Large solutions for an elliptic system of competitive type: Existence, uniqueness and asymptotic behavior

Nonlinear Analysis 71 (2009) 4544–4552

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Large solutions for an elliptic system of competitive type: Existence, uniqueness and asymptotic behaviorI Chunlai Mu a , Shuibo Huang a,b,∗ , Qiaoyu Tian b , Limin Liu a a

College of Mathematics and Physics, Chongqing University, Chongqing 400030, PR China

b

Department of Mathematics, Hezuo Minorities Teacher College, Hezuo, Gansu 747000, PR China

article

abstract

info

Article history: Received 2 July 2007 Accepted 11 March 2009

This paper deals with the existence, uniqueness and asymptotic behavior of boundary blow-up solutions for an elliptic system of competitive type, where the weight functions a(x), b(x) vanish on the boundary of the underlying domain at different rates according to the point of the boundary. The proof relies on subsolutions and supersolutions, and the localization method. Crown Copyright © 2009 Published by Elsevier Ltd. All rights reserved.

Keywords: Elliptic system Subsolution and supersolution Localization method

1. Introduction In this paper, we consider the existence, uniqueness and asymptotic behavior of positive solutions for the following elliptic system:

( 1u = a(x)up v q , 1v = b(x)ur v s , u = v = ∞,

in Ω , in Ω , on ∂ Ω ,

(1.1)

where p, s > 1, q, r > 0, (p − 1)(s − 1) − qr > 0, Ω is a bounded C 2 domain of RN , N ≥ 1, and there exist C (x), D(x) ∈ C (Ω ; R+ ), γ (x), η(x) ∈ C (Ω ; R+ ) such that lim

x→x0

a( x ) C (x0

)d(x)γ (x0 )

= 1,

lim

x→x0

b( x) D(x0 )d(x)η(x0 )

= 1,

(1.2)

where x0 ∈ ∂ Ω , d(x) = dist(x, ∂ Ω ). We must emphasize that the weight functions a(x), b(x) are allowed to decay to zero on Ω with arbitrary rate, depending upon the particular point of ∂ Ω . The boundary condition is to be understood as u(x) → ∞, v(x) → ∞ as d(x) → 0+, where d(x) stands for the distance function dist(x, ∂ Ω ). Problems like (1.1) are usually known in the literature as boundary blow-up problems, and their solutions are also named large solutions or boundary blow-up solutions. Boundary blow-up problems for single elliptic equations have received much attention during the last few years. Gómez [1] proved some optimal uniqueness results for large solutions for a canonical class of semilinear equations under minimal regularity conditions with the localization method, which was introduced in [2]. Du [3] refined these results, revealing some more interesting features. These problems with more general nonlinearities and weight functions have been discussed by many authors; see, for example, [4–14] and the references therein.

I This work is supported in part by NSF of China (10771226) and in part by Natural Science Foundation Project of CQ CSTC (2007BB0124).

∗

Corresponding author. Tel.: +86 0940 15109404502. E-mail address: [email protected] (S. Huang).

0362-546X/$ – see front matter Crown Copyright © 2009 Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.03.012

C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

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Problem (1.1) is considered a special case. In [15], problem (1.1) was analyzed with a(x) = 1, b(x) = 1 subject to Dirichlet boundary conditions of three different types. In the same paper, some existence, uniqueness and boundary behaviors of solutions were obtained under the assumption b(x) ∼ C2 d(x)κ2 ,

a(x) ∼ C1 d(x)κ1 ,

(1.3)

as d(x) → 0+ for some positive constants C1 , C2 and real numbers κ1 , κ2 > −2. This problem was later studied in [16] with the general form C3 d(x)κ2 ≤ b(x) ≤ C4 d(x)κ2 ,

C1 d(x)κ1 ≤ a(x) ≤ C2 d(x)κ1 ,

for x ∈ Ω , where a(x), b(x) ∈ C θ (Ω ) for some θ ∈ (0, 1), κ1 , κ2 > −2, and where the Ci (i = 1, 2, 3, 4) are positive constants. Recently, in [17] it was assumed that the weight functions a(x), b(x) satisfy C3 d(x)κ3 ≤ b(x) ≤ C4 d(x)κ4

C1 d(x)κ1 ≤ a(x) ≤ C2 d(x)κ2 ,

for x ∈ Ω ,

where a(x), b(x) ∈ C θ (Ω ) for some θ ∈ (0, 1), κ1 ≥ κ2 > −2, κ3 ≥ κ4 > −2, and the Ci (i = 1, 2, 3, 4) are positive constants, and it was proved that problem (1.1) admits a unique positive solution (u, v) with u = v = ∞ on ∂ Ω if and only if κi ∈ R, κ1 ≥ κ2 > −2, κ3 ≥ κ4 > −2, and q

<

s−1

2 + κ1 2 + κ4

2 + κ2

,

<

2 + κ3

p−1

.

r

At the same time, they also obtained boundary behavior of the solution (u, v). For more results for systems with boundary blow-up, we refer the reader to [18,19,1,17,9–11,13] and the references therein. Throughout of this paper, we set C1 = min C (x),

C2 = max C (x),

D1 = min D(x),

D2 = max D(x),

γ1 = max γ (x),

γ2 = min γ (x),

η1 = max η(x),

η2 = min η(x),

x∈Ω

x∈Ω

x∈Ω

x∈Ω

x∈Ω

x∈Ω

x∈Ω

x∈Ω

(2 + x)(s − 1) − (2 + y)q (2 + y)(p − 1) − (2 + x)r α(x, y) = , β(x, y) = , (p − 1)(s − 1) − qr (p − 1)(s − 1) − qr 1 1     (α(α + 1))s−1 xq (p−1)(s−1)−qr (β(β + 1))p−1 yr (p−1)(s−1)−qr E (x, y) = , F (x, y) = . (β(β + 1))q ys−1 (β(β + 1))r xp−1 nx0 stands for the outward unit normal at x0 ∈ ∂ Ω . Our main result is stated as follows. Theorem 1.1. Assume that Ω is a bounded C 2 domain of RN , with a(x), b(x) ∈ C θ (Ω ) for some θ ∈ (0, 1), a(x), b(x) > 0 in Ω , verifying (1.2), (p − 1)(s − 1) − qr > 0, p, s > 1, q, r > 0, γ (x), η(x) ∈ C (Ω , R+ ), and satisfying q s−1

<

2 + γ (x0 ) 2 + η(x0 )

<

p−1 r

for x0 ∈ ∂ Ω .

Then problem (1.1) has a unique solution (u, v) if and only if q s−1

<

2 + γ1 2 + η2

,

2 + γ2 2 + η1

<

p−1 r

.

(1.4)

= 1,

(1.5)

= 1,

(1.6)

And we have lim

x→ x0

lim

x→ x0

u(x) d(x)−α(γ (x0 ),η(x0 )) E (D(x

0

), C (x0 ))

v(x) d(x)−β(γ (x0 ),η(x0 )) F (D(x0 ), C (x0 ))

for each x0 ∈ ∂ Ω . This paper is organized as follows. In the next section, we give some propositions which will be used in the proof of Theorem 1.1. The main theorem will be proved in Section 3.

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C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

2. Preliminaries In this section, we will introduce some propositions. Definition 2.1. (u, v ) is a subsolution of



1u = a(x)up v q , 1v = b(x)ur v s ,

in Ω , in Ω ,

 1u ≥ a(x)up v q , provided 1v ≤ b(x)ur v s ,

in Ω , in Ω .

A supersolution (¯u, v¯ ) is defined by reversing the inequalities. First, by modifications of the arguments in the proof of Theorem A.2 in [15], we easily obtain the following proposition, and we omit the proof. Proposition 2.1. Assume (u, v) is a subsolution and (u, v) is a supersolution to problem (1.1), with u = v = u = v = ∞ on ∂ Ω , and u ≤ u, v ≥ v in Ω . Then problem (1.1) has at least a solution (u, v) with u ≤ u ≤ u, v ≥ v ≥ v in Ω . In particular u = v = ∞ on ∂ Ω . Proposition 2.2 (see [15]). Assuming a(x), b(x) satisfy (1.3), then problem (1.1) admits a positive solution (u, v) with u = v = ∞ on ∂ Ω if and only if κ1 , κ2 > −2 and q s−1

<

2 + κ1 2 + κ2

<

p−1 r

.

This solution is unique and satisfies lim

x→x0

u(x) d(x)−α(κ1 ,κ2 ) E (C

2

, C1 )

= 1,

lim

x→x0

v(x) d(x)−β(κ1 ,κ2 ) F (C

2

, C1 )

= 1,

(2.1)

for each x0 ∈ ∂ Ω . Next, we are ready to study two auxiliary problems in a ball and an annulus. To this end, for given 0 < R1 < R and x0 ∈ RN , N ≥ 1, set BR (x0 ) = {x ∈ RN : |x − x0 | < R},

AR1 ,R (x0 ) = {x ∈ RN : R1 < |x − x0 | < R}.

Proposition 2.3. Assume that Ω = BR (x0 ), (p − 1)(s − 1) − qr > 0, p, s > 1, q, r > 0. C (r ), D(r ) ∈ C ([0, R], R+ ) and γ , η > 0 satisfy q s−1

<

2+γ 2+η

<

p−1 r

.

(2.2)

Then the following system:

( 1Φ = C (r )(R − r )γ Φ p Ψ q , 1Ψ = D(r )(R − r )η Φ r Ψ s , Φ = Ψ = ∞,

in BR (x0 ), in BR (x0 ), on ∂ BR (x0 ),

(2.3)

possesses a unique radially symmetric positive solution (Φ (r ), Ψ (r )) satisfying

Φ (r )

lim

E (D(R), C (R))(R − r )−α(γ ,η)

lim

F (D(R), C (R))(R − r )−β(γ ,η)

r →R

r →R

Ψ (r )

= 1,

(2.4)

= 1,

(2.5)

where r = |x − x0 |. Proof. First, we consider the following systems:

 N −1 0 00 γ p q   Φ + r Φ = C (r )(R − r ) Φ Ψ , N −1 0 Ψ 00 + Ψ = D(r )(R − r )η Φ r Ψ s ,    r Φ (R) = Ψ (R) = ∞, Φ 0 (0) = Ψ 0 (0) = 0,

in (0, R), in (0, R),

(2.6)

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and we will show that problem (2.6) has a solution (Φ (r ), Ψ (r )), which provides a positive radially symmetric solution to problem (2.3). Indeed, any positive solution (Φ (r ), Ψ (r )) of the integral equation system

 Z t Z r   sN −1 C (s)(R − s)γ Φ p Ψ q dsdt , t 1−N Φ ( r ) = l + 0 0 Z t Z r   sN −1 D(s)(R − s)η Φ r Ψ s dsdt , t 1−N Ψ (r ) = m +

0 < r < R, (2.7) 0 < r < R,

0

0

provides a solution of (2.6), where Φ (0) = l, Ψ (0) = m, Φ (R) = ∞, Ψ (R) = ∞. Define Φ0 (r ) = l, Ψ0 (r ) = m, for all 0 < r < R, Let {Φk }, {Ψk } be the sequence functions given by

 Z t Z r  p q  sN −1 C (s)(R − s)γ Φk−1 Ψk−1 dsdt , t 1−N Φk (r ) = l + Z0 t Z0 r   sN −1 D(s)(R − s)η Φkr Ψks−1 dsdt , t 1 −N  Ψk ( r ) = m +

0 < r < R, 0 < r < R,

0

0

subject to Φk (0) = l, Ψk (0) = m, Φk (R) = Ψk (R) = k. We remark that {Φk }, {Ψk } are nondecreasing sequences. In fact,

Φ1 = l + lp mq

r

Z

t 1−N

t

Z

0

sN −1 C (s)(R − s)γ dsdt

0

= l + lp mq A(r ) ≥ l = Φ0 , and

Ψ1 ≥ m + lr ms

r

Z

t 1 −N

t

Z

0

sN −1 D(s)(R − s)η dsdt

0

= m + lr ms B(r ) ≥ m = Ψ0 , Rr Rt Rr Rt where A(r ) = 0 t 1−N 0 sN −1 C (s)(R − s)γ dsdt and B(r ) = 0 t 1−N 0 sN −1 D(s)(R − s)η dsdt. Proceeding in the same manner, we conclude that

l ≤ Φk ≤ Φk+1 ,

m ≤ Ψk ≤ Ψk+1 .

We now prove that {Φk }, {Ψk } are bounded in (0, R). To prove this, we consider

1Υ = (C (r )(R − r )γ + D(r )(R − r )η )(Υ p+q + Υ r +s ).

(2.8)

By Lemma 1 in [5], the problem (2.8) has a large positive radially symmetric solution Υ (r ), and r

Z

Υ (r ) = Υ (0) +

t

t

Z

1 −N

0

sN −1 (C (s)(R − s)γ + D(s)(R − s)η )(Υ p+q + Υ r +s )dsdt

0

where Υ (0) = l + m. It follows that

Φ1 = l + lp mq

r

Z

t 1−N 0

≤ Υ (0) +

sN −1 C (s)(R − s)γ dsdt

0 r

Z

t

Z

t 1−N

t

Z

0

sN −1 (C (s)(R − s)γ + D(s)(R − s)η )(Υ p+q + Υ r +s )dsdt

0

= Υ (r ). Similarly, we have Ψ1 ≤ Υ (r ). Arguing as above, we obtain Φk ≤ Υ (r ), Ψk ≤ Υ (r ). Therefore, we show that {Φk }, {Ψk } are nondecreasing and bounded sequences in (0, R), which implies that the following limit holds:

(Φ , Ψ ) := lim (Φk , Ψk ). k→∞

By standard elliptic regularity theory, we deduce that (Φ , Ψ ) is a positive solution of (2.6). Then (Φ (x), Ψ (x)) = (Φ (r ), Ψ (r )) is a positive radially symmetric solution to problem (2.3), and

Φ (R) := lim Φ (r ) = ∞, r →R

Ψ (R) := lim Ψ (r ) = ∞. r →R

Secondly, it is clear that C1 (R − r )γ ≤ C (r )(R − r )γ ≤ C2 (R − r )γ ,

D1 (R − r )η ≤ D(r )(R − r )η ≤ D2 (R − r )η .

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C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

By (2.2) and Proposition 2.2, we have

Φ (r ) ≤ E (D2 , C1 ), (R − r )−α(γ ,η) Ψ (r ) ≤ F (D1 , C2 ). F (D2 , C1 ) ≤ lim r →R (R − r )−β(γ ,η) E (D1 , C2 ) ≤ lim

r →R

Then we write l := lim

r →R

Φ (r ) , (R − r )−α(γ ,η)

k := lim

r →R

Ψ (r ) . (R − r )−β(γ ,η)

By using γ + 2 = (p − 1)α(γ , η) + qβ(γ , η), η + 2 = (s − 1)β(γ , η) + r α(γ , η), and the l’Hôpital rule, we obtain l = lim

Φ (0) +

Rr 0

t 1−N

Rt 0

sN −1 C (s)(R − s)γ Φ p Ψ q dsdt

(R − r )−α r t C (t )(R − t )γ Φ p Ψ q dt 0 = lim r →R α(R − r )−α−1 Rr C (r )(R − r )γ Φ p Ψ q + (1 − N )r −N 0 t N −1 C (t )(R − t )γ Φ p Ψ q dt = lim r →R α(α + 1)(R − r )−α−2 Rr (N − 1)r −N 0 t N −1 C (t )(R − t )γ Φ p Ψ q dt C (R) p q l k − lim . = r →R α(α + 1) α(α + 1)(R − r )−α−2 r →R

1 −N

Rr

r −N

Rr

N −1

We note that 0 ≤ lim

0

t N −1 C (t )(R − t )γ Φ p Ψ q dt

(R − r )−α−2 r C (t )(R − t )γ Φ p Ψ q dt 0 ≤ lim r →R (R − r )−α−2 C (r )(R − r )γ Φ p Ψ q = lim r →R R(α + 2)(R − r )−α−3 C (R)lp kq = lim (R − r ) = 0. R(α + 2) r →R r →R

R −1 r

This implies that l=

C (R) p q l k . α(α + 1)

(2.9)

Similarly, we obtain k=

D(R) lr ks . β(β + 1)

(2.10)

Since

α(α + 1) = E (D(R), C (R))p−1 F (D(R), C (R))q , C (R) β(β + 1) = E (D(R), C (R))r F (D(R), C (R))s−1 , D(R) by (2.9) and (2.10), we conclude that l = E (D(R), C (R)), k = F (D(R), C (R)); this is the desired result.



Proposition 2.4. Assume (p − 1)(s − 1) − qr > 0, p, s > 1, q, r > 0, γ > 0, η > 0 and q s−1

<

2+γ 2+η

<

p−1 r

,

D(r ), C (r ) ∈ C ([R1 , R], R+ ) are the reflections around R0 = following system:

( 1Φ = C (r )d(x)γ Φ p Ψ q , 1Ψ = D(r )d(x)η Φ r Ψ s , Φ = Ψ = ∞,

in AR1 ,R (x0 ), in AR1 ,R (x0 ), on ∂ AR1 ,R (x0 ),

R 1 +R 2

of some functions C (r ), D(r ) ∈ C ([R0 , R], R+ ). Then the

C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

4549

has a unique radially symmetric positive solution (Φ (r ), Ψ (r )) such that lim

d(x)→0

lim

d(x)→0

Φ (r ) E (D(R), C (R))d(x)−α(γ ,η) Ψ (r ) F (D(R), C (R))d(x)−β(γ ,η)

= 1, = 1,

where



R − |x − x0 |, | x − x0 | − R 1 ,

d(x) = d(x, ∂ AR1 ,R (x0 )) =

if R0 ≤ |x − x0 | < R, if R1 ≤ |x − x0 | < R0 ,

Proof. The proof is similar to the proof of Proposition 2.3; we omit it here.



3. Proof of Theorem 1.1 We are now ready to prove Theorem 1.1; the proof will be split into the following lemmas.

¯ ), a(x), b(x) > 0 in Ω and (1.2) holds, Lemma 3.1. Assume that (p − 1)(s − 1) − qr > 0, p, s > 1, q, r > 0.C (x), D(x) ∈ C (Ω γ (x) > 0, η(x) > 0 and satisfy q s−1

<

2 + γ (x0 ) 2 + η(x0 )

p−1

<

for each x0 ∈ ∂ Ω .

r

Then problem (1.1) has a solution (u, v) if q s−1

<

2 + γ1 2 + η2

2 + γ2

,

2 + η1

<

p−1 r

.

(3.1)

Proof. By (3.1) and Proposition 2.2, the following system:

( 1Φ = C1 d(x)γ1 Φ p Ψ q , 1Ψ = D2 d(x)η2 Φ r Ψ s , Φ = Ψ = ∞,

in Ω , in Ω , on ∂ Ω .

possesses a positive solution (u1 , v1 ). Next we will show that

(u, v) =

m+n q

(m − n) p−1

u1 ,

!

m−n r

(m + n) s−1

v1

is a supersolution of (1.1), if m is sufficiently large and 0 < m − n < 1, where m, n ∈ R+ and m > n. In fact, by

γ1 + 2 = (p − 1)α(γ1 , η2 ) + qβ(γ1 , η2 ),

η2 + 2 = (s − 1)β(γ1 , η2 ) + r α(γ1 , η2 ),

we have that (u, v) is a supersolution of (1.1) provided C1 d(x)γ1 ≤

m+n q

(m − n) p−1

a(x),

D2 d(x)η2 ≥

m−n r

(m + n) s−1

b(x).

(3.2)

¯ ), choosing m large enough, and m − n > 0 sufficiently small, it is easy to see that (3.2) holds over Since a(x), b(x) ∈ C (Ω the whole of Ω . The concludes the proof of the claim above. Similarly, if m is large enough, and m − n > 0 is sufficiently small, we can also prove that (u, v) =

m−n q

(m + n) p−1

u2 ,

!

m+n r

(m − n) s−1

v2

is a subsolution to (1.1), where (u2 , v2 ) is a solution of the following problem:

( 1Φ = C2 d(x)γ2 Φ p Ψ q , 1Ψ = D1 d(x)η1 Φ r Ψ s , Φ = Ψ = ∞,

in Ω , in Ω , on ∂ Ω .

Then by Proposition 2.1, problem (1.1) has a solution. Lemma 3.2. Assume that problem (1.1) has a solution (u, v); then (1.4) holds.

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C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

Proof. In fact, if (1.4) does not hold, this is a contradiction. From Lemma 3.1, we find that, if m is large enough and m − n > 0 is sufficiently small, u≤u=

m+n q

(m − n) p−1

u1 ,

v≤v=

m+n r

(m − n) s−1

v2 .

On the other hand, by (2.1), there exists a ε such that for x ∈ Ωε = {x ∈ Ω : d(x, ∂ Ω ) ≤ ε}, we get m+n

u≤

(m − n)

u1 ≤

q p−1

m+n

( m − n)

E (D2 , C1 )d(x)−α(γ1 ,η2 ) .

q p−1

(3.3)

Thus, if q s−1

≥

2 + γ1 2 + η2

,

by the definition of α(γ1 , η2 ), we obtain α(γ1 , η2 ) ≤ 0. This implies from (3.3) that u is bounded for x ∈ Ωε , which is impossible since u(x) = ∞ as d(x) = dist(x, ∂ Ω ) → 0+. If 2 + γ2 2 + η1

≥

p−1 r

,

it is similarly proved that v is bounded near ∂ Ω , which is also a contradiction. The proof of Lemma 3.2 is complete.



Lemma 3.3. Let (u, v) be a positive solution to (1.1); then (1.5) and (1.6) hold. Proof. Fix τ ∈ (0, 1); by (1.2), there exists δ ∈ (0, 1) such that, if d(x, x0 ) < δ , a(x) ≥ (1 − τ )C (x0 )d(x)γ (x0 ) ,

b(x) ≤ (1 + τ )D(x0 )d(x)η(x0 ) ,

where x0 ∈ ∂ Ω . For a fixed x0 ∈ ∂ Ω , set

Σ := B δ (x0 ) ∩ ∂ Ω , 2

and choose R > 0 small enough that K :=

[

BR (y − Rny ) ⊂ Bδ (x0 ) ∩ Ω ,

y∈Σ

where ny stands for the outward unit normal at y ∈ ∂ Ω . For x ∈ Bδ (x0 ) ∩ Ω , we get a(x) ≥ (1 − τ )C (x0 )d(x)γ (x0 ) ,

b(x) ≤ (1 + τ )D(x0 )d(x)η(x0 ) .

Since Ω is C 2 bounded, there exist R > 0 and δ0 > 0 such that BR (x0 − (R + δ)nx0 ) ⊂ Ω and BR (x0 − Rnx0 ) ∩ ∂ Ω = {x0 } for each δ ∈ (0, δ0 ). Let (uB,δ , vB,δ ) be any positive radially symmetric solution of the following problem:

 1u = (1 − τ )C (x0 )(R − |x − x0 |)γ (x0 ) up v q , 1v = (1 + τ )D(x0 )(R − |x − x0 |)η(x0 ) ur v s ,  u = v = ∞,

in BR (x0 − (R + δ)nx0 ), in BR (x0 − (R + δ)nx0 ), on ∂ BR (x0 − (R + δ)nx0 ).

(3.4)

It is easy to see that (uδ , v δ ) := (u, v)|BR (x0 −(R+δ)nx0 ) is a positive smooth subsolution of (3.4), where (u, v) is a positive solution of (1.1). Then we get uδ := u|BR (x0 −(R+δ)nx0 ) ≤ uB,δ ,

v δ := v|BR (x0 −(R+δ)nx0 ) ≥ vB,δ .

(3.5)

Let(uB , vB ) be any positive solution of the following problem:

 1u = (1 − τ )C (x0 )(R − |x − x0 |)γ (x0 ) up v q , 1v = (1 + τ )D(x0 )(R − |x − x0 |)η(x0 ) ur v s ,  u = v = ∞,

in BR (x0 − Rnx0 ), in BR (x0 − Rnx0 ), on ∂ BR (x0 − Rnx0 ),

By Proposition 2.2, (uB , vB ) satisfies lim

r →R

uB E ((1 + τ )D(x0 ), (1 − τ )C (x0 ))(R − r )−α(γ (x0 ),η(x0 ))

= 1,

vB = 1, F ((1 + τ )D(x0 ), (1 − τ )C (x0 ))(R − r )−β(γ (x0 ),η(x0 )) where r = |x − x0 |. lim

r →R

(3.6) (3.7)

C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

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Taking into account that, for x ∈ BR (x0 − (R + δ)nx0 ),

vB,δ (x) := vB (x + δ nx0 ).

uB,δ (x) := uB (x + δ nx0 ),

By (3.5), for each x ∈ BR (x0 − (R + δ)nx0 ) and δ ∈ (0, δ0 ), we have u(x) ≤ uB (x + δ nx0 ),

v(x) ≥ vB (x + δ nx0 ).

Letting δ → 0, u(x) ≤ uB (x),

in B(x + Rnx0 ).

v(x) ≥ vB (x),

It follows immediately from (3.6) and (3.7) that u

uB

lim

E (R − r )−α

≤ lim

E (R − r )−α

lim

F (R − r )−β

≥ lim

F (R − r )−β

r →R

v

r →R

r →R

r →R

vB

= 1,

(3.8)

= 1,

(3.9)

where E = E ((1 + τ )D(R), (1 − τ )C (R)), F = F ((1 + τ )D(R), (1 − τ )C (R)). T Next we want to prove the inverse inequalities. Similarly, there exist R > R1 > 0 and δ0 > 0 such that Ω ⊂ 0<δ<δo AR1 ,R (x0 + (R + δ)nx0 ) and AR0 ,R (x0 + R1 nx0 ) ∩ ∂ Ω = {x0 }. Fix a sufficiently small τ ; there exist radially symmetric functions a : AR1 ,R (x0 + R1 nx0 ) → R+ and b : AR1 ,R (x0 + R1 nx0 ) → R+ such that a ≥ a, b ≤ b in Ω , and max

AR ,R (x0 +R1 nx0 ) 1

a ≤ max a + 1, Ω

max b + 1 ≤ Ω

max

AR ,R (x0 +R1 nx0 ) 1

b,

and for each x ∈ AR1 ,R (x0 + R1 nx0 ), a(x) = a1 (|x − x0 − R1 nx0 |)[d(x, ∂ AR1 ,R (x0 + R1 nx0 ))]γ (x0 ) , b(x) = b1 (|x − x0 − R1 nx0 |)[d(x, ∂ AR1 ,R (x0 + R1 nx0 ))]η(x0 ) . where a1 , b1 ∈ C ([R1 , R], R+ ), satisfying a1 (R1 ) = C (x0 ) + τ ,

b1 (R1 ) = D(x0 ) − τ .

We now consider the system

( 1u = a(x)up v q , 1u = b(x)up v q , u = v = ∞,

in AR1 ,R (x0 + R1 nx0 ), in AR1 ,R (x0 + R1 nx0 ), on ∂ AR1 ,R (x0 + R1 nx0 ).

(3.10)

By Proposition 2.4, problem (3.10) possesses a solution (uA , vA ). But for the system

( 1u = a(x)up v q , 1u = b(x)up v q , u = v = ∞,

in AR1 ,R (x0 + (R1 + δ)nx0 ), in AR1 ,R (x0 + (R1 + δ)nx0 ), on ∂ AR1 ,R (x0 + (R1 + δ)nx0 ),

it has a solution (uA,δ , vA,δ ), and for each x ∈ AR1 ,R (x0 + (R1 + δ)nx0 ), we have




(uA,δ (x), vA,δ (x)) = uA x − δ nx0 , vA x − δ nx0 . It is also clear that (uA , v A ) := (uA,δ , vA,δ )|Ω is a subsolution of problem (1.1). Thus for each x ∈ AR1 ,R (x0 + (R1 + δ)nx0 ), we get uA (x − δ nx0 ) ≤ u(x), vA (x − δ nx0 ) ≥ v(x). Letting δ → 0, we have uA (x) ≤ u(x), v A ≥ v(x). Thus for x ∈ K , we get 1 = lim

|x|→R

≤ 1 =

≥

lim

uA ( x ) E (a(x), b(x))(R − |x|)−α(γ (x0 ),η(x0 )) u(x)

d(x)→0

lim

|x|→R,

lim

d(x)→0

E (a(x), b(x))(R − |x|)−α(γ (x0 ),η(x0 ))

,

(3.11)

.

(3.12)

v A (x) F (a(x), b(x))(R − |x|)−β(γ (x0 ),η(x0 )) v(x) F (a(x), b(x))(R − |x|)−β(γ (x0 ),η(x0 ))

But we have limτ →0 K = {x0 }. Therefore, by (3.8), (3.9), (3.11) and (3.12), we get (1.5) and (1.6). The proof of Lemma 3.3 is complete. 

4552

C. Mu et al. / Nonlinear Analysis 71 (2009) 4544–4552

Lemma 3.4. The solution to (1.1) is unique. Proof. Let (u1 , v1 ) and (u2 , v2 ) be positive solutions to problem (1.1); then for each ε > 0, there exists δ0 > 0 such that, for δ ∈ (0, δ0 ), x ∈ Ωδ , have 1−ε q

(1 + ε) p−1

u2 ≤ u1 ≤

1+ε q

(1 − ε) p−1

u2 ,

1−ε r

(1 + ε) s−1

v2 ≤ v1 ≤

1+ε r

(1 − ε) s−1

v2 .

(3.13)

Considering the following system:

( 1u = a(x)up v q , 1u = a(x)up v q , u = u1 , v = v 1 ,

in Ωδ , in Ωδ , on ∂ Ωδ ,

(3.14)

then 1−ε q

(1 + ε) p−1

u2 ,

!

1+ε r

(1 − ε) s−1

v2

is a subsolution to (3.14), and 1+ε q

(1 − ε) p−1

u2 ,

!

1−ε r

(1 + ε) s−1

v2

is a supersolution to (3.14). It follows that (3.14) has a solution (uδ , vδ ), and for x ∈ Ωδ we have 1−ε q

(1 + ε) p−1

u2 ≤ uδ ≤

1+ε q

(1 − ε) p−1

u2 ,

1−ε r

(1 + ε) s−1

v2 ≤ vδ ≤

1+ε r

v2 .

r

v2 .

(1 − ε) s−1

(3.15)

By the uniqueness of (3.14), we have uδ = u1 |Ωδ ,

vδ = v1 |Ωδ .

Combining (3.13) and (3.15), for each x ∈ Ω , we have 1−ε q

(1 + ε) p−1

u2 ≤ uδ ≤

1+ε q

(1 − ε) p−1

u2 ,

1−ε

(1 + ε)

r s−1

v2 ≤ vδ ≤

1+ε

(1 − ε) s−1

Letting ε → 0 gives u1 = u2 , v1 = v2 , as we wanted to prove. The proof of Lemma 3.4 is complete.



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