On uniform decay of wave equation of carrier model subject to memory condition at the boundary

J. Korean Math. Soc. 44 (2007), No. 4, pp. 1013–1024

ON UNIFORM DECAY OF WAVE EQUATION OF CARRIER MODEL SUBJECT TO MEMORY CONDITION AT THE

BOUNDARY

Jeong Ja Bae and Suk Bong Yoon

Reprinted from the

Journal of the Korean Mathematical Society Vol. 44, No. 4, July 2007

c

°2007 The Korean Mathematical Society

J. Korean Math. Soc. 44 (2007), No. 4, pp. 1013–1024

ON UNIFORM DECAY OF WAVE EQUATION OF CARRIER MODEL SUBJECT TO MEMORY CONDITION AT THE

BOUNDARY

Jeong Ja Bae and Suk Bong Yoon

Abstract. In this paper we consider the uniform decay for the wave equation of Carrier model subject to memory condition at the bound- ary. We prove that if the kernel of the memory decays exponentially or polynomially, then the solutions for the problems have same decay rates.

1. Introduction

In this paper, we are concerned with the mixed problem for the Carrier model subject to memory condition at the boundary given by

u ⁰⁰ − M ( Z

Ω

|u| ² dx)∆u + |u| ^α u = 0 in Ω × (0, ∞), (1.1)

u = ∂u

∂ν = 0 on Γ 0 × (0, ∞), (1.2)

u + Z _t

0

g(t − s)M ( Z

Ω

|u(s)| ² dx) ∂u

∂ν (s)ds = 0 on Γ 1 × (0, ∞) (1.3)

u(0) = u 0 , u t (0) = u 1 on Ω, (1.4)

where Ω is a bounded domain in R ⁿ with C ² boundary Γ := ∂Ω such that Γ = Γ 0 ∪ Γ 1 , Γ 0 ∩ Γ 1 = ∅ and Γ 0 , Γ 1 have positive measures, ν denotes the unit outer normal vector pointing towards the exterior of Ω and g ∈ W ^1,2 (0, ∞) is a non-increasing, positive function.

We note that the integral equation (1.3) describes the memory effect which can be caused by the interaction with another viscoelastic element. Indeed, boundary condition (1.3) means that Ω is clamped in a rigid body in the portion Γ 0 of its boundary and in a body in the portion Γ 1 with viscoelastic properties.

Carrier model u ⁰⁰ −M ( R

Ω u ² dx)∆u = f was derived in [2] to model vibrations of an elastic string with fixed ends when the changes in tension are not small.

For the global solvability of Carrier model, we refer [4, 5, 8]. When Γ ₁ = ∅,

Received March 14, 2006; Revised December 15, 2006.

2000 Mathematics Subject Classification. 35B35, 35B40, 35L70.

Key words and phrases. existence of solution, uniform decay, wave equation, carrier model, boundary value problem, a priori estimates.

c

°2007 The Korean Mathematical Society

1013

author [8] considered the global solvability for the Carrier model with nonlinear damping |u ⁰ | ^α u ⁰ and source term |u| ^ρ u.

On the other hand, when Γ 1 6= ∅, Cavalcanti et al. [3] have studied the uniform decay of solutions of coupled linear wave equations with M (s) = 1 and boundary conditions (1.2)-(1.3). Santos et al. [10] considered the decay rate for Kirchhoff type wave equation with strong damping u ⁰⁰ − M (k∇uk ² )∆u −

∆u t +f (u) = 0 on Ω×(0, ∞) subject to memory condition at the boundary.

Author [1] has proved the decay rates for the coupled wave equation of Kirchhoff type. For the global existence of Kirchhoff type model, we refer [1, 6, 9].

In this paper, we will study the existence of solutions for the Carrier model (1.1) subject to memory condition on the boundary. Moreover, we consider if the memory terms g decay exponentially or polynomially, then the solutions for the Carrier model with nonlinear damping have same decay rates.

2. Statement of results Throughout this paper we define

V := {u ∈ H ¹ (Ω); u = 0 on Γ 0 }, (u, v) :=

Z

Ω

u(x)v(x)dx,

(u, v) Γ = Z

Γ

u(x)v(x)dΓ, kuk ^p _p,Γ = Z

Γ

|u(x)| ^p dx and kuk ∞ = kuk L

(Ω) . Now, we shall assume that Γ 0 = {x ∈ Γ | (x − x 0 ) · ν(x) ≤ 0} and Γ 1 = {x ∈ Γ | (x − x 0 ) · ν(x) > 0}. Denoting by (g ∗ φ)(t) = R _t

0 g(t − s)φ(s)ds the convolution product operator and differentiating the equation (1.3), and then applying the Volterra’s inverse operator, we get M (kuk ² ) ^∂u _∂ν = − _g(0) ¹ (u ⁰ (t) + (k ∗ u ⁰ )(t)) on Γ 1 × (0, ∞). Here, the resolvent kernel satisfies k(t) + _g(0) ¹ (g ⁰ ∗ k)(t) = − _g(0) ¹ g ⁰ (t). Denoting by η 1 = _g(0) ¹ , we obtain on Γ 1 × (0, ∞),

M (kuk ² ) ∂u

∂ν = −η 1 [u ⁰ (t) + k(0)u(t) − k(t)u 0 + (k ⁰ ∗ u)(t)].

(2.1)

Now, we need the following assumptions:

(A 1 ) Let us consider u 0 , u 1 ∈ V ∩ H ² (Ω) verifying the compatibility condi- tions

M (ku 0 k ² ) ∂u 0

∂ν + 1

g(0) u 1 = 0 on Γ 1 .

(A 2 ) The function k ∈ W ^1,∞ (0, ∞) ∩ W ^2,1 (0, ∞) satisfies that there exist positive constants m i , i = 1, 2, such that for all t ≥ 0 and p > 1

k(0) > 0, k ⁰ (t) ≤ −m 1 k(t), k ⁰⁰ (t) ≥ −m 2 k ⁰ (t), (2.2)

or k(0) > 0, k ⁰ (t) ≤ −m 1 k(t) ¹⁺

, k ⁰⁰ (t) ≥ m 2 (−k ⁰ (t)) ¹⁺

. (2.3)

(A 3 ) M (·) is a nondecreasing C ¹ (0, ∞) function with 0 < m 0 ≤ M (s) for

every s ≥ 0.

Let us denote by (g¤φ)(t) := R _t

0 g(t − s)|φ(t) − φ(s)| ² ds. Then we state our main result.

Theorem 2.1. Under the assumptions (A 1 )-(A 3 ), if 0 < α ≤ _n−2 ¹ if n ≥ 3, or α > 0 if n = 1, 2, then problem (1.1)-(1.3) has a unique solution u : Ω → R such that u ∈ L ^∞ (0, ∞; V ∩ H ² (Ω)), u ⁰ ∈ L ^∞ (0, ∞; V ), u ⁰⁰ ∈ L ^∞ (0, ∞; L ² (Ω)).

Moreover, there exist positive constants C 1 and C 2 such that

E(t) ≤ C 1 E(0) exp(−C 2 t) or E(t) ≤ C 1 E(0)(1 + t) ^−(p+1) , where

E(t) = 1

2 {ku ⁰ (t)k ² + M (ku(t)k ² )k∇u(t)k ² + 2

α + 2 ku(t)k ^α+2 _α+2 − η 1 (k ⁰ ¤u)(t)}

+ η 1

2 k(t)ku(t)k ² _Γ

.

3. Proof of Theorem 2.1

Without loss of generality, we consider η 1 = 1. For each M > 0, we put (3.1)

W (M, T ) = {v ∈ L ^∞ (0, T ; V ∩ H ² (Ω)); v t ∈ L ^∞ (0, T ; V ), v tt ∈ L ² (Ω × (0, T )), kvk _L

_{(0,T ;V ∩H}

_(Ω)) + kv t k _L

_{(0,T ;V )} + kv tt k _L

_{(Ω×(0,T ))} ≤ M }, W 1 (M, T ) = {v ∈ W (M, T ); v tt ∈ L ^∞ (0, T ; L ² (Ω))},

K 0 = max

0≤s≤M

|M (s)| and K 1 = max

0≤s≤M

|M ⁰ (s)|.

Suppose that u m−1 ∈ W 1 (M, T ), then we associate the problem (1.1)-(1.3) with the following problem:

(u ⁰⁰ _m (t), w) + b m (t)(∇u m (t), ∇w) + (|u m (t)| ^α u m (t), w) + (u ⁰ _m (t), w) Γ

+ Z _t

0

k ⁰ (t − s)(u m (s), w) Γ

ds + k(0)(u m (t), w) Γ

(3.2)

= (k(t)u 0m , w) Γ

, w ∈ V m , u m (0) = u 0 , u ⁰ _m (0) = u 1 , (3.3)

where b _m (t) = M (ku _m−1 (t)k ² ), then we find u _m ∈ W ₁ (M, T ) which satisfies the problem (3.2)-(3.3). Using usual Faedo-Galerkin’s approximation and mul- tiplier method, we can obtain the following proposition.

Proposition 3.1. Under the assumption of Theorem 2.1, there exist positive constants M and T and the recurrent sequence {u m } ⊂ W 1 (M, T ) defined by (3.2)-(3.3).

Proposition 3.2. Under the assumption of Theorem 2.1, there exist positive constants M and T such that the problem (1.1)-(1.3) has a unique weak solution u ∈ W 1 (M, T ). On the other hand, the linear recurrent sequence {u m } defined by (3.2)-(3.3) converges to the solution u strongly in the space W 1 (T ) = {v ∈ L ^∞ (0, T ; V ) | v ⁰ ∈ L ^∞ (0, T ; L ² (Ω))}. Furthermore, we have

ku m − uk L

(0,T ;V ) + ku ⁰ _m − u ⁰ k L

(0,T ;L

(Ω)) ≤ Ck ^m _T , for all m,

where k T < 1 is some positive constant and C is a constant depending only on T , u 0 , u 1 .

Proof. First, we note that W 1 (T ) is a Banach space with respect to the norm kvk _W

_{(T )} = kvk _L

_{(0,T ;V )} + kv ⁰ k _L

_{(0,T ;L}

_(Ω)) . [7]. We shall prove that {u m } is a Cauchy sequence in W 1 (T ). Consider z m = u m+1 − u m . Then z m satisfies the variational problem

(z _m ⁰⁰ (t), w) + b m+1 (t)(∇z m (t), ∇w) + (b m+1 (t) − b m (t))(∇u m (t), ∇w) +(|u m+1 (t)| ^α u m+1 (t) − |u m (t)| ^α u m (t), w) + (z _m ⁰ (t), w) Γ

(3.4) +

Z _t

0

k ⁰ (t − s)(z m (s), w) Γ

ds + k(0)(z m (t), w) Γ

= 0, w ∈ V, z m (0) = z ⁰ _m (0) = 0 in V.

(3.5)

Now, we can write d

dt E 2m (t) + kz ⁰ _m (t)k ² _Γ

(3.6)

= 1

2 b ⁰ _m+1 (t)k∇z m (t)k ² + (|u m (t)| ^α u m (t) − |u m+1 (t)| ^α u m+1 (t), z _m ⁰ (t)) +(b m (t) − b m+1 (t))(∇u m (t), ∇z _m ⁰ (t)) −

Z _t

0

k ⁰ (t − s)(z m (s), z ⁰ _m (t)) Γ

ds, where E 2m (t) = ¹ ₂ {kz _m ⁰ (t)k ² +k(0)kz m (t)k ² _Γ

+b m+1 (t)k∇z m (t)k ² }. From mean value theorem, we obtain

|b m+1 (t) − b m (t)| = |M (ku m (t)k ² ) − M (ku m−1 (t)k ² )|

≤

Z _ku

_(t)k

ku

(t)k

|M ⁰ (ξ)|dξ

≤ K 1 (ku m−1 (t)k + ku m (t)k)kz m−1 (t)k _W

_{(T )}

≤ 2M K 1 kz m−1 (t)k W

(T ) , and so Green identity and boundary condition (2.1) imply

|(b m+1 (t) − b m (t))(∇u m (t), ∇z ⁰ _m (t))|

≤ 2M K 1 kz m−1 (t)k _W

_{(T )} [k∆u m (t)kkz _m ⁰ (t)k + k ∂u m

∂ν (t)k Γ

kz ⁰ _m (t)k Γ

]

≤ M ² K 1 [kz m−1 (t)k ² _W

_{(T )} + kz _m ⁰ (t)k ² ] + 2² ⁻¹ M ² K ₁ ² kz _m−1 (t)k ² _W

_{(T )} k ∂u m

∂ν (t)k ² _Γ

+ ²kz _m ⁰ (t)k ² _Γ

≤ M ² K 1 [kz m−1 (t)k ² _W

_{(T )} + kz _m ⁰ (t)k ² ] + ²kz ⁰ _m (t)k ² _Γ

+ 4(²m ² ₀ ) ⁻¹ M ⁴ K ₁ ² kz m−1 (t)k ² _W

_{(T )}

× (|k(0)| ² + kk(t)k ² _L

(0,∞) + kk(t)k ² _L

(0,∞) ).

Also, we have

(|u m (t)| ^α u m (t) − |u m+1 (t)| ^α u m+1 (t), z _m ⁰ (t))

≤ C 1 (ku m (t)k ^α _αn + ku m+1 (t)k ^α _αn )kz m (t)k

n−2

kz ⁰ _m (t)k (3.7)

≤ C 2 M ^α (k∇z m (t)k ² + kz _m ⁰ (t)k ² ).

Thus we arrive at d

dt E 2m (t) + (1 − 2²)kz _m ⁰ (t)k ² _Γ

≤ C M,K

kz m−1 (t)k ² _W

_{(T )} + 1

4² kk ⁰ k _L

_(0,∞) Z _t

0

|k ⁰ (t − s)|kz m (s)k ² _Γ

ds +[( 1

2 + M )M K 1 + C 2 M ^α ][kz _m ⁰ (t)k ² + k∇z m (t)k ² ], (3.8)

here

C M,K

= M ² K 1 [1 + 4(²m ² ₀ ) ⁻¹ M ² K 1 (|k(0)| ² + kk(t)k ² _L

(0,∞) + kk(t)k ² _L

(0,∞) )].

Integrating (3.8) over [0, t], choosing ² > 0 sufficiently small and employing Gronwall’s lemma we obtain

E _2m (t) + Z _t

0

kz ⁰ _m (s)k ² _Γ

ds ≤ ˜ k _T kz _m−1 (t)k ² _W

_{(T )} , (3.9)

where ˜ k T = C M,K

T e ^{[2 max{1,m}

^}((

^{+M )M K}

^+C

^M

⁾⁺

^kk

^k

^]T , that is,

kz m (t)k _W

_{(T )} ≤ k T kz m−1 (t)k _W

_{(T )} , (3.10)

where k T = √

2[min{1, m 0 , k(0)}] ⁻

(˜ k T )

< 1 Hence (3.11) ku m+p − u m k W

(T ) ≤ ku 1 − u 0 k W

(T ) k ^m _T

1 − k T for all m, p.

It follows from (3.11) that {u m } is a Cauchy sequence in W 1 (T ). Therefore there exists u ∈ W 1 (T ) such that

(3.12) u m → u strongly in W 1 (T ).

We also note that u m ∈ W 1 (M, T ), then from the sequence {u m } we can deduce a subsequence {u m

} such that

u m

→ u weak star in L ^∞ (0, T ; V ), (3.13)

u ⁰ _m

→ u ⁰ weak star in L ^∞ (0, T ; V ), u ⁰⁰ _m

→ u ⁰⁰ weak star in L ^∞ (0, T ; L ² (Ω)),

where u ∈ W (M, T ). The above convergence is sufficient to pass to the limit in the linear terms of (3.2). Taking into account the continuity of trace operator γ 0 : H ¹ (Ω) → H

(Γ), we have

u m

→ u in L ² (0, T ; L ² (Γ 1 )),

u ⁰ _m

→ u ⁰ in L ² (0, T ; L ² (Γ 1 )).

Sobolev imbedding and (3.13) imply that for every φ ∈ L ² (Ω),

|(|u m

(t)| ^α u m

(t) − |u(t)| ^α u(t), φ)|

≤ C(ku m

(t)k ^α _αn + ku(t)k ^α _αn )ku m

(t) − u(t)k

n−2

kφ(t)k

≤ 2CM ^α ku m

(t) − u(t)k _W

_{(T )} kφ(t)k → 0 as j → ∞.

Now, we notice that Z _T

0

b m (t)(∇u m (t), ∇v(t)) − b(t)(∇u(t), ∇v(t))dt

= Z _T

0

b m (t)(∇u m (t) − ∇u(t), ∇v(t)) + (b m (t) − b(t))(∇u(t), ∇v(t))dt

≤ C 1 [K 0 + 2K 1 M ]ku m − uk _W

_{(T )} kvk _L

_{(0,T ;H}

₎ for all v ∈ L ¹ (0, T ; H ¹ ).

Thus we can pass to the limit with m = m j → ∞ to obtain u ⁰⁰ − M (kuk ² )∆u + |u| ^α u = 0 in D ⁰ (0, ∞; L ² (Ω)).

Since u, u ⁰⁰ ∈ L ² _loc (0, ∞; L ² (Ω)),

u ⁰⁰ − M (kuk ² )∆u + |u| ^α u = 0 in L ² _loc (0, ∞; L ² (Ω)).

Returning to the approximate problem, making use of Green formula, we have M (kuk ² ) ∂u

∂ν + u ⁰ (t) + k(0)u(t) − k(t)u 0 + (k ⁰ ∗ u)(t) = 0 on D ⁰ (0, ∞; H ⁻

(Γ 1 )).

Since u ⁰ , u, k ⁰ ∗ u ∈ L ² _loc (0, ∞; H

(Γ 1 )), M (kuk ² ) ∂u

∂ν + u ⁰ (t) + k(0)u(t) − k(t)u 0 + (k ⁰ ∗ u)(t) = 0 on L ² _loc (0, ∞; H

(Γ 1 )).

This completes the proof of existence of solutions in Theorem 2.1. The unique- ness of solution is proved by applying the similar course to Proposition 3.2. We

omit it. ¤

4. Uniform decay 4.1. Exponentially decay

In this section we shall show the asymptotic behavior of solutions for the problem (1.1)-(1.3) when the resolvent kernel k decays exponentially. Let us assume k satisfies the condition (2.2). Note that the condition (2.2) implies k(t) ≤ k(0)e ^−m

^t for all t > 0. At first, we begin with the following Lemma.

Lemma 4.1 ([3]). Let f be a real positive function of class C ¹ . If there exists positive constants γ 0 , γ 1 and c 0 such that f ⁰ (t) ≤ −γ 0 f (t) + c 0 e ^−γ

^t , then there exist positive constants γ and c such that f (t) ≤ (f (0) + c)e ^−γt .

Now, we define

(k¤u)(t) :=

Z _t

0

k(t − r)ku(t) − u(r)k ² _Γ

dr,

(4.1)

and define energy by E(t) = 1

2 [ku ⁰ (t)k ² + b(t)k∇u(t)k ² − η 1 (k ⁰ ¤u)(t) + η 1 k(t)ku(t)k ² _Γ

]

+ 1

α + 2 ku(t)k ^α+2 _α+2 , (4.2)

where b(t) = M (ku(t)k ² ). Then we have d

dt E(t) ≤ b ⁰ (t)k∇u(t)k ² − η 1

2 ku ⁰ (t)k ² _Γ

+ η 1

2 |k(t)| ² ku 0 k ² _Γ

− η 1

2 (k ⁰⁰ ¤u)(t) + η 1

2 k ⁰ (t)ku(t)k ² _Γ

. (4.3)

For u ∈ W 1 (M, T ), we get d

dt E(t) ≤ 2K 1 M ² m ⁻¹ ₀ b(t)k∇u(t)k ² − η 1

2 ku ⁰ (t)k ² _Γ

+ η 1

2 |k(t)| ² ku 0 k ² _Γ

+ η ₁ m ₂

2 (k ⁰ ¤u)(t) − η ₁ m ₁

2 k(t)ku(t)k ² _Γ

. (4.4)

Let us define the perturbed modified energy by

N(t) = ku ⁰ (t)k ² + b(t)k∇u(t)k ² + ku(t)k ^α+2 _α+2 , ψ(t) =

Z

Ω

[m · ∇u(t) + ( n

2 − θ)u(t)]u ⁰ (t)dx, where θ is a small positive constant.

Consider the following operator (k♦h)(t) =

Z _t

0

k(t − s)(h(t) − h(s))ds.

Lemma 4.2. For any strong solution of the problem (1.1)-(1.3), we get for small ² > 0

d

dt ψ(t) ≤ C Z

Γ

(|u ⁰ (t)| ² + |k(t)u(t)| ² + |(k ⁰ ♦u)(t)| ² + |k(t)u 0 | ² )dΓ

− θ

2 N(t) − (1 − 2θ − ²m ⁻¹ ₀ )b(t) Z

Ω

|∇u(t)| ² dx.

(4.5)

Proof. From the equation (1.1) and integration by parts, we have d

dt ψ(t) ≤ 1 2 Z

Γ

(m · ν)|u ⁰ (t)| ² dΓ − θ Z

Ω

|u ⁰ (t)| ² dx

−(1 − θ)b(t) Z

Ω

|∇u(t)| ² dx − 1 2 b(t)

Z

Γ

(m · ν)|∇u(t)| ² dΓ +b(t)

Z

Γ

∂u

∂ν {m · ∇u(t) + ( n

2 − θ)u(t)}dΓ

− Z

Ω

|u(t)| ^α u(t)[m · ∇u(t) + ( n

2 − θ)u(t)]dx,

where we have used R

Γ

(m·ν)| ^∂u _∂ν | ² dΓ ≤ 0 and u| Γ

= 0. Taking θ small enough, we get

d dt ψ(t)

≤ − θN (t) − (1 − 2θ)b(t) Z

Ω

|∇u(t)| ² dx + 1 2

Z

Γ

(m · ν)|u ⁰ (t)| ² dΓ

− 1 2 b(t)

Z

Γ

(m · ν)|∇u(t)| ² dΓ + b(t) Z

Γ

∂u

∂ν {m · ∇u(t) + ( n

2 − θ)u(t)}dΓ

− Z

Ω

|u(t)| ^α u(t)[m · ∇u(t) + ( n

2 − 2θ)u(t)]dx.

Applying Young and Poincar´e inequality, we have for ² > 0, b(t)

Z

Γ

∂u

∂ν {m · ∇u(t) + ( n

2 − θ)u(t)}dΓ (4.6)

≤ ²b(t) Z

Γ

{|m · ∇u(t)| ² + ( n

2 − θ) ² |u(t)| ² }dΓ + C 1 (²)b(t) Z

Γ

| ∂u

∂ν | ² dΓ

≤ ²C 2 [b(t) Z

Γ

(m · ν)|∇u(t)| ² dΓ + N (t)] + C 1 (²)b(t) Z

Γ

| ∂u

∂ν | ² dΓ.

Also, we get Z

Ω

|u(t)| ^α u(t)m · ∇u(t)dx ≤ max

x∈ ¯ Ω kx − x 0 kk∇u(t)kku(t)k ^α+1 _2(α+1) . Gagliardo-Nirenberg inequality implies

ku(t)k ^α+1 _2(α+1) ≤ C 3 k∇u(t)k ^(α+1)θ

ku(t)k ^{(α+1)(1−θ} _α+2

⁾ , where θ 1 = (α+1)(4−α(n−2)) ^αn . Thus we have

Z

Ω

|u(t)| ^α u(t)m · ∇u(t)dx (4.7)

≤ C 3 max

x∈ ¯ Ω kx − x 0 kk∇u(t)k ^1+(α+1)θ

ku(t)k ^{(α+1)(1−θ} _α+2

⁾

≤ ²k∇u(t)k ² + C ₃ ² 4² [max

x∈ ¯ Ω kx − x 0 k] ² k∇u(t)k ^2(α+1)θ

ku(t)k ^{2(α+1)(1−θ} _α+2

⁾

≤ ²k∇u(t)k ² + C ₃ ² 4² [max

x∈ ¯ Ω kx − x 0 k] ² k∇u(t)k ^2(α+1)θ

ku(t)k ^α−2θ _α+2

^(α+1) N (t)

≤ ²k∇u(t)k ² + C 4 (²)E(0)

N (t).

Thus we obtain d

dt ψ(t) ≤ −θ(1 − C(², θ))N (t) − (1 − 2θ − ²m ⁻¹ ₀ )b(t) Z

Ω

|∇u(t)| ² dx +

Z

Γ

(m · ν)|u ⁰ (t)| ² dΓ − 1 2 b(t)

Z

Γ

(m · ν)|∇u(t)| ² dΓ +²C 2 b(t)

Z

Γ

(m · ν)|∇u(t)| ² dΓ + C 1 (²)b(t) Z

Γ

| ∂u

∂ν | ² dΓ,

where C(², θ) = ¹ _θ {²C 2 + C 4 (²)E(0)

}. Choosing small initial data with C(², θ)

< ¹ ₂ in the above inequality. Then we have d

dt ψ(t) ≤ − θ

2 N (t) − (1 − 2θ − ²m ⁻¹ ₀ )b(t) Z

Ω

|∇u(t)| ² dx +C ₅

Z

Γ

|u ⁰ (t)| ² + | ∂u

∂ν | ² dΓ.

(4.8)

Now, we note that the boundary condition (2.1) can be written as (4.9) b(t) ∂u

∂ν = −[u ⁰ (t) + k(t)u(t) − k(t)u 0 − (k ⁰ ♦u)(t)] on Γ 1 × (0, ∞).

Taking into account the above estimates with boundary conditions, we get

Lemma. ¤

On the other hand, using H¨older’s inequality, we get

|(k ⁰ ♦u)(t)| ² ≤ (k(t) − k(0))(k ⁰ ¤u)(t) ≤ −k(0)(k ⁰ ¤u)(t).

(4.10)

Thus we have

(4.11) d

dt ψ(t) ≤ C Z

Γ

(|u ⁰ (t)| ² + |k(t)u(t)| ² − k(0)(k ⁰ ¤u)(t) + |k(t)u 0 | ² )dΓ

− θ

2 N (t) − (1 − 2θ − ²m ⁻¹ ₀ )b(t) Z

Ω

|∇u(t)| ² dx.

We choose a positive ² and ¯ n with K 1 M ² < ^m

^{(1−2θ−²m} _{2¯ n}

⁾ and ² < m 0 (1−2θ).

Then let us introduce the Lyapunov functional L(t) := N E(t) + ψ(t) with N > 0. Then we have

d

dt L(t) ≤ − θ

2 E(t) − [1 − 2θ − ²m ⁻¹ ₀ − 2K 1 M ² m ⁻¹ ₀ N ]b(t) Z

Ω

|∇u(t)| ² dx

− (α + 1)θ

2(α + 2) ku(t)k ^α+2 _α+2 + ( η 1 m 2 N 2 − η 1 θ

4 − Ck(0))(k ⁰ ¤u)(t) (4.12)

+( η 1 θ

4 − m 1 N η 1

2 + C)k(t)ku(t)k ² _Γ

+ (C − η 1 N

2 )ku ⁰ (t)k ² _Γ

+( η 1 N

2 + C)|k(t)| ² ku 0 k ² _Γ

.

Taking N large with N > ¯ n, then from (2.3), we have d

dt L(t) ≤ − θ

2 E(t) + 2N ( η 1

2 |k(t)| ² ku 0 k ² _Γ

)

≤ − θ

2 E(t) + 2N ( η 1

2 |k(0)| ² ku 0 k ² _Γ

)e ^−2m

^t

≤ − θ

2 E(t) + 2N E(0)e ^−2m

^t . Using Young inequality and taking N large we find that

N

2 E(t) ≤ L(t) ≤ 2N E(t),

(4.13)

and so we get

d

dt L(t) ≤ − θ

4N L(t) + 2N E(0)e ^−2m

^t . Thus from Lemma 4.1, we have

L(t) ≤ (L(0) + C)e ^−γt for some positive constants C, γ.

Considering the inequality (4.13), we get the decay estimates of energy.

4.2. Polynomially decay

Now, we consider the uniform decay rates when the resolvent kernel k decay polynomially like (1 + t) ^−p . Let us assume that k satisfies the condition (2.3).

To obtain our result, we use the following Lemma.

Lemma 4.3 ([3]). Let f ≥ 0 be a differentiable function satisfying f ⁰ (t) ≤ − c 1

f (0)

f (t) ¹⁺

+ c 2

(1 + t) ^β f (0), t ≥ 0,

for some positive constants c 1 , c 2 , α and β such that β ≥ α + 1, then there exist a positive constant c such that

f (t) ≤ c

(1 + t) ^α f (0), t ≥ 0.

Lemma 4.4 ([3]). Let u = φ be a solution of problem (1.1)-(1.3). Then for p > 1, 0 < r < 1 and t ≥ 0, we have

( Z

Γ

(|k ⁰ |¤φ)(t)dΓ)

≤ 2(

Z _t

0

|k ⁰ (s)| ^r dskφk ² _L

(0,t;L

(Γ

)) )

Z

Γ

(|k ⁰ |

¤φ _i )(t)dΓ, (

Z

Γ

(|k ⁰ |¤φ i )(t)dΓ)

≤ 2(

Z _t

0

|φ(s)| ² _L

(Γ

) ds + tkφk ² _L

(Γ

) )

Z

Γ

(|k ⁰ |

¤φ)(t)dΓ.

Using (4.4) and conditions on k, we get d

dt E(t) ≤ − η 1

2 ku ⁰ (t)k ² _Γ

+ η 1

2 |k(t)| ² ku 0 k ² _Γ

− η 1 m 2

2 ((−k ⁰ ) ¹⁺

¤u)(t)

− η 1 m 1

2 k(t) ¹⁺

ku(t)k ² _Γ

+ 2K ₁ M ² m ⁻¹ ₀ b(t)k∇u(t)k ² . Using H¨older’s inequality and condition on k, we get

(4.14)

|(k ⁰ ♦u)(t)| ² ≤ Z _t

0

(−k ⁰ (s))

ds · ((−k ⁰ )

¤u)(t) ≤ C((−k ⁰ )

¤u)(t).

Thus we have d dt ψ(t)

≤ − θ

2 N(t) + C Z

Γ

(|u ⁰ (t)| ² + k(t)|u(t)| ² + |k(t)u 0 | ² + ((−k ⁰ )

¤u)(t))dΓ

−(1 − 2θ − ²m ⁻¹ ₀ )b(t) Z

Ω

|∇u(t)| ² dx.

Also, for large N > 0, Lyapunov functional L(t) satisfies d

dt L(t) ≤ − θ

2 N(t) + 2N |k(t)| ² E(0) − c 1 N 2

Z

Γ

((−k ⁰ )

¤u)(t)dΓ.

Condition on k and Lemma 4.4 imply d

dt L(t) ≤ −cL(0) ⁻

L(t) ¹⁺

+ 2N |k(t)| ² E(0).

Thus from Lemma 4.3, we have L(t) ≤ C

(1 + t) ^p+1 L(0) for some C > 0.

Therefore, (4.13) yields

E(t) ≤ C

(1 + t) ^p+1 E(0).

This completes the proof of Theorem. ¤

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Jeong Ja Bae

Department of Mathematics University of Ulsan Ulsan 680-749, Korea

E-mail address: [email protected] Suk Bong Yoon

Department of Mathematics Dongeui University

Pusan 614-714, Korea

E-mail address: [email protected]