Critical Exponent for a Porous Medium Equation with Inhomogeneous Density.pdf

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Mathematical Computation

December 2013, Volume 2, Issue 4, PP.98-104

Critical Exponent for a Porous Medium

Equation with Inhomogeneous DensityXie Li 1, 21. School of Mathematical Science, University of Electronic Science & Technology of China, Chengdu, 610054, China.

2. College of Mathematic and Information, China West Normal University, Nanchong, 637002, China.

Email: [email protected]

Abstract

In this paper, the critical exponent for a porous medium equation with inhomogeneous density has been investigated. Then it was

proved that there has a critical exponent cp , for cm p p no solutions that are global in time, while if cp p there have

global solutions for the initial value being sufficiently small, as well, blow-up solutions for the large data.

Keywords: Cr iti cal Exponent; Bl ow up; Porous Medium Equation; Cauchy Problem

1 INTRODUCTIONIn 1966, Fujita [1] proved the initial-value problem of the semilinear equation

, , 0,p Ntu u u x t that if1 1 2 /cp p N , then the problem does not have non-trivial non-negative global solutions, whereas

if ,cp p there have both global ( with small initial data ) and non-global (with large initial data) solutions. The

critical situation of cp p belongs to the blow-up case, proved later by Weissler[2]. cp is called the critical Fujitaexponent.

The problem of determining critical Fujita exponent is an interesting one in the general theory of blowing-up

solutions to different nonlinear evolution equations of mathematic physics. See the survey [3] where a full list of

references is given. Also see [1],[2],[9] etc.

In this paper, the following Cauchy problems which take the form were investigated

0

( ) ( ) , ( , ) (0, ),

( ,0) ( ),

m p N

t

N

x u u q x u x t T

u x u x x

(1)where 1p m , ( )x bounded and smooth, has a power-like decay at infinity, i.e.

1 2(1 | |) ( ) (1 | |)c x x c x (2)

with some positive constants 1 2,c c and 0 2 .

0 ( ) ( )Nq x C , satisfies

1 2(1 | |) ( ) (1 | |)s sc x q x c x

(3)

2 PRELIMINARIESIn this paper, the Cauchy problem of (1) was discussed with the initial value 0( ,0) ( ) 0,u x u x where

0 ( )Nu L with compact support. The solution to (1) is the weak solution in the following sense.

Definition 1

It was said that a function ( , )u x t is a weak solution to problem (1). If


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1) ( , )u x t is non-negative and continuous in (0, );N T

2) ([0, ) : ( ))m Nlocu C T L and1 1( ) ((0, ) : ( ))p Nlocq x u L T L ;

3) The identity

0 0 0( ) ( ) 0

N N N

T T Tm p

tx u dxdt u dxdt q x u dxdt holds every test function 0( (0, ));

NC T

4)0

( ,0) .u x u

A supersolution [or subsolution] is similarly defined with equality 3) replaced by [or ].

The local existence theorem and comparison principle are as follows

Theorem 1(local existence theorem) Let (2), (3) hold with 0 2, .s Then for 0 ( )Nu L with compact

support, there has at least one local solution to (1).

Theorem 2 (comparison principle) Let 1( , ),u x t 2( , )u x t be two solutions of (1) in 0, ,T and the corresponding

initial value satisfies1 2

0 ( ,0) ( ,0) ( )Nu x u x L with compact support. Then

1 20 ( , ) ( , ), . .( , ) (0, ).Nu x t u x t a e x t T

The proof is standard, and the details were omitted here.

For the solution ,u x t of (1), we using to denote the maximum existence time, that is

[0, )

sup{ 0; sup ( , ) }max

t T

T T u t

.

If 0 ,maxT it is said that the solution u blows up in finite time. If maxT , it is indicated that the problemadmits a global solution.

The main result of this paper is as follows

Theorem 3 Let 1m , (2) and (3) hold with 0 2, (2 ) / ( )cp m s N . Then, if cm p p , allsolutions of (1) blow up in finite time, whereas if cp p , there have non-trivial global solutions as well as non-global solutions.

3 THE PROOFOFTHEOREM 3In this section, the proof of Theorem 3 is available. First of all, the following lemma is iterated.

Lemma 1 Let be a positive continous function which satisfies the following inequality ( ) ,pt C in thedistributional sense, where 0C is a constant and 1p . Then is an increasing function, and there has a finite

0T such that ( )t as .t T

Proof. See [5].

Next two important lemmas were proved.

Lemma 2Let ,cm p p (2) and (3) hold with 0 2, 2,s then every solution to problem (1) blows up infinite time.

Proof. We consider a cut-off function2[0, )C , satisfying

0 10 1, 0, ( )

1

0 2,

zz

z

and we put, for 0 1 and Nz , ( ) ( | |).z z

It is easy to see that


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12( ) [ln(1 )]

p pt C

satisfies

2| | , | | .C C Define now

( ) ( ) ( , ) ( ) .N

t x u x t x dx

From the definition of weak solution, we have the following identity in the distribution sense

1 2( ) | | .m p m pt u qu u qu V V (4)

(Here and below g stands for ( )N g x dx ).

Applying Holders inequality we have

1 | | ( ) .

p mp mm p

p p m p mp

V qu q

(5)Consequently, it was derived from (4) and (5)

2 2( ) ( , ) ,

m

pt C V V where

( , ) [ | | ( ) ] .

p m p m

p m p m pC q

By the assumptions on q and , it can be computed that

2 ( ) /

( , ) ,N N s m p

C C

where Cis a positive constant independent of .

Set1

2 ( ) / pN N s m , then we get

1

2 2( ) ( ).

m p m

p pt V C V

(6)

Using Holders inequality again we get

1

1

1 1

2 .

pp

pp pV u q

Owing to the conditions on , q and , we have1

1 1 1 1

| | 2( | |) ,

p p s s pN

p p p p

xq C x dx

using polar coordinate, the above integral can be estimated as follows

121 1

| | 2 0( | |) ( )

s p s pN

p p

xx dx C r dr

1 1

2ln(1 ), 0

1

1[(2 ) ], 0.

1

1

s p s pN N

p p

s pC N

p

s pC N

s p pN

p

Hence


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2

11

1

1 1

2[ln(1 )] , 0

1

, 0,1

pp

p

p p

s pC N

pq

s pC N

p

where 2 [ ( 1) ] ,N p p s 0C independent of .

Therefore,

2

1

2

2[ln(1 )] , ( 1) 0

, ( 1) 0,

p p

p

C s p N pV

C s p N p

(7)

First we consider the case ( 1) 0s p N p .

It follows from (7)} that

21 2

( )

2

( ).

p mm p m

p mp p pCV C

(8)

Now we have

1 2 0

p m

p

, (9)

provided2 s

m p mN

, and consequently,

1 2

0 0.

p m

p as

Since 0u is non-trivial, we can choose small in such a way that (0) 0. Since (0) is a non-increasingfunction of, we can pick smaller, if necessary, in order to have

1 2

(0) 2 .

p m

p m pC

(10)Thus (8), (9) and (10) imply

2 ,pC

(11)for some ( maybe different ) positive constant .C Consequently, blows up in finite time.

Next we consider the case ( 1) 0.s p N p

Combining (6) and (7), it was found out that

1

( )(1 )

2 1 22( ) ( [ln(1 )] )

m p m p

p mp pt V C C

(12)

Based on the fact that is an non-increasing function of, it can be pointed out from (12) that there has 0,C 1 such that

12( ) [ln(1 )]p pt C

(13)

for 0.t Thus, blows up in finite time.

This completes the proof.

Remark Assume that 0u is large enough such that for some ,1 2

(0) 2

p m

p m pC

and


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1

( )( 1)2

(0) 2 [ln(1 )] ,

p m p

p m pC

then (11) and (13) hold simultaneously. Thus it can be concluded that

ublows up in finite time for sufficiently large 0u provided that .p m

It remains to check that blow-up occurs also in the borderline case2

1 .cm

p p

N

In this case 1 2 0p m

p

, (10) cannot be achieved by choosing small. It is proved next that is not

bounded in time. This will imply (10) at some 0t

large and (11) for0

t t , yielding the same conclusion.

Lemma 3Let ,cp p 2and3hold with 0 2, 2,s then every solution to problem (1) blows upin finite time.

Proof. In this case

[ ( 1)]( ) ( 2) ( ) 0.p s N p p m p N N s m

Suppose that0

lim ( )t

is unbounded in time, then the conclusion holds apparently. Otherwise, 1u L for all

0t , and there has an 0M that satisfies

( ) ( , )x u x t dx M (14)for all 0.t It was proved that (14) is impossible.

Let us return to (5). It is cleat that

1 ( , ) | ( ) | 0 0.V u x t x dx as

If ( ) ( , )pq x u x t dx , then, by (5) and (6), we can choose small enough, such that

1( ) ( ) ( , ) ( ) .

2

pt q x u x t x dx

If, on the contrary, ( ) ( , )pq x u x t dx , then by taking, if necessary, a smaller , we have ( ) 1.t

In any case,

1( ) min{1, ( ) ( , ) ( ) }

2

pt q x u x t x dx

for any 0t and small. Passing to the limit as 0, we get

1( ) ( ) : min{1, ( ) ( , ) }

2pt g t q x u x t dx

which, upon integration, gives

0( ) (0) ( ) .

t

t g t dt (15)Recalling some facts about the purely diffusive problem

0

( ) , , (0, ),

( ,0) ( ), ,

m N

t

N

x G G x t

G x G x x

(16)It is know that under the assumption (2), problem (16) has a unique bounded solution for any given

0 0) 0,( ,N

G L G

cf. [8] and [9]. If, moreover,1

0 ( , ( ) ),N

G L x dx then (see[10]})


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( , ) ( , ) 0, ,E L

t G t U t as t

(17)where

EU is a member of the explicit family

( , ) ( ) (| | )EU t C E t F x t (18)

with similarity exponents1

,

( 1) 2N m m

( )N . Where

2 1/( 1)( ) ( ) ,mF C 1

.(2 )[ ( 1) 2 ]

m

m N m m

The constant ( ) 0C E above is chosen in such a way that

1 1( ,| | ) ( , ( ) )

.N NE L x dx L x dxE U G While EU solves the singular problem

| | , (0, ),

| | ( ,0) ( ) .

,

,

m N

t

N

x U U x t

x U x E x x

Given a nontrivial 0 0,u we choose a nontrivial 0 0G with

10 0 0, ( , ( ) ).NG u G L x dx Clearly, ( , )u x t

is a super solution of problem (16) and by comparison u G in [0, ).N T On the other hand, it is know by (17)that

( , ) [ (| | ) ] , 0G x t F x t t c for 0( ).t t Choosing 1, we have

0

( ) ( , ) ( )[ (| | ) ] ( | |) [ (| |) ]

( | |) [ (| |) ] , 0.

p p p p N m m p

p N m m p p N m

q x u x t dx q x F x t t dx ct t x F x dx

ct t x F x dx ct c

It is ended by observation that this last exponent is 1p N m when ,cp p so by (15) we have ( )t

unbounded.

At the end of this section, it is shown that above the Fujita exponent cp there have global solutions.

Lemma 4 Let 1, ,cm p p (2) and (3) hold. If the initial value is small enough then the solution is global in time.

The global existence of (1) can be studied by using the similarity solution, and the method is similar to that in [5],

however, the details are omitted here.

Combining the above four lemmas, the proof of Theorem 3 is complete.

REFERENCES

[1] de Pablo, G. Reyes and A. Sanchez, The Cauchy problem for a nonhomogeneous heat equation with reaction, Discrete andContinuous Dynamical Systems, 33:2 (2013), 643-662.

[2] V. Martynenko and A. F. Tedeev, The Cauchy problem for a quasilinear parabolic equation with a souce and nonhomogeneousdensity, Zh. Vychisl. Mat. Fiz., 47 (2007), 245-255. (Russian); translation in Comput. Math. Phys., 47(2007), 238-248.

[3] F. B. Weissler, Existence and nonexistence of global solutions for a semilinear heat equation, Israel J.Math, 38 (1981), 29-40.[4] G. Reyes and J. L. Vazquez, Long time behavior for the inhomogeneous PME in a medium with slowly decaying density,

Commun. Pure Appl., 8 (2009), 493-508.

[5] G. Reyes and J. L. Vazquez, The inhomogeneous PME in several space dimensions. Existence and uniquence of finite energysolutions, Commun. Pure Appl. Anal., 7 (2008), 1275-1294.

[6] H. A. Levine, The role of critical exponents in blowup theorems, SIAM Rev., 32 (1990), 262-288.[7] H. Fujita, On the blowing up of solutions of the Cauchy problem for 1tu u u , J. Fac. Sci. Univ. Tokyo Sect. I 13

(1966), 109-124.[8] S. Kamin, R. Kersner and A. Tessi, On the Cauchy problem for a class of parabolic equations with variable density, Atti Accad.


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Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur., 9 (1998),279-298.

[9] V. A. Galaktionov, Blow-up for quasilinear heat equations with critical Fujita's exponents, Proc. R. Soc. Edinb. A 124(1994), 517-525.

[10] Y. W. Qi, The critical exponents of parabolic equations and blow-up in N , Proc. R. Soc. Edinb. A 128(1998), 123-136.AUTHORS

Xie Liwas born in Sichuan province in 1978. PhD in School of Mathematical Sciences, University of Electronic

Science and Technology of China, Chengdu. The authors major field is partial differential equation.

She once worked as a teacher in China West Normal University in Sichuan province.

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Critical Exponent for a Porous Medium Equation with Inhomogeneous Density.pdf