Determining nodes for partly dissipative reaction diffusion systems

Nonlinear Analysis 54 (2003) 873 – 884

www.elsevier.com/locate/na

Determining nodes for partly dissipative reaction di"usion systems Youmin Lua , Zhoude Shaob;∗ a Department

of Mathematics, Computer Science and Statistics Bloomsburg University, Bloomsburg, PA 17815, USA b Department of Mathematics, Millersville University, Millersville, PA 17551, USA Received 18 December 2001; accepted 27 October 2002

Abstract In this paper, we study the long-time dynamical properties of the solutions of partly dissipative reaction di"usion systems of the form
ut − d1u + f(x; u) + g(x; v) = 0; x ∈ ⊂ Rn ; t ¿ 0; vt + v + h(x; u) = 0: The main result is that, under appropriate assumptions, the large time behavior of a solution is completely determined in certain sense if the large time behavior of the solution is known on a densely chosen yet 6nite set of points which are called determining nodes, a concept initially introduced by Foias and Temam for the Navier–Stokes equations (Math. Comput. 43 (1984) 117). ? 2003 Elsevier Science Ltd. All rights reserved. MSC: 35B40; 35K57 Keywords: Partly dissipative systems; Determining nodes; Nodal values; Long-time behavior

1. Introduction We consider partly dissipative reaction di"usion systems of the form
ut − d1u + f(x; u) + g(x; v) = 0; x ∈ ⊂ Rn ; t ¿ 0; vt + v + h(x; u) = 0 ∗

Corresponding author. Tel.:+1-717-872-3109. E-mail address: [email protected] (Z. Shao).

0362-546X/03/$ - see front matter ? 2003 Elsevier Science Ltd. All rights reserved. doi:10.1016/S0362-546X(03)00112-3

(1.1)

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Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

along with initial conditions u(x; 0) = u0 (x);

v(x; 0) = v0 (x)

and one of the following boundary conditions for u:  Dirichlet : u(x; t) = 0; t ¿ 0; x ∈ @ ;     @u Neumann : (x; t) = 0; t ¿ 0; x ∈ @ ;  @n    Periodic : = (0; L)n and u is -periodic:

(1.2)

(1.3)

A typical example of system (1.1) is the FitzHugh–Nagumo equations. Under appropriate assumptions, it is proved that the dynamical system induced by the initial boundary value problem (1.1)– (1.3) possesses a global attractor with 6nite Hausdor" and fractal dimensions [10] and an (in6nite dimensional) inertial manifold [11,14]. As a consequence, the semiLow induced by problem (1.1) – (1.3) is reduced to a Low on the inertial manifold. See [13,15] for details on the theory of global attractors and inertial manifolds. Since the inertial manifold for problem (1.1) – (1.3) is in6nite dimensional, the reduced Low is still in6nite dimensional. However, the 6nite dimensionality of the global attractor suggests that the long-time asymptotic behavior of the solutions be determined by 6nitely many degrees of freedom. In this paper, we study the long-time behavior of the solutions using the approach of determining nodes and nodal values introduced for the Navier–Stokes equations by Foias and Temam [4]. It is proved in [4] that the di"erence of any two solutions of the Navier–Stokes equations goes to zero as time t goes to in6nity if the di"erence of these two solutions goes to zero as t goes to in6nity on a 6nite set of points, densely chosen from the underlying physical domain. These points are referred to as determining nodes and the values of a solution at the determining nodes are called nodal values. The concept of determining nodes is important from the practical point of view since experimental data, in general, can only be obtained from measurements at a 6nite number of points. In this paper, we show that similar result holds true for system (1.1), that is, there exists a 6nite set of determining nodes. This fact demonstrates that the long-time asymptotic behavior of the solutions of problem (1.1) – (1.3) is determined by 6nitely many degrees of freedom as supported by the 6nite dimensionality of the global attractor. The idea of the proof is similar to that used in [4]. However, due to the lack of dissipativity in the v-equation, extra care is needed in the relevant estimates. It is also noteworthy to point out that, when a system has a suMciently smooth 6nite dimensional inertial manifold, there certainly exists a 6nite set of determining nodes, see [1,5]. Research has also been done on estimating the number of determining nodes needed for the Navier–Stokes equations, see [6,7,12] and the references therein. Similar results have been obtained for other evolutionary equations such as the complex Ginzburg– Landau equation and the Kuramoto–Sivanshinsky equation [2,9,12]. It is also noted that similar concepts such as determining modes and determining volume elements, referred to as asymptotic determining degrees of freedom by [1], have been studied, see [1,3,8] and the references therein. Actually, Foias and Prodi [3] is the pioneer paper

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that initiated the study of determining degrees of freedom for evolutionary equations. Also, Foias and Titi [5] used nodal values to parameterize the inertial manifolds and obtained interesting results. In Section 2, we present some preliminary results related to problem (1.1) – (1.3) which are needed later on. The main results are stated and proved in Section 3. 2. Preliminaries We consider the initial-boundary value problem (1.1) – (1.3). It is assumed that f; g, and h satisfy the following conditions:  f; g; and h are all C 2 on O ⊂ Rn ;        ¿ 0 is a constant;    (2.1) 1 |u|k − 3 6 f(x; u)u 6 2 |u|k + 3 k ¿ 2;      |gv (x; v)| 6 4 ;      |hu (x; u)| 6 5 ; |hxi (x; u)| 6 6 (1 + |u|); i = 1; 2; : : : ; n; where k ¿ 2 and all the i ’s are positive constants. De6ne H = L2 ( ) and the subspace of V1 of H 1 ( ) as V1 = {u ∈ H 1 ( ): the corresponding boundary condition holds}: Under the assumptions stated in (2.1), it is shown [10,14] that, for any (u0 ; v0 ) ∈ H ×H , there exists a unique solution (u(t); v(t)) on [0; ∞) of problem (1.1) – (1.3) satisfying the following: (u; v) ∈ C([0; ∞); H × H ); u ∈ L2 (0; T ; V1 ) ∩ Lk (0; T ; Lk ( ));

T ¿ 0;

(u(0); v(0)) = (u0 ; v0 ) and (u(t); v(t)) is continuous with respect to (u0 ; v0 ). Therefore, problem (1.1) – (1.3) induces a (nonlinear) semiLow S(t) on H × H de6ned by S(t)(u0 ; v0 ) = (u(t); v(t));

t ¿ 0:

Furthermore, problem (1.1)– (1.3) can be written as an abstract initial value problem as follows. De6ne the operator A : D(A) → H as Au = −d1u + I; where I is the identity operator and D(A) = {u ∈ H 2 ( ): the corresponding boundary condition holds}:

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˜ g, The operator A is linear, selfadjoint, positive and has compact resolvent. De6ne f, ˜ and h˜ as, for any u ∈ L2 ( ), ˜ f(u)(x) = −f(x; u(x)) + u(x); g(u)(x) ˜ = −g(x; u(x)); ˜ h(u)(x) = −h(x; u(x)): Then problem (1.1)–(1.3) can be written as  d  ˜  u = −Au + f(u) + g(v); ˜   dt   d ˜ v = − v + h(u);   dt     (u(0); v(0)) = (u0 ; v0 ):

(2.2)

The conditions stated in (2.1) are suMcient to guarantee the existence of the global attractor of the semiLow S(t) induced by problem (1.1) – (1.3). See Marion [10] for the proof of the theorem below. Theorem 1. Under the assumptions stated in (2.1), the semi4ow S(t) associated with problem (2.2) possesses a global attractor A with 5nite Hausdor7 and fractal dimension. Moreover, A is bounded in L∞ ( ) × L∞ ( ). Based on Marion’s work, stronger regularity of the global attractor A can be obtained. Let b ¿ 0 be any constant, de6ne B2; 0 (b) = {(u; v) ∈ H 2 ( ) × L2 ( ): |u|H 2 ( ) ; |v| 6 b}; B2; 2 (b) = {(u; v) ∈ H 2 ( ) × H 2 ( ): |u|H 2 ( ) ; |v|H 2 ( ) 6 b}; B∞; ∞ (b) = {(u; v) ∈ L∞ ( ) × L∞ ( ): |u|L∞ ( ) ; |v|L∞ ( ) 6 b}: The following lemmas state some results that will be used later on. Lemma 2.1. Assume that the assumptions stated in (2.1) hold and the space dimension n 6 3. Then there exists a constant b0 such that (i) B2; 0 (b0 ) is an absorbing set for problem (2.2), i.e., for any bounded set B in H × H , there exits a T = T (B) ¿ 0 such that, the solution (u(t); v(t)) of (2.2) satis5es |u(t)|H 2 ( ) 6 b0 ;

|v(t)| 6 b0 ;

t ¿ T:

(ii) Let A is the global attractor of the semi4ow induced by problem (2.2), then A ⊂ B2; 2 (b0 ) ∩ B∞; ∞ (b0 ):

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Lemma 2.2. Assume that the assumptions stated in (2.1) hold, and (u(t); v(t)) is the solution of problem (2.2). Then (i) For any (u0 ; v0 ) ∈ H × H , t0 ¿ 0 and T ¿ t0 ¿ 0, (u; v) ∈ C([0; ∞); H × H ); u ∈ C((0; ∞); V1 ) ∩ L2 ((t0 ; T ); H 2 ( )): (ii) If n 6 3, for any (u0 ; v0 ) ∈ D(A) × D(A), (u(t); v(t)) ∈ H 2 ( ) × H 2 ( ) and is bounded in H 2 ( ) × H 2 ( ) for all t ¿ 0. For the proof of Lemma 2.1, see [10,14]. The conclusions of Lemma 2.2 is contained in [14] except the boundedness part of (u(t); v(t)) in H 2 ( ) × H 2 ( ) which is proved below. Assume (u0 ; v0 ) ∈ D(A) × D(A). There exists a constant M such that |u(t)|H 2 ( ) 6 M

for all t ¿ 0:

From (1.1), we have − t

v(t; x) = e



v0 (x) +

0

t

(2.3)

e− (t−s) h(x; u(s; x)) ds

and, therefore, for any i; j: 1 6 i; j 6 3, we have  t − t vxi xj (t; x) = e v{0; xi xj } + e− (t−s) (hxi xj + huxj uxi + huxi uxj ) ds  +

0

0

t

e− (t−s) (huu uxi uxj + hu uxi xj ) ds:

(2.4)

Inequality (2.3) and the embedding of H 2 ( ) into L∞ ( ) in case of n 6 3 imply that u(t; x) is bounded in L∞ ( ). Since h is C 2 on O ×R, h and all its partial derivatives of O s ¿ 0}. This implies that order 2 or lower are bounded on the set S ={(x; u(s; x)): x ∈ ; 2 2 vxi xj (t; x) ∈ L ( ) and is bounded in L ( ) for t ¿ 0. For example, if we let huu ∞ denote the bound of huu (x; u(s; x)) on S, by HSolder inequality, we have 2   t   t  t   − (t−s) 2  e− (t−s) huu ux ux ds d x 6 e |h u u | ds d x e− (t−s) ds uu xi xj i j  

0



0

6

1 huu 2∞

6

1 huu 2∞

 

0

 

0

t

0

t

e− (t−s) |uxi uxj |2 ds d x 1 e− (t−s) (|uxi |4 + |uxj |4 ) ds d x 2

1 Chuu 2∞ ; 2 where we have used the embedding of H 1 ( ) into L4 ( ) for n 6 3 and C is a constant. Similarly, we can show that all the other terms in the expression (2.4) of vxi xj are bounded in L2 ( ). Therefore, v(t; x) ∈ H 2 ( ) and remains bounded for all t ¿ 0. 6

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3. Determining nodes and nodal values For any positive integer N and any set of points E = {xj }N1 ⊂ , de6ne d = dN = max{min{|xi − xj |; 1 6 j 6 N; j = i}; 1 6 i 6 N }: For any w ∈ L2 ( ), de6ne $(w) = max{|w(xi )|: 0 6 i 6 N } provided that w is such that w(xi ) is well de6ned for all i = 1; 2; : : : ; N . Lemma 3.1. Assume the space dimension n 6 3. There exist positive constants c1 and c2 such that, for any w ∈ H 2 ( ), the following inequalities hold: |w|2∞ 6 $2 (w) + c2 dN1=2 |Aw|2 ; |w|2 6 c1 $2 (w) + c2 dN |Aw|2 ; |A1=2 w|2 6 c1 dN−1=2 $2 (w) + c2 dN1=2 |Aw|2 : The proof of this lemma can be found in [4]. Although it is possible to obtain sharper estimates than the ones stated in the lemma (see [6,7] for example), we will not pursue in that direction since we are mainly concerned with the 6niteness of the determining node set. Theorem 2. Under the conditions of Lemma 2:1, there exits a  ¿ 0 such that, for any E = {xj }N1 ⊂ with dN ¡ , if (u; v) and (u; O v) O are two equilibrium solutions of problem (1.1) –(1.3) satisfying u(xj ) = u(x O j );

j = 1; 2; : : : ; N

then, u = uO

v = v: O

Proof. Since u and uO are equilibrium solutions, we have  1 −d1u + f(x; u) + g x; h(x; u(x)) = 0;  1 −d1uO + f(x; u) O + g x; h(x; u(x)) = 0: Therefore, u − uO satis6es



(3.1)

 1 1 −d((u − u) O + f(x; u) − f(x; u) O + g x; h(x; u(x)) − g x; h(x; u(x)) O = 0:

Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

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Since u; uO ∈ D(A), u; uO ∈ L∞ ( ). The assumption that f, g and h are C 2 -functions on O × R implies that |A(u − u)(x)| O 6 |f(x; u) − f(x; u)| O + |u(x) − u(x)| O       1 1 + g x; h(x; u) − g x; h(x; u) O  6 C|u(x) − u(x)|; O where C is a constant independent of u and u. O Therefore, by Lemma 3.1, one has |A(u − u)| O 2 6 C 2 |u − u| O 2 6 C 2 c2 dN |A(u − u)| O 2: Clearly, if dN is suMciently small, we have A(u − u) O = 0 which implies u = uO and v = v. O Theorem 3. Under the assumptions of Lemma 2:1, there exists a  ¿ 0 such that, for any E ={xj }N1 ⊂ with dN ¡ , if (u(t; x); v(t; x)) is a solution of problem (1.1) – (1.3) satisfying u(t; xj ) → )j ;

j = 1; 2; : : : ; N;

(3.2)

as t → ∞ for some constant )j , then there exists a unique solution u∞ (x) of (3.1) that satis5es u∞ (xj ) = )j ;

for all j = 1; 2; : : : ; N:

(3.3)

Moreover, if v∞ (x) = (1= )h(x; u∞ (x)), (u∞ (x); v∞ (x)) is an equilibrium solution of (1.1) and (u(t; x); v(t; x)) → (u∞ (x); v∞ (x));

as t → ∞:

(3.4)

The convergence in (3.4) is in the metric of V × V and L∞ ( ) × L∞ ( ). Proof. First we assume that (u0 ; v0 ) ∈ D(A) × D(A). By Lemma 2.2, we have (u(t); v(t)) ∈ D(A) × D(A) and there exists a constant M ¿ 0 such that |u(t)| 6 M

|v(t)| 6 M;

t ∈ [0; ∞) × :

(3.5)

For any s ¿ 0, let (u(t); ˆ v(t)) ˆ = (u(t + s); v(t + s)). Then (u(t); ˆ v(t)) ˆ ∈ D(A)) × D(A) satis6es (3.5) and (u − u; ˆ v − v) ˆ satis6es the following equations:  d  ˜ ˜ u)  ˆ = −A(u − u) ˆ + f(u) − f( ˆ + g(v) ˜ − g( ˜ v); ˆ  dt (u − u) (3.6)    d (v − v) ˜ − h( ˜ u): ˆ = − (v − v) ˆ + h(u) ˆ dt

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Taking the inner product of the 6rst equation of (3.6) with A(u − u), ˆ one has 1 d 1=2 ˜ ˜ u); ˆ 2 = −|A(u − u)| ˆ 2 + f(u) − f( ˆ A(u − u) ˆ |A (u − u)| 2 dt + g(v) ˜ − g( ˜ v); ˆ A(u − u): ˆ Since f, g and h are C 2 functions on O × R, by (3.5), there exists a constant K such that ˜ ˜ u)| |f(u) − f( ˆ 6 K|u − u|; ˆ |g(u) ˜ − g( ˜ u)| ˆ 6 K|u − u|; ˆ

|A1=2 g(u) ˜ − (g( ˜ u)| ˆ 6 K|A1=2 (u − u))|; ˆ

˜ − h( ˜ u)| |h(u) ˆ 6 K|u − u|; ˆ

˜ − (˜u)| |A1=2 (h(u) ˆ 6 K|A1=2 (u − u))|: ˆ

(3.7)

It follows from (2.1), (3.7) and Lemma 3.1 that 1 d 1=2 |A (u − u)| ˆ 2 6 −|A(u − u)| ˆ − u)| ˆ ˆ 2 + K|u − uA(u 2 dt + K|A1=2 (v − v)A ˆ 1=2 (u − u)| ˆ 1 6 −|A(u − u)| ˆ 2 + K 2 |u − u| ˆ 2 + |A(u − u)| ˆ 2 4 K 2 1=2 |A (u − u)| + |A1=2 (v − v)| ˆ 2+ ˆ 2 4 3 6 − |A(u − u)| ˆ 2 ˆ 2 + |A1=2 (v − v)| 4 4 + K 2 (c1 $2 (u − u) ˆ + c2 dN |A(u − u)| ˆ 2) K2 ˆ + c2 dN1=2 |A(u − u)| ˆ 2) (c1 dN−1=2 $2 (u − u)  c2 K 2 1=2 3 2 |A(u − u)| ˆ 2 6− − c 2 K dN − d 4 N  1 −1=2 2 1=2 2 2 $ (u − u): ˆ + c1 K 1 + dN ˆ + |A (v − v)| 4 +

(3.8)

Next, taking the inner product of the second equation in (3.6) with A(v − v), ˆ one has 1 d 1=2 ˜ − h( ˜ u); ˆ 2 = − |A1=2 (v − v)| ˆ 2 + h(u) ˆ A(v − v): ˆ |A (v − v)| 2 dt

(3.9)

By Lemma 3.1 and (3.7), one has 1 d 1=2 1=2 ˜ − h( ˜ u))A |A (v − v)| ˆ 2 6 − |A1=2 (v − v)| ˆ 2 + |A1=2 (h(u) ˆ (v − v)| ˆ 2 dt ˆ 2+ 6 − |A1=2 (v − v)|

1 1=2 ˜ ˜ u))| ˆ 2 + |A1=2 (v − v)| ˆ 2 |A (h(u) − h( 4

Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

6−

3 1=2 K 2 1=2 |A (u− u)| ˆ 2+ ˆ 2 |A (v− v)| 4

6−

3 1=2 c2 K 2 1=2 ˆ 2+ ˆ 2 |A (v − v)| dN |A(u − u)| 4

+

881

c1 K 2 −1=2 2 ˆ dN $ (u − u):

(3.10)

Adding (3.8) and (3.10), one obtains 1 d ˆ 2 + |A1=2 (v − v)| ˆ 2) (|A1=2 (u − u)| 2 dt  2c2 K 2 1=2 3 − c 2 K 2 dN − dN |A(u − u)| ˆ 2 − |A1=2 (v − v)| ˆ 2 6− 4 2  2 + c1 K 2 1 + dN−1=2 $2 (u − u): ˆ

(3.11)

Let  ¿ 0 be chosen so small that 3 2c2 K 2 1=2 1 − c2 K 2  −  ¿ : 4 2

(3.12)

If dN ¡ , setting y(t) = |A1=2 (u − u)| ˆ 2 + |A1=2 (v − v)| ˆ 2 , one arrives at  d 2 ˆ 2 + 2c1 K 2 1 + dN−1=2 $2 (u − u) ˆ y(t) 6 −|A(u − u)| ˆ 2 − |A1=2 (v − v)| dt ˆ 2) 6 −(-1 |A1=2 (u − u)| ˆ 2 + |A1=2 (v − v)|  2 −1=2 2 2 + 2c1 K 1 + dN $ (u − u) ˆ  2 −1=2 2 2 6 −.y(t) + 2c1 K 1 + dN $ (u − u); ˆ

(3.13)

where . = min(-1 ; ) and -1 ¿ 0 is the 6rst eigenvalue of A. Assumption (3.2) implies that $(u − u) ˆ → 0 as t → ∞ and thus, for any / ¿ 0, T ¿ 0 can be chosen so large that  2 −1=2 2 2 2c1 K 1 + dN $ (u − u) ˆ ¡ / for all t ¿ T: This yields d y(t) 6 − .y(t) + /: dt An application of the Gronwall inequality yields that / y(t) 6 y(T )e−.(t−T ) + (1 − e−.(t−T ) ): .

(3.14)

(3.15)

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Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

This says that y(t)=|A1=2 (u− u)| ˆ 2 +|A1=2 (v− v)| ˆ 2 is a Cauchy sequence, or equivalently, that (u(t); v(t)) is a Cauchy Sequence in V as t → ∞. Therefore (u(t); v(t)) has a limit, which is denoted as (u∞ ; v∞ ). By (3.5), we can easily see that (u(t; x); v(t; x)) converges to (u∞ (x); v∞ (x)) uniformly as t → ∞ and thus (3.3) holds. Using standard arguments, one can show that (u∞ (x); v∞ (x)) is the unique equilibrium solution of (1.1). Now, we deal with general initial data (u0 ; v0 ) ∈ H × H . By Lemma 2.2, we know that there is a t0 such that u(t0 ) ∈ D(A). Without loss of generality, we will assume that u0 ∈ D(A), and therefore, u(t) ∈ D(A) for all t ¿ 0. We will write v(t; x)=v1 (t; x)+ v2 (t; x) where v1 (t; x) = v0 (x)e− t ;  t v2 (t; x) = e− (t−s) h(X; u(s; x)) ds: 0

It can be shown that v2 (t) ∈ H 2 ( ) is bounded and satis6es the following equation: d ˜ (3.16) v2 = − v2 + h(u): dt If (u; ˆ v) ˆ = (u; ˆ vˆ1 + vˆ2 ) is de6ned the same way as before, corresponding to (3.8), one has 1 d 1=2 ˆ 2 6 −|A(u − u)| ˆ 2 + K|u − uA(u ˆ − u)| ˆ |A (u − u)| 2 dt + K|A1=2 (v2 − vˆ2 )A1=2 (u − u)| ˆ ˜ v) ˆ − g( ˜ vˆ2 )|)|A(u − u)| ˆ + (|g(v) ˜ − g(v ˜ 2 )| + |g( 3 6 − |A(u − u)| ˆ 2 + |A1=2 (v2 − vˆ2 )|2 ˆ 2 + K 2 |u − u| 4 4 K 2 1=2 |A (u − u)| ˆ + 2K(|v1 | + |vˆ1 |)|A(u − u)| ˆ 3 ˆ + c2 dN |A(u − u)| ˆ 2) 6 − |A(u − u)| ˆ 2 + K 2 (c1 $2 (u − u) 4 ˆ + |A1=2 (v2 − vˆ2 )|2 + 2K(|v1 | + |vˆ1 |)|A(u − u)| 4 +

K2 ˆ + c2 dN1=2 |A(u − u)| ˆ 2) (c1 dN−1=2 $2 (u − u)  c2 K 2 1=2 3 2 |A(u − u)| ˆ 2 6− − c 2 K dN − d 4 N  1 −1=2 2 1=2 2 2 $ (u − u) + |A (v2 − vˆ2 )| + K c1 1 + dN ˆ 4 +

+ 2K(|v1 | + |vˆ1 |)|A(u − u)|: ˆ

(3.17)

Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

883

Further, Eq. (3.16) implies that 1 d 1=2 ˜ − h( ˜ u); ˆ A(v2 − vˆ2 ) A (v2 − vˆ2 )2 = − |A1=2 (v2 − vˆ2 )|2 + h(u) 2 dt 6−

3 1=2 K 2 1=2 |A (v2 − vˆ2 )|2 + ˆ 2 |A (u − u)| 4

6−

3 1=2 |A (v2 − vˆ2 )|2 4

K2 ˆ + c2 dN1=2 |A(u − u)| ˆ 2 ): (c1 dN−1=2 $2 (u − u) Adding (3.17) and (3.18), one obtains 1 d ˆ 2 + |A1=2 (v2 − vˆ2 )|2 ) (|A1=2 (u − u)| 2 dt  2c2 K 2 1=2 3 ˆ 2 − |A1=2 (v2 − vˆ2 )|2 6− − c 2 K 2 dN − dN |A(u − u)| 4 2  2 + K 2 c1 1 + dN−1=2 $2 (u − u) ˆ + 2K(|v1 | + |vˆ1 |)|A(u − u)|: ˆ +

(3.18)

(3.19)

By choosing dN ¡  with  ¿ 0 determined by (3.12), we have d (|A1=2 (u − u)| ˆ 2 + |A1=2 (v2 − vˆ2 )|2 ) dt ˆ 2 + |A1=2 (v2 − vˆ2 )|2 ) + 1(t); 6 − (-1 |A1=2 (u − u)| where -1 ¿ 0 is the 6rst eigenvalue of A and 2

ˆ + 2K c1 1(t) = 4K(|v1 | + |vˆ1 |)|A(u − u)|



(3.20)

2 −1=2 2 1 + dN $ (u − u): ˆ

Since |A(u − u)| ˆ is bounded on [0; ∞) and |v1 |; |vˆ1 | → 0 as t → ∞, one can show 1(t) → 0

as t → ∞:

Therefore, similarly as before, we can show that (u(t; x); v2 (t; x)) is a Cauchy sequence in V and, therefore, has a limit (u∞ (x); v∞ (x)), which is also the limit of (u(t; x); v(t; x)) as t → ∞. Theorem 4. Under the assumptions of Lemma 2.2, there exists a  ¿ 0 such that, for any E = {xj } ⊂ with dN ¡ , if (u1 (t; x); v1 (t; x)) and (u2 (t; x); v2 (t; x)) are two solutions of problem (1.1)– (1.3) satisfying u1 (t; xj ) − u2 (t; xj ) → 0;

j = 1; 2; : : : ; N;

(3.21)

as t → ∞, then (u1 (t; ·); v1 (t; ·)) − (u2 (t; ·); v2 (t; ·)) → 0

as t → ∞:

(3.22)

The convergence is in the metric of V × V and L∞ ( ) × L∞ ( ). The Proof of Theorem 4 is similar to the proof of Theorem 3, we omit the details.

884

Y. Lu, Z. Shao / Nonlinear Analysis 54 (2003) 873 – 884

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