The solution structure of the O(3) sigma model in a Maxwell-Chern-Simons theory

theory

Sze-Guang Yang, Zhi-You Chen, and Jann-Long Chern

Citation: Journal of Mathematical Physics 58, 071503 (2017); doi: 10.1063/1.4994060 View online: http://dx.doi.org/10.1063/1.4994060

View Table of Contents: http://aip.scitation.org/toc/jmp/58/7 Published by the American Institute of Physics

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The solution structure of the O(3) sigma model in a Maxwell-Chern-Simons theory

Sze-Guang Yang,^1,a)Zhi-You Chen,^2,b)and Jann-Long Chern^1,c)

1Department of Mathematics, National Central University, Chung-Li 32001, Taiwan

2Department of Mathematics, National Changhua University of Education, Changhua 500, Taiwan

(Received 24 December 2015; accepted 2 July 2017; published online 14 July 2017)

In this paper, a system of semilinear elliptic equations arising from a relativistic self-dual Maxwell-Chern-Simons O(3) sigma model is considered. We reveal the uniqueness aspect of the topological solutions for the model. The uniqueness result is associated with a clear solution structure of the equations of the radially symmet- ric case. We locate each solution set denoted by a planar diagram.Published by AIP Publishing.[http://dx.doi.org/10.1063/1.4994060]

I. INTRODUCTION

In the paper, we consider the following system of semilinear elliptic equations which comes from the self-dual equations arising in the Maxwell-Chern-Simons gauged O(3) sigma model:



















∆U=2q

−N+s−1−e^U 1 + e^U

+ 4πdδ₀,

∆N=−κ²q²

−N+s−1− e^U 1 + e^U

+ 4q e^U (1 + e^U)²N,

(1)

whereU,Nare unknown functions,δ₀denotes the Dirac distribution concentrated at the origin,dis a positive integer,κ,qare positive constants, andsis a parameter with 0≤s<1.

The Maxwell-Chern-Simons gaugedO(3) sigma model is described by the Lagrangian L=−1

4qF_αβF^αβ+ κ

2ε^αβγA_α∂_βA_γ+ 1

2|D_αφ|²+ 1

2q(∂_αN)²−V(φ,N), (2) whereA_α:R^1,2→R³ (α=0, 1, 2) is the gauge field,F_αβ=∂_αA_β−∂_βA_α is the corresponding cur- vature tensor,D_αφ=∂_αφ+A_αn×φis the gauge covariant derivative,n= (0, 0, 1),φ=(φ₁φ₂,φ₃) : R^1,2→S²is a vector field,N:R^1,2→Ris a scalar field,q,κ >0 are parameters,ε_αβγ,α,β,γ=0, 1, 2, is the totally skew-symmetric tensor withε₀₁₂=1, andV(φ,N) is a potential given by

V(φ,N)=q

2(κN+s−n·φ)²+1

2N²(n×φ)²,

in whichs∈Ris a constant. The Gauss equation (as a variational equation forA₀) of (2) is given by

1

q∂_αF_0α+κF₁₂+n·φ×D₀φ=0. (3) By virtue of (3), the static energy corresponding to (2) can be written as

a)E-mail:[email protected].

b)E-mail:[email protected].

c)Author to whom correspondence should be addressed:[email protected]

0022-2488/2017/58(7)/071503/15/$30.00 58, 071503-1 Published by AIP Publishing.

E=

R²

d²x 1

2q(|∂_jA₀|²+|F₁₂|²) +1

2(|A₀|²|n×φ|²+|D_jφ|²) + 1

2q|∂_jN|²+V(φ,N)

=

R²

d²x 1

2q|∂_jA₀∓∂N|²+ 1

2q|F₁₂∓q(κN+s−n·φ)|² +1

2|n×φ|²|A₀∓N|²

+ 1

2|D₁φ±φ×D₂φ|²

±φ·D₁φ×D₂φ+ (s−n·φ)F₁₂ . We choose the upper sign and thus obtain the lower energy bound

E≥

R²

d²x

φ·D₁φ×D₂φ+ (s−n·φ)F₁₂ .

When the field configurations saturate the energy bound, we have the Bogomol’nyi equations











 A₀=N,

F₁₂=q(κN+s−n·φ), D₁φ=−φ×D₂φ.

(4)

Finite energy condition indicates the following boundary conditions:

N→0, s−n·φ→0, as|x| → ∞, (5)

|n×φ| →0, κN+s−n·φ→0, as|x| → ∞. (6) Letu=u₁+ iu2be the projection ofφonto the complex plane through the south polen, which is given by

u₁= φ₁

1 +φ₃, u₂= φ₂ 1 +φ₃. Equation (4) is rewritten as













 A₀=N,

F₁₂=q(κN+s−n·φ),

∂u=−iαu,

(7)

where∂=(∂₁+ i∂₂)/2 andα=(A1−iA2)/2. SetU=logφandV=−κN. Equation (7) is transformed into



















∆U=2q

−V+s−1− e^U 1 + e^U

+ 4π

m

X

j=1

δ_pj −4π

n

X

j=1

δ_qj,

∆V=−κ²q²

−V+s−1−e^U 1 + e^U

+ 4q e^U (1 + e^U)²V,

(8)

whereδ_pis the Dirac distribution concentrated atp∈R²andp_j,q_jare the locations of vortices and antivortices. Conditions (5) and (6) indicate three kinds of boundary conditions for solutions of (8) as follows:

(a) Topological:U(x)→log¹_{1 +}⁻_s^sandN(x)→0, as|x| → ∞. (b) Non-topological I:U(x)→ −∞andN(x)→s−1, as|x| → ∞. (c) Non-topological II:U(x)→+∞andN(x)→s+ 1, as|x| → ∞.

Ifm=d≥0,n=0 withp₁=· · ·=p_d=0, Eq. (1) is a special case of Eq. (8). For the details of the theory, we refer the readers to Refs.11and13, for example.

Existence and asymptotic structure of those solutions of (8) which satisfy the topological bound- ary condition^2,3,10,15have been studied in Refs.9and11. Recently, there have been papers concerning the solution structure of the elliptic equations coming from the CS (Chern-Simons) gauge theory. For example, one may refer to Refs.7and8for the CS gaugedO(3) sigma model,⁵for the gravitational sigma model,^4,6for the self-dualU(1)×U(1) two-particle system, and¹²for a perturbation of the SU(3) Toda system. Classification of radially symmetric solutions is associated with the uniqueness of the topological solutions of the related models and can probably be an approach to realize more complicated solutions.

In this paper, we prove the uniqueness of the topological solutions and identify the solution sets of all solution types for the radially symmetric case of (1). We portray our main theorems as follows.

Theorem 1.1. Equation (1) possesses exactly one solution satisfying the topological boundary condition for each d≥0. Moreover, the topological solutions continuously deform with respect to d in the sense that for any d₀≥0, there exist a number ε >0 and a continuous mapping P: ρ 7→(Uρ,N_ρ)∈C⁰_loc(R²)×C_loc⁰ (R²)such that U_ρ,N_ρsolve Eq. (1) with d=ρand satisfy the topological boundary condition whenever ρ∈(d0−ε,d₀+ε)∩(0,∞).

The radially symmetric case of Eq. (1) is characterized by



























U⁰⁰+r⁻¹U⁰=2q

−N+s−1−e^U 1 + e^U

, r>0, N⁰⁰+r⁻¹N⁰=−κ²q²

−N+s−1−e^U 1 + e^U

+ 4q e^U

(1 + e^U)²N, r>0, U(r;c₁,c₂)=c₁+ 2dlogr+o(1), asr→0,

N(r;c₁,c₂)=c₂+o(1) asr→0

(9)

forc₁,c₂∈R. Define

E_I={(c1,c₂) : U(r;c₁,c₂)→log1−s

1 +s, N(r;c₁,c₂)→0, asr→ ∞}, E_II={(c1,c₂) : U(r;c₁,c₂)→ −∞, N(r;c₁,c₂)→s−1, asr→ ∞}, E_III={(c1,c₂) : U(r;c₁,c₂)→+∞, N(r;c₁,c₂)→s+ 1, asr→ ∞}, E_IV={(c1,c₂) : U(r;c₁,c₂)→ −∞, N(r;c₁,c₂)→+∞, asr→ ∞}, E_V={(c1,c₂) : U(r;c₁,c₂)→+∞, N(r;c₁,c₂)→ −∞, asr→ ∞}. Theorem 1.2. Assumeκ²q≥4.The sets EI,E_II,E_III,E_IV,E_Vcan be identified as follows.

(i) E_I∪E_II ∪E_III∪E_IV∪E_V=R².

(ii) There exist c^∗∈Rand a continuous functionΓ:R→Rsuch that E_I={(c,Γ(c)) :c=c^∗},

E_II={(c,Γ(c)) :−∞<c<c^∗}, E_III={(c,Γ(c)) :c^∗<c<∞}, E_IV={(c1,c₂) :c₁∈R, c₂<Γ(c1)},

E_V={(c1,c₂) :c₁∈R, c₂>Γ(c1)}.

Moreover, Γ(c)→s + 1 as c→ ∞ and Γ(c)→s − 1 as c→ −∞ (please see Fig. 1 for an illustration).

Preliminaries.With the substitution s=a−1

a+ 1, u=U+ loga, v=− 2 κ²qN,

FIG. 1. The location of each solution set.

Equation (9) can be transformed into the system

































u⁰⁰+r⁻¹u⁰=2qf

λv+ 2a(e^u −1) (a + 1)(e^u + a)

g, r>0, v⁰⁰+r⁻¹v⁰=2qf

λv+ 2a(e^u −1) (a + 1)(e^u + a)

g+ 4q ae^u

(e^u + a)²v, r>0, u(r)=α₁+ 2dlogr+o(1), asr→0,

v(r)=α₂+o(1) asr→0, α₁,α₂∈R,

(10)

wherea≥1 (by 0≤s<1) andλ=κ²q/2. The topological and non-topological solutions boundary conditions associated with Eq. (10) are given by

(Topological) u(x)→0,v(x)→0 as|x| → ∞, (11)

(Non-topological I) u(x)→ −∞,v(x)→ 2

λ(a+ 1) as|x| → ∞, (12) (Non-topological II) u(x)→+∞,v(x)→ −2a

λ(a+ 1) as|x| → ∞. (13) By the maximum principle, it is clear that the solutions satisfying either (11) or (12) have the following property:

u(x)<0, v(x)>0, for allx∈R², (14) and thus, lettingw=λv+ _(a^2a(e_{+ 1)(e}^u−u¹⁾+a), we have

∆w−2qf

λ+ 2ae^u (e^u+a)³

gw=4qλ ae^u

(e^u+a)²v+ 2ae^u(a−e^u)

(e^u+a)³ |∇u|²>0.

So that

λv+ 2a(e^u−1)

(a+ 1)(e^u+a)<0 (15)

for topological and non-topological I solutions. From (10),∆v=2q[f(e^u)v+g(e^u)], where f(t)=λ+ 2at

(t+a)², g(t)= 2a a+ 1

t−1 t+a

are bounded functions. Thus, the solutions of (10) are all entire solutions. Furthermore, by Lemma 2.3., u,vcannot oscillate infinitely many times. Especially, sinceg(0) =2/(a+ 1),g(∞)=2a/(a+ 1), and g⁰>0, it follows that

v⁰(r)>0, for allr>0, providedv(0)>2/λ(a+ 1),

v⁰(r)<0, for allr>0, providedv(0)<−2a/λ(a+ 1). (16) By the way, from Lemma 2.1. and Lemma 2.2., we have

−2a/λ(a+ 1)≤v≤2/λ(a+ 1) (17) wheneveru,vsatisfy one of the boundary conditions (11)–(13). It is not difficult to observe that all the possible solution types are included as follows:

Type I: u(r)→0 and v(r)→0 asr→ ∞,

type II: u(r)→ −∞ and v(r)→2/λ(a+ 1) asr→ ∞, type III: u(r)→+∞ and v(r)→ −2a/λ(a+ 1) asr→ ∞, type IV: u(r)→ −∞ and v(r)→ −∞ asr→ ∞, type V: u(r)→+∞ and v(r)→+∞ asr→ ∞. Letβ=(1/2π)∫_R²F₁₂, whereF₁₂is given in (7). We have

β=β(u,v)=

∞

0

−2qf

λv+ 2a(e^u−1) (a+ 1)(e^u+a)

gr dr.

Assume |β|<∞. Clearly,u(r)=(2d − β) logr+O(1) as r→ ∞. Since v(r)→constant, we have rv⁰(r)→0 asr→ ∞; thus,βhas an alternative expression

β=

∞

0

4q ae^u

(e^u+a)²vr dr. (18)

Ifu,vare of type I, thenβ=2d. Ifu,vare of type II, we have

∞

0

r^2d−β+1dr<c

∞

0

e^u

a+ 1r dr<c⁰

∞

0

e^u

(e^u+a)²vr dr<∞

and thus β >2d+ 2. Letu,v belong to type III, in whichv(r)<0 in (R,∞) for some largeR>0.

Then ∞

R

r^β−2d+1dr<c

∞

R

e^−ur dr<c⁰

∞

R

e^u

(e^u+a)²(−v)r dr<∞.

Soβ <2d−2. Besides, type IV and V solutions are characterized by β=+∞and−∞, respectively.

We summarize the above remark as follows:





















β=2d, for type I solutions, β >2d+ 2, for type II solutions, β <2d−2, for type III solutions, β=+∞, for type IV solutions, β=−∞, for type V solutions.

(19)

The paper is organized as follows. In Sec.II, we establish the non-degeneracy for the linearized equation of Eq. (10) and show the uniqueness of the topological solution. In Sec.III, we locate the regions of the initial data which account for each solution type we introduce as above and carry out a classification of the radially symmetric solutions of Eq. (9).

II. NON-DEGENERACY AND UNIQUENESS

In this section, we will establish the non-degeneracy property for the linearized equations of (10).

By introducing a functional deformation method and the non-degeneracy property, we can prove the uniqueness of the topological solutions.

Consider the linearized equation of (10)

(φ⁰⁰+r⁻¹φ⁰=F₁(φ,ψ),

ψ⁰⁰+r⁻¹ψ⁰=F₁(φ,ψ) +F₂(φ,ψ), (20)

where

F₁(φ,ψ)=4q ae^u

(e^u + a)²φ+ 2qλψ, F₂(φ,ψ)=4qae^u(a −e^u)

(e^u + a)³ vφ+ 4q ae^u (e^u + a)²ψ.

Then the non-degeneracy property of (20) is in the following:

Theorem 2.1. If (u,v)is a solution of (10) and (11), then(φ,ψ)≡(0, 0)is the only bounded solution of (20).

Remark 2.1. If (u,v) is a solution of (10), then, by∆v(0)=2qf

λv(0)− ^2a

a(a+1)

g,r= 0 cannot be a non-positive local minimum point ofv(r).

In the following, we first prove the monotone property of the solution of type I.

Lemma 2.1. Let(u(r),v(r))be a solution of (10).Then the following conditions are valid.

(i) Both u(r)andv(r)cannot attain any non-negative local maximum on(0,∞).

(ii) Both u(r)andv(r)cannot attain any non-positive local minimum on(0,∞).

(iii) If (u(r),v(r))is a solution of type I, then u⁰(r)>0and v⁰(r)<0on(0,∞).

Proof. Let (u(r),v(r)) be a solution of (10). First, we show the result of part (i). On the contrary, we assume thatr_u^∗andr^∗_vare the respective first non-negative local maximum points ofu(r) andv(r) on [0,∞). By Remark 2.1, we haver^∗_v,0. From (10) and

2a(e^x−1) (a+ 1)(e^x+a)

0

>0∀x∈R, we easily obtain thatu(r) [respectively,v(r)] does not possess a non-negative local minimum on (0,r^∗_v] (respectively,

0,r_u^∗

), and the following equations hold:



















λv(r_u^∗) + 2a(e^u(ru∗) −1) (a + 1)(e^u(r∗u) + a)≤0, λv(r_v^∗) + 2a(e^u(r∗v) − 1)

(a + 1)(e^u(r∗v) + a)≥ − 2ae^u(rv∗)

(a + 1)(e^u(r∗v) + a)²v(r_v^∗)≥0.

By the first inequality and

2a(e^x−1) (a+ 1)(e^x+a)

⁰

>0∀x∈R, we easily get λv(r^∗_v) + 2a(e^u(rv∗)−1)

(a+ 1)(e^u(rv∗)+a)<λv(r_u^∗) + 2a(e^u(r∗u)−1) (a+ 1)(e^u(ru∗)+a)≤0

which contradicts with the second inequality. This completes the proof of (i). Similarly, we can show the result of part (ii).

Next, in order to show the result of part (iii), we need the following claim.

Claim. If (u(r),v(r))is a solution of type I, then u(r)<0and−v(r)<0∀r∈(0,∞).

Proof of Claim.On the contrary, since (u(r),v(r))→(0, 0) asr→ ∞and by the boundary condi- tions of (10), we easily obtain thatu(r) andv(r) possess a respective non-negative local maximum point which contradicts to the result of part (i).

This proves ourClaim.Now, suppose the result of part (iii) is not true. Then by aboveClaim, we obtain thatu(r) andv(r) possess a respective non-positive local minimum point which contradicts to the result of part (ii).

This shows part (iii), and the proof of Lemma 2.1. is complete.

Next, we establish the range ofv(r) for solutions of type II and III as follows.

Lemma 2.2. Let(u(r),v(r))be a solution of (10).Then the following conditions are valid.

(i) If (u(r),v(r))is a solution of type II, then

u(r)<0 and 0≤v(r)≤2/λ(a+ 1)∀r∈(0,∞).

(ii) If (u(r),v(r))is a solution of type III, then−2a/λ(a+ 1)≤v(r)≤2/λ(a+ 1)∀r∈[0,∞).

Proof. (i) Let (u,v) be a solution of type II for Eq.(10). Thenu(r)→ −∞andv(r)→2/λ(a+ 1) asr→ ∞. By (i) and (ii) of Lemma 2.1. and the boundary conditions of (10), we easily obtainu(r)<0 andv(r)≥0∀r∈(0,∞). From (10), we have

(*) ∆v=2q[f(e^u)v+g(e^u)], wheref(t)=λ+ ^2at

(t+a)² andg(t)= _2a

a+1 t−1 t+a

are bounded functions.

Ifv(0)>_λ(a+1)² then, sinceg⁰(t)>0, lim

t→0g(t)=− ²

a+1andv(r)→2/λ(a+ 1) asr→ ∞, by (10) and (*), we obtain thatv(r) possesses a non-negative local maximum which contradicts with (ii) of Lemma 2.1.

Hence, we have 0< v(0)≤ ²

λ(a+1) andv(r)→ ²

λ(a+1) asr→ ∞. By using (ii) of Lemma 2.1. again, we finally obtain 0≤v(r)≤2/λ(a+ 1) on (0,∞). This proves part (i) of this lemma.

(ii) Let (u, v) be a solution of type III for Eq. (10). Then, we have u(r)→+∞ and v(r)

→ −2a/λ(a+ 1) asr→ ∞. Similar to the proof of above part (i), by above (*),g⁰(t)>0, lim

t→0g(t)

=− ²

a+1, lim

t→∞g(t)=_a+1^2a, the boundary conditions of (10) and (i) and (ii) of Lemma 2.1., we obtain that −2a/λ(a + 1)≤v(0)≤2/λ(a+ 1), and v(r) does not possess a local minimum (respectively, local maximum) at r_v∈(0,∞) such that v(r_v)≥ ²

λ(a+1) respectively, v(r_v)≤ − ^2a

λ(a+1)

. These imply that

−2a/λ(a+ 1)≤v(r)≤2/λ(a+ 1)∀r∈[0,∞).

Thus, we get the result of part (ii). The proof of Lemma 2.2. is complete.

Now for each solution of (10), we have the following non-oscillation result.

Lemma 2.3. Let(u(r), v(r))be a solution of (10).Then u, v cannot oscillate infinitely many times.

Proof. By the results of (i) and (ii) in Lemma 2.1., we easily obtain the conclusion.

Letϕ₁₁,ϕ₁₂andϕ₂₁,ϕ₂₂solve (20), respectively, with the initial data (ϕ11(0)=0, ϕ⁰₁₁(0)=0,

ϕ₂₁(0)=1, ϕ⁰₂₁(0)=0, and

(ϕ12(0)=1, ϕ₁₂⁰ (0)=0,

ϕ₂₂(0)=0, ϕ₂₂⁰ (0)=0. (21) Lemma 2.4. ϕ_ij>0andϕ_ij⁰>0for all i,j= 1, 2.

Proof. Since F₁(φ,ψ),F₂(φ,ψ)>0 whenever φ,ψ >0, it follows readily that each ϕ_ij is an

increasing function.

We introduce the auxiliary functionsA=ru⁰andB(r)=rv⁰. Then A⁰⁰+r⁻¹A⁰=F₁(A,B) +G₁,

B⁰⁰+r⁻¹B⁰=F₁(A,B) +F₂(A,B) +G₂, in whichG₁,G₂are given by

G₁=4qf

λv+ 2a(e^u−1) (a+ 1)(e^u+a)

g, G₂=4qf

λv+ 2a(e^u−1) (a+ 1)(e^u+a)

g+ 8q ae^u (e^u+a)²v.

Lemma 2.5. Letφ,ψsolve(20). SetΛ=(φ−ψ)A⁰−A(φ⁰−ψ⁰)−(φB⁰−Bφ⁰). Then Λ(τ)=1

τ

_τ

0

rf

(G1−G₂)φ−G₁ψg

dr, τ >0. (22)

Proof. Letξ=φ−ψ. From (20), (rξ⁰)⁰=−rF₂(φ,ψ). Making use of the expression f(rg⁰)⁰−g(rf⁰)⁰= d

dr[r(fg⁰−gf⁰)], we have

τ[ξA⁰−Aξ⁰](τ)=

_τ

0

ξ(rA⁰)⁰−A(rξ⁰)⁰dr

=

_τ

0

rf

(φ−ψ)F₁(A,B) +AF₂(φ,ψ) + (φ−ψ)G₁g dr

=

_τ

0

rf

F₁(Aφ−Aψ,Bφ−Bψ) +F₂(Aφ,Aψ) +G₁(φ−ψ)g dr, τ[φB⁰−Bφ⁰](τ)=

_τ

0

φ(rB⁰)⁰−B(rφ⁰)⁰dr

=

_τ

0

rf

φ(F₁+F₂)(A,B)−BF₁(φ,ψ) +G₂φg dr

=

_τ

0

rf

F₁(Aφ−Bφ,Bφ−Bψ) +F₂(Aφ,Bφ) +G₂φg dr.

Hence

τΛ(τ)=

_τ

0

rf

F₁(Bφ−Aψ, 0) +F₂(0,Aψ−Bφ) +G₁(φ−ψ)−G₂φg dr

=

_τ

0

rf

(G1−G₂)φ−G₁ψg dr.

We use the notationQ₀,Q₁to denote the following functions:

Q₀=ϕ11

ϕ₁₂ and Q₁=ϕ21

ϕ₂₂. Clearly,Q₀(0) = 0,Q₁(0) = 1, andQ₀,Q₁>0 (Lemma 2.4.).

Lemma 2.6. Let u,vsatisfy(10) and (11). Then Q₀⁰(r)>0, Q₁⁰(r)<0, and Q0(r)<Q₁(r)for all r∈(0,∞).

Proof. We construct the proof by two steps.

(A1)If there is M>0such that Q₀(r)<Q₁(r)for r∈(0,M), then Q⁰₀(r)>0and Q⁰₁(r)<0whenever r∈(0,M).

Assume there isr₁∈(0,M) such thatQ₀⁰(r1)=0 andQ₀(r)<Q₁(r) for allr∈(0,r₁). Without loss of generality, we may assume

Q₀(r)<Q₀(r1), r∈(0,r₁). (23) SetQ₀(r1) =c. Consider the solutionφ,ψof (20) given by

φ ψ

!

= ϕ₁₁ ϕ₂₁

!

−c ϕ₁₂ ϕ₂₂

!

. (24)

Note that

φ(r₁)=0, φ⁰(r1)=0, and ψ(r₁)>0.

Soφ⁰⁰(r1)>0 by virtue of (20). This contradicts (23). Therefore,Q₀⁰>0.

Since lim

r→0⁺Q₁(r)=∞,Q₁⁰(r)<0 for a sufficiently smallr>0. Hence, we haveQ₁⁰<0 according to the similar argument of the proof as in the caseQ₀⁰>0.

(A2)Q₁and Q₀cannot meet.

Suppose (A2) fails. SelectM>0 such thatQ₀(M)=Q₁(M)=c^∗andQ₁(r)>c^∗>Q₀(r) whenever r∈(0,M). Consider the solutionφ,ψgiven by (24) withc=c^∗. Note that

(φ(r)<0 and ψ(r)>0, r∈(0,M),

φ(M)=ψ(M)=0. (25)

Applying (15) and(22), 0>−Mf

−A(φ⁰−ψ⁰) +Bφ⁰g (M)=

M

0

rf

(G1−G₂)φ−G₁ψg dr>0,

which is a contradiction. The proof is completed.

Proof of Theorem 2.1. Letφ,ψbe a nontrivial solution of (20) which is characterized by (24) with somec>0. Ifφ,ψare bounded, thenrφ⁰(r),rψ⁰(r)→0 asr→ ∞. Sinceϕ₁₂,ϕ₂₂→ ∞asr→ ∞, it follows that

Q₀−c=φ/ϕ₁₂→0, r→ ∞, Q₁−c=ψ/ϕ₂₂→0, r→ ∞.

By Lemma 2.6.,Q₀(r)%candQ₁(r)&casr→ ∞. In particular,φ <0 andψ >0. On the other hand, sinceA,Bare bounded, we haverA⁰,rB⁰→0 asr→ ∞. Therefore,

0=lim

τ→0τ[(φ−ψ)A⁰−A(φ⁰−ψ⁰)−(φB⁰−Bφ⁰)]

=

∞

0

rf

(G1−G₂)φ−G₁ψg dr>0.

This yields a contradiction. So Eq. (24) cannot possess any non-trivial bounded solution.

Lemma 2.7. Assume d= 0. Let u,vsatisfy(10) and (11). Then u=v≡0.

Proof. ∆u≤0 inR²andu(x)→0 as|x| → ∞. Sou,v≡0 by the maximum principle.

Function Spaces.Define the scalar productsh,i_X

α andh,i_Y

α, 0< α <1, for functions in the spaces L_loc² (R²) andW_loc^2,2(R²),

hu,vi_Xα=

R²

(1 +|x|^2+α)uvdx, u,v∈L²_loc(R²), hu,vi_Yα=h∆u,∆vi_Xα+

R²

uv

1 +|x|^2+αdx, u,v∈W_loc^2,2(R²).

LetX_α,Y_αbe the Hilbert spaces given by X_α=(

u∈L²_loc(R²) :hu,ui_Xα<+∞)

, Y_α=( u∈W^2,2

loc(R²) :hu,ui_Yα<+∞) . TheX_α-norm andY_α-norm are induced by

kuk_Xα=q

hu,ui_Xα, kuk_Yα=q

hu,ui_Yα. We observe thatX_α,→L¹(R²) andY_α⊂C⁰

loc(R²) by H¨older’s inequality and the local regularity of the Laplace operator, respectively. Let the product spacesX_α ×X_α andY_α×Y_α be equipped with inner productshu,vi_Xα×Xα andhu,vi_Yα×Yαwhich are given by

hf,gi_Xα×Xα=hf₁,g₁i_Xα+hf₂,g₂i_Xα, hu,vi_Yα×Yα=hu₁,v₁i_Yα+hu₂,v₂i_Yα,

wheref=(f1,f₂),g=(g1,g₂)∈X_α×X_α andu=(u1,u₂),v=(v₁,v₂)∈Y_α×Y_α. In the following con- tent, we denote the subspaces consisting of radially symmetric functions byX_α^r andY_α^r.

We introduce the mappingP(d,ξ,η)=(P₁,P₂)(d,ξ,η) ford≥0,ξ,η∈Y_α^r, P₁=ρ⁻¹(

∆ξ−2qf

λη+ 2a(|x|^2de^ξ−1) (a+ 1)(|x|^2de^ξ+a)

g ), P2=ρ⁻¹(

∆η−2qf

λη+ 2a(|x|^2de^ξ−1) (a+ 1)(|x|^2de^ξ+a)

g−4q a|x|^2de^ξ (|x|^2de^ξ+a)²η)

,

where ρ=1 +|x|^4d+4. The triplet (d,ξ + 2dlogr,η) solving (10) is equivalent to P(d,ξ,η)=0.

According to Ref.1, Lemma 1.1, there existsC>0 such that

|u(x)| ≤Ckuk_Yα(log⁺|x|+ 1), x∈R², (26) for all u∈Y_α, where log⁺|x|=max{0, log|x|}. Then

P(d,ξ,η)∈X_α^r ×X_α^r

wheneverd≥0 andξ,η∈Y_α^r. Consider the partial derivativeL=∂P(d,ξ,η)/∂(ξ,η) at a fixedd>0 and (ξ,η)∈Y_α^r ×Y_α^r,

L: φ ψ

!

7→ρ⁻¹ ∆φ−L₁(φ,ψ)

∆ψ−L₁(φ,ψ)−L₂(φ,ψ)

!

, φ,ψ∈Y_α^r, (27)

in which

L₁(φ,ψ)=4q a|x|^2de^ξ

(|x|^2de^ξ+a)²φ+ 2qλψ, L₂(φ,ψ)=4qa|x|^2de^ξ(a− |x|^2de^ξ)

(|x|^2de^ξ+a)³ ηφ+ 4q a|x|^2de^ξ (|x|^2de^ξ+a)²ψ.

Lemma 2.8. Let d>0and ξ,η∈Y_α^r. If (ξ,η)=(u−2dlogr,v), where(u,v)solves(10) and (11)with d, then L is a surjective mapping from Y_α^r ×Y_α^r to X_α^r ×X_α^r.

Proof. Let ImL={g∈X_α^r ×X_α^r:Lf =g, ∃f∈Y_α^r×Y_α^r}. We claim that ImL=X_α^r ×X_α^r. Suppose the assertion fails. Letζ=(ζ₁,ζ₂) be a non-zero element ofX_α^r×X_α^r andζ<ImL. Since ImLis closed (via Ref.1, Proposition 2.1), we may decomposeX_α^r ×X_α^r=ImL⊕(ImL)^⊥. Assume (without loss of generality) thatζ⊥ImL. We have

0=hL(φ,ψ),ζi=

R²

(∆φ−L₁(φ,ψ)f +∆ψ−L₁(φ,ψ)−L₂(φ,ψ)g)

dx (28)

wheneverφ,ψ∈Y_α^r, in whichf=ρ⁻¹(1 +|x|^2+α)ζ₁≥0 andg=ρ⁻¹(1 +|x|^2+α)ζ₂≥0. SinceC^∞

0 (R²)

⊂Y_α, by elliptic regularity,f,gareC²-functions. Forφ,ψ∈C^∞₀ (R²), 0=

R²

φ∆f +ψ∆g−L₁(fφ+gφ,fψ+gψ)−L₂(gφ,gψ)dx

=

R²

f∆f −L₁(f +g, 0)−L₂(g, 0)g φ+f

∆g−L₁(0,f +g)−L₂(0,g)g ψdx.

Sof,gsolve

(∆f −L₁(f +g, 0)−L₂(g, 0)=0,

∆g−L₁(0,f +g)−L₂(0,g)=0. (29) Letθ=f +g. Note thatθ→0 as |x| → ∞. SinceX_α⊂L¹(R²), we getf,g→0 as|x| → ∞. By (29), (g,θ) satisfies (20). Thus by Theorem 2.1., we haveg≡0 andθ≡0; hencef=g= 0, that is,ζ₁=ζ₂=0, which contradicts our assumption. Therefore, ImL=X_α^r ×X_α^r. Assumeu₀,v₀solve (10) and (11) withd=d₀>0. Letu₀=ξ₀+ 2dlograndv₀=η₀. Clearly, ξ₀,η₀satisfy the assumption of Lemma 2.8. andP(d₀,ξ₀,η₀)=0. By applying a specific version of the implicit function theorem ([Ref.14, Theorem 2.7.5]) toP, we find a continuous mapping

h:d7→(ξ(d),η(d))

from a neighborhoodN(d₀) ofd₀toY_α^r ×Y_α^r such thatξ(d₀)=ξ₀,η(d₀)=η₀, and

P(d,ξ(d),η(d))=0 for all d∈N. (30)

This establishes the existence of a family of solutionsu_d,v_d to (10) which are given by

u_d=ξ(d) + 2dlogr, v_d=η(d), d∈N(d₀). (31) Lemma 2.9. Assume u₀,v₀solve(10) and (11)with d=d₀>0. Then there existsδ >0such that the solutions u_d,v_dgiven by(31)satisfy the boundary condition(11)whenever|d−d₀|< δ.

Proof. SinceX_α,→L¹(R²), by the continuity ofh, we have 2π

β(d) 2 −d₀

≤ k∆(ξ(d)−ξ₀)k_L1(R²)≤Ck∆(ξ(d)−ξ₀)k_Xα≤Ckξ(d)−ξ₀k_Yα→0

asd→d₀for some constantC>0. From (19),u_d,v_dmust belong to type I solutions asdis sufficiently

close tod₀.

Proof of Theorem 1.1. If d = 0, the only type I solution of (10) is the trivial solution by Lemma 2.7. Assume that there isd₁∈(0,∞) such that Eq. (10) withd =d₁ possesses at least two solutions satisfying the boundary condition (11). Let

J={σ≥0 : ∀d∈[0,σ],∃a unique (u,v) satisfying (10) and (11)}.

Setd^∗=supJ. Then 0≤d^∗≤d₁. We claim thatd^∗∈J. In fact, ifd^∗<J, there exist two (different) solutions (u,v), (¯u, ¯v) of (10) and (11) withd=d^∗. By virtue of Lemma 2.9., there are two sequences (ud,v_d) and (¯u_d, ¯v_d) of type I solutions which converge to (u,v), (¯u, ¯v) asd→d^∗, which indicates supJ<d^∗, a contradiction. Let (u,v) be the unique solution of (10) and (11) withd=d^∗. We select two sequences (dj,u_j,v_j) and (dj, ¯u_j, ¯v_j) such that

(i) d_j>d^∗andd_j→d^∗asj→ ∞;

(ii) for eachj, (uj, v_j) and (¯u_j, ¯v_j) are two different solutions which solve (10) and (11) with d=d_j.

Let

φ_j=(uj−u¯_j)/m_j and ψ_j=(v_j−v¯_j)/m_j, wherem_j=max{ku_j−u¯_jk_L∞,kv_j−v¯_jk_L∞}. Then

1<kφ_jk_L∞+kψ_jk_L∞<2 (32) andφ_j,ψ_jsatisfy

∆φ_j=2qf

λψ_j+ 2ae^aj (e^aj +a)²φ_jg

,

∆ψ_j=2qf

λψ_j+ 2ae^aj (e^aj +a)²φ_jg + 4qf ae^uj

(e^uj +a)²ψ_j+ae^bj(a−e^bj) (e^bj+a)³ v¯_jφ_jg

,

wherea_j,b_jare located betweenu_jand ¯u_j. Letx_j,y_jbe the maximum points given by (uj−u¯_j)(xj)=ku_j−u¯_jk_L^∞ and (v_j−v¯_j)(yj)=kv_j−¯v_jk_L^∞.

By Ref.4, Lemma 3.2,x_j,y_j are bounded sequences. Soφ_j,ψ_j converge. Letφ_j→φ,ψ_j→ψ. Note thatφ,ψare bounded functions which solve (20). Henceφ=ψ=0 by Theorem 2.1. This contradicts (32). Sod^∗=∞. Moreover, the continuous deformation follows readily from (31) and Lemma 2.9.

Theorem 1.1 is concluded.

III. LOCATE THE SOLUTION SETS

Using the uniqueness result shown in Sec.II, we are going to identify each solution set in the present section.

Letu(·,α1,α2),v(·,α1,α2) satisfy Eq. (10) with the initial dataα1,α2∈R. Consider the functions φ=∂u/∂α₂,ψ=∂v/∂α₂. (φ,ψ) solve Eq. (20) with (φ,ψ)(0)=(0, 1). From Lemma 2.4.,φ,ψ >0 and thus, we have

u(·,α₁,α₂)<u(·,α₁,α₂⁰), v(·,α₁,α₂)< v(·,α₁,α₂⁰) (33) wheneverα₂< α⁰₂. Let

E_i={(α₁,α₂) :u(·,α₁,α₂),v(·,α₁,α₂) solve (10) and satisfy typeiboundary condition}, fori=I,II,III,IV,V as given in Sec.I. Note thatS

iE_i=R²andE_i∩E_j=∅fori,j.

Theorem 3.1. For eachα₁∈R, there exists exactly oneα₂∈Rsuch that (α₁,α₂)∈E_I∪E_II∪E_III.

Lemma 3.1. E_IV ∩E_V=∅and E_V∩E_IV=∅, where E denotes the closure of E.

Proof. Assume (α₁,α₂)∈E_IV∩E_V. Note thatv(r,α₁,α₂)→ −∞asr→ ∞. In particular, there is R>0 so large thatv(R,α₁,α₂)<−2a/λ(a+1). By continuity, we may extract (α⁰₁,α⁰₂)∈E_Vsufficiently close to (α₁,α₂) such that v(R,α⁰₁,α⁰₂)<−2a/λ(a+ 1). However, v(r,α⁰₁,α⁰₂)→ ∞as r→ ∞and v(0,α₁⁰,α₂⁰)≥ −2a/λ(a+ 1) [by (16)], which indicates thatv(·,α₁⁰,α₂⁰) has a negative minimum which is less than−2a/λ(a+ 1) at somer=R⁰. Thus,

0≤v⁰⁰(R⁰,α₁⁰,α₂⁰)=2qf

λv+ 2a(e^u−1)

(a+ 1)(e^u+a)+ 2ae^u (e^u+a)²vg

(R⁰,α⁰₁,α⁰₂)<0.

That is a contradiction. So E_IV ∩E_V must be empty. Similarly, E_V ∩E_IV=∅. We omit the

details.

Lemma 3.2. E_II ∩E_III=∅andE_III∩E_II=∅.

Proof. E_III ∩E_II=∅ is evident becausev >0 [by (14)] for type II solutions but v(r)<0 as r→ ∞from type III boundary condition. To show thatE_II ∩E_III=∅, we let (α₁,α₂)∈E_II ∩E_III. Note that u(r;α1,α2)→ −∞ as r→0,∞ and u(r;α1,α2)<0,v(r;α1,α2)>0 for all r>0 from (14). By continuity, we may pick an interval [r1,r₂]⊂(0,∞) and (α⁰₁,α⁰₂)∈E_III in a sufficiently small neighborhood of (α₁,α₂) such that u(r;α₁⁰,α₂⁰) has a local maximum at r=r⁰∈(r1,r₂) and

u(r⁰;α⁰₁,α⁰₂)>u(r1;α⁰₁,α⁰₂)>u(r2;α₁⁰,α₂⁰), v(r2;α₁⁰,α₂⁰)>0.

Sinceu(r;α₁⁰,α₂⁰)→ ∞asr→ ∞, there isR≥r₂such that

u(R;α₁⁰,α₂⁰)=inf{u(r;α⁰₁,α⁰₂) :r∈[r1,∞)}.

Note that u⁰⁰(r⁰;α⁰₁,α⁰₂)≤0 and u⁰⁰(R;α⁰₁,α⁰₂)≥0. From (10), we deduce that v(r⁰;α₁⁰,α₂⁰)

< v(R;α₁⁰,α₂⁰). Since v(r;α⁰₁,α⁰₂)→ −2a/λ(a+ 1)<0 asr→ ∞, it is possible to pickR⁰>r⁰such that

v(R⁰;α⁰₁,α⁰₂)=sup{v(r;α⁰₁,α⁰₂) :r∈[r⁰,∞)}.

Hence v⁰⁰(R⁰;α₁⁰,α₂⁰)≤0. On the other hand, since u(R⁰;α₁⁰,α₂⁰)≥u(R;α⁰₁,α⁰₂) and v(R⁰;α₁⁰,α₂⁰)

≥v(R;α⁰

1,α⁰

2), it follows from (10) that λv+ 2a(e^u−1)

(a+ 1)(e^u+a)

(R⁰;α⁰₁,α⁰₂)≥

λv+ 2a(e^u−1) (a+ 1)(e^u+a)

(R;α⁰₁,α⁰₂)

=u⁰⁰(R;α⁰₁,α⁰₂)≥0.