Existence of Solutions for $p(x)$ Laplacian Equations Without Ambrosetti–Rabinowitz Type Condition

Abstract

This paper investigates the existence and multiplicity of solutions for superlinear $p(x)$-Laplacian equations with Dirichlet boundary conditions. Under no Ambrosetti–Rabinowitz’s superquadraticity conditions, we obtain the existence and multiplicity of solutions using a variant Fountain theorem without Palais-Smale type assumptions.

1 Introduction

We consider the following superlinear elliptic problem

and obtain infinitely many solutions, where $\Omega$ is a bounded smooth domain of $\mathbb {R}^{N}$ ($N\ge 3$) and $p\in C(\overline{\Omega })$ with $1<p(x)<N$ for all $x\in \overline{\Omega }$.

Generally, in order to search the existence of solutions for Dirichlet problems which is superlinear, it is essential to assume the following superquadraticity condition, which is known as Ambrosetti–Rabinowitz type condition [2]:

$$\begin{aligned} (AR) \exists M>0,\tau >p^{+}\,\,\,\,\, \hbox {such that}\,\,\,\,\, 0<\tau \,\, F(x,s) \le f(x,s) s, \,\, \vert s \vert \ge \,\,\, M, \,\, x\in \Omega , \end{aligned}$$

where $f$ is nonlinear term such that $F(x,t)=\int _{0}^{t}f(x,s)ds$.

There are many papers dealing with superlinear Dirichlet problems involving $p(x)$-Laplace operator $\Delta _{p(x) }u:={div}(\vert \nabla u \vert ^{p(x)-2}\nabla u)$, in which $(AR)$ is the main assumption to get the existence and multiplicity of solutions [8, 9]. However, as far as we are concerned, there are many functions which are superlinear but not satisfy $(AR)$ [3, 17].

It is well known that the main aim of using $(AR)$ is to ensure the boundedness of the Palais-Smale type sequences of the corresponding functional. In the present paper, we do not use $(AR)$. Instead, we use a variant Fountain theorem not including Palais-Smale type assumptions (see Theorem 2.1).

The study of differential equations and variational problems involving $p(x)$-growth conditions has attracted a special interest in recent years and a lot of researchers have devoted their work to this area [11, 13, 15, 16] since there are some physical phenomena which can be modeled by such kind of equations. In particular, we may mention some applications related to the study of elastic mechanics and electrorheological fluids [1, 4, 10, 14, 19]. The appearance of such physical models was facilitated by the development of variable exponent Lebesgue $L^{p(x)}$ and Sobolev spaces $W^{1,p(x)}$.

2 Preliminaries

At first, we shall mention some definitions and basic properties of generalized Lebesgue–Sobolev spaces $L^{p(x)}(\Omega )$, $W^{1,p(x)}(\Omega )$, and $W_{0}^{1,p(x)}\Omega$. We refer the reader to [5–7, 12] for the fundamental properties of these spaces.

Set

$$\begin{aligned} C_{+}(\overline{\Omega }) =\left\{ p;p\in C( \overline{\Omega }) , \inf p(X) >1,\,\forall x\in \overline{\Omega }\right\} . \end{aligned}$$

Let $p\in C_{+}(\overline{\Omega })$ and denote

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

For any $p\in C_{+}(\overline{\Omega })$, we define the variable exponent Lebesgue space by

$$\begin{aligned} L^{p\left( x\right) }(\Omega ) =\left\{ u\mid u:\Omega \rightarrow \mathbb {R}\,\,\,\,\, \hbox {is measurable},\,\,\, \int \limits _{\Omega }\vert u(x)\vert ^{p(x)}dx<\infty \right\} , \end{aligned}$$

then $L^{p(x)}(\Omega )$ endowed with the norm

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

becomes a Banach space.

The modular of the $L^{p(x)}(\Omega )$ space, which is the mapping $\rho :L^{p(x)}(\Omega ) \rightarrow \mathbb {R}$ defined by

$$\begin{aligned} \rho (u)=\int \limits _{\Omega }\left| u(x)\right| ^{\ p(x)} dx,\quad \forall u\in L^{p(x)}(\Omega ). \end{aligned}$$

(2.1)

Proposition 2.1

[6, 12] If $u,u_{n}\in L^{p(x)}(\Omega )\,\,\,(n=1,2,...)$, we have

• (i) $|u|_{p(x)}<1\left( =1;>1\right) \Leftrightarrow \rho \left( u\right) <1\left( =1;>1\right) ;$

• (ii) $|u|_{p(x)}>1\implies |u|_{p\left( x\right) }^{p^{-}}\le \rho \left( u\right) \le |u|_{p\left( x\right) }^{p^{+}};$

• (iii) $|u|_{p(x)}<1\implies |u|_{p\left( x\right) }^{p^{+}}\le \rho \left( u\right) \le |u|_{p\left( x\right) }^{p^{-}};$

Proposition 2.2

[6, 12] If $u,u_{n}\in L^{p(x)}( \Omega )$ ($n=1,2,...$), then the following statements are equivalent:

• (i) $\lim \limits _{n\rightarrow \infty }|u_{n}-u|_{p(x)}=0;$

• (ii) $\lim \limits _{n\rightarrow \infty }\rho (u_{n}-u)=0;$

• (iii) $u_{n}\rightarrow u$ in measure in $\Omega$ and $\lim \limits _{n\rightarrow \infty }\rho (u_{n})=\rho \left( u\right)$.

The variable exponent Sobolev space $W^{1,p(x)}\left( \Omega \right)$ is defined by

$$\begin{aligned} W^{1,p(x)}(\Omega )=\{u\in L^{p(x)}(\Omega ):|\nabla u|\in L^{p(x)}(\Omega ) \}, \end{aligned}$$

with the norm

$$\begin{aligned} \Vert u\Vert _{1,p(x)}=|u|_{p(x) }+|\nabla u|_{p(x)},\,\, \forall u\in W^{1,p(x)}(\Omega ) . \end{aligned}$$

Then $(W^{1,p(x)}(\Omega ),\Vert \cdot \Vert _{1,p(x)})$ becomes a Banach space. The space $W_{0}^{1,p(x)}(\Omega )$ is defined as the closure of $C_{0}^{\infty }(\Omega )$ in $W^{1,p(x)}(\Omega )$ with respect to the norm $\Vert \cdot \Vert _{1,p(x)}$. For $u\in$ $W_{0}^{1,p(x)}(\Omega )$, we can define an equivalent norm

$$\begin{aligned} \left\| u\right\| =|\nabla u|_{p(x)}, \end{aligned}$$

since Poincaré inequality

$$\begin{aligned} |u|_{p(x)}\le C|\nabla u|_{p(x)},\,\,\forall u\in W_{0}^{1,p(x)}(\Omega ) \end{aligned}$$

holds, where $C$ is a positive constant [8].

Proposition 2.3

[6, 12] If $1<p^{-}$ and $p^{+}<\infty$, then the spaces $L^{p(x)}(\Omega )$, $W^{1,p(x)}(\Omega )$, and $W_{0}^{1,p(x)}(\Omega )$ are separable and reflexive Banach spaces.

Proposition 2.4

[6, 12] Assume that $\Omega$ is bounded, the boundary of $\Omega$ possesses the cone property and $p\in C_{+}(\overline{\Omega })$ . If $q\in C_{+}(\overline{\Omega })$ and $q\left( x\right) <p^{*}\left( x\right) :=\frac{Np\left( x\right) }{N-p\left( x\right) }$ for all $x\in \overline{\Omega }$ , then the embedding $W^{1,p(x)}\left( \Omega \right) \hookrightarrow L^{q\left( x\right) }\left( \Omega \right)$ is compact and continuous.

From [18], let $X$ be a reflexive and separable Banach space, then there are $e_{j}\subset$ $X$ and $e_{j}^{*}\subset$ $X^{*}$ such that

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

and

$$\begin{aligned} \left\langle e_{i}^{*},e_{j}\right\rangle =\left\{ \begin{array}{ll} 1&{}\quad i=j, \\ 0&{}\quad i\ne j, \end{array} \right. \end{aligned}$$

where $\left\langle .,.\right\rangle$ denotes the duality product between $X$ and $X^{*}.$ For convenience, we write

$$\begin{aligned} X_{j}=span\left\{ e_{j}\right\} ,{\ \ \ \ \ }Y_{k}=\oplus _{j=1}^{k}X_{j},{\ \ \ \ \ }Z_{k}=\overline{\oplus _{j=k}^{\infty } X_{j}}. \end{aligned}$$

And let

$$\begin{aligned} B_{k}=\left\{ u\in Y_{k}:\left\| u\right\| \le \rho _{k}\right\} , \,\, N_{k}=\left\{ u\in Z_{k}:\left\| u\right\| =r_{k}\right\} , \hbox { for }\rho _{k}>r_{k}>0. \end{aligned}$$

Let consider the $C^{1}$-functional $I_{\lambda }:X\rightarrow \mathbb {R}$ defined by

$$\begin{aligned} I_{\lambda }\left( u\right) :=A(u)-\lambda B(u),\,\, \lambda \in \left[ 1,2\right] . \end{aligned}$$

Now we give the following variant Fountain theorem (see [20], Theorem 2.2), which we use in the proof of the main results of the present paper:

Theorem 2.1

(Variant Fountain Theorem) Assume the functional $I_{\lambda }$ satisfies the followings:

$(T_{1})$ :

$I_{\lambda }$ maps bounded sets to bounded sets uniformly for $\lambda \in \left[ 1,2\right]$. Moreover, $I_{\lambda }(-u)=I_{\lambda }(u)$ for all $(\lambda ,u)\in [1,2] \times X$.

$(T_{2})$ :

$B(u)\ge 0$; $B(u)\rightarrow \infty$ as $\left\| u\right\| \rightarrow \infty$ on any finite dimensional subspace of $X$.

$(T_{3})$ :

There exists $\rho _{k}>r_{k}>0$ such that

$$\begin{aligned} a_{k}(\lambda ):=\inf \limits _{u\in Z_{k},\Vert u \Vert =\rho _{k}}I_{\lambda }(u)\ge 0>b_{k}(\lambda ):=\max \limits _{u\in Y_{k},\Vert u \Vert =r_{k}}I_{\lambda }(u), \end{aligned}$$

for all $\lambda \in \left[ 1,2\right]$ and

$$\begin{aligned} d_{k}(\lambda ):=\inf \limits _{u\in Z_{k},\left\| u\right\| \le \rho _{k}}I_{\lambda }(u)\rightarrow 0\,\, as\,\, k\rightarrow \infty \ uniformly\, for \ \lambda \in \left[ 1,2\right] . \end{aligned}$$

Then there exists $\lambda _{n}\rightarrow 1$, $u(\lambda _{n})\in Y_{n}$ such that

$$\begin{aligned} I_{\lambda _{n}}^{\prime }\left| _{Y_{n}}(u(\lambda _{n}))\right. =0,\,\, I_{\lambda _{n}}(u(\lambda _{n}))\rightarrow c_{k}\in \left[ d_{k}(2),b_{k}(1)\right] \,\,\hbox {as}\,\,n\rightarrow \infty . \end{aligned}$$

Particularly, if $\left\{ u(\lambda _{n})\right\}$ has a convergent subsequence for every $k$, then $I_{1}$ has infinitely many nontrivial critical points $\left\{ u_{k}\right\} \in X\backslash \left\{ 0\right\}$ satisfying $I_{1}\left( u_{k}\right) \rightarrow 0^{-}$ as $k\rightarrow \infty$.

3 Main Results

For problem $\left( \mathbf {P}\right)$, we make the following assumptions

$\left( \mathbf {P}_{1}\right)$ $f\left( x,-t\right) =-f\left( x,t\right)$ and $g\left( x,-t\right) =-g\left( x,t\right)$ for any $x\in \Omega$, $t\in \mathbb {R}$.

$\left( \mathbf {P}_{2}\right)$ Assume that $f:\Omega \times \mathbb {R} \rightarrow \mathbb {R}$ is a Carathéodory function and there exist $1<\sigma \le \delta <p^{-}$ and $c_{1}>0,c_{2}>0,c_{3}>0$ such that

$$\begin{aligned} c_{1}\left| t\right| ^{\sigma }\le f\left( x,t\right) t\le c_{2}\left| t\right| ^{\delta }+c_{3}\left| t\right| ^{\sigma },\quad \hbox {for a.e.}x\in \Omega \hbox { and }t\in \mathbb {R}. \end{aligned}$$

$\left( \mathbf {P}_{3}\right)$ Assume that $g:\Omega \times \mathbb {R} \rightarrow \mathbb {R}$ is a Carathéodory function and $p,q\in C_{+}\left( \overline{\Omega }\right)$ with $p(x) \le p^{+}<q^{-}\le q (x) <p^{*}(x)$ such that

$$\begin{aligned} \vert g(x,t)\vert \le c\left( 1+\vert {t} \vert ^{q(x)-1}\right) , \quad \hbox {for a.e.}x\in \Omega \hbox { and }t\in \mathbb {R} , \end{aligned}$$

and $g(x,t) t\ge 0,$ for a.e. $x\in \Omega$ and $t\in \mathbb {R}$. Moreover, $\lim \limits _{t\rightarrow 0}\frac{g(x,t) }{t^{p^{-}-1}}=0$ uniformly for $x\in \Omega$.

$\left( \mathbf {P}_{4}\right)$ Assume one of the following conditions holds

$$\begin{aligned}&(\mathbf {1}) \lim \limits _{\vert t\vert \rightarrow \infty }\frac{g( x,t) }{t^{p^{-}-1}}=0\hbox { uniformly for } x\in \Omega .\\&(\mathbf {2}) \lim \limits _{\vert t \vert \rightarrow \infty }\frac{g(x,t) }{t^{p^{-}-1}}=-\infty \hbox {uniformly for }x\in \Omega . \end{aligned}$$

Moreover, $\frac{f(x,t)}{t^{p^{-}-1}}$ and $\frac{g(x,t)}{t^{p^{-}-1}}$ are decreasing in $t\in \mathbb {R}$ for $t$ large enough.

$$\begin{aligned} \left( \mathbf {3}\right) \liminf \limits _{\left| t\right| \rightarrow \infty }\frac{g\left( x,t\right) t-\epsilon G\left( x,t\right) }{\left| t\right| ^{\alpha }}\ge c>0\hbox { uniformly for }x\in \Omega , \end{aligned}$$

where $\alpha >\delta$ and $\epsilon >0$. Moreover, $\lim \limits _{\left| t\right| \rightarrow \infty }\frac{g\left( x,t\right) }{t^{p^{-}-1} }=\infty$ uniformly for $x\in \Omega$; $\frac{g\left( x,t\right) }{ t^{p^{-}-1}}$ is increasing in $t\in \mathbb {R}$ for $t$ large enough.

Theorem 3.1

Assume that $\left( \mathbf {P}_{1}\right) \!{-}\!\left( \mathbf {P}_{4}\right)$ hold, then problem $\left( \mathbf {P} \right)$ has infinitely many solutions $\{u_{k}\}$ satisfying

$$\begin{aligned} \Phi \left( u_{k}\right) \!:=\!\int \limits _{\Omega }\frac{1}{p(x)}\left| \nabla u_{k}\right| ^{p(x) }dx-\int \limits _{\Omega }G(x,u_{k})dx-\int \limits _{\Omega }F(x,u_{k})dx\rightarrow 0^{-}\ \hbox { as }k\!\rightarrow \! \infty , \end{aligned}$$

where $\Phi :W_{0}^{1,p(x)}(\Omega ) \rightarrow \mathbb {R}$ is the functional corresponding to problem $\left( \mathbf {P} \right)$ and $G(x,t) =\int _{0}^{t}g(x,s)ds$, $F(x,t)=\int _{0}^{t}f(x,s)ds$.

Remark 3.1

The conditions $\left( \mathbf {P}_{2}\right)$ and $\left( \mathbf {P} _{3}\right)$ imply the functional $\Phi$ is well defined and of class $C^{1}$. It is well known that the critical points of $\Phi$ are weak solutions of $(\mathbf {P})$. Moreover, the derivative of $\Phi$ is given by

$$\begin{aligned} \langle \Phi ^{\prime }(u) ,\upsilon \rangle =\int \limits _{\Omega }\left| \nabla u\right| ^{p(x) -2}\nabla u\nabla \upsilon dx-\int \limits _{\Omega }g(x,u)\upsilon dx-\int \limits _{\Omega }f(x,u)\upsilon dx, \end{aligned}$$

for any $u,\upsilon \in W_{0}^{1,p(x)}( \Omega )$.

Let us consider $C^{1}$-functional $\Phi _{\lambda }:W_{0}^{1,p(x)}(\Omega ) \rightarrow \mathbb {R}$ defined by

$$\begin{aligned} \Phi _{\lambda }(u)&= \int \limits _{\Omega }\frac{1}{p(x)}\left| \nabla u\right| ^{p(x)}dx-\int \limits _{\Omega }G(x,u)dx-\lambda \int \limits _{\Omega }F(x,u)dx\\&:= A(u) -K(u) -\lambda B(u) , \end{aligned}$$

where $\lambda \in [1,2]$. Then $B(u) \ge 0$ and $B(u)\rightarrow \infty$ as $\left\| u\right\| \rightarrow \infty$ on any finite dimensional subspace, where $n > k > 2$. To get the proof of Theorem 3.1, we will apply Theorem 2.1. Therefore, it is enough to obtain the results of Lemma 3.1 and Lemma 3.2.

Lemma 3.1

Under the assumptions of Theorem 3.1, there exist $\lambda _{n}\rightarrow 1$, $u_{n}(\lambda ) \in Y_{n}$ such that

$$\begin{aligned} \Phi _{\lambda _{n}}^{\prime }\mid _{Y_{n}}\left( u_{n}(\lambda )\right) =0,\,\, \Phi _{\lambda _{n}}\left( u_{n}( \lambda ) \right) \rightarrow c_{k}\in \left[ d_{k}(2),b_{k}\left( 1\right) \right] \,\, as\,\, n\rightarrow \infty . \end{aligned}$$

Proof

First, we prove that for some $r_{k}\in \left( 0,\rho _{k}\right)$ such that

$$\begin{aligned} b_{k}(\lambda ):=\max \limits _{u\in Y_{k,}\left\| u\right\| =r_{k}}\Phi _{\lambda }(u)<0, \end{aligned}$$

for $\lambda \in [1,2]$, $u\in Y_{k}$. The norms $\left| \cdot \right| _{\sigma }$ and $\left\| \cdot \right\|$ are equivalent on the finite dimensional subspace $Y_{k}$. Therefore, there is a constant $c>0$ such that

$$\begin{aligned} \left| u\right| _{\sigma }\ge c\left\| u\right\| ,\ \ \forall u\in Y_{k}. \end{aligned}$$

Moreover, by $\left( \mathbf {P}_{3}\right)$, for any $\varepsilon >0$ there exists $C_{\varepsilon }>0$ such that $\left| G(x,u)\right| \le \varepsilon \left| u\right| ^{p^{-}}+C_{\varepsilon }\left| u\right| ^{q(x) }$. Then, by $\left( \mathbf {P}_{2}\right)$ and Proposition 2.1, we have

$$\begin{aligned} \Phi _{\lambda }(u) \le&\,\frac{1}{p^{-}}\left\| u\right\| ^{p^{-}}-K(u) -\lambda B(u) \\ \le&\,\frac{1}{p^{-}}\left\| u\right\| ^{p^{-}}-\varepsilon \int \limits _{\Omega }\left| u\right| ^{p^{-}}dx-C_{\varepsilon }\int \limits _{\Omega }\left| u\right| ^{q(x)}dx-\lambda c_{1}\int \limits _{\Omega }\left| u\right| ^{\sigma }dx\\ \le&\,\frac{1}{p^{-}}\left\| u\right\| ^{p^{-}}-\varepsilon c^{p^{-}}\left\| u\right\| ^{p^{-}}-C_{\varepsilon }\left\| u\right\| ^{q^{+}}-c_{4}^{\sigma }\left\| u\right\| ^{\sigma }. \end{aligned}$$

Since $\sigma <p^{-}<q^{+}$, for $\left\| u\right\|$ small enough we get $b_{k}(\lambda ):=\max \limits _{u\in Y_{k,}\left\| u\right\| =r_{k}}\Phi _{\lambda }(u) <0$ for all $u\in Y_{k}$. $\square$

Second, we shall show that for some $0<r_{k}<\rho _{k}$ such that

$$\begin{aligned} a_{k}(\lambda ):=\inf \limits _{u\in Z_{k,}\left\| u\right\| =\rho _{k}}\Phi _{\lambda }(u) \ge 0 \end{aligned}$$

for $\lambda \in [ 1,2]$, and $u\in Z_{k}$.

Let

$$\begin{aligned} \beta _{k}\left( q(x) \right) :&=\sup \limits _{u\in Z_{k},\left\| u\right\| =1}\left| u\right| _{q(x)}, { \ \ \ }\beta _{k}( p^{-}) :=\sup \limits _{u\in Z_{k},\left\| u\right\| =1}\left| u\right| _{p^{-}}, \\ \beta _{k}\left( \delta \right) :&=\sup \limits _{u\in Z_{k},\left\| u\right\| =1}\left| u\right| _{\delta }, { \ \ \ \ \ \ }\beta _{k}(\sigma ) :=\sup \limits _{u\in Z_{k},\left\| u\right\| =1}\left| u\right| _{\sigma }. \end{aligned}$$

Then $\beta _{k}(q(x)) \rightarrow 0$, $\beta _{k}(p^{-}) \rightarrow 0,$ $\beta _{k}(\delta )\rightarrow 0$ and $\beta _{k}(\sigma ) \rightarrow 0$ as $k\rightarrow \infty$ see, [9]. Therefore, by $\left( \mathbf {P}_{2}\right)$ and Proposition 2.1, we have

$$\begin{aligned} \Phi _{\lambda }(u)&= A(u) -K(u)-\lambda B(u) \ge \frac{1}{p^{+}}\left\| u\right\| ^{p^{+}}-K(u) -\lambda B(u) \\&\ge \frac{1}{p^{+}}\left\| u\right\| ^{p^{+}}-{\varepsilon } \int \limits _{\Omega }\left| u\right| ^{p^{-}}dx-C_{\varepsilon } \int \limits _{\Omega }\left| u\right| ^{q(x)}dx-\lambda c_{2} \int \limits _{\Omega }\left| u\right| ^{\delta }dx\\&\qquad -\lambda c_{3} \int \limits _{\Omega }\left| u\right| ^{\sigma }dx \\&\ge \frac{1}{p^{+}}\left\| u\right\| ^{p^{+}}-c\left| u\right| _{p^{-}}^{p^{-}}-c\left| u\right| _{q(x)}^{q^{-}}-c\left| u\right| _{\delta }^{\delta }-c\left| u\right| _{\sigma }^{\sigma } \\&\ge \frac{1}{p^{+}}\left\| u\right\| ^{p^{+}}-c\beta _{k}^{p^{-}}(p^{-}) \left\| u\right\| ^{p^{-}}-c\beta _{k}^{q^{-}}(q(x)) \left\| u\right\| ^{q^{-}}-c\beta _{k}^{\delta }(\delta ) \left\| u\right\| ^{\delta }\\&\qquad -c\beta _{k}^{\sigma }(\sigma ) \left\| u\right\| ^{\sigma } \end{aligned}$$

where $c=\max \ \{ \varepsilon ,C_{\varepsilon },2c_{2},2c_{3}\}$. Let $\varphi \in Z_{k}$, $\left\| \varphi \right\| =1$ and $0<t<1,$ then it follows

$$\begin{aligned} \Phi _{\lambda }(t\varphi )&\ge \frac{1}{p^{+}}t^{p^{+}}-c\beta _{k}^{p^{-}}(p^{-}) t^{p^{-}}-c\beta _{k}^{q^{-}}(q(x)) t^{q^{-}}-c\beta _{k}^{\delta }(\delta ) t^{\delta }-c\beta _{k}^{\sigma }(\sigma ) t^{\sigma } \\&\ge \frac{1}{p^{+}}t^{q^{-}}-c\beta _{k}^{q^{-}}(q(x)) t^{q^{-}}-\left( c\beta _{k}^{p^{-}}(p^{-})+c\beta _{k}^{\delta }(\delta )+c\beta _{k}^{\sigma }(\sigma )\right) t^{\sigma }, \end{aligned}$$

since $\sigma <$ $\delta <p^{-}<p^{+}<q^{-}$ for sufficiently large $k$, by choosing $c\beta _{k}^{q^{-}}(q(x))<\frac{1}{2p^{+}}$, we get

$$\begin{aligned} \Phi _{\lambda }( t\varphi ) {\ge }\frac{1}{2p^{+}}t^{q^{-}}-( c\beta _{k}^{p^{-}}(p^{-}) +c\beta _{k}^{\delta }(\delta ) +c\beta _{k}^{\sigma }(\sigma )) t^{\sigma }. \end{aligned}$$

(3.1)

Put $\rho _{k}:=\left( 2cp^{+}\beta _{k}^{p^{-}}(p^{-})+2cp^{+}\beta _{k}^{\delta }(\delta ) +2cp^{+}\beta _{k}^{\sigma }( \sigma ) \right) ^{\frac{1}{q^{-}-\sigma }}$, then, for sufficiently large $k$, $\rho _{k}<1$. When $t=\rho _{k}$, $\varphi \in Z_{k}$ with $\left\| \varphi \right\| =1$, we have $\Phi _{\lambda }( t\varphi ) \ge 0$. So, for sufficiently large $k$, we obtain $a_{k}(\lambda ):=\inf \limits _{u\in Z_{k,}\Vert u\Vert =\rho _{k}}\Phi _{\lambda }( u) \ge 0$.

Finally, we prove

$$\begin{aligned} d_{k}(\lambda ):=\inf \limits _{u\in Z_{k,}\Vert u\Vert \le \rho _{k}}\Phi _{\lambda }( u) \rightarrow 0 \end{aligned}$$

as $k\rightarrow \infty$ uniformly. Indeed, since $Y_{k}\cap Z_{k}\ne \varnothing$ and $r_{k}<$ $\rho _{k}$, we have

$$\begin{aligned} d_{k}(\lambda ):=\inf \limits _{u\in Z_{k,}\left\| u\right\| \le \rho _{k}}\Phi _{\lambda }(u) \le b_{k}(\lambda ):=\max \limits _{u\in Y_{k,}\left\| u\right\| =r_{k}}\Phi _{\lambda }(u) <0. \end{aligned}$$

By $(3.1)$, for $\varphi \in Z_{k}$, $\left\| \varphi \right\| =1$, $0\le t\le \rho _{k}$ and $u=t\varphi$ it follows that

$$\begin{aligned} \Phi _{\lambda }(u)&=\Phi _{\lambda }(t\varphi )\ge \frac{1}{2p^{+}}t^{q^{-}}-\left( c\beta _{k}^{p^{-}}(p^{-}) +c\beta _{k}^{\delta }(\delta ) +c\beta _{k}^{\sigma }(\sigma ) \right) t^{\sigma } \\&\ge -\left( c\beta _{k}^{p^{-}}(p^{-}) +c\beta _{k}^{\delta }(\delta ) +c\beta _{k}^{\sigma }(\sigma )\right) t^{\sigma } \\&\ge -\left( c\beta _{k}^{p^{-}}(p^{-}) +c\beta _{k}^{\delta }(\delta ) +c\beta _{k}^{\sigma }(\sigma )\right) \rho _{k}^{\sigma } \\&\ge -\left( c\beta _{k}^{p^{-}}(p^{-}) +c\beta _{k}^{\delta }(\delta ) +c\beta _{k}^{\sigma }(\sigma )\right) , \end{aligned}$$

therefore, $d_{k}(\lambda )\rightarrow 0$ as $k\rightarrow \infty$. Hence, by Theorem 2.1, we can find $\lambda _{n}\rightarrow 1$ and $u_{n}(\lambda ) \in Y_{n}$ desired as the claim. The proof is completed.$\square$

Lemma 3.2

$\{u_{n}(\lambda )\} _{n=1}^{\infty }$ is bounded in $W_{0}^{1,p(x)}(\Omega )$.

Proof

Since $\Phi _{\lambda _{n}}^{\prime }\mid _{Y_{n}}\left( u\left( \lambda _{n}\right) \right) =0$, then we have

$$\begin{aligned} \Phi _{\lambda }^{\prime }\left( u\left( \lambda _{n}\right) \right) = \mathbf {A}^{\prime }\left( u\left( \lambda _{n}\right) \right) -K^{\prime }\left( u\left( \lambda _{n}\right) \right) -\lambda _{n}B^{\prime }\left( u\left( \lambda _{n}\right) \right) =o(1) \left\| u\left( \lambda _{n}\right) \right\| , \end{aligned}$$

or, by Proposition 2.1,

$$\begin{aligned} 1-o\left( 1\right)&=\lambda _{n}\int \limits _{\Omega }\frac{f(x,u\left( \lambda _{n}\right) )u\left( \lambda _{n}\right) }{\rho \left( u\left( \lambda _{n}\right) \right) }dx+\int \limits _{\Omega }\frac{g(x,u\left( \lambda _{n}\right) )u\left( \lambda _{n}\right) }{\rho \left( u\left( \lambda _{n}\right) \right) }dx \\&\le \lambda _{n}\int \limits _{\Omega }\frac{f(x,u\left( \lambda _{n}\right) )u\left( \lambda _{n}\right) }{\left\| u\left( \lambda _{n}\right) \right\| ^{p^{-}}}dx+ \int \limits _{\Omega }\frac{g(x,u\left( \lambda _{n}\right) )u\left( \lambda _{n}\right) }{\left\| u\left( \lambda _{n}\right) \right\| ^{p^{-}}}dx \end{aligned}$$

where $\rho \left( u(\lambda _{n}) \right)$ is defined as in (2.1). Passing to a subsequence, if necessary, $\left\| u(\lambda _{n}) \right\| \rightarrow \infty$ as $n\rightarrow \infty$, and using $\left( \mathbf {P}_{2}\right)$ it follows

$$\begin{aligned} 1-o(1) \le \int \limits _{\Omega }\frac{g(x,u(\lambda _{n})) u(\lambda _{n}) }{\left\| u\left( \lambda _{n}\right) \right\| ^{p^{-}}}dx, \end{aligned}$$

where $o(1) \rightarrow 0$ as $n\rightarrow \infty$. This is a contradiction providing that $\left( \mathbf {P}_{4}\right) \left( \mathbf {1}\right)$ holds. $\square$

Let $\{\omega _{n}\} \subset W_{0}^{1,p(x)}(\Omega )$ and put $\omega _{n}:=\frac{u(\lambda _{n})}{\left\| u\left( \lambda _{n}\right) \right\| }$. Since $\Vert \omega _{n}\Vert =1$, up to subsequences, from Proposition 2.4 we get

$$\begin{aligned} \omega _{n}&\rightharpoonup \omega \quad \hbox { in } W_{0}^{1,p(x)}(\Omega ) , \\ \omega _{n}&\rightarrow \omega \quad \hbox { in }L^{\gamma (x)}(\Omega ) ,\,\,p(x) \le \gamma (x)<p^{*}(x) , \\ \omega _{n}(x)&\rightarrow \omega (x)\quad \hbox { a.e. }x\in \Omega . \end{aligned}$$

Then the main concern is that either $\left\{ \omega _{n}\right\} \subset W_{0}^{1,p(x)}\left( \Omega \right)$ vanish or it does not vanish. We shall prove that none of these alternatives can occur and this contradiction will prove that $\left\{ \omega _{n}\right\} \subset W_{0}^{1,p(x)}\left( \Omega \right)$ is bounded.

If $\omega \ne 0$, from Proposition 2.1, Fatou’s Lemma, $\left( \mathbf {P} _{2}\right) ,\left( \mathbf {P}_{3}\right)$ and for $n$ large enough, we have

$$\begin{aligned} \Phi _{\lambda }^{\prime }\left( u\left( \lambda _{n}\right) \right) = \mathbf {A}^{\prime }\left( u\left( \lambda _{n}\right) \right) -K^{\prime }\left( u\left( \lambda _{n}\right) \right) -\lambda _{n}B^{\prime }\left( u\left( \lambda _{n}\right) \right) =o\left( 1\right) \left\| u\left( \lambda _{n}\right) \right\| , \end{aligned}$$

or

$$\begin{aligned} -1+o(1)&=\lambda _{n}\int \limits _{\Omega }\frac{-f(x,u(\lambda _{n}))u(\lambda _{n})}{\rho (u(\lambda _{n}))}dx+\int \limits _{\Omega }\frac{-g(x,u(\lambda _{n}))u(\lambda _{n})}{\rho (u(\lambda _{n}))}dx\\&\ge \lambda _{n}\int \limits _{\Omega }\frac{-f(x,u(\lambda _{n}))u(\lambda _{n})}{\Vert u(\lambda _{n})\Vert ^{p^{-}}}dx+\int \limits _{\Omega }\frac{-g(x,u(\lambda _{n}))u(\lambda _{n})}{\Vert u(\lambda _{n})\Vert ^{p^{-}}}dx. \end{aligned}$$

Using $\lim \limits _{\left| u\right| \rightarrow \infty } \frac{g(x,u)}{\left| u\right| ^{p^{-}-1}}=-\infty$ in $\left( \mathbf {P}_{4}\right)$ $\left( \mathbf {2}\right)$, we get

$$\begin{aligned} -1+o(1)&\ge \int \limits _{\Omega }\frac{-g(x,u(\lambda _{n}))u(\lambda _{n})}{\Vert u(\lambda _{n})\Vert ^{p^{-}}}dx=\int \limits _{\Omega }\frac{-g(x,u(\lambda _{n}))u(\lambda _{n})}{\vert u(\lambda _{n})\vert ^{p^{-}}}\vert \omega _{n}\vert ^{p^{-}}dx \\&\ge c+\int \limits _{\{\omega \ne 0\} \cap \{\vert u(\lambda _{n})\vert \ge c\} }\frac{-g(x,u(\lambda _{n})) u(\lambda _{n})}{\vert u(\lambda _{n})\vert ^{p^{-}}}\vert \omega _{n}\vert ^{p^{-}}dx\rightarrow \infty , \end{aligned}$$

which is a contradiction. Moreover, we can get the similar result if $\lim \limits _{\vert u\vert \rightarrow \infty }\frac{g(x,u)}{\vert u\vert ^{p^{-}-1}}=\infty$ in $(\mathbf {P}_{4})(\mathbf {3})$.

If $\omega \equiv 0$, we can define a sequence ${ t_{n}} \subset \mathbb {R}$ as in see, [17] such that

$$\begin{aligned} \Phi _{\lambda _{n}}(t_{n}u(\lambda _{n})):=\max \limits _{t\in [0,1]}\Phi _{\lambda _{n}}(tu(\lambda _{n})) . \end{aligned}$$

Let $\overline{\omega }_{n}:=(2p^{+}c)^{\frac{1}{p^{-}}}\omega _{n}$ with $c>0$. Then for $n$ large enough, we have

$$\begin{aligned} \Phi _{\lambda _{n}}(t_{n}u_{n})&\ge \Phi _{\lambda _{n}}(\overline{\omega }_{n}) \\&\ge A\left( (2p^{+}c)^{\frac{1}{p^{-}}}\omega _{n}\right) -K\left( (2p^{+}c)^{\frac{1}{p^{-}}}\omega _{n}\right) -\lambda _{n}B\left( (2p^{+}c)^{\frac{1}{p^{-}}}\omega _{n}\right) \\&\ge \frac{1}{p^{+}}(2p^{+}c)A(\omega _{n})-K(\overline{\omega }_{n})-\lambda _{n}B(\overline{\omega }_{n})\ge 2c-K(\overline{\omega }_{n})-\lambda _{n}B(\overline{\omega }_{n}) \\&\ge c, \end{aligned}$$

which implies that $\lim \limits _{n\rightarrow \infty }\Phi _{\lambda _{n}}( t_{n}u_{n}) \rightarrow \infty$ by the fact $c>0$ can be large arbitrarily. Noting that $\Phi _{\lambda _{n}}(0) =0$ and $\Phi _{\lambda _{n}}(u_{n}) \rightarrow c$, so $0<t_{n}<1$ when $n$ large enough. Hence we have $\langle \Phi _{\lambda _{n}}^{\prime }(t_{n}u(\lambda _{n})),t_{n}u(\lambda _{n}) \rangle =0$. Thus, it follows

$$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \Phi _{\lambda _{n}}( t_{n}u(\lambda _{n}))-\frac{1}{\overline{p}_{t_{n}}}\langle \Phi _{\lambda _{n}}^{\prime }( t_{n}u( \lambda _{n})),t_{n}u( \lambda _{n})t\rangle \right] \rightarrow \infty , \end{aligned}$$

where $\overline{p}_{t_{n}}=\frac{A^{\prime }( t_{n}u( \lambda _{n}) ) }{A( t_{n}u( \lambda _{n}) ) }$. Therefore,

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }\Bigg [(A(t_{n}u(\lambda _{n})) -K( t_{n}u(\lambda _{n}))-\lambda _{n}B( t_{n}u(\lambda _{n}))\\&\qquad -\frac{1}{\overline{p}_{t_{n}}}(A^{\prime }( t_{n}u(\lambda _{n})) +\frac{1}{\overline{p}_{t_{n}}}K^{\prime }(t_{n}u( \lambda _{n})) +\lambda _{n}\frac{1}{\overline{p} _{t_{n}}}B^{\prime }(t_{n}u(\lambda _{n}))\Bigg ]\rightarrow \infty , \end{aligned}$$

that is,

$$\begin{aligned} \lim _{n\rightarrow \infty }\left[ \lambda _{n}\frac{1}{\overline{p}_{t_{n}}}B^{\prime }(t_{n}u(\lambda _{n}))-\lambda _{n}B(t_{n}u(\lambda _{n}))+\frac{1}{\overline{p}_{t_{n}}}K^{\prime }( t_{n}u(\lambda _{n}))-K(t_{n}u(\lambda _{n})) \right] \rightarrow \infty . \end{aligned}$$

Moreover, if $(\mathbf {P}_{4}) (2)$ holds, we have

$$\begin{aligned} \frac{1}{\overline{p}_{t_{n}}}f\left( x,su\right) su-F\left( x,su\right) + \frac{1}{\overline{p}_{t_{n}}}g\left( x,su\right) su-G\left( x,su\right) \le c, \end{aligned}$$

for all $s>0$ and $u\in \mathbb {R}$, so we get a contradiction.

If $\left( \mathbf {P}_{4}\right)$ $\left( \mathbf {3}\right)$ holds, by $\left( \mathbf {P}_{2}\right)$, we get

$$\begin{aligned} \infty \le \frac{c_{2}}{\overline{p}_{n}}\int \limits _{\Omega }\vert u( \lambda _{n}) \vert ^{\delta }dx+\frac{1}{\overline{p}_{n}}K^{\prime }( u( \lambda _{n}) ) -K( u(\lambda _{n})) . \end{aligned}$$

Thus,

$$\begin{aligned} \frac{1}{\overline{p}_{n}}K^{\prime }( u( \lambda _{n})) -K( u(\lambda _{n})) \rightarrow \infty . \end{aligned}$$

(3.2)

Furthermore, using the property of $u\left( \lambda _{n}\right)$ (see Lemma 3.1), it follows that

$$\begin{aligned} b_{k}\left( 1\right) \ge&\,\lambda _{n}\left( \frac{1}{\overline{p}_{n}}B^{\prime }(u(\lambda _{n})) -B(u(\lambda _{n}) ) \right) +\frac{1}{\overline{p}_{n}}K^{\prime }( u( \lambda _{n})) -K (u( \lambda _{n})) \\ \ge&\,\frac{1}{\overline{p}_{n}}\left( \frac{1}{\overline{p}_{n}}K^{\prime }( u( \lambda _{n})) -K( u( \lambda _{n}) )\right) -\frac{c_{2}}{\overline{p}_{n}} \int \limits _{\Omega }\vert u(\lambda _{n})\vert ^{\delta }dx \\ \ge&\,c\frac{1}{\overline{p}_{n}}K^{\prime }(u(\lambda _{n})) -K(u(\lambda _{n})) -c, \end{aligned}$$

which contradicts (3.2). Therefore, ${(\lambda _{n})}$ is bounded. The proof is completed.