Nonlinear Analysis 73 (2010) 1739–1746

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Nonlinear Analysis

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The existence of large solutions of semilinear elliptic equations withnegative exponentsI

Lei WeiSchool of Mathematical Science, Xuzhou Normal University, Xuzhou 221116, PR China

a r t i c l e i n f o

Article history:Received 21 January 2010Accepted 4 May 2010

MSC:35J5535B40

Keywords:Boundary blow-upEigenvalueTopological degreeMaximal positive solution

a b s t r a c t

In this work, we consider semilinear elliptic equationswith boundary blow-upwhose non-linearities involve a negative exponent. Combining sub- and super-solution arguments,comparison principles and topological degree theory, we establish the existence of largesolutions. Furthermore, we show the existence of a maximal large positive solution.

© 2010 Elsevier Ltd. All rights reserved.

1. Introduction

In this paper we study the existence of positive solutions and the maximal positive solution of elliptic equations withboundary blow-up

−1u = a(x)u−m − b(x)up, x ∈ Ω,u = +∞, x ∈ ∂Ω, (1.1)

whereΩ ⊂ RN is a bounded and smooth domain, and the function b ∈ Cη(Ω) is a positive function. Throughout this paper,we always suppose p > 1,m > 0, a ∈ Cη(Ω).In (1.1), u = +∞ on ∂Ω means that u(x)→+∞ as x→ ∂Ω . Generally, the solutions of (1.1) are often said to be large

solutions. Problems related to boundary blow-up have been studied for a long time. Motivated by a geometric problem,Bieberbach in [1], for the cases N = 2 and f (u) = eu, proved that the following problem has a unique solution u ∈ C2(Ω)

−1u = f (u), x ∈ Ω,u = +∞, x ∈ ∂Ω.

So far, there is a large amount of literature on elliptic equations related with boundary blow-up, apart from the followingmentioned papers, the reader can also refer to [1–15]. In [16], Du introduced systematically

−1u = a(x)u− b(x)up, x ∈ Ω,u = +∞, x ∈ ∂Ω

I This work was supported by the National Natural Science Foundation of China 10771032,10771212.E-mail address:wleducn@yahoo.com.cn.

0362-546X/$ – see front matter© 2010 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2010.05.011

1740 L. Wei / Nonlinear Analysis 73 (2010) 1739–1746

and gave some brilliant results involving the existence, uniqueness and asymptotic behaviour of positive solutions. In [17],Delgado, López-Gómez and Suárez studied the following problem

−1u = λu1/m − a(x)up/m, x ∈ Ω,u = +∞, x ∈ ∂Ω,

whereΩ is a bounded and smooth domain and 1 < m < p, a(x) ≥ 0. Simultaneously, in [17] the open setΩ+ := x ∈ Ω :a(x) > 0 is connected with boundary ∂Ω+ of class C2, and the open setΩ0 := Ω \ Ω+ satisfiesΩ0 ⊂ Ω . In [18], Delgado,López-Gómez and Suárez studied the existence of positive solutions of the problem

−1u = W (x)uq − a(x)f (u), x ∈ Ω,u = +∞, x ∈ ∂Ω,

with 0 < q < 1, for a rather general class of functions f (u) (see Theorem 1.1 in [18]).Few papers involving boundary blow-up deal with such nonlinearities with negative exponents. Since nonlinearities

with negative exponents are singular at 0, 0 cannot be a sub-solution of the corresponding equations. Therefore, if a(x) is anegative function or sign-changing function, we have difficulty in finding a sub-solution of the auxiliary problem

−1u = a(x)u−m − b(x)up, x ∈ Ω,u = φ, x ∈ ∂Ω,

where φ is a positive function in C1+α(∂Ω). In order to overcome the difficulty, we will show the existence of positivesolutions of the following auxiliary problem by using topological degree theory under some conditions, and show theuniqueness of solutions which are larger than a constant γ

−1u = λu−m − µup, x ∈ Ω,u = l, x ∈ ∂Ω,

where−λ,µ are positive constants.In this paper, we also need the following lemma for operators involving the Laplacian, which goes back to [19,20].

Lemma 1. Assume that Ω ⊂ RN is a bounded domain of class C2+η , 0 < η < 1, and suppose that a(x) ∈ Cη(Ω). Then thefollowing assertions are equivalent:(i) L = −∆+ a satisfies the maximum principle,(ii) L = −∆+ a satisfies the strong maximum principle,(iii) σ1[∆+a,Ω] > 0, where σ1[∆+a,Ω] denotes the first eigenvalue of L with a homogeneous Dirichlet boundary condition,(iv) there exists a positive strict super-solution φ ∈ C2(Ω) ∩ C1(Ω) such that −1φ + a(x)φ ≥ 0 and φ|∂Ω ≥ 0 and φ is not a

solution.

In Section 2, we will give themain results involving the existence of positive solutions and themaximal positive solutionof (1.1).

2. Existence of positive solutions of (1.1)

We need to give some necessary results. Denote γ =(mp

)1/(m+p). Firstly, we show that the following problem has a

unique positive solutionwφ satisfyingwφ > γ−1u = λu−m − b(x)up, x ∈ Ω,u = φ, x ∈ ∂Ω, (2.1)

where λ ∈ R, φ ∈ C2+η(∂Ω), φ(x) > 0 on ∂Ω ,m > 0 and p > 1. The following proposition plays an important role in orderto prove the existence of positive solutions of (2.1).

Proposition 1. There exist two positive constants µ0 and L0 such that for any −λ ≤ µ ≤ µ0 (where λ < 0) and l ≥ L0, thefollowing problem has a unique positive solutionwl satisfying wl > γ

−1u = λu−m − µup, x ∈ Ω,u = l, x ∈ ∂Ω. (2.2)

Proof. Due to the strongmaximumprinciple, it holds that any positive solution of (2.2) is between 0 and l. Denote v = l−u.For the existence of positive solutions of (2.2), we only need to investigate the following problem

−1v = µ(l− v)p − λ(l− v)−m, x ∈ Ω,v = 0, x ∈ ∂Ω. (2.3)

Denote C0(Ω) = u ∈ C(Ω) : u|∂Ω = 0 and Dl,γ = C0(Ω) : 0 ≤ v < l− γ . Then, Dl,γ is a bounded and relatively opensubset of K , where K = u ∈ C0(Ω) : u ≥ 0 is the naturally positive cone of C0(Ω). Define a map by

N (t, w) = (−∆)−1[µ(l− w)p − λt(l− w)−m].

L. Wei / Nonlinear Analysis 73 (2010) 1739–1746 1741

It is clear thatN : [0, 1]×Dl,γ → K is a completely continuous operator by comparison principles and the standard ellipticarguments. Let R0 = supx,y∈Ω |x − y|/2, then there exists x0 ∈ Ω such that Ω ⊂ BR0(x0), where BR0(x0) denotes the openball of radius R0 centered at x0. Take R > R0, by the standard sub- and super-solution arguments,

−1u = −µup, x ∈ BR(x0),u = l, x ∈ ∂BR(x0)

(2.4)

has a non-negative solution. Due to the monotone property of the right hand side of the equation in (2.4), we see that (2.4)has a unique non-negative solution, denoted by ul. This implies that ul is radially symmetric for otherwise a different solutioncould be obtained by rotating ul. Therefore, denote zl(r) = ul(x), r = |x− x0|, thenz ′′l +

(N − 1)r

z ′l = µzpl , 0 < r < R,

z ′l (0) = 0, zl(R) = l.(2.5)

It is convenient to rewrite (2.5) in the form

(rN−1z ′l )′= µrN−1zpl .

Integrating this identity from 0 to r yields

z ′l (r) = r1−N

∫ r

0sN−1µzpl (s)ds ≥ 0.

Therefore, zl(r) is a non-decreasing function and

z ′l (r) ≤ µr1−Nzpl (r)

∫ r

0sN−1ds = µ

rNzpl (r).

Similarly to the proof of Theorem 6.1 in [16], it follows that∫ l

zl(0)H(τ , zl(0))−1/2dτ ≤ R ≤ N1/2

∫ l

zl(0)H(τ , zl(0))−1/2dτ ,

where H(τ , s) = 2∫ τs µξ

pdξ . From the left hand side of the above inequality, we have zl(0) > 0 since the integral in the lefthand side is not integrable if zl(0) = 0. Furthermore, zl is a positive function. By the uniqueness of positive solution of (2.4)and the standard sub- and super-solution arguments, zl is non-decreasing in l. It is well known that the following problemhas a unique positive solution, denoted by u (see [16])

−1u = −µup, x ∈ BR(x0),u = ∞, x ∈ ∂BR(x0).

(2.6)

We can easily prove zl(|x − x0|) ≤ u(x), x ∈ BR(x0). Denote liml→∞ zl(r) = z(r) and uµ(x) = z(r) as r = |x − x0|, by theuniqueness of positive solution of (2.6), then uµ = u. Thus, we have∫

∞

uµ(x0)H(τ , uµ(x0))−1/2dτ ≤ R ≤ N1/2

∫∞

uµ(x0)H(τ , uµ(x0))−1/2dτ . (2.7)

Denote f (β) =∫∞

β

(∫ τβ2ξ pdξ

)−1/2dτ , and we claim that f (β) is decreasing in (0,∞). Indeed, let 0 < β2 < β1, it is clear

that

f (β1)− f (β2) =∫∞

β1

(2∫ τ

β1

ξ pdξ)−1/2

dτ −∫∞

β2

(2∫ τ

β2

ξ pdξ)−1/2

dτ

=

∫∞

β1

(2∫ τ

β1

ξ pdξ)−1/2

dτ −∫∞

β1

(2∫ τ+β2−β1

β2

ξ pdξ)−1/2

dτ

< 0.

Denote µ1 =(1R

∫∞

γ

(2∫ τγξ pdξ

)−1/2dτ)2. We can see

∫∞

uµ(x0)H(τ , uµ(x0))−1/2dτ ≤ R H⇒

1R

∫∞

uµ(x0)

(2∫ τ

uµ(x0)ξ pdξ

)−1/2dτ

2 ≤ µ.So, ifµ ≤ µ1

2 , then theminimum uµ(x0) of the unique positive solution uµ of (2.6) is larger than γ . Denote L0 = max∂Ω u0(x),where u0 is the unique positive solution of (2.6) with µ =

µ12 . We firstly consider a special case l = L0. Suppose µ ≤

µ12 .

1742 L. Wei / Nonlinear Analysis 73 (2010) 1739–1746

Obviously, u0 is a sub-solution of the following problem−1u = −µup, x ∈ Ω,u = L0, x ∈ ∂Ω. (2.8)

It is clear that (u0, L0) is a pair ordered sub- and super-solution of (2.8), which implies that there exists a positive solutionof (2.8), denoted byWL0 such that u0 ≤ WL0 ≤ L0 in Ω . Obviously,WL0 is a unique positive solution of (2.8), and we haveWL0 > γ . Therefore, the following problem has only one solution belonging to DL0,γ

−1v = µ(L0 − v)p, x ∈ Ω,v = 0, x ∈ ∂Ω.

We claim

deg(I −N (0, ·),DL0,γ , 0) = 1.

Indeed, define an operator by N (s, v) = (−∆)−1[sµ(L0 − v)p], then it is clear that N : [0, 1] × DL0,γ → K is completelycontinuous. Assume that v ∈ DL0,γ and s ∈ [0, 1] satisfy v = N (s, v). Clearly,WL0 is a sub-solution of the following problem

−1u = −sµup, x ∈ Ω,u = L0, x ∈ ∂Ω.

By sub- and super-solution arguments, we haveWL0 ≤ L0 − v, which implies v < L0 − γ . So, deg(I − N (s, ·),DL0,γ , 0) iswell defined. By homotopy invariant property, we have

deg(I − N (1, ·),DL0,γ , 0) = deg(I − N (0, ·),DL0,γ , 0) = 1.

So, we have

deg(I −N (0, ·),DL0,γ , 0) = 1.

Next, we prove that deg(I −N (t, ·),DL0,γ , 0) is well defined. Define a set T ⊂ [0, 1] by

T =t ∈ [0, 1] : ∃vt ∈ DL0,γ s.t. vt = N (t, vt)

.

For any t ∈ T , letwt = L0 − vt , thenwt(x) ≥ γ inΩ and clearlywt is a positive solution of the following problem−1u = λtu−m − µup, x ∈ Ω,u = L0, x ∈ ∂Ω. (2.9)

Since 0 ∈ T , T is not empty, and let t0 = supt∈T t . By the standard elliptic arguments, we have t0 ∈ T . For any t ∈ T ,assume u1 and u2 are both positive solutions of (2.9) and u1, u2 ≥ γ , then we claim u1 = u2. Indeed, suppose Ω0 = x ∈Ω : u1(x) > u2(x) 6= ∅. By Green’s formula, we have∫

Ω0

|∇u2|2 −∫Ω0

∇u1∇u2 =∫Ω0

−1u2u2 +∫∂Ω0

u2∂u2∂ν+

∫Ω0

1u1u2 −∫∂Ω0

u2∂u1∂ν

=

∫Ω0

µu2(up1 − u

p2)− tλu2(u

−m1 − u

−m2 )+

∫∂Ω0

(u2∂u2∂ν− u2

∂u1∂ν

)(2.10)

and ∫Ω0

|∇u1|2 −∫Ω0

∇u1∇u2 =∫Ω0

−1u1u1 +∫∂Ω0

u1∂u1∂ν+

∫Ω0

1u2u1 −∫∂Ω0

u1∂u2∂ν

=

∫Ω0

µu1(up2 − u

p1)− tλu1(u

−m2 − u

−m1 )+

∫∂Ω0

(u1∂u1∂ν− u1

∂u2∂ν

). (2.11)

By adding (2.10) to (2.11), we have∫Ω0

|∇u1 −∇u2|2 =∫Ω0

(|∇u1|2 + |∇u2|2 − 2∇u1∇u2)

=

∫Ω0

[µ(up1 − up2)(u2 − u1)− tλ(u1 − u2)(u

−m2 − u

−m1 )]. (2.12)

This implies that

µ

∫Ω0

(u1 − u2)(up1 − u

p2) ≤ −tλ

∫Ω0

(u1 − u2)(u−m2 − u−m1 ). (2.13)

L. Wei / Nonlinear Analysis 73 (2010) 1739–1746 1743

Let h(s) = sp + s−m, then h′(s) = psp−1 − ms−m−1. It is clear that h′(s) > 0 as s >(mp

)1/(p+m)= γ . Due to u1, u2 ≥ γ , it

holds that up1 − up2 > u

−m2 − u

−m1 inΩ0. By (2.13) and together with−λ ≤ µ, we attain a contradiction. So we have u1 ≤ u2

inΩ . Similarly, we have u2 ≤ u1 inΩ . Therefore, for any t ∈ T , (2.9) has a unique positive solution wt satisfying wt ≥ γ .It is clear that wt0 is a sub-solution of (2.9) as t ∈ T , so (wt0 , L0) is a pair ordered sub and super-solution of (2.9). By theuniqueness of positive solution of (2.9) in L0 − DL0,γ (where L0 − DL0,γ = L0 − v : v ∈ DL0,γ ), we havewt ≥ wt0 . Assumex ∈ Ω such thatwt0(x) = minx∈Ω wt0(x), then we have

wt0(x) =∫Ω

G(x, y)(λt0w−mt0 (y)− µwpt0(y))dy−

∫∂Ω

L0∂G∂νdSy

=

∫Ω

G(x, y)(λt0w−mt0 (y)− µwpt0(y))dy+ L0

> L0 − c1µLp0 + c2λγ

−m, (2.14)

where G is the corresponding Green function, c1, c2 are two positive constants. By L0 > γ , there exists µ0 ≤µ12 such that it

holds that wt0(x) > γ as−λ ≤ µ ≤ µ0, which implies L0 − wt = vt ∈ DL0,γ for any t ∈ T . So deg(I − N (t, ·),DL0,γ , 0) iswell defined. By the homotopy invariant property of topological degree, it holds that

deg(I −N (1, ·),DL0,γ , 0) = deg(I −N (0, ·),DL0,γ , 0) = 1.

So, if 0 < −λ ≤ µ ≤ µ0, then (2.3) with l = L0 has a positive solution belonging to DL0,γ , i.e., (2.2) with l = L0 has a positivesolution wL0 such that wL0 > γ . If l ≥ L0, then (wL0 , l) is a pair ordered sub- and super-solution of (2.2). By the standardsub- and super-solution arguments, there exists at least one positive solution wl of (2.2) such that wL0 ≤ wl ≤ l. Clearly,wl(x) > γ inΩ , and similarly we can prove that (2.2) has a unique solution in l− Dl,γ .

For the positive function b in (2.1), denote b0 = supx∈Ω b(x) and b1 = infx∈Ω b(x).

Proposition 2. Suppose 0 < −λ ≤ b1 ≤ b0 ≤ µ0, infx∈∂Ω φ(x) ≥ L0 (L0, µ0 are given in Proposition 1). Then (2.1) has aunique positive solutionwφ satisfying wφ > γ .

Proof. Denote l = inf∂Ω φ(x), and take µ satisfying b0 ≤ µ ≤ µ0. By Proposition 1, (2.2) has a unique positive solution wlsatisfying wl > γ . Clearly, we can choose a large constant L such that (wl, L) is a pair ordered sub- and super-solution of(2.1). So, (2.1) has at least one positive solutionwφ such thatwl ≤ wφ ≤ L. Suppose thatw1, w2 are both positive solutionsof (2.1) and satisfyw1, w2 > γ . IfΩ1 := x ∈ Ω : w1(x) > w2(x) 6= ∅. Similarly to the proof of Proposition 1, we have∫

Ω1

|∇w1 −∇w2|2=

∫Ω1

(|∇w1|2+ |∇w2|

2− 2∇w1∇w2)

=

∫Ω1

b(x)(wp1 − wp2)(w2 − w1)− λ(w1 − w2)(w

−m2 − w

−m1 ),

which implies that∫Ω1

b(x)(w1 − w2)(wp1 − w

p2) ≤ −λ

∫Ω1

(w1 − w2)(w−m2 − w

−m1 ). (2.15)

Bywp1−wp2 > w−m2 −w

−m1 and b(x) ≥ −λ inΩ1, (2.15) is impossible. Therefore, (2.1) has a unique solution, which is larger

than γ .

Proposition 3. If λ ≥ 0, then (2.1) has a unique positive solution.

Proof. If λ = 0, it is obvious. We assume λ > 0. Since b is a positive function, we can take a sufficiently small positiveconstant ε and a sufficiently large positive constant L such that (ε, L) is a pair ordered sub- and super-solution of (2.1).Thus, there exists a positive solution wφ satisfying ε ≤ wφ ≤ L. Let wφ be an arbitrary positive solution of (2.1). SupposeΩ ′ := x ∈ Ω : wφ(x) > wφ(x) 6= ∅. We suppose Ω ′ is regular, otherwise we can make use of perturbation method tocomplete proof. By Lemma 1, we have

σ1[−∆+ b(x)wp−1φ − λw−m−1φ ,Ω ′] > 0.

It is not difficult to see that inΩ ′

−1wφ − λw−m−1φ wφ + b(x)w

p−1φ wφ = λwφ(w

−m−1φ − w−m−1φ )+ b(x)wφ(w

p−1φ − w

p−1φ ) > 0.

It holds that−1wφ − λw

−m−1φ wφ + b(x)w

p−1φ wφ > −1wφ − λw

−mφ + b(x)w

pφ, x ∈ Ω ′,

wφ = wφ, x ∈ ∂Ω ′.

1744 L. Wei / Nonlinear Analysis 73 (2010) 1739–1746

By Lemma 1, we have wφ ≥ wφ inΩ ′, so we obtain a contradiction. Thus, wφ ≥ wφ in Ω , and similarly, we have wφ ≥ wφin Ω . Therefore, wφ = wφ , i.e., the uniqueness is proved.

Theorem 1. If 0 < −λ ≤ b1 ≤ b0 ≤ µ0, then the following problem has a positive solution w such that w > γ−1u = λu−m − b(x)up, x ∈ Ω,u = ∞, x ∈ ∂Ω. (2.16)

If λ ≥ 0, b ∈ Cη(Ω) be a positive function, the problem (2.16) has a positive solution.

Proof. Suppose 0 < −λ ≤ b1 ≤ b0 ≤ µ0. By Proposition 2, as l ≥ L0, the following problem has a unique solution wl suchthat wl > γ

−1u = λu−m − b(x)up, x ∈ Ω,u = l, x ∈ ∂Ω. (2.17)

We claim that if l1 > l2 ≥ L0, then wl1 ≥ wl2 holds, where wl1 , wl2 are solutions of (2.17) with l = l1, l2, respectively, andsatisfy wl1 , wl2 > γ . Indeed, it is clear that (wl2 , l1) is a pair ordered sub- and super-solution of the following problem

−1u = λu−m − b(x)up, x ∈ Ω,u = l1, x ∈ ∂Ω. (2.18)

Since (2.18) has a unique positive solution wl1 satisfying wl1 > γ , and by the standard sub- and super-solution arguments,we have wl1 ≥ wl2 . Denote the unique positive solution of the following problem by ul

−1u = −b1up, x ∈ Ω,u = l, x ∈ ∂Ω. (2.19)

Clearly, ul is non-decreasing in l, and without loss of generality, assume ul → u in C2(Ω) as l→∞, and it is well knownthat u is the unique positive solution of the following problem (refer to [16])

−1u = −b1up, x ∈ Ω,u = ∞, x ∈ ∂Ω.

By (wl, l) being a pair ordered sub- and super-solution of (2.19), it holds that wl ≤ ul, furthermore we have γ < wl ≤ u.By the standard elliptic arguments, wl has a convergent subsequence in C2(Ω), without loss of generality, we assumewl → w in C2(Ω) as l→∞. So,w is a positive solution of (2.16) and w > γ .Suppose λ ≥ 0. Since we can refer to [16] when λ = 0, we only need to show the case λ > 0. By Proposition 3, we

denote the unique positive solution of (2.17) by wl. It is well known that wl is non-decreasing in l by the standard sub-and super-solution arguments and together with the uniqueness of positive solutions of (2.17). Denote the unique positivesolution of the following problem by u

−1u = −b1up/2, x ∈ Ω,u = ∞, x ∈ ∂Ω.

Denote ε0 = infx∈Ω u(x). For any compact set K ⊂ Ω , there exists δ0 > 0 such that for any δ ∈ (0, δ0] it holds that K ⊂ Ωδ ,where Ωδ = x ∈ Ω : d(x, ∂Ω) > δ. Without loss of generality, let l be large enough such that l can be a super-solutionof (2.17). Furthermore, we have wl ≤ l since a sufficiently small positive constant can be a sub-solution of (2.17). Takeδ ∈ (0, δ0] sufficiently small such that u(x) ≥ l for any x ∈ ∂Ωδ . So, we obtain u(x) ≥ wl(x) for each x ∈ ∂Ωδ . ChooseM = max2λε0−m−p/b1, 1, thenMb1up/2 ≥ λ/εm0 . Hence, we have

Mpb(x)up −Mb1up/2 ≥ λ/εm0 ,

we also have

b(x)Mpup −Mb1up/2 ≥ λ/(Mmum).

So, it holds that inΩ

−∆(Mu) = −Mb1up/2 ≥ λ/(Mu)m − b(x)(Mu)p.

So,Mu is a super-solution of the following problem−1u = λu−m − b(x)up, x ∈ Ωδ,u = wl, x ∈ ∂Ωδ.

(2.20)

We can take small enough ε′ (ε′ ≤ ε0) such that ε′ is a sub-solution of (2.20). By Proposition 3, we have wl ≤ Mu in Ωδ .From K ⊂ Ωδ it follows that for any sufficiently large l, we have wl(x) ≤ Mu(x) in K . By the arbitrariness of K and the

L. Wei / Nonlinear Analysis 73 (2010) 1739–1746 1745

standard elliptic arguments, wl has a convergent subsequence in C2(Ω), so without loss of generality, assume that wlitself is convergent to w in C2(Ω). Therefore, w is a positive solution of (2.16).

Theorem 2. Suppose a ∈ Cη(Ω) andmaxx∈Ω |a(x)| = a0 6= 0 (for the case a0 = 0, the problem (1.1) has been studied in somereferences). Then there existsµ0 > 0 such that if a0 < b1 ≤ b0 < µ0, the problem (1.1) has at least one positive solution, whichis larger than γ . Furthermore, (1.1) has a maximal positive solution.

Proof. We consider the following two auxiliary problems−1u = −a0u−m − b(x)up, x ∈ Ω,u = φ, x ∈ ∂Ω (2.21)

and −1u = a0u−m − b(x)up, x ∈ Ω,u = φ, x ∈ ∂Ω. (2.22)

By Proposition 2, if infx∈∂Ω φ(x) ≥ L0, then (2.21) has a unique positive solution vφ satisfying vφ > γ . By Proposition 3,it holds that (2.22) has a unique positive solution, denoted by ωφ . Choose sufficiently large L such that L ≥ maxx∈Ω vφ(x)and (vφ, L) can be a pair ordered sub- and super-solution of the problem (2.22). By the standard sub- and super-solutionarguments and the uniqueness of the positive solution of (2.22), it is well known that vφ ≤ ωφ . Clearly, (vφ, ωφ) is a pairordered sub- and super-solution of

−1u = a(x)u−m − b(x)up, x ∈ Ω,u = φ, x ∈ ∂Ω, (2.23)

so, (2.23) has at least one positive solution θφ such that ωφ ≥ θφ ≥ vφ > γ . For convenience, the corresponding positivesolutions of the problems (2.21), (2.22) and (2.23) with φ = l are denoted by vl, ωl and θl satisfying vl ≤ θl ≤ ωl inΩ , andassume vl > γ as l ≥ L0. By ωl being non-decreasing in l, denote ω = liml→∞ ωl, and by Theorem 1 we have that ω is apositive solution of (2.22) with φ = ∞. So, we have θl ≤ ω in Ω . So, if l ≥ L0, it holds that γ < vl ≤ θl ≤ ω in Ω . By thestandard elliptic arguments, θl has a convergent subsequence in C2(Ω), without loss of generality, we assume θl → θ inC2(Ω) as l→∞. Then, it is clear that θ is a positive solution of (1.1) and satisfies θ > γ .Now,we show the existence of amaximal positive solution of (1.1). Since (vl, ωl) is a pair ordered sub- and super-solution

of the following problem−1u = a(x)u−m − b(x)up, x ∈ Ω,u = l, x ∈ ∂Ω. (2.24)

Then, there exists amaximal positive solution θ∗l of (2.24) in the order interval [vl, ωl]. For any positive solution θ′

l of (2.24), itis clear that θ ′l ≤ ωl, so (maxvl, θ

′

l , ωl) is a pair ordered weak sub- and super-solution of (2.24). Thus, (2.24) has a maximalweak solution θl in the order interval [maxvl, θ ′l , ωl]. By regularity arguments, we θl is a classical solution. Obviously, wehave vl ≤ θl ≤ ωl, furthermore, we have θl ≤ θ∗l . By the maximality of θl, we have θ

′

l ≤ θ∗

l , that is, θ∗

l is a maximal positivesolution of (2.24). Similarly, for sufficiently small δ > 0 and sufficiently large constant l we can prove that the followingproblem has a maximal positive solution, denoted by θ∗l,δ

−1u = a(x)u−m − b(x)up, x ∈ Ωδ,u = l, x ∈ ∂Ωδ.

(2.25)

Denote the unique positive solution of the following problem by ωl,δ−1u = a0u−m − b(x)up, x ∈ Ωδ,u = l, x ∈ ∂Ωδ.

(2.26)

It is easy to prove θ∗l,δ ≤ ωl,δ . It is well known that ωl,δ is non-decreasing in l and denote ωδ = liml→∞ ωl,δ . Similarly tothe proof of Theorem 1, it follows that ωδ is a positive solution of

−1u = a0u−m − b(x)up, x ∈ Ωδ,u = ∞, x ∈ ∂Ωδ.

(2.27)

So, we have θ∗l,δ ≤ ωδ for any sufficiently large l, and from the standard elliptic theory it follows that θ∗

l,δ has a convergentsubsequence in C2(Ω), without loss of generality, we assume θ∗l,δ itself is convergent to θ

∗

δ in C2(Ω) as l → ∞. Assume

that θ is an arbitrary positive solution of (1.1). For any given compact set K ⊂ Ω , there exists δ′ > 0 such that K ⊂ Ωδ

for each δ ∈ (0, δ′]. Take l > 0 large enough such that l ≥ maxx∈∂Ωδ θ(x), then θ is a sub-solution of the problem (2.25).Since θ can be a sub-solution of (2.26), it holds θ ≤ ωl,δ . Hence, (θ, ωl,δ) is a pair ordered sub and super-solution of (2.25),furthermore, it holds θ ≤ θ∗l,δ inΩδ . So, we also have θ ≤ θ∗δ inΩδ . We claim that θ∗δ is non-decreasing about δ ∈ (0, δ

′] in

1746 L. Wei / Nonlinear Analysis 73 (2010) 1739–1746

the compact set K . Indeed, we assume δ′ ≥ δ1 > δ2 > 0, and we can choose l ≥ maxx∈∂Ωδ1 θ∗

δ2(x). Take L > l large enough

such that (θ∗δ2 , L) is a pair ordered sub- and super-solution of the following problem−1u = a(x)u−m − b(x)up, x ∈ Ωδ1 ,u = l, x ∈ ∂Ωδ1 .

Togetherwith the standard sub- and super-solution arguments, we have θ∗δ2 ≤ θ∗

δ1,linΩδ1 , so, θ

∗

δ2≤ θ∗δ1 inΩδ1 . Furthermore,

we have θ∗δ2 ≤ θ∗

δ1in K . By the arbitrariness of K , θ (x) = limδ→0+ θ∗δ (x) is well defined inΩ . By the standard elliptic theory,

θ is a positive solution of (1.1). Due to θ ≤ θ∗δ inΩδ , it is clear that θ is a maximal positive solution of (1.1).

Remark 1. If a, b are positive functions in Cη(Ω) (without additional conditions), then a sufficiently small positive constantε can be a sub-solution of (1.1), then (1.1) has at least one positive solution. Furthermore, (1.1) has a maximal positivesolution and a minimal positive solution.

For Remark 1, similarly to the proof of Theorem 2, the existence of a maximal positive solution can proved. Here, we onlygo to show that (1.1) has a minimal positive solution. Denote a1 = supx∈Ω a(x). Similarly to the proof of Proposition 3, wecan see that the problem (2.23) has a unique positive solution, denoted by θφ . The unique positive solution of (2.23) withφ = l is denoted by θl. By the uniqueness of the positive solution, it is not difficult to see that θl is non-decreasing in l.Denote the unique positive solution of the following problem by ξl

−1u = a1u−m − b(x)up, x ∈ Ω,u = l, x ∈ ∂Ω.

Similarly to the proof of Theorem 1, denote ξ := liml→∞ ξl(x), then ξ is a positive solution of the following problem−1u = a1u−m − b(x)up, x ∈ Ω,u = ∞, x ∈ ∂Ω.

Hence, we have θl ≤ ξ in Ω . By the standard elliptic theory, θl has a convergent subsequence in C2(Ω), without loss ofgenerality, we assume that θl itself is convergent to θ in C2(Ω) as l → ∞. Now, we show that θ is a minimal positivesolution of (1.1). For an arbitrary positive solution ζ of (1.1), in order to prove θ ≤ ζ , it is sufficient to prove that for anycompact set K ⊂ Ω it holds θ ≤ ζ in K . Therefore, we only need to prove that for any large enough constant l it holds θl ≤ ζin K . For the given K , there exists δ > 0 such that K ⊂ Ωδ for each δ ∈ (0, δ]. For any large enough l, there exists δ ∈ (0, δ]such that θl ≤ ζ on ∂Ωδ . So, ζ is a super-solution of the following problem

−1u = a(x)u−m − b(x)up, x ∈ Ωδ,u = θl, x ∈ ∂Ωδ.

At the same time, sufficiently small ε can be a sub-solution. So, by the standard sub- and super-solution arguments and theuniqueness of positive solution of the above problem, we have θl ≤ ζ inΩδ , furthermore, θl ≤ ζ in K .

Acknowledgements

The author is grateful to the referee for helpful comments. He is grateful to Professor Mingxin Wang and Professor JiangZhu for their help and suggestions.

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