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Finite Dimensional Global Attractor for Dissipative Schrödinger–Boussinesq Equations
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
205, 107]132 Ž1997.
AY965148
Finite Dimensional Global Attractor for Dissipative Schrodinger]Boussinesq Equations ¨ Li Yongsheng and Chen Qingyi Department of Mathematics, Huazhong Uni¨ ersity of Science and Technology, Wuhan, 430074, China Submitted by Colin Rogers Received April 19, 1995
In this paper the authors consider the initial boundary value problems of dissipative Schrodinger]Boussinesq equations and prove the existence of global ¨ attractors and the finiteness of the Hausdorff and the fractal dimensions of the attractors. Q 1997 Academic Press
1. INTRODUCTION In the laser and plasma physics under interaction of a nonlinear complex Schrodinger field and a real Boussinesq field, the dynamics is de¨ scribed by the equations ŽSBq., i c t q D c s fc ,
Ž 1.1.
f t t y D f q g D2f y D F Ž f . s D < c < 2 ,
Ž 1.2.
where c represents the complex Schrodinger field and f represents the ¨ real Boussinesq field Žsee w15]17, 19x.. The existence and the uniqueness of a global solution of Ž1.1., Ž1.2. have been studied in w8, 9, 11, 12x. In w10x the authors considered the one-dimensional dissipative ŽSBq. which involves a three-order nonlinear term in the Schrodinger equation and ¨ proved the existence of the global attractor and the finiteness of dimension of the attractor. In this paper, we consider the dissipative SBq equations i c t q D c q i ac y fc s f ,
Ž 1.3.
f t t y D f q g D2f y b D f t y D F Ž f . y D < c < 2 s g ,
Ž 1.4.
107 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
108
LI AND CHEN
c Ž 0, x . s c 0 Ž x . , f Ž 0, x . s f 0 Ž x . , f t Ž 0, x . s f 1 Ž x . ,
x g V,
Ž 1.5. c Ž t , x . s 0, f Ž t , x . s 0, D f Ž t , x . s 0,
t g R, x g V , Ž 1.6.
where a , b , and g are positive constants, F is a sufficiently smooth real function with F Ž0. s 0, and V ; R n Ž n F 3. is a bounded domain with a smooth boundary V. We arrange this paper as follows. First we establish time-uniform a priori estimates which imply the existence of solutions and bounded absorbing sets. Then we construct the global attractor of the system as the v-limit of one of those bounded absorbing sets Žsee w18x.. Finally, we show that the global attractor has finite Hausdorff and fractal dimensions. We introduce the following standard notations. We denote the spaces of complex-valued and the real-valued functions by the same symbols. W m, p Ž V . and W0m, p Ž V . are the usual Sobolev spaces. H m Ž V . s W m, 2 Ž V ., H0m Ž V . s W0m, 2 Ž V .. We denote by 5 ? 5 p the norm in L p Ž V ., especially 5 ? 5 s 5 ? 5 2 . We denote by Ž?, ? . the inner product in L2 Ž V ., by ² ? , ? : the dual product of Hy1 Ž V . and H01 Ž V ., and by 5 ? 5 m, p the norm of W m, p Ž V .. C b Ž I; E . denotes the space of continuous and bounded functions on I ; R with values in the Banach space E. For u g C b ŽRq; E ., or u g L`ŽRq; E ., A u A E denotes its norm, especially A u A s A u A L2 , A u A m, p s A u A W m , p . C is a generic constant and may assume different values in different formulae. We shall frequently use the Gargliardo]Nirenberg inequality Žsee w4x.: l 5 D j u 5 p F C 5 u 5 1y 5 D m u 5 lr , q
u g Lq l W m , r Ž V . ,
where 1 p
s
j n
ql
ž
1 r
y
m n
/
q
1yl q
,
1 F q, r F `, j is an integer, 0 F j F m, and jrm F l F 1. If m y j y nrr is a nonnegative integer, then the inequality holds for jrm F l - 1. We also use the short norm 5 =u 5 of u in H01 Ž V . and the equivalent norm 5 D u 5 of u in H 2 l H01 Ž V .. 2. THE TIME-UNIFORM A PRIORI ESTIMATES It is convenient to introduce a transformation u s f t q df , where d is a positive constant to be chosen later. The problems Ž1.3. ] Ž1.6. are
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
equivalent to i c t q D c q i ac y fc s f ,
Ž 2.1.
f t q df s u ,
Ž 2.2.
t g R, x g V ,
u t y du y b D u q d 2f y Ž 1 y db . D f q g D2f yD F Ž f . y D < c < 2 s g ,
Ž c , f , u . Ž 0, x . s Ž c 0 , f 0 , u 0 . Ž x . ,
Ž 2.3.
x g V,
Ž 2.4.
c s 0, f s D f s 0, u s 0, t g R, x g V ,
Ž 2.5.
where u 0 s df 0 q f 1. We set E0 s H01 = H01 = Hy1 Ž V . , E1 s H 2 l H01 = H 2 l H01 = L2 Ž V . . E2 s H 3 l H01 = H 3 l H01 = H01 Ž V . . Then E2 ¨ E1 ¨ E0 with compact imbedding. In this section we are going to establish time-uniform a priori estimates of solutions Ž c , f , u . in E0 and then in E1. 2.1. The Time-Uniform Estimates in E0 LEMMA 2.1. satisfies
Suppose that f g L`ŽRq; L2 Ž V ... Then c g L`ŽRq; L2 Ž V ..
5 c Ž t . 5 F 5 c 0 5 exp Ž ya t . q 2
2
A f A2
a2
Ž 1 y exp Ž ya t . . .
Ž 2.6.
Proof. Taking the inner product of Ž2.1. with c , then taking imaginary parts we obtain 1 d 2 dt
5 c 5 2 q a 5 c 5 2 s Im Ž f , c . F 5 f 5 5 c 5 F
By Gronwall inequality we have the lemma.
a 2
5c 52 q
5 f 52 2a
.
110
LI AND CHEN
We assume that in the following F satisfies the conditions lim inf < f <ª`
lim inf
H0f F Ž z . dz
f2
f F Ž f . y c 0 H0f F Ž z . dz f2
< f <ª`
G 0,
G 0,
Ž 2.7. for some c 0 ) 0.
Ž 2.8.
Under the above assumptions, ;« 1 , « 2 ) 0, there are constants C Ž « 1 ., C Ž « 2 . ) 0 such that f
H0 F Ž z . dz G y« f 1
f F Ž f . y c0
2
y CŽ «1 . ,
f
H0 F Ž z . dz G y«
2f
2
y CŽ «2 . .
Ž 2.9. Ž 2.10.
LEMMA 2.2. Suppose that F satisfies Ž2.7., Ž2.8., f g L`ŽRq; L2 Ž V .., f t , g g L`ŽRq; Hy1 Ž V ... If Ž c 0 , f 0 , u 0 . g E0 , then the solution Ž c , f , u . g L`ŽRq; E0 .. Proof. Taking the inner product of Ž2.1. with yŽ c t q ac ., then taking real parts we obtain 1 d 2 dt
5 =c 5 2 q a 5 =c 5 2 q Re Ž fc , c t . q a Ž f , < c < 2 . s Re Ž f , yc t y ac . ,
Noting that Re Ž fc , c t . s
1 d 2
Re Ž f , yc t y ac . s y
1
Ž f , < c < 2 . y 2 Ž ft , < c < 2 . , dt d dt
Re Ž f , c . y a Re Ž f , c . q Re² f t , c : ,
we obtain 1 d 2 dt
Ž 5 =c 5 2 q 2 Ž f , < c < 2 . q 2 Re Ž f , c . . q a 5 =c 5 2 q a Ž f , < c < 2 . q a Re Ž f , c . s
1 2
Ž ft , < c < 2 . q Re² ft , c : .
Ž 2.11.
111
FINITE DIMENSIONAL GLOBAL ATTRACTOR
Now taking the inner product of Ž2.3. with ŽyD .y1u , by inserting Ž2.2. we obtain 1 d 2 dt
ž
2 2 2 2 5 u 5y1 , 2 q d 5 f 5y1, 2 q Ž 1 y db . 5 f 5
f
qg 5 =f 5 2 q 2
HVH0 F Ž z . dz dx
/
2 2 2 2 2 3 y d 5 u 5y1 , 2 q b 5 u 5 q d 5 f 5y1, 2 q d Ž 1 y bd . 5 f 5 q dg 5 =f 5
HVf F Ž f . dx q d Ž f , < c < .
qd
2
s Ž Ž yD .
y1 r2
g , Ž yD .
y1r2
u . y Ž ft , < c < 2 . .
Ž 2.12.
Let 2 2 2 H0 Ž t . s 2 5 =c 5 2 q 4 Ž f , < c < 2 . q 4 Re Ž f , c . q 5 u 5y1 , 2 q d 5 f 5y1, 2
f
q Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 q 2
HVH0 F Ž z . dz dx,
Ž 2.13.
then 2Ž2.11. q Ž2.12. implies 1 d 2 dt
2 H0 Ž t . q 2 a 5 =c 5 2 q Ž 2 a q d . Ž f , < c < 2 . q 2 a Re Ž f , c . y d 5 u 5y1 ,2 2 2 q b 5 u 5 2 q d 3 5 f 5y1 , 2 q d Ž 1 y db . 5 f 5
q dg 5 =f 5 2 q d
HVf F Ž f . dx
s 2 Re² f t , c : q Ž Ž yD .
y1 r2
g , Ž yD .
y1r2
u ..
Ž 2.14.
Let l1 be the first eigenvalue of yD with a homogeneous Dirichlet boundary condition on V, and choose
d F min
½
1 2b
,
bl1 4
5
,
112
LI AND CHEN
2 then 5 =u 5 2 G l1 5 u 5 2 G l12 5 u 5y1 , 2 . We estimate the indefinite sign terms as
<4 Re Ž f , c . < F 2 5 f 5 2 q 2 5 c 5 2 , 2 Re² f t , c :< F
Ž Ž yD . y1 r2 g , Ž yD . y1r2 u . Žf,
2
a
a
2 5 f 5y1 ,2 q
2 F d 5 u 5y1 ,2 q
2 1 4d
5 =c 5 2 , 2 5 g 5y1, 2,
F C 5 f 5 4 5 c 5 4 5 c 5 F C 5 =f 5 5 c 5 5r4 5 =c 5 3r4 F « 3 5 =c 5 2 q « 4 5 =f 5 2 q C Ž « 3 , « 4 . 5 c 5 10 , ;« 3 ) 0, « 4 ) 0,
Ž 2.15.
where we take « 3 s arŽ2Ž2 a q d .., «4 s dgrŽ2Ž2 a q d ..; in Ž2.9., Ž2.10. we take « 1 s Ž1 y db .rŽ4 c 0 ., « 2 s Ž1 y db .r4, then f
HVH0 F Ž z . dz dx G y
1 y db 4 c0
f
5 f 5 2 y C1 ,
HVf F Ž f . dx G c HVH0 F Ž z . dz dx G y
1 y db
0
4
Ž 2.16.
5 f 5 2 y C2 . Ž 2.17.
From Ž2.14. and the above estimates we obtain d dt
H0 Ž t . q b 0 H0 Ž t . F K 0
2 10 where b 0 s min a , d , c 0 d 4 , K 0 s Ž2ra .5 f t 5y q Ž1r 1, 2 q C 5 c 5 2 q 4d .5 g 5y1 , 2 q C is bounded in R , and we denote its bound also by K 0 . By Gronwall inequality we obtain
H0 Ž t . F H0 Ž 0 . exp Ž yb 0 t . q
K0
b0
Ž 1 y exp Ž yb 0 t . . .
Ž 2.18.
Now we take « 3 s 1r8, « 4 s 1r16 in Ž2.15., « 1 s Ž1 y db .r4 in Ž2.9., and then we obtain 2 2 2 5 =c 5 2 q 5 u 5y1 , 2 q d 5 f 5y1, 2 q
F H0 Ž 0 . exp Ž yb 0 t . q
K0
b0
2
1
Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 2
Ž 1 y exp Ž yb 0 t . .
q C 5 c 5 10 q 2 5 f 5 2 q C, for all t G 0. So we have the lemma.
1
Ž 2.19.
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
2.2. The Time-Uniform Estimates in E1 LEMMA 2.3. Suppose that f g L`ŽRq; H01 Ž V .., f t g L`ŽRq; L2 Ž V .., and g g L`ŽRq; Hy1 Ž V ... Moreo¨ er, suppose that F satisfies Ž2.7., Ž2.8. and < F X Ž z . < F C1 < z < r q C 2 ,
1 F r - 2.
Ž 2.20.
If Ž c 0 , f 0 , u 0 . g E1 , then the solution Ž c , f , u . g L`ŽRq; E1 .. Moreo¨ er, t
H0 5 =u 5
2
dt F Ct q C.
Proof. First, taking the inner product of Ž2.1. with D c t q a D c , we obtain 1 d 2 dt
5 D c 5 2 q a 5 D c 5 2 y Re Ž fc , D c t . y a Re Ž fc , D c . s Re Ž f , D c t . q a Re Ž f , D c . ,
Ž 2.21.
Since yRe Ž fc , D c t . s y
d dt
Re Ž fc , D c . q Re Ž f t c , D c . q Re Ž fc t , D c . ,
while by Ž2.2. and Ž2.1., ReŽ f t c , D c . s ReŽ uc , D c . y d ReŽ fc , D c ., Re Ž fc t , D c . s Re Ž yi f Ž f y D c y i ac q fc . , D c . s Im Ž f f , D c . y a Re Ž fc , D c . q Im Ž f 2c , D c . , and therefore yRe Ž fc , D c t . s y
d dt
Re Ž fc , D c .
q Re Ž uc , D c . y Ž a q d . Re Ž fc , D c . q Im Ž f f , D c . q Im Ž f 2c , D c . , the first term in the right-hand side of Ž2.21. equaling Re Ž f , D c t . s
d dt
Re Ž f , D c . y Re Ž f t , D c . .
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LI AND CHEN
Thus Ž2.21. can be rewritten as 1 d 2 dt
Ž 5 D c 5 2 y 2 Re Ž fc , D c . y 2 Re Ž f , D c . . q a 5 D c 5 2 y Ž 2 a q d . Re Ž fc , D c . y a Re Ž f , D c . q Re Ž uc , D c . q Im Ž ff , D c . q Im Ž f 2c , D c . q Re Ž f t , D c . s 0.
Ž 2.22.
Next, taking the inner product of Ž2.3. with u , using Ž2.2., we obtain 1 d 2 dt
Ž 5 =u < 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f I 2 q g 5 D f 5 2 . y d 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 q Ž =F Ž f . , =u . q Ž = < c < 2 , =u . s ² g , u : .
Ž 2.23.
Define H1 Ž t . s 5 D c 5 2 y 2 Re Ž fc , D c . y 2 Re Ž f , D c . q 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f 5 2 q g 5 D f 5 2 . Then Ž2.22. q Ž2.23. implies 1 d 2 dt
H1 Ž t . q a 5 D c 5 2 y Ž 2 a q d . Re Ž fc , D c . y a Re Ž f , D c . y 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 q Re Ž uc , D c . q Im Ž f f , D c . q Im Ž f 2c , D c . q Re Ž f t , D c . q Ž =F Ž f . , =u . q Ž = < c < 2 , =u . s ² g , u : .
Ž 2.24.
Since
Ž 2.26.
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
here we take « 5 s arŽ14Ž2 a q d .., «6 s ar14,
a 14
5Dc 5 2 q
b 8
5 =u 5 2 q C 5 c 5 2 5 =c 5 8
Ž = < c < 2 , =u .
a 14
a 14 a 14
5 D c 5 2 q C 5 f 5 24 5 f 5 24 5 D c 5 2 q C 5 f 5 2 5 =c 5 8 ,
5 D c 5 2 q C 5 ft 5 2
F 2 5 c 5 6 5 =c 5 3 5 =u 5 F C 5 c 5 6 5 c 5 1r4 5 D c 5 3r4 5 =u 5 F
<² g , u :< F
a 14
b 8
5Dc 5 2 q
b 8
5 =u 5 2 q C 5 c 5 2 5 =c 5 8 ,
2 5 =u 5 2 q C 5 g 5y1 ,2,
and since F Ž z . satisfies Ž2.20.,
Ž =F Ž f . , =u . F Ž C1 5 < f < r =f 5 q C2 5 =f 5 . 5 =u 5 F Ž C 5 f 5 6r 5 =f 5 6rŽ3y r . q C2 5 =f 5 . 5 =u 5 F C Ž 5 f 5 6r 5 f 5 Ž2y r .r4 5 D f 5 Ž rq2.r4 q 5 =f 5 . 5 =u 5 F
dg 2
5Df 5 2 q
b 8
5 =u 5 2 q C 5 =f 5 2 q C 5 =f 5 Ž6 rq4.rŽ6y r . ,
taking b 1 s min a , d 4 , noting that 5 =u 5 2 G l1 5 u 5 2 . Inserting the above estimates into Ž2.24. we have d dt
H1 Ž t . q
1 4
b 5 =u 5 2 q b 1 H1 Ž t . F K 1 ,
Ž 2.27.
where K 1 s K 1Ž5 =c 5, 5 =f 5, 5 =f 5, 5 f t 5, 5 g 5y1 , 2 . is a constant which is independent of t. By Gronwall inequality we obtain H1 Ž t . F H1 Ž 0 . exp Ž yb 1 t . q
K1
b1
Ž 1 y exp Ž yb1 t . . .
116
LI AND CHEN
Now we take « 5 s « 6 s 1r4 in Ž2.25. and Ž2.26., then we obtain 1 2
5 D c 5 2 q 5 u 5 2 q g 5 D f 5 2 F H1 Ž 0 . ey b 1 t q
K1
b1
q C Ž 5 f 5 , 5 =f 5 , 5 =c 5 . .
Moreover, < H1Ž t .< F C. From Ž2.27. and the above estimates we have 5 =u 5 2 F
4 K1
b
y
4b 1
b
H1 Ž t . y
4 d
b dt
H1 Ž t . F C y
4 d
b dt
H1 Ž t . . Ž 2.28.
Integrating Ž2.28. over w0, t x we have the lemma. Remark. In the cases of n s 1 and 2, the restriction r - 2 can be omitted. In fact, when n s 1, 5 f 5 ` F C 5 =f 5, the term <Ž =F Ž f ., =u .< can be estimated as
Ž =F Ž f . , =u . F Ž C1 5 < f < r =f 5 q C2 5 =f 5 . 5 =u 5 F
b 2
5 =u 5 2 q C Ž 5 f 5 `r q 1 . 2 5 =f 5 2 .
And when n s 2, 5 f 5 p F C 5 =f 5 for all 2 F p - `,
Ž =F Ž f . , =u . F Ž C1 5 < f < r D f 5 q C2 5 =f 5 . 5 =u 5 F Ž C 5 f 5 4rr 5 =f 5 4 q C2 5 =f 5 . 5 =u 5 F C Ž 5 f 5 4rr 5 f 5 1r4 5 D f 5 3r4 q 5 =f 5 . 5 =u 5 F
dg 2
5Df 5 2 q
b 8
5 =u < 2 q C 5 =f 5 2 q C 5 f 5 84 rr r5 5 f 5 2r5 .
3. EXISTENCE OF SOLUTIONS AND BOUNDED ABSORBING SETS In Section 2 we have established the time-independent a priori estimates of solutions Ž c , f , u . in E0 and in E1. We can apply the obtained results and Galerkin method to prove the existence of a global solution of problems Ž2.1. ] Ž2.5.. The proof is in the standard manner and so it is omitted. Therefore we have THEOREM 3.1. Suppose that F, f, g satisfy the same conditions as in Lemma 2.2. Then for e¨ ery Ž c 0 , f 0 , u 0 . g E0 , problems Ž2.1. ] Ž2.5. ha¨ e a solution Ž c , f , u . g L`ŽRq; E0 .. Moreo¨ er, if f g Cb ŽRq; L2 Ž V .., f t , g g Cb ŽRq; Hy1 Ž V .., then Ž c , f , u . g Cb ŽRq; E0 ...
117
FINITE DIMENSIONAL GLOBAL ATTRACTOR
Accordingly, problems Ž2.1. ] Ž2.5. have a bounded absorbing set in E0 . THEOREM 3.2. Suppose that F, f, g satisfy the same conditions as in Lemma 2.2. Then problems Ž2.1. ] Ž2.5. possess a bounded absorbing set B0 in E0 . Proof. Let B Ž0, R . be the ball in E0 of radius R G 1 centered at 0. For any initial Ž c 0 , f 0 , u 0 . g B Ž0, R ., 5 c Ž0.5 2 F Ž1rl1 .5 =c Ž0.5 2 F Ž1rl1 . R 2 . By Lemma 2.1 we see that there exists a t 1Ž R . ) 0 such that 5 c Ž t .5 2 F 1 q Ž1ra 2 .A f A 2 for all t ) t 1Ž R .. By Ž2.13., H0 Ž0. F CR10 . From Ž2.19., when t G t 1Ž R ., 1 1 2 2 2 5 =c 5 2 q 5 u 5y1 Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 , 2 q d 5 f 5y1, 2 q 2 2 F CR10 exp Ž yb 0 t . q C 1 q
ž
5
1
a
A f A2 2
/
q
K0
b0
,
where K0 s
2
5
2 2 A f t Ay1 , 2 q C Ž1 q A f A . q
1
2 A g Ay1, 2 q C.
a 4d Therefore there exists a t 0 Ž R . G t 1Ž R . such that 1 1 2 2 2 5 =c 5 2 q 5 u 5y1 Ž 1 y db . 5 c 5 2 q g 5 =f 5 2 , 2 q d 5 f 5y1, 2 q 2 2 FC 2q
ž
1
a
A f A2 2
5
/
q
K0
b0
' R 02 ,
for all t G t 0 Ž R . .
Hence B0 s B Ž0, R 0 . is a bounded absorbing set in E0 . However, we are unable to prove the uniqueness of solution in E0 . But we can show that the solution exists uniquely in E1. Indeed, we have THEOREM 3.3. Suppose that F, f, g satisfy the conditions in Lemma 2.3. Suppose further that F Y is continuous in R. Then for e¨ ery Ž c 0 , f 0 , u 0 . g E1 , problems Ž2.1. ] Ž2.5. ha¨ e a unique solution Ž c , f , u . g L`ŽRq; E1 .. Moreo¨ er, if f g Cb ŽRq; H01 Ž V .., f t g Cb ŽRq; L2 Ž V .., g g Cb ŽRq; Hy1 Ž V .., then Ž c , f , u . g Cb ŽRq; E1 .. Proof. The proof of the existence of a solution is also in the standard manner. So we need only prove the uniqueness of a solution, which is a consequence of the following theorem. THEOREM 3.4. Let F, f, g be the same as in Theorem 3.3 and let SŽ t . be the solution operator, i.e., let Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . be the solution of Ž2.1. ] Ž2.5. with initial data Ž c 0 , f 0 , u 0 . g E1. Then SŽ t . is continuous from E1 into itself, uniformly on any compact inter¨ al w0, T x.
118
LI AND CHEN
Proof. Let Ž c k , f k , u k . Ž k s 1, 2. be two solutions with initial data Ž c k 0 , f k 0 , u k 0 . g E1. Set Ž c , f , u . s Ž c 1 y c 2 , f 1 y f 2 , u 1 y u 2 ., Ž c 0 , f 0 , u 0 . s Ž c 10 y c 20 , f 10 y f 20 , u 10 y u 20 .. Due to the time-uniform a priori estimates, Ž c k , f k , u k . Ž k s 1, 2. is in L`ŽRq; E1 ., and so is Ž c , c , u .. Moreover, Ž c , f , u . satisfies the equations i c t q D c q i ac s fc 1 q f 2 c ,
Ž 3.1.
f t q df s u ,
Ž 3.2.
u t y du y b D u q d 2f y Ž 1 y db . D f q g D2f sD
1 X
žH
0
F Ž f 2 q tf . f dt q D Ž cc 1 q c 2 c .
/
Ž 3.3.
with homogeneous boundary conditions and initial data Ž c , f , u .< ts0 s Ž c 0 , f 0 , u 0 .. Taking the inner product of Ž3.1. with D2c , then taking imaginary parts we obtain 1 d 5 D c 5 2 q a 5 D c 5 2 s Im Ž D Ž fc 1 q f 2 c . , D c . 2 dt F C Ž 5 c 1 5 22, 2 q 5 f 2 5 22, 2 . Ž 5 D c 5 2 q 5 D f 5 2 . . Ž 3.4. Taking the inner product of Ž3.3. with u and using Ž3.2., we have 1 d Ž 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =u 5 2 q g 5 D f 5 2 . 2 dt y d 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 s y Ž = Ž F X Ž f 2 q tf . f dt . , =u . q D Ž cc 1 q cc 2 . , u .
ž
/
Ž 3.5.
Let M0 s sup Rq 5 f 1 5 ` q 5 f 2 5 `4 , M1 s sup < z < F M 0< F X Ž z .<, < F Y Ž z .<4 , then =
1 X
ž žH
0
F Ž f 2 q tf . f dt , =u
/ /
F f
1 Y
ž žH Ž ž žH F
0
q =f
f 2 q tf . = Ž f 2 q tf . dt , =u
/ /
1 X
0
F Ž f 2 q tf . dt , =u
/ /
F M1 Ž 5 f 5 ` Ž 5 =f 1 5 q 5 =f 2 5 . q 5 =f 5 . 5 =u 5 F
1 4
ž D Ž cc
b 5 =u 5 2 q CM12 5 =f 5 2 q CM12 Ž 5 =f 1 5 2 q 5 =f 2 5 2 . 5 D f 5 2 , 1
q cc 2 . , u
/
2 2 F d 5 u 5 2 q C Ž 5 c 1 5 22 , 2 q 5 c 2 5 2, 2 . 5Dc 5 .
119
FINITE DIMENSIONAL GLOBAL ATTRACTOR
From Ž3.5. we have 1 d 2 dt
Ž 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =u 5 2 q g 5 D f 5 2 . 1 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 4 F C 5Dc 5 2 q C 5Df 5 2.
Ž 3.6.
Define H Ž t . s 5 D c 5 2 q 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f 5 2 q g 5 D f 5 2 , then Ž3.4. q Ž3.6. implies d dt
H Ž t . F KH Ž t . ,
where K is a constant which does not depend on t. By the Gronwall inequality we have H Ž t . F H Ž 0. e K t ,
t G 0.
Hence SŽ t . is uniformly continuous on t g w0, T x. THEOREM 3.5. Suppose that F, f, g satisfy the same conditions as those in Theorem 3.3. Then Ž2.1. ] Ž2.5. possess a bounded absorbing set B1 in E1. The proof is analogous to that of Theorem 3.2 and so it is omitted.
4. EXISTENCE OF GLOBAL ATTRACTOR 4.1. The Nonlinear Group Let f g H01 Ž V ., g g Hy1 Ž V . be independent of t. As shown in Theorem 3.3, problem Ž2.1. ] Ž2.5. has a unique solution Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . g Cb ŽRq, E1 .. Since now the system in discussion is autonomous, SŽ t . Ž t g R. forms a group. In fact, we can define SŽ t . on the whole real line, since the changing of t into yt in the system does not change the local existence of the solution. Therefore SŽ t . Ž t g R. is a group on E1 and is uniformly continuous on any compact set in R. To construct the global attractor, we treat SŽ t . as a compact perturbation of a contraction mapping. More precisely, we decompose S Ž t . into two parts: SŽ t . s S1Ž t . q S2 Ž t . for all t G 0, where S1Ž t . is continuous in E1 and decays uniformly on a bounded subset of E1 , i.e., for any bounded set
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B ; E1 , supj g B 5 SŽ t . j 5 E1 tends to zero as t ª `, and where S2 Ž t . is uniformly compact in E1 for t large, i.e., for any bounded set B ; E1 , there exists a t 0 G 0 such that D t G t 0 SŽ t . B is relatively compact. If it is the case, then for a bounded absorbing set B1 ; E1 Žas in Theorem 3.5., the v-limit set of B1 , A s v Ž B1 . s F s G 0D t G s S Ž t . B1 Žwhere the closure is with respect to the norm topology of E1 ., is the global attractor in E1 Žsee w18x.. For this purpose, we define S1Ž t . and S2 Ž t . in the following way. Let Ž c 0 , f 0 , u 0 . g E1 , and Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . be the solution of problems Ž2.1. ] Ž2.5.. Define S1Ž t .Ž c 0 , f 0 , u 0 . s Ž c 1 , f 1 , u 1 . to be the solution of the problems i c 1 t q D c 1 q i ac 1 s fc 1 ,
Ž 4.1.
f 1 t q df 1 s u 1 ,
Ž 4.2.
u 1 t y du 1 y b D u 1 q d 2f 1 y Ž 1 y db . D f 1 q g D2f 1 s 0, Ž 4.3. with initial data Ž c 1 , f 1 , u 1 .< ts0 s Ž c 0 , f 0 , u 0 . and homogeneous boundary conditions, then Ž c 2 , f 2 , u 2 . s S2 Ž t .Ž c 0 , f 0 , u 0 . s SŽ t .Ž c 0 , f 0 , u 0 . y S1Ž t .Ž c 0 , f 0 , u 0 . is the solution of problem i c 2 t q D c 2 q i ac 2 s fc 2 q f ,
Ž 4.4.
f 2 t q df 2 s u 2 ,
Ž 4.5.
u 2 t y du 2 y b D u 2 q d 2f 2 y Ž 1 y db . D f 2 q g D2f 2 s D FŽ f . q D
Ž 4.6.
with homogeneous initial data and homogeneous boundary conditions. Since the two systems are nonautonomous, S1Ž t . and S2 Ž t . are no longer Žsemi-.groups. 4.2. Decaying of S1Ž t . Uniformly on Bounded Sets of E1 LEMMA 4.1.
Let f g H01 Ž V ., g g Hy1 Ž V ., and Ž c 0 , f 0 , u 0 . g E1. Then
Ž1. 5 c 1Ž t .5 2 s 5 c 1Ž0.5 2 ey2 a t , Ž2. 5 =c 1Ž t .5 2 F Ž5 =c 1Ž0.5 2 q C 5 c 1Ž0.5 2 . eya t , where C s const. A f A 82.2 . Ž3. 5 D c 1Ž t .5 2 F Ž5 D c 1Ž0.5 2 q C 5 c 1Ž0.5 2 . eya t , where C s const. A f A 82.2 . Ž4. 5 u 1 5 2 q d 2 5 f 1 5 2 q Ž1 y db .5 =f 1 5 2 q g 5 D f 1 5 2 F Ž5 u 0 5 2 q d 2 5 f 0 5 2 q Ž1 y db .5 =f 0 5 2 q g 5 D f 1 5 2 . ey2 d t. Therefore S1Ž t . decays on bounded sets in E1.
FINITE DIMENSIONAL GLOBAL ATTRACTOR
121
Proof. Taking the inner product of Ž4.1. with c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 c 1 5 2 q a 5 c 1 5 2 s 0,
so we have Ž1.. Taking the inner product of Ž4.1. with yD c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 =c 1 5 2 q a 5 =c 1 5 2 s Im Ž =fc 1 , =c 1 . .
Since
a 2
5 =c 1 5 2 q C 5 f 5 82 , 2 5 c 1 5 2 ,
by Ž1. we have d dt
5 =c 1 5 2 q a 5 =c 1 5 2 F C 5 f 5 82, 2 5 c 1 5 2 F const. A f A 82, 2 5 c 1 Ž 0 . 5 ey2 a t .
By the Gronwall inequality we obtain 5 =c 1 5 2 F 5 =c 1 Ž 0 . 5 2 ey a t q
1
a
const. A f A 82, 2 5 c 1 Ž 0 . 5 ey a t Ž 1 y ey a t .
F Ž 5 =c 1 Ž 0 . 5 2 q C 5 c 1 Ž 0 . 5 2 . ey a t , so we have Ž2.. Taking the inner product of Ž4.1. with D2c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 D c 1 5 2 q a 5 D c 1 5 2 s Im Ž D fc 1 , D c 1 . q 2 Im Ž =f =c 1 , D c 1 . .
Since Im Ž D fc 1 , D c 1 . F 5 D f 5 5 c 1 5 ` 5 D c 1 5 F C 5 D f 5 5 c 1 5 1r4 5 D c 1 5 7r4 F
a 4
5 D c 1 5 2 q C 5 f 5 82, 2 5 c 1 5 2 ,
2 Im Ž =f =c 1 , D c 1 . F 2 5 =f 5 6 5 =c 1 5 3 5 D c 1 5 F C 5 f 5 2, 2 5 c 1 5 1r4 5 D c 1 5 7r4 F
a 4
5 D c 1 5 2 q C 5 f 5 82, 2 5 c 1 5 2 ,
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we obtain d dt
5 D c 1 5 2 q a 5 D c 1 5 2 F C 5 f 5 82, 2 5 c 1 5 2 F const. A f A 82, 2 5 c 1 Ž 0 . 5 ey2 a t .
Hence we have Equation Ž4. wave equations
Ž3.. follows from the classical results on the damped linear Žsee w18x..
4.3. Uniform Compactness of S2 Ž t . on Bounded Sets in E1 Since Ž c , f , u . and Ž c 1 , f 1 , u 1 . are uniformly bounded in E1 for t g Rq and any initial data Ž c 0 , f 0 , u 0 . in a bounded set B ; E1 , so is Ž c 2 , f 2 , u 2 . s Ž c , f , u . y Ž c 1 , f 1 , u 1 .. If we show that Ž c 2 , f 2 , u 2 . is uniformly bounded in E2 , for t g Rq, then by the compact imbedding of E2 into E1 , S2 Ž t . is uniformly compact for t g Rq in E1. LEMMA 4.2. Let f g H01 Ž V ., g g L2 Ž V ., B ; E1 be bounded. Then there exists a C that is independent of t such that 5 Ž c 2 , f 2 , u 2 . 5 E 2 F C,
; Ž c 0 , f 0 , u 0 . g B.
Proof. We need only show that 5 =D c 2 5 2 q 5 D u 2 5 2 q 5 =D f 2 5 2 F C. Taking the inner product of Ž4.4. with yD2 Ž c 2 t q ac 2 ., then by making use of a procedure analogous to that of Ž2.22. we obtain 1 d 2 dt
Ž 5 =D c 2 5 2 q 2 Re Ž D Ž fc 2 . , D c 2 . y 2 Re Ž =f , =D c 2 . . q a 5 =D c 5 2 q Ž 2 a q d . Re Ž D Ž fc 2 . , D c 2 . y a Re Ž =f , =D c 2 . q Re Ž = Ž uc 2 . , =D c 2 . q Im Ž = Ž f f . , =D c 2 . y Im Ž =f D c 2 , =D c 2 . q Im Ž = Ž f 2c 2 . , =D c 2 . s 0.
Ž 4.7.
The indefinite-sign terms in Ž4.7. can be estimated as Re Ž D Ž fc 2 . , D c 2 . F C 5 f 5 2, 2 5 c 2 5 22, 2 ,
a Re Ž =F , =D c 2 . F
a 10
5 =D c 2 5 2 q
5a 2
5 =f 5 2 ,
Re Ž = Ž uc 2 . , =D c 2 . F Ž 5 =u 5 5 c 2 5 ` q 5 u 5 4 5 =c 2 5 4 . 5 =D c 2 5 F
a 10
5 =D c 2 5 2 q C 5 =u 5 2 5 c 2 5 22, 2
FINITE DIMENSIONAL GLOBAL ATTRACTOR
123
Im Ž = Ž f f . , =D c 2 . F Ž 5 f 5 4 5 f 5 4 q 5 f 5 ` 5 =f 5 . 5 =D c 2 5 F
a 10
5 =D c 2 5 2 q C 5 f 5 2, 2 5 f 5 1, 2 ,
Im Ž =f D c 2 , =D c 2 . F 5 =f 5 4 5 D c 2 5 4 5 =D c 2 5 7r4 F C 5 f 5 2, 2 5 c 2 5 1r4 2, 2 5 =D c 2 5
F
a 10
2 5 =D c 2 5 2 q C 5 f 5 82, 2 5 c 2 5 2, 2,
Im Ž = Ž f 2c 2 . , =D c 2 . F C 5 f 5 22, 2 5 c 2 5 2, 2 5 =D c 2 5 F
a 10
4 2 5 =D c 2 5 2 q C 5 f 5 2, 2 5 c 2 5 2, 2 .
Let H2 Ž t . s 5 =D c 2 5 2 q 2 Re Ž D Ž fc 2 . , D c 2 . y 2 Re Ž =f , =D c 2 . . Then from Ž4.7. and the above estimates we get d dt
H2 Ž t . q a H2 Ž t . F K 2 q C 5 c 2 5 22, 2 5 =u 5 2 ,
where K 2 is a polynomial of 5 c 2 5 2, 2 , 5 f 5 2, 2 , and 5 f 5 1, 2 . We note that K 2 is bounded in Rq and denote its bound also by K 2 . However, we have been unable to show that 5 =u 5 2 g L`ŽRq. , but we have Ž2.28. instead. Since H2 Ž0. s 0, by using the Gronwall inequality, inserting Ž2.28., and noting that H1Ž t . is bounded we obtain H2 Ž t . F F s
K2
a K2
a
ž
t
Ž 1 y ey a t . q CH 5 =u 5 2 ey a Ž tys. ds 0
t
Ž 1 y ey a t . q CH C y 0
K2 q C
a
q
4C
b
/
ž
4
b
H1X Ž s . ey a Ž tys. ds
/
Ž 1 y ey a t . q
4
b
Ž H1Ž 0. ey a t y H1Ž t . . ,
and therefore we obtain 5 =D c 2 5 2 F
ž
2 K 2 q 2C
a
q
8C
b
/
Ž 1 y ey a t .
2 q C 5 f 5 22, 2 5 c 2 5 2, 2q
8
b
Ž H1Ž 0. ey a t y H1Ž t . . .
Therefore c 2 is uniformly bounded in H 3 Ž V ..
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Taking the inner product of Ž4.6. with yD u 2 and using Ž4.5. we get 1 d 2 dt
Ž 5 =u 2 5 2 q d 2 5 =f 2 5 2 q Ž 1 y db . 5 D f 2 5 2 q g 5 =D f 2 5 2 . q b 5 D u 2 5 2 y d 5 =u 2 5 2 q d 3 5 D f 2 5 2 q d Ž 1 y db . 5 D f 2 5 2 q dg 5 =D c 2 5 2 s Ž D F Ž f . , Du 2 . q Ž D < c < 2 , Du 2 . y Ž g , Du 2 . .
Let H3 Ž t . s 5 =u 2 5 2 q d 2 5 =f 2 5 2 q Ž 1 y db . 5 D f 2 5 2 q g 5 =D f 2 5 2 . Take b 3 s 2 d , note that 5 D u 2 5 2 G l1 5 =u 2 5 2 , and set
Ž D < c < 2 , Du 2 .
F
1 6
b 5 D u 2 5 2 q C 5 c 5 42, 2 ,
1
3
6
2b
Ž g , Du 2 . F b 5 Du 2 5 2 q
Ž D F Ž f . , Du 2 .
5 g52,
F Ž 5 F Y Ž f . < =f < 2 5 q 5 F X Ž f . D f 5 . 5 D u 2 5 F
1 6
2
b 5 D u 2 5 2 q C Ž F . Ž 5 =f 5 2 q 5 D f 5 . ,
then we have d dt
H3 Ž t . q b 3 H3 Ž t . F K 3 ,
where K 3 s C 5 c 5 42, 2 q Ž3r2 b .5 g 5 2 q C Ž F .Ž5 =f 5 2 q 5 D f 5. 2 is bounded in Rq. We denote its bound also by K 3 . Since H3 Ž0. s 0, by the Gronwall inequality we obtain H3 Ž t . F
K3
b3
Ž 1 y ey b 3 t . .
Therefore Ž c 2 , f 2 , u 2 . is uniformly bounded in t on bounded sets in E2 . 4.4. Construction of Global Attractor As discussed in Subsection 4.1, the uniform decaying of S1Ž t . and the uniform compactness of S2 Ž t . on bounded sets in E1 yield the existence of the global attractor.
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
THEOREM 4.3. Let f g H01 Ž V ., g g L2 Ž V ., F g C 2 ŽR. satisfy Ž2.7., Ž2.8., Ž2.20. and let B1 ; E1 be a bounded absorbing set which absorbs bounded sets in E1. Then A s v Ž B1 ., the v-limit set of B1 , is a global attractor of problems Ž2.1. ] Ž2.5..
5. FINITE DIMENSION OF THE GLOBAL ATTRACTOR In this section we are going to prove that the global attractor A has finite Hausdorff dimension and fractal dimension in the norm topology of E1. First we show that SŽ t . is Frechet differentiable in E1. From now on ´ we assume that f g H01, g g L2 Ž V ., and F g C 3 ŽR.. 5.1. The Uniform Differentiability of SŽ t . We consider the Žnonautonomous. linearized system i¨ t q D¨ q i a ¨ s f ¨ q u c ,
Ž 5.1.
u t q d u s w,
Ž 5.2.
wt y d w y b Dw q d 2 u y Ž 1 y db . D u q g D2 u s D Ž F X Ž f . u . q 2 Re Ž D c ¨ . q 4 Re Ž =c =¨ . ,
Ž ¨ , u, w . < ts0 s Ž ¨ 0 , u 0 , w 0 . ,
Ž 5.3.
x g V,
Ž 5.4.
¨ s 0, u s D u s 0, w s 0, x g V ,
Ž 5.5.
where Ž ¨ 0 , u 0 , w 0 . g E1 , Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . is the solution of problems Ž2.1. ] Ž2.5. with initial data Ž c 0 , f 0 , u 0 . g E1. It is easy to see that problems Ž5.1. ] Ž5.5. possess a unique solution Ž ¨ , u, w . g Cb ŽRq; E1 ., which means that SŽ t . is Frechet differentiable and that Ž ¨ , u, w . s ´ DS Ž t .Ž c 0 , f 0 , u 0 .Ž ¨ 0 , u 0 , w 0 . where DS Ž t .Ž c 0 , f 0 , u 0 . is the Frechet ´ derivative of SŽ t . at Ž c 0 , f 0 , u 0 . g E1. THEOREM 5.1. For any R ) 0, T ) 0, there exists a constant C s C Ž R, T . ) 0 such that for any j k 0 s Ž c k 0 , f k 0 , u k 0 . g E1 with 5 j k 0 5 E1 F R, k s 1, 2, S Ž t . j 20 y S Ž t . j 10 y Ž DS Ž t . j 10 . Ž j 20 y j 10 .
E1
F C 5 j 20 y j 10 5 2E1 , t g w 0, T x .
Consequently, supj 0 g X 5 DS Ž t . j 0 5 L Ž E1 . F C ŽT . bounded.
for t g w0, T x,
X ; E1
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The proof is similar to that of the continuity of S Ž t . on E1 , so it is omitted. 5.2. Transformation of m-Dimensional Volume by DS Ž t . Let X be invariant set of SŽ t . in E1 , j 0 g X. For any m linearly independent vectors j 1 , . . . , j m in E1 , we set Uk Ž t . s Ž DS Ž t . j 0 . j k Ž1 F k F m.. We are going to show that 5 U1Ž t . n ??? n UmŽ t .5 H m E1 , the m!-times of the volume of a polyhedron with vertices 0, U1Ž t ., . . . , UmŽ t ., decays exponentially as t ª q`, if m is sufficiently large. Since the solution of Ž5.1. ] Ž5.5. is expected to be exponentially damped, it is suitable to put Ž q, p, r . s e s t Ž ¨ , u, w ., where s is a positive constant to be determined later. It is easy to see that Ž q, p, r . satisfies the equations iqt q D q q i Ž a y s . q s qf q c p
Ž 5.6.
pt q Ž d y s . p s r ,
Ž 5.7.
rt y Ž d q s . r y b D r q d 2 p y Ž 1 y db . D p q g D2 p s D Ž F X Ž f . p . q 2 Re Ž D c q . q 4 Re Ž =c =q . ,
Ž 5.8.
where Ž c , f , u . s S Ž t .Ž c 0 , f 0 , u 0 ., Ž c 0 , f 0 , u 0 . g X. Note that Ž c , f , u . g Cb ŽRq; E1 .. Since 5 U1Ž t . n ??? n UmŽ t .5 2H m E1 s detŽUj Ž t ., Uk Ž t .., we may construct a bounded coercive quadratic form associated to Ž5.6. ] Ž5.8. and use it to study the development of 5 U1Ž t . n ??? n UmŽ t .5 2H m E1. Taking the inner product of Ž5.6. with q, then taking imaginary parts we obtain 1 d 2 dt
5 q 5 2 q Ž a y s . 5 q 5 2 s Im Ž c p, q . .
Ž 5.9.
Taking the inner product of Ž5.6. with D qt q Ž a y s . D q, making use of a procedure analogous to that of Ž2.22., we can obtain 1 d 2 dt
Ž 5 D q 5 2 y 2 Re Ž f q q c p, D q . . q Ž a y s . 5 D q 5 2 s Ž 2 a y 2 s q d . Re Ž f q, D q . q Ž a y 2 s q d . Re Ž c p, D q . q Im Ž f 2 q, D q . q Im Ž fc p, D q . y Re Ž u q, D q . qRe Ž c r , D q . y Re Ž c t p, D q . s 0.
Ž 5.10.
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
Taking the inner product of Ž5.8. with r, using Ž5.7. we get 1 d 2 dt
Ž 5 r 5 2 q d 2 5 p 5 2 q Ž 1 y db . 5 =p 5 2 q g 5 D p 5 2 . y Ž d q s . 5 r 5 2 q b 5 =r 5 2 q d 2 Ž d y s . 5 p 5 2 q Ž d y s . Ž 1 y db . 5 =p 5 2 qŽ d y s . g 5 D p 5 2 s y Ž =F X Ž f . p, =r . q 2 Re Ž D c q, r . q 4 Re Ž =c =q, r . .
Ž 5.11.
Let m , n be large constants Žto be chosen.. Define some quadratic forms on E1: for any h s Ž q, p, r . g E1 , G Ž t , h . s m 5 q 5 2 q 5 D q 5 2 y 2 Re Ž f q, D q . y 2 Re Ž c p, D q . q n 5 r 5 2 q nd 2 5 p 5 2 q n Ž 1 y db . 5 =p 5 2 q ng 5 D p 5 2 , Ž 5.12. G1 Ž h . s m Ž a y s . 5 q 5 2 q Ž a y s . 5 D q 5 2 y n Ž d q s . 5 r 5 2 q nb 5 =r 5 2 q nd 2 Ž d y s . 5 p 5 2 q n Ž d y s . Ž 1 y db . 5 =p 5 2 q n Ž d y s . g 5 D p5 2 ,
Ž 5.13.
G 2 Ž t , h . s m Im Ž c p, q . q Ž 2 a y 2 s q d . Re Ž f q, D q . q Ž a y 2 s q d . Re Ž c p, D q . q Im Ž f q2 , D q . y Im Ž fc p, D q . y Re Ž u q, D q . y Re Ž c r , D q . y Re Ž c t p, D q . q n Ž = Ž F X Ž f . p . , =r . q 2 n Re Ž D c q, r . q 4n Re Ž =c =q, r . .
Ž 5.14.
Then m Ž5.9. q Ž5.10. q n Ž5.11. implies 1 d 2 dt
G Ž t , h . q G1 Ž h . s G 2 Ž t , h . .
Ž 5.15.
Choose m G 8 A f A 20, ` , n G Ž8rd 2 .A c A 20, ` , and s F 12 min d , a 4 . Then <2 Re Ž f q, D q . < F 2 5 f 5 ` 5 q 5 5 D q 5 F 21 m 5 q 5 2 q 14 5 D q 5 2 , <2 Re Ž c p, D q . < F 2 5 c 5 ` 5 p 5 5 D q 5 F 21 nd 2 5 p 5 2 q 14 5 D q 5 2 ,
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LI AND CHEN
therefore GŽ t, h . is a bounded quadratic form on E1 , uniformly in t g Rq, i.e., < G Ž t , h . < F C 5h 5 2E1 ,
for h s Ž q, p, r . g E1 ,
Ž 5.16.
and coercive, uniformly in t g Rq: G Ž t , h . G 12 m 5 q 5 2 q 12 5 D q 5 2 q n 5 r 5 2 q 21 nd 2 5 p 5 2 q n Ž 1 y db . 5 =p 5 2 q ng 5 D p 5 2 G C 5h 5 2E1 .
Ž 5.17.
Obviously G1Žh . is also bounded and coercive quadratic form on E1. Noting that 5 c 5 2, 2 , 5 f 5 2, 2 , 5 u 5, 5 c t 5 are all in L`ŽRq. , by estimating G 2 Ž t, h . term by term we have 2 < G 2 Ž t , h . < F 12 G1 Ž h . q C Ž 5 q 5 2 q 5 p 5 2 q 5 r 5y1 ,2..
Ž 5.18.
Let J Ž t, h . s G 2 Ž t, h . y G1Žh ., then from the above analysis there exists an « ) 0 such that 1 d 2 dt
2 G Ž t , h . s J Ž t , h . F y« G Ž t , h . q C Ž 5 q 5 2 q 5 p 5 2 q 5 r 5y1 ,2..
Ž 5.19. Define Kh s ŽŽyD .y1 q, ŽyD .y1 p, ŽyD .y1 r . for any h s Ž p, q, r . g E1. Then K is a linear, self-adjoint, positive, and compact operator on E1. So it possesses a sequence of eigenvalues k l Ž l s 1, 2, . . . ., which is decreasing in l and tends to zero as l ª `. Moreover, 2 Ž Kh , h . E1 s 5 =q 5 2 q 5 =p 5 2 q 5 r 5y1, 2.
Ž 5.20.
Therefore there exists a constant C ) 0 such that J Ž t , h . F C Ž Kh , h . E 1 .
Ž 5.21.
˜Ž t, ? , ? . and J˜Ž t, ? , ? . be the polar forms associated to GŽ t, ? . and Let G ˜Ž t, ? , ? . is bounded and coercive, uniformly in J Ž t, ? . respectively. Then G ˜Ž t, ? , ? . in the sense of t g Rq. Moreover, J˜Ž t, ? , ? . is dominated by G Ž5.21.. Now let j 0 , j k Ž1 F k F m. and Uk Ž t . Ž1 F k F m. be the same as at the beginning of this subsection. Then hk Ž t . ' e s t Uk Ž t . is the solution of Ž5.6. ] Ž5.8. corresponding to initial data j k , and 5 U1Ž t . n ??? n Um Ž t .5 2H m E1
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
s ey2 m s t 5h1Ž t . n ??? n hmŽ t .5 2H m E1. By the boundedness and coercivity of ˜ there are two constants C1 , C2 ) 0 such that G,
˜ Ž t , hj Ž t . , h k Ž t . . C1m det Ž hj Ž t . , hk Ž t . . E1 F det G F C2m det Ž hj Ž t . , hk Ž t . . E1
Ž 5.22.
˜Ž t, hj Ž t ., hk Ž t .. s J˜Ž t, hj Ž t ., hk Ž t .., differentiating Žsee w5x.. Since Ž drdt .G ˜ Ž Ž . Ž .. det G t, hj t , hk t column by column we obtain d dt
˜ Ž t , hj Ž t . , h k Ž t . . det G m
s
Ý det ž Ž 1 y d k l . G˜Ž t , hj Ž t . , hk Ž t . . q d k l J˜Ž t , hj Ž t . , hk Ž t . . / , ls1
Ž 5.23. where d jk is the Kronecker deltas. By a linear algebraic lemma Žsee w5x., m
Ý det ž Ž 1 y d k l . G˜Ž t , hj Ž t . , hk Ž t . . q d k l J˜Ž t , hj Ž t . , hk Ž t . . / ls1 m
s
˜ Ž t , hj Ž t . , h k Ž t . . , v l det G
žÝ /
Ž 5.24.
ls1
where
v l s maxm min xgF F;R dim Fsl x/0
J Ž t , Ý mjs1 x jhj Ž t . .
Ž t , Ýmjs1 x jhj Ž t . .
.
Using Ž5.21. and coercivity of G,
v l F C maxm min
Ž K Ýmjs1 x jhj Ž t . , Ýmjs1 x jhj Ž t . . E 5Ý mjs1 x jhj
xgF F;R dim Fsl x/0
F C max min F;E1 h gF dim Fsl h/0
F Ck l .
Ž Kh , h . E 1 5h 5 2E1
Ž t.
5 2E1
1
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LI AND CHEN
Therefore Ž5.23., Ž5.24. yield m
˜Ž t , hj Ž t . , hk Ž t . . F det G˜Ž 0, j j , j k . exp C Ý k l t . det G
ž
/
ls1
Hence, by Ž5.22., 5h1 Ž t . n ??? n hm Ž t . 5 H m E1 s det Ž hj Ž t . , hk Ž t . . E 1 F
C2
m
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp C Ý k l t .
ž
ls1
/
Ž 5.25. So we obtain the following theorem. THEOREM 5.2. Let X be a bounded in¨ ariant set and j 0 g X. For linearly independent ¨ ectors j k g E1 Ž1 F k F m., Uk Ž t . s Ž DS Ž t . j 0 . j k . Then for any t G 0, 5 U1 Ž t . n ??? n Um Ž t . 5 2H m E1 F
C2
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp
m
žž
//
C Ý kl y 2 ms t . ls1
5.3. The Dimension of the Global Attractor We are now going to show the finiteness of the Hausdorff dimension and the fractal dimension of the global attractor A in the norm topology of E1. For the notions of the Hausdorff dimension and the fractal dimension we refer the reader to w2, 6, 18x. We shall use the theory developed in w6, 18x to show the finiteness of the dimension of the attractor. Since k l , the eigenvalues of K, tend to zero as l ª `, Ý m ls1 k lrm ª 0 as m ª `. So there exists an m 0 G 1 such that C m
m
Ý kl F s
for m G m 0 .
Ž 5.26.
ls1
Therefore, from Theorem 5.2, 5 U1 Ž t . n ??? n Um Ž t . 5 2H m E1 F
C2
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp Ž ym s t .
FINITE DIMENSIONAL GLOBAL ATTRACTOR
131
for t G 0, m G m 0 . From this inequality we see that there exists a t 0 G 0 such that the conditions of Theorem V.3.1 and V.3.2 in w18x Žor of Theorem 4.1 in w6x. hold. Therefore we have the following theorem. THEOREM 5.3.
Let m 0 be the same as in Ž5.26.. Then
Ž1. the Hausdorff dimension of the global attractor A is less than or equal to m 0 ; Ž2. the fractal dimension of the global attractor A is less than or equal to 2 m 0 .
REFERENCES 1. R. A. Adams, ‘‘Sobolev Spaces,’’ Academic Press, New York, 1975. 2. P. Constantin, C. Foias, and R. Temam, Attractors representing turbulent flows, Memoirs Amer. Math. Soc. 53, No. 314 Ž1985.. 3. I. Flahaut, Attractors for the dissipative Zakharov system, Nonlinear Anal. 16 Ž1991., 599]633. 4. A. Friedman, ‘‘Partial Differential Equations,’’ Holt, Reinhart, and Winston, 1969. 5. J. M. Ghidaglia, Weakly damped forced KdV equation behave as a finite dimensional dynamical system in the long time, J. Differential Equations 74 Ž1988., 369]390. 6. J. M. Ghidaglia and R. Temam, Attractors for damped nonlinear hyperbolic equations, J. Math. Pures Appl. 66 Ž1987., 273]319. 7. J. M. Ghidaglia, Finite dimensional behavior for weakly damped Schrodinger equations, ¨ Ann. Inst. H. Poincare´ Anal. 5 Ž1988., 365]405. 8. B. Guo, Global solutions of Schrodinger equation interacting with Bq-type self-consistent ¨ field, Acta Math. Sinica 26, No. 3 Ž1983., 295]306. 9. B. Guo, Initial boundary value problems for one class of system of multidimensional nonlinear Schrodinger]Boussinesq type equations, J. Math. Res. Exposition 8, No. 1 ¨ Ž1988., 61]71. 10. B. Guo and F. Chen, Finite dimensional behavior of global attractors for weakly damped nonlinear Schrodinger]Boussinesq equation, Phys. D 93 Ž1996., 101]118. ¨ 11. B. Guo and L. Shen, The global solution for the initial boundary value problem of the system of Schrodinger]Boussinesq type equations in three dimensions, Acta Math. Appl. ¨ Sinica 6, No. 1 Ž1990.; see also Symposium of DD7 Tianjin, China, 1986, 26]27. 12. Y. Li, On the initial boundary value problems for two dimensional systems of Zakharov equations and of complex-Schrodinger-real-Boussinesq equations, J. Partial Differential ¨ Equations 5, No. 2 Ž1992., 81]93. 13. Y. Li, Finite dimension of global attractor of dissipative Klein]Gordon]Schrodinger ¨ equations, Appl. Math. 8 Ž1995., 174]179. 14. J. L. Lions and E. Magenes, ‘‘Nonhomogeneous Boundary value Problem and Applications,’’ Vol. 1, Springer-Verlag, BerlinrHeidelbergrNew York, 1972; transl. from French by P. Kenneth. 15. V. G. Makhankov, On stationary solutions of Schrodinger equations with a self-consistent ¨ potential satisfying Boussinesq equations, Phys. Lett. A 50 Ž1974., 42]44.
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16. V. G. Makhankov, Dynamics of classical solitons Žin Nonintegrable systems., Phys. Rep., Phys. Lett. C 35, No. 1 Ž1978.. 17. K. Nishikawa, H. Hijo, K. Mima, and H. Ikezi, Coupled nonlinear electron-plasma and ion-acoustic waves, Phys. Lett. 33 Ž1974., 148]150. 18. R. Temam, ‘‘Infinite Dimensional Dynamical Systems in Mechanics and Physics,’’ Springer-Verlag, Berlin, 1988. 19. V. E. Zakharov and A. B. Shabat, Integrable system of nonlinear equations in mathematical physics, Funct. Anal. Appl. 83 Ž1974., 43]53.
205, 107]132 Ž1997.
AY965148
Finite Dimensional Global Attractor for Dissipative Schrodinger]Boussinesq Equations ¨ Li Yongsheng and Chen Qingyi Department of Mathematics, Huazhong Uni¨ ersity of Science and Technology, Wuhan, 430074, China Submitted by Colin Rogers Received April 19, 1995
In this paper the authors consider the initial boundary value problems of dissipative Schrodinger]Boussinesq equations and prove the existence of global ¨ attractors and the finiteness of the Hausdorff and the fractal dimensions of the attractors. Q 1997 Academic Press
1. INTRODUCTION In the laser and plasma physics under interaction of a nonlinear complex Schrodinger field and a real Boussinesq field, the dynamics is de¨ scribed by the equations ŽSBq., i c t q D c s fc ,
Ž 1.1.
f t t y D f q g D2f y D F Ž f . s D < c < 2 ,
Ž 1.2.
where c represents the complex Schrodinger field and f represents the ¨ real Boussinesq field Žsee w15]17, 19x.. The existence and the uniqueness of a global solution of Ž1.1., Ž1.2. have been studied in w8, 9, 11, 12x. In w10x the authors considered the one-dimensional dissipative ŽSBq. which involves a three-order nonlinear term in the Schrodinger equation and ¨ proved the existence of the global attractor and the finiteness of dimension of the attractor. In this paper, we consider the dissipative SBq equations i c t q D c q i ac y fc s f ,
Ž 1.3.
f t t y D f q g D2f y b D f t y D F Ž f . y D < c < 2 s g ,
Ž 1.4.
107 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
108
LI AND CHEN
c Ž 0, x . s c 0 Ž x . , f Ž 0, x . s f 0 Ž x . , f t Ž 0, x . s f 1 Ž x . ,
x g V,
Ž 1.5. c Ž t , x . s 0, f Ž t , x . s 0, D f Ž t , x . s 0,
t g R, x g V , Ž 1.6.
where a , b , and g are positive constants, F is a sufficiently smooth real function with F Ž0. s 0, and V ; R n Ž n F 3. is a bounded domain with a smooth boundary V. We arrange this paper as follows. First we establish time-uniform a priori estimates which imply the existence of solutions and bounded absorbing sets. Then we construct the global attractor of the system as the v-limit of one of those bounded absorbing sets Žsee w18x.. Finally, we show that the global attractor has finite Hausdorff and fractal dimensions. We introduce the following standard notations. We denote the spaces of complex-valued and the real-valued functions by the same symbols. W m, p Ž V . and W0m, p Ž V . are the usual Sobolev spaces. H m Ž V . s W m, 2 Ž V ., H0m Ž V . s W0m, 2 Ž V .. We denote by 5 ? 5 p the norm in L p Ž V ., especially 5 ? 5 s 5 ? 5 2 . We denote by Ž?, ? . the inner product in L2 Ž V ., by ² ? , ? : the dual product of Hy1 Ž V . and H01 Ž V ., and by 5 ? 5 m, p the norm of W m, p Ž V .. C b Ž I; E . denotes the space of continuous and bounded functions on I ; R with values in the Banach space E. For u g C b ŽRq; E ., or u g L`ŽRq; E ., A u A E denotes its norm, especially A u A s A u A L2 , A u A m, p s A u A W m , p . C is a generic constant and may assume different values in different formulae. We shall frequently use the Gargliardo]Nirenberg inequality Žsee w4x.: l 5 D j u 5 p F C 5 u 5 1y 5 D m u 5 lr , q
u g Lq l W m , r Ž V . ,
where 1 p
s
j n
ql
ž
1 r
y
m n
/
q
1yl q
,
1 F q, r F `, j is an integer, 0 F j F m, and jrm F l F 1. If m y j y nrr is a nonnegative integer, then the inequality holds for jrm F l - 1. We also use the short norm 5 =u 5 of u in H01 Ž V . and the equivalent norm 5 D u 5 of u in H 2 l H01 Ž V .. 2. THE TIME-UNIFORM A PRIORI ESTIMATES It is convenient to introduce a transformation u s f t q df , where d is a positive constant to be chosen later. The problems Ž1.3. ] Ž1.6. are
109
FINITE DIMENSIONAL GLOBAL ATTRACTOR
equivalent to i c t q D c q i ac y fc s f ,
Ž 2.1.
f t q df s u ,
Ž 2.2.
t g R, x g V ,
u t y du y b D u q d 2f y Ž 1 y db . D f q g D2f yD F Ž f . y D < c < 2 s g ,
Ž c , f , u . Ž 0, x . s Ž c 0 , f 0 , u 0 . Ž x . ,
Ž 2.3.
x g V,
Ž 2.4.
c s 0, f s D f s 0, u s 0, t g R, x g V ,
Ž 2.5.
where u 0 s df 0 q f 1. We set E0 s H01 = H01 = Hy1 Ž V . , E1 s H 2 l H01 = H 2 l H01 = L2 Ž V . . E2 s H 3 l H01 = H 3 l H01 = H01 Ž V . . Then E2 ¨ E1 ¨ E0 with compact imbedding. In this section we are going to establish time-uniform a priori estimates of solutions Ž c , f , u . in E0 and then in E1. 2.1. The Time-Uniform Estimates in E0 LEMMA 2.1. satisfies
Suppose that f g L`ŽRq; L2 Ž V ... Then c g L`ŽRq; L2 Ž V ..
5 c Ž t . 5 F 5 c 0 5 exp Ž ya t . q 2
2
A f A2
a2
Ž 1 y exp Ž ya t . . .
Ž 2.6.
Proof. Taking the inner product of Ž2.1. with c , then taking imaginary parts we obtain 1 d 2 dt
5 c 5 2 q a 5 c 5 2 s Im Ž f , c . F 5 f 5 5 c 5 F
By Gronwall inequality we have the lemma.
a 2
5c 52 q
5 f 52 2a
.
110
LI AND CHEN
We assume that in the following F satisfies the conditions lim inf < f <ª`
lim inf
H0f F Ž z . dz
f2
f F Ž f . y c 0 H0f F Ž z . dz f2
< f <ª`
G 0,
G 0,
Ž 2.7. for some c 0 ) 0.
Ž 2.8.
Under the above assumptions, ;« 1 , « 2 ) 0, there are constants C Ž « 1 ., C Ž « 2 . ) 0 such that f
H0 F Ž z . dz G y« f 1
f F Ž f . y c0
2
y CŽ «1 . ,
f
H0 F Ž z . dz G y«
2f
2
y CŽ «2 . .
Ž 2.9. Ž 2.10.
LEMMA 2.2. Suppose that F satisfies Ž2.7., Ž2.8., f g L`ŽRq; L2 Ž V .., f t , g g L`ŽRq; Hy1 Ž V ... If Ž c 0 , f 0 , u 0 . g E0 , then the solution Ž c , f , u . g L`ŽRq; E0 .. Proof. Taking the inner product of Ž2.1. with yŽ c t q ac ., then taking real parts we obtain 1 d 2 dt
5 =c 5 2 q a 5 =c 5 2 q Re Ž fc , c t . q a Ž f , < c < 2 . s Re Ž f , yc t y ac . ,
Noting that Re Ž fc , c t . s
1 d 2
Re Ž f , yc t y ac . s y
1
Ž f , < c < 2 . y 2 Ž ft , < c < 2 . , dt d dt
Re Ž f , c . y a Re Ž f , c . q Re² f t , c : ,
we obtain 1 d 2 dt
Ž 5 =c 5 2 q 2 Ž f , < c < 2 . q 2 Re Ž f , c . . q a 5 =c 5 2 q a Ž f , < c < 2 . q a Re Ž f , c . s
1 2
Ž ft , < c < 2 . q Re² ft , c : .
Ž 2.11.
111
FINITE DIMENSIONAL GLOBAL ATTRACTOR
Now taking the inner product of Ž2.3. with ŽyD .y1u , by inserting Ž2.2. we obtain 1 d 2 dt
ž
2 2 2 2 5 u 5y1 , 2 q d 5 f 5y1, 2 q Ž 1 y db . 5 f 5
f
qg 5 =f 5 2 q 2
HVH0 F Ž z . dz dx
/
2 2 2 2 2 3 y d 5 u 5y1 , 2 q b 5 u 5 q d 5 f 5y1, 2 q d Ž 1 y bd . 5 f 5 q dg 5 =f 5
HVf F Ž f . dx q d Ž f , < c < .
qd
2
s Ž Ž yD .
y1 r2
g , Ž yD .
y1r2
u . y Ž ft , < c < 2 . .
Ž 2.12.
Let 2 2 2 H0 Ž t . s 2 5 =c 5 2 q 4 Ž f , < c < 2 . q 4 Re Ž f , c . q 5 u 5y1 , 2 q d 5 f 5y1, 2
f
q Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 q 2
HVH0 F Ž z . dz dx,
Ž 2.13.
then 2Ž2.11. q Ž2.12. implies 1 d 2 dt
2 H0 Ž t . q 2 a 5 =c 5 2 q Ž 2 a q d . Ž f , < c < 2 . q 2 a Re Ž f , c . y d 5 u 5y1 ,2 2 2 q b 5 u 5 2 q d 3 5 f 5y1 , 2 q d Ž 1 y db . 5 f 5
q dg 5 =f 5 2 q d
HVf F Ž f . dx
s 2 Re² f t , c : q Ž Ž yD .
y1 r2
g , Ž yD .
y1r2
u ..
Ž 2.14.
Let l1 be the first eigenvalue of yD with a homogeneous Dirichlet boundary condition on V, and choose
d F min
½
1 2b
,
bl1 4
5
,
112
LI AND CHEN
2 then 5 =u 5 2 G l1 5 u 5 2 G l12 5 u 5y1 , 2 . We estimate the indefinite sign terms as
<4 Re Ž f , c . < F 2 5 f 5 2 q 2 5 c 5 2 , 2 Re² f t , c :< F
Ž Ž yD . y1 r2 g , Ž yD . y1r2 u . Žf,
2
a
a
2 5 f 5y1 ,2 q
2 F d 5 u 5y1 ,2 q
2 1 4d
5 =c 5 2 , 2 5 g 5y1, 2,
F C 5 f 5 4 5 c 5 4 5 c 5 F C 5 =f 5 5 c 5 5r4 5 =c 5 3r4 F « 3 5 =c 5 2 q « 4 5 =f 5 2 q C Ž « 3 , « 4 . 5 c 5 10 , ;« 3 ) 0, « 4 ) 0,
Ž 2.15.
where we take « 3 s arŽ2Ž2 a q d .., «4 s dgrŽ2Ž2 a q d ..; in Ž2.9., Ž2.10. we take « 1 s Ž1 y db .rŽ4 c 0 ., « 2 s Ž1 y db .r4, then f
HVH0 F Ž z . dz dx G y
1 y db 4 c0
f
5 f 5 2 y C1 ,
HVf F Ž f . dx G c HVH0 F Ž z . dz dx G y
1 y db
0
4
Ž 2.16.
5 f 5 2 y C2 . Ž 2.17.
From Ž2.14. and the above estimates we obtain d dt
H0 Ž t . q b 0 H0 Ž t . F K 0
2 10 where b 0 s min a , d , c 0 d 4 , K 0 s Ž2ra .5 f t 5y q Ž1r 1, 2 q C 5 c 5 2 q 4d .5 g 5y1 , 2 q C is bounded in R , and we denote its bound also by K 0 . By Gronwall inequality we obtain
H0 Ž t . F H0 Ž 0 . exp Ž yb 0 t . q
K0
b0
Ž 1 y exp Ž yb 0 t . . .
Ž 2.18.
Now we take « 3 s 1r8, « 4 s 1r16 in Ž2.15., « 1 s Ž1 y db .r4 in Ž2.9., and then we obtain 2 2 2 5 =c 5 2 q 5 u 5y1 , 2 q d 5 f 5y1, 2 q
F H0 Ž 0 . exp Ž yb 0 t . q
K0
b0
2
1
Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 2
Ž 1 y exp Ž yb 0 t . .
q C 5 c 5 10 q 2 5 f 5 2 q C, for all t G 0. So we have the lemma.
1
Ž 2.19.
113
FINITE DIMENSIONAL GLOBAL ATTRACTOR
2.2. The Time-Uniform Estimates in E1 LEMMA 2.3. Suppose that f g L`ŽRq; H01 Ž V .., f t g L`ŽRq; L2 Ž V .., and g g L`ŽRq; Hy1 Ž V ... Moreo¨ er, suppose that F satisfies Ž2.7., Ž2.8. and < F X Ž z . < F C1 < z < r q C 2 ,
1 F r - 2.
Ž 2.20.
If Ž c 0 , f 0 , u 0 . g E1 , then the solution Ž c , f , u . g L`ŽRq; E1 .. Moreo¨ er, t
H0 5 =u 5
2
dt F Ct q C.
Proof. First, taking the inner product of Ž2.1. with D c t q a D c , we obtain 1 d 2 dt
5 D c 5 2 q a 5 D c 5 2 y Re Ž fc , D c t . y a Re Ž fc , D c . s Re Ž f , D c t . q a Re Ž f , D c . ,
Ž 2.21.
Since yRe Ž fc , D c t . s y
d dt
Re Ž fc , D c . q Re Ž f t c , D c . q Re Ž fc t , D c . ,
while by Ž2.2. and Ž2.1., ReŽ f t c , D c . s ReŽ uc , D c . y d ReŽ fc , D c ., Re Ž fc t , D c . s Re Ž yi f Ž f y D c y i ac q fc . , D c . s Im Ž f f , D c . y a Re Ž fc , D c . q Im Ž f 2c , D c . , and therefore yRe Ž fc , D c t . s y
d dt
Re Ž fc , D c .
q Re Ž uc , D c . y Ž a q d . Re Ž fc , D c . q Im Ž f f , D c . q Im Ž f 2c , D c . , the first term in the right-hand side of Ž2.21. equaling Re Ž f , D c t . s
d dt
Re Ž f , D c . y Re Ž f t , D c . .
114
LI AND CHEN
Thus Ž2.21. can be rewritten as 1 d 2 dt
Ž 5 D c 5 2 y 2 Re Ž fc , D c . y 2 Re Ž f , D c . . q a 5 D c 5 2 y Ž 2 a q d . Re Ž fc , D c . y a Re Ž f , D c . q Re Ž uc , D c . q Im Ž ff , D c . q Im Ž f 2c , D c . q Re Ž f t , D c . s 0.
Ž 2.22.
Next, taking the inner product of Ž2.3. with u , using Ž2.2., we obtain 1 d 2 dt
Ž 5 =u < 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f I 2 q g 5 D f 5 2 . y d 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 q Ž =F Ž f . , =u . q Ž = < c < 2 , =u . s ² g , u : .
Ž 2.23.
Define H1 Ž t . s 5 D c 5 2 y 2 Re Ž fc , D c . y 2 Re Ž f , D c . q 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f 5 2 q g 5 D f 5 2 . Then Ž2.22. q Ž2.23. implies 1 d 2 dt
H1 Ž t . q a 5 D c 5 2 y Ž 2 a q d . Re Ž fc , D c . y a Re Ž f , D c . y 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 q Re Ž uc , D c . q Im Ž f f , D c . q Im Ž f 2c , D c . q Re Ž f t , D c . q Ž =F Ž f . , =u . q Ž = < c < 2 , =u . s ² g , u : .
Ž 2.24.
Since
Ž 2.26.
115
FINITE DIMENSIONAL GLOBAL ATTRACTOR
here we take « 5 s arŽ14Ž2 a q d .., «6 s ar14,
a 14
5Dc 5 2 q
b 8
5 =u 5 2 q C 5 c 5 2 5 =c 5 8
Ž = < c < 2 , =u .
a 14
a 14 a 14
5 D c 5 2 q C 5 f 5 24 5 f 5 24 5 D c 5 2 q C 5 f 5 2 5 =c 5 8 ,
5 D c 5 2 q C 5 ft 5 2
F 2 5 c 5 6 5 =c 5 3 5 =u 5 F C 5 c 5 6 5 c 5 1r4 5 D c 5 3r4 5 =u 5 F
<² g , u :< F
a 14
b 8
5Dc 5 2 q
b 8
5 =u 5 2 q C 5 c 5 2 5 =c 5 8 ,
2 5 =u 5 2 q C 5 g 5y1 ,2,
and since F Ž z . satisfies Ž2.20.,
Ž =F Ž f . , =u . F Ž C1 5 < f < r =f 5 q C2 5 =f 5 . 5 =u 5 F Ž C 5 f 5 6r 5 =f 5 6rŽ3y r . q C2 5 =f 5 . 5 =u 5 F C Ž 5 f 5 6r 5 f 5 Ž2y r .r4 5 D f 5 Ž rq2.r4 q 5 =f 5 . 5 =u 5 F
dg 2
5Df 5 2 q
b 8
5 =u 5 2 q C 5 =f 5 2 q C 5 =f 5 Ž6 rq4.rŽ6y r . ,
taking b 1 s min a , d 4 , noting that 5 =u 5 2 G l1 5 u 5 2 . Inserting the above estimates into Ž2.24. we have d dt
H1 Ž t . q
1 4
b 5 =u 5 2 q b 1 H1 Ž t . F K 1 ,
Ž 2.27.
where K 1 s K 1Ž5 =c 5, 5 =f 5, 5 =f 5, 5 f t 5, 5 g 5y1 , 2 . is a constant which is independent of t. By Gronwall inequality we obtain H1 Ž t . F H1 Ž 0 . exp Ž yb 1 t . q
K1
b1
Ž 1 y exp Ž yb1 t . . .
116
LI AND CHEN
Now we take « 5 s « 6 s 1r4 in Ž2.25. and Ž2.26., then we obtain 1 2
5 D c 5 2 q 5 u 5 2 q g 5 D f 5 2 F H1 Ž 0 . ey b 1 t q
K1
b1
q C Ž 5 f 5 , 5 =f 5 , 5 =c 5 . .
Moreover, < H1Ž t .< F C. From Ž2.27. and the above estimates we have 5 =u 5 2 F
4 K1
b
y
4b 1
b
H1 Ž t . y
4 d
b dt
H1 Ž t . F C y
4 d
b dt
H1 Ž t . . Ž 2.28.
Integrating Ž2.28. over w0, t x we have the lemma. Remark. In the cases of n s 1 and 2, the restriction r - 2 can be omitted. In fact, when n s 1, 5 f 5 ` F C 5 =f 5, the term <Ž =F Ž f ., =u .< can be estimated as
Ž =F Ž f . , =u . F Ž C1 5 < f < r =f 5 q C2 5 =f 5 . 5 =u 5 F
b 2
5 =u 5 2 q C Ž 5 f 5 `r q 1 . 2 5 =f 5 2 .
And when n s 2, 5 f 5 p F C 5 =f 5 for all 2 F p - `,
Ž =F Ž f . , =u . F Ž C1 5 < f < r D f 5 q C2 5 =f 5 . 5 =u 5 F Ž C 5 f 5 4rr 5 =f 5 4 q C2 5 =f 5 . 5 =u 5 F C Ž 5 f 5 4rr 5 f 5 1r4 5 D f 5 3r4 q 5 =f 5 . 5 =u 5 F
dg 2
5Df 5 2 q
b 8
5 =u < 2 q C 5 =f 5 2 q C 5 f 5 84 rr r5 5 f 5 2r5 .
3. EXISTENCE OF SOLUTIONS AND BOUNDED ABSORBING SETS In Section 2 we have established the time-independent a priori estimates of solutions Ž c , f , u . in E0 and in E1. We can apply the obtained results and Galerkin method to prove the existence of a global solution of problems Ž2.1. ] Ž2.5.. The proof is in the standard manner and so it is omitted. Therefore we have THEOREM 3.1. Suppose that F, f, g satisfy the same conditions as in Lemma 2.2. Then for e¨ ery Ž c 0 , f 0 , u 0 . g E0 , problems Ž2.1. ] Ž2.5. ha¨ e a solution Ž c , f , u . g L`ŽRq; E0 .. Moreo¨ er, if f g Cb ŽRq; L2 Ž V .., f t , g g Cb ŽRq; Hy1 Ž V .., then Ž c , f , u . g Cb ŽRq; E0 ...
117
FINITE DIMENSIONAL GLOBAL ATTRACTOR
Accordingly, problems Ž2.1. ] Ž2.5. have a bounded absorbing set in E0 . THEOREM 3.2. Suppose that F, f, g satisfy the same conditions as in Lemma 2.2. Then problems Ž2.1. ] Ž2.5. possess a bounded absorbing set B0 in E0 . Proof. Let B Ž0, R . be the ball in E0 of radius R G 1 centered at 0. For any initial Ž c 0 , f 0 , u 0 . g B Ž0, R ., 5 c Ž0.5 2 F Ž1rl1 .5 =c Ž0.5 2 F Ž1rl1 . R 2 . By Lemma 2.1 we see that there exists a t 1Ž R . ) 0 such that 5 c Ž t .5 2 F 1 q Ž1ra 2 .A f A 2 for all t ) t 1Ž R .. By Ž2.13., H0 Ž0. F CR10 . From Ž2.19., when t G t 1Ž R ., 1 1 2 2 2 5 =c 5 2 q 5 u 5y1 Ž 1 y db . 5 f 5 2 q g 5 =f 5 2 , 2 q d 5 f 5y1, 2 q 2 2 F CR10 exp Ž yb 0 t . q C 1 q
ž
5
1
a
A f A2 2
/
q
K0
b0
,
where K0 s
2
5
2 2 A f t Ay1 , 2 q C Ž1 q A f A . q
1
2 A g Ay1, 2 q C.
a 4d Therefore there exists a t 0 Ž R . G t 1Ž R . such that 1 1 2 2 2 5 =c 5 2 q 5 u 5y1 Ž 1 y db . 5 c 5 2 q g 5 =f 5 2 , 2 q d 5 f 5y1, 2 q 2 2 FC 2q
ž
1
a
A f A2 2
5
/
q
K0
b0
' R 02 ,
for all t G t 0 Ž R . .
Hence B0 s B Ž0, R 0 . is a bounded absorbing set in E0 . However, we are unable to prove the uniqueness of solution in E0 . But we can show that the solution exists uniquely in E1. Indeed, we have THEOREM 3.3. Suppose that F, f, g satisfy the conditions in Lemma 2.3. Suppose further that F Y is continuous in R. Then for e¨ ery Ž c 0 , f 0 , u 0 . g E1 , problems Ž2.1. ] Ž2.5. ha¨ e a unique solution Ž c , f , u . g L`ŽRq; E1 .. Moreo¨ er, if f g Cb ŽRq; H01 Ž V .., f t g Cb ŽRq; L2 Ž V .., g g Cb ŽRq; Hy1 Ž V .., then Ž c , f , u . g Cb ŽRq; E1 .. Proof. The proof of the existence of a solution is also in the standard manner. So we need only prove the uniqueness of a solution, which is a consequence of the following theorem. THEOREM 3.4. Let F, f, g be the same as in Theorem 3.3 and let SŽ t . be the solution operator, i.e., let Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . be the solution of Ž2.1. ] Ž2.5. with initial data Ž c 0 , f 0 , u 0 . g E1. Then SŽ t . is continuous from E1 into itself, uniformly on any compact inter¨ al w0, T x.
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LI AND CHEN
Proof. Let Ž c k , f k , u k . Ž k s 1, 2. be two solutions with initial data Ž c k 0 , f k 0 , u k 0 . g E1. Set Ž c , f , u . s Ž c 1 y c 2 , f 1 y f 2 , u 1 y u 2 ., Ž c 0 , f 0 , u 0 . s Ž c 10 y c 20 , f 10 y f 20 , u 10 y u 20 .. Due to the time-uniform a priori estimates, Ž c k , f k , u k . Ž k s 1, 2. is in L`ŽRq; E1 ., and so is Ž c , c , u .. Moreover, Ž c , f , u . satisfies the equations i c t q D c q i ac s fc 1 q f 2 c ,
Ž 3.1.
f t q df s u ,
Ž 3.2.
u t y du y b D u q d 2f y Ž 1 y db . D f q g D2f sD
1 X
žH
0
F Ž f 2 q tf . f dt q D Ž cc 1 q c 2 c .
/
Ž 3.3.
with homogeneous boundary conditions and initial data Ž c , f , u .< ts0 s Ž c 0 , f 0 , u 0 .. Taking the inner product of Ž3.1. with D2c , then taking imaginary parts we obtain 1 d 5 D c 5 2 q a 5 D c 5 2 s Im Ž D Ž fc 1 q f 2 c . , D c . 2 dt F C Ž 5 c 1 5 22, 2 q 5 f 2 5 22, 2 . Ž 5 D c 5 2 q 5 D f 5 2 . . Ž 3.4. Taking the inner product of Ž3.3. with u and using Ž3.2., we have 1 d Ž 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =u 5 2 q g 5 D f 5 2 . 2 dt y d 5 u 5 2 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 s y Ž = Ž F X Ž f 2 q tf . f dt . , =u . q D Ž cc 1 q cc 2 . , u .
ž
/
Ž 3.5.
Let M0 s sup Rq 5 f 1 5 ` q 5 f 2 5 `4 , M1 s sup < z < F M 0< F X Ž z .<, < F Y Ž z .<4 , then =
1 X
ž žH
0
F Ž f 2 q tf . f dt , =u
/ /
F f
1 Y
ž žH Ž ž žH F
0
q =f
f 2 q tf . = Ž f 2 q tf . dt , =u
/ /
1 X
0
F Ž f 2 q tf . dt , =u
/ /
F M1 Ž 5 f 5 ` Ž 5 =f 1 5 q 5 =f 2 5 . q 5 =f 5 . 5 =u 5 F
1 4
ž D Ž cc
b 5 =u 5 2 q CM12 5 =f 5 2 q CM12 Ž 5 =f 1 5 2 q 5 =f 2 5 2 . 5 D f 5 2 , 1
q cc 2 . , u
/
2 2 F d 5 u 5 2 q C Ž 5 c 1 5 22 , 2 q 5 c 2 5 2, 2 . 5Dc 5 .
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
From Ž3.5. we have 1 d 2 dt
Ž 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =u 5 2 q g 5 D f 5 2 . 1 q b 5 =u 5 2 q d 3 5 f 5 2 q d Ž 1 y db . 5 =f 5 2 q dg 5 D f 5 2 4 F C 5Dc 5 2 q C 5Df 5 2.
Ž 3.6.
Define H Ž t . s 5 D c 5 2 q 5 u 5 2 q d 2 5 f 5 2 q Ž 1 y db . 5 =f 5 2 q g 5 D f 5 2 , then Ž3.4. q Ž3.6. implies d dt
H Ž t . F KH Ž t . ,
where K is a constant which does not depend on t. By the Gronwall inequality we have H Ž t . F H Ž 0. e K t ,
t G 0.
Hence SŽ t . is uniformly continuous on t g w0, T x. THEOREM 3.5. Suppose that F, f, g satisfy the same conditions as those in Theorem 3.3. Then Ž2.1. ] Ž2.5. possess a bounded absorbing set B1 in E1. The proof is analogous to that of Theorem 3.2 and so it is omitted.
4. EXISTENCE OF GLOBAL ATTRACTOR 4.1. The Nonlinear Group Let f g H01 Ž V ., g g Hy1 Ž V . be independent of t. As shown in Theorem 3.3, problem Ž2.1. ] Ž2.5. has a unique solution Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . g Cb ŽRq, E1 .. Since now the system in discussion is autonomous, SŽ t . Ž t g R. forms a group. In fact, we can define SŽ t . on the whole real line, since the changing of t into yt in the system does not change the local existence of the solution. Therefore SŽ t . Ž t g R. is a group on E1 and is uniformly continuous on any compact set in R. To construct the global attractor, we treat SŽ t . as a compact perturbation of a contraction mapping. More precisely, we decompose S Ž t . into two parts: SŽ t . s S1Ž t . q S2 Ž t . for all t G 0, where S1Ž t . is continuous in E1 and decays uniformly on a bounded subset of E1 , i.e., for any bounded set
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LI AND CHEN
B ; E1 , supj g B 5 SŽ t . j 5 E1 tends to zero as t ª `, and where S2 Ž t . is uniformly compact in E1 for t large, i.e., for any bounded set B ; E1 , there exists a t 0 G 0 such that D t G t 0 SŽ t . B is relatively compact. If it is the case, then for a bounded absorbing set B1 ; E1 Žas in Theorem 3.5., the v-limit set of B1 , A s v Ž B1 . s F s G 0D t G s S Ž t . B1 Žwhere the closure is with respect to the norm topology of E1 ., is the global attractor in E1 Žsee w18x.. For this purpose, we define S1Ž t . and S2 Ž t . in the following way. Let Ž c 0 , f 0 , u 0 . g E1 , and Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . be the solution of problems Ž2.1. ] Ž2.5.. Define S1Ž t .Ž c 0 , f 0 , u 0 . s Ž c 1 , f 1 , u 1 . to be the solution of the problems i c 1 t q D c 1 q i ac 1 s fc 1 ,
Ž 4.1.
f 1 t q df 1 s u 1 ,
Ž 4.2.
u 1 t y du 1 y b D u 1 q d 2f 1 y Ž 1 y db . D f 1 q g D2f 1 s 0, Ž 4.3. with initial data Ž c 1 , f 1 , u 1 .< ts0 s Ž c 0 , f 0 , u 0 . and homogeneous boundary conditions, then Ž c 2 , f 2 , u 2 . s S2 Ž t .Ž c 0 , f 0 , u 0 . s SŽ t .Ž c 0 , f 0 , u 0 . y S1Ž t .Ž c 0 , f 0 , u 0 . is the solution of problem i c 2 t q D c 2 q i ac 2 s fc 2 q f ,
Ž 4.4.
f 2 t q df 2 s u 2 ,
Ž 4.5.
u 2 t y du 2 y b D u 2 q d 2f 2 y Ž 1 y db . D f 2 q g D2f 2 s D FŽ f . q D
Ž 4.6.
with homogeneous initial data and homogeneous boundary conditions. Since the two systems are nonautonomous, S1Ž t . and S2 Ž t . are no longer Žsemi-.groups. 4.2. Decaying of S1Ž t . Uniformly on Bounded Sets of E1 LEMMA 4.1.
Let f g H01 Ž V ., g g Hy1 Ž V ., and Ž c 0 , f 0 , u 0 . g E1. Then
Ž1. 5 c 1Ž t .5 2 s 5 c 1Ž0.5 2 ey2 a t , Ž2. 5 =c 1Ž t .5 2 F Ž5 =c 1Ž0.5 2 q C 5 c 1Ž0.5 2 . eya t , where C s const. A f A 82.2 . Ž3. 5 D c 1Ž t .5 2 F Ž5 D c 1Ž0.5 2 q C 5 c 1Ž0.5 2 . eya t , where C s const. A f A 82.2 . Ž4. 5 u 1 5 2 q d 2 5 f 1 5 2 q Ž1 y db .5 =f 1 5 2 q g 5 D f 1 5 2 F Ž5 u 0 5 2 q d 2 5 f 0 5 2 q Ž1 y db .5 =f 0 5 2 q g 5 D f 1 5 2 . ey2 d t. Therefore S1Ž t . decays on bounded sets in E1.
FINITE DIMENSIONAL GLOBAL ATTRACTOR
121
Proof. Taking the inner product of Ž4.1. with c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 c 1 5 2 q a 5 c 1 5 2 s 0,
so we have Ž1.. Taking the inner product of Ž4.1. with yD c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 =c 1 5 2 q a 5 =c 1 5 2 s Im Ž =fc 1 , =c 1 . .
Since
a 2
5 =c 1 5 2 q C 5 f 5 82 , 2 5 c 1 5 2 ,
by Ž1. we have d dt
5 =c 1 5 2 q a 5 =c 1 5 2 F C 5 f 5 82, 2 5 c 1 5 2 F const. A f A 82, 2 5 c 1 Ž 0 . 5 ey2 a t .
By the Gronwall inequality we obtain 5 =c 1 5 2 F 5 =c 1 Ž 0 . 5 2 ey a t q
1
a
const. A f A 82, 2 5 c 1 Ž 0 . 5 ey a t Ž 1 y ey a t .
F Ž 5 =c 1 Ž 0 . 5 2 q C 5 c 1 Ž 0 . 5 2 . ey a t , so we have Ž2.. Taking the inner product of Ž4.1. with D2c 1 , then taking imaginary parts we obtain 1 d 2 dt
5 D c 1 5 2 q a 5 D c 1 5 2 s Im Ž D fc 1 , D c 1 . q 2 Im Ž =f =c 1 , D c 1 . .
Since Im Ž D fc 1 , D c 1 . F 5 D f 5 5 c 1 5 ` 5 D c 1 5 F C 5 D f 5 5 c 1 5 1r4 5 D c 1 5 7r4 F
a 4
5 D c 1 5 2 q C 5 f 5 82, 2 5 c 1 5 2 ,
2 Im Ž =f =c 1 , D c 1 . F 2 5 =f 5 6 5 =c 1 5 3 5 D c 1 5 F C 5 f 5 2, 2 5 c 1 5 1r4 5 D c 1 5 7r4 F
a 4
5 D c 1 5 2 q C 5 f 5 82, 2 5 c 1 5 2 ,
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LI AND CHEN
we obtain d dt
5 D c 1 5 2 q a 5 D c 1 5 2 F C 5 f 5 82, 2 5 c 1 5 2 F const. A f A 82, 2 5 c 1 Ž 0 . 5 ey2 a t .
Hence we have Equation Ž4. wave equations
Ž3.. follows from the classical results on the damped linear Žsee w18x..
4.3. Uniform Compactness of S2 Ž t . on Bounded Sets in E1 Since Ž c , f , u . and Ž c 1 , f 1 , u 1 . are uniformly bounded in E1 for t g Rq and any initial data Ž c 0 , f 0 , u 0 . in a bounded set B ; E1 , so is Ž c 2 , f 2 , u 2 . s Ž c , f , u . y Ž c 1 , f 1 , u 1 .. If we show that Ž c 2 , f 2 , u 2 . is uniformly bounded in E2 , for t g Rq, then by the compact imbedding of E2 into E1 , S2 Ž t . is uniformly compact for t g Rq in E1. LEMMA 4.2. Let f g H01 Ž V ., g g L2 Ž V ., B ; E1 be bounded. Then there exists a C that is independent of t such that 5 Ž c 2 , f 2 , u 2 . 5 E 2 F C,
; Ž c 0 , f 0 , u 0 . g B.
Proof. We need only show that 5 =D c 2 5 2 q 5 D u 2 5 2 q 5 =D f 2 5 2 F C. Taking the inner product of Ž4.4. with yD2 Ž c 2 t q ac 2 ., then by making use of a procedure analogous to that of Ž2.22. we obtain 1 d 2 dt
Ž 5 =D c 2 5 2 q 2 Re Ž D Ž fc 2 . , D c 2 . y 2 Re Ž =f , =D c 2 . . q a 5 =D c 5 2 q Ž 2 a q d . Re Ž D Ž fc 2 . , D c 2 . y a Re Ž =f , =D c 2 . q Re Ž = Ž uc 2 . , =D c 2 . q Im Ž = Ž f f . , =D c 2 . y Im Ž =f D c 2 , =D c 2 . q Im Ž = Ž f 2c 2 . , =D c 2 . s 0.
Ž 4.7.
The indefinite-sign terms in Ž4.7. can be estimated as Re Ž D Ž fc 2 . , D c 2 . F C 5 f 5 2, 2 5 c 2 5 22, 2 ,
a Re Ž =F , =D c 2 . F
a 10
5 =D c 2 5 2 q
5a 2
5 =f 5 2 ,
Re Ž = Ž uc 2 . , =D c 2 . F Ž 5 =u 5 5 c 2 5 ` q 5 u 5 4 5 =c 2 5 4 . 5 =D c 2 5 F
a 10
5 =D c 2 5 2 q C 5 =u 5 2 5 c 2 5 22, 2
FINITE DIMENSIONAL GLOBAL ATTRACTOR
123
Im Ž = Ž f f . , =D c 2 . F Ž 5 f 5 4 5 f 5 4 q 5 f 5 ` 5 =f 5 . 5 =D c 2 5 F
a 10
5 =D c 2 5 2 q C 5 f 5 2, 2 5 f 5 1, 2 ,
Im Ž =f D c 2 , =D c 2 . F 5 =f 5 4 5 D c 2 5 4 5 =D c 2 5 7r4 F C 5 f 5 2, 2 5 c 2 5 1r4 2, 2 5 =D c 2 5
F
a 10
2 5 =D c 2 5 2 q C 5 f 5 82, 2 5 c 2 5 2, 2,
Im Ž = Ž f 2c 2 . , =D c 2 . F C 5 f 5 22, 2 5 c 2 5 2, 2 5 =D c 2 5 F
a 10
4 2 5 =D c 2 5 2 q C 5 f 5 2, 2 5 c 2 5 2, 2 .
Let H2 Ž t . s 5 =D c 2 5 2 q 2 Re Ž D Ž fc 2 . , D c 2 . y 2 Re Ž =f , =D c 2 . . Then from Ž4.7. and the above estimates we get d dt
H2 Ž t . q a H2 Ž t . F K 2 q C 5 c 2 5 22, 2 5 =u 5 2 ,
where K 2 is a polynomial of 5 c 2 5 2, 2 , 5 f 5 2, 2 , and 5 f 5 1, 2 . We note that K 2 is bounded in Rq and denote its bound also by K 2 . However, we have been unable to show that 5 =u 5 2 g L`ŽRq. , but we have Ž2.28. instead. Since H2 Ž0. s 0, by using the Gronwall inequality, inserting Ž2.28., and noting that H1Ž t . is bounded we obtain H2 Ž t . F F s
K2
a K2
a
ž
t
Ž 1 y ey a t . q CH 5 =u 5 2 ey a Ž tys. ds 0
t
Ž 1 y ey a t . q CH C y 0
K2 q C
a
q
4C
b
/
ž
4
b
H1X Ž s . ey a Ž tys. ds
/
Ž 1 y ey a t . q
4
b
Ž H1Ž 0. ey a t y H1Ž t . . ,
and therefore we obtain 5 =D c 2 5 2 F
ž
2 K 2 q 2C
a
q
8C
b
/
Ž 1 y ey a t .
2 q C 5 f 5 22, 2 5 c 2 5 2, 2q
8
b
Ž H1Ž 0. ey a t y H1Ž t . . .
Therefore c 2 is uniformly bounded in H 3 Ž V ..
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LI AND CHEN
Taking the inner product of Ž4.6. with yD u 2 and using Ž4.5. we get 1 d 2 dt
Ž 5 =u 2 5 2 q d 2 5 =f 2 5 2 q Ž 1 y db . 5 D f 2 5 2 q g 5 =D f 2 5 2 . q b 5 D u 2 5 2 y d 5 =u 2 5 2 q d 3 5 D f 2 5 2 q d Ž 1 y db . 5 D f 2 5 2 q dg 5 =D c 2 5 2 s Ž D F Ž f . , Du 2 . q Ž D < c < 2 , Du 2 . y Ž g , Du 2 . .
Let H3 Ž t . s 5 =u 2 5 2 q d 2 5 =f 2 5 2 q Ž 1 y db . 5 D f 2 5 2 q g 5 =D f 2 5 2 . Take b 3 s 2 d , note that 5 D u 2 5 2 G l1 5 =u 2 5 2 , and set
Ž D < c < 2 , Du 2 .
F
1 6
b 5 D u 2 5 2 q C 5 c 5 42, 2 ,
1
3
6
2b
Ž g , Du 2 . F b 5 Du 2 5 2 q
Ž D F Ž f . , Du 2 .
5 g52,
F Ž 5 F Y Ž f . < =f < 2 5 q 5 F X Ž f . D f 5 . 5 D u 2 5 F
1 6
2
b 5 D u 2 5 2 q C Ž F . Ž 5 =f 5 2 q 5 D f 5 . ,
then we have d dt
H3 Ž t . q b 3 H3 Ž t . F K 3 ,
where K 3 s C 5 c 5 42, 2 q Ž3r2 b .5 g 5 2 q C Ž F .Ž5 =f 5 2 q 5 D f 5. 2 is bounded in Rq. We denote its bound also by K 3 . Since H3 Ž0. s 0, by the Gronwall inequality we obtain H3 Ž t . F
K3
b3
Ž 1 y ey b 3 t . .
Therefore Ž c 2 , f 2 , u 2 . is uniformly bounded in t on bounded sets in E2 . 4.4. Construction of Global Attractor As discussed in Subsection 4.1, the uniform decaying of S1Ž t . and the uniform compactness of S2 Ž t . on bounded sets in E1 yield the existence of the global attractor.
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
THEOREM 4.3. Let f g H01 Ž V ., g g L2 Ž V ., F g C 2 ŽR. satisfy Ž2.7., Ž2.8., Ž2.20. and let B1 ; E1 be a bounded absorbing set which absorbs bounded sets in E1. Then A s v Ž B1 ., the v-limit set of B1 , is a global attractor of problems Ž2.1. ] Ž2.5..
5. FINITE DIMENSION OF THE GLOBAL ATTRACTOR In this section we are going to prove that the global attractor A has finite Hausdorff dimension and fractal dimension in the norm topology of E1. First we show that SŽ t . is Frechet differentiable in E1. From now on ´ we assume that f g H01, g g L2 Ž V ., and F g C 3 ŽR.. 5.1. The Uniform Differentiability of SŽ t . We consider the Žnonautonomous. linearized system i¨ t q D¨ q i a ¨ s f ¨ q u c ,
Ž 5.1.
u t q d u s w,
Ž 5.2.
wt y d w y b Dw q d 2 u y Ž 1 y db . D u q g D2 u s D Ž F X Ž f . u . q 2 Re Ž D c ¨ . q 4 Re Ž =c =¨ . ,
Ž ¨ , u, w . < ts0 s Ž ¨ 0 , u 0 , w 0 . ,
Ž 5.3.
x g V,
Ž 5.4.
¨ s 0, u s D u s 0, w s 0, x g V ,
Ž 5.5.
where Ž ¨ 0 , u 0 , w 0 . g E1 , Ž c , f , u . s SŽ t .Ž c 0 , f 0 , u 0 . is the solution of problems Ž2.1. ] Ž2.5. with initial data Ž c 0 , f 0 , u 0 . g E1. It is easy to see that problems Ž5.1. ] Ž5.5. possess a unique solution Ž ¨ , u, w . g Cb ŽRq; E1 ., which means that SŽ t . is Frechet differentiable and that Ž ¨ , u, w . s ´ DS Ž t .Ž c 0 , f 0 , u 0 .Ž ¨ 0 , u 0 , w 0 . where DS Ž t .Ž c 0 , f 0 , u 0 . is the Frechet ´ derivative of SŽ t . at Ž c 0 , f 0 , u 0 . g E1. THEOREM 5.1. For any R ) 0, T ) 0, there exists a constant C s C Ž R, T . ) 0 such that for any j k 0 s Ž c k 0 , f k 0 , u k 0 . g E1 with 5 j k 0 5 E1 F R, k s 1, 2, S Ž t . j 20 y S Ž t . j 10 y Ž DS Ž t . j 10 . Ž j 20 y j 10 .
E1
F C 5 j 20 y j 10 5 2E1 , t g w 0, T x .
Consequently, supj 0 g X 5 DS Ž t . j 0 5 L Ž E1 . F C ŽT . bounded.
for t g w0, T x,
X ; E1
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LI AND CHEN
The proof is similar to that of the continuity of S Ž t . on E1 , so it is omitted. 5.2. Transformation of m-Dimensional Volume by DS Ž t . Let X be invariant set of SŽ t . in E1 , j 0 g X. For any m linearly independent vectors j 1 , . . . , j m in E1 , we set Uk Ž t . s Ž DS Ž t . j 0 . j k Ž1 F k F m.. We are going to show that 5 U1Ž t . n ??? n UmŽ t .5 H m E1 , the m!-times of the volume of a polyhedron with vertices 0, U1Ž t ., . . . , UmŽ t ., decays exponentially as t ª q`, if m is sufficiently large. Since the solution of Ž5.1. ] Ž5.5. is expected to be exponentially damped, it is suitable to put Ž q, p, r . s e s t Ž ¨ , u, w ., where s is a positive constant to be determined later. It is easy to see that Ž q, p, r . satisfies the equations iqt q D q q i Ž a y s . q s qf q c p
Ž 5.6.
pt q Ž d y s . p s r ,
Ž 5.7.
rt y Ž d q s . r y b D r q d 2 p y Ž 1 y db . D p q g D2 p s D Ž F X Ž f . p . q 2 Re Ž D c q . q 4 Re Ž =c =q . ,
Ž 5.8.
where Ž c , f , u . s S Ž t .Ž c 0 , f 0 , u 0 ., Ž c 0 , f 0 , u 0 . g X. Note that Ž c , f , u . g Cb ŽRq; E1 .. Since 5 U1Ž t . n ??? n UmŽ t .5 2H m E1 s detŽUj Ž t ., Uk Ž t .., we may construct a bounded coercive quadratic form associated to Ž5.6. ] Ž5.8. and use it to study the development of 5 U1Ž t . n ??? n UmŽ t .5 2H m E1. Taking the inner product of Ž5.6. with q, then taking imaginary parts we obtain 1 d 2 dt
5 q 5 2 q Ž a y s . 5 q 5 2 s Im Ž c p, q . .
Ž 5.9.
Taking the inner product of Ž5.6. with D qt q Ž a y s . D q, making use of a procedure analogous to that of Ž2.22., we can obtain 1 d 2 dt
Ž 5 D q 5 2 y 2 Re Ž f q q c p, D q . . q Ž a y s . 5 D q 5 2 s Ž 2 a y 2 s q d . Re Ž f q, D q . q Ž a y 2 s q d . Re Ž c p, D q . q Im Ž f 2 q, D q . q Im Ž fc p, D q . y Re Ž u q, D q . qRe Ž c r , D q . y Re Ž c t p, D q . s 0.
Ž 5.10.
127
FINITE DIMENSIONAL GLOBAL ATTRACTOR
Taking the inner product of Ž5.8. with r, using Ž5.7. we get 1 d 2 dt
Ž 5 r 5 2 q d 2 5 p 5 2 q Ž 1 y db . 5 =p 5 2 q g 5 D p 5 2 . y Ž d q s . 5 r 5 2 q b 5 =r 5 2 q d 2 Ž d y s . 5 p 5 2 q Ž d y s . Ž 1 y db . 5 =p 5 2 qŽ d y s . g 5 D p 5 2 s y Ž =F X Ž f . p, =r . q 2 Re Ž D c q, r . q 4 Re Ž =c =q, r . .
Ž 5.11.
Let m , n be large constants Žto be chosen.. Define some quadratic forms on E1: for any h s Ž q, p, r . g E1 , G Ž t , h . s m 5 q 5 2 q 5 D q 5 2 y 2 Re Ž f q, D q . y 2 Re Ž c p, D q . q n 5 r 5 2 q nd 2 5 p 5 2 q n Ž 1 y db . 5 =p 5 2 q ng 5 D p 5 2 , Ž 5.12. G1 Ž h . s m Ž a y s . 5 q 5 2 q Ž a y s . 5 D q 5 2 y n Ž d q s . 5 r 5 2 q nb 5 =r 5 2 q nd 2 Ž d y s . 5 p 5 2 q n Ž d y s . Ž 1 y db . 5 =p 5 2 q n Ž d y s . g 5 D p5 2 ,
Ž 5.13.
G 2 Ž t , h . s m Im Ž c p, q . q Ž 2 a y 2 s q d . Re Ž f q, D q . q Ž a y 2 s q d . Re Ž c p, D q . q Im Ž f q2 , D q . y Im Ž fc p, D q . y Re Ž u q, D q . y Re Ž c r , D q . y Re Ž c t p, D q . q n Ž = Ž F X Ž f . p . , =r . q 2 n Re Ž D c q, r . q 4n Re Ž =c =q, r . .
Ž 5.14.
Then m Ž5.9. q Ž5.10. q n Ž5.11. implies 1 d 2 dt
G Ž t , h . q G1 Ž h . s G 2 Ž t , h . .
Ž 5.15.
Choose m G 8 A f A 20, ` , n G Ž8rd 2 .A c A 20, ` , and s F 12 min d , a 4 . Then <2 Re Ž f q, D q . < F 2 5 f 5 ` 5 q 5 5 D q 5 F 21 m 5 q 5 2 q 14 5 D q 5 2 , <2 Re Ž c p, D q . < F 2 5 c 5 ` 5 p 5 5 D q 5 F 21 nd 2 5 p 5 2 q 14 5 D q 5 2 ,
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therefore GŽ t, h . is a bounded quadratic form on E1 , uniformly in t g Rq, i.e., < G Ž t , h . < F C 5h 5 2E1 ,
for h s Ž q, p, r . g E1 ,
Ž 5.16.
and coercive, uniformly in t g Rq: G Ž t , h . G 12 m 5 q 5 2 q 12 5 D q 5 2 q n 5 r 5 2 q 21 nd 2 5 p 5 2 q n Ž 1 y db . 5 =p 5 2 q ng 5 D p 5 2 G C 5h 5 2E1 .
Ž 5.17.
Obviously G1Žh . is also bounded and coercive quadratic form on E1. Noting that 5 c 5 2, 2 , 5 f 5 2, 2 , 5 u 5, 5 c t 5 are all in L`ŽRq. , by estimating G 2 Ž t, h . term by term we have 2 < G 2 Ž t , h . < F 12 G1 Ž h . q C Ž 5 q 5 2 q 5 p 5 2 q 5 r 5y1 ,2..
Ž 5.18.
Let J Ž t, h . s G 2 Ž t, h . y G1Žh ., then from the above analysis there exists an « ) 0 such that 1 d 2 dt
2 G Ž t , h . s J Ž t , h . F y« G Ž t , h . q C Ž 5 q 5 2 q 5 p 5 2 q 5 r 5y1 ,2..
Ž 5.19. Define Kh s ŽŽyD .y1 q, ŽyD .y1 p, ŽyD .y1 r . for any h s Ž p, q, r . g E1. Then K is a linear, self-adjoint, positive, and compact operator on E1. So it possesses a sequence of eigenvalues k l Ž l s 1, 2, . . . ., which is decreasing in l and tends to zero as l ª `. Moreover, 2 Ž Kh , h . E1 s 5 =q 5 2 q 5 =p 5 2 q 5 r 5y1, 2.
Ž 5.20.
Therefore there exists a constant C ) 0 such that J Ž t , h . F C Ž Kh , h . E 1 .
Ž 5.21.
˜Ž t, ? , ? . and J˜Ž t, ? , ? . be the polar forms associated to GŽ t, ? . and Let G ˜Ž t, ? , ? . is bounded and coercive, uniformly in J Ž t, ? . respectively. Then G ˜Ž t, ? , ? . in the sense of t g Rq. Moreover, J˜Ž t, ? , ? . is dominated by G Ž5.21.. Now let j 0 , j k Ž1 F k F m. and Uk Ž t . Ž1 F k F m. be the same as at the beginning of this subsection. Then hk Ž t . ' e s t Uk Ž t . is the solution of Ž5.6. ] Ž5.8. corresponding to initial data j k , and 5 U1Ž t . n ??? n Um Ž t .5 2H m E1
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FINITE DIMENSIONAL GLOBAL ATTRACTOR
s ey2 m s t 5h1Ž t . n ??? n hmŽ t .5 2H m E1. By the boundedness and coercivity of ˜ there are two constants C1 , C2 ) 0 such that G,
˜ Ž t , hj Ž t . , h k Ž t . . C1m det Ž hj Ž t . , hk Ž t . . E1 F det G F C2m det Ž hj Ž t . , hk Ž t . . E1
Ž 5.22.
˜Ž t, hj Ž t ., hk Ž t .. s J˜Ž t, hj Ž t ., hk Ž t .., differentiating Žsee w5x.. Since Ž drdt .G ˜ Ž Ž . Ž .. det G t, hj t , hk t column by column we obtain d dt
˜ Ž t , hj Ž t . , h k Ž t . . det G m
s
Ý det ž Ž 1 y d k l . G˜Ž t , hj Ž t . , hk Ž t . . q d k l J˜Ž t , hj Ž t . , hk Ž t . . / , ls1
Ž 5.23. where d jk is the Kronecker deltas. By a linear algebraic lemma Žsee w5x., m
Ý det ž Ž 1 y d k l . G˜Ž t , hj Ž t . , hk Ž t . . q d k l J˜Ž t , hj Ž t . , hk Ž t . . / ls1 m
s
˜ Ž t , hj Ž t . , h k Ž t . . , v l det G
žÝ /
Ž 5.24.
ls1
where
v l s maxm min xgF F;R dim Fsl x/0
J Ž t , Ý mjs1 x jhj Ž t . .
Ž t , Ýmjs1 x jhj Ž t . .
.
Using Ž5.21. and coercivity of G,
v l F C maxm min
Ž K Ýmjs1 x jhj Ž t . , Ýmjs1 x jhj Ž t . . E 5Ý mjs1 x jhj
xgF F;R dim Fsl x/0
F C max min F;E1 h gF dim Fsl h/0
F Ck l .
Ž Kh , h . E 1 5h 5 2E1
Ž t.
5 2E1
1
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Therefore Ž5.23., Ž5.24. yield m
˜Ž t , hj Ž t . , hk Ž t . . F det G˜Ž 0, j j , j k . exp C Ý k l t . det G
ž
/
ls1
Hence, by Ž5.22., 5h1 Ž t . n ??? n hm Ž t . 5 H m E1 s det Ž hj Ž t . , hk Ž t . . E 1 F
C2
m
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp C Ý k l t .
ž
ls1
/
Ž 5.25. So we obtain the following theorem. THEOREM 5.2. Let X be a bounded in¨ ariant set and j 0 g X. For linearly independent ¨ ectors j k g E1 Ž1 F k F m., Uk Ž t . s Ž DS Ž t . j 0 . j k . Then for any t G 0, 5 U1 Ž t . n ??? n Um Ž t . 5 2H m E1 F
C2
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp
m
žž
//
C Ý kl y 2 ms t . ls1
5.3. The Dimension of the Global Attractor We are now going to show the finiteness of the Hausdorff dimension and the fractal dimension of the global attractor A in the norm topology of E1. For the notions of the Hausdorff dimension and the fractal dimension we refer the reader to w2, 6, 18x. We shall use the theory developed in w6, 18x to show the finiteness of the dimension of the attractor. Since k l , the eigenvalues of K, tend to zero as l ª `, Ý m ls1 k lrm ª 0 as m ª `. So there exists an m 0 G 1 such that C m
m
Ý kl F s
for m G m 0 .
Ž 5.26.
ls1
Therefore, from Theorem 5.2, 5 U1 Ž t . n ??? n Um Ž t . 5 2H m E1 F
C2
ž / C1
m
5 j 1 n ??? n j m 5 2H m E1exp Ž ym s t .
FINITE DIMENSIONAL GLOBAL ATTRACTOR
131
for t G 0, m G m 0 . From this inequality we see that there exists a t 0 G 0 such that the conditions of Theorem V.3.1 and V.3.2 in w18x Žor of Theorem 4.1 in w6x. hold. Therefore we have the following theorem. THEOREM 5.3.
Let m 0 be the same as in Ž5.26.. Then
Ž1. the Hausdorff dimension of the global attractor A is less than or equal to m 0 ; Ž2. the fractal dimension of the global attractor A is less than or equal to 2 m 0 .
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