# Large Solutions of Quasilinear Elliptic System of Competitive Type: Existence and Asymptotic Behavior

Authors: Lin Wei; Zuodong Yang
Journal: International Journal of Differential Equations
 (2010)

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## Abstract

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.


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## Body

## 1. Introduction

This paper is concerned with the study of positive boundary blow-up solutions to a quasilinear elliptic system of competitive type:(1.1)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
(1.2)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 functionsa(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 thep-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].Whenp=2, system (1.1) becomes
(1.3)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(1.4)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(1.5)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(1.6)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.Whenp≠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(1.7)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 Section2 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 Theorem1.1 in [12], we obtain the following main results.Theorem 1.1.
Assume thatΩ 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
(1.8)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
(1.9)be-p+1<p+γ1p+η2,p+γ2p+η1<a-p+1c.
And one has
(1.10)limx→x0u(x)d(x)-α(γ(x0),η(x0))E(D(x0),C(x0))=1,(1.11)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
(2.1){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 thata(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
(2.2)be-p+1<p+k1p+k2<a-p+1c.
This solution is unique and satisfies
(2.3)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 given0<R1<R and x0∈RN,N≥1, set(2.4)BR(x0)={x∈RN:|x-x0|<R},AR1,R(x0)={x∈RN:R1<|x-x0|<R}.Proposition 2.4.
AssumeΩ=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
(2.5)be-p+1<p+γp+η<a-p+1c.
Then the following systems
(2.6)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
(2.7)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(2.8)(|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
(2.9)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)=+∞.
DefineΦ0(r)=l,Ψ0(r)=m for all 0<r<R, let {Φk},{Ψk} be the function sequences given by
(2.10)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,
(2.11)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
(2.12)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
(2.13)l≤Φk≤Φk+1,m≤Ψk≤Ψk+1.
We now prove that{Φk},{Ψk} are bounded in (0,R). To prove this, we consider
(2.14)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
(2.15)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
(2.16)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 obtainΦk≤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
(2.17)(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
(2.18)Φ(R)=limr→RΦ(r)=∞,Ψ(R)=limr→RΨ(r)=∞.
Secondly, it is clear that
(2.19)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
(2.20)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
(2.21)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
(2.22)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
(2.23)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
(2.24)lp-1=C(R)α(α+1)(p-1)lakb.
Similarly, we obtain
(2.25)kp-1=D(R)β(β+1)(p-1)lcke.
Since
(2.26)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
(2.27)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, whenα≥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
(2.28)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
(2.29)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
(2.30)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
(2.31)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 Proposition2.4, so we omit it here.


## 3. Proof of Theorem1.1

We are now ready to prove Theorem1.1, 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
(3.1)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
(3.2)be-p+1<p+γ1p+η2,p+γ2p+η1<a-p+1c.Proof.
By (3.2) and Proposition 2.3, the following system
(3.3)ΔpΦ=C1d(x)γ1ΦaΨbinΩ,ΔpΨ=D2d(x)η2ΦcΨeinΩ,Φ=Ψ=+∞on∂Ω
possesses a positive solution (u1,v1).
Next we will show that(3.4)(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
(3.5)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
(3.6)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
(3.7)(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:
(3.8)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
(3.9)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
(3.10)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
(3.11)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
(3.12)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.
Fixτ∈(0,1), by (1.2), there exits σ∈(0,1) such that, if d(x,x0)<σ,
(3.13)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
(3.14)Σ=B̅σ/2⋂∂Ω
and choose R>0 small enough such that
(3.15)K=⋃y∈ΣB̅R(y-Rny)⊂Bσ(x0)⋂Ω,
where ny stands for the outward unit normal at y∈∂Ω.
Forx∈Bσ(x0)∩Ω, we get
(3.16)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
(3.17)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:
(3.18)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(3.19)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:
(3.20)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
(3.21)limr→RuBE((1+τ)D(x0),(1-τ)C(x0))(R-r)-α(γ(x0),η(x0))=1,(3.22)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, forx∈BR(x0-(R+σ)nx0),
(3.23)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
(3.24)u(x)≤uB(x+σnx0),v(x)≥vB(x+σnx0).
Let σ→0, we have
(3.25)u(x)≤uB(x),v(x)≥vB(x).
It follows immediately from (3.21), (3.22) that
(3.26)limr→RuE(R-r)-α≤limr→RuBE(R-r)-α=1,(3.27)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 exitsR>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
(3.28)maxA̅R1,R(x0+R1nx0)a̅≤maxΩa+1,maxΩb+1≤maxA̅R1,R(x0+R1nx0)b̲,
and for each x∈AR1,R(x0+R1nx0)(3.29)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
(3.30)a1(R1)=C(x0)+τ,b1(R1)=D(x0)-τ.
We now consider the system(3.31)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(3.32)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
(3.33)(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
(3.34)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)),(3.35)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.



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*Source: 104625-2009-12-02.xml*