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Article

# Existence of Solutions to a Nonlinear Parabolic Equation of Fourth-Order in Variable Exponent Spaces

by
Bo Liang
1,*,
Xiting Peng
1 and
Chengyuan Qu
2
1
School of Science, Dalian Jiaotong University, Dalian 116028, China
2
Department of Mathematics, Dalian Nationalities University, Dalian 116600, China
*
Author to whom correspondence should be addressed.
Entropy 2016, 18(11), 413; https://doi.org/10.3390/e18110413
Submission received: 17 September 2016 / Revised: 12 November 2016 / Accepted: 15 November 2016 / Published: 18 November 2016

## Abstract

:
This paper is devoted to studying the existence and uniqueness of weak solutions for an initial boundary problem of a nonlinear fourth-order parabolic equation with variable exponent $v t + div ( | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ) − | ▵ v | q ( x ) − 2 ▵ v = g ( x , v )$. By applying Leray-Schauder’s fixed point theorem, the existence of weak solutions of the elliptic problem is given. Furthermore, the semi-discrete method yields the existence of weak solutions of the corresponding parabolic problem by constructing two approximate solutions.

## 1. Introduction

We mainly study the following fourth-order parabolic equations with variable exponents:
$v t + div ( | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ) − | ▵ v | q ( x ) − 2 ▵ v = g ( x , v ) , ( x , t ) ∈ Ω T ,$
$v ( x , t ) = ▵ v ( x , t ) = 0 , ( x , t ) ∈ Γ T ,$
$v ( x , 0 ) = v 0 ( x ) , x ∈ Ω ,$
where Ω is an open, bounded domain in $R N$, $∂ Ω ∈ C 1$. Define $Ω T = Ω × ( 0 , T )$ and $Γ T = ∂ Ω × ( 0 , T )$. If p is a constant (especially $p ≡ 2$ and $q ≡ 2$), the Equation (1) has the structure of the classical Cahn–Hilliard problem, which is often used to describe the evolution of a conserved concentration field during phase separation in physics. It is also related to the thin-film equation if $| ∇ ▵ v | p ( x ) − 2$ becomes $v p$, which can analyze the motion of a very thin layer of viscous incompressible fluids along an include plane.
There have been some results related to the existence, uniqueness and properties of solutions to the fourth-order degenerate parabolic equations (see [1,2]). The paper [3] has studied the existence of the Cahn–Hilliard equation and the reader may refer to [4] to obtain its physical background. For the constant exponent case of (1), the paper [5] has given the existence and uniqueness of weak solutions. For the problems in variable exponent spaces, the papers [6,7,8] have studied the existence of some fourth-order parabolic equations with a variable exponent, and [9] has given the Fujita type conditions for fast diffusion equation.
For the research of the existence and long-time behavior of the fourth-order partial differential equations, the entropy functional method is often applied in order to obtain the necessary estimates and to show the entropy dissipation. The large time behavior of solutions of the thin film equation $u t + ( u n u x x x ) x = 0$ was addressed in [10,11] by the entropy function method. For $0 < n < 3$, [12] proved the existence of (1) in the distributional sense and obtained the exponentially fast convergence in $L ∞$-norm via the entropy method of a regularized problem. We apply the idea of the entropy method to deal with the corresponding problems with variable exponents.
In this paper, we apply the Leray-Schauder’s fixed point theorem to prove the existence of weak solutions of the corresponding elliptic problem of (1)–(3) in order to deal with the nonlinear source. Furthermore, the semi-discrete method yields the existence of weak solutions of the parabolic problem by constructing two approximate solutions. We will show the effect of the variable exponents and the second-order nonlinear diffusion to the degenerate parabolic Equation (1).

#### 1.1. Preliminaries

We introduce some elementary concepts and lemmas related to the variable exponent spaces in this part.
Let $p ( x ) ≥ 1$ be a continuous function in $Ω ¯$ and we define the variable exponent space as follows:
with the norm
$∥ v ∥ p ( x ) = inf γ > 0 : T p ( . ) v γ ≤ 1 .$
It is easy to check that the variable exponent space $L p ( x ) ( Ω )$ becomes the classical Lebesgue space $L p ( Ω )$ when $p ( x )$ is a positive constant.
For convenience, we list some definitions and notations of the generalized Lebesgue–Sobolev space $W k , p ( x ) ( Ω )$:
Moreover, $W 0 k , p ( x )$ denotes the closure of $C 0 ∞ ( Ω )$ in $W k , p ( x ) ( Ω )$-norm, $W 0 − 1 , p ′ ( x ) ( Ω )$ denotes the dual space of $W 0 1 , p ( x ) ( Ω )$. For any positive continuous function $θ ( x )$, we define
$θ − = inf x ∈ Ω ¯ θ ( x ) , θ + = sup x ∈ Ω ¯ θ ( x ) .$
Throughout the paper, C and $C i ( i = 1 , 2 , 3 , . . . )$ denote the general positive constants independent of solutions and may change from line to line.
In the following, we list some known results for the variable exponent spaces (see [13,14]).
Lemma 1.
Letting $f ∈ L p ( x ) ( Ω )$, one has
$( 1 ) ∥ f ∥ p ( x ) < 1 ( = 1 ; > 1 ) ⟺ T p ( . ) ( f ) < 1 ( = 1 ; > 1 ) ; ( 2 ) ∥ f ∥ p ( x ) < 1 ⟹ ∥ f ∥ p ( x ) p + ≤ T p ( . ) ( f ) ≤ ∥ f ∥ p ( x ) p − ; ∥ f ∥ p ( x ) ≥ 1 ⟹ ∥ f ∥ p ( x ) p − ≤ T p ( . ) ( f ) ≤ ∥ f ∥ p ( x ) p + ; ( 3 ) ∥ f ∥ p ( x ) → 0 ⟺ T p ( . ) ( f ) → 0 ; ∥ f ∥ p ( x ) → ∞ ⟺ T p ( . ) ( f ) → ∞ .$
Lemma 2.
(Poincaré’s inequality) Letting $f ∈ W 0 1 , p ( x ) ( Ω )$, there exists a positive constant C such that $∥ f ∥ p ( x ) ≤ C ∥ ∇ f ∥ p ( x ) .$
Lemma 3.
(Hölder’s inequality) Letting $f ∈ L p ( x ) ( Ω )$ and $g ∈ L p ′ ( x ) ( Ω )$, one has $∫ Ω f g d x ≤ 1 p − + 1 p ′ − ∥ f ∥ p ( x ) ∥ g ∥ p ′ ( x ) ≤ 2 ∥ f ∥ p ( x ) ∥ g ∥ p ′ ( x ) .$

#### 1.2. Results

In (1), we require that $p ( x )$ and $q ( x )$ are two continuous functions in $Ω ¯$ and $p − , q − > 1$. Besides, the nonlinear source term $g ( x , v ) ∈ C 1 ( Ω × R )$ satisfies the growth condition:
$| g ( x , v ) | ≤ K | v | l ( x ) + s ( x ) , v ∈ ( − ∞ , + ∞ ) , x ∈ Ω ,$
where K is a positive constant, $l ( x )$ is a continuous function in $Ω ¯$ and $s ( x ) ∈ L p ′ ( x ) ( Ω )$. Furthermore, by letting $π ( x ) = : l ( x ) p ′ ( x )$, we require that
$π + p − < 1 , 0 ≤ l ( x ) ≤ N p ( x ) ( N − p ( x ) ) p ′ ( x ) .$
The corresponding steady-state problem of (1)–(3) has the form:
$div ( | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ) − | ▵ v | q ( x ) − 2 ▵ v = g ( x , v ) in Ω ,$
$v = ▵ v = 0 on ∂ Ω .$
The weak solution is defined in the following sense.
Definition 1.
A function $v ∈ E 1$ is said to be a weak solution of (6) and (7) provided that
$− ∫ Ω | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ∇ ϕ d x − ∫ Ω | ▵ v | q ( x ) − 2 ▵ v ϕ d x = ∫ Ω g ( x , v ) ϕ d x$
for each $ϕ ∈ W 0 1 , p ( x ) ( Ω ) ∩ L q ( x ) ( Ω )$ and $g ( x , v ( x ) ) ∈ L p ′ ( x ) ( Ω )$.
The following theorem gives the existence of solutions.
Theorem 1.
Let $v 0 ∈ W 0 1 , p ( x ) ( Ω )$. There exists at least a weak solution of (6) and (7) satisfying Definition 1.
For the evolution equation case, we define the weak solution of (1)–(3) as following.
Definition 2.
A function v is said to be a weak solution of (1)–(3) provided that
(i)
$v ∈ C ( [ 0 , T ] ; L 2 ( Ω T ) ) ∩ L p − ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) ∩ L q − ( 0 , T ; W 2 , q ( x ) ( Ω ) )$, $▵ v ∈ L p ′ − ( 0 , T ; W 0 1 , p ( x ) ( Ω ) )$, $∂ v ∂ t ∈ L p ′ − ( 0 , T ; W − 1 , p ( x ) ( Ω ) )$, $v ( x , 0 ) = v 0 ( x )$ a.e. in Ω;
(ii)
For any $ϕ ∈ L p + ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) ∩ L q + ( 0 , T ; W 2 , q ( x ) ( Ω ) )$, one has
$∫ 0 T ∂ v ∂ t , ϕ d t = ∫ 0 T ∫ Ω | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ∇ ϕ d x d t + ∫ 0 T ∫ Ω | ▵ v | q ( x ) − 2 ▵ v ϕ d x d t + ∫ 0 T ∫ Ω g ( x , v ) ϕ d x d t .$
The existence of solutions is the following theorem.
Theorem 2.
Let $p ′ − > 1$, $q ′ − > 1$, $v 0 ∈ H 0 1 ( Ω ) ∩ W 0 1 , p ( x ) ( Ω )$ and $▵ v 0 ∈ W 0 1 , p ( x ) ( Ω )$. There exists at least a weak solution of (1)–(3).
Moreover, the solution of (1)–(3) is unique when $g ( x , v ) = μ ( v ( x , t ) − b ( x ) )$ where μ is a constant and $b ( x ) ∈ L p ′ ( x ) ( Ω )$.
This paper is organized as follows. In Section 2, we prove the existence and uniqueness of weak solution to the steady-state problem by using Leray-Schauder’s fixed point theorem. In Section 3, we prove the existence of the solution to an evolution equation by applying the semi-discrete method with necessary uniform estimates.

In order to apply the fixed point theorem, we consider a steady-state problem with the source $g ( x )$:
$div ( | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ) − | ▵ v | q ( x ) − 2 ▵ v = g ( x ) in Ω ,$
$v = ▵ v = 0 on ∂ Ω .$
By constructing an energy functional and obtaining its minimizer, we have the following existence of weak solutions.
Lemma 4.
Let $g ∈ L p ′ ( x ) ( Ω )$. There exists a unique weak solution $v ∈ E 1$ of (9) and (10) satisfying
$− ∫ Ω | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ∇ ▵ ϕ d x − ∫ Ω | ▵ v | q ( x ) − 2 ▵ v ▵ ϕ d x = ∫ Ω g ( x ) ▵ ϕ d x$
for any $ϕ ∈ E 1$.
Proof.
Introduce a functional
$ϝ ( v ) = ∫ Ω 1 p ( x ) | ∇ ▵ v | p ( x ) d x + ∫ Ω 1 q ( x ) | ▵ v | q ( x ) d x − ∫ Ω g ▵ v d x .$
For the last term, Hölder’s inequality, the Young inequality, the Sobolev embedding theorem (see [15]) and the $L p$-theory of the second-order elliptic equation (see [16]) gives
$| ∫ Ω g ▵ v d x | ≤ ∥ g ∥ p ′ ( x ) ∥ ▵ v ∥ p ( x ) ) ≤ C 1 ε max ∥ g ∥ p ′ ( x ) p ′ − , ∥ g ∥ p ′ ( x ) p ′ + + ε ∫ Ω | ▵ v | p ( x ) d x ≤ C max ∥ g ∥ p ′ ( x ) p ′ − , ∥ g ∥ p ′ ( x ) p ′ + + 1 2 ∫ Ω 1 p ( x ) | ∇ ▵ v | p ( x ) d x .$
On the other hand, (12) implies
$− C ≤ inf ν ∈ E 1 ϝ ( ν ) ≤ ϝ ( 0 ) = 0 .$
Hence there exists a sequence ${ v k } k = 1 ∞ ∈ E 1$ such that
$ϝ ( v k ) → inf ν ∈ E 1 ϝ ( ν ) ( k → ∞ ) .$
Equations (13) and (14) give
$∥ ∇ ▵ v k ∥ p ( x ) ≤ C , ∥ ▵ v k ∥ q ( x ) ≤ C ,$
which implies that $ϝ ( v k )$ is bounded and thus Lemmas 1–3 yield
$∥ v k ∥ 2 , p ( x ) ≤ ∥ ▵ v k ∥ p ( x ) ≤ C ∥ ∇ ▵ v k ∥ p ( x ) ≤ C 1 max ∫ Ω | ∇ ▵ v k | p ( x ) 1 p − , ∫ Ω | ∇ ▵ v k | p ( x ) 1 p + ≤ C 2 ,$
and
$∥ v k ∥ 2 , q ( x ) ≤ ∥ ▵ v k ∥ q ( x ) ≤ C 3 max ∫ Ω | ▵ v k | q ( x ) 1 q − , ∫ Ω | ▵ v k | q ( x ) 1 q + ≤ C 4 .$
It shows that $v k$ belongs to the space $W 0 1 , p ( x ) ⋂ W 2 , p ( x ) ⋂ W 2 , q ( x )$ uniformly, and then there exists a function $v ∈ E 1$ such that
Furthermore, since $ϝ ( v )$ is weakly lower semi-continuous on $E 1$, we have
$inf ν ∈ E 1 ϝ ( ν ) ≤ ϝ ( v ) ≤ lim k → ∞ inf ϝ ( v k ) = inf ν ∈ E 1 ϝ ( ν ) ,$
i.e., v is a minimizer of $ϝ ( · )$ and $ϝ ( v ) = inf ν ∈ E 1 ϝ ( ν ) .$ It guarantees that v is a weak solution of (9) and (10).
The uniqueness is obvious and we omit the details. ☐
Now, we consider the problem (6) and (7) with the nonlinear source $g ( x , v )$.
Lemma 5.
Letting $v ( x ) ∈ E 1$ be a weak solution of (6) and (7), one has $∥ v ∥ E 1 ≤ C$.
Proof.
Multiplying (8) by v gives
$∫ Ω | ∇ ▵ v | p ( x ) d x + ∫ Ω | ▵ v | q ( x ) d x = − ∫ Ω g ( x , v ) ▵ v d x ≤ 2 ∥ g ∥ p ′ ( x ) ∥ ▵ v ∥ p ( x ) ≤ C ∥ g ∥ p ′ ( x ) ∥ ∇ ▵ v ∥ p ( x ) ≤ 1 4 ∫ Ω | ∇ ▵ v | p ( x ) d x + C max { ∥ g ∥ p ′ ( x ) p ′ − , ∥ g ∥ p ′ ( x ) p ′ + } .$
By Lemmas 2 and 3 and $L p$-estimate (see [16]), we conclude that
$∫ Ω | g ( x , v ) | p ′ ( x ) d x ≤ C ∫ Ω ( K | v | l ( x ) + s ( x ) ) p ′ ( x ) d x ≤ C M p ′ + ∫ Ω | v | l ( x ) p ′ ( x ) d x + C ∫ Ω | s ( x ) | p ′ ( x ) d x + C ≤ C ∥ v ∥ W 1 , p ( x ) l ( x ) p ′ ( x ) + C ≤ C ∥ ▵ v ∥ p ( x ) l ( x ) p ′ ( x ) + C ≤ C ∥ ∇ ▵ v ∥ p ( x ) l ( x ) p ′ ( x ) + C ,$
and thus
$max { ∥ g ∥ p ′ ( x ) p ′ − , ∥ g ∥ p ′ ( x ) p ′ + } ≤ C max { ∥ ∇ ▵ v ∥ p ( x ) π + , ∥ ∇ ▵ v ∥ p ( x ) π − } + C ≤ 1 4 ∫ Ω | ∇ ▵ v | p ( x ) d x + C .$
Equations (15)–(17) yield
$∫ Ω | ∇ ▵ v | p ( x ) d x + ∫ Ω | ▵ v | q ( x ) d x ≤ C .$
It completes the proof of Lemma 5. ☐
Proof of Theorem 1.
Letting $ω ∈ L p * ( x ) ( Ω )$ and $δ ∈ [ 0 , 1 ]$ where we choose $p * ( x )$ such that $E 1 ↪ L p * ( x ) ( Ω )$ is compact, we consider the auxiliary problem
$div ( | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v ) − | ▵ v | q ( x ) − 2 ▵ v = δ g ( x , ω ) in Ω , v = ▵ v = 0 on ∂ Ω .$
Lemma 4 ensures its existence and so we can define the fixed point operator
$T : [ 0 , 1 ] × L p * ( x ) ( Ω ) ⟶ L p * ( x ) ( Ω ) , ( δ , ω ) ⟼ v$
and $T ( ω , 0 ) = 0$.
If $ω ∈ L p * ( x ) ( Ω )$ satisfies $T ( ω , δ ) = ω$, we can check that $∥ ω ∥ E 1 ≤ C$ where $C > 0$ is independence of ω and δ from the idea of Lemma 5. The compact embedding $E 1 ↪ L p * ( x ) ( Ω )$ can ensure that T is a continuous and compact operator. Leray-Schauder’s fixed point theorem yields the existence of solutions of (6) and (7). ☐

## 3. Evolution Equation

In this section, we study the existence solutions of (1)–(3). For this purpose, we establish a semi-discrete problem at first:
$1 h ( v k − v k − 1 ) + div ( | ∇ ▵ v k | p ( x ) − 2 ∇ ▵ v k ) − | ▵ v k | q ( x ) − 2 ▵ v k = g ( x , v k − 1 ) , x ∈ Ω ,$
$v k ( x , t ) = ▵ v k ( x , t ) = 0 , x ∈ ∂ Ω ,$
where $v k = v ( x , k h )$, $h = T n , k = 1 , 2 , ⋯ , n$ and $n ∈ N$.
Lemma 6.
Assume $v 0 ∈ E 2$. (19) and (20) admits a unique weak solution $v k ∈ E 2$ satisfying
$∑ k = 1 i ∫ Ω | ∇ v k | 2 d x + h ∑ k = 1 i ∫ Ω | ∇ ▵ v k | p ( x ) d x + h ∑ k = 1 i ∫ Ω | ▵ v k | q ( x ) d x ≤ ∑ k = 1 i ∫ Ω | ∇ v k − 1 | 2 d x + h 2 ∑ k = 1 i ∫ Ω | ∇ ▵ v k − 1 | p ( x ) d x + C T ,$
and
$∫ Ω | ∇ v i | 2 d x + h 2 ∑ k = 1 i ∫ Ω | ∇ ▵ v i | p ( x ) d x + h 2 ∫ Ω | ∇ ▵ v i | p ( x ) d x + h ∑ k = 1 i ∫ Ω | ▵ v k | q ( x ) d x ≤ ∫ Ω | ∇ v 0 | 2 d x + h 2 ∫ Ω | ∇ ▵ v 0 | p ( x ) d x + C T .$
Proof.
According to the argument of the Section 2, we conclude that the problem (19) and (20) has a unique weak solution $v k ∈ E 2$ satisfying
$1 h ∫ Ω ∇ ( v k − v k − 1 ) · ∇ ϕ d x + ∫ Ω | ∇ ▵ v k | p ( x ) − 2 | ∇ ▵ v k | · ∇ ▵ ϕ d x + ∫ Ω | ▵ v k | q ( x ) − 2 ▵ v k ▵ ϕ d x = − ∫ Ω g ( x , v k − 1 ) ▵ ϕ d x$
for any $ϕ ∈ C 0 ∞ ( Ω )$. Letting $ϕ = v k$ in (23), we have
$1 h ∫ Ω | ∇ v k | 2 d x + ∫ Ω | ∇ ▵ v k | p ( x ) d x + ∫ Ω | ▵ v k | q ( x ) d x = 1 h ∫ Ω ∇ v k · ∇ v k − 1 d x − ∫ Ω g ( x , v k − 1 ) ▵ v k d x ≤ 1 2 h ∫ Ω | ∇ v k − 1 | 2 d x + 1 2 h ∫ Ω | ∇ v k | 2 d x − ∫ Ω g ( x , v k − 1 ) ▵ v k d x .$
Similar to the proof of (16) and (17), we get
$− ∫ Ω g ( x , v k − 1 ) ▵ v k d x ≤ 1 2 ∫ Ω | ∇ ▵ v k | p ( x ) d x + C ∫ Ω | g ( x , v k − 1 ) | p ′ ( x ) d x ≤ 1 2 ∫ Ω | ∇ ▵ v k | p ( x ) d x + 1 4 ∫ Ω | ∇ ▵ v k − 1 | p ( x ) d x + C .$
By (24) and (25), one has
$1 2 h ∫ Ω | ∇ v k | 2 d x + 1 2 ∫ Ω | ∇ ▵ v k | p ( x ) d x + ∫ Ω | ▵ v k | q ( x ) d x ≤ 1 2 h ∫ Ω | ∇ v k − 1 | 2 d x + 1 4 ∫ Ω | ∇ ▵ v k − 1 | p ( x ) d x + C .$
Hence, for any $1 ≤ i ≤ n$, we obtain
$∑ k = 1 i ∫ Ω | ∇ v k | 2 d x + h ∑ k = 1 i ∫ Ω | ∇ ▵ v k | p ( x ) d x + h ∑ k = 1 i ∫ Ω | ▵ v k | q ( x ) d x ≤ ∑ k = 1 i ∫ Ω | ∇ v k − 1 | 2 d x + h 2 ∑ k = 1 i ∫ Ω | ∇ ▵ v k − 1 | p ( x ) d x + C T .$
It completes the proof of (21) and (22) obtained from (21). ☐
Now, we are in the position to define the first approximate solution of (1)–(3)
$ϖ ( n ) ( x , t ) = ∑ k = 1 n χ k ( t ) v k ( t )$
where $χ k ( t )$ is the characteristic function over the interval $( ( k − 1 ) h , k h ]$ for $k = 1 , 2 , . . . , n$. For this approximate solution, we have the following uniform estimates.
Lemma 7.
One has
$∥ ϖ ( n ) ∥ L ∞ ( 0 , T ; H 0 1 ( Ω ) ) + ∥ ▵ ϖ ( n ) ∥ L p − ( 0 , T ; W 0 1 , p ( x ) ( Ω ) ) + ∥ | ∇ ▵ ϖ ( n ) | p ( x ) − 2 ∇ ▵ ϖ ( n ) ∥ L p ′ − ( Ω T ) + ∥ | ▵ ϖ ( n ) | q ( x ) − 2 ▵ ϖ ( n ) ∥ L q ′ ( Ω T ) + ∥ g ( · , ϖ ( n ) ) ∥ L p ′ − ( Ω T ) ≤ C .$
Proof.
By Lemma 6 and
$∥ ∇ ϖ ( n ) ∥ L 2 ( Ω ) 2 = ∥ ∇ v k ( x ) ∥ L 2 ( Ω ) 2 ≤ C ,$
we have the estimate
$∥ ∇ ϖ ( n ) ∥ L ∞ ( 0 , T ; L 2 ( Ω ) ) ≤ C .$
On the other hand, we have
$∥ ▵ ϖ ( n ) ∥ L p − ( 0 , T ; W 0 1 , p ( x ) ) = ∫ 0 T ∥ ▵ ϖ ( n ) ∥ 1 , p ( x ) p − d x 1 p − = h ∑ k = 1 n ∥ ▵ v k ∥ 1 , p ( x ) p − d x 1 p − ≤ C h ∑ k = 1 n ∥ ∇ ▵ v k ∥ p ( x ) p − d x 1 p − ≤ C h ∑ k = 1 n ( ∫ Ω | ∇ ▵ v k | p ( x ) d x + 1 ) 1 p − ≤ C ( C + T ) 1 p − .$
Letting $i = n$ in (22), we get
$∫ 0 T ∫ Ω | ∇ ▵ ϖ ( n ) | p ( x ) d x d t + ∫ 0 T ∫ Ω | ▵ ϖ ( n ) | q ( x ) d x d t = h ∑ k = 1 n ∫ Ω | ∇ ▵ v k | p ( x ) d x + h ∑ k = 1 n ∫ Ω | ▵ v k | q ( x ) d x ≤ C .$
Another approximate solution is defined as follows:
$v ( n ) ( x , t ) = ∑ k = 1 n χ k ( t ) [ ϑ k ( t ) v k ( x ) + ( 1 − ϑ k ( t ) ) v k − 1 ( x ) ] ,$
where
We also obtain some uniform estimates for this approximate solution.
Lemma 8.
One has
$∥ ∂ v ( n ) ∂ t ∥ L p ′ − ( 0 , T ; W − 1 , p ′ ( x ) ( Ω ) ) + ∥ v ( n ) ∥ L ∞ ( 0 , T ; H 0 1 ( Ω ) ) ≤ C .$
Proof.
By (33), we get
$∂ v ( n ) ∂ t = 1 h ∑ k = 1 n χ k ( v k − v k − 1 )$
and then
$∫ 0 T ∥ ∂ v ( n ) ∂ t ∥ W − 1 , p ′ ( x ) ( Ω ) p ′ − d t ≤ ∫ 0 T | ∂ v ( n ) ∂ t , ψ | p ′ − d t ≤ C ∫ 0 T | ∑ k = 1 n χ k ( t ) ∫ Ω | ∇ ▵ v k | p ( x ) − 2 ∇ ▵ v k ∇ ψ d x | p ′ − d t + C ∫ 0 T | ∑ k = 1 n χ k ( t ) ∫ Ω g ( x , v k − 1 ) ψ d x | p ′ − d t + C ∫ 0 T | ∑ k = 1 n χ k ( t ) ∫ Ω | ▵ v k | q ( x ) − 2 ▵ v k ψ d x | p ′ − d t ≤ C ∥ ∇ ▵ ϖ ( n ) ∥ L p ′ − ( Ω T ) p ′ − + C ∥ g ( x , v k − 1 ) ∥ L p ′ − ( Ω T ) p ′ − ≤ C ,$
for any $ψ ∈ W 0 1 , p ( x ) ( Ω )$ with $∥ ψ ∥ W 0 1 , p ( x ) ( Ω ) ≤ 1$. It follows from (22) and (33) that
$∥ v ( n ) ∥ L m ( 0 , T ; H 0 1 ( Ω ) ) m ≤ C m ∫ 0 T ∫ Ω | ∇ v ( n ) | 2 d x m 2 d t = C m ∫ 0 T ∫ Ω | ∑ k = 1 n χ k ( t ) [ ϑ k ( t ) ∇ v k ( x ) + ( 1 − ϑ k ( t ) ) ∇ v k − 1 ( x ) ] | 2 d x m 2 d t = C m ∑ k = 1 n ∫ ( k − 1 ) h ( k h ) ∫ Ω | [ ϑ k ( t ) ∇ v k ( x ) + ( 1 − ϑ k ( t ) ) ∇ v k − 1 ( x ) ] | 2 d x m 2 d t ≤ C m ∑ k = 1 n h ∫ Ω ( | ∇ v k ( x ) | 2 + | ∇ v k − 1 ( x ) | 2 ) d x m 2 ≤ C m + m 2 T$
with $C > 0$ independent of m. Perform the limit $m → ∞$ to get
$∥ v ( n ) ∥ L ∞ ( 0 , T ; H 0 1 ( Ω ) ) = lim m → ∞ ∥ v ( n ) ∥ L m ( 0 , T ; H 0 1 ( Ω ) ) ≤ C .$
Proof of Theorem 2.
By (29), we can seek a subsequence of $ϖ ( n )$ (still denoted by itself) and two functions $ν ∈ L p ′ ( x ) ( Ω T ) , ν ′ ∈ L q ′ ( x ) ( Ω T )$ such that
$ϖ ( n ) ⇀ v weakly * in L ∞ ( 0 , T ; H 0 1 ( Ω ) ) , ▵ ϖ ( n ) ⇀ ▵ v weakly in L p − ( 0 , T ; W 0 1 , p ( Ω ) ) , | ∇ ▵ ϖ ( n ) | p ( x ) − 2 ∇ ▵ ϖ ( n ) ⇀ ν weakly in L p ′ ( x ) ( Ω T ) , | ▵ ϖ ( n ) | q ( x ) − 2 ▵ ϖ ( n ) ⇀ ν ′ weakly in L q ′ ( x ) ( Ω T ) ,$
as $n → ∞$.
It is easy to check that there exists a positive integer r such that $W − 1 , p ′ ( Ω ) ↪ H − r ( Ω )$ and thus the embedding $H 0 1 ( Ω ) ↪ compact L 2 ( Ω ) ↪ H − r ( Ω )$, the uniform estimate (34) and the Aubin lemma [17] yield the existence of a subsequence of $v ( n )$ and a function ϱ such that, as $n → ∞$,
$∂ v ( n ) ∂ t ⇀ ∂ ϱ ∂ t weakly in L p ′ − ( 0 , T ; W − 1 , p ( x ) ( Ω ) ) , v ( n ) ⇀ ϱ weakly * in L ∞ ( 0 , T ; H 0 1 ( Ω ) ) , v ( n ) → ϱ strongly in C ( [ 0 , T ] ; L 2 ( Ω ) ) , v ( n ) → ϱ a . e . in Ω T .$
Moreover, (23) gives, for any $ϕ ∈ C 0 ∞ ( Ω T )$,
$| ∫ 0 T ∫ Ω ( ϖ ( n ) − v ( n ) ) ϕ d x d t | = | ∫ 0 T ∫ Ω ∑ k = 1 n χ k ( t ) ( 1 − ϑ k ( t ) ) ( v k − v k − 1 ) ϕ d x d t | = | ∫ 0 T h ∑ k = 1 n χ k ( 1 − ϑ k ) ∫ Ω ( | ∇ ▵ v k | p ( x ) − 2 ∇ ▵ v k ∇ ϕ + | ▵ v k | q ( x ) − 2 ▵ v k ϕ + g ( x , v k − 1 ) ϕ ) d x d t | ≤ h ∫ 0 T ( | ∫ Ω | ∇ ▵ ϖ ( n ) | p ( x ) − 2 ∇ ▵ ϖ ( n ) ∇ ϕ d x | + | ∫ Ω | ▵ ϖ ( n ) | q ( x ) − 2 ▵ ϖ ( n ) ϕ d x | + | ∫ Ω g ( x , ϖ ( n ) ) ϕ d x | + | ∫ Ω g ( x , v 0 ) ϕ d x | ) d t ≤ C h → 0 ,$
as $n → ∞$.
By the continuity of g and $ϱ = v$ a.e. in $Ω T$, we have $g ( x , ϖ ( n ) ) → g ( x , v )$ a.e. in $Ω T$. Furthermore, the estimate (see Lemma 7) $∥ g ( x , ϖ ( n ) ) ∥ L p ′ ( x ) ( Ω T ) < ∞$ gives
$g ( x , ϖ ( n ) ) ⇀ g ( x , v ) weakly in L p ′ ( x ) ( Ω T ) .$
Applying (23) and (35), we obtain, for any test function ϕ,
$∫ 0 T ∂ v ( n ) ∂ t , ϕ d t = ∫ 0 T ∫ Ω | ∇ ▵ ϖ ( n ) | p ( x ) − 2 ∇ ▵ ϖ ( n ) ∇ ϕ d x d t + ∫ 0 T ∫ Ω | ▵ ϖ ( n ) | q ( x ) − 2 ▵ ϖ ( n ) ϕ d x d t + ∫ 0 T ∫ Ω g ( x , ϖ ( n ) ) ϕ d x d t + h ∫ Ω g ( x , v 0 ) ϕ ( x , 0 ) d x − ∫ ( n − 1 ) h T ∫ Ω g ( x , v n ) ϕ d x d t .$
By taking $n → ∞$, we have
$∫ 0 T ∂ v ∂ t , ϕ d t = ∫ 0 T ∫ Ω ν ∇ ϕ d x d t + ∫ 0 T ∫ Ω ν ′ ϕ d x d t + ∫ 0 T ∫ Ω g ( x , v ) ϕ d x d t .$
It remains to prove $ν = | ∇ ▵ v | p ( x ) − 2 ∇ ▵ v$ and $ν ′ = | ▵ v | q ( x ) − 2 ▵ v$. By using v as a test function in (23) and integrating by part, we get
$1 2 ∫ Ω | ∇ v ( x , T ) | 2 d x − 1 2 ∫ Ω | ∇ v 0 | 2 d x + ∫ 0 T ∫ Ω ν ∇ ▵ v d x d t + ∫ 0 T ∫ Ω ν ′ ▵ v d x d t = − ∫ 0 T ∫ Ω g ( x , v ) ▵ v d x d t .$
On the other hand, (23) implies
$1 2 ∫ Ω | ∇ ϖ ( n ) ( x , T ) | 2 d x − 1 2 ∫ Ω | ∇ v 0 | 2 d x + ∫ 0 T ∫ Ω | ∇ ▵ ϖ ( n ) | p ( x ) d x d t + ∫ 0 T ∫ Ω | ▵ ϖ ( n ) | q ( x ) d x d t ≤ − ∫ 0 T ∫ Ω g ( x , ϖ ( n ) ) ▵ ϖ ( n ) d x d t − h ∫ Ω g ( x , v 0 ) v 0 d x + ∫ ( n − 1 ) h T ∫ Ω g ( x , v n ) v n d x d t .$
For any test functions $ϕ , ϕ 1$ and constant $ε > 0$, we have
$∫ 0 T ∫ Ω ( | ρ ( n ) | p ( x ) − 2 ρ ( n ) − | η ε | p ( x ) − 2 η ε ) ( ρ ( n ) − η ε ) d x d t ≥ 0 ,$
and
$∫ 0 T ∫ Ω ( | ρ 1 ( n ) | q ( x ) − 2 ρ 1 ( n ) − | η ε | q ( x ) − 2 η 1 ε ) ( ρ 1 ( n ) − η 1 ε ) d x d t ≥ 0 ,$
where $ρ ( n ) = ∇ ▵ ϖ ( n )$, $η ε = ∇ ▵ ( v − ε ϕ ) )$ and $ρ 1 ( n ) = ▵ ϖ ( n )$, $η 1 ε = ▵ ( v − ε ϕ 1 )$.
By (37)–(39), we arrive at
$1 2 ∫ Ω | ∇ v ( x , T ) | 2 d x − 1 2 ∫ Ω | ∇ v 0 | 2 d x − ∫ 0 T ∫ Ω | ∇ ▵ ( v − ε ϕ ) | p ( x ) d x d t + ∫ 0 T ∫ Ω | ∇ ▵ ( v − ε ϕ ) | p ( x ) − 2 ∇ ▵ ( v − ε ϕ ) ∇ ▵ v d x d t + ∫ 0 T ∫ Ω ν ∇ ▵ ( v − ε ϕ ) d x d t − ∫ 0 T ∫ Ω | ▵ ( v − ε ϕ 1 ) | q ( x ) d x d t + ∫ 0 T ∫ Ω | ▵ ( v − ε ϕ 1 ) | q ( x ) − 2 ▵ ( v − ε ϕ 1 ) ▵ v d x d t + ∫ 0 T ∫ Ω ν ′ ▵ ( v − ε ϕ 1 ) d x d t ≤ − ∫ 0 T ∫ Ω g ( x , v ) ▵ v d x d t .$
Moreover, (36) gives
$∫ 0 T ∫ Ω [ | ∇ ▵ ( v − ε ϕ ) | p ( x ) − 2 ] ∇ ▵ ( v − ε ϕ ) − ν ] ∇ ▵ ϕ d x d t + ∫ 0 T ∫ Ω [ | ▵ ( v − ε ϕ 1 ) | q ( x ) − 2 ] ▵ ( v − ε ϕ ) − ν ′ ] ▵ ϕ d x d t ≤ 0 .$
By letting $ε → 0$, we obtain
$∫ 0 T ∫ Ω [ | ∇ ▵ v | p ( x ) − 2 ] ∇ ▵ v − ν ] ∇ ▵ ϕ d x d t + ∫ 0 T ∫ Ω [ | ▵ v | q ( x ) − 2 ▵ v − ν ′ ] ▵ ϕ 1 d x d t ≤ 0 .$
By letting $ϕ 1 = 0$, we have
$∫ 0 T ∫ Ω [ | ∇ ▵ v | p ( x ) − 2 ] ∇ ▵ v − ν ] ∇ ▵ ϕ d x d t ≤ 0 .$
The arbitrariness of ϕ yields $| ∇ ▵ v | p ( x ) − 2 ∇ ▵ v = ν$, a.e. in $Ω T$. Similarly, we can obtain $| ▵ v | q ( x ) − 2 ▵ v = ν ′$, a.e. in $Ω T$. ☐
Proof of Uniqueness.
Let $v 1 ( x )$ and $v 2 ( x )$ be two weak solutions to (1)–(3) and $ϕ = v 1 − v 2$. By taking $▵ ϕ$ as the test function, we get
$1 2 ∫ 0 t ∫ Ω d d t | ∇ ϕ | 2 d x d t + ∫ 0 t ∫ Ω [ | ∇ ▵ v 1 | p ( x ) − 2 ] ∇ ▵ v 1 − | ∇ ▵ v 2 | p ( x ) − 2 ] ∇ ▵ v 2 ] ∇ ▵ ϕ d x d t + ∫ 0 t ∫ Ω [ | ▵ v 1 | q ( x ) − 2 ▵ v 1 − | ▵ v 2 | q ( x ) − 2 ▵ v 2 ] ▵ ϕ d x d t = μ ∫ 0 t ∫ Ω | ∇ ϕ | 2 d x d t .$
It implies
$∫ Ω | ∇ ϕ | 2 d x ≤ 2 μ ∫ 0 t ∫ Ω | ∇ ϕ | 2 d x d t ,$
where we have used the fact $( | x | m − 2 x − | y | m − 2 y ) ( x − y ) ≥ 0$ for $m > 1$ and $x , y ∈ R$ (or $R N$). By Gronwall’s inequality, we obtain $v 1 ( x , t ) = v 2 ( x , t )$ a.e. in $Ω T$. ☐

## Acknowledgments

This work was supported by NSFC (No. 11201045, 11401078) and the Education Department Science Foundation of Liaoning Province of China(No. JDL2016029).

## Author Contributions

Bo Liang completed the main study. Xiting Peng carried out the results of this article. Chengyuan Qu verified the calculation. All authors have read and approved the final manuscript.

## Conflicts of Interest

The authors declare no conflict of interest.

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Liang, B.; Peng, X.; Qu, C. Existence of Solutions to a Nonlinear Parabolic Equation of Fourth-Order in Variable Exponent Spaces. Entropy 2016, 18, 413. https://doi.org/10.3390/e18110413

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Liang B, Peng X, Qu C. Existence of Solutions to a Nonlinear Parabolic Equation of Fourth-Order in Variable Exponent Spaces. Entropy. 2016; 18(11):413. https://doi.org/10.3390/e18110413

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Liang, Bo, Xiting Peng, and Chengyuan Qu. 2016. "Existence of Solutions to a Nonlinear Parabolic Equation of Fourth-Order in Variable Exponent Spaces" Entropy 18, no. 11: 413. https://doi.org/10.3390/e18110413

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