THE GREEN FUNCTION OF THE INTERIOR TRANSMISSION PROBLEM AND ITS APPLICATIONS

Inverse Problems and Imaging doi:10.3934/ipi.2012.6.487

Volume 6, No. 3, 2012, 487–521

THE GREEN FUNCTION OF THE INTERIOR TRANSMISSION

PROBLEM AND ITS APPLICATIONS

Kyoungsun Kim

Department of Mathematics of Inha UniversityIncheon 402-751, South Korea

Gen Nakamura

Department of Mathematics, Hokkaido University

Sapporo 060-0810, Japan

Mourad Sini

Johann Radon Institute for Computational and Applied Mathematics (RICAM)

Linz A-4040, Austria

(Communicated by Mikko Salo)

Abstract. The interior transmission problem appears naturally in the scat-

tering theory. In this paper, we construct the Green function associated to this

problem. In addition, we provide point-wise estimates of this Green functionsimilar to those known for the Green function related to the classical transmis-

sion problems. These estimates are, in particular, useful to the study of variousinverse scattering problems. Here, we apply them to justify some asymptotic

formulas already used for detecting partially coated dielectric mediums from

far field measurements.

1. Introduction. Let D be a C∞−smooth, bounded and connected domain ofRd, d ≥ 2 is an integer. Let ρ1, ρ2 ∈ H−1(D) := u ∈ H−1(Rd) : supp(u) ⊂ D1,

f ∈ H 12 (∂D) and g ∈ H− 1

2 (∂D). Assume that the matrix A is a C∞−smooth realsymmetric matrix valued function in D and n is a C∞−smooth and positive scalarfunction in D satisfying the condition 2

(1.1) ξ ·A(x)ξ ≥ γ|ξ|2, ∀ξ ∈ Rd and n(x) ≥ γ,∀x ∈ Dwith some constant γ > 1 independent of x and ξ.

We deal with the following boundary value problem called interior transmissionproblem:

(1.2)

∇ ·A∇v + κ2nv = ρ1 in D

∆w + κ2w = ρ2 in Dv − w = f on ∂D

∂νAv − ∂νw = g on ∂D,

where ∂νAv := A∇v · ν and ∂νw = ∇w · ν with the unit outer normal vector ν of∂D.

2000 Mathematics Subject Classification. Primary: 35J08, 35J25; Secondary: 35S15.Key words and phrases. Green function, interior transmission problem, pseudodifferential

operators.1The space H−1(D) is the dual of H1(D), i.e. H−1(D) = (H1(D))′, see [15].2A discussion of other complementary conditions can be found in [2]. The results proved in

this paper are also valid for those conditions.

487 c©2012 American Institute of Mathematical Sciences

http://dx.doi.org/10.3934/ipi.2012.6.487

488 Kyoungsun Kim, Gen Nakamura and Mourad Sini

This problem is naturally related to some studies of classical scattering by pen-etrable obstacles, see [2]. For instance we have the following two links:

1.) The far field operator related to the scattering problem by penetrable obsta-cles has a dense range if and only if the used frequency is not an eigenvalue of theoperator related to (1.2), called a transmission eigenvalue, see [7] and also [4, 17]for more information about the study of these eigenvalues.

2.) The second link is related to the study of the linear sampling method in-troduced in [6] to reconstruct obstacles using the far field map generated by theincident plane waves, see [2, 10] for more details.

One of our motivations is related to this second link. To explain this, let usrecall that the linear sampling method is based on the study of the solvability ofthe so-called far field equation, see section 5. It is shown that with the approximatesolutions of the far field equation, we define Herglotz wave functions which approx-imate a particular solution of (1.2) generated by point sources at the boundary, see[2]. The singular behavior of this last solution plays an important role in justifyingthe linear sampling method. In addition, the precise analysis of this solution pro-vides more qualitative and quantitative information about the unknown scatterer,see [5] where the asymptotic expansion of this solution is used for the identificationof a partially coated dielectric from far field measurements and [1, 13, 14, 16, 21, 22]for results related to impenetrable obstacles.

In the present work, we give a justification of the asymptotic analysis used in [5].A key ingredient in this analysis is the point-wise estimates of the Green function ofthe problem (1.2) similar to those known for classical transmission problems and thiscan be done if we can construct a local parametrix for (1.2). Recently, it is shownin [12] that the interior transmission problem including (1.2) satisfies the Shapiro-Lopatinskii condition. Thus, by the general theory of elliptic boundary value prob-lem for properly elliptic operators, it is possible to construct a local parametrix.There are several arguments for the construction of the local parametrix. They useeither the Calderon projector (see [8], [19], [23]) or the reduction to boundary argu-ment via the Poisson operator for the boundary value problem with inhomogeneousDirichlet data (see [11]). All of these arguments are quite involved and need to haveenough knowledge on the theory of pseudo-differential operators. In this paper,we present a new argument for constructing a local parametrix in a more simplerand concise way. The argument has its root in R. Seeley’s old and famous paper[18]. Furthermore, our new argument can easily give the dominant part of the localparametrix, which not only yields the estimate of the Green function from abovebut also the estimate of the difference of the Green function and the fundamentalsolution from below.

In applying our argument of construction of local parametrix, we observed somedifference between the classical transmission problem and interior transmissionproblem which could have some relation to the discussion in [7] on the ellipticityof the problem (1.2). Indeed, in our construction of the local parametrix for (1.2),see Theorem 4.3 in Section 4.1, its symbols with the same order are determined bysolving some linear algebraic system consisting of 6 equations with 12 unknowns.This system is apparently an under determined system. However, some unknownscan be combined together so that we only have 6 substantial unknowns which canbe uniquely determined by the 6 equations. If we apply the same argument for theconstruction of the local parametrix for the classical transmission problem, we justhave a determined linear algebraic system.

Inverse Problems and Imaging Volume 6, No. 3 (2012), 487–521

The Green function of the interior transmission problem 489

The interior transmission eigenvalue problem is now one of the growing hot topicsin inverse problems. In relation with our work, we would like to draw readersattention to [9] where a local parametrix has been constructed for an associatedboundary value problem for the 4 th order scalar elliptic operator. Although, theneed of the local parametrix is different, this paper and our paper show severalapplications of local parametrix for the interior transmission problem.

We would like to finish this introduction by saying that the construction of theGreen function for the interior transmission problem with its point-wise estimatesis important for itself. In particular, this enables us to use layer potential methodsand provide point-wise estimates for the solutions of (1.2).

The paper is organized as follows. In Section 2, we start by recalling the well-posedness of the interior transmission problem and prove its infinite smoothingproperty in Section 3. In Section 4, we construct the Green function and prove theclassical point-wise estimates. In Section 5, we justify the asymptotic analysis usedin [5].

2. The interior transmission problem is well-posed. The well-posedness ofthis problem has been studied by several authors, see [2]. For technical reasons, the

proof given in [3] allows ρ1 to be in H−1(D) but ρ2 needs to be in L2(D)3. To allow

ρ2 to be in H−1(D), we proceed as follows. We introduce two auxiliary functionsv1 and w1 solutions of the following problems:

(2.1)

∇ ·A∇v1 + κ2nv1 = ρ1 in D∂νAv1 + iλv1 = 0 on ∂D

and

(2.2)

∆w1 + κ2w1 = ρ2 in D∂νw1 + iλw1 = 0 on ∂D,

where λ is a positive real number.Integration by parts, the use of the unique continuation property and the fact

that λ is a positive real number show that the operators related to the last twoproblems have no eigenvalues. Using, for instance, integral equation methods, onecan show that these two problems are well-posed for every κ2. Hence v1 and w1

exist and satisfy the properties:

(2.3) ‖v1‖H1(D) ≤ C‖ρ1‖H−1(D) and ‖v1‖H1(D) ≤ C‖ρ2‖H−1(D)

with some constant C > 0 which is independent of ρ1, ρ2. We set V := v − v1 andW := w − w1, then

(2.4)

∇ ·A∇V + κ2nV = 0 in D

∆W + κ2W = 0 in DV −W = −(v1 − w1) + f on ∂D

∂νAV − ∂νW = −(∂νAv1 − ∂νw1) + g on ∂D,

where −(v1 − w1) + f ∈ H 12 (∂D) and −(∂νAv1 − ∂νw1) + g ∈ H− 1

2 (∂D).With the condition on the coefficients A and n as discussed before and assuming

that κ2 is not a transmission eigenvalue, this last problem is well-posed as it isshown in [3], see also [2] and we have the estimate:

‖W‖H1(D)+‖V ‖H1(D) ≤ C[‖−(v1−w1)+f‖H

12 (∂D)

+‖−(∂νAv1−∂νw1)+g‖H−

12 (∂D)

]

3In that work [3], d = 2. However, the arguments work for any dimension d ≥ 2.



with some constant C > 0 independent of ρi (i = 1, 2), f, g. Combining theseestimates with the ones in (2.3) along with the trace theorems, we deduce thefollowing theorem:

Theorem 2.1. Let D, n and A be C∞ smooth. Assume, in addition, that Aand n satisfy the estimate (1.1) and that κ2 is not a transmission eigenvalue, thenthe interior transmission problem (1.2) has one and only one solution (v, w) in(H1(D))2 and it satisfies the following estimate:(2.5)‖v‖H1(D) + ‖w‖H1(D) ≤ C[‖ρ1‖H−1(D) + ‖ρ2‖H−1(D) + ‖f‖

H12 (∂D)

+ ‖g‖H−

12 (∂D)

]

with some constant C > 0 independent of ρi (i = 1, 2), f, g.

3. The solution-operator of the interior transmission problem is smooth-ing. The object of this section is to prove the following theorem on the regularityproperty of the solution operator of the interior transmission problem (1.2).

Theorem 3.1. Let D, n and A be C∞ smooth. Assume, in addition, that A andn satisfy the estimate (1.1) and that κ2 is not an interior transmission eigenvalue.

Let m ∈ N ∪ 0 and assume also that ρi ∈ Hm−1(D) for m ≥ 1 or ρi ∈ H−1(D)

for m = 0, i = 1, 2, f ∈ Hm+ 12 (∂D) and g ∈ Hm− 1

2 (∂D). Then the solution (v, w)of the interior transmission problem (1.2) is in (Hm+1(Ω))2 and it satisfies (2.5)for m = 0 and the following estimates:

‖v‖Hm+1(D) + ‖w‖Hm+1(D) ≤

(3.1) C[‖ρ1‖Hm−1(D) + ‖ρ2‖Hm−1(D) + ‖f‖Hm+1

2 (∂D)+ ‖g‖

Hm−12 (∂D)

]

for m ≥ 1 with some constant C > 0 independent of ρi (i = 1, 2), f, g.

Proof. Assume that ρi ∈ Hm−1(D), i = 1, 2, f ∈ Hm+ 12 (∂D) and g ∈ Hm− 1

2 (∂D).The aim is to prove that the solution (v, w) of the problem (1.2) is in (Hm+1(D))2.We prove our theorem by induction on m. We already showed the proof for m = 0.We assume that it is true for m− 1 and we will prove it for m.

We use a partition of unity θi (i = 0, · · · , I) subordinated to an open covering

ϑi (i = 0, · · · , I) of D and write v =∑Ii=0 θiv, w =

∑Ii=0 θiv. Here, we take ϑ0 and

ϑi (i = 1, · · · , I) such that ϑ0 ⊂ D, ∂D ⊂ ∪Ii=1ϑi and D ⊂ ∪Ii=0ϑi.From the interior regularity results of the equations ∇·A∇+κ2n and ∆ +κ2, it

is clear that θ0v and θ0w are in Hm+1(D). The main special feature of the interiortransmission problem is the coupling at the boundary. So, we need to consider theterms θiv and θiw, i = 1, 2, · · · , n, with more care.

Let Q := x = (x1, x2, · · · , xd) ∈ Rd : |x1| < 1, |x2| < 1, · · · , |xd| < 1 andQ− := Q ∩ Rd−. Let Φi : Q → ϑi be the corresponding bijection, Φi ∈ C∞(Q)

and Ψi := Φ−1i ∈ C∞(ϑi). We write x = Φi(y) and y = Ψi(x). We set also

(vi, wi) := (θiv, θiw). Then we can show that (vi, wi) satisfies

(3.2)

∇ ·A∇vi + κ2nvi = (∇ ·A∇θi)v + 2A∇θi · ∇v + θiρ1 in D ∩ ϑi

∆wi + κ2wi = (∆θi)w + 2∇θi · ∇w + θiρ2 in D ∩ ϑivi − wi = θif on ∂(D ∩ ϑi)

∂νAvi − ∂νwi = (∂νAθi)v − (∂νθi)w + θig on ∂(D ∩ ϑi).



We set now vi(y) := vi(x) and wi(y) := wi(x). Hence, we show that

(3.3)

∇ · A∇vi + κ2nvi = Fi(y) in Q−∇ · B∇wi + κ2wi = Gi(y) in Q−

vi − wi = Hi(y) on ∂Q−∂νA vi − ∂νwi = Li(y) on ∂Q−.

with new coefficients related to those of (3.2), having the following regularity prop-erties:

(3.4) Fi, Gi ∈ Hm−1(Q−), Hi ∈ Hm+ 12 (∂Q−) and Li ∈ Hm− 1

2 (∂Q−)

Taking the tangential derivatives of vi and wi which we denote by vi,t and wi,t, weobtain

(3.5)

∇ · A∇vi,t + κ2nvi,t = Fi,t(y)−∇ · At∇vi − κ2ntvi in Q−∇ · B∇wi,t + κ2wi,t = Gi,t(y)−∇ · Bt∇wi in Q−

vi,t − wi,t = Hi,t(y) on ∂Q−∂νA vi,t − ∂νB wi,t = Li,t(y)− At∇vi · ν − Bt∇wi · ν on ∂Q−.

Remark that the tangential direction in our case is the (x1, x2, · · · , xd−1) direction.The problem (3.3) is issued from the problem (3.2) by the bijection Ψi. Hence the

well- posedness of the problem (3.2) implies the one of the problem (3.3) . From thehypothesis of our induction proof we deduce that the solution of the problem (3.3)vi and wi are in Hm(Q−). Hence the left hand sides of the equations in (3.5) are

in Hm−2(Q−), Hm− 12 (∂Q−) and Hm− 3

2 (∂Q−). Hence, again from the hypothesisof the induction, we deduce that vi,t and wi,t are in Hm(Q−).

From the first equation of the problem (3.3), we deduce that ∂2xdvi is also in

Hm−1(Q−). Hence ∂xk∂xl vi is in Hm−1(Q−) for k, l = 1, 2, · · · , d. This meansthat vi ∈ Hm+1(Q−). Similarly using the second equation of the problem (3.3), wededuce that wi ∈ Hm+1(Q−).

Coming back to the original coordinates, we have (vi, wi) ∈ (Hm+1(ϑi))2. Fi-

nally, we deduce that (v, w) ∈ (Hm+1(D))2 with the needed estimate.

4. The Green function of the interior transmission problem. Let D, A, nand κ2 satisfy the conditions in Theorem 3.1. The object of this section is to showthat there exists two functions G and H which satisfy the problem:

(4.1)

∇ ·A∇G+ κ2nG = δ(· − z) in D

∆H + κ2H = δ(· − z) in DG−H = 0 on ∂D

∂νAG− ∂νH = 0 on ∂D,

where δ(·, z) is the Dirac delta function with singularity at z. In addition, thefunctions G and H are C∞ smooth for (x, z) ∈ D ×D : x 6= z and satisfy someestimates near the diagonal (x, z) ∈ D × D : x = z. Precisely, we have thefollowing point-wise estimates of G and H:

Theorem 4.1. For d = 2, we have the following estimates:

|G(x, z)| .∣∣∣ ln(|x− z|−1)

∣∣∣, |H(x, z)| .∣∣∣ ln(|x− z|−1)

∣∣∣, x 6= z

and

|∂αxG(x, z)| . |x− z|−|α|, |∂αz G(x, z)| . |x− z|−|α|, x 6= z



while for d ≥ 3, we have:

|G(x, z)| . |x− z|2−d, |H(x, z)| . |x− z|2−d, x 6= z

and

|∂αxG(x, z)| . |x− z|2−d−|α|, |∂αz G(x, z)| . |x− z|2−d−|α|, x 6= z

for every multi-index α = (α1, α2, · · · , αd) ∈ Zd+ := (N ∪ 0)d such that |α| ≥ 1,where we set ∂αx = ∂α1

x1...∂αdxd and x = (x1, x2, · · · , xd). In the above estimates . is

≤ with some multiplication by a general positive constant.

Proof of Theorem 4.1. Using a partition of unity, we need to construct G and Hsatisfying (4.1) locally inside D and near the boundary. For the former case, G andH need to satisfy the equation ∇ · A∇G + κ2n(x)G = δ(x − z) and ∆H + κ2H =δ(x − z). However, the construction of such solution is already known and thesefunctions satisfy the desired estimates for those interior points. Hence, we need toconstruct G and H satisfying (4.1) and prove the estimates in Theorem 4.1 whenx and z are near ∂D. By flattening the boundary of D near a point x0 ∈ ∂D, sayx0 = 0, we need to look for G and H which satisfy the problem:

PG = δ(x− y) in xd < 0

QH = δ(x− y) in xd < 0

G = H on xd = 0

A∇G · ed = B∇H · ed on xd = 0,

where ed = (0, · · · , 0, 1) ∈ Rd. Note that we do the flattening so that the exteriorunit normal to the domain D becomes ed.

Here we used the two differential operators P := ∇ · (A∇) + k and Q := ∇ ·(B∇) + m. Without loss of generality, we can assume that A,B, k and m to beC∞-smooth with bounded derivatives. Also, we abused to use the same notationsA, B, G, H in the above problem.

Let us denote

G =

G+ for yd < xd < 0

G− for xd < yd, H =

H+ for yd < xd < 0

H− for xd < yd.

Then, we have the following problem :

(4.2)

PG± = 0 in ± (xd − yd) > 0 xd < 0

QH± = 0 in ± (xd − yd) > 0 xd < 0

G+ = G−, A∇(G+ −G−) · ed = δ(x′ − y′) on xd = yd

H+ = H−, B∇(H+ −H−) · ed = δ(x′ − y′) on xd = yd

G+ = H+, A∇G+ · ed = B∇H+ · ed on xd = 0,

where x′ = (x1, · · · , xd−1), y′ = (y1, · · · , yd−1).

4.1. Construction of the parametrix for the problem (4.2). We want tofind G and H by using pseudo-differential operators. We recall the definition ofthe symbol class Sm1,0(Rd−1

x′ × Rd−1ξ′ ). Assuming ρ, δ ∈ [0, 1], m ∈ R, we define

Smρ,δ(Rd−1x′ × Rd−1

ξ′ ) consisting of C∞-functions p(x′, ξ′) satisfying

|Dβx′D

αξ′p(x

′, ξ′)| ≤ Cα,β〈ξ′〉m−ρ|α|+δ|β| (x′, ξ′ ∈ Rd−1)



for all α, β, where Dx′ = −i∂x′ = −i(∂x1 , · · · , ∂xd−1), ξ′ = (ξ1, · · · , ξd−1), 〈ξ′〉 =

(1+ |ξ′|2)1/2, see [20]. We also define a family S(m) of symbols depending smoothlyon xd, for xd ≤ 0. Precisely, we have

p(x, ξ′) ∈ S(m)⇔ ∀α ∈ 0, 1, 2, .., 〈ξ′〉−α∂αxdp(x, ξ′)xd≤0 ⊂ Sm1,0(Rd−1

x′ × Rd−1ξ′ )

is bounded.The pseudo-differential operator p(x,Dx′) associated with this symbolis defined by

p(x,Dx′)φ(x) = (2π)d−1

∫R2d−2

ei(x′−y′)·ξ′p(x, ξ)φ(y′)dy′ dξ′

for any φ ∈ C∞0 (Rd−1). p(x, ξ′) is called the total symbol of p(x,Dx′) and it isdenoted by σ(p(x,Dx′)) = p(x, ξ′). We also abuse the notation to write p(x,Dx′) ∈S(m) which indicates that this pseudo-differential operator’s total symbol is inS(m). We further consider an ordinary differential operator

p(x,Dx′ , Dxd) =

m∑j=0

pj(x,Dx′)Dm−jxd

with pj(x,Dx′) ∈ S(j) (0 ≤ j ≤ m), Dxd = −i∂xd and denote the set formed by allsuch ordinary differential operators by S(m) by abusing the notation. Consider theoperator

S(2) 3 P = ∇ · (A(x)∇) + k = −D · (A(x)D) + k,

where D = −i∇. The total symbol σ(P) of P is denoted by p(x, ξ′, Dxd). Weexpand p(x, ξ′, Dxd) into its Taylor series around xd = yd. That is

p(x, ξ′, Dxd) =

∞∑j=0

(j!)−1(xd − yd)j(∂jxdp)(x′, yd, ξ

′, Dxd).

For our further argument, it is convenient to introduce the concept of order:

Definition 4.2. The multiplications by ξj (1 ≤ j ≤ d − 1) and Dxd = −i∂xd areconsidered as operators of order 1, and the multiplication by xd − yd is consideredas an operator of order −1. Further, Dx′ is considered as an operator of order 0.

We will see later in Remark 1 stated after the coming Theorem 4.3 the meaningof the order here.

We further look for the total symbol a =∑∞L=0 a−1−L, a−1−L ∈ S(−1−L), (L ≥

0) of the pseudo-differential operator G+ satisfying

0 = (PG+)ϕ = (2π)−(d−1)

∫Rd−1

eix′·ξ′A(x, ξ′)(Fϕ)(ξ′)dξ′,

whereA(x, ξ′) =

∑α(α!)−1∂αξ′p(x, ξ

′)Dαx′a(x, ξ′) and (Fϕ)(ξ′) =

∫Rd−1 e

−ix′·ξ′ϕ(x′)dx′

is the Fourier transform of ϕ(x′). Then, the total symbol σ(P) of P can be de-composed into pi (i = 0, 1, 2) of order i. That is,

σ(P) = p2 + p1 + p0,



where

p2 = −AddD2xd− 2

d−1∑j=1

Adj(x)ξjDxd −d−1∑i,j=1

Aij(x)ξiξj ,

p1 = i

d∑j=1

∂xjAdj(x)Dxd + i

d∑j=1

d−1∑l=1

∂xjAjlξl, p0 = k(x),

where i =√−1. Now, we arrange A(x, ξ′) in terms of order:

A =

∞∑l=0

A1−l, ordA1−l = 1− l,

where ordA1−l denotes the order of A1−l. For example, A1 and A0 are given as

A1 = p(0)2,0a−1,

A0 =∑

j+k+|α|=1

(xd − yd)jp(α)2,jD

αx′a−1−k + p

(0)1,0a−1

= p(0)2,0a−2 +

∑j+|α|=1

(xd − yd)jp(α)2,jD

αx′a−1 + p

(0)1,0a−1,

where p(α)i,j (i = 0, 1, 2, j ∈ Z+ := N ∪ 0, α ∈ Zd−1

+ ) are defined by p(α)i,j =

∂jxd∂αξ′pi(x

′, yd, ξ′). We also do the same manipulation to the amplitudes b, d and

e of G−, H+ and H−, respectively. For this, we will use B,D and E for b, d ande, respectively, instead of A. That is B, D and E are associated to PG−, QH+

and QH−, respectively. Then, from (4.2), we have the following representations ofamplitudes.

Theorem 4.3. Let us define A1 = A|xd=yd , B1 = B|xd=yd . We also define λ± and

µ± by

λ± = (A1dd)−1

[− i

d−1∑j=1

A1djξj ±

√√√√A1dd

d−1∑i,j=1

A1ijξiξj − (

d−1∑j=1

A1djξj)

2

],

µ± = (B1dd)−1

[− i

d−1∑j=1

B1djξj ±

√√√√B1dd

d−1∑i,j=1

B1ijξiξj − (

d−1∑j=1

B1djξj)

2

],

where the branch of the square roots are taken to satisfy their real parts are positivein each λ± and µ±. Then, the total symbols a, b, d, e of G+, G−, H+, H− aregiven as follows.

a =

∞∑L=0

a−1−L, b =

∞∑L=0

b−1−L, d =

∞∑L=0

d−1−L, e =

∞∑L=0

e−1−L,

where



a−L =

2L−2∑l=0

fLl,1(xd − yd)leλ+xd−λ−yd +

2L−2∑l=0

fLl,2(xd − yd)leλ+xd−µ−yd

+

2L−2∑l=0

fLl,3(xd − yd)leλ−xd−λ−yd ,

b−L =

2L−2∑l=0

fLl,1(xd − yd)leλ+xd−λ−yd +

2L−2∑l=0

fLl,2(xd − yd)leλ+xd−µ−yd

+

2L−2∑l=0

fLl,4(xd − yd)leλ+xd−λ+yd ,(4.3)

d−L =

2L−2∑l=0

fLl,5(xd − yd)leµ+xd−λ−yd +

2L−2∑l=0

fLl,6(xd − yd)leµ+xd−µ−yd

+

2L−2∑l=0

fLl,7(xd − yd)leµ−xd−µ−yd ,

e−L =

2L−2∑l=0

fLl,5(xd − yd)leµ+xd−λ−yd +

2L−2∑l=0

fLl,6(xd − yd)leµ+xd−µ−yd

+

2L−2∑l=0

fLl,8(xd − yd)leµ+xd−µ+yd

and ord fLl,j = l − L for j = 1, · · · , 8.

Remark 1. Looking, for example, at fLl,1(xd − yd)leλ+xd−λ−yd in the first sum of

a−L, we can estimate it as O(|ξ′|−L) as |ξ′| → ∞. Further, if we apply the threetypes of operators given in Definition 4.2 to a−L, · · · , e−L, we can easily see howtheir estimates change. For example, ξja−L = O(|ξ′|−L+1), Dxda−L = O(|ξ′|−L+1)and (xd − yd)a−L = O(|ξ′|−L−1) as |ξ′| → ∞.

Proof. We prove the result by induction. At first, let us find a−1, b−1, d−1 and e−1.A1 = B1 = D1 = E1 = 0 imply

(4.4) p(0)2,0a−1 = p

(0)2,0b−1 = q

(0)2,0d−1 = q

(0)2,0e−1 = 0,

where

p(0)2,0 = A1

dd∂2xd

+ 2i

d−1∑j=1

A1djξj∂xd −

d−1∑i,j=1

A1ijξiξj ,

q(0)2,0 = B1

dd∂2xd

+ 2i

d−1∑j=1

B1djξj∂xd −

d−1∑i,j=1

B1ijξiξj .



Then, the ordinary differential equations (4.4) give

a−1 = C1eλ+xd + C2e

λ−xd ,

b−1 = C3eλ+xd ,

d−1 = C4eµ+xd + C5e

µ−xd ,

e−1 = C6eµ+xd .

Here, we used limxd→−∞

b−1 = limxd→−∞

e−1 = 0. These four functions have to satisfy

the followings:

a−1 − b−1 = 0, ied ·A1

(ξ′

Dxd

)(a−1 − b−1) = e−iy

′·ξ′ on xd = yd,

d−1 − e−1 = 0, ied ·B1

(ξ′

Dxd

)(d−1 − e−1) = e−iy

′·ξ′ on xd = yd,

a−1 − d−1 = 0, ied ·A0

(ξ′

Dxd

)a−1 = ied ·B0

(ξ′

Dxd

)d−1 on xd = 0,

where A0 = A∣∣∣xd=0

and B0 = B∣∣∣xd=0

. Thus, we have the system of equations for

constants Cj , 1 ≤ j ≤ 6.

(4.5)

C1eλ+yd + C2e

λ−yd − C3eλ+yd = 0

λ+C1eλ+yd + λ−C2e

λ−yd − λ+C3eλ+yd = 1

A1dde−iy

′·ξ′

C4eµ+yd + C5e

µ−yd − C6eµ+yd = 0

µ+C4eµ+yd + µ−C5e

µ−yd − µ+C6eµ+yd = 1

B1nne−iy

′·ξ′

C1 + C2 − C4 − C5 = 0

(i∑d−1j=1 A

0djξj + λ+A

0dd)C1 + (i

∑d−1j=1 A

0djξj + λ−A

0dd)C2

−(i∑d−1j=1 B

0djξj + µ+B

0dd)C4 − (i

∑d−1j=1 B

0djξj + µ−B

0dd)C5 = 0.

From the first four equations, we have

C2 = − e−iy′·ξ′

A1dd(λ+ − λ−)

e−λ−yd , C5 = − e−iy′·ξ′

B1dd(µ+ − µ−)

e−µ−yd .

Then, the fifth and sixth equations are changed to

C1 − C4 = −C2 + C5 =: M1e−λ−yd +N1e

−µ−yd ,(4.6)

(i

d−1∑j=1

A0djξj + λ+A

0dd)C1 − (i

d−1∑j=1

B0djξj + µ+B

0dd)C4

= −(i

d−1∑j=1

A0djξj + λ−A

0dd)C2 + (i

d−1∑j=1

B0djξj + µ−B

0dd)C5(4.7)

=: M2e−λ−yd +N2e

−µ−yd .



From (4.6) and (4.7), we have

C4 =

(i

d−1∑j=1

B0djξj + µ+B

0dd)− (i

d−1∑j=1

A0djξj + λ+A

0dd)−1

×[(λ− − λ+)A0

ddC2 + (id−1∑j=1

A0djξj + λ+A

0dd)− (i

d−1∑j=1

B0djξj + µ−B

0dd)C5

]=: M3e

−λ−yd +N3e−µ−yd ,

and so

C1 = C4 +M1e−λ−yd +N1e

−µ−yd

= (M1 +M3)e−λ−yd + (N1 +N3)e−µ−yd .

Thus, we have

C3 = (M1 +M3)e−λ−yd + (N1 +N3)e−µ−yd − e−iy′·ξ′

A1dd(λ+ − λ−)

e−λ+yd ,

C6 = M3e−λ−yd +N3e

−µ−yd − e−iy′·ξ′

B1dd(µ+ − µ−)

e−µ+yd .

Therefore, we have

a−1 = (M1 +M3)eλ+xd−λ−yd + (N1 +N3)eλ+xd−µ−yd

− e−iy′·ξ′

A1dd(λ+ − λ−)

eλ−xd−λ−yd ,

b−1 = (M1 +M3)eλ+xd−λ−yd + (N1 +N3)eλ+xd−µ−yd


A1dd(λ+ − λ−)

eλ+xd−λ+yd ,

d−1 = M3eµ+xd−λ−yd +N3e

µ+xd−µ−yd


B1dd(µ+ − µ−)

eµ−xd−µ−yd ,

e−1 = M3eµ+xd−λ−yd +N3e

µ+xd−µ−yd − e−iy′·ξ′

B1dd(µ+ − µ−)

eµ+xd−µ+yd ,

where ordMj = ordNj = −1 for j = 1, 3. This shows (4.3) is true for L = 1. Now,let us show the case for L = 2. From A0 = B0 = D0 = E0 = 0, we have

p2,0a−2 = −∑

j+|α|=1

(xd − yd)jp(α)2,jDx′a−1 − p(0)

1,0a−1 =: Θa,

p2,0b−2 = −∑

j+|α|=1

(xd − yd)jp(α)2,jDx′b−1 − p(0)

1,0b−1 =: Θb,

q2,0d−2 = −∑

j+|α|=1

(xd − yd)jq(α)2,j Dx′d−1 − q(0)

1,0d−1 =: Θd,

q2,0e−2 = −∑

j+|α|=1

(xd − yd)jq(α)2,j Dx′e−1 − q(0)

1,0e−1 =: Θe.



From the forms of a−1, b−1, d−1, e−1, we have

Θa =

1∑l=0

El,1(xd − yd)leλ+xd−λ−yd +

1∑l=0

El,2(xd − yd)leλ+xd−µ−yd

+

1∑l=0

El,3(xd − yd)leλ−xd−λ−yd ,

Θb =

1∑l=0


1∑l=0


+

1∑l=0

El,4(xd − yd)leλ+xd−λ+yd ,

Θd =

1∑l=0

El,5(xd − yd)leµ+xd−λ−yd +

1∑l=0

El,6(xd − yd)leµ+xd−µ−yd

+

1∑l=0

El,7(xd − yd)leµ−xd−µ−yd ,

Θe =

1∑l=0

El,5(xd − yd)leµ+xd−λ−yd +

1∑l=0

El,6(xd − yd)leµ+xd−µ−yd

+

1∑l=0

El,8(xd − yd)leµ+xd−µ+yd ,

where ordEl,j = i for l = 0, 1, j = 1, · · · , 8. At first, let us consider a−2. From theform of Θa, we know

a−2 =

3∑j=1

aj−2, aj−2 =

2∑l=0

Fl,j(xd − yd)leβjxd−δjyd for j = 1, 2, 3,

where β1 = β2 = λ+, β3 = δ1 = δ3 = λ− and δ2 = µ− and satisfies

p2,0aj−2 =

1∑l=0

El,j(xd − yd)leβjxd−δjyd .

From the above equation, we can write Fl,j (l = 1, 2) in terms of El,j (l = 0, 1, 2).

F2,j =E1,j

4(βjA1dd + i

∑d−1j=1 A

1djξj)

F1,j =E0,j − 2A1

ddF2,j

2(βjA1dd + i

∑d−1j=1 A

1djξj)

(4.8)



where ordFl,j = l − 2 (l = 1, 2). We do the same for b−2, d−2 and e−2. Then, wehave

a−2

= 2∑l=1

Fl,1(xd − yd)l + C1

eλ+xd−λ−yd +

2∑l=1


eλ+xd−µ−yd

+ 2∑l=1


eλ−xd−λ−yd

b−2

= 2∑l=1


eλ+xd−λ−yd +

2∑l=1


eλ+xd−µ−yd

+ 2∑l=1


eλ+xd−λ+yd

d−2

= 2∑l=1


eµ+xd−λ−yd +

2∑l=1


eµ+xd−µ−yd

+ 2∑l=1


eµ−xd−µ−yd

e−2

= 2∑l=1


eµ+xd−λ−yd +

2∑l=1


eµ+xd−µ−yd

+ 2∑l=1


eµ+xd−µ+yd

where Fl,j (l = 1, 2, j = 1, · · · , 8) satisfies (4.8), ordFl,j = l − 2 and Cj , j =1, · · · , 12, are constants with respect to xd which we are going to seek. Thesefunctions have relations:

a−2 − b−2 = 0, ied ·A1

(ξ′

Dxd

)(a−2 − b−2) = 0 on xd = yd,

d−2 − e−2 = 0, ied ·B1

(ξ′

Dxd

)(d−2 − e−2) = 0 on xd = yd,

a−2 − d−2 = 0, ied ·A0

(ξ′

Dxd

)a−2 = ied ·B0

(ξ′

Dxd

)d−2 on xd = 0.



Thus, we have the system of equations for constants Cj , j = 1, · · · , 12.

(C1 − C4)e(λ+−λ−)yd + (C2 − C5)e(λ+−µ−)yd + C3 − C6 = 0,

λ+(C1 − C4)e(λ+−λ−)yd + λ+(C2 − C5)e(λ+−µ−)yd + λ−C3 − λ+C6

= −F1,3 + F1,4,

(C7 − C10)e(µ+−λ−)yd + (C8 − C11)e(µ+−µ−)yd + C9 − C12 = 0,

µ+(C7 − C10)e(µ+−λ−)yd + µ+(C8 − C11)e(µ+−µ−)yd + µ−C9 − µ+C12

= −F1,7 + F1,8,

(C1 + C3 − C7)e−λ−yd + (C2 − C8 − C9)e−µ−yd

= −2∑l=1

(Fl,1 + Fl,3 − Fl,5)(−yd)le−λ−yd −2∑l=1

(Fl,2 − Fl,6 − Fl,7)(−yd)le−µ−yd


−µ−yd ,

(i

d−1∑j=1

A0djξj + λ+A

0dd)(C1e

−λ−yd + C2e−µ−yd)

+ (i

d−1∑j=1

A0djξj + λ−A

0dd)C3e

−λ−yd

− (i

d−1∑j=1

B0djξj + µ+B

0dd)(C7e


− (i

d−1∑j=1

B0djξj + µ−B

0dd)C9e

−µ−yd

= −(i

d−1∑j=1

A0djξj)[

2∑l=1

(Fl,1 + Fl,3)(−yd)le−λ−yd +

2∑l=1

Fl,2(−yd)le−µ−yd ]

−A0dd[

2∑l=1

l(Fl,1 + Fl,3)(−yd)l−1 +2∑l=1

(λ+Fl,1 + λ−Fl,3)(−yd)le−λ−yd

+ 2∑l=1

lFl,2(−yd)l−1 + λ+

2∑l=1

Fl,2(−yd)le−µ−yd ]

+ (i

d−1∑j=1

B0djξj)[

2∑l=1

Fl,5(−yd)le−λ−yd +

2∑l=1

(Fl,6 + Fl,7)(−yd)le−µ−yd ]

+B0dd[

2∑l=1

lFl,5(−yd)l−1 + µ+

2∑l=1

Fl,5(−yd)le−λ−yd

+ 2∑l=1

l(Fl,6 + Fl,7)(−yd)l−1 +

2∑l=1

(µ+Fl,6 + µ−Fl,7)(−yd)le−µ−yd ]


−µ−yd ,

(4.9)

where ordM1 = ordN1 = −2 and ordM2 = ordN2 = −1.



Obviously, this system is under determined in contrast to the one we can getfor the usual transmission problem. However, the special structure of the systemenables us to solve it. From the first four equations, we have

C3 =F1,3 − F1,4

λ+ − λ−, C9 =

F1,7 − F1,8

µ+ − µ−,

where ordC3 = ordC9 = −2. Then, the fifth and sixth equation become

(4.10)

C1e−λ−yd + C2e

−µ−yd − C7e−λ−yd − C8e

−µ−yd

= (M1 − C3)e−λ−yd + (N1 + C9)e−µ−yd


−µ−yd ,

(i∑d−1j=1 A

0djξj + λ+A

0dd)(C1e


−(i∑d−1j=1 B

0djξj + µ+B

0dd)(C7e


= M2 − (i∑d−1j=1 A

0djξj + λ−A

0dd)C3e−λ−yd+

N2 + (i∑d−1j=1 B

0djξj + µ−B

0dd)C9e−µ−yd


−µ−yd ,

where ordM3 = ordN3 = −2 and ordM4 = ordN4 = −1. Then, we have

C7e−λ−yd + C8e

−µ−yd

=

(i

d−1∑j=1

B0djξj + µ+B

0dd)− (i

d−1∑j=1

A0djξj + λ+A

0dd)−1

×[

(i

d−1∑j=1

A0djξj + λ+A

0dd)M3 −M4

e−λ−yd+

(i

d−1∑j=1

A0djξj + λ+A

0dd)N3 −N4

e−µ−yd

]=: M5e

−λ−yd +N5e−µ−yd ,

where ordM5 = ordN5 = −2. From this and (4.10), we have

C1e−λ−yd + C2e

−µ−yd = (C7 +M3)e−λ−yd + (C8 +N3)e−µ−yd

= (M3 +M5)e−λ−yd + (N3 +N5)e−µ−yd .



From what we have obtained, we have:

(i) C4e−λ−yd + C5e

−µ−yd + C6e−λ+yd

= e−λ+yd [C4e(λ+−λ−)yd + C5e

(λ+−µ−)yd + C6]

= e−λ+yd [C1e(λ+−λ−)yd + C2e

(λ+−µ−)yd + C3]

= (M3 +M5)e−λ−yd + (N3 +N5)e−µ−yd +F1,3 − F1,4

λ+ − λ−e−λ+yd ,

(ii) C10e−λ−yd + C11e

−µ−yd + C12e−µ+yd

= e−µ+yd [C10e(µ+−λ−)yd + C11e

(µ+−µ−)yd + C12]

= e−µ+yd [C7e(µ+−λ−)yd + C8e

(µ+−µ−)yd + C9]

= M5e−λ−yd +N5e

−µ−yd +F1,7 − F1,8

µ+ − µ−e−µ+yd .

It is very important to remark here that we have observed that C1, C3 and the fourlinear combinations of the rest of Cj ’s given by C7e

−λ−yd +C5e−µ−yd , C1e

−λ−yd +C2e

−µ−yd , C4e−λ−yd+C5e

−µ−yd+C6e−λ+yd and C10e

−λ−yd+C11e−µ−yd+C12e

−µ+yd

are the substantial unknowns for the six equations (4.9). Hence, a−2, b−2, d−2, e−2

are uniquely determined and they are given as follows:

a−2 = 2∑l=1

Fl,1(xd − yd)l + (M3 +M5)eλ+xd−λ−yd

+ 2∑l=1

Fl,2(xd − yd)l + (N3 +N5)eλ+xd−µ−yd

+ 2∑

l=1

Fl,3(xd − yd)l +F1,3 − F1,4

λ+ − λ−


b−2 = 2∑l=1

Fl,1(xd − yd)l + (M3 +M5)eλ+xd−λ−yd

+ 2∑l=1

Fl,2(xd − yd)l + (N3 +N5)eλ+xd−µ−yd

+ 2∑l=1

Fl,4(xd − yd)l +F1,3 − F1,4

λ+ − λ−

eλ+xd−λ+yd ,

d−2 = 2∑l=1

Fl,5(xd − yd)l +M5

eµ+xd−λ−yd

+ 2∑l=1

Fl,6(xd − yd)l +N5

eµ+xd−µ−yd

+ 2∑l=1

Fl,7(xd − yd)l +F1,7 − F1,8

µ+ − µ−

eµ−xd−µ−yd



and

e−2 = 2∑l=1

Fl,5(xd − yd)l +M5

eµ+xd−λ−yd

+ 2∑l=1

Fl,6(xd − yd)l +N5

eµ+xd−µ−yd

+ 2∑l=1

Fl,8(xd − yd)l +F1,7 − F1,8

µ+ − µ−

eµ+xd−µ+yd .

These are the forms that we want for L = 2 in (4.3). Suppose that (4.3) is true forL ≥ 2. Let us show that it is true for L+ 1. Note that

A1−L = p(0)2,0a−1−L +

∑j+k+|α|=Lk≤L−1

(α!)−1(j!)−1(xd − yd)jp(α)2,jD

αx′a−1−k

+∑

j+k+|α|=L−1

(α!)−1(j!)−1(xd − yd)jp(α)1,jD

αx′a−1−k

+∑

j+k=L−2

(j!)−1(xd − yd)jp(0)0,ja−1−k.

Thus, A1−L = 0 implies that

p(0)2,0a−1−L = −

L−1∑k=0

[ ∑j+k+|α|=L

(α!)−1(j!)−1(xd − yd)jp(α)2,jD

αx′a−1−k

+∑

j+k+|α|=L−1

(α!)−1(j!)−1(xd − yd)jp(α)1,jD

αx′a−1−k

]−

∑j+k=L−2

(j!)−1(xd − yd)jp(0)0,ja−1−k =: Θ1−L

a .

From the form of a−1−k (k = 0, 1, · · · , L − 1), we see that Θ1−La is the sum of

eλ+xd−λ−yd , eλ+xd−µ−yd and eλ−xd−λ−yd with (2L− 1)-th degree’s polynomials inxd − yd as the coefficients of exponential. That is,

Θ1−La =

2L−1∑l=0


2L−1∑l=0


+

2L−1∑l=0

El,3(xd − yd)leλ−xd−λ−yd ,

where ordEl,j = l + 1− L for j = 1, 2, 3. So, we can see that

a−1−L =

3∑j=1

aj−2, aj−2 =

2L∑l=0

Fl,j(xd − yd)leβjxd−δjyd , j = 1, 2, 3,

where β1 = β2 = λ+, β3 = δ1 = δ3 = λ− and δ2 = µ− and satisfies

p(0)2,0a

j−1−L =

2L−1∑l=0

El,j(xd − yd)leβjxd−δjyd .



Thus, we can write Fl,j in terms of El,j .

F2L,j =E2L−1,j

4L(βjA1dd + i

∑d−1j=1 A

1djξj)

Fl+1,j =El,j − (l + 2)(l + 1)A1

ddFl+2,j

2(l + 1)(βjA1dd + i

∑d−1j=1 A

1djξj)

(4.11)

where ordFl+1,j = l − L for l = 0, 1, · · · , 2L− 1. Therefore, we have

a−1−L = 2L∑l=1

fL+1l,1 (xd − yd)l + C1

eλ+xd−λ−yd

+ 2L∑l=1

fL+1l,2 (xd − yd)l + C2

eλ+xd−µ−yd +

2L∑l=1

fL+1l,3 (xd − yd)l + C3


where fL+1l,j = Fl,j for l = 1, · · · , 2L, j = 1, 2, 3. By the same method, we can get

the similar expressions for b−1−L, d−1−L and e−1−L. We also note here that a−land b−l (l = 1, · · · , L) have the same coefficients for eλ+xd−λ−yd . Thus eλ+xd−λ−yd

in a−1−L and b−1−L have the same coefficients possibly except 0-th degree. It isthe same for eλ+xd−µ−yd in a−1−L and b−1−L, for eµ+xd−λ−yd in d−1−L and e−1−L,and for eµ+xd−µ−yd in d−1−L and e−1−L. Therefore, we have

b−1−L = 2L∑l=1

fL+1l,1 (xd − yd)l + C4

eλ+xd−λ−yd

+ 2L∑l=1

fL+1l,2 (xd − yd)l + C5

eλ+xd−µ−yd

+ 2L∑l=1

fL+1l,4 (xd − yd)l + C6

eλ+xd−λ+yd ,

d−1−L = 2L∑l=1

fL+1l,5 (xd − yd)l + C7

eµ+xd−λ−yd

+ 2L∑l=1

fL+1l,6 (xd − yd)l + C8

eµ+xd−µ−yd

+ 2L∑l=1

fL+1l,7 (xd − yd)l + C9

eµ−xd−µ−yd ,

e−1−L = 2L∑l=1

fL+1l,5 (xd − yd)l + C10

eµ+xd−λ−yd

+ 2L∑l=1

fL+1l,6 (xd − yd)l + C11

eµ+xd−µ−yd

+ 2L∑l=1

fL+1l,8 (xd − yd)l + C12

eµ+xd−µ+yd ,



where fL+1l,j are decided from Ei,j (l = 1, · · · , 2L, i = 0, 1, · · · , 2L−1, j = 1, · · · , 8)

in Θm, m = b, d, e like (4.11). To determine constants Cj , j = 1, · · · , 12, we usethe boundary conditions on xd = yd and xd = 0. That is,

a−1−L − b−1−L = 0, ied ·A1

(ξ′

Dxd

)(a−1−L − b−1−L) = 0 on xd = yd,

d−1−L − e−1−L = 0, ied ·B1

(ξ′

Dxd

)(d−1−L − e−1−L) = 0 on xd = yd,

a−1−L − d−1−L = 0, ied ·A0

(ξ′

Dxd

)a−1−L = ied ·B0

(ξ′

Dxd

)d−1−L on xd = 0.

These equations make the system of equations for Cj , j = 1, · · · , 12, and it is solvedthrough the same process as we did when L = 2. Then, we have the following results:

a−1−L = 2L∑l=1

fL+1l,1 (xd − yd)l + (M3 +M5)

eλ+xd−λ−yd

+ 2L∑l=1

fL+1l,2 (xd − yd)l + (N3 +N5)

eλ+xd−µ−yd

+ 2L∑l=1

fL+1l,3 (xd − yd)l +

fL+11,3 − f

L+11,4

λ+ − λ−

eλ−xd−λ−yd

b−1−L = 2L∑l=1

fL+1l,1 (xd − yd)l + (M3 +M5)

eλ+xd−λ−yd

+ 2L∑l=1

fL+1l,2 (xd − yd)l + (N3 +N5)

eλ+xd−µ−yd

+ 2L∑l=1


fL+11,3 − f

L+11,4

λ+ − λ−

eλ+xd−λ+yd

d−1−L = 2L∑l=1

fL+1l,5 (xd − yd)l +M5

eµ+xd−λ−yd

+ 2L∑l=1

fL+1l,6 (xd − yd)l +N5

eµ+xd−µ−yd

+ 2L∑l=1


fL+11,7 − f

L+11,8

µ+ − µ−

eµ−xd−µ−yd



and

e−1−L = 2L∑l=1

fL+1l,5 (xd − yd)l +M5

eµ+xd−λ−yd

+ 2L∑l=1

fL+1l,6 (xd − yd)l +N5

eµ+xd−µ−yd

+ 2L∑l=1


fL+11,7 − f

L+11,8

µ+ − µ−

eµ+xd−µ+yd

where ordMi = ordNi = −1−L for i = 3, 5 and ordfL+1l,j = l−1−L for j = 1, · · · , 8.

We have shown (4.3) for L+ 1. This completes the proof.

Remark 2. We conclude this section by pointing out the estimates for PG and QHif we take aM :=

∑ML=0 a−1−L instead of a =

∑∞L=0 a−1−L in PG+ and the same

for b, d and e. For these truncated aM , bM , dM and eM , we write the associatedG,H by GM and HM . Since a−1, b−1, d−1 and d−1 take care to produce δ(· − z)both in PGM and QHM , and for example aM =

∑ML=0 a−1−L enables to have

Al = 0 (−M ≤ l ≤ 1), it is not hard to see that (PGM−δ)(·, z) and (QHM−δ)(·, z)are in Cm(Rd−) with bounded derivatives uniformly with respect to y ∈ Rd− ifM > m+ d− 2.

4.2. Estimate of the Schwartz Kernel. The following proposition is given in[20].

Proposition 1. Let d ∈ N and suppose that x, ξ ∈ Rd. If p(x, ξ) ∈ Sm1,0(Rdx × Rdξ),then the Schwartz kernel K of p(x,D) satisfies estimates

(4.12) |Dβx,yK(x, y)| ≤ C|x− y|−m−d−|β|

provided m+ |β| > −d, and

(4.13) |Dβx,yK(x, y)| ≤ C| log |x− y||

provided m+ |β| = −d for some constant C > 0.

Now, let us consider the total symbols a, b, d, e computed in the previous sectionand we denote p = p(x′, xd, ξ

′) for them. From the form of p, we can see that

p(x′, xd, ξ′) ∈ S(−1)

and

(4.14) ∂αxd∂βx′∂

γξ′p(x

′, xd, ξ′) = O(〈ξ′〉α−1−|γ|e−c|xd||ξ

′|)

for some c > 0 and for arbitrary α ∈ R, β, γ ∈ Zd−1+ uniformly with respect to x′

and xd. We want to show the following:

(4.15) K(x) := (2π)−d+1

∫Rd−1

eix′·ξ′p(x′, xd, ξ

′)dξ′ .

|x|−(d−2) for d ≥ 3,∣∣∣ log |x|

∣∣∣ for n = 2

and

(4.16)∣∣∣∂βxK(x)

∣∣∣ . |x|−(d−2)−|β|.

We start by the case β = 0. In this case we have two situations.



Case 1. d ≥ 3.We want to see K(x) satisfies |K(x)| . |x|−(d−2). To apply Proposition 1, put

m = −1, d = d− 1. Then, m > −d⇔ −1 > −(d− 1)⇔ d > 2⇔ d ≥ 3. Thus, theresult of Proposition 1 gives us

|K(x)| . |x′|−(−1)−(n−1) = |x′|−d+2.

If we can show that |K(x)| . |xd|−(d−2), then it is done.

I(xd) :=

∫Rn−1

〈ξ′〉−1e−c|xd||ξ′|dξ′

=

∫Sn−2

dw

∫ ∞0

(1 + r2)−1/2e−c|xd|rrn−2dr

.∫ ∞

0

(1 + r)−1e−c|xd|rrd−2dr =: J(xd).

By substituting s = |xd|r, we see that

J(xd) =

∫ ∞0

e−cs(1 + |xd|−1s)−1(|xd|−1s)d−2|xd|−1ds

= |xd|−(d−2)

∫ ∞0

e−cs(|xd|+ s)−1sd−2ds

. |xd|−(d−2)

∫ ∞0

e−css−1sd−2ds

. |xd|−(d−2) (since d ≥ 3).

Case 2. d=2By Proposition 1 (m = −1, d = 1), we see that

|K(x)| .∣∣∣ log |x′|

∣∣∣.By the property of p, we have

|K(x)| .∫Re−c|xd||ξ1|(1 + |ξ1|)−1dξ1.

The following computation of integration gives us the estimation we want. Fory > 0, we have∫

Re−y|ξ|(1 + |ξ|)−1dξ = 2

∫ ∞0

e−yξ(1 + ξ)−1dξ

= 2y−1

∫ ∞0

e−η(1 + y−1η)−1dη

= 2

∫ ∞0

e−η(y + η)−1dη

= 2(∫ 1

0

+

∫ ∞1

)e−η(y + η)−1dη.

It is easy to see that the integration on (1,∞) is finite. Since∫ 1

0

e−η(y + η)−1dη ≤∫ 1

0

(y + η)−1dη = log(y + 1)− log y .∣∣∣ log |y|

∣∣∣,we have |K(x)| .

∣∣∣ log |x2|∣∣∣. Hence, we have |K(x)| .

∣∣∣ log |x|∣∣∣. Therefore, we have

(4.15).



To deal with the case where β 6= 0, we write

∂xβK(x) =

∫eix′·ξ′pβ(x′, xd, ξ

′)dξ′.

Due to (4.14), we have pβ ∈ S(|β| − 1). Repeating the same steps for m = |β| − 1,we obtain (4.16).

4.3. End of the proof of Theorem 4.1. Let z ∈ D and write the problem (4.1)in the form

P E(·, z) = δ(· − z)(1, 1)t

with E(·, z) := (G(·, z), H(·, z))t satisfying the boundary conditions of (4.1) where

P :=

(∇ ·A∇+ k2n 0

0 ∆ + k2

).

Further, let ΩjJj=0 be an open covering of D such that Ω0 ⊂ D, Ωj ∩∂D 6= ∅ (1 ≤j ≤ J). We have shown previously the existence of local parametrices Ej(·, z) satis-fying (4.1) in each Ωj . Of course, E0(·, z) does not have to satisfy the boundary con-ditions in (4.1). We denote by φjJj=0 ⊂ C∞0 (Rn) a partition of unity subordinated

to ΩjJj=0 and let ψj ∈ C∞0 (Ωj) (0 ≤ j ≤ J) to satisfy ψj = 1 on supp φj (0 ≤j ≤ J). Then, it is easy to see that E′(x, z) =

∑Jj=0 ψj(x)Ej(x, z)φj(z) gives a

parametrix for (4.1).The Green function E(x, z) of (4.1) is given by

E(x, z) = E′(x, z) +R(x, z),

where R(x, z) is the solution of

(4.17) PR(·, z) ∈ Cm(D) with the boundary conditions of (4.1),

where m ∈ N describes the smoothness of PEj(·, z) − δ(· − z)(1, 1)t (0 ≤ j ≤ J).From Remark 2, we know that ‖P R(·, z)‖Cm(D) is uniformly bounded for z ∈D. The infinitely smoothing property of the problem (4.1), see section 3, and theestimates of E′j(·, z) give the desired estimate for the Green function E(·, z).

5. Application to the identification of a partially coated dielectric mediumfrom far field measurements. In this section, we use the same notations as in[5] whenever it is possible. Let the bounded domain D ⊂ R2 with C∞-boundarysuch that the exterior domain De := R2 \ D is connected. We denote by ν theoutward unit normal to ∂D. We split the boundary into two open disjoint parts∂Dt and ∂Dc, i.e. ∂D = ∂Dt ∪ ∂Dc, where ∂Dt corresponds to the uncoated partand ∂Dc corresponds to the coated part. We take a function η defined on ∂Dc andC∞-smooth and a function a defined in D and C∞-smooth also. We are interestedin the following transmission problem.

(5.1)

∇·a(x)∇V + κ2 V = 0 in D∆U + κ2 U = 0 in De

V − U = 0 on ∂Dt

V − U = −iη(x)∂νU on ∂Dc

∂νaV − ∂νU = 0 on ∂DU = Us + U i

limr→∞√r

(∂Us

∂r− iκUs

)= 0,



r = |x|, where Us is the scattered field and U i is the plane wave incident field givenby U i := eikx·d, d ∈ Ω := x : |x| = 1,

∂νaV (x) := ν(x) · a(x)∇V (x), x ∈ Γ,

and the radiation condition in (5.1) holds uniformly with respect to x = x/|x|.In the following we assume, in addition, that a(x) ≥ a0 > 0 for x ∈ D and η(x) >η0 > 0 for x ∈ ∂Dc.

Let H1(D), H1loc(De) denote the usual Sobolev spaces and H

12 (∂D) the corre-

sponding trace space. For ∂Dc ⊂ ∂D we define

H12 (∂Dc) := u|∂Dc : u ∈ H 1

2 (∂D)H

12 (∂Dc) := u|∂Dc : u ∈ H 1

2 (∂D) with suppu ⊆ ∂Dc,

and denote by H−12 (∂Dc) and H−

12 (∂Dc) the dual spaces

(H

12 (∂Dc)

)′and(

H12 (∂Dc)

)′, respectively, with L2 as a pivot space (for details, see [15]). We also

define the Hilbert space4

H1(D, ∂Dc) :=u ∈ H1(D) such that ∂νu ∈ L2(∂Dc)

equipped with the usual graph norm

‖u‖2H1(D,∂Dc):= ‖u‖2H1(D) + ‖∂νu‖2L2(∂Dc)

.

The forward scattering problem reads: Given D, a, η and U i as above, findV ∈ H1(D) and U ∈ H1

loc(De) that satisfy (5.1) where the boundary conditions areassumed in the sense of the trace operator. In [2] it is shown that there exists aunique scattered field Us have the asymptotic behavior

(5.2) Us(x) =eiκr√rU∞(x, d) +O(r−3/2), r →∞

where U∞(x, d) is the far field pattern of the radiating solution Us which dependson the incident direction d ∈ Ω and the observation direction x ∈ Ω, where Ω is theunit circle.

The inverse scattering problem we are concernd with is to determine the shapeD, identify the uncoated part ∂Dt and coated part ∂Dc and reconstruct η anda|∂Dt from a knowledge of the far field pattern U∞(x, d) of the scattered field Us

for x, d ∈ Ω.

5.1. Solution of the inverse problem.

4If u ∈ H1(Ω), then ∂νu ∈ H−12 (∂D) (= (H

12 (∂D))′). Since every function in H

12 (∂Dc) can

be extended, by zero in ∂D \ ∂Dc, as a function in H12 (∂D), then ∂νu|∂Dc has a meaning as an

element of H−12 (∂Dc). Obviously L2(∂Dc) ⊂ H−

12 (∂Dc), then the space H1(D, ∂Dc) makes a

sense.



5.1.1. The far field equation and statement of the results. We now define the farfield operator F : L2(Ω)→ L2(Ω) by

(5.3) Fg(x) :=

∫Ω

U∞(x, d)g(d) ds(d).

The solution of the inverse problem via the linear sampling method, as describedin [6], is based on solving the far field equation

(5.4) (Fg)(x) = G(j)∞ (x, z), g ∈ L2(Ω), z ∈ R2, j = 1, 2

where G(j)∞ (x, z) is the far field pattern of

G(j)(x, z) = ∂xjΦ(x, z), j = 1, 2, x = (x1, x2)

with Φ(x, z) being the fundamental solution i4H

(1)0 (κ|x−z|) to the Helmholtz equa-

tion in R2 with H(1)0 being the Hankel function of the first kind of order zero.

To study the far field equation (5.4) we introduce the interior transmission prob-lem

∇·a∇V (j)z + κ2 V

(j)z = 0 in D

∆W(j)z + κ2W

(j)z = 0 in D

V(j)z − (W

(j)z +G(j)(·, z)) = 0 on ∂Dt(5.5)

V(j)z − (W

(j)z +G(j)(·, z)) = −iη ∂ν(W (j)

z +G(j)(·, z)) on ∂Dc

∂νaV(j)z − ∂ν(W (j)

z +G(j)(·, z)) = 0 on ∂D

for z ∈ D and j = 1, 2. The values of κ for which (5.5) is not uniquely solvable arecalled transmission eigenvalues. In [2], section 8.4, it is shown that assuming thatthere exists a constant γ > 0 such that a ≥ 1 + γ or 0 < a0 < a < 1− γ in D andκ is not a transmission eigenvalue, then there exists a unique solution of (5.5) such

that V(j)z ∈ H1(D) and W

(j)z ∈ H1(D, ∂Dc).

We recall that the Herglotz wave function with density g ∈ L2(Ω) is an entiresolution of the Helmholtz equation defined by

(5.6) vg(x) =

∫Ω

eiκx·dg(d)ds(d), x ∈ R2.

The following result, which we need to justify in this section, has been stated andused in [5].

Theorem 5.1. Assume that κ is not a transmission eigenvalue, a, η and ∂D satisfythe assumptions stated above and there exists constants a1 and γ > 0 such that

a ≥ 1 + γ or 0 < a1 < a < 1 − γ in D. Furthermore let W(j)z , V

(j)z be the solution

of (5.5). ThenI. Reconstruction of the shape of the obstacle:

1. For z ∈ D and a given ε > 0 there exists a g(j)z,ε ∈ L2(Ω) such that

‖Fg(j)z,ε −G(j)

∞ (· , z)‖L2(Ω) < ε

and the corresponding Herglotz function vg(j)z,ε

converges in H1(D, ∂Dc) to W(j)z

as ε→ 0.



2. For a fixed ε > 0, we have that

limz→z0

‖vg(j)z,ε‖H1(D,∂Dc) =∞ and lim

z→z0‖g(j)z,ε‖L2(Ω) →∞

for every z0 ∈ ∂D.

3. For z ∈ R2 \D and a given ε > 0, every g(j)z,ε ∈ L2(Ω) that satisfies

‖Fg(j)z,ε −G(j)

∞ (· , z)‖L2(Ω) < ε

is such that

limε→0‖vg(j)z,ε‖H1(D,∂Dc) =∞ and lim

ε→0‖g(j)z,ε‖L2(Ω) →∞.

II. Distinguishing the coated part from the uncoated part of the obstacle. Thecoated portion ∂Dc can be distinguished from ∂Dt using the following criteria, forz ∈ D

(5.7) limz→z0

limε→0

|Im(vgj,εz (z))|| ln |(z − z0) · ν||s

=

0 if z0 ∈ ∂Dt

∞ if z0 ∈ ∂Dc,

for every s ∈ (0, 1).III. Reconstruction of the boundary terms a and η on ∂Dt and ∂Dc, respectively:

(5.8) η(z0) = limz→z0

−νj(z0) ln |(z − z0) · ν(z0)|π limε→0 Im(v

g(j)z,ε

(z)), for z0 ∈ ∂Dc

whereas the index of refraction a on the uncoated part ∂Dt can be computed by

(5.9)a(z0)− 1

a(z0) + 1=

νj(z0)

4π limz→z0 limε→0 Re(vg(j)z,ε

(z))(z − z0) · ν(z0), z0 ∈ ∂Dt

where the normal vector ν(z0) is reconstructed by the following formula(5.10)

ν(z0) = limε→0

limz→z0

±Re(vg(1)z,ε

(z))

Re(vg(2)z,ε

(z))

√√√√√ 1

1 +

[Re(v

g(1)z,ε

(z))

Re(vg(2)z,ε

(z))

]2 ,±√√√√√ 1

1 +

[Re(v

g(1)z,ε

(z))

Re(vg(2)z,ε

(z))

]2

and the sign is chosen so that ν(z0) is oriented outside D.

5.1.2. Proof of Theorem 5.1. As it is explained in [5], the proof of part I of thetheorem can be done exactly in the same way as in Section 8.5 in [2]. In particular,

for z ∈ D, vg(j)z,ε

converges in H1(D, ∂Dc) to W(j)z as ε → 0. Therefore, since v

g(j)z,ε

and W(j)z satisfy the same Helmholtz equation in D, then interior estimates imply

that vg(j)z,ε

(z) converges to W(j)z (z) pointwisely for every z ∈ D.

In addition, the justification of part II and part III, is based on the study of the

asymptotic behavior of W(j)z for z near z0 ∈ ∂D. Precisely, we first show that the

dominant part of W(j)z with respect to the singularity created by the fundamental

solution near ∂D is a function satisfying a problem stated in the half plane wherea(z0) and η(z0) are the only ’unknown’ parameters. Then, as a second step, theexplicit form of the solution of this problem in the half plane derive explicit formulasto compute η(z0) if z0 ∈ ∂Dc and a(z0) if z0 ∈ ∂Dt. This second step has beenalready shown in [5], so we omit the calculations. The main result of this section isto justify the first step, i.e. Theorem 5.2.



For a sake of clarity, we state the problems satisfied by wjz and W jz . Let us fix j =

2. The other case is similar. From the assumption that ∂D is a C∞-closed curve, wehave that for every point z0 ∈ ∂D there exists a rigid transformation of coordinatesunder which z0 is transformed to 0 = (0, 0) and a function f ∈ C∞(−p, p) such thatf(0) = f ′(0) = 0 and D ∩ B(0, p) = (x, y) ∈ B(0, p); y < f(x). We consider thenew coordinate system by introducing the local coordinates transformation whichtakes x to x = T (x) = R(x) + Mz0 , where R is the rotation such that R(ν(z0)) =(0, 1), R(τ(z0)) = (1, 0) with the normal vector ν(z0) = (ν1(z0), ν2(z0)) of ∂D atz0 keeping z0 fixed and Mz0 is the translation such that Mz0 + R(z0) = (0, 0). Weset τ(z0) = R((0, 1)) = (−ν1(z0), ν2(z0)). In the new coordinate system we denoteby a(x) = a(x) and η(x) = η(x) for x where they are defined. Let a0 = a(z0)and η0 = η(z0). Denoting by Γ(x, z) = 1

2π ln 1|x−z| the fundamental solution of the

Laplace equation, we consider the following problems

(5.11)

∆v(·, z) = 0 in R2

−∆w(·, z) = 0 in R2

−v(·, z)− w(·, z) = ∇Γ(·, z) · τ on ∂R2

−a0∂x2

v(·, z)− ∂x2w(·, z) = ∂x2

∇Γ(·, z) · τ on ∂R2−

and(5.12)

∆v(·, z) = 0 in R2−

∆w(·, z) = 0 in R2−

v(·, z)− (w(·, z) +∇Γ(·, z) · τ) = −iη0∂x2(w(·, z) +∇Γ(·, z) · τ) on ∂R2−

a0∂x2v(·, z)− ∂x2

w(·, z) = ∂x2(∇Γ(·, z) · τ) on ∂R2

−

where x = (x1, x2) and R2− is the half plane x ∈ R2 : x2 < 0.

The explicit forms of these functions are computed in [5]. In particular, for theproblem (5.11), we have

(5.13) w(z, z) =(a0 + 1)ν2(z0)

4π(a0 − 1)z2,

For the problem (5.12), we have the asymptotic behaviors

(5.14)

Re w(z, z) =

ν2(z0)

4πz2+O(ln |z|), for z near 0

Im w(z, z) = −ν2(z0)

πηln(|z2|) +O(1) for z near 0.

Now, for x and z near the point z0 ∈ ∂D, define

(5.15) wz(x) := w(Tx, Tz) and vz(x) = v(Tx, Tz).

The next theorem which is the main result of this section, shows that wz is thedominant part of Wz with respect to the singularity near z0 where Wz, Vz solve(5.5).

Theorem 5.2. Let z0 ∈ ∂D and x, z ∈ D are in the vicinity of z0. Let Wz, Vzbe the unique solution of the interior transmission problem (5.5) for j = 2. Letv(·, z), w(·, z) satisfy (5.11) if z0 ∈ ∂Dt and (5.12) if z0 ∈ ∂Dc and wz is defined by(5.15). Then there exists a positive constant C such that:

1. |Re(Wz − wz)(x)| ≤ C| ln |z − z0||2. |Im(Wz − wz)(x)| ≤ C



for any x, z in the cone Cz0(θ) included in D with vertex z0 and angle θ0 ∈ (0, π/2).

Remark 3. We stated the results if G2∞(·, z) is used in the far field equation. The

other case, i.e. where G1∞(·, z) is used in the far field equation, is similar.

Combining Theorem 5.2 and the formulas (5.13), (5.14) and (5.15) gives a justi-fication of the parts II and III of Theorem 5.1.

5.2. Proof of Theorem 5.2. To prove Theorem 5.2, we proceed in two mainsteps, freezing the coefficients a and η first and then flattening the boundary ∂Dnear any point z0 ∈ ∂D. We divide the proof into several lemmas and we introducethe necessary notations for each step and lemma. For the sake of presentation weset ∂xj = ∂x2

and omit the index (j) in the solutions of (5.5). We also consider onlythe case z0 ∈ ∂Dc. The case z0 ∈ ∂Dt is similar with some appropriate changes butwith no substantial modification of the arguments.

We introduce the following function (Gη, Hη) which is the solution of the problem

(5.16)

∇ · a∇Gη + κ2Gη = 0 in D

∆Hη + κ2Hη = δ(· − z) in DGη −Hη + iη∂νHη = 0 on ∂D∂νaGη − ∂νHη = 0 on ∂D.

The results in Theorem 4.1 are also valid for this problem. We only remark herethat the computation of total symbols of the parametrix for this problem becomessimpler than that given in the proof of Theorem 4.3, because the term in the thirdequation with the highest order derivative is iη∂νHη and the fourth equation alsohas the term −∂νHη.

We denote also by (Vz, Wz) the solution of (5.5), in the case where ∂Dt = ∅.

Lemma 5.3. Let z0 ∈ ∂Dc. There exists a positive constant C such that:

|(Wz − Wz)(x)| ≤ Cfor x, z ∈ D near z0 ∈ ∂D.

Proof of Lemma 5.3. The function (Vz − Vz,Wz − Wz) satisfies:

(5.17)

∇ · a∇(Vz − Vz) + κ2(Vz − Vz) = 0 in D

∆(Wz − Wz) + κ2(Wz − Wz) = 0 in D

(Vz − Vz)− (Wz − Wz) = Wz − Vz +G(·, z)) on ∂Dt

(Vz − Vz)− (Wz − Wz) + iη∂ν(Wz − Wz) = 0 on ∂Dc

∂νa(Vz − Vz)− ∂ν(Wz − Wz) = 0 on ∂D.

Since the function (Vz, Wz) satisfies

(5.18)

∇ · a∇Vz + k2Vz = 0 in D

∆Wz + κ2Wz = 0 in D

Vz − Wz + iη∂νWz = fz on ∂D

∂νa Vz − ∂νWz = gz on ∂D.

where fz(x) := ∂x2Φ(·, z)− iη∂ν(∂x2

Φ)(·, z) and gz(·, z) := ∂ν(∂x2Φ)(·, z), then

(5.19) (Vz, Wz + ∂x2Φ(·, z)) = (Vz, Wz − ∂z2Φ(·, z)) = (∂z2Gη, ∂z2Hη).

Integrating by parts in (5.17), we obtain

(5.20) (Wz − Wz)(x) =

∫∂D

(Wz − Wz)∂νHηds(y)−∫∂D

∂ν(Wz − Wz)Hηds(y).



and

(5.21)

∫∂D

(Vz − Vz)∂νaGηds(y) =

∫∂D

∂νa(Vz − Vz)Gηds(y).

On the other hand, by using the boundary conditions, we have∫∂D

(Wz − Wz)∂νHηds(y) =∫∂D

(Vz − Vz)∂νaGηds(y)

−∫∂Dt

(Wz − Vz +G)∂νaGηds(y) + i∫∂Dc

∂ν(Wz − Wz)∂νaGηds(y),∫∂D

∂ν(Wz − Wz)Hηds(y)

=

∫∂D

∂νa(Vz − Vz)Gηds(y) + i

∫∂D

η∂νHη∂νa(Vz − Vz)ds(y)

Using the boundary conditions once again, we have

(Wz − Wz)(x)

= −∫∂Dt

∂νaGη(Wz − Vz +G)ds(y)− i∫∂Dt


= −∫∂Dt

∂νaGη(∂z2Hη − ∂z2Gη)ds(y)− i∫∂Dt


since Wz − Vz +G = ∂z2Hη − ∂z2Gη, see (5.19).From the well-poseness of the problem (5.17) and Theorem 4.1, we deduce that

there exists a positive constant c such that

|(Wz − Wz)(x)| ≤ c, for x, z near z0 ∈ ∂Dc

since ∂Dc ∩ ∂Dt = ∅. We denote by (V 0, W 0) the solution of (5.5) setting ∂Dt = ∅ and replacing η by

η0 = η(z0). Then we have the following lemma:

Lemma 5.4. There exists a positive constant c such that:

|(Wz − W 0z )(x)| ≤ c

for x ∈ D near z0 and z in a small cone Cz0,θ ⊂ D of center z0 ∈ ∂Dc and angle θ.

Proof of Lemma 5.4. The function Fz(x) := (F 1z (x), F 2

z (x)) := (Vz − V 0z , Wz − W 0

z )satisfies the problem:

(5.22)

∇ · a∇F 1

z + κ2F 1z = 0 in D

∆F 2z + κ2F 2

z = 0 in D

F 1z − F 2

z + iη∂νF2z = −i(η − η0)∂ν(W 0

z + ∂x2Φ) on ∂D

∂νaF1z − ∂νF 2

z = 0 on ∂D.

From the integral representation of solutions we obtain

(5.23) F 2z (x) = i

∫∂D

∂νHη(η − η0)∂ν(W 0z + ∂x2Φ)ds(y).

But we know that −iη∂ν(Wz + ∂x2Φ) = Vz − (Wz + ∂x2

Φ) = ∂z2Hη − ∂z2Gη andGη −Hη = −iη∂νHη, hence from Theorem 4.1, we have that

|∂ν(W 0z + ∂x2Φ)|(x, z) ≤ c|x− z|−1 for x ∈ ∂D.

Therefore

|F 2z (x)| ≤ c

∫∂D

| ln |y − x|||y − z0||y − z|−1ds(y).



Since y ∈ ∂D then it is easy to see that |y − z0| ≤ c(θ)|y − z| where c(θ) dependson the angle θ. Hence, we have

|F 2z (x)| ≤

∫∂D

| ln |y − x||ds(y)

which implies that |F 2z (x)| ≤ c.

We next denote by (V 0z ,W

0z ) the solution of (5.5) setting ∂Dt = ∅, k = 0, η = η0

and replacing Φ(·, z) by Γ(·, z). Using integral representation of solutions in termsof the Green’s function and the estimates for Φ, Γ, Gη and Hη, we can prove thefollowing lemma.

Lemma 5.5. There exists a positive constant c such that:

|(W 0z −W 0

z )(x)| ≤ c for x, z in D.

We further need the following last lemma.

Lemma 5.6. Let z0 ∈ Dc. There exists a positive constant C such that:

|Re(W 0z −Wz)(x)| ≤ C| ln |z − z0||

|Im(W 0z −Wz)(x)| ≤ C

for x ∈ D near z0 and z in a small cone Cz0,θ as in Lemma 5.4.

Proof of Lemma 5.6. The function (V 0z ,W

0z ) satisfies

(5.24)

∇ · a(z)∇V 0

z = 0 in D∆W 0

z = 0 in DV 0z −W 0

z + iη0∂νW0z = ∂2Γ− iη0∂ν(∂2Γ) on ∂D

∂νaV0z − ∂νW 0

z = ∂ν(∂2Γ) on ∂D.

We can assume without loss of generality that z0 = (0, 0) by using a rigid trans-formation of coordinates. Let ξ = F (x) be the following local change of variables

(5.25)

ξ1 = x1

ξ2 = x2 − f(x1),

where f is the function used for the local parametrization of ∂D near z0. Due tothe regularity of the boundary ∂D the following properties hold true

(5.26)

c1|x− z| ≤ |F (x)− F (z)| ≤ c2|x− z||F (x)− x| ≤ c3|x|2|DF (x)− I| ≤ c4|x|

for x, z near the point z0, where ci, i = 1, ..., 4 are positive constants.The vectors (V 0

z , W0z )(ξ, ζ) = (V 0

z ,W0z )(x, z), where ξ := F (x) and ζ := F (z),

satisfy(5.27)

∇ξ ·B1(ξ)∇V 0z = 0 in R2

−

∇ξ ·B2(ξ)∇W 0z = 0 in R2

−

V 0z − W 0

z + iη0|J−>ν|B2(ξ)∇ξW 0z · ν = (∂ξ2Γ)(F−1(ξ), F−1(ζ))

−iη0|J−>ν| ∇ξ(∂ξ2Γ)(F−1(ξ), F−1(ζ)) · J>ν(ξ) on ∂R2−

B1(ξ)∇ξV 0z · ν −B2(ξ)∇ξW 0

z · ν = ∇ξ(∂ξ2Γ)(F−1(ξ)F−1(ζ)) · J>ν(ξ) on ∂R2−



near F (z0), where J(ξ) := ∂xξ(F−1(ξ)), B1(ξ) = Ja(z)J>, B2(ξ) = JJ> and

ν := (0, 1).

We denote by R(ξ, ζ) := (V 0z , W

0z )(ξ, ζ)−(v, w)(ξ, ζ) with v(ξ, ζ) = vz(x), w(ξ, ζ)

= wz(x). It satisfies the following properties(5.28)

∇ξ ·B1(ξ)∇R1 = ∇ξ · (A(z0)−B1(ξ))∇ξv in R2−

∇ξ ·B2(ξ)∇R2 = ∇ξ · (I −B2)∇ξw in R2−

R1 − R2 + iη0|J−>ν| B2(ξ)∇R2 · ν =

−(∂ξ2Γ)(ξ, ζ) + (∂ξ2Γ)(F−1(ξ), F−1(ζ)) + iη0(∂ξ2w(ξ, ζ) + ∂ξ2(∂ξ2Γ))(ξ, ζ)

−iη0|J−>ν|[B2∇w · ν(ξ, ζ) +∇(∂ξ2Γ)(F−1(ξ), F−1(ζ)) · J>ν(ξ)] on ∂R2−

B1(ξ)∇ξR1 · ν −B2(ξ)∇ξR2 · ν = (∇∂ξ2Γ)(F−1(ξ), F−1(ζ)) · J>ν−∂ξ2(∂ξ2Γ)(ξ, ζ) + (I −B2)∇ξw · ν + (A(z0)−B1)∇ξv · ν on ∂R2

−

near F (z0), where we set A(z0) := a(z0)I and I is the identity matrix.We set also

(5.29)l(ξ, ζ):= −(∂ξ2Γ)(ξ, ζ) + (∂ξ2Γ)(F−1(ξ), F−1(ζ)) + iη0(∂ξ2w(ξ, ζ) + ∂ξ2(∂ξ2Γ))(ξ, ζ)−iη0|J−>ν|[B2∇w · ν +∇ξ(∂ξ2Γ)(F−1(ξ), F−1(ζ)) · J>ν(ξ)]

and

(5.30) h(ξ, ζ) := (∇∂ξ2Γ)(F−1(ξ), F−1(ζ)) · J>ν − ∂ξ2(∂ξ2Γ)(ξ, ζ).

Let Dr by a C∞-smooth domain containing B−r := Br ∩ R2− and contained in

B−R := BR∩R2− for some disks Br and BR of center 0 and radius r and R such that

R > r > 0. We denote νr to be the exterior unit normal of ∂B−r . Let (Gr, Hr) bethe function satisfying the problem:

(5.31)

∇ξ ·B1(ξ)∇Gr + κ2

rGr = 0 in Dr

∇ξ ·B2(ξ)∇Hr + κ2rHr = J−1δ(· − ζ) in Dr

Gr −Hr + iη0|J−>ν|B2(ξ)∇ξHr · ν = 0 on ∂Dr

B1(ξ)∇ξGr · ν −B2(ξ)∇ξHr · ν = 0 on ∂Dr,

where J = detJ and the coefficient κ2r is chosen so that it is not a transmission

eigenvalue. As in the sections 2, 3 and 4, we have existence of such a function andthe estimates in Theorem 4.1 are also valid.

Multiplying the first equation of (5.28) by Gr and integrating by parts, we havethat

−κ2r

∫Dr

Gr(z, ξ)R1(z, ζ)dz +

∫∂Dr

(B1(z)∇zR1 · νr)(z, ζ)Gr(z, ξ)ds(z)

−∫∂Dr

(B1(z)∇zGr · νr)(z, ξ)R1(z, ζ)ds(z)

= −∫Dr

(A(z0)−B1(z))∇zv(z, ζ)∇zGr(z, ξ)dz(5.32)

+

∫∂Dr

(A(z0)−B1(z))∇zv(z, ξ) · νrGr(z, ξ)ds(z).



Similarly, multiplying the second equation by Hr and integrating by parts, we havethat

−κ2r

∫Dr

Hr(z, ξ)R2(z, ζ)dz +

∫∂Dr

(B2(z)∇zR2 · νr)(z, ζ)Hr(z, ξ)ds(z)

−∫∂Dr

(B2(z)∇zHr · νr)(z, ξ)R2(z, ζ)ds(z)

= −J (ξ)−1R2(ξ, ζ)−∫Dr

(I −B2(z))∇zw(x, ζ)∇zHr(z, ξ)ds(z)(5.33)

+

∫∂Dr

(I −B2(z))∇zw(z, ζ) · νrHr(z, ξ)ds(z)

Next writing ∂Dr = Sr ∪ Scr , where Sr := ∂Dr ∩ ∂F (D) we see that (5.32) and(5.33) become∫

Sr

(B1(z)∇zR1 · νr)(z, ζ)Gr(z, ξ)ds(z)−∫Sr

(B1(z)∇zGr · νr)(z, ξ)R1(z, ζ)ds(z)

= −∫Dr

(A(z0)−B1(z))∇zv(z, ζ)∇zGr(z, ξ)dz + κ2r

∫Dr

Gr(z, ξ)R1(z, ζ)dz

+

∫Sr

(A(z0)−B1(z))∇zv(z, ξ) · νrGr(z, ξ)ds(z) + L1(ξ, ζ)(5.34)

and ∫Sr

(B2(z)∇zR2 · νr)(z, ζ)Hr(z, ξ)ds(z)

−∫Sr

(B2(z)∇zHr · νr)(z, ξ)R2(z, ζ)ds(z)

= −J (ξ)−1R2(ξ, ζ)−∫Dr

(I −B2(z))∇zw(z, ζ)∇zHr(z, ξ)ds(z)(5.35)

+ κ2r

∫Dr

Hr(z, ξ)R2(z, ζ)dz +

∫Sr

(I −B2(z))∇zw(z, ζ) · νrHr(z, ξ)ds(z)

+ L2(ξ, ζ),

where L1(ξ, ζ) and L2(ξ, ζ) are the terms integrating over Scr . Using the boundaryconditions from (5.28) and subtracting (5.35) from (5.34) we finally obtain

− J (ξ)−1R2(ξ, ζ) = −∫Dr

(A(z0)−B1(z))∇zv(z, ζ)∇zGr(z, ξ)dz(5.36)

+

∫Dr

(I −B2(z))∇zw(z, ζ)∇zHr(z, ξ)ds(z)

+L1(ξ, ζ)− L2(ξ, ζ) + L3(ξ, ζ) + L4(ξ, ζ)

where

L3(ξ, ζ) := −∫Sr

B2(ξ)∇Hr(z, ξ) · νrl(z, ζ)ds(z)−∫Sr

Gr(z, ξ)h(z, ζ)ds(z)

+

∫Sr

(Gr −Hr)(z, ξ)(I −B2)∇w(z, ζ) · νds(z)(5.37)

and

(5.38) L4(ξ, ζ) := κ2r

∫Dr

Gr(z, ξ)R1(z, ζ)dz − κ2

r

∫Dr

Hr(z, ξ)R2(z, ζ)dz



To finish the proof we make use of the following lemma.

Lemma 5.7. For every s > 0 there exists C := C(s) > 0 such that:

|∇(Im v(ξ, ζ))|, |∇(Imw(ξ, ζ))| ≤ C|ξ − ζ∗|−1

|ImGr(ξ, ζ)|, |ImHr(ξ, ζ)| ≤ C

|∇ξ(ImGr(ξ, ζ))|, |∇ξ(ImGr(ξ, ζ))| ≤ Cd−s(ξ, ∂Dr)

for ξ, ζ near the origin, where ζ∗ := (ζ1,−ζ2).

Proof of Lemma 5.7. The first estimate is due to the explicit form vz and wz, see(5.15) and the expressions before. We give here the proof for the last two estimates.To this end remark that, see (5.31), (ImGr, ImHr) satisfies the following problem

(5.39)

∇ξ ·B1(ξ)∇ImGr + κ2

rImGr = 0 in Dr

∇ξ ·B2(ξ)∇ImHr + κ2rImHr = 0 in Dr

ImGr − ImHr = −η0|J−>ν|B2(ξ)∇ξReHr · ν on ∂Dr

B1(ξ)∇ξImGr · ν −B2(ξ)∇ξImHr · ν = 0 on ∂Dr,

Let GN be the Neumann Green function of the Helmholtz (∆ +κ2r) in Dr assuming

that κ2r is not an eigenvalue of the Neumann-Laplace operator in Dr. Integrating

by parts we get:

ImHr(ξ, ζ) =

∫∂Dr

GN (ξ, y)∂νImHr(ζ, y)ds(y)

=

∫∂Dr

(|J−T ν|η0)−1GN (ξ, y)[ReGr(ζ, y)− ReG2r(ζ, y)]ds(y)

Hence

∇ξImHr(ξ, ζ) =

∫∂Dr

(|J−T ν|η0)−1∇ξGN (ξ, y)[ReGr(ζ, y)− ReHr(ζ, y)]ds(y)

which implies

|∇ξImHr(ξ, ζ)| ≤ C∫∂Dr

|ξ − y|−1| ln |ζ − y||ds(y)

≤ C∫∂Dr

|ξ − y|−1+s|ξ − y|−s|ζ − y|−tds(y).

Then (cf.[24])

|∇ξImHr(ξ, ζ)| ≤ Cd−s(ξ, ∂Dr)

∫∂Dr

|ξ − y|−1+s|ζ − y|−tds(y)

≤ Cα,s|ξ − ζ|−t+sd−s(ξ, ∂Dr)

Taking t = s we have

|∇ξImHr(ξ, ζ)| ≤ C(s)d−s(ξ, ∂Dr).

Similarly we have the estimates for ImGr(ξ, ζ).



Continuing now the proof, from (5.37), we have

− J (ξ)−1Im R2(ξ, ζ) = −∫Dr

(A(z0)−B1(z))∇zIm v(z, ζ)∇zReGr(z, ξ)dz

−∫Dr

(A(z0)−B1(z))∇zRe v(z, ζ)∇zImGr(z, ξ)dz

+

∫Dr

(I −B2(z))∇zImw(z, ζ)∇zReHr(z, ξ)dz

+

∫Dr

(I −B2(z))∇zReWz(z, ζ)∇zImHr(z, ξ)dz

+ImL(ξ, ζ) + ImL3(ξ, ζ) + ImL4(ξ, ζ),

where L(ξ, ζ) = L1(ξ, ζ)− L2(ξ, ζ). Hence

|Im R2(ξ, ζ)| ≤ C

[∫Dr

|z − z0||z − ζ∗|−1|z − ξ|−1dz

+

∫Dr

|z − z0|d−s(z, ∂Dr)|z − ζ∗|−2dz

]+|ImL(ξ, ζ) + ImL3(ξ, ζ) + ImL4(ξ, ζ)|.

Taking ζ ∈ CF (z0),θ and ξ ∈ Sr away from Scr , we have that L(ξ, ζ) = O(1).

Since R1(ξ, ζ) = V 0z (ξ, ζ) − vz(ξ, ζ) and both V 0

z and v0z have singularities of

the type |ξ − ζ|−1 and Gr(ξ, ζ) has a singularity of the type log(|ξ − ζ|), then∫DrGr(z, ξ)R

1(z, ζ)dz ≈ O(1). Similarly∫DrHr(z, ξ)R

2(z, ζ)dz ≈ O(1). Hence

L4(ξ, ζ) = O(1). With quite lengthy but similar considerations and using the prop-erties of the change of variable F , i.e (5.26), we can show also that ImL3(ξ, ζ) =O(1). Then

|Im R2(ξ, ζ)| ≤ C

[∫Dr

|z − z0||z − ζ∗|−1|z − ξ|−1ds(z)

+

∫Dr

|z − z0||z2 − z0,2|−s|z − ζ∗|−2dz

].

Hence we have

(5.40) ImR2(ξ, ζ) = O(1) for ξ ∈ Sr and ζ ∈ CF (z0),θ.

Similarly, we have

|ReR2(ξ, ζ)| ≤ C∫Dr

|z − z0||z − ζ∗|−2|z − ξ|−1ds(z) +O(1).

Therefore (cf. [24])

(5.41) |ReR2(ξ, ζ)| ≤ C| ln |ξ − ζ∗|| for ξ ∈ Sr and ζ ∈ CF (z0),θ.

We go back to estimate R(x, z) := (V 0z ,W

0z )(x, z)− (vz, wz)(x). To this end

R2(x, z) = W 0z (x, z)− w(F (x), F (z)) + w(F (x), F (z))− w(x, z),

which can be rewritten as(5.42)

R2(x, z) = R2(F (x), F (z)) + [w(F (x), F (z))− w(F (x), z)] + [w(F (x), z)− w(x, z)].

We want to show that F (Cz0,θ ∩B(0, δ(z0))) ⊂ CF (z0),θ′ for some θ′ ∈ (0, π/2) andθ ≤ θ′ with δ(z0) small enough, where B(0, δ(z0)) is a disk of center 0 and radius



δ(z0). To this end we set M := maxz1∈(−δ(z0),δ(z0)) |f(z1)| = |f(z01)| and recall that

z0 = (0, 0). We draw the cone of center z0, axis ν(z0) and having the point (z01 ,M)

on its boundary and denote by φ the angle between this cone and the plane z2 = 0.Since |∇f(0)| = 0 and f ∈ C1(0, δ(z0)), if δ(z0) tends to zero then φ = φ(δ(z0))tends to zero as well. Hence, taking δ(z0) small enough, if necessary, we can assumethat 0 ≤ φ < π/2−θ. Finally we define the cone of center z0, axis (ν(z0)) and angleθ′ := φ+ θ. Then F (Cz0,θ) ⊂ CF (z0),θ′ . Let now x be near z0 such that F (x) ∈ Srand away from Scr and z ∈ Cz0,θ ∩B(z0, δ(z0)). Then (F (x), F (z)) ∈ Sr ×CF (z0),θ′

and hence Im R2(F (x), F (z)) is bounded, as shown before.The boundedness of the second and the third terms on the right hand side of

(5.42) are based on the explicit form of wz and the properties (5.26) of F .The above means that

(5.43) |Im R2(x, z)| ≤ c(θ)for x ∈ B(z0, δ(z0)) such that F (x) ∈ Sr and z ∈ Cz0,θ ∩B(z0, δ(z0)).

Now for z ∈ Cz0,θ ∩B(z0,δ(z0)

2 ) and x ∈ [∂B(z0, δ(z0))] ∩D, we have

(5.44) |Im R2(x, z)| < c

with some positive constant c, because Cz0,θ ∩ B(z0,δ(z0)

2 ) and [∂B(z0, δ(z0))] ∩Dare separated sets. Since ∆xIm R2(x, z) = 0 in B(z0, δ(z0)) ∩ D, using (5.43)and (5.44) from the maximum principle we have that |Im R2(x, z)| ≤ c(θ) for

x ∈ D ∩B(z0, δ(z0)) and z ∈ Cz0,θ ∩B(z0,δ(z0)

2 ).To prove the second part of Lemma 5.6, we need to take the real part in (5.37)

and use the corresponding estimates to get

|ReR2(ξ, η)| ≤ C| ln |ξ − η∗||for ξ ∈ Sr and η ∈ Cz0,θ. Arguing in the same way as for the imaginary part, wededuce that |ReR2(x, z)| ≤ C| ln |x−z|| for x ∈ B(z0, δ(z0)) such that F (x) ∈ Sr andz ∈ Cz0,θ ∩ B(z0, δ(z0)). Hence |Re R2(x, z)| ≤ C| ln d(z, ∂D)| for x ∈ B(z0, δ(z0))such that F (x) ∈ Sr and z ∈ Cz0,θ ∩B(z0, δ(z0)) which is the counterpart of (5.44)

for Re R2. The rest of the proof is the same as for the imaginary part.

Acknowledgments. The authors would like to acknowledge the referee’s valuablecomments for improving this paper. Also, they acknowledge the scientific fundswhich they received while they did the research on this paper. More precisely,the first author was supported by BK 21 at Department of Mathematics of EwhaWomans Univ., the second author was partially supported by the Grant-in-Aidfor Scientific Research (B) (22340023) of the Japan Society for the Promotion ofScience and the third author was supported by the Johann Radon Institute forComputational and Applied Mathematics (RICAM) of the Austrian Academy ofSciences and the Austrian Science Fund (FWF): P22341-N18.

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Received February 2012; revised May 2012.

E-mail address: [email protected] address: [email protected] address: [email protected]



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http://dx.doi.org/10.1080/00036810903437820

http://dx.doi.org/10.1080/00036810903437820


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THE GREEN FUNCTION OF THE INTERIOR TRANSMISSION PROBLEM AND ITS APPLICATIONS

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