Two Classes of Nonlinear Singular Dirichlet Problems with Natural Growth: Existence and Asymptotic Behavior

Zhijun Zhang

doi:10.1515/ans-2019-2054

Article Open Access

Two Classes of Nonlinear Singular Dirichlet Problems with Natural Growth: Existence and Asymptotic Behavior

Zhijun Zhang

Published/Copyright: August 27, 2019

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From the journal Advanced Nonlinear Studies Volume 20 Issue 1

Abstract

This paper is concerned with the existence, uniqueness and asymptotic behavior of classical solutions to two classes of models -△⁢u±λ⁢|∇⁡u|2uβ=b⁢(x)⁢u-α, u>0, x∈Ω, u|∂⁡Ω=0, where Ω is a bounded domain with smooth boundary in ℝN, λ>0, β>0, α>-1, and b∈Clocν⁢(Ω) for some ν∈(0,1), and b is positive in Ω but may be vanishing or singular on ∂⁡Ω. Our approach is largely based on nonlinear transformations and the construction of suitable sub- and super-solutions.

Keywords: Semilinear Elliptic Equations; Singular Dirichlet Problems; Classical Solutions; Existence; Uniqueness; Asymptotic Behavior

MSC 2010: 35J25; 35J65; 35J67

1 Introduction and Main Results

In this paper, we consider the following two classes of singular boundary value problems

(1.1) - Δ ⁢ u ± λ ⁢ | ∇ ⁡ u | 2 u β = b ⁢ ( x ) ⁢ u - α , u > 0 , x ∈ Ω , u | ∂ ⁡ Ω = 0 ,

where Ω is a bounded domain with smooth boundary in ℝN, and b has the following property:

b ∈ C loc ν ⁢ ( Ω ) for some ν∈(0,1), is positive in Ω, and is such that the Poisson problem

(1.2) - Δ ⁢ u = b ⁢ ( x ) , u > 0 , x ∈ Ω , u | ∂ ⁡ Ω = 0 ,

has a unique solution u0∈C2+ν⁢(Ω)∩C⁢(Ω¯).

We will refer to the corresponding classes by (1.1)+ and (1.1)-, respectively.

Throughout this article, we set

λ > 0 , β > 0 , α > - 1 .

Firstly, when β=1 and α=0, we see that problem (1.1)+ is a special case of the following problem:

(1.3) - Δ ⁢ u + h ⁢ ( u ) ⁢ | ∇ ⁡ u | 2 u = b ⁢ ( x ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

where h∈C⁢((0,∞),(0,∞)).

Problem (1.3) is closely related to the following two problems.

The Boundary Blow-up Elliptic Problem:

(1.4) △ ⁢ w = b ⁢ ( x ) ⁢ f ⁢ ( w ) in ⁢ Ω , w | ∂ ⁡ Ω = + ∞ ,

where the boundary condition means that w⁢(x)→+∞ as d⁢(x)=dist⁢(x,∂⁡Ω)→0, and such solution is called ‘a large solution’ or ‘an explosive solution’ if f satisfies

f ∈ C 1 ⁢ [ a , ∞ ) for some a∈ℝ, f⁢(a)=0, and f is increasing on [a,∞) (or f∈C1⁢(ℝ), f⁢(s)>0 for all s∈ℝ, and f is increasing on ℝ),
the Keller–Osserman condition (see [23, 33])

∫ a + 1 ∞ ( F ⁢ ( s ) ) - 1 / 2 ⁢ 𝑑 s < ∞ , F ⁢ ( s ) = ∫ a s f ⁢ ( ν ) ⁢ 𝑑 ν , s ≥ a .

We note that ∫a+1∞(F⁢(s))-1/2⁢𝑑s<∞ implies ∫a+1∞(f⁢(s))-1⁢𝑑s<∞.

Let

u = ∫ w ∞ ( f ⁢ ( s ) ) - 1 ⁢ 𝑑 s .

We see that problem (1.4) is equivalent to problem (1.3) with h⁢(u)=u⁢f′⁢(ψ1⁢(u)), where ψ1 satisfies

∫ ψ 1 ⁢ ( t ) ∞ ( f ⁢ ( s ) ) - 1 ⁢ 𝑑 s = t , t > 0 .

In particular,

when f⁢(w)=wp, p>1, w≥0, we have ψ1⁢(t)=(t⁢(p-1))-1/(p-1), t>0, and h⁢(u)≡pp-1>1,
when f⁢(w)=ew, w∈ℝ, we have ψ1⁢(t)=-ln⁡t, t>0, and h⁢(u)≡1.

Problem (1.4) arises from some branches of mathematics and applied mathematics and has been extensively studied by many authors; see, for instance, [7, 17, 16, 23, 26, 27, 28, 29, 30, 31, 33, 37, 38, 41] and the references therein.

The Singular Dirichlet Problem:

(1.5) - △ ⁢ v = b ⁢ ( x ) ⁢ g ⁢ ( v ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

where

g ∈ C 1 ⁢ ( ( 0 , ∞ ) , ( 0 , ∞ ) ) , lims→0+⁡g⁢(s)=+∞ and g is decreasing on (0,∞).

Let

u = ∫ 0 v ( g ⁢ ( s ) ) - 1 ⁢ 𝑑 s .

We see that problem (1.5) is equivalent to problem (1.3) with h⁢(u)=-u⁢g′⁢(ψ2⁢(u)), where ψ2 satisfies

∫ 0 ψ 2 ⁢ ( t ) ( g ⁢ ( s ) ) - 1 ⁢ 𝑑 s = t , t ≥ 0 .

In particular, when g⁢(v)=v-η, v>0, η>0, we have h⁢(u)≡ηη+1∈(0,1).

Problem (1.5) has been extensively studied in many contexts, see, for instance, [14, 15, 19, 24, 35, 39, 43, 42], and the references therein.

Secondly, when β=1, let q=λ1+λ and u-=∫0vd⁢ssq=v1-q1-q. Then we see that problem (1.1)- is equivalent to the following problem:

(1.6) - △ ⁢ v = ( 1 + λ ) - α ⁢ b ⁢ ( x ) ⁢ v ( λ - α ) / ( 1 + λ ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 .

Thus, for the case λ=α, one can see that v=(1+λ)-λ⁢u0 is the unique solution to problem (1.6), i.e., u-=(u0⁢(1+λ))1/(1+λ) is the unique solution to problem (1.1)-. For the case λ<α, problem (1.6) is a special case of problem (1.5). For the case λ>α, problem (1.6) is a sublinear problem, since λ-α1+λ∈(0,1).

So, as in the above analysis, the main questions to study problem (1.1) are the cases β≠1 and α≠0.

Thirdly, for problems (1.1), there are many papers which have been dedicated to resolving existence, nonexistence, uniqueness and multiplicity issues for weak solutions (in H01⁢(Ω)), see, for instance, [4, 3, 6, 5, 7, 8, 12, 11, 10, 20, 21, 22, 25, 32, 13, 44, 45] and the references therein. In particular,

for λ=1, α=0 and b∈Lq⁢(Ω) (q>N2), which is strictly positive on every compactly contained subset of Ω, Arcoya et al. [4] proved that problem (1.1)+ has a weak solution in H01⁢(Ω) if and only if β<2,
for β=1, α=0 and b∈L2⁢N/(N+2)⁢(Ω), which is non-negative in Ω, Arcoya et al. [3] showed that problem (1.1)- has a weak solution in H01⁢(Ω) if and only if λ<1.

Moreover, for the case β=1, some results on the existence, uniqueness and asymptotic behavior of classical solutions (in C2⁢(Ω)∩C⁢(Ω¯)) to problem (1.1)+ can be found in [9, 18, 34, 36, 39, 40] and the references therein. In particular, for the case b≡1 in Ω, Porru and Vitolo [34] showed some results on the existence and boundary behavior of classical solutions to problem (1.1)+.

Next, we introduce two classes of functions.

First of all, we denote K the set of all functions L^ defined on (0,δ0] for some δ0>0 by

L ^ ( t ) = c 0 exp ( ∫ t δ 0 z ⁢ ( s ) s d s ) , t ∈ ( 0 , δ 0 ] ,

where c0>0 and the function z∈C⁢[0,δ0], with z⁢(0)=0. Some basic functions in K are

( - ln ⁡ t ) σ and (ln⁡(-ln⁡t))σ, σ∈ℝ,
exp ⁡ ( ( - ln ⁡ t ) σ ) , 0<σ<1.

Then let Λ denote the set of all positive monotonic functions θ in C1⁢(0,δ0)∩L1⁢(0,δ0) satisfying

lim t → 0 ⁡ d d ⁢ t ⁢ ( Θ ⁢ ( t ) θ ⁢ ( t ) ) := C θ ∈ [ 0 , ∞ ) , Θ ⁢ ( t ) := ∫ 0 t θ ⁢ ( s ) ⁢ 𝑑 s .

Some basic examples of the non-increasing functions in Λ are

θ ≡ c 0 > 0 , Θ⁢(t)=c0⁢t, Cθ=1,
θ ⁢ ( t ) = ( - ln ⁡ t ) σ , σ>0, Cθ=1,
θ ⁢ ( t ) = t - σ , with σ∈(0,1), Θ⁢(t)=t1-σ1-σ, Cθ=11-σ.

In addition, some basic examples of the non-decreasing functions in Λ are

θ ⁢ ( t ) = exp ⁡ ( - t - σ ) , σ>0, Cθ=0,
θ ⁢ ( t ) = t σ , σ>0, Θ⁢(t)=t1+σ1+σ, Cθ=11+σ,
θ ⁢ ( t ) = ( - ln ⁡ t ) - σ , σ>0, Cθ=1.

For the convenience of discussion, we introduce the following assumptions:

There exist θ∈Λ and positive constants bi (i=1,2) such that

b 1 := lim d ⁢ ( x ) → 0 ⁡ inf ⁡ b ⁢ ( x ) θ 2 ⁢ ( d ⁢ ( x ) ) ≤ b 2 := lim d ⁢ ( x ) → 0 ⁡ sup ⁡ b ⁢ ( x ) θ 2 ⁢ ( d ⁢ ( x ) ) ,

where d⁢(x)=dist⁢(x,∂⁡Ω), x∈Ω.
There exist positive constants bi (i=1,2) and L^∈K, with

h ⁢ ( t ) := ∫ 0 t L ^ ⁢ ( s ) s ⁢ 𝑑 s < ∞ , t ∈ ( 0 , δ 0 ] ,

such that

b 1 := lim d ⁢ ( x ) → 0 ⁡ inf ⁡ b ⁢ ( x ) ( d ⁢ ( x ) ) - 2 ⁢ L ^ ⁢ ( d ⁢ ( x ) ) ≤ b 2 := lim d ⁢ ( x ) → 0 ⁡ sup ⁡ b ⁢ ( x ) ( d ⁢ ( x ) ) - 2 ⁢ L ^ ⁢ ( d ⁢ ( x ) ) .

We note from [41, Lemma 2.1] that (b2) is quite different from (b3). In this paper, we show the existence, uniqueness and asymptotic behavior of classical solutions to problems (1.1).

Our main results are summarized in the following theorems. For β>1, we have Theorems 1.1 and 1.2.

Theorem 1.1.

Let b satisfy (b1). If β<2+α, then problem (1.1)+ has one classical solution u+. Moreover, if b satisfies the additional condition (b2) with b1=b2=b0, then for any classical solution u+ to problem (1.1)+, we have

lim d ⁢ ( x ) → 0 ⁡ ϕ 1 ⁢ ( u + ⁢ ( x ) ) exp ⁡ ( C 0 ⁢ ( b 0 ⁢ Θ ⁢ ( d ⁢ ( x ) ) ) - 2 ⁢ ( β - 1 ) / ( 2 + α - β ) ) = 1 ,

with

ln ⁡ ( ϕ 1 ⁢ ( τ ) ) ≅ λ β - 1 ⁢ τ 1 - β as ⁢ τ → 0 ,

where

ϕ 1 ⁢ ( τ ) = ∫ τ 1 exp ⁡ ( G 1 ⁢ ( s ) ) ⁢ 𝑑 s , τ ∈ ( 0 , 1 ) , G 1 ⁢ ( s ) = λ β - 1 ⁢ s 1 - β , s > 0 ,

and

(1.7) C 0 = ( λ β - 1 ) ( β + α ) / ( 2 + α - β ) ⁢ ( 2 ⁢ ( β - 1 ) λ ⁢ ( 2 + α - β ) ) 2 ⁢ ( β - 1 ) / ( 2 + α - β ) .

Remark 1.1.

In Theorem 1.1, the asymptotic behavior of u+ is not known if b satisfies the additional condition (b3) with b1=b2=b0.

Theorem 1.2.

Let b satisfy (b1). Then problem (1.1)- has at least one classical solution u-.

Remark 1.2.

In Theorem 1.2, the asymptotic behavior of u- is not known.

For β∈(0,1), we have Theorems 1.3 and 1.4.

Theorem 1.3.

Let b satisfy (b1). Then problem (1.1)+ has a unique classical solution u+ if α≥0, and problem (1.1)+ has at least one classical solution u+, provided α∈(-1,0). Moreover, for the case α>0, the following hold:

If b satisfies the additional condition (b2) with b 1 = b 2 = b 0 and

(1.8) C θ ⁢ ( 1 + α ) > 2 ,

then u + satisfies

(1.9) lim d ⁢ ( x ) → 0 ⁡ u + ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) 2 2 ⁢ ( C θ ⁢ ( 1 + α ) - 2 ) ) 1 / ( 1 + α ) .
If b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then u + satisfies

(1.10) lim d ⁢ ( x ) → 0 ⁡ u + ⁢ ( x ) ( h ⁢ ( d ⁢ ( x ) ) ) 1 / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) ) 1 / ( 1 + α ) .

Theorem 1.4.

Let b satisfy (b1). Then (1.1)- has at least one classical solution. Moreover, for α>0, the following hold:

If b satisfies the additional condition (b2) with b 1 = b 2 = b 0 and ( 1.8 ) holds, then any classical solution u - to problem ( 1.1 ) - satisfies ( 1.9 ).
If b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then any classical solution u - to problem ( 1.1 ) - satisfies ( 1.10 ).

Remark 1.3.

The uniqueness of solutions is not known in Theorem 1.4.

For β=1, we have Theorems 1.5–1.6.

Theorem 1.5.

Let b satisfy (b1). Then problem (1.1)+ has at least one classical solution. Moreover, for λ=1,

if b satisfies the additional condition (b2) with C θ > 0 , then problem ( 1.1 ) + has a unique classical solution u + satisfying

lim d ⁢ ( x ) → 0 ⁡ ln ⁡ ( u + ⁢ ( x ) ) ln ⁡ ( Θ ⁢ ( d ⁢ ( x ) ) ) = 2 1 + α ,
if b satisfies the additional condition (b3) , then problem ( 1.1 ) + has a unique classical solution u + satisfying

lim d ⁢ ( x ) → 0 ⁡ ln ⁡ ( u + ⁢ ( x ) ) ln ⁡ ( h ⁢ ( d ⁢ ( x ) ) ) = 1 1 + α .

For λ>1,

if b satisfies the additional condition (b2) with b 1 = b 2 = b 0 , then problem ( 1.1 ) + has a unique classical solution u + satisfying

(1.11) lim d ⁢ ( x ) → 0 ⁡ u + ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) 2 2 ⁢ ( C θ ⁢ ( 1 + α ) + 2 ⁢ λ - 2 ) ) 1 / ( 1 + α ) ,
if b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then problem ( 1.1 ) + has a unique classical solution u + satisfying ( 1.10 ).

For λ∈(0,1),

if λ > - α and b satisfies the additional condition (b2) with b 1 = b 2 = b 0 and

C θ ⁢ ( 1 + α ) + 2 ⁢ λ > 2 ,

then problem ( 1.1 ) + has a unique classical solution u + satisfying ( 1.11 );
if λ > - α and b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then problem ( 1.1 ) + has a unique classical solution u + satisfying ( 1.10 ),
if λ = - α , then u + = ( u 0 ⁢ ( 1 - λ ) ) 1 / ( 1 - λ ) is the unique classical solution to problem ( 1.1 ) + ,
if λ < - α , then problem ( 1.1 ) + has a classical solution u + . In addition, if b ∈ L 1 ⁢ ( Ω ) , then problem ( 1.1 ) + has a unique classical solution.

Theorem 1.6.

Let b satisfy (b1).

If λ = α , then u - = ( u 0 ⁢ ( 1 + λ ) ) 1 / ( 1 + λ ) is the unique classical solution to problem ( 1.1 ) - , where u 0 is given as in (b1).
If λ < α , then problem ( 1.1 ) - has a unique classical solution u - . Moreover,
1. if b satisfies the additional condition (b2) with b 1 = b 2 = b 0 and
  
  C θ ⁢ ( 1 + α ) - 2 ⁢ λ > 2 ,
  
  then u - verifies
  
  lim d ⁢ ( x ) → 0 ⁡ u - ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) 2 2 ⁢ ( C θ ⁢ ( 1 + α ) - 2 ⁢ λ - 2 ) ) 1 / ( 1 + α ) ,
2. if b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then u - verifies ( 1.11 ).
If λ > α , then problem ( 1.1 ) - has one classical solution u - . Moreover, if b ∈ L 1 ⁢ ( Ω ) , then u - is the unique classical solution to problem ( 1.1 ) - .

The outline of this paper is as follows. In Section 2 we give some preliminary considerations. The proofs of Theorems 1.1–1.6 are given in Section 3.

2 Preliminaries

In this section, we present some preliminary considerations.

Firstly, we provide the following lemmas with regard to problems (1.4) and (1.5).

Lemma 2.1 ([36, Theorem 1]).

Let b satisfy (b1), and let f satisfy (f1) and (f2). Then problem (1.4) has a classical solution w∈C2⁢(Ω).

Lemma 2.2 ([42, Theorems 1.1 and 1.2]).

Let b satisfy (b1) and f⁢(w)=exp⁡(w).

If b satisfies the additional condition (b2) with C θ > 0 , then problem ( 1.4 ) has a unique classical solution w satisfying

lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) - ln ⁡ ( Θ ⁢ ( d ⁢ ( x ) ) ) = 2 .
If b satisfies the additional condition (b3) , then problem ( 1.4 ) has a unique classical solution w satisfying

lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) - ln ⁡ ( h ⁢ ( d ⁢ ( x ) ) ) = 1 .

Lemma 2.3 ([17, Theorems 2.8], [27, Theorems 1.1 and 1.3], [41, Theorems 1.1 and 1.2]).

Let b satisfy (b1) and f⁢(w)=wp, with p>1.

If b satisfies the additional condition (b2) with b 1 = b 2 = b 0 , then problem ( 1.4 ) has a unique classical solution w satisfying

lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) ⁢ ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( p - 1 ) = ( 2 ⁢ ( 2 + C θ ⁢ ( p - 1 ) ) b 0 ⁢ ( p - 1 ) 2 ) 1 / ( p - 1 ) .
If b satisfies the additional condition (b3) with b 1 = b 2 = b 0 , then problem ( 1.4 ) has a unique classical solution w satisfying

lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) ⁢ ( h ⁢ ( d ⁢ ( x ) ) ) 1 / ( p - 1 ) = ( b 0 ⁢ ( p - 1 ) ) - 1 / ( p - 1 ) .

Lemma 2.4 ([41, Theorem 1.1]).

Let f satisfy (f1) and (f2), and let b satisfy (b1) and (b2) with b1=b2=b0. If f satisfies the additional condition

lim s → ∞ ⁡ d d ⁢ s ⁢ ( 2 ⁢ F ⁢ ( s ) ) ⁢ ∫ s ∞ d ⁢ υ 2 ⁢ F ⁢ ( υ ) = ∞ ,

then for any classical solution w to problem (1.4), we have

lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) ψ ⁢ ( b 0 ⁢ Θ ⁢ ( d ⁢ ( x ) ) ) = 1 ,

where ψ satisfies

(2.1) ∫ ψ ⁢ ( t ) ∞ d ⁢ s 2 ⁢ F ⁢ ( s ) = t for all ⁢ t > 0 .

Lemma 2.5 ([40, Theorem 4.1], [41, Theorems 1.1 and 1.2]).

Let g satisfy (g1) . Then problem ( 1.5 ) has a unique classical solution v if and only if b satisfies (b1).
Let g satisfy the following conditions
1. g ∈ C 1 ⁢ ( ( 0 , ∞ ) , ( 0 , ∞ ) ) , limt→0⁡g⁢(t)=∞,
2. there exists t 0 > 0 such that g ′ ⁢ ( t ) < 0 for all t ∈ ( 0 , t 0 ) ,
3. there exists η > 0 such that lim t → 0 ⁡ g ⁢ ( t ) ⁢ t η = 1 .
Then the following hold:
1. If b satisfies the additional condition (b2) with b 1 = b 2 = b 0 and
  
  C θ ⁢ ( 1 + η ) > 2 ,
  
  then for any classical solution v to problem ( 1.5 ), we have
  
  lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( 1 + η ) = ( b 0 ⁢ ( 1 + η ) 2 2 ⁢ ( C θ ⁢ ( 1 + η ) - 2 ) ) 1 / ( 1 + η ) .
2. If b satisfies the additional condition (b3) , with b 1 = b 2 = b 0 , then for any classical solution v to problem ( 1.5 ), we have
  
  lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( h ⁢ ( d ⁢ ( x ) ) ) 1 / ( 1 + η ) = ( b 0 ⁢ ( 1 + η ) ) 1 / ( 1 + η ) .

Next, we consider the following more general problem

(2.2) - Δ ⁢ v = H ⁢ ( x , v ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

and introduce the comparison principle and sub-supersolution method with the boundary restriction.

Lemma 2.6 (The Comparison Principle [35, Lemma 2.3]).

Suppose that H:Ω×(0,∞)→R is a continuous function such that H⁢(x,s)s is strictly decreasing for s>0 at each x∈Ω. Let v1,v2∈C2⁢(Ω)∩C⁢(Ω¯) satisfy

Δ ⁢ v 2 + H ⁢ ( x , v 2 ) ≤ 0 ≤ Δ ⁢ v 1 + H ⁢ ( x , v 1 ) , x∈Ω,
v 2 , v 1 > 0 in Ω and v2≥v1 on ∂⁡Ω,
Δ ⁢ v 1 ∈ L 1 ⁢ ( Ω ) .

Then we have v2≥v1 in Ω.

Lemma 2.7 (The Existence Theorem [15, Lemma 3]).

Let H⁢(x,s) be locally Hölder continuous in Ω×(0,∞) and continuously differentiable with respect to the variable s. Suppose problem (2.2) has a supersolution v¯ and a subsolution v¯ such that v¯≤v¯ on Ω and v¯|∂⁡Ω=v¯|∂⁡Ω=0. Then problem (2.2) has a classical solution v in the order interval [v¯,v¯].

Remark 2.1.

For f∈Cν⁢(Ω¯×[0,∞)) (ν∈(0,1) is given as in b1), and f⁢(x,0)≥0, x∈Ω, Amann [1] first established Lemma 2.7.

For convenience, denote

| u | ∞ = max x ∈ Ω ¯ | u ( x ) | , u ∈ C ( Ω ¯ ) .

Lemma 2.8.

Let σ∈(0,1) and b satisfy (b1). Then the problem

(2.3) - △ ⁢ v = b ⁢ ( x ) ⁢ v σ , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

has a classical solution v0 satisfying

( ( 1 - σ ) ⁢ u 0 ⁢ ( x ) ) 1 / ( 1 - σ ) ≤ v 0 ⁢ ( x ) ≤ ( | u 0 | ∞ ) σ / ( 1 - σ ) ⁢ u 0 ⁢ ( x ) , x ∈ Ω ,

where u0 is given as in (b1). Moreover, if b∈L1⁢(Ω), then problem (2.3) has a unique classical solution.

Remark 2.2.

Ambrosetti, Brezis and Cerami [2] established some existence and multiplicity results to the following Dirichlet problem:

- △ ⁢ u = u σ + λ ⁢ u p , u > 0 , x ∈ Ω , u | ∂ ⁡ Ω = 0 ,

where 0<σ<1<p and λ∈ℝ.

Proof.

We note that v¯=M⁢u0 is a supersolution to problem (2.3), provided M≥(|u0|∞)σ/(1-σ). To construct a subsolution, let v¯=((1-σ)⁢u0)1/(1-σ). We have, by a direct computation, that

- Δ ⁢ v ¯ = - σ ⁢ ( ( 1 - σ ) ⁢ u 0 ) ( 2 ⁢ σ - 1 ) / ( 1 - σ ) ⁢ | ∇ ⁡ u 0 | 2 + b ⁢ ( x ) ⁢ ( ( 1 - σ ) ⁢ u 0 ) σ / ( 1 - σ )

≤ b ⁢ ( x ) ⁢ ( ( 1 - σ ) ⁢ u 0 ) σ / ( 1 - σ ) = b ⁢ ( x ) ⁢ v ¯ σ in ⁢ Ω ,

i.e., v¯ is a subsolution to problem (2.3).

Since supσ∈(0,1)⁡(1-σ)1/(1-σ)=1, one can see that v¯≥v¯ on Ω¯. Lemma 2.7 implies that problem (2.3) has a classical solution v0 in the ordered interval [v¯,v¯]. Moreover, when b∈L1⁢(Ω), Lemma 2.6 implies that v0 is the unique classical solution to problem (2.3). ∎

3 Proofs of Theorems 1.1–1.6

In this section we prove Theorems 1.1–1.6.

Proof of Theorem 1.1.

For the case β∈(1,2+α), recall that

ϕ 1 ⁢ ( τ ) = ∫ τ 1 exp ⁡ ( G 1 ⁢ ( s ) ) ⁢ 𝑑 s , τ ∈ ( 0 , 1 ) , G 1 ⁢ ( s ) = λ β - 1 ⁢ s 1 - β .

It follows that

lim s → ∞ ⁡ G 1 ⁢ ( s ) = 0 , lim s → 0 ⁡ G 1 ⁢ ( s ) = ∞ , lim s → 0 ⁡ exp ⁡ ( - G 1 ⁢ ( s ) ) = 0 , lim s → ∞ ⁡ exp ⁡ ( ± G 1 ⁢ ( s ) ) = 1 , lim τ → 0 ⁡ ϕ 1 ⁢ ( τ ) = ∞ .

Let Φ1 be the inverse of ϕ1, i.e., Φ1 satisfies

∫ Φ 1 ⁢ ( t ) 1 exp ⁡ ( G 1 ⁢ ( s ) ) ⁢ 𝑑 s = t , t > 0 .

We obtain, by a direct computation, that

Φ 1 ′ ⁢ ( t ) = - exp ⁡ ( - G 1 ⁢ ( Φ 1 ⁢ ( t ) ) ) , Φ 1 ⁢ ( t ) > 0 , t > 0 , lim t → ∞ ⁡ Φ 1 ⁢ ( t ) = 0 ,

Φ 1 ′′ ⁢ ( t ) = λ ⁢ ( Φ 1 ⁢ ( t ) ) - β ⁢ exp ⁡ ( - 2 ⁢ G 1 ⁢ ( Φ 1 ⁢ ( t ) ) ) , t > 0 , lim t → ∞ ⁡ Φ 1 ′ ⁢ ( t ) = 0 .

Let u+=Φ1⁢(w). We see that problem (1.1)+ is equivalent to problem (1.4) with f⁢(t)=(Φ1⁢(t))-α⁢exp⁡(G1⁢(Φ1⁢(t))). Moreover, f⁢(t) has the following properties:

f ′ ⁢ ( t ) = λ ⁢ ( Φ 1 ⁢ ( t ) ) - ( β + α ) + α ⁢ ( Φ 1 ⁢ ( t ) ) - ( 1 + α ) , t>0,
we have

lim t → ∞ ⁡ F ⁢ ( t ) t 2 ⁢ ( ln ⁡ t ) ( β + α ) / ( β - 1 ) = λ 2 ⁢ ( β - 1 λ ) ( β + α ) / ( β - 1 ) , where ⁢ F ⁢ ( t ) = ∫ 0 t f ⁢ ( ν ) ⁢ 𝑑 ν ,
f satisfies (f1)–(f3), where (f3) is given as in Lemma 2.4.

In fact, (1) is easily obtained. To prove (2) and (3), let y=Φ1⁢(t). We see that t=∫y1exp⁡(G1⁢(s))⁢𝑑s and y→0 if and only if t→∞. We also note that for an arbitrary ρ>1, we have lims→+∞⁡sρexp⁡(s)=0. So, we set s=λβ-1⁢y1-β, ρ=ββ-1, and obtain

(3.1) lim y → 0 y - β exp ⁡ ( λ β - 1 ⁢ y 1 - β ) = 0 , i.e., lim y → 0 y β exp ( λ β - 1 y 1 - β ) = ∞ .

It follows from l’Hospital’s rule that

lim t → ∞ ⁡ ( Φ 1 ⁢ ( t ) ) 1 - β ln ⁡ t = lim y → 0 ⁡ y 1 - β ln ( ∫ y 1 exp ( G 1 ( s ) ) d s )

= ( β - 1 ) ⁢ lim y → 0 ⁡ ∫ y 1 exp ⁡ ( G 1 ⁢ ( s ) ) ⁢ 𝑑 s y β ⁢ exp ⁡ ( G 1 ⁢ ( y ) )

= ( β - 1 ) ⁢ lim y → 0 ⁡ - exp ⁡ ( G 1 ⁢ ( y ) ) β ⁢ y β - 1 ⁢ exp ⁡ ( G 1 ⁢ ( y ) ) - λ ⁢ exp ⁡ ( G 1 ⁢ ( y ) )

= ( β - 1 ) ⁢ lim y → 0 ⁡ 1 λ - β ⁢ y β - 1 = β - 1 λ ,

and thus

lim t → ∞ ⁡ F ⁢ ( t ) t 2 ⁢ ( ln ⁡ t ) ( β + α ) / ( β - 1 ) = lim t → ∞ ⁡ f ⁢ ( t ) 2 ⁢ t ⁢ ( ln ⁡ t ) ( β + α ) / ( β - 1 ) + β + α β - 1 ⁢ t ⁢ ( ln ⁡ t ) ( 1 + α ) / ( β - 1 ) = lim t → ∞ ⁡ f ⁢ ( t ) 2 ⁢ t ⁢ ( ln ⁡ t ) ( β + α ) / ( β - 1 )

= 1 2 ⁢ lim t → ∞ ⁡ λ ⁢ ( Φ 1 ⁢ ( t ) ) - ( β + α ) + α ⁢ ( Φ 1 ⁢ ( t ) ) - ( 1 + α ) ( ln ⁡ t ) ( β + α ) / ( β - 1 ) + β + α β - 1 ⁢ ( ln ⁡ t ) ( 1 + α ) / ( β - 1 ) = λ 2 ⁢ lim t → ∞ ⁡ ( Φ 1 ⁢ ( t ) ) - ( β + α ) ( ln ⁡ t ) ( β + α ) / ( β - 1 )

= λ 2 ⁢ ( β - 1 λ ) ( β + α ) / ( β - 1 ) .

Since 1<β<2+α, we have

β + α 2 ⁢ ( β - 1 ) > 1 and ∫ 2 ∞ d ⁢ t t ⁢ ( ln ⁡ t ) ( β + α ) / 2 ⁢ ( β - 1 ) = 2 ⁢ ( β - 1 ) 2 + α - β ⁢ ( ln ⁡ 2 ) ( - 2 - α + β ) / 2 ⁢ ( β - 1 ) .

Thus, (f2) holds. Moreover,

ψ ⁢ ( t ) ≅ ψ 3 ⁢ ( t ) = exp ⁡ ( C 0 ⁢ t - 2 ⁢ ( β - 1 ) / ( 2 + α - β ) ) as ⁢ t → 0 ,

where ψ and C0 are given as in (2.1) and (1.7), respectively, and ψ3 satisfies

∫ ψ 3 ⁢ ( t ) ∞ d ⁢ s s ⁢ ( ln ⁡ s ) ( β + α ) / 2 ⁢ ( β - 1 ) = λ ⁢ ( β - 1 λ ) ( β + α ) / 2 ⁢ ( β - 1 ) ⁢ t for all ⁢ t > 0 .

In addition, using l’Hospital’s rule and (3.1) again, we obtain

lim t → ∞ ⁡ F ⁢ ( t ) ⁢ f ′ ⁢ ( t ) f 2 ⁢ ( t ) = lim t → ∞ ⁡ F ⁢ ( t ) ⁢ ( λ ⁢ ( Φ 1 ⁢ ( t ) ) - ( β + α ) + α ⁢ ( Φ 1 ⁢ ( t ) ) - ( 1 + α ) ) ( Φ 1 ⁢ ( t ) ) - 2 ⁢ α ⁢ exp ⁡ ( 2 ⁢ G 1 ⁢ ( Φ 1 ⁢ ( t ) ) )

= λ ⁢ lim t → ∞ ⁡ F ⁢ ( t ) ( Φ 1 ⁢ ( t ) ) β - α ⁢ exp ⁡ ( 2 ⁢ G 1 ⁢ ( Φ 1 ⁢ ( t ) ) )

= λ ⁢ lim t → ∞ ⁡ ( Φ 1 ⁢ ( t ) ) - α ⁢ exp ⁡ ( G 1 ⁢ ( Φ 1 ⁢ ( t ) ) ) exp ⁡ ( G 1 ⁢ ( Φ 1 ⁢ ( t ) ) ) ⁢ ( 2 ⁢ λ ⁢ ( Φ 1 ⁢ ( t ) ) - α - ( β - α ) ⁢ ( Φ 1 ⁢ ( t ) ) β - α - 1 )

= λ ⁢ lim t → ∞ ⁡ 1 2 ⁢ λ - ( β - α ) ⁢ ( Φ 1 ⁢ ( t ) ) β - 1 = 1 2 ,

and thus

lim t → ∞ ⁡ F ⁢ ( t ) f 2 ⁢ ( t ) = lim t → ∞ ⁡ f ⁢ ( t ) 2 ⁢ f ⁢ ( t ) ⁢ f ′ ⁢ ( t ) = lim t → ∞ ⁡ 1 2 ⁢ f ′ ⁢ ( t )

= 1 2 ⁢ lim t → ∞ ⁡ 1 λ ⁢ ( Φ 1 ⁢ ( t ) ) - ( β + α ) + α ⁢ ( Φ 1 ⁢ ( t ) ) - ( 1 + α )

= 1 2 ⁢ lim t → ∞ ⁡ ( Φ 1 ⁢ ( t ) ) β + α λ + α ⁢ ( Φ 1 ⁢ ( t ) ) β - 1 = 0 .

Consequently, limt→∞⁡F⁢(t)f⁢(t)=0 and

lim t → ∞ ⁡ d d ⁢ t ⁢ ( 2 ⁢ F ⁢ ( t ) ) ⁢ ∫ t ∞ d ⁢ υ 2 ⁢ F ⁢ ( υ ) = 1 2 ⁢ lim t → ∞ ⁡ ∫ t ∞ d ⁢ υ F ⁢ ( υ ) F ⁢ ( t ) f ⁢ ( t ) = 1 2 ⁢ lim t → ∞ ⁡ - 1 F ⁢ ( t ) f 2 ⁢ ( t ) 2 ⁢ F ⁢ ( t ) - F ⁢ ( t ) ⁢ f ′ ⁢ ( t ) f 2 ⁢ ( t )

= 1 2 ⁢ lim t → ∞ ⁡ 1 F ⁢ ( t ) ⁢ f ′ ⁢ ( t ) f 2 ⁢ ( t ) - 1 2 = + ∞ ,

i.e., (f3) holds. Thus, Theorem 1.1 follows from Lemmas 2.1 and 2.4 with w=ϕ1⁢(u+). ∎

Proof of Theorem 1.2.

For the case β>1, recall that G1⁢(s)=λβ-1⁢s1-β, s>0.

Let

ϕ 2 ⁢ ( τ ) = ∫ 0 τ exp ⁡ ( - G 1 ⁢ ( s ) ) ⁢ 𝑑 s , τ > 0 ,

and let Φ2 be the inverse of ϕ2, i.e., Φ2 satisfies

∫ 0 Φ 2 ⁢ ( t ) exp ⁡ ( - G 1 ⁢ ( s ) ) ⁢ 𝑑 s = t , t > 0 .

It follows that

(3.2) lim t → ∞ ⁡ Φ 2 ⁢ ( t ) = ∞ , lim t → ∞ ⁡ exp ⁡ ( ± G 1 ⁢ ( Φ 2 ⁢ ( t ) ) ) = 1 ,

(3.3) Φ 2 ′ ⁢ ( t ) = exp ⁡ ( G 1 ⁢ ( Φ 2 ⁢ ( t ) ) ) , Φ 2 ⁢ ( t ) > 0 , t > 0 , Φ 2 ⁢ ( 0 ) = 0 ,

(3.4) Φ 2 ′′ ⁢ ( t ) = - λ ⁢ ( Φ 2 ⁢ ( t ) ) - β ⁢ exp ⁡ ( 2 ⁢ G 1 ⁢ ( Φ 2 ⁢ ( t ) ) ) , t > 0 .

Let u-=Φ2⁢(v). We see that problem (1.1)- is equivalent to the following problem:

(3.5) - △ ⁢ v = b ⁢ ( x ) ⁢ ζ ⁢ ( v ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

with ζ⁢(t)=(Φ2⁢(t))-α⁢exp⁡(-G1⁢(Φ2⁢(t))). Moreover, ζ⁢(t) has the properties

ζ ′ ⁢ ( t ) = λ ⁢ ( Φ 2 ⁢ ( t ) ) - β - α - α ⁢ ( Φ 2 ⁢ ( t ) ) - 1 - α , t>0,
lim t → ∞ ⁡ ζ ⁢ ( t ) = 0 ,
lim t → 0 ⁡ ζ ⁢ ( t ) = 0 and ∫01d⁢tζ⁢(t)<∞.

In fact, (1) is easily obtained and (2) comes from (3.2). In order to prove (3), let y=Φ2⁢(t). We see that t=∫0yexp⁡(-G1⁢(s))⁢𝑑s and y→0 if and only if t→0. It follows from l’Hospital’s rule and (3.3) that

lim t → 0 ⁡ ( Φ 2 ⁢ ( t ) ) 1 - β - ln ⁡ t = lim y → 0 ⁡ y 1 - β - ln ( ∫ 0 y exp ( - G 1 ( s ) ) d s )

= ( β - 1 ) ⁢ lim y → 0 ⁡ ∫ 0 y exp ⁡ ( - G 1 ⁢ ( s ) ) ⁢ 𝑑 s y β ⁢ exp ⁡ ( - G 1 ⁢ ( y ) )

= ( β - 1 ) ⁢ lim y → 0 ⁡ exp ⁡ ( - G 1 ⁢ ( y ) ) β ⁢ y β - 1 ⁢ exp ⁡ ( - G 1 ⁢ ( y ) ) + λ ⁢ exp ⁡ ( - G 1 ⁢ ( y ) )

= ( β - 1 ) ⁢ lim y → 0 ⁡ 1 λ + β ⁢ y β - 1 = β - 1 λ ,

and thus

lim t → 0 ⁡ ζ ⁢ ( t ) t ⁢ ( - ln ⁡ t ) ( β + α ) / ( β - 1 ) = lim t → 0 ⁡ λ ⁢ ( Φ 2 ⁢ ( t ) ) - ( β + α ) - α ⁢ ( Φ 2 ⁢ ( t ) ) - ( 1 + α ) ( - ln ⁡ t ) ( β + α ) / ( β - 1 ) - β + α β - 1 ⁢ ( - ln ⁡ t ) ( 1 + α ) / ( β - 1 )

= λ ⁢ lim t → 0 ⁡ ( Φ 2 ⁢ ( t ) ) - ( β + α ) ( - ln ⁡ t ) ( β + α ) / ( β - 1 ) = λ ⁢ ( β - 1 λ ) ( β + α ) / ( β - 1 ) .

Consequently, (3) holds, since

∫ 0 1 / 2 d ⁢ t t ⁢ ( - ln ⁡ t ) ( β + α ) / ( β - 1 ) = β - 1 1 + α ⁢ ( ln ⁡ 2 ) - ( 1 + α ) / ( β - 1 ) .

In addition, one can see from (1) and (2) that M0:=maxt∈[0,∞)⁡ζ⁢(t) exists and it is positive. Let v¯=M⁢u0, where M≥M0. Then v¯=M⁢u0⁢(x) satisfies

- Δ ⁢ v ¯ = M ⁢ b ⁢ ( x ) ≥ M 0 ⁢ b ⁢ ( x ) ≥ b ⁢ ( x ) ⁢ ζ ⁢ ( M ⁢ u 0 ⁢ ( x ) ) , x ∈ Ω ,

i.e., v¯ is a supersolution to problem (3.5).

To construct a subsolution v¯, we see from Φ2⁢(0)=0 and (1) that there exists t0>0 sufficiently small such that

ζ ′ ⁢ ( t ) = ( Φ 2 ⁢ ( t ) ) - β - α ⁢ ( λ - α ⁢ ( Φ 2 ⁢ ( t ) ) β - 1 ) > 0 , t ∈ ( 0 , t 0 ) ,

i.e., ζ is strictly increasing on (0,t0).

Let

φ ⁢ ( τ ) = ∫ 0 τ d ⁢ s ζ ⁢ ( s ) , τ ≥ 0 .

We have φ′⁢(τ)=1ζ⁢(τ)>0, τ>0, φ⁢(0)=0. It follows that φ is strictly increasing on [0,∞). Let Ψ be the inverse of φ, i.e., Ψ satisfies

∫ 0 Ψ ⁢ ( t ) d ⁢ s ζ ⁢ ( s ) = t , t ≥ 0 .

It follows that

(3.6) Ψ ′ ⁢ ( t ) = ζ ⁢ ( Ψ ⁢ ( t ) ) , Ψ ⁢ ( t ) > 0 , t > 0 , Ψ ⁢ ( 0 ) = 0 .

Let v¯⁢(x)=Ψ⁢(c⁢u0⁢(x)), x∈Ω, where c∈(0,1) is sufficiently small and such that

c ⁢ | u 0 | ∞ ≤ φ ⁢ ( t 0 ) .

Thus, v¯⁢(x)∈(0,t0), x∈Ω. In addition, we obtain by a direct computation that

- Δ ⁢ ( c ⁢ u 0 ) = c ⁢ b ⁢ ( x ) = - Δ ⁢ v ¯ ζ ⁢ ( v ¯ ) + ζ ′ ⁢ ( v ¯ ) ζ 2 ⁢ ( v ¯ ) ⁢ | ∇ ⁡ v ¯ | 2 ≥ - Δ ⁢ v ¯ ζ ⁢ ( v ¯ ) , x ∈ Ω ,

and thus

- Δ ⁢ v ¯ ≤ c ⁢ b ⁢ ( x ) ⁢ ζ ⁢ ( v ¯ ) ≤ b ⁢ ( x ) ⁢ ζ ⁢ ( v ¯ ) , x ∈ Ω , v ¯ | ∂ ⁡ Ω = 0 ,

i.e., v¯ is a subsolution to problem (3.5).

Next, we assert that v¯≤M⁢u0 in Ω, provided M≥M0 is sufficiently large. In fact, by (3), (3.6) and l’Hospital’s rule, we have

lim t → 0 ⁡ Ψ ⁢ ( t ) t = lim t → 0 ⁡ Ψ ′ ⁢ ( t ) = lim t → 0 ⁡ ζ ⁢ ( Ψ ⁢ ( t ) ) = 0 .

Therefore, maxx∈Ω¯⁡Ψ⁢(c⁢u0⁢(x))u0⁢(x) exists and is positive.

We choose M≥maxx∈Ω¯⁡Ψ⁢(c⁢u0⁢(x))u0⁢(x), and thus obtain v¯≤M⁢u0 in Ω. Lemma 2.7 implies that problem (3.5) has a classical solution v in the ordered interval [Ψ⁢(c⁢u0),M⁢u0]. So problem (1.1)- has a classical solution in the ordered interval [Φ2⁢(Ψ⁢(c⁢u0)),Φ2⁢(M⁢u0)]. The proof is complete. ∎

Proof of Theorem 1.3.

For the case β∈(0,1), let

G 2 ⁢ ( s ) = λ 1 - β ⁢ s 1 - β , s ≥ 0 , ϕ 3 ⁢ ( τ ) = ∫ 0 τ exp ⁡ ( - G 2 ⁢ ( s ) ) ⁢ 𝑑 s , τ ≥ 0 ,

and let Φ3 be the inverse of ϕ3, i.e., Φ3 satisfies

∫ 0 Φ 3 ⁢ ( t ) exp ⁡ ( - G 2 ⁢ ( s ) ) ⁢ 𝑑 s = t , t ∈ [ 0 , T 0 ) ,

with

T 0 = ∫ 0 ∞ exp ⁡ ( - G 2 ⁢ ( s ) ) ⁢ 𝑑 s < ∞ .

It follows that

Φ 3 ′ ⁢ ( t ) = exp ⁡ ( G 2 ⁢ ( Φ 3 ⁢ ( t ) ) ) , Φ 3 ⁢ ( t ) > 0 , t ∈ [ 0 , T 0 ) ,

Φ 3 ⁢ ( 0 ) = 0 , Φ 3 ′ ⁢ ( 0 ) = 1 , lim t → T 0 ⁡ Φ 3 ⁢ ( t ) = ∞ ,

Φ 3 ′′ ⁢ ( t ) = λ ⁢ ( Φ 3 ⁢ ( t ) ) - β ⁢ exp ⁡ ( 2 ⁢ G 2 ⁢ ( Φ 3 ⁢ ( t ) ) ) , t ∈ ( 0 , T 0 ) .

Let u+=Φ3⁢(v). We see that problem (1.1)+ is equivalent to the following problem:

(3.7) - △ ⁢ v = b ⁢ ( x ) ⁢ g ⁢ ( v ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

with g⁢(t)=(Φ3⁢(t))-α⁢exp⁡(-G2⁢(Φ3⁢(t))). Moreover, g⁢(t) has the following properties:

g ′ ⁢ ( t ) = - λ ⁢ ( Φ 3 ⁢ ( t ) ) - ( β + α ) - α ⁢ ( Φ 3 ⁢ ( t ) ) - ( 1 + α ) = ( Φ 3 ⁢ ( t ) ) - ( 1 + α ) ⁢ ( - λ ⁢ ( Φ 3 ⁢ ( t ) ) 1 - β - α ) , t∈(0,T0),
lim t → T 0 ⁡ g ⁢ ( t ) = 0 and

lim t → 0 ⁡ g ⁢ ( t ) = { ∞ , α > 0 , 1 , α = 0 , 0 , α < 0 ,
lim t → 0 ⁡ t Φ 3 ⁢ ( t ) = lim τ → 0 ⁡ ϕ 3 ⁢ ( τ ) τ = 1 .

In fact, (1) and (2) are easily obtained. In addition, by using the properties of Φ3⁢(t) and (2), one can see that g is strictly decreasing on [0,T0), provided α≥0. Moreover, g is strictly increasing on [0,t0) and strictly decreasing on [t0,T0) for some t0∈(0,T0) when α∈(-1,0).

(3) Let y=Φ3⁢(t). We see that t=∫0yexp⁡(-G2⁢(s))⁢𝑑s and y→0 if and only if t→0. It follows from l’Hospital’s rule that

lim t → 0 ⁡ t Φ 3 ⁢ ( t ) = lim y → 0 ⁡ ∫ 0 y exp ⁡ ( - G 2 ⁢ ( s ) ) ⁢ 𝑑 s y = lim y → 0 ⁡ exp ⁡ ( - G 2 ⁢ ( y ) ) = 1 ,

i.e., (3) holds.

Moreover, for α>0, since limt→0⁡g⁢(t)⁢tα=1, we see from the proof of Lemma 2.5 with η=α that (i) and (ii) in Theorem 1.3 hold.

Next, we prove the existence of solutions. Case 1: α≥0. Let

φ 1 ⁢ ( τ ) = ∫ 0 τ d ⁢ s g ⁢ ( s ) , τ ∈ ( 0 , T 0 ) ,

and z=Φ3⁢(t). Then we obtain

φ 1 ⁢ ( τ ) = ∫ 0 Φ 3 ⁢ ( τ ) z α ⁢ 𝑑 z = 1 1 + α ⁢ ( Φ 3 ⁢ ( τ ) ) 1 + α → ∞ as ⁢ τ → T 0 .

It follows that φ1:[0,T0)→[0,∞) is strictly increasing and φ1′⁢(τ)=1g⁢(τ)>0 for τ∈(0,T0).

Let Ψ1 be the inverse of φ1, i.e., Ψ1⁢(t)=ϕ3⁢((t⁢(1+α))1/(1+α)) and

∫ 0 Ψ 1 ⁢ ( t ) d ⁢ s g ⁢ ( s ) = t , t > 0 .

It follows that

Ψ 1 ′ ⁢ ( t ) = g ⁢ ( Ψ 1 ⁢ ( t ) ) , Ψ 1 ⁢ ( t ) > 0 , t > 0 , Ψ 1 ⁢ ( 0 ) = 0 .

Let v¯⁢(x)=Ψ1⁢(u0⁢(x)), x∈Ω. We obtain by the monotonicity of g and a direct computation that

- Δ ⁢ u 0 = b ⁢ ( x ) = - Δ ⁢ v ¯ g ⁢ ( v ¯ ) + g ′ ⁢ ( v ¯ ) g 2 ⁢ ( v ¯ ) ⁢ | ∇ ⁡ v ¯ | 2 ≤ - Δ ⁢ v ¯ g ⁢ ( v ¯ ) , x ∈ Ω ,

i.e.,

- Δ ⁢ v ¯ ⁢ ( x ) ≥ b ⁢ ( x ) ⁢ g ⁢ ( v ¯ ⁢ ( x ) ) , x ∈ Ω , v ¯ | ∂ ⁡ Ω = 0 .

So, v¯ is a supersolution to problem (3.7).

On the other hand, we have from (2) that

lim t → 0 ⁡ g ⁢ ( t ) t = ∞ and lim t → 0 ⁡ Ψ 1 ⁢ ( t ) t = lim t → 0 ⁡ Ψ 1 ′ ⁢ ( t ) = lim t → 0 ⁡ g ⁢ ( Ψ 1 ⁢ ( t ) ) = { ∞ , α > 0 , 1 , α = 0 .

It follows that infx∈Ω¯⁡Ψ1⁢(u0⁢(x))u0⁢(x) exists and is positive. Moreover, we see that there exists sufficiently small c0∈(0,min⁡{1,infx∈Ω¯⁡Ψ1⁢(u0⁢(x))u0⁢(x)}) such that

c 0 ⁢ | u 0 | ∞ < T 0 and g ⁢ ( c 0 ⁢ | u 0 | ∞ ) ≥ c 0 .

Let v¯=c0⁢u0⁢(x). Then we obtain, by the monotonicity of g, that

- Δ ⁢ ( c 0 ⁢ u 0 ⁢ ( x ) ) = c 0 ⁢ b ⁢ ( x ) ≤ b ⁢ ( x ) ⁢ g ⁢ ( c 0 ⁢ | u 0 | ∞ ) ≤ b ⁢ ( x ) ⁢ g ⁢ ( c 0 ⁢ u 0 ⁢ ( x ) ) , x ∈ Ω ,

and c0⁢u0⁢(x)≤Ψ1⁢(u0⁢(x)), x∈Ω¯. By Lemma 2.7, we see that problem (3.7) has a solution. Case 2: α∈(-1,0). For convenience, let σ=-α. From the above analysis, we see that M0=max[0,T0)⁡g⁢(t)=g⁢(t0). Let v¯=M0⁢u0⁢(x), where u0 is the unique solution of problem (1.2) and M0 is a positive constant satisfying M0⁢|u0|∞≤t0. It follows that

- △ ⁢ v ¯ = M 0 ⁢ b ⁢ ( x ) ≥ b ⁢ ( x ) ⁢ g ⁢ ( M 0 ⁢ u 0 ⁢ ( x ) ) , x ∈ Ω ,

i.e., v¯ is a supersolution to problem (3.7).

In addition, from the above analysis, we see that limt→0⁡g⁢(t)tσ=1, i.e., there exists t1∈(0,T0) such that

g ⁢ ( t ) ≥ t σ 2 , t ∈ ( 0 , t 1 ) .

Let v¯=m⁢v0, where v0 is given as in Lemma 2.8 and m is a positive constant satisfying

m = min ⁡ { t 1 | v 0 | ∞ , 2 - 1 / ( 1 - σ ) , M 0 ⁢ inf x ∈ Ω ¯ ⁡ u 0 ⁢ ( x ) v 0 ⁢ ( x ) } .

We note from Lemma 2.8 that infx∈Ω¯⁡u0⁢(x)v0⁢(x) is positive. It follows that

- △ ⁢ v ¯ = m ⁢ b ⁢ ( x ) ⁢ v 0 σ ⁢ ( x ) = 2 ⁢ b ⁢ ( x ) ⁢ m 1 - σ ⁢ 1 2 ⁢ ( m ⁢ v 0 ⁢ ( x ) ) σ ≤ b ⁢ ( x ) ⁢ g ⁢ ( m ⁢ v 0 ) , x ∈ Ω ,

i.e., v¯ is a subsolution to problem (3.7) and v¯≤v¯ on Ω. Thus, Lemma 2.7 implies that problem (3.7) has a solution for α∈(-1,0). The proof is complete. ∎

Proof of Theorem 1.4.

For the case β∈(0,1), recall that G2⁢(s)=λ1-β⁢s1-β, s≥0. Let

ϕ 4 ⁢ ( τ ) = ∫ 0 τ exp ⁡ ( G 2 ⁢ ( s ) ) ⁢ 𝑑 s , τ > 0 ,

and let Φ4 be the inverse of ϕ4, i,e., Φ4 satisfies

∫ 0 Φ 4 ⁢ ( t ) exp ⁡ ( G 2 ⁢ ( s ) ) ⁢ 𝑑 s = t , t > 0 .

It follows that

Φ 4 ′ ⁢ ( t ) = exp ⁡ ( - G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) , Φ 4 ⁢ ( t ) > 0 , t > 0 ,

Φ 4 ⁢ ( 0 ) = 0 , Φ 4 ′ ⁢ ( 0 ) = 1 , lim t → ∞ ⁡ Φ 4 ⁢ ( t ) = ∞ ,

Φ 4 ′′ ⁢ ( t ) = - λ ⁢ ( Φ 4 ⁢ ( t ) ) - β ⁢ exp ⁡ ( - 2 ⁢ G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) , t > 0 .

Let u-=Φ4⁢(v). Then we see that problem (1.1)- is equivalent to the following problem:

(3.8) - △ ⁢ v = b ⁢ ( x ) ⁢ g ⁢ ( v ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

with g⁢(t)=(Φ4⁢(t))-α⁢exp⁡(G2⁢(Φ4⁢(t))). Moreover, g⁢(t) has the following properties:

g ′ ⁢ ( t ) = λ ⁢ ( Φ 4 ⁢ ( t ) ) - β - α - α ⁢ ( Φ 4 ⁢ ( t ) ) - 1 - α ,
lim t → ∞ ⁡ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) t = lim t → ∞ ⁡ g ⁢ ( t ) t = 0 ,
lim t → 0 ⁡ Φ 4 ⁢ ( t ) t = 1 .

In fact, (1) is easily obtained. For (2) and (3), it follows from l’Hospital’s rule that

lim t → ∞ ⁡ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) t = lim t → ∞ ⁡ d d ⁢ t ⁢ ( exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) ) = λ ⁢ lim t → ∞ ⁡ ( Φ 4 ⁢ ( t ) ) - β = 0 ,

lim t → ∞ ⁡ g ⁢ ( t ) t = lim t → ∞ ⁡ g ′ ⁢ ( t ) = lim t → ∞ ⁡ ( Φ 4 ⁢ ( t ) ) - β - α ⁢ lim t → ∞ ⁡ ( λ - α ⁢ ( Φ 4 ⁢ ( t ) ) - ( 1 - β ) ) = 0

and

lim t → 0 ⁡ Φ 4 ⁢ ( t ) t = lim t → 0 ⁡ Φ 4 ′ ⁢ ( t ) = lim t → 0 ⁡ exp ⁡ ( - G 2 ⁢ ( Φ 4 ⁢ ( t ) ) ) = 1 .

Consequently,

(3.9) lim τ → 0 ⁡ ϕ 4 ⁢ ( τ ) τ = 1 .

In addition, when α>0, we see from the above properties of Φ4⁢(t) and (1) that there exists t0>0 sufficiently small such that

g ′ ⁢ ( t ) = ( Φ 4 ⁢ ( t ) ) - 1 - α ⁢ ( λ ⁢ ( Φ 4 ⁢ ( t ) ) 1 - β - α ) < 0 , t ∈ ( 0 , t 0 ) ,

i.e., g is strictly decreasing on (0,t0). Thus, (i) and (ii) in Theorem 1.4 follow from Lemma 2.5, with η=α.

Next, we prove the existence of solutions. Case 1: α=0. We note that g⁢(t)=exp⁡(G2⁢(Φ4⁢(t))), which is increasing on [0,∞) with g⁢(0)=1. It follows that

- Δ ⁢ u 0 = b ⁢ ( x ) ≤ b ⁢ ( x ) ⁢ g ⁢ ( u 0 ⁢ ( x ) ) , x ∈ Ω ,

i.e., u0 is a subsolution to problem (3.8), where u0 is given as in (b1). Moreover, using (2), we see that there exists M>1 sufficiently large such that

(3.10) g ⁢ ( t ⁢ | u 0 | ∞ ) ≤ t for all ⁢ t ≥ M .

Since g is increasing on [0,∞), we have that v¯=M⁢u0⁢(x) satisfies

- Δ ⁢ v ¯ = M ⁢ b ⁢ ( x ) ≥ b ⁢ ( x ) ⁢ g ⁢ ( M ⁢ | u 0 | ∞ ) ≥ b ⁢ ( x ) ⁢ g ⁢ ( M ⁢ u 0 ⁢ ( x ) ) , x ∈ Ω ,

i.e., v¯ is a supersolution to problem (3.8) and v¯≤v¯ on Ω¯. Lemma 2.7 implies that problem (3.8) has a classical solution v in the ordered interval [v¯,v¯]. Thus, problem (1.1)- has a classical solution u- satisfying Φ4⁢(c⁢u0⁢(x))≤u-≤Φ4⁢(M⁢u0⁢(x)) on Ω¯. Case 2: α>0. Using the above properties of Φ4⁢(t) and Lemma 2.5, one can see that the problem

- △ ⁢ v = b ⁢ ( x ) ⁢ ( Φ 4 ⁢ ( v ) ) - α , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 ,

has a unique classical solution v¯, which is a subsolution to problem (3.8).

In addition, in a similar way as (3.10), we see that there exists M>1 sufficiently large such that

M ≥ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( M ⁢ | v ¯ | ∞ ) ) ) ≥ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) ) ) , x ∈ Ω .

It follows that v¯=M⁢v¯ satisfies

- Δ ⁢ v ¯ = M ⁢ b ⁢ ( x ) ⁢ ( Φ 4 ⁢ ( v ¯ ⁢ ( x ) ) ) - α

≥ M ⁢ b ⁢ ( x ) ⁢ ( Φ 4 ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) ) - α

≥ b ⁢ ( x ) ⁢ ( Φ 4 ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) ) - α ⁢ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( M ⁢ | v ¯ | ∞ ) ) )

≥ b ⁢ ( x ) ⁢ ( Φ 4 ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) ) - α ⁢ exp ⁡ ( G 2 ⁢ ( Φ 4 ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) ) )

= b ⁢ ( x ) ⁢ g ⁢ ( M ⁢ v ¯ ⁢ ( x ) ) , x ∈ Ω ,

i.e., v¯ is a supersolution to problem (3.8) and v¯≤v¯ on Ω¯. Lemma 2.7 implies that problem (3.8) has a classical solution v in the ordered interval [v¯,v¯]. Thus, problem (1.1)- has a classical solution u- satisfying Φ4⁢(v0⁢(x))≤u-≤Φ4⁢(M⁢v0⁢(x)) on Ω¯. Case 3: α∈(-1,0). One can see that g is increasing on [0,∞). For the case α<0, combining with (1) and (2), we can see that v¯=M⁢u0⁢(x) is a supersolution to problem (3.8) when M>1 is sufficiently large.

To construct a subsolution v¯, let

φ 1 ⁢ ( τ ) = ∫ 0 τ d ⁢ s g ⁢ ( s ) , τ ≥ 0 .

We have φ1′⁢(τ)=1g⁢(τ)>0, τ>0, φ⁢(0)=0. It follows that φ1 is strictly increasing on [0,∞). Let Ψ1 be the inverse of φ1, i.e., Ψ1 satisfies

∫ 0 Ψ 1 ⁢ ( t ) d ⁢ s g ⁢ ( s ) = t , t ≥ 0 .

It follows that

Ψ 1 ′ ⁢ ( t ) = g ⁢ ( Ψ 1 ⁢ ( t ) ) , Ψ 1 ⁢ ( t ) > 0 , t > 0 , Ψ 1 ⁢ ( 0 ) = 0 .

Let v¯⁢(x)=Ψ1⁢(u0⁢(x)), x∈Ω. We obtain by a direct computation that

- Δ ⁢ u 0 = b ⁢ ( x ) = - Δ ⁢ v ¯ g ⁢ ( v ¯ ) + g ′ ⁢ ( v ¯ ) g 2 ⁢ ( v ¯ ) ⁢ | ∇ ⁡ v ¯ | 2 ≥ - Δ ⁢ v ¯ g ⁢ ( v ¯ ) , x ∈ Ω

and

- Δ ⁢ v ¯ ≤ b ⁢ ( x ) ⁢ g ⁢ ( v ¯ ) , x ∈ Ω , v ¯ | ∂ ⁡ Ω = 0 .

So v¯ is a subsolution to problem (3.8).

Next, we assert that v¯≤M⁢u0 in Ω, provided M≥supx∈Ω¯⁡Ψ1⁢(u0⁢(x))u0⁢(x). In fact, by (3), (3.9) and l’Hospital’s rule, we have that

lim t → 0 ⁡ Ψ 1 ⁢ ( t ) t = lim t → 0 ⁡ Ψ 1 ′ ⁢ ( t ) = lim t → 0 ⁡ g ⁢ ( Ψ 1 ⁢ ( t ) ) = 0 .

Therefore, maxx∈Ω¯⁡Ψ1⁢(u0⁢(x))u0⁢(x) exists and it is positive.

It follows from Lemma 2.7 that problem (3.8) has a classical solution v in the ordered interval [Ψ1⁢(u0),M⁢u0], i.e., problem (1.1)- has a classical solution in the ordered interval [Φ4⁢(Ψ1⁢(u0)),Φ4⁢(M⁢u0)]. The proof is complete. ∎

Proof of Theorem 1.5.

(i) and (ii) Since β=λ=1, let u+=∫w∞d⁢sexp⁡(s)=exp⁡(-w), and we see that problem (1.1)+ is equivalent to problem (1.4) with f⁢(w)=exp⁡((1+α)⁢w). Thus, the results (i) and (ii) follow from Lemma 2.2.

(iii) and (iv) Since β=1 and λ>1, let μ=λλ-1 and u+=∫w∞d⁢ssμ=w1-μμ-1, and we see that problem (1.1)+ is equivalent to problem (1.4) with f⁢(w)=(μ-1)α⁢wμ⁢(1+α)-α, where μ⁢(1+α)-α>1. By using Lemma 2.3, we see that problem (1.4) has a unique classical solution w satisfying

(3.11) lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) ⁢ ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 / ( μ - 1 ) ⁢ ( 1 + α ) = ( 2 ⁢ ( 2 + C θ ⁢ ( μ - 1 ) ⁢ ( 1 + α ) ) b 0 ⁢ ( μ - 1 ) α ⁢ ( μ - 1 ) 2 ⁢ ( 1 + α ) 2 ) 1 / ( μ - 1 ) ⁢ ( 1 + α ) ,

if b satisfies (b2) with b1=b2=b0, and

(3.12) lim d ⁢ ( x ) → 0 ⁡ w ⁢ ( x ) ⁢ ( h ⁢ ( d ⁢ ( x ) ) ) 1 / ( μ - 1 ) ⁢ ( 1 + α ) = ( b 0 ⁢ ( μ - 1 ) α ⁢ ( μ - 1 ) ⁢ ( 1 + α ) ) 1 / ( μ - 1 ) ⁢ ( 1 + α ) ,

provided b satisfies (b3) with b1=b2=b0. Putting λ=μμ-1 and w=(u+λ-1)-(λ-1) into (3.11) and (3.12), respectively, one can see that (iii) and (iv) hold.

(v) and (vi) By β=1 and λ∈(0,1), let κ=λ1-λ and u+=∫0vsκ⁢𝑑s=v1+κ1+κ, and we see that problem (1.1)+ is equivalent to the following problem:

(3.13) - △ ⁢ v = ( 1 - λ ) - α ⁢ b ⁢ ( x ) ⁢ v - ( λ + α ) / ( 1 - λ ) , v > 0 , x ∈ Ω , v | ∂ ⁡ Ω = 0 .

For λ>-α, we have from Lemma 2.5 that problem (3.13) has a unique classical solution v. Moreover, when Cθ⁢(1+α)+2⁢λ>2,

(3.14) lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 ⁢ ( 1 - λ ) / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) 2 2 ⁢ ( 1 - λ ) 1 + α ⁢ ( C θ ⁢ ( 1 + α ) + 2 ⁢ λ - 2 ) ) ( 1 - λ ) / ( 1 + α ) ,

if b satisfies (b2) with b1=b2=b0, and

(3.15) lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( h ⁢ ( d ⁢ ( x ) ) ) ( 1 - λ ) / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) ( 1 - λ ) 1 + α ) ( 1 - λ ) / ( 1 + α ) ,

provided b satisfies (b3) with b1=b2=b0. Putting v=(u+1-λ)1-λ into (3.14) and (3.15), respectively, one can also see that (v) and (vi) hold.

(vii) For λ=-α, it is obvious that u+=(u0⁢(1-λ))1/(1-λ) is the unique classical solution to problem (1.1)+, where u0 is given as in (b1).

(viii) For λ<-α, i.e., -(λ+α)/(1-λ)∈(0,1), we see that (viii) follows from Lemma 2.8. ∎

Proof of Theorem 1.6.

By β=1 and λ>0, as stated in the introduction, let q=λ1+λ and u-=∫0vd⁢ssq=v1-q1-q, and we see that problem (1.1)- is equivalent to problem (1.6). Case 1: λ=α. One can see that v=(1+λ)-λ⁢u0 is the unique solution to problem (1.6), i.e., we have that u-=(u0⁢(1+λ))1/(1+λ) is the unique solution to problem (1.1)-. Case 2: λ<α. We see from Lemma 2.5 that if b satisfies (b2) with b1=b2=b0 and Cθ⁢(1+α)-2⁢λ>2, then problem (1.6) has a unique classical solution v satisfying

lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( Θ ⁢ ( d ⁢ ( x ) ) ) 2 ⁢ ( 1 + λ ) / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) 2 2 ⁢ ( 1 + λ ) 1 + α ⁢ ( C θ ⁢ ( 1 + α ) - 2 ⁢ λ - 2 ) ) ( 1 + λ ) / ( 1 + α ) .

In addition, if b satisfies (b3) with b1=b2=b0, then problem (1.6) has a unique classical solution v satisfying

lim d ⁢ ( x ) → 0 ⁡ v ⁢ ( x ) ( h ⁢ ( d ⁢ ( x ) ) ) ( 1 + λ ) / ( 1 + α ) = ( b 0 ⁢ ( 1 + α ) ( 1 + λ ) 1 + α ) ( 1 + λ ) / ( 1 + α ) .

Hence, the results follow from v=(u-1+λ)1+λ. Case 3: λ>α. Since λ-α1+λ∈(0,1), one can see that the result follows from Lemmas 2.6 and 2.8 directly.

The proof is complete. ∎

Communicated by Julian Lopez Gomez

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 11571295

Funding statement: This work is supported in part by NSF of P. R. China under grant 11571295.

Acknowledgements

The author is greatly indebted to the anonymous referees for the very valuable suggestions and comments which improved the quality of the presentation.

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Received: 2019-04-29

Revised: 2019-06-24

Accepted: 2019-06-26

Published Online: 2019-08-27

Published in Print: 2020-02-01

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/ans-2019-2054

Keywords for this article

Semilinear Elliptic Equations; Singular Dirichlet Problems; Classical Solutions; Existence; Uniqueness; Asymptotic Behavior

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