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Extinction in finite time of solutions to the nonlinear diffusion equations involving p(x,t)-Laplacian operator

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Abstract

The aim of this paper is to study the extinction of solutions of the initial boundary value problem for u t =div( | u | p ( x , t ) 2 u)+b(x,t) | u | q a 0 u. The authors discuss how the relations of p(x,t) and dimension N affect the properties of extinction in finite time.

MSC:35K35, 35K65, 35B40.

1 Introduction

In this paper, we consider the following nonlinear degenerate parabolic equation:

{ u t = div ( | u | p ( x , t ) 2 u ) + b ( x , t ) | u | q a 0 u , ( x , t ) Ω × ( 0 , T ) = Q T , u ( x , t ) = 0 , ( x , t ) Ω × ( 0 , T ) = Γ T , u ( x , 0 ) = u 0 ( x ) , x Ω ,
(1.1)

where Q T =Ω×(0,T], Γ T denotes the lateral boundary of the cylinder Q T , a 0 >0. It will be assumed throughout the paper that the exponent p(x,t) is continuous in Q= Q T ¯ with logarithmic module of continuity

1< p = inf ( x , t ) Q p(x,t)p(x,t) p + = sup ( x , t ) Q p(x,t)<,
(1.2)
z=(x,t) Q T ,ξ=(y,s) Q T ,|zξ|<1, | p ( z ) p ( ξ ) | ω ( | z ξ | ) ,
(1.3)

where

lim sup τ 0 + ω(τ)ln 1 τ =C<+.

Model (1.1) may describe some properties of electro-rheological fluids which change their mechanical properties dramatically when an external electric field is applied. The variable exponent p in model (1.1) is a function of the external electric field | E | 2 , which satisfies the quasi-static Maxwell equations div( ε 0 E + p )=0, Curl( E )=0, where ε 0 is the dielectric constant in vacuum and the electric polarization p is linear in E , i.e., p =λ E . For more complete physical background, the readers may refer to [13].

These models include parabolic or elliptic equations which are nonlinear with respect to gradient of the thought solution and with variable exponents of nonlinearity; see [48] and references therein. Besides, another important application is the image processing where the anisotropy and nonlinearity of the diffusion operator and convection terms are used to underline the borders of the distorted image and to eliminate the noise [911]. When p is a fixed constant, authors in [1214] studied extinctions in finite time, blowing-up in finite time of solutions. Due to the lack of homogeneity and the gap between norm and modular, some methods in [1214] fail in solving our problems. In order to overcome some difficulties, we have to search for some new methods and techniques. To the best of our knowledge, there are only a few works about parabolic equations with variable exponents of nonlinearity. In [4], applying Galerkin’s method, Antontsev and Shmarev obtained the existence and uniqueness of weak solutions with the assumption that the function a(u) in div(a(u) | u | p ( x ) 2 u) was bounded. In the case when the function a(u) in div(a(u) | u | p ( x ) 2 u) might be not upper bounded, the authors in [15, 16] applied the method of parabolic regularization and Galerkin’s method to prove the existence of weak solutions. In this paper, we apply energy methods and Gronwall inequalities to prove that the solution vanishes in finite time. Moreover, we obtain the critical exponent of extinction in finite time.

The outline of this paper is the following. In Section 2, we shall introduce the function spaces of Orlicz-Sobolev type, give the definition of a weak solution to the problem; Section 3 will be devoted to the proof of the extinction of the solution obtained in Section 2.

2 Preliminaries

We will state some properties of variable exponent spaces and give the definition of a weak solution to the problem. Let us introduce the Banach spaces.

L p ( x , t ) ( Q T ) = { u ( x , t ) | u  is measurable in  Q T , A p ( ) ( u ) = Q T | u | p ( x , t ) d x d t < } , u p ( ) = inf { λ > 0 , A p ( ) ( u / λ ) 1 } ; V t ( Ω ) = { u | u L 2 ( Ω ) W 0 1 , 1 ( Ω ) , | u | L p ( x , t ) ( Ω ) } , u V t ( Ω ) = u 2 , Ω + u p ( ) Ω ; W ( Q T ) = { u : [ 0 , T ] V t ( Ω ) | u L 2 ( Q T ) , | u | L p ( x , t ) ( Q T ) , u = 0  on  Γ T } , u W ( Q T ) = u 2 , Q T + u p ( x ) , Q T

and denote by W ( Q T ) the dual of W( Q T ) with respect to the inner product in L 2 ( Q T ).

For the sake of simplicity, we first state some results about the properties of the Luxemburg norm.

Lemma 2.1 [17, 18]

For any u L p ( x ) (Ω),

( 1 ) u p ( x ) < 1 ( = 1 ; > 1 ) A p ( ) ( u ) < 1 ( = 1 ; > 1 ) ; ( 2 ) u p ( x ) < 1 u p ( x ) p + A p ( ) ( u ) u p ( x ) p ; u p ( x ) 1 u p ( x ) p A p ( ) ( u ) u p ( x ) p + ; ( 3 ) u p ( x ) 0 A p ( ) ( u ) 0 ; u p ( x ) A p ( ) ( u ) .

Lemma 2.2 [17, 18]

If pC( Ω ¯ ), 1< p < p + <, then L p ( x ) (Ω) and W 1 , p ( x ) (Ω) are uniformly convex. Hence they are reflexive.

Lemma 2.3 ([17, 18], Poincaré’s inequality)

There exists a constant C=C( p ± ,|Ω|)>0 such that for any u W 0 1 , p ( x ) (Ω),

u p ( x ) C u p ( x ) .

Definition 2.1 A function uW( Q T ) L (0,T; L (Ω)) is called a weak solution of problem (1.1) if every test-function

ξZ= { η ( x , t ) | η W ( Q T ) L ( 0 , T ; L 2 ( Ω ) ) , η t W ( Q T ) }

and for every t 1 < t 2 (0,T], the following identity holds:

t 1 t 2 Ω [ u ξ t | u | p ( x , t ) 2 u ξ + b ( x , t ) | u | q ξ a 0 u ξ ] dxdt= Ω uξdx | t 1 t 2 .
(2.1)

3 Main results and their proofs

Our main results read as follows.

Theorem 3.1 Assume that p(x,t) satisfies conditions (1.2)-(1.3) and 0<q<1, b L 1 (0,T; L (Ω)), a 0 >0, then problem (1.1) has at lease one weak solution u(x,t) in the sense of Definition  2.1.

Proof Define the operator

Lu,φ= Ω | u | p ( x , t ) 2 u φ b ( x , t ) | u | q φ + a 0 u φ dx+ Ω u t φdx,φ V t (Ω).

According to [4, 15], we know that { φ k (x)} V + (Ω)={u(x)|u L 2 (Ω) W 0 1 , 1 (Ω),|u| L p + } such that n = 1 V n is dense in V + (Ω) with

V n =Span{ φ 1 , φ 2 ,, φ n }.

We construct the approximate solution u ( m ) (x,t) as follows:

u ( m ) (x,t)= i = 1 m C k ( m ) (t) φ k (x),

where the coefficient C k ( m ) (t) may be obtained by solving the following identities:

L u ( m ) , φ k =0,k=1,2,,m.
(3.1)

Problem (1.1) generates the systems of m ODE

{ ( C k m ( t ) ) = G k ( t , C 1 m ( t ) C m m ( t ) ) , C k m ( 0 ) = Ω u 0 ( x ) φ k d x , k = 1 , 2 , , m .
(3.2)

If all the conditions in Theorem 3.1 hold, the functions G k are continuous in all arguments. □

In order to prove this theorem, we need the following lemmas.

Lemma 3.1 The approximate solutions u ( m ) of problem (1.1) satisfy the following inequality on the interval:

u ( m ) ( , τ ) 2 , Ω 2 + Q τ | u ( m ) | p ( x , t ) dxdtC.

Proof By Peano’s theorem, for every finite m, system (3.1) has solutions C i m , i=1,2,,m, on the interval (0, T 0 ).

Multiplying each of equations (3.1) by C k m (t) and summing over k=1,2,,m, we arrive at the relations

1 2 u ( m ) L 2 ( Ω ) 2 | t = 0 t = τ + Q τ | u ( m ) | p ( x , t ) d x d t = Q τ b | u ( m ) | q u ( m ) d x d t a 0 Q τ | u ( m ) | 2 d x d t C 0 0 τ ( Ω | b | 2 1 q d x ) d t + a 0 0 τ Ω | u ( m ) | 2 d x d t a 0 Q τ | u ( m ) | 2 d x d t C 0 0 τ Ω | b | 2 1 q d x d t .

This completes the proof of Lemma 3.1. □

The rest of the argument is similar to that in [4], we omit it here. In order to prove the locally extinction of weak solutions, we need to prove that the solution remains bounded.

Theorem 3.2 Assume that the conditions in Theorem  3.1 hold, then the solution of problem (1.1) remains bounded, i.e.,

u L ( Q T ) K(T)= { ( 1 q ) 0 T b L ( Q T ) d t + u 0 L ( Ω ) 1 q } 1 1 q .

Proof For any fixed M>0, define

u M =max { M , min { u , M } } .

For any integer k N , we choose u M 2 k 1 as a test-function in (2.1). Letting t 2 =t+h, t 1 =t, with t,t+h(0,T) in (2.1), we have

1 2 k t t + h d d t ( Ω u M 2 k d x ) d t + t t + h Ω ( 2 k 1 ) u M 2 ( k 1 ) | u M | p ( x , t ) d x d t = t t + h Ω b ( x , t ) u M 2 k 1 | u M | q d x d t a 0 t t + h Ω u M 2 k d x d t .

Multiplying the above identity by 1 h , letting h 0 + and applying Lebesgue’s dominated convergence theorem, we have that for every t(0,T),

1 2 k d d t Ω u M 2 k d x + Ω ( 2 k 1 ) u M 2 ( k 1 ) | u M | p ( x , t ) d x = Ω b ( x , t ) u M 2 k + q 1 d x a 0 Ω u M 2 k d x u M 2 k 2 k + q 1 2 k b 2 k 1 q 2 k 1 q a 0 u M 2 k 2 k .

We have

d d t u M 2 k u M 2 k q 2 k b 2 k 1 q 2 k 1 q a 0 u M 2 k .

By Gronwall’s inequality, we get

u M 2 k { ( 1 q ) 0 T b 2 k 1 q 2 k 1 q d t + u 0 1 q } 1 1 q .
(3.3)

Letting k+ in (3.3), we have

u M , Ω { ( 1 q ) 0 T b ( , t ) , Ω d t + u 0 , Ω 1 q } 1 1 q K(T),

We choose M>K(T), then u M =u a.e. in Q T . □

Theorem 3.3 (Extinction) (N>2)

Suppose 2 N N + 2 < p < 2 N 1 N + 1 .

  1. (1)

    1< p < p + < ( N 1 ) p N p N + p , then there exists T 1 T 1 ( p ± ,N,q,|Ω|, | b | L 2 1 q , a 0 ) such that every nonnegative solution of problem (1.1) vanishes in finite time, i.e.,

    lim t T 1 u L 2 ( Ω ) =0;
  2. (2)

    1< p < ( N 1 ) p N p N + p < p + < N p N p + p N , then there exists T 2 T 2 ( p ± ,N,q,|Ω|, | b | L 2 1 q , a 0 ) such that every nonnegative solution of problem (1.1) vanishes in finite time, i.e.,

    lim t T 2 u L r ( Ω ) =0,withr= p + ( N p N p ) + ( N + 1 ) p p >2.

Theorem 3.4 (Extinction) (N>2)

Assume that 2 N 1 N + 1 < p < p + < N p N p + p N <2, then there exists T 3 T 3 ( p ± ,N,q,|Ω|, | b | L 2 1 q , a 0 ) such that every nonnegative solution of problem (1.1) vanishes in finite time, i.e.,

lim t T 3 u L r ( Ω ) =0,with1<r=1+ p + ( N p ) N p ( p + 1 ) p <2.

Proof Case 1. 2 N N + 2 < p < 2 N 1 N + 1 , 1< p < p + < ( N 1 ) p N p N + p .

Choosing u as a test-function in (2.1), we have

1 2 0 t Ω d d t | u | 2 d x d t + 0 t Ω | u | p ( x , t ) d x d t = 0 t Ω b | u | 1 + q d x a 0 0 t Ω u 2 d x , 0 t Ω | u | p ( x , τ ) d x d τ Lemma 2.1 C 0 min { | u | p ( , t ) p | u | p ( , t ) p + } 0 t Ω | u | p ( x , τ ) d x d τ 00 C 0 min { M p + p , 1 } | u | p p + 0 t Ω | u | p ( x , τ ) d x d τ 00 C 1 | u | p p + .
(3.4)

Applying the embedding theorem W 1 , p L 2 , we have

( Ω u 2 d x ) 2 C 0 ( Ω | u | p d x ) 1 p , 0 t Ω b | u q + 1 | d x ( 0 t Ω b 2 1 q d x d t ) 1 q 2 ( 0 t Ω | u | 2 d x d t ) q + 1 2 , F ( t ) = 0 t Ω | u | 2 d x d t .
(3.5)

Moreover, we have

F (t)+ C 0 F ( t ) p + 2 + a 0 F(t) C 1 F ( t ) q + 1 2 .

This conclusion follows from Gronwall’s inequality.

Case 2. 2 N N + 2 < p < 2 N 1 N + 1 , 1< p < ( N 1 ) p N p N + p + < p + < N p N p N + p .

Choosing u m as a test-function with m=1+ p + ( N p ) N p ( p + 1 ) p , we have

1 2 0 t Ω d d t | u | m + 1 d x d t + ( m 1 ) 0 t Ω u m | u | p ( x , t ) d x d t = 0 t Ω b | u | m + q d x a 0 0 t Ω u m + 1 d x .
(3.6)

Applying | u | M, m1(α1) p 0 and Lemma 2.1, we get that

( m 1 ) 0 t Ω u m 1 | u | p ( x , τ ) d x d τ C 0 0 t Ω | u α | p ( x , τ ) d x d τ C 0 min { | u α | p ( , t ) p | u α | p ( , t ) p + } C 0 min { M p + p , 1 } | u α | p p + C 1 | u α | p p + .

Applying the embedding theorem W 1 , p L N p N p , we have

( Ω u α N p N p d x ) N p N p C 0 ( Ω | u α | p d x ) 1 p , 0 t Ω b | u q + m | d x ( 0 t Ω b m + 1 1 q d x d t ) 1 q m + 1 ( 0 t Ω | u | m + 1 d x d t ) q + 1 m + q .
(3.7)

Similarly as Case 1, letting α= ( m + 1 ) ( N p ) N p , we have

F (t)+F ( t ) p + ( N p ) N p + a 0 F(t) C 1 F ( t ) q + m 1 + m ,

with F(t)= 0 T Ω | u | m + 1 dxdt.

Case 3. 2 N 1 N + 1 < p < p + < N p N p + p N <2.

We choose u m as a test-function. We have

m=1+ p + ( N p ) N p ( p + 1 ) p .

The rest of the argument is the same as Case 2, we omit it here. □

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Acknowledgements

The work was supported the Natural Science Foundation of Jilin Province (Grant No. 201215044). We are very grateful to the anonymous referees for their valuable suggestions that improved the article.

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Correspondence to Peng Sun.

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The authors declare that they have no competing interests.

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Three authors collaborated in all the steps concerning the research and achievements presented in the final manuscript.

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Sun, P., Liu, M. & Cao, C. Extinction in finite time of solutions to the nonlinear diffusion equations involving p(x,t)-Laplacian operator. Bound Value Probl 2013, 259 (2013) doi:10.1186/1687-2770-2013-259

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Keywords

  • nonlinear diffusion equations
  • p(x,t)-Laplacian operator
  • extinction