Energy Decay for a Dissipative Wave Equation with Compactly Supported Data

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Energy Decay for a Dissipative Wave Equation

with Compactly Supported Data

By Kosuke Ono

Department of Mathematical Sciences The University of Tokushima Tokushima 770-8502, JAPAN e-mail : [email protected]

(Received September 30, 2011)

Abstract

Consider the Cauchy problem for the dissipative wave equation :

utt− ∆u + u = 0, u = u(x, t) in RN × (0, ∞) with u(x, 0) = u0(x)

and ut(x, 0) = u1(x). If {u0, u1} are compactly supported data

from the energy space, then there exists a domain Xmin RN such that{x ∈ RN |_x_{| ≥ t}1/2+δ_{} $ X}

mfor large t≥ 0 and ∫

Xm(|ut|

2₊

|∇u|2_{) dx} _{≤ C(1 + t)}−m _{with m > 0 for t} _{≥ 0, and moreover, if}

u0+ u1= 0, then

∫ Xm|u|

2_dx_{≤ C(1 + t)}−m _{for t}_{≥ 0.}

2000 Mathematics Subject Classiﬁcation. 35B40, 35L15

1 Introduction

We are concerned with the Cauchy problem for the dissipative wave equa-tion :

utt− ∆u + ut= 0 , u = u(x, t) in RN × (0, ∞) (1.1) with the initial data

u(x, 0) = u0(x) and ut(x, 0) = u1(x) , (1.2)

where ∆ =∇ · ∇ =∑N_j=1∂2_/∂x2

j is the Laplacian in RN.

We assume that {u0, u1} are compactly supported data from the energy

space :

u0∈ H1(RN) , u1∈ L2(RN) (1.3)

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and

supp u0∪ supp u1⊂ B(K) (1.4)

with K > 0, where B(K) is an open ball with center 0 and radius K :

B(K)≡ {x ∈ RN |x| < K} .

Then, it is well known that the problem (1.1)–(1.2) with (1.3)–(1.4) admits a unique global solution u(t) on [0,∞) such that

u(t)∈ C([0, ∞); H1(RN))∩ C1([0,∞); L2(RN)) (see [2], [5]) and

supp u(t)⊂ B(t + K) for t≥ 0 . (1.5) By the standard energy method, we obtain the following energy estimate :

E(t)≤ CE(0)(1 + t)−1 for t≥ 0

where

E(t)≡ ∥ut(t)∥2+∥∇u(t)∥2= ∫ RN ( |ut(x, t)|2+|∇u(x, t)|2 ) dx

and E(0) =∥u1∥2+∥∇u0∥2, and∥ · ∥ is the norm of L2(RN) (see [1], [3], [4]).

On the other hand, Todorova and Yordanov [6] have been obtained the following decay estimate :

∫ B(t1/2+δ₎c ( |ut|2+|∇u|2 ) dx≤ CE(0) exp(−t2δ/2) (1.6) with δ > 0, under the assumptions (1.3) and (1.4). Here, B(K)c is the com-plement of B(K), that is,

B(K)c≡ RN \ B(K) = {x ∈ RN |x| ≥ K} .

We are interested in the decay estimate for larger domains than B(t1/2+δ₎c_. When m > 0 and δ > 0, it is easy to see that for large t > 0,

(t + K)1/2log(1 + t)m< t1/2+δ

and hence

B(t1/2+δ)c$ B((t + K)1/2log(1 + t)m)c.

The purpose of this paper is to derive the decay estimate for large domain of integral B((t + K)1/2log(1 + t)m)c than B(t1/2+δ)c in (1.6).

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Theorem 1.1 Let m > 0. Suppose that the initial data {u0, u1} satisfy the

conditions (1.3) and (1.4). Then the solution u of (1.1)–(1.2) satisfies

∫ B((t+K)1/2_log(1+t)m₎c ( |ut|2+|∇u|2 ) dx≤ eKE(0)(1 + t)−m (1.7)

for t≥ 0. Moreover, if u0+ u1= 0, then

∫

B((t+K)1/2_log(1+t)m₎c

|u|2_dx_{≤ e}K_∥u

0∥2(1 + t)−m (1.8)

for t≥ 0.

Theorem 1.1 follows from Theorem 2.2 and Theorem 2.3 in next section.

2 Decay Estimates

The function ψ(x, t)≡1₂ ( t + K−√(t + K)2− |x|2 ) given by [6] plays an important role through this paper. It is easy to see that

ψ = 1 2 |x|2 t + K +√(t + K)2_{− |x|}2, ψt= 1 2 ( 1−√ t + K (t + K)2_{− |x|}2 ) =−√ ψ (t + K)2_{− |x|}2, (2.1) ψ_t2=1 4 ( 1 + (t + K) 2 (t + K)2− |x|2 − 2 t + K √ (t + K)2− |x|2 ) , |∇ψ|2₌1 4 |x|2 (t + K)2− |x|2,

and then we obtain the following. Lemma 2.1 The function ψ(x, t)≡ 1₂

( t + K−√(t + K)2− |x|2) _for |x| < t + K satisfies ψ(x, t)≥ 0 , ψt(x, t) = ψt(x, t)2− |∇ψ(x, t)|2 (2.2) and 1 4 |x|2 t + K ≤ ψ(x, t) ≤ 1 2(t + K) . (2.3)

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∫ |x|≥(t+K)1/2_log(1+t)m ( |ut|2+|∇u|2 ) dx≤ I₁2(1 + t)−m (2.4) for t≥ 0, where I₁2≡ ∫ RN e2ψ(x,0)(|u1(x)|2+|∇u0(x)|2 ) dx≤ eKE(0) .

Proof. Multiplying (1.1) by 2e2ψ_u_{t, we have}

0 = e2ψ ( d dt(u 2 t+|∇u| 2 )− 2 div(ut∇u) + 2u2t ) = d dt ( e2ψ(u2t+|∇u| 2 ))− 2 div(e2ψut∇u) + 2e2ψ (−ψt)P (x, t) (2.5) where

P (x, t)≡ (ψ2_t− ψt)u2_t− 2ψtut∇ψ · ∇u + ψt2|∇u|

2_, _{by (2.1)}

= u2_t|∇ψ|2− 2ψtut∇ψ · ∇u + ψt2|∇u|

2

=|ut∇ψ − ψt∇u|2 (≥ 0) .

When x̸= 0, we see ψt< 0 (by (2.1)) and hence P/(−ψt)≥ 0. When x = 0, we see ψt= 0 and|∇ψ| = 0 and hence P/(−ψt) = u2

t ≥ 0. Moreover, we see from (1.5) that supp P (·, t) ⊂ B(t + K) for t ≥ 0.

Integrating (2.5) overRN_{, we have}

d dt ( ∥eψ_u t∥2+∥eψ∇u∥2 ) ≤ 0 and hence ∥eψ_u

t∥2+∥eψ∇u∥2≤ ∥eψ(·,0)u1∥2+∥eψ(·,0)∇u0∥2 (≡ I12) (2.6)

for t≥ 0. By (2.3), it is easy to see that I2

1 ≤ eKE(0).

On the other hand, we observe from (1.5) that for t > 0,

∥eψ_u t∥2+∥eψ∇u∥2= ∫ |x|<t+K e2ψ(|ut|2+|∇u|2) dx , by (2.3) ≥ ∫ |x|<t+K e12t+K|x|2₍|u_t|2₊|∇u|2_{) dx} ≥ ∫ |x|≥(t+K)1/2_log(1+t)m e12 |x|2 t+K₍|u_t|2₊|∇u|2_{) dx} ≥ (1 + t)m ∫ |x|≥(t+K)1/2_log(1+t)m (|ut|2+|∇u|2) dx . (2.7)

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Therefore, we obtain from (2.6) and (2.7) that ∫ |x|≥(t+K)1/2_log(1+t)m ( |ut|2+|∇u|2 ) dx≤ I₁2(1 + t)−m for t≥ 0, which implies the desired estimate (2.4).

The following decay estimate means (1.8).

∫ |x|≥(t+K)1/2_log(1+t)m |u|2_dx_{≤ I}2 0(1 + t)−m (2.8) for t≥ 0, where I₀2≡ ∫ RN e2ψ(x,0)u0(x)2dx≤ eK∥u0∥2. Proof. Putting w(x, t) = ∫ t 0 u(x, s) ds

for the solution u = u(x, t) of (1.1)–(1.2), we observe that wt= u, w(x, 0) = 0 and ut+ u− ∆w = u0+ u1 in RN × (0, ∞) . (2.9) Multiplying (1.1) by 2e2ψ_{u, we have} 2e2ψ(u0+ u1)u = e2ψ ( d dtu 2_{+ 2u}2_{− 2 div(u∇w) +} d dt|∇w| 2 ) (2.10) = d dt ( e2ψ(u2+|∇w|2))− 2 div(e2ψu∇w) + 2e 2ψ (−ψt) Q(x, t) (2.11) where Q(x, t)≡ (ψ2_t− ψt)u2− 2ψtu∇ψ · ∇w + ψ2t|∇w| 2_, _{by (2.1)} = u2|∇ψ|2− 2ψtu∇ψ · ∇w + ψ2t|∇w|2 =|u∇ψ − ψt∇w|2 (≥ 0) .

When x̸= 0, we see ψt< 0 (by (2.1)) and hence Q/(−ψt)≥ 0. When x = 0, we see ψt = 0 and|∇ψ| = 0 and hence Q/(−ψt) = u2 ≥ 0. Moreover, we see from (1.5) that supp Q(·, t) ⊂ B(t + K) for t ≥ 0.

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Integrating (2.11) overRN, we have d dt ( ∥eψ_u_∥2₊_∥eψ_∇w∥2)_{≤ 2} ∫ RN e2ψ(u0+ u1)u dx If u0+ u1= 0, then we observe ∥eψ_u_∥2_{≤ ∥e}ψ(·,0)_u 0∥2 (≡ I02) (2.12)

for t≥ 0. By (2.3), it is easy to see that I2

0 ≤ eK∥u0∥2.

On the other hand, we observe from (1.5) that for t > 0,

∥eψ_u_∥2₌ ∫ |x|<t+K e2ψ|u|2dx , by (2.3) ≥ ∫ |x|<t+K e12t+K|x|2|u|2_dx ≥ ∫ |x|≥(t+K)1/2_log(1+t)m e12 |x|2 t+K|u|2_dx ≥ (1 + t)m ∫ |x|≥(t+K)1/2_log(1+t)m |u|2_{dx .} _(2.13)

Therefore, we obtain from (2.12) and (2.13) that ∫

|x|≥(t+K)1/2_log(1+t)m|u|

2_dx_{≤ I}2

0(1 + t)−m

for t≥ 0, which implies the desired estimate (2.8).

Acknowledgment. This work was in part supported by Grant-in-Aid for Science

Research (C) of JSPS (Japan Society for the Promotion of Science).

References

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