Nonlinear Analysis 70 (2009) 1096–1104www.elsevier.com/locate/na

Existence and estimate of large solutions for an elliptic systemI

Lei Weia,b,∗, Mingxin Wanga

a Department of Mathematics, Southeast University, Nanjing 210018, PR Chinab School of Mathematical Science, Xuzhou Normal University, Xuzhou 221116, PR China

Received 13 November 2007; accepted 18 January 2008

Abstract

In this paper, an elliptic system with boundary blow-up is considered in a smooth bounded domain. By constructing certainupper solution and subsolution, we show the existence of positive solutions and give a global estimate. Furthermore, the boundarybehavior of positive solutions is also discussed.c© 2008 Elsevier Ltd. All rights reserved.

MSC: 35J55; 35B40

Keywords: An elliptic system; Boundary blow-up; Existence; Uniqueness

1. Introduction

In this paper, we are concerned with a system of semilinear elliptic equations:∆u = u(a1um1 + b1(x)um+ δ1v

n), x ∈ Ω ,∆v = v(a2v

p1 + b2(x)vp+ δ2uq), x ∈ Ω ,

u = v = +∞, x ∈ ∂Ω ,(1.1)

where Ω ⊂ RN is a bounded and smooth domain, and constants ai ≥ 0 (i = 1, 2). Functions b1, b2 ∈ Cη(Ω)are positive weight functions, which is singular on ∂Ω and 0 < η < 1. The boundary condition is interpreted asu(x), v(x)→+∞ as d(x) = dist(x, ∂Ω)→ 0+. Assume that m > m1 > 0, p > p1 > 0 and q, n > 0.

There is a large amount of literature on elliptic equations related to (1.1) (refer to [1–3,5–8,11,12]). A single ellipticequation with blow-up condition is of the general form

∆u = f (x, u), x ∈ Ω; u = +∞, x ∈ ∂Ω . (1.2)

I This work was supported by the National Natural Science Foundation of China 10771032 and the Natural Science Foundation of Jiangsuprovince BK2006088.∗ Corresponding author at: Department of Mathematics, Southeast University, Nanjing 210018, PR China.

E-mail addresses: wleducn@yahoo.com.cn (L. Wei), mxwang@seu.edu.cn (M. Wang).

0362-546X/$ - see front matter c© 2008 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2008.01.036

L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104 1097

Problems related to (1.2) have been extensively studied. The pioneering work may go back to [1] with f (x, u) = eu

in the plane. A particular example is the following elliptic equation:∆u = b(x)ur+1, x ∈ Ω ,u = +∞, x ∈ ∂Ω ,

(1.3)

where Ω ⊂ RN is a C2 bounded domain, r > 0, the weight function b(x) is continuous and positive in Ω . Itis well known from [13] that (1.3) has a unique positive solution if there exist C1,C2 > 0, γ < 2 such thatC2d(x)−γ < b(x) < C1d(x)−γ in Ω . Now, there is a huge amount of literature dealing with single equations withboundary blow-up conditions, but very little has been said for the moment on elliptic system. In paper [4], the authorconsidered positive solutions of the following elliptic system:∆u = u pvq , x ∈ Ω ,

∆v = urvs, x ∈ Ω ,u = v = +∞, x ∈ ∂Ω .

(1.4)

In paper [9], the authors studied the following elliptic systems:−∆u = u(a1 − b1um− c1v

n) x ∈ Ω ,−∆v = v(a2 − b2u p

− c2vq), x ∈ Ω ,

u = v = +∞, x ∈ ∂Ω ,(1.5)

where ai ≥ 0, bi , ci (i = 1, 2) are positive constants, and m, q > 0, n, p ≥ 0.Before stating our results, we firstly give two hypotheses:

D2d(x)−γ1 ≤ b1(x) ≤ D1d(x)−γ1 , x ∈ Ω , (1.6)

K2d(x)−γ2 ≤ b2(x) ≤ K1d(x)−γ2 , x ∈ Ω , (1.7)

where Di , Ki > 0, 0 < γi < 2(i = 1, 2) are constants.The main results of the present paper are the following.

Theorem 1. Assume that q(2 − γ1) < 2m and n(2 − γ2) < 2p, and (1.6) and (1.7) are satisfied. If δ1, δ2 > 0 aresufficiently small, then elliptic system (1.1) has at least a positive solution. If δ1, δ2 < 0 and m > n, p > q, thenelliptic system (1.1) has at least a positive solution.

Theorem 2. Assume that q(2 − γ1) < 2m and n(2 − γ2) < 2p, and (1.6) and (1.7) are satisfied. If (δ1, δ2) →

(0+, 0+), then there exists a subsequence (uk, vk) of solution sequence corresponding to (1.1) such that (uk, vk)→

(Um,γ1 ,U p,γ2) in C2loc(Ω), where Um,γ1 and U p,γ2 are the unique positive solutions of the problems

∆u = a1um1+1+ b1(x)u

m+1 x ∈ Ω ,u = +∞, x ∈ ∂Ω

(1.8)

and ∆v = a2v

p1+1+ b2(x)v

p+1 x ∈ Ω ,v = +∞, x ∈ ∂Ω

(1.9)

respectively.

Theorem 3. Assume that n(2 − γ2) ≤ pγ1, q(2 − γ1) ≤ mγ2, δ1, δ2 > 0, and (1.6) and (1.7) are satisfied. If (u, v)is a positive solution of elliptic system (1.1), then there exist C1,C2 > 0 such that

C2d(x)−α ≤ u ≤ C1d(x)−α, C2d(x)−β ≤ v ≤ C1d(x)−β ,

where α = 2−γ1m , β =

2−γ2p .

1098 L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104

Theorem 4. Assume that n(2− γ2) ≤ pγ1, q(2− γ1) ≤ mγ2, δ1, δ2 > 0, and b1, b2 satisfies (1.6) (1.7) and

limx→∂Ω

b1(x)d(x)γ1 = A, lim

x→∂Ωb2(x)d(x)

γ2 = B, (1.10)

where A, B are positive constants. Then for any positive solution (u, v) to system (1.1), it holds that

limx→∂Ω

d(x)αu(x) =

(α(α + 1)

A

)1/m

, limx→∂Ω

d(x)β(x)v(x) =

(β(β + 1)

B

)1/p

.

2. Preliminaries

Our tool is to explore the upper and lower solutions method. Since the right-hand sides of the system (1.1) are bothmonotone, we can give the following definition and we can refer to [10].

Definition 1. Assume that δ1, δ2 > 0, then (u, v), (u, v) ∈ (C2(Ω))2 are called upper solution and subsolution of(1.1), provided that

∆u ≤ u(a1um1 + b1(x)um+ δ1v

n), x ∈ Ω ,∆v ≥ v(a2v

p1 + b2(x)vp+ δ2uq), x ∈ Ω , (2.1)

∆u ≥ u(a1um1 + b1(x)um+ δ1v

n), x ∈ Ω ,∆v ≤ v(a1v

p1 + b2(x)vp+ δ2uq), x ∈ Ω . (2.2)

Similarly, assume that δ1, δ2 < 0, then (u, v), (u, v) ∈ (C2(Ω))2 are called upper solution and subsolution of (1.1),provided that

∆u ≤ u(a1um1 + b1(x)um+ δ1v

n), x ∈ Ω ,∆v ≤ v(a1v

p1 + b2(x)vp+ δ2uq), x ∈ Ω , (2.3)

∆u ≥ u(a1um1 + b1(x)um+ δ1v

n), x ∈ Ω ,∆v ≥ v(a2v

p1 + b2(x)vp+ δ2uq), x ∈ Ω . (2.4)

Lemma 1 ([13]). Assume that b ∈ Cη(Ω) and there exist C ′′,C ′ > 0, γ < 2 such that C ′′d(x)−γ ≤ b(x) ≤C ′d(x)−γ in Ω . Then problem (1.3) has a unique positive solution u ∈ C2

loc(Ω). Moreover, there exist positive

constants M1 and M2, such that M2d(x)−2−γ

r ≤ u(x) ≤ M1d(x)−2−γ

r in Ω .

Lemma 2 ([3]). Let u ∈ C2(Ω) verify ∆u ≤ Cd(x)−γ u1+r in Ω for some positive constant, and u = +∞ on ∂Ω .

Then u(x) ≥ C−1r Ur,γ , where Ur,γ denotes the unique positive solution of the problem

∆u = d(x)−γ ur+1, x ∈ Ω ,u = +∞, x ∈ ∂Ω .

Similarly, if ∆u ≥ Cd(x)−γ ur+1 in Ω , then u(x) ≤ C−1r Ur,γ .

3. Proof of theorem

Proposition 1. Assume that (u, v), and (u, v) are the positive upper and lower solutions of (1.1) and u = u = v =v = +∞ on ∂Ω , u ≤ u, v ≤ v in Ω . Then (1.1) has at least a positive solution (u, v) satisfying u ≤ u ≤ u, v ≤ v ≤ vin Ω .

Proof. Denote Ωk = x ∈ Ω : d(x) > 1k . Consider the problem∆u = u(a1um1 + b1(x)u

m+ δ1v

n), x ∈ Ωk,

∆v = v(a2vp1 + b2(x)v

p+ δ2uq), x ∈ Ωk,

u = u, v = v, x ∈ ∂Ωk .

L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104 1099

Hence, there exists a solution (uk, vk) such that u ≤ uk ≤ u, v ≤ vk ≤ v in Ωk . In turn, we get a sequence (uk, vk)

and also give their bounds. From standard regularity theory and compactly embedding theory, it follows that thereexists subsequence (ukn , vkn ) such that ukn → u, vkn → v in C2

loc(Ω). So (u, v) is a solution to system (1.1)satisfying u ≤ u ≤ u, v ≤ v ≤ v in Ω .

Proposition 2. Assume that a1 ≥ 0, m > m1 > 0, and b1 satisfies (1.6). Then (1.8) has a unique positive solution u,and there exist C1,C2 > 0 such that

C2d(x)−2−γ1

m ≤ u(x) ≤ C1d(x)−2−γ1

m , x ∈ Ω .

Proof. Let U denote a unique positive solution to the equation∆u = b1(x)u

m+1, x ∈ Ω ,u = +∞, x ∈ ∂Ω .

Set u = U , u = εU , where ε > 0. Now we consider the following equation:∆u = a1um1+1

+ b1(x)um+1, x ∈ Ωk,

u = u, x ∈ ∂Ωk,(3.1)

where Ωk = x ∈ Ω : d(x) > 1k . Clearly, we have ∆u = b1(x)um+1

≤ a1um1+1+ b1(x)um+1 in Ωk . By

0 < γ1 < 2 and condition (1.6), we have b0 = infΩ b1(x) > 0. U is a positive function with boundary blow-up, thusm0 = infΩ U (x) > 0. If ε is sufficient small, it follows from m > m1 that

U m+1

2≥

a1εm1U m1+1

b0, x ∈ Ω .

If εm < 12 , we have

U m+1≥

a1εm1

b1(x)U m1+1

+ εmU m+1, x ∈ Ωk .

It is easy to see that

b1(x)Um+1≥ a1ε

m1U m1+1+ b1(x)ε

mU m+1, x ∈ Ωk,

furthermore, we have

∆u = εb1(x)Um+1≥ a1(εU )

m1+1+ b1(x)(εU )

m+1, x ∈ Ωk .

By the standard upper solution and subsolution argument, there exists a solution uk such that u ≤ uk ≤ u. By standardregularity theory and embedding argument, without loss of generality there exists a sequence uk such that uk → u0in C2

loc(Ω) as k → ∞. Thus u0 is a positive solution to (1.8) satisfying εU ≤ u0 ≤ U . From Lemmas 1 and 2, itfollows that there exist C1,C2 > 0 such that

C2d(x)−2−γ1

m ≤ u0(x) ≤ C1d(x)−2−γ1

m , x ∈ Ω . (3.2)

Assume that u is an arbitrary positive solution to (1.8). We will show u = u0 i.e. uniqueness. Clearly, ∆u =a1um1+1

+ b1(x)um+1≥ b1(x)um+1

≥ D2d(x)−γ1um+1, thus it implies that there exists L1 > 0 such that

u ≤ L1d(x)−2−γ1

m in Ω by Lemma 2. By infΩ u(x) > 0, infΩ b1(x) > 0 and m > m1, we can take a sufficiently largeconstant M > 0 such that

∆(Mu) = M(a1um1+1+ b1(x)u

m+1) ≤ b1(x)Mm+1um+1

= b1(x)(Mu)m+1, x ∈ Ω .

Hence, there exists L2 > 0 such that u ≥ L2d(x)−2−γ1

m in Ω by Lemma 2. These imply that uu0

is bounded in Ω .Assume that there exist x0 ∈ Ω and k > 1 such that u(x0) > ku0(x0). Denote Ω0 = u > ku0

⋂Br (x0), where

r = 12 d(x0). Then

1100 L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104

∆(u − ku0) = a1um1+1+ b1(x)u

m+1− ka1um1+1

0 − kb1(x)um+10

> a1km1+1um1+10 + b1(x)k

m+1um+10 − ka1um1+1

0 − kb1(x)um+10

> k(km− 1)D2d(x)−γ1C2d(x)−

(2−γ1)(m+1)m

≥ kCd(x)−2m−γ1+2

m ≥ kCr−2m−γ1+2

m , x ∈ Ω0.

Now, setting h(x) = Ckr−2m−γ1+2

m (r2−|x−x0|

2)/2N , then we have ∆(u−ku0+h) > 0 and (u−ku0+h)(x) > 0 inΩ0. By the maximum principle, there exists a point x1 ∈ ∂Ω0 such that h(x0) < (u−ku0+h)(x0) < (u−ku0+h)(x1).It easily follows that x1 ∈ ∂Br (x0), since otherwise h(x0) < h(x1), which is impossible. Thus x1 ∈ ∂Br (x0). Hence,

we have u(x1) − ku0(x1) > h(x0) =Ck2N r−

2m−γ1+2m +2

=Ck2N r−

2−γ1m ≥ Cku0(x1) i.e. u(x1) ≥ k(C + 1)u0(x1). We

iterate this procedure to obtain a sequence of points xn ⊂ Ω such that u(xn) > (1+C)nku0(xn). But this contradictsthat u

u0is bounded in Ω . Thus u ≤ u0, and symmetrically u0 ≤ u, which imply uniqueness.

Proof of Theorem 1. The assumptions q(2− γ1) < 2m and n(2− γ2) < 2p imply

δ0 = min

infΩ

C−q1 C p

2 K2

2d(x)

q(2−γ1)m −2, inf

Ω

C−n1 Cm

2 D2

2d(x)

n(2−γ2)p −2

> 0.

Firstly, we will show that system (1.1) has at least a positive solution when 0 < δ1, δ2 < δ0. Our aim is to look foran ordered upper solution and subsolution. Let (u, v) = (Um,γ1 ,U p,γ2) and (u, v) = (εUm,γ1 , εU p,γ2), where Um,γ1 ,U p,γ2 are unique positive solutions of (1.8) and (1.9) respectively. By Proposition 2, there exist C1,C2 > 0 such that

C2d(x)−2−γ1

m ≤ Um,γ1(x) ≤ C1d(x)−2−γ1

m , x ∈ Ω ,

C2d(x)−2−γ2

p ≤ U p,γ2(x) ≤ C1d(x)−2−γ2

p , x ∈ Ω .

It is clear that

∆u = a1 U m1+1m,γ1

+ b1(x)U m+1m,γ1≤ Um,γ1(a1 U m1

m,γ1+ b1(x)U m

m,γ1+ δ1(εU p,γ2)

n)

= u(a1um1 + b1(x)um+ δ1v

n).

From δ2 ≤ δ0, it follows that δ2 ≤C−q

1 C p2 K2

2 d(x)q(2−γ1)

m −2. Therefore, we have

K2C p2 d(x)−2

2≥ δ2Cq

1 d(x)−q(2−γ1)

m , x ∈ Ω .

Since

b2(x)

2U p

p,γ2 ≥K2C p

2

2d(x)−2, U q

m,γ1 ≤ Cq1 d(x)−

q(2−γ1)m , in Ω ,

we have thatb2(x)

2U p

p,γ2 ≥ δ2 U qm,γ1 .

Take ε > 0 such that maxε p, εm < 1

2 . It follows that

a2 U p1p,γ2 + b2(x)U p

p,γ2 ≥ a2εp1 U p1

p,γ2 + εpb2(x)U p

p,γ2 + δ2 U qm,γ1 .

Furthermore, we have

∆v ≥ v(a2vp1 + b2v

p+ δ2uq).

It is clear that

∆v = a2 U p1+1p,γ2 + b2(x)U p+1

p,γ2

≤ U p,γ2(a2 U p1p,γ2 + b2(x)U p

p,γ2 + δ2(εUm,γ1)q)

= v(a2vp1 + b2(x)v

p+ δ2uq).

L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104 1101

From δ1 ≤ δ0, it follows that δ1 ≤C−n

1 Cm2 D2

2 d(x)n(2−γ2)

p −2, i.e.

D2Cm2 d(x)−2

2≥ δ1Cn

1 d(x)−n(2−γ2)

p , x ∈ Ω .

Since

b1(x)

2U m

m,γ1≥

D2Cm2

2d(x)−2, U n

p,γ2≤ Cn

1 d(x)−n(2−γ2)

p , in Ω ,

we have that

b2(x)

2U m

m,γ1≥ δ1 U n

p,γ2.

Thus, we have

a1 U m1m,γ1+ b1(x)U m

m,γ1≥ a1ε

m1 U m1m,γ1+ εmb1(x)U m

m,γ1+ δ1 U n

p,γ2.

Furthermore, we conclude

∆u ≥ u(a1um1 + b1(x)um+ δ1v

n).

By Proposition 1, system (1.1) at least has a positive solution (u, v).Assume that δ1, δ2 < 0. Set (u, v) = (Um,γ1 ,U p,γ2), (u, v) = (M Um,γ1 ,M U p,γ2). Clearly, we have

∆u = a1um1+1+ b1(x)u

m+1≥ u(a1um

+ b1(x)um1 + δ1v

n),

∆v = a2vp1+1+ b2(x)v

p+1≥ v(a1v

p1 + b2(x)vp+ δ1uq).

Denote

M = max

2,(−2δ1Cn

1

D2Cm2

supΩ

d(x)2−βn) 1

m−n

,

(−2δ2Cq

1

K2C p2

supΩ

d(x)2−αq

) 1p−q

.Then we have

Mm−n D2Cm2 d(x)−2

2≥ −δ1Cn

1 d(x)−n(2−γ2)

p ,

b1(x)Mm−n U mm,γ1

2≥ −δ1 U n

p,γ2.

Therefore, it holds that

Mm+1b1(x)U m+1m,γ1

2≥ −δ1 Mn+1 Um,γ1 U n

p,γ2.

It follows that

∆u = a1 M U m1+1m,γ1

+ Mb1(x)U m+1m,γ1

≤ a1 Mm1+1 U m1+1m,γ1

+ b1(x)Mm+1 U m+1

m,γ1+ δ1 Mn+1 Um,γ1 U n

p,γ2

= u(a1um1 + b1(x)um+ δ1v

n).

Similarly, we have

∆v ≤ v(a2vp1 + b1v

p+ δ2uq).

By Proposition 1, systems (1.1) at least has a positive solution (u, v).

Proof of Theorem 2. Let (u, v) = (Um,γ1 ,U p,γ2) and (u, v) = (εUm,γ1 , εU p,γ2), where Um,γ1 , U p,γ2 are uniquepositive solutions of (1.8) and (1.9) respectively, and maxε p, εm

< 12 . If limk→∞ δ

k1 = 0, limk→∞ δ

k2 = 0 and

0 < δk1, δ

k2 ≤ δ0, we may similarly prove that (u, v) and (u, v) are still a pair of upper solution and subsolution

1102 L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104

corresponding to the elliptic system (1.1) (ε in the pair upper solution and subsolution can have no relation to δk1, δ

k2).

It follows from Theorem 1 that there exists a positive solution denoted by (uk, vk) such that

u ≤ uk ≤ u, v ≤ vk ≤ v.

Thus (uk, vk) has a convergent subsequence in C2loc(Ω) which is still denoted by itself and the limits are denoted by

(u0, v0). For any compact E ⊂ Ω , it follows that uk, vk are uniformly bounded in E . Let∆uk = uk(a1um1

k + b1(x)umk + δ

k1v

nk ), x ∈ E,

∆vk = vk(a2vp1k + b2(x)v

pk + δ

k2uq

k ), x ∈ E,(3.3)

pass to the limit as k →∞, we have that∆u0 = u0(a1um1

0 + b1(x)um0 ), x ∈ E,

∆v0 = v0(a2vp10 + b2(x)v

p0 ), x ∈ E .

(3.4)

Thus it follows from the arbitrary property of E ⊂ Ω that∆u0 = u0(a1um1

0 + b1(x)um0 ), x ∈ Ω ,

∆v0 = v0(a2vp10 + b2(x)v

p0 ), x ∈ Ω ,

u0 = v0 = +∞.

(3.5)

By Proposition 2, we have u0 = Um,γ1 , v0 = U p,γ2 .

Proof of Theorem 3. Since (u, v) is a positive solution of elliptic system (1.1), without loss of generality, we canassume that u(x), v(x) ≥ 1 in Ω . From the elliptic system, it follows that

∆u ≥ b1(x)um+1, ∆v ≥ b2(x)v

p+1, x ∈ Ω .

Thus, there exists C > 0 such that

u(x) ≤ Cd(x)−2−γ1

m = Cd(x)−α, v(x) ≤ Cd(x)−2−γ2

p = Cd(x)−β , x ∈ Ω .

Hence, we have

∆u = u(a1um1 + b1(x)um+ δ1v

n)

≤ u(a1um1 + D1d(x)−γ1um+ δ1Cnd(x)−nβ),

∆v = v(a2vp1 + b2(x)v

p+ δ2uq)

≤ v(a2vm1 + K1d(x)−γ2v p

+ δ2Cqd(x)−qα).

From 0 < nβ ≤ γ1 < 2, 0 < qα ≤ γ2 < 2, it follows that there exists C∗ > 0 such that in Ω

D1d(x)−γ1um+ δ1Cnd(x)−nβ

≤ C∗d(x)−γ1um,

K1d(x)−γ2v p+ δ2Cqd(x)−qα

≤ C∗d(x)−γ2v p.

Furthermore, we have

∆u ≤ u(a1um1 + C∗d(x)−γ1um), x ∈ Ω ,

∆v ≤ v(a2vp1 + C∗d(x)

−γ2v p), x ∈ Ω .

Since u, v are positive and boundary blow-up functions, infΩ u(x) > 0 and infΩ v(x) > 0 hold. By m > m1, p > p1,infΩ d(x)−γ1 > 0 and infΩ d(x)−γ2 > 0, we can take sufficient large constant C > 0 such that

∆u ≤ Cd(x)−γ1um+1, ∆v ≤ Cd(x)−γ2v p+1, x ∈ Ω .

That implies that there exists C2 > 0 such that

u(x) ≥ C2d(x)−α, v(x) ≥ C2d(x)−β , x ∈ Ω .

Thus, we may take suitable C1,C2 > 0 such that in Ω

C2d(x)−α ≤ u(x) ≤ C1d(x)−α, C2d(x)−β ≤ v(x) ≤ C1d(x)−β .

L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104 1103

Proof of Theorem 4. Let (u, v) be any positive solution to the system (1.1). It follows from Theorem 3 that (u, v)has the global estimates. Take any arbitrary point x0 ∈ ∂Ω and xk ⊂ Ω such that xk → x0. Choose an openneighborhood D of x0 so that ∂Ω admits C2,η local coordinates ξ : D → RN , and x ∈ D

⋂Ω if and only if

ξ1(x) > 0(ξ = (ξ1, ξ2, . . . , ξN )). Assume that ξ(x0) = 0. Denoting u(x) = u(ξ(x)), v(x) = v(ξ(x)), it follows that(u, v) satisfies the system:

N∑i, j=1

ai, j (ξ)∂2u

∂ξi∂ξ j+

N∑i=1

ei (ξ)∂u

∂ξi= u(a1um1 + b1(ξ)u

m+ δ1v

n),

N∑i, j=1

ai, j (ξ)∂2v

∂ξi∂ξ j+

N∑i=1

ei (ξ)∂v

∂ξi= v(a2v

p1 + b2(ξ)vp+ δ2uq),

(3.6)

in ξ(D⋂

Ω), where ai, j , ei are Cη, and ai, j (0) = δi, j , bi (ξ(x)) = bi (x). Let tk be the projections of ξ(xk) on toξ(D

⋂∂Ω), and define

uk(y) = dαk u(tk + dk y), vk(y) = dβk v(tk + dk y),

where dk = d(ξ(xk)). Then (uk, vk) satisfies for y ∈ E

N∑i, j=1

ai, j (tk + dk y)∂2uk

∂yi∂y j+ dk

N∑i=1

ei (tk + dk y)∂uk

∂yi

= uk(a1d2−αm1k um1

k + b1(tk + dk y)dγ1k um

k + δ1d2−nβk vn

k ),N∑

i, j=1

ai, j (tk + dk y)∂2vk

∂yi∂y j+ dk

N∑i=1

ei (tk + dk y)∂vk

∂yi

= vk(a2d2−βp1k v

p1k + b2(tk + dk y)dγ2

k vpk + δ2d2−qα

k uqk ),

(3.7)

where E = y ∈ RN: y1 > 0. From the global estimates of positive solutions to the system (1.1) in Theorem 3, it

follows that

C2 y−α1 ≤ uk(y) ≤ C1 y−α1 , C2 y−β1 ≤ vk(y) ≤ C1 y−β1 .

In particular, we have estimates for uk and vk in compact subset of E . As a result, it is standard to conclude withoutloss of generality that uk → u0 and vk → v0 in C2

loc(E). It is clear that

limk→∞

b1(tk + dk y)dγ1k = Ay−γ1

1 , limk→∞

b2(tk + dk y)dγ2k = By−γ2

1 .

From 2− αm1 > 0, 2− nβ > 0, 2− βp1 > 0, 2− αq > 0 and the global estimates to uk, vk , it follows that (u0, v0)

is a positive solution to the equation∆u0 = Ay−γ1

1 um+10 ,

∆v0 = By−γ21 v

p+10 ,

C2 y−α1 ≤ u0 ≤ C1 y−α1 , C2 y−β1 ≤ v0 ≤ C1 y−β1 .

(3.8)

Since this problem has unique positive solutions, we have that

u0(y) =

(α(α + 1)

A

)1/m

y−α1 , v0(y) =

(β(β + 1)

B

)1/p

y−β1 .

Take y = (1, 0, . . . , 0), then we have ξ(xk) = tk + dk(1, 0, . . . , 0). Furthermore, it follows that

u0(y) =

(α(α + 1)

A

)1/m

, v0(y) =

(β(β + 1)

B

)1/p

.

Therefore, we have

1104 L. Wei, M. Wang / Nonlinear Analysis 70 (2009) 1096–1104

limk→∞

d(xk)αu(xk) = u0(y) =

(α(α + 1)

A

)1/m

,

limk→∞

d(xk)αv(xk) = v0(y) =

(β(β + 1)

B

)1/p

.

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