Startseite Existence of solutions for a double-phase variable exponent equation without the Ambrosetti-Rabinowitz condition
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Existence of solutions for a double-phase variable exponent equation without the Ambrosetti-Rabinowitz condition

  • Jingjing Liu EMAIL logo und Patrizia Pucci
Veröffentlicht/Copyright: 27. Februar 2023

Abstract

The article deals with the existence of a pair of nontrivial nonnegative and nonpositive solutions for a nonlinear weighted quasilinear equation in R N , which involves a double-phase general variable exponent elliptic operator A . More precisely, A has behaviors like ξ q ( x ) 2 ξ if ξ is small and like ξ p ( x ) 2 ξ if ξ is large. Existence is proved by the Cerami condition instead of the classical Palais-Smale condition, so that the nonlinear term f ( x , u ) does not necessarily have to satisfy the Ambrosetti-Rabinowitz condition.

1 Introduction

The double-phase functional of the calculus of variations

(1.1) Ω ( D w p + a ( x ) D w q ) d x

was introduced by Zhikov in [25] in connection with homogenization, strongly anisotropic materials, elasticity, Lavrentiev phenomenon, and so on. Following [25], for strongly anisotropic materials, the regulation of the mixture between two different materials, with p and q hardening, is modulated by the coefficient a = a ( x ) . Independently, even without the presence of the modulating coefficient a = a ( x ) , for the regularity of minimizers, Marcellini in [15] proves that a condition of the type q p < 1 + o ( n ) is sufficient, while Giaquinta in [14] establishes the necessary part.

The articles [3,4] present sharp regularity theorems for minimizers of different double-phase integral models (1.1). In [7], the minimizers are proved to be C 1 , β -regular under appropriate conditions and further regularity results for related minimizers are given in [8] and in the more recent paper [10] and in the references therein.

Concerning existence results involving double-phase operators in bounded domains, we refer to [2,16], and in the whole space to [12], as well as to the references cited there.

In recent years, variable exponential problems have become a research hotpot because of its wide applications, for instance, in image restoration, image denoising, and image enhancement of image processing, etc. More application background can be found in [6,13,1921]. Variable exponential double-phase nonlinear problems describe objects, which contain two different materials involving power hardening exponents p ( x ) and q ( x ) . Energy functionals of this kind of problems are double-phase functionals, either isotropic or anisotropic. Indeed, these problems have been widely studied, and we refer to [5,18,22,23] and the references therein. In [22], Shi et al. deal with the equation:

div A ( x , u ) + u α ( x ) 2 u = λ a ( x ) u δ ( x ) 2 u + μ w ( x ) g ( x , u ) ,

with λ > 0 and μ 0 . They use the weighting method to overcome the lack of compactness and prove that the aforementioned equation has two pairs of nontrivial nonnegative and nonpositive solutions as well as infinitely many nontrivial solutions. Since they prove that the corresponding energy functional satisfies the Palais-Smale condition, in short PS, the function g ( x , u ) is required to satisfy the A-R condition. The previous article [23] deals with the following equation:

div A ( x , u ) + V ( x ) u α ( x ) 2 u = f ( x , u ) ,

where the differential operator div A ( x , u ) satisfies condition (1.2), which means the p ( ) -material is present if ξ > 1 and the q ( ) -material acts if ξ 1 . In [23], the weighted variable exponent spaces associated with p ( x ) and q ( x ) are introduced to prove two existence theorems of weak solutions without the A-R condition. In fact, this article improves [23], because there is one more term h ( x ) u r ( x ) 2 u in the equation ( ) than in the equation studied in [23]. Clearly, the addition of this term requires to use a new solution function space. Of course, although the proof ideas are somehow similar, the subsequent analysis and calculations are also different.

From now on, for v : R N R , we define

v + = ess sup x R N v ( x ) , v = ess inf x R N v ( x ) ,

and write v 1 v 2 to represent

ess inf x R N { v 2 ( x ) v 1 ( x ) } > 0 .

Motivated by [22] and [23], in this article, we study the equation

div A ( x , u ) + V ( x ) u α ( x ) 2 u + h ( x ) u r ( x ) 2 u = f ( x , u ) , ( )

where A : R N × R N R N admits a potential A , with respect to its second variable ξ , satisfying the following assumptions.

( A 1 ) the potential A = A ( x , ξ ) is a continuous function in R N × R N , with continuous derivative with respect to ξ , A = ξ A ( x , ξ ) , and verifies:

  1. A ( x , 0 ) = 0 and A ( x , ξ ) = A ( x , ξ ) , for all ( x , ξ ) R N × R N ;

  2. A ( x , ) is strictly convex in R N for all x R N ;

  3. there exist positive constants C 1 , C 2 and variable exponents p and q such that for all ( x , ξ ) R N × R N

    (1.2) C 1 ξ p ( x ) , if ξ > 1 C 1 ξ q ( x ) , if ξ 1 A ( x , ξ ) ξ ,

    and

    (1.3) A ( x , ξ ) C 2 ξ p ( x ) 1 , if ξ > 1 C 2 ξ q ( x ) 1 , if ξ 1 ;

  4. 1 p q min { N , p } in R N , and p , q are Lipschitz continuous functions in R N ;

  5. A ( x , ξ ) ξ s ( x ) A ( x , ξ ) for any ( x , ξ ) R N × R N , where s is a Lipschitz continuous function in R N , with q s p in R N ;

( A 2 ) A is uniformly convex, that is, for any ε ( 0 , 1 ) , there exists δ ( ε ) ( 0 , 1 ) such that either u v ε max { u , v } or A x , u + v 2 1 2 ( 1 δ ( ε ) ) ( A ( x , u ) + A ( x , v ) ) for any x , u , v R N .

When A ( x , u ) = u p ( x ) 2 u , if u > 1 u q ( x ) 2 u , if u 1 , then

div A ( x , u ) = div ( u p ( x ) 2 u ) , if u > 1 div ( u q ( x ) 2 u ) , if u 1

is an operator of p ( x ) -Laplacian type and the corresponding potential is

A ( x , ξ ) = 1 p ( x ) ξ p ( x ) + 1 q ( x ) 1 p ( x ) , if ξ > 1 , 1 q ( x ) ξ q ( x ) if ξ 1 .

Lemma A.2 in [23], see also Lemma 2.4.7 of [9], guarantees that the potential A satisfies ( A 1 ) and ( A 2 ), provided that 1 p q N in R N .

In addition, we also require the validity of

( V )

  1. V L loc 1 ( R N ) ;

  2. V V 0 > 0 a.e in R N ;

  3. V ( x ) as x .

( h )

  1. h L loc 1 ( R N ) ;

  2. h h 0 > 0 a.e. in R N ;

  3. h ( x ) as x .

( f 1 ) α , r C ( R N ) , with α , r > 1 , and α + , r + < , 0 f ( x , t ) t , f ( x , t ) t = o ( t α ( x ) ) and f ( x , t ) t = o ( t r ( x ) ) as t 0 , and f ( x , t ) C ( 1 + t γ ( x ) ) , where γ is a Lipschitz continuous function, with max { α , r } γ p .

( f 2 ) There exist positive constants M , C 1 , C 2 , and a function a q in R N such that for all t , with t M , and for all x R N

C 1 t q ( x ) [ ln ( e + t ) ] a ( x ) 1 C 2 t f ( x , t ) ln ( e + t ) t f ( x , t ) s ( x ) F ( x , t ) ,

where F ( x , t ) = 0 t f ( x , τ ) d τ .

The application of the mountain pass lemma to obtain a critical point of a functional φ , which is a nontrivial solution of the underlying equation, requires that the nonlinear term f satisfies the A-R condition [1], which ensures that the energy functional φ satisfies the PS condition, i.e., any sequence ( u n ) n such that φ ( u n ) c and φ ( u n ) 0 has a convergent subsequence. In this article, we use the Cerami condition (see Lemma 3.8) to replace the PS condition, so that f only needs to satisfy conditions ( f 1 )–( f 2 ), which are weaker than the A-R condition. For example, f ( x , t ) = t s ( x ) 2 t [ ln ( 1 + t ) ] a ( x ) , with s q , a q in R N , satisfies assumptions ( f 1 )–( f 2 ), but it does not verify the A-R condition.

Equation ( ) has been studied in [17], where α ( x ) = p ( x ) , f ( x , u ) = λ w ( x ) u q ( x ) 2 u and where div A ( x , u ) is only associated with the p ( x ) -Laplacian. More precisely, [17] deals with existence of entire solutions for it under suitable assumptions. In the present article, we study existence of a pair of nontrivial nonnegative and nonpositive solutions of ( ). For this purpose, we need to introduce an appropriate Sobolev space that matches the double-phase structure of the elliptic operator and the weighted term V ( x ) u α ( x ) 2 u . Meanwhile, we also need to consider the effects of h ( x ) u r ( x ) 2 u in the space. This brings new difficulties to handle. In addition, there are many differences in calculation from the previous work. In the rest of this article, we first present the weighted variable exponent spaces, involving the weighted term h ( x ) and then the mountain pass lemma. Our main result can be stated as follows:

Theorem 1.1

Assume that ( A 1 ) ( A 2 ) , ( V ) , ( h ) , and ( f 1 )–( f 2 ) hold, with

1 max { α , r } p q min { N , p } , 1 α p q p .

Then ( ) has a pair of nontrivial nonnegative and nonpositive solutions.

This article is divided into four sections. Section 2 presents some properties of function spaces with variable exponent; Section 3 presents some properties of functionals and operators; Section 4 presents the proof of Theorem 1.1.

2 Variable exponent space theory

Throughout the article, the letters c , c i , C , C i , i = 1 , 2 , , denote positive constants, which may vary from line to line, but are independent of the terms which will take part in any limit process.

To discuss the equation ( ), we need some properties of the variable exponent Lebesgue and Sobolev spaces. Let Ω R N be an open domain. Let S ( Ω ) be the set of all measurable real-valued functions defined on Ω . We write as follows:

C + ( Ω ¯ ) = { v v C ( Ω ¯ ) , v ( x ) > 1 for x Ω ¯ } , L p ( ) ( Ω ) = u S ( Ω ) Ω u ( x ) p ( x ) d x < ,

where p C + ( Ω ¯ ) . The space L p ( ) ( Ω ) is equipped with the norm

u L p ( ) ( Ω ) = inf λ > 0 Ω u ( x ) λ p ( x ) d x 1 ,

and ( L p ( ) ( Ω ) , u L p ( ) ( Ω ) ) is a Banach space, called generalized Lebesgue space. If Ω = R N , then we drop R N in our notations if there is no ambiguity. For example, we simply denote the space L p ( ) ( R N ) and the norm on it as L p ( ) and u L p ( ) .

Proposition 2.1

(Theorem 1.15 in [11]) The space ( L p ( ) ( Ω ) , u L p ( ) ( Ω ) ) is a separable, uniformly convex Banach space, and its conjugate space is L p ( ) ( Ω ) , where 1 p ( x ) + 1 p ( x ) = 1 . For any u L p ( ) ( Ω ) and v L p ( ) ( Ω ) ,

Ω u v d x 1 p + 1 p u L p ( ) ( Ω ) v L p ( ) ( Ω )

holds.

Proposition 2.2

(Theorem 1.16 in [11]) If f : Ω × R R is a Carathéodory function and satisfies

f ( x , t ) d ( x ) + b t p 1 ( x ) p 2 ( x ) for a n y x Ω , t R ,

where p 1 , p 2 C + ( Ω ¯ ) , d L p 2 ( ) ( Ω ) , d 0 , and b is a nonnegative number, then the Nemytsky operator from L p 1 ( ) ( Ω ) to L p 2 ( ) ( Ω ) defined by ( N f u ) ( x ) = f ( x , u ( x ) ) is a continuous bounded operator.

Proposition 2.3

(Theorem 1.3 in [11]) Put

ρ p ( ) ( u ) = Ω u p ( x ) d x

for all u L p ( ) ( Ω ) . Then

  1. u L p ( ) ( Ω ) < 1 ( = 1 ; > 1 ) ρ p ( ) ( u ) < 1 ( = 1 ; > 1 ) ;

  2. u L p ( ) ( Ω ) > 1 u L p ( ) ( Ω ) p ρ p ( ) ( u ) u L p ( ) ( Ω ) p + ; u L p ( ) ( Ω ) < 1 u L p ( ) ( Ω ) p ρ p ( ) ( u ) u L p ( ) ( Ω ) p + ;

  3. u L p ( ) ( Ω ) ρ p ( ) ( u ) .

Proposition 2.4

(Theorem 1.14 in [11]) If u , u k L p ( ) ( Ω ) , k = 1 , 2 , , then the following statements are equivalent to each other:

  1. lim k u k u L p ( ) ( Ω ) = 0 ;

  2. lim k ρ p ( ) ( u k u ) = 0 ;

  3. u k u in measure in Ω and lim k ρ p ( ) ( u k ) = ρ p ( ) ( u ) .

The variable exponent Sobolev space W 1 , p ( ) ( Ω ) is defined by

W 1 , p ( ) ( Ω ) = { u L p ( ) ( Ω ) u [ L p ( ) ( Ω ) ] N } ,

and it is equipped with the norm

u W 1 , p ( ) ( Ω ) = u L p ( ) ( Ω ) + u L p ( ) ( Ω )

for all u W 1 , p ( ) ( Ω ) . While W 0 1 , p ( ) ( Ω ) is the closure of C 0 ( Ω ) in W 1 , p ( ) ( Ω ) .

Since equation ( ) is a double-phase problem, we need to define the following variable exponent Sobolev space.

Definition 2.1

Assume ( A 1 )-(iv). Let the linear space

L p ( ) ( Ω ) + L q ( ) ( Ω ) = { u u = v + w , v L p ( ) ( Ω ) , w L q ( ) ( Ω ) }

be endowed with the norm

(2.1) u L p ( ) ( Ω ) + L q ( ) ( Ω ) = inf { v L p ( ) ( Ω ) + w L q ( ) ( Ω ) v L p ( ) ( Ω ) , w L q ( ) ( Ω ) , v + w = u } .

While the linear space

L p ( ) ( Ω ) L q ( ) ( Ω ) = { u u L p ( ) ( Ω ) and u L q ( ) ( Ω ) }

is endowed with the norm

u L p ( ) ( Ω ) L q ( ) ( Ω ) = max { u L p ( ) ( Ω ) , u L q ( ) ( Ω ) } .

Throughout this article, we denote

Λ u = { x Ω u ( x ) > 1 } and Λ u c = { x Ω u ( x ) 1 } .

Lemma 2.5

(Proposition 3.2 in [23]) Assume ( A 1 ) -(iv). Let Ω R N and let u be in L p ( ) ( Ω ) + L q ( ) ( Ω ) . Then the following properties hold:

  1. if Ω Ω is such that Ω < , then u L p ( ) ( Ω ) ;

  2. if Ω Ω is such that u L ( Ω ) , then u L q ( ) ( Ω ) ;

  3. Λ u < ;

  4. u L p ( ) ( Λ u ) L q ( ) ( Λ u c ) ;

  5. the infimum in (2.1) is attained;

  6. if B Ω , then u L p ( ) ( Ω ) + L q ( ) ( Ω ) u L p ( ) ( B ) + L q ( ) ( B ) + u L p ( ) ( Ω B ) + L q ( ) ( Ω B ) ;

  7. there exist ξ R N and a small positive constant c such that

    max 1 1 + 2 Λ u 1 p ( ξ ) 1 q ( ξ ) u L p ( ) ( Λ u ) , c min { u L q ( ) ( Λ u c ) , u L q ( ) ( Λ u c ) q ( ξ ) p ( ξ ) } u L p ( ) ( Ω ) + L q ( ) ( Ω ) u L p ( ) ( Λ u ) + u L q ( ) ( Λ u c ) .

Conditions ( A 1 )-(i) and (iii) imply that for all ( x , ξ ) R N × R N

A ( x , ξ ) = 0 1 d d t A ( x , t ξ ) d t = 0 1 1 t A ( x , t ξ ) t ξ d t C 1 p + ξ p ( x ) , ξ > 1 C 1 q + ξ q ( x ) , ξ 1 .

This estimate, in combination with (1.3) and ( A 1 )-(i) and (ii), yields

(2.2) c 1 ξ p ( x ) , ξ > 1 c 1 ξ q ( x ) , ξ 1 A ( x , ξ ) A ( x , ξ ) ξ C 2 ξ p ( x ) , ξ > 1 C 2 ξ q ( x ) , ξ 1 .

In what follows, we discuss ( ) in the variable exponent spaces ( E , E ) and ( X , X ).

Let E = { u L V α ( ) u ( L p ( ) + L q ( ) ) N } with the norm

u E = u L V α ( ) + u L p ( ) + L q ( ) ,

where

L V α ( ) = u S ( R N ) R N V ( x ) u ( x ) α ( x ) d x <

is endowed with the norm

u L V α ( ) = inf λ > 0 R N V ( x ) u ( x ) λ α ( x ) d x 1 .

Let

X = u E R N h ( x ) u ( x ) r ( x ) d x <

be endowed with the norm

u X = u E + u L h r ( ) = u L V α ( ) + u L p ( ) + L q ( ) + u L h r ( ) .

Lemma 2.6

(Lemma 2.2 in [17]) Assume ( V ) -(i), α > 1 and α + < . Then the space L V α ( ) ( R N ) is a separable uniformly convex Banach space.

Theorem 2.7

Assume ( A 1 ) -(iv) and ( V ) -(i). Then E is a reflexive uniformly convex Banach space, and X is a reflexive Banach space.

Proof

Propositions 3.10 and 3.11 in [23] imply that E is a reflexive Banach space under the conditions ( A 1 )-(iv) and ( V )-(i). Let us now prove that E is uniformly convex. It follows from Lemma 2.6 that L V α ( ) is a uniformly convex Banach space. On the space L p ( ) + L q ( ) , we consider the norm

u 2 = inf { ( v L p ( ) 2 + w L q ( ) 2 ) 1 2 v L p ( ) , w L q ( ) , v + w = u } ,

which is an equivalent norm on L p ( ) + L q ( ) . Hence, ( L p ( ) + L q ( ) , 2 ) is uniformly convex thanks to Propositions 3.3 and 3.5 of [23]. Similarly, ( L p ( ) + L q ( ) , # ) , with the equivalent norm,

u # = ( u L V α ( ) 2 + u 2 2 ) 1 2 ,

is uniformly convex.

The linear operator T : E L V α ( ) × ( L p ( ) + L q ( ) ) N , defined by T ( u ) = ( u , u ) for u E , is isometric. Hence, E is a closed subspace of L V α ( ) × ( L p ( ) + L q ( ) ) N . Thus, E is uniformly convex by Proposition 1.4.4 in [9].

For the second part of the proof, let Y = E × L h r ( ) , with the norm u Y = u E + u L h r ( ) . Because E and L h r ( ) are reflexive Banach spaces, so Y is reflexive. The operator T : X Y , given by T ( u ) = ( u , u ) , is well defined and linear. Note that X with the norm u Y , so T is an isometry. Thereby T ( X ) is a closed subspace of reflexive space Y , i.e. T ( X ) is reflexive. Thus, also ( X , Y ) = ( X , X ) is reflexive, being isomorphic to a reflexive space.□

Theorem 2.8

Assume that ( A 1 ) -(iv) and ( V ) -(i) and (ii) hold, with

1 p q p , 1 α p N 1 N , 1 α p q p .

Then the embeddings X E L p ( ) are continuous.

Proof

Lemma 2.4 of [17] assures that the embedding X E is continuous. Theorem 3.12 of [23] guarantees that the embedding E L p ( ) is continuous. This completes the proof.□

Let ψ C ( R N ) , with ψ ( x ) = 1 if x R N and x 1 , while ψ ( x ) = 0 if x R N and x 3 , and 0 ψ 1 and ψ ( x ) 1 in R N .

Let n N and ψ n ( x ) = ψ ( x n ) . Set A n = { x R N n x 3 n } , A n c = R N A n , B n = { x R N x n } , and B n c = { x R N x > n } .

Let u L and define u n = ψ n u ; thus, u n L and u n L u L . Clearly, u n has compact support and u n L V α ( ) . Since u n = ψ n u + u ψ n , then u n L p ( ) + L q ( ) if both terms of the sum are in L p ( ) + L q ( ) . Now, ψ n L and u L p ( ) + L q ( ) , so that ψ n u L p ( ) + L q ( ) . Since ψ n vanishes in A n c , A n < , ψ n ( L ) N and u L , then u ψ n L p ( ) + L q ( ) . In conclusion, u n X loc L .

Corollary 2.9

Assume that all the assumptions of Theorem 2.8are satisfied. Then, for any u X and ε > 0

  1. ψ n u u in X ;

  2. u ε = u j ε u in X as ε 0 + , where j ε ( x ) = ε N j ( x ε ) and j : R N R + is in C c ( R N ) and induces a probability measure;

  3. there exists a sequence ( u n ) n C c ( R N ) such that u n u in X .

Proof

Fix u X and ε > 0 .

  1. Theorem 2.8 yields that u L p ( ) . Let u n = ψ n u = ψ ( x n ) u , so that we have to prove that u n u in X , i.e., u n u X 0 . Recall that u n u X = u n u E + u n u L h r ( ) . Now, u n u in E by Corollary 3.13-(i) of [23]. Moreover,

    R N h ( x ) r ( x ) u u n r ( x ) d x B n c h ( x ) r ( x ) u r ( x ) d x = o n ( 1 ) ,

    where o n ( 1 ) is a vanishing function as n . Thus, u n u L h r ( ) 0 as n . In conclusion, u n = ψ n u u in X , as stated.

  2. Clearly, u ε u in L h r ( ) as ε 0 + thanks to the standard theory of mollifiers. Combining with Corollary 3.13-(ii) in [23], that is, that u ε u in E , we finally obtain that u ε u in X .

  3. The conclusion follows from (i) and (ii).□

Note that the space E in this article is X ( Ω ) in [23]. By Theorem 3.14 of [23], we have

Theorem 2.10

(Theorem 3.14 of [23]) Assume that all the hypotheses of Theorem 2.8hold and that s C ( R N ) , with α s p in R N .

  1. The embedding X L s ( ) ( R N ) is continuous.

  2. If Ω R N is bounded and furthermore s p , then embedding X L s ( ) ( Ω ) is compact.

  3. If moreover ( V )-(iii) holds, s is Lipschitz continuous and

    α s p i n R N ,

    then the embedding X L s ( ) ( R N ) is compact.

3 Properties of functionals and operators

Before proving the main properties of the underlying functionals and operators associated to ( ), let us begin with the definition of a (weak) solution of ( ).

Definition 3.1

We say that u X is a solution of ( ) if

R N A ( x , u ) v d x + R N V ( x ) u α ( x ) 2 u v d x + R N h ( x ) u r ( x ) 2 u v d x = R N f ( x , u ) v d x

for all v X .

Fix u X . It follows from (2.2) that

R N A ( x , u ) u d x + R N V ( x ) u α ( x ) d x c 1 R N Λ u u p ( x ) d x + R N Λ u c u q ( x ) d x + R N V ( x ) u α ( x ) d x ,

and

R N A ( x , u ) u d x + R N V ( x ) u α ( x ) d x c 2 R N Λ u u p ( x ) d x + R N Λ u c u q ( x ) d x + R N V ( x ) u α ( x ) d x ,

where c 1 and c 2 are positive constants.

Similarly, (2.2) implies that

R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x c 1 R N Λ u c u q ( x ) d x + R N V ( x ) α ( x ) u α ( x ) d x

and

R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x c 2 R N Λ u u p ( x ) d x + R N Λ u c u q ( x ) d x + R N V ( x ) α ( x ) u α ( x ) d x .

According to Definition 3.1, the energy functional Φ : X R associated to equation ( ) is

Φ ( u ) = R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x + R N h ( x ) r ( x ) u r ( x ) d x R N F ( x , u ) d x .

For simplicity, we write Φ = Φ A + Φ α + Φ r Φ F , where

Φ A ( u ) = R N A ( x , u ) d x , Φ α ( u ) = R N V ( x ) α ( x ) u α ( x ) d x , Φ r ( u ) = R N h ( x ) r ( x ) u r ( x ) d x , Φ F ( u ) = R N F ( x , u ) d x .

Lemma 3.1

(Lemma 4.1 of [23]) Assume the structure conditions ( A 1 ) and ( V ) -(i). Then the functional Φ A is convex, of class C 1 ( X ) , and sequentially weakly lower semicontinuous in X. Moreover, Φ A : X X is bounded.

Similar to the proofs of Lemmas 4.2–4.7 in [23], we obtain the following series of results.

Lemma 3.2

Let assume that conditions ( A 1 ) , ( V ) -(i), and ( h ) -(i) hold. Then the functionals Φ α and Φ r are convex, of class C 1 ( X ) and sequentially weakly lower semicontinuous. Moreover, if u n , u X and u n u in X, then Φ α ( u n ) Φ α ( u ) in X and Φ r ( u n ) Φ r ( u ) in X .

Lemma 3.3

Assume all the conditions of Theorem 2.10-(iii), ( f 1 ), that γ is Lipschitz continuous and α γ p in R N . Then the Nemytsky operator ( N f u ) ( x ) = f ( x , u ( x ) ) is weakly continuous from X to X , that is, N f u n N f u in X if u n u .

Lemma 3.4

Assume all the hypotheses of Theorem 2.10-(iii) and ( f 1 ) . Then Φ F is of class C 1 ( X ) and sequentially weakly continuous, that is, if u n u in X, then Φ F ( u n ) Φ F ( u ) and Φ F ( u n ) Φ F ( u ) in X .

Lemma 3.5

Assume conditions of Theorem 2.10-(iii) and ( f 1 ) . Then the functional Φ is of class C 1 ( X ) and sequentially weakly lower semicontinuous in X, that is, if u n u in X, then

Φ ( u ) lim inf n Φ ( u n ) .

Lemma 3.6

Suppose that A satisfies ( A 1 ) and ( A 2 ) . Then the next properties hold.

  1. Φ A : X R is uniformly convex, that is, for any ε ( 0 , 1 ) , there exists δ ( ε ) ( 0 , 1 ) such that for all u , v X either

    Φ A u v 2 ε Φ A ( u ) + Φ A ( v ) 2 or Φ A u + v 2 ( 1 δ ( ε ) ) Φ A ( u ) + Φ A ( v ) 2 ;

  2. if u n u in X and lim ¯ n ( Φ A ( u n ) Φ A ( u ) , u n u ) 0 , then Φ A ( u n u ) 0 and u n u L p ( ) + L q ( ) 0 .

Let ρ : X R be defined as follows:

ρ ( u ) = R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x + R N h ( x ) r ( x ) u r ( x ) d x ,

and let us denote its derivative operator by L , that is, L = ρ : X X and

( L ( u ) , v ) = R N A ( x , u ) v d x + R N V ( x ) u α ( x ) 2 u v d x + R N h ( x ) u r ( x ) 2 u v d x ,

for all u , v X .

Lemma 3.7

Under the structure assumptions ( A 1 ) and ( V ) -(i), the following properties hold.

  1. L : X X is a continuous, bounded strictly monotone operator. If ( A 2 ) is also satisfied, then

  2. L is a mapping of type ( S + ) , that is, if u n u in X and lim ¯ n ( L ( u n ) L ( u ) , u n u ) 0 , then u n u in X ;

  3. L : X X is a homeomorphism.

Let us recall that a functional φ satisfies the Cerami condition at a level c R , if ( u n ) n X , with φ ( u n ) c and φ ( u n ) X ( 1 + u n X ) 0 , has a strongly convergent subsequence.

Lemma 3.8

Assume the structure conditions ( A 1 ) ( A 2 ) , ( V ) , ( h ) , ( f 1 ) ( f 2 ) and that

1 α p q p a n d max { α , r } p

hold. Then Φ satisfies the Cerami compactness condition.

Proof

Let ( u n ) n X be a Cerami sequence at level c , that is,

Φ ( u n ) c , Φ ( u n ) X ( 1 + u n X ) 0 .

If ( u n ) n is bounded in X , then there exists a subsequence, still denoted by ( u n ) n such that u n u in X . By Theorem 2.10 and ( f 1 ) the embedding X L γ ( ) is compact. Hence, Φ f ( u n ) Φ f ( u ) in X . Lemma 3.7 guarantees that L 1 is continuous from X to X . From Φ ( u n ) = L ( u n ) Φ F ( u n ) 0 , it follows that u n L 1 Φ F ( u ) in X .

Thus, it just remains to prove that ( u n ) n is bounded in X . Otherwise, there exists a subsequence, still called ( u n ) n for simplicity such that u n X . Note that

1 s ( x ) u n L s ( ) 1 s u n L s ( ) , 1 s ( x ) u n L s ( ) 1 s u n L s ( ) + C u n L s ( ) .

Thus, 1 s ( x ) u n X C u n X . It follows that ( Φ ( u n ) , 1 s ( x ) u n ) 0 . Therefore, we have

c + 1 Φ ( u n ) Φ ( u n ) , 1 s ( x ) u n = R N A ( x , u n ) d x + R N V ( x ) α ( x ) u n α ( x ) d x + R N h ( x ) r ( x ) u r ( x ) d x R N F ( x , u n ) d x R N 1 s ( x ) A ( x , u n ) u n d x + R N V ( x ) s ( x ) u n α ( x ) d x + R N h ( x ) s ( x ) u n r ( x ) d x R N 1 s ( x ) f ( x , u n ) u n d x R N 1 s 2 ( x ) u n A ( x , u n ) s d x R N 1 s 2 ( x ) u n A ( x , u n ) s d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x + R N 1 s ( x ) f ( x , u n ) u n F ( x , u n ) d x .

Now, for σ ( 0 , 1 ] sufficiently small, there exists C ( σ ) > 0 such that

(3.1) R N f ( x , u n ) u n s ( x ) F ( x , u n ) d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C 1 R N u n A ( x , u n ) d x + 1 σ R N Λ u n A ( x , u n ) p ( x ) p ( x ) 1 ln ( e + u n ) d x + C ( σ ) R N Λ u n u n p ( x ) [ ln ( e + u n ) ] p ( x ) 1 d x + σ R N Λ u n c A ( x , u n ) q ( x ) q ( x ) 1 ln ( e + u n ) d x + C ( σ ) R N Λ u n c u n q ( x ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 1 σ C R N A ( x , u n ) u n ln ( e + u n ) d x + C ( σ ) R N Λ u n u n p ( x ) [ ln ( e + u n ) ] p ( x ) 1 d x + C ( σ ) R N Λ u n c u n q ( x ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 1 .

Moreover,

u n ln ( e + u n ) X and u n ln ( e + u n ) X C 2 u n X .

Therefore, taking u n ln ( e + u n ) as a test function, we obtain as n

R N f ( x , u n ) u n ln ( e + u n ) d x = R N A ( x , u n ) u n ln ( e + u n ) d x + R N V ( x ) u n α ( x ) ln ( e + u n ) d x + R N h ( x ) u n r ( x ) ln ( e + u n ) d x + o ( 1 ) = R N A ( x , u n ) u n ln ( e + u n ) d x R N u n A ( x , u n ) u n ( e + u n ) [ ln ( e + u n ) ] 2 d x + R N V ( x ) u n α ( x ) ln ( e + u n ) d x + R N h ( x ) u n r ( x ) ln ( e + u n ) d x + o ( 1 ) .

From

1 2 1 ln ( e + u n ) u n ( e + u n ) [ ln ( e + u n ) ] 2 1 2 1 ln ( e + u n ) ,

we have

(3.2) 1 2 R N A ( x , u n ) u n ln ( e + u n ) d x + R N V ( x ) u n α ( x ) ln ( e + u n ) d x + R N h ( x ) u n r ( x ) ln ( e + u n ) d x C 4 R N f ( x , u n ) u n ln ( e + u n ) d x 3 2 R N A ( x , u n ) u n ln ( e + u n ) d x + R N V ( x ) u n α ( x ) ln ( e + u n ) d x + R N h ( x ) u n r ( x ) ln ( e + u n ) d x + C 6 .

Combining (3.1)–(3.2) and condition ( f 2 ), we obtain

R N f ( x , u n ) u n ln ( e + u n ) d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C u n M ( u n α ( x ) + u n r ( x ) + u n γ ( x ) ) d x ( f 2 ) C 7 R N f ( x , u n ) u n s ( x ) F ( x , u n ) d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x ( 3.1 ) C 7 σ R N A ( x , u n ) u n ln ( e + u n ) d x + C ( σ ) R N Λ u n u n p ( x ) [ ln ( e + u n ) ] p ( x ) 1 d x + C ( σ ) R N Λ u n c u n q ( x ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 8 ( 3.2 ) 1 2 R N f ( x , u n ) u n ln ( e + u n ) d x R N V ( x ) u n α ( x ) ln ( e + u n ) d x R N h ( x ) u n r ( x ) ln ( e + u n ) d x + C ( σ ) R N Λ u n u n p ( x ) [ ln ( e + u n ) ] p ( x ) 1 d x + C ( σ ) R N Λ u n c u n q ( x ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 10 .

By this inequality and condition ( f 2 ), we have

R N u n q ( x ) [ ln ( e + u n ) ] a ( x ) 1 d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C u n M ( u n α ( x ) + u n r ( x ) + u n γ ( x ) ) d x C 11 R N f ( x , u n ) u n ln ( e + u n ) d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C u n M ( u n α ( x ) + u n r ( x ) + u n γ ( x ) ) d x C 12 R N Λ u n u n p ( x ) [ ln ( e + u n ) ] p ( x ) 1 d x + R N Λ u n c u n q ( x ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 12 C 12 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 12 .

Since V ( x ) and h ( x ) as x , we deduce that

R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C u n M ( u n α ( x ) + u n r ( x ) + u n γ ( x ) ) d x 1 2 R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + 1 2 R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C 0 .

Hence,

(3.3) R N u n q ( x ) [ ln ( e + u n ) ] a ( x ) 1 d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C 11 R N f ( x , u n ) u n ln ( e + u n ) d x + R N 1 α ( x ) 1 s ( x ) V ( x ) u n α ( x ) d x + R N 1 r ( x ) 1 s ( x ) h ( x ) u n r ( x ) d x C 12 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 12 .

Let us divide the remaining part of the proof into two steps.

Step 1. The sequence R N f ( x , u n ) u n ln ( e + u n ) d x n is unbounded.

If not, R N f ( x , u n ) u n ln ( e + u n ) d x n is bounded. Fix ε , with

0 < ε < min { 1 , α 1 , r 1 , 1 p + } .

Since Φ ( u n ) X ( 1 + u n X ) 0 , we obtain as n

R N A ( x , u n ) u n d x + R N V ( x ) u n α ( x ) d x + R N h ( x ) u n r ( x ) d x = R N f ( x , u n ) u n d x + o ( 1 ) = R N [ f ( x , u n ) u n ] ε [ ln ( e + u n ) ] 1 ε f ( x , u n ) u n ln ( e + u n ) 1 ε d x + o ( 1 ) C 9 ( 1 + u n X ) 1 + ε R N [ f ( x , u n ) u n ] 1 + ε + C 10 ( 1 + u n ) 1 + ε ε ε f ( x , u n ) u n ln ( e + u n ) 1 ε d x + o ( 1 ) C 11 ( 1 + u n X ) 1 + ε + C 12 .

On the other hand, we have

R N A ( x , u n ) u n d x + R N V ( x ) u n α ( x ) d x + R N h ( x ) u n r ( x ) d x C Λ u n u n p ( x ) d x + Λ u n c u n q ( x ) d x + R N V ( x ) u n α ( x ) d x + R N h ( x ) u n r ( x ) d x C ( u n L p ( ) + L q ( ) p + u n L V α ( ) α + u n L h r ( ) r 2 ) C min { u n X α , u n X r } 3 C .

Hence, the sequence ( u n ) n is bounded, which is an obvious contradiction.

Consequently, the sequence R N f ( x , u n ) u n ln ( e + u n ) d x n is unbounded as stated. This implies that also

(3.4) R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x n is unbounded.

Since a q in R N , there exists a positive constant M # > 1 + M , where M is defined in ( f 2 ), such that

[ ln ( e + t ) ] a ( x ) q ( x ) 14 C 12 ,

for all t , with t M # and all x R N .

Step 2. We claim that

lim ¯ n u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x < 1 2 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x .

If not, up to a subsequence, still denoted by ( u n ) n , we obtain

(3.5) u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x 1 3 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x .

Combining (3.4) and (3.5), we obtain

(3.6) u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x as n .

The definition of M # and (3.5) imply

14 C 12 u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] a ( x ) 1 d x R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] a ( x ) 1 d x 2 C 12 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 12 12 C 12 u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x + C 12 .

Hence, u n M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x C 12 for all n . This contradicts (3.6) and proves Step 2.

Consequently, for n large enough, we have

u n < M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x 1 2 R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x .

The aforementioned inequality and (3.3) give

R N [ u n < M # ] 1 α ( x ) 1 s ( x ) V ( x ) u n M # p ( x ) d x + R N [ u n < M # ] 1 r ( x ) 1 s ( x ) h ( x ) u n M # p ( x ) d x R N [ u n < M # ] 1 α ( x ) 1 s ( x ) V ( x ) u n M # α ( x ) d x + R N [ u n < M # ] 1 r ( x ) 1 s ( x ) h ( x ) u n M # p ( x ) d x 2 C 12 R N [ u n < M # ] ( M # ) p ( x ) u n M # p ( x ) + ( M # ) q ( x ) u n M # q ( x ) [ ln ( e + u n M # ) ] q ( x ) 1 d x + C 12 2 C 12 R N [ u n < M # ] ( M # ) p + u n M # p ( x ) + ( M # ) q + u n M # q ( x ) [ ln ( e + u n M # ) ] q ( x ) 1 d x + C 12 4 ( M # ) 2 q + + 1 C 12 R N [ u n < M # ] u n M # p ( x ) d x + C 12 .

Since V ( x ) and h ( x ) as x , there exists D > 0 such that

1 α ( x ) 1 s ( x ) V ( x ) > 5 ( M # ) 2 q + C 12 and 1 r ( x ) 1 s ( x ) h ( x ) > 5 ( M # ) 2 q + C 12 ,

for all x , with x > D . Hence,

( M # ) 2 q + C 12 R N [ u n < M # ] ( M # ) q + u n M # p ( x ) d x C 12 .

Therefore, the sequence R N [ u n < M # ] ( M # ) q + u n M # q ( x ) d x n is bounded. Thus, also the sequence ( u n < M # ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x ) n is bounded. Finally, the sequence ( R N ( u n p ( x ) + u n q ( x ) ) [ ln ( e + u n ) ] q ( x ) 1 d x ) n is bounded. This contradicts (3.4) and completes the proof of the lemma.□

For any x 0 R N , ε , δ , θ , with 0 < δ < ε and 0 < θ < π 2 , put

B ( x 0 , ε , δ , θ ) = x R N δ x x 0 ε , x x 0 x x 0 p ( x 0 ) p ( x 0 ) cos θ .

Lemma 3.9

(Lemma 2.8 in [24]) If p C 1 ( Ω ¯ ) , x 0 Ω satisfies p ( x 0 ) 0 , then there exists a small positive ε such that

( x x 0 ) p ( x ) > 0 for a l l x B ( x 0 , ε , δ , θ ) a n d max { p ( x ) x B ( x 0 , ε ) ¯ } = max { p ( x ) x B ( x 0 , ε , δ , θ ) , x x 0 = ε } .

Lemma 3.10

Suppose that F = F ( x , t ) satisfies

C 1 t q ( x ) [ ln ( e + t ) ] a ( x ) F ( x , t ) for a l l x R N a n d t , w i t h t M ,

where a q in R N , and x 0 R N with p ( x 0 ) 0 . Let

v ( x ) = 0 , x x 0 > ε ε x x 0 , x x 0 ε ,

where ε is defined in Lemma 3.9. Then v X and

R N A ( x , t v ) d x + R N V ( x ) α ( x ) t v α ( x ) d x + R N h ( x ) r ( x ) t v r ( x ) d x R N F ( x , t v ) d x

as t .

Proof

Since max { p , α , r } q and a q ,

C B ( x 0 , ε ) t p ( x ) d x + B ( x 0 , ε ) V ( x ) α ( x ) t v α ( x ) d x + R N h ( x ) r ( x ) t v r ( x ) d x B ( x 0 , ε ) C 1 t v q ( x ) [ ln ( e + t v ) ] a ( x ) d x

as t . Moreover,

Ω A ( x , t v ) d x C B ( x 0 , ε ) t p ( x ) d x for all t 1 .

This completed the proof of the lemma.□

4 Proof of Theorem 1.1

To prove Theorem 1.1, we first consider the following auxiliary problem:

div A ( x , u ) + V ( x ) u α ( x ) 2 u + h ( x ) u r ( x ) 2 u = f + ( x , u ) , ( + )

where

f + ( x , u ) = f ( x , u ) , if f ( x , u ) 0 0 , if f ( x , u ) < 0 .

Then, the corresponding energy functional in X is

Φ + ( u ) = R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x + R N h ( x ) r ( x ) u r ( x ) d x R N F + ( x , u ) d x ,

where F + ( x , u ) = 0 u f + ( x , t ) d t . With a proof similar to the that of Lemma 3.8, the functional Φ + is shown to satisfy the Cerami condition. Then, we shall prove that Φ + verifies the assumptions of the mountain pass lemma. Indeed, a direct computation implies that Φ + ( 0 ) = 0 . Lemma 3.10 shows that Φ + ( t v ) as t . Next, we only need to prove that there exist ϱ > 0 and δ > 0 such that Φ + ( u ) δ > 0 for every u X and u X = ϱ .

Since max { α , r } γ p and γ C ( R N ) is Lipschitz continuous, the embedding X L γ ( ) is compact by Theorem 2.10. Consequently, there exists C 0 > 0 such that

u L γ ( ) C 0 u X , for all u X .

Assumption ( f 1 ) ensures the existence of σ , ς ( 0 , min { V 0 4 α + , h 0 4 r + } ) , where V 0 and h 0 are given in conditions ( V )-(ii) and ( h )-(ii), such that

F + ( x , t ) σ 1 α ( x ) t α ( x ) + ς 1 r ( x ) t r ( x ) + C ( σ , ς ) t γ ( x )

for all ( x , t ) R N × R . A direct computation shows that

R N A ( x , u ) d x + R N V ( x ) α ( x ) u α ( x ) d x + R N h ( x ) r ( x ) u r ( x ) d x σ R N 1 α ( x ) u α ( x ) d x ς R N 1 r ( x ) u r ( x ) d x 3 4 R N A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x .

Now, we choose a sequence of equal small open disjoint cubes Ω i , i N , such that R N = i = 1 Ω i ¯ and

sup Ω i q ( x ) < inf Ω i γ ( x ) sup Ω i γ ( x ) < inf Ω i p ( x ) .

Let

ε inf i N { inf Ω i γ ( x ) sup Ω i q ( x ) } .

Clearly, ε > 0 as long as the side length of Ω i is made sufficiently small.

We denote u X ( Ω i ) as the norm of u X restricted to Ω i , that is,

u X ( Ω i ) = u E ( Ω i ) + u L h r ( ) ( Ω i ) = u L V α ( ) ( Ω i ) + u L p ( ) ( Ω i ) + L q ( ) ( Ω i ) + u L h r ( ) ( Ω i ) .

Obviously, u X ( Ω i ) u X . Moreover, there exists ξ i Ω i ¯ such that

u L p ( ) ( Ω i ) γ ( ξ i ) = Ω i u γ ( x ) d x

and

u X ( Ω i ) q + Ω i A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x .

If u X is small enough, then we can obtain

C ( σ ) R N u γ ( x ) d x = C ( σ ) i = 1 Ω i u γ ( x ) d x = C ( σ ) i = 1 u L γ ( ) ( Ω i ) γ ( ξ i ) ( where ξ i Ω i ¯ ) C i = 1 u X ( Ω i ) γ ( ξ i ) ( see Corollary 8.3.2 of [9] ) C u X ε i = 1 u X ( Ω i ) q + = C u X ε i = 1 Ω i A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x = C u X ε R N A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x 1 4 R N A ( x , u ) + V ( x ) p ( x ) u p ( x ) + h ( x ) r ( x ) u r ( x ) d x .

Thus, when u X is small enough, we have

Φ + ( u ) R N A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x σ R N 1 α ( x ) u α ( x ) d x C ( σ ) R N u γ ( x ) d x 1 2 R N A ( x , u ) + V ( x ) α ( x ) u α ( x ) + h ( x ) r ( x ) u r ( x ) d x .

This inequality implies that there exist r > 0 and δ > 0 such that Φ + ( u ) δ > 0 for every u X and u X = r . In conclusion, the functional Φ + satisfies all assumptions of the mountain pass lemma. Hence, ( + ) has a solution u 0 , and so ( ) has a nontrivial nonnegative solution.

Moreover, if f + ( x , u ) is replaced by f ( x , u ) in ( + ) , with

f ( x , u ) = f ( x , u ) , if f ( x , u ) 0 0 , if f ( x , u ) > 0 ,

then, in a similar way, we obtain a nontrivial nonpositive solution of ( ). This complete the proof of Theorem 1.1.

Acknowledgments

J. Liu is partly supported by the National Science Foundation of China (11701525 and 11971446). P. Pucci is a member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM) and is partly supported by the INdAM – GNAMPA Project 2022 Equazioni differenziali alle derivate parziali in fenomeni non lineari (CUP_E55F22000270001). P. Pucci was also partly supported by the Fondo Ricerca di Base di Ateneo – Esercizio 2017–2019 of the University of Perugia, named PDEs and Nonlinear Analysis.

  1. Conflict of interest: Prof. Pucci, who is the author of this article, is a current Editorial Board member of the journal Advances in Nonlinear Analysis. This fact did not affect the peer-review process. The author declare no other conflict of interest.

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Received: 2022-08-25
Revised: 2023-01-09
Accepted: 2023-01-19
Published Online: 2023-02-27

© 2023 the author(s), published by De Gruyter

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

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