# Finding and Representing Solutions to Transmission Problems for Open
Channels
Version 3.3

Charles L. Epstein111 Center for Computational Mathematics, Flatiron
Institute, 162 Fifth Avenue, New York, NY 10010. E-mail:
<EMAIL_ADDRESS>
(January 26, 2023)

###### Abstract

We introduce a layer potential representation for the solution of the
transmission problem defined by two dielectric channels, or open waveguides,
meeting along the straight-line interface, $\\{x_{1}=0\\}.$ The main
observation is that the outgoing fundamental solution for the operator
$\Delta+k_{1}^{2}+q(x_{2}),$ acting on functions defined in $\mathbb{R}^{2},$
is easily constructed using the Fourier transform in the $x_{1}$-variable and
the elementary theory of ordinary differential equations. These fundamental
solutions can then be used to represent the solution to the transmission
problem in half planes. The transmission boundary conditions then lead to
integral equations along the intersection of the half planes, which, in our
normalization, is the $x_{2}$-axis. We show that these integral equations are
solvable when the contrast between the two channels is small, and more
generally that they are Fredholm equations of second kind, which are therefore
generically solvable. This formulation suggests practicable numerical methods
to solve this class of problems.

###### Contents

  1. 1 Introduction
  2. 2 A Layer Potential Approach
  3. 3 The Structure of the Perturbed Green’s Function
  4. 4 Estimates for the Boundary Kernel
    1. 4.1 Estimates on the Wronskian and Basic Solutions of $L_{\xi}$
    2. 4.2 Asymptotics for $|x_{2}|,|y_{2}|>d$
    3. 4.3 Asymptotics for $|x_{2}|$ or $|y_{2}|<d$
    4. 4.4 The Diagonal Singularity
    5. 4.5 Summary
  5. 5 Estimates in the Right Half Plane
  6. 6 The Integral Equations
  7. 7 Outgoing Estimates for $\sigma,$ $\tau$ and $u^{r,l}$
  8. 8 The Projections onto Wave Guide Modes
  9. 9 Some Concluding Remarks
  10. A The Planar Case
    1. A.1 Proof of Theorem 2
    2. A.2 Proof of Theorem 3
  11. B The Outgoing Condition, I
  12. C A Second Representation for the Outgoing Resolvent
    1. C.1 Spectral Representation of $\partial_{x_{2}}^{2}+k_{1}^{2}+q(x_{2})$
    2. C.2 The Outgoing Condition, II
  13. D Stationary Phase Results

##  1 Introduction

In this paper we present a new analysis of the transmission problem for
dielectric, or open waveguides. This problem has already received a lot of
attention in works of many authors, for example, see [2, 3, 4, 5]. The goal of
this paper is to produce a layer potential representation for the solution of
the transmission problem defined by two semi-infinite, dielectric channels
meeting along a straight line interface. We assume that the refractive index
inside the wave-guide is larger than the surrounding space, which indicates
that one should expect wave guide modes to exist. Using this layer potential
representation, which is itself practically computable, we can derive integral
equations that lend themselves to numerical implementation for the approximate
solution of this basic, model problem.

As explained in [5], the fundamental electro-magnetic problem we are
considering is that of scattering in $\mathbb{R}^{3}$ by a non-magnetic,
dielectric sheet, which is independent of the $x_{3}$-variable in the TM, or
transverse magnetic mode. The electric field then takes the form
$\boldsymbol{E}=(0,0,u(x_{1},x_{2}))$ where $u$ satisfies an appropriate
Helmholtz-like equation

$(\Delta+k(x))u=0.$ (1)

For the case of two semi-infinite, dielectric channels meeting along a common
perpendicular line, the precise formulation is in terms of a pair operators:

$\Delta+k_{1}^{2}+q_{l,r}(x_{2}),$ (2)

see Figure 1. Here, and throughout the paper, $l$ refers to $\\{x_{1}<0\\},$
and $r$ refers to $\\{x_{1}>0\\}.$ Though the method applies to piecewise
continuous $q(x_{2})$ with bounded support, for simplicity we usually assume
that

$q_{l,r}(x_{2})=(k^{2}_{2;l,r}-k^{2}_{1})\chi_{[y_{l,r}-d_{l,r},y_{l,r}+d_{l,r}]}(x_{2}),$
(3)

with $k_{2;l,r}\geq k_{1}.$ This is essentially the same problem as that
considered in [2], but rather different from that considered in [5]. Chandler-
Wilde and Zhang consider more general potentials, but make assumptions that
preclude the existence of wave guide modes. They also focus on incoming data
similar to plane waves, whereas we consider incoming fields that decay as
$|x_{2}|\to\infty,$ which includes wave guide modes as well as smooth wave
packets formed from plane waves, see Example 1. Nonetheless, there are many
points of contact between our analysis and theirs.

Figure 1: Two dielectric channels meeting along a straight interface. The
$x_{3}$-axis is orthogonal to the plane of the image.

If we think of the waves as time harmonic, with time dependence
$e^{-it\omega},$ then conventionally $k_{1}=\omega\sqrt{\epsilon_{1}},$
$k_{2;l,r}=\omega\sqrt{\epsilon_{2;l,r}},$ with $\epsilon$ denoting the
permittivity of the material, which we assume is a positive real number. We
suppose that there is an incoming solution $u^{\operatorname{in}},$ which
satisfies

$(\Delta+k_{1}^{2}+q_{l}(x_{2}))u^{\operatorname{in}}=0,$ (4)

and then look for “outgoing” solutions $u^{l},u^{r},$ to the equations

$(\Delta+k^{2}_{1}+q_{l,r})u^{l,r}=0\text{ where }x_{1}<0,\text{ resp.
}x_{1}>0,$ (5)

with $u^{l}\in H^{2}_{\operatorname{loc}}((-\infty,0]\times\mathbb{R})$ and
$u^{r}\in H^{2}_{\operatorname{loc}}([0,\infty)\times\mathbb{R}),$ satisfying
the transmission boundary conditions

$\begin{split}u^{r}(0,x_{2})-u^{l}(0,x_{2})&=u^{\operatorname{in}}(0,x_{2})=g(x_{2}),\\\
\partial_{x_{1}}u^{r}(0,x_{2})-\partial_{x_{1}}u^{l}(0,x_{2})&=\partial_{x_{1}}u^{\operatorname{in}}(0,x_{2})=h(x_{2}).\end{split}$
(6)

The regularity and boundary conditions then imply that the field

$u^{\operatorname{tot}}(x_{1},x_{2})=\begin{cases}&u^{l}(x_{1},x_{2})+u^{\operatorname{in}}(x_{1},x_{2})\text{
for }x_{1}\leq 0,\\\ &u^{r}(x_{1},x_{2})\text{ for }x_{1}>0,\end{cases}$ (7)

is a weak solution, belonging $H^{2}_{\operatorname{loc}}(\mathbb{R}^{2}),$ to

$(\Delta+q(x_{1},x_{2})+k_{1}^{2})u^{\operatorname{tot}}=0,\\\ \text{ where
}q(x_{1},x_{2})=q_{l}(x_{2})\chi_{(-\infty,0]}(x_{1})+q_{r}(x_{2})\chi_{(0,\infty)}(x_{1}).$
(8)

The existence and uniqueness (under a certain definition of “outgoing”) of
solutions to this problem is established in [2]. We give a very different
formulation of the solution that we believe is better suited to numerical
approximation. That said, we have only proved existence generically, and have
considered neither the question of uniqueness, nor the relationship between
our solution and that found in [2].

For the applications of most immediate interest, the incoming field is a sum
of _wave guide modes_. A single wave guide mode for
$\Delta+k_{1}^{2}+q_{l}(x_{2}),$ is given by
$u^{\operatorname{in}}=e^{i\xi_{0}x_{1}}v_{0}(x_{2}),$ where $v_{0}(x_{2})$ is
an $H^{2}(\mathbb{R})$–solution to

$[\partial_{x_{2}}^{2}-\xi_{0}^{2}+k_{1}^{2}+q_{l}(x_{2})]v_{0}(x_{2})=0$ (9)

that is exponentially decaying outside the channel, which is the
$\operatorname{supp}q_{l}.$ If $\xi_{0}>0$ then this mode is rightward moving,
and if $\xi_{0}<0,$ leftward moving. For such a single mode, the jump data is
given by

$g(x_{2})=v_{0}(x_{2}),\quad h(x_{2})=i\xi_{0}v_{0}(x_{2}).$ (10)

While we do not give an explicit definition of “outgoing,” we represent the
scattered field in each half plane using the outgoing fundamental solutions
for the operators $(\Delta+k_{1}^{2}+q_{l,r}(x_{2})).$ These fundamental
solutions are, in turn, obtained using a limiting absorption principle, i.e.
as limits of the resolvents $\lim_{\delta\to
0^{+}}(\Delta+k_{1}^{2}+q_{l,r}(x_{2})+i\delta)^{-1}.$ In Appendices B and C
we show that these limits are outgoing in a sense very similar to the
classical Sommerfield radiation condition. Our representation is therefore
closely related to the _upper propagating radiation condition_ (UPRC) used in
[5]. As noted above, we have left the detailed analysis of our implicit
outgoing condition and questions of uniqueness to a later publication.

Section 2 introduces our approach, which uses the fundamental solutions for
the perturbed operators, $\Delta+k_{1}^{2}+q_{r,l}(x_{2}).$ In Sections 3 and
4 we provide details for the construction of, and estimates for the kernels
introduced in Section 2. In Section 5 we show how these estimates extend to a
half plane. In Section 6 we examine the integral equations we need to solve
and prove mapping results on the operators defined by these kernels in Banach
spaces of continuous functions with specified rates of decay: For
$0\leq\alpha,$ let $\mathcal{C}_{\alpha}(\mathbb{R})$ denote the subspace of
functions $f\in\mathcal{C}^{0}(\mathbb{R})$ with

$|f|_{\alpha}=\sup\\{(1+|x|)^{\alpha}|f(x)|:\>x\in\mathbb{R}\\}<\infty.$ (11)

We first show that if $q_{l}-q_{r}$ is sufficiently small, then the integral
equations obtained in Section 6 are solvable in
$\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R})$
for $0<\alpha<\frac{1}{2}.$ We next show that, on these spaces, these
equations are of the form $\operatorname{Id}+$ compact. They are therefore
Fredholm equations of second kind, which are generically solvable. In Section
7 we give refined estimates on the solutions obtained in Section 6, which
imply that, along rays $r(\eta_{1},\eta_{2}),$ with
$\eta_{1}^{2}+\eta_{2}^{2}=1,$ the scattered fields
$u^{r,l}(r(\eta_{1},\eta_{2}))$ are bounded as $r\to\infty,$ provided
$\eta_{1}\neq 0.$ We also show that the scattered waves decay in directions
orthogonal to the channels. In a final section we show that the projections of
the solutions onto the wave guide modes are determined by the projections of
the source terms onto these modes. We then derive equations that allow for the
approximate determination of the scattering relation from incoming wave guide
modes to transmitted and reflected modes, without solving the full system of
equations. Several appendices give background material and proofs of ancillary
results used at various points in the paper.

Acknowledgments

I would like to thank Leslie Greengard for suggesting this problem and for
many interesting conversations along the way. I would also like to thank
Shidong Jiang, Felipe Vico, and Alex Barnett for many very helpful discussions
of this material and pointers to the literature on this problem, and Mike
O’Neil for Figure 1. I am very grateful to Manas Rachh for carefully reading
this manuscript and providing very useful comments. I would like to thank
David Jerison for useful discussions about these issues. I am finally very
grateful for the support of the Flatiron Institute of the Simons Foundation.

##  2 A Layer Potential Approach

The starting point for our approach is the formula for the solution of the
classical transmission problem: find a function $u$ that solves

$(\Delta+k^{2})u=0,\text{ in }\mathbb{R}^{2}\setminus\\{x_{1}=0\\},$

and the transmission boundary condition

$\begin{split}u(0^{+},x_{2})-u(0^{-},x_{2})&=g(x_{2})\\\
\partial_{x_{1}}u(0^{+},x_{2})-\partial_{x_{1}}u(0^{-},x_{2})&=h(x_{2}).\end{split}$
(12)

The solution to this problem is given by

$u(x)=-\mathcal{D}_{k}g+\mathcal{S}_{k}h,$ (13)

where the single and double layers are given by

$\mathcal{S}_{k}f(x)=\int_{y_{1}=0}g_{k}(x-y)f(y_{2})dy_{2},\text{ and
}\mathcal{D}_{k}f(x)=\int_{y_{1}=0}\partial_{y_{1}}g_{k}(x-y)f(y_{2})dy_{2},$
(14)

with

$g_{k}(x-y)=\frac{i}{4}H^{(1)}_{0}(k|x-y|),$

the outgoing fundamental solution to $\Delta+k^{2}.$ That this gives a
solution follows from the classical jump formulæ for layer potentials, see
[6].

Our method for solving the transmission problem for 2 wave guides (5), and
(6), uses this general approach. Let $\mathfrak{E}^{l,r}(x;y)$ denote the
outgoing fundamental solutions for the operators
$\Delta+k_{1}^{2}+q_{l,r}(x_{2}),$ acting on the _whole plane_. The kernels of
these operators take a rather special form:

$\mathfrak{E}^{l,r}(x;y)=\frac{i}{4}H^{(1)}_{0}(k_{1}|x-y|)+w^{l,r}(x;y),$
(15)

where $w^{l,r}$ satisfies the equation

$(\Delta_{x}+k_{1}^{2}+q_{l,r}(x_{2}))w^{l,r}(x;y)=-\frac{i}{4}q_{l,r}(x_{2})H^{(1)}_{0}(k_{1}|x-y|).$
(16)

The right hand sides of (16) are compactly supported in the $x_{2}$-variable.
As we shall see, these equations can be solved quite explicitly by taking the
Fourier transform in the $x_{1}$-variable, and solving frequency-by-frequency.
The correction terms, $w^{l,r}(x;y),$ are smooth away from the diagonal, and 2
orders smoother along the diagonal, as distributions, than
$H^{(1)}_{0}(k|x-y|).$ Indeed, where $x_{1}=y_{1}=0,$ their singularities are
contained within the compact set $B_{d},$ where, for $d>0,$

$B_{d}\overset{d}{=}[-d,d]\times[-d,d].$ (17)

That $\mathfrak{E}^{l,r}$ give classically outgoing solutions to
$(\Delta+q_{l,r}(x_{2})+k_{1}^{2})u=f,$ for compactly supported $f,$ is proved
in Appendices B and C.

Once the fundamental solutions are constructed we can express the right and
left portions of the solution as

$u^{l,r}=-\mathcal{E}^{l,r\,^{\prime}}\sigma+\mathcal{E}^{l,r}\tau,$ (18)

where

$\begin{split}&\mathcal{E}^{l,r}f(x)=\int_{-\infty}^{\infty}\mathfrak{E}^{l,r}(x;0,y_{2})f(y_{2})dy_{2}=\mathcal{S}_{k_{1}}f+\mathcal{W}^{l,r}f(x)\\\
&\mathcal{E}^{l,r\,^{\prime}}f(x)=\int_{-\infty}^{\infty}[\partial_{y_{1}}\mathfrak{E}^{l,r}](x;0,y_{2})f(y_{2})dy_{2}=\mathcal{D}_{k_{1}}f+\mathcal{W}^{l,r\,^{\prime}}f(x),\end{split}$
(19)

with

$\begin{split}\mathcal{W}^{l,r}f(x)=&\int_{-\infty}^{\infty}w^{l,r}(x;0,y_{2})f(y_{2})dy_{2},\\\
\mathcal{W}^{l,r\,^{\prime}}f(x)=&\int_{-\infty}^{\infty}[\partial_{y_{1}}w^{l,r}](x;0,y_{2})f(y_{2})dy_{2}.\end{split}$
(20)

Using this representation we can derive a system of equations for
$(\sigma,\tau),$ which are somewhat better behaved than usual in so far as the
singularities of $w^{l,r}$ behave like $|x-y|^{2}\log|x-y|.$ The restrictions
to the boundary, $x_{1}=y_{1}=0,$ are given by

$\begin{split}u^{r}(0,x_{2})-u^{l}(0,x_{2})&=\sigma(x_{2})+\\\
&\int_{-\infty}^{\infty}[\partial_{y_{1}}w^{l}-\partial_{y_{1}}w^{r}]{(0,x_{2};0;y_{2})}\sigma(y_{2})dy_{2}+\\\
&\int_{-\infty}^{\infty}[w^{r}-w^{l}]{(0,x_{2};0;y_{2})}\tau(y_{2})dy_{2}=g(x_{2}),\\\
\partial_{x_{1}}u^{r}(0,x_{2})-\partial_{x_{1}}u^{l}(0,x_{2})&=\tau(x_{2})+\\\
&\int_{-\infty}^{\infty}[\partial_{x_{1}}w^{r}-\partial_{x_{1}}w^{l}]{(0,x_{2};0;y_{2})}\tau(y_{2})dy_{2}+\\\
&\int_{-\infty}^{\infty}[\partial^{2}_{x_{1}y_{1}}w^{l}-\partial^{2}_{x_{1}y_{1}}w^{r}]{(0,x_{2};0;y_{2})}\sigma(y_{2})dy_{2}=h(x_{2}).\end{split}$
(21)

Only the $H^{(1)}_{0}$-terms have jumps across $\\{x_{1}=0\\},$ and

$\partial_{y_{1}}w^{l,r}(0,x_{2};0,y_{2})=\partial_{x_{1}}w^{l,r}(0,x_{2};0,y_{2})=0.$
(22)

The behavior of these equations hinges on the analytic properties of the
$w^{l,r}$ along the plane where $x_{1}=y_{1}=0,$ which we now consider. It is
an interesting feature of this approach, via the fundamental solutions of
$\Delta+k_{1}^{2}+q_{l,r}(x_{2}),$ that if $q_{l}=q_{r},$ then these equations
reduce to

$\begin{split}u^{r}(0,x_{2})-u^{l}(0,x_{2})&=\sigma(x_{2})\\\
\partial_{x_{1}}u^{r}(0,x_{2})-\partial_{x_{1}}u^{l}(0,x_{2})&=\tau(x_{2}),\end{split}$
(23)

which is solved exactly as in (13).

The method presented below allows for the evaluation of the fundamental
solution $\mathfrak{E}(x;y)$ for an operator of the form
$\Delta+q(x_{2})+k_{1}^{2},$ at any pair of points in the
$\mathbb{R}^{2}\times\mathbb{R}^{2}.$ In the discussion below we show how this
leads to integral equations for describing the scattered field produced by two
open wave guides meeting along a common perpendicular line.

##  3 The Structure of the Perturbed Green’s Function

In Sections 3–5 we simplify notation by dropping the $l,r$ sub- and
superscripts. We use the resolvent kernel for
$\Delta+k_{1}^{2}+q(x_{2})+i\delta$ to find the kernel functions needed to
solve the transmission problem above; these functions are found by solving the
equation

$(\Delta+k_{1}^{2}+q(x_{2})+i\delta)w_{\delta}(x;y)=-\frac{i}{4}q(x_{2})H^{(1)}_{0}(\sqrt{k^{2}_{1}+i\delta}|x-y|),\text{
for }\delta>0.$ (24)

If $\delta>0,$ then, for fixed $y,$ the right hand side belongs to
$L^{2}(\mathbb{R}^{2}).$ We then let $\delta\to 0^{+},$ and denote this
“limiting absorption solution” by $w_{0+}(x;y).$ This insures that we get the
desired outgoing solution. To solve the limiting equation we simply take the
Fourier transform in the $x_{1}$-variable, and use the fact that, as
$\delta\to 0^{+},$ we get

$\mathcal{F}_{x_{1}}[(i/4)H^{(1)}_{0}(k_{1}|x-y|)](\xi)=\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}e^{-iy_{1}\xi}}{2\sqrt{k_{1}^{2}-\xi^{2}}}.$
(25)

Let $\tilde{w}_{0+}(\xi,x_{2};y)$ denote the Fourier transform of $w_{0^{+}}$
in the $x_{1}$-variable. For $\xi\in\mathbb{R},$ it is the outgoing solution
to the ordinary differential equation

$L_{\xi}\tilde{w}_{0+}=(\partial_{x_{2}}^{2}-\xi^{2}+k_{1}^{2}+q(x_{2}))\tilde{w}_{0+}=-q(x_{2})\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}e^{-iy_{1}\xi}}{2\sqrt{k_{1}^{2}-\xi^{2}}}.$
(26)

The spectral theory of $\Delta+q$ is reviewed in Appendix A, and we assume the
reader is familiar with this material. Let $R_{\xi,0+}(x_{2},z_{2})$ denote
the “outgoing” resolvent kernels for the 1-dimensional operators $L_{\xi}.$
They are the limits of the resolvent kernels for $(L_{\xi}+i\delta)^{-1}$ as
$\delta\to 0^{+},$ constructed out of the basic solutions,
$\widetilde{u}_{\pm}(\xi,0+;x_{2}),$ of $L_{\xi}$ and their Wronskian,
$W(\xi),$ see (252), (253), (259). Using this kernel we can write:

$\tilde{w}_{0+}(\xi,x_{2};y)=-\frac{ie^{-iy_{1}\xi}}{2\sqrt{k_{1}^{2}-\xi^{2}}}\int_{-d}^{d}R_{\xi,0+}(x_{2},z_{2})q(z_{2})e^{i|z_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}dz_{2},$
(27)

where $\sqrt{k_{1}^{2}-\xi^{2}}=i\sqrt{\xi^{2}-k_{1}^{2}},$ if $|\xi|>k_{1}.$

Note that the integral in (27) extends over the _finite_ interval $[-d,d],$
which is extremely useful from the perspective of numerical solutions. Away
from the diagonal in $[-d,d]\times[-d,d],$ it also decays exponentially as
$|\xi|\to\infty.$ We reconstruct $w_{0+}$ as a contribution from the
continuous spectrum of $L_{\xi},$ and a contribution from the wave guide
modes. The continuous spectrum contributes

$w^{c}_{0+}(x;y)=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}\tilde{w}_{0+}(\xi,x_{2};y)e^{ix_{1}\xi}d\xi\text{
for }x_{1}>0.$ (28)

The integral in (28) is over the contour $\Gamma_{\nu}^{+},$ which is defined
below, see Figure 2. In order to be able to deform the contour of integration
and use (28) to represent $w_{0+},$ it is necessary to assume that $\pm k_{1}$
are not roots of Wronskian, $W(\xi),$ of $L_{\xi},$ see (260) and (266)–(268).
If

$q(x_{2})=(k_{2}^{2}-k_{1}^{2})\chi_{[-d,d]}(x_{2}),$ (29)

then this amounts to the requirement that

$2d\sqrt{k_{2}^{2}-k_{1}^{2}}\neq n\pi\text{ for }n\in\mathbb{N}.$ (30)

In this case, as shown in Theorem 2, the roots of the Wronskian,
$\\{\pm\xi_{n}:\>n=1,\dots n\\},$ lie in $(-k_{2},-k_{1})\cup(k_{1},k_{2}).$
The contour $\Gamma_{\nu}^{+}$ is defined by replacing the intervals
$\\{[\pm\xi_{n}-\nu,\pm\xi_{n}+\nu]:\>n=1,\dots,N\\}$ in $\mathbb{R}$ with the
semi-circles in the upper half plane

$\\{\pm\xi_{n}+\nu e^{i\theta}:\>\theta\in[\pi,0]\\},$ (31)

with a clockwise orientation. We assume that $\nu>0$ is small enough so these
semi-circles are disjoint and intersect $\mathbb{R}$ within
$(-k_{2},-k_{1})\cup(k_{1},k_{2}).$ The contour $\Gamma_{\nu}^{-}$ is obtained
by reflecting $\Gamma_{\nu}^{+}$ in the real axis.

Figure 2: The portion of $\Gamma^{+}_{\nu}$ near to the positive wave guide
frequencies

To $w^{c}_{0+}(x;y)$ we add a contribution from the wave guide modes:

$w^{g}_{0+}(x;y)=\sum_{n=1}^{N}v_{n}(x_{2})a_{n}(y_{2})e^{i\xi_{n}(x_{1}-y_{1})}\text{
for }x_{1}>0,$ (32)

which is $i$ times the sum of the residues of
$\tilde{w}_{0+}(\xi,x_{2};y)e^{ix_{1}\xi}$ at the $\\{\xi_{n}\\}.$ The
$\\{v_{n}(x_{2})\\}$ are normalized to have $L^{2}$-norm 1. The coefficients
of the wave guide terms are given by

$a_{n}(y_{2})=-\frac{1}{2}\int_{-d}^{d}\frac{e^{-\sqrt{\xi_{n}^{2}-k_{1}^{2}}|y_{2}-z_{2}|}q(z_{2})v_{n}(z_{2})dz_{2}}{\sqrt{\xi_{n}^{2}-k_{1}^{2}}}.$
(33)

These contributions and their $\partial_{x_{1}}$-derivatives are in
$\mathcal{C}^{1}(\mathbb{R}).$ Using the fact that
$(\partial_{x_{2}}^{2}+k_{1}^{2}-\xi_{n}^{2})v_{n}(x_{2})=-q(x_{2})v_{n}(x_{2}),$
and integration by parts, we show that

$a_{n}(y_{2})=v_{n}(y_{2}),$ (34)

and therefore

$w^{g}_{0+}(x;y)=\sum_{n=1}^{N}v_{n}(x_{2})v_{n}(y_{2})e^{i\xi_{n}(x_{1}-y_{1})}\text{
for }x_{1}>0.$ (35)

Hence this term and its $x_{1}$-derivatives decay exponentially as
$|x_{2}|+|y_{2}|\to\infty.$ The fundamental solution for
$\Delta+k_{1}^{2}+q(x_{2})$ is given by

$\mathfrak{E}(x;y)=g_{k_{1}}(|x-y|)+w_{0+}^{c}(x;y)+w_{0+}^{g}(x;y);$ (36)

the guided modes are entirely captured by $w^{g}_{0+}.$ In Section 8 we show
that the continuous spectral terms,
$g_{k_{1}}(|(x_{1},x_{2}-y_{2})|)+w_{0+}^{c}(x_{1},x_{2};0,y_{2}),$ are
orthogonal to the wave guide modes.

This section describes the construction in the right half plane; the left half
plane is obtained by replacing $(x_{1},y_{1})$ with $(-x_{1},-y_{1}),$ and
$\Gamma_{\nu}^{+}$ with $\Gamma_{\nu}^{-}.$

######  Remark 1.

The kernel, $R_{\xi,0+}(x_{2},z_{2}),$ is constructed and analyzed in Appendix
A. For each $\xi$ it gives the outgoing solution in $x_{2}$-direction in that
the propagating (non-evanescent) waves travel away from the channel. As shown
in Appendices B and C, the fundamental solutions, $\mathfrak{E}^{l,r},$ for
$\Delta+q_{l,r}(x_{2})+k_{1}^{2},$ obtained as the sum
$g_{k_{1}}(x-y)+w^{l,r}_{0+}(x,y),$ are outgoing in the strongest possible
sense.

We need to analyze $w_{0+},$ and certain of its derivatives, along the set
$x_{1}=y_{1}=0.$ First observe that outside the interval $|x_{2}|\leq d,$ this
function satisfies the homogeneous elliptic equation
$(\Delta_{x}+k_{1}^{2})w_{0+}=0,$ and is therefore a
$\mathcal{C}^{\infty}$-function of $x.$ From (25) and (28) it is clear that
$\partial_{x_{1}}w^{c}_{0+}(x;y)=-\partial_{y_{1}}w^{c}_{0+}(x;y),$ hence it
suffices to analyze the smoothness and decay properties of
$\partial_{x_{1}}^{j}w^{c}_{0+}(x;y),$ for $j=0,1,2.$

As noted, away from $x_{2}=y_{2},$ the functions
$\partial_{x_{1}}^{j}w_{0+}(0,x_{2};0,y_{2})$ are
$\mathcal{C}^{\infty}$-functions provided $x_{2}\neq\pm d$ or $y_{2}\neq\pm
d.$ The function $w_{0+}$ is $\mathcal{C}^{1}$ in a neighborhood of these
points with higher regularity determined by the regularity of $q.$ If $q$ were
a smooth function, then it would follow that

$(\Delta_{x}+k_{1}^{2}+q(x_{2}))q(x_{2})|x-y|^{2}H^{(1)}_{0}(k_{1}|x-y|)=\\\
4q(x_{2})H^{(1)}_{0}(k_{1}|x-y|)+O(1+|x-y|\log|x-y|),$ (37)

showing that the principal singularity along the diagonal would be given by

$w_{0+}(x;y)=-i\frac{q(x_{2})}{16}|x-y|^{2}H^{(1)}_{0}(k_{1}|x-y|)+O(|x-y|^{3}\log|x-y|).$
(38)

Even with $q$ given by (29) this is essentially correct.

##  4 Estimates for the Boundary Kernel

In this section we analyze the behavior of the functions

$\mathfrak{w}^{[j]}(x_{2},y_{2})=\left[\frac{1}{i}\partial_{x_{1}}\right]^{j}w^{c}_{0+}(x_{1},x_{2};y_{1},y_{2})\upharpoonright_{x_{1}=y_{1}=0},\,j=0,1,2.$
(39)

With these functions one can easily construct the kernel functions appearing
in the equations in (21). In the following section we explain why similar
estimates hold for $w_{0+}(x_{1},x_{2};0,y_{2}),$ and its derivatives

$\partial_{y_{1}}w_{0+}(x_{1},x_{2};0,y_{2}),\partial_{x_{1}}w_{0+}(x_{1},x_{2};0,y_{2}),\text{
and }\partial_{x_{1}}\partial_{y_{1}}w_{0+}(x_{1},x_{2};0,y_{2}),$

where $x_{1}>0.$ These estimates require tedious, but rather standard
analysis, using the properties of solutions to second order ODEs, stationary
phase and integration by parts. A reader only interested in the final results
can safely skip to Section 4.5.

###  4.1 Estimates on the Wronskian and Basic Solutions of $L_{\xi}$

We first spell out the properties of the basic solutions,
$\widetilde{u}_{\pm}(\xi,0+;x_{2}),$ and their Wronskian, $W(\xi),$ in greater
detail. For these calculations we assume that $q$ is given in (29). Similar
estimates hold for any piecewise continuous, bounded potential, which is
supported in an interval, though we can no longer rely on explicit formulæ
within the support of $q.$

The Wronskian is given by

$W(\xi)=\begin{cases}&-e^{2iBd}\left[(2\xi^{2}-k_{1}^{2}-k_{2}^{2})\frac{\sin
2dA}{A}-2iB\cos 2dA\right]\text{ for }|\xi|<k_{1},\\\
&-e^{-2\widetilde{B}d}\left[(2\xi^{2}-k_{1}^{2}-k_{2}^{2})\frac{\sin
2dA}{A}+2\widetilde{B}\cos 2dA\right]\text{ for }k_{1}<|\xi|<k_{2},\\\
&-e^{-2\widetilde{B}d}\left[(2\xi^{2}-k_{1}^{2}-k_{2}^{2})\frac{\sinh
2d\widetilde{A}}{\widetilde{A}}+2\widetilde{B}\cosh
2d\widetilde{A}\right]\text{ for }k_{2}<|\xi|,\end{cases}$ (40)

where

$\begin{split}A=\sqrt{k_{2}^{2}-\xi^{2}},\,&B=\sqrt{k_{1}^{2}-\xi^{2}},\\\
\widetilde{A}=\sqrt{\xi^{2}-k_{2}^{2}},\,&\widetilde{B}=\sqrt{\xi^{2}-k_{1}^{2}}.\end{split}$
(41)

These formulæ and those for $\widetilde{u}_{+}$ below are provided for the
convenience of the reader, as all can be derived from the formula for
$|\xi|<k_{1}$ by choosing the correct branch of the square root: $\sqrt{z}$ is
defined in $\mathbb{C}\setminus(-\infty,0],$ with $\sqrt{x}\in(0,\infty)$ for
$x\in(0,\infty).$

The Wronskian is analytic in a neighborhood of $\pm k_{2},$ (where
$A,\widetilde{A}$ vanish), but has square root singularities at $\pm k_{1}.$
It never vanishes at $\pm k_{2},$ and does not vanish at $\pm k_{1},$ provided
that

$2d\sqrt{k_{2}^{2}-k_{1}^{2}}\neq n\pi\text{ for any }n\in\mathbb{Z}.$ (42)

As $|\xi|\to\infty,$ we have the lower bound

$|W(\xi)|>M|\xi|.$ (43)

Since $q$ is supported in $[-d,d]$ the asymptotics of $\tilde{w}_{0+}$ as
$|x_{2}|\to\infty$ are given by

$\begin{split}\tilde{w}_{0+}(\xi,x_{2};y)&=-\frac{\widetilde{u}_{+}(\xi,0+;x_{2})e^{-iy_{1}\xi}}{2W(\xi)\sqrt{\xi^{2}-k_{1}^{2}}}\times\\\
&\int_{-d}^{d}\widetilde{u}_{+}(\xi,0+;-z_{2})q(z_{2})e^{-|z_{2}-y_{2}|\sqrt{\xi^{2}-k_{1}^{2}}}dz_{2},\text{
for }x_{2}>d,\\\
\tilde{w}_{0+}(\xi,x_{2};y)&=-\frac{\widetilde{u}_{+}(\xi,0+;-x_{2})e^{-iy_{1}\xi}}{2W(\xi)\sqrt{\xi^{2}-k_{1}^{2}}}\times\\\
&\int_{-d}^{d}\widetilde{u}_{+}(\xi,0+;z_{2})q(z_{2})e^{-|z_{2}-y_{2}|\sqrt{\xi^{2}-k_{1}^{2}}}dz_{2},\text{
for }x_{2}<-d,\end{split}$ (44)

which shows that we need formulæ for $\widetilde{u}_{+}(\xi,0+;z_{2})$ for all
$z_{2}\geq-d.$ We include the formula for $z_{2}<-d,$ with $|\xi|<k_{1},$ as
we need the analytic properties of $\widetilde{u}_{+}(\xi,0+;z_{2})$ in the
appendices. For $|\xi|<k_{1},$

$\widetilde{u}_{+}(\xi,0+;z_{2})=\begin{cases}&e^{iBz_{2}},\text{ for
}z_{2}>d,\\\ &e^{iBd}\left[\cos A(d-z_{2})-iB\frac{\sin
A(d-z_{2})}{A}\right]\text{ for }|z_{2}|<d,\\\ &e^{iBd}\Big{[}\left(\cos
2dA\cos B(d+z_{2})+A\sin 2dA\frac{\sin B(d+z_{2})}{B}\right)+\\\ &iB\left(\cos
2dA\frac{\sin B(d+z_{2})}{B}-\frac{\sin 2dA}{A}\cos
B(d+z_{2})\right)\Big{]}\text{ for }z_{2}<-d.\end{cases}$ (45)

Note that for all $z_{2}<d,$

$\widetilde{u}_{+}(\xi,0+;z_{2})=e^{i\sqrt{k_{1}^{2}-\xi^{2}}d}\left[\theta(\xi,z_{2})+\sqrt{k_{1}^{2}-\xi^{2}}\varphi(\xi,z_{2})\right],$

where $\theta$ and $\varphi$ are entire functions $\xi.$ For the convenience
of the reader we include certain formulæ for $k_{1}<|\xi|<k_{2},$

$\widetilde{u}_{+}(\xi,0+;z_{2})=\begin{cases}&e^{-\widetilde{B}z_{2}},\text{
for }z_{2}>d\\\ &e^{-\widetilde{B}d}\left[\cos
A(d-z_{2})+\widetilde{B}\frac{\sin A(d-z_{2})}{A}\right]\text{ for
}|z_{2}|<d;\end{cases}$ (46)

and for $k_{2}<|\xi|,$

$\widetilde{u}_{+}(\xi,0+;z_{2})=\begin{cases}&e^{-\widetilde{B}z_{2}},\text{
for }z_{2}>d\\\
&e^{-\widetilde{B}d}\left[\cosh\widetilde{A}(d-z_{2})+\widetilde{B}\frac{\sinh\widetilde{A}(d-z_{2})}{\widetilde{A}}\right]\text{
for }|z_{2}|<d.\end{cases}$ (47)

If $\\{\xi_{n}\\}$ are the positive roots of the Wronskian, then

$\gamma^{-}_{n}=\\{-\xi_{n}+\nu e^{i\theta}:\>\theta\in[0,2\pi)\\}\text{ and
}\gamma^{+}_{n}=\\{\xi_{n}+\nu e^{i\theta}:\>\theta\in[0,2\pi)\\}$ (48)

are circles of radius $\nu$ centered in these roots. We let
$-\gamma^{\pm}_{n+}$ denote the intersection of these circles with the upper
half plane, oriented from left to right. Before proceeding with this analysis,
we observe that
$\tilde{w}_{0+}(\xi,x_{2};0,y_{2})=\tilde{w}_{0+}(-\xi,x_{2};0,y_{2}),$ which
implies that, in the integral defining $\mathfrak{w}^{[1]},$ the contributions
from $\Gamma_{\nu}^{+}\cap\mathbb{R}$ cancel exactly, and therefore

$\mathfrak{w}^{[1]}=\frac{1}{2\pi}\sum_{n=1}^{N}\left[\int_{-\gamma^{-}_{n+}}\xi\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi+\int_{-\gamma^{+}_{n+}}\xi\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi\right].$
(49)

Using the symmetries of the integrand it is not difficult to show that

$\mathfrak{w}^{[1]}=\frac{1}{2\pi}\sum_{n=1}^{N}\int_{-\gamma^{+}_{n}}\xi\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi;$
(50)

where $-\gamma^{+}_{n}$ indicates that the circles about the $\\{\xi_{n}\\}$
are oriented in the clockwise direction. This gives $-i$ times the sum of the
residues of $\xi\tilde{w}_{0+}(\xi,x_{2};0,y_{2})$ at the $\\{\xi_{n}\\}.$ It
is not difficult to see that

$i\mathfrak{w}^{[1]}+\partial_{x_{1}}w_{0+}^{g}(0,x_{2};0,y_{2})=0,$ (51)

as the function $\partial_{x_{1}}w_{0+}^{g}$ is $i$ times the sum of the
residues of $i\xi\tilde{w}_{0+}(\xi,x_{2};0,y_{2})$ at the $\\{\xi_{n}\\}.$
Hence the kernel $\mathfrak{w}^{[1]}$ does not play any role in the integral
equations derived in Section 6, nonetheless we include estimates for its
behavior as we need these results for the representation formula, which
involves $\partial_{y_{1}}w^{c}_{0+}(x;0,y_{2}).$

###  4.2 Asymptotics for $|x_{2}|,|y_{2}|>d$

For $x_{2}>d,$ we write these functions in the form

$\mathfrak{w}^{[j]}(x_{2},y_{2})=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}\xi^{j}\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi\\\
=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}\xi^{j}\frac{\widetilde{u}_{+}(\xi,0+;x_{2})\mathfrak{A}(\xi,\sqrt{\xi^{2}-k_{1}^{2}};y_{2})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi,$
(52)

where, for $|\xi|<k_{1},$
$\sqrt{\xi^{2}-k_{1}^{2}}=-i\sqrt{k_{1}^{2}-\xi^{2}}$ and we write

$\mathfrak{A}(\xi,\eta;y_{2})=\int_{-d}^{d}q(z_{2})e^{-\eta|y_{2}-z_{2}|}e^{-\eta
d}\left[\cos A(d+z_{2})+\eta\frac{\sin A(d+z_{2})}{A}\right]dz_{2}.$ (53)

We rewrite the Wronskian as
$W(\xi)=\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}}),$ with

$\mathfrak{W}(\xi,\eta)=-e^{2i\eta
d}\left[(2\xi^{2}-k_{1}^{2}-k_{2}^{2})\frac{\sin 2dA}{A}-2i\eta\cos
2dA\right]\text{ for }|\xi|<k_{1}.$ (54)

These formulæ hold for other ranges of $\xi$ by using the correct branch of
$B=\sqrt{k_{1}^{2}-\xi^{2}};$ as noted these formulæ do not depend on the
choice of branch for $A=\sqrt{k_{2}^{2}-\xi^{2}}.$ We use this formulation, as
$\mathfrak{A}(\xi,\eta;y_{2}),$ and $\mathfrak{W}(\xi,\eta)$ are entire
functions of $(\xi,\eta),$ which allows us to better keep track of the square
root singularities at $\pm k_{1}$ in our subsequent computations.

To analyze $\mathfrak{w}^{[j]}$ we split the integral over $\xi$ into the
segment where $|\xi|<k_{1},$ a segment with $k_{1}<|\xi|<k_{2},$ and finally
the segment where $k_{2}<|\xi|.$ The leading order behavior is largely
determined, via stationary phase, by the integral over small neighborhood of
$\xi=0;$ whereas the diagonal singularities come from large $|\xi|.$

We begin with $\xi\in[-k_{1},k_{1}],$ and denote these contributions by
$\mathfrak{w}^{[j]}_{0}(x_{2},y_{2}):$

$\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})=\frac{1}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi\\\
=\frac{1}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\frac{\widetilde{u}_{+}(\xi,0+;x_{2})\mathfrak{A}(\xi,\sqrt{\xi^{2}-k_{1}^{2}};y_{2})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi.$
(55)

We first consider $x_{2},y_{2}>d,$ so that

$\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})=\frac{1}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\tilde{w}_{0+}(\xi,x_{2};0,y_{2})d\xi\\\
=\frac{i}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\frac{e^{i\sqrt{k_{1}^{2}-\xi^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,-i\sqrt{k_{1}^{2}-\xi^{2}})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi,$
(56)

where

$\mathfrak{A}_{0}(\xi,\eta)=\int_{-d}^{d}q(z_{2})e^{\eta z_{2}}e^{-\eta
d}\left[\cos A(d+z_{2})+\eta\frac{\sin A(d+z_{2})}{A}\right]dz_{2}.$ (57)

We split the integral in (56) into the part near to $\xi=0,$ denoted
$\mathfrak{w}^{[j]}_{00},$ where the phase is stationary and the part near to
$\xi=\pm k_{1},$ denoted $\mathfrak{w}^{[j]}_{01},$ where the integrand
appears to have a singularity. As we will see, this singularity is removable
by changing variables. Let
$\phi\in\mathcal{C}^{\infty}_{c}((-\frac{1}{2},\frac{1}{2})),$ be a non-
negative, even function with $\phi(t)=1$ for $|t|<\frac{1}{4}.$ Using a
standard stationary phase argument we see that there are constants
$M_{0},M_{1},M_{2},$ so that

$\mathfrak{w}^{[j]}_{00}(x_{2},y_{2})=\frac{i}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\frac{e^{i\sqrt{k_{1}^{2}-\xi^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,-i\sqrt{k_{1}^{2}-\xi^{2}})\phi(\xi/k_{1})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi=\\\
M_{j}\frac{e^{ik_{1}(x_{2}+y_{2})}}{(x_{2}+y_{2})^{\frac{j+1}{2}}}+O\left((x_{2}+y_{2})^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (58)

To treat the other part of the integral we let $s=\sqrt{k_{1}^{2}-\xi^{2}},$
to obtain a smooth integrand that can be integrated by parts giving

$\mathfrak{w}^{[j]}_{01}(x_{2},y_{2})=\\\
\frac{i}{\pi}\int_{0}^{k_{1}}(k_{1}^{2}-s^{2})^{\frac{j}{2}}\frac{e^{is(x_{2}+y_{2})}\mathfrak{A}_{0}(\sqrt{k_{1}^{2}-s^{2}},-is)[1-\phi(\sqrt{k_{1}^{2}-s^{2}}/k_{1})}{\mathfrak{W}(\sqrt{k_{1}^{2}-s^{2}},s)\sqrt{k_{1}^{2}-s^{2}}}ds=\\\
-\frac{k_{1}^{j-1}}{\pi(x_{2}+y_{2})}\frac{\mathfrak{A}_{0}(k_{1},0)}{\mathfrak{W}(k_{1},0)}+O\left((x_{2}+y_{2})^{-2}\right).$
(59)

The leading term here will cancel with the leading contribution from this
endpoint for the integrals over set $k_{1}<|\xi|<k_{2}.$ So overall, the
contribution of this endpoint will be $O\left((x_{2}+y_{2})^{-2}\right).$

We next consider, $\mathfrak{w}^{[j]}_{1},$ the contributions from
$k_{1}<|\xi|<k_{2},$ which includes small semi-circles around the zeros of the
Wronskian. We first consider the portions of these integrals from near to
$\xi=\pm k_{1},$ which we denote $\mathfrak{w}^{[j]}_{10}(x_{2},y_{2}).$ By
evenness of the integrand it suffices to consider $\xi\in[k_{1},k_{1}+\mu].$
Let $\phi\in\mathcal{C}^{\infty}_{c}([0,\mu)),$ for an
$0<\mu\ll\xi_{1}-k_{1},$ with $\phi(t)=1$ for $t\in[0,\mu/2].$ Set
$s=\sqrt{\xi^{2}-k_{1}^{2}},$ and integrate by parts to get

$\mathfrak{w}^{[j]}_{10}(x_{2},y_{2})=\frac{1}{\pi}\int_{k_{1}}^{k_{1}+\mu}\xi^{j}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})\phi(\sqrt{\xi^{2}-k^{2}_{1}})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi=\\\
\frac{1}{\pi}\int_{0}^{\sqrt{2k_{1}\mu+\mu^{2}}}(s^{2}+k_{1}^{2})^{\frac{j}{2}}\frac{e^{-s(x_{2}+y_{2})}\mathfrak{A}_{0}(\sqrt{s^{2}+k_{1}^{2}},s)\phi(s)}{\mathfrak{W}(\sqrt{s^{2}+k_{1}^{2}},is)\sqrt{s^{2}+k_{1}^{2}}}ds=\\\
\frac{k_{1}^{j-1}}{\pi(x_{2}+y_{2})}\frac{\mathfrak{A}_{0}(k_{1},0)}{\mathfrak{W}(k_{1},0)}+O\left((x_{2}+y_{2})^{-2}\right).$
(60)

As stated above,

$\mathfrak{w}^{[j]}_{10}(x_{2},y_{2})+\mathfrak{w}^{[j]}_{01}(x_{2},y_{2})=O\left((x_{2}+y_{2})^{-2}\right);$
(61)

in fact the sum of these terms vanish to all orders, which can be proved by
deforming the portions of the contour near to $\pm k_{1}.$

Now we need to estimate the contributions from the remainder of
$\Gamma_{\nu}^{+}$ lying over
$[k_{1}+\mu/2,k_{2}]\cup[-k_{2},-(k_{1}+\mu/2)].$ Other than the semi-circles
centered on the roots of the Wronskian,

$\bigcup_{n=1}^{N}\\{\pm\xi_{n}+\nu
e^{i\theta}:\>\theta\in[\pi,0]\\}=\bigcup_{n=1}^{N}C^{\pm}_{n,\nu},$ (62)

we use the evenness of the integrand to restrict our attention to
$[k_{1}+\mu/2,k_{2}].$

To estimates the contributions from the semi-circles we need to estimate both
$|\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})|,$ from above, and
$|\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})|,$ from below, on these arcs. As
the roots of the Wronskian are all simple, there is a constant $M$ so that

$|\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})|\geq M\nu\text{ for }\xi\in
C^{\pm}_{n,\nu}.$ (63)

On the other hand, for $\xi\in C^{\pm}_{n,\nu},$

$|\mathfrak{A}_{0}(\xi,\eta)|\leq\int_{-d}^{d}q(z_{2})e^{\alpha(z_{2}-d)}\left[\exp\beta(d-z_{2})+|\eta|\frac{\exp\beta(d-z_{2})}{|\sqrt{k_{2}^{2}-\xi^{2}}|}\right]dz_{2},$
(64)

where

$\alpha=\operatorname{Re}\left[\sqrt{(\pm\xi_{n}+\nu
e^{i\theta})^{2}-k_{1}^{2}}\right]\text{ and
}\beta=\left|\operatorname{Im}\left[\sqrt{k_{2}^{2}-(\pm\xi_{n}+\nu
e^{i\theta})^{2}}\right]\right|.$ (65)

There is a constant, $m$ so that, for $\theta\in[0,2\pi],$ we have

$\alpha>\sqrt{\xi_{n}^{2}-k_{1}^{2}}(1-m\nu)\text{ and }\beta<m\nu.$ (66)

These inequalities rely on the fact that $k_{1}<\xi_{j}<k_{2}.$ If we choose
$0<\nu$ sufficiently small so that

$m\nu-\sqrt{\xi_{n}^{2}-k_{1}^{2}}(1-m\nu)<0\text{ for }n=1,\dots,N,$ (67)

then it follows that there exists a constant, $M,$ so that

$|\mathfrak{A}_{0}(\xi,\eta)|\leq M\text{ for }\xi\in C^{\pm}_{n,\nu}\text{
for }n=1,\dots,N.$ (68)

For sufficiently small $\nu$ we therefore have the estimates

$\left|\sum_{n=1}^{N}\int_{C_{n,\nu}^{\pm}}\xi^{j}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi\right|\leq\\\
M\sum_{n=1}^{N}\frac{e^{-\sqrt{\xi_{n}^{2}-k_{1}^{2}}(1-m\nu)(x_{2}+y_{2})}}{\nu}.$
(69)

To treat $\Gamma_{\nu}^{+}\cap[k_{1}+\mu/2,k_{2}],$ we see that
$|\mathfrak{A}_{0}(\xi,\eta)|$ is bounded by a constant on this set and
therefore

$\left|\,\int\limits_{\Gamma_{\nu}^{+}\cap[k_{1}+\mu/2,k_{2}]}\xi^{j}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi\right|\leq\\\
M\frac{e^{-\mu k_{1}(x_{2}+y_{2})/2}}{\nu}.$ (70)

To summarize, we have shown that for $x_{2},y_{2}>d,$ the contributions to the
integrals defining $\mathfrak{w}^{[j]}(x_{2},y_{2}),$ from frequencies
$\xi\in[-k_{2},k_{2}]$ are

$\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})+\mathfrak{w}^{[j]}_{1}(x_{2},y_{2})=C_{j}\frac{e^{ik_{1}(x_{2}+y_{2})}}{(x_{2}+y_{2})^{\frac{j+1}{2}}}+O\left((x_{2}+y_{2})^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (71)

We conclude the discussion of this case by estimating the contribution from
$|\xi|>k_{2},$ which we denote by $\mathfrak{w}^{[j]}_{2}(x_{2},y_{2}).$ As
before, by evenness, it suffices to consider $[k_{2},\infty).$ Using the
formulæ above we see that

$\mathfrak{w}_{2}^{[j]}(x_{2},y_{2})=\int_{k_{2}}^{\infty}\xi^{j}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}+y_{2})}\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi,$
(72)

where

$\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})=\int_{-d}^{d}q(z_{2})e^{\sqrt{\xi^{2}-k_{1}^{2}}(z_{2}-d)}\times\\\
\left[\cosh\sqrt{\xi^{2}-k_{2}^{2}}(d+z_{2})+\sqrt{\xi^{2}-k_{1}^{2}}\frac{\sinh\sqrt{\xi^{2}-k_{2}^{2}}(d+z_{2})}{\sqrt{\xi^{2}-k_{2}^{2}}}\right]dz_{2}.$
(73)

From this formula we can easily show that there is a constant $M$ so that

$|\mathfrak{A}_{0}(\xi,\sqrt{\xi^{2}-k_{1}^{2}})|\leq\frac{Me^{2d\sqrt{\xi^{2}-k_{1}^{2}}}}{1+|\xi|}\text{
for }\xi\in[k_{2},\infty),$ (74)

and therefore, for $|x_{2}|,|y_{2}|>d,\,|\xi|>k_{2}$ we have the estimate

$|\tilde{w}_{0+}(\xi,x_{2};0,y_{2})|\leq
M\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(|x_{2}|+|y_{2}|-2d)}}{(1+|\xi|)^{3}}.$
(75)

Hence

$\begin{split}|\mathfrak{w}_{2}^{[j]}(x_{2},y_{2})|&\leq
M\int_{k_{2}}^{\infty}\frac{\xi^{j}e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}+y_{2}-2d)}}{(1+|\xi|)^{3}}d\xi\\\
&\leq\begin{cases}&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(x_{2}+y_{2})}\text{ for
}j=0,1\\\
&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(x_{2}+y_{2})}\log\left(\frac{2(x_{2}+y_{2})}{x_{2}+y_{2}-2d}\right)\text{
for }j=2.\end{cases}\end{split}$ (76)

The calculation above indicates that the diagonal singularity should be
$|x-y|^{2}\log|x-y|,$ and this implies that we should expect a
$\log$-singularity in $\mathfrak{w}^{[2]}$ where $x_{2}+y_{2}=2d.$

Altogether we have shown that, for $x_{2},y_{2}>d$ we have the asymptotic
formulæ

$\mathfrak{w}^{[j]}(x_{2},y_{2})=C_{j}\frac{e^{ik_{1}(x_{2}+y_{2})}}{(x_{2}+y_{2})^{\frac{j+1}{2}}}+O\left((x_{2}+y_{2})^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (77)

We can also analyze

$\partial_{x_{2}}\mathfrak{w}^{[j]}(x_{2},y_{2})-ik_{1}\mathfrak{w}^{[j]}(x_{2},y_{2}),\text{
and
}\partial_{y_{2}}\mathfrak{w}^{[j]}(x_{2},y_{2})-ik_{1}\mathfrak{w}^{[j]}(x_{2},y_{2}),$
(78)

which are only significantly different in the contribution from
$\xi\in[-k_{1},k_{1}].$ In this interval we get an additional factor of
$i(\sqrt{k_{1}^{2}-\xi^{2}}-k_{1})=O(\xi^{2})$ in the integrand. From the
usual stationary phase argument it will follow that

$\begin{split}\partial_{x_{2}}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})-ik_{1}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})&=O\left((x_{2}+y_{2})^{-\frac{j+3}{2}}\right)\\\
\partial_{y_{2}}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})-ik_{1}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})&=O\left((x_{2}+y_{2})^{-\frac{j+3}{2}}\right).\end{split}$
(79)

It is not difficult to see that the same arguments apply if
$|x_{2}|,|y_{2}|>d$ to show that

$\mathfrak{w}^{[j]}(x_{2},y_{2})=C_{j}\frac{e^{ik_{1}(|x_{2}|+|y_{2}|)}}{(|x_{2}|+|y_{2}|)^{\frac{j+1}{2}}}+O\left((|x_{2}|+|y_{2}|)^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (80)

The function $\mathfrak{w}^{[0]}(x_{2},y_{2})$ is continuous as
$(x_{2},y_{2})\to(\pm d,\pm d),$ and

$|\mathfrak{w}^{[2]}(x_{2},y_{2})|\leq M|\log(|x_{2}+y_{2}|-2d)|,\text{ as
}|x_{2}+y_{2}|\to 2d.$ (81)

There is no singularity as $(x_{2},y_{2})\to(\pm d,\mp d).$ We can also show

$\begin{split}&\partial_{x_{2}}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{j+3}{2}}\right)\text{
for }\pm x_{2}>d,\\\ &\partial_{y_{2}}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[j]}_{0}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{j+3}{2}}\right)\text{
for }\pm y_{2}>d.\end{split}$ (82)

Similar conclusions hold for $w_{0+}=w_{0+}^{c}+w_{0+}^{g},$ as the
contribution of the wave guide modes is exponentially decaying as
$|x_{2}|+|y_{2}|\to\infty.$

###  4.3 Asymptotics for $|x_{2}|$ or $|y_{2}|<d$

We now consider what happens if either $|x_{2}|<d,$ or $|y_{2}|<d.$ We start
by assuming that $y_{2}>d,$ but $|x_{2}|<d.$ For this case the integrals
defining $\mathfrak{w}^{[j]}$ take the form

$\mathfrak{w}^{[j]}(x_{2},y_{2})=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}\xi^{j}\frac{e^{i\sqrt{k_{1}^{2}-\xi^{2}}(y_{2}+2d)}\mathfrak{B}(\xi,\sqrt{k_{1}^{2}-\xi^{2}};x_{2})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi,$
(83)

where

$\mathfrak{B}(\xi,\eta;x_{2})=\int\limits_{-d}^{x_{2}}\Bigg{[}\cos
A(d-x_{2})\cos A(d+z_{2})-i\frac{\eta}{A}\sin A(2d+z_{2}-x_{2})-\\\
\frac{\eta^{2}}{A^{2}}\sin A(d-x_{2})\sin
A(d+z_{2})\Bigg{]}e^{-iz_{2}\eta}q(z_{2})dz_{2}+\\\
\int\limits_{x_{2}}^{d}\Bigg{[}\cos A(d+x_{2})\cos
A(d-z_{2})-i\frac{\eta}{A}\sin A(2d-z_{2}+x_{2})-\\\
\frac{\eta^{2}}{A^{2}}\sin A(d+x_{2})\sin
A(d-z_{2})\Bigg{]}e^{-iz_{2}\eta}q(z_{2})dz_{2}.$ (84)

Again $\mathfrak{B}(\xi,\eta;x_{2})$ is an entire function of $(\xi,\eta).$ In
the integral over $\Gamma_{\nu}^{+}$ we take
$\eta=\sqrt{k_{1}^{2}-\xi^{2}}=i\sqrt{\xi^{2}-k_{1}^{2}}$ for $k_{1}<|\xi|.$

As before the principal contribution to $\mathfrak{w}^{[j]}(x_{2},y_{2})$, as
$y_{2}\to\infty,$ comes from the stationary phase at $\xi=0.$ The function
$|\mathfrak{B}(\xi,\sqrt{k_{1}^{2}-\xi^{2}};x_{2})|$ is easily seen to be
uniformly bounded where $|\xi|<k_{1},|x_{2}|<d.$

We separate the contributions from the stationary point at zero, and the
contributions from endpoints $\pm k_{1}.$ The contributions from the
stationary point are

$\mathfrak{w}^{[j]}_{00}(x_{2},y_{2})=C^{\prime}_{j}k_{1}^{\frac{j-1}{2}}\frac{e^{ik_{1}y_{2}}\mathfrak{B}(0,k_{1};x_{2})}{\mathfrak{W}(0,k_{1})\,y_{2}^{\frac{j+1}{2}}}+O\left(|y_{2}|^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (85)

To obtain the contributions from near to $\pm k_{1}$ we again change
variables, letting $s=\sqrt{k_{1}^{2}-\xi^{2}},$ and use evenness, to obtain

$\mathfrak{w}^{[j]}_{01}(x_{2},y_{2})=\frac{1}{\pi}\int_{0}^{k_{1}/2}(k_{1}^{2}-s^{2})^{\frac{j}{2}}\frac{e^{is(y_{2}+2d)}\mathfrak{B}(\sqrt{k_{1}^{2}-s^{2}},s,;x_{2})\phi(s)}{\mathfrak{W}(\sqrt{k_{1}^{2}-s^{2}},s)\sqrt{k_{1}^{2}-s^{2}}}ds,$
(86)

for a suitable bump function, $\phi,$ supported near 0. Integrating by parts
we see that this equals

$\frac{ik_{1}^{j-1}}{\pi
y_{2}}\frac{\mathfrak{B}(k_{1},0;x_{2})}{\mathfrak{W}(k_{1},0)}+O\left(y_{2}^{-2}\right).$
(87)

We now consider the part of $\Gamma_{\nu}^{+}$ lying over $k_{1}<|\xi|<k_{2}.$
The contributions from near to $\pm k_{1}$ are computed as before in (60), and
are seen to cancel the leading term in (60), leaving a term of
$O(y_{2}^{-2}).$ We next consider the integral over $\Gamma_{\nu}^{+}$ lying
above $k_{1}+\mu<|\xi|<k_{2},$ which is of the form

$\mathfrak{w}^{[j]}_{11}(x_{2},y_{2})=\\\
\frac{1}{2\pi}\int\limits_{\\{k_{1}+\mu<|\xi|<k_{2}\\}\cap\Gamma_{\nu}^{+}}\xi^{j}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}(y_{2}+2d)}\mathfrak{B}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};x_{2})\widetilde{\phi}(\xi)}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi,$
(88)

for a suitable cut-off function $\widetilde{\phi}.$ As before we use the lower
bound on the Wronskian, $|\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})|>m\nu.$
We also need to bound $\mathfrak{B}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};x_{2})$ from
above.

For $|x_{2}|<d,$ we can easily show that there is a constant $M$ so that

$|\mathfrak{B}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};x_{2})|\leq
Me^{\rho(\xi)d},\text{ where
}\rho(\xi)=\operatorname{Re}\left(\sqrt{\xi^{2}-k_{1}^{2}}\right).$ (89)

It then follows that

$|\mathfrak{w}^{[j]}_{11}(x_{2},y_{2})|\leq\frac{M}{\nu}\int\limits_{k_{1}+\mu<|\xi|<k_{2}}e^{-\rho(\xi)(y_{2}+d)}d\xi$
(90)

If we set

$\alpha=\min\\{\rho(\xi):\>\xi\in\Gamma_{\nu}^{+}\text{ with
}k_{1}+\mu<|\xi|<k_{2}\\},$ (91)

then $\alpha>0,$ provided that $\nu$ is sufficiently small, from which it
follows easily that

$|\mathfrak{w}^{[j]}_{11}(x_{2},y_{2})|\leq\frac{Me^{-\alpha y_{2}}}{\nu}.$
(92)

This leaves the integral over the set $\\{|\xi|>k_{2}\\}.$ In this set we can
prove the estimate

$\left|\mathfrak{B}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};x_{2})\right|\leq
Me^{(2d+x_{2})\sqrt{\xi^{2}-k_{1}^{2}}}\left[\frac{1}{1+|\xi|}+(d-x_{2})\right],$
(93)

and therefore, if either $|x_{2}|<d$ or $|y_{2}|<d,$ and $|\xi|>k_{2},$ we
have the estimate

$|\tilde{w}_{0+}(\xi,x_{2};0,y_{2})|<M\frac{e^{-\sqrt{\xi^{2}-k_{2}^{2}}|x_{2}-y_{2}|}}{(1+|\xi|)^{2}}\left[\frac{1}{(1+|\xi|)}+|y_{2}-x_{2}|\right],$
(94)

and therefore

$\begin{split}|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|&=\left|\frac{1}{2\pi}\int_{k_{2}<|\xi|}\xi^{j}\frac{e^{\sqrt{\xi^{2}-k_{1}^{2}}(y_{2}+2d)}\mathfrak{B}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};x_{2})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi\right|\\\
&\leq
M\int_{k_{2}}^{\infty}e^{-\sqrt{\xi^{2}-k_{1}^{2}}(y_{2}-x_{2})}\left[\frac{1}{(1+|\xi|)^{3-j}}+\frac{y_{2}-x_{2}}{(1+|\xi|)^{2-j}}\right]d\xi.\end{split}$
(95)

An elementary calculation now shows that

$|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|\leq\begin{cases}&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(y_{2}-x_{2})}\text{
for }j=0,1\\\
&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(y_{2}-x_{2})}\log\left(\frac{2y_{2}}{y_{2}-x_{2}}\right)\text{
for }j=2.\end{cases}$ (96)

We can again show that this analysis applies equally well if
$|x_{2}|<d,y_{2}<-d,$ and therefore we have the estimates

$\mathfrak{w}^{[j]}(x_{2},y_{2})=C^{\prime}_{j}k_{1}^{\frac{j-1}{2}}\frac{e^{ik_{1}|y_{2}|}\mathfrak{B}(0,k_{1};x_{2})}{\mathfrak{W}(0,k_{1})\,|y_{2}|^{\frac{j+1}{2}}}+O\left(|y_{2}|^{-\frac{j+3}{2}}\right)\text{
for }|y_{2}|>d,\,|x_{2}|<d.$ (97)

The functions
$\mathfrak{w}^{[0]}(x_{2},y_{2}),\mathfrak{w}^{[1]}(x_{2},y_{2})$ are
continuous as $y_{2}\to x_{2},$ and

$|\mathfrak{w}^{[2]}(x_{2},y_{2})|\leq M|\log|y_{2}-x_{2}||,\text{ as
}y_{2}\to x_{2}.$ (98)

As before we can show that

$\partial_{y_{2}}\mathfrak{w}^{[j]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[j]}(x_{2},y_{2})=O\left(|y_{2}|^{-\frac{j+3}{2}}\right).$
(99)

The final case we need to treat for asymptotics is $|y_{2}|<d,|x_{2}|>d.$ We
begin with $x_{2}>d.$ For this case we write

$\mathfrak{w}^{[j]}(x_{2},y_{2})=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}\frac{\xi^{j}e^{i\sqrt{k_{1}^{2}-\xi^{2}}x_{2}}\mathfrak{C}(\xi,\sqrt{k_{1}^{2}-\xi^{2}};y_{2})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi,$
(100)

where

$\mathfrak{C}(\xi,\eta;y_{2})=e^{i\eta(y_{2}+d)}\int_{-d}^{y_{2}}e^{-i\eta
z_{2}}\left[\cos A(d+z_{2})-i\eta\frac{\sin
A(d+z_{2})}{A}\right]q(z_{2})dz_{2}+\\\
e^{i\eta(d-y_{2})}\int_{y_{2}}^{d}e^{i\eta z_{2}}\left[\cos
A(d+z_{2})-i\eta\frac{\sin A(d+z_{2})}{A}\right]q(z_{2})dz_{2},\text{ with
}A=\sqrt{k_{2}^{2}-\xi^{2}}.$ (101)

As in the earlier cases, there is a stationary phase contribution from
$\xi=0,$ which we denote by $\mathfrak{w}^{[j]}_{00}(x_{2},y_{2}).$ The
function $\mathfrak{C}(\xi,\sqrt{k_{1}^{2}-\xi^{2}};y_{2})$ is an analytic
function of $\xi$ in $|\xi|<k_{1}-\epsilon,$ for any $\epsilon>0.$ The
stationary phase contributions are

$\mathfrak{w}^{[j]}_{00}(x_{2},y_{2})=C_{j}\frac{e^{ik_{1}x_{2}}\mathfrak{C}(0,k_{1};y_{2})}{\mathfrak{W}(0,k_{1})\,x_{2}^{\frac{j+1}{2}}}+O\left(x_{2}^{-\frac{j+3}{2}}\right)\text{
for }j=0,1,2.$ (102)

The analysis for the contribution from $\xi$ near to $\pm k_{1},$ which we
denote by $\mathfrak{w}^{[j]}_{01}(x_{2},y_{2}),$ is as before: we change
variables setting $s=\sqrt{k_{1}^{2}-\xi^{2}},$ to obtain

$\mathfrak{w}^{[j]}_{01}(x_{2},y_{2})=\frac{1}{\pi}\int_{0}^{k_{1}/2}\frac{e^{isx_{2}}\,(k_{1}^{2}-s^{2})^{\frac{j-1}{2}}\mathfrak{C}(\sqrt{k_{1}^{2}-s^{2}},s;y_{2})\phi(s)}{\mathfrak{W}(\sqrt{k_{1}^{2}-s^{2}},s)}ds,$
(103)

with $\phi\in\mathcal{C}^{\infty}([0,k_{1}/2)),$ equal to $1$ for near $s=0.$
Integration by parts gives

$\mathfrak{w}^{[j]}_{01}(x_{2},y_{2})=-\frac{k_{1}^{j-1}}{i\pi
x_{2}}\cdot\frac{\mathfrak{C}(k_{1},0;y_{2})}{\mathfrak{W}(k_{1},0)}+O(x_{2}^{-2});$
(104)

the leading terms here will cancel with the leading terms from near to $\pm
k_{1}$ in the intervals $[-k_{2},-k_{1}]\cup[k_{1},k_{2}],$ leaving an
$O(x_{2}^{-2})$ contribution from this endpoint. The contributions from
$[k_{1},\sqrt{k_{1}^{2}+\epsilon^{2}}]$ take the form

$\begin{split}\mathfrak{w}^{[j]}_{10}(x_{2},y_{2})=&\frac{1}{i\pi}\int_{k_{1}}^{\sqrt{k_{1}^{2}+\epsilon^{2}}}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}x_{2}}\,\xi^{j}\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})\widetilde{\phi}(\sqrt{\xi^{2}-k_{1}^{2}})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi\\\
=&\frac{1}{i\pi}\int_{0}^{\epsilon}\frac{e^{-sx_{2}}\,(s^{2}+k_{1}^{2})^{\frac{j-1}{2}}\mathfrak{C}(\sqrt{s^{2}+k_{1}^{2}},is;y_{2})\widetilde{\phi}(s)}{\mathfrak{W}(\sqrt{s^{2}+k_{1}^{2}},is)}ds\end{split}$
(105)

with $\widetilde{\phi}\in\mathcal{C}^{\infty}_{c}([0,\epsilon)),$ equal to $1$
near 0. In the second line we let $s=\sqrt{k_{1}^{2}-\xi^{2}}.$ Integrating by
parts gives

$\mathfrak{w}^{[j]}_{10}(x_{2},y_{2})=\frac{k_{1}^{j-1}}{i\pi
x_{2}}\cdot\frac{\mathfrak{C}(k_{1},0;y_{2})}{\mathfrak{W}(k_{1},0)}+O(x_{2}^{-2});$
(106)

showing the cancellation between the leading term here and in (104).

The remaining portion for $\sqrt{k_{1}^{2}+\epsilon^{2}}<|\xi|<k_{2}$ is given
by

$\mathfrak{w}^{[j]}_{11}(x_{2},y_{2})=\\\
\frac{1}{i\pi}\int\limits_{\\{\sqrt{k_{1}^{2}+\epsilon^{2}}<|\xi|<k_{2}\\}\cap\Gamma_{\nu}^{+}}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}x_{2}}\,\xi^{j}\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})(1-\widetilde{\phi}(\sqrt{\xi^{2}-k_{1}^{2}}))}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi,$
(107)

where

$\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})=e^{-\sqrt{\xi^{2}-k_{1}^{2}}(y_{2}+d)}\times\\\
\int_{-d}^{y_{2}}e^{\sqrt{\xi^{2}-k_{1}^{2}}z_{2}}\left[\cos
A(d+z_{2})+\sqrt{\xi^{2}-k_{1}^{2}}\frac{\sin
A(d+z_{2})}{A}\right]q(z_{2})dz_{2}+\\\
e^{-\sqrt{\xi^{2}-k_{1}^{2}}(d-y_{2})}\times\\\
\int_{y_{2}}^{d}e^{-\sqrt{\xi^{2}-k_{1}^{2}}z_{2}}\left[\cos
A(d+z_{2})+\sqrt{\xi^{2}-k_{1}^{2}}\frac{\sin
A(d+z_{2})}{A}\right]q(z_{2})dz_{2}\\\ \text{ with
}A=\sqrt{k_{2}^{2}-\xi^{2}}.$ (108)

It is easy to see that for $|y_{2}|<d,$
$|\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})|$ is bounded, and the
Wronskian satisfies $|\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})|>M\nu,$ on
the part of $\Gamma_{\nu}^{+}$ lying over
$\sqrt{k_{1}^{2}+\epsilon^{2}}<|\xi|<k_{2}.$ Moreover
$\operatorname{Re}\sqrt{\xi^{2}-k_{1}^{2}}>\alpha>0,$ therefore

$|\mathfrak{w}^{[j]}_{11}(x_{2},y_{2})|\leq\frac{Me^{-\alpha x_{2}}}{\nu}.$
(109)

Where $|\xi|>k_{2}$ we have

$\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})=e^{-\sqrt{\xi^{2}-k_{1}^{2}}(y_{2}+d)}\times\\\
\int_{-d}^{y_{2}}e^{\sqrt{\xi^{2}-k_{1}^{2}}z_{2}}\left[\cosh
A(d+z_{2})+\sqrt{\xi^{2}-k_{1}^{2}}\frac{\sinh
A(d+z_{2})}{A}\right]q(z_{2})dz_{2}+\\\
e^{-\sqrt{\xi^{2}-k_{1}^{2}}(d-y_{2})}\times\\\
\int_{y_{2}}^{d}e^{-\sqrt{\xi^{2}-k_{1}^{2}}z_{2}}\left[\cosh
A(d+z_{2})+\sqrt{\xi^{2}-k_{1}^{2}}\frac{\sinh
A(d+z_{2})}{A}\right]q(z_{2})dz_{2}\\\ \text{ with
}A=\sqrt{\xi^{2}-k_{2}^{2}}.$ (110)

The first term in (110) is bounded by

$M\frac{e^{\sqrt{\xi^{2}-k_{2}^{2}}y_{2}}}{1+|\xi|};$ (111)

the other term requires somewhat more care. It is bounded by

$Me^{\sqrt{\xi^{2}-k_{1}^{2}}y_{2}}\int_{y_{2}}^{d}e^{(\sqrt{\xi^{2}-k_{2}^{2}}-\sqrt{\xi^{2}-k_{1}^{2}})z_{2}}dz_{2}=\\\
Me^{\sqrt{\xi^{2}-k_{1}^{2}}y_{2}}\left[\frac{e^{(\sqrt{\xi^{2}-k_{2}^{2}}-\sqrt{\xi^{2}-k_{1}^{2}})d}-e^{(\sqrt{\xi^{2}-k_{2}^{2}}-\sqrt{\xi^{2}-k_{1}^{2}})y_{2}}}{\sqrt{\xi^{2}-k_{2}^{2}}-\sqrt{\xi^{2}-k_{1}^{2}}}\right].$
(112)

An elementary argument then shows that the expression in the bracket is
bounded by $|d-y_{2}|$ and therefore

$|\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})|\leq
Me^{\sqrt{\xi^{2}-k_{1}^{2}}y_{2}}\left[\frac{1}{1+|\xi|}+|d-y_{2}|\right].$
(113)

These estimates show that

$\begin{split}|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|&=\frac{1}{\pi}\left|\int_{k_{2}<|\xi|}\frac{e^{-\sqrt{\xi^{2}-k_{1}^{2}}x_{2}}\,\xi^{j}\mathfrak{C}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}};y_{2})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi\right|\\\
&\leq
M\int_{k_{2}}^{\infty}e^{-\sqrt{\xi^{2}-k_{1}^{2}}(x_{2}-y_{2})}\xi^{j-2}\left[\frac{1}{\xi}+(x_{2}-y_{2})\right]d\xi.\end{split}$
(114)

In the second line we use the fact that $x_{2}-y_{2}>d-y_{2}.$ It is easy to
show that the second term is bounded by
$Me^{\sqrt{k_{2}^{2}-k_{1}^{2}}(x_{2}-y_{2})}$ for $j=0,2.$ We estimate the
contribution from the first term, as before, to obtain that for
$|y_{2}|<d,x_{2}>d,$ we have

$|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|\leq\begin{cases}&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(x_{2}-y_{2})}\text{
for }j=0,1\\\
&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}(x_{2}-y_{2})}\cdot\log\frac{2x_{2}}{x_{2}-y_{2}}\text{
for }j=2.\end{cases}$ (115)

It is not difficult to show that the analogous estimates hold with $x_{2}<-d,$
so that

$|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|\leq\begin{cases}&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}|x_{2}-y_{2}|}\text{
for }j=0,1\\\
&Me^{-\sqrt{k_{2}^{2}-k_{1}^{2}}|x_{2}-y_{2}|}\cdot\log\frac{2|x_{2}|}{|x_{2}-y_{2}|}\text{
for }j=2.\end{cases}$ (116)

Altogether we have, for $|y_{2}|<d,|x_{2}|>d,$

$\mathfrak{w}^{[j]}(x_{2},y_{2})=C_{j}\frac{e^{ik_{1}|x_{2}|}\mathfrak{C}(0,k_{1};y_{2})}{\mathfrak{W}(0,k_{1})|x_{2}|^{\frac{j+1}{2}}}+O(|x_{2}|^{-\frac{j+3}{2}})\text{
for }j=0,1,2.$ (117)

The function $\mathfrak{w}^{[0]}(x_{2},y_{2})$ is continuous as
$|x_{2}-y_{2}|\to 0,$ and

$|\mathfrak{w}^{[2]}(x_{2},y_{2})|\leq M|\log|x_{2}-y_{2}||.$ (118)

As before we can show that

$\partial_{x_{2}}\mathfrak{w}^{[j]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[j]}(x_{2},y_{2})=O\left(|x_{2}|^{-\frac{j+3}{2}}\right)\text{
as }\pm x_{2}\to\infty.$ (119)

###  4.4 The Diagonal Singularity

The singularities of the function $w(x;y)$ are confined to the set $x=y$ where
$|x_{2}|,|y_{2}|\leq d.$ In this section we use the Fourier representation to
study the nature of this singularity. As expected from the discussion
surrounding (37), the principal singularity behaves like
$(x_{2}-y_{2})^{2}\log|x_{2}-y_{2}|.$ Note that the previous discussion
assumed that the potential $q(x_{2})$ is smooth, whereas here we continue
working with $q$ as defined in (29).

We start by assuming that $-d<y_{2}<x_{2}<d,$ so that

$\mathfrak{w}^{[j]}(x_{2},y_{2})=\int_{\Gamma_{\nu}^{+}}\frac{\xi^{j}\mathfrak{D}(\xi,\sqrt{\xi^{2}-k_{1}^{2}};x_{2},y_{2})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi,$
(120)

where

$\mathfrak{D}(\xi,\eta;x_{2},y_{2})=\int_{-d}^{x_{2}}\widetilde{u}_{+}(x_{2},0+;\xi)\widetilde{u}_{+}(-z_{2},0+;\xi)e^{-\eta|z_{2}-y_{2}|}q(z_{2})dz_{2}+\\\
\int_{x_{2}}^{d}\widetilde{u}_{+}(-x_{2},0+;\xi)\widetilde{u}_{+}(z_{2},0+;\xi)e^{-\eta|z_{2}-y_{2}|}q(z_{2})dz_{2}\\\
=\int_{-d}^{y_{2}}e^{-2\eta d}\left[\cosh A(d-x_{2})+\eta\frac{\sinh
A(d-x_{2})}{A}\right]\times\\\ \left[\cosh A(d+z_{2})+\eta\frac{\sinh
A(d+z_{2})}{A}\right]e^{-\eta(y_{2}-z_{2})}q(z_{2})dz_{2}+\\\
\int_{y_{2}}^{x_{2}}e^{-2\eta d}\left[\cosh A(d-x_{2})+\eta\frac{\sinh
A(d-x_{2})}{A}\right]\times\\\ \left[\cosh A(d+z_{2})+\eta\frac{\sinh
A(d+z_{2})}{A}\right]e^{-\eta(z_{2}-y_{2})}q(z_{2})dz_{2}+\\\
\int_{x_{2}}^{d_{2}}e^{-2\eta d}\left[\cosh A(d+x_{2})+\eta\frac{\sinh
A(d+x_{2})}{A}\right]\times\\\ \left[\cosh A(d-z_{2})+\eta\frac{\sinh
A(d-z_{2})}{A}\right]e^{-\eta(z_{2}-y_{2})}q(z_{2})dz_{2},$ (121)

with $A=\sqrt{\xi^{2}-k_{2}^{2}}.$ In the second, more explicit expression, we
use the condition that $-d<y_{2}<x_{2}<d.$ As usual
$\mathfrak{D}(\xi,\eta;x_{2},y_{2})$ is an entire function of $(\xi,\eta);$ we
use $\eta=-i\sqrt{k_{1}^{2}-\xi^{2}},$ for $|\xi|<k_{1},$ and
$\eta=\sqrt{\xi^{2}-k_{1}^{2}},$ for $|\xi|>k_{1}.$ The part of the integral
in (120) where $|\xi|<k_{2},$ defines $\mathcal{C}^{2}$-functions of
$(x_{2},y_{2}),$ so we focus our attention on $|\xi|\geq k_{2}.$ In set we
have

$\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})=-i\int_{|\xi|>k_{2}}\frac{\xi^{j}\mathfrak{D}(\xi,\sqrt{\xi^{2}-k_{1}^{2}};x_{2},y_{2})}{\mathfrak{W}(\xi,i\sqrt{\xi^{2}-k_{1}^{2}})\sqrt{\xi^{2}-k_{1}^{2}}}d\xi,$
(122)

Using the definition of $\mathfrak{D}$ we see that, with
$B=\sqrt{\xi^{2}-k_{1}^{2}},$

$|\mathfrak{D}(\xi,B;x_{2},y_{2})|\leq
M\Bigg{[}e^{-(Ax_{2}+By_{2})}\int_{-d}^{y_{2}}e^{(A+B)z_{2}}dz_{2}+\\\
e^{-(Ax_{2}-By_{2})}\int_{y_{2}}^{x_{2}}e^{(A-B)z_{2}}dz_{2}+e^{Ax_{2}+By_{2}}\int_{x_{2}}^{d}e^{-(A+B)z_{2}}dz_{2}\Bigg{]}.$
(123)

Performing these integrals and using elementary estimates we see that

$|\mathfrak{D}(\xi,B;x_{2},y_{2})|\leq
Me^{-\sqrt{\xi^{2}-k_{2}^{2}}|x_{2}-y_{2}|}\left[\frac{1}{1+|\xi|}+|x_{2}-y_{2}|\right].$
(124)

and therefore (94) holds if both $|x_{2}|<d,$ and $|y_{2}|<d.$ Inserting this
into (122), we see that

$\begin{split}|\mathfrak{w}^{[j]}_{2}(x_{2},y_{2})|&\leq
M\int_{k_{2}}^{\infty}\xi^{j-2}e^{-\sqrt{\xi^{2}-k_{2}^{2}}(x_{2}-y_{2})}\left[\frac{1}{1+|\xi|}+(x_{2}-y_{2})\right]d\xi\\\
&\leq\begin{cases}&M\text{ for }j=0,1\\\ &M|\log(x_{2}-y_{2})|\text{ for
}j=2.\end{cases}\end{split}$ (125)

It is straightforward to see that $|\mathfrak{D}(\xi,B;x_{2},y_{2})|$
satisfies the same estimate if $-d<x_{2}<y_{2}<d,$ and therefore the estimates
in (125) hold in this case as well. It is easy to see that differentiating
$\mathfrak{w}^{[j]}(x_{2},y_{2})$ with respect to $x_{2}$ or $y_{2}$ has the
effect of increasing $j$ by 1. Thus we see that
$\partial_{x_{2}}\mathfrak{w}^{[0]}(x_{2},y_{2}),\partial_{y_{2}}\mathfrak{w}^{[0]}(x_{2},y_{2}),$
are uniformly bounded, and
$\partial_{x_{2}}\mathfrak{w}^{[1]}(x_{2},y_{2}),\partial_{y_{2}}\mathfrak{w}^{[1]}(x_{2},y_{2}),$
along with the second derivatives of $\mathfrak{w}^{[0]}(x_{2},y_{2})$ have a
$\log|x_{2}-y_{2}|$-singularity along the diagonal.

###  4.5 Summary

Thus we see that the kernels that appear in the integral equations in (21)
have a very particular form. These kernels are smooth outside of
$B_{d}=[-d,d]\times[-d,d],$ and

  1. 1.

The kernel $\mathfrak{w}^{[0]}(x_{2},y_{2})$ has an
$(x_{2}-y_{2})^{2}\log|x_{2}-y_{2}|$-singularity in $B_{d};$ asymptotically

$\mathfrak{w}^{[0]}(x_{2},y_{2})=C^{0}_{\pm,\pm}\frac{e^{ik_{1}(|x_{2}|+|y_{2}|)}}{\sqrt{|x_{2}|+|y_{2}|}}+O((|x_{2}|+|y_{2}|)^{-\frac{3}{2}}),\\\
\text{ where }\pm x_{2}>d\text{ and }\pm y_{2}>d;$ (126)

we also have

$\begin{split}\mathfrak{w}^{[0]}(x_{2},y_{2})&=\frac{e^{ik_{1}|y_{2}|}b^{0}_{\pm}(x_{2})}{\sqrt{|y_{2}|}}+O(|y_{2}|^{-\frac{3}{2}})\text{
where }\pm y_{2}>d,\,|x_{2}|<d,\\\
\mathfrak{w}^{[0]}(x_{2},y_{2})&=\frac{e^{ik_{1}|x_{2}|}c^{0}_{\pm}(y_{2})}{\sqrt{|x_{2}|}}+O(|x_{2}|^{-\frac{3}{2}})\text{
where }\pm x_{2}>d,\,|y_{2}|<d;\end{split}$ (127)

the kernel is outgoing with

$\partial_{x_{2}}\mathfrak{w}^{[0]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[0]}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{3}{2}}\right)\text{
as }\pm x_{2}\to\infty,$ (128)
$\partial_{y_{2}}\mathfrak{w}^{[0]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[0]}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{3}{2}}\right)\text{
as }\pm y_{2}\to\infty.$ (129)

  2. 2.

The kernel
$i\mathfrak{w}^{[1]}(x_{2},y_{2})+\partial_{x_{1}}w_{0+}^{g}(0,x_{2};0,y_{2})=0.$

  3. 3.

The kernel $\mathfrak{w}^{[2]}(x_{2},y_{2})$ has a
$\log|x_{2}-y_{2}|$-singularity in $B_{d};$ asymptotically

$\mathfrak{w}^{[2]}(x_{2},y_{2})=C^{2}_{\pm,\pm}\frac{e^{ik_{1}(|x_{2}|+|y_{2}|)}}{(|x_{2}|+|y_{2}|)^{\frac{3}{2}}}+O((|x_{2}|+|y_{2}|)^{-\frac{5}{2}}),\\\
\text{ where }\pm x_{2}>d\text{ and }\pm y_{2}>d;$ (130)

we also have

$\begin{split}\mathfrak{w}^{[2]}(x_{2},y_{2})&=\frac{e^{ik_{1}|y_{2}|}b^{2}_{\pm}(x_{2})}{|y_{2}|^{\frac{3}{2}}}+O(|y_{2}|^{-\frac{5}{2}})\text{
where }\pm y_{2}>d,\,|x_{2}|<d,\\\
\mathfrak{w}^{[2]}(x_{2},y_{2})&=\frac{e^{ik_{1}|x_{2}|}c^{2}_{\pm}(y_{2})}{|x_{2}|^{\frac{3}{2}}}+O(|x_{2}|^{-\frac{5}{2}})\text{
where }\pm x_{2}>d,\,|y_{2}|<d;\end{split}$ (131)

the kernel is also outgoing with

$\partial_{x_{2}}\mathfrak{w}^{[2]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[2]}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{5}{2}}\right)\text{
as }\pm x_{2}\to\infty,$ (132)
$\partial_{y_{2}}\mathfrak{w}^{[2]}(x_{2},y_{2})\mp
ik_{1}\mathfrak{w}^{[2]}(x_{2},y_{2})=O\left((|x_{2}|+|y_{2}|)^{-\frac{5}{2}}\right)\text{
as }\pm y_{2}\to\infty.$ (133)

##  5 Estimates in the Right Half Plane

In this section we extend the estimates derived in the previous section to
$x_{1}>0.$ If $x_{1}$ remains bounded, then these is really nothing further to
do, as the principal term is a stationary phase contribution arising from
$\xi=0.$ We can also derive estimates along rays $r(\eta_{1},\eta_{2})$ as
$r\to\infty;$ here $|\eta|=1.$ In this case, a little more analysis is needed
as the stationary phase occurs where $\xi=k_{1}\eta_{1},$ which lies in the
interval $(-k_{1},k_{1}).$ We indicate the necessary analysis, leaving some of
the details, which are essentially covered in Appendices B–D, to the reader.

We begin with estimates for bounded $x_{1};$ these suffice to show that the
representations of $u^{r,l}$ in terms of
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R})$
with $0<\alpha<\frac{1}{2},$ make sense, see (11). Using (28) we express the
derivatives of $w_{0+}^{c}$ as the contour integrals:

$\partial_{x_{1}}^{j}w^{c}_{0+}(x;0,y_{2})=\frac{1}{2\pi}\int_{\Gamma_{\nu}^{+}}(i\xi)^{j}\tilde{w}_{0+}(\xi,x_{2};0,y_{2})e^{ix_{1}\xi}d\xi\text{
for }x_{1}>0.$ (134)

The only difference between these integrals and those estimated in Section 4
is the factor of $e^{i\xi x_{1}}$ in the integrand. Recalling that
$\partial_{x_{1}}^{j}w^{c}_{0+}(x;y)=(-1)^{j}\partial_{y_{1}}^{j}w^{c}_{0+}(x;y);$
it suffices to consider these expressions to estimate
$\partial_{x_{1}}w^{c}_{0+},$ $\partial_{y_{1}}w^{c}_{0+}$ and
$\partial_{x_{1}}\partial_{y_{1}}w^{c}_{0+}.$

The contribution from the guided modes is clearly infinitely differentiable in
the $x_{1}$-variable for all $x_{1},$ and satisfies the same estimates as for
$x_{1}=0.$ It is easy to see that the estimates derived in Section 4.2 for
$\mathfrak{w}^{[j]},j=0,1,2$ hold equally well for bounded $x_{1}>0,$ as the
principal contribution comes from the stationary phase at $\xi=0.$ The
addition of the factor $e^{i\xi x_{1}}$ does not change the contribution from
the semi-circular components, $\\{C^{\pm}_{j,\nu}\\},$ as
$\operatorname{Im}\xi\geq 0,$ and therefore $|e^{ix_{1}\xi}|\leq 1$ on this
part of the contour. Where $|x_{2}|,|y_{2}|>d,$ for bounded $x_{1},$ the
extension to the right half plane satisfies the asymptotic expression

$\partial_{x_{1}}^{j}w^{c}_{0+}(x_{1},x_{2};0,y_{2})=C_{j}\frac{e^{ik_{1}(|x_{2}|+|y_{2}|)}}{(|x_{2}|+|y_{2}|)^{\frac{j+1}{2}}}+O\left((|x_{2}|+|y_{2}|)^{-{\frac{j+3}{2}}}\right)\text{
for }j=0,1,2.$ (135)

It is similarly straightforward to handle the estimates where either
$|x_{2}|<d,$ or $|y_{2}|<d.$ Again, since the principal contribution comes
from $\xi=0,$ for bounded $x_{1}>0,$ we have

$\begin{split}\partial_{x_{1}}^{j}w^{c}_{0+}(x_{1},x_{2};0,y_{2})&=\frac{e^{ik_{1}|y_{2}|}b^{[j]}_{\pm}(x_{2})}{|y_{2}|^{\frac{j+1}{2}}}+O(|y_{2}|^{-\frac{j+3}{2}})\text{
where }\pm y_{2}>d,\,|x_{2}|<d,\\\
\partial_{x_{1}}^{j}w^{c}_{0+}(x_{1},x_{2};0,y_{2})&=\frac{e^{ik_{1}|x_{2}|}c^{[j]}_{\pm}(y_{2})}{|x_{2}|^{\frac{j+1}{2}}}+O(|x_{2}|^{-\frac{j+3}{2}})\text{
where }\pm x_{2}>d,\,|y_{2}|<d;\end{split}$ (136)

From ellipticity it follows that $w^{c}_{0+}(x_{1},x_{2};0,y_{2})$ is a
$\mathcal{C}^{\infty}$-function if $x_{1}>0,$ away from $x_{2},y_{2}=\pm d.$
It clear that if $f(y_{2})$ is a bounded continuous function, then for any
finite $L$ we have

$\lim_{x_{1}\to
0^{+}}\partial_{x_{1}}^{j}\int_{-L}^{L}w^{c}_{0+}(x_{1},x_{2};0,y_{2})f(y_{2})dy_{2}=\\\
\int_{-L}^{L}\partial_{x_{1}}^{j}w^{c}_{0+}(0,x_{2};0,y_{2})f(y_{2})dy_{2}\text{
for }j=0,1,2.$ (137)

From the Fourier representations it is clear that the asymptotics, for
$|x_{2}|+|y_{2}|$ large, hold uniformly as $x_{1}\to 0^{+}.$ Hence if $f$ is
continuous and

$|f(y_{2})|\leq
M\frac{(1+|y_{2}|)^{\frac{j}{2}}}{(1+|y_{2}|)^{\frac{1}{2}+\epsilon}},$ (138)

for an $\epsilon>0,$ then, for $j=0,1,2,$

$\begin{split}\lim_{x_{1}\to
0^{+}}\partial_{x_{1}}^{j}\int_{-\infty}^{\infty}w^{c}_{0+}(x_{1},x_{2};0,y_{2})f(y_{2})dy_{2}&=\lim_{x_{1}\to
0^{+}}\int_{-\infty}^{\infty}\partial_{x_{1}}^{j}w^{c}_{0+}(x_{1},x_{2};0,y_{2})f(y_{2})dy_{2}\\\
&=\int_{-\infty}^{\infty}\partial_{x_{1}}^{j}w^{c}_{0+}(0,x_{2};0,y_{2})f(y_{2})dy_{2}\\\
&=\left[\partial_{x_{1}}^{j}\int_{-\infty}^{\infty}w^{c}_{0+}(x_{1},x_{2};0,y_{2})f(y_{2})dy_{2}\right]_{x_{1}=0}.\end{split}$
(139)

With these observations we can now show that the representation in (18) can be
used with solutions of the corresponding boundary integral equations to find
and represent the scattered fields, $u^{r,l}(x).$

We now consider the behavior of the kernels $w_{0+}^{c}(r\eta;y_{2})$ and
$\partial_{x_{1}}w_{0+}^{c}(r\eta;y_{2})$ as $r\to\infty,$ with $y_{2}$ fixed,
as well as the result of applying $\partial_{r}-ik_{1}$ to these kernels. The
principal contribution again comes from the stationary phase that now occurs
where $\xi=\eta_{1}k_{1}.$ In the following analysis we avoid the channel
where $\eta_{2}=0,$ as well as the $x_{2}$-axis, $\eta_{1}=0.$ Along the rays
$x=r\eta$ the second component $|x_{2}|$ is eventually larger than $d.$

We start with the integral in (56) with the $e^{i\xi x_{1}}$ factor added to
the integrand for $x_{2},y_{2}>d:$

$w_{00+}^{[j]}(r\eta;y_{2})=\frac{i}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\frac{e^{i[\sqrt{k_{1}^{2}-\xi^{2}}(r\eta_{2}+y_{2})+i\xi
r\eta_{1}]}\mathfrak{A}_{0}(\xi,-i\sqrt{k_{1}^{2}-\xi^{2}})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi.$
(140)

As noted above, the stationary phase, as $r\to\infty,$ occurs where

$\frac{-\xi}{\sqrt{k_{1}^{2}-\xi^{2}}}\eta_{2}+\eta_{1}=0,$ (141)

which implies that

$\xi=\eta_{1}k_{1}.$ (142)

So long as $\eta_{1}\neq 1,$ this stationary point remains away from the
endpoints of $[-k_{1},k_{1}],$ and the contributions of the integral away from
the stationary point is easily shown to be of lower order. A standard
calculation shows that the contribution of the stationary phase is given by

$w_{00+}^{[j]}(r\eta;y_{2})=\eta_{1}^{j}f_{+}(\eta)\frac{e^{ik_{1}r}e^{i\eta_{2}y_{2}}}{r^{\frac{1}{2}}}+O(r^{-\frac{3}{2}}).$
(143)

To consider what happens where $x_{2}<0,$ we need to replace $\eta_{2}$ in
(143) with $-\eta_{2},$ which gives the same asymptotic result. If $y_{2}<0$
then we replace $y_{2}$ with $-y_{2}.$ Altogether we get the result

$w_{00+}^{[j]}(r\eta;y_{2})=\eta_{1}^{j}f_{-}(\eta)\frac{e^{ik_{1}r}e^{i|\eta_{2}y_{2}|}}{r^{\frac{1}{2}}}+O(r^{-\frac{3}{2}}).$
(144)

For $y_{2}\in[-d,d]$ there is a more complicated coefficient $f(\eta,y_{2}),$
which interpolates between $f_{+}(\eta)e^{i|\eta_{2}|y_{2}}$ and
$f_{-}(\eta)e^{i|\eta_{2}y_{2}|}.$ We do not explicitly compute this function
as we will not make use of it.

Applying $\partial_{r}-ik_{1}$ get

$(\partial_{r}-ik_{1})w_{00+}^{[j]}(r\eta;y_{2})=\\\
\frac{-1}{2\pi}\int_{-k_{1}}^{k_{1}}\xi^{j}\frac{e^{i[\sqrt{k_{1}^{2}-\xi^{2}}(r\eta_{2}+y_{2})+i\xi
r\eta_{1}]}[\sqrt{k_{1}^{2}-\xi^{2}}\eta_{2}+\xi\eta_{1}-k_{1}]\mathfrak{A}_{0}(\xi,-i\sqrt{k_{1}^{2}-\xi^{2}})}{\mathfrak{W}(\xi,\sqrt{k_{1}^{2}-\xi^{2}})\sqrt{k_{1}^{2}-\xi^{2}}}d\xi.$
(145)

The new term in the bracket has a second order zero at $\xi=\eta_{1}k_{1},$ so
that the stationary point now contributes

$(\partial_{r}-ik_{1})w_{00+}^{[j]}(r\eta;y_{2})=\eta_{1}^{j}\tilde{f}(\eta)\frac{e^{ik_{1}r}e^{i|\eta_{2}y_{2}|}}{r^{\frac{3}{2}}}+O(r^{-\frac{5}{2}}).$
(146)

This shows that, for fixed $y_{2}$ the correction terms in the resolvent
kernel, $\partial_{x_{1}}^{j}w^{c}_{0+}(x;y_{2}),$ are outgoing in the usual
sense, at least away from the channel. A more detailed analysis of these
estimates will appear in a later publication.

##  6 The Integral Equations

With the computations we have completed we can now express the kernels
appearing in the integral equations, (21), in terms of $\mathfrak{w}^{[j]}$
and the guided modes for the relevant equations. We now reintroduce the $l,r$
sub- and superscripts, letting $\mathfrak{w}^{[j]}_{l,r}$ denote the kernels
defined by $q_{l,r}$ and $w^{g}_{l,r}$ the contributions of the guided modes
from (35). With this notation we have

$\begin{split}\mathcal{W}^{0}_{l,r}=w^{l,r}(0,x_{2};0,y_{2})&=\mathfrak{w}^{[0]}_{l,r}(x_{2},y_{2})+w^{g}_{l,r}(0,x_{2};0;y_{2}),\\\
\partial_{x_{1}}w^{l,r}(0,x_{2};0,y_{2})&=\partial_{y_{1}}w^{l,r}(0,x_{2};0,y_{2})=0,\\\
\mathcal{W}^{2}_{l,r}=\partial^{2}_{x_{1}y_{1}}w^{l,r}(0,x_{2};0,y_{2})&=\mathfrak{w}^{[2]}_{l,r}(x_{2},y_{2})+\partial^{2}_{x_{1}y_{1}}w^{g}_{l,r}(0,x_{2};0;y_{2}).\end{split}$
(147)

The integral equations therefore can be written

$\left(\begin{matrix}\operatorname{Id}&\mathcal{W}^{0}_{r}-\mathcal{W}^{0}_{l}\\\
\mathcal{W}^{2}_{l}-\mathcal{W}^{2}_{r}&\operatorname{Id}\end{matrix}\right)\left(\begin{matrix}\sigma\\\
\tau\end{matrix}\right)=\left(\begin{matrix}g\\\ h\end{matrix}\right),$ (148)

which, to simplify notation, we rewrite as

$\left(\begin{matrix}\operatorname{Id}&D\\\
C&\operatorname{Id}\end{matrix}\right)\left(\begin{matrix}\sigma\\\
\tau\end{matrix}\right)=\left(\begin{matrix}g\\\ h\end{matrix}\right).$ (149)

The analysis of $\mathfrak{w}^{[2]}_{l,r}$ shows that the operator
$\mathcal{W}^{2}_{l}-\mathcal{W}^{2}_{r}$ is compact on $L^{2}(\mathbb{R}),$
amongst other spaces. While $\mathcal{W}^{0}_{r}-\mathcal{W}^{0}_{l}$ is
smoothing, the $(|x_{2}|+|y_{2}|)^{-\frac{1}{2}}$ asymptotic behavior of
$\mathfrak{w}^{[0]}_{l,r}$ prevents it from being defined on
$L^{2}(\mathbb{R}),$ let alone compact.

In the sequel we work with subspaces of $\mathcal{C}^{0}(\mathbb{R}).$

######  Definition 1.

For $\alpha\in\mathbb{R},$ let $\mathcal{C}_{\alpha}(\mathbb{R})$ denote
continuous functions on $\mathbb{R}$ with

$|f|_{\alpha}=\sup\\{(1+|x|)^{\alpha}|f(x)|:\>x\in\mathbb{R}\\}<\infty.$ (150)

These spaces are somewhat like Hölder spaces, in that, for $\alpha>0,$
$\mathcal{C}^{\infty}_{c}(\mathbb{R})$ is not dense in
$\mathcal{C}_{\alpha}(\mathbb{R})$ with respect to the
$|\cdot|_{\alpha}$-norm. A usable replacement is the fact that
$\mathcal{C}^{\infty}_{c}(\mathbb{R})$ is dense in
$\mathcal{C}_{\alpha}(\mathbb{R})$ with respect to the
$|\cdot|_{\alpha^{\prime}}$-norm, for any $0<\alpha^{\prime}<\alpha.$ It is
important to have a criterion for when a bounded linear operator
$A:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})$ is
compact. We give a simple sufficient condition.

######  Proposition 1.

Let $0<\alpha<\beta,$ and let
$A:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\beta}(\mathbb{R})$ be a
bounded linear operator. Let
$B_{r}=\\{f\in\mathcal{C}_{\alpha}(\mathbb{R}):\>|f|_{\alpha}<r\\}.$ If, for
any $0<X,$ the image $AB_{r}$ restricted to $[-X,X]$ is a uniformly
equicontinuous family of functions, then
$A:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})$ is a
compact operator.

###### Proof.

To prove the proposition we need to show that if
$\\{f_{n}\\}\subset\mathcal{C}_{\alpha}(\mathbb{R})$ is a bounded sequence,
then $\\{Af_{n}\\}$ has a $\mathcal{C}_{\alpha}(\mathbb{R})$-convergent
subsequence. The hypotheses of the proposition imply that there are positive
constants $M,r$ so that $\\{f_{n}\\}\subset B_{r}$ and

$|Af|_{\beta}\leq M|f|_{\alpha}\text{ for all
}f\in\mathcal{C}_{\alpha}(\mathbb{R}).$ (151)

The restriction of $\\{Af_{n}\\}$ to any interval $[-X,X]$ is a bounded,
uniformly equicontinuous family. Hence a simple diagonal argument using the
Arzela-Ascoli theorem produces a subsequence $\\{f_{n_{j}}\\}$ that converges
to $g\in\mathcal{C}^{0}(\mathbb{R})$ uniformly on any interval $[-X,X].$ In
fact this sequence also converges in $\mathcal{C}_{\alpha}(\mathbb{R}).$

For the terms of the sequence we have the estimate

$|Af_{n_{j}}(x)|(1+|x|)^{\beta}\leq Mr.$ (152)

Letting $j\to\infty$ shows that

$|g(x)|(1+|x|)^{\beta}\leq Mr$ (153)

as well. These estimates and the triangle inequality show that

$|Af_{n_{j}}(x)-g(x)|(1+|x|)^{\alpha}\leq
2Mr\frac{(1+|x|)^{\alpha}}{(1+|x|)^{\beta}}.$ (154)

For an $\epsilon>0,$ we can therefore choose $X$ so that

$|Af_{n_{j}}(x)-g(x)|(1+|x|)^{\alpha}\leq\epsilon\text{ if }|x|>X.$ (155)

As $\\{Af_{n_{j}}\upharpoonright_{[-X,X]}\\}$ converges uniformly to
$g\upharpoonright_{[-X,X]},$ there is a $J$ so that if $j>J,$ then

$|Af_{n_{j}}(x)-g(x)|(1+|x|)^{\alpha}\leq\epsilon\text{ if }|x|\leq X.$ (156)

Together these estimates show that

$|Af_{n_{j}}(x)-g(x)|_{\alpha}<\epsilon\text{ if }j>J,$ (157)

which completes the proof of the proposition. ∎

In the estimates below the function $m(q_{l},q_{r})$ is a continuous function
of the norms $\|q_{l}\|_{L^{\infty}},\|q_{r}\|_{L^{\infty}},$ and
$\|q_{l}-q_{r}\|_{L^{\infty}},$ which satisfies:

$m(q,q)=0.$ (158)

The estimates on the kernels follow from the analysis in Section 4. The
kernel, $k_{D}(x_{2},y_{2}),$ of $D$ is at least $\mathcal{C}^{1}$ and
satisfies an estimate of the form,

$|k_{D}(x_{2},y_{2})|\leq\frac{m(q_{l},q_{r})}{(1+|x_{2}|+|y_{2}|)^{\frac{1}{2}}}.$
(159)

The kernel, $k_{C}(x_{2},y_{2}),$ of $C$ has a singularity of the form
$m(q_{l},q_{r})\log|x_{2}-y_{2}|\chi_{B_{d}}(x_{2},y_{2})$ and

$|(1-\varphi(x_{2},y_{2}))k_{C}(x_{2},y_{2})|\leq\frac{m(q_{l},q_{r})}{(1+|x_{2}|+|y_{2}|)^{\frac{3}{2}}}.$
(160)

Here $\varphi\in\mathcal{C}^{\infty}_{c}(B_{d+2\epsilon})$ for an
$\epsilon>0,$ with

$\varphi(x_{2},y_{2})=1\text{ for }(x_{2},y_{2})\in B_{d+\epsilon}.$ (161)

We begin with the following boundedness result for the operator appearing in
(149).

######  Proposition 2.

For $0<\alpha<\frac{1}{2},$ there is a constant $M_{\alpha}$ so that if
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$
then

$|D\tau|_{\alpha}+|C\sigma|_{\alpha+\frac{1}{2}}\leq
M_{\alpha}m(q_{l},q_{r})\left[|\tau|_{\alpha+\frac{1}{2}}+|\sigma|_{\alpha}\right].$
(162)

The proof relies on the following lemma

######  Lemma 1.

If $0<\alpha<1,$ and $\alpha+\beta>1,$ then, for $x_{2}>0,$ we have the
estimate

$\int_{0}^{\infty}\frac{dy_{2}}{y_{2}^{\alpha}(x_{2}+y_{2})^{\beta}}\leq\frac{M_{\alpha,\beta}}{x_{2}^{\alpha+\beta-1}}.$
(163)

###### Proof of Lemma.

If we let $y_{2}=x_{2}t,$ then the integral becomes

$\frac{1}{x_{2}^{\alpha+\beta-1}}\int_{0}^{\infty}\frac{dt}{t^{\alpha}(1+t)^{\beta}}\leq\frac{M_{\alpha,\beta}}{x_{2}^{\alpha+\beta-1}}.$
(164)

∎

###### Proof of Proposition.

We first consider $C\sigma,$ splitting it into a compactly supported part,
$C_{0},$ with kernel $\varphi(x_{2},y_{2})k_{C}(x_{2},y_{2}),$ and an
unbounded part, $C_{1},$ with kernel
$(1-\varphi(x_{2},y_{2}))k_{C}(x_{2},y_{2}).$ It is clear that the compactly
supported part satisfies

$|C_{0}f(x_{2})|\leq
m(q_{l},q_{r})|\sigma|_{\alpha}\chi_{[-(d+2\epsilon),d+2\epsilon]}(x_{2}).$
(165)

To estimate the other part we observe that, if $0<\alpha<1,$ then applying the
lemma gives

$\begin{split}|C_{1}\sigma(x_{2})|&\leq\int_{\mathbb{R}}(1-\varphi(x_{2},y_{2}))|k_{C}(x_{2},y_{2})||\sigma(y_{2})|dy_{2}\\\
&\leq\int_{\mathbb{R}}\frac{m(q_{l},q_{r})|\sigma|_{\alpha}dy_{2}}{|y_{2}|^{\alpha}(|x_{2}|+|y_{2}|)^{\frac{3}{2}}}\\\
&\leq
K^{\prime}_{\alpha}\frac{m(q_{l},q_{r})|\sigma|_{\alpha}}{|x_{2}|^{\alpha+\frac{1}{2}}}.\end{split}$
(166)

Which shows that

$|C\sigma|_{\alpha+\frac{1}{2}}\leq Mm(q_{l},q_{r})|\sigma|_{\alpha}.$ (167)

We now estimate $|D\tau|_{\alpha},$ assuming that $0<\alpha<\frac{1}{2};$
applying the lemma gives

$\begin{split}|D\tau(x_{2})|&\leq
m(q_{l},q_{r})|\tau|_{\alpha+\frac{1}{2}}\int_{\mathbb{R}}\frac{dy_{2}}{(1+|y_{2}|)^{\alpha+\frac{1}{2}}(1+|x_{2}|+|y_{2}|)^{\frac{1}{2}}}\\\
&\leq
K_{\alpha}\frac{m(q_{l},q_{r})|\tau|_{\alpha+\frac{1}{2}}}{|x_{2}|^{\alpha}}.\end{split}$
(168)

The estimate in (162) follows from this and (167). ∎

If $M_{\alpha}m(q_{l},q_{r})<1,$ then the Neumann series,

$\left(\begin{matrix}\operatorname{Id}&D\\\
C&\operatorname{Id}\end{matrix}\right)^{-1}=\operatorname{Id}+\sum_{j=1}^{\infty}(-1)^{j}\left(\begin{matrix}0&D\\\
C&0\end{matrix}\right)^{j}$ (169)

converges uniformly in the space of bounded operators on
$\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}).$
This proves:

######  Theorem 1.

For $0<\alpha<\frac{1}{2},$ if $M_{\alpha}m(q_{l},q_{r})<1,$ and
$(g,h)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$
then the equations in (148) (or (149)) have a unique solution
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}).$

######  Remark 2.

As $m(q,q)=0,$ the smallness hypothesis is satisfied if the contrast between
the left and right channels is sufficiently small.

In fact we do not need to require the channels to have small contrast to prove
the generic solvability of (149) . To prove this it is easier to use the
single equation formulation

$(\operatorname{Id}-DC)\sigma=(g-Dh)$ (170)

obtained by multiplying (149) by $\left(\begin{matrix}\operatorname{Id}&-D\\\
0&\operatorname{Id}\end{matrix}\right).$ If
$h\in\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$ then
$Dh\in\mathcal{C}_{\alpha}(\mathbb{R}).$ In the table below we show the
leading order asymptotics for the kernels of $\mathcal{W}_{l,r}^{j},$ for
$j=0,2,$ assuming the channel is centered at $x_{2}=\gamma.$ The differences
between $\mathcal{W}_{l,r}^{j},$ and the leading terms, shown in Table 1, are
$O((|x_{2}|+|y_{2}|)^{-\frac{j+3}{2}});$ these differences are smoothing and
improve decay and therefore Proposition 1 implies that they define compact
operators from
$\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R}),$
$\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R})\to\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$
resp. for any $\alpha<\frac{1}{2}.$

$c^{j}_{-+}\frac{e^{ik_{1}(|y_{2}-\gamma|+|x_{2}-\gamma|)}}{(|y_{2}-\gamma|+|x_{2}-\gamma|)^{\frac{j+1}{2}}}$ | $\frac{e^{ik_{1}|y_{2}-\gamma|}b^{j}_{+}(x_{2}-\gamma)}{|y_{2}-\gamma|^{\frac{j+1}{2}}}$ | $c^{j}_{++}\frac{e^{ik_{1}(|x_{2}-\gamma|+|x_{2}-\gamma|)}}{(y_{2}-\gamma+x_{2}-\gamma)^{\frac{j+1}{2}}}$
---|---|---
$\frac{e^{ik_{1}|x_{2}-\gamma|}c^{j}_{-}(y_{2}-\gamma)}{|x_{2}-\gamma|^{\frac{j+1}{2}}}$ | $c_{00}^{j}(x_{2}-\gamma,y_{2}-\gamma)$ | $\frac{e^{ik_{1}|x_{2}-\gamma|}c^{j}_{+}(y_{2}-\gamma)}{|x_{2}-\gamma|^{\frac{j+1}{2}}}$
$c^{j}_{--}\frac{e^{ik_{1}(|x_{2}-\gamma|+|y_{2}-\gamma|)}}{(|y_{2}-\gamma|+|x_{2}-\gamma|)^{\frac{j+1}{2}}}$ | $\frac{e^{ik_{1}|y_{2}-\gamma|}b^{j}_{-}(x_{2}-\gamma)}{|y_{2}-\gamma|^{\frac{j+1}{2}}}$ | $c^{j}_{+-}\frac{e^{ik_{1}(|x_{2}-\gamma|+|y_{2}-\gamma|)}}{(|x_{2}-\gamma|+|y_{2}-\gamma|)^{\frac{j+1}{2}}}$
Table 1: Schematic for the structure of the leading terms of
$\mathcal{W}^{j}_{l,r}$ assuming the channel is centered on $\gamma.$ If the
channel has width $2\delta,$ then $+$ is the requirement that a variable is
greater than $\gamma+\delta,$ and $-$ is the requirement that a variable is
less than $\gamma-\delta.$

To analyze the kernel of composition $DC,$ we choose $d>0$ so that the
supports of $q_{l}$ and $q_{r}$ are contained in $(-d,d).$ As above we choose
a function $\varphi\in\mathcal{C}^{\infty}_{c}(B_{d+4\epsilon}),$ which equals
$1$ in $B_{d+2\epsilon}.$ We let $C_{0}$ (resp. $D_{0}$) have the kernel
$\varphi(x_{2},y_{2})k_{C}(x_{2},y_{2}),$ (resp.
$\varphi(x_{2},y_{2})k_{D}(x_{2},y_{2})$), and $C_{1}$ (resp. $D_{1}$) have
kernel $(1-\varphi(x_{2},y_{2}))k_{C}(x_{2},y_{2}),$ (resp.
$(1-\varphi(x_{2},y_{2}))k_{D}(x_{2},y_{2})$). The composition then splits
into the terms

$DC=D_{0}C_{0}+D_{0}C_{1}+D_{1}C_{0}+D_{1}C_{1}.$ (171)

As the kernels of $D_{0}$ and $C_{0}$ are compactly supported and smoothing,
and the kernels of $D_{1},C_{1}$ are smooth and have specified rates of decay
it is not difficult to show that

$D_{0}C_{0}+D_{0}C_{1}+D_{1}C_{0}:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})\text{
is compact for any }0<\alpha<\frac{1}{2}.$ (172)

Using smooth cut-off functions, we now divide $D_{1},C_{1}$ into operators
$D_{11},C_{11},$ with kernels supported in the set
$\\{(x_{2},y_{2}):\>|x_{2}|>d+\epsilon,|y_{2}|>d+\epsilon\\},$ and the
remainders $D_{10}=D_{1}-D_{11},C_{10}=C_{1}-C_{11}.$ This additonal spliting
is useful as the kernels for $D_{11},C_{11}$ have simpler asymptotics outside
of the channels centered on the $x_{2}$ and $y_{2}$ axes. It is again not
difficult to show that

$D_{10}C_{10}+D_{10}C_{11}+D_{11}C_{10}:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})\text{
is compact for any }0<\alpha<\frac{1}{2}.$ (173)

This leaves just the $D_{11}C_{11}$ term. If we suppose that the left channel
is centered at $0$ and the right at $\gamma,$ then, from Table 1, it follows
that the leading terms of $k_{C_{11}},k_{D_{11}}$ are given by

$\tilde{k}^{0}_{j}=\sum_{\epsilon_{0},\epsilon_{1}\in\\{-,+\\}}\psi(\epsilon_{0}x_{2})\psi(\epsilon_{1}y_{2})\Bigg{[}c_{\epsilon_{0}\epsilon_{1}}^{l,j}\frac{e^{ik_{1}(|x_{2}|+|y_{2}|)}}{(|x_{2}|+|y_{2}|)^{\frac{j+1}{2}}}-\\\
c_{\epsilon_{0}\epsilon_{1}}^{r,j}\frac{e^{ik_{1}(|x_{2}-\gamma|+|y_{2}-\gamma|)}}{(|x_{2}-\gamma|+|y_{2}-\gamma|)^{\frac{j+1}{2}}}\Bigg{]},$
(174)

with $j=0$ for $D$ and $j=2$ for $C,$ and
$\psi(z)\in\mathcal{C}^{\infty}(\mathbb{R})$ is a non-negative, even function
supported where $|z|\geq d+\epsilon$ and equal to $1$ where $|z|>d+4\epsilon.$
Note that $d$ is selected so that
$\operatorname{supp}q_{l},\operatorname{supp}q_{r}$ are compact subsets of
$(-d,d).$ The very simple form of these kernels allows us to estimate the
leading order part of the kernel of the composition, $D_{11}C_{11}.$

######  Proposition 3.

There are positive constants $m,M$ so that

$\left|\int_{-\infty}^{\infty}\tilde{k}^{0}_{0}(x_{2},z)\tilde{k}^{0}_{2}(z,y_{2})dz\right|\leq\frac{M}{(|x_{2}|+m)^{\frac{1}{2}}(|y_{2}|+m)^{\frac{3}{2}}}.$
(175)

###### Proof.

From (174) it follows that the integral in (175) is a sum of integals over
either $(-\infty,-d]$ or $[d,\infty)$ of terms of the form

$Ce^{ik_{1}(|x_{2}-a|+|y_{2}-b|)}\frac{\psi(|z|)e^{ik_{1}(|z-a|+|z-b|)}}{(|x_{2}-a|+|z-a|)^{\frac{1}{2}}(|y_{2}-b|+|z-b|)^{\frac{3}{2}}},$
(176)

where $a$ and $b$ equal either $0$ or $\gamma.$ Within the domain of
integration neither $z-a,$ nor $z-b$ changes sign. All of the various terms
are estimated by integrating by parts. For example, in case $a=b=0,$ an
integration by parts shows that the integral over $[d,\infty)$ equals

$\int_{d}^{\infty}\frac{\psi(z)e^{2ik_{1}z}}{(|x_{2}|+z)^{\frac{1}{2}}(|y_{2}|+z)^{\frac{3}{2}}}=\frac{1}{2ik_{1}}\Bigg{[}\int_{d}^{\infty}\frac{-e^{2ik_{1}z}\psi^{\prime}(z)}{(|x_{2}|+z)^{\frac{1}{2}}(|y_{2}|+z)^{\frac{3}{2}}}+\\\
\frac{1}{2}\int_{d}^{\infty}\frac{e^{2ik_{1}z}\psi(z)}{(|x_{2}|+z)^{\frac{1}{2}}(|y_{2}|+z)^{\frac{3}{2}}}\left(\frac{1}{|x_{2}|+z}+\frac{3}{|y_{2}|+z}\right)dz\Bigg{]}.$
(177)

The first integral on the right is easily seen to satisfy an estimate like
that in (175). Integrating by parts one more time in the other terms gives a
similar formula from which the estimates follows easily. All other types of
terms appearing in (176) are estimated using the same integrations by parts. ∎

We have the following corollary.

######  Corollary 1.

For any $0<\alpha<\frac{1}{2},$ the operator
$DC:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\frac{1}{2}}(\mathbb{R})$
is bounded, and
$DC:\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})$ is a
compact operator.

###### Proof.

The first statement is an immediate consequence of the estimate in (175), and
the fact that the differences between $k_{D}$ and $k_{C}$ and their leading
parts are bounded by $\frac{M}{(1+|x_{2}|+|y_{2}|)^{\frac{3}{2}}}$ and
$\frac{M}{(1+|x_{2}|+|y_{2}|)^{\frac{5}{2}}},$ respectively. The
$x_{2}$-derivative of the kernel of $DC$ is easily seen to satisfy the same
type of estimates, hence Proposition 1 applies to show that the second
statement is true as well. ∎

This corollary immediately implies:

######  Corollary 2.

$(\operatorname{Id}-DC):\mathcal{C}_{\alpha}(\mathbb{R})\to\mathcal{C}_{\alpha}(\mathbb{R})$
is a Fredholm operator of index 0.

######  Remark 3.

To prove the solvability for arbitrary data we would still need to show that
the operator $(\operatorname{Id}-DC)$ has a trivial null-space, which we leave
to a later publication. One might try to use a homotopy $\lambda
q_{l}+(1-\lambda)q_{r}$ to conclude that the operator $(\operatorname{Id}-DC)$
is invertible for “generic pairs” $(q_{l},q_{r}).$ This is somewhat delicate
as our construction of the fundamental solutions $\mathfrak{C}^{l,r}$ requires
the assumption that the Wronskians not vanish at $\pm k_{1},$ see (30).

If
$(g,h)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$

$(\operatorname{Id}-DC)\sigma=g-Dh$ (178)

is solvable for $\sigma\in\mathcal{C}_{\alpha}(\mathbb{R}),$ and we set
$\tau=h-C\sigma\in\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$ then the pair
$(\sigma,\tau)$ solves (149). In this case the solution to the transmission
problem is given by (19),

$u^{l,r}=-\mathcal{E}^{l,r\,^{\prime}}\sigma+\mathcal{E}^{l,r}\tau,$ (179)

which implies that

$\begin{split}u^{l,r}(x)&=\mathcal{S}_{k_{1}}\tau(x)+\int_{-\infty}^{\infty}w^{l,r}(x;0,y_{2})\tau(y_{2})dy_{2}-\\\
&\mathcal{D}_{k_{1}}\sigma(x)-\int_{-\infty}^{\infty}\partial_{y_{1}}w^{l,r}(x;0,y_{2})\sigma(y_{2})dy_{2},\end{split}$
(180)

where $\mp x_{2}>0.$ The kernel for single layer satisfies the estimate

$\left|\frac{H^{(1)}_{0}(k_{1}|x-(0,y_{2})|)}{4}\right|\leq\frac{M}{[x_{1}^{2}+(x_{2}-y_{2})^{2}]^{\frac{1}{4}}}.$
(181)

The kernel of the double layer is given by

$\partial_{y_{1}}\frac{iH^{(1)}_{0}(k_{1}|x-y|)}{4}\upharpoonright_{y_{1}=0}=-i\frac{k_{1}x_{1}}{4|x-(0,y_{2})|}\partial_{z}H^{(1)}_{0}(k_{1}|x-(0,y_{2})|).$
(182)

For $x_{1}\neq 0,$ as $y_{2}\to\infty,$ it satisfies the estimate

$\left|\partial_{y_{1}}\frac{H^{(1)}_{0}(k_{1}|x-y|)}{4}\upharpoonright_{y_{1}=0}\right|\leq\frac{Mx_{1}}{[x_{1}^{2}+(x_{2}-y_{2})^{2}]^{\frac{3}{4}}}.$
(183)

The estimates proved for
$w^{l,r}(0,x_{2};0,y_{2}),\partial_{y_{1}}w^{l,r}(0,x_{2};0,y_{2})$ also hold
where $x_{1}\neq 0,$ which, along with Proposition 2, shows that
representations for $u^{l,r}(x_{1},x_{2})$ in (180) are given by absolutely
convergent integrals.

In general our method is applicable to data
$(g,h)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}).$
The following example shows that our method applies to an interesting class of
incoming fields well beyond those defined by wave guide modes.

######  Example 1.

In the case that the wave guide is a bi-infinite channel, with $k_{1}<k_{2},$
as described by (3), the scattering problem for an incoming plane wave “from
above” has an elementary solution: Let
$\boldsymbol{\kappa}=(\kappa_{1},\kappa_{2})$ satisfy

$\kappa_{1}^{2}+\kappa_{2}^{2}=k_{1}^{2},\quad\kappa_{2}<0,$ (184)

then the function
$v^{\operatorname{in}}=e^{i\boldsymbol{\kappa}\cdot\boldsymbol{x}}$ is a
reasonable incoming field “from above” for the single channel modeled by
$(\Delta+q(x_{2})+k_{1}^{2}),$ with
$q(x_{2})=\chi_{[-d,d]}(x_{2})(k_{2}^{2}-k_{1}^{2}).$ We would like to find
the outgoing scattered wave, $v^{\operatorname{sc}},$ produced by this
solution. We let $\boldsymbol{\kappa}^{\prime}=(\kappa_{1},-\kappa_{2}),$ and
$\tilde{\boldsymbol{\kappa}}=(\kappa_{1},\hat{\kappa}_{2}),\tilde{\boldsymbol{\kappa}}^{\prime}=(\kappa_{1},-\hat{\kappa}_{2}),$
where $\hat{\kappa}_{2}=\sqrt{k_{2}^{2}-\kappa_{1}^{2}}>0.$ The scattered
field can be found using the jump conditions directly and takes the form
predicted by the Fresnel relations

$v^{\operatorname{sc}}(\boldsymbol{\kappa};\boldsymbol{x})=\begin{cases}&\alpha_{+}(\boldsymbol{\kappa})e^{i\boldsymbol{\kappa}^{\prime}\cdot\boldsymbol{x}}\text{
where }x_{2}>d,\\\
&\alpha_{0}(\boldsymbol{\kappa})e^{i\tilde{\boldsymbol{\kappa}}\cdot\boldsymbol{x}}+\beta_{0}(\boldsymbol{\kappa})e^{i\tilde{\boldsymbol{\kappa}}^{\prime}\cdot\boldsymbol{x}}\text{
where }|x_{2}|<d,\\\
&\alpha_{-}(\boldsymbol{\kappa})e^{i\boldsymbol{\kappa}\cdot\boldsymbol{x}}\text{
where }x_{2}<-d.\end{cases}$ (185)

The determinant of the linear system that defines the coefficients,
$(\alpha_{+}(\boldsymbol{\kappa}),\alpha_{0}(\boldsymbol{\kappa}),$
$\beta_{0}(\boldsymbol{\kappa}),\alpha_{-}(\boldsymbol{\kappa})),$ is a non-
zero multiple of $2\kappa_{2}\hat{\kappa}_{2}\cos
2\hat{\kappa}_{2}d+i(k_{2}^{2}-k_{1}^{2})\sin 2\hat{\kappa}_{2}d,$ which does
not vanish provided that $\kappa_{2}\neq 0.$ Hence if
$\boldsymbol{\kappa}=k_{1}(\cos\theta,-\sin\theta),$ then these coefficients
depend smoothly on $\theta\in(0,\pi).$

From the exponentials that arise in
$v^{\operatorname{sc}}(\boldsymbol{\kappa};\boldsymbol{x})$ where $|x_{2}|>d,$
it is apparent that it is outgoing as the scattered waves move away from the
channel. Since the first component of the wave vector is continuous across
$x_{2}=\pm d,$ and $|\kappa_{1}|<k_{1},$ it is clear that the incoming field
cannot excite any waveguide modes, $e^{\pm i\xi_{n}x_{1}}v_{n}(x_{2})$ for
which $k_{1}<|\xi_{n}|<k_{2}.$

A general field incoming “from above” takes the form

$v^{\operatorname{in}}_{\mu}(x)=\int_{0}^{\pi}e^{ik_{1}(\cos\theta,-\sin\theta)\cdot\boldsymbol{x}}d\mu(\theta),$
(186)

with $d\mu$ a finite measure on $(0,\pi).$ By linearity, it produces an
outgoing scattered field of the form

$v^{\operatorname{out}}_{\mu}=\int_{0}^{\pi}v^{\operatorname{sc}}(k_{1}(\cos\theta,-\sin\theta);\boldsymbol{x})d\mu(\theta),$
(187)

which again involves no wave guide modes. While our method for analyzing a
pair of intersecting semi-infinite wave guides does not apply to incoming
fields, which do not decay as $|x_{2}|\to\infty,$ if
$d\mu(\theta)=m(\theta)d\theta,$ with $m\in\mathcal{C}^{2}_{c}((0,\pi)),$ then

$v^{\operatorname{tot}}_{\mu}(x_{1},x_{2})=v^{\operatorname{in}}_{\mu}(x_{1},x_{2})\chi_{[d,\infty)}(x_{2})+v^{\operatorname{out}}_{\mu}(x_{1},x_{2})$

satisfies the transmission boundary conditions and
$(\Delta+q(x_{2})+k_{1}^{2})v^{\operatorname{tot}}_{\mu}=0.$ A stationary
phase computation shows that

$|v^{\operatorname{tot}}_{\mu}(0,x_{2})|\leq\frac{M}{\sqrt{1+|x_{2}|}}\text{
and
}|\partial_{x_{1}}v^{\operatorname{tot}}_{\mu}(0,x_{2})|\leq\frac{M}{1+|x_{2}|},\text{
as }|x_{2}|\to\infty,$ (188)

and therefore our method of solution, with $q_{l}=q,$ and
$u^{\operatorname{in}}$ given by $v^{\operatorname{tot}}_{\mu}$ does apply to
this case. It is an interesting question to determine whether an incoming
field of this type, scattered by a pair of intersecting, semi-infinite wave
guides, can produce non-trivial wave guide modes.

######  Remark 4.

One can imagine other uses for the fundamental solution, $\mathfrak{E},$ of
operators like $(\Delta+q(x_{2})+k_{1}^{2})$ constructed above. A simple
example would be to change the electrical properties of the wave guide in a
compact set replacing $q(x_{2})$ by $q(x_{2})+Q(x_{1},x_{2})$, with $Q$ a
compactly supported function. Suppose that $u^{\operatorname{in}}$ is a
solution to $(\Delta+q+k_{1}^{2})u^{\operatorname{in}}=0,$ and we seek an
outgoing solution, $u^{\operatorname{out}},$ to

$(\Delta+q+k_{1}^{2}+Q)[u^{\operatorname{in}}+u^{\operatorname{out}}]=0.$
(189)

Using the fundamental solution this can be rewritten as a Lipmann-Schwinger
type equation:

$(\operatorname{Id}+\mathfrak{E}Q)u^{\operatorname{out}}=-\mathfrak{E}Qu^{\operatorname{in}}.$
(190)

At least for small $Q,$ this equation can be solved using a Neumann series

$u^{\operatorname{out}}=-\mathfrak{E}Q\sum_{j=0}^{\infty}(-1)^{j}(\chi_{Q}\mathfrak{E}Q)^{j}u^{\operatorname{in}},$
(191)

where $\chi_{Q}$ is the characteristic function of $\operatorname{supp}Q.$ To
compute the terms of the sum only requires a knowledge of the kernel of
$\mathfrak{E}$ on $\operatorname{supp}Q\times\operatorname{supp}Q.$

A similar approach can be used to study the effect of placing an non-
transparent obstacle in the channel. For these cases the scattered field can
be represented in terms of the sum of a single and double layer with respect
to the kernel of $\mathfrak{E}$ over the boundary of the obstacle. This will
lead to a second kind Fredholm equation on the boundary of the obstacle.

A more ambitious application might be to study a network of channels meeting
in a compact set. Using an idea similar to that employed in [1] one can
decompose $\mathbb{R}^{2}$ into a collection of truncated sectors,
$\\{S_{1},\dots,S_{N}\\}$ each containing a single semi-infinite channel

$\\{x:\>|\langle x,v_{j}\rangle-c_{j}|\leq d_{j},\,\langle
x,v^{\bot}_{j}\rangle>e_{j}\\}$

with electrical properties modeled by an operator of the form

$L_{j}=(\Delta+q_{j}(\langle x,v_{j}\rangle-c_{j})+k_{1}^{2}).$

Here $\langle\cdot,\cdot\rangle$ is the Euclidean inner product in
$\mathbb{R}^{2}.$ The channels meet in a compact interaction zone, $D.$ See
Figure 3. Using our construction we can build a fundamental solution,
$\mathfrak{E}_{j}=\mathcal{S}_{k_{1}}+\mathcal{W}_{j},$ for each operator
$L_{j}.$

Figure 3: Three dielectric channels meeting in a compact interaction zone,
$D,$ showing sectors $S_{1},S_{2},S_{3}.$

Solutions to $L_{j}u_{j}=0$ in $S_{j}$ can then be written as sums of single
and double layers w.r.t. $\mathfrak{E}_{j}$ integrated over $\partial S_{j}.$
Imposing jump conditions across the common boundaries of the sectors would
then lead to systems of integral equations over $\cup\partial S_{j},$
analogous to (21). These would be supplemented with boundary conditions on
$\partial D.$ As before, the kernels of the various $\mathcal{W}_{j}$ decay
like $(|s|+|t|)^{-\frac{1}{2}}$ as one goes out to infinity along components
of the $\partial S_{j}.$ Unfortunately, unless the $\partial S_{j}$ is
orthogonal to the channel lying in $S_{j},$ the normal derivatives of these
kernels will also decay at this rate. Thus it seems unlikely that these
integral equations will be well posed in any useful function space. Adding a
little dissipation does lead to tractable integral equations.

##  7 Outgoing Estimates for $\sigma,$ $\tau$ and $u^{r,l}$

The solution that we obtain in Section 6 to the transmission problem is
essentially the limiting absorption solution. In Appendices B and C we show
that, for a single channel, the limiting absorption solution is outgoing in a
simple, classical sense. So, de facto, our solution to the transmission
problem should be thought of as “outgoing.” On the other hand, as we shall
see, the resultant solutions, $u^{r,l}(r\eta),$ do not seem to decay as
$r\to\infty.$ A similar phenomenon is observed in [5]. This is not a result of
the guided modes, which do not decay within the channels, but rather a
consequence of the non-compactness of the domain over which we integrate the
layer potentials, and the decay rate of $\sigma.$

Before estimating $u^{r,l}(r\eta),$ we use the estimates in (126)–(133) to
first show that, if $(g,h)$ decrease rapidly enough, then the solution
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\frac{1}{2}+\alpha}(\mathbb{R})$
to (149) actually satisfies the essentially optimal estimates:

$|\sigma(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{1}{2}}}\text{ and
}|\tau(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{3}{2}}}.$ (192)

These estimates are optimal in the sense that the result of applying the
integral operator in (148) to compactly supported data typically has these
rates of decay. We then use these improved estimates to derive estimates on
$u^{r,l}(r\eta)$ as $r\to\infty.$ We show that

$u^{r,l}(r\eta)=O(1)\text{ and }(\partial_{r}-ik_{1})u^{r,l}(r\eta)=O(1),$
(193)

as $r\to\infty,$ provided that $\eta_{1}\neq 0.$ The asymptotic behavior of
$u^{r,l}(r\eta)$ is largely determined by the rate of decay of $\sigma.$

We begin with the relations

$\sigma(x_{2})=g(x_{2})-D\tau(x_{2}),\quad\tau(x_{2})=h(x_{2})-C\sigma(x_{2}),$
(194)

the basic estimates for $(\sigma,\tau),$ and the assumptions on $g,h:$

$\begin{split}|\tau(x_{2})|\leq\frac{|\tau|_{\alpha+\frac{1}{2}}}{(1+|x_{2}|)^{\alpha+\frac{1}{2}}},&\quad|\sigma(x_{2})|\leq\frac{|\sigma|_{\alpha}}{(1+|x_{2}|)^{\alpha}},\\\
|g(x_{2})|,|g^{\prime}(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{2}},&\quad|h(x_{2})|,|h^{\prime}(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{3}}.\end{split}$
(195)

Note that outside a neighborhood of $B_{d}$ the kernels
$k_{C}(x_{2},y_{2}),k_{D}(x_{2},y_{2})$ are infinitely differentiable. Using
the estimate in (128) we see that, for $\pm x_{2}\to\infty,$ we have

$\begin{split}|(\partial_{x_{2}}\mp
ik_{1})D\tau(x_{2})|&\leq\left|\int_{-\infty}^{\infty}\frac{Mdy_{2}}{|y_{2}|^{\alpha+\frac{1}{2}}(|x_{2}|+|y_{2}|)^{\frac{3}{2}}}\right|\\\
&=\frac{1}{|x_{2}|^{\alpha+1}}\left|\int_{-\infty}^{\infty}\frac{Mdt}{|t|^{\alpha+\frac{1}{2}}(1+|t|)^{\frac{3}{2}}}\right|\leq\frac{M_{\alpha}}{|x_{2}|^{\alpha+1}}.\end{split}$
(196)

Similarly we can show that, as $\pm x_{2}\to\infty,$

$\begin{split}|(\partial_{x_{2}}\mp
ik_{1})C\sigma(x_{2})|&\leq\left|\int_{-\infty}^{\infty}\frac{Mdy_{2}}{|y_{2}|^{\alpha}(|x_{2}|+|y_{2}|)^{\frac{5}{2}}}\right|\\\
&=\frac{1}{|x_{2}|^{\alpha+\frac{3}{2}}}\left|\int_{-\infty}^{\infty}\frac{Mdt}{|t|^{\alpha}(1+|t|)^{\frac{5}{2}}}\right|\leq\frac{M_{\alpha}}{|x_{2}|^{\alpha+\frac{3}{2}}}.\end{split}$
(197)

From these estimates and the equations in (194), which show that

$\partial_{x_{2}}\sigma=\partial_{x_{2}}g-\partial_{x_{2}}D\tau,\quad\partial_{x_{2}}\tau=\partial_{x_{2}}g=h-\partial_{x_{2}}C\sigma,$
(198)

we see that, as $\pm x_{2}\to\infty,$ we have the estimates

$\begin{split}&|\partial_{x_{2}}\sigma(x_{2})\mp
ik_{1}\sigma(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\alpha+1}}\\\
&|\partial_{x_{2}}\tau(x_{2})\mp
ik_{1}\tau(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\alpha+\frac{3}{2}}}.\end{split}$
(199)

Let $\psi_{+}\in\mathcal{C}^{\infty}(\mathbb{R}),$ be a non-negative monotone
function with $\psi_{+}(x)=0,$ for $x\leq d,$ and $\psi_{+}(x)=1,$ for
$x>d+\frac{1}{2}.$ Using integration parts and (129) we see that for
$|x_{2}|>d,$

$\begin{split}\Bigg{|}\int_{d}^{\infty}k_{D}(x_{2},y_{2})[\partial_{y_{2}}(\psi_{+}\tau)(y_{2})+&ik_{1}\psi_{+}\tau(y_{2})]dy_{2}\Bigg{|}\\\
&=\left|\int_{d}^{\infty}(\partial_{y_{2}}-ik_{1})k_{D}(x_{2},y_{2})\psi_{+}\tau(y_{2})dy_{2}\right|\\\
&\leq\int_{d}^{\infty}\frac{Mdy_{2}}{|y_{2}|^{\alpha+\frac{1}{2}}(|x_{2}|+|y_{2}|)^{\frac{3}{2}}}\leq\frac{M_{\alpha}}{|x_{2}|^{\alpha+1}}.\end{split}$
(200)

A similar argument, with $\psi_{-}(x_{2})=\psi_{+}(-x_{2}),$ shows that

$\left|\int_{-\infty}^{-d}k_{D}(x_{2},y_{2})[\partial_{y_{2}}(\psi_{-}\tau)(y_{2})-ik_{1}\psi_{-}\tau(y_{2})]dy_{2}\right|\leq\frac{M_{\alpha}}{|x_{2}|^{\alpha+1}}.$
(201)

The estimates in (199) show that, for $\pm x_{2}>d,$ we have

$\partial_{x_{2}}(\psi_{\pm}\tau)(x_{2})=(\partial_{x_{2}}\psi_{\pm}(x_{2}))\tau(x_{2})+\psi_{\pm}(x_{2})\partial_{x_{2}}\tau(x_{2})=\\\
(\partial_{x_{2}}\psi_{\pm}(x_{2}))\tau\pm
ik_{1}\psi_{\pm}\tau(x_{2})+O(|x_{2}|^{-(\alpha+\frac{3}{2})}).$ (202)

Using this in (200) we see that

$\left|2ik_{1}\int_{d}^{\infty}k_{D}(x_{2},y_{2})\psi_{+}\tau(y_{2})dy_{2}\right|\leq\frac{M}{|x_{2}|^{1+\alpha}}+O(|x_{2}|^{-\frac{1}{2}})+\int_{d}^{\infty}\frac{dy_{2}}{y_{2}^{\alpha+\frac{3}{2}}(x_{2}+y_{2})^{\frac{1}{2}}}.$
(203)

An elementary estimate shows that the integral on the r.h.s. is bounded by
$Mx_{2}^{-\frac{1}{2}}.$ There is a similar estimate arising from (201) for
the integral from $-\infty$ to $-d.$

As

$D\tau(x_{2})=\int_{-\infty}^{-d}k_{D}(x_{2},y_{2})\psi_{-}\tau(y_{2})dy_{2}+\int^{\infty}_{d}k_{D}(x_{2},y_{2})\psi_{+}\tau(y_{2})dy_{2}+\\\
\int_{-(d+1)}^{d+1}k_{D}(x_{2},y_{2})(1-\psi_{+}(y_{2})-\psi_{-}(y_{2}))\tau(y_{2})dy_{2},$
(204)

from (203) and the analogue for $\psi_{-}\tau$ we conclude that

$|D\tau(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{1}{2}}},$ (205)

and therefore (194) implies that

$|\sigma(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{1}{2}}}$ (206)

as well.

We can repeat this argument for the integrals appearing in $C\sigma$ to
conclude that

$\begin{split}&\left|\int_{d}^{\infty}k_{C}(x_{2},y_{2})[\partial_{y_{2}}(\psi_{+}\sigma)(y_{2})+ik_{1}\psi_{+}\sigma(y_{2})]dy_{2}\right|\leq\frac{M}{x_{2}^{2}},\\\
&\left|\int_{-\infty}^{-d}k_{C}(x_{2},y_{2})[\partial_{y_{2}}(\psi_{-}\sigma)(y_{2})-ik_{1}\psi_{-}\sigma(y_{2})]dy_{2}\right|\leq\frac{M}{x_{2}^{2}}.\end{split}$
(207)

Combining these estimates with the estimate in (199) we see that

$|C\sigma(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{3}{2}}},$ (208)

and therefore (194) implies that

$|\tau(x_{2})|\leq\frac{M}{(1+|x_{2}|)^{\frac{3}{2}}}$ (209)

as well.

We see that the outgoing conditions along the $x_{2}$-axis satisfied by the
kernels $k_{C},k_{D}$ force the source terms to satisfy much stronger
estimates than the mere existence theory imposes. Using these refined
estimates we now use the representation formula (180) to derive estimates for
$u^{r,l}$ along radial lines through the origin. Recall that

$\begin{split}u^{r,l}(x)=&-\mathcal{E}^{l,r\,^{\prime}}\sigma(x)+\mathcal{E}^{l,r}\tau(x)\\\
=&-\mathcal{D}_{k_{1}}\sigma(x)-\mathcal{W}^{l,r\,^{\prime}}\sigma(x)+\mathcal{S}_{k_{1}}\tau(x)+\mathcal{W}^{l,r}\tau(x).\end{split}$
(210)

For the purposes of analyzing the behavior of $u^{r,l}(r\eta),$ as
$r\to\infty,$ the term
$\mathcal{W}^{l,r}\tau(x)-\mathcal{W}^{l,r\,^{\prime}}\sigma(x)$ is a small
perturbation of $-\mathcal{D}_{k_{1}}\sigma(x)+\mathcal{S}_{k_{1}}\tau(x),$ at
least if $\eta_{2}\neq 0.$ We leave their detailed analysis to the interested
reader.

We begin with the single layer term, and focus on $u^{r};$ the analysis of
$u^{l}$ is essentially the same. Let $\eta\in\mathbb{R}^{2}$ be a unit vector
with $\eta_{1}>0$ non-zero. We let

$u^{r}_{0}(r\eta)=\mathcal{S}_{k_{1}}\tau(r\eta)=\frac{i}{4}\int_{-\infty}^{\infty}H^{(1)}_{0}(k_{1}|r\eta-(0,y_{2})|)\tau(y_{2})dy_{2}.$
(211)

As $\eta_{1}>0,$ the $|r\eta-(0,y_{2})|>r\eta_{1}$ and therefore we can
estimate this integral using the large $z$ asymptotics of $H^{(1)}_{0}(z).$
With this is mind we see that

$u^{r}_{0}(r\eta)=\frac{i}{4}\int_{-\infty}^{\infty}\frac{e^{ik_{1}\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}}{[r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{\frac{1}{4}}}\left[1+O\left([r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{-\frac{1}{2}}\right)\right]\tau(y_{2})dy_{2}.$
(212)

Using the estimate $|\tau(y_{2})|\leq M(1+|y_{2}|)^{-\frac{3}{2}}$ we easily
show that the error term is $O(r^{-\frac{3}{2}}).$ Letting $y_{2}=rt,$ we
obtain

$\int_{-\infty}^{\infty}\frac{e^{ik_{1}\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}}{[r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{\frac{1}{4}}}\tau(y_{2})dy_{2}=r^{\frac{1}{2}}\int_{-\infty}^{\infty}\frac{e^{ik_{1}r\sqrt{1-2\eta_{2}t+t^{2}}}}{[1-2\eta_{2}t+t^{2}]^{\frac{1}{4}}}\tau(rt)dt.$
(213)

Using the estimate on $\tau$ we can show that this term is
$O(r^{-\frac{1}{2}}).$

Applying $\partial_{r}-ik_{1}$ to $u^{r}_{0}$ we see that

$(\partial_{r}-ik_{1})u^{r}_{0}(r\eta)=-\frac{k_{1}}{4}\int_{-\infty}^{\infty}\frac{e^{ik_{1}\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}}{[r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{\frac{1}{4}}}\Bigg{[}\frac{r-\eta_{2}y_{2}}{\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}-1+\\\
O\left([r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{-\frac{1}{2}}\right)\Bigg{]}\tau(y_{2})dy_{2}.$
(214)

Once again the error term is $O(r^{-\frac{3}{2}}).$ In the leading term we set
$y_{2}=rt$ to obtain

$\begin{split}&\int_{-\infty}^{\infty}\frac{1}{[r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}]^{\frac{1}{4}}}\left|\frac{r-\eta_{2}y_{2}}{\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}-1\right||\tau(y_{2})|dy_{2}=\\\
&\int_{-\infty}^{\infty}\frac{r^{\frac{1}{2}}}{[1-2\eta_{2}t+t^{2}]^{\frac{1}{4}}}\left|\frac{1-\eta_{2}t-\sqrt{1-2\eta_{2}t+t^{2}}}{\sqrt{1-2\eta_{2}t+t^{2}}}\right||\tau(rt)|dt.\end{split}$
(215)

To get a better estimate than $O(r^{-\frac{1}{2}})$ we split this into the
integral from $0$ to $r^{-\frac{3}{7}},$ and the integral from
$r^{-\frac{3}{7}}$ to $\infty.$ In the bounded part we approximate the
integral by a constant times $\|\tau\|_{\infty}t^{2},$ and in the unbounded
part by $M|\tau|_{\frac{3}{2}}/(1+|rt|)^{\frac{3}{2}}.$ With this choice, the
two terms decay at the same rate, and we see that

$(\partial_{r}-ik_{1})u^{r}_{0}(r\eta)=O(r^{-(\frac{1}{2}+\frac{2}{7})}),$
(216)

which reflects some sort of outgoing property.

We now consider the double layer term, which, as $r\to\infty,$ asymptotically
takes the form

$u_{1}^{r}(r\eta)=\mathcal{D}_{k_{1}}\sigma(r\eta)=\frac{-ik_{1}}{4}\int_{-\infty}^{\infty}\frac{r\eta_{1}\sigma(y_{2})e^{ik_{1}\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}}{(r^{2}-2r\eta_{2}y_{2}+y_{2}^{2})^{\frac{3}{4}}}\times\\\
\Bigg{[}1+O\left((r^{2}-2r\eta_{2}y_{2}+y_{2}^{2})^{-\frac{1}{2}}\right)\Bigg{]}dy_{2}.$
(217)

A straightforward calculation shows that the error term is $O(r^{-1}).$
Arguing as above, letting $y_{2}=rt,$ we can show that the leading term is
uniformly bounded as $r\to\infty,$ though it does not seem to decay.

A calculation shows that

$(\partial_{r}-ik_{1})u^{r}_{1}(r\eta)=\frac{-k^{2}_{1}}{4}\int_{-\infty}^{\infty}\frac{r\eta_{1}\sigma(y_{2})e^{ik_{1}\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}}{(r^{2}-2r\eta_{2}y_{2}+y_{2}^{2})^{\frac{3}{4}}}\times\\\
\Bigg{[}\frac{r-\eta_{2}y_{2}}{\sqrt{r^{2}-2r\eta_{2}y_{2}+y_{2}^{2}}}-1\Bigg{]}dy_{2}+O(r^{-1}).$
(218)

Once again we can show that $(\partial_{r}-ik_{1})u^{r}_{1}(r\eta)$ remains
bounded as $r\to\infty,$ but it too does not appear to decay. The failure to
decay or to satisfy a standard radiation condition is a consequence of the
fact that $\sigma(x_{2})$ appears to decay like $|x_{2}|^{-\frac{1}{2}}.$

While this appears to be the best general result we can obtain for
$u^{r,l}(r\eta)$ as $r\to\infty,$ we can obtain better estimates for
$u^{r,l}(x_{1},x_{2})$ as $x_{2}\to\pm\infty,$ with $|x_{1}|>0$ fixed. We
examine the simpler expression
$-\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})+\mathcal{S}_{k_{1}}\tau(x_{1},x_{2}).$
Using the rate of decay for $\tau$ from (192), an analysis using Lemma 1 shows
that

$\begin{split}|\mathcal{S}_{k_{1}}\tau(x_{1},x_{2})|&\leq\frac{M}{1+|x_{2}|^{\frac{1}{2}}},\\\
|(\partial_{x_{2}}\mp
ik_{1})\mathcal{S}_{k_{1}}\tau(x_{1},x_{2})|&\leq\frac{M}{1+|x_{2}|},\text{
for }\pm x_{2}\to\infty.\end{split}$ (219)

On the other hand the contribution of $\mathcal{D}_{k_{1}}\sigma$ satisfies
the estimates

$\begin{split}|\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})|&\leq\frac{M}{1+|x_{2}|^{\frac{1}{2}}},\\\
|(\partial_{x_{2}}\mp
ik_{1})\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})|&\leq\frac{M}{1+|x_{2}|^{\frac{1}{2}}},\text{
for }\pm x_{2}\to\infty.\end{split}$ (220)

The details of these estimates are left to the reader.

The $|x_{2}|^{-\frac{1}{2}}$ rate-of-decay of $\sigma(x_{2})$ is what prevents
$(\partial_{r}\mp ik_{1})\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})$ from decaying
any faster than $\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2}),$ for
$(x_{1},x_{2})=r\eta.$ If $\sigma(x_{2})$ decayed like $|x_{2}|^{-\alpha}$ for
any $\frac{1}{2}\leq\alpha<1,$ then $\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})$
and $(\partial_{r}\mp ik_{1})\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})$ would
decay like $r^{\frac{1}{2}-\alpha}.$ If $\alpha>1,$ then
$\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})$ would decay like $r^{-\frac{1}{2}}$
and $(\partial_{r}\mp ik_{1})\mathcal{D}_{k_{1}}\sigma(x_{1},x_{2})$ would
decay at a faster rate, which depends on $\alpha.$

The vertical line separating the two half planes is an artificial boundary,
with the actual physical boundary between the two open wave guides a finite
segment. Outside of the wave guides the total field is a smooth solution to
$(\Delta+k_{1}^{2})u=0,$ so one would expect this solution to satisfy more
standard decay and radiation conditions, at least in directions other than
$(\pm 1,0).$

##  8 The Projections onto Wave Guide Modes

In the foregoing pages we have explained a method to find and represent
solutions to the transmission problem that results from two semi-infinite wave
guides meeting along a common perpendicular line. The solution is represented
in each half plane by layer potentials along this line, with sources
$(\sigma,\tau),$ see (180). The solutions in each half plane can be split into
a contribution from the wave guide modes and “radiation,”

$u^{l,r}(x)=u^{l,r}_{g}(x)+u^{l,r}_{\operatorname{rad}}(x).$ (221)

If $\\{(v_{n}^{l,r},\xi_{n}^{l,r}):\>n=1,\dots,N_{l,r}\\}$ are the guided
modes, normalized to have $L^{2}$-norms 1, then the projection into the guided
modes is given by

$u^{l,r}_{g}(x_{1},x_{2})=\sum_{n=1}^{N_{l,r}}v_{n}^{l,r}(x_{2})\int_{-\infty}^{\infty}u^{l,r}(x_{1},y_{2})v_{n}^{l,r}(y_{2})dy_{2}$
(222)

This has a very simple expression in terms of our representation, coming
entirely from the $w^{g}_{0+}$-term in the expression for
$\mathfrak{E}^{l,r},$ see (36).

We assume that $u^{l,r}$ is given by (180), with sources
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}).$
As noted in Section 6 the representations for $u^{l,r}$ and
$\partial_{x_{1}}u^{l,r}$ are in terms of absolutely convergent integrals. The
key observation is the following

######  Proposition 4.

Let
$(\sigma,\tau)\in\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R})$
and let $v_{n}^{l,r}(x_{2})e^{i\xi_{n}^{l,r}x_{1}}$ be a wave guide mode for
$\Delta+k_{1}^{2}+q_{l,r}(x_{2}),$ then, for all $\pm x_{1}>0,$

$\begin{split}&\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}[g_{k_{1}}(|(x_{1},x_{2}-y_{2})|)+w^{c;l,r}_{0+}(x_{1},x_{2};0,y_{2})]\tau(y_{2})v_{n}^{l,r}(x_{2})dy_{2}dx_{2}=0,\\\
&\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\partial_{y_{1}}[g_{k_{1}}(|(x_{1}-y_{1},x_{2}-y_{2})|)+w^{c;l,r}_{0+}(x_{1},x_{2};y_{1},y_{2})]_{y_{1}=0}\tau(y_{2})v_{n}^{l,r}(x_{2})dy_{2}dx_{2}=0,\end{split}$
(223)

###### Proof.

We give the details for the right half plane. The estimates proved in the
previous sections show that these integrals are absolutely convergent and
therefore we can change the order of the integrations. To prove the
proposition we show that, for each $y_{2}$ and $x_{1}>0,$ we have

$\begin{split}&\int\limits_{-\infty}^{\infty}[g_{k_{1}}(|(x_{1},x_{2}-y_{2})|)+w^{c;r}_{0+}(x_{1},x_{2};0,y_{2})]v_{n}^{r}(x_{2})dx_{2}=0,\\\
&\int\limits_{-\infty}^{\infty}\partial_{y_{1}}[g_{k_{1}}(|(x_{1}-y_{1},x_{2}-y_{2})|)+w^{c;r}_{0+}(x_{1},x_{2};y_{1},y_{2})]_{y_{1}=0}v_{n}^{r}(x_{2})dx_{2}=0.\end{split}$
(224)

To prove these statements we use the Sommerfeld integral representation for
the free space fundamental solution, see (25).

We begin with the single layer term, which can be written as the iterated
integral:

$\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\left[\int\limits_{-\infty}^{\infty}\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\int\limits_{\Gamma_{\nu}^{+}}\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]e^{ix_{1}\xi}d\xi\,v_{n}^{r}(x_{2})dx_{2}.$
(225)

We would like to change the order of integrations in this integral, which
would be easily justified if the integral in $\xi$ were over any finite
interval. Note that

$\int\limits_{k_{1}+1}^{\infty}\left|\frac{ie^{-|x_{2}-y_{2}|\sqrt{\xi^{2}-k_{1}^{2}}}e^{-iy_{1}\xi}}{2\sqrt{\xi^{2}-k_{1}^{2}}}\right|d\xi\leq
M[1+|\log|x_{2}-y_{2}|].$ (226)

Combining this with the estimates (75) and (94), and the fact that
$|v_{n}^{r}(x_{2})|\leq Me^{-|x_{2}|\sqrt{\xi_{n}^{r\,2}-k_{1}^{2}}}$ shows
that these integrals are absolutely convergent and therefore we can
interchange the order of the integrations. By analyticity we can also replace
the integral in the first term with an integral over $\Gamma_{\nu}^{+},$ to
obtain

$\frac{1}{2\pi}\int\limits_{\Gamma_{\nu}^{+}}\left[\int\limits_{-\infty}^{\infty}\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]e^{ix_{1}\xi}v_{n}^{r}(x_{2})dx_{2}d\xi.$
(227)

Using the fact that
$\xi^{r\,2}_{n}v_{n}^{r}=(\partial_{x_{2}}^{2}+k_{1}^{2}+q_{r}(x_{2}))v_{n}^{r}$
and integrating by parts it follows that

$\int\limits_{-\infty}^{\infty}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]e^{ix_{1}\xi}v_{n}^{r}(x_{2})dx_{2}=\frac{ie^{ix_{1}\xi}v_{n}^{r}(y_{2})}{\xi_{n}^{r\,2}-\xi^{2}}.$
(228)

Hence the double integral is

$\frac{1}{2\pi
i}\int\limits_{\Gamma_{\nu}^{+}}\frac{ie^{ix_{1}\xi}v_{n}^{r}(y_{2})d\xi}{\xi^{2}-\xi_{n}^{r\,2}},$
(229)

which, using Cauchy’s theorem, is easily seen to vanish for $x_{1}>0.$

The double layer is almost the same; the double integral in (225) is replaced
by

$\frac{i}{2\pi}\int\limits_{-\infty}^{\infty}\left[\,\int\limits_{\Gamma_{\nu}^{+}}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]\xi
e^{ix_{1}\xi}d\xi\right]v_{n}^{r}(x_{2})dx_{2}.$ (230)

The integral involving $\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})$ is again easily
seen to be absolutely convergent, but the additional factor of $\xi$ makes the
other term more subtle. We need an estimate for this term that takes account
of the fact that $x_{1}>0.$

######  Lemma 2.

For $x_{1}>0,\lambda>0$ and $R>k_{1}$ let

$f^{\pm}_{R}(x_{1},\lambda)=\pm\int\limits_{\pm
R}^{\pm\infty}\frac{e^{ix_{1}\xi-\lambda\sqrt{\xi^{2}-k_{1}^{2}}}\xi
d\xi}{\sqrt{\xi^{2}-k_{1}^{2}}}.$ (231)

These functions satisfy the estimates

$|f^{\pm}_{R}(x_{1},\lambda)|\leq\frac{Me^{-\lambda\sqrt{R^{2}-k_{1}^{2}}}}{x_{1}}$
(232)

###### Proof.

As $f^{-}_{R}(x_{1},\lambda)=\overline{f^{+}_{R}(x_{1},\lambda)}$ it suffices
to do the $+$-case. If we let $s=\sqrt{\xi^{2}-k_{1}},$ then

$f^{+}_{R}(x_{1},\lambda)=\int\limits_{\sqrt{R^{2}-k_{1}^{2}}}^{\infty}e^{ix_{1}\sqrt{s^{2}+k_{1}^{2}}-\lambda
s}ds.$ (233)

Noting that

$\partial_{s}\left[\frac{\sqrt{s^{2}+k_{1}^{2}}}{s}e^{ix_{1}\sqrt{s^{2}+k_{1}^{2}}}\right]=\\\
ix_{1}e^{ix_{1}\sqrt{s^{2}+k_{1}^{2}}}-\frac{k_{1}^{2}}{s^{2}\sqrt{s^{2}+k_{1}^{2}}}e^{ix_{1}\sqrt{s^{2}+k_{1}^{2}}},$
(234)

integration by parts shows that

$f^{+}_{R}(x_{1},\lambda)=\frac{1}{ix_{1}}\Bigg{[}\frac{k_{1}e^{-\lambda\sqrt{R^{2}-k_{1}^{2}}+ix_{1}R}}{\sqrt{R^{2}-k_{1}^{2}}}+\\\
\int\limits_{\sqrt{R^{2}-k_{1}^{2}}}^{\infty}\left[\frac{\lambda\sqrt{s^{2}+k_{1}^{2}}}{s}+\frac{k_{1}^{2}}{s^{2}\sqrt{s^{2}+k_{1}^{2}}}\right]e^{ix_{1}\sqrt{s^{2}+k_{1}^{2}}-\lambda
s}ds\Bigg{]}.$ (235)

The integral is easily seen to be $O(e^{-\lambda\sqrt{R^{2}-k_{1}^{2}}}),$
which completes the proof of the lemma. ∎

We rewrite the integral in (230) as

$\frac{i}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{\Gamma_{\nu}^{+}}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}\right]\xi
e^{ix_{1}\xi}d\xi\,v_{n}^{r}(x_{2})dx_{2}=\\\
\frac{i}{2\pi}\int\limits_{-\infty}^{\infty}\left[\,\int\limits_{\Gamma_{\nu}^{+}\cap
D_{R}}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}\right]\xi
e^{ix_{1}\xi}d\xi+f_{R}^{+}(x_{1},|x_{2}-y_{2}|)+f_{R}^{-}(x_{1},|x_{2}-y_{2}|)\right]v_{n}^{r}(x_{2})dx_{2}.$
(236)

In the part of the integral over $\Gamma_{\nu}^{+}\cap D_{R}$ we can
interchange the order of the integrations. We use the lemma to estimate the
other terms

$\begin{split}\Bigg{|}\int\limits_{-\infty}^{\infty}[f_{R}^{+}(x_{1},|x_{2}-y_{2}|)&+f_{R}^{-}(x_{1},|x_{2}-y_{2}|)]v_{n}^{r}(x_{2})dx_{2}\Bigg{|}\\\
&\leq\frac{M}{x_{1}}\int_{-\infty}^{\infty}e^{-\sqrt{R^{2}-k_{1}^{2}}|x_{2}-y_{2}|}e^{-\sqrt{\xi_{n}^{r\,2}-k_{1}^{2}}|x_{2}|}dx_{2}\\\
&\leq\frac{M}{x_{1}(\sqrt{R^{2}-k_{1}^{2}}-\sqrt{\xi_{n}^{r\,2}-k_{1}^{2}})}.\end{split}$
(237)

Thus we see that

$\int\limits_{-\infty}^{\infty}\left[\,\int\limits_{\Gamma_{\nu}^{+}}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]\xi
e^{ix_{1}\xi}d\xi\right]v_{n}^{r}(x_{2})dx_{2}=\\\
\lim_{R\to\infty}\int\limits_{\Gamma_{\nu}^{+}\cap
D_{R}}\int\limits_{-\infty}^{\infty}\left[\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]v_{n}^{r}(x_{2})dx_{2}\right]\,\xi
e^{ix_{1}\xi}d\xi.$ (238)

The $x_{2}$-integral is computed in (228), giving

$\int\limits_{-\infty}^{\infty}\left[\,\int\limits_{\Gamma_{\nu}^{+}}\left[\frac{ie^{i|x_{2}-y_{2}|\sqrt{k_{1}^{2}-\xi^{2}}}}{2\sqrt{k_{1}^{2}-\xi^{2}}}+\tilde{w}_{0+}^{c;r}(\xi,x_{2};y_{2})\right]\xi
e^{ix_{1}\xi}d\xi\right]v_{n}^{r}(x_{2})dx_{2}=\\\
\lim_{R\to\infty}v_{n}(y_{2})\int\limits_{\Gamma_{\nu}^{+}\cap
D_{R}}\frac{i\xi e^{ix_{1}\xi}}{\xi_{n}^{r\,2}-\xi^{2}}d\xi=0.$ (239)

The last equality follows from Cauchy’s theorem. It completes the proof of
(224) and also of the proposition. ∎

Using this proposition we can compute the projections of $u^{r,l}$ to the
respective right and left wave guide modes given in (222). In our
representation

$u^{r,l}(x)=\int_{-\infty}^{\infty}\mathfrak{E}^{r,l}(x;0,y_{2})\tau(y_{2})dy_{2}-\int_{-\infty}^{\infty}\partial_{y_{1}}\mathfrak{E}^{r,l}(x;0,y_{2})\sigma(y_{2})dy_{2}.$
(240)

We give the details for $\\{x_{1}>0\\};$ Proposition 4 and (35) show that

$u_{g}^{r}(x_{1},x_{2})=\sum_{n=1}^{N_{r}}[\langle\tau,v_{n}^{r}\rangle+i\xi_{n}^{r}\langle\sigma,v_{n}^{r}\rangle]v_{n}^{r}(x_{2})e^{i\xi_{n}^{r}x_{1}},$
(241)

where

$\langle f,g\rangle=\int\limits_{-\infty}^{\infty}f(x)\overline{g(x)}dx.$
(242)

Hence the projections of $u^{r,l}$ onto the wave guide modes are completely
determined by the projections,
$\\{\langle\tau,v_{n}^{l,r}\rangle,\langle\sigma,v_{n}^{l,r}\rangle:\>n=1,\dots,N_{l,r}\\},$
of the source terms onto these modes.

If these projections could be determined directly from the data, then we could
determine the scattering relation, from incoming wave guide modes to outgoing
modes, without having to solve the complete problem. Starting with the
equation (149) we can almost find equations for the coefficients in (241).
Projecting these equations into span of the wave guide modes we obtain

$\begin{split}P^{l,r}_{g}\sigma+P^{l,r}_{g}D\tau&=P^{l,r}_{g}g\\\
P^{l,r}_{g}C\sigma+P^{l,r}_{g}\tau&=P^{l,r}_{g}h,\end{split}$ (243)

where we let

$P^{l,r}_{g}f=\sum_{n=1}^{N_{l,r}}\langle f,v_{n}^{l,r}\rangle
v_{n}^{l,r}(x_{2}).$ (244)

These equations can be rewritten as

$\begin{split}P^{l,r}_{g}\sigma+P^{l,r}_{g}DP_{g}^{l,r}\tau&=P^{l,r}_{g}g-P^{l,r}_{g}D(\operatorname{Id}-P_{g}^{l,r})\tau\\\
P^{l,r}_{g}CP^{l,r}_{g}\sigma+P^{l,r}_{g}\tau&=P^{l,r}_{g}h-P^{l,r}_{g}C(\operatorname{Id}-P^{l,r}_{g})\sigma.\end{split}$
(245)

While this is not quite a system of equations for the projections
$(P^{l,r}_{g}\sigma,P^{l,r}_{g}\tau),$ the facts that
$P^{l,r}_{g}(\operatorname{Id}-P^{l,r}_{g})=0,$ and the norms of $\|D\|$ and
$\|C\|$ are proportional to $m(q_{l},q_{r})$ suggests that, at least for two
channels with small contrast, dropping these terms leads to equations

$\begin{split}P^{l,r}_{g}\widetilde{\sigma}+P^{l,r}_{g}DP_{g}^{l,r}\widetilde{\tau}&=P^{l,r}_{g}g\\\
P^{l,r}_{g}CP^{l,r}_{g}\widetilde{\sigma}+P^{l,r}_{g}\widetilde{\tau}&=P^{l,r}_{g}h,\end{split}$
(246)

whose solutions,
$(P^{l,r}_{g}\widetilde{\sigma},P_{g}^{l,r}\widetilde{\tau}),$ should be very
close to $(P^{l,r}_{g}\sigma,P_{g}^{l,r}\tau).$

If the data, $(g,h),$ comes from an incoming wave packet, like that defined in
Example 1, then computations like those in the proof of Proposition 4 show
that

$P^{l}_{g}(g)=P^{l}_{g}(h)=0.$ (247)

This indicates that the reflected guided modes in the left half plane can be
expected to be very small. On the other hand, while $P^{r}_{g}(g)$ and
$P^{r}_{g}(h)$ are likely to be small, at least if the contrast is small, they
are unlikely to be zero, and therefore one can expect the outgoing wave guide
modes to be larger in the right half plane.

##  9 Some Concluding Remarks

In the foregoing pages we have constructed outgoing fundamental solutions for
operators of the form $\Delta+k_{1}^{2}+q(x_{2}),$ and shown how to use them
to represent the solution to the transmission problem defined by two semi-
infinite wave guides meeting along a common perpindicular line. The
construction of the “outgoing” fundamental solution is in a form that should
lend itself to numerical implementation. We have shown that the resultant
system of integral equations is Fredholm of second kind on the spaces
$\mathcal{C}_{\alpha}(\mathbb{R})\oplus\mathcal{C}_{\alpha+\frac{1}{2}}(\mathbb{R}),$
with $0<\alpha<\frac{1}{2},$ and are therefore generically solvable.

We have only presented a detailed analysis of this problem for the case of
potentials given by (3), though it is clear that our approach will apply,
mutatis mutandis, if $q(x_{2})$ is a bounded function with bounded support.
The principal difference will be that the basic solutions,
$\widetilde{u}_{\pm}(\xi,0^{+};x_{2}),$ of the ODE, and their Wronskian are no
longer explicit elementary functions. Various formulæ will need to be replaced
by (standard) estimates. To implement the method, $\widetilde{u}_{\pm}$ will
have to be computed numerically within the support of $q.$ The problem of
having 2 open wave guides that are of the form considered here outside a
compact set, is a realtively compact perturbation which, while requiring
further analysis, should not pose serious additional difficulties.

We have considered neither the question of uniqueness for the solution of our
integral equation, nor that of the 2d-scattering problem. It seems quite
likely that the solution we obtain is the physically meaningful solution.
These solutions do not seem to satisfy a classical outgoing radiation
condition, but this may well be an artifact of our representation, which
entails integration over a non-compact set. These problems of uniqueness and
the correct notion of outgoing will be considered in subsequent publications.

Appendix

In these appendices we collect a variety of background results, and study the
outgoing properties of the limiting absorption solution in the case of a bi-
infinite channel.

##  Appendix A The Planar Case

In order to estimate the correction terms $w^{l,r}(x;y)$ we need to have a
good description of the limit of the kernels for $(D_{q}+i\delta)^{-1}$ as
$\delta\to 0^{+},$ where

$D_{q}=\Delta+k_{1}^{2}+q(x_{2}).$ (248)

We are employing the limiting absorption principle limit, which, more or less
by definition, gives the outgoing solution to

$(\Delta+k_{1}^{2}+q)u=f,$ (249)

for certain functions $f,$ which includes, but is not limited to compactly
supported functions.

In this section we use the Fourier transform in the $x_{1}$-variable and basic
ODE theory to construct the kernels for these operators where we usually take

$q(x_{2})=(k_{2}^{2}-k_{1}^{2})\chi_{[-d,d]}(x_{2}).$ (250)

Much of this material consists of well known results from the theory of second
order ODE.

It would be more standard to consider the spectral theory and resolvent of the
operator $\Delta+q(x_{2}).$ However our analysis relies on detailed
analyticity properties of the kernel of $(\Delta+q+k_{1}^{2}+i\delta)$ for
$\delta>0,$ which is why we consider the shifted operator. The substance of
these results generalizes easily to piecewise continuous functions,
$q(x_{2}),$ with support in $[-d,d],$ which we will consider in a later
publication.

With $H^{2}(\mathbb{R}^{2})$ as the domain, $D_{q}$ defines an unbounded self
adjoint operator on $L^{2}(\mathbb{R}^{2}).$ The spectrum of this operator,
$\sigma(D_{q}),$ is well known to lie in the interval $(-\infty,k_{2}^{2}].$
In this section we present a construction for the resolvent kernel of this
operator, which allows for the construction of the perturbation terms
$w^{l,r},$ by taking for $q$ either $q_{l}$ or $q_{r}.$ To that end we need
compute the kernel of the limit

$\lim_{\delta\to 0^{+}}(D_{q}+i\delta)^{-1},$

which we denote by $\mathcal{R}_{0+}(x;y).$ This is the operator arising from
limiting absorption principle applied to this case. By a small abuse of
terminology, in the sequel we refer to $\mathcal{R}_{0+}$ as the _resolvent
kernel_ , or outgoing resolvent kernel.

Our construction of the resolvent kernel uses the partial Fourier transform in
the $x_{1}$-variable, which we denote by

$\widetilde{u}(\xi,x_{2})=\int_{-\infty}^{\infty}u(x_{1},x_{2})e^{-i\xi
x_{1}}dx_{1}.$ (251)

To construct the resolvent kernel we the need kernels for inverses of the
operators

$L_{\xi}+i\delta=\partial^{2}_{x_{2}}+k_{1}^{2}+q(x_{2})-\xi^{2}+i\delta,\,\xi\in\mathbb{R},$

with domain $H^{2}(\mathbb{R}).$ These kernels are constructed from the _basic
solutions_ to

$\partial^{2}_{x_{2}}\widetilde{u}_{\pm}(\xi,\delta;x_{2})+(k_{1}^{2}+q(x_{2})-\xi^{2}+i\delta)\widetilde{u}_{\pm}(\xi,\delta;x_{2})=0,$
(252)

which satisfy

$\widetilde{u}_{\pm}(\xi,\delta;x_{2})=e^{\pm
ix_{2}\sqrt{k^{2}_{1}-\xi^{2}+i\delta}}\text{ for }\pm x_{2}>d.$ (253)

The $\sqrt{z}$ is defined on $\mathbb{C}\setminus(-\infty,0],$ to be positive
on $(0,\infty);$ for $\delta,\xi\in\mathbb{R},$

$\operatorname{sgn}\operatorname{Im}\sqrt{k^{2}-\xi^{2}+i\delta}=\operatorname{sgn}\delta,$

and therefore, for $\delta>0,$ and $\xi^{2}>k_{1}^{2},$

$\sqrt{k_{1}^{2}-\xi^{2}+i\delta}=i\sqrt{\xi^{2}-k_{1}^{2}-i\delta}.$ (254)

Hence, for $\delta>0$ $\widetilde{u}_{+}(\xi,\delta;x_{2})$ decays
exponentially as $x_{2}\to\infty,$ whereas
$\widetilde{u}_{-}(\xi,\delta;x_{2})$ decays exponentially as
$x_{2}\to-\infty.$ Taking $\delta\to 0^{+}$ we get the basic solutions that
satisfy

$\widetilde{u}_{\pm}(\xi,0^{+};x_{2})=\begin{cases}&e^{\pm