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Some new results on the SU(3) Toda system and Lin-Ni problem Yang, Wen 2015

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Some new results on the SU(3) Todasystem and Lin-Ni problembyWen YangB.Sc., Wuhan University, 2010M.Phil, The Chinese University of Hong Kong, 2012A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)July 2015c© Wen Yang 2015AbstractIn this thesis, we mainly consider two problems. First, we study the SU(3)Toda system. Let (M, g) be a compact Riemann surface with volume 1, h1and h2 be a C1 positive function on M and ρ1, ρ2 ∈ R+. The SU(3) Todasystem is the following one on the compact surface M∆u1 + 2ρ1(h1eu1∫M h1eu1− 1)− ρ2(h2eu2∫M h2eu2− 1) = 4pi∑q∈S1 αq(δq − 1),∆u2 − ρ1(h1eu1∫M h1eu1− 1) + 2ρ2(h2eu2∫M h2eu2− 1) = 4pi∑q∈S2 βq(δq − 1),where ∆ is the Beltrami-Laplace operator, αq ≥ 0 for every q ∈ S1, S1 ⊂M,βq ≥ 0 for every q ∈ S2, S2 ⊂M and δq is the Dirac measure at q ∈M . Weinitiate the program for computing the Leray-Schauder topological degreeof SU(3) Toda system and succeed in obtaining the degree formula for ρ1 ∈(0, 4pi) ∪ (4pi, 8pi), ρ2 /∈ 4piN when S1 = S2 = ∅.Second, we consider the following nonlinear elliptic Neumann problem∆u− µu+ uq = 0 in Ω,u > 0 in Ω,∂u∂ν = 0 on ∂Ω.where q = n+2n−2 , µ > 0 and Ω is a smooth and bounded domain in Rn. Linand Ni (1986) conjectured that for µ small, all solutions are constants. Inthe second part of this thesis, we will show that this conjecture is false fora general domain in n = 4, 6 by constructing a nonconstant solution.iiPrefaceThis dissertation is ultimately based on the original intellectual product ofthe author, Wen Yang under the guidance of his supervisor Prof. Jun-chengWei.A version of Chapter 2 has been submitted and put on the arXiv:http://arxiv.org/abs/1408.5802.A version of Chapter 3 is contained in a preprint paper joint with Prof.Jun-cheng Wei and Prof. Bin Xu.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . viDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The Degree Counting Formula For SU(3) Toda System . . . 21.1.1 Background And Main Results . . . . . . . . . . . . . 21.1.2 Sketch Of The Proof Of Theorem 1.1.1 . . . . . . . . 61.2 Lin-Ni Problem . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.1 Background . . . . . . . . . . . . . . . . . . . . . . . 121.2.2 Previous Results On Lin-Ni Problem . . . . . . . . . 151.2.3 Sketch Of The Proof Of Theorem 1.2.1 . . . . . . . . 181.3 Organization Of The Thesis . . . . . . . . . . . . . . . . . . 232 The SU(3) Toda System . . . . . . . . . . . . . . . . . . . . . . 252.1 Proof Of Proposition 1.1.1, Proposition 1.1.2 And ShadowSystem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 A-priori Estimate . . . . . . . . . . . . . . . . . . . . . . . . 402.3 Approximate Blow-up Solution . . . . . . . . . . . . . . . . . 532.4 Deformation And Degree Counting Formula . . . . . . . . . 622.5 Proof Of Theorem 1.1.1 . . . . . . . . . . . . . . . . . . . . . 71ivTable of Contents2.6 Proof Of Lemma 2.3.2 And (2.4.14) . . . . . . . . . . . . . . 722.7 The Leray-Schauder degree . . . . . . . . . . . . . . . . . . . 883 The Lin-Ni Problem . . . . . . . . . . . . . . . . . . . . . . . . 923.1 Approximate Solutions . . . . . . . . . . . . . . . . . . . . . 923.2 Finite Dimensional Reduction . . . . . . . . . . . . . . . . . 973.3 Finite Dimensional Reduction: A Nonlinear Problem . . . . 1063.4 Finite Dimensional Reduction: Reduced Energy . . . . . . . 1123.5 Proof Of Theorem 1.2.1 . . . . . . . . . . . . . . . . . . . . . 1163.6 Proof Of Lemma 3.1.1 . . . . . . . . . . . . . . . . . . . . . . 122Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128vAcknowledgementsI am greatly indebted to my supervisor Prof. Wei Juncheng for his guidancethroughout my PhD study. He always give me a lot of useful suggestionsand some wonderful ideas.I would also like to thank Prof. Lin Changshou, Prof. Matthias Winter,Prof. Li Dong and Prof. Zhang Lei for their help and discussions.Moveover, I would also like to thank my colleagues in the MathematicsDepartment in UBC, especially to Ao Weiwei, Chan Hon To Hardy, ChangYifan, Duan Xiaoyu, Bai Fan, Chen Tian, Liu Ye, Lu Hongliang, Luo Yuwen,Sui Yi, Wang Jiaxing, Wang Li, Ye Zichun, and Zhu Qingsan, for the happytime we study together.Last but not least, I want to show my gratitude to my parents for theirsupport.viDedicationTo my parents.viiChapter 1IntroductionThis thesis mainly concerns two problems. First, we consider the SU(3)Toda system. Let (M, g) be a compact Riemann surface with volume 1, h1and h2 be C1 positive functions on M and ρ1, ρ2 ∈ R+. The SU(3) Todasystem is the following one on the compact surface M,∆u1 + 2ρ1(h1eu1∫M h1eu1− 1)− ρ2(h2eu2∫M h2eu2− 1) = 4pi∑q∈S1 αq(δq − 1),∆u2 − ρ1(h1eu1∫M h1eu1− 1) + 2ρ2(h2eu2∫M h2eu2− 1) = 4pi∑q∈S2 βq(δq − 1),(1.0.1)where ∆ is the Beltrami-Laplace operator, αq ≥ 0 for every q ∈ S1, S1 ⊂M,βq ≥ 0 for every q ∈ S2, S2 ⊂ M and δq is the Dirac measure at q ∈ M . Apartial result for the degree counting formula of (1.0.1) is obtained.The second part of this thesis concerns the following nonlinear ellipticNeumann problem∆u− µu+ uq = 0 in Ω,u > 0 in Ω,∂u∂ν = 0 on ∂Ω,(1.0.2)where 1 < q < +∞, µ > 0 and Ω is a smooth and bounded domain in Rn.In 1986, Lin and Ni proposed the following conjectureLin-Ni’s Conjecture [41]. For µ small and q = n+2n−2 , problem (1.0.2) ad-mits only the constant solution.When n = 4 and 6, we prove the existence of nonconstant solution to (1.0.2)provided µ is sufficiently small. This gives a counterexample of the conjec-ture in dimensions n = 4 and 6.11.1. The Degree Counting Formula For SU(3) Toda System1.1 The Degree Counting Formula For SU(3)Toda System1.1.1 Background And Main ResultsLet (M, g) be a compact Riemann surface with volume 1, h1 and h2 be C1positive functions on M and ρ1, ρ2 ∈ R+. The SU(3) Toda system on thecompact surface M is the following∆u1 + 2ρ1(h1eu1∫M h1eu1− 1)− ρ2(h2eu2∫M h2eu2− 1)= 4pi∑q∈S1 αq(δq − 1),∆u2 − ρ1(h1eu1∫M h1eu1− 1)+ 2ρ2(h2eu2∫M h2eu2− 1)= 4pi∑q∈S2 βq(δq − 1),(1.1.1)where ∆ is the Beltrami-Laplace operator, αq ≥ 0 for every q ∈ S1, S1 ⊂M,βq ≥ 0 for every q ∈ S2, S2 ⊂M and δq is the Dirac measure at q ∈M .When the two equations in (1.1.1) are identical, i.e., S1 = S2, αq = βq,u1 = u2 = u, h1 = h2 = h and ρ1 = ρ2 = ρ, system (1.1.1) is reduced to thefollowing mean field equation∆u+ ρ( heu∫M heu − 1)= 4pi∑q∈S1αq(δq − 1). (1.1.2)Equations (1.1.1) and (1.1.2) arise in many physical and geometric prob-lems. In physics, (1.1.2) or (1.1.1) is one of the limiting equations of theabelian gauge field theory or non-abelian Chern-Simons gauge field theory,one can see [24, 25, 48, 59, 60, 73] and references therein. In conformal geom-etry, equation (1.1.2) without singular sources corresponds to the Nirenbergproblem of prescribing Gaussian curvature. In general, equation (1.1.2) isrelated to the existence of positive constant curvature metric with conicsingularities. As for the Toda system (1.1.1), it is closely related to the clas-sical Plu¨cker formula for a holomorphic curve from M to CP2, the vortexpoints and αq, βq are exactly the branch points and its ramification indexof this holomorphic curve. See [43] for more precise formulation and also[8, 9, 13, 18, 26, 35] for connection with different aspects of geometry. Forthe past decade, there are many studies for the SU(3) Toda system, or more21.1. The Degree Counting Formula For SU(3) Toda Systemgenerally, system of equations with exponential nonlinearity. We refer thereaders to [7, 29, 30, 36, 40, 44–47, 49–51, 54–56, 60, 71, 72] and referencestherein.When S1 = ∅, equation (1.1.1) becomes the following nonlinear ellipticequation∆u+ ρ( heu∫M heu − 1)= 0. (1.1.3)Clearly, (1.1.3) is the Euler-Lagrange equation of the nonlinear functionalJρJρ(φ) =12∫M|∇φ|2 − ρ log(∫Mheφ)for φ ∈ {f ∈ H1(M) |∫M φ = 0}, where H1(M) denotes the Sobolevspace of L2 functions with L2−integrable first derivatives. For ρ < 8pi,Jρ(φ) is bounded from below and the infinimum of Jρ(φ) can be achievedby the well-known inequality due to Moser and Trudinger. For ρ ≥ 8pi, theexistence of (1.1.3) is more difficult. Struwe and Tarantello [65] were ableto obtain nontrivial solutions of (1.1.3) for 8pi < ρ < 4pi2 when h∗ ≡ 1 andM is the flat torus with fundamental domain [0, 1] × [0, 1]. Also, by usinga similar approach, Ding, Jost, Li, and Wang [22] proved the existence ofsolutions to (1.1.3) for 8pi < ρ < 16pi when M is a compact Riemann surfacewith genus g ≥ 1. For the case M = S2 and 8pi < ρ < 16pi, Lin [39]proved the nonvanishing of the Leray-Schauder degree to equation (1.1.3),and consequently, the existence of solutions follows for the case of genus 0.For the convenience of the reader, we provide a short introduction of theLeray-Schauder degree in Section 7 of Chapter 2. When the value of theparameter satisfies ρ > 16pi, the existence results for (1.1.3) can be deducedby the degree formula obtained by Chen-Lin in [14, 15] (Malchiodi uses adifferent approach to get the same degree counting formula in [53]). Moreprecisely, Li [37] proposed the problem of studying the existence of solutionsto (1.1.3) by the Leray-Schauder topological degree. Obviously, equation(1.1.3) is invariant under adding a constant. Hence, we can seek solutionsin the class of functions that are normalized by∫M u = 0. By the results of31.1. The Degree Counting Formula For SU(3) Toda SystemBrezis and Merle [11] and Li and Shafrir [38], it follows that for any integerm > 0 and for any compact set I in (8pim, 8pi(m + 1)), the normalizedsolutions of (1.1.3) are uniformly bounded for any positive C1 function hand ρ ∈ I. Thus, the Leray-Schauder degree dρ at zero of the Fredholm mapI + T (ρ) withT (ρ) = ρ∆−1( heu∫M heu − 1)is well defined. Moreover, dρ is independent of both the function h(x) andthe parameter ρ whenever ρ ∈ (8pim, 8pi(m+1)). Subsequently, Chen-Lin in[14, 15] were able to complete Li’s analysis and they arrived at the followingformula:dρ ={1, if ρ ∈ (0, 8pi),(m−χ(M))···(1−χ(M))m! , if ρ ∈ (8mpi, 8(m+ 1)pi), m > 0,(1.1.4)where χ(M) = 2(1− g) is the Euler characteristic of M with genus g.Since the degree counting formula of (1.1.3) is obtained, it is natural toconsider the same problem for equation (1.1.2). For equation (1.1.2), we letthe set Σ of the critical parameters be defined byΣ : ={8piN + Σq∈A8pi(1 + αq) | A ⊆ S1, N ∈ N ∪ {0}}\ {0}= {8piak | k = 1, 2, 3, · · · .},where ak will be defined in (1.1.5). We note that if S1 = ∅, Σ = 8mpi,which is indeed the set of the critical parameters for (1.1.3). It was provedthat if ρ /∈ Σ, then the a-priori estimate for any solution of (1.1.2) holdsin C2loc(M \ S1). This a-priori bound was obtained by Li and Shafrir [38]for the case without singular sources, i.e., S1 = ∅, and by Bartolucci andTarantello [6] for the general case with singular sources. After establishingthe a-priori bound for a non-critical parameter ρ, the Leray Schauder degreefor the equation (1.1.2) in the general case is well-defined for ρ ∈ R+ \ Σ.Following the same idea in [14, 15], Chen and Lin in [16, 17] have derivedthe topological degree counting formula for (1.1.2) as described below.We denote the topological degree of (1.1.2) for ρ /∈ Σ by dρ. By the41.1. The Degree Counting Formula For SU(3) Toda Systemhomotopic invariant of the topological degree, dρ is a constant for 8piak <ρ < 8piak+1, k = 0, 1, 2, · · · , where a0 = 0. Set dm = dρ for 8piam < ρ <8piam+1. To state the result, we introduce the following generating functionΞ0 :Ξ0(x) =(1 + x+ x2 + x3 + · · · )−χ(M)+|S1|Πq∈S1(1− x1+αq)=1 + c1xa1 + c2xa2 + · · ·+ ckxak + · · · . (1.1.5)The degree dm can be written in terms of cj , as shown in the followingtheorem.Theorem A. ([17]) Let dρ be the Leray-Schauder degree for (1.1.2). Suppose8ampi < ρ < 8am+1pi. Thendρ =m∑j=0cj ,where d0 = 1.For the application, it often requires that αq ∈ N for all q ∈ S1. In thiscase, Σ = {8pim | m ∈ N} and let dm = dρ for ρ ∈ (8pim, 8pi(m+ 1)). Thenthe generating functionΞ1(x) =∞∑k=0dkxk = (1 + x+ x2 + · · · )−χ(M)+1+|S1|Πq∈S1(1− xαq+1)=(1 + x+ x2 + · · · )−χ(M)+1Πq∈S1(1 + x+ x2 + · · ·+ xαq). (1.1.6)Clearly, we have dm ≥ 1, ∀m provided χ(M) ≤ 0. Hence we can obtain theexistence of the solution to (1.1.2) when the genus of M is nonzero.In the first part of this thesis, we want to initiate the program for com-puting the Leray-Schauder degree formula for the system (1.1.1). However,it seems still a very challenging problem in full generality. Hence we shallconsider the simplest (but nontrivial) case, described below. We assume(i) S1, S2 = ∅,(ii) ρ1 ∈ (0, 4pi) ∪ (4pi, 8pi) and ρ2 /∈ Σ1 = {4piN | N ∈ N}.51.1. The Degree Counting Formula For SU(3) Toda SystemIn order to state our result on the degree formula for SU(3) Toda system(1.1.1), we first introduce the following generating functionΞ1(x) = (1 + x+ x2 + x3 · · · )−χ(M)+1 = b0 + b1x1 + b2x2 + · · ·+ bmxm + · · · ,which is (1.1.6) provided αq = 0, ∀q ∈ S1. It is easy to see thatbm =(m− χ(M)m), (1.1.7)where(m− χ(M)m)={(m−χ(M))···(1−χ(M))m! , if m ≥ 1,1, if m = 0.Under the assumption (i) − (ii), our result on computing the Leray-Schauder degree for system (1.1.1) is as follows.Theorem 1.1.1. Suppose S1 = S2 = ∅. Let d(2)ρ1,ρ2 denote the topologicaldegree for (1.1.1) when ρ2 ∈ (4pim, 4pi(m+ 1)). Thend(2)ρ1,ρ2 ={bm, ρ1 ∈ (0, 4pi),bm − χ(M)(bm + bm−1), ρ1 ∈ (4pi, 8pi).1.1.2 Sketch Of The Proof Of Theorem 1.1.1In the following, we shall sketch the proof for the Theorem 1.1.1.Step 1. Find the critical parameters of (1.1.1).In order to compute the Leray-Schauder degree of the system, we needto develop a complete understanding of the blow-up phenomena for (1.1.1).The first main issue for the system is to determine the set of critical param-eters, i.e., those ρ = (ρ1, ρ2) such that the a-priori bounds for solutions of(1.1.1) fail.61.1. The Degree Counting Formula For SU(3) Toda SystemBased on the assumption S1 = S2 = ∅, we can write (1.1.1) as∆u1 + 2ρ1(h1eu1∫M h1eu1− 1)− ρ2(h2eu2∫M h2eu2− 1)= 0,∆u2 − ρ1(h1eu1∫M h1eu1− 1)+ 2ρ2(h2eu2∫M h2eu2− 1)= 0.(1.1.8)For equation (1.1.8), if u1 = u2 = u, h1 = h2 = h, and ρ1 = ρ2 = ρ,equation (1.1.8) turns to be∆u+ ρ( heu∫M heu − 1)= 0. (1.1.9)It is known that for mean field equation (1.1.9), the blow up phenomena isclosely related to the concentration phenomena, i.e., heuk∫M heuktends to a sumof Dirac measures, where uk is a sequence of blow up solutions to (1.1.9).More precisely, let uk be a sequence of blow-up solutions to (1.1.9). Then,ρheuk∫M heuk→ 8pi∑p∈Bδp,where B is the set containing all the blow up points of uk.As a consequence, once the blow-up phenomena happens, ρ = ρ heuk∫M heuk→8Npi. Therefore, if ρ 6= 8Npi, we can get a-priori bound on the solutions of(1.1.9). While for system (1.1.8), we can not find the counterpart result insystem. In fact, recently D’Aprile, Pistoia and Ruiz [19] have constructed asequence of bubbling solutions (u1k, u2k) to (1.1.8) with one of them failingto have the concentration property. So, finding the critical parameter forsystem (1.1.8) is more difficult than for the single equation, especially for thegeneral system (1.1.1). However, for the case without singular source term,i.e., the equation (1.1.8), we are able to determine all the critical parametersand the result is stated as follows,Proposition 1.1.1. Suppose hi in (1.1.8) are positive functions and ρi 6=4piN, i = 1, 2. Then, there exists a positive constant c such that for any71.1. The Degree Counting Formula For SU(3) Toda Systemsolution of equation (1.1.8), there holds:|ui(x)| ≤ c, ∀x ∈M, i = 1, 2.Step 2. Find out all the blow up solutions of (1.1.8).By Proposition 1.1.1, the Leray-Schauder degree d(2)ρ1,ρ2 for (1.1.1), orequivalently (1.1.8), is well-defined for ρ1 ∈ (0, 4pi)∪ (4pi, 8pi) and ρ2 /∈ 4piN.Clearly, d(2)ρ1,ρ2 = d(1)ρ2 if 0 < ρ1 < 4pi, and ρ2 /∈ 4piN. Hence, the main result ofthe first part of the thesis is to compute the degree d(2)ρ1,ρ2 for 4pi < ρ1 < 8pi.By the homotopic invariant, for any fixed ρ2 /∈ 4piN, d(2)ρ1,ρ2 is a constant forρ1 ∈ (0, 4pi), and the same holds true for ρ1 ∈ (4pi, 8pi). For simplicity, wemight let d(2)− and d(2)+ denote d(2)ρ1,ρ2 for ρ1 ∈ (0, 4pi) and ρ1 ∈ (4pi, 8pi). Sinced(2)− is known by Theorem A, computing d(2)+ is equivalent to computingthe difference of d(2)+ − d(2)− , which might be not zero due to the bubblingphenomena of (1.1.8) at (4pi, ρ2). To calculate d(2)+ −d(2)− , we need to computethe topological degree of the bubbling solution of (1.1.8) when ρ1 crosses 4pi,ρ2 /∈ 4piN. For convenience, we rewrite (1.1.8) as∆v1 + ρ1(h1e2v1−v2∫M h1e2v1−v2− 1)= 0,∆v2 + ρ2(h2e2v2−v1∫M h2e2v2−v1− 1)= 0,(1.1.10)where v1 = 13(2u1 + u2), v2 =13(u1 + 2u2). It is known that the Leray-Schauder degree for (1.1.8) and (1.1.10) are the same. So, our aim is tocompute the degree contribution of the bubbling solution of (1.1.10) whenρ1 crosses 4pi, ρ2 /∈ 4piN. We consider (v1k, v2k) to be a sequence of solutionsof (1.1.10) with (ρ1k, ρ2k) → (4pi, ρ2), and assume maxM (v1k, v2k) → ∞.Then we have the following theoremProposition 1.1.2. Let (v1k, v2k) be described as above. Then, the follow-ings hold:81.1. The Degree Counting Formula For SU(3) Toda System(i)ρ1kh1e2v1k−v2k∫M h1e2v1k−v2k→ 4piδp for some p ∈M, (1.1.11)(ii) v2k → 12w in C2,α(M), where (p, w) satisfies∇(log(h1e− 12w)(x) + 4piR(x, x))|x=p= 0, (1.1.12)and∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p)− 1)= 0. (1.1.13)Here R(x, p) refers to the regular part of the Green function G(x, p).Step 3. Computing the degree contributed by the blow up solution of(1.1.10).We write (1.1.12) and (1.1.13) as{∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 1)= 0,∇(log(h1e−12w)(x) + 4piR(x, x))|x=p= 0.(1.1.14)The system (1.1.14) is called the shadow system of (1.1.10). We say (p, w)is called a non-degenerate solution of (1.1.14) if the linearized equation. i.e.,for (φ, ν), where ν ∈ R2∆φ+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)φ−2ρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)φ)−8piρ2h2ew−4piG(x,p)(∇G(x,p)ν)∫M h2ew−4piG(x,p)+8piρ2h2ew−4piG(x,p)∫M(h2ew−4piG(x,p)(∇G(x,p)ν))( ∫M h2ew−4piG(x,p))2 = 0,∇2(log(h1e−12w)(x) + 4piR(x, x))|x=p ν − 12∇φ(p) = 0,∫M φ = 0,admits only trivial solution, i.e., (φ, ν) = (0, 0). For a given solution (p, w)91.1. The Degree Counting Formula For SU(3) Toda Systemof equation (1.1.14), we say λ is an eigenvalue of the linearized equation(1.1.14) if there exists an nontrivial pair (φ, ν) such that∆φ+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)φ−2ρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)φ)−8piρ2h2ew−4piG(x,p)(∇G(x,p)ν)∫M h2ew−4piG(x,p)+8piρ2h2ew−4piG(x,p)∫M(h2ew−4piG(x,p)(∇G(x,p)ν))( ∫M h2ew−4piG(x,p))2 + λφ = 0,∇2(log(h1e−12w)(x) + 4piR(x, x))|x=p ν − 12∇φ(p) + λν = 0,∫M φ = 0.The Morse index of solution (p, w) to (1.1.14) is the total number (countingmultiplicity) of the negative eigenvalues of the linearized system, and theLeray-Schauder topological degree contributed by (p, w) is given by (−1)N ,where N is the Morse index. From Proposition 1.1.2, it is known thatany blow up solution of (1.1.10) is closely related to (1.1.14). Furthermore,we shall prove that the topological degree contributed by all the blow-upsolutions equals the topological degree of the shadow system (1.1.14) up tosome factor, i.e., we have the following result.Proposition 1.1.3. Let dT denote the topological degree contributed by allthe blow up solutions of (1.1.10) and dS denote the topological degree con-tributed by (1.1.14). ThendT = −dS .Step 4. Computing the degree contributed by the shadow system(1.1.14).The equation (1.1.14) is a coupled system, and we can not directly com-pute the topological degree of (1.1.14). In this step, we introduce a defor-mation to decouple the system (1.1.14)(St){∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 1)= 0,∇(log(h1e−12w·(1−t)) + 4piR(x, x))|x=p= 0.(1.1.15)During the deformation from (S1) to (S0), we have the following lemma.101.1. The Degree Counting Formula For SU(3) Toda SystemLemma 1.1.1. Let ρ2 /∈ 4piN. Then there is uniform constant Cρ2 suchthat for all solutions to (1.1.15), we have |w|L∞(M) < Cρ2 .By using Lemma 1.1.1, computing the topological degree for the coupledsystem (S0) is equivalent to computing the topological degree for the decou-pled system (S1). For system (S1), we are able to compute its topologicaldegree. Therefore, we can get the degree formula for (1.1.14) and the degreecontributed by all the blow-up solutions. Then, we are able to complete theproof of Theorem 1.1.1.Remark: Based on Theorem 1.1.1, a natural question is: what happenswhen ρ1 crosses 8pi? As the case ρ1 crosses 8pi, we can also get a corre-sponding shadow system and the equation is as follows,∆w + 2ρ2( h2ew−4piG(x,p1)−4piG(x,p2)∫M h2ew−4piG(x,p1)−4piG(x,p2)− 1)= 0,∇(log(h1e−12w)(x) + 4piR(x, x) + 8piG(x, p2))|x=p1= 0,∇(log(h1e−12w)(x) + 4piR(x, x) + 8piG(x, p1))|x=p2= 0.(1.1.16)For the equation (1.1.16), if we treat it by a similar way as we did for(1.1.14), i.e., replace −12w by −12w(1− t) for the second and third equationof (1.1.16), we can get a decoupled system by letting t = 1. However, duringthe deformation, we can not provide a uniform estimate for the solution wuntil now. Indeed, when t changes form 0 to 1, the points p2 and p1 maygo to a same point, which may cause blow up phenomena for the solution wand the degree may change after the deformation, which is the most difficultpoint when we consider ρ1 crossing 4mpi for m ≥ 2,m ∈ N.111.2. Lin-Ni Problem1.2 Lin-Ni Problem1.2.1 BackgroundIn the second part of my thesis, I consider the following nonlinear Neumannelliptic problem∆u− µu+ up = 0 in Ω,u > 0 in Ω,∂u∂ν = 0 on ∂Ω,(1.2.1)where 1 < p < +∞, µ > 0, ν denotes the outward unit normal vector of∂Ω, and Ω is a smooth and bounded domain in Rn.Problem (1.2.1) arises in many applied models concerning biological pat-tern formations. For example, it gives rise to steady states in the Keller-Segelmodel of the chemotactic aggregation of the cellular slime molds. Chemo-taxis is the oriented movement of cells in response to chemicals in theirenvironment. Cellular slime molds (amoebae) release a certain chemical,move toward places of its higher concentration, and eventually form aggre-gates. Keller and Segel [33] proposed a model to describe the chemotacticaggregation stage of cellular slime molds. Let u(x, t) denote the populationof amoebae at place x and at time t and v(x, t) be the concentration of thechemical. Then the simplified Keller-Segel system is written as(KS1) ut = D1∆u− χ∇ · (u∇φ(v)) in Ω× (0,+∞),(KS2) vt = D2∆v + k(u, v) in Ω× (0,+∞),(IC) u(x, 0) = u0 > 0, v(x, 0) = v0 > 0 in Ω,(BC) ∂u∂ν = 0 =∂v∂ν on ∂Ω,where D1, D2 and χ are positive constants; φ is so-called sensitivity functionwhich is a smooth function such that φ′(r) > 0 for r > 0; k is a smoothfunction with ku ≥ 0 and kv ≤ 0.We are concerned with stationary solutions to Keller-Segel system in thecase of logarithmic sensitivity φ(v) = ln v and k(u, v) = −av + bu, where aand b are positive constants. Since∫Ω u(x, t)dx =∫Ω u0(x, t)dx for all t > 0by virtue of (KS1) and (BC), we consider the following problem for positive121.2. Lin-Ni Problemfunctions u and v :D1∆u− χ∇ · (u∇ ln(v)) = 0 in Ω,D2∆v − av + bu = 0 in Ω,∂u∂ν = 0 =∂v∂ν on ∂Ω,|Ω|−1∫Ω u(x)dx = u,(1.2.2)where |Ω| denotes the volume of Ω, and u > 0 is a given constant. Obviously,(u, v) = (u, v) with v = a−1bu is a solution.It is not difficult to reduce system (1.2.2) to a single equation. Indeed,we write the first equation as∇ · {D1u∇[lnu− χD−11 ln v]} = 0and using boundary condition we see that u = λvχD1 for some positive con-stant λ. Thus (1.2.2) is equivalent to the following system for (v, λ):D2∆v − av + bλvχD1 = 0 in Ω,∂v∂ν = 0 on ∂Ω,|Ω|−1∫Ω v(x)dx = v.(1.2.3)Now we set p = χD1 , d =D2a , θ = (a−1bλ)1q−1 , and w(x) = θv(x), we haved∆w − w + wp = 0 in Ω,∂w∂ν= 0 on ∂Ω. (1.2.4)By abuse of notation, let w(x) = d1p−1u(x) and µ = 1d , then (1.2.4) turns tobe (1.2.1).Problem (1.2.1) can be also viewed as the steady-state equation for theGierer-Meinhardt system, which is a system of reaction-diffusion equationof the format = d∆a− a+ aphq in Ω× (0,+∞),τht = D∆h− h+ arhs in Ω× (0,∞),∂a∂ν =∂h∂ν = 0 on ∂Ω× (0,∞),(1.2.5)131.2. Lin-Ni Problemwhere d,D, p, q, r, s are all positive constants, s ≥ 0, and0 <p− 1q<rs+ 1.This system was motivated by biological experiments on hydra in morpho-genesis. Hydra, an animal of a few millimeters in length, is made up ofapproximately 100, 000 cells of about fifteen different types. It consists ofa ”head” region located at one end along its length. Typical experimentson hydra involve removing part of the ”head” region and transplanting itto other parts of the body column. Then, a new ”head” will form if andonly if the transplanted area is sufficiently far from the (old) head. Theseobservations have led to the assumption of the existence of two chemicalsubstances: a slowly diffusing (short-range) activator and a rapidly diffus-ing (long-range) inhibitor. Here a(x, t), h(x, t) represent the density of theactivator and inhibitor respectively.For the full Gierer-Meinhardt system (1.2.5), there are a lot of worksconcerning the existence and the stability of the solutions with specific con-figuration. For example, D. Iron, M. Ward, and J. Wei [28] studied thestability of the symmetric k-peaked solutions by using matched asymptoticanalysis. For more results in this direction, one can see [28], [64], [69] andreferences therein.In general, the full (GM) system (1.2.5) is still very difficult to study.A very useful idea, which goes back to Keener [32] and Nishiura [58], isto consider the so-called shadow system. Namely, we let D → +∞ first.Suppose that the quantity −h+ aphq remains bounded, then we obtain∆h→ 0,∂h∂ν= 0 on ∂Ω. (1.2.6)Thus h(x, t) → ξ(t), a constant. To derive the equation for ξ(t), weintegrate both sides of the equation for h over Ω and then we obtain the141.2. Lin-Ni Problemfollowing so-called shadow systemat = d∆a− a+ apξq in Ω,τξt = −ξ + 1|Ω|∫Ωarξs ,a > 0 in Ω and ∂a∂ν = 0 on ∂Ω.(1.2.7)The advantage of shadow system is that by a simple scalinga = ξqp−1w, ξ =( 1|Ω|∫Ωwr) p−1(p−1)(s+1)−qr, (1.2.8)the stationary shadow system can be reduced to a single equationd∆w − w + wp = 0 in Ω,∂w∂ν= 0 on ∂Ω,which is exactly the same form as (1.2.4). By a same transformation as wedid for (1.2.4), we can obtain (1.2.1).Equation (1.2.1) enjoys at least one solution, namely the constant solu-tion u = µ1p−1 . In a series of seminal works, Lin, Ni and Takagi [42] andNi and Takagi [57] initiated quantitative analysis of non-constant solutionto equation (1.2.1). In particular, it is proved in [42],[57] that for µ large,the least energy solution concentrates at the boundary point of maximummean curvature. On the other hand, in the subcritical case 1 < p < 2∗ − 1,blow up arguments and the compactness of embedding imply that for smallpositive µ, the constant solution is the only solution. This uniqueness resultmotivated Lin and Ni to raise the following conjecture, the extension of thisresult to the critical case p = 2∗ − 1.Lin-Ni’s Conjecture [41]. For µ small and p = n+2n−2 , problem (1.2.1) ad-mits only the constant solution.1.2.2 Previous Results On Lin-Ni ProblemIn the following, we recall the the main results towards proving or disprovingLin-Ni’s conjecture. Adimurthi-Yadava [2]-[3] and Budd-Knapp-Peletier [12]151.2. Lin-Ni Problemfirst considered the following problem∆u− µu+ un+2n−2 = 0 in BR(0),u > 0 in BR(0),∂u∂ν = 0 on ∂BR(0).(1.2.9)They proved the following result:Theorem B. ([2–4, 12]) For µ sufficiently small(1) if n = 3 or n ≥ 7, problem (1.2.9) admits only the constant solution,(2) if n = 4, 5, 6, problem (1.2.9) admits a nonconstant solution.The proof of Theorem B relies on the radial symmetry of the domain. Inthe asymmetric case, the complete answer is not known yet, but there are afew results. In the general three-dimension domain case, Zhu [74] provedTheorem C. ([70, 74]) The conjecture is true if n = 3 (p = 5) and Ω isconvex.Zhu’s proof relies on a priori estimate. Later, Wei and Xu [70] gave adirect proof of Theorem C by using a method based only on integration byparts only. In comparison with the strong convexity condition assumed onthe domain, under the assumption on the bound of the energy and a weakerconvexity condition (mean convex domains), Druet, Robert and Wei [23]showed the following result:Theorem D. ([23]) Let Ω be a smooth bounded domain of Rn, n = 3 orn ≥ 7. Assume that H(x) > 0 for all x ∈ ∂Ω, where H(x) is the meancurvature of ∂Ω at x ∈ ∂Ω. Then for all µ > 0, there exists µ0(Ω,Λ) > 0such that for all µ ∈ (0, µ0(Ω,Λ)) and for any u ∈ C2(Ω), we have that−∆u+ µu = n(n− 2)u2∗−1 in Ωu > 0 in Ω∂νu = 0 on ∂Ω∫Ω u2∗dx ≤ Λ⇒ u ≡ (µn(n− 2))n−24 .161.2. Lin-Ni ProblemIt should be mentioned that the assumption of the bounded energy isnecessary in obtaining Theorem D. Without this technical assumption, itwas proved that the solutions to (1.2.9) may accumulate with infinite en-ergy when the mean curvature is negative somewhere (see Wang-Wei-Yan[66]). More precisely, Wang, Wei and Yan gave a negative answer to Lin-Ni’s conjecture in all dimensions (n ≥ 3) for non-convex domain by assumingthat Ω is a smooth and bounded domain satisfying the following conditions:(H1) y ∈ Ω if and only if (y1, y2, y3, · · · ,−yi, · · · , yn) ∈ Ω, ∀i = 3, · · · , n.(H2) If (r, 0, y′′) ∈ Ω, then (r cos θ, r sin θ, y′′) ∈ Ω, ∀θ ∈ (0, 2pi), wherey′′ = (y3, · · · , yn).(H3) Let T := ∂Ω∩{y3 = · · · = yn = 0}, there exists a connected componentΓ of T such that H(x) ≡ γ < 0, ∀x ∈ Γ.Theorem E. ([66]) Suppose n ≥ 3, q = n+2n−2 and Ω is a bounded smoothdomain satisfying (H1)-(H3). Let µ be any fixed positive number. Thenproblem (1.2.9) has infinitely many positive solutions, whose energy can bemade arbitrarily large.Wang, Wei and Yan [67] also gave a negative answer to Lin-Ni’s conjec-ture in some convex domain including the balls for n ≥ 4.Theorem F. ([67]) Suppose n ≥ 4, q = n+2n−2 and Ω satisfies (H1)-(H2). Letµ be a any fixed positive number. Then problem (1.2.9) has infinitely manypositive solutions, whose energy can be made arbitrarily large.Theorem B-F reveal that Lin-Ni’s conjecture depends very sensitivelynot only on the dimensions, but also on the shape of the domain. A naturalquestion is: what about a general domain? Inspired by the result of TheoremB, we expect to give a negative answer to the case n = 4, 5, 6. The onlyapproach in this direction is given by Rey and Wei [63]. They disproved theconjecture in the five-dimensional case by establishing a nontrivial solutionwhich blows up at K interior points in Ω provided µ is sufficiently small.The second part of my thesis is to establish a result similar to (2) of TheoremB in general four, and six-dimensional domains by establishing a nontrivial171.2. Lin-Ni Problemsolution which blows up at a single point in Ω provided µ is sufficientlysmall. Namely, we consider the problem∆u− µu+ un+2n−2 = 0 in Ω, u > 0 in Ω,∂u∂ν= 0 on ∂Ω, (1.2.10)where n = 4, 6 and Ω is a smooth bounded domain in Rn and µ > 0 verysmall. Our main result is stated as followsTheorem 1.2.1. For problem (1.2.10) in n = 4, 6, there exists µ0 > 0 suchthat for all 0 < µ < µ0, equation (1.2.10) possesses a nontrivial solution.1.2.3 Sketch Of The Proof Of Theorem 1.2.1Our proof use the localized energy method, which was introduced in [27] and[52] in dealing with spikes. This method usually consists of five steps.Step 1. Find out good approximate solutions.We setΩε := Ω/ε = {z|εz ∈ Ω},andµ =( c1− ln ε) 12 , n = 4,ε , n = 6,(1.2.11)where c1 is some constant that depends on the domain only, to be determinedlater.For the reason of normalization, we consider the following equation∆u− µu+ n(n− 2)un+2n−2 = 0, u > 0 in Ω,∂u∂ν= 0 on ∂Ω. (1.2.12)Through the transformation u(x) 7−→ ε−n−22 u(x/ε), (1.2.12) yields the re-181.2. Lin-Ni Problemscaled problem written as∆u− µε2u+ n(n− 2)un+2n−2 = 0, u > 0 in Ωε,∂u∂ν= 0 on Ωε. (1.2.13)For any Q ∈ Ω, we use UΛ,Qεas an approximate solution of (1.2.13),whereUΛ,Q =( ΛΛ2 + |x−Q|2)n−22, Λ > 0, Q ∈ Rn.Because of the appearance of the additional term µε2u, we need to add anextra term to get a better approximation. To this end, for n = 4, we considerthe following equation∆Ψ¯ + U1,0 = 0 in R4, Ψ¯(0) = 1.LetΨΛ,Q =Λ2ln1Λε+ ΛΨ¯(y −QΛ).Then∆ΨΛ,Q + UΛ,Q = 0.For n = 6, we denote Ψ(|y|) as the radial solution of∆Ψ + U1,0 = 0 in R6, Ψ→ 0 as |y| → +∞.We set Ψ(Λ, Q)(y) = Ψ(y−QΛ ), then∆ΨΛ,Q(y) + UΛ,Q = 0 in R6.In order to match the boundary condition, we need an extra correctionterm. For this purpose, we first introduce the Green’s function∆xG(x,Q) + δQ −1|Ω|= 0 in Ω,∂G∂ν= 0 on ∂Ω,∫ΩG(x,Q) = 0.191.2. Lin-Ni ProblemWe decomposeG(x,Q) = K(|x−Q|)−H(x,Q), K(r) =1cnrn−2, cn = (n− 2)|Sn|.We defineUˆΛ,Qε(z) = −ΨΛ,Qε(z)− cnµ−1εn−4Λn−22 H(εz,Q) +Rε,Λ,Q(z)χ(εz),where Rε,Λ,Q is defined by ∆Rε,Λ,Q − ε2Rε,Λ,Q = 0 in Ωε andµε2∂Rε,Λ,Q∂ν= −∂∂ν[UQε− µε2ΨQε− cnεn−2Λn−22 H(εz,Q)]on ∂Ωε.Here χ(x) is a smooth cut-off function in Ω such that χ(x) = 1 for d(x, ∂Ω) <δ/4 and χ(x) = 0 for d(x, ∂Ω) > δ/2.Since in n = 4 and 6, the constructions of the approximate solutions aredifferent, we shall treat them differently in the following. For n = 4, letΛ4,1 ≤ Λ ≤ Λ4,2, Q ∈Mδ4 := {x ∈ Ω| d(x, ∂Ω) > δ4}, (1.2.14)where Λ4,1 and Λ4,2 are constants depend on the domain and δ4 is a smallconstant, to be determined in Chapter 3. We writeQ¯ =1εQ,and define our approximate solutions asWε,Λ,Q = UΛ,Q¯ + µε2UˆΛ,Q¯ +c4Λ|Ω|µ−1ε2. (1.2.15)201.2. Lin-Ni ProblemFor n = 6, let√|Ω|c6(196− Λ6ε23 ) ≤ Λ ≤√|Ω|c6(196+ Λ6ε23 ),Q ∈Mδ6 := {x ∈ Ω| d(x, ∂Ω) > δ6},148− η6ε13 ≤ η ≤148+ η6ε13 , (1.2.16)where Λ6 and η6 are some constants that may depend on the domain, δ6is a small constant, which are also given in Chapter 3. Our approximatesolution for n = 6 is the followingWε,Λ,Q,η = UΛ,Q¯ + µε2UˆΛ,Q¯ + ηµ−1ε4. (1.2.17)Step 2. A-priori estimate for a linear problem.This is the most important step in reducing an infinite-dimensional prob-lem to finite dimensional one. The key result we need is the non-degeneracyof the following solution u:∆u+ un+2n−2 = 0, u > 0 in Rn, u(y)→ 0 as |y| → ∞.Using this non-degeneracy condition, we can show the solvability of thefollowing linearized problem in suitable function space for n = 4,−∆φ+ µε2φ− 24W 2φ = h+ Σ4i=0ciZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 4,where Z0 = −∆∂W∂Λ + µε2 ∂W∂Λ ,Zi = −∆ ∂W∂Q¯i + µε2 ∂W∂Q¯i, 1 ≤ i ≤ 4.211.2. Lin-Ni Problemand for n = 6,−∆φ+ µε2φ− 48Wφ = h+ Σ7i=0diZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 7,where Z0 = −∆∂W∂Λ + µε2 ∂W∂Λ ,Zi = −∆ ∂W∂Q¯i + µε2 ∂W∂Q¯i, 1 ≤ i ≤ 6,Z7 = −∆∂W∂η + µε2 ∂W∂η .Step 3. The solvability of the nonlinear problem.Using Wε,Λ,Q as the approximate solution for n = 4, Wε,Λ,Q,η as theapproximate solution for n = 6 and the result of Step 2, we can find thatthere exists a constant ε0 such that for all ε ≤ ε0, there is φ such that−∆(W + φ) + µε2(W + φ)− 8(W + φ)3 =∑i ciZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 4for n = 4, and−∆(W + φ) + µε2(W + φ)− 24(W + φ)2 =∑i diZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 7for n = 6.Step 4. The reduction lemma.We defineIε(Λ, Q¯) ≡ Jε[Wε,Λ,Q + φε,Λ,Q]221.3. Organization Of The Thesisfor n = 4 andIε(Λ, η, Q¯) ≡ Jε[Wε,Λ,Q,η + φε,Λ,Q,η]for n = 6.The reduction lemma says that if (Λ, Q¯) and (Λ, η, Q¯) are the criticalpoints of Iε for n = 4 and n = 6 respectively, then u = Wε,Λ,Q + φε,Λ,Q¯ andu = Wε,Λ,Q,η + φε,Λ,η,Q¯ are solution to problem (1.2.13) for n = 4 and n = 6respectively.Step 5. Finding the critical points of Iε for n = 4 and n = 6 respectively.For n = 4, we find the maximal value of Iε in (1.2.14), and for n = 6, wefind the min-max value of Iε in (1.2.16).Once we find the critical points of Iε for n = 4 and n = 6. Using Step4, we get a solution to (1.2.13). Furthermore, it is obvious that the solutionwe construct is nontrivial. Hence, we get Theorem 1.2.1, thereby disprovingthe Lin-Ni’s conjecture in n = 4 and n = 6 for a general domain.1.3 Organization Of The ThesisIn Chapter 2, we prove Theorem 1.1.1. The proof of Proposition 1.1.1 ,Proposition 1.1.2 and the existence of smooth positive function h1 and h2such that any solution of the shadow system (1.1.14) is non-degenerate aregiven in Section 2.1. In Section 2.2, we get the a-priori estimate for solutionsof (1.1.10) when ρ1 → 4pi and ρ2 /∈ 4piN. In Section 2.3 and Section 2.4, weuse the solutions of the shadow system (1.1.14) to get a good approximationof some bubbling solutions of (1.1.10) and thereby prove Proposition 1.1.3except for some important estimates which are shown in Section 2.6. InSection 2.5, we derive the degree formula for the shadow system (1.1.14)and prove Theorem 1.1.1.Chapter 3 is devoted to the proof of Theorem 1.2.1. In Section 3.1, we231.3. Organization Of The Thesisconstruct suitable approximated bubble solution W , and list their propertiesand some important estimates with the proof given in Section 3.6. In Section3.2, we solve the linearized problem at W in a finite-codimensional space.Then, in Section 3.3, we are able to solve the nonlinear problem in thatspace. In Section 3.4, we study the remaining finite-dimensional problemand solve it in Section 3.5 by finding critical points of the reduced energyfunctional.24Chapter 2The SU(3) Toda System2.1 Proof Of Proposition 1.1.1, Proposition 1.1.2And Shadow SystemWe shall prove Proposition 1.1.1 and Proposition 1.1.2 in this section. Fora sequence of bubbling solutions (u1k, u2k) of (1.1.8), we setu˜ik = uik −∫Mhieuik , i = 1, 2.Then u˜ik satisfy{∆u˜1k + 2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1) = 0,∆u˜2k − ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1) = 0.(2.1.1)We define the blow up set for u˜ik as followsSi = {p ∈M | ∃{xk}, xk → p, u˜ik(xk)→ +∞} (2.1.2)and S = S1 ∪S2. We note thatuik = u˜ik +∫Mhieuik ≥ u˜ik + Ce∫M uik ≥ u˜ik + C,where we used the Jensen’s inequality and hi (here hi = h∗i ) is a positivefunction in M. So, if p is a blow up point of u˜ik, then p is also a blow uppoint of uik. For any p ∈ S, we define the local mass byσip = limδ→0limk→+∞12pi∫Bδ(p)ρihieu˜ik . (2.1.3)252.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemLemma 2.1.1. If σ1p, σ2p < 13 , we have p /∈ S.Proof. Since σip < 13 , we can choose small r0, such that in Br0(p), thefollowing holds ∫Br0 (p)ρihieu˜ik < pi, (2.1.4)which implies∫Br0 (p)u˜+ik ≤ C, where C is some constant independent of k.In the following, C always denotes some generic constant independent of k,and may depend on the domain Br0(p). For the first equation in (2.1.1), wedecompose u˜1k =∑3j=1 u˜1k,j , where u˜1k,j satisfy the following equation−∆u˜1k,1 = 2ρ1h1eu˜1k − ρ2h2eu˜2k in Br0(p), u˜1k,1 = 0 on ∂Br0(p),−∆u˜1k,2 = −2ρ1 + ρ2 in Br0(p), u˜1k,2 = 0 on ∂Br0(p),−∆u˜1k,3 = 0 in Br0(p), u˜1k,3 = u˜1k on ∂Br0(p).(2.1.5)For the first equation in (2.1.5), since∫Br0 (p)∣∣∣2ρ1h1eu˜1k − ρ2h2eu˜2k∣∣∣ < 3pi,by [11, Theorem 1], we have∫Br0 (p)exp((1 + δ)|u˜1k,1|)dx ≤ C, (2.1.6)where δ ∈ (0, 13). Therefore, we have∫Br0 (p)|u˜1k,1| ≤ C. (2.1.7)For the second equation in (2.1.5), we can easily get∫Br0 (p)|u˜1k,2| ≤ C, and |u˜1k,2| ≤ C. (2.1.8)For the third equation in (2.1.5). By the mean value theorem for harmonic262.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow Systemfunction we have‖u˜+1k,3‖L∞(B r02(p)) ≤ C‖u˜+1k,3‖L1(Br0 (p))≤ C[‖u˜+1k‖L1(Br0 (p)) + ‖u˜1k,1‖L1(Br0 (p)) + ‖u˜1k,2‖L1(Br0 (p))]≤ C. (2.1.9)From (2.1.8)-(2.1.9), we have2ρ1h1eu˜1k,2+u˜1k,3 ≤ C in B r02(p). (2.1.10)By (2.1.6), (2.1.10) and Ho¨lder inequality, we obtaineu˜1k ∈ L1+δ1(Br0(p))with δ1 > 0 independent of k. Similarly, we haveeu˜2k ∈ L1+δ2(Br0(p))with δ2 > 0 independent of k. By using the standard elliptic estimate for thefirst equation in (2.1.5), we get ‖u˜1k,1‖L∞(Br0/2(p)) is uniformly bounded.Combined with (2.1.8) and (2.1.9), we have u˜1k is uniformly bounded abovein B r02(p). Following a same process, we can also obtain u˜2k is uniformlybounded above in B r02(p). Hence, we finish the proof of the lemma.From Lemma 2.1.1, we get if p ∈ S, either σ1p ≥ 13 or σ2p ≥13 , whichimplies |S| < ∞ and S is discrete in M. In fact, in next lemma, we shallprove that if p ∈ Si, σip must be positive.Lemma 2.1.2. If p ∈ Si, σip > 0.Proof. We prove it by contradiction. Without loss of generality, we mayassume p ∈ S2 and σ2p = 0. First, we claim that there is a constant CK > 0that depends on the compact set K such that|uik(x)| ≤ CK , ∀x ∈ K ⊂⊂M \S, i = 1, 2. (2.1.11)272.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemWe only prove for i = 1, the other one can be obtained similarlyu1k(x) =∫MG(x, z)(2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1))=∫M1G(x, z)(2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1))+∫M\M1G(x, z)(2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1)),where M1 = ∪p∈SBr0(p) and r0 is small enough to make K ⊂⊂M \M1. Itis easy to see that∫M1G(x, z)(2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1))= O(1),because G(x, z) is bounded due to the distance d(x, z) ≥ δ0 > 0 for z ∈M1,and x ∈ K. In M \M1, we can see that u˜ik are bounded above by someconstant depends on r0, then it is not difficult to obtain that∫M\M1G(x, z)(2ρ1(h1eu˜1k − 1)− ρ2(h2eu˜2k − 1))= O(1).Therefore, we prove the claim.Since σ2p = 0, we can find some r0, such that∫Br0 (p)ρ2h2eu˜2k ≤ pi (2.1.12)for all k (passing to a subsequence if necessary) and r0 ≤ 12d(p,S\{p}). On∂Br0(p), by (2.1.11)|u1k|, |u2k| ≤ C on ∂Br0(p). (2.1.13)Let wk satisfy the following equation{∆wk = ρ1(h1eu1k∫M h1eu1k− 1)in Br0(p),wk = u1k on ∂Br0(p).(2.1.14)282.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemWe set wk = wk1 + wk2 where wk1, wk2 satisfy{∆wk1 = ρ1h1eu1k∫M h1eu1kin Br0(p), wk1 = u1k on ∂Br0(p),∆wk2 = −ρ1 in Br0(p), wk1 = 0 on ∂Br0(p).(2.1.15)By maximum principle and (2.1.13), we have wk1 ≤ max∂Br0 (p) u1k ≤ C forx ∈ Br0(p). By elliptic estimate, we can easily get |wk2| ≤ C. Therefore,wk ≤ C, ∀x ∈ Br0(p). (2.1.16)We set u2k = fk1 + fk2 + wk, where fk1 and fk2 satisfy{∆fk1 = −2ρ2h2eu2k∫M h2eu2kin Br0(p), fk1 = 0 on ∂Br0(p),∆fk2 = 2ρ2 in Br0(p), fk2 = u2k − wk on ∂Br0(p).(2.1.17)For the second equation in (2.1.17), we have|fk2| ≤ |u2k|+ |wk| = |u2k|+ |u1k| ≤ C on ∂Br0(p).Thus |fk2| ≤ C in Br0(p). We denote gk = efk2+wk , and the first equation in(2.1.17) can be written as∆fk1 + 2ρ2h2egk∫M h2eu2kefk1 = 0 in Br0(p), fk1 = 0 on ∂Br0(p). (2.1.18)By using the Jensen’s inequality, we have∫M h2eu2k ≥ Ce∫M u2k ≥ C > 0.We set Vk = 2ρ2h2egk∫M h2eu2k, and have Vk ≤ C, where C depends on r0. Using(2.1.12), we get∫Br0 (p)Vkefk1 ≤ 2pi. By [11, Corollary 3], we have |fk1| ≤ Candu2k ≤ fk1 + fk2 + wk ≤ C.This leads to u˜2k = u2k −∫M h2eu2k ≤ C, which contradicts to the assump-tion u˜2k blows up at p. Thus we finish the proof of this lemma.By these two lemmas, we now begin to prove Proposition 1.1.1.292.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemProof of Proposition 1.1.1. We note that it is enough for us to prove u˜ik isuniformly bounded above. We shall prove it by contradiction.First, we claim S1 6= ∅. If not, u˜1k is uniformly bounded above and u˜2kblows up. We decompose u2k = u2k,1 + u2k,2, where u2k,1 and u2k,2 satisfythe following∆u2k,1 − ρ1(h1u˜1k − 1) = 0,∫M u2k,1 = 0,∆u2k,2 + 2ρ2(h˜2keu2k,2∫M h˜2keu2k,2− 1) = 0,∫M u2k,2 = 0,where h˜2k = h2eu2k,1 . By the Lp estimate, u2k,1 is bounded in W 2,p forany p > 1. Thus u2k,1 is bounded in C1,α for any α ∈ (0, 1), after passingto a subsequence if necessary, we gain u2k,1 converges to u0 in C1,α. Asa consequence, h˜2k → h2eu0 in C1,α. Since u˜2k blows up, u2k and u2k,2both blow up. Then applying the result of Li and Shafrir in [38], we haveρ2 ∈ 4piN, which contradicts to our assumption. Thus S1 6= ∅. Similarly,we can prove that S2 6= ∅.We note that our argument above can be applied to the local case, whichyields S1∩S2 6= ∅. Suppose S1∩S2 = ∅. For any point p ∈ S2, we considerthe behavior of u1k and u2k in Br0(p), where r0 is small enough such thatBr0(p) ∩ (S \ {p}) = ∅. We decompose u˜2k = u2k,3 + u2k,4, where u2k,3 andu2k,4 satisfy{∆u2k,3 − ρ1(h1eu˜1k − 1) = 0 in Br0(p), u2k,3 = 0 on ∂Br0(p),∆u2k,4 + 2ρ2(h˜2,keu2k,4 − 1) = 0 in Br0(p), u2k,4 = u˜2k on ∂Br0(p),(2.1.19)where h˜2,k = h2eu2k,3 . By using u˜1k uniformly bounded from above in Br0(p),we have h˜2,k converges in C1,α(Br0(p)). Since u2k,4 blows up simply at p, wehave∣∣∣u2k,4 − log( eu2k,4(p(k))(1 + ρ2h˜2,k(p(k))eu2k,4(p(k))4 |x− p(k)|2)2)∣∣∣ ≤ C, (2.1.20)where u2k,4(p(k)) = maxBr0 (p) u2k,4. (2.1.20) is proved in [5] and [37]. From302.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow System(2.1.20), we haveu2k,4 → −∞ in Br0(p) \ {p} and ρ2h2eu˜2k → 4piδp in Br0(p), (2.1.21)which impliesρ2 = limk→∞∫Mρ2h2eu˜2k = 4pi|S2|, (2.1.22)a contradiction to our assumption ρ2 /∈ 4piN, so S1 ∩S2 6= ∅.Let p ∈ S1 ∩ S2, and σip, i = 1, 2 be the local masses of them at p.Applying the result of Jost-Lin-Wang (Proposition 2.4 in [29]), we have(σ1p, σ2p) is one of (2, 4), (4, 2) and (4, 4).In the following, we claim if σip = 4, then u˜ik concentrate, i.e., u˜ik → −∞uniformly in any compact set of Br0(p) \ {p}. This implies∫M hieuik → +∞and u˜ik → −∞ uniformly in any compact set of M \Si. Then,ρihieu˜ik → αq∑q∈Si\{p}δq + 8piδp with αq = 4pi or 8pi, (2.1.23)which implies ρi = 4piN and again yields a contradiction. This completesthe proof of Proposition 1.1.1. The proof of the claim is given in Lemma2.1.3 below. Lemma 2.1.3. Suppose u˜ik, i = 1, 2 both blow up at p and let 4 be the localmass of u˜ik as before. Then u˜ik → −∞ in Br0(p) \ {p}.Proof. If the claim is not true, we have u˜ik is bounded by some constant C inL∞(∂Br0(p)). Without loss of generality, we assume i = 2. Then σ1p = 0, 2or 4. In fact, the proof of these three cases are the same, so, we only givethe proof of the case σ1p = 4. Let f1k = −ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1)and zk be the solution of{−∆zk = f1k in Br0(p),zk = −C on ∂Br0(p).(2.1.24)Note that f1k → f1 uniformly in any compact set of Br0(p) \ {p} and theintegration of the right hand side of (2.1.24) over Br0(p) is 8pi + o(1) as312.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow Systemr0 → 0. By maximum principle, u˜2k ≥ zk in Br0(p). In particular∫Br0 (p)ezk ≤∫Br0 (p)eu˜2k <∞.On the other hand, using Green representation formula for zk, we havezk(x) = −∫Br0 (p)12piln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))+O(1),(2.1.25)where we used the regular part of the Green function is bounded. For anyx ∈ Br0(p) \ {p}, we denote the distance between x and p by 2r. From(2.1.25), we havezk(x) =−∫Br0 (p)12piln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))+O(1)=−∫Br0 (p)∩Br(x)12piln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))−∫Br0 (p)\Br(x)12piln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))+O(1).It is easy to see∣∣∣∫Br0 (p)∩Br(x)ln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))∣∣∣ ≤ C,where we used the fact that u˜ik is uniformly bounded above in Br(x), i = 1, 2and C depends only on x. For y ∈ Br0(p) \Br(x), we have |x− y| ≥ r and∫Br0 (p)\Br(x)ln |x− y|(− ρ1(h1eu˜1k − 1) + 2ρ2(h2eu˜2k − 1))= (8pi + o(1)) ln |x− p|+O(1).Therefore, we get zk(x) is uniformly bounded below by some constant thatdepends on x only. Thus, we have zk → z in C2loc(Br0(p) \ {p}), where z322.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow Systemsatisfies{−∆z = f1 in Br0(p) \ {p},z = −C on ∂Br0(P ).For ϕ ∈ C∞0 (Br0(p)),− limk→+∞∫Br0 (p)ϕ∆zk =−∫Br0 (p)(ϕ(x)− ϕ(p))∆zk + ϕ(p)(∫Br0 (p)f1 + 8pi)=∫Br0 (p)ϕ(x)f1 + 8piϕ(p).Thus −∆z = f1 + 8piδp. Therefore, we have z(x) ≥ 4 log 1|x−p| + O(1) asx→ p, which implies∫Br0 (p)ez =∞, a contradiction. Henceu˜2k → −∞ in Br0(p) \ {p}. (2.1.26)Thus, Lemma 2.1.3 holds.Next, we prove Proposition 1.1.2 and derive the shadow system (1.1.14).Proof of Proposition 1.1.2. As ρ1k → 4pi, ρ2k → ρ2 and ρ2 /∈ 4piN, we con-sider a sequence of solutions (v1k, v2k) to (1.1.10) such that maxM (v1k, v2k)→+∞. We claim maxM (u˜1k, u˜2k) → +∞. Otherwise, u˜1k, u˜2k are uniformlybounded above. From Green representation theorem and Lp estimate, wecan get u1k, u2k are uniformly bounded. This implies v1k, v2k are uniformlybounded, which contradicts to our assumption. Let Si denote the blow uppoint of u˜ik, i = 1, 2 as before.We claim S2 = ∅ and S1 consists of one point only. Suppose firstS2 6= ∅. From the proof of Proposition 1.1.1, if S1 ∩ S2 = ∅, then u˜2kwould concentrate, i.e., u˜2k → −∞, ∀x ∈ M \ S2, which implies ρ2 =limk→+∞∫M h2eu˜2k ∈ 4piN, a contradiction. Thus S1 ∩ S2 6= ∅. Supposeq ∈ S1 ∩ S2, from Proposition 2.4 in [29] and the condition ρ1k < 8pi, weconclude σ1q = 2 and σ2q = 4. By Lemma 2.1.3, we have u˜2k concentrate,which implies ρ2 ∈ 4piN, a contradiction again. Hence S2 = ∅. By Lemma332.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow System2.1.2, u˜2k is uniformly bounded from above in M. Since maxM (u˜1k, u˜2k)→+∞, we get S1 6= ∅. By the fact ρ1k → 4pi, we have S1 contains only onepoint.We write the equation for vik, i = 1, 2 as∆v1k + ρ1k(h1e2v1k−v2k∫M h1e2v1k−v2k− 1)= 0,∆v2k + ρ2k(h2e2v2k−v1k∫M h2e2v2k−v1k− 1)= 0.(2.1.27)Since u˜2k is uniformly bounded above, the second equation of (2.1.27) impliesthat v2k is uniformly bounded in M and converges to some function 12w inC1,α(M). From the first equation of (2.1.27) and ρ1k → 4pi, v1k blows up atonly one point, say p ∈M.We write the first equation in (2.1.27) as∆v1k + ρ1k( h˜ke2v1k∫M h˜ke2v1k− 1)= 0, (2.1.28)where h˜k = h1e−v2k . We define v˜1k = v1k− 12 log∫M h˜ke2v1k . Due to the C1,αconvergence of h˜k, v˜1k simply blows up at p by a result of Li [37] (one canalso see [5]), i.e., the following inequality holds:∣∣∣2v˜1k − logeλk(1 + ρ1kh˜k(p(k))eλk4 |x− p(k)|2)2∣∣∣ < c for |x− p(k)| < r0, (2.1.29)where λk = 2v˜1k(p(k)) = maxx∈Br0 (p) 2v˜1k. By using this sharp estimate, wegetv˜1k → −∞ in M \ {p}, ρ1kh1e2v1k−v2k∫M h1e2v1k−v2k→ 4piδp, (2.1.30)and∇(log(h1e− 12w) + 4piR(x, x))|x=p= 0, (2.1.31)which proves (1.1.11) and (1.1.12).342.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemIn the following, we claim v2k → 12w in C2,α(M). From this claim and(2.1.28), it is easy to getv1k → 8piG(x, p) in C2,α(M \ {p}).Combined with v2k → 12w in C2,α(M), we have w satisfies the followingequation∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p)− 1)= 0. (2.1.32)This proves (1.1.13). Therefore, we finish the proof of Proposition 1.1.2.The proof of the claim is given in the following Lemma 2.1.4. Lemma 2.1.4. Let v1k, v2k be a sequence of blow up solutions of (2.1.27),which v1k blows at p and v2k → 12w in C1,α(M). Then v2k → 12w inC2,α(M).Proof. By (2.1.29), we have|λk − log∫Mh˜ke2v1k | < c. (2.1.33)To prove v2k → 12w in C2,α, we need the following estimate∣∣∣2∇v˜1k −∇(logeλk(1 + ρ1kh(p)eλk4 |x− p|2)2)∣∣∣ < c for |x− p| < r0, (2.1.34)where (2.1.34) comes from the error estimate of [14, Lemma 4.1]. We writeh2e2v2k−v1k = h2e−v1ke2v2k .By (2.1.29) and (2.1.34), it is not difficult to show∇(h2e−v1k)∈ L∞(M).Therefore, by classical elliptic regularity theory and Sobolev inequality, we352.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow Systemcan show thatv2k →12w in C2,α for any α ∈ (0, 1). (2.1.35)Then we finished the proof of this lemma.After deriving the shadow system (1.1.14), we show the non-degeneracyof (1.1.14) by applying the well-known transversality theorem, which can befound in [1], [61] and references therein. First, we recall thatTheorem 2.1.1. Let F : H × B → E be a Ck map. H, B and E Ba-nach manifolds with H and E separable. If 0 is a regular value of F andFb = F (·, b) is a Fredholm map of index < k, then the set {b ∈ B :0 is a regular value of Fb} is residual in B.We say y ∈ E is a regular value if every point x ∈ F−1(y) is a regularpoint, where x ∈ H×B is a regular point of F if DxF : Tx(H×B)→ TF (x)Eis onto. We say a set A is a residual set if A is a countable intersection ofopen dense sets in B, see [1], which implies A is dense in B (B is a Banachspace), see [31].Following the notations in Theorem 2.1.1, we denoteH = M × W˚ 2,p(M), B = C2,α(M)× C2,α(M), E = R2 × W˚ 0,p(M),whereW˚ 2,p(M) := {f ∈W 2,p |∫Mf = 0}, W˚ 0,p(M) := {f ∈ Lp |∫Mf = 0},andC2,α(M) = {f ∈ C2,α(M)}.We consider the mapT (w, p, h1, h2) =[∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 1)∇ log(h1e−12w + 4piR(x, x))(p)]. (2.1.36)362.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemClearly, T is C1. Next, we claim(i) T (·, ·, h1, h2) is a Fredholm map of index 0,(ii) 0 is a regular value of T.For the first claim, after computation, we getT ′w,p(w, p, h1, h2)[φ, ν] =[T0(w, p, h1, h2)[φ, ν]T1(w, p, h1, h2)[φ, ν]], (2.1.37)whereT0(w, p, h1, h2)[φ, ν] =∆φ+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)φ− 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)φ− 8piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)∇G(x, p) · ν+ 8piρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)∇G(x, p) · ν,T1(w, p, h1, h2)[φ, ν] =∇2x(log h1e− 12w + 4piR(x, x))|x=p ·ν −12∇φ(p).We decomposeT ′w,p[φ, ν] =[T01T11][φ, ν] +[T02T12][φ, ν], (2.1.38)whereT11 = 0, T12 = T1,T01(w, p, h1, h2)[φ, ν] =∆φ+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)φ− 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)φ,372.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemandT02(w, p, h1, h2)[φ, ν] =− 8piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)∇G(x, p)ν+ 8piρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)∇G(x, p)ν.We define T1 =[T01T11]and T2 =[T02T12]. We can easily see that T1is symmetric, it follows from the basic theory of elliptic operators that T1is a Fredholm operator of index 0. Combining the Sobolev inequality andR2 is a finite Euclidean space, we can show that T2 is a compact operator.Therefore, by the standard linear operator theory [31], we get T1 +T2 is alsoa Fredholm linear operator with index 0. Hence, we prove the first claimthat T is a Fredholm map with index 0.It remains to show that 0 is a regular value. We derive the differentiationof the operator T with respect to h1 and h2,T ′h1(w, p, h1, h2)[H1] =[0∇H1h1(p)− ∇h1(h1)2H1(p)],andT ′h2(w, p, h1, h2)[H2]=[2ρ2H2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M H2ew−4piG(x,p)0].By choosing ν = 0, and H1 such that∇H1h1− ∇h1(h1)2H1 =12∇φ at p. We getT ′w,p(w, p, h1, h2)[φ, ν] + T′h1(w, p, h1, h2)[H1]=[∆φ+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)φ− 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M h2ew−4piG(x,p)φ0].Next, we claim that the vector space spanned by T ′w,p(w, p, h1, h2)[φ, ν],382.1. Proof Of Proposition 1.1.1, Proposition 1.1.2 And Shadow SystemT ′h1(w, p, h1, h2)[H1] and T′h2(w, p, h1, h2)[H2] containsf0...0for all f ∈W˚ 0,p. It is enough for us to prove that only φ = 0 can satisfyφ ∈ Ker{∆·+2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)·−2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)·}and〈φ, 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)H2h2−2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)H2h2〉= 0,for all H2 ∈ C2,α(M). We setL = ∆ ·+2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)·−2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p) · .Using φ ∈ Ker(L), we obtain that for any H2 ∈W 0,p(M),∫ML(φ) ·H2 = 0. (2.1.39)Since C2,α(M) is dense in W 0,p(M) and〈φ, 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)H2−2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫Mh2ew−4piG(x,p)H2〉= 0,we deduce ∫M∆φ ·H2 = 0, ∀ H2 ∈W0,p(M). (2.1.40)Thus∆φ = 0 in M,∫Mφ = 0. (2.1.41)So φ ≡ 0. Therefore the claim is proved.392.2. A-priori EstimateOn the other hand, we choose two functions, H1,1 and H1,2 such that∇H1,1h1(p)−∇h1(h1)2H1,1(p) = (1, 0),and∇H1,2h1(p)−∇h1(h1)2H1,2(p) = (0, 1).Then it is not difficult to see that (by setting φ = 0, ν = 0)[0c]⊂ DT (w, p, h1, h2)[φ, ν]for all c ∈ R2. Therefore, we have proved that the differential map is onto.As a consequence, 0 is a regular point of T. By Theorem 2.1.1,{(h1, h2) ∈ B : 0 is a regular value of T (·, ·, h1, h2)}is residual in B. Since T (w, p, h1, h2) is a Fredholm map of index 0 for fixedh1, h2, we have{(h1, h2) ∈ B : the solution (w, p) of T (·, ·, h1, h2) = 0 is nondegenerate}is residual in B. Thus, we can choose h1, h2 > 0 such that the solution of(1.1.14) is non-degenerate.2.2 A-priori EstimateIn this section, we shall prove that all the blow up solutions of (1.1.10) mustbe contained in the set Sρ1(p, w)×Sρ2(p, w) when ρ1 → 4pi, ρ2 /∈ 4piN, wherethe definition of Sρi(p, w), i = 1, 2 are given in (2.2.14) and (2.2.15) of thissection.To simplify our description, we may assume M has a flat metric near aneighborhood of each blow up point. Of course we can modify our arguments402.2. A-priori Estimatewithout any difficulty for the general case, as in [15].We start to define the set Sρi(p, w). For any given non-degenerate solu-tion (p, w) of (1.1.14), we set (by abuse of the notation)h = h1e− 12w. (2.2.1)By noting that∇x(log h+ 4piR(x, x))|x=p= ∇x(log h(x) + 8piR(x, p))|x=p= 0, (2.2.2)whenever (p, w) is a solution of shadow system (1.1.14). For q such that|q − p|  1 and large λ > 0, we setU(x) = λ− 2 log(1 +ρ1h(q)4eλ|x− q|2), (2.2.3)and U(x) satisfies the following equation∆U(x) + 2ρ1h(q)eU = 0 in R2, U(q) = maxR2U(x) = λ. (2.2.4)LetH(x) = exp{logh(x)h(q)+ 8piR(x, q)− 8piR(q, q)}− 1, (2.2.5)ands = λ+ 2 log(ρ1h(q)4)+ 8piR(q, q) +∆H(q)ρ1h(q)λ2eλ. (2.2.6)Let σ0(t) be a cut-off function:σ0(t) ={1, if |t| < r0,0, if |t| ≥ 2r0.Set σ(x) = σ0(|x− q|) andJ(x) ={ (H(x)−∇H(q) · (x− q))σ, x ∈ B2r0(q),0, x /∈ B2r0(q).412.2. A-priori EstimateLet η(x) satisfy{∆η + 2ρ1h(q)eU (η + J(x)) = 0 on R2,η(q) = 0,∇η(q) = 0.(2.2.7)The existence of η was proved in [15]. Furthermore, we have the followinglemmaLemma 2.2.1. Let R =√ρ1h(q)4 eλ. For h ∈ C2,α(M) and large λ. thereexists a solution η satisfying (2.2.7) and the following(i) η(x) = −4∆H(q)ρ1h(q) e−λ[log(R|x− q|+ 2)]2 +O(λe−λ) on B2r0(q),(ii) η,∇xη, ∂qη, ∂λη,∇x∂qη,∇x∂λη = O(λ2e−λ) on B2r0(q).The proof of Lemma 2.2.1 was given in [15].We setvq(x) =(U(x) + η(x) + 8pi(R(x, q)−R(q, q)) + s)σ(x)+8piG(x, q)(1− σ(x)),vq = 1|M |∫M vq,vq,λ,a = a(vq − vq).(2.2.8)Next, we define O(1)q,λ and O(2)q,λ:O(1)q,λ ={φ ∈ H˚1(M)∣∣∣∫M∇φ · ∇vq =∫M∇φ · ∇∂qvq =∫M∇φ · ∇∂λvq = 0},(2.2.9)andO(2)q,λ ={ψ ∈W 2,p(M)∣∣∣∫ψ = 0}, p > 2. (2.2.10)For each (q, λ), we sett = λ+ 8piR(q, q) + 2 logρ1h(q)4+∆H(q)ρ1h(q)λ2e−λ − vq. (2.2.11)422.2. A-priori EstimateFor ρ1 6= 4pi, we define λ(ρ1) such thatρ1 − 4pi =∆ log h(p) + 8pi − 2K(p)h(p)λ(ρ1)e−λ(ρ1), (2.2.12)where (p, w) is the non-degenerate solution of (1.1.14) and K(p) denotesthe Gaussian curvature of p. By using the equation (1.1.12), we havee−4piG(x,p) |x=p= 0 and ∆w(p) = 2ρ2. Thus∆ log h(p) + 8pi − 2K(p) =∆ log h1(p)− ρ2 + 8pi − 2K(p). (2.2.13)Obviously, λ(ρ1) can be well-defined only if∆ log h1(p)− ρ2 + 8pi − 2K(p) 6= 0.Let c1 be a positive constant, which will be chosen later. By using ρ1,we setSρ1(p, w) ={v1 =12vq,λ,a + φ∣∣∣ |q − p| ≤ c1λ(ρ1)e−λ(ρ1),|λ− λ(ρ1)| ≤ c1λ(ρ1)−1, |a− 1| ≤ c1λ(ρ1)− 12 e−λ(ρ1),φ ∈ O(1)q,λ and ‖φ‖H1(M) ≤ c1λ(ρ1)e−λ(ρ1)}, (2.2.14)andSρ2(p, w) ={v2 =12w+ψ∣∣∣ ψ ∈ O(2)q,λ and ‖ψ‖∗ ≤ c1λ(ρ1)e−λ(ρ1)}, (2.2.15)where ‖ψ‖∗ = ‖ψ‖W 2,p(M).Now suppose (v1k, v2k) is a sequence of bubbling solutions of (1.1.10) suchthat v1k blows up at p and weakly converges to 4piG(x, p), while v2k → 12win C2,α(M). Then we want to prove that(v1k, v2k) ∈ Sρ1(p, w)× Sρ2(p, w).First of all, we prove the following lemma.432.2. A-priori EstimateLemma 2.2.2. Let (v1k, v2k) be a sequence of blow up solutions of (1.1.10),which v1k blows up at p, weakly converges to 4piG(x, p) and v2k → 12w inC2,α(M). Suppose (p, w) is a non-degenerate solution of (1.1.14) and∆ log h1(p)− ρ2 + 8pi − 2K(p) 6= 0. (2.2.16)Then there exist q∗k, λ∗k, a∗k, φ∗k, ψ∗k such thatv1k =12vq∗k,λ∗k,a∗k + φ∗k, v2k =12w + ψ∗k, (2.2.17)and (v1k, v2k) ∈ Sρ1(p, w)× Sρ2(p, w).Remark 1. Because the proof of this lemma is very long, we describe theprocess briefly. First of all, we obtain a good approximation of v1k. Sincev2k converges to 12w in C2,α(M), this fine estimate can be obtained by thesame proof in [14]. Next, we substitute v1k into the second equation of v2k.Then we use the non-degeneracy of (1.1.14) to get the sharp estimates ofψk and |q˜k − p|, where ψk = v2k − 12w and q˜k is the point where v1k obtainsits maximal value. After that, we get the lemma. In the following proof, weuse the same notation as the proof of Proposition 1.1.2.Proof. Let v1k and v2k be a sequence of blow up solutions of (1.1.10),∆v1k + ρ1k(h1e2v1k−v2k∫M h1e2v1k−v2k− 1)= 0,∆v2k + ρ2k(h2e2v2k−v1k∫M h2e2v2k−v1k− 1)= 0.(2.2.18)For convenience, we write the first equation in (2.2.18) as,∆v1k + ρ1k( h˜ke2v1k∫Ω h˜ke2v1k− 1)= 0, (2.2.19)whereh˜k = h1e−v2k = he−ψk and ψk = v2k −12w. (2.2.20)Since h˜k → h in C2,α(M), all the estimates in [14] can be applied to ourcase here, although in [14] the coefficient h˜k is independent of k. In the442.2. A-priori Estimatefollowings (up to (2.2.28) below), we sketch the estimates in [14, 15] whichwill be used here. We denote q˜k to be the maximal point of v˜1k near p,where v˜1k = v1k − 12 log∫M h˜ke2v1k . Letλk = 2v˜1k(q˜k)− log∫Mh˜ke2v1k .In the local coordinate near q˜k, we setU˜k(x) = logeλk(1 + ρ1kh˜k(qk)4 eλk |x− qk|2)2,where qk is chosen such that∇U˜k(q˜k) = ∇ log h˜k(q˜k).Clearly, |qk − q˜k| = O(e−λk). Then the error term inside Br0(qk) is set byη˜k(x) = 2v˜1k − U˜k(y)− (8piR(x, qk)− 8piR(qk, qk)), (2.2.21)and the error term outside Br0(qk) is set byξk(x) = 2v1k(x)− 8piG(x, qk). (2.2.22)By Green’s representation for v1k, it is not difficult to obtainξk(x) = O(λke−λk) for x ∈M \Br0(qk). (2.2.23)By a straightforward computation, the error term η˜k satisfies∆η˜k + 2ρ1kh˜k(qk)eU˜kH˜k(x, η˜k) = 0, (2.2.24)452.2. A-priori EstimatewhereH˜k(x, t) = exp{logh˜k(x)h˜k(qk)+ 8pi(R(x, qk)−R(qk, qk))+ t}− 1=Hk(x) + t+O(|t|2),andHk(x) = exp{logh˜k(x)h˜k(qk)+ 8piR(x, qk)− 8piR(qk, qk)}− 1.We see that except for the higher-order term O(|η˜k|2), equation (2.2.24) isexactly like (2.2.7). By Lemma 2.2.1, we can proveη˜k(x) = −4ρ1kh˜k(qk)∆Hk(qk)e−λk [log(Rk|x−qk|+2)]2+O(λke−λk) (2.2.25)for x ∈ B2r0(qk), where Rk =√ρ1kh˜k(qk)4 eλk .From [14, Theorem 1.1, Theorem 1.4 and Lemma 5.4], we haveρ1k − 4pi =∆ log h˜k(qk) + 8pi − 2K(qk)h˜k(qk)λke−λk +O(e−λk), (2.2.26)2v˜1k + λk + 2 logρ1kh˜k(qk)4+ 8piR(qk, qk) +∆Hk(qk)ρ1kh˜k(qk)λ2ke−λk = O(λke−λk),(2.2.27)and|∇Hk(qk)| = O(λke−λk). (2.2.28)Now we let ηk be defined as in (2.2.7), vqk and vqk,λk,ak be defined as in(2.2.8) with q = qk, λ = λk and a = ak = 1. By Lemma 2.2.1, (2.2.25) and(2.2.28), we haveηk(x) = η˜k +O(λke−λk) for x ∈ B2r0(qk). (2.2.29)462.2. A-priori EstimateNote that for x ∈ Br0(qk),vqk,λk,ak =U˜k(x) + ηk(x) +(8piR(x, qk)− 8piR(qk, qk))+ λk + 2 logρ1kh˜k(qk)4+ 8piR(qk, qk) +∆Hk(qk)ρ1kh˜k(qk)λ2ke−λk − vqk ,where vqk denotes the average of vqk . From [15, Lemma 2.2 and Lemma2.3], we havevqk − 8piG(x, qk) = O(λke−λk) in M \B2r0(qk), and vqk = O(λke−λk).(2.2.30)By (2.2.21), (2.2.27), (2.2.29) and (2.2.30), we have2v1k − vqk,λk,ak =2v˜1k +∫Mh˜ke2v1k − vqk,λk,ak=2v˜1k − U˜k −(8piR(x, x)− 8piR(x, qk))− ηk(x) +O(λke−λk)=η˜k(x)− ηk(x) +O(λke−λk) = O(λke−λk) (2.2.31)for x ∈ Br0(qk). For x ∈M \B2r0(qk), by (2.2.22) and (2.2.30), we get2v1k − vqk,λk,ak = 2v1k − 8piG(x, qk)− (vqk − 8piG(x, qk)) + vqk = O(λke−λk).For the intermediate domain B2r0(qk) \ Br0(qk), following a similar way,we can obtain that 2v1k − vqk,λk,ak = O(λke−λk). Thus, we find a goodapproximation 12vqk,λk,ak for v1k. For convenience, we writev1k =12vqk,λk,ak + φk, where ‖φk‖L∞(M) < c˜λke−λk , (2.2.32)where c˜ is independent of ψk.Next, we substitute (2.2.32) and v2k = 12w+ψk into the second equationof (2.2.18), after computation, we obtainLψk = I1 + I2 + I3,∫Mψk = 0, (2.2.33)472.2. A-priori EstimatewhereLψk =∆ψk + 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)ψk− 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)ψk)− 4piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)(∇G(x, p)(qk − p))+ 4piρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)(∇G(x, p)(qk − p))),I1 =− ρ2h2ew+2ψk−v1k∫M h2ew+2ψk−v1k+ ρ2h2ew+2ψk−4piG(x,qk)∫M h2ew+2ψk−4piG(x,qk),I2 =ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)− ρ2h2ew+2ψk−4piG(x,p)∫M h2ew+2ψk−4piG(x,p)+ 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)ψk− 2ρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)ψk),andI3 =− ρ2h2ew+2ψk−4piG(x,qk)∫M h2ew+2ψk−4piG(x,qk)+ ρ2h2ew+2ψk−4piG(x,p)∫M h2ew+2ψk−4piG(x,p)− 4piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)(∇G(x, p)(qk − p))+ 4piρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)(∇G(x, p)(qk − p))).We shall analyze the right hand side of (2.2.33) term by term in the following.482.2. A-priori EstimateFor I1, we setE1 = exp(w + 2ψk − 4piG(x, qk))− exp(w + 2ψk −12vqk,λk,ak − φk).For x ∈M \Br0(qk). We see that the difference between 4piG(x, qk) andvqk,λk,ak is of order λke−λk . As a consequence, E1 = O(λke−λk).For x ∈ Br0(qk),4piG(x, qk)−12vqk,λk,ak =4piG(x, qk)− 4piR(x, qk)− log(ρ1h˜k(qk)4)+ log(1 +ρ1h˜k(qk)eλk4|x− qk|2)− λk−12(ηk +∆Hk(qk)ρ1h˜k(qk)λ2keλk)+O(λke−λk)= log( 4ρ1h˜k(qk)eλk |x− qk|2+ 1)−12(ηk +∆Hk(qk)ρ1h˜k(qk)λ2keλk)+O(λke−λk).Since φk = O(λke−λk),exp(w + 2ψk −12vqk,λk,ak − φk)= exp(w + 2ψk −12vqk,λk,ak)+O(λke−λk).Then, we haveexp(w + 2ψk − 4piG(x, qk))− exp(w + 2ψk −12vqk,λk,ak − φk)= exp(w + 2ψk − 4piG(x, qk))− exp(w + 2ψk −12vqk,λk,ak)+O(λke−λk)= exp(w + 2ψk − 4piG(x, qk))(1− exp(4piG(x, qk)−12vqk,λk,ak))+O(λke−λk)= exp(w + 2ψk − 4piG(x, qk))(1− exp[log(1 +4ρ1h˜keλk |x− qk|2)+O(ηk +∆Hk(qk)ρ1h˜k(qk)λ2keλk)+O(λke−λk)])+O(λke−λk)492.2. A-priori EstimateWhen |x− qk| = O(e−λk2 ), we haveexp(w + 2ψk − 4piG(x, qk))= O(|x− qk|2),andlog(1 +4ρ1h˜keλk |x− qk|2)+O(ηk +∆Hk(qk)ρ1h˜k(qk)λ2keλk)= O(log(e−λk |x− qk|−2)),henceE1 = O(λke−λk) for |x− qk| = O(e−λk2 ).When |x− qk|  e−λk2 ,1− exp(log(1 +4ρ1h˜keλk |x− qk|2)+O(ηk +∆Hk(qk)ρ1h˜k(qk)λ2keλk))+O(λke−λk)= O( 4ρ1h˜keλk |x− qk|2+ λke−λk),which givesE1 = O(λke−λk) for r0 ≥ |x− qk|  e−λk2 .Thus, ‖E1‖L∞(M) = O(λke−λk), which implies I1 = O(λke−λk).For the second term, it is easy to see that I2 = O(‖ψk‖2∗). It remains toestimate I3. We divide it into three parts. I3 = I31 + I32 + I33, whereI31 =− ρ2h2ew+2ψk−4piG(x,qk)∫M h2ew+2ψk−4piG(x,qk)+ ρ2h2ew+2ψk−4piG(x,p)∫M h2ew+2ψk−4piG(x,p)− 4piρ2h2ew+2ψk−4piG(x,p)∫M h2ew+2ψk−4piG(x,p)(∇G(x, p)(qk − p)),+ 4piρ2h2ew+2ψk−4piG(x,p)(∫M h2ew+2ψk−4piG(x,p))2∫M(h2ew+2ψk−4piG(x,p)(∇G(x, p)(qk − p))),502.2. A-priori EstimateI32 =4piρ2h2ew−4piG(x,p)(∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)(∇G(x, p)(qk − p)))− 4piρ2h2ew+2ψk−4piG(x,p)(∫M h2ew+2ψk−4piG(x,p))2∫M(h2ew+2ψk−4piG(x,p)(∇G(x, p)(qk − p))),andI33 =4piρ2h2ew+2ψk−4piG(x,p)∫M h2ew+2ψk−4piG(x,p)(∇G(x, p)(qk − p))− 4piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)(∇G(x, p)(qk − p)).It is not difficult to seeI31 = O(|qk − p|2), I32 = O(1)‖ψ‖∗|qk − p|, I33 = O(1)‖ψ‖∗|qk − p|.Then (2.2.33) can be written asL(ψk) = o(1)‖ψk‖∗ +O(‖ψk‖2∗ + λke−λk) +O(|p− qk|2). (2.2.34)By the definition of Hk and (2.2.28), we have∇Hk(qk) = ∇ log h(qk)−∇ψk(qk) + 8pi∇R(qk, qk) = O(λke−λk). (2.2.35)By (2.2.2) and (2.2.35), we have∇2(log h(p) + 8piR(p, p))(qk − p)−∇ψk(p)=∇ log h(qk)−∇ψk(qk) + 8pi∇R(qk, qk)−(∇ log h(p) + 8pi∇R(p, p))+∇ψk(qk)−∇ψk(p)+O(|p− qk|2)=∇Hk(qk)−∇H(p) +O(|p− qk|γ‖ψk‖∗)+O(|p− qk|2), (2.2.36)where γ depends on p. We note that ∇H(p) = 0. From (2.2.34)-(2.2.36)512.2. A-priori Estimateand the non-degeneracy of p, w, we obtain‖ψk‖∗ + |p− qk| ≤ C(λke−λk + o(1)‖ψk‖∗ + ‖ψk‖2∗ + |p− qk|2), (2.2.37)where C is a generic constant, independent of k and ψk. Therefore, we haveψk = O(λke−λk), |p− qk| = O(λke−λk). (2.2.38)As a conclusion of (2.2.12), (2.2.26) and (2.2.38), we haveλk−λ(ρ1) = O(λ(ρ1)−1), h˜k = h+O(λ(ρ1)e−λ(ρ1)), |qk−p| = O(λ(ρ1)e−λ(ρ1))(2.2.39)andv2k −12w = O(λ(ρ1)e−λ(ρ1)). (2.2.40)We replace h˜k by h in the definition of vq, we denote the new terms by vq.By (2.2.38), we havevqk − vq = O(λ(ρ1)e−λ(ρ1)).We setvq,λ,a = vq − vq. (2.2.41)By (2.2.32) and (2.2.41), we obtainv1k −12vq,λ,a = O(λ(ρ1)e−λ(ρ1)). (2.2.42)By [15, Lemma 3.2], if we choose c1 in Sρ1(p, w) big enough, there exists atriplet (q∗k, λ∗k, a∗k) and φ∗ ∈ O(1)q∗k,λ∗ksuch thatv1k =12vq∗k,λ∗k,a∗k + φ∗k, (2.2.43)522.3. Approximate Blow-up Solutionwhere q∗k, λ∗k, a∗k satisfy the condition in Sρ1(p, w). Thus, we have proved(v1k, v2k) ∈ Sρ1(p, w)× Sρ2(p, w).In conclusion, we have the following result,Theorem 2.2.1. Suppose h1, h2 are two positive C2,α function on M suchthat any solution (p, w) of (1.1.14) is non-degenerate and ∆ log h1(p)−ρ2 +8pi − 2K(p) 6= 0. Then there exists ε0 > 0 and C > 0 such that for anysolution of (1.1.10) with ρ1 ∈ (4pi− ε0, 4pi+ ε0), ρ2 /∈ 4piN, either |v1|, |v2| ≤C,∀x ∈ M or (v1, v2) ∈ Sρ1(p, w) × Sρ2(p, w) for some solution (p, w) of(1.1.14).2.3 Approximate Blow-up SolutionIn the following two sections, we shall construct the blow up solutions of(1.1.10) when ρ1 → 4pi. The construction of such bubbling solution is basedon a non-degenerate solution of (1.1.14). Our aim is to compute the degreeof the following nonlinear operator(v1v2)= (−∆)−1ρ1( h1e2v1−v2∫M h1e2v1−v2− 1)ρ2( h2e2v2−v1∫M h2e2v2−v1− 1)in the space Sρ1(p, w)× Sρ2(p, w).SetT (v1, v2) =(T1(v1, v2)T2(v1, v2))= ∆−12ρ1( h1e2v1−v2∫M h1e2v1−v2− 1)2ρ2( h2e2v2−v1∫M h2e2v2−v1− 1) .Since each solution v1 in Sρ1(p, w) can be represented by (q, λ, a, φ), and v2532.3. Approximate Blow-up Solutionin Sρ2(p, w) can be represented by w and ψ, therefore the nonlinear operator2v1 + T1(v1, v2) can be divided according to this representation.Let v1 = 12vq,λ,a+φ ∈ Sρ1(p, w). Recalling that t = s−vq. For x ∈ Br0(q),we havevq,λ,a(x) + logh(x)h(q)=U + t+H(x) + η + (a− 1)(U + s)+O(|a− 1|(|y|+ |η|+ |vq|)),where y = x− q. Then, we getρ1h1e2v1−v2−2φ+ψ =ρ1hevq,λ,a = ρ1h(q)eU+t[1 + (a− 1)(U + s) + η+H(x) + (a− 1)O(|y|) +O(β˜2)], (2.3.1)whereβ˜ = λ|a− 1|+ |η|+ |H(x)|+ vq.Therefore in Br0(q), we haveρ1h1e2v1−v2 =(1 + ϕ)ρ1hevq,λ,a + (eϕ − 1− ϕ)ρ1hevq,λ,a=ρ1h(q)eU+t[1 + (a− 1)(U + s) + η +H(x)+ (a− 1)O(|y|) + ϕ]+ E˜, (2.3.2)whereE˜ = (eϕ − 1− ϕ)ρ1hevq,λ,a + ρ1h(q)eU+tO(ϕ2 + β˜2), (2.3.3)and ϕ = 2φ− ψ.Let 2 > 0 be small. E˜ can be written into two partsE˜ = E˜+ + E˜−,542.3. Approximate Blow-up SolutionwhereE˜+ ={E˜ if |ϕ| ≥ 20 if |ϕ| < 2,E˜− ={0 if |ϕ| ≥ 2E˜ if |ϕ| < 2.ThenE˜+ = O(e|ϕ|+2λ) if |ϕ| ≥ 2, (2.3.4)andE˜− = ρ1heU+λO(ϕ2 + β˜2). (2.3.5)Using the expression for ρ1h1e2v1−v2 above, we obtain the following estimatefor∫M ρ1h1e2v1−v2 .Lemma 2.3.1. Let v1 = 12vq,λ,a+φ ∈ Sρ1(p, w) and v2 =12w+ψ ∈ Sρ2(p, w).Then as ρ1 → 4pi, ρ2 /∈ 4piN, we have∫Mρ1h1e2v1−v2 = 4piet(1− ψ(p)) +4piρ1h(q)∆H(q)λeλet+ 8piλ(a− 1)et +O(|a− 1|eλ + 1). (2.3.6)Proof. By (2.3.2)∫Mρ1h1e2v1−v2 =∫Br0 (q){ρ1h(q)eU+t[1 + · · · ] + E˜}dy+∫M\Br0 (q)ρ1h1e2v1−v2 .By the explicit expression of U, we have∫M\Br0 (q)ρ1h1e2v1−v2 = O(1), (2.3.7)∫Br0 (q)ρ1h(q)eU+tdy = 4piet +O(1), (2.3.8)552.3. Approximate Blow-up Solution∫Br0 (q)ρ1h(q)eU+t(a− 1)(U + s)dy = 8pi(a− 1)λet +O(1 + |a− 1|eλ),(2.3.9)where U + s = 2λ− 2 log(1 + ρ1h(q)4 eλ|y|2)+O(1) is used. By the equationof η and the fact that ∇H(q) · y is an odd functions, we have∫Br0 (q)ρ1h(q)eU+t(η +H(x))dy = −12et∫Br0 (0)∆ηdy = −12et∫∂Br0 (0)∂η∂ν=4piρ1h(q)∆H(q)λet−λ +O(1). (2.3.10)To estimate the terms involving φ and ψ, we use (2.2.8) to obtain∆vq = ∆U + ∆η + 8pi for x ∈ Br0(q),and∆vq = ∆(vq − 8piG(x, q)) + 8pi for x /∈ Br0(q).This together with Lemma 2.2.1 implies∫Br0 (q)2ρ1h(q)eUφ =−∫Mφ∆vq +∫Br0 (q)φ∆η + 8pi∫Mφ+∫M\Br0 (q)φ∆(vq − 8piG(x, q))=∫M∇φ∇vq + 8pi∫Mφ+∫Br0 (q)φ∆η+∫M\Br0 (q)φ∆(vq − 8piG(x, q))≤ e−λλ∫M|φ|, (2.3.11)where |∆η(y)| = O(e−λ) and Lemma 2.2.1 are used. By the Poincare´ in-562.3. Approximate Blow-up Solutionequality, we have∣∣∣∫Br0 (q)ρh(q)eUφ∣∣∣ ≤ e−λλ∫M|φ| ≤ ce−λλ‖φ‖H1 . (2.3.12)While, for the terms involving ψ, we have∫Br0 (q)ρ1h(q)eUψ =∫Br0 (q)ρ1h(q)eUψ(q) +∫Br0 (q)ρ1h(q)eU (ψ − ψ(q))=4piψ(q) +O(λe−32λ). (2.3.13)For E˜+, we have∫Br0 (q)|E˜+| =O(1)∫Br0 (q)∩{|ϕ−ϕ|≥2}e|ϕ|+2λ=O(1)∫Br0 (q)∩{|ϕ−ϕ|≥2}e|ϕ−ϕ|+2λ,where ϕ =∫Br0 (q)ϕvol(Br0 (q))= O(‖ϕ‖H1) = O(λe−λ) for λ is large. We writee|ϕ−ϕ| = e|ϕ−ϕ|(1− 4pi|ϕ−ϕ|‖ϕ−ϕ‖2)e4pi|ϕ−ϕ|2‖ϕ−ϕ‖2 .Since ‖ϕ− ϕ‖−2 = ‖ϕ− ϕ‖−2H1  2λ, we havee|ϕ−ϕ|(1− 4pi|ϕ−ϕ|‖ϕ−ϕ‖2)≤ e22 (1−2pi2‖ϕ−ϕ‖2) e−2λ for ‖ϕ− ϕ‖ ≥22.Hence, by Moser-Trudinger inequality∫Br0 (q)∩{|ϕ−ϕ|≥22 }e|ϕ−ϕ| ≤ e−2λj∫Br0 (q)exp(4pi|ϕ− ϕ|2‖ϕ− ϕ‖2)≤ O(1)e−2λ,(2.3.14)which implies∫Br0 (q)|E˜+| ≤ O(1).572.3. Approximate Blow-up SolutionFor E˜−, (2.3.5) gives∫Br0 (q)|E˜−| ≤ O(1)∫Br0 (q)(|ϕ|2 + β˜2)ρ1h(q)eU+t. (2.3.15)By (2.6.3) in section 2.6, we can estimate the first term on the right handside of (2.3.15) by∫Br0 (q)ρ1h(q)eU+t|ϕ|2 ≤ O(1)et(∫M|∇φ|2 + ‖ψ‖2∗∫Br0 (q)eU)= O(1)etλ2e−2λ = O(1)λ2e−λ.For β˜2∫Br0 (q)ρ1h(q)eU+tβ˜2 =O(1)et(|vq|2 + (λ|a− 1|λ)2 +∫Br(0)(q)(η2 +H(q, η))eU)≤c.Therefore, we have ∫Br0 (q)|E˜−| = O(1). (2.3.16)By (2.3.2) and (2.3.7)-(2.3.16), we obtain (2.3.6). Hence we finish the proofof Lemma 2.3.1.Now we want to express 2v1 + T1(v1, v2) in a formula similar to (2.3.2).By Lemma 2.3.1 and the Taylor expansion of the exponential function,e−t∫Mρ1h1e2v1−v2 =4pi − 4piψ(q) +4piρ1h(q)∆H(q)λe−λ + 8piλ(a− 1)+O(|a− 1|) +O(e−λ). (2.3.17)Henceet∫M h1e2v1−v2− 1 =1e−t∫M ρ1h1e2v1−v2(ρ1 −∫M ρ1h1e2v1−v2et)=θ +O(|a− 1|) +O(e−λ), (2.3.18)582.3. Approximate Blow-up Solutionwhere θ is defined byθ =14pi[(ρ1 − 4pi)−4piρ1h(q)∆H(q)λe−λ + 4piψ(q)− 8piλ(a− 1)]. (2.3.19)Letβ =∣∣∣et∫M h1e2v1−v2− 1∣∣∣+ β˜, (2.3.20)andE = 2(eϕ − 1− ϕ)ρ1h1e2v1−v2∫M h1e2v1−v2+ 2ρ1h(q)eU(O(ϕ2) +O(β2)). (2.3.21)Then in Br0(q), we have by (2.3.2),ρ1h1e2v1−v2∫M h1e2v1−v2=(1 + ϕ)ρ1hevq,λ,a∫M h1e2v1−v2+ (eϕ − 1− ϕ)ρ1hevq,λ,a∫M h1e2v1−v2=ρ1h(q)eU[1 +( et∫M h1e2v1−v2− 1)+ (a− 1)(U + s)+ (a− 1)O(|y|) + η +H + ϕ+O(β2)]+ E. (2.3.22)Thus, we have∆(2v1 + T1(v1, v2))=2∆v1 +2ρ1h1e2v1−v2∫M h1e2v1−v2− 2ρ1=a(∆U + ∆η) + 2∆φ+ 8pia+2ρ1h1e2v1−v2∫M h1e2v1−v2− 2ρ1=− 2aρ1h(q)eU[1 + η +H −∇yH · y]+ 8pi − 2ρ1+ 8pi(a− 1) +2ρ1h1e2v1−v2∫M h1e2v1−v2=2∆φ+ (8pi − 2ρ1) + 8pi(a− 1)+ 2ρ1h(q)eU[(a− 1)(U + s− 1) + (a− 1)O(|y|)y · ∇H+( et∫M h1e2v1−v2− 1)+ ϕ]+ E. (2.3.23)592.3. Approximate Blow-up SolutionLet ε2 > 0 be small, which will be chosen later, see in section 6. WriteE = E+ + E−withE+ ={E if |ϕ| ≥ ε20 if |ϕ| < ε2and E− ={0 if |ϕ| ≥ ε2E if |ϕ| < ε2.As λ→∞, we haveE+ = O(e|ϕ|+λ)andE− = ρ1h(q)eU (O(ϕ2) +O(β2)).In B2r0(q) \Br0(q), since vq − vq − 8piG(x, q) is small, see [15, Lemma 2.2],we write ∆(2v1 + T1(v1, v2))as∆(2v1 + T1(v1, v2))= 2∆φ+ a∆(vq − 8piG(x, q))+ 8pi − 2ρ1 + 8pi(a− 1)+2ρ1h∫M h1e2v1−v2ea(vq−vq−8piG(x,q))+8piaG(x,q)+ϕ.(2.3.24)In M \B2r0(q), we have∆(2v1 + T1(v1, v2))= 2∆φ+ 8pi − 2ρ1 + 8pi(a− 1)+2ρ1h∫M h1e2v1−v2e8piaG(x,q)+ϕ−avq . (2.3.25)From (2.3.23)-(2.3.25), we have the followingLemma 2.3.2. Let v1 = 12vq,λ,a + φ ∈ Sρ1(p, w), v2 =12w + φ ∈ Sρ2(p, w).Then as ρ1 → 4pi,602.3. Approximate Blow-up Solution1.〈∇(2v1 + T1(v1, v2)),∇φ1〉 = 2B(φ, φ1) +O(λe−λ)‖φ1‖H10 (M),(2.3.26)whereB(φ, φ1) :=∫M∇φ · ∇φ1 −∫Br0 (q)2ρ1h(q)eUφφ1,is a positive symmetric, bilinear form satisfying B(φ, φ) ≥ c0‖φ‖2H1(M)for some constant c0 > 0.2.〈∇(2v1 + T1(v1, v2)),∇∂qvq〉=− 8pi∇H(q) + 8pi∇ψ(q)+O(λ|a− 1|+∣∣∣et∫M h1e2v1−v2− 1− ψ(q)∣∣∣+ λe−λ), (2.3.27)3.〈∇(2v1 + T1(v1, v2)),∇∂λjvq〉=− 16pi(a− 1)(λ− 1 + logρ1h(q)4+ 4piR(q, q))− 8pi(θ − ψ(q))+O(|a− 1|+ λ2e−32λ), (2.3.28)4.〈∇(2v1 + T1(v1, v2)),∇vq〉=(2λ− 2 + 8piR(q, q) + 2 logρ1h(q)4)× 〈∇(2v1 + T1(v1, v2)),∇∂λvq〉+ 16pi(a− 1)λ+O(1)‖φ‖H1(M) +O(λe−λ), (2.3.29)612.4. Deformation And Degree Counting FormulaWe leave the proof of Lemma 2.3.2 in the section 6 because it containsa lot of computations.2.4 Deformation And Degree Counting FormulaIn this section, we want to deform 2vi + Ti(v1, v2) into a simple form whichcan be solvable. Obviously, v1 = 12vq,λ,a + φ, v2 =12w + ψ is a solution of2v1 + T1(v1, v2) = 0, if and only if the left hand sides of (2.3.26)-(2.3.29)vanish. To solve the system (2.3.26)-(2.3.29) and 2v2 + T2(v1, v2) = 0, werecallH˚1 = O(1)q,λ⊕the linear subspace spanned by vq, ∂λvq and ∂qvqand deform 2vi+Ti(v1, v2) to a simpler operator 2vi+T 0i (v1, v2) by definingthe operator 2I + T ti , 0 ≤ t ≤ 1, i = 1, 2 through the following relations:〈∇(2v1 + Tt1(v1, v2)),∇φ1〉 =t〈∇(2v1 + T1(v1, v2)),∇φ1〉+ 2(1− t)B(φ, φ1) for φ1 ∈ O(1)q,λ, (2.4.1)〈∇(2v1 + Tt1(v1, v2)),∇∂qvq〉 =t〈∇(2v1 + T1(v1, v2)),∇∂qvq〉+ (1− t)(− 8pi∇H + 8pi∇ψ(q)), (2.4.2)〈∇(2v1 + Tt1(v1, v2)),∇∂λvq〉 =t〈∇(2v1 + T1(v1, v2)),∇∂λvq〉− 8pi(1− t)[2(a− 1)λ+ (θ − ψ(q))],(2.4.3)622.4. Deformation And Degree Counting Formula〈∇(2v1 + Tt1(v1, v2)),∇vq〉 =t[(2λ+O(1))〈∇(2v1 + Tt1(v1, v2)),∇∂λvp〉+O(1)‖φ‖H1 +O(λe−λ)]+ 16pi(a− 1)λ,(2.4.4)2v2 + Tt2(v1, v2) = t(2v2 + T2(v1, v2))+ (1− t)(w + 2ψ − 2ρ2(−∆)−1( h2ew+2ψ−4piG(x,q)∫M h2ew+2ψ−4piG(x,q)− 1)), (2.4.5)where those coefficients O(1) are those terms appeared in (2.4.4) so thatT 11 (v1, v2) = T1(v1, v2). From the construction above, we have2vi + Ti(v1, v2) = 2vi + T1i (v1, v2), i = 1, 2.When t = 0, the operator T 0i is simpler than Ti, i = 1, 2. During thedeformation from T 1i to T0i , i = 1, 2 we haveLemma 2.4.1. Assume (ρ1− 4pi) 6= 0, and ρ2 /∈ 4piN. Then there is ε1 > 0such that (2v1 + T t1(v1, v2), 2v2 + Tt2(v1, v2)) 6= 0 for (v1, v2) ∈ ∂(Sρ1(p, w)×Sρ2(p, w))and 0 ≤ t ≤ 1 if |ρ1 − 4pi| < ε1 and ρ2 is fixed.Proof. Assume (v1, v2) ∈ Sρ1(p, w)× Sρ2(p, w), where Sρi(p, w) denotes theclosure of Sρi(p, w), and 2vi + Tti (v1, v2) = 0, i = 1, 2 for some 0 ≤ t ≤ 1.We will show that (v1, v2) /∈ ∂(Sρ1(p, w)× Sρ2(p, w)).From 〈∇(2v1 + T t1(v1, v2)),∇φ〉 = 0, we have by Lemma 2.3.2‖φ‖2H1 ≤ O(λe−λ)‖φ‖H1 .This implies‖φ‖H1 = O(λe−λ) ≤ c2λe−λ, (2.4.6)for some constant c2 1 independent of c1.Using 〈∇(2v1 +T t1(v1, v2)),∇∂λvq〉 = 0 and 〈∇(2v1 +Tt1(v1, v2)),∇vq〉 =1Here c2 is independent of ψ, it can be shown in the proof of Lemma 2.3.2.632.4. Deformation And Degree Counting Formula0, (2.4.4) and (2.4.6) imply16piλ(a− 1) = O(λe−λ), (2.4.7)that is, when ρ1 is close to 4pi,|a− 1| = O(e−λ) < c1λ1(ρ)12 e−λ1(ρ). (2.4.8)By 〈∇(2v1+T t1(v1, v2)),∇λvq〉 = 0, we conclude from (2.3.28) and (2.4.8)thatθ − ψ(q) + 2λ(a− 1) = O(|a− 1|+ e−λ) = O(e−λ), (2.4.9)andet(∫Mh1e2v1−v2)−1− 1− θ = O(|a− 1|+ e−λ) = O(e−λ). (2.4.10)Together with〈∇(2v1 + Tt1(v1, v2)),∇∂qvq〉 = 0,(2.4.8), (2.4.10) and part (2) of Lemma 2.3.2, we have|∇H(q)−∇ψ(q)| = O(λ|a− 1|+ |et∫M h1e2v1−v2− 1− ψ(q)|+ λe−λ)≤ O(1)λe−λ,which implies∣∣∣∇2yH(p) · (q − p)−∇ψ(p)∣∣∣≤ O(1)λe−λ +O(1)‖ψ‖∗|p− q|γ +O(1)|p− q|2, (2.4.11)where we used ∇xH(p) = 0 and p > 2.642.4. Deformation And Degree Counting FormulaFor the second component, by (2.4.5), we have0 =(1− t)(∆w + 2∆ψ + 2ρ2( h2ew+2ψ−4piG(x,q)∫M h2ew+2ψ−4piG(x,q)− 1))+ t(∆w + 2∆ψ + 2ρ2( h2ew+2ψ−12vq,λ,a−φ∫M h2ew+2ψ− 12vq,λ,a−φ− 1)). (2.4.12)We set Θ = 2ρ2h2ew+2ψ− 12 vq,λ,a−φ∫M h2ew+2ψ− 12 vq,λ,a−φ− 2ρ2h2ew+2ψ−4piG(x,q)∫M h2ew+2ψ−4piG(x,q) , and claim‖Θ‖Lp(M) ≤ c3λe−λ, (2.4.13)where c3 is a constant that independent of c and p is defined in O(2)q,λ. By(2.4.6), it is not difficult to getexp(w + 2ψ −12vq,λ,a − φ)= exp(w + 2ψ −12vq,λ,a)+ Θ1,where ‖Θ1‖p ≤ c4λe−λ. By noting (2.4.6), it is enough for us to prove thefollowing one to get (2.4.13)∥∥∥ exp(w + 2ψ −12vq,λ,a)− exp(w + 2ψ − 4piaG(x, q))∥∥∥L∞(M)≤ c5λe−λ.(2.4.14)We leave the proof it in section 6. By (2.4.13), (2.4.12) can be written as∆w + 2∆ψ + 2ρ2( h2ew+2ψ−4piG(x,q)∫M h2ew+2ψ−4piG(x,q)− 1)+ tΘ = 0. (2.4.15)652.4. Deformation And Degree Counting FormulaWe expand the above equation,R =∆ψ + 2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)ψ− 2ρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)ψ)− 4piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)(∇G(x, p)(q − p))+ 4piρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)(∇G(x, p)(q − p))),(2.4.16)where R = tΘ + o(1)‖ψ‖∗ + |q − p|2. By the non-degeneracy of (p, w) to(1.1.14), (2.4.11) and (2.4.16), we can get‖ψ‖∗ ≤ c6λe−λ and |q − p| ≤ c7λe−λ. (2.4.17)Recall thatθ =14pi((ρ1 − 4pi)−4piρ1h(q)∆H(q)λe−λ + 4piψ(q)− 8piλ(a− 1)).From (2.4.9), we obtainO(1)e−λ =(ρ1 − 4pi −4piρ1h(q)∆H(q)λe−λ), (2.4.18)which impliesO(e−λ) =14pi∆H(q)(λ1(ρ)λ1(ρ)−λeλ).Hence|λ− λ1(ρ)| = c8(λ1(ρ))−1 (2.4.19)for some constant c8 independent of c1.662.4. Deformation And Degree Counting FormulaUsing (2.4.17) and (2.4.19), we have‖ψ‖∗ ≤ c9λ1(ρ)e−λ1(ρ) and |p− q| ≤ c10λ1(ρ)e−λ1(ρ). (2.4.20)By choosing c1 > c2, c7, c8, c9, c10. From (2.4.6), (2.4.8), (2.4.19) and (2.4.20),we obtain(v1, v2) /∈ ∂(Sρ1(p, w), Sρ2(p, w)).The proof is completed.Then, we want to apply Lemma 2.3.2 and Lemma 2.4.1 to get the degreeof the linear operator in Sρ1(p, w)× Sρ2(p, w) when ρ1 crosses 4pi.To compute the termdeg((2v1 + T1(v1, v2), 2v2 + T2(v1, v2));Sρ1(p, w)× Sρ2(p, w), 0),we setS∗1(p, w) ={(q, λ, a) :12vq,λ,a + φ ∈ Sρ1(p, w), φ ∈ O(1)q,λ}and define the mapΦp = (Φp,1,Φp,2,Φp,3,Φp,4) :Φp,1 =18pi(〈∇(2v1 + T01 (v1, v2)),∇∂qvq〉+ 〈∇2v2 + T02 (v1, v2), 0〉),Φp,2 = 〈∇(2v1 + T01 (v1, v2)),∇∂λvq〉+ 〈∇(2v2 + T02 (v1, v2)), 0〉,Φp,3 = 〈∇(2v1 + T01 (v1, v2)),∇vq〉+ 〈∇(2v2 + T02 (v1, v2)), 0〉,Φp,4 = 〈∇(2v1 + T01 (v1, v2)), 0〉+ (2v2 + T02 (v1, v2)).Clearly, by Lemma 2.3.2 and Lemma 2.4.1, we havedeg((2v1+T1(v1, v2), 2v2 + T2(v1, v2));Sρ1(p, w)× Sρ2(p, w), 0)= deg(Φp;S∗1(p, w)× Sρ2(p, w), 0). (2.4.21)672.4. Deformation And Degree Counting FormulaNext, we study the right hand side of (2.4.21) and prove Proposition1.1.3.Proof of Proposition 1.1.3. To compute the degree, we note that,Φp,2 = −[2ρ1 − 8pi − 8pi∆H(q)ρ1h(q)λe−λ]. (2.4.22)Clearly, we have∂Φp,1∂λ=∂Φp,1∂a=∂Φp,2∂a=∂Φp,3∂ψ=∂Φp,3∂q=∂Φp,4∂a=∂Φp,4∂λ= 0, (2.4.23)It is easy to see Φp,1 = 0, Φp,3 = 0 and Φp,4 = 0 if and only ifq = p, a = 1, ψ = 0, (2.4.24)and Φˆp,2 = 0 if and only ifρ1 − 4pi =4piρ1h(q)∆H(q)λe−λ. (2.4.25)It is not difficult to see that if |ρ1−4pi| is sufficiently small, equation (2.4.25)possesses a unique solution λ Hence (p, λ1(ρ), a, 0) is the solution of Φp,where a = 1. The degree of Φp at (p, λ1(ρ), a, 0) depends on the number ofnegative eigenvalue for the following matrixM =∂Φp,1∂q ,∂Φp,1∂λ ,∂Φp,1∂a ,∂Φp,1∂ψ∂Φp,2∂q ,∂ΦQ,2∂λ ,∂Φp,2∂a ,∂Φp,2∂ψ∂Φp,3∂q ,∂ΦQ,3∂λ ,∂Φp,3∂a ,∂Φp,3∂ψ∂Φp,4∂q ,∂ΦQ,4∂λ ,∂Φp,4∂a ,∂Φp,4∂ψHere we say µM is an eigenvalue ofM, if there exists ν ∈ R2, λ ∈ R, a ∈ R,682.4. Deformation And Degree Counting Formulaand Ψ such thatMνaλΨ= µMνaλ1(−∆)−1Ψ,where ∂Φp,1∂ψ [Ψ] = ∇Ψ(p), and∂Φp,4∂ψ[Ψ] =Ψ− (−∆)−1(2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p))Ψ− 2ρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)Ψ)).We set N(T ) as the number of the negative eigenvalue of matrix T ,M1 =[∂Φp,1∂q ,∂Φp,1∂ψ∂Φp,4∂q ,∂Φp,4∂ψ]and M2 =[∂Φp,2∂λ ,∂Φp,2∂a∂Φp,3∂λ ,∂Φp,3∂a].By using (2.4.23),N(M) = N(M1) +N(M2) = N(M1) + sgn(∂Φp,2∂λ)+ sgn(∂Φp,3∂a),Therefore,deg(Φp;S∗1(p, w)×Sρ2(p, w), 0)= (−1)N(M) = (−1)N(M1) × (−1)N(M2)= (−1)N(M1) × sgn(∂Φp,2∂λ)× sgn(∂Φp,3∂a).We first consider the last two terms on the right hand side of the aboveequality. For ∂Φp,3∂a , it is easy to see that the sign of this value is positive.Thereforesgn(∂Φp,3∂a) = 1.692.4. Deformation And Degree Counting FormulaTo compute ∂Φp,2∂λ , we have∂Φp,2∂λ= −8pi∆H(p)ρ1h(p)λe−λ +O(e−λ).Thus∂Φp,2∂λ= −2(ρ1 − 4pi) +O(e−λ).It remains to compute N(M1). According to the definition, we have[∂(Φp,1,Φp,4)∂(q, ψ)](νΨ)=(−∇2H · ν +∇Ψ(p)−I0), (2.4.26)whereI0 =−Ψ + (−∆)−1(2ρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)Ψ− 2ρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M(h2ew−4piG(x,p)Ψ)− 4piρ2h2ew−4piG(x,p)∫M h2ew−4piG(x,p)(∇G(x, p) · ν)+ 4piρ2h2ew−4piG(x,p)( ∫M h2ew−4piG(x,p))2∫M[h2ew−4piG(x,p)(∇G(x, p) · ν)]).According to the definition of the eigenvalue for the linearized equationof (1.1.14), we can get (−1)N(M1) is exactly the number of the negativeeigenvalue of the linearized equation of (1.1.14) when ∆H(p) has the samesign as ρ1 − 4pi. Therefore,d(2)+ − (−1)N1 = d(2)− − (−1)N2,where N1, N2 represent the number of the negative eigenvalue for the lin-earized equation of (1.1.14) when ρ1− 4pi, i.e., ∆H(p) > 0 and ρ1− 4pi < 0,i.e., ∆H(p) < 0 respectively. It is easy to see that the summation of N1 andN2 is the total negative eigenvalues of the linearized equation of (1.1.14) for702.5. Proof Of Theorem 1.1.1a given solution (p, w). Thus, by the definition of the topological degree forthe solution to the shadow system (1.1.14), we have all the topological de-gree contributed by the bubbling solution of (1.1.10) equals to the negativeof the topological degree of the shadow system (1.1.14). Hence, we provedProposition 1.1.3.2.5 Proof Of Theorem 1.1.1This section is devoted to prove Theorem 1.1.1. We first study the shadowsystem and give a proof the Lemma 1.1.1. As we mentioned in the firstChapter, we introduce a deformation to decouple the system (1.1.14).(St){∆w + 2ρ2(h2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 1) = 0,∇(log(h1e−12w·(1−t)) + 4piR(x, x))|x=p= 0.(2.5.1)It is easy to see that the system (2.5.1) is exactly (1.1.14) when t = 0,and will be a decoupled system when t = 1. During the deformation from(S1) to (S0), we haveLemma 2.5.1. Let ρ2 /∈ 4piN. Then there is a uniform constant Cρ2 suchthat for all solutions to (2.5.1), we have |w|L∞(M) < Cρ2 .Proof. Since ρ2 /∈ 4piN, then we can see any solution for the following equa-tion∆w + 2ρ2( h2ew−4piG(x,p)∫M h2ew−4piG(x,p)− 1)= 0 (2.5.2)is uniformly bounded above. By using the classical elliptic estimate, we have|w|C1(M) < C, where the constant C depends on ρ2.Proof of Theorem 1.1.1. It is known that the topological degree is inde-pendent of h1 and h2 as long as they are positive C1 functions. So we canalways choose h1 and h2 such that the hypothesis of Theorem 2.2.1 holds.Let dS denote the Leray-Schauder degree for (1.1.14). By Lemma 2.5.1,computing the topological degree for (1.1.14) is reduced to computing the712.6. Proof Of Lemma 2.3.2 And (2.4.14)topological degree for system (2.5.1) when t = 1,{∆w + 2ρ2(h2ew−4piG(x,p)∫M h2ew−4piG(x,p) − 1) = 0,∇[log h1 + 4piR(x, x)] |x=p= 0.(2.5.3)Since this is a decoupled system, the topological degree of (2.5.3) equals theproduct of the degree of first equation and degree contributed by the secondequation. By the Poincare´-Hopf Theorem, the degree of the second equationis χ(M). On the other hand, by Theorem A, the topological degree for thefirst equation is bm + bm−1, where bk is given (1.1.7). Therefore,dS = χ(M) · (bm + bm−1). (2.5.4)Hence we get the topological degree of the shadow system (1.1.14), combinedwith Proposition 1.1.3, we can get Theorem 1.1.1. 2.6 Proof Of Lemma 2.3.2 And (2.4.14)This Section is devoted to prove Lemma 2.3.2. Letv :=∫Br0 (0)eλ(1+eλ|y|2)2 v(y)dy∫R2eλ(1+eλ|y|2)2 dy=1pi∫Br0 (0)eλ(1 + eλ|y|2)2v(y)dy.Then we have the following Poincare-type inequality:∫Br0 (0)eλ(1 + eλ|y|2)2φ2(y)dy ≤ c(‖φ‖2H1(Br0 (0))+ φ2) (2.6.1)for some constant c independent of λ. Using (2.6.1) we can prove the fol-lowing result.Lemma 2.6.1. Let U(x) be defined as in (2.2.3). Assume φ ∈ O(1)q,λ. Then722.6. Proof Of Lemma 2.3.2 And (2.4.14)there is a constant c > 0 such that for large λ∫Br0 (q)eUφdy = O(λ2e−λ‖φ‖H1), (2.6.2)and ∫M[|∇φ|2 − 2ρ1h(q)eUσ(x)2φ2]≥ c∫M|∇φ|2. (2.6.3)For a proof, see [15].Proof of Lemma 2.3.2. We start with part (1). Let φ ∈ O(1)q,λ and ψ ∈ O(2)q,λ.Recall 2v1 = vq,λ,a + 2φ, φ ∈ O(1)q,λ and v2 =12w + ψ, ψ ∈ O(2)q,λ. We compute〈∇(2v1 + T1(v1, v2)),∇φ1〉 = −〈∆(2v1 + T1(v1, v2)), φ1〉.Here we will use the decomposition of ∆(2v1+T1(v1, v2))in (2.3.23)-(2.3.25).〈∇(2v1 + T1(v1, v2)),∇φ1〉 =∫2∇φ · ∇φ1 −∫Br0 (q)4ρ1h(p)eUφφ1 + remainders:=2B(φ, φ1) + remainders. (2.6.4)Clearly, B is a symmetric bilinear form inO(1)q,λ and by Lemma 2.6.1, B(φ, φ) ≥c0‖φ‖2H1(Ω) for some c0 > 0. For the remainder terms, by φ1 ∈ O(1)q,λ and(2.6.2), we have∫M(8pi(a− 1) + (8pi − 2ρ1))φ1 = 0 (2.6.5)and ∣∣∣∫Br0 (q)ρ1h(p)eUφ1∣∣∣ = O(λ2e−λ)‖φ1‖H1(M). (2.6.6)Since |∇H(q)| ≤ Cλe−λ for v1 ∈ Sρ1(p, w),∫Br0 (q)∇H · (x− q)ρ1h(q)eUφ1 =O(λe−λ)∫Br0 (q)|x− q|eU |φ1|=O(λe−λ)‖φ1‖H1(M). (2.6.7)732.6. Proof Of Lemma 2.3.2 And (2.4.14)Also, by Lemma 2.6.1, we have∫Br0 (q)2ρ1h(q)eU (a− 1)(U + s− 1)φ1=4λ∫Br0 (q)ρ1h(q)eU (a− 1)φ1+ 2∫Br0 (q)ρ1h(q)eU (a− 1)(U − λ+O(1))φ1=(a− 1)O(λ2e−λ)‖φ1‖H1(M)+ (a− 1)(∫Br0 (q)eU(U − λ+O(1))2) 12(∫Br0 (q)eUφ21) 12=O(|a− 1|)‖φ1‖H1(M) = O(λe−λ)‖φ1‖H1(M). (2.6.8)For E+, we obtain∫Br0 (q)|E+φ1| ≤(∫Br0 (p)|E+|2) 12(∫Br0 (q)φ21) 12= O(λe−λ)‖φ1‖H1(M), (2.6.9)where we used (2.3.14).For E−, we see that E− = 2ρ1h(q)eU (O(ϕ2) +O(β2)), thus∫Br0 (q)|E−φ1| ≤∫Br0 (q)2ρ1h(q)eU (O(ϕ2) +O(β2))φ1=O(ε2)(∫Br0 (q)eUϕ2) 12(∫Br0 (q)eUφ21) 12+O(λ3e2λ)(∫Br0 (q)eUφ21) 12=O(ε2)(‖φ‖H1(M) + λe−λ)‖φ1‖H1(M) +O(λ3e2λ)‖φ1‖H1(M)=O(λe−λ)‖φ1‖H1(M), (2.6.10)provided ε2 is small.742.6. Proof Of Lemma 2.3.2 And (2.4.14)For the term∫Br0 (q)ρ1h(q)eUφ1ψ, we have∣∣∣∫Br0 (q)ρ1h(q)eUφ1ψ∣∣∣=∣∣∣∫Br0 (q)ρ1h(q)eUφ1(ψ − ψ(q))∣∣∣+∣∣∣∫Br0 (q)ρ1h(q)eUφ1ψ(q)∣∣∣≤C(∫Br0 (q)ρ1h(q)eUφ21) 12(∫Br0 (q)ρ1h(q)eU (ψ − ψ(q))2) 12+ |ψ(q)|∣∣∣∫Br0 (q)ρ1h(q)eUφ1∣∣∣=O(λe−λ)‖φ1‖H1(M), (2.6.11)where we used ψ ∈ O(2)q,λ and (2.6.2). This finished the estimate in Br0(q).By Lemma [15, Lemma 2.2],∫B2r0 (q)\Br0 (q)∆(vq − 8piG(x, q))φ1 = O(λeλ)‖φ1‖H1(M). (2.6.12)For the nonlinear term in ∆T1(v1, v2) in M \Br0(q), by (2.3.14), we have∫M\Br0 (q)|eϕφ1| = O(∫|ϕ|≥ε2|eϕφ1|+∫|ϕ|≤ε2|eε2φ1|) = O(1)‖φ1‖H1(M).Because∫M h1e2v1−v2 ∼ eλ,∫M\Br0 (q)2ρ1h1e2v1−v2∫M h1e2v1−v2|φ1| = O(e−λ)∫M\Br0 (q)eϕ|φ1| = O(e−λ)‖φ1‖H1(M).(2.6.13)Combining (2.6.4)-(2.6.13), we reach at〈∇(2v1 + T1(v1, v2)),∇φ1〉 =2〈∇φ,∇φ1〉 −∫Br0 (q)4ρ1h(q)eUφφ1+O(λe−λ)‖φ1‖H1(M).752.6. Proof Of Lemma 2.3.2 And (2.4.14)Next, we prove part (3). In B2r0(q), by the setting of vq, we have∂λvq =(2−ρ1h(q)2 eλ|x− q|21 + ρ1h(q)4 eλ|x− q|2+ ∂λ[η +∆H(q)ρ1h(q)λ2e−λ])σ=(1 + ∂λU)σ +O(λ2e−λ). (2.6.14)In M \ B2r0(q), ∂λvq = 0. We compute 〈∇(2v1 + T1(v1, v2)),∇∂λvq〉 =−〈∆(2v1 + T1(v1, v2)), ∂λvq〉 by using (2.3.23)-(2.3.25).Since φ ∈ O(1)q,λ, we have∫M∇φ · ∇∂λvq = 0. (2.6.15)Direct computation yields,∫Br0 (q)∂λvq =∫Br0 (q)(1 + ∂λU) +O(λ2e−λ) = O(λ2e−λ). (2.6.16)Hence,(8pi(a− 1) + 8pi − 2ρ1)∫Br0 (q)∂λvq = O(λ3e−2λ). (2.6.17)Again by (2.6.14), we have∫Br0 (q)2ρ1h(q)eU∂λvq =∫Br0 (q)2ρ1h(q)eU (1 + ∂λU +O(λ2eλ))=∫R28(1 + r2)2(1 +1− r21 + r2) +O(λ2e−λ)=8pi + λ2e−λ (2.6.18)762.6. Proof Of Lemma 2.3.2 And (2.4.14)and∫Br0 (q)2ρ1h(q)eU[− 2 log(1 +ρ1h(q)4eλ|x− q|2)]∂λvq=∫R28(1 + r2)2[−2 log(1 + r2)](1 +1− r21 + r2+O(λ2eλ))+O(λeλ)= −8pi +O(λ2eλ). (2.6.19)Combining (2.6.18) and (2.6.19), we get∫Br0 (q)2ρ1h(q)eU (U + s− 1)∂λvq= 16piλ− 16pi + 16pi logρ1h(q)4+ 64pi2R(q, q) +O(λe−λ). (2.6.20)By scaling, we compute∫Br0 (q)2ρ1h(q)eUO(|x− q|)∂λvq= (a− 1)[O(R−1)∫|z|≤r0R16r(1 + r2)3dz +O(λ2eλ)]= O(e−12λ)|a− 1| = O(λe−32λ). (2.6.21)Since ∇H(q) · (x− q) is symmetry with respect to q∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)∂λvq=∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)(1 + ∂λU +O(λ2eλ))= O(λ2eλ)∫Br0 (q)eU |x− q| = O(λ2e−32λ). (2.6.22)772.6. Proof Of Lemma 2.3.2 And (2.4.14)Next, we estimate the term φ∂λvq and ψ∂λvq. Since0 =∫M∇φ · ∇∂λvq =−∫Mφ∆(∂λvq)=−∫Br0 (q)φ∆(∂λU) +O(λe−λ‖φ‖H1(M))=2ρ1h(q)∫Br0 (q)eUφ∂λU +O(λe−λ‖φ‖H1(M)).(2.6.23)Hence, by Lemma 2.6.1 and (2.6.23), we have∫Br0 (q)2ρ1h(p)eUφ∂λvq =∫Br0 (q)2ρ1h(q)eUφ(1 + ∂λU +O(λ2e−λ))=O(λ2e−λ)‖φ‖H1(M) = O(λ3e−2λ), (2.6.24)and∫Br0 (q)2ρ1h(q)eUψ∂λvq =∫Br0 (q)2ρ1h(q)eUψ(q)∂λvq+∫Br0 (q)2ρ1h(q)eU (ψ − ψ(q))∂λvq=8piψ(q) +O(λe−32λ), (2.6.25)where we have used |ψ − ψ(q)| ∼ O(λe−λ)|x − q|. By (2.6.14), (2.6.2) andthe Moser-Trudinger inequality,∫Br0 (q)|E+∂λvq| ≤ O(e−2λ) (2.6.26)and∫Br0 (q)|E−∂λvq| =∫Br0 (q)∩{|ϕ≤ε2|}2ρ1h(q)eU (O(ϕ2) +O(β2))=O(λ3e−2λ). (2.6.27)782.6. Proof Of Lemma 2.3.2 And (2.4.14)In M \Br0(q),e2v1−v2 = O(eφ),2ρ1h1e2v1−v2∫M h1e2v1−v2= O(e−λ)eφ, ∂λvq = O(λ2eλ).Hence, by the Moser-Trudinger inequality,∫M\Br0(q)2ρ1h1e2v1−v2∫M h1e2v1−v2∂λvq = O(λ3e−2λ). (2.6.28)By [15, Lemma 2.2] and ∂λvq = O(λ2e−λ),∫B2r0 (q)\Br0 (q)∆(vq − vq − 8piG(x, q)) · ∂λvq = O(λ3e−2λ). (2.6.29)Combining (2.6.14) to (2.6.29), we obtain〈∇(2v1 + T1(v1, v2)),∇∂λvq〉=− (a− 1)(16piλ− 16pi + 16pi logρ1h(q)4+ 64pi2R(q, q))− 8pi( et∫M h1e2v1−v2− 1)+ 8piψ(q) +O(λe−32λ). (2.6.30)This proves part (3).For the proof of part (4), we write〈∇(2v1 + T1(v1, v2)),∇vq〉 = 〈∇(2v1 + T1(v1, v2)),∇(vq − vq)〉= −〈∆(2v1 + T1(v1, v2)), (vq − vq)〉.First we note that∫M[8pi(a− 1) + 8pi − 2ρ1](vq − vq) = 0.By φ ∈ O(1)q,λ, ∫M∇φ · ∇(vq − vq) = 0.792.6. Proof Of Lemma 2.3.2 And (2.4.14)In Br0(p),vq − vq =2λ− 2 log(1 +ρ1h(q)4eλ|x− q|2) + 8piR(q, q) +O(|y|)+ 2 logρ1h(q)4+O(λ2eλ). (2.6.31)We use (2.6.31) to compute∫M 2ρ1heU (U + s− 1)(vq− vq), after scaling, wehave∫Br0 (q)2ρ1h(q)eU log(1 +ρ1h(q)4eλ|x− q|2)=∫R28(1 + r2)2log(1 + r2) +O(λeλ)=8pi +O(λeλ), (2.6.32)∫Br0 (q)2ρ1h(q)eU[log(1 +ρ1h(q)4eλ|x− q|2)]2=∫ ∞08(1 + r2)2[log(1 + r2)]22pirdr +O(λ2eλ)=8pi +O(λ2eλ), (2.6.33)and∫Br0 (q)2ρ1heU[log(1 +ρ1h(q)4eλ|x− q|2)]O(|x− q|) = O(e−12λ). (2.6.34)802.6. Proof Of Lemma 2.3.2 And (2.4.14)Therefore, by (2.6.32)-(2.6.34),∫Br0 (q)2ρ1heU (U + s− 1)(vq − vq)=∫Br0 (q)2ρ1heU[4λ2 − 8 log(1 +ρ1h(q)4eλ|x− q|2)λ+ λ(32piR(q, q) + 8 logρ1h(q)4− 2)]+O(1)=[256pi2R(q, q) + 64pi logρ1h(q)4− 16pi]λ+ 32piλ2 − 64piλ+O(1). (2.6.35)Similarly, we have∫Br0 (q)2ρ1h(q)eU (vq − vq) =16piλ− 16pi + 64pi2R(q, q)+ 16pi logρ1h(q)4+O(e−12λ), (2.6.36)and∫Br0 (q)2ρ1h(q)eUO(|x− q|)(vq − vq) = O(λe− 12λ). (2.6.37)Since ∇H(q) = O(λe−λ) and ∇H(q) · (x− q) is symmetry with respect to qin Br0(q), by (2.6.31), we have∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)(vq − vq)=∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)O(|x− q|) +O(λ2e−2λ)=O(λ2e−2λ). (2.6.38)812.6. Proof Of Lemma 2.3.2 And (2.4.14)By (2.6.2), and Lemma 2.6.1,∫Br0 (q)2ρ1h(q)eUϕ(vq − vq)=∫Br0 (q)4ρ1h(q)eUφ[λ+ s− 2 log(1 +ρ1h(q)4eλ|x− q|2) +O(|x− q|)]−∫Br0 (q)2ρ1h(q)eUψ[λ+ s− 2 log(1 +ρ1h(q)4eλ|x− q|2) +O(|x− q|)]=O(e−12λ)‖φ‖H1(M) + (∫Br0 (q)2ρ1h(q)eUφ2)12× (∫Br0 (q)2ρ1h(q)eU[log(1 +ρ1h(q)4eλ|x− q|2) +O(|x− q|)]2)12−∫Br0 (q)2ρ1h(q)eUψ(q)[λ+ s− 2 log(1 +ρ1h(q)4eλ|x− q|2) +O(|x− q|)]−∫Br0 (q)2ρ1h(q)eU (ψ − ψ(q))[λ+ s− 2 log(1 +ρ1h(q)4eλ|x− q|2) +O(|x− q|)]=O(1)‖φ‖H1(M) + o(1)‖ψ‖∗ − 16piλjψ(q). (2.6.39)By a similar argument as in the proof of part (3),∫Br0 (q)E(vq − vq) = O(λ4e−2λ). (2.6.40)Since vq = O(1) in M \Br0(q), by [15, Lemma 2.2],∫B2r0 (q)\Br0 (q)∆(vq − 8piG(x, q))(vq − vq) = O(λe−λ). (2.6.41)For the integral outside of Br0(q), we have (∫M h1e2v1−v2)−1 = O(e−λ)and∫B2r0 (q)\Br0 (q)2ρ1h∫M he2v1−v2e2v1−ψ(vq − vq)=∫B2r0 (q)\Br0 (q)O(e−λ)e2φ−ψ = O(e−λ). (2.6.42)822.6. Proof Of Lemma 2.3.2 And (2.4.14)Similarly∫M\B2r0 (q)2ρ1h∫M h1e2v1−v2e2v1−ψ(vq − vq) = O(e−λ). (2.6.43)By (2.6.35)-(2.6.43), we have〈∇(2v1 + T1(v1, v2),∇(vq − vq)〉=−(32piλ2 − 64piλ+ 256pi2R(q, q)λ+ 64pi logρ1h(q)4λ)(a− 1)+(16piλ− 16pi + 64pi2R(q, q) + 16pi logρ1h(q)4)( et∫M h1e2v1−v2− 1− ψ(q))+O(1)‖φ‖H1(M) + o(1)‖ψ‖∗ + 16pi(a− 1)λ+O(λe−λ)=(2λ− 2 + 8piR(q, q) + 2 logρ1h(q)8)〈∇(v1 + T1(v1, v2)),∇∂λvq〉+ 16pi(a− 1)λ+O(1)‖φ‖H1(M) + o(1)‖ψ‖∗ +O(λe−λ). (2.6.44)Finally, we prove part (2). We note that〈∇(2v1 + T1(v1, v2)),∇vq〉 = 〈∇(2v1 + T1(v1, v2)),∇(vq − vq)〉.From φ ∈ O(1)q,λ, we have〈∇φ,∇∂q(vq − vq)〉 = 0.Since∫M (vq − vq) = 0, we have∫M ∂q(vq − vq) = 0 and∫M(8pi(a− 1) + 8pi − 2ρ1)∂q(vq − vq) = 0.In Br0(q), by [14, Lemma 2.1]∂qvq =−∇xU +∂qh(q)h(q)∂λU + ∂q(2 log h(q) +∆H(q)ρ1h(q)λ2eλ)+ 8pi∂qR(x, q) |x=q +O(|x− q|) +O(λ2e−λ). (2.6.45)832.6. Proof Of Lemma 2.3.2 And (2.4.14)Since ∇xU is symmetry with respect to q in Br0(q),∫Br0 (q)2ρ1h(q)eU (U + s− 1)∇xU=∫Br0 (q)2ρ1h(q)eUO(|x− q|+ λ2e−λ)∇xU = 0. (2.6.46)Hence by the fact ∂λU is bounded,∫Br0 (q)2ρ1h(q)(U + s− 1)∂q(vq − vq) = O(λ). (2.6.47)For the other terms in (2.3.23), we need to estimate the following quantities:∫Br0 (q)2ρ1h(q)eU∂q(vq − vq) = O(1), (2.6.48)∫Br0 (q)2ρ1h(q)eUO(|x− q|)∂q(vq − vq) = O(1), (2.6.49)∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)∇xU=∇H(q)(∫R28(1 + r2)22r21 + r2+O(e−λ))=(8pi +O(λe−λ))∇H(q), (2.6.50)and∫Br0 (q)2ρ1h(q)eU∇H(q) · (x− q)∂q(vq − vq)= (8pi +O(e−λ))∇H(q) + |∇H(q)|∫Br0 (q)2ρ1h(q)eUO(|x− q|)= 8pi∇H(q) +O(λe−32λ), (2.6.51)where we have used ∇H(q) = O(λe−λ) for v1 ∈ Sρ1(p, w).842.6. Proof Of Lemma 2.3.2 And (2.4.14)For the term ρ1h(q)eUφ∂q(vq − vq),∫Br0 (q)2ρ1h(q)eUφ∂q(vq − vq)=4∫Br0 (q)ρ1h(q)eUφ∂q(vq − vq)=4∫Br0 (q)ρ1h(q)eUφ∂qU+∫Br0 (q)ρ1h(q)eUφ(O(λ2eλ) +O(|x− q|)). (2.6.52)Using 〈∇φ,∇∂q(vq − vq)〉 = 0, it holds that0 =∫M∇φ∇∂qvq = −∫Mφ∆(∂qvq)=∫Br0 (q)2ρ1h(q)eU∂qUφ+ ∂q log h(q)∫Br0 (q)2ρ1h(q)eUφ+O(λ2e−λ)‖φ‖H1(M).By (2.6.2) and the above equality, we have2∫Br0 (q)2ρ1h(q)eU∂qUφ = O(λ2e−λ)‖φ‖H1(Ω).While for the term et∫M h1e2v1−v2− 1 and ψ, we have∫Br0 (q)2ρ1h(q)eU( et∫M h1e2v1−v2− 1− ψ)∂q(vq − vq)=∫Br0 (q)2ρ1h(q)eU( et∫M h1e2v1−v2− 1− ψ(q))∂q(vq − vq)−∫Br0 (q)2ρ1h(q)eU (ψ − ψ(q))∂q(vq − vq)=− 8pi∇ψ(q) +O( et∫M h1e2v1−v2− 1− ψ(q)),852.6. Proof Of Lemma 2.3.2 And (2.4.14)where we used∫Br0 (q)2ρ1h(q)eU∇ψ(q)(x− q)∇yU = (8pi +O(e−λ))∇ψ(q),and (2.6.48). Since ∂q(vq − vq) = O(e12λ), as in the proof of part (3), wehave∫Br0 (q)E∂λq(vq − vq) = O(λ3e−32λ).In M \Br0(q), ∂q(vq − vq) = O(1). Hence by [15, Lemma 2.2],∫B2r0 (q)\Br0 (q)∆(vq − 8piG(x, q)) · ∂q(vq − vq) = O(λe−λ).Since∫M h1e2v1−v2 = O(e−λ), the integral of the products of ∂qvq and thenonlinear terms in (2.3.24) and (2.3.25) are of orderO(e−λ)∫Meϕ = O(e−λ).The estimates above imply〈∇(2v1 + T1(v1, v2)),∇∂q(vq − vq)〉=− 8pi∇H(q) + 8pi∇ψ(q)+O(λ|a− 1|+∣∣∣et∫M h1e2v1−v2− 1− ψ(q)∣∣∣+λeλ). (2.6.53)This proves the part (2) and hence the proof of Lemma is completed. Next, we give a proof of (2.4.14).Proof of (2.4.14): For convenience, we denoteE2 = exp(w + 2ψ −12vq,λ,a)− exp(w + 2ψ − 4piaG(x, q)).862.6. Proof Of Lemma 2.3.2 And (2.4.14)For x ∈M \Br0(q). By [15, Lemma 2.2], we have∣∣12vq,λ,a − 4piaG(x, q)∣∣ ≤ c˜λe−λfor some c˜ independent of c1. Thus |E2| ≤ c5λe−λ in M \Br0(q).For x ∈ Br0(q), we note4piaG(x, q)−12vq,λ,a =4piaG(x, q)− 4piaR(x, q)− a log(ρ1h(q)4)+ a log(1 +ρ1h(q)eλ4|x− q|2)− aλ−a2(η +∆H(q)ρ1h(qk)λ2eλ)+O(λe−λ)=a log( 4ρ1h(q)eλ|x− q|2+ 1)−a2(η +∆H(q)ρ1h(qk)λ2eλ)+O(λe−λ),where we have used vq = O(λe−λ). Then, we haveexp(w + 2ψ − 4piaG(x, q))− exp(w + 2ψ −12vq,λ,a)= O(1)|x− q|2a(1− exp(4piaG(x, p)−12vq,λ,a))= O(1)|x− q|2a(1− exp[a log(1 +4ρ1h(q)eλ|x− q|2)+O(η +∆H(q)ρ1h(q)λ2eλ)+O(λe−λ)]+O(λe−λ)). (2.6.54)When |x− q| = O(e−12λ), we haveexp(w + 2ψ − 4piaG(x, q))= O(|x− q|2a),and1− exp(a log(1 +4ρ1h(q)eλ|x− q|2)+O(λ2e−λ))= O(e−aλ|x− q|−2a),872.7. The Leray-Schauder degreewhich impliesE2 = O(λe−λ) for |x− q| = O(e−12λ).When |x− q|  e−12λ, then1− exp(a log(1 +4ρ1h(q)eλ|x− q|2) +O(η +∆H(q)ρ1h(q)λ2eλ))+O(λe−λ)= O( 1eaλ|x− q|2a+ λe−λ).As a result, we have the right hand side of (2.6.54) are of order O(λe−λ).ThereforeE2 = O(λe−λ) for x ∈ Br0(q).Thus, we get (2.4.14). 2.7 The Leray-Schauder degreeIn this section, we provide a short introduction of the Leray-Schauder degree,namely the degree for the maps T ∈ C(B,B), where B is a Banach spaceand T is a compact perturbation of the identity I = IB.Let D be an open bounded subset of the Banach space B. We shalldeal with compact perturbations of the identity, namely with operators T ∈C(D,B) such that T = I −K, where K is a compact.Let p /∈ T (∂D). It is easy to check that T (∂D) is closed and hencer := dist(p, T (∂D)) > 0.It is known, see [10], that there exists a sequence Kn ∈ C(D,B) such thatKn → K uniformly in D andKn(D) ⊂ En ⊂ B, with dim(En) <∞. (2.7.1)We shall define the degree of I−K as the limit of the degrees of I−Kn, which882.7. The Leray-Schauder degreewe are going to introduce. Before that, we make the following preparations.Let us consider a map φ ∈ C(Ω,Rm1), where Ω ⊂ Rm2 and m1 ≤ m2.We can regard Rm1 as the subset of Rm2 whose points have the last m2−m1components equal to zero:Rm1 = {x ∈ Rm2 : xm1+1 = · · · = xm2 = 0}.The above function φ can be considered as a map with values on Rm2 byunderstanding that the last m2 −m1 components are zero: φm1+1 = · · · =φm2 = 0. Let g(x) = x − φ(x) and let gm1 ∈ C(Ω ∩ Rm1 ,Rm1) denote therestriction of g to Ω ∩ Rm1 . Let us show that if p ∈ Rm1 \ g(∂Ω) thendeg(g,Ω, p) = deg(gm1 ,Ω ∩ Rm1 , p). (2.7.2)Let x ∈ Ω be such that g(x) = p. This means that x = φ(x) + p. Thusx ∈ Ω ∩ Rm1 and so gm1(x) = g(x) = p. This shows that g−1(p) ⊂ g−1m1(p).Since the converse is trivially true, it follows that g−1(p) = g−1m1(p). We cansuppose that Ω∩Rm1 6= ∅, otherwise, g−1m1(p) = ∅ and g−1(p) = ∅. As usual,we can suppose that φ is of class C1 and, moreover, that p is a regular valueof gm1 . Then according to the definition of degree, we getdeg(g,Ω, p) =∑x∈g−1(p)sgn[Jg(x)].Now the Jacobian matrix g′(x) is in triangular form(g′m1(x) ·0 IRm2−m1),and hence sgn[Jg(x)] = sgn[Jgm1 (x)]. As a consequence, we havedeg(g,Ω, p) =∑x∈g−1(p)sgn[Jg(x)] =∑x∈g−1m1 (p)sgn[Jgm1 (x)] = deg(gm1 ,Ω ∩ Rm1 , p),which proves (2.7.2), provided p is a regular value. In the general case, we892.7. The Leray-Schauder degreeuse the Sard Lemma to get the same conclusion.The above discussion allows us to define the degree for a map g such thatg(x) = x − φ(x), where φ(D) is contained in a finite dimensional subspaceE of B. Let p ∈ B, p 6= g(D). Let E1 be a subspace of B containing E andp. We get g1 = g |D∩E1 and definedeg(g,D, p) = deg(g1, D ∩ E1, p). (2.7.3)Let us show that the definition is independent of E1. Let E2 be anothersubspace of B such that E ⊂ E2 and p ∈ E2. Then E ∩ E1 ∩ E2 andp ∈ E1 ∩ E2. Applying (2.7.2) we obtaindeg(gi, D ∩ Ei, p) = deg(g |D∩E1∩E2 , D ∩ E1 ∩ E2, p), i = 1, 2.This justifies the definition given in (2.7.3).Now, let us come back to the map T = I − K, with K compact. LetKn → K satisfy (2.7.1) and set Tn = I −Kn. Taking n such thatsupx∈D‖Kn(x)−K(x)‖ ≤ε2, (2.7.4)we know that p /∈ Tn(D) and hence it makes sense to consider the degreedeg(Tn, D, p) defined in (2.7.3).Definition 2.7.1. Let p /∈ T (∂D), where T = I −K with K compact. Wesetdeg(T,D, p) = deg(I −Kn, D, p),for any Kn satisfying (2.7.1) and (2.7.4).Once more, we have to verify the definition, by showing that the degreedoes not depend on the approximation Kn. To prove this claim, let Ti, i =1, 2, be such (2.7.1)-(2.7.4) hold. Let Ei be finite dimensional spaces suchthat Ki(D) ⊂ Ei. If E is the space spanned by E1 and E2, we use the902.7. The Leray-Schauder degreedefinition (2.7.3) to getdeg(Ti, D, p) = deg((Ti) |D∩E , D ∩ E, p), i = 1, 2. (2.7.5)Consider the homotopyh(λ, ·) = λ(T1) |D∩E +(1− λ)(T2)D∩E .It is easy to check that h is admissible on D ∩ E, i.e., h(λ, ·) 6= p for all(λ, x) ∈ [0, 1]× ∂(D ∩ E). Thus,deg((T1) |D∩E , D ∩ E, p) = deg((T2) |D∩E , D ∩ E, p).This together with (2.7.5) proved that Definition 2.7.1 is well defined.91Chapter 3The Lin-Ni Problem3.1 Approximate SolutionsIn this section, we construct suitable approximate solution, in the neigh-bourhood of which solutions in Theorem 1.2.1 will be found.Let ε be as defined in (1.2.11). For any Q ∈ Ωε with d(Q, ∂Ωε) large,UΛ,Q/ε =(ΛΛ2+|x−Qε |2)n−22provides an approximate solution of (1.2.12). Be-cause of the appearance of the additional linear term µε2u, we need to addan extra term to get a better approximation. To this end, for n = 4, weconsider the following equation∆Ψ¯ + U1,0 = 0 in R4, Ψ¯(0) = 1. (3.1.1)ThenΨ¯(|y|) = −12ln |y|+I+O( 1|y|), Ψ¯′= −12|y|(1+O( ln(1 + |y|)|y|2))as |y| → ∞,(3.1.2)where I is a constant. LetΨΛ,Q =Λ2ln1Λε+ ΛΨ¯(y −QΛ). (3.1.3)Then∆ΨΛ,Q + UΛ,Q = 0.We note that|ΨΛ,Q(y)|, |∂ΛΨΛ,Q(y)| ≤ C∣∣∣ ln1ε(1 + |y −Q|)∣∣∣, |∂QiΨΛ,Q(y)| ≤C1 + |y −Q|.(3.1.4)923.1. Approximate SolutionsFor n = 6, let Ψ(|y|) be the radial solution of∆Ψ + U1,0 = 0 in Rn, Ψ→ 0 as |y| → +∞. (3.1.5)Then, it is easy to check thatΨ(y) =14|y|2(1 +O(1|y|2)) as |y| → +∞. (3.1.6)For Q ∈ Ωε, we setΨΛ,Q(y) = Ψ(y −QΛ).Then∆ΨΛ,Q(y) + UΛ,Q = 0 in R6.It is easy to see that|ΨΛ,Q(y)|, |∂ΛΨΛ,Q(y)| ≤C(1 + |y −Q|)2, |∂QiΨΛ,Q(y)| ≤C(1 + |y −Q|)3.(3.1.7)In order to obtain approximate solutions which satisfy the boundarycondition, we defineUˆΛ,Q/ε(z) = −ΨΛ,Q/ε(z)− cnµ−1εn−4Λn−22 H(εz,Q) +Rε,Λ,Q(z)χ(εz),(3.1.8)where Rε,Λ,Q is defined by ∆Rε,Λ,Q − ε2Rε,Λ,Q = 0 in Ωε andµε2∂Rε,Λ,Q∂ν= −∂∂ν[UΛ,Q/ε − µε2ΨΛ,Q/ε − cnεn−2Λn−22 H(εz,Q)]on ∂Ωε,(3.1.9)where χ(x) is a smooth cut-off function in Ω such thatχ(x) =1 for d(x, ∂Ω) < δ4 ,0 for d(x, ∂Ω) > δ2 .We note (3.1.2) and (3.1.6), an expansion of UΛ,Q/ε and the definition of Himply that the normal derivative of Rε,Q is of order εn−3 on the boundary933.1. Approximate Solutionsof Ωε, from which we deduce that 2|Rε,Λ,Q|+ |ε−1∇zRε,Λ,Q|+ |ε−2∇2zRε,Λ,Q| ≤CΛ, n = 4,Cε2, n = 6.(3.1.10)A similar estimate also holds for the derivatives of Rε,Λ,Q with respect toΛ, Q.Now we are able to define the approximate bubble solutions. Since it isdifferent in constructing the approximate solution for n = 4 and n = 6, weshall tact them respectively. For n = 4, letΛ4,1 ≤ Λ ≤ Λ4,2, Q ∈Mδ4 := {x ∈ Ω| d(x, ∂Ω) > δ4}, (3.1.11)where Λ4,1 = exp(−12)εβ, Λ4,2 = exp(−12)ε−β, β is a small constant with ageneric constant δ4, to be determined later. We writeQ¯ =1εQ,and define our approximate solutions asWε,Λ,Q = UΛ,Q/ε + µε2UˆΛ,Q/ε +c4Λ|Ω|µ−1ε2. (3.1.12)For n = 6, let√|Ω|c6(196− Λ6ε23 ) ≤Λ ≤√|Ω|c6(196+ Λ6ε23 ),Q ∈Mδ6 := {x ∈ Ω| d(x, ∂Ω) > δ6},148− η6ε13 ≤η ≤148+ η6ε13 , (3.1.13)where Λ6 and η6 are some constants that may depend on the domain, δ6 isa small constant, which are determined later. Our approximate solution for2For n = 4, we set the parameter Λ in a range that depends on ε, we have to takeΛ into consideration, and we note that each component on the right hand side of (3.1.9)carry Λ as a factor.943.1. Approximate Solutionsn = 6 is the followingWε,Λ,Q,η = UΛ,Q/ε + µε2UˆΛ,Q/ε + ηµ−1ε4. (3.1.14)For convenience, in the following, we write W, U, Uˆ , R, and Ψ insteadof Wε,Λ,Q, Uε,Q/ε, UˆΛ,Q/ε, Rε,Λ,Q and ΨΛ,Q/ε respectively in the following.By construction, the normal derivative of W vanishes on the boundary ofΩε, and W satisfies−∆W + µε2W ={8U3 + µ2ε4Uˆ − µε2∆(Rε,Λ,Qχ), n = 4,24U2 + µ2ε4Uˆ − µε2∆(Rε,Λ,Qχ) + ε6(η −c6Λ2|Ω| ), n = 6.(3.1.15)We note that W depends smoothly on Λ, Q¯. Setting, for z ∈ Ωε,〈z − Q¯〉 = (1 + |z − Q¯|2)12 .A simple computation shows|W (z)| ≤{C(ε2(− ln ε)12 + 〈z − Q¯〉−2), n = 4,C(ε3 + 〈z − Q¯〉−4), n = 6,(3.1.16)|DΛW (z)| ≤{C(ε2(− ln ε)12 + 〈z − Q¯〉−2), n = 4,C〈z − Q¯〉−4, n = 6,(3.1.17)|DQ¯W (z)| ≤{C(〈z − Q¯〉−3), n = 4,C(〈z − Q¯〉−5), n = 6,(3.1.18)and|DηW (z)| = O(ε3), n = 6. (3.1.19)According to the choice of W, we have the following error and energyestimates, we leave the proof in Section 6 of this Chapter.953.1. Approximate SolutionsLemma 3.1.1. We setSε[u] := −∆u+ µε2u− n(n− 2)un+2n−2+ , u+ = max(u, 0),and introduce the following functionalJε[u] :=12∫Ωε|∇u|2 +12µε2∫Ωεu2 −(n− 2)22∫Ωε|u|2nn−2 , u ∈ H1(Ωε).For n = 4, we have|Sε[W ](z)| ≤ C(〈z − Q¯〉−4ε2(− ln ε)12 + 〈z − Q¯〉−2ε4(− ln ε)+ε4(− ln ε)12+ε4(− ln ε)∣∣ ln( 1ε(1 + |z − Q¯|))∣∣), (3.1.20)|DΛSε[W ](z)| ≤ C(〈z − Q¯〉−4ε2(− ln ε)12 + 〈z − Q¯〉−2ε4(− ln ε)+ε4(− ln ε)12+ε4(− ln ε)∣∣ ln( 1ε(1 + |z − Q¯|))∣∣), (3.1.21)|DQ¯Sε[W ](z)| ≤ C(〈z − Q¯〉−5ε2(− ln ε)12 + 〈z − Q¯〉−3ε4(− ln ε)+ 〈z − Q¯〉−1ε4(− ln ε)), (3.1.22)andJε[W ] = 2∫R4U41,0 +c4Λ24ε2(c1− ln ε)12 ln1Λε−c24Λ22|Ω|ε2(c1− ln ε)−12+12c24ε2Λ2H(Q,Q) +O(ε2( c1− ln ε) 12 Λ2)+O(ε4(− ln ε)2).(3.1.23)963.2. Finite Dimensional ReductionFor n = 6, we haveSε[W ](z) = −ε6(24η2 − η +c6Λ2|Ω|)+O(1)ε3〈z − Q¯〉−4, (3.1.24)|DΛSε[W ](z)| = O(1)(〈z − Q¯〉−4ε3 + ε6), (3.1.25)|DηSε[W ](z)| = O(1)(〈z − Q¯〉−4ε3 + ε613), (3.1.26)|DQ¯Sε[W ](z)| ≤ C〈z − Q¯〉−5ε3, (3.1.27)andJε[W ] =4∫R6U31,0 +(12η2|Ω| − c6ηΛ2 +148c6Λ2 − 8η3|Ω|)ε3 +12c26Λ4ε4H(Q,Q)+12(η −c6Λ2|Ω|)ε4∫ΩΛ2|x−Q|4+O(ε5). (3.1.28)3.2 Finite Dimensional ReductionAccording to the general strategy used in Lyapunov-Schmidt reduction method,we first consider the linearized problem at W, and solve it in a finite-codimensional space, i.e., the orthogonal space to the finite-dimensional sub-space generated by the derivatives of W with respect to the parameters Λand Q¯i in the case n = 4, and the orthogonal space to the finite-dimensionalsubspace generated by the derivatives of W with respect to the parametersΛ, Q¯i and η in the case n = 6. Equipping H1(Ωε) with the scalar product(u, v)ε =∫Ωε(∇u · ∇v + µε2uv). (3.2.1)973.2. Finite Dimensional ReductionFor the case n = 4. Orthogonality to the functionsY0 =∂W∂Λ, Yi =∂W∂Q¯i, 1 ≤ i ≤ 4, (3.2.2)in that space is equivalent to the orthogonality in L2(Ωε), equipped withthe usual scalar product 〈·, ·〉, to the functions Zi, 0 ≤ i ≤ 4, defined asZ0 = −∆∂W∂Λ + µε2 ∂W∂Λ ,Zi = −∆ ∂W∂Q¯i + µε2 ∂W∂Q¯i, 1 ≤ i ≤ 4.(3.2.3)Straightforward computations provide us with the estimate:|Zi(z)| ≤ C(ε4 + 〈z − Q¯〉−6). (3.2.4)Then, we consider the following problem: given h, finding a solution φwhich satisfies−∆φ+ µε2φ− 24W 2φ = h+ Σ4i=0ciZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 4,(3.2.5)for some numbers ci.While for the case n = 6. Orthogonality to the functionsY0 =∂W∂Λ, Yi =∂W∂Q¯i, 1 ≤ i ≤ 6, Y7 =∂W∂η, (3.2.6)in that space is equivalent to the orthogonality in L2(Ωε), equipped withthe usual scalar product 〈·, ·〉, to the functions Zi, 0 ≤ i ≤ 7, defined asZ0 = −∆∂W∂Λ + µε2 ∂W∂Λ ,Zi = −∆ ∂W∂Q¯i + µε2 ∂W∂Q¯i, 1 ≤ i ≤ 6,Z7 = −∆∂W∂η + µε2 ∂W∂η .(3.2.7)983.2. Finite Dimensional ReductionDirect computations provide us the following estimate:|Zi(z)| ≤ C(ε6 + 〈z − Q¯〉−8), 0 ≤ i ≤ 6, Z7(z) = O(ε6). (3.2.8)Then, we consider the following problem: given h, finding a solution φwhich satisfies−∆φ+ µε2φ− 48Wφ = h+ Σ7i=0diZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 7,(3.2.9)for some numbers di.Existence and uniqueness of φ will follow from an inversion procedure insuitable weighted function space. To this end, we define‖φ‖∗ = ‖〈z − Q¯〉φ(z)‖∞, ‖f‖∗∗ = ε−3(− ln ε)12 |f |+ ‖〈z − Q¯〉3f(z)‖∞, n = 4,‖φ‖∗∗∗ = ‖〈z − Q¯〉2φ(z)‖∞, ‖f‖∗∗∗∗ = ‖〈z − Q¯〉4f(z)‖∞, n = 6,(3.2.10)where ‖f‖∞ = maxz∈Ωε |f(z)| and f = |Ωε|−1∫Ωεf(z)dz denotes the aver-age of f in Ωε.Before stating an existence result for φ in (3.2.5) and (3.2.9), we needthe following lemma:Lemma 3.2.1. Let u and f satisfy−∆u = f,∂u∂ν= 0, u¯ = f¯ = 0.Then|u(x)| ≤ C∫Ωε|f(y)||x− y|n−2dy. (3.2.11)Proof. The proof is similar to [63, Lemma 3.1], we omit it here.As a consequence, we have993.2. Finite Dimensional ReductionCorollary 3.2.1. For n = 4, suppose u and f satisfy−∆u+ µε2u = f in Ωε,∂u∂ν= 0 on ∂Ωε.Then‖u‖∗ ≤ C‖f‖∗∗. (3.2.12)For n = 6, suppose u and f satisfy−∆u+ cµε2u = f in Ωε,∂u∂ν= 0 on ∂Ωε, u = f = 0,where c is an arbitrary constant. Then‖u‖∗∗∗ ≤ C‖f‖∗∗∗∗. (3.2.13)Proof. For n = 4, integrating the equation yields f¯ = µε2u¯, we may rewritethe original equation as∆(u− u¯) = µε2(u− u¯)− (f − f¯).With the help of Lemma 3.2.1, we get|u(y)− u¯| ≤ Cµε2∫Ωε|u(x)− u¯||x− y|2dx+ C∫Ωε|f(x)− f¯ ||x− y|2dx.Since〈y − Q¯〉∫R41|x− y|2〈x− Q¯〉−3dx <∞,we obtain‖〈y − Q¯〉|u− u¯|‖∞ ≤ Cµε2‖〈y − Q¯〉3|u− u¯|‖∞ + C‖〈y − Q¯〉3|f − f¯ |‖∞≤ Cµ‖〈y − Q¯〉|u− u¯|‖∞ + C‖〈y − Q¯〉3|f − f¯ |‖∞,which gives‖〈y − Q¯〉|u− u¯|‖∞ ≤ C‖〈y − Q¯〉3|f − f¯ |‖∞,1003.2. Finite Dimensional Reductionwhence‖〈y − Q¯〉u‖∞ ≤ C‖〈y − Q¯〉‖∞|u¯|+ Cε−3|f¯ |+ ‖〈y − Q¯〉3f‖∞ ≤ C‖f‖∗∗.Hence we finish the proof of the case n = 4.For n = 6, by the help of Lemma 3.2.1,|〈y − Q¯〉2u| ≤ C∫Ωε〈y − Q¯〉2(|µε2u|+ |f |)|x− y|4dx ≤ C(|µ ln ε|‖u‖∗∗∗ + ‖f‖∗∗∗∗),where we used some similar estimates appeared in n = 4. From the aboveinequality, we obtain ‖u‖∗∗∗ ≤ ‖f‖∗∗∗∗. Hence we finish the proof.We now state the main result of this sectionProposition 3.2.1. There exist ε0 > 0 and a constant C > 0, independentof ε, Λ, Q¯ satisfying (3.1.11) and independent of ε, η, Λ, Q¯ satisfying(3.1.13), such that for all 0 < ε < ε0 and all h ∈ L∞(Ωε), problem (3.2.5),(3.2.9) has a unique solution φ = Lε(h) and the following estimates hold‖Lε(h)‖∗ ≤ C‖h‖∗∗, |ci| ≤ C‖h‖∗∗ for 0 ≤ i ≤ 4,‖Lε(h)‖∗∗∗ ≤ C‖h‖∗∗∗∗, |di| ≤ C‖h‖∗∗∗∗ for 0 ≤ i ≤ 6. (3.2.14)Moreover, the map Lε(h) is C1 with respect to Λ, Q¯ of the L∞∗ -norm inn = 4 and with respect to Λ, Q¯, η of the L∞∗∗∗-norm in n = 6, i.e.,‖D(Λ,Q¯)Lε(h)‖∗ ≤ C‖h‖∗∗ in n = 4, ‖D(η,Λ,Q¯)Lε(h)‖∗∗∗ ≤ Cε−1‖h‖∗∗∗∗ in n = 6.(3.2.15)The argument goes the same as the Proposition 3.1 in [63], and we listthe proof here. First, we need the following lemmaLemma 3.2.2. For n = 4, assuming that φε solves (3.2.5) for h = hε.If ‖hε‖∗∗ goes to zero as ε goes to zero, so does ‖φε‖∗. While for n = 6,assuming that φε solves (3.2.9) for h = hε. If ‖hε‖∗∗∗∗ goes to zero as ε goesto zero, so does ‖φε‖∗∗∗.1013.2. Finite Dimensional ReductionProof. We prove this lemma by contradiction and first consider n = 4.Assuming ‖φε‖∗ = 1. Multiplying the first equation in (3.2.5) by Yj andintegrating in Ωε we find∑ici〈Zi, Yj〉 = 〈−∆Yj + µε2Yj − 24W2Yj , φε〉 − 〈hε, Yj〉.We can easily get the following equalities from the definition of Zi, Yj〈Z0, Y0〉 = ‖Y0‖2ε = γ0 + o(1),〈Zi, Yi〉 = ‖Yi‖2ε = γ1 + o(1), 1 ≤ i ≤ 4, (3.2.16)where γ0, γ1 are strictly positive constants, and〈Zi, Yj〉 = o(1), i 6= j. (3.2.17)On the other hand, in view of the definition of Yj and W , straightforwardcomputations yield〈−∆Yj + µε2Yj − 24W2Yj , φε〉 = o(‖φε‖∗)and〈hε, Yj〉 = O(‖hε‖∗∗).Consequently, inverting the quasi diagonal linear system solved by the ci’swe findci = O(‖hε‖∗∗) + o(‖φε‖∗). (3.2.18)In particular, ci = o(1) as ε goes to zero.Since ‖φε‖∗ = 1, elliptic theory shows that along some subsequence, thefunctions φε,0 = φε(y − Q¯) converge uniformly in any compact subset of R4to a nontrivial solution of−∆φ0 = 24U2Λ,0φ0.A bootstrap argument (see e.g. Proposition 2.2 of [68]) implies |φ0(y)| ≤1023.2. Finite Dimensional ReductionC(1 + |y|)−2. As consequence, φ0 can be written asφ0 = α0∂UΛ,0∂Λ+∑iαi∂UΛ,0∂yi,(see [62]). On the other hand, equalities 〈Zi, φε〉 = 0 yield∫R4−∆∂UΛ,0∂Λφ0 =∫R4U2Λ,0∂UΛ,0∂Λφ0 = 0,∫R4−∆∂UΛ,0∂yiφ0 =∫R4U2Λ,0∂UΛ,0∂yiφ0 = 0, 1 ≤ i ≤ 4.As we also have∫R4∣∣∣∇∂UΛ,0∂Λ∣∣∣2= γ0 > 0,∫R4∣∣∣∇∂UΛ,0∂yi∣∣∣2= γ1 > 0, 1 ≤ i ≤ 4,and ∫R4∇∂UΛ,0∂Λ∇∂UΛ,0∂yi=∫R4∇∂UΛ,0∂yi∇∂UΛ,0∂yj= 0, i 6= j,the α′is solve a homogeneous quasi diagonal linear system, yielding αi =0, 0 ≤ i ≤ 4, and φ0 = 0. So φε(z − Q¯)→ 0 in C1loc(Ωε). Next, we will show‖φε‖∗ = o(1) by using the equation (3.2.5).Using (3.2.5) and Corollary 3.2.1, we have‖φε‖∗ ≤ C(‖W2φε‖∗∗ + ‖h‖∗∗ +∑i|ci|‖Zi‖∗∗). (3.2.19)Then we estimate the right hand side of (3.2.19) term by term. By the helpof (3.1.16), we deduce that|〈z − Q¯〉3W 2φε| ≤ C(ε4(− ln ε)〈z − Q¯〉2‖φε‖∗ + 〈z − Q¯〉−1|φε|). (3.2.20)Since ‖φε‖∗ = 1, the first term on the right hand side of (3.2.20) is dominatedby ε2(− ln ε). The last term goes uniformly to zero in any ball BR(Q¯), andis dominated by 〈z − Q¯〉−2‖φε‖∗ = 〈z − Q¯〉−2, which, through the choice of1033.2. Finite Dimensional ReductionR, can be made as small as possible in Ωε\BR(Q¯). Consequently,|〈z − Q¯〉3W 2φε| = o(1) (3.2.21)as ε goes to zero, uniformly in Ωε. On the other hand, we can also getε−3(− ln ε)12W 2φε ≤ Cε(− ln ε)12∫Ωε(〈z − Q¯〉−4 + ε4(− ln ε))|φε|≤ Cε(− ln ε)12∫Ωε(〈z − Q¯〉−5 + ε4(− ln ε)〈z − Q¯〉−1)‖φε‖∗= o(1).Finally, we obtain‖W 2φε‖∗∗ = o(1).In view of the formula (3.2.4), we have〈z − Q¯〉3|Zi| ≤ C(〈z − Q¯〉3ε4 + 〈z − Q¯〉−3)= O(1).andε−3(− ln ε)12Zi ≤ Cε(− ln ε)12∫Ωε∣∣〈z − Q¯〉−6 + ε4∣∣dx = o(1).Hence, ‖Zi‖∗∗ = O(1). Therefore, we have‖φε‖∗ ≤ C(‖W 2φε‖∗∗ + ‖h‖∗∗ +∑i|ci|‖Zi‖∗∗)= o(1), (3.2.22)which contradicts our assumption that ‖φε‖∗ = 1.For n = 6. We still assume that ‖φε‖∗∗∗ = 1. Using the similar argumentsin previous case, we obtain the followingdi = O(‖h‖∗∗∗∗) + o(‖φ‖∗∗∗) for 0 ≤ i ≤ 6,d7 = O(ε−2‖h‖∗∗∗∗) +O(ε−1‖φε‖∗∗∗) (3.2.23)and φε(z − Q¯)→ 0 in C1loc(Ωε). Then, we will show ‖φε‖∗∗∗ = o(1) by using1043.2. Finite Dimensional Reductionthe equation (3.2.9). At first, we write the equation (3.2.9) into the following−∆φε + µε2(1− 48η)φε = h+∑idiZi + 48Uφε + 48ε3Uˆφε. (3.2.24)Using Corollary 3.2.1 again, we have‖φε‖∗∗∗ ≤ C(‖(U + ε3Uˆ)φε‖∗∗∗∗ + ‖h‖∗∗∗∗ +∑i|di|‖Zi‖∗∗∗∗). (3.2.25)From the formula of U and Uˆ , it is not difficult to showU + ε3Uˆ ≤ C〈z − Q¯〉−4.Similar to the case n = 4, we could show ‖〈z − Q¯〉−4φε‖∗∗∗∗ = o(1),‖Zi‖∗∗∗∗ = O(1), 0 ≤ i ≤ 6 and ‖Z7‖∗∗∗∗ = O(ε2).Therefore, by the above facts and (3.2.23), we conclude‖φε‖∗∗∗ ≤ o(1) + C‖h‖∗∗∗∗ + o(1)‖φε‖∗∗∗ = o(1)which contradicts the previous assumption that ‖φε‖∗∗∗ = 1. Hence, wefinish the proof.Proof of Proposition 3.2.1. Since the proof of the case n = 4 and n = 6 arealmost the same, we only give the proof for the former one. We setH = {φ ∈ H1(Ωε) | 〈Zi, φ〉 = 0, 0 ≤ i ≤ 4},equipped with the scalar product (·, ·)ε. Problem (3.2.5) is equivalent tofinding φ ∈ H such that(φ, θ)ε = 〈24W2φ+ h, θ〉, ∀θ ∈ H,1053.3. Finite Dimensional Reduction: A Nonlinear Problemthat isφ = Tε(φ) + h˜, (3.2.26)where h˜ depends on h linearly, and Tε is a compact operator inH. Fredholm’salternative ensures the existence of a unique solution, provided that thekernel of Id−Tε is reduced to 0. We notice that any φε ∈ Ker(Id−Tε) solves(3.2.5) with h = 0. Thus, we deduce from Lemma 3.2.2 that ‖φε‖∗ = o(1) asε goes to zero. As Ker(Id−Tε) is a vector space and is {0}. The inequalities(3.2.14) follows from Lemma 3.2.2 and (3.2.18). This completes the proofof the first part of Proposition 3.2.1.The smoothness of Lε with respect to Λ and Q¯ is a consequence of thesmoothness of Tε and h˜, which occur in the implicit definition (3.2.26) ofφ ≡ Lε(h), with respect to these variables. Inequality (3.2.15) is obtainedby differentiating (3.2.5), writing the derivatives of φ with respect Λ and Q¯as linear combinations of the Zi’s and an orthogonal part, then we estimateeach term by using the first part of the proposition. One can see [20],[34]for detailed computations. 23.3 Finite Dimensional Reduction: A NonlinearProblemIn this section, we turn our attention to the nonlinear problem, which wesolve in the finite-dimensional subspace orthogonal to the Zi. Let Sε[u] beas defined in Lemma 3.1.1. Then (1.2.13) is equivalent toSε[u] = 0 in Ωε, u+ 6= 0,∂u∂ν= 0 on ∂Ωε. (3.3.1)Indeed, if u satisfies (3.3.1), the Maximal Principle ensures that u > 0 inΩε. Observing thatSε[W + φ] = −∆(W + φ) + µε2(W + φ)− n(n− 2)(W + φ)n+2n−21063.3. Finite Dimensional Reduction: A Nonlinear Problemmay be written asSε[W + φ] = −∆φ+ µε2φ− n(n+ 2)W4n−2φ+Rε − n(n− 2)Nε(φ) (3.3.2)withNε(φ) = (W + φ)n+2n−2 −Wn+2n−2 −n+ 2n− 2W4n−2φ (3.3.3)andRε = Sε[W ] = −∆W + µε2W − n(n− 2)Wn+2n−2 . (3.3.4)From Lemma 3.1.1 we get{‖Rε‖∗∗ ≤ CεΛ + ε2(− ln ε)12 , ‖D(Λ,Q¯)Rε‖∗∗ ≤ Cε, n = 4,‖Rε‖∗∗∗∗ ≤ Cε223 , ‖D(Λ,Q¯,η)Rε‖∗∗∗∗ ≤ Cε2, n = 6.(3.3.5)We now consider the following nonlinear problem: finding φ such that,for some numbers ci,−∆(W + φ) + µε2(W + φ)− 8(W + φ)3 =∑i ciZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 4(3.3.6)for n = 4, and finding φ such that, for some numbers di,−∆(W + φ) + µε2(W + φ)− 24(W + φ)2 =∑i diZi in Ωε,∂φ∂ν = 0 on ∂Ωε,〈Zi, φ〉 = 0, 0 ≤ i ≤ 7(3.3.7)for n = 6. The first equation in (3.3.6) and (3.3.7) reads−∆φ+ µε2φ− 24W 2φ = 8Nε(φ)−Rε +∑iciZi,−∆φ+ µε2φ− 48Wφ = 24Nε(φ)−Rε +∑idiZi. (3.3.8)In order to employ the contraction mapping theorem to prove that (3.3.6)1073.3. Finite Dimensional Reduction: A Nonlinear Problemand (3.3.7) are uniquely solvable in the set where ‖φ‖∗ and ‖φ‖∗∗∗ are smallrespectively, we need to estimate Nε in the following lemma.Lemma 3.3.1. There exists ε1 > 0, independent of Λ, Q¯, and C independentof ε,Λ, Q¯ such that for ε ≤ ε1 and‖φ‖∗ ≤ CεΛ for n = 4, ‖φ‖∗∗∗ ≤ Cε2 23 for n = 6.Then,‖Nε(φ)‖∗∗ ≤ CεΛ‖φ‖∗ for n = 4, ‖Nε(φ)‖∗∗∗∗ ≤ Cε‖φ‖∗∗∗ for n = 6.(3.3.9)For‖φi‖∗ ≤ CεΛ for n = 4, ‖φi‖∗∗∗ ≤ Cε2 23 for n = 6, i = 1, 2.Then,‖Nε(φ1)−Nε(φ2)‖∗∗ ≤ CεΛ‖φ1 − φ2‖∗ for n = 4,‖Nε(φ1)−Nε(φ2)‖∗∗∗∗ ≤ Cε‖φ1 − φ2‖∗∗∗ for n = 6. (3.3.10)Proof. Since the proof of these two cases are similar, we only consider n = 4here. From (3.3.3), we see|Nε(φ)| ≤ C(Wφ2 + |φ|3). (3.3.11)Using (3.1.16), we gainε−3(− ln ε)12Wφ2 + |φ|3 = ε(− ln ε)12∫Ωε(Wφ2 + |φ|3),where the integration term on the right hand side of the above equality can1083.3. Finite Dimensional Reduction: A Nonlinear Problembe estimated as|Wφ2 + |φ|3| ≤C((〈z − Q¯〉−2 + ε2(− ln ε)12 )|φ|2 + |φ|3)≤C(〈z − Q¯〉−4 + ε2(− ln ε)12 〈z − Q¯〉−2)‖φ‖2∗ + 〈z − Q¯〉−3‖φ‖3∗≤C((ε〈z − Q¯〉−4 + ε3(− ln ε)12 〈z − Q¯〉−2)Λ)‖φ‖∗.As a consequence,ε−3(− ln ε)12Wφ2 + |φ|3 ≤ Cε2(− ln ε)32 Λ‖φ‖∗ ≤ CεΛ‖φ‖∗.On the other hand,‖〈z − Q¯〉3(Wφ2 + |φ|3)‖∞ ≤ CεΛ‖φ‖∗.and (3.3.9) follows. Concerning (3.3.10), we writeNε(φ1)−Nε(φ2) = ∂ϑNε(ϑ)(φ1 − φ2)for some ϑ = xφ1 + (1− x)φ2, x ∈ [0, 1]. From∂ϑNε(ϑ) = 3[(W + ϑ)2 −W 2],we deduce that∂ϑNε(ϑ) ≤ C(|W ||ϑ|+ ϑ2) (3.3.12)and the proof of (3.3.10) is similar to the previous one.Proposition 3.3.1. For the case n = 4, there exists C, independent of εand Λ, Q satisfying (3.1.11), such that for small ε problem (3.3.6) has aunique solution φ = φ(Λ, Q¯, ε) with‖φ‖∗ ≤ CεΛ. (3.3.13)1093.3. Finite Dimensional Reduction: A Nonlinear ProblemMoreover, (Λ, Q¯)→ φ(Λ, Q¯, ε) is C1 with respect to the ∗-norm, and‖D(Λ,Q¯)φ‖∗ ≤ Cε. (3.3.14)For the case n = 6, there exists C, independent of ε and Λ, η, Q satisfying(3.1.13), such that for small ε problem (3.3.7) has a unique solution φ =φ(Λ, η, Q¯, ε) with‖φ‖∗∗∗ ≤ Cε83 . (3.3.15)Moreover, (Λ, η, Q¯)→ φ(Λ, η, Q¯, ε) is C1 with respect to the ∗ ∗ ∗-norm, and‖D(Λ,η,Q¯)φ‖∗∗∗ ≤ Cε53 . (3.3.16)Proof. We only give the proof of n = 4, the other case can be argued sim-ilarly. In the same spirit of [20], we consider the map Aε from F={φ ∈H1(Ωε)|‖φ‖∗ ≤ C′εΛ} to H1(Ωε) defined asAε(φ) = Lε(8Nε(φ) +Rε).Here C′is a large number, to be determined later, and Lε is given by Proposi-tion 3.2.1. We note that finding a solution φ to problem (3.3.6) is equivalentto finding a fixed point of Aε. On the one hand, we have for φ ∈ F . Then,using (3.3.5), Proposition 3.2.1 and Lemma 3.3.1,‖Aε(φ)‖∗ ≤ 8‖Lε(Nε(φ))||∗ + ‖Lε(Rε)‖∗ ≤ C1(‖Nε(φ)‖∗∗ + ε)≤ C2C′ε2Λ + C1εΛ ≤ C′εΛfor C′= 2C1 and ε small enough, which implies that Aε sends F into itself.On the other hand, Aε is a contraction. Indeed, for φ1 and φ2 in F , we write‖Aε(φ1)−Aε(φ2)‖∗ ≤ C‖Nε(φ1)−Nε(φ2)‖∗∗ ≤ CεΛ‖φ1−φ2‖∗ ≤12‖φ1−φ2‖∗for ε small enough. The contraction Mapping Theorem implies that Aε hasa unique fixed point in F , that is, problem (3.3.6) has a unique solution φ1103.3. Finite Dimensional Reduction: A Nonlinear Problemsuch that ‖φ‖∗ ≤ C′εΛ.In order to prove that (Λ, Q¯)→ φ(Λ, Q¯) is C1, we remark that if we setfor ψ ∈ F,B(Λ, Q¯, ψ) ≡ ψ − Lε(8Nε(ψ) +Rε),then φ is defined asB(Λ, Q¯, φ) = 0. (3.3.17)We have∂ψB(Λ, Q¯, ψ)[θ] = θ − 8Lε(θ(∂ψNε)(ψ)).Using Proposition 3.2.1 and (3.3.12) we write‖Lε(θ(∂ψNε)(ψ))‖∗ ≤ C‖θ(∂ψNε)(ψ)‖∗∗ ≤ ‖〈z − Q¯〉−1(∂ψNε)(ψ)‖∗∗‖θ‖∗≤ C‖〈z − Q¯〉−1(W+|ψ|+ |ψ|2)‖∗∗‖θ‖∗.Using (3.1.16), (3.2.10) and ψ ∈ F , we obtain‖Lε(θ(∂ψNε)(ψ))‖∗ ≤ ε‖θ‖∗.Consequently, ∂ψB(Λ, Q¯, φ) is invertible with uniformly bounded inverse.Then the fact that (Λ, Q¯) 7→ φ(Λ, Q¯) is C1 follows from the fact that(Λ, Q¯, ψ) 7→ Lε(Nε(ψ)) is C1 and the implicit function theorem.Finally, we consider (3.3.14). Differentiating (3.3.17) with respect to Λ,we find∂Λφ = (∂ψB(Λ, ξ, φ))−1((∂ΛLε)(Nε(φ)) + Lε((∂ΛNε)(φ)) + Lε(∂ΛRε)).Then by Proposition 3.2.1,‖∂Λφ‖∗ ≤ C(‖Nε(φ)‖∗∗ + ‖(∂ΛNε)(φ)‖∗∗ + ‖∂ΛRε‖∗∗).From Lemma 3.3.1 and (3.3.13), we know that ‖Nε(φ)‖∗∗ ≤ Cε2. Concerningthe next term, we notice that according to the definition of Nε,|∂ΛNε(φ)| = 3φ2|∂ΛW |.1113.4. Finite Dimensional Reduction: Reduced EnergyRecalling that|DΛW (z)| ≤ C(〈z − Q¯〉−2 + ε2(− ln ε)12 ),which gives‖∂ΛNε(φ)‖∗∗ ≤ Cε.Finally, using (3.3.5), we obtain‖∂Λφ‖∗ ≤ Cε.The derivative of φ with respect to Q¯ can be estimated in the same way.This concludes the proof.3.4 Finite Dimensional Reduction: ReducedEnergyLet us define the reduced energy functional asIε(Λ, Q) ≡ Jε[WΛ,Q¯ + φε,Λ,Q¯] (3.4.1)for n = 4 andIε(Λ, η,Q) ≡ Jε[WΛ,η,Q¯ + φε,Λ,η,Q¯] (3.4.2)for n = 6. Then, We haveProposition 3.4.1. The function u = WΛ,Q¯+φε,Λ,Q¯ is a solution to problem(1.2.13) for n = 4 if and only if (Λ, Q¯) is a critical point of Iε. The functionu = WΛ,η,Q¯ +φε,Λ,η,Q¯ is a solution to problem (1.2.13) for n = 6 if and onlyif (Λ, η, Q¯) is a critical point of Iε.Proof. Here we only give the proof for the case n = 6. We notice thatu = W +φ being a solution of (1.2.13) is equivalent to being a critical point1123.4. Finite Dimensional Reduction: Reduced Energyof Jε, which is also equivalent to the vanish of the di’s in (3.3.7) or, in viewof〈Z0, Y0〉 = ‖Y0‖2ε = γ0 + o(1),〈Zi, Yi〉 = ‖Yi‖2ε = γ1 + o(1), 1 ≤ i ≤ 6,〈Z7, Y7〉 = ‖Y7‖2ε = γ2ε3, (3.4.3)where γ0, γ1, γ2 are strictly positive constants, and〈Zi, Yj〉 = o(1), i 6= j, 0 ≤ i, j ≤ 6, 〈Zi, Yj〉 = O(ε3), i 6= j, i = 7 or j = 7.(3.4.4)We haveJ ′ε[W + φ][Yi] = 0, 0 ≤ i ≤ 7. (3.4.5)On the other hand, we deduce from (3.4.2) that I ′ε(Λ, η,Q) = 0 is equivalentto the cancellation of J ′ε(W + φ) applied to the derivative of W + φ withrespect to Λ, η and Q¯. By the definition of Yi’s and Proposition 3.3.1, wehave∂(W + φ)∂Λ= Y0 + y0,∂(W + φ)∂Q¯i= Yi + yi, 1 ≤ i ≤ 6,∂(W + φ)∂η= Y7 + y7with ‖yi‖∗∗∗ = O(ε2), 0 ≤ i ≤ 7. We write−∆(W + φ) + µε2(W + φ)− 24(W + φ)2 =7∑j=0djZjand denote aij = 〈yi, Zj〉. It turns out that I ′ε(Λ, η,Q) = 0 is equivalent,since J ′ε[W + φ][θ] = 0 for 〈θ, Zi〉 = (θ, Yi)ε = 0, 0 ≤ i ≤ 7, to([bij ] + [aij ])[dj ] = 0,where bij = 〈Yi, Zj〉. Using the estimate ‖yi‖∗∗∗ = O(ε2) and the expression1133.4. Finite Dimensional Reduction: Reduced Energyof Zi, Yi, 0 ≤ i ≤ 7, we directly obtainb00 = γ0 + o(1), bii = γ1 + o(1) for 1 ≤ i ≤ 6, b77 = γ2ε3,bij = o(1) for 0 ≤ i 6= j ≤ 6, bij = O(ε3) for i = 7 or j = 7, i 6= j,aij = O(ε2) for 0 ≤ i ≤ 7, 0 ≤ j ≤ 6, ai7 = O(ε4) for 0 ≤ i ≤ 7.Then it is easy to see the matrix [bij+aij ] is invertible by the above estimatesof each components, hence di = 0. We see that I ′ε(Λ, η,Q) = 0 means exactlythat (3.4.5) is satisfied.By Proposition 3.4.1, it remains to find critical points of Iε. First, weestablish an expansion of Iε.Proposition 3.4.2. In the case n = 4, for ε sufficiently small, we haveIε(Λ, η,Q) = Jε[W ] + ε2σε,4(Λ, Q) (3.4.6)where σε,4 = o(1) and DΛ(σε,4) = o(1) as ε goes to 0, uniformly with respectto Λ, Q satisfying (3.1.11).In the case n = 6, for ε sufficiently small, we haveIε(Λ, η,Q) = Jε[W ] + ε4σε,6(Λ, η,Q) (3.4.7)where σε,6 = o(1) and DΛ,η(σε,6) = o(1) as ε goes to 0, uniformly withrespect to Λ, η, Q satisfying (3.1.13).Proof. We only consider the case n = 4 here, the other case can be arguedsimilarly with minor changes. In view of (3.4.1), a Taylor expansion and thefact that J ′ε[W + φ][φ] = 0 yieldIε(Λ, Q)− Jε[W ] =Jε[W + φ]− Jε[W ] = −∫ 10J ′′ε (W + tφ)[φ, φ](t)dt=−∫ 10(∫Ωε(|∇φ|2 + µε2φ2 − 24(W + tφ)φ2))tdt,1143.4. Finite Dimensional Reduction: Reduced EnergywhenceIε(Λ, Q)− Jε[W ]= −∫ 10(8∫Ωε(Nε(φ)φ+ 3[W2 − (W + tφ)2]φ2))tdt−∫ΩεRεφ. (3.4.8)The first term on the right hand side of (3.4.8) can be estimated as∣∣∣∫ΩεNε(φ)φ∣∣∣ ≤ C∫Ωε|φ|4 + |Wφ3| = O(ε4 ln ε).Similarly, for the second term on the right hand side of (3.4.8), we obtain∣∣∣∫Ωε[W − (W + tφ)]φ2∣∣∣ ≤ C∫Ωε|φ|4 + |Wφ3| = O(ε4 ln ε).Concerning the last one, by using|Rε|∗ = |Sε[W ]| =O(ε4(− ln ε)〈z − Q¯〉−2 + ε2(− ln ε)12 〈z − Q¯〉−4)+O(Λ)( ε4(− ln ε)| ln1ε(1 + |z − Q¯|)|+ε4(− ln ε)),and ‖φ‖∗ = O(εΛ), we have∣∣∣∫ΩεRεφ∣∣∣ = O(ε2(− ln ε)12 Λ2 + ε3(− ln ε)12).This concludes the proof of the first part of Proposition (3.4.6).An estimate for the derivatives with respect to Λ is established exactlyin the same way, differentiating the right side in (3.4.8) and estimating eachterm separately, using (3.3.3), (3.3.5) and Lemma 3.1.1.1153.5. Proof Of Theorem 1.2.13.5 Proof Of Theorem 1.2.1In this section, we prove the existence of critical points for Iε(Λ, Q) andIε(Λ, η,Q), thereby proving Theorem 1.2.1 by Proposition 3.4.1. Accordingto the Proposition 3.4.2 and Lemma 3.1.1, we setKε(Λ, Q) =Iε(Λ, Q)− 2∫R4 U4(− ln εc1 )12 ε2(3.5.1)andKε(Λ, η,Q) =Iε(Λ, η,Q)− 4∫R6 U3ε3(3.5.2)When n = 4, we haveKε(Λ, Q) =14c4Λ2 ln1Λε(c1− ln ε)−c24Λ22|Ω|+12c24Λ2H(Q,Q)(c1− ln ε)12+O( Λ2− ln ε+ ε), (3.5.3)and when n = 6, we haveKε(Λ, η,Q) =(12η2|Ω| − c6Λ2η +148c6Λ2 − 8η3|Ω|)+12c26Λ4H(Q,Q)ε+12(η −c6Λ2|Ω|)ε∫ΩΛ2|x−Q|4+ o(ε). (3.5.4)Next, we consider Kε(Λ, Q), and find its critical point with respect toΛ, Q, and the critical point of Kε(Λ, η,Q) with respect to the parametersΛ, η,Q respectively.First, we consider Kε(Λ, Q) for n = 4. For the setting of the parametersΛ, Q, we see that Λ, Q are located in a compact set. As a consequence, wecan obtain a maximal value of Kε(Λ, Q). Then, we claim that:Claim: The maximal point of Kε(Λ, Q) with respect to Λ, Q can not happenon the boundary of the parameters.If we can prove this claim, then we could obtain an interior critical point1163.5. Proof Of Theorem 1.2.1of Kε(Λ, Q). Before proving the claim, we first considerFε(Λ) =14c4Λ2 ln1Λε(c1− ln ε)−c24Λ22|Ω|.Note that∂∂Λ[Fε(Λ)] =12c4Λ ln1Λε(c1− ln ε)−14c4Λ(c1− ln ε)−c24Λ|Ω|,Choosing c1 =2c4|Ω| , we could obtain that there existsΛ∗ = exp(−12) ∈ (exp(−12)εβ, exp(−12)ε−β)with some proper fixed constant β ∈ (0, 13), such that∂∂ΛFε |Λ=Λ∗= 0.It can be also found that such Λ∗ provides the maximal value of Fε(Λ) in[Λ4,1,Λ4,2], where Λ4,1 = exp(−12)εβ,Λ4,2 = exp(−12)ε−β. In order to provethe claim, we need to take Λ into consideration for the expansion of theenergy, going through the first part of the Appendix, we haveKε(Λ, Q) =14c4Λ2 ln1Λε(c1− ln ε)−c24Λ22|Ω|+12c24Λ2H(Q,Q)(c1− ln ε)12+O(Λ2− ln ε+ ε).Now, we go back to prove of the claim, choosing Λ = Λ∗ and Q = p.(Here p refers to the point where H(Q,Q) obtain its maximal value, it ispossible to find such a point. Indeed, we notice a fact H(Q,Q) → −∞as d(Q, ∂Ω) → 0 see [63] and references therein for a proof of this fact.Therefore we could find such p.)First, we prove that the maximal value can not happen on ∂Mδ4 . Wechoose δ4 such that d2 < max∂Mδ4 H < d1 for some proper constant d2, d1sufficiently negative, then we fixed Mδ4 . It is easy to see that Kε(Λ, Q) <1173.5. Proof Of Theorem 1.2.1Kε(Λ, p), where Q lies on the boundary of Mδ4 and Λ ∈ (Λ4,1,Λ4,2). ForΛ = Λ4,1 or Λ4,2, we go to the arguments below. Therefore, we prove thatthe maximal point can not lie on the boundary of Mδ4 × [Λ4,1,Λ4,2].Next, we show Kε(Λ∗, p) > Kε(Λ4,2, Q). It is easy to see thatFε[Λ4,2] ≤ cε−2β,where c < 0. Then we can find c1 < 0 such that Kε(Λ4,2, Q) ≤ c1ε−2β for anyQ ∈ Mδ4 , since the other terms compared to ε−2β are higher order term.On the other hand, for the choice of Λ∗, p, we see that Kε(Λ∗, p) = O(1).Therefore, we prove that Kε(Λ∗, p) > Kε(Λ4,2, Q) for any Q ∈Mδ4 .It remains to prove that the maximal value can not happen at Λ = Λ4,1.We choose Λ = εβ/2, Q = p, direct computation yields.Kε(εβ/2, p) =βc24εβ4|Ω|(1 + o(1)), Kε(Λ4,1, Q) =βc24ε2β2|Ω|(1 + o(1)).It is to see Kε(εβ/2, p) > Kε(Λ4,1, Q) for any Q ∈Mδ4 when ε is sufficientlysmall. Hence, we finish the proof of the claim. In other words, we couldobtain an interior maximal point in [Λ4,1,Λ4,2] ×Mδ4 . Therefore, we showthe existence of the critical points of Kε(Λ, Q) with respect to Λ, Q.For n = 6. We set η = 148 + aε13 , c6Λ2|Ω| =196 + bε23 , thenKε(a, b,Q) := Kε(Λ, η,Q) =16912|Ω|+[F (Q)− (8a3 + ab)|Ω|]ε+ o(ε),(3.5.5)whereF (x) =|Ω|18432(|Ω|H(x, x) +1c6∫Ω1|x− y|4dy),−η6 ≤ a ≤ η6 and −Λ6 ≤ b ≤ Λ6.We set C0 = F (p0), p0 refers to the point where F (x) obtains its maximalvalue. Indeed, we have H(Q,Q) → −∞ as d(Q, ∂Ω) → 0 and I(x) =∫Ω1|x−y|4 dy is uniformly bounded in Ω. Hence, we can always find such point1183.5. Proof Of Theorem 1.2.1p0. Let us introduce another five constants Ci, i = 1, 2, 3, 4, 5, with C2 <C1 < C0, 0 < C3 < C4 < η6 and 0 < C3 < C5 < Λ6, the value of these fiveconstants will be determined later.We setΣ0 ={− C4 ≤ a ≤ C4, − C5 ≤ b ≤ C5, Q ∈ NC2}, (3.5.6)whereNCi = {q : F (q) > Ci}, i = 1, 2 and δ6 is chosen such thatNC2 ⊂Mδ6 .We also defineB = {(a, b,Q) | (a, b) ∈ BC3(0), Q ∈ NC1},B0 = {(a, b) | (a, b) ∈ BC3(0)} × ∂NC1 , (3.5.7)where Br(0) := {0 ≤ a2 + b2 ≤ r}.It is trivial to see that B0 ⊂ B ⊂ Σ0, B is compact. Let Γ be the classof continuous functions ϕ : B → Σ0 with the property that ϕ(y) = y, y =(a, b,Q) for all y ∈ B0. Define the min-max value c asc = minϕ∈Γmaxy∈BKε(ϕ(y)).We now show that c defines a critical value. To this end, we just have toverify the following conditions(T1) maxy∈B0 Kε(ϕ(y)) < c, ∀ϕ ∈ Γ,(T2) For all y ∈ ∂Σ0 such that Kε(y) = c, there exists a vector τy tangentto ∂Σ0 at y such that∂τyKε(y) 6= 0.Suppose (T1) and (T2) hold. Then standard deformation argument en-sures that the min-max value c is a (topologically nontrivial) critical valuefor Kε(a, b,Q) in Σ0. (Similar notion has been introduced in [21] for degen-erate critical points of mean curvature.)To check (T1) and (T2), we define ϕ(y) = ϕ(a, b,Q) = (ϕa, ϕb, ϕQ)where (ϕa, ϕb) ∈ [−C4, C4]× [−C5, C5] and ϕQ ∈ NC2 .1193.5. Proof Of Theorem 1.2.1For any ϕ ∈ Γ and Q ∈ NC2 , the map Q → ϕQ(a, b,Q) is a continuousfunction from NC1 to NC2 such that ϕQ(a, b,Q) = Q for Q ∈ ∂NC1 . LetD be the smallest ball which contain NC1 , we extend ϕQ to a continuousfunction ϕ˜Q from D to D where ϕ˜(Q) is defined as follows:ϕ˜Q(x) = ϕ(x), x ∈ NC1 , ϕ˜Q(x) = Id, x ∈ D \ NC1 .Then we claim there exists Q′ ∈ D such that ϕ˜Q(Q′) = p0. Otherwiseϕ˜Q−p0|ϕ˜Q−p0|provides a continuous map from D to S5, which is impossible inalgebraic topology. Hence, there exists Q′ ∈ D such that ϕ˜Q(Q′) = p0. Bythe definition of ϕ˜, we can further conclude Q′ ∈ NC1 . Whencemaxy∈BKε(ϕ(y)) ≥Kε(ϕa(a, b,Q′), ϕb(a, b,Q′), p0)≥16912|Ω|+ (C0 − C6|Ω|)ε+ o(ε), (3.5.8)where C6 = 8C34 + C4C5 which stands for the maximal value of 8a3 + ab in[−C4, C4]× [−C5, C5]. As a consequencec ≥16912|Ω|+ (C0 − C6|Ω|)ε+ o(ε). (3.5.9)For (a, b,Q) ∈ B0, we have F (ϕQ(a, b,Q)) = C1. So,Kε(a, b,Q) ≤16912|Ω|+ (C1 + C7|Ω|)ε+ o(ε), (3.5.10)where C7 = max(a,b)∈BC3 (0) 8a3 + ab < 8C33 + C23 .If we choose C0 − C1 > 8C34 + C4C5 + 8C33 + C23 > C6 + C7, we havemaxy∈B0 Kε(ϕ(y)) < c holds. So (T1) is verified.To verify (T2), we observe that∂Σ0 =: {a, b,Q | a = −C4 or a = C4 or b = −C5 or b = C5 or Q ∈ ∂NC2}.Since C4, C5 are arbitrary, we choose 0 < 24C24 < C5. Then on a = −C4or a = C4, we choose τy = ∂∂b , on b = −C5 or b = C5, we choose τy =∂∂a .1203.5. Proof Of Theorem 1.2.1By our setting on C4, C5, we could show ∂τyKε(y) 6= 0. It only remains toconsider the case Q ∈ ∂NC2 . If Q ∈ ∂NC2 , thenKε(a, b,Q) ≤16912|Ω|+ (C2 + C7|Ω|)ε+ o(ε), (3.5.11)which is obviously less than c for C2 < C1. So (T2) is also verified.In conclusion, we proved that for ε sufficiently small, c is a critical value,i.e., a critical point (a, b,Q) ∈ Σ0 of Kε exists. Which means Kε indeed hascritical points respect to Λ, η,Q in (3.1.13).Proof of Theorem 1.2.1 completed. For n = 4, we proved that for ε smallenough, Iε has a critical point (Λε, Qε). Let uε = WΛε,Q¯ε,ε. Then uε is anontrivial solution to problem (1.2.13) for n = 4. The strong maximalprinciple shows uε > 0 in Ωε. Let uµ = ε−1uε(x/ε) and this is a nontrivialsolution of (1.2.12) for n = 4. Thus, we get Theorem 1.2.1 for n = 4.For n = 6, we proved that for ε small enough, Iε has a critical point(Λε, ηε, Qε). Let uε = WΛε,ηε,Q¯ε,ε. Then uε is a nontrivial solution to problem(1.2.13) for n = 6. The strong maximal principle shows uε > 0 in Ωε. Letuµ = ε−2uε(x/ε) and this is a nontrivial solution of (1.2.12) for n = 6. Thus,we get Theorem 1.2.1 for n = 6. Hence, we finish the proof. 1213.6. Proof Of Lemma 3.1.13.6 Proof Of Lemma 3.1.1We divide the proof into two parts. First, we study the case n = 4. Fromthe definition of W, (3.1.10) and (3.1.15), we know thatSε[W ] =−∆W + µε2W − 8W 3= 8U3 + ε4( c1− ln ε)Uˆ − ε2( c1− ln ε) 12 ∆(Rε,Λ,Qχ)− 8W3= O(ε4(− ln ε)〈z − Q¯〉−2 + ε2(− ln ε)12 〈z − Q¯〉−4)+O(Λ)( ε4(− ln ε)∣∣ ln1ε(1 + |z − Q¯|)∣∣+ε4(− ln ε)12).The estimates for DΛSε[W ] and DQ¯Sε[W ] can be computed in the sameway.We now turn to the proof of the energy estimate (3.1.23). From (3.1.15)and (3.1.16) we deduce that∫Ωε|∇W |2 + ε2( c1− ln ε) 12∫ΩεW 2 = 8∫ΩεU3W + ε4( c1− ln ε)∫ΩεUˆW− ε2( c1− ln ε) 12∫Ωε∆(Rχ)W. (3.6.1)Concerning the first term on the right hand side of (3.6.1), we have∫ΩεU3W =∫ΩεU4 + ε2( c1− ln ε) 12∫ΩεUˆU3+c4Λ|Ω|ε2( c1− ln ε)− 12∫ΩεU3. (3.6.2)By noting that∫ΩεU4 =∫R4U41,0 +O(ε4),∫ΩεU3 =c4Λ8+O(ε2),1223.6. Proof Of Lemma 3.1.1we get∫ΩεU3W =∫R4U41,0 +c24Λ28|Ω|ε2( c1− ln ε)− 12 + ε2( c1− ln ε) 12∫ΩεUˆU3+O(ε4( c1− ln ε)− 12).For the third term on the right hand side of the above equality, we have∫ΩεUˆU3 =−∫ΩεΨU3 − c4Λ( c1− ln ε)− 12∫ΩεH(x,Q)U3 +∫Ωε(Rχ)U3=−c4Λ216ln1Λε−c24Λ28( c1− ln ε)− 12H(Q,Q) +O(Λ2).Hence, we have∫ΩεU3W =∫R4U41,0 +c24Λ28|Ω|ε2( c1− ln ε)− 12 −c4Λ216ln1Λεε2( c1− ln ε) 12−c24Λ28ε2H(Q,Q) +O(ε2( c1− ln ε) 12 Λ2 + ε4( c1− ln ε)− 12).(3.6.3)For the second term on the right hand side of (3.6.1)∫ΩεUˆW =∫ΩεUˆU + ε2( c1− ln ε) 12∫ΩεUˆ2 +c4Λ|Ω|ε2( c1− ln ε)− 12∫ΩεUˆ ,by using∫ΩεUˆU = O(ε−2( c1− ln ε)− 12 Λ2),∫ΩεUˆ2 = O(ε−4(− ln ε)Λ2),∫ΩεUˆ = ε−4(c1− ln ε)−12∫ΩΛ|x−Q|2+O(ε−4Λ),and∫ΩG(x,Q) = 0, we obtainε4(c1− ln ε)∫ΩεUˆW =c4Λ2|Ω|ε2∫Ω1|x−Q|2+O(ε2( c1− ln ε) 12 Λ2). (3.6.4)1233.6. Proof Of Lemma 3.1.1For the last term on the right hand side of (3.6.1),∫Ωε∆(Rχ)W= ε2(c1− ln ε)−12c4Λ|Ω|∫Ωε∆(Rχ) +O(Λ2)=( c1− ln ε)−1 c4Λ|Ω|∫Ωε∆(U − ε2( c1− ln ε) 12 Ψ− c4Λε2H)+O(Λ2)=( c1− ln ε)−1 c4Λ|Ω|∫Ωε(− 8U3 + ε2( c1− ln ε) 12U + c4Λε4 1|Ω|)+O(Λ2)=( c1− ln ε)− 12 c4Λ2|Ω|∫Ω1(ε2Λ2 + |x−Q|2)+O(Λ2 + ε2(− ln ε)), (3.6.5)where we used (3.1.8). Using (3.6.3)-(3.6.5), we get12∫Ωε(|∇W |2 + ε2(c1− ln ε)12W 2)=4∫R4U41,0 + ε2(c1− ln ε)−12c24Λ22|Ω|−c24Λ22H(Q,Q)ε2−c4Λ24ε2(c1− ln ε)12 ln1Λε+O(ε2(c1− ln ε)12 Λ2) +O(ε4( c1− ln ε)− 12).(3.6.6)Next, we compute the term∫ΩεW 4.∫ΩεW 4 =∫ΩεU4 + 4ε2( c1− ln ε) 12∫ΩεU3Uˆ + 4ε2( c1− ln ε)− 12 c4Λ|Ω|∫ΩεU3+O(ε4( c1− ln ε)−2)=∫R4U41,0 −c4Λ24ε2( c1− ln ε) 12 ln1Λε−c24Λ22ε2H(Q,Q)+c24Λ22|Ω|ε2( c1− ln ε)− 12 +O(ε2( c1− ln ε) 12 Λ2)+O(ε4( c1− ln ε)−2).(3.6.7)1243.6. Proof Of Lemma 3.1.1Combining (3.6.6) and (3.6.7), we obtainJε[W ] =12∫Ωε|∇W |2 +µε22∫ΩεW 2 − 2∫ΩεW 4=2∫R4U41,0 +c4Λ24ε2( c1− ln ε) 12 ln1Λε−c24Λ22|Ω|ε2( c1− ln ε)− 12+12c24Λ2ε2H(Q,Q) +O(ε2( c1− ln ε) 12 Λ2)+O(ε4(− ln ε)2). (3.6.8)In the following, we prove (3.1.24)-(3.1.28). By the definition of W , (3.1.10)and (3.1.15), we know thatSε[W ] =−∆W + ε3W − 24W 2= 24U2 + ε6Uˆ − ε3∆(Rχ) + ε6(η −c6Λ2|Ω|)− 24U2 − 24η2ε6+O(ε3〈z − Q¯〉−4)We rearrange the right hand side of the above equality and obtainSε[W ] =− ε6(24η2 − η +c6Λ2|Ω|)+O(ε3〈z − Q¯〉−4)= O(〈z − Q¯〉−323 ε3).The estimates for DΛSε[W ], DQ¯Sε[W ] and DηSε[W ] can be derived in thesame way. Now we are in the position to compute the energy. From (3.1.15)and (3.1.16), we deduce that∫Ωε|∇W |2 + ε3∫ΩεW 2 =∫Ωε(−∆W + ε3W )W=∫Ωε(24U2 + ε6Uˆ − ε3∆(Rχ) + ε6(η −c6Λ2|Ω|))W.(3.6.9)1253.6. Proof Of Lemma 3.1.1Concerning the first term on the right hand side of (3.6.9), we have∫ΩεU2W =∫ΩεU3 + ε3∫ΩεUˆU2 + ηε3∫ΩεU2=∫R6U31,0 +124c6ηΛ2ε3 −124c26Λ4ε4H(Q,Q)−1576c6Λ2ε3 +O(ε5).(3.6.10)For the second, third and fourth term on the right hand side of (3.6.9),following the similar steps as we did in case n = 4.ε6∫ΩεUˆW = ε6∫ΩεUˆ(U + ε3Uˆ + ηε3) = −ηΛ2ε4∫Ω1|x−Q|4+O(ε5),(3.6.11)−ε3∫Ωε∆(Rχ)W =ε3η∫Ωε∆(U − ε3Ψ− c6ε4Λ2H) +O(ε5) = ε6η∫ΩεU +O(ε5)=ηΛ2ε4∫Ω1|x−Q|4+O(ε5), (3.6.12)andε6(η −c6Λ2|Ω|)∫ΩεW =(η2|Ω| − c6ηΛ2)ε3 +(η −c6Λ2|Ω|)ε4∫ΩΛ2|x−Q|4+O(ε5).(3.6.13)Using (3.6.10)-(3.6.13), we have12∫Ωε|∇W |2 +ε32∫ΩεW 2= 12∫R6U31,0 +(12η2|Ω| −148c6Λ2)ε3 −c26Λ42H(Q,Q)ε4+12(η −c6Λ2|Ω|)ε4∫ΩΛ2|x−Q|4+O(ε5). (3.6.14)1263.6. Proof Of Lemma 3.1.1Then,∫ΩεW 3 =∫R6U31,0 + 3ε3∫ΩεU2Uˆ + 3ε3∫ΩεU2η + 3ε6∫ΩεUη2 + 3ε9∫ΩεUˆη2+ ε9∫Ωεη3 +O(ε5)=∫R6U31,0 +18c6ηΛ2ε3 −1192c6Λ2ε3 + η3|Ω|ε3 −18c26Λ4H(Q,Q)ε4+O(ε5). 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