Self-dual radial non-topological solutions to a competitive Chern-Simons model ∗

Self-dual radial non-topological solutions to a competitive Chern-Simons model ^∗

Zhijie Chen

, Chang-Shou Lin

1Department of Mathematical Sciences, Yau Mathematical Sciences Center, Tsinghua University, Beijing 100084, China

2Taida Institute for Mathematical Sciences, Center for Advanced Study in Theoretical Sciences, National Taiwan University, Taipei 106, Taiwan

Abstract

We study a non-Abelian Chern-Simons system of rank 2:

∆u1

∆u2

+K

e^u1 e^u2

−K

e^u1 0 0 e^u2

K

e^u1 e^u2

=

4πN1δ0

4πN2δ0

in R², where N1, N2 ∈ N∪ {0}, δ0 is the Dirac measure at 0, and K = (aij) is a 2×2 matrix satisfyinga11, a22>0,a12, a21 <0 and detK >0, including the Cartan matrixB₂. The existence of non-topological solutions has remained a long-standing open problem. Here by applying the degree theory, we prove the existence of radial non-topological solutions (u₁, u₂) satisfying the prescribed asymptotic conditionu_k(x) =−2α_kln|x|+O(1) as|x| → ∞for some α_k >1.

We also construct bubbling solutions to show that the range of (α₁, α₂) is optimal in some sense. This generalizes a recent work by Choe, Kim and the second author, where theSU(3) case (i.e. K is the Cartan matrix A2) was investigated.

1 Introduction

In this paper, we study a non-Abelian Chern-Simons system of rank 2:

∆u₁

∆u2

+K

e^u1 e^u2

−K

e^u1 0 0 e^u2

K

e^u1 e^u2

=

4πN₁δ₀ 4πN2δ0

in R², (1.1) whereN1, N2∈N∪ {0},δ0is the Dirac measure at 0, andK= (aij) is a 2×2 matrix satisfying a11, a22>0, a12, a21<0 and a11a22−a12a21>0. (1.2) System (1.1) could be considered as a perturbation of the following Liouville system

∆u₁

∆u2

+

a₁₁ a₁₂ a21 a22

e^u1 e^u2

=

4πN₁δ₀ 4πN2δ0

inR². (1.3)

See [1, 2, 3]. Under the assumption (1.2), system (1.3) is also calledcompetitivein the literature, comparing to thecooperativecase wherea12, a21>0.

System (1.1) is motivated by various self-dual gauge field theories in physics. Our first motivation comes from the relativistic non-Abelian Chern-Simons model, which was proposed

∗E-mail address: [email protected], [email protected] (Chen); [email protected] (Lin)

by Kao and Lee [19] and Dunne [9, 10] to explain the physics of high critical temperature superconductivity. Following [9, 10], the relativistic non-Abelian Chern-Simons model is defined in the (2 + 1) Minkowski spaceR^1,2, and the gauge group is a compact Lie group with a semi- simple Lie algebra G. The Chern-Simons Lagrangian action density L in 2 + 1 dimensional spacetime involves the Higgs fieldφ and the gauge potential A = (A0, A1, A2). We consider the energy minimizers of the Lagrangian functional, which turn out to be the solutions of the following self-dual Chern-Simons equations:

D₋φ= 0, F₊₋= 1

κ²[φ−[[φ, φ^†], φ], φ^†], (1.4) where D₋ = D1−iD2, κ > 0, F₊₋ = ∂+A₋ −∂₋A+ + [A+, A₋] with A_± = A1 ±iA2,

∂_±=∂1±i∂2and [·,·] is the Lie bracket overG. In [10], Dunne considered a simplified form of the self-dual system (1.4), in which the fieldsφandA are algebraically restricted:

φ=

r

X

a=1

φ^aEa,

wherer is the rank of the gauge Lie algebra,E_a is the simple root step operator, and φ^a are complex-valued functions. Letu_a = ln|φ^a|, a= 1,· · · , r. Then system (1.4) can be reduced to the following system of nonlinear partial differential equations

∆ua+ 1 κ²

r

X

b=1

Kabe^ub−

r

X

b=1 r

X

c=1

e^ubKabe^ucKbc

!

= 4π

N_a

X

j=1

δp^a_j, 1≤a≤r, (1.5) whereK= (Kab) is the Cartan matrix of a semi-simple Lie algebra, {p^a_j}j=1,···,N_a are zeros of φ^a (a= 1,· · ·, r), and δ_p denotes the Dirac measure concentrated atpinR². See [27] for the derivation of (1.5) from (1.4). For example, there are three types of Cartan matrix of rank 2:

A2(i.e. SU(3)) =

2 −1

−1 2

, B2=

2 −1

−2 2

, G2=

2 −1

−3 2

. (1.6)

Let (K⁻¹)ab denote the inverse of the matrixK. Assume that

r

X

b=1

(K⁻¹)_ab>0, a= 1,2,· · · , r. (1.7) A solutionu= (u₁,· · ·, u_r) of (1.5) is called atopological solution if

ua(x)→ln

r

X

b=1

(K⁻¹)ab

!

as |x| →+∞, a= 1,· · ·, r, a solutionuis called anon-topological solutionif

ua(x)→ −∞ as |x| →+∞, a= 1,· · ·, r.

The existence of topological solutions of system (1.5) either inR² or a flat torus has been well investigated in the last decades. In 1997, Yang [27] proved the existence of topological solutions to (1.5) for any configuration p^a_j in R² via variational methods. Later, for a flat torus with doubly periodic boundary conditions, Nolasco and Tarantello [23] studied system (1.5) with the Cartan matrix A2 and obtained the existence of a topological solution and a mountain-pass type solution. Recently, Han and Tarantello [14] generalized the results of [23]

to the more general competitive case whereK is a 2×2 matrix satisfying (1.2). Their proof [14] is based on variational methods and does not seem to work in the cooperative case. This

indicates that the cooperative case is generally different from the competitive case. See also [13] for similar results to the Gudnason model.

On the other hand, the existence of non-topological solutions (and mixed-type solutions, see below) seems much more difficult than topological solutions to obtain, and there are very few results concerning the existence of non-topological solutions (and mixed-type solutions) in the literature. Very recently, some existence results of non-topological solutions to (1.5) with K = A₂,B₂ and G₂ have been proved by Ao, Lin and Wei [1, 2] by a perturbation from the corresponding Liouville system (1.3). However, their results are still very limited toward understanding the general theory of non-topological solutions.

In this paper, we focus on the radially symmetric solutions of (1.5) when all the vortices coincide at the origin. We only consider the rank 2 and competitive case, namelyKis a 2×2 matrix satisfying (1.2). Moreover, we may assume, without loss of generality, thatκ= 1. Then system (1.5) turns to be (1.1). In particular, whenK=A2, (1.1) becomes the followingSU(3) Chern-Simons system

(∆u1+ 2(e^u1−2e^2u1+e^u1+u2)−(e^u2−2e^2u2+e^u1+u2) = 4πN1δ0

∆u2+ 2(e^u2−2e^2u2+e^u1+u2)−(e^u1−2e^2u1+e^u1+u2) = 4πN2δ0

inR². (1.8) Recently, Huang and Lin made classifications of radially symmetric solutions for theSU(3) system (1.8) in [16] and for the more general system (1.1) in [17] respectively. As in [17], in order to simplify the expression of system (1.1), we consider the transformation

(u1, u2)→

u1+ ln a₂₂−a₁₂ a11a22−a12a21

, u2+ ln a₁₁−a₂₁ a11a22−a12a21

and let

(a1, a2) =

−a12(a11−a21) a11a22−a12a21

, −a21(a22−a12) a11a22−a12a21

.

Clearly the assumption (1.2) givesa1>0 anda2>0. Then system (1.1) becomes











∆u1+ (1 +a1)(e^u1−(1 +a1)e^2u1+a1e^u1+u2)

−a₁(e^u2−(1 +a₂)e^2u2+a₂e^u1+u2) = 4πN₁δ₀

∆u₂+ (1 +a₂)(e^u2−(1 +a₂)e^2u2+a₂e^u1+u2)

−a2(e^u1−(1 +a1)e^2u1+a1e^u1+u2) = 4πN2δ0

in R². (1.9)

Remark that the above transformation is just the identity for the caseK=A₂.

The main goal of this paper is to study the existence of non-topological solutions of the competitive Chern-Simons system (1.1) by seeking non-topological solutions of system (1.9).

Therefore, in the sequel, we only need to consider system (1.9). As in [17], we easily see that a solution (u1, u2) of (1.9) is a topological solution if (u1, u2) → (0,0) as |x| → +∞;

a non-topological solution if (u1, u2) → (−∞,−∞) as |x| → +∞; a mixed-type solution if (u1, u2)→(ln_1+a¹

1,−∞) or (u1, u2)→(−∞,ln_1+a¹

2) as|x| →+∞.

It is worth to point out that system (1.9) has also applications to the Lozano-Marqu´es- Moreno-Schaposnik model [22] of bosonic sector ofN = 2 supersymmetric Chern-Simons-Higgs theory, and the Gudnason model [11, 12] ofN = 2 supersymmetric Yang-Mills-Chern-Simons- Higgs theory. We refer the reader to [17] for details on these applications.

Define a continuous functionJ :R²→Rby J(x, y) = a2(1 +a2)

2 x²+a1a2xy+a1(1 +a1)

2 y². (1.10)

In [17], among other things, Huang and Lin proved the following interesting result.

Theorem A.[17]Leta1, a2>0. Suppose that(u1, u2)6= (0,0)is a radially symmetric solution of system(1.9). Then bothu1<0andu2<0inR², and one of the following conclusions holds.

(i) (u1, u2)is a topological solution.

(ii) (u1, u2)is a mixed-type solution.

(iii) (u1, u2)is a non-topological solution and there exist constantsα1, α2>1 such that uj(x) =−2αjln|x|+O(1) as|x| →+∞, j= 1,2. (1.11) Consequently,e^u1, e^u2 ∈L¹(R²). Moreover, (α₁, α₂)satisfies

J(α₁−1, α₂−1)> J(N₁+ 1, N₂+ 1). (1.12) Remark that the inequality (1.12) follows from the following Pohozaev identity (see [17] or Lemma 2.2 below):

J(α1−1, α2−1)−J(N1+ 1, N2+ 1)

=1 +a₁+a₂ 4

Z ∞ 0

r

a₂(1 +a₁)e^2u1+a₁(1 +a₂)e^2u2−2a₁a₂e^u1+u2

dr. (1.13)

Therefore, (1.12) is a necessary condition for the existence of radially symmetric non-topological solutions satisfying the asymptotic condition (1.11). After Theorem A, it is natural to consider the following long-standing open question (see [27, Section 8] for instance).

Open Question: Let a1, a2 >0. Given α1, α2 >1 satisfying (1.12). Is there a radial non- topological solution of system (1.9) subject to the asymptotic condition (1.11)?

If we letN1 =N2 =N, a1 =a2 and u1 =u2=uin (1.9), then system (1.9) turns to be the following Chern-Simons-Higgs equation

∆u+e^u(1−e^u) = 4πN δ0 inR². (1.14) Equation (1.14) is known as the SU(2) Chern-Simons equation for the Abelian case, which was proposed by Hong, Kim and Pac [15] and by Jakiw and Weinberg [18] independently. In the past twenty years, the topological solutions and non-topological solutions of (1.14) have been well studied; see [3, 4, 5, 6, 7, 25, 26] and references therein. Remark that the Pohozaev identity plays an important role in studying non-topological solutions of (1.14). Let u be a radially symmetric non-topological solution of (1.14) and satisfies u(x) = −2αln|x|+O(1) near∞. Then the Pohozaev identity gives

(α−1)²−(N+ 1)²=1 2

Z ∞ 0

re^2udr >0,

which impliesα > N+ 2. In 2002, Chan, Fu and Lin [4] proved that the inequality α > N+ 2 is also a sufficient condition for the existence of radial non-topological solutionsuwithu(x) =

−2αln|x|+O(1) near∞. However, as pointed out in [8], this might not hold for our problem (1.9) with the asymptotic condition (1.11). The reason is following: there might be a sequence of solutions (u1,n, u2,n) such that only one component blows up, but the other does not, i.e. the so-called phenomena ofpartial blowup; see Theorems C and 1.3 for instance. As a result, only one of theL¹norms ofe^2u1,n ande^2u2,n tends to 0 asn→ ∞, which implies that the quantity J(α₁−1, α₂−1)−J(N₁+ 1, N₂+ 1) might not converge to 0, namely it has a gap. Therefore, roughly speaking, the inequality (1.12) mightnotbe a sufficient condition for the existence of radial non-topological solutions satisfying (1.11).

DefineJ_A2(x, y) =x²+xy+y²for theSU(3) Chern-Simons system (1.8). Recently, Choe, Kim and the second author [8] gave a sufficient condition for the existence of non-topological solutions to system (1.8) subject to the asymptotic condition (1.11).

Theorem B.[8]Let N1, N2 be non-negative integers. Define

S_A2 =







(α₁, α₂)

−2N₁−N₂−3< α₂−α₁< N₁+ 2N₂+ 3 2α1+α2> N1+ 2N2+ 6

α1+ 2α2>2N1+N2+ 6







. (1.15)

Then

S_A2⊂Ω_A2 =

(α₁, α₂)

α₁, α₂>1

J_A2(α₁−1, α₂−1)> J_A2(N₁+ 1, N₂+ 1)

, (1.16)

and for each fixed(α1, α2)∈SA₂, system (1.8)has a radially symmetric non-topological solution (u1, u2)subject to the asymptotic condition (1.11).

In [8], Theorem B was established via the Leray-Schauder degree theory. To apply the degree theory, they made a deformation from (1.8) to theSU(3) Chern-Simons system without singular sources (i.e. N₁ = N₂ = 0 in (1.8)), and then proved a priori estimates for radial solutions satisfying (1.11) under the condition (α₁, α₂)∈S_A2. Furthermore, they also proved the following interesting result, which indicates that the phenomena of partial blowup occurs on some part of∂S_A2, and so S_A2 is anoptimalrange of (α₁, α₂) in view of the degree theory.

Theorem C.[8]LetN1, N2 be non-negative integers. Let(α1, α2)∈ΩA₂ satisfy α₂−α₁=N₁+ 2N₂+ 3.

Then there exists a sequence of radially symmetric bubbling solutions(u1,n, u2,n)to system(1.8) such thatsup_R2u2,n→ −∞asn→ ∞. Furthermore,

(i) there is a intersection pointR_1,n1ofu_1,nandu_2,nsuch thatu_1,n→UinC_loc² (B(0, R_1,n)) andlim_n→∞R∞

R1,nre^u1,ndr= 0, where U is the unique radial solution of (∆U+ 2e^U −4e^2U = 4πN₁δ₀ inR²,

U(x) =−2(α1+α2−1) ln|x|+O(1) as|x| → ∞.

(ii) there exists(α_1,n, α_2,n)∈Ωsuch that

uj,n(x) =−2αj,nln|x|+O(1) as|x| →+∞, j= 1,2, and(α_1,n, α_2,n)→(α₁, α₂)asn→ ∞.

The purpose of this paper is to generalize Theorems B and C to the more general system (1.9). Clearly, as pointed out in [14, 17] and seen also in the following proof, this more general situation poses new analytical difficulties compared to theSU(3) case. Define

Ω :={(α1, α2)|α1, α2>1 andJ(α1−1, α2−1)> J(N1+ 1, N2+ 1)}, (1.17) and

S:={(α1, α₂)|α₁, α₂>0 and (α₁, α₂) satisfies (1.19)−(1.22)}, (1.18) where

[(1 +a1)(1 +a2)−2a1a2]α2−a2(1 +a2)α1

< a2(1 +a2)N1+ (1 +a1)(1 +a2)N2+ 2(1 +a1+a2), (1.19) [(1 +a₁)(1 +a₂)−2a₁a₂]α₁−a₁(1 +a₁)α₂

< a1(1 +a1)N2+ (1 +a1)(1 +a2)N1+ 2(1 +a1+a2), (1.20) [3(1 +a1)(1 +a2)−4a1a2]α1+1 +a1

a2

[(1 +a1)(1 +a2)−2a1a2]α2

>(1 +a₁)(1 +a₂)N₁+(1 +a1)²(1 +a2)

a₂ N₂+

4 + 21 +a1

a₂

(1 +a₁+a₂), (1.21) [3(1 +a1)(1 +a2)−4a1a2]α2+1 +a2

a₁ [(1 +a1)(1 +a2)−2a1a2]α1

>(1 +a1)(1 +a2)N2+(1 +a2)²(1 +a1) a1

N1+

4 + 21 +a2

a1

(1 +a1+a2). (1.22) Then our first result is following.

Theorem 1.1. Let N1, N2 be non-negative integers. Suppose thata1, a2>0 satisfies (1 +a₁)(1 +a₂)>

6−2√ 5

a₁a₂. (1.23)

Let Ω and S be defined in (1.17)-(1.18). Then S∩Ω 6=∅, and for any fixed (α₁, α₂) ∈ S∩ Ω, system (1.9) admits a radially symmetric non-topological solution (u1, u2) satisfying the asymptotic condition (1.11).

Remark 1.1. Theorem 1.1 gives a partial answer to an open question raised by Yang [27, Section 8] in 1997. We will prove in Section 2 thatS6=∅if and only if(a₁, a₂)satisfies(1.23).

Hence (1.23) is a necessary condition for our result. For the SU(3) system (1.8), we have (a₁, a₂) = (1,1), which yields S = S_A2, Ω = Ω_A2 and J = J_A2, so Theorem 1.1 reduces to Theorem B. Differently from the SU(3) case, the statementS ⊂Ωmight not hold in general.

For example, we will prove in Section 2 that

(S⊂Ω if(1 +a1)(1 +a2)−4a1a2≤0,

S6⊂Ω if(1 +a₁)(1 +a₂)−4a₁a₂>0 small enough.

It is interesting that, when (1 +a1)(1 +a2)≤2a1a2, then (1.19)-(1.20) hold automatically for allα1, α2 >0. For example, let us consider the Cartan matrix B2 case. Let K =B2 in (1.1), then system (1.1) becomes the followingB₂ Chern-Simons system

(∆u1+ 2e^u1−e^u2−4e^2u1+ 2e^2u2 = 4πN1δ0

∆u2+ 2e^u2−2e^u1−4e^2u2+ 2e^u1+u2+ 4e^2u1 = 4πN2δ0

in R². (1.24) For thisB2system, we have (a1, a2) = (2,3) and so (1 +a1)(1 +a2) = 2a1a2. Then by applying Theorem 1.1 directly, we easily obtain

Theorem 1.2. Let N₁, N₂ be non-negative integers. Then for any(α₁, α₂)satisfying

α₁> N₁+N₂+ 3 and α₂>2N₁+N₂+ 4, (1.25) the B₂ Chern-Simons system (1.24) has a radially symmetric non-topological solution(u₁, u₂) subject to the asymptotic condition (1.11).

For the Cartan matrixK=G2case, we have (a1, a2) = (5,9), which does not satisfy (1.23).

Therefore, Theorem 1.1 can not be applied to theG2 case unfortunately. For the case (1 +a1)(1 +a2)≤

6−2√ 5

a1a2

(such as theG₂case), the existence of non-topological solutions of system (1.9) subject to the asymptotic condition (1.11) remains open.

As in [8], we will also prove Theorem 1.1 via the degree theory. Differently, here we make a continuous deformation from our original problem (1.9) to theSU(3) Chern-Simons system (1.8). By proving a priori estimates for this deformation problem, we will apply the homotopy invariance of the Leray-Schauder degree to prove Theorem 1.1. Since the variational method does not seem to work for the cooperative case, we believe that the degree theory is a feasible way to treat the cooperative case, which will be considered in a future project.

After Theorem 1.1, we may ask a natural question: Is the set S∩Ω an optimal range of (α1, α2)for the existence of radial solutions satisfying (1.11) in view of the degree theory? To answer this question, we prove the following result about the existence of bubbling solutions, which shows that the phenomena of partial blowup also occurs on some part of the boundary ofS∩Ω just as Theorem C.

Theorem 1.3. Let N₁, N₂ be non-negative integers. Suppose thata₁, a₂>0 satisfies

(1 +a₁)(1 +a₂)>2a₁a₂. (1.26) Let(α₁, α₂)∈Ω satisfyα₁6=α₂ and

[(1 +a1)(1 +a2)−2a1a2]α2−a2(1 +a2)α1

=a₂(1 +a₂)N₁+ (1 +a₁)(1 +a₂)N₂+ 2(1 +a₁+a₂). (1.27) Then system (1.9)admits a sequence of radially symmetric bubbling solutions(u_1,n, u_2,n) such thatsup_R2u_2,n→ −∞asn→ ∞. Furthermore,

(i) there exists a intersection point R_1,n 1 of u_1,n and u_2,n such that u_1,n → U in C_loc² (B(0, R_1,n)), whereU is the unique radial solution of

(∆U + (1 +a1)e^U −(1 +a1)²e^2U = 4πN1δ0 inR², U(x) =−2γln|x|+O(1) as|x| → ∞

withγ=α1+_1+a^2a1

2(α2−1). Besides,lim_n→∞R∞

R_1,nre^u1,ndr= 0.

(ii) there exists(α1,n, α2,n)∈Ωsuch that

uj,n(x) =−2αj,nln|x|+O(1) as|x| →+∞, j= 1,2, and(α1,n, α2,n)→(α1, α2)asn→ ∞.

Remark 1.2. Clearly Theorem 1.3 generalizes Theorem C. Observe that (1.26) is a neces- sary condition for Theorem 1.3, otherwise there are no (α1, α2) ∈ Ω satisfying (1.27). The assumption (α₁, α₂) ∈ Ω is also necessary in view of the Pohozaev identity. In the case (1 +a₁)(1 +a₂)≤4a₁a₂, we can prove that the assumption (α₁, α₂)∈ Ωholds automatically provided thatα₁, α₂>1 satisfy (1.27); see Remark 5.1.

Remark 1.3. The assumption α₁ 6= α₂ is a technical condition, since we can not treat the caseα1 =α2. In theSU(3) case(a1, a2) = (1,1) studied in [8], α2> α1 holds automatically in Theorem C. In Theorem 1.3 here, since we deal with generic(a1, a2), the relation α2> α1

holds automatically only when a2≥1 (i.e. (1 +a1)(1 +a2)−2a1a2≤a2(1 +a2)). Therefore, the assumptionα16=α2 in Theorem 1.3 is equivalent to

α26=α0:=a₂(1 +a₂)N₁+ (1 +a₁)(1 +a₂)N₂ (1−a2)(1 +a1+a2) + 2

1−a2

ifa2∈(0,1).

More precisely, when a₂ ∈ (0,1), we have α₁ < α₂ if and only if α₂ < α₀, and α₂ < α₁ if and only ifα₂> α₀. We will see in Section 5 that the situation is different with respect to the sign ofα1−α2. This phenomena provides an evidence that the general problem (1.9) is more involved than theSU(3)Chern-Simons system (1.8).

The rest of this paper is organized as follows. In Section 2, we give the deformation of the problem and prove some preliminary lemmas. In Section 3, by a delicate blow up analysis, we prove a priori estimates for radial non-topological solutions satisfying (1.11) under the assumption (α1, α2)∈S∩Ω. In Section 4, we complete the proof of Theorem 1.1 by applying the degree theory. In Section 5, we prove Theorem 1.3 via the shooting method.

2 Preliminaries and a deformation of the problem

In the sequel, we denote

A= (1 +a1)(1 +a2) and B=a1a2

for convenience. ThenA > B+ 1 since a1, a2>0. First we prove some facts aboutS and Ω mentioned in Remark 1.1.

Lemma 2.1. Let ΩandS be in (1.17)-(1.18).

(i) There holds

S6=∅ ⇐⇒(1 +a₁)(1 +a₂)>

6−2√ 5

a₁a₂. (2.1)

(ii) Let (1 +a₁)(1 +a₂)> 6−2√ 5

a₁a₂. Then

α₁>1, α₂>1, ∀(α₁, α₂)∈S. (2.2) Furthermore,S∩Ω6=∅ and

(S⊂Ω if (1 +a1)(1 +a2)−4a1a2≤0,

S6⊂Ω if (1 +a₁)(1 +a₂)−4a₁a₂>0small enough. (2.3) Proof. Recalling (1.19)-(1.22), we define

h1(α1, α2) =(A−2B)α2−a2(1 +a2)α1−a2(1 +a2)N1−AN2−2(A−B), (2.4) h₂(α₁, α₂) =(A−2B)α₁−a₁(1 +a₁)α₂−a₁(1 +a₁)N₂−AN₁−2(A−B), (2.5) h3(α1, α2) =(3A−4B)α1+1 +a₁

a2

(A−2B)α2

−AN1−1 +a1

a2

AN2−

4 + 21 +a1

a2

(A−B), (2.6)

h4(α1, α2) =(3A−4B)α2+1 +a2

a1

(A−2B)α1

−AN2−1 +a2

a1

AN1−

4 + 21 +a2

a1

(A−B). (2.7)

Clearly

S=

(α1, α2)

α1>0, h1(α1, α2)<0, h2(α1, α2)<0 α2>0, h3(α1, α2)>0, h4(α1, α2)>0

. Case 1. A−2B≤0.

In this case,h₁(α₁, α₂)<0 andh₂(α₁, α₂)<0 always hold for allα₁>0, α₂>0. Obviously, S=∅if 3A−4B≤0, andS6=∅ifA−2B = 0. So we assume 3A−4B >0> A−2B. Then it is easy to see thatS6=∅ if and only if the slopes of linesh3= 0 andh4= 0 satisfy

3A−4B

1+a₁

a₂ (2B−A) >

1+a₂

a1 (2B−A)

3A−4B . (2.8)

A direct calculation shows that (2.8) is equivalent to (A−B)(A²−12AB+ 16B²)<0. So A >(6−2√

5)B (note that 6−2√

5>4/3). (2.9)

This proves (2.1) in this case.

Now we let (2.9) holds and take any (α1, α2)∈ S. Then h3(α1, α2)>0 and A−2B ≤0 yield

(3A−4B)α1> AN1+ 4(A−B)≥(3A−4B)(N1+ 2),

namelyα1> N1+ 2. Similarly,α2> N2+ 2. Consequently, J(α₁−1, α₂−1)> J(N₁+ 1, N₂+ 1), namelyS⊂Ω in this case.

Case 2. A−2B >0 andA−4B≤0.

In this case, we have

a₂(1 +a₂)

A−2B ≥ A−2B a1(1 +a1)>0.

Consequently, the lineh1= 0 can not intersect with the lineh2= 0 in the first quadrant. Since the slopes ofh3= 0 andh4= 0 are both negative, so it is trivial to see thatS 6=∅.

To prove (2.2), we take any (α₁, α₂)∈S. Byh₁(α₁, α₂)<0 we have (A−2B)α2< a2(1 +a2)α1+a2(1 +a2)N1+AN2+ 2(A−B).

Substituting this inequality intoh₃(α₁, α₂)>0, we easily concludeα₁>1. Similarly,α₂>1.

Remark that this argument of proving (2.2) also works for Case 3 whereA−4B >0.

It remains to proveS⊂Ω. Note that the intersection point (β₁, β₂) ofh₃= 0 andh₄= 0 is







β1=^{A(A−4B)(N1+1)−}

2(1+a1 )

a2 AB(N₂+1) A²−12AB+16B² + 1, β2=^{A(A−4B)(N2+1)−}

2(1+a2 )

a1 AB(N₁+1) A²−12AB+16B² + 1.

(2.10)

SinceA²−12AB+ 16B²<0, we haveβ₁>1 and β₂>1. Take any (α₁, α₂)∈S. Then either α₁> β₁orα₂> β₂. Without loss of generality, we assumeα₂> β₂. In the following argument, we write

˜

αj:=αj−1, β˜j :=βj−1 and ˜Nj:=Nj+ 1 for convenience. Then ˜α₂>β˜₂. By h₃(α₁, α₂)>0 we have

(3A−4B) ˜α1>AN˜1+1 +a1

a2

AN˜2−1 +a1

a2

(A−2B) ˜α2

=:C−1 +a1

a₂ (A−2B) ˜α2. (2.11)

Recalling the definition (1.10) ofJ, a direct computation shows that J((3A−4B) ˜α1,(3A−4B) ˜α2)> J

C−1 +a1

a₂ (A−2B) ˜α2,(3A−4B) ˜α2

holds if we have 1 +a2

2a₁

(3A−4B) ˜α₁+AN˜₁+1 +a1

a₂ AN˜₂

+

3A−4B− A

2B(A−2B)

˜ α₂>0.

SinceA−4B≤0 gives 3A−4B−_2B^A (A−2B)>0, the above inequality holds. So J( ˜α₁,α˜₂)> 1

(3A−4B)²J

C−1 +a1

a₂ (A−2B) ˜α₂,(3A−4B) ˜α₂

=

a2(1+a2)

2 C²+^a1(1+a1_2B^)A2(A−B)α˜²₂−(A−B)(A−4B)Cα˜2

(3A−4B)²

>

a₂(1+a₂)

2 C²+^a1(1+a1_2B^)A2(A−B)β˜₂²−(A−B)(A−4B)Cβ˜₂

(3A−4B)² .

Recalling from (2.10)-(2.11) that

β˜2=A(A−4B) ˜N₂−^2(1+a_a ²⁾

1 ABN˜₁

A²−12AB+ 16B² , C=AN˜1+1 +a1

a2

AN˜2.

By substituting these two expressions into the last inequality and by a direct calculation, we conclude fromA−4B ≤0 that

J( ˜α1,α˜2)>a2(1 +a2) 2

1 + 16B(A−B)(A−4B)² (A²−12AB+ 16B²)²

N˜₁² +a1a2

1−4(A−B)(A−4B)(A²−4AB+ 16B²) (A²−12AB+ 16B²)²

N˜1N˜2

+a1(1 +a1) 2

1 + 16B(A−B)(A−4B)² (A²−12AB+ 16B²)²

N˜₂² (2.12)

≥J( ˜N1,N˜2).

Consequently, (α₁, α₂)∈Ω. The caseα₁> β₁is similar. Therefore,S ⊂Ω.

Case 3. A−4B >0.

Then the intersection point (γ₁, γ₂) ofh₁= 0 andh₂= 0 is (γ1= ^A(N1+1)+2a_A−4B^{1(1+a1)(N2+1)}+ 1>1,

γ2= ^A(N2+1)+2a_A−4B^{2(1+a2)(N1+1)}+ 1>1.

A direct calculation shows that hj(γ1, γ2) > 0 holds for j = 3,4. Therefore, S 6= ∅ with (γ1, γ2)∈∂S. We also write ˜γj =γj−1. Then

J(˜γ1,˜γ2)−J( ˜N1,N˜2)

=J(AN˜₁+ 2a₁(1 +a₁) ˜N₂, AN˜₂+ 2a₂(1 +a₂) ˜N₁)−(A−4B)²J( ˜N₁,N˜₂) (A−4B)²

=4B2a2(1 +a2)(A−B) ˜N₁²+ (A²+ 3AB−4B²) ˜N1N˜2+ 2a1(1 +a1)(A−B) ˜N₂² (A−4B)²

>0,

namely (γ₁, γ₂)∈Ω∩∂S. Since Ω is an open set, we conclude S∩Ω6=∅.

Finally, we need to prove thatS 6⊂Ω providedA−4B >0 small enough. Let 4B < A <5B, thenA²−12AB+16B²<0 anda1, a2are uniformly bounded. Note thatA= (1+a1)(1+a2)>1 and soB >1/5. Recalling (2.10), we can prove that bothh1(β1, β2)<0 andh2(β1, β2)<0 if and only if

2ABN˜₂> a₂(1 +a₂)(A−4B) ˜N₁ and 2ABN˜₁> a₁(1 +a₁)(A−4B) ˜N₂,

which are always true providedA−4B >0 small enough. So (β1, β2)∈∂S ifA−4B >0 is small enough. Also by (2.10), we can compute (the expression is the same as (2.12))

J( ˜β₁,β˜₂)−J( ˜N₁,N˜₂)

=J(A(A−4B) ˜N1−^2(1+a_a ¹⁾

2 ABN˜2, A(A−4B) ˜N2−^2(1+a_a ²⁾

1 ABN˜1)

(A²−12AB+ 16B²)² −J( ˜N₁,N˜₂)

= B(A−B)(A−4B) (A²−12AB+ 16B²)²

h8a₂(1 +a₂)(A−4B) ˜N₁²

−4[(A−4B)²+ 4AB)] ˜N1N˜2+ 8a1(1 +a1)(A−4B) ˜N₂²i

<0 ifA−4B >0 small enough,

namelyJ(β₁−1, β₂−1)< J(N₁+ 1, N₂+ 1) ifA−4B >0 small enough. Since (β₁, β₂)∈∂S, we conclude thatS 6⊂Ω forA−4B >0 small enough. This prove the lemma.

From now on, we always assume that (1.23) holds. As in [8], we will employ the Leray- Schauder degree theory to prove Theorem 1.1. Fort∈[0,1], we define

bk =bk(t) := 1 +t(ak−1), k= 1,2. (2.13)

Then

0<min{1, a1, a2} ≤b1, b2≤max{1, a1, a2}, ∀t∈[0,1], (2.14) and a straightforward computation gives

(1 +b₁)(1 +b₂)>

6−2√ 5

b₁b₂, ∀t∈[0,1]. (2.15) Consider the following deformation of system (1.9)











∆u1+ (1 +b1)(e^u1−(1 +b1)e^2u1+b1e^u1+u2)

−b1(e^u2−(1 +b2)e^2u2+b2e^u1+u2) = 4πN1δ0

∆u₂+ (1 +b₂)(e^u2−(1 +b₂)e^2u2+b₂e^u1+u2)

−b₂(e^u1−(1 +b₁)e^2u1+b₁e^u1+u2) = 4πN₂δ₀

inR². (2.16)

Clearly, ift= 1, thenbk=ak and system (2.16) is just our original problem (1.9); ift= 0, then bk = 1 and system (2.16) is just theSU(3) Chern-Simons system (1.8). As before, we define

Jt(x, y) := b2(1 +b2)

2 x²+b1b2xy+b1(1 +b1)

2 y², (2.17)

Ωt:={(α1, α2)|α1, α2>1 andJt(α1−1, α2−1)> Jt(N1+ 1, N2+ 1)}, (2.18) and

S_t:={(α1, α₂)|α₁, α₂>0 and (α₁, α₂) satisfies (2.20)−(2.23)}, (2.19) where

[(1 +b₁)(1 +b₂)−2b₁b₂]α₂−b₂(1 +b₂)α₁

< b2(1 +b2)N1+ (1 +b1)(1 +b2)N2+ 2(1 +b1+b2), (2.20) [(1 +b₁)(1 +b₂)−2b₁b₂]α₁−b₁(1 +b₁)α₂

< b1(1 +b1)N2+ (1 +b1)(1 +b2)N1+ 2(1 +b1+b2), (2.21) [3(1 +b₁)(1 +b₂)−4b₁b₂]α₁+1 +b₁

b2

[(1 +b₁)(1 +b₂)−2b₁b₂]α₂

>(1 +b₁)(1 +b₂)N₁+(1 +b1)²(1 +b2)

b₂ N₂+

4 + 21 +b1

b₂

(1 +b₁+b₂), (2.22) [3(1 +b1)(1 +b2)−4b1b2]α2+1 +b2

b1

[(1 +b1)(1 +b2)−2b1b2]α1

>(1 +b1)(1 +b2)N2+(1 +b₂)²(1 +b₁) b1

N1+

4 + 21 +b₂ b1

(1 +b1+b2). (2.23) Then Lemma 2.1 tell us thatSt∩Ωt6=∅for allt∈[0,1]. Remark thatJ1=J,S1=S, Ω1= Ω andJ0=JA₂,S0=SA₂, Ω0= ΩA₂. Moreover,JtandSt∩Ωtare both continuous with respect tot∈[0,1].

From now on, we fix any (α1, α2)∈S∩Ω. Our goal is to prove the existence of a radial non- topological solution of system (1.9) subject to the asymptotic condition (1.11). SinceSt∩Ωt

is open and continuous, we can take a continuous function (β1, β2) : [0,1]→R² such that (β1, β2)(1) = (α1, α2) and (β1, β2)(t)∈St∩Ωt ∀t∈[0,1]. (2.24) Obviously, (2.24) implies the existence of a constantc₀>0 such that

min

t∈[0,1]βk(t)≥1 +c0, k= 1,2. (2.25) Then we turn to study the existence of radially symmetric solutions to system (2.16) subject to the following asymptotic condition

uk(x) =−2βkln|x|+O(1) as|x| →+∞, k= 1,2. (2.26)

By Theorem B we know that, fort= 0, system (2.16) has a radial solution satisfying (2.26).

In fact, Choe, Kim and Lin [8] proved that the Leray-Schauder degree is not 0. Therefore, our strategy is to prove the uniform boundedness of radially symmetric solutions satisfying (2.26) whent varies in [0,1]; see Section 3. Consequently, the degree theory can be applied and the degree is invariant under this deformation, so Theorem 1.1 follows; see Section 4.

In the sequel, we always denote positive constants independent oft∈[0,1] (possibly different in different places) byC, C₀, C₁,· · ·. Before ending this section, we prove some useful results that are needed in Section 3. First we prove the Pohozaev identity of system (2.16). Clearly, any radially symmetric solution (u₁, u₂) depends on r = |x| as well as t. Also, b_k(t) and β_k(t) are continuous functions oft. When there is no confusion arising, we denote (u₁, u₂) by (u₁(r), u₂(r)), b_k(t) by b_k and β_k(t) by β_k for convenience (i.e. we omit the notation t). We will use ∆u=u⁰⁰+¹_ru⁰ frequently, whereu⁰(r) =^du_dr andu⁰⁰(r) =^d_dr^2u2.

By (2.24), it is easy to see that inf

t∈[0,1]

J_t(β₁(t)−1, β₂(t)−1)−J_t(N₁+ 1, N₂+ 1)

>0. (2.27)

Lemma 2.2. Let (u₁, u₂) be radially symmetric solutions of system (2.16) satisfying (2.26).

Thenu₁<0 andu₂<0 inR², and

Jt(β1−1, β2−1)−Jt(N1+ 1, N2+ 1)

=1 +b1+b2

4

Z ∞ 0

r

b2(1 +b1)e^2u1+b1(1 +b2)e^2u2−2b1b2e^u1+u2

dr. (2.28) Furthermore, there exist positive constantsC1 andC2 independent oft∈[0,1]such that

Z ∞ 0

r(e^2u1+e^2u2)dr≥C1, (2.29)

Z ∞ 0

r(e^u1+e^u2+e^2u1+e^2u2+e^u1+u2)dr≤C2. (2.30) Proof. The fact u1, u2<0 inR² was proved in Theorem A. By (2.16) and (2.26), we have

r→0limru⁰_k(r) = 2Nk, lim

r→∞ru⁰_k(r) =−2βk, k= 1,2. (2.31) Since (u1, u2) is radially symmetric, system (2.16) can be written as

(1 +b2)(ru⁰₁)⁰+b1(ru⁰₂)⁰=−(1 +b1+b2)r(e^u1−(1 +b1)e^2u1+b1e^u1+u2), (2.32) b2(ru⁰₁)⁰+ (1 +b1)(ru⁰₂)⁰=−(1 +b1+b2)r(e^u2−(1 +b2)e^2u2+b2e^u1+u2), (2.33) for 0< r <∞. We multiply (2.32) by b2ru⁰₁, (2.33) byb1ru⁰₂ and integrate them over [ε, R], which yields

Jt(ru⁰₁,ru⁰₂)

R ε =

b2(1 +b2)

2 (ru⁰₁)²+b1b2r²u⁰₁u⁰₂+b1(1 +b1) 2 (ru⁰₂)²

R

ε

=−(1 +b1+b2)r²

b2

e^u1−1 +b1

2 e^2u1

+b1

e^u2−1 +b2

2 e^2u2

+b1b2e^u1+u2

R

ε

+ 2(1 +b1+b2) Z R

ε

r

b2

e^u1−1 +b₁ 2 e^2u1

+b1

e^u2−1 +b2

2 e^2u2

+b1b2e^u1+u2

dr. (2.34)

By lettingε→0 andR→ ∞, it follows from (2.26) and (2.31) that J_t(2β₁,2β₂)−J_t(2N₁,2N₂) = 2(1 +b₁+b₂)

Z ∞ 0

r

b₂

e^u1−1 +b1

2 e^2u1

+b1

e^u2−1 +b₂ 2 e^2u2

+b1b2e^u1+u2

dr. (2.35) On the other hand, by integrating (2.32) and (2.33) over (0,∞), we obtain

(1 +b1+b2) Z ∞

0

r(e^u1−(1 +b1)e^2u1+b1e^u1+u2)dr

= 2(1 +b2)(β1+N1) + 2b1(β2+N2), (2.36) (1 +b₁+b₂)

Z ∞ 0

r(e^u2−(1 +b₂)e^2u2+b₂e^u1+u2)dr

= 2b₂(β₁+N₁) + 2(1 +b₁)(β₂+N₂). (2.37) By eliminating the termsR∞

0 re^ukdrin (2.35) via (2.36)-(2.37), we obtain the Pohozaev identity (2.28), and (2.29) follows directly from (2.14), (2.27) and (2.28). Since 2e^u1+u2 ≤e^2u1+e^2u2, (2.28) also implies

Z ∞ 0

r(e^2u1+e^2u2+e^u1+u2)dr≤C

for someC >0 independent oft. This, together with (2.36)-(2.37), gives (2.30).

Lemma 2.3. Let (u1, u2) be radially symmetric solutions of system (2.16) satisfying (2.26).

Then there exist constantsC1, C2>2 + 2N1+ 2N2 independent of t such that

|ru⁰_k(r)−2Nk| ≤

(C1r² if0< r <1,

C₁ ifr≥1, k= 1,2, (2.38)

uk(r)≤ −2 lnr+C2 ∀r≥1, k= 1,2. (2.39) Proof. Note that e^uk < 1 in R². Obviously, (2.38) follows directly from (2.16), (2.30) and (2.31). By (2.38), we have

uk(s) =uk(r) + Z s

r

u⁰_k(ρ)dρ≥uk(r)−(C1−2Nk) lns

r, fors≥r≥1.

Consequently, C≥

Z ∞ r

se^uk(s)ds≥e^uk(r) Z ∞

r

sr s

C1−2Nk

ds= e^uk(r)r²

C₁−2−2N_k, ∀r≥1, which implies (2.39).

Lemma 2.4. Recall Jtin (2.17). There holds Jt(x, y) =Jt(−x,−y)

=Jt

x,− 2b2

1 +b1

x−y

=Jt

−x, 2b2

1 +b1

x+y

=Jt

−x− 2b1

1 +b2

y, y

=Jt

x+ 2b1

1 +b2

y,−y

.

Proof. These formulae can be proved via direct calculations, and we omit the details.

3 A priori estimates

In this section, in order to apply the degree theory, we prove a priori estimates of radially symmetric solutions of (2.16) satisfying (2.26). Assume that (u₁, u₂) are any radially symmetric solutions of (2.16) satisfying (2.26). Define

fk(x) =fk(x;t) := 2Nkln|x| −(βk(t) +Nk) ln(1 +|x|²), k= 1,2. (3.1) Clearlyfk<0 inR². The main result of this section is following