Classification of the entire radial self-dual solutions to non-Abelian Chern–Simons systems

Journal of Functional Analysis 266 (2014) 6796–6841

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Journal of Functional Analysis

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Classification of the entire radial self-dual solutionsto non-Abelian Chern–Simons systems

Hsin-Yuan Huang a, Chang-Shou Lin b,∗

a Department of Applied Mathematics, National Sun Yat-sen University,Kaoshiung 804, Taiwanb Taida Institute for Mathematical Sciences, Center for Advanced Studyin Theoretical Science, National Taiwan University, Taipei 106, Taiwan

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

Article history:Received 14 October 2013Accepted 10 March 2014Available online 31 March 2014Communicated by S. Brendle

Keywords:Non-Abelian Chern–Simons systemClassificationNon-topological solutionsTopological solutions

We consider the entire radial solutions to the non-AbelianChern–Simons systems of rank 2(

ΔuΔv

)= −K

(eu 00 ev

(4πN1δ04πN2δ0

2, (0.1)

where Ni � 0, i = 1, 2 and K = (aij) is a 2 × 2 matrix satis-fying a11, a22 > 0, a12, a21 < 0 and a11a22 −a12a21 > 0. Thissystem is motivated by the relativistic non-Abelian Chern–Simons model, Lozano–Marqués–Moreno–Schaposnik modelof bosonic sector of N = 2 supersymmetric Chern–Simons–Higgs theory, and Gudnason model of N = 2 supersymmetricYang–Mills–Chern–Simons–Higgs theory. Understanding thestructure of entire radial solutions is one of fundamental is-sues for the system of nonlinear equations. We prove that anyentire radial solutions of (0.1) must be one of topological, non-topological and mixed type solutions, and completely classifythe asymptotic behaviors at infinity of these solutions. As anapplication of this classification, we prove that the two com-ponents u and v have intersection at most finite times.

* Corresponding author.E-mail addresses: hyhuang@math.nsysu.edu.tw (H.-Y. Huang), cslin@tims.ntu.edu.tw (C.-S. Lin).

H.-Y. Huang, C.-S. Lin / Journal of Functional Analysis 266 (2014) 6796–6841 6797

1. Introduction

In this paper, we consider the non-Abelian Chern–Simons systems of rank 2(Δu

)= −K

(eu 00 ev

4πN1δ04πN2δ0

)(1.1)

where K = (aij) is a 2 × 2 matrix satisfying

aii > 0, aij < 0 (i �= j), i, j ∈ {1, 2} and detK > 0, (1.2)

N1, N2 � 0, and δ0 is Dirac measure at the origin in R2. Obviously, (1.2) implies∑2

j=1(K−1)ij > 0, i = 1, 2, where K−1 is the inverse matrix of K. This system ismotivated by many non-Abelian gauged field models in physics, including the relativis-tic non-Abelian Chern–Simons model proposed by Kao and Lee [13] and Dunne [3,4],Lozano–Marqués–Moreno–Schaposnik model [20] of bosonic sector of N = 2 supersym-metric Chern–Simons–Higgs theory and Gudnason model [5,6] of N = 2 supersymmetricYang–Mills–Chern–Simons–Higgs theory. A typical example of K is the SU(3) Cartanmatrix

(2 −1−1 2

), where (1.1) is a perturbation of the well-known SU(3) Toda system

2 −1−1 2

). (1.3)

Recently, the SU(3)-Toda system has been extensively studied from the analytic pointof view. We refer the reader to [10–12,14,16–18,21,24,26] and reference therein for re-cent developments for the analytic studies of the Toda system and [1,19,22,27] for thecorresponding Chern–Simons SU(3) system.

Our first motivation is the relativistic non-Abelian Chern–Simons model proposed byKao and Lee [13] and Dunne [3,4]. This 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. TheChern–Simons Lagrangian density L in 2 + 1 dimensional spacetime involves the Higgsfield φ and the gauge potential A = (A0, A1, A2). We consider the energy minimizers ofthe Lagrangian functional, and thus obtain the self-dual Chern–Simons equations

D−φ = 0

F+− = 1κ2

[φ−

[[φ, φ†], φ], φ†] (1.4)

where D− = D1−iD2, κ2 > 0, and F+− = ∂+A−−∂−A++[A+, A−] with A± = A1±iA2and ∂± = ∂1 ± i∂2. Dunne considered a simplified form of the self-dual system (1.4), inwhich the fields φ and A are algebraically restricted:

φ =r∑

6798 H.-Y. Huang, C.-S. Lin / Journal of Functional Analysis 266 (2014) 6796–6841

where r is the rank of the gauge Lie algebra, Ea is the simple root step operator, andφa are complex-valued functions. Let

ua = log∣∣φa

∣∣, a = 1, 2, . . . , r.

Then Eqs. (1.4) can be reduced to the following system of equations

Δua + 1k2

Kabeub −

r∑b=1

r∑c=1

eubKbceucKac

)= 4π

Na∑j=1

δpaj,

a = 1, . . . , r, (1.5)

where K = (Kab) is the Cartan matrix of a semi-simple Lie algebra, {paj }j=1,...,Naare

zeros of φa (a = 1, . . . , r), and δp is the Dirac measure concentrated at p in R2. For

example, there are three types of Cartan matrix of rank 2, which are given by

A2(which is SU(3)

(2 −1−1 2

), B2 =

(2 −1−2 2

), G2 =

(2 −1−3 2

Let (K−1)ab be the inverse of the matrix K, and assume

r∑b=1

(K−1)

ab> 0, a = 1, 2, . . . , r. (1.7)

A solution u = (ua) of (1.5) is called topological solution if

ua(x) → log(

r∑b=1

(K−1)

)as |x| → +∞, a = 1, 2, . . . , r;

is called non-topological solution if

ua(x) → −∞ as |x| → +∞, a = 1, 2, . . . , r.

The existence of topological solutions either in R2 or a flat torus has been studied

in recent twenty years. In R2, Yang [27] used variational methods and a Cholesky de-

composition technique to obtain the existence of topological solutions to (1.5) for anyconfiguration when K satisfies (1.7). Nevertheless, for a flat tours, Nolasco and Tarantello[23] studied (1.5) with the Cartan matrix SU(3) and obtained the topological solutionsas well as mountain-pass solutions. Similar results for the solutions in the doubly peri-odic domain with K satisfying (1.2) can be founded in [7] and [8]. Roughly speaking, theexistence theory of topological solutions is satisfying to some extent. However, findingnon-topological solutions (or mixed-type solutions, see below) is far more difficult than

topological solutions. Up to now, there are very few results concerning the existence ofnon-topological solutions. Only recently, some existence of non-topological solutions to(1.5) with cases of A2 and B2 have been studied by Ao, Lin and Wei [1] by a perturba-tion from the A2 and B2 Toda system with singular sources. However, their result is stillvery limited toward understanding the general theory of non-topological solutions (andmixed-type solutions). Our study of entire radial solutions would play a significant rolefor this purpose. The following classification is the first major result in this direction.

Theorem 1.1. Suppose (u(r), v(r)) is an entire radial solution to (1.1). One of the fol-lowing holds.

(i) (u, v) is a topological solution.(ii) (u, v) is a mixed-type solution. More precisely, either eu ∈ L1(R2) and v(r) → some

constant as r → ∞, or ev ∈ L1(R2) and u(r) → some constant as r → ∞,(iii) (u, v) is a non-topological solution, and eu, ev ∈ L1(R2).

The strategy of our proof of Theorem 1.1 is to show that both nonlinear terms on theright hand side of (1.1) are L1 functions in R

2. For the purpose, we should simplify theexpression of (1.1) first. For simplicity, we let K =

(α −β−γ δ

), where α, β, γ and δ are

positive constants. By considering the transformation

(u, v) →(u + log β + δ

αδ − βγ, v + log α + γ

αδ − βγ

)and letting (

β(α + γ)αδ − βγ

,γ(β + δ)αδ − βγ

)= (a, b).

Then (1.1) becomes{Δu = −(1 + a)

(eu − (1 + a)e2u + aeu+v

(ev − (1 + b)e2v + beu+v

)+ 4πN1δ0

Δv = −(1 + b)(ev − (1 + b)e2v + beu+v

(eu − (1 + a)e2u + aeu+v

)+ 4πN2δ0

Note that a, b > 0 because K satisfies (1.2). By using (1.8), we see that (u, v) is a topologi-cal solution if (u, v) → (0, 0) as r → +∞; a mixed-type solution if (u, v) → (log 1

1+a ,−∞)or (−∞, log 1

1+b ) as r → +∞; a non-topological solution if (u, v) → (−∞,∞) as r → +∞.It is interesting to note that our system (1.8) is related to Lozano–Marqués–Moreno–

Schaposnik model [20] of bosonic sector of N = 2 supersymmetric Chern–Simons–Higgstheory and the Gunderson recent works [5,6] in the N = 2 supersymmetric Yang–Mills–Chern–Simons–Higgs theory.

Lozano, Marqués, Moreno, and Schaposnik [20] considered bosonic sector of N = 2supersymmetric Chern–Simons–Higgs theory when the gauge group is U(1) × SU(N)and has Nf flavors of fundamental matter fields. They investigated so-called local ZN

string-type solutions when Nf = N and obtained the following equations.

r∂rφ = ε

(f − fN2−1)φ, r∂rφN = ε

(f + (N − 1)fN2−1)φN ,

r∂rf = 14Nκ1

[f0((N − 1)φ2 + φ2

)+ fN2−1

0 (N − 1)(φ2 − φ2

)]r∂rf

N2−1 = 14Nκ2

[f0(φ2 − φ2

)+ fN2−1

0 (N − 1)(φ2 + (N − 1)φ2

)]f0 = ε

((N − 1)φ2 + φ2

N −N), fN2−1

0 = ε

(φ2 − φ2

), (1.9)

where N is positive integer, ε = ±1, and κ1, κ2 > 0. By letting φ = e12u, φN = e

12v and

k = κ1κ2

and using dimensional change of variables, we obtain

Δu = −N − 1 + k

(eu − N − 1 + k

Ne2u + k − 1

)+ k − 1

(ev − 1 + (N − 1)k

Ne2v + (N − 1)(k − 1)

)Δv = −1 + (N − 1)k

(ev − 1 + (N − 1)k

Ne2v + (N − 1)(k − 1)

)+ (N − 1)(k − 1)

(eu − N − 1 + k

Ne2u + k − 1

)(1.10)

In (1.10), we let a = k−1N and b = (N−1)(k−1)

N . Then a, b > 0 if k > 1.In Gudnason model, when the gauge group are SO(2M) and USp(2M), the so-called

master equations are written as follows:

ΔU = α∗M2

(M∑i=1

[eU+ui + eU−ui − 2

])(M∑j=1

[eU+uj + eU−uj

+ α∗β∗M

M∑i=1

(eU+ui − eU−ui

)+ 4π

M∑i=1

ni∑s=1

δpi,s(x),

Δuj = α∗β∗M

(M∑i=1

[eU+ui + eU−ui − 2

])(eU+uj − eU−uj

)+ β2

∗(e2U+2uj − e2U−2uj

)+ 4π

nj∑δpj ,s(x), j = 1, . . . ,M, (1.11)

where α∗ > 0 and β∗ > 0 are constant and {pj , s}s=1,...,nj

j=1,...,M ∈ R2. We refer to [5,6] for

more details on Gudnason’s model. When M = 1 and all vortices at the origin, we letu = U + u1 and v = U − u1 and rescale the dimensional variable, then (1.11) becomes⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Δu = −β∗ + α∗2α∗

(eu − β∗ + α∗

2α∗e2u + β∗ − α∗

2α∗eu+v

)+ β∗ − α∗

2α∗

(ev − β∗ + α∗

2α∗e2v + β∗ − α∗

2α∗eu+v

)+ 8π

n∑j=1

Δv = β∗ + α∗2α∗

(ev − β∗ + α∗

2α∗e2v + β∗ − α∗

2α∗eu+v

)+ β∗ − α∗

2α∗

(eu − β∗ + α∗

2α∗e2u + β∗ − α∗

2α∗eu+v

)(1.12)

In (1.12), we let a = b = β∗−α∗2α∗

. Then a, b > 0 if β∗ > α∗.We will prove Theorem 1.1 by using the system (1.8) instead of (1.1). To begin with

our approach, we apply the maximum principle and obtain the following theorem.

Theorem 1.2. Let (u(r), v(r)) be an entire radial solutions of (1.8). Assume a > 0 andb > 0. Then

u(r) < 0 and v(r) < 0 for r ∈ (0,∞),

unless u(r) = v(r) = 0 for r ∈ (0,∞).

Like in our previous work [9] on the SU(3) case, our strategy is to show both f1 andf2 are positive for large r if (u, v) is not a topological solution, where

f1 = eu − (1 + a)e2u + aeu+v, f2 = ev − (1 + b)e2v + beu+v.

We note that

f1(r) > 0 as long as u(r) < v(r) < 0. (1.13)

Similarly,

f2(r) > 0 as long as v(r) < u(r) < 0. (1.14)

This is a simple, but useful observation in our proof. In this paper, the positivity ofboth f1 and f2 will be established only for large r, but we conjecture that f1 and f2are positive for r > 0. We believe that the positivity of both f1 and f2, will play animportant role for the further study of uniqueness of solutions of the system (1.8). Weshall return this issue in our further works.

If (u, v) is not a topological solution, we show the positivity of f1 and f2 for large r

in the following theorem.

Theorem 1.3. Let (u, v) be an entire radial solution of (1.8). If (u, v) is not a topologicalsolution, then there exists R0 > 0 such that

fi > 0, i = 1, 2 for r > R0.

To prove Theorem 1.3, we will establish the a priori bound for not topological solutionsto (1.8): ⎧⎪⎨⎪⎩

u(r) < log 11 + a

if v(r) � u(r)

v(r) < log 11 + b

if u(r) � v(r)(1.15)

for r large enough. Then Theorem 1.3 follows. By applying the positivity of f1 and f2to the Pohozaev identity, we have

f1, f2 ∈ L1(R

2)and thus Theorem 1.1 follows.

For the topological solutions to (1.8), we have the following property for these solu-tions.

Theorem 1.4. (u, v) is a topological solution to (1.8) if and only if

(1 + 2b)ur(r) + (1 + 2a)vr(r) > 0 on (0,∞).

It is not difficult to see any topological solution (u, v) decays exponentially at infinity,because any topological solution (u, v) near infinity satisfies(

)+ higher order terms of (u, v) =

), (1.16)

where M =(

−(1+a)2−ab a(2+a+b)b(2+a+b) −(1+b)2−ab

)whose eigenvalues are negative. From here, it is

easy to see that the topological solutions decay as fast as√|x|e−λ1|x| at infinity, where

λ1 is the least eigenvalue of −M .By the positivity of f1 and f2, we obtain the asymptotic behaviors for non-topological

solutions and mixed-type solutions at infinity.

Corollary 1.5. Let (u(r), v(r)) be an entire radial solution of (1.8). Then the followinghold.

(1) If (u(r), v(r)) is a non-topological solution, then

u(r) = −β1 log r + O(1) and v(r) = −β2 log r + O(1) (1.17)

at infinity for some β1 > 2 and β2 > 2. Thus,

eu, ev ∈ L1(R

2).Furthermore, β1 and β2 satisfy

b(1 + b)2 β2

1 + abβ1β2 + a(1 + a)2 β2

2 − 4(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)− 2(1 + a + b)

(b(2N1 + β1) + a(2N2 + β2)

)= 2(1 + a + b)

∞∫0

((1 + a)b

2 e2u(s) + (1 + b)a2 e2v(s) − abe(u+v)(s)

> 0 (1.18)

(2) (u(r), v(r)) is a mixed-type solution, then either

u(r) → log 11 + a

and v(r) = −β log r + O(1) for some β > 2,

v(r) → log 11 + b

and u(r) = −β log r + O(1) for some β > 2,

as r → ∞.

Remark 1.6. From (1.18), it is easy to see that (β1, β2) satisfies

b(1 + b)2 β2

1 + abβ1β2 + a(1 + a)2 β2

2 − 4(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)> 2(1 + a + b)

(b(2N1 + β1) + a(2N2 + β2)

)(1.19)

The question whether (1.19) is also a sufficient condition for the existence of (1.8) satis-fying (1.17) is an issue which will be addressed in our coming paper.

To appreciate the result of Corollary 1.5, we could compare it with the followingsystem of equations: {Δu + ev(1 − eu) = 4πN1δ0

Δv + eu(1 − ev) = 4πN2δ0in R

2. (1.20)

The system (1.20) is related to Chern–Simons–Higgs model with two Higgs particles (see[2,15]) and corresponds to K =

( 0 11 0

)in (1.1) (or (a, b) = (−1,−1) in (1.8)). Chern, Chen

and Lin [2] classified all radial solutions of (1.20) according to their behaviors at ∞. In

(1.20), if limr→∞(u(r), v(r)) = (−∞,−∞), then the decay rate of (u, v) may be slow sothat eu and ev are not both in L1(R2). See [2] for more details of statements. It will bean interesting problem to investigate how generic matrix K without the condition (1.2)affects the classification of radial solutions to (1.1).

By Theorem 1.1 and Corollary 1.5, we obtain:

Corollary 1.7. Suppose (u(r), v(r)) be an entire radial solution of (1.8). Then u and v

have intersection finite times.

Finally, we want to discuss the existence problem of mixed-type solutions. We denote(u(r;α1, α2), v(r;α1, α2)) be a radial solution of (1.8) with the initial value

{u(r) = 2N1 log r + α1 + o(1)v(r) = 2N2 log r + α2 + o(1)

as r → 0+. (1.21)

We define the region of initial data of the non-topological solutions of (1.8).

Ω ={(α1, α2)

∣∣ (u(r;α1, α2), v(r;α1, α2))

is a non-topological solution of (1.8)}.

(1.22)

As in our previous work on the SU(3) case, one can obtain the following theorem byusing the same argument of Theorem 1.8 in [9].

Theorem 1.8. Ω is an open set. Furthermore, if (α1, α2) ∈ ∂Ω, then (u(r;α1, α2),v(r;α1, α2)) is either a topological solution or a mixed-type solution.

When N1 = N2 and u = v, solutions of (1.8) give Ω �= ∅ (see [25]). Furthermore, as acorollary of the existence result in [1], we have Ω �= ∅ for the cases of A2 and B2 with allnon-negative N1 and N2. It is not difficult to prove that Ω �= R

2, hence, ∂Ω �= ∅ if Ω �= ∅.However, we do not address the question whether Ω is a non-empty set for K satisfying(1.2) in this paper. Another question is whether (1.8) possesses a unique topologicalsolution. It is generally expected so. For instance, when N1 = N2 = 0, we have theuniqueness of topological solutions. With this uniqueness, we should be able to provethe existence of the mixed-type solutions. We will discuss these questions completely inour further works.

The paper is organized as follows. Section 2 is devoted to the proof of Theorem 1.2.In Sections 3, 4 and 5, we show some a priori estimates on the behaviors of solutions.Theorem 1.3 and Theorem 1.4 are proved in Section 6. The integrability of f1 and f2

in R2 will be discussed in Section 7 and Theorem 1.1 follows. The asymptotic behaviors

at infinity of solutions are shown in Section 8.

2. Proof of Theorem 1.2

In this section, we show that if (u, v) is an entire radial solution of (1.8), then u < 0and v < 0 for r ∈ (0,∞) unless u ≡ v ≡ 0.

If u(r0) = v(r0) and ur(r0) = vr(r0), by the uniqueness, u(r) = v(r) for r > 0. Thesystem of Eqs. (1.8) can be reduce to the single equation

urr + 1rur = e2u − eu on (0,∞).

Then it is known that limr→∞(u(r), v(r)) = (0, 0) or limr→∞(u(r), v(r)) = (−∞,−∞)(see [28]). Hence, if ur(r0) = vr(r0), we assume that u(r0) �= v(r0); if u(r0) = v(r0), weassume that ur(r0) �= vr(r0).

The following lemma gives a sufficient condition that (u, v) cannot be an entire solu-tion.

Lemma 2.1 (Finite time blow-up condition).

(1) Suppose u(r0) = v(r0) > 0, ur(r0) > 0 and (u− v)r(r0) > 0. Then u(r) blows up infinite time.

(2) Suppose u(r0) > v(r0), u(r0) > 0 and ur(r0) > 0. Then u(r) or v(r) blows up infinite time.

Proof. (1) Let (0, T ) be the maximal interval of existence for (u, v).Step 1. Let I = (r0, r1) be the interval such that u(r) > v(r) on I. We claim that

ur(r) > 0 on [r0, r1].Suppose it is not true, so there exits r2 ∈ [r0, r1] such that

ur(r2) = 0 and ur(r) > 0 on [r0, r2].

Hence,

0 � urr(r2) = a(f2(r2) − f1(r2)

)− f1(r2) (2.1)

From the assumptions u(r0) = v(r0) > 0, ur(r) > 0 on [r0, r2) and u(r) > v(r) on(r0, r1), we have

u(r2) > 0 and u(r2) � v(r2).

It follows that

(f2 − f1)(r2) =(ev(r2) − eu(r2)

)(1 − (1 + a)eu(r2) − (1 + b)ev(r2)

)� 0 (2.2)

−f1(r2) = −eu(r2) + (1 + a)e2u(r2) − aeu(r2)+v(r2)

= eu(r2)(eu(r2) − 1

)+ aeu(r2)

(eu(r2) − ev(r2)

)> 0, (2.3)

which contradict to (2.1). Hence, ur(r) > 0 on [r0, r1].Step 2. We show that u(r) > v(r) on (r0, T ).Suppose for the sake of contradiction that there exits r3 ∈ (r0, T ) such that

u(r3) = v(r3) and u(r) > v(r) on (r0, r3).

Since ur(r) > 0 on (r0, r3) and u(r0) > 0, we have

u(r) > 0 on [r0, r3] and u(r) > v(r) on (r0, r3).

(f2 − f1)(r) =(ev(r) − eu(r))(1 − (1 + a)eu(r) − (1 + b)ev(r)) > 0 for r ∈ (r0, r3).

By this and (u− v)r(r0) > 0, we get

r(u− v)r(r) = r0(u− v)r(r) + (1 + a + b)r∫

s(f2 − f1)(s) ds > 0 for r ∈ [r0, r3].

Obviously, it is a contradiction at r3.Step 3. We show that T < ∞.We introduce the change of variable

t = ln r, t0 = ln r0 and T0 = lnT

The first equation of (1.8) becomes

u′′ = e2t(−(1 + a)f1 + af2), −∞ < t < T0 (2.4)

where ′ denotes the differentiation with respect to t. Set u = u + 2t. On (t0, T ), byapplying u(r) > v(r) on (r0, T ) and u(r) > 0 on (r0, T ), we have

u′′ > eu(eu − 1

)� δeu, (2.5)

for δ ∈ (0, eu(r0) − 1). We further restrict

δ ∈(

0,min{eu(r0) − 1, 1

2 u′ 2(t0)e−u(t0)

Multiplying (2.5) by u′ and integrating on (t0, t), we have

12(u′ 2(t) − u′ 2(t0)

)� δ

(eu(t) − eu(t0)

It follows that

u′(t) �√

2(t) + 1)

because u′(t) > 0 on (t0, T ) and u(t) on (t0, T ). It leads u to blow up in finite time andthus T is finite.

(2) Since we do not assume (u− v)r(r0) > 0, u(r) may intersect v(r) on (r0, T ). Thuswe consider the following possible cases:

(1) u(r) > v(r) on (r0, T ).(2) u(r) intersects v(r) on (r0, T ).

Step 1. For the first case, we show that T < ∞.Repeating the arguments in the step 1 of this lemma (1), we find

ur(r) > 0 on (r0, T ),

because we suppose u(r) > v(r) on (r0, T ). As in the step 3 of this lemma (1), weconclude that T < ∞.

Step 2. For the second case, let r1 be the first intersection point of u and v on (r0, T ).We show that T < ∞.

As in the step 1 of this lemma (1) again, we know that

ur(r) > 0 on [r0, r1].

Thus, we have

v(r1) = u(r1) > 0, vr(r1) > 0 and (v − u)r(r1) > 0

and the second case follows from Lemma 2.1 (1). �Proof of Theorem 1.2. Step 1. For r > 0, we denote

FU (r) = max{u(r), v(r)

GU (r) ={max{ur(r), vr(r)} if u(r) = v(r)

ddrFU (r) if u(r) �= v(r)

Step 2. Suppose FU (r) attains its positive maximum value at some point rM ∈ [0,∞).By symmetry, we may assume that u(rM ) � v(rM ). Thus we obtain

0 � −(1 + a)f1(rM ) + af2(rM ). (2.6)

Note that

af2(rM ) − af1(rM ) � 0 if u(rM ) > 0 and u(rM ) � v(rM ),

and −f1(rM ) = eu(rM )[a(eu(rM ) − 1) + (eu(rM ) − ev(rM ))] > 0 if u(rM ) > 0 and u(rM ) �v(rM ). It is a contradiction to (2.6). We conclude that either FU (r) attains non-positivemaximum values on [0,∞) or FU (r) never attains a maximum on [0,∞).

Step 3. Suppose FU (r) never attains a maximum on [0,∞), we show that FU (r) � 0.Suppose otherwise that FU > 0 at some point r0 ∈ (0,∞). Since we assume that (u, v)

is an entire solution, by Lemma 2.1, GU (r0) � 0. In view of the boundary conditions at0 and the step 2, it is a contradiction that FU attains positive maximum value on [0, r0].Hence, FU (r) � 0 on (0,∞) if it never attains a maximum on [0,∞).

Step 4. It is easy to see that FU (r0) = 0 if only if u(r0) = v(r0) = 0. By this and step3, we have

ur(r0) = vr(r0),

and thus u(r) = v(r) = 0 for r ∈ (0,∞). We conclude that

u(r), v(r) < 0 on (0,∞),

unless u ≡ v ≡ 0. �3. A priori estimates

In this section, we prove two a priori estimate lemmas: Lemma 3.4 and Lemma 3.5which are important in the proof of Theorem 1.3. The proof of these two lemmas aresimilar but not identical to the proof of Lemma 3.5 and Lemma 2.6 in [9] because we needto consider generic (a, b) instead of (a, b) = (1, 1). Hence, we need to carry out more sharpanalysis to yield the conclusion. The most important result is to show that if u (resp.v) attains maximum at r0, then v(r0) < u(r0) < log 1

1+a (resp. u(r0) < v(r0) < log 11+b )

under some conditions. This estimate is a crucial step toward proving the positivity offi, i = 1, 2. We first have the following simple observation.

Lemma 3.1. Let (u, v) be the solution of (1.8). Suppose that limr→∞ eu(r) andlimr→∞ ev(r) exist. Then limr→∞(eu(r), ev(r)) must be one of the following:

(a) (1, 1). (b)(

11 + a

0, 11 + b

). (d) (0, 0).

The following lemma will be used to prove Lemma 3.4 and Lemma 3.5.

Lemma 3.2.

(1) Assume that u(r0) > v(r0) and ur(r0) � 0 � vr(r0). Then f1(r0) � 0.(2) Assume that v(r0) > u(r0) and vr(r0) � 0 � ur(r0). Then f2(r0) � 0.

Proof. The proof is essentially the same as Lemma 3.3 in [9], and we omit it. �If the derivative of one of u(r) and v(r) must be negative on an interval I, then u(r)

and v(r) cannot increase simultaneously on I and we say u(r) and v(r) satisfy non-simultaneous increasing condition (for brevity, nsi-condition) on I. If the derivative ofone of u(r) and v(r) must be positive on an interval I, then u(r) and v(r) cannot de-crease simultaneously on I and we say u(r) and v(r) satisfy non-simultaneous decreasingcondition (for brevity, nsd-condition) on I. In the proof of Theorem 1.3, we will showthat (u, v) satisfies nsi-condition for r large enough provided (u, v) is not a topologicalsolution.

Definition 3.3. (1) We say that a function f(r) has an S[x,y]-profile if

fr(x) = fr(y) = 0 and fr(r) � 0 on (x, y),

and a function f(r) has a reversive S[x,y]-profile if

fr(x) = fr(y) = 0 and fr(r) � 0 on (x, y).

(2) f(r) has an S-profile on [x, y] if f(r) has an S[x,y]-profile, where [x, y] ⊂ [x, y].Similarly, f(r) has a reversive S-profile on [x, y] if f(r) has a reversive S[x,y]-profile,where [x, y] ⊂ [x, y].

The following lemmas will be used to prove Theorem 1.3. We obtain the upper boundof maximum of u and v when one of them has S[a,b]-profile or reversive S[a,b]-profile, andthey satisfy nsi-condition on (a, b) and do not intersect on (a, b).

Lemma 3.4. Assume that a > 0 and b > 0.

(1) Assume that u(r) has an S[r0,r1]-profile, v(r) < u(r) on (r0, r1) and vr(r0) � 0. Thenu(r1) < log 1

1+a .(2) Assume that v(r) has an S[r0,r1]-profile, u(r) < v(r) on (r0, r1) and ur(r0) � 0. Then

v(r1) < log 11+b .

(3) Assume that u(r) has a reversive S[r0,r1]-profile, v(r) < u(r) on (r0, r1) andvr(r1) > 0. Then u(r0) < log 1 .

(4) Assume that v(r) has a reversive S[r0,r1]-profile, u(r) < v(r) on (r0, r1) andur(r1) > 0. Then u(r0) < log 1

Proof. We only prove the first part. The second part is similar to the first part. Heuris-tically, the third part and fourth part can be viewed as the reflection version of the firstpart and the second part respectively.

Suppose, for the sake of contradiction, that u(r1) � 11+a . We split the proof into the

following steps.Step 1. Integrating Δ(bu+(1+a)v) = −(1+a+b)f2, and using v(r) < u(r) on (r0, r1)

and vr(r0) � 0 = ur(r0), we obtain

(bu + (1 + a)v

)r(r) < 0 on (r0, r1].

It follows that vr(r) < 0 on (r0, r1].Step 2. We claim there is r2 ∈ (r0, r1] such that

(−(1 + a)f1 + af2

)(r) < 0 on (r2, r1) and

(−(1 + a)f1 + af2

)(r2) = 0.

Recall that (−(1+a)f1 +af2)(r1) = Δu(r1) � 0. Hence, if (−(1+a)f1 +af2)(r1) < 0,we are done. If (−(1 + a)f1 + af2)(r1) = 0, by computing d

dr (−(1 + a)f1 + af2)(r1), wehave

(−(1 + a)f1 + af2

)(r1) =

((1 + a)eu(r1) − (1 + a)2e2u(r1) − (1 + b)ae2v(r1)

)vr(r1)

> 0 (3.1)

where ur(r1) = 0, u(r1) � log 11+a and (−(1 + a)f1 + af2)(r1) = 0 are used. Thus, there

exists r2 with

(−(1 + a)f1 + af2

)(r) < 0 on (r2, r1) and

(−(1 + a)f1 + af2

)(r2) = 0.

Consequently,

(−(1 + a)f1 + af2

)(r2) � 0. (3.2)

The calculation of ddr (−(1+a)f1 +af2)(r2) will lead to a contradiction. We consider the

following two possible cases:

1. u(r2) � log 11+a .

2. u(r2) < log 1 .

Step 3. case 1: u(r2) � log 11+a

By this, (−(1 + a)f1 + af2)(r2) = 0 and vr(r2) < 0 � ur(r2), we obtain

(−(1 + a)f1 + af2

=(−(1 + a)eu(r2) + 2(1 + a)2e2u(r2) +

(ab− a(1 + a)

)e(u+v)(r2)

)ur(r2)

+((1 + a)eu(r2) − (1 + a)2e2u(r2) − a(1 + b)e(u+v)(r2)

)vr(r2) (3.3)

If (ab− a(1 + a)) � 0, then

−(1 + a)eu(r2) + 2(1 + a)2e2u(r2) +(ab− a(1 + a)

)e(u+v)(r2) > 0, (3.4)

where u(r2) � log 11+a is used. If (ab− a(1 + a)) < 0, then

−(1 + a)eu(r2) + 2(1 + a)2e2u(r2) +(ab− a(1 + a)

)e(u+v)(r2)

� −(1 + a)eu(r2) + 2(1 + a)2e2u(r2) +(ab− a(1 + a)

)e2u(r2)

= −(1 + a)eu(r2) +((2 + a)(1 + a) + ab

)e2u(r2) > 0 (3.5)

where u(r2) > v(r2) and u(r2) > log 11+a are used. As in (3.1), we have

(1 + a)eu(r2) − (1 + a)2e2u(r2) − a(1 + b)e(u+v)(r2) < 0.

By this, (3.4) and (3.5), we have ddr (−(1 + a)f1 + af2)(r2) > 0, which contradicts to

(3.2).Step 4. case 2: u(r2) < log 1

1+a .If u(r2) < log 1

1+a , there exists r3 ∈ (r2, r1], such that u(r3) = log 11+a . By the steps

1 and 2, we know that

−(1 + a)f1(r3) + af2(r3) < 0 and vr(r) < 0 on (r0, r1].

Thus, we obtain

v(r2) > v(r3) > log b

(1 + a)(1 + b) .

By (−(1 + a)f1 + af2)(r2) = 0, we obtain

(−(1 + a)f1 + af2

=(−aev(r2) + a(1 + b)e2v(r2) + (1 + a)2e2u(r2)

)ur(r2)

+ a(ev(r2) − 2(1 + a)e2v(r2) +

(b− (1 + a)

)e(u+v)(r2)

)vr(r2) (3.6)

Since (−(1 + a)f1 + af2)(r2) = 0, we have

a(eu(r2) − ev(r2)

)(1 − (1 + a)eu(r2) − (1 + b)ev(r2)

)= a(f1 − f2)(r2) = −f1(r2) < 0 (3.7)

which implies

1 < (1 + a)eu(r2) + (1 + b)ev(r2). (3.8)

If v(r2) � log 11+b , then

−aev(r2) + a(1 + b)e2v(r2) + (1 + a)2e2u(r2) > 0. (3.9)

If v(r2) < log 11+b , then

−aev(r2) + a(1 + b)e2v(r2) + (1 + a)2e2u(r2)

> −aev(r2) + a(1 + b)e2v(r2) + (1 + a)eu(r2)(1 − (1 + b)ev(r2)

((1 + a)eu(r2) − aev(r2)

)((1 − (1 + b)ev(r2)

))> 0 (3.10)

where (3.8) is used.If b− (1 + a) � 0, then

1 − 2(1 + b)ev(r2) +(b− (1 + a)

)eu(r2) < 0 (3.11)

where u(r2) < log 11+a and v(r2) > log b

(1+a)(1+b) are used.If b− (1 + a) < 0 and v(r2) � log 1

2(1+b) , then

1 − 2(1 + b)ev(r2) +(b− (1 + a)

)eu(r2) < 0. (3.12)

If b− (1 + a) < 0 and v(r2) < log 12(1+b) , then

1 − 2(1 + b)ev(r2) +(b− (1 + a)

)eu(r2)

< 1 − 2(1 + b)ev(r2) +(b− (1 + a)

)1 − (1 + b)ev(r2)

1 + a−

(b(1 + b)1 + a

+ (1 + b))ev(r2)

< − b2

(1 + a)2 (3.13)

where v(r2) > log b is used. By (3.9), (3.10), (3.11), (3.12) and (3.13), we obtain

(−(1 + a)f1 + af2

)(r2) > 0,

which contradicts to (3.2).We conclude that u(r2) < log 1

1+a . �Similar to above lemma, we obtain the upper bound of maximum of u and v when one

of them has S[x,y]-profile or reversive S[x,y]-profile and they intersect once on (x, y). Toprove the following lemma, we need the technical conditions (3.14) and (3.15) for (x, y).

Lemma 3.5. Assume that u(r) has an S[r0,r2]-profile, u(r1) = v(r1) < log 11+a for some

r1 ∈ (r0, r2) and v(r) is decreasing on (r0, r2). Then u(r2) < log 11+a provided

(1 + b) log (1 + b)a(1 + a)(1 + a + b) + a log a(1 + a)

b(1 + a + b) < 0. (3.14)

Similarly, assume that v(r) has an S[r0,r2]-profile, u(r1) = v(r1) < log 11+b for some

r1 ∈ (r0, r2) and u(r) is decreasing on (r0, r2). Then v(r2) < log 11+b provided

(1 + a) log (1 + a)b(1 + b)(1 + a + b) + b log b(1 + b)

a(1 + a + b) < 0. (3.15)

Proof. By symmetry, we only prove the first part. Suppose, for the sake of contradiction,that u(r1) � 1

1+a .Step 1. We claim that ((1 + b)u + av)r(r) < 0 on (r0, r2].Note that

Δu(r0) = −(1 + a)f1(r0) + af2(r0) � 0.

By this and u > v on [r0, r1), we obtain

f2(r0) = ev(r0)(1 − (1 + b)ev(r0) + beu(r0)

)� a

1 + af1(r0) > 0.

By this and ur � 0 and vr � 0 on (r0, r2), we have f2(r) > 0. By integratingΔ((1 + b)u + av) from r0 to r � r2, this claim follows.

Step 2. As in the steps 2, 3 and 4 of Lemma 3.4, it is not difficult to show that

(−(1 + a)f1 + af2

)(r) < 0 on [r1, r2)

and there is r3 ∈ (r1, r2] such that

u(r3) = log 11 + a

and v(r3) > log b

(1 + a)(1 + b) . (3.16)

Since 0 � Δu(r0) = (−(1 + a)f1 + af2)(r0), there is r4 ∈ (r0, r1) such that(−(1 + a)f1 + af2

)(r4) = 0.

After a simple calculation, we obtain that (−(1 + a)f1 + af2)(r4) = 0 which leads to theinequality

(1 + a)2

aeu(r4) + (1 + b)ev(r4) < 1. (3.17)

Step 3. By the step 1, (3.16) and (3.17), we obtain

(1 + b) log 11 + a

+ a log b

(1 + a)(1 + b)< (1 + b)u(r3) + av(r3)

< (1 + b)u(r4) + av(r4)

< (1 + b)u(r4) + a log1 − (1+a)2

a eu(r4)

� (1 + b) log (1 + b)a(1 + a)2(1 + a + b) + a log a

(1 + a + b)(1 + b) . (3.18)

It contradicts to (3.14). Thus, u(r2) < log 11+a provided (3.14) holds. �

Heuristically, the following lemma can be viewed as the reflection of Lemma 3.5. Weomit the proof.

Lemma 3.6. Assume that u(r) has a reversive S[r0,r2]-profile, u(r1) = v(r1) < log 11+a for

some r1 ∈ (r0, r2) and v(r) is increasing on (r0, r2). Then u(r0) < log 11+a provided

(1 + b) log (1 + b)a(1 + a)(1 + a + b) + a log a(1 + a)

b(1 + a + b) < 0. (3.19)

Similarly, assume that v(r) has a reversive S[r0,r2]-profile, u(r1) = v(r1) < log 11+b for

some r1 ∈ (r0, r2) and u(r) is increasing on (r0, r2). Then v(r0) < log 11+b provided

(1 + a) log (1 + a)b(1 + b)(1 + a + b) + b log b(1 + b)

a(1 + a + b) < 0. (3.20)

4. Asymptotic behavior (1): without intersection

In this section, under certain conditions, we show that (u, v) satisfies (1.15) whenthey do not have intersection for r sufficiently large. We mainly apply Lemma 3.1 andLemma 3.4. We first exclude one special case:

(∗∗) There exists R0 > 0, so that either

u(r) is decreasing to log 11 + a

, and v(r) is decreasing to −∞ for r � R0

v(r) is decreasing to log 11 + b

, and u(r) is decreasing to −∞ for r � R0.

Lemma 4.1. There is no solution of (1.8) which satisfies condition (∗∗).

Proof. Step 1. We only exclude the case that u and v are decreasing to log 11+a and −∞

on (R0,∞) respectively. We write u = log 11+a + u (here, u > 0 on (R0,∞)). Thus, there

exists R1 � R0 so that

Δu � ab

2(1 + a)ev + u on (R0,∞). (4.1)

We will show there is no such solution u = log 11+a + u which satisfies (4.1).

Step 2. We show that f2 ∈ L1(R2) which is important to the estimate of limr→∞ rur(r)and limr→∞ rvr(r).

We may assume that v(r) < 2 log 11+b on (R1,∞) which implies f2 > 0 on (R1,∞).

Thus, we only need to show that∫∞R1

sf2(s) ds < ∞. For r � R1,

rvr(r) = 2N2 +r∫

s(−(1 + b)f2 + bf1

)(s) ds

= 2N2 +R1∫0

s(−(1 + b)f2 + bf1

)(s) ds +

r∫R1

s(−(1 + b)f2 + bf1

)(s) ds

� 2N2 +R1∫0

s(−(1 + b)f2 + bf1

)(s) ds−

r∫R1

sf2(s) ds (4.2)

where (f1−f2)(s) = (eu(r)−ev(r))(1− (1+a)eu(r)− (1+ b)ev(r)) < 0 on (R1,∞) is used.Suppose that

∫∞R1

sf2(s) ds = ∞, then

rvr(r) → −∞ as r → ∞.

It implies that∫∞R1

sev(s) ds < ∞ and thus∫∞R1

sf2(s) ds < ∞ where u < 0 is used. Weconclude that f2 ∈ L1(R2).

Step 3. We show that limr→∞ rv(r) < −2.One can see that rvr(r) is a decreasing function on (R1,∞) from (4.2). Hence, if

limr→∞ rv(r) � −2, then

rvr(r) � −2 on (R1,∞),

which makes the L1-integrability of f2 fail. Therefore, limr→∞ rv(r) < −2.Step 4. We show that f1 ∈ L1(R2).We split the nonlinear term −(1 + a)f1 + af2 into

a(ev − (1 + b)e2v +

(b− (1 + a)

)− (1 + a)

(eu − (1 + a)e2u). (4.3)

Note that the first term of (4.3) is in L1(R2) by the step 2, and the second term of (4.3)is negative for r > R1. Hence, by the existence of limr→∞ u(r), we have

−(1 + a)∞∫

s(eu(s) − (1 + a)e2u(s)) ds > −∞.

It follows that f1 is L1-integrable in R2.

Step 5. By the steps 2, 3 and 4, we have limr→∞ ru(r) = 0 and limr→∞ rv(r) = −β

for some constant β > 2. Thus, for R2 > R1 large enough, we have

Δu � ab

2(1 + a)r−β + u, r > R2. (4.4)

Since u > 0 on (R2,∞),

Δu � ab

2(1 + a)r−β , r > R2. (4.5)

By integrating it, we obtain

u(r) � ab

2(1 + a)(2 − β)2 r2−β , r > R2.

Plugging it into (4.4) and integrating it, we obtain

u(r) � ab

2(1 + a)(2 − β)2(4 − β)2 r4−β , r > R2,

provided β > 4. Note that if β � 4, we obtain a contradiction because

lim u(r) = lim ru(r) = 0.

Since β is finite, if β > 4, one can repeat this argument at most a finite number of timesto get a contradiction. �

In the following lemma, we show that if

u(r) > v(r) on (r0,∞), ur(r0) � vr(r0), and(bu + (1 + a)v

)r(r0) � 0,

v(r) > u(r) on (r0,∞), vr(r0) � ur(r0), and((1 + b)u + av

)r(r0) � 0,

then (u, v) satisfies (1.15). We yield that (u, v) satisfies nsi-condition on (r0,∞) fromthese conditions, and then we are able to apply Lemma 3.4(1) and (2). In the secondpart of the following lemma, we show that if

u(r) > v(r) on (r0,∞) and((1 + 2b)u + (1 + 2a)v

)r(r) > 0 on (r0,∞), (4.6)

(u, v) must be a topological solution. The later condition of (4.6) implies (u, v) satisfiesnsd-condition on (r0,∞).

Lemma 4.2. (1) Suppose that

v(r) < u(r) on (r0,∞), vr(r0) � ur(r0) and(bu + (1 + a)v

)r(r0) � 0.

v(r) < u(r) < log 11 + a

on some interval (R1,∞) ⊆ (r0,∞).

Similarly, suppose that u(r) < v(r) on (r0,∞), ur(r0)� vr(r0) and ((1+b)u+av)r(r0)� 0.Then

u(r) < v(r) < log 11 + b

on some interval (R1,∞) ⊆ (r0,∞).

(2) Suppose that u(r) > v(r) on (r0,∞), and ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on(r0,∞). Then

u(r) and v(r) are increasing to 0 as r → ∞,

i.e. (u, v) is a topological solution to (1.8).

Proof. (1) We only prove the first part because the proof of the second part is similar.Step 1. We show that u(r) and v(r) satisfy the nsi-condition on (r0,∞).

Since we suppose u(r) > v(r) on (r0,∞) and (bu + (1 + a)v)r(r0) � 0, then

r(bu + (1 + a)v

)r(r) = r0

(bu + (1 + a)v

)r(r0) − (1 + a + b)

r∫r0

sf2(s) ds

< 0 for r ∈ (r0,∞). (4.7)

Consequently, u(r) and v(r) satisfy the nsi-condition on (r0,∞).Step 2. We consider the following possible cases:

(a) u(r) oscillates on (r0,∞).(b) u(r) is increasing on some interval (r1,∞) ⊆ (r0,∞).(c) u(r) is decreasing on some interval (r2,∞) ⊆ (r0,∞).

Step 2.1. Suppose u(r) oscillates on (r0,∞), which implies u(r) has infinitely manyS-profile on (r0,∞). Let S[α,β] be its first S-profile on (r0,∞). Combining the nsi-condition on (r0,∞) and Lemma 3.4(1), the local maximum values of u(r) on (α,∞) areless than log 1

1+a . Since we suppose u oscillates on (r0,∞), then

u(r) < log 11 + a

for r ∈ (α,∞).

Step 2.2. Suppose that u(r) is increasing on some interval (r1,∞) ⊂ (r0,∞). By thensi-condition on (r0,∞), v(r) is strictly decreasing on (r1,∞). Lemma 3.1 suggests

u(r) is increasing to log 11 + a

and v(r) is decreasing to −∞ as r → ∞.

Step 2.3. Suppose u(r) is decreasing on some interval (r2,∞) ⊆ (r0,∞). Then eitheru(r) is decreasing on (r0,∞) or there exits r3 ∈ (r2,∞) such that ur(r3) = 0 and u isdecreasing on (r3,∞). Note that vr(r3) < 0 because of (4.7). Let r4 = r0 if the first caseholds, and r4 = r3 if the second case holds.

u(r) is decreasing on (r4,∞) implies the existence of limr→∞ eu(r). Thus, iflimr→∞ eu(r) < 1

1+a , then u(r) is less than log 11+a on some interval (r5,∞) ⊆ (r4,∞).

We now show that if limr→∞ eu(r) � 11+a , then v(r) is decreasing on (r4,∞). Since

we suppose u(r) > v(r) and u(r) � log 11+a for r ∈ (r4,∞), then

f1(r) − f2(r) > 0 on (r4,∞).

Hence,

r(u− v)r(r) = r4(u− v)r(r4) − (1 + a + b)r∫s(f1 − f2) ds > 0 on (r4,∞).

It follows that vr(r) < ur(r) � 0 on (r4,∞), and thus limr→∞ eu(r) and limr→∞ ev(r)

exist. It is a contradiction by Lemma 3.1 and Lemma 4.1.(2) Step 1. As in the proof of (1), we consider the following possible cases:

(1) u(r) oscillates on (r0,∞).(2) u(r) is increasing on some interval (r1,∞) ⊆ (r0,∞).(3) u(r) is decreasing on some interval (r2,∞) ⊆ (r0,∞).

It will be shown that only the case (2) is possible.Step 2. We show u(r) cannot oscillate on (r0,∞).Suppose that u oscillates on (r0,∞). Then u(r) has infinitely many reversive S-profile

on (r0,∞). Let S[α,β] be its first reversive S-profile on (r0,∞). The assumption that((1+2b)u+(1+2a)v)r(r) > 0 on (r0,∞) implies the nsd-condition for (u, v) on (r0,∞).Combining the nsd-condition on (r0,∞) and Lemma 3.4 (3), the local maximum valuesof u(r) on (α,∞) are less than log 1

1+a . Thus,

v(r) < u(r) < log 11 + a

on (α,∞).

By this and ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on (r0,∞),

α((1 + 2b)u + (1 + 2a)v

)r(α) � (1 + a + b)

∞∫α

s(f1 + f2)(s) ds

� a(1 + a + b)∞∫α

se(u+v)(s) ds

� a(1 + a + b)ec(u+v)(t1)∞∫α

= +∞(

where c = max{(1 + 2a), (1 + 2b)}min{(1 + 2a), (1 + 2b)}

)(4.8)

which is a contradiction. Hence, u(r) cannot oscillate on (r0,∞).Step 3. Suppose that u(r) is increasing on some interval (r1,∞) ⊂ (r0,∞), which

implies limr→∞ u(r) exits. By this and ((1 + 2b)u + (1 + 2a)v)r > 0 on (r0,∞),

(1 + 2a)(v(∞) − v(r1)

)> (1 + 2b)

(u(r1) − u(∞)

which implies v(∞) > −∞. Lemma 3.1 suggests that limr→∞ u = limr→∞ v = 0.Step 4. Suppose that u(r) is decreasing on some interval (r2,∞) ⊂ (r0,∞). By the

nsd-condition for (u, v) on (r0,∞), v is increasing on (r2,∞). Thus, limr→∞ u(r) andlimr→∞ v(r) exist and are not both equal to 0, a contradiction to Lemma 3.1.

Step 5. Finally, we show vr(r) > 0 for r sufficiently large.Since limr→∞ v = 0, there exists r3 ∈ (r0,∞) such that vr(r3) > 0 and u(r) > v(r) >

log 11+b on (r3,∞). By this, we have −(1 + b)f2 + bf1 < 0 on (r3,∞). Hence,

rvr(r) = r3vr(r3) +r∫

s(−(1 + b)f2 + bf1

)ds > 0. �

5. Asymptotic behavior (2): general case

In this section, we consider that u and v may intersect for r large. We will show that,under certain condition, either (u, v) satisfies (1.15) for r large or (u, v) is a topologicalsolution to (1.8). To establish a priori bound for (u, v), we need the following localestimates for (u, v).

Lemma 5.1. Let 0 � s1 < s2 < s3 < s4. Assume that s2, s3 and s4 are consecutiveintersection points of u(r) and v(r), with u(r) > v(r) on (s1, s2), u(r) < v(r) on (s2, s3)and u(r) > v(r) on (s3, s4).

(1) Suppose that vr(s1) � ur(s1), (bu + (1 + a)v)r(s1) � 0 and (a, b) satisfies (3.14).Then

v(r) < u(r) < log 11 + a

for r ∈ [s3, s4].

(2) Suppose that u(s1) = v(s1), ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on [s1, s4], and u(s4) =v(s4) < log 1

1+a . Then

v(r) < u(r) < log 11 + a

for r ∈ [s1, s2].

Proof. (1) Step 1. We show that (u, v) satisfies the nsi-condition on [s1, s4], and thusur(s2) < 0, vr(s3) < 0 and ur(s4) < 0.

Since we suppose (bu + (1 + a)v)r(s1) � 0 and v(r) < u(r) on (s1, s2), then

r(bu + (1 + a)v

)r(r) = s1

(bu + (1 + a)v

)r(s1) − (1 + a + b)

r∫s1

sf2(s) ds < 0

for r ∈ [s1, s2]. (5.1)

Thus, ur(s2) < 0 which implies ((1 + b)u + av)r(s2) < 0. By this and u(r) < v(r) on(s2, s3), we obtain

r((1 + b)u + av

)r(r) = s2

((1 + b)u + av

)r(s2) − (1 + a + b)

r∫s2

sf1(s) ds < 0

for r ∈ [s2, s3]. (5.2)

Thus, vr(r3) < 0. Repeating the above argument, we get(bu + (1 + a)v

)r(r) < 0 on [s3, s4] and ur(s4) < 0. (5.3)

Step 2. We show that u(s3) = v(s3) < log 11+a .

By (5.1), ur(s2) < 0 and Lemma 3.4(1), we know that if u(r) has an S-profile on(s1, s2), then u(s2) must be less than log 1

1+a . Thus, we need only consider the followingcases:

(1) u(r) is decreasing on (s1, s2).(2) u(r) is increasing on (s1, r1) and u(r) is decreasing on (r1, s2) for some r1 ∈ (s1, s2).

(By (5.1), vr(r1) < 0.)

Let s = s1 if the first case holds, and s = r1 if the second case holds. Now, assumeu(s2) = v(s2) � log 1

1+a . Then,

r(u− v)r(r) = s(u− v)r(s) + (1 + a + b)r∫

s(f2 − f1) ds > 0 on (s, s2],

because u(r) > v(r) on (s, s2) and u(r) > u(s2) � log 11+a . Obviously, it is a contradic-

tion. Therefore, we proved u(s2) = v(s2) < log 11+a . By this and (5.2), we obtain

u(s3) = v(s3) < u(s2) = v(s2) < log 11 + a

. (5.4)

Step 3. We show that u(r) < log 11+a on [s3, s4] if ur(s3) � 0.

Since ur(s3) � 0 and ur(s4) < 0, then either u is decreasing on (s3, s4), or u attainslocal maximum on (s3, s4). The first case implies u(r) < log 1

1+a on [s3, s4]. For thesecond case, by ur(s3) � 0 and ur(s4) < 0, u(r) attains local maximum at m1 ∈ (s3, s4)implies u(r) has an S[m2,m1]-profile on (s3, s4). By (5.3) and Lemma 3.4(1), the localmaximum values of u(r) on [s3, s4] are less than log 1

1+a . So, we proved u(r) < log 11+a

on [s3, s4] provided ur(s3) � 0.Step 4. We show u(r) < log 1

1+a on [s3, s4] if ur(s3) > 0.The difference between step 3 and step 4 is that, with ur(s3) > 0, u(r) attains its

first local maximum on (s3, s4) cannot guarantee u(r) has an S-profile on (s3, s4). Sinceur(s2) < 0 and ur(s3) > 0, there exits r1 ∈ (s2, s3) with

ur(r1) = 0 and ur(r) > 0 on (r1, s3].

Let r2 be the first local maximum point of u on (s3, s4) which implies u(r) has anS[r1,r2]-profile. By (5.3) and Lemma 3.5(1),

u(r2) < log 11 + a

Since ur(r2) = 0 and ur(s3) < 0, as in the step 3, it is easy to see that

u(r) < log 11 + a

on [s3, s4].

(2) Since we suppose ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on [s1, s4] and u(s4) = v(s5) <log 1

1+a , we have u(si) = v(si) < log 11+a , i = 1, 2, 3, and

ur(s1) > 0, vr(s2) > 0, ur(s3) > 0 and vr(s4) > 0.

Step 1. We show u(r) < log 11+a on [s1, s2] if ur(s2) � 0.

Since ur(s1) > 0 and ur(s2) � 0, then either ur(r) in increasing on (s1, s2), or u

attains local maximum on (s1, s2). The first case implies u(r) < log 11+a on [s1, s2]

because u(s2) < log 11+a . For the second case, with ur(s1) > 0 and ur(s2) � 0, u(r)

attains local maximum on (s1, s2] implies u has a reversive S-profile on (s1, s2]. By((1 + 2b)u + (1 + 2a)v)r(r) > 0 on [s1, s4] and Lemma 3.4(3), the maximum values ofu(r) on [s1, s2] are less than log 1

1+a . Thus,

u < log 11 + a

on [s1, s2] if ur(s2) � 0.

Step 2. We show that u < log 11+a on [s1, s2] if ur(s2) < 0.

Since we suppose ur(s2) < 0 and ur(s1) > 0, u(r) attains local maximum on (s1, s2).Unlike the step 1, u(r) attains local maximum on (s1, s2) does not necessarily implyu(r) has reversive S-profile on (s1, s2). Let u attain local maximum at r3 ∈ (s1, s2) andur � 0 on (r3, s2]. Since we suppose ur(s2) < 0 and ur(s3) > 0, there exists r4 ∈ (s2, s3)such that

ur(r) < 0 on [s2, r4) and ur(r4) = 0,

which implies u(r) has a reversive S[r3,r4]-profile. By ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on[s1, s4] and Lemma 3.5(3),

u(r3) < log 1.

With ur(s1) > 0 and ur(r3) = 0, u(r) attains local maximum on (s1, r3) implies thatu(r) has a reversive S-profile on (s1, r3). By (u+v)r(r) > 0 on [s1, s3] and Lemma 3.4(3),the local maximum values of u(r) on [s1, s2] are less than log 1

1+a . It follows that

u < log 11 + a

on [s1, s2]. �By symmetry of Lemma 5.1, we have the following lemma.

Lemma 5.2. Let 0 � s1 < s2 < s3 < s4. Assume that s2, s3 and s4 are consecutiveintersection points of u(r) and v(r), with v(r) > u(r) on (s1, s2), v(r) < u(r) on (s2, s3)and v(r) > u(r) on (s3, s4).

(1) Suppose that ur(s1) � vr(s1) and ((1 + b)u+ av)r(s1) � 0 and (a, b) satisfies (3.15).Then

u(r) < v(r) < log 11 + b

for r ∈ [s3, s4].

(2) Suppose that u(s1) = v(s1), ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on [s1, s4], and u(s4) =v(s4) < log 1

1+b . Then

u(r) < v(r) < log 11 + b

for r ∈ [s1, s2].

Under the certain conditions (see (5.5), (5.6) and (5.7) below), we show that either(u, v) satisfies (1.15) for r large or (u, v) is a topological solution.

Lemma 5.3.

(1) Suppose (u(r), v(r)) satisfies either

u(r0) � v(r0), ur(r0) � vr(r0) and(bu + (1 + a)v

)r(r0) � 0, (5.5)

v(r0) � u(r0), vr(r0) � ur(r0) and(av + (1 + b)u

)r(r0) � 0, (5.6)

then there exists R0 > r0, such that⎧⎪⎨⎪⎩u(r) < log 1

1 + aif v(r) � u(r)

v(r) < log 11 + b

if u(r) � v(r)

for r ∈ (R0,∞).

(2) Suppose

((1 + 2b)u + (1 + 2a)v

)r(r) > 0 on (0,∞). (5.7)

Then (u, v) is a topological solution.

Proof. (1) Assume (5.5) holds. We discuss the following possible cases.

(1) u(r) and v(r) have finite intersection points on (r0,∞).(2) u(r) and v(r) have infinitely many intersection points on (r0,∞).

Step 1. If u(r) and v(r) have finite intersection points on (r0,∞), say {ti}ni=1. Toapply Lemma 4.2 to (u, v) on (rn,∞), we need to show that

(bu + (1 + a)v

)r(rn) � 0 if u > v on (rn,∞) (5.8)

((1 + b)u + av

)r(rn) � 0 if v > u on (rn,∞). (5.9)

As in the step 1 of Lemma 5.1(1), we have

((1 + b)u + av

)r(r) < 0 on [t2i−1, t2i](

bu + (1 + a)v)r(r) < 0 on [t2i, t2i+1]

i = 1, 2, . . . .

Hence, (u, v) satisfies (1.15).Step 2. Assume that u(r) and v(r) have infinitely many intersection points, say {ti}∞i=1.

v(r) < u(r) on (t2i−1, t2i)

u(r) < v(r) on (t2i, t2i+1)i = 1, 2, . . . .

As in the steps 1 and 2 of Lemma 5.1(1), we have

((1 + b)u + av

)r(r) < 0 on [t2i−1, t2i](

bu + (1 + a)v)r(r) < 0 on [t2i, t2i+1]

i = 1, 2, . . . , (5.10)

u(ti) = v(ti) < min{

log 1, log 1

}, i = 4, . . . . (5.11)

We further assume that b � a > 0, which implies that (3.14) holds. Hence, byLemma 5.1(1), we have

v(r) < u(r) < log 11 + a

on [t2i−1, t2i], i = 2, 3, . . . . (5.12)

We need to show that

u(r) < v(r) < log 11 + b

on [t2i, t2i+1], i = 2, 3, . . . .

Suppose it is not true. Since vr(t2i) < 0 for i = 1, 2, . . . , we may assume that v attainslocal maximum at si ∈ (t2i, t2i+1), i = 2, 3, . . . (after subsequence), and

v(si) � log 11 + b

, i = 2, 3, . . . .

By this, we know that vr(t2i+1) > 0, i = 2, 3, . . . , otherwise, by (5.10) and Lemma 3.4,v(si) < log 1

1+b , i = 2, 3, . . . . Furthermore, vr(r) � 0 on [t2i, si] and v has S[pi,si]-profilefor some pi ∈ (t2i−1, t2i), i = 2, 3, . . . . As in the step 2 of Lemma 3.5, there existsqi ∈ (t2i, si), such that

v(qi) = log 11 + b

and u(qi) > log a

(1 + a)(1 + b) , i = 2, 3, . . . . (5.13)

By (5.10) and (5.12), we have

((1 + b)u + av

)r(r) =

((1 + b)u + av

)r(t3) − (1 + a + b)

r∫t3

sf1 ds < 0

on (t3,∞) (5.14)

and thus u(ti) = v(ti) tends to −∞ as i tends to ∞. By this, (5.13) and (5.14), we have

(1 + b) log 11 + b

+ a log a

(1 + a)(1 + b)<

((1 + b)u + av

)(qi) <

((1 + b)u + av

)(t2i) → −∞ as i → ∞.

Obviously, it is a contradiction.(2) Step 1. We claim that if

u(r1) = v(r1) � log 11 + a

, vr(r1) < ur(r1) and((1 + 2b)u + (1 + 2a)v

)r> 0 on [r1,∞),

then u(r) and v(r) do not intersect after r1.

Suppose r2 is the first intersection point of u(r) and v(r) after r1. By ((1 + 2b)u +(1 + 2a)v)r(r) > 0 on [r1,∞) and u(r1) = v(r1) � log 1

1+a , we have, for r ∈ (r1, r2],

(2 + 2a + 2b) log 11 + a

�((1 + 2b)u + (1 + 2a)v

)(r4) <

((1 + 2b)u + (1 + 2a)v

� (2 + 2a + 2b)u(r), (5.15)

which implies f1(r)−f2(r) = (eu(v)−ev(r))(1− (1+a)eu(r)− (1+b)ev(r)) < 0 on (r1, r2).Hence,

r(u− v)r(r) = r4(u− v)r(r4) − (1 + a + b)r∫

s(f1 − f2) ds > 0 on (r4, r5]. (5.16)

Obviously, it is a contradiction. Similarly,if

u(r1) = v(r1) � log 11 + b

, ur(r1) < vr(r1) and((1 + 2b)u + (1 + 2a)v

)r> 0 on [r1,∞),

then u(r) and v(r) do not intersect after r1.Step 2. We discuss the following possible cases.

(1) u(r) and v(r) have infinitely many intersection points on (0,∞).(2) u(r) and v(r) have finite intersection points on (0,∞).

Step 3. By the step 1, if u(r) and v(r) have infinitely many intersection points, say{ti}∞i=1, then u(ti) = v(ti) < min{log 1

1+a , log 11+b}, i = 1, 2, . . .. Heuristically, it can be

viewed as the reflection version of the step 2 of this lemma (1), hence, we omit the proof.Thus, we have either

v(r) � u(r) < 11 + a

u(r) � v(r) < 11 + b

By these and ((1 + 2b)u + (1 + 2a)v)r > 0 on [r1,∞),

t1((1 + 2b)u + (1 + 2a)v

)r(t1) � (1 + a + b)

∞∫s(f1 + f2)(s) ds

� min{a, b}(1 + a + b)∞∫

se(u+v)(s) ds

� min{a, b}(1 + a + b)ec(u+v)(t1)∞∫

= +∞(

where c = max{(1 + 2a), (1 + 2b)}min{(1 + 2a), (1 + 2b)}

)(5.17)

which is a contradiction. Hence, u(r) cannot intersect v(r) infinitely many times.Step 4. If u(r) and v(r) have finite intersection points, say {ti}ni=1. By applying

Lemma 4.2 (2) to (u(r), v(r)) on [tn,∞), (u(r), v(r)) is a topological solution. �6. Proof of Theorems 1.3 and 1.4

Now, we are ready to prove Theorems 1.3 and 1.4. We will show that any entire radialsolutions (u, v) to (1.8) either satisfies (1.15) for r large or is a topological solution. Thus,Theorem 1.3 and Theorem 1.4 follow.

Step 1. From the equation

r((1 + 2b)u + (1 + 2a)v

= 2((1 + 2a)N1 + (1 + 2b)N2

)− (1 + a + b)

s(f1 + f2) ds, (6.1)

we know that either

r((1 + 2b)u + (1 + 2a)v

)r(r) > 0 for r ∈ (0,∞),

or there is r1 such that r1((1 + 2b)u + (1 + 2a)v

)r(r1) = 0 and(

(1 + 2b)u + (1 + 2a)v)r> 0 on [0, r1).

For the first case, (u, v) is a topological solution by Lemma 5.3. For the second case, wewill show (u, v) satisfies (1.15) for r large. Thus, Theorem 1.4 follows.

There are three possibilities on the derivative of (u, v) at r1. (Here, we assume thatu(r) > v(r) on some interval (r1, r∗1)):

(A) ur(r1) = vr(r1) = 0.(B) ur(r1) = −(1+2a

1+2b )vr(r1) > 0.(C) ur(r1) = −(1+2a )vr(r1) < 0.

Step 2. We discuss the cases (A) and (B).For the cases (A) and (B), it is obvious that

ur(r1) � 0 � vr(r1),(bu + (1 + a)v

)r(r1) � 0,

which satisfies (5.5). Hence, by Lemma 5.3, (u, v) satisfies (1.15) after r1.Step 3. We discuss the case (C).Unlike the cases (A) and (B), the case (C) does not satisfy the condition (5.5) at r1.

We will show u and v satisfy (5.5) or (5.6) at some point on (r1,∞).We define

s1 = sup{s > r1

∣∣ ur < 0 on (r1, s)}

s2 = sup{s > r1

∣∣ vr > 0 on (r1, s)}. (6.3)

We will use these definitions throughout this step.Step 3.1. We consider the case that (u, v) has no intersection point after r1.Step 3.1.1. We show that s2 � s1.Since ((1 + 2b)u + (1 + 2a)v)r(r) > 0 on (0, r1) and ur(r1) < 0, then there exits

r2 ∈ (0, r1) such that

ur(r2) = 0 and ur(r) < 0 on (r2, r1].

Hence, (−(1+a)f1 +af2)(r2) = urr(r2) � 0. By ((1+2b)u+(1+2a)v)r(r) > 0 on (0, r1)and ur(r) < 0 on (r2, r1], we have vr(r) > 0 on [r2, r1]. It follows that u(r) > v(r) on[r2, r1] and thus

f1(r2) �a

1 + af2(r2) > 0. (6.4)

Suppose for the sake of contradiction that s2 > s1. Then

ur < 0 < vr on (r2, s1), and ur(s1) = 0 < vr(s1). (6.5)

Note that f1(r2) = eu(r2)(1 − (1 + a)eu(r2) + aev(r2)) > 0. By this and (6.5),

1 − (1 + a)eu(r) + aev(r) > 0 on [r2, s1]. (6.6)

Consequently, f1(r) = eu(r)(1 − (1 + a)eu(r) + aev(r)) > 0 on [r2, s1] and

r((1 + 2b)u + (1 + 2a)v

)r(r) = −(1 + a + b)

r∫s(f1 + f2) ds < 0 on (r1, s1]. (6.7)

By this, ((1 + 2b)u + (1 + 2a)v)r(s1) < 0, which is a contradiction to (6.5). Hence, weproved

s2 � s1.

Step 3.1.2. By Lemma 3.1, u(r) and v(r) cannot be monotone after r1 which impliess2 is finite. Since we suppose

u(r) > v(r) on [r1,∞) and vr(s2) = 0 � ur(s2),

r(bu + (1 + a)v

)r(r) = s2

(bu + (1 + a)v

)r(s2) − (1 + a + b)

r∫s2

sf2(s) ds < 0

for r ∈ (s2,∞). (6.8)

If s1 is finite, by (6.8),

ur(s1) = 0 � vr(s1) and(bu + (1 + a)v

)r(s1) � 0.

Thus, by applying Lemma 5.3 to u(r) and v(r) at s1, we proved Theorem 1.3.If s1 = ∞, we consider the following possible cases:

(1) vr(r) > ur(r) on (s2,∞).(2) There exists r∗ ∈ (s2,∞) such that vr(r∗) � ur(r∗).

For the second case, Theorem 1.3 is proved by applying Lemma 5.3 again. For the firstcase, if limr→∞ u(r) < log 1

1+a , then we know there exists r∗∗ such that

v(r) < u(r) < log 11 + a

for r ∈(r∗∗,∞

which implies (1) of Theorem 1.3. Therefore, we may assume limr→∞ u(r) � log 11+a .

Note that s1 = ∞ implies ur(r) < 0 on [r1,∞). Hence,

vr(r) > ur(r) on (r1,∞.)

It follows that

v(∞) − v(r1) > u(∞) − u(r1),

which implies v(∞) > −∞. But it yields a contradiction to Lemma 3.1.Step 3.2. Suppose u(r) and v(r) have at least one intersection point after r1. Here, we

assume r3 is the first intersection point of u(r) and v(r) after r1.

To obtain Theorem 1.1, we want to show that

ur(r3) � vr(r3) and(av + (1 + b)u

)r(r3) � 0. (6.9)

Then Theorem 1.3 follows by applying Lemma 5.3. Recall the definition of s1 and s2in (6.2) and (6.3). By the step 3.1.1., if s1 < r3, then s2 � s1. Hence, we consider thefollowing possible cases to verify (5.6) at r3.

(1) s1, s2 ∈ [r3,∞).(2) s2 � s1 ∈ (r1, r3).(3) s2 ∈ (r1, r3) and s1 ∈ [r3,∞).

The case (1) implies that ur(r) < 0 < vr(r) on [r1, r3). As in (6.7),

((1 + 2b)u + (1 + 2a)v

)r(r) < 0 on (r1, r3],

which implies ur(r3) < 0 and thus (av + (1 + b)u)r(r3) < 0. Then (5.6) holds at r3.For the case (2), we have ur(s2) � 0 = vr(s2). Hence, as in (6.7),

((1 + 2b)u + (1 + 2a)v

)r(r) < 0 on (s2, r3],

which implies (5.5) holds at r3 again.The case (3) implies

ur(s2) < 0 = vr(s2).

Consequently,

r(bu + (1 + a)v

)r(r) = s2

(bu + (1 + a)v

)r(s2) − (1 + a + b)

r∫s2

sf2(s) ds < 0

on [s2, r3]. (6.10)

Since (u−v)r(r3) < 0, we have ur(r3) < vr(r3) and (bu+(1+a)v)r(r3) < 0 which imply(5.6).

7. Classification of radial solutions

In this section, we classify the radial solutions according to their boundary conditionsat ∞. To show this, we will study the L1(R2)-integrability of f1 and f2. We have thefollowing simple observation.

Lemma 7.1. Suppose (u, v) is not a topological solution to (1.8). Then the following hold.

(1)∫∞0 sf1(s) ds and

∫∞0 sf2(s) ds are either both finite or both infinite.

(2) If∫∞0 sf1(s) ds =

∫∞0 sf2(s) ds = ∞, then

∫∞0 se(u+v)(s) ds < +∞.

Next, we introduce the Pohozaev identity for (1.8).

Lemma 7.2 (The Pohozaev identity). Suppose (u, v) solves (1.8). Then (u, v) satisfies

r2(b(1 + b)

2 u2r(r) + abur(r)vr(r) + a(1 + a)

2 v2r(r)

)− 4

(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)(1 + a + b)r2

(beu(r) − b(1 + a)

2 e2u(r) + aev(r) − a(1 + b)2 e2v(r) + abe(u+v)(r)

= 2(1 + a + b)r∫

(beu(s) − b(1 + a)

2 e2u(s) + aev(s)

− a(1 + b)2 e2v(s) + abe(u+v)(s)

)ds (7.1)

for r > 0.

Using the Pohozaev identity, we show that f1 and f2 are in L1(R2) when (u, v) is nota topological solution to (1.8).

Lemma 7.3. Suppose (u, v) is not a topological solution. Then

f1 and f2 are in L1(R

2).Proof. Step 1. By the proof of Theorem 1.3, there exits R0 > 0 such that

⎧⎪⎨⎪⎩u(r) < log 1

1 + aif v(r) � u(r)

v(r) < log 11 + b

if u(r) � v(r)

for r ∈ (R0,∞), which implies f1(r) > 0 and f2(r) > 0 if r > R0. Hence, we only needto show that

∞∫sf1(s) ds < +∞ and

∞∫sf2(s) ds < +∞.

Suppose for the sake of contradiction that

∞∫0

sf1(s) ds = +∞ and∞∫0

sf2(s) ds = +∞. (7.2)

It implies that

∞∫R0

seu(s) ds = +∞ and∞∫

sev(s) ds = +∞ (7.3)

because of Lemma 7.1(2).Step 2. Let Γ{v<u} = {r ∈ (R0,∞) | v(r) < u(r)} and Γ{u<v} = {r ∈ (R0,∞) | u(r) <

v(r)}. Since we assume that

(1 + 2b)rur(r) + (1 + 2a)rvr(r)

= 2((1 + 2b)N1 + (1 + 2a)N2

)− (1 + a + b)

s(f1 + f2)(s) ds → −∞

as r → ∞. Hence, there exists R1 � R0 such that

(u + v)(r) < −6 log r for r > R1,

which implies

u(r) < −3 log r on Γ{u<v}

v(r) < −3 log r on Γ{v<u}.

It follows that for r � R1

r∫R1

seu ds =∫

(R1,r)∩Γ{v<u}

seu(s) ds + O(1) and

r∫R1

sev ds =∫

(R1,r)∩Γ{u<v}

sev(s) ds + O(1). (7.4)

We consider the following possible cases for (u, v) on (R1,∞):

(a) sup(R1,∞) u(r) < log 11+a and sup(R1,∞) v(r) < log 1

1+b .(b) sup(R1,∞) u(r) = log 1

1+a and sup(R1,∞) v(r) = log 11+b .

(c) sup(R1,∞) u(r) = log 11+a and sup(R1,∞) v(r) < log 1

1+b .(d) sup(R1,∞) u(r) < log 1

Step 3. Case (a): sup(R1,∞) u(r) < log 11+a and sup(R1,∞) v(r) < log 1

1+b .By this, there exists ε > 0 such that

eu(r) <1 − ε

1 + aand ev(r) <

1 − ε

1 + bfor r > R1. (7.5)

Thus, for r � R1,

s(f1 + f2)(s) ds

s(eu(s) − (1 + a)e2u(s) + ev(s) − (1 + b)e2v(s)) ds + O(1)

� ε

( r∫R1

seu(s) ds +r∫

sev(s) ds

)+ O(1) (7.6)

where Lemma 7.1(2) and (7.5) are used. It is easy to see from Pohozaev identity (7.1)that

min{a, b}4(1 + 2b)2 + 4(1 + 2a)2

((1 + 2b)rur(r) + (1 + 2a)rvr(r)

)2+ (1 + a + b)r2

(beu(r) − b(1 + a)

(b(1 + b)

2 r2u2r(r) + abr2ur(r)vr(r) + a(1 + a)

2 r2v2r(r)

)+ (1 + a + b)r2

(beu(r) − b(1 + a)

= 2(1 + a + b)r∫

(beu(s) − b(1 + a)

2 e2u(s) + aev(s)

− a(1 + b)2 e2v(s) + abe(u+v)(s)

+ 4(b(1 + b)

N21 + abN1N2 + a(1 + a)

)(7.7)

Note that

(1 + 2b)rur(r) + (1 + 2a)rvr(r) = 2((1 + 2b)N1 + (1 + 2a)N2

)− (1 + a + b)

s(f1 + f2)(s) ds + O(1).

Thus, for r � R1,

((1 + 2b)rur(r) + (1 + 2a)rvr(r)

)2 � 4(1 + a + b)2ε2

( r∫R1

s(eu(s) + ev(s)) ds)2

Therefore, (7.7) implies

min{a, b}4(1 + 2b)2 + 4(1 + 2a)2 ε

( r∫R1

s(eu(s) + ev(s)) ds)2

� 2(1 + a + b)( r∫

s(eu(s) + ev(s)) ds) + O(1). (7.8)

Due to (7.3), it is a contradiction.Step 4. Case (b): sup(R1,∞) u(r) = log 1

We denote

w1(r) = max(u(r), v(r)

)and w2(r) = min

(u(r), v(r)

We assume that log 11+b � log 1

1+a and there exists {rn}∞n=1 → +∞ such that

u(rn) = supr∈[R1,rn]

w1(r) tends to log 11 + a

as rn tends to + ∞.

Assume that r1 � R1, then we obtain

rn∫0

(beu(s) − b(1 + a)

2 e2u(s) + aev(s) − a(1 + b)2 e2v(s) + abe(u+v)(s)

=rn∫

(beu(s) − b(1 + a)

)ds + O(1)

(R1,rn)∩Γ1

(beu(s) − b(1 + a)

+ O(1)

(R1,rn)∩Γ2

(beu(s) − b(1 + a)

2e2u(s) + aev(s) − a(1 + b)

2e2v(s) + abe(u+v)(s)

+ O(1) (7.9)

Γ1 ={r ∈ [R1,∞)

∣∣ w1(r) � −3 log r}

Γ2 ={r ∈ [R1,∞)

∣∣ w1(r) < −3 log r}.

Therefore, we have

∫(R1,rn)

⋂Γ1

(beu(s) − b(1 + a)

�(beu(rn) − b(1 + a)

2 e2u(rn)) rn∫R1

s ds + O(1)

(beu(rn) − b(1 + a)

2 e2u(rn))

+ O(1). (7.10)

Moreover, one can easily see that

∫(R1,rn)

⋂Γ2

(beu(s) − b(1 + a)

= O(1). (7.11)

Combining (7.10) and (7.11), (7.7) implies that

min{a, b}4(1 + 2b)2 + 4(1 + 2a)2

((1 + 2b)rnur(rn) + (1 + 2a)rnvr(rn)

)2 � O(1). (7.12)

We conclude that f1 and f2 are in L1(R2). �Step 5. Case (c): sup(R1,∞) u(r) = log 1

1+a and sup(R1,∞) v(r) < log 11+b .

As in the step 3, there exists ε > 0 such that

ev(r) <1 − ε for r > R1. (7.13)

Since supΓ{v<u}u(r) = log 1

1+a , there here exists {rn}∞n=1 → +∞ such that

u(rn) = supr∈[R0,rn]

u(r) tends to log 11 + a

as rn tends to + ∞.

As in (7.7), (7.8) and step 4, we have

min{a, b}4(1 + 2b)2 + 4(1 + 2a)2 ε

( rn∫R0

sev(s) ds

� 2(1 + a + b)( rn∫

sev(s) ds

)+ O(1).

(7.14)

It is a contradiction.The case (d) is similar to the case (c). We omit the proof.

Proof of Theorem 1.1. We need to show that if the entire radial solution (u, v) of (1.8)is not a topological solution, then it must be either a non-topological solution or amixed-type solution. Note that Lemma 7.3 implies limr→∞ rur(r) and limr→∞ rvr(r)exist if (u, v) is not a topological solution. Hence, it is easy to see that limr→∞(u(r), v(r))exists and must be one of the following:

log 11 + a

,−∞)

(2)(−∞, log 1

)(3) (−∞,−∞).

Thus, we prove Theorem 1.1. �8. Proof of Corollary 1.5

We denote these two quantities

−β1 = 2N1 +∞∫0

s(−(1 + a)f1 + af2

)(s) ds

−β2 = 2N2 +∞∫0

s(−(1 + b)f2 + bf1

)(s) ds

to characterize the behavior of solutions of (1.8). By integrating Eqs. (1.8), it is easy tosee that

lim rur(r) = −β1

limr→∞

rvr(r) = −β2.

Proof of Corollary 1.5. We split the proof of Corollary 1.5 into the following two lemmas:Lemma 8.1 and Lemma 8.2.

Lemma 8.1. Suppose (u, v) is a non-topological solution of (1.8). Then

β1 > 2, β2 > 2,

b(1 + b)2 β2

1 + abβ1β2 + a(1 + a)2 β2

2 − 4(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)− 2(1 + a + b)

(b(2N1 + β1) + a(2N2 + β2)

)= 2(1 + a + b)

∞∫0

((1 + a)b

2 e2u(s) + (1 + b)a2 e2v(s) − abe(u+v)(s)

)ds > 0. (8.1)

Proof. Step 1. We first prove the inequality (8.1) with the assumptions β1 > 2 and β2 > 2.Applying

r(bu + av)r(r) = 2(bN1 + aN2) −r∫

s(bf1 + af2)(s) ds

andr∫

(beu(s) − b(1 + a)

s(bf1 + af2)(s) ds

2 + ab

)e2u(s) +

2 + ab

)e2v(s) − abe(u+v)(s)

)ds (8.2)

to the Pohozaev identity, we obtain

b(1 + b)2 β2

1 + abβ1β2 + a(1 + a)2 β2

2 − 4(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)+ (1 + a + b)r2

(beu(r) − b(1 + a)

e2u(r) + aev(r) − a(1 + b)e2v(r) + abe(u+v)(r)

− 2(1 + a + b)(b(2N1 + β1) + a(2N2 + β2)

)= 2(1 + a + b)

∞∫0

((1 + a)b

2 e2u(s) + (1 + b)a2 e2v(s) − abe(u+v)(s)

)ds (8.3)

Since we suppose β1 > 2 and β2 > 2, for r sufficiently large, we have

u(r) = −β1 log r + O(1) and v(r) = −β2 log r + O(1).

Plugging these into (8.3) and letting r tend to ∞, (8.1) follows.Step 2. Since (u, v) is a non-topological solution, there exists R1 > 0 such that

u(r), v(r) < min{

log 12(1 + a) , log 1

2(1 + b)

}for r � R1.

By this and Lemma 7.3, we obtain

∞∫R1

u(s) + ae(u+v)(s))ds �

∞∫R1

sf1(s) ds < ∞

and∞∫

v(s) + be(u+v)(s))ds �

∞∫R1

sf2(s) ds < ∞.

We conclude that

eu, ev and eu+v are in L1(R

2). (8.4)

Step 3. We show that β1 � 2 and β2 � 2.By the symmetry of the equation, we only show that β1 � 2. Suppose for the sake of

contradiction that β1 < 2, then

rur(r) � −2 for r sufficiently large,

and thus u(r) � −2 log r + O(1) for r sufficiently large. It contradicts to (8.4).Step 4. To show β1 > 2 and β2 > 2, we need to exclude the following other possible

cases:

(1) β1 = 2, β1 < β2.(2) β2 = 2, β2 < β1.(3) β1 = 2, β2 = 2.

The first two cases are symmetric, hence, we only consider the first case.

Step 4.1. Suppose β1 = 2 and β1 < β2. Then there exists R2 such that

u(r) < log 12(1 + a) and u(r) > v(r) for r � R2. (8.5)

Recall that

rur(r) = 2N1 +r∫

s(a(f2 − f1)(s) − f1(s)

Note that (f2 − f1)(r) < 0 and −f1(r) < 0 on (R2,∞) where (8.5) is used. Thus, rur(r)is a decreasing function on (R2,∞), which implies rur(r) > −2 on (R2,∞). It makes theL1(R2)-integrability of eu fail. Hence, this case is impossible.

Step 4.2. Suppose β1 = 2 and β2 = 2.For r sufficiently large, (8.3) becomes

4(b(1 + b)

2 + ab + a(1 + a)2

)− 4

(b(1 + b)

2 N21 + abN1N2 + a(1 + a)

)+ (1 + a + b)r2

(beu(r) − b(1 + a)

)> 2(1 + a + b)

(b(2N1 + 2) + a(2N2 + 2)

)(8.6)

which implies

r(beu(r) + aev(r)) � C

for some positive constant C. It contradicts to (8.4).We conclude that β1 > 2 and β2 > 2. �The mixed-type solutions of (1.8) will be discussed as follows.

Lemma 8.2. Suppose (u, v) is a mixed-type solution. Then

(1) β1 = 0 and β2 > 2 if (u, v) → (log 11+a ,−∞) as r → ∞.

(2) β2 = 0 and β1 > 2 if (u, v) → (−∞, log 11+b ) as r → ∞.

Proof. By the symmetry of the equations, we only need to prove the first part.Step 1. We show that b

1+aβ1 + β2 > 2.By Theorem 1.1, v(r) < u(r) < log 1

1+a on (R0,∞). Thus,

1 + au + v

(r) = 2(

1 + aN1 + N2

)− 1 + a + b

r∫sf2 ds

is a decreasing function on (R0,∞). If b1+aβ1 + β2 � 2, then

1 + au + v

(r) � −2 on (R0,∞).

It follows that∫∞R0

se( b1+au+v)(s) ds = ∞, which implies

∫∞R0

sev(s) ds = ∞. It is a contra-diction to Lemma 7.3.

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