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Local energy decay estimate of solutions to the thermoelastic plate equations in two- and three-dimensional exterior domains

Robert Denk, Reinhard Racke and Yoshihiro Shibata

Abstract

In this paper we prove frequency expansions of the resolvent and local energy decay estimates for the linear thermoelastic plate equations:

utt+ ∆2u+ ∆θ= 0 and θt∆θ∆ut= 0 in Ω×(0,∞),

subject to Dirichlet boundary conditions: u|Γ = Dνu|Γ = θ|Γ = 0 and initial conditions (u, ut, θ)|t=0 = (u0, v0, θ0). Here Ω is an exterior domain (domain with bounded comple- ment) in Rn with n = 2 or n = 3, the boundary Γ of which is assumed to be a C4- hypersurface.

1 Introduction and main results

Let Ω be an exterior domain (domain with bounded complement) in Rn with n= 2 or n= 3, the boundary Γ of which is assumed to be a C4-hypersurface. In this paper, we consider the linear thermoelastic plate equations

utt+ ∆2u+ ∆θ= 0 and θt−∆θ−∆ut= 0 in Ω×R+ (1.1) subject to the initial conditions

u(x,0) =u0(x), ut(x,0) =v0(x), θ(x,0) =θ0(x) (x∈Ω) (1.2) and Dirichlet boundary conditions

u|Γ=Dνu|Γ =θ|Γ = 0. (1.3)

Here Dν =Pn

j=1νjDj (Dj =∂/∂xj), and ν= (ν1, . . . , νn) denotes the unit outer normal to Γ.

In (1.1),u stands for a mechanical variable denoting the vertical displacement of the plate, whileθstands for a thermal variable describing the temperature relative to a constant reference temperature ¯θ. The thermal effect introduces a damping. In fact, when Ω is a bounded reference configuration, the exponential stability of the associated semigroup under several different kind of boundary conditions have been proved by Kim [5], Mun˜oz Rivera and Racke [18], Liu and Zheng [14], Avalos and Lasiecka [1], Lasiecka and Triggiani [7, 8, 9, 10] and Shibata [22]. Also, the analyticity of the semigroup has been shown, cf. Liu and Renardy [12] and then it has been studied by Russell [20], Liu and Liu [11], Liu and Yong [13], Mun˜oz Rivera and Racke [19] in

Department of Mathematics and Statistics, University of Konstanz, 78457 Konstanz, Germany

Department of Mathematical Sciences, School of Science and Engineering, Waseda University, Ohkubo 3-4-1, Shinjuku-ku, Tokyo 169-8555, Japan

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the L2 or Hilbert space setting (see also the book of Liu and Zheng [15] for a survey). In the Lp-setting this was investigated in our paper [4], where sufficiently strong a priori estimates for the resolvent in Lp-spaces have been proved. Before [4], Denk and Racke [3] studied the Cauchy problem for (1.1) in the whole spaceRn, also giving decay rates of solutions, and Naito and Shibata [16] studied the initial boundary value problem for (1.1) with Dirichlet boundary condition in the half-spaceRn+.

There were not yet any decay estimates for exterior domains. The purpose of this paper is to study the local energy decay of solutions to problem (1.1) – (1.3). To formulate the problem (1.1) – (1.3) in the semigroup setting, introducing the unknown functionv =ut, we rewrite it in matrix form:

Ut=AU in Ω×R+, U|t=0=U0, BU|Γ = 0, (1.4) where we have set

U =

 u v θ

, U0 =

 u0 v0 θ0

, A=

0 1 0

−∆2 0 −∆

0 ∆ ∆

, BU =

 u Dνu

θ

. (1.5)

To study the initial boundary value problem (1.4), we consider the corresponding resolvent problem:

(λI−A)U =F in Ω, BU|Γ= 0, (1.6)

whereIdenotes the 3×3 unit matrix. We shall give an expansion of the resolvent with respect to the frequency parameter λ(Theorem 1.3). Then, representing the semigroup via the resolvents (essentially: Laplace transform) will give the local energy decay result (Theorem 1.4).

To state our main results precisely, we introduce several spaces and some symbols at this point. Throughout this paper, let n ∈ {2,3}. For a general domain O ⊂ Rn, p ∈ (1,∞) and any integerm,Lp(O) and Wpm(O) stand for the usual Lebesgue space and Sobolev space, respectively. Letk · kLp(O) andk · kWm

p (O)denote their norms. For a general domainOwithC1 boundary ∂O, we introduce the spacesWp,02 (O) and Wp,Dm (O) (m= 2,4) as follows:

Wp,02 (O) ={u∈Wp2(O)|u|∂O = 0},

Wp,Dm (O) ={u∈Wpm(O)|u|∂O =Dνu|∂O= 0} (m= 2,4), (1.7) where ν = (ν1, . . . , νn) denotes the unit outer normal to ∂O. Let Hp(O) and Dp(O) be the spaces defined by the following formulas:

Hp(O) ={F =T(f, g, h)|f ∈Wp,D2 (O), g ∈Lp(O), h∈Lp(O)},

Dp(O) ={U =T(u, v, θ)|u∈Wp,D4 (O), v ∈Wp,D2 (O), θ∈Wp,02 (O)}. (1.8) Here and hereafter, TM denotes the transposed of M. We define the norms k · kHp(O) and k · kDp(O) by the following formulas:

kFkHp(O)=kfkW2

p(O)+k(g, h)kLp(O) (F =T(f, g, h)∈ Hp(O)), kUkDp(O)=kukW4

p(O)+k(v, θ)k

W2

p(O) (U =T(u, v, θ)∈ Dp(O)). (1.9) LetAO be the operator whose domain isDp(O) and whose operation is defined by the formula:

AOU =AU forU ∈ Dp(O). (1.10)

In [4] we proved the following theorem.

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Theorem 1.1. Let 1< p <∞. Let ρ(A) be the resolvent set of A. Let C+={λ∈C|Reλ≥0}

where C denotes the set of all complex numbers. Then, ρ(A)⊃C+\ {0}.

Moreover, for any λ0 >0 there exists a constant C depending onλ0, p and Ωsuch that for anyλ∈C+ with |λ| ≥λ0 andF ∈ Hp(Ω) there holds the estimate:

|λ|k(λI− Ap)−1FkHp(Ω)+k(λI− Ap)−1FkDp(Ω) ≤CkFkHp(Ω).

In view of Theorem 1.1, by standard arguments in the theory of analytic semigroups (cf.

Vrabie [24]) we know that for anyσ >0 there exists aθσ ∈(0, π/2) such that

ρ(A)⊃ {λ∈Σθσ | |λ|> σ}, (1.11) where we have set

Σ={λ∈C\ {0} | |argλ|< π−}. (1.12) Moreover, there exists a constantCσ depending onσ such that

|λ|k(λI− A)−1FkHp(Ω)+k(λI− A)−1FkDp(Ω)≤CσkFkHp(Ω) (1.13) for any λ∈Σθσ with|λ|> σ and F ∈ Hp(Ω). Let us define a setU by the formula

U = [

σ>0

{λ∈Σθσ | |λ|> σ}. (1.14) From (1.11) we see that

ρ(A)⊃ U. (1.15)

By (1.13), we have the following theorem.

Theorem 1.2. Let 1 < p < ∞. Then, A generates an analytic semigroup {T(t)}t≥0 in Hp(Ω).

Let bbe a number such thatBb ⊃Rn\Ω, whereBb ={x∈Rn| |x|< b}. Set Ωb =Bb∩Ω.

We introduce the following spaces:

Lp,b(Ω) ={f ∈Lp(Ω)|f(x) = 0 for |x|> b}, Hp,b(Ω) =Hp(Ω)∩(Lp,b(Ω))3

={F =T(f, g, h)|f ∈Wp,D2 (Ω)∩Lp,b(Ω), g, h∈Lp,b(Ω)}.

(1.16)

Replacing Ω byRn, we defineLp,b(Rn) andHp,b(Rn). For functionsU =T(u, v, θ) we will write kUkDp,loc(Ωb):=kU|bkDp(Ωb).

For Banach spaces X and Y, L(X, Y) denotes the set of all bounded linear operators from X into Y and L(X) = L(X, X). For any domain ω in C, Anal (ω, X) denotes the set of all holomorphic functions defined onω with their values inX. We set

ωτ :={λ∈C| |λ|< τ}, ω˙τ :=ωτ\(−∞,0].

The following two theorems are our main results.

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Theorem 1.3. Let n∈ {2,3},1< p <∞ and letb be a number such that Bb−3 ⊃Rn\Ω. Let U be the same set as in (1.14). Set Lp,b(Ω) =L(Hp,b(Ω),Dp,loc(Ωb)).

(a) In the case n = 2 there exist a constant τ > 0 and an operator-valued function G ∈ Anal( ˙ωτ,Lp,b(Ω)) such that for any F ∈ Hp,b(Ω)and λ∈ω˙τ∩ U there holds the equality:

(λI− A)−1F =G(λ)F in Ωb.

Moreover, there exist operators G1, G2∈ Lp,b(Ω) and an operator-valued function G3∈Anal( ˙ωτ,Lp,b(Ω))

such that

G(λ) =G1+ (logλ)−1G2+G3(λ) for any λ∈ω˙τ, kG3(λ)FkD

p,loc(Ωb) ≤C|logλ|−2kFkH

p(Ω) for any λ∈ω˙τ and F ∈ Hp,b(Ω). (1.17) (b) In the case n = 3 there exist a constant τ > 0 and operator-valued functions Gj ∈ Anal(ωτ,Lp,b(Ω)) (j = 1,2) such that for any F ∈ Hp,b(Ω) and λ ∈ ωτ ∩ U there holds the equality:

(λI− A)−1F =λ12G1(λ)F+G2(λ)F in Ωb. (1.18) For wave equations, elasticity or Maxwell equations, a collection of references for results on low frequency asymptotics is given in the work of Pauly [17].

With the expansion of the resolvent in terms of the frequency parameter above, we shall obtain the following local energy decay result.

Theorem 1.4. Let1< p <∞and letbbe the same constant as in Theorem 1.3. Let{T(t)}t≥0

be the semigroup associated with problem (1.1) – (1.3) which is given in Theorem 1.2. Then, we have

kT(t)FkDp,loc(Ωb)

(Cp,bt−1(logt)−2kFkHp(Ω) if n= 2,

Cp,bt32kFkHp(Ω) if n= 3 (1.19) for anyt≥1 and F ∈ Hp,b(Ω).

The difficulty in proving Theorem 1.3 arises from the facts that the expansion formula of the resolvent operator (λ−∆)−1 inR2 has the singularity logλand that of (λ−∆2)−1 inRn has the singularities λ−1logλ when n = 2 and λ12 when n = 3, respectively. Therefore, we can not use the usual compact perturbation method to obtain the expansion formula in the exterior domain. To prove Theorem 1.3, first of all employing the Seeley argument [21] about the invertibility ofI+Kλ,Kλ being a compact operator valued holomorphic function inλ, we shall show that (λI− A)−1 has an expansion formula nearλ= 0 which starts fromλs(logλ)β in two dimensional case and λs2 in three dimensional case for some integers s and β. Then, by a contradiction argument based on the uniqueness theorem we shall show that s = 0 and β = 0. Our strategy of the proof of Theorem 1.3 follows R. Kleinmann and B. Vainberg [6] and W. Dan and Y. Shibata [2], where the low frequency expansions of the Laplace operator and Stokes operator in the two dimensional case were obtained.

We will prove Theorems 1.3 and 1.4 in Sections 2–3 for the (somewhat simpler) case n= 3.

Modifications for the case n= 2 are indicated in Sections 4 and 5.

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2 Expansion formulas in three dimensions

We start with the three-dimensional case by showing an expansion formula of the resolvent in the whole-space.

Theorem 2.1. Let 1 < p < ∞ and b > 0. Let Lp,b(R3) be the set of all bounded linear operators from Hp,b(R3) into Dp,loc(Bb) and ρ(AR3) the resolvent set of AR3. Then, there exist constants ∈(0, π/2) and operator-valued functions Hj(λ)∈Anal (C,Lp,b(R3)) (j = 1,2) such thatρ(AR3)⊃Σ and

(λI− AR3)−1F =λ12E0F+E1F +λ12H1(λ)F+λH2(λ)F in Bb (2.1) for anyλ∈Σ andF ∈ Hp,b(R3). Here,Σ is the set defined in (1.12),

E0F =

 αR

R3g dx+βR

R3h dx 0

0

, E1F =

E32∗(−∆f +g+h)

−f E31∗(h−∆f)

, E31(x) = 1

4π|x|, E32(x) =−|x|

8π,

(2.2)

∗ stands for the convolution operator, is given in (2.6), and α and β are non-zero constants given in (2.11) in the proof below.

Remark 2.2. E31(x) and E32(x) are fundamental solutions to−∆ and ∆2 in R3, respectively.

Proof. For F ∈ Hp(R3), we set U(λ) = (λI− AR3)−1F. Let ˆU(λ)(ξ) = T(ˆuλ(ξ),vˆλ(ξ),θˆλ(ξ)) be the Fourier transform of U(λ). Then, from Naito and Shibata [16], we have the following formulas:

ˆ uλ(ξ) =

3

X

j=1

h A0j +A1j +A2j

(λ+γj|ξ|2)|ξ|2|ξ|2fˆ(ξ) + A0j +A1j

(λ+γj|ξ|2)|ξ|2g(ξ) +ˆ A0j

(λ+γj|ξ|2)|ξ|2ˆh(ξ) i

,

ˆ vλ(ξ) =

3

X

j=1

h−(A0j +A1j)|ξ|2

λ+γj|ξ|2 fˆ(ξ) + A1j+A12

λ+γj|ξ|2ˆg(ξ) + A1j

λ+γj|ξ|2ˆh(ξ)i , θˆλ(ξ) =

3

X

j=1

h A0j|ξ|2

λ+γj|ξ|2f(ξ)ˆ − A1j

λ+γj|ξ|2g(ξ) +ˆ A0j +A2j λ+γj|ξ|2ˆh(ξ)

i .

(2.3)

Here,γj (j= 1,2,3) are numbers such that

3

Y

j=1

(t+γj) =t3+t2+ 2t+ 1 for any t∈C, (2.4) 0< γ1 <1,γ3 is the complex conjugate ofγ2 and Reγ2 = (1−γ1)/2>0; andA0j,A1j and A2j (j = 1,2,3) are complex numbers such that

λk Q3

j=1(λ+γj|ξ|2) =

3

X

j=1

Akj

(λ+γj|ξ|2)|ξ|4−2k (k= 1,2,3)

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for any ξ∈R3 and λ∈Cwithλ+γj|ξ|26= 0 (j= 1,2,3). We have the following formulas:

3

X

j=1

A0j =

3

X

j=1

A1j = 0,

3

X

j=1

A2j = 1,

3

X

j=1

A0j γj

= 1,

3

X

j=1

A1j γj

=

3

X

j=1

A2j γj

= 0. (2.5)

Since γ2 and γ3 are complex conjugate and Reγ2 >0, we may assume that 0 <argγ2 < π/2.

Let us define by the formula:

= argγ2. (2.6)

Since λ+γj|ξ|2 6= 0 for any λ ∈ Σ and ξ ∈ R3, by Fourier multiplier theorem we have U(λ) =T(uλ, vλ, θλ)∈ Dp(R3). Moreover, for any 0 with < 0 < π/2 there exists a constant C depending on 0 such that

2

X

j=0

|λ|2−j2 k∇j(∇2uλ, vλ, θλ)kLp(R3) ≤CkFk

Hp(R3),

|λ|k∇uλkLp(R3)+|λ|2kuλkLp(R3) ≤Ck(|λ|f, g, h)kLp(R3)

(2.7)

for anyλ∈Σ0 (cf. Naito-Shibata [16]), where∇jw= (Dαw| |α|=j). From these observations, we see thatρ(AR3)⊃Σ.

Now, restricting ourselves to the case where F ∈ Hp,b(R3), we shall derive an expansion formula of (λI − A

R3)−1F by using the formula (2.3). Let Fξ−1 denote the Fourier inverse transform, and then we have

Fξ−1[(λ+|ξ|2)−1](x) = e

λ|x|

4π|x| , Fξ−1[(λ+|ξ|2)−1|ξ|−2](x) =−λ−1e

λ|x|

4π|x| − 1 4π|x|

(2.8)

for any λ∈C\(−∞,0]. Since we havee

λ|x|=P j=0(−√

λ|x|)j/(j!), we have Fξ−1[(λ+|ξ|2)−1](x) = 1

4π|x|−λ12

4πH11(λ|x|2) +λ|x|

8π H21(λ|x|2), (2.9) Fξ−1[(λ+|ξ|2)−1|ξ|−2](x) = λ12

4π − |x|

8π + λ12|x|2

4π H12(λ|x|2)−λ|x|3

4π H22(λ|x|2), (2.10) where we have set

H12(z) =

X

j=0

zj

(2j+ 3)!, H22(z) =

X

j=0

zj (2j+ 4)!, H11(z) = 1 +zH12(z), H21(z) = 1 + 2zH22(z).

Now, we assume thatF ∈ Hp,b(R3). Sinceλ+γj|ξ|2j(λγj−1+|ξ|2), using (2.10) and (2.5), from (2.3) we have

uλ(x) =hX3

j=1

A0j +A1j

√γj

1 4π

Z

R3

g dx+X3

j=1

A0j

√γj

1 4π

Z

R3

h dxi

λ12 +E32∗(−∆f +g+h)

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12

hnX3

j=1

A0j+A1j +A2j

γj3/2 H12j−1λ|x|2) |x|2

4π o

∗(−∆f)

+ nX3

j=1

A0j+A1j γj3/2

H12j−1λ|x|2) |x|2

4π o

∗g+ nX3

j=1

A0j γj3/2

H12j−1λ|x|2) |x|2

4π o

∗h i

+λhnX3

j=1

A0j +A1j +A2j

γj2 H22j−1λ|x|2)|x|3

o∗(−∆f)

+ nX3

j=1

A0j+A1j

γj2 H12j−1λ|x|2) |x|3

4π o

∗g+ nX3

j=1

A0j

γ2j H12j−1λ|x|2) |x|3

4π o

∗h i

. Setting

α=

3

X

j=1

A0j+A1j

√γj , β =

3

X

j=1

A0j

√γj, (2.11)

we have the first line of the formula (2.1) with (2.2). Using the fact that E31 ∗(−∆f) = f to obtain the formula forvλ(x), by (2.3), (2.5) and (2.9) we have

vλ(x) =−f +λ12hX3

j=1

A0j +A1j 4πγj3/2

H11j−1λ|x|2)

∗(−∆f)

−X3

j=1

A1j +A2j 4πγj3/2

H11j−1λ|x|2)

∗g−X3

j=1

A1j 4πγj3/2

H11j−1λ|x|2)

∗h i

−λhnX3

j=1

A0j +A1j

γj2 H21j−1λ|x|2)|x|

o∗(−∆f)

−nX3

j=1

A1j+A2j

γj2 H21j−1λ|x|2)|x|

8π o

∗g−nX3

j=1

A1j

γ2j H21j−1λ|x|2)|x|

8π o

∗hi ,

θλ(x) =E13∗(h−∆f)−λ12 hX3

j=1

A0j 4πγj3/2

H11−1j λ|x|2)

∗(−∆f)

−X3

j=1

A1j 4πγj3/2

H11−1j λ|x|2)

∗g+X3

j=1

A0j+A1j 4πγj3/2

H11j−1λ|x|2)

∗hi

hnX3

j=1

A0j

γj2H21j−1λ|x|2) |x|

8π o

∗(−∆f)

−nX3

j=1

A1j

γj2H21−1j λ|x|2) |x|

8π o

∗g+ nX3

j=1

A0j+A1j

γj2 H21j−1λ|x|2) |x|

8π o

∗h i

. This completes the proof of Theorem 2.1.

The next step in the proof of our main results consists in an expansion formula for the resolvent operator in Ω near λ= 0. We will show the following theorem.

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Theorem 2.3. Let 1 < p < ∞ and b be a positive number such that Bb−3 ⊃ R3\Ω. Let U and Lp,b(Ω) be the same sets as in (1.14) and Theorem 1.3, respectively. Then, there exist a constant τ >0, an integer sand operators Gj(λ)∈Anal (ωτ,Lp,b(Ω)) (j= 1,2)such that

(λI− A)−1F =λs2G1(λ)F +λs+12 G2(λ)F in Ωb for anyλ∈ωτ ∩ U and F ∈ Hp,b(Ω).

In what follows, we shall prove Theorem 2.3. For a given functionf defined on Ω,ιf denotes the zero extension off to the whole space R3 andrf denotes the restriction of f to the domain Ωb = Ω∩Bb. From Denk, Racke and Shibata [4] (also Simader [23]), we know the unique existence of a solution U0 =T(u0, v0, θ0)∈ Dp(Ωb) of the equation:

−AU0=F in Ωb, BU0|∂Ωb = 0 (2.12) for any F ∈ Hp(Ωb), Here, ∂Ωb = Γ∪Sb,Sb={x∈R3 | |x|=b} and BU0|∂Ωb = 0 means that

u0 =Dνu00 = 0 on Γ andSb,

where Dν = (x/|x|)· ∇ on Sb. Let us define the operatorSb by the formula: SbF =U0 and write SbF = (ub, vb, θb) as long as no confusion occurs. Let E0, E1, H1(λ) and H2(λ) be the same operator as in Theorem 2.1 and set

H(λ) =λ12E0+E112H1(λ) +λH2(λ). (2.13) In what follows, we write H(λ)F = (uλ,R3, vλ,R3, θλ,R3). Let ϕ be a function in C0(R3) such thatϕ(x) = 1 for|x|< b−2 andϕ(x) = 0 for|x|> b−1. With these preparations, we introduce the operator Φ as follows:

Φ(λ)F = (1−ϕ)H(λ)ιF +ϕSbrF. (2.14) By Theorem 2.1, we have

Φ(λ)F = (1−ϕ)(λI− AR2)−1ιF +ϕSbrF (2.15) when λ∈Σ. And therefore, applying λI−A to Φ(λ)F, we have

(λI−A)Φ(λ)F =F+T(λ)F in Ω, BΦ(λ)F|Γ= 0 (2.16) for any λ∈Σ, where T(λ)F is defined by the formula:

T(λ)F =

0

−L3ϕ(uλ,R2 −ub)−L1ϕλ,R2 −θb) L1ϕλ,R2−θb) +L1ϕ(vλ,R2 −vb)

, (2.17)

L3ϕ(w) = ∆2(ϕw)−ϕ∆2w, andL1ϕ(w) = ∆(ϕw)−ϕ∆w. If we consider (2.16) only on Ωb, the operators in both sides of (2.16) are analytic with respect toλ∈C\(−∞,0], and therefore by analytic continuation we have

(λI−A)Φ(λ)F =F +T(λ)F in Ωb, BΦ(λ)F|Γ= 0 (2.18) for anyλ∈C\(−∞,0]. If (I+T(λ))−1 exists, then Φ(λ)(I+T(λ))−1F solves equations (2.16) and (2.18).

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Lemma 2.4. Let U and Σ be the same sets as in (1.14)and Theorem 2.1, respectively. Then, (I+T(λ))−1 exists as a bounded linear operator on Hp,b(Ω) for anyλ∈ U ∩Σ.

Proof. Letλ∈Σ∩ U. Since the second and third components ofT(λ)F belong toWp1(Ω) and suppT(λ)F ⊂ Db−2,b−1 = Bb−1 \Bb−2, by Rellich’s compactness theorem T(λ) is a compact operator onHp,b(Ω). Therefore, to prove the lemma it suffices to show thatI+T(λ) is injective.

LetF be an element ofHp,b(Ω) such that (I+T(λ))F = 0. SetU = Φ(λ)F, and then by (2.18) we have

(λI−A)U = 0 in Ω, BU|Γ= 0.

Since SbrF ∈ Dp(Ωb) and (λI− AR3)−1ιF ∈ Dp(R3) forλ∈Σ (cf. (2.7)), by (2.15) we have U ∈ Dp(Ω). SinceU ⊂ρ(A) as follows from (1.15), we have U = 0, which implies that

(1−ϕ)(λI− AR3)−1ιF +ϕSbrF = 0 in Ω. (2.19) Recalling that ϕ(x) = 1 for |x|< b−2 and ϕ(x) = 0 for |x|> b−1, by (2.19) we have

(λI− AR3)−1ιF = 0 for|x|> b−1, SbrF = 0 for|x|< b−2.

If we setV(x) = (SbrF)(x) forx∈Ωb and V(x) = 0 forx6∈Ω, thenV(x) belongs to Dp(Bb) and satisfies the equation:

(λI−A)V =ιF inBb, BV|Sb = 0.

Since (λI−AR3)−1ιF also satisfies the above equation, by the uniqueness of solutions we have V = (λI−AR3)−1ιF inBb, and therefore SbF = (λI−AR3)−1ιF in Ωb, which inserted into (2.19) implies that

0 = (λI− AR3)−1ιF +ϕ(SbF −(λI− AR3)−1ιF) = (λI− AR3)−1ιF in Ω.

Therefore,F = (λI−A)(λI− AR3)−1ιF = 0 in Ω, which completes the proof of the lemma.

By Lemma 2.4 we have

(λI− A)−1= Φ(λ)(I+T(λ))−1 (2.20) forλ∈Σ∩ U.

Now, we shall discuss the invertibility of (I+T(λ)) for λ∈ω˙σ with someσ > 0, where we have set

˙

ωσ ={λ∈C\ {0} | |λ|< σ and |argλ|< π}.

For this purpose, we introduce an auxiliary operator:

Φ0F = (1−ϕ)E1ιF +ϕSbrF

forF ∈ Hp,b(Ω), where E1 is the same operator as in Theorem 2.1. Note that

−AE1ιF =ιF in R3.

We writeE1ιF =T(u0,R3, v0,R3, θ0,R3) unless any confusion may occur. ApplyingA to Φ0F, we have

−AΦ0F =F+T0F in Ω, BΦ0F|Γ= 0, (2.21)

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where

T0F =

0

−L3ϕ(u0,R3 −ub)−L1ϕ0,R3 −θb) L1ϕ0,R3−θb) +L1ϕ(v0,R3 −vb)

.

Since the second and third members of T0F belong to Wp1(Ω) and suppT0F ⊂ Db−2,b−1, by Rellich’s compactness theorem T0 is a compact operator on Hp,b(Ω). According to Theorem 2.1, we set

uλ,R3 =u0,R312T(αg+βh) +Uλ,R3, vλ,R3 =v0,R3 +Vλ,R3,

θλ,R30,R3+ Θλ,R3, whereT a=R

R3a dxand

T(Uλ,R3, Vλ,R3λ,R3) =λ12H1(λ)ιF +λH2(λ)ιF. (2.22) Then, we have

(I+T(λ))F = (I+T0)F +λ12(∆2ϕ)T(0, T(αg+βh),0) +R(λ)F (2.23) where

R(λ)F =

0

−L3ϕ(Uλ,R3)−L1ϕλ,R3) L1ϕλ,R3) +L1ϕ(Vλ,R3)

. (2.24)

In view of (2.22) and (2.24), there exist operatorsRj(λ)∈Anal (C,L(Hp,b(Ω))) (j= 1,2) such that

R(λ)F =λ12R1(λ)F +λR2(λ)F (2.25) for any λ∈C\(−∞,0]. In particular, we have

λ→0limkR(λ)kL(H

p,b(Ω))= 0. (2.26)

Here,k · kL(H

p,b(Ω)) denotes the operator norm ofL(Hp,b(Ω)). SinceT0 is a compact operator on Hp,b(Ω), by Seeley’s lemma [21] there exists a finite range operatorBsuch thatI+T0−Bhas an inverse operator (I+T0−B)−1 ∈ L(Hp,b(Ω)). SetGλ =I+T0−B+R(λ) andG0 =I+T0−B, and then

(I+T(λ))F =GλF+BF+λ12(∆2ϕ)T(0, T(αg+βh),0) (2.27)

Gλ= (I+R(λ)G−10 )G0. (2.28)

By (2.26) there exists aτ0>0 such thatkR(λ)G−10 kL(H

p,b(Ω) ≤1/2 for anyλ∈ω˙τ0, and therefore by Neumann series expansion we have

G−1λ =G−10 (I+R(λ)G−10 )−1 =G−10

X

j=0

(−R(λ)G−10 )j (λ∈ω˙τ0). (2.29) In view of (2.25), we see that there exist aτ1 >0 and operators Gj(λ)∈Anal (ωτ1,L(Hp,b(Ω)) (j = 1,2) such that

G−1λ12G1(λ) +G2(λ) for any λ∈ω˙τ1. (2.30)

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We define the operator ˜B by the formula ˜BF = (∆2ϕ)T(0,R

R3(αg+βh)dx,0). As both operators B and ˜B are finite range operators, we can choose h1, . . . ,hm ∈ Hp,b(Ω) which are linearly independent overC in such a way that

BF =

m

X

j=1

βj(F)hj, BF˜ =

m

X

j=1

β˜j(F)hj

with βj(F), ˜βj(F)∈ C. To represent βj(F), ˜βj(F) ∈C in more convenient way, we introduce h1, . . ., hm ∈ Hp,b(Ω) such that < hj,hk >=δjk, where < ·,·> is the dual paring between Hp,b(Ω) and its dual space Hp,b(Ω) and δjk denote the Kronecker delta symbols. By using these symbols, we write

βj(F) =< BF,hj >=< F, Bhj >, β˜j(F) =<BF,˜ hj >=< F,B˜hj > . Setting`aj =Bhj and`bj = ˜Bhj, we have

BF +λ12(∆2ϕ)T(0, T(αg+βh),0) =

m

X

j=1

< F, `aj12`bj>hj,

and therefore we have

(I+T(λ))F =GλF +

m

X

j=1

< F, `aj12`bj>hj. (2.31)

Applying G−1λ to the both side of (2.31), we have G−1λ (I+T(λ))F =F +

m

X

j=1

< F, `aj12`bj > G−1λ hj = (I+Nλ)F (2.32)

where we have defined the operator Nλ by the formula:

NλF =

m

X

j=1

< F, `aj12`bj > G−1λ hj. (2.33) Now, we shall show the existence of the inverse operator ofI+Nλ. For the notational simplicity, we set G−1λ hj = vλ,j and `aj12`bj = Aλ,j. Since {hj}mj=1 is linearly independent, so is {vλ,j}mj=1. Let us consider the m×m matrix: M(λ) = (δjk+ <vλ,k, Aλ,j >). By (2.30) the (j, k) component δjk+<vλ,k, Aλ,j >is of the form: λ12m1jk(λ) +m2jk(λ), wherem1jk(λ) and m2jk(λ) are complex valued holomorphic functions defined onωτ1. LetD(λ) be the determinant of M(λ). In particular, we can say that D(λ) ≡ 0 on ωτ1 or there exist an integer q1, and functions Dj(λ) (j= 1,2) such that

D(λ) =λq21D1(λ) +λq1+12 D2(λ) forλ∈ω˙τ1, (2.34) D1(0)6= 0, andDj(λ) (j= 1,2) are both holomorphic inωτ1. We shall show that

D(λ)6≡0 inωτ1. (2.35)

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In fact, let λ ∈ U ∩Σ ∩ωτ1 and assume that D(λ) = 0. Then there exists a vector xλ =

T(xλ1, . . . , xλm)∈Rm\ {0} such that 0 =

m

X

k=1

jk+<vλ,k, Aλ,j >)xλ,k=xλ,j+

m

X

k=1

<vλ,k, Aλ,j > xλ,k (2.36) for j = 1, . . . , m. Set Fλ = Pm

k=1xλ,kvλ,k ∈ Hp,b(Ω), and then Fλ 6= 0, because {vλ,k}mk=1 is linearly independent. On the other hand, by (2.33) and (2.36)

NλFλ =

m

X

j=1

< Fλ, Aλ,j >vλ,j =

m

X

j,k=1

xλ,k<vλ,k, Aλ,j >vλ,j =−

m

X

j=1

xλ,jvλ,j =−Fλ, which implies that (I+Nλ)Fλ = 0. And therefore, by (2.32) and (2.31) (I+T(λ))Fλ = 0. On the other hand, by Lemma 2.4 I+T(λ) is invertible when λ∈ U ∩Σ, and therefore we have Fλ= 0. This leads to a contradiction. Therefore, we have (2.35), and then (2.34) holds.

From (2.34), there exist a constant τ2 (0 < τ2 ≤ τ1) and holomorphic functions Ej(λ) (j = 1,2) defined onωτ2 such that

D−1(λ) =λq21E1(λ) +λq21+12E2(λ) forλ∈ω˙τ2. (2.37) By using this fact, we shall show the existence of (I+Nλ)−1. We may assume thatD−1(λ)6= 0 whenλ∈ωτ2\ {0}. Let us denote the (j, k) cofactor ofM(λ) byMjk(λ), which has the similar formula toD−1(λ) in (2.37). We observe that

(I+Nλ)[G−D(λ)−1

m

X

j=1 m

X

k=1

< G, Aλ,k> Mjk(λ)vλ,j]

=G−D(λ)−1

m

X

j,k=1

< G, Aλ,k> Mjk(λ)vλ,j

+NλG−D(λ)−1

m

X

j,k=1

< G, Aλ,k > Mj,k(λ)Nλvλ,j = (∗).

Since Nλvλ,j =Pm

`=1 <vλ,j, Aλ,` >vλ,` as follows from (2.33) and our short notation: `aj + λ12`bj =Aλ,j, we can proceed as follows:

(∗) =G−D(λ)−1

m

X

j,k=1

< G, Aλ,k > Mjk(λ)vλ,j +

m

X

k=1

< G, Aλ,k>vλ,k

−D(λ)−1

m

X

j,k,`=1

< G, Aλ,k> Mjk(λ)<vλ,j, Aλ,` >vλ,`

=G+

m

X

k=1

< G, Aλ,k>vλ,k−D(λ)−1

m

X

j,k,`=1

`j+<vλ,j, Aλ,` >)Mjk(λ)< G, Aλ,k>

vλ,`

=G+

m

X

k=1

< G, Aλ,k>vλ,k

m

X

k,`=1

δ`k < G, Aλ,k>vλ,`

=G.

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From this observation and our short notations: G−1λ hj =vλ,j and`aj12`bj =Aλ,j, we have (I+N(λ))−1G=G−D(λ)−1

m

X

j,k=1

< G, `ak12`bk > Mjk(λ)G−1λ hk forλ∈ωτ2 \ {0}. By (2.35), we see that

(I+T(λ))−1 = (I +Nλ)−1G−1λ

which combined with (2.30) and (2.37) implies that there exist an integer q2 and operators Tj(λ)∈Anal (ωτ2,L(Hp,b(Ω))) (j = 1,2) such that

(I+T(λ))−1q22T1(λ) +λq2+12 T2(λ)

for anyλ∈ωτ2\ {0}. Combining this fact with (2.20), (2.14) and Theorem 2.1 implies Theorem 2.3.

3 The proofs of Theorems 1.3 and 1.4 in the three-dimensional case

In what follows,b denotes a large number such thatBb−3 ⊃R3\Ω. To prove Theorem 1.3, we start with the following lemmas.

Lemma 3.1. Let `be a positive integer and n∈ {2,3}. If u∈ S0(Rn)∩L1,loc(Rn) satisfies the homogeneous equation:

`u= 0 in Rn (3.1)

and the radiation condition:

u(x) =O(|x|m) as |x| → ∞, (3.2)

for some non-negative integer m, then u is a polynomial of orderm.

Proof. Since u ∈ S0(Rn), applying the Fourier transform to (3.1) we have |ξ|2`u(ξ) = 0, whichˆ implies that supp ˆu(ξ) ⊂ {0}. By the structure theorem of distributions, ˆu(ξ) is represented as follows: ˆu(ξ) =P

|α|≤kcαδ(α)(ξ) for some non-negative integer k, whereδ denotes the Dirac delta function and cα are complex numbers. By the Fourier inverse transform, we have

u(x) = X

|α|≤k

cα(−ix)α,

which combined with (3.2) implies that u = u(x) should be a polynomial of order m. This completes the proof of the lemma.

Lemma 3.2. Let E1 be the same operator as in Theorem 2.1. Given F = T(f, g, h), we set U =E1F =T(u, v, θ). If F ∈ Hp,b(R3) and

Z

R3

(g(x) +h(x))dx= 0, (3.3)

then

u(x) =O(1), ∇u(x) =O(|x|−1), (3.4)

θ(x) =O(|x|−1) (3.5)

as |x| → ∞.

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Proof. Since R

R3(g(y) +h(y)−∆f(y))dy= 0 as follows from (3.3), by (2.2) we have u(x) = −1

8π Z

R3

(|x−y| − |x|)(g(y) +h(y)−∆f(y))dy.

By Taylor’s formula we have

|x−y| − |x|= Z 1

0

d

dθ|x−θy|dθ=−

3

X

i=1

Z 1 0

(xi−θyi)yi|x−θy|−1dθ,

and therefore

u(x) =

3

X

i=1

Z 1 0

nZ

R3

(xi−θyi)yi

|x−θy| (g(y) +h(y)−∆f(y))dyo dθ,

which combined with the fact that g(y) +h(y)−∆f(y) = 0 vanishes for|y| ≥b implies (3.4).

Since

θ=E31∗(h−∆f) = 1

4π|x|∗(h−∆f)

and since h(y)−∆f(y) vanishes for |y| ≥ b, we have (3.5), which completes the proof of the lemma.

Lemma 3.3. Let 1< p <∞. (1)If θ∈Wp,loc2 (Ω)satisfies the homogeneous equation:

∆θ= 0 in Ω, θ|Γ= 0 (3.6)

and the radiation condition:

θ(x) =O(|x|−1) (3.7)

as |x| → ∞, then θ= 0.

(2) If u∈Wp,loc4 (Ω) satisfies the homogeneous equation:

2u= 0 in Ω, u|Γ =Dνu|Γ= 0 (3.8) and the radiation condition:

u(x) =O(1) (3.9)

as |x| → ∞, then u= 0.

Proof. (1) By Lp (1< p <∞) solvability in any C2 bounded domain for the Dirichlet problem of the Laplace operator (cf. Simader [23]) and Sobolev’s imbedding theorem, we see that θ∈W2,loc2 (Ω). Let ρ be a function inC0(R3) such that ρ(x) = 1 for |x| ≤1 and ρ(x) = 0 for

|x| ≥2. SetρL(x) =ρ(x/L) for L > b. Then, we have

0 = (∆θ, ρLθ)=−(∇θ, ρL∇θ)+ (1/2)(θ,(∆ρL)θ) (3.10) where (a, b)=R

a(x)b(x)dx. Since

|(θ,(∆ρL)θ)| ≤ k∆ρk

L∞(R3)L−2 Z

L≤|x|≤2L

|θ(x)|2dx,

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