ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

ALMOST GLOBAL EXISTENCE FOR THE NEUMANN PROBLEM OF QUASILINEAR WAVE EQUATIONS OUTSIDE

STAR-SHAPED DOMAINS IN 3D

LULU REN, JIE XIN Communicated by Goong Chen

Abstract. In this article, we prove the almost global existence of solutions for quasilinear wave equations in the complement of star-shaped domains in three dimensions, with a Neumann boundary condition.

1. Introduction

Assume the obstacleK ⊂R^{3}be a smooth, closed and strictly star-shaped domain
with respect to the origin. Then consider the Neumann problem for the quasilinear
wave equation

^{c}u=F(du, d^{2}u), (t, x)∈R+×R^{3}\K,

∂νu
_{∂K} = 0,

u(0, x) =f(x), ∂tu(0, x) =g(x).

(1.1)
Herec = (c1,c2,· · ·,cN) is a vector-value multiple-speed D’Alembertian with
^{c}I = ∂^{2}_{t} −c^{2}_{I}∆, and we suppose that all cI’s are positive but not necessarily
distinct.

∂νu=~n· ∇xu=

3

X

j=1

∂u

∂xj

nj

denotes differentiation with respect to the outward normal toK. If we set∂t=∂0, then

F^{I}(du, d^{2}u) =

3

X

0≤j,k,l≤3,0≤J,K≤N

C_{K,l}^{IJ,jk}∂lu^{K}∂j∂ku^{J}, 1≤I≤N,
whereC_{K,l}^{IJ,jk} are real constants satisfying the symmetry conditions

C_{K,l}^{IJ,jk}=C_{K,l}^{J I,jk}=C_{K,l}^{IJ,kj}.

Let ∂ = (∂t, ∂1, ∂2, ∂3) = (∂0,∇) Denote the time-space gradient, and ∂u = u^{0}.
We write Ω = {Ωij}, where Ωij =xi∂j−xj∂i, 1≤i ≤j ≤3, are the Euclidean

2010Mathematics Subject Classification. 35L70, 74H20, 74B20.

Key words and phrases. Quasilinear wave equations; exterior problem; almost global existence;

Neumann boundary condition.

c

2018 Texas State University.

Submitted August 8, 2017. Published December 31, 2017.

1

R^{3} rotation operators. Set Z ={∂t, ∂j,Ωij}, S =t∂t+x· ∇x=t∂t+r∂r, hxi=
(1 +|x|^{2})^{1/2}.

To simplify the notation, we let

=∂^{2}_{t}−∆

be the scale unit-speed D’Alembertian. Since the estimates foryield ones for^{c},
we will state most of our estimates in terms ofinstead of^{c}.

We suppose that the Cauchy data satisfies the relevant compatibility conditions.

LetJ_{k}u={∂_{x}^{α}u: 0≤ |α| ≤k}. Ifmis fixed anduis a formalH^{m}solution of (1.1),
then we write∂_{t}^{k}u(0,·) =ψ_{k}(J_{k}f, J_{k−1}g)(0≤k≤m). The compatibility condition
for (1.1) with (f, g)∈H^{m}×H^{m−1} is just the requirement that ψ_{k} vanish on ∂K
for 0≤k≤m. Furthermore, (f, g)∈C^{∞} satisfies the compatibility conditions to
infinite order if these conditions hold for allm.

There have been many results on the almost global existence of wave equations, mostly with Dirichlet boundary condition. The almost global existence for non- linear wave equations was proved in [1] on Minkowski space by using the Lorentz invariance of the wave operator. In [3], the authors gave the same result without relying on Lorentz invariance. The exterior problem of nonlinear wave equation was considered in [2]. Mitsuru Ikawa [4] studied some mixed problems for hyperbolic system of second order. The almost global existence for the Dirichlet problem of quasilinear, semilinear wave equations in three space dimensions were proved in [5, 8] and [7], respectively. Also [12, 13, 14] give the global existence for Dirich- let problem of nonlinear wave equations in exterior domains. The nonexistence of global solutions for exterior problem to critical semilinear wave equations in high dimensions was obtained in [9].

There are also some results on the almost existence to Neumann problem for wave equations. The Neumann problem for the wave equation in wedge was considered in [15]. [16] considered the Neumann exterior problem for wave equation in 2D and studied the asymptotic behavior of the solutions for large times. Katayama et al [10] proved the almost global existence of solutions to exterior problem for semilin- ear wave equations with Neumann condition. Metcalfe et al [11] gave the almost global existence for quasilinear Neumann wave equations on infinite homogeneous waveguides.

To our acknowledge there are very few results on the almost global existence
or lifespan estimate of exterior Neumann problem for quasilinear wave equations
in 3D. In this paper, we study the almost global existence of solutions to the
exterior problem for quasilinear wave equations with Neumann condition by using
the estimates similar to Dirichlet problem in [5]. Compared with the Dirichlet
problem,u= 0 changes into∂_{ν}u= 0 on∂Ω. So the estimates on the boundary, we
decompose the estimated terms into the terms which contain ∂_{ν}u. The key steps
in this paper are the piontwise estimates and weighted L^{2} estimates. At last, we
proof the almost global existence to this problem and give a lower bound for the
lifespan of the solutions. To study this problem conveniently, we need some known
lemmas (see [5]).

Lemma 1.1. Suppose that u∈C^{5} solves the Cauchy problem
u=F(s, x), (s, x)∈[0, t]×R^{3}

u(0, x) =∂_{t}u(0, x) = 0. (1.2)

Then

(1 +t)|u(t, x)| ≤C Z t

0

Z

R^{3}

X

|α|+j≤3,j≤1

|S^{j}Z^{α}F(s, y)| 1

|y|dy ds. (1.3)
Lemma 1.2. Let u∈C^{5} solve (1.2), and fix x∈R^{3} with |x|=r. Then

|x| |u(t, x)| ≤ 1 2

Z t

0

Z r+t−s

|r−(t−s)|

sup

|θ|=1

|F(s, ρθ)|ρdρds. (1.4) Lemma 1.3. Suppose that usolves the Cauchy problem

u=F,

u(0, x) =f, ∂_{t}u(0, x) =g. (1.5)
Then

(ln(2 +t))^{−1/2}khxi^{−1/2}u^{0}k_{L}2(R^{3})

≤Cku^{0}(0, x)kL^{2}(R^{3})ds+C
Z t

0

kF(s,·)kL^{2}(R^{3})ds,

(1.6)

ku^{0}k_{L}2([0,t]×{|x|<1}≤Cku^{0}(0, x)k_{L}2(R^{3})ds+C
Z t

0

kF(s,·)k_{L}2(R^{3})ds. (1.7)
Lemma 1.4. Suppose that h∈C^{∞}(R^{3}). Then forR >1,

khk_{L}^{∞}(R/2<|x|<R)≤CR^{−1} X

|α|+|γ|≤2

kΩ^{α}∂_{x}^{γ}hk_{L}2(R/4<|x|<2R).

2. Pointwise estimates outside of obstacles

In this section, we shall consider the exterior problem of Neumann wave equations
u=F(t, x), (t, x)∈R+×R^{3}\K,

∂νu(t, x) = 0, x∈∂K, u(t, x) = 0, t≤0.

(2.1)

Any of the following estimates for extend to estimates for ^{c} after applying
straightforward scaling argument. We will prove the following pointwise estimate.

Theorem 2.1. Suppose thatu=u(t, x)∈C^{∞} is the solution of (2.1). Then for
each |α|=N >1,

t|Z^{α}u(t, x)| ≤C
Z t

0

X

|γ|+j≤N+3, j≤1

kS^{j}∂^{γ}F(s,·)k_{L}2(R^{3}\K)ds

+C Z t

0

Z

R^{3}\K

X

|β|+j≤N+6j≤1

|S^{j}Z^{β}F(s, y)| 1

|y|dy ds.

(2.2)

We assume, without loss of generality, that K ⊂ {x∈ R^{3} :|x|<1}. As a first
step, we prove the following lemma.

Lemma 2.2. Suppose that u= u(t, x)∈ C^{∞} is the solution of (2.1). Then for
each |α|=N >1,

t|Z^{α}u(t, x)| ≤C
Z t

0

Z

R^{3}\K

X

|γ|+j≤3,j≤1

|S^{j}Z^{α+γ}F(s, y)| 1

|y|dy ds

+C sup

|y|≤2,0≤s≤t

(1 +s)(|Z^{α}u^{0}(s, y)|+|Z^{α}u(s, y)|).

(2.3)

Proof. Inequality (2.3) obviously holds for|x|<2. Letρ∈C^{∞}(R) be a cut function
satisfying

ρ(r) =

(1, r≥2, 0, r≤1.

Thenω(t, x) =ρ(|x|)∂^{α}u(t, x), solves the following problem inR^{3},
ω=ρ∂^{α}F+G,

ω(t, x) = 0, t≤0, where

G=−2∇ρ(|x|)· ∇∂^{α}u−(∆ρ(|x|))u.

Splitω=ω1+ω2, whereω1 andω2 solve the following problems:

ω1=ρ∂^{α}F,
ω1(t, x) = 0, t≤0,
and

ω_{2}=G,
ω2(t, x) = 0, t≤0,
respectively. Applying Lemma 1.1, we conclude that

t|ω1(t, x)| ≤C Z t

0

Z

R^{3}\K

X

|γ|+j≤3,j≤1

|S^{j}Z^{γ}∂^{α}F(s, y)| 1

|y|dy ds.

By Lemma 1.2,

|ω2(t, x)| ≤C 1

|x|

Z t

0

Z |x|+t−s

||x|−(t−s)|

sup

|θ|=1

|G(s, rθ)|r dr ds. (2.4) For|x| ≤1 and|x| ≥2,G(t, x) = 0. Hence the right-hand side of (2.4) is nonzero only when

−2≤ |x| −(t−s)≤2, namely,

(t− |x|)−2≤s≤(t− |x|) + 2.

We conclude that

|ω2(t, x)|

≤C 1

|x|

1

1 +|t− |x|| sup

(t−|x|)−2≤s≤(t−|x|)+2,

|y|≤2

(1 +s)(|Z^{α}u^{0}(s, y)|+|Z^{α}u(s, y)|). (2.5)

This implies that (2.3) still holds for|x| ≥2.

Lemma 2.3. Suppose that u∈C^{∞} solves (2.1)andF(t, x) = 0 for|x|>4. Then
there exists a constant c >0 such that

ku^{0}(t,·)k_{L}2(R^{3}\K:|x|<4)≤C
Z t

0

e^{−c(t−s)}kF(s,·)k_{L}2(R^{3}\K)ds. (2.6)
Consequently, for any fixed nonnegative integer M, we have

X

|α|+j≤M, j≤1

k(t∂t)^{j}∂^{α}u^{0}(t,·)kL^{2}(R^{3}\K:|x|<4)

≤C X

|α|+j≤M−1, j≤1

k(t∂t)^{j}∂^{α}F(t,·)k_{L}2(R^{3}\K)

+C Z t

0

e^{−}^{c}^{2}^{(t−s)} X

|α|+j≤M, j≤1

k(s∂s)^{j}∂^{α}F(s,·)kL^{2}(R^{3}\K)ds,

(2.7)

X

|α|+j≤M, j≤1

k(t∂t)^{j}∂^{α}u^{0}(t,·)k_{L}2(R^{3}\K:|x|<4)

≤C X

|α|+j≤M−1, j≤1

kS^{j}∂^{α}f(t,·)k_{L}2(R^{3}\K)

+C Z t

0

e^{−}^{c}^{2}^{(t−s)} X

|α|+j≤M, j≤1

kS^{j}∂^{α}F(s,·)k_{L}2(R^{3}\K)ds.

(2.8)

Proof. First, we provide the exponential energy decay [17, Theorem III, p. 480]

and [18, (iii), p. 230]: Suppose thatω is the solution to the problem ω= 0,

∂νω= 0, x∈∂K. (2.9)

Let

E(ω, D, t) = 1 2 Z

D

|∂tω|^{2}+|∇ω|^{2}
dx.

Then there exist positive constantsC, c, such that
E(ω, D, t)≤Ce^{−ct}E(ω, D,0).

Next, homogenizing (2.1), we have

ω= 0,

∂νω
_{∂K}= 0,

ω|t=s= 0, ∂tω|t=s=F(s, x).

(2.10)

Suppose thatω solves problem (2.10), then u=Rt

0ω(x, t, s)dssolves (2.1). Thus we derive

ku^{0}k^{2}_{L}2(R^{3}\K:|x|<4)≤
Z t

0

kω^{0}(x, t, s)k^{2}_{L}2(R^{3}\K:|x|<4)ds

≤C Z t

0

E(ω,(R^{3}\K:|x|<4), t−s)ds

≤C Z t

0

e^{−c(t−s)}E(ω,(R^{3}\K:|x|<4), s)ds

≤C Z t

0

e^{−c(t−s)}kF(s,·)k^{2}_{L}2(R^{3}\K:|x|<4)ds,

which implies

ku^{0}k_{L}2(R^{3}\K:|x|<4)≤C
Z t

0

e^{−c(t−s)}kF(s,·)k_{L}2(R^{3}\K)ds.

Therefore, estimate (2.6) holds.

Estimate (2.8) follows from (2.7). Using induction and elliptic regularity we can

prove the estimate (2.7).

Proof of Theorem 2.1. By Lemma 2.2, we need only to proof that the last term on the right-hand side of (2.3) can be dominated by the right-hand side of (2.2), namely prove

t sup

|x|<2

|∂^{α}u(t, x)| ≤ right-hand side of (2.3),
holds for each|α|=N. We have

|t∂^{α}u(t, x)| ≤
Z t

0

X

j≤1

|(s∂_{s})^{j}∂^{α}u(s, x)|ds. (2.11)
First we discuss the case: F(s, y)≡0 when |y|>4.

By Sobolev Lemma, from (2.8), we obtain that for|α|=N, t sup

|x|<2

|∂^{α}u(t, x)|

≤C Z t

0

X

|γ|+j≤N+2, j≤1

kS^{j}∂^{γ}F(s,·)kL^{2}(R^{3}\K:|x|≤4)ds

+C Z t

0

Z s

0

e^{−}^{c}^{2}^{(s−τ)} X

|γ|+j≤N+2, j≤1

kS^{j}∂^{γ}F(τ,·)kL^{2}(R^{3}\K:|x|≤4)dτ ds.

(2.12)

Therefore, t sup

|x|<2

|∂^{α}u(t, x)| ≤first term on the right-hand side of (2.3).

Now we deal with the second case: F(s, y)≡0 when|y| <3. Suppose that u0

solves the Cauchy problem

u0=F(t, x), (t, x)∈R^{+}×R^{3},

u_{0}(t, x) = 0, t≤0. (2.13)

Letη∈C_{0}^{∞}(R^{3}) be a cut function satisfying
η(x) =

(1, |x|<2, 0, |x| ≥3.

If we set ˜u= (η−1)u_{0}+u, then ˜usolves the problem
u˜=G(t, x), (t, x)∈R+×R^{3}\K,

∂νu˜
_{∂K} = 0,

˜

u(t, x) = 0, t≤0,

(2.14)

where

G=−2∇η· ∇u0−(∆η)u_{0}

vanishes unless 2≤ |x| ≤4. Hence by the first case, t sup

|x|<2

|∂^{α}u(t, x)|=t sup

|x|<2

|∂^{α}u(t, x)|˜

≤C Z t

0

X

|γ|≤N+2, j≤1

kS^{j}∂^{γ}G(s,·)k_{L}2(R^{3}\K)ds

≤C Z t

0

X

|γ|≤N+3, j≤1

kS^{j}∂^{γ}u_{0}(s,·)kL^{∞}(R^{3}\K:2≤|x|≤4)ds.

(2.15)

Setω=S^{j}∂^{γ}u_{0} withj= 0,1. By (1.4), we obtain
kS^{j}∂^{γ}u0(s,·)k_{L}^{∞}_{(2≤|x|≤4)}

≤C Z s

0

Z

|s−τ−ρ|≤4

sup

|θ|=1

|S^{j}∂^{γ}F(τ, ρθ)|ρdρdτ

≤C X

|µ|≤2

Z s

0

Z

|s−τ−ρ|≤4

|S^{j}∂^{γ}Ω^{µ}F(τ, ρθ)|ρdρdθdτ

=C X

|µ|≤2

Z s

0

Z

|s−τ−|y||≤4

|S^{j}∂^{γ}Ω^{µ}F(τ, y)|dydτ

|y| .

(2.16)

Set Λs={(τ, y) : 0≤τ≤s,|s−τ− |y|| ≤4}satisfying Λs∩Λs^{0} =∅if|s−s^{0}|>20.

Therefore, by (2.15) and (2.16), we conclude that t sup

|x|<2

|∂^{α}u(t, x)| ≤C X

γ≤N+3,|µ|≤2,j≤1

Z t

0

Z

R^{3}\K

|S^{j}Ω^{µ}∂^{γ}F(τ, y)|dydτ

|y| .

The proof is complete.

3. WeightedL^{2}_{t,x} estimates for D’Alembertian outside of star-shaped
obstacles

In this section, we prove the following theorem.

Theorem 3.1. Suppose that u=u(t, x)solves problem (2.1). Then ifN is fixed, we have

(ln(2 +t))^{−1/2} X

|α|≤N

khxi^{−1/2}∂^{α}u^{0}kL^{2}([0,t]×R^{3}\K)

≤C Z t

0

X

|α|≤N

k∂^{α}u(s,·)k_{L}2(R^{3}\K)ds

+C X

|α|≤N−1

k∂^{α}ukL^{2}([0,t]×R^{3}\K), ∀t≥0.

(3.1)

Additionally,

(ln(2 +t))^{−1/2} X

|α|+m≤N, m≤1

khxi^{−1/2}S^{m}∂^{α}u^{0}kL^{2}([0,t]×R^{3}\K)

≤C Z t

0

X

|α|+m≤N, m≤1

kS^{m}∂^{α}u(s,·)kL^{2}(R^{3}\K)ds

+C X

|α|+m≤N−1, m≤1

kS^{m}∂^{α}uk_{L}2([0,t]×R^{3}\K), ∀t≥0,

(3.2)

and

(ln(2 +t))^{−1/2} X

|α|+m≤N, m≤1

khxi^{−1/2}S^{m}Z^{α}u^{0}kL^{2}([0,t]×R^{3}\K)

≤C Z t

0

X

|α|+m≤N, m≤1

kS^{m}Z^{α}u(s,·)k_{L}2(R^{3}\K)ds

+C X

|α|+m≤N−1, m≤1

kS^{m}Z^{α}ukL^{2}([0,t]×R^{3}\K), ∀t≥0.

(3.3)

Proposition 3.2. Suppose that usolves problem (2.1). Then we have
ku^{0}kL^{2}([0,t]×R^{3}\K:|x|<2)≤C

Z t

0

ku(s,·)kL^{2}(R^{3}\K)ds, ∀t≥0 (3.4)
and for any given positive integer N,

X

|α≤N

k∂^{α}u^{0}k_{L}2([0,t]×R^{3}\K:|x|<2)

≤C Z t

0

X

|m≤N

k∂_{s}^{m}u(s,·)k_{L}2(R^{3}\K)ds+C X

|α≤N−1

k∂^{α}uk_{L}2([0,t]×R^{3}\K),

∀t≥0.

(3.5)

Proof. Using the elliptic regularity argument, we know that (3.5) is a consequence of (3.4). To prove (3.4), we discuss the first case: F(s, y)≡0 for|y|>6.

By (2.6) and the Schwarz inequality, we have
ku^{0}(τ,·)k^{2}_{L}2(R^{3}\K:|x|<2)

≤C Z τ

0

e^{−c(τ−s)}kF(s,·)k_{L}2(R^{3}\K)ds
Z τ

0

kF(s,·)k_{L}2(R^{3}\K)ds,
for allτ≥0. Integratingτ from 0 toton the above inequality,

Z t

0

ku^{0}(τ,·)k^{2}_{L}2(R^{3}\K:|x|<2)dτ

≤C Z t

0

Z τ

0

e^{−c(τ−s)}kF(s,·)k_{L}2(R^{3}\K)ds
Z τ

0

kF(s,·)k_{L}2(R^{3}\K)dsdτ

≤C Z t

0

Z τ

0

e^{−c(τ−s)}kF(s,·)k_{L}2(R^{3}\K)dsdτ
Z t

0

kF(s,·)k_{L}2(R^{3}\K)ds

=C Z t

0

Z t

s

e^{−c(τ−s)}kF(s,·)kL^{2}(R^{3}\K)dτ ds
Z t

0

kF(s,·)kL^{2}(R^{3}\K)ds

≤CZ t 0

kF(s,·)kL^{2}(R^{3}\K)ds^{2}

, ∀t≥0 therefore, (3.4) holds.

Now we consider the second case: F(s, y)≡0 for |y|<4. By (3.4), we have
ku^{0}kL^{2}([0,t]×R^{3}\K:|x|<2)≤CkFkL^{2}([0,t]×R^{3}\K:|x|<2), if F(s, y)≡0, |y|>4. (3.6)
Letη∈C^{∞}(R^{3}) be a cut function satisfying

η(x) =

(1, |x| ≤2, 0, |x| ≥4.

Suppose thatu0 solves the Cauchy problem (2.13). Set ˜u= (η−1)u0+u, then ˜u solves the following problem

u˜= ˜F ,

∂νu˜ ∂K= 0,

˜

u(0, x) = 0, t≤0, where

F˜=−2∇η· ∇u0−(∆η)u0.

Note that ˜u=ufor|x|<2, and ˜F(s, y) = 0 for|y|>4. Then by (3.6) and (1.7), we obtain

ku^{0}kL^{2}([0,t]×R^{3}\K:|x|<2)≤Cku^{0}_{0}kL^{2}([0,t]×R^{3}\K:|x|<4)+Cku0kL^{2}([0,t]×R^{3}\K:|x|<4)

≤C Z t

0

kuk_{L}2(R^{3}\K)ds, ∀t≥0.

Repeating the proof of Proposition 3.2 and using (2.8), we have the following proposition.

Proposition 3.3. Suppose that usolves problem (2.1). Then X

|α|+m≤N, m≤1

kS^{m}∂^{α}u^{0}k_{L}2([0,t]×R^{3}\K:|x|<2)

≤C Z t

0

X

|α|+m≤N, m≤1

kS^{m}∂^{m}u(s,·)kL^{2}(R^{3}\K)ds

+C X

|α|+m≤N−1, m≤1

kS^{m}∂^{α}uk_{L}2([0,t]×R^{3}\K), ∀t≥0.

(3.7)

Additionally, X

|α|+|γ|+m≤N, m≤1

kS^{m}Ω^{γ}∂^{α}u^{0}kL^{2}([0,t]×R^{3}\K:|x|<2)

≤C Z t

0

X

|α|+|γ|+m≤N, m≤1

kS^{m}Ω^{γ}∂^{m}u(s,·)k_{L}2(R^{3}\K)ds

+C X

|α|+|γ|+m≤N−1, m≤1

kS^{m}Ω^{γ}∂^{α}ukL^{2}([0,t]×R^{3}\K), ∀t≥0.

(3.8)

Proof of Theorem 3.1. Let us first proof estimate (3.1). By Proposition 3.2 it suf- fices to prove that

(ln(2 +t))^{−1/2} X

|α|≤N

khxi^{−1/2}∂^{α}u^{0}kL^{2}([0,t]×R^{3}\K:|x|>2)

≤C Z t

0

X

|α|≤N

k∂^{α}u(s,·)k_{L}2(R^{3}\K)ds+C X

|α|≤N−1

k∂^{α}uk_{L}2([0,t]×R^{3}\K).
(3.9)

Letβ∈C^{∞}(R^{3}) be a cut function satisfying
β(x) =

(1, |x| ≥2, 0, |x| ≤1.

Thenω=βusolves the Cauchy problem

ω=βu−2∇β· ∇u−(∆β)u, ω(t, x) = 0, t≤0,

We splitω=ω1+ω2, whereω1=βuandω2=−2∇β· ∇u−(∆β)u. By (1.6), we have

(ln(2 +t))^{−1/2} X

|α|≤N

khxi^{−1/2}∂^{α}ω^{0}_{1}kL^{2}([0,t]×R^{3}\K:|x|>2)

≤C X

|α|≤N

Z t

0

k∂^{α}(βu)k_{L}2(R^{3}\K)ds≤C X

|α|≤N

Z t

0

k∂^{α}uk_{L}2(R^{3}\K)ds

To bound the left of (3.9) it suffices to proof
(ln(2 +t))^{−1/2} X

|α|≤N

khxi^{−1/2}∂^{α}ω_{2}^{0}k_{L}2([0,t]×R^{3}\K:|x|>2)

≤C Z t

0

X

|α|≤N

k∂^{α}u(s,·)kL^{2}(R^{3}\K)ds+C X

|α|≤N−1

k∂^{α}ukL^{2}([0,t]×R^{3}\K).

(3.10)

Note that G =−2∇β· ∇u−(∆β)u= ω2 vanishes unless 1 <|x| <2. To use
this, letχ ∈C_{0}^{∞}(R) satisfying χ(s) = 0, |s|>2, and P

jχ(s−j) = 1. Then we splitG=P

jGj, whereGj(s, x) =χ(s−j)G(s, x), and letω2,j solvesω2,j =Gjon Minkowski space with zero initial data. By the sharp Huygens principle, we have

|∂^{α}ω2(t, x)|^{2}≤CP

j|∂^{α}ω2,j(t, x)|^{2}. Therefore, by (1.6) it follows that

(ln(2 +t))^{−1/2} X

|α|≤N

khxi^{−1/2}∂^{α}ω^{0}_{2}k_{L}2([0,t]×R^{3}\K:|x|>2)

^{2}

≤ X

|α|≤N

X

j

Z t

0

k∂^{α}Gj(s,·)kL^{2}(R^{3})ds^{2}

≤C X

|α|≤N

k∂^{α}Gk^{2}_{L}2([0,t]×R^{3})

≤C X

|α|≤N

k∂^{α}u^{0}k^{2}_{L}2([0,t]×{1<|x|<2})+C X

|α|≤N

k∂^{α}uk^{2}_{L}2([0,t]×{1<|x|<2})

≤C X

|α|≤N

k∂^{α}u^{0}k^{2}_{L}2([0,t]×{|x|<2})

≤CZ t 0

X

|α|≤N

k∂^{α}u(s,·)k_{L}2(R^{3}\K)ds+C X

|α|≤N−1

k∂^{α}uk_{L}2([0,t]×R^{3}\K)

^{2}
,

which completes the proof of (3.1). Estimates (3.2) and (3.3) follow by a similar

argument.

4. L^{2}_{x} estimates outside of obstacles
Suppose thatv is a sufficiently smooth function such that

k∇vk_{L}∞([0,T]×R^{3}\K)≤δ, (4.1)

k∂∇vk_{L}1

tL^{∞}_{x}([0,T]×R^{3}\K)≤C0, (4.2)

where δ > 0 is a sufficiently small constant, C0 is a positive constant. Let ^{γ}
denote a second order operator given by

^{γ}=^{c}−X

l,m

C^{lm}(∇v)∂l∂m. (4.3)

Consider the Neumann wave equations

^{γ}ω=G, (t, x)∈R+×R^{3}\K,

∂νω
_{∂K}= 0,
ω(t, x) = 0, t≤0.

(4.4)

Let

E0=|∂0ω|^{2}+c^{2}_{I}|∇ω|^{2}+

3

X

l,m=1

(∂lω)^{T}C^{lm}(∇v)∂mω,

Ej=−2c^{2}_{I}(∂0ω)^{T}(∂jω)−2

3

X

k=1

(∂0ω)^{T}C^{jk}(∇v)∂kω, j = 1,2,3,

e=

3

X

l,m=1

(∂_{l}ω)^{T}∂_{0}C^{lm}(∇v)∂_{m}ω−2(∂_{l}ω)^{T}∂_{l}C^{lm}(∇v)∂_{m}ω
.

Noting the symmetry condition ofC^{lm}(∇v), we have

∂_{0}E_{0}+

3

X

j=1

∂_{j}E_{j} = 2(∂_{0}ω)^{T}^{γ}ω+e. (4.5)
By (4.1), there exist positive constantsλ, µdepending only onc1, c2, δ, such that

λ|ω^{0}|^{2}≤E0≤µ|ω^{0}|^{2}. (4.6)
Integrating (4.6) over [0, t]×R^{3}\K, we obtain

Z

R^{3}\K

E_{0}(t, x)dx−
Z

R^{3}\K

E_{0}(0, x)dx−
Z

[0,t]×∂K 3

X

j=1

E_{j}n_{j}dσ ds

= 2 Z

[0,t]×R^{3}\K

(∂0ω)^{T}^{γ}ω ds dx+
Z

[0,t]×R^{3}\K

e ds dx.

(4.7)

Noticing the Neumann condition∂νω =P3

j=1∂jωnj = 0 when ω ∈∂K, we have P3

j=1Ejnj= 0 on∂K, andE0(0, x) = 0. Therefore, Z

R^{3}\K

E0(t, x)dx= 2 Z

[0,t]×R^{3}K

(∂0ω)^{T}^{γ}ω ds dx+
Z

[0,t]×R^{3}K

e ds dx. (4.8) Using (4.6) and (4.8), we have

kω^{0}k^{2}_{L}2(R^{3}\K)≤C
Z t

0

kω^{0}k_{L}2(R^{3}\K)kGk_{L}2(R^{3}\K)ds

+C Z t

0

X

l,m

k∂C^{lm}(∇v)k_{L}^{∞}

x(R^{3}\K)kω^{0}k^{2}_{L}2(R^{3}\K)ds.

(4.9)

From assumption (4.2) and applying Gronwall inequality, we obtain
kω^{0}k^{2}_{L}2(R^{3}\K)≤C

Z t

0

kω^{0}k_{L}2(R^{3}\K)kGk_{L}2(R^{3}\K)ds

≤C sup

0≤s≤t

kω^{0}k_{L}2(R^{3}\K)

Z t

0

kGk_{L}2(R^{3}\K)ds.

Therefore,

kω^{0}kL^{2}(R^{3}\K)≤C
Z t

0

kGkL^{2}(R^{3}\K)ds, 0≤t≤T. (4.10)
In general, we have the following theorem.

Theorem 4.1. Assume that (4.1) and (4.2) hold, andω = ω(t, x) ∈ C^{∞} solves
problem (4.4). Then for any nonnegative integerN, there is a positive constantC,
such that

X

|α|≤N

k∂^{α}ω^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

X

|α|≤N

k^{γ}∂_{s}^{m}ω(s,·)kL^{2}(R^{3}\K)ds

+C X

|α|≤N−1

k^{c}∂^{α}ω(t,·)k_{L}2(R^{3}\K), 0≤t≤T.

(4.11)

The second term on the right-hand side of (4.11) vanishes when N = 0.

Proof. Proof by induction. WhenN = 0, (4.10) shows that (4.11) holds.

We suppose that (4.11) is valid ifN is replaced byN−1, then we proof it is valid forN. We first notice that∂tω satisfies (4.4), then by the assumption of induction,

X

|α|≤N−1

k∂^{α}(∂tω)^{0}(t,·)k_{L}2(R^{3}\K)≤ right-hand side of (4.11).

Hence it suffices to show that, forN ≥1 X

|α|≤N

k∂^{α}_{x}∇_{x}ω(t,·)k_{L}2(R^{3}\K)≤the right side of (4.11).

However, X

|α|≤N−1

k∆∂_{x}^{α}ω(t,·)kL^{2}(R^{3}\K)

≤C X

|α|≤N−1

k∂_{x}^{α}∂_{t}^{2}ω(t,·)kL^{2}(R^{3}\K)+C X

|α|≤N−1

k^{c}∂_{x}^{α}ω(t,·)kL^{2}(R^{3}\K),

(4.12)

whereC depends only on the wave speedscI.

The first term on the right-hand side of (4.12) is bounded by the right-hand side of (4.11), thus the right-hand side of (4.12) is similarly bounded. By elliptic regularity, so isP

|α|=Nk∂_{x}^{α}∇_{x}ω(t,·)k_{L}2(R^{3}\K), which completes the proof.

5. L^{2}_{x} estimates involving operators S^{j}Z^{α} outside of obstacles
We suppose that ω solves problem (4.4). Let P = P(t, x, D) be differential
operator and ∂_{ν}P ω not necessarily vanishes on ∂K. We will give some rough L^{2}
estimates forP ω. In this section, we assume thatv satisfies (4.1) and (4.2).

Proposition 5.1. Suppose thatP ω(0,·) =∂tP ω(0,·) = 0and there exist an integer M and a constant C0 such that

|(P ω)^{0}(t, x)| ≤C0t X

|α|≤M−1

|∂t∂^{α}ω^{0}(t, x)|+C0

X

|α|≤M

|∂^{α}ω^{0}(t, x)|, x∈∂K. (5.1)

Then,

k(P ω)^{0}(t,·)k_{L}2(R^{3}\K)≤C
Z t

0

k^{γ}P ω(s,·)k_{L}2(R^{3}\K)ds

+C Z t

0

X

|α|+j≤M+1, j≤1

k^{c}S^{j}∂^{α}ω(s,·)kL^{2}(R^{3}\K)ds

+ X

|α|+j≤M, j≤1

k^{c}S^{j}∂^{α}ω(s,·)k_{L}2([0,t]×R^{3}\K).

(5.2)

Proof. We will use the analogue of (4.7) where ω is replaced by P ω. Then we obtain

Z

R^{3}\K

E0(t, x)dx− Z

R^{3}\K

E0(0, x)dx− Z

[0,t]×∂K 3

X

j=1

Ejnjdσds

= 2 Z

[0,t]×R^{3}\K

(∂_{0}P ω)^{T}γP ω ds dx+
Z

[0,t]×R^{3}\K

e ds dx,

(5.3)

where

E0=|∂0P ω|^{2}+c^{2}_{I}|∇P ω|^{2}+

3

X

l,m=1

(∂lP ω)^{T}C^{lm}(∇v)∂mP ω,

E_{j}=−2c^{2}_{I}(∂_{0}P ω)^{T}(∂_{j}P ω)−2

3

X

k=1

(∂_{0}P ω)^{T}C^{jk}(∇v)∂_{k}P ω, j= 1,2,3,

e=

3

X

l,m=1

(∂lP ω)^{T}∂0C^{lm}(∇v)∂mP ω−2(∂lP ω)^{T}∂lC^{lm}(∇v)∂mP ω
.

It is obvious thatE_{0}(0, x) = 0. Use (4.1) and (4.2) and apply Gronwall’s inequality,
we obtain that ifδ >0 is small enough, then

k(P ω)^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

kγP ω(s,·)k_{L}2(R^{3}\K)ds

+CZ

[0,t]×∂K

(|∂_{t}P ω(s, x)|^{2}+|∇_{x}P ω(s, x)|^{2})dσ1/2

.

(5.4)

Recall thatK ⊂ {|x|<1}. By (5.1) and trace inequality, we have Z

[0,t]×∂K

(|∂tP ω(s, x)|^{2}+|∇xP ω(s, x)|^{2})dσ

≤C Z

[0,t]×∂K

X

|α|+j≤M, j≤1

|S^{j}∂^{α}ω^{0}|^{2}dσ

≤C X

|α|+j≤M+1, j≤1

kS^{j}∂^{α}ω^{0}k^{2}_{L}2([0,t]×∂K:|x|<2), ∀t≥0.

(5.5)

Therefore, by (5.4), (5.5) and (3.7), we obtain
k(P ω)^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

k^{γ}P ω(s,·)k_{L}2(R^{3}\K)ds+C X

|α|+j≤M+1, j≤1

kS^{j}∂^{α}ω^{0}k_{L}2([0,t]×∂K:|x|<2)

≤C Z t

0

k^{γ}P ω(s,·)kL^{2}(R^{3}\K)ds+C
Z t

0

X

|α|+j≤M+1, j≤1

k^{c}S^{j}∂^{α}ω(s,·)kL^{2}(R^{3}\K)ds

+C X

|α|+j≤M, j≤1

k^{c}S^{j}∂^{α}ωk_{L}2([0,t]×R^{3}\K), ∀t≥0.

Obviously,P =S^{j}Z^{α}(j≤1) satisfies (5.1), then we have the following theorem.

Theorem 5.2. Suppose that ω =ω(t, x)∈C^{∞} solves (4.4). If M = 1,2, . . ., we
have

X

|α|+j≤M, j≤1

k(S^{j}Z^{α}ω)^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

X

|α|+j≤M, j≤1

k^{γ}S^{j}Z^{α}ω(s,·)kL^{2}(R^{3}\K)ds

+C Z t

0

X

|α|+j≤M+1, j≤1

kcS^{j}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds

+ X

|α|+j≤M, j≤1

k^{c}S^{j}∂^{α}ω(s,·)kL^{2}([0,t]×R^{3}\K).

(5.6)

6. L^{2}_{x} estimates involving S^{m}∂^{α} outside of star-shaped obstacles
In this section, we shall assume furthermore that

k∇vk_{L}^{∞}_{(}_{R}3\K)≤ δ

1 +t, (6.1)

with δ small enough. Assume that ω solves problem (4.4). Using that K is a star-shaped obstacle, we will obtain a better estimate forSω.

Proposition 6.1. Suppose that (6.1) holds and ω=ω(t, x)∈C^{∞} solves problem
(4.4), then

k(Sω)^{0}(t,·)k_{L}2(R^{3}\K)≤C
Z t

0

kγSω(s,·)k_{L}2(R^{3}\K)ds

+C Z t

0

X

|α|≤2

k^{c}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds

+C X

|α|≤1

kc∂^{α}ωk_{L}2([0,t]×R^{3}\K).

(6.2)

Proof. Using the analogue of (4.7) whereω is replaced bySω, we have Z

R^{3}\K

E0(t, x)dx− Z

R^{3}\K

E0(0, x)dx− Z

[0,t]×∂K 3

X

j=1

Ejnjdσds

= 2 Z

[0,t]×R^{3}\K

(∂0Sω)^{T}^{γ}Sω ds dx+
Z

[0,t]×R^{3}\K

e ds dx,

(6.3)

where

E_{0}=|∂_{0}Sω|^{2}+c^{2}_{I}|∇Sω|^{2}+

3

X

l,m=1

(∂_{l}Sω)^{T}C^{lm}(∇v)∂_{m}Sω,

Ej=−2c^{2}_{I}(∂0Sω)^{T}(∂jSω)−2

3

X

k=1

(∂0Sω)^{T}C^{jk}(∇v)∂kSω, j = 1,2,3,

e=

3

X

l,m=1

(∂lSω)^{T}∂0C^{lm}(∇v)∂mSω−2(∂lSω)^{T}∂lC^{lm}(∇v)∂mSω
.

First we consider the right most term on the left-hand side of (6.3). When (s, x)∈
R+×∂K, the Neumann condition∂_{ν}ω=h~n,∇xiω= 0 gives us

∂sSω=s∂^{2}_{s}ω+∂sω+∂shx,∇xiω=s∂_{s}^{2}ω+∂sω+hx, ~ni∂s∂νω=s∂^{2}_{s}ω+∂sω.

Similarly,

3

X

j=1

nj∂jSω=

3

X

j=1

snj∂j∂sω+

3

X

j=1

nj∂jhx,∇xiω=s∂ν∂sω+∂νhx,∇xiω= 0 onR+×∂K. Noticing the assumption (6.1), we have

−

3

X

j=1

E_{j}n_{j}= 2(s∂_{s}^{2}ω+∂_{s}ω)^{T}

3

X

j,k=1

C^{jk}(∇v)(s∂k∂_{s}ω+∂_{k}(hx,∇iω))nj

≤C X

1≤|α|≤2

|∂^{α}ω|^{2}.

Hence, identity (6.3) yields Z

R^{3}\K

E_{0}(t, x)dx≤2
Z

[0,t]×R^{3}K

(∂_{0}Sω)^{T}γSω ds dx
+

Z

[0,t]×R^{3}K

e ds dx+C Z

[0,t]×∂K

X

1≤|α|≤2

|∂^{α}ω|^{2}dσ.

(6.4)

Applying Gronwall’s inequality, we obtain
k(Sω)^{0}(t,·)k_{R}^{3}_{\K}

≤ Z t

0

k^{γ}Sω(t,·)k_{R}3\Kds+CZ

[0,t]×∂K

X

1≤|α|≤2

|∂^{α}ω|^{2}dσ^{1/2}

. (6.5)

By the trace inequality and (3.5), we obtain Z

[0,t]×∂K

X

1≤|α|≤2

|∂^{α}ω|^{2}dσ^{1/2}

≤ X

|α|≤2

k∂^{α}ω^{0}(s,·)k_{L}2([0,t]×R^{3}\K:|x|<2)

≤C Z t

0

X

|α|≤2

k^{c}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds+C X

|α|≤1

k^{c}∂^{α}ωk_{L}2([0,t]×R^{3}\K).

(6.6)

Inequalities (6.5) and (6.6) complete the proof of (6.2).

Applying Proposition 6.1 and repeating the procedure of Theorem 4.1, we have the following theorem.

Theorem 6.2. Suppose that (6.1) holds and ω = ω(t, x) ∈ C^{∞} solves problem
(4.4). Then for any nonnegative integerN,

X

|α|+m≤N, m≤1

kS^{m}∂^{α}ω^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

X

|α|+m≤N, m≤1

k^{γ}S^{m}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds

+ X

|α|+m≤N−1, m≤1

kcS^{m}∂^{α}ω(s,·)k_{L}2(R^{3}\K)

+C Z t

0

X

|α|≤N+1

k^{c}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds

+C X

|α|≤N

kc∂^{α}ωkL^{2}([0,t]×R^{3}\K), ∀t≥0.

(6.7)

7. MainL^{2} estimates outside of star-shaped obstacles
We assume thatvsatisfies (4.2) and (6.1), then we have the following result.

Proposition 7.1. Suppose thatω =ω(t, x)∈C^{∞} solves problem (4.4). Then for
any fixed nonnegative integer N, we have

X

|α|≤N+4

k∂^{α}ω^{0}(t,·)kL^{2}(R^{3}\K)+ X

|α|+m≤N+2, m≤1

kS^{m}∂^{α}ω^{0}(t,·)kL^{2}(R^{3}\K)

+ X

|α|+m≤N, m≤1

kS^{m}Z^{α}ω^{0}(t,·)k_{L}2(R^{3}\K)

≤C Z t

0

X

|α|≤N+4

k^{γ}∂^{α}ω(s,·)kL^{2}(R^{3}\K)

+ X

|α|+m≤N+2, m≤1

k^{γ}S^{m}∂^{α}ω(s,·)k_{L}2(R^{3}\K)

+ X

|α|+m≤N, m≤1

k^{γ}S^{m}Z^{α}ω(s,·)k_{L}2(R^{3}\K)

ds

+C X

|α|≤N+3

k^{γ}∂^{α}ω(t,·)k_{L}2(R^{3}\K)

+C X

|α|+m≤N+1, m≤1

kγS^{m}∂^{α}ω(t,·)kL^{2}(R^{3}\K)

+C X

|α|≤N+2

k^{c}∂^{α}ωk_{L}2([0,t]×R^{3}\K)

+C X

|α|+m≤N, m≤1

k^{c}S^{m}∂^{α}ωk_{L}2([0,t]×R^{3}\K).

(7.1)

Proof. We denote the left side of (7.1) byI+II+III, and the right-hand side
side of (7.1) by RHS. Noticing that ^{c} = ^{γ} +P3

l,m=1C^{lm}(∇v)∂l∂m, then by
Theorem 4.1, we have

I≤RHS+C

3

X

l,m=1

X

|α|≤N+3

kC^{lm}(∇v)∂l∂_{m}∂^{α}ω(t,·)kL^{2}(R^{3}\K). (7.2)
Similarly, by Theorem 6.2, we obtain

II ≤RHS+C Z t

0 3

X

l,m=1

X

|α|≤N+3

kC^{lm}(∇v)∂l∂m∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds

+C

3

X

j,k=1

X

|α|+m≤N+1, m≤1

kC^{jk}(∇v)∂j∂kS^{m}∂^{α}ω(t,·)kL^{2}(R^{3}\K).

(7.3)

Similarly, by Theorem 5.2, we obtain III≤RHS+C

Z t

0 3

X

j,k=1

X

|α|+m≤N+1, m≤1

kC^{jk}(∇v)∂j∂kS^{m}∂^{α}ω(s,·)k_{L}2(R^{3}\K)ds.

Applying assumption (6.1), the last term on the right-hand side of (7.2) is domi- nated by

C sup

x∈R^{3}, l,m

|C^{lm}(∇v)| X

|α|≤N+4

k∂^{α}ω^{0}(t,·)k_{L}2(R^{3}\K)≤CδI.