The Existence of Entropy Solutions for a Class of Parabolic Equations

Abstract

The existence and uniqueness of entropy solutions for a class of parabolic equations involving a $p\left(x\right)$-Laplace operator are investigated. We first prove existence of the global weak solution for the $p\left(x\right)$-Laplacian equations with regular initial data via the difference and variation methods as well as the standard domain expansion technique. Then, by constructing and solving a related approximation problem, the entropy solution for the $p\left(x\right)$-Laplacian equations with irregular initial data in whole space is also obtained.

1. Introduction

In this paper, we consider the following nonlinear parabolic $p\left(x\right)$-Laplacian equations:

$$\left\{\begin{array}{cc}{u}_t-{\mathrm{div}(|\nabla u|}^{p\left(x\right)-2}{\nabla u)+\lambda |u|}^{q\left(x\right)-2}u=f(x,u)+g\left(x\right),\hfill & \mathrm{in}{\mathbb{R}}^N\times {\mathbb{R}}^{+},\hfill \\ u(x,0)=u_0\left(x\right),\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \end{array}\right.$$

(1)

where $u_0,g\left(x\right)\in L^1\left({\mathbb{R}}^N\right)(N\ge 2)$, $\lambda >0$, $q\left(x\right)\ge 2$. The exponent $p\left(x\right)$ is a so-called Log-Hölder continuous function, namely, there exists a positive constant $c_0$ such that

$$|p\left(x\right)-p\left(y\right)|\le -\frac{c_0}{log|x-y|}\mathrm{for} \mathrm{every}x,y\in {\mathbb{R}}^N\mathrm{with}|x-y|\le \frac{1}{2},$$

(2)

which satisfies

$$1<p^{-}=\underset{x\in {\mathbb{R}}^N}{inf}p\left(x\right)\le \underset{x\in {\mathbb{R}}^N}{sup}p\left(x\right)=p^{+}<+\infty .$$

The study of partial differential equations referring to $p\left(x\right)$-Laplacian attracts much attention due to the extensive application in the study of electrorheological fluid, image processing and other various fields with significant physical backgrounds (see [1,2,3]). As we know, electrorheological fluid is mainly used to make liquids and solids transform in electric fields, such as the automobile transmission principle, and the image processing is often used to remove noise, enhance image quality, restore, segment and extract features. The results on both elliptic and parabolic equations involving a variable exponent have kept growing in recent years. We refer to the references [4,5,6,7,8,9,10,11,12].

One of the interesting topics in the study of $p\left(x\right)$-Laplacian equations is the existence of solutions for the differential equations with irregular data (such as $L^1$-data, measure data). Because of the irregularity of the initial value and the forced term, the normal technique of getting weak solutions for the equation is no longer feasible. People may transfer to find the “very weak” solution, such as, DiPerna [13] who first introduced a renormalized solution into the Boltzmann equations, which overcame the deficiency of a strong a priori estimation and obtained the weak solutions. After that, Bénilan and Boccardo introduced the concept of entropy solution and obtained a unique entropy solution for the nonlinear elliptic problem in [14]. Then, more and more scholars studied renormalized and entropy solutions in elliptic equations (see [15,16,17]) and parabolic equations (see [18,19,20,21,22]). At the same time, many people have made more general equations about Leray–Lions-type operator equations [23,24,25,26,27], anisotropic equations [26,28] and fractional equations [29]. When ${\mathbb{R}}^N$ is replaced by a bounded domain $\mathsf{\Omega }\subset {\mathbb{R}}^N$ in (1), Zhang and Zhou in [22] obtained the existence and uniqueness of the renormalized and entropy solutions in (1) with $\lambda =0, f(x,u)=0$, and then Bendahmane and Wittbold proved the existence and uniqueness of the renormalized solution in [18]. Recently, under the assumptions: f is a Carathéodory mapping and there exist positive constants $l,C,c_1,c_2,{\sigma }_0$ and $b\left(x\right)\in L^1\left(\mathsf{\Omega }\right)$ such that

($H_1$)

$(f(x,s_1)-f(x,s_2))(s_1-s_2)\ge l{|s_1-s_2|}^2$ for a.e. $x\in \mathsf{\Omega }$ and any $s_1,s_2\in \mathbb{R}$;

($H_2$)

$c_2{|s|}^{q+1}-C\le f(x,s)s\le c_1{|s|}^{q+1}+C,q\ge 1$ for a.e. $x\in \mathsf{\Omega }$ and any $s\in \mathbb{R}$;

($H_3$)

$|f(x,s)|\le b\left(x\right)$ for a.e. $x\in \mathsf{\Omega }$ and all $s\in \mathbb{R}$ with $|s|<{\sigma }_0$,

Niu and Chai in [21] obtained a unique entropy solution and the global attractor in $L^q\left(\mathsf{\Omega }\right)$ in the case of $\lambda =0, f(x,u)=-f(x,u)$ in (1). The generalized equations involving a variable exponent are also getting more attractive than before. Li and Gao in [20] considered the existence of a renormalized solution when the equation contains the gradient term $g(x,t,u,\nabla u)$. Jamea, Lamrani and El Hachimi in [30] obtained the existence of the entropy solutions of the $p\left(x\right)$-Laplace equations with the Neumann boundary condition. In case of $\mathsf{\Omega }\subset {\mathbb{R}}^N$ being an arbitrary domain, the uniqueness of the entropy solution or renormalized solution is investigated by [24,31].

In view of what is mentioned above, we intend to get not only the existence of the entropy solution but also the uniqueness of the solution for the equation (1) in the whole space. To this end, we assume that external force $f(x,u)$ is a Carathéodory function satisfying $f(x,s)s\le 0$ for almost all $x\in {\mathbb{R}}^N$, and

$$\begin{array}{c}\hfill -l_1|s_1-s_2{|}^2\le (f(x,s_1)-f(x,s_2))(s_1-s_2)\le l_2{|s_1-s_2|}^2,\end{array}$$

(3)

for any $s_1,s_2\in \mathbb{R}$ and positive constants $l_1,l_2$. It is notable that there are more difficulties to be solved when we locate the problem in an unbounded domain, such as no compactness Sobolev embedding, invalidity of Poincaré inequality, and so on. One can see that the double boundedness in (3) is different from the conditions ($H_1$) and ($H_2$) in [21]. Actually, the authors in [21] depended on the Vitali’s convergence theorem and Marcinkiewicz space to get the important convergence of $f(x,u_n)$ in $L^1$. In our situation, we cannot use this technique again, and we refer to the global Lipschitz condition of $f(x,u)$ to get the corresponding convergence. Moreover, in order to overcome the limitation brought by the unbounded region, we firstly obtain the global weak solution in ${\mathbb{R}}^N$ by the difference and variation methods and the standard region expansion technique, and then prove the existence and uniqueness of the entropy solution in whole space by constructing and solving the approximation problem which corresponds to the $p\left(x\right)$-Laplacian equation with irregular initial data.

This paper is organized as follows. In Section 2, we provide the definitions and relevant lemmas, and list our results. Section 3 is devoted to obtaining a unique weak solution in the bounded domain, then extend to ${\mathbb{R}}^N$. In Section 4, we prove the existence and uniqueness of the entropy solution in ${\mathbb{R}}^N$.

For the convenience, throughout the paper, we denote $\mathsf{\Omega }\times (0,T)$ by ${\mathsf{\Omega }}_T$ and ${\mathbb{R}}^N\times (0,T)$ by $Q_T$ for any $T>0$. C is a positive constant, which may be different from line to line and even in the same line.

2. Notations, Definitions and Preliminary Results

In this section, we recall the definitions and some basic properties of the generalized Lebesuge and Sobolev spaces (see [32,33] for more details).

Let $\mathsf{\Omega }$ be an open domain of ${\mathbb{R}}^N$, $N\ge 2$. For a variable exponent $p\left(x\right)\in C\left(\mathsf{\Omega }\right)$ satisfying the log-Hölder continuity condition (2), we define the Lebesgue space $L^{p(·)}\left(\mathsf{\Omega }\right)$ as

$$L^{p(·)}\left(\mathsf{\Omega }\right)=\{u:\mathsf{\Omega }⟶\mathbb{R},\mathrm{u} \mathrm{is} \mathrm{measurable} \mathrm{and}{\int }_{\mathsf{\Omega }}{|u|}^{p\left(x\right)}<\infty \}$$

with the norm

$${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}=inf\{\lambda >0:{\int }_{\mathsf{\Omega }}|\frac{u\left(x\right)}{\lambda }{|}^{p\left(x\right)}\mathrm{d}x\le 1\},$$

which is a separable and reflexive Banach space. The dual space of $L^{p\left(x\right)}\left(\mathsf{\Omega }\right)$ is $L^{p^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)$, where $\frac{1}{p\left(x\right)}+\frac{1}{p^{\prime }\left(x\right)}=1$. The Sobolev space ${\mathcal{D}}_0^{1,p\left(x\right)}\left(\mathsf{\Omega }\right)$ is defined as the closure of $C_0^{\infty }\left(\mathsf{\Omega }\right)$ in the norm

$$\begin{array}{c}\hfill {\parallel u\parallel }_{{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right)}={\parallel \nabla u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}.\end{array}$$

For a positive integer m, the generalized Sobolev space $W^{m,p(·)}\left(\mathsf{\Omega }\right)$ is defined as

$$\begin{array}{c}\hfill W^{m,p(·)}\left(\mathsf{\Omega }\right)=\{u\in L^{p(·)}\left(\mathsf{\Omega }\right):D^{\alpha }u\in L^{p(·)}\left(\mathsf{\Omega }\right),|\alpha |\le m\}\end{array}$$

with norm

$$\begin{array}{c}\hfill {\parallel u\parallel }_{W^{m,p(·)}\left(\mathsf{\Omega }\right)}=\sum _{|\alpha |\le m}{\parallel D^{\alpha }u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}.\end{array}$$

Under the assumption (2), smooth functions are dense in it, and the space $W_0^{m,p(·)}\left(\mathsf{\Omega }\right)$ can be defined as the completion of $C_0^{\infty }\left(\mathsf{\Omega }\right)$ in $W^{m,p(·)}\left(\mathsf{\Omega }\right)$ with respect to the norm ${\parallel ·\parallel }_{W^{m,p(·)}\left(\mathsf{\Omega }\right)}$.

Let $T_k,k\ge 0$ be the truncation function:

$$T_k\left(r\right)=min\{k,max\{r,-k\}\}=\left\{\begin{array}{cc}k,\hfill & \mathrm{if}r\ge k,\hfill \\ r,\hfill & \mathrm{if}|r|<k,\hfill \\ -k,\hfill & \mathrm{if}r\le -k,\hfill \end{array}\right.$$

and its primitive function ${\mathsf{\Phi }}_k:\mathbb{R}⟶{\mathbb{R}}^{+}$

$${\mathsf{\Phi }}_k\left(r\right)={\int }_0^r{T}_k\left(s\right)\mathrm{ds}=\left\{\begin{array}{cc}\frac{r^2}{2},\hfill & \mathrm{if}|r|\le k,\hfill \\ k|r|-\frac{k^2}{2},\hfill & \mathrm{if}|r|\ge k.\hfill \end{array}\right.$$

It is obvious that ${\mathsf{\Phi }}_k\left(r\right)\ge 0$ and ${\mathsf{\Phi }}_k\left(r\right)\le k|r|$.

Lemma 1

(see [33,34]). The space $L^{p(·)}\left(\mathsf{\Omega }\right)$ is a separable, uniformly convex Banach space, and its conjugate space is $L^{p^{\prime }(·)}\left(\mathsf{\Omega }\right)$, where $\frac{1}{p\left(x\right)}+\frac{1}{p^{\prime }\left(x\right)}=1$. For any $u\in L^{p\left(x\right)}\left(\mathsf{\Omega }\right)$ and $v\in L^{p^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)$, we have

$${\int }_{\mathsf{\Omega }}|uv|dx\le \left(\frac{1}{p^{-}}+\frac{1}{p^{\prime -}}\right){\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}{\parallel v\parallel }_{L^{p^{\prime }(·)}\left(\mathsf{\Omega }\right)}\le {2\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}{\parallel v\parallel }_{L^{p^{\prime }(·)}\left(\mathsf{\Omega }\right)}.$$

Lemma 2

(see [33,34]). Let

$$\rho \left(u\right)={\int }_{\mathsf{\Omega }}{|u|}^{p\left(x\right)}dx for all u\in L^{p(·)}\left(\mathsf{\Omega }\right).$$

Then,

(1) 

$\rho \left(u\right)>1(=1,<1)$ if and only if ${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}>1(=1,<1,respectively);$

(2) 

if ${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}>1$, then ${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}^{p^{-}}\le \rho \left(u\right)\le {\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}^{p^{+}}$;

(3) 

if ${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}<1$, then ${\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}^{p^{+}}\le \rho \left(u\right)\le {\parallel u\parallel }_{L^{p(·)}\left(\mathsf{\Omega }\right)}^{p^{-}}$.

We are now in a position to give the main results of this paper.

Theorem 1.

Let $u_0,g\in L^1\left({\mathbb{R}}^N\right)$ and (2) and (3) hold, then the problem (1) admits a unique entropy solution u for any $T>0$.

3. Proof of the Weak Solution

In this section, we first consider the existence of weak solutions in a bounded, smooth domain $\mathsf{\Omega }\subset {\mathbb{R}}^N$ with the boundary condition:

$$\begin{array}{c}\hfill {u\left(t\right)|}_{\partial \mathsf{\Omega }}=0\mathrm{for} \mathrm{all} \mathrm{values} \mathrm{of}t>0.\end{array}$$

(4)

Definition 1.

A function $u(x,t)$ is called a weak solution of (1) and (4) on  $[0,T],$ if for any $T>0$, 

$$u\in C([0,T];L^2\left(\mathsf{\Omega }\right))\cap L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p\left(x\right)}\left(\mathsf{\Omega }\right))\cap L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right)$$

with $\nabla u\in {\left(L^{p(·)}\left({\mathsf{\Omega }}_T\right)\right)}^N$,  $u_t\in L^{{\left(p^{-}\right)}^{\prime }}(0,T;{\mathcal{D}}^{-1,p^{\prime }(·)}\left(\mathsf{\Omega }\right))+L^{q^{\prime }(·)}\left({\mathsf{\Omega }}_T\right)$ and  ${u|}_{t=0}=u_0$ almost everywhere in Ω such that 

$${\int }_{{\mathsf{\Omega }}_T}(-u{\phi }_t+{|\nabla u|}^{p\left(x\right)-2}\nabla u·\nabla \phi +{\lambda |u|}^{q\left(x\right)-2}u\phi )dxdt={\int }_{{\mathsf{\Omega }}_T}(f(x,u)\phi +g\phi )dxdt+{\int }_{\mathsf{\Omega }}{u}_0\phi \left(0\right)dx$$

 for any  $\phi \in C^1\left({\overline{\mathsf{\Omega }}}_T\right)$ with  $\phi =0on\partial \mathsf{\Omega }\times (0,T]$.

Next, we prove the existence and uniqueness of weak solution to (1) and (4) by the difference and variation methods (see [8]).

Proposition 1.

Let $\mathsf{\Omega }\subset {\mathbb{R}}^N$ be a bounded smooth domain and assumptions (2) and (3) hold. Then for any $u_0\in L^2\left(\mathsf{\Omega }\right)$ and a smooth function g, there exists a unique weak solution u to (1) and (4) for any $T>0$.

Proof. 

We denote $W=\{u\in L^2\left(\mathsf{\Omega }\right)\cap {\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right)\cap L^{q(·)}\left(\mathsf{\Omega }\right)\}$ with the norm

$$\begin{array}{c}\hfill {\parallel u\parallel }_W={\parallel u\parallel }_{L^2\left(\mathsf{\Omega }\right)}+{\parallel u\parallel }_{{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right)}+{\parallel u\parallel }_{L^{q(·)}\left(\mathsf{\Omega }\right)}.\end{array}$$

Then, it is obvious that W is a Banach space. Now we construct an approximation sequence of solution for problems (1) and (4). Let n be a positive integer. Denote $h=\frac{T}{n}$, we consider the following elliptic problems:

$$\left\{\begin{array}{c}\frac{u_k-u_{k-1}}{h}-\mathrm{div}(|\nabla u_k{|}^{p\left(x\right)-2}\nabla u_k)+\lambda |u_k{|}^{q\left(x\right)-2}{u}_k=f(x,u_k)+g,\hfill \\ u_k{|}_{\partial \mathsf{\Omega }}=0,k=1,2,\dots ,n.\hfill \end{array}\right.$$

(5)

We divide the proof of this proposition into four steps.

Step 1. Find solution $u_k$ of (5) for every $k\in \{1,2,\dots ,n\}$. For $k=1$, we introduce a functional J in W as follows

$$\begin{array}{cc}\hfill J\left(u\right)=& \frac{1}{2h}{\int }_{\mathsf{\Omega }}{|u|}^2\mathrm{d}x+{\int }_{\mathsf{\Omega }}\frac{1}{p\left(x\right)}{|\nabla u|}^{p\left(x\right)}\mathrm{d}x+\lambda {\int }_{\mathsf{\Omega }}\frac{1}{q\left(x\right)}{|u|}^{q\left(x\right)}\mathrm{d}x\hfill \\ \hfill & -\frac{1}{h}{\int }_{\mathsf{\Omega }}{u}_{0}u\mathrm{d}x-{\int }_{\mathsf{\Omega }}F(x,u)\mathrm{d}x-{\int }_{\mathsf{\Omega }}gu\mathrm{d}x,\hfill \end{array}$$

(6)

where $F(x,u)={\int }_0^{u}f(x,s)\mathrm{d}s$. Firstly, by condition of $f(x,u)$, we have

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}F(x,u)\mathrm{d}x\le 0,\end{array}$$

(7)

and by Hölder and Young inequalities, it is easy to see that

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}{u}_{0}u\mathrm{d}x\le C\left(\epsilon \right){\int }_{\mathsf{\Omega }}|u_0{|}^2\mathrm{d}x+\epsilon {\int }_{\mathsf{\Omega }}{|u|}^2\mathrm{d}x,\end{array}$$

(8)

and

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}gu\mathrm{d}x\le C\left(\epsilon \right){\int }_{\mathsf{\Omega }}{|g|}^2\mathrm{d}x+\epsilon {\int }_{\mathsf{\Omega }}{|u|}^2\mathrm{d}x,\end{array}$$

(9)

where $\epsilon >0$ is a small positive number. Thus, applying (7)–(9) and choosing $\epsilon$ sufficiently small, we have from (6) that

$$\begin{array}{cc}\hfill J\left(u\right)& \ge \frac{1}{4h}{\int }_{\mathsf{\Omega }}{|u|}^2\mathrm{d}x+\frac{1}{p^{+}}{\int }_{\mathsf{\Omega }}{|\nabla u|}^{p\left(x\right)}\mathrm{d}x+\frac{\lambda }{q^{+}}{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)}\mathrm{d}x\hfill \\ \hfill & -C({\int }_{\mathsf{\Omega }}|u_0{|}^2\mathrm{d}x+{\int }_{\mathsf{\Omega }}|g{|}^2\mathrm{d}x)\hfill \\ \hfill & \ge -C(\parallel u_0{\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2,\parallel g{\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2).\hfill \end{array}$$

This immediately implies that

$$-C\le \underset{w\in W}{inf}J\left(w\right)\le J\left(0\right)\le C.$$

Now, we choose a minimizing sequence ${\left\{v_m\right\}}_{m=1}^{\infty }\subset W$ such that

$$\begin{array}{c}\hfill J\left(v_m\right)⟶\underset{w\in W}{inf}J\left(w\right).\end{array}$$

Then we have $J\left(v_m\right)\le C$ for $m=1,2,\dots$, and by the argument as (7)–(9), it reduces to

$$\begin{array}{cc}\hfill & \frac{1}{4h}{\int }_{\mathsf{\Omega }}|v_m{|}^2\mathrm{d}x+\frac{1}{p^{+}}{\int }_{\mathsf{\Omega }}|\nabla v_m{|}^{p\left(x\right)}\mathrm{d}x+\frac{\lambda }{q^{+}}{\int }_{\mathsf{\Omega }}{|v_m|}^{q\left(x\right)}\mathrm{d}x\hfill \\ \hfill & \le C+C(\parallel v_{0m}{\parallel }_{L^2\left(\mathsf{\Omega }\right)},\parallel g{\parallel }_{L^2\left(\mathsf{\Omega }\right)}).\hfill \end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill \parallel v_m{\parallel }_{L^2\left(\mathsf{\Omega }\right)}\le C,\parallel v_m{\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}\le C,{\parallel \nabla v_m\parallel }_{L^{p\left(x\right)}\left(\mathsf{\Omega }\right)}\le C,\end{array}$$

and there exists a subsequence ${\left\{v_{m_j}\right\}}_{j=1}^{\infty }\subset {\left\{v_m\right\}}_{m=1}^{\infty }$ and a function $u_1\in W$, such that

$$\begin{array}{c}\hfill v_{m_j}\rightharpoonup u_1\mathrm{weakly} \mathrm{in}{L}^2\left(\mathsf{\Omega }\right)\cap L^{q\left(x\right)}\left(\mathsf{\Omega }\right),\\ \hfill \nabla v_{m_j}\rightharpoonup \nabla u_1\mathrm{weakly} \mathrm{in}{\left(L^{p\left(x\right)}\left(\mathsf{\Omega }\right)\right)}^N.\end{array}$$

Since $W^{1,p\left(x\right)}\left(\mathsf{\Omega }\right)\hookrightarrow \hookrightarrow L^{\alpha \left(x\right)}\left(\mathsf{\Omega }\right),1\le \alpha \left(x\right)<p^{*}\left(x\right)=\frac{Np\left(x\right)}{N-p\left(x\right)}$, we get

$$\begin{array}{c}\hfill v_{m_j}\rightharpoonup u_1\mathrm{strongly} \mathrm{in}{L}^{\alpha \left(x\right)}\left(\mathsf{\Omega }\right).\end{array}$$

By means of f condition, we have $|f(x,s)|\le l|s|$, where $l=max\{l_1,l_2\}$. Then,

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}|F(x,v_{m_j})-F(x,u_1)|\mathrm{d}x={\int }_{\mathsf{\Omega }}|{\int }_{u_1}^{v_{m_j}}f(x,s)\mathrm{d}s|\mathrm{d}x\le {\int }_{\mathsf{\Omega }}{\int }_{u_1}^{v_{m_j}}|f(x,s)|\mathrm{d}s\mathrm{d}x\hfill \\ \hfill & \le \frac{1}{2}{\int }_{\mathsf{\Omega }}|(u_1-v_{m_j})(u_1+v_{m_j})|\mathrm{d}x\le \frac{1}{2}\parallel u_1-v_{m_j}{\parallel }_{L^{\alpha \left(x\right)}\left(\mathsf{\Omega }\right)}{\parallel u_1+v_{m_j}\parallel }_{L^{{\alpha }^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)}<\epsilon ,\hfill \end{array}$$

as $j⟶+\infty$, where $\frac{1}{\alpha \left(x\right)}+\frac{1}{{\alpha }^{\prime }\left(x\right)}=1$. Thus, we conclude that $F(x,v_{m_j})⟶F(x,u_1)$ strongly in $L^1\left(\mathsf{\Omega }\right)$, $J\left(u\right)$ is weakly lower semi-continuous on W, that is,

$$\begin{array}{c}\hfill J\left(u_1\right)\le \underset{j⟶\infty }{lim\; inf}J(v_{m_j})=\underset{w\in W}{inf}J\left(w\right).\end{array}$$

This implies that $u_1\in W$ is a minimizer of the functional $J\left(u\right)$ in W, i.e.,

$$\begin{array}{c}\hfill J\left(u_1\right)=\underset{w\in W}{inf}J\left(w\right).\end{array}$$

With $u_1$, by using the same argument as before, we can find weak solution $u_k$ of (5) for $k=2,\dots ,n$.

Step 2. Find the limit u of $u_h$ with $h=\frac{T}{n}$ based on $u_k$. Taking $L^2$ inner product to (5) by $u_k$, one has

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}\frac{u_k-u_{k-1}}{h}{u}_k\mathrm{d}x+{\int }_{\mathsf{\Omega }}|\nabla u_k{|}^{p\left(x\right)}\mathrm{d}x+\lambda {\int }_{\mathsf{\Omega }}{|u_k|}^{q\left(x\right)}\mathrm{d}x\hfill \\ \hfill & ={\int }_{\mathsf{\Omega }}f(x,u_k)u_k\mathrm{d}x+{\int }_{\mathsf{\Omega }}g{u}_k\mathrm{d}x.\hfill \end{array}$$

(10)

By Hölder and Young inequalities, we have

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}\frac{u_k-u_{k-1}}{h}{u}_k\mathrm{d}x\ge \frac{1}{2h}{\int }_{\mathsf{\Omega }}|u_k{|}^2\mathrm{d}x-\frac{1}{2h}{\int }_{\mathsf{\Omega }}{|u_{k-1}|}^2\mathrm{d}x,\end{array}$$

(11)

and

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}g{u}_k\mathrm{d}x\le 2{\parallel u_k\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}{\parallel g\parallel }_{L^{q^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)}\le \epsilon {\parallel u_k\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}^{\beta }+C\left(\epsilon \right){\parallel g\parallel }_{L^{q^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)}^{{\beta }^{\prime }}\hfill \\ \hfill & \le \epsilon {\int }_{\mathsf{\Omega }}|u_k{|}^{q\left(x\right)}\mathrm{d}x+C(\epsilon ,{\parallel g\parallel }_{L^{q^{\prime }\left(x\right)}\left(\mathsf{\Omega }\right)}),\hfill \end{array}$$

(12)

where $\frac{1}{\beta }+\frac{1}{{\beta }^{\prime }}=1$,

$$\begin{array}{c}\hfill \beta =\left\{\begin{array}{cc}{q}^{-},\hfill & \parallel u_k{\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}\ge 1;\hfill \\ q^{+},\hfill & \parallel u_k{\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}\le 1.\hfill \end{array}\right.\end{array}$$

Then, combining (10)–(12) and $f(x,s)s\le 0$, we have

$$\begin{array}{c}\hfill \frac{1}{2h}{\int }_{\mathsf{\Omega }}|u_k{|}^2\mathrm{d}x+{\int }_{\mathsf{\Omega }}|\nabla u_k{|}^{p\left(x\right)}\mathrm{d}x+\frac{\lambda }{2}{\int }_{\mathsf{\Omega }}|u_k{|}^{q\left(x\right)}\mathrm{d}x\le \frac{1}{2h}{\int }_{\mathsf{\Omega }}{|u_{k-1}|}^2\mathrm{d}x+C\left(\epsilon \right).\end{array}$$

(13)

Integrating above inequality with respect to time on $\left(\right(k-1)h,kh]$, one has

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}|u_k{|}^2\mathrm{d}x+2{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}|\nabla u_k{|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t+\lambda {\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}{|u_k|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{\mathsf{\Omega }}{|u_{k-1}|}^2\mathrm{d}x+2hC\left(\lambda \right).\hfill \end{array}$$

(14)

Summing up all the equalities of (14) for $k=1,2,\dots ,n$, we have

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}|u_n{|}^2\mathrm{d}x+2\sum _{i=1}^n{\int }_{(i-1)h}^{ih}{\int }_{\mathsf{\Omega }}|\nabla u_i{|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t+\lambda \sum _{i=1}^n{\int }_{(i-1)h}^{ih}{\int }_{\mathsf{\Omega }}{|u_i|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{\mathsf{\Omega }}{|u_0|}^2\mathrm{d}x+CT.\hfill \end{array}$$

Now, for every $h=\frac{T}{n}$, we define

$$u_h(x,t)=\left\{\begin{array}{cc}{u}_0\left(x\right),\hfill & t=0,\hfill \\ u_1\left(x\right),\hfill & 0<t\le h,\hfill \\ \dots \hfill \\ u_j\left(x\right),\hfill & (j-1)h<t\le jh,\hfill \\ \dots \hfill \\ u_n\left(x\right),\hfill & (n-1)h<t\le nh=T.\hfill \end{array}\right.$$

(15)

Then we have

$$\begin{array}{cc}\hfill & {\int }_{\mathsf{\Omega }}|u_h{(x,t)|}^2\mathrm{d}x+2{\int }_0^T{\int }_{\mathsf{\Omega }}|\nabla u_h{(x,t)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t+\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u_h(x,t)|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{\mathsf{\Omega }}{|u_0\left(x\right)|}^2\mathrm{d}x+CT.\hfill \end{array}$$

It implies that

$$\begin{array}{cc}\hfill {\int }_0^{T}min\{\parallel \nabla u_h{\parallel }_{L^{p\left(x\right)}\left(\mathsf{\Omega }\right)}^{p^{-}},\parallel \nabla u_h{\parallel }_{L^{p\left(x\right)}\left(\mathsf{\Omega }\right)}^{p^{+}}\}\mathrm{d}t& \le {\int }_0^T{\int }_{\mathsf{\Omega }}{|\nabla u_h(x,t)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\le CT,\hfill \\ \hfill {\int }_0^{T}min\{\parallel u_h{\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}^{q^{-}},\parallel u_h{\parallel }_{L^{q\left(x\right)}\left(\mathsf{\Omega }\right)}^{q^{+}}\}\mathrm{d}t& \le {\int }_0^T{\int }_{\mathsf{\Omega }}{|u_h(x,t)|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}t\le CT.\hfill \end{array}$$

Thus, we obtain

$$\begin{array}{c}\hfill \parallel u_h{\parallel }_{L^{\infty }(0,T;L^2\left(\mathsf{\Omega }\right))}+\parallel u_h{\parallel }_{L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))}+\parallel u_h{\parallel }_{L^{q(·)}\left({\mathsf{\Omega }}_T\right)}+{\parallel \nabla u_h\parallel }_{L^{p(·)}\left({\mathsf{\Omega }}_T\right)}\le CT,\end{array}$$

(16)

and we may choose a subsequence (we also denote it by the original sequence for simplicity) such that

$$\begin{array}{c}\hfill u_h\rightharpoonup u\mathrm{weakly} * \mathrm{in}{L}^{\infty }(0,T;L^2\left(\mathsf{\Omega }\right));\end{array}$$

$$\begin{array}{c}\hfill u_h\rightharpoonup u\mathrm{weakly} \mathrm{in}{L}^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))\cap L^{q(·)}\left({\mathsf{\Omega }}_T\right);\end{array}$$

$$\begin{array}{c}\hfill |u_h{|}^{q\left(x\right)-2}{u}_h\rightharpoonup \eta \mathrm{weakly} \mathrm{in}{L}^{q^{\prime }(·)}\left({\mathsf{\Omega }}_T\right);\end{array}$$

(17)

$$\begin{array}{c}\hfill |\nabla u_h{|}^{p\left(x\right)-2}\nabla u_h\rightharpoonup \xi \mathrm{weakly} \mathrm{in}{\left(L^{p^{\prime }(·)}\left({\mathsf{\Omega }}_T\right)\right)}^N,\end{array}$$

(18)

which follows that

$$\begin{array}{c}\hfill {\parallel u\parallel }_{L^{\infty }(0,T;L^2\left(\mathsf{\Omega }\right))}+{\parallel u\parallel }_{L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))}+{\parallel u\parallel }_{L^{q(·)}\left({\mathsf{\Omega }}_T\right)}\le CT.\end{array}$$

(19)

For each $\phi \in C^1\left({\overline{\mathsf{\Omega }}}_T\right)$ with $\phi (·,T)=0$, ${\phi (x,t)|}_{\partial \mathsf{\Omega }}=0$ and for every $k\in \{1,2,\dots ,n\}$, let $\phi (x,kh)$ be a test function in (5), we get

$$\begin{array}{cc}\hfill & \frac{1}{h}{\int }_{\mathsf{\Omega }}{u}_k\left(x\right)\phi (x,kh)\mathrm{d}x-\frac{1}{h}{\int }_{\mathsf{\Omega }}{u}_{k-1}\left(x\right)\phi (x,kh)\mathrm{d}x\hfill \\ \hfill & +{\int }_{\mathsf{\Omega }}|\nabla u_k{|}^{p\left(x\right)-2}\nabla u_k·\nabla \phi (x,kh)\mathrm{d}x+\lambda {\int }_{\mathsf{\Omega }}{|u_k|}^{q\left(x\right)-2}{u}_k\phi (x,kh)\mathrm{d}x\hfill \\ \hfill & ={\int }_{\mathsf{\Omega }}f(x,u_k)\phi (x,kh)\mathrm{d}x+{\int }_{\mathsf{\Omega }}g\left(x\right)\phi (x,kh)\mathrm{d}x.\hfill \end{array}$$

Summing up above equalities for $k=1,2,\dots ,n-1,$ recalling the definition of $u_h$ in (15) and $\phi (·,T)=\phi (·,nh)=0$, we have

$$\begin{array}{cc}\hfill & h\sum _{k=1}^{n-1}{\int }_{\mathsf{\Omega }}{u}_h(x,kh)\frac{\phi (x,kh)-\phi (x,(k+1\left)h\right)}{h}\mathrm{d}x-{\int }_{\mathsf{\Omega }}{u}_0\left(x\right)\phi (x,h)\mathrm{d}x\hfill \\ \hfill & +h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}(|\nabla u_h{|}^{p\left(x\right)-2}\nabla u_h)(x,kh)·\nabla \phi (x,kh)\mathrm{d}x\hfill \\ \hfill & +\lambda h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}(|u_h{|}^{q\left(x\right)-2}{u}_h)(x,kh)\phi (x,kh)\mathrm{d}x\hfill \\ \hfill & =h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}f(x,u_h(x,kh))\phi (x,kh)\mathrm{d}x+h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}g\left(x\right)\phi (x,kh)\mathrm{d}x.\hfill \end{array}$$

(20)

In what follows, we deal with every term in above equation in turn. In view of (18), we obtain

$$\begin{array}{cc}\hfill & h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}(|\nabla u_h{|}^{p\left(x\right)-2}\nabla u_h)(x,kh)·\nabla \phi (x,kh)\mathrm{d}x\hfill \\ \hfill & ={\int }_0^T{\int }_{\mathsf{\Omega }}(|\nabla u_h{|}^{p\left(x\right)-2}\nabla u_h)(x,\tau )·\nabla \phi (x,\tau )\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\sum _{k=1}^n{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}(|\nabla u_h{|}^{p\left(x\right)-2}\nabla u_h)(x,\tau )·(\nabla \phi (x,kh)-\nabla \phi (x,\tau ))\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ⟶{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·\nabla \phi (x,\tau )\mathrm{d}x\mathrm{d}\tau ,\mathrm{as}h⟶0.\hfill \end{array}$$

Since ${\parallel u\parallel }_{L^{p\left(x\right)}}\le C{\parallel \nabla u\parallel }_{L^{p\left(x\right)}}$, $L^{p^{-}}(0,T;W_0^{1,p\left(x\right)}\left(\mathsf{\Omega }\right))\hookrightarrow \hookrightarrow L^{p^{-}}(0,T;L^{\gamma \left(x\right)}\left(\mathsf{\Omega }\right))\hookrightarrow L^{s\left(x\right)}\left({\mathsf{\Omega }}_T\right)$, where $1\le \gamma \left(x\right)<p^{*}\left(x\right)=\frac{Np\left(x\right)}{N-p\left(x\right)}$, $s\left(x\right)\le \min \{p^{-},\gamma \left(x\right)\}$, then we obtain

$$\begin{array}{c}\hfill u_h⟶u\mathrm{strongly} \mathrm{in}{L}^{s\left(x\right)}\left({\mathsf{\Omega }}_T\right),\end{array}$$

and then

$$\begin{array}{cc}\hfill u_h& ⟶u \mathrm{a}.\mathrm{e}.\mathrm{in} {\mathsf{\Omega }}_T,\hfill \\ \hfill |u_h{|}^{q\left(x\right)-2}{u}_h& ⟶{|u|}^{q\left(x\right)-2}u \mathrm{a}.\mathrm{e}.\mathrm{in} {\mathsf{\Omega }}_T.\hfill \end{array}$$

(21)

From (17), we have

$$\begin{array}{c}\hfill \parallel |u_h{|}^{q\left(x\right)-2}{u}_h{\parallel }_{L^{q^{\prime }\left(x\right)}\left({\mathsf{\Omega }}_T\right)}\le \parallel u_h{\parallel }_{L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right)}^{q^{+}-1}+{\parallel u_h\parallel }_{L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right)}^{q^{-}-1}\le C.\end{array}$$

(22)

According to the well-known results of Lemma 1.3 in [35], we conclude that $\eta ={|u|}^{q\left(x\right)-2}u$. Hence

$$\begin{array}{cc}\hfill & h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}(|u_h{|}^{q\left(x\right)-2}{u}_h)(x,kh)\phi (x,kh)\mathrm{d}x={\int }_0^T{\int }_{\mathsf{\Omega }}(|u_h{|}^{q\left(x\right)-2}{u}_h)(x,\tau )\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\sum _{k=1}^n{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}(|u_h{|}^{q\left(x\right)-2}{u}_h)(x,\tau )(\phi (x,kh)-\phi (x,\tau ))\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ⟶{\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)-2}u\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau ,\mathrm{as}h⟶0.\hfill \end{array}$$

For the term of $f(x,u)$, we first deduce from (3) that

$$\begin{array}{c}\hfill |f(x,u_h)|\le l|u_h|,\end{array}$$

(23)

where $l=max\{l_1,l_2\}$. Since for a.e. $x\in {\mathbb{R}}^N$, the map $u⟶f(x,u)$ is continuous, then combining (16), (21) and (23), it implies that

$$\begin{array}{c} f(x,u_h)⟶f(x,u)\mathrm{a}.\mathrm{e}.\mathrm{in} {\mathsf{\Omega }}_T,\hfill \\ \parallel f(x,u_h){\parallel }_{L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right)}\le C_1{\parallel u_h\parallel }_{L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right)}\le C.\hfill \end{array}$$

By using Lemma 1.3 of [35] again, we obtain

$$\begin{array}{c}\hfill f(x,u_h)\rightharpoonup f(x,u)\mathrm{weakly} \mathrm{in} L^{q\left(x\right)}\left({\mathsf{\Omega }}_T\right),\end{array}$$

it follows that

$$\begin{array}{cc}\hfill & h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}f(x,u_h(x,kh))\phi (x,kh)\mathrm{d}x={\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u_h(x,\tau ))\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\sum _{k=1}^n{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}f(x,u_h(x,\tau ))(\phi (x,kh)-\phi (x,\tau ))\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ⟶{\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u)\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau ,\mathrm{as}h⟶0.\hfill \end{array}$$

Moreover, it is obvious that

$$\begin{array}{cc}\hfill & h\sum _{k=1}^n{\int }_{\mathsf{\Omega }}g\left(x\right)\phi (x,kh)\mathrm{d}x={\int }_0^T{\int }_{\mathsf{\Omega }}g\left(x\right)\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\sum _{k=1}^n{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}g\left(x\right)(\phi (x,kh)-\phi (x,\tau ))\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ⟶{\int }_0^T{\int }_{\mathsf{\Omega }}g\phi (x,\tau )\mathrm{d}x\mathrm{d}\tau ,\mathrm{as}h⟶0.\hfill \end{array}$$

Therefore, let h converge to 0 and use above mentioned convergence result, we obtain from (20) that

$$\begin{array}{cc}\hfill & -{\int }_0^T{\int }_{\mathsf{\Omega }}u\frac{\partial \phi }{\partial t}\mathrm{d}x\mathrm{d}\tau -{\int }_{\mathsf{\Omega }}{u}_0\left(x\right)\phi (x,0)\mathrm{d}x+{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·\nabla \phi \mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)-2}u\phi \mathrm{d}x\mathrm{d}\tau ={\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u)\phi \mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}g\phi \mathrm{d}x\mathrm{d}\tau .\hfill \end{array}$$

(24)

Now, we choose $\phi \in C_0^{\infty }\left({\mathsf{\Omega }}_T\right)$, then (24) reduces to

$$\begin{array}{cc}\hfill & -{\int }_0^T{\int }_{\mathsf{\Omega }}u\frac{\partial \phi }{\partial t}\mathrm{d}x\mathrm{d}\tau =-{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·\nabla \phi \mathrm{d}x\mathrm{d}\tau -\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)-2}u\phi \mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +{\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u)\phi \mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}g\phi \mathrm{d}x\mathrm{d}\tau ,\hfill \end{array}$$

(25)

and it immediately shows

$$\begin{array}{c}\hfill u_t\in L^{{\left(p^{-}\right)}^{\prime }}(0,T;{\mathcal{D}}^{-1,p^{\prime }(·)}\left(\mathsf{\Omega }\right))+L^{q^{\prime }(·)}\left({\mathsf{\Omega }}_T\right).\end{array}$$

(26)

Finally, by combining (19) and (26), we obtain

$$\begin{array}{c}\hfill u\in C([0,T];L^2\left(\mathsf{\Omega }\right)).\end{array}$$

Step 3. The function u is a weak solution. It is enough for us to prove $\xi ={|\nabla u|}^{p\left(x\right)-2}\nabla u$ a.e. in ${\mathsf{\Omega }}_T$. By taking u as a test function in (24), we have

$$\begin{array}{cc}\hfill & \frac{-\parallel u_0{\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2-{\parallel u\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2}{2}+{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·\nabla u\mathrm{d}x\mathrm{d}\tau +\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ={\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u)u\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}gu\mathrm{d}x\mathrm{d}\tau .\hfill \end{array}$$

(27)

Denote $Au={|\nabla u|}^{p\left(x\right)-2}\nabla u$ and recall the following elementary inequality

$$\begin{array}{c}\hfill {(|\xi |}^{p-2}\xi -{|\eta |}^{p-2}\eta )(\xi -\eta )\ge 0,\mathrm{for} \mathrm{all}\xi ,\eta \in {\mathbb{R}}^N,\end{array}$$

(28)

we see that

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}(A{u}_k-Av\left(\tau \right))·(\nabla u_k-\nabla v\left(\tau \right))\mathrm{d}x\ge 0,\end{array}$$

(29)

for each $k=1,2,\dots ,n$, and every $v\in L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))$. From (10), we obtain

$$\begin{array}{cc}\hfill & \frac{1}{h}{\int }_{\mathsf{\Omega }}(u_k-u_{k-1})u_k\mathrm{d}x+{\int }_{\mathsf{\Omega }}A{u}_k·\nabla v\left(\tau \right)\mathrm{d}x+{\int }_{\mathsf{\Omega }}Av\left(\tau \right)·(\nabla u_k-\nabla v\left(\tau \right))\mathrm{d}x\hfill \\ \hfill & +\lambda {\int }_{\mathsf{\Omega }}{|u_k|}^{q\left(x\right)}\mathrm{d}x-{\int }_{\mathsf{\Omega }}f(x,u_k)u_k\mathrm{d}x-{\int }_{\mathsf{\Omega }}g{u}_k\mathrm{d}x\le 0.\hfill \end{array}$$

(30)

Applying the estimates

$$\begin{array}{c}\hfill {\int }_{\mathsf{\Omega }}(u_k-u_{k-1})u_k\mathrm{d}x={\int }_{\mathsf{\Omega }}{|u_k|}^2\mathrm{d}x-{\int }_{\mathsf{\Omega }}{u}_k{u}_{k-1}\mathrm{d}x\ge \frac{\parallel u_k{\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2-{\parallel u_{k-1}\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2}{2},\end{array}$$

and then integrating (30) with respect to time over $\left(\right(k-1)h,kh]$, we have

$$\begin{array}{cc}\hfill & \frac{\parallel u_k{\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2-{\parallel u_{k-1}\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2}{2}+{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}A{u}_k·\nabla v\left(\tau \right)\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}Av\left(\tau \right)·(\nabla u_k-\nabla v\left(\tau \right))\mathrm{d}x\mathrm{d}\tau +\lambda {\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}{|u_k|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & -{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}f(x,u_k)u_k\mathrm{d}x\mathrm{d}\tau -{\int }_{(k-1)h}^{kh}{\int }_{\mathsf{\Omega }}g{u}_k\mathrm{d}x\mathrm{d}\tau \le 0.\hfill \end{array}$$

Summing up the above inequalities for $k=1,2,\dots ,n$, we obtain

$$\begin{array}{cc}\hfill & \frac{\parallel u_h{\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2-{\parallel u_0\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2}{2}+{\int }_0^T{\int }_{\mathsf{\Omega }}A{u}_h·\nabla v\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}Av·(\nabla u_h-\nabla v)\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u_h|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau \le {\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u_h)u_h\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}g{u}_h\mathrm{d}x\mathrm{d}\tau .\hfill \end{array}$$

Passing to limits as $h⟶0$ and noting

$$\begin{array}{c}\hfill {\parallel u\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}\le \underset{h⟶0}{lim\; inf}{\parallel u_h\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)},\\ \hfill {\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau \le \underset{h⟶0}{lim\; inf}{\int }_0^T{\int }_{\mathsf{\Omega }}{|u_h|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau ,\end{array}$$

we obtain that

$$\begin{array}{cc}\hfill & \frac{{\parallel u\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2-{\parallel u_0\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2}{2}+{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·\nabla v\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}Av·(\nabla u-\nabla v)\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{\mathsf{\Omega }}{|u|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}\tau \le {\int }_0^T{\int }_{\mathsf{\Omega }}f(x,u)u\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}gu\mathrm{d}x\mathrm{d}\tau .\hfill \end{array}$$

(31)

Let’s substitute (27) into (31), we have

$$\begin{array}{c}\hfill {\parallel u\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2+{\int }_0^T{\int }_{\mathsf{\Omega }}\xi ·(\nabla v-\nabla u)\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{\mathsf{\Omega }}Av·(\nabla u-\nabla v)\mathrm{d}x\mathrm{d}\tau \le 0,\end{array}$$

and further that

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{\mathsf{\Omega }}(\xi -Av)·(\nabla u-\nabla v)\mathrm{d}x\mathrm{d}\tau \ge {\parallel u\left(T\right)\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2\ge 0.\end{array}$$

By choosing $v=u-\zeta w$ for any $\zeta >0,w\in L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))$ in the above inequality, one has

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{\mathsf{\Omega }}(\xi -A(u-\zeta w))·\nabla w\mathrm{d}x\mathrm{d}\tau \ge 0.\end{array}$$

Passing to limits as $\zeta ⟶0$ and using Lebesgue’s dominated convergence theorem, we obtain

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{\mathsf{\Omega }}(\xi -Au)·\nabla \psi \mathrm{d}x\mathrm{d}\tau \ge 0,\end{array}$$

for every $\psi \in L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(\mathsf{\Omega }\right))$. Thus, we conclude that $\xi =Au$ a.e. in ${\mathsf{\Omega }}_T$. Therefore, we finish the proof of the existence of weak solutions.

Step 4. The uniqueness of the weak solution. Suppose there exist two weak solutions u and v to problem (1) with (4), then $u-v$ satisfies the following problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{(u-v)}_t-{\mathrm{div}(|\nabla u|}^{p\left(x\right)-2}\nabla u-{|\nabla v|}^{p\left(x\right)-2}\nabla v)\hfill \\ +\lambda (|u{|}^{q\left(x\right)-2}u-|v{|}^{q\left(x\right)-2}v)=f(x,u)-f(x,v),\hfill & \mathrm{in}\mathsf{\Omega }\times {\mathbb{R}}^{+},\hfill \\ u-v=0,\hfill & \mathrm{on}\partial \mathsf{\Omega }\times {\mathbb{R}}^{+},\hfill \\ u(x,0)-v(x,0)=0,\hfill & \mathrm{in}\mathsf{\Omega }.\hfill \end{array}\right.\end{array}$$

Choosing $u-v$ as a test function in the above problem, we obtain from condition (3) that

$$\begin{array}{cc}\hfill & \frac{1}{2}\frac{d}{dt}{\parallel u-v\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2+{\int }_{\mathsf{\Omega }}{(|\nabla u|}^{p\left(x\right)-2}\nabla u-{|\nabla v|}^{p\left(x\right)-2}\nabla v)·\nabla (u-v)\mathrm{d}x\hfill \\ \hfill & +\lambda {\int }_{\mathsf{\Omega }}{(|u|}^{q\left(x\right)-2}u-{|v|}^{q\left(x\right)-2}v)(u-v)\mathrm{d}x\hfill \\ \hfill & ={\int }_{\mathsf{\Omega }}(f(x,u)-f(x,v))(u-v)\mathrm{d}x\hfill \\ \hfill & \le l_2{\int }_{\mathsf{\Omega }}{|u-v|}^2\mathrm{d}x\hfill \\ \hfill & =l_2{\parallel u-v\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2.\hfill \end{array}$$

Discarding the nonnegative term and using Gronwall inequality, we have

$$\begin{array}{c}\hfill {\parallel u-v\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2\left(t\right)\le e^{2l_{2}t}{\parallel u_0-v_0\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2=0\end{array}$$

for every $t\in (0,T)$, which implies $u=v$ a.e. in ${\mathsf{\Omega }}_T$. Therefore, we obtain the uniqueness of weak solutions. □

Proposition 2.

If the conditions (2) and (3) are satisfied, then for any $u_0\in L^2\left({\mathbb{R}}^N\right)$ and a smooth function g, there exists a unique weak solution u to (1) for any fixed $T>0$.

Proof. 

We use the standard domain expansion technique to get the desired result. The proof is along with the idea of Babin–Vishik (see [36,37]). For the convenience of the readers, we give the detail below. Let $u_R,R\to +\infty$ be the sequence of solutions of the (1) on the $B_R=\{x\in {\mathbb{R}}^N,|x|\le R\}$ with the initial data $u_{0,R}=\psi (|x|)u_0\left(x\right)$, where $\psi \left(\xi \right)$ is a smooth function and satisfies

$$\psi \left(\xi \right)=1\mathrm{as}|\xi |\le R-1;\psi \left(\xi \right)=0\mathrm{as}|\xi |\ge R;0\le \psi \left(\xi \right)\le 1\mathrm{and}|{\psi }^{\left(k\right)}\left(\xi \right)|\le C_k.$$

Note that

$$\begin{array}{c}\hfill \parallel u_0-u_{0,R}{\parallel }_{L^2\left({\mathbb{R}}^N\right)}^2={\int }_{{\mathbb{R}}^N}{(1-\psi (|x|\left)\right)}^2|u_0{|}^2\mathrm{d}x\le {\int }_{\{|x|>R-1\}}{|u_0|}^2\mathrm{d}x\to 0\mathrm{as}R\to \infty .\end{array}$$

(32)

We use the method of difference variation to obtain the weak solution, where the original equation is replaced by $\frac{u_k-u_{k-1}}{h}$ with $u_t$ and u with $u_k$, which satisfies (10)–(14). For $u_{0,R}\in L^2\left(B_R\right)$, we see the estimates (10)–(14) are true uniformly with respect to R. Define these functions as zero for $|x|\ge R$ and denote functions so extended multiplied by $\psi (|x|)$, by $u_R$. From (10)–(14), it follows that there exists a subsequence $u_j=u_{R_j}$ of $u_R$, which is weakly convergent in $L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p(·)}\left(B_R\right))\cap L^{q(·)}(0,T;L^{q(·)}\left(B_R\right))$ and ∗-weakly convergent in $L^{\infty }(0,T;L^2\left(B_R\right))$.

Denote the limit of $u_j$ by $u_{\infty }=u_{\infty }(x,t)$. For every fixed $R=R_k$, denote by $u_{kj}=L_k{u}_j$ the restriction on $B_{R_k}\times [0,T]$ of $u_j$, $L_k$ is the operator of restriction. For $L_k{u}_j$, the estimates (10)–(14) are valid, where the norms are on the domain $B_{R_k}$. From estimates (10)–(14), it follows that there exists a subsequence $u_{j1}$, such that $L_k{u}_{j1}$ is weakly convergent in the above mentioned spaces. Denote the limit of $L_k{u}_{j1}$ by $u_{k\infty }$. Since

$${\int }_0^T{\int }_{B_R}{L}_k{u}_{j1}v(x,t)\mathrm{d}x\mathrm{d}t={\int }_0^T{\int }_{{\mathbb{R}}^N}{u}_{j1}v(x,t)\mathrm{d}x\mathrm{d}t,$$

for any $v\in C_0^{\infty }(B_R\times [0,T]),$ we have $u_{k\infty }=L_k{u}_{\infty }$. As in the case of a bounded domain, we prove that $L_k{u}_{\infty }$ is a solution of (1) in $B_R\times [0,T]$ in the class of distributions, so that we have

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_{{\mathbb{R}}^N}\frac{\partial u_{\infty }}{\partial t}v(x,t)\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla u_{\infty }|}^{p\left(x\right)-2}\nabla u_{\infty }·\nabla v(x,t)\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{|u_{\infty }|}^{q\left(x\right)-2}{u}_{\infty }v(x,t)\mathrm{d}x\mathrm{d}\tau \hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u_{\infty })v(x,t)\mathrm{d}x\mathrm{d}\tau +{\int }_0^T{\int }_{{\mathbb{R}}^N}gv(x,t)\mathrm{d}x\mathrm{d}\tau ,\hfill \end{array}$$

(33)

for any $v\in C_0^{\infty }(B_{R-1}\times [0,T])$.

Since R is arbitrary, we deduce that (33) is fulfilled for arbitrary $v\in C_0^{\infty }([0,T]\times {\mathbb{R}}^N)$, and $u_{\infty }$ is a solution of (1). From (32), we see that the solution satisfies (1). In estimates (10)–(14) we pass to the limit as $R\to \infty$ and we find that these estimates are true for $u=u_{\infty }$.

The proof of this proposition is finished. □

4. Proof of the Entropy Solution

In this section, we consider the following related approximate problem

$$\left\{\begin{array}{cc}{u}_t^n-\mathrm{div}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n)+\lambda |u^n{|}^{q\left(x\right)-2}{u}^n=f(x,u^n)+g^n,\hfill & \mathrm{in}{\mathbb{R}}^N\times {\mathbb{R}}^{+},\hfill \\ u^n(x,0)=u_0^n,\hfill & \mathrm{in}{\mathbb{R}}^N,\hfill \end{array}\right.$$

(34)

where ${\left\{g^n\right\}}_{n\in N},{\left\{u_0^n\right\}}_{n\in N}$ are smooth approximations of g and $u_0$, respectively, satisfying

$$\begin{array}{c}\hfill \parallel u_0^n{\parallel }_{L^1\left({\mathbb{R}}^N\right)}\le \parallel u_0{\parallel }_{L^1\left({\mathbb{R}}^N\right)},\parallel g^n{\parallel }_{L^1\left({\mathbb{R}}^N\right)}\le {\parallel g\parallel }_{L^1\left({\mathbb{R}}^N\right)}.\end{array}$$

By Propositions 1 and 2, we obtain the approximate solutions $u^n$ when the initial value $u_0^n\in L^2\left({\mathbb{R}}^N\right)$, $g^n$ is a smooth function. Due to the initial value $u_0^n$ converges to $u_0$, the $g^n$ converges to g in $L^1$-norm, we verify u is an entropy solution of (1). Next, we give the definition of entropy solutions firstly.

Definition 2.

A function $u(x,t)$ is called an entropy solution to problem (1), if for any $T>0$,

(1) 

$u\in C([0,T];L^1\left({\mathbb{R}}^N\right))$,

(2) 

$f(x,u)\in L^1\left(Q_T\right)$,

(3) 

$T_k\left(u\right)\in L^{p^{-}}(0,T;W_0^{1,p(·)}\left({\mathbb{R}}^N\right))\cap L^{q(·)}\left(Q_T\right)$,

(4) 

$\nabla T_k\left(u\right)\in {\left(L^{p(·)}\left(Q_T\right)\right)}^N$,

and

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u-\phi )\left(T\right)dx-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0-\phi \left(0\right))dx+{\int }_0^T\langle {\phi }_t,T_k(u-\phi )\rangle dt\hfill \\ \hfill & +{\int }_{Q_T}{|\nabla u|}^{p\left(x\right)-2}\nabla u·\nabla T_k(u-\phi )dxdt+\lambda {\int }_{Q_T}{|u|}^{q\left(x\right)-2}u{T}_k(u-\phi )dxdt\hfill \\ \hfill & ={\int }_{Q_T}f(x,u)T_k(u-\phi )dxdt+{\int }_{Q_T}g{T}_k(u-\phi )dxdt,\hfill \end{array}$$

holds for all $k>0$ and $\phi \in C_0^1\left(Q_T\right)$.

Proof of Theorem 1.

We divide the proof into several steps.

Step 1.  $Theconvergenceof\left\{u^n\right\}inC([0,T];L^1\left({\mathbb{R}}^N\right))$

Let n and m be two integers, then by (34), we have

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{(u^n-u^m)}_t\phi \mathrm{d}x+{\int }_{{\mathbb{R}}^N}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n-|\nabla u^m{|}^{p\left(x\right)-2}\nabla u^m)·\nabla \phi \mathrm{d}x\hfill \\ \hfill & +\lambda {\int }_{{\mathbb{R}}^N}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)\phi \mathrm{d}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^N}(f(x,u^n)-f(x,u^m))\phi \mathrm{d}x+{\int }_{{\mathbb{R}}^N}(g^n-g^m)\phi \mathrm{d}x,\hfill \end{array}$$

(35)

for all $\phi \in L^{p^{-}}(0,T;{\mathcal{D}}_0^{1,p\left(x\right)}\left(\mathsf{\Omega }\right))\cap L^{\infty }\left(Q_T\right)$ with $\nabla \phi \in {\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N$. Choosing $\phi =\frac{1}{k}{T}_k(u^n-u^m)$ in (35), we get

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{(u^n-u^m)}_t\frac{1}{k}{T}_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & +\frac{1}{k}{\int }_{{\mathbb{R}}^N}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n-|\nabla u^m{|}^{p\left(x\right)-2}\nabla u^m)·\nabla T_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & +\frac{\lambda }{k}{\int }_{{\mathbb{R}}^N}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)T_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & =\frac{1}{k}{\int }_{{\mathbb{R}}^N}(f(x,u^n)-f(x,u^m))T_k(u^n-u^m)\mathrm{d}x+\frac{1}{k}{\int }_{{\mathbb{R}}^N}(g^n-g^m)T_k(u^n-u^m)\mathrm{d}x.\hfill \end{array}$$

(36)

By elementary inequality (28), the second term on the left-hand side of (36) estimates

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n-|\nabla u^m{|}^{p\left(x\right)-2}\nabla u^m)·\nabla T_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & ={\int }_{\{x\in {\mathbb{R}}^N,|u^n-u^m|<k\}}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n-|\nabla u^m{|}^{p\left(x\right)-2}\nabla u^m)·(\nabla u^n-\nabla u^m)\mathrm{d}x\ge 0.\hfill \end{array}$$

(37)

Due to $T_k(u^n-u^m)=sgn(u^n-u^m)|T_k(u^n-u^m)|$, the third term on the left-hand side of (36) reads

$$\begin{array}{cc}\hfill & \phantom{==}{\int }_{{\mathbb{R}}^N}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)T_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^N}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)sgn(u^n-u^m)|T_k(u^n-u^m)|\mathrm{d}x\ge 0.\hfill \end{array}$$

(38)

By virtue of condition (3), the terms on the right-hand side of (36) shows

$$\begin{array}{cc}\hfill & \phantom{==}\frac{1}{k}{\int }_{{\mathbb{R}}^N}(f(x,u^n)-f(x,u^m))T_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & =\frac{1}{k}{\int }_{{\mathbb{R}}^N}(f(x,u^n)-f(x,u^m))sgn(u^n-u^m)|T_k(u^n-u^m)|\mathrm{d}x\hfill \\ \hfill & \le \frac{l_2}{k}{\int }_{{\mathbb{R}}^N}|u^n-u^m||T_k(u^n-u^m)|\mathrm{d}x,\hfill \end{array}$$

(39)

$$\frac{1}{k}{\int }_{{\mathbb{R}}^N}(g^n-g^m)T_k(u^n-u^m)\mathrm{d}x\le {\int }_{{\mathbb{R}}^N}|g^n-g^m|\mathrm{d}x=\parallel g^n-g^m{\parallel }_{L^1\left({\mathbb{R}}^N\right)}.$$

(40)

Inserting (37)–(40) into (36) and then after discarding, we obtain

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{(u^n-u^m)}_t\frac{1}{k}{T}_k(u^n-u^m)\mathrm{d}x\hfill \\ \hfill & \le \frac{l_2}{k}{\int }_{{\mathbb{R}}^N}|u^n-u^m||T_k(u^n-u^m)|\mathrm{d}x+\parallel g^n-g^m{\parallel }_{L^1\left({\mathbb{R}}^N\right)},\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill & \frac{\mathrm{d}}{\mathrm{d}t}({\int }_{\{|u^n-u^m|<k\}}\frac{1}{2k}|u^n-u^m{|}^2+{\int }_{\{|u^n-u^m|\ge k\}}|u^n-u^m|\mathrm{d}x)\hfill \\ \hfill & \le 2l_2{\int }_{\{|u^n-u^m|<k\}}\frac{1}{2k}|u^n-u^m{|}^2+l_2{\int }_{\{|u^n-u^m|\ge k\}}|u^n-u^m|\mathrm{d}x+\parallel g^n-g^m{\parallel }_{L^1\left({\mathbb{R}}^N\right)}\hfill \\ \hfill & \le 2l_2({\int }_{\{|u^n-u^m|<k\}}\frac{1}{2k}|u^n-u^m{|}^2+{\int }_{\{|u^n-u^m|\ge k\}}|u^n-u^m|\mathrm{d}x)+\parallel g^n-g^m{\parallel }_{L^1\left({\mathbb{R}}^N\right)}.\hfill \end{array}$$

By Gronwall’s inequality, we obtain

$$\begin{array}{cc}\hfill & {\int }_{\{|u^n-u^m|<k\}}\frac{1}{2k}|u^n-u^m{|}^2+{\int }_{\{|u^n-u^m|\ge k\}}|u^n-u^m|\mathrm{d}x\hfill \\ \hfill & \le e^{2l_{2}t}({\int }_{\{|u_0^n-u_0^m|<k\}}\frac{1}{2k}|u_0^n-u_0^m{|}^2+{\int }_{\{|u_0^n-u_0^m|\ge k\}}|u_0^n-u_0^m|\mathrm{d}x)\hfill \\ \hfill & +\frac{1}{2l_2}{e}^{2l_{2}t}{\parallel g^n-g^m\parallel }_{L^1\left({\mathbb{R}}^N\right)}\hfill \\ \hfill & \le e^{2l_{2}t}({\int }_{\{|u_0^n-u_0^m|<k\}}|u_0^n-u_0^m|+{\int }_{\{|u_0^n-u_0^m|\ge k\}}|u_0^n-u_0^m|\mathrm{d}x)\hfill \\ \hfill & +\frac{1}{2l_2}{e}^{2l_{2}t}{\parallel g^n-g^m\parallel }_{L^1\left({\mathbb{R}}^N\right)}\hfill \\ \hfill & =e^{2l_{2}t}\parallel u_0^n-u_0^m{\parallel }_{L^1\left({\mathbb{R}}^N\right)}+\frac{1}{2l_2}{e}^{2l_{2}t}{\parallel g^n-g^m\parallel }_{L^1\left({\mathbb{R}}^N\right)}:=a_{n,m}.\hfill \end{array}$$

Let k converge to 0, we then conclude that

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}|u^n-u^m|\mathrm{d}x\le a_{n,m}.\end{array}$$

Since $\left\{g^n\right\},\left\{u_0^n\right\}$ are convergent in $L^1$, we have $a_{n,m}⟶0$ for $n,m⟶+\infty$. Thus, $\left\{u^n\right\}$ is a Cauchy sequence in $C([0,T];L^1\left({\mathbb{R}}^N\right))$ and $u^n$ converges to u in $C([0,T];L^1\left({\mathbb{R}}^N\right))$. Then, we get the first item in the definition of entropy solution. Moreover, we find an a.e. convergent subsequence (still denoted by $\left\{u^n\right\}$ ) in $Q_T$ such that

$$\begin{array}{c}\hfill u^n⟶ua.e.\mathrm{in}{Q}_T,\end{array}$$

(41)

and

$$\begin{array}{c}\hfill T_k\left(u^n\right)⟶T_k\left(u\right)a.e.\mathrm{in}{Q}_T.\end{array}$$

(42)

Step 2.  $Uptoasubsequence,f(x,u^n)convergestof(x,u)in{L}^1\left(Q_T\right)foranyT>0.$

Since for a.e. $x\in {\mathbb{R}}^N$, the map $u⟼f(x,u)$ is continuous. Then, we have from (41) that

$$\begin{array}{c}\hfill f(x,u^n)⟶f(x,u)\mathrm{a}.\mathrm{e}.\mathrm{in}{Q}_T.\end{array}$$

(43)

From condition (3), $l=\max \{l_1,l_2\}$, and $u^n⟶u\mathrm{strongly} \mathrm{in} L^1\left(Q_T\right)$, we have

$$\begin{array}{c}\hfill {\int }_{Q_T}|f(x,u^n)-f(x,u)|\mathrm{d}x\mathrm{d}t\le l{\int }_{Q_T}|u^n-u|\mathrm{d}x\mathrm{d}t<\epsilon ,\end{array}$$

as $n⟶\infty$. Then, we obtain that

$$\begin{array}{c}\hfill f(x,u^n)⟶f(x,u)\mathrm{strongly} \mathrm{in} L^1\left(Q_T\right).\end{array}$$

(44)

Therefore, we obtain the second item in the definition of entropy solution.

Step 3.  $Theconvergenceof|u^n{|}^{q\left(x\right)-2}{u}^n⟶{|u|}^{q\left(x\right)-2}u\mathrm{in}{L}^1\left(Q_T\right).$

On one hand, by choosing $T_k\left(u^n\right)$ as a test function in (34), we have

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k\left(u^n\right)\left(T\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k\left(u_0^n\right)\mathrm{d}x+{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla T_k\left(u^n\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n{|}^{q\left(x\right)-1}|T_k\left(u^n\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u^n)T_k\left(u^n\right)\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}{g}^n{T}_k\left(u^n\right)\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(45)

Applying the fact ${\mathsf{\Phi }}_k\left(r\right)\ge 0,{\mathsf{\Phi }}_k\left(r\right)\le k|r|$ and $f(x,s)s\le 0$, we have

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}|\nabla T_k\left(u^n\right){|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\le {kT\parallel g\parallel }_{L^1\left({\mathbb{R}}^N\right)}+k{\parallel u_0\parallel }_{L^1\left({\mathbb{R}}^N\right)}\le Ck,\end{array}$$

(46)

$$\begin{array}{c}\hfill \lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n{|}^{q\left(x\right)-1}|T_k\left(u^n\right){|\mathrm{d}x\mathrm{d}t\le kT\parallel g\parallel }_{L^1\left({\mathbb{R}}^N\right)}+k{\parallel u_0\parallel }_{L^1\left({\mathbb{R}}^N\right)}\le Ck.\end{array}$$

(47)

By above inequalities, we have the boundedness

$$\begin{array}{cc}\hfill & {\int }_{Q_T}|u^n{|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t={\int }_{\{|u^n|\le k\}}|u^n{|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t+{\int }_{\{|u^n|\ge k\}}{|u^n|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \max \{k^{q^{+}-2},k^{q^{-}-2}\}{\int }_{\{|u^n|\le k\}}|u^n|\mathrm{d}x\mathrm{d}t+\frac{1}{k}{\int }_{\{|u^n|\ge k\}}|u^n{|}^{q\left(x\right)-1}|T_k\left(u^n\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le C.\hfill \end{array}$$

In view of (41), we conclude that

$$\begin{array}{c}\hfill u^n\rightharpoonup u\mathrm{weakly} \mathrm{in}{L}^{q\left(x\right)-1}\left(Q_T\right).\end{array}$$

(48)

On the other hand, by integrating (36) with respect to time, we get

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}\frac{1}{k}{\mathsf{\Phi }}_k(u^n-u^m)\left(T\right)\mathrm{d}x\hfill \\ \hfill & +\frac{1}{k}{\int }_{Q_T}(|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n-|\nabla u^m{|}^{p\left(x\right)-2}\nabla u^m)·\nabla T_k(u^n-u^m)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\frac{\lambda }{k}{\int }_{Q_T}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)T_k(u^n-u^m)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{Q_T}|f(x,u^n)-f(x,u^m)|\mathrm{d}x\mathrm{d}t+T{\int }_{{\mathbb{R}}^N}|g^n-g^m|\mathrm{d}x+{\int }_{{\mathbb{R}}^N}|u_0^n-u_0^m|\mathrm{d}x.\hfill \end{array}$$

Similar to before, we obtain

$$\begin{array}{cc}\hfill & E\left(n\right)=\frac{1}{k}{\int }_{Q_T}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)T_k(u^n-u^m)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \frac{1}{\lambda }({\int }_{Q_T}|f(x,u^n)-f(x,u^m)|\mathrm{d}x\mathrm{d}t+T{\int }_{{\mathbb{R}}^N}|g^n-g^m|\mathrm{d}x+{\int }_{{\mathbb{R}}^N}|u_0^n-u_0^m|\mathrm{d}x)\hfill \\ \hfill & :=b_{n,m}.\hfill \end{array}$$

Since $f(x,u^n)⟶f(x,u)\mathrm{in} L^1\left(Q_T\right),g^n⟶g\mathrm{in} L^1\left({\mathbb{R}}^N\right),u_0^n⟶u_0\mathrm{in} L^1\left({\mathbb{R}}^N\right)$, we have $b_{n,m}⟶0,\mathrm{as}n,m⟶+\infty$, i.e., $E\left(n\right)⟶0,\mathrm{as}n,m⟶+\infty$.

Moreover, we also note that

$$\begin{array}{cc}\hfill & {\int }_{Q_T}{|u^n-u^m|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{\{|u^n-u^m|\le k\}}|u^n-u^m{|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t+{\int }_{\{|u^n-u^m|\ge k\}}{|u^n-u^m|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \max \{k^{q^{+}-2},k^{q^{-}-2}\}{\int }_{Q_T}|u^n-u^m|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\frac{1}{k}{\int }_{\{|u^n-u^m|\ge k\}}|u^n-u^m{|}^{q\left(x\right)-1}|T_k(u^n-u^m)|\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Next, we deal with the second term on the right side of the above inequality. We first recall the well-known inequalities:

$$\begin{array}{c}\hfill {c\left(p\right)|a-b|}^p\le {(|a|}^{p-2}a-{|b|}^{p-2}b)(a-b),\mathrm{for} \mathrm{any} a,b\in {\mathbb{R}}^N,p\ge 2,\end{array}$$

(49)

and for every $\epsilon \in (0,1],1<p<2$,

$$\begin{array}{c}\hfill {|a-b|}^p\le c\left(p\right){\epsilon }^{(p-2)/p}{(|a|}^{p-2}a-{|b|}^{p-2}{b)(a-b)+\epsilon |b|}^p,\mathrm{for} \mathrm{any} a,b\in {\mathbb{R}}^N,\end{array}$$

(50)

where $c\left(p\right)=\frac{2^{1-p}}{p-1}$ when $p\ge 2$ and $c\left(p\right)=\frac{3^{2-p}}{p-1}$ when $1<p<2$. Since $q\left(x\right)\ge 2$, we have

$$\begin{array}{cc}\hfill & \frac{1}{k}{\int }_{\{|u^n-u^m|\ge k:q\left(x\right)\ge 2\}}|u^n-u^m{|}^{q\left(x\right)-1}|T_k(u^n-u^m)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \frac{q^{+}-1}{2^{1-q^{+}}}{\int }_E\frac{1}{k}(|u^n{|}^{q\left(x\right)-2}{u}^n-|u^m{|}^{q\left(x\right)-2}{u}^m)\mathrm{sgn}(u^n-u^m)|T_k(u^n-u^m)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \left(2^{q^{+}-1}\right)(q^{+}-1)E\left(n\right)⟶0,\mathrm{as}n,m⟶\infty ,\hfill \end{array}$$

(51)

where $E=\{|u^n-u^m|\ge k:q\left(x\right)\ge 2\}$. Since $\left\{u^n\right\}$ is a Cauchy sequence in $C([0,T];L^1\left({\mathbb{R}}^N\right))$ and (51) hold true, we then conclude that

$$\begin{array}{c}\hfill {\int }_{Q_T}{|u^n-u^m|}^{q\left(x\right)-1}\mathrm{d}x\mathrm{d}t⟶0,\mathrm{as}n,m⟶+\infty .\end{array}$$

Therefore, $\left\{u^n\right\}$ is a Cauchy sequence in $L^{q\left(x\right)-1}\left(Q_T\right)$. In view of (48), we obtain

$$\begin{array}{c}\hfill |u^n{|}^{q\left(x\right)-2}{u}^n⟶{|u|}^{q\left(x\right)-2}u\mathrm{strongly} \mathrm{in} L^1\left(Q_T\right).\end{array}$$

(52)

Step 4.  $Theconvergenceof\nabla T_k\left(u^n\right)⟶\nabla T_k\left(u\right)\mathrm{strongly} \mathrm{in}{\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N.$

We deduce from (46), (47) and $|T_k\left(u^n\right)|\le |u^n|$ that

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}|T_k\left(u^n\right){|}^{q\left(x\right)}\mathrm{d}x\mathrm{d}t\le {\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n{|}^{q\left(x\right)-1}|T_k\left(u^n\right)|\mathrm{d}x\mathrm{d}t\le Ck,\end{array}$$

(53)

$$\begin{array}{cc}\hfill & {\int }_0^{T}min\{\parallel \nabla T_k\left(u^n\right){\parallel }_{L^{p\left(x\right)}\left({\mathbb{R}}^N\right)}^{p^{+}},\parallel \nabla T_k\left(u^n\right){\parallel }_{L^{p\left(x\right)}\left({\mathbb{R}}^N\right)}^{p^{-}}\}\mathrm{d}t\hfill \\ \hfill & \le {\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla T_k\left(u^n\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\le Ck,\hfill \end{array}$$

(54)

and

$$\begin{array}{cc}\hfill & {\int }_0^{T}min\{\parallel T_k\left(u^n\right){\parallel }_{L^{p\left(x\right)}\left({\mathbb{R}}^N\right)}^{p^{+}},\parallel T_k\left(u^n\right){\parallel }_{L^{p\left(x\right)}\left({\mathbb{R}}^N\right)}^{p^{-}}\}\mathrm{d}t\le {\int }_0^T{\int }_{{\mathbb{R}}^N}{|T_k\left(u^n\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \max \{k^{p^{+}-1},k^{p^{-}-1}\}{\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n|\mathrm{d}x\mathrm{d}t\le C\left(k\right).\hfill \end{array}$$

(55)

Then, we conclude that $T_k\left(u^n\right)$ is bounded in $L^{p^{-}}(0,T;W_0^{1,p\left(x\right)}\left({\mathbb{R}}^N\right))\cap L^{q\left(x\right)}\left(Q_T\right)$. From (42), the third item of entropy solution is obviously.

Next, we prove the limit of $u^n$ meets the fourth item of entropy solution.

Since $\left\{u^n\right\}\subset C([0,T];L^1\left({\mathbb{R}}^N\right))$ and $|T_k\left(u^n\right)|\le |u^n|$, $|T_k\left(u^n\right)|\le k$, we have

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_{{\mathbb{R}}^N}|T_k\left(u^n\right){|}^{p^{*}\left(x\right)}\mathrm{d}x\mathrm{d}t\le \max \{k^{p^{*+}-1},k^{p^{*-}-1}\}{\int }_0^T{\int }_{{\mathbb{R}}^N}|T_k\left(u^n\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \max \{k^{p^{*+}-1},k^{p^{*-}-1}\}{\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n|\mathrm{d}x\mathrm{d}t\le C\left(k\right),\hfill \end{array}$$

(56)

where $p^{*}\left(x\right)=\frac{Np\left(x\right)}{N-p\left(x\right)}$. By virtue of boundedness of $T_k\left(u^n\right)$ in $L^{p^{*}\left(x\right)}\left(Q_T\right)$ and $\nabla T_k\left(u^n\right)$ in ${\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N$, there is a subsequence ( still denoted by $\left\{u^n\right\}$ ) such that

$$\begin{array}{c}\hfill T_k\left(u^n\right)\rightharpoonup T_k\left(u\right)\mathrm{weakly} \mathrm{in}{L}^{p^{*}\left(x\right)}\left(Q_T\right),\\ \hfill \nabla T_k\left(u^n\right)\rightharpoonup \xi \mathrm{weakly} \mathrm{in}{\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N.\end{array}$$

Then, we get

$$\begin{array}{c}\hfill {\int }_{Q_T}\nabla T_k\left(u^n\right)·\phi \mathrm{d}x\mathrm{d}t⟶{\int }_{Q_T}\xi ·\phi \mathrm{d}x\mathrm{d}t,\end{array}$$

and

$$\begin{array}{cc}\hfill & {\int }_{Q_T}\nabla T_k\left(u^n\right)·\phi \mathrm{d}x\mathrm{d}t=-{\int }_{Q_T}{T}_k\left(u^n\right)·\nabla \phi \mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ⟶-{\int }_{Q_T}{T}_k\left(u\right)·\nabla \phi \mathrm{d}x\mathrm{d}t={\int }_{Q_T}\nabla T_k\left(u\right)·\phi \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Thus, $\xi =\nabla T_k\left(u\right)$ a.e. in ${\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N$, i.e.,

$$\begin{array}{c}\hfill \nabla T_k\left(u^n\right)\rightharpoonup \nabla T_k\left(u\right)\mathrm{weakly} \mathrm{in}{\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N,\end{array}$$

(57)

which means that we obtain the fourth item of entropy solution.

In order to get the result of this step, we refer to the regularization method which is given by Landes [38]. We introduce a time-regularization of function $T_k\left(u\right)$, which is defined

$$\begin{array}{c}\hfill {\left(T_k\left(u\right)\right)}_{\mu }(x,t)=\mu {\int }_{-\infty }^t{e}^{\mu (s-t)}{T}_k\left(u(x,s)\right)\mathrm{d}s,\end{array}$$

(58)

here, $T_k\left(u\right)$ is extended by 0 for $s<0$. It satisfies the following properties:

$$\begin{array}{c}\hfill {\left(T_k\left(u\right)\right)}_{\mu }\in L^{p^{-}}(0,T;W_0^{1,p\left(x\right)}\left({\mathbb{R}}^N\right))\cap L^{\infty }\left(Q_T\right),\nabla {\left(T_k\left(u\right)\right)}_{\mu }\in {\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N,\\ \hfill |{\left(T_k\left(u\right)\right)}_{\mu }(x,t)|\le k(1-e^{-\mu t})<k,\frac{\partial \left({\left(T_k\left(u\right)\right)}_{\mu }\right)}{\partial t}=\mu (T_k\left(u\right)-{\left(T_k\left(u\right)\right)}_{\mu }),\end{array}$$

and

$$\begin{array}{c}\hfill \nabla {\left(T_k\left(u\right)\right)}_{\mu }⟶\nabla T_k\left(u\right)\mathrm{strongly} \mathrm{in} {\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N.\end{array}$$

(59)

Let us take a sequence $\left\{{\psi }_j\right\} \mathrm{of} C_0^{\infty }\left({\mathbb{R}}^N\right)$ functions that strongly converge to $u_0\mathrm{in}{L}^1\left({\mathbb{R}}^N\right)$, and set

$$\begin{array}{c}\hfill {\eta }_{\mu ,j}\left(u\right)={\left(T_k\left(u\right)\right)}_{\mu }+e^{-\mu t}{T}_k\left({\psi }_j\right).\end{array}$$

The definition of ${\eta }_{\mu ,j}$, which is a smooth approximation of $T_k\left(u\right)$, is needed to deal with a nonzero initial datum (see also [22,39]). Note that this function has the following properties:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\left({\eta }_{\mu ,j}\left(u\right)\right)}_t=\mu (T_k\left(u\right)+{\eta }_{\mu ,j}\left(u\right)),{\eta }_{\mu ,j}\left(u\right)\left(0\right)=T_k\left({\psi }_j\right),|{\eta }_{\mu ,j}\left(u\right)|\le k,\hfill \\ \nabla {\eta }_{\mu ,j}\left(u\right)⟶\nabla T_k\left(u\right)\mathrm{strongly} \mathrm{in} {\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N,\mathrm{as} \mu ⟶+\infty .\hfill \end{array}\right.\end{array}$$

(60)

We set

$$\begin{array}{c}\hfill {\omega }^n=T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right)),\end{array}$$

and it is easy to see that $\nabla {\omega }^n=0$ for $|u^n|>M=4k+h$. Now, we choose ${\omega }^n$ as test function in (34) with $h>k>0$, then it follows that

$$\begin{array}{cc}\hfill & {\int }_0^T\langle \frac{\partial u^n}{\partial t},{\omega }^n\rangle \mathrm{d}t+{\int }_{Q_T}{|\nabla T_M\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)·\nabla {\omega }^n\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\lambda {\int }_{Q_T}{|u^n|}^{q\left(x\right)-2}{u}^n{\omega }^n\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{Q_T}f(x,u^n){\omega }^n\mathrm{d}x\mathrm{d}t+{\int }_{Q_T}{g}^n{\omega }^n\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(61)

For the first term of (61), we have (see the detail proof of Lemma 2.1 in [39])

$$\begin{array}{c}\hfill {\int }_0^T\langle \frac{\partial u^n}{\partial t},{\omega }^n\rangle \mathrm{d}t\ge \omega (n,\mu ,j,h),\end{array}$$

(62)

where $\omega (n,\mu ,j,h)$ satisfies

$$\begin{array}{c}\hfill \underset{h\to +\infty }{lim}\underset{j\to +\infty }{lim}\underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}\omega (n,\mu ,j,h)=0.\end{array}$$

Combining (61) with (62), we have

$$\begin{array}{cc}\hfill & {\int }_{Q_T}{|\nabla T_M\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)·\nabla {\omega }^n\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{Q_T}f(x,u^n){\omega }^n\mathrm{d}x\mathrm{d}t+{\int }_{Q_T}{g}^n{\omega }^n\mathrm{d}x\mathrm{d}t-\lambda {\int }_{Q_T}{|u^n|}^{q\left(x\right)-2}{u}^n{\omega }^n\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\omega (n,\mu ,j,h).\hfill \end{array}$$

(63)

Splitting the integral on the left-hand side on the set where $|u^n|<k$ and where $|u^n|\ge k$ and discarding some nonnegative terms, we find

$$\begin{array}{cc}\hfill & {\int }_{Q_T}{|\nabla T_M\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)·\nabla T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \ge {\int }_{Q_T}{|\nabla T_k\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_k\left(u^n\right)·\nabla (T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & -{\int }_{\{|u^n|\ge k\}}||\nabla T_M\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)|·|\nabla {\eta }_{\mu ,j}\left(u\right)|\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Inserting above inequality into (63), one has

$$\begin{array}{cc}\hfill & {\int }_{Q_T}{|\nabla T_k\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_k\left(u^n\right)·\nabla (T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{\{|u^n|\ge k\}}||\nabla T_M\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)|·|\nabla {\eta }_{\mu ,j}\left(u\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +{\int }_{Q_T}f(x,u^n){\omega }^n\mathrm{d}x\mathrm{d}t+{\int }_{Q_T}{g}^n{\omega }^n\mathrm{d}x\mathrm{d}t-\lambda {\int }_{Q_T}{|u^n|}^{q\left(x\right)-2}{u}^n{\omega }^n\mathrm{d}x\mathrm{d}t+\omega (n,\mu ,j,h).\hfill \end{array}$$

Moreover, we get further

$$\begin{array}{cc}\hfill & I\left(n\right)\hfill \\ \hfill & \stackrel{\mathrm{def}}{=}{\int }_{Q_T}(|\nabla T_k\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_k\left(u^n\right)-|\nabla T_k\left(u\right){|}^{p\left(x\right)-2}\nabla T_k\left(u\right))·\nabla (T_k\left(u^n\right)-T_k\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le {\int }_{\{|u^n|\ge k\}}||\nabla T_M\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)|·|\nabla {\eta }_{\mu ,j}\left(u\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & -{\int }_{Q_T}{|\nabla T_k\left(u\right)|}^{p\left(x\right)-2}\nabla T_k\left(u\right)·\nabla (T_k\left(u^n\right)-T_k\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +{\int }_{Q_T}f(x,u^n)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +{\int }_{Q_T}{g}^n{T}_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & -\lambda {\int }_{Q_T}{|u^n|}^{q\left(x\right)-2}{u}^n{T}_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{d}x\mathrm{d}t+\omega (n,\mu ,j,h).\hfill \\ \hfill & =I_1+I_2+I_3+I_4+I_5+\omega (n,\mu ,j,h).\hfill \end{array}$$

(64)

Now, we show the limits of $I_1,I_2,I_3,I_4 \mathrm{and} I_5$ are zero when $n,\mu ,h \mathrm{and} j$ tend to infinity, respectively.

$Limitof{I}_1$. We know that $|\nabla T_M\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)$ is bounded in ${\left(L^{p^{\prime }\left(x\right)}\left(Q_T\right)\right)}^N$, and by the dominated convergence theorem ${\chi }_{\{|u^n|\ge k\}}|\nabla {\eta }_{\mu ,j}\left(u\right)|$ converges strongly in $L^{p\left(x\right)}\left(Q_T\right)$ to ${\chi }_{\{|u|\ge k\}}|\nabla T_k\left(u\right)|$, which is zero, as $n,\mu$ tends to infinity. Thus, we obtain

$$\begin{array}{cc}\hfill \underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}{I}_1& =\underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}{\int }_{\{|u^n|\ge k\}}||\nabla T_M\left(u^n\right){|}^{p\left(x\right)-2}\nabla T_M\left(u^n\right)||\nabla T_k\left(u\right)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & =0.\hfill \end{array}$$

(65)

$Limitof{I}_2$. From (57), we have $\nabla T_k\left(u^n\right)\rightharpoonup \nabla T_k\left(u\right)\mathrm{weakly} \mathrm{in}{\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N$, then

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}{I}_2=0.\end{array}$$

(66)

$Limitof{I}_3$. Notice that

$$\begin{array}{cc}\hfill & I_3\le {\int }_{Q_T}|f(x,u^n)-f(x,u)||T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +{\int }_{Q_T}|f(x,u)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le 2k{\int }_{Q_T}|f(x,u^n)-f(x,u)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +{\int }_{Q_T}|f(x,u)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))|\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(67)

For the first term on the right-hand side of the last inequality of (67), according to (43) we have

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}2k{\int }_{Q_T}|f(x,u^n)-f(x,u)|\mathrm{d}x\mathrm{d}t=0.\end{array}$$

For the second term on the right-hand side of the last inequality of (67), and (41) and (42) hold, we get

$$\begin{array}{cc}\hfill & f(x,u)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\hfill \\ \hfill & ⟶f(x,u)T_{2k}(u-T_h\left(u\right)+T_k\left(u\right)-{\eta }_{\mu ,j}\left(u\right))\mathrm{a}.\mathrm{e}.\mathrm{in} Q_T,\hfill \end{array}$$

and

$$\begin{array}{c}\hfill f(x,u)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))\le 2kf(x,u).\end{array}$$

By Lebesgue dominated convergence theorem, we have

$$\begin{array}{cc}\hfill & \underset{n\to +\infty }{lim}{\int }_{Q_T}|f(x,u)T_{2k}(u^n-T_h\left(u^n\right)+T_k\left(u^n\right)-{\eta }_{\mu ,j}\left(u\right))|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{Q_T}|f(x,u)T_{2k}(u-T_h\left(u\right)+T_k\left(u\right)-{\eta }_{\mu ,j}\left(u\right))|\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Since ${\eta }_{\mu ,j}$ is a smooth approximation of $T_k\left(u\right)$, we obtain that

$$\begin{array}{c}\hfill \underset{h\to +\infty }{lim}\underset{j\to +\infty }{lim}\underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}|I_3|\le \underset{h\to +\infty }{lim}{\int }_{Q_T}|f(x,u)T_{2k}(u-T_h\left(u\right))|\mathrm{d}x\mathrm{d}t=0.\end{array}$$

(68)

$Limitof{I}_4,I_5$. Similarly to $Limitof{I}_3$, for $g^n⟶g \mathrm{in} L^1\left({\mathbb{R}}^N\right)$ and (52) hold, we can conclude

$$\begin{array}{c}\hfill \underset{h\to +\infty }{lim}\underset{j\to +\infty }{lim}\underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}|I_4|\le \underset{h\to +\infty }{lim}{\int }_{Q_T}|g{T}_{2k}(u-T_h\left(u\right))|\mathrm{d}x\mathrm{d}t=0,\end{array}$$

(69)

and

$$\begin{array}{c}\hfill \underset{h\to +\infty }{lim}\underset{j\to +\infty }{lim}\underset{\mu \to +\infty }{lim}\underset{n\to +\infty }{lim}|I_5|\le \underset{h\to +\infty }{lim}{\int }_{Q_T}{||u|}^{q\left(x\right)-2}u{T}_{2k}(u-T_h\left(u\right))|\mathrm{d}x\mathrm{d}t=0.\end{array}$$

(70)

Therefore, passing to the limits in (64) as $n,\mu ,j,h$ tend to infinity, by means of (65)–(70), we deduce that

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}I\left(n\right)=0.\end{array}$$

Recalling the well-known inequalities (49) and (50), we have

$$\begin{array}{c}\hfill {\int }_{\{(x,t)\in Q_T:p\left(x\right)\ge 2\}}{|\nabla T_k\left(u^n\right)-\nabla T_k\left(u\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\le \left(2^{p^{+}-1}\right)(p^{+}-1)I\left(n\right),\end{array}$$

and

$$\begin{array}{cc}\hfill & {\int }_{\{(x,t)\in Q_T:1<p\left(x\right)<2\}}{|\nabla T_k\left(u^n\right)-\nabla T_k\left(u\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le \frac{3^{2-p^{-}}}{p^{-}-1}{\epsilon }^{(p^{-}-2)/p^{-}}I\left(n\right)+\epsilon {\int }_{Q_T}{|\nabla T_k\left(u\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Since $I\left(n\right)⟶0 \mathrm{as} n⟶+\infty$, then using the arbitrariness of $\epsilon$ and $\nabla T_k\left(u\right)$ is bounded in ${\left(L^{p\left(x\right)}\left(Q_T\right)\right)}^N$, we conclude that

$$\begin{array}{c}\hfill \underset{n\to +\infty }{lim}{\int }_{Q_T}{|\nabla T_k\left(u^n\right)-\nabla T_k\left(u\right)|}^{p\left(x\right)}\mathrm{d}x\mathrm{d}t=0,\end{array}$$

which implies that, for every $k>0$,

$$\begin{array}{c}\hfill \nabla T_k\left(u^n\right)⟶\nabla T_k\left(u\right)\mathrm{strongly} \mathrm{in} {(L^{p\left(x\right)}\left(Q_T\right))}^N.\end{array}$$

(71)

Step 5.  $Theintegralin\mathrm{d}entity.$

Taking $T_k(u^n-\phi )$ as a test function in (34), we have

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u^n-\phi )\left(T\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0^n-\phi \left(0\right))\mathrm{d}x+{\int }_0^T\langle {\phi }_t,T_k(u^n-\phi )\rangle \mathrm{d}t\hfill \\ \hfill & +{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla u^n|}^{p\left(x\right)-2}\nabla u^n·\nabla T_k(u^n-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{|u^n|}^{q\left(x\right)-2}{u}^n{T}_k(u^n-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u^n)T_k(u^n-\phi )\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}{g}^n{T}_k(u^n-\phi )\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(72)

We will prove the convergence term by term in (72). Because ${\mathsf{\Phi }}_k$ is Lipschitz continuous and $u^n$ converges to u in $C([0,T];L^1\left({\mathbb{R}}^N\right))$, then

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u^n-\phi )\left(T\right)\mathrm{d}x⟶{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u-\phi )\left(T\right)\mathrm{d}x,\\ \hfill {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u^n-\phi )\left(0\right)\mathrm{d}x⟶{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0-\phi \left(0\right))\mathrm{d}x,\end{array}$$

as $n⟶+\infty$. Since (71) hold, then there exists a constant L such that

$$\begin{array}{cc}\hfill & {\int }_{Q_T}{|\nabla u^n|}^{p\left(x\right)-2}\nabla u^n·\nabla T_k(u^n-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{Q_T}{|\nabla T_L\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_L\left(u^n\right)·\nabla T_k(T_L\left(u^n\right)-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{\{|T_L\left(u^n\right)-\phi |\le k\}}{|\nabla T_L\left(u^n\right)|}^{p\left(x\right)-2}\nabla T_L\left(u^n\right)·(\nabla T_L\left(u^n\right)-\nabla \phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ⟶{\int }_{\{|T_L\left(u\right)-\phi |\le k\}}{|\nabla T_L\left(u\right)|}^{p\left(x\right)-2}\nabla T_L\left(u\right)·(\nabla T_L\left(u\right)-\nabla \phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{Q_T}{|\nabla T_L\left(u\right)|}^{p\left(x\right)-2}\nabla T_L\left(u\right)·\nabla T_k(T_L\left(u\right)-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_{Q_T}{|\nabla u|}^{p\left(x\right)-2}\nabla u·\nabla T_k(u-\phi )\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

From (44) and (52), as $n⟶+\infty$, we conclude that

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u^n)T_k(u^n-\phi )\mathrm{d}x\mathrm{d}t⟶{\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u)T_k(u-\phi )\mathrm{d}x\mathrm{d}t,\end{array}$$

and

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}|u^n{|}^{q\left(x\right)-2}{u}^n{T}_k(u^n-\phi )\mathrm{d}x\mathrm{d}t⟶{\int }_0^T{\int }_{{\mathbb{R}}^N}{|u|}^{q\left(x\right)-2}u{T}_k(u-\phi )\mathrm{d}x\mathrm{d}t.\end{array}$$

Then, let $n\to +\infty$ in (72), we obtain

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u-\phi )\left(T\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0-\phi \left(0\right))\mathrm{d}x+{\int }_0^T\langle {\phi }_t,T_k(u-\phi )\rangle \mathrm{d}t\hfill \\ \hfill & +{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla u|}^{p\left(x\right)-2}\nabla u·\nabla T_k(u-\phi )\mathrm{d}x\mathrm{d}t+\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{|u|}^{q\left(x\right)-2}u{T}_k(u-\phi )\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u)T_k(u-\phi )\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}g{T}_k(u-\phi )\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(73)

Therefore, we prove that u is an entropy solution to problem (1).

Step 6. The uniqueness of the entropy solution.

Assume that there is another entropy solution v to the problem (1). For any $T>0$, setting $\phi =u^n$ in (73), we have

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u^n)\left(T\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0-u^n\left(0\right))\mathrm{d}x+{\int }_0^T\langle u_t^n,T_k(v-u^n)\rangle \mathrm{d}t\hfill \\ \hfill & +{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla v|}^{p\left(x\right)-2}\nabla v·\nabla T_k(v-u^n)\mathrm{d}x\mathrm{d}t+\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{|v|}^{q\left(x\right)-2}v{T}_k(v-u^n)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,v)T_k(v-u^n)\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}g{T}_k(v-u^n)\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(74)

At the same time, we take $T_k(v-u^n)$ as a test function in (34) to obtain

$$\begin{array}{cc}\hfill & {\int }_0^T\langle u_t^n,T_k(v-u^n)\rangle \mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}{|\nabla u^n|}^{p\left(x\right)-2}\nabla u^n·\nabla T_k(v-u^n)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{|u^n|}^{q\left(x\right)-2}{u}^n{T}_k(v-u^n)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}f(x,u^n)T_k(v-u^n)\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}{g}^n{T}_k(v-u^n)\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(75)

Subtracting (75) from (74), it yields

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u^n)\left(T\right)\mathrm{d}x-{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(u_0-u^n\left(0\right))\mathrm{d}x\hfill \\ \hfill & +{\int }_0^T{\int }_{{\mathbb{R}}^N}{(|\nabla v|}^{p\left(x\right)-2}\nabla v-|\nabla u^n{|}^{p\left(x\right)-2}\nabla u^n)·\nabla T_k(v-u^n)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & +\lambda {\int }_0^T{\int }_{{\mathbb{R}}^N}{(|v|}^{q\left(x\right)-2}v-|u^n{|}^{q\left(x\right)-2}{u}^n)T_k(v-u^n)\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & ={\int }_0^T{\int }_{{\mathbb{R}}^N}(f(x,v)-f(x,u^n))T_k(v-u^n)\mathrm{d}x\mathrm{d}t+{\int }_0^T{\int }_{{\mathbb{R}}^N}(g-g^n)T_k(v-u^n)\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

(76)

Due to the properties of ${\mathsf{\Phi }}_k$ and the same initial condition for $u^n$ and v, as $n⟶\infty$, we get

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u^n)\left(T\right)\mathrm{d}x⟶{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u)\left(T\right)\mathrm{d}x,\hfill \\ \hfill & {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u^n)\left(0\right)\mathrm{d}x⟶0.\hfill \end{array}$$

Moreover, the fact $g^n⟶g \mathrm{in} L^1\left({\mathbb{R}}^N\right)$, $|T_k\left(r\right)|\le k$ process

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}(g-g^n)T_k(v-u^n)\mathrm{d}x\mathrm{d}t⟶0,\end{array}$$

and from the assumptions on f, we obtain

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_{{\mathbb{R}}^N}(f(x,v)-f(x,u^n))T_k(v-u^n)\mathrm{d}x\mathrm{d}t\le {\int }_0^T{\int }_{{\mathbb{R}}^N}{l}_2|v-u^n||T_k(v-u^n)|\mathrm{d}x\mathrm{d}t\hfill \\ \hfill & \le 2l_2{\int }_0^T{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u^n)\mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

Taking the limit $n⟶\infty$ in (76) and discarding the nonnegative term, we deduce that

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u)\left(T\right)\mathrm{d}x\le 2l_2{\int }_0^T{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u)\mathrm{d}x\mathrm{d}t.\end{array}$$

(77)

Using the Gronwall’s inequality, we obtain

$$\begin{array}{c}\hfill {\int }_0^T{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u)\mathrm{d}x\mathrm{d}t\le e^{2l_{2}t}{\int }_0^T{\int }_{{\mathbb{R}}^N}{\mathsf{\Phi }}_k(v-u)\left(0\right)\mathrm{d}x\mathrm{d}t=0,\end{array}$$

(78)

which implies that ${\mathsf{\Phi }}_k(v-u)\left(t\right)=0$ for all $t\in [0,T]$. Thus, we have $u=v$ a.e. in $Q_T$. Therefore, we obtain the uniqueness of entropy solutions. This finishes the proof. □