Energy Decay Result In A Quasilinear Parabolic System With Viscoelastic Term
Mohamed Ferhat
y, Ali Hakem
zReceived 27 June 2015
Abstract
In this paper we consider a quasilinear parabolic system of the form:
A(t)jutjm 2ut Lu+ Zt
0
g(t s)Lu(s)ds= 0
in a bounded domain , A(t) is a bounded and positive de…nite matrix, Rn(n 1), initial data (u0; u1)are given functions belonging to suitable spaces andga continuously di¤erentiable decaying function. We use the lemma of Mar- tinez to establish a general decay result. This improves the result obtained by Messaoudi and Tellab [3].
1 Introduction
In this paper we consider
A(t)jutjm 2ut Lu+ Z t
0
g(t s)Lu(s)ds= 0; m >2; (1) subjected to the following boundary conditions
u(x; t) = 0; x2@ ; t 0; (2)
and initial conditions
u(x;0) =u0(x); ut(x;0) =u1(x); x2 ; (3) where is a bounded open subset ofRn(n 1),A(t)is a bounded and positive de…nite matrix,
Lu= div(Mru) = XN i;j=1
@
@xi ai;j(x)@u
@xi :
Mathematics Sub ject Classi…cations: 35L05, 35L15, 35L70, 93D15.
yDepartment of Mathematics, USTO University, 31000 Oran, Algeria
zLaboratory ACEDP, Djillali Liabes University, 22000 Sidi Bel Abbes, Algeria
56
The matrix M = (ai;j(x)), where ai;j 2 C1( ), is symmetric and there exists a constant a0 > 0 such that for all x 2 and = ( 1; 2; :::::; N) 2 RN, we have PN
i;j=1ai;j(x) j i a0j j2:Also, a(u(t); (t)) =
XN i;j=1
ai;j(x)@u(t)
@xj
@ (t)
@xi
dx= Z
Mru(t):r (t)dx;
and
a1= max XN
i=1
kai;jk21
! :
The values ofuare taken inRn andA2C(R+)is a bounded square matrix satisfying c0j j2 (A(t) ; ) c1j j2; 8t2R+; 2Rn: (4) The initial data (u0; u1)are given functions belonging to suitable spaces andg a con- tinuously di¤erentiable decaying function. To motivate our work, let us recall some results regarding quasilinear parabolic system. This type of equation arises from a variety of mathematical models in engineering and physical sciences. For example, in the study of a heat conduction in materials with memory, the classical Fourier’s law is replaced by the following form (cf. [7]):
q= dru Z t
1
k(x; t)u(x; s)ds;
where u is the temperature, d is the di¤usion coe¢ cient and the integral term rep- resents the memory e¤ect in the material. The study of this type of equations has drawn a considerable attention and many results have been obtained, see ([2, 3, 6, 9]).
From a mathematical point of view one would expect that the integral term should be dominated by the leading term in the equation. Messaoudi and Tellab [3] studied the following system
A(t)jutjm 2ut u+ Z t
0
g(t s) u(s)ds= 0;
with the same conditions in (2)–(3) and obtained an energy decay result although the memory term makes more complex situation. Berrimi and Messaoudi [6] showed that ifAsatis…es
(A(t) ; ) c0j j2; 8t2R+; 2Rn
then the solutions with small initial energy decay exponentially form= 2and polyno- mially ifm >2. Very recently for a framework of blow-up in …nite time Liu and Chen [2] studied the following system
A(t)jutjm 2ut u+ Z t
0
g(t s) u(s)ds=jujp 2u;
with the same conditions in (2)–(3) and proved a blow-up result for both positive and negative initial energy under suitable conditions ongandp. Motivated by the previous
works, in the present paper we investigate problem (1) in which we generalize the results obtained in [3], supposing new conditions with which the stability is assured, by using the lemma of Martinez.
Our work is organized as follows. In section 2, we present some preliminaries and some lemmas. In section 3, the decay property is derived. Our result improves the one in Messaoudi and Tellab [3].
2 Preliminary Results
In this section, we present some material needed for the proof of our main result. For the relaxation function gwe assume
(A0) g:R+!R+ is a boundedC1 function satisfying g(0)>0; 1
Z 1
0
g(s)ds=l <1;
and there exists a nonincreasing di¤erentiable function :R+!R+ satisfying g0(t) (t)g(t); t 0;
Z +1 0
(s)ds= +1:
(A1) We also assume that
2 m 2n
n 2 if n 3; m 2; if n= 1;2:
REMARK 1. The same as in [3], there are many functions satisfying (A0). Exam- ples of such functions are
g1(t) =e (t+1) ; 0< 1 and g2(t) = (1 +t) ; < 1:
We will also be using the embedding
H01( ),!Lp( ); H01( ),!Lm( ):
LEMMA 1 ([2]). LetE :R+!R+ be a nonincreasing function and :R+ !R+
be aC2 increasing function with (0) = 0andlimt!1 (t) = +1. Assume that there exists c >0for which
Z T S
E(t) 0(t)dt cE(S); 8S 0:
Then
E(t) E(0)e (R0t (s)ds); 8t 0;
where and! are positive constants.
LEMMA 2 (Sobolev-Poincaré’s inequality). Let2 m n2n2. The inequality kukm cskruk2 for u2H01( )
holds with some positive constantcs.
LEMMA 3 ([3]). Foru(:; t)2H01( ), we have Z Z t
0
g(t s)(u(x; t) u(x; s))ds
2
dx (1 l)cs2
a0(g u)(t);
where csis the Poincaré constant and lis given in (A1), and (g u)(t) =
Z t 0
g(t s) Z
a(u(x; t) u(x; s); u(x; t) u(x; s))dxds:
3 Asymptotic Behavior
In this section, we consider the energy decay of solutions associated to the system (1)–(3). Similarly as in [7] we give a de…nition of a weak solution of the system (1)–(3):
DEFINITION 1. A weak solution of (1)–(3) is a functionu(x; t)such that u(x; t)2C [0; T); [H01( )]n \C1((0; T); [Lm( )]n);
which satis…es Z t 0
Z
A(s)jus(s)jm 2us(x; s) (x; s)dsdx+ Z t
0
Z
Lu(x; s) (x; s)dsdx +
Z s 0
Z t 0
Z
g(t ) (x; s)Lu(x; )d dxds= 0;
for allt2[0; T]and 2C [0; T); [H01( )]n :
REMARK 2. Similar to ([3, 9]), we assume the existence of a solution. For the linear case (m = 2), one can easily establish the existence of a weak solution by the Galerkin method. In the one-dimensional case(n= 1), the existence is established, in a more general setting, by Yin [9].
Now we de…ne the "modi…ed" energy equation related with problem (1)–(3) by E(t) = 1
2(g u)(t) +1 2 1
Z t 0
g(s)ds a(u(t); u(t)):
LEMMA 4. Letu(x; t)be the solution of (1)–(3). Then the energy equation satis…es E0(t) 1
2(g0 u)(t) 1
2g(t)a(u(t); u(t)) Z
A(t)jut(t)jmdx:
PROOF. By multiplying (1) byut(t), and integrating over we get Z
A(t)jut(t)jmdx 1 2
d
dta(u(t); u(t)) + Z Z t
0
g(t s)Mru(s)rut(t)dsdx= 0: (5) Note that
a(u(t); ut(t)) = 1 2
d
dta(u(t); u(t));
following the ideas of [10], we can obtain Z t
0
g(t s) Z
Mru(s)rut(t)dxds
= XN i;j=1
Z t 0
Z
g(t s)ai;j(x)@u(s)
@xj
@ut(t)
@xi dxds
= XN i;j=1
Z t 0
Z
g(t s)ai;j(x)@u(t)
@xi
@ut(t)
@xi
dxds XN
i;j=1
Z t 0
Z
g(t s)ai;j(x) @u(t)
@xi
@ut(s)
@xj
@ut(t)
@xi dxds
= 1
2 Z t
0
g(t s) d
dta(u(t); u(t)) ds 1
2 Z t
0
g(t s) d
dta(u(t) u(s); u(t) u(s)) ds
= 1
2 d dt
Z t 0
g(t s)a(u(t); u(t) ds 1
2 d dt
Z t 0
g(t s)a(u(t) u(s); u(t) u(s)) ds 1
2g(t)a(u(t); u(t)) +1 2
Z t 0
g0(t s)a(u(t) u(s); u(t) u(s))ds
= 1
2 d
dt(g u)(t) +1
2(g0 u)(t) +1 2
d
dt a(u(t); u(t)) Z t
0
g(s)ds 1
2g(t)a(u(t); u(t)); (6)
where
(g u)(t) = Z t
0
g(t s)a(u(x; t) u(x; s); u(x; t) u(x; s))ds:
By (5) and (6), we obtain E0(t) 1
2(g0 u)(t) 1
2g(t)a(u(t); u(t)) Z
A(t)jut(t)jmdx 0:
THEOREM 1. Let(u0; u1)2(H01( ))2be given. Suppose that (A0)–(A1) and (4) hold. Then there exist two positive constants wandK, depending on the initial data andc0 for which the solution of (1)–(3) satis…es
E(t) Ke wR0t (s)ds:
PROOF. From now on, we denote by ci various positive constants which may be di¤erent at di¤erent occurrences. We multiply the equation (1) by (t)u, integrate over
(S; T), and use the boundary conditions to get Z T
S
Z
(t)A(t)jut(t)jm 2ut(t)u(t)dxdt Z T
S
(t)a(u(t); u(t))dtdx +
Z T S
(t) Z
Mru(t) Z t
0
g(t s)ru(s)dsdx= 0: (7)
We then estimate
Z
(t):Mru(t) Z t
0
g(t s)ru(s)dsdx
= Z
(t):M Z t
0
g(t s)(ru(t) ru(s))ru(t)dsdx Z t
0
g(s)ds: (t)a(u(t); u(t)); (8)
by substituting (8) in (7) and adding the following term in (7) 1
2 Z T
S
(t)(g u)(t) 1 2
Z T S
(t)(g u)(t): (9)
So (7) becomes Z T
S
(t)E(t)dt =
Z T S
Z
(t)A(t)jut(t)jm 2ut(t)u(t)dxdt Z T
S
Z
(t)M:
Z t 0
g(t s)(ru(t) ru(s)):ru(t)dsdxdt +
Z T S
(t)(g u)(t)dt: (10)
By Lemma 4, equation (4), the boundedness of A(t)and condition (A1), we see that, 8 >0;
Z
A(t)jut(t)jm 2ut(t)u(t)dx
Z
ju(t)jmdx+c Z
jut(t)jmdx cms kru(t)km+c
Z
jut(t)jmdx cms 2E(0)
l
m 2 2
E(t) c c0
E0(t):
Then Z T
S
Z
A(t)jut(t)jm 2ut(t)u(t)dxdt cms 2E(0) l
m 2 2 ! Z T
S
E(t) (t)dt c
c0 Z T
S
E0(t) (t)dt; 8 >0: (11) We also have
Z Z t 0
M g(t s)(ru(t) ru(s))ru(t)dsdx
= XN i;j=1
Z t 0
g(t s) Z
aij(x)@u(t)
@xj
@u(t)
@xi
@u(s)
@xi dxds XN
i;j=1
Z Z t 0
aij(x)@u(t)
@xj ds
2
dx
+1 XN
i;j=1
Z Z t 0
g(t s) @u(s)
@xi
@u(t)
@xi
ds
2
dx
a0
0
@max XN i;j=1
kaijk21 1
Aa(u(t); u(t)) + N 4a0
(1 l)(g u)(t)
a0 0
@ XN i;j=1
kaijk21 1
AE(t) + N
4a0 (1 l)(g u)(t); (12)
and using the fact that
j (t)(g u)(t)j= (t)(g u)(t) (g u)(t) ( E0(t));
we obtain
Z T S
(t)(g u)(t)dt Z T
S
( E0(t))dt CE(S): (13)
By combining (10)–(13), we easily deduce the following 8<
:1 cms 2E(0) l
m 2 2
2 4( + 1)
0
@max
i i N
XN i;j=1
kaijk21 1 A+ N
4a0 (1 l) 3 5
9=
; Z T
S
(t)E(t)dt c
c0
(0) +e E(S):
Finally, we get
Z T S
(t)E(t)dt CE(S); 8S 0;
by choosing , 2, small enough and by the hypothesisl < 1. By letting T go to in…nity, one can easily see that (A0) is satis…ed with (t) =Rt
0 (s)ds. This completes the proof.
Acknowledgment. The author thanks the editor and the referees for their remarks and suggestions to improve this paper.
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