the Maxwell–Chern–Simons model Uniqueness of topological multivortex solutions in Journal of Functional Analysis

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Journal of Functional Analysis

www.elsevier.com/locate/jfa

Uniqueness of topological multivortex solutions in the Maxwell–Chern–Simons model

Zhi-You Chen^a,1, Jann-Long Chern^b,c,∗,2

aDepartmentofMathematics,NationalChanghuaUniversityofEducation, Changhua500,Taiwan

bDepartmentofMathematics,NationalCentralUniversity,Chung-Li32001, Taiwan

cNationalCenterforTheoreticalSciencesofTaiwan,Taiwan

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received29March2014 Accepted27January2016 Availableonline5February2016 CommunicatedbyF.Otto

MSC:

primary35J47,35J60 secondary35A02

Keywords:

Maxwell–Chern–Simonsmodel Multivortexsolutions

Fluxfornon-topologicalsolutions Classificationofsolutions

Inthispaper,weprovetheuniquenessoftopologicalmulti- vortexsolutionsfortheself-dualMaxwell–Chern–SimonsU(1) modeliftheChern–Simonscouplingparameterissufficiently largeandthechargeofelectronissufficientlysmallorlarge.

Ontheotherhand,wealsoestablishthesharpregionofthe fluxfornon-topologicalsolutionsandprovidetheclassification ofradialsolutionsofalltypesinthecaseofonevortexpoint.

© 2016ElsevierInc.All rights reserved.

* Correspondingauthorat: DepartmentofMathematics,National CentralUniversity,Chung-Li32001, Taiwan.

E-mailaddresses:[email protected](Z.-Y. Chen),[email protected](J.-L. Chern).

1 Work partially supported by Ministry of Science and Technology, Taiwan under Grant MOST- 103-2115-M-018-002-MY3.

2 Work partially supported by Ministry of Science and Technology, Taiwan under Grant MOST- 101-2115-M-008-009-MY3.

http://dx.doi.org/10.1016/j.jfa.2016.01.024 0022-1236/© 2016ElsevierInc.All rights reserved.

1. Introduction

Inthisarticle,we studythesolutionsof thefollowing coupledequations

⎧⎪

⎪⎨

⎪⎪

⎩

Δu= 2q

e^u−1−κv + 4π

i=1

n_iδ_pi inR², Δv=−κq²

e^u−1−κv

+ 2qe^uv inR²,

(1.1)

where Δ=

2 i=1

∂²

∂x²_i,n1,. . . ,n,q,κarepositiveconstantsand≥0 iscalledthenumber ofvortexpointsandδ_pistheDiracmeasureatp.Inparticular,if= 0,thenSystem(1.1) hasnovortices involved.System(1.1)arisesfrom theself-dualMaxwell–Chern–Simons U(1) modelon(2+1)-dimensionalMinkowskispaceR^2,1withthemetricdiag(1,−1,−1).

Wereferthereadersto[5,16,17,19]andthereferencestherein.Herewebrieflyintroduce theself-dualMaxwell–Chern–SimonsU(1) modelasfollows.

Theself-dualMaxwell–Chern–SimonsU(1) modelisdefinedbytheLagrangian:

L(A, φ, N) =

(D^q)αφ

(D^q)^αφ

−1

4FαβF^αβ+μ

4ε^αβγAαFβγ+1

2∂αN ∂^αN

−V(|φ|, N) (1.2)

with self-dualpotential

V(|φ|, N) =q²|φ|²N+1 2

q|φ|²+μN−q2

,

where μ>0 isthe Chern–Simonscoupling parameter,q isthe chargeof electron, ^{ςτ ρ} is a totally skew-symmetric tensor with ⁰¹² = 1, A = (A₁,A₂), A_α : R×R² → R is the gaugefield, α = 0,1,2,φ : R×R² → C is the Higgs field,N : R×R² → R is the neutralscalarfield,Fαβ =∂αAβ−∂βAα is thefieldstrength, andthe covariant derivativeDAis definedas follows:

D_Aφ=

(D^q)₁φ,(D^q)₂φ

=∇φ−iqAφ, i=√

−1.

We saythat(φ,Aα) is gaugeequivalent to (ψ,Bα), ifthere exists afunctionχ such that

(ψ, Bα) = (e^iχφ, Aα+∂αχ).

ThentheLanditsEuler–Lagrangeequationsareinvariantunderthegaugetransforma- tion.

We consider thestationary solutionsof theEuler–Lagrangeequations for(1.2). The Gauss lawconstraintobtainedfrom thevariationofA0 isgivenby

μFA= ΔA0−2q²|φ|²A0 in R², whereFA=∂1A2−∂2A1isthemagneticfield.

Inthisarticle,westudythestaticconfigurationforself-dualMaxwell–Chern–Simons U(1) model,andwecanwrite theenergyfunctionalas

Eq,μ(A, φ, N) =

R²

D^q₁φ±iD₂^qφ²+q²|φ|²|A0±N|²+1

2|∇A0± ∇N|²

+1

2FA±

q|φ|²+μN−q² dx±q

R²

FAdx.

Thus the field configurations saturating the energy bounded by Eq,μ(A,φ,N) =

±q

R²F_AdxsatisfytheGauss lawconstraintandfollowingself-dualequations,

⎧⎪

⎨

⎪⎩

D^q₁φ±iD₂^qφ= 0, A₀±N = 0, FA±

q|φ|²+μN−q

= 0.

(1.3)

If(A,φ,N) isafinite energysolutionof(1.3),then |φ|andN satisfy q|φ|²+μN−q

→0 as |x| → ∞.

Inthispaper,weareinterestedinlookingformultivortexsolutionssothatφvanishes atp_i withcorrespondingorders n_i, i.e.,

|φ(y)|=|y−pi|ⁿⁱ fory nearpi, i= 1,· · ·, .

Withoutlossofgenerality,wewillonlytaketheupper(plus)signintoaccount.Theflux correspondingtothemagneticfieldF_AisgivenbyΦ=q

R²F_Adx,whereF_A=q|φ|²+ μN−q.Withthesubstitution(u,v,κ,q)=

ln|φ|²,−κq³²N,^√_q2^μ,q²

,orequivalently,

φ(x) = exp u

2+ i

i=1

n_iArg(x−p_i)

and N(x) =−κ⁻¹q⁻³²v(x),

where x = (x1,x2) ∈ R², (1.3) can be reduced to (1.1). Conversely, once we find a solution(u,v) of(1.1),wemayrecoverA andN from (1.3)bytheformula

μ=κ²q², q=√

q, qA₁+iqA₂=−2i∂¯lnφ and N =− v q√

μ, where∂¯= ∂₁+i∂₂

2 .

For(1.1),there arenaturalboundaryconditionsforsolutionsat infinity,namely,

|xlim|→∞u(x) = 0, lim

|x|→∞v(x) = 0, (1.4)

|xlim|→∞u(x) =−∞, lim

|x|→∞v(x) =−1

κ. (1.5)

We note that if (u,v) is a solution with the boundary condition (1.4), then, by the maximum principle, we have u(x),v(x) < 0 on R²\{p₁,. . . ,p}. In physics literature, a solution (u,v) satisfying boundary condition (1.4) is called the topological solution.

Correspondingly, a solution (u,v) is called the non-topological solution if it satisfies boundary condition (1.5). Note that one can also consider (1.3) on the ’t Hooft type periodicdomain,ofwhichsolutionsarecallcondensatesolutions.Therehavebeenseveral resultsforcondensatesolutionsobtainedin[5,18,19].

From[4],weobtaintheexistenceoftopologicalsolutionsto(1.1)forallκ,q >0.Then it is naturalto askthe questionabout theuniquenessof topologicalsolutions to (1.1).

Forthecaseofκ= 0,q=¹₂ andv≡0,(1.1)reducesto

Δu=e^u−1 + 4π i=1

n_iδ_pi inR², (1.6)

whichistheself-dualequationoftheAbelian–Higgsmodelandtheuniquenessoftopo- logicalsolutionsof(1.6)wasprovedin[21].Similarly,ifwesetμ=κ²q,v=κ⁻¹

e^u−1 , q→ ∞,thenby[5],(1.1)with(1.4)tendsformallyto

Δu= 4

κ²e^u(e^u−1) + 4π i=1

n_iδ_pi inR², (1.7) whichistheself-dualequationoftheChern–Simons–Higgsmodel.For(1.7),theunique- nessoftopologicalsolutionswasobtainedin[12]and[8].Now,westateourmain result abouttheuniquenessoftopologicalsolutionsof(1.1)asfollows.

Theorem 1.1. Letp1,. . . ,p ∈R² and n1,. . . ,n∈R⁺ be given. Thenthere existthree positive constants κ0 = κ0(pi,ni), q0 = q0(pi,ni) and q1 = q1(pi,ni) < q0 such that (1.1)possesses oneandonlyonetopological solutionin R² forany (κ,q)∈

(κ,q): 0<

q < q₁ orq > q₀ or κ > κ₀

. Moreover, if = 0 or 1, then (1.1) possesses a unique topological solution inR² forany κ,q >0.

Moreover, we alsodiscuss the existenceof solutionsto (1.1)satisfying(1.5). In[15], wegetthatif= 0,1,then(1.1)possessesaradiallysymmetricnon-topologicalsolution inR²foranyκ,q >0.Inaddition,forany(,κ,q)∈Z⁺×(0,∞)×(0,∞),wealsohave that there exist infinitely many non-topological solutionsof (1.1) inR² and their flux

Φ∈ 4π

i=1

n_i+ 1 ,4π

i=1

n_i+ 1 +ε₀

for someε₀ =ε₀(n_i,p_i)> 0,i = 1,. . . ,, whichisderivedin[3].Hence,there isan intersectingproblem aboutfinding thesharp rangeof fluxΦ forall non-topologicalsolutions of(1.1). Wenote thatthere havebeen many well-knownresults about such aproblem for Chern–Simons limit equation (1.7) in,forexample,[2,6,13].

To answer the problem about the range of flux for (1.1), we consider the case of = 0,1 andsetp1 tobetheoriginO.Then(1.1)reducesto thefollowing equations

Δu= 2q

e^u−1−κv

+ 4πN δO inR², Δv=−κq²

e^u−1−κv

+ 2qe^uv inR²,

(1.8)

whereN ≥0.Wegiveaconsequenceabouttherangeoffluxfor(1.8)mentionedbelow.

Theorem1.2.If(u(x),v(x))isanon-topologicalsolutionof (1.8),thenΘ1(u,v)∈(4N+ 4,∞)andΘ2(u,v)= 0,where

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

Θ1(u, v) = q π

R²

1 +κv(x)−e^u(x) dx,

Θ2(u, v) = 1 2π

R²

κq²

e^u(x)−κv(x)−1

−2qe^u(x)v(x) dx.

Moreover, (1.8)possesses a non-topologicalsolution (u(x),v(x))such that Θ₁(u,v)=θ andΘ2(u,v)= 0 forany θ∈(4N+ 4,∞).

From[3,4,15],weknowthat(1.8)possessatleasttwodifferenttypesofsolutionsand hencewewanttoask:Howmanydifferenttypesofsolutionsof(1.8)arethere?Inorder tosolvethisproblem,wewillinvestigatethestructureofallradialsolutionsof(1.8)and considerthefollowingODE system

⎧⎪

⎨

⎪⎩

u(r) +1

ru(r) = 2q

e^u−1−κv

for r >0, v(r) +1

rv(r) =−κq²

e^u−1−κv

+ 2qe^uv for r >0

(1.9)

withtheinitialvalue

u(r) = 2Nlnr+α1+o(1), v(r) =α2+o(1)

asr→0⁺. (1.10)

Accordingtothebehaviorsat∞,allsolutionsof(1.9)definedin(0,∞) canbeclassified intothefollowingfour types.

Type (I): lim

r→∞(u(r),v(r))= (0,0).

Type (II): lim

r→∞(u(r),v(r))= (−∞,−_κ¹).

Type (III): lim

r→∞(u(r),v(r))= (−∞,∞).

Type (IV): lim

r→∞(u(r),v(r))= (∞,∞).

Exceptfourtypesofentiresolutionstatedabove,weseethatothersolutionsmustblow upat finiteR >0.Indeed,suchsolutionsadmit threetypes,i.e.,

Type (V): lim

r→R⁻(u(r),v(r))= (∞,∞).

Type (VI): lim

r→R⁻(u(r),v(r))= (∞,−∞).

Type (VII): lim

r→R⁻(u(r),v(r))= (∞,^κq₂).

Let α= (α1,α2)∈ R² and (u(r,α),v(r,α)) denote thesolution of(1.9)–(1.10). Ac- cording to the behavior of (u,v), the set of initial data could be classified into the following regions.

T ={α|(u(r,α),v(r,α)) is a Type (I) solution}, Λ={α|(u(r,α),v(r,α)) is a Type (II) solution}, B∞={α|(u(r,α),v(r,α)) is a Type (IV) solution}, B={α|(u(r,α),v(r,α)) is a Type (V) or (VII) solution}, Bv={α|(u(r,α),v(r,α)) is a Type (III) solution}, Bu={α|(u(r,α),v(r,α)) is a Type (VI) solution}.

Then thestructureofsolutionssetsaredescribedas follows.

Theorem 1.3.The followingstatements arevalid.

(i) R²=Bv Λ

T Bu

B B∞.

(ii) Thereexistapoint(α⁰₁,α₂⁰)∈R² andastrictly increasingfunctionγ: (−∞,∞)→ (−∞,∞)suchthat lim

α1→−∞γ(α1)=−_κ¹,γ(α⁰₁)=α⁰₂ and

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎩

T ={(α⁰₁, α⁰₂)}, Λ ={(α1, α2) :α1< α⁰₁, α2=γ(α1)}, {(α1, α2) :α1∈(−∞, α⁰₁), γ(α1)< α2<∞} ⊇ Bv, {(α1, α2) :α1>−∞,−∞< α2< γ(α1)} ⊇ Bu, {(α1, α2) :α1∈(α⁰₁,∞), α2=γ(α1)} ⊆ B

B∞.

Now we keep ondiscussing entire solutionsof (1.9)–(1.10) definedfor allr >0.Let (u(r),v(r))= (u(r;α1,α2),v(r;α1,α2)) beasolutionof(1.9)–(1.10).Wedefine

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎩

β1(α1, α2) = 2q ∞ 0

r

1 +κv(r)−e^u(r) dr,

β2(α1, α2) = ∞ 0

r κq²

e^u(r)−1−κv(r)

−2qe^u(r)v(r) dr,

(1.11)

and sometimes denote it by (β₁,β₂) if no confusion arises. Here we call β₁(u,v) and β₂(u,v) theu-flux andv-flux withrespect tosolution(u,v),respectively.

Remark 1.1. Let (u(r),v(r)) be an entire (1.9)–(1.10). Then we obtain the following statementsfrom (1.9)–(1.11).

(i) (u(r;α₁,α₂),v(r;α₁,α₂)) is asolution of Type (I) ifand onlyif β₁(α₁,α₂) = 2N andβ2(α1,α2)= 0.

(ii) (u(r;α1,α2),v(r;α1,α2)) is a solution of Type (II) if and only if 4N + 4 <

β₁(α₁,α₂)<∞andβ₂(α₁,α₂)= 0.

(iii) (u(r;α₁,α₂),v(r;α₁,α₂)) is asolutionof Type(III) ifand onlyif β₁(α₁,α₂)=∞ andβ2(α1,α2)=−∞.

(iv) If α1 < α₁⁰, then lim

α1→α⁰1

β1(α1,γ(α1)) = ∞, lim

α1→−∞β1(α1,γ(α1)) = 4N + 4 and β2(α1,γ(α1))= 0.

This paper is organized as follows. First we investigate the monotone and non- degeneracypropertiesofthelinearizedequationsatthesolutionsof(1.9)inSection2.We applytheImplicitFunctionTheoremtoprovetheuniquenessofthetopologicalsolutions for(1.1)withatmostonevortexpointinSection3.Theexistencesandclassificationof radialsolutionsof alltypeswill be showninSection4. InSection5,we findthe sharp rangeof fluxΦ for all non-topological solutionsof (1.8). Finally, theuniquenessof the topological multivortexsolutionsfor (1.1)will bederived inSection6. Roughlyspeak- ing,Sections 1–5aredevoted todealing with(1.1)with atmost onevortex point. The situationfor multivortex pointswill be treatedby applyingthe consequencesobtained fromthecasewithatmostonevortexpoint.

2. Linearizedequations

In this section, we give the proof about the non-degeneracy of linearized equations on thesolution of Type(I) to (1.8). Before going to our proof, we need to statesome propertiesconcerningsolutions.First,wehavethefollowingpropertyforsolutionswith zeroboundaryvalues.

Lemma 2.1. Let(u(r),v(r))be a solution of (1.9)–(1.10) satisfying u(R₀)=v(R₀)= 0 forsomeR0>0 (orR0= +∞).Thenthefollowingarevalid.

(i) u(r) < 0, −_κ¹ ≤ v(r) < 0, u(r) > 0 and v(r) > 0 on (0,R0). Furthermore, if R0=∞,i.e.,(u,v)isatopologicalsolutionof (1.8),thenthecorresponding(β1,β2) satisfiesβ1= 2N andβ2= 0.

(ii) 1+κv(r)−e^u(r)≥0on [0,R₀].

Proof. We note thatr = 0 is not anon-negative local maximum point of v(r) due to Δv(0)=κq²

1+κv(0)

.Toproveit,wedividetheproofinto twosteps.

Step 1. Both u(r)and v(r) do not attain non-negative local maximum on (0,R₀). Fur- thermore, one of u(r)and v(r)is negativeon (0,R0).Onthe contrary,withoutloss of generality,wemayassumethatr¹_uandr_v¹arebothnon-negativemaximumpointsofu(r) and v(r) on(0,R0),respectively.Thenwesee

−κv(r_u¹)−1 +e^u(r1u)≤0 and κq²

−κv(r_v¹)−1 +e^u(r1v)

≥2qe^u(r1v)v(r_v¹)≥0, whichcontradicts−κv(r¹_u)−1+e^u(r1u)≥ −κv(r_v¹)−1+e^u(r1v)whenv(r_v¹)>0.Ifv(r¹_v)= 0, thenv(r¹_u)<0 ande^u(r1u)−1<0,whichcontradictsu(r¹_u)≥0.Therefore,wefinishthis step.

Step2. Bothu(r)andv(r)arenegativeon(0,R0).Moreover,−¹_κ ≤v(r)<0on[0,R0).

Onthecontrary,wefirstassumethatu(r) possessesalocalmaximumpointru∈(0,R0) suchthatu(r_u)≥0 and−κv(r_u)−1+e^u(ru)≤0.Thenwehaveκv(r_u)≥e^u(ru)−1≥0, whichcontradictsStep 1.Ifv(r) possessesalocalmaximumpointrv∈(0,R0) suchthat

v(rv)≥0 and κq²

e^u(rv)−1

≥

κ²q²+ 2qe^u(rv) v(rv), then u(r_v)≥0.However,itisimpossiblebyStep1.Wecompletethisstep.

Moreover, we can see that one of u(r) and v(r) does not attain non-positive local minimum on(0,R0) bythesimilarproofs inStep 1andStep 2.Thusu(r) andv(r) are both strictly increasing on (0,R0) and tend to zero as r → R0. Furthermore, we have v(0)≥ −¹_κ;otherwise,Δv(0)<0 and v(R₀)= 0,whichisincontradiction tov(r)>0 on(0,R0).Consequently,−¹_κ ≤v(r)<0 on [0,R0).

LetR0=∞. Combiningwith(1.9)andStep2,weget

⎧⎪

⎪⎨

⎪⎪

⎩

Δu(r)≥2qu(r) on [0,∞),

Δv(r)≥(κ²q+ 2)qv(r) on [0,∞),

r→∞lim(u(r), v(r)) = (0,0).

It follows that |u(r)| = O(e^−√2qr) and |v(r)| = O(e^−κ2q2+2qr) as r → ∞, which imply 1+κv(r)−e^u(r) ∈ L¹([0,∞)) ande^u(r)v(r) ∈L¹([0,∞)).Then we obtain that

rlim→∞ru(r)= lim

r→∞rv(r)= 0,thatis,β₁= 2N andβ₂= 0.Hence(i)holds.

Nowwewanttoshowthat−κv(r)−1+e^u(r)≤0on[0,R0].Onthecontrary,wemay assumethatthere existthreeconstants r0,r1,r2∈(0,R0] suchthatr0< r1< r2 and

⎧⎪

⎨

⎪⎩

−κv(r0)−1 +e^u(r0)=−κv(r2)−1 +e^u(r2)= 0,

−κv(r)−1 +e^u(r)>0 on (r₀, r₂),

−κv(r1) +e^u(r1)u(r1) = 0

(2.1)

fromu(R0)=v(R0)= 0.Itiseasyto seethat ru(r)

= 2q

−κv(r)−1 +e^u(r)

≥0 on (r0, r2), rv(r)

=−κq²

−κv(r)−1 +e^u(r)

+ 2qe^u(r)v(r)≤0 on (r0, r2), whichimplies

−κv(r) +e^u(r)u(r)≥

−κv(r₁) +e^u(r1)u(r₁)r₁

r >0 on [r₁, r₂] by

e^u(r)

>0 on (0,R₀). Then it follows that −κv(r₂)−1+e^u(r2) ≥ −κv(r₁)−1+ e^u(r1)>0,whichcontradicts(2.1).Hencetheresult(ii)is proved. 2

Secondly,wehavethefollowinguseful identity.

Lemma 2.2. Let (u(r),v(r)) be a solution of (1.9)–(1.10) in (0,R] for some R > 0 or R=∞.Thenwe havethefollowingidentity

κq 4

ru(r)2

+r²uv(r) + 2qr²

v(r) +κv²(r)

2 −e^u(r)v(r)

=κqN²+ 2q r

0

t

κv(t) + 2−2e^u(t)

v(t)dt (2.2)

forr∈(0,R].

Proof. Bymultiplying ^κq₂ru,rvandruonbothsidesofthefirstandsecondequations of(1.9),andusingtheinitial value(1.10),wecaneasily obtain(2.2). 2

In the following, we will apply the similar method used in [11,9,10] to prove the non-degeneracypropertiesofthelinearizedequations.Firstweinvestigatethemonotone property of solutions of (1.9)–(1.10) and let, for i = 1,2, φi(r)= ∂U

∂αi

(r) and ψi(r) =

∂v

∂α_i(r), where U(r) = u(r)−2Nlnr. Then (φ_i,ψ_i), i = 1,2, satisfy the linearized equations

⎧⎪

⎪⎨

⎪⎪

⎩

Δφ_i−2qe^uφ_i =−2κqψ_i on [0, R₀), Δψi−

κ²q²+ 2qe^u

ψi=−qe^u

κq−2v

φi on [0, R0),

φ1(0) = 1 =ψ2(0), φ2(0) = 0 =ψ1(0), φ_i(0) = 0 =ψ_i(0), i= 1,2,

(2.3)

where u(r) andv(r) are both finite on [0,R0). Themonotone property of φi and ψi is as follows.

Lemma2.3.If(u(r),v(r))beasolutionof (1.9)–(1.10)on[0,Rv),thenthecorresponding (φi(r),ψi(r))satisfies

φ1(r)>1, φ₁(r)>0, φ2(r)<0, φ₂(r)<0, ψ1(r)<0, ψ₁(r)<0, ψ2(r)>1, ψ₂(r)>0

(2.4)

on (0,R_v), whereR_v = sup

R:v(r)≤0on (0,R]

.Moreover, if R_v =∞,thenφ₁(r), ψ₁(r),φ₂(r) andψ₂(r) areallunbounded on[0,∞).

Proof. Sinceφ1(0)= 1,ψ1(0)= 0 and(2.3),wehave φ₁(r)>0 and Δψ₁(r)−

κ²q²+ 2qe^u(r)

ψ₁(r)≤0 on [0, r₀]

for somer0∈(0,Rv),and henceψ1(r)<0 on(0,r0] bythemaximum principle.From (2.3),we alsoget

rψ₁(r) = r 0

s

κ²q²+ 2qe^u(s)

ψ1(s)−qe^u(s)

κq−2v(s) φ1(s)

ds <0 on (0, r0] (2.5) and

φ₁(r)>1 and rφ₁(r) = 2q r 0

s

e^u(s)φ₁(s)−κψ₁(s)

ds >0 on (0, r₀]. (2.6)

Since the inequalities in (2.5) and (2.6) hold whenever φ1(r) > 0 and ψ1(r) < 0, we deduce that φ1(r) > 1, φ₁(r) > 0 and ψ1(r) < 0, ψ₁(r) < 0 for all r ∈ (0,Rv]. The situationsforφ2(r) andψ2(r) aresimilar,and wecompletethisproof. 2

Finally,westateandprovethenon-degeneracypropertyofthelinearizedequationat atopologicalsolutioninthefollowing lemma.

Lemma2.4.Let(u(r),v(r))beasolutionof (1.9)–(1.10)satisfyingu(R₀)= 0,v(R₀)= 0 forsomeR0>0 (orR0= +∞).If (φi(r),ψi(r)),i= 1,2,istherespective solutionpair

of (2.3),thenthecorrespondinglinearizedequation (2.7)isnon-degenerate,thatis,there doesnotexistanonzerobounded solutionpair(Φ(r),Ψ(r))of

ΔΦ−2qe^uΦ + 2κqΨ = 0 on [0, R₀), ΔΨ−

κ²q²+ 2qe^u

Ψ +qe^u

κq−2v

Φ = 0 on [0, R₀). (2.7) Proof. Let

MΦ(r) =−φ₁(r)

φ2(r) andMΨ(r) =−ψ₁(r) ψ2(r).

Thenwe see thatifMΦ(r)> MΨ(r) on(0,r0) for somer0≤R0, thenM_Φ(r)<0 and M_Ψ(r)>0 on (0,r₀).Toprovethislemmaweneedthefollowingthefact.

Claim.Theredoesnot existR∈(0,R₀)suchthat M_Φ(R)=M_Ψ(R).

ProofofClaim. Weprovethisclaimbyacontradictionandhencethereexistsasmallest R∈(0,R₀] suchthatM_Φ(R)=M_Ψ(R)(≡C) andM_Φ(r)> M_Ψ(r)>0∀r∈(0,R).Let Φc(r)=ϕ1(r)+cϕ2(r) andΨc(r)=ψ1(r)+cψ2(r) foranyc∈R.ThenΦc(r) andΨc(r) satisfy(2.7)anditfollowsthat

⎧⎪

⎪⎨

⎪⎪

⎩

Φ_C(r)>0,Ψ_C(r)>0∀r∈(0, R), ΦC(R) = ΨC(R) = 0,

Φ_C(R)<0,Ψ_C(R)<0 ifR <∞.

(2.8)

Taking thedifferentiation with respect to αi, i = 1,2, onthe both sides of (2.2), then foranyc>0 andr∈(0,R₀],weobtain

r²κq

2 Φ_c(r)u(r) + Φ_c(r)v(r) + Ψ_c(r)u(r)

+ 2qr²

κv(r) + 1−e^u(r) Ψc(r)

= 2qr²v(r)e^u(r)Φc(r) + 4q r

0

t

κv(t) + 1−e^u(t)

Ψc(t)−2e^u(t)v(t)Φc(t)

dt. (2.9)

If R < ∞ then, by replacing c and r with C and R in (2.9)respectively, we easily have

R² κq

2 Φ_c(R)u(R) + Φ_c(R)v(R) + Ψ_c(R)u(R)

= 4q R

0

t

κv(t) + 1−e^u(t)

Ψc(t)−2e^u(t)v(t)Φc(t)

dt. (2.10)

Then, combiningwithLemma 2.1,(2.8)and(2.10),we getthat 0> R²κq

2Φ_c(R)u(R) + Φ_c(R)v(R) + Ψ_c(R)u(R)

= 4q R

0

t

κv(t) + 1−e^u(t)

Ψ_c(t)−2e^u(t)v(t)Φ_c(t) dt >0,

whichyieldsacontradiction.Thisshowstheclaim.Therefore,weeasilyobtainM_Φ(r)>

MΨ(r)>0 on[0,R0] (resp.,[0,∞) ifR0=∞)andM_Φ(r)<0,M_Ψ(r)>0 on (0,R0).

Now weshowthatif R0=∞,thenforany C >0,

either ΦC(r)or ΨC(r)is unbounded on [0,∞). (†) Suppose that (†) is not true. Then Φ_C(r) and Ψ_C(r) are bounded on [0,∞) for some C >0. Note that M_Φ(r) (resp., M_Ψ(r)) is strictly decreasing (resp., increasing) to

r→∞lim MΦ(r) (resp., lim

r→∞MΨ(r)) and lim

r→∞MΨ(r) ≤ lim

r→∞MΦ(r) by maximum principle and Claim,respectively.Hence,we haveC∈

r→∞lim M_Ψ(r), lim

r→∞M_Φ(r)

.Indeed,onthe contrary, without loss of generality, we may assume that 0 < C < lim

r→∞MΨ(r). Then we obtain thatΦC(r)>0 on [0,∞) andΨC(r) has only oneroot r_∗ on (0,∞),which impliesthat lim

r→∞Ψ_C(r)=−∞from(2.7).Ityieldsacontradiction.

It followsfromC∈

rlim→∞MΨ(r), lim

r→∞MΦ(r) that

Φ_C(r)>0 and Ψ_C(r)>0 on (0,∞). (2.11) Since |u| and|v|decayexponentiallyat∞, wegetthat

r→∞lim ru(r) = 0 = lim

r→∞rv(r) and

rlim→∞r²Φ_C(r)u(r) = 0 = lim

r→∞r²Ψ_C(r)v(r). (2.12) Moreover,

rlim→∞r²

κv(r) + 1−e^u(r)

= 0 = lim

r→∞r²v(r)e^u(r). (2.13) Hence,byvirtueof(2.10)–(2.13),weobtain

0 = 4q ∞ 0

t

κv(t) + 1−e^u(t)

Ψc(t)−2e^u(t)v(t)Φc(t)

dt >0.

This contradictionshowsthatΦC orΨC isunbounded,and (†)isproved.

Finally,let(u,v) beatopologicalsolutionof(1.8).Thenanysolutionpair(Φ(r),Ψ(r)) ofthelinearizedequations(2.7)canbewrittenas

Φ(r) =c1φ1(r) +c2ψ1(r) and Ψ(r) =c1φ2(r) +c2ψ2(r) for somec1, c2∈R.

By (†), we easily obtain the non-degeneracy result if c = c₂/c₁ > 0. When c ≤ 0 or c₁= 0, byLemma 2.3,wealsogetthatbothΦ_c(r) andΨ_c(r) areunboundedon[0,∞).

Theproofofthistheorem iscompleted. 2

3. Uniquenessof topologicalsolutionswithonevortex point

Inthissection,wewilluseLemma 2.4,ImplicitFunctionTheoremandacontinuation argument to establish the existence and uniqueness of topological solutions with one vortex point which, without loss of generality, maybe assumed to be the origin point O= (0,0).First,wewillgiveauniquenessresultforsmallenoughκ>0 asfollows.

Theorem 3.1. Letq > 0and N ≥0be fixed. Thenthere exists a constant κ^∗ >0 such that (1.8)possessesaunique topologicalsolution inR² forany 0< κ< κ^∗.

Proof. Let (u¹_κ(x),v¹_κ(x)) and (u²_κ(x),v²_κ(x)) be topological solutions of (1.8). Both (u¹_κ(x),v_κ¹(x)) and (u²_κ(x),v_κ²(x)) are radially symmetric with respect to O by [14]. It is easy to see thatboth (u¹_κ(x),v_κ¹(x)) and (u²_κ(x),v_κ²(x)) converge to (u^∗(x),0) point- wiseas κ→0,where

Δu^∗= 2q

e^u∗−1

+ 4πN δO inR². (3.1)

Since

R²

1−e^u∗(x)

dx= lim

κ→0

R²

1−e^uiκ(x)−κvⁱ_κ(x)

dx= 2πN q

fori= 1,2 dueto u^∗(x)→0 as|x|→ ∞, thatis,u^∗(x) isauniquetopologicalsolution of(3.1)andu^∗(x)=u^∗(r),wherer=|x|.

Without loss of generality, we may assume |(u¹_κ−u²_κ)(x_κ)| = u¹_κ −u²_κL^∞(R²) ≥ v¹_κ−v_κ²L^∞(R²) for all κ. Set Φ_κ = _u1^{(u1κ−u2κ)}

κ−u²κL∞(R2 ) and Ψ_κ = _u1^{(v1κ−v2κ)}

κ−u²κL∞(R2 ). Then (Φ_κ,Ψ_κ) satisfies

⎧⎨

⎩

ΔΦ_κ−2qe^ξκΦ_κ+ 2κqΨ_κ= 0 inR², ΔΨκ−

κ²q²+ 2qe^u1κ

Ψκ+qe^ξκ

κq−2v_κ²

Φκ= 0 inR²,

(3.2) where

ξκ(|x|)∈ min

u¹_κ(|x|), u²_κ(|x|) ,max

u¹_κ(|x|), u²_κ(|x|) .

First,weclaimthat{xκ}isbounded. LetP bea 2×2 matrixsuchthatP⁻¹MP = Λ, where

M=

−2q 2κq κq² −κ²q²−2q

, Λ =

λ₁ 0 0 λ₂

, (3.3)

and λ1,λ2 are distinct negative eigenvalue of M with λ1 > λ2. Introduce a new set variables Aκ, Bκ suchthat

Aκ

Bκ

=P⁻¹

Φκ

Ψκ

. From(3.2),weobtain

⎧⎪

⎨

⎪⎩

ΔAκ+λ1Aκ+a11(|x|)Aκ+a12(|x|)Bκ= 0 on [0,∞), ΔBκ+λ2Bκ+a21(|x|)Aκ+a22(|x|)Bκ= 0 on [0,∞), Aκ(|x|)→0, Bκ(|x|)→0 as |x| → ∞,

(3.4)

where

a11(|x|) a12(|x|) a21(|x|) a22(|x|)

=P⁻¹

2q

1−e^ξκ

0 qe^ξκ

κq−2v_κ²

−κq² 2q

1−e^u1κ

P.

Then wehave

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

Aκ(xκ) =−

R²

K1(xκ, y)

a11(|y|)Aκ(y) +a12(|y|)Bκ(y)

dy, Bκ(xκ) =−

R²

K2(xκ, y)

a21(|y|)Aκ(y) +a22(|y|)Bκ(y)

dy,

(3.5)

where 0 < Ki(x,y)≤ Ce^{−√−λi|x−y|} for |x−y| ≥1, i = 1,2. Since uⁱ_κ(|x|)→ u^∗(|x|), vⁱ_κ(|x|) →0 pointwise as κ →0 for i= 1,2 andu^∗(|x|) decay exponentiallyat ∞, by (3.5),we have

|Aκ(xκ)|+|Bκ(xκ)| ≤o(1)

AκL^∞+BκL^∞

as κ→0, whichcontradicts toΦκL^∞ = 1.Thuswecomplete thisclaim.

By claim,Φ_k and Ψ_k converges to Φ andΨ (passingto asubsequence ifnecessary) inC²([0,∞)),respectively,where (Φ(r),Ψ(r)) satisfies

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

Φ(r) +1

rΦ(r)−2qe^u∗Φ(r) = 0 on [0,∞), Ψ(r) +1

rΨ(r)−2qe^u∗Ψ(r) = 0 on [0,∞), Φ(0) =C₁, Φ(0) =C₂, Φ(0) = Ψ(0) = 0

forsome C1,C2 ∈ R. SinceΦ(r) and Ψ(r) are bounded on [0,∞), wehave Φ≡0 and Ψ≡0.Thentheproofof thistheoremiscompleted. 2

Lemma3.1.Suppose(u(r),v(r))isasolutionofType(I)of (1.9)withrespectto(κ,q, N).

Then

|U(r)| ≤max

2q ,2N

and |v(r)| ≤max

κq²+2q κ , κqN

forr∈[0,∞),whereu(r)=U(r)+ 2Nln_1+r^r22.

Proof. Without lossof generality,weonlyprovethecaseofU(r) and thecaseofv(r) issimilarly.Since(u,v) isa topologicalsolutionof(1.9)with(κ,q, N),we have

∞ 0

r

e^u(r)−1−κv(r)

dr=−N q and

∞ 0

re^u(r)v(r)dr=−κN

2 . (3.6)

CombiningwithLemma 2.1and(3.6),weobtain

−rU(r)≤

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

2qr² forr∈[0,1], 2q

r 0

t

1 +κv(t)−e^u(t)

dt forr∈(1,∞)

and

−rv(r)≤

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎩

κq²+^2q_κ

r² forr∈[0,1], κq²

r 0

t

e^u(t)−1−κv(t) dt−2q

r 0

te^u(t)v(t)dt forr∈(1,∞).

Thenwecompletethis result. 2

Proposition 3.1. Suppose (u(x),v(x)) is a topological solution of (1.8) with respect to (κ,q, N). Then the linearized operatorL : W_r^2,2(R²)×W_r^2,2(R²) to L²_r(R²)×L²_r(R²) definedby

L:=

Δ−2qe^u 2κq qe^u(κq−2v) Δ−

κ²q²+ 2qe^u

(3.7) is invertible and L⁻¹ is a linear bounded operator where L²_r(R²) =

z(x) = z(r)|z ∈ L²(R²)

andW_r^2,2(R²)={z(x)=z(r)|z,z,z∈L²(R²)}.