SCIENCE IN CHINA (Series A)

Vol. 46 No. 5

September 2003

Analyticity of thermoelastic plates with dynamical boundary conditions ZHANG Qiong (

 ) & HUANG Falun (
)

Department of Mathematics, Sichuan University, Chengdu 610064, China Correspondence should be addressed to Zhang Qiong (email: [email protected]) Received July 16, 2002 Abstract We consider a thermoelastic plate with dynamical boundary conditions. Using the contradiction argument of Pazy’s well-known analyticity criterion and P.D.E. estimates, we prove that the corresponding C0 semigroup is analytic, hence exponentially stable. Keywords:

thermoelasticity, C0 semigroup, analyticity, exponential stability.

DOI: 10.1360/02ys0222

1

Introduction

Let Ω be a bounded, open set in R2 with C 2 -smooth boundary Γ . We assume that Γ =  Γ0 Γ1 , Γ0 and Γ1 are relatively open subsets of Γ , Γ0 = ∅ has positive boundary measure, and  Γ0 Γ1 = ∅. We consider the following hybrid thermoelastic system: ⎧  u + Δ2 u + Δθ = 0 in Ω × R+ , ⎪ ⎪ ⎪ ⎪ ⎪ θ − Δθ + θ − Δu = 0 in Ω × R+ , ⎪ ⎪ ⎪ ⎪ ⎪ on Γ0 × R+ , ⎪ ⎨ u = ∂ν u = 0 (1.1) J∂ν u + B1 u + θ = 0 on Γ1 × R+ , ⎪ ⎪  + ⎪ ⎪ ρu − B u − ∂ θ = 0 on Γ × R , 2 ν 1 ⎪ ⎪ ⎪ ⎪ ⎪ θ + λθ = 0, λ  0 on Γ × R+ , ∂ ν ⎪ ⎪ ⎩ u(0) = u0 , u (0) = u1 , θ(0) = θ0 in Ω , where ρ is the linear boundary density and J is the bending moment of inertia per unit length of the boundary. The boundary operators B1 , B2 are given by  2 2 ∂2u 2∂ u 2∂ u − ν1 2 − ν2 2 , B1 u = Δu + (1 − μ) 2ν1 ν2 ∂x1 ∂x2 ∂x2 ∂x1

 2  2 ∂ u ∂2u ∂ u 2 2 B2 u = ∂ν Δu + (1 − μ)∂τ (ν1 − ν2 ) + ν1 ν2 − 2 . ∂x1 ∂x2 ∂x22 ∂x1 ν = (ν1 , ν2 ) is the unit outer normal vector and τ = (−ν2 , ν1 ) is the unit tangent vector. 0 < μ <

1 2

denotes the elastic Poisson’s ratio. The PDE model (1.1) describes a thermoelastic thin plate, which is clamped on one part of the boundary and rimmed along the other free part with a flange that has mass and moment of inertia of the boundary. The purpose of this paper is to investigate the analyticity of the C0 semigroup associated with (1.1).

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There have been many papers on analyticity of thermoelastic system. Russell[1] considered the problem with boundary conditions u = Δu = θ = 0. Liu and Renardy[2] studied the problem with the clamped edge and zero temperature boundary conditions. Liu and Liu[3] and Lasiceka and Triggiani[4] both proved analyticity of abstract thermoelastic systems, which include several models with different boundary conditions. Analyticity of the thermoelastic plate with the Neumann temperature boundary conditions and the hinged/free edge was showed by Lasiceka and Triggiani[5,6]. As for the dynamical mechanical boundary conditions[7] , to our knowledge, similar results are not available in the literature. In this paper, we prove that the semigroup associated with closed-loop system (1.1) is analytic. The proof employs the contradiction argument of Pazy’s analyticity criterion[8,9], properties of boundary operators B1 , B2 , and estimates associated with the partial differential equations. Once analyticity is established, exponential stability can be obtained by excluding the possibility that the generator of the C0 contraction semigroup has spectrum on the imaginary axis.

2

Abstract formulation and main results

We will consider system (1.1) as an abstract evolution equation in a certain Hilbert space, for which we introduce some definitions and notations. First, we introduce the following Hilbert space: V = {u ∈ H 2 (Ω ) | u|Γ0 = ∂ν u|Γ0 = 0}, . u2V = a(u), ∀u ∈ V, where a(u) = a(u, u), and a(·, ·) is the sesquilinear form defined by   2 ∂ 2 u1 ∂ 2 u2 ∂ 2 u1 ∂ 2 u2 ∂ 2 u1 ∂ 2 u2 ∂ u1 ∂ 2 u2 a(u1 , u2 ) = + +μ + ∂x21 ∂x21 ∂x22 ∂x22 ∂x21 ∂x22 ∂x22 ∂x21 Ω  ∂ 2 u1 ∂ 2 u2 + 2(1 − μ) dx, ∀u1 , u2 ∈ V. ∂x1 ∂x2 ∂x1 ∂x2 It is well known that a(·, ·) is an equivalent norm on V since Γ0 is non-empty and has positive boundary measure (see, e.g. ref. [10]). Next, we denote the Hilbert space H to be . H = V × L2 (Ω ) × L2 (Ω ) × L2 (Γ1 ) × L2 (Γ1 ), 1

zH = [u2V + v2L2 (Ω ) + θ2L2 (Ω ) + Jξ2L2 (Γ1 ) + ρη2L2 (Γ1 ) ] 2 , ∀z = (u, v, θ, ξ, η) ∈ H. Finally, let A be the following operator on H:     u ∈ W, v ∈ V, θ ∈ H 2 (Ω ), ξ = ∂ν v|Γ1 ,  D(A) = (u, v, θ, ξ, η) ∈ H  ,  η = v|Γ1 , (∂ν θ + λθ)|Γ1 = 0  1 A(u, v, θ, ξ, η) = v, −Δ2 u − Δθ, Δv + Δθ − θ, − (B1 u + θ) , J 1 (B2 u + ∂ν θ) ρ

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633

where . W = {u ∈ V | Δ2 u ∈ L2 (Ω ), B1 u, B2 u ∈ L2 (Γ1 )}. Hence, we can rewrite system (1.1) in the following abstract form  z  (t) = Az(t), z = (u, u , θ, ∂ν u |Γ1 , u |Γ1 ),

z(0) = z0 ∈ H.

We shall begin by giving preliminary results regarding the well-posedness of system (1.1) and spectrum properties of the operator A. Theorem 2.1.

Suppose the boundary Γ is C 2 -smooth. Then

(i) A generates a C0 contraction semigroup etA on H. (ii) The operator A has compact resolvent on H, and {iω | ω ∈ R} ∈ ρ(A). Our main result is as follows. Theorem 2.2.

C0 semigroup etA is analytic on H.

We know that the analytic C0 semigroup satisfies spectrum determining growth assumption. Hence by Theorem 2.1(ii) and Theorem 2.2 we have: Corollary 2.1.

C0 semigroup etA is exponentially stable, i.e. there exist positive constants

M and  such that the energy of system (1.1), defined by

1  2 E(u, θ, t) = a(u(t)) + |u (t)| dx + |θ(t)|2 dx 2 Ω Ω   2 |∂ν u (t)| dΓ + ρ |u (t)|2 dΓ , +J Γ1

(2.1)

Γ1

satisfies E(u, θ, t)  M e−t ,

∀t  0.

3 Proofs The following integration-by-parts formula is needed in the analysis below. Δ2 uvdx =a(u, v) + B2 uvdΓ − B1 u∂ν vdΓ Ω Γ1 Γ1 ΔuΔvdx + ∂ν ΔuvdΓ − Δu∂ν vdΓ = Ω

for all u ∈ W and v ∈ V .

Γ1

(3.1)

Γ1

3.1 Proof of Theorem 2.1 Proof of (i). In establishing the semigroup generation of A, we will show that the conditions of the Lumer-Phillips Theorem[8] are satisfied. To show the dissipativity of A, for z = (u, v, θ, ξ, η) ∈ D(A) we have Re(Az, z)H = − (|∇θ|2 + |θ|2 )dx − λ |θ|2 dΓ  0. Ω

(3.2)

Γ

To show the maximality of A, for arbitrary f = (f1 , f2 , f3 , f4 , f5 ) ∈ H, we solve the equation Az = f,

z = (u, v, θ, ξ, η) ∈ D(A),

(3.3)

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which implies v = f1 in V,

(3.4)

2

Δ u + Δθ = −f2 in L2 (Ω ),

(3.5)

Δv + Δθ − θ = f3 in L2 (Ω ),

(3.6)

B1 u + θ = −Jf4 in L2 (Γ1 ),

(3.7)

B2 u + ∂ν θ = ρf5 in L2 (Γ1 ).

(3.8)

By (3.4), we have ξ = ∂ν v|Γ1 = ∂ν f |Γ1 ,

η = v|Γ1 = f |Γ1 .

Using (3.4) and (3.6) yields that θ satisfies  Δθ − θ = −Δf1 + f3 ∂ν θ + λθ = 0

in

Ω,

on

Γ.

(3.9)

(3.10)

Hence we have that for any f1 ∈ V and f3 ∈ L2 (Ω ) there exists a unique θ ∈ H 2 (Ω ) satisfying (3.10)(see, e.g. ref. [11]). Furthermore, replacing (3.6) into (3.5), we obtain that u satisfies ⎧ 2 in Ω, ⎪ ⎨ Δ u = −θ + Δf1 − f2 − f3 B1 u = −θ − Jf4 on Γ1 , ⎪ ⎩ B2 u = λθ + ρf5 on Γ1 , which is equivalent to the following variational equation: a(u, φ) = − (θ − Δf1 + f2 + f3 )φdx Ω − ((ρf5 + λθ)φ + (Jf4 + θ)∂ν φ)dΓ

(3.11)

(3.12)

Γ1

for all φ ∈ V . Thanks to the Lax-Milgram theorem, eq. (3.12) admits a unique solution u ∈ V 1 for any f ∈ H and θ ∈ H 2 (Ω ). In summary, we get that for any f ∈ H there exists a unique z ∈ D(A) such that Az = f . The closed graph theorem implies that the inverse A−1 is bounded, and hence 0 ∈ ρ(A). Therefore, (ωI − A) = A(ωA−1 − I) is boundedly invertible for any ω satisfying |ω| < A−1 −1 . Applying Theorem 1.4.6 in ref. [8] yields D(A) = H. The proof of (i) is complete. Proof of (ii).

5

Thanks to the elliptic theory[12] , we deduce from (3.11) that u ∈ H 2 (Ω )

and u

5 2

H (Ω )

  C1 f H + θ

 1 2

H (Ω )

,

C1 > 0.

(3.13)

Similarly, by (3.10) θH 2 (Ω )  C2 f H ,

C2 > 0.

(3.14)

Therefore, from (3.9), (3.13) and (3.14), we have (u, v, θ, ξ, η)

5

1

3

H 2 (Ω )×V ×H 2 (Ω )×H 2 (Γ1 )×H 2 (Γ1 )

 Cf H ,

C > 0.

(3.15)

−1

It follows that A is compact. Finally, since we have proved that 0 ∈ ρ(A) and A−1 is compact, it follows that the spectrum of A consists of its eigenvalues. Thus, if {iω | ω ∈ R} ∈ ρ(A) is not true, there is a real

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635

number ω = 0 such that iω is an eigenvalue of A. It turns out that there is a vector function z = (u, v, θ, ξ, η) ∈ D(A) with zH = 1 such that (iωI − A)z = 0.

(3.16)

This means that iωu − v = 0 in V,

(3.17)

2

iωv + Δ u + Δθ = 0 in L2 (Ω ),

(3.18)

iωθ − Δv − Δθ + θ = 0 in L2 (Ω ),

(3.19)

iωJ∂ν v + B1 u + θ = 0 in L2 (Γ1 ),

(3.20)

iωρv − B2 u + λθ = 0 in L2 (Γ1 ).

(3.21)

Taking the real part of the inner product of (3.16) with z in H and applying (3.2) yields |∇θ|2 dx = |θ|2 dx = |θ|2 dΓ = 0. Ω

Ω

(3.22)

Γ

Consequently, θ=0

in Ω .

(3.23)

Therefore, by (3.17)—(3.19) and (3.23) we obtain Δ2 u − ω 2 u = 0 in Ω , Δu = 0 in Ω .

(3.24)

It is clear that u = 0 in V (see, e.g. ref. [10]). It turns out from (3.17) that v = 0 in V , and then we have v = 0 in L2 (Ω ) and ξ = ∂ν v|Γ1 = 0, η = v|Γ1 = 0 in L2 (Γ1 ). Thus we get a contradiction. 3.2 Proof of Theorem 2.2 From Theorem 2.1 (ii) and Pazy’s analyticity criterion (Theorem 5.2 in ref. [8] or Theorem 1.3.3 in ref. [9]), our goal is to verify the following estimate: lim ω(iωI − A)−1   ∞.

(3.25)

|ω|→∞

If (3.25) is not true, then there exist {ωn } ⊂ R with |ωn | → ∞ and a sequence of vector functions zn = (un , vn , θn , ξn , ηn ) ∈ D(A2 ) such that zn H = 1, n = 1, 2, · · · ,     1  lim A zn  i−  = 0. n→∞  ωn

(3.26) (3.27)

H

(3.27) implies 1 . f1,n = iun − vn −→ 0 in ωn 1 2 1 . f2,n = ivn + Δ un + Δθn −→ 0 ωn ωn

V,

(3.28)

in L2 (Ω ),

(3.29)

1 1 1 . f3,n = iθn − Δvn − Δθn + θn −→ 0 in L2 (Ω ), ωn ωn ωn

(3.30)

1 1 . f4,n = iJ∂ν vn + B1 un + θn −→ 0 in L2 (Γ1 ), ωn ωn

(3.31)

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1 λ . f5,n = iρvn − B2 un + θn −→ 0 ωn ωn

Vol. 46

in L2 (Γ1 ).

(3.32)

1 Re(Azn , zn )H = 0. Thus, we obtain via (3.2) ωn 1 1 lim 1 ∇θn L2 (Ω ) = lim 1 θn L2 (Ω ) = 0, n→∞ |ω | 2 n→∞ |ω | 2 n n (3.33) 1 1 lim ∂ θ  = lim θ  = 0. ν n L2 (Γ ) n L2 (Γ ) 1 1 n→∞ |ω | 2 n→∞ |ω | 2 n n We shall complete the proof by several lemmas. In fact, we have the following results for

Moreover, it follows from (3.27) that lim

n→∞

zn = (un , vn , θn , ξn , ηn ) ∈ D(A2 ) satisfy (3.26) and (3.27). Lemma 3.1. Lemma 3.2. Lemma 3.3. Lemma 3.4.

(i) lim θn L2 (Ω ) = 0; (ii) lim ∂ν un L2 (Γ1 ) = lim un L2 (Γ1 ) = 0. n→∞

n→∞

n→∞

lim un 2V = lim (vn 2L2 (Ω ) + J∂ν vn 2L2 (Γ1 ) + ρvn 2L2 (Γ1 ) ) =

n→∞

n→∞

lim Δun L2 (Ω ) = 0.

1 . 2

n→∞

lim (vn 2L2 (Ω ) + J∂ν vn 2L2 (Γ1 ) + ρvn 2L2 (Γ1 ) ) = 0.

n→∞

It is clear that Lemma 3.2 is in contradiction to Lemma 3.4. Thus Theorem 2.2 is proved. 3.3 Proof of Lemmas 3.1—3.4 Proof of Lemma 3.1.

(i) We take the inner product of (3.30) with θn in L2 (Ω ), and use

(3.33), iθn 2L2 (Ω ) −

1 ωn

Ω

Δvn θn dx −→ 0.

(3.34)

By Gagliardo-Nirenberg inequality[5,6,9] and (3.26), (3.28), we have 1 1 1 1 2 2 1 vn H 1 (Ω )  1 vn V vn L2 (Ω )  1, n = 1, 2, · · · . 2 2 |ωn | |ωn | Combining (3.33) with (3.35) yields     1 1 1  ∇vn ∇θn dx  1 vn H 1 (Ω ) 1 ∇θn L2 (Ω ) −→ 0.  ωn 2 |ωn | |ωn | 2 Ω Similarly,

  1   ωn

  1 1 ∂ν vn θn dΓ   1 ∂ν vn L2 (Γ1 ) 1 θn L2 (Γ ) −→ 0. 2 |ωn | |ωn | 2 Γ1

Thus, by (3.36) and (3.37), we obtain    1    −→ 0. Δv θ dx n n  ωn 

(3.35)

(3.36)

(3.37)

(3.38)

Ω

Substituting (3.38) into (3.34) yields lim θn L2 (Ω ) = 0.

(3.39)

n→∞

(ii) It is clear from (3.28) that 1 ∂ν vn L2 (Γ1 ) . (3.40) |ωn |  1, we can deduce from (3.40) that lim ∂ν un L2 (Γ1 ) = 0.

lim ∂ν un L2 (Γ1 ) = lim

n→∞

Since |ωn | → ∞ and ∂ν vn L2 (Γ1 )

Similarly, we have lim un L2 (Γ1 ) = 0. n→∞

n→∞

n→∞

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637

Proof of Lemma 3.2.

We take the V -inner product of (3.28) with un , 1 a(un , vn ) −→ 0. (3.41) ia(un ) + ωn Furthermore, we take the inner product of (3.29) with vn in L2 (Ω ), take the inner product of (3.31), (3.32) with ∂ν vn , vn in L2 (Γ1 ), respectively, and then add them up,  |vn |2 dx + J |∂ν vn |2 dΓ + ρ |vn |2 dΓ 0 ←−i Ω Γ1 Γ1 1 1 1 + a(un , vn ) − ∇θn ∇vn dx + θn ∂ν vn dΓ ωn ωn Ω ωn Γ1  ←−i |vn |2 dx + J |∂ν vn |2 dΓ + ρ |vn |2 dΓ Ω

Γ1

Γ1

1 + a(un , vn ) (by (3.33) and (3.36)). ωn Hence combining (3.26), Lemma 3.1(i) with (3.41), (3.42), we get the desired result. Proof of Lemma 3.3.

(3.42)

We combine (3.28) with (3.30), and take the L2 (Ω )-inner product

with Δun :

 1 1 − (Δθn , Δun )L2 (Ω ) + i + (3.43) (θn , Δun )L2 (Ω ) −→ 0. ωn ωn Applying Lemma 3.1(i) and Lemma 3.2, we have lim (θn , Δun )L2 (Ω ) = 0. Hence it follows from −iΔun 2L2 (Ω )

n→∞

(3.43) that lim Δun 2L2 (Ω )

n→∞

1 (∇Δun L2 (Ω ) ∇θn L2 (Ω ) + Δun L2 (Γ ) ∂ν θn L2 (Γ ) ). |ωn | Furthermore, by (3.28)—(3.30), 1 Δ2 un L2 (Ω ) lim n→∞ |ωn |  1  lim 1+ θn L2 (Ω ) + vn L2 (Ω ) + Δun L2 (Ω ) n→∞ |ωn | 1 (by (3.26)).  lim

n→∞

Applying the interpolation theorem, we can deduce from (3.32) that 1 lim B2 un  12 H (Γ1 ) n→∞ |ωn |

 λ θn  12  lim + ρvn  12 H (Γ ) H (Γ1 ) n→∞ |ωn |

 1 1 1 1 λ ∂ν θn L2 2 (Γ ) θn L2 2 (Γ ) + ρ∂ν vn L2 2 (Γ1 ) vn L2 2 (Γ1 ) C  lim n→∞ |ωn |  C (by (3.33) and Lemma 3.2, C  , C  are positive constants). In summary, we know that

1 ωn u n

(3.44)

(3.45)

(3.46)

is the solution of the elliptic boundary value problem (3.29),

(3.31) and (3.32), and satisfies estimates (3.45), (3.46). Thus the elliptic regularity theory and the trace theorem lead to[6,9,12] 1 un H 4 (Ω )  C  , C  > 0. lim (3.47) n→∞ |ωn |

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Moreover, by Gagliardo-Nirenberg inequality and (3.47), Lemma 3.2,

√ C  √ . lim 1 un H 3 (Ω )  lim 1 un H 4 (Ω ) un V  4 n→∞ |ω | 2 n→∞ |ω | 2 2 n n Hence, applying trace theorem and (3.48) yields 1 1 1 1  2 2 , C  > 0. lim 1 Δun L2 (Γ )  lim 1 un H 3 (Ω ) un V  C n→∞ |ω | 4 n→∞ |ω | 4 n n We obtain the desired result by substituting (3.33), (3.48) and (3.49) into (3.44). 1

1

1 2

1 2

(3.48)

(3.49)

Proof of Lemma 3.4.

We take the inner product of (3.29) with vn in L2 (Ω ) to get 1 1 ivn 2L2 (Ω ) + (Δun , Δvn )L2 (Ω ) + (∂ν Δun vn − Δun vn )dΓ ωn ωn Γ1 1 + Δθn vn dx −→ 0. (3.50) ωn Ω By using the same argument as (3.35)—(3.38), we have 1 Δθn vn dx −→ 0. (3.51) ωn Ω And from (3.28) and Lemma 3.3, 1 (Δun , Δvn )L2 (Ω ) = −i lim Δun 2L2 (Ω ) = 0. (3.52) lim n→∞ ωn n→∞ Therefore, substituting (3.51) and (3.52) into (3.50) yields 1 2 ivn L2 (Ω ) + (∂ν Δun vn − Δun ∂ν vn )dΓ −→ 0. (3.53) ωn Γ1 Moreover, taking the inner product of (3.31) with ∂ν vn in L2 (Γ1 ) and using (3.33), we have 1 2 iJ∂ν vn L2 (Γ1 ) + B1 un ∂ν vn −→ 0. (3.54) ωn Γ1 Similarly, by (3.32) and (3.33), 1 2 iρvn L2 (Γ1 ) − B2 un vn −→ 0. (3.55) ωn Γ1 Thus, combining (3.53)—(3.55), we obtain ivn 2L2 (Ω ) + iJ∂ν vn 2L2 (Γ1 ) + iρvn 2L2 (Γ1 ) 1 1 − (B2 un − ∂ν Δun )vn dΓ + (B1 un − Δun )∂ν vn dΓ −→ 0. ωn Γ1 ωn Γ1 From Proposition 3C.1 and Proposition 3C.8 in ref. [6], we have

(3.56)

B1 un − Δun = −∂τ τ un − ( div ν)∂ν un ; B2 un − ∂ν Δun = ∂τ ντ un .

(3.57)

Substituting (3.57) and Lemma 3.1 (ii) into (3.56), we obtain i lim (vn 2L2 (Ω ) + J∂ν vn 2L2 (Γ1 ) + ρvn 2L2 (Γ1 ) ) n→∞ 1 (∂τ ντ un vn + ∂τ τ un ∂ν vn )dΓ . (3.58) = lim n→∞ ωn Γ 1 Furthermore, by interpolation theory and trace theorem, we have via (3.47) and (3.48) that 1 1 1 1 2 2 lim 3 ∂τ ντ un L2 (Γ1 )  lim 3 un H 4 (Ω ) un H 3 (Ω ) n→∞ |ω | 4 n→∞ |ω | 4 n n κ1 , κ1 > 0. (3.59)

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Thus, by the fact that |ωn | → ∞,     1 1  lim  ∂τ ντ un vn dΓ   lim ∂τ ντ un L2 (Γ1 ) vn L2 (Γ1 ) = 0. n→∞ ωn Γ n→∞ |ωn | 1 Similarly, we have from (3.48)     1 1 ∂τ τ un L2 (Γ1 ) ∂ν vn L2 (Γ1 ) = 0. ∂τ τ un ∂ν vn dΓ   lim lim  n→∞ ωn Γ n→∞ |ωn | 1 Therefore, Lemma 3.4 is established by substituting (3.60) and (3.61) into (3.58).

639

(3.60)

(3.61)

4 The case Γ0 =∅ When Γ0 = ∅, the sesquilinear form a(·, ·) is not coercive and so it is not the equivalent norm on H 2 (Ω ). In this section, we deal with this difficulty by applying a control on one of the boundary conditions. The resulting model is described as u + Δ2 u + Δθ = 0 



θ − Δθ + θ − Δu = 0 

J∂ν u + B1 u + θ = 0 

ρu − B2 u + u − ∂ν θ = 0 ∂ν θ + λθ = 0,

λ>0



u(0) = u0 , u (0) = u1 , θ(0) = θ0

in Ω × R+ , +

in Ω × R , +

on Γ × R , +

(4.1) (4.2) (4.3)

on Γ × R ,

(4.4)

on Γ × R+ ,

(4.5)

in Ω .

(4.6)

Similar to section 2, we first induce the energy space . H = H 2 (Ω ) × L2 (Ω ) × L2 (Ω ) × L2 (Γ1 ) × L2 (Γ1 ), where the equivalent norm on H 2 (Ω ) is defined by . u2H 2 (Ω ) = a(u) + u2L2(Γ ) , Remark 4.1.

∀u ∈ H 2 (Ω ).

(4.7)

The term u|Γ in the boundary condition (4.4) is critical since it is the 1

presence of the term u|Γ that makes [a(u) + u2L2(Γ ) ] 2 an equivalent norm on H 2 (Ω ) (see ref. [10] or refs. [5, 6]). In fact, a direct computation gives    2 2   2 2  2 2     ∂ u  ∂ ∂ u u  dx. a(u) = μ|Δu|2 + (1 − μ)  2  +  2  + 2(1 − μ)  ∂x1 ∂x2 ∂x1 x2  Ω

(4.8)

Thus it follows from a(u) + u2L2 (Γ ) = 0 that u is a harmonic function satisfying u|Γ = 0. Therefore, u = 0 in Ω .

Define the unbounded operator A0 on H:     u ∈ W , v ∈ H 2 (Ω ), θ ∈ H 2 (Ω ), ξ = ∂ v| , 0 ν Γ  , D(A0 ) = (u, v, θ, ξ, η) ∈ H   η = v|Γ , (∂ν θ + λθ)|Γ = 0  1 A(u, v, θ, ξ, η) = v, −Δ2 u − Δθ, Δv + Δθ − θ, − (B1 u + θ), J 1 (B2 u − u + ∂ν θ) , ρ where . W0 = {u ∈ H 2 (Ω ) | Δ2 u ∈ L2 (Ω ), B1 u, B2 u ∈ L2 (Γ )}.

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By using the same argument as in section 3, we can prove that: Theorem 4.1. Corollary 4.1.

A0 generates an analytic C0 semigroup etA0 on H. The energy of system (4.1)—(4.6) is exponentially stable, i.e. there exist

positive constants M0 and 0 such that the energy

1 2  2 E0 (u, θ, t) = a(u(t)) + |u(t)| dΓ + |u (t)| dx + |θ(t)|2 dx 2 Γ Ω Ω   2  2 +J |∂ν u (t)| dΓ + ρ |u (t)| dΓ Γ

Γ

satisfies E0 (u, θ, t)  M0 e−0 t ,

∀t  0.

Acknowledgements The first author would like to thank Prof. Liu Kangsheng, Prof. Liu Zhuangyi, Prof. Lasiecka Irena, Prof. Zhang Xu. This work was supported by the National Natural Science Foundation of China (Grant No. 10071079).

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