Z. Angew. Math. Phys. 61 (2010), 235–265 c 2009 Birkh¨
auser Verlag Basel/Switzerland 0044-2275/10/020235-31 published online June 4, 2009 DOI 10.1007/s00033-009-0009-z

Zeitschrift f¨ ur angewandte Mathematik und Physik ZAMP

Energy decay rates for solutions of the wave equation with boundary damping and source term Jong Yeoul Park, Tae Gab Ha and Yong Han Kang Abstract. In this paper, we consider the wave equation u

− ∆u = |u|ρ u with the following nonlinear boundary condition ∂u + ∂ν


t

k(t − s, x)u
(s)ds + a(x)g(u
) = 0.

0

We show energy decay rates for solutions of the wave equation in bounded domain with nonlinear boundary damping and source term. Mathematics Subject Classification (2000). 35L05 · 35L20 · 35B37 · 35B40. Keywords. Wave equation · Boundary damping term · Source term.

1. Introduction In this paper, we are concerned with the energy decay rate of the wave equation with ⎧

u − ∆u = |u|ρ u in Ω × R+ , ⎪ ⎪ ⎪ ⎨u = 0 on Γ × R , 0 + t ∂u
⎪ + 0 k(t − s, x)u (s)ds + a(x)g(u
) = 0 on Γ1 × R+ , ⎪ ⎪ ⎩ ∂ν u(x, 0) = u0 (x), u
(x, 0) = u1 (x), x ∈ Ω,

(1.1)

where Ω ⊂ Rn is an open bounded domain, n ≥ 1, with boundary Γ = Γ0 ∪ Γ1 of class C 2 , where Γ0 and Γ1 are closed and disjoint. a : Γ1 → R+ ∈ L∞ (Γ1 ) is such that a(x) ≥ a0 > 0, k : R+ × Γ1 → R+ ∈ C 2 (R+ , L∞ (Γ1 )) and g is a continuous function. We shall denote by ν the unit outward normal vector to Γ. The problem of proving uniform decay rates for the solutions to the wave equation with a boundary dissipation has attracted a lot of attention in recent years. Indeed, the linear problem has been treated by several authors; see for instance [10,11,21]. When the boundary conditions are nonlinear, Aassila et al. [1], Komornik [9] and Zuazua [24] studied. In the case of nonlinear wave equation, it has been studied more energetically (see [2–7,12,14–16,23]). The boundary condition (1.1)3 describes the reflection of sound at surfaces of some materials with memory of interest in engineering practice. It is quite general and covers a fairly large variety of physical configurations. Problem containing the boundary condition (1.1)3 was studied by many authors, e.g., Aassila et al. [1], Guesmia [7], Nicaise and Pignotti [14], Pr¨ uss [17], Rivera [18], Tadayuki and Rinko [20],

236

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

and the references therein. In the case of [7], Guesmia shows the exponential stability (or polynomial decay) of the system when u0 is zero on Γ1 . The extension to a nonlinear boundary condition of memory type is made in this paper with the same assumption on k but without the above assumption on u0 . Another difference is that this paper has a memory term |u|ρ u. The memory term |u|ρ u is very important in this paper. Because of the memory term |u|ρ u, we can not guarantee that solutions exist globally. Furthermore, energy can blow up in finite time. To overcome this problem, we have to find a condition  T p−1  that solutions exist and energy decays. Also , the term S E 2 (t) Ω |u|2 dxdt appears in the process of energy decay. Using Lasiecka and Tataru’s method [12], we solve this term since it is difficult that we induce the result from this term directly. Moreover, we consider more general case of g (see (H3 )). (H3 ) means that g has not a polynomial behavior near the origin. In order to solve energy decay rates using Lemma 3.2 in this paper, we need a polynomial growth of the function g near the origin. Hence, we can not use Lemma 3.2 when we assume (H3 ). To overcome this problem we use the Martinez’s method [13], however this paper generalizes Martinez results. This paper is organized as follows: In Sect. 2, we recall the hypotheses to prove our main result and introduce a main result. In Sect. 3, under hypotheses (H3 ), we prove boundedness of energy and then we prove the exponential or polynomial decay rate to our purpose using Komornik’s method and Lasiecka and Tataru’s method. In Sect. 4, using Martinez’s method and Lasiecka and Tataru’s method, we prove the energy decay rate having more general case of g than condition in Sect. 3.

2. Hypotheses and main results We consider the Hilbert space HΓ10 (Ω) := {u ∈ H 1 (Ω); u = 0onΓ0 }. The following hypotheses are made on Ω and on the functions k and g: (H1 ) Hypotheses on Ω. Let Ω ⊂ Rn be an open bounded domain, n ≥ 1, with boundary Γ = Γ0 ∪ Γ1 of class C 2 , where Γ0 and Γ1 are closed and disjoint satisfying the following condition: Γ0 = ∅ or m·ν ≤0

m · ν ≥ δ > 0 on Γ1 ,

inf Γ1 ×R+ k = 0, on Γ0 ,

(2.1) 0

0

n

m(x) = x − x (x ∈ R ).

(2.2)

(H2 ) Hypotheses on k. Let k : R+ × Γ1 → R+ ∈ C 2 (R+ , L∞ (Γ1 )) such that k ≥ 0 on Γ1 × R+ , k
≤ 0 on Γ1 × R+ , k

≥ −αk
on Γ1 × R+ , for some α > 0,

(2.3) (2.4) (2.5)

α inf Γ1 k(0) > −2 inf Γ1 k
(0).

(2.6)

The idea of hypotheses (H1 ) and (H2 ) are taken from [7]. (H3 ) Hypotheses on g. Let g : R → R be a continuous nondecreasing function such that g(0) = 0,

|g(x)| ≤ 1 + C1 |x| and p

1 p

C2 |x| ≤ |g(x)| ≤ C3 |x| C4 |x| ≤ |g(x)| ≤ C5 |x|

where p ≥ 1 and Ci (1 ≤ i ≤ 5) are five positive constants.

g(s)s > 0 if |x| ≤ 1, if |x| ≥ 1,

for s = 0,

(2.7) (2.8) (2.9)

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Energy decay rates for solutions of the wave equation

237

And suppose that a : Γ1 → R+ ∈ L∞ (Γ1 ) is such that a(x) ≥ a0 > 0 and 2 if n ≥ 3 and ρ > 0 n−2 Next lemma is used to estimate energy identity. 0<ρ<

if n = 1, 2.

(2.10)

Lemma 2.1. For g, ϕ ∈ C 1 ([0, ∞) : R), we have 
t 0


t 1 1
g(t − s)ϕ(s)ds ϕ = g
(t − s)|ϕ(t) − ϕ(s)|2 ds − g(t)|ϕ|2 2 2 0

1 d − 2 dt Proof. Differentiating the term

t 0


t


t 

2 g(t − s)|ϕ(t) − ϕ(s)| ds− g(s)ds |ϕ| . 2

0

0

g(t − s)|ϕ(t) − ϕ(s)|2 ds, we arrive at the above inequality.




We define the energy of the solution by the following formula

1 1 (|u
(t, x)|2 + |∇u(t, x)|2 )dx + k(t, x)|u(t, x) − u0 (x)|2 dΓ E(t) : = 2 2 Ω

−

1 2

Γ1



t

k
(t − s, x)|u(t, x) − u(s, x)|2 ds dΓ −

Γ1 0

1 ρ+2




|u(t, x)|ρ+2 dx.

(2.11)

Ω

According to (2.10), we have the imbedding: HΓ10 (Ω) → L2(ρ+1) (Ω) → Lρ+2 (Ω). Let B1 > 0 be the optimal constant of Sobolev imbedding which satisfies the inequality for all v ∈ HΓ10 (Ω).

||v||ρ+2 ≤ B1 ||∇v||2 , From the above inequality one has  K0 :=

sup

1 (Ω), v=0 v∈HΓ 0

ρ+2  1 ρ+2 ||v||ρ+2 ||∇v||ρ+2 2

≤

B1ρ+2 , ρ+2

for all v ∈ HΓ10 (Ω), v = 0.

Note that K0 > 0 and 1 ρ+2 ||v||ρ+2 , ρ+2 ≤ K0 ||∇v||2 ρ+2

for all v ∈ HΓ10 (Ω)

(2.12)

Let us consider the functional J(u) =

1 1 ||∇u||22 − ||u||ρ+2 ρ+2 , 2 ρ+2

We define the positive number d :=

inf

 sup J(λv) .

1 (Ω),v=0 v∈HΓ 0

u ∈ HΓ10 (Ω).

λ>0

Setting f (λ) =

1 2 λ − K0 λρ+2 , 2

λ > 0,

(2.13)

238

J. Y. Park, T. G. Ha and Y. H. Kang

then

 λ1 =

1 K0 (ρ + 2)

ZAMP

 ρ1

is the absolute maximum point of f and d = f (λ1 ) > 0. It is well known that the number d is the Mountain Pass level associated with the elliptic problem ⎧ ρ ⎪ ⎨−∆u = |u| u in Ω, u=0 on Γ0 , ⎪ ⎩ ∂u = 0 on Γ1 . ∂ν In fact (see [22]), d = inf sup J(γ(t)), γ∈Λ t∈[0,1]

where Λ = {γ ∈ C([0, 1];

HΓ10 (Ω));

γ(0) = 0,

J(γ(1)) < 0}.

Furthermore, we get d = f (λ1 ) =

ρ λ2 . 2(ρ + 2) 1

The energy associated with problem (1.1) is given by

1 1
2 E(t) = |u (t, x)| dx + k(t, x)|u(t, x) − u0 (x)|2 dΓ 2 2 Ω

−

1 2

Γ1



t

k
(t − s, x)|u(t, x) − u(s, x)|2 ds dΓ + J(u(t)),

u ∈ HΓ10 (Ω).

Γ1 0

By (2.4) and (2.12), we have E(t) ≥ J(u(t)) ≥

1 ||∇u||22 − K0 ||∇u||ρ+2 = f (||∇u(t)||2 ). 2 2

(2.14)

Now, if one consider ||∇u(t)||2 < λ1 . By (2.14) and (2.15), we have E(t) ≥ J(u(t)) >

||∇u(t)||22



1 − λρ1 K0 2

(2.15)

 =

||∇u(t)||22



 1 1 − . 2 ρ+2

(2.16)

So, if (2.15) is satisfied, from the above inequality we deduce that J(t) ≥ 0 and

||∇u(t)||22 ≤

2(ρ + 2) E(t). ρ

(2.17)

Moreover, we suppose that E(0) < d

and

||∇u0 ||2 < λ1 .

(2.18)

Before stating the main result of this paper, we briefly recall the following result on existence of solution of (1.1). According to the previous results existing in the literature, it can be summarized as follows:

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Energy decay rates for solutions of the wave equation

239

Theorem 2.1. Under assumptions initial data (u0 , u1 ) ∈ HΓ10 (Ω)×L2 (Ω) and (2.18), problem (1.1) admits a unique global weak solution u ∈ C(R+ , HΓ10 (Ω)) ∩ C 1 (R+ , L2 (Ω)). Furthermore, if (u0 , u1 ) ∈ (H 2 (Ω) ∩ HΓ10 (Ω)) × HΓ10 (Ω) satisfying (2.18) and g is globally Lipschitz continuous, then the solution has the following regularity: u ∈ L∞ (0, ∞, HΓ10 (Ω)),

1 u
∈ L∞ loc (0, ∞, HΓ0 (Ω)),

2 u

∈ L∞ loc (0, ∞, L (Ω)).

The existence of solutions may be proven by Galerkin method. Our main result is the following theorem. Theorem 2.2. Assume that hypotheses (H1 ), (H2 ), (H3 ), (2.10) and (2.18) hold. Then we have E(t) ≤ CE(0)e−ωt if p = 1, C > 0, ω > 0, t > 0, CE(0) if p > 1, C > 0, t > 0 E(t) ≤ (1 + t)2/(p−1)

(2.19) (2.20)

for every weak solution to (1.1) and initial data (u0 , u1 ) ∈ H01 (Ω) × L2 (Ω). Now, we consider more general case of g. (H3 ). Let g : R → R be a nondecreasing C 1 function such that g(0) = 0 and suppose that there exists a strictly increasing and odd function β of C 1 class on [−1, 1] such that |β(s)| ≤ |g(s)| ≤ |β −1 (s)| if |s| ≤ 1, C6 |s| ≤ |g(s)| ≤ C7 |s| if |s| > 1,

(2.21) (2.22)

where β −1 denotes the inverse function of β and C6 , C7 are positive constants. Theorem 2.3. Assume that hypotheses (H1 ), (H2 ), (H3 ), (2.10) and (2.18) hold. Then we have

 1  2 E(t) ≤ C F −1 for all t, t where F (s) := sβ(s) and C is a positive constant. Moreover, if the function G(s) := on [0, η] for some η > 0 and G(0) = 0, then we have

1 2 E(t) ≤ C β −1 ( ) for all t. t

β(s) s

(2.23)

is nondecreasing

(2.24)

3. Proof of Theorem 2.2 We will transform the boundary condition in another more practical one considering u0 = 0 on Γ1 . This idea was introduced from Aassila et al. [1]. On the other hand, Nicaise and Pignotti [14] considering u0 = 0 on Γ1 obtained the same results. However, computations are more complicated to apply our problem. So, by [1] we can consider u0 = 0 on Γ1 to simplify the computation and without loss of generality. So, the problem (1.1) is now transformed into ⎧

u − ∆u = |u|ρ u in Ω × R+ , ⎪ ⎪ ⎪ ⎨u = 0 on Γ × R , + t
0 (3.1) ∂u ⎪ + k (t − s, x)u(s, x)ds + k(0)u + a(x)g(u
) = 0 on Γ1 × R+ , ⎪ ∂ν 0 ⎪ ⎩ u(x, 0) = u0 (x), u
(x, 0) = u1 (x), x ∈ Ω. In order to solve the energy decay of (3.1), we use the following lemma.

240

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Lemma 3.1. (cf. [8]) Let E : R+ → R+ be a nonincreasing function and assume that there exist two constants α > 0 and T > 0 such that +∞
E α+1 (s)ds ≤ T E(0)α E(t) for all t ∈ R+ . t

Then we have

T + αt −1/α E(t) ≤ E(0) T + αT

for all t ≥ T.

In particular, assume that +∞
E(s)ds ≤ T E(t)

for all t ∈ R+ .

t

Then t

E(t) ≤ E(0)e1− T ,

for all t ≥ T.

And we include the following lemma,which will be used to estimate an integral involving the ρ. Lemma 3.2. (Gagliardo–Nirenberg). Let 1 ≤ r < p < ∞, 1 ≤ q ≤ p and 0 ≤ m. Then, ||v||W k,q ≤ C||v||θW m,p ||v||1−θ Lr for v ∈ W m,p (Ω) ∩ Lr (Ω), Ω ⊂ RN, where C is a positive constant and   −1 k 1 1 m 1 1 + − + − θ= N r q N r p provided that 0 < θ ≤ 1. Now, we begin the estimates for the energy of (3.1). In the following section, the symbol C indicates positive constants, which may be different. Lemma 3.3. The energy is nonincreasing and it holds that
T

T

t 1 a(x)g(u
)u
dΓdt + k

(t − s)(u(t) − u(s))2 dsdΓdt 2 S Γ1 0

S Γ1

−

1 2

for every 0 ≤ S < T < +∞. In particular,


T




k
|u|2 dΓdt = E(S) − E(T ) ≤ E(S)

S Γ1



g(u
)u
≤ c|E
(0)| for a suitable constant c.

Proof.
T E(T ) − E(S) =


T





E (t)dt = S

S Ω


T
+

kuu
dΓdt −

S Γ1

1 − 2

1 (u u + ∇u · ∇u )dxdt + 2


T

t













T

S





Γ1


0

t


T
S Γ1

k
(t − s)(u(t) − u(s))u
(t)dsdΓdt

2


T


k (t − s)(u(t) − u(s)) dsdΓdt − S Γ1 0

k
|u|2 dΓdt

S

Ω

|u|ρ uu
dxdt.

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Energy decay rates for solutions of the wave equation

241

From (3.1) and Green formula, we have
T





E(T ) − E(S) = −


T





S Γ1

−

k
|u|2 dΓdt

a(x)g(u )u dΓdt +

1 2

S Γ1


T

t

k

(t − s)(u(t) − u(s))2 dsdΓdt

S Γ1 0


T








t

uu k − k(0) −

+

 k
(t − s)ds dΓdt.

0

S Γ1

Hence
T


a(x)g(u
)u
dΓdt

E(S) − E(T ) = S Γ1

1 + 2


T

t






T


2

k (t − s)(u(t) − u(s)) dsdΓdt − S Γ1 0

k
|u|2 dΓdt.

S Γ1


Remark 3.1. Another method of proof of Lemma 3.3 is E
(t) is negative. Using the Lemma 2.1, we can easily show that
 1

d 1 E(t) = − k (t − s, x)|u(t, x) − u(s, x)|2 − k
(t, x)|u|2 + a(x)g(u
)u
dΓ. dt 2 2 Γ1

By hypotheses of k and g, we get energy is nonincreasing. Lemma 3.4. Let M u := 2(m · ∇u) + (n − 1)u. Then
T E

p−1 2


(t) Ω

S

(|u
|2 + |∇u|2 −

2 |u|ρ+2 )dxdt ρ+2





T


T ≤ C( )

E

p−1 2

(t) Ω

S


T −λ

E

p−1 2

S

E S

E

p+1 2

p−1 2

(t)dt + CE(S) + C( )E(S)

S




2


T

k(0)|u| dΓdt − (2λ + 1 − n)

(t) Γ1


T +2

|u|2 dxdt + C

S


t


(t)

γ Γ1

E

p−1 2


(t)

u Γ1

∂u dΓdt ∂ν

2 k
(t − s, x)u(s, x)ds dΓdt

0

where 0 ≤ S < T < +∞, R = maxx∈Ω ||x − x0 ||, is an arbitrary small real number, γ = γ(x) = 2

R ∞ with λ > max{ n−1 2 , δ ||k(0)||L (Γ1 ) } and δ comes from (2.2).

λ k(0)

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J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

We note that the idea of constructions of γ and λ are taken from [7]. Proof.
T 0=

E

p−1 2




(M u)(u

− ∆u − |u|ρ u)dxdt

(t)

S

Ω


T =

E

p−1 2




(2(m · ∇u) + (n − 1)u)(u

− ∆u − |u|ρ u)dxdt

(t)

S

Ω


T E

=

p−1 2





T





2u (m · ∇u)dxdt +

(t)

S

Ω


T −

E

p−1 2

(t)

2 Γ


T + (n − 1)

E

∂u (m · ∇u)dΓdt − ∂ν

p−1 2




S

Ω


T




− (n − 1)

E

p−1 2

E

p−1 2




2|u|ρ u(m · ∇u)dxdt

(t)

S

Ω


T E

p−1 2




Γ1

∂u udΓdt − (n − 1) ∂ν

|∇u|2 dxdt

(t)

S

(t)

S

Ω


T

uu

dxdt + (n − 1)

(t)

2∇u · ∇(m · ∇u)dxdt

(t)

S




S

E




p−1 2

Ω


T E

p−1 2




|u|ρ+2 dxdt

(t)

S

Ω

Note that
T E

p−1 2




2u (m · ∇u)dxdt = E

(t)

S







p−1 2


(t) Ω

Ω

p−1 − 2
T E

p−1 2


T E

p−3 2







S


T




2u (m · ∇u)dxdt −

(t)E (t)

S

T 2u (m · ∇u)dx


Ω

E

p−1 2


(t)

S

2u
(m · ∇u
)dxdt,

Ω


2∇u · ∇(m · ∇u)dxdt

(t)

S

Ω


T = (2 − n)

E

p−1 2




|∇u| dxdt +

(t)

S


T

2

Ω

E

p−1 2


(t)

S

(m · ν)|∇u|2 dΓdt

Γ

and
T (n − 1)

E S

p−1 2


(t)



T
p−1 uu

dxdt = (n − 1) E 2 (t) u
udx

Ω

(n − 1)(p − 1) − 2

Ω


T E S

p−3 2










S


T

u udxdt − (n − 1)

(t)E (t) Ω

E S

p−1 2


(t) Ω

|u
|2 dxdt.

(3.2)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

243

Replacing above calculations in (3.2), we also obtain

T


T p−1 p−3 p−1
E 2 (t)E
(t) 2u
(m · ∇u)dxdt 0 = E 2 (t) 2u (m · ∇u)dx − 2 S Ω


T −

E

p−1 2

S


Ω


T




E

p−1 2

S

Γ




E

p−1 2

(m · ν)|∇u|2 dΓdt −

(t)

E

p−1 2

+ (n − 1)

E


(t)

2 Γ

∂u (m · ∇u)dΓdt ∂ν


T E

p−3 2

p−1 2


(t)


p−1 2





T




u udxdt − (n − 1)

(t)E (t) Ω

Ω

S




S

S

E

S

Ω


T − (n − 1)

|∇u|2 dxdt

Ω

Ω


T

∂u ∂ν ν






T
p−1 2|u|ρ u(m · ∇u)dxdt + (n − 1) E 2 (t) u
udx

(n − 1)(p − 1) − 2

Using ∇u =


T S

(t)

S

E

p−1 2

S

(t)


T −




2u (m · ∇u )dxdt + (2 − n)

(t)

S

+




Ω


T

(t)

E

p−1 2


(t)

S


T

|∇u|2 dxdt − (n − 1)

E

p−1 2


(t)

S

Γ1

|u
|2 dxdt

Ω

∂u udΓdt ∂ν

|u|ρ+2 dxdt.

Ω

on Γ0 and definition of M u, it follows that


T E

p−1 2


 (t) |u
|2 + |∇u|2 −

S

Ω


T E

=2

p−1 2


Ω

 + (n − 1) −
T +

p−1 |u| u(m · ∇u)dxdt + 2 ρ

(t)

S

E

p−1 2

2 ρ+2




S

Γ0




E S

p−1 2


T E

p−1 2




ρ+2

(t) Ω

S 2

|u|
T

(m · ν)|∇u| dΓdt +

(t)


T +

 2 ρ+2 |u| dxdt ρ+2

(t)

E


T E

p−3 2

(t)E
(t)




u
(M u)dxdt Ω

S



T
p−1
2 dxdt − E (t) u (M u)dx Ω

p−1 2

S


(t) Γ1

 (m · ν) |u
|2 − |∇u|2 dΓdt.

Γ1

Next, we will estimate the terms of the right-hand side of (3.3).

S

∂u (M u)dΓdt ∂ν

(3.3)

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J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

T p−1  Estimates for I1 := 2 E 2 (t) |u|ρ u(m · ∇u)dxdt: S

Ω


T |I1 | ≤ 2R

E

p−1 2




|∇u||u|

(t)

S

ρ+1


T dxdt ≤ 2R

Ω

E

p−1 2

(t)||∇u||2 ||u||ρ+1 2(ρ+1) dt.

(3.4)

S

By Lemma 3.1, we obtain that ||u||θp , ||u||q ≤ C||u||1−θ 2 Let us consider 0 < ρ < then

2 n−2 ,

if n ≥ 3, 0 < s <

2n n−2

θ=

p(q − 2) . q(p − 2)

(3.5)

− 2(ρ + 1) and (3.5) with q = 2(ρ + 1), p = 2(ρ + 1) + s,

||u(t)||θ2(ρ+1)+s ||u(t)||2(ρ+1) ≤ C||u(t)||1−θ 2

for all t ≥ 0,

(3.6)

where θ = ρ(2(ρ+1)+s) (ρ+1)(2ρ+s) . From (3.6), we have (1−θ)(ρ+1)

||u||ρ+1 2(ρ+1) ≤ C||u||2

Considering the choice of s we have that 2(ρ + 1) + s ≤ we obtain that

θ(ρ+1)

||u||2(ρ+1)+s .

2n n−2

(1−θ)(ρ+1)

||u||ρ+1 2(ρ+1) ≤ C||u||2

which implies that HΓ10 (Ω) → L2(ρ+1)+s . So θ(ρ+1)

||∇u||2

.

(3.7)

From (3.4) and (3.7), we have
T |I1 | ≤ 2RC

E

p−1 2

(1−θ)(ρ+1)

(t)||u||2

θ(ρ+1)+1

||∇u||2

dt.

(3.8)

S

Using Young’s inequality with p = (1−θ)(ρ+1)

||u||2

2 (1−θ)(ρ+1) θ(ρ+1)+1

||∇u||2

and q =

2 2−(1−θ)(ρ+1) ,

we get 2{θ(ρ+1)+1}

≤ C
( )||u||22 + ||∇u||22−(1−θ)(ρ+1) ,

(3.9)




where > 0 and C (ε) is a for some constant. By the inequality (2.17), we can easily check that ||∇u||22 ≤

2(ρ + 2) E(0). ρ

(3.10)

To calculate the (3.8) with (3.9) and (3.10), we obtain that I1 ≤ C ∗ ( )


T E

p−1 2

S


(t)

|u|2 dxdt + η

Ω

where

Estimates for I2 :=

p−1 2

T S

E

p−3 2

E

p+1 2

(t)dt,

(3.11)

S



C ∗ ( ) = 2RCC
( ),


T

η = 4RC

ρ 2−(1−θ)(ρ+1) 2(ρ + 2) E(0) . ρ

 (t)E
(t) u
(M u)dxdt: Ω

From Young’s inequality and Poincare’s inequality, we obtain that 
    u
(M u)dx ≤ CE.   Ω

(3.12)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

245

And as using above inequality with the nonincreasingness of the energy, we deduce that    

  p−3 E 2 (t)E
(t) u
(M u)dx ≤ −C E p+1 2 (t) .   Ω

Hence, I2 ≤ CE(S).

(3.13)

  T p−1   2 Estimates for I3 := (n − 1) − ρ+2 E 2 (t) Ω |u|ρ+2 dxdt: S Using the inequality (3.5) with q = ρ + 2 and p = 2(ρ + 1), we deduce that ρ+1

1

ρ+2 ||u||ρ+2 ≤ C||u||2ρ+2 ||u||2(ρ+1) ,

where θ =

ρ+1 ρ+2 .

Also using the fact HΓ10 (Ω) → L2(ρ+1) (Ω) and Cauchy’s inequality, ρ+1 ≤ C

( )||u||22 + C E(t), ||u||ρ+2 ρ+2 ≤ C||u||2 ||∇u||2

where > 0 and C

( ) is a for some constant. Hence,
T I3 ≤ C

E

p+1 2


T

∗∗

(t)dt + C ( )

S

E

p−1 2


(t) Ω

S

|u|2 dxdt,

(3.14)

  2 . where C ∗∗ ( ) = C

( ) (n − 1) − ρ+2

T 
p−1 Estimates for I4 := − E 2 (t) u (M u)dx : Ω

S

We can easily calculate I4 by the inequality (3.12). We substitute (3.12) for I4 , then we have I4 ≤ −C E

Estimates for I5 :=

T

E

p−1 2

(t)

 Γ1

S

By (2.8) and (2.9), we have


p−1 2

T (t)E(t) ≤ CE(S).

(3.15)

S

(m · ν)|u
|2 dΓdt:


(m · ν)|u
|2 dΓ ≤ C

|u |≤1


≤C

2

(m · ν)(u
g(u
)) p+1 dΓ

|u |≤1

2  p+1 2 (m · ν)u g(u )dΓ ≤ C(−E
) p+1







Γ1

and





2




(m · ν)|u | dΓ ≤ |u |≥1

|u |≥1

(m · ν)u
g(u
)dΓ ≤ −CE
.

246

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Hence
T E




p−1 2

(m · ν)|u | dΓdt ≤

(t)

S


T


2

E

p+1 2

(t) − C( )E
(t)dt

S

|u |≤1


T ≤

E

p+1 2

(t)dt + C ∗∗∗ ( )E(S)

S

and
T

p−1 2

E




(m · ν)|u
|2 dΓdt ≤ CE(S),

(t)

S

|u |≥1

where C ∗∗∗ ( ) is a for some constant. Replacing above inequalities, (3.11), (3.13), (3.14) and (3.15) in (3.3) and using the fact m · ν ≤ 0 on Γ0 and Young’s inequality, we obtain that
T E

p−1 2


(t)

S

(|u
|2 + |∇u|2 −

2 |u|ρ+2 )dxdt ρ+2





T

Ω


T ≤ C( )

E

p−1 2

|u| dxdt + C

(t)

S

R2 + δ

2

Ω




T

E

p−1 2

p+1 2

(t)dt + CE(S) + C( )E(S)

S




(t) Γ1

S

E

∂u ∂ν

2


T dΓdt + (n − 1)

E

p−1 2


(t)

S

u Γ1

∂u dΓdt, ∂ν

where C( ) is a constant what dependent for . Now, if we estimate the term of the right-hand side of (3.16), we obtain the result we want.   T p−1  R2 ∂u 2 ∂u 2 Estimates for I6 := S E (t) Γ1 δ ( ∂ν ) + (n − 1)u ∂ν dΓdt : From (3.1) and the Young’s inequality, we have  2  2 R2 ∂u ∂u ∂u ∂u ≤γ + k(0)u − γk 2 (0)|u|2 + (n − 1 − 2γk(0))u + (n − 1)u δ ∂ν ∂ν ∂ν ∂ν 
t 2 k
(t − s, x)u(s, x)ds ≤ 2γa2 (x)g 2 (u
) + 2γ 0

2

−λk(0)|u| + (n − 1 − 2γk(0))u

∂u . ∂ν

By calculating I5 in similarly way we obtain
T 2

E

p−1 2




2





T

γa (x)g (u )dΓdt ≤

(t)

S

2

E

p+1 2

(t)dt + C( )E(S)

S

|u |≤1

and
T 2

E S

p−1 2


(t) |u |≥1

γa2 (x)g 2 (u
)dΓdt ≤ CE(S).

(3.16)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

247

Hence,
T I6 ≤

E

p+1 2

(t)dt + C( )E(S) + CE(S)

S


T −λ

E

p−1 2




S

Γ1


T E

+2

k(0)|u|2 dΓdt

(t)

p−1 2


t


(t)

S

γ 0

Γ1


T + (n − 1 − 2λ)

2 k (t − s, x)u(s, x)ds dΓdt


E

p−1 2


(t)

S

u Γ1

∂u dΓdt. ∂ν

(3.17)


Now, we estimate the last two terms of the result of Lemma 3.4.

t 2  T p−1  Estimates for I7 := 2 S E 2 (t) Γ1 γ k
(t − s, x)u(s, x)ds dΓdt; 0

Let ξ > 0 be satisfying

ξ inf k
(0) + 1 > 0

(3.18)

Γ1

and we define (we use the idea of [7]) j := j(x) =

k(0) α(1 + ξk
(0))

on Γ1 .

By (3.18), j ≥ 0 and j ∈ L∞ (Γ1 ). On the other hand we have 
t 0

2
t k (t − s, x)u(s, x)ds − j k

(t − s, x)(u(t, x) − u(s, x))2 ds + jk
u2



t

≤

0


t 
t

2 k (t − s, x)ds k (t − s, x)u (s, x)ds − j k

(t − s, x)u2 (s, x)ds

0

0


t + 2ju

0

k

(t − s, x)u(s, x)ds − jk
(0)u2 − jk
u2 + jk
u2

0


t ≤ (k − k(0)) 




2


t

k (t − s, x)u (s, x)ds − j 0

+ j k
(0) +

1 ξ



u2 + ξj


t

k

(t − s, x)u2 (s, x)ds

0

2 k

(t − s, x)u(s, x)ds

0

1 
t

1 2 1 k(0) − j(1 + ξk
(0)) |u| ≤ k(0)|u|2 . ≤ k

(t − s, x)u2 (s, x)ds + j k
(0) + α ξ ξα 0

248

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

As j ∈ L∞ (Γ1 ),
T E

p−1 2


t


(t)

S

2 k
(t − s, x)u(s, x)ds

2γ 0

Γ1


t −j 0


T
p−1 2λ 2 k (t − s, x)(u(t, x) − u(s, x)) ds + jk u dΓdt ≤ E (t) |u|2 dΓdt ξα



2


2

S

Γ1

By the above inequality and Lemma 3.3, we obtain
T E

2

p−1 2


t


(t)

S

0

Γ1

2λ ≤ CE(S) + ξα T

p−1 2

2 k
(t − s, x)u(s, x)ds dΓdt

γ


T E

p−1 2


(t)

S

|u|2 dΓdt

(3.19)

Γ1



Estimates for I8 := − S E (t) Γ1 ∂u ∂ν udΓdt ; Let ϕ be a solution of −∆ϕ = |u|ρ u in Ω, ϕ = u on Γ. By the classical results of elliptic partial differential equations theory we have



∇ϕ · ∇udx − |u|ρ+2 dx = |∇ϕ|2 dx − |u|ρ uϕdx, Ω







2

Ω

|ϕ| dx ≤ C Ω

−

E

p−1 2


(t)

S

|u| dΓ

ϕ Γ

p−1 2

Ω





2




|ϕ | dx ≤ C

and

Γ

Multiplying (3.1) with E
T

Ω 2

Ω

|u
|2 dΓ.

Γ

ϕ and integrating from S to T we obtain

∂u dΓdt ∂ν



T

T


T p−1 p−3 p−1 p−1
=− E 2 u ϕdx + E 2 (t)E
(t) u
ϕdxdt + E 2 (t) u
ϕ
dxdt 2 S S Ω


T −

E

p−1 2

S

∇u∇ϕdxdt +

(t)

S

Ω


T


Ω

E

p−1 2


(t)

S

Ω

|u|ρ uϕdxdt

Ω

T



T
T p−1 p−3 p−1 p−1


2 2 2 u ϕdx + E (t)E (t) u ϕdxdt + E (t) u
ϕ
dxdt =− E 2 S Ω


T −

E

p−1 2

S


(t)

S

|∇ϕ|2 dxdt −

Ω


T E

+2 S

p−1 2

E S


(t) Ω

Ω


T

|u|ρ uϕdxdt.

p−1 2


(t)

S

Ω

|u|ρ+2 dxdt

Ω

(3.20)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

Note that

249



T
p−3
2 − E u ϕdx ≤ CE(S), S

Ω


T E

p−3 2

(t)E
(t)


Ω

S


T E

p−1 2




u
ϕ
dxdt

(t)

S

u
ϕdxdt ≤ CE(S),

Ω


T ≤

E

p−1 2





T


2

|u | dxdt + C( )

(t)

E

S

Ω

S


T





T

≤

E

p−1 2

|u
|2 dxdt + C( )

(t)

S

Ω

p−1 2


Ω

E

p−1 2




≤ CE(S) + C( )E(S) + 3

E

p+1 2

|u
|2 dxdt

(t)

S


T

|ϕ
|2 dxdt

(t)

Γ1

(t)dt,

S

and using the imbedding HΓ10 (Ω) → L2(ρ+1) (Ω) and Poincare’s inequality, we get
T E

2

p−1 2




|u|ρ uϕdxdt

(t)

S

Ω


T ≤

E

p−1 2




|u|

(t)

S

2(ρ+1)


T dxdt +

Ω


T ≤C

E

p+1 2

E S

p−1 2


(t)

|ϕ|2 dxdt

Ω

(t)dt.

S

Replacing above calculations in (3.20), we obtain
T −

E

p−1 2


(t)

S

Γ

∂u ϕ dΓdt ≤ CE(S) + C( )E(S) + C ∂ν


T E

p+1 2

(t)dt

(3.21)

S

p−1 2

Multiplying (2.11) by E (t) and then integrating from S to T and using Lemma 3.4, (3.19) and (3.21) and then we choose > 0 such that is small enough, we can obtain that
T E S

p+1 2


T (t)dt ≤ C

E

p−1 2


(t)

S

Ω


T + CE(S) + C

E S

|u|2 dxdt

p−1 2


 (1 − λ)k(0) +

(t) Γ1

2λ ξα



|u|2 dΓdt.

(3.22)

250

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Now, we estimate the last term of right-hand side of (3.22). we would like to show that  p−1 2λ E 2 (t) Γ1 ((1 − λ)k(0) + ξα )|u|2 dΓdt ≤ 0. This inequality is sufficient if we can choose λ such S 2λ that (1 − λ)k(0) + ξα ≤ 0. The condition (2.6) implies that there exist ξ
> 0 such that α inf Γ1 k(0) > −(2 + ξ
) inf Γ1 k
(0) (see [7]). We choose ξ > 0 such that

T

ξ
+ 4 . 2(ξ
+ 2)

−ξ inf k
(0) = Γ1

Then, we get  2λ λ ξ
≤ 2 + ξ(2 + ξ
) inf k
(0) + k(0) = − λ + k(0). Γ1 ξα ξα 2ξα

(1 − λ)k(0) + Therefore, if we choose

 λ = max n − 1,

R2

2ξα  +
||k(0)||L∞ (Γ1 ) δ ξ



then, we have (1 − λ)k(0) +

2λ ≤ 0. ξα

(3.23)

Next lemma is estimated the first term of the fight-hand side of (3.22). Lemma 3.5. If u(x, t) is a solution of (3.1), then it holds that
T E

p−1 2


(t)

S

|u|2 dxdt

Ω


T ≤C

E S

p−1 2



t  1 1 (t) k

(t − s, x)|u(t, x) − u(s, x)|2 ds − k
(t, x)|u|2 + a2 (x)g 2 (u
) dΓdt. 2 2 0

Γ1

(3.24) Proof. We will argue by contradiction. If (3.24) was false and let {ui (0), u
i (0)} be a sequence of initial data where the corresponding solutions {ui } of (3.1) with Ei (0) uniform bounded in i, verifies ⎡ lim ⎣ i→∞  T S

T p−1 2

Ei

(t)

 Γ1

S  1 t 2 0

T + T S

p−1 2

Ei

S

(t)

  Γ1

p−1 2

Ei

(t)

 Ω

|ui |2 dxdt

k

(t − s, x)|ui (t, x) − ui (s, x)|2 dsdΓdt p−1 2

Ei

(t)



|ui |2 dxdt Ω

 − 12 k
(t, x)|ui |2 + a2 (x)g 2 (u
i ) dΓdt

where Ei (t)is defined by E(t) with u replaced by ui .

⎤ ⎦ = ∞,

(3.25)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

251

Since Ei (0) is uniformly bounded in i, we have Ei (t) ≤ C for all i ∈ N, for all t ≥ 0. Then, we get a subsequence, still denoted by {ui }, that satisfies following properties ui → u weakly in H 1 (Q), Q = Ω × (0, T ), ui → u weak star in L∞ (0, T ; HΓ10 (Ω)), u
i → u
weak star in L∞ (0, T ; L2 (Ω)), ui → u weakly in L2 (0, T ; L2 (Γ)). By compactness results we have that ui → u strongly in L2 (0, T ; L2 (Ω))

(3.26)

ui → u strongly in L2 (0, T ; L2 (Γ)).

(3.27)

and

In what follows we are going to use the ideas contained in Lasiecka and Tataru [12], applied to our context. Assume that u = 0. According to (3.26) we obtain that |ui |ρ ui → |u|ρ u

a.e. in Q.

From the above convergence and the sequence {|ui |ρ ui } is bounded in L2 (0, T ; L2 (Ω)) we conclude by Lions’ Lemma that |ui |ρ ui → |u|ρ u Also, since

T S

p−1 2

Ei

(t)




T

Ω

weakly in L2 (0, T ; L2 (Ω)).

|ui |2 dxdt is bounded,



t 1 (t) k

(t − s, x)|ui (t, x) − ui (s, x)|2 dsdΓ 2 0 Γ1

1
k (t, x)|ui |2 dΓ + a2 (x)g 2 (u
i )dΓ dt → 0 as i → ∞. − 2 p−1 2

Ei

S

Γ1

(3.28)

Γ1

As S is chosen in the interval [0, T ) and by assumption of k
, k

, a(x) and g(x), we can obtain that u(s, x) = u(x) a.e. in Γ1 × (0, T ), ui → 0 strongly in L2 (0, T ; L2 (Γ1 )), a(x)g(u
i ) → 0 strongly in L2 (0, T ; L2 (Γ1 )). Passing to the limit in the equation, when i → ∞, we get for u, ⎧

ρ ⎪ ⎨u − ∆u = |u| u in Q, u = 0 on Γ0 × (0, T ), ⎪ ⎩ ∂u u
= 0 on Γ1 × (0, T ), ∂ν = 0, and for u
= v, ⎧

ρ ⎪ ⎨v − ∆v = (ρ + 1)|u| v in Q, v = 0 on Γ0 × (0, T ), ⎪ ⎩ ∂v ∂ν = 0, v = 0 on Γ1 × (0, T ).

(3.29)

252

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Since u ∈ L∞ (0, T ; HΓ10 (Ω)), (ρ + 1)|u|ρ ∈ L∞ (0, T ; Ln (Ω)). Then, by the result of [12], we conclude that v = u
≡ 0 for sufficiently large T . Returning to (3.29) we obtain the following elliptic equation: ⎧ ρ ⎪ ⎨−∆u = |u| u in Ω, u = 0 on Γ0 , ⎪ ⎩ ∂u on Γ1 , ∂ν = 0, Multiplying the equation by u, we get


|∇u|2 dx −

Ω




|u|ρ+2 dx = 0,

Ω

hence J(u) =

ρ ||∇u||22 . 2(ρ + 2)

J(u) >

ρ ||∇u||22 , 2(ρ + 2)

But if u = 0, we obtain that

because of (2.16). This is a contradiction. Assume that u ≡ 0. Setting

χ2i


T =

p−1 2

Ei


(t)

S

|ui |2 dxdt and ui (t) =

Ω

ui (t) . χi

Then, we get
T

p−1 2

Ei


(t)

0

Ω

1 |ui | dxdt = 2 χi 2


T

p−1 2

Ei S


(t)

|ui |2 dxdt = 1.

(3.30)

Ω

In addition, as u = 0, we have that χi → 0 as i → ∞. As S is chosen in the interval [0, T ), from (3.25) we have
T

p−1 2

Ei 0


− Γ1



t 1 (t) k

(t − s, x)|ui (t, x) − ui (s, x)|2 dsdΓ 2 Γ1

0

1
k (t, x)|ui |2 dΓ + 2


 Γ1

2 1 a(x)g(u
i ) dΓ dt → 0 as i → ∞. χi

(3.31)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

253

On the other hand, 1 E i (t) = 2 −


|u
Ω

1 2




1 i (t, x)| dx + 2 2


t




Ω

|u
i (t)|2 dx +



t −

Ω




1 |∇ui (t, x)| dx + 2 2

k(t, x)|ui (t, x)|2 dΓ

Γ1

k
(t − s, x)|ui (t, x) − ui (s, x)|2 dsdΓ −

Γ1 0

1 ≤ 2χ2i







|∇ui (t)|2 dx +

Ω




1 ρ+2




|ui (t, x)|ρ+2 dx

Ω

k(t, x)|ui (t, x)|2 dΓ

Γ1

k (t − s, x)|ui (t, x) − ui (s, x)| dsdΓ .


2

Γ1 0

From (2.16), we have  1 ρ + 2 1 1 ||∇ui (t)||22 ≤ ||∇ui (t)||22 − ||ui (t)||ρ+2 ρ+2 . 2 ρ 2 ρ+2 Hence E i (t) ≤

ρ+2 Ei (t), ρχ2i

it implies that ||u
i (t)||22 and ||∇ui (t)||22 are bounded. Then, in particular for a subsequence {ui }, from boundedness of ||u
i (t)||22 and ||∇ui (t)||22 and (3.31) we obtain ui → u u
i → u


weak star in L∞ (0, T ; HΓ10 (Ω)), weak star in L∞ (0, T ; L2 (Ω)),

ui → u strongly in L2 (0, T ; L2 (Ω)), u(s, x) = u(x) a.e. in Γ1 × (0, T ), ui → 0 strongly in L2 (0, T ; L2 (Γ)), 1
2 2 χi a(x)g(ui ) → 0 strongly in L (0, T ; L (Γ1 )). In addition, ui satisfies the equation ⎧ ⎪ u

i − ∆ui = |ui |ρ ui in Ω × (0, T ), ⎪ ⎪ ⎨ ui = 0 on Γ0 × (0, T ), t ⎪ ⎪ ⎪ ∂ui + k
(t − s, x)ui (s, x)ds + k(0)ui + ⎩ ∂ν 0

(3.32) 1
χi a(x)g(ui )

= 0 on Γ1 × (0, T ).

We note that
T


||ui |ρ ui |2 dxdt =

0 Ω


Q

|ui |2 |ui |2ρ dxdt


2

2ρ




|ui | |ui | dxdt +

= |ui |≤

|ui |>

|ui |2 |ui |2ρ dxdt,

(3.33)

254

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

where Q = Ω × (0, T ). According to the fact that the function F (s) = |s|ρ is continuous in R and S = sup|x|≤ |F (x)| is well defined, from (3.33) we have
T
0

Ω

2ρ+2 ||ui |ρ ui |2 dxdt ≤ S 2 ||ui ||2L2 (Q) + χ2ρ i ||ui ||L2ρ+2 (Q) .

Since {ui } is bounded in L∞ (0, T ; HΓ10 (Ω)) → L∞ (0, T ; L2ρ+2 (Ω)), there exists C > 0 such that
T
Ω

0

||ui |ρ ui |2 dxdt ≤ C(S 2 + χ2ρ i ).

Then, taking ε → 0 and i → ∞ we conclude that |ui |ρ ui → 0

in

L2 (0, T ; L2 (Ω))

as

i → ∞.

(3.34)

Passing to the limit in (3.32) as i → ∞, we obtain ⎧

⎪ ⎨u − ∆u = 0 in Ω × (0, T ), u = 0 on Γ0 × (0, T ), ⎪ ⎩ ∂u ∂ν = 0 on Γ1 × (0, T ).

(3.35)

Furthermore
T E 0

p−1 2


(t)

|u|2 dxdt = 1

(3.36)

Ω

Then v = u
satisfies (differentiating problem (3.35) with respect to t) ⎧

⎪ ⎨v − ∆v = 0 in Ω × (0, T ), v = 0 on Γ0 × (0, T ), ⎪ ⎩ ∂v ∂ν = 0 on Γ1 × (0, T ). Hence, we can obtain that v = u
= 0 (see [19]). We are able to rewrite (3.35) as below ⎧ ⎪ ⎨−∆u = 0 in Ω, u = 0 on Γ0 , ⎪ ⎩ ∂u ∂ν = 0 on Γ1 . Since u ∈ HΓ10 (Ω) we conclude from this that u ≡ 0 in Ω. This is a contradiction to (3.36).




Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

255

We replace (3.23) and (3.24) in (3.22), we have
T E

p+1 2

(t)dt

S


T ≤C

p−1 2

E S



t  1 1 (t) k

(t − s, x)|u(t, x) − u(s, x)|2 ds − k
(t, x)|u|2 dΓdt 2 2 Γ1




T

+C

E

p−1 2




(t)

S


T ≤C

E

0

a2 (x)g 2 (u
)dΓdt + CE(S)

Γ1 p−1 2

  (t) −E
(t) dt + C

S

E

p−1 2




+C

E

p−1 2

S


(t)

a2 (x)g 2 (u
)dΓdt

(t)

S


T

|u |≤1

a2 (x)g 2 (u
)dΓdt + CE(S)

|u |≥1


T ≤ C


T

E

p+1 2

(t)dt + CE(S) + C( )E(S).

S

Hence we can choose that is satisfied the following inequality
T E

p+1 2

(t)dt ≤ CE(S).

S

Consequently, we can arrive at the result of Theorem 2.2 using the Lemma 3.1. Remark 3.2. If we except (2.9) from hypotheses, p ≥ n − 1 and g is bounded and Lipschitz continuous,  p−1 p+1 we shall estimate E 2 (t) Γ1 |u
|2 dΓ ≤ E 2 (t) − C( )E
for every > 0 to obtain same result of Theo p−1 rem 2.2 (cf. estimation of I5 and I6 ). In fact, we can calculate the inequality E 2 (t) Γ1 ,|u |≥1 |u
|2 dΓ ≤

E

p+1 2

2p

2p


1 p−1 (Γ), we get the formula that we want (t)||u
|| p−1 2p + C( )E . Using the trace theorem H (Ω) → L p−1

(cf. [8] Theorem 9.10 and Remark 9.11).

4. Proof of Theorem 2.3 In the Sect. 3, using a polynomial growth of the function g near the origin, we proved the energy decay rates for solution of (3.1). However, having hypotheses (H3 ), we can not prove the energy decay rates like Sect. 3 since (H3 ) means that g has not a polynomial behavior near the origin. It implies that we can not use Lemma 3.1 in this section. So, we present two technical lemmas which will play an essential role when establishing the asymptotic behavior. Lemma 4.1. ([13]) Let E : R+ → R+ be a nonincreasing function and φ : R+ → R+ a strictly increasing function of class C 1 such that φ(0) = 0

and

φ(t) → +∞

as t → +∞.

256

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Assume that there exists σ > 0, σ
≥ 0 and C > 0 such that +∞
E 1+σ (t)φ
(t)dt ≤ CE 1+σ (S) + S

C E σ (0)E(S), (1 + φ(S))σ

0 ≤ S < +∞.

Then, there exists C > 0 such that E(t) ≤ E(0)

C , (1 + φ(t))(1+σ )/σ

for all t > 0.

Lemma 4.2. ([13]) There exists a function φ : R+ → R+ of class C 2 increasing and such that φ is concave, φ(t) → +∞ as t → +∞, φ
(t) → 0 as t → +∞ and +∞
φ
(t)(β −1 (φ
(t)))2 dt < +∞, 1

where β is defined on (H3 ). Now, we begin the proof of our main result. In the following section the symbol C indicates positive constants, which may be different. Let us multiply (3.1) by E(t)φ
(t)M u, where φ(t) is a function under the hypotheses of Lemmas 4.1 and 4.2. Then we have
T 0=

E(t)φ
(t)

S







E(t)φ (t) S

(2(m · ∇u) + (n − 1)u)(u

− ∆u − |u|ρ u)dxdt

Ω


T =

(M u)(u

− ∆u − |u|ρ u)dxdt

Ω


T =










2u (m · ∇u)dxdt +

E(t)φ (t) S


T





Ω


T −

E(t)φ
(t)

S

S


2 Γ


T + (n − 1)

∂u (m · ∇u)dΓdt − ∂ν





S

Ω


T




S

E(t)φ
(t)





Γ1


2∇u · ∇(m · ∇u)dxdt Ω


T

E(t)φ
(t)

S




2|u|ρ u(m · ∇u)dxdt

Ω


T

uu dxdt + (n − 1)

E(t)φ (t)

− (n − 1)

E(t)φ
(t)







E(t)φ (t) S

∂u udΓdt − (n − 1) ∂ν


T S

|∇u|2 dxdt

Ω

E(t)φ
(t)


Ω

|u|ρ+2 dxdt

(4.1)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

257

Note that
T







2u

(m · ∇u)dxdt

E(t)φ (t) S

Ω



T
T




E (t)φ
(t) + E(t)φ

(t) = E(t)φ (t) 2u (m · ∇u)dx − 2u
(m · ∇u)dxdt S

Ω


T −

E(t)φ
(t)

S


T




S

Ω

2u
(m · ∇u
)dxdt,

Ω

E(t)φ
(t)

S


2∇u · ∇(m · ∇u)dxdt Ω


T = (2 − n)







|∇u| dxdt +

E(t)φ (t) S


T

2

Ω

E(t)φ
(t)

S




(m · ν)|∇u|2 dΓdt

Γ

and
T (n − 1)

E(t)φ
(t)




uu

dxdt Ω

S



T


T 


E (t)φ
(t) + E(t)φ

(t) = (n − 1) E(t)φ (t) u udx − (n − 1) u
udxdt S

Ω


T −(n − 1)

E(t)φ
(t)

S




S

Ω

|u
|2 dxdt.

Ω

Replacing above calculations in (4.1), we also obtain
T

E(t)φ
(t)

S




|u
|2 + |∇u|2 −

Ω


T =2

E(t)φ
(t)

S

 + (n − 1) −




 2 |u|ρ+2 dxdt ρ+2

|u|ρ u(m · ∇u)dxdt +

Ω

2 ρ+2


T

:=J1





 E
(t)φ
(t) + E(t)φ

(t)







ρ+2

|u|

E(t)φ (t) S

Ω

 u
(M u)dxdt

Ω

S


T






T


dxdt − E(t)φ (t) u (M u)dx Ω

S

258

J. Y. Park, T. G. Ha and Y. H. Kang


T +

E(t)φ
(t)

S


T +




(m · ν)|∇u|2 dΓdt +

Γ0


T

ZAMP

:=J2




E(t)φ
(t)

S





(m · ν)|u
|2 dΓdt

Γ1

:=J3

E(t)φ
(t)

S


Γ1

 

T ∂u (M u)dΓdt − E(t)φ
(t) (m · ν)|∇u|2 dΓdt . ∂ν S

(4.2)

Γ1

Next, we will estimate the terms of the right-hand side of (4.2).  T Estimates for J1 := S E
(t)φ
(t) + E(t)φ

(t) Ω u
(M u)dxdt; From (3.12), we have
T J1 ≤








T





|E (t)φ (t) + E(t)φ (t)|E(t)dt ≤ C S




2


T

−E (t)E(t)dt + CE (S) S

−φ

(t)dt

S

≤ C(E 2 (S) − E 2 (T )) − CE 2 (S)(φ
(S) − φ
(T )) ≤ CE 2 (S). T  Estimates for J2 := S E(t)φ
(t) Γ1 (m · ν)|u
|2 dΓdt ; For every t > 1 let us define

(4.3)

Γ1,1 = {x ∈ Γ1 ; |u
| ≤ h(t)}, Γ1,2 = {x ∈ Γ1 ; h(t) < |u
| ≤ h(1)}, Γ1,3 = {x ∈ Γ1 ; |u
| > h(1)}, where each Γ1,i depends on t and h(t) = β −1 (φ
(t)) such that h is a decreasing positive function which satisfies h(t) → 0 as t → +∞. Now, let us consider these three cases. Case 1 Part on Γ1,3 . First, we claim that h(1) > 0. Indeed, we assume that h(1) = 0. Then β −1 (φ
(1)) = h(1) = 0, i.e.,
φ (1) = β(0) = 0. It implies that φ
(t) ≤ φ
(1) for all t ≥ 1, consequently, φ
(t) = 0 for all t ≥ 1. This contradicts the fact that φ is strictly increasing. Thus, h(1) > 0. If h(1) > 1, then from (2.22) |g(u
)| ≥ C6 |u
|. If h(1) ≤ 1, we note that the function H : s → g(s) s is positive and continuous on [−1, −h(1)] ∪ [h(1), 1]
which implies that there exists a positive constant C8 satisfying g(s) s ≥ C8 for |s| ∈ [h(1), 1], i.e., g(u ) ≥ C8 |u
|. Hence, for C9 = min{C6 , C8 }, |u
| ≤ C19 |g(u
)|. Then we have
T S

E(t)φ
(t)


Γ1,3

(m · ν)|u
|2 dΓdt ≤ RE(S)


T

φ
(t)

S

≤

Rφ
(S) E(S) a0 C9




|u
| |g(u
)|dΓdt

Γ1,3


T




a(x)u
g(u
)dΓdt

S Γ1

Rφ
(S) 2 ≤ E (S). a0 C9 Case 2 Part on Γ1,2 . Since β is increasing, if x ∈ Γ1,2 , then φ
(t) = β(h(t)) ≤ β(|u
|) = |β(u
)|. If h(1) < 1, then |u
| < 1, consequently, by (2.21), |u
|2 |β(u
)| ≤ u
g(u
).

(4.4)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

259

If h(1) ≥ 1 and |u
| ∈ [1, h(1)], then −h(1) ≤ u
≤ h(1). Since β is an increasing and odd function, β(h(1))

|β(u
)| ≤ β(h(1)). Hence, from (2.22) |u
|2 |β(u
)| ≤ β(h(1)) C6 u g(u ). Therefore, for C10 = max{1, C6 },
2


|u | |β(u )| ≤ C10 u g(u ). Consequently, we obtain
T







|u
|2 |β(u
)|dΓdt

(m · ν)|u | dΓdt ≤ RE(S)

E(t)φ (t) S


T



2

Γ1,2

S Γ1,2

RC10 ≤ E(S) a0


T


a(x)u
g(u
)dΓdt

S Γ1

RC10 2 E (S). a0

≤

(4.5)

Case 3 Part on Γ1,1 . By the definition of the boundary of this paper, we have
T







(m · ν)|u
|2 dΓdt

E(t)φ (t) S

Γ1,1


T ≤ RE(S)







h (t)dΓdt ≤ meas(Γ)RE(S)

φ (t) S


T

2

Γ1,1

φ
(t)(β −1 (φ
(t)))2 dt.

(4.6)

S

From (4.4)–(4.6), we get
T

2

J2 ≤ CE (S) + CE(S)

φ
(t)(β −1 (φ
(t)))2 dt.

(4.7)

S

Estimates for J3 :=

T S

E(t)φ
(t)



∂u (M u)dΓdt Γ1 ∂ν

−

T



E(t)φ
(t)

S

Γ1

(m · ν)|∇u|2 dΓdt;

From (3.1) and Young’s inequality and by same arguments as I7 and I8 , we have 
 2  2
T ∂u ∂u R
dΓdt + (n − 1)u J3 ≤ E(t)φ (t) δ ∂ν ∂ν S

Γ1


T ≤2







2

E(t)φ (t) S

2


T




γa (x)g (u )dΓdt + 2 Γ1


T −λ





t




E(t)φ (t) S

E(t)φ
(t)

S




γ Γ1

k(0)|u|2 dΓdt + (n − 1 − 2λ)

Γ1


T

≤ CE 2 (S) + C( )E 2 (S) + C

0

E(t)φ
(t)

S


T

2 k (t − s, x)u(s, x)ds dΓdt



u Γ1

∂u dΓdt ∂ν

E 2 (t)φ
(t)dt

S


T +2







E(t)φ (t) S

2

2





T

γa (x)g (u )dΓdt + Γ1

S

E(t)φ
(t)


 −λk(0) + Γ1

2λ ξα



|u|2 dΓdt.

(4.8)

260

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

Furthermore, other terms of the right-hand side of (4.2) can be obtained easily by same arguments as I1 –I3 , that is,
T

 |u| u(m · ∇u)dxdt + (n − 1) −







ρ

E(t)φ (t)

2 S

Ω

2 ρ+2


T







E(t)φ (t) S

|u|ρ+2 dxdt

Ω



T
T

− E(t)φ
(t) u
(M u)dx + E(t)φ
(t) (m · ν)|∇u|2 dΓdt S

Ω


T

2

2

≤ CE (S) + C

S

Γ0


T




E (t)φ (t)dt + C( ) S







E(t)φ (t) S

|u|2 dxdt.

(4.9)

Ω

Replacing (4.3), (4.7)–(4.9) in (4.2), we obtain
T

E(t)φ
(t)

S




|u
|2 + |∇u|2 −

Ω

2

 2 |u|ρ+2 dxdt ρ+2
T

2

2

≤ CE (S) + C( )E (S) + C


T




E (t)φ (t)dt + C( ) S


T +2







E(t)φ (t) S

2

2

S


T




γa (x)g (u )dΓdt + Γ1


T + CE(S)

E(t)φ
(t)







E(t)φ (t) S

Γ1




|u|2 dxdt

Ω

2λ −λk(0) + ξα



|u|2 dΓdt

φ
(t)(β −1 (φ
(t)))2 dt.

(4.10)

S

Multiplying (2.11) by E(t)φ
(t) and then integrating from S to T and using (4.10) and choosing > 0 sufficiently small, we can obtain that
T

E 2 (t)φ
(t)dt ≤ CE 2 (S) + C

S


T

E(t)φ
(t)

S


T +C





Γ1


T + CE(S) S

|u|2 dxdt

Ω

E(t)φ (t) S




1 a (x)g (u )dΓdt + 2 2

2




φ
(t)(β −1 (φ
(t)))2 dt.


T S

E(t)φ
(t)


 (1 − λ)k(0) + Γ1

2λ ξα



|u|2 dΓdt

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

261

From the fact (3.23) and by the same arguments as Lemma 3.5 we rewrite above inequality that
T

2


T




2


T


1 2

E (t)φ (t)dt ≤ CE (S) + C S

E(t)φ (t) S

+C

E(t)φ
(t)

S

Γ1


T

2


t

2

2


T




a (x)g (u )dΓdt + CE(S) Γ1

φ
(t)(β −1 (φ
(t)))2 dt

S

 1 k

(t − s, x)|u(t, x) − u(s, x)|2 ds − k
(t, x)|u|2 + a2 (x)g 2 (u
) dΓdt 2

0


≤ CE (S) + CE(S)







φ (t)(β

−1





T

2

(φ (t))) dt + C

S







E(t)φ (t) S

a(x)g 2 (u
)dΓdt.

(4.11)

Γ1

Now, we will estimate the last term of the right-hand side of (4.11). T  Estimates for J4 := S E(t)φ
(t) Γ1 a(x)g 2 (u
)dΓdt ; For every t ≥ 1, let us define Γ1,4 = {x ∈ Γ1 ; |u
| ≤ φ
(t)}, Γ1,5 = {x ∈ Γ1 ; φ
(t) < |u
| ≤ φ
(1)}, Γ1,6 = {x ∈ Γ1 ; |u
| > φ
(1)}. Let us consider these three cases. Case 1 Part on Γ1,6 . We can easily see that
T







2

a(x)g (u )dΓdt ≤ C

E(t)φ (t) S


T




Γ1,6







E(t)φ (t) S

a(x)u
g(u
)dΓdt ≤ CE 2 (S).

(4.12)

a(x)u
g(u
)dΓdt ≤ CE 2 (S).

(4.13)

Γ1,6

Case 2 Part on Γ1,5 . By monotonicity of g, we have
T







E(t)φ (t) S

a(x)g 2 (u
)dΓdt

Γ1,5





T ≤

2





T




a(x)|u |g (u )dΓdt ≤ C

E(t) S




Γ1,5

E(t) S

Γ1,5

Case 3 Part on Γ1,4 .
T

E(t)φ
(t)

S




a(x)g 2 (u
)dΓdt

Γ1,4


T ≤C







E(t)φ (t) S

Γ1,4

(β

−1




2


T

(|u |)) dΓdt ≤ C(meas(Γ))E(S) S

φ
(t)(β −1 (φ
(t)))2 dt.

(4.14)

262

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

From (4.12)–(4.14), we get
T

2

J4 ≤ CE (S) + CE(S)

φ
(t)(β −1 (φ
(t)))2 dt.

(4.15)

S

Replacing (4.15) in (4.11), we obtain
T

2





T

2

E (t)φ (t)dt ≤ CE (S) + CE(S) S

φ
(t)(β −1 (φ
(t)))2 dt.

(4.16)

S

Finally, we will estimate the second term of the right hand side of (4.16). By condition of φ, we can consider, without loss of generality that φ(1) = 1. Considering the change of variables s = φ(t), one has +∞ +∞ +∞ 



−1
2 −1
−1 2 −1 φ (t)(β (φ (t))) dt = (β (φ (φ (s)))) ds = β 1

1

1

1 (φ−1 )
(s)

2 ds.

Let us consider the function ψ by
t ψ(t) = 1 + 1

1 ds, β( 1s )

t ≥ 1.

Then ψ is a strictly increasing function of class C 2 that satisfies 1 ψ
(t) = → +∞ ψ(t) → +∞ as t → +∞ β( 1t ) and +∞


β

−1

1



1 ψ
(s)

2

+∞


ds = 1

1 ds < +∞. s2





A simple computation shows that ψ ≥ 0 which implies that ψ
is nondecreasing and ψ is convex. Moreover, it is easy to verify that ψ −1 is concave. Setting φ(t) = ψ −1 (t), then we can rewrite (4.16) as
T

E 2 (t)φ
(t)dt ≤ CE 2 (S) + CE(S)

S 2

+∞
φ
(t)(β −1 (φ
(t)))2 dt S +∞ 


≤ CE (S) + CE(S)

β

φ(S)

−1

2 1 (
) ds ψ (s)

C E(S), ≤ CE 2 (S) + φ(S) and applying Lemma 4.1 with σ = σ
= 1, we deduce C for all t > 0. E(t) ≤ 2 φ (t) Let s0 be a number such that β( s10 ) ≤ 1. Since β is nondecreasing, we have ψ(s) ≤ 1 + (s − 1)

1 1 ≤ , β( 1s ) F ( 1s )

for all s ≥ s0 .

(4.17)

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

263

Consequently, having in mind that φ = ψ −1 , the last inequality yields

1  1 = φ(t) with t = s≤φ . F ( 1s ) F ( 1s ) Then 1 1 ≤ F −1 ( ). φ(t) t

(4.18)

Combining (4.17) and (4.18) we can prove (2.23). It remains to prove (2.24). Assume that G(0) = 0 and G is nondecreasing on [0, η] for some η > 0. Let T1 be a number such that G( 1t ) ≤ η for all t ≥ T1 and set T2 = sup{T1 , η1 }. Assume that φ is an increasing and concave function such that for all t ≥ T2 , φ
(t) ≤ η and φ
(t) ≤ G(η) and φ
(t) → 0 as t → +∞. Then thanks to Martinez’s idea ([13], p. 438), we obtain
T J2 =








T

2

(m · ν)|u | dΓdt ≤ CE (S) + CE(S)

E(t)φ (t) S


2

Γ1

2 φ
(t) G−1 (φ
(t)) dt

S

and
T J4 =







2


T

2

a(x)g (u )dΓdt ≤ CE (S) + CE(S)

E(t)φ (t) S




Γ1

2 φ
(t) G−1 (φ
(t)) dt.

S

Therefore, by the same arguments as (4.16) we have
T

2





T

2

E (t)φ (t)dt ≤ CE (S) + CE(S) S

φ
(t)(G−1 (φ
(t)))2 dt.

S

Define φ˜−1 (t) = T2 +




t

T2

1 ds, G( 1s )

for all t ≥ T2 .

Then ˜ ≥ T2 ≥ 1 , φ(t) η so φ˜
(t) = G



1 ˜ φ(t)

 ≤ G(η)

and φ˜
(t) ≤ φ˜
(T2 ) = G



1 ˜ 2) φ(T

 ≤ G(

1 )≤G T2



1 T1

 ≤η

for all t ≥ T2 .

Then φ˜ satisfies all the required properties. Thus from the same argument as (4.17) we obtain C for all t > T2 . E(t) ≤ 2 ˜ φ (t) Since for all s ∈ [0, 1], β(s) ≤ s i.e., G(s) ≤ 1, we see that t − T2 t 1 φ˜−1 (t) ≤ T2 + ≤ = . G( 1t ) G( 1t ) β( 1t )

(4.19)

(4.20)

Combining (4.19) and (4.20) we can prove (2.24). Thus the proof of Theorem 2.3 is completed.

Acknowledgments The authors are thankful to the referee of this paper for valuable suggestions which improved this paper.

264

J. Y. Park, T. G. Ha and Y. H. Kang

ZAMP

References 1. Aassila, M., Cavalcanti, M.M., Soriano, J.A.: Asymptotic stability and energy decay rates for solutions of the wave equation with memory in a star-shaped domain. SIAM J. Control Optim. 38(5), 1581–1602 (2000) 2. Cavalcanti, M.M., Cavalcanti, V.N., Lasiecka, I.: Well-posedness and optimal decay rates for the wave equation with nonlinear boundary damping-source interation. J. Differ. Equ. 236, 407–459 (2007) 3. Cavalcanti, M.M., Cavalcanti, V.N., Martinez, P.: Existence and decay rate estimates for the wave equation with nonlinear boundary damping and source term. J. Differ. Equ. 203, 119–158 (2004) 4. Cavalcanti, M.M., Guesmia, A.: General decay rates of solutions to a nonlinear wave equation with boundary condition of memory type. Differ. Integral Equ. 18(5), 583–600 (2005) 5. Cavalcanti, M.M., Oquendo, H.P.: Frictional versus viscoelastic damping in a semilinear wave equation. SIAM J. Control Optim. 42(4), 1310–1324 (2003) 6. Guesmia, A.: A new approach of stabilization of nondissipative distributed systems. SIAM J. Control Optim. 42(1), 24–52 (2003) 7. Guesmia, A.: Satabilisation de l’´equation des ondes avec condition aux limites de type m´ emoire. Afr. Matematika 3(10), 14–25 (1999) 8. Komornik, V.: Exact Controllability and Stabilization. The Multiplier Method. Masson, Chicester (1994) 9. Komornik, V.: On the nonlinear boundary stabilization of the wave equation. China Ann. Math. 14B(2), 153–164 (1993) 10. Komornik, V., Zuazua, E.: A direct method for the boundary stabilization of the wave equation. J. Math. Pure Appl. 69, 33–54 (1990) 11. Lagnese, J.: Decay of solutions of the wave equation in a bounded region with boundary dissipation. J. Differ. Equ. 50, 163–182 (1983) 12. Lasiecka, I., Tataru, D.: Uniform boundary stabilization of semilinear wave equation with nonlinear boundary damping. Differ. Integral Equ. 6(3), 507–533 (1993) 13. Martinez, P.: A new method to obtain decay rate estimates for dissipative systems. ESAIM Control Optim. Calc. Var. 4, 419–444 (1999) 14. Nicaise, S., Pignotti, C.: Stabilization of the wave equation with variable coefficients and boundary condition of memory type. Asymptot. Anal. 50, 31–67 (2006) 15. Park, J.Y., Ha, T.G.: Energy decay for nondissipative distributed systems with boundary damping and source term. Nonlinear Anal. 70, 2416–2434 (2009) 16. Park, J.Y., Ha, T.G.: Existence and asymptotic stability for the semilinear wave equation with boundary damping and source term. J. Math. Phys. 49, 053511 (2008) 17. Pr¨ uss, J.: Evolutionary Integral Equations and Applications. Birkh¨ auser, Basel (1993) 18. Rivera, J.E.M.: Global solution of a quasilinear wave equation with memory. Boll. Un. Math. Ital. B 8, 289–303 (1994) 19. Ruiz, A.: Unique continuation for weak solutions of the wave equation plus a potential. J. Math. Pures Appl. 71, 455–467 (1992) 20. Tadayuki, H., Rinko, M.: Equivalent condition for stability of Volterra integro-differential equation. J. Math. Anal. Appl. 174, 298–316 (1993) 21. Triggiani, R.: Wave equation on a bounded domain with boundary dissipation: an operator approach. J. Math. Anal. Appl. 137, 438–461 (1989) 22. Vitillaro, E.: A potential well method for the wave equation with nonlinear source and boundary damping terms. Glasgow Math. J. 44, 375–395 (2002) 23. Zuazua, E.: Exponential decay for the semilinear wave equation with locally distributed damping. Comm. P. D. E. 15, 205–235 (1990) 24. Zuazua, E.: Uniform stabilization of the wave equation by nonlinear boundary feedback. SIAM J. Control optim. 28(2), 466–477 (1990) Jong Yeoul Park and Tae Gab Ha Department of Mathematics Pusan National University Pusan 609-735 Korea e-mail: [email protected] Tae Gab Ha e-mail: [email protected]

Vol. 61 (2010)

Energy decay rates for solutions of the wave equation

Yong Han Kang Department of Mathematics University of Ulsan Ulsan 680-749 Korea e-mail: [email protected] (Received: November 8, 2007; revised: January 21, 2009)

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