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Large solutions for a cooperative elliptic system of p -Laplacian equations
Nonlinear Analysis 73 (2010) 450–457
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Large solutions for a cooperative elliptic system of p-Laplacian equationsI Yujuan Chen ∗ , Yueping Zhu Department of Mathematics, Nantong University, Nantong 226007, PR China
article
abstract
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This paper deals with the quasilinear elliptic systems ∆p u = ua v b , ∆p v = uc v e in a smooth bounded domain Ω ⊂ RN , with the boundary conditions u = v = +∞ on ∂ Ω . The operator ∆p stands for the p-Laplacian operator defined by ∆p u = div(|∇ u|p−2 ∇ u), p > 1, and the exponents verify a, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. We prove the existence and uniqueness of the positive solution, and obtain the exact blow-up rate near the boundary of the solution. © 2010 Elsevier Ltd. All rights reserved.
Article history: Received 31 May 2009 Accepted 24 March 2010 MSC: 35J55 35B40 Keywords: Large solutions Boundary blow-up Cooperative Quasilinear elliptic system
1. Introduction In this paper we study the positive boundary blow-up solutions to a quasilinear elliptic system of cooperative type:
∆p u = ua v b , ∆ v = uc v e , u p= v = +∞,
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω,
(1.1)
where Ω is a bounded domain in RN with smooth boundary ∂ Ω , and ∆p stands for the p-Laplacian operator defined by ∆p u = div(|∇ u|p−2 ∇ u), p > 1, a, e > p − 1 and b, c < 0. The boundary condition is assumed in the sense u(x), v(x) → +∞ when d(x) → 0, where d(x) stands for the distance function dist(x, ∂ Ω ). We will focus our attention on positive weak 1 ,p 2 solutions, that is, pairs of positive functions (u, v) ∈ [Wloc (Ω ) ∩ L∞ loc (Ω )] verifying (1.1) in the weak sense. However, 1,η
according to standard regularity for the p-Laplacian, we know that positive weak solutions verify u, v ∈ Cloc (Ω ) for some η ∈ (0, 1) (cf. [1–4]). There is a vast amount of literature on elliptic problems related to (1.1), see for example, [5–7,1,8–13,4,14,15] for semilinear problems, and [16–19] for problems with the p-Laplacian. Moreover, we would like to quote some references such as [20–25,2,26–29], in which systems of boundary blow-up solutions related to (1.1) were analyzed. Our motivation mainly comes from [22], where problem (1.1) was analyzed for the competitive case, i.e. a, e > p − 1, and b, c > 0. Under the condition (a − p + 1)(e − p + 1) > bc, García-Melián proved that problem (1.1) has positive solutions
I This work was supported by the foundation of NSFC 10971228, the foundation of Jiangsu Education Commission 07KJD110166, and the doctoral project in Nantong University 08B02. ∗ Corresponding author. Tel.: +86 0513 85015889. E-mail address: [email protected] (Y. Chen).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.03.036
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if and only if c < a − p + 1, b < e − p + 1, and the positive solution is unique when it exists. Moreover, the exact blow-up rate near the boundary of the positive solution was obtained. In the critical case (a − p + 1)(e − p + 1) = bc, infinitely many positive solutions were constructed. We want to obtain similar results as those in [22] for the problem (1.1) under some conditions on the parameters. However, we find that we can only get the existence, uniqueness and boundary behavior for the positive solution in the subcritical case. We remark that though most of the proofs in our paper are an adaptation of the corresponding ones in [22], there are some important differences. For instance, concerning the proof of existence of solutions, the construction of sub- and supersolutions involves more complicated technology. Furthermore, in order to get rough estimates for positive solutions (see Lemma 4 in Section 4), the iterative method in [22] cannot be used here, we shall use a different method. We now state our main results. Theorem 1. Assume that the parameters satisfy a > p − 1,
e > p − 1,
b < 0,
c < 0,
(a − p + 1)(e − p + 1) > bc .
(1.2)
Then the problem (1.1) admits a unique positive solution (u, v). Moreover, (u, v) verifies u(x) ∼ Ad−α (x),
v(x) ∼ Bd−β (x)
(1.3)
as d(x) → 0, where
α=
p(e − p + 1 − b) , (a − p + 1)(e − p + 1) − bc
β=
p(a − p + 1 − c ) , (a − p + 1)(e − p + 1) − bc
(1.4)
and
e−p+1 ! (a−p+1)(e1−p+1)−bc (p − 1)α p−1 (α + 1) , b (p − 1)β p−1 (β + 1) a−p+1 ! (a−p+1)(e1−p+1)−bc (p − 1)β p−1 (β + 1) . c (p − 1)α p−1 (α + 1)
A = B =
(1.5)
The paper is organized as follows. In Section 2 we collect some results on a single equation and the method of sub- and supersolutions which are instrumental in our proofs. In Section 3 the two uniqueness results for problems related to (1.1) will be proved. In Section 4 we prove Theorem 1. 2. Preliminaries The main results of this paper are established by applying the method of sub- and supersolutions and by making use of some results on a single equation with singular boundary conditions. 2.1. Some results on a single equation with singular boundary condition To understand our problem more clearly, we collect some properties of positive solutions to the following problem
∆p u = d−γ (x)uq , u = +∞,
x ∈ Ω, x ∈ ∂Ω.
(2.1)
This problem has been recently considered in [17], and the results concerning existence, uniqueness and asymptotic behavior near the boundary of positive solutions were used in [22]. We cite some results from the papers [22,17]. Lemma 1. Let q > p − 1 and γ ∈ [0, p). Then the problem (2.1) admits a unique positive solution denoted by Uq,γ . Moreover, Uq,γ ∼ (p − 1)%p−1 (% + 1)
q−p1+1
d−% (x)
as d(x) → 0, where % = q−p+1 . p−γ
1,η
Lemma 2. Let u ∈ Cloc (Ω ) for some η ∈ (0, 1). If u verifies
∆p u ≥ Cd−γ (x)uq in Ω , Then u ≤ C
− q−p1+1
∆p u ≤ Cd
Uq,γ in Ω . Similarly, if u verifies
−γ
Then u ≥ C
− q−p1+1
u = +∞ on ∂ Ω .
(x)uq in Ω ,
Uq,γ in Ω .
u = +∞ on ∂ Ω .
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2.2. Sub- and supersolutions This subsection collects some results about the method of sub- and supersolutions for the system:
∆p u = ua v b , ∆p v = uc v e ,
x ∈ Ω, x ∈ Ω,
(2.2)
where a, e > p − 1 and b, c < 0. 1 ,p
2 Definition 1. A pair of positive functions (u, v) ∈ [Wloc (Ω ) ∩ L∞ loc (Ω )] is called a subsolution to the system (2.2), provided a b c e that in the weak sense (u, v) satisfies ∆p u ≥ u v and ∆p v ≥ u v in Ω . As always, a supersolution (u, v) is defined by revising these inequalities.
Proposition 1. Assume that (u, v) is a subsolution and (u, v) is a supersolution to the problem (2.2) with u = u = v = v = +∞ on ∂ Ω , and u ≤ u, v ≤ v in Ω . Then the problem (2.2) admits at least one weak solution (u, v) with u ≤ u ≤ u, v ≤ v ≤ v in Ω and u = v = +∞ on ∂ Ω . The proof of this proposition is similar to Theorem A.2 of [22]. We omit the details here. 3. Two uniqueness results Our first uniqueness result is concerned with problem (1.1) with finite boundary conditions. To this end, Lemma 8 in [22] will be used. For the readers’ convenience, we state it below. Lemma 3. Let f , g ∈ C (Ω ) and u, v ∈ C 1,η (Ω ) be weak solutions to ∆p u = f , ∆p v = g in Ω with u ≤ v and u = v at some point of Ω . Assume moreover that u < v on ∂ Ω . Then there exists x0 ∈ Ω such that u(x0 ) = v(x0 ) and f (x0 ) ≤ g (x0 ). Proposition 2. Assume that the condition (1.2) holds. Let (u1 , v1 ) and (u2 , v2 ) be positive solutions to the problem
∆p u = ua v b , x ∈ Ω , ∆ v = uc v e , x ∈ Ω , u p= f (x), v = g (x),
(3.1)
x ∈ ∂Ω,
where f > 0, g > 0 on ∂ Ω . Then u1 = u2 , v1 = v2 . Proof. The scale argument in this proof has been used by many author, see, for example, [22,30] and the references therein. ¯ and c < 0, we can select a large k such that Since u2 , v2 are positive on Ω u1 ≤ ku2 ,
c
¯. v1 ≤ k− e−p+1 v2 on Ω
(3.2)
Choose the least k with this property, and assume k > 1. Then one of the two inequalities in (3.2) is not strict. Assume it is − e−pc+1
the second one. We can apply Lemma 3 to obtain a point x0 ∈ Ω , such that v1 (x0 ) = k (p−1) − ce− p+1
uc1 (x0 )v1e (x0 ) ≤ k
v2 (x0 ) and
uc2 (x0 )v2e (x0 ).
It follows that uc1 (x0 ) ≤ kc uc2 (x0 ). By c < 0, we have u1 (x0 ) ≥ ku2 (x0 ), which shows that the first inequality in (3.2) is not strict. Thus we may apply Lemma 3 again to obtain a point x0 ∈ Ω , such that u1 (x0 ) = ku2 (x0 ) and ua1 (x0 )v1b (x0 ) ≤ kp−1 ua2 (x0 )v2b (x0 ). It follows that ka−p+1 v1b (x0 ) ≤ v2b (x0 ),
and k
a−p+1 b
v1 (x0 ) ≥ v2 (x0 ).
In view of the second inequality of (3.2) we have k
a−p+1 b
c
v1 (x0 ) ≥ v2 (x0 ) ≥ k e−p+1 v1 (x0 ),
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which implies that k
c
a−p+1 b
≥ k e−p+1 ,
i.e. k
bc −(a−p+1)(e−p+1) b(e−p+1)
≤ 1.
Note that bc − (a − p + 1)(e − p + 2) < 0, b(e − p + 1) < 0 and k > 1, the above inequality is impossible. So, we have k ≤ 1. That is u1 ≤ u2 and v1 ≤ v2 . Similarly, we can prove u2 ≤ u1 and v2 ≤ v1 . We conclude the proof. Our second uniqueness result is concerned with problem (1.1) in a half-space D := {x ∈ RN : x1 > 0}, where for a point x in RN we write x = (x1 , x0 ), with x0 ∈ RN −1 , i.e. we prove the unique existence result for the following problem:
∆ p u = ua v b , ∆ v = uc v e , u p= v = +∞,
x ∈ D, x ∈ D, x ∈ ∂ D.
(3.3)
Theorem 2. Assume that the condition (1.2) holds. Let α and β be given by (1.4) and (u, v) be a positive weak solution to (3.3) verifying −β
C1 x−α ≤ u ≤ C2 x−α 1 1 ,
−β
≤ v ≤ C2 x1
C 1 x1
in D
(3.4)
for some positive constants C1 , C2 . Then u(x) = Ax−α 1 ,
−β
v(x) = Bx1 ,
where A and B are given by (1.5). −β
Proof. We are going to show that u ≤ Ax−α 1 , v ≤ Bx1 , since the complementary inequality can be proved in a similar way. Thanks to (3.4) we can choose K 1 such that u(x) ≤ Kx−α 1 ,
c
−β
v(x) ≤ K − e−p+1 x1 .
Take K = max{λ, µ}, where
− e−pc+1 β . µ = sup(x1 v(x))
λ = sup(xα1 u(x)),
D
D
Assume that λ ≥ µ, the other case being treated similarly. We may also assume K ≥ A, since otherwise there is nothing to prove. According to the definition of K , there exists a sequence {xn } ⊂ D such that xαn,1 u(xn ) → K , where to simplify the notation xn,1 stands for the first component of xn . Let ξn be the projection of xn onto ∂ D and introduce the functions β
Un (y) = xαn,1 u(ξn + xn,1 y),
Vn (y) = xn,1 v(ξn + xn,1 y).
It is not hard to see that Un , Vn verify the equations
∆p Un = Una Vnb , ∆p Vn = Unc Vne ,
y ∈ D, y ∈ D,
and the inequalities C1 y−α ≤ Un ≤ Ky−α 1 1 ,
−β
C1 y1
c
−β
≤ Vn ≤ K − e−p+1 y1
in D.
(3.5)
Notice that (3.5) gives in particular uniform local bounds for Un , Vn , so that we can use the standard C 1 to obtain that (up to a subsequence) Un → U, Vn → V in Cloc (D), where U , V is a solution to
∆p U = U a V b , ∆p V = U c V e ,
1,η
interior estimates
y ∈ D, y ∈ D,
(3.6)
which in addition verifies U ≤ Ky−α 1 ,V ≤ K
− e−pc+1 −β
y1 , U (e1 ) = K , where e1 is the first vector in the canonical basis of RN . −
c
−β
e−p+1 y On the other hand, it is easily seen that if K ≥ A, then (u, v) given by u = Ky−α 1 ,v = K 1 , is a supersolution to (3.6). We choose M > 0 such that the function h(ζ ) = ζ a v(x)b − M ζ is decreasing in a neighborhood of K for fixed x in a neighborhood of e1 . Then
∆p U − MU = U a V b − MU ≥ U a v b − MU ≥ ua v b − Mu ≥ ∆p u − Mu in a neighborhood of e1 . Since U ≤ u and U (e1 ) = K = u(e1 ), the strong comparison principle [31, Theorem 1.4] gives U ≡ u in a neighborhood of e1 . Notice that it is important here that |∇ u(e1 )| = K α > 0. We also obtain then that V ≡ v in the same neighborhood, and this gives that (u, v) is in fact a solution to (3.6) in a neighborhood of e1 . Thus K = A, K
− e−p1+1
−β
= B, which finally implies u ≤ Ax−α 1 , v ≤ Bx1 . The proof is finished.
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4. Proof of Theorem 1 Proof of the existence. The main tool of this proof is the method of sub- and supersolutions. We look for a subsolution of the form (u(x), v(x)) = (ε Uµ,γ (x), ε δ Uν,σ (x)), where µ, ν, γ , σ and δ are to be chosen and ε is small enough. It is not hard to see that (u(x), v(x)) will be a subsolution provided that
( p−1 µ a b ε d(x)−γ Uµ,γ ≥ εa+δb Uµ,γ Uν,σ ,
(4.1)
ν c e ε δ(p−1) d(x)−σ Uν,σ ≥ ε c +δe Uµ,γ Uν,σ ,
in Ω . We may choose δ in such a way that p − 1 < a + δ b,
δ(p − 1) < c + δ e,
that is
−c a−p+1 <δ< e−p+1 −b since a > p − 1, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. Thus (4.1) will hold for small enough ε provided that µ−a −b −c ν−e d−γ Uµ,γ Uν,σ and d−σ Uµ,γ Uν,σ are bounded from above and from below in Ω . According to the boundary behavior of Uµ,γ and Uν,σ given by Lemma 1, it is enough to have
µ, ν > p − 1, γ , σ ∈ [0, p), p−σ p−γ (µ − a) − b = 0, γ+ µ−p+1 ν−p+1 p−σ p−γ σ+ (ν − e) − c=0 ν−p+1 µ−p+1
(4.2)
which is also possible thanks to the condition (1.2), i.e. a > p − 1, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. In fact, by the condition (1.2), we have p−1<
a(e − p + 1) + b(1 − c − p) e−p+1−b
,
p−1<
e(a − p + 1) + c (1 − b − p) a−p+1−c
.
If we choose µ and ν such that p−1<µ<
a(e − p + 1) + b(1 − c − p) e−p+1−b
,
p−1<ν <
e(a − p + 1) + c (1 − b − p) a−p+1−c
,
and set
γ =p−
p(µ − p + 1)(−b + e − p + 1)
(a − p + 1)(e − p + 1) − bc
,
σ =p−
p(ν − p + 1)(−c + a − p + 1)
(a − p + 1)(e − p + 1) − bc
,
it is not difficult to verify that µ, ν, γ and σ satisfy (4.2). Thus we have a subsolution (u(x), v(x)) = (ε Uµ,γ (x), ε δ Uν,σ (x)). In a similar way, we can show that (¯u(x), v¯ (x)) = (MUµ,γ (x), M δ Uν,σ (x)) is a supersolution for the same choice of µ, ν, γ , σ and δ , provided that M is large enough. Since the sub- and supersolutions are ordered, it follows from Proposition 1 that the problem (1.1) has at least one positive solution (u, v) and satisfies
(u(x), v(x)) ≤ (u, v) ≤ (¯u(x), u¯ (x)), The existence is proved.
x ∈ Ω.
Note that Uµ,γ (x) ∼ d−α (x),
Uν,σ (x) ∼ d−β (x),
we see that such a solution obtained in the above satisfies cˆ1 d−α (x) ≤ u(x) ≤ cˆ2 d−α (x),
cˆ1 d−β (x) ≤ v(x) ≤ cˆ2 (x)d−β (x),
x∈Ω
for some positive constants cˆ1 and cˆ2 which are independent of the domain Ω , where α and β are given by (1.4). To obtain the boundary behavior of solutions we use a blow-up argument. For this sake we need some rough estimates of all possible positive solutions. Lemma 4. Let the condition of Theorem 1 hold and (u, v) be a positive solution to (1.1). Then there exist positive constants C1 , C2 such that C1 d−α (x) ≤ u(x) ≤ C2 d−α (x), where α and β are given by (1.4).
C1 d−β (x) ≤ v(x) ≤ C2 d−β (x),
(4.3)
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Before giving the proof of Lemma 4, we first give a comparison principle which proof is standard and the details is omitted. Lemma 5. Assume that the condition (1.2) holds. Let (u, v) and (w, z ) be positive solutions of the following problems
∆ p u = ua v b , ∆ v = uc v e , u p= v = ∞,
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω
and
∆p w = w a z b , x ∈ Ω , ∆ z = wc z e , x ∈ Ω , wp< ∞, z < ∞, x ∈ ∂ Ω , respectively. Then (u, v) ≥ (w, z ) in Ω . Proof of Lemma 4. We first prove the upper bound. Let εn & 0 and Ωn = {x ∈ Ω : dist(x, ∂ Ω ) > εn }. From the above proof for the existence result we see that the problem
∆p un = uan vnb , c e ∆ p v n = un v n , un = vn = +∞,
x ∈ Ωn , x ∈ Ωn , x ∈ ∂ Ωn
has a positive solution (un , vn ) satisfying −α c1 d−α n (x) ≤ un (x) ≤ c2 dn (x),
−β c1 d−β n (x) ≤ vn (x) ≤ c2 dn (x),
(4.4)
where c1 and c2 are independent of n, and dn (x) = dist(x, ∂ Ωn ). By Lemma 5, u(x) ≤ un (x),
v(x) ≤ vn (x),
x ∈ Ωn .
(4.5)
Note that (4.4) gives in particular the uniform local bounds for un , vn when n is large enough. One can use the standard C 1,η interior estimate [1,4] and a diagonal process to find a subsequence of {(un , vn )}, denoted by itself, such that (un , vn ) → 1 (u∗ , v ∗ ) in Cloc (Ω ) and (u∗ , v ∗ ) satisfies the differential equations of the problem (1.1) in the weak sense. It follows from (4.5) that u(x) ≤ u∗ (x),
v(x) ≤ v ∗ (x),
∀x ∈ Ω ,
(4.6)
which implies that (u , v ) satisfies the boundary condition of (1.1). For any fixed x ∈ Ω , there is a N1 1 such that d(x) < 2dn (x) for all n > N1 . Hence ∗
∗
−α un (x) ≤ c2 d−α , n (x) < c2 (d(x)/2)
−β vn (x) ≤ c2 d−β . n (x) < c2 (d(x)/2)
Letting n → ∞ it yields u∗ (x) ≤ 2α c2 d−α (x),
v ∗ (x) ≤ 2β c2 d−β (x).
This combines with (4.6) gives the upper bound in (4.3). ∗ Now we prove the lower bound. Similar to the above, let εm & 0 and Ωm = {x ∈ RN : dist(x, Ω ) < εm }. Then Ω ⊂ Ωm∗ . From the above proof for the existence result we see that the problem
∆p um = uam vmb , c e ∆ p v m = um v m , um = vm = +∞,
x ∈ Ωm , x ∈ Ωm , x ∈ ∂ Ωm
has a positive solution (um , vm ) satisfying
ˆ2 d−α cˆ1 d−α m (x) ≤ um (x) ≤ c m (x),
ˆ2 d−β cˆ1 d−β m (x) ≤ vm (x) ≤ c m (x),
x ∈ Ωm ,
(4.7)
where cˆ1 and cˆ2 are independent of m, and dm (x) = dist(x, ∂ Ωm ). Thanks to Lemma 5 we have u(x) ≥ um (x),
v(x) ≥ vm (x) for all x ∈ Ω .
The estimate (4.7) gives in particular the uniform local bounds of um and vm in Ω . We can use the standard C 1,η interior estimate [1,32] and a diagonal process to find a subsequence of {(um , vm )}, denoted by itself, such that (um , vm ) → (u∗ , v∗ ) 1 in Cloc (Ω ) and (u∗ , v∗ ) satisfies the differential equations of the problem (1.1) in weak sense. It is obvious that u(x) ≥ u∗ (x),
v(x) ≥ v∗ (x),
∀x ∈ Ω .
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For any fixed x ∈ Ω , there is a N2 1 such that d(x) ≥
ˆ1 (2d(x))−α , um (x) ≥ cˆ1 d−α m (x) ≥ c
1 d 2 m
(x) for all m > N2 . Hence
ˆ1 (2d(x))−β . vm (x) ≥ cˆ1 d−β m (x) ≥ c
Letting m → ∞ it yields
v(x) ≥ v∗ (x) ≥ 2−β cˆ1 d−β (x).
u(x) ≥ u∗ (x) ≥ 2−α cˆ1 d−α (x), The proof is completed.
Proof of the boundary behavior (1.3). Let (u, v) be an arbitrary positive solution to (1.1). Fix a point x0 ∈ ∂ Ω . Without loss of generality we may assume x0 = 0 and ν(x0 ) = −e1 , the first vector in the canonical basis of RN . Take an arbitrary sequence {xn } ⊂ Ω with xn → 0, and denote dn = d(xn ) and let ξn be the projection of xn onto ∂ Ω . Introducing the functions:
wn (y) = dαn u(ξn + dn y),
zn (y) = dβn v(ξn + dn y),
where y ∈ Ωn := {y ∈ RN : ξn + dn y ∈ Ω }. Observe that Ωn → D as n → ∞, while d(ξn + dn y) ∼ dn y1 , where y = (y1 , y0 ). It is not hard to see that (wn , zn ) satisfies the system
∆p w = w a z b , ∆p z = w c z e ,
y ∈ Ωn , y ∈ Ωn ,
and thanks to Lemma 4, also the inequalities C1 dαn d−α (ξn + dn y) ≤ wn (y) ≤ C2 dαn d−α (ξn + dn y), C1 dβn d−α (ξn + dn y) ≤ zn (y) ≤ C2 dβn d−α (ξn + dn y). It follows that (wn , zn ) is locally uniformly bounded. Applying the C 1,η interior estimate, we obtain that, passing to a 1 subsequence if necessary, (wn , zn ) converges in Cloc (D) to a pair (w, z ) which is a weak solution to
∆p w = w a z b , ∆p z = w c z e ,
y ∈ D, y ∈ D,
and verifies C1 y−α ≤ w ≤ C2 y−α 1 1 ,
−β
C1 y1
−β
≤ z ≤ C2 y1
in D. β
α According to Theorem 2, w = Ay−α 1 , z = By1 , and setting y = e1 , we obtain that dn u(xn ) → A, dn v(xn ) → B. Since the sequence {xn } is arbitrary, the estimate (1.3) is proved. −β
Proof of the uniqueness. Let (u1 , v1 ) and (u2 , v2 ) be two positive solutions of the problem (1.1). According to the limit (1.3), we have lim
d(x)→0
u1 (x) u2 (x)
= lim
d(x)→0
v1 (x) = 1. v2 (x)
Thus, for every 0 < ε < 1, there exists δ > 0 such that when d(x) ≤ δ , there holds:
(1 − ε)u2 ≤ u1 ≤ (1 + ε)u2 ,
c
c
(1 − ε)− e−p+1 v2 ≤ v1 ≤ (1 + ε)− e−p+1 v2 .
(4.8)
Now we set Ωδ = {x ∈ Ω : d(x) > δ}, and consider the problem
∆p w = w a z b , x ∈ Ωδ , (4.9) ∆ z = w c z e , x ∈ Ωδ , wp= u , z = v 1 , x ∈ ∂ Ωδ . 1 − c − c It is not difficult to see that the pairs (1 − ε)u2 , (1 − ε) e−p+1 v2 and ((1 + ε)u2 , (1 + ε) e−p+1 v2 ) are the ordered suband supersolutions to the problem (4.9). So, the problem (4.9) has at least one positive solution (u, v) and satisfies (1 − ε)u2 ≤ u ≤ (1 + ε)u2 ,
c
c
(1 − ε)− e−p+1 v2 ≤ v ≤ (1 + ε)− e−p+1 v2 in Ωδ .
On the other hand, thanks to Proposition 2, the problem (4.9) has a unique positive solution, which is precisely (u1 , v1 ). Thus, (4.8) is valid in Ωδ . Therefore, (4.8) holds for all x ∈ Ω . Letting ε go to zero in (4.8) we arrive at u1 = u2 , v1 = v2 , which proves the uniqueness.
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Acknowledgements The authors are grateful to Prof. Yihong Du and Prof. Mingxin Wang for pointing out this issue and for helpful suggestions on the manuscript. They would also like to thank the referees for their careful reading and for suggesting valuable references [20,21,28,12,29]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32]
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Large solutions for a cooperative elliptic system of p-Laplacian equationsI Yujuan Chen ∗ , Yueping Zhu Department of Mathematics, Nantong University, Nantong 226007, PR China
article
abstract
info
This paper deals with the quasilinear elliptic systems ∆p u = ua v b , ∆p v = uc v e in a smooth bounded domain Ω ⊂ RN , with the boundary conditions u = v = +∞ on ∂ Ω . The operator ∆p stands for the p-Laplacian operator defined by ∆p u = div(|∇ u|p−2 ∇ u), p > 1, and the exponents verify a, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. We prove the existence and uniqueness of the positive solution, and obtain the exact blow-up rate near the boundary of the solution. © 2010 Elsevier Ltd. All rights reserved.
Article history: Received 31 May 2009 Accepted 24 March 2010 MSC: 35J55 35B40 Keywords: Large solutions Boundary blow-up Cooperative Quasilinear elliptic system
1. Introduction In this paper we study the positive boundary blow-up solutions to a quasilinear elliptic system of cooperative type:
∆p u = ua v b , ∆ v = uc v e , u p= v = +∞,
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω,
(1.1)
where Ω is a bounded domain in RN with smooth boundary ∂ Ω , and ∆p stands for the p-Laplacian operator defined by ∆p u = div(|∇ u|p−2 ∇ u), p > 1, a, e > p − 1 and b, c < 0. The boundary condition is assumed in the sense u(x), v(x) → +∞ when d(x) → 0, where d(x) stands for the distance function dist(x, ∂ Ω ). We will focus our attention on positive weak 1 ,p 2 solutions, that is, pairs of positive functions (u, v) ∈ [Wloc (Ω ) ∩ L∞ loc (Ω )] verifying (1.1) in the weak sense. However, 1,η
according to standard regularity for the p-Laplacian, we know that positive weak solutions verify u, v ∈ Cloc (Ω ) for some η ∈ (0, 1) (cf. [1–4]). There is a vast amount of literature on elliptic problems related to (1.1), see for example, [5–7,1,8–13,4,14,15] for semilinear problems, and [16–19] for problems with the p-Laplacian. Moreover, we would like to quote some references such as [20–25,2,26–29], in which systems of boundary blow-up solutions related to (1.1) were analyzed. Our motivation mainly comes from [22], where problem (1.1) was analyzed for the competitive case, i.e. a, e > p − 1, and b, c > 0. Under the condition (a − p + 1)(e − p + 1) > bc, García-Melián proved that problem (1.1) has positive solutions
I This work was supported by the foundation of NSFC 10971228, the foundation of Jiangsu Education Commission 07KJD110166, and the doctoral project in Nantong University 08B02. ∗ Corresponding author. Tel.: +86 0513 85015889. E-mail address: [email protected] (Y. Chen).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.03.036
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if and only if c < a − p + 1, b < e − p + 1, and the positive solution is unique when it exists. Moreover, the exact blow-up rate near the boundary of the positive solution was obtained. In the critical case (a − p + 1)(e − p + 1) = bc, infinitely many positive solutions were constructed. We want to obtain similar results as those in [22] for the problem (1.1) under some conditions on the parameters. However, we find that we can only get the existence, uniqueness and boundary behavior for the positive solution in the subcritical case. We remark that though most of the proofs in our paper are an adaptation of the corresponding ones in [22], there are some important differences. For instance, concerning the proof of existence of solutions, the construction of sub- and supersolutions involves more complicated technology. Furthermore, in order to get rough estimates for positive solutions (see Lemma 4 in Section 4), the iterative method in [22] cannot be used here, we shall use a different method. We now state our main results. Theorem 1. Assume that the parameters satisfy a > p − 1,
e > p − 1,
b < 0,
c < 0,
(a − p + 1)(e − p + 1) > bc .
(1.2)
Then the problem (1.1) admits a unique positive solution (u, v). Moreover, (u, v) verifies u(x) ∼ Ad−α (x),
v(x) ∼ Bd−β (x)
(1.3)
as d(x) → 0, where
α=
p(e − p + 1 − b) , (a − p + 1)(e − p + 1) − bc
β=
p(a − p + 1 − c ) , (a − p + 1)(e − p + 1) − bc
(1.4)
and
e−p+1 ! (a−p+1)(e1−p+1)−bc (p − 1)α p−1 (α + 1) , b (p − 1)β p−1 (β + 1) a−p+1 ! (a−p+1)(e1−p+1)−bc (p − 1)β p−1 (β + 1) . c (p − 1)α p−1 (α + 1)
A = B =
(1.5)
The paper is organized as follows. In Section 2 we collect some results on a single equation and the method of sub- and supersolutions which are instrumental in our proofs. In Section 3 the two uniqueness results for problems related to (1.1) will be proved. In Section 4 we prove Theorem 1. 2. Preliminaries The main results of this paper are established by applying the method of sub- and supersolutions and by making use of some results on a single equation with singular boundary conditions. 2.1. Some results on a single equation with singular boundary condition To understand our problem more clearly, we collect some properties of positive solutions to the following problem
∆p u = d−γ (x)uq , u = +∞,
x ∈ Ω, x ∈ ∂Ω.
(2.1)
This problem has been recently considered in [17], and the results concerning existence, uniqueness and asymptotic behavior near the boundary of positive solutions were used in [22]. We cite some results from the papers [22,17]. Lemma 1. Let q > p − 1 and γ ∈ [0, p). Then the problem (2.1) admits a unique positive solution denoted by Uq,γ . Moreover, Uq,γ ∼ (p − 1)%p−1 (% + 1)
q−p1+1
d−% (x)
as d(x) → 0, where % = q−p+1 . p−γ
1,η
Lemma 2. Let u ∈ Cloc (Ω ) for some η ∈ (0, 1). If u verifies
∆p u ≥ Cd−γ (x)uq in Ω , Then u ≤ C
− q−p1+1
∆p u ≤ Cd
Uq,γ in Ω . Similarly, if u verifies
−γ
Then u ≥ C
− q−p1+1
u = +∞ on ∂ Ω .
(x)uq in Ω ,
Uq,γ in Ω .
u = +∞ on ∂ Ω .
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2.2. Sub- and supersolutions This subsection collects some results about the method of sub- and supersolutions for the system:
∆p u = ua v b , ∆p v = uc v e ,
x ∈ Ω, x ∈ Ω,
(2.2)
where a, e > p − 1 and b, c < 0. 1 ,p
2 Definition 1. A pair of positive functions (u, v) ∈ [Wloc (Ω ) ∩ L∞ loc (Ω )] is called a subsolution to the system (2.2), provided a b c e that in the weak sense (u, v) satisfies ∆p u ≥ u v and ∆p v ≥ u v in Ω . As always, a supersolution (u, v) is defined by revising these inequalities.
Proposition 1. Assume that (u, v) is a subsolution and (u, v) is a supersolution to the problem (2.2) with u = u = v = v = +∞ on ∂ Ω , and u ≤ u, v ≤ v in Ω . Then the problem (2.2) admits at least one weak solution (u, v) with u ≤ u ≤ u, v ≤ v ≤ v in Ω and u = v = +∞ on ∂ Ω . The proof of this proposition is similar to Theorem A.2 of [22]. We omit the details here. 3. Two uniqueness results Our first uniqueness result is concerned with problem (1.1) with finite boundary conditions. To this end, Lemma 8 in [22] will be used. For the readers’ convenience, we state it below. Lemma 3. Let f , g ∈ C (Ω ) and u, v ∈ C 1,η (Ω ) be weak solutions to ∆p u = f , ∆p v = g in Ω with u ≤ v and u = v at some point of Ω . Assume moreover that u < v on ∂ Ω . Then there exists x0 ∈ Ω such that u(x0 ) = v(x0 ) and f (x0 ) ≤ g (x0 ). Proposition 2. Assume that the condition (1.2) holds. Let (u1 , v1 ) and (u2 , v2 ) be positive solutions to the problem
∆p u = ua v b , x ∈ Ω , ∆ v = uc v e , x ∈ Ω , u p= f (x), v = g (x),
(3.1)
x ∈ ∂Ω,
where f > 0, g > 0 on ∂ Ω . Then u1 = u2 , v1 = v2 . Proof. The scale argument in this proof has been used by many author, see, for example, [22,30] and the references therein. ¯ and c < 0, we can select a large k such that Since u2 , v2 are positive on Ω u1 ≤ ku2 ,
c
¯. v1 ≤ k− e−p+1 v2 on Ω
(3.2)
Choose the least k with this property, and assume k > 1. Then one of the two inequalities in (3.2) is not strict. Assume it is − e−pc+1
the second one. We can apply Lemma 3 to obtain a point x0 ∈ Ω , such that v1 (x0 ) = k (p−1) − ce− p+1
uc1 (x0 )v1e (x0 ) ≤ k
v2 (x0 ) and
uc2 (x0 )v2e (x0 ).
It follows that uc1 (x0 ) ≤ kc uc2 (x0 ). By c < 0, we have u1 (x0 ) ≥ ku2 (x0 ), which shows that the first inequality in (3.2) is not strict. Thus we may apply Lemma 3 again to obtain a point x0 ∈ Ω , such that u1 (x0 ) = ku2 (x0 ) and ua1 (x0 )v1b (x0 ) ≤ kp−1 ua2 (x0 )v2b (x0 ). It follows that ka−p+1 v1b (x0 ) ≤ v2b (x0 ),
and k
a−p+1 b
v1 (x0 ) ≥ v2 (x0 ).
In view of the second inequality of (3.2) we have k
a−p+1 b
c
v1 (x0 ) ≥ v2 (x0 ) ≥ k e−p+1 v1 (x0 ),
Y. Chen, Y. Zhu / Nonlinear Analysis 73 (2010) 450–457
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which implies that k
c
a−p+1 b
≥ k e−p+1 ,
i.e. k
bc −(a−p+1)(e−p+1) b(e−p+1)
≤ 1.
Note that bc − (a − p + 1)(e − p + 2) < 0, b(e − p + 1) < 0 and k > 1, the above inequality is impossible. So, we have k ≤ 1. That is u1 ≤ u2 and v1 ≤ v2 . Similarly, we can prove u2 ≤ u1 and v2 ≤ v1 . We conclude the proof. Our second uniqueness result is concerned with problem (1.1) in a half-space D := {x ∈ RN : x1 > 0}, where for a point x in RN we write x = (x1 , x0 ), with x0 ∈ RN −1 , i.e. we prove the unique existence result for the following problem:
∆ p u = ua v b , ∆ v = uc v e , u p= v = +∞,
x ∈ D, x ∈ D, x ∈ ∂ D.
(3.3)
Theorem 2. Assume that the condition (1.2) holds. Let α and β be given by (1.4) and (u, v) be a positive weak solution to (3.3) verifying −β
C1 x−α ≤ u ≤ C2 x−α 1 1 ,
−β
≤ v ≤ C2 x1
C 1 x1
in D
(3.4)
for some positive constants C1 , C2 . Then u(x) = Ax−α 1 ,
−β
v(x) = Bx1 ,
where A and B are given by (1.5). −β
Proof. We are going to show that u ≤ Ax−α 1 , v ≤ Bx1 , since the complementary inequality can be proved in a similar way. Thanks to (3.4) we can choose K 1 such that u(x) ≤ Kx−α 1 ,
c
−β
v(x) ≤ K − e−p+1 x1 .
Take K = max{λ, µ}, where
− e−pc+1 β . µ = sup(x1 v(x))
λ = sup(xα1 u(x)),
D
D
Assume that λ ≥ µ, the other case being treated similarly. We may also assume K ≥ A, since otherwise there is nothing to prove. According to the definition of K , there exists a sequence {xn } ⊂ D such that xαn,1 u(xn ) → K , where to simplify the notation xn,1 stands for the first component of xn . Let ξn be the projection of xn onto ∂ D and introduce the functions β
Un (y) = xαn,1 u(ξn + xn,1 y),
Vn (y) = xn,1 v(ξn + xn,1 y).
It is not hard to see that Un , Vn verify the equations
∆p Un = Una Vnb , ∆p Vn = Unc Vne ,
y ∈ D, y ∈ D,
and the inequalities C1 y−α ≤ Un ≤ Ky−α 1 1 ,
−β
C1 y1
c
−β
≤ Vn ≤ K − e−p+1 y1
in D.
(3.5)
Notice that (3.5) gives in particular uniform local bounds for Un , Vn , so that we can use the standard C 1 to obtain that (up to a subsequence) Un → U, Vn → V in Cloc (D), where U , V is a solution to
∆p U = U a V b , ∆p V = U c V e ,
1,η
interior estimates
y ∈ D, y ∈ D,
(3.6)
which in addition verifies U ≤ Ky−α 1 ,V ≤ K
− e−pc+1 −β
y1 , U (e1 ) = K , where e1 is the first vector in the canonical basis of RN . −
c
−β
e−p+1 y On the other hand, it is easily seen that if K ≥ A, then (u, v) given by u = Ky−α 1 ,v = K 1 , is a supersolution to (3.6). We choose M > 0 such that the function h(ζ ) = ζ a v(x)b − M ζ is decreasing in a neighborhood of K for fixed x in a neighborhood of e1 . Then
∆p U − MU = U a V b − MU ≥ U a v b − MU ≥ ua v b − Mu ≥ ∆p u − Mu in a neighborhood of e1 . Since U ≤ u and U (e1 ) = K = u(e1 ), the strong comparison principle [31, Theorem 1.4] gives U ≡ u in a neighborhood of e1 . Notice that it is important here that |∇ u(e1 )| = K α > 0. We also obtain then that V ≡ v in the same neighborhood, and this gives that (u, v) is in fact a solution to (3.6) in a neighborhood of e1 . Thus K = A, K
− e−p1+1
−β
= B, which finally implies u ≤ Ax−α 1 , v ≤ Bx1 . The proof is finished.
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4. Proof of Theorem 1 Proof of the existence. The main tool of this proof is the method of sub- and supersolutions. We look for a subsolution of the form (u(x), v(x)) = (ε Uµ,γ (x), ε δ Uν,σ (x)), where µ, ν, γ , σ and δ are to be chosen and ε is small enough. It is not hard to see that (u(x), v(x)) will be a subsolution provided that
( p−1 µ a b ε d(x)−γ Uµ,γ ≥ εa+δb Uµ,γ Uν,σ ,
(4.1)
ν c e ε δ(p−1) d(x)−σ Uν,σ ≥ ε c +δe Uµ,γ Uν,σ ,
in Ω . We may choose δ in such a way that p − 1 < a + δ b,
δ(p − 1) < c + δ e,
that is
−c a−p+1 <δ< e−p+1 −b since a > p − 1, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. Thus (4.1) will hold for small enough ε provided that µ−a −b −c ν−e d−γ Uµ,γ Uν,σ and d−σ Uµ,γ Uν,σ are bounded from above and from below in Ω . According to the boundary behavior of Uµ,γ and Uν,σ given by Lemma 1, it is enough to have
µ, ν > p − 1, γ , σ ∈ [0, p), p−σ p−γ (µ − a) − b = 0, γ+ µ−p+1 ν−p+1 p−σ p−γ σ+ (ν − e) − c=0 ν−p+1 µ−p+1
(4.2)
which is also possible thanks to the condition (1.2), i.e. a > p − 1, e > p − 1, b, c < 0 and (a − p + 1)(e − p + 1) > bc. In fact, by the condition (1.2), we have p−1<
a(e − p + 1) + b(1 − c − p) e−p+1−b
,
p−1<
e(a − p + 1) + c (1 − b − p) a−p+1−c
.
If we choose µ and ν such that p−1<µ<
a(e − p + 1) + b(1 − c − p) e−p+1−b
,
p−1<ν <
e(a − p + 1) + c (1 − b − p) a−p+1−c
,
and set
γ =p−
p(µ − p + 1)(−b + e − p + 1)
(a − p + 1)(e − p + 1) − bc
,
σ =p−
p(ν − p + 1)(−c + a − p + 1)
(a − p + 1)(e − p + 1) − bc
,
it is not difficult to verify that µ, ν, γ and σ satisfy (4.2). Thus we have a subsolution (u(x), v(x)) = (ε Uµ,γ (x), ε δ Uν,σ (x)). In a similar way, we can show that (¯u(x), v¯ (x)) = (MUµ,γ (x), M δ Uν,σ (x)) is a supersolution for the same choice of µ, ν, γ , σ and δ , provided that M is large enough. Since the sub- and supersolutions are ordered, it follows from Proposition 1 that the problem (1.1) has at least one positive solution (u, v) and satisfies
(u(x), v(x)) ≤ (u, v) ≤ (¯u(x), u¯ (x)), The existence is proved.
x ∈ Ω.
Note that Uµ,γ (x) ∼ d−α (x),
Uν,σ (x) ∼ d−β (x),
we see that such a solution obtained in the above satisfies cˆ1 d−α (x) ≤ u(x) ≤ cˆ2 d−α (x),
cˆ1 d−β (x) ≤ v(x) ≤ cˆ2 (x)d−β (x),
x∈Ω
for some positive constants cˆ1 and cˆ2 which are independent of the domain Ω , where α and β are given by (1.4). To obtain the boundary behavior of solutions we use a blow-up argument. For this sake we need some rough estimates of all possible positive solutions. Lemma 4. Let the condition of Theorem 1 hold and (u, v) be a positive solution to (1.1). Then there exist positive constants C1 , C2 such that C1 d−α (x) ≤ u(x) ≤ C2 d−α (x), where α and β are given by (1.4).
C1 d−β (x) ≤ v(x) ≤ C2 d−β (x),
(4.3)
Y. Chen, Y. Zhu / Nonlinear Analysis 73 (2010) 450–457
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Before giving the proof of Lemma 4, we first give a comparison principle which proof is standard and the details is omitted. Lemma 5. Assume that the condition (1.2) holds. Let (u, v) and (w, z ) be positive solutions of the following problems
∆ p u = ua v b , ∆ v = uc v e , u p= v = ∞,
x ∈ Ω, x ∈ Ω, x ∈ ∂Ω
and
∆p w = w a z b , x ∈ Ω , ∆ z = wc z e , x ∈ Ω , wp< ∞, z < ∞, x ∈ ∂ Ω , respectively. Then (u, v) ≥ (w, z ) in Ω . Proof of Lemma 4. We first prove the upper bound. Let εn & 0 and Ωn = {x ∈ Ω : dist(x, ∂ Ω ) > εn }. From the above proof for the existence result we see that the problem
∆p un = uan vnb , c e ∆ p v n = un v n , un = vn = +∞,
x ∈ Ωn , x ∈ Ωn , x ∈ ∂ Ωn
has a positive solution (un , vn ) satisfying −α c1 d−α n (x) ≤ un (x) ≤ c2 dn (x),
−β c1 d−β n (x) ≤ vn (x) ≤ c2 dn (x),
(4.4)
where c1 and c2 are independent of n, and dn (x) = dist(x, ∂ Ωn ). By Lemma 5, u(x) ≤ un (x),
v(x) ≤ vn (x),
x ∈ Ωn .
(4.5)
Note that (4.4) gives in particular the uniform local bounds for un , vn when n is large enough. One can use the standard C 1,η interior estimate [1,4] and a diagonal process to find a subsequence of {(un , vn )}, denoted by itself, such that (un , vn ) → 1 (u∗ , v ∗ ) in Cloc (Ω ) and (u∗ , v ∗ ) satisfies the differential equations of the problem (1.1) in the weak sense. It follows from (4.5) that u(x) ≤ u∗ (x),
v(x) ≤ v ∗ (x),
∀x ∈ Ω ,
(4.6)
which implies that (u , v ) satisfies the boundary condition of (1.1). For any fixed x ∈ Ω , there is a N1 1 such that d(x) < 2dn (x) for all n > N1 . Hence ∗
∗
−α un (x) ≤ c2 d−α , n (x) < c2 (d(x)/2)
−β vn (x) ≤ c2 d−β . n (x) < c2 (d(x)/2)
Letting n → ∞ it yields u∗ (x) ≤ 2α c2 d−α (x),
v ∗ (x) ≤ 2β c2 d−β (x).
This combines with (4.6) gives the upper bound in (4.3). ∗ Now we prove the lower bound. Similar to the above, let εm & 0 and Ωm = {x ∈ RN : dist(x, Ω ) < εm }. Then Ω ⊂ Ωm∗ . From the above proof for the existence result we see that the problem
∆p um = uam vmb , c e ∆ p v m = um v m , um = vm = +∞,
x ∈ Ωm , x ∈ Ωm , x ∈ ∂ Ωm
has a positive solution (um , vm ) satisfying
ˆ2 d−α cˆ1 d−α m (x) ≤ um (x) ≤ c m (x),
ˆ2 d−β cˆ1 d−β m (x) ≤ vm (x) ≤ c m (x),
x ∈ Ωm ,
(4.7)
where cˆ1 and cˆ2 are independent of m, and dm (x) = dist(x, ∂ Ωm ). Thanks to Lemma 5 we have u(x) ≥ um (x),
v(x) ≥ vm (x) for all x ∈ Ω .
The estimate (4.7) gives in particular the uniform local bounds of um and vm in Ω . We can use the standard C 1,η interior estimate [1,32] and a diagonal process to find a subsequence of {(um , vm )}, denoted by itself, such that (um , vm ) → (u∗ , v∗ ) 1 in Cloc (Ω ) and (u∗ , v∗ ) satisfies the differential equations of the problem (1.1) in weak sense. It is obvious that u(x) ≥ u∗ (x),
v(x) ≥ v∗ (x),
∀x ∈ Ω .
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Y. Chen, Y. Zhu / Nonlinear Analysis 73 (2010) 450–457
For any fixed x ∈ Ω , there is a N2 1 such that d(x) ≥
ˆ1 (2d(x))−α , um (x) ≥ cˆ1 d−α m (x) ≥ c
1 d 2 m
(x) for all m > N2 . Hence
ˆ1 (2d(x))−β . vm (x) ≥ cˆ1 d−β m (x) ≥ c
Letting m → ∞ it yields
v(x) ≥ v∗ (x) ≥ 2−β cˆ1 d−β (x).
u(x) ≥ u∗ (x) ≥ 2−α cˆ1 d−α (x), The proof is completed.
Proof of the boundary behavior (1.3). Let (u, v) be an arbitrary positive solution to (1.1). Fix a point x0 ∈ ∂ Ω . Without loss of generality we may assume x0 = 0 and ν(x0 ) = −e1 , the first vector in the canonical basis of RN . Take an arbitrary sequence {xn } ⊂ Ω with xn → 0, and denote dn = d(xn ) and let ξn be the projection of xn onto ∂ Ω . Introducing the functions:
wn (y) = dαn u(ξn + dn y),
zn (y) = dβn v(ξn + dn y),
where y ∈ Ωn := {y ∈ RN : ξn + dn y ∈ Ω }. Observe that Ωn → D as n → ∞, while d(ξn + dn y) ∼ dn y1 , where y = (y1 , y0 ). It is not hard to see that (wn , zn ) satisfies the system
∆p w = w a z b , ∆p z = w c z e ,
y ∈ Ωn , y ∈ Ωn ,
and thanks to Lemma 4, also the inequalities C1 dαn d−α (ξn + dn y) ≤ wn (y) ≤ C2 dαn d−α (ξn + dn y), C1 dβn d−α (ξn + dn y) ≤ zn (y) ≤ C2 dβn d−α (ξn + dn y). It follows that (wn , zn ) is locally uniformly bounded. Applying the C 1,η interior estimate, we obtain that, passing to a 1 subsequence if necessary, (wn , zn ) converges in Cloc (D) to a pair (w, z ) which is a weak solution to
∆p w = w a z b , ∆p z = w c z e ,
y ∈ D, y ∈ D,
and verifies C1 y−α ≤ w ≤ C2 y−α 1 1 ,
−β
C1 y1
−β
≤ z ≤ C2 y1
in D. β
α According to Theorem 2, w = Ay−α 1 , z = By1 , and setting y = e1 , we obtain that dn u(xn ) → A, dn v(xn ) → B. Since the sequence {xn } is arbitrary, the estimate (1.3) is proved. −β
Proof of the uniqueness. Let (u1 , v1 ) and (u2 , v2 ) be two positive solutions of the problem (1.1). According to the limit (1.3), we have lim
d(x)→0
u1 (x) u2 (x)
= lim
d(x)→0
v1 (x) = 1. v2 (x)
Thus, for every 0 < ε < 1, there exists δ > 0 such that when d(x) ≤ δ , there holds:
(1 − ε)u2 ≤ u1 ≤ (1 + ε)u2 ,
c
c
(1 − ε)− e−p+1 v2 ≤ v1 ≤ (1 + ε)− e−p+1 v2 .
(4.8)
Now we set Ωδ = {x ∈ Ω : d(x) > δ}, and consider the problem
∆p w = w a z b , x ∈ Ωδ , (4.9) ∆ z = w c z e , x ∈ Ωδ , wp= u , z = v 1 , x ∈ ∂ Ωδ . 1 − c − c It is not difficult to see that the pairs (1 − ε)u2 , (1 − ε) e−p+1 v2 and ((1 + ε)u2 , (1 + ε) e−p+1 v2 ) are the ordered suband supersolutions to the problem (4.9). So, the problem (4.9) has at least one positive solution (u, v) and satisfies (1 − ε)u2 ≤ u ≤ (1 + ε)u2 ,
c
c
(1 − ε)− e−p+1 v2 ≤ v ≤ (1 + ε)− e−p+1 v2 in Ωδ .
On the other hand, thanks to Proposition 2, the problem (4.9) has a unique positive solution, which is precisely (u1 , v1 ). Thus, (4.8) is valid in Ωδ . Therefore, (4.8) holds for all x ∈ Ω . Letting ε go to zero in (4.8) we arrive at u1 = u2 , v1 = v2 , which proves the uniqueness.
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Acknowledgements The authors are grateful to Prof. Yihong Du and Prof. Mingxin Wang for pointing out this issue and for helpful suggestions on the manuscript. They would also like to thank the referees for their careful reading and for suggesting valuable references [20,21,28,12,29]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32]
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