1 Introduction

In this paper, we study the following nonlinear problem of p(x)-triharmonic type with Navier boundary conditions:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(x)}^{3} u=\lambda f(x,u), & \hbox { in } \Omega,\\ u=\Delta u=\Delta ^{2}u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(1)

where \(N \ge 3\)\(\Delta _{p(x)}^{3}\) is defined as

$$\begin{aligned} \Delta _{p(x)}^{3}u:=div\left( \Delta \left( \left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right) \right), \end{aligned}$$

and is called a p(x)-triharmonic operator. Here, \(p(x)\in C\left( {\overline{\Omega }}\right)\) satisfies

$$\begin{aligned} 3<p^-:=\inf _{x\in {\overline{\Omega }}}p(x)\le p^+:=\sup _{x\in {\overline{\Omega }}}p(x). \end{aligned}$$

\(\lambda\) is a real number, f satisfies \(Carath\acute{e}odory\) condition, and

$$\begin{aligned} F\left( x,t\right) =\int _{0}^{t}f\left( x,s\right) ds. \end{aligned}$$
(2)

The analysis of problem (1) involves the p(x)-triharmonic operator is a natural extension of the p-triharmonic operator and has garnered considerable interest in recent years due to its more complex nonlinear characteristics compared to the p-triharmonic operator. The triharmonic problem has a wide range of applications in science and engineering, particularly requiring precise numerical solutions. The triharmonic equation is typically used to describe the slow rotational flow of highly viscous fluids within small cavities [1] and is also applied to geometric modeling in phase-field crystal models [2], such as surface optimization [3], mixing, and pore filling [4, 5]. Additionally, these equations can be viewed as a linearization of the Euler-Lagrange equations, used to minimize curvature variations [6]. In surface generation and forming, controlling boundary conditions to ensure curvature smoothness is crucial for designing high-quality surfaces. The variable exponent triharmonic equation further extends these applications by accurately describing high-order nonlinear effects and spatial variations in complex physical systems. It is widely used in fields such as material mechanics, fluid mechanics, nonlinear wave and vibration analysis, heat conduction and diffusion, biophysics, and biomedical engineering. This equation effectively models stress distribution in heterogeneous materials, the flow behavior of non-uniform fluids, the vibration modes of complex structures, and heat transfer in functionally graded materials, providing powerful tools for studying and predicting the behavior of these complex systems.

The investigation of Lebesgue and Sobolev spaces with variable exponent has been the subject of intensive research. \(Kov\acute{a}{\check{c}}ik\) and \(R\acute{a}kosn\acute{i}k\) [7] introduced the generalized Lebesgue space \(L^{p(x)}\) and generalized Sobolev space \(W^{k,p(x)}\), which serve as counterparts to the traditional Lebesgue space \(L^p\) and Sobolev space \(W^{k,p}\), respectively. Fan and Zhao [8] presented fundamental results regarding the generalized Lebesgue spaces \(L^{p(x)}\left( \Omega \right)\) and generalized Lebesgue-Sobolev spaces \(W^{k,p(x)}(\Omega )\). MarekGalewski [9] provided a generalized sufficient condition for the continuity of the Nemytskij operator between \(L^{p_1(x)}\) and \(L^{p_2(x)}\), among other results. The study of these spaces has laid a foundation for exploring partial differential equations and variational problems in the context of variable exponent p(x) growth, which has found extensive applications in various areas such as the modeling of electrorheological fluids (see [10,11,12,13]), mathematical representation of stationary thermo-rheological viscous flows in non-Newtonian fluids (see [14]), phenomena related to image processing (see [15,16,17]), elasticity (see [18]), and flow in porous media (see [19]).

A representative elliptic Dirichlet boundary value problem featuring variable exponent growth conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(x)}u=\lambda f(x,u), & \hbox { in } \Omega,\\ u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(3)

has been studied extensively. For instance, numerous researchers have investigated the following nonhomogeneous eigenvalue problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\operatorname {div}(|\nabla u|^{p(x)-2}\nabla u)=\lambda {\left| u\right| }^{q(x)-2}u, & \hbox { in } \Omega,\\ u=0, & \hbox { on }\partial \Omega.\\ \end{array}\right. } \end{aligned}$$
(4)

The scenario where \(p(x) = q(x)\) has been explored by Fan, Zhang, and Zhao in [20]. Utilizing an argument based on the Ljusternik-Schnirelmann critical point theory, the findings in [20] demonstrate that problem (4) possesses infinitely many eigenvalues. By defining \(\Lambda\) as the set of all nonnegative eigenvalues, the authors establish that \(\Lambda\) is a nonempty infinite set and \(\sup \Lambda = +\infty\). They further note that, in general, \(\inf \Lambda = 0\) can occur, but \(\inf \Lambda> 0\) holds under certain specific conditions, which are somewhat related to a type of monotonicity in the function p(x). It is also noted that for the p-Laplace operator (where \(p(x) \equiv p\)), \(\inf \Lambda> 0\) always applies. \(M. Mih\breve{a}ilescu\) and \(V. R\breve{a}dulescu\) [21] investigate problem (4) under a fundamental assumption:

$$\begin{aligned} \begin{aligned}1<\min _{x\in {\overline{\Omega }}}q(x)<\min _{x\in {\overline{\Omega }}}p(x)<\max _{x\in {\overline{\Omega }}}q(x).\end{aligned} \end{aligned}$$

The authors establish the existence of a continuous family of eigenvalues for problem (4) in a neighborhood of the origin. Specifically, they demonstrate the existence of \({\lambda }^{\star }\) such that every \(\lambda \in (0, {\lambda }^{\star })\) is an eigenvalue of problem (4). The proof relies on simple variational arguments based on Ekeland’s variational principle. This finding was later extended by X.Fan in [22]. The authors further investigate the eigenvalues of problem (4) through a constrained variational approach. They prove that for each \(t> 0\), problem (4) admits at least one sequence of solutions satisfying \(\int _{\Omega }{\frac{1}{p(x)}}|\nabla {u_{n,t}}|^{p(x)} dx=t\) and \(\lambda _{n,t} \rightarrow \infty\) as \(n \rightarrow \infty\). They also examine the principal eigenvalues of problem (4) in the general case, as well as in the cases where \(\Omega ={\Omega }_{-}=\{x\in \Omega:q(x)<p(x)\}\) and \(\Omega ={\Omega }_{+}=\{x\in \Omega:q(x)>p(x)\}\), and reveal the similarities and differences between the variable exponent case and the constant exponent case in Problem (4). Additionally, it is noted that the established results for problem (4) can be generalized to the more comprehensive form (3) under suitable conditions on f(xu).                                        

Similarly, the following eigenvalue problem involving p(x)-biharmonic:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta ^2_{p(x)}u=\lambda f(x,u), & \hbox { in } \Omega,\\ u=\Delta u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(5)

has also been studied in depth. For instance, the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta ^2_{p(x)}u=\lambda {\left| u\right| }^{q(x)-2}u, & \hbox { in } \Omega,\\ u=\Delta u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(6)

has been studied by A.Ayoujil and A.R.ElAmrouss in [23] for the case \(p(x) = q(x)\) and in [24] for the case \(p(x) \ne q(x)\). Notably, in [23], using Ljusternik-Schnirelmann theory on \(C^1\)-manifolds, the authors demonstrated the existence of a sequence of eigenvalues and showed that \(\sup \Lambda = \infty\), where \(\Lambda\) represents the set of all nonnegative eigenvalues. For the variable exponent p(x), unlike the constant exponent scenario, they provided sufficient conditions under which \(\inf \Lambda = 0\). In [24], employing the mountain pass lemma and Ekeland’s variational principle, the authors further established various existence criteria for eigenvalues.

As research progresses, numerous scholars are now focusing on elliptic boundary value problems involving the p(x)-triharmonic operator. For further details, refer to [25,26,27,28,29]. Notably, BelgacemRahal [25] was the first to extend the triharmonic problem from a constant exponent framework to a variable exponent setting, marking a significant starting point for the study of variable exponent triharmonic problems. The eigenvalue equation:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(\cdot )}^3u=\lambda|u|^{q(\cdot )-2}u, & \hbox {in }\Omega,\\ u=\Delta u=\Delta ^2u=0,\quad & \hbox {on }\partial \Omega, \end{array}\right. } \end{aligned}$$
(7)

similar to the above has also been studied by \({\dot{I}}sma{\dot{i}}l\)Aydin in [28]. The author demonstrated the existence of multiple weak solutions for problem (7) using the Fountain theorem under suitable assumptions.  

Inspired by the above research process, this paper will supplement and extend the results of [28], and explore more general forms (1) under appropriate assumptions.

We assume the following hypotheses:

(H1):

\(\alpha (x),\ \beta (x),\ \omega \left( x\right),\ \eta \left( x\right),\ p(x),\ q(x)\in C_+\left( {\overline{\Omega }}\right)\) and \(p(x)<\frac{N}{3}\) such that

$$\begin{aligned} & 1<\alpha ^{-}\le \alpha ^{+}<\beta ^{-}\le\beta ^{+}<p^{-},\\ & 1<\omega ^{-}\le \omega ^{+}<p^{-}<p^{+}<q^{-}\le q^{+},\\ & 1<\eta ^{-}\le\eta ^{+}<p^{-}<p^{+}<l^{-}\le l^{+}<q^{-}, \end{aligned}$$

and

$$\begin{aligned} 3<p^-\le p^+<q^-\leq q^+<p_3^*(x). \end{aligned}$$
(H2):

Denote

$$\begin{aligned} \gamma _1\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -\omega \left( x\right) },\quad&\gamma _2\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -q\left( x\right) },\\ \gamma _3\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -\alpha \left( x\right) },\quad&\gamma _4\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -\beta \left( x\right) },\\ \gamma _5\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -\eta \left( x\right) },\quad&\gamma _6\left( x \right) =\frac{r\left( x\right) }{r\left( x\right) -l\left( x\right) }. \end{aligned}$$

Then \(a_i\left( x\right) \in L^{\gamma _{2i-1}\left( x \right) }\left( \Omega \right) \bigcap L^{\infty }\left( \Omega \right)\) and \(b_i\left( x\right) \in L^{\gamma _{2i}\left( x \right) }\left( \Omega \right) \bigcap L^{\infty }\left( \Omega \right),\) for \(i=1,2,3\), where \(r\big (x\big )\in C_{+}\big ({\overline{\Omega }}\big )\) with \(q(x)<r(x)<p_3^*(x)\), \(meas\left\{ x\in \Omega:a_i\left( x\right)>0\right\}>0\) for \(i=1,2,3\), and \(meas\left\{ x\in \Omega:b_i\left( x\right)>0\right\}>0\) for \(i=1,2,3\).                

(F1):

\(f:\Omega \times {\mathbb {R}}\rightarrow {\mathbb {R}}\) satisfies the \(Carath\acute{e}odory\) condition, which states that \(f\left( \cdot,t\right)\) is measurable for all \(t\in {\mathbb {R}}\) and \(f\left( x,\cdot \right)\) is continuous for almost all \(x\in \Omega\).

(F2):

f satisfies the following condition:

$$\begin{aligned} \left| f\left( x,t\right) \right| \le \left| a_1\left( x\right) \right||t|^{\omega \left( x\right) -1}+\left| b_1\left( x\right) \right||t|^{q\left( x\right) -1},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$

where \(\omega \left( x\right)\), \(q\left( x\right)\), \(a_1\left( x\right)\) and \(b_1\left( x\right)\) are given in (H1) and (H2).

(F3):

There are positive constants \(M_{1}\) and \(\theta _{1}\) such that \(\theta _{1}>p^{+}\) and

$$\begin{aligned} 0<\theta _1F\left( x,t\right) \le f\left( x,t\right) t,\ \left| t\right| \ge M_1. \end{aligned}$$
(F4):

\(f\left( x,t\right) =o\left( \left| t\right| ^{p^+-1}\right)\), as \(t\rightarrow 0\) for \(x\in \Omega\) uniformly.

(F5):

f satisfies the following condition:

$$\begin{aligned} \left| f\left( x,t\right) \right| \ge \left| a_3\left( x\right) \right||t|^{\eta \left( x\right) -1}+\left| b_3\left( x\right) \right||t|^{l\left( x\right) -1},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$

where \(\eta \left( x\right)\), \(l\left( x\right)\), \(a_3\left( x\right)\) and \(b_3\left( x\right)\) are given in (H1) and (H2).

(F6):

f satisfies the following condition:

$$\begin{aligned} \left| f\left( x,t\right) \right| \le \left| a_2\left( x\right) \right||t|^{\alpha \left( x\right) -1}+\left| b_2\left( x\right) \right||t|^{\beta \left( x\right) -1},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$

where \(\alpha \left( x\right)\), \(\beta \left( x\right)\), \(a_2\left( x\right)\) and \(b_2\left( x\right)\) are given in (H1) and (H2). It follows from (F2) that:

\((F2')\):

F satisfies the following growth condition:

$$\begin{aligned} \left| F(x,t)\right| \le \frac{\left| a_1(x)\right| }{\omega (x)}|t|^{\omega (x)}+\frac{\left| b_1(x)\right| }{q(x)}|t|^{q(x)},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}. \end{aligned}$$

Our main result is given by the following theorem:

Theorem 1.1

Assume (H1), (H2), (F1) and (F6) hold. Then problem (1) has a weak solution for any \(\lambda> 0\).

Theorem 1.2

Assume (H1),(H2),(F1)-(F4) hold, then the problem (1) has a non-trivial weak solution for all \(\lambda>0\).

Theorem 1.3

Assume (H1), (H2), (F1)-(F3) hold. If

$$\begin{aligned} f\left( x,-t\right) =-f\left( x,t\right),\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$
(8)

then for all \(\lambda>0\), \(I_{\lambda }\) has a list of critical points \(\left\{ \pm u_n\right\}\) in X such that, as \(n\rightarrow \infty\).

Theorem 1.4

Assume (H1), (H2), (F1)-(F3) and (F5) hold. If

$$\begin{aligned} f\left( x,-t\right) =-f\left( x,t\right),\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$

then for all \(\lambda>0\), \(I_{\lambda }\) has a list of critical points \(\left\{ \pm u_n\right\}\) in X such that \(I_{\lambda }\left( \pm u_{n}\right) <0\) and \(I_{\lambda }\left( \pm u_{n}\right) \rightarrow 0\), as \(n\rightarrow \infty\).

The remainder of this paper is organized as follows: In Section 2, we review some fundamental theories and results related to variable exponent Lebesgue-Sobolev spaces. Then, in Section 3, we establish the existence of weak solutions to problem (1) by employing different methods based on various assumptions on f(xu).

2 Preliminaries

In order to discuss the problem (1), we need some theories about the spaces \(L^{p(x)}(\Omega )\) and \(W^{k,p(x)}\left( \Omega \right)\). Below, we first give some basic definitions and properties relevant to this paper.

For any \(p(x)\in C_{+}\big ({\overline{\Omega }}\big )\), we introduce the Lebesgue space with variable exponents:

$$\begin{aligned} L^{p(x)}\left( \Omega \right) =\left\{ u: u\ \text {is a measurable real-valued function in}\ \Omega, \int _{\Omega }|u(x)|^{p(x)}dx<\infty \right\}, \end{aligned}$$

with the corresponding Luxemburg norm:

$$\begin{aligned} \left\| u\right\| _{L^{p(x)}(\Omega )}=\inf \left\{ \lambda>0:\int _{\Omega }\left| \frac{u\left( x\right) }{\lambda }\right| ^{p(x)}dx\le 1\right\}. \end{aligned}$$

It is worth noting that if \(p\left( x\right) \equiv p,\ p\ge 1\) is a constant, the norm of \(L^{p(x)}(\Omega )\) is equivalent to the standard norm of \(L^{p}(\Omega )\), namely,

$$\begin{aligned} \left\| u\right\| _{L^p(\Omega )}=\left( \int _{\Omega }\left| u\left( x\right) \right| ^pdx\right) ^{\frac{1}{p}}. \end{aligned}$$

Proposition 2.1

[7, 8] The space \(\left( L^{p(x)}\left( \Omega \right),\left\| \bullet \right\| _{L^{p(x)}(\Omega )}\right)\) is a separable, uniformly convex, reflexive Banach space. Its conjugate space is \(L^{p^{\prime }(x)}(\Omega )\), where \(p^{\prime }(x)\) is conjugate function of \(p\left( x\right)\), namely,

$$\begin{aligned}\begin{aligned} \frac{1}{p(x)}+\frac{1}{p^{\prime }(x)}&=1.\end{aligned}\end{aligned}$$

For any \(u\in L^{p(x)}\left( \Omega \right)\) and \(v\in L^{p^{\prime }(x)}(\Omega )\), we have

$$\begin{aligned} \left| \int _{\Omega }uvdx\right| \le \left( \frac{1}{p^-}+\frac{1}{(p^{\prime })^-}\right) \Vert u\Vert _{L^{p(x)}(\Omega )}\Vert v\Vert _{L^{p^{\prime }(x)}(\Omega )}\le 2\Vert u\Vert _{L^{p(x)}(\Omega )}\Vert v\Vert _{L^{p^{\prime }(x)}(\Omega )}. \end{aligned}$$

Proposition 2.2

[8] Let \(\ell _{p(x)}(u)=\int _{\Omega }|u(x)|^{p(x)}dx\), for all \(u\in L^{p\left( x \right)}\left( \varOmega \right).\)We have

$$\begin{aligned} \left\| u\right\| _{L^{p(x)}(\Omega )}^{{\check{p}}}\le \ell _{p(x)}\left( u\right) \le \left\| u\right\| _{L^{p(x)}(\Omega )}^{\hat{p}}. \end{aligned}$$

Next we introduce the variable exponent Sobolev space:

$$\begin{aligned} W^{k,p(x)}\left( \Omega \right) =\left\{ u\in L^{p(x)}\left( \Omega \right) {:}D^{\alpha }u\in L^{p(x)}\left( \Omega \right),\left| \alpha \right| {\le }k\right\}, \end{aligned}$$

with the norm:

$$\begin{aligned} \left\| u\right\| _{k,p(x)}=\sum _{|\alpha|\le k}\left\| D^\alpha u\right\| _{L^{p(x)}(\Omega )}, \end{aligned}$$

where

$$\begin{aligned} D^\alpha u=\frac{\partial ^{|\alpha|}}{\partial {x_1}^{\alpha _1}\partial {x_2}^{\alpha _2}\cdots \partial {x_N}^{\alpha _N}}u, \end{aligned}$$

\(\alpha =\left( \alpha _{1},\cdots,\alpha _{N}\right)\) is a multi-index and \(\left| \alpha \right| =\sum _{i=1}^N\alpha _i\).

In addition, \(W_0^{k,p(x)}\left( \Omega \right)\) is the closure of \(C_0^\infty (\Omega )\) in \(W^{k,p(x)}(\Omega )\).

Proposition 2.3

[7, 8] Let \(p(x)\in C_+\left( {\overline{\Omega }}\right)\). Then the Space \(\left( W^{k,p(x)}\left( \Omega \right),\left\| \bullet \right\| _{k,p(x)}\right)\) is a reflexive and separable Banach space.

Proposition 2.4

[8] Let \(p(x),q(x)\in C_+\left( {\overline{\Omega }}\right)\) such that: \(q\left( x\right) \le p_{k}^{*}(x)\). Then there is continuous embedding:

$$\begin{aligned} W^{k,p(x)}(\Omega )\hookrightarrow L^{q(x)}(\Omega ). \end{aligned}$$

If \(\le\) is replaced by <, then the embedding is compact.

The problem (1) studied in this paper is discussed in the following Sobolev space:

$$\begin{aligned} X:=W^{3,p(x)}\left( \Omega \right) \cap W_{0}^{1,p(x)}\left( \Omega \right), \end{aligned}$$

with the corresponding norm:

$$\begin{aligned} \left\| u\right\| _X=\left\| u\right\| _{1,p(x)}+\left\| u\right\| _{2,p(x)}+\left\| u\right\| _{3,p(x)}. \end{aligned}$$

\(\left\| u\right\| _{X}\) and \(\Vert \nabla \Delta u\Vert _{L^{p(x)}(\Omega )}\) are two equivalent norms in X [25]. For any \(u\in X\), define

$$\begin{aligned} \Vert u\Vert =\inf \left\{ \lambda>0:\int _{\Omega }\left| \frac{\nabla \Delta u\left( x\right) }{\lambda }\right| ^{p(x)}dx\le 1\right\}. \end{aligned}$$

Thus, \(\Vert u\Vert\) in X is equivalent to \(\left\| u\right\| _X\).

Proposition 2.5

[25]If \(1<p^{-}\le p^{+}<\infty\), the space \(\left( X,\left\| \bullet \right\| _{X}\right)\) is a reflexive and separable Banach space.

According to Proposition 2.2, there is the following proposition

Proposition 2.6

Let \(\zeta \left( u\right) =\int _{\Omega }\left| \nabla \Delta u\right| ^{p(x)}dx\), for all \(u\in X\). We have

$$\begin{aligned} \Vert u\Vert ^{{\check{p}}}\le \zeta \left( u\right) \le \Vert u\Vert ^{\hat{p}}. \end{aligned}$$

Proposition 2.7

[25] Assume \(q(x)\in C_{+}\left( {\overline{\Omega }}\right)\) and \(q\left( x\right) <p_{3}^{*}\left( x\right)\). Then there is a continuous and compact embedding:

$$\begin{aligned} X\hookrightarrow L^{q(x)}(\Omega ). \end{aligned}$$

3 Existence of Solution

Definition 3.1

If

$$\begin{aligned} \int _{\Omega }|\nabla \Delta u|^{p(x)-2}\nabla \Delta u\nabla \Delta vdx=\lambda \int _{\Omega }f\left( x,u\right) v dx \end{aligned}$$

for all \(v\in X\), then \(u\in X\) is said to be a weak solution of the problem (1).

Define the functionals \(\varphi\), \(\psi\) and \(I_{\lambda }:X\rightarrow {\mathbb {R}}\) by:

$$\begin{aligned} \varphi (u)&=\int _{\Omega }\frac{1}{p\left( x\right) }|\nabla \Delta u|^{p(x)}dx, \end{aligned}$$
(9)
$$\begin{aligned} \psi (u)&=\int _\Omega F(x,u)dx, \end{aligned}$$
(10)
$$\begin{aligned} I_{\lambda }\left( u\right)&=\varphi \left( u\right) -\lambda \psi \left( u\right). \end{aligned}$$
(11)

3.1 The proof of Theorem 1.1

Lemma 3.1

\(\varphi \in C^{1}\left( X,{\mathbb {R}}\right)\) and its \(Fr\acute{e}chet\) derivative is given by:

$$\begin{aligned} \left\langle \varphi ^{\prime }(u),v\right\rangle =\int _{\Omega }\left| \nabla \Delta u(x)\right| ^{p(x)-2}\nabla \Delta u(x)\nabla \Delta v(x)dx. \end{aligned}$$

Proof

Let \(u(x),\ v(x)\in X\), \(x\in \Omega\) and \(0<\left| t\right| <1\). By the mean value theorem, there exists \(s\in \left[ 0,1\right]\) such that

$$\begin{aligned}&\left| \frac{\left| \nabla \Delta \left( u\left( x\right) +tv\left( x\right) \right) \right| ^{p\left( x\right) }-\left| \nabla \Delta u\left( x\right) \right| ^{p\left( x\right) }}{p\left( x\right) t}\right| \\&\quad =\left| \nabla \Delta \left( u\left( x\right) +tsv\left( x\right) \right) \right| ^{p\left( x\right) -1}\left| \nabla \Delta v\left( x\right) \right| \\&\quad \le \left( \left| \nabla \Delta u\left( x\right) \right| +\left| \nabla \Delta v(x)\right| \right) ^{p(x)-1}\left| \nabla \Delta v(x)\right|. \end{aligned}$$

Using the inequality [8]

$$\begin{aligned} \left| u(x)+v(x)\right| ^{p(x)}\le 2^{p^{+}-1}\left( \left| u(x)\right| ^{p(x)}+\left| v(x)\right| ^{p(x)}\right), \end{aligned}$$

Proposition 2.1 and Proposition 2.6, We can obtain

$$\begin{aligned}&\int _{\Omega }\left( \left| \nabla \Delta u\left( x\right) \right| +\left| \nabla \Delta v\left( x\right) \right| \right) ^{p(x)-1}\left| \nabla \Delta v(x)\right| dx\\&\quad \le 2^{p^+-2}\int _{\Omega }\big (\left| \nabla \Delta u\left( x\right) \right| ^{p\left( x\right) -1}\left| \nabla \Delta v\left( x\right) \right| +\left| \nabla \Delta v\left( x\right) \right| ^{p\left( x\right) }\big )dx\\&\quad \le 2^{p^+-1}\left\| \left| \nabla \Delta u\left( x\right) \right| ^{p\left( x\right) -1}\right\| _{L^{\frac{p(x)}{p(x)-1}}(\Omega )}\left\| \nabla \Delta v\left( x\right) \right\| _{L^{p(x)}(\Omega )}+2^{p^+-2}\left\| v\left( x\right) \right\| ^{\hat{p}}\\&\quad \le 2^{p^+-1}\left\| u\left( x\right) \right\| ^{\hat{p}-1}\left\| v\left( x\right) \right\| +2^{p^+-2}\left\| v\left( x\right) \right\| ^{\hat{p}}. \end{aligned}$$

Thus,

$$\begin{aligned} \left( \left| \nabla \Delta u(x)\right| +\left| \nabla \Delta v(x)\right| \right) ^{p(x)-1}\left| \nabla \Delta v(x)\right| \in L^1\left( \Omega \right). \end{aligned}$$

By the Lebesgue dominated convergence theorem, we obtain

$$\begin{aligned} \left\langle \varphi ^{\prime }(u\left( x\right) ),v\left( x\right) \right\rangle&=\lim _{t\rightarrow 0}\frac{\varphi (u\left( x\right) +tv\left( x\right) )-\varphi (u\left( x\right) )}{t}\\&=\lim _{t\rightarrow 0}\int _{\Omega }\frac{\left| \nabla \Delta \left( u(x)+tv(x)\right) \right| ^{p(x)}-\left| \nabla \Delta u\left( x\right) \right| ^{p(x)}}{tp(x)}dx\\&=\int _{\Omega }\lim _{t\rightarrow 0}\left| \nabla \Delta \left( u\left( x\right) +tsv(x)\right) \right| ^{p(x)-2}\nabla \Delta \left( u\left( x\right) +tsv\left( x\right) \right) \nabla \Delta v(x)dx\\&=\int _{\Omega }\left| \nabla \Delta u(x)\right| ^{p(x)-2}\nabla \Delta u(x)\nabla \Delta v(x)dx. \end{aligned}$$

Let \(u_{n}\left( x\right) \rightarrow u\left( x\right)\) in X, i.e., \(\nabla \Delta u_{n}\left( x\right) \rightarrow \nabla \Delta u\left( x\right)\) in \(L^{p(x)}\big (\Omega \big )\). Then,

$$\begin{aligned} \left| \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),v\right\rangle \right|&=\left| \int _{\Omega }\left( \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right) \nabla \Delta vdx\right| \\&\le \int _{\Omega }\left| \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right| \left| \nabla \Delta v\right| dx\\&\le 2\left\| \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right\| _{L^{\frac{p(x)}{p(x)-1}}(\Omega )}\left\| \nabla \Delta v\right\| _{L^{p(x)}(\Omega )}\\&=2\left\| \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right\| _{L^{\frac{p(x)}{p(x)-1}}(\Omega )}\Vert v\Vert. \end{aligned}$$

Let \(P\big (x,\nabla \Delta u\big )=\big|\nabla \Delta u\big|^{p(x)-2}\nabla \Delta u\). We deduce from theorem 1.16 of [8] or theorem 1.1 of [9] that \(P\left( x,\bullet \right):L^{p(x)}\left( \Omega \right) \rightarrow L^{\frac{p(x)}{p(x)-1}}\left( \Omega \right)\) is continuous,which shows that

$$\begin{aligned} P\left( x,\nabla \Delta u_n\right) \rightarrow P\left( x,\nabla \Delta u\right) \ \text {in}\ L^{\frac{p(x)}{p(x)-1}}\left( \Omega \right). \end{aligned}$$

Therefore,

$$\begin{aligned} \left\| \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u)\right\|&=\sup _{0\ne v\in X}\frac{\left| \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),v\right\rangle \right| }{\Vert v\Vert }\\&\le \sup _{0\ne v\in X}\frac{2\left\| \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right\| _{L^{\frac{p(x)}{p(x)-1}}(\Omega )}\Vert v\Vert }{\Vert v\Vert }\\&\left. =2\left\| P\left( x,\nabla \Delta u_n\right. \right) -P\left( x,\nabla \Delta u\right) \right\| _{L^{\frac{p(x)}{p(x)-1}}(\Omega )}\rightarrow 0,\ n\rightarrow \infty. \end{aligned}$$

To sum up, we can conclude that \(\varphi \in C^{1}\left( X,\mathbb R\right)\). \(\square\)

Lemma 3.2

\(\varphi ^{\prime }\) is a mapping of type \((S_{+})\), i.e., if \(u_{n}\rightharpoonup u\) in X and

$$\begin{aligned} \limsup \limits _{n\rightarrow \infty }\left<\varphi ^{\prime }\left( u_n\right) -\varphi ^{\prime }\left( u\right),u_n-u\right>\le 0, \end{aligned}$$

then \(u_{n}\rightarrow u\) in X, as \(n\rightarrow \infty\).

Proof

By Lemma 3.1, it follows that

$$\begin{aligned} & \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),u_n-u\right\rangle \\ & \quad =\int _{\Omega }\left( \left| \nabla \Delta u_n\right| ^{p(x)-2}\nabla \Delta u_n-\left| \nabla \Delta u\right| ^{p(x)-2}\nabla \Delta u\right) \nabla \Delta (u_n-u)dx. \end{aligned}$$

Using the following elementary inequality [30]:

$$\begin{aligned} \left| x-y\right| ^\gamma \le 2^\gamma \left( \left| x\right| ^{\gamma -2}x-\left| y\right| ^{\gamma -2}y\right) \cdot \left( x-y\right),\ \gamma \ge 2, \end{aligned}$$
(12)

we can conclude that for all \(u,\ v\in X\) with \(u\ne v\),

$$\begin{aligned} \left\langle \varphi ^{\prime }(u)-\varphi ^{\prime }(v),u-v\right\rangle>0. \end{aligned}$$

This implies \(\varphi ^{\prime }\) is strictly monotone. Let \(\{u_n\}\subset X\) be a sequence such that \(u_{n}\rightharpoonup u,\) and

$$\begin{aligned} \limsup \limits _{n\rightarrow \infty }\left<\varphi ^{\prime }\left( u_n\right) -\varphi ^{\prime }\left( u\right),u_n-u\right>\le 0. \end{aligned}$$

By Proposition 2.6, it suffices to show that \(\int _{\Omega }|\nabla \Delta u_n-\nabla \Delta u|^{p(x)}dx\rightarrow 0\). By the monotonicity of \(\varphi ^{\prime }\), we have

$$\begin{aligned} \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),u_n-u\right\rangle \ge 0. \end{aligned}$$

Thus,

$$\begin{aligned} \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),u_n-u\right\rangle \rightarrow 0,\ n\rightarrow \infty. \end{aligned}$$
(13)

From (12) and (13), we deduce that

$$\begin{aligned} \left\langle \varphi ^{\prime }(u_n)-\varphi ^{\prime }(u),u_n-u\right\rangle&\ge \int _{\Omega }\frac{1}{2^{p(x)}}\left| \nabla \Delta (u_n-u)\right| ^{p(x)}\\&\ge \frac{1}{2^{p^+}}\int _{\Omega }\left| \nabla \Delta (u_n-u)\right| ^{p(x)}\rightarrow 0,\ n\rightarrow \infty. \end{aligned}$$

Therefore, \(u_{n}\rightarrow u\) in X, which implies that \(\varphi ^{\prime }\) is a mapping of type \((S_{+})\). \(\square\)

Lemma 3.3

Assume that (H1) holds. Then \(\varphi\) is weakly lower semi-continuous, meaning that if \({{u}_{n}}\rightharpoonup u\) as \(n\rightarrow \infty\) in X, then \(\varphi \left( u \right) \le \underset{n\rightarrow \infty }{\mathop {\lim \inf }}\,\varphi \left( {{u}_{n}} \right)\).

Proof

Let \(u_{n}\rightharpoonup u\) in X. From Lemma 3.1 and Lemma 3.2, we know that \(\varphi\) is convex functional [31] and

$$\begin{aligned} \varphi \left( {{u}_{n}} \right) \ge \varphi \left( u \right) +\left\langle {{\varphi }^{'}}\left( u \right),{{u}_{n}}-u \right\rangle,\ \text {for any}\ n. \end{aligned}$$

According to the definition of weak convergence we can obtain

$$\begin{aligned} \left\langle \varphi'\left( u \right),{{u}_{n}}-u \right\rangle \rightarrow 0. \end{aligned}$$

Thus,

$$\begin{aligned} \underset{n\rightarrow \infty }{\mathop {\lim \inf }}\,\varphi \left( {{u}_{n}} \right) \ge \varphi \left( u \right) +\underset{n\rightarrow \infty }{\mathop {\lim \inf }}\,\left\langle {{\varphi }^{'}}\left( u \right),{{u}_{n}}-u \right\rangle =\varphi \left( u \right). \end{aligned}$$

\(\square\)

Lemma 3.4

Assume

$$\begin{aligned} \left| f\left( x,t\right) \right| \le \left| a_2\left( x\right) \right||t|^{\alpha \left( x\right) -1}+\left| b_2\left( x\right) \right||t|^{\beta \left( x\right) -1},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$
(14)

and that conditions (H1), (H2) and (F1) hold. Then for almost all \(x\in \Omega\) and all \(t\in {\mathbb {R}},\) the following estimate can be obtained:

$$\begin{aligned} F\left( x,t \right) \le \frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}}{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}}{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }+2L{\left| t \right| }^{r\left( x \right) }, \end{aligned}$$

where \(L=max\left\{ \frac{1}{{{\alpha }^{-}{\left( {{\gamma _3 }^{+}} \right) }^{'}}},\ \frac{1}{{{\beta }^{-}{\left( {{\gamma _4 }^{+}} \right) }^{'}}}\right\}\). Also, the Nemytskij operator

$$\begin{aligned} u\mapsto F\left( x,u \right) \end{aligned}$$

is continuous from \({{L}^{r\left( x \right) }}\left( \Omega \right)\) to \({{L}^{1}}\left( \Omega \right)\).

Proof

According to (14) and the Young inequality, we can deduce that

$$\begin{aligned} F\left( x,t \right)&\le \frac{\left| a_2\left( x \right) \right| }{\alpha \left( x\right) }{{\left| t \right| }^{\alpha \left( x\right) }}+\frac{\left| b_2\left( x \right) \right| }{\beta \left( x \right) }{{\left| t \right| }^{\beta \left( x \right) }}\\&\le \frac{1}{{{\alpha }^{-}}}\left| a_2\left( x \right) \right| {{\left| t \right| }^{\alpha \left( x \right) }}+\frac{1}{{{\beta }^{-}}}\left| b_2\left( x \right) \right| {{\left| t \right| }^{\beta \left( x \right) }} \\&\le \frac{1}{{{\alpha }^{-}}}\left( \frac{{{\left| a_2\left( x \right) \right| }^{\gamma _3\left( x \right) }}}{{{\gamma _3 }^{-}}}+\frac{{{\left| t \right| }^{\alpha \left( x \right) {{\left( \gamma _3\left( x \right) \right) }^{'}}}}}{{{\left( {{\gamma _3 }^{+}} \right) }^{'}}} \right) +\frac{1}{{{\beta }^{-}}}\left( \frac{{{\left| b_2\left( x \right) \right| }^{\gamma _4\left( x \right) }}}{{{\gamma _4 }^{-}}}+\frac{{{\left| t \right| }^{\beta \left( x \right) {{\left( \gamma _4\left( x \right) \right) }^{'}}}}}{{{\left( {{\gamma _4 }^{+}} \right) }^{'}}} \right) \\&=\frac{1}{{{\alpha }^{-}}}\left( \frac{{{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }}}{{{\gamma _3 }^{-}}}+\frac{{{\left| t \right| }^{r\left( x \right) }}}{{{\left( {{\gamma _3 }^{+}} \right) }^{'}}} \right) +\frac{1}{{{\beta }^{-}}}\left( \frac{{{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }}}{{{\gamma _4 }^{-}}}+\frac{{{\left| t \right| }^{r\left( x \right) }}}{{{\left( {{\gamma _4 }^{+}} \right) }^{'}}} \right) \\&\le \frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}}{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}}{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }+2L{\left| t \right| }^{r\left( x \right) }. \end{aligned}$$

Let \(u\in {{L}^{r\left( x \right) }}\). By Proposition 2.2, we have

$$\begin{aligned} \int _{\Omega }{F\left( x,u \right) }dx&\le \frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}} \int _{\Omega }{{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }}dx+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}} \int _{\Omega }{{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }}dx\\&\quad +2L \int _{\Omega }{{\left| t \right| }^{r\left( x \right) }}dx\\&\le \frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}}\left\| a_2\left( x \right) \right\| _{{{L}^{\gamma _3 \left( x \right) }}\left( \Omega \right) }^{{\hat{\gamma _3 }}}+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}} \left\| b_2\left( x \right) \right\| _{{{L}^{\gamma _4 \left( x \right) }}\left( \Omega \right) }^{{\hat{\gamma _4 }}}\\&\quad +2L\left\| u \right\| _{{{L}^{r\left( x \right) }}\left( \Omega \right) }^{{\hat{r}}} \\&<\infty. \end{aligned}$$

So \(F\left( x,u \right) \in {{L}^{1}}\left( \Omega \right)\). Let \({{u}_{n}}\rightarrow u\) in \({{L}^{r\left( x \right) }}\left( \Omega \right)\). By Lemma 2.2 of [9], in the subsequence sense, there is \(g\in {{L}^{r\left( x \right) }}\) such that

$$\begin{aligned} \left| F\left( x,{{u}_{n}} \right) -F\left( x,u \right) \right|&\le \frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}}{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}}{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }+2L{\left| {u}_{n} \right| }^{r\left( x \right) }\\&\quad +\frac{1}{{{\alpha }^{-}{\gamma _3 }^{-}}}{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }+\frac{1}{{{\beta }^{-}{\gamma _4 }^{-}}}{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }+2L{\left| u \right| }^{r\left( x \right) }\\&\le \frac{2}{{{\alpha }^{-}{\gamma _3 }^{-}}}{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }+\frac{2}{{{\beta }^{-}{\gamma _4 }^{-}}}{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }\\&\quad +2L{\left| g \right| }^{r\left( x \right) }+2L{\left| u \right| }^{r\left( x \right) }. \end{aligned}$$

Also using Proposition 2.2, we get

$$\begin{aligned} \int _{\Omega }|F(x,u_{n})-F(x,u)|dx&\le \frac{2}{{{\alpha }^{-}{\gamma _3 }^{-}}} \int _{\Omega }{{\left| a_2\left( x \right) \right| }^{\gamma _3 \left( x \right) }}dx+\frac{2}{{{\beta }^{-}{\gamma _4 }^{-}}} \int _{\Omega }{{\left| b_2\left( x \right) \right| }^{\gamma _4 \left( x \right) }}dx\\&\quad +2L \int _{\Omega }{{\left| g \right| }^{r\left( x \right) }}dx+2L \int _{\Omega }{{\left| u \right| }^{r\left( x \right) }}dx\\&\le \frac{2}{{{\alpha }^{-}{\gamma _3 }^{-}}}\left\| a_2\left( x \right) \right\| _{{{L}^{\gamma _3 \left( x \right) }}\left( \Omega \right) }^{{\hat{\gamma _3 }}}+\frac{2}{{{\beta }^{-}{\gamma _4 }^{-}}} \left\| b_2\left( x \right) \right\| _{{{L}^{\gamma _4 \left( x \right) }}\left( \Omega \right) }^{{\hat{\gamma _4 }}}\\&\quad +2L\left\| g \right\| _{{{L}^{r\left( x \right) }}\left( \Omega \right) }^{{\hat{r}}}+2L\left\| u \right\| _{{{L}^{r\left( x \right) }}\left( \Omega \right) }^{{\hat{r}}} \\&<\infty. \end{aligned}$$

Thus, \(\left| F\left( x,{{u}_{n}} \right) -F\left( x,u \right) \right| \in {{L}^{1}}\left( \Omega \right)\). It is clear that F satisfies the Caratheodory condition, so by the Lebesgue dominated convergence theorem, we have \(F\left( x,{{u}_{n}} \right) \rightarrow F\left( x,u \right)\) in \({{L}^{1}}\left( \Omega \right).\) This concludes the proof of the theorem. \({\hfill}{\square}\)

Lemma 3.5

Assuming that (14), (H1), (H2) and (F1) hold, then \(\psi\) is weak-strong continuous, which means that if \(u_{n}\rightharpoonup u\) in X, then \(\psi \left( {{u}_{n}} \right) \rightarrow \psi \left( u \right)\).

Proof

Let \(\left\{ {{u}_{n}} \right\} \subset X\) such that \(u_{n}\rightharpoonup u\). By proposition 2.7, we can deduce that \({{u}_{n}}\rightarrow u\) in \({{L}^{r\left( x \right) }}\left( \Omega \right)\). Then, \(\psi \left( {{u}_{n}} \right) \rightarrow \psi \left( u \right)\) follows from Lemma 3.4. \(\hfill{\square}\)

The Proof of Theorem 1.1

Let \(u \in X\) with \(\Vert u\Vert> 1\). From Proposition 2.1, Proposition 2.6 and Proposition 2.7, we can deduce that

$$\begin{aligned} I_{\lambda }\left( u\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta u\right| ^{p(x)}dx-\lambda \int _{\Omega }F\left( x,u\right) dx\\&\ge \frac{1}{p^+}\int _{\Omega }|\nabla \Delta u|^{p(x)}dx-\lambda \int _{\Omega }\left( \frac{\left| a_2\left( x \right) \right| }{\alpha \left( x\right) }{{\left| u \right| }^{\alpha \left( x\right) }}+\frac{\left| b_2\left( x \right) \right| }{\beta \left( x \right) }{{\left| u \right| }^{\beta \left( x \right) }}\right) dx\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-\frac{2\lambda }{\alpha ^-}\Vert a_2(x)\Vert _{L^{\gamma _3}(\Omega )}\left\| \left| u\right| ^{\alpha (x)}\right\| _{L^{{(\gamma _3)}^{\prime }}(\Omega )}\\&\quad -\frac{2\lambda }{\beta ^-}\Vert b_2(x)\Vert _{L^{\gamma _4}(\Omega )}\left\| \left| u\right| ^{\beta (x)}\right\| _{L^{{(\gamma _4)}^{\prime }}(\Omega )}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-\frac{2\lambda }{\alpha ^-}\Vert a_2(x)\Vert _{L^{\gamma _3}(\Omega )}\left\| u\right\| ^{\hat{\alpha }}_{L^{r(x)}(\Omega )}-\frac{2\lambda }{\beta ^-} \Vert b_2(x)\Vert _{L^{\gamma _4} (\Omega )}\left\| u\right\| ^{\hat{\beta }}_{L^{r(x)}(\Omega )}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-\frac{2\lambda C}{\alpha ^-}\Vert a_2(x)\Vert _{L^{\gamma _3}(\Omega )}\left\| u\right\| ^{\hat{\alpha }}-\frac{2\lambda C}{\beta ^-} \Vert b_2(x)\Vert _{L^{\gamma _4} (\Omega )}\left\| u\right\| ^{\hat{\beta }}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-2L_{1}\left\| u\right\| ^{\hat{\beta }}, \end{aligned}$$

where \(L_{1}=max\left\{ \frac{2\lambda C}{\alpha ^-}\Vert a_2(x)\Vert _{L^{\gamma _3}(\Omega )},\frac{2\lambda C}{\beta ^-} \Vert b_2(x)\Vert _{L^{\gamma _4} (\Omega )}\right\}\). Since \(1\le \alpha ^{-}\le \alpha ^{+}<\beta ^{-}\le \beta ^{+}<p^{-}\) implies that \(I_\lambda (u)\rightarrow \infty\) as \(\Vert u\Vert \rightarrow \infty\) for \(\lambda>0,\) namely, \(I_\lambda\) is coercive. By Lemma 3.3 and Lemma 3.5, the functional \(I_\lambda\) is weakly lower semi-continuous, so there exists a global minimizer, denoted as u, of \(I_\lambda\) in the function space X. Moreover, u is a weak solution to problem (1). This completes the proof.

\(\hfill\square\)

Example 3.1

Assume that

$$\begin{aligned} \left| f\left( x,t\right) \right| \le C_1(1+|t|^{\beta \left( x\right) -1}),\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}, \end{aligned}$$
(15)

where \(C_1\) is a positive constant. If conditions (H1) and (F1) hold, then problem (1) has a weak solution for any \(\lambda> 0\).

Proof

According to (15), we get

$$\begin{aligned} F\left( x,t \right)&\le C_1\left( \left| t \right| +\frac{1}{\beta \left( x \right) }{{\left| t \right| }^{\beta \left( x \right) }}\right) \\&\le C_1\left( \left| t \right| +\frac{1}{{{\beta }^{-}}}{{\left| t \right| }^{\beta \left( x \right) }}\right). \end{aligned}$$

Let \(u\in {{L}^{\beta \left( x \right) }}\). By Theorem 2.8 of [7] and Proposition 2.2, we have

$$\begin{aligned} \int _{\Omega }{F\left( x,u \right) }dx&\le C_1\int _{\Omega }{\left( \left| u \right| +\frac{1}{{{\beta }^{-}}}{{\left| u \right| }^{\beta \left( x \right) }}\right) }dx \\&\le C_1\left\| u \right\| _{{L}^{1}\left( \Omega \right) }+\frac{C_1}{{{\beta }^{-}}}\left\| u \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}} \\&\le C\left\| u \right\| _{{L}^{\beta \left( x \right) }\left( \Omega \right) }+\frac{C_1}{{{\beta }^{-}}}\left\| u \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}} \\&<\infty. \end{aligned}$$

So, \(F\left( x,u \right) \in {{L}^{1}}\left( \Omega \right)\). Let \({{u}_{n}}\rightarrow u\) in \({{L}^{\beta \left( x \right) }}\left( \Omega \right)\). By Lemma 2.2 of [9], in the subsequence sense, there is \(g\in {{L}^{\beta \left( x \right) }}\) such that

$$\begin{aligned} \left| F\left( x,{{u}_{n}} \right) -F\left( x,u \right) \right|&\le C_1\left( \left| {u}_{n} \right| +\frac{1}{{{\beta }^{-}}}{{\left| {u}_{n} \right| }^{\beta \left( x \right) }}+\left| u \right| +\frac{1}{{{\beta }^{-}}}{{\left| u \right| }^{\beta \left( x \right) }}\right) \\&\le C_1\left( \left| g \right| +\left| u \right| +\frac{1}{{{\beta }^{-}}}{{\left| g \right| }^{\beta \left( x \right) }}+\frac{1}{{{\beta }^{-}}}{{\left| u \right| }^{\beta \left( x \right) }}\right). \end{aligned}$$

Using Theorem 2.8 from [7] and Proposition 2.2, we can derive

$$\begin{aligned} \int _{\Omega }|F(x,u_{n})-F(x,u)|dx&\le C_1\int _{\Omega }{\left( \left| g \right| +\left| u \right| +\frac{1}{{{\beta }^{-}}}{{\left| g \right| }^{\beta \left( x \right) }}++\frac{1}{{{\beta }^{-}}}{{\left| u \right| }^{\beta \left( x \right) }}\right) }dx\\&\le C_1\left\| g \right\| _{{L}^{1}\left( \Omega \right) }+C_1\left\| u \right\| _{{L}^{1}\left( \Omega \right) }+\frac{C_1}{{{\beta }^{-}}}\left\| g \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}}\\&\quad +\frac{C_1}{{{\beta }^{-}}}\left\| u \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}} \\&\le C\left\| g \right\| _{{L}^{\beta \left( x \right) }\left( \Omega \right) }+C\left\| u \right\| _{{L}^{\beta \left( x \right) }\left( \Omega \right) }+\frac{C_1}{{{\beta }^{-}}}\left\| g \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}}\\&\quad +\frac{C_1}{{{\beta }^{-}}}\left\| u \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}} \\&<\infty. \end{aligned}$$

Thus, \(\left| F\left( x,{{u}_{n}} \right) -F\left( x,u \right) \right| \in {{L}^{1}}\left( \Omega \right)\). It is clear that F satisfies the Caratheodory condition since condition (F1), so by the Lebesgue dominated convergence theorem, we have \(F\left( x,{{u}_{n}} \right) \rightarrow F\left( x,u \right)\) in \({{L}^{1}}\left( \Omega \right).\) Let \(u_{n}\rightharpoonup u\) in X. By Proposition 2.7, we deduce that \({{u}_{n}}\rightarrow u\) in \({{L}^{\beta \left( x \right) }}\left( \Omega \right)\), which implies \(\psi \left( {{u}_{n}} \right) \rightarrow \psi \left( u \right)\). Hence, \(\psi\) is weak-strong continuous. Combining this with Lemma 3.3, we know that the functional \(I_\lambda\) is weakly lower semi-continuous. Let \(u \in X\) with \(\Vert u\Vert> 1\). From Proposition 2.6 and Proposition 2.7, we can conclude

$$\begin{aligned} I_{\lambda }\left( u\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta u\right| ^{p(x)}dx-\lambda \int _{\Omega }F\left( x,u\right) dx\\&\ge \frac{1}{p^+}\int _{\Omega }|\nabla \Delta u|^{p(x)}dx-C_1\lambda \int _{\Omega }{\left( \left| u \right| +\frac{1}{{{\beta \left( x\right) }}}{{\left| u \right| }^{\beta \left( x \right) }}\right) }dx \\&\ge \frac{1}{p^+}\int _{\Omega }|\nabla \Delta u|^{p(x)}dx-C_1\lambda \int _{\Omega }{\left| u \right| dx-\frac{C_1\lambda }{{{\beta }^{-}}}\int _{\Omega }{{\left| u \right| }^{\beta \left( x \right) }}}dx \\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-C_1\lambda \left\| u \right\| _{{L}^{1}\left( \Omega \right) }-\frac{C_1\lambda }{{{\beta }^{-}}}\left\| u \right\| _{{{L}^{\beta \left( x \right) }}\left( \Omega \right) }^{{\hat{\beta }}} \\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-C\lambda \left\| u \right\| -\frac{C\lambda }{{{\beta }^{-}}}\left\| u \right\| ^{{\hat{\beta }}} \\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^-}-2L_2\left\| u \right\| ^{{\hat{\beta }}}, \end{aligned}$$

where \(L_2=max\{C\lambda, \frac{C\lambda }{{{\beta }^{-}}}\}\). Since \(1\le \beta ^{-}\le \beta ^{+}<p^{-}\), this implies \(I_\lambda (u)\rightarrow \infty\) as \(\Vert u\Vert \rightarrow \infty\) for \(\lambda>0\). Thus, \(I_\lambda\) is coercive. Therefore, there exists a global minimizer u of \(I_\lambda\) in the function space X. Moreover, u is a weak solution to the problem (1). This completes the proof. \(\square\)

Example 3.2

Assume that conditions (H1), (H2), (F1) hold. Then problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(x)}^{3} u=\lambda \left| b_2\left( x\right) \right| \left| t\right| ^{\beta \big (x\big )-2}t, & \hbox { in } \Omega,\\ u=\Delta u=\Delta ^{2}u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(16)

has a nontrivial weak solution for any \(\lambda> 0\).

Proof

The same proof technique used in Theorem 1.1 and Example 3.1 can be applied to derive the following result: there exists a global minimizer, denoted as \(u_0,\) of \(I_\lambda\) in the function space X. Moreover, \(u_0\) is a weak solution to problem (16). Let \(v\in X\) such that \(\left\| v\right\| \ne 0\) and \(0<t<1\). Then Since \(\beta ^+<p^-\), we can conclude that \(I_{\lambda }\left( tv\right) <0\) for sufficiently small t. Thus, \(I_{\lambda }\left( u_0\right) <0\). Hence, the weak solution \(u_0\) is non-trivial. \(\hfill\square\)

3.2 The Proof of Theorem 1.2

Lemma 3.6

Assume that (F1) and (F3) hold. Then \(I_{\lambda }\) satisfies \(\left( PS\right)\)-condition for all \(\lambda>0\).

Proof

Since \(\psi ^{\prime }\) is compact, it’s clear that \(\psi ^{\prime }\) is a mapping of type \((S_{+})\).

Let \(\{u_n\}\subset X\) be a \(\left( PS\right)\)-sequence, i.e., there exists a constant M such that

$$\begin{aligned} \sup _{n}\left| I_{\lambda }\left( u_{n}\right) \right| \le M\quad \text {and}\quad I_\lambda ^{\prime }\left( u_n\right) \rightarrow 0,\ \text {as}\ n\rightarrow \infty. \end{aligned}$$

Because \(I_{\lambda }^{\prime }\) is a mapping of type \((S_{+})\) and X is reflexive, it suffices to prove that \(\left\{ u_{n}\right\}\) is bounded in X.

Suppose \(\left\| u_n\right\| \rightarrow \infty\), in the subsequence sense. For n large enough, assume \(\left\| u_n\right\|>1.\) From (F3), Proposition 2.6 and \(\left\langle I_{\lambda }^{\prime }\left( u_{n}\right),u_{n}\right\rangle \ge -\left\| I_{\lambda }^{'}\left( u_{n}\right) \right\| _{X^{*}}\left\| u_{n}\right\| _{X}\), we have

$$\begin{aligned} M&\ge I_{\lambda }\left( u_{n}\right) \\&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta u_{n}\right| ^{p(x)}dx-\lambda \int _{\Omega }F\left( x,u_{n}\right) dx\\&\ge \frac{1}{p^+}\int _{\Omega }|\nabla \Delta u_n|^{p(x)}dx-\lambda \int _{\Omega }\frac{1}{\theta _1}f(x,u_n)u_ndx\\&=\left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \int _{\Omega }|\nabla \Delta u_n|^{p(x)}dx+\frac{1}{\theta _1}\int _{\Omega }(|\nabla \Delta u_n|^{p(x)}-\lambda f(x,u_n)u_n)dx\\&=\left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \int _{\Omega }\left| \nabla \Delta u_n\right| ^{p(x)}dx+\frac{1}{\theta _1}\left\langle I_\lambda ^{\prime }(u_n),u_n\right\rangle \\&\ge \left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \int _{\Omega }|\nabla \Delta u_n|^{p(x)}dx-\frac{1}{\theta _1}\Vert I_\lambda ^{^{\prime }}(u_n)\Vert _{X^*}\Vert u_n\Vert _X\\&\ge \left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \left\| u_n\right\| ^{p^-}-\frac{1}{\theta _1}\left\| I_\lambda ^{\prime }\left( u_n\right) \right\| _{X^*}\left\| u_n\right\| _X. \end{aligned}$$

Due to \(p^{+}<\theta _{1}\) and \(p^{-}>3\), we have \(I_\lambda \big (u_n\big )\rightarrow \infty\) as \(\left\| u_n\right\| \rightarrow \infty\). This leads to a contradiction. \(\hfill\square\)

The Proof of Theorem 1.2

It’s clear from (2) and (11) that \(I_{\lambda }(0)=0\). It follows from Lemma 3.6 that \(I_{\lambda }\) satisfies \(\left( PS\right)\)-condition, so it suffices to prove the geometric conditions in the Mountain pass theorem.

By (F2) and (F4), for any \(\varepsilon>0\), there exists a positive constant \(C(\varepsilon )\) such that for all \((x,t)\in \Omega \times {\mathbb {R}}\),

$$\begin{aligned} \left| F(x,t)\right| \le \varepsilon \left| t\right| ^{p^+}+C(\varepsilon )|b_1(x)||t|^{q(x)}. \end{aligned}$$
(17)

Assume \(\left\| u\right\| <1\). Then, by Proposition 2.1, Proposition 2.6 and Proposition 2.7, we have

$$\begin{aligned} I_{\lambda }\left( u\right)&=\int _{\Omega }\frac{1}{p(x)}|\nabla \Delta u|^{p(x)}dx-\lambda \int _{\Omega }F\left( x,u\right) dx\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^+}-\lambda \int _{\Omega }\left( \varepsilon \left| u\right| ^{p^+}+C\left( \varepsilon \right)|b_1\left( x\right)||u|^{q\left( x\right) }\right) dx\\&=\frac{1}{p^+}\Vert u\Vert ^{p^+}-\lambda \varepsilon \int _{\Omega }|u|^{p^+}dx-\lambda C(\varepsilon )\int _{\Omega }|b_1(x)||u|^{q(x)}dx\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^+}-\lambda \varepsilon \Vert u\Vert _{L^{p^+}(\Omega )}^{p^+}-2\lambda C(\varepsilon )\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\Vert \left| u\right| ^{q(x)}\Vert _{L^{{\left( \gamma _2\left( x\right) \right) }^{\prime }}(\Omega )}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^+}-\lambda \varepsilon C\Vert u\Vert ^{p^+}-2\lambda C(\left. \varepsilon \right) C_2\Vert u\Vert _{L^{r(x)}(\Omega )}^{\hat{q}}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{p^+}-\lambda \varepsilon C\Vert u\Vert ^{p^+}-2\lambda C(\varepsilon )CC_2\Vert u\Vert ^{q^-}, \end{aligned}$$

where \(C_2=\left\| b_1\left( x\right) \right\| _{L^{\gamma _2\left( x\right) }(\Omega )}\). Choosing \(\varepsilon>0\) small enough such that \(0<\lambda \varepsilon C_1<\frac{1}{2p^+}\). Then

$$\begin{aligned} I_\lambda (u)\ge \frac{1}{2p^+}\Vert u\Vert ^{p^+}-C(\lambda,\varepsilon )CC_2\Vert u\Vert ^{q^-}. \end{aligned}$$

Since \(q^{-}>p^{+}\), there exist \(R>0\) small enough and \(\delta>0\) such that

$$\begin{aligned} \inf _{\Vert u\Vert =R}I_{\lambda }\left( u\right) \ge \delta>I_{\lambda }\left( 0\right) =0. \end{aligned}$$

It follows from (F3) that for all \(\left( x,t\right) \in \Omega \times {\mathbb {R}}\) and some constants \(C_3,\ C_4>0\),

$$\begin{aligned} F(x,t)\ge C_3\left| t\right| ^{\theta _1}-C_4. \end{aligned}$$
(18)

Thus, by (18), for \(v\in X\setminus \{0\}\), we have

$$\begin{aligned} I_\lambda \left( tv\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta tv\right| ^{p(x)}dx-\lambda \int _{\Omega }F\left( x,tv\right) dx\\&\le \frac{t^{p^+}}{p^-}\int _{\Omega }\left| \nabla \Delta v\right| ^{p(x)}dx-\lambda C_3t^{\theta _1}\int _{\Omega }\left| v\right| ^{\theta _1} dx+\lambda C_4\left| \Omega \right|, \end{aligned}$$

where \(t\ge 1\) and \(\left| \Omega \right|\) denotes the Lebesgue measure of \(\Omega\). Since \(\theta _1>p^+\), we have \(I_\lambda \left( tv\right) \rightarrow -\infty\), as \(t\rightarrow \infty\). That is, there exists \(e:=tv\) such that \(\Vert e\Vert>R\) and \(I_{\lambda }\left( e\right) \le 0\). Therefore, \(I_{\lambda }\) satisfies the geometric structure of the Mountain pass theorem, and the theorem is proved. \(\square\)

Example 3.3

Assume \(\left| f\left( x,t\right) \right| \le \left| b_1\left( x\right) \right||t|^{q\left( x\right) -1},\ \text {for all}\ \left( x,t\right) \in \Omega \times {\mathbb {R}}\), and that conditions (H1), (H2), (F1), (F3), (F4) hold. Then the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(x)}^{3} u=f(x,u), & \hbox { in } \Omega,\\ u=\Delta u=\Delta ^{2}u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(19)

has a non-trivial weak solution.

3.3 The Proof of Theorem 1.3

Lemma 3.7

[32] Let X be a reflexive and separable Banach space. Then there exist \(\left\{ e_{j}\right\} \subset X\) and \(\left\{ e_j^*\right\} \subset X^*\) such that

$$\begin{aligned} X=\overline{span\left\{ e_j|j=1,2,\cdots \right\} },\ X^*=\overline{span\left\{ e_j^*|j=1,2,\cdots \right\} }, \end{aligned}$$

and

$$\begin{aligned} \left. \langle e_i^*,e_j\rangle =\left\{ \begin{array}{ccc}1,& i=j,\\ 0,& i\ne j.\end{array}\right. \right. \end{aligned}$$

For convenience, we write \(X_{j}=\textrm{span}\{e_{j}\}\), \(Y_k=\bigoplus _{j=1}^kX_j\), \(Z_k=\overline{\bigoplus _{j=k}^\infty X_j}\).

Lemma 3.8

Denote

$$\begin{aligned} \alpha _k:=\sup \left\{ \left\| u\right\| _{L^{r(x)}\left( \Omega \right) }|\left\| u\right\| =1,u\in Z_{_k}\right\}. \end{aligned}$$

Then \(\lim \limits _{k\rightarrow \infty }\alpha _k\rightarrow 0\), for all \(x\in {\overline{\Omega }}\).

Proof

It’s clear that \(0<\alpha _{k+1}\le \alpha _{k}\), so \(\alpha _k\rightarrow \alpha \ge 0\).

Let \(u_{k}\in Z_{k}\) be such that \(\left\| u_{k}\right\| =1\) and \(0\le \alpha _{k}-\left\| u_{k}\right\| _{L^{r(x)}(\Omega )}<\frac{1}{k}\). Then there exist subsequences \(\{u_{k}\}\), which we also denote by \(\{u_{k}\}\), such that \(u_{k}\rightharpoonup u\) and

$$\begin{aligned} \left\langle e_j^*,u\right\rangle =\lim _{k\rightarrow +\infty }\left\langle e_j^*,u_k\right\rangle =0,j=1,2,\cdots. \end{aligned}$$

This implies that \(u=0\), so \(u_{k}\rightharpoonup 0\), as \(k\rightarrow \infty\). It is obtainable from Proposition 2.7 that \(u_{k}\rightarrow 0\) in \(L^{r(x)}(\Omega )\). Therefore we get \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). \(\square\)

Lemma 3.9

Assume (F1) and (F3) hold. Then \(I_{\lambda }\) satisfies \(\left( PS\right) _c^*\)-condition for all \(\lambda>0\).

Proof

Let \(\{u_n\}\subset X\) be a \(\left( PS\right) _c^*\)-sequence, meaning \(u_n\in Y_n,\ \ I_\lambda \left( u_n\right) \rightarrow c,\ \left( I_\lambda|_{Y_n}\right) ^{\prime }\left( u_n\right) \rightarrow 0,\ n\rightarrow \infty\). May as well set \(\left\| u_n\right\|>1\), for sufficiently large n. Using (F3) and Proposition 2.6 we have

$$\begin{aligned} \text {c+1}&\ge I_{\lambda }\left( u_{n}\right) =\int _{\Omega }\frac{1}{p(x)}\Big|\nabla \Delta u_{n}\Big|^{p(x)}dx-\lambda \int _{\Omega }F\big (x,u_{n}\big )dx \\&\ge \frac{1}{p^+}\int _\Omega \left| \nabla \Delta u_n\right| ^{p(x)}dx-\lambda \int _\Omega \frac{1}{\theta _1}f\left( x,u_n\right) u_ndx \\&=\left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \int _\Omega \left| \nabla \Delta u_n\right| ^{p(x)}dx+\frac{1}{\theta _1}\int _\Omega (\left| \nabla \Delta u_n\right| ^{p(x)}-\lambda f\left( x,u_n\right) u_n)dx \\&\ge \Bigg (\frac{1}{p^+}-\frac{1}{\theta _1}\Bigg )\Vert u_n\Vert ^{p^-}+\frac{1}{\theta _1}\Big \langle I_\lambda ^{'}\big (u_n\big ),u_n\Big \rangle \\&=\Big (\frac{1}{p^+}-\frac{1}{\theta _1}\Big )\left\| u_n\right\| ^{p^-}+\frac{1}{\theta _1}\Big <\Big (I_\lambda \mid _{Y_n}\Big )'(u_n),u_n\Big> \\&\ge \left( \frac{1}{p^+}-\frac{1}{\theta _1}\right) \left\| u_n\right\| ^{p^-}-\frac{1}{\theta _1}\left\| \left( I_\lambda \mid _{Y_n}\right)'\left( u_n\right) \right\| _{X^*}\left\| u_n\right\| _X. \end{aligned}$$

Because \(\left( I_\lambda \mid _{Y_{n}}\right) ^{\prime }\left( u_{n}\right) \rightarrow 0,\ \text {as}\ n\rightarrow \infty\), the sequence \(\left\{ u_{n}\right\}\) is bounded. In a subsequence sense, we have \(u_{n}\rightharpoonup u\). Denoted \(X=\overline{\bigcup \limits _ {n}Y_{n}}\) and take a sequence \(\left\{ v_{n}\right\} \subset Y_{n}\) such that \(v_{n}\rightarrow u\). Therefore,

$$\begin{aligned} \operatorname {lim}_{n\rightarrow \infty }\left\langle I_{\lambda }^{'}\left( u_{n}\right),u_{n}-u\right\rangle&=\lim _{n\rightarrow \infty }\left\langle I_\lambda ^{'}\left( u_{n}\right),u_{n}-v_{n}\right\rangle +\lim _{n\rightarrow \infty }\left\langle I_\lambda ^{'}\left( u_{n}\right),v_{n}-u\right\rangle \\&=\lim _{n\rightarrow \infty }\left\langle \left( I_\lambda \mid _{Y_{n}}\right) ^{'}\left( u_{n}\right),u_{n}-v_{n}\right\rangle \\&=0. \end{aligned}$$

By Lemma 3.6, \(I_{\lambda }^{\prime }\) is a mapping of type \((S_{+})\), so \(u_{n}\rightarrow u\). From \(I_{\lambda }\left( u\right) \in C^{1}\left( X,R\right)\), it follows that \(I_\lambda ^{'}\left( u_{n}\right) \rightarrow I_\lambda ^{'}\left( u\right)\). For any \(\omega _k\in Y_k\), when \(n\ge k\), we have

$$\begin{aligned} \left\langle I_{\lambda }^{'}\big (u\big ),\omega _{k}\right\rangle&=\left\langle I_\lambda ^{'}\left( u\right) -I_\lambda ^{'}\left( u_{n}\right),\omega _k\right\rangle +\left\langle I_\lambda ^{'}\left( u_{n}\right),\omega _k\right\rangle \\&=\left\langle I_\lambda ^{'}\left( u\right) -I_\lambda ^{'}\left( u_{n}\right),\omega _k\right\rangle +\left\langle \left( I_\lambda \mid _{Y_{n}}\right) ^{\prime }\left( u_{n}\right),\omega _k\right\rangle. \end{aligned}$$

In the limit of n, we have \(\left\langle I_\lambda ^{\prime }(u),\omega _k\right\rangle =0,\ \text {for any}\ \omega _k\in Y_k\). Since \(X=\overline{\bigcup \limits _ {k}Y_{k}}\), it follows that \(I_\lambda ^{\prime }\left( u\right) =0\). Hence, we have proven that \(I_{\lambda }\) satisfies \(\left( PS\right) _c^*\)-condition for any \(c\in {\mathbb {R}}\). \(\square\)

The Proof of Theorem 1.3

It follows from (2) and (8) that \(I_{\lambda }\) is even functional. From Lemma 3.9 we know that \(I_{\lambda }\) satisfies \(\left( PS\right)\)-condition for all \(\lambda>0\). Thus, it suffices to prove that for sufficiently large k, there exists \(\rho _{k}>\delta _{k}>0\) such that

  1. (1)

    \(b_k:=\inf \left\{ I_\lambda (u)|u\in Z_k,\ \Vert u\Vert =\delta _k\right\} \rightarrow \infty,\ \text {as}\ k\rightarrow \infty,\)

  2. (2)

    \(a_k:=\max \left\{ I_\lambda (u)|u\in Y_k,\ \Vert u\Vert =\rho _k\right\} \le 0.\)

Suppose \(u\in Z_{_k}\) with \(\left\| u\right\| \ge 1\). By \((F2')\) and Propositions 2.1, Propositions 2.6 and Propositions 2.7, we have

$$\begin{aligned} I_{\lambda }\left( u\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta u\right| ^{p(x)}dx-\lambda \int _{\Omega }F\left( x,u\right) dx\\&\ge \frac{1}{p^+}\int _{\Omega }|\nabla \Delta u|^{p(x)}dx-\lambda \int _{\Omega }{\left( \frac{|a_1(x)|}{\omega (x)}|u|^{\omega (x)}+\frac{|b_1(x)|}{q(x)}|u|^{q(x)}\right) }dx\\&\ge \frac{1}{p^+}\Vert u\Vert ^{{p}^{-}}-\frac{2\lambda }{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\left\| \left| u\right| ^{\omega (x)}\right\| _{L^{\left( \gamma _1\left( x\right) \right) ^{\prime }}(\Omega )}\\&\quad -\frac{2\lambda }{q^-}\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\left\| \left| u\right| ^{q(x)}\right\| _{L^{\left( \gamma _2\left( x\right) \right) ^{\prime }}(\Omega )}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{{p}^{-}}-\frac{2\lambda }{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\Vert u\Vert _{L^{r(x)}(\Omega )}^{\hat{\omega }}- \frac{2\lambda }{q^-}\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\Vert u\Vert _{L^{r(x)}(\Omega )}^{\hat{q}}\\&\ge \frac{1}{p^+}\Vert u\Vert ^{{p}^{-}}-\frac{2\lambda }{\omega ^-}C_4\big (\alpha _k\Vert u\Vert \big )^{\hat{\omega }}-\frac{2\lambda }{q^-}C_2\big (\alpha _k\Vert u\Vert \big )^{\hat{q}}, \end{aligned}$$

where \(C_4=\left\| a_1(x)\right\| _{L^{\gamma _1\left( x\right) }(\Omega )}\), \(C_2=\left\| b_1(x)\right\| _{L^{\gamma _2\left( x\right) }(\Omega )}\). We can choose \(\frac{2}{\omega ^-}C_4{\alpha _k}^{\hat{\omega }}<\frac{1}{2\lambda p^+}\) since \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). Therefore,

$$\begin{aligned} I_{\lambda }\left( u\right)&\ge \frac{1}{2p^+}\Vert u\Vert ^{{p}^{-}}-\frac{2\lambda }{q^-}C_2\big (\alpha _k\Vert u\Vert \big )^{\hat{q}}. \end{aligned}$$

Choosing \(\delta _{k}=\left( \frac{2\lambda }{q^{-}}C_{2}2{q}^{+}\alpha _{k}^{\hat{q}}\right) ^{\frac{1}{{p}^{-}-\hat{q}}}\). Then \(\delta _k\rightarrow \infty\) since \({p}^{-}<\hat{q}\) and \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). If \(u\in Z_{_k}\) with \(\left\| u\right\| =\delta _{k}\rightarrow \infty\), as \(k\rightarrow \infty\) and (H1) holds, we can deduce that

$$\begin{aligned} I_{\lambda }\left( u\right) \ge \left( \frac{1}{2p^{+}}-\frac{1}{2q^{+}}\right) {\delta _{k}}^{{p}^{-}}\rightarrow \infty,\ \text {as}\ k\rightarrow \infty. \end{aligned}$$

This completes the proof of (1).            

Let \(u\in Y_{k}\) such that \(\rho _{k}>\delta _{k}>1\). According to (18), we have

$$\begin{aligned} I_\lambda (u)&=\int _{\Omega }\frac{1}{p(x)}|\nabla \Delta u|^{p(x)}dx-\lambda \int _{\Omega }F(x,u)dx\\&\le \frac{1}{p^-}\Vert u\Vert ^{{p}^{+}}-\lambda C_3\int _{\Omega }|u|^{\theta _1}dx+\lambda C_4|\Omega|. \end{aligned}$$

Since \(\dim Y_k<\infty\) (this implies that all norms are equivalent) and \( \theta _{1}>p^{+}\)

$$\begin{aligned} I_{\lambda }\left( u\right) {\le }\frac{1}{p^{-}}\left\| u\right\| ^{{p}^{+}}-\lambda C_{3}\left\| u\right\| ^{\theta _{1}}+\lambda C_{4}\left| \Omega \right|. \end{aligned}$$

For \(\left\| u\right\| =\rho _{k}\), we have \(I_{\lambda }\left( u\right) \le 0\). This completes the proof of (2). Therefore, we can choose \(\rho _{k}>\delta _{k}>0\), which completes the proof of the theorem. \(\square\)

3.4 The Proof of Theorem 1.4

The proof of Theorem 1.4

It follows from the assumptions of the theorem and Lemma 3.9 that \(I_{\lambda }\) is even functional and satisfies the \(\left( PS\right) _c^*\)-condition. Therefore, it suffices to show that for any \(k\ge {k}_{0}\), there exists \(\rho _{k}>\delta _{k}>0\) such that

  1. (1)

    \(c_{k}:=\inf \left\{ I_{\lambda }(u):\ u\in Z_{k},\ \Vert u\Vert =\rho _{k}\right\} \ge 0,\)

  2. (2)

    \(d_{k}:=\max \left\{ I_{\lambda }\left( u\right):\ u\in Y_{k},\ \left\| u\right\| =\delta _{k}\right\} <0,\)

  3. (3)

    \(f_{k}:=\inf \left\{ I_{\lambda }\big (u\big ):\ u{\in }Z_{k},\ \Vert u\Vert {\le }\rho _{k}\right\} {\rightarrow }0,\ k{\rightarrow }\infty.\)

Let \(v\in Z_{k}\) be such that \(\left\| v\right\| =1\). Taking \(u=tv,\ 0<t<1\), and using \((F2')\) as well as Proposition 2.1, Proposition 2.6 and Proposition 2.7, we have

$$\begin{aligned} I_{\lambda }(tv)&=\int _{\Omega }\frac{1}{p(x)}\Big|\nabla \Delta tv\Big|^{p(x)}dx-\lambda \int _{\Omega }F\big (x,tv\big )dx \\&\ge \frac{t^{p^{+}}}{p^{+}}\int _{\Omega }\left| \nabla \Delta v\right| ^{p(x)}dx-\frac{\lambda {t}^{\omega ^-}}{\omega ^-} \int _{\Omega }\left| a_1(x)\right||v|^{\omega (x)}dx\\&\quad -\frac{\lambda {t}^{q^-}}{q^-} \int _{\Omega }\left| b_1(x)\right||v|^{q(x)}dx \\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\left\| \left| v\right| ^{\omega (x)}\right\| _{L^{\left( \gamma _1\left( x\right) \right) ^{\prime }}(\Omega )}\\&\quad -\frac{2\lambda {t}^{q^-}}{q^-}\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\left\| \left| v\right| ^{q(x)}\right\| _{L^{\left( \gamma _2\left( x\right) \right) ^{\prime }}(\Omega )}\\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\left\| v\right\| ^{{\hat{\omega }}}_{L^{r(x)}(\Omega )}\\&\quad -\frac{2\lambda {t}^{q^-}}{q^-}\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\left\| v\right\| ^{{\hat{q}}}_{L^{r(x)}(\Omega )}\\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5\big (\alpha _k\Vert v\Vert \big )^{\hat{\omega }}-\frac{2\lambda {t}^{q^-}}{q^-}C_2\big (\alpha _k\Vert v\Vert \big )^{\hat{q}}\\&=\frac{t^{p^{+}}}{p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}-\frac{2\lambda {t}^{q^-}}{q^-}C_2{\alpha _k}^{\hat{q}}, \end{aligned}$$

where \(C_5=\left\| a_1(x)\right\| _{L^{\gamma _1\left( x\right) }(\Omega )}\), \(C_2=\left\| b_1(x)\right\| _{L^{\gamma _2\left( x\right) }(\Omega )}\). For sufficiently large k, if we take \(\frac{2 C_2{\alpha _k}^{\hat{q}}}{q^-}<\frac{1}{2\lambda {p^+}}\),we obtain

$$\begin{aligned} I_{\lambda }(tv)\ge \frac{t^{p^{+}}}{2p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}. \end{aligned}$$
(20)

Choosing \(\rho _k=\left( \frac{2\lambda }{\omega ^-}C_{5}2{q}^{+}{\alpha _k}^{\hat{\omega }}\right) ^{\frac{1}{p^+-\omega ^-}}\), we note that \(\rho _k\rightarrow 0\) since \({p}^{+}>\omega ^-\) and \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). When \(t=\rho _k\), we have

$$\begin{aligned} I_{\lambda }(tv)\ge \left( \frac{1}{2p^{+}}-\frac{1}{2q^{+}}\right) \rho _k^{p^{+}}. \end{aligned}$$

Since \(p^{+}<q^{+}\), this implies \(I_{\lambda }\left( u\right) \ge 0\), namely, \(\inf \limits _{u\in Z_{k},\left\| u\right\| =\rho _{k}}I_{\lambda }\left( u\right) \ge 0\). Thus, (1) is proved. Let \(v\in Y_{k}\) with \(\left\| v\right\| =1\). Taking \(u=tv,\ 0<t<1\), and using (F5) and Proposition 2.6, we get

$$\begin{aligned} I_{\lambda }(tv)&=\int _{\Omega }\frac{1}{p(x)}\Big|\nabla \Delta tv\Big|^{p(x)}dx-\lambda \int _{\Omega }F\big (x,tv\big )dx \\&\le \frac{t^{p^{-}}}{p^{-}}\int _{\Omega }\left| \nabla \Delta v\right| ^{p(x)}dx-\frac{\lambda {t}^{\eta ^+}}{{\eta }^{+}} \int _{\Omega }\left| a_3(x)\right||v|^{\eta (x)}dx-\frac{\lambda {t}^{l^+}}{{l}^{+}} \int _{\Omega }\left| b_3(x)\right||v|^{l(x)}dx \\&\le \frac{t^{p^{-}}}{p^{-}}\left\| v\right\| ^{{p}^{-}}-\frac{\lambda {t}^{\eta ^+}}{{\eta }^{+}} \inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }\left| a_3(x)\right||v|^{\eta (x)}dx\\&\quad -\frac{\lambda {t}^{l^+}}{{l}^{+}} \inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }\left| b_3(x)\right||v|^{l(x)}dx\\&=\frac{t^{p^{-}}}{p^{-}}-\frac{\lambda {t}^{\eta ^+}}{{\eta }^{+}} \inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }\left| a_3(x)\right||v|^{\eta (x)}dx\\&\quad -\frac{\lambda {t}^{l^+}}{{l}^{+}} \inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }\left| b_3(x)\right||v|^{l(x)}dx. \end{aligned}$$

From \(1<\eta ^{-}\le \eta ^{+}<p^{-}<p^{+}<l^{-}\le l^{+}<q^{-}\), we can derive that there exists a \(\delta _{k}\in \left( 0,\rho _{k}\right)\) such that when \(t=\delta _{k}\), \(I_{\lambda }(tv)<0\). Thus, \(d_{k}=\max \limits _{u\in Y_{k}, \left\| u\right\| =\delta _{k}}I_{\lambda }\left( u\right) <0.\) In summary, (2) is proved.

Since \(Y_k\cap Z_k\ne \emptyset\) and \(\delta _{k}<\rho _{k}\), we have

$$\begin{aligned} f_k:=\inf _{u\in Z_k,\Vert u\Vert \le \rho _k}I_{\lambda }\left( u\right) \le d_k:=\max _{u\in Y_k,\Vert u\Vert =\delta _k}I_{\lambda }\left( u\right) <0. \end{aligned}$$

By (20), for \(v\in Z_{k}\) with \(\left\| v\right\| =1\), let \(u=tv\) and \(0\le t\le {\rho }_{k}\), we can deduce

$$\begin{aligned} I_{\lambda }(u)= & I_{\lambda }(tv)\ge \frac{t^{p^{+}}}{2p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5\big (\alpha _k\big )^{\hat{\omega }} \ge -\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5\big (\alpha _k\big )^{\hat{\omega }}\\\ge & -\frac{2\lambda {\rho _k}^{\omega ^-}}{\omega ^-}C_5\big (\alpha _k\big )^{\hat{\omega }} \ge -\frac{2\lambda }{\omega ^-}C_5\big (\alpha _k\big )^{\hat{\omega }}. \end{aligned}$$

So \(\inf \limits _{u\in Z_{k}, \left\| u\right\| \le \rho _{k}}I_{\lambda }\left( u\right) \rightarrow 0,k \rightarrow \infty\). Hence, (3) is proved. \(\hfill\square\)

Example 3.4

Denote the functional

$$\begin{aligned} I\left( u\right) =\int _{\Omega }\frac{1}{p(x)}\Big|\nabla \Delta u\Big|^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|u|^{\omega \left( x\right) }}dx -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|u|^{q\left( x\right) }}dx. \end{aligned}$$

Suppose (H1), (H2) and (F1) hold. Then the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{p(x)}^{3} u=\lambda \left| a_1\left( x\right) \right||u|^{\omega \left( x\right) -2}u+\mu \left| b_1\left( x\right) \right||u|^{q\left( x\right) -2}u, & \hbox { in } \Omega,\\ u=\Delta u=\Delta ^{2}u=0, & \hbox { on }\partial \Omega,\\ \end{array}\right. } \end{aligned}$$
(21)

has the following result:

  1. (1)

    If \(\mu>0\), then for any \(\lambda \in {\mathbb {R}}\), the equation (21) possesses a set of solutions \(\left( \pm {u}_{k}\right) ^{\infty }_{k=1}\) such that \(I_{\lambda }(u_k)\rightarrow \infty\), as \(k\rightarrow \infty\).

  2. (2)

    If \(\lambda>0\), then for any \(\mu \in {\mathbb {R}}\), the equation (21) possesses a set of solutions \(\left( \pm {v}_{k}\right) ^{\infty }_{k=1}\) such that \(I_{\lambda }(v_k)<0\), when \(k\rightarrow \infty\), \(I_{\lambda }(v_k)\rightarrow 0\).

Proof

Step one. \(I\left( u\right)\) satisfies \(\left( PS\right) _c^*\)-condition for all \(c>0\).

(i) Let \(\{u_n\}\subset X\) be a \(\left( PS\right) _c^*\)-sequence, i.e., \(u_n\in Y_n,\ \ I\left( u_n\right) \rightarrow c,\ \left( I|_{Y_n}\right) ^{\prime }\left( u_n\right) \rightarrow 0,\ n\rightarrow \infty\). Without loss of generality, we assume that \(\lambda>0\), \(\mu>0\). Let’s assume that \(\left\| u_n\right\|>1\), for sufficiently large n. Using Proposition 2.1, Proposition 2.6, Proposition 2.7 and \(\left<\left( I\mid _{Y_n}\right)'(u_n),u_n\right>\ge -\left\| \left( I\mid _{Y_n}\right)'\left( u_n\right) \right\| _{X^*}\left\| u_n\right\| _X\), when \(\lambda \ge 0\), we have

$$\begin{aligned} c+1&\ge I\left( u_{n}\right) =\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta u_{n}\right| ^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|u_{n}|^{\omega \left( x\right) }}dx\\&\quad -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|u_{n}|^{q\left( x\right) }}dx \\&\ge \frac{1}{p^+}\int _{\Omega }\left| \nabla \Delta u_{n}\right| ^{p(x)}dx-\frac{\lambda }{\omega ^-}\int _{\Omega }{\left| a_1\left( x\right) \right||u_{n}|^{\omega \left( x\right) }}dx -\frac{\mu }{q^-}\int _{\Omega }{\left| b_1\left( x\right) \right||u_{n}|^{q\left( x\right) }}dx\\&=\left( \frac{1}{p^{+}}-\frac{1}{q^{-}}\right) \int _{\Omega }\left| \nabla \Delta u_{n}\right| ^{p(x)}dx-\lambda \left( \frac{1}{\omega ^{-}}-\frac{1}{q^{-}}\right) \int _{\Omega }{\left| a_1\left( x\right) \right||u_{n}|^{\omega \left( x\right) }}dx\\&\quad +\frac{1}{q^-}\left\langle I^{\prime }\left( u_n\right),u_n\right\rangle \\&\ge \left( \frac{1}{p^{+}}-\frac{1}{q^{-}}\right) \left\| u_n\right\| ^{p^-}-2\lambda \left( \frac{1}{\omega ^{-}}-\frac{1}{q^{-}}\right) {\left\| a_1\left( x\right) \right\| } _{L^{\gamma _{1}\left( x\right) }\left( \Omega \right) } \left\||u_{n}|^{\omega \left( x\right) }\right\| _{L^{{\left( \gamma _{1}\left( x\right) \right) }^{\prime }}\left( \Omega \right) }\\&\quad +\frac{1}{q^-}\left<\left( I\mid _{Y_n}\right)'(u_n),u_n\right>\\&\ge \left( \frac{1}{p^{+}}-\frac{1}{q^{-}}\right) \left\| u_n\right\| ^{p^-}-2\lambda \left( \frac{1}{\omega ^{-}}-\frac{1}{q^{-}}\right) {\left\| a_1\left( x\right) \right\| } _{L^{\gamma _{1}\left( x\right) }\left( \Omega \right) }\left\| u_{n}\right\| ^{\hat{\omega }}_{L^{r\left( x\right) }\left( \Omega \right) }\\&\quad +\frac{1}{q^-}\left<\left( I\mid _{Y_n}\right)'(u_n),u_n\right>\\&\ge \left( \frac{1}{p^{+}}-\frac{1}{q^{-}}\right) \left\| u_n\right\| ^{p^-}-2C\lambda \left( \frac{1}{\omega ^{-}}-\frac{1}{q^{-}}\right) {\left\| a_1\left( x\right) \right\| } _{L^{\gamma _{1}\left( x\right) }\left( \Omega \right) }\left\| u_{n}\right\| ^{{\omega }^{+}}\\&\quad -\frac{1}{q^-}\left\| \left( I\mid _{Y_n}\right)'\left( u_n\right) \right\| _{X^*}\left\| u_n\right\| _X. \end{aligned}$$

Since \(\omega ^{+}<p{-}\), \(\left( I|_{Y_n}\right) ^{\prime }\left( u_n\right) \rightarrow 0\), as \(n\rightarrow \infty\), we prove the boundedness of \(\{{u}_{n}\}\).

(ii) Consider a subsequence of \(\{u_n\}\), still denoted by \(\{u_n\}\), where \(u_{n}\rightharpoonup u\). Denoted \(X=\overline{\bigcup \limits _ {n}Y_{n}}\), and take \(\left\{ v_{n}\right\} \subset Y_{n}\) such that \(v_{n}\rightarrow u\). Then

$$\begin{aligned} \operatorname {lim}_{n\rightarrow \infty }\left\langle I^{\prime }\left( u_{n}\right),u_{n}-u\right\rangle&=\lim _{n\rightarrow \infty }\left\langle I^{\prime }\left( u_{n}\right),u_{n}-v_{n}\right\rangle +\lim _{n\rightarrow \infty }\left\langle I^{\prime }\left( u_{n}\right),v_{n}-u\right\rangle \\&=\lim _{n\rightarrow \infty }\left\langle \left( I\mid _{Y_{n}}\right) ^{\prime }\left( u_{n}\right),u_{n}-v_{n}\right\rangle \\&=0. \end{aligned}$$

By Lemma 3.6, it follows that \(I^{\prime }\) is a mapping of type \((S_{+})\), so \(u_{n}\rightarrow u\). From \(I\left( u\right) \in C^{1}\left( X,R\right)\), we have \(I^{\prime }\left( u_{n}\right) \rightarrow I^{\prime }\left( u\right)\). For any \(\omega _k\in Y_k\), when \(n\ge k\), we have

$$\begin{aligned} \left\langle I^{\prime }\big (u\big ),\omega _{k}\right\rangle&=\left\langle I^{\prime }\left( u\right) -I^{\prime }\left( u_{n}\right),\omega _k\right\rangle +\left\langle I^{\prime }\left( u_{n}\right),\omega _k\right\rangle \\&=\left\langle I^{\prime }\left( u\right) -I^{\prime }\left( u_{n}\right),\omega _k\right\rangle +\left\langle \left( I\mid _{Y_{n}}\right) ^{\prime }\left( u_{n}\right),\omega _k\right\rangle. \end{aligned}$$

In the limit of n, we have \(\left\langle I^{\prime }(u),\omega _k\right\rangle =0,\ \text {for any}\ \omega _k\in Y_k\). Since\(X=\overline{\bigcup \limits _ {k}Y_{k}}\), \(I^{\prime }\left( u\right) =0\). In this way we prove that I satisfies \(\left( PS\right) _c^*\)-condition for any \(c\in {\mathbb {R}}\).

step two. If \(\mu>0\), then for any \(\lambda \in {\mathbb {R}}\), \(I\left( u\right)\) satisfies the remaining conditions of the Fountain theorem, i.e., for sufficiently large k there exists \(\rho _{k}>\delta _{k}>0\) such that

(i):

\(b_k:=\inf \left\{ I(u)|u\in Z_k,\ \Vert u\Vert =\delta _k\right\} \rightarrow \infty,\ \text {as}\ k\rightarrow \infty,\)

(ii):

\(a_k:=\max \left\{ I(u)|u\in Y_k,\ \Vert u\Vert =\rho _k\right\} \le 0.\)

(i)Let \(u\in Z_{_k}\) with \(\left\| u\right\| \ge 1\), By Propositions 2.1, Propositions 2.6 and Propositions 2.7, we have

$$\begin{aligned} I\left( u\right)&=\int _{\Omega }\frac{1}{p(x)}\Big|\nabla \Delta u\Big|^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|u|^{\omega \left( x\right) }}dx -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|u|^{q\left( x\right) }}dx \\&\ge \frac{1}{p^+}\left\| u\right\| ^{{p}^{-}}-\frac{\left| \lambda \right| }{\omega ^-}\int _{\Omega }{\left| a_1\left( x\right) \right||u|^{\omega \left( x\right) }}dx -\frac{\mu }{q^-}\int _{\Omega }{\left| b_1\left( x\right) \right||u|^{q\left( x\right) }}dx\\&\ge \frac{1}{p^+}\left\| u\right\| ^{{p}^{-}}-\frac{2\left| \lambda \right| }{\omega ^-}{\left\| a_1\left( x\right) \right\| } _{L^{\gamma _{1}\left( x\right) }\left( \Omega \right) } \left\||u|^{\omega \left( x\right) }\right\| _{L^{{\left( \gamma _{1}\left( x\right) \right) }^{\prime }}\left( \Omega \right) }\\&\quad -\frac{2\mu }{q^-}{\left\| b_1\left( x\right) \right\| } _{L^{\gamma _{2}\left( x\right) }\left( \Omega \right) } \left\||u|^{q\left( x\right) }\right\| _{L^{{\left( \gamma _{2}\left( x\right) \right) }^{\prime }}\left( \Omega \right) }\\&\ge \frac{1}{p^+}\left\| u\right\| ^{{p}^{-}}-\frac{2\left| \lambda \right| }{\omega ^-}{\left\| a_1\left( x\right) \right\| } _{L^{\gamma _{1}\left( x\right) }\left( \Omega \right) }\left\| u\right\| ^{\hat{\omega }}_{L^{r\left( x\right) }\left( \Omega \right) } -\frac{2\mu }{q^-}{\left\| b_1\left( x\right) \right\| } _{L^{\gamma _{2}\left( x\right) }\left( \Omega \right) }\left\| u\right\| ^{\hat{q}}_{L^{r\left( x\right) }\left( \Omega \right) }\\&\ge \frac{1}{p^+}\left\| u\right\| ^{{p}^{-}}-\frac{2\left| \lambda \right| }{\omega ^-}C_5\left( \alpha _k\Vert u\Vert \right) ^{\hat{\omega }} -\frac{2\mu }{q^-}C_2\left( \alpha _k\Vert u\Vert \right) ^{\hat{q}}, \end{aligned}$$

where \(C_5=\left\| a_1(x)\right\| _{L^{\gamma _1\left( x\right) }(\Omega )}\), \(C_2=\left\| b_1(x)\right\| _{L^{\gamma _2\left( x\right) }(\Omega )}\). We can choose \(\frac{2}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}<\frac{1}{2\left| \lambda \right| p^+}\) since \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). Thus,

$$\begin{aligned} I\left( u\right)&\ge \frac{1}{2p^+}\Vert u\Vert ^{{p}^{-}}-\frac{2\mu }{q^-}C_2\left( \alpha _k\Vert u\Vert \right) ^{\hat{q}}. \end{aligned}$$

Choosing \(\delta _{k}=\left( \frac{2\mu }{q^{-}}C_{2}2q^{+}\alpha _{k}^{\hat{q}}\right) ^{\frac{1}{{p}^{-}-\hat{q}}}\). Then \(\delta _k\rightarrow \infty\) since \({p}^{-}<\hat{q}\) and \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). If \(u\in Z_{_k}\) with \(\left\| u\right\| =\delta _{k}\rightarrow \infty\), as \(k\rightarrow \infty\) and (H1) holds, we deduce that

$$\begin{aligned} I\left( u\right) \ge \left( \frac{1}{2p^{+}}-\frac{1}{2q^{+}}\right) {\delta _{k}}^{{p}^{-}}\rightarrow \infty,\ \text {as}\ k\rightarrow \infty. \end{aligned}$$

This completes the proof of (i).

(ii) Let \(v\in Y_{k}\) with \(\left\| v\right\| =1\), and \(t>1\), we have

$$\begin{aligned} I\left( tv\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta tv\right| ^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|tv|^{\omega \left( x\right) }}dx -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|tv|^{q\left( x\right) }}dx \\&\le \frac{{t^{p^+}}}{p^-}{\left\| v\right\| }^{{p}^{+}}+\left| \lambda \right| t^{\omega ^+}\int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|v|^{\omega \left( x\right) }}dx -\mu t^{q^-}\int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|v|^{q\left( x\right) }}dx\\&\le \frac{{t^{p^+}}}{p^-}+\left| \lambda \right| t^{\omega ^+}\sup \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|v|^{\omega \left( x\right) }}dx\\&\quad -\mu t^{q^-}\inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|v|^{q\left( x\right) }}dx. \end{aligned}$$

Since \(1<\omega ^{-}\le \omega ^{+}<p^{-}<p^{+}<q^{-}\le q^{+}\), there exists \(\rho _{k}>\delta _{k}\) such that for \(t=\rho _{k}\) and \(u=tv\), we have \(I\left( u\right) =I\left( tv\right) \le 0\). This completes the proof of (ii).

Step three. If \(\lambda>0\), then for any \(\mu \in {\mathbb {R}}\), \(I\left( u\right)\) satisfies the remaining conditions of the dual fountain theorem, i.e., for any \(k\ge {k}_{0}\) there exists \(\rho _{k}>\delta _{k}>0\) such that

(i):

\(c_{k}:=\inf \left\{ I(u):\ u\in Z_{k},\ \Vert u\Vert =\rho _{k}\right\} \ge 0,\)

(ii):

\(d_{k}:=\max \left\{ I\left( u\right):\ u\in Y_{k},\ \left\| u\right\| =\delta _{k}\right\} <0,\)

(iii):

\(f_{k}:=\inf \left\{ I\left( u\right):\ u{\in }Z_{k},\ \Vert u\Vert {\le }\rho _{k}\right\} {\rightarrow }0,\ k{\rightarrow }\infty.\)

            

(i)Let \(v\in Z_{k}\) be such that \(\left\| v\right\| =1\). Taking \(u=tv,\ 0<t<1\), and applying Proposition 2.1, Proposition 2.6 and Proposition 2.7, we have

$$\begin{aligned} I\left( tv\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta tv\right| ^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|tv|^{\omega \left( x\right) }}dx -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|tv|^{q\left( x\right) }}dx \\&\ge \frac{t^{p^{+}}}{p^{+}}\int _{\Omega }\left| \nabla \Delta v\right| ^{p(x)}dx-\frac{\lambda {t}^{\omega ^-}}{\omega ^-} \int _{\Omega }\left| a_1(x)\right||v|^{\omega (x)}dx\\&\quad -\frac{\left| \mu \right| {t}^{q^-}}{q^-} \int _{\Omega }\left| b_1(x)\right||v|^{q(x)}dx \\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\left\| \left| v\right| ^{\omega (x)}\right\| _{L^{\left( \gamma _1\left( x\right) \right) ^{\prime }}(\Omega )}\\&\quad -\frac{2\left| \mu \right| {t}^{q^-}}{q^-}\Vert b_1(x)\Vert _{L^{\gamma _2\left( x\right) }(\Omega )}\left\| \left| v\right| ^{q(x)}\right\| _{L^{\left( \gamma _2\left( x\right) \right) ^{\prime }}(\Omega )}\\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}\Vert a_1(x)\Vert _{L^{\gamma _1\left( x\right) }(\Omega )}\left\| v\right\| ^{{\hat{\omega }}}_{L^{r(x)}(\Omega )}\\&\quad -\frac{2\left| \mu \right| {t}^{q^-}}{q^-}\Vert b_1(x)\Vert _{L^{\left| \mu \right| _2\left( x\right) }(\Omega )}\left\| v\right\| ^{{\hat{q}}}_{L^{r(x)}(\Omega )}\\&\ge \frac{t^{p^{+}}}{p^{+}}{\left\| v\right\| }^{{p}^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_4\left( \alpha _k\Vert v\Vert \right) ^{\hat{\omega }}-\frac{2\left| \mu \right| {t}^{q^-}}{q^-}C_2\left( \alpha _k\Vert v\Vert \right) ^{\hat{q}}\\&=\frac{t^{p^{+}}}{p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}-\frac{2\left| \mu \right| {t}^{q^-}}{q^-}C_2{\alpha _k}^{\hat{q}}, \end{aligned}$$

where \(C_5=\left\| a_1(x)\right\| _{L^{\gamma _1\left( x\right) }(\Omega )}\), \(C_2=\left\| b_1(x)\right\| _{L^{\gamma _2\left( x\right) }(\Omega )}\). When k is large enough, we take \(\frac{2 C_2{(\alpha _k)}^{\hat{q}}}{q^-}<\frac{1}{2\left| \mu \right| {p^+}}\) to obtain

$$\begin{aligned} I(tv)\ge \frac{t^{p^{+}}}{2p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}. \end{aligned}$$
(22)

Choosing \(\rho _k=\left( \frac{2\lambda }{\omega ^-}C_{5}2{q}^{+}{\alpha _k}^{\hat{\omega }}\right) ^{\frac{1}{p^+-\omega ^-}}\), then \(\rho _k\rightarrow 0\) since \({p}^{+}>\omega ^-\) and \(\alpha _{k}\rightarrow 0\), as \(k\rightarrow \infty\). When \(t=\rho _k\), we have

$$\begin{aligned} I(tv)\ge \left( \frac{1}{2p^{+}}-\frac{1}{2q^{+}}\right) \rho _k^{p^{+}}. \end{aligned}$$

Thus, \(I_{\lambda }\left( u\right) \ge 0\) since \(p^{+}<q^{+}\), namely, \(\inf \limits _{u\in Z_{k},\left\| u\right\| =\rho _{k}}I_{\lambda }\left( u\right) \ge 0\). In summary, (i) is proved.                

(ii) Let \(v\in Y_{k}\) with \(\left\| v\right\| =1\). Taking \(u=tv,\ 0<t<1\), by Proposition 2.6, we get

$$\begin{aligned} I\left( tv\right)&=\int _{\Omega }\frac{1}{p(x)}\left| \nabla \Delta tv\right| ^{p(x)}dx-\lambda \int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|tv|^{\omega \left( x\right) }}dx -\mu \int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|tv|^{q\left( x\right) }}dx \\&\le \frac{{t^{p^-}}}{p^-}{\left\| v\right\| }^{{p}^{-}}-\lambda t^{\omega ^+}\int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|v|^{\omega \left( x\right) }}dx +\left| \mu \right| t^{q^-}\int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|v|^{q\left( x\right) }}dx\\&\le \frac{{t^{p^-}}}{p^-}-\lambda t^{\omega ^+}\inf \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }{\frac{\left| a_1\left( x\right) \right| }{\omega \left( x\right) }|v|^{\omega \left( x\right) }}dx\\&\quad +\left| \mu \right| t^{q^-}\sup \limits _{v\in Y_{k}, \left\| v\right\| =1}\int _{\Omega }{\frac{\left| b_1\left( x\right) \right| }{q\left( x\right) }|v|^{q\left( x\right) }}dx. \end{aligned}$$

From \(1<\omega ^{-}\le \omega ^{+}<p^{-}<p^{+}<q^{-}\le q^{+}\), we can derive that there exists a \(\delta _{k}\in \left( 0,\rho _{k}\right)\) such that when \(t=\delta _{k}\), \(I(tv)<0\). So \(d_{k}=\max \limits _{u\in Y_{k}, \left\| u\right\| =\delta _{k}}I\left( u\right) <0\). In summary, (ii) is proved.

(iii) Since \(Y_k\cap Z_k\ne \emptyset\) and \(\delta _{k}<\rho _{k}\), we have

$$\begin{aligned} f_k:=\inf _{u\in Z_k,\Vert u\Vert \le \rho _k}I\left( u\right) \le d_k:=\max _{u\in Y_k,\Vert u\Vert =\delta _k}I\left( u\right) <0. \end{aligned}$$

By (22), for \(v\in Z_{k}\), \(\left\| v\right\| =1\). \(u=tv,\text {and}\ 0\le t\le {\rho }_{k}\), we can deduce

$$\begin{aligned} I(u)= & I(tv)\ge \frac{t^{p^{+}}}{2p^{+}}-\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }} \ge -\frac{2\lambda {t}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}\\\ge & -\frac{2\lambda {\rho _k}^{\omega ^-}}{\omega ^-}C_5{\alpha _k}^{\hat{\omega }} \ge -\frac{2\lambda }{\omega ^-}C_5{\alpha _k}^{\hat{\omega }}. \end{aligned}$$

Thus, \(\inf \limits _{u\in Z_{k}, \left\| u\right\| \le \rho _{k}}I\left( u\right) \rightarrow 0,k \rightarrow \infty\). (iii) is proved. \(\square\)

4 Index of Notations

We use the following notations:

  • \(\Omega\): bounded domain of \(R^N\) with smooth boundary \(\partial \Omega\),

  • \(\Delta _{p(x)}u:=div(|\nabla u|^{p(x)-2}\nabla u)\) is called the p(x)-Laplacian operator,

  • \(\Delta _{p(x)}^2:=\Delta \left(|\Delta u|^{p(x)-2}\Delta u\right)\) is called the p(x)-biharmonic operator,

  • \(C_+\left( {\overline{\Omega }}\right):=\left\{ h(x):h(x)\in C\left( {\overline{\Omega }}\right),\min _{x\in {\overline{\Omega }}}h(x)>1\right\},\)

  • \(h^+:=\sup _{x\in {\overline{\Omega }}}h(x)\), \(h(x)\in C_{+}\big ({\overline{\Omega }}\big ),\)\(h^-:=\inf _{x\in {\overline{\Omega }}}h(x)\), \(h(x)\in C_{+}\big ({\overline{\Omega }}\big ),\)

  • \({\mathcal {K}}^{{\check{p}}}:=\left\{ \begin{array}{ll}{\mathcal {K}}^{p^+},0<{\mathcal {K}}\leqslant 1,\\ {\mathcal {K}}^{p^-},{\mathcal {K}}>1,\end{array}\right.\)

  • \({\mathcal {K}}^{\hat{p}}:=\left\{ \begin{array}{ll}{\mathcal {K}}^{p^-},0<{\mathcal {K}}\leqslant 1,\\ {\mathcal {K}}^{p^+},{\mathcal {K}}>1,\end{array}\right.\)

  • \(p_{k}^{*}(x):=\left\{ \begin{array}{ll}\frac{Np(x)}{N-kp(x)},& p(x)<\frac{N}{k},\\ +\infty,& p(x)\geqslant \frac{N}{k}.\end{array}\right.\)

We denote by \(\rightarrow\) (resp. \(\rightharpoonup\)) the strong (resp. weak) convergence.