Science, Faculty of
Mathematics, Department of
DSpace
UBCV
Zhang, Qidi
Doctor of Philosophy - PhD
University of British Columbia
In this dissertation, we construct blow-up solutions for the critical heat equations and the two-dimensional Landau-Lifshitz-Gilbert equation. In Chapter 2, we construct a radial smooth positive ancient solution for the energy critical semi-linear heat equation in the Euclidean space with a dimension greater or equal to seven. It blows up at the origin with the profile of multiple Aubin-Talenti bubbles in the backward time infinity. In Chapter 3, we consider the Cauchy problem for the four-dimensional energy critical heat equation. We construct a positive infinite time blow-up solution with the blow-up rate ln t as t goes to infinity and study the stability of the infinite time blow-up solution. This gives rigorous proof of the infinite time blow-up predicted by Fila and King. In Chapter 4, we construct finite time blow-up solutions to the Landau-Lifshitz-Gilbert (LLG) equation from the two-dimensional Euclidean space into the unit sphere with dimension two. Given any prescribed N points in the two-dimensional Euclidean space and a small positive constant T, we prove that there exists smooth initial data such that the solution blows up precisely at these points at finite time T, taking around each point the profile of sharply scaled degree 1 harmonic map.
The construction of blow-up solutions for some evolution equationsbyQidi ZhangA THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFDoctor of PhilosophyinTHE FACULTY OF GRADUATE AND POSTDOCTORAL STUDIES(Mathematics)The University of British Columbia(Vancouver)May 2023© Qidi Zhang, 2023The following individuals certify that they have read, and recommend to the Faculty of Graduate andPostdoctoral Studies for acceptance, the thesis entitled:The construction of blow-up solutions for some evolution equationssubmitted by Qidi Zhang in partial fulfillment of the requirements for the degree of Doctor of Phi-losophy inMathematics.Examining Committee:Juncheng Wei, Professor, Department of Mathematics, UBCSupervisorStephen Gustafson, Professor, Department of Mathematics, UBCSupervisory Committee MemberTai-Peng Tsai, Professor, Department of Mathematics, UBCSupervisory Committee MemberMichael Jeffrey Ward, Professor, Department of Mathematics, UBCSupervisory Committee MemberYoung-Heon Kim, Professor, Department of Mathematics, UBCUniversity ExaminerJiahua Chen, Professor, Department of Statistics, UBCUniversity ExamineriiAbstractIn this dissertation, we construct blow-up solutions for the critical heat equations and the two-dimensionalLandau-Lifshitz-Gilbert equation.In Chapter 2, we construct a radial smooth positive ancient solution for the energy critical semi-linear heat equation in the Euclidean space with a dimension greater or equal to seven. It blows up atthe origin with the profile of multiple Aubin-Talenti bubbles in the backward time infinity.In Chapter 3, we consider the Cauchy problem for the four-dimensional energy critical heat equa-tion. We construct a positive infinite time blow-up solution with the blow-up rate ln t as t goes toinfinity and study the stability of the infinite time blow-up solution. This gives rigorous proof of theinfinite time blow-up predicted by Fila and King.In Chapter 4, we construct finite time blow-up solutions to the Landau-Lifshitz-Gilbert (LLG)equation from the two-dimensional Euclidean space into the unit sphere with dimension two. Givenany prescribed N points in the two-dimensional Euclidean space and a small positive constant T, weprove that there exists smooth initial data such that the solution blows up precisely at these points atfinite time T, taking around each point the profile of sharply scaled degree 1 harmonic map.iiiLay SummaryIn this dissertation, we use the gluing method to construct ancient solutions and infinite time blow-up solutions for the critical heat equation and finite time blow-up solutions for the two-dimensionalLandau-Lifshitz-Gilbert (LLG) equation. In the outer problem, we use the convolution estimate toprovide uniform estimates for the heat equation, which can be applied directly to the other parabolicequations. For the inner part, we use regluing method to facilitate the inner linear theory, in somesense, almost to the optimal. For the mode −1 in the LLG equation, the distorted Fourier transformhas a good performance.ivPrefaceThis dissertation is based on the author’s original research articles.Chapter 2 is based on [96] joint work with Liming Sun and Juncheng Wei, which is published inCalculus of Variations and Partial Differential Equations. I contribute 33% of the methodology andcomputation.Chapter 3 is based on [108] joint with Juncheng Wei and Yifu Zhou. I contribute 40% of themethodology and computation.Chapter 4 is based on [107] joint work with Juncheng Wei and Yifu Zhou. I contribute 40% ofthe methodology and computation.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix1 Introduction and main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.0.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Bubble towers in the ancient solution of the energy-critical heat equation . . . . . . . 162.1 Sketch of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 A first approximation and the ansatz . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 The inner-outer gluing system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 The linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.4.1 The linear inner problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.4.2 The linear outer problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Orthogonal equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.6 The Schauder fixed point argument . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.7 Additional information 1: Estimates for the data in the outer problem . . . . . . . . . 392.8 Additional information 2: Some estimates for the outer problem . . . . . . . . . . . 523 Existence and stability of infinite time blow-up for the four dimensional energy criticalheat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.1 Approximate solution and improvement . . . . . . . . . . . . . . . . . . . . . . . . 653.1.1 First approximate solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.1.2 Transferring slow decaying terms by heat equations . . . . . . . . . . . . . . 663.1.3 Further improvement by solving an elliptic equation . . . . . . . . . . . . . 78vi3.2 Gluing system and solving the outer problem . . . . . . . . . . . . . . . . . . . . . 923.3 Orthogonal equations for µ1, ξ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.3.1 Solving µ1 and ξ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.3.2 Hölder continuity of µ1t and estimate for µEν [µ1] . . . . . . . . . . . . . . . 1053.4 Solving the inner problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.5 Stability of blow-up: proof of Theorem 1.0.6 . . . . . . . . . . . . . . . . . . . . . 1123.6 Linear theory for the inner problem . . . . . . . . . . . . . . . . . . . . . . . . . . 1133.6.1 Mode 0 without orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . 1223.6.2 Modes 1 to n without orthogonality . . . . . . . . . . . . . . . . . . . . . . 1253.6.3 Higher modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.6.4 Mode 0 with orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.6.5 Modes 1 to n with orthogonality . . . . . . . . . . . . . . . . . . . . . . . . 1313.7 Additional information 1: Estimates for heat equations . . . . . . . . . . . . . . . . 1323.7.1 Heat equation with right hand side v(t)|x|−b1{l1(t)≤|x|≤l2(t)} . . . . . . . . . . 1323.7.2 Heat equation with right hand side v(t)|x|−b1{|x|≥t1/2} . . . . . . . . . . . . . 1363.7.3 Cauchy problem with initial value 〈y〉−b . . . . . . . . . . . . . . . . . . . . 1373.8 Additional information 2: Proof of Proposition 3.2.1: solving the outer problem . . . 1393.9 Additional information 3: Estimates for ∇x¯φ[µ¯0] and ∂tφ[µ¯0] . . . . . . . . . . . . . 1474 Finite time blow-up solution of the two-dimensional Landau-Lifshitz-Gilbert equation 1514.1 Sketch of the construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1514.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1544.3 Approximation and improvement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1604.3.1 First approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1604.3.2 Global correction by parabolic systems . . . . . . . . . . . . . . . . . . . . 1614.4 Gluing system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1804.4.1 Error calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1804.4.2 Simplification ofN [Φ] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1894.4.3 Inner-outer gluing system . . . . . . . . . . . . . . . . . . . . . . . . . . . 1924.4.4 Weighted topologies for the inner and outer problems . . . . . . . . . . . . . 1964.4.5 Strategy for solving the inner problem . . . . . . . . . . . . . . . . . . . . . 1984.5 Orthogonal equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2004.5.1 Mode 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2014.5.2 Mode 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2054.5.3 Solving the orthogonal equations . . . . . . . . . . . . . . . . . . . . . . . 2064.6 Linear theory for the outer problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 2094.6.1 Fundamental solution for the outer problem . . . . . . . . . . . . . . . . . . 2094.6.2 Properties about the leading coefficients . . . . . . . . . . . . . . . . . . . . 213vii4.7 Solving the gluing system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2174.8 Linear theory for the inner problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 2184.8.1 Complex-valued form of the inner linear equation . . . . . . . . . . . . . . . 2214.8.2 Energy estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2244.8.3 Higher modes |k| ≥ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2324.8.4 Mode 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2424.8.5 Mode 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2484.8.6 Mode −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.9 Additional information 1: Pointwise estimate for heat equations . . . . . . . . . . . 2694.10 Additional information 2: Integral estimate for distorted Fourier . . . . . . . . . . . 2764.11 Additional information 3: Convolution estimates in finite time . . . . . . . . . . . . 2794.11.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2794.11.2 Convolution with v(t)|x−q|−b1{l1(t)≤|x−q|≤l2(t)} . . . . . . . . . . . . . . . 2904.11.3 Convolution with v(t)|x−q|−b1{|x−q|≥(T−t) 12 } . . . . . . . . . . . . . . . . . 3024.12 Additional information 4: Derivation of the weighted topology for the outer problem 3074.13 Additional information 5: Estimates of G andH j . . . . . . . . . . . . . . . . . . . 3144.13.1 Estimates for terms involvingU∗, Φout, Φ[ j]in, Φ∗[ j]0 . . . . . . . . . . . . . . . 3144.13.2 Estimate ∇xA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3194.13.3 Estimate G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3234.13.4 Pointwise and Hölder estimate forH [ j] . . . . . . . . . . . . . . . . . . . . 3504.14 Additional information 6: Convergence . . . . . . . . . . . . . . . . . . . . . . . . 360Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371viiiAcknowledgmentsFirst and foremost, I would like to express my sincere gratitude to my advisor Professor JunchengWei for his continuous guidance and patience over the years.I would like to thank the members of my supervisory committee Professors Stephen Gustafson,Tai-Peng Tsai, and Michael Ward, university examiners Young-Heon Kim, Jiahua Chen, and externalexaminers Pengfei Guan, Yu Yuan, Robert L. Jerrard for their time and help.Last but not least, I would like to express my deepest gratitude to my parents Yuecong Zhang,Junlian Wu, and grandparents Hesheng Zhang, Gen-e Jiang, Sanmei Zhang, for their constant supportand love.ixChapter 1Introduction and main resultsIt is a general belief that energy concentration will give rise to the blow-up phenomenon for manynonlinear partial differential equations. In this dissertation, we will use the gluing method to constructinfinite time blow-up solutions backward and forward for the critical heat equation and the two-dimensional Landau-Lifshitz-Gilbert equation.This dissertation is divided into three parts and the introduction is also split into three parts re-spectively.• For the first part, we deal with the analysis of ancient solutions that exhibit infinite time blow-upin the energy critical semi-linear heat equationut = ∆u+ |u|p−1u in Rn× (−∞,0) (1.0.1)where n ≥ 3 and p is the critical Sobolev exponent pS := n+2n−2 . We are interested in the positivesolutions u(x, t) globally defined for the ancient time such thatlimt→−∞‖u(·, t)‖L∞(Rn) =+∞. (1.0.2)Problem (1.0.1) has a more well-known counterpart in the forward direction, namely, the nonlin-ear Fujita equation ut = ∆u+ |u|p−1u in Rn× (0,T ),u(x,0) = u0(x) in Rn. (1.0.3)The energy functional associated to (1.0.3) isJ(u) :=ˆRn12|∇u|2− 1p+1|u|p+1. (1.0.4)The scaling u(x, t) 7→ λ 2/(p−1)u(λx,λ 2t) keeps the equation invariant and transforms J(u) to λ 4p−1+2−nJ(u).1Evidently, (1.0.3) is energy critical when p= pS.Problem (1.0.3) has been extensively studied in the literature. It is well-known that for a largeclass of initial data, say bounded continuous, there is a unique maximal classical solution u(x, t) fort ∈ (0,T ). If T is finite, then u will blow up at T . There are two types of blow-ups depending on therateType I : limsupt→T(T − t) 1p−1 ‖u(·, t)‖L∞(Rn) < ∞, (1.0.5)Type II : limsupt→T(T − t) 1p−1 ‖u(·, t)‖L∞(Rn) = ∞. (1.0.6)The blow-up is almost completely understood in the sub-critical range p < pS, for instance, by [36,43–45, 85, 102]. The solution always blows up in type I in this range. The existence of type IIblow-up has been established in various settings, for instance by [55, 56, 73] when p> pJL, wherepJL =∞ if n≤ 10,1+ 4n−4−2√n−1 if n≥ 11.(1.0.7)There are active researches in the energy critical case p = pS by [16, 21–23, 37, 52, 53, 92]. Theseworks found that u can exhibit type II blow-up in finite time in lower dimensions, while Wang andWei [105] precluded this fast blow-up for n≥ 7.Ancient solutions play an important role in studies of singularities and long-time behavior ofsolutions of many evolution problems, for instance in the mean curvature flow [10], Ricci flow [9] andYamabe flow [17]. Compared to the forward direction, the studies of ancient solutions of semi-linearheat equation (1.0.1) are quite limited. In the sub-critical case, Merle and Zaag [71] first establishedthe following result.Theorem 1.0.1 ([71]). Let 1< p< pS and u be a positive solution of (1.0.1) satisfying‖u(·, t)‖L∞(Rn) ≤C(−t)−1/(p−1) for t ∈ (−∞,0). (1.0.8)Then there exists T ∗ ≥ 0 such that u(x, t) = (p−1)−1/(p−1) (T ∗− t)−1/(p−1).The above result about ancient solutions has some interesting and important consequences in thestudy of the (forward) blow-up behavior of solutions of (1.0.3) when p < pS. See [71] for details.Recently, Quittner [87] improves the above theorem by removing the decay constraint (1.0.8).For the super-critical case, one knows that there exist one-parameter radial positive steady states{ϕα} for each α > 0. Furthermore, if p> pJL, then these solutions are ordered as ϕα < ϕβ for α < βand ϕβ → ϕ∞ as β → ∞, where (see [88, 106])ϕ∞(x) := L|x|−2/(p−1), L :=(2(p−1)2 ((n−2)p−n)) 1p−1.2The following Liouville-type results are known by Fila and Yanagida [33, Theorem 2.4] and Polácˇikand Yanagida [78, Theorem 1.2].Theorem 1.0.2 ([33, 78]). Let u be a non-negative radial solution of (1.0.1).1. Assume pS ≤ p< pJL and u(·, t)≤ ϕ∞ for all t ≤ 0. Then u≡ 0.2. Assume p> pJL and ϕα ≤ u(·, t)≤ ϕ∞ for some α > 0 and all t ≤ 0. Then u(·, t)≡ ϕγ for someγ ≥ α .Without this ϕ∞ bound, [33] also constructed some radial positive bounded solutions which dodepend on time. Polácˇik and Quittner [77] classified all radial positive ancient solutions under somefurther conditions for the super-critical regime.We are interested in the energy-critical case p = pS. The steady states of the equation (1.0.3)satisfy∆u+ |u| 4n−2 u= 0 in Rn. (1.0.9)We recall that all positive entire solutions of the equation are given by the family of Aubin-Talentisolitons [2, 42, 98]Uµ,ξ (x) = µ−n−22 U(x−ξµ)(1.0.10)whereU(y) is the standard bubble solitonU(y) = αn(11+ |y|2) n−22, αn = [n(n−2)] n−24 . (1.0.11)This family of solitons is also called Aubin-Talenti ground state solitary wave of the energy functionalJ. Collot et al. [14] classified the ancient solutions near the ground states.Theorem 1.0.3 ([14]). Let n≥ 7 and p= pS. There exist two strictly positive, C∞ radial solutions of(1.0.1), Q+ and Q− such that limt→−∞ ‖Q±−U‖H˙1 = 0. Conversely, there exists 0< δ 1 such thatif u is a solution of (1.0.1) withsupt≤0infµ>0,ξ∈Rn‖u(t)−Uµ,ξ‖H˙1 ≤ δ ,then u= Q± or u=U up to the symmetry of the flow.They also pointed out the forward behavior: Q+ explodes according to type I blow-up in finitetime, and Q− is global and dissipates Q−→ 0 as t→ ∞ in H˙1(Rn).A natural question is whether we have multiple Aubin-Talenti solitons in the backward limit. Inthe forward direction, del Pino et al. [24] constructed an initial condition u0 such that (1.0.3) blowsup in infinite time exactly at the origin. The solutions constructed in [24] consist of sign-changing3bubbling towers in the forward limit t → +∞. We investigate the possibility of such a phenomenonin the backward direction.Recall that for any Palais-Smale sequence {u(x, tn)}∞n=1 ≥ 0 of the energy functional J, Struwe’sprofile decomposition [93] tells us that passing to a subsequence, there are positive scalars {µ j(tn)}kj=1and points {ξ j(tn)}kj=1 such thatµi(tn)µ j(tn)+µ j(tn)µi(tn)+|ξi−ξ j|2µiµ j(tn)→ ∞ as n→ ∞ (1.0.12)andu(x, tn) =k∑j=11µ j(tn)n−22U(x−ξ j(tn)µ j(tn))+o(1) as n→ ∞ (1.0.13)where (after some permutation) µk(t) ≤ ·· · ≤ µ1(t). Our main result is the following existence of abubbling-tower solution in the backward limit t→−∞:Theorem 1.0.4. Let n≥ 7, k≥ 2. There exists a radial smooth positive solution of (1.0.1) that blowsup backward in infinite time exactly at 0 with a profile of the formu(x, t) =(1+O(|t|−ε)) k∑j=1µ j(t)−n−22 U(xµ j(t))for all t ≤ t0 (1.0.14)where O(|t|−ε) denotes some function g(x, t) satisfying ‖g(·, t)‖L∞(Rn) ≲ |t|−ε . Furthermore, we have‖u(x, t)−k∑j=1µ j(t)−n−22 U(xµ j(t))‖H1(Rn) ≲ |t|−ε for all t ≤ t0. (1.0.15)Here ε > 0 is small,µ j(t) = β j(−t)−α j(1+O(|t|−σ )), α j = 12(n−2n−6) j−1− 12, j = 1, . . . ,k,where β j, σ are certain positive constants.One interesting question is the forward behavior of the ancient solution we construct. Eitheru(x, t) is an eternal solution, or it will blow up in type I at some later time.There are some other related results. Del Pino and Gkikas [20] studied the ancient solution in theAllen-Cahn equation. Daskalopoulos et al. [17] constructed the ancient bubbling-tower solution forYamabe flow. For the construction of radial symmetric bubbling-towers in NLS and energy-criticalwave equations, we refer to [58–60].• For the second part, we consider the infinite time blow-up for the four-dimensional critical heat4equation. For the critical heat equationut = ∆u+ |u|p−1u in Rn× (0,∞),u(x,0) = u0(x) in Rn, (1.0.16)with p> 1, n≥ 3. Regarding infinite time blow-up for p= ps, Galaktionov and King [39] investigatedpositive, radially symmetric, global unbounded solutions for problem (1.0.16) in the case of a unitball with Dirichlet boundary condition in dimensions n≥ 3. See also [97, Theorem 1.4] for the casethat the domain is symmetric and convex. In the non-radial setting, positive infinite time blow-upsolutions for problem (1.0.16) with Dirichlet boundary condition and n≥ 5 were constructed in [16].The solution constructed in [16] takes the profile of sharply scaled Aubin-Talenti bubbles.Moreover, the blow-up location for the solution is determined by the Green’s function of −∆in Ω, while for elliptic problems, the role of the Green’s function in bubbling phenomena has beenknown for a long time since the works [3] and [4]. In [25], the non-radial and sign-changing solutionwhich blows up at infinite time has been constructed. In a very interesting paper [32], Fila and Kingstudied problem (1.0.16) in the whole space Rn with the critical exponent p= ps and gave insight onthe infinite time blow-up in the case of a radially symmetric, positive initial condition with an exactpower decay rate. By formal matched asymptotic analysis, they demonstrated that the blow-up rate isdetermined by the power decay in a precise manner. Intriguingly enough, their analysis leads them toconjecture that infinite time blow-ups should only happen in low dimensions 3 and 4, see Conjecture1.1 in [32]. Recently, this has been confirmed and rigorously proved in [22] for n = 3, where theleading part of the scaling parameter is achieved by asymptotic analysis. For the case, n= 4, Fila andKing conjectured that infinite time blow-up only exists when ℓ > 2 for radial solutions, wherelim|x|→∞|x|ℓu0(|x|) = Afor some A> 0.We are concerned with the following Cauchy problem for the Fujita equation with critical expo-nent in dimension four ut = ∆u+u3 in R4× (t0,∞),u(x, t0) = u0(x) in R4. (1.0.17)The aim is to construct an infinite time blow-up solution, confirming the conjecture by [32, Conjecture1.1], and further investigate the stability of the infinite time blow-up.Assume η is a smooth cut-off function which satisfies that η(s) = 1 for s ≤ 1 and η(s) = 0 fors≥ 32 . Our main results are stated as follows.Theorem 1.0.5. For t0 sufficiently large, there exists an initial value u0 > 0 with exponential decaysuch that the positive solution u(x, t) to (1.0.17) blows up at the infinite time. More precisely, the5solution takes the form of the sharply scaled bubbleu(x, t) = η(x−ξ√t)µ−1(t)U(x−ξ (t)µ(t))+O((ln t)−1min{t−1, |x|−2})where U(y) = 232 11+|y|2 is the Aubin-Talenti solution in dimension four. The blow-up rate and locationare given byµ(t) =1ln t(1+O( ln ln tln t)), ξ (t) = O(t−1).More precisely, the positive initial value of the solution constructed above isu(x, t0) = µ−1(t0)U(x−ξ (t0)µ(t0))η(x−ξ (t0)√t0)+232 µ(t0)|x−ξ (t0)|−2(e−|x−ξ (t0)|24t0 −η(x−ξ (t0)√t0))+ µ¯−10 (t0)Φ0(x−ξ (t0)µ¯0(t0), t0)η(4(x−ξ (t0))√t0)+η( x−ξ (t0)µ0(t0)R(t0))e0µ−1(t0)Z0(x−ξ (t0)µ(t0)),where µ0, µ¯0 are the leading order of µ , and µ¯0 ∼ µ0 = (ln t)−1; Φ0 is a global correction functiongiven in Section 3.1.3; e0 is a constant and Z0 is the eigenfunction with respect to the first eigenvaluefor the linearized operator, which has exponential decay, see (3.6.3).We further investigate the stability of the blow-up solution constructed in Theorem 1.0.5 andobtain the stability in the following sense.Theorem 1.0.6. For any g0, not necessarily radially symmetric, satisfying |g0(x)| ≤Cgt−min{ℓ,4}20 〈x〉−ℓ,ℓ > 3, and for t0 sufficiently large, there exists a solution u[g0](x, t) to (1.0.17) blowing up at infinitetime with the rateµ[g0](t) =1ln t(1+O( ln ln tln t)), ξ [g0](t) = O(t−1).The initial value is given byu(x, t0) =(µ[g0](t0))−1U(x−ξ [g0](t0)µ[g0](t0))η(x−ξ [g0](t0)√t0)+232 µ[g0](t0)∣∣x−ξ [g0](t0)∣∣−2(e− |x−ξ [g0](t0)|24t0 −η(x−ξ [g0](t0)√t0 ))+(µ¯0(t0))−1Φ0(x−ξ [g0](t0)µ¯0(t0) , t0)η(4(x−ξ [g0](t0))√t0 )+η(x−ξ [g0](t0)µ0(t0)R(t0))e0[g0](µ[g0](t0))−1Z0(x−ξ [g0](t0)µ[g0](t0) )+g0where µ[g0]→ µ under the norm (3.2.9), ξ [g0]→ ξ under the norm (3.2.10), e0[g0]→ e0 as Cg→ 0.In the radial setting, the same conclusion holds for ℓ > 2 with ξ [g0] ≡ 0 and µ¯0[g0]→ µ¯0 under thenorm (3.1.22) as Cg → 0 additionally.Remark 1.0.6.1.6• Indeed, the initial value of the infinite time blow-up solution in Theorem 1.0.5 has exponentialdecay at space infinity. By Theorem 1.0.6, we can add suitable perturbation for the initial valueto achieve thatlim|x|→∞|x|ℓu(x, t0) = Afor any |A| small enough, recovering the assumption on the initial value in the conjecture byFila and King [32].• It is very possible to generalize the stability result for all ℓ > 2 in the non-radial setting, seeRemark 3.5.0.1.• The infinite time blow-up (grow-up) threshold solution can be constructed by the same methodin Remark 2(i) [86].Some essential new features and difficulties are listed below.One key feature and difficulty is the non-local dynamics for the scaling parameter µ(t). It turnsout that the dynamics for µ(t) is governed by an integrodifferential operator, which is a natural conse-quence of the fact that the linear generator of dilations of the Aubin-Talenti bubble is of slow decay inlower dimensions. This non-local phenomenon has also been observed in [18, 19, 22, 23] for lower di-mensional problems. In our case here, neither the usual Laplace transform nor the Riemann-Louvilletype method is applicable since the integrodifferential equation is not in the class of Abel-type inte-gral equations. The non-local operator here is the threshold/endpoint case in a certain sense, and oneneeds to carry out a much more delicate analysis to investigate its solvability.Our strategy is to decompose the non-local equation for µ(t) into two parts: the dominating termand the remainder term. The dominating term will be solved by the contraction mapping theorem,while the remaining term will leave a much smaller error. To be more precise, the desired blow-uprate is determined at the leading order. However, due to the way that we handle the non-local operator,the time decay is not fast enough for the remainder in the gluing procedure, and we will iterate thisprocess finitely many times to make the remainder term have a faster time decay than the one providedby the outer problem. This smaller remainder will be handled when solving the next order of µ(t).After getting the leading order of µ(t), we need to solve the corresponding linearized ellipticequation to improve the time decay of the error term, which is essential for finding suitably weightedtopologies ensuring the implementation of the gluing procedure. When solving the next order µ1(t),we still need to decompose the non-local equation into two parts. The main difference is that theinvolved outer problem in the equation of µ1(t) only has Hölder continuity in t variable. The deriva-tive of µ1(t) will inherit Hölder continuity from the outer problem, which will be used to control theremainder term.On the other hand, the rather slow logarithmic blow-up rate produces the following difficulties.Several slow-decaying linear terms involving the inner part cannot be controlled as the right hand side7of the inner or outer problem. Instead, we regard these slow-decaying terms as part of the linearizationof the inner problem and develop a new linear theory. See Remark 3.2.0.1. The dealing of these termsis in a similar spirit as in [18], where the logarithmic blow-up speed also appears.Thanks to the generality of the gluing method, we can study the stability of the solution con-structed in Theorem 1.0.5 with both radial and non-radial perturbations, and non-radial infinite timeblow-up solutions are easily found by suitable perturbation for the initial value.• For the last part, we study the finite time blow-up solutions for the two-dimensional Landau-Lifshitz-Gilbert equation.LetM be am-dimensional Riemannian manifold of the metric g and S2 be the 2-sphere embeddedin R3. The Landau-Lifshitz-Gilbert (LLG) equation onM is given byut =−au∧ (u∧∆M u)−bu∧∆M u in M × (0,T )u(·,0) = u0 ∈ S2 in M , (1.0.18)where a2+ b2 = 1, a ≥ 0, b ∈ R, ∆M u = 1√g∂xβ (gαβ√g∂xαu) is the Laplace-Beltrami operator andu = [u1,u2,u3]tr is a 3-vector with normalized length which is a mapping u(x, t) :M × (0,T )→ S2.First formulated by Landau and Lifshitz [66] in 1935, the LLG equation (1.0.18) is an importantsystem modeling the effects of a magnetic field on ferromagnetic materials in micromagnetics, and itdescribes the evolution of spin fields in continuum ferromagnetism. See also Gilbert [46]. The LLGequation (1.0.18) can be viewed as a bridge between the harmonic map flow (HMF) when a= 1,b= 0and the Schrödinger map flow (SMF) when a= 0,b=−1.In the context of HMF, Struwe [94] proved the existence and uniqueness of weak solution with atmost finitely many singular points whenM is a Riemann surface. See [38] for further generalizationsand [13, 95] for higher dimensional cases. Chang, Ding, and Ye [11] first proved the existence of finitetime blow-up solutions for HMF from disk into S2. See also [12, 15, 27, 67, 83, 84, 99, 103] and thereferences therein. In [101], van den Berg, Hulshof, and King used formal analysis to predict theexistence of blow-up solutions with quantitative ratesλk(t)∼ |T − t|k| ln(T − t)| 2k2k−1, k ∈ N+ (1.0.19)for the two-dimensional HMF into S2. For the caseM = R2 and the target manifold is a revolutionsurface, using degree 1 harmonic map Q1 as the building block, Raphaël and Schweyer [89, 90]constructed finite time blow-up solutions with rates (1.0.19) for all k ≥ 1 in the equivariant class,where the initial data can be taken arbitrarily close to Q1 in the energy-critical topology. For the casethatM ⊂R2 is a general bounded domain, Dávila, del Pino andWei [19] constructed solutions whichblow up at finitely many points with the type II rate (1.0.19) for k = 1, and they further investigatedthe stability and reverse bubbling phenomena. The construction in [19] can be generalized to the case8M = R2.On the other hand, for SMF withM = R2, Merle, Raphaël and Rodnianski [72] constructed thefinite time blow-up solution with the rate (1.0.19) for k = 1 in the 1-equivariant class. Analogous tothe results in Krieger, Schlag, and Tataru [63] for wave maps, Perelman [76] constructed finite timeblow-up solutions with continuous rates. See also [5–8, 57] and the references therein for the globalwell-posedness results and the dynamics of SMF near ground state.For the LLG equation, in the caseM =R3,a> 0, Alouges and Soyeur [1] proved the existence ofweak solutions for (1.0.18) and constructed infinitely many weak solutions. The existence of the weaksolution to the LLG equation has been established by Guo and Hong [47, Theorem 4.2] whenM is aclosed Riemannian manifold with m ≥ 3, while for the case thatM is a closed Riemannian surface,the weak solution was shown to be unique and regular except for at most finitely many points [47,Theorem 3.13]. WhenM = R2 and the target manifold is a smooth closed surface embedded in R3,approximation by discretization was used in [61] to construct a solution of the LLG equation whichis smooth away from a two-dimensional locally finite Hausdorff measure. In general, one cannotexpect good partial regularity results for weak solutions in the higher dimensional case m≥ 3 withoutfurther regularity or energy minimizing assumptions. See the famous example by Rivière [91], whereweakly harmonic maps from the ball B3 ⊂ R3 into S2 were constructed for which the singular setSing u is the closed ball B3, and this result can be generalized to higher dimensions. In a similarspirit to the existence results for partially regular solution for HMF in higher dimensions of Chen andStruwe [13], Melcher [69] proved that forM =Rm (m= 3) there exists a global weak solution whosesingular set has finite 3-dimensional parabolic Hausdorff measure. Later, this result was generalizedto m ≤ 4 by Wang [104]. With the additional stability assumption for the weak solution, for m ≤ 4,Moser [74] proved a better estimate for the singular set. The partial regularity of the LLG equation(1.0.18) for m≥ 5 remains open.ForM = Rm, the global existence, uniqueness, and decay properties for the solution of (1.0.18)were established by Melcher [70] for m ≥ 3 with initial data u0 close to a fixed point in S2 in the Lmnorm. Lin, Lai, and Wang [68] generalized the result to Morrey space and m≥ 2. For u0 away froma fixed point in S2 with BMO semi-norm sufficiently small, Gutiérrez and de Laire [51] proved theglobal existence, uniqueness, and regularity results for the LLG equation.The study of the dynamics for the LLG equation with initial data close to harmonic maps is ofspecial significance and can provide hints on the mechanism of singularity formation. A series ofworks by Gustafson, Kang, Nakanishi, and Tsai [48–50] are devoted to the behavior of the solutionsto the LLG equation withM =R2 with initial data u0 close to the harmonic map in the n-equivariantclass. They found, among other things, that there is no finite time blow-up for the LLG equationand HMF with u0 close to n-equivariant harmonic maps for n ≥ 3 and n = 2, respectively. See [50,Theorem 1.1], [50, Theorem 1.2].9The singularity formation for the LLG equation is an important and challenging topic. For thecase that M is a compact manifold with or without boundary in dimensions m = 3,4, Ding andWang [26] obtained the existence of a smooth finite time blow-up solution for the LLG equation. ForM ⊂ R2, as an analogue of Qing [83] for HMF, Harpes [54] gave descriptions of solutions to theLLG equation (1.0.18) near the singular points. For the energy critical case thatM is a disk in R2, inan interesting paper [100], van den Berg and Williams predicted the existence of finite time blow-upby formal asymptotic analysis supported with numerical simulations. For M = R2, Xu and Zhao[109] rigorously constructed a finite time blow-up solution to (1.0.18) in the 1-equivariant class.We consider the case with target manifold S2,M = R2, and positive damping parameter a > 0.The LLG equation (1.0.18) can then be written asut = a(∆u+ |∇u|2u)−bu∧∆u in R2× (0,T ),u(·,0) = u0 ∈ S2 in R2. (1.0.20)We are interested in the case of multiple bubbles to the LLG equation (1.0.20) in the general non-radially symmetric setting. Our construction is based on the following degree 1 profileW (y) =1|y|2+1 2y12y2|y|2−1 , y= (y1,y2) ∈ R2,and clearlyQγW(x−ξλ)solves the stationary equation of the LLG equation (1.0.20) for any ξ ∈R2, λ > 0, and any γ-rotationmatrix around z-axisQγ :=cosγ −sinγ 0sinγ cosγ 00 0 1 .Let us defineU∞ = [0,0,1]tr, andU∞ =W (∞). Our main result is stated as follows.Theorem 1.0.7. For any prescribed N distinct points q[ j] ∈ R2, j = 1,2, . . . ,N, N ∈ Z+ and T suf-ficiently small, there exists a smooth initial data u0 such that the gradient of the solution u to theLLG equation (1.0.20) with a> 0 blows up at these N points at finite time t = T . More precisely, thesolution u takes the sharply scaled degree 1 profile around each point q[ j]u(x, t) =−(N−1)U∞+N∑j=1Qγ jW(x−ξ [ j]λ j)+Φper10withλ j(t) = κ∗j λ∗(t)(1+o(1)), λ∗(t) =| lnT |(T − t)| ln(T − t)|2 , ξ[ j](t) = q[ j]+o(1), γ j(t) = γ∗j (1+o(1))where o(1)→ 0 as t → T; κ∗j > 0, γ∗j ∈ R; the leading part of κ∗j is determined by ∇u0 and inde-pendent of a,b; the leading part of γ∗j is determined by ∇u0 and a,b; ‖Φper‖L∞(R2×(0,T )) 1, andλ∗(t)‖∇Φper(·, t)‖L∞(R2) ≲ λ ε∗ (t) for ε > 0 small. Moreover,u(x, t)−u∗(x)−N∑j=1Qγ j[W(x−ξ [ j]λ j)−W (∞)]→ 0 as t→ Tin H1loc(R2)∩L∞(R2) for some u∗(x) ∈ H1loc(R2)∩L∞(R2).Remark 1.0.7.1.• The positivity of the damping term (a) plays a crucial role in our construction.• Obviously, for j = 1, . . . ,N,|∇u(ξ [ j](t), t)| ∼ | ln(T − t)|2κ∗j | lnT |(T − t)as t→ T.• Due to the method employed, the construction works equally well for the case of smooth,bounded domain Ω ⊂ R2 as long as the technical ingredient in [31] can be generalized togeneral domains with proper boundary conditions.Our study of the singularity formation for the LLG equation is motivated by the one for HMF[19]. However, substantial difficulties arise due to the coupling between HMF and SMF in the LLGequation (1.0.20), and such coupling produces both dissipative (a> 0) and dispersive (b 6= 0) features.A direct consequence of the presence of dispersion is the lack of maximum principle for suitablequantities, which makes the analysis more delicate even at the linearized level. The dispersion cannotbe treated perturbatively even in the dissipation-dominating case a/|b| 1, and one has to includethis as part of the leading order. In our inner-outer gluing construction, new linear theories for bothinner and outer problems need to be developed, taking into account the dispersion.The new linear theory for the inner problem is developed by analyzing each Fourier mode, whichis the Fourier expansion of the complex form on each tangent plane of the bubble on S2. Due tothe absence of the maximum principle, we first employ energy methods to get rough upper boundsfor each mode. In order to refine the bounds and get better pointwise decay estimates, we performanother gluing procedure, called the re-gluing process, at all the modes except mode −1. The re-gluing process was first used in the analysis of linearization of HMF at mode 0 in [19], and here wegeneralize this technique to all other modes except mode −1. For mode −1, using the above method11does give a solution, but this solution gets deteriorated in the innermost region and is not sufficientfor the gluing to implement. Instead, we utilize the techniques of distorted Fourier transform for thedealing of mode −1. The use of distorted Fourier transform is motivated by a recent work of Krieger,Miao, and Schlag [65] on the stability of blow-up for wave maps beyond the equivariant class. Seealso Krieger, Miao, Schlag, and Tataru [62–64] and the references therein and Gesztesy, Zinchenko[40] for the general theory. Using the distorted Fourier transform, we develop linear theory at mode−1 with or without orthogonality conditions. The version with orthogonality removes the logarithmicloss compared to the one without orthogonality. See Section 4.8.6 for more details. For mode −1, weonly use the one without orthogonality since the introduction of two new modulation parameters willfurther complicate the interactions, and we control the logarithmic loss by Hölder properties inheritedfrom the equations.The outer problem turns out to be a quasi-linear parabolic system in non-divergence form. Dif-ferent from the linearized outer problem in HMF, the one in the LLG equation is a coupled system,and thus cannot be solved componentwisely. The coefficients of the coupled system for the linearizedouter problem are part of the blow-up profile. So one cannot expect good Hölder continuity in thecoefficients and has to work in the certain weaker class. The linear theory for the outer problemis achieved by using the sub-Gaussian estimate for the fundamental solution of the parabolic sys-tem in non-divergence form with coefficients of Dini mean oscillation in space (DMOx), which wasproved by Dong, Kim, and Lee [31] recently. We introduce Dini mean absolute oscillation in space(|DMO|x), which is a subspace of DMOx. Under some weak assumptions, the functions in |DMO|xare closed under arithmetic (see Lemma 4.6.1). This property makes it easier to achieve that theleading coefficients of the outer problem belong to |DMO|x.Another aspect of the construction is to deal with slow decaying errors, usually present in lower di-mensional problems. The improvement of these slow decaying errors involves finding good global/non-local corrections, which in turn make the dynamics for the parameters in the corresponding modenon-local. In the context of the LLG equation, the mode with slow decaying error that we shall dealwith is mode 0, which corresponds to the invariance of scaling and rotation around z-axis. To capturethe precise blow-up dynamics, the non-local correction at mode 0 should be rather explicit. But dueto the aforementioned structure of the outer problem, one cannot improve the error by solving thelinearized system directly and has to extract part of the parabolic system instead. It turns out that thecombination of the new error produced by the non-local correction and the remainder in the parabolicsystem together make the non-local equations for the scaling parameter λ and rotational parameter γa well-structured complex system. See Section 4.5.The construction of multiple bubbles also involves analyzing complicated interactions, and thisreflects in the analysis of the tangent plane of each bubble. On the other hand, the ansatz for thesolution u with multiple bubbles needs to be carefully chosen as the unit length of the map |u| = 1should be kept for all space-times. This further produces delicate interactions. Fortunately, we can12control these in some well-designed topologies thanks to subtle cancellations as well as a trick thatwe call U∗-operation (see (4.4.6)), which simplifies analysis. This trick first appeared in [19] in thecase of the single bubble for HMF.1.0.1 Notations• We assume c1 ≲ c2 (c1 ≳ c2) if there exists a positive constantC such that c1 ≤Cc2 (c1 ≥Cc2).Denote c1 ∼ c2 if c1 ≲ c2 ≲ c1. All constants stated in the dissertation are independent of timevariables. For x ∈ Rd , denote 〈x〉 :=√|x|2+1.• In general, the letter C(a,b, . . .) stands for a positive constant depending on parameters a,b . . .that might change its actual value at each occurrence.• The symbol f [g1,g2, . . . ] means that the function f depends on some functions g1, g2, . . . .• For a set Ω ⊂ Rd , denote 1{x∈Ω} as the characteristic function with 1{x∈Ω} = 1 if x ∈ Ω and1{x∈Ω} = 0 if x 6∈Ω.• The symbol O( f (x)) is used to denote a real-valued function that satisfies |O( f (x))| ≲ | f (x)|in a domain of x that is either specified explicitly or follows from the context.• For any c ∈ R, we use the notation c− to denote a constant less than c and can be chosen closeto c arbitrarily.• For f ∈ Cm(Rd), the symbol Dkx f with k ∈ {0,1, . . .m} is used to denote ∂ k1x1 ∂ k2x2 · · ·∂ kdxd f forsome nonnegative intergers k j, j= 1,2, . . . ,d andd∑j=1k j = k. We will omit the subscript “x” andadopt Dk f if no ambiguity.• η(x) = χ(x) denotes a smooth cut-off function inRd such that 0≤ χ(x)≤ 1, χ(x) = 1 if |x| ≤ 1and χ(x) = 0 if |x| ≥ 2.• f ≈ g means that | f −g| → 0 as t→ ∞.• The notations in this part are devoted to Chapter 2.For j = 1, . . . ,k, µ j, µ0 j are some positive functions about t. We will use the notationµ⃗ = (µ1, . . . ,µk) , U¯ :=k∑j=1U j (1.0.21)whereU j(x, t) = µ j(t)−n−22 U(xµ j(t))(1.0.22)13andU(y) is given by (1.0.11). We denoteµ¯ j :=√µ jµ j−1, µ¯0 j :=√µ0 jµ0, j−1, j = 2, · · · ,k (1.0.23)and make the convention thatµ¯1 = µ¯01 = (−t)δ , µ¯k+1 = µ¯0,k+1 = 0. (1.0.24)where δ > 0 is a small constant.All the following notations are devoted to Chapter 4.• Denote Γ♮d as the fundamental solution of∂tu= (a− ib)∆u in Rdand Γ♮d is given byΓ♮d(x, t) = (a− ib)−d2 (4πt)−d2 e−|x|24(a−ib)t (1.0.25)where (a− ib)− d2 = e−iθ d2 if a− ib= eiθ .It is easy to have|Γ♮d(x, t)|≲ t−d2 e−a|x|24t .Given a fundamental solution Γ(x,y, t,s) for a parabolic system in Rd and some admissiblefunctions f (x), h(x, t), denote(Γ∗ f )(x, t, t0) :=ˆRdΓ(x,y, t, t0) f (y)dy, (Γ∗∗h)(x, t, t0) :=ˆ tt0ˆRdΓ(x,y, t,s)h(y,s)dyds.We usually omit the initial time t0 if no ambiguity.• For any vector a⃗ = [a1,a2,a3]tr ∈ R3, where “[· · · ]tr” means the transpose of a matrix, thereis a natural one-to-one correspondence between [a1,a2,a3]tr and [a1+ ia2,a3]tr and we alwaysregard them as the same thing with the notation [a1,a2,a3]tr := [a1+ ia2,a3]tr if no ambiguity.For b⃗= [b1,b2,b3]tr ∈ R3, it is easy to see that a⃗ · b⃗= Re [(a1+ ia2)(b1− ib2)]+a3b3.• For any matrix A= (ai j)n×m, denote |A|= (n∑i=1m∑j=1|ai j|2) 12 . Specially, for a ∈C, |a| is the usualabsolute value.• For any smooth function f (x, t) and x = x(t) depending on t smoothly, denote ∂t f (x(t), t) =(∂t f )(x(t), t) and ∂t( f (x(t), t)) = (∂t f )(x(t), t)+ x′(t) · (∇x f )(x(t), t).14• For v⃗= [v1,v2,v3]tr ∈C1(R2,R3), w⃗ ∈ R2 denotew⃗ · ∇⃗v :=[w⃗ ·∇v1, w⃗ ·∇v2, w⃗ ·∇v3]tr.Denotea1a2a3 ·b11 b12b21 b22b31 b32 := [ 3∑k=1akbk1,3∑k=1akbk2]; v⃗∧ := 0 −v3 v2v3 0 −v1−v2 v1 0 . (1.0.26)15Chapter 2Bubble towers in the ancient solution ofthe energy-critical heat equationIn this chapter, we study the bubble tower solutions for the critical heat equation.2.1 Sketch of the proofThe method of this chapter is close in spirit to the analysis in the works [16, 24], where the inner-outer gluing method is employed. That approach consists of reducing the original problem to solvinga basically uncoupled system, which depends in subtle ways on the parameter choices (which aregoverned by relatively simple ODE systems).We start with the ansatz solution U¯ =∑kj=1U j =∑kj=1 µ j(t)−n−22 U(x/µ j(t)) and search for φ(x, t)such that U¯+φ is a solution forS[u] :=−ut +∆u+ |u|p−1u= 0 in Rn× (−∞, t0) . (2.1.1)Because of the specific form of U¯ , we anticipate φ ∼ ∑kj=1 µ−n−22j ϕ(x/µ j(t))χ j with some cut-offfunction χ j supporting in the region where U j dominates other Ul , l 6= j. Plugging in U¯ +φ , wefound that in the support of χ j, the linearized operator of φ is ∆+ pU p−1j and the leading error is−∂tU j+ pU p−1j U j−1(0). Making a change of variable y= x/µ j(t), we will choose the ϕ satisfying∆yϕ + pU(y)p−1ϕ +h j(y, t) = 0 in Rn, ϕ(y)→ 0 as |y| → ∞ (2.1.2)whereh j(y, t) = µ j µ˙ jZn+1(y)+ pU(y)p−1(µ jµ j−1) n−22U(0). (2.1.3)One knows that (2.1.2) is solvable if and only if´Rn h j(y, t)Zn+1(y)dy= 0. Using the above expression16of h j, it implies that µ j µ˙ j = c∗(µ j/µ j−1)n−22 (see (2.2.17)). This implies µ j(t) = β j(−t)−α j . We willdenote it as µ0 j, because the above process is the first approximation.Next we will start with u∗ = U¯+∑kj=1 µ0 j(t)−n−22 ϕ0 j(x/µ0 j(t))χ j and search for φ with the formφ =k∑j=1µ−n−22j ϕ j(xµ j(t), t)η j+Ψ(x, t)such that u∗+φ is a solution of (2.1.1) and µ j(t) = µ0 j(t)+µ1 j(t). Plugging in u∗+φ to (2.1.1) candeduce the following equations of ϕ j and Ψµ2j ∂tϕ j = ∆yϕ j+ pU(y)p−1ϕ j+H j[Ψ, µ⃗1](y, t) in B8R× (−∞, t0), j = 1, . . . ,k, (2.1.4)∂tΨ= ∆xΨ+G [ϕ⃗ ,Ψ, µ⃗1](x, t) in Rn× (−∞, t0), (2.1.5)whereH j is defined in (2.3.13) and G is defined in (2.3.14), µ⃗1(t)= (µ11, · · · ,µ1k) and ϕ⃗ =(ϕ1, · · · ,ϕk).(2.1.4) is the so-called inner problem and (2.1.5) is the so-called outer problem. One will see that thesetwo problems are weakly coupled in the sense that the dependence ofH j on Ψ and G on ϕ⃗ is smallin an appropriate norm. The strategy to solve (2.1.4) and (2.1.5) is: for each fixed ϕ⃗ and µ⃗1, one cansolve (2.1.5) forΨ=Ψ[ϕ⃗ , µ⃗1]. Next, inserting suchΨ to (2.1.4) and using fixed point theorem to findϕ⃗ and µ⃗1.The foundation of this process lies in a clear understanding of the linearized problem of (2.1.4)and (2.1.5) respectively. The study of the linearized equation of inner problem (2.1.4) has been donein [16] for the forward direction. For the backward direction, one encounters new difficulties whentaking subsequences. We establish a uniqueness statement to make sure that different subsequenceswill give the same limit function. More details are given in the proof of Lemma 2.4.3. The linearizedequation of the outer problem (2.1.5) occupies the bulk of this chapter. Notice (2.1.5) actually canbe thought of nonhomogenous heat equation, we leverage Duhamel’s formula to get a solution Ψ.The main difficulty is finding a suitable topology for the outer problem due to the bubble towerphenomenon. Great efforts are needed to find suitable topological space for G . Check Remark 2.4.3.1and 2.4.7.1 for further explanation. Having set up the right space, we apply the Schauder fixed pointtheorem to prove the existence of ancient solution of (1.0.1).Here is the structure of the chapter. In Section 2.2, we derive the first approximation from theansatz solution. Section 2.3 is devoted to splitting the flow equation into a system of the inner problemand the outer problem. In Section 2.4, we study the linear problem of the inner one and outer onerespectively. We put off some tedious computations to Appendices in this chapter. Section 2.5 is usedto derive the orthogonal equations µ⃗1 should satisfy. In the last section, we put everything togetherand solve the problem by using Schauder’s fixed point theorem.172.2 A first approximation and the ansatzProblem (1.0.1) is equivalent toS[u] :=−ut +∆u+ |u|p−1u= 0 in Rn× (−∞, t0) (2.2.1)where t0 is a very negative constant. After some translation in time, we can assume the solution livesup to t = 0.For any integer k ≥ 2, let us consider k positive functionsµk(t)< µk−1(t)< · · ·< µ1(t) in (−∞, t0)which will be chosen later, such that as t→−∞,µ1(t)→ 1, µ j+1(t)µ j(t) → 0 for all j = 1, . . . ,k−1. (2.2.2)We assume that for j= 1, . . . ,k, µ0 j is the leading order of µ j and has the similar property of µ j above.µ0 j will be determined later. We will get an accurate first approximation to a solution of (2.2.1) ofthe form U¯ +φ0 that reduces the part of the error S[U¯ ] created by the interaction of the bubbles U jand U j−1, j = 2, · · · ,k. To get the correction φ0, we will need to fix the parameters µ j at main orderaround certain explicit values.Let us introduce the cut-off functionsχ j(x, t) =χ (2 |x|/µ¯0 j)−χ (2|x|/µ¯0, j+1) j = 2, . . . ,k−1,χ (2 |x|/µ¯0k) j = k. (2.2.3)One readily sees thatχ j(x, t) =0 if |x| ≤ 12 µ¯0, j+1,1 if µ¯0, j+1 ≤ |x| ≤ 12 µ¯0 j,0 if |x| ≥ µ¯0 j.(2.2.4)We define our approximate solution to be given byu∗ = U¯+φ0. (2.2.5)The correction φ0 has the formφ0 =k∑j=2φ0 jχ j (2.2.6)18whereφ0 j(x, t) = µ j(t)−n−22 ϕ0 j(xµ j(t), t)(2.2.7)for certain functions ϕ0 j(y, t) defined in entire y ∈ Rn which we will suitably determine. Let us writeS (u∗) = E¯1+LU¯ [φ0]+NU¯ [φ0] (2.2.8)whereLU¯ [φ0] =−∂tφ0+∆xφ0+ pU¯ p−1φ0, (2.2.9)NU¯ [φ0] = |U¯+φ0|p−1 (U¯+φ0)− pU¯ p−1φ0−U¯ p, (2.2.10)E¯1 =−k∑j=1∂tU j+U¯ p−k∑j=1U pj . (2.2.11)Next we writeLU¯ [φ0] using the form of φ0 in (2.2.6) as followsLU¯ [φ0] =k∑j=2(∆xφ0 j+ pU p−1j φ0 j)χ j+k∑j=2p(U¯ p−1−U p−1j)φ0 jχ j+k∑j=2(2∇xφ0 j ·∇xχ j+∆x (χ j)φ0 j)−k∑j=2∂t (φ0 jχ j) .In the end, we have the error expansionS (u∗) =−∂tU1+k∑j=2(∆xφ0 j+ pU p−1j φ0 j−∂tU j+ pU p−1j U j−1(0))χ j+ E¯11+k∑j=2p(U¯ p−1−U p−1j)φ0 jχ j+k∑j=2(2∇xφ0 j ·∇xχ j+∆xχ jφ0 j)−k∑j=2∂t (φ0 jχ j)+NU¯ [φ0](2.2.12)whereE¯11 = U¯ p−k∑j=1U pj −k∑j=2pU p−1j U j−1(0)χ j−k∑j=2(1−χ j)∂tU j. (2.2.13)The function φ0 j is chosen to eliminate at main order the terms in the first line of (2.2.12), after19conveniently restricting the range of variation of µ⃗ ,E j[φ0 j, µ⃗] := ∆xφ0 j+ pU p−1j φ0 j−∂tU j+ pU p−1j U j−1(0)=µ−n+22j[∆yϕ0 j+ pU(y)p−1ϕ0 j+µ j µ˙ jZn+1(y) + pU p−1(y)(µ jµ j−1) n−22U(0)]y= xµ j(2.2.14)where Zn+1(y) = n−22 U(y)+y ·∇U(y). The elliptic equation (for a radially symmetric function ϕ(y))∆yϕ + pU(y)p−1ϕ +h j(y, t) = 0 in Rn (2.2.15)whereh j(y, t) = µ j µ˙ jZn+1(y)+ pU(y)p−1(µ jµ j−1) n−22U(0) (2.2.16)has a solution with ϕ(y)→ 0 as |y| → ∞ if and only if h j satisfies the solvability conditionˆRnh j(y, t)Zn+1(y)dy= 0.The latter conditions hold if the parameters µ j(t) satisfy the following relations:µ1 = 1, µ j µ˙ j = c∗λn−22j , λ j =µ jµ j−1for all j = 2, . . . ,k (2.2.17)wherec∗ =−U(0)p´RnUp−1Zn+1dy´Rn Z2n+1dy=U(0)n−22´RnUpdy´Rn Z2n+1dy> 0. (2.2.18)Let µ⃗0 = (µ01, . . .µ0k) be the solution of (2.2.17) in (−∞, t0) given byµ0 j(t) = β j(−t)−α j , t ∈ (−∞, t0) (2.2.19)whereα j =12(n−2n−6) j−1− 12, j = 1, . . . ,kand the numbers β j are determined by the recursive relationsβ1 = 1, β j = (α jc−1∗ )2n−6βn−2n−6j−1 . (2.2.20)From (2.2.17), we setλ0 j(t) =µ0 jµ0, j−1(t). (2.2.21)20We haveh j (y, t) = λn−220 j h¯(y), h¯(y) = h¯(|y|) = pU(0)U(y)p−1+ c∗Zn+1(y).Since´Rn h¯Zn+1dy= 0, there exists a radially symmetric solution ϕ¯(y) to the equation∆ϕ¯ + pU(y)p−1ϕ¯ + h¯(|y|) = 0 in Rnsuch that ϕ¯(y) = O(|y|−2) as |y| →+∞.Then we define ϕ0 j(y, t) asϕ0 j(y, t) = λn−220 j ϕ¯(y). (2.2.22)In what follows we let the parameters µ j(t) in (2.2.2) have the form µ⃗ = µ⃗0+ µ⃗1, namelyµ j(t) = µ0 j(t)+µ1 j(t), (2.2.23)where the parameters µ1 j(t) to be determined satisfy∣∣µ1 j(t)∣∣≲ µ0 j(t)(−t)−σ (2.2.24)for some small and fixed constant 0< σ < 1. We ansatz 34 ≤|µ j||µ0 j| ≤ 43 for j = 1, . . . ,k.We observe that for some positive number c j we haveλ0 j(t) = c j(−t)− 2n−6( n−2n−6)j−2.With these choices, the expression E j[φ0 j; µ⃗] in (2.2.14) can be decomposed asE j[φ0 j, µ⃗0+ µ⃗1] = µ− n+22j[(µ j µ˙ j−µ0 j µ˙0 j)Zn+1 (y j)+(λn−22j −λn−220 j)pU p−1 (y j)U(0)]= µ−n+22j D j [⃗µ1] (y j, t)+µ− n+22j Θ j [⃗µ1] (y j, t), y j =xµ j(t)where j = 2, · · · ,k andD j [⃗µ1](y j, t) =(µ˙0 jµ1 j+µ0 j µ˙1 j)Zn+1 (y j)+n−22pU p−1 (y j)U(0)λn−220 j(µ1 jµ0 j− µ1, j−1µ0, j−1),Θ j [⃗µ1](y j, t) =µ1 j µ˙1 jZn+1 (y j)+ pU p−1 (y j)λn−220 j O( |µ1 j|µ0 j+|µ1, j−1|µ0, j−1)2.(2.2.25)21Here we have used the fact thatλn−22j −λn−220 j =n−22λn−220 j(µ1 jµ0 j− µ1, j−1µ0, j−1)+λn−220 j O( |µ1 j|µ0 j+|µ1, j−1|µ0, j−1)2. (2.2.26)We also introduce the notationD1 [⃗µ1](y1, t) = (1+µ11)µ˙11Zn+1 (y1) , y1 =xµ1, (2.2.27)which is derived from−∂tU1 = µ−n+221 D1 [⃗µ1]. (2.2.28)2.3 The inner-outer gluing systemWe consider the approximation u∗ = u∗ [⃗µ1] in (2.2.5) built in the previous section and want to find asolution of equation (2.2.1) in the form u= u∗+φ. By Lemma 2.7.2, we have u∗ > 0 when t0 is verynegative. The problem becomesS [u∗+φ] =−φt +∆φ+ pup−1∗ φ+Nu∗ [φ]+S [u∗] = 0 in Rn× (−∞, t0) (2.3.1)whereNu∗ [φ] = |u∗+φ|p−1 (u∗+φ)−up∗ − pup−1∗ φ.We consider the cut-off functions η j,ζ j, j = 1, . . . ,k, defined asη j(x, t) = χ( |x|2Rµ0 j(t))(2.3.2)andζ j(x, t) =χ( |x|Rµ0 j(t))−χ(R |x|µ0 j(t))j = 1, · · · ,k−1,χ( |x|Rµ0k(t))j = k.(2.3.3)We observe that ηiζi = ζi, becauseη j(x, t) ={1 for |x| ≤ 2Rµ0 j(t),0 for |x| ≥ 4Rµ0 j(t),22andζ j(x, t) ={1 for 2R−1µ0 j(t)≤ |x| ≤ Rµ0 j(t),0 for |x| ≥ 2Rµ0 j(t) or |x| ≤ R−1µ0 j(t),j = 1, . . . ,k−1, (2.3.4)ζk(x, t) ={1 for |x| ≤ Rµ0k(t),0 for |x| ≥ 2Rµ0k(t).(2.3.5)Here R is a large constant to be determined later. In fact, we fix R first, then take t0 very negative.We consider functions ϕ j(y, t), j= 1, · · · ,k defined in B8R×(−∞, t0) and a functionΨ(x, t) definedin Rn× (−∞, t0). We look for the φ(x, t) in (2.3.1) of the formφ(x, t) =k∑j=1φ jη j(x, t)+Ψ(x, t) (2.3.6)whereφ j(x, t) = µ− n−22j ϕ j(xµ j(t), t). (2.3.7)Let us substitute φ given by (2.3.6) into equation (2.3.1). We getS [u∗+φ] =k∑j=1η jµ− n+22j(−µ2j ∂tϕ j(y j, t)+∆yϕ j(y j, t)+ pU(y j)p−1ϕ j(y j, t)+µn−22j ζ jpU(y j)p−1Ψ+D j [⃗µ1])−Ψt +∆xΨ+VΨ+B[ϕ⃗ ]+N [ϕ⃗ ,Ψ, µ⃗1]+Eout .Here we denote for ϕ⃗ = (ϕ1, . . . ,ϕk), µ⃗ = (µ1, . . . ,µk) andB[ϕ⃗ ] =k∑j=12∇xη j ·∇xφ j+(−∂tη j+∆xη j)φ j+ p(up−1∗ −U p−1j)φ jη j− µ˙ j ∂∂µ jφ jη j, (2.3.8)N [ϕ⃗ ,Ψ, µ⃗1] = Nu∗(k∑j=1φ jη j+Ψ), V = pup−1∗ −k∑j=1ζ jpU p−1j , (2.3.9)Eout = S [u∗]−k∑j=1µ−n+22j D j [⃗µ1]η j, (2.3.10)where D j [⃗µ1] are defined in (2.2.25) and (2.2.27). We will have that S [u∗+φ] = 0 if the followingsystem of k+1 equations is satisfied.µ2j ∂tϕ j = ∆yϕ j+ pU(y)p−1ϕ j+H j[Ψ, µ⃗1](y, t) in B8R× (−∞, t0), j = 1, . . . ,k, (2.3.11)∂tΨ= ∆xΨ+G [ϕ⃗ ,Ψ, µ⃗1](x, t) in Rn× (−∞, t0), (2.3.12)23whereH j[Ψ, µ⃗1](y, t) = µn−22j ζ j(µ jy)pU(y)p−1Ψ(µ jy, t)+D j [⃗µ1](y, t), (2.3.13)G [ϕ⃗ ,Ψ, µ⃗1](x, t) = VΨ+B[ϕ⃗ ]+N [ϕ⃗ ,Ψ, µ⃗1]+Eout . (2.3.14)In the next sections, we will solve this system in a well-designed topology with a suitable choice ofparameters µ⃗1.2.4 The linear equationsIn order to solve the system (2.3.11)-(2.3.12), we need to study their linear equations respectively.The linear estimates of this section are crucial to the fixed point argument.2.4.1 The linear inner problemFirst, we consider the linear theory of (2.3.11).µ j(t)2∂tϕ = ∆yϕ + pU(y)p−1ϕ +h(y, t), B8R× (−∞, t0). (2.4.1)where µ j(t) ∼ (−t)−α j and R is a sufficiently large constant. We aim to solve (2.4.1) by finding alinear mapping ϕ = ϕ [h] that keeps the spatial decay property of h, provided that certain solvabilitycondition for h is satisfied. Making the change of variablesτ(t) = τ0+ˆ tt0µ j(s)−2ds∼−(−t)2α j+1,where τ0 is a suitably chosen that τ0 ∼−(−t0)2α j+1, transforms (2.4.1) into∂τϕ = ∆yϕ + pU(y)p−1ϕ +h(y,τ), B8R× (−∞,τ0). (2.4.2)In order to solve this equation, we need to know the space h(y,τ) belongs to. This amounts toexamining the decay of H j in (2.3.13). Inspired by the estimate of |H j(y, t)| in Lemma 2.6.1, wedefine the following norms‖h‖inν(τ),2+a := sups<τ0supy∈B8Rν−1(s)〈y〉2+a|h(y,s)|, (2.4.3)‖ϕ‖in,∗ν(τ),a := sups<τ0supy∈B8RR−(n+1−a)ν−1(s)〈y〉n+1|ϕ(y,s)|. (2.4.4)24where 0< a< 1 and ν(τ) : (−∞,τ0)→ R+ is a positiveC1 function satisfyinglimτ→−∞ν(τ) = 0, and ∂τν ∼ν(τ)−τ for τ ≤ τ0. (2.4.5)Lemma 2.4.1. Consider∂τϕ = ∆ϕ + pU p−1ϕ +h(y,τ) in B8R× (−∞,τ0). (2.4.6)For all sufficiently large R> 0, if τ0 = τ0(R) is very negative, ‖h‖inν(τ),2+a <+∞, and h(y,τ) satisfiesˆB8Rh(y,τ)Z j(y)dy= 0 for all τ ∈ (−∞,τ0) (2.4.7)j= 1, . . . ,n+1, where Z j(y) = ∂y jU(y) and Zn+1(y) = n−22 U(y)+y ·∇U(y). Then there exists a linearmappingϕ =T inν(τ)[h] (2.4.8)which solves (2.4.6) and satisfies the estimate‖〈y〉∇yϕ‖in,∗ν(τ),a+‖ϕ‖in,∗ν(τ),a ≤Cinν(τ),a‖h‖inν(τ),2+a, (2.4.9)where Cinν(τ),a is a constant depending on ν(τ) and a.Remark 2.4.1.1. Since we consider radial scheme throughout this chapter,´B8Rh(y,τ)Z j(y)dy = 0,j = 1, . . . ,n, are satisfied automatically.The proof inherits the spirit of [16]. First, we consider the linear problem (2.4.6) in a finite timeregion (s,τ0) and get a uniform estimate independent of the initial time s. Second, we make s go to−∞ and get an ancient solution by the compaction argument, like [17]. We need to use some Liouvilletype theorem to guarantee the uniqueness of the ancient solution derived from this operation, whichdeduces the existence of the desired linear mapping.Since the proof is very similar to the linear theory in [16], we only stress the difference due totaking subsequence as s→−∞. We have to prove no matter what convergent subsequence we choose,the limit function is the same.First, we need the following preparation lemma.Lemma 2.4.2. Given L(u) = ∆u+c(x)u defined in a bounded domain Ω, if L has a positive superso-lution w ∈C2(Ω)∩C(Ω¯), that is L(w)≤ 0 in Ω and w> 0 in Ω¯, then for all ϕ ∈C2(Ω)∩C(Ω¯) with25ϕ = 0 on ∂Ω, the corresponding energyQ(ϕ ,ϕ) :=ˆΩ(|∇ϕ |2− c(x)ϕ 2) dy≥ 0.Proof. Since w> 0 in Ω¯, ∃ψ ∈C2(Ω)∩C0(Ω¯) such that ϕ = wψ . ThenQ(ϕ ,ϕ) =ˆΩ(−wψ2∆w−2wψ∇w ·∇ψ−w2ψ∆ψ− c(x)w2ψ2) dy=ˆΩ[−(∆w+ c(x)w)wψ2−2wψ∇w ·∇ψ−w2ψ∆ψ] dy. (2.4.10)Using the assumption Lw≤ 0 and w> 0, we haveQ(ϕ ,ϕ)≥ˆΩ(−2wψ∇w ·∇ψ−w2ψ∆ψ) dy=ˆΩ[−2wψ∇w ·∇ψ+∇ψ ·∇(w2ψ)] dy= ˆΩ|∇ψ|2w2 dy≥ 0.We take the following typical lemma, whose counterpart is given in Lemma 7.3 in [16], to il-lustrate the difference with the linear theory in [16] due to taking subsequence. Define χM(y) =χ(|y|−M).Lemma 2.4.3. Considerϕτ = ∆ϕ + pU p−1(1−χM)ϕ +h(y,τ) in B8R× (−∞,τ0),ϕ = 0 on ∂B8R× (−∞,τ0), (2.4.11)where ‖h‖ν ,a < +∞, 0 ≤ a < n. If M is a large constant, there exists a very negative constant τ˜0. Ifτ0 ≤ τ˜0, there exists a linear map ϕ∗[h] satisfying (2.4.11) and the following estimate:|ϕ∗[h]|≲M ν(τ)Θ0Ra(|y|)‖h‖ν ,a, (2.4.12)whereΘ0Ra(r) =(1+ r)2−a if 2< a< nlnR if a= 2R2−a if 0≤ a< 2.26Proof. First, we considerϕ sτ = ∆ϕ s+ pU p−1(1−χM)ϕ s+h(y,τ) in B8R× (−∞,τ0),ϕ s = 0 on ∂B8R× (−∞,τ0),ϕ s(·,s) = 0 in B8R.By the same method in Lemma 7.3 in [16], we have|ϕ s∗[h]|≲M ν(τ)Θ0Ra(|y|)‖h‖ν ,a. (2.4.13)Notice this estimate is independent of s. By parabolic estimate, Arzelà-Ascoli theorem and diago-nalization argument, taking s→ −∞, we find a weak solution ϕ∗[h] to (2.4.11) with the followingestimate|ϕ∗[h]|≲M ν(τ)Θ0Ra(|y|)‖h‖ν ,a. (2.4.14)Next, we need to demonstrate this operation is really a mapping. That is, if the operation gives twofunctions ϕ 1∗ [h], ϕ 2∗ [h] due to the different choices of subsequences, we need to prove ϕ 1∗ [h] = ϕ 2∗ [h].In fact, set Φ∗ = ϕ 1∗ [h]−ϕ 2∗ [h]. By (2.4.11), (2.4.14), Φ∗ satisfies∂τΦ∗ = ∆Φ∗+ pU p−1(1−χM)Φ∗ in B8R× (−∞,τ0),Φ∗ = 0 on ∂B8R× (−∞,τ0),|Φ∗|≲M ν(τ)Θ0Ra(|y|)‖h‖ν ,a.(2.4.15)By parabolic regularity theory, Φ∗ is smooth. Multiplying Φ∗ for both sides and integrating by part,we have12∂τˆB8R|Φ∗|2 dx=ˆB8R[−|∇Φ∗|2+ pU p−1(1−χM)Φ2∗] dx≤ 0. (2.4.16)The inequality is due to Lemma 2.4.2 since LM(ϕ) = ∆ϕ + pU p−1(y)(1−χM)ϕ has a positive kernelg2(|y|) given in Lemma 7.3 of [16].By the upper bound of |Φ∗|,ˆB8R|Φ∗|2 dx≲ ‖h‖2ν ,aν2(τ)Rn1 if 2< a< n,ln2(R) if a= 2,R4−2a if 0≤ a< 2.Thus ˆB8R|Φ∗|2 dx→ 0 as τ →−∞, (2.4.17)we have´B8R|Φ∗|2 dx≡ 0, which implies Φ∗ ≡ 0.27By the same argument, we can prove that ϕ∗[h] is a linear mapping. That is, for all functions f , gsatisfying ‖ f‖ν ,a, ‖g‖ν ,a < ∞, we have ϕ∗[ f +g] = ϕ∗[ f ]+ϕ∗[g].Since we aim to find ancient solutions. The initial value given in the linear theory of [16] willdisappear as s→−∞.2.4.2 The linear outer problemWe consider the solution ofψt = ∆xψ+g(x, t), in Rn× (−∞, t0). (2.4.18)It is well-known that the above equation has a solution which is given by Duhamel’s formulaψ(x, t) =T out [g](x, t) :=1(4π) n2ˆ t−∞ds(t− s) n2ˆRne−|x−y|24(t−s) g(y,s)dy (2.4.19)whenever the integral is well-defined.In order to design a topology to solve the outer problem (2.3.12), we define three types of weights.w11(x, t) =|t|−1−σ1+ |x|2+α 1{|x|≤µ¯01}+ |t|−1−σ µ¯n−2−α01 |x|−1−n1{µ¯01≤|x|≤|t| 12 }∼ |t|γ11{|x|≤1}+ |t|γ1 |x|−2−α1{1≤|x|≤µ¯01}+ |t|γ1 µ¯n−2−α01 |x|−1−n1{µ¯01≤|x|≤|t| 12 },w1 j(x, t) =|t|−σµn+220 jλn−220 j1+( |x|µ0 j )2+α1{|x|≤µ¯0 j}∼ µ−20 j |t|γ j1{|x|≤µ0 j}+µα0 j|t|γ j |x|−2−α1{µ0 j≤|x|≤µ¯0 j},(2.4.20)where γ1 =−1−σ and γ j = n−22 α j−1−σ for j = 2, . . . ,k.w21(x, t) = |t|−2σµn2−202 µ−101 |x|2−n1{µ¯02≤|x|≤1},w2 j(x, t) = |t|−2σµn2−20, j+1µ−10 j |x|2−n1{µ¯0, j+1≤|x|≤µ¯0 j}, for j = 2, . . . ,k−1.(2.4.21)andw3(x, t) = R|t|−1−σ |x|2−n1{|x|≥µ¯01}, (2.4.22)where δ > 0 is a small constant.Remark 2.4.3.1. These ad hoc weights are used to control the behavior of G in (2.3.14). Thereare four terms in G , namely B[ϕ⃗ ] (the influence of inner problem), VΨ (linear term on Ψ), Eout(error comes from ansatz U¯ and mainly depends on µ⃗1),N (higher order nonlinear term). Roughly28speaking, w1 j will be used to control B[ϕ⃗ ] in the support of χ j. Specially, w11 is also designed tocontrol the influence of w∗11 in {|x| ≥ µ¯01}. The regions between the support of χ j of B[ϕ⃗ ] is controlledby w2 j. Also notice the support of B[ϕ⃗ ] is contained in {|x| ≤ 4Rµ01}. w3 is designed for controllingEout in {|x| ≥ µ¯01}. See Remark 2.4.7.1 how to control the other three terms.Lemma 2.4.4. For j = 1, . . . ,k, we have the following estimate:T out [w1 j]≲ w∗1 j :=|t|γ j if |x| ≤ µ0 j,|t|γ jµα0 j|x|−α if µ0 j ≤ |x| ≤ µ¯0 j,|t|γ jµα0 j µ¯n−2−α0 j |x|2−n if µ¯0 j ≤ |x| ≤ |t|12 ,|x|2γ∗j+2−n if |x| ≥ |t| 12 .(2.4.23)where γ∗j = (1− n2)α j− α2 (α j−α j−1)−σ for j = 2, . . . ,k and γ∗1 = γ1+(n−2−α)δ .Here γ∗j satisfies |t|γ jµα0 j µ¯n−2−α0 j ∼ |t|γ∗j for simplicity. Approximately, w∗1 j is like a radially non-increasing function about |x| for every fixed t up to a constant multiplicity, that isw∗1 j ∼min{|t|γ j , |t|γ j−αα j |x|−α , |t|γ∗j |x|2−n, |x|2γ∗j+2−n}. (2.4.24)Similarly, we have the following fact.Lemma 2.4.5. We have the following estimates:T out [w21]≲ w∗21 :=|t|−2σ if |x| ≤ µ¯02,|t|−2σ µ¯n−402 |x|4−n if µ¯02 ≤ |x| ≤ 1,|t|−2σ µ¯n−402 |x|2−n if 1≤ |x| ≤ |t|12 ,(|x|2)−2σ−( n2−2)α2 |x|2−n if |x| ≥ |t| 12 .(2.4.25)and for j = 2, . . . ,k−1,T out [w2 j]≲ w∗2 j :=|t|−2σµ1−n20 j if |x| ≤ µ¯0, j+1,|t|−2σµn2−20, j+1µ−10 j |x|4−n if µ¯0, j+1 ≤ |x| ≤ µ¯0 j,|t|−2σµn2−20, j+1µ0, j−1|x|2−n if µ¯0 j ≤ |x| ≤ |t|12 ,(|x|2)−2σ−( n2−2)α j+1−α j−1 |x|2−n if |x| ≥ |t| 12 .(2.4.26)29Lemma 2.4.6. For δ ≤ 12 , we have the following estimate:T out [w3]≲ w∗3 := R|t|−1−σ µ¯4−n01 if |x| ≤ µ¯01,|t|−1−σ |x|4−n if µ¯01 ≤ |x| ≤ |t| 12 ,|t|−σ |x|2−n if |x| ≥ |t| 12 .(2.4.27)Remark 2.4.6.1. Just like (2.4.24), w∗2 j, j = 1, . . . ,k−1, w∗3 are approximate to some non-increasingfunctions about |x| for every fixed t.The proofs of Lemma 2.4.4, 2.4.5 and 2.4.6 are deferred to subsection 2.8 in the section Additionalinformation.For a function h= h(x, t), we define the weighted L∞ norm ‖h‖outα,σ , ‖h‖out,∗α,σ as the following formrespectively.‖h‖outα,σ := inf{K∣∣∣ |h(x, t)| ≤ K( k∑j=1w1 j+k−1∑j=1w2 j+w3)(x, t), Rn× (−∞, t0)}. (2.4.28)‖h‖out,∗α,σ := inf{K∣∣∣ |h(x, t)| ≤ K( k∑j=1w∗1 j+k−1∑j=1w∗2 j+w∗3)(x, t), Rn× (−∞, t0)}. (2.4.29)Using Lemma 2.4.4, 2.4.5 2.4.6 and Lemma 2.7.3-2.7.8, we get the following proposition:Proposition 2.4.7. Suppose that σ ,ε > 0 is chosen small enough, ‖⃗µ1‖σ ≤ 1 and t0 is negativeenough. Then there exists a constant Cout > 0, independent of R and t0, such that ∂tψ = ∆xψ +G [ϕ⃗ ,Ψ, µ⃗1] has a solution T out [G [ϕ⃗ ,Ψ, µ⃗1]] in Rn× (−∞, t0) satisfying‖T out [G [ϕ⃗ ,Ψ, µ⃗1]]‖out,∗α,σ ≤CoutRα−a(1+‖ϕ⃗‖in,∗a,σ +‖Ψ‖out,∗α,σ +(‖ϕ⃗‖in,∗a,σ )p+(‖Ψ‖out,∗α,σ )p)where G is defined in (2.3.14) and T out [g] is given by (2.4.19).Remark 2.4.7.1. There are some subtleties to bound VΨ. Some term in VΨ can not be much smallerthan w1 j in the sense of L∞(see (2.7.37)). Thanks to its narrow support, we could still get the smallnesswhen applying T out to it. The estimate ofN and Eout are straightforward.2.5 Orthogonal equationsIn this section, we deal with the orthogonal equationsˆB8RH j[Ψ, µ⃗1](y, t)Zn+1(y)dy= 0, for j = 1, . . . ,k. (2.5.1)30Lemma 2.5.1. (2.5.1) is equivalent toµ˙11 =M1[Ψ, µ⃗1](t),µ˙1 j+ n−42 α jt µ1 j− n−22 α jt λ0 jµ1, j−1 =M j[Ψ, µ⃗1](t), for j = 2, . . . ,k, (2.5.2)where M j are given in (2.5.4) and (2.5.7).Proof. For j = 1, using (2.3.13) and (2.2.27), (2.5.1) is equivalent toµ˙11 =M1[Ψ, µ⃗1](t), (2.5.3)whereM1[Ψ, µ⃗1](t) =− µn−2211+µ11´B8Rζ1(µ1y)pU(y)p−1Zn+1(y)Ψ(µ1y, t)dy´B8RZ2n+1(y)dy. (2.5.4)For j = 2, . . . ,k, by (2.3.13) and (2.2.25), (2.5.1) is equivalent toµ˙0 jµ1 j(t)+µ0 j µ˙1 j(t)+n−22U(0)´B8RpU p−1(y)Zn+1(y)dy´B8RZ2n+1(y)dyλn−220 j(µ1 jµ0 j− µ1, j−1µ0, j−1)= −µn−22j´B8Rζ j(µ jy)pU(y)p−1Zn+1(y)Ψ(µ jy, t)dy´B8RZ2n+1(y)dy. (2.5.5)Since |Zn+1(y)|≲ 〈y〉2−n,ˆB8RpU p−1(y)Zn+1(y)dy=ˆRnpU p−1(y)Zn+1(y)dy+O(R−2),ˆB8RZ2n+1(y)dy=ˆRnZ2n+1(y)dy+O(R4−n).It follows that−U(0)´B8RpU p−1(y)Zn+1(y)dy´B8RZ2n+1(y)dy= c∗+O(R−2),where c∗ is the positive constant defined in (2.2.18). Notice the fact thatµ˙0 jµ0 j=α j−t ,c∗λn−220 jµ20 j=µ˙0 jµ0 jfor j = 2, . . . ,k.We can simplify (2.5.5) toµ˙1 j+n−42α jtµ1 j− n−22α jtλ0 jµ1, j−1 =M j[Ψ, µ⃗1](t), (2.5.6)31whereM j[Ψ, µ⃗1](t) =−µn−22jµ0 j´B8Rζ j(µ jy)pU(y)p−1Zn+1(y)Ψ(µ jy, t)dy´B8RZ2n+1(y)dy+ µ˙0 j(µ1 jµ0 j− µ1, j−1µ0, j−1)O(R−2).(2.5.7)In order to solve (2.5.2) by the fixed point theorem, we reformulate (2.5.2) as the followingmapping. Let us define S⃗ [Ψ, µ⃗1] = (S1[Ψ, µ⃗1], . . . ,Sk[Ψ, µ⃗1]) whereS1[Ψ, µ⃗1](t) =ˆ t−∞M1[Ψ, µ⃗1](s)ds,S j[Ψ, µ⃗1](t) = (−t)− n−42 α jˆ tt0(−s) n−42 α j(n−22α jsλ0 j(s)S j−1[Ψ, µ⃗1](s)+M j[Ψ, µ⃗1](s))ds,(2.5.8)for j = 2, . . . ,k.For a constant b and a function g(t), we define‖g‖#b := supt≤t0|(−t)bg(t)|. (2.5.9)We introduce the norm about µ⃗1:‖⃗µ1‖σ :=k∑i=1(‖µ˙1i‖#1+αi+σ +‖µ1i‖#αi+σ) , (2.5.10)where σ > 0.Lemma 2.5.2. Suppose Ψ and µ⃗1 satisfy ‖Ψ‖out,∗α,σ < ∞, ‖⃗µ1‖σ ≤ 1, 0 < σ < 1 respectively, when t0is very negative, there exists CS such that‖S⃗ [Ψ, µ⃗1]‖σ ≤CS (‖Ψ‖out,∗α,σ +O(R−2)). (2.5.11)Proof. Note that the support of ζ1 is contained in {R−1µ01 ≤ |x| ≤ 2Rµ01}. By Lemma 2.7.5 and2.7.6, we have |Ψ|≲ (w∗11+w∗12+w∗21)‖Ψ‖out,∗α,σ ≲ |t|−1−σ‖Ψ‖out,∗α,σ . Then using (2.5.4), we have|M1[Ψ, µ⃗1]|≲ |t|−1−σ‖Ψ‖out,∗α,σ . (2.5.12)By (2.5.8), we have‖S˙1[Ψ, µ⃗1]‖#1+σ +‖S1[Ψ, µ⃗1]‖#σ ≲ ‖Ψ‖out,∗α,σ . (2.5.13)Similarly, for j= 2, . . . ,k, the support of ζ j is contained in {R−1µ0 j ≤ |x| ≤ 2Rµ0 j}. By Lemma 2.7.532and 2.7.6, we have|Ψ|≲ (w∗1 j+w∗1, j+1+w∗2 j+w∗2, j−1)‖Ψ‖out,∗α,σ ≲ |t|γ j‖Ψ‖out,∗α,σ ,where w∗1, j+1, w∗2 j are vacuum if j = k.Then using (2.5.7), we have|M j[Ψ, µ⃗1]|≲ |t|−σµ−10 j λn−220 j ‖Ψ‖out,∗α,σ + |t|−1−α j−σO(R−2)≲ |t|−α j−1−σ (‖Ψ‖out,∗α,σ +O(R−2)),(2.5.14)where we have used that µ0 j µ˙0 j = c∗λn−220 j .We will prove‖S˙ j[Ψ, µ⃗1]‖#1+α j+σ +‖S j[Ψ, µ⃗1]‖#α j+σ ≲ ‖Ψ‖out,∗α,σ +O(R−2), (2.5.15)by induction. The case j = 1 has been proved.Suppose we have proved ‖S˙ j−1[Ψ, µ⃗1]‖#1+α j−1+σ + ‖S j−1[Ψ, µ⃗1]‖#α j−1+σ ≲ ‖Ψ‖out,∗α,σ +O(R−2)by induction. Consequently∣∣s−1λ0 j(s)S j−1(s)∣∣ ≲ (−s)−α j−1−σ (‖Ψ‖out,∗α,σ +O(R−2)). Now using(2.5.8) and (2.5.14),|S j[Ψ, µ⃗1]|≲ (−t)− n−42 α jˆ tt0(−s) n−42 α j(−s)−α j−1−σds(‖Ψ‖out,∗α,σ +O(R−2))≲ (−t)−α j−σ (‖Ψ‖out,∗α,σ +O(R−2)),(2.5.16)where we have used σ < 1=min2≤ j≤k{n−62 α j}. Similarly, we can get|S˙ j[Ψ, µ⃗1]|≲ (−t)−1−α j−σ (‖Ψ‖out,∗α,σ +O(R−2)).This completes the induction.2.6 The Schauder fixed point argumentIn this section, we will solve the system (2.3.11)-(2.3.12) by fixed point argument. We need to setup appropriate topologies and operators. Recall (2.4.4), (2.4.3) and (2.4.8) in the previous section.When ν(τ) = (−t)γ jµn−220 j , we write‖〈y〉∇ϕ‖in,∗ν(τ),a+‖ϕ‖in,∗ν(τ),a = ‖ϕ‖in,∗j,a,σ , ‖h‖inν(τ),a = ‖h‖inj,a,σ , T inν(τ) =T inj (2.6.1)for short, where 0< a< 1, γ1 =−1−σ and γ j = n−22 α j−1−σ , j = 2, . . . ,k.33Now we state precisely the topology we are going to use. Suppose σ is small enough and 0 <α < a< 1. Define‖ϕ⃗‖in,∗a,σ :=k∑j=1‖ϕ j‖in,∗j,a,σ . (2.6.2)We will cope with ϕ⃗ , Ψ, µ⃗1 in the topology (2.6.2), (2.4.29) and (2.5.10) respectively.The following lemma justifies why we choose ν(τ) = (−t)γ jµn−220 j .Lemma 2.6.1. For any R> 0 large, there exists t0 negative enough such that for t < t0 one has|H j(x, t)| ≤CH µn−220 j (−t)γ j〈y j〉−4(‖⃗µ1‖σ +‖Ψ‖out,∗α,a ), j = 1, · · · ,k. (2.6.3)Proof. By (2.5.10), we have|D1 [⃗µ1]|≲|µ˙11Zn+1(y1)|≲ (−t)γ1〈y1〉2−n‖⃗µ1‖σ ,|D j [⃗µ1]|≲λ j(t) n−22 (−t)−σ (|Zn+1(y j)|+ 〈y j〉−4)‖⃗µ1‖σ ≲ µn−220 j (−t)γ j〈y j〉−4‖⃗µ1‖σ ,(2.6.4)for j = 2, . . . ,k. By the same estimate in Lemma 2.5.2, we have|ζ jU(y j)p−1µn−22j Ψ|≲ µn−220 j (−t)γ j〈y j〉−4‖Ψ‖out,∗α,a , j = 1, . . . ,k. (2.6.5)We reformulate the inner-outer gluing system and the orthogonal equation into the mapping T⃗ :(ϕ⃗ ,Ψ, µ⃗1) = T⃗ [ϕ⃗ ,Ψ, µ⃗1], (2.6.6)where T⃗ = (T⃗ 1,T 2, T⃗ 3), T⃗ 1 = (T⃗ 11 , . . . , T⃗1k ), T⃗3 = (T⃗ 31 , . . . , T⃗3k ), with the following expressions,T⃗ 1j [Ψ, µ⃗1] = T inj [H j[Ψ, µ⃗1]− c j [Ψ, µ⃗1]Zn+1], j = 1, . . . ,k, (2.6.7)T 2[ϕ⃗ ,Ψ, µ⃗1] = T out [G [ϕ⃗ ,Ψ, µ⃗1]], (2.6.8)T⃗ 3j [Ψ, µ⃗1] =S j[Ψ, µ⃗1], j = 1, . . . ,k. (2.6.9)where c j [Ψ, µ⃗1] (t) = ‖Zn+1‖−2L2(B8R)´B8RH j [Ψ, µ⃗1] (y, t)Zn+1(y)dy. Here T inj in (2.6.7) is obtainedfrom (2.4.8). It is well-defined becauseH j− c jZn+1 satisfies (2.4.7). DenoteBout = {Ψ | ‖Ψ‖out,∗α,σ ≤ R−2ρ}, Bin = {ϕ⃗ | ‖ϕ⃗‖in,∗a,σ ≤ 1}, Bµ1 = {⃗µ1 | ‖⃗µ1‖σ ≤ R−ρ}, (2.6.10)where 0< 2ρ ≤ a−α2 is a small constant.34Proof of Theorem 1.0.4: • Existence part. Fix 0 < α < a < 1. We choose σ ,ε > 0 small enoughsuch that Proposition 2.4.7 holds. Let B =Bin×Bout ×Bµ1 , then we claim that T⃗ maps B to Bprovided taking R large enough and t0 negative enough.First, for any fixed Ψ ∈Bout , µ⃗1 ∈Bµ1 . Applying (2.4.9) with h =H j and ν(τ) = (−t)γ jµn−220 j ,Lemma 2.6.1 implies that‖T⃗ 1[Ψ, µ⃗1]‖in,∗a,σ ≤k∑j=1Cinj,aCH (R−ρ +R−2ρ)≤ 1 (2.6.11)provided taking R large enough.Second, for any fixed ϕ⃗ ∈Bin, µ⃗1 ∈Bµ1 , by Proposition 2.4.7, there exists t0 = t0(R) negativeenough such that‖T 2[ϕ⃗ ,Ψ, µ⃗1]‖out,∗α,σ ≤CoutRα−a(3+2e−ρ)≤ R−2ρ .Third, for any fixed ϕ⃗ ∈Bin, Ψ ∈Bout . It follows from Lemma 2.5.2 that‖T⃗ 3[Ψ, µ⃗1]‖σ ≤CS (R−2ρ +O(R−2))≤ R−ρ . (2.6.12)Therefore T⃗ :B→B.Next, we need to show T⃗ is a compact mapping. Thus for any sequence (ϕ⃗m,Ψm, µ⃗m1 ) ∈ B,where ϕ⃗m = (ϕm1 , . . . ,ϕmk ), µ⃗m1 = (µm1 , . . . ,µmk ), we have to show T⃗ [ϕ⃗m,Ψm, µ⃗m1 ] has a convergentsubsequence. Let us consider first the sequence ϕ˜mj =T inj [H j[Ψm, µ⃗m1 ]−c j [Ψm, µ⃗m1 ]Zn+1]. It satisfies∂τ ϕ˜mj = ∆yϕ˜mj +hmj (y,τ), hmj =H j[Ψm, µ⃗m1 ]− c j [Ψm, µ⃗m1 ]Zn+1.Using Lemma 2.6.1 and interior estimate of parabolic equations, we get that in any compact setK ⊂ B8R × (−∞, t0), we have ϕ˜mj ∈ C1+γ,1+γ2 in K for each fixed γ ∈ (0,1). Thus ϕ˜mj and ∇yϕ˜mjare equi-continuous in K. By Arzelà-Ascoli theorem, going to a subsequence if necessary, ϕ˜mj willconverge uniformly in compact sets of B8R×(−∞, t0). Since ϕ˜mj ∈Bin, then the limit will also belongtoBin.Second, consider Ψ˜m = T out [G [ϕ⃗m,Ψm, µ⃗m1 ]]. Since G [ϕ⃗m,Ψm, µ⃗m1 ] are uniformly bounded, Ψ˜mhave a uniform C1+γ,1+γ2 bound in compact sets of Rn× (−∞, t0). By Arzelà-Ascoli theorem, Ψ˜m (upto a subsequence) converges uniformly to a function Ψ˜ ∈Bout .Third, consider S [Ψm, µ⃗m1 ]. Note that (2.5.4) and (2.5.7) imply M1[Ψm, µ⃗m1 ] and M j[Ψm, µ⃗m1 ] areC1(−∞, t0). ThusS [Ψm, µ⃗m1 ] ∈C2(−∞, t0). ConsequentlyS [Ψm, µ⃗m1 ] has a convergent subsequenceinBµ1 .By Schauder’s fixed point theorem, T⃗ :B→B has a fixed point (ϕ⃗ ,Ψ, µ⃗1). Then (2.6.9) impliesc j[Ψ, µ⃗1] = 0 and consequently (ϕ⃗ ,Ψ, µ⃗1) makes (2.3.11) and (2.3.12) hold.35We have constructed a bubble tower solution for (1.0.1). Recall that u = U¯ +φ0+∑kj=1φ jη j +Ψ. One can see (1.0.21), (2.2.6) and (2.3.6) for their respective definitions. We shall prove that U¯dominates in the sum in the sense of L∞ and H1 when t0 is negative enough.• Convergence in L∞(Rn) and positiveness. It is easy to see the first approximation and innersolutions are smaller than U¯ . Namely, by Lemma 2.7.2, we know |φ0|≲ |t|−εU¯ . For j = 1, . . . ,k, by(2.7.12) and (2.6.10),∣∣∣∣µ− n−22j ϕ j( xµ j , t)η j∣∣∣∣≲ |t|γ jRn+1−a〈y j〉−n−11{|x|≤4Rµ0 j}∼ µn−22j |t|γ jRn+1−a〈y j〉−31{|x|≤4Rµ0 j}U j(x, t)≲ |t|−σU j(x, t).The solution Ψ in the outer problem is more involved to estimate. First, by (2.6.10), we have|Ψ|≲k∑j=1w∗1 j+k−1∑j=1w∗2 j+w∗3.We will estimate it in several regions. We will use Lemma 2.7.5 and 2.7.6 repeatedly in the followingargument.In {|x| ≤ µ¯0k}, we have|Ψ|≲ w∗1k+w∗2,k−1 ≲ |t|γk ∼ |t|γkµn−22k 〈yk〉n−2Uk(x, t)≲ |t|−σUk(x, t).In {µ¯0,i+1 ≤ |x| ≤ µ¯0i}, i= 2, . . .k−1, we have|Ψ|≲ w∗1i+w∗1,i+1+w∗2i+w∗2,i−1∼ (w∗1i+w∗1,i+1+w∗2i+w∗2,i−1)µn−22i 〈yi〉n−2Ui(x, t)≲ |t|−σUi(x, t).This is becausew∗1iµn−22i 〈yi〉n−2 ≲ |t|γiµn−22i (µ¯0iµi)n−2 ≲ |t|−σ ,w∗1,i+1µn−22i 〈yi〉n−2 ≲ |t|γi+1 µ¯n−2i+1 |x|2−nµn−22i 〈yi〉n−2 ≲ |t|γi+1 µ¯n−2i+1 µn−22i µ¯2−n0,i+1 ≲ |t|−σ ,w∗2iµn−22i 〈yi〉n−2 ∼ |t|−2σµn2−2i+1 µn2−2i |x|4−n〈yi〉n−2≲ |t|−2σµn2−2i+1 µn2−2i(µ¯4−n0,i+1+ µ¯20iµ2−ni)≲ |t|−2σ ,w∗2,i−1µn−22i 〈yi〉n−2 ≲ |t|−2σµ1− n2i−1 µn−22i (µ¯0iµi)n−2 ≲ |t|−2σ .In {µ¯02 ≤ |x| ≤ µ¯01}, similarly, we have|Ψ|≲ (w∗11+w∗12+w∗21)(1+ |x|)n−2U1(x, t)≲ |t|−σU1(x, t),36since for δ ≤ (n−2−α)−1,w∗11(1+ |x|)n−2 ≲ |t|γ1〈x〉−α(1+ |x|)n−2 ≲ |t|γ1 |t|δ (n−2−α) = |t|−σ ,w∗12(1+ |x|)n−2 ≲ |t|γ2 µ¯n−22 |x|2−n(1+ |x|)n−2 ≲ |t|−σ ,w∗21(1+ |x|)n−2 ≲ |t|−2σ µ¯n−42 |x|2−nmin{1, |x|2}(1+ |x|)n−2 ≲ |t|−2σ .In {µ¯01 ≤ |x| ≤ |t| 12 },|Ψ|≲ w∗11+w∗3 ≲ |t|−σ2U1(x, t).In {|x| ≥ |t| 12 },|Ψ|≲ w∗3 ≲ |t|−σ2U1(x, t).Therefore |Ψ|≲ |t|− σ2 U¯ .Combining the above analysis, we have u = U¯(1+O(|t|−ε)) > 0. By the parabolic regularitytheory, we improve the regularity of u to be smooth.• Convergence in H1(Rn). The solution we construct is u= U¯+φ0+φ , where φ0 is from (2.2.6)and φ is from (2.3.6). Set φ¯ = φ0 +φ . We have already proved |φ¯| ≲ |t|−εU¯ . Formally, we canexpect |∇xφ¯|≲ |t|−ε |∇xU¯ |. Note that φ¯ satisfies∂t φ¯ = ∆xφ¯+ f (x, t) (2.6.13)where f (x, t) = (U¯+ φ¯)p−∑kj=1U pj −∑kj=1 ∂tU j. It follows that| f (x, t)|≲|t|−εµ−n+220k 〈y0k〉−n−2+µ− n20k |µ˙0k|〈y0k〉2−n if |x| ≤ µ¯0k,|t|−εµ−n+220 j 〈y0 j〉−n−2+µ− n20 j |µ˙0 j|〈y0 j〉2−n if µ¯0, j+1 ≤ |x| ≤ µ¯0 j, j = 2, . . . ,k−1,|t|−ε〈x〉−n−2+ |t|−1−σ 〈x〉2−n if |x| ≥ µ¯02,∼|t|−εµ−n+220k if |x| ≤ µ0k,|t|−εµ−n+220 j if µ¯0, j+1 ≤ |x| ≤ µ0 j, j = 2, . . . ,k−1,|t|−εµn+220 j |x|−n−2 if µ0 j ≤ |x| ≤ µ¯0 j, j = 2, . . . ,k,|t|−ε if µ¯02 ≤ |x| ≤ 1,|t|−ε |x|−n−2+ |t|−1−σ |x|2−n if 1≤ |x|.since |t|−εµ−n+220 j 〈y0 j〉−n−2 ≳ µ− n20 j |µ˙0 j|〈y0 j〉2−n in {µ¯0, j+1 ≤ |x| ≤ µ¯0 j}, j = 2, . . . ,k when ε is small.By the similar argument about uniqueness in Corollary 2.4.3, we knowφ¯ =T out [ f ].37Then|∇xφ¯|≲T d [| f |], (2.6.14)whereT d [g] :=ˆ t−∞ds(t− s) n2+1ˆRne−|x−y|24(t−s) |x− y|g(y,s)dyds.Claim:|∇xφ¯|≲ |t|−εn∑j=1µ−n20 j 〈y0 j〉1−n. (2.6.15)Notice µ−n20 j 〈y0 j〉1−n is approximate to |∇xU j(x, t)|. Once we complete the proof of (2.6.15), it isstraightforward to have ‖u(·, t)−U¯‖H1(Rn) = O(|t|−ε).By Lemma 2.8.6, for j = 2, . . . ,k−1, we getT d [|t|−εµ−n+220 j 1{µ¯0, j+1≤|x|≤µ0 j}]≲|t|−εµ−n20 j if |x| ≤ µ0 j,|t|−εµn2−10 j |x|1−n if µ0 j ≤ |x| ≤ |t|12 ,(|x|2)−ε−α j( n2−1)|x|1−n if |t| 12 ≤ |x|.≲ |t|−εµ−n20 j 〈y0 j〉1−n.Similarly,T d [|t|−εµ−n+220k 1{|x|≤µ0k}]≲ |t|−εµ− n20k 〈y0k〉1−n,T d [|t|−ε1{µ¯02≤|x|≤1}]≲ |t|−εµ− n201 〈y01〉1−n.For j = 2, . . . ,k,T d [|t|−εµn+220 j |x|−n−21{µ0 j≤|x|≤µ¯0 j}]≲|t|−εµ−n20 j if |x| ≤ µ0 j,|t|−εµn2−10 j |x|1−n if µ0 j ≤ |x| ≤ |t|12 ,(|x|2)−ε−α j( n2−1)|x|1−n if |x| ≥ |t| 12 ,≲ |t|−εµ−n20 j 〈y0 j〉1−n.Similarly,T d [|t|−ε |x|−n−21{1≤|x|≤|t| 12 }]≲ |t|−εµ−n201 〈y01〉1−n.38The left part can be transformed into the estimate in Appendix.T d [[|t|−ε |x|−n−21{|t| 12≤|x|}+ |t|−1−σ |x|2−n1{1≤|x|}]]≲ˆ t−∞ds(t− s) n+12ˆRne−|x−y|28(t−s)[(−s)−ε |y|−n−31{(−s) 12≤|y|}+(−s)−1−σ |y|1−n1{1≤|y|}]|y|dyds≲(|t|−1−ε + |t|−σ)µ− n201 〈y01〉1−n,whose estimate process is similar to the convolution of Gaussian kernel in Rn+1.This completes the proof of (2.6.15).2.7 Additional information 1: Estimates for the data in the outerproblemWe will prove Proposition 2.4.7 in this section. Throughout this section, we assume ‖Ψ‖out,∗α,σ +‖ϕ⃗‖in,∗a,σ < ∞, ‖⃗µ1‖σ ≤ 1.The parameters are determined in the following order. First, we choose R as a large fixed positiveconstant. Second, we choose σ > 0 small. Third, we choose δ > 0 small. Fourth, we choose ε > 0small. Finally, we take t0 very negative such that µ j ∼ µ0 j, for j = 1, . . . ,k, µ˙ j ∼ µ˙0 j for j = 2, . . . ,k.We introduce the notation y j = x/µ j, y¯ j = x/µ¯ j, y0 j = x/µ0 j, y¯0 j = x/µ¯0 j for j = 1, . . . ,k. Onereadily sees that |y j| ∼ |y0 j|, |y¯ j| ∼ |y¯0 j| for j = 1, . . . ,k.Lemma 2.7.1. Consider the U j defined in (1.0.22). For j = 1, · · · ,k−1, one hasU j <U j+1 in {|x|< µ¯ j+1} and U j >U j+1 in {|x|> µ¯ j+1}. (2.7.1)In {|x| ≤ µ¯0k},Uk ≳Uk−1 >Uk−2 > · · ·>U1. (2.7.2)In {|x| ≥ µ¯02},U1 ≳U2 >U3 > · · ·>Uk. (2.7.3)In {µ¯0, j+1 ≤ |x| ≤ µ¯0 j}, j = 2, . . . ,k−1,U j ≳U j+1 >U j+2 > · · ·>Uk, U j ≳U j−1 >U j−2 > · · ·>U1. (2.7.4)MoreoverU j+1U j∼ λ−n−22j+1 〈y j+1〉−(n−2)1{|x|≤µ0 j}+λn−22j+11{|x|>µ0 j} for j = 1, . . . ,k−1, (2.7.5)39U j−1U j∼ λn−22j 〈y j〉n−21{|x|≤µ0, j−1}+λ− n−22j 1{|x|>µ0, j−1} for j = 2, . . . ,k. (2.7.6)Proof. (2.7.1)-(2.7.4) follow from that U j+1U j is strictly decreasing about |x| andU j+1U j(µ¯ j+1) = 1. Up toa multiplicity of the constant αn,U j = µ2−n2j (1+ |y j|2)2−n2 andU j+1 =µn−22j+1(µ2j+1+ |x|2)n−22=µn−22j+1µ2−nj(λ 2j+1+ |y j|2)n−22, (2.7.7)thenU j+1U j= λn−22j+1(1+ |y j|2) n−22(λ 2j+1+ |y j|2)n−22∼ λ−n−22j+1 〈y j+1〉−(n−2)1{|x|≤µ0 j}+λn−22j+11{|x|>µ0 j}, (2.7.8)for j = 1, . . . ,k−1. This finishes the proof of (2.7.5). Similarly,U j−1U j= λ−n−22j(λ 2j + |y j−1|2)n−22(1+ |y j−1|2) n−22∼ λn−22j 〈y j〉n−21{|x|≤µ0, j−1}+λ− n−22j 1{|x|>µ0, j−1} (2.7.9)for j = 2, . . . ,k. Then (2.7.6) holds.Lemma 2.7.2. Consider φ0 defined in (2.2.6). One has |φ0|≲k∑i=2λiUiχi.Proof. By (2.2.6) and (2.2.22), we have|φ0|≲k∑i=2µ−n−22i−1 〈yi〉−2χi. (2.7.10)It follows from (2.2.4) that the support of χi are disjoint. More precisely, the support of χi is containedin {λ12i+1 ≤ |yi| ≤ λ− 12i }. It is easy to verify that µ− n−22i−1 〈yi〉−2 ≲ λiUi in this set.Lemma 2.7.3. For 0< α < a< 1, there exists R large enough and t0 negative enough such that B[ϕ⃗ ]defined in (2.3.8) satisfies‖B[ϕ⃗ ]‖outα,σ ≲ Rα−a‖ϕ⃗‖in,∗a,σ (2.7.11)Proof. By the definition (2.6.1) and (2.4.4),〈y j〉|∇y jϕ j(y j, t)|+ |ϕ j(y j, t)|≲ |t|γ jµn−220 j Rn+1−a〈y j〉−n−1‖ϕ j‖in,∗j,a,σ . (2.7.12)40• First, using (2.3.7)|µ˙ j ∂φ j∂µ jη j|=∣∣∣∣µ˙ jµ− n2j (n−22 ϕ j(y j, t)+ y j ·∇y jϕ j(y j, t))η j∣∣∣∣≲ |µ˙ j|µ−1j (−t)γ jRn+1−a〈y j〉−n−1η j‖ϕ j‖in,∗j,a,σ≲ |t0|−εRn+1−aw1 j‖ϕ j‖in,∗j,a,σ .(2.7.13)Here we choose ε small such that |µ˙ jµ j| ≲ |t|−ε for j = 1, . . . ,k. We have used (2.4.20) in the laststep.• Second, (2.7.12) implies that|φ j(x, t)|≲ |t|γ jR−a‖ϕ j‖in,∗j,a,σ for 2R≤ |y j| ≤ 4R. (2.7.14)Using (2.3.2), we obtain|∆η jφ j|≲ (Rµ j)−2|t|γ jR−a‖ϕ j‖in,∗j,a,σ1{2R≤|y j|≤4R} ≲ Rα−aw1 j‖ϕ j‖in,∗j,a,σ . (2.7.15)Similarly, we have|∇xη j ·∇xφ j|≲ R−1µ−1j |t|γ jµ−1j R−1−a‖ϕ j‖in,∗j,a,σ1{2Rµ j≤|x|≤4Rµ j} ≲ Rα−aw1 j‖ϕ j‖in,∗j,a,σ (2.7.16)and|∂tη jφ j|≲ R−1µ−2j |µ˙ j||t|γ jR−a‖ϕ j‖in,∗j,a,σ1{2R≤|y j|≤4R} ≲ R1+α−a|µ˙ j|w1 j‖ϕ j‖in,∗j,a,σ≲ R1+α−a|t0|−εw1 j‖ϕ j‖in,∗j,a,σ ,(2.7.17)when |µ˙ j|≲ |t|−ε for j = 1, . . . ,k.• Third, to estimate∣∣∣p(up−1∗ −U p−1j )φ jη j∣∣∣. We only give calculation details for j = 2, . . . ,k−1since the case j = 1 and j = k can be dealt with similarly. Consider it in {µ¯0, j+1 ≤ |x| ≤ 4Rµ0 j}. ByLemma 2.7.1 and Lemma 2.7.2,U−1j(∑i 6= jUi+φ0)≥ 2−1U−1j(∑i6= jUi)−λ12j >−12,when t0 is very negative. Consequently u∗ > 12U j. Thus by the mean value theorem and (2.7.5) and(2.7.6),|up−1∗ −U p−1j |≲U p−1j(U j+1U j+U j−1U j+λ j)≲ µ−2j 〈y j〉−4(λ−n−22j+1 〈y j+1〉2−n1{µ¯0, j+1≤|x|≤µ0 j}+λn−22j+1 +λn−22j 〈y j〉n−2+λ j).(2.7.18)41Therefore, using |φ j|≲ |t|γ jRn+1−a‖ϕ j‖in,∗j,a,σ ,∣∣∣p(up−1∗ −U p−1j )φ jη j∣∣∣1{µ¯0, j+1≤|x|≤4Rµ j}≲ Rn+1−aµ−2j |t|γ jλ− n−22j+1 〈y j+1〉2−n1{µ¯0, j+1≤|x|≤µ0 j}‖ϕ j‖in,∗j,a,σ+Rn+1−a(λn−22j+1 +λn−22j Rn−2+λ j)w1 j‖ϕ j‖in,∗j,a,σ≲ Rn+1−a|t0|−ε(w2 j+w1 j)‖ϕ j‖in,∗j,a,σ .(2.7.19)Here we have used the following fact.Rn+1−aµ−2j |t|γ jλ− n−22j+1 〈y j+1〉2−n1{µ¯0, j+1≤|x|≤µ0 j}∼ Rn+1−aλ j+1λn2−1j |t|σ |t|−2σµn2−2j+1 µ−1j |x|2−n1{µ¯0, j+1≤|x|≤µ0 j}≲ Rn+1−a|t0|−εw2 j.In {µ¯0,m+1 ≤ |x| ≤ µ¯0m}, m = j+ 1, · · · ,k, one has |up−1∗ −U p−1j | ≲ U p−1m ∼ µ−2m 〈ym〉−4 and|φ j|≲ (−t)γ jRn+1−a‖ϕ j‖in,∗j,a,σ . Then it is easy to see∣∣∣p(up−1∗ −U p−1j )φ jη j∣∣∣≲ Rn+1−a|t|γ j−γmw1m‖ϕ j‖in,∗j,a,σ ≲ Rn+1−a|t0|−εw1m‖ϕ j‖in,∗j,a,σ .Taking t0 very negative such that |t0|−ε < Rα−n−1, we obtain (2.7.11).Recall Eout defined in (2.3.10). We reorganize the terms as the following.Eout = E¯11+ E¯2+ E¯3+ E¯4+ E¯5 (2.7.20)where E¯11 is defined in (2.2.13), andE¯2 =µ− n+221 D1 [⃗µ1](1−η1)+k∑j=2µ−n+22j D j [⃗µ1](χ j−η j)+k∑j=2µ−n+22j Θ j [⃗µ1]χ j (2.7.21)E¯3 =k∑j=2p(U¯ p−1−U p−1j )φ0 jχ j (2.7.22)E¯4 =k∑j=2(2∇xφ0 j ·∇xχ j+∆xχ jφ0 j)−k∑j=2∂t (φ0 jχ j) (2.7.23)E¯5 =NU¯ [φ0]. (2.7.24)Lemma 2.7.4. There exist σ ,ε > 0 small and t0 negative enough, such thatEout ≲ R−1(k∑j=1w1 j+k−1∑j=1w2 j+w3).42Proof. • Estimate of E¯2. Consider the first term in E¯2. The support of 1−η1 is {|y01| ≥ 2R}. Sincewe assume ‖⃗µ‖σ < 1, one has |µ˙11| ≤ ‖⃗µ1‖σ |t|−1−σ ≤ |t|−1−σ . Then using (2.2.27)|µ−n−221 D1 [⃗µ1](1−η1)|≲ |t|−1−σ |x|2−n1{|x|≥2R} ≲ R4+α−nw11+R−1w3 ≲ R−1 (w11+w3) .For j ≥ 2, the support of χ j−η j is contained in {|x| ≤ µ¯0, j+1}∪{2Rµ0 j ≤ |x| ≤ µ¯0 j}. In the firstset (it is vacuum if j = k), one has |χ j−η j| ≤ χ(2y¯0, j+1) and 〈y0 j〉 ∼ 1. It follows from (2.6.4) that∣∣∣∣µ− n+22j D j [⃗µ1](χ j−η j)∣∣∣∣≲ µ−2j |t|γ jχ(2y¯0, j+1)≲ |t|−εw1, j+1.In the second set, since |y0 j| ≥ 2R, then∣∣∣∣µ− n+22j D j [⃗µ1](χ j−η j)∣∣∣∣≲ µ−2j |t|γ j |y j|−4 ≲ R−1w1 j.It is straightforward to have|µ−n+22j Θ j [⃗µ1]χ j|≲ |t|−σµ−2j |t|γ j〈y j〉−4χ j ≲ |t|−σw1 j.• Estimate of E¯3. In the support of χ j, by Lemma (2.7.1), we have|U¯ p−1−U p−1j |≲U p−1j (U j+1U j+U j−1U j).It follows from (2.2.7) and (2.2.22) that φ0 j ≤ |t|γ j+σ 〈yi〉−2. Using (2.7.18), similar to (2.7.19), weget|E¯3|≲k∑j=2λn−22j |t|σw1 j+k−1∑j=1λ j+1λn−22j |t|2σw2 j ≲ |t0|−ε(k∑j=2w1 j+k−1∑j=1w2 j).• Estimate of E¯4. Notice∣∣∇xχ j∣∣≲ µ¯−10 j 1{ 12 µ¯0 j<|x|<µ¯0 j}+ µ¯−10, j+11{ 12 µ¯0, j+1<|x|<µ¯0, j+1}.∣∣∆xχ j∣∣≲ µ¯−20 j 1{ 12 µ¯0 j<|x|<µ¯0 j}+ µ¯−20, j+11{ 12 µ¯0, j+1<|x|<µ¯0, j+1}.By (2.2.22), one has |φ0 j|≲ |t|γ j+σ 〈y j〉−2, |∇xφ0 j|≲ λn−220 j µ− n2j 〈y j〉−3 ∼ µ−1j |t|γ j+σ 〈y j〉−3, then|∇xφ0 j ·∇xχ j|≲ |t|σ µ jµ¯ jw1 j+ |t|γ j−γ j+1+σ(µ j+1µ j) 1−α2w1, j+1 ≲ |t0|−ε(w1 j+w1, j+1).43|φ0 j∆xχ j|≲ |t|σ µ jµ¯ jw1 j+ |t|n−22 (α j−1−α j)+ α2 (α j+1−α j)+σw1, j+1 ≲ |t0|−ε(w1 j+w1, j+1).where we have used that n−22 (α j−1−α j)+ α2 (α j+1−α j) = ( αn−6 −1)(n−2n−6) j−1 ≤ αn−6 −1.For ∂t(φ0 jχ j), we have|∂t(φ0 j)χ j|+ |φ0 j∂tχ j|≲ |t|γ j+σ−1〈y j〉−2 (χ j+ |∇xχ j|)≲ |t|σ−1µ2−α2j µα2j−1w1 j ≲ |t0|−εw1 j.• Estimate of E¯5. It follows from (2.2.10), (2.7.10) and p ∈ (1,2) that|NU¯ [φ0]|≲ |φ0|p ≲k∑j=2|t|(γ j+σ)p〈y j〉−2pχ pj ∼k∑j=2|t|2(α j−1−α j)+σ 〈y j〉2+α−2pµ−2j |t|γ j〈y j〉−2−αχ pj .If 2p≥ 2+α , it is easy to see |NU¯ [φ0]|≲ |t0|−ε ∑kj=2w1 j.If 2p< 2+α ,|NU¯ [φ0]|≲k∑j=2|t| 2−α+2p2 (α j−1−α j)+σµ−2j |t|γ j〈y j〉−2−αχ pj ≲ |t0|−εk∑j=2w1 j.• Estimate of E¯11. Regrouping the terms in (2.2.13), one obtainE¯11 =k∑j=2pU p−1j[∑l 6= jUl−U j−1(0)]χ j+k∑j=2[U¯ p−k∑i=1U pi − pU p−1j ∑l 6= jUl]χ j+(−k∑j=2(1−χ j)∂tU j)+[U¯ p−k∑j=1U pj](1−k∑i=2χi):=J1+ J2+ J3+ J4.(2.7.25)Claim:E¯11(x, t)≲ |t0|−ε(k∑j=1w1 j+k−1∑j=1w2 j+w3).1. Estimate of J1.J1 =k∑j=2pU p−1j(∑l 6= j, j−1Ul)χ j+k∑j=2pU p−1j (U j−1−U j−1(0))χ j.We will bound each term in the above equation. Fix j ≥ 2. If i≤ j−2,|pU p−1j Uiχ j|≲ µ−2j 〈y j〉−4µ− n−22j−2 χ j ≲ |t|−εw1 j.44If i≥ j+1, by Lemma 2.7.1|pU p−1j Uiχ j|≲U pjU j+1U jχ j≲ µ−n+22j 〈y j〉−n−2(λ−n−22j+1 〈y j+1〉2−n1{µ¯0, j+1≤|x|≤µ0 j}+λn−22j+1)χ j≲ µn−22j+1µ−2j |x|2−n1{µ¯0, j+1≤|x|≤µ0 j}+(λ j+1λ j)n−22 |t|σw1 j≲ |t|−ε(w2 j+w1 j)(2.7.26)when we choose σ small first and then chose ε small enough. Using |U j−1−U j−1(0)|χ j ≲µ−n−22j−1 λ j, we have|pU p−1j (U j−1−U j−1(0))χ j|≲ µ−2j 〈y j〉−4µ− n−22j−1 λ jχ j ≲ |t|−εw1 j. (2.7.27)2. Estimate of J2. By Lemma (2.7.1), we have∣∣∣∣∣U¯ p− k∑i=1U pi − pU p−1j ∑l 6= jUl∣∣∣∣∣χ j ≲ (U pj−1+U pj+1)χ j.Therein,U pj−1χ j ∼ |t|n+22 α j−1χ j ≲ |t| n+22 α j−1µ2j |t|−γ j(µ¯ jµ j)2+αw1 jχ j∼ ( µ jµ j−1)2−α2 |t|σw1 jχ j ≲ |t|−εw1 j,andU pj+1χ j ∼ µn+22j+1|x|−2−nχ j ≲ |t|2σµ3j+1µ j|x|−4w2 jχ j ≲ |t|−εw2 j,when we take σ small first and then take ε small enough.3. Estimate of J3. For j = 2, . . . ,k, notice that|∂tU j|= |µ˙ jµ−n2j Zn+1(y j)|≲ µ−2j µ− n−22j−1 〈y j〉2−n. (2.7.28)The support of 1−χ j is contained in {|x| ≤ µ¯0, j+1}∪{ 12 µ¯0 j ≤ |x|< µ¯0 j}∪{µ¯0 j ≤ |x|}. In thefirst set, it is easy to see 1−χ j = χ(2|x|/µ¯0, j+1), then|(1−χ j)∂tU j|≲ (µ j+1µ j )2−α2 (µ jµ j−1)n−22 |t|σw1, j+1χ(2|x|/µ¯0, j+1)≲ |t|−εw1, j+1.45In the second set,|∂tU j|1{ 12 µ¯0 j≤|x|<µ¯0 j} ≲(µ jµ j−1) n−4−α2|t|σw1 j1{ 12 µ¯0 j≤|x|<µ¯0 j} ≲ |t0|−εw1 j.In the third set, we split it further to be {µ¯0 j ≤ |x|}= ∪ jm=2{µ¯0m ≤ |x| ≤ µ¯0,m−1}∪{µ¯01 ≤ |x|}.Since |y j| is very large in the third set, (2.7.28) implies |∂tU j| ≲ µn−4j µ− n−22j−1 |x|2−n. Note thatµn−4j µ− n−22j−1 decreases about j up to some constant multiplicity. Then in {µ¯0m ≤ |x| ≤ µ¯0,m−1},m= 2 . . . , j,|∂tU j|≲ µn−4m µ−n−22m−1 |x|2−n ≲ |t|−εw2,m−1.In {µ¯01 ≤ |x|}, we have|∂tU j|≲ µn−42 |x|2−n ≲ |t|−εw3.4. Estimate of J4. Recall the definition of χi in (2.2.4), we have the support of J4 is contained inthe set ∪km=3{12 µ¯0m ≤ |x| ≤ µ¯0m}∪{ 12 µ¯02 ≤ |x|}.In {12 µ¯0m ≤ |x| ≤ µ¯0m}, for m = 3, . . . ,k, by Lemma 2.7.1, one has Um ∼Um−1 ∼ µ− n−22m−1 Uifor i 6= m,m−1. Therefore|J4|1{ 12 µ¯0m≤|x|≤µ¯0m} ≲ µ− n+22m−1 1{ 12 µ¯0m≤|x|≤µ¯0m} ≲ (µm−1µm)α−22 |t|σw1m ≲ |t0|−εw1m.In {12 µ¯02 ≤ |x|}, by Lemma 2.7.1, when σ < (n−6)−1,|J4|≲U p−11 U2∼ µn−222(|x|2−n1{ 12 µ¯02≤|x|≤µ¯02}+ |x|2−n1{µ¯02≤|x|≤1}+ |x|−2−n1{1≤|x|≤µ¯01}+ |x|−2−n1{µ¯01≤|x|})≲ |t0|−ε (w12+w21+w11+w3) .Lemma 2.7.5. There exist σ > 0 small enough and t0 negative enough, such that for t < t0,1. In {|x| ≤ µ¯0i}, i= 1, . . . ,k, we have w∗1 j ≲ w∗1i for j = 1, . . . , i−1 (it is vacuum if i= 1)2. In {µ¯0i ≤ |x|}, i= 1, . . . ,k, we have w∗1 j ≲ w∗1i for j = i+1, . . . ,k (it is vacuum if i= k).3. In {|x| ≤ µ¯0 j}, w∗3 ≲ w∗1 j for j = 1, . . . ,k. In {|x| ≥ |t|12 }, w∗3 ≳ w∗1 j for j = 1, . . . ,k whenδ ≤ (n−2−α)−1.46Consequently,k∑j=1w∗1 j+w∗3 ≲w∗1k if |x| ≤ µ¯0k,w∗1i+w∗1,i+1 if µ¯0,i+1 ≤ |x| ≤ µ¯0i, i= 1, · · · ,k−1,w∗11+w∗3 if µ¯01 ≤ |x| ≤ |t|12 ,w∗3 if |x| ≥ |t|12 .(2.7.29)Proof. (1) For j= 1, . . . , i−1, in {|x| ≤ µ¯0i}, we have w∗1 j(x, t) = |t|γ j ≤ |t|γi−1 and w∗1i ≥ |t|γiµαi µ¯−αi .It is easy to verify |t|γi−1 ≲ |t|γiµαi µ¯−αi if α < n−6.(2) For j = i+ 1, . . . ,k, in {µ¯0i ≤ |x| ≤ |t| 12 }, w∗1 j(x, t) = |t|γ jµαj µ¯n−2−αj |x|2−n ∼ |t|γ∗j |x|2−n ≤|t|γ∗i |x|2−n ∼ w∗1i(x, t), because γ∗j is strictly decreasing on j, i.e.γ∗1 > γ∗2 > · · ·> γ∗k .In {|x| ≥ |t| 12 }, we have w∗1 j ≲ w∗1i by the same reason.(3) Due to (1), we only need to check w∗3 ≲ w∗11 in {|x| ≤ µ¯01}. It is straightforward to haveR|t|−1−σ+δ (4−n) ≲ |t|−1−σ 〈x〉−α in {|x| ≤ µ¯01}. Due to (2), in {|x| ≥ |t| 12 }, we only need to checkw∗3 ≳ w∗11, which is easy to get when δ ≤ (n−2−α)−1.Lemma 2.7.6. There exists t0 negative enough such that1. In {µ¯0,i+1 ≤ |x|}, we have w∗2 j ≲ w∗2i for j = i+ 1, . . . ,k− 1 (it is vacuum if i = k,k− 1). In{|x| ≤ µ¯0i}, we have w∗2,i−1 ≳ w∗2 j for j = 1, . . . , i−2 (it is vacuum if i= 1,2).2. In {|x| ≥ µ¯01}, w∗2 j ≲ w∗3 for j = 1, . . . ,k−1.Consequentlyk−1∑j=1w∗2 j ≲w∗2,k−1 if |x| ≤ µ¯0k,w∗2i+w∗2,i−1 if µ¯0,i+1 ≤ |x| ≤ µ¯0i, for i= 2, . . . ,k−1,w∗21 if µ¯02 ≤ |x| ≤ µ¯01,w∗3 if |x| ≥ µ¯01.(2.7.30)Proof. (1) In {µ¯0,i+1 ≤ |x| ≤ µ¯0i}, it follows from (2.4.26) that w∗2 j = |t|−2σµn2−20, j+1µ0, j−1|x|2−n forj = i+1, . . . ,k−1 and w∗2i = |t|−2σµn2−20,i+1µ−10i |x|4−n.w∗2iw∗2 j≳ (µ0,i+1µ0,i+2)n2−2 µ0,i+1µ0i∼ |t|( n2−2)(αi+2−αi+1)−(αi+1−αi) ≳ 147where we have used that (n2 −2)(αi+2−αi+1)− (αi+1−αi) = (n−4)2+4(n−6)2 (n−2n−6)i−1. Thus w∗2 j ≲ w∗2i. Itis easy to see that w∗2 j ≲ w∗2i also holds in {|x| > µ¯0i}. This deduces the first part. The second partholds obviously by (2.4.26).(2) In {µ¯01 ≤ |x| ≤ |t| 12 }, by the definition (2.4.26) and (2.4.27), it is easy to see that w∗2 j ≲ w∗3 forj = 1, . . . ,k− 1 since w2 j∗ have more time decay than w∗3. In {|x| ≥ |t|12 }, we have w∗2 j ≲ w∗3 by thesimilar reason.Remark 2.7.6.1. Lemma 2.7.5 and 2.7.6 help us consider much fewer terms in the topology of theouter problem in some special domains.Lemma 2.7.7. There exist σ ,ε > 0 small and t0 negative enough such that‖T out [VΨ]‖out,∗α,σ ≲ R−1‖Ψ‖out,∗α,σ . (2.7.31)Proof. Without loss of generality, we assume ‖Ψ‖out,∗α,σ = 1. By (2.3.9), we rewrite V asV = pup−1∗ (1−k∑j=1ζ j)+k∑j=1ζ jp(up−1∗ −U p−1j ). (2.7.32)We shall handle terms respectively.Consider the first term in (2.7.32). Using (2.3.3), the support of 1−∑kj=1 ζ j is ∪ki=2{Rµ0i ≤ |x| ≤2R−1µ0,i−1}∪{Rµ01 ≤ |x|}.• In {Rµ01 ≤ |x|}, we have pup−1∗ ≲ µ21 |x|−4 ≤ R−1|x|−3 by Lemma 2.7.1 and 2.7.2. Split theregion into {Rµ01 ≤ |x| ≤ µ¯01} ∪ {µ¯01 ≤ |x| ≤ |t| 12 } ∪ {|x| ≥ |t| 12 }. In the first set, one has |Ψ| ≲w∗11+w∗12+w∗21 by (2.7.29) and (2.7.30). Notice w∗12+w∗21 ≲ w∗11 in {Rµ01 ≤ |x| ≤ µ¯01}. Therefore∣∣pup−1∗ (1− k∑j=1ζ j)Ψ∣∣≲ R−1|x|−3w∗11 ≲ R−1w11. (2.7.33)In the second set, by (2.7.29) and (2.7.30), one has |Ψ|≲ w∗11+w∗3. Then∣∣pup−1∗ (1− k∑j=1ζ j)Ψ∣∣≲ R−1|x|−3(w∗11+w∗3)≲ R−1(w11+w3). (2.7.34)In the third set, similarly, we have∣∣pup−1∗ (1− k∑j=1ζ j)Ψ∣∣≲ R−1|x|−3w∗3 ≲ R−1w3. (2.7.35)• Consider the region {Rµ0i ≤ |x| ≤ 2R−1µ0,i−1}, i= 2, . . . ,k. We divide it further into two parts,{Rµ0i ≤ |x| ≤ µ¯0i}∪{µ¯0i ≤ |x| ≤ 2R−1µ0,i−1}. In {Rµ0i ≤ |x| ≤ µ¯0i}, we have pup−1∗ ≲ µ2i |x|−4 by48Lemma 2.7.1 and 2.7.2. Moreover, one has |Ψ| ≤ w∗1i+w∗1,i+1+w∗2,i−1+w∗2i by Lemma 2.7.5 and2.7.6. One readily has w∗1,i+1 ≲ w∗1i in {Rµ0i ≤ |x| ≤ µ¯0i}. Thus∣∣pup−1∗ (1− k∑j=1ζ j)Ψ∣∣≲ µ2i |x|−4 (w∗1i+w∗1,i+1+w∗2,i−1+w∗2i)≲ µ2i |x|−4(w∗1i+w∗2,i−1+w∗2i)≲ R−1 (w1i+w2i) ,(2.7.36)where we have used the fact that in {Rµ0i ≤ |x| ≤ µ¯0i},µ2i |x|−4w∗1i ≲ |t|γiµ2+αi |x|−4−α ≲ R−2w1i,µ2i |x|−4w∗2,i−1 ≲ µ2i |x|−4|t|−2σµ1−n2i−1 ≲ Rα−2|t|−σw1i,µ2i |x|−4w∗2,i ≲ µ2i |x|−4|t|−2σµn2−2i+1 µ−1i |x|4−n ≲ R−2w2i.In the other part {µ¯0i ≤ |x| ≤ 2R−1µ0,i−1}, we have pup−1∗ ≲U p−1i−1 ≲ µ−2i−1 and w∗2,i−2 ≲ w∗1,i−1(which is vacuum if i= 2). Then∣∣pup−1∗ (1− k∑j=1ζ j)Ψ∣∣≲ µ−2i−1 (w∗1,i−1+w∗1i+w∗2,i−2+w∗2,i−1)≲ µ−2i−1(w∗1,i−1+w∗1i+w∗2,i−1)≲(µ−2i−1|t|γi−11{|x|≤2R−1µi−1}+R−2w2,i−1),(2.7.37)where we have used the fact that in {µ¯0i ≤ |x| ≤ 2R−1µ0,i−1},µ−2i−1w∗2,i−1 ≲ µ−2i−1|t|−2σµn2−2i µ−1i−1|x|4−n ≲ R−2|t|−2σµn2−2i µ−1i−1|x|2−n ≲ R−2w2,i−1,µ−2i−1w∗1i ≤ µ−2i−1|t|γiµαi µ¯n−2−αi |x|2−n ≲ |t|γi+2σµα2 +1i µn−4−α2i−1 w2,i−1 ≲ |t|−εw2,i−1.By Lemma 2.8.2,T out [µ−2i−1|t|γi−11{|x|≤2R−1µ0,i−1}]≲|t|γi−1R−2 if |x| ≤ R−1µ0,i−1,µ−2i−1|t|γi−1(R−1µi−1)n|x|2−n if R−1µ0,i−1 ≤ |x| ≤ |t|12 ,R−n(|x|2)(2−n)αi−1+γi−1 |x|2−n if |x| ≥ |t| 12 .≲ R−1w∗1,i−1.(2.7.38)Next, we consider the second term in (2.7.32). Recall the support of ζ j (2.3.4) is contained in{R−1µ0 j ≤ |x| ≤ 2Rµ0 j}, which are mutually disjoint.49• For j = 1, we have |ζ1(up−1∗ −U p−11 )| ≲U p−12 ζ1 ≲ µ22 |x|−41{R−1µ01≤|x|≤2Rµ01} since φ0 = 0 inthe support of ζ1. Then∣∣∣ζ1(up−1∗ −U p−1j )Ψ∣∣∣≲ µ22 |x|−41{R−1µ01≤|x|≤2Rµ01} (w∗11+w∗12+w∗21)≲ |t|−εw11, (2.7.39)where we have used the fact that in {R−1µ01 ≤ |x| ≤ 2Rµ01},µ22 |x|−4w∗11 ≲ µ22 |x|−4|t|γ1(1+ |x|)−α ≤ R4µ22w11 ≤ R4|t|−2εw11,µ22 |x|−4w∗21 ≲ µ22 |x|−4|t|−2σµn2−22 |x|2−n ≲ Rn+2|t|−2εw11,µ22 |x|−4w∗12 ≲ µ22 |x|−4|t|γ2µα2 µ¯n−2−α2 |x|2−n ≲ Rn+2|t|−2εw11.• For j = 2, . . . ,k, we have |ζ j(up−1∗ −U p−1j )|≲ (U p−1j−1 +U p−1j+1 +φ p−10 j )|ζ j|≲ µ−2j−1|ζ j| where wehave used the fact that φ0 = φ0 jχ j in {R−1µ0 j ≤ |x| ≤ 2Rµ0 j} andU j+1 is vacuum if j = k. Then∣∣∣ζ j(up−1∗ −U p−1j )Ψ∣∣∣≲ µ−2j−1 (w∗1 j+w∗1, j+1+w∗2, j−1+w∗2 j) |ζ j|≲ |t|−ε (w1 j+w2 j) , (2.7.40)where we have used the fact that in {R−1µ0 j ≤ |x| ≤ 2Rµ0 j}, w∗2, j−1 ≲ w∗1 j andµ−2j−1w∗1 j ≲ µ−2j−1|t|γ j ≲ λ 2j 〈y j〉2+αw1 j ≲ R2+α |t|−2εw1 j,µ−2j−1w∗1, j+1 ≲ µ−2j−1|t|γ j+1µαj+1µ¯n−2−αj+1 |x|2−n ≲ µ−2j−1µ1− α2j µ1+ α2j+1 |t|σw2 j ≲ |t|−εw2 j,µ−2j−1w∗2 j ≲ µ−2j−1|t|−2σµn2−2j+1 µ−1j |x|4−n ≲ µ−2j−1(Rµi)2w2 j ≲ R2|t|−2εw2 j.Combining the above calculations of the two terms in (2.7.32), we get the conclusion.Lemma 2.7.8. There exist σ ,ε > 0 small and t0 negative enough such that‖N [ϕ⃗ ,Ψ, µ⃗1]‖outα,σ ≲ |t0|−ε(‖ϕ⃗‖in,∗a,σ +‖Ψ‖out,∗α,σ)p.Proof. By (2.3.9) and some elementary inequality|N [ϕ⃗ ,Ψ, µ⃗1]|≲k∑j=1µ−n+22j |ϕ j|pη j+ |Ψ|p. (2.7.41)For the first part on the RHS, recalling (2.7.12), we obtainµ−n+22j |ϕ j|pη j ≲ |t|γ j pR(n+1−a)p〈y j〉(n+1)p1{|x|≤4Rµ0 j}(‖ϕ j‖in,∗j,a,σ)p≲ R(n+1−a)p|t|−2εw1 j(‖ϕ j‖in,∗j,a,σ)p.(2.7.42)For the second part on the RHS of (2.7.41),• In {|x| ≥ µ¯01}, by (2.7.29), we have |Ψ|≲ (w∗11+w∗3)‖Ψ‖out,∗α,σ in {µ¯01 ≤ |x| ≤ |t|12 } and |Ψ|≲50w∗3‖Ψ‖out,∗α,σ in {|x|> |t|12 }. Notice(w∗3)p = Rp−1|t|1+(1−p)σ |x|−4w3 ≲ |t|−1w3 if |x|> |t| 12 ,(w∗3)p = Rp−1|t|−(1+σ)(p−1)|x|− 2n−12n−2 w3 ≲ |t|−εw3 if µ¯01 ≤ |x| ≤ |t| 12 ,(w∗11)p = R−1|t|− 4n−2 (1+σ)+δ (n−2−pα)w3 ≲ |t|−εw3 if µ¯01 ≤ |x| ≤ |t| 12 ,where we take δ ≤ 2(n−2)−2 in the last formula. Thus|Ψ|p ≲ |t|−εw3(‖Ψ‖out,∗α,σ )p . (2.7.43)• In {1≤ |x| ≤ µ¯01}, we have|Ψ|≲ (w∗11+w∗12+w∗21)‖Ψ‖out,∗α,σ ≲ w∗11‖Ψ‖out,∗α,σsince(w∗12+w∗21)1{1≤|x|≤µ¯01} ≲ w∗11.Therefore|Ψ|p ≲ (w∗11)p(‖Ψ‖out,∗α,σ )p ≲ |t|−(1+σ) 4n−2+2δw11 (‖Ψ‖out,∗α,σ )p ≲ |t|−εw11 (‖Ψ‖out,∗α,σ )p , (2.7.44)for δ ≤ (n−2)−1.• In {µ¯02 ≤ |x| ≤ 1}, we have(w∗11)p ≲ |t|−(1+σ)p ≲ |t|−εw11,(w∗12)p ≲ |t|γ2pµ¯n+22 |x|−n−2 ≲ |t|γ2pµ¯n−22 |x|2−n ≲ |t|−εw21,(w∗21)p ≲ |t|−2σ pµ(n2−2)p2 |x|(4−n)p ≲ |t|−2σ pµn2−12 |x|2−n ≲ |t|−εw21,where we have used n−2≤ (n−4)p when n≥ 6 in the last inequality. Then|Ψ|p ≲ [(w∗11)p+(w∗12)p+(w∗21)p](‖Ψ‖out,∗α,σ )p ≲ |t|−ε(w11+w21)(‖Ψ‖out,∗α,σ )p. (2.7.45)• In {µ¯0,i+1 ≤ |x| ≤ µ¯0i}, i= 2, . . . ,k, we have(w∗1i)p ≲ |t|γip〈yi〉−α p ≲ µ2i |t|γi(p−1)〈yi〉2+α−α pw1i ≲ |t|−σ(p−1)µiµ−1i−1w1i ≲ |t|−εw1iprovided ε < 2n−6 .(w∗1,i+1)p ≲ |t|pγi+1 µ¯n+2i+1 |x|−n−2 ≲ µ− n+22i |t|−σ pµ¯n−2i+1 |x|2−n51∼ λi+1µn2−2i+1 µ−1i |t|−σ p|x|2−n ≲ |t|−εw2iprovided ε <−2σ +σ p+ 2n−6 .(w∗2,i−1)p ≲ |t|−2σ pµ−n+22i−1 ≲ |t|−2σ pµ− n+22i−1 (µ¯iµi)2+α〈yi〉−2−α (2.7.46)≲ |t|−2σ p( µiµi−1)1−α2 µ−2i µ1− n2i−1 〈yi〉−2−α ≲ |t|−εw1i (2.7.47)provided ε < 1n−6 .(w∗2i)p ≲ |t|−2σ pµ(n2−2)pi+1 µ−pi |x|(4−n)p ≲ |t|−2σ pµ( n2−2)pi+1 µ−pi |x|2−nµ¯(4−n)p−(2−n)i+1 (2.7.48)∼ λ j+1|t|−2σ pµn2−2i+1 µ−1i |x|2−n ≲ |t|−εw2i (2.7.49)provide ε < 2n−6 . Here we have used (4−n)p− (2−n)≤ 0 when n≥ 6.Therefore, for i= 2, . . . ,k,|Ψ|p ≲(w∗1i+w∗1,i+1+w∗2,i−1+w∗2i)p (‖Ψ‖out,∗α,σ )p ≲ |t|−ε (w1i+w2i)(‖Ψ‖out,∗α,σ )p ,where w∗1,i+1,w∗2i are vacuum if i= k and w∗1,i+1 are vacuum if i= k,k−1.Proof of Proposition 2.4.7. This is a combination of results in Lemma 2.7.3, 2.7.4, 2.7.7, 2.7.8 andLemma 2.4.4, 2.4.5 2.4.6.2.8 Additional information 2: Some estimates for the outer problemBasic estimatesLet G(x, t) denote the standard heat kernel on Rn, that isG(x, t) =1(4πt)n/2e−|x|24t . (2.8.1)Recall the T out [g](x, t) defined by (2.4.19).Lemma 2.8.1. Suppose n> 2, a≥ 0, d1 ≤ d2 ≤ 12 and b satisfiesn2 −b+d2(a−n)> 1 if a< n,n2 −b+d1(a−n)> 1 if a≥ n,(2.8.2)520≤ c1,c2 ≤ c∗∗. Then there exists C depending on n,a,b,d1,d2,c∗∗ such that for t <−1T out[ |t|b|x|a 1{c1|t|d1≤|x|≤c2|t|d2}](0, t)≤C|t|b(c1|t|d1)2−a if a ∈ (2,∞),|t|b ln(c2|t|d2/(c1|t|d1)) if a= 2,|t|b(c2|t|d2)2−a if a ∈ [0,2).Proof. Using (2.8.1), we obtainˆ t−∞ˆRn1(t− s)n/2 e− |y|24(t−s) |s|b|y|a 1{c1|s|d1≤|y|≤c2|s|d2}dyds∼ˆ t−∞ˆ c2|s|d2c1|s|d1|s|b(t− s)n/2 e− r24(t−s) rn−1−adrds∼ˆ t−∞ˆ c22 |s|2d24(t−s)c21 |s|2d14(t−s)e−zzn−a−22|s|b(t− s)a/2 dzds=ˆ t−∞|s|b(t− s)a/2F(c21|s|2d14(t− s) ,c22|s|2d24(t− s))ds, (2.8.3)whereF(A,B) :=ˆ BAe−zzn−a−22 dz.We shall split (2.8.3) into four integrals J1,J2,J3,J4 according to the regions of s. First, in the regions ∈ [t− c21|t|2d1 , t], one has |s− t| ≤ c21|t| when t <−1. ThereforeJ1 =ˆ tt−c21|t|2d1|s|b(t− s)a/2F(c21|s|2d14(t− s) ,c22|s|2d24(t− s))ds≲ |t|bˆ tt−c21|t|2d11(t− s)a/2 e− c21 |t|2d14(t−s) (1+c2∗∗)2(d1)−ds∼ c2−a1 |t|b+d1(2−a)ˆ ∞(1+c2∗∗)2(d1)−4e−s˜s˜a2−2ds˜≲ c2−a1 |t|b+d1(2−a),where (d1)− =min{0,d1}.53Second, in the region s ∈ [t− c22|t|2d2 , t− c21|t|2d1 ],J2 =ˆ t−c21|t|2d1t−c22|t|2d2|s|b(t− s)a/2F(c21|s|2d14(t− s) ,c22|s|2d24(t− s))ds≲ˆ t−c21|t|2d1t−c22|t|2d2|s|b(t− s)a/2F(c21|t|2d1(1+ c2∗∗)2(d1)−4(t− s) ,c22|t|2d2(1+ c2∗∗)2(d2)+4(t− s))ds≲ˆ t−c21|t|2d1t−c22|t|2d2|t|b(t− s)a/21 if a< n,∣∣∣∣ln( c21|t|2d1 (1+c2∗∗)2(d1)−4(t−s) )∣∣∣∣ if a= n,(c21|t|2d1t−s) n−a2if a> n,ds≲c2−a1 |t|b+d1(2−a) if 2< a,|t|b log(c2/c1|t|2(d2−d1)) if a= 2,c2−a2 |t|b+d2(2−a) if 0≤ a< 2,(2.8.4)where (d2)+ =max{0,d2}.Third, when s ∈ [2t− c22|t|2d2 , t− c22|t|2d2 ],J3 =ˆ t−c22|t|2d22t−c22|t|2d2|s|b(t− s)a/2F(c21|s|2d14(t− s) ,c22|s|2d24(t− s))ds≲ˆ t−c22|t|2d22t−c22|t|2d2|t|b(t− s)a/2F(c21|t|2d1(2+ c2∗∗)2(d2)−4(t− s) ,c22|t|2d2(2+ c2∗∗)2(d2)+4(t− s))ds≲ˆ t−c22|t|2d22t−c22|t|2d2|t|b(t− s)a/2(c22|t|2d2t−s) n−a2if a< n,ln(c22|t|2d2 (2+c2∗∗)2(d2)+c21|t|2d1 (2+c2∗∗)2(d2)−)if a= n,(c21|t|2d1t−s) n−a2if a> n,ds≲c2−a2 |t|b+d2(2−a) if a< n,c2−n2 |t|b+d2(2−n) ln(c22|t|2d2 (2+c2∗∗)2(d2)+c21|t|2d1 (2+c2∗∗)2(d2)−)if a= n,|t|bcn−a1 |t|d1(n−a)c2−n2 |t|d2(2−n) if a> n.54Fourth, when s ∈ (−∞,2t− c22|t|2d2 ], we have −s2 ≤ t− s≤−s. For a< n,J4 =ˆ 2t−c22|t|2d2−∞|s|b(t− s)a/2F(c21|s|2d14(t− s) ,c22|s|2d24(t− s))ds≲ˆ 2t−c22|t|2d2−∞|s|b− a2F(c21|s|2d14|s| ,c22|s|2d22|s|)ds≲ˆ 2t−c22|t|2d2−∞|s|b− a2 cn−a2 |s|(2d2−1)n−a2 ds≲cn−a2 |t|b+d2(n−a)−n2+1 ≲ c2−a2 |t|b+d2(2−a),where (2.8.2) is needed to guarantee the integrability and the last step is using d2 ≤ 1/2, c2 ≤ c∗∗ andn> 2. For a≥ n, similarly we haveJ4 ≲cn−a1 |t|b+d1(n−a)+1−n2 if a> n,|t|b+1− a2 ln(2c2|t|d2c1|t|d1)if a= n,≲c2−a1 |t|b+d1(2−a).Collecting the estimates of J1 to J4, we get the conclusion.Remark 2.8.1.1. After close examination of the proof, only (2.8.4) needs the comparison between aand 2. In fact, if a< 2, one can let c1 = 0 to getT out[ |t|b|x|a 1{|x|≤c2|t|d2}](0, t)≤Cc2−a2 |t|b+d2(2−a).Lemma 2.8.2. Suppose n> 2, a≥ 0, d1 ≤ d2 ≤ 12 and b satisfies (2.8.2), 0≤ c1,c2 ≤ c∗∗. Denoteu(x, t) =T out[ |t|b|x|a 1{c1|t|d1≤|x|≤c2|t|d2}](x, t).Then there exists C depending on n,a,b,d1,d2,c∗∗ such that for t <−1u(x, t)≤Cc2−a1 |t|b+d1(2−a) if a ∈ (2,∞),c2−a2 |t|b+d2(2−a) if a ∈ [0,2). (2.8.5)Moreover, when |x|> 2c2|2t|d2 , a< n,u(x, t)≤Ccn−a2|t|b+d2(n−a)|x|2−n if 2c2|2t|d2 ≤ |x| ≤ |t|12 ,|x|2b+2d2(n−a)+2−n if |x| ≥ |t| 12 .(2.8.6)55When |x| ≥ 2c1|2t|d1 , a> n,u(x, t)≤Ccn−a1|t|b+d1(n−a)|x|2−n if 2c1|2t|d1 ≤ |x| ≤ |t|12 ,|x|2b+2d1(n−a)|x|2−n if |x| ≥ |t| 12 .(2.8.7)Proof. • Since|t|b|x|a 1{c1|t|d1≤|x|≤c2|t|d2} ≤ |t|bmin{1ca1|t|d1a,1|x|a}1{|x|≤c2|t|d2} = f (x, t),thenu(x, t)≲ˆ t−∞ˆRnG(x− y, t− s) f (y,s)dyds.Since G and f are both decreasing functions for each time slice, using Hardy-Littlewood rearrange-ment inequality, thenu(x, t)≤ u(0, t) = J1+ J2,whereJ1 =ˆ t−∞ˆRnG(y, t− s)|s|bmin{1ca1|s|d1a,1|y|a}1{|y|≤c1|s|d1}dyds, (2.8.8)J2 =ˆ t−∞ˆRnG(y, t− s)|s|bmin{1ca1|s|d1a,1|y|a}1{c1|s|d1≤|y|≤c2|s|d2}dyds. (2.8.9)Applying Lemma 2.8.1 and Remark 2.8.1.1, we obtainJ1 ≲ c2−a1 |t|b+d1(2−a), J2 ≲c2−a1 |t|b+d1(2−a) if a ∈ (2,∞),c2−a2 |t|b+d2(2−a) if a ∈ (0,2).Therefore (2.8.5) is established.• Next we will establish (2.8.6) when d2 ≤ 0. In this case, for x> 2c2|t|d2 , one has12|x| ≤ |x− y| ≤ 2|x| for y with |y| ≤ |s|d2 and s≤ t. (2.8.10)Thenu(x, t)≲ˆ t−∞ˆRn1(t− s)n/2 e− |x|216(t−s) |s|b|y|a 1{c1|s|d1≤|y|≤c2|s|d2}dyds∼ˆ t−∞ˆ c2|s|d2c1|s|d11(t− s)n/2 e− |x|216(t−s) |s|brn−1−adrds56≲cn−a2ˆ t−∞1(t− s)n/2 e− |x|216(t−s) |s|b+d2(n−a)ds≤ cn−a2(max{|t|, |x|2})b+d2(n−a) |x|2−n.The last step follows from the following two factsˆ t2t1(t− s)n/2 e− |x|216(t−s) |s|b+d2(n−a)ds∼ |t|b+d2(n−a)ˆ t2t1(t− s)n/2 e− |x|216(t−s) ds∼|t|b+d2(n−a)|x|2−nˆ ∞|x|216|t|e−zzn2−2dz≲|t|b+d2(n−a)|x|2−n if |x|< |t| 12 ,|t|b+d2(n−a) ´ ∞|x|216|t|e−zzn2−2dz if |x| ≥ |t| 12 .andˆ 2t−∞1(t− s)n/2 e− |x|216(t−s) |s|b+d2(n−a)ds∼ˆ 2t−∞e−|x|216|s| |s|b+d2(n−a)− n2 ds∼|x|2b+2d2(n−a)−n+2ˆ |x|232|t|0e−zz−b−d2(n−a)+n2−2dz∼|t|b+d2(n−a)−n2+1 if |x|< |t| 12 ,|x|2b+2d2(n−a)−n+2 if |x| ≥ |t| 12 ,where −b− d2(n− a)+ n2 − 2 > −1 is needed to guarantee the integrability. Thus (2.8.6) is estab-lished.• Next we will establish (2.8.6) when d2 > 0. We do not have (2.8.10) anymore. In this case,|x| ≥ 2c2|2t|d2 is equivalent to −( |x|2c2 )1/d2 ≤ 2t. We write(−∞, t) = (−( |x|2c2)1/d2 , t)∪ (−(2|x|c2)1/d2 ,−( |x|2c2)1/d2)∪ (−∞,−(2|x|c2)1/d2)and thusu≲ˆ t−∞ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds= u1+u2+u3 (2.8.11)where u1,u2,u3 are the integrations according to the three intervals respectively. We shall verify thatu1,u2,u3 all satisfy (2.8.6). For u1, one has c2|s|d2 ≤ |x|/2 for such s. Thenu1 =ˆ t−(|x|/2c2)1/d2ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds≲ˆ t−(|x|/2c2)1/d2ˆRn1(t− s)n/2 e− |x|216(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds≲(ˆ t2t+ˆ 2t−(|x|/2c2)1/d2)ˆRn1(t− s)n/2 e− |x|216(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds≲cn−a2(max{|t|, |x|2})b+d2(n−a) |x|2−n.57The last step follows from some easy integration which has been done many times in this section. Foru2,u2 ≲ˆ −(|x|/2c2)1/d2−(2|x|/c2)1/d2ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds≲(|x|/c2)b/d2ˆ −(|x|/2c2)1/d2−(2|x|/c2)1/d2ˆRn1(t− s)n/2 e− |x−y|24(t−s) 1|y|a 1{|y|≤2|x|}dyds≲c−b/d22 |x|b/d2+n−aˆ −(|x|/2c2)1/d2−(2|x|/c2)1/d21(t− s)n/2 ds≲ c(−b+ n2−1)/d22 (|x|1/d2)b+d2(n−a)+1−n2≲cn−a2|t|b+d2(n−a)|x|2−n if 2c2|2t|d2 ≤ |x|< |t|12 ,|x|2b+2d2(n−a)+2−n if |x| ≥ |t| 12 .The last step follows from 0< d2 ≤ 12 and b+d2(n−a)+1− n2 < 0. Similarly, for u3,u3 ≲ˆ −(2|x|/c2)1/d2−∞ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{|y|≤c2|s|d2}dyds≲(|x|/c2)(b+1− n2 )/d2 |x|n−a+ˆ −(2|x|)1/d2−∞1|s| n2 e− |y|216(t−s) |s|a|y|b 1{2|x|≤|y|≤c2|s|d2}dyds≲c(−b+n2−1)/d22 (|x|1/d2)b+d2(n−a)+1−n2≲cn−a2|t|b+d2(n−a)|x|2−n if 2c2|2t|d2 ≤ |x|< |t|12 ,|x|2b+2d2(n−a)+2−n if |x| ≥ |t| 12 .Collecting the results of u1,u2,u3, we can get (2.8.6).• By the similar calculation like (2.8.6), we will get (2.8.7).Lemma 2.8.3. Suppose 2 < a < n, 0 ≤ d2 ≤ 12 , n2 − b > 1 and 0 < c2 ≤ c∗∗. Then there exists Cdepending on n,a,b,d2,c∗∗ such that for t <−1,T out[ |t|b|x|a 1{|x|≤c2|t|d2}]≤C|t|b|x|2−a for |x|< c2|t|d2 . (2.8.12)Proof. We divide u into three partsu(x, t) =ˆ t−∞ˆRnG(x− y, t− s) |s|b|y|a 1{|y|≤c2|s|d2}dyds= u1+u2+u3,where u1 is the term with 1{|y|≤ 12 |x|} inside the integrand, u2 is the one with 1{ 12 |x|≤|y|≤2|x|} and u3 is theone with {2|x| ≤ |y| ≤ c2|s|d2}. Since most of the calculations are similar to the proof of the previous58lemma. We omit some details here. For u1, we proceed asu1 ≲ˆ t−∞ˆRn1(t− s)n/2 e− |x|216(t−s) |s|b|y|a 1{|y|≤ 12 |x|}dyds∼|x|n−aˆ t−∞1(t− s)n/2 e− |x|216(t−s) |s|bds≲ |t|b|x|2−aˆ ∞|x|216(−t)e−zzn2−2dz+ |x|2b+2−aˆ |x|232(−t)0e−zz−b+n2−2dz≲|t|b|x|2−a+ |x|n−a|t|1− n2+b,where we have used the fact that n2 −b> 1. For u2, we haveu2 ≲ˆ t−∞ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{ 12 |x|≤|y|≤2|x|}dyds≲|x|−aˆ t−∞ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b1{|x−y|≤3|x|}dydx∼|x|−aˆ t−∞ˆ 3|x|01(t− s)n/2 e− r24(t−s) |s|brn−1dyds∼ |x|−aˆ t−∞ˆ 9|x|24(t−s)0|s|be−zz n2−1dzds∼|x|−aˆ t−∞|s|bmin{1,( |x|2t− s) n2}ds∼ |t|b|x|2−a+ |t|1+b− n2 |x|n−a.For u3, we haveu3 ≲ˆ t−∞ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|a 1{2|x|≤|y|≤c2|s|d2}dyds≲ˆ t−∞ˆRn1(t− s)n/2 e− |y|216(t−s) |s|b|y|a 1{2|x|≤|y|≤c2|s|d2}dyds∼ˆ t−∞ˆ c2|s|d22|x|1(t− s)n/2 e− r216(t−s) |s|brn−1−adrds∼ˆ t−∞ˆ c2 |s|d216(t−s)|x|24(t−s)(t− s)− a2 |s|be−zz n−a2 −1dzds≲ |t|b|x|2−a.Combining the estimate of u1,u2,u3 and using the fact that |x|n−a|t|1− n2+b ≤ |t|b|x|2−a because |x| ≤c2|t|d2 , we get (2.8.12).Corollary 2.8.4. Suppose that n> 2, d1 ≤ d2 ≤ 12 , b satisfies (2.8.2), 0≤ c1,c2 ≤ c∗∗. Denoteu(x, t) =T out[|t|b|x|−a1{c1|t|d1≤|x|≤c2|t|d2}].Then there exists C depending on n,a,b,d1,d2,c∗∗ such that for t <−1,59if 0≤ a< 2,u(x, t)≤Cc2−a2 |t|b+d2(2−a) if |x| ≤ c2|t|d2 ,cn−a2 |t|b+d2(n−a)|x|2−n if c2|t|d2 ≤ |x| ≤ |t|12 ,cn−a2 |x|2b+2d2(n−a)+2−n if |x| ≥ |t|12 .(2.8.13)If 2< a< n,u(x, t)≤Cc2−a1 |t|b+d1(2−a) if |x| ≤ c1|t|d1 ,|t|b|x|2−a if c1|t|d1 ≤ |x| ≤ c2|t|d2 ,cn−a2 |t|b+d2(n−a)|x|2−n if c2|t|d2 ≤ |x| ≤ |t|12 ,cn−a2 |x|2b+2d2(n−a)+2−n if |x| ≥ |t|12 .(2.8.14)If a> n,u(x, t)≤Cc2−a1 |t|b+d1(2−a) if |x| ≤ c1|t|d1 ,cn−a1 |t|b+d1(n−a)|x|2−n if c1|t|d1 ≤ |x| ≤ |t|12 ,cn−a1 |x|2b+2d1(n−a)+2−n if |x| ≥ |t|12 .(2.8.15)Proof. (2.8.13) and (2.8.15) follow Lemma 2.8.2 directly. (2.8.14) follows from (2.8.6) and (2.8.12).Lemma 2.8.5. Suppose that a2 −b> 1. ThenT out[|t|b|x|−a1{|x|≥|t| 12 }]≲ |t|1+b− a2 1{|x|≤|t| 12 }+1{|x|≥|t| 12 }|x|−a|t|1+b, if b<−1,1+ ln( |x|2|t|)if b=−1,|x|2+2b if b>−1.(2.8.16)Proof. Denote u(x, t) =T out[|t|b|x|−a1[|x|≥|t| 12 ]].• For |x| ≤ 12 |t|12 , we have |x− y| ≥ 12 |y| for |y| ≥ |s|12 ≥ |t| 12 . Thenu(x, t)≲ˆ t−∞ˆRn1(t− s)n/2 e− |y|216(t−s) |s|b|y|−a1{|y|≥|s| 12 } dξ ds≲(ˆ 2t−∞+ˆ t2t)|s|b(t− s)−a/2e−(−s)32(t−s) ds≲ |t|1+b− a2 ,(2.8.17)where a2 −b> 1 is used to guarantee the integrability in the last step.60• Consider |x| ≥ 4|t| 12 . We make the following decomposition.u(x, t) =(ˆ t− 14 |x|2+ˆ − 14 |x|2−4|x|2+ˆ −4|x|2−∞)ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|−a1{|y|≥|s| 12 } dyds:= P1+P2+P3.For P1, we divide it further to beP1 =ˆ t− 14 |x|2ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|−a1{|y|≥|s| 12 }dyds= P11+P12+P13where P11 is the term with 1{|s| 12≤|y|≤ 12 |x|}in the integrand, P12 is the one with 1{ 12 |x|≤|y|≤2|x|} and P13 isthe one with 1{2|x|≤|y|}. For P11, when a< n,P11 ≲ˆ t− 14 |x|2ˆRn1(t− s)n/2 e− |x|216(t−s) |s|b|y|−a1{|s| 12≤|y|≤ |x|2 } dyds≲ˆ t− 14 |x|21(t− s)n/2 e− |x|216(t−s) |s|b|x|n−a ds= |x|n−aˆ t− 14 |x|21(t− s)n/2 e− |x|216(t−s) |s|b ds≲ |x|2−a+2b.(2.8.18)When a≥ n, by similar calculation, P11 ≲ |x|2−a+2b still holds.For P12,P12 ≲ |x|−aˆ t− 14 |x|2ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b1{ |x|2 ≤|y|≤2|x|} dyds≲ |x|−a(ˆ t2t+ˆ 2t− 14 |x|2)ˆ 3|x|01(t− s)n/2 e− r24(t−s) |s|brn−1 drds≲ |x|−a|t|1+b+ |x|−a|t|1+b if b<−1,1+ ln( |x|2−t)if b=−1,(|x|2)1+b if b>−1.(2.8.19)For P13,P13 ≲ˆ t− 14 |x|2ˆRn1(t− s)n/2 e− |y|216(t−s) |s|b|y|−a1{2|x|≤|y|} dyds∼ˆ t− 14 |x|2ˆ ∞2|x|(−s)b(t− s)n/2 e− r216(t−s) rn−1−a drds=(ˆ t2t+ˆ 2t− 14 |x|2)ˆ ∞2|x|(−s)b(t− s)n/2 e− r216(t−s) rn−1−a drds≲ (−t)b|x|2−ae−|x|216(−t) + |x|2+2b−a.(2.8.20)61For P2, since −14 |x|2 ≤ 4t in this case,P2 ≲ˆ − 14 |x|2−4|x|2ˆRn1|s|n/2 e− |x−y|24(−s) |s|b|y|−a1{|y|≥ |x|2 } dyds∼ |x|−n+2bˆ − 14 |x|2−4|x|2ˆRne−|x−y|24(−s) |y|−a(1{ |x|2 ≤|y|≤2|x|}+1{2|x|≤|y|})dyds≲ |x|2+2b−a.(2.8.21)For P3, in this case, |x| ≤ 12 |s|12 ,P3 ≲ˆ −4|x|2−∞ˆRn1(t− s)n/2 e− |y|216(t−s) |s|b|y|−a1{|y|≥|s| 12 } dyds≲ˆ −4|x|2−∞ˆRn(−s)b− n2 e−|y|216(−s) |y|−a1{|y|≥|s| 12 } dyds≲ |x|2+2b−a,(2.8.22)where a2 − b > 1 is required to guarantee the integrability. Combining the above estimates of P1, P2and P3, we get, when |x| ≥ 4|t| 12u(x, t)≲ |x|−a|t|1+b if b<−1,1+ ln( |x|2−t)if b=−1,|x|2+2b if b>−1.(2.8.23)• Consider the case 12 |t|12 ≤ |x| ≤ 4|t| 12 ,u(x, t) =ˆ t−∞ˆRn1(4π(t− s))n/2 e− |x−y|24(t−s) |s|b|y|−a1[|y|≥|s| 12 ] dyds≲(ˆ t64t+ˆ 64t−∞)ˆRn1(t− s)n/2 e− |x−y|24(t−s) |s|b|y|−a1{|y|≥|s| 12 } dyds≲ |t|1+b− a2 .(2.8.24)In order to get the gradient estimate of φ¯ , we need the following lemma.Lemma 2.8.6. For d1 ≤ d2 ≤ 12 , n2 −b−d2n> 0, c1,c2 ∼ 1, we haveT d [|t|b1[c1|t|d1≤|x|≤c2|t|d2 ]]≲|t|b+d2 if |x| ≤ |t|d2 ,|t|b+d2n|x|1−n if |t|d2 ≤ |x| ≤ |t| 12 ,(|x|2)b+d2n|x|1−n if |x| ≥ |t| 12 .(2.8.25)62For d1 ≤ d2 ≤ 12 , a> n, n2 −b−d1(n−a)> 0, c1,c2 ∼ 1, we haveT d [|t|b|x|a 1[c1|t|d1≤|x|≤c2|t|d2 ]]≲|t|b+d1(1−a) if |x| ≤ |t|d1 ,|t|b+d1(n−a)|x|1−n if |t|d1 ≤ |x| ≤ |t| 12 ,(|x|2)b+d1(n−a)|x|1−n if |x| ≥ |t| 12 .(2.8.26)We omit the proof since it relies on splitting integral domain like above.Proofs of three lemmas in the outer problemProof of Lemma 2.4.4. For j = 2, . . . ,k, by Corollary 2.8.4,T out [µ−20 j (t)|t|γ j1{|x|≤µ0 j}]≲|t|γ j if |x| ≤ µ0 j,|t|γ jµn−20 j |x|2−n if µ0 j ≤ |x| ≤ |t|12 ,|x|2γ j+(4−2n)α j+2−n if |x| ≥ |t| 12 ,T out [µα0 j(t)|t|γ j |x|−2−α1{µ0 j≤|x|≤µ¯0 j}]≲|t|γ j if |x| ≤ µ0 j,|t|γ jµα0 j|x|−α if µ0 j ≤ |x| ≤ µ¯0 j,|t|γ jµα0 j µ¯n−2−α0 j |x|2−n if µ¯0 j ≤ |x| ≤ |t|12 ,|x|2γ∗j+2−n if |x| ≥ |t| 12 .Thus T out [w1 j]≲ w∗1 j.For j = 1, the first part of w11 can be dealt with by the same method above. For the rest part, byCorollary 2.8.4,T out [|t|γ1 µ¯n−2−α01 |x|−1−n1{µ¯01≤|x|≤|t| 12 }]≲|t|γ1 µ¯−1−α01 if |x| ≤ µ¯01,|t|γ1 µ¯n−3−α01 |x|2−n if µ¯01 ≤ |x| ≤ |t|12 ,|x|2(γ1+δ (n−3−α))+2−n if |x| ≥ |t| 12 .≲ w∗11.Proof of Lemma 2.4.5. This just follows from (2.8.14). b=−2σ − (n2 −2)α j+1+α j, d2 =−12(α j+α j−1), a= n−2.63Proof of Lemma 2.4.6. By Corollary 2.8.4, we haveT out [|t|−1−σ |x|2−n1{µ¯01≤|x|≤|t| 12 }]≲|t|−1−σ µ¯4−n01 if |x| ≤ µ¯01,|t|−1−σ |x|4−n if µ¯01 ≤ |x| ≤ |t| 12 ,|x|2−2σ−n if |x| ≥ |t| 12 .By Lemma 2.8.5, we haveT out [|t|−1−σ |x|2−n1{|x|≥|t| 12 }]≲|t|1−σ−n2 if |x| ≤ |t| 12 ,|t|−σ |x|2−n if |x| ≥ |t| 12 .Then (2.4.27) follows when δ ≤ 12 .64Chapter 3Existence and stability of infinite timeblow-up for the four dimensional energycritical heat equationIn this chapter, we construct the infinite time blow-up solution for the critical heat equation in dimen-sion four.3.1 Approximate solution and improvement3.1.1 First approximate solutionWe consider the energy critical heat equation in dimension 4ut = ∆u+u3 in R4× (0,∞),u(x,0) = u0(x) in R4. (3.1.1)Since changing the initial time will not change the structure of the nonlinear heat equation, weassume the initial time is t = t0 and t0 is sufficiently large.In this chapter, we use the Aubin-Talenti solution in dimension fourU(y) = 23211+ |y|2as the building block of construction. It is known that all the bounded kernels of the corresponding65linearized operator ∆+3U2 are given byZi(y) = ∂yiU =−252yi(1+ |y|2)2 , for i= 1, . . . ,4, Z5(y) =U+ y ·∇U = 2321−|y|2(1+ |y|2)2 .We take the leading profile of the infinite time blow-up solution asu1(x, t) = µ−1(t)U(x−ξ (t)µ(t))η(x−ξ (t)√t)where µ(t), ξ (t) ∈C1(t0,∞). Throughout this chapter, we make the following ansatz1Cµ ln t≤ |µ|+ t ln t|µt | ≤ Cµln t , (3.1.2)ξ (t)→ 0 as t→ ∞whereCµ ≥ 1 is a large constant. Later we shall rigorously justify the above ansatz about the asymp-totics for the scaling and translation parameters.Denote the error function asS[g] :=−∂tg+∆g+g3.Then the error produced by the first approximate solution u1 is given byS[u1] = µ−2µtZ5(x−ξµ)η(x−ξ√t)+E[µ]+µ−2ξt ·∇U(x−ξµ)η(x−ξ√t)+µ−1t−12 ξt ·∇η(x−ξ√t)U(x−ξµ)whereE[µ] := 2−1µ−1t−1U(x−ξµ)∇η(x−ξ√t)· x−ξ√t+2µ−2t−12∇U(x−ξµ)·∇η(x−ξ√t)+µ−1t−1U(x−ξµ)∆η(x−ξ√t)+µ−3U3(x−ξµ)[η3(x−ξ√t)−η(x−ξ√t)].In the next section, we shall add two global corrections to improve the slow decaying error.3.1.2 Transferring slow decaying terms by heat equationsFor some admissible function f (x, t), denoteT outn [ f ](x, t) :=ˆ tt0ˆRn(4π(t− s))− n2 e−|x−z|24(t−s) f (z,s)dzds. (3.1.3)66In the rest of the chapter, we will use Lemma 3.7.1 and Lemma 3.7.2 in the additional information toestimate T outn frequently and sometimes will not state repeatedly.Set y= x−ξµ . A term is said to be of slow decay if its spatial decay is equal to or slower than 〈y〉−2( 6∈ L2(R4)). Otherwise, it is of fast decay. Fast decay is usually necessary for the gluing procedure.For this reason, we will transfer the slow decaying terms in S[u1] by heat equations. We now introducethe correction function φ to improve the error. ForS[u1+φ] =−∂tφ+∆φ+S[u1]+ (u1+φ)3−u31,we set x¯= x−ξ and choose φ(x¯, t) = φ1(x¯, t)+φ2(x¯, t) such that∂tφ1 = ∆x¯φ1+E[µ], ∂tφ2 = ∆x¯φ2+µ−2µtZ5(x¯µ)η(x¯√t).The properties of φ1 and φ2 are given in the following two lemmas.Lemma 3.1.1. Assume that µ satisfies (3.1.2) and µ1 satisfies |µ1| ≤ µ2 . Consider∂tφ1 = ∆x¯φ1+E[µ]. (3.1.4)There exists a solution φ1 = φ1[µ] satisfying the following pointwise estimates|φ1[µ]|≲ (t ln t)−11{|x¯|≤2t 12 }+ t2(ln t)−1|x¯|−61{|x¯|>2t 12 },|∇x¯φ1|≲ t− 32 (ln t)−11{|x¯|≤2t 12 }+ t32 (ln t)−1|x¯|−61{|x¯|>2t 12 }.More precisely,φ1[µ] =[−2− 12 µt−1+O(µt−2|x¯|2)+O(t−2ˆ tt0/2(s−1µ3(s)+ s|µt(s)|)ds)]1{|x¯|≤2t 12 }+O(µ|x¯|−2e− |x¯|216t + |x¯|−6ˆ tt0/2s2|µt(s)|ds+ t−2e−|x¯|216tˆ t/2t0/2(s−1µ3(s)+ s|µt(s)|)ds)1{|x¯|>2t 12 },φ1[µ+µ1]−φ1[µ] =[−2− 12 µ1t−1+O(|µ1|t−2|x¯|2)+O(t−2µ2 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|+ t−2ˆ t/2t0/2(s−1|µ1(s)|µ2(s)+ s|µ1t(s)|)ds)]1{|x¯|≤2t 12 }+O(supt1∈[t/2,t]|µ1(t1)||x¯|−2e−|x¯|216t + |x¯|−6(t3 supt1∈[t/2,t]|µ1t(t1)|+ˆ t/2t0/2s2|µ1t(s)|ds)+ t−2e−|x¯|216tˆ t/2t0/2(s−1|µ1(s)|µ2(s)+ s|µ1t(s)|)ds)1{|x¯|>2t 12 }.67Proof. The support of E is in {t 12 ≤ |x¯| ≤ 2t 12 }. In this region, by (3.1.2), µ−1|x¯| 1, which impliesU(|x¯|µ) = 232 µ2|x¯|−2+O(µ4|x¯|−4), |x¯|µU ′(|x¯|µ) =−2 52 µ2|x¯|−2+O(µ4|x¯|−4).Then the leading term of E denoted by E˜ is given byE˜ = 232 µt−2(2−1ζ−1η ′(ζ )−ζ−3η ′(ζ )+ζ−2η ′′(ζ )) , ζ = |x¯|√t.Take φ˜1 as the approximate solution to (3.1.4). Set φ˜1 = µφˆ1, E˜ = µEˆ and φˆ1 satisfies∂t φˆ1 = ∆x¯φˆ1+ Eˆ.We take φˆ1 = t−1A( |x¯|√t)in the self-similar form. ThenA′′+(3ζ+ζ2)A′+A+h(ζ ) = 0, (3.1.5)whereh(ζ ) = 232 ζ−2(η ′′(ζ )− 1ζη ′(ζ )+ζ2η ′(ζ )).Observe that ζ−2, ζ−2(1− e− ζ24 ) are linearly independent kernels to the homogeneous part of(3.1.5). And (3.1.5) has a particular solutionAp(ζ ) =−ζ−2ˆ ζ0ae−a24ˆ a0h(b)beb24 dbda=−ζ−2ˆ ζ0232η ′(a)da= 232 ζ−2(1−η(ζ )),where we have used h(b)beb24 = 232 (b−1eb24 η ′(b))′.In order to find a solution with fast spatial decay, we takeA(ζ ) = Ap(ζ )−2 32 ζ−2(1− e−ζ24 ) = 232 ζ−2(e−ζ24 −η(ζ ))which implies thatφˆ1(x¯, t) = 232 |x¯|−2(e−|x¯|24t −η( |x¯|√t)), φ˜1[µ] = 232 µ|x¯|−2(e−|x¯|24t −η( |x¯|√t)). (3.1.6)It is straightforward to see φ˜1[µ](0, t) =−2− 12 µt−1,φ˜1[µ] =(−2− 12 µt−1+O(µt−2|x¯|2))1{|x¯|≤2t 12 }+O(µ|x¯|−2e− |x¯|24t)1{|x¯|>2t 12 },68φ˜1[µ+µ1]− φ˜1[µ] =(−2− 12 µ1t−1+O(|µ1|t−2|x¯|2))1{|x¯|≤2t 12 }+O(|µ1||x¯|−2e−|x¯|24t)1{|x¯|>2t 12 },and∂|x¯|φˆ1 = −252 |x¯|−3(e−|x¯|24t −η( |x¯|√t))−2 32 |x¯|−2(e−|x¯|24t|x¯|2t+η ′(|x¯|√t)t−12)= O(|x¯|t−21{|x¯|≤t 12 }+ |x¯|−1t−1e−|x¯|24t 1{|x¯|>t 12 }),|∇x¯φ˜1[µ]|= |µ∇x¯φˆ1|≲ |x¯||µ|t−21{|x¯|≤2t 12 }+ |x¯|−1|µ|t−1e− |x¯|24t 1{|x¯|>2t 12 }. (3.1.7)Take φ1 = φ˜1+ φ˜1b. Then φ˜1b satisfies∂t φ˜1b = ∆x¯φ˜1b−µt φˆ1+E− E˜,where φ˜1b is given byφ˜1b[µ](x¯, t) =T out4 [−µt φˆ1+E− E˜](x¯, t) (3.1.8)withE− E˜ = 12µ−1t−1(U(|x¯|µ)−2 32 µ2|x¯|−2)η ′(|x¯|√t)|x¯|√t+2µ−1t−1( |x¯|µU ′(|x¯|µ)+252 µ2|x¯|−2) 1|x¯|t− 12η ′(|x¯|√t)+µ−1t−1(U(|x¯|µ)−2 32 µ2|x¯|−2)∆η(|x¯|√t)+µ−3U3(|x¯|µ)(η3(|x¯|√t)−η( |x¯|√t))= O(µ3t−31{√t≤|x¯|≤2√t}).(3.1.9)Similarly, we evaluate(E− E˜)[µ+µ1]− (E− E˜)[µ] = O(|µ1|µ2t−31{√t≤|x¯|≤2√t}).By Lemma 3.7.1, one hasT out4[µ3t−31{√t≤|x¯|≤2√t}]≲ t−2e−|x¯|216tˆ t/2t0/2µ3(s)s−1ds+µ3t−2 if |x¯| ≤ t12µ3t−1|x¯|−2e− |x¯|216t if |x¯|> t 12,T out4[|µ1|µ2t−31{√t≤|x¯|≤2√t}]≲ t−2e−|x¯|216tˆ t/2t0/2|µ1(s)|µ2(s)s−1ds+supt1∈[t/2,t]|µ1(t1)|µ2t−2 if |x¯| ≤ t 12supt1∈[t/2,t]|µ1(t1)|µ2t−1|x¯|−2e−|x¯|216t if |x¯|> t 12.Notice that|µt φˆ1|≲ |µt |t−11{|x¯|≤t 12 }+ |µt ||x¯|−2e−|x¯|24t 1{|x¯|>t 12 }. (3.1.10)69Therefore, by Lemma 3.7.1, we obtainT out4[|µt |t−11{|x¯|≤t 12 }]≲ t−2e−|x¯|216tˆ t/2t0/2s|µt(s)|ds+|µt | if |x¯| ≤ t12|µt |t|x|−2e−|x¯|216t if |x¯|> t 12.By Lemma 3.7.2, we haveT out4[|µt ||x¯|−2e−|x¯|24t 1{|x¯|>t 12 }]≲T out4[|µt |t2|x¯|−61{|x¯|>t 12 }]≲ t−2e−|x¯|216tˆ t/2t0/2s|µt(s)|ds+|µt | if |x¯| ≤ t12|x¯|−6(t3|µt |+´ t/2t0/2|µt(s)|s2ds) if |x¯|> t 12,and thus|T out4 [µt φˆ1]|≲ t−2e−|x¯|216tˆ t/2t0/2|µt(s)|sds+|µt | if |x¯| ≤ t12|x¯|−6 ´ tt0/2 |µt(s)|s2ds if |x¯|> t 12.It then follows that|φ˜1b[µ]|≲ t−2e−|x¯|216tˆ t/2t0/2(s−1µ3(s)+ s|µt(s)|)ds+µ3t−2+ |µt | if |x¯| ≤ t12µ3t−1|x¯|−2e− |x¯|216t + |x¯|−6 ´ tt0/2 s2|µt(s)|ds if |x¯|> t 12,∣∣φ˜1b[µ+µ1]− φ˜1b[µ]∣∣≲ t−2e− |x¯|216t ˆ t/2t0/2(s−1|µ1(s)|µ2(s)+ s|µ1t(s)|)ds+t−2µ2 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)| if |x¯| ≤ t 12supt1∈[t/2,t]|µ1(t1)|µ2t−1|x¯|−2e−|x¯|216t + |x¯|−6(t3 supt1∈[t/2,t]|µ1t(t1)|+´ t/2t0/2s2|µ1t(s)|ds)if |x¯|> t 12.(3.1.11)In particular, for |µ|≲ (ln t)−1, |µt |≲ t−1(ln t)−2, one has|φ˜1b[µ]|≲ t−1(ln t)−21{|x¯|≤t 12 }+ t2(ln t)−2|x¯|−61{|x¯|>t 12 }, (3.1.12)|E− E˜|≲ (t ln t)−31{√t≤|x¯|≤2√t}, |µt φˆ1|≲ (t ln t)−21{|x¯|≤t 12 }+ t−1(ln t)−2|x¯|−2e− |x¯|24t 1{|x¯|>t 12 }.Then by scaling argument, we have|∇x¯φ˜1b[µ]|≲ t− 32 (ln t)−21{|x¯|≤t 12 }+ t32 (ln t)−2|x¯|−61{|x¯|>t 12 }.70Combining above estimates with (3.1.7), we have|∇x¯φ1|≲ t− 32 (ln t)−11{|x¯|≤2t 12 }+ t32 (ln t)−1|x¯|−61{|x¯|>2t 12 }.Lemma 3.1.2. Assume that µ satisfies (3.1.2) and µ1 satisfies |µ1| ≤ µ2 , |µ1t | ≤ |µt |2 . Consider∂tφ2 = ∆x¯φ2+µ−2µtZ5(x¯µ)η(x¯√t),where φ2 is given by φ2 = φ2[µ] =T out4[µ−2µtZ5( x¯µ )η(x¯√t )]. Then the following estimates hold|φ2[µ]|≲ t−2e−|x¯|216tˆ t/2t0/2s|µt(s)|ds+|µt |(ln(µ−1t 12 )+1) if |x¯| ≤ µ|µt |(ln(|x¯|−1t 12 )+1) if µ < |x¯| ≤ t 12|µt |t|x¯|−2e−|x¯|216t if |x¯|> t 12≲(t ln t)−1 if |x¯| ≤ µt−1(ln t)−2(ln(|x¯|−1t 12 )+1) if µ < |x¯| ≤ t 12t−1(ln t)−2e−|x¯|216t if |x¯|> t 12,(3.1.13)|∇x¯φ2[µ]|≲t−1 if |x¯| ≤ µt−1(ln t)−2(ln(|x¯|−1t 12 )+1)|x¯|−1 if µ < |x¯| ≤ t 12t−32 (ln t)−2e−|x¯|230t if |x¯|> t 12, (3.1.14)∣∣φ2[µ+µ1]−φ2[µ]∣∣≲ t−2e− |x¯|216t ˆ t/2t0/2s|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|)ds+ |µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)ln(µ−1t 12 )+1 if |x¯| ≤ µln(|x¯|−1t 12 )+1 if µ < |x¯| ≤ t 12t|x¯|−2e− |x¯|216t if |x¯|> t 12.(3.1.15)More precisely,φ2[µ] =[−2− 12ˆ t−µ20t/2µt(s)t− s ds+O(t−2ˆ tt0/2s|µt(s)|ds+min{µ−1|µt ||x¯|, ln t|µt |})]1{|x¯|≤2t 12 }+O(t−2ˆ tt0/2s|µt(s)|ds)e−|x¯|216t 1{|x¯|>2t 12 },71φ2[µ+µ1]−φ2[µ]=[−2− 12ˆ t−µ20t/2µ1t(s)t− s ds+O(|µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)+ t−2ˆ t/2t0/2s|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|)ds+ |µt | ln t supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)2+µ−1|µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)|x¯|)]1{|x¯|≤2t 12 }+O(|µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)+ t−2ˆ t/2t0/2s|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|)ds)e−|x¯|216t 1{|x¯|>2t 12 }.Proof. Since|µ−2µtZ5( x¯µ )η(x¯√t)|≲ µ−2µt1{|x¯|≤µ}+µt |x¯|−21{µ<|x¯|≤2t 12 }, (3.1.16)by Lemma 3.7.1 and (3.1.2), we conclude the validity of (3.1.13). By scaling argument, (3.1.14)follows.For µ1 satisfying |µ1| ≤ µ2 and |µ1t | ≤ |µt |2 , we have(µ+µ1)−2(µt +µ1t)Z5(x¯µ+µ1)−µ−2µtZ5( x¯µ )= µ−2µ1tZ5(x¯µ)−µ−3µ1µt(2Z5(x¯µ)+x¯µ·∇Z5( x¯µ ))+(µ−3µ1µt +µ−2µ1t)〈 x¯µ〉−2O( |µ1|µ+|µ1t ||µt |)= O(µ−2|µt |( |µ1|µ+|µ1t ||µt |)〈 x¯µ〉−2).(3.1.17)Then by Lemma 3.7.1, one gets (3.1.15).In order to extract the dominating part of φ2 for the preparation of solving the orthogonal equation,we split φ2 into several parts to estimate. Set µ0(t) = (ln t)−1 and considerφ2 =(ˆ t/2t0+ˆ t−µ20 (t)t/2+ˆ tt−µ20 (t))ˆR4(4π(t− s))−2e−|x−z|24(t−s) µ−2(s)µt(s)Z5(|z|µ(s))η(|z|√s)dzds:= I1+ I2+ I3.For I1, by rearrangement inequality, we have|I1|≲ˆ t/2t0/2ˆR4(t− s)−2e−|x−z|24(t−s) µ−2(s)|µt(s)|〈 |z|µ(s)〉−2η(|z|√s)dzds≲ t−2ˆ t/2t0/2ˆR4e−|z|24t µ−2(s)|µt(s)|〈 |z|µ(s)〉−2η(|z|√s)dzds≲ t−2ˆ t/2t0/2µ−2(s)|µt(s)|ˆ 2√s0〈 rµ(s)〉−2r3drds≲ t−2ˆ t/2t0/2s|µt(s)|dssince´ µ(s)0 〈 rµ(s)〉−2r3dr ∼ µ4(s),´ 2√sµ(s) 〈 rµ(s)〉−2r3dr ≲ sµ2(s).72Using (3.1.17) and similar calculations above, one has∣∣I1[µ+µ1]− I1[µ]∣∣≲ t−2ˆ t/2t0/2s|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|)ds.For I3, we have|I3|≲ˆ tt−µ20 (t)ˆR4(t− s)−2e−|z|24(t−s) µ−2(s)|µt(s)|〈 |z|µ(s)〉−2η(|z|√s)dzds≲ µ−2|µt |ˆ tt−µ20 (t)(t− s)−2ˆ 2√t0e−r24(t−s) 〈 rµ(t)〉−2r3drds≲ |µt |since for s ∈ (t−µ20 (t), t),ˆ µ(t)0e−r24(t−s) 〈 rµ(t)〉−2r3dr ∼ˆ µ(t)0e−r24(t−s) r3dr ∼ (t− s)2ˆ µ2(t)4(t−s)0e−zzdz∼ (t− s)2,ˆ 2√tµ(t)e−r24(t−s) 〈 rµ(t)〉−2r3dr ∼ (t− s)µ2(t)ˆ tt−sµ2(t)4(t−s)e−zdz≲ (t− s)µ2(t)e−µ2(t)4(t−s) ≲ (t− s)2.Similarly, using (3.1.17), one has∣∣I3[µ+µ1]− I3[µ]∣∣≲ |µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|).For I2, more delicate calculations are needed to single out the leading term. SetI2 = I02+(I2− I02),whereI02 =ˆ t−µ20 (t)t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)µt(s)Z5(|z|µ(s))η(|z|√s)dzds := I∗+ I021+ I022andI∗ = −2 32ˆ t−µ20 (t)t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)µt(s)µ2(s)|z|2 dzds,I021 = 232ˆ t−µ20 (t)t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)µt(s)µ2(s)|z|2(1−η( |z|√s))dzds,I022 =ˆ t−µ20 (t)t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)µt(s)(Z5(|z|µ(s))+232µ2(s)|z|2)η(|z|√s)dzds.73For I∗, we evaluateI∗ =−2 32 |S3|ˆ t−µ20 (t)t/2µt(s)ˆ ∞0(4π(t− s))−2e− r24(t−s) rdrds=−2− 12ˆ t−µ20 (t)t/2µt(s)t− s ds.In the same way, one hasI∗[µ+µ1]− I∗[µ] =−2− 12ˆ t−µ20 (t)t/2µ1t(s)t− s ds.For I021, we get|I021|≲ |µt |ˆ t−µ20 (t)t2ˆ ∞√t2(t− s)−2e− r24(t−s) rdrds≲ |µt |,∣∣I021[µ+µ1]− I021[µ]∣∣≲ supt1∈[t/2,t]|µ1t(t1)|.For I022, we have|I022|≲ˆ t−µ20t/2ˆR4(t− s)−2e−|z|24(t−s) µ−2(s)|µt(s)|( |z|µ(s))−2〈 |z|µ(s)〉−2η( |z|√s)dzds≲ |µt |ˆ t−µ20t/2ˆR4(t− s)−2e−|z|24(t−s) |z|−2 〈 |z|µ(t)〉−2η( |z|2√t)dzds≲ |µt |ˆ t−µ20t/2(t− s)−2ˆ 4√t0e−r24(t−s) 〈 rµ(t)〉−2rdrds≲ |µt |µ2ˆ t−µ20t/2(t− s)−2(1− ln( µ2(t)4(t− s)))ds≲ |µt |since for t2 ≤ s≤ t−µ2(t), ˆ µ(t)0e−r24(t−s) 〈 rµ(t)〉−2rdr ≲ µ2(t),ˆ 4√tµ(t)e−r24(t−s) 〈 rµ(t)〉−2rdr ∼ µ2(t)ˆ 4√tµ(t)e−r24(t−s) r−1dr ∼ µ2(t)ˆ 4tt−sµ2(t)4(t−s)e−zz−1dz≲ µ2(t)(1− ln( µ2(t)4(t− s))).74Next, we estimate I022[µ+µ1]− I022[µ]. By (3.1.17), we have(µ+µ1)−2(µt +µ1t)(Z5(|z|µ+µ1)+232(µ+µ1)2|z|2)−µ−2µt(Z5(|z|µ)+232µ2|z|2)= µ−2µ1tZ5(zµ)+232µ1t|z|2 −µ−3µ1µt(2Z5(zµ)+zµ·∇Z5( zµ ))+(µ−3µ1µt +µ−2µ1t)〈 zµ〉−2O( |µ1|µ+|µ1t ||µt |)= O(µ−2|µ1t | |z|−2µ−2〈 zµ〉−2)−µ−3µ1µt(2Z5(zµ)+zµ·∇Z5( zµ ))+(µ−3µ1µt +µ−2µ1t)〈 zµ〉−2O( |µ1|µ+|µ1t ||µt |)= O(µ−2|µt |( |µ1|µ+|µ1t ||µt |) |z|−2µ−2〈 zµ〉−2)+O(µ−2|µt |( |µ1|µ+|µ1t ||µt |)2〈 zµ〉−2).Similar to the estimates of I022, we then haveˆ t−µ20t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|) |z|−2µ−2(s)〈 zµ(s)〉−2η( |z|√s)dzds≲ |µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|),ˆ t−µ20t/2ˆR4(4π(t− s))−2e−|z|24(t−s) µ−2(s)|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|)2〈 zµ(s)〉−2η( |z|√s)dzds≲ µ−2|µt |( |µ1|µ+|µ1t ||µt |)2ˆ t−µ20t/2(t− s)−2ˆ 2t 120e−r24(t−s) 〈 rµ(t)〉−2r3drds≲ |µt | ln t supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)2sinceˆ µ(t)0e−r24(t−s) 〈 rµ(t)〉−2r3dr ≲ µ4,ˆ 2t 12µ(t)e−r24(t−s) 〈 rµ(t)〉−2r3dr ∼ µ2ˆ 2t 12µ(t)e−r24(t−s) rdr ≲ µ2(t− s).(3.1.18)Therefore, one has∣∣I022[µ+µ1]− I022[µ]∣∣≲ |µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)+ |µt | ln t supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)2.75Let us now estimate I2− I02|I2− I02|=∣∣∣∣ˆ t−µ20 (t)t/2ˆR4ˆ 10(4π(t− s))−2e−|θx−z|24(t−s) θx− z2(t− s) · xµ−2(s)µt(s)Z5(|z|µ(s))η(|z|√s)dθdzds∣∣∣∣≲ µ−2|µt ||x|ˆ t−µ20 (t)t/2ˆR4ˆ 10(t− s)− 52 e−|θx−z|28(t−s) 〈 |z|µ(s)〉−2η( |z|√s)dθdzds≲ µ−2|µt ||x|ˆ t−µ20 (t)t/2ˆR4(t− s)− 52 e−|z|28(t−s) 〈 |z|µ(s)〉−2η( |z|√s)dzds≲ µ−2|µt ||x|ˆ t−µ20 (t)t/2(t− s)− 52ˆ 2√t0e−r28(t−s) 〈 rµ(t)〉−2r3drds≲ µ−1|µt ||x¯|since´ 2√t0 e− r28(t−s) 〈 rµ(t)〉−2r3dr ≲ µ2(t− s) by similar estimate in (4.10.3).Using rearrangement inequality, one has another upper bound for |I2− I02|,|I2− I02|≲ˆ t−µ20 (t)t/2ˆR4(t− s)−2e−|z|24(t−s) µ−2(s)|µt(s)|〈 |z|µ(s)〉−2η(|z|√s)dzds≲ µ−2|µt |ˆ t−µ20 (t)t/2(t− s)−2ˆ 2t 120e−r24(t−s) 〈 rµ(t)〉−2r3drds≲ |µt(t)|ˆ t−µ20 (t)t/2(t− s)−1ds≲ ln t|µt |.Thus|I2− I02|≲min{µ−1|µt ||x¯|, ln t|µt |}.Using (3.1.17) and similar calculations, one has∣∣(I2− I02)[µ+µ1]− (I2− I02)[µ]∣∣≲ µ−1|µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)|x¯|.Combining all the estimates above, we conclude the validity of Lemma 3.1.2.Recalling φ[µ] = φ1[µ]+φ2[µ] and combining Lemma 3.1.1 and Lemma 3.1.2, one has76Corollary 3.1.3. Assume that µ satisfies (3.1.2) and µ1 satisfies |µ1| ≤ µ2 , |µ1t | ≤ |µt |2 . We have|φ[µ]|≲ (µt−1+g[µ])1{|x¯|≤2t 12 }+|µt |(ln(µ−1t 12 )+1) if |x¯| ≤ µ|µt |(ln(|x¯|−1t 12 )+1) if µ < |x¯| ≤ t 12t|µt ||x¯|−2e−|x¯|216t if |x¯|> t 12+O(µ|x¯|−2e− |x¯|216t + |x¯|−6ˆ tt0/2s2|µt(s)|ds+g[µ]e−|x¯|216t)1{|x¯|>2t 12 }≲ (t ln t)−11{|x¯|≤2t 12 }+O(t2(ln t)−1|x¯|−6)1{|x¯|>2t 12 }whereg[µ] = O(t−2ˆ tt0/2(s−1µ3(s)+ s|µt(s)|)ds).|∇x¯φ[µ]|≲t−1 if |x¯| ≤ µt−1(ln t)−2(ln(|x¯|−1t 12 )+1)|x¯|−1+ t− 32 (ln t)−1 if µ < |x¯| ≤ t 12t32 (ln t)−1|x¯|−6 if |x¯|> t 12.∣∣φ[µ+µ1]−φ[µ]∣∣≲ (O(|µ1|t−1)+ g˜[µ,µ1])1{|x¯|≤2t 12 }+ supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)|µt |(ln(µ−1t 12 )+1) if |x¯| ≤ µ|µt |(ln(|x¯|−1t 12 )+1) if µ < |x¯| ≤ t 12t|µt ||x¯|−2e−|x¯|216t if |x¯|> t 12+O(supt1∈[t/2,t]|µ1(t1)||x¯|−2e−|x¯|216t + |x¯|−6(t3 supt1∈[t/2,t]|µ1t(t1)|+ˆ tt0/2s2|µ1t(s)|ds)+ g˜[µ,µ1]e−|x¯|216t)1{|x¯|>2t 12 }whereg˜[µ,µ1] = O(|µt | ln t supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)2)+O(|µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|)+ t−2ˆ tt0/2(s−1|µ1(s)|µ2(s)+ s|µt(s)|( |µ1(s)|µ(s)+|µ1t(s)||µt(s)|))ds).More precisely,φ[µ] =[−2− 12(µt−1+ˆ t−µ20t/2µt(s)t− s ds)+O(µt−2|x¯|2+ |µt |min{|x¯|µ , ln t})+g[µ]]1{|x¯|≤2t 12 }+O(µ|x¯|−2e− |x¯|216t + |x¯|−6ˆ tt0/2s2|µt(s)|ds+g[µ]e−|x¯|216t)1{|x¯|>2t 12 },77φ[µ+µ1]−φ[µ] =[−2− 12(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)+O(|µ1|t−2|x¯|2+ |µt | supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|) |x¯|µ)+ g˜[µ,µ1]]1{|x¯|≤2t 12 }+O(supt1∈[t/2,t]|µ1(t1)||x¯|−2e−|x¯|216t + |x¯|−6(t3 supt1∈[t/2,t]|µ1t(t1)|+ˆ t/2t0/2s2|µ1t(s)|ds)+ e−|x¯|216t g˜[µ,µ1])1{|x¯|>2t 12 }.In order to extract the leading term, we will use the precise version of φ[µ] and φ[µ+µ1]−φ[µ]when calculating the orthogonal equation. In other cases, we are inclined to adopt the rougher upperbound.With the introduction of the correction term φ , the new error is given byS[u1+φ[µ]] = 3u21φ[µ]+3u1φ2[µ]+φ3[µ]+ξt ·∇x¯φ[µ](x−ξ , t)+µ−2ξt ·∇U(x−ξµ )η(x−ξ√t)+µ−1t−12 ξt ·∇η(x−ξ√t )U(x−ξµ).3.1.3 Further improvement by solving an elliptic equationIn order to find suitable parameters to design the topology for solving the inner-outer gluing systemand the orthogonal equation, we will use the corresponding linearized elliptic equation to cut off theerror term so that the time decay rate will be improved.Set the correction term asµ¯−10 Φ0(x−ξµ¯0, t)where µ¯0 is the leading order of µ to be determined later. Formally speaking, Φ0 will be chosen tosatisfy the following equation∆yΦ0+3w2(y)Φ0 ≈ −µ3(3u21φ[µ](x¯, t)+3u1φ2[µ](x¯, t))= −3µ(U2(y)η2(µy√t)φ[µ](µy, t)+µU(y)η(µy√t)φ2[µ](µy, t)).(3.1.19)SetM [µ] :=ˆR4(U2(y)η2(µy√t)φ[µ](µy, t)+µU(y)η(µy√t)φ2[µ](µy, t))Z5(y)dy= µ−4ˆR4(U2(x¯µ)Z5(x¯µ)η2(x¯√t)φ[µ](x¯, t)+µU(x¯µ)Z5(x¯µ)η(x¯√t)φ2[µ](x¯, t))dx¯.In order to find Φ0 with fast spatial decay, we aim to find µ¯0 as the leading order of µ such thatM [µ]≈ 0. In other words, the above orthogonality condition is satisfied at leading order for careful78choice of µ¯0, which will be adjusted and corrected several times in order to further improve the timedecay, and we shall see thatµ¯0 ∼ (ln t)−1.The iteration of finding proper µ¯0 consists of three steps:• the first step is to single out the leading part in the above orthogonal equation, and this resultsin the blow-up rate predicted in [32],• the second step is to add next-order correction of the scaling parameter,• the last step is to iterate the second step finitely many times such that the new error has suffi-ciently fast time decay.We now start the iteration.Step 1. Finding the leading part µ0.Using the precise expression of φ[µ] in Corollary 3.1.3, one hasˆR4U2(y)Z5(y)η2(µy√t)φ(µy, t)dy= −2− 12(µt−1+ˆ t−µ20t/2µt(s)t− s ds)ˆR4U2(y)Z5(y)η2(µy√t)dy+O(µ3t−2 ln(µ−1t12 ))+O(|µt |)+g[µ],andµˆR4U(y)η(µy√t)φ2(µy, t)Z5(y)dy= µˆR4U(y)η(µy√t)Z5(y)O(µ2t−2+g2[µ]+ |µt |2(ln(µ−1t 12 ))2)dy= µ ln(µ−1t12 )O(µ2t−2+g2[µ]+ |µt |2(ln(µ−1t 12 ))2).Therefore, we obtainM [µ] = −2− 12ˆR4U2(y)Z5(y)η2(µy√t)dy(µt−1+ˆ t−µ20t/2µt(s)t− s ds+O(|µt |)+g[µ]+µ ln(µ−1t 12 )O(µ2t−2+g2[µ]+ |µt |2(ln(µ−1t 12 ))2))where´R4U2(y)Z5(y)η2( µy√t )dy< 0 when t is large. Balancing the following two leading termsµt−1+ˆ t−µ20t/2µt(s)t− s ds∼ µt−1+µt(t)ˆ t−µ20t/21t− sds∼ µt−1+µt ln t = 0,79one gets µ0 = (ln t)−1 as the leading order of µ . Notice thatµ0t−1+ˆ t−µ20t/2µ0t(s)t− s ds= (t ln t)−1− t−1ˆ 1−t−1(ln t)−21/2(ln t+ lnz)−2z(1− z) dz= (t ln t)−1− t−1(ln t)−2(1+O((ln t)−1))ˆ 1−t−1(ln t)−21/21z(1− z)dz= (t ln t)−1− t−1(ln t)−2(1+O((ln t)−1)) ln(t(ln t)2−1) = O(t−1(ln t)−2 ln ln t),O(|µ0t |)+g[µ0]+µ0 ln(µ−10 t12 )O(µ20 t−2+ |µ0t |2(ln(µ−10 t12 ))2+g2[µ0])= O(t−1(ln t)−2),and thusM [µ0] = O(t−1(ln t)−2 ln ln t).Step 2. Finding the corrected term µ01.In order to improve the time decay of the error, we introduce the next order term µ01 and makethe ansatz |µ01| µ0, |µ01t | |µ0t |.Then by Corollary 3.1.3, we estimateM [µ0+µ01]=ˆR4((µ0+µ01)−4U2(x¯µ0+µ01)Z5(x¯µ0+µ01)η2(x¯√t)φ[µ0+µ01](x¯, t)+(µ0+µ01)−3U(x¯µ0+µ01)Z5(x¯µ0+µ01)η(x¯√t)φ2[µ0+µ01](x¯, t))dx¯=ˆR4{[(µ0+µ01)−4U2(x¯µ0+µ01)Z5(x¯µ0+µ01)−µ−40 U2(x¯µ0)Z5(x¯µ0)]η2(x¯√t)φ[µ0+µ01](x¯, t)+µ−40 U2(x¯µ0)Z5(x¯µ0)η2(x¯√t)(φ[µ0+µ01]−φ[µ0])(x¯, t)+µ−40 U2(x¯µ0)Z5(x¯µ0)η2(x¯√t)φ[µ0](x¯, t)+[(µ0+µ01)−3U(x¯µ0+µ01)Z5(x¯µ0+µ01)−µ−30 U(x¯µ0)Z5(x¯µ0)]η(x¯√t)φ2[µ0+µ01](x¯, t)+µ−30 U(x¯µ0)Z5(x¯µ0)η(x¯√t)(φ2[µ0+µ01]−φ2[µ0])(x¯, t)+µ−30 U(x¯µ0)Z5(x¯µ0)η(x¯√t)φ2[µ0](x¯, t)}dx¯=ˆR4{O( |µ01|µ0µ−40 〈x¯µ0〉−6)η2(x¯√t)[−2− 12((µ0+µ01)t−1+ˆ t−µ20t/2µ0t(s)+µ01t(s)t− s ds)+O((µ0+µ01)t−2|x¯|2+ |µ0t +µ01t | |x¯|µ0+µ01)+g[µ0+µ01]]+µ−40 U2(x¯µ0)Z5(x¯µ0)η2(x¯√t)[−2− 12(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)+O(|µ01|t−2|x¯|2+ |µ0t | supt1∈[t/2,t]( |µ01(t1)|µ0+|µ01t(t1)||µ0t |) |x¯|µ0)+ g˜[µ0,µ01]]80+|µ01|µ0µ−30 〈x¯µ0〉−4η( x¯√t)O((t ln t)−2)+µ−30 U(x¯µ0)Z5(x¯µ0)η(x¯√t)O((t ln t)−2)}dx¯+M [µ0]=ˆR4{O( |µ01|µ0〈 x¯µ0〉−6)η2(x¯√t)[−2− 12(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)+O(t−1(ln t)−2 ln ln t)+O(µ30 t−2|x¯|2µ20+ |µ0t | |x¯|µ0)]+U2(x¯µ0)Z5(x¯µ0)η2(x¯√t)[−2− 12(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)+O(µ20 |µ01|t−2|x¯|2µ20+ |µ0t | supt1∈[t/2,t]( |µ01(t1)|µ0+|µ01t(t1)||µ0t |) |x¯|µ0)+ g˜[µ0,µ01]]+|µ01|µ20〈 x¯µ0〉−4η( x¯√t)O((t ln t)−2)+µ0〈 x¯µ0 〉−4η(x¯√t)O((t ln t)−2)}d(x¯µ0)+M [µ0]= O(|µ01|µ0)(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)−2− 12ˆR4U2(y)Z5(y)η2(µ0y√t)dy(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)+O(t−1(ln t)−2 ln ln t)|µ01|µ0+O(µ20 |µ01|t−2 ln t+ |µ0t | supt1∈[t/2,t]( |µ01(t1)|µ0+|µ01t(t1)||µ0t |))+ g˜[µ0,µ01]+|µ01|µ20ln tO((t ln t)−2)+µ0 ln tO((t ln t)−2)+M [µ0]=(O(|µ01|µ0)−2− 12ˆR4U2(y)Z5(y)η2(µ0y√t)dy)(µ01t−1+ˆ t−µ20t/2µ01t(s)t− s ds)+O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)+O(t−1(ln t)−1 ln ln t)|µ01|+ g˜[µ0,µ01]+O((t ln t)−2)+M [µ0]=(O(|µ01|µ0)−2− 12ˆR4U2(y)Z5(y)η2(µ0y√t)dy)[µ01t−1(1+O((ln t)−12 ))+ˆ t−t1−ν1t/2µ01t(s)t− s ds+ˆ t−µ20 (t)t−t1−ν1µ01t(t)t− s ds+Eν1 [µ01]+O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)+ g˜[µ01,µ0]+O((t ln t)−2)+M [µ0]]=(O(|µ01|µ0)−2− 12ˆR4U2(y)Z5(y)η2(µ0y√t)dy)[µ01t−1(1+O((ln t)−12 ))+ˆ t−t1−ν1t/2µ01t(s)t− s ds+µ01t((1−ν1) ln t+2ln ln t)+O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)+Eν1 [µ01]+ g˜[µ01,µ0]+O((t ln t)−2)+M [µ0]],whereEν1 [µ01] =ˆ t−µ20 (t)t−t1−ν1µ01t(s)−µ01t(t)t− s ds.Since it is too difficult to solve the nonlocal equation about µ01 thoroughly, we put Eν1 [µ01] aside81as the new error term and consider the following equationµ01t +βν1(t)µ01 = fν1 [µ01], (3.1.20)whereβν1(t) = t−1(1+O((ln t)−12 ))[(1−ν1) ln t+2ln ln t]−1,fν1 [µ01] = χ(t)[(1−ν1) ln t+2ln ln t]−1(−ˆ t−t1−ν1t/2µ01t(s)t− s ds− g˜[µ01,µ0]+O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)+O((t ln t)−2)−M [µ0]),χ(t) is a smooth cut-off function such that χ(t) = 0 for t < 34 t0 and χ(t) = 1 for t ≥ t0. Since µ01(t)will be defined in ( t04 ,∞), the introduction of χ(t) is used to avoid the occurrence of µ01(t) for tbeyond ( t04 ,∞) in the terms like´ t−t1−ν1t/2µ01t(s)t−s ds. After all, the original orthogonal equation is onlyrequired to hold in (t0,∞). For technical reasons, we extend the domain of µ01 to ( t04 ,∞).It then suffices to consider the following fixed point problem:Aν1 [µ01](t) = −ˆ ∞t∂tAν1 [µ01](s)ds=−e−´ t βν1 (u)duˆ ∞te´ s βν1 (u)du fν1 [µ01](s)ds,∂tAν1 [µ01](t) = βν1(t)e−´ t βν1 (u)duˆ ∞te´ s βν1 (u)du fν1 [µ01](s)ds+ fν1 [µ01](t),(3.1.21)where ν1 ∈ (0, 12) will be determined later.SinceO((t ln t)−2)+ |M [µ0]| ≤C0t−1(ln t)−2 ln ln twhereC0 ≥ 1 is a large constant independent of t0, we have∣∣[(1−ν1) ln t+2ln ln t]−1(O((t ln t)−2)+ |M [µ0]|)∣∣≤ (1−ν1)−1C0t−1(ln t)−3 ln ln t.By L’Hôpital’s rule,limt→∞−´ ∞t e´ s βν1 (u)dus−1(lns)−3 ln lnsdse´ t βν1 (u)du(ln t)−2 ln ln t= limt→∞e´ t βν1 (u)dut−1(ln t)−3 ln ln tβν1(t)e´ t βν1 (u)du(ln t)−2 ln ln t+ e´ t βν1 (u)du(−2)t−1(ln t)−3 ln ln t+ e´ t βν1 (u)dut−1(ln t)−3= [(1−ν1)−1−2]−1.Notice ν1 < 12 implies e´ t βν1 (u)du (ln t)2 so that ´ ∞t e´ s βν1 (u)dus−1(lns)−3 ln lnsds is well defined.82Thus we have−´ ∞t e´ s βν1 (u)dus−1(lns)−3 ln lnsdse´ t βν1 (u)du(ln t)−2 ln ln t= [(1−ν1)−1−2]−1+o(1)where o(1)→ 0 as t0 → ∞. Then∣∣∣∣e−´ t βν1 (u)duˆ t∞e´ s βν1 (u)du[(1−ν1) lns+2ln lns]−1s−1(lns)−2 ln lnsds∣∣∣∣≤ (1−ν1)−1∣∣∣∣e−´ t βν1 (u)duˆ t∞e´ s βν1 (u)dus−1(lns)−3 ln lnsds∣∣∣∣= |(2ν1−1)−1+o(1)|(ln t)−2 ln ln t,∣∣∣∣βν1(t)e−´ t βν1 (u)duˆ t∞e´ s βν1 (u)du[(1−ν1) lns+2ln lns]−1s−1(lns)−2 ln lnsds∣∣∣∣+∣∣∣[(1−ν1) ln t+2ln ln t]−1t−1(ln t)−2 ln ln t∣∣∣≤ t−1(1−ν1)−1(ln t)−1|(2ν1−1)−1+o(1)|(ln t)−2 ln ln t+(1−ν1)−1t−1(ln t)−3 ln ln t= (1−ν1)−1(1+ |(2ν1−1)−1+o(1)|)t−1(ln t)−3 ln ln t.From the estimates above, for µ01 ∈C1(t0/4,∞) and µ01(t)→ 0 as t→ ∞, we set the norm as‖µ01‖01 = supt≥t0/4t(ln t)3(ln ln t)−1|µ01t(t)| (3.1.22)and will solve the fixed point problem (3.1.21) in the spaceB01 ={g(t) ∈C1(t0/4,∞), g(t)→ 0 as t→ ∞ : ‖g‖01 ≤ 2C0C(ν1)}whereC(ν1) = (1−ν1)−1(1+ |(2ν1−1)−1+o(1)|). We take ν1 < 12 and t0 large enough to guaranteeC(ν1)< ∞. Let us estimate other terms for ∂tAµ1 [µ01] in (3.1.21).For any µ¯01 ∈ B01,χ(t)∣∣∣ˆ t−t1−ν1t/2µ¯01t(s)t− s ds∣∣∣≤ ‖µ¯01‖01ˆ t−t1−ν1t/2s−1(lns)−3 ln lnst− s ds= ‖µ¯01‖01t−1ˆ 1−t−ν11/2(ln t+ lnz)−3 ln(ln t+ lnz)z(1− z) dz≤ ‖µ¯01‖01(1+O((ln t)− 12 ))t−1(ln t)−3 ln ln tˆ 1−t−ν11/21z(1− z)dz= ‖µ¯01‖01(1+O((ln t)− 12 ))t−1(ln t)−3 ln ln t ln(tν1−1)≤ ‖µ¯01‖01ν1(1+O((ln t)− 12 ))t−1(ln t)−2 ln ln t≤ 2C0C(ν1)ν1(1+O((ln t)− 12 ))t−1(ln t)−2 ln ln t(3.1.23)83which implies ∣∣∣∣[(1−ν1) ln t+2ln ln t+O(1)]−1χ(t)ˆ t−t1−ν1t/2µ¯01t(s)t− s ds∣∣∣∣≤ 2C0C(ν1)ν1(1−ν1)−1(1+O((ln t)− 12 ))t−1(ln t)−3 ln ln t.We take ν1 ∈ (0, 14) to make ν1(1−ν1)−1(1+ |(2ν1−1)−1|)< 1.Since µ¯01 ∈ B01, one has |µ¯01(t)|= |´ ∞t µ¯01t(s)ds|≲ ‖µ¯01‖01(ln t)−2 ln ln t. Thenχ(t)g˜[µ¯01,µ0] = O(t−2ˆ tt0/2(s−1‖µ¯01‖01(lns)−4 ln lns+(lns)−2‖µ¯01‖01(lns)−1 ln lns)ds+(t ln t)−1‖µ¯01‖201((ln t)−1 ln ln t)2)= (ln t0)−12O((‖µ¯01‖01+‖µ¯01‖201)t−1(ln t)−2 ln ln t),χ(t)O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)≲ t−1(ln t)−3 ln ln t ≲ (ln t0)−1t−1(ln t)−2 ln ln t.Then for any fixed ν1 ∈ (0, 14) and t0 large enough, one sees that Aν1 [µ¯01] ∈ B01.The contraction property can be derived similarly. Indeed, for any µ01a,µ01b ∈ B01, similar to(3.1.23), we haveχ(t)∣∣∣∣ˆ t−t1−ν1t/2∂t µ¯01a(s)t− s ds−ˆ t−t1−ν1t/2∂t µ¯01b(s)t− s ds∣∣∣∣≤ ‖µ¯01a− µ¯01b‖01ˆ t−t1−ν1t/2s−1(lns)−3 ln lnst− s ds≤ ‖µ¯01a− µ¯01b‖01ν1(1+O((ln t)− 12 ))t−1(ln t)−2 ln ln t,χ(t)|g˜[µ01a,µ0]− g˜[µ01b,µ0]|= (ln t0)− 12O(C0C(ν1)t−1(ln t)−2 ln ln t)‖µ¯01a− µ¯01b‖01,χ(t)∣∣∣O((t ln t)−1 supt1∈[t/2,t]|µ01a(t1)|+ supt1∈[t/2,t]|∂tµ01a(t1)|)−O((t ln t)−1 supt1∈[t/2,t]|µ01b(t1)|+ supt1∈[t/2,t]|∂tµ01b(t1)|)∣∣∣≲ χ(t)∣∣O((t ln t)−1 supt1∈[t/2,t]|µ01a(t1)−µ01b(t1)|+ supt1∈[t/2,t]|∂tµ01a(t1)−∂tµ01b(t1)|)∣∣≲ ‖µ¯01a− µ¯01b‖01t−1(ln t)−3 ln ln t ≲ (ln t0)−1‖µ¯01a− µ¯01b‖01t−1(ln t)−2 ln ln t(3.1.24)by the estimate of φ[µ0+µ01a]−φ[µ0+µ01b] in Corollary 3.1.3.Due to the choice of ν1 and t0 above, the contraction property is achieved. By contraction mappingtheorem, there exists a unique solution µ01 ∈ B01 for (3.1.21).From now on, ν1 will be regarded as a general constant unless otherwise stated. For notationalsimplicity, ∂t is denoted by “ ′ ”. Once we have solved µ01, the regularity of µ01 can be improved bythe equation of µ01 and µ ′′01 decays to 0 as t → ∞. For the purpose of finding a better decay estimate84of µ ′′01, we take derivative on both sides of (3.1.20). Thenµ ′′01+(βν1(t))′µ01+βν1(t)µ′01 = ( fν1 [µ01])′,where we can evaluate|(βν1(t))′µ01+βν1(t)µ ′01|≲ t−2(ln t)−3 ln ln t,(χ(t)[(1−ν1) ln t+2ln ln t]−1)′(−ˆ t−t1−ν1t/2µ01t(s)t− s ds− g˜[µ01,µ0]+O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|)+O((t ln t)−2)−M [µ0])≲ t−2(ln t)−3 ln ln t,χ(t)[(1−ν1) ln t+2ln ln t]−1(− g˜[µ01,µ0]+O((t ln t)−2)−M [µ0])′ ≲ t−2(ln t)−3 ln ln t,where we used similar calculation in (3.9.2) for (g˜[µ01,µ0])′.χ(t)[(1−ν1) ln t+2ln ln t]−1(−ˆ t−t1−ν1t/2µ01t(s)t− s ds)′= χ(t)[(1−ν1) ln t+2ln ln t]−1[− µ ′01(t− t1−ν1)t1−ν1(1− (1−ν1)t−ν1)+µ ′01(t2)t+ˆ t−t1−ν1t/2µ ′01(s)(t− s)2 ds)]= χ(t)[(1−ν1) ln t+2ln ln t]−1[(1−ν1)µ′01(t− t1−ν1)t− µ′01(t2)t−ˆ t−t1−ν1t/2µ ′′01(s)t− s ds]= −χ(t)[(1−ν1) ln t+2ln ln t]−1ˆ t−t1−ν1t/2µ ′′01(s)t− s ds+O(t−2(ln t)−4 ln ln t).Revisiting the process of proving Corollary 3.1.3, we haveχ(t)[(1−ν1) ln t+2ln ln t]−1(O((t ln t)−1 supt1∈[t/2,t]|µ01(t1)|+ supt1∈[t/2,t]|µ01t(t1)|))′= χ(t)[(1−ν1) ln t+2ln ln t]−1O( supt1∈[t/2,t]|µ ′′01(t1)|)+O(t−2(ln t)−4 ln ln t).(3.1.25)Using the estimates above, we haveµ ′′01 = χ(t)[(1−ν1) ln t+2ln ln t]−1(−ˆ t−t1−ν1t/2µ ′′01(s)t− s ds+O( supt1∈[t/2,t]|µ ′′01(t1)|))+ f˜ (t)where | f˜ (t)| ≤C1t−2(ln t)−3 ln ln t. For this reason, we solve µ ′′01 in the following spaceB2,2 = {g ∈C(t0/4,∞),g(t)→ 0 as t→ ∞ : ‖g‖2,2 ≤ 2C1},85where for g ∈C(t0/4,∞), we define‖g‖a,b := supt≥t0/4ta(ln t)b|g(t)|.For any g1,g2 ∈ B2,2, similar to (3.1.23), we haveχ(t)[(1−ν1) ln t+2ln ln t]−1∣∣∣∣ˆ t−t1−ν1t/2g1(s)t− s ds−ˆ t−t1−ν1t/2g2(s)t− s ds∣∣∣∣≤ (1−ν1)−1(ln t)−1‖g1−g2‖2,2ˆ t−t1−ν1t/2s−2(lns)−2t− s ds≤ ν1(1−ν1)−1‖g1−g2‖2,2t−2(ln t)−2(1+(ln t)−1)(1+(ν1 ln t)−1).Similar to (3.1.25), one hasχ(t)[(1−ν1) ln t+2ln ln t]−1∣∣∣O( supt1∈[t/2,t]|g1(t1)|)−O( supt1∈[t/2,t]|g2(t1)|)∣∣∣≲ χ(t)[(1−ν1) ln t+2ln ln t]−1O( supt1∈[t/2,t]|g1(t1)−g2(t1)|)≲ t−2(ln t)−3‖g1−g2‖2,2.For any g ∈ B2,2, we have∣∣∣χ(t)[(1−ν1) ln t+2ln ln t]−1ˆ t−t1−ν1t/2g(s)t− sds∣∣∣≤ (1−ν1)−1(ln t)−1‖g‖2,2ˆ t−t1−ν1t/2s−2(lns)−2t− s ds≤ ν1(1−ν1)−1‖g‖2,2t−2(ln t)−2(1+(ln t)−1)(1+(ν1 ln t)−1).Since ν1 ∈ (0, 14), when t0 is large enough, the contraction property follows and then µ ′′01 ∈ B2,2. Thusthe improved error is given byM [µ0+µ01] = Eν1 [µ01] =ˆ t−µ20 (t)t−t1−ν1µ ′01(s)−µ ′01(t)t− s ds= O(t−1−ν1(ln t)−2).Step 3. Further improvement by iteration.Repeating Step 2 finitely many times, we can find µ0i, i= 1, . . . ,k0 such thatM[µ0+k0∑i=1µ0i]= O(t−2). (3.1.26)Denoteµ¯0 = µ0+k0∑i=1µ0i.From the construction above, we see that µ¯0 ∼ µ0 = (ln t)−1, µ¯0t ∼ µ0t .86Since µ¯0 is determined, we are now able to describe Φ0 rigorously. Set y¯= x¯µ¯0 and consider∆y¯Φ0+3U2(y¯)Φ0 = H˜(|y¯|, t),whereH˜(|y¯|, t) = −3µ¯0(U2(y¯)η2(µ¯0y¯√t)φ[µ¯0](µ¯0y¯, t)+ µ¯0U(y¯)η(µ¯0y¯√t)φ2[µ¯0](µ¯0y¯, t))+3µ¯0M [µ¯0]η(y¯)Z5(y¯)´B2η(z)Z25(z)dz.Then Φ0(y¯, t) is given byΦ0(y¯, t) = Z˜5(y¯)ˆ |y¯|0H˜(s, t)Z5(s)s3ds−Z5(y¯)ˆ |y¯|0H˜(s, t)Z˜5(s)s3ds,where Z˜5(r) is the other linearly independent kernel of the homogeneous equation, which satisfiesthat the WronskianW [Z5, Z˜5] = r−3, so Z˜5(r)∼ r−2 if r→ 0 and Z˜5(r)∼ 1 if r→ ∞.By the definition ofM [µ¯0], it is easy to haveˆR4H˜(z, t)Z5(z)dz= 0. (3.1.27)By Corollary 3.1.3, |H˜| ≲ t−1(ln t)−2〈y¯〉−4. Due to the special choice of µ¯0, one can get bettertime decay for H˜. Indeed, we haveU2(y¯)η2(µ¯0y¯√t)φ[µ¯0](µ¯0y¯, t)=U2(y¯)η2(µ¯0y¯√t)[−2− 12(µ¯0t−1+ˆ t−µ20t/2µ¯0t(s)t− s ds)+O(µ¯30 t−2|y¯|2+ |µ¯0t ||y¯|)+O(t−1(ln t)−2)]=[O(t−1(ln t)−2|y¯|)+O(t−1(ln t)−2 ln ln t)]〈y¯〉−41{|y¯|≤2µ¯−10 t 12 }= O(t−1(ln t)−2 ln ln t)〈y¯〉−31{|y¯|≤2µ¯−10 t 12 },µ¯0U(y¯)η(µ¯0y¯√t)φ2[µ¯0](µ¯0y¯, t)=O(t−2(ln t)−3)〈y¯〉−21{|y¯|≤2µ¯−10 t 12 }=O(t−1(ln t)−2 ln ln t)〈y¯〉−31{|y¯|≤2µ¯−10 t 12 },which implies |H˜|≲ t−1(ln t)−3 ln ln t〈y¯〉−3. As a result, one has|H˜|≲min{t−1(ln t)−2〈y¯〉−4, t−1(ln t)−3 ln ln t〈y¯〉−3}. (3.1.28)87Claim:|Φ0(y¯, t)|≲ min{t−1(ln t)−2〈y¯〉−2 ln(2+ |y¯|), t−1(ln t)−3 ln ln t〈y¯〉−1},|∇y¯Φ0(y¯, t)|≲ min{t−1(ln t)−2〈y¯〉−3 ln(2+ |y¯|), t−1(ln t)−3 ln ln t〈y¯〉−2},|∂tΦ0(y¯, t)|≲ t−2(ln t)−1〈y¯〉−2 ln(2+ |y¯|).(3.1.29)Indeed, the estimate aboutΦ0(y¯, t) is derived from (3.1.27) and (3.1.28). The upper bound of∇y¯Φ0(y¯, t)follows by scaling argument. In order to estimate ∂tΦ0(y¯, t), we need to take a closer look at ∂tH˜(z, t).By the definition of H˜, it is straightforward to haveˆR4∂tH˜(z, t)Z5(z)dz= 0.∂tH˜(|y¯|, t) =−3µ¯0t(U2(y¯)η2(µ¯0y¯√t)φ[µ¯0](µ¯0y¯, t)+ µ¯0U(y¯)η(µ¯0y¯√t)φ2[µ¯0](µ¯0y¯, t)−M [µ¯0]η(y¯)Z5(y¯)´B2ηZ25dz)−3µ¯0[U2(y¯)2η(µ¯0y¯√t)∇η(µ¯0y¯√t) · µ¯0y¯√t(µ¯0√t)−1∂t(µ¯0√t)φ[µ¯0](µ¯0y¯, t)+U2(y¯)η2(µ¯0y¯√t)(∇x¯φ[µ¯0](µ¯0y¯, t) · µ¯0t y¯+∂tφ[µ¯0](µ¯0y¯, t))+ µ¯0tU(y¯)η(µ¯0y¯√t)φ2[µ¯0](µ¯0y¯, t)+ µ¯0U(y¯)∇η(µ¯0y¯√t) · µ¯0y¯√t(µ¯0√t)−1∂t(µ¯0√t)φ2[µ¯0](µ¯0y¯, t)+ µ¯0U(y¯)η(µ¯0y¯√t)2φ[µ¯0](µ¯0y¯, t)(∇x¯φ[µ¯0](µ¯0y¯, t) · µ¯0t y¯+∂tφ[µ¯0](µ¯0y¯, t))−∂t(M [µ¯0])η(y¯)Z5(y¯)´B2ηZ25dz].Using (3.9.1) in Appendix, one has∣∣∣∇x¯φ[µ¯0](µ¯0y¯, t) · µ¯0t y¯+∂tφ[µ¯0](µ¯0y¯, t)∣∣∣1{µ¯0|y¯|≤2t 12 }≲∣∣∣[(t ln t)−11{|x¯|≤µ0}+(t−1(ln t)−2|x¯|−1+ t− 32 (ln t)−1)1{|x¯|>µ0}]µ¯−10 |µ¯0t ||x¯|+ t−2∣∣∣1{|x¯|≤2t 12 } ≲ t−2.Thus|∂tH˜(|y¯|, t)|≲ t−2(ln t)−3〈y¯〉−4+(ln t)−1[t−2(ln t)−1〈y¯〉−4+ t−2〈y¯〉−4+ t−3(ln t)−4〈y¯〉−2+ t−3(ln t)−3〈y¯〉−2+ t−3(ln t)−2〈y¯〉−2+ t−31{|y¯|≤3}]1{µ¯0|y¯|≤2t12 }≲ t−2(ln t)−3〈y¯〉−4+(ln t)−1(t−2〈y¯〉−4+ t−3(ln t)−2〈y¯〉−2)1{µ¯0|y¯|≤2t12 } ≲ t−2(ln t)−1〈y¯〉−4.Therefore, we have the estimate about ∂tΦ0(y¯, t) in (3.1.29).In order to avoid the influence in the remote region |x¯| ≳ t 12 , we add cut-off function and set88µ¯−10 Φ0(x−ξµ¯0 , t)η(4(x−ξ )√t ) as the correction term. It is easy to check∂t(µ¯−10 Φ0(x−ξµ¯0, t))=−µ¯−20 µ¯0tΦ0(x−ξµ¯0, t)− µ¯−20 ∇y¯Φ0(x−ξµ¯0, t)·(µ¯0t x−ξµ¯0 +ξt)+ µ¯−10 ∂tΦ0(x−ξµ¯0 , t).Set µ = µ¯0+µ1 where |µ1| ≤ µ¯02 , |µ1t | ≤ |µ¯0t |2 . Let us estimate the new errorS[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)](3.1.30)= −∂t(µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))+∆x(µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))+S[u1+φ[µ]]+(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))3− (u1+φ[µ])3= µ¯−10 Φ0(x−ξµ¯0, t)∇η(4(x−ξ )√t) · (4t− 12 ξt +2t− 32 (x−ξ ))−η(4(x−ξ )√t)(− µ¯−20 µ¯0tΦ0(x−ξµ¯0, t)− µ¯−20 ∇y¯Φ0(x−ξµ¯0, t) · (µ¯0t x−ξµ¯0 +ξt)+ µ¯−10 ∂tΦ0(x−ξµ¯0, t))+ µ¯−30 ∆y¯Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+8t−12 µ¯−20 ∇y¯Φ0(x−ξµ¯0, t) ·∇η(4(x−ξ )√t)+16t−1µ¯−10 Φ0(x−ξµ¯0, t)∆η(4(x−ξ )√t)+S[u1+φ[µ]]+3µ¯−30 U2(x−ξµ¯0)Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))3− (u1+φ[µ])3−3µ¯−30 U2(x−ξµ¯0)Φ0(x−ξµ¯0, t)η(4(x−ξ )√t) (3.1.31)= µ¯−10 Φ0(x¯µ¯0, t)∇η(4x¯√t) · (4t− 12 ξt +2t− 32 x¯)+8t−12 µ¯−20 ∇y¯Φ0(x¯µ¯0, t) ·∇η( 4x¯√t)+16t−1µ¯−10 Φ0(x¯µ¯0, t)∆η(4x¯√t)−η( 4x¯√t)[− µ¯−20 µ¯0tΦ0(x¯µ¯0, t)− µ¯−20 ∇y¯Φ0(x¯µ¯0, t) · (µ¯0t x¯µ¯0 +ξt)+ µ¯−10 ∂tΦ0(x¯µ¯0, t)]+3(η2(x¯√t)−η( 4x¯√t))µ−2U2(x¯µ)φ[µ](x¯, t)+3(η(x¯√t)−η( 4x¯√t))µ−1U(x¯µ)φ2[µ](x¯, t)+3η(4x¯√t)(µ−2U2(x¯µ)φ[µ](x¯, t)− µ¯−20 U2(x¯µ¯0)φ[µ¯0](x¯, t))+3η(4x¯√t)(µ−1U(x¯µ)φ2[µ](x¯, t)− µ¯−10 U(x¯µ¯0)φ2[µ¯0](x¯, t))+ξt ·∇x¯φ[µ](x¯, t)+µ−2ξt ·∇U( x¯µ )η(x¯√t)+µ−1t−12U(x¯µ)ξt ·∇η( x¯√t )+φ3[µ]+3µ¯−20 M [µ¯0](ˆB2η(z)Z25(z)dz)−1η( x¯µ¯0)Z5(x¯µ¯0)+(u1+φ[µ]+ µ¯−10 Φ0(x¯µ¯0, t)η(4x¯√t))3− (u1+φ[µ])3−3µ¯−30 U2(x¯µ¯0)Φ0(x¯µ¯0, t)η(4x¯√t).89Claim:∣∣∣∣S[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0 , t)η(4(x−ξ )√t )]∣∣∣∣≲ t−2〈y¯〉−2 ln(2+ |y¯|)1{|x|≤9t 12 }+[t−1(ln t)2|µ1|+(ln t)2(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))]〈y〉−41{|x|≤9t 12 }+ |ξt |(ln t)2〈y〉−31{|x|≤2t 12 }+[|ξt |t 32 (ln t)−1|x|−6+(t2(ln t)−1|x|−6)3]1{|x|>2t 12 }+ t−2(ln t)2η(y¯).(3.1.32)We need to estimate term by term. Indeed, by (3.1.29), one has∣∣∣µ¯−10 Φ0( x¯µ¯0 , t)∇η( 4x¯√t ) · (4t− 12 ξt +2t− 32 x¯)∣∣∣≲ (t ln t)−1〈y¯〉−2 ln(2+ |y¯|)(t− 12 |ξt |+ t−1)1{9−1t 12≤|x|≤9t 12 }∼[t−52 (ln t)−2|ξt |+ t−3(ln t)−2]1{9−1t 12≤|x|≤9t 12 }.Also, we have∣∣∣8t− 12 µ¯−20 ∇y¯Φ0( x¯µ¯0 , t) ·∇η( 4x¯√t )+16t−1µ¯−10 Φ0( x¯µ¯0 , t)∆η( 4x¯√t )∣∣∣≲∣∣∣t− 32 〈y¯〉−3 ln(2+ |y¯|)+ t−2(ln t)−1〈y¯〉−2 ln(2+ |y¯|)∣∣∣1{9−1t 12≤|x|≤9t 12 } ∼ t−3(ln t)−21{9−1t 12≤|x|≤9t 12 },∣∣∣µ¯−20 µ¯0tΦ0( x¯µ¯0 , t)+ µ¯−20 ∇y¯Φ0( x¯µ¯0 , t) · (µ¯0t x¯µ¯0 +ξt)− µ¯−10 ∂tΦ0( x¯µ¯0 , t)∣∣∣η( 4x¯√t)≲[(t ln t)−2〈y¯〉−2 ln(2+ |y¯|)+ t−1|ξt |〈y¯〉−3 ln(2+ |y¯|)+ t−2〈y¯〉−2 ln(2+ |y¯|)]η(4x¯√t)∼(t−1|ξt |〈y¯〉−3 ln(2+ |y¯|)+ t−2〈y¯〉−2 ln(2+ |y¯|))η(4x¯√t).By Corollary 3.1.3, we have the following estimates∣∣∣∣3(η2( x¯√t )−η( 4x¯√t ))µ−2U2( x¯µ )φ[µ](x¯, t)+3(η( x¯√t )−η( 4x¯√t ))µ−1U( x¯µ )φ2[µ](x¯, t)∣∣∣∣≲[(ln t)2〈y〉−4(t ln t)−1+ ln t〈y〉−2(t ln t)−2]1{9−1t 12≤|x¯|≤9t 12 } ∼ (t ln t)−31{9−1t 12≤|x¯|≤9t 12 },∣∣∣∣η( 4x¯√t )(µ−2U2( x¯µ )φ[µ](x¯, t)− µ¯−20 U2( x¯µ¯0 )φ[µ¯0](x¯, t))∣∣∣∣=∣∣∣∣η( 4x¯√t )[(µ−2U2( x¯µ )− µ¯−20 U2( x¯µ¯0 ))φ[µ]+ µ¯−20 U2( x¯µ¯0 )(φ[µ]−φ[µ¯0])]∣∣∣∣≲ η( 4x¯√t)[t−1(ln t)2|µ1|〈y〉−4+(ln t)2〈y〉−4(|g˜[µ¯0,µ1]|+(t ln t)−1 supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))],90η(4x¯√t)∣∣∣(µ−1U( x¯µ )φ2[µ](x¯, t)− µ¯−10 U( x¯µ¯0 )φ2[µ¯0](x¯, t))∣∣∣= η(4x¯√t)∣∣∣(µ−1U( x¯µ )− µ¯−10 U( x¯µ¯0 ))φ2[µ]+ µ¯−10 U( x¯µ¯0 )(φ[µ]−φ[µ¯0])(φ[µ]+φ[µ¯0])∣∣∣≲ η( 4x¯√t)[µ¯−20 |µ1|〈y〉−2(t ln t)−2+ µ¯−10 〈y〉−2(t ln t)−1(|µ1|t−1+ |g˜[µ¯0,µ1]|+ |µt | ln t supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|))]∼ η( 4x¯√t)[t−2|µ1|〈y〉−2+ t−1〈y〉−2(|g˜[µ¯0,µ1]|+(t ln t)−1 supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|))]≲ η( 4x¯√t)[t−1(ln t)2|µ1|〈y〉−4+(ln t)2〈y〉−4(|g˜[µ¯0,µ1]|+(t ln t)−1 supt1∈[t/2,t]( |µ1(t1)|µ(t)+|µ1t(t1)||µt(t)|))].Using Corollary 3.1.3, one has∣∣∣ξt ·∇x¯φ[µ](x¯, t)+µ−2ξt ·∇U( x¯µ )η( x¯√t )+µ−1t− 12U( x¯µ )ξt ·∇η( x¯√t )∣∣∣≲ |ξt |(t−1〈y〉−11{|x¯|≤2t 12 }+ t32 (ln t)−1|x¯|−61{|x¯|>2t 12 }+(ln t)2〈y〉−31{|x¯|≤2t 12 })∼ |ξt |((ln t)2〈y〉−31{|x|≤2t 12 }+ t32 (ln t)−1|x|−61{|x|>2t 12 }).|φ3[µ]|≲ (t ln t)−31{|x¯|≤2t 12 }+(t2(ln t)−1|x¯|−6)31{|x¯|>2t 12 },∣∣∣µ¯−20 M [µ¯0] η(y¯)Z5(y¯)´B2η(z)Z25(z)dz∣∣∣≲ t−2(ln t)2η(y¯).∣∣∣∣(u1+φ[µ]+ µ¯−10 Φ0( x¯µ¯0 , t)η( 4x¯√t ))3− (u1+φ[µ])3−3µ¯−30 U2(x¯µ¯0)Φ0(x¯µ¯0, t)η(4x¯√t)∣∣∣∣= 3∣∣∣∣(µ−1U( x¯µ )− µ¯−10 U( x¯µ¯0 )+φ[µ]+θ µ¯−10 Φ0( x¯µ¯0 , t)η( 4x¯√t ))×(u1+φ[µ]+θ µ¯−10 Φ0(x¯µ¯0, t)η(4x¯√t)+ µ¯−10 U(x¯µ¯0))µ¯−10 Φ0(x¯µ¯0, t)η(4x¯√t)∣∣∣∣≲(|µ1|µ−20 〈y〉−2+(t ln t)−1+ t−1(ln t)−1〈y¯〉−2 ln(2+ |y¯|))× ((t ln t)−1+ ln t〈y〉−2)(t ln t)−1〈y¯〉−2 ln(2+ |y¯|)η( 4x¯√t)≲(|µ1|(ln t)2〈y〉−2+(t ln t)−1)t−1〈y〉−4 ln(2+ |y|)η( 4x¯√t )= |µ1|t−1(ln t)2〈y〉−6 ln(2+ |y|)η( 4x¯√t )+ t−2(ln t)−1〈y〉−4 ln(2+ |y|)η( 4x¯√t).We have completed the proof of claim (3.1.32).913.2 Gluing system and solving the outer problemIn this section, we formulate the inner–outer gluing system such that an infinite time blow-up solutionto (3.1.1) with desired asymptotics can be found. We look for the solution of the formu(x, t) = u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+ψ(x, t)+ηRµ−1ϕ(x−ξµ, t)withηR(x, t) = η(x−ξµ0R(t)), R(t) = tγ , 0< γ <12,where ψ , ϕ are perturbations in the outer region and inner region, respectively. In order for thefollowing to hold0= S[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+ψ+ηRµ−1ϕ(x−ξµ, t)]= −∂tψ−∂tηRµ−1ϕ(x−ξµ , t)+ηRµ−2µt(ϕ(x−ξµ, t)+x−ξµ·∇yϕ(x−ξµ , t))+ηRµ−2ξt ·∇yϕ(x−ξµ , t)−ηRµ−1∂tϕ(x−ξµ, t)+∆xψ+∆xηRµ−1ϕ(x−ξµ, t)+2∇xηR ·µ−2∇yϕ(x−ξµ , t)+ηRµ−3∆yϕ(x−ξµ, t)+3(µ−1U(x−ξµ))2(ψ+ηRµ−1ϕ(x−ξµ , t))+S[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0 , t)η(4(x−ξ )√t )]+(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+ψ+ηRµ−1ϕ(x−ξµ, t))3−(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))3−3(µ−1U(x−ξµ))2(ψ+ηRµ−1ϕ(x−ξµ , t)),it suffices to solve the following inner-outer gluing system for (ψ,ϕ).The outer problem for ψ:∂tψ = ∆xψ+G [ψ,ϕ ,µ1,ξ ] in R4× (t0,∞),92whereG [ψ,ϕ ,µ1,ξ ] := 3µ−2U2(x−ξµ)ψ(1−ηR)+ηRµ−2ξt ·∇yϕ(x−ξµ , t)+∆xηRµ−1ϕ(x−ξµ, t)+2∇xηR ·µ−2∇yϕ(x−ξµ , t)−∂tηRµ−1ϕ(x−ξµ, t)+(1−ηR)S[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)]+(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+ψ+ηRµ−1ϕ(x−ξµ, t))3−(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))3−3(µ−1U(x−ξµ))2(ψ+ηRµ−1ϕ(x−ξµ , t))−[3(u1+φ[µ]−µ−1U(x−ξµ ))(u1+φ[µ]+µ−1U(x−ξµ))+6(u1+φ[µ])µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)]ηRµ−1ϕ(x−ξµ, t).(3.2.1)The inner problem for ϕ :µ2∂tϕ(y, t)=∆yϕ(y, t)+3U2(y)ϕ(y, t)+ f1(y, t)ϕ(y, t)+ f2(t)y ·∇yϕ(y, t)+H [ψ,µ1,ξ ](y, t) in D4R(3.2.2)where y= x−ξµ , D4R = {(y, t) : y ∈ B4R(t), t ∈ (t0,∞)}, andf1(y, t) = µµt +µ2[3(u1+φ[µ]−µ−1U(y))(u1+φ[µ]+µ−1U(y))+6(u1+φ[µ])µ¯−10 Φ0(µyµ¯0, t)],f2(t) = µµt ,H [ψ,µ1,ξ ](y, t) := µ3(3µ−2U2(y)ψ(µy+ξ , t)+S[u1+φ[µ]+ µ¯−10 Φ0(µyµ¯0, t)η(4µy√t)]). (3.2.3)By (3.1.32), one has∣∣H [ψ,µ1,ξ ](y, t)∣∣≲ (ln t)−1〈y〉−4|ψ(µy+ξ , t)|+ t−2(ln t)−3〈y¯〉−2 ln(2+ |y¯|)+(t ln t)−1|µ1|〈y〉−4+(ln t)−1(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))〈y〉−4+ |ξt |(ln t)−1〈y〉−3+ t−2(ln t)−1η(y¯)≲ (ln t)−1〈y〉−4|ψ(µy+ξ , t)|+ ta1γ−2(ln t)−2〈y〉−2−a1+(t ln t)−1|µ1|〈y〉−4+(ln t)−1(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))〈y〉−4+ |ξt |(ln t)−1〈y〉−3 (3.2.4)93where we have used |y| ≤ 4R= 4tγ , and for later purpose, we require thata1γ−2< 5δ −κ−aγ, 0< a1 ≤ 1. (3.2.5)Here above constants are those which measure the weighted topology for the inner problem (see(3.2.8)). Notice in D4R, we have| f1(y, t)|+ | f2(t)|≲ t−1(ln t)−3+[(t ln t)−1 ln t〈y〉−2+ ln t〈y〉−2t−1(ln t)−1〈y¯〉−2 ln(2+ |y¯|)](ln t)−2∼ t−1(ln t)−3+ t−1(ln t)−2〈y〉−2.Remark 3.2.0.1. Due to the slow time decay rate of f1(y, t), f2(t), the terms f1(y, t)ϕ(y, t)+ f2(t)y ·∇yϕ(y, t) have to be regarded as the linear part of the inner problem. We can not put this term on theright hand side of the outer problem since this will influence the Hölder continuity of ψ in t variable.Besides, we can not use the inner linear theory in [16] since f1(y, t)ϕ(y, t)+ f2(t)y ·∇yϕ(y, t) willinfluence the Hölder property about µ1t through the orthogonal equation, which will result in failureto choose suitable topology for solving the inner–outer gluing system. Instead, we rebuild a new innerlinear theory in Section 3.6 to avoid including f1(y, t)ϕ(y, t) + f2(t)y ·∇yϕ(y, t) in the orthogonalequation about µ1.We decompose the inner problem (3.2.2) into two parts. Set ϕ = ϕ1 + ϕ2, then it suffices toconsiderµ2∂tϕ1(y, t) = ∆yϕ1(y, t)+3U2(y)ϕ1(y, t)+ f1(y, t)ϕ1(y, t)+ f2(t)y ·∇yϕ1(y, t)+H [ψ,µ1,ξ ](y, t)+(ˆB2η(z)Z25(z)dz)−1(−2− 12 3ˆB2R0U2(z)Z5(z)dz+O((t ln t)−1))µEν [µ1]η(y)Z5(y) in D4R,(3.2.6)µ2∂tϕ2(y, t) = ∆yϕ2(y, t)+3U2(y)ϕ2(y, t)+ f1(y, t)ϕ2(y, t)+ f2(t)y ·∇yϕ2(y, t)− (ˆB2η(z)Z25(z)dz)−1(−2− 12 3ˆB2R0U2(z)Z5(z)dz+O((t ln t)−1))µEν [µ1]η(y)Z5(y) in D4R,(3.2.7)where R0(τ) = τδ with δ > 0 very small andEν [µ1] =ˆ t−µ20 (t)t−t1−νµ1t(s)−µ1t(t)t− s ds.Setτ(t) =ˆ tt0µ−2(s)ds+ t0(ln t0)2, τ0 = t0(ln t0)2.Then τ(t)∼ t(ln t)2 for all t ≥ t0. In τ variable, D4R = {(y,τ) : y ∈ B4R(t(τ)), τ ∈ (τ0,∞)}. It is easyto rewrite (3.2.6) and (3.2.7) in the form as in Proposition 3.6.1 and Lemma 3.6.5, respectively.The reason for decomposing the inner problem into the above two parts is that the orthogonal94equation involving µ1 is too difficult to solve. More detailed explanations will be given in Section3.3.1.Before starting the solvability of the outer problem, let us first fix the inner solution ϕ to the innerproblem, the next order of scaling parameter µ1 and translating parameter ξ in the spaces with thefollowing norms‖ϕ‖i,κ−5δ ,a := sup(y,τ)∈D4Rτκ−5δ 〈y〉a(〈y〉|∇ϕ(y, t(τ))|+ |ϕ(y, t(τ))|) (3.2.8)where κ , a are some positive constants to be determined later. For µ1(t) ∈ C1( t04 ,∞), µ1(t)→ 0 ast→ ∞, denote‖µ1‖∗1 := supt≥t0/4[ln t(t(ln t)2)5δ−κR(t)−a]−1|µ1t |. (3.2.9)For ξ (t) = (ξ1(t), . . . ,ξ4(t)) ∈C1(t0,∞), ξ (t)→ 0 as t→ ∞, denote‖ξ‖∗2 := max1≤ j≤4supt≥t0[(ln t)2(t(ln t)2)5δ−κR(t)−a]−1|ξ jt |. (3.2.10)The outer problem is solved in the following Proposition.Proposition 3.2.1. Consider∂tψ(x, t) = ∆ψ(x, t)+G [ψ,ϕ ,µ1,ξ ] in R4× (t0,∞), ψ(x, t0) = 0 in R4 (3.2.11)where G [ψ,ϕ ,µ1,ξ ] is given in (3.2.1). Assume ϕ , µ1, ξ satisfy ‖ϕ‖i,κ−5δ ,a,‖µ1‖∗1,‖ξ‖∗2 < Λ1where Λ1 > 1 is a constant and the parameters satisfy5δ −κ−aγ >−2, 5δ −κ <−1, 0< a< 2, 0< γ < 12, (3.2.12)then there exists a solution ψ = ψ[ϕ ,µ1,ξ ] with the following estimates:|ψ(x, t)| ≤C(Λ1) ln t(t(ln t)2)5δ−κR−a(1{|x|≤t 12 }+ t|x|−21{|x|>t 12 }),|∇ψ(x, t)| ≤C(Λ1) ln t(t(ln t)2)5δ−κR−a,sups1,s2∈(t− λ2(t)4 ,t)|ψ(x,s1)−ψ(x,s2)||s1− s2|α ≤C(Λ1,α){λ−2α(t) ln t(t(ln t)2)5δ−κR−a+λ 2−2α(t)[(µ0R)−2 ln t(t(ln t)2)5δ−κR−a+(ln t)3(t(ln t)2)10δ−2κ]}where 0< λ (t)≤ t 12 .95The proof is postponed to Appendix 3.8.3.3 Orthogonal equations for µ1, ξ3.3.1 Solving µ1 and ξIn order to utilize Proposition 3.6.1 with R0 = τδ ∼ (t(ln t)2)δ where δ > 0 is small, one needs toadjust µ1, ξ such that ci[H ] = 0, i= 1, . . . ,5 in Proposition 3.6.1 withH given in (3.2.3).However, for i = 5, it is too difficult to solve c5[H ] = 0 thoroughly. We are only able to makec5[H ] ≈ 0 and leave a smaller remainder to be solved by the non-orthogonal linear theory of theinner problem.In this section, we only care about the estimate in |y| ≤ 4R since this is served for the innerproblem. SetMi[ψ,µ1,ξ ] =ˆB2R0H [ψ,µ1,ξ ](y, t)Zi(y)dy, i= 1, . . . ,5,H5(|y|, t) =ˆS3H [ψ,µ1,ξ ](|y|θ , t)ϒ0(θ)dθ , Hi(|y|, t) =ˆS3H [ψ,µ1,ξ ](|y|θ , t)ϒi(θ)dθ , i= 1, . . . ,4where ϒi are spherical harmonic functions, which are given in Section 3.6.Using (3.1.30), for i= 5, since Z5 is radially symmetric, one hasM5[ψ,µ1,ξ ] =ˆB2R03µU2(y)ψ(µy+ξ , t)Z5(y)dy+ˆB2R0µ3(µ¯−20 µ¯0tΦ0(µyµ¯0, t)+ µ¯−20 ∇y¯Φ0(µyµ¯0, t) · µ¯0t µyµ¯0 − µ¯−10 ∂tΦ0(µyµ¯0, t))Z5(y)dy+ˆB2R03µ3(µ−2U2(y)φ[µ](µy, t)− µ¯−20 U2(µyµ¯0)φ[µ¯0](µy, t))Z5(y)dy+ˆB2R03µ3(µ−1U(y)φ2[µ](µy, t)− µ¯−10 U(µyµ¯0)φ2[µ¯0](µy, t))Z5(y)dy+ˆB2R0µ3φ3[µ](µy, t)Z5(y)dy+ˆB4R3µ3µ¯−20 M [µ¯0]η(µ¯−10 µy)Z5(µ¯−10 µy)´B2η(z)Z25(z)dzZ5(y)dy+ˆB2R0µ3[(u1+φ[µ]+ µ¯−10 Φ0(µyµ¯0, t)η(4µy√t))3− (u1+φ[µ])3−3µ¯−30 U2(µyµ¯0 )Φ0(µyµ¯0 , t)η(4µy√t )]Z5(y)dy,96andH5(|y|, t) =ˆS33µU2(|y|θ)ψ(µ|y|θ +ξ , t)ϒ0(θ)dθ+ˆS3µ3(µ¯−20 µ¯0tΦ0(µ|y|θµ¯0, t)+ µ¯−20 ∇y¯Φ0(µ|y|θµ¯0, t) · µ¯0t µ|y|θµ¯0 − µ¯−10 ∂tΦ0(µ|y|θµ¯0, t))ϒ0(θ)dθ+ˆS33µ3(µ−2U2(|y|θ)φ[µ](µ|y|θ , t)− µ¯−20 U2(µ|y|θµ¯0)φ[µ¯0](µ|y|θ , t))ϒ0(θ)dθ+ˆS33µ3(µ−1U(|y|θ)φ2[µ](µ|y|θ , t)− µ¯−10 U(µ|y|θµ¯0)φ2[µ¯0](µ|y|θ , t))ϒ0(θ)dθ+ˆS3µ3φ3[µ](µ|y|θ , t)ϒ0(θ)dθ +ˆS33µ3µ¯−20 M [µ¯0]η(µ¯−10 µ|y|θ)Z5(µ¯−10 µ|y|θ)´B2η(z)Z25(z)dzϒ0(θ)dθ+ˆS3µ3[(u1+φ[µ]+ µ¯−10 Φ0(µ|y|θµ¯0, t)η(4µ|y|θ√t))3− (u1+φ[µ])3−3µ¯−30 U2(µ|y|θµ¯0)Φ0(µ|y|θµ¯0, t)η(4µ|y|θ√t)]ϒ0(θ)dθ .For i= 1, . . . ,4, we haveMi[ψ,µ1,ξ ] =ˆB2R03µU2(y)ψ(µy+ξ , t)Zi(y)dy+ξitˆB2R0µ3µ¯−20 ∂y¯iΦ0(µyµ¯0, t)Zi(y)dy+ξitˆB2R0(µ3∂x¯iφ(µy, t)Zi(y)+µZ2i (y))dy=ˆB2R03µU2(y)ψ(µy+ξ , t)Zi(y)dy+µξit(ˆB2R0Z2i (y)dy+O(t− 12 ))by Corollary 3.1.3 and (3.1.29). Also,Hi(|y|, t) =ˆS33µU2(|y|θ)ψ(µ|y|θ +ξ , t)ϒi(θ)dθ +ξitˆS3µ3µ¯−20 ∂y¯iΦ0(µ|y|θµ¯0, t)ϒi(θ)dθ+ξitˆS3µ3(∂x¯iφ(µ|y|θ , t)+µ−2∂ziU(|y|θ)η(µ|y|θ√t))ϒi(θ)dθ .Using similar calculations as in (3.1.32), one has|H5(|y|, t)|≲ (ln t)−1〈y〉−3 supz∈B4R(t)〈z〉−1|ψ(µz+ξ , t)|+(ln t)−3{t−2〈y¯〉−2 ln(2+ |y¯|)+[t−1(ln t)2|µ1|+(ln t)2(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))]〈y〉−4+ t−2(ln t)21{|y|≤4}}≲ (ln t)−1〈y〉−3 supz∈B4R(t)〈z〉−1|ψ(µz+ξ , t)|+ ta1γ−2(ln t)−2〈y〉−2−a1+[(t ln t)−1|µ1|+(ln t)−1(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|))]〈y〉−497where a1 > 0 is chosen such that a1γ−2< 5δ −κ−aγ . It then follows that|Hi(|y|, t)|≲ (ln t)−1〈y〉−3 supz∈B4R(t)〈z〉−1|ψ(µz+ξ , t)|+ |ξit |(ln t)2〈y〉−3, i= 1, . . .4.By Proposition 3.6.1, the orthogonal equation ci[H ][τ] = 0 (i= 1, . . . ,4) is equivalent to solvingMi[ψ,µ1,ξ ]+ (t(ln t)2)−δε0O(supy∈B4R(t)〈y〉3|Hi(y, t)|)=ˆB2R03µU2(y)ψ(µy+ξ , t)Zi(y)dy+µξit(ˆB2R0Z2i (y)dy+O(t− 12 ))+(t(ln t)2)−δε0O((ln t)−1 supz∈B4R(t)〈z〉−1|ψ(µz+ξ , t)|+ |ξit |(ln t)2)= 0where ε0 > 0 is given in Proposition 3.6.1. One can write the above equation asξit =Πi[µ1,ξ ] (3.3.1)whereΠi[µ1,ξ ] =(ˆB2R0Z2i (y)dy+O(t(ln t)2)−δε02)−1[−ˆB2R03U2(y)ψ(µy+ξ , t)Zi(y)dy− (t(ln t)2)−δε0O( supz∈B4R(t)〈z〉−1|ψ(µz+ξ , t)|)].Let us estimateM5 term by term. By (3.1.29), one hasˆB2R0µ3[µ¯−20 µ¯0tΦ0(µyµ¯0, t)+ µ¯−20 ∇y¯Φ0(µyµ¯0, t) · µ¯0t µyµ¯0 − µ¯−10 ∂tΦ0(µyµ¯0, t)]Z5(y)dy= O(t−2(ln t)−1).By Corollary 3.1.3 and the special choice of µ¯0, we have for |x¯| ≤ 2t 12φ[µ¯0] = −2− 12(µ¯0t−1+ˆ t−µ20t/2µ¯0t(s)t− s ds)+O(µ¯0t−2|x¯|2+ |µ¯0t | |x¯|µ¯0)+O(t−1(ln t)−2)= O(t−1(ln t)−2 ln ln t)+O(µ¯0t−2|x¯|2+ |µ¯0t | |x¯|µ¯0).Notice thatµ−2U2(y)− µ¯−20 U2(µyµ¯0) = −2µ1(θµ+(1−θ)µ¯0)−3(U2(yθ )+U(yθ )∇U(yθ ) · yθ )∣∣∣yθ=x−ξθµ+(1−θ)µ¯0= −2µ1µ¯−30 (U2(y)+U(y)∇U(y) · y)+O(µ21 µ¯−40 〈y〉−4).98Then by Corollary 3.1.3, it follows thatˆB2R03µ3(µ−2U2(y)φ[µ](µy, t)− µ¯−20 U2(µyµ¯0)φ[µ¯0](µy, t))Z5(y)dy= 3µ3ˆB2R0[µ−2U2(y)(φ[µ]−φ[µ¯0])+(µ−2U2(y)− µ¯−20 U2(µyµ¯0))φ[µ¯0]]Z5(y)dy= 3µ3ˆB2R0{µ−2U2(y)Z5(y)[−2− 12(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)+O(|µ1|µ¯20 t−2|x¯|2µ¯20+ |µ¯0t | supt1∈[t/2,t]( |µ1(t1)|µ¯0+|µ1t(t1)||µ¯0t |) |x¯|µ¯0)+ g˜[µ¯0,µ1]]+[−2µ1µ¯−30 (U2(y)+U(y)∇U(y) · y)+O(µ21 µ¯−40 〈y〉−4)]Z5(y)×(O(t−1(ln t)−2 ln ln t)+O(µ¯0t−2|x¯|2+ |µ¯0t | |x¯|µ¯0 ))}dy= µ[−2− 12 3ˆB2R0U2(y)Z5(y)dy(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+ g˜[µ¯0,µ1]+O(|µ1|t−1(ln t)−1 ln ln t)].By Corollary 3.1.3, we haveˆB2R03µ3(µ−1U(y)φ2[µ](µy, t)− µ¯−10 U(µyµ¯0)φ2[µ¯0](µy, t))Z5(y)dy= 3µ3ˆB2R0[µ−1U(y)(φ[µ]−φ[µ¯0])(φ[µ]+φ[µ¯0])+(µ−1U(y)− µ¯−10 U(µyµ¯0))φ2[µ¯0]]Z5(y)dy= 3µ3ˆB2R0{µ−1U(y)Z5(y)[−2− 12(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)+O(|µ1|µ¯20 t−2|x¯|2µ¯20+ |µ¯0t | supt1∈[t/2,t]( |µ1(t1)|µ¯0+|µ1t(t1)||µ¯0t |) |x¯|µ¯0)+ g˜[µ¯0,µ1]](t ln t)−1+(−µ1µ¯−20 (U(y)+ y ·∇U(y))+O(µ21 µ¯−30 )〈y〉−2)Z5(y)(t ln t)−2}dy= µ3{[(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)O(ln t)+O(|µ1|µ¯20 t−2R20+ |µ¯0t | supt1∈[t/2,t]( |µ1(t1)|µ¯0+|µ1t(t1)||µ¯0t |)R0)+O(ln t)g˜[µ¯0,µ1]]O(t−1)+µ1µ¯−20 O(t−2(ln t)−1)}= µ[(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)O((t ln t)−1)99+O(t−32 supt1∈[t/2,t]|µ1(t1)|+ t− 12 supt1∈[t/2,t]|µ1t(t1)|)+O((t ln t)−1)g˜[µ¯0,µ1]]since δ > 0 is very small and µ−1U(y)− µ¯−10 U( µyµ¯0 ) =−µ1µ¯−20 (U(y)+y ·∇U(y))+O(µ21 µ¯−30 )〈y〉−2.Similarly, the following estimates holdˆB2R0µ3φ3[µ](µy, t)Z5(y)dy= O(µ3(t ln t)−3R20) = O(t−52 )when δ is small enough.∣∣∣∣ˆB2R0µ3µ¯−20 M [µ¯0]η(y¯)Z5(y¯)´B2η(z)Z25(z)dzZ5(y)dy∣∣∣∣≲ O(t−2(ln t)−1),ˆB2R0µ3[(u1+φ[µ]+ µ¯−10 Φ0(µyµ¯0, t)η(4µy√t))3− (u1+φ[µ])3−3µ¯−30 U2(µyµ¯0 )Φ0(µyµ¯0 , t)η(4µy√t )]Z5(y)dy=ˆB2R0µ3[(u1+φ[µ]+ µ¯−10 Φ0(µyµ¯0, t))3− (u1+φ[µ])3−3µ−2U2(y)µ¯−10 Φ0(µyµ¯0 , t)+3(µ−2U2(y)− µ¯−20 U2(µyµ¯0))µ¯−10 Φ0(µyµ¯0, t)]Z5(y)dy= µO(|µ1|(t ln t)−1 ln ln t)+O(t−2(ln t)−4)since by (3.1.29),ˆB2R03µ3[(µ−1U(y))2− (µ¯−10 U(µyµ¯0))2]µ¯−10 Φ0(µyµ¯0, t)Z5(y)dy= 3µ3µ¯−10ˆB2R0[−2µ1µ¯−30 (U2(y)+U(y)∇U(y) · y)+O(µ21 µ¯−40 〈y〉−4)]O(t−1(ln t)−3 ln ln t〈y〉−1)Z5(y)dy= µO(|µ1|(t ln t)−1 ln ln t),∣∣∣∣∣ˆB2R0µ3[(u1+φ[µ]+ µ¯−10 Φ0(µyµ¯0, t))3− (u1+φ[µ])3−3µ−2U2(y)µ¯−10 Φ0(µyµ¯0 , t)]Z5(y)dy∣∣∣∣∣=∣∣∣∣∣µ3ˆB2R0[3(u1+φ[µ])(µ¯−10 Φ0(µyµ¯0, t))2+(µ¯−10 Φ0(µyµ¯0, t))3+3(u1+φ[µ]−µ−1U(y))(u1+φ[µ]+µ−1U(y))µ¯−10 Φ0(µyµ¯0, t)]Z5(y)dy∣∣∣∣∣≲ µ(ln t)−2ˆB2R0[(ln t〈y〉−2+(t ln t)−1)((t ln t)−1〈y¯〉−2 ln(2+ |y¯|))2+((t ln t)−1〈y¯〉−2 ln(2+ |y¯|))3+(ln t〈y〉−21{|x¯|≥t 12 }+(t ln t)−1)(ln t〈y〉−2+(t ln t)−1)(t ln t)−1〈y¯〉−2 ln(2+ |y¯|)]〈y〉−2dy100≲ µ(ln t)−2ˆB2R0[ln t〈y〉−2((t ln t)−1〈y¯〉−2 ln(2+ |y¯|))2+((t ln t)−1〈y¯〉−2 ln(2+ |y¯|))3+(t ln t)−1 ln t〈y〉−2(t ln t)−1〈y¯〉−2 ln(2+ |y¯|)]〈y〉−2dy≲ t−2(ln t)−4.Finally, we getM5[ψ,µ1,ξ ] =ˆB2R03µU2(y)ψ(µy+ξ , t)Z5(y)dy+O(t−2(ln t)−1)+µ[−2− 12 3ˆB2R0U2(y)Z5(y)dy(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+ g˜[µ¯0,µ1]+O(|µ1|t−1(ln t)−1 ln ln t)]+µ[(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)O((t ln t)−1)+O(t−32 supt1∈[t/2,t]|µ1(t1)|+ t− 12 supt1∈[t/2,t]|µ1t(t1)|)+O((t ln t)−1)g˜[µ¯0,µ1]]+µO(|µ1|(t ln t)−1 ln ln t)= µ{ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy+O(t−2)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+O(g˜[µ¯0,µ1])+O(|µ1|(t ln t)−1 ln ln t)+(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))(µ1t−1+ˆ t−µ20t/2µ1t(s)t− s ds)}= µ(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))×{(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))−1ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy+O(t−2)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+O(g˜[µ¯0,µ1])+µ1t−1(1+O((ln t)−1 ln ln t))+ˆ t−t1−νt/2µ1t(s)t− s ds+µ1t((1−ν) ln t+2ln ln t)+Eν [µ1]}whereEν [µ1] =ˆ t−µ20 (t)t−t1−νµ1t(s)−µ1t(t)t− s ds.101By Proposition 3.6.1, c5[H ] = 0 is equivalent toM5[ψ,µ1,ξ ]+ (t(ln t)2)−δε0O(supy∈B4R(t)〈y〉2+a1 |H5(y, t)|)= µ(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))×{(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))−1ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy+O(t−2)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+O(g˜[µ¯0,µ1])+µ1t−1(1+O((ln t)−1 ln ln t))+ˆ t−t1−νt/2µ1t(s)t− s ds+µ1t((1−ν) ln t+2ln ln t)+Eν [µ1]}+(t(ln t)2)−δε0O((ln t)−1 supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|+ ta1γ−2(ln t)−2+(t ln t)−1|µ1|+(ln t)−1(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|)))= µ(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))×{(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))−1ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy+O(t−2)+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+O(g˜[µ¯0,µ1])+µ1t−1(1+O((ln t)−1 ln ln t))+ˆ t−t1−νt/2µ1t(s)t− s ds+µ1t((1−ν) ln t+2ln ln t)+Eν [µ1]+ (t(ln t)2)−δε0O( supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|+ ta1γ−2(ln t)−1)}= 0where we have used similar calculations as in (3.2.4), andsupy∈B4R(t)〈y〉2+a1 |H5(y, t)|≲ (ln t)−1 supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|+ ta1γ−2(ln t)−2+(t ln t)−1|µ1|+(ln t)−1(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|)).Similar to the methodology in Section 3.1.3, we leave Eν [µ1] as the remainder term and consider102the following equation about µ1(−2− 12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))−1ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy+O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)+O(g˜[µ¯0,µ1])+µ1t−1(1+O((ln t)−1 ln ln t))+ˆ t−t1−νt/2µ1t(s)t− s ds+µ1t((1−ν) ln t+2ln ln t)+(t(ln t)2)−δε0O( supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|)+(t(ln t)2)−δε0O(ta1γ−2(ln t)−1) = 0when a1γ > δε0. That is,µ1t +βv(t)µ1 =Π5[µ1,ξ ] (3.3.2)whereβν(t) = t−1 [(1−ν) ln t+2ln ln t]−1 (1+O((ln t)−1 ln ln t)),Π5[µ1,ξ ] = χ(t)((1−ν) ln t+2ln ln t)−1[−ˆ t−t1−νt/2µ1t(s)t− s ds−O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)−O(g˜[µ¯0,µ1])+(2−12 3ˆB2R0U2(y)Z5(y)dy+O((t ln t)−1))−1ˆB2R03U2(y)ψ(µy+ξ , t)Z5(y)dy− (t(ln t)2)−δε0O( supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|)− (t(ln t)2)−δε0O(ta1γ−2(ln t)−1)].(3.3.3)Similar to (3.1.21), in order to solve (3.3.2) and (3.3.1), it is sufficient to consider the followingfixed point problem:S5[µ1,ξ ](t) = Π5[µ1,ξ ](t)+βν(t)e−´ t βν (u)duˆ ∞te´ s βν (u)duΠ5[µ1,ξ ](s)ds,Si[µ1,ξ ](t) = Πi[µ1,ξ ], i= 1, . . .4.(3.3.4)Notice that |ψ|≲ ln t(t(ln t)2)5δ−κR−a and recall the norms (3.2.9), (3.2.10) for µ1, ξ . We will solve(3.3.4) in the following spacesBµ1 = {µ1 ∈C1(t0/4,∞) : ‖µ1‖∗1 ≤ 2}, Bξ = {ξ ∈C1(t0,∞) : ‖ξ‖∗2 ≤ 2}. (3.3.5)For any µ1a,µ1b ∈ Bµ1 and ξa,ξb ∈ Bξ , similar to (3.1.23), one hasχ(t)∣∣∣∣ˆ t−t1−νt/2µ1at(s)−µ1bt(s)t− s ds∣∣∣∣≤ ‖µ1a−µ1b‖∗1ˆ t−t1−νt/2lns(s(lns)2)5δ−κR−a(s)t− s ds= ‖µ1a−µ1b‖∗1(1+O((ln t)−1))ν(ln t)2(t(ln t)2)5δ−κR−a.103By gradient estimate in Proposition 3.2.1, we have∣∣∣∣ˆB2R0U2(y)(ψ(µ1ay+ξa, t)−ψ(µ1by+ξb, t))Z5(y)dy∣∣∣∣≤C ln t(t(ln t)2)5δ−κR−a(|µ1a−µ1b|+ |ξa−ξb|)≤Ct ln t[ln t(t(ln t)2)5δ−κR−a]2(‖µ1a−µ1b‖∗1+‖ξa−ξb‖∗2).The estimate for´B2R03U2(y)ψ(µy+ξ , t)Zi(y)dy is the same.(t(ln t)2)−δε0∣∣∣∣O( supy∈B4R(t)〈y〉−1|ψ(µ1ay+ξa, t)|)−O( supy∈B4R(t)〈y〉−1|ψ(µ1by+ξb, t)|)∣∣∣∣≤ (t(ln t)2)−δε0∣∣∣∣O( supy∈B4R(t)〈y〉−1|ψ(µ1ay+ξa, t)−ψ(µ1by+ξb, t)|)∣∣∣∣≤C(t(ln t)2)−δε0t ln t[ln t(t(ln t)2)5δ−κR−a]2(‖µ1a−µ1b‖∗1+‖ξa−ξb‖∗2)since O( supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|) depends on ψ linearly.From the same calculations as in (3.8.1), one hasχ(t)|g˜[µ¯0,µ1a]− g˜[µ¯0,µ1b]| ≤C ln t(t(ln t)2)5δ−κR−a‖µ1a−µ1b‖∗1when 5δ −κ−aγ >−2. Similar to (3.1.24), one hasχ(t)∣∣∣O((t ln t)−1 supt1∈[t/2,t]|µ1a(t1)|+ supt1∈[t/2,t]|µ1at(t1)|)−O((t ln t)−1 supt1∈[t/2,t]|µ1b(t1)|+ supt1∈[t/2,t]|µ1bt(t1)|)∣∣∣≲ χ(t)O((t ln t)−1 supt1∈[t/2,t]|µ1a(t1)−µ1b(t1)|+ supt1∈[t/2,t]|µ1at(t1)−µ1bt(t1)|)≲ ln t(t(ln t)2)5δ−κR−a‖µ1a−µ1b‖∗1.In conclusion, under the following restrictionsa1 > 0, a1γ−2< 5δ −κ−aγ, a1γ > δε0, 5δ −κ−aγ >−2, 0< ν < 12 , (3.3.6)for t0 is sufficiently large, (S5,Si) is a contraction mapping in Bµ1×Bξ .Similarly, for (µ1,ξ ) ∈ Bµ1×Bξ , we haveχ(t)∣∣∣ˆ t−t1−νt/2µ1t(s)t− s ds∣∣∣≤ ‖µ1‖∗1(1+O((ln t)−1))v(ln t)2(t(ln t)2)5δ−κR−a,|g˜[µ¯0,µ1]| ≤C ln t(t(ln t)2)5δ−κR−a‖µ1‖∗1,χ(t)∣∣∣O((t ln t)−1 supt1∈[t/2,t]|µ1(t1)|+ supt1∈[t/2,t]|µ1t(t1)|)∣∣∣≲ ln t(t(ln t)2)5δ−κR−a‖µ1‖∗1.104Then(S5,Si) : Bµ1×Bξ → Bµ1×Bξ .Consequently, by the contraction mapping theorem, we find a unique solution (µ1,ξ ) in Bµ1×Bξ .3.3.2 Hölder continuity of µ1t and estimate for µEν [µ1]In order to estimate the left errorEν [µ1] =ˆ t−µ20 (t)t−t1−νµ1t(s)−µ1t(t)t− s ds,we need Hölder estimate of µ1t , which satisfiesµ1t =Π5[µ1,ξ ](t)+βν(t)e−´ t βν (u)duˆ ∞te´ s βν (u)duΠ5[µ1,ξ ](s)ds.Assume 3t4 ≤ t2 < t1 ≤ t, 89 < A < 1. A will be chosen to be close to 1 later depending on ν andindependent of t0. We revisit (3.3.3) term by term.Notice that ψ only has Hölder continuity in t variable, which restricts the regularity for µ1t . UsingProposition 3.2.1 with λ 2(t) = t 12 , one has|ψ(µ(t1)y+ξ (t1), t1)−ψ(µ(t2)y+ξ (t2), t2)|≤ |ψ(µ(t1)y+ξ (t1), t1)−ψ(µ(t2)y+ξ (t2), t1)|+ |ψ(µ(t2)y+ξ (t2), t1)−ψ(µ(t2)y+ξ (t2), t2)|≲[ln t(t(ln t)2)5δ−κR−a(t)|y|+(ln t)2(t(ln t)2)5δ−κR−a(t)] ln t(t(ln t)2)5δ−κR−a(t)|t1− t2|+C(α){λ−2α(t) ln t(t(ln t)2)5δ−κR−a(t)+λ 2−2α(t)[(µ0R)−2(t) ln t(t(ln t)2)5δ−κR−a(t)+(ln t)3(t(ln t)2)10δ−2κ ]}|t1− t2|αwhich implies(ln t1)−1∣∣∣ˆB2R0U2(y)Z5(y)(ψ(µ(t1)y+ξ (t1), t1)−ψ(µ(t2)y+ξ (t2), t2))dy∣∣∣≲ [ln t(t(ln t)2)5δ−κR−a(t)]2|t1− t2|+C(α){λ−2α(t)(t(ln t)2)5δ−κR−a(t)+λ 2−2α(t)[(µ0R)−2(t)(t(ln t)2)5δ−κR−a(t)+(ln t)2(t(ln t)2)10δ−2κ ]}|t1− t2|α .105Similarly, (t(ln t)2)−δε0O( supy∈B4R(t)〈y〉−1|ψ(µy+ξ , t)|) provides the same Hölder estimate as above.Reviewing the analysis details in solving (3.3.4), one has|Π5[µ1,ξ ]|≲ ln t(t(ln t)2)5δ−κR−aχ(t).Then∣∣∣(βν(t)e−´ t βν (u)duˆ ∞te´ s βν (u)duΠ5[µ1,ξ ](s)ds)′∣∣∣= ∣∣∣∣∣β ′ν(t)e−´ t βν (u)duˆ ∞te´ s βν (u)duΠ5[µ1,ξ ](s)ds−β 2ν (t)e−´ t βν (u)duˆ ∞te´ s βν (u)duΠ5[µ1,ξ ](s)ds−βν(t)Π5[µ1,ξ ]∣∣∣∣∣≲ t−2(ln t)−1e−´ t βν (u)duˆ ∞te´ s βν (u)du lns(s(lns)2)5δ−κR−a(s)ds+(t ln t)−2e−´ t βν (u)duˆ ∞te´ s βν (u)du lns(s(lns)2)5δ−κR−a(s)ds+(t ln t)−1 ln t(t(ln t)2)5δ−κR−a≲ (t ln t)−1 ln t(t(ln t)2)5δ−κR−awhere in the last inequality, we have used e−´ t βν (u)du ´ ∞t e´ s βν (u)du lns(s(lns)2)5δ−κR−a(s)ds≲ t ln t(t(ln t)2)5δ−κR−awhen t0 is sufficiently large. Also∣∣[((1−ν) ln t+2ln ln t)−1χ(t)]′[((1−ν) ln t+2ln ln t)Π5[µ1,ξ ]]∣∣≲ t−1 ln t(t(ln t)2)5δ−κR−a1{t≥ 3t04 },∣∣∣((1−ν) ln t+2ln ln t)−1[−O((t ln t)−1 supτ1∈[t/2,t]|µ1(τ1)|)−O(g˜[µ¯0,µ1])− (t(ln t)2)−δε0O(ta1γ−2(ln t)−1)]′∣∣∣≲ (ln t)−1(t−1 ln t(t(ln t)2)5δ−κR−a+ t−1(t(ln t)2)−δε0O(ta1γ−2(ln t)−1))≲ (t ln t)−1 ln t(t(ln t)2)5δ−κR−a.In order to get the estimate((1−ν) ln t+2ln ln t)−1∣∣∣O( supτ1∈[t1/2,t1]|µ1t(τ1)|)−O( supτ1∈[t2/2,t2]|µ1t(τ1)|)∣∣∣≲ (ln t)−1(C(A)t−1 ln t(t(ln t)2)5δ−κR−a|t1− t2|+[µ1t ]Cα ( 3At4 ,t)|t1− t2|α),rigorously speaking, we need to estimate all the terms that appeared in the proof of Lemma 3.1.1 andLemma 3.1.2 except the leading term. For simplicity, we take φ˜1b[µ+µ1]− φ˜1b[µ] as an example toillustrate the key idea. We decompose φ˜1b[µ+µ1]− φ˜1b[µ] into two parts to estimate.(φ˜1b[µ+µ1]− φ˜1b[µ])(x¯, t) =T out4 [−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ]](x¯, t)=(ˆ Att0+ˆ tAt)ˆR4(4π(t− s))−2e−|x¯−z|24(t−s) (−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)dzds.106Here x¯ is regarded to be independent of t. Then∣∣∣∣∂t(ˆ Att0ˆR4(4π(t− s))−2e−|x¯−z|24(t−s) (−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)dzds)∣∣∣∣=∣∣∣∣∣AˆR4(4π(t−At))−2e−|x¯−z|24(t−At) (−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,At)dz+ˆ Att0ˆR4∂t((4π(t− s))−2e−|x¯−z|24(t−s) )(−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)dzds∣∣∣∣∣≲C(A)t−2ˆR4e−|x¯−z|24(t−At)(|µ1t(At)|t−11{|z|≤t 12 }+ |µ1t(At)||z|−2e−|z|24t 1{|z|>t 12 }+ |µ1(At)|µ2(At)t−31{2−1√t≤|z|≤4√t})dz+ˆ Att0ˆR4(t− s)−3e−|x¯−z|28(t−s)∣∣−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)∣∣dzds≲C(A)t−2ˆR4e−|z|24(t−At)(|µ1t(At)|t−11{|z|≤t 12 }+ |µ1t(At)||z|−2e−|z|24t 1{|z|>t 12 }+ |µ1(At)|µ2(At)t−31{|z|≤4√t})dz+C(A)t−1ˆ Att0ˆR4(t− s)−2e−|x¯−z|28(t−s)∣∣∣∣(−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)∣∣∣∣dzds≲C(A)[t−1|µ1t(At)|+ t−3|µ1(At)|µ2(At)+ t−1ˆ Att0ˆR4(t− s)−2e−|x¯−z|28(t−s)∣∣∣∣(−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)∣∣∣∣dzds]≲C(A)t−1 ln t(t(ln t)2)5δ−κR−awhere the last inequality follows from the same calculations as in (3.1.11).For the other part, we haveˆ tAtˆR4(4π(t− s))−2e−|x¯−z|24(t−s)(−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z,s)dzds=ˆ 1AˆR4t(t− ta)−2e−|x¯−z|24(t−ta)(−µ1t φˆ1+(E− E˜)[µ+µ1]− (E− E˜)[µ])(z, ta)dzda.The terms independent of µ1t are C1 in time variable t. We only need to focus on the terms107including µ1t . By similar calculations in (3.1.11), we have∣∣∣∣ˆ 1AˆR4∂t(t(t− ta)−2e−|x¯−z|24(t−ta) )(µ1t φˆ1)(z, ta)dzda∣∣∣∣≲ ˆ 1AˆR4(t− ta)−2e−|x¯−z|28(t−ta) )|µ1t φˆ1(z, ta)|dzda≲ t−1 ln t(t(ln t)2)5δ−κR−a,∣∣∣∣ˆ 1AˆR4t(t− ta)−2e−|x¯−z|24(t−ta) (µ1t(t1a)−µ1t(t2a))φˆ1(z, ta)dzda∣∣∣∣≲ˆ 1AˆR4t(t− ta)−2e−|x¯−z|24(t−ta) [µ1t ]Cα ( 3At4 ,t)|t1− t2|αaα |φˆ1(z, ta)|dzda≲ [µ1t ]Cα ( 3At4 ,t)|t1− t2|α .Next, for´ t−t1−νt/2µ1t(s)t−s ds= (´ Att/2+´ t−t1−νAt )µ1t(s)t−s ds, we have∣∣∣∣(ˆ Att/2µ1t(s)t− s ds)′∣∣∣∣= ∣∣∣∣Aµ1t(At)t−At − µ1t( t2)t −ˆ Att/2µ1t(s)(t− s)2 ds∣∣∣∣≲C(A)t−1 ln t(t(ln t)2)5δ−κR−a,and ∣∣∣∣∣ˆ t1−t1−ν1At1µ1t(s)t1− s ds−ˆ t2−t1−ν2At2µ1t(s)t2− s ds∣∣∣∣∣=∣∣∣∣∣ˆ 1−t−ν1Aµ1t(t1z)1− z dz−ˆ 1−t−ν2Aµ1t(t2z)1− z dz∣∣∣∣∣=∣∣∣∣∣ˆ 1−t−ν1Aµ1t(t1z)−µ1t(t2z)1− z dz+ˆ 1−t−ν11−t−ν2µ1t(t2z)1− z dz∣∣∣∣∣≤ |t1− t2|α [µ1t ]Cα ( 3At4 ,t)ˆ 1−t−ν1Azα1− zdz+Cˆ 1−t−ν11−t−ν2(t2z)5δ−κ−aγ(ln(t2z))1+2(5δ−κ)1− z dz= |t1− t2|α [µ1t ]Cα ( 3At4 ,t)ν ln t1(1+O(| ln(1−A)|(ln t1)−1))+C(1+O((ln t2)−1))t5δ−κ−aγ2 (ln t2)1+2(5δ−κ)ˆ 1−t−ν11−t−ν2z5δ−κ−aγ1− z dz≤ |t1− t2|α [µ1t ]Cα ( 3At4 ,t)ν ln t1(1+O(| ln(1−A)|(ln t1)−1))+C(1+O((ln t2)−1))t5δ−κ−aγ2 (ln t2)1+2(5δ−κ)(1+O(t−1))ν | ln t1− ln t2|≤ |t1− t2|α [µ1t ]Cα ( 3At4 ,t)ν ln t1(1+O(| ln(1−A)|(ln t1)−1))+C(1+O((ln t2)−1))t5δ−κ−aγ2 (ln t2)1+2(5δ−κ)(1+O(t−1))νt−12 |t1− t2|.108Combining the estimates above, one gets that|µ1t(t1)−µ1t(t2)|≤C(A)[(t ln t)−1 ln t(t(ln t)2)5δ−κR−a(t)+ t−1 ln t(t(ln t)2)5δ−κR−a(t)1{t≥ 3t04 }]|t1− t2|+C(α){λ−2α(t)(t(ln t)2)5δ−κR−a(t)+λ 2−2α(t)[(µ0R)−2(t)(t(ln t)2)5δ−κR−a(t)+(ln t)2(t(ln t)2)10δ−2κ ]}|t1− t2|α+ |t1− t2|α [µ1t ]Cα ( 3At4 ,t)[ν(1−ν)−1(1+O(| ln(1−A)|(ln t)−1)+O((ln t)−1))]where we have used 3t4 ≤ t2 < t1 ≤ t. Thus one has[µ1t ]Cα ( 3t4 ,t)≤C(A,α)ρ(t)+[µ1t ]Cα ( 3At4 ,t)[ν(1−ν)−1(1+O(| ln(1−A)|(ln t0)−1)+O((ln t)−1)])1{t≥ 3t04 }whereρ(t) = t−α ln t(t(ln t)2)5δ−κR−a+ t1−α [(µ0R)−2(t(ln t)2)5δ−κR−a+(ln t)2(t(ln t)2)10δ−2κ ].Thussupt02 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3t4 ,t)≤C(A,α)+ [ν(1−ν)−1(1+O(| ln(1−A)|(ln t0)−1))+O((ln t0)−1)] supt02 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3At4 ,t)1{t≥ 3t04 }=C(A,α)+ [ν(1−ν)−1(1+O(| ln(1−A)|(ln t0)−1))+O((ln t0)−1)] sup3t04 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3At4 ,t).Noticeρ−1(t)[µ1t ]Cα ( 3At4 ,t) = ρ−1(t) sups1,s2∈( 3At4 ,t)|µ1t(s1)−µ1t(s2)||s1− s2|α= max{ρ−1(t) sups1,s2∈( 3t4 ,t)|µ1t(s1)−µ1t(s2)||s1− s2|α ,ρ−1(t) sups1,s2∈( 3At4 ,At)|µ1t(s1)−µ1t(s2)||s1− s2|α ,ρ−1(t) sups2∈( 3At4 , 3t4 ),s1∈(At,t)|µ1t(s1)−µ1t(s2)||s1− s2|α}≤ max{ρ−1(t) sups1,s2∈( 3t4 ,t)|µ1t(s1)−µ1t(s2)||s1− s2|α ,ρ−1(t)ρ(At)ρ−1(At) sups1,s2∈( 3At4 ,At)|µ1t(s1)−µ1t(s2)||s1− s2|α}+C,109then one hassup3t04 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3At4 ,t) ≤ AC(α,δ ,κ,a)(1+O((ln t0)−1)) supt02 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3t4 ,t)+C(A,α).Thus, when ν < 12 , taking A close to 1 sufficiently, which depends on ν , and then making t0 largeenough, one has supt02 ≤t≤Tρ−1(t)[µ1t ]Cα ( 3t4 ,t) ≤C(ν ,α). Letting T → ∞, one finally getssupt≥ t02ρ−1(t)[µ1t ]Cα ( 3t4 ,t) ≤C(ν ,α). (3.3.7)Finally, we estimate µEν [µ1] as follows|µEν [µ1]|≲C(ν ,α){(ln t)−1t−να ln t(t(ln t)2)5δ−κR−a+(ln t)−1t1−να(µ0R)−2(t(ln t)2)5δ−κR−a+(ln t)−1t1−να(ln t)2(t(ln t)2)10δ−2κ}.(3.3.8)Although Cν ,α goes to ∞ as ν → 12 and α → 1, the smallness is given by t−ε0 where ε > 0 whensolving (3.2.7). Once ν and α are fixed, we take t0 large enough.3.4 Solving the inner problemRecalling (3.2.8), for any fixed ϕ ∈ Bi withBi = {ϕ : ‖ϕ‖i,κ−5δ ,a ≤ 2Ci} (3.4.1)where Ci > 1 is a constant, we have found ψ[ϕ ] ∈ Bo, µ1[ϕ ] ∈ Bµ1 and ξ [ϕ ] ∈ Bξ . We abbre-viate H [ϕ ] =H [ψ[ϕ ],µ1[ϕ ],ξ [ϕ ]]. By (3.2.4), (3.2.5) and (3.3.5), we obtain |H [ϕ ](y, t(τ))| ≲τ5δ−κ(lnτ)4R−a(t(τ))〈y〉−2−a1 . The orthogonal equations of µ1 and ξ have been solved in Section3.3.1, then by Proposition 3.6.1, one finds a solution for (3.2.6) satisfying〈y〉|∇ϕ1(y,τ)|+ |ϕ1(y,τ)|≲ τ5δ−κ(lnτ)4R−a(t(τ))R50〈y〉−a1 ≲ τ−ε0 τ5δ−κ〈y〉−awith ε > 0 sufficiently small providedγmin{a,a1}> 5δ . (3.4.2)Combining (3.3.8) and Lemma 3.6.5, one can find a solution for (3.2.7) with the estimate〈y〉|∇ϕ2(y,τ)|+ |ϕ2(y,τ)|≲ τ−ε0 τ5δ−κ〈y〉−a110if−να+(2−a)γ < 0, 1−να−aγ < 0, 1−να+5δ −κ+2γ < 0, 0< a< 2. (3.4.3)Combining (3.2.5), (3.2.12), (3.3.6), (3.4.2), (3.4.3) and the assumption about parameters inProposition 3.6.1 and Lemma 3.6.5, one needs to choose parameters such that all the inequalitiesbelow hold5δ −κ−aγ >−2, 5δ −κ <−1, 0< a< 2, 0< γ < 12, 0< α < 1, 0< ν <12,a1γ−2< 5δ −κ−aγ, 0< a1 ≤ 1, γmin{a,a1}> 5δ , 6δ < 1,−να+(2−a)γ < 0, 1−να−aγ < 0, 1−να+5δ −κ+2γ < 0.(3.4.4)There exists a solution given by1< κ ≤ 54,2−2αν−1+κ+ν < a< 2,1−ανa< γ <−1+κ+ν2,2−κ2< αν <12, 0< α < 1, 0< ν <12, 6δ < 1,0< 5δ < κ−1, a1γ < 5δ +2−κ−aγ, 5δ < γmin{a,a1}, 0< a1 ≤ 1.(3.4.5)Indeed, one may take for example κ = 98 ,ν =49100 ,α =375392 ,a=3619 ,γ =932 ,5δ =164 ,a1 =19 .Thanks to (3.4.4), the desired ϕ1, ϕ2 can really be found and then ϕ1+ϕ2 ∈ Bi when τ0 is largeenough. The compactness is a consequence of parabolic estimates, so we can find a solution for theinner problem (3.2.2). Making more efforts to calculate the Lipschitz continuity of H [ϕ ] about ϕ ,one can prove the existence of the inner problem (3.2.2) by the contraction mapping theorem.Collecting the estimates in Proposition 3.2.1, Corollary 3.1.3, (3.1.29) and (3.4.1), one gets∣∣∣∣φ[µ]+ µ¯−10 Φ0(x−ξµ¯0 , t)η(4(x−ξ )√t )+ψ+ηRµ−1ϕ(x−ξµ , t)∣∣∣∣≲ (t ln t)−11{|x¯|≤2t 12 }+O(t2(ln t)−1|x¯|−6)1{|x¯|>2t 12 }+(ln t)t−1(ln t)−2〈y¯〉−2 ln(2+ |y¯|)1{|x¯|≤t 12 }+ ln t(t(ln t)2)5δ−κR−a(1{|x|≤t 12 }+ t|x|−21{|x|>t 12 })+(t(ln t)2)−κ+5δ 〈y〉−a1{|y|≤4R}≲ (t ln t)−11{|x¯|≤2t 12 }+O((ln t)−1|x¯|−2)1{|x¯|>2t 12 } ≲ (ln t)−1min{t−1, |x|−2}.Positivity of the solution u. We will demonstrate that the initial value u(x, t0) that we take in theconstruction is positive. For simplicity, we abuse the symbols µ = µ(t0), µ¯0 = µ¯0(t0) in the remainder111of this section. Indeed, recalling (3.1.6), (3.1.29) and (3.4.1), we haveu(x, t0) = µ−1U(x¯µ)η(x¯√t0)+ φ˜1(x¯, t0)+ µ¯−10 Φ0(x¯µ¯0, t0)η(4x¯√t0)+ηRµ−1ϕ(x¯µ, t0)= 232 µ|x¯|−2{e− |x¯|24t0 −(1+( |x¯|µ)2)−1+2−32 µ−1|x¯|2µ¯−10 Φ0(x¯µ¯0, t0)η(4x¯√t0)η( |x¯|√t0)+ e−|x¯|24t0(1−η( |x¯|√t0))}+ηRµ−1ϕ(x¯µ, t0)≥ 2 32 µ|x¯|−21− |x¯|24t0−(1+( |x¯|µ)2)−1−C|x¯|2(t0 ln t0)−1 ln ln t0η(4x¯√t0)η( |x¯|√t0)+ηRµ−1ϕ(x¯µ, t0)= 232 µ|x¯|−2|x¯|2[(µ2+ |x¯|2)−1− (4t0)−1−C(t0 ln t0)−1 ln ln t0η(4x¯√t0)]η( |x¯|√t0)+ηRµ−1ϕ(x¯µ, t0)≥ 2− 12 µ−1(1+ |y|2)−1η( |x¯|√t0)−C1µ−1(t0(ln t0)2)5δ−κ〈y〉−aηR > 0where we have used η(s)= 0 for s≥ 32 to make [58(µ2+|x¯|2)−1−(4t0)−1−C(t0 ln t0)−1 ln ln t0η( 4x¯√t0 )]η(|x¯|√t0)≥0 when t0 is large, and 5δ −κ + γ(2− a) < 0 is used in the last inequality. Therefore, the solutionu(x, t) is positive by the maximum principle.3.5 Stability of blow-up: proof of Theorem 1.0.6In this section, we will analyze the stability of the blow-up solution constructed in Theorem 1.0.5.Proof of Theorem 1.0.6. Consider any perturbation g0(x) satisfying |g0(x)| ≲ t−min{ℓ,4}20 〈x〉−ℓ, ℓ > 2.Setψ0(x, t) = (4π(t− t0))−2ˆR4e−|x−z|24(t−t0) g0(z)dzwhich satisfies ∂tψ0 = ∆ψ0 in R4×(t0,∞), ψ(x, t0) = g0(x) in R4. Without loss of generality, we onlyconsider the case 2< ℓ < 4. By Lemma 3.7.3, one has|ψ0(x, t)|≲ t−ℓ20(〈t− t0〉− ℓ2 1{|x|≤〈t−t0〉 12 }+ |x|−ℓ1{|x|>〈t−t0〉12 })≲ t− ℓ2 1{|x|≤t 12 }+ |x|−ℓ1{|x|>t 12 }.We modify the proof of Proposition 3.2.1 slightly in order to match the perturbation g0. Indeed,we split ψ = ψ¯+ψ0 and consider∂tψ¯(x, t) = ∆ψ¯(x, t)+G [ψ¯+ψ0,ϕ ,µ1,ξ ] in R4× (t0,∞), ψ¯(x, t0) = 0 in R4.When ℓ > 2(κ + aγ), by (3.4.5), one has |ψ0| ≲ t−ε0 wo, and thus ψ¯ can be solved in Bo by the same112method in Proposition 3.2.1. Repeating the rest procedures in the construction of Theorem 1.0.5,(µ1,ξ ,ϕ ,e0) = (µ1[g0],ξ [g0],ϕ [g0],e[g0]) can be solved in the same topology that we have used be-fore, and the leading order of blowup rate µ¯0 ∼ (ln t)−1 remains the same. The perturbed initial valueis then given by[(µ¯0+µ1[g0])−1U(x−ξ [g0]µ¯0+µ1[g0])η(x−ξ [g0]√t)+232 (µ¯0+µ1[g0])|x−ξ [g0]|−2(e−|x−ξ [g0]|24t −η(x−ξ [g0]√t))+ µ¯−10 Φ0(x−ξ [g0]µ¯0, t)η(4(x−ξ [g0])√t)+η(x−ξ [g0]µ0R)e0[g0](µ¯0+µ1[g0])−1Z0(x−ξ [g0]µ¯0+µ1[g0])]∣∣∣∣∣t=t0+g0.From (3.4.5), κ > 1 and aγ > 1−αν . So all ℓ > 3 is permitted for κ and αν close to 1 and 12 ,respectively.In the radial setting, the translation parameter ξ ≡ 0 automatically in (1.0.17). Then for 2< ℓ≤ 3,we put 3u21ψ0 into the right hand side in the equation (3.1.19). Since |ψ0(x, t)|≲ t−ℓ2 , ℓ > 2, the extraterm involving 3u21ψ0 will not influence the leading order µ0 and will be absorbed into Φ0. Butrecalling the construction of µ¯0 in Section 3.1.3, µ¯0 depends on g0, namely, µ¯0 = µ¯0[g0].We omit the tedious calculations about the Lipschitz continuity with respect to g0 for ψ,ϕ ,µ1,ξhere.Remark 3.5.0.1.• In general nonradial case and ℓ > 2, since ψ0 is not radial about x¯= x−ξ , the previous ODEsolution about (3.1.19) is not allowed. Instead, we can expand (3.1.19) by modes similar to themanipulation in section 3.6 and solve the leading order of µ and ξ . Since this involves moretechnicalities, given the length of this chapter, we refrain from considering such a generalityhere.• The borderline ℓ > 2 is also provided in [32].• The stability result can be expected for |g0(x)| ≲ t−10 (ln t0)−b1〈x〉−2(ln(|x|+ 2))−b2 for someb1,b2 > 0. The proof can be in fact achieved by similar computations as in the proof of Theorem1.0.6.3.6 Linear theory for the inner problemIn this section, we develop a linear theory for the associated inner problem. Since the constructionis independent of the spatial dimension n, we assume n ≥ 3 in this section unless specifically stated113otherwise. SetDR = {(y,τ) | y ∈ BR(τ), τ ∈ (τ0,∞)}, ∂DR = {(y,τ) | y ∈ ∂BR(τ), τ ∈ (τ0,∞)}.We consider the associated linear problem∂τϕ = ∆ϕ + pU p−1ϕ + f1(y,τ)ϕ + f2(y,τ)y ·∇ϕ +h(y,τ) in DR (3.6.1)wherep=n+2n−2 , U(y) = (n(n−2))n−24(1+ |y|2)− n−22 .Throughout this section, we always assume that f1, f2 satisfyfi(y,τ) = fi(|y|,τ) are radial in space, i= 1,2, | f1|, | f2|, |y||∇ f2| ≤C f τ−d , d > 0, C f ≥ 0. (3.6.2)It is easier to make mode expansion by spherical harmonic functions when f1 and f2 are radial.And it is very possible to generalize the linear theory without the assumption that f1 and f2 are radiallysymmetric.Recall that the linearized operator ∆+ pU p−1 has only one positive eigenvalue γ0 > 0 such that∆Z0+ pU p−1Z0 = γ0Z0, (3.6.3)where the corresponding eigenfunction Z0 ∈ L∞(Rn) is radially symmetric with the asymptotic be-haviorZ0(y)∼ |y|− n−12 e−√γ0|y| as |y| → ∞.The bounded kernels of ∆+ pU p−1 are given byZi(y) = ∂yiU(y), i= 1,2, . . . ,n, Zn+1(y) = y ·∇U(y)+n−22U(y).Define the weighted L∞ norm‖h‖v,a := sup(y,τ)∈DRv−1(τ)〈y〉a|h(y,τ)|where a≥ 0 is a constant. Throughout this section, we assume R(τ), v(τ) ∈C1(τ0,∞) with the formv(τ) = a0τa1(lnτ)a2(ln lnτ)a3 · · · , R(τ) = b0τb1(lnτ)b2(ln lnτ)b3 · · · , v(τ)> 0, 1 R(τ) τ 12 ,v′(τ) = O(τ−1v(τ)), R′(τ) = O(τ−1R(τ))where a0,b0 > 0, ai,bi ∈ R, i= 1,2, . . . . For brevity, we write v= v(τ), R= R(τ).114We impose a linear constraint on the initial value ϕ(y,τ0) to handle the instability caused by Z0.Consider the associated Cauchy problem∂τϕ = ∆ϕ + pU p−1(y)ϕ + f1(y,τ)ϕ + f2(y,τ)y ·∇ϕ +h, in DR,ϕ(y,τ0) = e0Z0(y), in BR(τ0), (3.6.4)where τ0 is sufficiently large. Formally speaking, when R τ d2−ε for some ε > 0, we can expect thatf1ϕ + f2y ·∇ϕ is a small perturbation since | fi| τ−2εR−2 ≲ τ−2ε〈y〉−2 in DR.The construction of a solution to (3.6.4) is achieved by decomposing the equation into differentspherical harmonic modes. Consider an orthonormal basis {ϒi}∞i=0 made up of spherical harmonicfunctions in L2(Sn−1), namely eigenvalues of the problem∆Sn−1ϒ j+ ι jϒ j = 0 in Sn−1.where 0= ι0 < ι1 = ι2 = · · ·= ιn = n−1< ιn+1 ≤ . . . and´Sn−1 ϒi(θ)ϒ j(θ)dθ = δi j. More precisely,ϒ0(y) = a0, ϒi(y) = a1yi, i= 1, · · · ,n for two constants a0, a1 and the eigenvalue ιl = l(n−2+ l) hasmultiplicity (n+ l−1l)−(n+ l−3l−2)for l ≥ 2.For h(·,τ) ∈ L2(BR(τ)), we decompose h into the formh(y,τ) =∞∑j=0h j(r,τ)ϒ j(y/r), r = |y|, h j(r,τ) =ˆSn−1h(rθ ,τ)ϒ j(θ)dθ .Write h= h0+h1+h⊥ withh0 = h0(r,τ)ϒ0, h1 =n∑j=1h j(r,τ)ϒ j, h⊥ =∞∑j=n+1h j(r,τ)ϒ j.Also, we decompose ϕ = ϕ 0+ϕ 1+ϕ⊥ in a similar form. Then looking for a solution to problem(3.6.4) is equivalent to finding the pairs (ϕ 0,h0), (ϕ 1,h1), (ϕ⊥,h⊥) in each mode.The key linear theory for the inner problem is stated as follows.Proposition 3.6.1. Consider∂τϕ = ∆ϕ + pUp−1ϕ + f1ϕ + f2y ·∇ϕ +h(y,τ)+n+1∑i=1ci(τ)η(y)Zi(y) in DRϕ(y,τ0) = e0Z0(y) in BR(τ0)where n ≥ 4, ‖h‖v,2+a < ∞, 0 < a < 2. Suppose that R2 τd−, R0 =Cτδ 1, δ ≥ 0 and Rn+20 115τmin{1,d}−, then for τ0 sufficiently large, there exists (ϕ ,e0,ci) solving above equation, and (ϕ ,e0,ci)=(T3i[h],T3e[h],ci[h]) defines a linear mapping of h with the estimates〈y〉|∇ϕ |+ |ϕ |≲ Rn+10 v〈y〉−a‖h‖v,2+a, |e0|≲ v(τ0)R2−a0 (τ0)‖h‖v,2+a,ci[h](τ) = −(ˆB2η(y)Z2i (y)dy)−1(ˆB2R0h(y,τ)Zi(y)dy+R−ε00 O(v‖hi‖v,2+a)), i= 1, . . . ,n,cn+1[h](τ) = −(ˆB2η(y)Z2n+1(y)dy)−1(ˆB2R0h(y,τ)Zn+1(y)dy+R−ε00 O(v‖h0‖v,2+a)),where 0< ε0 < min{a,1}2 is a small constant,h(y,τ) =∞∑j=0ϒ j(y|y|)h j(|y|,τ), h j(|y|,τ) =ˆSn−1h(|y|θ ,τ)ϒ j(θ)dθ ,O(v‖hi‖v,2+a) linearly depends on hi for i= 0,1, . . . ,n.The proof of Proposition 3.6.1 is achieved by the following Proposition and by another gluingprocedure (re-gluing).Proposition 3.6.2. Consider∂τϕ = ∆ϕ + pU p−1ϕ + f1ϕ + f2y ·∇ϕ +h(y,τ) in DRϕ(y,τ0) = e0Z0(y) in BRwhere ‖h‖v,2+a < ∞, a> 0 and h satisfies the orthogonal conditionˆBR(τ)h(y,τ)Zi(y)dy= 0 for all τ > τ0, i= 1, . . . ,n+1.Assume Rnθ 1Ra τmin{1,d}. Then for τ0 sufficiently large, there exists a solution (ϕ ,e0)= (T2i[h],T2e[h])which is a linear mapping of h with the estimates〈y〉|∇ϕ |+ |ϕ |≲ vmin{τ 12 ,λ−12R }λ− 12R θ0Raˆ0(〈y〉−n+C f τ−dmin{τ 12 ,λ−12R }λ− 12R lnR〈y〉2−n)‖h0‖v,2+a+ v(Θ0Raˆ0(|y|)〈y〉−2+C f τ−dmin{τ12 ,λ−12R }λ− 12R θ0Raˆ0 lnR)‖h0‖v,2+a+ vθ 1Raˆ1Rn(〈y〉−1−n+C f τ−dRn〈y〉1−n)‖h1‖v,2+a+ v(Θ0R,2+a(|y|)+R〈y〉2−n)‖h⊥‖v,2+a,|e0|≲ v(τ0)θ 0R(τ0)aˆ0(1+C f τ−d0 min{τ120 ,λ− 12R(τ0)}λ− 12R(τ0) lnR(τ0))‖h0‖v,2+a116whereaˆ0 =a if a 6= n−2(n−2)− if a= n−2 , aˆ1 =a if a 6= n−1(n−1)− if a= n−1 , (3.6.5)θ 0Ra =R2−a if a< 2lnR if a= 21 if a> 2, θ 1Ra =R1−a if a< 1lnR if a= 11 if a> 1, (3.6.6)λR =R−2 if n= 3(R2 lnR)−1 if n= 4R2−n if n≥ 5, Θ0Ra(r) =R2−a if a< 2lnR if a= 2〈r〉2−min{a,n−} if a> 2. (3.6.7)Before we prove Proposition 3.6.1, we first use Proposition 3.6.2 to prove Proposition 3.6.1.Proof of Proposition 3.6.1. Set ϕ(y,τ) = ηR0(y)ϕi(y,τ)+ϕo(y,τ), where ηR0(y) = η(yR0). In order tofind a solution ϕ , it suffices to consider the following inner–outer gluing system for (ϕi,ϕo)∂τϕo = ∆ϕo+ J[ϕo,ϕi] in DR,ϕo = 0 on ∂DR, ϕo = 0 in BR(τ0), (3.6.8)∂τϕi = ∆ϕi+ pUp−1ϕi+ f1ϕi+ f2y ·∇ϕi+ pU p−1ϕo+h+n+1∑i=1ci(τ)η(y)Zi(y) in D2R0 ,ϕi = e0Z0(y) in B2R(τ0),(3.6.9)whereJ[ϕo,ϕi] = f1ϕo+ f2y ·∇ϕo+ pU p−1ϕo(1−ηR0)+A[ϕi]+h(1−ηR0),A[ϕi] = ∆ηR0ϕi+2∇ηR0 ·∇ϕi+ f2y ·∇ηR0ϕi−∂τηR0ϕi.Here ci(τ) is given byci(τ) = ci[ϕo](τ) =CiˆB2R0(pU p−1(z)ϕo(z,τ)+h(z,τ))Zi(z)dz, Ci =−(ˆB2η(y)Z2i (y)dy)−1such that the orthogonal conditionsˆB2R0(pU p−1(z)ϕo(z,τ)+h(z,τ)+n+1∑i=1ci(τ)η(y)Zi(z))Z j(z)dy= 0 for j = 1, . . . ,n+1are satisfied.117We reformulate (4.8.56) and (4.8.57) into the following formϕo(y,τ) = To[J[ϕo,ϕi]], ϕi(y,τ) =T2i[pU p−1(y)ϕo+h+n+1∑i=1ci(τ)η(y)Zi(y)],e0 = T2e[pU p−1(y)ϕo+h+n+1∑i=1ci(τ)η(y)Zi(y)],(3.6.10)where To is a linear mapping given by the standard parabolic theory, and T2i, T2e are given byProposition 3.6.2. We will solve the system (4.8.72) by the contraction mapping theorem.Denote the leading term of the right hand side of (4.8.57) asH1 := h+n+1∑i=1Ciη(y)Zi(y)´B2R0h(z,τ)Zi(z)dz.It is easy to check ‖H1‖v,2+a ≲ ‖h‖v,2+a. If H1 satisfies the orthogonal condition in D2R0 , under theassumption Rn+20 τmin{1,d}−, Proposition 3.6.2 gives following a priori estimates〈y〉|∇T2i[H1]|+ |T2i[H1]| ≤ Diwi(y,τ), |T2e[H1]| ≤ Div(τ0)R2−a0 ‖h‖v,2+a,where Di ≥ 1 is a constant andwi(y,τ) = v(λ−1R0 R2−a0 〈y〉−n+θ 1R0aRn0〈y〉−1−n+ 〈y〉−a+R0〈y〉2−n)‖h‖v,2+awhere θ 1R0a is given in (3.6.6). For this reason, we will solve the inner part in the spaceBi = {g(y,τ) : 〈y〉|∇yg(y,τ)|+ |g(y,τ)| ≤ 2Diwi(y,τ)} .For any ϕ˜i ∈Bi, we will find a solution ϕo = ϕo[ϕ˜i] of (4.8.56) by the contraction mapping theorem.Let us estimate J[0, ϕ˜i] term by term. For n≥ 4,|A[ϕ˜i]|≲ Div(R−20 + τ−d)(R−a0 lnR0+R−10 )1{R0≤|y|≤2R0}‖h‖v,2+a ≲ DivR−ε00 〈y〉−2−a1‖h‖v,2+awhere constants 0< a1 <min{a,1} and ε0 = min{a,1}−a12 . Also, we have|h(1−ηR0)|≲ 1{y≥R0}v〈y〉−2−a‖h‖v,2+a ≲ vR−ε00 〈y〉−2−a1‖h‖v,2+a.Consider (4.8.56) with the right hand side J[0, ϕ˜i]. Using Cv(−∆)−1(〈y〉−2−a1)R−ε00 ‖h‖v,2+a as thebarrier function with a large constantC and then scaling argument, we have〈y〉|∇To[J[0, ϕ˜i]](y,τ)|+ |To[J[0, ϕ˜i]](y,τ)| ≤ wo(y,τ) = DoDivR−ε00 〈y〉−a1‖h‖v,2+awith a large constant Do ≥ 1. This suggests us solve ϕo in the following space:Bo = { f (y,τ) : 〈y〉|∇ f (y,τ)|+ | f (y,τ)| ≤ 2wo(y,τ)} .118For any ϕ˜o ∈Bo, due to |y| ≤ 2R(τ), we have|pU p−1ϕ˜o(1−ηR0)|≲ R−20 DoDivR−ε00 〈y〉−2−a1‖h‖v,2+a,| f1ϕo+ f2y ·∇ϕo|≲ τ−dR2(τ)DoDivR−ε00 〈y〉−2−a1‖h‖v,2+a.Since τ−dR2, R−20 provide smallness, by comparison principle, we haveTo[J[ϕ˜o, ϕ˜i]] ∈Bo.The contraction mapping property can be deduced in the same way.Now we have found a solution ϕo = ϕo[ϕ˜i] ∈Bo. It follows that∥∥∥pU p−1(y)ϕo[ϕ˜i]+ n+1∑i=1CiˆB2R0pU p−1(z)ϕo[ϕ˜i](z,τ)Zi(z)dzη(y)Zi(y)∥∥∥v,2+a≲ DoDiR−ε00 ‖h‖v,2+a.Due to the choice of ci(τ), H2 := pU p−1(y)ϕo[ϕ˜i]+h+n+1∑i=1ci[ϕo[ϕ˜i]](τ)η(y)Zi(y) satisfies the orthog-onal condition in D2R0 . By Proposition 3.6.2, since R−ε00 provides smallness, we haveT2i[h2] ∈Bi.The contraction property can be deduced in the same way. Thus we find a solutionϕi = ϕi[h] ∈Bi. (3.6.11)Finally, we obtain a solution (ϕo,ϕi) for (4.8.56) and (4.8.57).From the construction above and the topology ofBi, ϕi[h] = 0 if h= 0, which deduces that ϕi[h]is a linear mapping of h. By the similar argument, ϕo[h] and ci[h] are also linear mappings of h, andso does ϕ .We will regard Do, Di as general constants hereafter. Then by Proposition 3.6.2 and (4.8.72), wehave|e0|≲ v(τ0)R2−a0 ‖h‖v,2+a.Since ϕo[h] ∈Bo, one hasci[h](τ) =CiˆB2R0h(y,τ)Zi(y)dy+R−ε00 O(v)‖h‖v,2+a.119Since the above operation is linear about h, we are able to decompose h intoh(y,τ) =∞∑j=0ϒ j(y|y|)h j(|y|,τ), h j(|y|,τ) =ˆSn−1h(|y|θ ,τ)ϒ j(θ)dθand repeat the construction about ϒ j( y|y|)h j(|y|,τ) separately. Thencn+1[h](τ) =Cn+1ˆB2R0h(y,τ)Zn+1(y)dy+R−ε00 O(v)‖h0‖v,2+a,ci[h](τ) =CiˆB2R0h(y,τ)Zi(y)dy+R−ε00 O(v)‖hi‖v,2+a for i= 1, . . . ,n.Reviewing the re-gluing procedure, we have|J[0,ϕi]|≲ R0v〈y〉−2−a‖h‖v,2+a.Using comparison principle to (4.8.56) several times, the upper bound of ϕo can be improved to|ϕo|≲ R0v〈y〉−a‖h‖v,2+a. (3.6.12)Combining (3.6.11), (3.6.12) and then using scaling argument, we conclude〈y〉|∇ϕ |+ |ϕ |≲ Rn+10 v〈y〉−a‖h‖v,2+a.The rest of this section is devoted to the proof of Proposition 3.6.2. We first invoke a coerciveestimate for the linearized operatorLemma 3.6.3. [16, Lemma 7.2] There exists a constant c0 > 0 such that for all sufficiently large Rand all radially symmetric functions ϕ ∈ H10 (BR) with´BRϕZ0 = 0, we havec0λRˆBR|ϕ |2 ≤ Q(ϕ ,ϕ),where λR is given in (3.6.7) and Q(ϕ ,ϕ) :=´BR(|∇ϕ |2− pU p−1|ϕ |2).Note that in [16, Lemma 7.2], there is the above coercive estimate only for higher dimensionsn≥ 5. The proof in lower dimensions n= 3, 4 is similar and by slight modifications.120Lemma 3.6.4. Consider∂τϕ = ∆ϕ + pU p−1(1−χM)ϕ + f1ϕ + f2y ·∇ϕ +h in DRϕ = 0 on ∂DR, ϕ(·,τ0) = 0 in BR(τ0)where χM(y) = η( yM ), M> 0 is a large constant, R2 lnR τmin{1,d}, ‖h‖v,a <∞, a≥ 0. Then when τ0sufficiently large, for Θ0Ra(|y|) given in (3.6.7), the unique solution ϕ∗[h] has the following estimate:|ϕ∗[h]|≲C(M,a,n)vΘ0Ra(|y|)‖h‖v,a.Proof. Set a¯ = min{a,n−}, r = |y|, LMϕ = ∆ϕ + pU p−1(y)(1− χM)ϕ . Set a barrier function asϕ¯(r,τ) =Cvg(r,R), whereLMg(r,R) =−〈r〉−a¯, g(r,R) = g2(r)ˆ Rrdρg22(ρ)ρn−1ˆ ρ0g2(s)sn−1〈s〉−a¯dsand g2(r)> 0 is the positive kernel of LM and g2(r)∼ 1 for r ∈ (0,∞). By direct calculation, one has〈r〉a¯g(r,R)≲ 〈r〉a¯Θ0Ra¯(|y|)≲ R2 lnR.By scaling argument, one has 〈r〉a¯|r∂rg(r,R)|≲ R2 lnR. ThenP(ϕ¯) := LM(Cvg(r,R))+h(y,τ)+Cv( f1g(r,R)+ f2r∂rg(r,R))−∂τ(Cvg(r,R))= −Cv〈r〉−a¯+h(y,τ)+Cv( f1g(r,R)+ f2r∂rg(r,R))−Cv′g(r,R)− Cvg2(r)R′g22(R)Rn−1ˆ R0g2(s)sn−1〈s〉−a¯ds≤Cv〈r〉−a¯[−1+ 〈r〉a¯( f1g(r,R)+ f2r∂rg(r,R))− v′v−1g(r,R)〈r〉a¯− 〈r〉a¯g2(r)R′g22(R)Rn−1ˆ R0g2(s)sn−1〈s〉−a¯ds]+ v〈r〉−a‖h‖v,a ≤−34Cv〈r〉−a¯+ v〈r〉−a¯‖h‖v,awhere we have used|〈r〉a¯( f1g(r,R)+ f2r∂rg(r,R))|≲ (| f1|+ | f2|)R2 lnR≲C f τ−dR2 lnR 1,|v′v−1g(r,R)〈r〉a¯|≲ |v′|v−1R2 lnR≲ τ−1R2 lnR 1,〈r〉a¯g2(r)|R′|g22(R)Rn−1ˆ R0g2(s)sn−1〈s〉−a¯ds∼ 〈r〉a¯|R′|Rn−1ˆ R0sn−1〈s〉−a¯ds≲ R|R′|≲ τ−1R2 1.SetC = 2‖h‖v,a, then P(ϕ¯)≤ 0.1213.6.1 Mode 0 without orthogonalityLemma 3.6.5. Consider∂τϕ 0 = ∆ϕ 0+ pU p−1ϕ 0+ f1ϕ 0+ f2y ·∇ϕ 0+h0 in DR,ϕ(·,τ0) = e0Z0(y) in BR(τ0) (3.6.13)where ‖h0‖v,a < ∞, a≥ 0. Assume λRτd 1, R2 lnR τmin{1,d}. Then for τ0 sufficiently large, thereexists a linear mapping (ϕ 0,e0) = (T1i[h0],T1e[h0]) solving (3.6.13) with the following estimates〈y〉|∇ϕ 0|+ |ϕ 0|≲ v(min{τ 12 ,λ−12R }λ− 12R θ0Ra〈y〉2−n+Θ0Ra(|y|))‖h0‖v,a,|e0|≲ v(τ0)θ 0R(τ0)a‖h0‖v,a.Proof. First, we decompose ϕ 0 into two partsϕ 0 = ϕ∗[h0]+ ϕ˜ ,where ϕ∗[h0] is the solution derived from Lemma 3.6.4 with the following estimate|ϕ∗[h0]|≲ vΘ0Ra(|y|)‖h0‖v,a. (3.6.14)Then∂τϕ∗[h0]+∂τ ϕ˜ = ∆ϕ∗[h0]+∆ϕ˜ + pU p−1(y)χMϕ∗[h0]+ pU p−1(y)(1−χM)ϕ∗[h0]+ pU p−1(y)ϕ˜ + f1(ϕ∗[h0]+ ϕ˜)+ f2y ·∇(ϕ∗[h0]+ ϕ˜)+h0 in DR,which implies that∂τ ϕ˜ = ∆ϕ˜ + pU p−1(y)ϕ˜ + f1ϕ˜ + f2y ·∇ϕ˜ + pU p−1(y)χMϕ∗[h0] in DR.We will construct a linear mapping ϕ˜ = ϕ˜ [h0]. Take ϕ˜ = ϕ˜1+ e(τ)Z0(y) and consider the followingequation∂τ ϕ˜1 = ∆ϕ˜1+ pU p−1ϕ˜1+ f1ϕ˜1+ f2y ·∇ϕ˜1−∂τe(τ)Z0+ γ0e(τ)Z0+pU p−1χMϕ∗[h0]+ e(τ)( f1Z0(y)+ f2y ·∇Z0(y)) in DR,ϕ˜1 = 0 on ∂DR, ϕ˜1(·,τ0) = 0 in BR(τ0),´BR(τ)ϕ˜1(y,τ)Z0(y)dy= 0 ∀τ > τ0.(3.6.15)Here e(τ)will be chosen to make´BR(τ)ϕ˜1(y,τ)Z0(y)dy= 0 for all τ > τ0. Indeed, multiplying (3.6.15)122by Z0 and integrating by parts, one has∂τˆBR(τ)ϕ˜1Z0(y)dy=ˆBR(τ)∂τ ϕ˜1Z0(y)dy= γ0ˆBR(τ)ϕ˜1Z0(y)dy+ˆ∂BR(τ)Z0(y)∂nϕ˜1dy+ˆBR(τ)( f1ϕ˜1+ f2y ·∇ϕ˜1)Z0(y)dy− (∂τe(τ)− γ0e(τ))ˆBR(τ)Z20(y)dy+ˆBR(τ)pU p−1χMϕ∗[h0]Z0(y)dy+ e(τ)ˆBR(τ)( f1Z0(y)+ f2y ·∇Z0(y))Z0(y)dy.By ϕ˜1(·,τ0) = 0, the orthogonality´BR(τ)ϕ˜1(y,τ)Z0(y)dy= 0 holds for all τ > τ0 if and only if∂τe(τ)− γ˜0(τ)e(τ) =(ˆBR(τ)Z20(y)dy)−1[ˆ∂BR(τ)Z0(y)∂nϕ˜1dy+ˆBR(τ)( f1ϕ˜1+ f2y ·∇ϕ˜1)Z0(y)dy+ˆBR(τ)pU p−1χMϕ∗[h0]Z0(y)dy],where γ˜0(τ) = γ0+(´BR(τ)Z20(y)dy)−1 ´BR(τ)( f1Z0(y)+ f2y ·∇Z0(y))Z0(y)dy. By (3.6.2), limτ→∞ γ˜0(τ) =γ0 as τ → ∞. We take e(τ) ase(τ) =−e´ τ γ˜0(u)duˆ ∞τe−´ s γ˜0(u)du(ˆBR(s)Z20(y)dy)−1[ˆ∂BR(s)Z0(y)∂nϕ˜1(y,s)dy+ˆBR(s)( f1(y,s)ϕ˜1(y,s)+ f2(y,s)y ·∇ϕ˜1(y,s))Z0(y)dy+ˆBR(s)pU p−1(y)χM(y)ϕ∗[h0](y,s)Z0(y)dy]ds.Set‖ϕ˜1‖w = supτ>τ0(min{τ 12 ,λ−12R }λ− 12R vθ0Ra)−1(‖ϕ˜1(·,τ)‖L∞(BR(τ))+‖〈·〉∇ϕ˜1(·,τ)‖L∞(BR(τ))).By (3.6.14), it is straightforward to get|e(τ)|≲ e´ τ γ˜0(u)duˆ ∞τe−´ s γ˜0(u)du(e−cR(s)‖∇ϕ˜1‖L∞(BR(s))+‖| f1ϕ˜1|+ | f2∇ϕ˜1|‖L∞(BR(s))+ v(s)θ 0Ra(s)‖h0‖v,a)ds≲ e´ τ γ˜0(u)duˆ ∞τe−´ s γ˜0(u)du[(s−d + e−cR(s))min{s 12 ,λ−12R (s)}λ− 12R (s)v(s)θ0Ra(s)‖ϕ˜1‖w+ v(s)θ 0Ra(s)‖h0‖v,a]ds≲ (τ−d0 + e−cR(τ0))min{τ12 ,λ−12R }λ− 12R vθ0Ra‖ϕ˜1‖w+ vθ 0Ra‖h0‖v,a(3.6.16)for some constant c> 0, and θ 0Ra is given in (3.6.6). It follows that|∂τe(τ)|≲ (τ−d0 + e−cR(τ0))min{τ12 ,λ−12R }λ− 12R vθ0Ra‖ϕ˜1‖w+ vθ 0Ra‖h0‖v,a.123With the above choice of e(τ), the global existence of (3.6.15) can be deduced by the local exis-tence.Multiplying equation (3.6.15) by ϕ˜1 and integrating by parts, one has12∂τˆBR(τ)(ϕ˜1)2dy+ˆBR(τ)(|∇ϕ˜1|2− pU p−1(ϕ˜1)2)dy=ˆBR(τ)( f1ϕ˜1+ f2y ·∇ϕ˜1)ϕ˜1dy+ˆBR(τ)pU p−1χMϕ∗[h0]ϕ˜1dy+ e(τ)ˆBR(τ)( f1Z0(y)+ f2y ·∇Z0(y))ϕ˜1dy.Then by Lemma 3.6.3 and (3.6.2), we get12∂τˆBR(τ)(ϕ˜1)2dy+ cλRˆBR(τ)(ϕ˜1)2dy≤Cτ−dˆBR(τ)(ϕ˜1)2dy+ˆBR(τ)4cλR(pU p−1χMϕ∗[h0])2dy+ˆBR(τ)cλR4(ϕ˜1)2dy+ˆBR(τ)4cλRe2(τ)( f1Z0(y)+ f2y ·∇Z0(y))2dy+ˆBR(τ)cλR4(ϕ˜1)2dyfor some constant c> 0. By (3.6.2), (3.6.14), (3.6.16) and the assumption λRτd 1, we get12∂τˆBR(τ)(ϕ˜1)2dy+cλR4ˆBR(τ)(ϕ˜1)2dy≲ λ−1R [(vθ 0Ra‖h0‖v,a)2+ τ−2de2(τ)]≲ λ−1R (vθ 0Ra)2[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w]2.Since ϕ˜1(·,τ0) = 0, one hasˆBR(τ)(ϕ˜1)2dy≲ e−´ τ cλR(u)2 duˆ ττ0e´ s cλR(u)2 duλ−1R (s)(v(s)θ0Ra(s))2[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w]2ds≲ min{τ,λ−1R }λ−1R (vθ 0Ra)2[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w]2.Applying parabolic estimate to (3.6.15), one has‖ϕ˜1(·,τ)‖L∞(BR(τ)) ≲min{τ12 ,λ−12R }λ− 12R vθ0Ra[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w]+ τ−d(‖ϕ˜1(·,τ)‖L∞(BR(τ))+‖y ·∇ϕ˜1‖L∞(BR(τ)))+ |∂τe(τ)|+ |e(τ)|+ vθ 0Ra‖h0‖v,a≲ min{τ 12 ,λ−12R }λ− 12R vθ0Ra[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w].By comparison principle, the spatial decay of ϕ˜1 can be improved and the scaling argument willgive the spatial decay about ∇ϕ˜1. Then one has〈y〉|∇ϕ˜1|+ |ϕ˜1|≲min{τ 12 ,λ−12R }λ− 12R vθ0Ra[‖h0‖v,a+(τ−d0 + e−cR(τ0))‖ϕ˜1‖w]〈y〉2−n,124which implies〈y〉|∇ϕ˜1|+ |ϕ˜1|≲min{τ 12 ,λ−12R }λ− 12R vθ0Ra〈y〉2−n‖h0‖v,a.Reviewing the computations in (3.6.16) and using λRτd 1, one has |e(τ)|≲ vθ 0Ra‖h0‖v,a and then|ϕ˜ |= |ϕ˜1+ e(τ)Z0(y)|≲min{τ 12 ,λ−12R }λ− 12R vθ0Ra〈y〉2−n‖h0‖v,a. (3.6.17)Finally, we take e0 = e(τ0). Combining (3.6.14) and (3.6.17), we complete the proof of this Lemma.3.6.2 Modes 1 to n without orthogonalityLemma 3.6.6. Consider∂τϕ 1 = ∆ϕ 1+ pU p−1ϕ 1+ f1ϕ 1+ f2y ·∇ϕ 1+h1(y,τ) in DRϕ 1 = 0 on ∂DR ϕ 1(·,τ0) = 0 in BR(τ0) (3.6.18)where h1(y,τ) =n∑j=1h j(|y|,τ)ϒ j. Assume Rnθ 1Ra τmin{1,d}, where θ 1Ra is given in (3.6.6). Then forτ0 sufficiently large, there exists a unique linear mapping ϕ 1 = ϕ 1[h1] solving (3.6.18) of the formϕ 1 =n∑j=1ϕ j(|y|,τ)ϒ j with the following estimate〈y〉|∇ϕ 1|+ |ϕ 1|≲ vθ 1RaRn〈y〉1−n‖h1‖v,a.Proof. Set r = |y|. Notice y ·∇(ϕ j(r,τ)ϒ j) = r∂rϕ j(r,τ)ϒ j. It is equivalent to considering∂τϕ j =L1[ϕ j]+ f1ϕ j(r,τ)+ f2r∂rϕ j(r,τ)+h j(r,τ) for r ∈ (0,R(τ)), τ ∈ (τ0,∞)∂rϕ j(0,τ) = 0= ϕ j(R(τ),τ) for τ ∈ (τ0,∞), ϕ j(r,τ0) = 0 for r ∈ (0,R(τ0)) (3.6.19)whereL1[ϕ j] := ∂rrϕ j+ n−1r ∂rϕ j− n−1r2 ϕ j+ pU(r)p−1ϕ j, |h j| ≤ v〈y〉−a‖h j‖v,a, ‖h j‖v,a ≲ ‖h1‖v,a.One positive kernel of L1 is given by Z(r) := −Ur = (n(n− 2)) n−24 (n− 2)r(1+ r2)− n2 . Set abarrier function of (3.6.19) as ϕs =Cvϕ¯(r,R), whereL1[ϕ¯ ] =−〈r〉−a¯, a¯=min{a,n−1}with ϕ¯ given by the variation of parameter formulaϕ¯(r,R) = Z(r)ˆ Rr1ρn−1Z2(ρ)ˆ ρ0〈s〉−a¯Z(s)sn−1dsdρ.125Thenϕ¯ ≲ Rnθ 1Ra¯r〈r〉−n, |∂Rϕ¯ |=∣∣∣∣ Z(r)Rn−1Z2(R)ˆ R0〈s〉−a¯Z(s)sn−1ds∣∣∣∣≲ Rn−1θ 1Ra¯r〈r〉−nfor all r > 0. This estimate holds for all n> 2, and a≤ 0 is also allowed here. Next, we computeP(ϕs) :=L1ϕs+ f1ϕs+ f2r∂rϕs−∂τϕs+h j =−Cv〈r〉−a¯+Cv( f1ϕ¯ + f2r∂rϕ¯)−Cv′ϕ¯ −Cv∂Rϕ¯R′+h j≤Cv〈r〉−a¯ [−1+ 〈r〉a¯( f1ϕ¯ + f2r∂rϕ¯)− v−1v′〈r〉a¯ϕ¯ −〈r〉a¯∂Rϕ¯R′+C−1〈r〉a¯−a‖h j‖v,a]≤Cv〈r〉−a¯(−34+C−1‖h j‖v,a)where we have used|〈r〉a¯( f1ϕ¯ + f2r∂rϕ¯)|≲ τ−dRnθ 1Ra¯〈r〉a¯+1−n ≲ τ−dRnθ 1Ra 1,|v−1v′〈r〉a¯ϕ¯ |≲ τ−1Rnθ 1Ra¯〈r〉a¯+1−n ≲ τ−1Rnθ 1Ra 1,|〈r〉a¯∂Rϕ¯R′|≲ 〈r〉a¯Rn−1θ 1Ra¯r〈r〉−n|R′|≲ τ−1Rnθ 1Ra 1by (3.6.2), θ 1Ra¯ = θ 1Ra and the assumption Rnθ 1Ra τmin{1,d}. TakingC= 4‖h j‖v,a, one has P(ϕs)< 0.3.6.3 Higher modesLemma 3.6.7. Consider∂τϕ⊥ = ∆ϕ⊥+ pU p−1ϕ⊥+ f1ϕ⊥+ f2y ·∇ϕ⊥+h⊥ in DRϕ⊥ = 0 on ∂DR, ϕ⊥(·,τ0) = 0 in BR(τ0)where ‖h⊥‖v,a < ∞, a ≥ 0. Assume R2 lnR τmin{1,d}. Then there exists a unique linear mappingϕ⊥ = ϕ⊥[h⊥] of the formϕ⊥ =∞∑j=n+1ϕ⊥j (|y|,τ)ϒ j (3.6.20)with the following estimate〈y〉|∇ϕ⊥|+ |ϕ⊥|≲ v(Θ0Ra(|y|)+θ 0RaR〈y〉2−n)‖h⊥‖v,a.First, we give the following technical lemma.Lemma 3.6.8. For f ∈ C2(BR)∩C0(BR), by the expansion of spherical harmonic functions, f =126∞∑j=0f j(r)ϒ j, where r = |y|, f j(r) =´Sn−1 f (rθ)ϒ j(θ)dθ ∈C2[0,R]. ThenQ( f , f ) =ˆBR(|∇ f |2− pU p−1 f 2)dy= |Sn−1|∞∑j=0Q j( f j, f j),where |Sn−1| is the volume of the unit (n−1)-sphere andQ j( f j, f j) =ˆ R0(f ′2j +ι jr2f 2j − pU p−1 f 2j)rn−1dr.Specially, if f j = 0 for j = 0,1, . . . ,n, it holds thatQ( f , f )≥ (n+1)ˆBR| f |2|y|2 dy. (3.6.21)Proof. Since ∆Sn−1ϒi =−ιiϒi, ιi = i(n−2+ i) for a nonnegative integer i, we have∆( fiϒi) = ( f ′′i +n−1rf ′i −ιir2fi)ϒi.f |∂BR = 0 implies f j(R) = 0, j = 0,1, . . . . ThenQ( f , f ) =ˆBR|∇ f |2− pU p−1 f 2dy=−ˆBR( f∆ f + pU p−1 f 2)dy= −|Sn−1|ˆ R0[∞∑i=0fi( f ′′i +n−1rf ′i −ιir2fi)+ pU p−1∞∑i=0f 2i]rn−1dr = |Sn−1|∞∑i=0Qi( fi, fi).For i≥ n+1, ιi ≥ 2n, we haveQi( fi, fi)≥ Q1( fi, fi)+(n+1)ˆ R0f 2ir2rn−1dr ≥ (n+1)ˆ R0f 2ir2rn−1drsince ∆u− n−1r2 u+ pU p−1(y)u = 0 has a positive kernel −Ur, and by [96, Lemma 4.2], one hasQ1( fi, fi)≥ 0. Specially, if f j = 0 for j = 0,1, . . . ,n, we have (3.6.21).Proof of Lemma 3.6.7. The existence and uniqueness of the linear mapping ϕ⊥ = ϕ⊥[h⊥] are guar-anteed by the classical parabolic theory. The form (3.6.20) is derived from the existence of everycomponent ϕ j with∂τϕ j = ∂rrϕ j+ n−1r ∂rϕ j−ι jr2ϕ j+ pUp−1ϕ j+ f1ϕ j+ f2r∂rϕ j+h j for r ∈ (0,R(τ)),τ ∈ (τ0,∞)∂rϕ j(0,τ) = ϕ j(R(τ),τ) = 0 for τ ∈ (τ0,∞), ϕ j(r,τ0) = 0 for r ∈ (0,R(τ0)).127By similar operation in mode 0, we set ϕ⊥ = ϕ∗[h⊥]+ ϕ˜ [h⊥], where ϕ∗[h⊥] satisfies∂τϕ∗ = ∆ϕ∗+ pU p−1(1−χM)ϕ∗+ f1ϕ∗+ f2y ·∇ϕ∗+h⊥ in DR,ϕ∗ = 0 on ∂DR, ϕ∗(·,τ0) = 0 in BR(τ0),and ϕ˜ [h⊥] satisfies∂τ ϕ˜ = ∆ϕ˜ + pU p−1ϕ˜ + f1ϕ˜ + f2y ·∇ϕ˜ + pU p−1χMϕ∗[h⊥] in DR,ϕ˜ = 0 on ∂DR, ϕ˜(·,τ0) = 0 in BR(τ0). (3.6.22)Under the assumption R2 lnR τmin{1,d}, by Lemma 3.6.4, we have|ϕ∗[h⊥]|≲ vΘ0Ra(|y|)‖h⊥‖v,a. (3.6.23)ϕ∗[h⊥] has the form ϕ∗[h⊥] =∞∑j=n+1ϕ∗ j(r,τ)ϒ j by the same reason as (3.6.20). By the same argument,there exists a linear mapping ϕ˜ = ϕ˜ [h⊥] and ϕ˜ has the same form as (3.6.20). Thus we are able toapply (3.6.21) to ϕ˜ .Multiplying (3.6.22) by ϕ˜ and integrating both sides, we have12∂τˆBRϕ˜ 2dy+Q(ϕ˜ , ϕ˜) =ˆBR( f1ϕ˜ + f2y ·∇ϕ˜)ϕ˜dy+ˆBRpU p−1(y)χMϕ∗[h⊥]ϕ˜dy.By (3.6.21), (3.6.2) and Hölder inequality, one has12∂τˆBRϕ˜ 2dy+(n+12)ˆBRϕ˜ 2|y|2 dy≤Cτ−dˆBRϕ˜ 2dy+12ˆBR(pU p−1(y)χMϕ∗[h⊥]|y|)2dy.Then, by (3.6.23) and the assumption R2 lnR τmin{1,d}, we have∂τˆBRϕ˜ 2dy+R−2ˆBR|ϕ˜ |2dy≲ (vθ 0Ra)2‖h⊥‖2v,a.Since ϕ˜(·,τ0) = 0 and the assumption R2 lnR τmin{1,d}, we haveˆBRϕ˜ 2dy≲ e−´ τ R−2(u)duˆ ττ0e´ sR−2(u)du(v(s)θ 0Ra(s))2ds‖h⊥‖2v,a ≲ (vθ 0RaR‖h⊥‖v,a)2.Using the same argument in Lemma 3.6.5, one has|ϕ˜(y,τ)|≲ vθ 0RaR〈y〉2−n‖h⊥‖v,a. (3.6.24)128Combining (3.6.23), (3.6.24) and scaling argument, we get〈y〉|∇ϕ⊥|+ |ϕ⊥|≲ v(Θ0Ra(|y|)+θ 0RaR〈y〉2−n)‖h⊥‖v,a.Proof of Proposition 3.6.2. The case for higher modes has been given in Lemma 3.6.7. Since the fastspatial decay of the right hand side, h cannot be recovered in the non-orthogonal case in lower modesi, 0 ≤ i ≤ n, we transform the fast decay right hand side into a slower decay function by solving thecorresponding elliptic equation.3.6.4 Mode 0 with orthogonalityConsider∆H0+ pU p−1H0 = h˜0(r,τ) in Rnwhere h˜0 is the extension of h0 as zero outside DR. The orthogonal condition is reformulated asˆ R0h0(r,τ)Zn+1(r)rn−1dr = 0 for all τ > τ0.Take H0(r,τ) as in the following formH0(r,τ) = Z˜n+1(r)ˆ r0h˜0(s,τ)Zn+1(s)sn−1ds−Zn+1(r)ˆ r0h˜0(s,τ)Z˜n+1(s)sn−1ds, if a≤ n−2,H0(r,τ) = Z˜n+1(r)ˆ r0h˜0(s,τ)Zn+1(s)sn−1ds+Zn+1(r)ˆ ∞rh˜0(s,τ)Z˜n+1(s)sn−1ds, if a> n−2,where Z˜n+1(r) is the other linearly independent kernel of the homogeneous equation, which satisfiesthat the Wronskian W [Zn+1, Z˜n+1] = r1−n, Z˜n+1(r) ∼ r2−n if r→ 0 and Z˜n+1(r) ∼ 1 if r→ ∞. It isstraightforward to check‖H0‖v,aˆ0 ≲ ‖h0‖v,2+a,where aˆ0 is given in (3.6.5) and a > 0 is used to ensure that the spatial decay of h˜0(s,τ)Zn+1(s)sn−1is faster than s−1 for s≥ 1. Next, consider∂τΦ0 = ∆Φ0+ pU p−1Φ0+H0 in D2R,Φ0(·,τ0) = e¯0Z0 in B2R(τ0), (3.6.25)129where (Φ0, e¯0) is given by Lemma 3.6.5 under the condition f1 = f2 = 0. By scaling argument, onehas〈y〉2|∇2Φ0|+ 〈y〉|∇Φ0|+ |Φ0|≲ v(min{τ 12 ,λ−12R }λ− 12R θ0Raˆ0〈y〉2−n+Θ0Raˆ0(|y|))‖h0‖v,2+a,|e¯0|≲ v(τ0)θ 0R(τ0)aˆ0‖h0‖v,2+a.Acting the operator L := ∆+ pU p−1 on both sides of (3.6.25) and denoting ϕ 01 = LΦ0, we obtain∂τϕ 01 = ∆ϕ 01 + pU p−1ϕ 01 +h0 in DRϕ 01 (·,τ0) = γ0e¯0Z0 in BRwith the following estimate〈y〉|∇ϕ 01 |+ |ϕ 01 |≲ v(min{τ 12 ,λ−12R }λ− 12R θ0Raˆ0〈y〉−n+Θ0Raˆ0(|y|)〈y〉−2)‖h0‖v,2+a.Taking into account f1ϕ 0+ f2y ·∇ϕ 0, we consider∂τϕ 02 = ∆ϕ 02 + pU p−1ϕ 02 + f1ϕ 02 + f2y ·∇ϕ 02 + f1ϕ 01 + f2y ·∇ϕ 01 in DRϕ 02 (·,τ0) = e02Z0 in BRwhere ∣∣ f1ϕ 01 + f2y ·∇ϕ 01 ∣∣≲C f τ−dmin{τ 12 ,λ− 12R }λ− 12R vθ 0Raˆ0〈y〉−2‖h0‖v,2+a.Using Lemma 3.6.5 again, one can find a solution (ϕ 02 ,e02) with the following estimates|ϕ 02 |≲C f τ−dmin{τ12 ,λ−12R }λ− 12R vθ0Raˆ0(min{τ 12 ,λ−12R }λ− 12R lnR〈y〉2−n+ lnR)‖h0‖v,2+a,|e02|≲C f τ−d0 min{τ120 ,λ− 12R(τ0)}λ− 12R(τ0)v(τ0)θ0R(τ0)aˆ0 lnR(τ0)‖h0‖v,2+a.Finally, we take (ϕ 0,e0) = (ϕ 01 +ϕ 02 ,γ0e¯0+ e02) and conclude the result for mode 0:|ϕ 0|≲ vmin{τ 12 ,λ−12R }λ− 12R θ0Raˆ0(〈y〉−n+C f τ−dmin{τ 12 ,λ−12R }λ− 12R lnR〈y〉2−n)‖h0‖v,2+a+ v(Θ0Raˆ0(|y|)〈y〉−2+C f τ−dmin{τ12 ,λ−12R }λ− 12R θ0Raˆ0 lnR)‖h0‖v,2+a,|e0|≲ v(τ0)θ 0R(τ0)aˆ0(1+C f τ−d0 min{τ120 ,λ− 12R(τ0)}λ− 12R(τ0) lnR(τ0))‖h0‖v,2+a.1303.6.5 Modes 1 to n with orthogonalitySet r = |y|. Consider h1(y,τ) =n∑j=1h j(r,τ)ϒ j satisfying´B2Rh1Z j = 0 for all j = 1, . . . ,n, τ ∈ (τ0,∞).Then ˆ 2R0h j(r,τ)Ur(r)rn−1dr = 0 for all τ ∈ (τ0,∞) (3.6.26)whereUr(r) = (n(n−2)) n−24 (2−n)r(1+ r2)− n2 . Let H = H j(r,τ)ϒ j satisfyingL1H j+ h˜ j = 0 in Rn,where h˜ j is the extension of h j as zero outside DR. H j is given byH j(r,τ) =Ur(r)ˆ r01ρn−1Ur(ρ)2ˆ ∞ρh˜ j(s,τ)Ur(s)sn−1dsdρ for −1< a≤ n−1,H j(r,τ) = −Ur(r)ˆ ∞r1ρn−1Ur(ρ)2ˆ ∞ρh˜ j(s,τ)Ur(s)sn−1dsdρ for a> n−1where a>−1 is used to guarantee that the spatial decay of h˜ j(s,τ)Ur(s)sn−1 is faster than s−1. Using(3.6.26), one has the following estimate‖H j‖v,aˆ1 ≲ ‖h j‖v,2+a,where aˆ1 is given in (3.6.5). Next, consider∂τΦ= ∆Φ+ pU p−1Φ+H j(r,τ)ϒ j in D2R,Φ= 0 on ∂D2R Φ(·,τ0) = 0 in B2R(τ0).By Lemma 3.6.6, we find a solution Φ j with the estimate|Φ j|≲ vθ 1Raˆ1Rn〈y〉1−n‖h1‖v,2+a.It follows thatϕ j1 = LΦ j with |ϕ j1|≲ vθ 1Raˆ1Rn〈y〉−1−n‖h1‖v,2+a.Consider∂τϕ j2 =L1ϕ j2+ f1ϕ j2+ f2r∂rϕ j2+ f1ϕ j1+ f2r∂rϕ j1 for r ∈ (0,R(τ)), τ ∈ (τ0,∞)∂rϕ j2(0,τ) = 0= ϕ j2(R(τ),τ) for τ ∈ (τ0,∞), ϕ j2(r,τ0) = 0 for r ∈ (0,R(τ0))where| f1ϕ j1+ f2r∂rϕ j1|≲C f τ−dvθ 1Raˆ1Rn〈y〉−1−n‖h1‖v,2+a.Using Lemma 3.6.6 again, we get ϕ j2 with the following estimate|ϕ j2|≲C f τ−dvθ 1Raˆ1R2n〈y〉1−n‖h1‖v,2+a,131Set ϕ j[h j] = ϕ j1+ϕ j2. Then ϕ 1[h1] =n∑j=1ϕ j[h j]ϒ j with the following estimate〈y〉|∇ϕ 1|+ |ϕ 1|≲ vθ 1Raˆ1Rn(〈y〉−1−n+C f τ−dRn〈y〉1−n)‖h1‖v,2+a.3.7 Additional information 1: Estimates for heat equationsRecalling T outn defined in (3.1.3), we only require t0 ≥ 0 in Lemma 3.7.1 and Lemma 3.7.2.3.7.1 Heat equation with right hand side v(t)|x|−b1{l1(t)≤|x|≤l2(t)}Lemma 3.7.1. Assume n > 2, v(t) ≥ 0, b ∈ R, 0 ≤ l1(t) ≤ l2(t) ≤ C∗t 12 , C−1l li(t) ≤ li(s) ≤ Clli(t),i= 1,2, for all t2 ≤ s≤ t, t ≥ t0 ≥ 0, where C∗ > 0,Cl ≥ 1. ThenT outn[v(t)|x|−b1{l1(t)≤|x|≤l2(t)}]≲ t− n2 e−|x|216tˆ t2t02v(s)ln−b2 (s) if b< nln( l2(s)l1(s)) if b= nln−b1 (s) if b> nds+ supt1∈[t/2,t]v(t1)l2−b2 (t) if b< 2〈ln( l2(t)l1(t))〉 if b= 2l2−b1 (t) if b> 2for |x| ≤ l1(t)l2−b2 (t) if b< 2〈ln( l2(t)|x| )〉 if b= 2|x|2−b if 2< b< n|x|2−n〈ln( |x|l1(t))〉 if b= n|x|2−nln−b1 (t) if b> nfor l1(t)< |x| ≤ l2(t)|x|2−ne− |x|216tln−b2 (t) if b< n〈ln( l2(t)l1(t))〉 if b= nln−b1 (t) if b> nfor |x|> l2(t).132Proof.T outn[v(t)|x|−b1{l1(t)≤|x|≤l2(t)}]≲ t− n2ˆ t2t02ˆRne−|x−y|24t v(s)|y|−b1{l1(s)≤|y|≤l2(s)}dyds+ supt1∈[t/2,t]v(t1)ˆ tt2ˆRn(t− s)− n2 e−|x−y|24(t−s) |y|−b1{C−1l l1(t)≤|y|≤Cl l2(t)}dyds := u1+ supt1∈[t/2,t]v(t1)u2.For u1, notice |y| ≤C∗t 12 . For |x| ≤ 2C∗t 12 , we haveu1 ≲ t−n2ˆ t2t02ˆRnv(s)|y|−b1{l1(s)≤|y|≤l2(s)}dyds≲ t−n2ˆ t2t02v(s)ln−b2 (s) if b< nln( l2(s)l1(s)) if b= nln−b1 (s) if b> nds.For |x|> 2C∗t 12 , one has |x− y| ≥ |x|2 . Thenu1 ≲ t−n2 e−|x|216tˆ t2t02v(s)ln−b2 (s) if b< nln( l2(s)l1(s)) if b= nln−b1 (s) if b> nds.Let us estimate u2 in different regions.For |x| ≤ (2Cl)−1l1(t), since |y|2 ≤ |x− y| ≤ 2|y|, thenu2 ≤ˆ tt2ˆRn(t− s)− n2 e−|y|216(t−s) |y|−b1{C−1l l1(t)≤|y|≤Cl l2(t)}dyds∼ˆ tt2(t− s)− b2ˆ C2l l22 (t)16(t−s)l21 (t)16C2l (t−s)e−zzn−b2 −1dzds=(ˆ t−l22(t)t2+ˆ t−l21(t)t−l22(t)+ˆ tt−l21(t)· · ·):= u21+u22+u23 ≲l2−b2 (t) if b< 2〈ln( l2(t)l1(t))〉 if b= 2l2−b1 (t) if b> 2.In order to get the last inequality above, we need the following estimates. For u21, since n > 2, wehaveu21 ∼ˆ t−l22(t)t2(t− s)− b2ˆ C2l l22 (t)16(t−s)l21 (t)16C2l (t−s)zn−b2 −1dzds≲l2−b2 (t) if b< n〈ln( l2(t)l1(t))〉l2−n2 (t) if b= nln−b1 (t)l2−n2 (t) if b> n.133For u22, since l21(t)≲ t− s≲ l22(t), thenu22 ≲ˆ t−l21(t)t−l22(t)(t− s)− b21 if b< n〈ln( t−sl21(t))〉 if b= n(l21(t)t−s )n−b2 if b> nds≲l2−b2 (t) if b< 2〈ln( l2(t)l1(t))〉 if b= 2l2−b1 (t) if b> 2.For u23, we haveu23 ≲ˆ tt−l21(t)(t− s)− b2ˆ C2l l22 (t)16(t−s)l21 (t)16C2l (t−s)e−z2 dzds≲ˆ tt−l21(t)(t− s)− b2 e−l21 (t)32C2l (t−s) ds∼ l2−b1 (t).For (2Cl)−1l1(t)≤ |x| ≤ 2Cll2(t), thenu2 ≤ˆ tt2ˆRn(t− s)− n2 e−|x−y|24(t−s) |y|−b(1{(4Cl)−1l1(t)≤|y|≤ |x|2 }+1{ |x|2 ≤|y|≤2|x|}+1{2|x|≤|y|≤4Cl l2(t)})dyds:= u21+u22+u23 ≲l2−b2 (t) if b< 2〈ln( l2(t)|x| )〉 if b= 2|x|2−b if 2< b< n|x|2−n〈ln( |x|l1(t))〉 if b= n|x|2−nln−b1 (t) if b> n.For the last inequality above, we need to estimate u21, u22 and u23. For u21, since n> 2, one hasu21 ≤ˆ tt2ˆRn(t− s)− n2 e−|x|216(t−s) |y|−b1{(4Cl)−1l1(t)≤|y|≤ |x|2 }dyds≲|x|2−b if b< n|x|2−n〈ln( |x|l1(t))〉 if b= n|x|2−nln−b1 (t) if b> n.For u22, we haveu22 ≲ |x|−bˆ tt2ˆRn(t− s)− n2 e−|x−y|24(t−s) 1{|x−y|≤3|x|}dyds∼ |x|−bˆ tt2ˆ 9|x|24(t−s)0e−zzn2−1dzds∼ |x|2−b.134For u23, we haveu23 ≤ˆ tt2ˆRn(t− s)− n2 e−|y|216(t−s) |y|−b1{2|x|≤|y|≤4Cl l2(t)}dyds∼ˆ tt2ˆ C2l l22 (t)t−s|x|24(t−s)(t− s)− b2 e−zz n−b2 −1dzds=ˆ t− l22 (t)2C2∗t2+ˆ t− |x|28C2l C2∗t− l22 (t)2C2∗+ˆ tt− |x|28C2l C2∗· · · := u231+u232+u233 ≲l2−b2 (t) if b< 2〈ln( l2(t)|x| )〉 if b= 2|x|2−b if b> 2.In order to get the last inequality above, we need the following estimates. For u231, we haveu231 ∼ˆ t− l22 (t)2C2∗t2ˆ C2l l22 (t)t−s|x|24(t−s)(t− s)− b2 z n−b2 −1dzds≲l2−b2 (t) if b< nl2−n2 (t)〈ln( l2(t)|x| )〉 if b= n|x|n−bl2−n2 (t) if b> n.For u232, since n> 2, we estimateu232 ≲ˆ t− |x|28C2l C2∗t− l22 (t)2C2∗(t− s)− b21 if b< n〈ln( |x|2t−s)〉 if b= n( |x|2t−s)n−b2 if b> nds≲l2−b2 (t) if b< 2〈ln( l2(t)|x| )〉 if b= 2|x|2−b if b> 2.For u233, one hasu233 ≲ˆ tt− |x|28C2l C2∗(t− s)− b2 e−|x|28(t−s) ds∼ |x|2−bˆ ∞C2l C2∗e−zzb2−2dz∼ |x|2−b.For |x| ≥ 2Cll2(t), since |x|2 ≤ |x− y| ≤ 2|x|, then for n> 2, it follows thatu2 ≲ˆ tt2(t− s)− n2 e−|x|216(t−s) dsln−b2 (t) if b< n〈ln( l2(t)l1(t))〉 if b= nln−b1 (t) if b> n≲ |x|2−ne− |x|216tln−b2 (t) if b< n〈ln( l2(t)l1(t))〉 if b= nln−b1 (t) if b> n.1353.7.2 Heat equation with right hand side v(t)|x|−b1{|x|≥t1/2}Lemma 3.7.2. Assume n> 0, v(t)≥ 0, b ∈ R, t0 ≥ 0, thenT outn[v(t)|x|−b1{|x|≥t 12 }]≲t−n2´ t/2t0/2v(s)tn−b2 if b< n〈ln(ts−1)〉 if b= nsn−b2 if b> nds+ t1−b2 supt1∈[t/2,t]v(t1) if |x| ≤ t 12|x|−b(t supt1∈[t/2,t]v(t1)+´ t/2t0/2v(s)ds)+ t−n2 e−|x|216t´ t/2t0/2v(s)0 if b< n〈ln(|x|s− 12 )〉 if b= nsn−b2 if b> nds if |x|> t 12.Proof. By definition, we writeT outn[v(t)|x|−b1{|x|≥t 12 }]≲ t− n2ˆ t/2t0/2ˆRne−|x−y|24t v(s)|y|−b1{|y|≥s 12 }dyds+ supt1∈[t/2,t]v(t1)ˆ tt/2ˆRn(t− s)− n2 e−|x−y|24(t−s) |y|−b1{|y|≥2− 12 t 12 }dyds := t− n2 u1+ supt1∈[t/2,t]v(t1)u2.For u1, when |x| ≤ 2t 12 , we haveu1 ≲ˆ t/2t0/2ˆRnv(s)|y|−b1{s 12≤|y|≤4t 12 }dyds+ˆ t/2t0/2ˆRne−|y|216t v(s)|y|−b1{4t 12≤|y|}dyds≲ˆ t/2t0/2v(s)tn−b2 if b< n〈ln(ts−1)〉 if b= nsn−b2 if b> nds+ tn−b2ˆ t/2t0/2v(s)ds∼ˆ t/2t0/2v(s)tn−b2 if b< n〈ln(ts−1)〉 if b= nsn−b2 if b> nds.For u1, when |x|> 2t 12 , we haveu1 =ˆ t/2t0/2ˆRne−|x−y|24t v(s)|y|−b(1{s 12≤|y|≤ |x|2 }+1{ |x|2 ≤|y|≤2|x|}+1{2|x|≤|y|})dyds≲ˆ t/2t0/2v(s)ˆRn(e−|x|216t |y|−b1{s 12≤|y|≤ |x|2 }+ |x|−be−|x−y|24t 1{|x−y|≤3|x|}+ e−|y|216t |y|−b1{2|x|≤|y|})dyds≲ e−|x|216tˆ t/2t0/2v(s)|x|n−b if b< n〈ln(|x|s− 12 )〉 if b= nsn−b2 if b> nds+ tn2 |x|−bˆ t/2t0/2v(s)ds+ tn−b2 e−|x|28tˆ t/2t0/2v(s)ds136≲ e−|x|216tˆ t/2t0/2v(s)0 if b< n〈ln(|x|s− 12 )〉 if b= nsn−b2 if b> nds+ tn2 |x|−bˆ t/2t0/2v(s)ds.For u2, when |x| ≤ 2− 32 t 12 , we have |y| ≥ 2|x|. Thenu2 ≤ˆ tt/2ˆRn(t− s)− n2 e−|y|216(t−s) |y|−b1{|y|≥2− 12 t 12 }dyds≲ˆ tt/2(t− s)− b2 e− t64(t−s) ds∼ t1− b2 .For u2, when |x| ≥ 2− 32 t 12 , one hasu2 ≤ˆ tt/2ˆRn(t− s)− n2 e−|x−y|24(t−s) |y|−b(1{9−1t 12≤|y|≤ |x|2 }+1{ |x|2 ≤|y|≤4|x|}+1{4|x|≤|y|})dyds≲ˆ tt/2ˆRn(t− s)− n2(e−|x|216(t−s) |y|−b1{9−1t 12≤|y|≤ |x|2 }+ |x|−be−|x−y|24(t−s) 1{|x−y|≤5|x|}+ e− |y|216(t−s) |y|−b1{4|x|≤|y|})dyds≲ |x|2−ne− |x|216t|x|n−b if b< n〈ln(|x|t− 12 )〉 if b= ntn−b2 if b> n+ t|x|−b+ |x|2−be− |x|22t ∼ t|x|−b.3.7.3 Cauchy problem with initial value 〈y〉−bLemma 3.7.3. For n≥ 1, b ∈ R and t > 0, it holds that(4πt)−n2ˆRne−|x−y|24t 〈y〉−bdy≲〈t〉− b2 1{|x|≤〈t〉 12 }+ |x|−b1{|x|>〈t〉 12 } if b< n〈t〉− n2 ln(t+2)1{|x|≤〈t〉 12 }+(|x|−n+ t− n2 e− |x|216t ln(|x|+2))1{|x|>〈t〉 12 } if b= n〈t〉− n2 1{|x|≤〈t〉 12 }+(|x|−b+ t− n2 e− |x|216t)1{|x|>〈t〉 12 } if b> n.Proof. Setu(x, t) = (4πt)−n2ˆRne−|x−y|24t 〈y〉−bdy∼ t− n2(ˆ|y|≤ |x|2+ˆ|x|2 ≤|y|≤2|x|+ˆ2|x|≤|y|)e−|x−y|24t 〈y〉−bdy.137We estimate term by term:ˆ|y|≤ |x|2e−|x−y|24t 〈y〉−bdy≲ e− |x|216tˆ|y|≤ |x|2〈y〉−bdy∼e−|x|216t |x|n if |x| ≤ 1e−|x|216t |x|n−b if b< ne−|x|216t ln(|x|+2) if b= ne−|x|216t if b> nif |x|> 1,ˆ|x|2 ≤|y|≤2|x|e−|x−y|24t 〈y〉−bdy≲ 〈x〉−bˆ|x−y|≤3|x|e−|x−y|24t dy∼〈x〉−b|x|n if |x| ≤ t12〈x〉−bt n2 if |x|> t 12,and ˆ2|x|≤|y|e−|x−y|24t 〈y〉−bdy≤ˆ2|x|≤|y|e−|y|216t 〈y〉−bdy.For |x| ≥ 1, we haveˆ2|x|≤|y|e−|y|216t 〈y〉−bdy∼ t n−b2ˆ ∞|x|24te−zzn−b2 −1dz≲tn−b2 if |x| ≤ t 12 ,b< nln( t|x|2 )+1 if |x| ≤ t12 ,b= n|x|n−b if |x| ≤ t 12 ,b> ntn−b2 e−|x|28t if |x|> t 12.For |x|< 1, we haveˆ2|x|≤|y|e−|y|216t 〈y〉−bdy∼ˆ 22|x|e−r216t rn−1dr+ˆ ∞2e−r216t rn−1−bdr≲tn2 e−|x|28t if t ≤ |x|2tn2 if |x|2 < t ≤ 11 if t > 1+tn−b2 e−18t if t < 1tn−b2 if t ≥ 1,b< nln(t+2) if t ≥ 1,b= n1 if t ≥ 1,b> n≲tn2 e−|x|216t if t ≤ |x|2tn2 if |x|2 < t ≤ 1tn−b2 if t ≥ 1,b< nln(t+2) if t ≥ 1,b= n1 if t ≥ 1,b> n.138Combining the above estimates, one hasu(x, t)≲1 if t ≤ 1t−b2 if t ≥ 1,b< nt−n2 ln(t+2) if t ≥ 1,b= nt−n2 if t ≥ 1,b> nif |x|< 1t−b2 if |x| ≤ t 12 ,b< nt−n2 ln(t+2) if |x| ≤ t 12 ,b= nt−n2 if |x| ≤ t 12 ,b> n〈x〉−b if |x|> t 12 ,b< n〈x〉−n+ t− n2 e− |x|216t ln(|x|+2) if |x|> t 12 ,b= n〈x〉−b+ t− n2 e− |x|216t if |x|> t 12 ,b> nif |x| ≥ 1≲〈t〉− b2 if |x| ≤max{1, t 12 },b< n〈t〉− n2 ln(t+2) if |x| ≤max{1, t 12 },b= n〈t〉− n2 if |x| ≤max{1, t 12 },b> n〈x〉−b if |x|>max{1, t 12 },b< n〈x〉−n+ t− n2 e− |x|216t ln(|x|+2) if |x|>max{1, t 12 },b= n〈x〉−b+ t− n2 e− |x|216t if |x|>max{1, t 12 },b> n.This completes the proof of Lemma 3.7.3.3.8 Additional information 2: Proof of Proposition 3.2.1: solving theouter problemProof. It suffices to find a fixed point for ψ =T out4 [G [ψ,ϕ ,µ1,ξ ]]. Setwo(x, t) = ln t(t(ln t)2)5δ−κR−a(1{|x|≤t 12 }+ t|x|−21{|x|>t 12 }),‖g‖o = sup(x,t)∈R4×(t0,∞)w−1o (x, t)|g(x, t)|, Bo = {g(x, t) : ‖g‖o ≤ Do},where Do ≥ 1 will be determined later. For any ψ1 ∈ Bo, let us estimate G [ψ1,ϕ ,µ1,ξ ] term by term.In this proof, we will apply Lemma 3.7.1 and Lemma 3.7.2 multiple times to estimate convolutionT out4 and will not state them repetitively.139By the definitions of the norms (3.2.8), (3.2.9), (3.2.10), one has|ϕ(y, t)|+ 〈y〉|∇ϕ(y, t)|≲ (t(ln t)2)5δ−κ〈y〉−a‖ϕ‖i,κ−5δ ,a,|µ1|+ t|µ1t |≲ t ln t(t(ln t)2)5δ−κR−a‖µ1‖∗1, |ξ |+ t|ξt |≲ t(ln t)2(t(ln t)2)5δ−κR−a‖ξ‖∗2.Then ∣∣∣∣∆xηRµ−1ϕ(x−ξµ , t)∣∣∣∣≲ (µ0R)−21{µ0R≤|x−ξ |≤2µ0R}µ−1(t(ln t)2)5δ−κ〈y〉−a‖ϕ‖i,κ−5δ ,a∼ Λ1(µ0R)−2 ln t(t(ln t)2)5δ−κR−a1{µ0R≤|x−ξ |≤2µ0R},∣∣∣∣2∇xηR ·µ−2∇yϕ(x−ξµ , t)∣∣∣∣≲ (µ0R)−11{µ0R≤|x−ξ |≤2µ0R}µ−2(t(ln t)2)5δ−κ〈y〉−1−a‖ϕ‖i,κ−5δ ,a≲ Λ1(µ0R)−2 ln t(t(ln t)2)5δ−κR−a1{µ0R≤|x−ξ |≤2µ0R},∣∣∣∣∂tηRµ−1ϕ(x−ξµ , t)∣∣∣∣= ∣∣∣∣∇η(x−ξµ0R ) · (−ξtµ0R − x−ξµ0R (µ0R)tµ0R )µ−1ϕ(x−ξµ , t)∣∣∣∣≲ Λ21(µ0R)−2 ln t(t(ln t)2)5δ−κR−a1{µ0R≤|x−ξ |≤2µ0R},where we have used γ < 12 and 5δ −κ <−1 in the last inequality. Then one hasT out4[(µ0R)−2 ln t(t(ln t)2)5δ−κR−a1{µ0R≤|x−ξ |≤2µ0R}]≲ T out4[(µ0R)−2 ln t(t(ln t)2)5δ−κR−a1{µ0R/2≤|x|≤4µ0R}]≲ t−2e−|x|216tˆ t2t02(lns)−1(s(lns)2)5δ−κR2−a(s)ds+ln t(t(ln t)2)5δ−κR−a if |x| ≤ µ0Rln t(t(ln t)2)5δ−κR−a(µ0R)2|x|−2e− |x|216t if |x|> µ0R ≲ woprovided 5δ −κ−aγ >−2. Also,∣∣∣∣ηRµ−2ξt ·∇yϕ(x−ξµ , t)∣∣∣∣≲ Λ211{|x|≤3µ0R}(ln t)2(ln t)2(t(ln t)2)5δ−κR−a(t(ln t)2)5δ−κ〈y〉−1−a∼ Λ21(ln t)4(t(ln t)2)10δ−2κR−a(1{|x|≤µ0}+(ln t)−1−a|x|−1−a1{µ0<|x|≤3µ0R}),140andT out4[(ln t)4(t(ln t)2)10δ−2κR−a(ln t)−1−a|x|−1−a1{µ0<|x|≤3µ0R}]≤ T out4[(ln t)3(t(ln t)2)10δ−2κR−a|x|−11{µ0<|x|≤3µ0R}]≲ t−2e−|x|216tˆ t2t02(lns)3(s(lns)2)10δ−2κR−a(s)(µ0R)3(s)ds+µ0R(ln t)3(t(ln t)2)10δ−2κR−a if |x| ≤ µ0R(ln t)3(t(ln t)2)10δ−2κR−a|x|−2e− |x|216t (µ0R)3 if |x|> µ0R ≲ t−ε0 wo,T out4 [(ln t)4(t(ln t)2)10δ−2κR−a1{|x|≤µ0}]≲ t−ε0 woprovided 5δ −κ−aγ >−2 and ε > 0 is sufficiently small.Using (3.1.32), one has∣∣∣∣(1−ηR)S[u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0 , t)η(4(x−ξ )√t )]∣∣∣∣≲[t−2(ln t)−1|x|−2+ t−1(ln t)−2|µ1||x|−4+(ln t)−2(|g˜[µ¯0,µ1]|+ |µ¯0t | ln t supt1∈[t/2,t](|µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)| ))|x|−4+ |ξt |(ln t)−1|x|−3]1{ µ0R2 ≤|x|≤9t12 }+ |ξt |t 32 (ln t)−1|x|−61{|x|>2t 12 }+(t2(ln t)−1|x|−6)31{|x|>2t 12 }≲(t−2(ln t)−1|x|−2+Λ21 ln t(t(ln t)2)5δ−κR−a|x|−3)1{ µ0R2 ≤|x|≤9t12 }+Λ1t32 ln t(t(ln t)2)5δ−κR−a|x|−61{|x|>2t 12 }+(t2(ln t)−1|x|−6)31{|x|>2t 12 }since|µ¯0t | ln t supt1∈[t/2,t]( |µ1(t1)|µ¯0(t)+|µ1t(t1)||µ¯0t(t)|)≲ Λ1(ln t)2(t(ln t)2)5δ−κR−a,g˜[µ¯0,µ1]≲ Λ21t−2ˆ tt0/2(lns(s(lns)2)5δ−κR−a(s)(lns)−2+ s lns(s(lns)2)5δ−κR−a(s))ds+Λ21(t ln t)−1[t(ln t)3(t(ln t)2)5δ−κR−a]2 ≲ Λ21 ln t(t(ln t)2)5δ−κR−a(3.8.1)when 5δ −κ−aγ >−2.141Then we estimateT out4[t−2(ln t)−1|x|−2]1{ µ0R2 ≤|x|≤9t 12 }≲ t−2 ln ln te−|x|216t +t−2 if |x| ≤ µ0Rt−2(ln t)−1(ln(|x|−1t 12 )+1) if µ0R< |x| ≤ t 12(t ln t)−1|x|−2e− |x|216t if |x|> t 12≲ t−ε00 wosince 5δ −κ−aγ >−2.T out4[ln t(t(ln t)2)5δ−κR−a|x|−31{ µ0R2 ≤|x|≤9t 12 }]≲ t−2ˆ t2t02lns(s(lns)2)5δ−κR−a(s)s12 ds+ln t(t(ln t)2)5δ−κR−a(µ0R)−1 if |x| ≤ µ0Rln t(t(ln t)2)5δ−κR−a|x|−1 if µ0R< |x| ≤ t 12ln t(t(ln t)2)5δ−κR−at12 |x|−2e− |x|216t if |x|> t 12≲ t−ε0 wosince 5δ −κ−aγ >−2.T out4[t32 ln t(t(ln t)2)5δ−κR−a|x|−61{|x|>2t 12 }]≲t−2 ´ tt02s12 lns(s(lns)2)5δ−κR−a(s)ds if |x| ≤ t 12|x|−6 ´ tt02s32 lns(s(lns)2)5δ−κR−a(s)ds+ t−2e−|x|216t´ t2t02s12 lns(s(lns)2)5δ−κR−a(s)ds if |x|> t 12≲ t−ε0 wo.T out4[(t2(ln t)−1|x|−6)31{|x|>2t 12 }]≲ t−21{|x|≤t 12 }+(t7(ln t)−3|x|−18+ t−2e− |x|216t)1{|x|>t 12 } ≲ t−ε0 wowhen 5δ −κ−aγ >−2.(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)+ψ1+ηRµ−1ϕ(x−ξµ, t))3−(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))3−3(µ−1U(x−ξµ))2(ψ1+ηRµ−1ϕ(x−ξµ, t))−[3(u1+φ[µ]−µ−1U(x−ξµ ))(u1+φ[µ]+µ−1U(x−ξµ))+6(u1+φ[µ])µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)]ηRµ−1ϕ(x−ξµ, t)= 3(u1+φ[µ]−µ−1U(x−ξµ ))(u1+φ[µ]+µ−1U(x−ξµ))ψ1142+6(u1+φ[µ])µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t)ψ1+3(µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))2(ψ1+ηRµ−1ϕ(x−ξµ, t))+3(u1+φ[µ]+ µ¯−10 Φ0(x−ξµ¯0, t)η(4(x−ξ )√t))(ψ1+ηRµ−1ϕ(x−ξµ, t))2+(ψ1+ηRµ−1ϕ(x−ξµ, t))3≲((t ln t)−11{|x¯|≤2t 12 }+ t2(ln t)−1|x¯|−61{|x¯|>2t 12 }+ ln t〈y〉−21{|x¯|≥t 12 })×((t ln t)−11{|x¯|≤2t 12 }+ t2(ln t)−1|x¯|−61{|x¯|>2t 12 }+ ln t〈y〉−2)|ψ1|+(ln t〈y〉−21{|x¯|≤2t 12 }+(t ln t)−11{|x¯|≤2t 12 }+ t2(ln t)−1|x¯|−61{|x¯|>2t 12 })(t ln t)−1〈y¯〉−2 ln(2+ |y¯|)1{|x¯|≤8t 12 }|ψ1|+(t ln t)−2〈y¯〉−4 ln2(2+ |y¯|)1{|x¯|≤8t 12 }(|ψ1|+ |ηRµ−1ϕ(x−ξµ , t)|)+(ln t〈y〉−21{|x¯|≤2t 12 }+(t ln t)−11{|x¯|≤2t 12 }+O(t2(ln t)−1|x¯|−6)1{|x¯|>2t 12 }+(t ln t)−1〈y¯〉−2 ln(2+ |y¯|)1{|x¯|≤8t 12 })∣∣∣ψ1+ηRµ−1ϕ(x−ξµ , t)∣∣∣2+ ∣∣∣ψ1+ηRµ−1ϕ(x−ξµ , t)∣∣∣3≲(t−1〈y〉−21{|x|≤t 12 }+(ln t)−2|x|−41{|x|>t 12 })|ψ1|+(t ln t)−2〈y¯〉−4 ln2(2+ |y¯|)1{|x¯|≤8t 12 }∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣+(ln t〈y〉−21{|x|≤t 12 }+ t2(ln t)−1|x|−61{|x|>t 12 })∣∣∣ψ1+ηRµ−1ϕ(x−ξµ , t)∣∣∣2+ ∣∣∣ψ1+ηRµ−1ϕ(x−ξµ , t)∣∣∣3,where we have used Corollary 3.1.3.Consider the terms involving ϕ :T out4[(t ln t)−2〈y¯〉−4 ln2(2+ |y¯|)1{|x¯|≤8t 12 }∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣]≲ Λ1T out4 [t−2〈y¯〉−41{|x¯|≤2µ0R} ln t(t(ln t)2)5δ−κ〈y〉−a]∼ Λ1T out4 [t−2 ln t(t(ln t)2)5δ−κ〈y〉−4−a1{|x¯|≤2µ0R}]≲ Λ1t−2e−|x|216tsinceT out4 [t−2 ln t(t(ln t)2)5δ−κ〈y〉−4−a1{µ0≤|x¯|≤2µ0R}]≲ T out4 [t−2(ln t)−3−a(t(ln t)2)5δ−κ |x|−4−a1{ µ02 ≤|x|≤4µ0R}]≲ t−2e−|x|216t +t−2(ln t)−1(t(ln t)2)5δ−κ if |x| ≤ µ0t−2(ln t)−3(t(ln t)2)5δ−κ |x|−2e− |x|216t if |x|> µ0 ≲ t−2e−|x|216t ≲ t−ε0 wo,143T out4 [t−2 ln t(t(ln t)2)5δ−κ〈y〉−4−a1{|x¯|<µ0}]≲T out4 [t−2 ln t(t(ln t)2)5δ−κ1{|x|<2µ0}]≲ t−2e−|x|216t +t−2(ln t)−1(t(ln t)2)5δ−κ if |x| ≤ µ0t−2(ln t)−3(t(ln t)2)5δ−κ |x|−2e− |x|216t if |x|> µ0 ≲ t−2e−|x|216t .Next, we haveT out4[(ln t〈y〉−21{|x|≤t 12 }+ t2(ln t)−1|x|−61{|x|>t 12 })∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣2]≲ Λ21T out4 [(ln t1{|x|≤µ0}+(ln t)−1|x|−21{µ0<|x|≤4µ0R})(ln t)2(t(ln t)2)10δ−2κ〈y〉−2a]∼ Λ21T out4[(ln t)3(t(ln t)2)10δ−2κ1{|x|≤µ0}+(t(ln t)2)10δ−2κ(ln t)1−2a|x|−2−2a1{µ0<|x|≤4µ0R}],andT out4[(t(ln t)2)10δ−2κ(ln t)1−2a|x|−2−2a1{µ0<|x|≤4µ0R}]≲T out4[(t(ln t)2)10δ−2κ ln t|x|−21{µ0<|x|≤4µ0R}]≲ t−2e−|x|216tˆ t2t02(s(lns)2)10δ−2κ lns(µ0R)2(s)ds+(t(ln t)2)10δ−2κ(ln t)2 if |x| ≤ µ0(t(ln t)2)10δ−2κ ln t(ln( µ0R|x| )+1) if µ0 < |x| ≤ µ0R(t(ln t)2)10δ−2κ ln t(µ0R)2|x|−2e−|x|216t if |x|> µ0R≲ t−ε0 wowhen 5δ −κ <−1 and 5δ −κ−aγ >−2. Also,T out4[(ln t)3(t(ln t)2)10δ−2κ1{|x|≤µ0}]≲ t−ε0 wo.When 5δ −κ+(2−a)γ < 0, one has∣∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣∣3≲Λ1 ln t(t(ln t)2)5δ−κ〈y〉−a ∣∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣∣2≲ ln t〈y〉−2 ∣∣∣∣ηRµ−1ϕ(x−ξµ , t)∣∣∣∣2 .Let us now estimate terms involving ψ1.∣∣∣∣µ−2U2(x−ξµ )ψ1(1−ηR)∣∣∣∣≲ (ln t)−2|x|−4|ψ1|1{|x|≥ µ0R2 }≲ Do(ln t)−2|x|−4 ln t(t(ln t)2)5δ−κR−a(1{ µ0R2 ≤|x|≤t12 }+ t|x|−21{|x|>t 12 })= Do(ln t)−1(t(ln t)2)5δ−κR−a|x|−41{ µ0R2 ≤|x|≤t 12 }+Dot(ln t)−1(t(ln t)2)5δ−κR−a|x|−61{|x|>t 12 }.144For the first term, we haveT out4[(ln t)−1(t(ln t)2)5δ−κR−a|x|−41{ µ0R2 ≤|x|≤t 12 }]≲ t−2e−|x|216t +(ln t)−1(t(ln t)2)5δ−κR−a(µ0R)−2 if |x| ≤ µ0R(ln t)−1(t(ln t)2)5δ−κR−a|x|−2(ln( |x|µ0R)+1) if µ0R< |x| ≤ t12(t(ln t)2)5δ−κR−a|x|−2e− |x|216t if |x|> t 12≲ t−ε0 wowhen 5δ −κ−aγ >−2. For the second term, one hasT out4[t(ln t)−1(t(ln t)2)5δ−κR−a|x|−61{|x|>t 12 }]≲ t−21{|x|≤t 12 }+(|x|−6+ t−2e− |x|216t)1{|x|>t 12 } ≲ t−ε0 wo.Thus ∣∣∣∣T out4 [µ−2U2(x−ξµ )ψ1(1−ηR)]∣∣∣∣≲ t−ε0 wo.Notice (ln t〈y〉−21{|x|≤t 12 }+ t2(ln t)−1|x|−61{|x|>t 12 })|ψ1|2+ |ψ1|3≲ D2o(t−1〈y〉−21{|x|≤t 12 }+(ln t)−2|x|−41{|x|>t 12 })|ψ1|when 5δ −κ−aγ <−1. And(ln t)−2|x|−41{|x|>t 12 }|ψ1|≲ (ln t)−2|x|−4|ψ1|1{|x|≥ µ0R2 },where the last term has been estimated above. So we only need to estimate the following termt−1〈y〉−21{|x|≤t 12 }|ψ1|≲ Do ln t(t(ln t)2)5δ−κR−at−1〈y〉−21{|x|≤t 12 }≲ Do ln t(t(ln t)2)5δ−κR−at−1(1{|x|≤µ0}+(ln t)−2|x|−21{µ0<|x|≤t 12 }),andT out4[(ln t)−1(t(ln t)2)5δ−κR−at−1|x|−21{µ0<|x|≤t 12 }]≲ t−2e−|x|216t +(t(ln t)2)5δ−κR−at−1 if |x| ≤ µ0(ln t)−1(t(ln t)2)5δ−κR−at−1(ln(|x|−1t 12 )+1) if µ0 < |x| ≤ t 12(ln t)−1(t(ln t)2)5δ−κR−a|x|−2e− |x|216t if |x|> t 12≲ t−ε0 wo,T out4 [ln t(t(ln t)2)5δ−κR−at−11{|x|≤µ0}]≲ t−ε0 wo.145These implyT out4[t−1〈y〉−21{|x|≤t 12 }|ψ1|]≲ t−ε0 wo.Taking Do = Do(Λ1) large depending on Λ1 and then choosing t0 large enough, we haveT out4 [G [ψ1,ϕ ,µ1,ξ ]] ∈ Bo.The contraction property is given by a similar method used in dealing with terms including ψ1.Then the unique solution ψ is found in Bo for (3.2.11) by the contraction mapping theorem.From now on, we also regard Do(Λ1) as a constant depending on Λ1. Reviewing the estimatesabove and utilizing 1{|x|≥ µ0R2 }to transform the spatial decay to time decay, one has|G [ψ,ϕ ,µ1,ξ ]|≲C(Λ1)[(µ0R)−2 ln t(t(ln t)2)5δ−κR−a+(ln t)3(t(ln t)2)10δ−2κ ]whereC(Λ1) is a constant depending on Λ1 which changes from line to line.By gradient estimate, we have|∇ψ|≲C(Λ1) ln t(t(ln t)2)5δ−κR−a.Next, we will use the scaling argument to deduce the Hölder estimate of ψ(x, t) in time variablet. For x1 ∈ R4, t1 > 4t0, setψ˜(z,s) = ψ(x1+λ (t1)z, t1+λ 2(t1)s)where 0< λ (t1)≤ t121 . Then∂sψ˜ = ∆zψ˜+ G˜ (z,s)where G˜ (z,s) = λ 2(t1)G [ψ,ϕ ,µ1,ξ ](x1+λ (t1)z, t1+λ 2(t1)s), and standard parabolic regularity the-ory implies‖ψ˜‖C2α,α (B(0, 116 )×(− 14 ,0)) ≤C(α)(‖ψ˜‖L∞(B(0, 14 )×(− 12 ,0))+‖G˜ ‖L∞(B(0, 14 )×(− 12 ,0)))where α can be chosen as any constant in (0,1) and C(α) is a constant depending on α . Moreover,one has‖ψ˜‖L∞(B(0, 14 )×(− 12 ,0)) ≲C(Λ1) ln t1(t1(ln t1)2)5δ−κR−a(t1),‖G˜ ‖L∞(B(0, 14 )×(− 12 ,0)) ≲C(Λ1)λ2(t1)[(µ0R)−2(t1) ln t1(t1(ln t1)2)5δ−κR−a(t1)+(ln t1)3(t1(ln t1)2)10δ−2κ],146and‖ψ˜‖C2α,α (B(0, 116 )×(− 14 ,0)) ≥ sups1,s2∈(−1/4,0)|ψ(x1, t1+λ (t1)2s1)−ψ(x1, t1+λ (t1)2s2)||s1− s2|α= λ 2α(t1) sups1,s2∈(−1/4,0)|ψ(x1, t1+λ (t1)2s1)−ψ(x1, t1+λ (t1)2s2)||(t1+λ (t1)2s1)− (t1+λ (t1)2s2)|α= λ 2α(t1) sups1,s2∈(t1− λ (t1)24 ,t1)|ψ(x1,s1)−ψ(x1,s2)||s1− s2|α .Thussups1,s2∈(t1− λ (t1)24 ,t1)|ψ(x1,s1)−ψ(x1,s2)||s1− s2|α ≲C(Λ1,α){λ−2α(t1) ln t1(t1(ln t1)2)5δ−κR−a(t1)+λ 2−2α(t1)[(µ0R)−2(t1) ln t1(t1(ln t1)2)5δ−κR−a(t1)+(ln t1)3(t1(ln t1)2)10δ−2κ ]}.3.9 Additional information 3: Estimates for ∇x¯φ[µ¯0] and ∂tφ[µ¯0]In this section, we will revisit the calculations in Section 3.1.2 and derive the following estimates|∂tφ[µ¯0]|≲ t−2(ln t)−1, |∇x¯φ[µ¯0]|≲(t ln t)−1 if |x¯| ≤ µ0t−1(ln t)−2|x¯|−1+ t− 32 (ln t)−1 if µ0 < |x¯| ≤ t 12t−32 (ln t)−1e−|x¯|216t + t(ln t)−2|x¯|−5 if |x¯|> t 12.(3.9.1)Proof. Notice µ¯0 ∼ (ln t)−1 and |µ¯0t | ∼ t−1(ln t)−2. By (3.1.7), we have|∇x¯φ˜1[µ¯0]|≲ |x¯|t−2(ln t)−11{|x¯|≤2t 12 }+ |x¯|−1(t ln t)−1e−|x¯|24t 1{|x¯|>2t 12 }.For ∇x¯φ˜1b[µ¯0], we abbreviate ∇x¯φ˜1b[µ¯0] as ∇x¯φ˜1b. By (3.1.8), then∇x¯φ˜1b(x¯, t) =T out4 [∇x¯(−µ¯0t φˆ1+(E− E˜)[µ¯0])](x¯, t).Notice by (3.1.7), we have|µ¯0t∇x¯φˆ1(x¯, t)|≲ |x¯|t−3(ln t)−21{|x¯|≤t 12 }+ |x¯|−1t−2(ln t)−2e−|x¯|24t 1{|x¯|>t 12 },147|∇x¯(E− E˜)[µ¯0]|≲ t− 72 (ln t)−31{t 12≤|x¯|≤2t 12 },andT out4[t−72 (ln t)−31{t 12≤|x¯|≤2t 12 }+ |x¯|t−3(ln t)−21{|x¯|≤t 12 }]≲T out4[t−52 (ln t)−21{|x¯|≤2t 12 }]≲ t− 32 (ln t)−2e−|x¯|216t ,T out4[|x¯|−1t−2(ln t)−2e− |x¯|24t 1{|x¯|>t 12 }]≲T out4[(ln t)−2|x¯|−51{|x¯|>t 12 }]≲t−32 (ln t)−2 if |x¯| ≤ t 12t(ln t)−2|x¯|−5 if |x¯|> t 12.Thus|∇x¯φ˜1b|≲ t− 32 (ln t)−21{|x¯|≤t 12 }+ t(ln t)−2|x¯|−51{|x¯|>t 12 }.Next, let us consider ∇x¯φ2[µ¯0]. Recall the definition of φ2 in Lemma 3.1.2, then∇x¯φ2 =T out4[µ¯−20 µ¯0t∇x¯(Z5(x¯µ¯0)η(x¯√t))].Notice∣∣∣∣µ¯−20 µ¯0t∇x¯(Z5( x¯µ¯0 )η( x¯√t ))∣∣∣∣= ∣∣∣∣µ¯−20 µ¯0t(µ¯−10 ∇Z5( x¯µ¯0 )η( x¯√t )+ t− 12Z5( x¯µ¯0 )∇η( x¯√t ))∣∣∣∣≲ t−1[ln t(1+ | x¯µ¯0|)−31{|x¯|≤2t 12 }+ t− 12 (1+ | x¯µ¯0|)−21{t 12≤|x¯|≤2t 12 }]∼ t−1 ln t(1+ | x¯µ¯0|)−31{|x¯|≤2t 12 }∼ t−1 ln t1{|x¯|≤µ0}+ t−1(ln t)−2|x¯|−31{µ0<|x¯|≤2t 12 }.Therefore, we obtain|∇x¯φ2|≲T out4[t−1 ln t1{|x¯|≤µ0}+ t−1(ln t)−2|x¯|−31{µ0<|x¯|≤2t 12 }]≲(t ln t)−1 if |x¯| ≤ µ0t−1(ln t)−2|x¯|−1 if µ0 < |x¯| ≤ t 12t−32 (ln t)−2e−|x¯|216t if |x¯|> t 12.Since φ = φ˜1+ φ˜1b+φ2, one concludes the upper bound of |∇x¯φ[µ¯0]| in (3.9.1).The left part is devoted to estimating ∂tφ[µ¯0]. For ∂t φ˜1[µ¯0], by (3.1.6),|∂t φ˜1[µ¯0]|=∣∣∣∣2 32 µ¯0t |x¯|−2(e− |x¯|24t −η( |x¯|√t ))+232 µ¯0|x¯|−2( |x¯|24t2e−|x¯|24t +|x¯|2t32η ′(|x¯|√t))∣∣∣∣≲ t−2(ln t)−1e− |x¯|24t .148Next, we estimate ∂t φ˜1b[µ¯0]. For any integer n≥ 1 and f (x, t) ∈C1(Rn× (t0,∞)),∂t(ˆ tt0ˆRn[4π(t− s)]− n2 e−|x−y|24(t−s) f (y,s)dyds)= ∂t(ˆ t2t0ˆRn[4π(t− s)]− n2 e−|x−y|24(t−s) f (y,s)dyds+ˆ t20ˆRn(4πa)−n2 e−|x−y|24a f (y, t−a)dyda)=12ˆRn(2πt)−n2 e−|x−y|22t f (y,t2)dy+ˆ t2t0ˆRn∂t{[4π(t− s)]− n2 e−|x−y|24(t−s)}f (y,s)dyds+ˆ tt2ˆRn[4π(t− s)]− n2 e−|x−y|24(t−s) (∂t f )(y,s)dyds.(3.9.2)As a consequence of (3.1.8) and (3.9.2), we have∂t φ˜1b =12ˆR4(2πt)−2e−|x−y|22t(−µ¯0t φˆ1+(E− E˜)[µ¯0])(y, t2)dy+ˆ t2t0ˆR4∂t{[4π(t− s)]−2 e−|x−y|24(t−s)}(−µ¯0t φˆ1+(E− E˜)[µ¯0])(y,s)dyds+ˆ tt2ˆR4[4π(t− s)]−2 e−|x−y|24(t−s)[∂t(−µ¯0t φˆ1+(E− E˜)[µ¯0])](y,s)dyds,where by (3.1.9) and (3.1.10), it follows that∣∣∣(−µ¯0t φˆ1+(E− E˜)[µ¯0])(x¯, t2)∣∣∣+ ∣∣(−µ¯0t φˆ1+(E− E˜)[µ¯0])(x¯, t)∣∣≲ (t ln t)−21{|x¯|≤t 12 }+ t−1(ln t)−2|x¯|−2e− |x¯|24t 1{|x¯|>t 12 }+(t ln t)−31{t 12≤|x¯|≤2t 12 } ≲ (t ln t)−2e−|x¯|24t ,and|∂t(µ¯0t φˆ1)|= |µ¯0tt φˆ1+ µ¯0t∂t φˆ1|≲∣∣∣∣(t ln t)−2(t−11{|x¯|≤2t 12 }+ |x¯|−2e− |x¯|24t 1{|x¯|>2t 12 })+ t−3(ln t)−2e−|x¯|24t∣∣∣∣∼ t−3(ln t)−2e− |x¯|24t ,and|∂t(E− E˜)[µ¯0]|≲ t−4(ln t)−31{√t≤|x¯|≤2√t}.Thus by Lemma 3.7.3 and same calculation for deducing (3.1.12), we have|∂t φ˜1b|≲ (t ln t)−2ˆR4t−2e−|x−y|22t dy+ t−1ˆ tt0ˆR4(t− s)−2e−|x−y|28(t−s) (s lns)−2e−|y|24s dyds≲ (t ln t)−2.149Finally, we consider ∂tφ2[µ¯0]. By (3.9.2) and the definition of φ2 in Lemma 3.1.2, we have∂tφ2 =12ˆR4(2πt)−2e−|x−y|22t µ¯−20 (t2)µ¯0t(t2)Z5(yµ¯0( t2))η(√2y√t)dy+ˆ t2t0ˆR4∂t{[4π(t− s)]−2 e−|x−y|24(t−s)}µ¯−20 (s)µ¯0t(s)Z5(yµ¯0(s))η(y√s)dyds+ˆ tt2ˆR4[4π(t− s)]−2 e−|x−y|24(t−s)[∂s(µ¯−20 (s)µ¯0t(s)Z5(yµ¯0(s))η(y√s))]dyds,and by (3.1.16), ∣∣∣∣∣µ¯−20 ( t2)µ¯0t( t2)Z5( x¯µ¯0( t2))η(√2x¯√t)∣∣∣∣∣+∣∣∣∣µ¯−20 (t)µ¯0t(t)Z5( x¯µ¯0(t))η( x¯√t )∣∣∣∣≲ t−11{|x¯|≤(ln t)−1}+ t−1(ln t)−2|x¯|−21{(ln t)−1<|x¯|≤2t 12 },and ∣∣∣∣∂t(µ¯−20 µ¯0tZ5( x¯µ¯0 )η( x¯√t ))∣∣∣∣=∣∣∣∣∂t(µ¯−20 µ¯0t)Z5( x¯µ¯0 )η( x¯√t )− µ¯−20 µ¯0t x¯µ¯0 ·∇Z5( x¯µ¯0 ) µ¯0tµ¯0 η( x¯√t )− µ¯−20 µ¯0tZ5( x¯µ¯0 ) x¯2t 32 ·∇η( x¯√t )∣∣∣∣≲ t−2〈 x¯µ¯0〉−21{|x¯|≤2t 12 }.Thus, by similar calculation for Lemma 3.7.3 and the upper bound of φ2 in Lemma 3.1.2, we have|∂tφ2|≲ˆR4t−2e−|x−y|22t(t−11{|y|≤(ln t)−1}+ t−1(ln t)−2|y|−21{(ln t)−1<|y|≤2t 12 })dy+ t−1ˆ tt0ˆR4(t− s)−2e−|x−y|28(t−s)(s−11{|y|≤(lns)−1}+ s−1(lns)−2|y|−21{(lns)−1<|y|≤2s 12 })dyds≲ t−2(ln t)−1.Collecting the above estimates, we obtain |∂tφ[µ¯0]|≲ t−2(ln t)−1.150Chapter 4Finite time blow-up solution of thetwo-dimensional Landau-Lifshitz-GilbertequationThis chapter will construct the finite time blow-up solutions for the two-dimensional Landau-Lifshitz-Gilbert equation.4.1 Sketch of the constructionDue to the complexities and technicalities in the construction, we sketch a roadmap of the significantsteps in this section.• Ansatz of multiple bubblesThe construction begins with a careful choice of first approximation. Since the target is the 2-sphere, one has to choose some profile for multiple bubbles, which is relatively reasonable to analyze.Here we chooseU∗ =−(N−1)U∞+N∑j=1Qγ jW(x−ξ [ j]λ j)as the first approximation. Notice that |U∗| ≈ 1 at any space-times as those bubbles are essentiallyseparated. Based onU∗, we then look for solutions to the LLG equation in the formu(x, t) = (1+A)U∗+Φ− (Φ ·U∗)U∗ (4.1.1)151for some perturbation term Φ and scalar function A. Here the purpose of the scalar function A,depending on Φ, is to preserve the unit length of the map u(x, t) for any (x, t) ∈ R2× (0,T ). So here,part of the interactions between bubbles get encoded in the scalar function A. Let us denote the errorof u asS[u] :=−ut +a(∆xu+ |∇xu|2u)−bu∧∆xu.An important observation here is that instead of solving S[u] = 0, we only need to solveS[u] = Ξ(x, t)U∗for some scalar function Ξ. Indeed, since |u| = 1 is kept for all t ∈ (0,T ) and u =U∗+ w˜ where theperturbation w˜ is uniformly small, thenΞ(U∗ ·u) = S[u] ·u=−12∂t(|u|2)+a2∆|u|2 = 0.Thus Ξ ≡ 0 follows from U∗ · u ≥ δ0 > 0. This provides us with the flexibility to adjust the errorterms in U∗ direction, and we will call this U∗-operation throughout this chapter. This operation cansimplify analysis for the dealing of multiple bubbles especially.• Slow decaying errors and non-local corrections by approximate parabolic systemThe error S(U∗) contains slowing decaying termsN∑j=1E [ j]0 /∈ L2(R2)which correspond to the rescaling and rotation around the z-axis. To improve the spatial decay of theerror in the remote region, we add well-designed global/non-local corrections around each bubble.Since the operator−∂t +(a−bU [ j]∧)∆xdepends on the blow-up profileU j as well as the parameters λ j, γ j and ξ [ j], one cannot expect explicitrepresentation formula apriori without knowing the blow-up dynamics. We consider an approximateparabolic operator instead−∂t +(a−bU∞∧)∆xand add correction Φ∗[ j]0 around each bubbleU[ j] with−∂tΦ∗[ j]0 +(a−bU∞∧)∆xΦ∗[ j]0 +E [ j]0 ≈ 0.Then the new error with corrections is given by those created byΦ∗[ j]0 and the remainder b(U∞−U [ j])∧∆xΦ∗[ j]0 . This is rather important in the analysis of the non-local reduced problems.152• Formulation of the inner-outer gluing systemWe next look for the perturbation Φ in (4.1.1) consisting of inner and outer parts with non-localcorrections addedΦ(x, t) =N∑j=1(η [ j]R (x, t)Qγ jΦ[ j]in(y[ j], t)+η [ j]dq(x, t)Φ∗[ j]0)+Φout(x, t)where Φ[ j]in is on the tangent plane ofW[ j], η [ j]R and η[ j]dq are suitable cut-off functions near q[ j]. Then usolving the LLG equation implies a coupled inner-outer gluing system for Φ[ j]in and Φout, j = 1, . . . ,Nλ 2j ∂tΦ[ j]in = (a−bW [ j]∧)[∆y[ j]Φ[ j]in+ |∇y[ j]W [ j]|2Φ[ j]in+2(∇y[ j]W [ j] ·∇y[ j]Φ[ j]in)W [ j]]+H [ j][Φ[ j]in,Φout,λ j,γ j,ξ[ j]] in D2R,∂tΦout = BΦ,U∗∆xΦout+G [Φ[ j]in,Φout,λ j,γ j,ξ[ j]] in R2× (0,T ),whereW [ j] is the j-th bubble expressed in the rescaled variable y[ j] = x−ξ[ j]λ j , the right hand sidesH[ j],G consists of the error terms, couplings and nonlinear terms depending on the parameters λ j, γ j, ξ [ j],and BΦ,U∗ is a matrix, defined in (4.4.26), that involves the perturbationΦ and the blow-up profileU∗.For the whole system above, finding blow-up of the LLG equation at multiple points is reducedto finding well-behaved inner and outer profiles such that gluing procedure can be implemented. Inother words, we need to devise appropriately weighted topologies in which the gluing system becomesweakly coupled and thus can be solved by fixed point arguments. For the outer problem, we makeuse of the sub-Guassian estimate recently proved in [31]. For the inner problem, reasonable solutionswith sufficient decay in space and time can only be captured with careful choices of the parametersλ j, γ j, ξ [ j]. We shall develop the linear theory for the inner problems with orthogonality conditions,and these orthogonalities determine the blow-up dynamics.• Solving the inner problemThe linear theory for the inner problem is established by analyzing each Fourier mode. Decom-posing the complex form in Fourier modes, one obtains the linearized operator at mode k of the formλ 2j ∂t − (a− ib)(∂ρρ +∂ρρ− (k+1)2ρ4+(2k2−6)ρ2+(k−1)2(ρ2+1)21ρ2), (4.1.2)where ρ j = |y[ j]|. Then for all the modes k ∈Z\{−1}, good inner solutions are found by the followingStep 1: We first use energy methods to get a rough pointwise upper bound for the inner solutions ϕk;Step 2: Next, we use Duhamel’s formula to refine the pointwise bounds and further gain decay esti-mates;153Step 3: Finally, we perform the re-gluing procedure to obtain better estimates in the innermost region.As mentioned earlier, the treatment for mode k = −1 differs from the techniques we employ for allthe other modes. The reason is the following: as one can see from (4.1.2), mode −1 can be roughlyviewed as a problem in 2D, which is worse than any other mode as one cannot gain spatial decay inStep 2 above. Fortunately, distorted Fourier transform can give us almost the optimal bound.• Non-local reduced problemsThe development of the linear theory for the inner problem relies on orthogonalities achievedby adjusting modulation parameters λ j, γ j, ξ [ j]. The dynamics for ξ [ j] are governed by an ODE,which is relatively straightforward to solve. However, the non-local feature in the corrections Φ∗[ j]0gets inherited by the mode 0 (λ j and γ j) of each bubble as the corrections are essentially for mode0. Here one might expect the complex system involving both λ j and γ j to be a rather sophisticatedform due to dispersion. Fortunately, it turns out that the contribution of both Φ∗[ j]0 and the remainderb(U∞−U [ j])∧∆xΦ∗[ j]0 in the orthogonal equation at mode 0 results in the following well-ordered non-local problem ˆ t−λ 2j (t)0p˙ j(s)t− s ds∼ O(1),where p j(t) = λ j(t)eiγ j(t). This λ j-γ j system was first found and handled in the context of HMF [19].Surprisingly, this comes with a similar form in the LLG equation with dispersion.• Solving the outer problemThe linear theory for the outer problem is done by using the sub-Gaussian estimate for the lin-earized parabolic system proved in [31], where DMOx-regularity is required for the entries in thematrix BΦ,U∗ . The dependence on Φ in the matrix shall be dealt with via Schauder fixed point the-orem, and the critical thing here is the dependence on the blow-up profile U∗. In fact, to ensureDMOx-regularity inU∗, the leading blow-up rate λ∗ plays an important role. See Section 4.6.2.4.2 PreliminariesWe consider the Landau-Lifshitz-Gilbert equation given in (1.0.20). The harmonic map equation isthe steady state equation of (1.0.20). LetW be the least energy harmonic mapW (y) =1|y|2+1 2y12y2|y|2−1 , y= (y1,y2) ∈ R2,154which solves the harmonic map equation. Since we shall consider the case of multiple bubbles,subscript “ j” or superscript “[ j]” will be used to distinguish different bubbles and their associatedtangent planes. There is a natural correspondence between the vectors in R2 and complex numbers.In the (rescaled) polar coordinates around ξ [ j] = (ξ [ j]1 ,ξ[ j]2 ) ∈ R2, denotey[ j] =x−ξ [ j]λ j= ρ jeiθ j , x= ξ [ j]+λ jρ jeiθ j , ρ j = |y[ j]|, r j = |x−ξ [ j]|= λ jρ j, θ j = arctan(x2−ξ [ j]2x1−ξ [ j]1).(4.2.1)We can write for j = 1,2, . . . ,N,W [ j] :=W (y[ j]) =cosθ j sinw(ρ j)sinθ j sinw(ρ j)cosw(ρ j) := [eiθ j sinw(ρ j)cosw(ρ j)](4.2.2)withw(ρ j) = π−2arctan(ρ j),and we havewρ j =−2ρ2j +1, sinw(ρ j)=−ρ jwρ j =2ρ jρ2j +1, cosw(ρ j)=ρ2j −1ρ2j +1, |∇y[ j]W (y[ j])|2 = 2w2ρ j =8(ρ2j +1)2.(4.2.3)We denote the Frenet basis associated toW [ j] asE [ j]1 =cosθ j cosw(ρ j)sinθ j cosw(ρ j)−sinw(ρ j) := [eiθ j cosw(ρ j)−sinw(ρ j)], E [ j]2 =−sinθ jcosθ j0 := [ieiθ j0], (4.2.4)soW [ j]∧E [ j]1 = E [ j]2 , W [ j]∧E [ j]2 =−E [ j]1 , E [ j]1 ∧E [ j]2 =W [ j]. (4.2.5)It is direct to check that in the polar coordinates (4.2.1)∂ρ jW[ j] = wρ jE[ j]1 , ∂ρ jρ jW[ j] = wρ jρ jE[ j]1 −w2ρ jW [ j], ∂θ jW [ j] = sinw(ρ j)E [ j]2 ,∂θ jθ jW[ j] =−sinw(ρ j)(sinw(ρ j)W [ j]+ cosw(ρ j)E [ j]1),∂ρ jE[ j]1 =−wρ jW [ j], ∂ρ jρ jE [ j]1 =−wρ jρ jW [ j]−w2ρ jE [ j]1 , ∂θ jE [ j]1 = cosw(ρ j)E [ j]2 ,∂θ jθ jE[ j]1 =−cosw(ρ j)(sinw(ρ j)W [ j]+ cosw(ρ j)E [ j]1),∂ρ jE[ j]2 = ∂ρ jρ jE[ j]2 = 0, ∂θ jE[ j]2 =−sinw(ρ j)W [ j]− cosw(ρ j)E [ j]1 , ∂θ jθ jE [ j]2 =−E [ j]2 .(4.2.6)155The linearization of the harmonic map equation aroundW [ j] is the elliptic operatorLW [ j] [ϕ ] := ∆y[ j]ϕ + |∇y[ j]W [ j]|2ϕ +2(∇y[ j]W[ j] ·∇y[ j]ϕ)W [ j]. (4.2.7)Denote the s-rotation matrices around z-axis, x-axis, y-axis respectively asQs :=coss −sins 0sins coss 00 0 1 , Qxs :=1 0 00 coss −sins0 sins coss , Qys := coss 0 sins0 1 0−sins 0 coss . (4.2.8)Due to the invariance of group action for harmonic map equation, we haveZ [ j]0,1(y[ j]) =−∂s(W (s−1y[ j]))∣∣∣s=1= ρ jwρ j(ρ j)E[ j]1 (y[ j]),Z [ j]0,2(y[ j]) =−∂s(QsW (y[ j]))∣∣∣s=0= ρ jwρ j(ρ j)E[ j]2 (y[ j]),Z [ j]1,1(y[ j]) = ∂y[ j]1W (y[ j]) = wρ j(ρ j)[cosθ jE[ j]1 (y[ j])+ sinθ jE [ j]2 (y[ j])],Z [ j]1,2(y[ j]) = ∂y[ j]2W (y[ j]) = wρ j(ρ j)[sinθ jE[ j]1 (y[ j])− cosθ jE [ j]2 (y[ j])],Z [ j]−1,1(y[ j]) =−Z [ j]1,1(y[ j])−2∂s(QysW (y[ j]))∣∣∣s=0= ρ2jwρ j(ρ j)[cosθ jE[ j]1 (y[ j])− sinθ jE [ j]2 (y[ j])],Z [ j]−1,2(y[ j]) =−Z [ j]1,2(y[ j])+2∂s(QxsW (y[ j]))∣∣∣s=0= ρ2jwρ j(ρ j)[sinθ jE[ j]1 (y[ j])+ cosθ jE [ j]2 (y[ j])](4.2.9)which satisfyL[ j]W [Z[ j]p,q] = 0 for p=±1,0, q= 1,2.SetU [ j](x, t) := Qγ jW(x−ξ [ j]λ j). (4.2.10)For f= [ f1, f2, f3]tr ∈ R3,Qγ j f=[Re(eiγ j( f1+ i f2)), Im(eiγ j( f1+ i f2)), f3]tr:=[eiγ j( f1+ i f2)f3]. (4.2.11)For f,g ∈ R3, by basic linear algebra, (Qγ j f)∧ (Qγ jg) = Qγ j(f∧g). Combining (4.2.5), we haveU [ j]∧ (Qγ jE [ j]1 ) = Qγ jE [ j]2 , U [ j]∧ (Qγ jE [ j]2 ) =−Qγ jE [ j]1 , (Qγ jE [ j]1 )∧ (Qγ jE [ j]2 ) =U [ j]. (4.2.12)It is convenient to use complex notations to deal with linearization near concentration zones, asall the analysis will be done on the associated tangent plane.156For any f ∈ R3 satisfying f ·U [ j] = 0, we define the equivalent complex form of f asfC j := f · (Qγ jE [ j]1 )+ if · (Qγ jE [ j]2 ).For any complex-valued function f , we definefC−1j := (Re f )Qγ jE[ j]1 +(Im f )Qγ jE[ j]2 . (4.2.13)By (4.2.12),U [ j]∧ fC−1j = (Re f )Qγ jE[ j]2 − (Im f )Qγ jE [ j]1 = (i f )C−1j . (4.2.14)Similarly, for any g ∈ R3 satisfying g ·W [ j] = 0, the equivalent complex form of g is defined asgC j := g ·E [ j]1 + ig ·E [ j]2 . (4.2.15)For any complex-valued function g, we definegC−1j := (Reg)E[ j]1 +(Img)E[ j]2 . (4.2.16)For any g(y[ j],τ) ∈ R3 satisfying g ·W [ j] = 0, we denote the mode k component for g asgC j,k(ρ j,τ) := (2π)−1ˆ 2π0gC j(ρ jeis,τ)e−iksds. (4.2.17)For any f,g ∈ R3, we defineΠg⊥f := f− (f ·g)g. (4.2.18)Specially, when |g|= 1, Πg⊥ is the usual orthogonal projection on the plane orthogonal to g.Notice for any f= [ f1, f2, f3]tr ∈ R3, by (4.2.11), we have(ΠU [ j]⊥f)C j =(1− 2ρ2j +1Re)[( f1+ i f2)e−i(θ j+γ j)]− 2ρ jρ2j +1f3,f ·U [ j] = 2ρ jρ2j +1Re[( f1+ i f2)e−i(θ j+γ j)]+ρ2j −1ρ2j +1f3.(4.2.19)The linearization of the harmonic map equation aroundU [ j] is given byLU [ j] [ϕ ] := ∆xϕ + |∇xU [ j]|2ϕ +2(∇xU [ j] ·∇xϕ)U [ j]. (4.2.20)157It is easy to haveLU [ j] [Qγ j f(y[ j])] = λ−2j Qγ jLW [ j] [f(y[ j])], where y[ j] =x−ξ [ j]λ j.We now give several useful formulas with proof similar to that of [19, Section 3]. For any vector-valued function f : R2 → R3, we setL˜U [ j] [f] := |∇xU [ j]|2ΠU [ j]⊥f−2∇x(f ·U [ j]) ·∇xU [ j], (4.2.21)with∇x(f ·U [ j]) ·∇xU [ j] =2∑k=1∂xk(f ·U [ j])∂xkU [ j].It is easy to see thatU [ j] · L˜U [ j] [f] = 0. Similarly, we setL˜W [ j] [f] := |∇y[ j]W [ j]|2ΠW [ j]⊥f−2∇y[ j](f ·W [ j]) ·∇y[ j]W [ j].Then it is straightforward to getL˜U [ j] [Qγ j f(y[ j])] = λ−2j Qγ j L˜W [ j] [f(y[ j])].LU [ j] [ΠU [ j]⊥f] =ΠU [ j]⊥∆xf+ L˜U [ j] [f].For f= [ f1, f2, f3]tr, we haveL˜U [ j] [Qγ j f] = λ−1j{ρ jw2ρ j(ρ j)(∂x1 f1+∂x2 f2)−2wρ j(ρ j)cosw(ρ j)(cosθ j∂x1 f3+ sinθ j∂x2 f3)+ρ jw2ρ j(ρ j) [cos(2θ j)(∂x1 f1−∂x2 f2)+ sin(2θ j)(∂x2 f1+∂x1 f2)]}Qγ jE[ j]1+λ−1j{−ρ jw2ρ j(ρ j)(∂x2 f1−∂x1 f2)−2wρ j(ρ j)cosw(ρ j)(sinθ j∂x1 f3− cosθ j∂x2 f3)+ρ jw2ρ j(ρ j) [sin(2θ j)(∂x1 f1−∂x2 f2)− cos(2θ j)(∂x2 f1+∂x1 f2)]}Qγ jE[ j]2= λ−1j{ρ jw2ρ j(ρ j)(∂x1 f1+∂x2 f2)− eiθ jwρ j(ρ j)cosw(ρ j)(∂x1 f3− i∂x2 f3)− e−iθ jwρ j(ρ j)cosw(ρ j)(∂x1 f3+ i∂x2 f3)+ e2iθ j12ρ jw2ρ j(ρ j) [(∂x1 f1−∂x2 f2)− i(∂x2 f1+∂x1 f2)]+ e−2iθ j12ρ jw2ρ j(ρ j) [(∂x1 f1−∂x2 f2)+ i(∂x2 f1+∂x1 f2)]}Qγ jE[ j]1+λ−1j{−ρ jw2ρ j(ρ j)(∂x2 f1−∂x1 f2)+ eiθ jwρ j(ρ j)cosw(ρ j)(∂x2 f3+ i∂x1 f3)+ e−iθ jwρ j(ρ j)cosw(ρ j)(∂x2 f3− i∂x1 f3)− e2iθ j12ρ jw2ρ j(ρ j) [(∂x2 f1+∂x1 f2)+ i(∂x1 f1−∂x2 f2)]+ e−2iθ j12ρ jw2ρ j(ρ j) [−(∂x2 f1+∂x1 f2)+ i(∂x1 f1−∂x2 f2)]}Qγ jE[ j]2 .158Then the corresponding complex form is given by(L˜U [ j] [Qγ j f])C j = λ−1j{ρ jw2ρ j(ρ j) [(∂x1 f1+∂x2 f2)− i(∂x2 f1−∂x1 f2)]+ eiθ j2wρ j(ρ j)cosw(ρ j)(−∂x1 f3+ i∂x2 f3)+ e2iθ jρ jw2ρ j(ρ j) [(∂x1 f1−∂x2 f2)− i(∂x2 f1+∂x1 f2)]}.(4.2.22)Specially,(L˜U [ j] [f])C j = (L˜U [ j] [f])C j0 + eiθ j(L˜U [ j] [f])C j1 + e2iθ j(L˜U [ j] [f])C j2 , (4.2.23)where(L˜U [ j] [f])C j0 := λ−1j ρ jw2ρ j(ρ j)[∂x1(Q−γ j f)1+∂x2(Q−γ j f)2− i(∂x2(Q−γ j f)1−∂x1(Q−γ j f)2)]= λ−1j ρ jw2ρ j(ρ j)e−iγ j [∂x1 f1+∂x2 f2+ i(∂x1 f2−∂x2 f1)] ,(L˜U [ j] [f])C j1 := 2λ−1j wρ j(ρ j)cosw(ρ j)(−∂x1(Q−γ j f)3+ i∂x2(Q−γ j f)3)= 2λ−1j wρ j(ρ j)cosw(ρ j)(−∂x1 f3+ i∂x2 f3) ,(L˜U [ j] [f])C j2 := λ−1j ρ jw2ρ j(ρ j)[∂x1(Q−γ j f)1−∂x2(Q−γ j f)2− i(∂x2(Q−γ j f)1+∂x1(Q−γ j f)2)]= λ−1j ρ jw2ρ j(ρ j)eiγ j [∂x1 f1−∂x2 f2− i(∂x1 f2+∂x2 f1)] ,(4.2.24)since by (4.2.11), we have∂x1(Q−γ j f)1+∂x2(Q−γ j f)2− i(∂x2(Q−γ j f)1−∂x1(Q−γ j f)2)= ∂x1Re[e−iγ j ( f1+ i f2)]+∂x2Im[e−iγ j ( f1+ i f2)]− i∂x2Re[e−iγ j ( f1+ i f2)]+ i∂x1Im[e−iγ j ( f1+ i f2)]= e−iγ j [∂x1 ( f1+ i f2)− i∂x2 ( f1+ i f2)] = e−iγ j [∂x1 f1+∂x2 f2+ i(∂x1 f2−∂x2 f1)] ,∂x1(Q−γ j f)1−∂x2(Q−γ j f)2− i(∂x2(Q−γ j f)1+∂x1(Q−γ j f)2)= ∂x1Re[e−iγ j ( f1+ i f2)]−∂x2Im[e−iγ j ( f1+ i f2)]− i∂x2Re[e−iγ j ( f1+ i f2)]− i∂x1Im[e−iγ j ( f1+ i f2)]= eiγ j [∂x1 f1−∂x2 f2− i(∂x1 f2+∂x2 f1)] .By (4.2.12),Q−γ j[(a−bU [ j]∧) L˜U [ j] [f]]= Q−γ j[(a−bU [ j]∧){Re[(L˜U [ j] [f])C j]Qγ jE [ j]1 + Im[(L˜U [ j] [f])C j]Qγ jE [ j]2 }]= Re[(L˜U [ j] [f])C j](aE [ j]1 −bE [ j]2)+ Im[(L˜U [ j] [f])C j](aE [ j]2 +bE[ j]1).Thus, {Q−γ j[(a−bU [ j]∧) L˜U [ j] [f]]}C j = (a− ib)(L˜U [ j] [f])C j . (4.2.25)1594.3 Approximation and improvement4.3.1 First approximationFix arbitrary N different points q[ j] ∈ R2, j = 1,2, . . . ,N, where N is an arbitrary positive integer.Denotedq :=19mink 6=m|q[k]−q[m]|> 0, p j(t) := λ j(t)eiγ j(t). (4.3.1)Throughout this chapter, we ansatz that for all j = 1,2, . . . ,N,C−1λ λ∗(t)≤ |p j(t)|= λ j(t)≤Cλλ∗(t) with λ∗(t) =| lnT |(T − t)ln2(T − t) ,C−1λ| lnT |ln2(T − t) ≤ |p˙ j(t)| ≤Cλ| lnT |ln2(T − t) , |γ˙ j(t)|≲Cγ(T − t)−1, |ξ˙ [ j](t)| ≤Cξλ εξ∗ (t), ξ [ j](T ) = q[ j](4.3.2)whereCλ ≥ 1,Cξ > 0,Cγ > 0; εξ > 0 is small.We will construct blow-up solutions which blow up simultaneously at these prescribed points q[ j].We take the approximationU∗(x, t) :=−(N−1)U∞+N∑j=1U [ j](x, t), (4.3.3)whereU [ j] are given in (4.2.10) andU∞ = [0,0,1]tr.Take T sufficiently small, thenmink 6=m|ξ [k](t)−ξ [m](t)|> 8dq > 0 for all t ∈ [0,T ] and |U∗|= 1+O(N∑j=1λ j). (4.3.4)Let us denote the error functionS [u] :=−ut +a(∆xu+ |∇xu|2u)−bu∧∆xu.The error of the approximate solution isS [U∗] =−N∑j=1∂tU [ j]+a(∆xU∗+ |∇xU∗|2U∗)−bU∗∧∆xU∗.Notice−∂tU [ j] = E [ j]0 +E [ j]1 , E [ j]0 :=−λ˙ j∂λ jU [ j]− γ˙ j∂γ jU [ j], E [ j]1 :=−ξ˙ [ j]1 ∂ξ [ j]1 U[ j]− ξ˙ [ j]2 ∂ξ [ j]2 U[ j] (4.3.5)160where ∂λ jU[ j](x) =−λ−1j Qγ jZ [ j]0,1(y[ j]), ∂γ jU [ j](x) =−Qγ jZ [ j]0,2(y[ j]),∂ξ [ j]1U [ j](x) =−λ−1j Qγ jZ [ j]1,1(y[ j]), ∂ξ [ j]2 U[ j](x) =−λ−1j Qγ jZ [ j]1,2(y[ j])with Z [ j]p,q, E[ j]1 , E[ j]2 given in (4.2.9), (4.2.4), respectively. It is straightforward to haveE [ j]0 = Qγ j(λ−1j λ˙ jZ[ j]0,1(y[ j])+ γ˙ jZ [ j]0,2(y[ j]))= ρ jwρ jQγ j(λ−1j λ˙ jE[ j]1 + γ˙ jE[ j]2)=−2ρ jρ2j +1[(λ−1j λ˙ j cosw(ρ j)+ iγ˙ j)ei(θ j+γ j)−λ−1j λ˙ j sinw(ρ j)],(4.3.6)(E [ j]0 )C j =−2ρ jρ2j +1(λ−1j λ˙ j+ iγ˙ j), (4.3.7)E [ j]1 = ξ˙[ j]1 λ−1j Qγ jZ[ j]1,1(y[ j])+ ξ˙ [ j]2 λ−1j Qγ jZ[ j]1,2(y[ j])= ξ˙ [ j]1 λ−1j Qγ jwρ j(cosθ jE [ j]1 + sinθ jE[ j]2)+ ξ˙ [ j]2 λ−1j Qγ jwρ j(sinθ jE [ j]1 − cosθ jE [ j]2)= λ−1j wρ j(ξ˙ [ j]1 cosθ j+ ξ˙[ j]2 sinθ j)Qγ jE[ j]1 +λ−1j wρ j(ξ˙ [ j]1 sinθ j− ξ˙ [ j]2 cosθ j)Qγ jE[ j]2=−2λ−1jρ2j +1Re[(ξ˙ [ j]1 − iξ˙ [ j]2)eiθ j]Qγ jE[ j]1 −2λ−1jρ2j +1Im[(ξ˙ [ j]1 − iξ˙ [ j]2)eiθ j]Qγ jE[ j]2 ,(4.3.8)(E [ j]1 )C j =−2λ−1j (ξ˙ [ j]1 − iξ˙ [ j]2 )ρ2j +1eiθ j . (4.3.9)Combining (4.3.6) and (4.3.8), we have|∂tU [ j]|≲(λ−1j |λ˙ j|+ |γ˙ j|)〈ρ j〉−1+λ−1j |ξ˙ [ j]|〈ρ j〉−2. (4.3.10)Notice that S(U∗) contains errors E [ j]0 with slow decay in space, which makes the gluing processbreak down. We shall introduce global corrections to improve the error.4.3.2 Global correction by parabolic systemsIn this part, we will transfer slow decay terms by parabolic systems. Around each bubble, the slowdecaying term in (4.3.6) is given byE [ j]0 ≈−2z j[p˙ j(t)eiθ j0],wherez j = (λ 2j (t)+ r2j )1/2, r j = |x[ j]|, x[ j] = x−ξ [ j](t), p j(t) = λ j(t)eiγ j(t). (4.3.11)161We aim to find global corrections Φ∗[ j]0 (r j, t) to make−∂t(Φ∗[ j]0 )+(a−bU∞∧)∆xΦ∗[ j]0 −2z j[p˙ j(t)eiθ j0]≈ 0with the formΦ∗[ j]0 (r j, t) :=rµjrµj +λµjΦ[ j]0 (√r2j +λ 2j , t)eiθ j0= ρµjρµj +1Φ[ j]0 (z j, t)eiθ j0 (4.3.12)where µ ≥ 2 is a constant to be determined later. The term rµjrµj +λµjis used to avoid singular terms whencalculating new errors.Formally, the approximate calculation is displayed here.∆x[Φ[ j]0 eiθ j0]≈[(∂z jz jΦ[ j]0 + z−1j ∂z jΦ[ j]0 − z−2j Φ[ j]0)eiθ j0].Since for any v1,v2 ∈ R,(a−bU∞∧)v1v20=Re [(a− ib)(v1+ iv2)]Im [(a− ib)(v1+ iv2)]0 := [(a− ib)(v1+ iv2)0], (4.3.13)then−∂t(Φ∗[ j]0 )+(a−bU∞∧)∆xΦ∗[ j]0 −2z j[p˙ j(t)eiθ j0]≈[−∂tΦ[ j]0 eiθ j +(a− ib)(∂z jz jΦ[ j]0 + z−1j ∂z jΦ[ j]0 − z−2j Φ[ j]0)eiθ j0]− 2z j[p˙ j(t)eiθ j0].For this reason, we choose Φ[ j]0 (z j, t) to solve(a+ ib)∂tΦ[ j]0 = ∂z jz jΦ[ j]0 +1z j∂z jΦ[ j]0 −1z2jΦ[ j]0 −2(a+ ib)p˙ j(t)z j. (4.3.14)The analysis of (4.3.14) is same as [19, (4.7)]. Consider the more general equation(a+ ib)∂t f = ∂zz f +1z∂z f − 1z2 f +g(t)z. (4.3.15)162First we look for the self-similar profile to(a+ ib)∂t f1 = ∂zz f1+1z∂z f1− 1z2 f1+1zwithf1(z, t) = t1/2 f2(zt1/2).Then f2 satisfiesf ′′2 (ξ )+(1ξ+a+ ib2ξ)f ′2(ξ )−(1ξ 2+a+ ib2)f2(ξ )+1ξ= 0, ξ =zt1/2.Observing that ξ is a homogeneous solution, so we have a solutionf2(ξ ) = ξˆ ∞ξe−a+ib4 η2η3dηˆ η0sea+ib4 s2ds=2ξa+ ibˆ ∞ξ1− e− a+ib4 η2η3dη ,and| f2(ξ )|≲ ξ 〈lnξ 〉1{0≤ξ≤1}+ξ−11{ξ>1}.Then for g(t) ∈C1([0,T ]), by Duhamel’s formula, one has a solution to (4.3.15)f (z, t)=ˆ t0g˙(s) f1(z, t−s)ds+g(0) f1(z, t)=ˆ t0g(s)∂t f1(z, t−s)ds=ˆ t0g(s)1a+ ib1− e− a+ib4 z2t−szds.(4.3.16)If g ∈ L∞([0,T ]), then (4.3.16) satisfies (4.3.15) in weak sense. Thus for (4.3.14), we haveΦ[ j]0 (z j, t) =−z jˆ t0p˙ j(s)t− s K0(z2jt− s)ds, K0(ζ j) := 21− e− a+ib4 ζ jζ j, (4.3.17)whereζ j =z2jt− s = ι j(ρ2j +1), ι j :=λ 2j (t)t− s . (4.3.18)Since a> 0, it is straightforward to haveK0(ζ j) =(a+ ib2+O(ζ j))1{ζ j≤1}+O(ζ−1j)1{ζ j>1}, ζ jK0ζ j(ζ j) = O(ζ j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1},ζ 2j K0ζ jζ j(ζ j) = O(ζ 2j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1}.(4.3.19)It is easy to get∂z jΦ[ j]0 =−ˆ t0p˙ j(s)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds,∂z jz jΦ[ j]0 =−z−1jˆ t0p˙ j(s)t− s (6ζ jK0ζ j(ζ j)+4ζ2j K0ζ jζ j(ζ j))ds.(4.3.20)163The upper bound of the nonlocal termsBy (4.3.19) and |p˙ j|≲ |λ˙∗| in (4.3.2), then|Φ[ j]0 |+ z j|∂z jΦ[ j]0 |+ z2j |∂z jz jΦ[ j]0 |≲ z jˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})ds. (4.3.21)Claim:ˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})∣∣ζ j=z2j(t−s)−1ds≲t| lnT |−1z−2j , z2j ≥ t| lnT |−1〈ln( tz2j )〉, t ≤T2 ,1− | lnT || ln(2(T−t))| + |λ˙∗(t)|〈ln(T−tz2j )〉, t >T2 ,z2j < T − t,1− | lnT || ln(T−t+z2j)| + | lnT |(lnz j)−2, t > T2 ,z2j ≥ T − t,z2j < t≲ 1{z2j<t}+ t| lnT |−1z−2j 1{z2j≥t}.(4.3.22)By (4.3.21) and (4.3.22), we have|Φ[ j]0 |+ z j|∂z jΦ[ j]0 |+ z2j |∂z jz jΦ[ j]0 |≲ z j1{z2j<t}+ t| lnT |−1z−1j 1{z2j≥t}. (4.3.23)Using (4.3.12), (4.3.27), (4.3.30) and (4.3.23), we have|Φ∗[ j]0 |+ z j|∇xΦ∗[ j]0 |+ z2j |∆xΦ∗[ j]0 |≲ z j1{z2j<t}+ t| lnT |−1z−1j 1{z2j≥t}. (4.3.24)Proof of Claim (4.3.22). Denoteg(z j, t) :=ˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})∣∣ζ j=z2j(t−s)−1ds.• For z2j ≥ t,g(z j, t) = z−2jˆ t0|λ˙∗(s)|ds∼ z−2j | lnT |ˆ t0| ln(T − s)|−2ds∼ t| lnT |−1z−2jsince if t ≤ T2 , ˆ t0| ln(T − s)|−2ds∼ t| lnT |−2;164if T2 < t ≤ T ,ˆ t0| ln(T−s)|−2ds=(ˆ T20+ˆ tT2)| ln(T−s)|−2ds∼T | lnT |−2+ˆ T2T−t(lnz)−2dz∼T | lnT |−2∼ t| lnT |−2.• For z2j < t,g(z j, t) =ˆ t−z2j0|λ˙∗(s)|t− s ds+ z−2jˆ tt−z2j|λ˙∗(s)|ds.If t ≤ T2 ,g(z j, t)∼ | lnT |−1〈ln( tz2j)〉.If t > T2 and z2j < T − t,g(z j, t)∼ 1− | lnT || ln(2(T − t))| + |λ˙∗(t)|〈ln(T − tz2j)〉sinceˆ t−z2j0|λ˙∗(s)|t− s ds=(ˆ t−(T−t)0+ˆ t−z2jt−(T−t))|λ˙∗(s)|t− s ds∼ˆ t−(T−t)0|λ˙∗(s)|T − s ds+ |λ˙∗(t)|ˆ t−z2jt−(T−t)1t− sds∼ | lnT |ˆ t−(T−t)01(T − s)| ln(T − s)|2 ds+ |λ˙∗(t)| ln(T − tz2j) = 1− | lnT || ln(2(T − t))| + |λ˙∗(t)| ln(T − tz2j),z−2jˆ tt−z2j|λ˙∗(s)|ds∼ |λ˙∗(t)|.If t > T2 and T − t ≤ z2j < t,g(z j, t)≲ 1− | lnT || ln(T − t+ z2j)|+ | lnT |(lnz j)−2sinceˆ t−z2j0|λ˙∗(s)|t− s ds∼ˆ t−z2j0|λ˙∗(s)|T − s ds∼ | lnT |ˆ t−z2j01(T − s)| ln(T − s)|2 ds= 1−| lnT || ln(T − t+ z2j)|,z−2jˆ tt−z2j|λ˙∗(s)|ds∼ z−2j | lnT |ˆ tt−z2j(ln(T − s))−2ds= z−2j | lnT |ˆ T−t+z2jT−t(lnv)−2dv≲ z−2j | lnT |(T − t+ z2j)(ln(T − t+ z2j))−2 ∼ | lnT |(lnz j)−2.In sum, we get the first part of (4.3.22).165Specially, for z2j < t, t ≤ T2 ,t > z2j ≳ (| lnT |(T − t)| ln(T − t)|2 )2 ∼ T 2| lnT |−2.Thus| lnT |−1〈ln( tz2j)〉≲ 1.For z2j < t, t >T2 ,z2j < T − t,|λ˙∗(t)|〈ln(T − tz2j)〉≲ | lnT || ln(T − t)|2 〈ln(T − tλ 2∗ (t))〉≲ 1.Thus we have the second part of (4.3.22).New errors produced by the global correctionsIn this part, we will calculate the new errors produced by the introduction of Φ∗[ j]0 defined in (4.3.12),that is,S [ j] := −∂t(Φ∗[ j]0 )+(a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]= −∂t(Φ∗[ j]0 )+(a−bU∞∧)∆xΦ∗[ j]0 −∂tU [ j]−b(U [ j]−U∞)∧∆xΦ∗[ j]0+a|∇xU [ j]|2Φ∗[ j]0 +b|∇xU [ j]|2Φ∗[ j]0 ∧U [ j]+(a−bU [ j]∧)[−2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]] .We present some formulas we will use frequently. Recall (4.3.11), thenz j =√r2j +λ 2j =√|x−ξ [ j](t)|2+λ 2j (t), ∂r j(√r2j +λ 2j )=r j√r2j +λ 2j, ∂r jr j(√r2j +λ 2j )=λ 2j(r2j +λ 2j )32,∂t(√r2j +λ 2j ) =λ˙ jλ j− ξ˙ [ j] · (x−ξ [ j])√r2j +λ 2j, ∂tr j =−r−1j (x−ξ [ j]) · ξ˙ [ j].∂r j(rµjrµj +λµj) =µλ µj rµ−1j(rµj +λµj )2, ∂r jr j(rµjrµj +λµj) =(µ−1)µλ 2µj rµ−2j − (µ+1)µλ µj r2µ−2j(rµj +λµj )3,∂t(rµjrµj +λµj) =−µλµj rµ−2j ξ˙ [ j] · (x−ξ [ j])+µλ˙ jλ µ−1j rµj(rµj +λµj )2,166θ j = arctan(x2−ξ [ j]2x1−ξ [ j]1), ∂tθ j =−ξ˙ [ j]2 (x1−ξ [ j]1 )+ ξ˙ [ j]1 (x2−ξ [ j]2 )r2j.∂t(rµjrµj +λµjΦ[ j]0 (z j, t)) =rµjrµj +λµj[∂tΦ[ j]0 +λ˙ jλ j− ξ˙ [ j] · (x−ξ [ j])√r2j +λ 2j∂z jΦ[ j]0]− µλµj rµ−2j ξ˙ [ j] · (x−ξ [ j])+µλ˙ jλ µ−1j rµj(rµj +λµj )2Φ[ j]0 ,∂r j(rµjrµj +λµjΦ[ j]0 (z j, t)) =rµ+1j(rµj +λµj )(r2j +λ 2j )12∂z jΦ[ j]0 +µλ µj rµ−1j(rµj +λµj )2Φ[ j]0 ,∂r jr j(rµjrµj +λµjΦ[ j]0 (z j, t)) =rµ+2j(rµj +λµj )(r2j +λ 2j )∂z jz jΦ[ j]0+[ 2µλ µj rµj(rµj +λµj )2(r2j +λ 2j )12+λ 2j rµj(rµj +λµj )(r2j +λ 2j )32]∂z jΦ[ j]0 +(µ−1)µλ 2µj rµ−2j − (µ+1)µλ µj r2µ−2j(rµj +λµj )3Φ[ j]0 .Then recalling (4.3.12), we have∂r jΦ∗[ j]0 =[[ rµ+1j(rµj +λµj )(r2j +λ 2j )12∂z jΦ[ j]0 +µλ µj rµ−1j(rµj +λµj )2Φ[ j]0]eiθ j ,0]tr=[[ ρµ+1j(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j ρµ−1j(ρµj +1)2Φ[ j]0]eiθ j ,0]tr,∂θ jΦ∗[ j]0 =[ rµjrµj +λµjΦ[ j]0 ieiθ j ,0]tr=[ ρµjρµj +1Φ[ j]0 ieiθ j ,0]tr.(4.3.25)It follows that|∇xΦ∗[ j]0 |2 = |∇x[ j]Φ∗[ j]0 |2 = |∂r jΦ∗[ j]0 |2+ r−2j |∂θ jΦ∗[ j]0 |2=∣∣∣∣∣ ρµ+1j(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j ρµ−1j(ρµj +1)2Φ[ j]0∣∣∣∣∣2+λ−2j ρ−2j∣∣∣∣∣ ρµjρµj +1Φ[ j]0∣∣∣∣∣2.(4.3.26)Since µ ≥ 2, it follows that|∇xΦ∗[ j]0 |≲ |∂z jΦ[ j]0 |+ z−1j |Φ[ j]0 |. (4.3.27)•−∂t(Φ∗[ j]0 ) =[−∂t( rµjrµj +λµjΦ[ j]0 (z j, t)eiθ j),0]tr167=[ −rµjrµj +λµj[∂tΦ[ j]0 +λ˙ jλ j− ξ˙ [ j] · (x−ξ [ j])√r2j +λ 2j∂z jΦ[ j]0]eiθ j +µλ µj rµ−2j ξ˙ [ j] · (x−ξ [ j])+µλ˙ jλ µ−1j rµj(rµj +λµj )2Φ[ j]0 eiθ j− rµjrµj +λµjΦ[ j]0 i−ξ˙ [ j]2 (x1−ξ [ j]1 )+ ξ˙ [ j]1 (x2−ξ [ j]2 )r2jeiθ j ,0]tr=[{ −rµjrµj +λµj∂tΦ[ j]0 −rµj[λ˙ jλ j− ξ˙ [ j] · (x−ξ [ j])](rµj +λµj )(r2j +λ 2j )12∂z jΦ[ j]0 +[µλ µj rµ−2j ξ˙ [ j] · (x−ξ [ j])+µλ˙ jλ µ−1j rµj(rµj +λµj )2+irµ−2j[ξ˙ [ j]2 (x1−ξ [ j]1 )− ξ˙ [ j]1 (x2−ξ [ j]2 )]rµj +λµj]Φ[ j]0}eiθ j ,0]tr=[{ −ρµjρµj +1∂tΦ[ j]0 +ρµj (ξ˙ [ j] · y[ j]− λ˙ j)(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +[µλ−1j ρµ−2j ξ˙ [ j] · y[ j]+µλ˙ jλ−1j ρµj(ρµj +1)2+iλ−1j ρµ−2j (ξ˙[ j]2 y[ j]1 − ξ˙ [ j]1 y[ j]2 )ρµj +1]Φ[ j]0}eiθ j ,0]tr. (4.3.28)•∆xΦ∗[ j]0 = ∆x[ j]Φ∗[ j]0 =[(∂r jr j +1r j∂r j +1r2j∂θ jθ j)( rµjrµj +λµjΦ[ j]0 eiθ j),0]tr(4.3.29)=[{ rµ+2j(rµj +λµj )(r2j +λ 2j )∂z jz jΦ[ j]0 +[ 2µλ µj rµj(rµj +λµj )2(r2j +λ 2j )12+λ 2j rµj(rµj +λµj )(r2j +λ 2j )32]∂z jΦ[ j]0+(µ−1)µλ 2µj rµ−2j − (µ+1)µλ µj r2µ−2j(rµj +λµj )3Φ[ j]0+rµj(rµj +λµj )(r2j +λ 2j )12∂z jΦ[ j]0 +µλ µj rµ−2j(rµj +λµj )2Φ[ j]0 −rµ−2jrµj +λµjΦ[ j]0}eiθ j ,0]tr=[{ rµ+2j(rµj +λµj )(r2j +λ 2j )∂z jz jΦ[ j]0+[ 2µλ µj rµj(rµj +λµj )2(r2j +λ 2j )12+λ 2j rµj(rµj +λµj )(r2j +λ 2j )32+rµj(rµj +λµj )(r2j +λ 2j )12]∂z jΦ[ j]0+[µ2λ 2µj rµ−2j −µ2λ µj r2µ−2j(rµj +λµj )3− rµ−2jrµj +λµj]Φ[ j]0}eiθ j ,0]tr=[{ ρµ+2j(ρµj +1)(ρ2j +1)∂z jz jΦ[ j]0 +[ 2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32+λ−1j ρµj(ρµj +1)(ρ2j +1)12]∂z jΦ[ j]0+[µ2λ−2jρµ−2j −ρ2µ−2j(ρµj +1)3− λ−2j ρµ−2jρµj +1]Φ[ j]0}eiθ j ,0]tr.168Since µ ≥ 2, we have|∆xΦ∗[ j]0 |≲ |∂z jz jΦ[ j]0 |+ z−1j |∂z jΦ[ j]0 |+ z−2j |Φ[ j]0 |. (4.3.30)• By (4.3.5), (4.3.28), (4.3.29) and (4.3.13), we have−∂t(Φ∗[ j]0 )+(a−bU∞∧)∆xΦ∗[ j]0 −∂tU [ j]=[{ −ρµjρµj +1∂tΦ[ j]0 +ρµj (ξ˙ [ j] · y[ j]− λ˙ j)(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +[µλ−1j ρµ−2j ξ˙ [ j] · y[ j]+µλ˙ jλ−1j ρµj(ρµj +1)2+iλ−1j ρµ−2j (ξ˙[ j]2 y[ j]1 − ξ˙ [ j]1 y[ j]2 )ρµj +1]Φ[ j]0}eiθ j ,0]tr+[(a− ib){[ ρµjρµj +1− ρµj(ρµj +1)(ρ2j +1)]∂z jz jΦ[ j]0+[ 2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32+λ−1j ρµj(ρµj +1)(ρ2j +1)12]∂z jΦ[ j]0+[µ2λ−2jρµ−2j −ρ2µ−2j(ρµj +1)3− λ−2j ρµ−2j(ρµj +1)(ρ2j +1)− λ−2j ρµj(ρµj +1)(ρ2j +1)]Φ[ j]0}eiθ j ,0]tr+E [ j]0 +E[ j]1=[{ ξ˙ [ j] · y[ j]ρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +[µλ−1j ξ˙ [ j] · y[ j]ρµ−2j(ρµj +1)2+iλ−1j (ξ˙[ j]2 y[ j]1 − ξ˙ [ j]1 y[ j]2 )ρµ−2jρµj +1]Φ[ j]0}eiθ j ,0]tr+[[ −λ˙ jρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j λ˙ jρµj(ρµj +1)2Φ[ j]0]eiθ j ,0]tr+[(a− ib){ −ρµj(ρµj +1)(ρ2j +1)∂z jz jΦ[ j]0 +[ 2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0+λ−2j ρµ−2j[µ21−ρµj(ρµj +1)3− 1(ρµj +1)(ρ2j +1)]Φ[ j]0}eiθ j ,0]tr+E [ j]0 +[ 2λ−1j p˙ jρµj(ρµj +1)(ρ2j +1)12eiθ j ,0]tr+E [ j]1(4.3.31)where we have used (4.3.14) in the last equality. Therein,ξ˙ [ j] · y[ j] = 2−1ρ j[(ξ˙ [ j]1 − iξ˙ [ j]2 )eiθ j +(ξ˙ [ j]1 + iξ˙ [ j]2 )e−iθ j],ξ˙ [ j]2 y[ j]1 − ξ˙ [ j]1 y[ j]2 = 2−1ρ j[(ξ˙ [ j]2 + iξ˙[ j]1 )eiθ j +(ξ˙ [ j]2 − iξ˙ [ j]1 )e−iθ j].(4.3.32)169By (4.3.6),E [ j]0 +[ 2λ−1j p˙ jρµj(ρµj +1)(ρ2j +1)12eiθ j ,0]tr=−2ρ jρ2j +1[[λ−1j λ˙ j(1− 2ρ2j +1)+ iγ˙ j]ei(θ j+γ j),−λ−1j λ˙ j2ρ jρ2j +1]tr+[ 2(λ−1j λ˙ j+ iγ˙ j)ρµj(ρµj +1)(ρ2j +1)12ei(θ j+γ j),0]tr=[{(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)+4λ−1j λ˙ jρ j(ρ2j +1)2}ei(θ j+γ j),4λ−1j λ˙ jρ2j(ρ2j +1)2]tr.Then by (4.2.19),(ΠU [ j]⊥(E [ j]0 +[ 2λ−1j p˙ jρµj(ρµj +1)(ρ2j +1)12eiθ j ,0]tr))C j=(1− 2ρ2j +1Re)[(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)+4λ−1j λ˙ jρ j(ρ2j +1)2]− 8λ−1j λ˙ jρ3j(ρ2j +1)3= (λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)−λ−1j λ˙ j4ρµj [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2,(4.3.33)(E [ j]0 +[ 2λ−1j p˙ jρµj(ρµj +1)(ρ2j +1)12eiθ j ,0]tr)·U [ j] = 4λ−1j λ˙ jρµ+1j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2. (4.3.34)•−b(U [ j]−U∞)∧∆xΦ∗[ j]0=−2b|y[ j]|2+1[y[ j]1 cosγ j− y[ j]2 sinγ j,y[ j]1 sinγ j+ y[ j]2 cosγ j,−1]tr∧∆xΦ∗[ j]0=−2bρ2j +1[(∆xΦ∗[ j]0 )2,−(∆xΦ∗[ j]0 )1,(y[ j]1 cosγ j− y[ j]2 sinγ j)(∆xΦ∗[ j]0 )2− (y[ j]1 sinγ j+ y[ j]2 cosγ j)(∆xΦ∗[ j]0 )1]tr=−2bρ2j +1[(∆xΦ∗[ j]0 )2,−(∆xΦ∗[ j]0 )1,ρ jRe[((∆xΦ∗[ j]0 )2− i(∆xΦ∗[ j]0 )1)e−i(θ j+γ j)]]tr(4.3.35)where in the last equality, we use the following formula. For any a1,a2 ∈ R,(y[ j]1 cosγ j− y[ j]2 sinγ j)a1−(y[ j]1 sinγ j+ y[ j]2 cosγ j)a2 = ρ j (a1 cos(θ j+ γ j)−a2 sin(θ j+ γ j))= ρ jRe[(a1− ia2)e−i(θ j+γ j)]= ρ jIm[(a2+ ia1)e−i(θ j+γ j)].(4.3.36)170Then by (4.2.19), we have(ΠU [ j]⊥[−b(U [ j]−U∞)∧∆xΦ∗[ j]0 ])C j=−2bρ2j +1{(1− 2ρ2j +1Re)[((∆xΦ∗[ j]0 )2− i(∆xΦ∗[ j]0 )1)e−i(θ j+γ j)]− 2ρ jρ2j +1ρ jRe[((∆xΦ∗[ j]0 )2− i(∆xΦ∗[ j]0 )1)e−i(θ j+γ j)]}=2ibρ2j +1((∆xΦ∗[ j]0 )1+ i(∆xΦ∗[ j]0 )2)e−i(θ j+γ j)={2ibρµ+2j(ρµj +1)(ρ2j +1)2∂z jz jΦ[ j]0 +2ibλ−1j ρµjρ2j +1[ 2µ(ρµj +1)2(ρ2j +1)12+ρ2j +2(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0− 2ibλ−2j ρµ−2jρ2j +1[µ2ρµj −1(ρµj +1)3+1ρµj +1]Φ[ j]0}eiγ j(4.3.37)where we have used (4.3.29) in the last equality.[−b(U [ j]−U∞)∧∆xΦ∗[ j]0 ] ·U [ j]=−2bρ2j +1{2ρ jρ2j +1Re[((∆xΦ∗[ j]0 )2− i(∆xΦ∗[ j]0 )1)e−i(θ j+γ j)]+ρ2j −1ρ2j +1ρ jRe[((∆xΦ∗[ j]0 )2− i(∆xΦ∗[ j]0 )1)e−i(θ j+γ j)]}=−2bρ jρ2j +1Im[((∆xΦ∗[ j]0 )1+ i(∆xΦ∗[ j]0 )2)e−i(θ j+γ j)].(4.3.38)• By (4.2.3) and (4.3.12),a|∇xU [ j]|2Φ∗[ j]0 = aλ−2j |∇y[ j]U [ j]|2Φ∗[ j]0 =[ 8aλ−2j ρµj(ρ2j +1)2(ρµj +1)Φ[ j]0 eiθ j ,0]tr.Then by (4.2.19),(ΠU [ j]⊥(a|∇xU [ j]|2Φ∗[ j]0))C j=8aλ−2j ρµj(ρ2j +1)2(ρµj +1)(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j) ,(a|∇xU [ j]|2Φ∗[ j]0) ·U [ j] = 16aλ−2j ρ1+µj(ρ2j +1)3(ρµj +1)Re(Φ[ j]0 e−iγ j) . (4.3.39)• Notice (b|∇xU [ j]|2Φ∗[ j]0 ∧U [ j]) ·U [ j] = 0. (4.3.40)171By (4.2.14) and (4.3.39), we have(b|∇xU [ j]|2Φ∗[ j]0 ∧U [ j])C j= b([ΠU [ j]⊥(|∇xU [ j]|2Φ∗[ j]0 )]∧U [ j])C j=−8ibλ−2j ρµj(ρ2j +1)2(ρµj +1)(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j) . (4.3.41)• Notice∂r jU[ j] = λ−1j ∂ρ jU[ j] = λ−1j wρ jQγ jE[ j]1 , ∂θ jU[ j] = sinw(ρ j)Qγ jE[ j]2 .By (4.3.12) and (4.3.25),−2∇x(Φ∗[ j]0 ·U [ j]) ·∇xU [ j] =−2∇x[ j](Φ∗[ j]0 ·U [ j]) ·∇x[ j]U [ j]= −2∂r j(Φ∗[ j]0 ·U [ j])∂r jU [ j]−2r−2j ∂θ j(Φ∗[ j]0 ·U [ j])∂θ jU [ j]= −2(∂r jΦ∗[ j]0 ·U [ j]+Φ∗[ j]0 ·∂r jU [ j])∂r jU [ j]−2r−2j (∂θ jΦ∗[ j]0 ·U [ j]+Φ∗[ j]0 ·∂θ jU [ j])∂θ jU [ j]= −2(∂r jΦ∗[ j]0 ·U [ j]+Φ∗[ j]0 ·λ−1j wρ jQγ jE [ j]1)λ−1j wρ jQγ jE[ j]1−2r−2j(∂θ jΦ∗[ j]0 ·U [ j]+Φ∗[ j]0 · sinw(ρ j)Qγ jE [ j]2)sinw(ρ j)Qγ jE[ j]2= −2{[[ ρµ+1j(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j ρµ−1j(ρµj +1)2Φ[ j]0]eiθ j ,0]tr·U [ j]+[ ρµjρµj +1Φ[ j]0 eiθ j ,0]tr·λ−1j−2ρ2j +1Qγ jE[ j]1}λ−1j−2ρ2j +1Qγ jE[ j]1−2r−2j{[ ρµjρµj +1Φ[ j]0 ieiθ j ,0]tr·U [ j]+[ ρµjρµj +1Φ[ j]0 eiθ j ,0]tr· 2ρ jρ2j +1Qγ jE[ j]2}2ρ jρ2j +1Qγ jE[ j]2={Re{[ρµ+1j(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j ρµ−1j(ρµj +1)2Φ[ j]0]sinw(ρ j)e−iγ j}+−2λ−1j ρµj(ρµj +1)(ρ2j +1)cosw(ρ j)Re(Φ[ j]0 e−iγ j)} 4λ−1jρ2j +1Qγ jE[ j]1−2r−2j{ρµjρµj +1sinw(ρ j)Re(Φ[ j]0 ie−iγ j)− 2ρµ+1j(ρµj +1)(ρ2j +1)Re(Φ[ j]0 ie−iγ j)} 2ρ jρ2j +1Qγ jE[ j]2={8λ−1j ρµ+2j(ρµj +1)(ρ2j +1)52Re(∂z jΦ[ j]0 e−iγ j)+[8µλ−2j ρµj(ρµj +1)2(ρ2j +1)2− 8λ−2j ρµj (ρ2j −1)(ρµj +1)(ρ2j +1)3]Re(Φ[ j]0 e−iγ j)}Qγ jE [ j]1 .(4.3.42)172Then it is easy to see{(a−bU [ j]∧)[−2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]} ·U [ j] = 0. (4.3.43)By (4.2.12),(a−bU [ j]∧)[−2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]={8λ−1j ρµ+2j(ρµj +1)(ρ2j +1)52Re(∂z jΦ[ j]0 e−iγ j)+[8µλ−2j ρµj(ρµj +1)2(ρ2j +1)2− 8λ−2j ρµj (ρ2j −1)(ρµj +1)(ρ2j +1)3]Re(Φ[ j]0 e−iγ j)}(aQγ jE [ j]1 −bQγ jE [ j]2 ) .(4.3.44)In sum, by (4.2.19), (4.3.31), (4.3.34), (4.3.8), (4.3.38), (4.3.39), (4.3.40), (4.3.43),S [ j] ·U [ j] = 2ρ jρ2j +1Re{[ξ˙ [ j] · y[ j]ρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0+[µλ−1j ξ˙ [ j] · y[ j]ρµ−2j(ρµj +1)2+iλ−1j (ξ˙[ j]2 y[ j]1 − ξ˙ [ j]1 y[ j]2 )ρµ−2jρµj +1]Φ[ j]0 +−λ˙ jρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j λ˙ jρµj(ρµj +1)2Φ[ j]0+(a− ib){ −ρµj(ρµj +1)(ρ2j +1)∂z jz jΦ[ j]0 +[ 2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0+λ−2j ρµ−2j[µ21−ρµj(ρµj +1)3− 1(ρµj +1)(ρ2j +1)]Φ[ j]0}]e−iγ j}+4λ−1j λ˙ jρµ+1j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2− 2bρ jρ2j +1Im[((∆xΦ∗[ j]0 )1+ i(∆xΦ∗[ j]0 )2)e−i(θ j+γ j)]+16aλ−2j ρ1+µj(ρ2j +1)3(ρµj +1)Re(Φ[ j]0 e−iγ j).(4.3.45)By µ ≥ 2, (4.3.2), (4.3.23) and (4.3.24), we have|S [ j] ·U [ j]|≲ |ξ˙ [ j]|〈ρ j〉−1+ |λ j|−1〈ρ j〉−2. (4.3.46)By (4.2.19), (4.3.32) and (4.3.31), (4.3.33), (4.3.8), (4.3.37), (4.3.39), (4.3.41), (4.3.44),(ΠU [ j]⊥S[ j])C j=[(ξ˙ [ j]1 − iξ˙ [ j]2 )eiθ j +(ξ˙ [ j]1 + iξ˙ [ j]2 )e−iθ j]×[ρµ+1j2(ρµj +1)(ρ2j +1)12(1− 2ρ2j +1Re)(∂z jΦ[ j]0 e−iγ j)+ µλ−1j ρµ−1j2(ρµj +1)2(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j)]+[(ξ˙ [ j]2 + iξ˙[ j]1 )eiθ j +(ξ˙ [ j]2 − iξ˙ [ j]1 )e−iθ j] λ−1j ρµ−1j2(ρµj +1)(i+2ρ2j +1Im)(Φ[ j]0 e−iγ j)173+(1− 2ρ2j +1Re)[{−λ˙ jρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j λ˙ jρµj(ρµj +1)2Φ[ j]0+(a− ib){ −ρµj(ρµj +1)(ρ2j +1)∂z jz jΦ[ j]0 +[2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0+λ−2j ρµ−2j[µ21−ρµj(ρµj +1)3− 1(ρµj +1)(ρ2j +1)]Φ[ j]0}}e−iγ j]+(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)−λ−1j λ˙ j4ρµj [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2− 2λ−1j (ξ˙[ j]1 − iξ˙ [ j]2 )ρ2j +1eiθ j +{2ibρµ+2j(ρµj +1)(ρ2j +1)2∂z jz jΦ[ j]0+2ibλ−1j ρµjρ2j +1[ 2µ(ρµj +1)2(ρ2j +1)12+ρ2j +2(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0− 2ibλ−2j ρµ−2jρ2j +1[µ2ρµj −1(ρµj +1)3+1ρµj +1]Φ[ j]0}eiγ j +8(a− ib)λ−2j ρµj(ρ2j +1)2(ρµj +1)(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j)+(a− ib){8λ−1j ρµ+2j(ρµj +1)(ρ2j +1)52Re(∂z jΦ[ j]0 e−iγ j)+[8µλ−2j ρµj(ρµj +1)2(ρ2j +1)2− 8λ−2j ρµj (ρ2j −1)(ρµj +1)(ρ2j +1)3]Re(Φ[ j]0 e−iγ j)}=(1− 2ρ2j +1Re)[{−λ˙ jρµj(ρµj +1)(ρ2j +1)12∂z jΦ[ j]0 +µλ−1j λ˙ jρµj(ρµj +1)2Φ[ j]0+(a− ib){ −ρµj(ρµj +1)(ρ2j +1)∂z jz jΦ[ j]0 +[2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0+λ−2j ρµ−2j[µ21−ρµj(ρµj +1)3− 1(ρµj +1)(ρ2j +1)]Φ[ j]0}}e−iγ j]+(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)−λ−1j λ˙ j4ρµj [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2+{2ibρµ+2j(ρµj +1)(ρ2j +1)2∂z jz jΦ[ j]0 +2ibλ−1j ρµjρ2j +1[ 2µ(ρµj +1)2(ρ2j +1)12+ρ2j +2(ρµj +1)(ρ2j +1)32]∂z jΦ[ j]0− 2ibλ−2j ρµ−2jρ2j +1[µ2ρµj −1(ρµj +1)3+1ρµj +1]Φ[ j]0}eiγ j +8(a− ib)λ−2j ρµj(ρ2j +1)2(ρµj +1)(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j)+(a− ib){8λ−1j ρµ+2j(ρµj +1)(ρ2j +1)52Re(∂z jΦ[ j]0 e−iγ j)174+[8µλ−2j ρµj(ρµj +1)2(ρ2j +1)2− 8λ−2j ρµj (ρ2j −1)(ρµj +1)(ρ2j +1)3]Re(Φ[ j]0 e−iγ j)}+ eiθ j{−2λ−1j (ξ˙ [ j]1 − iξ˙ [ j]2 )ρ2j +1+(ξ˙ [ j]1 − iξ˙ [ j]2 )[ρµ+1j2(ρµj +1)(ρ2j +1)12(1− 2ρ2j +1Re)(∂z jΦ[ j]0 e−iγ j)+µλ−1j ρµ−1j2(ρµj +1)2(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j)]+(ξ˙ [ j]2 + iξ˙ [ j]1 ) λ−1j ρµ−1j2(ρµj +1)(i+2ρ2j +1Im)(Φ[ j]0 e−iγ j)}+ e−iθ j{(ξ˙ [ j]1 + iξ˙[ j]2 )[ρµ+1j2(ρµj +1)(ρ2j +1)12(1− 2ρ2j +1Re)(∂z jΦ[ j]0 e−iγ j)+µλ−1j ρµ−1j2(ρµj +1)2(1− 2ρ2j +1Re)(Φ[ j]0 e−iγ j)]+(ξ˙ [ j]2 − iξ˙ [ j]1 ) λ−1j ρµ−1j2(ρµj +1)(i+2ρ2j +1Im)(Φ[ j]0 e−iγ j)}.Using (4.3.17), (4.3.20), we have(ΠU [ j]⊥S[ j])C j=(1− 2ρ2j +1Re){λ˙ jρµj(ρµj +1)(ρ2j +1)12ˆ t0p˙ j(s)e−iγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds− µλ˙ jρµj (ρ2j +1)12(ρµj +1)2ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds+(a− ib){ λ−1j ρµj(ρµj +1)(ρ2j +1)32ˆ t0p˙ j(s)e−iγ j(t)t− s(6ζ jK0ζ j(ζ j)+4ζ2j K0ζ jζ j(ζ j))ds−[2µλ−1j ρµj(ρµj +1)2(ρ2j +1)12+λ−1j ρµj(ρµj +1)(ρ2j +1)32]ˆ t0p˙ j(s)e−iγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds−λ−1j ρµ−2j[µ2(1−ρµj )(ρ2j +1)12(ρµj +1)3− 1(ρµj +1)(ρ2j +1)12]ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds}}+(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)−λ−1j λ˙ j4ρµj [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2+−2ibλ−1j ρµ+2j(ρµj +1)(ρ2j +1)52ˆ t0p˙ j(s)eiγ j(t)t− s(6ζ jK0ζ j(ζ j)+4ζ2j K0ζ jζ j(ζ j))ds− 2ibλ−1j ρµjρ2j +1[ 2µ(ρµj +1)2(ρ2j +1)12+ρ2j +2(ρµj +1)(ρ2j +1)32]ˆ t0p˙ j(s)eiγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds+2ibλ−1j ρµ−2j(ρ2j +1)12[µ2ρµj −1(ρµj +1)3+1ρµj +1]ˆ t0p˙ j(s)eiγ j(t)t− s K0(ζ j)ds− 8(a− ib)λ−1j ρµj(ρ2j +1)32 (ρµj +1)(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)175+(a− ib){−8λ−1j ρµ+2j(ρµj +1)(ρ2j +1)52Re(ˆ t0p˙ j(s)e−iγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds)−[8µλ−1j ρµj(ρµj +1)2(ρ2j +1)32− 8λ−1j ρµj (ρ2j −1)(ρµj +1)(ρ2j +1)52]Re(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)}+ eiθ j{−2λ−1j (ξ˙ [ j]1 − iξ˙ [ j]2 )ρ2j +1− (ξ˙ [ j]1 − iξ˙ [ j]2 )[ρµ+1j2(ρµj +1)(ρ2j +1)12(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds)+µρµ−1j (ρ2j +1)122(ρµj +1)2(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)]− (ξ˙ [ j]2 + iξ˙ [ j]1 )ρµ−1j (ρ2j +1)122(ρµj +1)(i+2ρ2j +1Im)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)}+ e−iθ j{(ξ˙ [ j]1 + iξ˙[ j]2 )[−ρµ+1j2(ρµj +1)(ρ2j +1)12×(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s(K0(ζ j)+2ζ jK0ζ j(ζ j))ds)− µρµ−1j (ρ2j +1)122(ρµj +1)2(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)]− (ξ˙ [ j]2 − iξ˙ [ j]1 )ρµ−1j (ρ2j +1)122(ρµj +1)(i+2ρ2j +1Im)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds)}= M[ j]0 (ρ j, t)+ M˜[ j]0 (ρ j, t)+ eiθ j(M[ j]1 (ρ j, t)+ M˜[ j]1 (ρ j, t))+ e−iθ jM[ j]−1(ρ j, t)where M[ j]0 (ρ j, t),M˜[ j]0 (ρ j, t),M[ j]1 (ρ j, t),M˜[ j]1 (ρ j, t),M[ j]−1(ρ j, t) are given as follows.M[ j]0 (ρ j, t) := λ−1j(1− 2ρ2j +1Re){(a− ib)[(4.3.47)ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρµ−2j(ρµj +1)(ρ2j +1)32− 2µρµj(ρµj +1)2(ρ2j +1)12−µ2ρµ−2j (1−ρµj )(ρ2j +1)12(ρµj +1)3]K0(ζ j)+[4ρµj(ρµj +1)(ρ2j +1)32− 4µρµj(ρµj +1)2(ρ2j +1)12]ζ jK0ζ j(ζ j)+4ρµj(ρµj +1)(ρ2j +1)32ζ 2j K0ζ jζ j(ζ j)}ds]}− ibλ−1jˆ t0p˙ j(s)eiγ j(t)t− s{[2ρµj [2µ(ρ2j +1)+(ρ2j +2)(ρµj +1)](ρµj +1)2(ρ2j +1)52− 2ρµ−2j [µ2(ρµj −1)+(ρµj +1)2](ρµj +1)3(ρ2j +1)12]K0(ζ j)+[8µρµj(ρµj +1)2(ρ2j +1)32+16ρµ+2j +8ρµj(ρµj +1)(ρ2j +1)52]ζ jK0ζ j(ζ j)+8ρµ+2j(ρµj +1)(ρ2j +1)52ζ 2j K0ζ jζ j(ζ j)}ds176− (a− ib)λ−1j(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s8ρµj(ρ2j +1)32 (ρµj +1)K0(ζ j)ds)− (a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− s{[8µρµj(ρµj +1)2(ρ2j +1)32+8ρµj(ρµj +1)(ρ2j +1)52]K0(ζ j)+16ρµ+2j(ρµj +1)(ρ2j +1)52ζ jK0ζ j(ζ j)}ds]+(λ−1j λ˙ j+ iγ˙ j)2ρ j[ρµ−1j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)−λ−1j λ˙ j4ρµj [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρµj +1)(ρ2j +1)2,M˜[ j]0 (ρ j, t) := λ˙ j(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρµj(ρµj +1)(ρ2j +1)12− µρµj (ρ2j +1)12(ρµj +1)2]K0(ζ j)+2ρµj(ρµj +1)(ρ2j +1)12ζ jK0ζ j(ζ j)}ds],(4.3.48)M[ j]1 (ρ j, t) :=−2λ−1j (ξ˙ [ j]1 − iξ˙ [ j]2 )ρ2j +1, (4.3.49)M˜[ j]1 (ρ j, t) :=−(ξ˙ [ j]1 − iξ˙ [ j]2 )(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρµ+1j2(ρµj +1)(ρ2j +1)12+µρµ−1j (ρ2j +1)122(ρµj +1)2]K0(ζ j)+ρµ+1j(ρµj +1)(ρ2j +1)12ζ jK0ζ j(ζ j)}ds]− (ξ˙ [ j]2 + iξ˙ [ j]1 )ρµ−1j (ρ2j +1)122(ρµj +1)(i+2ρ2j +1Im)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds),(4.3.50)M[ j]−1(ρ j, t) :=−(ξ˙ [ j]1 + iξ˙ [ j]2 )(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρµ+1j2(ρµj +1)(ρ2j +1)12+µρµ−1j (ρ2j +1)122(ρµj +1)2]K0(ζ j)+ρµ+1j(ρµj +1)(ρ2j +1)12ζ jK0ζ j(ζ j)}ds]− (ξ˙ [ j]2 − iξ˙ [ j]1 )ρµ−1j (ρ2j +1)122(ρµj +1)(i+2ρ2j +1Im)(ˆ t0p˙ j(s)e−iγ j(t)t− s K0(ζ j)ds).(4.3.51)In order to avoid the discontinuity due to terms like eiθ j , we need to take µ large to reserve somepower of ρ j as ρ j → 0 in the new error. From now on, we take µ = 3. ThenM[ j]0 = λ−1j(1− 2ρ2j +1Re)[(a− ib){ˆ t0p˙ j(s)e−iγ j(t)t− s[ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)32 (ρ3j +1)3177×K0(ζ j)+4ρ3j (ρ3j −3ρ2j −2)(ρ2j +1)32 (ρ3j +1)2ζ jK0ζ j(ζ j)+4ρ3j(ρ3j +1)(ρ2j +1)32ζ 2j K0ζ jζ j(ζ j)]ds}]− ibλ−1jˆ t0p˙ j(s)eiγ j(t)t− s[− 2ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)52 (ρ3j +1)3K0(ζ j)+8ρ3j (2ρ5j +ρ3j +5ρ2j +4)(ρ2j +1)52 (ρ3j +1)2ζ jK0ζ j(ζ j)+8ρ5j(ρ3j +1)(ρ2j +1)52ζ 2j K0ζ jζ j(ζ j)]ds− (a− ib)λ−1j(1− 2ρ2j +1Re)(ˆ t0p˙ j(s)e−iγ j(t)t− s8ρ3j(ρ2j +1)32 (ρ3j +1)K0(ζ j)ds)− (a− ib)λ−1j Re{ˆ t0p˙ j(s)e−iγ j(t)t− s[8ρ3j (ρ3j +3ρ2j +4)(ρ2j +1)52 (ρ3j +1)2K0(ζ j)+16ρ5j(ρ3j +1)(ρ2j +1)52ζ jK0ζ j(ζ j)]ds}+ p−1j p˙ j2ρ j[ρ2j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)−λ−1j λ˙ j4ρ3j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)2. (4.3.52)By (4.3.19), we haveM[ j]0 = λ−1j(1− 2ρ2j +1Re){(a− ib)[ˆ t0p˙ j(s)e−iγ j(t)t− s{ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)32 (ρ3j +1)3[(a+ ib2+O(ζ j))1{ζ j≤1}+O(ζ−1j)1{ζ j>1}]+4ρ3j (ρ3j −3ρ2j −2)(ρ2j +1)32 (ρ3j +1)2(O(ζ j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1})+4ρ3j(ρ3j +1)(ρ2j +1)32(O(ζ 2j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1})}ds]}− ibλ−1jˆ t0p˙ j(s)eiγ j(t)t− s{− 2ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)52 (ρ3j +1)3×[(a− ib2+O(ζ j))1{ζ j≤1}+O(ζ−1j)1{ζ j>1}]+8ρ3j (2ρ5j +ρ3j +5ρ2j +4)(ρ2j +1)52 (ρ3j +1)2(O(ζ j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1})+8ρ5j(ρ3j +1)(ρ2j +1)52(O(ζ 2j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1})}ds− (a− ib)λ−1j(1− 2ρ2j +1Re){ˆ t0p˙ j(s)e−iγ j(t)t− s8ρ3j(ρ2j +1)32 (ρ3j +1)×[(a+ ib2+O(ζ j))1{ζ j≤1}+O(ζ−1j)1{ζ j>1}]ds}178− (a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− s{8ρ3j (ρ3j +3ρ2j +4)(ρ2j +1)52 (ρ3j +1)2[(a+ ib2+O(ζ j))1{ζ j≤1}+O(ζ−1j)1{ζ j>1}]+16ρ5j(ρ3j +1)(ρ2j +1)52(O(ζ j)1{ζ j≤1}+O(ζ−1j)1{ζ j>1})}ds]+ p−1j p˙ j2ρ j[ρ2j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)−λ−1j λ˙ j4ρ3j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)2= λ−1j(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)2(ρ2j +1)32 (ρ3j +1)3+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]+b(ia+b)λ−1jˆ t0p˙ j(s)eiγ j(t)t− s{[ρ j(3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)52 (ρ3j +1)3+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds− (a− ib)λ−1j(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s×{[4(a+ ib)ρ3j(ρ2j +1)32 (ρ3j +1)+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]− (a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− s{[(a+ ib)4ρ3j (ρ3j +3ρ2j +4)(ρ2j +1)52 (ρ3j +1)2+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]+ p−1j p˙ j2ρ j[ρ2j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)−λ−1j λ˙ j4ρ3j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)2. (4.3.53)Then by (4.3.2), (4.3.19), (4.3.22), we obtain|M[ j]0 |≲ λ−1∗ 〈ρ j〉−3ˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})ds+λ−1∗ |λ˙∗|〈ρ j〉−3 ≲ λ−1∗ 〈ρ j〉−3,|M˜[ j]0 |≲ |λ˙∗|〈ρ j〉−1ˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})ds≲ |λ˙∗|〈ρ j〉−1,|M[ j]1 |≲ λ−1∗ |ξ˙ [ j]|〈ρ j〉−2, |M˜[ j]1 |≲ |ξ˙ [ j]|ˆ t0|λ˙∗(s)|t− s(1{ζ j≤1}+ζ−1j 1{ζ j>1})ds≲ |ξ˙ [ j]|, |M[ j]−1|≲ |ξ˙ [ j]|.(4.3.54)By (4.3.54), we have∣∣∣(ΠU [ j]⊥S [ j])C j ∣∣∣≲ λ−1∗ 〈ρ j〉−3+ |λ˙∗|〈ρ j〉−1+ |ξ˙ [ j]|(λ−1∗ 〈ρ j〉−2+1) . (4.3.55)179Integrating (4.3.46) and (4.3.55), we have∣∣S [ j]∣∣≲ λ−1∗ 〈ρ j〉−2+ |λ˙∗|〈ρ j〉−1+ |ξ˙ [ j]|. (4.3.56)4.4 Gluing system4.4.1 Error calculationIn this section, we formulate the inner–outer gluing system such that a solution with desired asymp-totics can be found. We look for the solution u of the formu= (1+A)U∗+Φ− (Φ ·U∗)U∗,Φ(x, t) :=N∑j=1(η [ j]R (x, t)Qγ jΦ[ j]in(y[ j], t)+η [ j]dq(x, t)Φ∗[ j]0 (r j, t))+Φout(x, t),Φ[ j]in(y[ j], t) ·W [ j] = 0 for all t ∈ (0,T ), j = 1,2, . . . ,N,(4.4.1)where η is a smooth cut-off function and0≤ η ≤ 1, η(s) =1, for s≤ 10, for s≥ 2, η [ j]R (x, t) = η(x−ξ [ j](t)λ∗(t)R(t)), η [ j]dq(x, t) = η(x−ξ [ j](t)dq),(4.4.2)where A is a real-valued function to to make |u| = 1; Φ[ j]in and Φout will be solved in the inner-outersystem, whereΦ[ j]in solves the inner problem near each bubbleU[ j], whileΦout handles the region awayfrom the concentration zones; Φ∗[ j]0 (r j, t) is defined in (4.3.12). Throughout this chapter, we ansatzR(t) = λ−β∗ (t), |Φ| 1 (4.4.3)where 0< β < 1 will be chosen later. Notice thatη [ j]dq ≡ 1 in |x−ξ [ j](t)| ≤ 2λ j(t)R(t).A will be chosen in (4.4.1) to make |u|= 1. Indeed,|u|2 = 1⇐⇒ (1+A)2|U∗|2+2(1+A)(Φ ·U∗)(1−|U∗|2)+ |Φ− (Φ ·U∗)U∗|2 = 1⇐⇒[1+A+(Φ ·U∗)(1−|U∗|2)|U∗|2]2=1−|Φ− (Φ ·U∗)U∗|2|U∗|2 +[(Φ ·U∗)(1−|U∗|2)|U∗|2]2.180We takeA={1+1−|U∗|2−|Φ− (Φ ·U∗)U∗|2|U∗|2 +[(Φ ·U∗)(1−|U∗|2)|U∗|2]2} 12−1− (Φ ·U∗)(1−|U∗|2)|U∗|2 .(4.4.4)By (4.3.4), (4.3.2), (4.4.3), we haveA=(1+O(λ∗+ |Φ|2)+O(λ 2∗ |Φ|2)) 12 −1+O(λ∗|Φ|) = O(λ∗+λ∗|Φ|+ |Φ|2) = O(λ∗+ |Φ|2).(4.4.5)One important insight is that we only need to solveS [u] = Ξ(x, t)U∗ (4.4.6)for some scalar function Ξ. Indeed, since |u|= 1 is kept for all t ∈ (0,T ) and u−U∗ as the perturbationis uniformly small, then(U∗ ·u)Ξ= S[u] ·u=−12∂t(|u|2)+a2∆(|u|2) = 0.IfU∗ ·u≥ δ0 > 0, then Ξ≡ 0. (4.4.6) provides us the flexibility to adjust the error terms inU∗ directionand we will call thisU∗-operation throughout this chapter.We compute−∂tΦ= −∂tΦout−N∑j=1∂t(η [ j]dqΦ∗[ j]0)+N∑j=1η [ j]R Qγ j[−∂tΦ[ j]in+(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]−N∑j=1Qγ jΦ[ j]in∂tη[ j]R ,∆xΦ= ∆xΦout+N∑j=1∆x(η [ j]dqΦ∗[ j]0)+N∑j=1η [ j]R Qγ j∆xΦ[ j]in+N∑j=1Qγ j(Φ[ j]in∆xη[ j]R +2∇xη[ j]R ·∇xΦ[ j]in)(4.4.7)where we used ∂t(Qγ j) = γ˙ jJQγ j = γ˙ jQγ jJ,J :=0 −1 01 0 00 0 0 . (4.4.8)Notice thatU∗∆xA+(1+A)∆xU∗+2∇xA ·∇xU∗+∆x [Φ− (Φ ·U∗)U∗]+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 [(1+A)U∗+Φ− (Φ ·U∗)U∗]181= ∆x [Φ− (Φ ·U∗)U∗]+ |∇xU∗|2 [Φ− (Φ ·U∗)U∗]+ |∇x [(1+A)U∗]|2 [Φ− (Φ ·U∗)U∗]−|∇xU∗|2 [Φ− (Φ ·U∗)U∗]+{2∇x [(1+A)U∗] ·∇x [Φ− (Φ ·U∗)U∗]+ |∇x [Φ− (Φ ·U∗)U∗]|2}[Φ− (Φ ·U∗)U∗]+2∇xA ·∇xU∗+(1+A)∆xU∗+U∗[∆xA+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 (1+A)]= ∆xΦ−2∇x(Φ ·U∗) ·∇xU∗+ |∇xU∗|2Φ+ |∇x [(1+A)U∗]|2 [Φ− (Φ ·U∗)U∗]−|∇xU∗|2 [Φ− (Φ ·U∗)U∗]+{2∇x [(1+A)U∗] ·∇x [Φ− (Φ ·U∗)U∗]+ |∇x [Φ− (Φ ·U∗)U∗]|2}[Φ− (Φ ·U∗)U∗]+2∇xA ·∇xU∗+[1+A− (Φ ·U∗)]∆xU∗+U∗{∆xA+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 (1+A)−|∇xU∗|2 (Φ ·U∗)−∆x(Φ ·U∗)},and[(1+A)U∗+Φ− (Φ ·U∗)U∗]∧∆x [(1+A)U∗+Φ− (Φ ·U∗)U∗]= [Φ− (Φ ·U∗)U∗]∧∆x [(1+A)U∗]+ [(1+A)U∗]∧∆x [Φ− (Φ ·U∗)U∗]+ [Φ− (Φ ·U∗)U∗]∧∆x [Φ− (Φ ·U∗)U∗]+ (1+A)U∗∧∆x [(1+A)U∗]= [Φ− (Φ ·U∗)U∗]∧∆xU∗+U∗∧∆x [Φ− (Φ ·U∗)U∗]+ [Φ− (Φ ·U∗)U∗]∧∆x (AU∗)+AU∗∧∆x [Φ− (Φ ·U∗)U∗]+ [Φ− (Φ ·U∗)U∗]∧∆x [Φ− (Φ ·U∗)U∗]+ (1+A)U∗∧∆x [(1+A)U∗]= Φ∧∆xU∗+U∗∧ [∆xΦ−2∇x(Φ ·U∗) ·∇xU∗]+ [Φ− (Φ ·U∗)U∗]∧∆x (AU∗)+AU∗∧∆x [Φ− (Φ ·U∗)U∗]+ [Φ− (Φ ·U∗)U∗]∧∆x [Φ− (Φ ·U∗)U∗]+ (1+A)U∗∧∆x [(1+A)U∗]−2(Φ ·U∗)U∗∧∆xU∗.182ThenS[u] =−U∗∂tA− (1+A)∂tU∗−∂tΦ+(Φ ·U∗)∂tU∗+U∗∂t(Φ ·U∗)+a{U∗∆xA+(1+A)∆xU∗+2∇xA ·∇xU∗+∆x[Φ− (Φ ·U∗)U∗]+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 [(1+A)U∗+Φ− (Φ ·U∗)U∗]}−b [(1+A)U∗+Φ− (Φ ·U∗)U∗]∧∆x [(1+A)U∗+Φ− (Φ ·U∗)U∗]= − (1+A)∂tU∗−∂tΦ+(Φ ·U∗)∂tU∗+U∗ [∂t(Φ ·U∗)−∂tA]+a[∆xΦ−2∇x(Φ ·U∗) ·∇xU∗+ |∇xU∗|2Φ+ |∇x [(1+A)U∗]|2 [Φ− (Φ ·U∗)U∗]−|∇xU∗|2 [Φ− (Φ ·U∗)U∗]+{2∇x [(1+A)U∗] ·∇x [Φ− (Φ ·U∗)U∗]+ |∇x [Φ− (Φ ·U∗)U∗]|2}[Φ− (Φ ·U∗)U∗]+2∇xA ·∇xU∗+[1+A− (Φ ·U∗)]∆xU∗+U∗{∆xA+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 (1+A)−|∇xU∗|2 (Φ ·U∗)−∆x(Φ ·U∗)}]−b{Φ∧∆xU∗+U∗∧ [∆xΦ−2∇x(Φ ·U∗) ·∇xU∗]+ [Φ− (Φ ·U∗)U∗]∧∆x (AU∗)+AU∗∧∆x [Φ− (Φ ·U∗)U∗]+ [Φ− (Φ ·U∗)U∗]∧∆x [Φ− (Φ ·U∗)U∗]+ (1+A)U∗∧∆x [(1+A)U∗]−2(Φ ·U∗)U∗∧∆xU∗}= −∂tΦ+a[∆xΦ−2∇x(Φ ·U∗) ·∇xU∗+ |∇xU∗|2Φ]−b{Φ∧∆xU∗+U∗∧ [∆xΦ−2∇x(Φ ·U∗) ·∇xU∗]}−∂tU∗+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗(4.4.9)whereN [Φ] := a[{|∇x [(1+A)U∗]|2−|∇xU∗|2+2∇x [(1+A)U∗] ·∇xΠU⊥∗ Φ+∣∣∣∇xΠU⊥∗ Φ∣∣∣2}ΠU⊥∗ Φ+2∇xA ·∇xU∗+[1+A− (Φ ·U∗)]∆xU∗]−b{ΠU⊥∗ Φ∧∆x (AU∗)+AU∗∧∆xΠU⊥∗ Φ+ΠU⊥∗ Φ∧∆xΠU⊥∗ Φ+(1+A)U∗∧∆x [(1+A)U∗]−2(Φ ·U∗)U∗∧∆xU∗},(4.4.10)Ξ[Φ] := ∂t [(Φ ·U∗)−A]+a{∆xA+ |∇x [(1+A)U∗+Φ− (Φ ·U∗)U∗]|2 (1+A)−|∇xU∗|2 (Φ ·U∗)−∆x(Φ ·U∗)}.183ThenS[u] =−∂tΦ+(a−bU∗∧) [∆xΦ−2∇x(Φ ·U∗) ·∇xU∗]+aΦN∑j=1|∇xU [ j]|2+bΦ∧N∑j=1|∇xU [ j]|2U [ j]−∂tU∗+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗= −∂tΦ+(a−bU∗∧)[∆xΦ−2N∑j=1∇x(Φ ·U [ j]) ·∇xU [ j]]+ N∑j=1|∇xU [ j]|2(a−bU [ j]∧)Φ−∂tU∗+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗.Using (4.4.1), (4.4.7), we haveS[u] =−∂tΦout−N∑j=1∂t(η [ j]dqΦ∗[ j]0 )+N∑j=1η [ j]R Qγ j[−∂tΦ[ j]in+(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]−N∑j=1Qγ jΦ[ j]in∂tη[ j]R +(a−bU∗∧){∆xΦout+N∑j=1∆x(η [ j]dqΦ∗[ j]0 )+N∑j=1η [ j]R Qγ j∆xΦ[ j]in+N∑j=1Qγ j(Φ[ j]in∆xη[ j]R +2∇xη[ j]R ·∇xΦ[ j]in)−2N∑j=1∇x(U [ j] ·Φout) ·∇xU [ j]−2 N∑j=1∇x[U [ j] ·(η [ j]R Qγ jΦ[ j]in+η[ j]dqΦ∗[ j]0)]·∇xU [ j]−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧)Φout+ N∑j=1|∇xU [ j]|2(a−bU [ j]∧)(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)−∂tU∗+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗184= −∂tΦout+(a−bU∗∧)∆xΦout+N∑j=1|∇xU [ j]|2(a−bU [ j]∧)Φout−N∑j=1∂t(η [ j]dqΦ∗[ j]0 )+N∑j=1|∇xU [ j]|2(a−bU [ j]∧)η [ j]dqΦ∗[ j]0−N∑j=1η [ j]R Qγ j∂tΦ[ j]in+N∑j=1|∇xU [ j]|2(a−bU [ j]∧)η [ j]R Qγ jΦ[ j]in+N∑j=1η [ j]R Qγ j[(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]−N∑j=1Qγ jΦ[ j]in∂tη[ j]R+N∑j=1[a−b(U [ j]+U∗−U [ j])∧]{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)−2∇x(U [ j] ·Φout) ·∇xU [ j]−2∇x [U [ j] ·(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )] ·∇xU [ j]}−∂tU∗+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗185= −∂tΦout+(a−bU∗∧)∆xΦout+N∑j=1|∇xU [ j]|2(a−bU [ j]∧)Φout+N∑j=1(a−bU [ j]∧)[−2∇x (U [ j] ·Φout) ·∇xU [ j]]−N∑j=1∂t(η [ j]dqΦ∗[ j]0 )+N∑j=1(a−bU [ j]∧)∆x(η [ j]dqΦ∗[ j]0 )+ N∑j=1|∇xU [ j]|2(a−bU [ j]∧)η [ j]dqΦ∗[ j]0+N∑j=1(a−bU [ j]∧)[−2∇x(U [ j] ·η [ j]dqΦ∗[ j]0 ) ·∇xU [ j]]−N∑j=1η [ j]R Qγ j∂tΦ[ j]in+N∑j=1(a−bU [ j]∧)η [ j]R Qγ j∆xΦ[ j]in+ N∑j=1|∇xU [ j]|2(a−bU [ j]∧)η [ j]R Qγ jΦ[ j]in+N∑j=1(a−bU [ j]∧){−2∇x [U [ j] · (η [ j]R Qγ jΦ[ j]in)] ·∇xU [ j]}+N∑j=1η [ j]R Qγ j[(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]−N∑j=1Qγ jΦ[ j]in∂tη[ j]R +N∑j=1(a−bU [ j]∧)[Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)]−∂tU∗−N∑j=1b(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)−2∇x(U [ j] ·Φout) ·∇xU [ j]−2∇x [U [ j] ·(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗186= −∂tΦout+(a−bU∗∧)∆xΦout+N∑j=1(a−bU [ j]∧)[|∇xU [ j]|2Φout−2∇x (U [ j] ·Φout) ·∇xU [ j]]+N∑j=1η [ j]R Qγ j[−∂tΦ[ j]in+ (a−bW [ j]∧)(∆xΦ[ j]in+ |∇xW [ j]|2Φ[ j]in−2∇x (W [ j] ·Φ[ j]in) ·∇xW [ j])]+N∑j=1{−∂t(η [ j]dqΦ∗[ j]0 )+(a−bU [ j]∧)[∆x(η [ j]dqΦ∗[ j]0 )+ |∇xU [ j]|2η [ j]dqΦ∗[ j]0 −2∇x(U [ j] ·η [ j]dqΦ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]}+N∑j=1η [ j]R Qγ j[(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]+N∑j=1Qγ j[−Φ[ j]in∂tη [ j]R +(a−bW [ j]∧)(Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)+(W [ j] ·Φ[ j]in)(a−bW [ j]∧)(−2∇xη [ j]R ·∇xW [ j])]−N∑j=1b(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)−2∇x(U [ j] ·Φout) ·∇xU [ j]−2∇x [U [ j] ·(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗187= −∂tΦout+(a−bU∗∧)∆xΦout+N∑j=1(1−η [ j]R)(a−bU [ j]∧)[|∇xU [ j]|2Φout−2∇x (U [ j] ·Φout) ·∇xU [ j]]+N∑j=1(1−η [ j]R){−∂t(η [ j]dqΦ∗[ j]0 )+ (a−bU [ j]∧)[∆x(η [ j]dqΦ∗[ j]0 )+ |∇xU [ j]|2η [ j]dqΦ∗[ j]0−2∇x(U [ j] ·η [ j]dqΦ∗[ j]0)·∇xU [ j]]−∂tU [ j]}+N∑j=1η [ j]R Qγ j{−∂tΦ[ j]in+λ−2j(a−bW [ j]∧)[∆y[ j]Φ[ j]in+ |∇y[ j]W [ j]|2Φ[ j]in−2∇y[ j] (W [ j] ·Φ[ j]in) ·∇y[ j]W [ j]+2(∇y[ j]W[ j] ·∇y[ j]Φ[ j]in)W [ j]]+Q−γ j{(a−bU [ j]∧)[|∇xU [ j]|2ΠU [ j]⊥Φout−2∇x (U [ j] ·Φout) ·∇xU [ j]]}+Q−γ jΠU [ j]⊥{−∂t(Φ∗[ j]0 )+(a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]}}+N∑j=1η [ j]R Qγ j[(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]+N∑j=1Qγ j{−Φ[ j]in∂tη [ j]R + (a−bW [ j]∧)[Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in− (W [ j] ·Φ[ j]in)(2∇xη [ j]R ·∇xW [ j])]}−N∑j=1b(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)−2∇x(U [ j] ·Φout) ·∇xU [ j]−2∇x [U [ j] ·(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N [Φ]+Ξ[Φ]U∗+N∑j=1η [ j]R(U [ j]−U∗+U∗){−2a(∇xW [ j] ·∇xΦ[ j]in)+a|∇xU [ j]|2 (U [ j] ·Φout)+{−∂t(Φ∗[ j]0 )+(a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]} ·U [ j]}.1884.4.2 Simplification ofN [Φ]In this part, we will extract terms about Φ and its derivatives inN [Φ].ΠU⊥∗ Φ∧∆x (AU∗)+AU∗∧∆xΠU⊥∗ Φ+ΠU⊥∗ Φ∧∆xΠU⊥∗ Φ= (Φ∧U∗)∆xA+ΠU⊥∗ Φ∧ [A∆xU∗+2∇xA ·∇xU∗]+AU∗∧∆xΦ−AU∗∧ [(Φ ·U∗)∆xU∗+2∇x(Φ ·U∗) ·∇xU∗]+ΠU⊥∗ Φ∧∆xΦ− (Φ∧U∗)∆x(Φ ·U∗)−ΠU⊥∗ Φ∧ [(Φ ·U∗)∆xU∗+2∇x(Φ ·U∗) ·∇xU∗]= (Φ∧U∗)∆xA+AU∗∧∆xΦ+ΠU⊥∗ Φ∧∆xΦ− (Φ∧U∗)∆x(Φ ·U∗)− (ΠU⊥∗ Φ+AU∗)∧ [2∇x(Φ ·U∗) ·∇xU∗]+ [A− (Φ ·U∗)]Φ∧∆xU∗+ΠU⊥∗ Φ∧ (2∇xA ·∇xU∗)+ [(Φ ·U∗)2−2A(Φ ·U∗)]U∗∧∆xU∗.(4.4.11)Next, we give explicit formulas for∇xA and ∆xA. Due to the choice of (4.4.1), |u|= 1 is equivalentto(1+A)2|U∗|2+2(1+A)(U∗ ·ΠU⊥∗ Φ)+ |ΠU⊥∗ Φ|2 = 1. (4.4.12)Taking ∇x for (4.4.12), we get2(1+A)|U∗|2∇xA+(1+A)2∇x(|U∗|2)+∇x(|ΠU⊥∗ Φ|2)+2(1+A)∇x(U∗ ·ΠU⊥∗ Φ)+2(U∗ ·ΠU⊥∗ Φ)∇xA= 0.So∇xA=−(1+A)2∇x(|U∗|2)+∇x(|ΠU⊥∗ Φ|2)+2(1+A)∇x(U∗ ·ΠU⊥∗ Φ)2(1+A)|U∗|2+2(U∗ ·ΠU⊥∗ Φ). (4.4.13)Taking ∆x for (4.4.12), we have(1+A)2∆x(|U∗|2)+ |U∗|2[2(1+A)∆xA+2|∇xA|2]+4(1+A)∇x(|U∗|2) ·∇xA+2(1+A)∆x(U∗ ·ΠU⊥∗ Φ)+2(U∗ ·ΠU⊥∗ Φ)∆xA+4∇x(U∗ ·ΠU⊥∗ Φ)·∇xA+∆x(|ΠU⊥∗ Φ|2)= 0.Thus, we have∆xA= −2−1[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1 [∆x(|ΠU⊥∗ Φ|2)+2(1+A)∆x(U∗ ·ΠU⊥∗ Φ)+4∇x(U∗ ·ΠU⊥∗ Φ)·∇xA+2|U∗|2|∇xA|2+4(1+A)∇x(|U∗|2) ·∇xA+(1+A)2∆x(|U∗|2)].(4.4.14)Notice that∆x(|ΠU⊥∗ Φ|2) = 2Φ ·∆xΦ+2(|U∗|2−2)[(Φ ·U∗)∆x(Φ ·U∗)+ |∇x(Φ ·U∗)|2]+2∇x(|U∗|2) ·∇x[(Φ ·U∗)2]+2|∇xΦ|2+(Φ ·U∗)2∆x(|U∗|2)189and∆x(U∗ ·ΠU⊥∗ Φ) = (1−|U∗|2)∆x(Φ ·U∗)− (Φ ·U∗)∆x(|U∗|2)−2∇x(|U∗|2) ·∇x(Φ ·U∗).Then (4.4.14) is rephrased into∆xA=−2−1[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1×[2Φ ·∆xΦ+[2(|U∗|2−2)(Φ ·U∗)+2(1+A)(1−|U∗|2)]∆x(Φ ·U∗)+2(|U∗|2−2)|∇x(Φ ·U∗)|2+2|∇xΦ|2+4[(Φ ·U∗)− (1+A)]∇x(|U∗|2) ·∇x(Φ ·U∗)+2|U∗|2|∇xA|2+4∇x(U∗ ·ΠU⊥∗ Φ) ·∇xA+4(1+A)∇x(|U∗|2) ·∇xA+[(Φ ·U∗)− (1+A)]2∆x(|U∗|2)].(4.4.15)By (4.4.15), then partial terms in (4.4.11) can be rewritten as(Φ∧U∗)∆xA+AU∗∧∆xΦ+ΠU⊥∗ Φ∧∆xΦ− (Φ∧U∗)∆x(Φ ·U∗)= −2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1×{2Φ ·∆xΦ+ [2(|U∗|2−2)(Φ ·U∗)+2(1+A)(1−|U∗|2)](U∗ ·∆xΦ)}+AU∗∧∆xΦ+ΠU⊥∗ Φ∧∆xΦ− (Φ∧U∗)(U∗ ·∆xΦ)−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1×{[2(|U∗|2−2)(Φ ·U∗)+2(1+A)(1−|U∗|2)](2∇xΦ ·∇xU∗+Φ ·∆xU∗)+2(|U∗|2−2)|∇x(Φ ·U∗)|2+2|∇xΦ|2+4[(Φ ·U∗)− (1+A)]∇x(|U∗|2) ·∇x(Φ ·U∗)+2|U∗|2|∇xA|2+4∇x(U∗ ·ΠU⊥∗ Φ) ·∇xA+4(1+A)∇x(|U∗|2) ·∇xA+[(Φ ·U∗)− (1+A)]2∆x(|U∗|2)}− (Φ∧U∗)(2∇xΦ ·∇xU∗+Φ ·∆xU∗)= −2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1[2Φ ·∆xΦ+2(1+A−Φ ·U∗)(U∗ ·∆xΦ)]+AU∗∧∆xΦ+ΠU⊥∗ Φ∧∆xΦ−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1{2(1+A−Φ ·U∗)(2∇xΦ ·∇xU∗+Φ ·∆xU∗)+2(|U∗|2−2)|∇x(Φ ·U∗)|2+2|∇xΦ|2+4[(Φ ·U∗)− (1+A)]∇x(|U∗|2) ·∇x(Φ ·U∗)+2|U∗|2|∇xA|2+4∇x(U∗ ·ΠU⊥∗ Φ) ·∇xA+4(1+A)∇x(|U∗|2) ·∇xA+[(Φ ·U∗)− (1+A)]2∆x(|U∗|2)}.(4.4.16)190Combining (4.4.10), (4.4.11) and (4.4.16), we getN [Φ] = b{(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1[Φ ·∆xΦ+(1+A−Φ ·U∗)(U∗ ·∆xΦ)]−AU∗∧∆xΦ−(ΠU⊥∗ Φ)∧∆xΦ}+a[{|∇x [(1+A)U∗]|2−|∇xU∗|2+2∇x [(1+A)U∗] ·∇x(ΠU⊥∗ Φ)+∣∣∣∇x(ΠU⊥∗ Φ)∣∣∣2}ΠU⊥∗ Φ+2∇xA ·∇xU∗+(1+A−Φ ·U∗)∆xU∗]−b[−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1{2(1+A−Φ ·U∗)(2∇xΦ ·∇xU∗+Φ ·∆xU∗)+2(|U∗|2−2)|∇x(Φ ·U∗)|2+2|∇xΦ|2+4[(Φ ·U∗)− (1+A)]∇x(|U∗|2) ·∇x(Φ ·U∗)+2|U∗|2|∇xA|2+4∇x(U∗ ·ΠU⊥∗ Φ) ·∇xA+4(1+A)∇x(|U∗|2) ·∇xA+[(Φ ·U∗)− (1+A)]2∆x(|U∗|2)}− (ΠU⊥∗ Φ+AU∗)∧ [2∇x(Φ ·U∗) ·∇xU∗]+ [A− (Φ ·U∗)]Φ∧∆xU∗+ΠU⊥∗ Φ∧ (2∇xA ·∇xU∗)+ [(Φ ·U∗)2−2A(Φ ·U∗)]U∗∧∆xU∗+(1+A)U∗∧ [(1+A)∆xU∗+2∇xA ·∇xU∗]−2(Φ ·U∗)U∗∧∆xU∗].(4.4.17)Since2∇x [(1+A)U∗] ·∇x(ΠU⊥∗ Φ)= 22∑k=1{[(∂xkA)U∗ ·∂xkΦ+(1+A)∂xkU∗ ·∂xkΦ]−∂xk (U∗ ·Φ)[|U∗|2∂xkA+(1+A)U∗ ·∂xkU∗]− (U∗ ·Φ)[(∂xkA)U∗ ·∂xkU∗+(1+A) |∂xkU∗|2]},thenN [Φ] is rephrased asN [Φ] = b{(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1 [Φ ·∆xΦ+(1+A−Φ ·U∗)(U∗ ·∆xΦ)]−AU∗∧∆xΦ−(ΠU⊥∗ Φ)∧∆xΦ}+a{[|∇xA|2|U∗|2+2(1+A)∇xA · (U∗ ·∇xU∗)+A(2+A) |∇xU∗|2+22∑k=1{[(∂xkA)U∗ ·∂xkΦ+A∂xkU∗ ·∂xkΦ]−∂xk (U∗ ·Φ)[|U∗|2∂xkA+(1+A)U∗ ·∂xkU∗]− (U∗ ·Φ)[(∂xkA)U∗ ·∂xkU∗+(1+A) |∂xkU∗|2]}+2∑k=1|∂xkΦ−U∗∂xk (Φ ·U∗)− (Φ ·U∗)∂xkU∗|2]ΠU⊥∗ Φ+2(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ) ·∇xU∗+∆xU∗−2(U∗ ·∇xU∗) ·∇xU∗+(A−Φ ·U∗)∆xU∗}+2a [(∇xU∗ ·∇xΦ)Φ− (Φ ·∇xΦ) ·∇xU∗]−2a(∇xU∗ ·∇xΦ)(U∗ ·Φ)U∗191−2bU∗∧ [(∇xU∗ ·∇xΦ)Φ− (Φ ·∇xΦ) ·∇xU∗]+b(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1(1+A−Φ ·U∗)(2∇xΦ ·∇xU∗)−b(Φ∧U∗)(2∇xΦ ·∇xU∗)−b[−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1{2(1+A−Φ ·U∗)(Φ ·∆xU∗)+2(|U∗|2−2)|∇x (Φ ·U∗) |2+2|∇xΦ|2+8[(Φ ·U∗)− (1+A)](U∗ ·∇xU∗) ·∇x (Φ ·U∗)+2|U∗|2|∇xA|2+4[−2(Φ ·U∗)U∗ ·∇xU∗+(1−|U∗|2)∇x (Φ ·U∗)] ·∇xA+8(1+A)(U∗ ·∇xU∗) ·∇xA+2 [(Φ ·U∗)− (1+A)]2(|∇xU∗|2+U∗ ·∆xU∗)}− (ΠU⊥∗ Φ+AU∗)∧ [2∇x (Φ ·U∗) ·∇xU∗]+ [A− (Φ ·U∗)]Φ∧∆xU∗+ΠU⊥∗ Φ∧ (2∇xA ·∇xU∗)+[(Φ ·U∗)2−2A(Φ ·U∗)−2(Φ ·U∗)]U∗∧∆xU∗+(1+A)U∗∧ [A∆xU∗+2(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ) ·∇xU∗+∆xU∗−2(U∗ ·∇xU∗) ·∇xU∗]]+2bAU∗∧ [(Φ ·∇xΦ) ·∇xU∗] (4.4.18)whereU∗ ·∇xU∗,Φ ·∇xΦ are defined in (1.0.26).4.4.3 Inner-outer gluing systemBy U∗-operation (4.4.6), we can adjust the terms in the U∗ direction flexibly. Then using (4.4.18), asufficient condition for S[u] = 0 is that (Φ[ j]in,Φout) solve the following gluing system∂tΦout = BΦ,U∗∆xΦout+G in R2× (0,T ),Φout(x,0) = Z∗(x)+N∑m=13∑n=1cmnϑmn(x) in R2.(4.4.19)λ 2j ∂tΦ[ j]in =(a−bW [ j]∧)[∆y[ j]Φ[ j]in+ |∇y[ j]W [ j]|2Φ[ j]in−2∇y[ j] (W [ j] ·Φ[ j]in) ·∇y[ j]W [ j]+2(∇y[ j]W[ j] ·∇y[ j]Φ[ j]in)W [ j]]+H [ j] in D2R,(4.4.20)whereD2R := {(y, t) | |y| ≤ 2R(t), t ∈ (0,T )} ; (4.4.21)H [ j] :=H [ j]1 +H[ j]in , (4.4.22)H [ j]1 := λ2j Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]+(M[ j]0 + eiθ jM[ j]1 )C−1j]=(a−bW [ j]∧) L˜W [ j] [Q−γ jΦout]+λ 2j (M[ j]0 + eiθ jM[ j]1 )C−1j ,(4.4.23)192H [ j]in := λ2j Q−γ j[2(a−bU [ j]∧){[∇xU [ j] ·∇x (η [ j]R Qγ jΦ[ j]in)](Qγ jΦ[ j]in)− [(Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)] ·∇xU [ j]}]= 2(a−bW [ j]∧){[∇y[ j]W [ j] ·∇y[ j] (η [ j]R Φ[ j]in)]Φ[ j]in−[Φ[ j]in ·∇y[ j] (η [ j]R Φ[ j]in)] ·∇y[ j]W [ j]} ;(4.4.24)G :=N∑j=1(1−η [ j]R)(a−bU [ j]∧)[|∇xU [ j]|2Φout−2∇x (U [ j] ·Φout) ·∇xU [ j]]+N∑j=1(1−η [ j]R){−∂t(η [ j]dqΦ∗[ j]0 )+ (a−bU [ j]∧)[∆x(η [ j]dqΦ∗[ j]0 )+ |∇xU [ j]|2η [ j]dqΦ∗[ j]0 −2∇x(U [ j] ·η [ j]dqΦ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]}+N∑j=1η [ j]R(M˜[ j]0 + eiθ jM˜[ j]1 + e−iθ jM[ j]−1)C−1j+N∑j=1η [ j]R Qγ j[(λ−1j λ˙ jy[ j]+λ−1j ξ˙[ j])·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]+N∑j=1Qγ j{−Φ[ j]in∂tη [ j]R + (a−bW [ j]∧)[Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in− (W [ j] ·Φ[ j]in)(2∇xη [ j]R ·∇xW [ j])]}−N∑j=1b(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ·∇xΦ[ j]in)−2∇x(U [ j] ·Φout) ·∇xU [ j]−2∇x [U [ j] ·(η [ j]R Qγ jΦ[ j]in+η [ j]dqΦ∗[ j]0 )] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])] ·∇xU [ j]}+(a−bU∗∧){−2N∑j=1∇x[U [ j] ·N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]·∇xU [ j]}+N∑j=1|∇xU [ j]|2(a−bU [ j]∧) N∑k=1,k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)+aΦN∑j,k=1, j 6=k∇xU [ j] ·∇xU [k]+[(Φ ·U∗)−A]∂tU∗+N∑j=1η [ j]R(U [ j]−U∗)[−2a(∇xW [ j] ·∇xΦ[ j]in)+a|∇xU [ j]|2 (U [ j] ·Φout)+{−∂t(Φ∗[ j]0 )+ (a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 ) ·∇xU [ j]]−∂tU [ j]} ·U [ j]]+b{(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1[Φ+(1+A−Φ ·U∗)U∗] ·∆x (Φ−Φout)−(AU∗+ΠU⊥∗ Φ)∧∆x (Φ−Φout)}+a{[|∇xA|2|U∗|2+2(1+A)∇xA · (U∗ ·∇xU∗)+A(2+A) |∇xU∗|2+22∑k=1{[(∂xkA)U∗ ·∂xkΦ+A∂xkU∗ ·∂xkΦ]−∂xk (U∗ ·Φ)[|U∗|2∂xkA+(1+A)U∗ ·∂xkU∗]− (U∗ ·Φ)[(∂xkA)U∗ ·∂xkU∗+(1+A) |∂xkU∗|2]}+2∑k=1|∂xkΦ−U∗∂xk (Φ ·U∗)− (Φ ·U∗)∂xkU∗|2]ΠU⊥∗ Φ193+2(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ) ·∇xU∗+∆xU∗−2(U∗ ·∇xU∗) ·∇xU∗+(A−Φ ·U∗)∆xU∗}+2(a−bU∗∧) [(∇xU∗ ·∇xΦ)Φ− (Φ ·∇xΦ) ·∇xU∗]−2a(∇xU∗ ·∇xΦ)(U∗ ·Φ)U∗−N∑j=12(a−bU [ j]∧){[∇xU [ j] ·∇x (η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)] ·∇xU [ j]}+b(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1(1+A−Φ ·U∗)(2∇xΦ ·∇xU∗)−b(Φ∧U∗)(2∇xΦ ·∇xU∗)−b[−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1{2(1+A−Φ ·U∗)(Φ ·∆xU∗)+2(|U∗|2−2)|∇x (Φ ·U∗) |2+2|∇xΦ|2+8[(Φ ·U∗)− (1+A)](U∗ ·∇xU∗) ·∇x (Φ ·U∗)+2|U∗|2|∇xA|2+4[−2(Φ ·U∗)U∗ ·∇xU∗+(1−|U∗|2)∇x (Φ ·U∗)] ·∇xA+8(1+A)(U∗ ·∇xU∗) ·∇xA+2 [(Φ ·U∗)− (1+A)]2(|∇xU∗|2+U∗ ·∆xU∗)}− (ΠU⊥∗ Φ+AU∗)∧ [2∇x (Φ ·U∗) ·∇xU∗]+ [A− (Φ ·U∗)]Φ∧∆xU∗+ΠU⊥∗ Φ∧ (2∇xA ·∇xU∗)+[(Φ ·U∗)2−2A(Φ ·U∗)−2(Φ ·U∗)]U∗∧∆xU∗+(1+A)U∗∧ [A∆xU∗+2(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ) ·∇xU∗+∆xU∗−2(U∗ ·∇xU∗) ·∇xU∗]]+2bAU∗∧ [(Φ ·∇xΦ) ·∇xU∗]+ΞG (x, t)U∗ (4.4.25)where ΞG (x, t) is some scalar function due to U∗-operation; M[ j]0 ,M˜[ j]0 ,M[ j]1 ,M˜[ j]1 ,M[ j]−1 are given in(4.3.52), (4.3.48), (4.3.49), (4.3.50), (4.3.51) respectively with µ = 3;BΦ,U∗ := aI3−bU∗∧+B˜Φ,U∗ , (4.4.26)I3 is the 3×3 identity matrix,B˜Φ,U∗ := b[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1(Φ∧U∗)1 [Φ+(1+A−Φ ·U∗)U∗]tr(Φ∧U∗)2 [Φ+(1+A−Φ ·U∗)U∗]tr(Φ∧U∗)3 [Φ+(1+A−Φ ·U∗)U∗]tr−b(AU∗+ΠU⊥∗ Φ)∧,(4.4.27)Z∗(x)∈C∞0 (R2), OuppZ∗⊂BCq ‖Z∗‖C3(R2)<CZ∗ , [∂x1Z∗1+∂x2Z∗2+ i(∂x1Z∗2−∂x2Z∗1)] (q[ j]) 6= 0(4.4.28)whereCZ∗ 1, j = 1,2, . . . ,N andCq = 9 maxj=1,2,...,N |q[ j]|;ϑmn ∈C∞0 (R2), Ouppϑmn ⊂ BCq ‖ϑmn‖C3(R2) ≤ 2,ϑmn(qk) = δmken, ∇ϑmn(qk) = 0 for m,k = 1,2, . . . ,N, n= 1,2,3,(4.4.29)e1 = [1,0,0]tr, e2 = [0,1,0]tr, e3 = [0,0,1]tr.194Under the assumption in Propostion 4.6.3 and using Propostion 4.6.2, there exists a fundamentalsolution ΓΦ,U∗(x, t,y,s) for∂tf= BΦ,U∗∆xf in R2× (0,T ).cmn will be chosen to make Φout(qk,T ) = 0 for k = 1,2, . . . ,N, that is,(ΓΦ,U∗ ∗∗G )(qk,T )+(ΓΦ,U∗ ∗Z∗)(qk,T )+N∑m=13∑n=1cmn (ΓΦ,U∗ ∗ϑmn)(qk,T ) = 0 for k = 1,2, . . . ,N.(4.4.30)By Propostion 4.6.3 and Propostion 4.6.2, for f= Z∗ or ϑmn, we have|ΓΦ,U∗ ∗ f|+ |Dx (ΓΦ,U∗ ∗ f)|+∣∣D2x (ΓΦ,U∗ ∗ f)∣∣+ |∂t (ΓΦ,U∗ ∗ f)|≲ ‖f‖C3(R2) in R2× (0,T ), (4.4.31)|Dx (ΓΦ,U∗ ∗ f)(x, t)−Dx (ΓΦ,U∗ ∗ f)(x∗, t∗)|(|x− x∗|+√|t− t∗|)α ≲C(α)‖f‖C3(R2) for 0<α < 1, (x, t),(x∗, t∗)∈R2×(0,T ).(4.4.32)By (4.4.31), for m,k = 1,2, . . . ,N, n= 1,2,3, we have|(ΓΦ,U∗ ∗ϑmn)(qk,T )−ϑmn(qk)|= |(ΓΦ,U∗ ∗ϑmn)(qk,T )−δmken|≲ T.Thus we can find unique cmn = cmn[Φ,U∗,G ,Z∗] = cmn1+cmn2 for m= 1,2, . . . ,N, n= 1,2,3 solving(4.4.30) where cmn1 = cmn1[Φ,U∗,G ] and cmn2 = cmn2[Φ,U∗,Z∗] satisfy(ΓΦ,U∗ ∗∗G )(qk,T )+N∑m=13∑n=1cmn1 (ΓΦ,U∗ ∗ϑmn)(qk,T ) = 0,(ΓΦ,U∗ ∗Z∗)(qk,T )+N∑m=13∑n=1cmn2 (ΓΦ,U∗ ∗ϑmn)(qk,T ) = 0 for k = 1,2, . . . ,Nwith the following upper bound|cmn1|≲N∑k=1|(ΓΦ,U∗ ∗∗G )(qk,T )| , |cmn2|≲N∑k=1|(ΓΦ,U∗ ∗Z∗)(qk,T )|≲ ‖Z∗‖C3(R2). (4.4.33)In order to find a solution for (4.4.19), it suffices to solve the following fixed point problem:To[Φout] := ΓΦ,U∗ ∗∗G +ΓΦ,U∗ ∗Z∗+N∑m=13∑n=1(cmn1[Φ,U∗,G ]+ cmn2[Φ,U∗,Z∗]) (ΓΦ,U∗ ∗ϑmn) .(4.4.34)DenoteΦ(1)out := ΓΦ,U∗ ∗Z∗+N∑m=13∑n=1cmn2[Φ,U∗,Z∗] (ΓΦ,U∗ ∗ϑmn) . (4.4.35)195By (4.4.31), (4.4.32) and (4.4.33), we havesupR2×(0,T )(∣∣Φ(1)out∣∣+ ∣∣DxΦ(1)out∣∣+ ∣∣D2xΦ(1)out∣∣+ ∣∣∂tΦ(1)out∣∣)+ sup(x,t),(x∗,t∗)∈R2×(0,T )∣∣DxΦ(1)out(x, t)−DxΦ(1)out(x∗, t∗)∣∣(|x− x∗|+√|t− t∗|)α ≤ 9−1Λo1‖Z∗‖C3(R2)(4.4.36)for a large constant Λo1 ≥ 1 depending on α .4.4.4 Weighted topologies for the inner and outer problemsThe topologies for the inner and outer problems are listed in this section. Recall (4.3.2) and the formof (4.4.20). It is natural to introduce the new time variableτ j = τ j(t) :=ˆ t0λ−2j (s)ds+CτTλ−2∗ (0), τ j(0) = τ0 :=CτTλ−2∗ (0) (4.4.37)with a constantCτ > 0 sufficiently large to make λ 2j ∂tΦ[ j]in = ∂τΦ[ j]in. It follows thatτ j(t)∼ | lnT |−2(T − t)−1| ln(T − t)|4, ln(τ j(t))∼ | ln(T − t)|, λ∗(t(τ j))∼ | lnT |−1τ−1j (lnτ j)2,T − t(τ j)∼ | lnT |−2τ−1j (lnτ j)4, ∂τ jt(τ j) = λ 2j (t(τ j)),∂τ jλ∗(t(τ j)) = (∂tλ∗)(t(τ j))∂τ jt(τ j)∼−| lnT |−1τ−2j (lnτ j)2.(4.4.38)• For the inner problems, we endow the right hand sides and solutions with the following normsrespectively‖H‖ν ,2+l := supD2Rλ−ν∗ (t(τ j))〈y〉2+l∣∣H(y,τ j)∣∣ ,‖H‖ν ,2+l,ςH := supD2R[λ−ν∗ (t(τ j))〈y〉2+l(∣∣H(y,τ j)∣∣+ 〈y〉ςH [H]CςH ,ςH2 (Q−((y,τ j), |y|2 )))],(4.4.39)where 0< ςH < 1 is small,[H]CςH ,ςH2 (Q−((y,τ j), |y|2 )):= sup(y∗,τ∗)∈Q−((y,τ j), |y|2 )|H(y,τ j)−H(y∗,τ∗)|(|y− y∗|+ |τ j− τ∗| 12)ςH ,D2R :={(y,τ j) | |y|< 2R, τ j > τ0},Q−((y,τ j),|y|2) :={(z,s) ∈D2R | |z− y|< |y|2 , τ j− (|y|2)2 < s< τ j}.(4.4.40)196Denote‖Φ[ j]in‖in,ν−δ0,l := supD2R[λ−ν+δ0∗ (t(τ j))〈y〉l(∣∣Φ[ j]in(y,τ j)∣∣+ 〈y〉 ∣∣DyΦ[ j]in(y,τ j)∣∣+ 〈y〉2 ∣∣D2yΦ[ j]in(y,τ j)∣∣)],[Φ[ j]in]in,ν−δ0,l,ςin := supD2R[λ−ν+δ0∗ (t(τ j))〈y〉l×(〈y〉ςin [Φ[ j]in]Cςin , ςin2 (Q−((y,τ j), |y|2 ))+ 〈y〉ςin+1 [DyΦ[ j]in]Cςin , ςin2 (Q−((y,τ j), |y|2 )))],‖Φ[ j]in‖in,ν−δ0,l,ςin := ‖Φ[ j]in‖in,ν−δ0,l +[Φ[ j]in]in,ν−δ0,l,ςin ,(4.4.41)where0< ςin < 1, 0< δ0 < ν < 1. (4.4.42)Set R0(t) = λ−δ0/6∗ (t), which will be used in the inner problems and orthogonal equations.The inner problem will be solved in the following spaceB[ j]in :={f | ‖f‖in,ν−δ0,l,ςin ≤ Λin, f ·W [ j] = 0}(4.4.43)for a constant Λin ≥ 1 to be determined later.• For the outer problem, we modify the weights used in [19, Section 6.6] to control the right handside of the outer problemρ [ j]1 := λΘ∗ (λ∗R)−11{|x−q[ j]|≤3λ∗R}, ρ[ j]2 := T−σ0 λ1−σ0∗|x−q[ j]|2 1{λ∗R/2≤|x−q[ j]|≤dq}, ρ3 := T−σ0 (4.4.44)where dq is given in (4.3.1),Θ+β −1< 0, 0<Θ< 1, 0< σ0 < 1. (4.4.45)For a function f (x, t), we define the L∞-weighted norm‖ f‖∗∗ := supR2×(0,T )[N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3]−1| f (x, t)|. (4.4.46)197Also, we define the L∞-weighted norm for Φout:‖Φout‖♯,Θ,α :=(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))−1 ‖Φout‖L∞(R2×(0,T ))+(λΘ∗ (0)+‖Z∗‖C3(R2))−1 ‖∇xΦout‖L∞(R2×(0,T ))+ supR2×(0,T )[| ln(T − t)|λΘ+1∗ (t)R(t)+(T − t)‖Z∗‖C3(R2)]−1 |Φout(x, t)−Φout(x,T )|+ supR2×(0,T )[λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2)]−1 |∇xΦout(x, t)−∇xΦout(x,T )|+ sup[λΘ∗ (t)(λ∗(t)R(t))−α +‖Z∗‖C3(R2)]−1 |∇xΦout(x, t)−∇xΦout(x∗, t∗)|(|x− x∗|+√|t− t∗|)α(4.4.47)under assumptions (4.12.1) for the parameters, where the last supremum is taken in the regionx, x∗ ∈ R2, t, t∗ ∈ (0,T ), |t− t∗|< 14(T − t).The outer problem will be solved inBout :={f : ‖f‖♯,Θ,α ≤ Λo, f(q[ j],T ) = 0 for j = 1,2, . . . ,N}(4.4.48)where Λo ≥ 1 will be determined later.4.4.5 Strategy for solving the inner problemIn order to find inner solutions with sufficient space-time decay, we need to impose orthogonalityconditions forH [ j] given in (4.4.22).Due to the non-local feature at mode 0, we will only solve the non-local problem at the leadingorder and leave the remainder to another piece of the inner problem without orthogonality at mode 0.The smallness in the remainder is crucial to control the solution to the non-orthogonal inner problem.Proposition 4.8.1 and Proposition 4.8.8 will deal with the orthogonal inner problem and the non-orthogonal inner problem in mode 0 respectively.By (4.2.25), (4.2.24), we reformulate(Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]])C jinto(Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]])C j (x, t) = L˜#j [Φout](y[ j], t)+ l˜#j [Φout](x, t) (4.4.49)198where the leading term L˜#j [Φout] is given byL˜#j [Φout](y[ j], t) := (a− ib){λ−1j ρ jw2ρ j(ρ j)e−iγ j [∂x1 (Φout)1+∂x2 (Φout)2+ i(∂x1 (Φout)2−∂x2 (Φout)1)] (q[ j], t)+ eiθ j2λ−1j wρ j(ρ j)cosw(ρ j) [−∂x1 (Φout)3+ i∂x2 (Φout)3] (q[ j], t)+ e2iθ jλ−1j ρ jw2ρ j(ρ j)eiγ j [∂x1 (Φout)1−∂x2 (Φout)2− i(∂x1 (Φout)2+∂x2 (Φout)1)] (q[ j], t)},(4.4.50)and the smaller term l˜#j [Φout] is given byl˜#j [Φout](x, t) := (a− ib)[λ−1j ρ jw2ρ j(ρ j)e−iγ j{[∂x1 (Φout)1+∂x2 (Φout)2+ i(∂x1 (Φout)2−∂x2 (Φout)1)] (x, t)− [∂x1 (Φout)1+∂x2 (Φout)2+ i(∂x1 (Φout)2−∂x2 (Φout)1)] (q[ j], t)}+ eiθ j2λ−1j wρ j(ρ j)cosw(ρ j){[−∂x1 (Φout)3+ i∂x2 (Φout)3] (x, t)− [−∂x1 (Φout)3+ i∂x2 (Φout)3] (q[ j], t)}+ e2iθ jλ−1j ρ jw2ρ j(ρ j)eiγ j{[∂x1 (Φout)1−∂x2 (Φout)2− i(∂x1 (Φout)2+∂x2 (Φout)1)] (x, t)− [∂x1 (Φout)1−∂x2 (Φout)2− i(∂x1 (Φout)2+∂x2 (Φout)1)] (q[ j], t)}].(4.4.51)By (4.4.47), we have|l˜#j [Φout](x, t)|≲ λ−1j 〈ρ j〉−2‖Φout‖♯,Θ,α(λΘ∗ (t)(λ∗R)−α +‖Z∗‖C3(R2))∣∣x−ξ [ j]+ξ [ j]−q[ j]∣∣α≲ λ−1j (T − t)α(λΘ∗ (t)(λ∗R)−α +‖Z∗‖C3(R2))〈ρ j〉α−2‖Φout‖♯,Θ,α(4.4.52)We will not put the mode 0 component of l˜#j [Φout], which is defined asl˜#j,0[Φout](ρ j, t) := (2π)−1ˆ 2π0l˜#j [Φout](ρ jeis+ξ [ j](t), t)ds. (4.4.53)into the orthogonal inner problem due to Proposition 4.5.1. L˜#j,0[Φout](ρ j, t) is defined by the sameway.In order to simplify the calculation, we also put the mode 0 component (H [ j]in )C j,0 (see (4.2.17))into the non-orthogonal version of the inner problem.More precisely, for j ∈ {1, . . . ,N}, (4.4.20) is split into the following two parts.199Orthogonal inner problem:λ 2j ∂tΦ[ j1]in =(a−bW [ j]∧)[∆y[ j]Φ[ j1]in + |∇y[ j]W [ j]|2Φ[ j1]in −2∇y[ j] (W [ j] ·Φ[ j1]in ) ·∇y[ j]W [ j]+2(∇y[ j]W[ j] ·∇y[ j]Φ[ j1]in)W [ j]]+H [ j]1 −λ 2j(l˜#j,0[Φout])C−1j+R0[Φout]+H [ j]in −((H [ j]in )C j,0)C−1j+(1∑k=0eikθ jc[ j]k (τ j(t))η(|y[ j]|)Zk,1(|y[ j]|))C−1jin D2R,(4.4.54)Non-orthogonal inner problem:λ 2j ∂tΦ[ j2]in =(a−bW [ j]∧)[∆y[ j]Φ[ j2]in + |∇y[ j]W [ j]|2Φ[ j2]in −2∇y[ j] (W [ j] ·Φ[ j2]in ) ·∇y[ j]W [ j]+2(∇y[ j]W[ j] ·∇y[ j]Φ[ j2]in)W [ j]]+λ 2j(l˜#j,0[Φout])C−1j−R0[Φout]+((H [ j]in )C j,0)C−1jin D2R,(4.4.55)whereR0[Φout](y, t) := λ j(ˆB2η(y)Z 20,1(y)dy)−1η(|y[ j]|)Z0,1(|y[ j]|)×(e−iγ j(t)2πR0[eiγ j(t)ˆ 2R00λ jL˜#j,0(ρ j, t)Z0,1(ρ j)ρ jdρ j](t))C−1j,(4.4.56)the operatorR0 is given in Proposition 4.5.1 with m=Θ−α(1−β ) given in (4.5.16);c[ j]0 (t) = c[ j]0 [(H[ j]1 +R0)C j,0−λ 2j l˜#j,0](t), c[ j]∗0(t) = c[ j]∗0[(H [ j]1 +R0)C j,0−λ 2j l˜#j,0](t),c[ j]1 (t) = c[ j]1 [(H[ j]1 +H[ j]in )C j,1](t), c[ j]∗1(t) = c[ j]∗1[(H[ j]1 +H[ j]in )C j,1](t)are given by Proposition 4.8.1 andc[ j]0 (t) =−(ˆB2η(y)Z 20,1(y)dy)−1ˆB2R0[(H [ j]1 +R0)C j,0−λ 2j l˜#j,0](y, t)Z0,1(y)dy+ c[ j]∗0(t),c[ j]1 (t) =−(ˆB2η(y)Z 21,1(y)dy)−1ˆB2R0(H [ j]1 +H[ j]in)C j,1(y, t)Z1,1(y)dy+ c[ j]∗1(t).4.5 Orthogonal equationsIn this section, we will use Proposition 4.5.1 to find λ j,γ j,ξ [ j] satisfying (4.3.2) to makec[ j]0 (t) = 0, c[ j]1 (t) = 0 for all t ∈ (0,T ) (4.5.1)200which is equivalent toλ−1j eiγ j(t)ˆB2R0[(H [ j]1 +R0)C j,0−λ 2j l˜#j,0](y, t)Z0,1(y)dy−(ˆB2η(y)Z 20,1(y)dy)λ−1j eiγ j(t)c[ j]∗0(t) = 0,(4.5.2)λ−1jˆB2R0(H [ j]1 +H[ j]in)C j,1(y, t)Z1,1(y)dy−(ˆB2η(y)Z 21,1(y)dy)λ−1j c[ j]∗1(t) = 0. (4.5.3)4.5.1 Mode 0There exists the correspondence between (4.2.9) and (4.8.19) through (4.8.20) and (4.8.21). See(4.8.19), Z0,1(ρ j)ρ j =ρ2jρ2j+1. By (4.4.22) and (4.4.49),ˆB2R0[(H [ j]1 +R0)C j,0−λ 2j l˜#j,0](y, t)Z0,1(y)dy= λ j{2πˆ 2R00[λ jL˜#j,0+λ jM[ j]0](ρ j, t)Z0,1(ρ j)ρ jdρ j+λ−1jˆB2R0(R0)C j,0(y, t)Z0,1(y)dy}.(4.5.4)Recall (4.3.18), ζ j = ι j(ρ2j +1), ι j =λ 2j (t)t−s . Then by (4.3.53),ˆ 2R00M0(ρ j, t)Z0,1(ρ j)ρ jdρ j= λ−1jˆ 2R00(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[ρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)2(ρ2j +1)52 (ρ3j +1)3+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]dρ j+b(ia+b)λ−1jˆ 2R00ˆ t0p˙ j(s)eiγ j(t)t− s{[ρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)72 (ρ3j +1)3+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}dsdρ j− (a− ib)λ−1jˆ 2R00(1− 2ρ2j +1Re)[ˆ t0p˙ j(s)e−iγ j(t)t− s{[4(a+ ib)ρ5j(ρ2j +1)52 (ρ3j +1)+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]dρ j− (a− ib)λ−1jˆ 2R00Re[ˆ t0p˙ j(s)e−iγ j(t)t− s{201[(a+ ib)4ρ5j (ρ3j +3ρ2j +4)(ρ2j +1)72 (ρ3j +1)2+O(ζ j〈ρ j〉−3)]1{ζ j≤1}+O(ζ−1j 〈ρ j〉−3)1{ζ j>1}}ds]dρ j+ p−1j p˙ jˆ 2R002ρ3j [ρ2j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)2dρ j−λ−1j λ˙ jˆ 2R004ρ5j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)3dρ j= λ−1jˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[ρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)2(ρ2j +1)52 (ρ3j +1)3+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds−λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[ρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)72 (ρ3j +1)3+O(ι j〈ρ j〉−3)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−7)1{ι j(ρ2j+1)>1}}dρ jds]+b(ia+b)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[ρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)72 (ρ3j +1)3+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds]+b(ib−a)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[iρ3j (3ρ7j +ρ6j +12ρ5j −15ρ4j +11ρ3j −24ρ2j −8)(ρ2j +1)72 (ρ3j +1)3+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds]− (a− ib)λ−1jˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[4(a+ ib)ρ5j(ρ2j +1)52 (ρ3j +1)+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds+(a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[8(a+ ib)ρ5j(ρ2j +1)72 (ρ3j +1)+O(ι j〈ρ j〉−3)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−7)1{ι j(ρ2j+1)>1}}dρ jds]− (a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{202[(a+ ib)4ρ5j (ρ3j +3ρ2j +4)(ρ2j +1)72 (ρ3j +1)2+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds]+Cp1(R0)p−1j p˙ j+Cp2(R0)λ−1j λ˙ j= λ−1jˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[−8ρ11j +3ρ10j +ρ9j −4ρ8j −15ρ7j +11ρ6j −32ρ5j −8ρ3j2(ρ2j +1)52 (ρ3j +1)3+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds+(a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− sˆ 2R00{[(a+ ib)4ρ11j −15ρ10j −ρ9j −16ρ8j +3ρ7j −11ρ6j +16ρ5j +8ρ3j(ρ2j +1)72 (ρ3j +1)3+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ jds]+Cp1(R0)p−1j p˙ j+Cp2(R0)λ−1j λ˙ jwhereCp1(R0) :=ˆ 2R002ρ3j [ρ2j −ρ j− (ρ2j +1)12 ][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)2dρ j ∼−1,Cp2(R0) :=−ˆ 2R004ρ5j [ρ2j +ρ j(ρ2j +1)12 +1][ρ j+(ρ2j +1)12 ](ρ3j +1)(ρ2j +1)3dρ j ∼−1.For brevity, we denotef1(ρ j) =−8ρ11j +3ρ10j +ρ9j −4ρ8j −15ρ7j +11ρ6j −32ρ5j −8ρ3j2(ρ2j +1)52 (ρ3j +1)3,f2(ρ j) =4ρ11j −15ρ10j −ρ9j −16ρ8j +3ρ7j −11ρ6j +16ρ5j +8ρ3j(ρ2j +1)72 (ρ3j +1)3.Notice ˆ ∞0f1(ρ j)dρ j =−1,ˆ ∞0f2(ρ j)dρ j = 0.• ˆ 2R00{[f1(ρ j)+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ j=(−1+O(R−20 + ι j〈ln ι j〉))1{ι j≤1}+O(ι−1j )1{ι j>1}since for ι j > 1, ˆ 2R00O(ι−1j 〈ρ j〉−5)dρ j = O(ι−1j);203for ι j < (4R20+1)−1,ˆ 2R00(f1(ρ j)+O(ι j〈ρ j〉−1))dρ j =−1+O(R−20 + ι j lnR0);for (4R20+1)−1 ≤ ι j ≤ 1,ˆ (2ι−1j −1) 120[f1(ρ j)+O(ι j〈ρ j〉−1)]dρ j+ˆ 2R0(2ι−1j −1)12O(ι−1j 〈ρ j〉−5)dρ j =−1+O(ι j〈ln ι j〉).•ˆ 2R00{[(a+ ib) f2(ρ j)+O(ι j〈ρ j〉−1)]1{ι j(ρ2j+1)≤1}+O(ι−1j 〈ρ j〉−5)1{ι j(ρ2j+1)>1}}dρ j= O(R−40 + ι j〈ln ι j〉)1{ι j≤1}+O(ι−1j )1{ι j>1}since for ι j > 1,ˆ 2R00O(ι−1j 〈ρ j〉−5)dρ j = O(ι−1j);for ι j < (4R20+1)−1,ˆ 2R00[(a+ ib) f2(ρ j)+O(ι j〈ρ j〉−1)]dρ j = O(R−40 + ι j lnR0);for (4R20+1)−1 ≤ ι j ≤ 1,ˆ (2ι−1j −1) 120[(a+ ib) f2(ρ j)+O(ι j〈ρ j〉−1)]dρ j+ˆ 2R0(2ι−1j −1)12O(ι−1j 〈ρ j〉−5)dρ j = O(ι j〈ln ι j〉).Thus, we arrive atˆ 2R00M0(ρ j, t)Z0,1(ρ j)ρ jdρ j = λ−1jˆ t0p˙ j(s)e−iγ j(t)t− s×[(−1+O(R−20 + ι j〈ln ι j〉))1{ι j≤1}+O(ι−1j )1{ι j>1}]ds+(a− ib)λ−1j Re[ˆ t0p˙ j(s)e−iγ j(t)t− s(O(R−40 + ι j〈ln ι j〉)1{ι j≤1}+O(ι−1j )1{ι j>1})ds]+Cp1(R0)p−1j p˙ j+Cp2(R0)λ−1j λ˙ j.(4.5.5)Next, we handle the influence of the outer problem in the orthogonal equations. For |x− ξ [ j]| ≤2042λ jR, by (4.4.47), we have∣∣∣DxΦout(x, t)−DxΦout(q[ j],T )∣∣∣= ∣∣∣DxΦout(x, t)−DxΦout(q[ j], t)+DxΦout(q[ j], t)−DxΦout(q[ j],T )∣∣∣≲ ‖Φout‖♯,Θ,α{[λΘ∗ (t)(λ∗(t)R(t))−α +‖Z∗‖C3(R2)]|x−q[ j]|α +[λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2)]}≲ ‖Φout‖♯,Θ,α[λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2)].(4.5.6)By (4.4.28), (4.4.29) and (4.5.6), when T is sufficiently small, for all 0≤ t ≤ T , we have[∂x1 (Φout)1+∂x2 (Φout)2+ i(∂x1 (Φout)2−∂x2 (Φout)1)] (q[ j], t)= [∂x1Z∗1+∂x2Z∗2+ i(∂x1Z∗2−∂x2Z∗1)] (q[ j])+O(T ε)where 0< ε <min{Θ, α2 }. Thenˆ 2R00L˜#j,0(ρ j, t)Z0,1(ρ j)ρ jdρ j= (a− ib)λ−1j e−iγ j(t)(1+O(R−20 )) [∂x1(Φout)1+∂x2(Φout)2+ i(∂x1(Φout)2−∂x2(Φout)1)] (q[ j], t)where we used´ 2R00 ρ jw2ρ j(ρ j)Z0,1(ρ j)ρ jdρ j =´ 2R004ρ3j(ρ2j+1)3dρ j = 1+O(R−20 ) and O(R−20 ) is smoothabout R0.4.5.2 Mode 1See (4.8.19), Z1,1(ρ j) = 1ρ2j+1 . ThenˆB2R0(H [ j]1 +H[ j]in)C j,1(y, t)Z1,1(y)dy= 2πˆ 2R00{λ 2j(Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]])C j,1+λ 2j M[ j]1 +(H [ j]in )C j,1}(ρ j, t)Z1,1(ρ j)ρ jdρ j.(4.5.7)Therein, recall M[ j]1 given in (4.3.49),ˆ 2R00M[ j]1 (ρ j, t)Z1,1(ρ j)ρ jdρ j =−(ξ˙ [ j]1 −iξ˙ [ j]2 )λ−1jˆ 2R002ρ j(ρ2j +1)2dρ j =−(ξ˙ [ j]1 −iξ˙ [ j]2 )λ−1j (1+O(R−20 )).(4.5.8)For the influence of the outer problem, by (4.4.52),ˆ 2R00(Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]])C j,1 (ρ j, t)Z1,1(ρ j)ρ jdρ j= O(λ δ0/3∗ )(a− ib)λ−1j (−∂x1(Φout)3+ i∂x2(Φout)3)(q[ j], t)+h∗1[l˜#j [Φout]](4.5.9)205where we used´ 2R00 wρ j(ρ j)cosw(ρ j)Z1,1(ρ j)ρ jdρ j =´ 2R00−2ρ j(ρ2j−1)(ρ2j+1)3dρ j = O(R−20 ) = O(λδ0/3∗ );h∗1[l˜#j [Φout]] is a linear mapping about l˜#j [Φout] and∣∣h∗1[l˜#j [Φout]]∣∣≲ λ−1j ‖Φout‖♯,Θ,α [λΘ∗ (t)+λα(1−β )∗ (t)‖Z∗‖C3(R2)] . (4.5.10)The effect from the nonlinear part of the inner problem. By (4.13.86),∣∣∣ˆB2R0(H [ j]in )C j,1(y, t)Z1,1(y)dy∣∣∣≲ ‖Φ[ j]in‖2in,ν−δ0,lλ 2ν−2δ0∗ . (4.5.11)4.5.3 Solving the orthogonal equationsTo introduce the space for the parameter function p(t), we recall the non-local operator B0 appearsat mode 0 is of the approximate formB0[p] =ˆ t−λ 2∗−Tp˙(s)t− sds +O(|p˙(t)|).For Θ ∈ (0,1), ϖ ∈ R and a continuous function g : [−T,T ]→ C, we define the norm‖g‖Θ,ϖ = supt∈[−T,T ](T − t)−Θ| ln(T − t)|ϖ |g(t)|,and for α ∈ (0,1), m, ϖ ∈ R, we define the semi-norm[g] α2 ,m,ϖ = sup(T − t)−m| ln(T − t)|ϖ |g(t)−g(s)|(t− s) α2 ,where the supremum is taken over s≤ t in [−T,T ] such that t− s≤ 14(T − t).The following proposition proved in [19, Proposition 6.5, Proposition 6.6] gives an approximateinverse of the non-local operatorB0 with a small remainderR0.Proposition 4.5.1. Let α0, α2 ∈ (0, 12), ϖ ∈ R. There is ♭ > 0 such that if Θ ∈ (0, ♭) and m ≤ Θ− α2 ,then for h(t) : [0,T ]→ C satisfying{|h(T )|> 0,TΘ| lnT |1+σ−ϖ‖h(·)−h(T )‖Θ,ϖ−1+[h] α2 ,m,ϖ−1 ≤C1,(4.5.12)for some C1 > 0, σ ∈ (0,1), then, for T > 0 sufficiently small there exist two operatorsP andR0 so206that p=P[h] : [−T,T ]→ C satisfiesB0[p](t) = h(t)+R0[h](t), t ∈ [0,T ]with|R0[h](t)| ≤C(Tσ +TΘln | lnT || lnT | ‖h(·)−h(T )‖Θ,ϖ−1+[h] α2 ,m,ϖ−1)(T − t)m+ (1+α0)α2| ln(T − t)|ϖ .Moreover,P[h] = p0,κ +P1[h]+P2[h],withp0,κ(t) = κ| lnT |ˆ Tt1| ln(T − s)|2 ds, t ≤ T,where κ = κ[h]. Moreover, the following bounds hold:κ = h(T )(1+O(| lnT |−1)) , |∂tP1[h](t)| ≤C | lnT |1−σ (ln(| lnT |))2| ln(T − t)|3−σ ,|∂ 2t P1[h](t)| ≤C| lnT || ln(T − t)|3(T − t) , ‖∂tP2[h]‖Θ,ϖ ≤C(T12+σ−Θ+‖h(·)−h(T )‖Θ,ϖ−1),[∂tP2[h]] α2 ,m,ϖ ≤C(| lnT |ϖ−3T ♭−m− α2 +TΘ ln | lnT || lnT | ‖h(·)−h(T )‖Θ,ϖ−1+[h] α2 ,m,ϖ−1).(4.5.13)One plugging (4.4.56), (4.5.4), (4.5.5) into (4.5.2), then (4.5.2) is rewritten asˆ t0p˙ j(s)t− s[(−1+O(R−20 + ι j〈ln ι j〉))1{ι j≤1}+O(ι−1j )1{ι j>1}]ds+(a− ib)eiγ j(t)Re[ˆ t0p˙ j(s)e−iγ j(t)t− s(O(R−40 + ι j〈ln ι j〉)1{ι j≤1}+O(ι−1j )1{ι j>1})ds]+Cp1(R0)p˙ j+Cp2(R0)eiγ j(t)λ˙ j+(a− ib)(1+O(R−20 )) [∂x1(Φout)1+∂x2(Φout)2+ i(∂x1(Φout)2−∂x2(Φout)1)] (q[ j], t)+R0[(a− ib)(1+O(R−20 )) [∂x1(Φout)1+∂x2(Φout)2+ i(∂x1(Φout)2−∂x2(Φout)1)] (q[ j], t)]− (2π)−1(ˆB2η(y)Z 20,1(y)dy)λ−1j eiγ j(t)c[ j]∗0(t) = 0.(4.5.14)207One plugging (4.5.7), (4.5.8) into (4.5.3), then (4.5.3) is rewritten as(ξ˙ [ j]1 − iξ˙ [ j]2 )(1+O(R−20 )) = λ jˆ 2R00(Q−γ j[(a−bU [ j]∧) L˜U [ j] [Φout]])C j,1 (ρ j, t)Z1,1(ρ j)ρ jdρ j+λ−1jˆ 2R00(H [ j]in )C j,1(ρ j, t)Z1,1(ρ j)ρ jdρ j− (2π)−1(ˆB2η(y)Z 21,1(y)dy)λ−1j c[ j]∗1(t).(4.5.15)We now impose constraints on the parameters such that contraction property in the non-localproblem can be obtained. We apply Proposition 4.5.1 to (4.5.14) withh[Ψout] := (a− ib) [∂x1(Φout)1+∂x2(Φout)2+ i(∂x1(Φout)2−∂x2(Φout)1)] (q[ j], t).The vanishing and Hölder properties (4.5.12) are exactly the ones inherited from the weighted topol-ogy (4.4.47) for the outer problem, namely|h[Ψout](t)−h[Ψout](T )|≲ λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2),|h[Ψout](t)−h[Ψout](s)||t− s|α/2 ≲ λ∗(t)Θ−α(1−β )+‖Z∗‖C3(R2) for |t− s|<T − t4.So it is then natural to choose in the [·] α2 ,m,ϖ−1-seminormm :=min{Θ−α(1−β ),0}=Θ−α(1−β ), (4.5.16)where we needΘ−α(1−β )< 0. (4.5.17)Then in the last line of (4.5.13), ♭−m− α2 > 0. In order for both ‖h[Ψout](·)− h[Ψout](T )‖Θ,ϖ−1,[h[Ψout]] α2 ,m,ϖ−1 to be finite, we needΘ<α2, ϖ −1−2Θ< 0, ϖ −1−2m< 0. (4.5.18)Also the assumption m≤Θ− α2 in Proposition 4.5.1 impliesβ ≤ 1/2, (4.5.19)which is in the desired self-similar regime as we require before. Recall the estimate of R0[h[Ψout]].We requirem+(1+α0)α2>Θ,namely,0< α0 < 1/2, 2β −1+α0 > 0, (4.5.20)208so that the vanishing order ofR0[a] as t→ T is faster than the leading part h[Ψout] itself.Under the parameter assumption above, (4.5.14) will be solved by Proposition 4.5.1 and the solu-tion p j meets the estimate in (4.5.13). Combining estimate (4.5.9), (4.5.10) and (4.5.11), we can finda solution ξ [ j] with the estimate|ξ˙ [ j]|≲ λ εξ∗ with 0< εξ <min{δ03,Θ,α(1−β ),2ν−2δ0−1}(4.5.21)providedν−δ0− 12 > 0. (4.5.22)We then conclude that with above choices of m, α0, α2 , ϖ , the remainder gains smallness|R0[h[Ψout]](t)|≲ λΘ+σ1∗ (4.5.23)compare to the leading part h[Ψout] itself, where0< σ1 < m+(1+α0)α2−Θ. (4.5.24)We put the remainder R0[h[Ψout]] in the non-orthogonal inner problem (4.4.55), where the extrasmallness measured above by σ1 is crucial to control the non-orthogonal part.In summary, the restrictions on the constants needed when dealing with the reduced problem aregiven by0< β <12, 2Θ< α, 0< α0 <12, 2β −1+α0 > 0, ν−δ0− 12 > 0. (4.5.25)4.6 Linear theory for the outer problem4.6.1 Fundamental solution for the outer problemConsiderLu= ∑|α|≤1,|β |≤1Aαβ (x, t)DαDβu in RdwhereDα = Dα1x1 · · ·Dαdxd , α = (α1, · · · ,αd), u= (u1, · · · ,un)trand, for each α , β , Aαβ (x, t) = [Aαβi j (x, t)]ni, j=1 is an n×n real matrix-valued function.209For X = (x, t) ∈ Rd+1 and r > 0, we define the (d+ 1)-dimensional cylinders Qr(X), Q+r (X),Q−r (X) byQr(X) = (t− r2, t+ r2)×Br(x), Q+r (X) = (t, t+ r2)×Br(x), Q−r (X) = (t− r2, t)×Br(x).DefineωxA(r,X) := Q−r (X)|A(y,s)− A¯xx,r(s)|dyds, where A¯xx,r(s) := Br(x)A(z,s)dz.Then for a set Q⊂ Rd+1, we defineωxA(r,Q) := sup{ωxA(r,X) : X ∈ Q} and ωxA(r) := ωxA(r,Rd+1).We say that A is of Dini mean oscillation in x over Q and write A ∈ DMOx(Q) if ωxA(r,Q) satisfiesthe Dini condition ˆ 10ωxA(r,Q)rdr <+∞. (4.6.1)Similarly, we define|ω|xA(r,X) := Q−r (X) Br(x)|A(y,s)−A(z,s)| dzdydsThen for a set Q⊂ Rd+1, we define|ω|xA(r,Q) := sup{|ω|xA(r,X) : X ∈ Q} and |ω|xA(r) := |ω|xA(r,Rd+1).We say that A is of Dini mean absolute oscillation in x over Q and write A ∈ |DMO|x(Q) if|ω|xA(r,Q) satisfies the Dini conditionˆ 10|ω|xA(r,Q)rdr <+∞. (4.6.2)We present some basic properties about DMOx(Q) and |DMO|x(Q) in the following Lemma.Lemma 4.6.1.1. |DMO|x(Q)⊂ DMOx(Q).2. If | f (x, t)− f (y, t)| ≤ C|x− y|α f for constants C > 0 and 0 < α f ≤ 1 and all (x, t),(y, t) ∈ Q,then f (x, t) ∈ |DMO|x(Q).3. For all f ,g ∈ DMOx(Q)(|DMO|x(Q)), c ∈ R, then f +g,c f ∈ DMOx(Q)(|DMO|x(Q)).2104. For f ∈ |DMO|x(Q), then | f | ∈ |DMO|x(Q).5. For f ∈ |DMO|x(Q) and | f | ≥ ε f > 0 uniformly in Q, then 1f , | f |θ ∈ |DMO|x(Q)with 0< θ < 1.6. For all f ,g ∈ |DMO|x(Q)∩L∞(Q), then f g, | f |θ ∈ |DMO|x(Q)∩L∞(Q) with θ > 1.Proof. The proof is straightforward by the definition.Compared with DMOx(Q), |DMO|x(Q) has the advantage that the functions in |DMO|x(Q) areclosed under arithmetic under some weak assumptions.Proposition 4.6.2. Assume the principal coefficients Aαβ ∈ DMOx(Rd+1) and satisfy |Aαβ | ≤ Λ fora constant Λ> 0 and Legendre-Hadamard ellipticity∑|α|=|β |=1θ trξαξ βAαβ (x, t)θ ≥ λ |ξ |2|θ |2 (4.6.3)for a constant λ > 0 and all (x, t) ∈ Rd+1, ξ ∈ Rd , and θ ∈ Rn. Then the parabolic systemut = Lu (4.6.4)has a fundamental solution Γ(x, t,y,s), which satisfies(∂t −L)Γ(·, ·,y,s) = 0 in Rd× (s,∞), limt→s+Γ(·, t,y,s) = δy(·) on Rd .Moreover, for any δ ∈ (0,1), there exists a universal constant c> 0 such that for 0< t− s≤ 1,(t− s)(|∂tΓ(x, t,y,s)|+ |D2xΓ(x, t,y,s)|)+(t− s) 12 |DxΓ(x, t,y,s)|+ |Γ(x, t,y,s)|≤C(d,λ ,Λ,ωxA,δ )(t− s)−d2 e−c( |x−y|√t−s)2−δ;(4.6.5)for s< t1 < t2 ≤ s+1, x1,x2 ∈ Rd ,|Γ(x1, t1,y,s)−Γ(x2, t2,y,s)|(|x1− x2|+√|t1− t2|)α≤C(α,d,λ ,Λ,ωxA,δ )(t2− s)−α2[(t1− s)− d2 e−c( |x1−y|√t1−s)2−δ+(t2− s)− d2 e−c( |x2−y|√t2−s)2−δ];(4.6.6)|(DxΓ)(x1, t1,y,s)− (DxΓ)(x2, t2,y,s)|(|x1− x2|+√|t1− t2|)α≤C(α,d,λ ,Λ,ωxA,δ )(t2− s)−α2[(t1− s)− d+12 e−c( |x1−y|√t1−s)2−δ+(t2− s)− d+12 e−c( |x2−y|√t2−s)2−δ];(4.6.7)211additionally, for f ∈C30(Rd) and Oupp f ⊂BC1 with a constantC1> 0, then (Γ∗ f )(x, t) :=´Rd Γ(x, t,y,s) f (y)dysatisfies for 0< t− s≤ 1, s< t1 < t2 ≤ s+1, x1,x2 ∈ Rd ,|Γ∗ f |+ |Dx (Γ∗ f )|+∣∣D2x (Γ∗ f )∣∣+ |∂t (Γ∗ f )|≲C(C1,d,λ ,Λ,ωxA,δ )‖ f‖C3(Rd), (4.6.8)|Dx (Γ∗ f)(x1, t1)−Dx (Γ∗ f)(x2, t2)|(|x1− x2|+√|t1− t2|)α ≲C(α,C1,d,λ ,Λ,ωxA,δ )‖f‖C3(Rd) (4.6.9)where α is an arbitrary number in (0,1) and constants depending on α could be unbounded asα → 1−.Proof. The existence of the fundamental solution is given by the generalization of [31, Theorem 1.3]to parabolic systems (4.6.4). Indeed, the authors in [31] also claimed that their methods apply toparabolic systems of the second order. For further details, W 2,p estimate for parabolic systems isgiven in [28], which can be used to generalize [31, Lemma 2.2] to parabolic systems. [31, Lemma2.3], which is given in [30, Theorem 3.3] can be generalized to parabolic systems.By [30, Theorem 3.2] and generalized version of [31, Theorem 1.3], we have (4.6.5), whereC(d,λ ,Λ,ωxA,δ ),c > 0 will vary from line to line in the following calculation. The “C” in the proofdepends on d,λ ,Λ,ωxA,δ and for brevity, we will not stress this dependence by writing these explic-itly.For any (x∗, t∗),(y,s) ∈ Rd+1, s < t∗, set ρ∗ = (t∗− s) 12 . Consider Γ(x∗+ρ∗z, t∗+ρ2∗τ,y,s) as afunction of z,τ . For p> n+2, set α1 = 1− n+2p . Thenρ−1−α1∗ ‖Γ(x∗+ρ∗z, t∗+ρ2∗τ,y,s)‖Lp(B(0, 12 )×(− 14 ,0))≲ (t∗− s)−1+α12 ‖(t∗+ρ2∗τ− s)−d2 e−c( |x∗+ρ∗z−y|√t∗+ρ2∗ τ−s)2−δ‖Lp(B(0, 12 )×(− 14 ,0))≲(t∗− s)−d+1+α12 if |x∗− y| ≤ (t∗− s) 12(t∗− s)−d+1+α12 e−c(|x∗−y|√t∗−s )2−δif |x∗− y|> (t∗− s) 12∼ (t∗− s)−d+1+α12 e−c(|x∗−y|√t∗−s )2−δ,where we have used34(t∗− s)≤ t∗+ρ2∗τ− s≤ t∗− s,|x∗+ρ∗z− y|√t∗+ρ2∗τ− s≲ 1 if |x∗− y| ≤ (t∗− s)12∼ |x∗−y|√t∗−s if |x∗− y|> (t∗− s)12 .212Similarly,ρ−α1∗ ‖(DxΓ)(x∗+ρ∗z, t∗+ρ2∗τ,y,s)‖Lp(B(0, 12 )×(− 14 ,0))+ρ1−α1∗ ‖(D2xΓ)(x∗+ρ∗z, t∗+ρ2∗τ,y,s)‖Lp(B(0, 12 )×(− 14 ,0))+ρ1−α1∗ ‖(∂tΓ)(x∗+ρ∗z, t∗+ρ2∗τ,y,s)‖Lp(B(0, 12 )×(− 14 ,0)) ≲ (t∗− s)− d+1+α12 e−c(|x∗−y|√t∗−s )2−δ.By the Sobolev embedding theorem (see [30, Lemma 2.1]),supx1,x2∈B(x∗, (t∗−s)122 ),t1,t2∈(t∗− t∗−s4 ,t∗)|(DxΓ)(x1, t1,y,s)− (DxΓ)(x2, t2,y,s)|(|x1− x2|+√|t1− t2|)α1 ≲C(α1)(t∗−s)− d+1+α12 e−c(|x∗−y|√t∗−s )2−δ.For (x1, t1) /∈ B(x∗, (t∗−s)122 )× (t∗− t∗−s4 , t∗), by (4.6.5), we have|(DxΓ)(x1, t1,y,s)− (DxΓ)(x∗, t∗,y,s)|(|x1− x∗|+√|t1− t∗|)α1 ≲ (t∗−s)− α12 {(t1−s)− d+12 e−c(|x1−y|√t1−s )2−δ+(t∗−s)− d+12 e−c(|x∗−y|√t∗−s )2−δ}.Similarly, we have|Γ(x1, t1,y,s)−Γ(x∗, t∗,y,s)|(|x1− x∗|+√|t1− t∗|)α2 ≲C(α2)(t∗− s)− d+α22 e−c(|x∗−y|√t∗−s )2−δfor (x1, t1) ∈ B(x∗, (t∗−s)122 )× (t∗− t∗−s4 , t∗), whereα2 =2− n+2p if p< n+21− ε for any ε ∈ (0,1), if p≥ n+2.For (x1, t1) /∈ B(x∗, (t∗−s)122 )× (t∗− t∗−s4 , t∗), by (4.6.5), we get|Γ(x1, t1,y,s)−Γ(x∗, t∗,y,s)|(|x1− x∗|+√|t1− t∗|)α2 ≲ (t∗− s)− α22 {(t1− s)− d2 e−c(|x1−y|√t1−s )2−δ+(t∗− s)− d2 e−c(|x∗−y|√t∗−s )2−δ}.By (4.6.5), (4.11.2), then |Γ∗ f |≲ ‖ f‖L∞ . Combining [28, Theorem 2] (where we used Oupp f ⊂BC1), [30, Theorem 3.2], we conclude (4.6.8). ByW1,2p estimate, it follows (4.4.32).4.6.2 Properties about the leading coefficientsIn this section, we will prove the following Proposition.213Proposition 4.6.3. Suppose T sufficiently small, λ j given in (4.3.2), |Φ| a, Φ[ j]in ∈ B[ j]in defined in(4.4.43) and parameter assumption (4.4.42), Φout ∈ Bout defined in (4.4.48), andν−δ0− 12 > 0, (4.6.10)then BΦ,U∗ defined in (4.4.26) satifies (4.6.3) and BΦ,U∗ ∈ |DMO|x(R2+1)∩L∞(R2+1)Proof. SetA(1,0)(1,0) = A(0,1)(0,1) = aI3−bU∗∧,and Aαβ = 0 for all (α,β ) 6= ((1,0),(1,0)),((0,1),(0,1)). Then for any f ∈ R2, g ∈ R3,∑|α|=|β |=1gtrfα fβAαβg= gtr[|f|2(aI3−bU∗∧)g] = |f|2(a|g|2−bgtr[U∗∧g])= a|f|2|g|2.By (4.4.27), (4.4.5), then |B˜Φ,U∗ |=O(λ∗+ |Φ|). One taking T small and |Φ| a, then BΦ,U∗ satifies(4.6.3).Next we will prove the coefficients in BΦ,U∗ belong to |DMO|x(R2+1)∩L∞(R2+1). RecallU [ j] =1|y[ j]|2+1[2y[ j]|y[ j]|2−1],where2y[ j]|y[ j]|2+1 =2λ j(t)(x−ξ [ j](t))|x−ξ [ j](t)|2+λ 2j (t),|y[ j]|2−1|y[ j]|2+1 =|x−ξ [ j](t)|2−λ 2j (t)|x−ξ [ j](t)|2+λ 2j (t)= 1− 2λ2j (t)|x−ξ [ j](t)|2+λ 2j (t).ObviouslyU [ j] ∈ L∞(R2+1). In order to getU [ j] ∈ |DMO|x(R2+1), it suffices to proveλ 2j (t)|x−ξ [ j](t)|2+λ 2j (t),λ j(t)(x−ξ [ j](t))|x−ξ [ j](t)|2+λ 2j (t)∈ |DMO|x(R2+1). (4.6.11)Proof of (4.6.11):∣∣∣∣∣ λ 2j (s)|w−ξ [ j](s)|2+λ 2j (s) − λ2j (s)|z−ξ [ j](s)|2+λ 2j (s)∣∣∣∣∣= λ 2j (s)∣∣|z−ξ [ j](s)|2−|w−ξ [ j](s)|2∣∣(|w−ξ [ j](s)|2+λ 2j (s))(|z−ξ [ j](s)|2+λ 2j (s))≤ |w− z|λ 2j (s)|z−ξ [ j](s)|+ |w−ξ [ j](s)|(|w−ξ [ j](s)|2+λ 2j (s))(|z−ξ [ j](s)|2+λ 2j (s))214≲ |w− z|λ j(s)[(∣∣w−ξ [ j](s)∣∣2+λ 2j (s))−1+(∣∣z−ξ [ j](s)∣∣2+λ 2j (s))−1] .Then Q−r (X) Br(x)∣∣∣∣∣ λ 2j (s)|w−ξ [ j](s)|2+λ 2j (s) −λ 2j (s)|z−ξ [ j](s)|2+λ 2j (s)∣∣∣∣∣dzdwds≲ r Q−r (X) Br(x)λ j(s)[(∣∣w−ξ [ j](s)∣∣2+λ 2j (s))−1+(∣∣z−ξ [ j](s)∣∣2+λ 2j (s))−1]dzdwds∼ r tt−r2 Br(x)λ j(s)(∣∣z−ξ [ j](s)∣∣2+λ 2j (s))−1 dzds≤ r tt−r2 Br(0)λ j(s)(|z|2+λ 2j (s))−1dzds∼ r−3ˆ tt−r2λ j(s)ˆ r0(v2+λ 2j (s))−1vdvds∼ r−3ˆ tt−r2λ j(s) ln(1+λ−2j (s)r2)ds.(4.6.12)Then it suffices to prove the following integral is bounded.ˆ 10r−4ˆ tt−r2λ j(s) ln(1+λ−2j (s)r2)dsdr =ˆ 10r−4ˆ t0λ j(s) ln(1+λ−2j (s)r2)1{s≥t−r2}dsdr=ˆ t0ˆ 10r−4λ j(s) ln(1+λ−2j (s)r2)1{r≥(t−s) 12 }drds=(ˆ tt−(T−t)+ˆ t−(T−t)0)λ−2j (s)ˆ 1λ j(s)(t−s)12λ j(s)z−4 ln(1+ z2)dzds.Recall λ j in (4.3.2). For the first part, since T − t ≤ T − s≤ 2(T − t), there exist constants c1,c2 > 0such thatˆ tt−(T−t)λ−2j (s)ˆ 1λ j(s)(t−s)12λ j(s)z−4 ln(1+ z2)dzds≲(ˆ t−λ 2j (t)t−(T−t)+ˆ tt−λ 2j (t))λ−2j (t)ˆ c1λ j(t)c2(t−s)12λ j(t)z−4 ln(1+ z2)dzds≲ˆ t−λ 2j (t)t−(T−t)λ−2j (t)((t− s) 12λ j(t))−3 ln(1+((t− s) 12λ j(t))2)ds+1≲ˆ (T−t) 12λ j(t)1y−2 ln(1+ y2)dy+1≲ 1.For the second part, since T−s2 ≤ t− s≤ T − s,ˆ t−(T−t)0λ−2j (s)ˆ 1λ j(s)(t−s)12λ j(s)z−4 ln(1+ z2)dzds≤ˆ t−(T−t)0λ−2j (s)ˆ 1λ j(s)(T−s)12√2λ j(s)z−4 ln(1+ z2)dzds≲ˆ t−(T−t)0λ j(s)(T − s)− 32 ln(1+((T − s) 12√2λ j(s))2)ds≲ 1.215Next for i= 1,2,∣∣∣∣∣ λ j(s)(wi−ξ [ j]i (s))|w−ξ [ j](s)|2+λ 2j (s)− λ j(s)(zi−ξ [ j]i (s))|z−ξ [ j](s)|2+λ 2j (s)∣∣∣∣∣= λ j(s)∣∣∣∣∣∣(wi−ξ [ j]i (s))(|z−ξ [ j](s)|2+λ 2j (s))− (zi−ξ [ j]i (s))(|w−ξ [ j](s)|2+λ 2j (s))(|w−ξ [ j](s)|2+λ 2j (s))(|z−ξ [ j](s)|2+λ 2j (s))∣∣∣∣∣∣≤ |w− z|λ j(s)|z−ξ [ j](s)|2+λ 2j (s)+ |z−ξ [ j](s)|(|w−ξ [ j](s)|+ |z−ξ [ j](s)|)(|w−ξ [ j](s)|2+λ 2j (s))(|z−ξ [ j](s)|2+λ 2j (s))≲ |w− z|λ j(s)[(∣∣w−ξ [ j](s)∣∣2+λ 2j (s))−1+(∣∣z−ξ [ j](s)∣∣2+λ 2j (s))−1] .Thus we conclude that λ j(t)(x−ξ[ j](t))|x−ξ [ j](t)|2+λ 2j (t)∈ |DMO|x(R2+1) by the same reason as (4.6.12). Thus, wecomplete (4.6.11). It follows thatU∗ ∈ |DMO|x(R2+1)∩L∞(R2+1).By Lemma 4.6.1, the multiplicity of finite many terms of the component of U∗ also belongsto |DMO|x(R2+1)∩L∞(R2+1). By Lemma 4.6.1(2) and (4.3.24), thenN∑j=1Φ∗[ j]0 ∈ |DMO|x(R2+1)∩L∞(R2+1); similarly forΦout ∈Bout defined in (4.4.48), we haveΦout,η [ j]dq ∈ |DMO|x(R2+1)∩L∞(R2+1).For Φ[ j]in ∈ B[ j]in defined in (4.4.43) and parameter assumption (4.4.42), we have Q−r (X) Br(x)∣∣∣∣η(x−ξ [ j](s)λ j(s)R(s))Qγ jΦ[ j]in(x−ξ [ j](s)λ j(s),s)−η(z−ξ [ j](s)λ j(s)R(s))Qγ jΦ[ j]in(z−ξ [ j](s)λ j(s),s)∣∣∣∣dxdzds≲ r−2ˆ tt−r2rλ ν−δ0−1∗ (s)ds∼ r−1| lnT |ν−δ0−1ˆ T−t+r2T−tzν−δ0−1| lnz|2ν−2δ0−2 dz≲ | lnT |ν−δ0−1 r2ν−2δ0−1| lnr|2ν−2δ0−2which is a Dini function under the assumption (4.6.10). In this case,Φin ∈ |DMO|x(R2+1)∩L∞(R2+1).In sum, by Lemma 4.6.1 and (4.4.1), under the assumption (4.4.42), (4.6.10), we haveΦ ∈ |DMO|x(R2+1)∩L∞(R2+1). (4.6.13)Recalling the definition (4.4.4) for A, by (4.3.4), (4.6.13), |Φ| 1 and Lemma 4.6.1, we haveA ∈ |DMO|x(R2+1)∩L∞(R2+1). By (4.4.5), λ∗ in (4.3.2) and |Φ| 1, we have |A| 1. By a similarargument, BΦ,U∗ ∈ |DMO|x(R2+1)∩L∞(R2+1).2164.7 Solving the gluing system• Recall the space Bout defined in (4.4.48) for the outer problem, we choose Λo = 9Λo1 with Λo1 givenin (4.4.36). In order to find a solution for the outer problem (4.4.19), it suffices to find a fixed point for(4.4.34). One part of (4.4.34) (see Φ(1)out in (4.4.35)) has been handled in (4.4.36). For the remainingpart, we denoteΦ(2)out := ΓΦ,U∗ ∗∗G [Φout]+N∑m=13∑n=1cmn1[Φ,U∗,G [Φout]] (ΓΦ,U∗ ∗ϑmn) . (4.7.1)For Φ[ j]in ∈ B[ j]in, j = 1,2, . . . ,N, using Lemma 4.13.1, Proposition 4.12.1 to estimate ΓΦ,U∗ ∗ ∗G [Φout]and (4.4.33), (4.4.31), (4.4.32) to estimate cmn1[Φ,U∗,G [Φout]] (ΓΦ,U∗ ∗ϑmn) and the estimate (4.4.36)for Φ(1)out, we finally haveTo[Φout] =Φ(1)out+Φ(2)out ∈ Bout. (4.7.2)The regularity of the parabolic system with DMOx coefficients is guaranteed by [30, Theorem 1.2].Then Schauder fixed point theorem implies the existence of (4.4.19).• For the orthgonal inner problem (4.4.54), under the parameter assumption (4.13.84), by (4.13.85)and Proposition 4.8.1, we have ‖Φ[ j1]in ‖in,ν−δ0,l ≲ T ε .For the non-orthogonal inner problem (4.4.55), by (4.4.52),∣∣∣∣λ 2j (l˜#j,0[Φout])C−1j∣∣∣∣≲ λ∗(| lnT |−2τ−1j (lnτ j)4)α (λΘ−α(1−β )∗ +‖Z∗‖C3(R2))〈ρ j〉α−2‖Φout‖♯,Θ,α .By (4.13.86), ∣∣((H [ j]in )C j,0)C−1j ∣∣≲ ‖Φ[ j]in‖2in,ν−δ0,lλ 2ν−2δ0∗ 〈ρ j〉−2l−3Recall R0[Φout] given in (4.4.56) involving R0[h[Ψout]] with the estimate (4.5.23). By Proposition4.8.8, in order to find a solution Φ[ j2]in of (4.4.55) with the estimate ‖Φ[ j2]in ‖in,ν−δ0,l ≲ T ε , we require1+α+min{Θ−α(1−β ),0}−2β > ν−δ0, 2β +δ0−ν < 0,1+Θ−α(1−β )+ (1+α0)α2−2β > ν−δ0.(4.7.3)Plus (4.4.54) and (4.4.55) together and regard Φ[ j1]in +Φ[ j2]in as a whole. Suppose (4.13.95), by(4.13.96) and regularity results in [28], [30, Theorem 1.2], we have ‖Φ[ j1]in +Φ[ j2]in ‖in,ν−δ0,l,ςin ≲ T ε .Thus for T sufficiently small, we haveΦ[ j1]in +Φ[ j2]in ∈ B[ j]in.The compactness of this mapping is guaranteed by the similar reason as solving the outer problem.217Combining restrictions for the outer problem (4.4.45), (4.6.10), (4.13.39), (4.12.1), for the innerand nonlocal problems (4.4.42), (4.7.3), (4.13.84), (4.13.95), (4.5.25), we need to solve the followingparameter inequality:ν−δ0− 12 > 0, Θ+β +δ0−ν < 0, 3β < 1+Θ,0< δ0 < β < 12 , β (l+1)−1+ν−δ0−Θ> 0, Θ+2β −1< 0, 2β +δ0−ν < 0,0<Θ< β , 0< α < 1, Θ+ 12 −β − α2 < 0, 0< σ0,β −σ0− α2 < 0, 1−σ0− (1+α)(1−β )< 0, Θ+2σ0−β < 0,0< ν < 1, 0< l < 1, ν+β l−1< 0, Θ+1−β −ν > 0, ν−2δ0−β (ςin+ l)> 0,ςH ≤min{ςin,α} , 0< α0 < 12 , 2β −1+α0 > 0, 1+Θ−α(1−β )+ (1+α0)α2 −2β > ν−δ0.(4.7.4)This can be solved by using Mathematica. Indeed, sound choices satisfying all the restrictions aregiven by0<Θ<14,14< β <1+Θ4, 0< σ0 <β −Θ2, 0< δ0 <1−4Θ4,1−2β +δ0+Θ< ν < 14 (3−4β +4δ0+4Θ) ,1−β +δ0−ν+Θβ< l < 1max{1−2β ,2ν−2δ0+2β −1−2Θ}< α0 < 12 ,max{2Θ+1−2β ,2β −2σ0, β −σ01−β ,2(ν−δ0+2β −1−Θ)2β +α0−1}< α < 1,ςin <min{1,β−1 (ν−2δ0−β l)}, ςH ≤min{ςin,α} .(4.7.5)Finally, we have solved the inner-outer gluing system (4.4.19) and (4.4.20).4.8 Linear theory for the inner problemIn this section, we study the key linear theory for the inner problem (4.4.20). Since this section ispretty independent of the other parts, we abuse R for more general cases. Recall (4.4.37), in the timevariable τ j, (4.4.20) is the usual parabolic system. Since the inner problems for j = 1,2, . . . ,N allhave the same structure and they are “localized”, we omit the subscripts or superscripts “ j”, “[ j]” inthis section for brevity and all spatial derivative is about y. Consider ∂τΨ= (a−bW∧)(LinΨ)+H in DRΨ(y,τ) ·W (y) = H(y,τ) ·W (y) = 0 in DR, (4.8.1)whereLinf := ∆f+ |∇W |2f−2∇(W · f) ·∇W +2(∇W ·∇f)W, (4.8.2)218DR :={(y,τ) | τ ∈ (τ0,∞),y ∈ BR(τ)}, BR :={y ∈ R2 | |y| ≤ R(τ)} .Throughout this section, if no special statement, we always assume that v(τ),R(τ),R0(τ) ∈C1(τ0,∞) with the formv(τ) = a0τa1(lnτ)a2(ln lnτ)a3 · · · , R(τ) = b0τb1(lnτ)b2(ln lnτ)b3 · · · ,R0(τ) = c0τc1(lnτ)c2(ln lnτ)c3 · · · , for finite multiplicity,a0,b0,c0 > 0, a1 >−1, ai,bi,ci ∈ R, i= 1,2, . . . , R(τ),R0(τ) are non-decreasing,v(τ)> 0, 1 R0(τ) R(τ) τ 12−,v′ = O(τ−1v), R′ = O(τ−1R), R′0 = O(τ−1R0),ˆ ττ0v(s)ds= O(τv(τ))(4.8.3)Obviously, for f (τ) = v(τ),R(τ) or R0(τ), C−1f f (τ) ≤ f (s) ≤ C f f (τ) for all τ ≤ s ≤ 2τ , with aconstantC f > 1. For brevity, we write v= v(τ), R= R(τ), R0 = R0(τ).Suppose thatΨC(y,τ), HC(y,τ) defined by (4.2.15) have the folllowing Fourier expansion respec-tively,ΨC(y,τ) = ∑k∈Zψk(ρ,τ)eikθ , ψk(ρ,τ) :=ΨC,k(ρ,τ) = (2π)−1ˆ 2π0ΨC(ρeis,τ)e−iksds,HC(y,τ) = ∑k∈Zhk(ρ,τ)eikθ , hk(ρ,τ) := HC,k(ρ,τ) = (2π)−1ˆ 2π0HC(ρeis,τ)e−iksds(4.8.4)where ΨC,k and HC,k are defined in (4.2.17) andy= ρeiθ , ρ = |y|, θ = arctan(y2/y1).Using (4.2.16), we denoteΨk(y,τ) :=(ψk(ρ,τ)eikθ)C−1, Hk(y,τ) :=(hk(ρ,τ)eikθ)C−1for k∈Z; h>(y,τ) := ∑|k|≥2hk(ρ,τ)eikθ .(4.8.5)It is easy to see|Ψk|= |ψk|, |Hk|= |hk|. (4.8.6)For ℓ∈R and v(τ)> 0 and vectorial complex-valued function f, the weighted topology are defined219by‖f‖Rv,ℓ := sup(y,τ)∈DRv−1(τ)〈y〉ℓ|f(y,τ)|,[f]Rς ,v,ℓ := sup(y,τ)∈DRv−1(τ)〈y〉ℓ+ς sup(y1,τ1),(y2,τ2)∈B(y, |y|2 )×(τ− |y|24 ,τ)|f(y1,τ1)− f(y2,τ2)|(|y1− y2|+ |τ1− τ2| 12)ς , 0< ς < 1whereR =R(τ) is a scalar function which could depend on τ . By (4.8.4), (4.8.6), we have‖ψk‖Rv,ℓ = ‖Ψk‖Rv,ℓ, ‖hk‖Rv,ℓ = ‖Hk‖Rv,ℓ ≲ ‖H‖Rv,ℓ; for |k| ≥ 2, ‖hk‖Rv,ℓ ≲ ‖h>‖Rv,ℓ;|Ψ(y,τ)|≲ ∑k∈Z|Ψk(y,τ)| (4.8.7)In order to ensure the convergence for the summation in the last inequality, we have to make thedependence on k very clear in the estimate for the higher mode case. The key inner linear theory isstated as follows.Proposition 4.8.1. Consider∂τΨ= (a−bW∧)(LinΨ)+H+(1∑k=0eikθck(τ)η(|y|)Zk,1(|y|))C−1in DR,Ψ(y,τ) ·W (y) = H(y,τ) ·W (y) = 0 in DR,(4.8.8)where the Fourier expansion of H is given in (4.8.4). Suppose (4.8.3),‖h0‖Rv,ℓ,‖h1‖Rv,ℓ,‖h>‖Rv,ℓ,‖h−1‖Rv,ℓ−1 < ∞ with 1< ℓ < 3, ℓ−1 >32,then there exists linear mappings (Ψin,c0(τ),c1(τ))= (Tin[H],Tc0 [h0](τ),Tc1 [h1](τ)) solving (4.8.8),whereck(τ) =−(ˆB2η(y)Z 2k,1(y)dy)−1ˆB2R0hk(y,τ)Zk,1(y)dy+ c∗k[hk] for k = 0,1and c∗k[hk] depends on hk linearly and satisfies |c∗k[hk]|≲ R−ε00 v(τ)‖hk‖Rv,ℓ with 0< ε0 < ℓ−1 suffi-ciently small; Ψ satisfies the following estimate in DR/2〈y〉 |DΨ|+ |Ψ|≲ R50v(τ)〈y〉2−ℓ‖h0‖Rv,ℓ+R60v(τ)〈y〉2−ℓ‖h1‖Rv,ℓ+ v(τ)〈y〉2−ℓ‖h>‖Rv,ℓ+‖h−1‖Rv,ℓ−1v(τ)τ1−ℓ−12 + τ−ℓ−12´ τ2τ02v(s)ds if 32 < ℓ−1 < 2v(τ)(lnτ)2+ τ−1 lnτ´ τ2τ02v(s)ds if ℓ−1 = 2v(τ) lnτ+ τ−1´ τ2τ02v(s)ds if ℓ−1 > 2.Remark 4.8.1.1. We emphasize that the constants in “≲” above depend on ℓ and can be unbounded220as ℓ→ 1 or 3.The rest of this section is devoted to the proof of Proposition 4.8.1.4.8.1 Complex-valued form of the inner linear equationThe following Lemma is applied for transforming the inner problem from the parabolic system intothe complex-scalar form.Lemma 4.8.2. For Lin defined in (4.8.2) and Ψ ·W = 0, we have[(a−bW∧)(LinΨ)] ·W = 0 (4.8.9)and[(a−bW∧)(LinΨ)]C = (a− ib)LinΨC, (4.8.10)whereLinΨC :=[∂ρρ +1ρ∂ρ +1ρ2∂θθ − 1ρ2 + i2cosw(ρ)ρ2∂θ +8(ρ2+1)2]ΨC. (4.8.11)Then (4.8.1) is equivalent to the complex equation∂τΨC = (a− ib)LinΨC+HC in DR. (4.8.12)Under the Fourier expansion (4.8.4), thenLin[eikθψk]= eikθLkψk, (4.8.13)whereLk f := ∂ρρ f +∂ρ fρ+Vk(ρ) f , Vk(ρ) :=−(k+1)2ρ4+(2k2−6)ρ2+(k−1)2(ρ2+1)21ρ2. (4.8.14)It follows that∂τΨk = (a−bW∧)(LinΨk)+Hk (4.8.15)is equivalent to∂τψk = (a− ib)Lkψk+hk. (4.8.16)Proof. SetΨ(y,τ) = φ1(y,τ)E1(y)+φ2(y,τ)E2(y), that is, ΨC = φ1+ iφ2.221By (4.2.6), one has∆(φ1E1) = (∆φ1)E1+2(∂ρφ1∂ρE1+1ρ2∂θφ1∂θE1)+φ1(∂ρρ +∂ρρ+∂θθρ2)E1= (∆φ1)E1−2∂ρφ1wρW + 2ρ2 ∂θφ1 coswE2+φ1[−wρρW −w2ρE1−1ρwρW − 1ρ2 cosw(sinwW + coswE1)]=[∆φ1−φ1(w2ρ +cos2wρ2)]E1+2coswρ2∂θφ1E2+[−2wρ∂ρφ1−φ1(wρρ +wρρ+sinwcoswρ2)]W=(∂ρρφ1+1ρ∂ρφ1+1ρ2∂θθφ1− 1ρ2φ1)E1+2coswρ2∂θφ1E2+(−2wρ∂ρφ1− 2sinwcoswρ2 φ1)Wwhere we used wρρ +wρρ − sinwcoswρ2 = 0 in the last equality. Similarly,∆(φ2E2) = ∆(φ2)E2−2 1ρ2 ∂θφ2(sinwW + coswE1)−φ21ρ2E2= − 2coswρ2∂θφ2E1+(∂ρρφ2+1ρ∂ρφ2+1ρ2∂θθφ2− 1ρ2φ2)E2− 2sinwρ2 ∂θφ2W.Thus∆Ψ=(∂ρρφ1+1ρ∂ρφ1+1ρ2∂θθφ1− 1ρ2φ1−2coswρ2∂θφ2)E1+(∂ρρφ2+1ρ∂ρφ2+1ρ2∂θθφ2− 1ρ2φ2+2coswρ2∂θφ1)E2+(−2wρ∂ρφ1− 2sinwρ2 ∂θφ2−2sinwcoswρ2φ1)W.(4.8.17)By (4.2.6),∂ρW = wρE1, ∂θW = sinwE2, ∂ρΨ= (∂ρφ1)E1+(∂ρφ2)E2−φ1wρW,∂θΨ= (∂θφ1)E1+(∂θφ2)E2+φ1 coswE2−φ2(sinwW + coswE1),then we have2(∇W ·∇Ψ)W = 2(∂ρW ·∂ρΨ+ 1ρ2 ∂θW ·∂θΨ)W =(2sinwcoswρ2φ1+2wρ∂ρφ1+2sinwρ2∂θφ2)W.(4.8.18)Then plugging (4.2.3), (4.8.17), (4.8.18) into (4.8.2) and using (4.2.5), we have(a−bW∧)(LinΨ) ={∂ρρφ1+1ρ∂ρφ1+1ρ2∂θθφ1+[8(ρ2+1)2− 1ρ2]φ1− 2coswρ2 ∂θφ2}(aE1−bE2)+{∂ρρφ2+1ρ∂ρφ2+1ρ2∂θθφ2+[8(ρ2+1)2− 1ρ2]φ2+2coswρ2∂θφ1}(aE2+bE1)222which implies (4.8.9), (4.8.10) and the equivalence between (4.8.1) and (4.8.12). Finally, (4.8.13) and(4.8.16) are derived directly.The linearly independent kernelsZk,1,Zk,2 ofLk in (4.8.14) satisfying theWronskianW [Zk,1,Zk,2] =ρ−1 are given as follows:Z−1,1(ρ) = ρ2ρ2+1 , Z−1,2(ρ) =4ρ2(ρ2 ln(ρ)−1)−14ρ2(ρ2+1) , k =−1,Z0,1(ρ) = ρρ2+1 , Z0,2(ρ) =ρ4+4ρ2 ln(ρ)−12ρ(ρ2+1) , k = 0,Z1,1(ρ) = 1ρ2+1 , Z1,2(ρ) =ρ4+4ρ2+4ln(ρ)4(ρ2+1) , k = 1,Zk,1(ρ) = ρ1−kρ2+1 , Zk,2(ρ) =ρk−1ρ2+1(ρ42k+2 +ρ2k +12k−2), k 6=−1,0,1.(4.8.19)It is easy to see thatZk,1(ρ)∼ρ1−k if ρ → 0ρ−1−k if ρ → ∞ , Zk,2(ρ)∼ρk−12k−2 if ρ → 0ρk+12k+2 if ρ → ∞for k 6=−1,0,1.Recall (4.2.9) and (4.8.4) and notice Z0,1(ρ) =−12ρwρ . Then for mode 0,(h0(ρ,τ))C−1 ·Z0,1+ i(h0(ρ,τ))C−1 ·Z0,2 = ρwρh0(ρ,τ) =−2Z0,1(ρ)h0(ρ,τ). (4.8.20)Notice Z1,1(ρ) =−12wρ . For mode 1,(h1(ρ,τ)eiθ)C−1·Z1,1 = Re(h1(ρ,τ)eiθ )wρ cosθ + Im(h1(ρ,τ)eiθ )wρ sinθ ,(h1(ρ,τ)eiθ)C−1·Z1,2 = Re(h1(ρ,τ)eiθ )wρ sinθ − Im(h1(ρ,τ)eiθ )wρ cosθ ,whose equivalent complex form is given by(h1(ρ,τ)eiθ)C−1·Z1,1− i(h1(ρ,τ)eiθ)C−1·Z1,2 = wρh1(ρ,τ) =−2Z1,1(ρ)h1(ρ,τ). (4.8.21)The quadratic form corresponding toLk in BR is defined asQR,k( f , f ) = 2πˆ R0[|∂ρ f |2+ (k+1)2ρ4+(2k2−6)ρ2+(k−1)2(ρ2+1)2| f |2ρ2]ρdρ. (4.8.22)Specially,QR,0( f , f )= 2πˆ R0[|∂ρ f |2+ ρ4−6ρ2+1(ρ2+1)2| f |2ρ2]ρdρ = 2πˆ R0[|∂ρ f |2+ | f |2ρ2− 8(ρ2+1)2| f |2]ρdρ,223QR,1( f , f ) = 2πˆ R0[|∂ρ f |2+ 4(ρ2−1)(ρ2+1)2| f |2]ρdρ,QR,−1( f , f )= 2πˆ R0[|∂ρ f |2+ −4ρ2+4(ρ2+1)2| f |2ρ2]ρdρ = 2πˆ R0[|∂ρ f |2+4 | f |2ρ2− 4ρ2+12(ρ2+1)2| f |2]ρdρ.Define the following norms‖ f‖X(BR) =[2πˆ R0(|∂ρ f |2+ | f |2ρ2)ρdρ] 12,‖ f‖H10 (BR) =(2πˆ R0|∂ρ f |2ρdρ) 12, ‖ f‖L2(BR) =(2πˆ R0| f |2ρdρ) 12.SetX0(BR) ={f (ρ)∣∣∣ f (R) = 0, ‖ f‖X(BR) < ∞} .There exists the Sobolev embedding X0 → L∞ in [50, p 216]:‖ f‖2L∞(BR) ≤ ‖fρ‖L2(BR)‖∂ρ f‖L2(BR) ≤ ‖ f‖2X(BR) for f ∈ X0. (4.8.23)For |k| ≥ 2, it is easy to see‖ f‖2X(BR) ≤ QR,k( f , f ). (4.8.24)By [96, Lemma 4.2], QR,k( f , f )≥ 0 for all f ∈C2(BR)∩C(B¯R) with f = 0 on ∂BR and QR,k( f , f ) = 0implies f ≡ 0.4.8.2 Energy estimatesWe use the method in [16, Lemma 7.2] to analyze the first eigenvalue of QR,k.Lemma 4.8.3. LetλR,k = inff∈X0(BR)\{0}QR,k( f , f )‖ f‖2L2(BR)for k 6= 1, λR,1 = inff∈H10 (BR)\{0}QR,1( f , f )‖ f‖2L2(BR)for k = 1.λR,k is attained by a real-valued function in X0(BR) for k 6= 1 and H10 (BR) for k= 1. When R is large,λR,0 ∼ (R2 lnR)−1, λR,1 ∼ R−4, λR,−1 ≳ (R2 lnR)−1, λR,k ≳ |k|2R−2 for |k| ≥ 2.Proof. For any complex-valued function f = f1+ i f2 where f1 = Re f , f2 = Im f ,QR,k( f , f )‖ f‖2L2(BR)=QR,k( f1, f1)+QR,k( f2, f2)‖ f1‖2L2(BR)+‖ f2‖2L2(BR)≥min{QR,k( f1, f1)‖ f1‖2L2(BR),QR,k( f2, f2)‖ f2‖2L2(BR)}.224Thus for k 6= 1,λR,k = inf{QR,k( f , f )‖ f‖2L2(BR)| f ∈ X0(BR)\{0}, f is real-valued}.The same argument can be applied to λR,1.Hereafter, we focus on real-valued functions. We choose a sequence fn ∈ X0(BR) if k 6= 1 ( fn ∈H10 (BR) if k = 1) with ‖ fn‖L2(BR) = 1 and λR,k+1≥ QR,k( fn, fn)→ λR,k. By the definition (4.8.22),ˆ R0(∂ρ fn)2ρdρ ≲ λR,k+1.The Sobolev compact embedding theorem implies fn → f∞ in L2(BR) up to a subsequence.For k 6= 1,QR,k( fn, fn) = 2πˆ R0[(∂ρ fn)2+(k−1)2(ρ2+1)2f 2nρ2+(k+1)2ρ2+(2k2−6)(ρ2+1)2f 2n]ρdρ.Up to a subsequence, we haveˆ R0[(∂ρ f∞)2+(k−1)2(ρ2+1)2f 2∞ρ2]ρdρ ≤ liminfn→∞ˆ R0[(∂ρ fn)2+(k−1)2(ρ2+1)2f 2nρ2]ρdρ,ˆ R0(k+1)2ρ2+(2k2−6)(ρ2+1)2f 2∞ρdρ = limn→∞ˆ R0(k+1)2ρ2+(2k2−6)(ρ2+1)2f 2n ρdρ.Moreover,ˆ R0[(∂ρ f∞)2+(k−1)2(ρ2+1)2f 2∞ρ2]ρdρ ∼C(R,k)ˆ R0[(∂ρ f∞)2+f 2∞ρ2]ρdρ.ThusQR,k( f∞, f∞)≤ λR,k, ‖ f∞‖L2(BR) = 1, f∞ ∈ X0(BR),which implies the minima λR,k is attained by f∞.For k = 1, similarly, we choose a subsequence such that fn ⇀ f∞ in H10 (BR), fn → f∞ in L2(BR).ˆ R0(∂ρ f∞)2ρdρ ≤ liminfn→∞ˆ R0(∂ρ fn)2ρdρ,ˆ R04(ρ2−1)(ρ2+1)2f 2∞ρdρ = limn→∞ˆ R04(ρ2−1)(ρ2+1)2f 2n ρdρ.ThenQR,1( f∞, f∞)≤ λR,1, ‖ f∞‖L2(BR) = 1, f∞ ∈ H10 (BR).Thus f∞ attains λR,1.225Next, we will use the Lagrange multiplier for the real-valued minimum function f∞ to estimateλR,k, k = −1,0,1. In order to avoid confusion, we denote wk as the eigenfunction corresponding tothe eigenvalue λR,k for every mode k with the normalization ‖wk‖L2(BR) = 1.For k = 0,L0w0 =−λR,0w0 in BR, w0 = 0 on ∂BR.w0 is given byw0(ρ) =Z0,2(ρ)ˆ ρ0(−λR,0w0(s))Z0,1(s)sds+Z0,1(ρ)ˆ Rρ(−λR,0w0(s))Z0,2(s)sds−AR,0Z0,1(ρ)whereAR,0 = (Z0,1(R))−1Z0,2(R)ˆ R0(−λR,0w0(s))Z0,1(s)sds.For 0≤ ρ ≤ 1,|Z0,2(ρ)ˆ ρ0w0(s)Z0,1(s)sds|≲ ρ−1‖w0‖L2(Bρ )‖Z0,1‖L2(Bρ ) ≲ 1,|Z0,1(ρ)ˆ Rρw0(s)Z0,2(s)sds| ≤ |Z0,1(ρ)ˆ R1w0(s)Z0,2(s)sds|+|Z0,1(ρ)ˆ 1ρw0(s)Z0,2(s)sds|≲R2,|AR,0Z0,1(ρ)|≲ |AR,0|≲ λR,0R2(lnR) 12 .Thus ‖w0‖L2(B1) ≲ λR,0R2(lnR)12 .For ρ ≥ 1,‖Z0,2(ρ)ˆ ρ0w0(s)Z0,1(s)sds‖L2(BR\B1) ≤ ‖Z0,2‖L2(BR\B1)‖w0‖L2(BR)‖Z0,1‖L2(BR) ≲ R2(lnR)12 ,‖Z0,1(ρ)ˆ Rρw0(s)Z0,2(s)sds‖L2(BR\B1) ≲ ‖Z0,1‖L2(BR\B1)‖w0‖L2(BR\B1)‖Z0,2‖L2(BR\B1) ≲ R2(lnR)12 ,‖AR,0Z0,1(ρ)‖L2(BR\B1) ≲ λR,0R2 lnR.In sum, when R is large, we have1= ‖w0‖L2(BR) ≲ λR,0R2 lnR.On the other hand, when R is large,‖η R2Z0,1‖2L2(BR) ∼ lnR,226QR,0(η R2Z0,1,η R2Z0,1)∼(ˆ R20+ˆ RR2)[(∂ρ(η R2Z0,1))2+ρ4−6ρ2+1(ρ2+1)2(η R2Z0,1)2ρ2]ρdρ= −ˆ ∞R2[(∂ρZ0,1)2+ρ4−6ρ2+1(ρ2+1)2(Z0,1)2ρ2]ρdρ+ˆ RR2[(∂ρ(η R2Z0,1))2+ρ4−6ρ2+1(ρ2+1)2(η R2Z0,1)2ρ2]ρdρ ∼ R−2,where we usedL0Z0,1 = 0. Then we haveλR,0 ≲ (R2 lnR)−1.For k = 1,L1w1 =−λR,1w1 in BR, w1 = 0 on ∂BR.w1 is written asw1(ρ) =Z1,2(ρ)ˆ ρ0(−λR,1w1(s))Z1,1(s)sds+Z1,1(ρ)ˆ Rρ(−λR,1w1(s))Z1,2(s)sds−AR,1Z1,1(ρ),(4.8.25)whereAR,1 = (Z1,1(R))−1Z1,2(R)ˆ R0(−λR,1w1(s))Z1,1(s)sds.For 0≤ ρ ≤ 1,|Z1,2(ρ)ˆ ρ0w1(s)Z1,1(s)sds|≲ (| lnρ|+1)‖w1‖L2(Bρ )‖Z1,1‖L2(Bρ ) ≲ 1,|Z1,1(ρ)ˆ Rρw1(s)Z1,2(s)sds| ≤ |Z1,1(ρ)ˆ R1w1(s)Z1,2(s)sds|+|Z1,1(ρ)ˆ 1ρw1(s)Z1,2(s)sds|≲R3,|AR,1Z1,1(ρ)|≲ |AR,1|≲ λR,1R4.Thus ‖w1‖L2(B1) ≲ λR,1R4.For ρ ≥ 1,‖Z1,2(ρ)ˆ ρ0w1(s)Z1,1(s)sds‖L2(BR\B1) ≤ ‖Z1,2‖L2(BR\B1)‖w1‖L2(BR)‖Z1,1‖L2(BR) ≲ R3,‖Z1,1(ρ)ˆ Rρw1(s)Z1,2(s)sds‖L2(BR\B1) ≲ ‖Z1,1‖L2(BR\B1)‖w1‖L2(BR\B1)‖Z1,2‖L2(BR\B1) ≲ R3,‖AR,1Z1,1(ρ)‖L2(BR\B1) ≲ λR,1R4.227In sum, when R is large, we have1= ‖w1‖L2(BR) ≲ λR,1R4.On the other hand, when R is large,‖η R2Z1,1‖2L2(BR) ∼ 1,QR,1(η R2Z1,1,η R2Z1,1)∼(ˆ R20+ˆ RR2)[(∂ρ(η R2Z1,1))2+4(ρ2−1)(ρ2+1)2(η R2Z1,1)2]ρdρ= −ˆ ∞R2[(∂ρZ1,1)2+4(ρ2−1)(ρ2+1)2(Z1,1)2]ρdρ+ˆ RR2[(∂ρ(η R2Z1,1))2+4(ρ2−1)(ρ2+1)2(η R2Z1,1)2]ρdρ ∼ R−4,where we usedL1Z1,1 = 0. ThusλR,1 ≲ R−4.For k =−1,L−1w−1 =−λR,−1w−1 in BR, w−1 = 0 on ∂BR.w−1(ρ) is written asw−1(ρ) = Z−1,2(ρ)ˆ ρ0(−λR,−1w−1(s))Z−1,1(s)sds+Z−1,1(ρ)ˆ Rρ(−λR,−1w−1(s))Z−1,2(s)sds−AR,−1Z−1,1(ρ)whereAR,−1 = (Z−1,1(R))−1Z−1,2(R)ˆ R0(−λR,−1w−1(s))Z−1,1(s)sds.For 0≤ ρ ≤ 1,|Z−1,2(ρ)ˆ ρ0w−1(s)Z−1,1(s)sds|≲ ρ−2‖w−1‖L2(Bρ )‖Z−1,1‖L2(Bρ ) ≲ 1,|Z−1,1(ρ)ˆ Rρw−1(s)Z−1,2(s)sds|≤ |Z−1,1(ρ)ˆ R1w−1(s)Z−1,2(s)sds|+ |Z−1,1(ρ)ˆ 1ρw−1(s)Z−1,2(s)sds|≲ R lnR,|AR,−1Z−1,1(ρ)|≲ |AR|≲ λR,−1R lnR.228Thus ‖w−1‖L2(B1) ≲ λR,−1R lnR.For ρ ≥ 1,‖Z−1,2(ρ)ˆ ρ0w−1(s)Z−1,1(s)sds‖L2(BR\B1) ≤ ‖Z−1,2‖L2(BR\B1)‖w−1‖L2(BR)‖Z−1,1‖L2(BR) ≲ R2 lnR,‖Z−1,1(ρ)ˆ Rρw−1(s)Z−1,2(s)sds‖L2(BR\B1)≲ ‖Z−1,1‖L2(BR\B1)‖w−1‖L2(BR\B1)‖Z−1,2‖L2(BR\B1) ≲ R2 lnR,‖ARZ−1,1(ρ)‖L2(BR\B1) ≲ λR,−1R2 lnR.Thus when R is large, we have1= ‖w−1‖L2(BR) ≲ λR,−1R2 lnR.For |k| ≥ 2,QR,k( f , f )≳ 2π|k|2ˆ R0| f |2ρ2ρdρ ≥ |k|2R−2‖ f‖2L2(BR).Lemma 4.8.4. Consider ∂τϕk = (a− ib)Lkϕk+h(ρ,τ) in DR,ϕk = 0 on ∂DR, ϕk(·,τ0) = 0 in BR(τ0) (4.8.26)with R given in (4.8.3). Assume ‖h(·,τ)‖2L2(BR) ≲ g(τ), g(s)∼ g(τ) for τ ≤ s≤ τ+1,ˆ ττ0ec´ s λ˜R,k(λ˜R,k)−1g(s)ds≲C(c)ec´ τ λ˜R,k min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ) (4.8.27)for any fixed constant c> 0, a constant C(c)> 0 depending on c andλ˜R,0 = (R2 lnR)−1, λ˜R,1 = R−4, λ˜R,−1 = (R2 lnR)−1, λ˜R,k = |k|2R−2 for |k| ≥ 2.Then we have the following estimates‖ϕk(·,τ)‖L∞(BR) ≲[min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ)]1/2for k 6= 1, (4.8.28)‖ϕ1(·,τ)‖L2(BR)+‖ϕ1(·,τ)‖H10 (BR) ≲[min{τ,(λ˜R,1)−1}(λ˜R,1)−1g(τ)]1/2. (4.8.29)229In particular,‖ϕ0(·,τ)‖L∞(BR) ≲ R2 lnRθR,ℓv(τ)‖h‖Rv,ℓ, ‖ϕ−1(·,τ)‖L∞(BR) ≲ R2 lnRθR,ℓv(τ)‖h‖Rv,ℓ,‖ϕk(·,τ)‖L∞(BR) ≲ |k|−2R2θR,ℓv(τ)‖h‖Rv,ℓ for |k| ≥ 2,‖ϕ1(·,τ)‖L2(BR)+‖ϕ1(·,τ)‖H10 (BR) ≲min{τ12 ,R2}R2θR,ℓv(τ)‖h‖Rv,ℓwhereθR,ℓ :=C(ℓ) if ℓ > 1(lnR)12 if ℓ= 1C(ℓ)R1−ℓ if ℓ < 1,(4.8.30)C(ℓ) can be unbounded as ℓ→ 1.Proof. By continuity argument, it suffices to assume that h is smooth. For k = 1, by the parabolictheory, there exists a unique solution ϕ1 ∈ H10 (BR). For k 6= 1, setting ϕk = ρ |k−1|ϕk1, then we canprove there exists a unique solution ϕk ∈ X0(BR). Multiplying ϕ¯k and integrating by parts, we haveˆBR∂τϕkϕ¯k+(a− ib)QR,k(ϕk,ϕk) =ˆBRhϕ¯k.We take the real part for both parts.12∂τˆBR|ϕk|2+aQR,k(ϕk,ϕk) =ˆBRRe(hϕ¯k). (4.8.31)By Lemma 4.8.3, we have∂τˆBR|ϕk|2+ cλ˜R,kˆBR|ϕk|2 ≤ 2ˆBR|h||ϕk|for a fixed constant c> 0. By Young inequality, we have∂τˆBR|ϕk|2+ cλ˜R,kˆBR|ϕk|2 ≲ (λ˜R,k)−1ˆBR|h|2 ≲ (λ˜R,k)−1g(τ).Since ϕk(·,τ0) = 0 in BR(τ0), by (4.8.27), we haveˆBR|ϕk|2 ≲min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ). (4.8.32)Next we will estimate QR,k(ϕk,ϕk). Integrating (4.8.31) from τ to τ+1, we have12(ˆBR|ϕk|2(τ+1)−ˆBR|ϕk|2(τ))+aˆ τ+1τQR,k(ϕk,ϕk) =ˆ τ+1τˆBRRe(hϕ¯k),230which derives ˆ τ+1τQR,k(ϕk,ϕk)≲min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ).So there exists τ0 ∈ (τ,τ+1) such thatQR,k(ϕk,ϕk)(τ0)≲min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ). (4.8.33)We multiplyLkϕk to (4.8.26) and integrate by parts.−ˆBR∇∂τϕk ·∇ϕ¯k−ˆBR(k+1)2|y|4+(2k2−6)|y|2+(k−1)2(|y|2+1)2ϕ¯k∂τϕk|y|2 =(a−ib)ˆBR|Lkϕk|2+ˆBRLkϕkh.Taking the real part of the equation, we have−12∂τˆBR|∇ϕk|2− 12∂τˆBR(k+1)2|y|4+(2k2−6)|y|2+(k−1)2(|y|2+1)2|ϕk|2|y|2 = aˆBR|Lkϕk|2+ˆBRRe(Lkϕkh).That is∂τQR,k(ϕk,ϕk) =−2aˆBR|Lkϕk|2−2ˆBRRe(Lkϕkh). (4.8.34)By Young inequality, we derive∂τQR,k(ϕk,ϕk)≲ˆBR|h|2 ≲ g(τ). (4.8.35)Combing (4.8.33) and (4.8.35), we haveQR,k(ϕk,ϕk)(τ+1)≲min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ).Since τ is arbitrary here, we getQR,k(ϕk,ϕk)(τ)≲min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ) if τ > τ0+1. (4.8.36)For τ0 ≤ τ ≤ τ0+1, by (4.8.35) and ϕk(·,τ0) = 0 in BR(τ0), we haveQR,k(ϕk,ϕk)(τ)≲ g(τ). (4.8.37)Combining (4.8.23), (4.8.32), (4.8.36) and (4.8.37), we have‖ϕk(·,τ)‖L∞(BR) ≲ ‖ϕk(·,τ)‖X(BR) ≲[min{τ,(λ˜R,k)−1}(λ˜R,k)−1g(τ)]1/2for k 6= 1,‖ϕ1(·,τ)‖H10 (BR) ≲[min{τ,(λ˜R,1)−1}(λ˜R,1)−1g(τ)]1/2.231Lemma 4.8.5. Consider∂τϕ = (a− ib)(∂ρρϕ +1ρ∂ρϕ − k2ρ2ϕ)+h(ρ,τ), ϕ(ρ,τ0) = g(ρ)where k ≥ 0. Then using Γ♮d defined in (1.0.25), we can find a solution ϕ given byϕ(ρ,τ) = ρkΓ♮2k+2 ∗∗(|y|−kh(y,s))(ρ,τ,τ0)+ρkΓ♮2k+2 ∗(|y|−kg(y))(ρ,τ,τ0). (4.8.38)Proof. Set ϕ(ρ,τ) = ρkψ(ρ,τ). Then∂τψ = (a− ib)(∂ρρψ+2k+1ρ∂ρψ)+ρ−kh(ρ,τ), ψ(ρ,τ0) = ρ−kg(ρ) (4.8.39)which can be regarded as the heat equation in R2k+2. Then ψ is given byψ(ρ,τ) = Γ♮2k+2 ∗∗(|y|−kh(y,s))(ρ,τ,τ0)+Γ♮2k+2 ∗(|y|−kg(y))(ρ,τ,τ0),which satisfies (4.8.39) in weak sense and pointwise sense except at ρ = 0. (4.8.38) follows.4.8.3 Higher modes |k| ≥ 2In order to analyze the case that the right hand side of the equation has singularity at y = 0, givenR =R(τ), we introduce the norm‖h‖Rv,ℓ1,ℓ := sup(y,τ)∈DRv(τ)−1(1{|y|≤1}|y|ℓ1 +1{|y|>1}|y|ℓ)|h(y,τ)|.We use the notation ‖h‖∞v,ℓ1,ℓ if R(τ) = ∞. Obviously, ‖h‖Rv,0,ℓ ∼ ‖h‖Rv,ℓ; ‖h‖Rv,ℓ1,ℓ ≲ ‖h‖Rv,ℓ if ℓ1 > 0;‖h‖Rv,ℓ1,ℓ ≳ ‖h‖Rv,ℓ if ℓ1 < 0.Lemma 4.8.6. Consider∂τΨk = (a−bW∧)(LinΨk)+Hk in DR, Ψk(·,τ0) = 0 in BR(τ0)where Hk =(hk(ρ,τ)eikθ)C−1 , ‖Hk‖Rv,ℓ1,ℓ < ∞, 0 ≤ ℓ1 ≤ 1.9, 1 < ℓ < 3. Then there exists a solutionΨk =T Rkr [Hk] which is a linear mapping about Hk with the following estimate〈y〉|∇Ψk(y,τ)|+ |Ψk(y,τ)|≲C(ℓ)|k|−2‖Hk‖Rv,ℓ1,ℓv(τ)R5−ℓ(τ)〈y〉−3 ln(|y|+2), (4.8.40)232where “≲” is independent of k and C(ℓ) can be unbounded as ℓ→ 1 or 3. Moreover, Ψk ·W = 0 ande−ikθ (Ψk)C is radial in space.Proof. For brevity, denote ‖hk‖= ‖hk‖Rv,ℓ1,ℓ in this proof. Assume hk(ρ,τ) = 0 in DcR. Consider(a−bW∧)(LinGk) = Hk where Gk =(gk(ρ,τ)eikθ)C−1.By Lemma 4.8.2, it is equivalent to considering(a− ib)Lkgk = hk,where gk is given bygk(ρ,τ) = (a+ ib)Zk,2(ρ)´ ρ0 Zk,1(r)hk(r,τ)rdr+Zk,1(ρ)´ ∞ρ Zk,2(r)hk(r,τ)rdr if k ≤−2−Zk,2(ρ)´ ∞ρ Zk,1(r)hk(r,τ)rdr−Zk,1(ρ)´ ρ0 Zk,2(r)hk(r,τ)rdr if k ≥ 2.We will estimate the upper bound of gk. For k ≤−2, ρ ≤ 1,∣∣∣∣Zk,2(ρ)ˆ ρ0Zk,1(r)hk(r,τ)rdr∣∣∣∣≲ |k|−1‖hk‖v(τ)ρk−1ˆ ρ0r2−k−ℓ1dr ∼ |k|−2‖hk‖v(τ)ρ2−ℓ1for 0≤ ℓ1 ≤ 4.∣∣∣∣Zk,1(ρ)ˆ ∞ρZk,2(r)hk(r,τ)rdr∣∣∣∣≲ ρ1−k‖hk‖v(τ)(ˆ 1ρ|k|−1rk−ℓ1dr+ˆ ∞1|k|−1rk+2−ℓdr)≲ |k|−1ρ1−k‖hk‖v(τ)[ρk−ℓ1+1ℓ1− (k+1) +1ℓ− (k+3)]≲C1(ℓ)|k|−2‖hk‖v(τ)ρ2−ℓ1for 0≤ ℓ1 ≤ 4 and 1< ℓ≤ 5 whereC1(ℓ)→ ∞ as ℓ→ 1 due to k =−2.For k ≤−2, ρ ≥ 1,∣∣∣∣Zk,2(ρ)ˆ ρ0Zk,1(r)hk(r,τ)rdr∣∣∣∣≲ |k|−1‖hk‖v(τ)ρk+1(ˆ 10r2−k−ℓ1dr+ˆ ρ1r−ℓ−kdr)≲ |k|−1‖hk‖v(τ)ρk+1(13− k− ℓ1 +ρ1−k−ℓ1− k− ℓ)≲C2(ℓ)|k|−2‖hk‖v(τ)ρ2−ℓfor 0≤ ℓ1 ≤ 4 and 0≤ ℓ < 3 whereC2(ℓ)→ ∞ as ℓ→ 3 due to k =−2.|Zk,1(ρ)ˆ ∞ρZk,2(r)hk(r,τ)rdr|≲ ρ−1−k‖hk‖v(τ)ˆ ∞ρ|k|−1rk+2−ℓdr ≲C3(ℓ)|k|−2‖hk‖v(τ)ρ2−ℓfor 1< ℓ≤ 4 whereC3(ℓ)→ ∞ as ℓ→ 1 due to k =−2.233In sum, for 0≤ ℓ1 ≤ 4, 1< l < 3, k ≤−2,‖gk‖∞v,ℓ1−2,ℓ−2 ≲C4(ℓ)|k|−2‖hk‖ (4.8.41)whereC4(ℓ)→ ∞ as ℓ→ 1 or 3.For k ≥ 2, ρ ≤ 1, 0≤ ℓ1 ≤ 3, 0≤ ℓ≤ 4,∣∣∣∣Zk,2(ρ)ˆ ∞ρZk,1(r)hk(r,τ)rdr∣∣∣∣≲ k−1‖hk‖v(τ)ρk−1(ˆ ∞1r−k−ℓdr+ˆ 1ρr2−k−ℓ1dr)= k−1‖hk‖v(τ)ρk−1(1k+ ℓ−1 +1−ρ3−k−ℓ13− k− ℓ1 1{ℓ1 6=3−k}+(− lnρ)1{ℓ1=3−k})≲k−1‖hk‖v(τ)ρk−1(k−1+ ρ3−k−ℓ1k+ℓ1−3)∼ k−2‖hk‖v(τ)ρ2−ℓ1 , k ≥ 4‖hk‖v(τ)ρ2〈lnρ〉, k = 3, ℓ1 = 0‖hk‖v(τ)ρ2(3−1+ ρ−ℓ1−1ℓ1)≲ ‖hk‖v(τ)ρ2−ℓ1〈lnρ〉, k = 3,0< ℓ1 ≤ 3‖hk‖v(τ)ρ(2−1+ 1−ρ1−ℓ11−ℓ1)≲ ‖hk‖v(τ)ρ〈lnρ〉, k = 2,0≤ ℓ1 < 1‖hk‖v(τ)ρ〈lnρ〉, k = 2, ℓ1 = 1‖hk‖v(τ)ρ(2−1+ ρ1−ℓ1−1ℓ1−1)≲ ‖hk‖v(τ)ρ2−ℓ1〈lnρ〉, k = 2,1< ℓ1 ≤ 3where we used ρ t −1= tρθ t lnρ for some 0≤ θ ≤ 1.For the other part,∣∣∣∣Zk,1(ρ)ˆ ρ0Zk,2(r)hk(r,τ)rdr∣∣∣∣≲ ‖hk‖v(τ)ρ1−k ˆ ρ0k−1rk−ℓ1dr ∼ k−2‖hk‖v(τ)ρ2−ℓ1for 0≤ ℓ1 ≤ 2.9.For k ≥ 2, ρ ≥ 1,∣∣∣∣Zk,2(ρ)ˆ ∞ρZk,1(r)hk(r,τ)rdr∣∣∣∣≲ k−1‖hk‖v(τ)ρk+1ˆ ∞ρr−k−ℓdr ∼ k−2‖hk‖v(τ)ρ2−ℓ,when 0≤ ℓ≤ 4.∣∣∣∣Zk,1(ρ)ˆ ρ0Zk,2(r)hk(r,τ)rdr∣∣∣∣≲ ‖hk‖v(τ)ρ−1−k(ˆ 10k−1rk−ℓ1dr+ˆ ρ1k−1rk+2−ℓdr)≲ k−1‖hk‖v(τ)ρ−1−k(1k+1− ℓ1 +ρk+3−ℓk+3− ℓ)∼ k−2‖hk‖v(τ)ρ2−ℓ,when 0≤ ℓ1 ≤ 2.9, 0≤ ℓ≤ 4.234In sum, for 0≤ ℓ1 ≤ 2.9, 0≤ ℓ≤ 4,‖gk‖∞v,ℓ1−2,ℓ−2 ≲ k−2‖hk‖, k ≥ 4‖g3‖∞v,ε+ℓ1−2,ℓ−2 ≲C(ε)‖hk‖, k = 3‖g2‖∞v,ε+(ℓ1−1)+−1,ℓ−2 ≲C(ε)‖hk‖, k = 2(4.8.42)where ε > 0 is arbitrarily small andC(ε) is a constant depending on ε .Combining (4.8.41), (4.8.42), for 0≤ ℓ1 ≤ 1.9, 1< ℓ < 3, we have‖Gk‖∞v,ℓ−2 = ‖gk‖∞v,ℓ−2 ≲C(ℓ)|k|−2‖hk‖ for |k| ≥ 2 (4.8.43)whereC(ℓ)→ ∞ as ℓ→ 1 or 3.Consider ∂τΦk = (a−bW∧)(LinΦk)+Gk in D2R,Φk = 0 on ∂D2R, Φk(·,τ0) = 0 in B2R(τ0). (4.8.44)In order to find a solution Φk with the form Φk =(ϕk(ρ,τ)eikθ)C−1 , by Lemma 4.8.2, it suffices toconsider ∂τϕk = (a− ib)Lkϕk+gk in D2R,ϕk = 0 on ∂D2R, ϕk(·,τ0) = 0 in B2R(τ0). (4.8.45)Recall (4.8.14). Set ϕk(ρ,τ) = ρ |k−1|ϕ˜k(ρ,τ). Then∂τ ϕ˜k = (a− ib)[∂ρρ ϕ˜k+(2|k−1|+1) ∂ρ ϕ˜kρ + −4kρ2+8−4k(ρ2+1)2 ϕ˜k]+ gkρ |k−1| in D2R,ϕ˜k = 0 on ∂D2R, ϕ˜k(·,τ0) = 0 in B2R(τ0)(4.8.46)where the existence follows the parabolic theory.Applying Lemma 4.8.4 to (4.8.45), we have‖ϕk(·,τ)‖L∞(B2R(τ)) ≲C(ℓ)|k|−2v(τ)R5−ℓ(τ)‖gk‖∞v,ℓ−2. (4.8.47)In order to improve the spatial decay of ϕk, we reformulate the equation (4.8.45) into the followingform ∂τϕk = (a− ib)[∂ρρϕk+∂ρϕkρ − (k+1)2ρ2 ϕk]+ g˜k in D2R,ϕk = 0 on ∂D2R, ϕk(·,τ0) = 0 in B2R(τ0),(4.8.48)where g˜k(ρ,τ) := (a− ib) (4k+8)ρ2+4k(ρ2+1)21ρ2ϕk + gk. Set ϕ∗k(y,τ) = ei(k+1)θϕk(ρ,τ). Then (4.8.48) is235equivalent to ∂τϕ∗k = (a− ib)∆R2ϕ∗k+ ei(k+1)θ g˜k in D2R,ϕ∗k = 0 on ∂D2R, ϕ∗k(·,τ0) = 0 in B2R(τ0). (4.8.49)(4.8.49) can be regarded as a real-valued parabolic system in the varying time domain. Combin-ing [29, Theorem 3.2] and [75, Lemma 2.26 and Remark 2.27], there exists a fundamental solutionΓ2(x,y,τ,s) for the homogeneous part of (4.8.49) with the estimate|Γ2(x,y,τ,s)| ≤C (τ− s)−1 e−κ|x−y|2τ−sfor constantsC,κ > 0. Then by scaling argument, we have|∇yΓ2(x,y,τ,s)|≲ (τ− s)−32 e−κ|x−y|2τ−s . (4.8.50)Then ϕ∗k can be written asϕ∗k(y,τ) =ˆ ττ0ˆB2R(s)Γ2(y,z,τ,s)ei(k+1)θ(z)g˜k(|z|,s)dzds (4.8.51)where θ(z) = arctan( z2z1 ).In order to utilize the special form of ei(k+1)θ g˜k, we set g˜k = 0 in Dc2R and want to find P˜k(y,τ)satisfying∆R2P˜k(y,τ) = ei(k+1)θ g˜k in R2. (4.8.52)Set P˜k(y,τ) = ei(k+1)θ p˜k(ρ,τ).∂ρρ p˜k+1ρ∂ρ p˜k− (k+1)2ρ2p˜k = g˜k.Set p˜k = ρ |k+1| p˜k,1(ρ,τ). It is equivalent to considering∂ρρ p˜k,1+(2|k+1|+1)∂ρ p˜k,1ρ= ρ−|k+1|g˜k.p˜k,1 is given byp˜k,1(ρ,τ) =−ρ−2|k+1|ˆ ρ0u2|k+1|−1ˆ ∞urr−|k+1|g˜k(r,τ)drdu.Notice|g˜k|≲ 1{r≤2R(τ)}[|k|(ρ−21{ρ≤1}+ρ−41{ρ>1}) |ϕk|+ v(τ)〈ρ〉2−ℓ‖gk‖∞v,ℓ−2] .236Then|p˜k,1|≲ ρ−2|k+1|ˆ ρ0u2|k+1|−1ˆ ∞ur1{r≤2R(τ)}×[|k|(r−2−|k+1|1{r≤1}+ r−4−|k+1|1{r>1})|ϕk|+ v(τ)(r−|k+1|1{r≤1}+ r2−ℓ−|k+1|1{r>1})‖gk‖∞v,ℓ−2]drdu.Therein, by Lemma 4.9.1,ρ−2|k+1|ˆ ρ0u2|k+1|−1ˆ ∞u1{r≤2R(τ)}rv(τ)(r−|k+1|1{r≤1}+ r2−ℓ−|k+1|1{r>1})‖gk‖∞v,ℓ−2drdu≲C(ℓ)v(τ)‖gk‖∞v,ℓ−2|k|−2 (ρ2−|k+1|1{ρ≤1}+ρ4−ℓ−|k+1|1{ρ>1}) for k ≤−4, k ≥ 2R(τ)(〈lnρ〉1{ρ≤1}+ρ1−ℓ1{ρ>1}) for k =−3R2(τ)(1{ρ≤1}+ρ1−ℓ1{ρ>1})for k =−2where for the cases k =−3,−2, we used1{r≤2R(τ)}r2−ℓ−|k+1|1{r>1} ≲R(τ)r−1−ℓ for k =−3R2(τ)r−1−ℓ for k =−2.By (4.8.47) and Lemma 4.9.1,ρ−2|k+1|ˆ ρ0u2|k+1|−1ˆ ∞ur1{r≤2R(τ)}|k|(r−2−|k+1|1{r≤1}+ r−4−|k+1|1{r>1})|ϕk|drdu≲C(ℓ)|k|−1v(τ)R5−ℓ(τ)‖gk‖∞v,ℓ−2|k|−2 (ρ−|k+1|1{ρ≤1}+ρ−2−|k+1|1{ρ>1}) for k ≤−4, k ≥ 2ρ−21{ρ≤1}+ρ−4〈lnρ〉1{ρ>1} for k =−3ρ−11{ρ≤1}+ρ−21{ρ>1} for k =−2.|∂ρ p˜k,1| can also be bounded by (4.9.1) in Lemma 4.9.1 similarly. As a result, for ρ ≤ 2R(τ),|k|−1ρ|∂ρ p˜k,1|+ |p˜k,1|≲C(ℓ)v(τ)R5−ℓ(τ)‖gk‖∞v,ℓ−2|k|−2 (ρ−|k+1|1{ρ≤1}+ρ−1−|k+1|1{ρ>1}) , k ≤−4, k ≥ 2ρ−21{ρ≤1}+ρ−31{ρ>1}, k =−3ρ−11{ρ≤1}+ρ−21{ρ>1}, k =−2.237Notice P˜k(y,τ) = ei(k+1)θρ |k+1| p˜k,1(ρ,τ). Then∣∣∇P˜k∣∣= (∣∣∂ρ P˜k∣∣2+ρ−2 ∣∣∂θ P˜k∣∣2) 12 = (∣∣∣|k+1|ρ |k+1|−1 p˜k,1+ρ |k+1|∂ρ p˜k,1∣∣∣2+ρ−2|k+1|2 ∣∣∣ρ |k+1| p˜k,1∣∣∣2) 12≲ |k+1|ρ |k+1|−1 (|p˜k,1|+ |k+1|−1ρ|∂ρ p˜k,1|)≲C(ℓ)v(τ)R5−ℓ(τ)‖gk‖∞v,ℓ−2|k|−1 (ρ−11{ρ≤1}+ρ−21{ρ>1}) , k ≤−4, k ≥ 2ρ−11{ρ≤1}+ρ−21{ρ>1}, k =−3ρ−11{ρ≤1}+ρ−21{ρ>1}, k =−2≲ |k|−1C(ℓ)v(τ)R5−ℓ(τ)(ρ−11{ρ≤1}+ρ−21{ρ>1})‖gk‖∞v,ℓ−2.By ϕ∗k(y,τ) = ei(k+1)θϕk(ρ,τ), (4.8.51) and (4.8.52), we haveϕk(y,τ) =−e−i(k+1)θˆ ττ0ˆB2R(s)∇zΓ2(y,z,τ,s) ·∇P˜k(z,s)dzds.Combining (4.8.50), then|ϕk|≲ˆ ττ0ˆB2R(s)(τ− s)− 32 e− κ|y−z|2τ−s∣∣∇P˜k(z,s)∣∣dzds≲ |k|−1C(ℓ)‖gk‖∞v,ℓ−2ˆ ττ0ˆB2R(s)(τ− s)− 32 e− κ|y−z|2τ−s v(s)R5−ℓ(s)(|z|−21{|z|≤1}+ |z|−31{|z|>1}) |z|dzds,which can be estimated by a similar convolution estimate in R3. By the same argument in [108,Lemma A.1], we have|ϕk|≲ |k|−1C(ℓ)v(τ)R5−ℓ(τ)|y|−1 ln(|y|+2)‖gk‖∞v,ℓ−2 for 1≤ |y| ≤ τ12 . (4.8.53)Combining (4.8.47) and (4.8.53), we have|Φk|= |ϕk|≲ |k|−1C(ℓ)v(τ)R5−ℓ(τ)〈y〉−1 ln(|y|+2)‖gk‖∞v,ℓ−2.Applying [30, Theorem 1.2], (4.8.43) and scaling argument to (4.8.44), we have〈y〉2|D2Φk|+ 〈y〉|DΦk|+ |Φk|≲C(ℓ)|k|−2v(τ)R5−ℓ(τ)〈y〉−1 ln(|y|+2)‖hk‖ in D3R/2. (4.8.54)We take Ψk = (a−bW∧)(LinΦk) and manipulate (a−bW∧)Lin to (4.8.44). Combining (4.8.54)and scaling argument, we conclude (4.8.40). Recalling Φk =(ϕk(ρ,τ)eikθ)C−1 and applying Lemma4.8.2, we have (Ψk)C = eikθ (a− ib)Lkϕk.238Since theΨk given in Lemma 4.8.6 loses some power of R(τ) when |y| is small, we will constructa Ψk with a better estimate by another gluing procedure.Proposition 4.8.7. Consider∂τΨk = (a−bW∧)(LinΨk)+Hk in DR, Ψk(·,τ0) = 0 in BR(τ0)where Hk =(hk(ρ,τ)eikθ)C−1 , ‖Hk‖Rv,ℓ < ∞, 1 < ℓ < 3. There exists a solution Ψk = T Rk [Hk] as alinear mapping about Hk with the following estimate〈y〉|∇Ψk(y,τ)|+ |Ψk(y,τ)|≲C(ℓ)|k|−2v(τ)〈y〉2−ℓ‖Hk‖Rv,ℓwhereC(ℓ) can be unbounded as ℓ→ 1 or 3. Moreover,Ψk ·W = 0 and e−ikθ (Ψk)C is radial in space.Proof. Denote ‖hk‖ = ‖hk‖Rv,ℓ and take hk = 0 in DcR. In order to find a solution Ψk with the formΨk =(ψk(ρ,τ)eikθ)C−1 , by Lemma 4.8.2, it is equivalent to considering∂τψk = (a− ib)Lkψk+hk in DR. (4.8.55)Set ψk(ρ,τ) = ηR0(ρ)ψi,k(ρ,τ)+ψo,k(ρ,τ), where ηR0(ρ) = η(ρR0) and R0 is a large fixed constantindependent of τ0, τ , k. In order to find a solution for (4.8.55), it suffices to consider the followinginner-outer system ∂τψo,k = (a− ib)[∂ρρψo,k+ 1ρ ∂ρψo,k− (k+1)2ρ2 ψo,k]+ J[ψo,k,ψi,k]1{ρ≤4R(τ)} in R2× (τ0,∞),ψo,k(·,τ0) = 0 in R2.(4.8.56)∂τψi,k = (a− ib)Lkψi,k+K[ψo,k] in D2R0 , ψi,k(·,τ0) = 0 in B2R(τ0) (4.8.57)whereJ[ψo,k,ψi,k] := (a− ib)(1−ηR0)(4k+8)ρ2+4k(ρ2+1)2ρ2ψo,k+A0[ψi,k]+ (1−ηR0)hk,K[ψo,k] := (a− ib)(4k+8)ρ2+4k(ρ2+1)2ρ2ψo,k+hk,A0[ψi,k] := (a− ib)[(∂ρρηR0 +1ρ∂ρηR0)ψi,k+2∂ρηR0∂ρψi,k].Set Ψi,k(y,τ) =(ψi,k(ρ,τ)eikθ)C−1 , that is, ψi,k = e−ikθ (Ψi,k ·E1+ iΨi,k ·E2). By Lemma 4.8.2,(4.8.57) is equivalent to∂τΨi,k = (a−bW∧)LinΨi,k+(K[ψo,k]eikθ)C−1in D2R0 , Ψi,k(·,τ0) = 0 in B2R(τ0). (4.8.58)239The linear theories of (4.8.56) and (4.8.58) are given by Lemma 4.8.5 and Lemma 4.8.6 respec-tively and we reformulate (4.8.56) and (4.8.58) into the following formψo,k(ρ,τ) = ρ |k+1|[Γ♮2|k+1|+2 ∗∗(|z|−|k+1|J[ψo,k,ψi,k])1{|z|≤4R(s)}](ρ,τ,τ0),Ψi,k(y,τ) = T 2R0kr[(K[ψo,k]eikθ)C−1].(4.8.59)We will solve (ψo,k,Ψi,k) for (4.8.59) by the contraction mapping theorem.By Lemma 4.8.6,〈y〉∣∣∣∇T 2R0kr [(hkeikθ)C−1]∣∣∣+ ∣∣∣T 2R0kr [(hkeikθ)C−1]∣∣∣≤ Diwi,k(ρ,τ)‖hk‖,where the constant Di ≥ 1 is independent of k; for C(ℓ) given in Lemma 4.8.6,wi,k(ρ,τ) :=C(ℓ)|k|−2v(τ)R5−ℓ0 lnR0〈ρ〉−3.DenoteBi,k :={F(y,τ) ∈C1 (B2R0 ,R3) | F(y,τ) = (eikθ f (ρ,τ))C−1 for some radial scalar functionf (ρ,τ) and 〈y〉|∇yF(y,τ)|+ |F(y,τ)| ≤ 2Diwi,k(ρ,τ)‖hk‖}.For any Ψ˜i,k ∈Bi,k, denote ψ˜i,k = e−ikθ(Ψ˜i,k ·E1+ iΨ˜i,k ·E2). We will find a solutionψo,k =ψo,k[ψ˜i,k]of (4.8.56) by the contraction mapping theorem. Let us estimate J[ψo,k, ψ˜i,k] term by term. Notice by(4.2.6), ∣∣∂ρ ψ˜i,k∣∣= ∣∣e−ikθ (Ψ˜i,k ·∂ρE1+E1 ·∂ρΨ˜i,k+ iΨ˜i,k ·∂ρE2+ iE2 ·∂ρΨ˜i,k)∣∣≲ |Ψ˜i,k|〈ρ〉−2+ |∂ρΨ˜i,k|≲ DiC(ℓ)|k|−2v(τ)R5−ℓ0 lnR0〈ρ〉−4‖hk‖.Then for 1< ℓ˜ < ℓ,|A0[ψ˜i,k]|≲ DiC(ℓ)1{R0≤ρ≤2R0}|k|−2v(τ)R−ℓ0 lnR0‖hk‖≲ DiC(ℓ)|k|−2Rℓ˜−ℓ0 lnR0v(τ)〈ρ〉−ℓ˜‖hk‖,|(1−ηR0)hk|≲ 1{ρ≥R0}v(τ)〈ρ〉−ℓ‖hk‖≲ Rℓ˜−ℓ0 v(τ)〈ρ〉−ℓ˜‖hk‖.By (4.8.3), Lemma 4.9.2 (for k ≤−3 and k ≥ 2) and [108, Lemma A.1, Lemma A.2] (for k =−2),ρ |k+1|∣∣∣Γ♮2|k+1|+2∣∣∣∗∗(v(s)|z|−|k+1|〈z〉−ℓ˜)≲ wo,k(ρ,τ) := |k|−2v(τ)(ρ1{ρ≤1}+ρ2−ℓ˜1{1<ρ≤C1τ12 })+ρ−ℓ˜τv(τ)1{ρ>C1τ12 }where C1 is a constant independent of k. The spatial decay rate near ρ = 0 is restricted by the case240k =−2. It follows that∣∣∣ρ |k+1|Γ♮2|k+1|+2 ∗∗{|z|−|k+1| [A0[ψ˜i,k]+ (1−ηR0)hk]}∣∣∣≤ DoDiC(ℓ)R(ℓ˜−ℓ)/20 wo,k(ρ,τ)‖hk‖where the constant Do ≥ 1 is independent of k. DenoteBo,k :={f (ρ,τ) | | f (ρ,τ)| ≤ 2DoDiC(ℓ)R(ℓ˜−ℓ)/20 wo,k(ρ,τ)‖hk‖}.For any ψ˜o,k ∈Bo,k∣∣∣∣(1−ηR0) (4k+8)ρ2+4k(ρ2+1)2ρ2 ψ˜o,k1{ρ≤4R(τ)}∣∣∣∣≲ |k|−1DoDiC(ℓ)R(ℓ˜−ℓ)/20 v(τ)〈ρ〉−2−ℓ˜1{R0≤ρ≤4R(τ)}‖hk‖≲ R−20 DoDiC(ℓ)R(ℓ˜−ℓ)/20 v(τ)〈ρ〉−ℓ˜‖hk‖.By the same convolution estimate above, with the small quantity R−20 when R0 is large,ρ |k+1|Γ♮2|k+1|+2 ∗∗(|z|−|k+1|J[ψ˜o,k, ψ˜i,k])∈Bo,k.We can deduce the contraction mapping property in the same way.Now we have found a solution ψo,k = ψo,k[ψ˜i,k] ∈Bo,k. Let us estimate the following term inD2R0 :∣∣∣∣(4k+8)ρ2+4k(ρ2+1)2ρ2 ψo,k∣∣∣∣≲ |k|(1+ρ)−2ρ−2DoDiC(ℓ)R(ℓ˜−ℓ)/20 |k|−2v(τ)(ρ1{ρ≤1}+ρ2−ℓ˜1{1<ρ≤C1τ 12 })‖hk‖≲ |k|−1DoDiC(ℓ)R(ℓ˜−ℓ)/20 v(τ)(ρ−11{ρ≤1}+ρ−ℓ1{1<ρ≤C1τ12 })‖hk‖.By Lemma 4.8.6,〈y〉∣∣∣∣∇yT 2R0kr {[eikθ (a− ib)(4k+8)ρ2+4k(ρ2+1)2ρ2 ψo,k]C−1}∣∣∣∣+ ∣∣∣∣T 2R0kr {[eikθ (a− ib)(4k+8)ρ2+4k(ρ2+1)2ρ2 ψo,k]C−1}∣∣∣∣≲ |k|−1DoDiC(ℓ)R(ℓ˜−ℓ)/20 wi,k(ρ,τ)‖hk‖.Since R(ℓ˜−ℓ)/20 provides small quantity when R0 is large, we haveT 2R0kr[(eikθK[ψo,k[ψ˜i,k]])C−1]∈Bi,k.The contraction property can be deduced in the same way. Thus we find a solution Ψi,k =Ψi,k[hk] ∈Bi,k. Finally we find a solution (ψo,k,Ψi,k) for (4.8.56) and (4.8.58).Substituting the right hand side hk by c1h(1)k , c2h(2)k , c1h(1)k + c2h(2)k respectively, where c1,c2 arearbitrary constants and h(1)k , h(2)k are in the same topology as hk, then making subtraction and repeating241process above, we deduce that Ψi,k[hk] and ψo,k[hk] are linear mappings about hk. So does ψk.We will regard Do, Di and R0 as general constants hereafter. Reviewing the calculation process,we have|J[ψo,k,ψi,k]|1{ρ≤4R(τ)} ≲C(ℓ)v(τ)〈ρ〉−ℓ‖hk‖.Using (4.8.59) again, the upper bound of ψo,k can be improved to|ψo,k|≲C(ℓ)[|k|−2v(τ)(ρ1{ρ≤1}+ρ2−ℓ1{1<ρ≤C1τ12 })+ρ−ℓτv(τ)1{ρ>C1τ12 }]‖hk‖.Combining the upper bound of ψo,k and Ψi,k, we have|Ψk|≲C(ℓ)|k|−2v(τ)(R5−ℓ0 lnR0〈ρ〉−31{ρ≤2R0}+ρ2−ℓ1{2R0<ρ≤2R(τ)})‖hk‖≲C(ℓ)|k|−2v(τ)〈ρ〉2−ℓ‖hk‖in DR. By the scaling argument, the proof of the proposition is concluded.4.8.4 Mode 0In next proposition, we do not need a1 > −1,´ ττ0 v(s)ds = O(τv(τ)) in (4.8.3). Instead, we requireR2 lnRθR,ℓv(τ)≳ τ−(2−).Proposition 4.8.8. Consider∂τΨ0 = (a−bW∧)(LinΨ0)+H0 in DR,Ψ0 = 0 on ∂DR, Ψ0(·,τ0) = 0 in BR(τ0),where H0 = (h0(ρ,τ))C−1 , ‖H0‖Rv,ℓ < ∞. Then there exists a linear mapping Ψ0 = T R00[H0] with theestimate|Ψ0|≲ ‖H0‖Rv,ℓv(τ)〈y〉−1C(ℓ)R2 lnR if ℓ > 1R2(lnR)32 if ℓ= 1C(ℓ)R3−ℓ lnR if ℓ < 1.where C(ℓ) can be unbounded as ℓ→ 1. Moreover, Ψ0 ·W = 0 and (Ψ0)C is radial in space.Proof. Denote ‖h0‖ = ‖h0‖Rv,ℓ. In order to find a solution with the form Ψ0 = (ψ0(ρ,τ))C−1 , byLemma 4.8.2, it is equivalent to considering∂τψ0 = (a− ib)L0ψ0+h0 in DR,ψ0 = 0 on ∂DR, ψ0(·,τ0) = 0 in BR(τ0). (4.8.60)242The existence of (4.8.60) is deduced by the same argument as (4.8.45). By Lemma 4.8.4,‖ψ0(·,τ)‖L∞(BR) ≲ R2 lnRθR,ℓv(τ)‖h0‖.In order to improve the spatial decay, we reformulate (4.8.60) into the following form∂τψ0 = (a− ib)(∂ρρψ0+ 1ρ ∂ρψ0− 1ρ2ψ0)+ h˜0 in DR,ψ0 = 0 on ∂DR, ψ0(·,τ0) = 0 in BR(τ0),(4.8.61)where h˜0 := (a− ib) 8(ρ2+1)2ψ0+h0. Set ψ0 = ρψ∗0. Then (4.8.61) is equivalent to∂τψ∗0 = (a− ib)∆R4ψ∗0+ |y|−1h˜0 in DR,ψ∗0 = 0 on ∂DR, ψ∗0(·,τ0) = 0 in BR(τ0) (4.8.62)where we abuse the symbol DR ={(y,τ) | y ∈ R4, |y| ≤ R(τ)} as the corresponding time-varyingdomain inR4 and similarly ∂DR, BR(τ0). By the same argument for deducing (4.8.51), the fundamentalsolution for (4.8.62) is given by Γ4(x,y, t,s) with the bound|Γ4(x,y,τ,s)|≲ (τ− s)−2 e−κ|x−y|2τ−s for a constant κ > 0.Then by [108, Lemma A.1, Lemma A.2],|ψ0|= ρ|ψ∗0|≲ ρ∣∣Γ4 ∗∗(|z|−1|h˜0|1{|z|≤R(s)})∣∣≲ R2 lnRθR,ℓv(τ)〈ρ〉−1‖h0‖+‖h0‖v(τ)〈ρ〉2−ℓ if ℓ > 1(R(τ))1−ℓ+ε〈ρ〉1−ε if ℓ≤ 1∼ ‖h0‖v(τ)〈ρ〉−1C(ℓ)R2 lnR if ℓ > 1R2(lnR)32 if ℓ= 1C(ℓ)R3−ℓ lnR if ℓ < 1(4.8.63)whereC(ℓ) can be unbounded as ℓ→ 1 and we used|z|−1〈z〉−ℓ1{|z|≤R(s)} ≲ (R(s))1−ℓ+ε |z|−1〈z〉−1−εfor ℓ≤ 1 with a small fixed constant ε > 0.Next, we will give the linear theory with the orthogonal condition. In this proposition, we do notneed a1 >−1,´ ττ0 v(s)ds= O(τv(τ)) in (4.8.3). Instead, we require v(τ)R5−ℓ lnR≳ τ−(2−).243Lemma 4.8.9. Consider∂τΨ0 = (a−bW∧)(LinΨ0)+H0 in DR, Ψ0(·,τ0) = 0 in BR(τ0)where H0 = (h0(ρ,τ))C−1 , ‖H0‖Rv,ℓ < ∞ with 1< ℓ < 3 and the orthogonal conditionˆBR(τ)h0(y,τ)Z0,1(y)dy= 0 for all τ ∈ (τ0,∞). (4.8.64)Then there exists a solution Ψ0 =T R0r [H0] as a linear mapping about H0 with the following estimate〈y〉|∇Ψ0(y,τ)|+ |Ψ0(y,τ)|≲C(ℓ)v(τ)R5−ℓ lnR〈y〉−3‖H0‖Rv,ℓ (4.8.65)where C(ℓ) can be unbounded as ℓ→ 1 or 3. Moreover, Ψ0 ·W = 0 and (Ψ0)C is radial in space.Proof. This proof follows Lemma 4.8.6. Denote ‖h0‖ = ‖h0‖Rv,ℓ and assume h0 = 0 in DcR. Weconsider(a−bW∧)(LinG0) = H0 where G0 = (g0(ρ,τ))C−1 .By Lemma 4.8.2, it is equivalent to considering(a− ib)L0g0 = h0,where g0 is given byg0(ρ,τ) = (a+ ib)(Z0,2(ρ)ˆ ρ0h0(r,τ)Z0,1(r)rdr−Z0,1(ρ)ˆ ρ0h0(r,τ)Z0,2(r)rdr).Then under the orthogonal condition (4.8.64), if 1< ℓ < 3, we have‖G0‖∞v,ℓ−2 = ‖g0‖∞v,ℓ−2 ≲C(ℓ)‖h0‖ (4.8.66)whereC(ℓ) can be unbounded as ℓ→ 1 or 3. Next, let us consider∂τΦ0 = (a−bW∧)(LinΦ0)+G0 in D2R,Φ0 = 0 on ∂D2R, Φ0(·,τ0) = 0 in B2R(τ0). (4.8.67)By Proposition 4.8.8, there exists a solution Φ0 =Φ0[G0] with the form Φ0 = (ϕ0(ρ,τ))C−1 for somescalar function ϕ0 and the estimate|Φ0(y,τ)|≲C(ℓ)v(τ)R5−ℓ lnR〈y〉−1‖G0‖∞v,ℓ−2.244Applying [30, Theorem 1.2] and scaling argument to (4.8.67), we have〈y〉2|D2Φ0|+ 〈y〉|DΦ0|+ |Φ0|≲C(ℓ)v(τ)R5−ℓ lnR〈y〉−1‖h0‖ in D3R/2. (4.8.68)We takeΨ0 = (a−bW∧)(LinΦ0) and manipulate (a−bW∧)Lin to (4.8.67). Combining (4.8.68) andscaling argument, we conclude (4.8.65). Applying Lemma 4.8.2, we have (Ψ0)C = (a− ib)L0ϕ0.In next proposition, we do not need a1 >−1,´ ττ0 v(s)ds=O(τv(τ)) in (4.8.3). Instead, we requirea1 >−2.Proposition 4.8.10. Consider∂τΨ0 = (a−bW∧)(LinΨ0)+H0+(c0(τ)η(ρ)Z0,1(ρ))C−1 in DR, Ψ0(·,τ0) = 0 in BR(τ0)where H0 = (h0(ρ,τ))C−1 , ‖H0‖Rv,ℓ < ∞ with 1 < ℓ < 3. Then there exists a solution (Ψ0,c0) =(T R0 [H0],c0[H0](τ)) which is a linear mapping about H0 with the following estimate〈y〉|∇Ψ0|+ |Ψ0|≲C(ℓ) lnR0v(τ)(R5−ℓ0 〈y〉−31{|y|≤2R0}+ 〈y〉2−ℓ1{|y|>2R0})‖H0‖Rv,ℓ,c0[H0](τ) =−(ˆB2η(y)Z 20,1(y)dy)−1ˆB2R0h(y,τ)Z0,1(y)dy+ c∗0[H0],where R0 is given in (4.8.3), C(ℓ) can be unbounded as ℓ→ 1 or 3; c∗0[H0] is a scalar functionlinearly depending on H0 and satisfies |c∗0[H0]|≲C(ℓ)R−ε00 v(τ)‖H0‖Rv,ℓ and 0< ε0 < ℓ−1 is a smallconstant independent of τ0.Proof. Denote ‖h0‖= ‖h0‖Rv,ℓ and take h0 = 0 in DcR. By Lemma 4.8.2, in order to find a solution Ψ0with the form Ψ0 = (ψ0(ρ,τ))C−1 , it is equivalent to considering∂τψ0 = (a− ib)L0ψ0+h0+ c0(τ)η(ρ)Z0,1(ρ) in DR, ψ0(·,τ0) = 0 in BR(τ0)Set ψ0 = ηR0(ρ)ψi,0(ρ,τ) +ψo,0(ρ,τ), where ηR0(ρ) = η(ρR0). In order to find a solution ψ0, itsuffices to consider the following inner-outer system ∂τψo,0 = (a− ib)(∂ρρψo,0+ 1ρ ∂ρψo,0− 1ρ2ψo,0)+ J[ψo,0,ψi,0]1{ρ≤4R(τ)} in R2× (τ0,∞),ψo,0(·,τ0) = 0 in R2,(4.8.69)∂τψi,0 = (a− ib)L0ψi,0+K[ψo,0]+ c0(τ)η(ρ)Z0,1(ρ) in D2R0 , ψi,0(·,τ0) = 0 in B2R0(τ0)(4.8.70)245whereJ[ψo,0,ψi,0] = (1−ηR0)8(a− ib)ψo,0(1+ρ2)2+A0[ψi,0]+ (1−ηR0)h0, K[ψo,0] =8(a− ib)ψo,0(1+ρ2)2+h0,A0[ψi,0] = (a− ib)[(∂ρρηR0 +1ρ∂ρηR0)ψi,0+2∂ρηR0∂ρψi,0]−ψi,0∂τηR0 .Denote Ψi,0(y,τ) = (ψi,0)C−1 , that is, ψi,0 =Ψi,0 ·E1+ iΨi,0 ·E2. By Lemma 4.8.2, (4.8.70) is equiv-alent to∂τΨi,0 =(a−bW∧)LinΨi,0+(K[ψo,0]+ c0(τ)η(ρ)Z0,1(ρ))C−1 in D2R0 , Ψi,0(·,τ0)= 0 in B2R0(τ0)(4.8.71)In order to meet the orthogonal condition (4.8.64) in D2R0 , we takec0(τ) = c0[ψo,0](τ) :=C0,1ˆB2R0[8(a− ib)ψo,0(y,τ)(1+ |y|2)2 +h0(y,τ)]Z0,1(y)dywhereC0,1 :=−(´B2η(y)Z 20,1(y)dy)−1.By Lemma 4.8.5 and Lemma 4.8.9, we reformulate (4.8.69) and (4.8.71) into the following formψo,0(ρ,τ) = ρ[Γ♮4 ∗∗(|z|−1J[ψo,0,ψi,0])1{|z|≤4R(s)}](ρ,τ,τ0),Ψi,0(y,τ) = T 2R00r[(K[ψo,0]+ c0[ψo,0](τ)η(ρ)Z0,1(ρ))C−1].(4.8.72)We will solve this system by the contraction mapping theorem.Denote HI :=[h0+C0,1(´B2R0h0(y,τ)Z0,1(y)dy)η(ρ)Z0,1(ρ)]C−1. It is easy to have ‖HI‖2R0v,ℓ ≲‖h0‖. Inspired Lemma 4.8.9, if (HI)C satisfies the orthogonal condition (4.8.64) in D2R0 , we have thefollowing estimate〈y〉|∇T 2R00r [HI](y,τ)|+ |T 2R00r [HI](y,τ)| ≤ Diwi,0(ρ,τ)‖h0‖,where Di ≥ 1 is a constant andwi,0(ρ,τ) :=C(ℓ)v(τ)R5−ℓ0 lnR0〈ρ〉−3.DenoteBi,0 :={F(y,τ) ∈C1 (B2R0 ,R3) | F(y,τ) = ( f (ρ,τ))C−1 for some radial scalar functionf (ρ,τ) and 〈y〉|∇yF(y,τ)|+ |F(y,τ)| ≤ 2Diwi,0(ρ,τ)‖h0‖}.For any Ψ˜i,0 ∈Bi,0, denote ψ˜i,0 = Ψ˜i,0 ·E1+ iΨ˜i,0 ·E2. We will find a solution ψo,0 = ψo,0[ψ˜i,0] of246(4.8.69) by the contraction mapping theorem. Let us estimate J[ψo,0, ψ˜i,0] term by term. By (4.2.6),∣∣∂ρ ψ˜i,0∣∣= ∣∣Ψ˜i,0 ·∂ρE1+E1 ·∂ρΨ˜i,0+ iΨ˜i,0 ·∂ρE2+ iE2 ·∂ρΨ˜i,0∣∣≲ |Ψ˜i,0|〈ρ〉−2+ |∇yΨ˜i,0|≲ DiC(ℓ)v(τ)R5−ℓ0 lnR0〈ρ〉−4‖h0‖.For R0 is given in (4.8.3), we have |R′0|= O(R−10 ). Then, for 1< ℓ˜ < ℓ and ℓ˜ very close to ℓ,|A0[ψ˜i,0]|≲ DiC(ℓ)1{R0≤ρ≤2R0}v(τ)R−ℓ0 lnR0‖h0‖≲ DiC(ℓ)Rℓ˜−ℓ0 lnR0v(τ)〈ρ〉−ℓ˜‖h0‖,|(1−ηR0)h0|≲ 1{ρ≥R0}v(τ)〈ρ〉−ℓ‖h0‖≲ Rℓ˜−ℓ0 v(τ)〈ρ〉−ℓ˜‖h0‖.By (4.8.3), [108, Lemma A.1, Lemma A.2], we have∣∣∣ρΓ♮4 ∗∗{|z|−1 [A0[ψ˜i,0]+ (1−ηR0)h0]}∣∣∣≤ DoDiRℓ˜−ℓ0 lnR0wo,0(ρ,τ)‖h0‖,where Do ≥ 1 is a large constant,wo,0(ρ,τ) :=C(ℓ)(v(τ)〈ρ〉2−ℓ˜1{ρ≤τ 12 }+ρ−ℓ˜τv(τ)1{ρ>τ 12 }).DenoteBo,0 :={f (ρ,τ) | | f (ρ,τ)| ≤ 2DoDiRℓ˜−ℓ0 lnR0wo,0(ρ,τ)‖h0‖}.For any ψ˜o,0 ∈Bo,0,∣∣∣∣8(1−ηR0)(ρ2+1)2 ψ˜o,01{ρ≤4R(τ)}∣∣∣∣≲ DoDiC(ℓ)Rℓ˜−ℓ0 lnR0v(τ)〈ρ〉−2−ℓ˜1{R0≤ρ≤4R(τ)}‖h0‖≲ R−20 (τ0)DoDiC(ℓ)Rℓ˜−ℓ0 lnR0v(τ)〈ρ〉−ℓ˜‖h0‖.By the same convolution estimate above, with the small quantity R−20 (τ0) when R0(τ0) is large,ρΓ♮4 ∗∗(|z|−1J[ψ˜o,0, ψ˜i,0]1{|z|≤4R(s)}) ∈Bo,0.We can deduce the contraction mapping property in the same way.Now we have found a solution ψo,0 = ψo,0[ψ˜i,0] ∈Bo,0. It follows that‖8(1+ρ2)−2ψo,0[ψ˜i,0]+C0,1ˆB2R08ψo,0[ψ˜i,0](y,τ)(1+ |y|2)2 Z0,1(y)dyη(ρ)Z0,1(ρ)‖2R0v,ℓ ≲DoDiC(ℓ)Rℓ˜−ℓ0 lnR0‖h‖.Due to the choice of c0(τ), hII :=K[ψo,0[ψ˜i,0]]+c0[ψo,0[ψ˜i,0]](τ)η(ρ)Z0,1(ρ) satisfies the orthogonalcondition (4.8.64) in D2R0 . By Lemma 4.8.9, we haveT 2R00r [(hII)C−1 ] ∈Bi,0247since Rℓ˜−ℓ0 lnR0 provides small quantity.The contraction property can be deduced in the same way. Thus we find a solution Ψi,0 =Ψi,0[h0] ∈Bi,0. Finally we find a solution (ψo,0,Ψi,0) for (4.8.69) and (4.8.71).From the construction process, ψi,0[h0], ψo,0[h0] and c0[h0] are linear mappings about h0. So doesψ0.We will regard Do, and Di as general constants hereafter. Since ψo,0[h0] ∈Bo,0, thenc0[h0](τ) =C0,1ˆB2R0h0(y,τ)Z0,1(y)dy+ c∗0[h0]where c∗0[h0](τ) is a linear mapping about h0 and |c∗0[h0]|≲C(ℓ)Rℓ˜−ℓ0 lnR0v(τ)‖h0‖.Reviewing the calculation process, we have|J[ψo,0,ψi,0]|1{ρ≤4R(τ)} ≲C(ℓ) lnR0v(τ)〈ρ〉−ℓ‖h0‖.Using (4.8.72) again, the upper bound of ψo,0 can be improved to|ψo,0|≲C(ℓ) lnR0(v(τ)〈ρ〉2−ℓ1{ρ≤τ 12 }+ρ−ℓτv(τ)1{ρ>τ 12 })‖h0‖.Combining the upper bound of ψo,0 and Ψi,0, we have|Ψ0(y,τ)|≲C(ℓ) lnR0v(τ)(R5−ℓ0 〈y〉−31{|y|≤2R0}+ 〈y〉2−ℓ1{|y|>2R0})‖h0‖ in DR.By the scaling argument again, we conclude this proposition.4.8.5 Mode 1Proposition 4.8.11. Consider∂τΨ1 = (a−bW∧)(LinΨ1)+H1 in DR,Ψ1 = 0 on ∂DR, Ψ1(·,τ0) = 0 in BR(τ0),where H1 =(h1(ρ,τ)eiθ)C−1 , ‖H1‖Rv,ℓ < ∞. Then there exists a linear mapping Ψ1 = T R10[H1] withthe estimate|Ψ1(y,τ)|≲min{τ 12 ,R2}R2θR,ℓv(τ)〈y〉−2‖H1‖Rv,ℓ.Moreover, Ψ1 ·W = 0 and e−iθ (Ψ1)C is radial in space.248Proof. In order to find a solution Ψ1 with the form Ψ1 =(ψ1(ρ,τ)eiθ)C−1 , by Lemma 4.8.2, it isequivalent to considering ∂τψ1 = (a− ib)L1ψ1+h1 in DR,ψ1 = 0 on ∂DR, ψ1(·,τ0) = 0 in BR(τ0).For brevity, denote ‖h1‖= ‖h1‖Rv,ℓ. By Lemma 4.8.4,‖ψ1(·,τ)‖L2(BR) ≲min{τ12 ,R2}R2θR,ℓv(τ)‖h1‖. (4.8.73)In order to get the L∞ estimate, we reformulate the equation into the following form:∂τψ1 = (a− ib)(∂ρρψ1+ 1ρ ∂ρψ1)+ hˆ1 in DR,ψ1 = 0 on ∂DR, ψ1(·,τ0) = 0 in BR(τ0),(4.8.74)where hˆ1 =−4ρ2+4(ρ2+1)2ψ1+h1. By the similar argument for deducing (4.8.51), denote Γ2(x,y,τ,s) as thefundamental solution of the homogeneous part of (4.8.74) with the estimate|Γ2(x,y,τ,s)| ≤C(τ− s)−1e−κ|x−y|2τ−swith positive constantsC, κ .ψ1(y,τ) =ˆBR(τ−1)Γ2(y,z,τ,τ−1)ψ1(z,τ−1)dz+ˆ ττ−1ˆBR(s)Γ2(y,z,τ,s)hˆ1(z,s)dzdsThen by (4.8.73),|ˆBR(τ−1)Γ2(y,z,τ,τ−1)ψ1(z,τ−1)dz|≲ˆBR(τ−1)e−κ|y−z|2 |ψ1(z,τ−1)|dz≲min{τ 12 ,R2}R2θR,ℓv(τ)‖h1‖,‖hˆ1(·,τ)‖L2(BR) ≲min{τ12 ,R2}R2θR,ℓv(τ)‖h1‖, and then∣∣∣∣∣ˆ ττ−1ˆBR(s)Γ2(y,z,τ,s)hˆ(z,s)dzds∣∣∣∣∣≲ˆ ττ−1(τ− s)−1(ˆBR(s)e−2κ|y−z|2τ−s dz) 12‖hˆ1(·,s)‖L2(BR(s))ds≲ min{τ 12 ,R2}R2θR,ℓv(τ)‖h1‖ˆ ττ−1(τ− s)− 12 ∼min{τ 12 ,R2}R2θR,ℓv(τ)‖h1‖.Therefore, we get‖ψ1(·,τ)‖L∞ ≲min{τ 12 ,R2}R2θR,ℓv(τ)‖h1‖.249In order to get the spatial decay, we rewrite the equation into the following form:(a+ ib)∂τψ1 = ∂ρρψ1+ 1ρ ∂ρψ1− 4ρ2ψ1+ h˜1 in DR,ψ1 = 0 on ∂DR, ψ1(·,τ0) = 0 in BR(τ0),where h˜1 =12ρ2+4(ρ2+1)21ρ2ψ1+h1. By similar argument in (4.8.63),|ψ1(ρ,τ)|≲ ρ2∣∣Γ6 ∗∗(|y|−2|h˜1|1{|y|≤R(s)})∣∣ .where|Γ6(x,y, t,s)|≲ (t− s)−3 e−κ|x−y|2t−s for a constant κ > 0.Notice in DR, by (4.8.3), [108, Lemma A.1, Lemma A.2],∣∣∣∣Γ6 ∗∗(|z|−2 12|z|2+4(|z|2+1)2 1|z|2 |ψ1(z,s)|1{|z|≤R(s)})∣∣∣∣≲ (ρ−21{ρ≤1}+ρ−3.91{ρ>1})min{τ12 ,R2}R2θR,ℓv(τ)‖h1‖,∣∣Γ6 ∗∗(|z|−2|h1(z,s)|1{|z|≤R(s)})∣∣≲ v(τ)‖h1‖(〈lnρ〉1{ρ≤1}+〈ρ〉−4〈lnρ〉 if ℓ≥ 4〈ρ〉−ℓ if 0< ℓ < 4R−ℓ+ε〈ρ〉−ε if ℓ≤ 0)where we used |y|−2〈|y|〉−ℓ1{1<|y|≤R(s)} ≲ R−ℓ+ε〈|y|〉−2−ε for ℓ ≤ 0 where the constant ε > 0 can bechosen arbitrarily small. Then we have|ψ1(ρ,τ)|≲ (1{ρ≤1}+ρ−1.91{ρ>1})min{τ12 ,R2}R2θR,ℓv(τ)‖h1‖.By the iteration of the above estimate, we gain|ψ1(ρ,τ)|≲min{τ 12 ,R2}R2θR,ℓv(τ)〈ρ〉−2‖h1‖.Lemma 4.8.12. Consider∂τΨ1 = (a−bW∧)(LinΨ1)+H1 in DR, Ψ1(·,τ0) = 0 in BR(τ0)where H1 =(h1(ρ,τ)eiθ)C−1 , ‖H1‖Rv,ℓ < ∞ with 0< ℓ < 3 and the orthogonal conditionˆBR(τ)h1(y,τ)Z1,1(y)dy= 0 for all τ ≥ τ0. (4.8.75)250Then there exists a solution Ψ1 = T R1r [H1] which is a linear mapping about H1 with the followingestimate〈y〉|∇Ψ1(y,τ)|+ |Ψ1(y,τ)|≲C(ℓ)min{τ 12 ,R2}R5−ℓv(τ)〈y〉−4‖H1‖Rv,ℓ in DRwhere C(ℓ) can be unbounded as ℓ→ 0 or 3. Moreover,Ψ1 ·W = 0 and e−iθ (Ψ1)C is radial in space.Proof. Denote ‖h1‖= ‖h1‖Rv,ℓ and set h1 = 0 in DcR. Consider(a−bW∧)(LinG1) = H1 where G1 =(g1(ρ,τ)eiθ)C−1.By Lemma 4.8.2, it is equivalent to considering(a− ib)L1g1 = h1,where g1 is given byg1(ρ,τ) = (a+ ib)(Z1,2(ρ)ˆ ρ0h1(r,τ)Z1,1(r)rdr−Z1,1(ρ)ˆ ρ0h1(r,τ)Z1,2(r)rdr).By the orthogonal condition (4.8.75), we have the following estimate‖g1‖∞v,ℓ−2 ≲C(ℓ)‖h1‖ for 0< ℓ < 4 (4.8.76)whereC(ℓ) can be unbounded as ℓ→ 0 or 4. In fact,Z1,1(ρ)ˆ ρ0h1(r,τ)Z1,2(r)rdr ≲C(ℓ)v(τ)〈ρ〉2−ℓ‖h1‖ if ℓ < 4.For ρ ≤ 1,|Z1,2(ρ)ˆ ρ0h1(r,τ)Z1,1(r)rdr|≲ v(τ)‖h1‖.For ρ ≥ 1, by the orthogonal condition (4.8.75),|Z1,2(ρ)ˆ ρ0h1(r,τ)Z1,1(r)rdr|= |Z1,2(ρ)ˆ ∞ρh1(r,τ)Z1,1(r)rdr|≲C(ℓ)v(τ)〈ρ〉2−ℓ‖h1‖ for ℓ> 0.Next, let us consider∂τΦ1 = (a−bW∧)(LinΦ1)+G1 in D2R,Φ1 = 0 on ∂D2R, Φ1(·,τ0) = 0 in B2R(τ0).251Here Φ1 is given by proposition 4.8.11. Using (4.8.76), we have the estimate|Φ1(ρ,τ)|≲C(ℓ)min{τ 12 ,R2}R5−ℓv(τ)〈ρ〉−2‖h1‖when ℓ < 3. Take Ψ1 = (a−bW∧)(LinΦ1). Combining the scaling argument, we have〈y〉|∇Ψ1(y,τ)|+ |Ψ1(y,τ)|≲C(ℓ)min{τ 12 ,R2}R5−ℓv(τ)〈y〉−4‖h1‖ in DR.Proposition 4.8.13. Consider∂τΨ1 = (a−bW∧)(LinΨ1)+H1+(c1(τ)η(ρ)Z1,1(ρ)eiθ)C−1in DR, Ψ1(·,τ0) = 0 in BR(τ0)where H1 =(h1(ρ,τ)eiθ)C−1 , ‖H1‖Rv,ℓ < ∞ with 1 < ℓ < 3. Then there exists a solution (Ψ1,c1) =(T R1 [H1],c1[H1](τ)) which is a linear mapping about H1 with the following estimate〈y〉|∇yΨ1(y,τ)|+ |Ψ1(y,τ)|≲C(ℓ)R0v(τ)(R6−ℓ0 〈y〉−41{|y|≤2R0}+ 〈y〉2−ℓ1{|y|>2R0})‖H1‖Rv,ℓ,c1[H1](τ) =−(ˆB2η(y)Z 21,1(y)dy)−1ˆB2R0h1(y,τ)Z1,1(y)dy+ c∗1[H1](τ)where R0 is given in (4.8.3), C(ℓ) can be unbounded as ℓ→ 1 or 3; c∗1[H1] is a scalar functionlinearly depending on H1 and satisfies |c∗1[H1]| ≲ C(ℓ)R−ε00 v(τ)‖H1‖Rv,ℓ where 0 < ε0 < ℓ− 1 is asmall constant independent of τ0. Moreover, Ψ1 ·W = 0 and e−iθ (Ψ1)C is radial in space.Proof. The proof is similar to Proposition 4.8.10. In order to find a solution Ψ1 with the form Ψ1 =(ψ1(ρ,τ)eiθ)C−1 , by Lemma 4.8.2, it is equivalent to considering∂τψ1 = (a− ib)L1ψ1+h1+ c1(τ)η(ρ)Z1,1(ρ) in DR, ψ1(·,τ0) = 0 in BR(τ0).Denote ‖h1‖ = ‖h1‖v,ℓ and take h1 = 0 in DcR. Set ψ1(ρ,τ) = ηR0ψi,1(ρ,τ) +ψo,1(ρ,τ), whereηR0 = η(ρR0). In order to find a solution ψ1, it suffices to consider the following inner-outer system ∂τψo,1 = (a− ib)(∂ρρψo,1+ 1ρ ∂ρψo,1− 4ρ2ψo,1)+ J[ψo,1,ψi,1]1{ρ≤4R(τ)} in R2× (τ0,∞),ψo,1(·,τ0) = 0 in R2.(4.8.77)∂τψi,1 = (a− ib)L1ψi,1+K[ψo,1]+ c1(τ)η(ρ)Z1,1(ρ) in D2R0 , ψi,1(·,τ0) = 0 in B2R0(τ0)(4.8.78)252whereJ[ψo,1,ψi,1] = (1−ηR0)(a− ib)(12ρ2+4)(ρ2+1)2ρ2ψo,1+A0[ψi,1]+ (1−ηR0)h1,K[ψo,1] =(a− ib)(12ρ2+4)(ρ2+1)2ρ2ψo,1+h1,A0[ψi,1] = (a− ib)[(∂ρρηR0 +1ρ∂ρηR0)ψi,1(ρ,τ)+2∂ρηR0∂ρψi,1(ρ,τ)]−ψi,1∂τηR0 .SetΨi,1(y,τ)=(eiθψi,1(ρ,τ))C−1 , that is,ψi,1 = e−iθ (Ψi,1 ·E1+ iΨi,1 ·E2). By Lemma 4.8.2, (4.8.78)is equivalent to ∂τΨi,1 = (a−bW∧)LinΨi,1+[(K[ψo,1]+ c1(τ)η(ρ)Z1,1(ρ))eiθ]C−1 in D2R0 ,Ψi,1(·,τ0) = 0 in B2R0(τ0).(4.8.79)In order to meet the orthogonal condition (4.8.75) in D2R0 , we takec1(τ) = c1[ψo,1](τ) :=C1,1ˆB2R0[(a− ib)(12|y|2+4)(|y|2+1)2|y|2 ψo,1(y,τ)+h1(y,τ)]Z1,1(y)dywhereC1,1 :=−(´B2η(y)Z 21,1(y)dy)−1.By Lemma 4.8.5 and Lemma 4.8.12, we reformulate (4.8.77) and (4.8.79) into the following formψo,1(ρ,τ) = ρ2[Γ♮6 ∗∗(|z|−2J[ψo,1,ψi,1]1{|z|≤4R(s)})](ρ,τ,τ0),Ψi,1(y,τ) = T 2R01r[[(K[ψo,1]+ c1(τ)η(ρ)Z1,1(ρ))eiθ]C−1].(4.8.80)We will solve (ψo,1,Ψi,1) by the contraction mapping theorem.DenoteHI :={[h1+C1,1(´B2R0h(y,τ)Z1,1(y)dy)η(ρ)Z1,1(ρ)]eiθ}C−1. It is easy to have ‖HI‖2R0v,ℓ ≲‖h1‖. Inspired Lemma 4.8.12, if e−iθ (HI)C satisfies the orthogonal condition (4.8.75) in D2R0 , wehave the estimate〈y〉|∇y(T 2R01r [HI])(y,τ)|+ |T 2R01r [HI](y,τ)| ≤ Diwi,1(ρ,τ)‖h1‖,where the constant Di ≥ 1 is large andwi,1(ρ,τ) :=C(ℓ)R7−ℓ0 v(τ)〈ρ〉−4.253DenoteBi,1 :={F(y,τ) ∈C1 (B2R0 ,R3) | F(y,τ) = (eiθ f (ρ,τ))C−1 for some radial scalar functionf (ρ,τ) and 〈y〉|∇yF(y,τ)|+ |F(y,τ)| ≤ 2Diwi,1(ρ,τ)‖h1‖}.For any Ψ˜i,1 ∈Bi,1, denote ψ˜i,1 = e−iθ(Ψ˜i,1 ·E1+ iΨ˜i,1 ·E2). We will find a solutionψo,1 =ψo,1[Ψ˜i,1]of (4.8.77) by the contraction mapping theorem.Let us estimate J[ψo,1, ψ˜i,1] term by term. By (4.2.6),∣∣∂ρ ψ˜i,1∣∣≲ |Ψ˜i,1|〈ρ〉−2+ |∇yΨ˜i,1|≲ DiC(ℓ)R7−ℓ0 v(τ)〈ρ〉−5‖h1‖.For R0 is given in (4.8.3), we have |R′0| = O(R−10 ). Then for max{1, ℓ−1} < ℓ˜ < ℓ and ℓ˜ very closeto ℓ,|A0[ψ˜i,1]|≲ DiC(ℓ)1{R0≤ρ≤2R0}v(τ)R1−ℓ0 ‖h1‖≲ 1{ρ≥R0}DiC(ℓ)Rℓ˜−ℓ0 v(τ)〈ρ〉1−ℓ˜‖h1‖,|(1−ηR0)h1|≲ 1{ρ≥R0}v(t)〈ρ〉−ℓ‖h1‖≲ 1{ρ≥R0}Rℓ˜−ℓ0 v(t)〈ρ〉−ℓ˜‖h1‖.By (4.8.3) and Lemma 4.9.3, for 1< ℓ˜ < 5 and ℓ˜ very close to ℓ,C(ℓ)ρ2∣∣∣Γ♮6∣∣∣∗∗(Rℓ˜−ℓ0 (s)v(s)|z|−2〈z〉1−ℓ˜1{|z|≥R0(s)})≲C(ℓ)ρ2∣∣∣Γ♮6∣∣∣∗∗(Rℓ˜−ℓ0 (s)v(s)〈z〉−1−ℓ˜)≲ Rℓ˜−ℓ0 wo,1(ρ,τ)wherewo,1(ρ,τ) :=C(ℓ)v(τ)(ρ21{ρ≤1}+ρ3−ℓ˜1{1<ρ≤τ 12 }+ τρ1−ℓ˜1{ρ>τ 12 }).It follows that∣∣∣ρ2Γ♮6 ∗∗[|z|−2(A0[ψ˜i,1]+ (1−ηR0)h1)1{|z|≤4R(s)}]∣∣∣≤ DoDiRℓ˜−ℓ0 wo,1(ρ,τ)‖h1‖for a large constant Do ≥ 1. Then we denoteBo,1 :={f (ρ,τ) | | f (ρ,τ)| ≤ 2DoDiRℓ˜−ℓ0 wo,1(ρ,τ)‖h1‖}.For any ψ˜o,1 ∈Bo,1,∣∣∣∣(1−ηR0) 12ρ2+4(ρ2+1)2ρ2 ψ˜o,11{ρ≤4R(τ)}∣∣∣∣≲ DoDiC(ℓ)Rℓ˜−ℓ0 v(τ)〈ρ〉−1−ℓ˜1{R0≤ρ≤4R(τ)}‖h1‖≲ R−20 DoDiC(ℓ)Rℓ˜−ℓ0 v(τ)〈ρ〉1−ℓ˜‖h1‖.254Using Lemma 4.9.3 again, due to the small quantity provided by R−20 , we haveρ2Γ♮6 ∗∗(|z|−2J[ψ˜o,1, ψ˜i,1]) ∈Bo,1.We can deduce the contraction mapping property in the same way. Thus we find a solution ψo,1 =ψo,1[ψ˜i,1] ∈Bo,1. Then in D2R0 , due to ℓ≤ 1+ l˜, we have the following estimate∣∣∣∣(12ρ2+4)ψo,1[ψ˜i,1](ρ2+1)2ρ2∣∣∣∣≲ (ρ〈ρ〉)−2DoDiC(ℓ)Rℓ˜−ℓ0 v(τ)(ρ21{ρ≤1}+ρ3−ℓ˜1{1<ρ≤τ 12 })‖h1‖≲ DoDiC(ℓ)Rℓ˜−ℓ0 v(τ)〈ρ〉−ℓ‖h1‖,∣∣∣∣∣C1,1(ˆB2R012|y|2+4(|y|2+1)2|y|2ψo,1[ψ˜i,1](y,τ)Z1,1(y)dy)η(ρ)Z1,1(ρ)∣∣∣∣∣≲ DoDiC(ℓ)Rℓ˜−ℓ0 v(τ)〈ρ〉−ℓ‖h1‖.Due to the choice of c1(τ), hII := K[ψo,1[ψ˜i,1]] + c1[ψo,1[ψ˜i,1]](τ)η(ρ)Z1,1(ρ) satisfies the or-thogonal condition (4.8.75) in D2R0 . By Lemma 4.8.12, we haveT 2R01r [(hIIeiθ )C−1 ] ∈Bi,1due to the small quantity provided by Rℓ˜−ℓ0 . The contraction property can be deduced in the same way.Thus we find a solution Ψi,1 =Ψi,1[h1] ∈Bi,1. Finally we find a solution (ψo,1,Ψi,1) for (4.8.77) and(4.8.79). By similar argument in Proposition 4.8.7, we have that ψo,1, Ψi,1 c1(τ) linearly depends onh1.We will regard Do, and Di as general constants hereafter. Since ψo,1[h1] ∈Bo,1, thenc1[h1](τ) =C1,1ˆB2R0h1(y,τ)Z1,1(y)dy+ c∗1[h1]where c∗1[h1] depends on h1 linearly and |c∗1[h1]|≲C(ℓ)Rℓ˜−ℓ0 v(τ)‖h1‖.Reviewing the calculation process, we have|J[ψo,1,ψi,1]|1{ρ≤4R(τ)} ≲C(ℓ)R0v(τ)〈ρ〉−ℓ‖h1‖.Iterating Lemma 4.9.3 to (4.8.80), the upper bound of ψo,1 can be improved to|ψo,1|≲C(ℓ)R0v(τ)(〈ρ〉2−ℓ1{ρ≤τ 12 }+ τρ−ℓ1{ρ>τ 12 })‖h1‖.Combining the upper bound of ψo,1 and Ψi,1, we have|Ψ1(y,τ)|≲C(ℓ)R0v(τ)(R6−ℓ0 〈ρ〉−41{ρ≤2R0}+ 〈ρ〉2−ℓ1{ρ>2R0})‖h1‖ in DR.255By the scaling argument, the proof of the proposition is concluded.4.8.6 Mode −1For τ0 ≥ 2, first we consider a general case(a+ ib)∂τφk(ρ,τ) =Lkφk(ρ,τ),φk(ρ,τ0) = g(ρ),whereLk is given in (4.8.14) and g is a Schwartz function. Set φk = ρ−12Ak(ρ,τ), then(a+ ib)∂τAk(ρ,τ) = L˜kAk(ρ,τ),Ak(ρ,τ0) = ρ 12 g(ρ). (4.8.81)where L˜k := ∂ρρ + 14ρ−2− (k−1)2ρ2 − 4kρ2+1 + 8(ρ2+1)2 . The Fourier basis Φk(ρ,ξ ) with respect to −L˜kis given by−L˜kΦk(ρ,ξ ) = ξΦk(ρ,ξ ).We multiply Φk(ρ,ξ ) to (4.8.81) and integrate by parts. Then(a+ ib)∂τ Aˆk(ξ ,τ) =−ξ Aˆk(ξ ,τ),Aˆk(ξ ,τ0) = ´ ∞0 ρ 12 g(ρ)Φk(ρ,ξ )dρ,where Aˆk(ξ ,τ) =´ ∞0 Ak(ρ,τ)Φk(ρ,ξ )dρ . ThusAˆk(ξ ,τ) = e−(a−ib)ξτ Aˆk(ξ ,τ0).By the distorted Fourier transform,Ak(ρ,τ) =ˆ ∞0Aˆk(ξ ,τ)Φk(ρ,ξ )ρk(dξ ) =ˆ ∞0e−(a−ib)ξτ Aˆk(ξ ,τ0)Φk(ρ,ξ )ρk(dξ )=ˆ ∞0e−(a−ib)ξτΦk(ρ,ξ )ˆ ∞0x12 g(x)Φk(x,ξ )dxρk(dξ )=ˆ ∞0ˆ ∞0e−(a−ib)ξτΦk(ρ,ξ )Φk(x,ξ )ρk(dξ )x12 g(x)dx.Thus, by Duhamel’s principle,ϕk(ρ,τ) =ˆ ττ0ˆ ∞0ˆ ∞0e−(a−ib)ξ (τ−s)ρ−12Φk(ρ,ξ )Φk(x,ξ )x12 h(x,s)ρk(dξ )dxds (4.8.82)256is a solution for (a+ ib)∂τϕk(ρ,τ) =Lkϕk(ρ,τ)+h(ρ,τ) in (0,∞)× (τ0,∞),ϕk(ρ,τ0) = 0 in (0,∞).For k =−1, we summarize the results in [62, Section 4.3.2]. See also [40] for a general theory.Proposition 4.8.14 ( [62]). For all ρ ≥ 0, ξ ≥ 0, we have∣∣Φ−1(ρ,ξ )∣∣≲ρ52 〈ρ〉−2 if ρ2ξ ≤ 1ξ− 14 〈ξ 〉−1 if ρ2ξ > 1.Φ−1(ρ,ξ ) has the following expansion:Φ−1(ρ,ξ ) =Φ−10 (ρ)+ρ12∞∑j=1(−ρ2ξ ) jΦ j(ρ2),which converges absolutely, where Φ−10 (ρ) =ρ521+ρ2 . It converges uniformly if ρξ12 remains bounded.Here Φ j(u)≥ 0 are smooth functions of u≥ 0 satisfyingΦ j(u)≤ 1j!u1+u, for all u≥ 0, j ≥ 1,and Φ1(u)≥ c1 u1+u for all u≥ 0 with some absolute constant c1 > 0.The spectrum measure ρ−1(dξ ) of −L˜−1 is absolutely continuous on ξ ≥ 0 with densitydρ−1(ξ )dξ∼ 〈ξ 〉2.We emphasize that we do not require the assumption about v(τ) in (4.8.3) in the next proposition.Proposition 4.8.15. Consider(a+ ib)∂τϕ−1(ρ,τ) =L−1ϕ−1(ρ,τ)+h(ρ,τ) in (0,∞)× (τ0,∞),ϕ−1(ρ,τ0) = 0 in (0,∞).where τ0 ≥ 2, ‖h‖∞v,ℓ < ∞, where v(τ) ≥ 0, ℓ > 32 . Then the solution ϕ−1 = T−1[h], where T−1[h] is257given by the linear mapping (4.8.82) with k =−1, satisfies the following estimate|ϕ−1(ρ,τ)|≲ ‖h‖∞v,ℓ1{ρ≤τ 12 }τ1− ℓ2 sups∈[τ/2,τ]v(s)+ τ− ℓ2´ τ2τ02v(s)ds if ℓ < 2(lnτ)2 sups∈[τ/2,τ]v(s)+ τ−1 lnτ´ τ2τ02v(s)ds if ℓ= 2lnτ sups∈[τ/2,τ]v(s)+ τ−1´ τ2τ02v(s)ds if ℓ > 2+‖h‖∞v,ℓ1{ρ>τ 12 }ρ− 12τ 54− ℓ2 sups∈[τ/2,τ]v(s)+ τ 14− ℓ2´ τ2τ02v(s)ds if ℓ < 2τ 14 〈lnτ〉 sups∈[τ/2,τ]v(s)+ τ− 34 〈lnτ〉´ τ2τ02v(s)ds if ℓ= 2τ 14 sups∈[τ/2,τ]v(s)+ τ− 34´ τ2τ02v(s)ds if ℓ > 2.Moreover, assuming 2< ℓ < 52 and the orthogonality conditionˆR2h(y,τ)Z−1,1(y)dy= 0 for all τ > τ0, (4.8.83)we have the following estimate|ϕ−1(ρ,τ)|≲ ‖h‖∞v,ℓ〈ρ〉2−ℓ sups∈[τ/2,τ]v(s)+ τ− ℓ2´ τ2τ02v(s)ds if ρ ≤ τ 12ρ− 12(τ 54− ℓ2 sups∈[τ/2,τ]v(s)+ τ 14− ℓ2´ τ2τ02v(s)ds)if ρ > τ 12 .Proof. Without loss of generality, we assume ‖h‖∞v,ℓ = 1.Estimate without orthogonality.|ϕ−1(ρ,τ)|≲ ρ− 12ˆ ττ0v(s)ˆ ∞0ˆ ∞0e−aξ (τ−s)|Φ−1(ρ,ξ )||Φ−1(x,ξ )|x 12 〈x〉−ℓ〈ξ 〉2dxdξds.First, we considerF(ξ ) :=ˆ ∞0|Φ−1(x,ξ )|x 12 〈x〉−ℓdx=ˆ ξ− 120+ˆ ∞ξ−12· · · := F1+F2.For F1,F1 ≲ˆ ξ− 120x52 〈x〉−2x 12 〈x〉−ℓdx≲ξ ℓ2−1 if ℓ < 2〈lnξ 〉 if ℓ= 21 if ℓ > 2for ξ ≤ 1ξ−2 for ξ > 1.258For F2,F2 ≲ˆ ∞ξ−12ξ−14 〈ξ 〉−1x 12 〈x〉−ℓdx≲ξℓ2−1 if ξ ≤ 1ξ− 54 if ξ > 1,where we have used ℓ > 32 .ThusF(ξ )≲ξ ℓ2−1 if ℓ < 2〈lnξ 〉 if ℓ= 21 if ℓ > 2for ξ ≤ 1ξ− 54 for ξ > 1.Next, let us estimateP(ρ,τ,s) :=ˆ ∞0e−aξ (τ−s)|Φ−1(ρ,ξ )|F(ξ )〈ξ 〉2dξ =ˆ 1ρ20+ˆ ∞1ρ2· · · := P1+P2.First, let us estimate P1,P1 ≲ ρ52 〈ρ〉−2ˆ 1ρ20e−aξ (τ−s)F(ξ )〈ξ 〉2dξ .For ρ ≥ 1,P1 ≲ ρ12ˆ 1ρ20e−aξ (τ−s)ξ ℓ2−1 if ℓ < 2〈lnξ 〉 if ℓ= 21 if ℓ > 2dξ≲ρ12−ℓ if τ− s≤ ρ2ρ 12 (τ− s)− ℓ2 if τ− s> ρ2if ℓ < 2ρ−32 〈lnρ〉 if τ− s≤ ρ2ρ 12 (τ− s)−1〈ln(a(τ− s))〉 if τ− s> ρ2if ℓ= 2ρ−32 if τ− s≤ ρ2ρ 12 (τ− s)−1 if τ− s> ρ2if ℓ > 2by Lemma 4.10.1.For ρ < 1,P1 ≲ ρ52 (ˆ 10+ˆ 1ρ21)e−aξ (τ−s)F(ξ )〈ξ 〉2dξ ,259whereˆ 10e−aξ (τ−s)F(ξ )〈ξ 〉2dξ ≲1 if τ− s≤ 1(τ− s)− ℓ2 if τ− s> 1 for ℓ < 21 if τ− s≤ 1(τ− s)−1〈ln(a(τ− s))〉 if τ− s> 1 for ℓ= 21 if τ− s≤ 1(τ− s)−1 if τ− s> 1 for ℓ > 2by the same estimate above.ˆ 1ρ21e−aξ (τ−s)F(ξ )〈ξ 〉2dξ ≲ˆ 1ρ21e−aξ (τ−s)ξ34 dξ ∼ (τ− s)− 74ˆ a(τ−s)ρ2a(τ−s)e−zz34 dz≲ρ− 72 if τ− s≤ ρ2(τ− s)− 74 if ρ2 < τ− s≤ 1(τ− s)− 74 e− a(τ−s)2 if τ− s> 1.Thus for ρ < 1,P1 ≲ρ−1 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)− ℓ2 if τ− s> 1for ℓ < 2ρ−1 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)−1〈ln(a(τ− s))〉 if τ− s> 1for ℓ= 2ρ−1 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)−1 if τ− s> 1for ℓ > 2.Next, let us estimate P2.P2 ≲ˆ ∞1ρ2e−aξ (τ−s)ξ−14 〈ξ 〉−1F(ξ )〈ξ 〉2dξ .260For ρ ≤ 1,P2 ≲ˆ ∞1ρ2e−aξ (τ−s)ξ−14 〈ξ 〉−1ξ− 54 〈ξ 〉2dξ ∼ˆ ∞1ρ2e−aξ (τ−s)ξ−12 dξ ∼ (τ− s)− 12ˆ ∞a(τ−s)ρ2e−zz−12 dz≲(τ− s)− 12 if τ− s≤ ρ2(τ− s)− 12 e−a(τ−s)2ρ2 if τ− s> ρ2.For ρ > 1,P2 ≲ (ˆ ∞1+ˆ 11ρ2)e−aξ (τ−s)ξ−14 〈ξ 〉−1F(ξ )〈ξ 〉2dξ .We estimateˆ ∞1e−aξ (τ−s)ξ−14 〈ξ 〉−1F(ξ )〈ξ 〉2dξ ≲(τ− s)−12 if τ− s≤ 1(τ− s)− 12 e− a(τ−s)2 if τ− s> 1for the same reason as above.ˆ 11ρ2e−aξ (τ−s)ξ−14 〈ξ 〉−1F(ξ )〈ξ 〉2dξ≲ˆ 11ρ2e−aξ (τ−s)ξ−14ξ ℓ2−1 if ℓ < 2〈lnξ 〉 if ℓ= 21 if ℓ > 2dξ≲1 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2(τ− s) 14− ℓ2 e−a(τ−s)2ρ2 if τ− s> ρ2for ℓ < 21 if τ− s≤ 1(τ− s)− 34 〈ln(a(τ− s))〉 if 1< τ− s≤ ρ2(τ− s)− 34 〈ln(a(τ− s))〉e−a(τ−s)2ρ2 if τ− s> ρ2for ℓ= 21 if τ− s≤ 1(τ− s)− 34 if 1< τ− s≤ ρ2(τ− s)− 34 e−a(τ−s)2ρ2 if τ− s> ρ2for ℓ > 2by Lemma 4.10.1.261Thus, for ρ > 1,P2 ≲(τ− s)− 12 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2(τ− s) 14− ℓ2 e−a(τ−s)4ρ2 if τ− s> ρ2for ℓ < 2(τ− s)− 12 if τ− s≤ 1(τ− s)− 34 〈ln(a(τ− s))〉 if 1< τ− s≤ ρ2(τ− s)− 34 〈ln(a(τ− s))〉e−a(τ−s)4ρ2 if τ− s> ρ2for ℓ= 2(τ− s)− 12 if τ− s≤ 1(τ− s)− 34 if 1< τ− s≤ ρ2(τ− s)− 34 e−a(τ−s)4ρ2 if τ− s> ρ2for ℓ > 2.In sum, for ρ ≤ 1,P≲(τ− s)− 12 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)− ℓ2 if τ− s> 1for ℓ < 2(τ− s)− 12 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)−1〈ln(a(τ− s))〉 if τ− s> 1for ℓ= 2(τ− s)− 12 if τ− s≤ ρ2ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1ρ 52 (τ− s)−1 if τ− s> 1for ℓ > 2.262For ρ > 1,P≲(τ− s)− 12 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2ρ 12 (τ− s)− ℓ2 if τ− s> ρ2for ℓ < 2(τ− s)− 12 if τ− s≤ 1(τ− s)− 34 〈ln(a(τ− s))〉 if 1< τ− s≤ ρ2ρ 12 (τ− s)−1〈ln(a(τ− s))〉 if τ− s> ρ2for ℓ= 2(τ− s)− 12 if τ− s≤ 1(τ− s)− 34 if 1< τ− s≤ ρ2ρ 12 (τ− s)−1 if τ− s> ρ2for ℓ > 2.Now let us estimate ϕ−1. For ρ ≤ 1,|ϕ−1(ρ,τ)|≲ ρ− 12 (ˆ ττ−ρ2+ˆ τ−ρ2τ−1+ˆ τ−1τ02)v(s)P(ρ,τ,s)ds≲ ρ− 12[v˜(τ)ˆ ττ−ρ2(τ− s)− 12 ds+ v˜(τ)ρ 52ˆ τ−ρ2τ−1(τ− s)− 74 ds+ρ52´ τ−1τ02v(s)(τ− s)− ℓ2 ds if ℓ < 2´ τ−1τ02v(s)(τ− s)−1〈ln(a(τ− s))〉ds if ℓ= 2´ τ−1τ02v(s)(τ− s)−1ds if ℓ > 2]≲ v˜(τ)ρ 12 +ρ2v˜(τ)τ1− ℓ2 + τ− ℓ2´ τ2τ02v(s)ds if ℓ < 2v˜(τ)(lnτ)2+ τ−1 lnτ´ τ2τ02v(s)ds if ℓ= 2v˜(τ) lnτ+ τ−1´ τ2τ02v(s)ds if ℓ > 2where v˜(τ) := sups∈[τ/2,τ]v(s).For 1< ρ ≤ ( τ2 )12 ,|ϕ−1(ρ,τ)|≲ ρ− 12 (ˆ ττ−1+ˆ τ−1τ−ρ2+ˆ τ−ρ2τ02)v(s)P(s)ds≲ ρ− 12[v˜(τ)ˆ ττ−1(τ− s)− 12 ds+ v˜(τ)ˆ τ−1τ−ρ2(τ− s) 14− ℓ2 if ℓ < 2(τ− s)− 34 〈ln(a(τ− s))〉 if ℓ= 2(τ− s)− 34 if ℓ > 2ds263+ρ12ˆ τ−ρ2τ02v(s)(τ− s)− ℓ2 if ℓ < 2(τ− s)−1〈ln(a(τ− s))〉 if ℓ= 2(τ− s)−1 if ℓ > 2ds]≲ ρ− 12 v˜(τ)+ v˜(τ)ρ2−ℓ if ℓ < 2〈lnρ〉 if ℓ= 21 if ℓ > 2+v˜(τ)τ1− ℓ2 + τ− ℓ2´ τ2τ02v(s)ds if ℓ < 2v˜(τ)´ τ2ρ2〈lnz〉z−1dz+ τ−1 lnτ´ τ2τ02v(s)ds if ℓ= 2v˜(τ) ln( τ2ρ2 )+ τ−1 ´ τ2τ02v(s)ds if ℓ > 2≲v˜(τ)τ1− ℓ2 + τ− ℓ2´ τ2τ02v(s)ds if ℓ < 2v˜(τ)(〈lnρ〉+´ τ2ρ2〈lnz〉z−1dz)+ τ−1 lnτ´ τ2τ02v(s)ds if ℓ= 2v˜(τ)〈ln( τ2ρ2 )〉+ τ−1´ τ2τ02v(s)ds if ℓ > 2.For ( τ2 )12 ≤ ρ ≤ τ 12 ,|ϕ−1(ρ,τ)|≲ ρ− 12 (ˆ ττ−1+ˆ τ−1τ−ρ2+ˆ τ−ρ2τ02)v(s)P(s)ds≲ ρ− 12[v˜(τ)ˆ ττ−1(τ− s)− 12 ds+ˆ τ−1τ−ρ2v(s)(τ− s) 14− ℓ2 if ℓ < 2(τ− s)− 34 〈ln(a(τ− s))〉 if ℓ= 2(τ− s)− 34 if ℓ > 2ds+ρ12ˆ τ−ρ2τ02v(s)(τ− s)− ℓ2 if ℓ < 2(τ− s)−1〈ln(a(τ− s))〉 if ℓ= 2(τ− s)−1 if ℓ > 2ds]≲ ρ− 12[v˜(τ)+ v˜(τ)τ 54− ℓ2 if ℓ < 2τ 14 〈lnτ〉 if ℓ= 2τ 14 if ℓ > 2+ˆ τ2τ−ρ2v(s)dsτ 14− ℓ2 if ℓ < 2τ− 34 〈lnτ〉 if ℓ= 2τ− 34 if ℓ > 2+ρ12ˆ τ−ρ2τ02v(s)dsτ− ℓ2 if ℓ < 2τ−1〈lnτ〉 if ℓ= 2τ−1 if ℓ > 2]≲ ρ− 12[v˜(τ)τ 54− ℓ2 if ℓ < 2τ 14 〈lnτ〉 if ℓ= 2τ 14 if ℓ > 2+ˆ τ2τ02v(s)dsτ 14− ℓ2 if ℓ < 2τ− 34 〈lnτ〉 if ℓ= 2τ− 34 if ℓ > 2]264∼τ1− ℓ2 v˜(τ)+ τ− ℓ2´ τ2τ02v(s)ds if ℓ < 2〈lnτ〉v˜(τ)+ τ−1〈lnτ〉´ τ2τ02v(s)ds if ℓ= 2v˜(τ)+ τ−1´ τ2τ02v(s)ds if ℓ > 2.For ρ ≥ τ 12 ,|ϕ−1(ρ,τ)|≲ ρ− 12 (ˆ ττ−1+ˆ τ−1τ02)v(s)P(s)ds≲ ρ− 12 v˜(τ)ˆ ττ−1(τ− s)− 12 ds+ρ− 12ˆ τ−1τ02v(s)(τ− s) 14− ℓ2 if ℓ < 2(τ− s)− 34 〈ln(a(τ− s))〉 if ℓ= 2(τ− s)− 34 if ℓ > 2ds≲ v˜(τ)ρ− 12 +ρ− 12v˜(τ)τ 54− ℓ2 + τ 14− ℓ2´ τ2τ02v(s)ds if ℓ < 2v˜(τ)τ 14 〈lnτ〉+ τ− 34 〈lnτ〉´ τ2τ02v(s)ds if ℓ= 2v˜(τ)τ 14 + τ− 34´ τ2τ02v(s)ds if ℓ > 2≲ ρ− 12v˜(τ)τ 54− ℓ2 + τ 14− ℓ2´ τ2τ02v(s)ds if ℓ < 2v˜(τ)τ 14 〈lnτ〉+ τ− 34 〈lnτ〉´ τ2τ02v(s)ds if ℓ= 2v˜(τ)τ 14 + τ− 34´ τ2τ02v(s)ds if ℓ > 2.Estimate with orthogonality.For one part,ρ−12 |ˆ ττ−1ˆ ∞0ˆ ∞0e−(a−ib)ξ (τ−s)Φ−1(ρ,ξ )Φ−1(x,ξ )x12 h(x,s)ρ−1(dξ )dxds|≲ v˜(τ)(ρ 12 1{ρ≤1}+ρ−12 1{ρ>1}).(4.8.84)For the other part,ϕ˜−1 := ρ−12 |ˆ τ−1τ0ˆ ∞0ˆ ∞0e−(a−ib)ξ (τ−s)Φ−1(ρ,ξ )Φ−1(x,ξ )x12 h(x,s)ρ−1(dξ )dxds|.By the orthogonal condition (4.8.83), we haveF(ξ ,s) := |ˆ ∞0Φ−1(x,ξ )x12 h(x,s)dx|= |(ˆ ξ− 120+ˆ ∞ξ−12)(Φ−1(x,ξ )− x521+ x2)x12 h(x,s)dx|265Firstly, by proposition 4.8.14, we have|ˆ ξ− 120(Φ−1(x,ξ )− x521+ x2)x12 h(x,s)dx|≲ v(s)ˆ ξ− 120x521+ x2x2ξx12 〈x〉−ℓdx≲ v(s)ξℓ2−1 if ξ ≤ 1ξ−2 if ξ ≥ 1when ℓ < 4.Secondly,|ˆ ∞ξ−12Φ−1(x,ξ )x12 h(x,s)dx|≲ v(s)ξ− 14 〈ξ 〉−1ˆ ∞ξ−12x12 〈x〉−ℓdx∼ v(s)ξℓ2−1 if ξ ≤ 1ξ− 54 if ξ ≥ 1where we require ℓ > 32 to guarantee the integrability.Thirdly, by (4.8.83),ˆ ∞ξ−12x521+ x2x12 h(x,s)dx=−ˆ ξ− 120x521+ x2x12 h(x,s)dx,where we require ℓ > 2. Then we have∣∣∣∣∣ˆ ∞ξ−12x521+ x2x12 h(x,s)dx∣∣∣∣∣≲ v(s)ξℓ2−1 if ξ ≤ 1ξ−2 if ξ ≥ 1.ThusF(ξ ,s)≲ v(s)(ξ ℓ2−11{ξ≤1}+ξ−54 1{ξ>1}).Next, let us estimateP(ρ,τ,s) :=ˆ ∞0e−aξ (τ−s)|Φ−1(ρ,ξ )|F(ξ ,s)〈ξ 〉2dξ =ˆ 1ρ20+ˆ ∞1ρ2· · · := P1+P2.Let us estimate P1. For ρ ≥ 1,P1 ≲ v(s)ˆ 1ρ20e−aξ (τ−s)ρ52 〈ρ〉−2ξ ℓ2−1〈ξ 〉2dξ ∼ v(s)ρ 12ˆ 1ρ20e−aξ (τ−s)ξℓ2−1dξ≲ v(s)ρ12−ℓ if τ− s≤ ρ2ρ 12 (τ− s)− ℓ2 if τ− s> ρ2by Lemma 4.10.1.266For ρ < 1,P1 ≲ˆ 1ρ20e−aξ (τ−s)ρ52 〈ρ〉−2F(ξ ,s)〈ξ 〉2dξ∼ ρ 52 (ˆ 10+ˆ 1ρ21)e−aξ (τ−s)F(ξ ,s)〈ξ 〉2dξ≲ v(s)ρ 52 (ˆ 10e−aξ (τ−s)ξℓ2−1dξ +ˆ 1ρ21e−aξ (τ−s)ξ34 dξ )≲ v(s)ρ 52ρ− 72 if τ− s≤ ρ2(τ− s)− 74 if ρ2 < τ− s≤ 1(τ− s)− ℓ2 if τ− s> 1since ˆ 10e−aξ (τ−s)ξℓ2−1dξ ≲1 if τ− s≤ 1(τ− s)− ℓ2 if τ− s> 1by Lemma 4.10.1 andˆ 1ρ21e−aξ (τ−s)ξ34 dξ ∼ (τ− s)− 74ˆ a(τ−s)ρ2a(τ−s)e−zz34 dz∼ρ− 72 if τ− s≤ ρ2(τ− s)− 74 if ρ2 < τ− s≤ 1(τ− s)− 74 e− a(τ−s)2 if τ− s> 1.Next, we will estimate P2. For ρ ≤ 1,P2 ≲ v(s)ˆ ∞1ρ2e−aξ (τ−s)ξ−14 〈ξ 〉−1ξ− 54 〈ξ 〉2dξ ∼ v(s)ˆ ∞1ρ2e−aξ (τ−s)ξ−12 dξ∼ v(s)(τ− s)− 12ˆ ∞a(τ−s)ρ2e−zz−12 dz≲v(s)(τ− s)− 12 if τ− s≤ ρ2v(s)(τ− s)− 12 e−a(τ−s)2ρ2 if τ− s> ρ2.For ρ > 1,P2 ≲ˆ ∞1ρ2e−aξ (τ−s)ξ−14 〈ξ 〉−1F(ξ ,s)〈ξ 〉2dξ =ˆ 11ρ2+ˆ ∞1· · ·≲ v(s)(ˆ 11ρ2e−aξ (τ−s)ξℓ2− 54 dξ +ˆ ∞1e−aξ (τ−s)ξ−12 dξ )≲ v(s)(τ− s)− 12 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2(τ− s) 14− ℓ2 e−a(τ−s)4ρ2 if τ− s> ρ2267since ˆ ∞1e−aξ (τ−s)ξ−12 dξ ≲(τ− s)−12 if τ− s≤ 1(τ− s)− 12 e− a(τ−s)2 if τ− s> 1,andˆ 11ρ2e−aξ (τ−s)ξℓ2− 54 dξ ≲1 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2(τ− s) 14− ℓ2 e−a(τ−s)2ρ2 if τ− s> ρ2by Lemma 4.10.1.In sum, for ρ ≤ 1,P(ρ,τ,s)≲v(s)(τ− s)− 12 if τ− s≤ ρ2v(s)ρ 52 (τ− s)− 74 if ρ2 < τ− s≤ 1v(s)ρ 52 (τ− s)− ℓ2 if τ− s> 1.For ρ > 1,P(ρ,τ,s)≲ v(s)(τ− s)− 12 if τ− s≤ 1(τ− s) 14− ℓ2 if 1< τ− s≤ ρ2ρ 12 (τ− s)− ℓ2 if τ− s> ρ2.Finally, we will estimate ϕ˜−1. Obviously,ϕ˜−1 ≲ ρ−12ˆ τ−1τ02P(τ,s,ρ)ds.For ρ ≤ 1,ϕ˜−1 ≲ ρ−12ˆ τ−1τ02v(s)ρ52 (τ− s)− ℓ2 ds≲ ρ2(v˜(τ)+ τ−ℓ2ˆ τ2τ02v(s)ds).For 1< ρ ≤ ( τ2 )12 ,ϕ˜−1 ≲ ρ−12 (ˆ τ−1τ−ρ2+ˆ τ−ρ2τ02)P(τ,s,ρ)ds≲ ρ− 12[v˜(τ)ˆ τ−1τ−ρ2(τ− s) 14− ℓ2 ds+(ˆ τ−ρ2τ2+ˆ τ2τ02)v(s)ρ12 (τ− s)− ℓ2 ds]≲ v˜(τ)ρ2−ℓ+ τ− ℓ2ˆ τ2τ02v(s)dswhere we have used ℓ < 52 .268For ( τ2 )12 < ρ ≤ τ 12 ,ϕ˜−1 ≲ ρ−12 (ˆ τ−1τ−ρ2+ˆ τ−ρ2τ02)P(τ,s,ρ)ds≲ ρ− 12[ˆ τ−1τ−ρ2v(s)(τ− s) 14− ℓ2 ds+ˆ τ−ρ2τ02v(s)ρ12 (τ− s)− ℓ2 ds]≲ ρ− 12[v˜(τ)τ54− ℓ2 + τ14− ℓ2ˆ τ2τ−ρ2v(s)ds+ρ12 τ−ℓ2ˆ τ−ρ2τ02v(s)ds]∼ v˜(τ)ρ2−ℓ+ τ− ℓ2ˆ τ2τ02v(s)dswhere we have used ℓ < 52 .For ρ > τ 12 ,ϕ˜−1 ≲ ρ−12ˆ τ−1τ02v(s)(τ− s) 14− ℓ2 ds≲ ρ− 12(v˜(τ)τ54− ℓ2 + τ14− ℓ2ˆ τ2τ02v(s)ds)where we have used ℓ < 52 .Combining (4.8.84), we conclude the estimate.4.9 Additional information 1: Pointwise estimate for heat equationsLemma 4.9.1. Consider−∆u= f (x) in Rn,where n≥ 3, f (x) = f (|x|) is radial with the upper bound | f (x)|≲ |x|−l11{|x|≤1}+ |x|−l1{|x|>1}, l1 < n,l > 2. u is given byu(x) =1(n−2)|Sn−1|ˆRn|x− y|2−n f (y)dy,where |Sn−1| is the volume of the unit sphere Sn−1. Thenu(x) = u(|x|) = |x|2−nˆ |x|0an−3ˆ ∞ab f (b)dbda, ∂|x|u=−|x|1−nˆ |x|0f (a)an−1da.Moreover,|∂|x|u|≲|x|1−l1n− l1 1{|x|≤1}+( |x|1−nn− l1 +|x|1−ln− l)1{|x|>1}. (4.9.1)269Specially, if f (x) = |x|−l11{|x|≤1}+ |x|−l1{|x|>1}, then for |x| ≤ 1,u(x) =1(l−2)(n−2) +1(2−l1)(n−2) −|x|2−l1(2−l1)(n−l1) if l1 < 2− ln |x|n−2 +1(n−2)2 if l1 = 2|x|2−l1(l1−2)(n−l1) − 1(l1−2)(n−2) if 2< l1 < n,for |x| ≥ 1,u(x) =|x|2−nn−2 (1l−2 +1n− l1 )+|x|2−l−|x|2−n(l−2)(n−l) if 2< l < n|x|2−n ln |x|n−2 if l = n|x|2−n−|x|2−l(l−2)(l−n) if l > n=|x|2−l(l−2)(n−l) +|x|2−n(n−2)(n−l1) −|x|2−n(n−2)(n−l) if 2< l < n|x|2−n ln |x|n−2 +|x|2−nn−2 (1n−2 +1n−l1 ) if l = n|x|2−n(n−2)(l−n) +|x|2−n(n−2)(n−l1) −|x|2−l(l−2)(l−n) if l > n=|x|2−l(l−2)(n−l) +(l1−l)|x|2−n(n−2)(n−l1)(n−l) if 2< l < n|x|2−n ln |x|n−2 +|x|2−nn−2 (1n−2 +1n−l1 ) if l = n(l−l1)|x|2−n(n−2)(n−l1)(l−n) −|x|2−l(l−2)(l−n) if l > n.Specially, for |x| ≥ 1,u(x)≥|x|2−l(l−2)(n−2) +|x|2−n(n−2)(n−l1) if 2< l < n|x|2−n ln |x|n−2 +|x|2−nn−2 (1n−2 +1n−l1 ) if l = n|x|2−n(l−2)(n−2) +|x|2−n(n−2)(n−l1) if l > n.Proof. It is easy to see that u is radial. By the removable singularity theorem for harmonic function(It is used for the case 2≤ l1 < n) and maximum principle, we haveu(x) = u(|x|) = |x|2−nˆ |x|0an−3ˆ ∞ab f (b)dbda.Denote r = |x|. It is straightforward to have∂ru= (2−n)r1−nˆ r0an−3ˆ ∞ab f (b)dbda+ r−1ˆ ∞rb f (b)db=−r1−nˆ r0f (a)an−1dafor | f (x)|≲ |x|−l11{|x|≤1}+ |x|−l1{|x|>1}, l1 < n, l > 2.For r ≤ 1, ∣∣∣∣ˆ r0f (a)an−1da∣∣∣∣≲ rn−l1n− l1 .270For r > 1, ∣∣∣∣ˆ r0f (a)an−1da∣∣∣∣= ∣∣∣∣(ˆ 10+ˆ r1)f (a)an−1da∣∣∣∣≲ 1n− l1 + rn−ln− l .Thus we have (4.9.1).Hereafter, we assume f (r) = r−l11{r≤1}+ r−l1{r>1}. For a≥ 1, under the assumption l > 2, thenˆ ∞ab f (b)db=a2−ll−2 ,For a≤ 1,ˆ ∞ab f (b)db=ˆ ∞1b1−ldb+ˆ 1ab1−l1db=1l−2 +12−l1 (1−a2−l1) if l1 < 2− lna if l1 = 21l1−2(a2−l1−1) if l1 > 2.For r ≤ 1,ˆ r0an−3ˆ ∞ab f (b)dbda=rn−2(l−2)(n−2) +rn−2(2−l1)(n−2) − rn−l1(2−l1)(n−l1) if l1 < 2rn−2(− lnr)n−2 +rn−2(n−2)2 if l1 = 2rn−l1(l1−2)(n−l1) − rn−2(l1−2)(n−2) if 2< l1 < n(4.9.2)where we require l1 < n to guarantee the integrability around 0.For r ≥ 1, sinceˆ r1an−3ˆ ∞ab f (b)dbda=ˆ r1an−1−ll−2 da=rn−l−1(l−2)(n−l) if 2< l < nlnrn−2 if l = n1−rn−l(l−2)(l−n) if l > n,then combining (4.9.2), we haveˆ r0an−3ˆ ∞ab f (b)dbda=1n−2(1l−2 +1n− l1 )+rn−l−1(l−2)(n−l) if 2< l < nlnrn−2 if l = n1−rn−l(l−2)(l−n) if l > n.In the next lemma, we build the upper bound of the heat equation with a clear dependence onparameters.271Lemma 4.9.2. Consider ∂tu= c0∆u+ f (x, t) in Rn× (t0,∞)u(·, t0) = 0 in Rnwhere c0 is a complex constant and Re(c0)> 0, n> 4, | f (x, t)| ≤C f v(t)(|x|−l11{|x|≤1}+ |x|−l1{|x|>1}),l1 < n, l > 2 and u is given byu(x, t) =ˆ tt0ˆRn[4πc0(t− s)]−n2 e−|x−y|24c0(t−s) f (y,s)dydswhere c−n20 = |c0|−n2 e−iθn2 if c0 = |c0|eiθ .Suppose that 0< v(t) ∈C1(t0,∞), then for 2< l ≤ n−2, l1 ≤ l, v′(t)≥ 0, we have|u(r, t)| ≤C f 2|c0|n2−2(Re(c0))n2−1min{v(t)g(r),|c0|2Re(c0)r−lˆ tt0v(s)ds}in Rn× (t0,∞);for 2< l ≤ n−2, 0≤ l1 ≤ l, −(v(t))−1v′(t)≤Cvt−1, (v(t))−1´ tt0v(s)ds≤Cvt,t0 ≥ 2Re(c0)|c0|2 Cvmax{C1(n, l),C2(n, l1, l)}, we have|u(r, t)|≤ 2C f |c0|n2−2(Re(c0))n2−1min{max{1,2(Cv)2n−2n− l}v(t)g(r)1{r≤(2CvC1(n,l))−12 (|c0 |2Re(c0))12 t12 },|c0|2Re(c0)r−lˆ tt0v(s)ds}where g satisfies −∆g= r−l11{r≤1}+ r−l1{r>1} and is given by Lemma 4.9.1,C1(n, l) :=1(l−2)(n− l) , C2(n, l1, l) :=1(l−2)(n−2) +1(2−l1)(n−2) if l1 < 21n−2 +1(n−2)2 if l1 = 21(l1−2)(n−l1) if 2< l1 < n.(4.9.3)Proof. Since|u(x, t)| ≤C fˆ tt0ˆRn[4π|c0|(t− s)]−n2 exp(−Re(c0)|x− y|24|c0|2(t− s))× v(s)(|y|−l11{|y|≤1}+ |y|−l1{|y|>1})dyds=C f( |c0|Re(c0)) n2ˆ tt0ˆRn[4π|c0|2Re(c0)(t− s)]− n2exp− |x− y|24 |c0|2Re(c0)(t− s)× v(s)(|y|−l11{|y|≤1}+ |y|−l1{|y|>1})dyds272=C f|c0| n2−2(Re(c0))n2−1ˆ t˜t˜0ˆRn[4π(t˜− s˜)]− n2 exp(− |x− y|24(t˜− s˜))v˜(s˜)(|y|−l11{|y|≤1}+ |y|−l1{|y|>1})dyds˜where t˜ := |c0|2Re(c0)t, t˜0 :=|c0|2Re(c0)t0, v˜(s˜) := v(Re(c0)|c0|2 s˜). Thus we only need to consider the case c0 = 1and f (x, t) = v(t)(|x|−l11{|x|≤1}+ |x|−l1{|x|>1}). With this radial right hand side, it is ready to get thatu(x, t) = u(|x|, t) is radial about the spatial variable.Denote r = |x|, Lg= ∆g−∂tg+ v(t)(r−l11{r≤1}+ r−l1{r>1}). Set ϕ1(r, t) = 2r−l´ tt0v(s)ds, thenLϕ1 = 2l(l+2−n)r−l−2ˆ tt0v(s)ds−2r−lv(t)+ v(t)(r−l11{r≤1}+ r−l1{r>1})≤ 2l(l+2−n)r−l−2ˆ tt0v(s)ds− r−lv(t),when l1 ≤ l. If 0≤ l ≤ n−2, then Lϕ1 ≤ 0 in Rn× (t0,∞). It followsu(r, t)≤ 2r−lˆ tt0v(s)ds in Rn× (t0,∞).Next, we want to improve the estimate of u in the region |x| ≲ t 12 . Set −∆g = r−l11{r≤1} +r−l1{r>1}, where l1 < n, l > 2 and g> 0 is given by Lemma 4.9.1. For r ≥ 1, l1 ≤ l, we haveg(r)≤C1(n, l)r2−l;for r ≤ 1,g(r)≤C2(n, l1, l)1 if l1 < 21− lnr if l1 = 2r2−l1 if 2< l1 < nwhereC1(n, l),C2(n, l1, l) are given in (4.9.3).Set ϕ2 = D2v(t)g(r) with D2 ≥ 2. ThenLϕ2 = −D2v(t)(r−l11{r≤1}+ r−l1{r>1})−D2v′(t)g(r)+ v(t)(r−l11{r≤1}+ r−l1{r>1})≤ − D22v(t)(r−l11{r≤1}+ r−l1{r>1})−D2v′(t)g(r)= D2g(r)v(t)[− 12g(r)(r−l11{r≤1}+ r−l1{r>1})−v′(t)v(t)].If v′(t)≥ 0, then Lϕ2 ≤ 0 in Rn× (t0,∞). Take D2 = 2. Thenu(r, t)≤ 2v(t)g(r) in Rn× (t0,∞).273Thus for 2< l ≤ n−2, l1 ≤ l, v′(t)≥ 0, we haveu(r, t)≤min{2v(t)g(r),2r−lˆ tt0v(s)ds} in Rn× (t0,∞).If v′(t) ≥ 0 does not hold for all t ≥ t0, we assume − v′(t)v(t) ≤ Cvt−1. Then, for r > 1, r−l2g(r) ≥(2C1(n, l))−1r−2. So in r ≤ r0 := (2CvC1(n, l))− 12 t 12 ,− r−l2g(r)− v′(t)v(t)≤−(2C1(n, l))−1r−2+Cvt−1 = (2C1(n, l))−1r−2(−1+2C1(n, l)Cvr2t−1)≤ 0.For r ≤ 1, when l1 ≥ 0, t0 ≥ 2CvC2(n, l1, l), we have− r−l12g(r)− v′(t)v(t)≤−(2C2(n, l1, l))−1+Cvt−1 ≤ 0.We still need to find D2 to guarantee ϕ2(r0, t)≥ u(r0, t). It suffices to ensure ϕ2(r0, t)≥ ϕ1(r0, t).That is D2rl0g(r0)≥ 2(v(t))−1´ tt0v(s)ds. By Lemma 4.9.1, we knowg(r)≥ r2−l(l−2)(n−2) in r ≥ 1.Since (v(t))−1´ tt0v(s)ds≤Cvt, when t0 ≥ 2CvC1(n, l), i.e. r0 ≥ 1, it suffices to ensureD2rl0r2−l0(l−2)(n−2) ≥ 2Cvt.That is,D2 ≥ 4(Cv)2(l−2)(n−2)C1(n, l) = 4(Cv)2(n− l)−1(n−2).Take D2 =max{2,4(Cv)2(n− l)−1(n−2)}. So ϕ2 is a supersolution of u in r ≤ (2CvC1(n, l))− 12 t 12 .Thus for 2< l≤ n−2, 0≤ l1≤ l,−(v(t))−1v′(t)≤Cvt−1, (v(t))−1´ tt0v(s)ds≤Cvt, t0≥ 2Cvmax{C1(n, l),C2(n, l1, l)},u(r, t)≤min{max{2,4(Cv)2n−2n− l}v(t)g(r)1{r≤(2CvC1(n,l))−12 t12 },2r−lˆ tt0v(s)ds}.The next lemma gives the convolution estimate, where the relationship between the constant in-volved in “≲” and parameters are not explicit. This lemma is efficient if we do not need the accuratedependence on k.274Lemma 4.9.3. For d > 2, 2< ℓ∗ < d and v1(τ)≥ 0,(|Γ♮d | ∗ ∗(v1(s)〈z〉−ℓ∗))(y,τ,τ0)≲ 1{|y|≤τ 12 }(supτ1∈[τ/2,τ]v1(τ1)〈y〉2−ℓ∗ + τ−ℓ∗2ˆ τ2τ02v1(s)ds)+1{|y|>τ 12 }(τ supτ1∈[τ/2,τ]v1(τ1)+ˆ τ2τ02v1(s)ds)|y|−ℓ∗ .(4.9.4)Proof. Noticev1(s)〈z〉−ℓ∗ ∼ v1(s)(1{|z|≤1}+ |z|−ℓ∗1{1<|z|≤τ 12 }+ |z|−ℓ∗1{|z|>τ 12 }).By [108, Lemma A.1],|Γ♮d | ∗ ∗(v1(s)1{|z|≤1})≲ τ− d2 e−c|y|2τˆ τ2τ02v1(s)ds+ supτ1∈[τ/2,τ]v1(τ1)(1{|y|≤1}+1{|y|>1}|y|2−de−c|y|2τ)where c> 0 depends on a. And|Γ♮d | ∗ ∗(v1(s)|z|−ℓ∗1{1<|z|≤s 12 })≲ τ− d2 e−c|y|2τˆ τ2τ02v1(s)sd−ℓ∗2 ds+ supτ1∈[τ/2,τ]v1(τ1)(1{|y|≤τ 12 }〈y〉2−ℓ∗ +1{|y|>τ 12 }|y|2−de−c|y|2τ τd−ℓ∗2)∼ 1{|y|≤τ 12 }(supτ1∈[τ/2,τ]v1(τ1)〈y〉2−ℓ∗ + τ− d2ˆ τ2τ02v1(s)sd−ℓ∗2 ds)+1{|y|>τ 12 }(supτ1∈[τ/2,τ]v1(τ1)|y|2−dτd−ℓ∗2 + τ−d2ˆ τ2τ02v1(s)sd−ℓ∗2 ds)e−c|y|2τ .By [108, Lemma A.2], we have|Γ♮d |∗∗(v1(s)〈z〉−ℓ∗1{|z|>s 12 })≲(τ supτ1∈[τ/2,τ]v1(τ1)+ˆ τ2τ02v1(s)ds)(1{|y|≤τ 12 }τ− ℓ∗2 +1{|y|>τ 12 }|y|−ℓ∗).Collecting the above estimates, we conclude the Lemma.2754.10 Additional information 2: Integral estimate for distorted FourierLemma 4.10.1. Assume´ 10 xa〈lnx〉bdx<∞, that is, a, b satisfy either a>−1 or a=−1 and b<−1.For 0≤ x0 ≤ x1 ≤ 12 , we haveˆ x1x0e−λxxa(− lnx)bdx≲xa+11 (− lnx1)b if a>−1(− lnx1)b+1− (− lnx0)b+1 if a=−1,b<−1 for 0≤ λ ≤ x−11(lnλ )bλ a+1 +0 if a>−1(lnλ )b+1− (− lnx0)b+1 if a=−1,b<−1 for x−11 ≤ λ ≤ x−10(lnλ )bλ a+1 e− x0λ2 for λ ≥ x−10.(4.10.1)Specially, for a constant C∗ > 0, 0≤ x0 ≤ c1 where c1 =min{12 ,C∗}, we haveˆ C∗x0e−λxxa〈lnx〉bdx≲1 for 0≤ λ ≤ 2(lnλ )bλ a+1 +0 if a>−1(lnλ )b+1− (− lnx0)b+1 if a=−1,b<−1 for 2≤ λ ≤ x−10(lnλ )bλ a+1 e− x0λ2 for λ ≥ x−10.For x0 = 0,ˆ C∗0e−λxxa〈lnx〉bdx≲1 for 0≤ λ ≤ 2(lnλ )bλ a+1 if a>−1(lnλ )b+1 if a=−1,b<−1for 2≤ λ < ∞.Proof. For 0≤ λ ≤ x−11 ,ˆ x1x0e−λxxa(− lnx)bdx∼ˆ x1x0xa(− lnx)bdx≲xa+11 (− lnx1)b if a>−1(− lnx1)b+1− (− lnx0)b+1 if a=−1,b<−1 .(4.10.2)276For λ ≥ x−10 ,ˆ x1x0e−λxxa(− lnx)bdx= 1λ a+1ˆ x1λx0λe−zza(lnλ − lnz)bdz≲(lnλ )bλ a+1 e− x0λ2 if x−10 ≤ λ ≤ x−201λ a+1 e− 3x0λ4 if λ ≥ x−20≲ (lnλ )bλ a+1e−x0λ2 .(4.10.3)In order to get the first "≲", one needs the following estimate.If x1λ ≤ λ 12 , that is λ ≤ x−21 ,1λ a+1ˆ x1λx0λe−zza(lnλ − lnz)bdz∼ (lnλ )bλ a+1ˆ x1λx0λe−zzadz≲ (lnλ )bλ a+1e−x0λ2 . (4.10.4)If x0λ ≥ λ 12 , that is λ ≥ x−20 ,1λ a+1ˆ x1λx0λe−zza(lnλ − lnz)bdz≲ 1λ a+1e−3x0λ4 , (4.10.5)since − lnx1 ≤ lnλ − lnz≤ lnλ − ln(x0λ )≤ lnλ2 ,za ≲1 if a≤ 0λ a if a> 0 , (lnλ − lnz)b ≲1 if b≤ 0(lnλ )b if b> 0 .If x0λ ≤ λ 12 ≤ x1λ , that is x−21 ≤ λ ≤ x−20 ,1λ a+1ˆ x1λx0λe−zza(lnλ − lnz)bdz≲ (lnλ )bλ a+1e−x0λ2 ,since by (4.10.4),1λ a+1ˆ λ 12x0λe−zza(lnλ − lnz)bdz≲ (lnλ )bλ a+1e−x0λ2 ,and by (4.10.5),1λ a+1ˆ x1λλ12e−zza(lnλ − lnz)bdz≲ 1λ a+1e−3λ124 .With the restriction x−10 ≤ λ ≤ x−20 , one has 1λ a+1 e−3λ124 ≲ (lnλ )bλ a+1 e− x0λ2 .277For x−11 ≤ λ ≤ x−10 , we haveˆ x1x0e−λxxa(− lnx)bdx≲ (lnλ )bλ a+1+0 if a>−1(lnλ )b+1− (− lnx0)b+1 if a=−1,b<−1 .Indeed, by (4.10.2),ˆ 1λx0e−λxxa(− lnx)bdx≲(lnλ )bλ a+1 if a>−1(lnλ )1+b− (− lnx0)1+b if a=−1,b<−1,and by (4.10.3), ˆ x11λe−λxxa(− lnx)bdx≲ (lnλ )bλ a+1.This complete the proof of general case (4.10.1).For the special cases, ifC∗ ≤ 12 , by (4.10.1),ˆ C∗x0e−λxxa〈lnx〉bdx≲1 for 0≤ λ ≤C−1∗(lnλ )bλ a+1 +0 if a>−1(lnλ )b+1− (− lnx0)b+1 if a=−1,b<−1 for C−1∗ ≤ λ ≤ x−10(lnλ )bλ a+1 e− x0λ2 for λ ≥ x−10≲1 for 0≤ λ ≤ 2(lnλ )bλ a+1 +0 if a>−1(lnλ )b+1− (− lnx0)b+1 if a=−1,b<−1 for 2≤ λ ≤ x−10(lnλ )bλ a+1 e− x0λ2 for λ ≥ x−10.IfC∗ ≥ 12 , ˆ C∗x0e−λxxa〈lnx〉bdx= (ˆ 12x0+ˆ C∗12)e−λxxa〈lnx〉bdx.The first part can be estimated in the same way as above. For the second part,ˆ C∗12e−λxxa〈lnx〉bdx≲ e− λ2 .This concludes the proof.2784.11 Additional information 3: Convolution estimates in finite time4.11.1 PreliminariesFor s≤ t and t ≤ t∗ ≤ T ,T−s2 ≤ t− s≤ T − s for s≤ t− (T − t)T − t ≤ T − s≤ 2(T − t) for s≥ t− (T − t)t∗−s2 ≤ t− s≤ t∗− s for s≤ t− (t∗− t)t∗− t ≤ t∗− s≤ 2(t∗− t) for s≥ t− (t∗− t).(4.11.1)For any x,q ∈ Rd , p> 0, b≥ 0 and L> 0, 0≤ L1 ≤ L2 ≤ ∞,ˆRde−c(|x−y|√L)p |y−q|−b1{L1≤|y−q|≤L2}dy= Ld2− b2ˆRde−c|x˜−z|p |z|−b1{L1L− 12≤|z|≤L2L− 12 }dz≤ L d2− b2ˆRde−c|x˜−z|pmin{|z|−b,(L1L− 12 )−b}1{|z|≤L2L− 12 }dz≤ L d2− b2ˆRde−c|z|pmin{|z|−b,(L1L− 12 )−b}1{|z|≤L2L− 12 }dz= Ld2− b2{ˆRde−c|z|p[(L1L−12 )−b1{|z|≤L1L−12 }+ |z|−b1{L1L−12<|z|≤L2L−12 }]dz}≲Ld2 L−b1 if L≤ L21Ld2− b2 if b< d〈ln( LL21 )〉 if b= dLd−b1 if b> dif L21 < L≤ L22Ld−b2 if b< d〈ln(L2L1 )〉 if b= dLd−b1 if b> dif L> L22(4.11.2)where x˜= (x−q)L− 12 .Specially, for b≥ 0, L3 ≥CL> 0 whereC > 0 is a constant,ˆRde−c(|x−y|√L)p |y−q|−b1{|y−q|≥√L3}dy≲ Ld2 L−b23 . (4.11.3)279Next, we want to establish the basic calculation of the time variable. Setg(s) =(t− s) d2−d∗L−b1 if t− s≤ L21(t− s) d2− b2−d∗ if b< d(t− s)−d∗〈ln( t−sL21 )〉 if b= d(t− s)−d∗Ld−b1 if b> dif L21 < t− s≤ L22(t− s)−d∗Ld−b2 if b< d(t− s)−d∗〈ln(L2L1 )〉 if b= d(t− s)−d∗Ld−b1 if b> dif t− s> L22.Claim: for δ > 0, if d2 +1> d∗ >d2 +1−δ ,x−δˆ tt−xg(s)ds≤ ∞,and for δ = 0, d∗ < d2 +1,ˆ tt−xg(s)ds≲(max{x,L22})1−d∗Ld−b2 if d∗ < 1〈ln(max{x,L22}L22 )〉Ld−b2 if d∗ = 1Ld+2−b−2d∗2 if 1< d∗ < 1+d−b2〈ln(L2L1 )〉 if d∗ = 1+ d−b2Ld+2−b−2d∗1 if d∗ > 1+d−b2if b< d(max{x,L22})1−d∗〈ln(L2L1 )〉 if d∗ < 1〈ln(max{x,L22}L21 )〉〈ln(L2L1)〉 if d∗ = 1L2−2d∗1 if d∗ > 1if b= d(max{x,L22})1−d∗Ld−b1 if d∗ < 1〈ln(max{x,L22}L21 )〉Ld−b1 if d∗ = 1Ld+2−b−2d∗1 if d∗ > 1if b> d.(4.11.4)280and if δ > 0, d∗ ≤ d2 +1−δ ,x−δˆ tt−xg(s)ds≲(max{x,L22})1−d∗−δLd−b2 if d∗ ≤ 1−δLd+2−b−2d∗−2δ2 if 1−δ < d∗ ≤ 1+ d−b2 −δLd+2−b−2d∗−2δ1 if d∗ > 1+d−b2 −δif b< d(max{x,L22})1−d∗−δ 〈ln(L2L1)〉 if d∗ ≤ 1−δL2−2d∗−2δ1 if d∗ > 1−δif b= d(max{x,L22})1−d∗−δLd−b1 if d∗ ≤ 1−δLd+2−b−2d∗−2δ1 if d∗ > 1−δ if b> d.(4.11.5)Proof. For x≤ L21, due to the assumption d∗ < d2 +1, thenˆ tt−xg(s)ds≲ x d2+1−d∗L−b1 .For L21 < x≤ L22,ˆ tt−xg(s)ds= (ˆ tt−L21+ˆ t−L21t−x)g(s)ds≲ Ld+2−b−2d∗1+xd2+1− b2−d∗ , b< d+2−2d∗ln( xL21), b= d+2−2d∗Ld+2−b−2d∗1 , b> d+2−2d∗if b< dx1−d∗〈ln( xL21 )〉, d∗ < 1〈ln( xL21 )〉2, d∗ = 1L2−2d∗1 , d∗ > 1if b= dx1−d∗Ld−b1 , d∗ < 1Ld−b1 ln(xL21), d∗ = 1Ld+2−b−2d∗1 , d∗ > 1if b> d∼xd2+1− b2−d∗ , b< d+2−2d∗〈ln( xL21 )〉, b= d+2−2d∗Ld+2−b−2d∗1 , b> d+2−2d∗if b< dx1−d∗〈ln( xL21 )〉, d∗ < 1〈ln( xL21 )〉2, d∗ = 1L2−2d∗1 , d∗ > 1if b= dx1−d∗Ld−b1 , d∗ < 1Ld−b1 〈ln( xL21 )〉, d∗ = 1Ld+2−b−2d∗1 , d∗ > 1if b> d.281For x> L22,ˆ tt−xg(s)ds= (ˆ tt−L22+ˆ t−L22t−x)g(s)ds≲Ld+2−b−2d∗2 , b< d+2−2d∗〈ln(L2L1 )〉, b= d+2−2d∗Ld+2−b−2d∗1 , b> d+2−2d∗if b< dL2−2d∗2 〈ln(L2L1 )〉, d∗ < 1〈ln(L2L1 )〉2, d∗ = 1L2−2d∗1 , d∗ > 1if b= dL2−2d∗2 Ld−b1 , d∗ < 1Ld−b1 〈ln(L2L1 )〉, d∗ = 1Ld+2−b−2d∗1 , d∗ > 1if b> d+x1−d∗Ld−b2 , d∗ < 1ln( xL22)Ld−b2 , d∗ = 1Ld+2−b−2d∗2 , d∗ > 1if b< dx1−d∗〈ln(L2L1 )〉, d∗ < 1ln( xL22)〈ln(L2L1 )〉, d∗ = 1L2−2d∗2 〈ln(L2L1 )〉, d∗ > 1if b= dx1−d∗Ld−b1 , d∗ < 1ln( xL22)Ld−b1 , d∗ = 1L2−2d∗2 Ld−b1 , d∗ > 1if b> d∼x1−d∗Ld−b2 , d∗ < 1Ld−b2 〈ln( xL22 )〉, d∗ = 1Ld+2−b−2d∗2 , 1< d∗ < 1+d−b2〈ln(L2L1 )〉, d∗ = 1+ d−b2Ld+2−b−2d∗1 , d∗ > 1+d−b2if b< dx1−d∗〈ln(L2L1 )〉, d∗ < 1〈ln( xL21 )〉〈ln(L2L1)〉, d∗ = 1L2−2d∗1 , d∗ > 1if b= dx1−d∗Ld−b1 , d∗ < 1Ld−b1 〈ln( xL21 )〉, d∗ = 1Ld+2−b−2d∗1 , d∗ > 1if b> d.282Thus, for d∗ < d2 +1,ˆ tt−xg(s)ds≲xd2+1−d∗L−b1 if x≤ L21xd2+1− b2−d∗ b< d+2−2d∗〈ln( xL21 )〉 b= d+2−2d∗Ld+2−b−2d∗1 b> d+2−2d∗if b< dx1−d∗〈ln( xL21 )〉 if d∗ < 1〈ln( xL21 )〉2 if d∗ = 1L2−2d∗1 if d∗ > 1if b= dx1−d∗Ld−b1 if d∗ < 1Ld−b1 〈ln( xL21 )〉 if d∗ = 1Ld+2−b−2d∗1 if d∗ > 1if b> dif L21 < x≤ L22x1−d∗Ld−b2 if d∗ < 1Ld−b2 〈ln( xL22 )〉 if d∗ = 1Ld+2−b−2d∗2 if 1< d∗ < 1+d−b2〈ln(L2L1 )〉 if d∗ = 1+ d−b2Ld+2−b−2d∗1 if d∗ > 1+d−b2if b< dx1−d∗〈ln(L2L1 )〉 if d∗ < 1〈ln( xL21 )〉〈ln(L2L1)〉 if d∗ = 1L2−2d∗1 if d∗ > 1if b= dx1−d∗Ld−b1 if d∗ < 1Ld−b1 〈ln( xL21 )〉 if d∗ = 1Ld+2−b−2d∗1 if d∗ > 1if b> dif x> L22.283Thenx−δˆ tt−xg(s)ds≲Ld+2−b−2d∗−2δ1 if δ ≤ d2 +1−d∗∞ if δ > d2 +1−d∗if x≤ L21Ld+2−b−2d∗−2δ2 if b≤ d+2−2d∗−2δLd+2−b−2d∗−2δ1 if b> d+2−2d∗−2δb< d+2−2d∗L−2δ2 〈ln(L2L1)〉 if δ ≤ 0L−2δ1 if δ > 0b= d+2−2d∗L−2δ2 Ld+2−b−2d∗1 if δ ≤ 0Ld+2−b−2d∗−2δ1 if δ > 0b> d+2−2d∗if b< dL2−2d∗−2δ2 〈ln(L2L1 )〉 if δ ≤ 1−d∗L2−2d∗−2δ1 if δ > 1−d∗if d∗ < 1L−2δ2 〈ln(L2L1)〉2 if δ ≤ 0L−2δ1 if δ > 0if d∗ = 1L−2δ2 L2−2d∗1 if δ ≤ 0L2−2d∗−2δ1 if δ > 0if d∗ > 1if b= dL2−2d∗−2δ2 Ld−b1 if δ ≤ 1−d∗Ld+2−b−2d∗−2δ1 if δ > 1−d∗if d∗ < 1L−2δ2 Ld−b1 〈ln(L2L1)〉 if δ ≤ 0Ld−b−2δ1 if δ > 0if d∗ = 1L−2δ2 Ld+2−b−2d∗1 if δ ≤ 0Ld+2−b−2d∗−2δ1 if δ > 0if d∗ > 1if b> dif L21 < x≤ L22.284For x> L22,x−δˆ tt−xg(s)ds≲x1−d∗−δLd−b2 if δ ≤ 1−d∗Ld+2−b−2d∗−2δ2 if δ > 1−d∗ if d∗ < 1x−δ 〈ln(xL22)〉Ld−b2 if δ ≤ 0Ld−b−2δ2 if δ > 0if d∗ = 1x−δLd+2−b−2d∗2 if δ ≤ 0Ld+2−b−2d∗−2δ2 if δ > 0if 1< d∗ < 1+ d−b2x−δ 〈ln(L2L1)〉 if δ ≤ 0L−2δ2 〈ln(L2L1 )〉 if δ > 0if d∗ = 1+ d−b2x−δLd+2−b−2d∗1 if δ ≤ 0L−2δ2 Ld+2−b−2d∗1 if δ > 0if d∗ > 1+ d−b2if b< dx1−d∗−δ 〈ln(L2L1)〉 if δ ≤ 1−d∗L2−2d∗−2δ2 〈ln(L2L1 )〉 if δ > 1−d∗if d∗ < 1x−δ 〈ln(xL21)〉〈ln(L2L1 )〉 if δ ≤ 0L−2δ2 〈ln(L2L1 )〉2 if δ > 0if d∗ = 1x−δL2−2d∗1 if δ ≤ 0L−2δ2 L2−2d∗1 if δ > 0if d∗ > 1if b= dx1−d∗−δLd−b1 if δ ≤ 1−d∗L2−2d∗−2δ2 Ld−b1 if δ > 1−d∗ if d∗ < 1x−δ 〈ln(xL21)〉Ld−b1 if δ ≤ 0L−2δ2 Ld−b1 〈ln(L2L1 )〉 if δ > 0if d∗ = 1x−δLd+2−b−2d∗1 if δ ≤ 0L−2δ2 Ld+2−b−2d∗1 if δ > 0if d∗ > 1if b> d.285Specially, for δ = 0,ˆ tt−xg(s)ds≲Ld+2−b−2d∗1 if d∗ ≤ d2 +1∞ if d∗ > d2 +1if x≤ L21Ld+2−b−2d∗2 if d∗ < 1+d−b2〈ln(L2L1 )〉 if d∗ = 1+ d−b2Ld+2−b−2d∗1 if d∗ > 1+d−b2if b< dL2−2d∗2 〈ln(L2L1 )〉 if d∗ < 1〈ln(L2L1 )〉2 if d∗ = 1L2−2d∗1 if d∗ > 1if b= dL2−2d∗2 Ld−b1 if d∗ < 1Ld−b1 〈ln(L2L1 )〉 if d∗ = 1Ld+2−b−2d∗1 if d∗ > 1if b> dif L21 < x≤ L22,For x> L22,ˆ tt−xg(s)ds≲x1−d∗Ld−b2 if d∗ < 1〈ln( xL22 )〉Ld−b2 if d∗ = 1Ld+2−b−2d∗2 if 1< d∗ < 1+d−b2〈ln(L2L1 )〉 if d∗ = 1+ d−b2Ld+2−b−2d∗1 if d∗ > 1+d−b2if b< dx1−d∗〈ln(L2L1 )〉 if d∗ < 1〈ln( xL21 )〉〈ln(L2L1)〉 if d∗ = 1L2−2d∗1 if d∗ > 1if b= dx1−d∗Ld−b1 if d∗ < 1〈ln( xL21 )〉Ld−b1 if d∗ = 1Ld+2−b−2d∗1 if d∗ > 1if b> d.286For δ > 0,x−δˆ tt−xg(s)ds≲Ld+2−b−2d∗−2δ1 if δ ≤ d2 +1−d∗∞ if δ > d2 +1−d∗if x≤ L21Ld+2−b−2d∗−2δ2 if d∗ ≤ 1+ d−b2 −δLd+2−b−2d∗−2δ1 if d∗ > 1+d−b2 −δif b< dL2−2d∗−2δ2 〈ln(L2L1 )〉 if δ ≤ 1−d∗L2−2d∗−2δ1 if δ > 1−d∗if b= dL2−2d∗−2δ2 Ld−b1 if δ ≤ 1−d∗Ld+2−b−2d∗−2δ1 if δ > 1−d∗if b> dif L21 < x≤ L22.For x> L22,x−δˆ tt−xg(s)ds≲x1−d∗−δLd−b2 if δ ≤ 1−d∗Ld+2−b−2d∗−2δ2 if δ > 1−d∗ if d∗ < 1+ d−b2L−2δ2 〈ln(L2L1 )〉 if d∗ = 1+ d−b2L−2δ2 Ld+2−b−2d∗1 if d∗ > 1+d−b2if b< dx1−d∗−δ 〈ln(L2L1)〉 if δ ≤ 1−d∗L2−2d∗−2δ2 〈ln(L2L1 )〉 if δ > 1−d∗if d∗ < 1L−2δ2 〈ln(L2L1 )〉2 if d∗ = 1L−2δ2 L2−2d∗1 if d∗ > 1if b= dx1−d∗−δLd−b1 if δ ≤ 1−d∗L2−2d∗−2δ2 Ld−b1 if δ > 1−d∗ if d∗ < 1L−2δ2 Ld−b1 〈ln(L2L1 )〉 if d∗ = 1L−2δ2 Ld+2−b−2d∗1 if d∗ > 1if b> d.• Throughout this section, we assumev(t)=Cv(T )(T−t)m1(ln(T−t))m2(ln ln(T−t))m3 · · · , li(t)=C2(T )(T−t)ki1(ln(T−t))ki2(ln ln(T−t))ki3 · · ·with finite terms multiplication and m j,ki j ∈R for i= 1,2, j= 1,2, . . . . And l2(t)≤C(T − t) 12 wherethe general constant C is independent of T .287Setψ(x, t) =ˆ t0v(s)(t− s)−d∗ˆRde−c(|x−y|√t−s )2−δ |y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds.Claim: for b≥ 0, d∗ < d2 +1,ψ(x, t)≲ˆ t−(T−t)0v(s)(T − s)−d∗ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)(T − t)1−d∗ ld−b2 (t) if d∗ < 1〈ln( T−tl22(t))〉ld−b2 (t) if d∗ = 1ld+2−b−2d∗2 (t) if 1< d∗ < 1+d−b2〈ln( l2(t)l1(t))〉 if d∗ = 1+ d−b2ld+2−b−2d∗1 (t) if d∗ > 1+d−b2if b< d(T − t)1−d∗〈ln( l2(t)l1(t))〉 if d∗ < 1〈ln( T−tl21(t))〉〈ln(l2(t)l1(t))〉 if d∗ = 1l2−2d∗1 (t) if d∗ > 1if b= d(T − t)1−d∗ ld−b1 (t) if d∗ < 1〈ln( T−tl21(t))〉ld−b1 (t) if d∗ = 1ld+2−b−2d∗1 (t) if d∗ > 1if b> d.(4.11.6)Remark 4.11.0.1. When b= 0< d, the cases d∗ = d2 − b2 +1 and d∗ > d2 − b2 +1 are vacuum.Proof. Since b≥ 0, by (4.11.2),ψ(x, t)≲ˆ t0v(s)(t− s) d2−d∗ l−b1 (s) if t− s≤ l21(s)(t− s) d2− b2−d∗ if b< d(t− s)−d∗〈ln( t−sl21(s))〉 if b= d(t− s)−d∗ ld−b1 (s) if b> dif l21(s)< t− s≤ l22(s)(t− s)−d∗ ld−b2 (s) if b< d(t− s)−d∗〈ln( l2(s)l1(s))〉 if b= d(t− s)−d∗ ld−b1 (s) if b> dif t− s> l22(s)ds=ˆ t−(T−t)0+ˆ tt−(T−t)· · ·=: P1+P2where we have used b≥ 0 in the first “≤”.288For P1, since T−s2 ≤ t− s≤ T − s, l22(s)≤C(T − s),P1 ∼ˆ t−(T−t)0v(s)(T − s)−d∗ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds. (4.11.7)For P2, since T − t ≤ T − s≤ 2(T − t),P2 ∼ v(t)ˆ tt−(T−t)(t− s) d2−d∗ l−b1 (t) if t− s≤ l21(t)(t− s) d2− b2−d∗ if b< d(t− s)−d∗〈ln( t−sl21(t))〉 if b= d(t− s)−d∗ ld−b1 (t) if b> dif l21(t)< t− s≤ l22(t)(t− s)−d∗ ld−b2 (t) if b< d(t− s)−d∗〈ln( l2(t)l1(t))〉 if b= d(t− s)−d∗ ld−b1 (t) if b> dif t− s> l22(t)ds≲ v(t)(T − t)1−d∗ ld−b2 (t) if d∗ < 1〈ln( T−tl22(t))〉ld−b2 (t) if d∗ = 1ld+2−b−2d∗2 (t) if 1< d∗ < 1+d−b2〈ln( l2(t)l1(t))〉 if d∗ = 1+ d−b2ld+2−b−2d∗1 (t) if d∗ > 1+d−b2if b< d(T − t)1−d∗〈ln( l2(t)l1(t))〉 if d∗ < 1〈ln( T−tl21(t))〉〈ln(l2(t)l1(t))〉 if d∗ = 1l2−2d∗1 (t) if d∗ > 1if b= d(T − t)1−d∗ ld−b1 (t) if d∗ < 1〈ln( T−tl21(t))〉ld−b1 (t) if d∗ = 1ld+2−b−2d∗1 (t) if d∗ > 1if b> dby (4.11.4) and l22(t)≤C(T − t) when d∗ < d2 +1.• For b≥ 0, d∗ < d2 +1, by (4.11.3),ˆ t0v(s)(t− s)−d∗ˆRde−c(|x−y|√t−s )2−δ |y−q|−b1{|y−q|≥(T−s) 12 }dyds289≲ˆ t0v(s)(t− s) d2−d∗(T − s)− b2 ds=ˆ t−(T−t)0+ˆ tt−(T−t)· · ·≲ˆ t−(T−t)0v(s)(T − s) d2−d∗− b2 ds+ v(t)(T − t)− b2ˆ tt−(T−t)(t− s) d2−d∗ds∼ˆ t−(T−t)0v(s)(T − s) d2−d∗− b2 ds+ v(t)(T − t)1−d∗+ d2− b2∼ˆ t0v(s)(T − s) d2−d∗− b2 ds (4.11.8)for d∗ < d2 +1.4.11.2 Convolution with v(t)|x−q|−b1{l1(t)≤|x−q|≤l2(t)}For| f (x, t)| ≤ v(t)|x−q|−b1{l1(t)≤|x−q|≤l2(t)},considerψ(x, t) =T outd [ f ] :=ˆ t0ˆRdΓ(x, t,y,s) f (y,s)dyds.For d ≥ 1,|ψ(x, t)|≲ˆ t−(T−t)0v(s)(T − s)− d2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)(T − t)1− d2 ld−b2 (t) if d < 2l2−b2 (t)〈ln( T−tl22(t))〉 if d = 2l2−b2 (t) if d > 2,b< 2〈ln( l2(t)l1(t))〉 if b= 2l2−b1 (t) if b> 2if b< d(T − t)1− d2 〈ln( l2(t)l1(t))〉 if d < 2〈ln( T−tl21(t))〉〈ln(l2(t)l1(t))〉 if d = 2l2−d1 (t) if d > 2if b= d(T − t)1− d2 ld−b1 (t) if d < 2l2−b1 (t)〈ln( T−tl21(t))〉 if d = 2l2−b1 (t) if d > 2if b> d.(4.11.9)290|∇ψ(x, t)|≲ˆ t−(T−t)0v(s)(T − s)− d+12ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)l1−b2 (t)〈ln( T−tl22(t))〉 if d = 1l1−b2 (t) if d > 1,b< 1〈ln( l2(t)l1(t))〉 if b= 1l1−b1 (t) if b> 1if b< d〈ln(T−tl21(t))〉〈ln( l2(t)l1(t))〉 if d = 1l1−d1 (t) if d > 1if b= dl1−b1 (t)〈ln( T−tl21(t))〉 if d = 1l1−b1 (t) if d > 1if b> d.(4.11.10)|ψ(x, t)−ψ(x,T )|≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− d2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)ˆ tt−(T−t)l−b1 (t) if t− s≤ l21(t)(t− s)− b2 if b< d(t− s)− d2 〈ln( t−sl21(t))〉 if b= dld−b1 (t)(t− s)−d2 if b> dif l21(t)< t− s≤ l22(t)ld−b2 (t)(t− s)−d2 if b< d〈ln( l2(t)l1(t))〉(t− s)−d2 if b= dld−b1 (t)(t− s)−d2 if b> dif t− s> l22(t)ds+ v(t)(T − t)1− d2ld−b2 (t) if b< d〈ln( l2(t)l1(t))〉 if b= dld−b1 (t) if b> d+ˆ Tt(T − s)− d2 v(s)ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds.(4.11.11)291For b≥ 0, 0< α < 1,|∇ψ(x, t)−∇ψ(x,T )|≲C(α)[(T − t) α2ˆ t−(T−t)0v(s)(T − s)− d+1+α2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)〈ln( T−tl22(t))〉l1−b2 (t) if d = 1l1−b2 (t) if d > 1,b< 1〈ln( l2(t)l1(t))〉 if b= 1l1−b1 (t) if b> 1if b< d〈ln(T−tl21(t))〉〈ln( l2(t)l1(t))〉 if d = 1l1−d1 (t) if d > 1if b= d〈ln(T−tl21(t))〉l1−b1 (t) if d = 1l1−b1 (t) if d > 1if b> d.]+ˆ Ttv(s)(T − s)− d+12ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds.(4.11.12)For b≥ 0, 0< α < 1 and t < t∗ ≤ T ,|∇ψ(x, t)−∇ψ(x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0v(s)(T − s)− d+1+α2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds+ v(t)l1−α−b2 (t) if b≤ 1−αl1−α−b1 (t) if b> 1−α]+1{t∗> T+t2 }[(T − t∗) α2 v(t∗)l1−α−b2 (t∗) if b≤ 1−αl1−α−b1 (t∗) if b> 1−α+ˆ t∗−(T−t∗)tv(s)(T − s)− d+12ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds].(4.11.13)292(4.11.9) and (4.11.10) are derived by (4.6.5) and (4.11.6).Proof of (4.11.11).ψ(x, t)−ψ(x,T ) =ˆ t−(T−t)0ˆRd(Γ(x, t,y,s)−Γ(x,T,y,s)) f (y,s)dyds+ˆ tt−(T−t)ˆRd(Γ(x, t,y,s)−Γ(x,T,y,s)) f (y,s)dyds−ˆ TtˆRdΓ(x,T,y,s) f (y,s)dyds := I1+ I2+ I3.By (4.6.5),|I1|≲ (T − t)ˆ t−(T−t)0ˆRdˆ 10|(∂tΓ)(x,θ t+(1−θ)T,y,s)| | f (y,s)|dyds≲ (T − t)ˆ t−(T−t)0ˆRdˆ 10[θ t+(1−θ)T − s]−1− d2 e−c(|x−y|√θ t+(1−θ)T−s)2−δ× v(s)|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dθdyds≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− d2ˆRde−c( |x−y|√T−s)2−δ|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− d2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> ddswhere we have used (4.11.2) in the last "≲" and l22(s)≤C(T − s).By (4.6.5),|I2|≲ˆ tt−(T−t)ˆRd(|Γ(x, t,y,s)|+ |Γ(x,T,y,s)|)| f (y,s)|dyds≲ˆ tt−(T−t)v(s)(t− s)− d2ˆRde−c( |x−y|√t−s)2−δ|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds+ˆ tt−(T−t)v(s)(T − s)− d2ˆRde−c( |x−y|√T−s)2−δ|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds293≲ˆ tt−(T−t)v(s)(t− s)− d2(t− s) d2 l−b1 (s) if t− s≤ l21(s)(t− s) d2− b2 if b< d〈ln( t−sl21(s))〉 if b= dld−b1 (s) if b> dif l21(s)< t− s≤ l22(s)ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dif t− s> l22(s)ds+ˆ tt−(T−t)v(s)(T − s)− d2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds≲ v(t)ˆ tt−(T−t)l−b1 (t) if t− s≤ l21(t)(t− s)− b2 if b< d(t− s)− d2 〈ln( t−sl21(t))〉 if b= dld−b1 (t)(t− s)−d2 if b> dif l21(t)< t− s≤ l22(t)ld−b2 (t)(t− s)−d2 if b< d〈ln( l2(t)l1(t))〉(t− s)−d2 if b= dld−b1 (t)(t− s)−d2 if b> dif t− s> l22(t)ds+ v(t)(T − t)1− d2ld−b2 (t) if b< d〈ln( l2(t)l1(t))〉 if b= dld−b1 (t) if b> dwhere we have used (4.11.2) and l22(s)≤C(T − s).By (4.6.5) and (4.11.2),|I3|≲ˆ TtˆRd(T − s)− d2 e−c( |x−y|√T−s)2−δv(s)|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲ˆ Tt(T − s)− d2 v(s)ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> ddswhere we have used l22(s)≤C(T − s).294Proof of (4.11.12).∂xiψ(x, t)−∂xiψ(x,T )=ˆ t0ˆRd(∂xiΓ(x, t,y,s)−∂xiΓ(x,T,y,s)) f (y,s)dyds−ˆ TtˆRd∂xiΓ(x,T,y,s) f (y,s)dyds=: I1+ I2.For I1, by (4.6.7) and (4.11.2),|I1|≲ (T − t) α2ˆ t0ˆRd(T − s)− α2 [(t− s)− d+12 e−β ( |x−y|√t−s )2−δ +(T − s)− d+12 e−β ( |x−y|√T−s )2−δ ]∗ v(s)|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲C(α)(T − t) α2[ˆ t0v(s)(T − s)− α2(t− s)− 12 l−b1 (s) if t− s≤ l21(s)(t− s)− b+12 if b< d(t− s)− d+12 〈ln( t−sl21(s))〉 if b= d(t− s)− d+12 ld−b1 (s) if b> dif l21(s)< t− s≤ l22(s)(t− s)− d+12 ld−b2 (s) if b< d(t− s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(t− s)− d+12 ld−b1 (s) if b> dif t− s> l22(s)ds+ˆ t0v(s)(T − s)− d+1+α2 ld−b2 (s) if b< d(T − s)− d+1+α2 〈ln( l2(s)l1(s))〉 if b= d(T − s)− d+1+α2 ld−b1 (s) if b> dds]=C(α)(T − t) α2 (ˆ t−(T−t)0+ˆ tt−(T−t)· · ·)=:C(α)(T − t) α2 (I11+ I12).For I11, by (4.11.1),I11 ≲ˆ t−(T−t)0v(s)(T − s)− d+1+α2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> ddssince t− s∼ t∗− s∼ T − s≳ l22(s).295For I12,I12 ≲ v(t)(T − t)− α2ˆ tt−(T−t)(t− s)− 12 l−b1 (t) if t− s≤ l21(t)(t− s)− 12− b2 if b< d(t− s)− 12− b2 〈ln( t−sl21(t))〉 if b= d(t− s)− 12− d2 ld−b1 (t) if b> dif l21(t)< t− s≤ l22(t)(t− s)− 12− d2 ld−b2 (t) if b< d(t− s)− 12− d2 〈ln( l2(t)l1(t))〉 if b= d(t− s)− 12− d2 ld−b1 (t) if b> dif t− s> l22(t)ds+ v(t)(T − t)− α2(T − t) 12− d2 ld−b2 (t) if b< d(T − t) 12− d2 〈ln( l2(t)l1(t))〉 if b= d(T − t) 12− d2 ld−b1 (t) if b> d≲ v(t)(T − t)− α2〈ln( T−tl22(t))〉l1−b2 (t) if d = 1l1−b2 (t) if d > 1,b< 1〈ln( l2(t)l1(t))〉 if b= 1l1−b1 (t) if b> 1if b< d〈ln(T−tl21(t))〉〈ln( l2(t)l1(t))〉 if d = 1l1−d1 (t) if d > 1if b= d〈ln(T−tl21(t))〉l1−b1 (t) if d = 1l1−b1 (t) if d > 1if b> d.where we have used (4.11.4) in the last “≲”.For I2,|I2|≲ˆ Ttv(s)(T − s)− d+12ˆRde−β (|x∗−y|√T−s )2−δ |y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲ˆ Ttv(s)(T − s)− d+12 ld−b2 (s) if b< d(T − s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(T − s)− d+12 ld−b1 (s) if b> ddsby (4.11.2).296Proof of (4.11.13).∂xiψ(x, t)−∂xiψ(x∗, t∗)=ˆ t0ˆRd(∂xiΓ(x, t,y,s)−∂xiΓ(x∗, t∗,y,s)) f (y,s)dyds−ˆ t∗tˆRd∂xiΓ(x∗, t∗,y,s) f (y,s)dyds=: I1+ I2.For I1, by (4.6.7) and (4.11.2),|I1|≲ (|x− x∗|+√|t− t∗|)αˆ t0ˆRd(t∗− s)− α2[(t− s)− d+12 e−β (|x−y|√t−s )2−δ+(t∗− s)− d+12 e−β (|x∗−y|√t∗−s )2−δ ]∗ v(s)|y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲C(α)(|x− x∗|+√|t− t∗|)α[ˆ t0v(s)(t∗− s)− α2(t− s)− 12 l−b1 (s) if t− s≤ l21(s)(t− s)− b+12 if b< d(t− s)− d+12 〈ln( t−sl21(s))〉 if b= d(t− s)− d+12 ld−b1 (s) if b> dif l21(s)< t− s≤ l22(s)(t− s)− d+12 ld−b2 (s) if b< d(t− s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(t− s)− d+12 ld−b1 (s) if b> dif t− s> l22(s)ds+ˆ t0v(s)(t∗− s)− 1+α2 l−b1 (s) if t∗− s≤ l21(s)(t∗− s)− b+1+α2 if b< d(t∗− s)− d+1+α2 〈ln( t∗−sl21(s))〉 if b= d(t∗− s)− d+1+α2 ld−b1 (s) if b> dif l21(s)< t∗− s≤ l22(s)(t∗− s)− d+1+α2 ld−b2 (s) if b< d(t∗− s)− d+1+α2 〈ln( l2(s)l1(s))〉 if b= d(t∗− s)− d+1+α2 ld−b1 (s) if b> dif t∗− s> l22(s)ds]=C(α)(|x− x∗|+√|t− t∗|)α(ˆ t−(T−t)0+ˆ t−(t∗−t)t−(T−t)+ˆ tt−(t∗−t)· · ·)=:C(α)(|x− x∗|+√|t− t∗|)α(I11+ I12+ I13).For I11,I11 ≲ˆ t−(T−t)0v(s)(T − s)− d+1+α2ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> dds297since t− s∼ t∗− s∼ T − s≳ l22(s).For I12, by (4.11.2),I12 ≲ v(t)ˆ t−(t∗−t)t−(T−t)(t− s)− 1+α2 l−b1 (t) if t− s≤ l21(t)(t− s)− b+1+α2 if b< d(t− s)− d+1+α2 〈ln( t−sl21(t))〉 if b= d(t− s)− d+1+α2 ld−b1 (t) if b> dif l21(t)< t− s≤ l22(t)(t− s)− d+1+α2 ld−b2 (t) if b< d(t− s)− d+1+α2 〈ln( l2(t)l1(t))〉 if b= d(t− s)− d+1+α2 ld−b1 (t) if b> dif t− s> l22(t)ds≲ v(t)l1−b−α2 (t) if b≤ 1−αl1−b−α1 (t) if b> 1−α if t∗− t ≤ l21(t)l1−b−α2 (t) if b≤ 1−α(t∗− t) 1−b−α2 if b> 1−α if b< d(t∗− t) 1−d−α2 〈ln( t∗−tl21(t))〉 if b= d(t∗− t) 1−d−α2 ld−b1 (t) if b> dif l21(t)< t∗− t ≤ l22(t)(t∗− t) 1−d−α2ld−b2 (t) if b< d〈ln( l2(t)l1(t))〉 if b= dld−b1 (t) if b> dif t∗− t ≥ l22(t)≲ v(t)l1−b−α2 (t) if b≤ 1−αl1−b−α1 (t) if b> 1−α if t∗− t ≤ l21(t)l1−b−α2 (t) if b≤ 1−αl1−b−α1 (t) if b> 1−α if b< dl1−d−α1 (t) if b= dl1−b−α1 (t) if b> dif l21(t)< t∗− t ≤ l22(t)l1−b−α2 (t) if b< dl1−d−α2 (t)〈ln( l2(t)l1(t))〉 if b= dl1−d−α2 (t)ld−b1 (t) if b> dif t∗− t ≥ l22(t)∼ v(t)l1−b−α2 (t) if b≤ 1−αl1−b−α1 (t) if b> 1−α298where we have used the following calculation in the second “≲”. For l21(t)< t∗− t ≤ l22(t),l1−d−α2 (t)ld−b2 (t) if b< d〈ln( l2(t)l1(t))〉 if b= dld−b1 (t) if b> d+ˆ t−(t∗−t)t−l22(t)(t− s)− b+1+α2 if b< d(t− s)− d+1+α2 〈ln( t−sl21(t))〉 if b= d(t− s)− d+1+α2 ld−b1 (t) if b> dds≲ l1−d−α2 (t)ld−b2 (t) if b< d〈ln( l2(t)l1(t))〉 if b= dld−b1 (t) if b> d+l1−b−α2 (t) if b≤ 1−α(t∗− t) 1−b−α2 if b> 1−α if b< d(t∗− t) 1−d−α2 〈ln( t∗−tl21(t))〉 if b= d(t∗− t) 1−d−α2 ld−b1 (t) if b> d∼l1−b−α2 (t) if b≤ 1−α(t∗− t) 1−b−α2 if b> 1−α if b< d(t∗− t) 1−d−α2 〈ln( t∗−tl21(t))〉 if b= d(t∗− t) 1−d−α2 ld−b1 (t) if b> dwhere for b= d, we have usedˆ t−(t∗−t)t−l22(t)(t− s)− d+1+α2 〈ln( t− sl21(t))〉ds= l1−d−α1 (t)ˆ l22 (t)l21 (t)t∗−tl21 (t)z−d+1+α2 〈lnz〉dz≲ (t∗− t) 1−d−α2 〈ln( t∗− tl21(t))〉.For t∗− t ≤ l21(t), l1−b−α2 (t) if b≤ 1−αl1−b−α1 (t) if b> 1−α.For I13,I13 ≲ v(t)(t∗− t)− α2ˆ tt−(t∗−t)(t− s)− 12 l−b1 (t) if t− s≤ l21(t)(t− s)− 12− b2 if b< d(t− s)− 12− b2 〈ln( t−sl21(t))〉 if b= d(t− s)− 12− d2 ld−b1 (t) if b> dif l21(t)< t− s≤ l22(t)(t− s)− 12− d2 ld−b2 (t) if b< d(t− s)− 12− d2 〈ln( l2(t)l1(t))〉 if b= d(t− s)− 12− d2 ld−b1 (t) if b> dif t− s> l22(t)ds299+ v(t)(t∗− t)− α2(t∗− t) 12 l−b1 (t) if t∗− t ≤ l21(t)(t∗− t) 12− b2 if b< d(t∗− t) 12− b2 〈ln( t∗−tl21(t))〉 if b= d(t∗− t) 12− d2 ld−b1 (t) if b> dif l21(t)< t∗− t ≤ l22(t)(t∗− t) 12− d2 ld−b2 (t) if b< d(t∗− t) 12− d2 〈ln( l2(t)l1(t))〉 if b= d(t∗− t) 12− d2 ld−b1 (t) if b> dif t∗− t > l22(t)≲ v(t)l1−α−b2 (t) b≤ 1−αl1−α−b1 (t) b> 1−αwhere we have used (4.11.5) in the last “≲”.For I2,|I2|≲ˆ t∗tv(s)(t∗− s)− d+12ˆRde−β (|x∗−y|√t∗−s )2−δ |y−q|−b1{l1(s)≤|y−q|≤l2(s)}dyds≲ˆ t∗tv(s)(t∗− s)− 12 l−b1 (s) if t∗− s≤ l21(s)(t∗− s)− 12− b2 if b< d(t∗− s)− d+12 〈ln( t∗−sl21(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif l21(s)< t∗− s≤ l22(s)(t∗− s)− d+12 ld−b2 (s) if b< d(t∗− s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif t∗− s> l22(s)dsby (4.11.2).When t∗− t ≤ (T − t)/2, then for s ∈ (t, t∗),T − t2≤ T − t∗ ≤ T − s≤ 2(T − t∗)≤ 2(T − t).It follows that300|I2|≲ v(t)ˆ t∗t∗−(t∗−t)(t∗− s)− 12 l−b1 (t) if t∗− s≤ l21(t)(t∗− s)− 12− b2 if b< d(t∗− s)− d+12 〈ln( t∗−sl21(t))〉 if b= d(t∗− s)− d+12 ld−b1 (t) if b> dif l21(t)< t∗− s≤ l22(t)(t∗− s)− d+12 ld−b2 (t) if b< d(t∗− s)− d+12 〈ln( l2(t)l1(t))〉 if b= d(t∗− s)− d+12 ld−b1 (t) if b> dif t∗− s> l22(t)ds≲ (t∗− t) α2 v(t)l1−α−b2 (t) b≤ 1−αl1−α−b1 (t) b> 1−αby (4.11.5).When t∗− t > (T − t)/2,|I2|≲ (T − t∗) α2 v(t∗)l1−α−b2 (t∗) b≤ 1−αl1−α−b1 (t∗) b> 1−α +ˆ t∗−(T−t∗)tv(s)(T − s)− d+12ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> ddssinceˆ t∗t∗−(T−t∗)v(s)(t∗− s)− 12 l−b1 (s) if t∗− s≤ l21(s)(t∗− s)− 12− b2 if b< d(t∗− s)− d+12 〈ln( t∗−sl21(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif l21(s)< t∗− s≤ l22(s)(t∗− s)− d+12 ld−b2 (s) if b< d(t∗− s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif t∗− s> l22(s)ds≲ (T − t∗) α2 v(2t∗−T )l1−α−b2 (2t∗−T ) b≤ 1−αl1−α−b1 (2t∗−T ) b> 1−α∼ (T − t∗) α2 v(t∗)l1−α−b2 (t∗) b≤ 1−αl1−α−b1 (t∗) b> 1−α ,301andˆ t∗−(T−t∗)tv(s)(t∗− s)− 12 l−b1 (s) if t∗− s≤ l21(s)(t∗− s)− 12− b2 if b< d(t∗− s)− d+12 〈ln( t∗−sl21(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif l21(s)< t∗− s≤ l22(s)(t∗− s)− d+12 ld−b2 (s) if b< d(t∗− s)− d+12 〈ln( l2(s)l1(s))〉 if b= d(t∗− s)− d+12 ld−b1 (s) if b> dif t∗− s> l22(s)ds≲ˆ t∗−(T−t∗)tv(s)(T − s)− d+12ld−b2 (s) if b< d〈ln( l2(s)l1(s))〉 if b= dld−b1 (s) if b> ddssince T−s2 ≤ t∗− s≤ T − s and l22(s)≲ T − s.4.11.3 Convolution with v(t)|x−q|−b1{|x−q|≥(T−t) 12 }For| f (x, t)| ≤ v(t)|x−q|−b1{|x−q|≥(T−t) 12 }considerψ(x, t) =ˆ t0ˆRdΓ(x, t,y,s) f (y,s)dyds.Claim: for b≥ 0, 0< α < 1, 0≤ t < t∗ ≤ T ,|ψ(x, t)|≲ˆ t0v(s)(T − s)− b2 ds,|∇ψ(x, t)|≲ˆ t0v(s)(T − s)− b+12 ds,|ψ(x, t)−ψ(x,T )|≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− b2 ds+ v(t)(T − t)1− b2 +ˆ Ttv(s)(T − s)− b2 ds,(4.11.14)302|∇ψ(x, t)−∇ψ(x,T )|≲C(α)(T − t) α2[ˆ t−(T−t)0v(s)(T − s)− 1+b+α2 ds+ v(t)(T − t) 1−b−α2]+ˆ Ttv(s)(T − s)− 1+b2 ds.(4.11.15)|∇xψ(x, t)−∇xψ(x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0v(s)(T − s)− 1+b+α2 ds+ v(t)(T − t) 1−b−α2]+1{t∗> T+t2 }ˆ t∗tv(s)(T − s)− 1+b2 ds.(4.11.16)Proof of (4.11.14).ψ(x, t)−ψ(x,T ) =ˆ t−(T−t)0ˆRd(Γ(x, t,y,s)−Γ(x,T,y,s)) f (y,s)dyds+ˆ tt−(T−t)ˆRd(Γ(x, t,y,s)−Γ(x,T,y,s)) f (y,s)dyds−ˆ TtˆRdΓ(x,T,y,s) f (y,s)dyds := I1+ I2+ I3.By (4.6.5) and (4.11.3),|I1|≲ (T − t)ˆ t−(T−t)0ˆRdˆ 10|(∂tΓ)(x,θ t+(1−θ)T,y,s)|| f (y,s)|dθdyds≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− d2ˆRde−c( |x−y|√T−s)2−δ|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ (T − t)ˆ t−(T−t)0v(s)(T − s)−1− b2 ds,|I2|≲ˆ tt−(T−t)v(s)(t− s)− d2ˆRde−c( |x−y|√t−s)2−δ|y−q|−b1{|y−q|≥(T−s) 12 }dyds+ˆ tt−(T−t)v(s)(T − s)− d2ˆRde−c( |x−y|√T−s)2−δ|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ v(t)(T − t)1− b2 ,|I3|≲ˆ TtˆRd(T − s)− d2 e−c( |x−y|√T−s)2−δv(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ˆ Ttv(s)(T − s)− b2 ds.This concludes (4.11.14).303Proof of (4.11.15).∂xiψ(x, t)−∂xiψ(x,T )=ˆ t0ˆRd(∂xiΓ(x, t,y,s)−∂xiΓ(x,T,y,s)) f (y,s)dyds−ˆ TtˆRd∂xiΓ(x,T,y,s) f (y,s)dyds=: I1+ I2.For I1, by (4.6.7),|I1|≲C(α)(T − t) α2ˆ t0ˆRd(T − s)− α2[(t− s)− d+12 e−c( |x−y|√t−s)2−δ+(T − s)− d+12 e−c( |x−y|√T−s)2−δ]∗ v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds=C(α)(T − t) α2(ˆ t−(T−t)0+ˆ tt−(T−t)· · ·)=:C(α)(T − t) α2 (I11+ I12).For I11,I11 ≲ˆ t−(T−t)0ˆRd(t− s)− d+1+α2[e−c( |x−y|√t−s)2−δ+ e−c( |x∗−y|√t−s)2−δ]v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ˆ t−(T−t)0v(s)(T − s)− 1+b+α2 dsby the same calculation in (4.11.8).For I12, since T − t ≤ T − s≤ 2(T − t),I12 =ˆ tt−(T−t)ˆRd(T − s)− α2[(t− s)− d+12 e−c( |x−y|√t−s)2−δ+(T − s)− d+12 e−c( |x∗−y|√T−s)2−δ]∗ v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ v(t)(T − t)− α2ˆ tt−(T−t)(t− s)− d+12ˆRde−c( |x−y|√t−s)2−δ|y−q|−b1{|y−q|≥(T−t) 12 }dyds+ v(t)(T − t)− d+1+α2ˆ tt−(T−t)ˆRde−c( |x∗−y|√T−t)2−δ|y−q|−b1{|y−q|≥(T−t) 12 }dyds≲ v(t)(T − t) 1−b−α2where we have used (4.11.3) in the last “≲”.304For I2, by (4.11.3),|I2|≲ˆ Ttv(s)(T − s)− d+12ˆRde−c( |x∗−y|√T−s)2−δ|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ˆ Ttv(s)(T − s)− 1+b2 ds.Proof of (4.11.16).∂xiψ(x, t)−∂xiψ(x∗, t∗)=ˆ t0ˆRd(∂xiΓ(x, t,y,s)−∂xiΓ(x∗, t∗,y,s)) f (y,s)dyds−ˆ t∗tˆRd∂xiΓ(x∗, t∗,y,s) f (y,s)dyds=: I1+ I2.For I1, by (4.6.7),|I1|≲C(α)(|x− x∗|+√|t− t∗|)α×ˆ t0ˆRd(t∗− s)− α2[(t− s)− d+12 e−c( |x−y|√t−s)2−δ+(t∗− s)− d+12 e−c( |x∗−y|√t∗−s)2−δ]∗ v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds=C(α)(|x− x∗|+√|t− t∗|)α(ˆ t−(T−t)0+ˆ t−(t∗−t)t−(T−t)+ˆ tt−(t∗−t)· · ·)=:C(α)(|x− x∗|+√|t− t∗|)α(I11+ I12+ I13).For I11,I11 ≲ˆ t−(T−t)0ˆRd(t− s)− d+1+α2[e−c( |x−y|√t−s)2−δ+ e−c( |x∗−y|√t−s)2−δ]v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ˆ t−(T−t)0v(s)(T − s)− 1+b+α2 dsby the same calculation in (4.11.8).For I12,I12 ≲ v(t)ˆ t−(t∗−t)t−(T−t)ˆRd(t− s)− d+1+α2[e−c( |x−y|√t−s)2−δ+ e−c( |x∗−y|√t−s)2−δ]|y−q|−b1{|y−q|≥(T−t) 12 }dyds≤ v(t)ˆ tt−(T−t)ˆRd(t− s)− d+1+α2[e−c( |x−y|√t−s)2−δ+ e−c( |x∗−y|√t−s)2−δ]|y−q|−b1{|y−q|≥(T−t) 12 }dyds305≲ v(t)(T − t) 1−b−α2by the same calculation in (4.11.8) where we have used α < 1.For I13, since t∗− t ≤ t∗− s≤ 2(t∗− t), T − t ≤ T − s≤ 2(T − t),I13 =ˆ tt−(t∗−t)ˆRd(t∗− s)− α2[(t− s)− d+12 e−c( |x−y|√t−s)2−δ+(t∗− s)− d+12 e−c( |x∗−y|√t∗−s)2−δ]∗ v(s)|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ v(t)(t∗− t)− α2ˆ tt−(t∗−t)(t− s)− d+12ˆRde−c( |x−y|√t−s)2−δ|y−q|−b1{|y−q|≥(T−t) 12 }dyds+ v(t)(t∗− t)− d+1+α2ˆ tt−(t∗−t)ˆRde−c( |x∗−y|√t∗−t)2−δ|y−q|−b1{|y−q|≥(T−t) 12 }dyds≲ v(t)(T − t)− b2 (t∗− t) 1−α2where we have used (4.11.3) in the last “≲”.For I2, by (4.11.3),|I2|≲ˆ t∗tv(s)(t∗− s)− d+12ˆRde−c( |x∗−y|√t∗−s)2−δ|y−q|−b1{|y−q|≥(T−s) 12 }dyds≲ˆ t∗tv(s)(t∗− s)− 12 (T − s)− b2 ds≲ v(t)(T − t)− b2 (t∗− t) 12 1{t∗≤ T+t2 }+ˆ t∗tv(s)(T − s)− 1+b2 ds1{t∗> T+t2 }.Indeed, when t∗− t ≤ (T − t)/2, then for s ∈ (t, t∗),T − t2≤ T − t∗ ≤ T − s≤ 2(T − t∗)≤ 2(T − t).It follows that|I2|≲ v(t)(T − t)− b2 (t∗− t) 12 .When t∗− t > (T − t)/2,ˆ t∗t∗−(T−t∗)v(s)(t∗− s)− 12 (T − s)− b2 ds≲ v(2t∗−T )(T − t∗) 12− b2 ∼ v(t∗)(T − t∗) 12− b2 .ˆ t∗−(T−t∗)tv(s)(t∗− s)− 12 (T − s)− b2 ds∼ˆ t∗−(T−t∗)tv(s)(T − s)− 1+b2 ds306since T−s2 ≤ t∗− s≤ T − s.4.12 Additional information 4: Derivation of the weighted topologyfor the outer problemProposition 4.12.1. For| f |≲N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3,suppose that0<Θ< β <12, 0< α < 1, Θ+12−β − α2< 0,0< σ0 < β , β −σ0− α2 < 0, 1−σ0− (1+α)(1−β )< 0,Θ+2σ0−β < 0,(4.12.1)then we have|T out2 [ f ]|≲ | lnT |λΘ+1∗ (0)R(0), (4.12.2)|∇xT out2 [ f ]|≲ λΘ∗ (0), (4.12.3)|T out2 [ f ](x, t)−T out2 [ f ](x,T )|≲ | ln(T − t)|λΘ+1∗ R, (4.12.4)|∇xT out2 [ f ](x, t)−∇xT out2 [ f ](x,T )|≲C(α)λΘ∗ , (4.12.5)for 0< t < t∗ ≤ T , t∗− t < 14(T − t),|∇xT out2 [ f ](x, t)−∇xT out2 [ f ](x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)αλΘ∗ (t)(λ∗R)−α(t). (4.12.6)Proof. Convolution estimate about ρ [ j]1 . For| f |≲ ρ [ j]1 := λΘ∗ (λ∗R)−11{|x−q[ j]|≤3λ∗R},T out2 [ f ]≲ˆ t−(T−t)0λΘ∗ (s)(λ∗R)−1(s)(T − s)−1(λ∗R)2(s)ds+λΘ∗ (λ∗R)−1(λ∗R)2| ln(T − t)|=ˆ t−(T−t)0λΘ∗ (s)λ∗R(T − s)−1ds+λΘ∗ λ∗R| ln(T − t)|≲ λΘ∗ (0)(λ∗R)(0)| lnT |(4.12.7)provided1+Θ−β > 0. (4.12.8)307|∇xT out2 [ f ]|≲ˆ t−(T−t)0λΘ∗ (s)(λ∗R)−1(s)(T − s)−32 (λ∗R)2(s)ds+λΘ∗ (λ∗R)−1λ∗R=ˆ t−(T−t)0λΘ∗ (s)(λ∗R)(s)(T − s)−32 ds+λΘ∗ ≲ λΘ∗ (0)(4.12.9)providedβ <12, Θ+12−β > 0. (4.12.10)|T out2 [ f ](x, t)−T out2 [ f ](x,T )|≲ (T − t)ˆ t−(T−t)0λΘ∗ (s)(λ∗R)−1(s)(T − s)−2(λ∗R)2(s)ds+λΘ∗ (λ∗R)−1ˆ tt−(T−t)1 if t− s≤ (λ∗R)2(t)(λ∗R)2(t)(t− s)−1 if t− s> (λ∗R)2(t) ds+λΘ∗ (λ∗R)−1(λ∗R)2+ˆ Tt(T − s)−1λΘ∗ (s)(λ∗R)−1(s)(λ∗R)2(s)ds≲ (T − t)ˆ t−(T−t)0λΘ∗ (s)(λ∗R)(s)(T − s)−2ds+λΘ∗ (λ∗R)| ln(T − t)|+ˆ Tt(T − s)−1λΘ∗ (s)(λ∗R)(s)ds≲ λΘ∗ (λ∗R)| ln(T − t)|(4.12.11)provided0< β −Θ< 1. (4.12.12)|∇T out2 [ f ](x, t)−∇T out2 [ f ](x,T )|≲C(α)[(T − t) α2ˆ t−(T−t)0λΘ∗ (s)(λ∗R)−1(s)(T − s)−3+α2 (λ∗R)2(s)ds+λΘ∗ (λ∗R)−1λ∗R]+ˆ TtλΘ∗ (s)(λ∗R)−1(s)(T − s)−32 (λ∗R)2(s)ds=C(α)[(T − t) α2ˆ t−(T−t)0λΘ∗ (s)(λ∗R)(s)(T − s)−3+α2 ds+λΘ∗]+ˆ TtλΘ∗ (s)(λ∗R)(s)(T − s)−32 ds≲C(α)λΘ∗(4.12.13)provided0< α < 1, β <12, 0<Θ+12−β < α2. (4.12.14)308For 0< α < 1 and 0< t < t∗ ≤ T ,|∇T out2 [ f ](x, t)−∇T out2 [ f ](x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0λΘ∗ (s)(λ∗R)−1(s)(T − s)−3+α2 (λ∗R)2(s)ds+λΘ∗ (t)(λ∗R)−1(t)(λ∗R)1−α(t)]+1{t∗> T+t2 }[(T − t∗) α2 λΘ∗ (t∗)(λ∗R)−1(t∗)(λ∗R)1−α(t∗)+ˆ t∗−(T−t∗)tλΘ∗ (s)(λ∗R)−1(s)(T − s)−32 (λ∗R)2(s)ds]=C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0λΘ∗ (s)(λ∗R)(s)(T − s)−3+α2 ds+λΘ∗ (t)(λ∗R)−α(t)]+1{t∗> T+t2 }[(T − t∗) α2 λΘ∗ (t∗)(λ∗R)−α(t∗)+ˆ t∗−(T−t∗)tλΘ∗ (s)(λ∗R)(s)(T − s)−32 ds]≲C(α)(|x− x∗|+√|t− t∗|)αλΘ∗ (t)(λ∗R)−α(t)+1{t∗> T+t2 }[(T − t∗) α2 λΘ∗ (t∗)(λ∗R)−α(t∗)+λΘ∗ (t)(λ∗R)(t)(T − t)−12]≲C(α)(|x− x∗|+√|t− t∗|)αλΘ∗ (t)(λ∗R)−α(t)+1{t∗> T+t2 }(T − t∗)α2 λΘ∗ (t∗)(λ∗R)−α(t∗)(4.12.15)providedΘ+12−β < α2, β <12. (4.12.16)Convolution estimate about ρ [ j]2 .Recall ρ [ j]2 = T−σ0 λ 1−σ0∗|x−q[ j]|2 1{λ∗R≤|x−q[ j]|≤dq}. Consider| f | ≤ λ 1−σ0∗ |x−q[ j]|−21{λ∗R≤|x−q[ j]|≤(T−t) 12 }+λ1−σ0∗ |x−q[ j]|−21{(T−t) 12<|x−q[ j]|≤dq}.ThenT out2 [ f ]≲ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−1| ln(T − s)|ds+λ 1−σ0∗ | ln(T − t)|2+ˆ t0λ 1−σ0∗ (s)(T − s)−1ds≲ λ 1−σ0∗ (0)(lnT )2.(4.12.17)309|∇T out2 [ f ]|≲ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−32 | ln(T − s)|ds+λ 1−σ0∗ (λ∗R)−1+ˆ t0λ 1−σ0∗ (s)(T − s)−32 ds≲ λ 1−σ0∗ (0)(λ∗R)−1(0)(4.12.18)providedσ0 < β <12. (4.12.19)|T out2 [ f ](x, t)−T out2 [ f ](x,T )|≲ (T − t)ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−2| ln(T − s)|ds+λ 1−σ0∗ˆ tt−(T−t)(λ∗R)−2 if t− s≤ (λ∗R)2(t)(t− s)−1〈ln( t−s(λ∗R)2(t))〉 if (λ∗R)2(t)< t− s≤ T − tds+λ 1−σ0∗ | ln(T − t)|+ˆ Tt(T − s)−1λ 1−σ0∗ (s)| ln(T − s)|ds+(T − t)ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−2ds+λ 1−σ0∗ +ˆ Ttλ 1−σ0∗ (s)(T − s)−1ds≲ λ 1−σ0∗ ln2(T − t)(4.12.20)provided0< σ0 < 1, β <12. (4.12.21)where we have usedˆ t−(λ∗R)2(t)t−(T−t)(t− s)−1〈ln( t− s(λ∗R)2(t))〉ds=ˆ T−t(λ∗R)2(t)1z−1〈lnz〉dz= O(ln2(T − t)).For 0< α < 1,|∇T out2 [ f ](x, t)−∇T out2 [ f ](x,T )|≲C(α)[(T − t) α2ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 | ln(T − s)|ds+λ 1−σ0∗ (λ∗R)−1]+ˆ Ttλ 1−σ0∗ (s)(T − s)−32 | ln(T − s)|ds+C(α)(T − t) α2[ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 ds+λ 1−σ0∗ (T − t)−1−α2]+ˆ Ttλ 1−σ0∗ (s)(T − s)−32 ds310≲C(α)[(T − t) α2ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 | ln(T − s)|ds+λ 1−σ0∗ (λ∗R)−1]≲C(α)λ 1−σ0∗ (λ∗R)−1(4.12.22)providedβ <12, σ0 <12, β −σ0 < α2 . (4.12.23)where we have used the following estimate in the last “≲”: If 1−σ0− 1+α2 ≤ 0, then(T−t) α2ˆ t−(T−t)0λ 1−σ0∗ (s)(T−s)−3+α2 | ln(T−s)|ds≲ λ 1−σ0∗ (T−t)−12−ε | ln(T−t)| λ 1−σ0∗ (λ∗R)−1where ε = 0 when 1−σ0− 1+α2 < 0 and 0< ε < 12 −β when 1−σ0− 1+α2 = 0;If 1−σ0− 1+α2 > 0, then(T − t) α2ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 | ln(T − s)|ds≲ (T − t) α2 λ 1−σ0∗ (0)T−1+α2 | lnT | λ 1−σ0∗ (λ∗R)−1when β < 12 ,β −σ0 < α2 .For 0< α < 1 and 0< t < t∗ ≤ T ,|∇T out2 [ f ](x, t)−∇T out2 [ f ](x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 | ln(T − s)|ds+λ 1−σ0∗ (t)(λ∗R)−1−α(t)]+1{t∗> T+t2 }[(T − t∗) α2 λ 1−σ0∗ (t∗)(λ∗R)−1−α(t∗)+ˆ t∗−(T−t∗)tλ 1−σ0∗ (s)(T − s)−32 | ln(T − s)|ds]+C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 ds+λ 1−σ0∗ (t)(T − t)−1−α2]+1{t∗> T+t2 }ˆ t∗tλ 1−σ0∗ (s)(T − s)−32 ds≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 | ln(T − s)|ds+λ 1−σ0∗ (t)(λ∗R)−1−α(t)]+1{t∗> T+t2 }[(T − t∗) α2 λ 1−σ0∗ (t∗)(λ∗R)−1−α(t∗)+ˆ t∗−(T−t∗)tλ 1−σ0∗ (s)(T − s)−32 | ln(T − s)|ds]+1{t∗> T+t2 }ˆ t∗tλ 1−σ0∗ (s)(T − s)−32 ds≲C(α)(|x− x∗|+√|t− t∗|)αλ 1−σ0∗ (t)(λ∗R)−1−α(t)311+1{t∗> T+t2 }[(T − t∗) α2 λ 1−σ0∗ (t∗)(λ∗R)−1−α(t∗)+ˆ t∗−(T−t∗)tλ 1−σ0∗ (s)(T − s)−32 | ln(T − s)|ds]+1{t∗> T+t2 }ˆ t∗tλ 1−σ0∗ (s)(T − s)−32 ds (4.12.24)providedβ <12, 1−σ0− (1+α)(1−β )< 0 (4.12.25)where we have used the following estimate in the last “≲”: If 1−σ0− 1+α2 ≤ 0,ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 ds≲ λ 1−σ0∗ (t)(T − t)−1+α2 −ε λ 1−σ0∗ (t)(λ∗R)−1−α(t)where ε = 0 when 1−σ0− 1+α2 < 0 and 0< ε < (1+α)(12 −β ) when 1−σ0− 1+α2 = 0;If 1−σ0− 1+α2 > 0,ˆ t−(T−t)0λ 1−σ0∗ (s)(T − s)−3+α2 ds≲ λ 1−σ0∗ (0)T−1+α2 λ 1−σ0∗ (t)(λ∗R)−1−α(t)when β < 12 ,1−σ0− (1+α)(1−β )< 0.Convolution estimate about ρ3.Recall ρ3 = T−σ0 . Consider| f | ≤ 1= 1{|x|≤√T−t}+1{|x|>√T−t}.Then|T outd [ f ]|≲ T. (4.12.26)|∇T outd [ f ]|≲ T12 . (4.12.27)|T outd [ f ](x, t)−T outd [ f ](x,T )|≲ (T − t)| ln(T − t)|. (4.12.28)For 0< α < 1,|∇T outd [ f ](x, t)−∇T outd [ f ](x,T )|≲C(α)[(T − t) α2ˆ t−(T−t)0(T − s)− 1+α2 ds+(T − t) 12]≲C(α)T 1−α2 (T − t) α2 .(4.12.29)312For 0< α < 1 and 0< t < t∗ ≤ T ,|∇T outd [ f ](x, t)−∇T outd [ f ](x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [ˆ t−(T−t)0(T − s)− 1+α2 ds+(T − t) 1−α2]+1{t∗> T+t2 }[(T − t∗) 12 +ˆ t∗t(T − s)− 12 ds]≲C(α)(|x− x∗|+√|t− t∗|)αT1−α2 +1{t∗> T+t2 }(T − t)12 .(4.12.30)In sum, for| f |≲N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3,combining (4.12.7), (4.12.17) and (4.12.26), we have|T outd [ f ]|≲ λΘ∗ (0)(λ∗R)(0)| lnT |+T−σ0λ 1−σ0∗ (0)(lnT )2+T 1−σ0 ≲ λΘ∗ (0)(λ∗R)(0)| lnT | (4.12.31)where in the last “≲”, we requireΘ+2σ0−β < 0, σ0 > 0. (4.12.32)Combining (4.12.9), (4.12.18) and (4.12.27), we have|∇xT out2 [ f ]|≲ λΘ∗ (0)+T−σ0λ 1−σ0∗ (0)(λ∗R)−1(0)+T12−σ0 ≲ λΘ∗ (0) (4.12.33)where in the last “≲” we have usedΘ+2σ0−β < 0, σ0 > 0, β < 12 . (4.12.34)Combining (4.12.11), (4.12.20) and (4.12.28), then|T out2 [ f ](x, t)−T out2 [ f ](x,T )|≲ λΘ∗ (λ∗R)| ln(T − t)|+T−σ0λ 1−σ0∗ ln2(T − t)+T−σ0(T − t)| ln(T − t)|≲ λΘ∗ (λ∗R)| ln(T − t)|(4.12.35)where in the last “≲” we have usedΘ+2σ0−β < 0, σ0 > 0. (4.12.36)313Combining (4.12.13), (4.12.22) and (4.12.29), then|∇T out2 [ f ](x, t)−∇T out2 [ f ](x,T )|≲C(α)[λΘ∗ +T−σ0λ 1−σ0∗ (λ∗R)−1+T−σ0T1−α2 (T − t) α2]=C(α)λΘ∗[1+T−σ0λ 1−σ0−Θ∗ (λ∗R)−1+T−σ0T1−α2 (T − t) α2−Θ]≲C(α)λΘ∗(4.12.37)where in the last “≲” we have usedΘ+2σ0−β < 0, Θ− α2 < 0, Θ+σ0−12< 0. (4.12.38)Combining (4.12.15), (4.12.24), (4.12.30), then for 0< t < t∗ ≤ T , t∗− t < 14(T − t),|∇T out2 [ f ](x, t)−∇T out2 [ f ](x∗, t∗)|≲C(α)(|x− x∗|+√|t− t∗|)α [λΘ∗ (t)(λ∗R)−α(t)+T−σ0λ 1−σ0∗ (t)(λ∗R)−1−α(t)+T−σ0T1−α2]=C(α)(|x− x∗|+√|t− t∗|)αλΘ∗ (t)(λ∗R)−α(t)×[1+T−σ0λ 1−σ0−Θ∗ (t)(λ∗R)−1(t)+T−σ0T1−α2 λ−Θ∗ (t)(λ∗R)α(t)]≲C(α)(|x− x∗|+√|t− t∗|)αλΘ∗ (t)(λ∗R)−α(t)(4.12.39)where in the last “≲” we have usedΘ+2σ0−β < 0, Θ−α(1−β )< 0, Θ+σ0− 12 −α(12−β )< 0. (4.12.40)Collecting (4.12.8), (4.12.10), (4.12.12), (4.12.14), (4.12.16), (4.12.19), (4.12.21), (4.12.23),(4.12.25), (4.12.32), (4.12.34), (4.12.36), (4.12.38) and (4.12.40), we conclude the restrictions (4.12.1)on the parameters.4.13 Additional information 5: Estimates of G andH j4.13.1 Estimates for terms involvingU∗, Φout, Φ[ j]in, Φ∗[ j]0First, we prepare some useful formulas. By (4.2.3),∣∣∇xU [k]∣∣≲ λ−1k 〈ρk〉−2, |∇xU∗|≲ N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗. (4.13.1)314|∆xU∗|≲N∑j=1λ−2∗ 〈ρ j〉−4 ≲N∑j=11{|x−q[ j]|<3dq}λ−2∗ 〈ρ j〉−4+1{∩Nj=1{|x−q[ j]|≥3dq}}λ2∗ . (4.13.2)|U∗ ·∇xU∗|=∣∣∣∣∣ N∑m=1 ∑k 6=m(U [k]−U∞) ·∇xU [m]∣∣∣∣∣≲ N∑m=1 ∑k 6=m〈ρk〉−1λ−1m 〈ρm〉−2≲N∑j=11{|x−q[ j]|<3dq}〈ρ j〉−2+1{∩Nj=1|x−q[ j]|≥3dq}λ2∗(4.13.3)where for the last step, we used that for any fixed j = 1,2, . . . ,N,1{|x−q[ j]|<3dq}[λ−1j 〈ρ j〉−2∑k 6= j〈ρk〉−1+ ∑m 6= j∑k 6=m〈ρk〉−1λ−1m 〈ρm〉−2]≲ 1{|x−q[ j]|<3dq}(〈ρ j〉−2+λ∗ ∑m 6= j∑k 6=m〈ρk〉−1)∼ 1{|x−q[ j]|<3dq}〈ρ j〉−2;1{∩Nj=1{|x−q[ j]|≥3dq}}N∑m=1∑k 6=m〈ρk〉−1λ−1m 〈ρm〉−2 ≲ λ 2∗ .By the same argument for (4.13.3), then∣∣∣∣∣ N∑j=1 |∇xU [ j]|2 (U [ j]−U∗)∣∣∣∣∣≲ N∑j=1λ−2j 〈ρ j〉−4∑k 6= j〈ρk〉−1≲N∑j=11{|x−q[ j]|<3dq}(λ−1∗ 〈ρ j〉−4+λ 2∗ 〈ρ j〉−1)+1{∩Nj=1{|x−q[ j]|≥3dq}}λ3∗ .(4.13.4)Notice|∇xU∗|2+U∗ ·∆xU∗ =∣∣∣∣∣ N∑j=1∇xU [ j]∣∣∣∣∣2−N∑j=1|∇xU [ j]|2−N∑j=1|∇xU [ j]|2(U∗−U [ j]) ·U [ j]=N∑j=1∑k 6= j∇xU [ j] ·∇xU [k]−N∑j=1|∇xU [ j]|2(U∗−U [ j]) ·U [ j].Then∣∣|∇xU∗|2+U∗ ·∆xU∗∣∣≲ N∑m=1∑k 6=mλ−1m λ−1k 〈ρm〉−2〈ρk〉−2+N∑m=1∑k 6=mλ−2m 〈ρm〉−4〈ρk〉−1≲N∑j=11{|x−q[ j]|<3dq}(〈ρ j〉−2+λ−1∗ 〈ρ j〉−4)+1{∩Nj=1{|x−q[ j]|≥3dq}}λ 2∗(4.13.5)315where for the last step, we used that for any fixed j = 1,2, . . . ,N,1{|x−q[ j]|<3dq}[λ−1j 〈ρ j〉−2∑k 6= jλ−1k 〈ρk〉−2+λ−2j 〈ρ j〉−4∑k 6= j〈ρk〉−1+ ∑m 6= j∑k 6=m(λ−1m λ−1k 〈ρm〉−2〈ρk〉−2+λ−2m 〈ρm〉−4〈ρk〉−1)]≲ 1{|x−q[ j]|<3dq}[〈ρ j〉−2+λ−1∗ 〈ρ j〉−4+ ∑m6= j∑k 6=m(〈ρk〉−2+λ 2∗ 〈ρk〉−1)]∼ 1{|x−q[ j]|<3dq} (〈ρ j〉−2+λ−1∗ 〈ρ j〉−4) ;1{∩Nj=1{|x−q[ j]|≥3dq}}( N∑m=1∑k 6=mλ−1m λ−1k 〈ρm〉−2〈ρk〉−2+N∑m=1∑k 6=mλ−2m 〈ρm〉−4〈ρk〉−1)≲ 1{∩Nj=1{|x−q[ j]|≥3dq}}λ2∗ .|U∗∧∆xU∗|=∣∣∣∣∣U∗∧ N∑j=1∣∣∇xU [ j]∣∣2U [ j]∣∣∣∣∣≲ N∑j=1λ−2j 〈ρ j〉−4∑k 6= j〈ρ j〉−1≲N∑j=11{|x−q[ j]|<3dq}(λ−1∗ 〈ρ j〉−4+λ 2∗ 〈ρ j〉−1)+1{∩Nj=1{|x−q[ j]|≥3dq}}λ3∗(4.13.6)where we used (4.13.4) in the last step.Next, we derive some estimates involving Φout, Φin, Φ∗[ j]0 that will be used frequently in theestimate of G .For Φout ∈ Bout defined in (4.4.48), since Φout(q[ j],T ) = 0 for all j = 1,2, . . . ,N, then|Φout(x, t)|=∣∣Φout(x, t)−Φout(x,T )+Φout(x,T )−Φout(q[ j],T )∣∣≲ ‖Φout‖♯,Θ,α[| ln(T − t)|λΘ+1∗ R+(T − t)‖Z∗‖C3(R2)+ |x−q[ j]|(λΘ∗ (0)+‖Z∗‖C3(R2))].Thus|Φout|≲ ‖Φout‖♯,Θ,αmin{| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2),| ln(T − t)|λΘ+1∗ R+(T − t)‖Z∗‖C3(R2)+ infj=1,...,N |x−q[ j]|(λΘ∗ (0)+‖Z∗‖C3(R2))}.(4.13.7)Combining (4.3.2) and the parameter restriction Θ< β , we have|Φout|≲N∑j=11{|x−q[ j]|<3dq}‖Φout‖♯,Θ,α(| ln(T − t)|λΘ+1∗ R+λ jρ j)+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.8)By (4.4.47), we have|∇xΦout| ≤ ‖Φout‖♯,Θ,α(λΘ∗ (0)+‖Z∗‖C3(R2)). (4.13.9)316For |x−ξ [ j](t)| ≤ 2λ∗R, by (4.3.2), then for T sufficiently small, we have |x−q[ j]| ≤ 3λ∗R. RecallΦ given in (4.4.1), thenΦ−Φout =N∑j=1(η [ j]R Qγ jΦ[ j]in(y[ j], t)+η [ j]dqΦ∗[ j]0 (r j, t)).By (4.4.41) and (4.3.24), we have|Φ−Φout|≲N∑j=1[η [ j]R ‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l +η [ j]dq(z j1{z2j<t}+ t| lnT |−1z−1j 1{z2j≥t})]≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq}λ∗〈ρ j〉].(4.13.10)By (4.4.41), we get∣∣∇x (η [ j]R Qγ jΦ[ j]in(y[ j], t))∣∣= ∣∣η [ j]R ∇x (Qγ jΦ[ j]in(y[ j], t))+Qγ jΦ[ j]in(y[ j], t)∇xη [ j]R ∣∣≲ η [ j]R λ−1j ‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈y[ j]〉−l−1+(λ∗R)−11{λ∗R≤|x−ξ [ j]|≤2λ∗R}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈y[ j]〉−l≲ 1{|x−q[ j]|≤3λ∗R}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−1∗ 〈ρ j〉−l−1.(4.13.11)By (4.3.24), we have∣∣∣∇x(η [ j]dqΦ∗[ j]0 (r j, t))∣∣∣= ∣∣∣η [ j]dq∇xΦ∗[ j]0 (r j, t)+Φ∗[ j]0 (r j, t)∇xη [ j]dq∣∣∣≲η [ j]dq+t 12 1{dq≤|x−ξ [ j]|≤2dq}≲ 1{|x−q[ j]|<3dq}.(4.13.12)Combining (4.13.11) and (4.13.12), we have|∇x (Φ−Φout)|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−1∗ 〈ρ j〉−l−1+1)+1{3λ∗R<|x−q[ j]|<3dq}].(4.13.13)By (4.3.24) and (4.4.41), it holds that|∆x (Φ−Φout) |=∣∣∣ N∑j=1[η [ j]R ∆x(Qγ jΦ[ j]in(y[ j], t))+2∇xη [ j]R ∇x(Qγ jΦ[ j]in(y[ j], t))+Qγ jΦ[ j]in(y[ j], t)∆xη [ j]R+η [ j]dq∆xΦ∗[ j]0 (r j, t)+2∇xη[ j]dq∇xΦ∗[ j]0 (r j, t)+Φ∗[ j]0 (r j, t)∆xη[ j]dq]∣∣∣≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−2∗ 〈ρ j〉−l−2+λ−1∗ 〈ρ j〉−1)+1{3λ∗R<|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−1].(4.13.14)317Combining (4.13.8) and (4.13.10), we have|Φ|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.15)Integrating (4.13.9), (4.13.13), we have|∇xΦ|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0−1∗ 〈ρ j〉−l−1+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.16)By (4.13.15), (4.13.16), then|Φ||∇xΦ|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2×(λ 2ν−2δ0−1∗ 〈ρ j〉−2l−1+λ ν−δ0∗ 〈ρ j〉−l + | ln(T − t)|λ ν−δ0+Θ∗ R〈ρ j〉−l−1+λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α .(4.13.17)Recalling Φ[ j]in ·W [ j] = 0 in (4.4.1) yields(N∑j=1η [ j]R Qγ jΦ[ j]in)·U∗ =N∑j=1η [ j]R Qγ jΦ[ j]in ·(U∗−U [ j]), (4.13.18)which implies∣∣∣∣∣(N∑j=1η [ j]R Qγ jΦ[ j]in)·U∗∣∣∣∣∣≲ N∑j=1η [ j]R∣∣Φ[ j]in∣∣λ∗ ≲ N∑j=11{|x−q[ j]|≤3λ∗R}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0+1∗ 〈ρ j〉−l.(4.13.19)By (4.13.19), (4.3.24) and (4.13.8), we obtain|Φ ·U∗|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0+1∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(| ln(T − t)|λΘ+1∗ R+λ∗〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.20)318Using (4.4.41), (4.13.11), (4.13.1), we have∣∣∇x [η [ j]R Qγ jΦ[ j]in · (U∗−U [ j])]∣∣=∣∣∣∣∣(U∗−U [ j]) ·∇x (η [ j]R Qγ jΦ[ j]in)+(∑k 6= j∇xU [k])·η [ j]R Qγ jΦ[ j]in∣∣∣∣∣≲ 1{|x−q[ j]|≤3λ∗R}(λ∗‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−1∗ 〈ρ j〉−l−1+λ∗‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l)≲ 1{|x−q[ j]|≤3λ∗R}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l−1.(4.13.21)(4.3.24), (4.13.12), and (4.13.1) imply∣∣∣∇x(η [ j]dqΦ∗[ j]0 ·U∗)∣∣∣=∣∣∣∣U∗ ·∇x(η [ j]dqΦ∗[ j]0 )+η [ j]dqΦ∗[ j]0 ·∇xU∗∣∣∣∣≲ 1{|x−q[ j]|<3dq}(1+λ j〈ρ j〉N∑k=1λ−1k 〈ρk〉−2)≲ 1{|x−q[ j]|<3dq}.(4.13.22)By (4.13.8), (4.13.9) and (4.13.1), we have|∇x (Φout ·U∗)|≲N∑j=11{|x−q[ j]|<3dq}‖Φout‖♯,Θ,α(| ln(T − t)|λΘ∗ R〈ρ j〉−2+1)+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.23)Combining (4.13.21), (4.13.22) and (4.13.23), we have|∇x (Φ ·U∗)|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(| ln(T − t)|λΘ∗ R〈ρ j〉−2+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.24)4.13.2 Estimate ∇xAClaim: Suppose (4.4.42),Θ> 0, Θ+β+2δ0−2ν < 0, Θ+β−1< 0, β < 1/2, Θ+β+4δ0−4ν+1< 0, 3β <Θ+1,(4.13.25)319then for ∇xA given in (4.4.13) and ε > 0 sufficiently small, we have∇xA=−U∗ ·∇xU∗−Φ ·∇xΦ+O( N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)4× (λ ε+1∗ λΘ∗ (λ∗R)−1+λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)4 (| ln(T − t)|λΘ+1∗ R+λ∗〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)4).(4.13.26)Using (4.13.3), (4.13.17), then∇xA= O( N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)4×(〈ρ j〉−2+λ 2ν−2δ0−1∗ 〈ρ j〉−2l−1+λ ν−δ0∗ 〈ρ j〉−l + | ln(T − t)|λ ν−δ0+Θ∗ R〈ρ j〉−l−1+λ ε+1∗ λΘ∗ (λ∗R)−1+λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)4(〈ρ j〉−2+ | ln(T − t)|λΘ+1∗ R+λ∗〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)4).(4.13.27)Proof of Claim. Let us simplify (4.4.13) first|ΠU⊥∗ Φ|2 = |Φ|2+(|U∗|2−2)(Φ ·U∗)2 , U∗ ·ΠU⊥∗ Φ=(1−|U∗|2)(Φ ·U∗) ,∇x(|ΠU⊥∗ Φ|2)= 2Φ ·∇xΦ+2(Φ ·U∗)2U∗ ·∇xU∗+2(|U∗|2−2)(Φ ·U∗)∇x (Φ ·U∗) ,∇x(U∗ ·ΠU⊥∗ Φ)=(1−|U∗|2)∇x (Φ ·U∗)−2(Φ ·U∗)U∗ ·∇xU∗.By (4.4.5), (4.3.4), (4.3.2) and (4.4.3), we have(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)= 1+O(λ∗+ |Φ|2), then[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1= 1+O(λ∗+ |Φ|2).(4.13.28)Thus we obtain∇xA=−(1+O(λ∗+ |Φ|2)){(1+A)2U∗ ·∇xU∗+Φ ·∇xΦ+(Φ ·U∗)2U∗ ·∇xU∗+(|U∗|2−2)(Φ ·U∗)∇x (Φ ·U∗)+(1+A)[(1−|U∗|2)∇x (Φ ·U∗)−2(Φ ·U∗)U∗ ·∇xU∗]}= − (1+O(λ∗+ |Φ|2)){[1+A(2+A)−2(1+A)(Φ ·U∗)+(Φ ·U∗)2]U∗ ·∇xU∗+Φ ·∇xΦ+(|U∗|2−2)(Φ ·U∗)∇x (Φ ·U∗)+(1+A)(1−|U∗|2)∇x (Φ ·U∗)}= −U∗ ·∇xU∗−Φ ·∇xΦ+(2Φ ·U∗+O(λ∗+ |Φ|2))U∗ ·∇xU∗−O(λ∗+ |Φ|2)Φ ·∇xΦ320− (1+O(λ∗+ |Φ|2))[(|U∗|2−2)(Φ ·U∗)∇x (Φ ·U∗)+(1+A)(1−|U∗|2)∇x (Φ ·U∗)](4.13.29)where we used A= O(λ∗+ |Φ|2) 1 by (4.4.5) and (4.4.3).By (4.13.15), (4.13.20), and parameter assumption (4.4.42), we have|Φ ·U∗|+λ∗+ |Φ|2≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2(λ 2ν−2δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2 .(4.13.30)Notice λ ε+1∗ λΘ∗ (λ∗R)−1 = λε+Θ+β∗ . Then using (4.13.3) and (4.13.30), we get∣∣(2Φ ·U∗+O(λ∗+ |Φ|2))U∗ ·∇xU∗∣∣≲ N∑j=1{1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2×(λ 2ν−2δ0∗ 〈ρ j〉−l−2+λ∗〈ρ j〉−1+ | ln(T − t)|λΘ+1∗ R〈ρ j〉−2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉−1+ | ln(T − t)|λΘ+1∗ R〈ρ j〉−2)}+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2λ 2∗≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2λ ε+1∗ λΘ∗ (λ∗R)−1+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2λ∗〈ρ j〉−1]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2λ 2∗(4.13.31)for some ε > 0 where for the last “≲”, we requireΘ> 0, Θ+β +2δ0−2ν < 0, Θ+β −1< 0, β < 1/2. (4.13.32)By (4.13.15) and (4.13.16), it follows that∣∣O(λ∗+ |Φ|2)Φ ·∇xΦ∣∣≲ (|Φ|3+λ∗ |Φ|) |∇xΦ|≲{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)3(λ 3ν−3δ0∗ 〈ρ j〉−3l +λ 3∗ 〈ρ j〉3+ | ln(T − t)|3λ 3Θ+3∗ R3+λ ν−δ0+1∗ 〈ρ j〉−l +λ 2∗ 〈ρ j〉+ | ln(T − t)|λΘ+2∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)3(λ 3∗ 〈ρ j〉3+ | ln(T − t)|3λ 3Θ+3∗ R3+λ 2∗ 〈ρ j〉+ | ln(T − t)|λΘ+2∗ R)]321+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)3}×{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0−1∗ 〈ρ j〉−l−1+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)4×(λ 4ν−4δ0−1∗ +λ 3∗ 〈ρ j〉3+λ ν−δ0+2∗ 〈ρ j〉2−l + | ln(T − t)|3λ ν−δ0+3Θ+2∗ R3+λ 2ν−2δ0∗ + | ln(T − t)|λ ν−δ0+Θ+1∗ R+λ 2∗ 〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)4(λ 3∗ 〈ρ j〉3+ | ln(T − t)|3λ 3Θ+3∗ R3+λ 2∗ 〈ρ j〉+ | ln(T − t)|λΘ+2∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)4≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)4λ ε+1∗ λΘ∗ (λ∗R)−1+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)4(λ 3∗ 〈ρ j〉3+ | ln(T − t)|3λ 3Θ+3∗ R3+λ 2∗ 〈ρ j〉+ | ln(T − t)|λΘ+2∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)4 (4.13.33)where we require (4.4.42) and additional patameters restriction for the last “≲”β < 1/2, Θ+β +4δ0−4ν+1< 0, Θ+β +2δ0−2ν < 0, Θ+2β < 2. (4.13.34)Combining (4.13.20) and (4.13.24), we have|(Φ ·U∗)∇x (Φ ·U∗)|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2×(| ln(T − t)|λ ν−δ0+1+Θ∗ R+ | ln(T − t)|λΘ+1∗ R+ | ln(T − t)|2λ 2Θ+1∗ R2+λ ν−δ0+1∗ +λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(| ln(T − t)|λΘ+1∗ R+λ∗〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 (λ ε+1∗ λΘ∗ (λ∗R)−1+λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(| ln(T − t)|λΘ+1∗ R+λ∗〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α(4.13.35)where for the last step, we requireβ < 1/2, 3β <Θ+1, Θ+β < ν−δ0+1. (4.13.36)322By (4.3.4), (4.13.24), we have∣∣(1−|U∗|2)∇x (Φ ·U∗)∣∣≲ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)λ ε+1∗ λΘ∗ (λ∗R)−1+1{3λ∗R<|x−q[ j]|<3dq}λ∗ (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗‖Φout‖♯,Θ,α(4.13.37)providedΘ+β < 1, β < 1/2. (4.13.38)Under the parameters restriction (4.13.25), which is the combination of (4.13.32), (4.13.34),(4.13.36), and (4.13.38), we conclude the validity of (4.13.26) from (4.13.31), (4.13.33), (4.13.35),and (4.13.37).4.13.3 Estimate GLemma 4.13.1. Suppose ‖Φ[ j]in‖in,ν−δ0,l ≤Cin for j = 1,2, . . . ,N, under the parameter assumptionsΘ< β , Θ+β +δ0−ν < 0, 3β < 1+Θ, δ0 < β < 12 , β (l+1)−1+ν−δ0−Θ> 0,Θ+2β −1< 0, δ0 < ν , 2β +δ0−ν < 0, Θ+β +1+3δ0−3ν < 0,Θ+β +4δ0−4ν+1< 0.(4.13.39)Then for ε > 0 sufficiently small,‖G ‖∗∗ ≲ T ε . (4.13.40)where ‖ · ‖∗∗ is defined in (4.4.46).323Proof. • By (4.13.7) and (4.13.9), we have∣∣(1−η [ j]R )(a−bU [ j]∧)[|∇xU [ j]|2Φout−2∇x (U [ j] ·Φout)∇xU [ j]]∣∣≲ 1{|x−q[ j]|≥λ∗R/2}(λ−2j 〈ρ j〉−4 |Φout|+ |∇xΦout|λ−1j 〈ρ j〉−2)≲ 1{|x−q[ j]|≥λ∗R/2}‖Φout‖♯,Θ,α×{λ−2j 〈ρ j〉−4[| ln(T − t)|λΘ+1∗ R(t)+(T − t)‖Z∗‖C3(R2)+λ jρ j(λΘ∗ (0)+‖Z∗‖C3(R2))]+(λΘ∗ (0)+‖Z∗‖C3(R2))λ−1j 〈ρ j〉−2}≲ 1{|x−q[ j]|≥λ∗R/2}‖Φout‖♯,Θ,α×{〈ρ j〉−2[| ln(T − t)|λΘ−1∗ R−1+(T − t)λ−2j R−2‖Z∗‖C3(R2)+λ−1j R−1(λΘ∗ (0)+‖Z∗‖C3(R2))]+(λΘ∗ (0)+‖Z∗‖C3(R2))λ−1j 〈ρ j〉−2}≲ T ε(ρ [ j]2 +ρ3).(4.13.41)• ∣∣∣(1−η [ j]R ){−∂t(η [ j]dqΦ∗[ j]0 )+ (a−bU [ j]∧)[∆x(η [ j]dqΦ∗[ j]0 )+ |∇xU [ j]|2η [ j]dqΦ∗[ j]0−2∇x(U [ j] ·η [ j]dqΦ∗[ j]0)∇xU [ j]]−∂tU [ j]}∣∣∣=∣∣∣(1−η [ j]R ){−η [ j]dq∂tΦ∗[ j]0 −Φ∗[ j]0 ∂tη [ j]dq + (a−bU [ j]∧)[η [ j]dq∆xΦ∗[ j]0 +2∇xη [ j]dq∇xΦ∗[ j]0 +Φ∗[ j]0 ∆xη [ j]dq+ |∇xU [ j]|2η [ j]dqΦ∗[ j]0 −2(U [ j] ·Φ∗[ j]0)∇xη [ j]dq∇xU[ j]−2η [ j]dq∇x(U [ j] ·Φ∗[ j]0)∇xU [ j]]−∂tU [ j]}∣∣∣=∣∣∣(1−η [ j]R )η [ j]dq{−∂tΦ∗[ j]0 + (a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 )∇xU [ j]]−∂tU [ j]}− (1−η [ j]dq)∂tU [ j]+(1−η [ j]R){−Φ∗[ j]0 ∂tη [ j]dq + (a−bU [ j]∧)[2∇xη [ j]dq∇xΦ∗[ j]0 +Φ∗[ j]0 ∆xη [ j]dq−2(U [ j] ·Φ∗[ j]0 )∇xη [ j]dq∇xU [ j]]}∣∣∣≲ 1{λ jR≤|x−ξ [ j]|≤2dq}(λ−1∗ 〈ρ j〉−2+ |λ˙∗|〈ρ j〉−1+ |ξ˙ [ j]|)+1{|x−ξ [ j]|≥dq}[(λ−1j |λ˙ j|+ |γ˙ j|)〈ρ j〉−1+λ−1j |ξ˙ [ j]|〈ρ j〉−2]+1{dq≤|x−ξ [ j]|≤2dq}≲(1{λ jR/2≤|x−q[ j]|≤λ jR}+1{λ jR<|x−q[ j]|≤3dq})λ∗|x−q[ j]|−2+1{λ jR/2≤|x−q[ j]|≤3dq}(|λ˙∗|R−1+ |ξ˙ [ j]|)+1{|x−ξ [ j]|≥dq}[(|λ˙ j|+λ j|γ˙ j|)+λ j|ξ˙ [ j]|]+1{dq≤|x−ξ [ j]|≤2dq}≲ T ε(ρ [ j]1 +ρ[ j]2 +ρ3)providedΘ< β (4.13.42)where we have used (4.3.56), (4.3.10) and (4.3.24) in the first “≲”.324• By (4.3.54), we have ∣∣∣∣η [ j]R (eiθ jM˜[ j]1 + e−iθ jM[ j]−1)C−1j∣∣∣∣≲ η [ j]R |ξ˙ [ j]|≲ T ερ1.• Since |γ˙ j|≲ (T − t)−1, |ξ˙ [ j]|≲ O(R−20 ), one has∣∣∣η [ j]R Qγ j [(λ−1j λ˙ jy[ j]+λ−1j ξ˙ [ j]) ·∇y[ j]Φ[ j]in− γ˙ jJΦ[ j]in]∣∣∣≲ η [ j]R (T − t)−1‖Φin‖[ j]in,ν−δ0,lλ ν−δ0∗ ≲ T ερ [ j]1for ε > 0 sufficiently small providedΘ+δ0+β −ν < 0. (4.13.43)• ∣∣Qγ j {−Φ[ j]in∂tη [ j]R + (a−bW [ j]∧)[Φ[ j]in∆xη [ j]R +2∇xη [ j]R ∇xΦ[ j]in+W [ j] ·Φ[ j]in (−2∇xη [ j]R ∇xW [ j])]}∣∣=∣∣−Φ[ j]in∂tη [ j]R + (a−bW [ j]∧)[Φ[ j]in∆xη [ j]R +2∇xη [ j]R ∇xΦ[ j]in+W [ j] ·Φ[ j]in (−2∇xη [ j]R ∇xW [ j])]∣∣=∣∣∣Φ[ j]in (∇η)(x−ξ [ j]λ jR)·(ξ˙ [ j]λ jR+x−ξ [ j]λ jR(λ jR)′λ jR)+(a−bW [ j]∧)[Φ[ j]in(λ jR)−2 (∆η)(x−ξ [ j]λ jR)+2(λ jR)−1 (∇η)(x−ξ [ j]λ jR)λ−1j ∇y[ j]Φ[ j]in−2(W [ j] ·Φ[ j]in)(λ jR)−1 (∇η)(x−ξ [ j]λ jR)λ−1j ∇y[ j]W[ j]]∣∣∣≲ ‖Φ[ j]in‖in,ν−δ0,l1{λ jR≤|x−ξ [ j]|≤2λ jR}[(T − t)−1λ ν−δ0∗ 〈y[ j]〉−l +(λ jR)−2λ ν−δ0∗ 〈y[ j]〉−l+(λ jR)−1λ−1j λν−δ0∗ 〈y[ j]〉−l−1+(λ jR)−1λ−1j 〈y[ j]〉−2λ ν−δ0∗ 〈y[ j]〉−l]∼ ‖Φ[ j]in‖in,ν−δ0,l1{λ jR≤|x−ξ [ j]|≤2λ jR}(λ jR)−2λ ν−δ0∗ R−l ≲ T ερ1(4.13.44)providedδ0 < β <12, ν−δ0+β l− (1−β )>Θ. (4.13.45)•(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )+η [ j]R Qγ j∆xΦ[ j]in+Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ∇xΦ[ j]in)−2∇x(U [ j] ·Φout)∇xU [ j]−2∇x[U [ j] ·(η [ j]R Qγ jΦ[ j]in+η[ j]dqΦ∗[ j]0)]∇xU [ j]}=(U∗−U [ j])∧{∆x(η [ j]dqΦ∗[ j]0 )−2∇x(U [ j] ·η [ j]dqΦ∗[ j]0 )∇xU [ j]325+Qγ j(Φ[ j]in∆xη[ j]R +2∇xη[ j]R ∇xΦ[ j]in)−2(W [ j] ·Φ[ j]in)∇xη [ j]R ∇xU [ j]+η [ j]R Qγ j∆xΦ[ j]in−2η [ j]R ∇x(W [ j] ·Φ[ j]in)∇xU [ j]−2∇x(U [ j] ·Φout)∇xU [ j]},where by (4.3.24), it follows that∣∣∣(U∗−U [ j])∧[∆x(η [ j]dqΦ∗[ j]0 )−2∇x(U [ j] ·η [ j]dqΦ∗[ j]0 )∇xU [ j]]∣∣∣=∣∣∣(U∗−U [ j])∧[Φ∗[ j]0 ∆xη [ j]dq +2∇xη [ j]dq∇xΦ∗[ j]0 +η [ j]dq∆xΦ∗[ j]0−2(U [ j] ·Φ∗[ j]0 )∇xη [ j]dq∇xU [ j]−2η [ j]dq (U [ j] ·∇xΦ∗[ j]0 )∇xU [ j]−2η [ j]dq (Φ∗[ j]0 ·∇xU [ j])∇xU [ j]]∣∣∣≲ λ∗(1+λ−1j 〈ρ j〉−1+λ−1∗ 〈ρ j〉−2+λ j〈ρ j〉λ−2∗ 〈ρ j〉−4)1{|x−q j|≤3dq} ≲ T ερ3.• ∣∣(U∗−U [ j])∧ [Qγ j (Φ[ j]in∆xη [ j]R +2∇xη [ j]R ∇xΦ[ j]in)−2(W [ j] ·Φ[ j]in)∇xη [ j]R ∇xU [ j]]∣∣≲ λ∗∣∣Φ[ j]in∆xη [ j]R +2∇xη [ j]R ∇xΦ[ j]in−2(W [ j] ·Φ[ j]in)∇xη [ j]R ∇xW [ j]∣∣≲ T ερ1by the similar estimate in (4.13.44).• ∣∣(U∗−U [ j])∧η [ j]R [Qγ j∆xΦ[ j]in−2∇x (W [ j] ·Φ[ j]in)∇xU [ j]]∣∣= λ−2j∣∣∣(U∗−U [ j])∧η [ j]R [Qγ j∆y[ j]Φ[ j]in−2∇y[ j] (W [ j] ·Φ[ j]in)∇y[ j]U [ j]]∣∣∣= λ−2j∣∣∣(U∗−U [ j])∧η [ j]R [Qγ j∆y[ j]Φ[ j]in−2(W [ j] ·∇y[ j]Φ[ j]in+Φ[ j]in ·∇y[ j]W [ j])∇y[ j]U [ j]]∣∣∣≲ 1{|x−ξ [ j]|≤2λ jR}λ−1∗ ‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈y[ j]〉−l−2 ≲ T ερ1providedΘ+β +δ0−ν < 0. (4.13.46)• (U∗−U [ j])∧ [∇x (U [ j] ·Φout)∇xU [ j]]= (U∗−U [ j])∧ [(U [ j] ·∇xΦout+Φout ·∇xU [ j])∇xU [ j]]≲ ∑k 6= j〈ρk〉−1(|∇xΦout|+ |Φout|λ−2j 〈ρ j〉−4),where∑k 6= j〈ρk〉−1|∇xΦout|≲ ∑k 6= j〈ρk〉−1‖Φout‖♯,Θ,α(λΘ∗ (0)+‖Z∗‖C3(R2))≲ T ερ3.326By (4.13.7), we haveη [ j]R ∑k 6= j〈ρk〉−1|Φout|λ−2j 〈ρ j〉−4 ≲ ‖Φout‖♯,Θ,αη [ j]R ∑k 6= jλkλ−2j 〈ρ j〉−4×[| ln(T − t)|λΘ+1∗ (t)R(t)+(T − t)‖Z∗‖C3(R2)+ infm=1,...,N |x−q[m]|(λΘ∗ (0)+‖Z∗‖C3(R2))]≲ ‖Φout‖♯,Θ,αη [ j]R ∑k 6= jλkλ−2j 〈ρ j〉−4×[| ln(T − t)|λΘ+1∗ (t)R(t)+(T − t)‖Z∗‖C3(R2)+(λ jR+ |ξ [ j]−q[ j]|)(λΘ∗ (0)+‖Z∗‖C3(R2))]≲ ‖Φout‖♯,Θ,αη [ j]R λ−1∗ 〈ρ j〉−4λ∗R(| lnT |λΘ∗ (0)+‖Z∗‖C3(R2))≲ T ερ [ j]1provided2β +Θ−1< 0. (4.13.47)For all k 6= j,η [k]R(1−η [ j]R)〈ρk〉−1|Φout|λ−2j 〈ρ j〉−4≲ ‖Φout‖♯,Θ,αη [k]R(1−η [ j]R)〈ρk〉−1λ 2j (| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ T ερ [k]1 .(1−η [k]R)(1−η [ j]R)〈ρk〉−1|Φout|λ−2j 〈ρ j〉−4≲ ‖Φout‖♯,Θ,α(1−η [k]R)(1−η [ j]R)λ 3∗ |x−q[k]|−1|x−q[ j]|−4(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ T ε(ρ [ j]2 +ρ3)since1{|x−q[k]|≤dq}‖Φout‖♯,Θ,α(1−η [k]R)(1−η [ j]R)λ 3∗ |x−q[k]|−1|x−q[ j]|−4(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ ‖Φout‖♯,Θ,α1{|x−q[k]|≤dq}(1−η [k]R)(1−η [ j]R)λ 3∗ (λkR)−1(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ T ερ3,1{|x−q[ j]|≤dq}‖Φout‖♯,Θ,α(1−η [k]R)(1−η [ j]R)λ 3∗ |x−q[k]|−1|x−q[ j]|−4(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ 1{|x−q[ j]|≤dq}‖Φout‖♯,Θ,α(1−η [k]R)(1−η [ j]R)λ∗R−2|x−q[ j]|−2(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ T ερ [ j]2 , (1−1{|x−q[k]|≤dq}−1{|x−q[ j]|≤dq})‖Φout‖♯,Θ,α(1−η [k]R)(1−η [ j]R)×λ 3∗ |x−q[k]|−1|x−q[ j]|−4(| lnT |λΘ+1∗ (0)R(0)+‖Z∗‖C3(R2))≲ T ερ3.327• For(a−bU∗∧){−2N∑j=1∇x[Φ · (U∗−U [ j])]∇xU [ j]} ,it suffices to estimate∇x[Φ · (U∗−U [ j])]∇xU [ j] = [(U∗−U [ j]) ·∇xΦ+Φ ·∇x (U∗−U [ j])]∇xU [ j].Then for any fixed j,∣∣∇x [Φ · (U∗−U [ j])]∇xU [ j]∣∣≲(|∇xΦ|∑k 6= j〈ρk〉−1+ |Φ|∑k 6= jλ−1∗ 〈ρk〉−2)λ−1∗ 〈ρ j〉−2≲ |∇xΦ| 〈ρ j〉−1∑k 6= j(min{〈ρ j〉,〈ρk〉})−1+ |Φ|∑k 6= j(min{〈ρ j〉,〈ρk〉})−2since〈ρ j〉〈ρk〉≳ λ−1∗ min{〈ρ j〉,〈ρk〉} for j 6= k. (4.13.48)Proof of (4.13.48). For |x−ξ [ j]| ≤ |ξ [ j]−ξ [k]|2 , then 〈ρk〉 ∼ λ−1∗ , which implies〈ρ j〉〈ρk〉 ∼ λ−1∗ 〈ρ j〉 ∼ λ−1∗ min{〈ρ j〉,〈ρk〉} .For |x−ξ [k]| ≤ |ξ [ j]−ξ [k]|2 , similarly, we have〈ρ j〉〈ρk〉 ∼ λ−1∗ min{〈ρ j〉,〈ρk〉} .For |x−ξ [ j]|> |ξ [ j]−ξ [k]|2 and |x−ξ [k]|> |ξ[ j]−ξ [k]|2 ,〈ρ j〉〈ρk〉 ∼ λ−2∗ |x−ξ [ j]||x−ξ [k]|≳ λ−1∗ min{〈ρ j〉,〈ρk〉} .For k 6= j, by (4.13.7) and (4.13.10), we have|Φ|(min{〈ρ j〉,〈ρk〉})−2 ≲ T ερ3.328By (4.13.9) and (4.3.24), we have|∇xΦ| 〈ρ j〉−1(min{〈ρ j〉,〈ρk〉})−1≲ 〈ρ j〉−1(min{〈ρ j〉,〈ρk〉})−1 [‖Φout‖♯,Θ,α (λΘ∗ (0)+‖Z∗‖C3(R2))+N∑m=1(1{|x−ξ [m]|≤2λmR}λ−1m ‖Φ[m]in ‖in,ν−δ0,lλ ν−δ0∗ 〈ρm〉−l−1+1{|x−ξ [m]|≤2dq})≲ T ε(N∑m=1ρ [m]1 +ρ3)providedΘ+β +δ0−ν < 0. (4.13.49)•(a−bU∗∧){−2N∑j=1∇x[U [ j] ·∑k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0)]∇xU [ j]}.For k 6= j, we get∇x[U [ j] · (η [k]R QγkΦ[k]in)]∇xU [ j]=[η [k]R(W [ j] ·∇xΦ[k]in+Φ[k]in ·∇xW [ j])+(W [ j] ·Φ[k]in)∇xη [k]R]∇xU [ j]=[η [k]R(W [ j] ·λ−1k ∇y[k]Φ[k]in+Φ[k]in ·λ−1j ∇y[ j]W [ j])+(W [ j] ·Φ[k]in)∇xη [k]R]λ−1j ∇y[ j]U[ j]≲ ‖Φ[k]in‖in,ν−δ0,lλ ν−δ0∗[1{|x−ξ [k]|≤2λkR}(λ−1k 〈ρk〉−l−1+ 〈ρk〉−lλ−1j 〈ρ j〉−2)+ 〈ρk〉−l(λkR)−11{λkR≤|x−ξ [k]|≤2λkR}]λ−1j 〈ρ j〉−2∼ 1{|x−ξ [k]|≤2λkR}‖Φ[k]in‖in,ν−δ0,lλ ν−δ0∗(λ−1k 〈ρk〉−l−1+ 〈ρk〉−lλ j)λ j≲ 1{|x−ξ [k]|≤2λkR}‖Φ[k]in‖in,ν−δ0,lλ ν−δ0∗(〈ρk〉−l−1+ 〈ρk〉−lλ 2∗)≲ T ερ [k]1whenδ0 < ν ; (4.13.50)and by (4.3.24),∣∣∣∇x(U [ j] ·η [k]dqΦ∗[k]0 )∇xU [ j]∣∣∣= ∣∣∣[(U [ j] ·Φ∗[k]0 )∇xη [k]dq +η [k]dq (U [ j] ·∇xΦ∗[k]0 +Φ∗[k]0 ·∇xU [ j])]∇xU [ j]∣∣∣≲ λ j1{|x−ξ [k]|≤2dq} ≲ Tερ3.•N∑j=1|∇xU [ j]|2(a−bU [ j]∧)∑k 6= j(η [k]R QγkΦ[k]in+η[k]dqΦ∗[k]0).For k 6= j, ∣∣|∇xU [ j]|2η [k]R QγkΦ[k]in∣∣≲ λ 2j η [k]R ‖Φ[k]in‖in,ν−δ ,lλ ν−δ0∗ 〈ρk〉−l ≲ T ερ [k]1329and by (4.3.24),|∇xU [ j]|2|η [k]dqΦ∗[k]0 |≲ λ 2j ≲ T ερ3.• By (4.13.7), (4.13.10) and (4.13.48),∣∣∣∣∣aΦ∑j 6=k∇xU [ j] ·∇xU [k]∣∣∣∣∣≲ |Φ|∑j 6=kλ−1j 〈ρ j〉−2λ−1k 〈ρk〉−2 ≲ |Φ|∑j 6=k(min{〈ρ j〉,〈ρk〉})−2 ≲ T ερ3.• By (4.4.5), (4.3.10) and (4.13.30),|[(Φ ·U∗)−A]∂tU∗|≲(|Φ ·U∗|+λ∗+ |Φ|2) N∑j=1[(λ−1j |λ˙ j|+ |γ˙ j|)〈ρ j〉−1+λ−1j |ξ˙ [ j]|〈ρ j〉−2]≲{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2×1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2}N∑j=1(λ−1j |λ˙ j|+ |γ˙ j|+λ−1j |ξ˙ [ j]|)〈ρ j〉−1≲ T ε(N∑j=1ρ [ j]1 +ρ3)providedΘ+β +2δ0−2ν < 0. (4.13.51)•2∑j=1η [ j]R(U [ j]−U∗){−2a(∇xW [ j] ·∇xΦ[ j]in)+a|∇xU [ j]|2 (U [ j] ·Φout)+{−∂t(Φ∗[ j]0 )+(a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 )∇xU [ j]]−∂tU [ j]} ·U [ j]}.For above terms, we first estimate∣∣η [ j]R (U [ j]−U∗)(∇xW [ j] ·∇xΦ[ j]in)∣∣=∣∣∣λ−2j η [ j]R (U [ j]−U∗)(∇y[ j]W [ j] ·∇y[ j]Φ[ j]in)∣∣∣≲ λ−2j 1{|x−ξ [ j]|≤2λ jR}λ∗〈ρ j〉−2‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l−1∼ 1{|x−ξ [ j]|≤2λ jR}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−1∗ 〈ρ j〉−l−3 ≲ T ερ [ j]1330whenΘ+δ0+β −ν < 0. (4.13.52)Secondly, by (4.13.8),∣∣η [ j]R (U [ j]−U∗) |∇xU [ j]|2 (U [ j] ·Φout)∣∣≲ ‖Φout‖♯,Θ,αη [ j]R λ−1∗ 〈ρ j〉−4λ∗R(| lnT |λΘ∗ (0)+‖Z∗‖C3(R2))≲ T ερ [ j]1providedΘ+2β −1< 0. (4.13.53)Thirdly, by (4.3.46), we obtain∣∣∣∣∣η [ j]R (U [ j]−U∗){{−∂t(Φ∗[ j]0 )+(a−bU [ j]∧)[∆xΦ∗[ j]0 + |∇xU [ j]|2Φ∗[ j]0 −2∇x (U [ j] ·Φ∗[ j]0 )∇xU [ j]]−∂tU [ j]}·U [ j]}∣∣∣∣∣≲ η [ j]R λ∗(|ξ˙ [ j]|〈ρ j〉−1+ |λ j|−1〈ρ j〉−2)≲ T ερ3.• By (4.13.28), we have(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1[Φ+(1+A−Φ ·U∗)U∗] ·∆x (Φ−Φout)−(AU∗+ΠU⊥∗ Φ)∧∆x (Φ−Φout)= (Φ∧U∗)(1+O(λ∗+ |Φ|2))[Φ+(1+A−Φ ·U∗)U∗] ·∆x (Φ−Φout)− [AU∗+Φ− (Φ ·U∗)U∗]∧∆x (Φ−Φout)= (Φ∧U∗) [U∗ ·∆x (Φ−Φout)]−Φ∧∆x (Φ−Φout)+(Φ∧U∗) [Φ+(A−Φ ·U∗)U∗] ·∆x (Φ−Φout)+(Φ∧U∗)O(λ∗+ |Φ|2)[Φ+(1+A−Φ ·U∗)U∗] ·∆x (Φ−Φout)− [AU∗− (Φ ·U∗)U∗]∧∆x (Φ−Φout) .For above terms, we estimate by (4.4.5)∣∣∣(Φ∧U∗) [Φ+(A−Φ ·U∗)U∗] ·∆x (Φ−Φout)+(Φ∧U∗)O(λ∗+ |Φ|2)[Φ+(1+A−Φ ·U∗)U∗] ·∆x (Φ−Φout)− [AU∗− (Φ ·U∗)U∗]∧∆x (Φ−Φout)∣∣∣≲(λ∗+ |Φ|2+ |Φ ·U∗|) |∆x (Φ−Φout)| .331Using (4.13.30) and (4.13.14), we have(λ∗+ |Φ|2+ |Φ ·U∗|) |∆x (Φ−Φout)|≲{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2}×N∑j=1[1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−2∗ 〈ρ j〉−l−2+λ−1∗ 〈ρ j〉−1)+λ−1∗ 〈ρ j〉−11{3λ∗R<|x−q[ j]|<3dq}]∼N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)3 1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0+1∗ 〈ρ j〉1−l +λ 2∗ 〈ρ j〉2+ | ln(T − t)|λΘ+2∗ R〈ρ j〉+ | ln(T − t)|λ 2ν−2δ0+Θ+1∗ R〈ρ j〉−l + | ln(T − t)|λΘ+2∗ R〈ρ j〉+ | ln(T − t)|2λ 2Θ+2∗ R2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2]≲ T ερ3.We need more refined estimates for the other part. Recalling (4.4.1), we have(Φ∧U∗) [U∗ ·∆x (Φ−Φout)]−Φ∧∆x (Φ−Φout)= −Φ∧{∆x (Φ−Φout)− [U∗ ·∆x (Φ−Φout)]U∗}= −(N∑j=1η [ j]dqΦ∗[ j]0 +Φout)∧{∆x (Φ−Φout)− [U∗ ·∆x (Φ−Φout)]U∗}−N∑j=1η [ j]R Qγ jΦ[ j]in∧{[U [ j] ·∆x (Φ−Φout)]U [ j]− [U∗ ·∆x (Φ−Φout)]U∗}−N∑j=1η [ j]R Qγ jΦ[ j]in∧{∆x (Φ−Φout)−[U [ j] ·∆x (Φ−Φout)]U [ j]}.By (4.3.24) and (4.13.8), one has∣∣∣∣∣ N∑j=1η [ j]dqΦ∗[ j]0 +Φout∣∣∣∣∣≲N∑j=11{|x−q[ j]|<3dq}[λ j〈ρ j〉+‖Φout‖♯,Θ,α(| ln(T − t)|λΘ+1∗ R+λ jρ j)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α .(4.13.54)332Then using (4.13.14), we get∣∣∣∣∣(N∑j=1η [ j]dqΦ∗[ j]0 +Φout)∧{∆x (Φ−Φout)− [U∗ ·∆x (Φ−Φout)]U∗}∣∣∣∣∣≲{N∑j=11{|x−q[ j]|<3dq}[λ j〈ρ j〉+‖Φout‖♯,Θ,α(| ln(T − t)|λΘ+1∗ R+λ jρ j)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}×N∑j=1[1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0−2∗ 〈ρ j〉−l−2+λ−1∗ 〈ρ j〉−1)+λ−1∗ 〈ρ j〉−11{3λ∗R<|x−q[ j]|<3dq}]≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2×(λ ν−δ0−1∗ 〈ρ j〉−l−1+1+ | ln(T − t)|λ ν−δ0+Θ−1∗ R〈ρ j〉−l−2+ | ln(T − t)|λΘ∗ R〈ρ j〉−1)+(1+‖Φout‖♯,Θ,α)1{3λ∗R<|x−q[ j]|<3dq}]≲ T ε(N∑j=1ρ [ j]1 +ρ3)providedΘ+β +δ0−ν < 0, 2β +δ0−ν < 0. (4.13.55)By (4.13.14) and (4.4.41), we have∣∣η [ j]R Qγ jΦ[ j]in∧{[U [ j] ·∆x (Φ−Φout)]U [ j]− [U∗ ·∆x (Φ−Φout)]U∗}∣∣=∣∣η [ j]R Qγ jΦ[ j]in∧{[(U [ j]−U∗) ·∆x (Φ−Φout)]U [ j]+[U∗ ·∆x (Φ−Φout)](U [ j]−U∗)}∣∣≲ η [ j]R λ∗∣∣Φ[ j]in∣∣ |∆x (Φ−Φout)|≲ 1{|x−q[ j]|≤3λ∗R}λ∗‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l×N∑k=1[1{|x−q[k]|≤3λ∗R}(‖Φ[k]in‖in,ν−δ0,lλ ν−δ0−2∗ 〈ρk〉−l−2+λ−1∗ 〈ρk〉−1)+λ−1∗ 〈ρk〉−11{3λ∗R<|x−q[k]|<3dq}]≲ 1{|x−q[ j]|≤3λ∗R}(1+‖Φ[ j]in‖in,ν−δ0,l)2(λ 2ν−2δ0−1∗ 〈ρ j〉−2l−2+λ ν−δ0∗ 〈ρ j〉−l−1)≲ T ερ [ j]1providedΘ+β +2δ0−2ν < 0. (4.13.56)Noticeη [ j]R Qγ jΦ[ j]in∧{∆x (Φ−Φout)−[U [ j] ·∆x (Φ−Φout)]U [ j]}is a vector parallel withU [ j]. That is,η [ j]R Qγ jΦ[ j]in∧{∆x (Φ−Φout)−[U [ j] ·∆x (Φ−Φout)]U [ j]}= η [ j]R f j(x, t)U[ j]333where ∣∣ f j(x, t)∣∣= ∣∣Qγ jΦ[ j]in∧{∆x (Φ−Φout)− [U [ j] ·∆x (Φ−Φout)]U [ j]}∣∣ .ByU∗-operation, we only need to estimate the following term.∣∣η [ j]R f j(x, t)(U [ j]−U∗)∣∣≲ η [ j]R λ∗ ∣∣Φ[ j]in∣∣ |∆x (Φ−Φout)|≲ T ερ [ j]1by the same calculations as the above terms under the assumptions (4.13.56) on the parameters.Under the parameter assumptions (4.13.25), by (4.13.26) and (4.13.1), we have|(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗|≲{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)4 1{|x−q[ j]|≤3λ∗R} (λ ε+1∗ λΘ∗ (λ∗R)−1+λ∗〈ρ j〉)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)4(| ln(T − t)|λΘ+1∗ R+λ j〈ρ j〉)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)4}(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)≲ T ε(N∑j=1ρ [ j]1 +ρ3).(4.13.57)•(A−Φ ·U∗)∆xU∗ =−(A−Φ ·U∗)N∑j=1|∇xU [ j]|2U [ j].ByU∗-operation and (4.4.5), it suffices to estimate∣∣∣∣∣(A−Φ ·U∗) N∑j=1 |∇xU [ j]|2 (U [ j]−U∗)∣∣∣∣∣≲ (λ∗+ |Φ|2+ |Φ ·U∗|) N∑j=1λ−2j 〈ρ j〉−4∑k 6= j〈ρk〉−1which will be dealt with uniformly in (4.13.58) later.• By (4.13.28), one has∣∣∣∣(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1 (1+A−Φ ·U∗)(2∇xΦ ·∇xU∗)− (Φ∧U∗)(2∇xΦ ·∇xU∗)∣∣∣∣=∣∣∣(Φ∧U∗)(1+O(λ∗+ |Φ|2+ |Φ ·U∗|))2 (2∇xΦ ·∇xU∗)− (Φ∧U∗)(2∇xΦ ·∇xU∗)∣∣∣≲ |Φ|(λ∗+ |Φ|2+ |Φ ·U∗|) |∇xΦ ·∇xU∗|≲(λ∗+ |Φ|2+ |Φ ·U∗|) |Φ| |∇xU∗|2+ (λ∗+ |Φ|2+ |Φ ·U∗|) |Φ| |∇xΦ|2which will be controlled by (4.13.58) and (4.13.60) later.334• ∣∣∣∣∣[|∇xA|2|U∗|2+2(1+A)∇xA · (U∗ ·∇xU∗)+A(2+A) |∇xU∗|2+22∑k=1{[(∂xkA)U∗ ·∂xkΦ+A∂xkU∗ ·∂xkΦ]−∂xk (U∗ ·Φ)[|U∗|2∂xkA+(1+A)U∗ ·∂xkU∗]− (U∗ ·Φ)[(∂xkA)U∗ ·∂xkU∗+(1+A) |∂xkU∗|2]}+2∑k=1|∂xkΦ−U∗∂xk (Φ ·U∗)− (Φ ·U∗)∂xkU∗|2]ΠU⊥∗ Φ∣∣∣∣∣+∣∣∣∣∣−b[−2−1(Φ∧U∗)[(1+A)|U∗|2+(U∗ ·ΠU⊥∗ Φ)]−1{2(1+A−Φ ·U∗)(Φ ·∆xU∗)+2(|U∗|2−2)|∇x (Φ ·U∗) |2+2|∇xΦ|2+8[(Φ ·U∗)− (1+A)](U∗ ·∇xU∗) ·∇x (Φ ·U∗)+2|U∗|2|∇xA|2+4[−2(Φ ·U∗)U∗ ·∇xU∗+(1−|U∗|2)∇x (Φ ·U∗)] ·∇xA+8(1+A)(U∗ ·∇xU∗) ·∇xA+2 [(Φ ·U∗)− (1+A)]2(|∇xU∗|2+U∗ ·∆xU∗)}− (ΠU⊥∗ Φ+AU∗)∧ [2∇x (Φ ·U∗)∇xU∗]+ [A− (Φ ·U∗)]Φ∧∆xU∗+ΠU⊥∗ Φ∧ (2∇xA∇xU∗)+[(Φ ·U∗)2−2A(Φ ·U∗)−2(Φ ·U∗)]U∗∧∆xU∗+(1+A)U∗∧ [A∆xU∗+2(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗+∆xU∗−2(U∗ ·∇xU∗)∇xU∗]]+2bAU∗∧ [(Φ ·∇xΦ)∇xU∗]∣∣∣∣∣≲(|∇xA|2+ |∇xA| |U∗ ·∇xU∗|+(λ∗+ |Φ|2) |∇xU∗|2+ |∇xA||∇xΦ|+(λ∗+ |Φ|2)|∇xU∗||∇xΦ|+ |∇xA| |∇x (U∗ ·Φ)|+ |U∗ ·∇xU∗| |∇x (U∗ ·Φ)|+ |∇xA||U∗ ·∇xU∗| |U∗ ·Φ|+ |∇xU∗|2 |U∗ ·Φ|+ |∇xΦ|2+ |∇x (Φ ·U∗) |2+ |Φ ·U∗|2|∇xU∗|2)|Φ|+ |Φ|(∣∣∣∣∣Φ · N∑j=1 |∇xU [ j]|2U [ j]∣∣∣∣∣+ |∇x (Φ ·U∗) |2+ |∇xΦ|2+ |U∗ ·∇xU∗||∇x (Φ ·U∗) |+ |∇xA|2+ |Φ ·U∗||U∗ ·∇xU∗||∇xA|+λ∗|∇x (Φ ·U∗) ||∇xA|+ |U∗ ·∇xU∗||∇xA|+∣∣|∇xU∗|2+U∗ ·∆xU∗∣∣)+(λ∗+ |Φ|) |∇x (Φ ·U∗)| |∇xU∗|+(λ∗+ |Φ|2+ |Φ ·U∗|) |Φ∧∆xU∗|+ |Φ| |∇xA∇xU∗|+ |Φ ·U∗| |U∗∧∆xU∗|+(λ∗+ |Φ|2) |U∗∧∆xU∗|+ |U∗∧ [∆xU∗−2(U∗ ·∇xU∗)∇xU∗]|335+ |(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗|+(λ∗+ |Φ|2) |Φ||∇xΦ||∇xU∗|≲ |∇xA|2|Φ|+(λ∗+ |Φ|2+ |U∗ ·Φ|) |Φ|(|∇xU∗|2+ |∆xU∗|)+ |U∗ ·∇xU∗|2 |Φ|+ |∇xΦ|2|Φ|+∣∣∣∣∣Φ · N∑j=1 |∇xU [ j]|2U [ j]∣∣∣∣∣ |Φ|+ ∣∣|∇xU∗|2+U∗ ·∆xU∗∣∣ |Φ|+(λ∗+ |Φ|) |∇x (Φ ·U∗)| |∇xU∗|+ |Φ| |∇xA∇xU∗|+(λ∗+ |Φ|2+ |U∗ ·Φ|) |U∗∧∆xU∗|+ |U∗∧ [∆xU∗−2(U∗ ·∇xU∗)∇xU∗]|+ |(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗|≲(λ∗+ |Φ|2+ |U∗ ·Φ|)[|Φ|(|∇xU∗|2+ |∆xU∗|)+ |U∗∧∆xU∗|]+ |U∗ ·∇xU∗|2 |Φ|+ |∇xΦ|2|Φ|+∣∣∣∣∣Φ · N∑j=1 |∇xU [ j]|2U [ j]∣∣∣∣∣ |Φ|+ ∣∣|∇xU∗|2+U∗ ·∆xU∗∣∣ |Φ|+(λ∗+ |Φ|) |∇x (Φ ·U∗)| |∇xU∗|+ |Φ| |(U∗ ·∇xU∗)∇xU∗|+ |Φ| |(Φ ·∇xΦ)∇xU∗|+ |U∗∧ [∆xU∗−2(U∗ ·∇xU∗)∇xU∗]|+O(T ε)(N∑j=1ρ [ j]1 +ρ3),where in the last “≲”, we require (4.13.25) and then by (4.13.26),|∇xA|2|Φ|≲ |U∗ ·∇xU∗|2 |Φ|+ |Φ|3|∇xΦ|2+T ερ3,|Φ| |∇xA∇xU∗|≲ |Φ| |(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗|+ |Φ| |(U∗ ·∇xU∗)∇xU∗|+ |Φ| |(Φ ·∇xΦ)∇xU∗| ,and |(∇xA+U∗ ·∇xU∗+Φ ·∇xΦ)∇xU∗| has been controlled by (4.13.57).• Combining (4.13.1), (4.13.2), (4.13.6), (4.13.30) and (4.13.15), we then obtain(λ∗+ |Φ|2+ |U∗ ·Φ|)[|Φ|(|∇xU∗|2+ |∆xU∗|)+ |U∗∧∆xU∗|]≲(λ∗+ |Φ|2+ |U∗ ·Φ|)(|Φ| N∑j=1λ−2∗ 〈ρ j〉−4+N∑j=1λ−2j 〈ρ j〉−4∑k 6= j〈ρ j〉−1)≲(λ∗+ |Φ|2+ |U∗ ·Φ|)×[N∑j=11{|x−q[ j]|<3dq}(|Φ|λ−2∗ 〈ρ j〉−4+λ−1∗ 〈ρ j〉−4+λ 2∗ 〈ρ j〉−1)+1{∩Nj=1{|x−q[ j]|≥3dq}} (|Φ|λ 2∗ +λ 3∗ )]≲{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}} (1+‖Φout‖♯,Θ,α)2}336×{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)×(λ ν−δ0−2∗ 〈ρ j〉−l−4+λ−1∗ 〈ρ j〉−3+ | ln(T − t)|λΘ−1∗ R〈ρ j〉−4)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)λ−1∗ 〈ρ j〉−3]+1{∩Nj=1{|x−q[ j]|≥3dq}}(‖Φout‖♯,Θ,αλ 2∗ +λ 3∗ )}≲N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)3 1{|x−q[ j]|≤3λ∗R}×(λ 3ν−3δ0−2∗ + | ln(T − t)|λ ν−δ0+Θ−1∗ R+ | ln(T − t)|2λ 2Θ∗ R2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)3 〈ρ j〉−2]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ2∗ (1+‖Φout‖♯,Θ,α)3 ≲ T ε(N∑j=1ρ [ j]1 +ρ3)(4.13.58)providedΘ+β +1+3δ0−3ν < 0, 2β +δ0−ν < 0, 3β < 1+Θ. (4.13.59)• By (4.13.3), we get|U∗ ·∇xU∗|2 |Φ|≲ T ερ3.•|∇xΦ|2|Φ|≲{N∑j=1(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 [1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0−2∗ 〈ρ j〉−2l−2+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α}×{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)1{|x−q[ j]|≤3λ∗R}(λ ν−δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}≲N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)3 1{|x−q[ j]|≤3λ∗R}(λ∗〈ρ j〉+λ 3ν−3δ0−2∗ + | ln(T − t)|λ 2ν−2δ0+Θ−1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)3(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖3♯,Θ,α≲ T ε(N∑j=1ρ [ j]1 +ρ3)(4.13.60)337providedΘ+β +1+3δ0−3ν < 0, β +δ0−ν < 0. (4.13.61)• By (4.13.4), we have∣∣∣∣∣Φ · N∑j=1 |∇xU [ j]|2U [ j]∣∣∣∣∣ |Φ| ≤∣∣∣∣∣ N∑j=1 |∇xU [ j]|2Φ · (U [ j]−U∗)∣∣∣∣∣ |Φ|+∣∣∣∣∣ N∑j=1 |∇xU [ j]|2Φ ·U∗∣∣∣∣∣ |Φ|≲ |Φ|2N∑j=1λ−2j 〈ρ j〉−4∑k 6= j〈ρk〉−1+ |Φ ·U∗| |Φ|N∑j=1λ−2∗ 〈ρ j〉−4 ≲ T ε(N∑j=1ρ [ j]1 +ρ3)where the last step is derived by the same way as (4.13.58) under the parameter assumption (4.13.59).• By (4.13.5) and (4.13.15), we get∣∣|∇xU∗|2+U∗ ·∆xU∗∣∣ |Φ|≲(N∑j=11{|x−q[ j]|<3dq}(〈ρ j〉−2+λ−1∗ 〈ρ j〉−4)+λ 2∗ 1{∩Nj=1{|x−q[ j]|≥3dq}})×{ N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)1{|x−q[ j]|≤3λ∗R}(λ ν−δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}≲N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)1{|x−q[ j]|≤3λ∗R}(λ ν−δ0−1∗ + | ln(T − t)|λΘ∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)× (λ∗〈ρ j〉−1+ | ln(T − t)|λΘ+1∗ R〈ρ j〉−2+ 〈ρ j〉−3+ | ln(T − t)|λΘ∗ R〈ρ j〉−4)]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ2∗ ‖Φout‖♯,Θ,α ≲ T ε(N∑j=1ρ [ j]1 +ρ3)(4.13.62)providedΘ+β +δ0−ν < 0. (4.13.63)• Combining (4.13.15), (4.13.24) and (4.13.1), one has(λ∗+ |Φ|) |∇x (Φ ·U∗)| |∇xU∗|≲{ N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)1{|x−q[ j]|≤3λ∗R}(λ ν−δ0∗ 〈ρ j〉−l +λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]338+1{∩Nj=1{|x−q[ j]|≥3dq}} (λ∗+‖Φout‖♯,Θ,α)}×{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(| ln(T − t)|λΘ∗ R〈ρ j〉−2+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α(λΘ∗ (0)+‖Z∗‖C3(R2))}×(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)≲N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 1{|x−q[ j]|≤3λ∗R}(| ln(T − t)|λ ν−δ0+Θ−1∗ R+ | ln(T − t)|2λ 2Θ∗ R2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(〈ρ j〉−1+ | ln(T − t)|λΘ∗ R〈ρ j〉−2)]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗ (1+‖Φout‖♯,Θ,α)2 ≲ T ε(N∑j=1ρ [ j]1 +ρ3)provided2β +δ0−ν < 0, 3β < 1+Θ. (4.13.64)• By (4.13.1) and (4.13.3), we have|Φ| |(U∗ ·∇xU∗)∇xU∗|≲ |Φ|(N∑j=11{|x−q[ j]|<3dq}〈ρ j〉−2+1{∩Nj=1|x−q[ j]|≥3dq}λ2∗)×(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)= |Φ|(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−4+1{∩Nj=1{|x−q[ j]|≥3dq}}λ3∗)≲ T ε(N∑j=1ρ [ j]1 +ρ3)which is obtained by the same calculation as in (4.13.62) under the parameter assumption (4.13.63).• By (4.13.15), (4.13.16) and (4.13.1), we get|Φ| |(Φ ·∇xΦ)∇xU∗|≲{ N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 1{|x−q[ j]|≤3λ∗R}×(λ 2ν−2δ0∗ 〈ρ j〉−2l +λ 2∗ 〈ρ j〉2+ | ln(T − t)|2λ 2Θ+2∗ R2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ 2∗ 〈ρ j〉2+ | ln(T − t)|2λ 2Θ+2∗ R2)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α}339×{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0−1∗ 〈ρ j〉−l−1+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}×(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)≲N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)3 1{|x−q[ j]|≤3λ∗R}(λ 3ν−3δ0−2∗ + | ln(T − t)|2λ 2Θ+ν−δ0∗ R2)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)3λ∗]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗‖Φout‖3♯,Θ,α≲ T ε(N∑j=1ρ [ j]1 +ρ3)providedΘ+β +1+3δ0−3ν < 0, 3β <Θ+1+ν−δ0. (4.13.65)• By (4.13.83), we have|U∗∧ [∆xU∗−2(U∗ ·∇xU∗)∇xU∗]|≲ T ε[N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3].• Consider2(a−bU∗∧)[(∇xU∗ ·∇xΦ)Φ− (Φ ·∇xΦ)∇xU∗−N∑j=1{[∇xU∗ ·∇x(η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)]∇xU∗}+N∑j=1{[∇xU∗ ·∇x(η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)]∇xU∗}−N∑j=1{[∇xU [ j] ·∇x(η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)]∇xU [ j]}]+2(a−bU∗∧)N∑j=1{[∇xU [ j] ·∇x(η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)]∇xU [ j]}−N∑j=12(a−bU [ j]∧){[∇xU [ j] ·∇x (η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [(η [ j]R Qγ jΦ[ j]in) ·∇x (η [ j]R Qγ jΦ[ j]in)]∇xU [ j]}.340We estimate by (4.13.11) that∣∣[∇xU∗ ·∇x (η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)− [∇xU [ j] ·∇x (η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)∣∣≲ λ∗∣∣∇x (η [ j]R Qγ jΦ[ j]in)∣∣ ∣∣η [ j]R Qγ jΦ[ j]in∣∣≲ 1{|x−q[ j]|≤3λ∗R}(‖Φ[ j]in‖in,ν−δ0,l)2λ 2ν−2δ0∗ 〈ρ j〉−2l−1 ≲ T ερ [ j]1 .By (4.13.11), (U∗−U [ j])∧ [∇xU [ j] ·∇x (η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)≲ 1{|x−q[ j]|≤3λ∗R}‖Φ[ j]in‖2in,ν−δ0,lλ 2ν−2δ0−1∗ 〈ρ j〉−2l−3 ≲ T ερ[ j]1providedΘ+β +2δ0−2ν < 0. (4.13.66)By the property of cut-off function,N∑j=1[∇xU∗ ·∇x(η [ j]R Qγ jΦ[ j]in)](η [ j]R Qγ jΦ[ j]in)=[∇xU∗ ·∇x(N∑j=1η [ j]R Qγ jΦ[ j]in)](N∑j=1η [ j]R Qγ jΦ[ j]in).Then by (4.13.1) and (4.13.16), it follows that∣∣∣∣∣(∇xU∗ ·∇xΦ)Φ−[∇xU∗ ·∇x(N∑j=1η [ j]R Qγ jΦ[ j]in)](N∑j=1η [ j]R Qγ jΦ[ j]in)∣∣∣∣∣=∣∣∣∣∣(∇xU∗ ·∇xΦ)(Φ−N∑j=1η [ j]R Qγ jΦ[ j]in)+[∇xU∗ ·∇x(Φ−N∑j=1η [ j]R Qγ jΦ[ j]in)](N∑j=1η [ j]R Qγ jΦ[ j]in)∣∣∣∣∣≲(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)×[{ N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)(λ ν−δ0−1∗ 〈ρ j〉−l−1+1)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}×{N∑j=1[(1+‖Φout‖♯,Θ,α)1{|x−q[ j]|≤3λ∗R}(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}+{ N∑j=1[1{|x−q[ j]|≤3λ∗R} (1+‖Φout‖♯,Θ,α)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖♯,Θ,α}341×(N∑j=11{|x−q[ j]|≤3λ∗R}‖Φ[ j]in‖in,ν−δ0,lλ ν−δ0∗ 〈ρ j〉−l)]≲(N∑j=11{|x−q[ j]|<3dq}λ−1∗ 〈ρ j〉−2+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗)×{N∑j=1[(1+‖Φout‖♯,Θ,α +‖Φ[ j]in‖in,ν−δ0,l)2 1{|x−q[ j]|≤3λ∗R}(λ∗〈ρ j〉+ | ln(T − t)|λ ν−δ0+Θ∗ R)+1{3λ∗R<|x−q[ j]|<3dq} (1+‖Φout‖♯,Θ,α)2(λ∗〈ρ j〉+ | ln(T − t)|λΘ+1∗ R)]+1{∩Nj=1{|x−q[ j]|≥3dq}}‖Φout‖2♯,Θ,α}≲ T ε(N∑j=1ρ [ j]1 +ρ3)provided2β +δ0−ν < 0. (4.13.67)The other terms in this collection can be dealt in the same way.• Before proceeding, we take a closer look at −∆xU∗ and ∇x(|U∗|2)∇xU∗.In the single bubble caseN= 1, then−∆xU∗ can be neglected by theU∗-operation and∇x(|U∗|2)∇xU∗vanishes automatically. But for case of multiple bubbles, the phenomenon is different, and interac-tions appear here.Recall (4.2.13). Claim:∆xU∗ =−N∑j=1|∇xU [ j]|2U [ j]=N∑j=1∑k 6= j{16η [ j]R λ−2j ρ2j (ρ2j +1)−3λk|q[ j]−q[k]|−2[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(γk−γ j)e−iθ j−16η [ j]R λ−2j (ρ2j +1)−3λk|q[ j]−q[k]|−2[q[ j]1 −q[k]1 − i(q[ j]2 −q[k]2)]e−i(γk−γ j)eiθ j}C−1j+O(T ε)[N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3]−Ξ1(x, t)U∗(4.13.68)for some scalar function Ξ1(x, t) whenΘ+2β −1< 0. (4.13.69)342Under the assumption (4.13.69), then∇x(|U∗|2)∇xU∗ = 2(U∗ ·∇xU∗)∇xU∗=N∑j=1∑m 6= j{16η [ j]R λ−2j ρ2j (ρ2j +1)−3λm|q[ j]−q[m]|−2[q[ j]1 −q[m]1 + i(q[ j]2 −q[m]2)]ei(γm−γ j)e−iθ j−16η [ j]R λ−2j (ρ2j +1)−3λm|q[ j]−q[m]|−2[q[ j]1 −q[m]1 − i(q[ j]2 −q[m]2)]e−i(γm−γ j)eiθ j}C−1j+O(T ε)[N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3].(4.13.70)Proof of (4.13.68).N∑j=1|∇xU [ j]|2U [ j] =N∑j=1|∇xU [ j]|2(U [ j]−U∗+U∗)=N∑j=1|∇xU [ j]|2(U [ j]−U∗)+N∑j=1|∇xU [ j]|2U∗,where ∣∣∣∣∣ N∑j=1 |∇xU [ j]|2 (U [ j]−U∗)∣∣∣∣∣≲ N∑j=1λ−2j 〈y[ j]〉−4∑k 6= j〈y[k]〉−1.Then (1−N∑j=1η [ j]R)∣∣∣∣∣ N∑j=1 |∇xU [ j]|2 (U [ j]−U∗)∣∣∣∣∣≲(1−N∑j=1η [ j]R)(N∑j=1λ−2j 〈y[ j]〉−4∑k 6= j〈y[k]〉−1)≲(1−N∑j=1η [ j]R)(N∑j=1∑k 6= jλ 3∗ |x−q[ j]|−4|x−q[k]|−1)≲(1−N∑j=1η [ j]R)(N∑j=1λ 3∗ (λ∗R)−2|x−q[ j]|−2)= λ∗R−2(1−N∑j=1η [ j]R)(N∑j=1|x−q[ j]|−2)≲ T ε(N∑j=1ρ [ j]2 +ρ3)where we have used|x−q[ j]|−1|x−q[k]|−1 ≲max{|x−q[ j]|−1, |x−q[k]|−1} for j 6= k. (4.13.71)For any fixed j = 1, . . . ,N,η [ j]R∣∣∣∣∣ N∑j=1 |∇xU [ j]|2 (U [ j]−U∗)∣∣∣∣∣≲ η [ j]R(λ−1∗ 〈y[ j]〉−4+ ∑m 6= jλ−2m 〈y[m]〉−4 ∑k 6=m〈y[k]〉−1)≲ η [ j]R(λ−1∗ 〈y[ j]〉−4+λ 2∗ (〈y[ j]〉−1+λ∗))≲ η [ j]R λ−1∗ 〈y[ j]〉−4.343This estimate is too rough that can not be controlled by the outer topology. More sophisticatedanalysis will be applied. Indeed,|∇xU [ j]|2(U [ j]−U∗)=−2λ−2j |∇y[ j]U [ j]|2∑k 6= j(|y[k]|2+1)−1Qγk[y[k]1 ,y[k]2 ,−1]tr= −16λ−2j (ρ2j +1)−2∑k 6= j(ρ2k +1)−1[eiγk(y[k]1 + iy[k]2),−1]tr.Thenη [ j]R ∑m 6= j|∇xU [m]|2(U [m]−U∗)= η [ j]R ∑m 6= jO(λ 2m)(〈ρ j〉−1+ ∑k 6=m, jλk)= η [ j]R O(λ2∗ )〈ρ j〉−1 ≲ T ερ [ j]1 .η [ j]R |∇xU [ j]|2(U [ j]−U∗)=−16η [ j]R λ−2j (ρ2j +1)−2∑k 6= j(ρ2k +1)−1[eiγk(y[k]1 + iy[k]2),−1]tr.and for k 6= j,ρ2k = λ−2k |x−ξ [k]|2 = λ−2k[|x−ξ [ j]|2+2(x−ξ [ j]) · (ξ [ j]−ξ [k])+ |ξ [ j]−ξ [k]|2]= λ−2k |ξ [ j]−ξ [k]|2{1+ |ξ [ j]−ξ [k]|−2 [|x−ξ [ j]|2+2(x−ξ [ j]) · (ξ [ j]−ξ [k])]} .Then(ρ2k +1)−1= λ 2k |ξ [ j]−ξ [k]|−2{1+ |ξ [ j]−ξ [k]|−2 [λ 2k + |x−ξ [ j]|2+2(x−ξ [ j]) · (ξ [ j]−ξ [k])]}−1 .Specially,η [ j]R(ρ2k +1)−1= η [ j]R λ2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR)), (4.13.72)which impliesη [ j]R |∇xU [ j]|2(U [ j]−U∗)= −16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλ 2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))[eiγk(y[k]1 + iy[k]2),−1]tr.(4.13.73)Noticey[k]1 + iy[k]2 =[λ jy[ j]1 +ξ[ j]1 −ξ [k]1 + i(λ jy[ j]2 +ξ[ j]2 −ξ [k]2)]λ−1k= λ jλ−1k ρ jeiθ j +λ−1k[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)].344Then by (4.2.19), we have(ΠU [ j]⊥[eiγk(y[k]1 + iy[k]2),−1]tr)C j=(1− 2ρ2j +1Re)[(y[k]1 + iy[k]2)ei(−θ j+γk−γ j)]+2ρ jρ2j +1= λ jλ−1k ρ j(1− 2ρ2j +1Re)[ei(γk−γ j)]+2ρ jρ2j +1+λ−1k(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)},(4.13.74)[eiγk(y[k]1 + iy[k]2),−1]tr·U [ j] = 2ρ jρ2j +1Re[(y[k]1 + iy[k]2)ei(−θ j+γk−γ j)]− ρ2j −1ρ2j +1= λ jλ−1k2ρ2jρ2j +1Re[ei(γk−γ j)]− ρ2j −1ρ2j +1+λ−1k2ρ jρ2j +1Re{[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)}.Thenη [ j]R{ΠU [ j]⊥[|∇xU [ j]|2 (U [ j]−U∗)]}C j= −16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλ 2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))×[λ jλ−1k ρ j(1− 2ρ2j +1Re)[ei(γk−γ j)]+2ρ jρ2j +1+λ−1k(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)}]= −16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλ 2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))×{λ jλ−1k ρ j(1− 2ρ2j +1Re)[ei(γk−γ j)]+2ρ jρ2j +1}−16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))×(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)},(4.13.75)where ∣∣∣16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλ 2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))×{λ jλ−1k ρ j(1− 2ρ2j +1Re)[ei(γk−γ j)]+2ρ jρ2j +1}∣∣∣≲ η [ j]R 〈ρ j〉−3 ≲ T ερ [ j]1 ,345∣∣∣16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|ξ [ j]−ξ [k]|−2(O(λ 2k +λ jR))×(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)}∣∣∣≲ T ερ [ j]1 ,providedΘ+2β −1< 0. (4.13.76)∣∣∣−16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|ξ [ j]−ξ [k]|−2(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)}+16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|q[ j]−q[k]|−2(1− 2ρ2j +1Re){[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(−θ j+γk−γ j)}∣∣∣≲ η [ j]R | lnT |−1 ln2(T − t)〈ρ j〉−4 ≲ T ερ [ j]1 .(4.13.77)−16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|q[ j]−q[k]|−2(1− 2ρ2j +1Re){[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(−θ j+γk−γ j)}= −16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλk|q[ j]−q[k]|−2[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(γk−γ j)e−iθ j+16η [ j]R λ−2j (ρ2j +1)−3∑k 6= jλk|q[ j]−q[k]|−2×{[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(γk−γ j)e−iθ j +[q[ j]1 −q[k]1 − i(q[ j]2 −q[k]2)]e−i(γk−γ j)eiθ j}= −16η [ j]R λ−2j ρ2j (ρ2j +1)−3∑k 6= jλk|q[ j]−q[k]|−2[q[ j]1 −q[k]1 + i(q[ j]2 −q[k]2)]ei(γk−γ j)e−iθ j+16η [ j]R λ−2j (ρ2j +1)−3∑k 6= jλk|q[ j]−q[k]|−2[q[ j]1 −q[k]1 − i(q[ j]2 −q[k]2)]e−i(γk−γ j)eiθ j(4.13.78)which will be put into mode 1 and mode −1, respectively.For the projection inU [ j], we calculateη [ j]R |∇xU [ j]|2(U [ j]−U∗) ·U [ j]= −16η [ j]R λ−2j (ρ2j +1)−2∑k 6= jλ 2k |ξ [ j]−ξ [k]|−2(1+O(λ 2k +λ jR))×[λ jλ−1k2ρ2jρ2j +1Re[ei(γk−γ j)]− ρ2j −1ρ2j +1+λ−1k2ρ jρ2j +1Re{[ξ [ j]1 −ξ [k]1 + i(ξ [ j]2 −ξ [k]2)]ei(−θ j+γk−γ j)}].(4.13.79)346Thus byU∗-operation,∣∣η [ j]R |∇xU [ j]|2 [(U [ j]−U∗) ·U [ j]](U [ j]−U∗)∣∣≲ η [ j]R 〈ρ j〉−4 ≲ T ερ [ j]1 .In sum, we conclude the validity of (4.13.68).Proof of (4.13.70).2(U∗ ·∇xU∗)∇xU∗ = 2(N∑k=1U∗ ·∇xU [k])N∑m=1∇xU [m]= 2[N∑k=1∑n6=k(U [n]−U∞) ·∇xU [k]] N∑m=1∇xU [m].By (4.13.71), we have(1−N∑j=1η [ j]R)∣∣∣∣∣[N∑k=1∑n6=k(U [n]−U∞) ·∇xU [k]] N∑m=1∇xU [m]∣∣∣∣∣≲(1−N∑j=1η [ j]R)N∑k=1∑n6=kN∑m=1λn|x−q[n]|λk|x−q[k]|2λm|x−q[m]|2≲(1−N∑j=1η [ j]R)N∑k=1∑n6=kN∑m=1λnmax{|x−q[k]|−1, |x−q[n]|−1} λk|x−q[k]| λm|x−q[m]|2≲(1−N∑j=1η [ j]R)N∑m=1R−2λm|x−q[m]|2 ≲ Tε(N∑j=1ρ [ j]2 +ρ3).For any fixed j = 1,2, . . . ,N,∣∣∣∣∣η [ j]R[N∑k=1∑n6=k(U [n]−U∞) ·∇xU [k]] ∑m6= j∇xU [m]∣∣∣∣∣≲ η [ j]R ≲ T ερ [ j]1 ,∣∣∣∣∣η [ j]R[∑k 6= j∑n6=k(U [n]−U∞) ·∇xU [k]]∇xU [ j]∣∣∣∣∣≲ η [ j]R ≲ T ερ [ j]1 .Thus we only need to focus on2η [ j]R[∑m 6= j(U [m]−U∞) ·∇xU [ j]]∇xU [ j].347For m 6= j,[(U [m]−U∞) ·∇xU [ j]]∇xU [ j] =2∑k=1[(U [m]−U∞) ·∂xkU [ j]]∂xk(U [ j])12∑k=1[(U [m]−U∞) ·∂xkU [ j]]∂xk(U [ j])22∑k=1[(U [m]−U∞) ·∂xkU [ j]]∂xk(U [ j])3= λ−2j2∑k=1[(U [m]−U∞) ·∂y[ j]kU [ j]]∂y[ j]kU [ j].Notice∂y[ j]1=∂ρ j∂y[ j]1∂ρ j +∂θ j∂y[ j]1∂θ j = cosθ j∂ρ j −sinθ jρ j∂θ j ,∂y[ j]2=∂ρ j∂y[ j]2∂ρ j +∂θ j∂y[ j]2∂θ j = sinθ j∂ρ j +cosθ jρ j∂θ j .(4.13.80)Then∂y[ j]1U [ j] = cosθ j∂ρ jU[ j]− sinθ jρ j∂θ jU[ j] = cosθ jwρ jQγ jE[ j]1 −sinθ jρ jsinw(ρ j)Qγ jE[ j]2= −2(ρ2j +1)−1(cosθ jQγ jE[ j]1 + sinθ jQγ jE[ j]2),∂y[ j]2U [ j] = sinθ j∂ρ jU[ j]+cosθ jρ j∂θ jU[ j] = sinθ jwρ jQγ jE[ j]1 +cosθ jρ jsinw(ρ j)Qγ jE[ j]2= −2(ρ2j +1)−1(sinθ jQγ jE[ j]1 − cosθ jQγ jE [ j]2).(4.13.81)Thusλ−2j2∑k=1[(U [m]−U∞) ·∂y[ j]kU [ j]]∂y[ j]kU [ j]= 4λ−2j (ρ2j +1)−2{[(U [m]−U∞) · (cosθ jQγ jE [ j]1 + sinθ jQγ jE [ j]2 )](cosθ jQγ jE [ j]1 + sinθ jQγ jE [ j]2 )+[(U [m]−U∞) · (sinθ jQγ jE [ j]1 − cosθ jQγ jE [ j]2 )](sinθ jQγ jE [ j]1 − cosθ jQγ jE [ j]2 )}= 4λ−2j (ρ2j +1)−2{[(U [m]−U∞) ·Qγ jE [ j]1 ]Qγ jE [ j]1 + [(U [m]−U∞) ·Qγ jE [ j]2 ]Qγ jE [ j]2 } .NoticeU [m]−U∞ = 2(|y[m]|2+1)−1Qγm[y[m]1 ,y[m]2 ,−1]tr= 2(ρ2m+1)−1[eiγm(y[m]1 + iy[m]2),−1]tr.348Then it follows that(λ−2j2∑k=1[(U [m]−U∞) ·∂y[ j]kU [ j]]∂y[ j]kU [ j])C j= 4λ−2j (ρ2j +1)−2 [(U [m]−U∞) ·Qγ jE [ j]1 + i(U [m]−U∞) ·Qγ jE [ j]2 ]= 8λ−2j (ρ2j +1)−2(ρ2m+1)−1(ΠU [ j]⊥[eiγm(y[m]1 + iy[m]2),−1]T)C j= 8λ−2j (ρ2j +1)−2(ρ2m+1)−1[λ jλ−1m ρ j(1− 2ρ2j +1Re)[ei(γm−γ j)]+2ρ jρ2j +1+λ−1m(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [m]1 + i(ξ [ j]2 −ξ [m]2)]ei(−θ j+γm−γ j)}]where we used (4.13.74) in the last equality.We estimate∣∣∣∣∣2η [ j]R 8λ−2j (ρ2j +1)−2(ρ2m+1)−1{λ jλ−1m ρ j(1− 2ρ2j +1Re)[ei(γm−γ j)]+2ρ jρ2j +1}∣∣∣∣∣≲η [ j]R 〈ρ j〉−3≲T ερ [ j]1 .By (4.13.72), one has2η [ j]R 8λ−2j (ρ2j +1)−2(ρ2m+1)−1λ−1m(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [m]1 + i(ξ [ j]2 −ξ [m]2)]ei(−θ j+γm−γ j)}= 16η [ j]R λ−2j λm(ρ2j +1)−2|ξ [ j]−ξ [m]|−2 (1+O(λ 2m+λ jR))×(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [m]1 + i(ξ [ j]2 −ξ [m]2)]ei(−θ j+γm−γ j)}.Notice that∣∣∣16η [ j]R λ−2j λm(ρ2j +1)−2|ξ [ j]−ξ [m]|−2O(λ 2m+λ jR)×(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [m]1 + i(ξ [ j]2 −ξ [m]2)]ei(−θ j+γm−γ j)}∣∣∣≲ η [ j]R R〈ρ j〉−4 ≲ T ερ [ j]1providedΘ+2β −1< 0. (4.13.82)16η [ j]R λ−2j λm(ρ2j +1)−2|ξ [ j]−ξ [m]|−2(1− 2ρ2j +1Re){[ξ [ j]1 −ξ [m]1 + i(ξ [ j]2 −ξ [m]2)]ei(−θ j+γm−γ j)}= 16η [ j]R λ−2j ρ2j (ρ2j +1)−3λm|q[ j]−q[m]|−2[q[ j]1 −q[m]1 + i(q[ j]2 −q[m]2)]ei(γm−γ j)e−iθ j−16η [ j]R λ−2j (ρ2j +1)−3λm|q[ j]−q[m]|−2[q[ j]1 −q[m]1 − i(q[ j]2 −q[m]2)]e−i(γm−γ j)eiθ j +O(T ε)ρ [ j]1 .349by the same estimate as in (4.13.77) and (4.13.78), which will be assigned into mode 1 and mode−1.Under the parameter assumption (4.13.69), combining (4.13.68) and (4.13.70), we have∆xU∗−2(U∗ ·∇xU∗) ·∇xU∗ = O(T ε)[N∑j=1(ρ [ j]1 +ρ[ j]2)+ρ3]−Ξ1(x, t)U∗. (4.13.83)Combining (4.13.25), (4.13.42), (4.13.43), (4.13.45), (4.13.46), (4.13.47), (4.13.49), (4.13.50),(4.13.51), (4.13.52), (4.13.53), (4.13.55), (4.13.56), (4.13.59), (4.13.61), (4.13.63), (4.13.64), (4.13.65),(4.13.66), (4.13.67) and (4.13.69), we get the parameter requirement (4.13.39).4.13.4 Pointwise and Hölder estimate forH [ j]Claim: suppose2δ0 < ν < 1, 0< l < 1, ν+β l−1< 0, (4.13.84)then forH [ j] defined in (4.4.22), there exists a small constant ε > 0 such that∣∣H [ j]∣∣≲ λ ν+ε∗ 〈ρ j〉−2−l in D2R. (4.13.85)Proof of (4.13.85). By (4.4.41),|H [ j]in |≲ ‖Φ[ j]in‖2in,ν−δ0,lλ 2ν−2δ0∗ 〈ρ j〉−2l−3 ≲ λ ν+ε∗ 〈ρ j〉−2−l (4.13.86)provided2δ0 < ν . (4.13.87)By (4.3.54), (4.2.24) and (4.4.52), we get|(H [ j]1 )C j,0|≲ λ 2∗{λ−1∗ 〈ρ j〉−3+λ−1j 〈ρ j〉−3∣∣∇xΦout(q[ j],T )∣∣+λ−1j 〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ +(T − t)α2 ‖Z∗‖C3(R2)]}≲ λ∗〈ρ j〉−3+λ∗〈ρ j〉−3‖Φout‖♯,Θ,α(λΘ∗ (0)+‖Z∗‖C3(R2))+λ∗〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ +(T − t)α2 ‖Z∗‖C3(R2)]≲ λ ν+ε∗ 〈ρ j〉−2−l(4.13.88)350for |ρ j| ≤ 3R and ε > 0 sufficiently small provided0< ν < 1, 0< l < 1, ν−Θ+β l−1< 0, ν− α2+β l−1< 0. (4.13.89)By (4.3.54), (4.2.24) and (4.4.52),|(H [ j]1 )C j,1|≲ λ 2∗{|ξ˙ [ j]|λ−1∗ 〈ρ j〉−2+λ−1j 〈ρ j〉−2∣∣∇xΦout(q[ j],T )∣∣+λ−1j 〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2)]}≲ |ξ˙ [ j]|λ∗〈ρ j〉−2+λ∗〈ρ j〉−2‖Φout‖♯,Θ,α(λΘ∗ (0)+‖Z∗‖C3(R2))+λ∗〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ +(T − t)α2 ‖Z∗‖C3(R2)]≲ λ ν+ε∗ 〈ρ j〉−2−l(4.13.90)providedν+β l−1< 0. (4.13.91)By (4.4.52),|(H [ j]1 )C j,−1|≲ λ∗〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ +(T − t)α2 ‖Z∗‖C3(R2)](4.13.92)By (4.4.37) and Proposition 4.8.1, Tin[(H [ j])C j,−1] ∈ Bin providedν+β l−δ0−1< 0. (4.13.93)By (4.2.24) and (4.4.52), we have∣∣∣(H [ j]1 )C j − ∑|k|≤1eikθ j(H [ j]1 )C j,k∣∣∣≲ λ 2∗{λ−1j 〈ρ j〉−3 ∣∣∇xΦout(q[ j],T )∣∣+λ−1j 〈ρ j〉−2‖Φout‖♯,Θ,α[λΘ∗ (t)+(T − t)α2 ‖Z∗‖C3(R2)]}≲ λ ν+ε∗ 〈ρ j〉−2−l(4.13.94)provided the same parameter assumption (4.13.89) holds.Hölder estimate forH [ j]. When we have solved the outer problem and the orthogonal equations, inorder to get the spatial decay of the second derivative of the inner solutions, we need the spatial-timedecay of Hölder seminorm in (y[ j],τ j) ofH [ j].Claim: Suppose2δ0 < ν , Θ+1−β −ν > 0, ν+β l−1< 0, ν−2δ0−β (ςin+ l)> 0, ςH ≤min{ςin,α} ,(4.13.95)351then there exists a small constant ε > 0 such thatsup(y,τ j)∈D2R[λ−ν∗ (t(τ j))〈y〉2+l+ςH[H [ j]]CςH ,ςH2 (Q−((y,τ j), |y|2 ))]≲ T ε (4.13.96)Proof of (4.13.96): For brevity, we denote y = y[ j], τ = τ j, Iτ := [τ − |y|24 ,τ], By := B(y,|y|2 ), Qy :=By× Iτ and λ∗ = λ∗(t(τ)). Notice that[ f ]CςH ,ςH2 (Qy)≤ sups∈Iτ[ f (·,s)]CςHy (By)+ supz∈By[ f (z, ·)]CςH2τ (Iτ )≤ 2[ f ]CςH ,ςH2 (Qy),[g]CςHy (By) := supz1,z2∈By|g(z1)−g(z2)||z1− z2|ςH , [g]CςH2τ (Iτ ):= sups1,s2∈Iτ|g(s1)−g(s2)||s1− s2|ςH2,[ f g]CςH ,ςH2 (Qy)≤ [ f ]CςH ,ςH2 (Qy)‖g‖L∞(Qy)+‖ f‖L∞(Qy)[g]CςH , ςH2 (Qy)(4.13.97)where the last inequality holds when CςH ,ςH2 (Qy), L∞(Qy) are replaced by CςHy (By) (CςH2τ (Iτ)) andL∞(By) (L∞(Iτ)) respectively. It is easy to have that in D2R,〈y〉ςH [a−bW [ j]∧]CςH ,ςH2 (Qy)+ |a−bW [ j]∧|≲ 1, 〈y〉ςH [∇y[ j]W [ j]]CςH , ςH2 (Qy)+ |∇y[ j]W[ j]|≲ 〈y〉−2,〈y〉1+ςH [∇y[ j]η [ j]R ]CςH , ςH2 (Qy)+ 〈y〉|∇y[ j]η[ j]R |+ 〈y〉ςH [η [ j]R ]CςH , ςH2 (Qy)+ |η[ j]R |≲ 1,〈y〉1+ςH [∇y[ j]Φ[ j]in]CςH , ςH2 (Qy)+ 〈y〉|∇y[ j]Φ[ j]in|+ 〈y〉ςH [Φ[ j]in]CςH , ςH2 (Qy)+ |Φ[ j]in|≲ λ ν−δ0∗ 〈y〉−l‖Φ[ j]in‖in,ν−δ0,l,ςinwhere in the last inequality, by (4.4.41), we requireςH ≤ ςin. (4.13.98)Thus by (4.13.97),[H [ j]in ]CςH ,ςH2 (Qy)≲ λ 2ν−2δ0∗ 〈y〉−3−2l−ςH‖Φ[ j]in‖2in,ν−δ0,l,ςin ≲ λ ν+ε∗ 〈y〉−2−l−ςH‖Φ[ j]in‖2in,ν−δ0,l,ςinprovided2δ0 < ν . (4.13.99)• Hölder continuity of the coupling terms from the outer problem inH [ j]1(a−bW [ j]∧) L˜W [ j] [Q−γ jΦout]=(a−bW [ j]∧)(|∇y[ j]W [ j]|2ΠW [ j]⊥(Q−γ jΦout)−2∇y[ j]((Q−γ jΦout) ·W [ j]) ·∇y[ j]W [ j]) . (4.13.100)352By (4.13.7), (4.4.38),|Φout|≲[lnτλΘ+1−β∗ +(| lnT |−2τ−1(lnτ)4+λ jρ j)(λΘ∗ (0)+‖Z∗‖C3(R2))]‖Φout‖♯,Θ,α .(4.13.101)For p j = p j(t(τ)) = λ jeiγ j , λ j = |p j|, γ j = arctan( ImpjRepj ) given in Proposition 4.5.1.|∂t(pc3j |p j|c4)|≲ λ∗(t)c3+c4(T − t)−1 for c3,c4 ∈ R,|∂tγ j|= |p j|−2∣∣Im(∂tpj)Repj− ImpjRe(∂tpj)∣∣≲ (T − t)−1. (4.13.102)By (4.4.37), (4.4.38), (4.5.21), for any max{0,c2} ≤ c1,sups1,s2∈Iτ|t(s1)− t(s2)|c1|s1− s2|c2 ≲ λ2c1∗ |y|2c1−2c2 ,sups1,s2∈Iτ|(λ j(t(s1))−λ j(t(s2))) |c1|s1− s2|c2 ≲(| lnT |(lnτ)−2λ 2∗ )c1 |y|2c1−2c2 ∼ (| lnT |−1τ−2(lnτ)2)c1 |y|2c1−2c2 ,sups1,s2∈Iτ|ξ [ j](t(s1))−ξ [ j](t(s2))|c1|s1− s2|c2 ≲ λc1(2+εξ )∗ |y|2c1−2c2 .(4.13.103)[λ j]CςH2τ (Iτ )+[p j]CςH2τ (Iτ )≲ | lnT |(lnτ)−2λ 2∗ |y|2−ςH ,[eicγ j(t(·))]CςH2τ (Iτ )≲ | lnT |(lnτ)−2λ∗|y|2−ςH for c ∈ R.(4.13.104)By same methodology for deriving (4.4.47), thensup[| ln(T − t)|λΘ∗ (t)(λ∗(t)R(t))1−α +(T − t)1−α2 ‖Z∗‖C3(R2)]−1 |Φout(x, t)−Φout(x, t∗)||t− t∗| α2≲ ‖Φout‖♯,Θ,α(4.13.105)where the last supremum is taken in the regionx ∈ R2, t, t∗ ∈ (0,T ), |t− t∗|< 14(T − t).Then suppose ςH ≤ α ,[Φout(zλ j(t(·))+ξ [ j](t(·)), t(·))]CςH2τ (Iτ )≤ sups∈Iτ[Φout(zλ j(t(·))+ξ [ j](t(·)), t(s))]CςH2τ (Iτ )+ sups∈Iτ[Φout(zλ j(t(s))+ξ [ j](t(s)), t(·))]CςH2τ (Iτ )≲[zλ j(t(·))+ξ [ j](t(·))]CςH2τ (Iτ )(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+[lnτλΘ∗ (λ∗R)1−α +(T − t(τ))1−α2 ‖Z∗‖C3(R2)]sups1,s2∈Iτ|t(s1)− t(s2)| α2|s1− s2|ςH2‖Φout‖♯,Θ,α≲ 〈z〉| lnT |−1τ−2(lnτ)2|y|2−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α353+[lnτλΘ+(1−β )(1−α)∗ +(| lnT |−2τ−1(lnτ)4)1− α2 ‖Z∗‖C3(R2)]λα∗ |y|α−ςH‖Φout‖♯,Θ,α .Then since β < 12 ,[Φout]CςH ,ςH2 (Qy)≤ sups∈Iτ[Φout(·λ j(t(s))+ξ [ j](t(s)), t(s))]CςHy (By)+ supz∈By[Φout(zλ j(t(·))+ξ [ j](t(·)), t(·))]CςH2τ (Iτ )≲ λ∗|y|1−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+ | lnT |−1τ−2(lnτ)2〈y〉|y|2−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+[lnτλΘ+(1−β )(1−α)∗ +(| lnT |−2τ−1(lnτ)4)1− α2 ‖Z∗‖C3(R2)]λα∗ |y|α−ςH‖Φout‖♯,Θ,α∼ λ∗|y|1−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+λα∗[lnτλΘ+(1−β )(1−α)∗ +(| lnT |−2τ−1(lnτ)4)1− α2 ‖Z∗‖C3(R2)]|y|α−ςH‖Φout‖♯,Θ,α .(4.13.106)Suppose ςH ≤ α , by (4.13.103), we have[(∇xΦout)(zλ j(t(·))+ξ [ j](t(·)), t(·))]CςH2τ (Iτ )≤ sups∈Iτ[(∇xΦout)(zλ j(t(·))+ξ [ j](t(·)), t(s))]CςH2τ (Iτ )+ sups∈Iτ[(∇xΦout)(zλ j(t(s))+ξ [ j](t(s)), t(·))]CςH2τ (Iτ )≲[λΘ∗ (λ∗R)−α +‖Z∗‖C3(R2)](sups1,s2∈Iτ|z(λ j(t(s1))−λ j(t(s2)))+ξ [ j](t(s1))−ξ [ j](t(s2))|α|s1− s2|ςH2+ sups1,s2∈Iτ|t(s1)− t(s2)| α2|s1− s2|ςH2)‖Φout‖♯,Θ,α≲(λΘ−α+αβ∗ +‖Z∗‖C3(R2))[〈z〉α (| lnT |−1τ−2(lnτ)2)α |y|2α−ςH +λα∗ |y|α−ςH]‖Φout‖♯,Θ,αsince by (4.5.13), for s1,s2 ∈ Iτ , |s1− s2|< c|T − s1| for a constant c< 1. Then since β < 12 ,[∇y[ j]Φout]CςH ,ςH2 (Qy)≤ sups∈Iτ[λ j(t(s))(∇xΦout)(·λ j(t(s))+ξ [ j](t(s)), t(s))]CςHy (By)+ supz∈By[λ j(t(·))(∇xΦout)(zλ j(t(·))+ξ [ j](t(·)), t(·))]CςH2τ (Iτ )≲(λ 1+Θ+αβ∗ +λ 1+α∗ ‖Z∗‖C3(R2))|y|α−ςH‖Φout‖♯,Θ,α+ | lnT |−1τ−2(lnτ)2|y|2−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+λ∗(λΘ−α+αβ∗ +‖Z∗‖C3(R2))[(| lnT |−1τ−2(lnτ)2)α 〈y〉α |y|2α−ςH +λα∗ |y|α−ςH]‖Φout‖♯,Θ,α∼| lnT |−1τ−2(lnτ)2|y|2−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+λ 1+α∗(λΘ−α+αβ∗ +‖Z∗‖C3(R2))|y|α−ςH‖Φout‖♯,Θ,α . (4.13.107)354We will give the Hölder estimate of two typical terms in (4.13.100) and the same other two termscan be handled similarly. By (4.13.101), (4.13.104), (4.13.106),[(a−bW [ j]∧) |∇y[ j]W [ j]|2 (W [ j] · (Q−γ jΦout))W [ j]]CςH , ςH2 (Qy)≲ 〈y〉−4[Q−γ jΦout]CςH , ςH2 (Qy)+ 〈y〉−4−ςH‖Φout‖L∞(Qy)≲ 〈y〉−4([Φout]CςH ,ςH2 (Qy)+[Q−γ j ]CςH ,ςH2 (Qy)‖Φout‖L∞(Qy))+ 〈y〉−4−ςH‖Φout‖L∞(Qy)≲ 〈y〉−4{λ∗|y|1−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+λα∗[lnτλΘ+(1−β )(1−α)∗ +(| lnT |−2τ−1(lnτ)4)1− α2 ‖Z∗‖C3(R2)]|y|α−ςH‖Φout‖♯,Θ,α+ | lnT |(lnτ)−2λ∗|y|2−ςH×[lnτλΘ+1−β∗ +(| lnT |−2τ−1(lnτ)4+λ j|y|)(λΘ∗ (0)+‖Z∗‖C3(R2))]‖Φout‖♯,Θ,α}+ 〈y〉−4−ςH[lnτλΘ+1−β∗ +(| lnT |−2τ−1(lnτ)4+λ j|y|)(λΘ∗ (0)+‖Z∗‖C3(R2))]‖Φout‖♯,Θ,α≲ λ ν+ε∗ 〈y〉−2−l−ςH‖Φout‖♯,Θ,αprovidedΘ+1−β > ν . (4.13.108)By (4.4.47), (4.13.104), (4.13.107),[(a−bW [ j]∧)[(W [ j] · (Q−γ j∇y[ j]Φout)) ·∇y[ j]W [ j]]]CςH , ςH2 (Qy)≲ 〈y〉−2−ςH‖∇y[ j]Φout‖L∞(Qy)+ 〈y〉−2([∇y[ j]Φout]CςH ,ςH2 (Qy)+[Q−γ j ]CςH ,ςH2 (Qy)‖∇y[ j]Φout‖L∞(Qy))≲ 〈y〉−2−ςHλ∗(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+ 〈y〉−2[| lnT |−1τ−2(lnτ)2|y|2−ςH(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+λ 1+α∗(λΘ−α+αβ∗ +‖Z∗‖C3(R2))|y|α−ςH‖Φout‖♯,Θ,α+ | lnT |(lnτ)−2λ∗|y|2−ςHλ∗(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α]≲ λ ν+ε∗ 〈y〉−2−l−ςH‖Φout‖♯,Θ,αprovided1−β l > ν . (4.13.109)• Hölder continuity aboutλ 2j Q−γ j[(M[ j]0 + eiθ jM[ j]1)C−1j],whereM[ j]0 ,M[ j]1 are defined in (4.3.52), (4.3.49) respectively. Back to the vector form, for k= 0,1, we355have (eikθ jM[ j]k)C−1j= Re[eikθ j(ReM[ j]k + iImM[ j]k )]Qγ jE[ j]1 + Im[eikθ j(ReM[ j]k + iImM[ j]k )]Qγ jE[ j]2=[cos(kθ j)ReM[ j]k − sin(kθ j)ImM[ j]k )]Qγ jE[ j]1 +[cos(kθ j)ImM[ j]k + sin(kθ j)ReM[ j]k )]Qγ jE[ j]2= ReM[ j]k Qγ j[cos(kθ j)E [ j]1 + sin(kθ j)E[ j]2]+ ImM[ j]k Qγ j[cos(kθ j)E [ j]2 − sin(kθ j)E [ j]1].Here, by (4.2.3), we computecos(kθ j)E [ j]1 + sin(kθ j)E[ j]2=cos(kθ j)cosθ j cosw(ρ j)− sin(kθ j)sinθ jcos(kθ j)sinθ j cosw(ρ j)+ sin(kθ j)cosθ j−cos(kθ j)sinw(ρ j)=−cos((k−1)θ j)+ 2ρ2jρ2j+1cos(kθ j)cosθ jsin((k−1)θ j)+ 2ρ2jρ2j+1cos(kθ j)sinθ j−cos(kθ j) 2ρ jρ2j+1 ,(4.13.110)cos(kθ j)E [ j]2 − sin(kθ j)E [ j]1=−cos(kθ j)sinθ j− sin(kθ j)cosθ j cosw(ρ j)cos(kθ j)cosθ j− sin(kθ j)sinθ j cosw(ρ j)sin(kθ j)sinw(ρ j)=sin((k−1)θ j)− 2ρ2jρ2j+1sin(kθ j)cosθ jcos((k−1)θ j)− 2ρ2jρ2j+1sin(kθ j)sinθ jsin(kθ j)2ρ jρ2j+1 .(4.13.111)Thus,λ 2j Q−γ j[(eiθ jM[ j]1)C−1j]=2λ jρ2j +1(− ξ˙ [ j]1−1+ 2ρ2jρ2j+1cos2 θ j2ρ2jρ2j+1cosθ j sinθ j−cosθ j 2ρ jρ2j+1+ ξ˙ [ j]2− 2ρ2jρ2j+1sinθ j cosθ j1− 2ρ2jρ2j+1sin2 θ jsinθ j2ρ jρ2j+1).We will give the Hölder estimate of ξ˙ [ j] given by (4.5.15). Notice for k = 0,±1, |t− s| ≤ T−t2 , by(4.4.47), (4.13.102),∣∣eikγ j(t)(∇xΦout)(λ j(t)y[ j]+ξ [ j](t), t)− eikγ j(s)(∇xΦout)(λ j(s)y[ j]+ξ [ j](s),s)∣∣≲ |eikγ j(t)− eikγ j(s)|(λΘ∗ (0)+‖Z∗‖C3(R2))‖Φout‖♯,Θ,α+[λΘ∗ (t)(λ∗(t)R(t))−α +‖Z∗‖C3(R2)]∣∣(λ j(t)−λ j(s))y[ j]+ξ [ j](t)−ξ [ j](s)∣∣α ‖Φout‖♯,Θ,α≲ (T − t)−1(t− s)+(λ∗(t))Θ−α(1−β )(|λ˙∗(t)|〈y[ j]〉+1)|t− s|α .356By (4.4.41), (4.13.103), for s1,s2 ∈ [τ− |y|24 ,τ],∣∣∣(η [ j]R Φ[ j]in∇y[ j]Φ[ j]in)(y[ j],s1)−(η [ j]R Φ[ j]in∇y[ j]Φ[ j]in)(y[ j],s2)∣∣∣≲(|t(s1)− t(s2)|(T − t(τ))−1λ 2(ν−δ0)∗ 〈y[ j]〉−2l−1+λ 2(ν−δ0)∗ 〈y[ j]〉−2l−1−ςin |s1− s2|ςin2)‖Φ[ j]in‖2in,ν−δ0,l,ςin∼ λ 2(ν−δ0)∗ 〈y[ j]〉−2l−1−ςin |s1− s2|ςin2 ‖Φ[ j]in‖2in,ν−δ0,l,ςin .Similarly,∣∣∣(∇y[ j](η [ j]R )(Φ[ j]in)2)(y[ j],s1)−(∇y[ j](η [ j]R )(Φ[ j]in)2)(y[ j],s2)∣∣∣ has the same estimate above.By (4.5.15), (4.2.24), (4.4.24), (4.13.103), for s1,s2 ∈ [τ− |y|24 ,τ],∣∣∣ξ˙ [ j](s1)− ξ˙ [ j](s2)∣∣∣≲ | lnT |2τ(lnτ)−4λ 2∗ |s1−s2|+λΘ+α(1+β )∗ |s1−s2|α+λ 2(ν−δ0)−1∗ |s1−s2| ςin2 ‖Φ[ j]in‖2in,ν−δ0,l,ςin ,which implies[ξ˙ [ j]]CςH2 (Iτ )≲ | lnT |2τ(lnτ)−4λ 2∗ |y|2−ςH +λΘ+α(1+β )∗ |y|2α−ςH +λ 2(ν−δ0)−1∗ |y|ςin−ςH‖Φ[ j]in‖2in,ν−δ0,l,ςinprovidedςH ≤min{ςin,2α} . (4.13.112)By (4.13.103) [λ 2j Q−γ j[(eiθ jM[ j]1 )C−1j]]CςH ,ςH2 (Qy)≲(λ∗[ξ˙ [ j]]CςH2 (Iτ )+λ εξ∗ | lnT |−1τ−2(lnτ)2|y|2−ςH)〈y〉−2+λ 1+εξ∗ 〈y〉−2−ςH≲ λ ν+ε∗ 〈y〉−2−l−ςH (‖Φ[ j]in‖2in,ν−δ0,l,ςin +1)providedν+β l−1< 0, ν−2δ0−β (ςin+ l)> 0. (4.13.113)For mode 0,λ 2j Q−γ j(M[ j]0 )C−1j= λ 2j(ReM[ j]0−cosθ j+ 2ρ2jρ2j+1cosθ j−sinθ j+ 2ρ2jρ2j+1sinθ j− 2ρ jρ2j+1+ ImM[ j]0−sinθ jcosθ j0).One recalling (4.3.52), the ρ j vanishing condition as ρ j → 0 make the above error smooth at the357origin. Thus[λ 2j Q−γ j(M[ j]0 )C−1j]CςH ,ςH2 (Qy)≲ 〈y〉−3−ςHλ∗m[ j]01(y,τ)+ 〈y〉−3([p j]CςH2τ (Iτ )m[ j]01(y,τ)+λ∗m[ j]02(y,τ))+ 〈y〉−3−ςHλ∗|λ˙∗|+ 〈y〉−3[([p j]CςH2τ (Iτ )+[λ j]CςH2τ (Iτ ))|λ˙∗|+λ∗([p˙ j]CςH2τ (Iτ )+[λ˙ j]CςH2τ (Iτ ))](4.13.114)where we used p j = λ jeiγ j and m[ j]01(y,τ), m[ j]02(y,τ) are defined asm[ j]01(y,τ) := ‖ˆ t(τ)0p˙ j(s)K0(ζ j)t(τ)− s ds‖L∞(Qy)+‖ˆ t(τ)0p˙ j(s)ζ jK0ζ j(ζ j)t(τ)− s ds‖L∞(Qy)+‖ˆ t(τ)0p˙ j(s)ζ 2j K0ζ jζ j(ζ j)t(τ)− s ds‖L∞(Qy),m[ j]02(y,τ) :=[ˆ t(τ)0p˙ j(s)K0(ζ j)t(τ)− s ds]CςH ,ςH2 (Qy)+[ˆ t(τ)0p˙ j(s)ζ jK0ζ j(ζ j)t(τ)− s ds]CςH ,ςH2 (Qy)+[ˆ t(τ)0p˙ j(s)ζ 2j K0ζ jζ j(ζ j)t(τ)− s ds]CςH ,ςH2 (Qy).Claim: ∣∣∣[λ 2j Q−γ j(M[ j]0 )C−1j ]CςH , ςH2 (Qy)∣∣∣≲ λ ν+ε∗ 〈y〉−2−l−ςH . (4.13.115)The rest part is devoted to the proof of Claim (4.13.115). By (4.3.19), (4.3.22),m[ j]01(y,τ)≲ ‖ˆ t(τ)0|p˙ j(s)|t(τ)− s(1{ζ j≤1}+O(ζ−1j)1{ζ j>1})ds‖L∞(Qy) ≲ 1.Then combining (4.13.104), we have〈y〉−3−ςHλ∗m[ j]01(y,τ)+ 〈y〉−3 [p j]CςH2τ (Iτ )m[ j]01(y,τ)≲ 〈y〉−3−ςHλ∗+ 〈y〉−3| lnT |(lnτ)−2λ 2∗ |y|2−ςH ≲ λ ν+ε∗ 〈y〉−2−l−ςH .〈y〉−3−ςHλ∗|λ˙∗|≲ λ ν+ε∗ 〈y〉−2−l−ςH .By (4.13.103), (4.13.104),〈y〉−3([p j]CςH2τ (Iτ )+[λ j]CςH2τ (Iτ ))|λ˙∗|≲ 〈y〉−3| lnT |(lnτ)−2λ 2∗ |y|2−ςH | lnT |(lnτ)−2≲ λ ν+ε∗ 〈y〉−2−l−ςH .358By (4.5.13), (4.13.103), (4.4.38), provided ςH ≤ α ,[p˙ j]CςH2τ (Iτ )≲ | lnT || ln(T − t(τ))|3(T − t(τ)) sups1,s2∈Iτ|t(s1)− t(s2)||s1− s2|ςH2+(T − t(τ))Θ−α(1−β )| ln(T − t(τ))|−ϖ sups1,s2∈Iτ|t(s1)− t(s2)| α2|s1− s2|ςH2≲ | lnT |3τ(lnτ)−7λ 2∗ |y|2−ςH +(| lnT |−2τ−1(lnτ)4)Θ−α(1−β )(lnτ)−ϖλα∗ |y|α−ςH .Notice λ˙ j = 2−1|p j|−1 (p j ¯˙p j+ p˙ j p¯ j),[λ˙ j]CςH2τ (Iτ )≲ [p˙ j]CςH2τ (Iτ )+λ−1∗ |λ˙∗|[p j]CςH2τ (Iτ )+λ∗|λ˙∗|[|p j|−1]CςH2τ (Iτ )≲ | lnT |3τ(lnτ)−7λ 2∗ |y|2−ςH +(| lnT |−2τ−1(lnτ)4)Θ−α(1−β )(lnτ)−ϖλα∗ |y|α−ςH+λ−1∗ |λ˙∗|| lnT |(lnτ)−2λ 2∗ |y|2−ςH +λ∗|λ˙∗|| lnT |(lnτ)−2|y|2−ςH∼ | lnT |3τ(lnτ)−6λ 2∗ |y|2−ςH +(| lnT |−2τ−1(lnτ)4)Θ−α(1−β )(lnτ)−ϖλα∗ |y|α−ςH .Thus〈y〉−3λ∗([p˙ j]CςH2τ (Iτ )+[λ˙ j]CςH2τ (Iτ ))≲ λ ν+ε∗ 〈y〉−2−l−ςH .Let us estimate 〈y〉−3λ∗m[ j]02(y,τ). In order to estimate m[ j]02(y,τ) uniformly, we consider a moregeneral form g0(y[ j], t), which covers all cases in m[ j]02(y,τ). Setg0(y[ j], t) :=ˆ t0p˙ j(s)K (ζ j)t− s ds, K (ζ j) := c0ζ−1j(1− e−d0ζ j)+n∑k=1ciζ k−1j e−dkζ j ,ζ j = ζ j(y[ j], t,s) =λ 2j (t)(1+ |y[ j]|2)t− swhere ck,dk are complex constants and Re(dk)> 0 for k = 0,1, . . . ,n. It is easy to see that∣∣K (ζ j)∣∣+ ∣∣∣ζ j∂ζ jK (ζ j)∣∣∣+ ∣∣∣ζ 2j ∂ζ jζ jK (ζ j)∣∣∣≲ 1{ζ j≤1}+ζ−1j 1{ζ j>1}. (4.13.116)Since∇y[ j]g0(y[ j], t) =ˆ t0p˙ j(s)t− s ∂ζ jK (ζ j)2λ 2j (t)y[ j]t− s ds=ˆ t0p˙ j(s)t− s ζ j∂ζ jK (ζ j)2y[ j]1+ |y[ j]|2 ds,by (4.3.22), we havesupτ1∈Iτ[g0(·, t(τ1))]CςHy (By) ≲ 〈y〉−ςH .359Noticeg0(y[ j], t) = λ−2j (t)(1+ |y[ j]|2)−1g01(y[ j], t), where g01(y[ j], t) :=ˆ t0p˙ j(s)ζ jK (ζ j)ds.By (4.13.103),[λ−2j (t(·))]CςH2τ (Iτ )≲ λ−3∗ | lnT |−1τ−2(lnτ)2|y|2−ςH . By (4.13.116), (4.3.22), |g01| ≲λ 2∗ 〈y[ j]〉2.For t < T ,∂tg01 = c0 p˙ j(t)+ˆ t0p˙ j(s)∂ζ j(ζ jK (ζ j))(1+ |y[ j]|2)2λ j(t)λ˙ j(t)(t− s)−λ 2j (t)(t− s)2 ds.By (4.3.22), it follows that|∂tg01|≲ |λ˙∗|+ˆ t0|λ˙∗(s)|t− s e−cdζ j(λ∗(t)|λ˙∗(t)|〈y[ j]〉2+1)ds≲ 1where cd := mink=0,1...,nRedk/9. It follows that [g01(y[ j], t(·))]CςH2τ (Iτ )≲ λ 2∗ |y|2−ςH . Thus we arrivesupz∈By[g0(z, t(·))]CςH2τ (Iτ )≲ 〈y〉−ςH .By (4.13.97),[g0]CςH ,ςH2 (Qy)∼ supτ1∈Iτ[g0(·, t(τ1))]CςHy (By)+ supz∈By[g0(z, t(·))]CςH2τ (Iτ )≲ 〈y〉−ςH .Thus we get〈y〉−3λ∗m[ j]02(y,τ)≲ λ∗〈y〉−3−ςH ≲ λ ν+ε∗ 〈y〉−2−l−ςH .4.14 Additional information 6: ConvergenceIn this section, based on the weighted spaces for the solution constructed, we shall show the conver-gence results in Theorem 1.0.7.Throughout this section, we adopt the usual Sobolev norm for the mapping between Euclideanspaces. M > 0 is an arbitrary constant. Given a function f (x, t), if there is no special declaration, weuse f (t) to denote f (x, t) for simplicity. A function f (t)→ 0 means that f (t) converges to 0 undersome norms as t→ T .360Recall (4.4.1). The solution to the LLG equation has the formu(x, t) =U∗+Φper, Φper := AU∗+Φ− (Φ ·U∗)U∗whereU∗ is the multi-bubble profile defined in (4.3.3), and A is given in (4.4.4).By (4.13.15) and the choice of Z∗ in (4.4.28), then |Φ| 1. Combining‖|U∗(·, t)|−1‖L∞(R2) = O(λ∗), |A|≲ λ∗+ |Φ|2by (4.3.4), (4.4.5) respectively, then we obtain‖Φper‖L∞(R2×(0,T )) 1. (4.14.1)It is straightforward to get∇xΦper =U∗∇xA+A∇xU∗+∇xΦ− (Φ ·U∗)∇xU∗−U∗∇x(Φ ·U∗)where we denoteU∗∇x f = (U∗∂x1 f ,U∗∂x2 f ) for a scalar function f . Therein, by (4.13.27),|U∗∇xA|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+λ 2ν−2δ0−1∗ 〈ρ j〉−2l−1+ | ln(T − t)|λ ν−δ0+Θ∗ R〈ρ j〉−l−1)+1{3λ∗R<|x−q[ j]|<3dq}]+1{∩Nj=1{|x−q[ j]|≥3dq}}.By (4.4.5), (4.13.1), (4.13.30),|A∇xU∗− (Φ ·U∗)∇xU∗|≲(|Φ ·U∗|+λ∗+ |Φ|2) |∇xU∗|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(λ 2ν−2δ0−1∗ 〈ρ j〉−l−2+ 〈ρ j〉−1+ | ln(T − t)|λΘ∗ R〈ρ j〉−2)+1{3λ∗R<|x−q[ j]|<3dq}(〈ρ j〉−1+ | ln(T − t)|λΘ∗ R〈ρ j〉−2)]+1{∩Nj=1{|x−q[ j]|≥3dq}}λ∗.By (4.13.16), (4.13.24),|∇xΦ−U∗∇x(Φ ·U∗)|≲N∑j=1[1{|x−q[ j]|≤3λ∗R}(1+λ ν−δ0−1∗ 〈ρ j〉−l−1+ | ln(T − t)|λΘ∗ R〈ρ j〉−2)+1{3λ∗R<|x−q[ j]|<3dq}]+1{∩Nj=1{|x−q[ j]|≥3dq}}.Thus, there exists a constant ε > 0 small such thatλ∗(t)‖∇xΦper(t)‖L∞(R2) ≲ λ ε∗ (t), Φper ∈ L∞((0,T );H1loc(R2)). (4.14.2)361We defineu∗(x) =Φper(x,T )+W (∞).By definition, we haveu(x, t)−u∗(x)−N∑j=1Qγ j[W(x−ξ [ j]λ j)−W (∞)]=Φper(x, t)−Φper(x,T ).Claim:Φ(x,T ) =N∑j=1η [ j]dq(x,T )Φ∗[ j]0 (|x−q[ j]|,T )+Φout(x,T ), (4.14.3)‖Φ(t)−Φ(T )‖L∞(R2) → 0, ‖∇xΦ(t)−∇x (Φ(T ))‖L2(BM) → 0. (4.14.4)Combining (4.13.7), (4.13.9), (4.3.24), we have‖Φ(T )‖L∞(R2)+‖∇x (Φ(T ))‖L∞(R2) ≲ 1. (4.14.5)Proof of (4.14.3) and (4.14.4). For Φout solved in Bout defined in (4.4.48), we have‖Φout(t)−Φout(T )‖L∞(R2) ≲ | ln(T − t)|λΘ+1∗ R+(T − t),‖∇xΦout(t)−∇xΦout(T )‖L∞(R2) ≲ λΘ∗ +(T − t)α/2.(4.14.6)For Φ[ j]in solved in B[ j]in defined in (4.4.43), since 0< l < 1, we have‖Φ[ j]in(t)‖L∞(R2) ≲ λ ν−δ0∗ ,ˆBM|Φ[ j]in(y[ j], t)|2dx≲ λ 2ν−2δ0∗ˆBM〈y[ j]〉−2ldx≲ λ 2ν−2δ0+2∗ˆ|y[ j]|≤C(M)λ−1∗〈y[ j]〉−2ldy[ j] ≲ λ 2ν−2δ0+2l∗ ,ˆBM∣∣∇x (η [ j]R (x, t)Φ[ j]in(y[ j], t))∣∣2 dx= ˆBM∣∣∣Φ[ j]in(y[ j], t)∇xη [ j]R (x, t)+η [ j]R (x, t)λ−1j ∇y[ j]Φ[ j]in(y[ j], t)∣∣∣2 dx≲ λ 2ν−2δ0−2∗ˆBM〈y[ j]〉−2l−2dx≲ λ 2ν−2δ0∗ˆ|y[ j]|≤C(M)λ−1∗〈y[ j]〉−2l−2dy[ j] ∼ λ 2ν−2δ0∗ .(4.14.7)Specially, Φ[ j]in(T ) = 0, which implies (4.14.3).It is easy to get‖η [ j]dq(t)−η[ j]dq(T )‖L∞(R2) → 0, ‖∇xη[ j]dq(t)−∇xη[ j]dq(T )‖L∞(R2) → 0. (4.14.8)We next consider the non-local correction terms Φ∗[ j]0 defined in (4.3.12). Recalling (4.3.18), wedenoteA :=a+ ib4, ζ j(t) :=|x−ξ [ j](t)|2+λ j(t)2t− s , ζ j(T ) :=|x−q[ j]|2T − s .362By (4.3.2), we have |ξ [ j]−q[ j]| λ∗.Recall (4.3.12), (4.3.17), (4.2.1) and µ = 3,Φ∗[ j]0 (|x−ξ [ j]|, t) =r2jr3j +λ 3j[x1−ξ [ j]1 + i(x2−ξ [ j]2)][Φ[ j]0 (√r2j +λ 2j , t),0]tr=−2|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )](|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2 [ˆ t0p˙ j(s)(1− e−Aζ j(t))ds,0]tr.(4.14.9)Then ∣∣Φ∗[ j]0 (|x−ξ [ j]|, t)−Φ∗[ j]0 (|x−q[ j]|,T )∣∣≲(|x−ξ [ j]|+λ j)−1 ∣∣∣∣ˆ t0p˙ j(s)(1− e−Aζ j(t))ds−ˆ T0p˙ j(s)(1− e−Aζ j(T ))ds∣∣∣∣+∣∣∣∣∣∣∣|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )](|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2 − |x−q[ j]|2[x1−q[ j]1 + i(x2−q[ j]2)]|x−q[ j]|3|x−q[ j]|∣∣∣∣∣∣∣×∣∣∣∣ˆ T0p˙ j(s)(1− e−Aζ j(T ))ds∣∣∣∣ .Denote r˜ j := |x−q[ j]|. Then∣∣∣∣∣∣∣|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )](|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2 − |x−q[ j]|2[x1−q[ j]1 + i(x2−q[ j]2)]|x−q[ j]|3|x−q[ j]|∣∣∣∣∣∣∣≲ˆ 10|x−θξ [ j]− (1−θ)q[ j]|2 (|ξ [ j]−q[ j]|+λ j)[|x−θξ [ j]− (1−θ)q[ j]|+θλ j]4dθ ≲(∣∣ξ [ j]−q[ j]∣∣+λ∗)ˆ 10(r˜ j+θλ∗)−2 dθ .(4.14.10)Using (4.3.22), we have∣∣∣∣ˆ T0p˙ j(s)(1− e−Aζ j(T ))ds∣∣∣∣= r˜2j ∣∣∣∣ˆ T0p˙ j(s)T − s(1− e−Aζ j(T ))ζ j(T )−1ds∣∣∣∣≲ r˜2j .ˆ t0p˙ j(s)(1− e−Aζ j(t))ds−ˆ T0p˙ j(s)(1− e−Aζ j(T ))ds=(ˆ (t−λ 2∗−r˜2j)+0+ˆ (t−r˜2j)+(t−λ 2∗−r˜2j)++ˆ t(t−r˜2j)+)p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds−ˆ Ttp˙ j(s)(1− e−Aζ j(T ))ds.363Therein, for the first part,∣∣∣∣∣ˆ (t−λ 2∗−r˜2j)+0p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds∣∣∣∣∣≲ˆ (t−λ 2∗−r˜2j)+0∣∣∣λ˙∗(s)∣∣∣ ∣∣ζ j(T )−ζ j(t)∣∣ds≲ˆ (t−λ 2∗−r˜2j)+0∣∣∣λ˙∗(s)∣∣∣[r˜2j( 1t− s − 1T − s)+ r˜ j |ξ [ j](t)−q[ j]|+λ∗(t)2t− s]ds≲ r˜2j ln( t[T −(t−λ 2∗ − r˜2j)+]T[t−(t−λ 2∗ − r˜2j)+])+ (r˜ j ∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( tt−(t−λ 2∗ − r˜2j)+).For the second part, ∣∣∣∣∣ˆ (t−r˜2j)+(t−λ 2∗−r˜2j)+p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds∣∣∣∣∣≲ λ 2∗ .For the third part, if r˜ j ≤√T − t, we have∣∣∣∣∣ˆ t(t−r˜2j)+p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds∣∣∣∣∣≲ r˜2j ;if r˜ j >√T − t, t > T/2, we have∣∣∣∣∣ˆ t(t−r˜2j )+p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds∣∣∣∣∣=∣∣∣∣∣(ˆ t−(T−t)(t−r˜2j )++ˆ tt−(T−t))p˙ j(s)(e−Aζ j(T )− e−Aζ j(t))ds∣∣∣∣∣≲ˆ t−(T−t)(t−r˜2j )+∣∣∣λ˙∗(s)∣∣∣[r˜2j( 1t− s − 1T − s)+ r˜ j |ξ [ j](t)−q[ j]|+λ∗(t)2t− s]ds+ˆ tt−(T−t)∣∣∣λ˙∗(s)∣∣∣ds≲ r˜2j| lnT |ln2(T − t) +(r˜ j∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( t− (t− r˜2j )+T − t)+λ∗where we used 1t−s − 1T−s ∼ (T − t)(T − s)−2 for s≤ t− (T − t).For the last part, by (4.3.2),∣∣∣∣ˆ Ttp˙(s)(1− e−Aζ j(T ))ds∣∣∣∣≲r˜2j´ T−r˜2jt|λ˙∗(s)|T−s ds+´ TT−r˜2j∣∣∣λ˙∗(s)∣∣∣ds, if r˜ j ≤√T − tλ∗, if r˜ j >√T − t≲ r˜2j1{r˜ j≤√T−t}+λ∗1{r˜ j>√T−t}.364Collecting the above estimates, we conclude that for t > T/2,∣∣Φ∗[ j]0 (|x−ξ [ j]|, t)−Φ∗[ j]0 (|x−q[ j]|,T )∣∣≲ (r˜ j+λ∗)−1{r˜2j ln( t[T −(t−λ 2∗ − r˜2j)+]T[t−(t−λ 2∗ − r˜2j)+])+ (r˜ j ∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( tt−(t−λ 2∗ − r˜2j)+)+λ 2∗ +1{r˜ j≤√T−t}r˜2j +1{r˜ j>√T−t}[r˜2j| lnT |ln2(T − t) +(r˜ j∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( t− (t− r˜2j )+T − t)+λ∗]}+∣∣ξ [ j]−q[ j]∣∣+λ∗≲ r˜ j[ln(t−1T)+ ln(T −(t−λ 2∗ − r˜2j)+t−(t−λ 2∗ − r˜2j)+)]+λ∗∣∣∣∣ ln( t−(t−λ 2∗ − r˜2j)+t)∣∣∣∣+λ∗+1{r˜ j≤√T−t}r˜ j+1{r˜ j>√T−t}[r˜ j| lnT |ln2(T − t) +λ∗ ln( t− (t− r˜2j )+T − t)+λ∗ (T − t)−1/2].For all r˜2j ≤M, we have∣∣Φ∗[ j]0 (|x−ξ [ j]|, t)−Φ∗[ j]0 (|x−q[ j]|,T )∣∣→ 0 uniformly as t→ T. (4.14.11)Indeed, for T/4< r˜2j ≤M, the conclusion is easy to derive; for r˜2j ≤ T/4 and t close to T sufficiently,we have∣∣Φ∗[ j]0 (|x−ξ [ j]|, t)−Φ∗[ j]0 (|x−q[ j]|,T )∣∣≲ r˜ j ln(t−1T)+√T − t |ln(T − t)|+λ∗ (|ln t|+ ∣∣ln(λ 2∗ + r˜2j)∣∣)+1{r˜ j≤√T−t}r˜ j+1{r˜ j>√T−t}[r˜ j| lnT |ln2(T − t) +λ∗ (T − t)−1/2],which converges to 0 uniformly as t→ T , where we used∣∣r˜ j ln(T − t+λ 2∗ + r˜2jλ 2∗ + r˜2j )∣∣≲ r˜ j |ln(T − t)|1{r˜ j≤√T−t}+(T − t)r˜−1j 1{r˜ j>√T−t}.From (4.14.6), (4.14.7), (4.14.8), and (4.14.11), we obtain the first part of (4.14.4).Using (4.14.9), we have∂x1Φ∗[ j]0 (|x−ξ [ j]|, t)=−2 f [ j]1 (t)[ˆ t0p˙ j(s)(1− e−Aζ j(t))ds,0]tr+ f [ j]2 (t)[ˆ t0p˙ j(s)ζ j(t)e−Aζ j(t)ds,0]trwheref [ j]1 (t) :=2(x1−ξ [ j]1 )[x1−ξ [ j]1 + i(x2−ξ [ j]2)](|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2 + |x−ξ [ j]|2(|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2365+|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )](|x−ξ [ j]|2+λ 2j)1/2 (−1)(|x−ξ [ j]|3+λ 3j )−2 3|x−ξ [ j]|(x1−ξ [ j]1 )+|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )]|x−ξ [ j]|3+λ 3j(−1)(|x−ξ [ j]|2+λ 2j )−3/2 (x1−ξ [ j]1 ),f [ j]2 (t) :=−2|x−ξ [ j]|2 [x1−ξ [ j]1 + i(x2−ξ [ j]2 )](|x−ξ [ j]|3+λ 3j)(|x−ξ [ j]|2+λ 2j)1/2 2A x1−ξ [ j]1|x−ξ [ j]|2+λ j(t)2 .It is easy to get ∣∣ f [ j]1 (t)∣∣+ ∣∣ f [ j]2 (t)∣∣≲ (r˜ j+λ∗)−2 .Similarly to (4.14.10), we have∣∣ f [ j]1 (t)− f [ j]1 (T )∣∣+ ∣∣ f [ j]2 (t)− f [ j]2 (T )∣∣≲ λ∗ˆ 10(r˜ j+θλ∗)−3 dθ .´ t0 p˙ j(s)ζ j(t)e−Aζ j(t)ds can be handled similarly to´ t0 p˙ j(s)(1− e−Aζ j(t))ds. In sum,∣∣∂x1Φ∗[ j]0 (|x−ξ [ j]|, t)−∂x1Φ∗[ j]0 (|x−q[ j]|,T )∣∣≲ (r˜ j+λ∗)−2{r˜2j ln( t[T −(t−λ 2∗ − r˜2j)+]T[t−(t−λ 2∗ − r˜2j)+])+ (r˜ j ∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( tt−(t−λ 2∗ − r˜2j)+)+λ 2∗ +1{r˜ j≤√T−t}r˜2j +1{r˜ j>√T−t}[r˜2j| lnT |ln2(T − t) +(r˜ j∣∣ξ [ j]−q[ j]∣∣+λ 2∗ ) ln( t− (t− r˜2j )+T − t)+λ∗]}+λ∗ˆ 10(r˜ j+θλ∗)−1 dθ≲ (r˜ j+λ∗)−1{r˜ j[ln(t−1T)+ ln(T −(t−λ 2∗ − r˜2j)+t−(t−λ 2∗ − r˜2j)+)]+λ∗∣∣∣∣ ln( t−(t−λ 2∗ − r˜2j)+t)∣∣∣∣+λ∗+1{r˜ j≤√T−t}r˜ j+1{r˜ j>√T−t}[r˜ j| lnT |ln2(T − t) +λ∗ ln( t− (t− r˜2j )+T − t)+λ∗ (T − t)−1/2]}+ ln(1+ r˜−1j λ∗).And ∂x2Φ∗[ j]0 (|x−ξ [ j]|, t) can be dealt with similarly. Thus we getˆBM∣∣∇xΦ∗[ j]0 (|x−ξ [ j]|, t)−∇xΦ∗[ j]0 (|x−q[ j]|,T )∣∣2 dx→ 0.Combining this with (4.14.6), (4.14.7), and (4.14.8), we conclude the second part of (4.14.4).366RecallU∗ given in (4.3.3). The pointwsie limitation as t goes to T is given byU∗(T ) =[0,0,1]tr =U∞, if x 6∈ {q[ j] | j = 1, . . . ,N}[0,0,−1]tr if x ∈ {q[ j] | j = 1, . . . ,N} .U∗(t) does not converge in L∞loc sinceU∗(T ) is not continuous. Instead,‖U∗(·, t)−U∞‖L∞(∩Nj=1{|x−q[ j]|>3λ∗(t)R(t)}) → 0 as t→ T. (4.14.12)By (4.13.15), Φ(q[ j],T ) = 0, j = 1, . . . ,N, which implies ((Φ ·U∗)U∗)(T ) = (Φ(T ) ·U∞)U∞.‖(Φ ·U∗)U∗− ((Φ ·U∗)U∗)(T )‖L∞(R2)≤ ‖(Φ · (U∗−U∞))U∗‖L∞(R2)+‖(Φ ·U∞)(U∗−U∞)‖L∞(R2)+‖ [(Φ−Φ(T )) ·U∞]U∞‖L∞(R2) → 0,(4.14.13)where we used (4.13.15), (4.14.12), (4.14.4) in the last step.‖∂xi [(Φ ·U∗)U∗]−∂xi [((Φ ·U∗)U∗)(T )]‖L2(BM)≤ ‖(Φ ·U∗)∂xiU∗‖L2(BM)+‖(Φ ·∂xiU∗)U∗‖L2(BM)+‖ [∂xiΦ · (U∗−U∞)]U∗‖L2(BM)+‖(∂xiΦ ·U∞)(U∗−U∞)‖L2(BM)+‖ [(∂xiΦ−∂xiΦ(T )) ·U∞]U∞‖L2(BM) → 0(4.14.14)where for the last step, we used the following analysis.By (4.13.1), (4.13.15), then ‖(Φ ·U∗)∂xiU∗‖L2(BM)+‖(Φ ·∂xiU∗)U∗‖L2(BM) → 0.By (4.13.16), (4.14.12), then ‖ [∂xiΦ · (U∗−U∞)]U∗‖L2(BM)+‖(∂xiΦ ·U∞)(U∗−U∞)‖L2(BM) → 0.By (4.14.4), then ‖ [(∂xiΦ−∂xiΦ(T )) ·U∞]U∞‖L2(BM) → 0.Recall A defined in (4.4.4). It is straightforward to getA(T ) =[1−|Φ(T )|2+(Φ(T ) ·U∞)2] 12 −1, A(q[ j],T ) = 0, j = 1, . . . ,N, (AU∗)(T ) = A(T )U∞,∇x (A(T )) =−∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)2(1+A(T )),∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2 = |Φ(T )|2− (Φ(T ) ·U∞)2 .(4.14.15)Note that|ΠU⊥∗ Φ|2 = |Φ− (Φ ·U∗)U∗|2 = |Φ|2− (Φ ·U∗)2+(Φ ·U∗)2(|U∗|2−1) .367By (4.13.15), (4.14.4), (4.14.12), we have‖(Φ ·U∗)(t)− (Φ ·U∗)(T )‖L∞(R2) ≤ ‖Φ(t) · (U∗(t)−U∗(T ))‖L∞(∩Nj=1{|x−q[ j]|>3λ∗(t)R(t)})+‖Φ(t) · (U∗(t)−U∗(T ))‖L∞(∪Nj=1{|x−q[ j]|≤3λ∗(t)R(t)})+‖(Φ(t)−Φ(T )) ·U∗(T )‖L∞(R2) → 0.We can handle the convergence of the other terms involved in A directly. Then‖A(t)−A(T )‖L∞(R2) → 0. (4.14.16)Similarly, using (4.4.5), (4.13.15), (4.14.12), and (4.14.16), we have‖(AU∗)(t)− (AU∗)(T )‖L∞(R2) ≤ ‖A(t)(U∗(t)−U∗(T ))‖L∞(∩Nj=1{|x−q[ j]|>3λ∗(t)R(t)})+‖A(t)(U∗(t)−U∗(T ))‖L∞(∪Nj=1{|x−q[ j]|≤3λ∗(t)R(t)})+‖(A(t)−A(T ))U∗(T )‖L∞(R2) → 0 as t→ T.(4.14.17)Using (AU∗)(T ) = A(T )U∞, we write∂xi ((AU∗)(t)− (AU∗)(T )) = (U∗(t)−U∞)∂xiA(t)+A(t)∂xiU∗(t)+U∞∂xi (A(t)−A(T )) , i= 1,2and consider the following three termsI1 :=ˆBM|∇xA(x, t)|2 |U∗(x, t)−U∞|2dx, I2 :=ˆBM|A(x, t)|2 |∇xU∗(x, t)|2 dx,I3 :=ˆBM|∇x (A(x, t)−A(x,T ))|2 dx.By (4.13.27), (4.14.12), and the parameters restriction (3.4.4), we have I1 → 0.By (4.4.5), (4.13.15), and (4.13.1), we have I2 → 0.For I3, using ∇x (A(T )) in (4.14.15) and ∇xA given in (4.4.13), we have|∇xA−∇x (A(T ))|≲∣∣∣[(1+A)2∇x(|U∗|2)+∇x(∣∣∣ΠU⊥∗ Φ∣∣∣2)+2(1+A)∇x(U∗ ·ΠU⊥∗ Φ)](1+A(T ))−∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)[(1+A)|U∗|2+U∗ ·ΠU⊥∗ Φ]∣∣∣=∣∣∣(1+A)2∇x(|U∗|2)(1+A(T ))+2(1+A)∇x(U∗ ·ΠU⊥∗ Φ)(1+A(T ))+[∇x(∣∣∣ΠU⊥∗ Φ∣∣∣2)−∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)](1+A(T ))+∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)(A(T )−A)+∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)(1+A)(1−|U∗|2)−∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)(1−|U∗|2)(Φ ·U∗) ∣∣∣.368• By (4.13.3), we have∇x(|U∗|2)→ 0 in L2loc.• By (4.3.4), (4.13.24), (4.13.20), and (4.13.3),∣∣∣∇x(U∗ ·ΠU⊥∗ Φ)∣∣∣= ∣∣(1−|U∗|2)∇x (Φ ·U∗)−2(Φ ·U∗)U∗ ·∇xU∗∣∣≲ λ∗+N∑j=11{|x−q[ j]|<3dq}| ln(T − t)|λΘ+1∗ R〈ρ j〉−2which converges to 0 in L2loc as t→ T .• Note that∇x(|ΠU⊥∗ Φ|2)= 2Φ ·∇xΦ+(Φ ·U∗)2∇x(|U∗|2)+2(|U∗|2−2)(Φ ·U∗)∇x (Φ ·U∗) .Combining (4.14.15), we have∣∣∣∣∂xi(∣∣∣ΠU⊥∗ Φ∣∣∣2)−∂xi(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)∣∣∣∣≤ 2 |Φ ·∂xiΦ−Φ(T ) ·∂xi (Φ(T ))|+ ∣∣∣(Φ ·U∗)2 ∂xi (|U∗|2)∣∣∣+2∣∣∣(|U∗|2−1)(Φ ·U∗)∂xi (Φ ·U∗)∣∣∣+2 |−(Φ ·U∗)∂xi (Φ ·U∗)+(Φ(T ) ·U∞) [∂xi (Φ(T )) ·U∞]| .By (4.14.4), (4.14.5),‖Φ ·∂xiΦ−Φ(T ) ·∂xi (Φ(T ))‖L2(BM) → 0.By (4.13.15), (4.13.3),‖(Φ ·U∗)2 ∂xi(|U∗|2)‖L2(BM) ≲ ‖|Φ|2∣∣∣∂xi (|U∗|2)∣∣∣‖L2(BM) → 0.By (4.13.20), (4.13.24),‖(|U∗|2−1)(Φ ·U∗)∂xi (Φ ·U∗)‖L∞(R2) ≲ λ∗‖(Φ ·U∗)∂xi (Φ ·U∗)‖L∞(R2) → 0.For the last term,‖− (Φ ·U∗)∂xi (Φ ·U∗)+(Φ(T ) ·U∞) [∂xi (Φ(T )) ·U∞]‖L2(BM)≤ ‖ [(Φ(T )−Φ) ·U∗]∂xi (Φ ·U∗)‖L2(BM)+‖ [Φ(T ) · (U∞−U∗)]∂xi (Φ ·U∗)‖L2(BM)+‖(Φ(T ) ·U∞)(Φ ·∂xiU∗)‖L2(BM)+‖(Φ(T ) ·U∞) [∂xiΦ · (U∞−U∗)]‖L2(BM)+‖(Φ(T ) ·U∞) [(∂xi(Φ(T ))−∂xiΦ) ·U∞]‖L2(BM) → 0where we used (4.14.4), (4.13.24), (4.14.5), (4.14.12), (4.13.15), (4.13.1), (4.13.16) in the last step.369In sum,‖∂xi(∣∣∣ΠU⊥∗ Φ∣∣∣2)−∂xi(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)‖L2(BM) → 0.• By (4.14.15), (4.14.5), ∣∣∣∣∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)∣∣∣∣≲ 1.Combining (4.14.16), we have‖∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)(A(T )−A)‖L∞(R2) → 0,‖∇x(∣∣∣(ΠU⊥∗ Φ)(T )∣∣∣2)(1+A−Φ ·U∗)(1−|U∗|2)‖L∞(R2) ≲ λ∗.As a result, we conclude I3 → 0. 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Thesis/Dissertation
eng
Mathematics
Vancouver : University of British Columbia Library
Attribution-NonCommercial-NoDerivatives 4.0 International
Graduate
Wei, Juncheng, 1968-
The construction of blow-up solutions for some evolution equations
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