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On singular double phase problem with variable exponents and convolution term

  • Rui He ORCID logo and Sihua Liang EMAIL logo
Published/Copyright: February 18, 2026

Abstract

In this article, we consider the following singular double phase problem with variable exponents and convolution term of the form:

T p ( x ) , q ( x ) α ( x ) ( u ) = λ s ( x ) u γ ( x ) + Ω | u ( x ) | h ( x ) | x y | μ ( x , y ) | u ( y ) | h ( x ) 2 u ( y ) in Ω , u = 0 on Ω ,

where the operator T p ( x ) , q ( x ) α ( x ) ( u ) : = div u p ( x ) 2 u + α ( x ) u q ( x ) 2 u is the double phase operator with variable exponents, Ω R N is a bounded domain with smooth boundary ∂Ω, 0 ≤ α(⋅) ∈ L (Ω), λ is a positive real parameter. The functions s ( x ) C ( Ω ̄ ) are positive with compact support in Ω, h : R N R and μ : R N × R N R are continuous functions. Under the suitable conditions, the existence of at least one weak solution is obtained for the above problem by using the Nehari manifold approach. The novelty of this paper is that this problem includes singular term and convolution term. Moreover, the emergence of p(x) and q(x) Laplacian operator makes the study of this problem more complicated and interesting.

MSC 2020: 35J30; 35J75; 35D30

1 Introduction

In this paper, we investigate some existence results for the following singular double phase problem with variable exponents and convolution term of the form:

(1.1) T p ( x ) , q ( x ) α ( x ) ( u ) = λ s ( x ) u γ ( x ) + Ω | u ( x ) | h ( x ) | x y | μ ( x , y ) | u ( y ) | h ( x ) 2 u ( y ) in Ω , u = 0 on Ω ,

where the operator T p ( x ) , q ( x ) α ( x ) ( u ) : = div u p ( x ) 2 u + α ( x ) u q ( x ) 2 u is the double phase operator with variable exponents, Ω R N is a bounded domain with smooth boundary ∂Ω, 0 ≤ α(⋅) ∈ L (Ω), λ is a positive real parameter. The function s C ( Ω ̄ ) is positive with compact support in Ω, μ : R N × R N R is continuous function specified later and C may denote a positive constant and the same C may represent different constant.

Furthermore, throughout this paper, we denote with r + : = max x Ω ̄ r ( x ) and r : = min x Ω ̄ r ( x ) for any function r C ( Ω ̄ ) , and we consider the following hypotheses:

  1. 1 < p(x) < N and 0 < 1 − γ(x) ≪ p(x) ≪ q(x) ≪ h(x) ≪ p*(x) ≔ Np(x)/(Np(x)) for a.e. x ∈ Ω, where p*(x) ≔ Np(x)/(Np(x)) is the critical Sobolev exponent for variable exponents, the notation 1(x) ≪ 2(x) means that inf x Ω ( 2 ( x ) 1 ( x ) ) > 0 . Moreover, we assume that

    q + 1 γ + < 2 h q + 2 h + + γ 1 2 h + γ + 1 2 h + p .

  2. μ : R N × R N R is a continuous function such that

    1 < μ μ + < N

    and there exists a function q C + R N such that

    1 p ( x ) + μ ( x , y ) N + 1 q ( y ) = 2 , x , y R N .

The study of this problem (1.1) is based on two reasons. On the one hand, double-phase problem has a wealthy physical background, on the grounds that the double phase operator has been utilized to describe steady-state solutions of reaction diffusion problems in biophysics, plasma physics, and chemical reaction analysis. The research on the double phase problems and applicable functionals originates from the seminal paper Zhikov [1]. Recently, Ambrosio and Rădulescu [2] had been concerned with a classification of fractional problems with uneven growth as follows:

( Δ ) p s u + ( Δ ) q s u + V ( ε x ) | u | p 2 u + | u | q 2 u = f ( u ) in R N , u W s , p R N W s , q R N , u > 0 in R N ,

where 0 < s < 1, ɛ > 0, ( Δ ) t s is the fractional t-Laplacian operator, 2 p < q < N s and t ∈ {p, q}. For ɛ > 0 small enough, the authors showed multiple positive solutions and associated concentration properties related to the set of potential V reaching its minimum by suitable variational and topological arguments. In particular, the authors derived that there are at least five nontrivial smooth solutions to superlinear (p, q)-equations with indefinite potential and a concave boundary term, and they are all sign-informative and linearly ordered for small values of all parameters in [3].

In recent years, we note that the (p, q)-Laplacian equation has been extended to the (p(x), q(x))-Laplacian problem. For example, Alves et al. [4] studied the following equation:

Δ p ( x ) u = λ u α 1 ( x ) v β 1 ( x ) in Ω , Δ q ( x ) v = λ u α 2 ( x ) v β 2 ( x ) in Ω , u , v > 0 in Ω , u , v = 0 on Ω ,

where Ω is a bounded domain in R N ( N > 2 ) with C 2 boundary ∂Ω and λ > 0 is a parameter. Here, Δ p(x) and Δ q(x) stand for the p(x)-Laplacian and q(x)-Laplacian operators, respectively. By using sub and supersolution methods, the authors derived the existence and regularity of solutions for the problem.

In addition, the variable exponential growth problem with singular nonlinear terms is widely used in non-Newtonian mechanics, cosmology, elasticity theory, electromagnetism and so on. This kind of problem has been widely studied in the past 30 years. In [5], Arora et al. proved the existence of two weak solutions for the singular double phase problem with variable exponents by the way of using the Nehari manifold technique primarily based on fibering maps. For some other interesting results can be found in [2],[6], [7], [8], [9], and the references therein.

On the other hand, the nonlinearity on the right-hand side of problem (1.1) is motivated by the equation

Δ u + V u = I α u p u p 1 in R N ,

where I α : R N R is a Riesz potential defined for x R N \ { 0 } by I α ( x ) = A α | x | N α . Choquard delivered this equation to describe the electrons trapped in their own holes. Such equations also appear in the theory of stationary polarons in [10] and in the simulation of self-gravitating matter in [11]. Edmunds and Rákosník in [12] studied the existence and uniqueness of the minimizing solution of the problem

Δ u + u = | x | μ F ( u ) f ( u ) in R N ,

where f has subcritical growth. For critical case, Gao and Yang in [13],14] considered the following Brézis-Nirenberg type problem of the nonlinear Choquard equation:

Δ u = Ω | u | 2 μ * | x y | μ d y u 2 μ * + λ u ,

where 2 μ * = 2 N μ N 2 , 0 < μ < N, the multiplicity results are obtained for this problem by using the variational method.

Alves et al. in [15] proved the Hardy-Littlewood-Sobolev inequality for variable exponents, and used it to study the quasilinear Choquard equation involving variable exponents. Zhang et al. in [16] obtained the existence of infinitely many solutions for the variable exponent critical reaction Choquard problem by using variational and analytical methods, including the variable exponent Hardy-Littlewood-Sobolev inequality and the concentration compactness principle for the variable growth problem. Some other results for the Choquard problem involving nonlocal operators can be founded in [8],9],[17], [18], [19], [20]. For the study of a class of partial differential equations with nonstandard growth, although many important results have been obtained at present, for Choquard equations involving (p(x), q(x))-Laplace operators and singular term, the results are still less, which is one of the motivations of this paper.

Inspired by the above literatures, this paper is devoted to the study of singular double phase problem with variable exponents and convolution term. The main difficulty is that due to the convolution term and the singular term, we adopt different technical means to study the existence of solutions to such problem. Inspired by [5], we use the Nehari manifold method based on the fibering mapping method to prove our existence results, which also supplement [5] to some extent.

Our main results are as follows:

Theorem 1.1.

Assume that the hypotheses (H 1)–(H 2) hold. Then, there exists λ * > 0, such that for each λ ∈ (0, λ *), the problem (1.1) has at least two non-trivial weak solutions.

The paper is organized as follows. First, Section 2 contains some definitions and properties of Musielak-Orlicz Sobolev spaces. Secondly, in Section 3 of this paper, we explore the properties of the fibering mapping corresponding to the energy functional, and prove some lemmas. Finally, in Section 4, the main theorem is proved by a series of lemmas.

2 Preliminaries

In this section, we introduce some symbols and related conclusions, which are very important to prove Theorem 1.1. Let C + ( Ω ̄ ) = { m C ( Ω ̄ ) : m ( x ) > 1 , x Ω ̄ } and S ( Ω ) represents the set of all measurable functions u : Ω R . The variable Lebesgue space L m(⋅)(Ω), for mC +(Ω) is given by

L m ( ) ( Ω ) : = u S ( Ω ) : Ω | u | m ( x ) d x <

and the norm is defined as

u m ( ) = inf ν > 0 : Ω u ( x ) ν m ( x ) d x 1 .

The space (L m(⋅)(Ω), ‖ ⋅‖ m(⋅)) is separable and reflexive Banach space [17],21]. The conjugate space of L m(⋅)(Ω) is L m (⋅)(Ω) with 1 m ( x ) + 1 m ( x ) = 1 . For uL m(⋅)(Ω) and hL m (⋅)(Ω), we have

(2.1) Ω u ( x ) h ( x ) d x 1 m + 1 ( m ) u m ( x ) h m ( x ) 2 u m ( x ) h m ( x ) .

Lemma 2.1

(see [21]). Let mC +(Ω). Define ρ m (u) = Ω|u| m(x)dx for uL m(x)(Ω). Then, the norm ‖ ⋅‖ m(x) and the modular function ρ m (⋅) satisfy the following relations:

  1. u m(⋅) > 1(=1, <1) ⇔ ρ m (u) > 1(=1, <1), respectively.

  2. u m ( ) > 1 u m ( ) m ρ m ( u ) u m ( ) m + .

  3. u m ( ) < 1 u m ( ) m + ρ m ( u ) u m ( ) m .

  4. For u n L m(x)(Ω), ‖u m(⋅) → 0 ⇔ ρ m (u) → 0.

Lemma 2.2

(see [12]). Let hL (Ω), mC +(Ω) and 1 ≤ h(x), m(x) ≤ ∞ for a.a. x ∈ Ω. Then, for any uL m(x)(Ω) with u ≠ 0 we have

u m ( ) h ( ) > 1 u m ( ) h ( ) h | u | h ( x ) m ( ) u m ( ) h ( ) h +

and

u m ( ) h ( ) < 1 u m ( ) h ( ) h + | u | h ( x ) m ( ) u m ( ) h ( ) h .

Moreover, if h(x) = h, i.e., a constant, then we have

(2.2) u m ( ) h h = | u | h m ( ) .

Definition 2.1.

If the function k: Ω → (0, ∞) is measurable and locally integrable, it is called a weight function. The weighted variable Lebesgue space L k ( ) v ( ) ( Ω ) is defined as

L k ( ) v ( ) ( Ω ) : = u S ( Ω ) : Ω k ( x ) u η v ( x ) < for some η > 0 ;

L k ( ) v ( ) ( Ω ) is equipped with the norm

u L k ( ) v ( ) ( Ω ) = inf η > 0 : Ω k ( x ) u η v ( x ) 1 .

Next, define M : Ω × [ 0 , ) by M ( x , s ) : = s p ( x ) + λ ( x ) s q ( x ) for all (x, s) ∈ Ω × [0, ∞). The Musielak space L M ( Ω ) is given by

L M ( Ω ) : = u S ( Ω ) : Ω M ( x , | u | ) < .

L M ( Ω ) is equipped with the norm

u M : = inf ϒ > 0 : Ω M x , u ϒ 1 .

The Musielak space L M ( Ω ) is a reflexive Banach space [10]. The Musielak-Orlicz Sobolev space W 1 , M ( Ω ) is defined as

W 1 , M ( Ω ) : = u L M ( Ω ) : u L M ( Ω ) ,

W 1 , M ( Ω ) is equipped with the norm

u 1 , M : = u M + u M

and W 0 1 , M ( Ω ) is defined as the completion of C c ( Ω ) in W 1 , M ( Ω ) . For u W 0 1 , M ( Ω ) , if (H 1) holds, then we have u H C u M , for some constant C > 0 [10]. The space W 0 1 , M ( Ω ) can be equipped with the norm  u = u M . W 0 1 , M ( Ω ) and W 1 , M ( Ω ) are separable reflexive Banach spaces.

Lemma 2.3

(see [10]). Define ρ M ( u ) = Ω | u | p ( x ) + λ ( x ) | u | q ( x ) d x for p, qC +(Ω). If u L M ( Ω ) , u ≠ 0. Then, the following holds:

  1. u M = ϒ if and only if ρ M ( u ϒ ) = 1 ,

  2. u M > 1 ( = 1 , < 1 ) if and only if ρ M ( u ) > 1 ( = 1 , < 1 ) ,

  3. u M > 1 , then u M p ρ M ( u ) u M q + ,

  4. u M < 1 , then u M q + ρ M ( u ) u M p ,

  5. u M 0 ρ M ( u ) 0 .

Lemma 2.4

(see [10]). Let p, qC +(Ω) and (H 1) holds. Then, we have the following embedding results:

  1. For w C ( Ω ̄ ) and 1 ≤ v(x) ≤ p*(x), W 0 1 , M ( Ω ) L v ( ) ( Ω ) .

  2. For w C ( Ω ̄ ) and 1 ≤ v(x) ≪ p*(x), W 0 1 , M ( Ω ) L v ( ) ( Ω ) .

  3. For continuous function v(x) and 1 ≤ v(x) ≤ p(x), W 0 1 , M ( Ω ) W 0 1 , v ( ) ( Ω ) .

Next, inspired by [22], Theorem 2.3, 2.4], we have the following results:

Theorem 2.1

(see [22]). Let hL w(x), w C + ( Ω ̄ ) and w 0 C + ( Ω ̄ ) be such that 1 w ( x ) + 1 w 0 ( x ) = 1 , γ C ( Ω ̄ ) with

(2.3) p ( x ) < 1 γ + w 0 ( x ) 1 γ w 0 ( x ) < p * ( x ) for a.a. x Ω .

Then, the space W 0 1 , H ( Ω ) is embedded into L h ( x ) 1 γ ( x ) ( Ω ) and the embedding is compactly. Moreover

(2.4) Ω h ( x ) | u | 1 γ ( x ) d x C 1 u 1 γ + u 1 γ + ,

where C 1 > 0 is a constant.

Theorem 2.2

(see [22]). For a given u W 0 1 , H ( Ω ) , the following inequalities hold

(2.5) Ω h ( x ) | u | 1 η ( x ) d x C 2 u 1 γ if u > 1 , C 3 u 1 γ + if u < 1 ,

where C 2, C 3 are positive constants.

Then we give the Hardy-Littlewood-Sobolev inequality for variable exponents, see [15].

Proposition 2.1.

Let z 1 , z 2 C + R N , h L z 1 + R N L z 1 R N , g L z 2 + R N L z 2 R N , and μ : R N × R N R be a continuous function such that

0 < μ : = inf x R N μ ( x ) μ + : = sup x R N μ ( x ) < N

and

1 z 1 ( x ) + μ ( x , y ) N + 1 z 2 ( y ) = 2 for any x , y R N .

Then, we have

R 2 N h ( x ) g ( y ) | x y | μ ( x , y ) d x d y C h L z 1 + R N g L z 2 + R N + h L z 1 R N g L z 2 R N ,

where C > 0 is a constant that does not depend on h and g.

Corollary 2.3.

In particular for h ( x ) = g ( x ) = | u ( x ) | β ( x ) L r + R N L r R N , we have

R N × R N | u ( x ) | β ( x ) | u ( y ) | β ( y ) | x y | μ ( x , y ) d x d y C | u | β ( ) L r + R N 2 + | u | β ( ) L r R N 2 ,

where β , r C + ( R N ̄ ) such that 1 < β r β(x)r β(x)r + < p*(x) for all x R N ̄ , C > 0 is a constant that does not depend on r.

3 Some useful lemmas

The energy functional J ( u ) : W 0 1 , H ( Ω ) R corresponding to problem (1.1) is given by

(3.1) J ( u ) = Ω u p ( x ) p ( x ) d x + Ω α ( x ) u q ( x ) q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

If u W 0 1 , H ( Ω ) is a weak solution of problem (1.1), it means that there hold s(x)|u|γ(x) φL 1(Ω) and

Ω u p ( x ) 2 u + α ( x ) u q ( x ) 2 u φ d x = λ Ω s ( x ) | u | γ ( x ) φ d x + Ω Ω | u ( y ) | h ( x ) | u ( x ) | h ( x ) 2 u ( x ) φ ( x ) | x y | μ ( x , y ) d x d y

for all φ W 0 1 , H ( Ω ) . The functional J ( u ) is not C 1 because of the singular term s ( x ) | u | 1 γ ( x ) 1 γ ( x ) . Hence, the fibering map ϕ u : [ 0 , ) R we defined by ϕ u ( t ) = J ( t u ) , ∀t ≥ 0. We have ϕ u C (0, ∞) and

ϕ u ( t ) = Ω t p ( x ) u p ( x ) p ( x ) + α ( x ) t q ( x ) u q ( x ) q ( x ) d x λ Ω s ( x ) t 1 γ ( x ) | u | 1 γ ( x ) 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) t 2 h ( x ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y ,

ϕ u ( t ) = Ω t p ( x ) 1 u p ( x ) + α ( x ) t q ( x ) 1 u q ( x ) d x λ Ω s ( x ) t γ ( x ) | u | 1 γ ( x ) d x Ω Ω t 2 h ( x ) 1 | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y ,

ϕ u ( t ) = Ω ( p ( x ) 1 ) t p ( x ) 2 u p ( x ) d x + Ω α ( x ) ( q ( x ) 1 ) t q ( x ) 2 u q ( x ) d x + λ Ω s ( x ) γ ( x ) t γ ( x ) 1 | u | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) t 2 h ( x ) 2 | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

To solve problem (1.1), the Nehari manifold N λ is defined as

N λ : = t W 0 1 , M ( Ω ) \ { 0 } : ϕ T ( 1 ) = 0 = s t W 0 1 , M ( Ω ) \ { 0 } : ϕ t ( s ) = 0 .

Next, we split N λ into three constrained subsets

N λ + : = t N λ : ϕ t ( 1 ) > 0 = s t W 0 1 , M ( Ω ) \ { 0 } : ϕ t ( s ) = 0 , ϕ t ( s ) > 0 ,

N λ : = t N λ : ϕ t ( 1 ) < 0 = s t W 0 1 , M ( Ω ) \ { 0 } : ϕ t ( s ) = 0 , ϕ t ( s ) < 0 ,

N λ 0 : = t N λ : ϕ t ( 1 ) = 0 = s t W 0 1 , M ( Ω ) \ { 0 } : ϕ t ( s ) = 0 , ϕ t ( s ) = 0 .

Lemma 3.1.

Suppose that the hypotheses (H 1)–(H 2) hold. Then, the functional J is coercive on N λ .

Proof.

For any u N λ with ‖u‖ > 1, we have

(3.2) Ω u p ( x ) + α ( x ) u q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) d x Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) = 0 .

Using (H 1), (3.2), Lemma 2.3 and Theorem 2.2, we have

J ( u ) = Ω u p ( x ) p ( x ) d x + Ω α ( x ) u q ( x ) q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y > 1 q + ρ M ( u ) λ 1 γ + Ω s ( x ) | u | 1 γ ( x ) d x 1 2 h Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 1 q + 1 2 h ρ M ( u ) λ 1 1 γ + 1 2 h Ω s ( x ) | u | 1 γ ( x ) d x 1 q + 1 2 h u p λ 1 1 γ + 1 2 h C 2 u 1 γ .

In view of (H 1), we derive that J ( u ) as ‖u‖ → ∞. Thus, J ( u ) is coercive on N λ . □

Next, we discuss the properties of fibering maps in all possible cases in [23]. Define

m u : = ρ M ( u ) , n u : = Ω s ( x ) | u | 1 γ ( x ) d x , r u : = Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

We have

(3.3) ϕ u ( t ) t q + 1 m u λ t γ + n u t 2 h 1 r u X ( 0,1 ) ( t ) + t p 1 m u λ t γ n u t 2 h + 1 r u X [ 1 , ) ( t )

and

(3.4) ϕ u ( t ) t q + 1 m u λ t γ + n u t 2 h 1 r u X [ 1 , ) ( t ) + t p 1 m u λ t γ n u t 2 h + 1 r u X ( 0,1 ) ( t ) .

For β > ζ > δ > 0 and λ, m, n, r > 0, choose the function l λ (t):= t ζ mλt δ nt β r. For t > 0, l λ (t) = 0 ⇔ t δ l λ (t) = 0, i.e., t ζ+δ mt β+δ r = λn. Let f(t) = t ζ+δ mt β+δ r and g(t) = λn. The function f(t) satisfies

lim t 0 f ( t ) = 0 and lim t f ( t ) = , f ( t ) > 0 for t < m r 1 β ζ and f ( t )  attains maxima a t t = ζ + δ m β + δ r 1 β ζ .

If we choose λ > 0 and sufficiently small then l λ (t) has exactly two solutions t 1, t 2 with t 1 < t 2, such that l λ ( t 1 ) > 0 and l λ ( t 2 ) < 0 .

In view of the inequalities (3.3) and (3.4), we note that the graph of ϕ u ( t ) lies between the graphs of ν λ , u = t q + 1 m u λ t γ + n u t 2 h 1 r u and ω λ , u ( t ) = t p 1 m u λ t γ n u t 2 h + 1 r u . Therefore, there exists at least two critical points of ϕ u (t), namely, t 1 < t 2, such that ϕ u (t) has local minima at t 1 = t 1(u) and at local maxima at t 2 = t 2(u). If u N λ + , t 1 = 1 < t 2, t 2 u N λ and for u N λ , t 1 < t 2 = 1, such that t 1 u N λ + . Also, ϕ u is decreasing in (0, t 1), increasing on (t 1, t 2), and again decreasing on (t 2, ∞). Thus, there exists λ * > 0, such that for 0 < λ < λ *, we have ϕ u ( t ) = 0 , i.e., t u N λ but t u N λ 0 .

Lemma 3.2.

There exists λ * > 0 such that N λ 0 = for λ ∈ (0, λ *).

Proof.

If possible, for any λ * > 0, there exists λ ∈ (0, λ *), such that N λ 0 . Let u N λ 0 with ‖u‖ > 1. As u N λ 0 N λ , we have

(3.5) Ω u p ( x ) + α ( x ) u q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) d x Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) = 0

and

(3.6) Ω ( p ( x ) 1 ) u p ( x ) d x + Ω α ( x ) ( q ( x ) 1 ) u q ( x ) d x + λ Ω s ( x ) γ ( x ) | u | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 0 .

From (H 1) and (3.6), we have that

(3.7) p 1 ρ M ( u ) + λ γ Ω s ( x ) | u | 1 γ ( x ) d x 2 h + 1 Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y 0

and

(3.8) q + 1 ρ M ( u ) + λ γ + Ω s ( x ) | u | 1 γ ( x ) d x 2 h 1 Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y 0 .

And in view of Corollary 2.3, we obtain

(3.9) Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y C u | h ( ) L p + ( Ω ) 2 + u | h ( ) L p ( Ω ) 2 .

Taking into account (3.5), (3.7) and (3.9), Lemma 2.3, and Theorem 2.2, we have

0 p 1 ρ M ( u ) + λ γ Ω s ( x ) | u | 1 γ ( x ) d x 2 h + 1 Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = p + γ 1 ρ M ( u ) 2 h + + γ 1 Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y p + γ 1 u p 2 h + + γ 1 C u 2 h + .

This fact implies that

u C p + γ 1 2 h + + γ 1 1 2 h + p .

Again, by using (3.5)(3.8), Theorem 2.2 and Lemma 2.3, we get

0 q + 1 ρ M ( u ) + λ γ + Ω s ( x ) | u | 1 γ ( x ) d x 2 h 1 Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = q + 2 h ρ M ( u ) + λ 2 h + γ + 1 Ω s ( x ) | u | 1 γ ( x ) d x q + 2 h u p + λ 2 h + γ + 1 C u 1 γ .

Moreover

u C λ 2 h + γ + 1 2 h q + 1 p + γ 1 .

Therefore, we have

C λ 2 h + γ + 1 2 h q + 1 p + γ 1 u C p + γ 1 2 h + + γ 1 1 2 h + p .

Furthermore

λ C p + γ 1 2 h q + 2 h + γ + 1 p + γ 1 2 h + + γ 1 p γ 1 2 h + p .

Choosing

λ 1 = C p + γ 1 2 h q + 2 h + γ + 1 p + γ 1 2 h + + γ 1 p γ 1 2 h + p ,

then for λ ∈ (0, λ 1), we have that u < C p + γ 1 2 h + + γ 1 , which is impossible. Similarly, for ‖u‖ < 1, we have λ 2, such that there exists a contradiction to our hypothesis. If we let λ * = min(λ 1, λ 2), then we have N λ 0 = for any λ ∈ (0, λ *).□

Therefore, for any λ ∈ (0, λ *), we have N λ = N λ + N λ .

4 Proof of main theorem

Lemma 4.1.

Assume that the hypotheses (H 1)–(H 2) are satisfied. Then, m λ + < 0 , where m λ + : = inf u N λ + J ( u ) .

Proof.

First, we note that u N λ + N λ . From the definitions of N λ and N λ + , respectively, it means that there hold

(4.1) Ω u p ( x ) + α ( x ) u q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) d x = Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y )

and

(4.2) Ω ( p ( x ) 1 ) u p ( x ) + α ( x ) ( q ( x ) 1 ) u q ( x ) d x + λ Ω s ( x ) η ( x ) | u | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y > 0 .

In view of (4.1) and (4.2), we can get

(4.3) q + 2 h ρ M ( u ) + λ 2 h + γ + 1 Ω s ( x ) | u | 1 γ ( x ) d x > 0 .

This fact implies that

2 h q + 2 h + γ + 1 ρ M ( u ) < λ Ω s ( x ) | u | 1 γ ( x ) d x .

From (H 1), (4.1)(4.3), and (H 2)

J ( u ) = Ω u p ( x ) p ( x ) d x + Ω α ( x ) u q ( x ) q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) 1 p Ω u p ( x ) + α ( x ) u q ( x ) d x λ 1 γ Ω s ( x ) | u | 1 γ ( x ) d x 1 2 h + Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) = 1 p 1 2 h + ρ M ( u ) λ 1 1 γ 1 2 h + Ω s ( x ) | u | 1 γ ( x ) d x < 1 p 1 2 h + 1 1 γ 1 2 h + 2 h q + 2 h + γ + 1 ρ M ( u ) = 2 h + p 2 h + 1 p 2 h + + γ 1 2 s q + ( 1 γ ) 2 h + γ + 1 2 h + p ρ M ( u ) < 0 .

Hence, m λ + < 0 . Now, the proof of Lemma 4.1 is completed.□

Lemma 4.2.

Assume that the hypotheses (H 1)–(H 2) are satiafied. Then, m λ > 0 , where m λ : = inf u N λ J ( u ) .

Proof.

Let u N λ N λ . From definition of N λ and N λ , respectively, it means that there hold

p 2 h + ρ H ( u ) + λ 2 h + + γ 1 Ω s ( x ) | u | 1 γ ( x ) d x < 0

and

(4.4) Ω s ( x ) | u | 1 γ ( x ) d x < 2 h + p 2 h + + γ 1 ρ H ( u ) .

Using definition of (H 1), (4.1)(4.4), and (H 2)

J ( u ) = Ω u p ( x ) p ( x ) d x + Ω α ( x ) u q ( x ) q ( x ) d x λ Ω s ( x ) | u | 1 γ ( x ) 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y )

1 q + ρ H ( u ) λ 1 γ + Ω s ( x ) | u | 1 γ ( x ) d x 1 2 h Ω Ω | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) = 1 q + 1 2 h ρ H ( u ) λ 1 1 γ + 1 2 h Ω s ( x ) | u | 1 γ ( x ) d x > 2 h q + 2 h 1 q + 2 h + γ + 1 2 h + p ( 1 γ + ) 2 h + + γ 1 2 h q + ρ H ( u ) > 0 .

Thus, for u N λ , we have J ( u ) > 0 , so m λ > 0 . Now, the proof of Lemma 4.2 is completed.□

Lemma 4.3.

For λ ∈ (0, λ *), there exists u 0 N λ + ( Ω ) satisfying J ( u 0 ) = m λ + and u 0 ≥ 0 for a.e. x ∈ Ω.

Proof.

Since J | N λ is bounded below and thus bounded on N λ + . Therefore, there exists a sequence { u n } N λ + ( Ω ) , such that J ( u n ) m λ + as n → ∞. Due to the coerciveness of I λ , the sequence {u n } is bounded in W 0 1 , M ( Ω ) . In view of the reflexivity of W 0 1 , M ( Ω ) and applying Theorem 2.2, we have

(4.5) u n u 0 in W 0 1 , M ( Ω ) , u n u 0 in L s ( x ) 1 γ ( x ) ( Ω ) and u n u 0 in L p ( x ) h ( x ) ( Ω ) .

Now, we claim that u n u 0 in W 0 1 , M ( Ω ) .

If possible, let u n does not converge to u 0 in W 0 1 , M ( Ω ) . Then

lim inf n ρ M ( u n ) > ρ M ( u 0 ) ,

that is

(4.6) Ω u 0 p ( x ) + α ( x ) u 0 q ( x ) d x < lim inf n Ω u n p ( x ) + α ( x ) u n q ( x ) d x .

We get J ( u 0 ) lim inf n J ( u n ) < 0 = J ( 0 ) , this means that u 0 ≠ 0. Moreover

ϕ u n ( t ) = Ω t p ( x ) 1 u n p ( x ) + α ( x ) t q ( x ) 1 u n q ( x ) d x λ Ω s ( x ) t γ ( x ) | u n | 1 γ ( x ) d x Ω Ω t 2 h ( x ) 1 | u n ( x ) | h ( x ) | u n ( y ) | h ( x ) | x y | μ ( x , y ) d x d y

and

ϕ u 0 ( t ) = Ω t p ( x ) 1 u 0 p ( x ) + α ( x ) t q ( x ) 1 u 0 q ( x ) d x λ Ω s ( x ) t γ ( x ) | u 0 | 1 γ ( x ) d x Ω Ω t 2 h ( x ) 1 | u 0 ( x ) | h ( x ) | u 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

In Section 3, though the analysis of fibering map, there exists t 1 = t 1(u 0), such that t 1 ( u 0 ) N λ + . From N λ + N λ , we have ϕ u 0 ( t 1 ) = 0

lim n ϕ u n ( t 1 ) = lim n Ω t 1 p ( x ) 1 u n p ( x ) + α ( x ) t 1 q ( x ) 1 u n q ( x ) d x

lim n λ Ω s ( x ) t 1 γ ( x ) | u n | 1 γ ( x ) d x + Ω Ω t 1 2 h ( x ) 1 | u n ( x ) | h ( x ) | u n ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = lim n Ω t 1 p ( x ) 1 u n p ( x ) + α ( x ) t 1 q ( x ) 1 u n q ( x ) d x λ Ω s ( x ) t 1 γ ( x ) | u 0 | 1 γ ( x ) d x Ω Ω t 1 2 h ( x ) 1 | u 0 ( x ) | h ( x ) | u 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y ( using ( 4.5 ) ) > 0 .

Therefore, as sufficiently large n, ϕ u n ( t 1 ) > 0 . Since the sequence u n N λ + , it is easy to obtain that ϕ u n ( 1 ) = 0 by the analysis of fibering maps, and for 0 < t < 1, ϕ u n ( t ) < 0 for all n. Hence, we get t 1 > 1. For 0 < t < 1

ϕ t 1 u 0 ( 1 ) < ϕ t 1 u 0 ( t ) ;

assume t = 1 t 1 , J ( t 1 u 0 ) = ϕ t 1 u 0 ( 1 ) < ϕ t 1 u 0 ( 1 t 1 ) = J ( u 0 ) . Thus, J ( t 1 u 0 ) < J ( u 0 ) < lim n J ( u 0 ) = inf u N λ + J ( u ) , which is a contradiction to t 1 u 0 N λ + . Therefore, ρ M ( u n ) ρ M ( u 0 ) , so by using (Preposition 2.19, [10]) u n u 0 in W 0 1 , M ( Ω ) . As { u n } N λ + N λ and the embedding W 0 1 , M ( Ω ) L s ( x ) 1 γ ( x ) ( Ω ) ( L p ( x ) h ( x ) ( Ω ) ) compact embedding. Thus

lim n Ω u n p ( x ) + α ( x ) u n q ( x ) d x λ Ω s ( x ) | u n | 1 γ ( x ) d x Ω Ω | u n ( x ) | h ( x ) | u n ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 0

and

Ω u 0 p ( x ) + α ( x ) u 0 q ( x ) d x λ Ω s ( x ) | u 0 | 1 γ ( x ) d x Ω Ω | u 0 ( x ) | h ( x ) | u 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 0 .

Due to the boundedness of p, q, h, γ, we have

Ω ( p ( x ) 1 ) u 0 p ( x ) + α ( x ) ( q ( x ) 1 ) u 0 q ( x ) d x + λ Ω s ( x ) η ( x ) | u 0 | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | u 0 ( x ) | h ( x ) | u 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y 0 .

Since N λ 0 = for λ ∈ (0, λ *), so the above equality is not possible. Thus, we get u 0 N λ + . We can also replace u 0 with |u 0|, so for a.a. x ∈ Ω with u 0 ≠ 0, we can have u 0(x) ≥ 0.□

Lemma 4.4.

Assume that hypotheses (H 1)–(H 2) are satisfied. Then, for u N λ + N λ , there exist ϵ > 0 and a function z: B ϵ (0) → (0, ∞), such that z is continuous and z(0) = 1, z ( r ) ( u + r ) N λ + N λ for every rB ϵ (0), with B ϵ ( 0 ) = r W 0 1 , M ( Ω ) : r < ϵ .

Proof.

Consider u N λ + and define T : W 0 1 , M ( Ω ) × ( 0 , ) R given by

T ( r , t ) = Ω t p ( x ) 1 ( u + r ) p ( x ) d x + Ω α ( x ) t q ( x ) 1 ( u + r ) q ( x ) d x λ Ω s ( x ) t γ ( x ) | u + r | 1 γ ( x ) d x Ω Ω t 2 h ( x ) 1 | ( u + r ) ( x ) | h ( x ) | ( u + r ) ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

We have T(0, 1) = 0 and

T t ( 0,1 ) = Ω ( p ( x ) 1 ) u p ( x ) + α ( x ) ( q ( x ) 1 ) u q ( x ) d x + λ Ω s ( x ) γ ( x ) | u | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | u ( x ) | h ( x ) | u ( y ) | h ( x ) | x y | μ ( x , y ) d x d y > 0 .

Therefore, by using implicit function theorem (see p. 115, [17]), there exists ϵ > 0 and a continuous function z: B ϵ (0) → (0, ∞), where z(0) = 1 and z ( y ) ( u + r ) N λ , ∀rB ϵ (0). If we choose small enough ϵ > 0, then we have z(0) = 1 and z ( r ) ( u + r ) N λ + .

For u N λ , we omit the details on similar lines.□

Lemma 4.5.

Suppose that the hypotheses (H 1)–(H 2) are satisfied. For λ ∈ (0, λ *) and g W 0 1 , H ( Ω ) , there exists a > 0 with J ( u 0 ) J ( u 0 + t g ) for all t ∈ [0, a].

Proof.

We define Γ g given by

Γ g ( t ) = Ω ( p ( x ) 1 ) ( u 0 + t g ) p ( x ) + α ( x ) ( q ( x ) 1 ) ( u 0 + t g ) q ( x ) d x + λ Ω s ( x ) γ ( x ) | u 0 + t g | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | ( u 0 + t g ) ( x ) | h ( x ) | ( u 0 + t g ) ( y ) | h ( x ) | x y | μ ( x , y ) d x d y .

Since u 0 N λ + N λ , we have Γ g (0) > 0. The function Γ g is continuous, so for all t ∈ [0, b], there exists b > 0 with Γ g (t) > 0. Combining Lemma 4.4, we obtain that for all t ∈ [0, b], g(t) > 0 satisfying

z ( t ) ( u 0 + t g ) N λ + , z ( t ) 1 as t 0 + .

In view of Lemma 4.1, we get

(4.7) m λ + = inf u N λ + J ( u ) = J ( u 0 ) J ( z ( t ) ( u + t ) ) for all  t [ 0 , b ] .

In addition, we have ϕ u 0 ( 1 ) > 0 , and due to the continuity of ϕ″, for any t ∈ [0, a], there exists a ∈ (0, b], such that ϕ u 0 + t g ( 1 ) > 0 . Applying (4.7)

m λ + = J ( u 0 ) J ( z ( t ) ( u + t ) ) = ϕ u 0 + t g ( z ( t ) ) ϕ u 0 + t g ( 1 ) = J ( u 0 + t g )

for every t ∈ [0, a].□

Lemma 4.6.

Assume that hypotheses (H 1)–(H 2) are satisfied and λ ∈ (0, λ *). Then, u 0 is weak solution of problem (1.1) with J ( u 0 ) < 0 .

Proof.

By using Lemmas 4.3 and 4.1, we obtain that u 0 ≥ 0 for a.a. x ∈ Ω with J ( u 0 ) < 0 . We claim that u 0 > 0 for a.a. x ∈ Ω. Suppose that there exists a set G , G > 0 with u 0 = 0 for a.a. x G . Let g W 0 1 , M ( Ω ) , such that g > 0. For t ∈ (0, a) where a comes from Lemma 4.5, then we have ( u 0 + t g ) 1 γ ( x ) > u 0 1 γ ( x ) for a.a. x Ω \ G . By using Lemma 4.5, we obtain that

0 J ( u 0 + t g ) J ( u 0 ) t = Ω 1 p ( x ) ( u 0 + t g ) p ( x ) u 0 p ( x ) t d x + Ω α ( x ) 1 q ( x ) ( u 0 + t g ) q ( x ) u 0 q ( x ) t d x λ G s ( x ) t γ ( x ) ( 1 γ ( x ) ) g 1 γ ( x ) d x λ Ω \ G s ( x ) 1 γ ( x ) | u 0 + t g | 1 γ ( x ) | u 0 | 1 γ ( x ) t d x 1 2 Ω Ω 1 h ( x ) | ( u 0 + t g ) ( x ) | h ( x ) | ( u 0 + t g ) ( y ) | h ( x ) | u 0 | h ( x ) | u 0 ( y ) | h ( x ) t | x y | μ ( x , y ) d x d y < Ω 1 p ( x ) ( u 0 + t g ) p ( x ) u 0 p ( x ) t d x + Ω α ( x ) 1 q ( x ) ( u 0 + t g ) q ( x ) u 0 q ( x ) t d x λ G s ( x ) t γ ( x ) ( 1 γ ( x ) ) g 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) | ( u 0 + t g ) ( x ) | h ( x ) | ( u 0 + t g ) ( y ) | h ( x ) | u 0 | h ( x ) | u 0 ( y ) | h ( x ) t | x y | μ ( x , y ) d x d y as t 0 +

which is a contradiction. Thus, u 0 ≥ 0 for a.a. x ∈ Ω. Next, we will prove the following conclusions:

(4.8) s ( x ) u 0 γ ( x ) g L 1 ( Ω ) for all g W 0 1 , M ( Ω )

and

(4.9) Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 z d x λ Ω s ( x ) ( u 0 ) γ ( x ) z d x + Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 z ( x ) | x y | μ ( x , y ) d x d y

for g W 0 1 , M ( Ω ) and g ≥ 0.

Suppose that g W 0 1 , M ( Ω ) with g ≥ 0 and { t n } n N ( 0,1 ] be decreasing sequence and lim n→∞ t n = 0. Consider the functions

z n ( x ) = s ( x ) 1 γ ( x ) ( u 0 + t n g ) ( x ) 1 γ ( x ) | u 0 ( x ) | 1 γ ( x ) t n , n N .

Then, for a.a. x ∈ Ω, z n (x) are non-negative measurable functions with lim n z n ( x ) = s ( x ) u 0 γ ( x ) g ( x ) . By applying Fatou’s Lemma, we obtain that

(4.10) Ω s ( x ) u 0 γ ( x ) g ( x ) d x lim inf n Ω z n ( x ) d x .

Using Lemma 4.5 and choosing n N is large enough, we get

0 J ( u 0 + t n z ) J ( u 0 ) t n = Ω 1 p ( x ) ( u 0 + t n g ) p ( x ) u 0 p ( x ) t n d x + Ω α ( x ) 1 q ( x ) ( u 0 + t n g ) q ( x ) u 0 q ( x ) t n d x λ Ω z n ( x ) d x 1 2 Ω Ω 1 h ( x ) ( u 0 + t n g ) ( x ) h ( x ) ( u 0 + t n g ) ( y ) h ( x ) | u 0 | h ( x ) | u 0 ( y ) | h ( x ) t n | x y | μ ( x , y ) d x d y .

Take the limit n → ∞ and by (4.10), we get

Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 g d x λ Ω s ( x ) | u 0 | γ ( x ) g ( x ) d x Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 g ( x ) | x y | μ ( x , y ) d x d y ,

i.e.

λ Ω s ( x ) | u 0 | γ ( x ) g ( x ) d x Ω ( u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ) g d x Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 g ( x ) | x y | μ ( x , y ) d x d y .

Thus, we have proved (4.8) and (4.9). Next, we will prove that u 0 is a weak solution of problem (1.1). Let w W 0 1 , M ( Ω ) and ϵ > 0. We substitute g = ( u 0 + ϵ w ) + in (4.9) and using u 0 ( > 0 ) N λ + N λ , we have

0 Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ( u 0 + ϵ w ) + d x λ Ω s ( x ) | u 0 | γ ( x ) ( u 0 + ϵ w ) + d x Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 ( u 0 + ϵ w ) + | x y | μ ( x , y ) d x d y = { u 0 + ϵ w 0 } u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ( u 0 + ϵ w ) d x λ { u 0 + ϵ w 0 } s ( x ) | u 0 | γ ( x ) ( u 0 + ϵ w ) d x { u 0 + ϵ w 0 } { u 0 + ϵ w 0 } | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 ( u 0 + ϵ w ) | x y | μ ( x , y ) d x d y = Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ( u 0 + ϵ w ) d x { u 0 + ϵ w < 0 } u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ( u 0 + ϵ w ) d x λ Ω s ( x ) | u 0 | γ ( x ) ( u 0 + ϵ w ) d x Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 ( u 0 + ϵ w ) | x y | μ ( x , y ) d x d y + λ { u 0 + ϵ w < 0 } s ( x ) | u 0 | γ ( x ) ( u 0 + ϵ w ) d x + { u 0 + ϵ w < 0 } { u 0 + ϵ w < 0 } | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 ( u 0 + ϵ w ) | x y | μ ( x , y ) d x d y = Ω u 0 p ( x ) + α ( x ) u 0 q ( x ) d x λ Ω s ( x ) | u 0 | 1 γ ( x ) d x Ω Ω | u 0 ( x ) | h ( x ) | u 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y + ϵ Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 w d x ϵ λ Ω s ( x ) | u 0 | γ ( x ) w d x ϵ Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 w | x y | μ ( x , y ) d x d y { u 0 + ϵ w < 0 } u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 ( u 0 + ϵ w ) d x + λ { u 0 + ϵ w < 0 } s ( x ) | u 0 | γ ( x ) ( u 0 + ϵ w ) d x + { u 0 + ϵ w < 0 } { u 0 + ϵ w < 0 } | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 ( u 0 + ϵ w ) | x y | μ ( x , y ) d x d y ϵ Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 w d x ϵ λ Ω s ( x ) | u 0 | γ ( x ) w d x ϵ Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 w | x y | μ ( x , y ) d x d y { u 0 + ϵ w < 0 } u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 w d x .

Divide the above inequality by ϵ and let ϵ → 0, also

{ u 0 + ϵ w < 0 } u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 w d x 0 as ϵ 0 .

We have that

Ω u 0 p ( x ) 2 u 0 + α ( x ) u 0 q ( x ) 2 u 0 w d x λ Ω s ( x ) | u 0 | γ ( x ) v d x + Ω Ω | u 0 ( y ) | h ( x ) | u 0 ( x ) | h ( x ) 1 w | x y | μ ( x , y ) d x d y .

Since the above inequality holds for each w W 0 1 , M ( Ω ) . Therefore, when J ( u 0 ) < 0 , u 0 is a weak solution of problem (1.1).□

Lemma 4.7.

Assume that the hypotheses (H 1)–(H 2) are satisfied. Then, for any λ ∈ (0, λ *), there exists w 0 ( 0 ) N λ with J ( w 0 ) = m λ .

Proof.

Let { w n } N λ N λ be a sequence of minimizations such that J ( w n ) m λ as n → ∞. The sequence {w n } is bounded in W 0 1 , M ( Ω ) as J | N λ is coercive. Therefore, we obtain that

w n w 0 in W 0 1 , H ( Ω ) , w n w 0 in L s ( x ) 1 γ ( x ) ( Ω ) and w n w 0 in L p ( x ) h ( x ) ( Ω ) .

Claim: w n w 0 in W 0 1 , M ( Ω ) .

If possible, let w n do not converge to w 0 in W 0 1 , M ( Ω ) . Then

lim inf n ρ M ( w n ) > ρ M ( w 0 ) ,

i.e.

(4.11) Ω w 0 p ( x ) + α ( x ) w 0 q ( x ) d x < lim inf n Ω w n p ( x ) + α ( x ) w n q ( x ) d x .

Through the analysis of fibering map in Section 3, this means that there exists t 2 = t 2(w 0) such that t 2 ( w 0 ) N λ hold. We know

lim n ϕ w n ( t 2 ) = lim n Ω t 2 p ( x ) 1 w n p ( x ) + α ( x ) t 2 q ( x ) 1 w n q ( x ) d x lim n λ Ω s ( x ) t 2 γ ( x ) | w n | 1 γ ( x ) d x + Ω Ω t 2 2 h ( x ) 1 | w n ( x ) | h ( x ) | w n ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = lim n Ω t 2 p ( x ) 1 w n p ( x ) + α ( x ) t 2 q ( x ) 1 w n q ( x ) d x λ Ω s ( x ) t 2 γ ( x ) | w 0 | 1 γ ( x ) d x Ω Ω t 2 2 h ( x ) 1 | w 0 ( x ) | h ( x ) | w 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y > 0 .

Therefore, as n is sufficiently large, ϕ w n ( t 2 ) > 0 . Through the analysis of fibering maps, as the sequence w n N λ , we can easily check that ϕ w n ( 1 ) = 0 for all n and we have ϕ w n ( t ) < 0 for t > 1. Hence, 0 < t 2 < 1

J ( t 2 w 0 ) < lim inf n J ( t 2 w n ) lim inf n J ( w n ) = m λ ,

which is a contradiction to t 2 w 0 N λ . Therefore, w n w 0 in W 0 1 , M ( Ω ) .

Since { w n } N λ N λ and W 0 1 , M ( Ω ) L s ( x ) 1 γ ( x ) ( Ω ) ; L p ( x ) h ( x ) ( Ω ) compactly, we have

lim n Ω w n p ( x ) + α ( x ) w n q ( x ) d x λ Ω s ( x ) | w n | 1 γ ( x ) d x Ω Ω | w n ( x ) | h ( x ) | w n ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 0

and

Ω w 0 p ( x ) + α ( x ) w 0 q ( x ) d x λ Ω s ( x ) | w 0 | 1 γ ( x ) d x Ω Ω | w 0 ( x ) | h ( x ) | w 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y = 0 .

Due to the boundedness of p, q, s, γ, we have

Ω ( p ( x ) 1 ) w 0 p ( x ) + α ( x ) ( q ( x ) 1 ) w 0 q ( x ) d x + λ Ω s ( x ) γ ( x ) | w 0 | 1 γ ( x ) d x Ω Ω ( 2 h ( x ) 1 ) | w 0 ( x ) | h ( x ) | w 0 ( y ) | h ( x ) | x y | μ ( x , y ) d x d y ) 0 .

Since N λ 0 = for λ ∈ (0, λ *), so we note that the equality is not possible. Therefore, we obtain w 0 N λ . We can also replace w 0 with |w 0|, so we can take w 0(x) ≥ 0 for a.a. x ∈ Ω with w 0 ≠ 0.□

Lemma 4.8.

Assume that (H 1)–(H 2) are satisfied and λ ∈ (0, λ *). Then, when J ( w 0 ) > 0 , w 0 is weak solution of problem (1.1).

Proof.

By using Lemma 4.5, u 0 = w 0 is substituted into the definition of Γ and by Lemma 4.4, it can be easily verified that for 0 ≤ ta 0, we have the function z(t) > 0 satisfying z ( t ) ( w 0 + t g ) N λ and z(t) → 1 as t → 0+. In view of Lemma 4.7, we get

(4.12) m λ = J ( w 0 ) J ( z ( t ) ( w 0 + t g ) ) , t [ 0 , λ * ] .

Now, for a.a. x ∈ Ω, there is w 0 > 0. Suppose there exists a set O with | O | > 0 and w 0 = 0 a.e. in O . We consider g W 0 1 , M ( Ω ) , such that g > 0. For t ∈ (0, a 0), we have

( z ( t ) ( w 0 + t g ) ) 1 γ ( x ) > z ( t ) w 0 1 γ ( x )

for a.e. x Ω \ O . We know ϕ w 0 ( 1 ) is the global maximum, so ϕ w 0 ( 1 ) ϕ w 0 ( z ( t ) )

0 J ( z ( t ) ( w 0 + t g ) ) J ( w 0 ) t J ( z ( t ) ( w 0 + t g ) ) J ( z ( t ) w 0 ) t = Ω 1 p ( x ) ( z ( t ) ( w 0 + t g ) ) p ( x ) ( z ( t ) w 0 ) p ( x ) t d x + Ω α ( x ) 1 q ( x ) ( z ( t ) ( w 0 + t g ) ) q ( x ) ( z ( t ) w 0 ) q ( x ) t d x λ O s ( x ) t γ ( x ) ( 1 γ ( x ) ) z ( t ) 1 γ ( x ) g 1 γ ( x ) d x λ Ω \ O s ( x ) 1 γ ( x ) | z ( t ) ( w 0 + t g ) | 1 γ ( x ) | ( z ( t ) w 0 ) | 1 γ ( x ) t d x 1 2 Ω Ω 1 h ( x ) | ( z ( t ) ( w 0 + t g ) ) ( x ) | h ( x ) | ( z ( t ) ( w 0 + t g ) ) ( y ) | g ( x ) | ( z ( t ) w 0 ) ( x ) | h ( x ) | ( z ( t ) w 0 ) ( y ) | h ( x ) t | x y | μ ( x , y ) d x d y < Ω 1 p ( x ) ( z ( t ) ( w 0 + t g ) ) p ( x ) ( z ( t ) w 0 ) p ( x ) t d x + Ω α ( x ) 1 q ( x ) ( z ( t ) ( w 0 + t g ) ) q ( x ) ( z ( t ) w 0 ) q ( x ) t d x λ O s ( x ) t γ ( x ) ( 1 γ ( x ) ) z ( t ) 1 γ ( x ) g 1 γ ( x ) d x 1 2 Ω Ω 1 h ( x ) ( z ( t ) ( w 0 + t g ) ) ( x ) h ( x ) | ( z ( t ) ( w 0 + t g ) ) ( y ) | h ( x ) | ( z ( t ) w 0 ) | h ( x ) | ( z ( t ) w 0 ) ( y ) | h ( x ) t | x y | μ ( x , y ) d x d y as t 0 +

which is a contradiction. Therefore, for a.e. x ∈ Ω, w 0 ≥ 0. The rest of the proof is similar to Lemma 4.6, which can be obtained by using ϕ w 0 ( 1 ) ϕ w 0 ( z ( t ) ) and (4.12).□

Proof of Theorem 1.1.

Therefore, according to Lemmas 4.6 and 4.8, when N λ + N λ = , the problem (1.1) has at least two weak solutions. The proof of Theorem 1.1 is completed.□


Corresponding author: Sihua Liang, College of Mathematics, Changchun Normal University, Changchun, 130032, P.R. China, E-mail: 

Acknowledgments

The authors wish to extend sincere gratitude to the referees and the editor for providing valuable comments and suggestions that will enhance the quality of this article.

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: The authors declare that they have contributed equally to this article. Both authors have read and approved the final version of the manuscript.

  4. Use of Large Language Models, AI and Machine Learning Tools: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  5. Conflict of interest: The author states no conflict of interest.

  6. Research funding: The authors were supported by the National Natural Science Foundation of China (No. 12571114), the Young outstanding talents project of Scientific Innovation and entrepreneurship in Jilin (No. 20240601048RC) and the Research Foundation of Department of Education of Jilin Province (No. JJKH20261207KJ).

  7. Data availability: Not applicable.

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Received: 2023-07-21
Accepted: 2026-01-12
Published Online: 2026-02-18

© 2026 the author(s), published by De Gruyter, Berlin/Boston

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