Local Splitting into Incoming and Outgoing Waves and the Integral Representation of Regular Scalar Waves

Abstract

The problem of the integral representation over a bounded surface of a regular field satisfying the Helmholtz equation in all space is investigated. This problem is equivalent to local splitting into an incoming field and an outgoing field. This splitting is not possible in general.

1. Introduction

Let us consider the following harmonic scattering problem, where a time dependence of $e^{-i\omega t}$ is implied, with $\omega =kc$, where $k=2\pi /\lambda$ is the wavenumber and $\lambda$ is the wavelength in a vacuum. Below, the results are given for a space dimension of 3 but they can be easily adapted for any other dimension $>1$.

Let $\mathrm{\Omega }$ be a bounded domain of ${\mathbb{R}}^3$. Throughout this work, it is assumed that $k^2$ does not belong to the discrete spectrum $\left\{k_j^2\right\}$ of the operator $-\mathrm{\Delta }$ defined over $\mathrm{\Omega }$ with homogeneous Dirichlet boundary condition on $\partial \mathrm{\Omega }$. This hypotheses is necessary to ensure the uniqueness of the scattering problem. Let us stress that since the spectrum is discrete, $k^2$ belonging to this spectrum is a non-generic situation, i.e., it can be removed by an infinitesimal variation of the parameters of the system.

Let $\kappa ,\rho \in L^{\infty }(\mathrm{\Omega };\mathbb{R})$ be such that $\kappa -1$ and $\rho -1$ have compact support in $\mathrm{\Omega }$. Let $u^{\mathrm{reg}}$ be a regular field satisfying $\mathrm{\Delta }{u}^{\mathrm{reg}}+k^2{u}^{\mathrm{reg}}=0$ in ${\mathbb{R}}^n$. This field is the “incident field”. The total field u satisfies $\mathrm{div}(\rho \nabla u)+k^2\kappa u=0$. The scattered field is defined as $u^s=u-u^{\mathrm{reg}}$ and it satisfies a radiation condition at infinity. This radiation condition expresses the fact that the energy of the scattered field flows away from the obstacle, towards infinity. The field is said to be outgoing. In scattering theory, an important object is the scattering matrix, which is defined in the following way. Let us assume that the incident field is of Herglotz type [1], [2] (p. 74), [3] (p. 6); that is, it can be written $u^{\mathrm{reg}}\left(\mathbf{x}\right)={\int }_{S^2}{e}^{ik\mathbf{x}·\widehat{y}}g\left(\widehat{y}\right)d\widehat{y}$ with a smooth enough kernel g. According to that hypothesis, at a large distance from the obstacle, the total field exhibits the following asymptotic behavior [1,3]:

$$u\left(\mathbf{x}\right)\sim {\displaystyle \frac{e^{ikx}}{kx}}{w}^{+}\left(\widehat{x}\right)+{\displaystyle \frac{e^{-ikx}}{kx}}{w}^{-}\left(\widehat{x}\right),$$

(1)

i.e., it is the sum of an outgoing field (the first term) and an incoming field (the second term). Indeed, the first term corresponds to a spherical wave with a Poynting vector directed towards infinity and the second one corresponds to a spherical wave with a Poynting vector directed towards the origin. The scattering matrix is then defined as the operator S relating $w^{-}$ to $w^{+}$: $w^{+}=S\left[w^{-}\right]$. The scattering matrix is thus operating the field at infinity. In view of applications in multiple scattering theory [4,5,6], it would be interesting to know whether the scattering matrix can be defined in the vicinity of an obstacle. Indeed, let us consider a situation involving scattering by two obstacles, as depicted in Figure 1. In order to compute the field diffracted by these obstacles according to the methodology of multiple scattering, one decomposes the total field around each obstacle ${\mathrm{\Omega }}_j$ into an outgoing field $u_j^{+}$ and an incoming field $u_j^{-}$. The incoming field is the sum of the true incident field and the field scattered by the other obstacle. Writing $u_1^{-}=u_1^{\mathrm{reg}}+T_{2\to 1}{u}_2^{+}$ and $u_2^{-}=u_2^{\mathrm{reg}}+T_{1\to 2}{u}_1^{+}$, we obtain the following (symbolic) relations:

$$u_1^{+}=S_1\left(u_1^{\mathrm{reg}}+T_{2\to 1}{u}_2^{+}\right),u_2^{+}=S_2\left(u_2^{\mathrm{reg}}+T_{1\to 2}{u}_1^{+}\right),$$

(2)

where $S_1$ and $S_2$ are the scattering matrices of the obstacles. This derivation is based on the hypotheses that there is a local decomposition of the total field into an outgoing field and an incoming field and that the scattering matrices are valid locally. This is true in the situation depicted in Figure 1a, where the obstacles are contained in two non-overlapping spheres, because one can use a modal expansion of the fields into spherical Hankel functions and apply the scattering matrices to the coefficients of these series. However, in the situation depicted in Figure 1b the situation is quite different, since the modal expansions are not valid. This raises the problem of applying multiple scattering analysis to this situation. A pioneering result in that direction was obtained in [7].

We have proven in [4] that, given any smooth bounded surface $\mathrm{\Gamma }$ enclosing $\mathrm{\Omega }$, there is a density ${\sigma }^{+}\in H^{1/2}\left(\mathrm{\Gamma }\right)$ such that $u^s\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}{\sigma }^{+}\left({\mathbf{x}}^{\prime }\right)g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })d{\mathbf{x}}^{\prime }$ (see also [7]), where $g^{+}$ is the fundamental solution to the Helmholtz equation with an outgoing radiation condition (cf. Equation (3)). Note also that several works have investigated the inverse problem of the existence of a scattered field given the far-field pattern [2,8]. In order to investigate this further, in this work, we address the question of the possibility of representing the regular (incident) field by an integral over $\mathrm{\Gamma }$. Indeed, if it were possible to find a density ${\sigma }^{\mathrm{reg}}$ allowing us to represent the incident field as an integral over a surface enclosing the obstacle, then we could construct a local scattering matrix relating ${\sigma }^{\mathrm{reg}}$ to ${\sigma }^{+}$. Assuming that $\mathrm{\Gamma }$ is the boundary of a domain $\mathrm{\Omega }$, we have proven in [4] that such a representation exists inside $\mathrm{\Omega }$. More precisely, there exists ${\sigma }^{\mathrm{reg}}\in H^{1/2}\left(\mathrm{\Gamma }\right)$ such that $u^{\mathrm{reg}}\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}{\sigma }^{\mathrm{reg}}\left({\mathbf{x}}^{\prime }\right)g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })d{\mathbf{x}}^{\prime }$, for $\mathbf{x}\in \mathrm{\Omega }$. However, the existence of a representation of the field outside $\mathrm{\Omega }$ is not obvious at all. In fact, the main result of the present work is that such a representation is not possible in general.

2. Setting for the Scattering Problem

Let us specify a few notations. The unit sphere of ${\mathbb{R}}^3$ is denoted $S^2$. For $\mathbf{x}\in {\mathbb{R}}^3\setminus \left\{0\right\}$, we denote $x=\left|\mathbf{x}\right|$ the norm of $\mathbf{x}$, and $\widehat{x}=\mathbf{x}/x$. The fundamental solution $g^{+}$ of the Helmholtz equation with outgoing wave condition is

$$g^{+}\left(\mathbf{x}\right)=-{\displaystyle \frac{1}{4\pi x}}{e}^{ikx}.$$

(3)

The Green function with the incoming wave condition is denoted as $g^{-}\left(\mathbf{x}\right)$. Explicitly, $g^{-}\left(\mathbf{x}\right)=-{\displaystyle \frac{1}{4\pi x}}{e}^{-ikx}$. The following expansion of the Green functions over the spherical harmonics holds:

$$g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })=-{\displaystyle \frac{e^{ik|\mathbf{x}-{\mathbf{x}}^{\prime }|}}{4\pi |\mathbf{x}-{\mathbf{x}}^{\prime }|}}=-ik\sum _{n,m}{j}_n\left(k{x}_{<}\right)h_n^{\left(1\right)}\left(k{x}_{>}\right)\overline{Y_n^m}\left({\widehat{x}}^{\prime }\right)Y_n^m\left(\widehat{x}\right),$$

(4)

where $x_{<}=min(x,x^{\prime })$ and $x_{>}=max(x,x^{\prime })$.

Let $\mathrm{\Omega }$ be a bounded domain of ${\mathbb{R}}^3$ with the boundary $\partial \mathrm{\Omega }=\mathrm{\Gamma }$. For $u\in H^1\left(\mathrm{\Omega }\right)$ (the Sobolev space of function of $L^2\left(\mathrm{\Omega }\right)$ with a gradient in $L^2\left(\mathrm{\Omega }\right)$; see [9] (chap. 2) for more results on Sobolev spaces), the interior traces [9] (chap. 2) of u and its normal derivative on $\mathrm{\Gamma }$ are denoted by

$${\gamma }^{-}\left(u\right)={\left.u\right|}_{\mathrm{\Gamma }},{\gamma }^{-}\left({\partial }_{n}u\right)={\left.{\partial }_{n}u\right|}_{\mathrm{\Gamma }}.$$

(5)

For fields belonging to $H_{\mathrm{loc}}^1(\mathrm{\Omega }\setminus {\mathbb{R}}^3)$, we denote the exterior traces by:

$${\gamma }^{+}\left(u\right)={\left.u\right|}_{\mathrm{\Gamma }},{\gamma }^{+}\left({\partial }_{n}u\right)={\left.{\partial }_{n}u\right|}_{\mathrm{\Gamma }}.$$

(6)

Given a field $u\in H_{\mathrm{loc}}^1\left({\mathbb{R}}^3\right)$, we denote ${\left[u\right]}_{\mathrm{\Gamma }}$ the jump of u across $\mathrm{\Gamma }$, i.e.,

$${\left[u\right]}_{\mathrm{\Gamma }}={\gamma }^{+}\left(u\right)-{\gamma }^{-}\left(u\right)\mathrm{and}{\left[{\partial }_{n}u\right]}_{\mathrm{\Gamma }}={\gamma }^{+}\left({\partial }_{n}u\right)-{\gamma }^{-}\left({\partial }_{n}u\right).$$

(7)

3. Warm Up: The Simple Case of a Circular Cylinder

For the sake of clarity, let us present the main problems in the simple situation of a 2D problem with a cylinder $\mathrm{\Omega }$ bounded by a circle of radius 1 as the curve $\mathrm{\Gamma }$ [6,10]. That is, we consider the scattering of a regular field by a circular cylinder. The regular field $u^{\mathrm{reg}}$ satisfies a Helmholtz equation $\mathrm{\Delta }{u}^{\mathrm{reg}}+k^2{u}^{\mathrm{reg}}=0$ in ${\mathbb{R}}^2$. The field can be expanded in Fourier series in the following form:

$$u^{\mathrm{reg}}(r,\theta )=\sum _n{i}_n{J}_n\left(kr\right)e^{in\theta },$$

(8)

where $J_n$ is a Bessel function [11]. The scattering field $u^s$ can be expanded in series of Hankel functions in the following form:

$$u^s(r,\theta )=\sum _n{s}_n{H}_n^{\left(1\right)}\left(kr\right)e^{in\theta },r>1.$$

(9)

Given these fields, we can define the scattering amplitude T as the operator relating $u^{reg}$ to $u^s$: $u^s=T\left[u^{reg}\right]$. It has an explicit representation as an infinite matrix relating the sequence ${\left(i_n\right)}_n$ to the sequence ${\left(s_n\right)}_n$: ${\left(s_n\right)}_n=T\left[{\left(i_n\right)}_n\right]$.

As far as the local splitting into an outgoing field and an incoming field is involved, one could think of using the decomposition into Hankel functions $J_n={\displaystyle \frac{1}{2}}\left(H_n^{\left(1\right)}+H_n^{\left(2\right)}\right)$ and write, for $r>0$,

$$u^{\mathrm{reg}}(r,\theta )=\sum _n\left({\displaystyle \frac{i_n}{2}}{H}_n^{\left(1\right)}\left(kr\right)+{\displaystyle \frac{i_n}{2}}{H}_n^{\left(2\right)}\left(kr\right)\right)e^{in\theta }.$$

(10)

For a fixed $r>0$, this series cannot be split into two converging Fourier series, in general, due to the asymptotic behavior: $H_n\left(kr\right)\simeq {\left({\displaystyle \frac{n}{kr}}\right)}^n$ as $n\to \infty$. Indeed, $\left(i_n\right)$ is not a rapidly decreasing sequence in general. For example, for a plane wave, it has constant modulus, since it holds that

$$e^{ikr sin\left(\theta \right)}=\sum _n{J}_n\left(kr\right)e^{in\theta }.$$

As a matter of fact, due to the exponential growth of the Hankel function, the series ${\sum }_n{\displaystyle \frac{i_n}{2}}{H}_n^{\left(1\right)}\left(kr\right)e^{in\theta }$ (for a fixed r) does not even define a tempered distribution. A possible way out is to consider the asymptotic behavior of the series as $r\to +\infty$ and to consider it in terms of the Schwartz distribution meaning. By this, we mean the following. Consider the asymptotic behavior of the Hankel functions [11]:

$$H_n^{\left(1\right)}\left(kr\right){\sim }_{r\to \infty }\sqrt{{\displaystyle \frac{2}{\pi }}}{e}^{-i\pi /4}{\displaystyle \frac{e^{ikr}}{\sqrt{kr}}}{(-i)}^n.$$

(11)

Then, at least formally, we can write a decomposition into outgoing and incoming fields:

$$\sum _n{i}_n{J}_n\left(kr\right)e^{in\theta }\simeq \sqrt{{\displaystyle \frac{2}{\pi }}}{e}^{-i\pi /4}\left({\displaystyle \frac{e^{ikr}}{\sqrt{kr}}}\sum _n{(-i)}^n{\displaystyle \frac{i_n}{2}}{e}^{in\theta }+{\displaystyle \frac{e^{-ikr}}{\sqrt{kr}}}\sum _n{(-i)}^n{\displaystyle \frac{i_n}{2}}{e}^{in\theta }\right).$$

Obviously, the series ${\displaystyle \frac{1}{2}}{\sum }_n{(-i)}^n{i}_n{e}^{in\theta }$ does not converge pointwise in general. However, provided the coefficients do not grow too rapidly, it has a meaning in the sense of Schwartz distributions. Indeed, given a function $\phi \left(\theta \right)={\sum }_n{\phi }_n{e}^{in\theta }\in \mathcal{D}\left(S^1\right)$ (the Schwartz space of $C^{\infty }$ functions with compact support in $S^1$) and $\mathrm{\Psi }\in {\mathcal{D}}^{\prime }\left(S^1\right)$ (the dual space of Schwartz distributions), the duality $(\mathcal{D},{\mathcal{D}}^{\prime })$ can be defined by $\langle \mathrm{\Psi },\phi \rangle ={\sum }_n{i}_n{\phi }_n$, and the coefficients $\left({\phi }_n\right)$ are such that ${\phi }_n=O(1/n^p),\forall p>0$. As said before, we can consider the space of sequences ${\mathbb{C}}^{\mathbb{Z}}$ and define the scattering amplitude T as a linear operator relating the sequence of coefficients of the incident field to the sequence of coefficients of the scattered field: ${\left(s_n\right)}_n=T\left[{\left(i_n\right)}_n\right]$. Then, the scattering matrix relates the “incoming” field to the “outgoing” field as follows. The incoming coefficients are $1/2{\left(i_n\right)}_n$ and the outgoing coefficients are ${\left(s_n\right)}_n+1/2{\left(i_n\right)}_n$. We determine directly that ${\left(s_n\right)}_n+1/2{\left(i_n\right)}_n=1/2{\left(i_n\right)}_n+T\left[{\left(i_n\right)}_n\right]=S[1/2\left(i_n\right)]$. Thus: $S=1+2T$. This gives a meaning to the splitting of the field and a definition of the scattering matrix in terms of bilateral complex sequences. However, the splitting does not hold in terms of fields, since it would require the series ${\sum }_n{i}_n{H}_n\left(kr\right)e^{in\theta }$ to be convergent and, thus, the coefficients ${\left(i_n\right)}_n$ to tend rapidly to 0. Therefore, the scattering matrix is only defined for the field at infinity. As for the integral representation, the question is whether it is possible to find $\sigma$ such that

$$u^{\mathrm{reg}}\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)\Re \left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\right)d{\mathbf{x}}^{\prime },\forall \mathbf{x}\in {\mathbb{R}}^3.$$

Passing to polar coordinates, we are led to considering the following integral equation:

$$K\left(\sigma \right)={\int }_0^{2\pi }\sigma \left({\theta }^{\prime }\right)J_0\left(k\right|e^{i\theta }-e^{i{\theta }^{\prime }}\left|\right)d{\theta }^{\prime }=u\left(\theta \right).$$

Lemma 1.

The operator K does not have a bounded inverse, since 0 belongs to its essential spectrum.

Proof. 

To see this, let us reformulate the equation in terms of Fourier coefficients by using the expansion

$$J_0\left(k\right|e^{i\theta }-e^{i{\theta }^{\prime }}\left|\right)=\sum _n{J}_n^2\left(k\right)e^{in(\theta -{\theta }^{\prime })}.$$

Then, upon inserting this expansion into the integral equation, we obtain

$$\begin{array}{c}\hfill K\left(\sigma \right)=\int \sum _n{J}_n^2\left(k\right)e^{in(\theta -{\theta }^{\prime })}\sigma \left({\theta }^{\prime }\right)d{\theta }^{\prime }=\\ \hfill \sum _n{J}_n^2\left(k\right)e^{in\theta }\left({\int }_0^{2\pi }\sigma \left({\theta }^{\prime }\right)e^{-in{\theta }^{\prime }}d{\theta }^{\prime }\right)=\sum _n{\sigma }^n{J}_n^2\left(k\right)e^{in\theta }.\end{array}$$

Consider now the sequence ${\sigma }_n\left(\theta \right)=e^{in\theta }$, then $\parallel {\sigma }_n{\parallel }_2=1$. Since ${\sigma }_n^m={\delta }_n^m$, we see that $\parallel K\left({\sigma }_n\right){\parallel }_2=J_n^2\left(k\right)\to 0$. Moreover, given $\phi \in L^2\left(S^1\right)$, the scalar product ${({\sigma }_n,\phi )}_{L^2\left(S^1\right)}$ tends to 0, by the Riemann–Lebesgue lemma. We thus see that $\left({\sigma }_n\right)$ is a Weyl sequence for K and 0 [12]. □

This lemma does not prevent the solving of the integral equation $K\left(\sigma \right)=u\left(\theta \right)$, which is a clue regarding the possibility of representing the field as an integral over $\mathrm{\Gamma }$. However, it is an ill-conditioned problem. Expanding the field u in the Fourier series, we obtain $u\left(\theta \right)={\sum }_n{u}_n{e}^{in\theta }$. Thus solving the integral equation gives

$$\sigma \left(\theta \right)=\sum _n{\displaystyle \frac{u_n}{J_n^2\left(k\right)}}{e}^{in\theta }.$$

Therefore, the $L^2$ convergence of the series requires

$$\sum _n{\left|{\displaystyle \frac{u_n}{J_n^2\left(k\right)}}\right|}^2<+\infty .$$

This is the so-called Picard condition [13] (th. 15.18, p. 311). Indeed, denoting $e_n\left(\theta \right)=e^{in\theta }$, we remark that $K\left(e_n\right)=J_n^2\left(k\right)e_n$. The integral operator K is self-adjoint with eigenvalues ${\mu }_n=J_n^2\left(k\right)$ and eigenvectors $e_n$. The eigenvalues are thus also the singular values of K.

Consider, as a counter-example, the possibility of an integral representation for $u(r,\theta )=e^{ikr sin\theta }$. Since $e^{ikr sin\theta }={\sum }_n{J}_n\left(kr\right)e^{in\theta }$, we find that $\sigma \left(\theta \right)={\sum }_n{\displaystyle \frac{1}{J_n\left(k\right)}}{e}^{in\theta }$, which diverges exponentially. This shows that a simple plane wave is not amenable to such an integral representation. More generally, consider the series expansion (8) and plug it into the integral equation, this gives

$$\sigma \left(\theta \right)=\sum _n{\displaystyle \frac{i_n}{J_n\left(k\right)}}{e}^{in\theta }.$$

(12)

Since the Bessel function behaves as $e^{-n log n}$, this shows that the coefficients $\left(i_n\right)$ should be such that $i_n{e}^{n log n}$ is of polynomial growth for the series to make sense as a tempered distribution.

4. Local Splitting and Integral Representation

4.1. The Far-Field Operator

The scattered field can be represented in the following integral form [4]

$$u^s\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)g^{+}\left(k\right|\mathbf{x}-{\mathbf{x}}^{\prime }\left|\right)ds\left({\mathbf{x}}^{\prime }\right).$$

Asymptotically, this reads as

$$u^s\left(\mathbf{x}\right)\sim {\displaystyle \frac{-k}{4\pi }}{\displaystyle \frac{e^{ikx}}{kx}}{\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)e^{-ik\widehat{x}·{\mathbf{x}}^{\prime }}ds\left({\mathbf{x}}^{\prime }\right).$$

Below, we denote $J_{\mathrm{\Gamma }}$ as the “far-field operator” [13] (p. 24) defined by

$$H^{-1/2}\left(\mathrm{\Gamma }\right)\ni \sigma \to \left(\widehat{x}\in S^2\to J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)e^{-ik\widehat{x}·{\mathbf{x}}^{\prime }}ds\left({\mathbf{x}}^{\prime }\right)\in L^2\left(S^2\right).\right)$$

We derive a few properties of this operator that will be useful later on.

Proposition 1.

The adjoint $J_{\mathrm{\Gamma }}^{+}$ of $J_{\mathrm{\Gamma }}$ is given by

$$J_{\mathrm{\Gamma }}^{+}\left[v\right]\left({\mathbf{x}}^{\prime }\right)={\int }_{S^2}{e}^{ik\widehat{x}·{\mathbf{x}}^{\prime }}v\left(\widehat{x}\right)d\widehat{x}.$$

(13)

The adjoint is therefore of the form $J_{\mathrm{\Gamma }}^{+}\left[v\right]\left({\mathbf{x}}^{\prime }\right)=J_{S^2}\left[v\right](-{\mathbf{x}}^{\prime })$.

Proof. 

The adjoint $J_{\mathrm{\Gamma }}^{+}$ of $J_{\mathrm{\Gamma }}$ is defined by:

$${(J_{\mathrm{\Gamma }}\left[\sigma \right],v)}_{S^2}={\langle \sigma ,J_{\mathrm{\Gamma }}^{+}\left[v\right]\rangle }_{\mathrm{\Gamma }},$$

where ${\langle ·,·\rangle }_{\mathrm{\Gamma }}$ denotes the duality $(H^{-1/2}\left(\mathrm{\Gamma }\right),H^{1/2}\left(\mathrm{\Gamma }\right))$. This gives

$${(J_{\mathrm{\Gamma }}\left[\sigma \right],v)}_{S^2}={\int }_{S^2}d\widehat{x}\left({\int }_{\mathrm{\Gamma }}d{\mathbf{x}}^{\prime }\overline{\sigma \left({\mathbf{x}}^{\prime }\right)}{e}^{ik\widehat{x}·{\mathbf{x}}^{\prime }}\right)v\left(\widehat{x}\right)={\int }_{\mathrm{\Gamma }}d{\mathbf{x}}^{\prime }\overline{\sigma \left({\mathbf{x}}^{\prime }\right)}{\int }_{S^2}d\widehat{x}{e}^{ik\widehat{x}·{\mathbf{x}}^{\prime }}v\left(\widehat{x}\right).$$

The proposition follows. □

Proposition 2.

The self-adjoint operator $J_{\mathrm{\Gamma }}^{+}{J}_{\mathrm{\Gamma }}$ has the following expression:

$$J_{\mathrm{\Gamma }}^{+}{J}_{\mathrm{\Gamma }}\left[\sigma \right]\left(\mathbf{x}\right)={\left(4\pi \right)}^{3/2}{\int }_{\mathrm{\Gamma }}d{\mathbf{x}}^{\prime }{j}_0\left(k\right|\mathbf{x}-{\mathbf{x}}^{\prime }\left|\right)\sigma \left({\mathbf{x}}^{\prime }\right)={\left(4\pi \right)}^{5/2}\sum _{nm}{\sigma }_{nm}{j}_n\left(kx\right)Y_n^m\left(\widehat{x}\right)$$

where

$${\sigma }_{nm}={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m(\widehat{x^{\prime }})}d{\mathbf{x}}^{\prime }.$$

Proof. 

By a direct computation, we find that

$$J_{\mathrm{\Gamma }}^{+}{J}_{\mathrm{\Gamma }}\left[\sigma \right]\left(\mathbf{x}\right)={\int }_{S^2}d\widehat{x}{e}^{ik\widehat{x}·\mathbf{x}}{\int }_{\mathrm{\Gamma }}d{\mathbf{x}}^{\prime }\sigma \left({\mathbf{x}}^{\prime }\right)e^{-ik\widehat{x}·{\mathbf{x}}^{\prime }}={\int }_{\mathrm{\Gamma }}d{\mathbf{x}}^{\prime }\sigma \left({\mathbf{x}}^{\prime }\right){\int }_{S^2}d\widehat{x}{e}^{ik\widehat{x}·(\mathbf{x}-{\mathbf{x}}^{\prime })}.$$

Inserting the following expansions:

$$e^{ik\widehat{x}·{\mathbf{x}}^{\prime }}=4\pi \sum _{n,m}{i}^n{j}_n\left(k{x}^{\prime }\right)\overline{Y_n^m\left(\widehat{x}\right)}{Y}_n^m\left(\widehat{x^{\prime }}\right),$$

(14)

$$j_0\left(k\right|\mathbf{x}-{\mathbf{x}}^{\prime }\left|\right)=4\pi \sum _{n,m}{j}_n\left(k{x}^{\prime }\right)j_n\left(kx\right)Y_n^m\left({\widehat{x}}^{\prime }\right)\overline{Y_n^m}\left(\widehat{x}\right),$$

the proposition follows. □

Two natural questions arise about the operator $J_{\mathrm{\Gamma }}$: is it injective, and how can we characterize its image? The first question is answered in the following proposition.

Proposition 3.

Assume that $k^2$ is not an eigenvalue of $-\mathrm{\Delta }$ inside Ω with homogeneous Dirichlet boundary conditions on Γ. If $J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)=0$ a.e., then $\sigma =0$.

Proof. 

Consider the function defined by $v\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}{g}^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\sigma \left({\mathbf{x}}^{\prime }\right)ds\left({\mathbf{x}}^{\prime }\right)$. Then, from potential theory [13], v is the unique function satisfying

$$\mathrm{\Delta }v+k^{2}v=0\mathrm{in}\mathrm{\Omega }\cup ({\mathbb{R}}^p\setminus \overline{\mathrm{\Omega }}),$$

with an outgoing wave condition and the boundary conditions:

$${\left[v\right]}_{\mathrm{\Gamma }}=0,{\left[{\partial }_{n}v\right]}_{\mathrm{\Gamma }}=\sigma .$$

The unicity follows from the following argument: assume there exists a function $u\ne v$ satisfying the same problem. Then, $w=u-v$ satisfies $\mathrm{\Delta }w+k^{2}w=0$ in ${\mathbb{R}}^p$ and a radiation condition. Therefore, it is null thanks to Rellich lemma. Outside $B_e$, the field can be expanded in spherical harmonics in the form

$$v\left(\mathbf{x}\right)=\sum _{nm}{v}_{nm}{h}_n^{\left(1\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right).$$

Explicitly, the coefficients $\left(v_{nm}\right)$ are obtained by use of formula (14)

$$v\left(\mathbf{x}\right)=-ik\sum _{nm}\left({\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m}\left({\widehat{x}}^{\prime }\right)d{\mathbf{x}}^{\prime }\right)h_n^{\left(1\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right).$$

Using the asymptotic forms of the Hankel functions (11), we obtain the following asymptotic behavior

$$v\left(\mathbf{x}\right)\sim {\displaystyle \frac{e^{ikx}}{kx}}w\left(\widehat{x}\right),$$

with:

$$w\left(\widehat{x}\right)=\sum _{nm}{v}_{nm}{e}^{-i(n+1)\pi /2}{Y}_n^m\left(\widehat{x}\right).$$

Besides, using the asymptotic form of the Green function

$$g^{\pm }(\mathbf{x}-{\mathbf{x}}^{\prime }){\sim }_{x\to \infty }-{\displaystyle \frac{e^{\pm ikx}}{4\pi x}}{e}^{\mp kx·x^{\prime }},$$

we obtain

$$v\left(x\right){\sim }_{x\to \infty }-{\displaystyle \frac{k}{4\pi }}{\displaystyle \frac{e^{ikx}}{4\pi x}}{J}_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right),$$

and thus $J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)=-{\displaystyle \frac{4\pi }{k}}w\left(\widehat{x}\right)$. Consequently, the nullity of $J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)=0$ implies that of w and thus that of v in ${\mathbb{R}}^3\setminus \mathrm{\Omega }$. Therefore, ${\gamma }^{-}\left(v\right)=0$ and ${\gamma }^{-}\left({\partial }_{n}v\right)$ is non-zero if v is a solution to the Dirichlet problem; therefore, $v=0$ in $\mathrm{\Omega }$ by the hypothesis on $k^2$, and thus $\sigma =0$. □

Corollary 1.

It holds that

$${\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m\left(\widehat{x^{\prime }}\right)}d{\mathbf{x}}^{\prime }=0\mathit{if}\mathit{and}\mathit{only}\mathit{if}\sigma =0.$$

Therefore, ${\left(\gamma \left(j_n\left(k{x}^{\prime }\right)Y_n^m\left(\widehat{x^{\prime }}\right)\right)\right)}_{(n,m)}$ is dense in $L^2\left(\mathrm{\Gamma }\right)$.

Proof. 

Note that

$$J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)=4\pi \sum _{n,m}{(-i)}^n{\sigma }_{nm}{Y}_n^m\left(\widehat{x}\right),$$

where

$${\sigma }_{nm}={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m\left(\widehat{x^{\prime }}\right)}d{\mathbf{x}}^{\prime }.$$

Let us proof the “only if” direction: assume ${\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m\left(\widehat{x^{\prime }}\right)}d{\mathbf{x}}^{\prime },\forall (n,m)$. Then $J\left[\sigma \right]\left(\widehat{x}\right)=0$ and thus $\sigma =0$ by the proposition above. The converse is obvious. □

Finally, we state the following:

Proposition 4.

Provided that $k^2$ is not an eigenvalue of $-\mathrm{\Delta }$ in the ball of radius 1, $J_{\mathrm{\Gamma }}$ has a dense range in $L^2\left(S^2\right)$.

Proof. 

Take $v\in L^2\left(S^2\right)$ in the range of $J_{\mathrm{\Gamma }}$; therefore, there is $\sigma \in H^{-1/2}\left(\mathrm{\Gamma }\right)$, such that $J_{\mathrm{\Gamma }}\left[\sigma \right]=v$. To show that the range is dense, we show that if ${\int }_{S^2}\overline{v\left(\widehat{x}\right)}\varphi \left(\widehat{x}\right)d\widehat{x}=0,\forall v\in \mathrm{Ran}\left(J_{\mathrm{\Gamma }}\right)$ then $\varphi =0$. Assuming that it holds, then we have

$$0={(J_{\mathrm{\Gamma }}\left[\sigma \right],\varphi )}_{S^2}={\langle \sigma ,J_{\mathrm{\Gamma }}^{*}\left[\varphi \right]\rangle }_{\mathrm{\Gamma }}.$$

Hence, explicitly

$${\int }_{S^2}d\widehat{x}{e}^{ik\widehat{x}·{\mathbf{x}}^{\prime }}\varphi \left(\widehat{x}\right)=0,\mathrm{for}\mathrm{a}.\mathrm{e}.{\mathbf{x}}^{\prime }\in \mathrm{\Gamma }.$$

Therefore, we are reduced to a special case of Proposition 3. We conclude that $\varphi \left(\mathbf{x}\right)=0\mathrm{a}.\mathrm{e}.$, provided $k^2$ is not an eigenvalue of $-\mathrm{\Delta }$ in the ball of radius 1. □

On this topic, see also [13] (p. 76).

We can now remark the following. When solving the equation $J_{\mathrm{\Gamma }}\left[\sigma \right]=\varphi$, it holds that $J_{\mathrm{\Gamma }}\left[\sigma \right]\left(\widehat{x}\right)=4\pi {\sum }_{n,m}{(-i)}^n{\sigma }_{nm}{Y}_n^m\left(\widehat{x}\right)$, where $\varphi ={\sum }_{nm}{\varphi }_{nm}{Y}_n^m$. Hence we conclude that $|{\sigma }_{nm}|=|{\varphi }_{nm}|$, hence ${\varphi }_{nm}=O\left(j_n\right)$, since ${\sigma }_{nm}={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m(\widehat{x^{\prime }})}d{\mathbf{x}}^{\prime }$.

4.2. Integral Representation of a Regular Field

At this stage, we address the possibility of representing a regular field as an integral over $\mathrm{\Gamma }$ that would be valid outside $\mathrm{\Omega }$. For an incident field satisfying the Helmholtz equation in all ${\mathbb{R}}^3$, we cannot impose an outgoing wave condition and the kernel of the integral representation should be regular. This leads to a natural representation in the following form:

$$u^{\mathrm{reg}}\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)\Im \left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\right)d{\mathbf{x}}^{\prime },\mathbf{x}\in {\mathbb{R}}^3.$$

As we shall see, it turns out that this representation imposes strong constraints on the field $u^{\mathrm{reg}}$. This is somewhat similar to the possibility of splitting, at infinity, a regular field defined over ${\mathbb{R}}^3$ into an incoming and an outgoing field. The possibility of this decomposition was analyzed in [1], where it was shown that this was possible provided $u^{\mathrm{reg}}\left(\mathbf{x}\right)=\int A\left(\mathbf{k}\right)e^{i\mathbf{k}·\mathbf{x}}d\mathbf{k}$ where $A\left(\mathbf{k}\right)$ is twice continuously differentiable. If the field $u^{\mathrm{reg}}\left(\mathbf{x}\right)$ admits the integral representation over $\mathrm{\Gamma }$ then it can be decomposed at infinity into an outgoing field and an incoming field, as can be inferred by noting that $\Im \left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\right)$ is the sum of two functions satisfying an outgoing wave condition and an incoming wave condition, respectively. However, the converse is not true, as the integral representation implies that the splitting is always true, i.e., for all $\mathbf{x}$, and not only at infinity. This is described in the following result.

Theorem 1.

Let $u^{\mathrm{reg}}$ satisfy $\mathrm{\Delta }{u}^{\mathrm{reg}}+k^2{u}^{\mathrm{reg}}=0,\mathbf{x}\in {\mathbb{R}}^3$. The following statements are equivalent.

1. 

The field $u^{\mathrm{reg}}$ can be represented in the form

$$u^{\mathrm{reg}}\left(\mathbf{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)\Im \left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\right)d{r}^{\prime },\mathbf{x}\in {\mathbb{R}}^3$$

(15)

where σ belongs to $H^{-1/2}\left(\mathrm{\Gamma }\right)$.

2. 

The field $u^{\mathrm{reg}}$ can be split locally into an outgoing field and an incoming field, i.e., it is the sum of two pointwise-convergent series in the form

$$u^{\mathrm{reg}}\left(\mathbf{x}\right)=\sum _{nm}{\displaystyle \frac{1}{2}}{i}_{nm}{h}_n^{\left(1\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right)+\sum _{nm}{\displaystyle \frac{1}{2}}{i}_{nm}{h}_n^{\left(2\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right).$$

(16)

3. 

At infinity, the field can be split into an outgoing field and an incoming field in the form

$$u^{\mathrm{reg}}\left(x\right)\sim {\displaystyle \frac{e^{ikx}}{kx}}{u}_{\infty }^{+}\left(\widehat{x}\right)-{\displaystyle \frac{e^{-ikx}}{kx}}{u}_{\infty }^{+}(-\widehat{x}),$$

(17)

where $u_{\infty }^{+}\left(\widehat{x}\right)$ is a function in $L^2\left(S^2\right)$ and such that $J\left[\sigma \right]\left(\widehat{x}\right)=u_{\infty }^{+}\left(\widehat{x}\right)$, and therefore it satisfies a Picard condition.

Proof. 

Assume the integral form. Writing that

$$\Im \left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })\right)=1/2\left(g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })+g^{-}(\mathbf{x}-{\mathbf{x}}^{\prime })\right),$$

it is obtained that

$$u^{\mathrm{reg}}\left(\mathbf{x}\right)=1/2{\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)g^{+}(\mathbf{x}-{\mathbf{x}}^{\prime })d{\mathbf{x}}^{\prime }+1/2{\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)g^{-}(\mathbf{x}-{\mathbf{x}}^{\prime })d{\mathbf{x}}^{\prime }.$$

Then, upon expanding $g^{\pm }(\mathbf{x}-{\mathbf{x}}^{\prime })$ in series using (4), it is obtained that

$$\begin{array}{c}\hfill u^{\mathrm{reg}}\left(\mathbf{x}\right)=k\sum _{n,m}{\displaystyle \frac{1}{2}}\left({\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m}\left({\widehat{x}}^{\prime }\right)\right)h_n^{\left(1\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right)+\\ \hfill k\sum _{n,m}{\displaystyle \frac{1}{2}}\left({\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)j_n\left(k{x}^{\prime }\right)\overline{Y_n^m}\left({\widehat{x}}^{\prime }\right)\right)h_n^{\left(2\right)}\left(kx\right)Y_n^m\left(\widehat{x}\right)\end{array}$$

Let us now assume the representation of $u^{\mathrm{reg}}$ as a sum of two series. Then, upon using the asymptotic forms of the spherical Hankel functions (11), we obtain the existence of two functions $u_{\infty }^{\pm }\left(\widehat{x}\right)$ defined on $S^2$ such that

$$u^{\mathrm{reg}}\left(x\right)\sim {\displaystyle \frac{e^{ikx}}{kx}}{u}_{\infty }^{+}\left(\widehat{x}\right)+{\displaystyle \frac{e^{-ikx}}{kx}}{u}_{\infty }^{-}\left(\widehat{x}\right).$$

Explicitly, these functions are given by

$$u_{\infty }^{+}\left(\widehat{x}\right)={\displaystyle \frac{1}{2}}\sum _{nm}{i}_{nm}{e}^{-i(n+1)\pi /2}{Y}_n^m\left(\widehat{x}\right),u_{\infty }^{-}\left(\widehat{x}\right)={\displaystyle \frac{1}{2}}\sum _{nm}{i}_{nm}{e}^{i(n+1)\pi /2}{Y}_n^m\left(\widehat{x}\right).$$

Since

$$e^{-i(n+1)\pi /2}={(-1)}^{n+1}{e}^{i(n+1)\pi /2}\mathrm{and}{Y}_n^m(-\widehat{x})={(-1)}^n{Y}_n^m\left(\widehat{x}\right),$$

we state that

$$\begin{array}{c}\hfill u_{\infty }^{-}(-\widehat{x})={\displaystyle \frac{1}{2}}\sum _{nm}{i}_{nm}{e}^{i(n+1)\pi /2}{Y}_n^m(-\widehat{x})={\displaystyle \frac{1}{2}}\sum _{nm}{i}_{nm}{(-1)}^n{e}^{i(n+1)\pi /2}{Y}_n^m\left(\widehat{x}\right)\\ \hfill =-{\displaystyle \frac{1}{2}}\sum _{nm}{i}_{nm}{e}^{-i(n+1)\pi /2}{Y}_n^m\left(\widehat{x}\right)=-u_{\infty }^{+}\left(\widehat{x}\right).\end{array}$$

The existence of $\sigma$ satisfying $u_{\infty }^{+}\left(\widehat{x}\right)={\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)e^{-ik\widehat{x}·{\mathbf{x}}^{\prime }}d{\mathbf{x}}^{\prime }$ follows from corollary (1) and lemma (4). The relation

$$u_{\infty }^{-}\left(\widehat{x}\right)=-{\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)e^{ik\widehat{x}·{\mathbf{x}}^{\prime }}d{\mathbf{x}}^{\prime }$$

is fulfilled thanks to

$$u_{\infty }^{-}\left(\widehat{x}\right)=-u_{\infty }^{+}(-\widehat{x})=-{\int }_{\mathrm{\Gamma }}\sigma \left({\mathbf{x}}^{\prime }\right)e^{ik\widehat{x}·{\mathbf{x}}^{\prime }}d{\mathbf{x}}^{\prime }.$$

Finally, the Picard condition is a consequence of [13] (th. 15.18, p. 311) and is in fact encoded in the convergence of the series (16). □

5. Conclusions

We have established in Theorem (1) that for a regular field, splitting into an incoming field and an outgoing field is, in general, not possible in the vicinity of a given surface. We have obtained a characterization of the fields that can be split that way in terms of equivalent properties (15)–(17). These results constitute a generalization of the results already obtained for the fields at infinity [1,13]. Since a local splitting is, in general, not possible, the definition of a local scattering matrix stumbles on this problem. However, we have shown in [4] that, locally, the total field can be split into a regular field, which is to be considered as a local incident field and a local scattered field. Therefore, despite the negative result that was obtained here, it is possible to define a local scattering amplitude as shown in [4].