We study the existence and asymptotic behavior of positive solutions for a class of quasilinear elliptic systems in a smooth boundary via the upper and lower solutions and the localization method. The main results of the present paper are new and extend some previous results in the literature.
1. Introduction
This paper is concerned with the study of positive boundary blow-up solutions to a quasilinear elliptic system of competitive type:
Δpu=a(x)uavbinΩ,Δpv=b(x)ucveinΩ,u=v=+∞on∂Ω,
where Ω is a bounded C2 domain of RN and Δp stands for the p-Laplacian operator defined by Δpu=div(|∇u|p-2∇u),p>1. The exponents a,b,c,e verify a,e>p-1,b,c>0,(a-p+1)(e-p+1)>bc. There exists C(x),D(x)∈C(Ω̅,R+),γ(x),η(x)∈C(Ω̅,R+) such that
limx→x0a(x)C(x0)d(x)γ(x0)=1,limx→x0b(x)D(x0)d(x)η(x0)=1,
where x0∈∂Ω,d(x)=dist(x,∂Ω).
We must emphasize that the weight functions a(x),b(x) are allowed decaying to zero on Ω with arbitrary rate, depending upon the particular point of ∂Ω. The boundary condition is to be understood u(x)→∞,v(x)→∞ as d(x)→0+. 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.
The problem of the previous form is mathematical models occuring in studies of the p-Laplace system, generalized reaction-diffusion theory, non-Newtonian fluid theory [1, 2], non-Newtonian filtration [3], and the turbulent flow of a gas in porous medium. In the non-Newtonian fluid theory, the quantity p is a characteristic of the medium. Media with p>2 are called dilatant fluids and those with p<2 are called pseudoplastics. If p=2, they are Newtonian fluids. When p≠2, the problem becomes more complicated since certain nice properties inherent to the case p=2 seem to be lost or at least difficult to verify. The main differences between p=2 and p≠2 can be founded in [4, 5].
When p=2, system (1.1) becomes
Δu=a(x)uavbinΩ,Δv=b(x)ucveinΩ,u=v=+∞on∂Ω,
for which the existence, uniqueness, and asymptotic behavior of large solutions have been investigated extensively. We list here, for example, [6–12].
This is a huge amount of literature dealing with single equation with infinite boundary conditions (see, e.g., [13–34]). This problem with more general nonlinearies and weight-function has been discussed by many authors recently [35–39].
Problem (1.1) is considered in special case. When p=2, in [40], problem (1.1) was analyzed with a(x)=1,b(x)=1. In the same paper, some existence, uniqueness, and boundary behavior of solutions were obtained under the assumptions
a(x)~C1d(x)k1,b(x)~C2d(x)k2
as d(x)→0+ for some positive constants C1,C2 and real numbers k1,k2>-2. This problem was later studied in [41] with general form, where
C1d(x)γ1≤a(x)≤C2d(x)γ1,C1′d(x)γ1≤b(x)≤C2′d(x)γ1,
for x∈Ω, γ1,γ2∈RN,C1,C2,C1′,C2′ are positive constants. The author also obtained uniqueness results.
In [42], Yang extended the quasilinear elliptic system to
Δpu=um1vn1inΩ,Δqv=um2vn2inΩ,u=v=+∞on∂Ω,
where m1>p-1,n2>q-1,m2,n1>0, and Ω⊆RN is a smooth bounded domain, subject to three different types of Dirichlet boundary conditions:u=λ,v=μ or u=v=+∞ or u=+∞,v=μ on ∂Ω, where λ,μ>0. Under several hypotheses on the parameters m1,n1,m2,n2, the author showed the existence of positive solutions and further provided the asymptotic behavior of the solutions near ∂Ω.
When p≠2, in [43], problem (1.1) was analyzed with a(x)=1,b(x)=1 under assumption (1.4). The author obtained the existence, uniqueness, and behavior of solutions to problem (1.1).
Very recently, Huang et al. [12] obtained existence, uniqueness, and asymptotic behavior of problem (1.1) when p=2, and a(x),b(x) satisfy condition (1.2). Motivated by the results of the papers [12, 40, 41, 43], we consider the quasilinear elliptic system (1.1). We modify the method developed by Huang et al. [12] and extend the results to a quasilinear elliptic system (1.1) under condition (1.2).
Throughout of this paper, set
C1=minx∈Ω̅C(x),C2=maxx∈Ω̅C(x),D1=minx∈Ω̅D(x),D2=maxx∈Ω̅D(x),γ1=maxx∈Ω̅γ(x),γ2=minx∈Ω̅γ(x),η1=maxx∈Ω̅η(x),η2=minx∈Ω̅η(x),α(x,y)=(p+x)(e-p+1)-(p+y)b(a-p+1)(e-p+1)-bc,β(x,y)=(p+y)(a-p+1)-(p+x)c(a-p+1)(e-p+1)-bc,E(x,y)=(((p-1)αp-1(α+1))e-p+1xb((p-1)βp-1(β+1))bye-p+1)1/((a-p+1)(e-p+1)-bc),F(x,y)=(((p-1)βp-1(β+1))a-p+1yc((p-1)αp-1(α+1))cxa-p+1)1/((a-p+1)(e-p+1)-bc),
nx0 stands for the outward unit normal at x0∈∂Ω.
The paper is organized as follows. In Section 2 we consider some preliminaries which will be used in proof of Theorem 1.1. In Section 3 we will give the proof of the main theorem.
By modifications of the arguments in the proof of Theorem 1.1 in [12], we obtain the following main results.
Theorem 1.1.
Assume that Ω is a bounded C2 domain of RN, a(x),b(x)∈Cθ(Ω) for some θ∈(0,1),a(x),b(x)>0inΩ and verify (1.2), (a-p+1)(e-p+1)>bc,a,e>p-1,b,c>0,γ(x),η(x)∈C(Ω̅,R+) and satisfy
be-p+1<p+γ(x0)p+η(x0)<a-p+1cforx0∈∂Ω.
Then problem (1.1) has a solution (u,v) if and only if
be-p+1<p+γ1p+η2,p+γ2p+η1<a-p+1c.
And one has
limx→x0u(x)d(x)-α(γ(x0),η(x0))E(D(x0),C(x0))=1,limx→x0v(x)d(x)-β(γ(x0),η(x0))F(D(x0),C(x0))=1.
2. Preliminaries
In this section, we will introduce some propositions.
Definition 2.1.
(u̲,v̲) is a subsolution of
{Δpu=a(x)uavbinΩ,Δpv=b(x)ucveinΩ,provided{Δpu̲≥a(x)u̲av̲binΩ,Δpv̲≤b(x)u̲cv̲einΩ.
A supersolution (u̅,v̅) is defined by reversing the inequalities.
Proposition 2.2.
Assume that (u̲,v̲) is a subsolution and (u̅,v̅) is a supersolution of problem (1.1), with u̲=v̲=u̅=v̅=+∞on∂Ω. Then problem (1.1) has at least a solution (u,v) with u̲≤u≤u̅,v̲≥v≥v̅inΩ. In particular u=v=+∞on∂Ω.
Proposition 2.3 (see [43]).
Assume that a(x),b(x) satisfy (1.4), then problem (1.1) admits a positive solution (u,v) with u=v=+∞on∂Ω if and only if k1,k2>-p and
be-p+1<p+k1p+k2<a-p+1c.
This solution is unique and satisfies
limx→x0u(x)d(x)-α(k1,k2)E(C2,C1)=1,limx→x0v(x)d(x)-β(k1,k2)F(C2,C1)=1
for each x0∈∂Ω.
Next, we are ready to study two auxiliary problems in a ball and an annuli. To this aim, for given 0<R1<R and x0∈RN,N≥1, set
Assume Ω=BR(x0),(a-p+1)(e-p+1)>bc,a,e>p-1,b,c>0,C(r),D(r)∈C([0,R],R+) and γ,η>0 satisfy
be-p+1<p+γp+η<a-p+1c.
Then the following systems
ΔpΦ=C(r)(R-r)γΦaΨbinBR(x0),ΔpΨ=D(r)(R-r)ηΦcΨeinBR(x0),Φ=Ψ=+∞on∂BR(x0)
possess a unique radially symmetric positive solution (Φ(r),Ψ(r)) satisfying
limx→x0Φ(r)E(D(R),C(R))(R-r)-α(γ,η)=1,limx→x0Ψ(r)F(D(R),C(R))(R-r)-β(γ,η)=1,
where r=|x-x0|.
Proof.
At first, we consider the following systems
(|Φ′|p-2Φ′)′+N-1r(|Φ′|p-2Φ′)=C(r)(R-r)γΦaΨbin(0,R),(|Ψ′|p-2Ψ′)′+N-1r(|Ψ′|p-2Ψ′)=D(r)(R-r)ηΦcΨein(0,R),Φ(R)=Ψ(R)=+∞,Φ′(0)=Ψ′(0)=0.
We will show that problem (2.8) has a solution (Φ(r),Ψ(r)), which provide a positive radially symmetric solution to problem (2.6). Indeed, any positive solution (Φ(r),Ψ(r)) of the integral equation system
Φ(r)=l+∫0r[t1-N∫0tsN-1C(s)(R-s)γΦaΨbds]1/(p-1)dt,0<r<R,Ψ(r)=m+∫0r[t1-N∫0tsN-1D(s)(R-s)ηΦcΨeds]1/(p-1)dt,0<r<R,
provides a solution of (2.8), 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 function sequences given by
Φk(r)=l+∫0r[t1-N∫0tsN-1C(s)(R-s)γΦk-1aΨk-1bds]1/(p-1)dt,0<r<R,Ψk(r)=m+∫0r[t1-N∫0tsN-1D(s)(R-s)ηΦk-1cΨk-1eds]1/(p-1)dt,0<r<R,
subject to Φk(0)=l,Ψk(0)=m,Φk(R)=Ψk(R)=k.
We remark that {Φk},{Ψk} are nondecreasing sequences. In fact,
Φ1(r)=l+(lamb)1/(p-1)∫0r[t1-N∫0tsN-1C(s)(R-s)γds]1/(p-1)dt=l+(lamb)1/(p-1)A(r)≥l=Φ0(r),Ψ1(r)=m+(lcme)1/(p-1)∫0r[t1-N∫0tsN-1D(s)(R-s)ηds]1/(p-1)dt=m+(lcme)1/(p-1)B(r)≥m=Ψ0(r),
where
A(r)=∫0r[t1-N∫0tsN-1C(s)(R-s)γds]1/(p-1)dt,B(r)=∫0r[t1-N∫0tsN-1D(s)(R-s)ηds]1/(p-1)dt.
Proceeding by 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
ΔpΥ=(C(r)(R-r)γ+D(r)(R-r)η)(Υa+b+Υc+e),
problem (2.14) has a large radially symmetric solution Υ(r), and
Υ(r)=Υ(0)+∫0r[t1-N∫0tsN-1(C(s)(R-s)γ+D(s)(R-s)η)(Υa+b+Υc+e)ds]1/(p-1)dt,
where Υ(0)=l+m. It follows that
Φ1(r)=l+(lamb)1/(p-1)∫0r[t1-N∫0tsN-1C(s)(R-s)γds]1/(p-1)dt≤Υ(0)+∫0r[t1-N∫0tsN-1(C(s)(R-s)γ+D(s)(R-s)η)(Υa+b+Υc+e)ds]1/(p-1)dt=Υ(r).
Similarly, we have Ψ1≤Υ(r).
Arguing as before, 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)=limk→∞(Φk,Ψk),
we deduce that (Φ,Ψ) is a positive solution of (2.8). Then (Φ(x),Ψ(x))=(Φ(r),Ψ(r)) is a positive radially symmetric solution to problem (2.6) and
Φ(R)=limr→RΦ(r)=∞,Ψ(R)=limr→RΨ(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.
By (2.5) and Proposition 2.3, we have
E(D1,C2)≤limr→RΦ(r)(R-r)-α(γ,η)≤E(D2,C1),F(D2,C1)≤limr→RΨ(r)(R-r)-β(γ,η)≤F(D1,C2).
Denote by
l=limr→RΦ(r)(R-r)-α(γ,η),k=limr→RΨ(r)(R-r)-β(γ,η).
By using γ+p=(a-p+1)α(γ,η)+bβ(γ,η),η+p=(e-p+1)β(γ,η)+cα(γ,η) and L′Hôpital rule, we obtain
l=limr→RΦ(0)+∫0r[t1-N∫0tsN-1C(s)(R-s)γΦaΨbds]1/(p-1)dt(R-r)-α=limr→R[r1-N∫0rtN-1C(t)(R-t)γΦaΨbdt]1/(p-1)α(R-r)-α-1=[limr→Rr1-N∫0rtN-1C(t)(R-t)γΦaΨbdtα(R-r)-(α+1)(p-1)]1/(p-1)=[limr→R(1-N)r-N∫0rtN-1C(t)(R-t)γΦaΨbdt+C(r)(R-r)γΦaΨbα(α+1)(p-1)(R-r)-αp+α-p]1/(p-1)=[C(R)α(α+1)(p-1)limr→R(R-r)aα+bβΦaΨb+1-Nα(α+1)(p-1)limr→Rr-N∫0rtN-1C(t)(R-t)γΦaΨbdt(R-r)-αp+α-p]1/(p-1)=[C(R)α(α+1)(p-1)lakb+1-Nα(α+1)(p-1)limr→Rr-N∫0rtN-1C(t)(R-t)γΦaΨbdt(R-r)-αp+α-p]1/(p-1).
We note that
0≤limr→Rr-N∫0rtN-1C(t)(R-t)γΦaΨbdt(R-r)-αp+α-p≤limr→Rr-1∫0rC(t)(R-t)γΦaΨbdt(R-r)-αp+α-p=limr→RC(r)(R-r)γΦaΨbdtR(-αp+α-p)(R-r)-αp+α-p+1=C(R)R(-αp+α-p)limr→R(R-r)γ+αp-α+p+1ΦaΨb=C(R)R(-αp+α-p)lakblimr→R(R-r)=0.
This implies that
lp-1=C(R)α(α+1)(p-1)lakb.
Similarly, we obtain
kp-1=D(R)β(β+1)(p-1)lcke.
Since
αp-1(α+1)(p-1)C(R)=E(D(R),C(R))a-p+1F(D(R),C(R))b,βp-1(β+1)(p-1)D(R)=E(D(R),C(R))cF(D(R),C(R))e-p+1.
If 0<α<1,1<p≤2, then
Ea-p+1Fb=αp-1(α+1)(p-1)C(R)≥α(α+1)(p-1)C(R)=la-p+1kb,
therefore, we get E≥l,F≥k. If 0<α<1,p>2, we get E≤l,F≤k. So, when 0<α<1, we get E=l,F=k.
Similarly, when α≥1, we also get E=l,F=k.
By (2.24) and (2.25), we conclude that l=E(D(R),C(R)),k=F(D(R),C(R)), this completes the proof.
Proposition 2.5.
Assume (a-p+1)(e-p+1)>bc,a,e>p-1,b,c>0,γ>0,η>0, and
be-p+1<p+γp+η<a-p+1c,D(r),C(r)∈C([R1,R],R+) are the reflection around R0=(R1+R)/2 of some functions C¯(r),D¯(r)∈C([R0,R],R+). Then the following system
ΔpΦ=C(r)d(x)γΦaΨbinAR1,R(x0),ΔpΨ=D(r)d(x)ηΦcΨeinAR1,R(x0),Φ=Ψ=+∞on∂AR1,R(x0)
has a unique radially symmetric positive solution (Φ(r),Ψ(r)) such that
limd(x)→0Φ(r)E(D(R),C(R))d(x)-α(γ,η)=1,limd(x)→0Ψ(r)F(D(R),C(R))d(x)-β(γ,η)=1,
where
d(x)=d(x,∂AR1,R(x0))={R-|x-x0|,ifR0≤|x-x0|≤R,|x-x0|-R1,ifR1≤|x-x0|≤R0.
Proof.
The proof is similarl to the proof of Proposition 2.4, so we omit it here.
3. Proof of Theorem 1.1
We are now ready to prove Theorem 1.1, whose proof will be split into the following several lemmas.
Lemma 3.1.
Assume (a-p+1)(e-p+1)>bc,a,e>p-1,b,c>0,C(x),D(x)∈C(Ω̅),a(x),b(x)>0inΩ and (1.2) holds, γ(x),η(x)>0 and satisfy
be-p+1<p+γ(x0)p+η(x0)<a-p+1c,
for each x0∈∂Ω, then problem (1.1) has a solution (u,v) if
be-p+1<p+γ1p+η2,p+γ2p+η1<a-p+1c.
Proof.
By (3.2) and Proposition 2.3, the following system
ΔpΦ=C1d(x)γ1ΦaΨbinΩ,ΔpΨ=D2d(x)η2ΦcΨeinΩ,Φ=Ψ=+∞on∂Ω
possesses a positive solution (u1,v1).
Next we will show that
(u̅,v̅)=((m+n(m-n)b/(a-p+1))1/(p-1)u1,(m-n(m+n)c/(e-p+1))1/(p-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+p=(a-p+1)α(γ1,η2)+bβ(γ1,η2),η2+p=(e-p+1)β(γ1,η2)+cα(γ1,η2).
We have (u̅,v̅) is a supersolution of (1.1) provided
C1d(x)γ1≤a(x)(m+n)((e-p+1)(a-p+1)-bc)/(e-p+1)(p-1),D2d(x)≥b(x)(m-n)((a-p+1)(e-p+1)-bc)/(a-p+1)(p-1).
Since a(x),b(x)∈C(Ω̅), choosing m is large enough, and m-n>0 is sufficiently small, we can prove that
(u̲,v̲)=((m-n(m+n)b/(a-p+1))1/(p-1)u2,(m+n(m-n)c/(e-p+1))1/(p-1)v2)
is a subsolution of (1.1), where (u2,v2) is a solution of the following problem:
ΔpΦ=C2d(x)γ2ΦaΨbinΩ,ΔpΨ=D1d(x)η1ΦcΨeinΩ,Φ=Ψ=+∞on∂Ω.
Then by Proposition 2.2, problem (1.1) has a solution.
Lemma 3.2.
Assume that problem (1.1) has a solution (u,v), then (1.9) holds.
Proof.
In fact, if (1.9) does not hold, it will lead to a contradiction. From Lemma 3.1, we find that if m is large enough and m-n>0 is sufficiently small, we have
u≤u̅=(m+n(m-n)b/(a-p+1))1/(p-1)u1,v≤v̲=(m+n(m-n)c/(e-p+1))1/(p-1)v2.
On the other hand, by (2.3), there exists ε>0 such that for x∈Ωε={x∈Ω:d(x,∂Ω)≤ε}, we get
u≤(m+n(m-n)b/(a-p+1))1/(p-1)u1≤(m+n(m-n)b/(a-p+1))1/(p-1)E(D2,C1)d(x)-α(γ1,η2).
Thus, if
be-p+1≥p+γ1p+η2,
by the definition of α(γ1,η2), we obtain α(γ1,η2)≤0. By (3.10), it implies that u is bounded for x∈Ωε, which is impossible since u(x)=+∞ as d(x)=dist(x,∂Ω)→0+. If
p+γ2p+η2≥a-p+1c,
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 of (1.1), then (1.10) and (1.11) hold.
Proof.
Fix τ∈(0,1), by (1.2), there exits σ∈(0,1) such that, if d(x,x0)<σ,
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=B̅σ/2⋂∂Ω
and choose R>0 small enough such that
K=⋃y∈ΣB̅R(y-Rny)⊂Bσ(x0)⋂Ω,
where ny stands for the outward unit normal at y∈∂Ω.
For x∈Bσ(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 of C2 bounded domain, there exit R>0 and σ0>0 such that
BR(x0-(R+σ)nx0)⊂Ω,BR(x0-Rnx0)⋂∂Ω={x0},
for each σ∈(0,σ0).
Let (uB,σ,vB,σ) be any positive radially symmetric solution to the following system:
Δpu=(1-τ)C(x0)(R-|x-x0|)γ(x0)uavbinBR(x0-(R+σ)nx0),Δpv=(1+τ)D(x0)(R-|x-x0|)η(x0)ucveinBR(x0-(R+σ)nx0),u=v=+∞on∂BR(x0-(R+σ)nx0).
It is easy to see that (u̲σ,v̲σ)=(u,v)|BR(x0-(R+σ)nx0) is a positive smooth subsolution of (3.18), where (u,v) is a positive solution of (1.1).
Then we get
u̲σ=u|BR(x0-(R+σ)nx0)≤uB,σ,v̲σ=v|BR(x0-(R+σ)nx0)≥vB,σ.
Let (uB,vB) be any positive solution to the following system:
Δpu=(1-τ)C(x0)(R-|x-x0|)γ(x0)uavbinBR(x0-Rnx0),Δpv=(1+τ)D(x0)(R-|x-x0|)η(x0)ucveinBR(x0-Rnx0),u=v=+∞on∂BR(x0-Rnx0).
By Proposition 2.3, (uB,vB) satisfies
limr→RuBE((1+τ)D(x0),(1-τ)C(x0))(R-r)-α(γ(x0),η(x0))=1,limr→RvBF((1+τ)D(x0),(1-τ)C(x0))(R-r)-β(γ(x0),η(x0))=1,
where r=|x-x0|.
Taking into account that, for x∈BR(x0-(R+σ)nx0),
uB,σ(x)=uB(x+σnx0),vB,σ(x)=vB(x+σnx0),
by (3.19), for each x∈BR(x0-(R+σ)nx0) and σ∈(0,σ0), we have
u(x)≤uB(x+σnx0),v(x)≥vB(x+σnx0).
Let σ→0, we have
u(x)≤uB(x),v(x)≥vB(x).
It follows immediately from (3.21), (3.22) that
limr→RuE(R-r)-α≤limr→RuBE(R-r)-α=1,limr→RvF(R-r)-β≥limr→RvBF(R-r)-β=1,
where E=E((1+τ)D(R),(1-τ)C(R)),F=F((1+τ)D(R),(1-τ)C(R)).
We next have to prove the inverse inequalities. Similarly, there exits R>R1>0 and σ0>0 such that Ω⊂⋂0<σ<σ0AR1,R(x0+(R+σ)nx0) and AR0,R(x0+R1nx0)⋂∂Ω={x0}. Fix a sufficiently small τ, there exit radially symmetric functions a̅:AR1,R(x0+R1nx0)→R+ and b̲:AR1,R(x0+R1nx0)→R+ such that a̅≥a,b̲≤b in Ω, and
maxA̅R1,R(x0+R1nx0)a̅≤maxΩa+1,maxΩb+1≤maxA̅R1,R(x0+R1nx0)b̲,
and for each x∈AR1,R(x0+R1nx0)a̅(x)=a1(|x-x0-R1nx0|)[d(x,∂AR1,R(x0+R1nx0))]γ(x0),b̲(x)=b1(|x-x0-R1nx0|)[d(x,∂AR1,R(x0+R1nx0))]η(x0),
where a1,b1∈C([R1,R],R+), satisfing
a1(R1)=C(x0)+τ,b1(R1)=D(x0)-τ.
We now consider the system
Δpu=a̅(x)uavbinAR1,R(x0+R1nx0),Δpv=b̲(x)ucveinAR1,R(x0+R1nx0),u=v=+∞on∂AR1,R(x0+R1nx0).
By Proposition 2.5, problem (3.31) possesses a solution (uA,vA).
But for the system
Δpu=a̅(x)uavbinAR1,R(x0+(R1+σ)nx0),Δpv=b̲(x)ucveinAR1,R(x0+(R1+σ)nx0),u=v=+∞on∂AR1,R(x0+(R1+σ)nx0),
it has a solution (uA,σ,vA,σ), and for each x∈AR1,R(x0+(R1+σ)nx0), we have
(uA,σ(x),vA,σ(x))=(uA(x-σnx0),vA(x-σnx0)).
It is also clear that (u̲A(x),v̲A(x))=(uA,σ(x),vA,σ(x))|Ω is a subsolution of problem (1.1). Thus for each x∈AR1,R(x0+(R1+σ)nx0), we get uA(x-σnx0)≤u(x),vA(x-σnx0)≥v(x). Let σ→0, we have u̲A(x)≤u(x),v̲A(x)≥v(x). Thus for x∈K, we get
1=lim|x|→Ru̲A(x)E(a̅(x),b̲(x))(R-|x|)-α(γ(x0),η(x0))≤limd(x)→0u(x)E(a̅(x),b̲(x))(R-|x|)-α(γ(x0),η(x0)),1=lim|x|→Rv̲A(x)F(a̅(x),b̲(x))(R-|x|)-β(γ(x0),η(x0))≥limd(x)→0v(x)F(a̅(x),b̲(x))(R-|x|)-β(γ(x0),η(x0))
but we have limτ→0K={x0}. Therefore, by (3.26), (3.27), (3.34), and (3.35), we finish (1.10) and (1.11). The proof of Lemma 3.3 is complete. From Lemma 3.1 to Lemma 3.3, we finish the proof of Theorem 1.1.
Acknowledgments
This paper was supported by the National Natural Science Foundation of China (Grant no. 10871060) by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (Grant no. 8KJB110005).
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