MATHEMATICAL CONTROL AND RELATED FIELDS Volume 1, Number 2, June 2011

doi:10.3934/mcrf.2011.1.251 pp. 251–265

DECAY OF SOLUTIONS OF THE WAVE EQUATION WITH LOCALIZED NONLINEAR DAMPING AND TRAPPED RAYS

Kim Dang Phung Yangtze Center of Mathematics, Sichuan University Chengdu 610064, China

(Communicated by Sylvain Ervedoza) Abstract. We prove some decay estimates of the energy of the wave equation governed by localized nonlinear dissipations in a bounded domain in which trapped rays may occur. The approach is based on a comparison with the linear damped wave equation and an interpolation argument. Our result extends to the nonlinear damped wave equation the well-known optimal logarithmic decay rate for the linear damped wave equation with regular initial data.

1. Introduction. We study a nonlinear wave equation in a bounded
connected open set Ω of Rn , n > 1 with a C 2 boundary ∂Ω. Let M = αij 1≤i,j≤n ∈
Therefore C ∞ Ω; Rn×n be a symmetric and uniformly positive P definite matrix. 
Pn n ij 1/2 1j nj β 1≤i,j≤n = M is well defined. Denote ∇ = j=1 β ∂xj , . . . , j=1 β ∂xj Pn and by ∆ = i,j=1 ∂xi (αij ∂xj ) the “Laplacian” associated to the matrix M. We deal with the following damped wave equation  2   ∂t u − ∆u + ag (∂t u) = 0 in Ω × (0, +∞) , u = 0 on ∂Ω × (0, +∞) , (1.1)   (u, ∂t u) (·, 0) = (u0 , u1 ) in Ω , where a = a (x) ∈ L∞ (Ω) is a non-negative bounded function, a (x) ≥ 0 for all x ∈ Ω, and g : R → R is a non-decreasing continuous function with g (0) = 0, sg (s) ≥ 0 and the additional conditions. r

1/r

(i) ∃r ∈ [1, ∞) , ∃c1 , c2 > 0, |s| ≤ 1 =⇒ c1 |s| ≤ |g (s)| ≤ c2 |s| . k p (ii) ∃k ∈ [0, 1] , ∃p ∈ [1, ∞) , ∃c3 , c4 > 0, |s| > 1 =⇒ c3 |s| ≤ |g (s)| ≤ c4 |s| . (iii) (n − 2) (1 − k) ≤ 4r and (n − 2) (p − 1) ≤ 1 . p−1

For example, when g (s) = |s| s, we can take k = 1 and 1 ≤ r = p ≤ n−1 n−2 . s When g (s) = √1+s2 , we can take k = 0 and r = p = 1. Standard arguments assure existence, uniqueness and regularity of the solution u of (1.1) (see [11], [10], [13]). For (u0 , u1 ) ∈ H01 (Ω) × L2 (Ω), there exists a 2000 Mathematics Subject Classification. Primary: 35L05, 35L71; Secondary: 35B40, 35B35. Key words and phrases. Wave equation, decay estimates, localized nonlinear damping, trapped rays. This work was supported by the “Sichuan Youth Science & Technology Foundation” 2010JQ0013 and by the NSF of China under grants 10771149 and 60974035.

251

252

KIM DANG PHUNG



unique weak solution u ∈ C [0, +∞) ; H01 (Ω) ∩ C 1 [0, +∞) ; L2 (Ω) . Introduce the energy at instant t ≥ 0 Z   1 2 2 |∂t u (x, t)| + |∇u (x, t)| dx . (1.2) E (u, t) = 2 Ω E (u, t) is a non-increasing function of time and for t2 > t1 ≥ 0 Z t2 Z E (u, t2 ) − E (u, t1 ) = − a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt ≤ 0 . t1

(1.3)

Ω

1 For more regular initial data (u0 , u1 ) ∈ H 2 (Ω)∩H01 (Ω)×H the solution u has
0 (Ω),
∞ 2 1 1,∞ the following regularity u ∈ L 0, +∞; H (Ω) ∩ H (Ω) ∩W 0, +∞; H01 (Ω) ∩ 0
W 2,∞ 0, +∞; L2 (Ω) . Moreover, there exists a constant c > 0 such that for a.e. t≥0 Z  2  2 (1.4) |∇∂t u (x, t)| + ∂t2 u (x, t) dx ≤ cE1 (u0 , u1 ) , Ω

where 2

E1 (u0 , u1 ) = k(∆u0 − ag (u1 ) , u1 )kL2 (Ω)×H 1 (Ω) . 0

(1.5)

Our main interest here is the decay rate of the energy when trapped ray occurs. The decay rate for solutions of the initial boundary value problem to the wave equation governed by a localized nonlinear dissipation is described in [17, Theorem 1.2, p.506] and [2, Theorem 2, p.307-308] (see also [5]). In [17] (see also [13]), a geometrical condition is added involving the sign of (x − xo ) · ν (x) for x ∈ ∂Ω (ν (x) being the outward normal at x ∈ ∂Ω). Here, no such condition is imposed and we consider the geometric case where the geometrical control condition (see [1]) may not be satisfied. When we deal with control of waves with localized damping, it is reasonable to choose the support of the damping large enough and to add a geometric assumption because of trapped rays. However, it is also natural to study what happen without such geometric assumption. We shall give two results in this context. Denote (2p−1) 1+ r−k X (u , u ) = E (u, 0)+E (u , u )+[E (u , u )] +[E (u , u )]( r+1 ) . (1.6) 0

1

1

0

1

1

0

1

1

0

1

Our main results are stated as follows. Theorem 1.1. Suppose that there exists ω ⊂ Ω a non-empty subdomain such that a (x) ≥ ao > 0 a.e. in ω. Then there is C > 0 such that for any t > 0 and (u0 , u1 ) ∈ H 2 (Ω) ∩ H01 (Ω) × H01 (Ω), the solution u of (1.1) satisfies E (u, t) ≤

C 2

[ln (2 + t)]

((r − 1) + X (u0 , u1 )) .

(1.7)

Theorem 1.2. Suppose that there exist c > 0 and γ > 0 such that for any t > 0 and (u0 , u1 ) ∈ H 2 (Ω) ∩ H01 (Ω) × H01 (Ω), the solution u` of  2   ∂t u` − ∆u` + a (x) ∂t u` = 0 in Ω × (0, +∞) , u` = 0 on ∂Ω × (0, +∞) ,   (u` , ∂t u` ) (·, 0) = (u0 , u1 ) in Ω , satisfies E (u` , t) ≤

c 2 k(∆u0 , u1 )kL2 (Ω)×H 1 (Ω) . 0 tγ

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

253

Then there is C > 0 such that for any t > 0 and (u0 , u1 ) ∈ H 2 (Ω)∩H01 (Ω)×H01 (Ω), the solution u of (1.1) satisfies E (u, t) ≤ with δ =

C ((r − 1) + X (u0 , u1 )) , tδ

(1.8)

γ 2r(γ+1)+γ+2 .

Remark 1. In Theorem 1.1, the damping in our nonlinear mechanism is effective in an arbitrary non-empty subdomain. Theorem 1.1 improves the logarithmic decay in [2] (see also [5]). In [2] (see also [5]), the logarithmic decay is obtained from an inequality of observation type for the linear wave equation deduced by a Carleman estimate (see [8] or [6]) and a Fourier-Bros-Iagolnitzer transform (see [9]). Here, our logarithmic decay, that is (1.7), comes directly from the optimal decay rate for the linear damped wave equation established by [7] and [3, Remarque 4.1, p.18]. In particular, we recover the optimal logarithmic decay when r = k = p = 1. Notice that following [6], the fact that a ∈ L∞ (Ω) and ∂Ω is only C 2 does not affect the getting of the resolvent estimate in [7]. Remark 2. Concerning Theorem 1.2, we set an assumption linked with the linear damped wave equation which of course may be fulfilled. Indeed, the polynomial decay estimate for E (u` , t) the energy of the damped linear wave, with respect to a stronger norm of the initial data, when trapped rays occur is described in different works (see [4], [12], [15], [14]). In particular, the conclusion of Theorem P2 2 2 1.2 holds when ∆ = i=1 ∂xi , Ω ⊂ R is a partially rectangular domain and
0 ≤ a ∈ C Ω is such that a > 0 in the closure of the non-rectangular part of Ω (see [4] and more recently [14]). Also, the conclusion of Theorem 1.2 holds when P3 ∆ = i=1 ∂x2i , Ω ⊂ R3 generates trapped rays bouncing up and down infinitely between two parallel parts Γ1 , Γ2 of the boundary ∂Ω and a ≥ ao > 0 a.e. on a neighborhood of ∂Ω \(Γ1 ∪ Γ2 ) in R3 (see [15]). The outline of the paper is the following. Next section gives technical estimates whose proof is based on H¨ older inequality and Sobolev’s imbedding theorem. In section 3, we recall some results for the linear damped wave equation and we establish interpolation estimates for the nonlinear damped wave equation. In section 4, we complete the proof of Theorem 1.1 (the logarithmic decay) and the proof of Theorem 1.2 (the polynomial decay). Throughout the remainder of this paper, C denotes different positive constants independent of the initial data. 2. Preliminary estimates. In this section, we prove some technical estimates that will be needed in the sequel. Theses estimates are very similar to the ones in [2]. We have the following results. Proposition 1. If (n − 2) (1 − k) ≤ 4r, then for a.e. t ≥ 0, Z 2 a (x) |∂t u (x, t)| dx Ω

Z ≤ Cε

(1+ r−k r+1 ) |∇∂t u (x, t)| dx 2

(2.1)

Ω 2  Z  r+1  1 + C+ a (x) g (∂t u (x, t)) ∂t u (x, t) dx ε Ω

∀ε > 0 .

254

KIM DANG PHUNG

Proposition 2. If (n − 2) (1 − k) ≤ 4r and (n − 2) (p − 1) ≤ 1, then for a.e. t ≥ 0, Z 2 a (x) |g (∂t u (x, t))| dx Ω "Z Z (2p−1) # (1+ r−k r+1 ) 2 2 ≤ Cε + |∇∂t u (x, t)| dx |∇∂t u (x, t)| dx (2.2) Ω

Ω



1 + C+ 3 ε

2  r+1 a (x) g (∂t u (x, t)) ∂t u (x, t) dx

 Z

∀ε > 0 .

Ω

Corollary 1. If (n − 2) (1 − k) ≤ 4r and (n − 2) (p − 1) ≤ 1, then for a.e. t ≥ 0, 2

2

k∆u (·, t)kL2 (Ω) + k∇∂t u (·, t)kL2 (Ω)   1 (2p−1) 1+ r−k ≤ C [E (u, 0)] r + C E1 (u0 , u1 ) + [E1 (u0 , u1 )] + [E1 (u0 , u1 )]( r+1 ) . (2.3) Proof of Corollary 1. First from the equation solved by u and (1.4), we have 2

2

k∆u (·, t)kL2 (Ω) + k∇∂t u (·, t)kL2 (Ω)

2

2 ≤ k∇∂t u (·, t)k 2 + 2 ∂t2 u (·, t) 2

+ 2 kag (∂t u) (·, t)kL2 (Ω)

≤ 2cE1 (u0 , u1 ) +

.

L (Ω)

L (Ω) 2 2 kag (∂t u) (·, t)kL2 (Ω)

2

Next, we check that Z 2 |a (x) g (∂t u (x, t))| dx Ω   1 (2p−1) 1+ r−k ≤ C [E (u, 0)] r + C [E1 (u0 , u1 )] + [E1 (u0 , u1 )]( r+1 ) .

(2.4)

(2.5)

Indeed, from Proposition 2 with ε = 1, we get Z 2 |a (x) g (∂t u (x, t))| dx Ω "Z Z (1+ r−k (2p−1) # r+1 ) 2 2 ≤ kakL∞ (Ω) C + |∇∂t u (x, t)| dx |∇∂t u (x, t)| dx Ω

Ω

2  r+1

Z + kakL∞ (Ω) (C + 1)

a (x) g (∂t u (x, t)) ∂t u (x, t) dx

.

Ω

(2.6) On the other hand, Z

2  r+1 a (x) g (∂t u (x, t)) ∂t u (x, t) dx

Ω 2

2

r+1 ≤ kag (∂t u)kLr+1 by Cauchy-Schwarz, 2 (Ω) k∂t ukL2 (Ω)    r1    r+1 r+1 r 2 2 1 r r+1 r+1 ≤ ε0 kag (∂t u)kL2 (Ω) + k∂t ukL2 (Ω) r + 1 ε0 (r + 1)

∀ε0 > 0 , (2.7)

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

by Young’s inequality (ab ≤ ε0 ap + 1q



1 ε0 p

 pq

255

∀a, b, ε0 > 0, p1 + 1q = 1, 1 < p, q
0 small enough, we obtain the desired estimate from (2.6) thanks to (1.3)-(1.4). This concludes the proof of Corollary 1. The proof of Proposition 1 and Proposition 2 can easily be deduced from the four lemmas below. Let us introduce the following two sets. Ω1 = {x ∈ Ω; |∂t u (x, t)| ≤ 1} ,

Ω2 = {x ∈ Ω; |∂t u (x, t)| > 1}

for a fixed t ≥ 0. We need to estimate the following four quantities. Z Z 2 2 a (x) |∂t u (x, t)| dx, a (x) |∂t u (x, t)| dx, Ω1 Ω Z Z 2 2 2 a (x) |g (∂t u (x, t))| dx, a (x) |g (∂t u (x, t))| dx. Ω1

Ω2

Lemma 2.1. There is C > 0 such that for a.e. t ≥ 0, Z

Z

2

a (x) |∂t u (x, t)| dx ≤ C Ω1

2  r+1 . a (x) g (∂t u (x, t)) ∂t u (x, t) dx

(2.8)

Ω r

1/r

Proof. Indeed,using c1 |s| ≤ |g (s)| ≤ c2 |s| |sg (s)| = sg (s) and Z

Z

2

2

Ω1

Ω1

 ≤

1 c1

2  r+1

≤ C kak because 0
0 such that for a.e. t ≥ 0, Z 2 a (x) |∂t u (x, t)| dx Ω2

Z ≤ Cε

2

( 2r−k+1 ) r+1

|∇∂t u (x, t)| dx

(2.10)

Ω

1 +C ε

2  r+1

Z a (x) g (∂t u (x, t)) ∂t u (x, t) dx Ω

∀ε > 0 .

256

KIM DANG PHUNG k

p

k+1

Proof. Indeed, using c3 |s| ≤ |g (s)| ≤ c4 |s| for |s| > 1, we deduce that c3 |s| |sg (s)| = sg (s) and Z

Z

2

2−α

a |∂t u| dx ≤ Ω2

a |∂t u|

α

|∂t u| dx

Ω2

Z

α 1− k+1

≤

a

a

α k+1

2−α



|∂t u|

Ω2

 ≤

1 c3

α  k+1

1− α kakL∞k+1 (Ω)

r

≤ C kakLr+1 ∞ (Ω) where α =

≤

k+1 r+1

Z

Z

Z

1 g (∂t u) ∂t u c3 2−α

|∂t u|

α  k+1

dx (2.11)

(ag (∂t u) ∂t u)

α k+1

dx

Ω

|∂t u|(

2r−k+1 r+1

1 ) (ag (∂ u) ∂ u) r+1 dx t t

Ω

∈ (0, k + 1), α < 2. Now, using H¨older inequality 2r−k+1 1 |∂t u|( r+1 ) (ag (∂t u) ∂t u) r+1 dx

Ω

1− δ1 Z  δ1 2r−k+1 δ 1 ( )( r+1 δ−1 ) r+1 δ |∂t u| dx dx (ag (∂t u) ∂t u)

Z ≤ Ω

(2.12)

Ω r 1  r+1 Z  r+1 2r−k+1 ( ) r , |∂t u| dx ag (∂t u) ∂t udx

Z ≤ Ω

Ω

with δ = r + 1. Recall that by Sobolev’s imbedding theorem,  2n 2≤q≤ ,if n > 2 n−2 =⇒ kf kLq (Ω) ≤ C k∇f kL2 (Ω) ∀f ∈ H01 (Ω) .  2 ≤ q < ∞,if n ≤ 2 Consequently, since 2r−k+1 ≥ 2, if (n − 2) (1 − k) ≤ 4r (that is r n > 2), we have by Sobolev’s imbedding theorem Z

2r−k+1 r

≤

 Z  21 ( 2r−k+1 ) r 2r−k+1 2 ) ( r |∂t u| dx ≤ C |∇∂t u| dx .

Ω

2n n−2

when

(2.13)

Ω

Finally,combining (2.13) and (2.12) with (2.11), we conclude that if (n − 2) (1 − k) ≤ 4r, Z

2

a |∂t u| dx ≤ C kak Ω2

r r+1 L∞ (Ω)

1  21 ( 2r−k+1  r+1 ) Z r+1 ag (∂t u) ∂t udx . |∇∂t u| dx

Z

2

Ω

Ω

(2.14)

Lemma 2.3. There is C > 0 such that for a.e. t ≥ 0, Z

2

a (x) |g (∂t u (x, t))| dx ≤ C Ω1

2  r+1

Z a (x) g (∂t u (x, t)) ∂t u (x, t) dx Ω

. (2.15)

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING r

1/r

Proof. Indeed,using c1 |s| ≤ |g (s)| ≤ c2 |s| r (c2 ) |s| |g (s)| and Z

Z

2

for |s| ≤ 1,we deduce that |g (s)|

2

2

Ω1

Ω1

Z

r−1

≤ C kakLr+1 ∞ (Ω)

r+1

≤

2

r

a1− r+1 a r+1 ((c2 ) g (∂t u) ∂t u) r+1 dx Ω1 Z r−1 2r 2 ≤ (c2 ) r+1 kakLr+1 (ag (∂t u) ∂t u) r+1 dx ∞ (Ω)

a |g (∂t u)| dx ≤

257

 ag (∂t u) ∂t udx

(2.16)

2 r+1

.

Ω

Lemma 2.4. If (n − 2) (1 − k) ≤ 4r and (n − 2) (p − 1) ≤ 1, then there is C > 0 such that for a.e. t ≥ 0, Z

2

a (x) |g (∂t u (x, t))| dx Ω2

(2p−1) # ( 2r−k+1 ) Z r+1 2 + |∇∂t u (x, t)| dx |∇∂t u (x, t)| dx (2.17)

"Z

2

≤ Cε

Ω

Ω

+C

1 ε3

2  r+1

Z

∀ε > 0 .

a (x) g (∂t u (x, t)) ∂t u (x, t) dx Ω k

p

Proof. Indeed, using c3 |s| ≤ |g (s)| ≤ c4 |s| for |s| > 1, we deduce that Z

Z

2

p 2

a |g (∂t u)| dx ≤

a (c4 |∂t u| ) dx Ω2 Z   1  1 2 2p−1 a 2 |∂t u| a 2 |∂t u| ≤ (c4 ) dx

Ω2

Ω2 2

Z

 12 

2

≤ (c4 )

a |∂t u| dx Ω2

Z kakL∞ (Ω)

2(2p−1)

|∂t u|

 12 dx

Ω2

(2.18) by Cauchy-Schwarz inequality. Remark that we already proved in Lemma 2.2 that if (n − 2) (1 − k) ≤ 4r, Z

2

a |∂t u| dx ≤ C kak Ω2

r r+1 L∞ (Ω)

Z

1  r+1  21 ( 2r−k+1 ) Z r+1 |∇∂t u| dx ag (∂t u) ∂t udx .

2

Ω

Ω

(2.19) On the other hand, since 2 (2p − 1) ≥ 2, if (n − 2) (p − 1) ≤ 1 (that is 2 (2p − 1) ≤ 2n n−2 when n > 2), we have by Sobolev’s imbedding theorem, Z

2(2p−1)

|∂t u| Ω

1  2(2p−1) Z  12 2 dx ≤C |∇∂t u| dx .

Ω

(2.20)

258

KIM DANG PHUNG

Consequently, inserting (2.19) and (2.20) into (2.18), we obtain Z 2 a |g (∂t u)| dx Ω2

 12 Z  12 2(2p−1) a |∂t u| dx |∂t u| dx

Z

2

≤C Ω2

"

Ω

r r+1 L∞ (Ω)

Z

≤ C kak

1 1 #2  12 ( 2r−k+1  r+1 ) Z r+1 2 |∇∂t u| dx ag (∂t u) ∂t udx

Ω

Ω

 2p−1 2 2 |∇∂t u| dx

Z × Ω

1 2 #2 Z  r+1 ( 2r−k+1 ) r+1 1 + 2 ≤ C kak ε |∇∂t u| dx ag (∂t u) ∂t udx ε Ω Ω (2.21) Z  2p−1 2 2 × |∇∂t u| dx r 2(r+1) L∞ (Ω)

"

2

Z

2

Ω 2 # Z ( 2r−k+1 Z  r+1 ) r+1 1 2 1 2 ε |∇∂t u| dx ag (∂t u) ∂t udx ≤C + 2 ε ε Ω Ω Z (2p−1) 2 +ε |∇∂t u| dx

"

Ω

( 2r−k+1 (2p−1) # ) Z r+1 2 |∇∂t u| dx + |∇∂t u| dx

"Z

2

≤ Cε Ω

+C

1 ε3

Ω 2  r+1

Z ag (∂t u) ∂t udx

.

Ω

3. Interpolation estimates. First, we consider the linear damped wave equation  ∂t2 u` − ∆u` + a (x) ∂t u` = 0 in Ω × (0, +∞) ,   u` = 0 on ∂Ω × (0, +∞) , (3.1)   2 1 1 (u` , ∂t u` ) (·, 0) ∈ H (Ω) ∩ H0 (Ω) × H0 (Ω) . Recall that E (u` , t) is a continuous decreasing function of time satisfying Z   1 2 2 E (u` , t) = |∂t u` (x, t)| + |∇u` (x, t)| dx 2 Ω Z tZ 2 = E (u` , 0) − a (x) |∂t u` (x, t)| dxdt . 0

(3.2)

Ω

We need the following assumption: (A): There exists a positive continuous decreasing function F : (0, +∞) → (0, +∞) with the properties lim F (h) = +∞ and lim F (h) = 0, such that h→0

h→+∞

for any t > 0, the solution u` of (3.1) satisfies 2

E (u` , t) ≤ F −1 (t) k(∆u` (·, 0) , ∂t u` (·, 0))kL2 (Ω)×H 1 (Ω) . 0

(3.3)

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

259

The existence of such function F holds in the following two cases. • Under the assumption of Theorem 1.2 (see also Remark 2), then we can choose
1/γ F −1 (t) = tcγ in (3.3) that is F (h) = hc . • Under the assumption of Theorem 1.1, it is known that a logarithmic decay holds for the linear damped wave equation (see [7] and [3, Remarque 4.1, C p.18]). Precisely, we can choose F −1 (t) = [ln(1+t)] 2 in (3.3) that is F (h) = √ e C/h − 1. Remark that under the assumption (A), by (3.2) and taking h = F −1 (t), the solution u` of (3.1) satisfies, for any h > 0, 2

E (u` , 0) ≤ h k(∆u` (·, 0) , ∂t u` (·, 0))kL2 (Ω)×H 1 (Ω) 0 Z F (h) Z 2 a (x) |∂t u` (x, t)| dxdt . +

(3.4)

Ω

0

Next, we derive a similar type of interpolation estimate for the solution u of (1.1) when a = a (x) ∈ L∞ (Ω) is a non-negative bounded function, a (x) ≥ 0 for all x ∈ Ω, and g : R → R is a non-decreasing continuous function with g (0) = 0, sg (s) ≥ 0 and satisfying (i) − (ii) − (iii). Proposition 3. Let (A) holds. There is C > 0 such that for any (u0 , u1 ) ∈ H 2 (Ω) ∩ H01 (Ω) × H01 (Ω), the solution u of (1.1) satisfies E (u, s) ≤ Ch ((r − 1) + X (u0 , u1 )) Z s+Y (h) Z +C a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt , s

(3.5)

Ω

for any h > 0 and any s ≥ 0 where  (2r+1) 1 2(r+1) (F (h)) . Y (h) = C h

(3.6)

Proof of Proposition 3. First, we consider the following linear
wave equation with a second member in which we suppose Θ ∈ L∞ 0, +∞; L2 (Ω) .  ∂t2 v − ∆v = a (x) Θ in Ω × (0, +∞) ,   v = 0 on ∂Ω × (0, +∞) , (3.7)   (v (·, 0) , ∂t v (·, 0)) = (u` , ∂t u` ) (·, 0) in Ω . Then, the solution w = u` − v solves  2   ∂t w − ∆w + a (x) ∂t w = −a (x) ∂t v − a (x) Θ in Ω × (0, +∞) , w = 0 on ∂Ω × (0, +∞) ,   (w, ∂t w) (·, 0) = (0, 0) in Ω ,

(3.8)

and by a classical energy multiplier method, we get Z F (h) Z Z F (h) Z 2 2 a (x) |∂t w (x, t)| dxdt ≤ a (x) |∂t v (x, t) + Θ (x, t)| dxdt . 0

Ω

0

Ω

(3.9)

260

KIM DANG PHUNG

Consequently, we obtain thanks to (3.4) 2

E (v, 0) = E (u` , 0) ≤ h k(∆v (·, 0) , ∂t v (·, 0))kL2 (Ω)×H 1 (Ω) 0 Z F (h) Z 2 a (x) |∂t v (x, t)| dxdt +6 0

Ω F (h)

Z

Z

(3.10)

2

a (x) |Θ (x, t)| dxdt .

+4 0

Ω


More generally, the following inequality holds for any Θ ∈ L∞ 0, +∞; L2 (Ω) , h > 0 and any s ≥ 0, 2

E (v, s) ≤ h k(∆v (·, s) , ∂t v (·, s))kL2 (Ω)×H 1 (Ω) 0 Z s+F (h) Z 2 a (x) |∂t v (x, t)| dxdt +6 s

Ω s+F (h)

Z

Z

(3.11)

2

a (x) |Θ (x, t)| dxdt .

+4 s

Ω

Next, applying (3.11) to the solution u with Θ = −g (∂t u), (2.1) in Proposition 1, (2.2) in Proposition 2 and (1.4), we obtain for any s ≥ 0 and any h, ε > 0, E (u, s) 2

≤ h k(∆u (·, s) , ∂t u (·, s))kL2 (Ω)×H 1 (Ω) 0 Z s+F (h) Z 2 +6 a (x) |∂t u (x, t)| dxdt s

Z

Ω s+F (h)

Z

2

a (x) |g (∂t u (x, t))| dxdt

+4 s

Ω

(3.12)

2

≤ h k(∆u (·, s) , ∂t u (·, s))kL2 (Ω)×H 1 (Ω) 0   (2p−1) 1+ r−k + CF (h) ε [E1 (u0 , u1 )] + [E1 (u0 , u1 )]( r+1 ) 2  Z  r+1 Z s+F (h)  1 1 +C 1+ + 3 a (x) g (∂t u (x, t)) ∂t u (x, t) dx dt . ε ε s Ω h By applying (2.3) in Corollary 1 and by taking ε = F (h) in (3.12), we get the existence of a constant C > 1 such that for any s ≥ 0 and any h > 0  3 !   1 F (h) F (h) + E (u, s) ≤ Ch X (u0 , u1 ) + [E (u, 0)] r + C 1 + h h (3.13) 2 Z  Z s+F (h) r+1 × a (x) g (∂t u (x, t)) ∂t u (x, t) dx dt . s

Ω

We claim that for any s ≥ 0 and any h > 0,  3   1 F (h) E (u, s) ≤ Ch X (u0 , u1 ) + [E (u, 0)] r + C h 2  r+1 Z s+F (h) Z × a (x) g (∂t u (x, t)) ∂t u (x, t) dx dt . s

(3.14)

Ω F (h) h

Indeed, if h ∈ (0, 1], we use that ∈ [F (1) , +∞) in (3.13) to get (3.14) for any 0 < h ≤ 1. When h > 1, (3.14) holds because E (u, s) ≤ E (u, 0) ≤ ChX (u0 , u1 )

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

261 1

if h ∈ [1, +∞). Now, we treat the power term in the expressions [E (u, 0)] r and
2 R s+F (h) R a (x) g (∂t u (x, t)) ∂t u (x, t) dx r+1 dt as follows. By Young’s inequality, s Ω we obtain 1

[E (u, 0)] r ≤ (r − 1) + q

(here, ab ≤ a q−1 + and

1 q



q−1 q

q−1

1 E (u, 0) rr

(3.15) 1

bq where a = 1, b = [E (u, 0)] r , q = r,  = r − 1)

2  r+1

Z a (x) g (∂t u (x, t)) ∂t u (x, t) dx Ω

   r−1 2 1 2 1 ≤ (r − 1) ε1 + r + 1 ε1 r + 1 Z × a (x) g (∂t u (x, t)) ∂t u (x, t) dx ∀ε1 > 0 .

(3.16)

Ω

Therefore, (3.16) easily implies that for any h > 0, 2 3 Z s+F (h) Z  r+1 F (h) a (x) g (∂t u (x, t)) ∂t u (x, t) dx dt h s Ω  3  3    r−1 2 F (h) F (h) 2 1 1 ≤ F (h) (r − 1) ε1 + h h r + 1 ε1 r + 1 Z s+F (h) Z × a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt ∀ε1 > 0 .



s

Ω



Choosing ε1 > 0 such that 

(3.17)

F (h) h

3 Z

s+F (h)

3

≤ h (r − 1) + s+F (h)

F (h) ε1 = h, we get that for any h > 0, 2  r+1 dt a (x) g (∂t u (x, t)) ∂t u (x, t) dx

Z

s

Ω



Z

F (h) h

F (h) h

3

2 r+1



1 r+1



F (h) h

4 ! r−1 2

(3.18)

Z

×

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s

Ω

Finally, inserting (3.15) and (3.18) into (3.14) we have that for any h > 0 and any s ≥ 0,  E (u, s) ≤ Ch ((r − 1) + X (u0 , u1 )) + C Z

s+F (h)

F (h) h

(2r+1)

Z

×

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s

Ω

(3.19)

262

KIM DANG PHUNG

In particular, there is C > 1 such that for any n ∈ {1, · · · , N − 1} with N > 1, E (u, s + nF (h))  ≤ Ch ((r − 1) + X (u0 , u1 )) + C Z

s+(n+1)F (h)

F (h) h

(2r+1) (3.20)

Z

×

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s+nF (h)

Ω

So for any N > 1, N E (u, s + N F (h)) X ≤ E (u, s + nF (h))

by (1.3),

n=0,..,N −1

 ≤ CN h ((r − 1) + X (u0 , u1 )) + C s+(n+1)F (h)

Z

X

×

(2r+1)

Z a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt

n=0,..,N −1

s+nF (h)

≤ CN h ((r − 1) + X (u0 , u1 )) + C s+N F (h)

(3.21)

Ω

 Z

F (h) h

F (h) h

(2r+1)

Z

×

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s

Ω

 (2r+1) 1 Choosing N > 1 such that C F (h) h N ' 1, we deduce from (1.3) and (3.21) that for any h > 0 and any s ≥ 0, Z s+N F (h) Z E (u, s) = E (u, s + N F (h)) + a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt s

Ω

≤ Ch ((r − 1) + X (u0 , u1 )) (2r+1) Z s+C ( F (h) F (h) Z h ) a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . +C s

(3.22)

Ω

This concludes the proof of Proposition 3. 4. End of the proofs. Based on Proposition 3 and assumption (A) in the previous section, we will distinguish two cases. C C • If F (h) = h1/γ in (3.3), then by (3.6) Y (h) ≤ h1/δ with 1δ = (2r + 1) + 2(r+1) . γ This case corresponds to the polynomial decay rate of Theorem 1.2. √ √ • If F (h) = eC/ h − 1 in (3.3), then by (3.6) Y (h) ≤ CeC/ h . This case corresponds to the logarithmic decay rate of Theorem 1.1 for the linear case, i.e. g (s) = s, (see [7] and [3, Remarque 4.1, p.18]).

4.1. Logarithmic decay. Under the assumption of Theorem 1.1, we can take √ √ C/ h C/ h F (h) = e − 1, then Y (h) ≤ Ce and we prove (1.7) in Theorem 1.1. Indeed, we get from (3.5) in Proposition 3, E (u, s) ≤ Ch ((r − 1) + X (u0 , u1 )) Z s+CeC/√h Z +C a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt s

Ω

(4.1) ∀h > 0 .

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

263

Choosing h=

1 E (u, s) , 2C ((r − 1) + X (u0 , u1 ))

(4.2)

it implies the existence of a constant c > 0 such that Z

 h  i ((r−1)+X(u0 ,u1 )) 1/2 s+exp c E(u,s)

Z

E (u, s) ≤ c

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s

Ω

(4.3) ≤ 1, we deduce from (4.3) and (1.3) that !! " !!# 1 1 G s + exp c ≤ G (s) ≤ c G (s) − G s + exp c 1/2 1/2 [G (s)] [G (s)] (4.4) which gives !! 1 c G s + exp c ≤ G (s) . (4.5) 1/2 1+c [G (s)] Denoting G (s) =

E(u,s) ((r−1)+X(u0 ,u1 ))

We distinguish two cases. Let n ∈ N, n ≥ 1.     1 1 • If s ≤ exp c [G(ns)] , then 2 + s ≤ exp (2 + c) and 1/2 [G(ns)]1/2 2 2+c . (4.6) ln (2 + s)     1 1 • If s > exp c [G(ns)] , then ns+exp c [G(ns)] < (n + 1) s and from (4.5), 1/2 1/2 

G ((n + 1) s) ≤ G (ns) ≤

!!

1

G ((n + 1) s) ≤ G ns + exp c

1/2

≤

[G (ns)]

c G (ns) . 1+c

Consequently, from (4.6) and (4.7), for any s > 0 and n ∈ N, n ≥ 1, ( ) 2 2+c c G ((n + 1) s) ≤ max , G (ns) . ln (2 + s) 1+c Then by induction, we deduce that ∀s > 0, ∀n ∈ N, n ≥ 1, ( ) 2  n 2+c c G ((n + 1) s) ≤ max , G (s) . ln (2 + s) 1+c

(4.7)

(4.8)

(4.9)

We choose n ≥ 1 such that n ≤ s + 1 < n + 1 and we obtain that for any s > 0   2       2 (2 + c)  2  , e−s ln(1+1/c) . (4.10) G (s + 2) ≤ G ((s + 2) s) ≤ max    ln (2 + s)2    We conclude that there is C > 0 such that E (u, t) C = G (t) ≤ 2 ((r − 1) + X (u0 , u1 )) [ln (2 + t)]

∀t > 0 .

(4.11)

264

KIM DANG PHUNG

4.2. Polynomial decay. Under the assumption of Theorem1.2,we can take F (h) = C C , and we prove (1.8) in Theorem , then Y (h) ≤ h1/δ with 1δ = (2r + 1) + 2(r+1) γ h1/γ 1.2 similarly than in [16]. Indeed, we get from (3.5) in Proposition 3, E (u, s) ≤ Ch ((r − 1) + X (u0 , u1 )) Z s+C/h1/δ Z +C a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt s

(4.12) ∀h > 0 .

Ω

Choosing h=

E (u, s) 1 , 2C ((r − 1) + X (u0 , u1 ))

(4.13)

it implies the existence of a constant c > 0 such that Z

i h ((r−1)+X(u0 ,u1 )) 1/δ s+c E(u,s)

Z

E (u, s) ≤ c

a (x) g (∂t u (x, t)) ∂t u (x, t) dxdt . s

(4.14)

Ω E(u,s) ((r−1)+X(u0 ,u1 ))

≤ 1, we deduce from (4.14) and (1.3) that "  1/δ !  1/δ !# 1 1 G s+c ≤ G (s) ≤ c G (s) − G s + c (4.15) G (s) G (s)

Denoting G (s) =

which gives 

1/δ !

c G (s) . (4.16) 1+c  δ
1/δ Let us introduce c1 = 1+c − 1 > 0 and denote d (s) = cc1 1s . We distinguish c two cases.   1/δ δ 1 • If c1 s ≤ c G(s) , then G (s) ≤ cc1 1s and G s+c

1 G (s)

≤

G ((1 + c1 ) s) ≤ G (s) ≤ d (s) .  1/δ  1/δ 1 1 • If c1 s > c G(s) , then s + c G(s) < (1 + c1 ) s and  G ((1 + c1 ) s) ≤ G s + c

1 G (s)

1/δ ! ≤

c G (s) . 1+c

(4.17)

(4.18)

Then by induction, we deduce from (4.17) and (4.18) that ∀s > 0, ∀n ∈ N, n ≥ 1,    c s G ((1 + c1 ) s) ≤ max d (s) , d , · · ·, 1+c (1 + c1 )  n    n+1  # c s c s , d , G . n n 1+c 1+c (1 + c1 ) (1 + c1 ) (4.19) Now, remark that with the above choice of c1 ,   c s d = d (s) 1+c (1 + c1 )

∀s > 0 .

(4.20)

DECAY OF THE WAVE EQUATION WITH NONLINEAR DAMPING

265

Consequently,  G ((1 + c1 ) s) ≤ max d (s) ,  ≤ max d (s) ,

c 1+c

n+1

c 1+c

n+1 !

 G

s n (1 + c1 )

! ∀n ≥ 1 (4.21)

∀n ≥ 1 .

and we conclude that E (u, s) = G (s) ≤ d ((r − 1) + X (u0 , u1 ))



s 1 + c1



 =

c (1 + c1 ) c1

δ

1 sδ

∀s > 0 . (4.22)

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Received November 2010; revised April 2011. E-mail address: kim dang [email protected]