Mathematical Computation December 2013, Volume 2, Issue 4, PP.98-104
Critical Exponent for a Porous Medium Equation with Inhomogeneous Density Xie Li 1, 2 1. 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: xielimath@gmail.com
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
pc , for m p pc no solutions that are global in time, while if p pc
there have
global solutions for the initial value being sufficiently small, as well, blow-up solutions for the large data. Keywords: Critical Exponent; Blow up; Porous Medium Equation; Cauchy Problem
1 INTRODUCTION In 1966, Fujita [1] proved the initial-value problem of the semilinear equation
ut u u p ,
x
N
,t 0,
that if 1 p pc 1 2 / N , then the problem does not have non-trivial non-negative global solutions, whereas if p pc , there have both global ( with small initial data ) and non-global (with large initial data) solutions. The critical situation of p pc belongs to the blow-up case, proved later by Weissler [2]. pc is called the critical Fujita exponent. 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 m p ( x)ut u q( x)u , ( x, t ) N u ( x, 0) u0 ( x), x
where p m 1 ,
N
(0, T )
N
(1)
( x) bounded and smooth, has a power-like decay at infinity, i.e.
c1 (1 | x |) ( x) c2 (1 | x |) with some positive constants c1 , c2 and 0 2 . 0 q ( x) C (
,
(2)
) , satisfies
c1 (1 | x |) s q( x) c2 (1 | x |) s
(3)
2 PRELIMINARIES In this paper, the Cauchy problem of (1) was discussed with the initial value u( x,0) u0 ( x) 0, where u0 L ( N ) 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 - 98 www.ivypub.org/MC
N
1) u ( x, t ) is non-negative and continuous in 2) u C ([0, T ) : Lloc ( m
N
(0, T );
)) and q( x)u p L1 ((0, T ) : L1loc (
N
)) ;
3) The identity T
0
N
( x)u t dxdt
holds every test function C0 (
T
0
N
(0, T ));
N
u m dxdt
T
0
N
q( x)u p dxdt 0
4) u ( x, 0) u0 . 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 u0 L ( support, there has at least one local solution to (1).
N
) with compact
Theorem 2 (comparison principle) Let u1 ( x, t ), u2 ( x, t ) be two solutions of (1) in 0, T , and the corresponding N initial value satisfies 0 u1 ( x,0) u2 ( x,0) L ( ) with compact support. Then
0 u1 ( x, t ) u2 ( x, t ),
a.e.( x, t )
N
(0, 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
Tmax sup{T 0; sup‖u(, t ‖ ) } t[0,T )
. If 0 Tmax , it is said that the solution u blows up in finite time. If Tmax , it is indicated that the problem admits a global solution. The main result of this paper is as follows Theorem 3 Let m 1 , (2) and (3) hold with 0 2, pc m (2 s) / ( N ) . Then, if m p pc , all solutions of (1) blow up in finite time, whereas if p pc , there have non-trivial global solutions as well as nonglobal solutions.
3 THE PROOF OF THEOREM 3 In 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 (t ) C p , in the distributional sense, where C 0 is a constant and p 1 . Then is an increasing function, and there has a finite T 0 such that (t ) as t T . Proof. See [5]. Next two important lemmas were proved. Lemma 2 Let m p pc , (2) and (3) hold with 0 2, s 2, then every solution to problem (1) blows up in finite time. Proof. We consider a cut-off function C 2 [0, ) , satisfying
1 0 z 1 0 1, 0, ( z ) 0 z 2, N ( z ) ( | z |). and we put, for 0 1 and z , It is easy to see that
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2
(t ) Cp [ln(1 )]1 p satisfies
| | C ,
| | C 2 .
Define now
(t ) ( x)u( x, t ) ( x)dx. N
From the definition of weak solution, we have the following identity in the distribution sense
(t ) u m qu p u m | | qu p V1 V2 .
(Here and below g stands for
N
(4)
g ( x)dx ).
Applying Holder’s inequality we have
V1
qu p
m p
p m pm p m (q ) | |
p m p
.
(5)
Consequently, it was derived from (4) and (5) m p 2
(t ) C ( , )V V2 , where
C ( , ) [ | |
p p m
By the assumptions on q and , it can be computed that
(q )
m p m
]
p m p
.
C ( , ) C 2 N ( N s ) m/ p , where C is a positive constant independent of . Set 1 2 N ( N s)m / p , then we get m
(t ) V2 p (C 1 V2
p m p
).
(6)
Using Holder’s inequality again we get
V2 Owing to the conditions on , q and
u
, we have p 1 p 1 1 p
q
C
1 p
p
p 1 p 1 1 p q
p s N p 1
| x| 2
( | x |)
.
s p p 1
dx,
using polar coordinate, the above integral can be estimated as follows
2 s p N 0 C ln(1 ), p 1 s p s p N 1 2 s p s p p 1 p 1 N N 1 s p ( | x |) dx C ( r ) dr C p 1 [(2 ) p 1 ], N 0. | x| 2 0 s p p 1 N p 1
Hence - 100 www.ivypub.org/MC
2 s p C[ln(1 )]1 p , N 0 p 1 q s p C 2 , N 0, p 1 where 2 [ N ( p 1) p s] , C 0 independent of . 1 p
p 1 p 1 1 p
Therefore,
2 1 p p C [ln(1 )] , s p N ( p 1) 0 V2 Cp 2 , s p N ( p 1) 0,
(7)
First we consider the case s p N ( p 1) 0 . It follows from (7)} that m
CV2 p
( p m ) 2 p
(
pm
C
1
p m 2 p
).
(8)
Now we have
1 provided m p m
pm 2 0 , p
(9)
2s , and consequently, N 1
p m
2
p 0 as 0. Since u0 is non-trivial, we can choose small in such a way that (0) 0. Since (0) is a non-increasing function of , we can pick smaller, if necessary, in order to have (0) 2C p m
1
p m 2 p
.
(10)
Thus (8), (9) and (10) imply
Cp , for some ( maybe different ) positive constant C. Consequently, blows up in finite time. Next we consider the case s p N ( p 1) 0. 2
(11)
Combining (6) and (7), it was found out that m p 2
1
(t ) V (C1 C2
p m
2 [ln(1 )]
( p m )(1 p ) p
)
(12)
Based on the fact that is an non-increasing function of , it can be pointed out from (12) that there has C 0, 1 such that
2
for t 0. Thus,
blows up in finite time.
(t ) Cp [ln(1 )]1 p
(13)
This completes the proof. Remark Assume that u0
is large enough such that for some - 101 www.ivypub.org/MC
, p m (0) 2C
1
p m 2 p
and
1
2
(0) 2C [ln(1 )] p m
( p m )( p 1) p
, then (11) and (13) hold simultaneously. Thus it can be concluded that
u blows up in finite time for sufficiently large u provided that p m. 0
It remains to check that blow-up occurs also in the borderline case p pc 1
2m . N
pm 2 0 , (10) cannot be achieved by choosing small. It is proved next that is not p t bounded in time. This will imply (10) at some 0 large and (11) for t t , yielding the same conclusion. In this case
1
0
Lemma 3 Let p pc , (2)and (3) hold with 0 2, s 2, then every solution to problem (1) blows up in finite time. Proof. In this case
[ p s N ( p 1)]( p m) p( N 2) ( N s)m 0. Suppose that lim (t ) is unbounded in time, then the conclusion holds apparently. Otherwise, u L for all 1
0
t 0 , and there has an M 0 that satisfies
( x)u( x, t )dx M for all t 0. It was proved that (14) is impossible.
(14)
Let us return to (5). It is cleat that
If
q( x)u
V1 u( x, t ) | ( x) | dx 0 as 0. p
( x, t )dx , then, by (5) and (6), we can choose small enough, such that
(t ) If, on the contrary,
q( x)u
p
1 q( x)u p ( x, t ) ( x)dx. 2
( x, t )dx , then by taking, if necessary, a smaller , we have (t ) 1.
In any case,
1 q( x)u p ( x, t ) ( x)dx} 2 for any t 0 and small. Passing to the limit as 0, we get
(t ) min{1,
(t ) g (t ) : min{1,
1 q( x)u p ( x, t )dx} 2
which, upon integration, gives t
(t ) (0) g (t )dt. 0
(15)
Recalling some facts about the purely diffusive problem
( x)Gt G m , x G( x, 0) G0 ( x), x
N
,t (0, ),
N
,
(16) It is know that under the assumption (2), problem (16) has a unique bounded solution for any given 1 N G0 L ( N ), G0 0, cf. [8] and [9]. If, moreover, G0 L ( , ( x)dx), then (see[10]}) - 102 www.ivypub.org/MC
t ‖G(, t ) U E (, t ‖ ) L 0,
as t ,
(17)
where U E is a member of the explicit family
U E (, t ) C ( E )t F (| x | t ) with similarity exponents
(18)
1 , ( N ) . Where N (m 1) 2 m
m1) F ( ) (C 2 )1/( ,
m 1 . m(2 )[ N (m 1) 2 m ]
The constant C ( E ) 0 above is chosen in such a way that
E ‖U E‖L1 (
N
,| x| dx )
‖G‖L1 (
N
, ( x ) dx )
.
While U E solves the singular problem
| x | U t U m , x N ,t (0, ), N | x | U ( x, 0) E ( x), x . 1 N Given a nontrivial u0 0, we choose a nontrivial G0 0 with G0 u0 , G0 L ( , ( x)dx). Clearly, u( x, t ) N is a super solution of problem (16) and by comparison u G in [0, T ). On the other hand, it is know by (17) that for t t0 ( ). Choosing
q( x)u
p
G( x, t ) [ F (| x | t ) ] t ,
c0
1, we have
( x, t )dx q( x)[ F (| x | t ) ]p t p dx ct p N m (t | x |) m [ F (| x |) ]p dx
ct p N m (t0 | x |) m [ F (| x |) ]p dx ct p N m ,
c 0.
It is ended by observation that this last exponent is p N m 1 when p pc , so by (15) we have (t ) unbounded. At the end of this section, it is shown that above the Fujita exponent pc there have global solutions. Lemma 4 Let m 1, p pc , (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.
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ut u u1
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AUTHORS Xie Li was born in Sichuan province in 1978. PhD in School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu. The author’s major field is partial differential equation. She once worked as a teacher in China West Normal University in Sichuan province.
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