Page 1

arXiv:0804.0298v3 [math.AP] 6 Jan 2010

The Pointwise Estimates of Solutions for

Semilinear Dissipative Wave Equation

Yongqin Liu∗

DepartmentofMathematics,FudanUniversity,Shanghai,China

DepartmentofMathematics,KyushuUniversity,Fukuoka,Japan

Abstract

In this paper we focus on the global-in-time existence and the

pointwise estimates of solutions to the initial value problem for the

semilinear dissipative wave equation in multi-dimensions. By using

the method of Green function combined with the energy estimates,

we obtain the pointwise decay estimates of solutions to the problem.

keywords: semilinear dissipative wave equation, pointwise esti-

mates, Green function.

MSC(2000): 35E15; 35L15.

1Introduction

In this paper we consider the initial value problem for the semilinear dissi-

pative wave equation in n(n ≥ 1) dimensions,

(? + ∂t)u(x,t) = f(u), x ∈ Rn, t > 0,

with initial condition

(1.1)

1a

(u,∂tu)(x,0) = (u0,u1)(x), x ∈ Rn,

t− △x+ ∂tis the dissipative wave operator with Lapla-

xj, f(u) = −|u|θu, θ > 0 is an integer. Equation (1a

often called the semilinear dissipative wave equation or semilinear telegraph

equation.

There have been many results on the equation (1a

responding to the different forms of f(u). By employing the weighted L2

(1.2)

IC

where ? + ∂t= ∂2

cian △x =

n ?

j=1∂2

1a

1.1) is

1a

1.1) and its variants cor-

∗email: yqliu2@yahoo.com.cn

1

Page 2

energy method and the explicit formula of solutions, Ikehata, Nishihara and

Zhao

pected to be same as that for the corresponding heat equation, Nishihara

Ni0

[15] studied the global asymptotic behaviors in three and four dimensions,

and Nishihara and Zhao[19] obtained the decay properties of solutions to

the problem (1a

1.2). Kawashima, Nakao and Ono

property of solutions to (1a

1.1) by using the energy method combined with

Lp− Lqestimates, and Ono

case of solutions to (1a

1.1) in unbounded domains in RNwithout the smallness

condition on initial data. Also, recently Nishihara, etc. in

ied the following semilinear damped wave equations with time or space-time

dependent damping term,

INZ

[7] obtained that the behavior of solutions to (1a1a

1.1) as t → ∞ is ex-

NZ

1a

1.1)(IC IC KNO

[9] studied the decay

1a

Ono1

[20] derived sharp decay rates in the subcritical

1a

Ni1, Ni2

[16, 17] stud-

utt− ∆u + b(t)ut+ |u|ρ−1u = 0,(1.3)

n1

and

utt− ∆u + b(t,x)ut+ |u|ρ−1u = 0,(1.4)

n2

where ρ > 1, b(t) = b0(1 + t)−βwith b0 > 0,−1 < β < 1, and b(t,x) =

b0(1 + |x|2)−α

obtained the global existence and the L2decay rate of the solution by using

the weighted energy method. (n1

yield (1a

1.1). For studies on the case f(u) = |u|θu, see

studies on the case f(u) = |u|θ+1, see

on the global attractors, see

[1, 10] and the references cited there.

The main purpose of this paper is to study the pointwise estimates of so-

lutions for (1a

1.2). In [11], Liu and Wang studied the corresponding linear

problem, i.e. (1a

mates of solutions. In this paper, we first obtain the global-in-time solutions

by energy method combined with the fixed point theorem of Banach, and

then obtain the optimal pointwise decay estimates of the solutions by using

the properties of the Green function proved in

analysis. One point worthy to be mentioned is that, different from that for

solutions to the corresponding linear problem, the order of derivatives with

respect to time variable t of solutions does not contribute to the decay rate

of solutions due to the presence of the semilinear term, which could be seen

from (pe1

2.4) in Theorem5.1) in Theorem

The rest of the paper is arranged as follows. In section 2, the main results

are stated. We give the proof of Proposition

in-time existence of solutions in section 3. In section 4 we give estimates on

the Green function by Fourier analysis which will be used in the last section

where the proof of Theorem

2.4 is given.

2(1 + t)−βwith b0 > 0,α ≥ 0,β ≥ 0,α + β ∈ [0,1), and

n1

1.3) and (n2

1.4) with the exponents α = β = 0

n2

1a HO, IMN, IO, Na, Ni, NO

[3, 6, 8, 13, 14, 18], for

Ik, LZ, Ono2, Ono3, TY, Z

[5, 12, 21, 22, 23, 25], and for studies

BP, KS

1a

1.1)(IC

1a

1.1) with f(u) = 0 and (IC

IC LW

IC

1.2), and obtained the pointwise esti-

LW

[11] combined with Fourier

pe1pepe

2.4 and (lr1lr1lrlr

5.1.

aeae

2.3 and then obtain the global-

pepe

2

Page 3

Before closing this section, we give some notations to be used below. Let

F[f] denote the Fourier transform of f defined by

F[f](ξ) =ˆf(ξ) :=

?

Rne−ix·ξf(x)dx,

and we denote its inverse transform by F−1.

For 1 ≤ p ≤ ∞, Lp= Lp(Rn) is the usual Lebesgue space with the norm

? · ?Lp. Let s be a nonnegative integer. Then Hs= Hs(Rn) denotes the

Sobolev space of L2functions, equipped with the norm

?f?Hs :=

?

s

?

k=0

?∂k

xf?2

L2

?1

2.

In particular, we use ? · ? = ? · ?L2, ? · ?s= ? · ?Hs. Here, for a multi-index

α, Dα

to x ∈ Rn. Also, Ck(I;Hs(Rn)) denotes the space of k-times continuously

differentiable functions on the interval I with values in the Sobolev space

Hs= Hs(Rn).

Finally, in this paper, we denote every positive constant by the same

symbol C or c without confusion. [·] is Gauss’ symbol.

xdenotes the totality of all the |α|-th order derivatives with respect

2Main theorems

The first main result is about the global existence of solutions to the initial

value problem (1a

1.2).

1a

1.1)(IC IC

ge

Theorem 2.1 (Global existence). Let θ > 0 be an integer. Assume that

(u0,u1) ∈ Hs+1(Rn) × Hs(Rn), s ≥ [n

E0:= ?u0?Hs+1 + ?u1?Hs.

Then if E0is suitably small, (1a

1.2) admits a unique solution

2] , put

1a

1.1) (ICIC

u ∈

s+1

?

i=0

Ci([0,∞);Hs+1−i(Rn)),

which satisfies

s+1

?

i=0

?∂i

tu(t)?2

s+1−i+

?t

0

(?∇u(τ)?2

s+

s+1

?

i=1

?∂i

τu(τ)?2

s+1−i)dτ ≤ CE2

0.(2.1)

ge1

3

Page 4

Theorem

in the following Theorem

sition

2.3.

gege

2.1 is proved by combining the local existence of solutions stated

le le

2.2 with a priori estimate in the following Propo-

ae ae

le

Theorem 2.2 (Local existence).

(u0,u1) ∈ Hs+1(Rn)×Hs(Rn), s ≥ [n

solution to (1a

1.2) satisfying

Let θ > 0 be an integer. Assume that

2], then there exists T > 0 and a unique

1a

1.1) (IC IC

u ∈

s+1

?

i=0

Ci([0,T);Hs+1−i(Rn)).

The proof of the local existence result is based on the fixed point theorem

of Banach and standard argument, so the detail is omitted.

Based on the a priori assumption

sup

0<t<T?u(t)?L∞ ≤¯δ,(2.2)

aa

where s > n is an integer and¯δ < 1 is a small constant, the following a priori

estimate is obtained.

ae

Proposition 2.3 (A priori estimate). Under the same assumptions as in The-

orem

and verifies (aa

2.2), then the following estimate holds,

gege

2.1, let u(x,t) be the solution to (1a

aa

1a

1.1)(ICIC

1.2) which is defined on [0,T]

sup

0<t<T{

s+1

?

i=0

?∂i

tu(t)?2

s+1−i} +

?T

0

(?∇u(τ)?2

s+

s+1

?

i=1

?∂i

τu(τ)?2

s+1−i)dτ ≤ CE2

0.

(2.3)

ae1

Remark 1. In (1a

makes it possible to close energy estimates. Otherwise, if f(u) = |u|θu, then

Theorem

2.1 does not hold, since the lower-order term present in the energy

estimates could not be controlled.

The second main result is about the pointwise estimate to the solution

obtained in Theorem

2.1.

1a

1.1), f(u) = −|u|θu is called absorption term which

gege

gege

pe

Theorem 2.4 (Pointwise estimate). Under the same assumptions as in The-

orem 2.1, if s > n, θ ≥ 2 + [1

there exists some constant r > max{n

|Dα

then for h ≥ 0 satisfying |α| + h < s − n, the solution to (1a

in Theorem

2.1 satisfies the following pointwise estimate,

ge ge

n], and for any multi-indexes α, |α| < s −n

2,1} such that

xu0(x)| + |Dα

2,

xu1(x)| ≤ C(1 + |x|2)−r,

1a

1.1)(ICIC

1.2) obtained

gege

|∂h

tDα

xu(x,t)| ≤ CE0(1 + t)−n+|α|

2 (1 +|x|2

1 + t)−r.(2.4)

pe1

4

Page 5

Remark 2.

of derivatives with respect to t of the solution obtained in Theorem

no effect on the decay rate of the solution, as is different from that for the

solution to the corresponding linear problem studied in

As a direct corollary of Theorem 2.4 we have

From the estimate in Theorem

pepe

2.4, we see that the order

ge ge

2.1 has

LW

[11].

pepe

44

Corollary 2.5. Assume that the same assumptions as in Theorem

then for p ∈ [1,∞], |α| + h < s − n, the solution to (1a

?∂h

pe pe

2.4 hold,

1a

1.1)(ICIC

1.2) satisfies,

tDα

xu(·,t)?Lp ≤ CE0(1 + t)−n

2(1−1

p)−|α|

2.

3The global existence of solutions

First we give a lemma which will be used in our next energy estimates.

L

Lemma 3.1. Let n ≥ 1, 1 ≤ p,q,r ≤ ∞ and1

estimate holds:

p=1

q+1

r. Then the following

?∂k

x(uv)?Lp ≤ C(?u?Lq?∂k

xv?Lr + ?v?Lq?∂k

xu?Lr)(3.1)

d1

for k ≥ 0.

Proof. The estimate (d1

proof. To prove (d1

k1+ k2= k, the following estimate holds:

d1

3.1) can be found in a literature but we give here a

3.1), it is enough to show that, for k1 ≥ 1, k2 ≥ 1 and

d1

?∂k1

xu∂k2

xv?Lp ≤ C(?u?Lq?∂k

xv?Lr + ?v?Lq?∂k

xu?Lr).

Let θj=

kj

k, j = 1,2, and define pj, j = 1,2, by

1

pj

−kj

n= (1 − θj)1

q+ θj(1

r−k

n).

Since θ1+ θ2= 1, we have

the Gagliardo-Nirenberg inequality, we have

1

p=

1

p1+

1

p2. By using the H¨ older inequality and

?∂k1

xu∂k2

xv?Lp ≤ ?∂k1

xu?Lp1?∂k2

≤ C(?u?1−θ1

≤ C(?u?Lq?∂k

≤ C(?u?Lq?∂k

xv?Lp2

xu?θ1

xv?Lr)θ2(?v?Lq?∂k

xv?Lr + ?v?Lq?∂k

Lq ?∂k

Lr)(?v?1−θ2

Lq ?∂k

xv?θ2

xu?Lr)θ1

xu?Lr).

Lr)

In the last inequality, we have used the Young inequality. Thus (d1

proved.

d1

3.1) is

5

Page 6

Now, let T > 0 and consider solutions to the problem (1a

are defined on the time interval [0,T] and verify the regularity mentioned in

Proposition

2.3. we derive energy estimates under the a priori assumption

(aa

2.2).

By multiplying (1a

1.1) with utand integrating on Rn× (0,t) with respect

to (x,t), we get

1a

1.1), (ICIC

1.2), which

aeae

aa

1a

?ut(t)?2+ ?∇u(t)?2+

?

Rn|u|θ+2(x,t)dx +

?t

0

?uτ(τ)?2dτ ≤ CE2

0. (3.2)

3a0

∀α, 1 ≤ |α| ≤ s, by multiplying Dα

Rn× (0,t) with respect to (x,t), in view of Lemma

?Dα

x(1a

1.1) with Dα

1a

xutand integrating on

L L3.1 and (aa

2.2) we get

aa

xut(t)?2+ ?Dα

x∇u(t)?2+?t

0?u(τ)?2θ

0?Dα

xuτ(τ)?2dτ

≤ CE2

0+ C?t

0+ C¯δ?t

L∞?Dα

xu(τ)?2dτ

≤ CE2

0?Dα

xu(τ)?2dτ.

By taking sum for α with 1 ≤ |α| ≤ s, it yields that

?∇ut(t)?2

s−1+ ?∇2u(t)?2

s−1+?t

0?∇uτ(τ)?2

s−1dτ

≤ CE2

0+ C¯δ?t

1a

1.1) with u and integrating on Rn× (0,t) with respect

to (x,t), by virtue of (3a0

3.2) we get

0?∇u(τ)?2

s−1dτ.

(3.3)

3a

By multiplying (1a

3a0

?u(t)?2+?t

≤ C(E2

0?∇u(τ)?2dτ +?t

0+ ?ut(t)?2+?t

0

?

Rn(|u|θ+2)(x,τ)dxdτ

0?uτ(τ)?2dτ) ≤ CE2

x(1a

1.1) with Dα

0.

(3.4)

3b

∀α, 1 ≤ |α| ≤ s, by multiplying Dα

Rn× (0,t) with respect to (x,t), by virtue of Lemma

?Dα

1a

xu and integrating on

L L3.1 we get

xu(t)?2+?t

0?Dα

x∇u(τ)?2dτ≤ C(E2

0+ ?Dα

xut(t)?2+?t

L∞?Dα

0?Dα

xuτ(τ)?2dτ)

+C?t

0?u(τ)?2θ

xu(τ)?2dτ.

3a

3.3) and (aa

By taking sum for α with 1 ≤ |α| ≤ s and in view of (3a

yields that

aa

2.2), it

?∇u(t)?2

s−1+

?t

0

?∇2u(τ)?2

s−1dτ ≤ CE2

0+ C¯δ

?t

0

?∇u(τ)?2

s−1dτ.(3.5)

3c

6

Page 7

(3a03a0

3.2), (3a3a

3.3), (3b3b

3.4) and (3c 3c

3.5) yield that

?u(t)?2

s+1+ ?ut(t)?2

s+

?t

0

(?∇u(τ)?2

s+ ?uτ(τ)?2

s)dτ ≤ CE2

0.(3.6)

3d

Proof of Proposition

prove that the following estimate holds for ∀h ∈ [1,s + 1], ∀t ∈ [0,T].

aeae

2.3. To prove (ae1 ae1

2.3) in Proposition

aeae

2.3, it is enough to

h

?

i=0

?∂i

tu(t)?2

s+1−i+

?t

0

(?∇u(τ)?2

s+

h

?

i=1

?∂i

τu(τ)?2

s+1−i)dτ ≤ CE2

0.(3.7)

3e

It is obvious that (3e

3.7) holds with h = j(1 ≤ j ≤ s), next we will prove that (3e

h = j + 1.

From (1a

1.1), by using induction argument we could prove that the following

two equalities hold for k ≥ 1,

∂2k

+P{∆iu(x,t),∆jut(x,t),0 ≤ i ≤ k − 1,0 ≤ j ≤ k − 2},

3e

3.7) holds with h = 1 by virtue of (3d3d

3.6). Assume that

3e

3.7) holds with

(3e3e

1a

tu(x,t) = a2k∆ku(x,t) + b2k∆k−1ut(x,t)

(3.8)

even

∂2k+1

t

u(x,t) = a2k+1∆ku(x,t) + b2k+1∆kut(x,t)

+P{∆iu(x,t),∆jut(x,t),0 ≤ i ≤ k − 1,0 ≤ j ≤ k − 1},

(3.9)

odd

where a2k, a2k+1, b2k, b2k+1 are constants, P{∆iu(x,t),∆ju(x,t),0 ≤ i ≤

k−1,0 ≤ j ≤ k−2} is a polynomial with arguments ∆iu(x,t),∆ju(x,t),0 ≤

i ≤ k − 1,0 ≤ j ≤ k − 2.

Let t = 0 in (even

3.9), we have for k ≥ 1,

∂2k

+P{∆iu0(x),∆ju1(x),0 ≤ i ≤ k − 1,0 ≤ j ≤ k − 2},

even

3.8)(oddodd

tu(x,0) = a2k∆ku0(x) + b2k∆k−1u1(x)

(3.10)

3h

∂2k+1

t

u(x,0) = a2k+1∆ku0(x) + b2k+1∆ku1(x)

+P{∆iu0(x),∆ju1(x),0 ≤ i ≤ k − 1,0 ≤ j ≤ k − 1}.

t(1a

on Rn× (0,t) with respect to (x,t), in view of (3e

(3i

3.11) we get

(3.11)

3i

∀α, |α| ≤ s − j, by multiplying Dα

x∂j

1a

1.1) with Dα

x∂j+1

t

u and integrating

3h

3.10) and

3e

3.7) with h = j, (3h

3i

?Dα

x∂j+1

t

u(t)?2+?t

0

?

Rn?Dα

x∂j+1

τ

u(τ)?2dτ ≤ CE2

0.

7

Page 8

By taking sum for α with 0 ≤ |α| ≤ s − j, it yields that

?∂j+1

t

u(t)?2

s−j+

?t

0

?∂j+1

τ

u(τ)?2

s−jdτ ≤ CE2

0.(3.12)

3f

(3e3e

3.7) with h = j and (3f3f

3.12) yield that

j+1

?

i=0

?∂i

tu(t)?2

s+1−i+

?t

0

(?∇u(τ)?2

s+

j+1

?

i=1

?∂i

τu(τ)?2

s+1−i)dτ ≤ CE2

0.(3.13)

3g

It means that (3e

complete the proof of Proposition

3e

3.7) holds with h = j + 1. Thus by induction method, we

aeae

2.3.

Now we give the proof of Theorem

2.1. By virtue of the a priori estimate (ae1

2.3, we can continue a unique solution obtained in Theorem

time, provided that E0is suitably small, say, E0< δ0, δ0depends only on

¯δ in (aa

finishes the proof of Theorem

2.1.

gege

2.1.

Proof of Theorem

aeae

gege

ae1

2.3) in Proposition

lele

2.2 globally in

aa

2.2). The global solution thus obtained satisfies (ge1

gege

ge1

2.1) and (aaaa

2.2). This

4 Estimates on Green function

In this section, we list some formulas and properties of the Green function

obtained in[11] to make preparation for the next section about the pointwise

estimates of solutions.

The Green function or the fundamental solution to the corresponding

linear dissipative wave equation (i.e. f(u) = 0 in (1a

LW

1a

1.1)) to (1a1a

1.1) satisfies

(? + ∂t)G(x,t) = 0,x ∈ Rn, t > 0,

G(x,0) = 0,x ∈ Rn,

∂tG(x,0) = δ(x),x ∈ Rn.

By Fourier transform we get that,

(∂2

t+ ∂t)ˆG(ξ,t) + |ξ|2ˆG(ξ,t) = 0,ξ ∈ Rn, t > 0,

ˆG(ξ,0) = 0,ξ ∈ Rn,

∂tˆG(ξ,0) = 1,ξ ∈ Rn.

8

Page 9

The symbol of the operator for equation (1a1a

1.1) is

σ(? + ∂t) = τ2+ τ + |ξ|2,

∂

∂t, and

2a

4.1) are τ = µ±(ξ) =1

(4.1)

2a

τ and ξ correspond to

that the eigenvalues of (2a

calculation we have that

1

√−1

∂

∂xj, j = 1,2,··· ,n. It is easy to see

2(−1 ±?1 − 4|ξ|2). By direct

ˆG(ξ,t) = (1 − 4|ξ|2)−1

2(eµ+(ξ)t− eµ−(ξ)t).

For convenience we decomposeˆG(ξ,t) =ˆG+(ξ,t) +ˆG−(ξ,t), where

ˆG±(ξ,t) = ±µ−1

0eµ±(ξ,t), µ0(ξ) = (1 − 4|ξ|2)

1

2.

Let

χ1(ξ) =

1,|ξ| < ε,

0,|ξ| > 2ε,

χ3(ξ) =

1,|ξ| > R,

0,|ξ| < R − 1,

be the smooth cut-off functions, where ε and R are any fixed positive num-

bers satisfying 2ε < R − 1.

Set

χ2(ξ) = 1 − χ1(ξ) − χ3(ξ),

and

ˆG±

i(ξ,t) = χi(ξ)ˆG±(ξ,t), i = 1,2,3.

We are going to study G±

responding toˆG±

Denote BN(|x|,t) = (1+|x|2

to G1(x,t) and G2(x,t), the proof can be seen in

i(x,t), which is the inverse Fourier transform cor-

i(ξ,t).

1+t)−N. First we give two propositions regarding

LW, WY

[11, 24].

21

Proposition 4.1. For sufficiently small ε, there exists constant C > 0, and

N > n such that

|∂h

tDα

xG1(x,t)| ≤ CNt−(n+|α|+2h)/2BN(|x|,t).

22

Proposition 4.2. For fixed ε and R, there exist positive numbers m , C and

N > n such that

|∂h

tDα

xG2(x,t)| ≤ Ce−

t

2mBN(|x|,t).

Next we will come to consider G3(x,t). Now we list some lemmas which

are useful in dealing with the higher frequency part.

9

Page 10

23

Lemma 4.3. If suppˆf(ξ) ⊂ OR:= {ξ; |ξ| > R}, andˆf(ξ) satisfies

|ˆf(ξ)| ≤ C, |Dβ

then there exist distributions f1(x), f2(x), and constant C0such that

ξˆf(ξ)| ≤ C|ξ|−1−|β|, |β| ≥ 1,

f(x) = f1(x) + f2(x) + C0δ(x),

where δ(x) is the Dirac function. Furthermore, for positive integer 2N >

n + |α|,

|Dα

?f2?L1 ≤ C,

with ε1being sufficiently small.

xf1(x)| ≤ C(1 + |x|2)−N,

supp f2(x) ⊂ {x; |x| < 2ε1},

24

Lemma 4.4. For any N > 0, τ ≥ 0, we have that

?

|z|=1

(1 +|x+tz|2

√1 + τ

2323

4.3 and Lemma

(1 +|x + tz|2

1 + τ

)−NdSz≤ C(1 + t)2N(1 +

|x|2

1 + τ)−N.

?

|z|≤1

1+τ)−N

dVz≤ C(1 + t)2N(1 +

|x|2

1 + τ)−N.

The proof of Lemma

2424

4.4 can be seen in

WY

[24].

The following Kirchhoff formulas can be seen in

E,HZ

[2, 4].

25

Lemma 4.5. Assume that w(x,t) is the fundamental solution of the follow-

ing wave equation with c = 1,

wtt− c2△w = 0,

w|t=0= 0,

∂tw|t=0= δ(x).

There are constants aα, bαdepending only on the spatial dimension n ≥ 1

such that, if h ∈ C∞(Rn), then

(w ∗ h)(x,t) =

?

0≤|α|≤n−3

2

aαt|α|+1

?

|z|=1

Dαh(x + tz)zαdSz,

(wt∗ h)(x,t) =

?

0≤|α|≤n−1

2

bαt|α|

?

|z|=1

Dαh(x + tz)zαdSz,

10

Page 11

for odd n, and

(w ∗ h)(x,t) =

?

0≤|α|≤n−2

2

aαt|α|+1

?

|z|≤1

Dαh(x + tz)zα

?1 − |z|2

Dαh(x + tz)zα

?1 − |z|2

dz,

(wt∗ h)(x,t) =

?

0≤|α|≤n

2

bαt|α|

?

|z|≤1

dz,

for even n. Here dSzdenotes surface measure on the unit sphere in Rn.

By denoting λη=√η − 4 and then taking the Taylor expansion for ληin

η, we have that

λη= 2√−1 +

j=1

m−1

?

ajηj+ O(ηm).

Since

µ±(ξ) =−1 ±?1 − 4|ξ|2

we have that, when ξ is sufficiently large,

2

=1

2(−1 ± |ξ|

?

|ξ|−2− 4),

µ±(ξ) =1

2(−1 ± 2√−1|ξ| ± (

m−1

?

j=1

aj|ξ|1−2j)) + O(|ξ|1−2m).

µ−1

0(ξ) =

1

?1 − 4|ξ|2= |ξ|−1(−

√−1

2

+ O(|ξ|−2)).

This implies that

eµ±(ξ)t

= e−t/2e±√−1|ξ|t(1 + (

m−1

?

j=1(±aj)|ξ|1−2j)t + ···

+1

m!(

m−1

?

j=1(±aj)|ξ|1−2j)mtm+ R±(ξ,t)),

where R±(ξ,t) ≤ (1 + t)m+1(1 + |ξ|)1−2m.

Denote

ˆ w(ξ,t) = (2π)−n/2sin(|ξ|t)/|ξ|,ˆ wt= (2π)−n/2cos(|ξ|t).

Since

∂h

tˆG+(ξ,t) =(µ+(ξ))h

µ0

eµ+(ξ)t, ∂h

tˆG−(ξ,t) = −(µ−(ξ))h

µ0

eµ−(ξ)t.

11

Page 12

By a direct and a little tedious calculation we get that,

∂h

tˆG3(ξ,t) = e−t/2ˆ wt(

h−1

?

j=0p1

h ?

1j(t)q1

1j(ξ) +

2m−2

?

j=1

p1

2j(t)q1

2j(ξ) +ˆR1(ξ,t))

+e−t/2ˆ w(

j=0p2

1j(t)q2

1j(ξ) +

2m−2

?

j=1

p2

2j(t)q2

2j(ξ) +ˆR2(ξ,t)),

here

p1

1j(t) ≤ C(1 + t)h−1−j, q1

1j(ξ) = χ3(ξ)|ξ|j, 0 ≤ j ≤ h − 1;

p1

2j(t) ≤ C(1 + t)h+j, q1

2j(ξ) = χ3(ξ)|ξ|−j, 1 ≤ j ≤ 2m − 2;

p2

1j(t) ≤ C(1 + t)h−j, q2

1j(ξ) = χ3(ξ)|ξ|j, 0 ≤ j ≤ h;

p2

2j(t) ≤ C(1 + t)h+j, q2

2j(ξ) = χ3(ξ)|ξ|−j, 1 ≤ j ≤ 2m − 2;

|ˆR1(ξ,t)|, |ˆR2(ξ,t)| ≤ C(1 + t)m+1(1 + |ξ|)h+1−2m.

In the following we denote q1

h), q1

R1(Dx,t), R2(Dx,t) the pseudo-differential operators with symbols q1

j ≤ h − 1), q2

2m − 2), ˆ w(ξ,t), ˆ wt(ξ,t), ˆR1(ξ,t), ˆR2(ξ,t) respectively. It is easy to get

that, for any multi-indexes β, |β| ≥ 1,

|Dβ

1j(Dx)(0 ≤ j ≤ h − 1), q2

2j(Dx)(0 ≤ j ≤ 2m−2), w(Dx,t), wt(Dx,t),

1j(Dx)(0 ≤ j ≤

2j(Dx)(0 ≤ j ≤ 2m−2), q2

1j(ξ)(0 ≤

1j(ξ)(0 ≤ j ≤ h), q1

2j(ξ)(0 ≤ j ≤ 2m − 2), q2

2j(ξ)(0 ≤ j ≤

ξq1

2j(ξ)| ≤ C|ξ|−1−|β|, |Dβ

ξq2

2j(ξ)| ≤ C|ξ|−1−|β|, 1 ≤ j ≤ 2m − 2;

|Dβ

ξχ3(ξ)| ≤ C|ξ|−1−|β|, |Dβ

ξ(|ξ|−1χ3(ξ))| ≤ C|ξ|−1−|β|;

supp q1

1j(0 ≤ j ≤ h − 1), supp q2

1j(0 ≤ j ≤ h) ⊂ OR−1= {ξ; |ξ| > R − 1};

supp q1

2j(1 ≤ j ≤ 2m − 2), supp q2

By Lemma

4.3 we have the following lemma.

2j(1 ≤ j ≤ 2m − 2) ⊂ OR−1.

2323

28

Lemma 4.6. For R being sufficiently large, there exist distributions ¯ q1

˜ q1

1j(x),

1j(x), 0 ≤ j ≤ h − 1; ¯ q2

1j(x), ˜ q2

1j(x), 0 ≤ j ≤ h; ¯ q1

2j(x), ˜ q1

2j(x), 1 ≤ j ≤

12

Page 13

2m − 2; ¯ q2

2j(x), ˜ q2

2j(x), 1 ≤ j ≤ 2m − 2 and constant C0such that

q1

1j(Dx)δ(x) = (−△)

j

2(¯ q1

1j+ ˜ q1

1j+ C0δ(x)), 0 ≤ j ≤ h − 1;

q2

1j(Dx)δ(x) = (−△)

j

2(¯ q2

1j+ ˜ q2

1j+ C0δ(x)), 0 ≤ j ≤ h;

q1

2j(Dx)δ(x) = ¯ q1

2j+ ˜ q1

2j+ C0δ(x), 1 ≤ j ≤ 2m − 2;

q2

2j(Dx)δ(x) = ¯ q2

2j+ ˜ q2

2j+ C0δ(x), 1 ≤ j ≤ 2m − 2;

and

|Dα

x¯ q1

1j|(0 ≤ j ≤ h − 1), |Dα

x¯ q2

1j|(0 ≤ j ≤ h) ≤ C(1 + |x|2)−N;

|Dα

x¯ q1

2j|(1 ≤ j ≤ 2m − 2), |Dα

x¯ q2

2j|(1 ≤ j ≤ 2m − 2) ≤ C(1 + |x|2)−N;

?˜ q1

1j?L1(0 ≤ j ≤ h − 1), ?˜ q2

1j?L1(0 ≤ j ≤ h) ≤ C;

?˜ q1

2j?L1(1 ≤ j ≤ 2m − 2), ?˜ q2

2j?L1(1 ≤ j ≤ 2m − 2) ≤ C;

supp ˜ q1

1j(0 ≤ j ≤ h − 1), supp ˜ q2

1j(0 ≤ j ≤ h) ⊂ {x; |x| < 2ε1};

supp ˜ q1

2j(1 ≤ j ≤ 2m − 2), supp ˜ q2

with ε1being sufficiently small.

2j(1 ≤ j ≤ 2m − 2) ⊂ {x; |x| < 2ε1},

Let

Q1

1j(x) = ˜ q1

1j(x) + C0δ(x), 0 ≤ j ≤ h − 1;

Q2

1j(x) = ˜ q2

1j(x) + C0δ(x), 0 ≤ j ≤ h;

2j(x) = ˜ q1

Q1

2j(x) + C0δ(x), 1 ≤ j ≤ 2m − 2;

Q2

2j(x) = ˜ q2

2j(x) + C0δ(x), 1 ≤ j ≤ 2m − 2,

and

L1

1j(x,t) = p1

1j(t)wt(Dx,t)(−△)

j

2Q1

1j(x), 0 ≤ j ≤ h − 1;

L2

1j(x,t) = p2

1j(t)w(Dx,t)(−△)

j

2Q2

1j(x), 0 ≤ j ≤ h;

L2j(x,t) = p1

2j(t)wt(Dx,t)Q1

2j(x) + p2

2j(t)w(Dx,t)Q2

2j(x), 1 ≤ j ≤ 2m − 2,

we have the following proposition.

13

Page 14

29

Proposition 4.7. For R sufficiently large, there exists distribution

Kh

m(x,t) = e−t/2(

h−1

?

j=0

L1

1j(x,t) +

h

?

j=0

L2

1j(x,t) +

2m−2

?

j=1

L2j(x,t))

such that for m ≥ [|α|+n+h+3

2

], we have that

|Dα

x(∂h

tG3− Kh

m)(x,t)| ≤ Ce−t/4BN(|x|,t).

4.7 can be seen in

4.1, Proposition4.2 and Proposition

The proof of Proposition

By Proposition

following proposition on the Green function.

2929 LW

[11].

21 2122 222929

4.7, we have the

210

Proposition 4.8. For any integer h ≥ 0, any multi-index α, and m ≥

[|α|+n+h+3

2

], we have that

|Dα

x(∂h

tG − Kh

m)(x,t)| ≤ C(1 + t)−(n+|α|+2h)/2BN(|x|,t),

where N > n can be big enough.

5Pointwise estimates

In this section, we aim at verifying that the solution obtained in Theorem

ge ge

2.1 satisfies the pointwise decay estimates expressed in Theorem

By Duhamel’s principle, the solution to (1a

following,

pepe

2.4.

1a

1.1)(ICIC

1.2) can be expressed as

u(x,t) = G(x − ·,t) ∗ (u0+ u1)(·) + ∂tG(x − ·,t) ∗ u0(·)

−?t

0G(x − ·,t − τ) ∗ (|u|θu)(·,τ)dτ.

We denote the solution to the corresponding linear dissipative wave equa-

tion as ¯ u, then

¯ u(x,t) := G(x − ·,t) ∗ (u0+ u1)(·) + ∂tG(x − ·,t) ∗ u0(·).

Denote

˜ u(x,t) :=

?t

0

G(x − ·,t − τ) ∗ (|u|θu)(·,τ)dτ,

then the solution u to (1a

In

[11], the following pointwise estimate of the solution ¯ u to the linear

problem is obtained.

1a

1.1) can be expressed as: u = ¯ u − ˜ u.

LW

14

Page 15

Theorem 5.1.

and for any multi-index α ∈ Zn, |α| < s −n

|Dα

|α| + h < s − n,

x¯ u(x,t)| ≤ C(1 + t)−(n+|α|+2h)/2(1 +|x|2

LW

[11]Assume that (u0,u1) ∈ Hs+1× Hs,s > n is an integer,

2, there exists r >

xu0(x)| + |Dα

lr

n

2such that

xu1(x)| ≤ C(1 + |x|2)−r, then the solution ¯ u satisfies, for

|∂h

tDα

1 + t)−r. (5.1)

lr1

Now we give some lemmas which will be used later.

41

Lemma 5.2. Assume n ≥ 1, then the following inequalities hold,

(1). If τ ∈ [0,t], and A2≥ t, then

A2

1 + τ)−n≤ 2n(1 + τ

(1 +

1 + t)n(1 +

A2

1 + t)−n.

(2). If A2≤ t, then 1 ≤ 2n(1 +

Lemma 5.3. Assume that 0 ≤ τ ≤ t and h(x,τ) satisfies

A2

1+t)−n.

42

Dα

xh(x,τ) ≤ C(1 + τ)−θn+|α|

2

(1 +

|x|2

1 + τ)−r,

then we have that,

(1).

?

|z|=1|Dα

xh(x + tz,τ)|dSz≤ C(1 + τ)−θn+|α|

2

(1 + t)2r(1 +

|x|2

1+τ)−r.

(2).

?

|z|≤1

|Dα

xh(x+tz,τ)|

√

1−|z|2

dVz≤ C(1 + τ)−θn+|α|−1

24

4.4)1,

2

(1 + t)2r(1 +

|x|2

1+τ)−r.

Proof. (1). By using Lemma (24

?

|z|=1|Dα

xh(x + tz,τ)|dSz

≤ C?t

≤ C(1 + τ)−θn+|α|

24

4.4)2,

0(1 + τ)−θn+|α|

2

(1 +|x+tz|2

(1 + t)2r(1 +

1+τ)−rdSz

|x|2

2

1+τ)−r.

(2). By H¨ older inequality and Lemma (24

?

≤ (?

≤ C(1 + τ)−θn+|α|−1

≤ C(1 + τ)−θn+|α|−1

Thus we complete the proof of Lemma

|z|≤1

|Dα

xh(x+tz,τ)|

√

1−|z|2

|z|≤1|Dα

dVz

xh(x + tz,τ)|3dVz)

1

3(?

1+τ

√1+τ

|z|≤1(

)−3r

1

√

1−|z|2)

dVz)

3

2dVz)

2

3

2

(?

(1 + t)2r(1 +

|z|≤1

(1+|x+tz|2

1

3(?1

0(1 − r2)−3

4rn−1dr)

2

3

2

|x|2

1+τ)−r.

4242

5.3.

15

Page 16

Proof of Theorem

pepe

2.4. For s > n, denote

ϕα(x,t) := (1 + t)

n+|α|

2 (Br(|x|,t))−1, r >n

2,

M(T) :=sup

(x,τ) ∈ Rn× [0,T)

|α| + h < s − n

|Dα

x∂h

τu(x,τ)|ϕα(x,τ).

Now we come to make estimates to ˜ u(x,t) under the assumption that s > n

and θ ≥ 2 + [1

By induction argument, we obtain the following expression,

n].

∂h

t˜ u(x,t) = ∂h

t

?t

0G(t − τ) ∗ (|u|θu)(τ)dτ

(h−1)+

?

=: J1+ J2,

=

j=0

∂j

tG(t) ∗ ∂(h−1)+−j

t

(|u|θu)(0) +?t

0G(t − τ) ∗ ∂h

τ(|u|θu)(τ)dτ

(5.2)

J

where (h − 1)+= max{h − 1,0}.

From (3h

mial with arguments ∆iu0(x) and ∆ku1(x), 0 ≤ i ≤ [(h−1)+−j

[(h−1)+−j−1

2

]. By using the similar estimates as that for ¯ u in

the following estimate for Dα

3h

3.10) and (3i3i

3.11), we know that ∂(h−1)+−j

t

(|u|θu)(x,0) is a polyno-

2

LW

[11], we obtain

], 0 ≤ k ≤

xJ1,

|Dα

xJ1| ≤ CE0(1 + t)−n+|α|

xJ2, we divide it as following ,

2 BN(|x|,t).

As for the estimate to Dα

Dα

xJ2 =? t

2

0

?

{y:|x|≤2|y|}Dα

x(G − K0

m)(x − y,t − τ)∂h

τ(|u|θu)(y,τ)dydτ

+? t

+?t

+?t

+?t

2

0

?

?

?

{y:|x|≥2|y|}Dα

x(G − K0

m)(x − y,t − τ)∂h

τ(|u|θu)(y,τ)dydτ

t

2

{y:|x|≤2|y|}(G − K0

m)(x − y,t − τ)Dα

y∂h

τ(|u|θu)(y,τ)dydτ

t

2

{y:|x|≥2|y|}(G − K0

m)(x − y,t − τ)Dα

y∂h

τ(|u|θu)(y,τ)dydτ

0

?

RnK0

m(x − y,t − τ)Dα

y∂h

τ(|u|θu)(y,τ)dydτ

=: J21+ J22+ J23+ J24+ J25.

Next we estimate J2i(i = 1,2,3,4,5) respectively by using Proposition

(with N ≥ r) and the fact that B(|x|,t) ≤ 1 and is an increasing function of

t and decreasing function of |x| .

210210

4.8

16

Page 17

By the definition of M(T), we have that

|J21| ≤ C? t

2

0

?

{y:|x|≤2|y|}(1 + t − τ)−n+|α|

M(T)θ+1(1 + τ)−(θ+1)n+2h

2 BN(|x − y|,t − τ)

Br(|y|,τ)dydτ.

2

Now we estimate J21in two cases.

Case 1. |x|2≥ t. We have

|J21| ≤ CM(T)θ+1? t

BN(|x − y|,t − τ)(1 + τ)−(θ+1)n

≤ CM(T)θ+1Br(|x|,t)? t

BN(|x − y|,t − τ)(1 + τ)−(θ+1)n

≤ CM(T)θ+1Br(|x|,t)? t

≤ CM(T)θ+1Br(|x|,t)(1 + t)−n+|α|

here in the second inequality we used Lemma

Case 2. |x|2≤ t. We have

|J21| ≤ CM(T)θ+1? t

(1 + τ)−(θ+1)n+2h

≤ CM(T)θ+1? t

≤ CM(T)θ+1(1 + t)−n+|α|

≤ CM(T)θ+1Br(|x|,t)(1 + t)−n+|α|

here in the last inequality we used Lemma

Combining the two cases, we have that

2

0

?

{y:|x|≤2|y|}(1 + t − τ)−n+|α|

2

2

Br(|x|,τ)dydτ

2

0

?

{y:|x|≤2|y|}(1 + t − τ)−n+|α|

2

(1+τ

0(1 + t − τ)−|α|

2 ,

4141

5.2 (1).

2

1+t)rdydτ

2 (1 + τ)−(θ+1)n

2

2

(1+τ

1+t)

n

2dτ

2

0

?

2

{y:|x|≤2|y|}(1 + t − τ)−n+|α|

Br(|y|,τ)dydτ

0(1 + t − τ)−n+|α|

2

2

2

2 (1 + τ)−θn+2h

2

dτ

2 ,

4141

5.2 (2).

|J21| ≤ CM(T)θ+1(ϕα(x,t))−1.

For J22, we have

|J22| ≤ C? t

2

0

?

{y:|x|≥2|y|}(1 + t − τ)−n+|α|

M(T)θ+1(1 + τ)−(θ+1)n+2h

2 BN(|x − y|,t − τ)

Br(|y|,τ)dydτ,

2

noticing that if |x| ≥ 2|y|, then |x − y| ≥

|J22| ≤ CM(T)θ+1? t

BN(|x|,t)(1 + τ)−(θ+1)n+2h

≤ CM(T)θ+1BN(|x|,t)? t

≤ CM(T)θ+1BN(|x|,t)(1 + t)−n+|α|

≤ CM(T)θ+1(ϕα(x,t))−1.

|x|

2, it yields that

2

0

?

{y:|x|≥2|y|}(1 + t − τ)−n+|α|

2

Br(|y|,τ)dydτ

2

0(1 + t − τ)−n+|α|

2

2 (1 + τ)−θn+2h

2

dτ

2

17

Page 18

For J23, by using the monotonic properties of B(|x|,t) with respect to |x|

and t, we have that

|J23| ≤ C?t

t

2

?

{y:|x|≤2|y|}(1 + t − τ)−n

M(T)θ+1(1 + τ)−(θ+1)n+|α|+2h

≤ CM(T)θ+1?t

BN(|x − y|,t− τ)(1 + τ)−(θ+1)n+|α|+2h

≤ CM(T)θ+1Br(|x|,t)?t

≤ CM(T)θ+1Br(|x|,t)(1 + t)−n+|α|

≤ CM(T)θ+1(ϕα,h(x,t))−1.

For J24, similar to J22we have that

2BN(|x − y|,t − τ)

Br(|y|,τ)dydτ

2

2

t

2

?

{y:|x|≤2|y|}(1 + t − τ)−n

2

Br(|x|,t)dydτ

dτ

t

2(1 + τ)−(θ+1)n+|α|+2h

2

2

|J24| ≤ C?t

t

2

?

{y:|x|≥2|y|}(1 + t − τ)−n

M(T)θ+1(1 + τ)−(θ+1)n+|α|+2h

≤ CM(T)θ+1?t

BN(|x|,t)(1 + τ)−(θ+1)n+|α|+2h

≤ CM(T)θ+1BN(|x|,t)?t

≤ CM(T)θ+1BN(|x|,t)(1 + t)−n+|α|+2h

≤ CM(T)θ+1(ϕα(x,t))−1.

Finally we come to estimate J25. From the definition of K0

2BN(|x − y|,t − τ)

Br(|y|,τ)dydτ

2

2

t

2

?

{y:|x|≥2|y|}(1 + t − τ)−n

2

Br(|y|,τ)dydτ

2(1 + τ)−θn+|α|+2h

t

2(1 + t − τ)−n

2

dτ

2

mwe have

|J25| = |?t

0

2m−2

?

+p2

≤ |?t

+

?

Rne−t−τ

2 p2

10(t − τ)wt(Dx,t − τ)(˜ q2

2j(t − τ)wt(Dx,t − τ)(˜ q1

2j(t − τ)w(Dx,t − τ)(˜ q2

0

?

?

+p2

h−1

?p2

+

?

C0δ(x − y)Dα

=: K1+ K2.

10+ C0δ)(x − y)

2j+ C0δ)(x − y)

2j+ C0δ)(x − y)]}Dα

10(t − τ)w(Dx,t − τ)˜ q2

2j(t − τ)wt(Dx,t − τ)˜ q1

2j(t − τ)w(Dx,t − τ)˜ q2

+|?t

2m−2

[p1

+

j=1

[p1

y∂h

τ(|u|θu)(y,τ)dydτ|

Rne−t−τ

2m−2

[p1

2 p2

10(x − y)

j=1

2j(x − y)

2j(x − y)]}Dα

10(t − τ)w(Dx,t − τ)

y(|u|θu)(y,τ)dydτ|

t

2

?

Rne−t−τ

2 {

j=1

2j(t − τ)wt(Dx,t − τ) + p2

y(|u|θu)(y,τ)dydτ|

2j(t − τ)w(Dx,t − τ)]}

(5.3)

j25

18

Page 19

By using Lemma

1+t)−1≤ C(1 +|x|2

2525

4.5, Lemma

1+t)−1, if |x − y| ≤ 2ε1, K1can be estimated as follows,

2828

4.6, Lemma

42 42

5.3 and the fact that (1 +

|y|2

K1 ≤ C?t

0

?

Rne−t−τ

4 (˜ q2

10| +

2m−2

?

2

j=1

(|˜ q1

2j| + |˜ q2

(1 +

2j|))(x − y)

|y|2

M(T)θ+1(1 + τ)−(θ+1)n+|α|

≤ CM(T)θ+1(1 + t)−n+|α|

2525

4.5 and Lemma

1+τ)−rdydτ

1+t)−r.

2 (1 +|x|2

(5.4)

k1

By using Lemma

4242

5.3, K2can be controlled by

K2 ≤ C?t

≤ CM(T)θ+1(1 + t)−n+|α|

5.3), (k1

5.5) yield that

0e−t−τ

4 M(T)θ+1(1 + τ)−(θ+1)n+|α|

2 (1 +|x|2

2

(1 +

|x|2

1+τ)−rdτ

1+t)−r.

(5.5)

k2

(j25 j25k1

5.4) and (k2k2

|J25| ≤ CM(T)θ+1(ϕα(x,t))−1.

Combining the estimates for J2i,i = 1,2,3,4,5, we have that

|Dα

x∂h

t˜ u(x,t)| ≤ CM(T)θ+1(ϕα(x,t))−1.

lr1

5.1) Theorem 5.1 yields that,

(5.6)

pe2

(pe2pe2

5.6) combined with (lr1 lrlr

M(T) ≤ C(E0+ M(T)θ+1).

Since θ ≥ 2, we have that M(T) ≤ CE0if E0is suitably small. It yields that

tu(x,t)| ≤ CE0(1 + t)−n+|α|

Thus Theorem

2.4 is proved.

|Dα

pepe

x∂h

2 Br(|x|,t).

Acknowledgement

This work was partially supported by Grant-in-Aid for JSPS Fellows. Also,

the author would give thanks to professor Weike Wang for the helpful dis-

cussion.

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