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On the existence, structure, and stability of chiral magnetic skyrmions Wang, Li 2020

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On the Existence, Structure and Stability of ChiralMagnetic SkyrmionsbyLi WangB.A. University of California, Berkeley, 2013M.Sc. University of Washington, 2014A THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFDoctor of PhilosophyinTHE FACULTY OF GRADUATE AND POSTDOCTORALSTUDIES(Mathematics)The University of British Columbia(Vancouver)November 2020c© Li Wang, 2020The following individuals certify that they have read, and recommend to the Fac-ulty of Graduate and Postdoctoral Studies for acceptance, the thesis entitled:On the Existence, Structure and Stability of Chiral MagneticSkyrmionssubmitted by Li Wang in partial fulfillment of the requirements for the degree ofDoctor of Philosophy in Mathematics.Examining Committee:Stephen Gustafson, MathematicsSupervisorTai-Peng Tsai, MathematicsSupervisory Committee MemberJun-cheng Wei, MathematicsSupervisory Committee MemberYoung-Heon Kim, MathematicsUniversity ExaminerMark Van Raamsdonk, PhysicsUniversity ExamineriiAbstractThis main focus of this thesis is on the existence and stability of topological soli-tons called chiral magenetic skyrmions, arising in planar ferromagnets includingcombinations of Zeeman, anisotropy, and chiral (Dzyoloshinskii-Moriya) interac-tions.The first part deals with the existence of skyrmions with co-rotational symme-try under small chiral interaction, which is a variational problem. By treating itas a perturbation problem, our results make precise the sense in which skyrmionsare perturbations of bubbles (degree-one harmonic maps). The perturbation ap-proach relies on delicate 2D resolvent expansion for a linearized operator with azero-energy resonance. We show the uniqueness and (in a special case) the mono-tonicity of the skyrmion solution profile. By applying our precise bounds of thedifferences between skyrmions and harmonic maps, we obtain precise asymptoticsof the skyrmion energy.In the second part, as an application of the first part, we study the spectrum ofthe second variation of the energy about the skyrmion, by perturbation theory inthe presence of a threshold resonance. We first prove the linear (spectral) stabilityof chiral magnetic skyrmions against arbitrary perturbations; thus the skyrmionsolution from the first part is a strict local minimizer (modulo translations) ofthe energy. We also prove the orbital stability of the skyrmions for the Landau-Lifshitz equation with dissipation, which is an alternate quantitative proof of therecent skyrmion stability result of Li-Melcher in Journal of Functional Analysis2018 . These results are consistent with observations in the physics literature.iiiLay SummaryThis thesis focuses on studying the existence and stability of special “topologi-cal soliton solutions” of nonlinear differential equations modelling ferromagnets.These so-called “chiral magnetic skyrmions” were predicted in the physics lit-erature, and have been observed in experiments on materials exhibiting. Theirlocalized vortex-like structures are shown to be the Dzyaloshinsky-Moriya inter-action, a special micromagnetic energy. Motivated by this observation, we providea rigorous mathematical proof of the existence of skyrmions under symmetry as-sumptions, and also examine their spectral and dynamical stability.ivPrefaceThis thesis is mostly adapted from two papers [21] and [55]. A version of Chapter2 has been submitted to publication. The research is joint work with Dr. StephenGustafson. I conducted much of the analysis and computation in Section 2.2 andSection 2.3.A version of Chapter 3 will be submitted for publication soon. Dr. StephenGustafson suggested the problem and supervised the research. I conducted theanalysis, and wrote the whole manuscript.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Micromagnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Micromagnetic energy . . . . . . . . . . . . . . . . . . . . . . . 41.3 Symmetric reduction . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Co-rotational harmonic maps . . . . . . . . . . . . . . . . . . . . 111.5 First main result: skyrmion existence and properties . . . . . . . . 121.6 Landau-Liftshitz Equations . . . . . . . . . . . . . . . . . . . . . 161.6.1 Preservation of equivariance by the Landau-Lifshitz dy-namics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.6.2 Non-perservation of co-rotational maps by Landau-Lifshitzdynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 201.7 Second main results: skyrmion stability . . . . . . . . . . . . . . 21vi2 Existence and Isotropic Limit of Co-rotational Skyrmions . . . . . . 262.1 Minimization problem and function space . . . . . . . . . . . . . 262.1.1 Lower bounds . . . . . . . . . . . . . . . . . . . . . . . . 322.1.2 Upper bounds . . . . . . . . . . . . . . . . . . . . . . . . 352.1.3 Existence of minimizer . . . . . . . . . . . . . . . . . . . 372.1.4 Convergence to Q, after rescaling . . . . . . . . . . . . . . 372.2 Convergence Rate Estimates . . . . . . . . . . . . . . . . . . . . 412.2.1 Equation for the difference . . . . . . . . . . . . . . . . . 422.2.2 Resolvent estimates . . . . . . . . . . . . . . . . . . . . . 432.2.3 Reparameterization . . . . . . . . . . . . . . . . . . . . . 442.2.4 Remainder estimate . . . . . . . . . . . . . . . . . . . . . 472.2.5 Parameter relation . . . . . . . . . . . . . . . . . . . . . . 492.2.6 Derivative estimates . . . . . . . . . . . . . . . . . . . . . 502.2.7 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . 572.2.8 Energy asymptotics and Lp estimates . . . . . . . . . . . . 652.3 Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692.3.1 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . 692.3.2 Existence of a minimizer . . . . . . . . . . . . . . . . . . 712.3.3 Monotonicity of the profile . . . . . . . . . . . . . . . . . 752.3.4 Resolvent estimates . . . . . . . . . . . . . . . . . . . . . 792.3.5 Computation of some integrals . . . . . . . . . . . . . . . 973 Stability of Chiral Magnetic Skyrmions . . . . . . . . . . . . . . . . 993.1 Equivariant perturbations . . . . . . . . . . . . . . . . . . . . . . 993.2 General perturbations . . . . . . . . . . . . . . . . . . . . . . . . 1163.2.1 The unperturbed operator . . . . . . . . . . . . . . . . . . 1173.2.2 The full linearized operator . . . . . . . . . . . . . . . . . 1193.3 Translation zero-modes j =±1 . . . . . . . . . . . . . . . . . . . 1203.3.1 Dealing with H1 . . . . . . . . . . . . . . . . . . . . . . . 1213.3.2 Dealing with H−1 . . . . . . . . . . . . . . . . . . . . . . 1233.4 Higher order modes | j| ≥ 2 . . . . . . . . . . . . . . . . . . . . . 127vii3.5 Dynamical Stability . . . . . . . . . . . . . . . . . . . . . . . . . 1313.6 Second variation of the energy . . . . . . . . . . . . . . . . . . . 1364 Directions for Future Research . . . . . . . . . . . . . . . . . . . . . 1414.1 Asymptotic stability . . . . . . . . . . . . . . . . . . . . . . . . . 1414.2 Dynamics of Schro¨dinger map flow . . . . . . . . . . . . . . . . . 142Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145viiiAcknowledgmentsFirst, I would like to express sincere gratitude to my supervisor Professor StephenGustafson for all his support in the past six years. This thesis would not be possi-ble without his constant guidance. His patience, and encouragement were invalu-able, and our stimulating discussions have truly shaped my thinking as a mathe-matician. I feel lucky and grateful to have Stephen as my supervisor.I would also like to thank Professor Tai-Peng Tsai and Professor Jun-ChengWei for their time as my supervisor committee members.I would like to thank Professor Maciej Zworski for introducing me to mathe-matical physics when I was an undergraduate, which inspired me to pursue furthergraduate studies in mathematics.I thank my fellow graduate students in Auditorium Annex, for making mytime enjoyable.Finally, I would like to thank my Aunt’s family and my parents for all thesupport since I started studying abroad. I am also thankful to my friends MochongDuan, Yun Li, and Peng Zheng for their kindness and support.ixChapter 1Introduction1.1 MicromagneticsMicromagnetics is a widely used continuum theory for describing the staticand dynamic behavior of ferromagnets [10, 33, 35]. Phenomena captured by thetheory include domain walls and vortices in ferromagnetic materials. Microscop-ically, within a small region of a ferromagnet, neighboring magnetic moments orspins tend to align with each other to minimize the energy. This small regionwith uniform moments can be simply regarded as a bigger magnetic dipole. Theboundary between such regions is called domain wall. A ferromagnet is dividedinto many of these domains. The size of each domain and the shape of the do-main wall are determined by the interplay of the short-range interaction betweenneighboring spins and the long-range magnetic dipolar force as well as the detailsof the material and external magnetic field. Boundary effects can cause domainorientations to curl around whirlpool-like structures, known as magnetic vortices,on the surface of the material.The state of the ferromagnet (at a fixed time) is given by its magnetization(density of magnetic moment), a constant-length (here normalized to one) vector1fieldmˆ(x) = m1(x)m2(x)m3(x) |mˆ(x)|2 = m21(x)+m22(x)+m23(x)≡ 1.As an idealization of a situation where the magnetization does not vary muchin one spatial direction, where boundary effects can be ignored, we consider aninfinite, two-dimensional magnet, so that mˆ(x) is defined for all x = (x1,x2) ∈R2.Geometrically, the magnetization is a map into the unit 2-spheremˆ : R2→ S2, S2 = {mˆ ∈ R3 | |mˆ|= 1} ⊂ R3 . (1.1.1)The behavior of the magnet is determined by a micromagnetic energy func-tional E(mˆ) of the magnetization mˆ. Equlibrium (static) configurations are criticalpoints of this energy, so satisfy the Euler-Lagrange equationE ′(mˆ) = 0, (1.1.2)that is,ddεE(mˆε)|ε=0 = 0for any smooth, one parameter variation of magnetizations mˆε with |mˆε(x)| ≡ 1,mˆ0(x) ≡ mˆ(x). Micromagnetics, and the form of the energy functional E can bederived from microscopic theory (see e.g., [27]).The contribution to this energy which reflects the ferromagnetic character ofthe material is the exchange energyEe(mˆ) := 12∫R2|∇mˆ(x)|2dx,which penalizes rapid spatial variation of mˆ(x).Configurations with finite exchange energy may be classified by topological2degree (or Skyrmion number) [49]N :=14pi∫R2mˆ · (∂ x2 mˆ×∂ x1 mˆ)dx =14pi∫R2∂ x1 mˆ · J ∂ x2 mˆ dx ∈ Z, (1.1.3)N is the degree of the map mˆ : R2 → S2, which counts how many times the mˆwraps around the unit sphere. The integrand in (1.1.3) is the pull-back by mˆ of thevolume form on S2. Here J denotes the pi2 rotation (i.e. complex structure) on thetangent planeTmˆS2 = {ξ ∈ R3 | mˆ ·ξ = 0} (1.1.4)to the sphere at mˆ:J = Jmˆ : TmˆS2→ TmˆS2ξ 7→ mˆ×ξ .(1.1.5)The degree N determines the minimum possible exchange energy by a classical“completing-the-square” computation [3, 33]:Ee(mˆ) = 12∫R2(|∂ x1 mˆ|2+ |∂ x2 mˆ|2)dx = 12∫R2|(∂ x1∓J ∂ x2)mˆ|2 dx±4piN≥ 4pi|N|.(1.1.6)The minimum of Ee(mˆ) is attained if ∂ x1 mˆ = ±J ∂ x2 mˆ. If we apply the stereo-graphic projection w = m1+im21+m3 ,∂ x1 mˆ =±J ∂ x2 mˆ ↔ (∂ x1±i∂ x2)w = 0,which is the Cauchy-Riemann equations.Hence, w(z) must be a holomorphic or anti-holomorphic function of z = x1+3ix2. If we fix N > 0, and a boundary conditionlim|x|→∞mˆ(x) = kˆ = 001the lower bound is attained by magnetizations given by a family of rational func-tionsw(z) =bN(z¯−b1)(z¯−b2) · · ·(z¯−bN−1)(z¯−a1)(z¯−a2) · · ·(z¯−aN) (1.1.7)Here a1 · · ·aN ,b1 · · ·bN are arbitrary complex numbers. Thus this is a 4N-dimensionalfamily of minimizers. Each one is a minimizer, but a degenerate one. They are(in an informal sense) each unstable, because it costs no energy to drift within thisfamily.1.2 Micromagnetic energyBesides the exchange energy, there are usually other affects contributing to themicromagnetic energy. An anisotropy in the direction orthogonal to the plane ofthe magnet is represented by a term of the formEa(mˆ) = 12∫R2(1− (mˆ · kˆ)2)dx = 12∫R2(1−m23)dx.The effect of a constant external magnetic field perpendicular to the plane of themagnet is given by Zeeman-type energy:E z(mˆ) =∫R2(1− mˆ · kˆ)dx =∫R2(1−m3)dx.If we change variable by scaling,λ > 0, mˆ(x) 7→ mˆ(λx), =⇒ Ee 7→ Ee, Ea 7→ λ−2Ea, E z 7→ λ−2E z .4In particular, Ee(mˆ) is scale invariant. Since |mˆ| = 1, both Ea(mˆ) and E z(mˆ) arepositive. If we consider a combination of exchange, anisotropy, and Zeeman en-ergies, then by sending λ → ∞,inf(Ee(mˆ)+aEa(mˆ)+bE z(mˆ)) = infEe(mˆ) = 4pi|N|, a,b≥ 0and there is no minimizer.The energy of Dzyaloshinskii-Moriya interaction is:EDM(mˆ) = 12∫R2mˆ · (∇× mˆ)dx =∫R2(m1∂x2m3−m2∂x1m3)dx(the second expression comes from a formal integration by parts). The D-M en-ergy can be negative (see example we constructed in Proposition 2.1.2 ), and so itmay be possible to find configurations with energy below 4piN, in which case thescaling argument does not prevent the existence of minimizers. The presence of aD-M term limits the energy gain under dilation. In this sense, the D-M term hasthe potential to stabilize the minimizers of Ee.A large physics literature (see, e.g., [5, 6, 9, 30]) has confirmed that in certainchiral magnets, an important spin-orbit coupling effect is well modelled by a D-M contribution to the energy, and that it has the effect of stabilizing the localizedstructure with fixed rotation [29].Chiral magnetic skyrmions are topological solitons, of significant physical in-terest, arising in ferromagnets described by a micromagnetic energy including achiral (D-M) interaction term. Chiral magnetic skyrmions were first predictedtheoretically [6, 9, 29], and then later observed experimentally in recent years asisolated structures, and forming lattices [30]. They are of potential technologicalimportance [11, 17, 26, 29, 43, 56].Remark 1.2.1. In general, there should also be a contribution to the energy fromthe demagnetizing (or stray) field of the form −12∫R2~H · mˆ dx, where ∇ · ~H =−∇ · mˆ,∇× ~H =~0. However, for the configurations we consider here (see (1.3.5)below), and ∇ · mˆ =~0, and so this contribution vanishes.5So the full micromagnetic energy we consider isEk,α,β (mˆ)= Ee(mˆ)+β k EDM(mˆ)+β 2α Ea(mˆ)+β 2(1−α) E z(mˆ), k∈R, β > 0, α ≤ 1.The form of the coefficients requires some explanation. First, we will combinethe anisotropy and Zeeman energies and writeE (α)(mˆ) := α Ea(mˆ)+(1−α)E z(mˆ) =∫R2(1−m3− 12 α(1−m3)2)dx.This includes the special casesE (1) = Ea, E (0) = E z,but in general allows both easy-axis (α > 0) and easy-plane (α < 0) type anisotropies(though, in the latter case, balanced by a sufficiently strong Zeeman term). Notethatα ≤ 1 =⇒ E (α) ≥ E (1) = Ea ≥ 0. (1.2.1)The full energy is then written asEk,α,β (mˆ) = Ee(mˆ)+β k EDM(mˆ)+β 2 E (α)(mˆ), k ∈ R, β > 0, α ≤ 1.(1.2.2)The sign of parameter k determines the direction of rotation of the skyrmion. Sec-ond, note that β is purely a scaling parameter: for λ > 0,Ek,α,β (mˆ(λ ·)) = Ee(mˆ)+βλk EDM(mˆ)+ β2λ 2E (α)(mˆ) = Ek,α, βλ(mˆ), (1.2.3)so in particular we may arrange β = 1 by taking λ = β to giveEk,α,β (mˆ(β ·)) = Ek,α,1(mˆ).Because of the unit length constraint (1.1.1) on mˆ, the Euler-Lagrange equa-6tion (1.1.2) for critical points of this energy functional is0 = E ′k,α,β (mˆ) = PTmˆS2[−∆mˆ+β k ∇× mˆ+β 2 (α−1−αm3) kˆ]=−∆mˆ−|∇mˆ|2mˆ+β k (∇× mˆ− (mˆ · (∇× mˆ))mˆ)+β 2 (α−1−αm3)(kˆ−m3mˆ)(1.2.4)wherePTmˆS2 : ξ 7→ ξ − (mˆ ·ξ )ξis the orthogonal projection from R3 onto the tangent plane (1.1.4). Thoughthe energy functional appears to be quadratic in the magnetization mˆ, the Euler-Lagrange equation (1.2.4) is clearly nonlinear, as a consequence of the geometricconstraint (1.1.1). In fact, this geometric constraint is the main reason the prob-lems from micromagnetics are mathematically challenging and interesting .Mathematically, it was shown in [15, 41] that the energy (1.2.2) has a globalminimizer under the constraint that the skyrmion number (1.1.3) is fixed at N =1, rigorously establishing the existence of chiral skyrmions. The main difficultyin showing the existence of minimizers is the possible lack of compactness ofminimizing sequences due to scaling; i.e. a minimizing sequence might have alength scale which goes to zero along the sequence and so (every sub-sequence ofit) would fail to converge. This is just like the scaling argument above showingthere is no minimizer without the D-M term. The essential key to preventing thisis showing that the minimum energy is below 4pi (the minimal exchange energy).The technical tool for implementing this is the concentration compactness theory.A simple version of this is done in Section 2.3.2.1.3 Symmetric reductionMuch of the physics literature (e.g. [6, 7, 29, 30] ) on chiral skyrmions con-figurations with co-rotational symmetry. Here we explain the reduction to thissymmetry class. It seems not to be known if the global minimizer of [15, 41] has7this symmetry.It is easily verified that the energy functionals Ee, Ea, and E z (hence also E (α))are each separately invariant under• spatial rotations: mˆ(x) 7→ mˆ(eφ R˜x), R˜=[0 −11 0], eφ R˜ =[cosφ −sinφsinφ cosφ],φ ∈ R;• target rotations fixing kˆ: mˆ(x) 7→ eφRmˆ(x), R = 0 −1 01 0 00 0 0,eφR = cosφ −sinφ 0sinφ cosφ 00 0 1;• spatial reflections: mˆ(x) 7→ mˆ(±F˜x), F˜ =[1 00 −1];• target reflections fixing kˆ: mˆ(x) 7→ ±Fmˆ(x), F = 1 0 00 −1 00 0 1.This follows from:• (eφRmˆ)3,(Fmˆ)3 = mˆ3;• ∇[(m j)(eφ R˜x)] = (e−φ R˜∇m j)(eφ R˜x);• e−φ R˜, F˜ preserve length in R2;• x 7→ eφ R˜x, x 7→ F˜x have unit Jacobian: d(eφ R˜x) = dx,d(F˜x) = dx.The chiral energy term EDM breaks each of these symmetries, but the fol-lowing lemma, proved in Section 2.3.1, shows that it retains invariance undercombined spatial and target rotations.mˆ(x) 7→ e−φRmˆ(eφ R˜x), φ ∈ R, (1.3.1)8and combined reflectionsmˆ(x) 7→ −Fmˆ(F˜ x). (1.3.2)Lemma 1.3.1. We haveEDM(e−φRmˆ(eφ R˜·))= EDM(mˆ) ∀φ ∈ R, (1.3.3)andEDM(−Fmˆ(F˜ ·))= EDM(mˆ). (1.3.4)Remark 1.3.1. For general maps mˆ : R3→ S2, the chiral energy term is invariantto combined transformations in SO(3) rotation, and sign reversing for combinedorientation reversing elements in O(3). When restricted to mˆ :R2→ S2 which pre-serve kˆ, only invariance (1.3.3) remains. The invariance (1.3.4) is an additionalone.We refer to magnetization vectors fields mˆ which are invariant under the trans-formation (1.3.1) as equivariant, and those which are additionally invariant un-der (1.3.2) as co-rotational. The latter take the special formmˆ(x) = [−sin(u(r))sin(θ), sin(u(r))cos(θ), cos(u(r)) ] , u : [0,∞)→ R,(1.3.5)where (r,θ) are polar coordinates on R2 (up to a reflection mˆ 7→ epiRmˆ, whichreverses the sign of EDM(mˆ) and so can be absorbed by change of coefficientk 7→ −k). By Lemma 1.3.1, the full energy Ek,α,β is invariant under both (1.3.1)and (1.3.2), and so a critical point of Ek,α,β restricted to co-rotational maps (1.3.5)will be a critical point of Ek,α,β , hence a solution of the Euler-Lagrange equa-tion (1.2.4) – see Proposition 2.1.1.A straightforward computation shows that the full energy (1.2.2) of a co-rotational magnetization (1.3.5) produces the following functional of the profileu:Ek,α,β (u) :=12piEk,α,β (mˆ) = Ee(u)+β k EDM(u)+β 2 E(α)(u) (1.3.6)9whereEe(u) =12∫ ∞0(u2r +1r2sin2(u(r)))rdrE(α)(u) =∫ ∞0(1− cos(u(r))− 12α(1− cos(u(r)))2)rdr.Since mˆ·(∇×mˆ)= ur+ 1r cos(u(r))sin(u(r))= 1r (r cos(u(r))sin(u(r))r+2sin2(u(r))ur,and 1r (r cosusinu)r is a “null Lagrangian”. So integration by parts givesEDM(u)=12piEDM(mˆ)= r cosu(r)sinu(r)|∞0 +∫ ∞0sin2(u(r))urrdr=∫ ∞0sin2(u(r))urrdr,if u(r)→ 0 sufficiently fast as r→ ∞.Another straightforward computation shows that for co-rotational magnetiza-tions∂ x1 mˆ · Jmˆ ∂ x2 mˆ =−1rsinu ur =1r(cosu)rHere we use ∂ x1 mˆ = cosθ mˆr− 1r sin(θ)mˆθ , ∂ x2 mˆ = sin(θ)mˆr + 1r cos(θ)mˆθ , andmˆr = ur~e, mˆθ = sin(u)J~e , and~e and J~e (see (3.6.1) in Section 3.6) are an orthonor-mal basis on TmˆS2. Thus, Jmˆ ∂ x2 mˆ = mˆ×∂ x2 mˆ = sin(θ)urJ~e− 1r sinucos(θ)~e.So by (1.1.3), the Skyrmion number of a co-rotational map isN =14pi∫R2∂ x1 mˆ · J ∂ x2 mˆ dx =14pi2pi∫ ∞0(cosu)r dr =12cosu(r)∣∣r=∞r=0 . (1.3.7)By the scaling law (1.2.3), we haveEk,α,β (u(λ ·)) = Ek,α, βλ (u). (1.3.8)101.4 Co-rotational harmonic mapsHarmonic maps are critical points of the Dirichlet energy(exchange energy) func-tionalEe(mˆ) := 12∫R2|∇mˆ(x)|2dx,and so solve the corresponding Euler-Lagrange equation0 =−E ′e(mˆ) = ∆mˆ+ |∇mˆ|2mˆ.Restricting the minimizers of Ee(mˆ) (1.1.7) to the equivariant class (1.3.1) givesw(z) =az¯N, a ∈ C,and to the co-rotational class (1.3.5) givesw(z) =iaz¯N, a ∈ R .The corresponding equivariant family ismˆ(x) = eφReNθRHˆN(rs), φ ∈ R,s > 0,where HˆN(r) = 0sinQ(rN)cosQ(rN) ,Q(r) = pi−2tan−1(r).The co-rotational family (1.3.5) ismˆ(x) = eθRHˆN(rs), s > 0.11The bubble Q corresponds, via (1.3.5), to a degree-one harmonic map R2→ S2.Note that this is a one-parameter family, due to the scale invariance of Ee.1.5 First main result: skyrmion existence andpropertiesWe are interested in minimizing profiles u of the reduced energy (1.3.6), especiallythose satisfying boundary conditionsu(0) = pi, u(∞) = 0 =⇒ N = 1 (1.5.1)by (1.3.7), which is corresponding to the configuration of the skyrmion point-ing downward (viewed from above the plane of magnet) at the origin r = 0,and pointing upward at infinity r = ∞. Such minimizers produce solutions ofthe form (1.3.5) to the full Euler-Lagrange equations (1.2.4). These are chiralmagnetic skyrmions solutions with co-rotational symmetry and Skyrmion num-ber N = 1.[37] showed the existence of minimizers of (1.3.6) in this topological class for0 < k < 1, in the case α = 0. They also showed that for k 1 the minimizers areunique and monotone. [19] shows existence of skyrmions for a related model ofliquid crystals.Our approach is to treat the minimization problem for (1.3.6), for k 1, as aperturbation of the minimization problem for the k = β = 0 limit energy E0,α,0 =Ee.Our first main result makes precise the sense in which the skyrmion profilesare perturbations of the bubble Q. To state it, we introduce the family of BanachspacesX p := {ξ : [0,∞)→ R | ‖ξ‖X p < ∞}, ‖ξ‖X p := ‖ξr‖Lp +∥∥∥∥ξr∥∥∥∥Lp, 1≤ p≤ ∞12where‖ f‖Lp :=(∫ ∞0| f (r)|p rdr) 1pdenotes the Lp norm of the radially-symmetric function f (|x|) on R2. So specifi-cally, when p = ∞,‖ξ‖X∞ = ‖ξr‖L∞+‖ξr ‖L∞ .The case p = 2,X = X2 = {ξ : [0,∞)→ R | ‖ξ‖X := ‖ξr‖L2 +∥∥∥∥ξr∥∥∥∥L2< ∞},plays a key role as our “energy space”.Theorem 1. Let A > 0 be given. There is k0 > 0 and C > 0 such that for each0< k≤ k0,−A≤ α ≤ 1, β > 0, there is a unique minimizer vk,α,β of Ek,α,β in theclass Q+X . The corresponding mapmˆk,α,β =[−sin(vk,α,β )sin(θ), sin(vk,α,β )cos(θ), cos(vk,α,β ) ]is a smooth solution of the Euler-Lagrange equations (1.2.4) with Skyrmion num-ber N = 1. Moreover, for a particular choice of scaling β = β (k)→ 0 as k→ 0,satisfyingk = β(2log(1β)+O(1)), (1.5.2)the skyrmion profile satisfies the estimates‖vk,α,β −Q‖X ≤Cβ , ‖vk,α,β −Q‖X∞ ≤Cβ 2 log(1β). (1.5.3)Finally, the skyrmion energy behaves asEk,α,β (vk,α,β ) = Ee(Q)−k22log(1k) (1+O( 1log(1k))) . (1.5.4)13Remark 1.5.1. By the scaling relation (1.3.8), for any β˜ > 0, vk,α,β˜(ββ˜r)=vk,α,β (r), so the estimates (1.5.3) give the k→ 0 behaviour of all the skyrmionprofiles.Compared to [37], we consider a more general energy functional, involvingE(α) versus just E(0). More importantly, our perturbative approach gives the pre-cise description (1.5.3) of the skyrmion profile for k 1, which has several advan-tages. For example, it allows us to compute the precise energy asymptotics (1.5.4)(see also [32], where energy asymptotics are found using formal asymptotic ar-guements). Also, in the forthcoming paper [55], this information is used to provethe stability of these skyrmion solutions (against general perturbations), with pre-cise estimates on the spectral gap. Stability is already obtained by [37] in the caseα = 0, but by a very different method. They use the monotonicity of the skyrmionprofile, which relies on an ODE argument. It is unclear how to adapt this argu-ment to the more complicated energy functionals we consider here, or if it couldbe used to compute the spectral gap. After submission of this thesis, we learned ofthe independent work [4] which considers the isotropic limit of skyrmions outsideof the co-rotational class. In particular, they prove a (non-sharp) version of thefirst estimate of (1.5.3)There are two main complications in treating minimizers of Ek,α,β for small kas perturbations of the minimizers Q(·/s) of E0,α,0 = Ee. Firstly, the unperturbedminimizers have slow spatial decay: Q 6∈ L2. Second, the unperturbed problem isscale invariant. As a consequence, the linear operatorH =−∆+ 1r2− 8(1+ r2)2which appears when linearizing the Euler-Lagrange equation for Ee around Q (seeSection 2.2.1) has a resonance at the threshold of its essential spectrum:Hh = 0, h(r) =ddsQ(rs)∣∣s=1 =2r1+ r26∈ L2,14while by Weyl’s Theorem [45], the essential spectrum of H is σess(H)=σess(−∆)=[0,∞). (Recall that the essential spectrum of an operator A is those λ ∈C for whichA−λ I is not invertible, but λ is not an isolated eigenvalue of finite multiplicity.)The equation for the correction to Q involves the resolvent (H + β 2)−1, and animportant ingredient in our analysis is various estimates of this resolvent actingon spaces of functions with slow spatial decay.In particular, the resolvent becomes singular as β → 0 due to the resonance.To prove the first estimate in (1.5.3), we must remove this singularity by care-fully adjusting the scaling s in Q(r/s), which results in the relation (1.5.2). Thiskind of Lyapunov-Schmidt reduction in the presence of a resonance is similar tothe analysis in [13] of solitons of perturbations of the energy-critical nonlinearSchro¨dinger equation in 3D, though the specific estimates are quite different forour 2D problem.After removing the effect of the resonance, we may expect our resolvent tobehave roughly like the free resolvent−∆+λ which in 2D is given by convolutionwith the (modified) Bessel function which has a small-argument expansion [28]12piK0(√λ |x|)= log 1λ· 1pi+12pi(− log |x|+ log2+ γ)+λ log 1λ· 14pi|x|2+O(λ |x|2).which is still singular for λ → 0. However, because of the equivariant symmetry,it effectively acts on functions of the form eiθ f (r). Since these have averagezero, convolution of such a function with a constant vanishes, and so we find thesingular term disappears: on radial functions,(−∆+ 1r2+λ )−1 = G˜0+λ log1λG˜1+λ Gˆ, Gˆ = O(1)whereG˜ j = e−iθG jeiθ , G0 =− 12pi log |x|∗= (−∆)−1, G1 =14pi|x|2 ∗ .Another key ingredient is to exploit the factorized structure of the linear oper-15atorH = F∗F, F = h∂r1h= ∂r +r2−1r(r2+1).Applying F to the equation for the correction ξ to Q produces an equation for Fξwhose linear operatorHˆ = FF∗ =−∆+ 4r2(r2+1)has no threshold resonance or eigenvalue, and so good estimates for its resolvent(Hˆ+β 2)−1, without singularity as β → 0, can be proved. This is how the secondestimate of (1.5.3) is established.1.6 Landau-Liftshitz EquationsThe dynamics of a time-dependent magnetization mˆ(x, t) is governed by the Landau-Lifshitz (or Landau-Lifshitz-Gilbert) equation. The Landau-Lifshitz equation as-sociated with energy (1.2.2) is∂ t mˆ = a mˆ×E ′(mˆ)−b E ′(mˆ) (1.6.1)where a∈R, b≥ 0 is a damping parameter. Landau-Lifshitz combines Schro¨dingerflow ∂tmˆ= mˆ×E(mˆ) (conserved Hamiltonian) with heat-flow (gradient flow, diss-pation) ∂tmˆ =−E ′(mˆ), satisfying the energy dissipation relationE(mˆ(t))+ ba2+b2∫ t0‖∂tmˆ‖2L2ds = E(mˆ(0))The Landau-Lifshitz equations incorporating only the “exchange energy”∂ mˆ∂ t=−amˆ×∆mˆ+b(∆mˆ+ |∇mˆ|2mˆ) (1.6.2)have been well-studied. Space dimension n= 2 is critical case for Ee(mˆ), becausethe energy Ee(mˆ) is invariant under scaling. The local theory and global theory16for critical problems is more difficult compared to subcritical problems. The casewith a = 0, b 6= 0 corresponds to the harmonic map heat-flow. Global existenceand characterization of blow up solutions for the harmonic map flow and Landau-Lifshitz flow on compact domains were studied in, e.g., [12, 14, 25, 44, 53, 54].On the other hand, the case with b= 0, a 6= 0 (i.e no dissipation term), correspondsto the Schro¨dinger map, the geometric generalization of the linear Scho¨dingerequation for maps into the Ka¨hler manifold S2. Questions of local well-posednessand global behavior of equivarant solutions with energy near the harmonic mapshave been studied in [20, 22–24]. Travelling wave solutions of the Landau-Lifshitz equation with easy-plane anisotropy were shown to exist in [38].1.6.1 Preservation of equivariance by the Landau-LifshitzdynamicsFirst, let’s observe that if an energy functional E is invariant under a unitary groupof symmetries, then its variational derivative E ′ is covariant, which is shown be-low.If eα T is a one-parameter group of transformations of maps from (in our case)R2 7→ S2 ⊂ R3 which acts unitarily on tangent vector fields: denoting as usual〈~f , ~g〉= ∫R2 ~f (x) ·~g(x) dx,~ξ (x),~η(x) ∈ Tmˆ(x)S2 =⇒ 〈~ξ , eα T~η〉 := 〈e−α T~ξ , ~η〉Recall the variational derivative E ′(mˆ) is defined in terms of one-parametervariations mˆε of mˆ (that is, mˆε : R2→ S2 ⊂ R3 depends smoothly on ε , and mˆ0 =mˆ) and the inner-product. The one-parameter variations of eα T mˆ are of the formeα T mˆε .If mˆε is a one-parameter variation of mˆ with ξ (x) := dd ε mˆε(x)|ε=0 (so ξ (x) ∈17Tmˆ(x)S2), then dd ε eα T mˆε |ε=0 = eα Tξ , so using the unitarity and the invariance,〈e−α T E ′(eα T mˆ), ξ 〉= 〈E ′(eα T mˆ), eα Tξ 〉= dd εE(eα T mˆε)|ε=0=dd εE(mˆε)|ε=0 = 〈E ′(mˆ), ξ 〉.This holds for any smooth tangent vector field ξ , soe−α T E ′(eα T mˆ) = E ′(mˆ). (1.6.3)For the micromagnetic energy E = Ek,α,β , we’d like to apply this to the trans-formation groupmˆ(x) 7→ (eα T mˆ)(x) = e−α Rmˆ(eα R˜x)which leaves E invariant. Note that eα T acts unitarily, because each of the targetand spatial rotation operators separately act unitarily. If mˆ is equivariant, thenby (1.6.3),eα T mˆ = mˆ =⇒ E ′(mˆ) = E ′(eα T mˆ) = eα T E ′(mˆ)and so we see E ′(~m) is also equivariant. Thus we expect that the solution of thegradient flow ∂t~m =−E ′(~m) with equivariant initial data should remain equivari-ant.We can be more precise. The generating operator T can be found simply bydifferentiating:T mˆ(x) = ∂α(eα T mˆ)(x)|α=0 =−Rmˆ(x)+(R˜x ·∇)mˆ(x) = (∂ θ −R)mˆ(x)so T = ∂ θ −R, just the sum of the generators of the target and spatial rotations.(The fact that eα T is unitary is reflected in the fact that its generator T is skew-adjoint: T ∗ = −T .) By differentiating with respect to α at α = 0, we find an18equivalent condition for mˆ to be equivariant:eα T mˆ = mˆ =⇒ 0 = T mˆ = (∂ θ −R)mˆ (∈ Tmˆ(x)S2),and a similar condition at the level of tangent vector fields which we should dif-ferentiate covariantly as shown below: Dµ~ξ = PTmˆS2 ∂ µ~ξ = ∂µ~ξ +(∂µmˆ ·~ξ )mˆ,so~ξ (x) ∈ Tmˆ(x)S2, eα T~ξ =~ξ =⇒ 0= PT~ξ = (Dθ −PR)~ξ (∈ Tmˆ(x)S2). (1.6.4)Suppose mˆ(x, t) solves the Landau-Lifshitz equation (1.6.1) where recall Jmˆ~ξ :=mˆ×~ξ for ~ξ ∈ TmˆS2. ThenDtT mˆ = Dt(∂ θ −R)mˆ = (Dθ −PR)∂ t mˆ = (Dθ −PR)(aJmˆE ′(mˆ)−bE ′(mˆ)),where we used the fact that usual and covariant derivatives commute: Dµ ∂ ν mˆ =∂µ∂νmˆ+(∂ µ mˆ ·∂ ν mˆ)mˆ= ∂ ν ∂ µ mˆ+(∂ ν mˆ ·∂ µ mˆ)mˆ=Dν ∂ µ mˆ. So using (1.6.4),Jmˆ commutes with covariant derivatives, and with PR: for ~ξ ∈ TmˆS2,DθJmˆ~ξ = ∂θ (m×~ξ )+ [∂θ mˆ · (mˆ×~ξ )]mˆ= ∂ θ mˆ×~ξ + mˆ×∂ θ ~ξ +(~ξ ×∂ θ mˆ)mˆ= mˆ×∂ θ ~ξ = JmˆDθ~ξ .And, JmˆPR~ξ = mˆ×R~ξ =−m3~ξ . Thus,PRJmˆ~ξ = R(mˆ×~ξ )− (mˆ ·R(mˆ×~ξ ))mˆ = R(mˆ×~ξ )− (Rmˆ×~ξ )mˆ= R(mˆ×~ξ )− (~ξ · kˆ)mˆ=−m3~ξ = JmˆPR~ξ .So we see that if mˆ is (initially) equivariant, thenDtT mˆ = 019and so we expect the equivariance condition T mˆ = 0 to be preserved by the dy-namics.1.6.2 Non-perservation of co-rotational maps byLandau-Lifshitz dynamicsConsider now the (discrete) transformationmˆ(x) 7→ (Fmˆ)(x) := −m1(x1,−x2)m2(x1,−x2)m3(x1,−x2)which is in fact a reflection: F2 = I (indeed the composition of a reflection on thedomain R2 and a reflection on the target S2 ⊂ R3). It is easily checked that(Fmˆ ·∇× (Fmˆ))(x) = (mˆ ·∇× mˆ)(x1,−x2)and so EDM(Fmˆ) = EDM(mˆ) (since the Jacobian of (x1,x2) 7→ (x1,−x2) is just 1).It is also clear the other terms in the energy are invariant under F : E(Fmˆ) = E(mˆ).As above this impliesE ′(Fmˆ) = F E ′(mˆ) and so: Fmˆ = mˆ =⇒ F E ′(mˆ) = E ′(mˆ).Maps which are both equivariant (as above) and invariant under F are exactly theco-rotational ones:eα T mˆ= mˆ and Fmˆ= mˆ =⇒ mˆ(r,θ)= eθR 0sin(u(r))cos(u(r))= −sin(u(r))sin(θ)sin(u(r))cos(θ)cos(u(r)) ,20(up to an overall rotation). Now for the gradient-flow, ∂ t mˆ =−E ′(mˆ), we expectthe class Fmˆ = mˆ to be preserved, since if this holds initially, then∂ t(Fmˆ− mˆ) = (F− I)∂ t mˆ =−(F− I)E ′(mˆ) = 0.For the Schro¨dinger flow ∂ t mˆ = JmˆE ′(mˆ), however,∂ t(Fmˆ− mˆ) = FJmˆE ′(mˆ)− JmˆE ′(mˆ),but F and J do not commute. In fact, they anticommute: for ~ξ ∈ TmˆS2,FJmˆξˆ =−JFmˆF~ξso if (initially) Fmˆ = mˆ,∂ t(Fmˆ− mˆ) =−2JmˆE ′(mˆ) 6= 0and so this condition is not preserved by the Schro¨dinger term.1.7 Second main results: skyrmion stabilityOnce we understand a special solution, a static solution in our case, what wewhat to know next is: is the solution stable? In practice, a “stable” solution isone that maintains its basic shape/form under small perturbations, so the stablesolutions can be observed physically or numerically. Mathematically, there arethree possible types of stability that could apply to our static skyrmion solutionssˆ(x). We say a sˆ is Lyapunov stable if every solution which starts “close” to, itstays “close” for all future times :∀ε > 0, ∃δ > 0 s.t ‖mˆ(·,0)− sˆ‖< δ =⇒ ‖mˆ(·, t)− sˆ‖< ε,∀t ≥ 021for some choice of norm ‖ · ‖. When there is some symmetry of the equationwhich is broken by the static solution e.g., translation in our case, we may expectLyapunov only up to symmetry, or orbital stability:∀ε > 0, ∃δ > 0 s.t ‖mˆ(·,0)− sˆ‖< δ =⇒ infy∈R2‖mˆ(·, t)− sˆ(·− y)‖< ε.Finally, asymptotic stability is when nearby solutions actually converge to thestatic solution (or its symmetry orbits) as t→ ∞:∃δ > 0 s.t ‖mˆ(·,0)− sˆ‖< δ =⇒ limt→∞ infy∈R2‖mˆ(·, t)− sˆ(·− y)‖= 0.Because of the translation invariance of the Landau-Liftshitz equation, orbitaland asymptotic stability are the possibilities for skyrmions. For a dissipative sys-tem, the energy dissipation can sometimes be used to show that the solutions mustalways stay close to the (translated) static solutions (skyrmions in our case), andeven converge to them asymptotically. For a conservative Hamiltonian system,asymptotic stability becomes a possibility for dispersive PDE only through themore subtle mechanism of dispersion.In the second part of the thesis (Chapter 3), we provide an alternate, quantita-tive proof of the recent skyrmion stability result of Li-Melcher [37]. We preciselyestimate the lowest eigenvalues of the second variation of the energy around theskyrmion by perturbing from the isotropic (k = β = 0) case, to prove the strictlocal minimality up to translations. In order to do the perturbation theory, we firstfix the scaling k = k(β ) = 2β(log( 1β +O(1))).We decompose the second variation in Fourier modes indexed by j ∈Z, whichis done in the Section 3.6. Since the full energy is invariant under spatial transla-tion, the second variation has a non-zero kernel:0 =∂∂ εE ′k,α,β (sˆk,α,β (·+ ε eˆl))|ε=0 = E ′′k,α,β (sˆk,α,β )∂ sˆk,α,β∂ xl, l = 1,2.Under the Fourier decomposition, these zero-modes are in the j =±1 subspaces.22The j = 0 case, which is corresponding to the equivariant perturbation, is themost delicate one. The unperturbed operator is:H =−∆˜−W, −∆˜=−∆+ 1r2, W = 2h2r2, h := sin(Q) =2rr2+1.Since ker(H) = 〈h〉. and h(r) ∼ 1r as r → ∞, so h is not in the L2 kernel: itis a resonance eigenfunction, sitting at the threshold of the essential spectrum.In Section 3.1, using spectral perturbation theory via resolvent expansion, andcareful estimation of the perturbed term, we show the j = 0 operator has positivespectrum, bounded away from zero by β 2−O( β 2log( 1β )). This is a contradictionargument. In the first step, we show that (normalized in X) eigenfunction φ foreigenvalues below this level would converge to h ∈ kerH. Then, studying theequation for the perturbation η = φ − h, using the resolvent expansion provedin Proposition 3.1.1, we get a contradiction after many careful estimates of thesource terms in this equation.In Section 3.3, for j = ±1, using the factorization of the linearized operator,we show that orthogonal to the translational zero-modes, the positive spectrumis bounded below by β 2−O(β 4 log4( 1β )) in Theorem 11. Here there are severaladditional difficulties. First, the quadratic form for the unperturbed operator H1acting on the first component does not control the X or L∞ norm (at r = 0) of aneigenfunction, so we have to use a weaker norm to normalize the eigenfunction.Second, the inner-product with the (first component of the) unperturbed transla-tion zero mode hr does not make sense in this space, so we first replace it with alocalized bump function. Third, the quadratic form for the unperturbed operatorH−1 of the second component does not control X or L∞ (at r = ∞). For this, weuse a factorization H−1 = F∗−1F−1 like (3.1.2) and a corresponding interpolationinequality (3.3.5).In Section 3.4, we can directly compute the quadratic form in the higher modes| j| ≥ 2 to obtain a uniform spectral gap of β 2−O(β 4 log3( 1β )). Combining all theFourier modes together at the end, we obtain the spectral stability. Orthogonal23to the translational zero modes ( j = ±1), we obtain a spectral gap of the secondvariation of the energy E ′′k,α,β (sˆk,α,β ) around the local minimizer sˆk,α,β .Theorem 2. For 0 < k 1,−A≤ α ≤ 1,β > 0E ′′k,α,β (sˆk,α,β )|T⊥ ≥ β 2 γk,γk ≥ 1− C1log( 1k ) > 0, and T := span{∂ sˆk,α,β∂ x1,∂ sˆk,α,β∂ x2}. -The dynamics of skyrmion is governed by the Landau-Lifshitz equation withdissipation:∂tmˆ =−bE ′(mˆ)+amˆ×E ′(mˆ), b > 0, a ∈ RFor simplicity, we write E ′(mˆ) = E ′k,α,β (mˆ). As shown in the Section 3.6E ′(mˆ) = E ′e(mˆ)+ kβ E ′DM(mˆ)+β 2Eα′(mˆ)=−∆mˆ−|∇mˆ|2mˆ+ kβ (∇× mˆ− (mˆ · (∇× mˆ)~m))+β 2(α(−m3kˆ+m23mˆ)+(1−α)(−kˆ+m3mˆ))Applying mˆ× on both sides of the equation, we rewrite it as:γ1∂t(mˆ)+ γ2mˆ×∂tmˆ =−E ′(mˆ) (1.7.1)where γ1 := ba2+b2 > 0,γ2 :=aa2+b2 .Multiplying (1.7.1) by ∂tmˆ produces the energy dissipation relation:γ1∫ T0∫R2|∂tmˆ|2dxdt =∫ T0∂t E(mˆ(·, t))) =⇒ E(mˆ(t))+γ1∫ t0‖∂tmˆ‖2L2ds= E(mˆ(0))(1.7.2)Dynamic stability of a spin-current driven chiral skyrmion has been studied in[15] with specific combination of Zeeman and in-plane anisotropy interaction.By following their argument, we obtain the orbital stability around the skyrmionsolution sˆk,α,β as follows:24Theorem 3. Fix k,β ,α , such that E ′′k,α,β (sˆk,α,β )|T⊥ > 0. For any ε > 0, there exista δ > 0 such that if ‖mˆ(0)− sˆk,α,β‖H1 ≤ δ , then infx∈R2 ‖mˆ(·−x, t)− sˆk,α,β‖H1 ≤ εfor all t ≥ 0.25Chapter 2Existence and Isotropic Limit ofCo-rotational SkyrmionsIn this chapter, we prove Theorem on the existence of minimizing skyrmion pro-files, and their convergence to the bubble Q as k→ 0. Much of the existence part issimilar to [15, 37, 41], but since we have a more general energy functional it is in-cluded, though briefly, with technical details left to appendix Sections 2.3.1- 2.3.3.The main argument establishing the perturbation estimates (1.5.2) and (1.5.3) isin Section 2.2. This argument relies heavily on various resolvent estimates whichare important, but for readability are left to the appendix Section 2.3.4. Finally,uniqueness is shown in Section 2.2.7, by a variant of the perturbation estimates inSection 2.2, and the energy asymptotics (1.5.4) are proved in Section 2.2.8.2.1 Minimization problem and function spaceTo produce a skyrmion solution, we wish to minimize the energy Ek,α,β (u) amongprofile functions u(r), in an appropriate function space, corresponding to topolog-ically non-trivial magnetization configurations.To identify the natural function space, first observe that in the co-rotationalsetting, we have the following localized version of the elementary topological26lower bound (1.1.6): for any 0≤ r1 < r2 ≤ ∞,E [r1,r2]e (u) :=12∫ r2r1(u2r +1r2sin2 u)rdr =∫ r2r1(±1r(cosu)r +12(ur± 1r sinu)2)rdr=±(cosu(r2)− cosu(r1))+ 12∫ r2r1(ur± 1r sinu)2rdr≥±(cosu(r2)− cosu(r1)).(2.1.1)Note that by one-dimensional Sobolev embedding, if Ee(u)<∞, then u∈C((0,∞)).Moreover, using (2.1.1), the limits limr→0+,∞ cosu(r), and hence u(0)= limr→0+ u(r),u(∞) = limr→∞ u(r) exist, and (by finite energy again) are multiples of pi:Ee(u)< ∞ =⇒ u ∈C([0,∞]), u(0),u(∞) ∈ piZ . (2.1.2)Finiteness of Ez further requires u(∞) ∈ 2piZ, and we may then assume u(∞) = 0by the shift u 7→ u− u(∞) which leaves energy unchanged. Further, by u 7→ −u,k 7→−k, is is enough to consider n≥ 0. This leads to the family of function spacesXn := {u : [0,∞)→R |Ee(u)<∞, limr→0+u(r)= npi, limr→∞u(r)= 0}⊂C([0,∞]), n= 0,1,2, . . .on which the exchange energy Ee is well-defined. By (1.3.7), the Skyrmion num-ber isu ∈ Xn =⇒ N = 12(1− cos(npi)) ={1 n odd0 n even.For n = 0 we denote X := X0. Since |sinu| ≤ |u| ≤C|sinu| for |u| ≤ pi2 ,X := X0 = {u : [0,∞)→ R | ur ∈ L2, ur ∈ L2},is a Banach space with norm‖u‖2X :=∫ ∞0(u2r +u2r2)rdr.27By writing u2(r) as the integral of its derivative and using Ho¨lder’s inequality, weobtain an elementary embedding inequalityu ∈ X =⇒ ‖u‖L∞ ≤ ‖u‖X (2.1.3)which is used frequently below.It is easy to check that Xn = u¯n +X0 for any fixed u¯n ∈ Xn, and so Xn is anaffine Banach space.To ensure finiteness of E(α) we add the condition u ∈ L2: since 1− cos(u) ≤min(12u2,2),0≤ E(α)(u)≤∫ ∞0(1+ |α |)12u2(r)r dr =12(1+ |α |)‖u‖2L2.This also makes EDM finite via Ho¨lder’s inequality:|EDM(u)| ≤∫ ∞0|sin(u(r))||ur|r dr ≤[∫ ∞0sin2(u(r))r dr] 12[∫ ∞0u2r (r)r dr] 12≤ 2√Ea(u)√Ee(u)≤ 2√E(α)(u)√Ee(u)< ∞.(2.1.4)So finally our variational problem is:e(n)k,α,β := inf{Ek,α,β (u) | u∈Xn∩L2}, k∈R, α ≤ 1, β > 0, n∈{0,1,2, . . .}.(2.1.5)X0∩L2 is a Banach space, while for n 6= 0, Xn∩L2 = v¯n +X0∩L2 for any fixedv¯n ∈ Xn ∩ L2 is an affine Banach space. A simple variant of (2.1.3) shows thatfunctions in Xn∩L2 (unlike functions in Xn) have a decay rate:u ∈ Xn∩L2 ⊂ H1 =⇒ u2(r)≤ 1r ‖u‖2H1 ≤1r(2Ee(u)+‖u‖2L2). (2.1.6)Proposition 2.1.1. Ek,α,β is a (Fre´chet) differentiable function on Xn∩L2. If v ∈28Xn∩L2 solves (2.1.5), then v ∈C∞((0,∞)) satisfies the Euler-Lagrange equation0=E ′k,α,β (v)=−∆rv+12r2sin(2v)−k β 1rsin2(v)+β 2 (1−α(1− cos(v)))sin(v).(2.1.7)Moreover, v(r) and vr(r) decay exponentially as r→ ∞, and the Pohozaev-typerelationβ kEDM(v)+2β 2 E(α)(v) = 0 (2.1.8)is satisfied. Finally, the map mˆ : R2→ S2 given bymˆ(x) = [−sin(v(r))sin(θ), sin(v(r))cos(θ), cos(v(r)) ] (2.1.9)satisfies the full Euler-Lagrange equation (1.2.4).Proof. Let u ∈ Xn∩L2 and ξ ∈ X ∩L2. Using simple trig identities and the ele-mentary bounds |1− cosξ |+ |ξ − sinξ |.ξ 2, we find:∣∣∣∣Ee(u+ξ )−Ee(u)−∫ ∞0(urξr +1r2sinucosu)rdr∣∣∣∣.‖ξ‖2X ;∣∣∣∣EDM(u+ξ )−EDM(u)+∫ ∞0 1r sin2 u ξ rdr−∫ ∞0(sin2 u ξ r)r dr∣∣∣∣.‖ur‖L2‖ξ‖L2‖ξ‖L∞+‖ξr‖L2‖ξ‖L2and so since u ∈ Xn∩L2, ξ ∈ X ∩L2 implies sin2 u ξ r→ 0 as r→ 0 and r→ ∞(using (2.1.6)),∣∣∣∣EDM(u+ξ )−EDM(u)+∫ ∞0 1r sin2 u ξ rdr∣∣∣∣.‖ξ‖2X∩L2;∣∣∣∣Ez(u+ξ )−Ez(u)−∫ ∞0 sinu ξ rdr∣∣∣∣.‖ξ‖2L2;∣∣∣∣Ea(u+ξ )−Ea(u)−∫ ∞0 sinucosu ξ rdr∣∣∣∣.‖ξ‖2L2 .29Combining these shows that Ek,α,β is differentiable on Xn∩L2 with Fre´chet deriva-tive at u given byX∩L2 3 ξ 7→∫ ∞0(urξr +[1r2sinucosu− kβ 1rsin2 u+β 2 (α sinucosu+(1−α)sinu)]ξ)rdr.If v minimizes Ek,α,β on Xn ∩L2, this derivative vanishes for all ξ ∈ X ∩L2. Inparticular, taking ξ ∈C∞0 ((0,∞)), we see that v satisfies the Euler-Lagrange equa-tion (2.1.7) in the sense of distributions. It follows that vrr ∈ L2loc((0,∞)), so infact v ∈ H2loc. Continuing in this way, v ∈ Hkloc for all k, and so v ∈ C∞((0,∞))satisfies (2.1.7) in the classical sense.The exponential decay of v is standard. Since v ∈ Xn ∩ L2, it has some de-cay by (2.1.6). The linear approximation to the Euler-Lagrange equation (2.1.7)around v = 0 is0≈−vrr− vrr +vr2+β 2 vwhose decaying fundamental solution is the modified Bessel function K1(β r),which decays like e−β r√β ras r→ ∞. See [37] for details.To obtain the Pohozaev-type identity (2.1.8), multiply the Euler-Lagrangeequation (2.1.7) by rvr and integrate over an interval [s R] with 0 < s < R < ∞to obtainkβ∫ RseDM(v)rdr+2β 2∫ Rse(α)(v)rdr=[−12(rvr)2+12sin2 v+β 2 r2e(α)(v)]∣∣∣Rs.(2.1.10)Since u ∈ Xn, we have limr→0v(r) = npi and so limr→0sinv = limr→0r2e(α)(v) = 0. Sincev ∈ Xn ∩ L2, vr ∈ L2, and by (2.1.7) and the exponential decay, (rvr)r ∈ L2. Sorvr ∈ X and in particular limr→0rvr = 0. Taking s→ 0 in (2.1.10) giveskβ∫ R0eDM(v)rdr+2β 2∫ R0e(α)(v)rdr=−12Rv2r (R)+12sin2 v(R)+β 2 R2e(α)(v)(R).(2.1.11)Taking R→ ∞ and using the exponential decay gives the desired relation (2.1.8).30Finally, we verify directly that (2.1.9) satisfies (1.2.4). Leteˆ(r,θ) := [−cos(v(r))sin(θ),cos(v(r))cos(θ),−sin(v(r))].We notice that |eˆ|= 1, and eˆ⊥ mˆ, so eˆ(r,θ) ∈ Tmˆ(r,θ)S2. Computemˆr = vreˆ, mˆrr = vrreˆ− v2r mˆ, mˆθθ =−sin2(v)mˆ− sin(v)cos(v)eˆ,and soPTmˆS2∆mˆ = PTmˆS2(mˆrr +1rmˆr +1r2mˆθθ)=(vrr +1rvr− sin(v)cos(v)r2)eˆ.Compute∇×mˆ= [−vr sin(v)sin(θ),vr sin(v)cos(θ),vr cos(v)+ 1r sin(v)]= vrmˆ+1rsin(v)kˆ,soPTmˆS2(∇× mˆ) =1rsin(v)PTmˆS2 kˆ,andPTmˆS2 kˆ = kˆ−m3mˆ =−sin(v)eˆ.Then by (1.2.4)E ′(mˆ) = PTmˆS2[−∆mˆ+β k ∇× mˆ+β 2(α−1−αm3)kˆ]=(−vrr− 1r vr +sin(v)cos(v)r2−β k 1rsin2(v)+β 2(1−α+α cos(v))sin(v))eˆ.So the full Euler-Lagrange equation (1.2.4) is satsified, since the reduced one (2.1.7)is.Taking λ = β in the scaling relation (1.3.8), we haveEk,α,β (u(β ·)) = Ek,α,1(u), e(n)k,α,β = e(n)k,α,1, (2.1.12)31and so we may fix β = 1 and consider the minimization probleme(n)k,α,1 := inf{ Eκ,α,1(u) | u ∈ Xn∩L2}, k ∈ R, α ≤ 1. (2.1.13)2.1.1 Lower boundsTo begin the study of (2.1.13), we investigate if the energy is bounded from below.First, using (2.1.4) and Young’s inequality yieldsEk,α,1(u)≥ Ee(u)+E(α)(u)−min(k2Ee(u)+E(α)(u), Ee(u)+ k2E(α)(u))= max((1− k2)Ee(u), (1− k2)E(α)(u)).(2.1.14)Second:Lemma 2.1.1. If u ∈ Xn satisfies u(r) = jpi for some r ∈ [0,∞), j ∈ Z, thenEe(u)≥ 2(|n− j|+ | j|). Therefore,u ∈ Xn =⇒ Ee(u)≥ 2|n| (2.1.15)andu ∈ Xn =⇒ ‖u‖L∞ ≤(12Ee(u)+1)pi. (2.1.16)Proof. If r 6= 0 and j 6= n, by continuity of u, there are non-intersecting sub-intervals (rm, rm+1) ⊂ [0, r], m = 1,2, . . . , |n− j| with |u(rm)− u(rm+1)| = pi .By (2.1.1), E [rm, rm+1]e (u)≥ 2, and soE [0, r]e (u)≥ 2|n− j|.Similarly, if j 6= 0, there are non-intersecting sub-intervals (rm, rm+1) ⊂ [r, ∞),m= 1,2, . . . , | j|, r| j|+1 =∞, with |u(rm)−u(rm+1)|= pi . By (2.1.1), E [rm, rm+1]e (u)≥2, and soE [r, ∞)e (u)≥ 2| j|.32The first statement of the lemma follows from summing the two bounds above.Then (2.1.15) follows from taking r = 0 and j = n. For (2.1.16): if ‖u‖L∞ >(12Ee(u)+1)pi , then by continuity of u, for some r∈ [0,∞), u(r)= jpi where j∈Zwith | j| ≤ 12Ee(u)+1< | j|+1. Then by the first statement, Ee(u)≥ 2| j|> Ee(u),a contradiction.Combining (2.1.14) and (2.1.15) gives:u ∈ Xn, |k| ≤ 1 =⇒ e(n)k,α,1 ≥ 2|n|(1− k2). (2.1.17)We are not sure if the condition |k| ≤ 1 is necessary for boundedness frombelow of the energy, but we do have the following partial converse to (2.1.17).Proposition 2.1.2.|k|> 4− 32α =⇒ e(n)k,α,1 =−∞. (2.1.18)Proof. Without loss of generality, we may take n≥ 0. First suppose k > 0. Con-sider, for M  t  1, M an odd integer, a piecewise linear test function of theformu(r) =npi+ tr 0≤ r ≤ M−nt pi(M+ M−nt )pi− r M−nt pi ≤ r ≤ (M−nt +M)pi0 r ≥ (M−nt +M)pi ∈ Xn.Begin with the exchange energy:12∫ ∞0u2r rdr =12t212(M−n)2t2pi2+1212pi2((M−nt+M)2− (M−n)2t2)=M2pi24((1−n/M)2+1+ 2t(1−n/M))=M2pi22(1+O(1t)).The other part of the exchange is (mainly) bounded by the anisotropy: using33sin2(npi+ tr) = sin2(tr)≤min(t2r2, 1),12∫ ∞0sin2(u)r2rdr ≤ 12∫ 1/t0t2 rdr+12∫ M/t1/tdrr+12t2M2∫ ∞M/tsin2(u) rdr= O(1+ logM+t2M2Ea(u)),and combining these givesEe(u) =M2pi22(1+O(1t+t2M4Ea(u))).Next the anisotropy: setting S(y) = y/2− sin(2y)/4, Sˆ(y) = y2/4− ysin(2y)/4+sin2(y)/4, so that S′ = sin2(y) and Sˆ′ = ysin2(y),Ea(u) =12∫ ∞0sin2(u(r)) rdr =12∫ M−nt pi0sin2(tr) rdr+12∫ (M−nt +M)piM−nt pisin2(r−M−ntpi) rdr=12t2Sˆ((M−n)pi)+ 12Sˆ(Mpi)+(M−n)pi2tS(Mpi)=M2pi28(1+O(1t)).The Zeeman term is similar: set S˜±(y) = y2/2±ysin(y)±cos(y)∓1 so that S˜′± =y(1± cos(y)),Ez(u) =∫ ∞0(1− cos(u(r))) rdr =∫ M−nt pi0(1± cos(tr)) rdr+∫ (M−nt +M)piM−nt pi(1+ cos(r−M−ntpi)) rdr=1t2S˜±((M−n)pi)+ S˜+(Mpi)+ (M−n)pit Mpi =M2pi22(1+O(1t)).34Finally the D-M term (which is the source of negative energy):EDM(u) =∫ ∞0ur sin2(u) rdr = 2tE[0,(M−n)pi/t]a −2E [(M−n)pi/t,(M+(M−n)/t)pi]a=−M2pi24(1+O(1t)).So in totalEk,α,1(u) =M2pi24(2− k+ α2+2(1−α)+O(1t+t2M2))=M2pi24(4− 3α2− k+O(1t+t2M2))which→−∞ as M t→ ∞, provided k > 4− 3α2 .The case k < −(4− 3α2 ) is handled by a similar test function with the slopesreversed:u(r) =npi− tr 0≤ r ≤ M+nt pi−(M+ M+nt )pi+ r M+nt pi ≤ r ≤ (M+nt +M)pi0 r ≥ (M+nt +M)pi ∈ Xn.We omit the details.2.1.2 Upper boundsHere we specialize to the case n = 1, and show an upper-bound which ensurescompactness of minimizing sequences. For this, we use the well-known ex-plicit minimizers of the exchange energy Ee on X1, which saturate the inequal-ity (2.1.15):Ee(u) = 2, u∈ X1 =⇒ u(r) =Q(r/s) for some s> 0, Q(r) = pi−2tan−1(r).35This follows from (2.1.1), since equality in Ee(u) ≥ 2 holds if and only if ur +1r sin(u) = 0 (almost everywhere) on (0, ∞), and the only solutions to this ode inX1 are the given ones.Proposition 2.1.3. e(1)k,α,1 ≤ 2. Moreover,k > 0 =⇒ e(1)k,α,1 < 2. (2.1.19)Proof. Q 6∈ L2, so to use it as a test function we cut it off: setuR(r) := Q(r)φR(r), φR(r) := φ(r/R), R > 0,where φ ≥ 0 is a standard smooth cut-off function with φ(r) = 1 for r ≤ 1 andφ(r) = 0 for r ≥ 2. Then uR ∈ X1∩L2, and as R→ ∞ it is easy to check thatEe(uR) = 2+O(R−2), E(α)(uR) = O(logR), EDM(uR) =−2+o(1),using (2.3.56). Rescaling, usR(r) := uR(r/s), s > 0, we getEk,α .1(usR) = 2−2ks+O(R−2)+ s o(1)+ s2 O(logR),so1s[Ek,α,1(usR)−2]=−2k+ 1sO(R−2)+o(1)+ s O(logR).Then if we take R→ ∞ and s = 1/R→ 0, we see1s[Eκ,α,1(usR)−2]→−2k ,which proves both statements of the proposition.Remark 2.1.1. For n≥ 2, a test function built of a sequence of n re-scaled, shifted,cut-off bubbles Q, connecting u= npi at r= 0 to u= 0 at r=∞, with (small) lengthscales whose ratios are diverging, shows that e(n)k,1,α ≤ 2n.362.1.3 Existence of minimizerWe now show that for n = 1 and 0 < k < 1, lower bound (2.1.17) and upperbound (2.1.19) imply existence of a minimizer for problem (2.1.13):Theorem 4. Let n = 1, k ∈ (0,1), α ≤ 1. There is v ∈ X1∩L2 with Ek,α,1(v) =e(1)k,α,1.Remark 2.1.2. Theorem 4, together with Proposition 2.1.1, prove the existencepart of Theorem 1.The proof is an elementary concentration-compactness-type argument, withcondition (2.1.19) used to exclude concentration. This in the spirit of [41], butmuch simpler because of the co-rotational symmetry. For this reason, we give thedetails in the appendix, Section 2.3.2.For the case α = 0 (no anisotropy), it is shown in [37] that for k 1, the min-imizing profile v(r) is monotone. For completeness, we prove in Section 2.3.3 ofthe appendix that monotonicity holds also in the opposite case α = 1 (no externalfield):Proposition 2.1.4. For α = 1 and 0 < k 1, a minimizing profile v(r) is mono-tonically decreasing.By applying the scaling (2.1.12) to the minimizer provided by Theorem 4, wealso have:Corollary 5. Let n = 1, k ∈ (0,1), α ≤ 1, β > 0. There is v ∈ X1 ∩ L2 withEk,α,β (v) = e(1)k,α,β .By Proposition 2.1.1, any such minimizer satisfies the Euler-Lagrange equa-tion (2.1.7), as well as the Pohozaev-type relation (2.1.8).2.1.4 Convergence to Q, after rescalingTheorem 6. Consider a family of minimizers {uk} ⊂ X1, Ek,αk,β k(uk) = e(1)k,αk,β k,indexed by (0,1) 3 k→ 0, with αk ≤ 1 and β k > 0. Then there are rk > 0 such37that vk(r) := uk(rk r) satisfies‖vk−Q‖X → 0 (2.1.20)as k→ 0. Moreover Ek,αk,βˆ k(vk) = e(1)k,αk,βˆ kwith βˆ k := rk β k, and (1+ |αk |)βˆ k→0.Remark 2.1.3. The existence of such minimizers is ensured by Corollary 5.Proof. By rescaling uk(r) 7→ uk(β k r)we may assume β k = 1. By (2.1.14), (2.1.15),and (2.1.17),2 > e(1)k,αk,1 = Ek,αk,1(uk)≥ (1− k2)Ee(uk)≥ 2(1− k2)→ 2,which together with (2.1.8) showsEk,αk,1(uk) = e(1)k,αk,1→ 2, Ee(uk)→ 2, E(αk)(uk)→ 0. (2.1.21)Since uk ∈ X1 is continuous, there is rk > 0 such that uk(rk) = pi2 . Setvk(r) := uk(rkr),sovk(1) =pi2, Ee(vk)→ 2, r2kE(αk)(vk)→ 0.The proof of (2.1.20) is a standard variational argument, together with theuse of (2.1.1). If (2.1.20) fails, then along some subsequence v j := vk j , k j → 0,we have ‖v j−Q‖X ≥ ε for some ε > 0. We will contradict this. Since ‖(v j)r‖2L2 ≤2Ee(v j).1, and using the 1D (compact) Sobolev embedding H1dr([r1,r2])⊂C([r1,r2])for any 0 < r1 < r2 < ∞, there is a further subsequence (still denoted v j) converg-ing uniformly on compact subintervals of (0,∞) to a continuous function v(r),with v(1) = pi2 . Moreover, (v j)r converges weakly in L2rdr(0,∞) to vr. By weak38lower-semi-continuity of the L2 norm, and Fatou’s lemmaEe(v)≤ liminfE(v j) = 2. (2.1.22)By (2.1.2), v ∈C([0,∞]) with v(0), v(∞) ∈ piZ.We next show v(0) = pi , v(∞) = 0, and v = Q. Apply (2.1.1) to v j on each ofthe intervals [0,1] and [1,∞], using (2.1.22), to conclude.1≤ E [0,1]e (v j)≤ 1+o(1), 1≤ E [1,∞)e (v j)≤ 1+o(1). (2.1.23)Then by weak lower-semi-continuity and Fatou,E [0,1]e (v)≤ liminfE [0,1]e (v j) = 1, E [1,∞)e (v)≤ liminfE [1,∞)e (v j) = 1. (2.1.24)Applying (2.1.1) for v j on [0,r] and [r,1] for r ≤ 1, using (2.1.23), givespi2−o(1)≤ v j(r)≤ pi+o(1), 0≤ r ≤ 1, (2.1.25)which impliespi2≤ v(r)≤ pi, 0≤ r ≤ 1,and in particular v(0) = pi . Similarly, applying (2.1.1) on [1,r] and [r,∞) for r ≥ 1gives−o(1)≤ v j(r)≤ pi2 +o(1), 1≤ r < ∞, (2.1.26)which implies0≤ v(r)≤ pi2, 1≤ r < ∞,and in particular v(∞) = 0. Then (2.1.1) on [0,∞), with (2.1.22) forces vr +1r sinv = 0 a.e. in (0,∞). The unique solution of this ODE with v(1) =pi2 isv = Q.Sow j := v j−Q→ 0 uniformly on compact subintervals of (0,∞), and (w j)r→ 0 weakly in L2.39It remains to show the convergence is strong in X . This will be a consequence ofconvergence of the energy: 2 = Ee(Q) = limEe(v j). The energy relationv j = Q+w j =⇒ Ee(v j) = Ee(Q)+Ee(w j)+∫ ∞0(Qr(w j)r +1r2cos(Q+w j)sinQsinw j)rdris an elementary consequence of trig identities. Letting j → ∞ here, using theweak convergence of (w j)r, the local uniform convergence of w j, and the factsthat Qr ∈ L2rdr, 1r2 sinQ ∈ L1rdr, we findlimj→∞Ee(w j) = 0.Now by (2.1.25) and (2.1.26),−pi2−o(1)≤ w j(r)≤ pi2 +o(1),we have w2j.sin2 w j and so‖w j‖2X.Ee(w j)→ 0,which completes the contradiction argument.By (2.1.12), vk ∈ X1 minimizes Ek,αk,βˆ k with βˆ k := rk. Finally,E(α)(u)≥∫ ∞0((1− cosu)+α− 12(1− cosu)2)rdr≥ 12(1+α−)∫ ∞0(1−cosu)2rdr,where α− := max(0,−α), so by (2.1.21),0← βˆ 2kE(αk)(vk)≥ βˆ 2k(1+(αk)−)12∫ ∞0(1− cosvk)2 rdr.40By Fatou and (2.3.55),liminfk→0∫ ∞0(1− cos(vk))2rdr ≥∫ ∞0(1− cos(Q))2rdr = 2 > 0,so βˆ 2k(1+(αk)−)→ 0, and since αk ≤ 1, βˆ 2k(1+ |αk |)→ 0.2.2 Convergence Rate EstimatesOur main result says that by a more refined re-scaling of a sequence of minimizersas in Theorem 6, we can sharpen the convergence (2.1.20) to give quantitativeestimates:Theorem 7. Fix A≥ 0. Let {vk} ⊂ X1 be a family of minimizers Ek j,αk,βˆ k(uk) =e(1)k,αk,βˆ k, indexed by (0,1) 3 k→ 0, with −A ≤ αk ≤ 1, 0 < βˆ k → 0, such that‖vk−Q‖X → 0 as k→ 0. Then for all k sufficiently small, there is µk = 1+o(1),withk = β k(2log(1β k)+O(1)), β k := µkβˆ k→ 0, (2.2.1)such that if we writevk(µkr) = Q(r)+ξk(r),then‖ξk‖X.β k, (2.2.2)and‖ξk‖X∞.β 2k log2(1β k). (2.2.3)Remark 2.2.1. The existence of such minimizers is ensured by Corollary 5 andTheorem 6.Remark 2.2.2. Theorem 7 proves (1.5.2) and (1.5.3) of Theorem 1.Proof. The proof occupies the next several subsections. It is based on a carefulanalysis of the Euler-Lagrange equation (2.1.7) with β  1 and k 1. Having41fixed a family vk, αk, βˆ k as in the statement of the theorem, we will drop thesubscript k from the notation for simplicity.2.2.1 Equation for the differenceWe writev(r) = Q(r)+ ξˆ (r), ‖ξˆ‖X = o(1), (2.2.4)and reorganize the Euler-Lagrange equation (2.1.7) as an equation for ξˆ :(H + βˆ 2) ξˆ = s+N, (2.2.5)where we denoteh := sin(Q) =2rr2+1, hˆ := cos(Q) =r2−1r2+1H =−∆r + 1r2 −W (r), W = 2h2r2,s := sk,α,βˆ (r) := kβˆ1rh2+ βˆ 2(αhr−1)h =−βˆ 2h+(k+ βˆ α)βˆ h2r(2.2.6)andN = Nk,α,βˆ (ξˆ ) =12r2(2hˆh(1− cos(2ξˆ ))+(2h2−1)(sin(2ξˆ )−2ξˆ ))+ kβˆ1r(sin2(Q+ ξˆ )− sin2(Q))+ βˆ 2((1−α+α cos(Q))sin(Q)+ ξˆ − (1−α+α cos(Q+ ξˆ ))sin(Q+ ξˆ ))(2.2.7)so that|N|. 1r2(hξˆ 2+|ξˆ |3)+kβˆ 1r(h|ξˆ |+ ξˆ 2)+ βˆ 2((1+ |α |)(hr|ξˆ |+ |ξˆ |3+hξˆ 2)+ |α |hξˆ 4).(2.2.8)422.2.2 Resolvent estimatesRe-writing equation (2.2.5) asξˆ =(H+ βˆ 2)−1 (s+N) , s= sk,α,βˆ =−βˆ 2h+(k+ βˆ α)βˆh2r, N =Nk,α,βˆ (ξˆ )(2.2.9)it is clear we need to understand the mapping properties of the resolvent (H +β 2)−1 as β → 0, which are dominated by the presence of a threshold resonancefor the β = 0 limit H:Hh = 0, h ∈ Lp, p > 2, h 6∈ L2.This means we need an orthogonality condition to avoid singular behaviour of(H + β 2)−1 as β → 0. With such a condition, we find (H + β 2)−1 satisfies thesame estimates as the free resolventR0(β ) :=(−∆r + 1r2 +β2)−1.Precisely,g⊥ R0(β )Wh =⇒ ‖(H +β 2)−1g‖X.‖R0(β )g‖X . (2.2.10)This estimate is a consequence of the factorization(H +β 2)−1 = (I−R0(β )W )−1R0(β ) (2.2.11)and the estimatef ⊥Wh =⇒ ‖(I−R0(β )W )−1 f‖X.‖ f‖X , (2.2.12)43which is proved in Section 2.3.4. The bound (2.2.10) will be used together withestimates on the free resolvent which are proved in Section 2.3.4:‖R0(β )g‖X.‖rg‖L2 , (2.2.13)‖R0(β )g‖X. 1β ‖g‖L2, (2.2.14)‖R0(β )g‖X. 1β 2‖g‖X . (2.2.15)In addition, we use a refinement of (2.2.14) for the case g = h 6∈ L2:‖R0(β )h‖X. 1β . (2.2.16)This is proved in Section 2.3.4.2.2.3 ReparameterizationTo apply estimate (2.2.10) to equation (2.2.9) would require the orthogonalitycondition0 = 〈sk,α,βˆ +Nk,α,βˆ (ξˆ ),R0(βˆ )Wh〉,which does not hold a priori. We first re-express this condition:Lemma 2.2.1. If ξˆ satisfies (2.2.9), then〈s+N,R0(βˆ )Wh〉= 0 ⇐⇒ 〈W ξˆ ,R0(βˆ )h〉= 0. (2.2.17)Proof. This is a computation using the equation (2.2.9) for ξˆ , the equation for theresonance eigenfunction0 = Hh = (−∆+ 1r2−W )h =⇒ R0(0)(Wh) = h,44and the resolvent identityR0(γ)−R0(β ) = (β 2−γ2)R0(β )R0(γ), β ,γ ≥ 0 (2.2.18)with γ = 0:〈s+N,R0(βˆ )Wh〉= 〈(H + βˆ 2)ξˆ ,R0(βˆ )Wh〉= 〈ξˆ ,(−∆+ 1r2 + βˆ2−W )R0(βˆ )Wh〉= 〈ξˆ ,Wh−WR0(0)Wh+W (R0(0)−R0(βˆ ))Wh〉= βˆ 2〈ξˆ ,WR0(βˆ )R0(0)Wh〉= βˆ 2〈W ξˆ ,R0(βˆ )h〉.To impose this condition we will reparameterize:v(r) = Q(r)+ ξˆ (r) = Q(r/µ)+ξ (r/µ), µ = 1+o(1).It follows that the new perturbation ξ solves the Euler-Lagrange equation (2.2.9)with rescaled parameter:(H +β 2)ξ = sk,α,β +Nk,α,β (ξ ), β := µβˆ → 0. (2.2.19)The idea is to choose µ so as to enforce the orthogonality condition correspondingto (2.2.17):〈Wξ ,R0(β )h〉= 0. (2.2.20)Proposition 2.2.1. For all k sufficiently small, there exists µ = 1+o(1) so that (2.2.20)holds.Proof. We haveξ (r) = v(µr)−Q(r) = Q(µr)−Q(r)+ ξˆ (µr).45Now since∂∂ µQ(µr) = rQ′(µr) =− 1µh(µr),∂ 2∂ µ2Q(µr) =1µrh2(µr),by Taylor’s theorem we haveQ(µr)−Q(r)= (1−µ)h(r)+(µ−1)2ζµ(r), ζµ(r)= 12µ∗ rh2(µ∗r), µ∗= µ∗(r)∈ (1,µ),and so we express condition (2.2.20) as0 = (1−µ)〈Wh,R0(β )h〉+(µ−1)2〈Wζµ ,R0(β )h〉+ 〈W ξˆ (µ·),R0(β )h〉.(2.2.21)To estimate these terms, we use a key estimate of inner-products involving the freeresolvent acting on the slowly decaying function h, proved in Section 2.3.4:Lemma 2.2.2. Let g ∈ L1 be a (radial) function:1. if rqg ∈ L∞ for some q > 3, then as β → 0+,∣∣∣∣〈g,R0(β )h〉−(∫ ∞0 rg(r) rdr)log(1β)∣∣∣∣.‖g‖L1∩r−qL∞ , (2.2.22)and in particular|〈g,R0(β )h〉|. log(1β)‖g‖L1∩r−qL∞ , (2.2.23)and〈Wh,R0(β )h〉= 4log(1β)+O(1); (2.2.24)2. if r3g ∈ L∞, then|〈g,R0(β )h〉|. log2(1β)‖g‖L1∩r−3L∞ . (2.2.25)46We apply (2.2.24), (2.2.23) for q = 4 with g = Wζµ , and with g = Wξ (µ·)to (2.2.21), noting that‖Wζµ‖L1∩r−4L∞ ≤ ‖W‖L1∩r−4L∞‖ζµ‖L∞.1(µ∗)2.1for µ near 1, and‖W ξˆ (µ·)‖L1∩r−4L∞.‖W‖L1∩r−4L∞‖ξˆ (µ·)‖L∞.‖ξˆ‖L∞.‖ξˆ‖X = o(1)by (2.2.4). After dividing through by log(1βˆ), this yields(4+o(1))(µ−1)+O((µ−1)2)+o(1) = 0.By the intermediate value theorem, there is a solution µ = 1+o(1) of this equationfor all k small enough.2.2.4 Remainder estimateBy Proposition 2.2.1 and Lemma 2.2.1 (for k small enough), the rescaled remain-derξ (r) = v(µr)−Q(r), ‖ξ‖X → 0satisfies equation (2.2.19) and the orthogonality condition〈sk,α,β +Nk,α,β (ξ ),R0(β )Wh〉= 0. (2.2.26)So using (2.2.10), we have‖ξ‖X.‖R0(β )(sk,α,β +Nk,α,β (ξ ))‖X . (2.2.27)We will use this bound to show estimate (2.2.2) for ‖ξ‖X .47We begin with the contributions fromsk,α,β =−β 2 h+(k+β α)βh2r. (2.2.28)We use (2.2.16) for the main term −β 2 h, and (2.2.13) for the second term, sincer h2r = h2 ∈ L2, to get‖R0(β )sk,α,β‖X.β +(k+β |α |)β .β . (2.2.29)Next we will bound the contributions coming from the nonlinear terms Nk,α,β (ξ ).From the pointwise estimate (2.2.8), and recalling that ‖ξ‖L∞.‖ξ‖X  1, wehaveNk,α,β (ξ ) = N1+N2+N3, (2.2.30)whereN1.hξ 2r2+|ξ |3r2+(k+β (1+ |α |))β h |ξ |rN2.kβξ 2r+β 2(1+ |α |)hξ 2N3.β 2(1+ |α |)|ξ |3.(2.2.31)Using (2.2.13):‖R0(β )N1‖X.‖r N1‖L2.‖rh‖L∞∥∥∥∥ξr∥∥∥∥2L2+‖ξ‖2L∞∥∥∥∥ξr∥∥∥∥L2+(k+β (1+ |α |))β ‖rh‖L∞∥∥∥∥ξr∥∥∥∥L2.(‖ξ‖X +(k+β (1+ |α |))β )‖ξ‖X.(‖ξ‖X +β )‖ξ‖X = o(1)‖ξ‖X .(2.2.32)48Using (2.2.14):‖R0(β )N2‖X. 1β ‖N2‖L2.1β(kβ ‖ξ‖L∞‖ξr ‖L2 +β2(1+ |α |)‖rh‖L2‖ξ‖L∞‖ξr‖L2).(k+β (1+ |α |))‖ξ‖2X ‖ξ‖2X = o(1)‖ξ‖X .(2.2.33)Using (2.2.15):‖R0(β )N3‖X. 1β 2‖N3‖X.(1+ |α |)‖ξ‖2L∞‖ξ‖X.‖ξ‖3X = o(1)‖ξ‖X , (2.2.34)since |α |.1+A.1.Combining (2.2.32)-(2.2.34) with (2.2.30), (2.2.29), and (2.2.27), we get‖ξ‖X.β +o(1)‖ξ‖X ,and (2.2.2) of Theorem 7 follows for k sufficiently small.2.2.5 Parameter relationHere we will use (2.2.26) to determine the leading-order relationship (2.2.1) be-tween k and β .Begin with the contributions from the source term sk,α,β . By (2.2.24) ofLemma 2.2.2, we have〈h,R0(β )Wh〉= 4log(1β)+O(1). (2.2.35)Using the resolvent identity (2.2.18) with γ = 0, and R0(0)(Wh) = h, we haveR0(β )Wh = h−β 2 R0(β )h,49so〈h2r, R0(β )Wh〉= 〈h2r, h〉−β 2〈h2r, R0(β )h〉= 2+O(β 2 log2(1β))(2.2.36)by using (2.3.56), as well as (2.2.25) of Lemma 2.2.2 with g = h2r ∈ L1∩ r−3L∞.Combining (2.2.35) and (2.2.36) yields〈sk,α,β , R0(β )Wh〉=−4β 2 log(1β)+2kβ +O(β 2). (2.2.37)To bound the contributions of the nonlinear terms we will use the decomposi-tion (2.2.30), the basic estimate∣∣〈N j, R0(β )Wh〉∣∣= ∣∣〈R0(β )N j, Wh〉∣∣.‖R0(β )N j‖L∞‖Wh‖L1.‖R0(β )N j‖X ,the estimates (2.2.32)-(2.2.34), and (2.2.2), to conclude:∣∣〈Nk,α,β (ξ ) R0(β )Wh〉∣∣.β 2 . (2.2.38)Inserting (2.2.37) and (2.2.38) into (2.2.26) shows (2.2.1) of Theorem 7, for ksufficiently small.2.2.6 Derivative estimatesIn this section, we complete the proof of Theorem 7 by establishing the higher-regularity estimate (2.2.3).First, we show:Lemma 2.2.3. ξr ∈ L∞, with ∥∥∥∥ξr∥∥∥∥L∞.β . (2.2.39)50Proof. Defineξ¯ (r) :=1βξ (r), M(δ ) := supδ<r<1|ξ¯ (r)|r, 0 < δ < 1.By (2.2.2), ‖ξ¯‖L∞.‖ξ¯‖X.1, so we have the crude boundM(δ ). 1δ, (2.2.40)and it suffices to show M(δ ).1 uniformly for δ < 1. Re-write the Euler-Lagrangeequation (2.2.19) as− ξ¯rr− 1r ξ¯r +1r2ξ¯ = g, g :=(2h2r2−β 2)ξ¯ +1β(s+N). (2.2.41)We will single out one term from the nonlinear terms N which needs to be treatedcarefully, and writeg = g1+g2, |g2|.β 2 |ξ¯ |3r2,where using (2.2.31) and (2.2.28) it is easily checked that‖ξ¯‖X.1 =⇒∥∥∥g1r∥∥∥L1≤1.1.We claim the representation formula2(ξ¯ (r)r− ξ¯ (1))=1r2∫ r0yg(y) ydy+∫ 1r1yg(y) ydy−∫ 10yg(y) ydy (2.2.42)holds. Indeed, since r and 1/r are fundamental solutions of (2.2.41) with g = 0,by the variation of parameters formula and ODE uniqueness,2ξ¯ (r)r=1r2(a−∫ 1ryg(y) ydy)+b+∫ 1r1yg(y) ydy (2.2.43)51for some a,b ∈ R. Now for r ≤ 1,∣∣∣∣∫ 1r 1y g(y) ydy∣∣∣∣.∥∥∥g1r ∥∥∥L1≤1 +β 2∫ 1r1y|ξ¯ |3y2ydy.1+β 2 ‖ξ¯‖2X M(r).1+β 2 M(r).1r(2.2.44)using (2.2.40), and so since ξ¯ ∈ L∞ we must havea =∫ 10yg(y) ydy, and so 2ξ¯ (r)r=1r2∫ r0yg(y) ydy+b+∫ 1r1yg(y) ydy,and then the formula (2.2.42) follows from plugging in r = 1 to find b.Now using (2.2.44) again, as well as∣∣∣∣∫ r0 yg(y) ydy∣∣∣∣.r2∥∥∥g1r ∥∥∥L1≤1 +β 2∫ r0y|ξ¯ (y)|3y2ydy.r2+β 2∫ r0|ξ¯ (y)|dy,in (2.2.42), we arrive at|ξ¯ (r)|r.1+β 2 M(r)+ β2r2∫ r0|ξ¯ (y)|dy.1+β 2 M(δ )+ β2r2∫ r0|ξ¯ (y)|dyif δ ≤ r ≤ 1. Taking supremum over such r givesM(δ ).1+β 2 M(δ )+β 2 supδ≤r≤11r2∫ r0|ξ¯ (y)|dy,and so, since β  1,|ξ¯ (δ )|δ≤M(δ )≤C(1+β 2 supδ≤r≤11r2∫ r0|ξ¯ (y)|yydy)(2.2.45)for some constant C. To finish the argument, we will iterate this relation, begin-52ning with the crude bound (2.2.40), i.e.|ξ¯ (r)|r≤ Cr=⇒ supδ≤r≤11r2∫ r0|ξ¯ (y)|yydy≤C supδ≤r≤11r2∫ r0dy =Cδ, (2.2.46)so by (2.2.45),|ξ¯ (δ )|δ≤C(1+β 2Cδ).We can use this improved estimate in place of the crude one in (2.2.46), and returnagain to (2.2.45) to even further improve the estimate. Iterating in this way k timesyields|ξ¯ (δ )|δ≤C(k−1∑j=0(Cβ 2) j +(Cβ 2)kδ).1+ (Cβ2)kδ,and the Lemma is proved by taking k→ ∞.The idea behind the proof of estimate (2.2.3) is to exploit the factorized struc-ture of the linearized operator H appearing in the Euler-Lagrange equation (2.2.19)for ξ ,H = F∗F, F = ∂ r+1rhˆ = h∂ r1h, F∗ =−∂r + 1r (hˆ−1),by applying the first-order factor F to (2.2.19), to obtain an equation for Fξ :(Hˆ +β 2)η = Fs+FN, η := Fξ , (2.2.47)whereHˆ = FF∗ =−∆r +Vˆ, Vˆ := 4r2(r2+1) .Equation (2.2.47) is better than equation (2.2.19) in two ways. First, since Fh= 0,the term −β 2 h in sk,α,β which was the main term in the bound ‖ξ‖X.β , is53completely absent from the source term Fsk,α,β in (2.2.47). More precisely,Fh = 0, Fh2r=−h2r2(1+ hˆ) =⇒ Fs =−(k+β α)β h2r2(1+ hˆ),which is well-localized. In particular, defining the space L1,log by the norm‖ f‖L1,log :=∫ ∞0log(2+ r)| f (r)| rdr,(this slight refinement of L1 is needed in Proposition 2.2.2 below), and observingthat h2r2 (1+ hˆ) ∈ L1,log,‖Fs‖L1,log.kβ .β 2 log(1β). (2.2.48)by (2.2.2). Second, Hˆ (unlike H) is a positive operator on L2 (with domain D(Hˆ)=D(−∆r + 4r2 )), and in particular has no zero-energy resonance or eigenvalue. Soa uniform in β bound for the resolvent (Hˆ + β 2)−1, without any orthogonalitycondition, is possible:Proposition 2.2.2. For any f ∈ L1,log∩L2∩C((0,∞)),‖(Hˆ +β 2)−1 f‖L∞∩r2L1≤1.‖ f‖L1,log . (2.2.49)This proved in Section 2.3.4.Combining (2.2.49) with (2.2.48) gives‖(Hˆ +β 2)−1Fs‖L∞∩r2L1≤1.β2 log(1β). (2.2.50)From the expression for N = Nβ ,κ,α(ξ ) we can derive a pointwise estimate of|FN|. |Nr|+ 1r |N| analogous to (2.2.8), using ‖ξ‖L∞  1 to simplify slightly:N = N(1)+N(2)54with|(N(1))r|+ 1r |N(1)|. |ξ |r(hr+|ξ |r)(|ξr|+ |ξ |r)+(k+β (1+|α |))β hr(|ξr|+ |ξ |r)and|(N(2))r|+ 1r |N(2)|. kβ |ξ |r(|ξr|+ ξr)+β 2(1+ |α |)|ξ |(h+ξ 2)(|ξr|+ |ξ |r).Using the pointwise bound above, FN(1) ∈ L1,log, with‖FN(1)‖L1,log.‖ξ‖2X(∥∥∥∥hr∥∥∥∥L∞,log+∥∥∥∥ξr∥∥∥∥L∞,log)+(k+β (1+ |α |))β ‖ξ‖X∥∥∥∥hr∥∥∥∥L2,log.β 2(1+∥∥∥∥ξr∥∥∥∥L∞+‖ξ‖L∞).β 2,using h/r ∈ L∞,log∩L2,log, ‖ξ‖L∞.‖ξ‖X.β , (2.2.39), and∥∥∥∥ξr∥∥∥∥L∞,log= suprlog(2+r)|ξ (r)|r.supr≤1|ξ (r)|r+(supr≥1log(2+ r)r)‖ξ‖L∞.∥∥∥∥ξr∥∥∥∥L∞+‖ξ‖L∞.Then by (2.2.49),‖(Hˆ +β 2)−1FN(1)‖L∞∩L1≤1.β2 . (2.2.51)Since the terms of FN(2) decay too slowly to lie in L1,log, we also consider theresolvent (Hˆ+β )−1 acting on L2 functions, and get a bound with some loss in β :Proposition 2.2.3. For f ∈ L2,‖(Hˆ +β 2)−1 f‖L∞∩r2L1≤1 .1βlog32 (1β)‖ f‖L2. (2.2.52)This is proved in Section 2.3.4.55Using the pointwise bound above, FN(2) ∈ L2, with‖FN(2)‖L2.kβ∥∥∥∥ξr∥∥∥∥L∞‖ξ‖X +β 2(1+|α |)‖ξ‖L∞(1+‖ξ‖2L∞)‖ξ‖X.β 4 log(1β),using (2.2.1), (2.2.2), and (2.2.39). Then by (2.2.52),‖(Hˆ +β 2)−1FN(2)‖L∞∩L1≤1.β3 log52(1β). (2.2.53)Using (2.2.50), (2.2.51) and (2.2.53) in (2.2.47) gives‖η‖L∞∩L1≤1.β2 log(1β). (2.2.54)Finally, we by recover the desired estimate (2.2.3) for ξ from the estimate (2.2.54)for η = Fξ using:Lemma 2.2.4.‖ξ‖X∞. log(1β)‖η‖L∞∩r2L1≤1. (2.2.55)Proof. Solving the first order equation η = Fξ = h(ηh)r givesξ = ch+ξ ], ξ ](r) = h(r)∫ r1η(s)h(s)ds, (2.2.56)for some c = c(η) ∈ R.The estimate‖ξ ]‖X∞.‖η‖L∞∩r2L1≤1, (2.2.57)follows from the asymptotic behaviour of hh(r)∼ r, h′(r)∼ 1 for r ≤ 1, h(r)∼ 1r, h′(r)∼ 1r2for r ≥ 1,56and the estimates: for r ≤ 1,∣∣∣∣∫ r1 η(s)h(s) ds∣∣∣∣.∫ 1r |η(s)|s2 sds.‖η‖r2L1≤1,and for r ≥ 1, ∣∣∣∣∫ r1 η(s)h(s) ds∣∣∣∣.‖η‖L∞ ∫ r1 sds.r2‖η‖L∞.To bound the constant c in (2.2.56), we use the orthogonality condition (2.2.26),in the form (2.2.17):0 = 〈ξ , WR0(β )h〉= c〈h, WR0(β )h〉+ 〈ξ ], WR0(β )h〉. (2.2.58)Since‖Wξ ]‖L1∩r−3L∞ ≤ ‖rW‖L1∩r−3L∞∥∥∥∥ξ ]r∥∥∥∥L∞.∥∥∥ξ ]∥∥∥X∞,we may apply (2.2.25), as well as (2.2.24) in (2.2.58) to get(4log(1β)+O(1))c = O(log2(1β)∥∥∥ξ ]∥∥∥X∞)and so using (2.2.57),|c|. log(1β)∥∥∥ξ ]∥∥∥X∞. log(1β)‖η‖L∞∩r2L1≤1 .This, together with (2.2.56) and (2.2.57), gives (2.2.55).Combining (2.2.55) and (2.2.54) shows (2.2.3) and completes the proof ofTheorem 7.2.2.7 UniquenessTheorem 8. Fix A ≥ 0. There is k0 > 0 such that for all k ∈ (0,k0], α ∈ [−A,1]and β > 0, there is a unique minimizer of Ek,α,β in X1.57Remark 2.2.3. Theorem 8 proves the uniqueness part of Theorem 1.Proof. The existence of minimizers is shown in Theorem 4. If the uniquenessfails, there are two families {u(ν)k } ⊂ X1, ν = 1,2, of minimizers Ek,αk,βˆ k(u(ν)k ) =ek,αk,βˆ k , αk ∈ [−A,1], βˆ k > 0, which disagree along a sequence (0,1) 3 k j → 0:u(1)k j 6= u(2)k j.Applying Theorems 6 and 7 to these families of minimizers we have, droppingthe subscripts k,u(ν)(λ (ν) r) = Q(r)+ξ (ν)(r), ‖ξ (ν)‖X.β (ν), ν = 1,2, (2.2.59)for some λ (ν) > 0 withβ (ν) = λ (ν) βˆ → 0, k = β (ν)(2log(1β (ν))+O(1))(2.2.60)where the orthogonality conditions (2.2.26)F (ν) := R0(β (ν))(sk,α,β (ν)+Nk,α,β (ν)(ξ(ν)))⊥ Wh (2.2.61)hold, and the Euler-Lagrange equations (2.2.19), using (2.2.11),ξ (ν)=(H+β (ν))−1(sk,α,β (ν)+Nk,α,β (ν)(ξ(ν)))= I(β (ν))F (ν), I(β ) :=(I−R0(β )W )−1(2.2.62)hold. We will use difference estimates for ξ (1)− ξ (2) and β (1)−β (2) to showthat for k small enough, ξ (1) = ξ (2) and β (1) = β (2), which shows u(1) = u(2) andproves the uniqueness theorem. Without loss of generality, we may assumeβ (2).β (1), (2.2.63)otherwise we may restrict to a subsequence for which β (1).β (2), and reverse theroles of ν = 1 and ν = 2 below.58From (2.2.62),‖ξ (1)−ξ (2)‖X.∥∥∥I(β (1))(F (1)−F (2))∥∥∥X+∥∥∥(I(β (1))− I(β (2)))F (2)∥∥∥X.‖F (1)−F (2) ‖X +∥∥∥(I(β (1))− I(β (2)))F (2)∥∥∥X,(2.2.64)where the second line follows from (2.2.12), since F (1)−F (2) ⊥Wh by (2.2.61).From (2.2.61),‖F (1)−F (2) ‖X.‖R0(β (1))(sk,α,β (1)− sk,α,β (2))‖X+∥∥∥R0(β (1))(Nk,α,β (1)(ξ (1))−Nk,α,β (2)(ξ (2)))∥∥∥X+∥∥∥(R0(β (1))−R0(β (2)))(sk,α,β (2)+Nk,α,β (2)(ξ (2)))∥∥∥X .(2.2.65)For the first term in (2.2.65), notesk,α,β (1)− sk,α,β (2) =((β (2))2− (β (1))2)h+α((β (1))2− (β (2))2) h2r,and so by (2.2.16) and (2.2.13),‖R0(β (1))(sk,α,β (1)− sk,α,β (2))‖X.∣∣∣β (1)−β (2)∣∣∣(1+ β (2)β (1)).∣∣∣β (1)−β (2)∣∣∣ .(2.2.66)For the third term in (2.2.65), the resolvent identity (2.2.18), (2.2.15), togetherwith the estimates (2.2.29) and (2.2.32)-(2.2.34) show∥∥∥(R0(β (1))−R0(β (2)))(sk,α,β (2)+Nk,α,β (2)(ξ (2)))∥∥∥X.∣∣∣β (1)−β (2)∣∣∣(β (1)+β (2)) 1(β (1))2β (2).∣∣∣β (1)−β (2)∣∣∣ .(2.2.67)59From the form (2.2.7) of the nonlinear terms, and using also (2.2.59), (2.2.60)and (2.2.63), we can get difference estimate versions of (2.2.31):Nk,α,β (1)(ξ(1))−Nk,α,β (2)(ξ (2))= (M1+M2+M3)(ξ (1)−ξ (2))+(L1+L2+L3)(β (1)−β (2)),(2.2.68)where|M1|. 1r2((h+ |ξ (1)|+ |ξ (2)|)(|ξ (1)|+ |ξ (2)|))+ kβ (1)hr|M2|.kβ (1)(1r+h)(|ξ (1)|+ |ξ (2)|)|M3|.(β (1))2((ξ (1))2+(ξ (2))2)|L1|.khr (|ξ(1)|+ |ξ (2)|)|L2|.k(1r+h)((ξ (1))2+(ξ (2))2)|L3|.β (1)(|ξ (1)|3+ |ξ (2)|3)60Then using (2.2.13), (2.2.14), and (2.2.15),‖R0(β (1))M1(ξ (1)−ξ (2))‖X.‖r2M1‖L∞‖ξ (1)−ξ (2)‖X.(‖ξ (1)‖X +‖ξ (2)‖X + kβ (1))‖ξ (1)−ξ (2)‖X.β (1) ‖ξ (1)−ξ (2)‖X ,‖R0(β (1))M2(ξ (1)−ξ (2))‖X. 1β (1)‖rM2‖L∞‖ξ (1)−ξ (2)‖X.kβ (1) ‖ξ (1)−ξ (2)‖X ,‖R0(β (1))M3(ξ (1)−ξ (2))‖X. 1(β (1))2‖M3‖L∞‖ξ (1)−ξ (2)‖X.(β (1))2‖ξ (1)−ξ (2)‖X ,‖R0(β (1))L1‖X.‖rL1‖L2.k(‖ξ (1)‖X +‖ξ (2)‖X).kβ (1),‖R0(β (1))L2‖X. 1β (1)‖L2‖L2. 1β (1)k(‖ξ (1)‖X +‖ξ (2)‖X)(‖ξ (1)‖L∞+‖ξ (2)‖L∞).kβ (1),‖R0(β (1))L3‖X. 1(β (1))2‖L3‖X.(β (1))2.Using all of these in (2.2.68), for the second term in (2.2.65) we get∥∥∥R0(β (1))(Nk,α,β (1)(ξ (1))−Nk,α,β (2)(ξ (2)))∥∥∥X.β (1) ‖ξ (1)−ξ (2)‖X + kβ (1) |β (1)−β (2) |.(2.2.69)Combining (2.2.66), (2.2.67), and (2.2.69) we get an estimate of the first termin (2.2.64):‖F (1)−F (2) ‖X. |β (1)−β (2) |+β (1) ‖ξ (1)−ξ (2)‖X . (2.2.70)61For the second term in (2.2.64), by resolvent identities(I(β (1))− I(β (2)))F (2) =((β (2))2− (β (1))2)I(β (1))R0(β (1))R0(β (2))WI(β (2))F (2)=((β (2))2− (β (1))2)I(β (1))R0(β (1))R0(β (2))Wξ (2),using (2.2.62) in the last step. The difficulty here is that I(β (1)) acts on a functionwhich is not ⊥Wh, and so bahaves badly as β (1)→ 0. Precisely,f ∈ X =⇒ ‖I(β ) f‖X. 1β 2 log(1β) |〈Wh, f 〉|+‖ f‖X (2.2.71)is proved in Section 2.3.4. Applying this,∥∥∥(I(β (1))− I(β (2)))F (2)∥∥∥X.∣∣∣β (1)−β (2)∣∣∣β (1)( 1(β (1))2 log(1β (1)) ∣∣∣(〈R0(β (1))Wh, R0(β (2))Wξ (2)〉)∣∣∣+‖R0(β (1))R0(β (2))Wξ (2)‖X).(2.2.72)Use (2.2.15) and (2.2.13) for the second term in (2.2.72):‖R0(β (1))R0(β (2))Wξ (2)‖X. 1(β (1))2‖rWξ (2)‖L2.1(β (1))2‖r2W‖L∞‖ξ (2)‖X. β(2)(β (1))2.For the first term, we use an L2 estimate for the free resolvent acting on well-localized functions, proved in Section 2.3.4,‖R0(β ) f‖L2. log12(1β)‖r f‖L1 +‖ f‖L1. log12(1β)‖(1+ r) f‖L1 (2.2.73)62to get ∣∣∣〈R0(β (1))Wh, R0(β (2))Wξ (2)〉∣∣∣. log 12(1β (1))log12(1β (2))‖(1+ r)Wh‖L1‖r(1+ r)W‖L2‖ξ (2)‖X. log 12(1β (1))log12(1β (2))β (2) .Using the last two estimates in (2.2.72) gives∥∥∥(I(β (1))− I(β (2)))F (2)∥∥∥X.∣∣∣β (1)−β (2)∣∣∣β (1) log12(1β (1))log12(1β (2))β (2)(β (1))2 log(1β (1)) + β (2)(β (1))2.∣∣∣β (1)−β (2)∣∣∣β(2) log(1β (2))β (1) log(1β (1)) + β (2)β (1). ∣∣∣β (1)−β (2)∣∣∣by (2.2.63). Combining this with (2.2.70), we complete the estimate of ξ (1)−ξ (2)in (2.2.65):‖ξ (1)−ξ (2)‖X.∣∣∣β (1)−β (2)∣∣∣+β (1) ‖ξ (1)−ξ (2)‖Xand so for sufficiently large j,‖ξ (1)−ξ (2)‖X.∣∣∣β (1)−β (2)∣∣∣ . (2.2.74)It remains to estimate β (1)−β (2). For this we use the orthogonality condi-63tions (2.2.61) re-written asγ(β (ν)) = S(ν), γ(β ) := β (R0(β )Wh, h) , andS(ν) := (k+β (ν)α)(R0(β (ν))Wh,h2r)+1β(R0(β (ν))Wh,Nk,α,β (ν)(ξ(ν))).We need a monotonicity estimate for γ(β ), proved in Section 2.3.4:Lemma 2.2.5.|γ(β (1))− γ(β (2))|& log(1β (1))|β (1)−β (2) |. We also need more difference estimates:Lemma 2.2.6.|S(1)−S(2)|. |β (1)−β (2) |+‖ξ (1)−ξ (2)‖X .The proof is below. Applying these two lemmas to the previous relation gives|β (1)−β (2) |. 1log(1β (1)) (|β (1)−β (2) |+‖ξ (1)−ξ (2)‖X) ,so for k sufficiently small,|β (1)−β (2) |. 1log(1β (1))‖ξ (1)−ξ (2)‖X .Combining this with (2.2.74), we see that ξ (1) = ξ (2) and β (1) = β (2) for ksufficiently small. It follows that u(1) = u(2) for k sufficiently small. This contra-diction proves Theorem 8.64Proof of Lemma 2.2.6: by resolvent identity, (2.2.68), (2.2.13), (2.2.14), (2.2.73),(2.2.15), the estimates of Section 2.2.4 and (2.2.69),|S(1)−S(2)|. |β (1)−β (2) |[ ∣∣∣∣(R0(β (1))Wh, h2r)∣∣∣∣+β (2)(β (1)+β (2)) ∣∣∣∣(R0(β (2))R0(β (1))Wh, h2r)∣∣∣∣+β (1)+β (2)β (1)∣∣∣(Wh,R0(β (1))R0(β (2))Nk,α,β (2)(ξ (2)))∣∣∣+ 1β (1)β (2)∣∣∣(Wh,R0(β (2))Nk,α,β (2)(ξ (2)))∣∣∣]+1β (1)∣∣∣(Wh,R0(β (1))Nk,α,β (1)(ξ (1))−Nk,α,β (2)(ξ (2)))∣∣∣. |β (1)−β (2) |[1+β (2)(β (1)+β (2))1β (2)log(1β (1))+β (1)+β (2)β (1)(β (2))2(β (1))2+(β (2))2β (1)β (2)]+1β (1)(β (1) ‖ξ (1)−ξ (2)‖X +β (1) |β (1)−β (2) |).. |β (1)−β (2) |+‖ξ (1)−ξ (2)‖Xas required. 2.2.8 Energy asymptotics and Lp estimatesTo complete the proof of Theorem 1, it remains to prove the energy asymp-totics (1.5.4). By (1.3.8), we may compute for the special casek = k(β ) = β(2log(1β)+O(1))so that by Theorem 7, the minimizer vk,α,β satisfiesvk,α,β = Q+ξ , ξ = ξk,α,β , ‖ξ‖X.β .This energy-space bound alone is not sufficient to estimate the D-M energy EDM(vk,α,β ),so at the same time, we obtain Lp estimates of ξ by interpolating between theenergy space estimate, and the L2 (and weak L2) information provided by the65anisotropy/Zeeman energy term E(α)(vk,α,β ).Begin with the D-M energy:|EDM(Q+ξ )−EDM(Q)|=∣∣∣∣∫ ∞0 (sin2(Q+ξ )(Qr +ξr)− sin2 Q Qr)rdr∣∣∣∣=∣∣∣∣∫ ∞0(h2ξr +((1−2h2)sin2 ξ +2hhˆcosξ sinξ )(−hr +ξr))rdr∣∣∣∣.∫ ∞0((h2+ξ 2)|ξr|+ hr (ξ2+h|ξ |))rdr.(‖h‖2L4 +‖ξ‖2L4)‖ξr‖L2 + (‖h‖2L4 +‖h‖L∞‖ξ‖L∞)‖ξr ‖L2.(1+‖ξ‖2L4 +‖ξ‖X)‖ξ‖X.(1+‖ξ‖2L4)β ,while by (2.3.56),EDM(Q) =∫ ∞0sin2 Q Qr rdr =−∫ ∞0h3rrdr =−2,so thatEDM(vk,α,β ) = EDM(Q+ξ ) =−2+O((1+‖ξ‖2L4)β ). (2.2.75)Next, the Zeeman-anisotropy energy: since vk,α,β is a critical point of Ek,α,β ,the Pohozaev relation (2.1.8) holds,β 2 E(α)(vk,α,β ) =−12β k(β )EDM(Q) = β k(β )(1+O((1+‖ξ‖2L4)β )).(2.2.76)We can extract L2 information from this:‖sin(Q+ξ )‖2L2 =∫ ∞0sin2(Q+ξ ) rdr = 2Ea(vk,α,β )≤ 2E(α)(vk,α,β ). log(1β)(1+β ‖ξ‖2L4).66By trig identity,sinξ +h = sin(Q+ξ )+h(1− cos(ξ ))+(1− hˆ)sin(ξ ),and so‖sinξ +h‖L2.‖sin(Q+ξ )‖L2 +‖rh‖L∞‖ξ‖L∞‖ξ‖X +‖1− hˆ‖L2‖ξ‖L∞. log 12(1β)(1+√β‖ξ‖L4)+β 2+β. log 12(1β)(1+√β‖ξ‖L4).(2.2.77)Since ‖ξ‖L∞.β  1,|ξ (r)|. |sinξ (r)| ≤ |sinξ (r)+h(r)|+h(r),and since h ∈ L2,w,‖ξ‖L2,w.‖sinξ +h‖L2 +‖h‖L2,w. log12(1β)(1+√β‖ξ‖L4)+1. log 12(1β)(1+√β‖ξ‖L4).Simple interpolation with ‖ξ‖L∞.β yields‖ξ‖Lp ≤(pp−2) 1p‖ξ‖2pL2,w‖ξ‖1− 2pL∞.(pp−2) 1plog1p(1β)(1+√β‖ξ‖L4) 2p β 1−2p , p > 2.(2.2.78)In particular, the case p = 4‖ξ‖L4. log14(1β)(1+√β‖ξ‖L4) 12 β12 . log 14(1β)β12 + log14(1β)β34 ‖ξ‖12L467shows that‖ξ‖L4. log14(1β)β12 . (2.2.79)With this, we can return to (2.2.78) to get‖ξ‖Lp.(pp−2) 1plog1p(1β)β 1−2p , p > 2,and (2.2.77) to get‖sinξ +h‖L2. log12(1β).Then‖ξ − sinξ‖L2.‖ξ 3‖L2 = ‖ξ‖3L6. log12(1β)β 2,so‖ξ +h‖L2 ≤ ‖sinξ +h‖L2 +‖ξ − sinξ‖L2. log12(1β)(1+β 2). log 12(1β).Finally, we can return to the energy computation. For the exchange energy:since Q minimizes Ee (among finite-energy configurations with the given bound-ary conditions),0≤ Ee(Q+ξ )−Ee(Q)= Ee(Q+ξ )−Ee(Q)−〈E ′e(Q),ξ 〉=12∫ ∞0(ξ 2r +1r2(sin2(Q+ξ )− sin2 Q−2sinQcosQ ξ ))rdr=12∫ ∞0(ξ 2r +1r2((1−2h2)sin2 ξ +2hhˆ(cosξ sinξ −ξ )))rdr.∫ ∞0(ξ 2r +1r2(ξ 2+ξ 3))rdr.‖ξ‖2X(1+‖ξ‖L∞).‖ξ‖2X(1+‖ξ‖X).β 2,so thatEe(vk,α,β ) = Ee(Q+ξ ) = Ee(Q)+O(β2) = 2+O(β 2). (2.2.80)68Combining (2.2.80), (2.2.75), (2.2.76) and (2.2.79):Ee(Q)−Ek,α,β (vk,α,β )=−Ee(vk,α,β )+2−β k(β )EDM(vk,α,β )−β 2 E(α)(vk,α,β )=−Ee(vk,α,β )+2−12β k(β )EDM(vk,α,β ) = O(β2)+β k(β )(1+O(β ))= β k(β )+O(β 2) = 2β 2 log(1β)+O(β 2)which implies (1.5.4).This completes the proof of Theorem 1. 2.3 Appendices2.3.1 SymmetriesHere we Prove Lemma 1.3.1:Proof. For (1.3.3), just compute:∇×(mˆ(eφ R˜x))= ∂ 2 m3(eφ R˜x)−∂ 1 m3(eφ R˜x)∂ 1 m2(eφ R˜x)−∂ 2 m1(eφ R˜x)= −sin(φ)(m3)1+ cos(φ)(m3)2−cos(φ)(m3)1− sin(φ)(m3)2cos(φ)(m2)1+ sin(φ)(m2)2+ sin(φ)(m1)1− cos(φ)(m1)2and somˆ(eφ R˜x) ·∇×(mˆ(eφ R˜x))= sin(φ)(−m1(m3)1−m2(m3)2+m3(m2)2+m3(m1)1)+ cos(φ)(m1(m3)2−m2(m3)1+m3(m2)1−m3(m1)2)Since the change of variables x 7→ eφ R˜x has unit Jacobian, after integration we69haveEDM(mˆ(eφ R˜·))= sin(φ)∫(−m1(m3)1−m2(m3)2+m3(m2)2+m3(m1)1) dx+ cos(φ)∫(m1(m3)2−m2(m3)1+m3(m2)1−m3(m1)2) dx(2.3.1)On the other hand, compute∇×(eφRmˆ)= ∂ 2 m3−∂ 1 m3∂ 1 (sin(φ)m1+ cos(φ)m2)−∂ 2 (cos(φ)m1− sin(φ)m2)so(eφRmˆ)·∇×(eφRmˆ)= (cos(φ)m1− sin(φ)m2)(m3)2− (sin(φ)m1+ cos(φ)m2)(m3)1+m3 (sin(φ)((m1)1+(m2)2)+ cos(φ)((m2)1− (m1)2))and so after integration,EDM(eφRmˆ)= sin(φ)∫(−m2(m3)2−m1(m3)1+m3(m1)1+m3(m2)2) dx+ cos(φ)∫(m1(m3)2−m2(m3)1+m3(m2)1−m3(m1)2) dx= EDM(mˆ(eφ R˜·))(2.3.2)using (2.3.1) in the last step. Now replacing mˆ with e−φRmˆ in (2.3.2) yields (1.3.3).For (1.3.4), it is easily checked that(Fmˆ ·∇× (Fmˆ))(x) = (mˆ ·∇× mˆ)(x1,−x2),and then (1.3.4) follows from the fact that (x1,x2) 7→ (x1,−x2) has unit Jacobian.702.3.2 Existence of a minimizerHere we prove Theorem 4.Proof. By (2.1.17) and (2.1.19),e(1)k,α,1 ∈ [2(1− k2),2)⊂ (0,2). (2.3.3)Let {u j}∞j=1 ⊂ X1 be a minimizing sequence: Ek,α,1(u j)→ ek,α,1. Since |k| < 1,by (2.1.14) we have uniform bounds: Ee(u j).1, E(α)(u j).1. We then also havea uniform pointwise bound, ‖u j‖∞.1 by (2.1.16).A useful observation is that the co-rotational symmetry allows for easy uni-form control of the r→ ∞ tail of minimizing sequences (c.f. the classical Strausslemma [52] giving compact embedding of H1 into a Lebesgue space for radialfunctions):Lemma 2.3.1. For u ∈ Xn,1− cos(u(r))+ 12sin2(u(r))≤ 4r√E(α)(u)√Ee(u). (2.3.4)Proof. Since 1− cos(u(r))+ 12 sin2(u(r))→ 0 as r→ ∞, by the fundamental the-orem of calculus, and Cauchy-Schwarz,1− cos(u(r))+ 12sin2(u(r)) =∫ ∞r1ssin(u(s))(1− cos(u(s))ur(s) sds≤ 2r‖sin(u)‖L2rdr‖ur‖L2rdr≤ 4r√Ea(u)√Ee(u)≤ 4r√E(α)(u)√Ee(u)where the last inequality is from (1.2.1).It follows from (2.3.4) (and continuity) that there is R such that |u j(r)| ≤ pi/271for all r ≥ R and j. Then since |u|. |sin(u)| when |u| ≤ pi/2,‖u j‖2L2 =∫ ∞0u2j(r) rdr =∫ R0u2j(r) rdr+∫ ∞Ru2j(r) rdr.R2‖u j‖2∞+E(α)(u j).1.(2.3.5)From here we also get uniform decay as r→ ∞, just as in (2.3.4):u2j(r) =−2∫ ∞r1su j(s)(u j)r(s) sds≤ 2r ‖u j‖2‖(u j)r‖2.1r. (2.3.6)Combining (2.3.5) with Ee(u).1, shows ‖u j‖H1rdr.1, and so by the standard ar-guments (theorems of Alaoglu and Rellich-Kondrachov) there is a subsequence(which we still denote {u j}) such that∃H1 3 v← u j weakly in H1rdr; strongly in Lp([0,R))rdr ∀R, 1≤ p<∞; and a.e.(2.3.7)Moreover, since Ee(u j).1, we have ‖sin(u j)/r‖L2rdr.1 and we may assume(passing to a further subsequence if needed) thatsin(u j)r→ sin(v)rweakly in L2rdr. (2.3.8)Additionally, we have uniform convergence away from the origin. This is becauseon compact intervals I ⊂ (0,∞), u j are uniformly bounded in H1(I)dr and so theone-dimensional Sobolev embedding gives compactness in L∞(I). Combining thisobservation with the small tail estimate (2.3.6):∀ R > 0, limj→∞(supr≥R|u j(r)− v(r)|)= 0. (2.3.9)As usual, we have weak lower semicontinuity of ‖ur‖2L2 , and so (by Fatou’sLemma) of Ee and E(α):Ee(v)≤ liminfEe(u j), E(α)(v)≤ liminfE(α)(u j). (2.3.10)72We have full convergence of the DM term, because of the extra sin(u) factor, andthe uniform decay (2.3.4):Lemma 2.3.2.EDM(v) = limj→∞EDM(u j). (2.3.11)Proof. By (2.3.4), and Ho¨lder, for any u ∈ H1rdr,∫ ∞Rsin2(u(r))|ur| rdr ≤supr≥R|sin(u(r)|[∫ ∞Rsin2 u rdr]1/2[∫ ∞Ru2r rdr]1/2. 1√R√E(α)(u)√Ee(u).So given ε > 0, choose R such that∫ ∞R sin2(u j)|(u j)r| rdr+∫ ∞R sin2(v)|vr| rdr < ε .Then1k[EDM(u j)−EDM(v)] =∫ R0(sin2(u j)− sin2(v))(u j)r rdr+∫ R0sin2(v)((u j)r− vr) rdr+∫ ∞R(sin2(u j)(u j)r− sin2(v)vr)rdr =: I+ II+ III.Since sin2 is a Lipshitz function, and u j → v strongly in L2 on BR, ‖sin2(u j)−sin2(v)‖L2(BR)→ 0, and so by Ho¨lder,|I|.‖sin2(u j)− sin2(v)‖L2(BR)E1/2e (u j)→ 0.Weak H1 convergence of u j shows II → 0. Finally, by choice of R, |III| < ε ,which was arbitrary.Combining (2.3.10) and (2.3.11), we have:Ek,α,1(v)≤ liminfj→∞Ek,α,1(u j) = e(1)k,α,1,and so it remains to show v ∈ X1.73Since Ee(v) < ∞, by (2.1.2) and (2.3.6), v ∈ C([0,∞]) with limr→∞ v(r) = 0,and sov ∈ Xm for some m ∈ Z .It remains to show m = 1.Lemma 2.3.3.liminfj→∞Ee(u j)≥ 2|m−1|+Ee(v). (2.3.12)Proof. Set w j = u j− v and write u j = v+w j so thatEe(u j)=Ee(v)+Ee(w j)+∫ ∞0vr(w j)r rdr+∫ ∞012r2(sin2(v+w j)−sin2(v)−sin2(w j)) rdr.(2.3.13)By the weak convergence, the first integral on the right → 0. For the secondintegral, we first claim thatsin(w j)r→ 0 weakly in L2rdr. (2.3.14)To see this, notesin(w j)r= cos(v)(sin(u j)r− sin(v)r)+sin(v)r(cos(v)− cos(u j)).The first term tends weakly to 0 by (2.3.8), and the fact that cos(v) is bounded.The second term tends to 0 strongly, since for any R > 0,‖sin(v)r(cos(v)−cos(u j))‖2≤ 2‖sin(v)r ‖L2(BR)+‖cos(u j)−cos(v)‖L∞(BcR)‖sin(v)r‖2,the first term → 0 as R→ 0 while the second → 0 (for any fixed R) as j → ∞by the uniform convergence (2.3.9). By trig identities, the second integral on theright of (2.3.13) can be written∫ ∞0cos(v)sin(v)rcos(w j)sin(w j)rrdr−∫ ∞0sin2(v)r2sin2(w j) rdr.74The first integral → 0 by (2.3.14) (and boundedness of cos(w j) and cos(v), andsin(v)/r ∈ L2), while the second integral is bounded by∫ R0sin2(v)r2rdr+ supr≥Rsin2(w j(r))‖sin(v)/r‖2,with the first term → 0 as R→ 0, and the second → 0 (for fixed R) as j → ∞by (2.3.9) (and trig identities). Thus from (2.3.13) we seeliminfj→∞Ee(u j) = Ee(v)+ liminfj→∞Ee(w j).Finally, since w j(0) = (1−m)pi , the lower bound (2.1.15) shows Ee(w j)≥ 2|m−1|, and the lemma follows.Combining (2.1.19), (2.3.10), (2.3.11), (2.3.12), and (2.1.17) we get2 > e(1)k,α,1 = liminfj→∞Ek,α,1(u j)≥ 2|m−1|+Ek,α,1(v)≥ 2|m−1|+ e(m)k,α,1 ≥ 2|m−1|+2(1− k2)|m| ≥ 2|m−1|from which m = 1 follows, completing the proof of the theorem.2.3.3 Monotonicity of the profileHere we prove Proposition 2.1.4Proof. With α = 1 and β = 1, the minimizer v(r)∈C∞((0,∞)) satisfies the Euler-Lagrange equation (2.1.7)− vrr− 1r vr +1r2sin(v)cos(v)− krsin2(v)+ sin(v)cos(v) = 0 (2.3.15)and the boundary conditionslimr→0+v(r) = pi, limr→∞v(r) = 0.75And by (2.1.19),Ek,1,1(v) = e(1)k,1,1 < 2. (2.3.16)Lemma 2.3.4. For k sufficiently small, 0≤ v(r)≤ pi .Proof. We use proof by contradiction, and a maximum-principle argument.First suppose v has a local minimum at r1 ∈ (0,∞) with v(r1) < 0. Thenvr(r1) = 0 and vrr(r1)≥ 0, and so(1r21+1)sin(v(r1))cos(v(r1))− kr1 sin2(v(r1))≥ 0. (2.3.17)Thus sin(v(r1))cos(v(r1)) ≥ 0, implying v(r1) ≤ −pi2 . Then by the energy lowerbounds (2.1.14) and (2.1.1),Ek,1,1(v)≥ (1− k2)Ee(v)≥ 4(1− k2)≥ 2for k ≤ 1√2, contradicting (2.3.16).On the other hand, suppose v has a local maximum at r2 ∈ (0,∞) with v(r2)>pi . Then vr(r2) = 0 and vrr(r2)≤ 0 so(1r22+1)sin(v(r2))cos(v(r2))− kr2 sin2(v(r2))≤ 0. (2.3.18)If v(r2)≥ 32pi , then again by the lower bounds (2.1.14) and (2.1.1),Ek,1,1(v)≥ (1− k2)Ee(v)≥ 4(1− k2)≥ 2for k ≤ 1√2, contradicting (2.3.16). So we may assume pi < v(r2) < 32pi and sosin(v(r2))< 0 and cos(v(r2))< 0. Then by (3.1.20),0≥ sin(v(r2))cos(v(r2))[1r22+1− kr2tan(v(r2))],76sotan(v(r2))≥ 1k(1r2+ r2)≥ 2kand so v(r2)≥ pi+ arctan(2k ). Then again by lower bounds (2.1.14) and (2.1.1),Ek,1,1(v)≥ (1− k2)Ee(v)≥ (1− k2)[2∣∣∣∣cos(pi+ arctan(2k ))+1∣∣∣∣+2]≥ 2(1− k2)[2− k√4+ k2]≥ 2provided k is small enough, contradicting (2.3.16).Lemma 2.3.5. If v has a local minimum at r1 ∈ (0,∞), then 0≤ v(r1)≤ pi2 .Proof. Inequality (2.3.17) holds, so sin(v(r1))cos(v(r1))≥ 0, and by Lemma 2.3.4,0≤ v(r1)< pi . So 0≤ v(r1)≤ pi2 .Lemma 2.3.6. If v has a local maximum at r2 ∈ (0,∞), then arctan(2k )≤ v(r2)≤ pi .Proof. By Lemma 2.3.4, 0 < v(r2) ≤ pi . If cos(v(r2)) ≤ 0, then pi2 ≤ v(r2) ≤ pi .Otherwise, cos(v(r2)) > 0 and sin(v(r2)) > 0, and since (3.1.20) holds, 1r22+ 1−kr2tan(v(r2))≤ 0. Sotan(v(r2))≥ 1k(1r2+ r2)≥ 2k,and v(r2)≥ arctan(2k ).Lemma 2.3.7. For k sufficiently small, if v has a local minimum at r1 ∈ (0,∞),thensin(v(r1)) = 1−O(k2). (2.3.19)Proof. v must have a local maximum at some r2 > r1. By (2.3.16) and the lower77bounds (2.1.14) and (2.1.1),2 > Ek,1,1(v)≥ (1− k2)Ee(v)≥ (1− k2)[cos(v(r1))+1+2(cos(v(r1))− cos(v(r2)))+1− cos(v(r1))]socos(v(r1))− cos(v(r2))< 11− k2 −1 =k21− k2 .By Lemmas 2.3.5 and 2.3.6,0≤ cos(v(r1))< k21− k2 + cos(arctan(2k)) =k21− k2 +k√4+ k2= O(k),andsin(v(r1)) =√1− cos2(v(r1)) = 1−O(k2).To complete the proof of Proposition 2.1.4, we suppose v has a local minimumat r1 ∈ (0,∞), and derive a contradiction. By the localized version (2.1.11) of thePohozaev-type identity,2E [0,r1]a (v)+ kE[0,r1]DM (v) =12(1+ r21)sin2(v(r1)), (2.3.20)since vr(r1) = 0. By Lemma 2.3.5, cos(u(r1))≥ 0. So using (2.3.20),E [0,r1]k,1,1 (v) = E[0,r1]e (v)+12(1+ r21)sin2(v(r1))−E [0,r1]a (v)≥ 1+ cos(v(r1))+ 12(1+ r21)sin2(v(r1))− 14r21where we used (2.1.1) and E [0,r1]a (v)≤ 12∫ r10 rdr =14r21. Then using again (2.1.14)78and (2.1.1), and that sin(v(r1)) = 1−O(k2) by Lemma 2.3.7,Ek,1,1(v) = E[0,r1]k,1,1 (v)+E[r1,∞)k,1,1 (v)≥ 1+ cos(v(r1))+ 12(1+ r21)sin2(v(r1))− 14r21 +(1− k2)(1− cos(v(r1))≥ 2− k2+ 12(1+ r21)(1−O(k2))−14r21≥ 52−O(k2)> 2for k sufficiently small, contradicting (2.3.16).2.3.4 Resolvent estimatesFree resolvent estimatesHere we record some simple estimates for the free resolvent R0(β ) = (−∆r +1r2 )−1, which include (2.2.13), (2.2.14), and (2.2.15):Lemma 2.3.8. For f radial:‖R0(β ) f‖X.β p−1 ‖rp f‖L2 , 0≤ p≤ 1; (2.3.21)‖R0(β ) f‖X. 1β 2‖ f‖X ; (2.3.22)‖R0(β ) f‖X.β p−1(‖rp+1 fr‖L2 +‖rp f‖L2) , −1≤ p≤ 0. (2.3.23)Proof. These estimates follow from an elementary relation for functions on R2:forg(x), k(x), with g = (−∆+β 2)−1k,(g, k) = (g, −∆g)+β 2(g, g) = ‖∇g‖2L2 +β 2 ‖g‖2L2 . (2.3.24)In particular,‖∇g‖2L2 ≤∣∣∣∣〈 g|x| , |x|k〉∣∣∣∣≤ ∥∥∥∥ g|x|∥∥∥∥L2‖|x|k‖L2 .79Then takingk(x) = eiθ f (r), so g(x) = eiθR0(β ) f ,‖R0(β ) f‖2X.‖∇g‖2L2 ≤∥∥∥∥ 1|x|R0(β ) f∥∥∥∥L2‖r f‖L2 ≤ ‖R0(β ) f‖X‖r f‖L2,from which the p = 1 case of (2.3.21) follows. Similarly from (2.3.24),β 2 ‖g‖2L2 +‖∇g‖2L2 ≤ |(g,k)| ≤ ‖g‖L2‖k‖L2, =⇒ ‖g‖L2 ≤1β 2‖k‖L2and then,‖R0(β ) f‖2X.‖∇g‖2L2 ≤1β 2‖ f‖2L2from which the p = 0 case of (2.3.21) follows. The 0 < p < 1 cases of (2.3.21)then follow from elementary interpolation.For (2.3.22), replacing g by ∇g and k by ∇k in (2.3.24) (since ∇ commuteswith −∆+β 2) yieldsβ 2 ‖∇g‖2L2 ≤ |〈∇g,∇k〉| ≤ ‖∇g‖L2‖∇k‖L2,from which (2.3.22) follows.Bound (2.3.23) follows from interpolation between the other two bounds.Inner products with the free resolventFirst we prove Lemma 2.2.2.Proof. We use the following Fourier-Bessel transformation: for a real-valued, ra-dial function f ,f˜ (ρ) :=−iêiθ f (r)(ξ )|ξ=(ρ,0) =∫ ∞0J1(ρr) f (r) r dr,80where J1 is the order-1 Bessel function of the first kind. In particular((−∆r + 1r2 ) f )˜(ρ) = ρ2 f˜ (ρ), (R0(β ) f )˜(ρ) =f˜ (ρ)ρ2+β 2.Then by Plancherel,〈g,R0(β )h〉= 〈g˜, h˜ρ2+β 2〉=∫ 10g˜h˜ρ2+β 2ρdρ+∫ ∞1g˜h˜ρ2+β 2ρdρ. (2.3.25)By standard estimates for the Fourier transform,‖R0(β )g‖L∞.‖g‖L1 , (2.3.26)andh ∈ X =⇒ ρ h˜ ∈ L2, (2.3.27)and we can easily bound the large ρ integral in (2.3.25):∣∣∣∣∫ ∞1 g˜h˜ρ2+β 2ρdρ∣∣∣∣≤ ∫ ∞1 |g˜||h˜|ρ2 ρdρ ≤ ‖g˜‖L∞‖ρ h˜‖L2(∫ ∞11ρ6ρdρ) 12.‖g‖L1.(2.3.28)Now we consider the small ρ contribution. Suppose first rqg ∈ L∞ for someq > 3. We assume q < 4 (which would imply the desired result for any higher q).Choosing 0 < ε < q−3, using the pointwise estimate|J1(y)− 12y|.yq−2−εfrom Taylor’s theorem, and settingc :=∫ ∞0rg rdr, |c| ≤ ‖rqg‖L∞∫ ∞0r1−q rdr.‖g‖r−qL∞,81we have∣∣∣g˜(ρ)− c2ρ∣∣∣= ∣∣∣∣∫ ∞0(J1(ρr)− 12ρr)g(r) rdr∣∣∣∣.ρq−2−ε ∫ ∞0 rq−2−ε |g(r)| rdr.ρq−2−ε(∫ 10|g(r)| rdr+‖rqg‖L∞∫ ∞1r−2−ε rdr).ρq−2−ε‖g‖L1∩r−qL∞.(2.3.29)Now writeh = 2χ≥1(r)1r+h#,and note thath# := h−2χ≥1(r)1r ∈ L1∩L2 =⇒ h˜# ∈ L2∩L∞.Moreover, (χ≥1(r)1r)˜(ρ) =∫ ∞1J1(ρr)dr =1ρ∫ ∞ρJ1(y)dy, (2.3.30)and since∫ ∞0 J1(y)dy = 1, and |J1(y)|.y,∣∣∣∣∣(χ≥1(r)1r)˜(ρ)− 1ρ∣∣∣∣∣.ρ.Then|h˜(ρ)− 2ρ| ≤ 2∣∣∣∣∣(χ≥1(r)1r)˜(ρ)− 1ρ∣∣∣∣∣+ |h˜#(ρ)|.1 for ρ ≤ 1, (2.3.31)and in particular|h˜(ρ)|. 1ρfor ρ ≤ 1. (2.3.32)We can now use estimates (2.3.29), (2.3.31) and (2.3.32) in the small ρ intergal82of (2.3.25) to get∣∣∣∣∫ 10 g˜h˜ρ2+β 2ρdρ− c∫ 10ρ dρρ2+β 2∣∣∣∣.∫ 10 ρ dρρ2(|g˜(ρ)− c2ρ||h˜(ρ)|+ |c|2ρ|h˜(ρ)− 2ρ|).∫ 10ρ dρρ2(ρq−2−ε1ρ+ρ)‖g‖L1∩r−qL∞.‖g‖L1∩r−qL∞Then since ∫ 10ρdρρ2+β 2= log(1β)+O(β 2),and recalling (2.3.25) and (2.3.28), we have established (2.2.22).The bound (2.2.23) is an immediate consequence of (2.2.22), while (2.2.24)follows from the computation, using (2.3.56):∫ ∞0rW (r)h(r) = 2∫ ∞0h3rrdr = 4.Finally, we turn to (2.2.25). So suppose now r3g∈L∞. Since |J1(y)|.min(y, 1√y),we have, for ρ ≤ 1,|g˜(ρ)|=∣∣∣∣∫ ∞0 J1(rρ)g(r) rdr∣∣∣∣.∫ 10 ρr|g(r)| rdr+∫ 1ρ1ρr|g(r)| rdr+∫ ∞1ρ1√ρ1√r|g(r)| rdr.ρ‖g‖L1 +ρ‖r3g‖L∞∫ 1ρ1drr+1√ρ ‖r3g‖L∞∫ ∞1ρdrr52.‖g‖L1 ρ+‖r3g‖L∞ ρ log(1ρ)+‖r3g‖L∞ ρ.‖g‖L1∩r−3L∞ ρ log(1ρ).Using this, together with (2.3.32), in the small ρ contribution to (2.3.25) gives∣∣∣∣∫ 10 g˜(ρ)h˜(ρ)ρ2+β 2 ρdρ∣∣∣∣.‖g‖L1∩r−3L∞ ∫ 10 log(1ρ)ρ2+β 2ρdρ.83By change of variable ρ = β y,∫ 10log(1ρ)ρ2+β 2ρdρ = log(1β)∫ 1β0y dyy2+1+∫ 1β0log(1y)y dyy2+1. log2(1β)which, together with (2.3.25) and (2.3.28), establishes (2.2.25).Next we prove the monotonicity estimate Lemma 2.2.5.Proof. We may assume β (2).β (1). We have, by resolvent identity,γ(β (1))− γ(β (2))=(β (1)−β (2))(R0(β (1))Wh,h)+β (2)((β (1))2− (β (2))2)(R0(β (2))R0(β (1))Wh,h)=(β (1)−β (2))(4log(1β (1))+O(1)+β (2)(β (1)+β (2))(R0(β (1))Wh,R0(β (2))h))(2.3.33)using (2.2.24). Using the Fourier-Bessel transform,(R0(β (1))Wh,R0(β (2))h)=∫ ∞0(Wh)˜(ρ)h˜(ρ) ρ dρ(ρ2+(β (1))2)(ρ2+(β (2))2)Since r5Wh ∈ L∞, we have |Wh˜(ρ)|.ρ from (2.3.26) and (2.3.29). And sinceρ h˜(ρ) =Wh˜(ρ), we have |h˜(ρ)| ≤ 1/ρ . Using these above, we have∣∣∣(R0(β (1))Wh,R0(β (2))h)∣∣∣.∫ ∞0ρ dρ(ρ2+(β (1))2)(ρ2+(β (2))2) .Using ρ2+(β (2))2&ρ β (2) and the change of variable ρ = β (1) y, we get∣∣∣(R0(β (1))Wh,R0(β (2))h)∣∣∣. 1β (1)1β (2)∫ ∞0dyy2+1. 1β (1)1β (2).84Inserting this in (2.3.33) we seeγ(β (1))− γ(β (2)) =(β (1)−β (2))(4log(1β (1))+O(1)+O(β (2)β (1)))&(β (1)−β (2))log(1β (1)).Refined free resolvent estimatesFirst we prove the refined version (2.2.16) of the p = 0 case of (2.3.21) for f =h 6∈ L2.Proof. Using (2.3.32) and (2.3.27),‖R0(β )h‖X.‖ρ (R0(β )h)˜‖L2 =∥∥∥∥ ρρ2+β 2 h˜∥∥∥∥L2.∥∥∥∥ 1ρ2+β 2∥∥∥∥L2≤1+∥∥∥∥ 1ρ2+β 2∥∥∥∥L∞≥1=(∫ 10ρdρ(ρ2+β 2)2) 12+11+β 2. 1β.Next we prove the L2 estimate (2.2.73) for the free resolvent acting on well-localized functions.Proof. Using the basic Fourier-Bessel estimates|J1(y)|.1 =⇒ | f˜ (ρ)|.‖ f‖L1, |J1(y)|.y =⇒ | f˜ (ρ)|.ρ‖r f‖L1 ,85we have‖R0(β ) f‖2L2 =∥∥∥∥ f˜ρ2+β 2∥∥∥∥2L2.‖r f‖2L1∫ 10ρ2 ρ dρ(ρ2+β 2)2+‖ f‖2L1∫ ∞1ρ dρ(ρ2+β 2)2. log(1β)‖r f‖2L1 +‖ f‖2L1,and (2.2.73) follows.Estimates for (H +β 2)−1Since H +β 2 =−∆˜+β 2−W ,(H+β 2)−1 =(−∆˜+β 2−W )−1 =((−∆˜+β 2)(I−R0(β )W ))−1=(I−R0(β )W )−1R0(β )(2.3.34)whereR0(β ) = (−∆˜+β 2)−1, ∆˜= ∆r− 1r2denotes the free resolvent.Let us consider first the β → 0 limit of I−R0(β )W(I−R0(β )W )|β=0 = I−G0Wwhere G0 = (−∆˜)−1 has the explicit formG0 f (r) =12r∫ r0s2 f (s)ds+r2∫ ∞rf (s)ds =∫ ∞012min(sr,rs)f (s) sds.We summarize its mapping properties on the space X . For this purpose, it isconvenient to consider X as a Hilbert space with inner-product〈 f , g〉X := 〈 fr, gr〉+ 〈1r f ,1rg〉=∫ ∞0(fr(r)gr(r)+1r2f (r)g(r))rdr,and to identify the dual space X∗ with X via this inner-product.86Lemma 2.3.9. We have:1. I−G0W : X → X is Fredholm;2. I−G0W is self-adjoint (with respect to 〈·, ·〉X );3. ker(I−G0W ) = 〈h〉;4. ran(I−G0W ) = X ∩ (Wh)⊥ := { f ∈ X | 〈Wh, f 〉= 0}.5. I−G0W : X ∩ (rW )⊥→ X ∩ (Wh)⊥ is bijective;6. ‖rp+1 ∂ r G0W f‖L2 +‖rpG0W f‖L2.‖ f‖X , −2 < p < 0.Proof. 1. The first statement will follow from the fact that G0W : rL2→ X iscompact. We first show G0W : rL2 → rL2 is compact. Equivalently, weshow 1r G0Wr : L2→ L2 is compact. Indeed, it is Hilbert-Schmidt, since itsintegral kernel is square integrable:∫ ∞0sds∫ ∞0rdr(1r12min(sr,rs)W (s)s)2=14∫ ∞0W 2(s)s2 sds(1s2∫ s0rdr+ s2∫ ∞s1r4rdr)=14∫ ∞0W 2(s)s2 sds < ∞.A very similar calculation shows that the operator∂ r G0W : f (r) 7→ − 12r2∫ r0sW (s) f (s) sds+12∫ ∞r1sW (s) f (s) sdsis compact (in fact Hilbert-Schmidt) from rL2 to L2.2. Compute, for f , g ∈C∞0 ((0,∞)):〈(I−G0W ) f , g〉X = 〈(I−G0W ) f , −∆˜g〉= 〈 f , (I−WG0(−∆˜)g〉= 〈 f , −∆˜(I−G0W )g〉= 〈 f , (I−G0W )g〉X .873. This is a computation. Clearly h ∈ ker(I−G0W ), since(I−G0W )h = h−h = 0.Conversely, if X 3 f = G0W f , then, since X ⊂ C((0,∞)), f ∈ C2((0,∞)),and then0 =−∆˜ f +W f = H f .Since h and k are independent solutions of this second-order ODE, whilek 6∈ X , it follows that f ∈ 〈h〉.4. Since I−G0W is Fredholm, its range is the orthogonal complement (withrespect to 〈·, ·〉X ) of the kernel of its adjoint. Since ker(I−G0W )∗= ker(I−G0W ) = 〈h〉,ran(I−G0W ) = { f ∈ X | 0 = 〈h, f 〉X = 〈−∆˜h, f 〉= 〈Wh, f 〉 }.5. Since we have identified the kernel and range, the last statement followsfrom the facts that rW ∈ X , and〈rW, h〉=∫ ∞0rh2(r)r2h(r) rdr =∫ ∞0h3rrdr = 2 6= 0. (2.3.35)6. Using the explicit form of G0, and W (r).(1+ r2)−2, we haver≤ 1 =⇒ |G0W f |. 1r ‖ f‖L∞(∫ r0s2ds)+r‖ f‖L∞(∫ ∞0W (s)ds).r‖ f‖L∞,whiler≥ 1 =⇒ |G0W f |. 1r ‖ f‖L∞(∫ ∞0s2W (s)ds)+r‖ f‖L∞(∫ ∞r1s4ds). 1r‖ f‖L∞,88and so for −2 < p < 0,‖rpG0W f‖L2.‖rpr(1+ r2)−1‖L2‖ f‖L∞.‖ f‖L∞.‖ f‖X .A similar calculation gives, again for −2 < p < 0,‖rp+1 ∂ r G0W‖L2.‖rp+1(1+ r2)−1‖L2‖ f‖L∞.‖ f‖L∞.‖ f‖X .We will therefore denote by (I−G0W )−1 the (bounded) inverse operator(I−G0W )−1 : X ∩ (hW )⊥→ X ∩ (rW )⊥. (2.3.36)The particular choice (rW )⊥ of subspace for the range of the inverse is what al-lows for estimate (2.3.37) below.We return now to consider I−R0(β )W for β > 0. We first confirm that it isindeed invertible on X – essentially just by manipulating the factorization (2.3.34),and exploiting the fact that H ≥ 0, and hence H +β 2 is invertible:Lemma 2.3.10. For β > 0, I−R0(β )W : X → X is bounded and bijective, hence(I−R0(β )W )−1 : X → Xexists (and is bounded).Proof. Boundedness follows fromf ∈ X =⇒ ‖R0(β )W f‖X.‖rW f‖L2.‖ f‖L∞.‖ f‖X ,using (2.3.21) with p = 1.For surjectivity: given f ∈ X , setg := f +(H +β 2)−1W f .89As a self-adjoint operator on L2, H ≥ 0, and so (H + β 2)−1 : L2 → D(H) ⊂ Xboundedly. Since f ∈ X ⊂ L∞, W f ∈ L2, and hence g ∈ X . And just compute(1−R0(β )W )g = f −R0(β )W f +(H +β 2)−1W f−R0(β )(−H−β 2−∆˜+β 2)(H +β 2)−1W f= f −R0(β )W f +(H +β 2)−1W f +R0(β )W f − (H +β 2)−1W f= f .For injectivity: if f ∈ X satisfies (1−R0(β )W ) f = 0, then since W f ∈ L2,f = R0(β )W f ∈ D(R0(β )) = D(H), and0 = f −R0(β )(−H−β 2−∆˜+β 2) f = R0(β )(H +β 2) f ,and so f = 0 by the invertibility of R0(β ) and H +β 2.The key result is:Proposition 2.3.1. If f ∈ X ∩ (Wh)⊥, then for β sufficiently small,‖(1−R0(β )W )−1 f − (1−G0W )−1 f‖X. 1log(1β)‖ f‖X , (2.3.37)and so in particular‖(1−R0(β )W )−1 f‖X.‖ f‖X (2.3.38)(uniformly in β ), and we have (2.2.10):g⊥ R0(β )Wh =⇒ ‖(H +β 2)−1g‖X.‖R0(β )g‖X .Proof. Setg := (1−R0(β )W )−1 f − (1−G0W )−1 f ∈ X90so that, using a resolvent identity,(1−G0W )g = (1−R0(β )W +(R0(β )−G0)W )(1−R0(β )W )−1 f − f=−β 2 R0(β )G0W (1−R0(β )W )−1 f=−β 2 R0(β )G0W(g+(1−G0W )−1 f).Evidently, the right hand side lies in Ran(1−G0W )=X∩(Wh)⊥, so using (2.3.36)and Lemma 2.3.9,g =−β 2(1−G0W )−1R0(β )G0W(g+(1−G0W )−1 f)+µ h (2.3.39)for some µ ∈ R, and so‖g‖X.β 2 ‖R0(β )G0W(g+(1−G0W )−1 f)‖X + |µ|.Then using (2.3.23) for any −1 < p < 0, the last statement of Lemma 2.3.9,and (2.3.36),‖g‖X.β 2 β p−1 (‖g‖X +‖ f‖X)+ |µ|and so for β small enough,‖g‖X.β 1+p ‖ f‖X + |µ|. (2.3.40)Finally, we bound µ by taking the (L2) inner-product of (2.3.39) with rW , andusing (2.3.36),〈rW, h〉 µ =−〈rW, g〉=−〈rW, (1−R0(β )W )−1 f 〉and so by (2.3.35),|µ|. |(rW, gˆ)|, gˆ := (1−R0(β )W )−1 f = g+(1−G0W )−1 f .Now since f ∈ X ∩ (Wh)⊥, using the relation G0(Wh) = h, a resolvent identity,91and estimate (2.2.22),0 = 〈Wh, f 〉= 〈Wh, (1−R0(β )W )gˆ〉= 〈(1−WR0(β ))Wh, gˆ〉= 〈(1−WG0)Wh+W (G0−R0(β ))Wh, gˆ〉= β 2〈WR0(β )G0Wh, gˆ〉= β 2〈R0(β )h, Wgˆ〉= β 2(〈rW, gˆ〉 log(1β)+O(‖gˆ‖L∞))from which follows|µ|. |〈rW, gˆ〉|. 1log(1/β )‖gˆ‖X. 1log(1/β ) (‖g‖X +‖ f‖X) .Returning to (2.3.40), we have (for any −1 < p < 0),‖g‖X.β 1+p ‖ f‖X + 1log(1/β ) (‖g‖X +‖ f‖X) ,and (2.3.37) follows.Finally, (2.3.38) is a consequence of (2.3.37) and (2.3.36), and then (2.2.10)follows from the factorization (2.3.34).Estimates for (Hˆ +β 2)−1Here we prove the estimates for the resolvent (Hˆ +β 2)−1 used in Section 2.2.6for the higher-regularity estimate.First we prove Proposition 2.2.2, by comparing (Hˆ +β 2)−1 to Hˆ−1.Proof. Since Hˆ ≥ 0, Hˆ +β 2 is an invertible operator on L2, soζ := (Hˆ +β 2)−1 f ∈ D(Hˆ)⊂ L2exists, and(Hˆ +β 2)ζ = f . (2.3.41)92Note that by second-order ODE regularity, f ∈ C((0,∞)) =⇒ ζ ∈ C2((0,∞)),so (2.3.41) holds for all r ∈ (0,∞), and D(Hˆ) ⊂ X ⊂ L∞ shows that ζ is alsobounded.We will show (2.2.49) by comparing (Hˆ +β 2)−1 to Hˆ−1, which has a simpleexplicit form. Indeed, it is easily checked thatu(r) =2rh(r)=r2+1r2, v(r) =(r2+1) ln(r2+1)r2−1are homogeneous solutions: Hˆu = Hˆv = 0. It is also easily checked that:u(r)> 0, v(r)> 0, |u(r)|.{1r2 r→ 01 r→ ∞ , |v(r)|.{r2 r→ 0logr r→ ∞ .(2.3.42)Since f ∈ L1∩C((0,∞)), by (2.3.42),(Hˆ−1 f )(r) := u(r)∫ r0v(s) f (s) sds+ v(r)∫ ∞ru(s) f (s) sds (2.3.43)is well-defined for r ∈ (0,∞), and by the variation of parameters formula, Hˆ−1 f ∈C2((0,∞)) satisfiesHˆ(Hˆ−1 f ) = f (2.3.44)for r ∈ (0,∞).Lemma 2.3.11. If 0 ≤ f ∈ L1 ∩ L2 ∩C((0,∞)), then Hˆ−1 f ≥ 0 and (Hˆ−1 +β 2)−1 f ≥ 0.Proof. The non-negativity of Hˆ−1 f follows immediately from the expression (2.3.43),and the positivity of u and v (see (2.3.42)). Non-negativity of ζ := (Hˆ +β 2)−1 ffollows from the maximum principle: since ζ ∈D(Hˆ)⊂ X , it is bounded and van-ishes as r→ 0+ and r→ ∞. Using the equation (2.3.41), if for some rˆ ∈ (0,∞),0 > minr ζ (r) = ζ (rˆ), then0≤ f (rˆ) = ((Hˆ +β 2)ζ)(rˆ) =−∆rζ (rˆ)+Vˆ (rˆ)ζ (rˆ)< 0,93since Vˆ > 0, a contradiction .Lemma 2.3.12. ∣∣((Hˆ +β 2)−1 f )(r)∣∣≤ (Hˆ−1| f |)(r). (2.3.45)Proof. Write f (r) = f+(r)− f−(r), f+ := max{ f ,0}, f− := max{− f ,0}, f± ∈L1∩L2∩C((0,∞)). Setζ± := (Hˆ +β 2)−1 f± ∈ D(Hˆ)∩C2((0,∞)), ζˆ± := H−1 f±.By Lemma 2.3.11, ζˆ±≥ 0 and ζ±≥ 0. Then using the equations (2.3.41) and (2.3.44),(Hˆ +β 2)(ζˆ±−ζ±) = f±+β 2 ζˆ±− f± = β 2 ζˆ± ≥ 0,and so by Lemma 2.3.11 again, ζˆ±−ζ± ≥ 0. That is,Hˆ−1 f± ≥ (Hˆ +β 2)−1 f± ≥ 0,and so|(Hˆ +β 2)−1 f |= |(Hˆ +β 2)−1( f+− f−)| ≤ (Hˆ +β 2)−1 f++(Hˆ +β 2)−1 f−≤ Hˆ−1 f++ Hˆ−1 f− = Hˆ−1( f++ f−) = Hˆ−1| f |,which is (2.3.45).It follows from the pointwise bound (2.3.45) that‖(Hˆ +β )−1 f‖L∞∩r2L1≤1 . ‖Hˆ−1 f‖L∞∩r2L1≤1, (2.3.46)and so it remains to consider Hˆ−1.Lemma 2.3.13.‖Hˆ−1 f‖L∞∩r2L1≤1 . ‖ f‖L1,log (2.3.47)94Proof. We estimate directly using the explicit form (2.3.43) and the properties (2.3.42).First the pointwise bound: for r ≤ 2,|Hˆ−1 f (r)|. 1r2∫ r0s2| f (s)|sds+r2∫ 2r1s2| f (s)|sds+r2∫ ∞2| f (s)|sds. ‖ f‖L1 ≤‖ f‖L1,log ,and for r ≥ 2,|Hˆ−1 f (r)|.∫ 20s2| f (s)|sds+∫ r2log(s)| f (s)|sds+ logr∫ ∞r| f (s)|sds.‖ f‖L1 +‖ f‖L1,log.‖ f‖L1,log ,which together give ‖Hˆ−1 f‖L∞.‖ f‖L1,log . For the r2L1≤1 bound: using Tonelli’stheorem,∫ 101r2|Hˆ−1 f (r)|rdr .∫ 10rdr1r4∫ r0s2| f (s)|sds+∫ 10rdr∫ 1r1s2| f (s)|sds+∫ 10rdr∫ ∞1| f (s)|sds=∫ 10s2| f (s)|sds∫ 1sdrr3+∫ 101s2| f (s)|sds∫ s0rdr+12∫ ∞1| f (s)|sds.∫ 10| f (s)|sds+∫ ∞1| f (s)|sds = ‖ f‖L1 ,and so ‖Hˆ−1 f‖r2L1≤1.‖ f‖L1.‖ f‖L1,log , giving (2.3.47).Finally, (2.2.49) follows from (2.3.47) and (2.3.46).Next, we prove Proposition 2.2.3 by direct integration.Proof. Given f ∈ L2, setη := (Hˆ +β 2)−1 f ∈ D(Hˆ).Taking the inner product of η with(Hˆ +β 2)η = (−∆r +Vˆ +β 2)η = f ,95we get ∫ ∞0(η2r +Vˆη2+β 2η2)rdr = ( f ,η). ‖η‖L2‖ f‖L2 ,which, since Vˆ > 0, gives the L2 estimate‖η‖L2 .1β 2‖ f‖L2 (2.3.48)and then also‖ηr‖L2 +‖√Vˆη‖L2 . ‖η‖12L2‖ f‖12L2 .1β‖ f‖L2. (2.3.49)We use (2.3.48) and (2.3.49) to estimate ‖η‖L∞ . For r ≤ 1, 1r .√Vˆ (r), andsoη2(r) = 2∫ r0ηr(s)η(s)ssds. ‖ηr‖L2‖√Vˆη‖L2 .1β 2‖ f‖2L2. (2.3.50)For r ≥ 1βη2(r) =−2∫ ∞r1sηr(s)η(s)sds.1r‖ηr‖L2‖η‖L2 .1β 2‖ f‖2L2. (2.3.51)For 1≤ r ≤ 1β ,|η(r)−η(1)|.∫ r1|ηr(s)|1s sds. ‖ηr‖L2 log12 (r). ‖ηr‖L2 log12 (1β).log12 ( 1β )β‖ f‖L2(2.3.52)Combining (2.3.50), (2.3.51) and (2.3.52) gives‖η‖L∞ . 1β log12 (1β)‖ f‖L2. (2.3.53)96Now we estimate ‖η‖r2L1≤1 . SinceHˆ =−∆r + 4r2 −V, with V =4r2+1bounded ,Using (2.3.48) again,∥∥∥ηr2∥∥∥L2.‖(−∆r + 4r2 )η‖L2 . ‖ f +Vη+β2η‖L2.‖ f‖L2 +‖η‖L2.1β 2‖ f‖L2,and then Ho¨lder, ∫ β0|η |r2rdr . β ‖ηr2‖L2 .1β‖ f‖L2 .For the remaining interval, use the estimate (2.3.53) of ‖η‖L∞:∫ 1β|η |r2rdr . ‖η‖L∞ log( 1β ).1βlog32 (1β)‖ f‖L2Combining the last two bounds gives‖η‖r2L1≤1 .1βlog32 (1β)‖ f‖L2. (2.3.54)Then (2.2.52) follows from (2.3.53) and (2.3.54).2.3.5 Computation of some integrals∫ ∞0h2r2rdr =∫ ∞04r(r2+1)2rdr =− 2r2+1∣∣∞0 = 2 (2.3.55)∫ ∞01rhˆh3rdr = 8∫ ∞0(r2−1)r3(r2+1)4dr = 4∫ ∞0(1(r2+1)2− 3(r2+1)3+2(r2+1)4)2rdr= 4(1− 32+23)=2397∫ ∞0h4rdr = 16∫ ∞0r4(r2+1)4rdr = 8∫ ∞0(1(r2+1)2− 2(r2+1)3+1(r2+1)4)2rdr= 8(1−1+ 13)=83∫ ∞01r2h4rdr = 16∫ ∞0r2(r2+1)4rdr = 8∫ ∞0(1(r2+1)3− 1(r2+1)4)2rdr= 8(12− 13)=43∫ ∞01r2h3rdr = 8∫ ∞0r2(r2+1)3dr = 2∫ pi20sin2(2θ)dθ =pi2∫ ∞01rh3rdr = 4∫ ∞0r2(r2+1)32rdr = 4∫ ∞0(1(r2+1)2− 1(r2+1)3)2rdr= 4(1− 12)= 2.(2.3.56)∫ ∞01r3h3rdr = 4∫ ∞01(r2+1)32rdr = 4 · 12=−2,∫ ∞01r3hˆh3rdr = 4∫ ∞0(r2−1)(r2+1)42rdr = 4∫ ∞0(1(r2+1)3− 2(r2+1)4)2rdr= 4(12− 23)=2398Chapter 3Stability of Chiral MagneticSkyrmionsIn this chapter, we prove Theorems 2, and 3 on the stability of skyrmions. Thespectral gap for equivariant perturbations is established in Section 3.1. Generalperturbations are considered in Section 3.2.1. The spectral gap in directions or-thogonal to the translation zero-modes is proved in Section 3.3. The higher Fouriermodes are dealt with in Section 3.4. The dynamical stability arguments appear inSection 3.5. Computations of the second variation of the energy in Fourier modesappear in Section 3.6.3.1 Equivariant perturbationsLet v = vk,α,β (r) be the energy minimizing(in the co-rotational class) skyrmionprofile of Theorem 1, and let sˆ = sˆk,α,β denote the corresponding skyrmion solu-tion.Consider a perturbation preserving equivariance (1.3.1), that produces a tan-99gent vector field of the formddε(−sin(u(r)+ ε a(r))sin(θ + ε b(r)),sin(u(r)+ ε a(r))cos(θ + ε b(r)),cos(u(r)+ ε a(r))) |ε=0= a(r)~e+b(r)J~ewhere the tangent orthonormal basis~e, J~e are defined as (3.6.1). When b(r) = 0,it is the case of perturbations preserving the co-rotational class (1.3.5). In thenotation of Section 3.6, the second variation of the energy at the skyrmion forsuch perturbations corresponds to the operator E ′′j=0(sˆ) acting on vector[a(r)b(r)].The operator E ′′j=0(sˆ) is given by a diagonal matrix whose entries are the op-erators acting on the Hilbert space L2 = L2([0,∞];rdr):L+ :=−∆r + 1r2 cos(2v)− kβ1rsin(2v)+β 2(1+(1−α)(cos(v)−1)−2α sin2(v)) ,L− :=−∆r + 1r2 cos(2v)− kβ(vr +12rsin(2v))+β 2(1+(1−α)(cos(v)−1)−α sin2(v)) .By Weyl’s theorem [45], σess(L+) = σess(L−) = [β 2,∞), and so in order to fix theessential spectrum at [0,∞), introduce shifted operators:Lˆ+ := L+−β 2, Lˆ− := L−−β 2 . (3.1.1)Since L+ = E ′′k,α,β (v) gives the second variation for perturbation within the co-rotational class, and v(r) is a minimizer within this class, we must have L+ ≥ 0.We are interested to know L±> 0, implying stability against (more general) equiv-ariant perturbations.With β = 0, k = 0, v = Q, in both cases, the ‘unperturbed’ operator isLˆ±|k=be=0 = H = F∗F ≥ 0, F = h∂ r 1h , ker(H) = ker(F) = 〈h〉, (3.1.2)100H =−∆˜−W, W = 2h2r2, −∆˜=−∆+ 1r2though note thath(r)∼ 1ras r→ ∞ =⇒ h 6∈ L2,so h is not in the L2 kernel: rather, it is a resonance eigenfunction, sitting at thethreshold of the essential spectrum.Remark 3.1.1. The resonance for Lˆ+|k=β=0 = H comes from scale invariance ofthe exchange energy, while the resonance for Lˆ−|k=β=0 = H comes from rotationinvariance of the exchange energy.WriteLˆ± =−∆˜+Z± (3.1.3)whereZ+=− 2r2sin2(v)−kβ 1rsin(2v)−2β 2 sin2(v)+β 2 ((1−α)(cos(v)−1)−2α sin2(v))Z−=− 2r2sin2(v)−kβ(vr +12rsin(2v))+β 2((1−α)(cos(v)−1)−α sin2(v)) .Sincev = Q+ξ , ξr,ξr∈ L∞,it follows that Z± ∈ L∞, and so by the Kato theory [45],Lˆ± are self-adjoint operators on L2r dr with domain D=D(Lˆ±)=D(−∆˜)⊂X∩L2.Moreover, sincevr,sin(v)r, sin(v) ∈ L2,it follows that Z± ∈ L2, hence that Z± are relatively compact perturbations of−∆˜,101and so by Weyl’s theorem [45],σess(Lˆ±) = σess(−∆˜) = [0,∞).To do perturbation theory, we can write the operators more explicitly asLˆ± = H +V±V+ :=−2kβ 1rhˆh−β 2((1−α)hr+2α h2)− 4r2hˆhξ+1r2((1−2h2)(cos(2ξ )−1)+2hˆh(2ξ − sin(2ξ )))− kβr(2hˆh(cos(2ξ )−1)+(1−h2)sin(2ξ ))−β 2 (−(1−α)[hˆ(cos(ξ )−1)−hsin(ξ )])−2β 2α ([h2(cos2(ξ )−1)hˆ2 sin2(ξ )+2hˆhcos(ξ )sin(ξ )])V− := kβh2r2−β 2((1−α)hr+α h2)− 4r2hˆhξ+1r2((1−2h2)(cos(2ξ )−1)+2hˆh(2ξ − sin(2ξ )))− kβ(ξr +12r[2hˆh(cos(2ξ )−1)+(1−h2)sin(2ξ )])−β 2 (−(1−α)[hˆ(cos(ξ )−1)−hsin(ξ )])−β 2α ([h2(cos2(ξ )−1)+ hˆ2 sin2(ξ )+2hˆhcos(ξ )sin(ξ )])and V± satisfies the following point wise bound|V±|.kβ hr+β 2 h2+hr2|ξ |+ ξ2r2+ kβ (|ξr|+ 1r |ξ |)+β2 h|ξ |+β 2 ξ 2. (3.1.4)We fix β = β (k) as in Theorem 1. Then we havek.β log(1β), ‖ξ‖L∞.‖ξ‖X.β , ‖|ξr|+ |ξ |r ‖L∞.β2 log2(1β),102and so for each of V =V± we have‖V‖L∞.β 2 log2( 1β ). (3.1.5)Moreover, we may writeV =V1+V2+V3, ‖r2V1‖L∞.β , ‖V2‖L2.β 3 log(1β), ‖V3‖L∞.β 4,leading to the following quadratic form estimate: for φ ∈ D,|〈φ ,V±φ〉|.‖r2V1‖L∞‖φ‖2X +‖V2‖L2‖φ‖X‖φ‖L2 +‖V3‖L∞‖φ‖2L2.β ‖φ‖2X +β 3 log(1β)‖φ‖X‖φ‖L2 +β 4 ‖φ‖2L2.β ‖φ‖2X +β 4 ‖φ‖2L2.(3.1.6)We will rewrite the eigenvalue equation in terms of the resolvent.(H +β 2)−1 = (−∆˜+β 2−W )−1 =((−∆˜+β 2)(I−R0(β )W ))−1= (I−R0(β )W )−1R0(β )(3.1.7)whereR0(β ) = (−∆˜+β 2)−1denotes the free resolvent. For readers’ convenience, we recall some estimates onthe free resolvent from Section 2.3.4. :‖R0(β )g‖X.‖rg‖L2 , (3.1.8)‖R0(β )g‖X. 1β ‖g‖L2, (3.1.9)‖R0(β )g‖X. 1β 2‖g‖X . (3.1.10)103In addition, we use a refinement of (2.2.14) for the case g = h 6∈ L2:‖R0(β )h‖X. 1β . (3.1.11)We have the following estimates for the inverse acting on a function in X whichdoes not satisfy the X orthogonal conditions (2.2.12).Proposition 3.1.1. For f ∈ X ,(1−R0(λ )W )−1 f = 〈Wh, f 〉2λ 2 log( 1λ) ([1+O( 1log( 1λ))]h+OX(λ ))+OX(‖ f‖X).Proof. Start with the X-orthogonal decompositionf =〈h, f 〉X‖h‖2Xh+ fˆ , 〈h, fˆ 〉X = 0,so‖(1−R0(λ )W )−1 f − 〈h, f 〉X‖h‖2X(1−R0(λ )W )−1h‖X.‖ fˆ‖X.‖ f‖X , (3.1.12)and so it remains to determineg := (1−R0(λ )W )−1h.Nowh = (1−R0(λ )W )g = (1−G0W )g+λ 2 G0R0(λ )Wg,and so(1−G0W )g = h−λ 2 G0R0(λ )Wg ∈ Ran(1−G0W ) = h⊥X = X ∩ (Wh)⊥.104So using our previous estimates of certain inner-products (2.2.22),0 = 〈h, h−λ 2 G0R0(λ )Wg〉X = ‖h‖2X −λ 2〈h, R0(λ )Wg〉= ‖h‖2X −λ 2(log(1λ)〈rW, g〉+O(‖g‖X))and so〈rW, g〉= ‖h‖2Xλ 2 log( 1λ) +O( ‖g‖Xlog( 1λ)) .Also, by Lemma 2.3.9(I−G0W ) : X ∩ (rW )⊥→ X ∩ (Wh)⊥is boundedly invertible,g−µh = (1−G0W )−1(h−λ 2 R0(λ )G0Wg)∈ X ∩ (rW )⊥for some µ ∈ R. Thus by Lemma 3.1.1 below,‖g−µh‖X.‖h‖X +λ 2λ−1 ‖g‖X.1+λ ‖g‖X ,and in particular‖g‖X. |µ|+1+λ ‖g‖X =⇒ ‖g‖X. |µ|+1.Moreover0 = 〈rW, g−µh〉= ‖h‖2Xλ 2 log( 1λ) +O( ‖g‖Xlog( 1λ))− (rW, h)µsoµ =1λ 2 log( 1λ) ‖h‖2X〈rW, h〉 +O(‖g‖Xlog( 1λ))= 1λ 2 log( 1λ) ‖h‖2X〈rW, h〉 +O(1+ |µ|log( 1λ)) .105and thusµ =1λ 2 log( 1λ) ‖h‖2X〈rW, h〉(1+O(1log( 1λ))) .Putting these few estimates together yieldsg =1λ 2 log( 1λ) ‖h‖2X〈rW, h〉(1+O(1log( 1λ)))h+OX( 1λ log( 1λ )),and inserting this into (3.1.12) yields the result.The following Lemma 3.1.1 is needed for Proposition 3.1.1Lemma 3.1.1.‖R0(λ )G0Wg‖X. 1λ ‖g‖X .Proof. 1. Defineζ (r) :=1rχ≥1(r),andcg :=12∫ ∞0s2W (s)g(s)ds, |cg| ≤ 12‖s2W‖L2‖gs‖L2.‖g‖X .2. We first show‖G0Wg− cgζ‖L2.‖g‖X .For this, use the explicit formulaγ(r) :=G0Wg(r)−cgζ = 12r(∫ r0− χ≥1(r)∫ ∞0)s2W (s)g(s)ds+r2∫ ∞rW (s)g(s)ds.Using Ho¨lder, for r ≤ 1,|γ(r)|. 1r∫ r0s2|g(s)|ssds+ r∫ ∞0W (s)|g(s)|ssds.(r2+ r)‖g‖X.r‖g‖X ,106while for r ≥ 1,|γ(r)|. 1r∫ ∞r1s2|g(s)|ssds+ r∫ ∞r1s4|g(s)|ssds. 1r2‖g‖X ,and the desired estimate follows.3. Then in light of the free resolvent estimate ‖R0(λ )‖L2→X. 1λ , it suffices toshow‖R0(λ )ζ‖X. 1λ .For this, use the estimate |ζ˜ (ρ)|. 1ρ (2.3.30), so that‖R0(λ )ζ‖2X.∫ ∞0ρ2(ρ2+λ 2)2ζ˜ 2(ρ)ρdρ.∫ ∞01(ρ2+λ 2)2ρdρ. 1λ 2.The following is the main theorem for equivariant perturbations.Theorem 9. There is a constant K > 0 such that for all β > 0 sufficiently small,the (shifted) linearized operatorsLˆ± = H +V±satisfyLˆ± ≥−K β2log(1β) .Proof. 1. Arguing by contradiction, if the statement of the Theorem is false,there must be a sequence β j→ 0+, and eigenvalues(Lˆ+λ 2j)ψ j = 0, Lˆ = Lˆ+ or Lˆ−, λ 2j β 2log(1β)107,andψ j ∈ D(−∆˜)⊂ X , ‖ψ j‖L2 = 1,where we have chosen to normalize the eigenfunctions in L2. We will showthis cannot happen. For future use, we consider also the X-normalizedeigenfunctionsφ j :=‖h‖X‖ψ j‖X ψ j, ‖φ j‖X = ‖h‖X , ‖φ j‖L2 =‖h‖X‖ψ j‖X .In what follows we drop the subscript j for simplicity of notation.2. Taking inner-product of the eigenvalue equation with φ , and using (3.1.6),λ 2 ‖φ‖2L2 + 〈φ , Hφ〉=−〈φ , V±φ〉.β +β 4 ‖φ‖2L2 = β +λ 2β 4λ 2‖φ‖2L2and so sinceλ 2 β2log(1β)  β 4,and using the positivity H ≥ 0,0 < λ 2 ‖φ‖2L2 + 〈φ , Hφ〉.β → 0.We concludeλ 2 ‖φ‖2L2.β → 0, 0≤ 〈φ , Hφ〉.β → 0,and so also in light of (3.1.6),λ 2+‖ψ‖2X = 〈ψ,(λ 2− ∆˜)ψ〉= 〈ψ,(W −V±)ψ〉. 1+β 2 ‖ψ‖2X +β 4and so,λ 2+‖ψ‖2X . 1.108Thus,λ 2.β 1‖φ‖2L2= β‖ψ‖2X‖h‖2X.β → 0.3. If {φ j}∞j=1 ∈X is normalized as ‖φ j‖X = ‖h‖X , and ‖Fφ j‖L2→ 0, by Lemma3.1.2 below, up to subsequence ‖φ j−±h‖X → 0.So we insertφ = h+η , ‖η‖X = o(1).Writing the eigenvalue equation as:0 = (H +V±+λ 2)(h+η) = (H +λ 2)η+V±η+(V±+λ 2)h.by resolvent identityη = (I−R0(λ )W )−1R0(λ )[(−V±−λ 2)h−V±η ]. (3.1.13)By Proposition 3.1.1,η =〈Wh, f 〉2λ 2 log( 1λ) ([1+O( 1log( 1λ))]h+OX(λ ))+OX(‖ f‖X),where f := R0(λ )((−V±−λ 2)h−V±η) .4. We first estimate the size of ‖ f‖X . Using the pointwise bounds on V±(3.1.4), and the free resolvent estimates (2.2.16), (2.2.13) and (2.2.2), wehave• ‖λ 2 R0(λ )h‖X.λ• ‖R0(λ )V±h‖X.‖rh‖L∞‖kβ hr +β 2 h2+ hr2 |ξ |+kβ (|ξr|+ 1r |ξ |)+β2 h|ξ |‖L2+‖h‖L∞‖r ξ2r2 ‖L2+ 1λ ‖rh‖L∞‖1r β 2 ξ 2‖L2• ‖R0(λ )V±η‖X.‖1rη‖L2 ‖kβ rh+β 2 r2h2+h|ξ |+ξ 2‖L∞109+ 1λ ‖η‖L∞‖kβ (|ξr|+ 1r |ξ |)+β 2 h|ξ |‖L2+ 1λ 2‖η‖X‖β 2 ξ 2‖XSince ‖η‖X = o(1)‖R0(λ )V±η‖X  β + 1λ β3 log(1β)+1λ 2β 4 (3.1.14)Putting the above estimates together, we have‖ f‖X = ‖R0(λ )((−V±−λ 2)h−V±η)‖X . λ +β = o(1).5. Returning then to step (3), since ‖η‖X = o(1), we must have|〈Wh, f 〉|  λ 2 log(1λ)(3.1.15)Now to estimate 〈Wh, f 〉. We first compute the case L+, and writeV+ := V¯+1 +V˜+V¯+1 :=−2kβ hˆhr−β 2((1−α)hr+2αh2)− 4r2hˆhξ+1r2((1−2h2)(cos(2ξ )−1)+2hˆh(2ξ − sin(2ξ )))and V˜ satisfying the bound|V˜+|. kβ (ξr + 1r ξ )+β2hξ +β 2ξ 2 (3.1.16)110〈Wh, f 〉= 〈R0(λ )((−V+−λ 2)h+V+η),Wh〉=−〈λ 2h,R0(λ )Wh〉−〈V+h,R0(λ )Wh〉+ 〈V+η ,R0(λ )Wh〉=−〈λ 2h,R0(λ )Wh〉−〈(V¯+1 +V˜+)h,R0(λ )Wh〉+ 〈V+η ,R0(λ )Wh〉=−4λ 2 log( 1λ)+O(λ 2)−〈V¯+1 h,h〉+λ 2〈V¯+1 h,R0(λ )h〉+ 〈V˜+h,R0(λ )Wh〉+ 〈V+η ,R0(λ )Wh〉(3.1.17)where in the last equality, we use (2.2.24), G0Wh = h, and the resolventidentity (2.2.18)〈V¯1h, R0(λ )Wh〉= 〈V¯1h, G0Wh〉−λ 2〈V¯1h,R0(λ )h〉= 〈V¯1h,h〉−λ 2〈V¯1h,R0(λ )h〉.Next, we estimate each inner product 〈V¯1h,h〉, λ 2〈V¯1h,R0(λ )h〉, 〈V˜ h,R0(λ )Wh〉,〈V+η ,R0(λ )Wh〉 and then show that the main contributions of 〈Wh, f 〉 arefrom −〈λ 2h,R0(λ )Wh〉−〈V¯1h,R0(λ )Wh〉.We first estimate the last term of (3.1.17), 〈V+η ,R0(λ )Wh〉. Recall that wehave the expression of η asη =〈Wh, f 〉2λ 2 log( 1λ) [(1+OX( 1log( 1λ)))h+OX(λ )]+OX(‖ f‖X).(3.1.18)Inserting (3.1.18) in the inner product, by ‖ f‖X = o(1), and the bounds(3.1.14) we have〈V+η ,R0(λ )Wh〉= o(1)〈V+h,R0(λ )Wh〉+o(λβ )+o(β 2+λβ ).And by the assumption λβ  λ 2 log( 1λ ),〈V+η ,R0(λ )Wh〉= o(1)〈V+h,R0(λ )Wh〉+o(β 2+λ 2 log( 1λ )). (3.1.19)So we can absorb 〈V+η ,R0(λ )Wh〉 into 〈V+h,R0(λ )Wh〉.111Next, we show that the size of 〈R0(λ )V˜ h,Wh〉 is O(β 3 log( 1β )).Since‖R0(λ )V˜+h‖X . ‖rh‖L∞kβ (‖ξr‖L2 +‖ξr‖L2)+β 2‖rh‖L∞‖ξ 2r‖L2. β 3 log 1β+β 4λ,|〈R0(λ )V˜+h,Wh〉|. ‖rWh‖L2‖R0(λ )V˜+h‖X . β 3 log1β+β 4λ= O(β 3 log(1β)) β 2.(3.1.20)Next we compute 〈h,V¯1h〉,〈h,V¯+1 h〉=−2kβ∫ ∞01rhˆh3rdr−β 2∫ ∞0((1−α)h3r+2αh4)rdr−∫ ∞04r2hˆhξ rdr+∫ ∞01r2(1−h2)(cos(2ξ )−1)h2rdr+O(β 3)= kβ(∫ ∞0−21rhˆh3rdr+43)−β 2∫ ∞0((1−α)h3r+2αh4)rdr+β 2(43α−2)−∫ ∞0hˆh3r2(cos(2ξ )−1)rdr+∫ ∞0h2(1−h2)r2(cos(2ξ )−1)rdr+O(β 3 log( 1β))= β 2∫ ∞0(−(1−α)h3r−2αh4)rdr+β 2(43α−2)+∫ ∞0(−hˆh+1−h2) h2r2(cos(2ξ )−1)rdr+O(β 3 log( 1β))=−(4+2α)β 2+∫ ∞0(−hˆh+1−h2) h2r2(cos(2ξ )−1)rdr+O(β 3 log( 1β)).Because |(cos(2ξ )−1)|= O(β 2), and |2ξ − sin(2ξ )|= O(β 3),〈h,V¯+1 h〉= O(β 2). (3.1.21)112In the second equality, we used H(h2r ) = −4 hˆh3r2 and Euler-Lagrange equa-tion (2.2.5) to evaluate the inner product 〈h,−4 hˆh2r2 ξ 〉 to cancel the leadingorder kβ = β 2 log( 1β )+O(β2) :〈h,−4 hˆh2r2ξ 〉= 〈H h2r,ξ 〉= 〈h2r,Hξ 〉= 〈h2r,s+N−β 2 ξ 〉=−β 2∫ ∞0rdrh3r+(kβ +αβ 2)∫ ∞0rdrh4r2−∫ ∞0rdrhˆh3r2(cos(2ξ )−1)+O(β 3 log(1β))=−2β 2+ 43(kβ +αβ 2)−∫ ∞0rdrhˆh3r2(cos(2ξ )−1)+O(β 3 log( 1β)).Next, we estimate λ 2(V¯+1 h,R0(λ )h). Since h3, h2r ,hr2 ∈ L1 ∩ r−3L∞ and‖ξ‖L∞ . β , by (2.2.25)• λ 2〈kβ h2r ,R0(λ )h〉= O(λ 2kβ log2( 1λ )) β 2• λ 2〈2β 2h3,R0(λ )h〉= O(β 2λ 2 log2( 1λ )) β 2• λ 2〈ξ 2 hr2 ,R0(λ )h〉. λ 2β 2 log2( 1λ) λ 2 log( 1λ )• λ 2〈 hr2ξ ,R0(λ )h〉. λ 2β log2( 1λ ) λ 2 log( 1λ )Adding the above estimates together, we haveλ 2〈V¯+1 h,R0(λ )h〉= o(λ 2 log(1λ)+β 2). (3.1.22)Therefore by (3.1.21) and (3.1.22),〈V¯+1 h,R0(λ )Wh〉= 〈h,V¯+1 h〉−λ 2〈V¯+1 h,R0(λ )h〉= O(β 2)+o(λ 2 log(1λ))113Returning to (3.1.19), and using (3.1.20)〈V+η ,R0(λ )Wh〉= o(1)〈V+h,R0(λ )Wh〉+o(β 2+λ 2 log( 1λ ))= o(β 2+λ 2 log(1λ)).(3.1.23)With this, we can return to (3.1.17), and adding all the estimates (3.1.20),(3.1.21), (3.1.22), and (3.1.23) togethero(λ 2 log(1λ)) = 〈Wh, f 〉=−4λ 2 log(1λ)+O(β 2).indicating λ 2 log 1λ . β 2, which contradicts β 2 λ 2 log( 1β ).For the case of V−, similarly as the L+, we write V− = V¯−1 + V˜−,whereV¯−1 := kβh2r2−β 2((1−α)hr+α h2)− 4r2hˆhξ+1r2((1−2h2)(cos(2ξ )−1)+2hˆh(2ξ − sin(2ξ )))V˜− satisfies the same bound as V˜+ (3.1.16), so 〈R0(λ )V˜+h,Wh〉=O(β 3 log 1β ).Similar to the estimates of (3.1.22),λ 2〈V¯−1 h,R0(λ )h〉= o(λ 2 log(1λ)+β 2).One of the leading orders of 〈Wh, f 〉, where f =R0(λ )((−V−−λ 2)h−V−η)is〈h,V¯−1 h〉= kβ(∫ ∞01r2h4rdr+43)+O(β 2) =83kβ +O(β 2).Combining the above and (3.1.20),〈Wh, f 〉=−4λ 2 log( 1λ)− 83kβ +O(β 2).114Recall that k > 0, and because of (3.1.15), this is a contradiction.Lemma 3.1.2. If {φ j}∞j=1 ⊂ X is normalized as ‖φ j‖X = ‖h‖X , and ‖Fφ j‖L2→ 0,then, up to subsequence, ‖φ j−±h‖X → 0.Remark 3.1.2. With a further condition, such as φ j ≥ 0, the conclusion is ‖φ j−h‖X → 0, and the convergence is for the full sequence (since the limit is thenunique).Proof. Since ‖φ j‖X.1, up to subsequence, we haveφ j→ φ ∈ Xweakly in X , and uniformly on compact subintervals of (0,∞) (the latter by the1D compact embedding H1([a,b])⊂C([a,b]) for compact intervals [a,b]⊂R). Itfollows that Fψ j→ 0 weakly in L2, and so‖Fφ‖L2 ≤ liminf‖Fφ j‖L2 = 0, =⇒ Fφ = 0, =⇒ φ = γhfor some γ ∈ R. By the local uniform convergence, and since∫ ε0Wφ2j rdr+∫ ∞1εWφ2j rdr ≤ 2 supr<ε, r> 1εr2W (r)∫ ∞0φ2jr2rdr. supr<ε, r> 1εr2W (r)→ 0 as ε → 0+,we have ∫ ∞0Wφ2j rdr → γ2∫ ∞0Wh2 rdr.Then by the identity‖Fφ j‖2L2 = 〈φ j, Hφ j〉= 〈φ j, −∆˜φ j−Wφ j〉= ‖ψ j‖2X −∫ ∞0Wφ2j rdr,115we have0←‖Fφ j‖2L2 = ‖φ j‖2X −∫ ∞0Wφ2j rdr = ‖h‖2X − γ2∫ ∞0Wh2 rdr+o(1)= (1− γ2)‖h‖2X +o(1),hence γ = ±1. In particular, ‖φ‖X = ‖h‖X = lim‖φ j‖X , and it follows that theconvergence is strong in X .Corollary 10. The original linearized operatorsL± = Lˆ±+β 2 ≥ β 2−K β2log(1β) > 0and so the co-rotational skyrmion solution is (energetically) stable against equiv-ariant perturbations.Proof. This follows directly from Theorem 9 and the definition of shifted operator(3.1.1).3.2 General perturbationsThe linearized operator governing general perturbations, acting on the j-th angularFourier mode (see Section 3.6) is represented by the operator valued matrixE′′j =[L++ j2r2i jr(2r cos(u)− kβ sin(u))− i jr(2r cos(u)− kβ sin(u))L−+ j2r2],acting on L2×L2 = L2([0,∞);rdr])×L2([0,∞);rdr]).To set up for perturbation theory, writeE′′j −β 2 I =H j +Vβ , j116where the “unperturbed” operator isH j =[H + j2r22i jr2 hˆ−2i jr2 hˆ H +j2r2].3.2.1 The unperturbed operatorNote we can explicitly diagonalize the unperturbed operator:U :=1√2[1 1−i i], Hˆ j :=U∗HU =[H j 00 H− j]where, denoting∆(m) := ∆r− m2r2so ∆(0) = ∆r and ∆(1) = ∆˜,H j :=−∆( j−1)+Wj, Wj := 1r2(2 j(1+ hˆ)−2h2) , so H0 = H.Since Wj is bounded, it follows from the standard Kato theory that H j is a self-adjoint operator on L2r dr with domainD(H j) = D(−∆( j−1))= f ∈ L2 ∣∣frr, 1r fr ∈ L2 j = 1frr, 1r fr− 1r2 f ∈ L2 j = 0, 2frr, 1r fr,1r2 f ∈ L2 | j−1| ≥ 2 .Remark 3.2.1. This characterization of the domain comes from determining theconditions on f (r) such that f (r)ei( j−1)θ ∈ H˙2(R2). Note also thatD(H j)⊂ X ⇐⇒ j 6= 1,117since‖ fr‖2L2 +( j−1)2‖fr‖2L2 = 〈 f , −∆( j−1) f 〉 ≤ ‖ f‖L2‖∆( j−1) f‖L2 , (3.2.1)while a radial function f which is (non-zero) constant in a neighbourhood of r = 0lies in D(−∆r), but f/r 6∈ L2.Observe also the factorizationH j = F∗j Fj, Fj = ∂ r+1r( j+ hˆ) = r− jh ∂ rr jh, so F0 = F, F1 = ∂ r+h.Remark 3.2.2. This factorization is a consequence of the energy relationE(mˆ) = 2pi+12∫R2|∂ x1 mˆ+ J ∂ x2 mˆ|2 dx.Indeed, for a one-parameter variationmˆε = mˆ+ ε~ξ + · · · , ξ ∈ TmˆS2,∂ x1 mˆε + J ∂ x2 mˆε = ε~η+ · · · , ~η ∈ TmˆS2,and Fj results from expressing the operator~ξ 7→~η : TmˆS2→ TmˆS2in the natural frame used in Section 3.6 (and applying the diagonalizing transfor-mation).Note thatFj : X → L2, F1 : D(H1)→ X .Lemma 3.2.1.kerX Fj ={〈h〉 j = 00 j 6= 0}, kerL2 Fj ={〈hr 〉 j = 10 j 6= 1}.118Proof. Solving the first order ODEs: Fjφ = 0,φ =Ch(r)r− j =C r1− jr2+1 ∼{r1− j, r→ 0r−1− j r→ ∞}, so φ ∈ L2, onlywhen j = 1. And φr = h′(r)r− j− jr− j−1h(r)∼{r− j, r→ 0r− j−2 r→ ∞}, so φ ∈ Xonly when j = 0.Next, F∗j = −∂r + 1r ( j+ hˆ− 1). Similarly, solving for F∗j φ = 0, φ =C(r2 +1)r j−2, φr /∈ L2 for any j, sokerX F∗j = kerL2 F∗j = 0.Because hr ∈ D(H1)⊂ L2, using the factorization H j = F∗F , it follows thatkerX H j ={〈h〉 j = 00 else}, kerD(H j)H j ={〈hr 〉 j = 10 else}.3.2.2 The full linearized operatorApplying the same diagonalization transformation to the full linearized operatorproducesE˜ ′′j :=U∗E ′′j U =[L¯+ j2r2 +jr(2r cos(v)− kβ sin(v))LL L¯+ j2r2 −jr(2r cos(v)− kβ sin(v)) ]whereL¯ =12(L++L−)=−∆r + 1r2 cos(2v)− kβ(12vr +34rsin(2v))+β 2(1+(1−α)(cos(v)−1)− 32α sin2(v)),andL =12(L+−L−) =−kβ(14rsin(2v)− 12vr)− 12β 2α sin2(v).119To set up for perturbation theory, we write the shifted second variationEˆ ′′j := E˜′′j −β 2 I = Hˆ j +VjVj =−kβ V (k)j −β 2V (β )j −V (ξ )j , (3.2.2)whereV (k)j =[12r h(3hˆ−1)+ jr h 12r h(1+ hˆ)12r h(1+ hˆ)12r h(3hˆ−1)− jr h],V (β )j =[(1−α)(1− hˆ)+ 32 α h2 12 α h212 α h2 (1−α)(1− hˆ)+ 32 α h2],andV (ξ )j =2r2(2hˆ+ j)hξ I+V˜ (ξ )j , ‖V˜ (ξ )j ‖.ξ 2r2+kβ (|ξr|+ |ξ |r )+β2 ξ 2+β 2 h|ξ |.By the same considerations as for L± above, we have: Eˆ ′′j is a self-adjoint operatoron L2×L2 with domainD j := D(Eˆ ′′j ) = D(Hˆ j) = D(H j)×D(H− j),andσess(Eˆ ′′j ) = [0,∞).3.3 Translation zero-modes j =±1Translation invariance of the energy produces the translation zero-modes. Asshown (3.6.2) in Section 3.6∂1mˆ+ i∂2mˆ = (ur(r)~e+ isin(u(r))rJ~e)eiθ120and∂2mˆ+ i∂1mˆ = (iur(r)~e+sin(u(r))rJ~e)e−iθ ,which corresponding to the j = 1, and j =−1 sectors. Without loss of generalitywe may assume j = 1. The goal is to prove:Theorem 11. For sufficiently small β , there exists K > 0 independent of β , suchthatE ′′j=1|~T⊥ ≥ β 2−K β 4 log4(1β)> 0, where ~T =[vri sinvr].After the diagonalization by U , each shifted Eˆ ′′j=±1 have −β 2 as a simpleeigenvalue, this reads(Eˆ ′′1 +β2 I)~T1 =~0,where ~T1 =[12(1r sin(v)− vr)12(−1r sin(v)− vr)].The corresponding “unperturbed” statement isHˆ1 =[H1 00 H−1][hr0]= 0.We first do the perturbation of the H1 (Section 3.3.1), and H−1 (Section 3.3.2)separately, and combine the results later.3.3.1 Dealing with H1Consider a model problem: suppose V is a potential satisfying a bound of thetype (3.1.4), and that the perturbed operator H1+V has an eigenvalue(H1+V +β 2)ψ = 0, ψ = ψβ ∈ D(H1), ‖ψ−hr‖L2 → 0, ‖ψ‖L2 = ‖hr‖L2.121The problem with trying to proceed as in Section 3.1 is that sinceH1 =−∆r +W1, W1 = 2r2(1+ hˆ−h2) ∼ { −4 r→ 04r2 r→ ∞,the quadratic form for H1, does not control ‖·‖2X , even on the subspace orthogonalto the zero-mode h/r. So we replace ‖ · ‖X here with the weaker norm‖φ‖2Y := ‖φr‖2L2 +∥∥∥∥ φ〈r〉∥∥∥∥2L2.We do not have Y ⊂ L∞ since Y allows a logarithmic singularity at the origin, butby R2 Sobolev embedding, and X ⊂ L∞, we still have‖φ‖L∞([1,∞)).‖φ‖Y , ‖φ‖Lp([0,1]).p ‖φr‖L2.‖φ‖Y , p < ∞. (3.3.1)Moreover, the orthogonality φ ⊥ hr is not well-defined for φ ∈ Y , and so we willreplace hr with a localized bump function:0 6= ζ ∈C∞0 ((0,∞)), ζ ≥ 0,and instead of directly considering possible eigenfunctions, we will work with thequadratic form:Proposition 3.3.1. There is K > 0 such that for every φ ∈D(H1) with 〈φ , ζ 〉= 0,〈φ , (H1+V )φ〉 ≥ −K β 4 ‖φ‖2L2 . (3.3.2)Proof. By way of contradiction, suppose that for any K > 0, there areβ = β j→ 0+, φ = φ j ∈D(H1), 〈ζ , φ j〉= 0, 〈φ , (H1+V )φ〉<−K β 4 ‖φ‖2L2.122Normalize the sequence as ‖φ‖Y = 1. The key is the estimate:|〈φ , Vφ〉|.kβ +β 2+β +β 2+(kβ 2+β 3)‖φ‖L2 +β 4 ‖φ‖2L2.β +β 4 ‖φ‖2L2 ,(3.3.3)which follows from (3.1.4), the facts that 〈r〉2 (hr +h2) is bounded and ‖ 〈r〉r ξ‖L∞.β ,and the Sobolev-type inequalities (3.3.1). Since H1 ≥ 0,0≤ K β 4 ‖φ‖2L2 + 〈φ , H1φ〉<−〈φ , Vφ〉.β +β 4 ‖φ‖2L2,and so for K chosen large enough,0≤ ‖F1φ‖2L2 = 〈φ , H1φ〉.β → 0.Up to subsequence,φ ⇀ φ∗ in Y, φ → φ∗ in L∞loc([1,∞)) and Lploc([0,1]), p < ∞. (3.3.4)It follows that F1φ∗ = 0, hence φ∗ = µ hr . But also 〈ζ , φ∗〉 = lim〈ζ ,φ〉 = 0, andhence µ = 0. Then using (3.3.1), we arrive at a contradiction:1= ‖φ‖2Y .〈φ , (−∆r+χ≥1(r)4r2)φ〉= 〈φ , H1φ〉+∫ ∞0(−W1+χ≥1 4r2)φ2 rdr→ 0,by the convergence (3.3.4), and |W1−χ≥1 4r2 |.〈r〉−4.3.3.2 Dealing with H−1SinceH−1 =−∆r +W−1, W−1 = 4r2(r2+1) ∼{4r2 r→ 04r4 r→ ∞,123The quadratic form of H−1 does not control ‖ · ‖2X . So we use the factorization:H−1 = F∗−1F−1, F−1 = rh∂r1rh〈φ ,H−1φ〉= ‖F−1φ‖2L2and the interpolation:(φrh)2(r) =−2∫ ∞rsdsφs3h2(φsh)ssh≤ ‖F−1φ‖L2‖φ‖L2 sups≥r1s3h2so,φ2(r)≤ ‖F−1φ‖L2‖φ‖L2r2h2 sups≥r1s3h2≤ 1r‖F−1φ‖L2‖φ‖L2 (3.3.5)Proposition 3.3.2. V is the lower diagonal component of V1 defined as (3.2.2).There is K > 0 such that〈φ ,(H−1+V )φ〉 ≥ −K β 4 log4( 1β )‖φ‖2L2Proof. Using young’s inequality and the interpolation (3.3.5),〈φ ,H−1+Vˆ1+Vˆ2)φ〉= ‖F−1φ‖2L2 +∫Vˆ1φ2rdr+∫Vˆ2φ2rdr≥ ‖F−1φ‖2L2−‖rφ‖L∞∫Vˆ1dr+∫Vˆ2φ2rdr≥ ‖F−1φ‖2L2−√2‖F−1φ‖L2‖φ‖L21√2∫|Vˆ1|dr−‖Vˆ2‖L∞‖φ‖2L2≥(−14(∫|Vˆ1|dr)2−‖Vˆ2‖L∞)‖φ‖2L2Here, we write V = Vˆ1+Vˆ2,Using the higher regularity (2.2.3) on Vˆ2 and (3.1.4)‖Vˆ1r‖2L1 . β 4 log4(1β), ‖Vˆ2‖L∞ . β 4 log4( 1β ) (3.3.6)The result follows.124We are now ready to prove Theorem 11Proof. Suppose that(H1+V1+λ 2I)~η = 0, ~η 6= 0, 〈~η , ~T1〉= 0, λ 2 β 4 log4( 1β ), ‖~η‖L2 = 1~ζ =[ζ0], ζ ∈C∞0 is a positive bump function.Because for the component of ~T1,sin(v)r−vr = sin(Q+ξ )r −Qr−ξr =hr(1−O(ξ 2))+ hr+O(|ξ |r+ξr)= 2hr+O(β 2 log2(1β)),and hr > 0, for small β , 〈~ζ , ~T1〉> 0. So we can define~φ :=~η+ γ~T1, γ :=−〈~η ,~ζ 〉〈~T,~ζ 〉so that 〈~φ ,~ζ 〉= 0, and 〈φ+,ζ 〉= 0.Since ~η ⊥ ~T1,‖~φ‖2L2 = 1+ γ2‖~T1‖2L2.Writing~φ =[φ+φ−]〈~φ , Eˆ ′′j=1~φ〉= 〈~φ ,(H1+Vj=1)~φ〉= 〈[φ+φ−],[H1+V1,1j=1 V1,2j=1V 2,1j=1 H−1+V2,2j=1][φ+φ−]〉where Vj=1 is defined as (3.2.2).For the off-diagonal terms, by Ho¨lder and elementary inequalities,125|〈φ+,V 1,2j=1φ−〉|≤ ‖φ+√|V 1,2j=1|‖L2‖φ−√|V 1,2j=1|‖L2 ≤12〈φ+, |V 1,2j=1|φ+〉+12〈φ−, |V 1,2j=1|φ−〉.Thus,〈~φ ,(H1+Vj=1〉~φ) = 〈φ+,(H1+V 1,1j=1)φ+〉+ 〈φ+,V 1,2j=1φ−〉+ 〈φ−,(H−1+V 2,2j=1〉φ−)+ 〈φ−,V 1,2j=1φ+〉≥ 〈φ+,(H1+V 1,1j=1〉φ+)−12〈φ+, |V 1,2j=1|φ+〉−12〈φ−, |V 1,2j=1|φ−〉− 12〈φ+, |V 2,1j=1|φ+〉−12〈φ−, |V 2,1j=1|φ−〉+ 〈φ+,(H−1+V 2,2j=1)φ−〉≥ 〈φ+,(H1+V 1,1j=1−12|V 1,2j=1|−12|V 2,1j=1|)φ+〉+ 〈φ−,(H−1+V 2,2j=1−12|V 1,2j=1|−12|V 2,1j=1|)φ−〉.By Proposition 3.3.1 and Proposition 3.3.2, each components of Vj=1 satisfyingthe same bounds of (3.3.3), and (3.3.6),〈~φ ,(H1+V1)~φ〉≥−K1β 4 ‖φ+‖2L2−K2β 4 log4(1β)‖φ−‖2L2 ≥−K β 4 log4(1β)‖~φ‖2L2.and K1,K2,K > 0. Inserting into the quadratic form,λ 2+β 2 γ2‖~T1‖L2 =−〈~φ ,(H1+V1)~φ〉≤ K β 4 log4( 1β)‖~φ‖L2= K β 4 log4(1β)+ γ2K β 4 log4(1β)‖~T1‖2L2contradicting λ 2 β 4 log4( 1β ).1263.4 Higher order modes | j| ≥ 2In this section, we study spectral gaps of the high order modes | j| ≥ 2, and obtaina uniform bound (in j) on the spectrum.We first writeE˜ ′′j = Hˆ j +β 2I− jkβsin(v)rI− kβW k +β 2Wβ , (3.4.1)whereW k =[12vr +34r sin(2v)14r sin(2v)− 12vr14r sin(2v)− 12vr 12vr + 34r sin(2v)],Wβ =[(1−α)(cos(v)−1)+ 32α sin2(v) −12α sin2(v)−12α sin2(v) (1−α)(cos(v)−1)− 32α sin2(v)]First, we obtain a lower bound of the unperturbed operator Hˆ j:Lemma 3.4.1. If | j| ≥ 2, the unperturbed operatorHˆ j ≥−∆r + (| j|−1)2r2,Proof. It suffices to show that the diagonals of Hˆ j, H j ≥−∆ j + (| j|−1)2r2 .We split it in two cases.For j ≥ 2, H j =−∆r + (| j|−1)2r2 +Wj, andWj =1r2(2 j(1+ hˆ)−2h2) = 1r24 jr2(r2+1)−8r2(r2+1)2≥ 0.For j ≤−2, H j =−∆r + ( j−1)2r2 +Wj =−∆r +(| j|−1)2r2 −4jr2 +Wj, andWj−4 jr2 =1r2(2 j(hˆ−1)−2h2) = 4| j|(r2+1)−8r2(r2+1)2r2≥ 0.127Theorem 12. For | j| ≥ 2, there is K > 0 (independent of j), such that for suffi-ciently small β ,E˜ ′′j ≥ β 2−K β 4 log3(1β)> 0.Proof. Using |sin(v)|. h+ |ξ |, and choose kβ  µ  1,kβ |〈φ , | j|sin(v)rφ〉. | j|kβ (‖rh‖L∞‖φr ‖2L2 +‖ξ‖L∞‖φ‖L2‖φr‖2L2). | j|kβ ‖φr‖2L2 +1µ| j|kβ ‖φr‖L2µβ‖φ‖L2. | j|kβ ‖φr‖2L2 +j2k2β 2µ2‖φr‖2L2 +µ2β 2‖φ‖2L2.Now each component of W k satisfies|Wκ |. |vr|+ |sin(v)|r .hr+ |ξr|+ |ξ |r .hr+β 2 log21β.Thus,|〈~φ ,W k~φ〉|. ‖rh‖L∞‖~φr‖2L2 +β 2 log21β‖~φ‖2L2.From the expression (3.2.2) and (2.2.2) each component of Wβ satisfies|Wβ |. h2+ hr+ξ 2 . h2+ hr+β 2,so|〈~φ ,Wβ~φ〉|. ‖r2h2‖L∞‖~φr‖2L2 +‖rh‖L∞‖~φr‖2L2 +β 2 ‖~φ‖2L2.Using the expression (3.4.1), the previous Lemma 3.4.1, and combining all theabove〈~φ , E˜ ′′j ~φ〉 ≥ ‖~φr‖2L2 +β 2(1− cµ2− cβ 2−ckβ log2(1β)− cβ 2)‖~φ‖2L2+((| j|−1)2− c j2k2β 2µ2− c| j|kβ −c| j|β −ckβ)‖~φr‖2L2128And for the last terms in the last equality, we claim it is positive. As | j| ≥ 2((| j|−1)2− c j2k2β 2µ2− c| j|kβ −c| j|β −ckβ −cβ 2)=(| j|−1)22+ | j|((12− ck2β 2µ2)| j|− ckβ −cβ −1)+12− ckβ −cβ 2 > 0.Since for | j| ≥ 2, (| j| − 1)2 ≥ j24 , then uniformly in | j| ≥ 2, for β sufficientlysmall, and with (2.2.1) k = k(β ) = β(2log( 1β )+O(1))E˜ ′′j ≥18(−∆r + j2r2)+β 2(1−O(β 2 log3( 1β))).Now we prove Theorem 2. For convenience, we restate Theorem 2 hereTheorem 13. For 0 < k 1,−A≤ α ≤ 1,β > 0E ′′k,α,β (sˆk,α,β )|T⊥ ≥ β 2 γk,γk ≥ 1− C1log( 1k ) > 0, and T := span{∂ sˆk,α,β∂ x1,∂ sˆk,α,β∂ x2}.Proof. For all k 1, β > 0, with ~φ ∈ H2,~φ ∈ T⊥. Let ~φλ := ~φ(λ ·), sˆλ := sˆ(λ ·).Choose λ such that it satisfies the relation (2.2.1) k = k(λ β ) = 2λ β (log( 1λ β ))+O(1)), and log( 1λ β )∼ log(1k ).It follows from Corollary 10, Theorem 11 and Theorem 12, and by Plancherel,〈~φ ,E ′′k,α,β (sˆ)~φ〉‖~φ‖2L2=〈~φλ ,E ′′k,α,β λ (sˆλ )~φλ 〉λ 2‖~φλ‖2L2≥ λ2β 2λ 2(1− C1log(1k ))= β 2(1− C1log(1k )).Corollary 14. • For j 6=±1, E ′′j ≥ εk,β (−∆+ 1r2 +1)129• For j =±1, E ′′±1|T⊥±1 ≥ εk,β (−∆+1)• E ′′|T⊥±1 ≥ εk,β (−∆+1), i.e(~φ ,E ′′~φ)≥ εk,β‖~φ‖2H10 < εk,β  β 2 γk. where γk is defined in Theorem 2.Proof. For j ≥ 2, because j2+1r2 ≤ 2j2r2 . It follows directly from Theorem 12,E˜ ′′j ≥116(−∆+ 1r2)+β 2 γkFor j = 0,E ′′0 =[−∆+ 1r2 00 −∆+ 1r2]+Vβ ,0and by (3.1.3), Vβ ,0 :=[Z+ 00 Z−]+β 2 I is bounded.By Theorem 2, and choose 0 < εk,β  β 2 γkE ′′0 = εk,βE′′0 +(1− εk,β )E ′′0≥ εk,β[−∆+ 1r2 00 −∆+ 1r2]− εk,β‖Vβ ,0‖L∞+(1− εk,β )β 2 γk≥ εk,β[−∆+ 1r2 00 −∆+ 1r2]+12β 2 γk≥ εk,β[−∆+ 1r2 00 −∆+ 1r2]+ εk,β .For j =±1 (without loss of generality, we take j = 1 here), because of the behav-ior of W1, and W−1 near the origin, we can write E˜ ′′1 as.130E˜ ′′1 = Hˆ1+V1 =[−∆ 00 −∆+ 4r2]+V1,βand V1,β is bounded. Thus similarly as for j = 0 above, on ~T⊥1 ,E˜ ′′1 ≥ εk,β[−∆ 00 −∆+ 4r2]+ εk,β .Now, we write the quadratic form for as ~φ =[φ+φ−]⊥ ~T1.〈~φ , E˜ ′′1~φ〉 ≥ εk,β(‖φ+‖2H1 +‖φ−‖2H1∩rL2) .Therefore, combining all the cases of j, for ~φ ⊥ T±1, and by Plancherel,~φ =∑j~φ j(r)ei jθ〈~φ , E˜ ′′~φ〉=∑j〈~φ j, E˜ ′′j ~φ j〉≥ εk,β 〈~φ ,(−∆+1)~φ〉≥ εk,β‖~φ‖2H1.Since the unitary transform preserves the inner product, the proof follows.Remark 3.4.1. for the j = 1 mode, φ− ∈ r1L2∩ H˙1 ⊂ L∞, while φ+ /∈ L∞. So inthis mode, we lose control of the L∞ norm.3.5 Dynamical StabilityWe denote sˆ = (s1,s2,s3) as the skyrmion solution. In this section , we show theorbital stability result around the skymion solution sˆ. As shown in the Section 3.6,131the Hessian of the full energy isE ′′(sˆ) := PTsˆS2H(sˆ)where〈~ξ ,H(sˆ)~ξ 〉= 〈~ξ , [He(sˆ)+ kβ HDM(sˆ)+β 2αHα(sˆ)+β 2(1−α)Hz(sˆ)]~ξ 〉.with Hα(sˆ)~ξ = s23ξ− kˆξ3, Hz(sˆ)~ξ = s3~ξ , HDM(sˆ)~ξ =∇×~ξ− sˆ ·(∇× sˆ)~ξ , He(sˆ)~ξ =−∆~ξ −|∇sˆ|2~ξ .Proposition 3.5.1. For 0< k 1, there exist µ > 0, and 0< δ0 . β 2 γk. such thatC2‖mˆ− sˆ‖2H1 ≥ E(mˆ)−E(sˆ)≥ µ infx∈R2‖mˆ(·+ x)− sˆ‖2H1when ‖mˆ− sˆ‖H1 ≤ δ0.Proof. We proceed as in Theorem 2 of [37].~ξ := mˆ− sˆ,and the only difference from [37] is that our energy has an extra anisotropy,β 2α(Ea(mˆ)−Ea(sˆ)) = 12 β2α∫(−2ξ3m3−ξ 23 )dx =12β 2α(2E ′a(sˆ)~ξ + |~ξ |2m23−ξ 23 )= β 2α E ′a(sˆ)~ξ +12〈~ξ ,Ha~ξ 〉Since sˆ is the critical point of the Euler-Lagrange equation,E ′e(sˆ)+β 2α E ′α(sˆ)+β 2(1−α)E ′z(sˆ)+ kβ E ′DM(sˆ) = 0,E(mˆ)−E(sˆ) = 12(~ξ ,H(sˆ)~ξ ).As the result of implicit function Lemma 10 in [37], by shifting mˆ to mˆ(·+a) forsome a ∈ R2 with |a|. δ0, we have 〈mˆ(·+a)− sˆ, ∂ sˆ∂x1 〉= 〈mˆ(·+a),∂ sˆ∂x2〉= 0.132Thus we can assume 〈~ξ , ∂ sˆ∂x1 〉= 0 = 〈~ξ ,∂ sˆ∂x2〉.Next, we write ~ξ = ξ>+ξ⊥, where ξ> = (~ξ · sˆ)sˆ =−12 |~ξ |2sˆ, ξ⊥ ⊥ sˆand ‖ξ>‖2L2 = 14‖~ξ‖4L4 ≤ c‖~ξ‖4H1 ;〈~ξ ,H(sˆ)~ξ 〉= 〈ξ⊥,H(sˆ)ξ⊥〉+ 〈ξ>,H(sˆ)ξ>〉+2〈ξ⊥,H(sˆ)ξ>〉.By the estimate of the leading order of the cross term in the proof of Theorem 2in [37],〈ξ⊥,H(sˆ)ξ>〉 ≤C‖~ξ‖3H1.And,〈ξ>,H(sˆ)ξ>〉=∫|∇ξ>|2+ (β2(1−α)s3−β 2αs23−|∇sˆ|2)4|~ξ |4+β 2α(ξ>3 )2≥ ‖ξ>‖2H1−C‖~ξ‖4H1So combining all and the above corollary,E(mˆ)−E(sˆ)≥ 12(εk,β −Cδ0)‖~ξ‖2H1.Using Ho¨lder and Sobolev embedding L4 ⊂H1, the upper bound follows directly.The proof follows from choosing δ0 < 1Cεk,β with sufficiently small δ0.For simplicity, we assume a = 1, γ = b is the damping parameter. We firstrewrite the Landau-Lifshitz equation (1.7.1) as :∂tmˆ = mˆ× (γ∂tmˆ+E ′(mˆ))Proposition 3.5.2. If E(mˆ(0))< 4pi , (1.7.1) has a unique global smooth solutionmˆ(x, t) ∈C([0,∞); H˙1)∩C∞(R2×(0,∞)).Proof. See the Proposition 3 of [15]. Suppose this is a finite-time singularity.After translation and dilation, we can assume (x, t) = (0,0) is a singularity. For asuitable sequence xk → 0, tk ↗ 0,rk ↘ 0 the blow-up solution mˆk(x, t) = mˆ(xk +133rkx, tk + r2kt) satisfies the perturbed Landau-Lifshitz equation:∂tmˆk = mˆk×(γ∂tmˆk +E ′e(mˆk)+ kβ rk E ′DM(mˆk)+β 2 r2k [α E ′a(mˆk)+(1−α)E ′z(mˆk)])= mˆk× (γ∂tmˆk−∆mˆk)+~fkand ~fk = mˆk× [kβ rk(∇×mˆk)+β 2r2k(−αm3kˆ− (1−α)kˆ)]. So fk⊥mk, and | fk|.rk|∇mˆk|+ r2k |mˆk− kˆ|2.The perturbed term is ‖ fk(t)‖L2 =O(rk). Because of the lower order perturba-tion, local higher sobolev bounds can be estimated, and the perturbed term of L-Lequation from the rescaled approximating solution is bounded uniformly. Fromthe energy inequality∫ 0−1∫R2 |∂tmˆk|2dxdt → 0, the rescaled solution converges toa non-constant harmonic map, and so.limsupt↗0E(mˆ(t))≥ 4pi.That is, it costs energy at least 4pi to form a bubble. However, our initial energy isbelow the ground state, and energy is dissipating:E(mˆ(t))+ γ1∫ t0‖∂tmˆ‖2L2ds = E(mˆ(0))< 4pi.Thus we have a global smooth solution.Together with Proposition 3.5.1, and a bootstrap argument, we have the orbitalstability.Theorem 15. Fix k,β ,α , such that E ′′k,α,β (sˆk,α,β )|T⊥ > 0. For any ε > 0, there ex-ist a δ > 0 such that if ‖mˆ(0)− sˆk,α,β‖H1 ≤ δ , then infx∈R2 ‖mˆ(·−x, t)− sˆk,α,β‖H1 ≤ε for all t ≥ 0.Proof. We choose δ such that δ < min{2δ0√µC2,√µC2ε}, where C2,µ defined asin the above Proposition 3.5.1.Let T = {supτ| infx∈R2‖mˆ(t)− sˆ(·+ x)‖H1 ≤ 2√C2µ δ < δ0,∀0 ≤ t ≤ τ}. Since134mˆ∈C([0,∞);H1) by Proposition 3.5.2, and we can assume 2√C2µ δ ≥ δ , so T > 0.We claim that T = ∞. By Proposition 3.5.1 , and the energy dissipation (1.7.2),infx∈R2‖mˆ(t)− sˆ(·+ x)‖2H1 ≤1µ(E(mˆ(t))−E(sˆ))≤ 1µ(E(mˆ(0))−E(sˆ))≤ C2µ‖mˆ(0)− sˆ‖2H1 ≤C2µδ 2for all 0≤ t ≤ T , so T = ∞.By the choice of δ , we have infx∈R2 ‖mˆ(t)− sˆ(·+ x)‖H1 ≤ ε .1353.6 Second variation of the energyIn this section, we compute the second variation of the energy, and decompose itin angular Fourier modes index by j ∈ Z . Working in the original representation,the tangent space to S2 at mˆ ∈ S2 isTmˆS2 = mˆ⊥,and the corresponding orthogonal projection isR3 3 ~η 7→ PTmˆS2 ~η :=~η− (~η · mˆ) mˆ ∈ TmˆS2.A one-parameter variationmˆε = mˆ+ ε ~ξ +12ε2 ~η+ · · · ∈ S2of magnetization mˆ ∈ S2 satisfies1 = |mˆε |2 = 1+ ε(2mˆ ·~ξ)+ ε2(|~ξ |2+ mˆ ·~η)+ · · ·and somˆ ·~ξ = 0 =⇒ ~ξ ∈ TmˆS2,andmˆ ·~η =−|~ξ |2 =⇒ ~η =−|~ξ |2mˆ+~ζ , ~ζ ∈ TmˆS2,thusmˆε = mˆ+ ε ~ξ +12ε2(−|~ξ |2mˆ+~ζ)+ · · · , ~ξ , ~ζ ∈ TmˆS2.We compute the first variation and second variation of each energy term at thecritical point:136• exchange, Ee(mˆ) = 12∫R2 |∇mˆ|2dx:Ee(mˆε)−Ee(mˆ) = ε∫∇mˆ ·∇~ξ + 12ε2∫ (|∇~ξ |2+∇mˆ ·∇(−|~ξ |2mˆ+~ζ ))+ · · ·= ε(−PTmˆS2∆mˆ, ~ξ +12ε~ζ)+12ε2(~ξ , −PTmˆS2∆~ξ −|∇mˆ|2~ξ)+ · · ·from whichE ′e(mˆ) =−PTmˆS2∆mˆ =−∆mˆ−|∇mˆ|2mˆ,E ′′e (mˆ) =−PTmˆS2∆ −|∇mˆ|2.• D-M, EDM(mˆ) = 12∫R2 mˆ · (∇× mˆ)dx:EDM(mˆε)−EDM(mˆ) = ε∫(∇× mˆ) ·~ξ+12ε2∫ (~ξ · (∇×~ξ )+(∇× mˆ) · (−|~ξ |2mˆ+~ζ ))+ · · ·= ε(PTmˆS2(∇× mˆ), ~ξ +12ε~ζ)+12ε2(~ξ , PTmˆS2∇×~ξ − (mˆ · (∇× mˆ))~ξ)+ · · ·from whichE ′DM(mˆ) = PTmˆS2∇× mˆ = ∇× mˆ− (mˆ · (∇× mˆ))mˆ,E ′′DM(mˆ) = PTmˆS2∇× −mˆ · (∇× mˆ).• Zeeman, E z(mˆ) =∫R2(1−m3)dx:E z(mˆε)−E z(mˆ) = ε∫(−ξ3)+ 12 ε2∫(|~ξ |2m3−ζ3)+ · · ·= ε(−PTmˆS2 kˆ, ~ξ +12ε~ζ)+12ε2(~ξ ,m3 ~ξ)+ · · ·137from whichE ′z(mˆ) =−PTmˆS2 kˆ =−kˆ+m3mˆ,E ′′z (mˆ) = m3.• Anisotropy, Ea(mˆ) = 12∫R2(1−m23)dx:Ea(mˆε)−Ea(mˆ) = ε∫(−m3ξ3)+ 12 ε2∫ (m3(|~ξ |2m3−ζ3)−ξ 23)+ · · ·= ε(−m3PTmˆS2 kˆ, ~ξ +12ε~ζ)+12ε2(~ξ ,m23−PTmˆS2 kˆkˆ · ~ξ)+ · · ·from whichE ′a(mˆ) =−m3PTmˆS2 kˆ = m3(−kˆ+m3mˆ),E ′′a(mˆ) = m23−PTmˆS2 kˆkˆ · .In the co-rotational class we have the solution as the following vector form:mˆ := (−sin(u(r))sin(θ),sin(u(r))cos(θ),cos(u(r)))~e := (−cos(u(r))sin(θ),cos(u(r))cos(θ),−sin(u(r)))J~e := mˆ×~e = (−cos(θ),−sin(θ),0)(3.6.1)Note~e and J~e are an orthonormal basis on TmˆS2 and mˆr = ur~e, mˆθ = sin(u)J~e,For ~ξ ∈ TmˆS2, we write~ξ (r,θ) = g(r,θ)~e(r,θ)+h(r,θ)J~e(r,θ).Each second variation operator maps tangent vector fields to tangent vector fields.Writing them as the vector form with respect to the basis~e and J~e givesE ′′e (mˆ)[gh]=[−∆+ 1r2 cos2(u)− 1r2 sin2(u) 2r2 cos(u)∂θ− 2r2 cos(u)∂θ −∆+ 1r2 cos2(u)−u2r][gh]138E ′′a (mˆ)[gh]=[cos2(u)− sin2(u) 00 cos2(u)][gh]E ′′z (mˆ)[gh]=[cos(u) 00 cos(u)][gh]E ′′DM(mˆ)[gh]=[−2r sin(u)cos(u) −1r sin(u)∂θ1r sin(u)∂θ −ur− 1r sin(u)cos(u)][gh]Thus the second variations of the energy terms are all self-adjoint operators.If we write ~ξ in angular fourier modes:~ξ (r,θ) =∞∑j=∞(g j(r)~e+h j(r)J~e)ei jθE ′′(mˆ)~ξ =∞∑j=−∞[E ′′j (mˆ)(g j(r)~e+h j(r)J~e)]ei jθE ′′j (mˆ) := E′′e j(mˆ)+β2αE ′′a j(mˆ)+β2(1−α)E ′′z j(mˆ)+ kβ E ′′DM j(mˆ),where the j-th second variations of the energy terms are:E ′′e j(mˆ) =[−∆r + 1r2 ( j2+ cos2(u)− sin2(u)) 2r2 i j cos(u)− 2r2 i j cos(u) −∆r + 1r2 ( j2+ cos2(u))−u2r]E ′′a j(mˆ) =[cos2(u)− sin2(u) 00 cos2(u)]E ′′z j(mˆ)[gh]=[cos(u) 00 cos(u)][gh]E ′′DM j(mˆ) =[−2r sin(u)cos(u) − i jr sin(u)i jr sin(u) −ur− 1r sin(u)cos(u)]139For the j = 1 case:E ′′e1(mˆ)[uri sin(u)r]=[∂r(−∆ru+ sin(2u)2r2 )i cos(u)r (−∆ru+ sin(2u)2r2 )]E ′′a1(mˆ)[uri sin(u)r]=[∂r(cos(u)sin(u))i cos(u)r (cos(u)sin(u))]E ′′z1(mˆ)[uri sin(u)r]=[∂r(sin(u))i cos(u)r sin(u)]E ′′DM1(mˆ)[uri sin(u)r]=[∂r(−1r sin2(u))i cos(u)r (−1r sin2(u))]By the Euler-Lagrange equation (2.1.7), we see[uri sinurr]∈ ker(E ′′1 ).As we have , ∂1mˆ= cos(θ)ur~e− sin(u)sin(θ)r J~e, and ∂2mˆ= sin(θ)ur~e+ sin(u)cos(θ)r J~e,we see that∂1mˆ+ i∂2mˆ = (ur~e+ isin(u)rJ~e)eiθ . (3.6.2)The fact that the spatial derivative ∂1mˆ+ i∂2mˆ is in the kernel of E ′′(mˆ) is becauseof the fact that the total energy is translation invariant. Similarly, ∂2mˆ+ i∂1mˆ ∈ker(E ′′(mˆ)) implies that [iursinur]∈ ker(E ′′−1).140Chapter 4Directions for Future ResearchIn this Chapter, we collect some suggestions for future problems related to Chap-ters 2 and 3.4.1 Asymptotic stabilityWe obtained the orbital stability in Theorem 3 for the Landau-Lifshitz equationwith dissipationProblem 1. Do we have asymptotic stability of the skyrmion solution, sˆ? To bemore precise, if the initial data mˆ(0) is close enough to the sˆ, does the solutionmˆ(t) converge to sˆ as t→ ∞, up to some translation in an appropriate norm?If we writemˆ(x, t) = sˆ(x−X(t))+~η(x, t)where X(t) is the space displacement depending on time t, then the orbital stabilitysays ‖~η(·,0)‖H1 < δ , =⇒ ‖~η‖H1 ≤ ε for all t. We would like to know:Question 1. ‖~η‖H1 → 0 as t→ ∞?andQuestion 2. X(t)→ X∞ ∈ R2?141These do not seem to follow immediately from the energy dissipation relation.If the skyrmion solution could drift to infinity, which is saying |X(t)| → ∞, oroscillate between the two positions, this can cause failure of asymptotic stability.Problem 2. Is the co-rotational minimizer the global minimizer (with N=1)?There are classical methods for showing minimizers are symmetric such as“symmetrization of minimizing sequences”, the “moving planes method” [18],and its more recent refinements such as [39] . But none of these apply to prob-lems involving a vector-valued map with point-wise constraint, like the skyrmionvariational problem.Most of our results require small chiral interaction k 1. We encounter someopen problems relating to the strength of chiral term k:Problem 3. 1. Is the energy Ek,α,1 bounded below for 1 < |k| ≤ 4− 32 α? Dowe have a Co-rotational minimizer in this case ?2. For n = 1 and −1 < k < 0, is e(1)k,α,1 = 2 or < 2?3. Is the minimizing profile v(r) monotone for a wide range of k?Note that (2.1.17), (2.1.18), and (2.1.19) still leave open some questions aboutthe minimal energy e(n)k,α,1. We are not sure |k| ≤ 1 is necessary for boundednessfrom below the energy, and construct counter examples in (2.1.18) for |k| ≥ 4−32α . The monotonicity of the profile is closely related to stability arguments: thestability argument in [37] relies on the monotonicity of the profile.4.2 Dynamics of Schro¨dinger map flowProblem 4. Orbital stability, or asymptotic stability of the skyrmion under Schro¨dingerevolution with (1.6.1) b = 0.The only obstacle to orbital stability for the Schro¨dinger case is a local ex-istence theory. If we had a local existence theory in the energy space, orbital142stability should follow from the same argument as in the dissipative case becausewe still have energy conservation.In the pure Schro¨dinger case ((1.6.2), b = 0), local well-possedness resultshave been studied in [40, 51] for general case, [2] for the small data in energyspace, and [22] and [50] for the equivariant maps with energy close to the har-monic map energy. There is so far no local well-posedness result including thechiral term.Asymptotic stability in the Schro¨dinger case is a much harder problem. [24],which shows asymptotic stability of harmonic maps for the equivariant (pure)Schro¨dinger flow provides a possible strategy. The problems for our cases areno local well-posedness theory, and the derived equation (for the “q”) see (4.2.1)below is more complicated.We consider the conservative Landau-Lifshitz equation with energy E = Ee+kβ EDMfor equivariant mapts mˆ(r,θ , t) = eθR~v(r, t). We first show that E ′DM is also equiv-ariant. By Lemma 1.3.1EDM(e−αR(mˆ(eαR˜·)) = EDM(mˆ) =⇒ E ′DM(eαRmˆ(eαR˜·)) = (eαRE ′DM(mˆ))(eαR˜·)so we can write E ′DM(mˆ)(r,θ) = eθR~w(r).~w = e−θRPmˆ∇× (eθR~v) = P~ve−θR∇× (eθR~v), where Pmˆ denotes the orthogonalprojection on mˆ. Now∇× eθR~v = sin(θ)v′3−cos(θ)v′3v′2+v2r ,Soe−θR(∇× eθR~v) = cos(θ) sin(θ) 0−sin(θ) cos(θ) 00 0 1 sin(θ)v′3−cos(θ)v′3v′2+v2r= 0−v′3v′2+v2r .143We writeq := qˆ1+ iqˆ2, and vr = qˆ1(r)~e+ qˆ2(r)J~e = q◦ eˆ.the coordinates ~e and J~e defined as (3.6.1), and convert the equation as in [24].Applying the covariant derivative Dr on P~ve−θR∇× (eθ Rˆ~v(r)),Pv 0−v′3v′2= PvΓ~Vr = Pv[q1Γeˆ+q2ΓJeˆ] = q1αJeˆ+q2(−α) = iαq◦ eˆwhere Γ=0 0 00 0 −10 1 0 , eˆ=~e+iJ~e, and a◦b, denotes as Re(a)Re(b)+Im(a)Im(b).Then DrP~v 0−v′3v′2= i(αq)r ◦ eˆ = i(αrq+αqr)◦ eˆ, and α is determined byΓ~v = µ1~e+µ2J~eΓ~e =−µ1~v+αJ~eΓJ~e =−µ2~v−α~e.The end result is a nonlinear Schro¨dinger-type equation of the formiqt = (−∆r +V )q±|q|2q+ ikβ (αr + iα∂r)q, (4.2.1)which could be the starting point for the analysis of this problem.144Bibliography[1] I. Bejenaru, A. D. Ionescu, C. E. Kenig, and D. Tataru. 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