Generalized Snell’s law and Maxwell equationsThanks: 2020 AMS Math Subject Classification: 78A40, 35Q61, 35D30.Thanks: C.E.G was partially supported by NSF Grant DMS-1600578, and A. S. was partially supported by the University Research Board, Grants 104107 and 104631 from the American University of Beirut. August 11, 2026

Cristian E. Gutiérrez and Ahmad Sabra
August 11, 2026 Address: Department of Mathematics
Temple University
Philadelphia, PA 19122 Email address: gutierre@temple.edu Address: Department of Mathematics
American University of Beirut
Beirut, Lebanon Email address: asabra@aub.edu.lb

Abstract.

This paper examines the Maxwell system of electrodynamics within the framework of distributions. A primary objective is to establish general boundary conditions for fields at interfaces when the charge and current densities are measures localized on the interface. From this analysis, the paper presents a derivation of the generalized Snell’s law, along with formulas for the amplitudes of the reflected and transmitted waves in terms of the incident amplitude.

Key words and phrases:

Electromagnetism, Generalized functions, Generalized Snell’s law, Metasurfaces

1. Introduction

Metasurfaces or metalenses are ultra-thin layers built with nano-materials that can steer light in unconventional ways. In beam shaping, the subject of metasurfaces is a rapidly growing area of research with diverse practical applications. Central to this field is the generalized Snell’s law of refraction and reflection, which explains how beams propagate across metasurfaces. The law was introduced in the influential works [16] and [1] for planar geometries. Its formulation involves a function defined in a small neighborhood of the metasurface, called the phase discontinuity, and is further discussed in [15]. A rigorous mathematical derivation of the law for non-planar geometries was first obtained using wave fronts in [7, Sect. 3] and later by applying the Fermat principle of least action in [8, Sect. 2]. These works also demonstrate the existence of phase discontinuities for various geometric configurations and multiple applications.

Let us recall precisely the generalized Snell’s law in vector form. Given a surface $\Gamma$ separating two media $I$ and $II$ with refractive indices $n_{1}$ and $n_{2}$ respectively, and $f$ a phase function defined on $\Gamma$ , the generalized Snell’s law of refraction states that if a wave in medium $I$ with unit direction vector ${\mathbf{k}}_{i}$ strikes $\Gamma$ at a point $P$ , then the wave is refracted into medium $II$ with unit direction vector ${\mathbf{k}}_{t}$ satisfying

(1.1)

n_{1}\,{\mathbf{k}}_{i}-n_{2}\,{\mathbf{k}}_{t}=\lambda\,{\mathbf{n}}(P)+\nabla f(P)

where ${\mathbf{n}}(P)$ is the unit normal to $\Gamma$ at $P$ , and $\lambda\in{\mathbb{R}}$ , see [8, Equation (2.4)]. On the other hand, a wave reflected back into medium $I$ has unit direction vector ${\mathbf{k}}_{r}$ and satisfies the generalized law of reflection

(1.2)

n_{1}\,{\mathbf{k}}_{i}-n_{1}\,{\mathbf{k}}_{r}=\lambda^{\prime}\,{\mathbf{n}}(P)+\nabla f(P).

A primary goal of this paper is to explore Maxwell’s equations from a distributional perspective and derive relationships between the electric and magnetic fields on either side of a boundary or interface—that is, boundary conditions—when the current and charge densities are measures concentrated on the interface. While Maxwell’s equations are well understood in the classical sense, analyzing them in the setting of generalized functions (or distributions) becomes essential when dealing with discontinuous fields across surfaces; see, for example, [10] and [9]. The main result of this part is Theorem 3.1, which we believe has independent interest, since the discontinuities of the fields are assumed to be measures concentrated on the interface, a novel assumption in this generality.

Using this analysis, the second objective of the paper is to derive the generalized laws of refraction and reflection directly from Maxwell’s equations, which is a novel approach in the metasurface setting. To achieve this, we propose representing the transmitted and reflected electric fields as nonlinear waves incorporating the phase discontinuity. This representation—detailed in equations (4.3) and (4.4)—enables us to deduce the generalized Snell’s laws from the boundary conditions established in Theorem 3.1. Because the electric fields must satisfy the Maxwell system, this imposes constraints on the phase discontinuity, and the main result of this part is Theorem 4.3. Using the boundary conditions obtained, a third objective is to calculate the amplitudes of the transmitted and reflected waves in terms of incident wave, Proposition 6.1. This is possible under the necessary and sufficient condition (6.3) between the amplitude of the incident wave and the wave vectors. This is also a new result in the paper.

To place the results in broader context, it is worth noting that metasurfaces that refract or reflect beams according to prescribed energy patterns are closely connected to Monge–Ampère type partial differential equations, as discussed in [5]. The analysis of chromatic aberration in metalenses is carried out in [8], with further insights available in [16]. For applications related to tunable metasurfaces using graphene, see [2]. Recent developments and applications in the field can also be found in [12], [11], and [14].

The paper is organized as follows. In Section 2, we recall results concerning distributions, outline the assumptions on the fields, and establish formulas needed later. These are applied in Section 3 to prove Theorem 3.1, which provides general boundary conditions.

The derivation of the generalized Snell law from the Maxwell’s system is the subject of Section 4, where we employ Theorem 3.1. In Section 5, we obtain boundary conditions for the magnetic fields, which, combined with the previously derived conditions, are used in Section 6 to derive explicit formulas for the wave amplitudes. Finally, the Appendix 7 contains a proof of the exponential Lemma 4.2.

2. Preliminaries

In this section, we revisit concepts related to distributions that are essential for analyzing the Maxwell system in this context. Specifically, we derive the formulas that are subsequently utilized to establish general boundary conditions in Theorem 3.1. These formulas are presented in Propositions 2.1 and 2.4.

Let $\Omega\subset{\mathbb{R}}^{3}$ be an open and bounded domain. A generalized function or distribution in $\Omega$ is a complex-valued continuous linear functional defined in the class of test functions $\mathcal{D}(\Omega)=C_{0}^{\infty}(\Omega)$ that are infinitely differentiable in $\Omega$ having compact support in $\Omega$ . As usual, $\mathcal{D}^{\prime}(\Omega)$ denotes the class of distributions in $\Omega$ [13]. If $g\in\mathcal{D}^{\prime}(\Omega)$ , then, as usual, $\langle g,\varphi\rangle$ denotes the value of the distribution $g$ on the test function $\varphi\in\mathcal{D}(\Omega)$ .

We say that ${\bf G}=(G_{1},G_{2},G_{3})$ is a vector valued distribution in $\Omega$ if each component $G_{i}\in\mathcal{D}^{\prime}(\Omega)$ , $1\leq i\leq 3$ . The divergence of ${\bf G}$ is the scalar distribution defined by

(2.3)

\langle\nabla\cdot{\bf G},\varphi\rangle=-\sum_{i=1}^{3}\langle G_{i},\varphi_{x_{i}}\rangle,

and the curl of ${\bf G}$ is the vector valued distribution in $\Omega$ defined by

(2.4)

\langle\nabla\times{\bf G},\varphi\rangle=\left(\langle G_{2},\varphi_{x_{3}}\rangle-\langle G_{3},\varphi_{x_{2}}\rangle\right){\bf i}-\left(\langle G_{1},\varphi_{x_{3}}\rangle-\langle G_{3},\varphi_{x_{1}}\rangle\right){\bf j}+\left(\langle G_{1},\varphi_{x_{2}}\rangle-\langle G_{2},\varphi_{x_{1}}\rangle\right){\bf k}.

Then it follows that

(2.5)

\nabla\cdot(\nabla\times{\bf G})=0,

in the sense of distributions. When the distribution ${\bf G}=(G_{1},G_{2},G_{3})$ is locally integrable in $\Omega$ we obtain from (2.3), and (2.4) that

\langle\nabla\cdot{\bf G},\varphi\rangle=-\int_{\Omega}{\bf G}\cdot\nabla\varphi\,dx,\qquad\langle\nabla\times{\bf G},\varphi\rangle=\int_{\Omega}{\bf G}\times\nabla\varphi\,dx.

We consider the following configuration. $\Omega$ is a smooth open and bounded domain in $\mathbb{R}^{3}$ and $\Gamma$ is a smooth surface that splits $\Omega$ into two disjoint open parts $\Omega_{+}$ and $\Omega_{-}$ , i.e., $\Omega=\Omega_{-}\cup\Gamma\cup\Omega_{+}$ , as follows: for every $x_{0}\in\Gamma$ there exists a ball $B(x_{0},r)\subset\Omega$ and $\gamma\in C^{1}\left(B(x_{0},r)\right)$ such that

	$\displaystyle\Omega_{-}\cap B(x_{0},r)$	$\displaystyle=\{(x_{1},x_{2},x_{3})\in B(x_{0},r):x_{3}<\gamma(x_{1},x_{2})\}$
	$\displaystyle\Omega_{+}\cap B(x_{0},r)$	$\displaystyle=\{(x_{1},x_{2},x_{3})\in B(x_{0},r):x_{3}>\gamma(x_{1},x_{2})\}.$

Figure 1.

\Omega=\Omega_{-}\cup\Gamma\cup\Omega_{+}.

We are given fields ${\bf G}_{-}$ in $\Omega_{-}$ and ${\bf G}_{+}$ in $\Omega_{+}$ satisfying the following properties

(F1)

${\bf G}_{-}\in C^{1}(\Omega_{-}),{\bf G}_{+}\in C^{1}(\Omega_{+})$ .
(F2)

The first order derivatives of ${\bf G}_{\pm}$ are in $L^{1}(\Omega_{\pm})$ , respectively.
(F3)

For every $x\in\Gamma$ , $\lim_{y\to x,y\in\Omega_{-}}{\bf G}_{-}(y)$ and $\lim_{y\to x,y\in\Omega_{+}}{\bf G}_{+}(y)$ exist and are finite.

As a consequence, each ${\mathbf{G}}_{-}$ and ${\mathbf{G}}_{+}$ can be extended continuously to $\Gamma$ by setting

{\mathbf{G}}_{-}(x)=\lim_{y\to x,y\in\Omega_{-}}{\bf G}_{-}(y);\qquad{\mathbf{G}}_{+}(x)=\lim_{y\to x,y\in\Omega_{+}}{\bf G}_{+}(y),

for each $x\in\Gamma$ . For such fields ${\bf G_{-}}$ and ${\bf G_{+}}$ , the linear functional ${\mathbf{G}}$ given by

(2.6)

\langle{\mathbf{G}},\varphi\rangle=\int_{\Omega_{-}}{\mathbf{G}}_{-}(x)\,\varphi(x)\,dx+\int_{\Omega_{+}}{\mathbf{G}}_{+}(x)\,\varphi(x)\,dx

is a well defined distribution for $\varphi\in\mathcal{D}(\Omega)$ . The jump of the fields in $\Gamma$ is defined by

[[{\bf G}(x)]]={\bf G_{+}}(x)-{\bf G_{-}}(x),\quad\text{for $x\in\Gamma$}.

We then have the following expressions for the curl and divergence of ${\mathbf{G}}$ .

Proposition 2.1.

If the field ${\mathbf{G}}$ satisfies (F1)–(F3), then for each $\varphi\in\mathcal{D}(\Omega)$ we have

(2.7)		$\displaystyle\langle\nabla\cdot{\bf G},\varphi\rangle$	$\displaystyle=\int_{\Gamma}\varphi(x)\,[[{\bf G}(x)]]\cdot{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\,\nabla\cdot{\bf G_{-}}\,dx+\int_{\Omega_{+}}\varphi\nabla\cdot{\bf G_{+}}\,dx$
(2.8)		$\displaystyle\langle\nabla\times{\bf G},\varphi\rangle$	$\displaystyle=-\int_{\Gamma}\varphi(x)\,[[{\bf G}(x)]]\times{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\nabla\times{\bf G_{-}}\,dx+\int_{\Omega_{+}}\varphi\nabla\times{\bf G_{+}}\,dx$

with ${\bf n}$ the unit normal to $\Gamma$ pointing toward $\Omega_{+}$ .

Proof.

Given $\varepsilon>0$ , define

\Omega_{-}^{\varepsilon}=\{x\in\Omega_{-}:\text{dist}(x,\Gamma)>\varepsilon\}\qquad\qquad\Omega_{+}^{\varepsilon}=\{x\in\Omega_{+}:\text{dist}(x,\Gamma)>\varepsilon\}.

For $\varphi\in\mathcal{D}(\Omega)$ , we have from (2.3) and the definition of ${\mathbf{G}}$ in (2.6) that

	$\displaystyle\langle\nabla\cdot{\bf G},\varphi\rangle$	$\displaystyle=-\int_{\Omega_{-}}{\bf G_{-}}\cdot\nabla\varphi\,dx-\int_{\Omega_{+}}{\bf G_{+}}\cdot\nabla\varphi\,dx$
		$\displaystyle=-\int_{\Omega_{-}^{\varepsilon}}{\bf G_{-}}\cdot\nabla\varphi\,dx-\int_{\Omega_{-}\setminus\Omega_{-}^{\varepsilon}}{\bf G_{-}}\cdot\nabla\varphi\,dx-\int_{\Omega_{+}^{\varepsilon}}{\bf G_{+}}\cdot\nabla\varphi\,dx-\int_{\Omega_{+}\setminus\Omega_{+}^{\varepsilon}}{\bf G_{+}}\cdot\nabla\varphi\,dx$
		$\displaystyle=-(I+II+III+IV).$

I=\int_{\partial\Omega_{-}^{\varepsilon}}\varphi{\bf G_{-}}\cdot{\bf n_{-}^{\varepsilon}}\,d\sigma-\int_{\Omega_{-}^{\varepsilon}}\varphi{\bf\nabla\cdot G_{-}}\,dx\\

with ${\bf n}^{\varepsilon}_{-}$ the outward unit normal to $\partial\Omega^{\varepsilon}_{-}$ . Since $\varphi$ has compact support in $\Omega$ , $\nabla\cdot{\bf G_{-}}\in L^{1}(\Omega_{-})$ , and ${\mathbf{G}}_{-}$ is locally bounded in $\Omega_{-}\cup\Gamma$ , it follows that

I\to\int_{\Gamma}\varphi{\bf G_{-}}\cdot{\bf n}\,d\sigma-\int_{\Omega_{-}}\varphi\nabla\cdot{\bf G_{-}}\,dx,

with ${\bf n}$ is the unit normal to $\Gamma$ toward $\Omega_{+}$ . Similarly we get

III\to\int_{\Gamma}\varphi{\bf G_{+}}\cdot({-\bf n})\,d\sigma-\int_{\Omega_{+}}\varphi\nabla\cdot{\bf G_{+}}\,dx,

with $-{\bf n}$ the unit normal to $\Gamma$ toward $\Omega_{-}$ . Hence (2.7) follows.

We next prove (2.8). Let $\varphi\in\mathcal{D}(\Omega)$ , we have from (2.4) and the definition of ${\bf G}$ in (2.6) that

	$\displaystyle\langle\nabla\times{\bf G},\varphi\rangle$	$\displaystyle=\int_{\Omega_{-}}{\bf G_{-}}\times\nabla\varphi\,dx+\int_{\Omega_{+}}{\bf G_{+}}\times\nabla\varphi\,dx$
		$\displaystyle=\int_{\Omega_{-}^{\varepsilon}}{\bf G_{-}}\times\nabla\varphi\,dx+\int_{\Omega_{-}\setminus\Omega_{-}^{\varepsilon}}{\bf G_{-}}\times\nabla\varphi\,dx+\int_{\Omega_{+}^{\varepsilon}}{\bf G_{+}}\times\nabla\varphi\,dx+\int_{\Omega_{+}\setminus\Omega_{+}^{\varepsilon}}{\bf G_{+}}\times\nabla\varphi\,dx$
		$\displaystyle=I+II+III+IV$

$II,IV\to 0$ as $\varepsilon\to 0^{+}$ because the extensions ${\mathbf{G}}_{\pm}$ are locally bounded in $\Omega_{\pm}\cup\Gamma$ and $\varphi$ is compactly supported in $\Omega$ . We write $I=(I_{1},I_{2},I_{3})$ , ${\bf G}_{-}=(G_{1}^{-},G_{2}^{-},G_{3}^{-})$ , and ${\bf n^{\varepsilon}_{-}}=(n_{1}^{\varepsilon},n_{2}^{\varepsilon},n_{3}^{\varepsilon})$ the outward unit normal to $\partial\Omega^{\varepsilon}_{-}$ . Using the divergence theorem

	$\displaystyle I_{1}$	$\displaystyle=\int_{\Omega_{-}^{\varepsilon}}G_{2}^{-}\varphi_{x_{3}}-G_{3}^{-}\varphi_{x_{2}}\,dx$
		$\displaystyle=\left(\int_{\partial\Omega_{-}^{\varepsilon}}\varphi G_{2}^{-}n_{3}^{\varepsilon}\,d\sigma-\int_{\Omega_{-}^{\varepsilon}}(G_{2}^{-})_{x_{3}}\varphi\,dx\right)-\left(\int_{\partial\Omega_{-}^{\varepsilon}}\varphi G_{3}^{-}n_{2}^{\varepsilon}\,d\sigma-\int_{\Omega_{-}^{\varepsilon}}(G_{3}^{-})_{x_{2}}\varphi\,dx\right)$
		$\displaystyle=\int_{\partial\Omega^{\varepsilon}_{-}}\varphi(G_{2}^{-}n_{3}^{\varepsilon}-G_{3}^{-}n_{2}^{\varepsilon})\,d\sigma+\int_{\Omega^{\varepsilon}_{-}}\varphi((G_{3}^{-})_{x_{2}}-(G_{2}^{-})_{x_{3}})\,dx.$

Similarly

\displaystyle I_{2}

\displaystyle=\int_{\Omega_{-}^{\varepsilon}}-G_{1}^{-}\varphi_{x_{3}}+G_{3}^{-}\varphi_{x_{1}}\,dx=\int_{\partial\Omega^{\varepsilon}_{-}}\varphi(G_{3}^{-}n_{1}^{\varepsilon}-G_{1}^{-}n_{3}^{\varepsilon})\,d\sigma+\int_{\Omega^{\varepsilon}_{-}}\varphi((G_{1}^{-})_{x_{3}}-(G_{3}^{-})_{x_{1}})\,dx.

and

\displaystyle I_{3}=\int_{\Omega_{-}^{\varepsilon}}G_{1}^{-}\varphi_{x_{2}}-G_{2}^{-}\varphi_{x_{1}}\,dx=\int_{\partial\Omega^{\varepsilon}_{-}}\varphi(G_{1}^{-}n_{2}^{\varepsilon}-G_{2}^{-}n_{1}^{\varepsilon})\,d\sigma+\int_{\Omega^{\varepsilon}_{-}}\varphi((G_{2}^{-})_{x_{1}}-(G_{1}^{-})_{x_{2}})\,dx.

Combining the above calculations, we deduce that

I=\int_{\partial\Omega_{-}^{\varepsilon}}\varphi{\bf G_{-}}\times{\bf n_{-}^{\varepsilon}}\,d\sigma+\int_{\Omega^{\varepsilon}_{-}}\varphi\nabla\times{\bf G_{-}}\,dx.

Since $\varphi$ is compactly supported in $\Omega$ , ${\bf G_{-}}$ is locally bounded in $\Omega_{-}\cup\Gamma$ , and $\nabla\times{\bf G_{-}}\in L^{1}(\Omega_{-})$ then

I\to\int_{\Gamma}\varphi{\bf G}_{-}\times{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\nabla\times{\bf G_{-}}\,dx.

Similarly we get

III=\int_{\partial\Omega_{+}^{\varepsilon}}\varphi{\bf G_{+}}\times{\bf n}^{\varepsilon}_{+}\,d\sigma+\int_{\Omega^{\varepsilon}_{+}}\varphi\nabla\times{\bf G_{+}}\,dx\to\int_{\Gamma}\varphi{\bf G}_{+}\times({\bf-n})\,d\sigma+\int_{\Omega_{+}}\varphi\nabla\times{\bf G_{+}}\,dx.

Hence (2.8) follows. ∎

2.1. Distributions depending on a parameter

Since the fields satisfying Maxwell’s equations depend on time, we consider vector-valued distributions in $\Omega$ depending on a parameter $t\in\mathbb{R}$ , that is, for each $t\in\mathbb{R}$ , $g(\cdot,t)\in\mathcal{D}^{\prime}(\Omega)$ , see [4, Appendix 2, p. 147]. We need the following.

Definition 2.2.

Let $g(\cdot,t)$ be a distribution in $\Omega\subseteq\mathbb{R}^{n}$ depending on the parameter $t\in{\mathbb{R}}$ . We say that the derivative of $g(\cdot,t)$ with respect to the parameter $t$ exists if for each test function $\varphi\in\mathcal{D}(\Omega)$ , the function $\langle g(\cdot,t),\varphi\rangle$ is differentiable in $t$ , and there exists a distribution $h(\cdot,t)$ depending on the parameter $t$ such that

\langle h(\cdot,t),\varphi\rangle=\dfrac{d}{dt}\langle g(\cdot,t),\varphi\rangle.

We write $h(x,t)=\dfrac{\partial g}{\partial t}(x,t).$

Proposition 2.3.

Given a distribution $g(\cdot,t)$ in $\Omega$ , $t\in\mathbb{R}$ , if $\dfrac{\partial g}{\partial t}(\cdot,t)$ exists for each $t\in\mathbb{R}$ , then for every multi-index $\alpha=(\alpha_{1},\cdots,\alpha_{n})$ with $\alpha_{i}\in{\mathbb{Z}}_{\geq 0}$ , the derivative with respect to $t$ of the distribution $D^{\alpha}g$ exists and we have

\dfrac{\partial(D^{\alpha}g)}{\partial t}=D^{\alpha}\left(\dfrac{\partial g}{\partial t}\right),

with $D^{\alpha}=\dfrac{\partial^{\alpha_{1}}\partial^{\alpha_{2}}\cdots\partial^{\alpha_{n}}}{\partial x_{1}^{\alpha_{1}}\partial x_{2}^{\alpha_{2}}\cdots\partial x_{n}^{\alpha_{n}}}$ .

Proof.

If $h(\cdot,t)=\dfrac{\partial g}{\partial t}$ and $\varphi\in\mathcal{D}(\Omega)$ , then $\langle g(\cdot,t),\varphi\rangle$ is differentiable in $t$ . Since

\langle D^{\alpha}g(\cdot,t),\varphi\rangle=(-1)^{|\alpha|}\langle g(\cdot,t),D^{\alpha}\varphi\rangle,

with $|\alpha|=\alpha_{1}+\cdots+\alpha_{n}$ , and $D^{\alpha}\varphi\in\mathcal{D}(\Omega)$ , then $\langle D^{\alpha}g(\cdot,t),\varphi\rangle$ is differentiable in $t$ and

\dfrac{d}{dt}\langle D^{\alpha}g(\cdot,t),\varphi\rangle=(-1)^{|\alpha|}\dfrac{d}{dt}\langle g(\cdot,t),D^{\alpha}\varphi\rangle=(-1)^{|\alpha|}\langle h(\cdot,t),D^{\alpha}\varphi\rangle=\langle D^{\alpha}h,\varphi\rangle.

∎

Recalling the set up at the beginning of this section in Figure 1, $\Omega$ is a smooth open and bounded domain in $\mathbb{R}^{3}$ , and $\Gamma$ is a smooth surface that splits $\Omega$ into two disjoint open parts $\Omega_{+}$ and $\Omega_{-}$ , i.e., $\Omega=\Omega_{-}\cup\Gamma\cup\Omega_{+}$ . For $t\in\mathbb{R}$ , we are given a function $g(\cdot,t)$ satisfying

(H1)

$g(\cdot,t)\in L^{1}_{loc}(\Omega)$ for every $t$ ,
(H2)

for each fixed $x\in\Omega_{\pm}$ the function $g(x,\cdot)$ is differentiable with respect to $t$ and there exists a function $\psi\in L^{1}(\Omega_{+}\cup\Omega_{-})$ such that $\left|\dfrac{\partial g}{\partial t}(x,t)\right|\leq\psi(x)$ for a.e. $x\in\Omega_{\pm}$ and for each $t$ .

For every $t$ , the linear functional $g(\cdot,t)$ given by

\langle g(\cdot,t),\varphi\rangle=\int_{\Omega}g(x,t)\varphi(x)\,dx,

is then a well defined distribution by Item (H1).

Proposition 2.4.

If ${\mathbf{G}}(\cdot,t)=(g_{1}(\cdot,t),g_{2}(\cdot,t),g_{3}(\cdot,t))$ is a vector-valued distribution with each $g_{i}$ satisfying (H1) and (H2), then the distribution ${\mathbf{G}}(\cdot,t)$ has a derivative with respect to the parameter $t$ , and

\left\langle\dfrac{\partial{\mathbf{G}}}{\partial t}(\cdot,t),\varphi\right\rangle=\int_{\Omega_{-}}\dfrac{\partial{\mathbf{G}}}{\partial t}(x,t)\varphi(x)\,dx+\int_{\Omega_{+}}\dfrac{\partial{\mathbf{G}}}{\partial t}(x,t)\varphi(x)\,dx.

Proof.

Let us denote by $g(\cdot,t)$ any of the components of ${\mathbf{G}}$ . We write for $\varphi\in\mathcal{D}(\Omega)$ and $t\in{\mathbb{R}}$

\displaystyle\langle g(\cdot,t),\varphi\rangle=\int_{\Omega}g(x,t)\,\varphi(x)\,dx=\int_{\Omega_{-}}g(x,t)\,\varphi(x)\,dx+\int_{\Omega_{+}}g(x,t)\,\varphi(x)\,dx,

since from (H1), the integral $\int_{\Gamma}g(x,t)\,\varphi(x)\,dx=0$ . Using condition (H2) and the Lebesgue dominated convergence theorem, we can justify differentiation under the integral sign and obtain that $\langle g(\cdot,t),\varphi\rangle$ is differentiable in $t$ , and that

\displaystyle\dfrac{d}{dt}\langle g(\cdot,t),\varphi\rangle=\int_{\Omega_{-}}\dfrac{\partial g}{\partial t}(x,t)\,\varphi(x)\,dx+\int_{\Omega_{+}}\dfrac{\partial g}{\partial t}(x,t)\,\varphi(x)\,dx.

From (H2), the linear functional $h(\cdot,t)$ given by

\langle h(\cdot,t),\varphi\rangle=\int_{\Omega_{-}}\dfrac{\partial g}{\partial t}(x,t)\,\varphi(x)\,dx+\int_{\Omega_{+}}\dfrac{\partial g}{\partial t}(x,t)\,\varphi(x)\,dx,

is a well defined distribution and hence we obtain $h(\cdot,t)=\dfrac{\partial g}{\partial t}(\cdot,t)$ . ∎

3. Maxwell equations in distributional sense and general boundary conditions

We are given $\Omega$ open and bounded domain in $\mathbb{R}^{3}$ , and $\Gamma$ a smooth surface separating $\Omega$ into two open parts $\Omega_{+}$ and $\Omega_{-}$ as in Section 2. We are interested in the Maxwell system [3, Sections 1.1 and 1.2] which written in Gaussian (or cgs) units has the form

(3.1)

\begin{cases}\nabla\times{\bf H}=\dfrac{4\pi}{c}{\bf J}+\dfrac{1}{c}\dfrac{\partial{\bf D}}{\partial t}\\ \nabla\cdot{\bf D}=4\pi\rho\\ \nabla\times{\bf E}=-\dfrac{1}{c}\dfrac{\partial{\bf B}}{\partial t}\\ \nabla\cdot{\bf B}=0\end{cases},

where the curl and divergence are understood in the sense of distributions as in Section 2, and the fields $\bf H,J,D,E,B$ are vector valued distributions in $\Omega$ depending on the parameter $t\in{\mathbb{R}}$ in the sense of Section 2.1, with ${\mathbf{J}}$ given, and $\rho$ is scalar distribution in $\Omega$ , also given, depending also on the parameter $t\in{\mathbb{R}}$ .

The purpose of this section is to show that under general assumptions on the current density field ${\mathbf{J}}$ and the charge density $\rho$ each equation in the Maxwell system (3.1), understood in distributional sense, implies a boundary condition at the interface $\Gamma$ and the solutions are classical solutions away from $\Gamma$ . Viceversa, classical solutions in $\Omega_{\pm}$ discontinuous across $\Gamma$ , give rise to distributions solutions in $\Omega$ . This is the contents of the following theorem.

Theorem 3.1.

Let us assume that ${\mathbf{J}}$ and $\rho$ satisfy

(a)

${\mathbf{J}}(x,t)={\mathbf{J}}_{0}(x,t)+\nu_{t}$ with ${\mathbf{J}}_{0}(x,t)$ a locally integrable ${\mathbb{C}}^{3}$ -valued function for $x\in\Omega$ for each $t$ ; and $\nu_{t}$ is a family of ${\mathbb{C}}^{3}$ -valued Borel measures in $\Omega$ depending on the parameter $t$ that are all concentrated on $\Gamma$ , that is, the support of $\nu_{t}$ is contained in $\Gamma$ ;
(b)

$\rho(x,t)=\rho_{0}(x,t)+\mu_{t}$ with $\rho_{0}(x,t)$ locally integrable in $\Omega$ for each $t$ , and $\mu_{t}$ are Borel measures in $\Omega$ depending on the parameter $t$ that are all concentrated on the surface $\Gamma$ .

Suppose also that ${\mathbf{B}}$ and ${\mathbf{D}}$ are given fields satisfying (F1), (F2), (F3), (H1), and (H2); and ${\mathbf{E}}$ and ${\mathbf{H}}$ are also given fields satisfying (F1), (F2), and (F3); ${\bf n}$ denotes the unit normal to $\Gamma$ toward $\Omega_{+}$ .

Then we have the following

(1)

If ${\mathbf{D}}$ satisfies $\nabla\cdot{\mathbf{D}}=4\pi\rho$ in $\Omega$ in the sense of distributions, then $\nabla\cdot{\mathbf{D}}_{\pm}(x,t)=4\pi\rho_{0}(x,t)$ for a.e. $x\in\Omega_{\pm}$ and for each $t$ , and

(3.2) $d\mu_{t}(x)=\frac{1}{4\pi}[[{\mathbf{D}}(x,t)]]\cdot{\bf n}(x)\,d\sigma(x),$

where $d\sigma$ denotes the surface measure on $\Gamma$ . Reciprocally, if $\nabla\cdot{\mathbf{D}}_{\pm}=4\pi\rho_{0}$ holds point-wise in $\Omega_{\pm}$ and (3.2) holds, then ${\mathbf{D}}=\chi_{\Omega_{-}}{\mathbf{D}}_{-}+\chi_{\Omega_{+}}{\mathbf{D}}_{+}$ satisfies the equation $\nabla\cdot{\mathbf{D}}=4\pi\rho$ in $\Omega$ in the sense of distributions; as usual, $\chi_{E}$ denotes the characteristic function of the set $E$ .

(2)

If ${\mathbf{B}}$ satisfies $\nabla\cdot{\mathbf{B}}=0$ in $\Omega$ in the sense of distributions, then $\nabla\cdot{\mathbf{B}}_{\pm}(x,t)=0$ point-wise $x\in\Omega_{\pm}$ and for each $t$ and

(3.3)

[[{\mathbf{B}}(x,t)]]\cdot{\bf n}(x)=0\quad\text{for a.e. $x\in\Gamma$ (with respect to surface measure) for all $t$.}

Reciprocally, if $\nabla\cdot{\mathbf{B}}_{\pm}=0$ holds point-wise in $\Omega_{\pm}$ and (3.3) holds, then ${\mathbf{B}}=\chi_{\Omega_{-}}{\mathbf{B}}_{-}+\chi_{\Omega_{+}}{\mathbf{B}}_{+}$ satisfies the equation $\nabla\cdot{\mathbf{B}}=0$ in the sense of distributions.

(3)

If ${\mathbf{H}}$ and ${\mathbf{D}}$ satisfy $\nabla\times{\mathbf{H}}=\dfrac{4\pi}{c}{\mathbf{J}}+\dfrac{1}{c}\dfrac{\partial{\mathbf{D}}}{\partial t}$ in $\Omega$ in the sense of distributions, then $\nabla\times{\mathbf{H}}_{\pm}(x,t)=\dfrac{4\pi}{c}{\mathbf{J}}_{0}(x,t)+\dfrac{1}{c}\dfrac{\partial{\mathbf{D}}_{\pm}}{\partial t}(x,t)$ point-wise for $x\in\Omega_{\pm}$ and

(3.4) $d\nu_{t}(x)=-\dfrac{c}{4\pi}[[{\mathbf{H}}(x,t)]]\times{\bf n}(x)\,d\sigma(x).$

Reciprocally, if the equation holds point-wise in $\Omega_{\pm}$ and (3.4) also holds, then the distributional equation holds for ${\mathbf{H}}=\chi_{\Omega_{-}}{\mathbf{H}}_{-}+\chi_{\Omega_{+}}{\mathbf{H}}_{+}$ and ${\mathbf{D}}=\chi_{\Omega_{-}}{\mathbf{D}}_{-}+\chi_{\Omega_{+}}{\mathbf{D}}_{+}$ .

(4)

If ${\mathbf{E}}$ and ${\mathbf{B}}$ satisfy $\nabla\times{\mathbf{E}}=-\dfrac{1}{c}\dfrac{\partial{\mathbf{B}}}{\partial t}$ in $\Omega$ in the sense of distributions, then $\nabla\times{\mathbf{E}}_{\pm}(x,t)=-\dfrac{1}{c}\dfrac{\partial{\mathbf{B}}_{\pm}}{\partial t}(x,t)$ point-wise in $\Omega_{\pm}$ and

(3.5)

[[{\mathbf{E}}(x,t)]]\times{\bf n}(x)={\bf 0}\quad\text{for a.e. $x\in\Gamma$ (with respect to surface measure) for all $t$.}

Reciprocally, if the equation holds point-wise in $\Omega_{\pm}$ and (3.5) also holds, then the distributional equation holds for ${\mathbf{E}}=\chi_{\Omega_{-}}{\mathbf{E}}_{-}+\chi_{\Omega_{+}}{\mathbf{E}}_{+}$ and ${\mathbf{B}}=\chi_{\Omega_{-}}{\mathbf{B}}_{-}+\chi_{\Omega_{+}}{\mathbf{B}}_{+}$ .

Proof.

(1) From $(b)$ , $\rho(\cdot,t)$ is a distribution depending on $t$ given by

\langle\rho(\cdot,t),\varphi\rangle=\int_{\Omega}\rho_{0}(x,t)\,\varphi(x)\,dx+\int_{\Omega}\varphi(x)\,d\mu_{t}(x)=\int_{\Omega}\rho_{0}(x,t)\,\varphi(x)dx+\int_{\Gamma}\varphi(x)\,d\mu_{t}(x),

for each $\varphi\in\mathcal{D}(\Omega)$ ; and $\nabla\cdot{\mathbf{D}}$ is a distribution that acting on a test function $\varphi$ is given by (2.7). We have

\langle\nabla\cdot{\mathbf{D}}(\cdot,t),\varphi\rangle=4\pi\langle\rho(\cdot,t),\varphi\rangle

for each $t$ . If $\text{supp}(\varphi)\subset\Omega_{-}$ or $\text{supp}(\varphi)\subset\Omega_{+}$ , then from (2.7)

\int_{\Omega_{-}}\varphi(x)\,\nabla\cdot{{\mathbf{D}}_{-}(x,t)}\,dx+\int_{\Omega_{+}}\varphi(x)\,\nabla\cdot{{\mathbf{D}}_{+}(x,t)}\,dx=4\pi\int_{\Omega}\rho_{0}(x,t)\,\varphi(x)\,dx,

which implies $\nabla\cdot{\mathbf{D}}_{\pm}(x,t)=4\pi\rho_{0}(x,t)\text{ for a.e. $x\in\Omega_{\pm}$ and for each $t$.}$ That is, the equation $\nabla\cdot{\mathbf{D}}=4\pi\rho$ is satisfied pointwise a.e. in $\Omega_{\pm}$ . If $\text{supp}(\varphi)\cap\Gamma\neq\emptyset$ , we then get again from (2.7) that

\int_{\Gamma}\varphi(x)\,[[{\mathbf{D}}(x,t)]]\cdot{\bf n}(x)\,d\sigma(x)=4\pi\int_{\Gamma}\varphi(x)\,d\mu_{t}(x),

that is, the measure $\mu_{t}$ has density $\dfrac{1}{4\pi}[[{\mathbf{D}}(x,t)]]\cdot{\bf n}(x)$ , and (3.2) follows. Notice that this part only uses that ${\mathbf{D}}$ satisfies (F1)–(F3).

For the converse, applying (2.7) to ${\mathbf{D}}$ yields

	$\displaystyle\langle\nabla\cdot{\mathbf{D}},\varphi\rangle$	$\displaystyle=\int_{\Gamma}\varphi(x)\,[[{\mathbf{D}}(x)]]\cdot{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\,\nabla\cdot{\mathbf{D}}_{-}\,dx+\int_{\Omega_{+}}\varphi\nabla\cdot{\mathbf{D}}_{+}\,dx$
		$\displaystyle=4\pi\,\int_{\Gamma}\varphi(x)\,d\mu_{t}+4\pi\,\int_{\Omega_{-}}\varphi\,\rho_{0}\,dx+\int_{\Omega_{+}}\varphi\,\rho_{0}\,dx\quad\text{from \eqref{eq:density of mut bis}}$
		$\displaystyle=4\pi\,\int_{\Gamma}\varphi(x)\,d\mu_{t}+4\pi\,\int_{\Omega}\varphi\,\rho_{0}\,dx=4\pi\,\langle\rho,\varphi\rangle.$

(2) We proceed as in the proof of (1) and in this way we obtain $\nabla\cdot{\mathbf{B}}_{\pm}(x,t)=0$ for a.e. $x\in\Omega_{\pm}$ and for each $t$ ; and (3.3). Reciprocally, ${\mathbf{B}}=\chi_{\Omega_{-}}{\mathbf{B}}_{-}+\chi_{\Omega_{+}}{\mathbf{B}}_{+}$ satisfies the equation $\nabla\cdot{\mathbf{B}}=0$ in the sense of distributions.

(3) The equation reads

\left\langle\nabla\times{\mathbf{H}},\varphi\right\rangle=\dfrac{4\pi}{c}\left\langle{\mathbf{J}},\varphi\right\rangle+\dfrac{1}{c}\left\langle\dfrac{\partial{\mathbf{D}}}{\partial t},\varphi\right\rangle,

for each $\varphi\in\mathcal{D}(\Omega)$ . From Proposition 2.4

\left\langle\dfrac{\partial{\mathbf{D}}}{\partial t},\varphi\right\rangle=\int_{\Omega_{-}}\dfrac{\partial{\mathbf{D}}_{-}(x,t)}{\partial t}\varphi(x)\,dx+\int_{\Omega_{+}}\dfrac{\partial{\mathbf{D}}_{+}(x,t)}{\partial t}\varphi(x)\,dx.

If $\text{supp}(\varphi)\subset\Omega_{-}$ or $\text{supp}(\varphi)\subset\Omega_{+}$ , then from (2.8)

		$\displaystyle\int_{\Omega_{-}}\varphi(x)\,\nabla\times{\mathbf{H}}_{-}(x,t)\,dx+\int_{\Omega_{+}}\varphi(x)\,\nabla\times{\mathbf{H}}_{+}(x,t)\,dx$
		$\displaystyle\qquad=\dfrac{4\pi}{c}\int_{\Omega}{\mathbf{J}}_{0}(x,t)\,\varphi(x)\,dx+\dfrac{1}{c}\int_{\Omega_{-}}\dfrac{\partial{\mathbf{D}}_{-}}{\partial t}(x,t)\varphi(x)\,dx+\dfrac{1}{c}\int_{\Omega_{+}}\dfrac{\partial{\mathbf{D}}_{+}}{\partial t}(x,t)\varphi(x)\,dx,$

which implies $\nabla\times{\mathbf{H}}_{\pm}(x,t)=\dfrac{4\pi}{c}{\mathbf{J}}_{0}(x,t)+\dfrac{1}{c}\dfrac{\partial{\mathbf{D}}_{\pm}}{\partial t}(x,t)\text{ for a.e. $x\in\Omega_{\pm}$ and all $t$}$ . If $\text{supp}(\varphi)\cap\Gamma\neq\emptyset$ , we then get again from (2.8) that

-\int_{\Gamma}\varphi(x)\,[[{\mathbf{H}}(x,t)]]\times{\bf n}(x)\,d\sigma(x)=\dfrac{4\pi}{c}\int_{\Gamma}\varphi(x)\,d\nu_{t}(x),

that is, the measure $\nu_{t}$ has density $-\dfrac{c}{4\pi}[[{\mathbf{H}}(x,t)]]\times{\bf n}(x)$ , so (3.4) follows.

If each equation $\nabla\times{\mathbf{H}}_{\pm}=\dfrac{4\pi}{c}{\mathbf{J}}_{0}+\dfrac{1}{c}\dfrac{\partial{\mathbf{D}}_{\pm}}{\partial t}$ holds in $\Omega_{\pm}$ in the classical sense and (3.4) holds, then the equation $\nabla\times{\mathbf{H}}=\dfrac{4\pi}{c}{\mathbf{J}}+\dfrac{1}{c}\dfrac{\partial{\mathbf{D}}}{\partial t}$ is satisfied in $\Omega$ in the sense of distributions where ${\mathbf{H}}$ is the distribution given by the locally integrable function $\chi_{\Omega_{-}}{\mathbf{H}}_{-}+\chi_{\Omega_{+}}{\mathbf{H}}_{+}$ . In fact, from (2.8) and (3.4)

	$\displaystyle\langle\nabla\times{\mathbf{H}},\varphi\rangle$	$\displaystyle=-\int_{\Gamma}\varphi(x)\,[[{\mathbf{H}}(x)]]\times{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\,\nabla\times{\mathbf{H}}_{-}\,dx+\int_{\Omega_{+}}\varphi\,\nabla\times{\mathbf{H}}_{+}\,dx$
		$\displaystyle=\dfrac{4\pi}{c}\int_{\Gamma}\varphi(x)\,d\nu_{t}+\dfrac{4\pi}{c}\int_{\Omega}\varphi\,{\mathbf{J}}_{0}\,dx+\dfrac{1}{c}\int_{\Omega_{-}}\varphi\,\dfrac{\partial{\mathbf{D}}_{-}}{\partial t}\,dx+\dfrac{1}{c}\int_{\Omega_{+}}\varphi\,\dfrac{\partial{\mathbf{D}}_{+}}{\partial t}\,dx$
		$\displaystyle=\dfrac{4\pi}{c}\langle{\mathbf{J}},\varphi\rangle+\dfrac{1}{c}\left\langle\dfrac{\partial{\mathbf{D}}}{\partial t},\varphi\right\rangle.$

(4) The equation reads

\left\langle\nabla\times{\mathbf{E}},\varphi\right\rangle=-\dfrac{1}{c}\left\langle\dfrac{\partial{\mathbf{B}}}{\partial t},\varphi\right\rangle,

for each $\varphi\in\mathcal{D}(\Omega)$ . From Proposition 2.4

\left\langle\dfrac{\partial{\mathbf{B}}}{\partial t},\varphi\right\rangle=\int_{\Omega_{-}}\dfrac{\partial{\mathbf{B}}(x,t)}{\partial t}\varphi(x)\,dx+\int_{\Omega_{+}}\dfrac{\partial{\mathbf{B}}(x,t)}{\partial t}\varphi(x)\,dx.

If $\text{supp}(\varphi)\subset\Omega_{-}$ or $\text{supp}(\varphi)\subset\Omega_{+}$ , then from (2.8)

		$\displaystyle\int_{\Omega_{-}}\varphi(x)\,\nabla\times{\mathbf{E}}_{-}(x,t)\,dx+\int_{\Omega_{+}}\varphi(x)\,\nabla\times{\mathbf{E}}_{+}(x,t)\,dx$
		$\displaystyle\qquad=-\dfrac{1}{c}\int_{\Omega_{-}}\dfrac{\partial{\mathbf{B}}_{-}}{\partial t}(x,t)\varphi(x)\,dx-\dfrac{1}{c}\int_{\Omega_{+}}\dfrac{\partial{\mathbf{B}}_{+}}{\partial t}(x,t)\varphi(x)\,dx,$

which implies $\nabla\times{\mathbf{E}}_{\pm}(x,t)=-\dfrac{1}{c}\dfrac{\partial{\mathbf{B}}_{\pm}}{\partial t}(x,t)$ for a.e. $x\in\Omega_{\pm}$ and all $t$ . If $\text{supp}(\varphi)\cap\Gamma\neq\emptyset$ , we then get again from (2.8) that

	$\displaystyle\langle\nabla\times{\mathbf{E}},\varphi\rangle$	$\displaystyle=-\int_{\Gamma}\varphi(x)\,[[{\mathbf{E}}(x,t)]]\times{\bf n}\,d\sigma+\int_{\Omega_{-}}\varphi\nabla\times{\mathbf{E}}_{-}\,dx+\int_{\Omega_{+}}\varphi\nabla\times{\mathbf{E}}_{+}\,dx$
		$\displaystyle=-\int_{\Gamma}\varphi(x)\,[[{\mathbf{E}}(x,t)]]\times{\bf n}\,d\sigma-\dfrac{1}{c}\int_{\Omega_{-}}\varphi\dfrac{\partial{\mathbf{B}}_{-}}{\partial t}(x,t)\,dx-\dfrac{1}{c}\int_{\Omega_{+}}\varphi\dfrac{\partial{\mathbf{B}}_{+}}{\partial t}(x,t)\,dx$
		$\displaystyle=-\int_{\Gamma}\varphi(x)\,[[{\mathbf{E}}(x,t)]]\times{\bf n}\,d\sigma-\dfrac{1}{c}\left\langle\dfrac{\partial{\mathbf{B}}}{\partial t},\varphi\right\rangle$

implying that

\int_{\Gamma}\varphi(x)\,[[{\mathbf{E}}(x,t)]]\times{\bf n}\,d\sigma=0

for all test functions $\varphi\in\mathcal{D}(\Omega)$ and all $t$ , therefore (3.5) holds.

Reciprocally, if (3.5) holds we obtain that $\nabla\times{\mathbf{E}}=-\dfrac{1}{c}\dfrac{\partial{\mathbf{B}}}{\partial t}$ holds in $\Omega$ in the sense of distributions.

∎

3.1. Compatibility condition

Let us assume that (3.1) holds with vector valued distributions ${\mathbf{E}},{\mathbf{H}},{\mathbf{D}},{\mathbf{B}},{\mathbf{J}}$ depending on the parameter $t$ , with ${\mathbf{B}},{\mathbf{D}},{\mathbf{J}}$ also differentiable with respect to this parameter. Hence from the first and second Maxwell equations in (3.1), (2.5), and Proposition 2.3 we have in the distributional sense the continuity equation

(3.6)

0=\nabla\cdot(\nabla\times{\bf H})=\dfrac{4\pi}{c}\nabla\cdot{\bf J}+\dfrac{1}{c}\nabla\cdot\dfrac{\partial{\bf D}}{\partial t}=\dfrac{4\pi}{c}\nabla\cdot{\bf J}+\dfrac{1}{c}\dfrac{\partial\rho}{\partial t}.

When ${\mathbf{J}}$ and $\rho$ satisfy the assumptions in Theorem 3.1, equation (3.6) leads to the following compatibility condition between the current ${\mathbf{J}}$ and the density $\rho$ :

\dfrac{4\pi}{c}\left(\nabla\cdot{\mathbf{J}}_{0}+\nabla\cdot\nu_{t}\right)+\dfrac{1}{c}\left(\dfrac{\partial\rho_{0}}{\partial t}+\dfrac{\partial\mu_{t}}{\partial t}\right)=0,

in the sense of distributions.

4. Generalized Snell’s law deduced from Maxwell equations

Letting as usual [3, Section 1.1.2] the material or constitutive equations

(4.1)

{\mathbf{D}}=\epsilon\,{\mathbf{E}},\qquad{\mathbf{B}}=\mu\,{\mathbf{H}},

we obtain from (3.1)

(M.1)	$\displaystyle\nabla\cdot\epsilon{\mathbf{E}}$	$\displaystyle=4\,\pi\,\rho,\quad$
(M.2)	$\displaystyle\nabla\cdot\mu{\mathbf{H}}$	$\displaystyle=0,\quad$
(M.3)	$\displaystyle\nabla\times{\mathbf{E}}$	$\displaystyle=-\dfrac{\mu}{c}\dfrac{\partial{\mathbf{H}}}{\partial t},\quad$
(M.4)	$\displaystyle\nabla\times{\mathbf{H}}$	$\displaystyle=\dfrac{4\pi}{c}\,{\mathbf{J}}+\dfrac{\epsilon}{c}\dfrac{\partial{\mathbf{E}}}{\partial t},$

where $\rho(x,t)$ is the charge density, ${\mathbf{J}}(x,t)$ is the current density vector, ${\mathbf{E}}$ is the electric field, ${\mathbf{H}}$ is the magnetic field, and $\epsilon,\mu$ are constants, the permittivity and permeability of the media (isotropic), respectively. If $\epsilon_{0},\mu_{0}$ are the permittivity and permeability of vacuum, then $\epsilon_{\text{rel}}=\epsilon/\epsilon_{0}$ and , $\mu_{\text{rel}}=\mu/\mu_{0}$ denote the relative permittivity and relative permeability, respectively, and $n=\sqrt{\epsilon_{\text{rel}}\mu_{\text{rel}}}$ is the refractive index of the media. Since the speed of light in vacuum is $c=1/\sqrt{\epsilon_{0}\mu_{0}}$ , and the phase velocity of light in the media is $v=1/\sqrt{\epsilon\mu}$ , then $v=c/n$ .

Let $\Gamma$ be the plane $x_{3}=0$ , and let $\Omega_{+}$ , $\Omega_{-}$ denote the regions above $\Gamma$ and below $\Gamma$ respectively, with $\Omega_{-}$ filled with medium $I$ and $\Omega_{+}$ filled with medium $II$ . The constants $\epsilon,\mu$ in the Maxwell system (M.1)–(M.4) may be different in media $I$ and $II$ , and they are denoted by $\epsilon_{-},\mu_{-}$ in medium $I$ , and $\epsilon_{+},\mu_{+}$ in medium $II$ . Suppose the incoming incident electric field in media $I$ is a plane wave with the form

(4.2)

{\mathbf{E}}_{i}(x,t)={\bf A}_{i}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot x}{v_{1}}-t\right)}

where ${\mathbf{k}}_{i}$ is the incident unit vector, $v_{1}$ is the velocity of propagation in medium $I$ , ${\bf A}_{i}$ is a three dimensional constant complex vector, the amplitude, and $\omega$ is a constant (the angular frequency). Here $x=(x_{1},x_{2},x_{3})$ . As usual, the Roman numeral $\rm i$ in the exponentials denotes the unit imaginary number. This wave is defined for $x_{3}<0$ , i.e., the field is incident to the plane $\Gamma$ from below and defined in $\Omega_{-}$ . This wave strikes the plane $\Gamma$ and it is then transmitted into medium $II$ as a nonlinear wave and the ansatz is to assume it has the form

(4.3)

{\mathbf{E}}_{t}(x,t)={\bf A}_{t}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot x}{v_{2}}+f(x)-t\right)}

where now ${\mathbf{k}}_{t}$ is the refracted unit vector, $v_{2}$ is the velocity of propagation in medium $II$ , ${\bf A}_{t}$ is the amplitude, a constant vector, the wave is defined for $x_{3}>0$ , i.e., on $\Omega_{+}$ , and $f(x)$ is a $C^{2}$ function defined in a neighborhood of the plane $\Gamma$ . There is also a wave reflected back into medium $I$ that will be assumed to have also a similar form

(4.4)

{\mathbf{E}}_{r}(x,t)={\bf A}_{r}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot x}{v_{1}}+f(x)-t\right)},

with ${\bf A}_{r}$ a constant vector, ${\mathbf{k}}_{r}$ is the reflected back unit vector, $v_{1}$ is the velocity of propagation in medium $I$ , the wave is defined for $x_{3}<0$ , i.e., on $\Omega_{-}$ . We are assuming that $f$ depends only on $x_{1},x_{2}$ , i.e., the gradient of $f$ is tangential to the plane $\Gamma$ ; we then denote $\nabla f=(f_{x_{1}},f_{x_{2}},0)$ . In addition, and without loss of generality, we assume that $f(0)=0$ , otherwise that simply changes the values of the amplitudes.

Figure 2. Reflection and transmission at the planar interface

\Gamma=\{x_{3}=0\}

separating medium

I

from medium

II

The plan of this section is the following:

(1)

The fields ${\mathbf{E}}_{i},{\mathbf{E}}_{t},{\mathbf{E}}_{r}$ have corresponding magnetic fields ${\mathbf{H}}_{i},{\mathbf{H}}_{t},{\mathbf{H}}_{r}$ so that the Maxwell system (M.1)–(M.4) is satisfied; these are calculated in Section 4.1. These magnetic fields are used in Section 5 to obtain boundary conditions for them.
(2)

Section 4.2 contains the proof of the main result, Theorem 4.3. The proof uses Theorem 3.1 to obtain the boundary conditions (4.10), (4.11), and (4.15) for the electric field (equations that will be used later in Section 6). As a consequence of these we obtain the generalized Snell law for the first two components of the wave vectors, Equation (4.8). It also contains the statement of Lemma 4.2 used to prove Theorem 4.3 . The proof of this lemma is given in the Appendix 7.
(3)

In Section 4.3 we deduce from the previous items the generalized Snell law for refraction (1.1) and the generalized law of reflection (1.2).
(4)

In Section 4.4 we show relationships for the third components of the wave vectors, Corollary 4.4.
(5)

In Section 4.5 we deduce orthogonality conditions for the amplitudes that are used later in Section 6 to calculate the amplitudes of the transmitted and reflected waves.

We use the following notation throughout the paper

(4.5)

{\mathbf{m}}_{i}=\omega\frac{{\mathbf{k}}_{i}}{v_{1}},\qquad{\mathbf{m}}_{r}=\omega\frac{{\mathbf{k}}_{r}}{v_{1}},\qquad{\mathbf{m}}_{t}=\omega\dfrac{{\mathbf{k}}_{t}}{v_{2}},

and write ${\mathbf{m}}_{\ell}=(m_{1}^{\ell},m_{2}^{\ell},m_{3}^{\ell})$ , with $\ell=i,r,t$ .

4.1. Calculation of the corresponding magnetic fields

The values of these magnetic fields are given in the following lemma.

Lemma 4.1.

Suppose the electric fields ${\mathbf{E}}_{i},{\mathbf{E}}_{t},{\mathbf{E}}_{r}$ are given by (4.2), (4.3), and (4.4), respectively. If the field ${\mathbf{H}}^{\prime}(x,t)$ solves

\nabla\times\left({\mathbf{E}}_{i}+{\mathbf{E}}_{r}\right)=-\dfrac{\mu_{-}}{c}\dfrac{\partial{\mathbf{H}}^{\prime}}{\partial t}

in $\Omega_{-}$ , and the field ${\mathbf{H}}_{t}(x,t)$ solves

\nabla\times{\mathbf{E}}_{t}=-\dfrac{\mu_{+}}{c}\dfrac{\partial{\mathbf{H}}_{t}}{\partial t}

in $\Omega_{+}$ , then ${\mathbf{H}}^{\prime}={\mathbf{H}}_{i}+{\mathbf{H}}_{r}$ with

{\mathbf{H}}_{i}=-\dfrac{c}{\mu_{-}}\,{\mathbf{E}}_{i}\times\dfrac{{\mathbf{k}}_{i}}{v_{1}};\quad{\mathbf{H}}_{r}=-\dfrac{c}{\mu_{-}}\,{\mathbf{E}}_{r}\times\left(\dfrac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(x)\right),

and

{\mathbf{H}}_{t}=-\dfrac{c}{\mu_{+}}\,{\mathbf{E}}_{t}\times\left(\dfrac{{\mathbf{k}}_{t}}{v_{2}}+\nabla f(x)\right),

modulo fields only depending on $x$ which we assume to be zero. We have denoted $\nabla f(x)=\left(f_{x_{1}}(x),f_{x_{2}}(x),0\right)$ .

Proof.

Calculation of ${\mathbf{H}}^{\prime}$ . We have from (4.2) and (4.4)

\nabla\times\left({\mathbf{E}}_{i}+{\mathbf{E}}_{r}\right)=-{\rm i}\,\omega\,\left({\bf A}_{i}\times\dfrac{{\mathbf{k}}_{i}}{v_{1}}\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot x}{v_{1}}-t\right)}-{\rm i}\,\omega\,{\bf A}_{r}\times\left(\dfrac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(x)\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot x}{v_{1}}+f(x)-t\right)}.

Integrating yields

	$\displaystyle{\mathbf{H}}^{\prime}$	$\displaystyle=-\dfrac{c}{\mu_{-}}\int\nabla\times\left({\mathbf{E}}_{i}+{\mathbf{E}}_{r}\right)\,dt$
		$\displaystyle=-\dfrac{c}{\mu_{-}}\,\left({\bf A}_{i}\times\dfrac{{\mathbf{k}}_{i}}{v_{1}}\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot x}{v_{1}}-t\right)}-\dfrac{c}{\mu_{-}}\,{\bf A}_{r}\times\left(\dfrac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(x)\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot x}{v_{1}}+f(x)-t\right)}$
		$\displaystyle=-\dfrac{c}{\mu_{-}}\,{\mathbf{E}}_{i}\times\dfrac{{\mathbf{k}}_{i}}{v_{1}}-\dfrac{c}{\mu_{-}}\,{\mathbf{E}}_{r}\times\left(\dfrac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(x)\right),$

modulo a field only depending on $x$ which we assume to be zero.

Similarly, the calculation of ${\mathbf{H}}_{t}$ follows by integration using (4.3).

∎

4.2. Main result and the generalized Snell law

In this section, we shall prove Theorem 4.3 showing that the phase function $f$ in the scattered waves (4.3) and (4.4) is necessarily affine and we prove relationships between the components of the wave vectors ${\mathbf{k}}_{i},{\mathbf{k}}_{r}$ and ${\mathbf{k}}_{t}$ and $\nabla f$ , that imply the generalized Snell law (1.1) and (1.2). The proof of Theorem 4.3 requires the following lemma whose proof is given in the Appendix 7.

Lemma 4.2.

Suppose that the equation

(4.6)

a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+b\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}+c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0

holds for all $X=(x_{1},x_{2},0)$ , where $a,b,c$ are fixed complex constants, ${\mathbf{m}}_{\ell}=\left(m_{1}^{\ell},m_{2}^{\ell},m_{3}^{\ell}\right)$ for $\ell=i,r,t$ , and $\psi$ is a real-valued $C^{2}$ -function; we use the notation $\psi(X)=\psi(x_{1},x_{2})$ . We have

(i)

If $a\neq 0$ , $b\neq 0$ , and $c\neq 0$ , then $\psi$ is an affine function and

$m_{j}^{i}-m_{j}^{r}=\psi_{x_{j}},\text{ and }m_{j}^{i}-m_{j}^{t}=\psi_{x_{j}},\quad j=1,2.$
(ii)

If $a=0$ , $b\neq 0$ , and $c\neq 0$ , then $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ .
(iii)

If $a\neq 0$ , $b=0$ , and $c\neq 0$ , then $\psi$ is affine and

$m_{j}^{i}-m_{j}^{t}=\psi_{x_{j}},\quad j=1,2.$
(iv)

If $a\neq 0$ , $b\neq 0$ , and $c=0$ , then $\psi$ is affine and

$m_{j}^{i}-m_{j}^{r}=\psi_{x_{j}},\quad j=1,2.$

Now, let us proceed to state and the prove the main result of the section.

Theorem 4.3.

Recall the definitions of the fields ${\mathbf{E}}_{i},{\mathbf{E}}_{r}$ and ${\mathbf{E}}_{t}$ from (4.2)–(4.4), and the corresponding magnetic fields ${\mathbf{H}}_{i},{\mathbf{H}}_{r}$ and ${\mathbf{H}}_{t}$ from Lemma 4.1. Let

(4.7)

{\mathbf{E}}^{\prime}(x,t)={\mathbf{E}}_{i}(x,t)+{\mathbf{E}}_{r}(x,t),\quad{\mathbf{H}}^{\prime}(x,t)={\mathbf{H}}_{i}(x,t)+{\mathbf{H}}_{r}(x,t),

and define ${\mathbf{E}}(x,t)=\chi_{\Omega_{-}}{\mathbf{E}}^{\prime}(x,t)+\chi_{\Omega_{+}}{\mathbf{E}}_{t}(x,t)$ , and ${\mathbf{H}}(x,t)=\chi_{\Omega_{-}}{\mathbf{H}}^{\prime}(x,t)+\chi_{\Omega_{+}}{\mathbf{H}}_{t}(x,t)$ .

Then the fields ${\mathbf{E}}^{\prime},{\mathbf{H}}^{\prime},{\mathbf{E}}_{t}$ and ${\mathbf{H}}_{t}$ satisfy conditions (F1), (F2), and (F3).

If ${\mathbf{E}}$ and ${\mathbf{H}}$ are distributional solutions to (M.3), and ${\mathbf{E}}$ is a distributional solution to (M.1) with $\rho$ non-singular, i.e. $\rho$ is as in Theorem 3.1(b) with $\mu_{t}=0$ , then the function $f$ must be affine, i.e., a linear function of $x_{1},x_{2}$ plus a constant, and

(4.8)

\frac{k_{j}^{i}}{v_{1}}-\frac{k_{j}^{r}}{v_{1}}=f_{x_{j}}\quad\text{and}\quad\frac{k_{j}^{i}}{v_{1}}-\frac{k_{j}^{t}}{v_{2}}=f_{x_{j}}\quad\quad j=1,2,

with ${\mathbf{k}}_{\ell}=\left(k_{1}^{\ell},k_{2}^{\ell},k_{3}^{\ell}\right)$ with $\ell=i,r,t$ .

Proof.

The proof uses Lemma 4.2 below and Theorem 3.1.

From the explicit form of the fields, it is clear that ${\mathbf{E}}^{\prime},{\mathbf{H}}^{\prime},{\mathbf{E}}_{t}$ and ${\mathbf{H}}_{t}$ satisfy conditions (F1), (F2), and (F3). Since we assume that ${\mathbf{E}}$ and ${\mathbf{H}}$ are distributional solutions to (M.3), it follows that Theorem 3.1, Part (4), is applicable.

Therefore, the jump of the electric field equals

	$\displaystyle[[{\mathbf{E}}(X,t)]]$	$\displaystyle=\lim_{x\to X,x\in\Omega_{+}}{\mathbf{E}}(x,t)-\lim_{x\to X,x\in\Omega_{-}}{\mathbf{E}}(x,t)$
		$\displaystyle={\bf A}_{t}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot X}{v_{2}}+f(X)-t\right)}-{\bf A}_{i}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot X}{v_{1}}-t\right)}-{\bf A}_{r}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)-t\right)}\,$

with $X=(x_{1},x_{2},0)$ , and from the boundary condition (3.5)

(4.9)

\left({\bf A}_{t}\times{\mathbf{n}}\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot X}{v_{2}}+f(X)-t\right)}-\left({\bf A}_{i}\times{\mathbf{n}}\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot X}{v_{1}}-t\right)}-\left({\bf A}_{r}\times{\mathbf{n}}\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)-t\right)}={\bf 0},

for all $X\in\Gamma$ .

If we write the components of $A_{\ell}=\left(A_{1}^{\ell},A_{2}^{\ell},A_{3}^{\ell}\right)$ with $\ell=i,r,t$ , then $A_{\ell}\times{\mathbf{n}}=\left(A_{2}^{\ell},-A_{1}^{\ell},0\right)$ . Also, if we set $\psi(X)=\omega f(X)$ and recall (4.5), then (4.9) is the system of two scalar equations

(4.10)

-A_{2}^{i}\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}-A_{2}^{r}\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}+A_{2}^{t}\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0,

(4.11)

A_{1}^{i}\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}+A_{1}^{r}\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}-A_{1}^{t}\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0.

To prove the desired result we shall use Lemma 4.2. Let us first assume that

(4.12)

{\bf A}_{i}\times{\mathbf{n}}\neq 0,\quad{\bf A}_{r}\times{\mathbf{n}}\neq 0,\quad\text{and }{\bf A}_{t}\times{\mathbf{n}}\neq 0.

From (4.12) we have that $A_{2}^{\ell}\neq 0$ or $A_{1}^{\ell}\neq 0$ for $\ell=i,r,t$ . Notice that in (4.10) if one coefficient is different from zero then at least one of the other two must be different from zero; and likewise in (4.11). If in (4.10) $A_{2}^{\ell}\neq 0$ for $\ell=i,r,t$ , then applying Lemma 4.2 (i) it follows that $\psi$ is affine and (4.8) holds. Likewise, if in (4.11) $A_{1}^{\ell}\neq 0$ for $\ell=i,r,t$ , then applying Lemma 4.2 (i) it follows that $\psi$ is affine and (4.8) holds. If in (4.10) $A_{2}^{i}\neq 0$ , then $A_{2}^{r}\neq 0$ or $A_{2}^{t}\neq 0$ . If $A_{2}^{r}\neq 0$ and $A_{2}^{t}=0$ , by Lemma 4.2 (iv) we have $\psi$ is affine and $m_{j}^{i}-m_{j}^{r}=\psi_{x_{j}}$ , $j=1,2$ . Since ${\bf A}_{t}\times{\mathbf{n}}\neq 0$ , if $A_{2}^{t}=0$ we then must have $A_{1}^{t}\neq 0$ and from (4.11) $A_{1}^{i}\neq 0$ or $A_{1}^{r}\neq 0$ . If $A_{1}^{i}\neq 0$ and $A_{1}^{r}=0$ , then by Lemma 4.2 (iii) $m_{j}^{i}-m_{j}^{t}=\psi_{x_{j}}$ for $j=1,2$ ; so (4.8) follows. On the other hand, if $A_{1}^{i}=0$ and $A_{1}^{r}\neq 0$ , then by Lemma 4.2 (ii) $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ and so also (4.8) follows. In general, all the possibilities for the values of the coefficients with their conclusions are summarized in the following table:

$A_{2}^{i}$	$A_{2}^{r}$	$A_{2}^{t}$	$A_{1}^{i}$	$A_{1}^{r}$	$A_{1}^{t}$	Conclusion
$\neq 0$	$\neq 0$	0	0	$\neq 0$	$\neq 0$	${\mathbf{m}}_{i}-{\mathbf{m}}_{r}=\nabla\psi,$ and ${\mathbf{m}}_{r}={\mathbf{m}}_{t}$ by Lemma 4.2 (iv) (ii)
0	$\neq 0$	$\neq 0$	$\neq 0$	0	$\neq 0$	${\mathbf{m}}_{r}={\mathbf{m}}_{t}$ and ${\mathbf{m}}_{i}-{\mathbf{m}}_{t}=\nabla\psi,$ by Lemma 4.2 (ii) (iii)
$\neq 0$	0	$\neq 0$	0	$\neq 0$	$\neq 0$	${\mathbf{m}}_{i}-{\mathbf{m}}_{t}=\nabla\psi,$ and ${\mathbf{m}}_{r}={\mathbf{m}}_{t}$ by Lemma 4.2 (iii) (ii)
$\neq 0$	0	$\neq 0$	$\neq 0$	$\neq 0$	0	${\mathbf{m}}_{i}-{\mathbf{m}}_{t}=\nabla\psi,$ and ${\mathbf{m}}_{i}-{\mathbf{m}}_{r}=\nabla\psi$ by Lemma 4.2 (iii) (iv)
0	$\neq 0$	$\neq 0$	$\neq 0$	$\neq 0$	0	${\mathbf{m}}_{r}={\mathbf{m}}_{t},$ and ${\mathbf{m}}_{i}-{\mathbf{m}}_{r}=\nabla\psi$ by Lemma 4.2 (ii) (iv)
0	$\neq 0$	$\neq 0$	$\neq 0$	0	$\neq 0$	${\mathbf{m}}_{r}={\mathbf{m}}_{t},$ and ${\mathbf{m}}_{i}-{\mathbf{m}}_{t}=\nabla\psi$ by Lemma 4.2 (ii) (iii)

Therefore $\psi$ is affine and (4.8) follows when (4.12) holds since $\psi=\omega f$ .

It remains to prove (4.8) when (4.12) does not hold. That is, suppose

(4.13)

{\bf A}_{i}\times{\mathbf{n}}=0,\text{ or }{\bf A}_{r}\times{\mathbf{n}}=0,\text{ or }{\bf A}_{t}\times{\mathbf{n}}=0.

Here we use the constitutive equations (4.1) and Part (1) of Theorem 3.1. Notice that the permittivity constant for $\Omega_{-}$ is $\epsilon_{-}$ and for $\Omega_{+}$ is $\epsilon_{+}$ . We recall the assumption that the field ${\bf D}=\epsilon_{-}\,\chi_{\Omega_{-}}\left({\bf E}_{i}+{\bf E}_{r}\right)+\epsilon_{+}\,\chi_{\Omega_{+}}{\bf E}_{t}=\epsilon_{-}\,\chi_{\Omega_{-}}{\mathbf{E}}^{\prime}+\epsilon_{+}\,\chi_{\Omega_{+}}{\bf E}_{t}$ is a distributional solution to the second equation in (3.1) with $\rho$ having singular part equals zero. Then Part (1) of Theorem 3.1 is applicable with $\mu_{t}=0$ and we have

(4.14)

[[{\bf D}(X,t)]]\cdot{\bf n}=0,

where

[[{\bf D}(X,t)]]=\lim_{x\to X,x\in\Omega_{+}}\epsilon_{+}{\mathbf{E}}_{t}(x,t)-\lim_{x\to X,x\in\Omega_{-}}\epsilon_{-}{\mathbf{E}}^{\prime}(x,t).

Since ${\bf n}=(0,0,1)$ , we then have from the form of the fields ${\mathbf{E}}_{t}$ and ${\mathbf{E}}^{\prime}$ that the equation (4.14) reads ( $\psi=\omega f$ )

(4.15)

\epsilon_{-}\,A_{3}^{i}e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+\epsilon_{-}\,A_{3}^{r}e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}-\epsilon_{+}\,A_{3}^{t}e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0,

which will be used to deal with the case (4.13). To begin with let us assume ${\bf A}_{i}\times{\mathbf{n}}=0$ , that is, $A_{1}^{i}=0$ and $A_{2}^{i}=0$ . Since ${\bf A}_{i}\neq 0$ , it follows that $A_{3}^{i}\neq 0$ . Hence from (4.15) it follows that $A_{3}^{r}\neq 0$ or $A_{3}^{t}\neq 0$ . If $A_{3}^{r}\neq 0$ and $A_{3}^{t}\neq 0$ , then by Lemma 4.2 (i) we get that $\psi$ is affine and (4.8) holds. If $A_{3}^{i}\neq 0$ , $A_{3}^{r}\neq 0$ and $A_{3}^{t}=0$ , by Lemma 4.2 (iv) it follows that $m_{j}^{i}-m_{j}^{r}=\psi_{x_{j}}$ for $j=1,2$ . But if $A_{3}^{t}=0$ , since the amplitude ${\bf A}_{t}\neq 0$ , we must have $A_{1}^{t}\neq 0$ or $A_{2}^{t}\neq 0$ . If $A_{1}^{t}\neq 0$ and $A_{2}^{t}=0$ , then using (4.11) we must have $A_{1}^{r}\neq 0$ , and by Lemma 4.2 (ii) $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ and so (4.8) follows. If $A_{1}^{t}=0$ and $A_{2}^{t}\neq 0$ , then from (4.10) $A_{2}^{r}\neq 0$ so by Lemma 4.2 (ii) $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ and so (4.8) follows.

If $A_{3}^{i}\neq 0$ , $A_{3}^{r}=0$ and $A_{3}^{t}\neq 0$ , by Lemma 4.2 (iii) it follows that $m_{j}^{i}-m_{j}^{t}=\psi_{x_{j}}$ for $j=1,2$ . But if $A_{3}^{r}=0$ , since the amplitude ${\bf A}_{r}\neq 0$ , we must have $A_{1}^{r}\neq 0$ or $A_{2}^{r}\neq 0$ . If $A_{1}^{r}\neq 0$ and $A_{2}^{r}=0$ , then using (4.11) we must have $A_{1}^{t}\neq 0$ , and by Lemma 4.2 (ii) $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ and so (4.8) follows. If $A_{1}^{r}=0$ and $A_{2}^{r}\neq 0$ , then from (4.10) $A_{2}^{t}\neq 0$ so by Lemma 4.2 (ii) $m_{j}^{r}=m_{j}^{t}$ for $j=1,2$ and so again (4.8) follows.

The remaining cases in (4.13) are treated similarly. ∎

4.3. Deduction of the generalized Snell law for a general phase discontinuity

Let us now consider a general phase discontinuity function $\phi$ defined on the plane $x_{3}=0$ and let ${\mathbf{E}}_{i}$ be the incident electric field given by (4.2). For a point $P$ on the plane $x_{3}=0$ , we model the scattering of the wave taking into account the value of the gradient of $\phi$ , in other words, the function $f$ in (4.3) and (4.4) is chosen to be $f(x)=\nabla\phi(P)\cdot x$ ; $x=(x_{1},x_{2})$ . We will prove in Section 6 that the amplitudes of the scattered waves (4.3) and (4.4) depend on the choice of the function $f$ and therefore in this case will depend of the point $P$ . Applying Theorem 4.2 with this choice of $f$ we then obtain from (4.8) the generalized Snell’s law for refraction (1.1) and the generalized law of reflection (1.2). In fact, since $n_{i}=c/v_{i}$ , from (4.8) we have $n_{1}k_{j}^{i}-n_{2}k_{j}^{t}=cf_{x_{j}}$ for $j=1,2$ and since ${\mathbf{n}}=(0,0,1)$ taking $\lambda=n_{1}k_{3}^{i}-n_{2}k_{3}^{t}$ , (1.1) follows with $f$ replaced by $cf$ . Similarly, (1.2) follows with $\lambda^{\prime}=n_{1}k_{3}^{i}-n_{1}k_{3}^{r}$ .

We can now extend this argument as follows. Suppose we take a finite number of points $P_{1},\cdots,P_{N}$ on the plane $x_{3}=0$ , and for each point $P_{j}$ we choose the phase function $\nabla\phi(P_{j})\cdot x$ . Each of these phases gives rise to a transmitted and a reflected back waves. Then by superposition, the scattered waves for all the points $P_{j}$ will have the form

\sum_{j=1}^{N}{\bf A}_{t}(P_{j})\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot x}{v_{2}}+\nabla\phi(P_{j})\cdot x-t\right)},

for the transmitted wave and

\sum_{j=1}^{N}{\bf A}_{r}(P_{j})\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot x}{v_{1}}+\nabla\phi(P_{j})\cdot x-t\right)}

for the wave reflected back.

4.4. Calculation of the third components of the wave vectors

Recalling the notation (4.5), and from (4.8) we have

m_{j}^{i}-m_{j}^{r}=\omega f_{x_{j}},\quad m_{j}^{i}-m_{j}^{t}=\omega f_{x_{j}}\qquad j=1,2.

The following corollary, shows formulas for the third component $m_{3}^{\ell}$ for $\ell=r,t$ as a function of $m_{1}^{i}$ , $m_{2}^{i}$ and $f$ .

Corollary 4.4.

Under the assumptions of Theorem 4.3, and if

(4.16)

\sqrt{\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}+\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}}\leq\min\left\{\frac{\omega}{v_{1}},\frac{\omega}{v_{2}}\right\},

then

(4.17)		$\displaystyle m_{3}^{r}=-\sqrt{\left(\frac{\omega}{v_{1}}\right)^{2}-\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}-\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}}$
(4.18)		$\displaystyle m_{3}^{t}=\sqrt{\left(\frac{\omega}{v_{2}}\right)^{2}-\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}-\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}}.$

Proof.

From (4.8)

(4.19)

{\mathbf{m}}_{i}-{\mathbf{m}}_{r}=\lambda\,{\mathbf{n}}+\omega\left(f_{x_{1}},f_{x_{2}},0\right),

where ${\bf n}$ is the vertical unit direction, and $\lambda=m_{3}^{i}-m_{3}^{r}$ . Dotting (4.19) with ${\mathbf{m}}_{i}$ and with ${\mathbf{m}}_{r}$ yields

	$\displaystyle{\mathbf{m}}_{i}\cdot{\mathbf{m}}_{i}-{\mathbf{m}}_{r}\cdot{\mathbf{m}}_{i}$	$\displaystyle=\lambda m_{3}^{i}+\omega\left(f_{x_{1}},f_{x_{2}},0\right)\cdot{\mathbf{m}}_{i}$
	$\displaystyle{\mathbf{m}}_{i}\cdot{\mathbf{m}}_{r}-{\mathbf{m}}_{r}\cdot{\mathbf{m}}_{r}$	$\displaystyle=\lambda m_{3}^{r}+\omega\left(f_{x_{1}},f_{x_{2}},0\right)\cdot{\mathbf{m}}_{r},$

and adding these equations, we obtain that

|{\mathbf{m}}_{i}|^{2}-|{\mathbf{m}}_{r}|^{2}=\lambda\,(m_{3}^{i}+m_{3}^{r})+\omega\left((m_{1}^{i}+m_{1}^{r})f_{x_{1}}+(m_{2}^{i}+m_{2}^{r})f_{x_{2}}\right).

Since $|{\mathbf{m}}_{i}|^{2}=|{\mathbf{m}}_{r}|^{2}=\omega^{2}/v_{1}^{2}$ , it follows that

(4.20)

\lambda\,\left(m_{3}^{i}+m_{3}^{r}\right)+\omega\left((m_{1}^{i}+m_{1}^{r})f_{x_{1}}+(m_{2}^{i}+m_{2}^{r})f_{x_{2}}\right)=0.

Hence, from (4.19)

\left(m_{3}^{i}\right)^{2}-\left(m_{3}^{r}\right)^{2}+\omega\left((m_{1}^{i}+m_{1}^{r})f_{x_{1}}+(m_{2}^{i}+m_{2}^{r})f_{x_{2}}\right)=0.

Since ${\mathbf{k}}_{r}$ is the unit direction of the ray reflected back in medium $I$ , we have $m_{3}^{r}=\dfrac{\omega}{v_{1}}k_{3}^{r}<0$ and so solving for $m_{3}^{r}$ gives

m_{3}^{r}=-\sqrt{\left(m_{3}^{i}\right)^{2}+\omega\left((m_{1}^{i}+m_{1}^{r})f_{x_{1}}+(m_{2}^{i}+m_{2}^{r})f_{x_{2}}\right)}.

Since ${\mathbf{k}}_{i}$ is a unit vector with $k_{3}^{i}>0$ , it follows that $m_{3}^{i}=\sqrt{\left(\omega/v_{1}\right)^{2}-(m_{1}^{i})^{2}-(m_{2}^{i})^{2}}$ and from the fact that $m_{j}^{r}=m_{j}^{i}-\omega f_{x_{j}}$ for $j=1,2$ we get

	$\displaystyle m_{3}^{r}$	$\displaystyle=-\sqrt{\left(\frac{\omega}{v_{1}}\right)^{2}-(m_{1}^{i})^{2}-(m_{2}^{i})^{2}+\omega\left((2m_{1}^{i}-\omega f_{x_{1}})f_{x_{1}}+((2m_{2}^{i}-\omega f_{x_{2}})f_{x_{2}}\right)}$
		$\displaystyle=-\sqrt{\left(\frac{\omega}{v_{1}}\right)^{2}-\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}-\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}},$

which proves (4.17).

To prove (4.18), we proceed similarly using (4.8) and that ${\mathbf{m}}_{t}=\dfrac{\omega}{v_{2}}{\mathbf{k}}_{t}$ , $|{\mathbf{k}}_{t}|=1$ , with $m_{3}^{t}>0$ . ∎

Remark 4.5.

In the standard case, i.e., when $f=0$ , (4.16) obviously reads $\sqrt{(m_{1}^{i})^{2}+(m_{2}^{i})^{2}}\leq\min\left\{\omega/v_{1},\omega/v_{2}\right\}$ . In particular, when the refractive index of medium $I$ is smaller than the one for medium $II$ , that is, when $v_{1}>v_{2}$ , this always happens since $|{\mathbf{m}}_{i}|=\omega/v_{1}$ . On the other hand, if $v_{1}<v_{2}$ , given an incident wave satisfying (4.16) to avoid total internal reflection, the third component must satisfy

{\mathbf{k}}_{i}\cdot{\bf n}=\dfrac{v_{1}}{\omega}m_{3}^{i}=\dfrac{v_{1}}{\omega}\sqrt{\left(\dfrac{\omega}{v_{1}}\right)^{2}-(m_{1}^{i})^{2}-(m_{2}^{i})^{2}}\geq\dfrac{v_{1}}{\omega}\sqrt{\left(\dfrac{\omega}{v_{1}}\right)^{2}-\left(\dfrac{\omega}{v_{2}}\right)^{2}}=\sqrt{1-\left(\dfrac{v_{1}}{v_{2}}\right)^{2}}.

This condition agrees with the one obtained for the standard Snell’s law, see [6, Section 2]. On the other hand, when $f\neq 0$ , the compatibility conditions for ${\mathbf{m}}_{i}$ and $f$ in (4.16) must be satisfied in order to have reflected and transmitted waves.

We also remark that from (4.17) and (4.18) the following relation between the third components of ${\mathbf{m}}_{r}$ and ${\mathbf{m}}_{t}$ holds:

(4.21)

m_{3}^{t}=\sqrt{\omega^{2}/v_{2}^{2}-\omega^{2}/v_{1}^{2}+\left(m_{3}^{r}\right)^{2}}.

4.5. Orthogonality conditions for the amplitudes

The following conditions must be satisfied by the amplitudes, which will be utilized in Section 6.

Lemma 4.6.

Recall from (4.7) the definitions of ${\mathbf{E}}^{\prime}$ and ${\mathbf{H}}^{\prime}$ and the fields ${\mathbf{E}}(x,t)=\chi_{\Omega_{-}}{\mathbf{E}}^{\prime}(x,t)+\chi_{\Omega_{+}}{\mathbf{E}}_{t}(x,t)$ , and ${\mathbf{H}}(x,t)=\chi_{\Omega_{-}}{\mathbf{H}}^{\prime}(x,t)+\chi_{\Omega_{+}}{\mathbf{H}}_{t}(x,t)$ .

If ${\mathbf{E}}$ and ${\mathbf{H}}$ are distributional solutions to (M.3), and ${\mathbf{E}}$ is a distributional solution to (M.1) with $\rho$ non-singular, then the amplitudes satisfy the following orthogonality conditions

(4.22)

A_{1}^{i}m_{1}^{i}+A_{2}^{i}m_{2}^{i}+A_{3}^{i}m_{3}^{i}=0,

(4.23)

A_{1}^{r}m_{1}^{i}+A_{2}^{r}m_{2}^{i}+A_{3}^{r}m_{3}^{r}=0,

and

(4.24)

A_{1}^{t}m_{1}^{i}+A_{2}^{t}m_{2}^{i}+A_{3}^{t}m_{3}^{t}=0.

Proof.

From Theorem 4.3, (4.8) holds and from the form of the fields (4.2), (4.3), (4.4), and the notation (4.5) it follows that

	$\displaystyle{\mathbf{E}}_{i}(x,t)$	$\displaystyle={\bf A}_{i}e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}+m_{3}^{i}x_{3}-\omega\,t\right)},\quad{\mathbf{E}}_{r}(x,t)={\bf A}_{r}e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}+m_{3}^{r}x_{3}-\omega\,t\right)}$
	$\displaystyle{\mathbf{E}}_{t}(x,t)$	$\displaystyle={\bf A}_{t}e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}+m_{3}^{t}x_{3}-\omega\,t\right)}.$

Since ${\mathbf{E}}$ is a distributional solution to (M.1), it follows that ${\mathbf{E}}^{\prime}$ satisfies (M.1) pointwise in $\Omega_{-}$ and ${\mathbf{E}}_{t}$ satisfies (M.1) pointwise in $\Omega_{+}$ . We then have for $x_{3}<0$ that

	$\displaystyle 0$	$\displaystyle=\text{\rm div}\,{\mathbf{E}}^{\prime}=\text{\rm div}\,{\mathbf{E}}_{i}+\text{\rm div}\,{\mathbf{E}}_{r}$
		$\displaystyle={\rm i}\,\left(m_{1}^{i},m_{2}^{i},m_{3}^{i}\right)\cdot{\bf A}_{i}\,e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}+m_{3}^{i}x_{3}-\omega\,t\right)}+{\rm i}\,\left(m_{1}^{i},m_{2}^{i},m_{3}^{r}\right)\cdot{\bf A}_{r}\,e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}+m_{3}^{r}x_{3}-\omega\,t\right)}$
		$\displaystyle={\rm i}\,e^{{\rm i}\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}-\omega\,t\right)}\left(\left(m_{1}^{i},m_{2}^{i},m_{3}^{i}\right)\cdot{\bf A}_{i}\,e^{{\rm i}m_{3}^{i}x_{3}}+\left(m_{1}^{i},m_{2}^{i},m_{3}^{r}\right)\cdot{\bf A}_{r}\,e^{{\rm i}m_{3}^{r}x_{3}}\right)$

which implies

\left(m_{1}^{i},m_{2}^{i},m_{3}^{i}\right)\cdot{\bf A}_{i}\,e^{{\rm i}\,m_{3}^{i}x_{3}}+\left(m_{1}^{i},m_{2}^{i},m_{3}^{r}\right)\cdot{\bf A}_{r}\,e^{{\rm i}\,m_{3}^{r}x_{3}}=0

for all $x_{3}<0$ . Since $m_{3}^{i}>0$ and $m_{3}^{r}<0$ , the exponentials in the last identity are linearly independent and therefore the coefficients must be zero, that is, (4.22) and (4.23) follow.

Since $\text{\rm div}\,{\mathbf{E}}_{t}=0$ for $x_{3}>0$ , (4.24) also follows.

∎

5. Boundary conditions for the magnetic fields

In this section, we derive the boundary conditions for the magnetic fields presented in Lemma 4.1 from Theorem 3.1. These boundary conditions will be utilized in Section 6 in the calculation of the amplitudes.

Proposition 5.1.

Under the assumptions of Theorem 4.3, if ${\mathbf{E}}$ and ${\mathbf{H}}$ are solutions to (M.4), with current density ${\bf J}$ satisfying the assumptions in Theorem 3.1 with $\nu_{t}=0$ , then

(5.1)		$\displaystyle-(1/\mu_{+})\left(A_{3}^{t}m_{1}^{i}-m_{3}^{t}A_{1}^{t}\right)+(1/\mu_{-})\left(A_{3}^{i}m_{1}^{i}-m_{3}^{i}A_{1}^{i}+A_{3}^{r}m_{1}^{i}-m_{3}^{r}A_{1}^{r}\right)$	$\displaystyle=0$
(5.2)		$\displaystyle-(1/\mu_{+})\left(A_{3}^{t}m_{2}^{i}-m_{3}^{t}A_{2}^{t}\right)+(1/\mu_{-})\left(A_{3}^{i}m_{2}^{i}-m_{3}^{i}A_{2}^{i}+A_{3}^{r}m_{2}^{i}-m_{3}^{r}A_{2}^{r}\right)$	$\displaystyle=0.$

Proof.

From (4.1) we recall that ${\mathbf{D}}=\epsilon_{-}\,{\mathbf{E}}$ in $\Omega_{-}$ , ${\mathbf{D}}=\epsilon_{+}\,{\mathbf{E}}$ in $\Omega_{+}$ , and ${\mathbf{B}}=\mu_{-}\,{\mathbf{H}}$ in $\Omega_{-}$ , ${\mathbf{B}}=\mu_{+}\,{\mathbf{H}}$ in $\Omega_{+}$ . Since ${\mathbf{J}}$ has not a singular part, then Theorem 3.1 Part (3) is applicable and we get that

(5.3)

[[{\mathbf{H}}(X,t)]]\times{\mathbf{n}}(X)=0

for $X$ in the interface plane $\Gamma=\{x_{3}=0\}$ .

From the expression of the electric fields in (4.2), (4.3), (4.4), and the corresponding magnetic fields obtained in Lemma 4.1, we have that for every $X=(x_{1},x_{2},0)\in\Gamma$

	$\displaystyle\lim_{x\to X,x\in\Omega_{-}}{\mathbf{H}}^{\prime}(x,t)$	$\displaystyle=-\dfrac{c}{\mu_{-}}\left({\bf A}_{i}\times\frac{{\mathbf{k}}_{i}}{v_{1}}\,\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot X}{v_{1}}-t\right)}+{\bf A}_{r}\times\left(\frac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(X)\right)\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)-t\right)}\right)$
	$\displaystyle\lim_{x\to X,x\in\Omega_{+}}{\mathbf{H}}_{t}(x,t)$	$\displaystyle=-\dfrac{c}{\mu_{+}}\,{\bf A}_{t}\times\left(\dfrac{{\mathbf{k}}_{t}}{v_{2}}+\nabla f(X)\right)\,e^{{\rm i}\omega\left(\frac{{\mathbf{k}}_{t}\cdot X}{v_{2}}+f(X)-t\right)},$

where $\nabla f(X)=\left(f_{x_{1}},f_{x_{2}},0\right)$ . Substituting these into (5.3), we get

	$\displaystyle 0$	$\displaystyle=-\dfrac{c}{\mu_{+}}\,\left({\bf A}_{t}\times\left(\dfrac{{\mathbf{k}}_{t}}{v_{2}}+\nabla f(X)\right)\right)\times{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot X}{v_{2}}+f(X)-t\right)}$
		$\displaystyle\qquad+\dfrac{c}{\mu_{-}}\,\left(\left({\bf A}_{i}\times\frac{{\mathbf{k}}_{i}}{v_{1}}\right)\times{\mathbf{n}}\,\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot X}{v_{1}}-t\right)}+\left({\bf A}_{r}\times\left(\frac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(X)\right)\right)\times{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)-t\right)}\right).$

Then from (4.8)

		$\displaystyle-\dfrac{c}{\mu_{+}}\,\left({\bf A}_{t}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{t}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)\right)}$
		$\displaystyle\quad+\dfrac{c}{\mu_{-}}\,\left(\left({\bf A}_{i}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{i}^{3}}{v_{1}}\right)\right)\times{\mathbf{n}}\,\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+\nabla f(X)\cdot X\right)}+\left({\bf A}_{r}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{r}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)\right)}\right)=0.$

From Theorem 4.3, $f(X)$ is affine, and since $f(0)=0$ we have $f(X)=\nabla f(X)\cdot X$ . So canceling the exponential and the constant $c$ in the last equation we obtain

(5.4)

-\dfrac{1}{\mu_{+}}\,\left({\bf A}_{t}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{t}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}+\dfrac{1}{\mu_{-}}\,\left(\left({\bf A}_{i}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{i}^{3}}{v_{1}}\right)\right)\times{\mathbf{n}}+\left({\bf A}_{r}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{r}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}\right)=0.

Let us calculate the triple cross products, we use the formula $(a\times b)\times c=b(c\cdot a)-a(b\cdot c)$ and obtain

	$\displaystyle\left({\bf A}_{t}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{t}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}$	$\displaystyle=\left({\bf A}_{t}\cdot{\mathbf{n}}\right)\,\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{t}^{3}}{v_{2}}\right)-\left(\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{t}^{3}}{v_{2}}\right)\cdot{\mathbf{n}}\right)\,{\bf A}_{t}$
		$\displaystyle=\left(A_{3}^{t}\frac{k_{1}^{i}}{v_{1}}-\frac{k_{3}^{t}}{v_{2}}A_{1}^{t},A_{3}^{t}\frac{k_{2}^{i}}{v_{1}}-\frac{k_{3}^{t}}{v_{2}}A_{2}^{t},0\right)$
	$\displaystyle\left({\bf A}_{r}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{r}^{3}}{v_{2}}\right)\right)\times{\mathbf{n}}$	$\displaystyle=\left(A_{3}^{r}\frac{k_{1}^{i}}{v_{1}}-\frac{k_{3}^{r}}{v_{1}}A_{1}^{r},A_{3}^{r}\frac{k_{2}^{i}}{v_{1}}-\frac{k_{3}^{r}}{v_{1}}A_{2}^{r},0\right)$
	$\displaystyle\left({\bf A}_{i}\times\left(\dfrac{k_{i}^{1}}{v_{1}},\dfrac{k_{i}^{2}}{v_{1}},\dfrac{k_{i}^{3}}{v_{1}}\right)\right)\times{\mathbf{n}}$	$\displaystyle=\left(A_{3}^{i}\frac{k_{1}^{i}}{v_{1}}-\frac{k_{3}^{i}}{v_{1}}A_{1}^{i},A_{3}^{i}\frac{k_{2}^{i}}{v_{1}}-\frac{k_{3}^{i}}{v_{1}}A_{2}^{i},0\right).$

Replacing the obtained formulas above in (5.4), and using the notation (4.5), equations (5.1) and (5.2) follow. ∎

Remark 5.2.

We observe that analyzing the boundary condition (3.3) does not provide any additional information. Specifically, assume that the magnetic field ${\bf H}$ presented in Theorem 4.3 is a distributional solution to (M.2). The jump on ${\mathbf{B}}$ is given by

[[{\mathbf{B}}(X,t)]]=\lim_{x\to X,x\in\Omega_{+}}\mu_{+}{\mathbf{H}}_{t}(x,t)-\lim_{x\to X,x\in\Omega_{-}}\mu_{-}{\mathbf{H}}^{\prime}(x,t),\quad X\in\Gamma,

and so from Lemma 4.1 and (3.3)

	$\displaystyle[[{\mathbf{B}}(X,t)]]\cdot{\mathbf{n}}$	$\displaystyle=-c\,\left({\bf A}_{t}\times\left(\dfrac{{\mathbf{k}}_{t}}{v_{2}}+\nabla f(X)\right)\right)\cdot{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{t}\cdot X}{v_{2}}+f(X)-t\right)}$
		$\displaystyle\qquad+c\,\left(\left({\bf A}_{i}\times\frac{{\mathbf{k}}_{i}}{v_{1}}\right)\cdot{\mathbf{n}}\,\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{i}\cdot X}{v_{1}}-t\right)}+\left({\bf A}_{r}\times\left(\frac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(X)\right)\right)\cdot{\mathbf{n}}\,e^{{\rm i}\,\omega\left(\frac{{\mathbf{k}}_{r}\cdot X}{v_{1}}+f(X)-t\right)}\right)=0.$

As in the proof of Proposition 5.1, the exponentials are all equal and so cancelling them yields

-\left({\bf A}_{t}\times\left(\dfrac{{\mathbf{k}}_{t}}{v_{2}}+\nabla f(X)\right)\right)\cdot{\mathbf{n}}+\left({\bf A}_{i}\times\frac{{\mathbf{k}}_{i}}{v_{1}}\right)\cdot{\mathbf{n}}+\left({\bf A}_{r}\times\left(\frac{{\mathbf{k}}_{r}}{v_{1}}+\nabla f(X)\right)\right)\cdot{\mathbf{n}}=0.

From the triple product formula $a\cdot(b\times c)=b\cdot(c\times a)$ and (4.8), it follows that

	$\displaystyle 0$	$\displaystyle=-{\bf A}_{t}\cdot\left(\left(\dfrac{k_{1}^{i}}{v_{1}},\dfrac{k_{2}^{i}}{v_{1}},\dfrac{k_{3}^{t}}{v_{2}}\right)\times{\mathbf{n}}\right)+{\bf A}_{i}\cdot\left(\left(\frac{k_{1}^{i}}{v_{1}},\frac{k_{2}^{i}}{v_{1}},\frac{k_{3}^{i}}{v_{1}}\right)\times{\mathbf{n}}\right)+{\bf A}_{r}\cdot\left(\left(\frac{k_{1}^{i}}{v_{1}},\frac{k_{2}^{i}}{v_{1}},\frac{k_{3}^{r}}{v_{1}}\right)\times{\mathbf{n}}\right)$
		$\displaystyle=-{\bf A}_{t}\cdot\left(\frac{k_{2}^{i}}{v_{1}},-\frac{k_{1}^{i}}{v_{1}},0\right)+{\bf A}_{i}\cdot\left(\frac{k_{2}^{i}}{v_{1}},-\frac{k_{1}^{i}}{v_{1}},0\right)+{\bf A}_{r}\cdot\left(\frac{k_{2}^{i}}{v_{1}},-\frac{k_{1}^{i}}{v_{1}},0\right)$
		$\displaystyle=\left(-{\bf A}_{t}+{\bf A}_{i}+{\bf A}_{r}\right)\left(\frac{k_{2}^{i}}{v_{1}},-\frac{k_{1}^{i}}{v_{1}},0\right)$

which written in terms of $m$ ’s is

(5.5)

m_{2}^{i}\,\left(-A_{1}^{t}+A_{1}^{i}+A_{1}^{r}\right)-m_{1}^{i}\,\left(-A_{2}^{t}+A_{2}^{i}+A_{2}^{r}\right)=0.

From equations (4.10) and (4.11), equation (5.5) is satisfied. Consequently, the boundary condition (3.3) does not provide any additional equations for the amplitudes.

6. Calculation of the amplitude coefficients

Recall these amplitudes are ${\bf A}_{i}=(A_{1}^{i},A_{2}^{i},A_{3}^{i}),{\bf A}_{r}=(A_{1}^{r},A_{2}^{r},A_{3}^{r})$ , and ${\bf A}_{t}=(A_{1}^{t},A_{2}^{t},A_{3}^{t})$ , where the unknowns are ${\bf A}_{r}$ and ${\bf A}_{t}$ . The purpose of the section is to find explicit formulas for ${\bf A}_{r}$ and ${\bf A}_{t}$ in terms of ${\bf A}_{i}$ and the wave vectors. Listing all the relationships obtained for the amplitudes in a table we get

\begin{matrix}A_{1}^{r}&A_{2}^{r}&A_{3}^{r}&A_{1}^{t}&A_{2}^{t}&A_{3}^{t}&A_{1}^{i}&A_{2}^{i}&A_{3}^{i}&\text{From equation}\\ \hline\cr 1&0&0&-1&0&0&1&0&0&\eqref{eq:equation first components}\\ 0&-1&0&0&1&0&0&-1&0&\eqref{eq:equation second components}\\ 0&0&\epsilon_{-}&0&0&-\epsilon_{+}&0&0&\epsilon_{-}&\eqref{eq:equation third components}\\ -m_{3}^{r}/\mu_{-}&0&m_{1}^{i}/\mu_{-}&m_{3}^{t}/\mu_{+}&0&-m_{1}^{i}/\mu_{+}&-m_{3}^{i}/\mu_{-}&0&m_{1}^{i}/\mu_{-}&\eqref{eq:relations with mupmA}\\ 0&-m_{3}^{r}/\mu_{-}&m_{2}^{i}/\mu_{-}&0&m_{3}^{t}/\mu_{+}&-m_{2}^{i}/\mu_{+}&0&-m_{3}^{i}/\mu_{-}&m_{2}^{i}/\mu_{-}&\eqref{eq:relations with mupmB}\\ 0&0&0&0&0&0&m_{1}^{i}&m_{2}^{i}&m_{3}^{i}&\eqref{eq:orthogonality for Ai}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{r}&0&0&0&0&0&0&\eqref{eq:orthogonality for Ar}\\ 0&0&0&m_{1}^{i}&m_{2}^{i}&m_{3}^{t}&0&0&0&\eqref{eq:orthogonality for At}\end{matrix}.

We remark that the first five rows in the table, (4.11), (4.10), (4.15), (5.1), and (5.2) are all consequences of the boundary conditions, and the last three rows in the table, (4.22), (4.23), and (4.24), are orthogonality type conditions derived in Lemma 4.6 from the fact that ${\mathbf{E}},{\mathbf{H}}$ satisfy the Maxwell system. Notice that we have six unknowns, the components of ${\bf A}_{t}$ and ${\bf A}_{r}$ , and eight equations. The system will be first reduced to solve (6.2) for ${\bf A}_{t}$ , and substitute this value in (6.1) to get ${\bf A}_{r}$ . A necessary and sufficient condition for the solvability of the system (6.2) is given in the following section. The coefficient matrix of the system is then

M=\begin{bmatrix}1&0&0&-1&0&0&1&0&0\\ 0&-1&0&0&1&0&0&-1&0\\ 0&0&\epsilon_{-}&0&0&-\epsilon_{+}&0&0&\epsilon_{-}\\ -m_{3}^{r}/\mu_{-}&0&m_{1}^{i}/\mu_{-}&m_{3}^{t}/\mu_{+}&0&-m_{1}^{i}/\mu_{+}&-m_{3}^{i}/\mu_{-}&0&m_{1}^{i}/\mu_{-}\\ 0&-m_{3}^{r}/\mu_{-}&m_{2}^{i}/\mu_{-}&0&m_{3}^{t}/\mu_{+}&-m_{2}^{i}/\mu_{+}&0&-m_{3}^{i}/\mu_{-}&m_{2}^{i}/\mu_{-}\\ 0&0&0&0&0&0&m_{1}^{i}&m_{2}^{i}&m_{3}^{i}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{r}&0&0&0&0&0&0\\ 0&0&0&m_{1}^{i}&m_{2}^{i}&m_{3}^{t}&0&0&0\end{bmatrix}.

For the calculation of the amplitudes it is convenient to write this matrix in terms of blocks. If we set

	$\displaystyle M_{1}$	$\displaystyle=\begin{bmatrix}1&0&0\\ 0&-1&0\\ 0&0&\epsilon_{-}\end{bmatrix},\qquad M_{2}=\begin{bmatrix}-1&0&0\\ 0&1&0\\ 0&0&-\epsilon_{+}\end{bmatrix}$
	$\displaystyle N_{1}$	$\displaystyle=\begin{bmatrix}-m_{3}^{r}/\mu_{-}&0&m_{1}^{i}/\mu_{-}\\ 0&-m_{3}^{r}/\mu_{-}&m_{2}^{i}/\mu_{-}\\ 0&0&0\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{r}\\ 0&0&0\end{bmatrix},\qquad N_{2}=\begin{bmatrix}m_{3}^{t}/\mu_{+}&0&-m_{1}^{i}/\mu_{+}\\ 0&m_{3}^{t}/\mu_{+}&-m_{2}^{i}/\mu_{+}\\ 0&0&0\\ 0&0&0\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{t}\end{bmatrix},$
	$\displaystyle N_{3}$	$\displaystyle=\begin{bmatrix}-m_{3}^{i}/\mu_{-}&0&m_{1}^{i}/\mu_{-}\\ 0&-m_{3}^{i}/\mu_{-}&m_{2}^{i}/\mu_{-}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{i}\\ 0&0&0\\ 0&0&0\end{bmatrix},$

then we can write $M=\begin{bmatrix}M_{1}&M_{2}&M_{1}\\ N_{1}&N_{2}&N_{3}\end{bmatrix}$ . Therefore the amplitudes must verify the equations, written with ${\bf A}_{r},{\bf A}_{t},{\bf A}_{i}$ column 3-vectors,

\begin{bmatrix}M_{1}&M_{2}&M_{1}\\ N_{1}&N_{2}&N_{3}\end{bmatrix}\begin{bmatrix}{\bf A}_{r}\\ {\bf A}_{t}\\ {\bf A}_{i}\end{bmatrix}=0\in{\mathbb{R}}^{8}

which means

M_{1}{\bf A}_{r}+M_{2}{\bf A}_{t}+M_{1}{\bf A}_{i}=0\in{\mathbb{R}}^{3},\qquad N_{1}{\bf A}_{r}+N_{2}{\bf A}_{t}+N_{3}{\bf A}_{i}=0\in{\mathbb{R}}^{5}.

From the first equation and since the matrix $M_{1}$ is invertible

(6.1)

{\bf A}_{r}=-M_{1}^{-1}M_{2}{\bf A}_{t}-{\bf A}_{i}

which substituted in the second equation yields

(6.2)

\left(N_{2}-N_{1}M_{1}^{-1}M_{2}\right){\bf A}_{t}=\left(N_{1}-N_{3}\right){\bf A}_{i}.

Now

P=N_{2}-N_{1}M_{1}^{-1}M_{2}=\begin{bmatrix}-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&0&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{1}^{i}\\ 0&-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{2}^{i}\\ 0&0&0\\ m_{1}^{i}&m_{2}^{i}&\dfrac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{t}\end{bmatrix}

and

Q=N_{1}-N_{3}=\begin{bmatrix}\dfrac{m_{3}^{i}-m_{3}^{r}}{\mu_{-}}&0&0\\ 0&\dfrac{m_{3}^{i}-m_{3}^{r}}{\mu_{-}}&0\\ -m_{1}^{i}&-m_{2}^{i}&-m_{3}^{i}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{r}\\ 0&0&0\end{bmatrix}.

Q{\bf A}_{i}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ -A_{1}^{i}m_{1}^{i}-A_{2}^{i}m_{2}^{i}-A_{3}^{i}m_{3}^{i}\\ A_{1}^{i}m_{1}^{i}+A_{2}^{i}m_{2}^{i}+A_{3}^{i}m_{3}^{r}\\ 0\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ 0\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\\ 0\end{bmatrix}.

Therefore, (6.2) has a solution ${\bf A}_{t}$ if the vector $QA_{i}$ is in the column space of the matrix $P$ , and hence ${\bf A}_{r}$ follows from (6.1).

6.1. Solvability of the system (6.2)

We shall prove the following proposition.

Proposition 6.1.

The system (6.2) is solvable if and only if

(6.3)

A_{3}^{i}m_{3}^{i}\dfrac{1}{\mu_{-}}=\left(-\delta\,m_{3}^{t}+\alpha_{1}m_{1}^{i}+\alpha_{2}m_{2}^{i}\right)A_{3}^{i}\dfrac{1}{\alpha_{3}-m_{3}^{t}}

holds, where

\displaystyle\delta=\frac{m_{3}^{t}}{\mu_{+}}-\frac{m_{3}^{r}}{\mu_{-}},\quad\alpha_{1}=\left(\frac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\frac{1}{\mu_{+}}\right)m_{1}^{i},\quad\alpha_{2}

\displaystyle=\left(\frac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\frac{1}{\mu_{+}}\right)m_{2}^{i},\quad\alpha_{3}=\frac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}};

notice that $\delta>0$ and $\alpha_{3}<0$ since $m_{3}^{t}>0$ and $m_{3}^{r}<0$ . Moreover, we obtain the following formula for the amplitude ${\bf A}_{t}$ :

(6.4)

\begin{cases}A_{1}^{t}&=\dfrac{m_{3}^{i}-m_{3}^{r}}{-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}}\left(\dfrac{A_{1}^{i}}{\mu_{-}}-\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{1}^{i}\dfrac{\epsilon_{-}}{\epsilon_{-}m_{3}^{t}-\epsilon_{+}m_{3}^{r}}A_{3}^{i}\right)\\ A_{2}^{t}&=\dfrac{m_{3}^{i}-m_{3}^{r}}{-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}}\left(\dfrac{A_{2}^{i}}{\mu_{-}}-\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{2}^{i}\dfrac{\epsilon_{-}}{\epsilon_{-}m_{3}^{t}-\epsilon_{+}m_{3}^{r}}A_{3}^{i}\right)\\ A_{3}^{t}&=\dfrac{\epsilon_{-}\left(m_{3}^{i}-m_{3}^{r}\right)}{\epsilon_{-}m_{3}^{t}-\epsilon_{+}m_{3}^{r}}A_{3}^{i}.\end{cases}

Inserting this value of ${\bf A}_{t}$ into (6.1) we obtain the amplitude ${\bf A}_{r}$ .

Proof.

Since the third row of the system (6.2) is zero, the system of equations to solve is

\begin{bmatrix}-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&0&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{1}^{i}\\ 0&-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{2}^{i}\\ m_{1}^{i}&m_{2}^{i}&\dfrac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}}\\ m_{1}^{i}&m_{2}^{i}&m_{3}^{t}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\\ 0\end{bmatrix}.

Subtracting the third equation from the fourth we need to solve

\begin{bmatrix}-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&0&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{1}^{i}\\ 0&-\dfrac{m_{3}^{r}}{\mu_{-}}+\dfrac{m_{3}^{t}}{\mu_{+}}&\left(\dfrac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\dfrac{1}{\mu_{+}}\right)m_{2}^{i}\\ m_{1}^{i}&m_{2}^{i}&\dfrac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}}\\ 0&0&m_{3}^{t}-\dfrac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\\ -A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\end{bmatrix},

which written in terms of $\delta,\alpha_{1},\alpha_{2},\alpha_{3}$ is the system

\begin{bmatrix}\delta&0&\alpha_{1}\\ 0&\delta&\alpha_{2}\\ m_{1}^{i}&m_{2}^{i}&\alpha_{3}\\ 0&0&m_{3}^{t}-\alpha_{3}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\\ -A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\end{bmatrix}.

Dividing rows one and two by $\delta$ gives

\begin{bmatrix}1&0&\alpha_{1}/\delta\\ 0&1&\alpha_{2}/\delta\\ m_{1}^{i}&m_{2}^{i}&\alpha_{3}\\ 0&0&m_{3}^{t}-\alpha_{3}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\\ -A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\end{bmatrix};

multiplying the first row by $-m_{1}^{i}$ and add in it to the third row gives

\begin{bmatrix}1&0&\alpha_{1}/\delta\\ 0&1&\alpha_{2}/\delta\\ 0&m_{2}^{i}&\alpha_{3}-m_{1}^{i}(\alpha_{1}/\delta)\\ 0&0&m_{3}^{t}-\alpha_{3}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)-m_{1}^{i}\dfrac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ -A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\end{bmatrix};

multiplying the second row by $-m_{2}^{i}$ and add in it to the third row gives

\begin{bmatrix}1&0&\alpha_{1}/\delta\\ 0&1&\alpha_{2}/\delta\\ 0&0&\alpha_{3}-m_{1}^{i}(\alpha_{1}/\delta)-m_{2}^{i}(\alpha_{2}/\delta)\\ 0&0&m_{3}^{t}-\alpha_{3}\end{bmatrix}\begin{bmatrix}A_{1}^{t}\\ A_{2}^{t}\\ A_{3}^{t}\end{bmatrix}=\begin{bmatrix}\displaystyle\frac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ \displaystyle\frac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)-m_{1}^{i}\dfrac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}-m_{2}^{i}\dfrac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}\\ -A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)\end{bmatrix}.

From the fourth equation, the value of $A_{3}^{t}$ is given by

A_{3}^{t}=\dfrac{-A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)}{m_{3}^{t}-\alpha_{3}}.

We shall verify that (6.3) implies that this value of $A_{3}^{t}$ satisfies the third equation, i.e., under (6.3) equations three and four are equivalent. In fact, substituting this value of $A_{3}^{t}$ on the left hand side of the third equation we need to prove the identity

\left(\alpha_{3}-m_{1}^{i}(\alpha_{1}/\delta)-m_{2}^{i}(\alpha_{2}/\delta)\right)\dfrac{A_{3}^{i}\left(m_{3}^{i}-m_{3}^{r}\right)}{m_{3}^{t}-\alpha_{3}}=A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)-m_{1}^{i}\dfrac{A_{1}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}-m_{2}^{i}\dfrac{A_{2}^{i}(m_{3}^{i}-m_{3}^{r})}{\delta\mu_{-}}.

Since $m_{1}^{i}A_{1}^{i}+m_{2}^{i}A_{2}^{i}=-m_{3}^{i}A_{3}^{i}$ , this identity is equivalent to

\left(\alpha_{3}-m_{1}^{i}(\alpha_{1}/\delta)-m_{2}^{i}(\alpha_{2}/\delta)\right)\dfrac{A_{3}^{i}\left(m_{3}^{i}-m_{3}^{r}\right)}{m_{3}^{t}-\alpha_{3}}=A_{3}^{i}\left(m_{3}^{r}-m_{3}^{i}\right)+m_{3}^{i}A_{3}^{i}\dfrac{m_{3}^{i}-m_{3}^{r}}{\delta\mu_{-}}.

Now notice that moving terms around and since $m_{3}^{i}-m_{3}^{r}\neq 0$ , the last identity is equivalent to (6.3). Hence (6.4) follows.

Notice that given ${\bf A}_{i}$ , the amplitude solution ${\bf A}_{t}$ , and consequently ${\bf A}_{r}$ , are unique since the null space of the matrix

\begin{bmatrix}1&0&\alpha_{1}/\delta\\ 0&1&\alpha_{2}/\delta\\ 0&0&\alpha_{3}-m_{1}^{i}(\alpha_{1}/\delta)-m_{2}^{i}(\alpha_{2}/\delta)\\ 0&0&m_{3}^{t}-\alpha_{3}\end{bmatrix}

is zero since $m_{3}^{r}<0$ and so $m_{3}^{t}-\alpha_{3}>0$ .

∎

6.2. Analysis of the condition (6.3)

In case $A_{3}^{i}=0$ (the incident wave is transverse electric (TE), that is, it is perpendicular to the normal to the interface), condition (6.3) obviously holds and from (6.4) the amplitudes are

\left(A_{1}^{t},A_{2}^{t},A_{3}^{t}\right)=\dfrac{\mu_{+}\left(m_{3}^{i}-m_{3}^{r}\right)}{\mu_{-}m_{3}^{t}-\mu_{+}m_{3}^{r}}\left(A_{1}^{i},A_{2}^{i},0\right),

and hence from (6.1)

\left(A_{1}^{r},A_{2}^{r},A_{3}^{r}\right)=\dfrac{\mu_{+}\left(m_{3}^{i}-m_{3}^{r}\right)}{\mu_{-}m_{3}^{t}-\mu_{+}m_{3}^{r}}\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&\epsilon_{+}/\epsilon_{-}\end{bmatrix}{\bf A}_{i}-{\bf A}_{i}=\left(\dfrac{\mu_{+}\left(m_{3}^{i}-m_{3}^{r}\right)}{\mu_{-}m_{3}^{t}-\mu_{+}m_{3}^{r}}-1\right)\left(A_{1}^{i},A_{2}^{i},0\right).

In case $A_{3}^{i}\neq 0$ , if (6.3) holds, then some relationships between the wave vectors $m^{\ell}$ , $\ell=i,r,t$ , must be satisfied. Indeed, cancelling $A_{3}^{i}$ in (6.3) for the solvability of the system we must have

	$\displaystyle m_{3}^{i}\dfrac{\frac{\epsilon_{+}m_{3}^{r}}{\epsilon_{-}}-m_{3}^{t}}{\mu_{-}}=m_{3}^{i}\dfrac{\alpha_{3}-m_{3}^{t}}{\mu_{-}}$	$\displaystyle=-\delta\,m_{3}^{t}+\alpha_{1}m_{1}^{i}+\alpha_{2}m_{2}^{i}$
		$\displaystyle=-\left(\frac{m_{3}^{t}}{\mu_{+}}-\frac{m_{3}^{r}}{\mu_{-}}\right)\,m_{3}^{t}+\left(\frac{\epsilon_{+}}{\epsilon_{-}\mu_{-}}-\frac{1}{\mu_{+}}\right)\left(\left(m_{1}^{i}\right)^{2}+\left(m_{2}^{i}\right)^{2}\right).$

Re writing this expression we get

(6.5)

m_{3}^{i}\dfrac{\epsilon_{+}m_{3}^{r}-\epsilon_{-}m_{3}^{t}}{\mu_{-}\epsilon_{-}}=-\left(\frac{m_{3}^{t}}{\mu_{+}}-\frac{m_{3}^{r}}{\mu_{-}}\right)\,m_{3}^{t}+\left(\frac{\epsilon_{+}\mu_{+}-\epsilon_{-}\mu_{-}}{\epsilon_{-}\mu_{-}\mu_{+}}\right)\left(\left(m_{1}^{i}\right)^{2}+\left(m_{2}^{i}\right)^{2}\right).

Since $(m_{1}^{i})^{2}+(m_{2}^{i})^{2}+(m_{3}^{i})^{2}=\omega^{2}/v_{1}^{2}=\omega^{2}\,\epsilon_{-}\mu_{-}$ , we have $(m_{1}^{i})^{2}+(m_{2}^{i})^{2}=\omega^{2}\,\epsilon_{-}\mu_{-}-(m_{3}^{i})^{2}$ which substituted in the last expression yields

m_{3}^{i}\dfrac{\epsilon_{+}m_{3}^{r}-\epsilon_{-}m_{3}^{t}}{\mu_{-}\epsilon_{-}}=-\left(\frac{m_{3}^{t}}{\mu_{+}}-\frac{m_{3}^{r}}{\mu_{-}}\right)\,m_{3}^{t}+\left(\frac{\epsilon_{+}\mu_{+}-\epsilon_{-}\mu_{-}}{\epsilon_{-}\mu_{-}\mu_{+}}\right)\left(\omega^{2}\,\epsilon_{-}\mu_{-}-(m_{3}^{i})^{2}\right).

From (4.21), $m_{3}^{t}=\sqrt{\omega^{2}(\epsilon_{+}\mu_{+}-\epsilon_{-}\mu_{-})+(m_{3}^{r})^{2}}$ , which substituted in the last expression we get after simplification that the difference between the left-hand and right-hand sides becomes:

(m_{3}^{i}+m_{3}^{r})\left(\epsilon_{+}\mu_{+}m_{3}^{i}-\epsilon_{-}\mu_{-}m_{3}^{i}+\epsilon_{-}\mu_{-}m_{3}^{r}-\epsilon_{-}\mu_{+}m_{3}^{t}\right)=0

Thus, (6.5) holds if and only if:

(6.6)

m_{3}^{i}+m_{3}^{r}=0

(6.7)

\left(\epsilon_{+}\mu_{+}-\epsilon_{-}\mu_{-}\right)m_{3}^{i}=\epsilon_{-}\left(\mu_{+}m_{3}^{t}-\mu_{-}m_{3}^{r}\right).

From Equation (4.17) $m_{3}^{r}=-\sqrt{\omega^{2}\mu_{-}\epsilon_{-}-\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}-\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}}$ , and since $m_{3}^{i}=\sqrt{\omega^{2}\mu_{-}\epsilon_{-}-\left(m_{1}^{i}\right)^{2}-\left(m_{2}^{i}\right)^{2}}$ , (6.6) means

\sqrt{\omega^{2}\mu_{-}\epsilon_{-}-\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}-\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}}=\sqrt{\omega^{2}\mu_{-}\epsilon_{-}-\left(m_{1}^{i}\right)^{2}-\left(m_{2}^{i}\right)^{2}}

which implies $\left(m_{1}^{i}-\omega f_{x_{1}}\right)^{2}+\left(m_{2}^{i}-\omega f_{x_{2}}\right)^{2}=\left(m_{1}^{i}\right)^{2}+\left(m_{2}^{i}\right)^{2}$ . So (6.6) holds if and only if the gradient of $f$ satisfies

\omega\left((f_{x_{1}})^{2}+(f_{x_{2}})^{2}\right)=2\left(m_{1}^{i}f_{x_{1}}+m_{2}^{i}f_{x_{2}}\right).

This is clearly satisfied for the standard refraction case when $f=0$ .

Suppose $m_{3}^{i}+m_{3}^{r}\neq 0$ . Notice that since $m_{3}^{i}>0,m_{3}^{r}<0$ , and $m_{3}^{t}>0$ , it follows that $\mu_{+}m_{3}^{t}-\mu_{-}m_{3}^{r}>0$ and so $\epsilon_{+}\mu_{+}>\epsilon_{-}\mu_{-}$ . Therefore if $\epsilon_{+}\mu_{+}\leq\epsilon_{-}\mu_{-}$ , then from (6.7) for the system (6.2) to be solvable we must have (6.6) or $A_{3}^{i}=0$ .

7. Appendix

Here we prove Lemma 4.2 used in the proof of Theorem 4.3.

Proof.

Differentiating (4.6) with respect to $x_{1}$ and dividing by ${\rm i}$ yields

(7.8)

a\,m_{1}^{i}\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}+b\,\left(m_{1}^{r}+\psi_{x_{1}}\right)\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}+c\,\left(m_{1}^{t}+\psi_{x_{1}}\right)\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0.

Next differentiate (7.8) with respect to $x_{1}$ to get

		$\displaystyle{\rm i}\,a\,(m_{1}^{i})^{2}\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}+b\,\left(\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{r}+\psi_{x_{1}}\right)^{2}\right)\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}$
		$\displaystyle\qquad+c\,\left(\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{t}+\psi_{x_{1}}\right)^{2}\right)\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0.$

Putting together (4.6), (7.8), and the last equation yields the following system

\left(\begin{matrix}1&1&1\\ m_{1}^{i}&m_{1}^{r}+\psi_{x_{1}}&m_{1}^{t}+\psi_{x_{1}}\\ {\rm i}\,(m_{1}^{i})^{2}&\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{r}+\psi_{x_{1}}\right)^{2}&\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{t}+\psi_{x_{1}}\right)^{2}\\ \end{matrix}\right)\left(\begin{matrix}a\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}\\ b\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}\\ c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}\\ \end{matrix}\right)=0.

If $a\neq 0$ , or $b\neq 0$ , or $c\neq 0$ , then the vector $\left(\begin{matrix}a\,e^{{\rm i}\,\left({\mathbf{m}}_{i}\cdot X\right)}\\ b\,e^{{\rm i}\,\left({\mathbf{m}}_{r}\cdot X+\psi(X)\right)}\\ c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}\\ \end{matrix}\right)$ is a non trivial solution to the system and therefore the matrix

\displaystyle M(x_{1},x_{2})=\left(\begin{matrix}1&1&1\\ m_{1}^{i}&(m_{1}^{r}+\psi_{x_{1}}&m_{1}^{t}+\psi_{x_{1}}\\ {\rm i}\,(m_{1}^{i})^{2}&\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{r}+\psi_{x_{1}}\right)^{2}&\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{t}+\psi_{x_{1}}\right)^{2}\\ \end{matrix}\right)

has determinant equals zero. Multiplying the first row of $M$ by $-m_{1}^{i}$ and adding it to the second row, and multiplying the first row by $-{\rm i}\,(m_{1}^{i})^{2}$ and adding it to the third row yields

\left(\begin{matrix}1&1&1\\ 0&-m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}&-m_{1}^{i}+m_{1}^{t}+\psi_{x_{1}}\\ 0&\psi_{x_{1}x_{1}}+{\rm i}\,\left(\left(m_{1}^{r}+\psi_{x_{1}}\right)^{2}-(m_{1}^{i})^{2}\right)&\psi_{x_{1}x_{1}}+{\rm i}\,\left(\left(m_{1}^{t}+\psi_{x_{1}}\right)^{2}-(m_{1}^{i})^{2}\right)\\ \end{matrix}\right).

So the determinant of the last matrix is zero and factoring the difference of squares yields

		$\displaystyle\left(-m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}\right)\left(\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{t}+\psi_{x_{1}}-m_{1}^{i}\right)\left(m_{1}^{t}+\psi_{x_{1}}+m_{1}^{i}\right)\right)$
		$\displaystyle\qquad-\left(-m_{1}^{i}+m_{1}^{t}+\psi_{x_{1}}\right)\left(\psi_{x_{1}x_{1}}+{\rm i}\,\left(m_{1}^{r}+\psi_{x_{1}}-m_{1}^{i}\right)\left(m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}\right)\right)=0$

that is,

(7.9)

\left(m_{1}^{r}-m_{1}^{t}\right)\,\left(\psi_{x_{1}x_{1}}-{\rm i}\,\left(-m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}\right)\,\left(m_{1}^{t}+\psi_{x_{1}}-m_{1}^{i}\right)\right)=0.

Proceeding in the same way differentiating (4.6) with respect to $x_{2}$ we obtain for the second components the equation

(7.10)

\left(m_{2}^{r}-m_{2}^{t}\right)\,\left(\psi_{x_{2}x_{2}}-{\rm i}\,\left(-m_{2}^{i}+m_{2}^{r}+\psi_{x_{2}}\right)\,\left(m_{2}^{t}+\psi_{x_{2}}-m_{2}^{i}\right)\right)=0.

Proof of (i). We shall prove first that $m_{1}^{t}=m_{1}^{r}$ , and a similar argument proves that $m_{2}^{t}=m_{2}^{r}$ . Suppose by contradiction that $m_{1}^{t}\neq m_{1}^{r}$ , then from (7.9) we must have

\psi_{x_{1}x_{1}}-{\rm i}\,\left(-m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}\right)\,\left(m_{1}^{t}+\psi_{x_{1}}-m_{1}^{i}\right)=0

and since the wave vectors $m^{\ell}$ have real components for $\ell=i,r,t$ and the function $\psi$ is real valued we obtain that

\psi_{x_{1}x_{1}}(x_{1},x_{2})=0\text{ and }\left(-m_{1}^{i}+m_{1}^{r}+\psi_{x_{1}}\right)\,\left(m_{1}^{t}+\psi_{x_{1}}-m_{1}^{i}\right)=0.

which implies that $\psi_{x_{1}}(x_{1},x_{2})=g(x_{2})$ , and

(7.11)

m_{1}^{i}-m_{1}^{r}=\psi_{x_{1}}(x_{1},x_{2}),\text{ or }m_{1}^{i}-m_{1}^{t}=\psi_{x_{1}}(x_{1},x_{2})

for all $x_{1},x_{2}$ . Since $\psi_{x_{1}}$ is continuous we get that $\psi_{x_{1}}$ is constant. Then $\psi(x_{1},x_{2})=c_{0}\,x_{1}+h(x_{2})$ with $c_{0}$ constant. Also at least one of the equations in (7.11) holds, so $m_{1}^{i}-m_{1}^{r}=c_{0}$ or $m_{1}^{i}-m_{1}^{t}=c_{0}$ . Suppose first that $m_{1}^{i}-m_{1}^{r}=c_{0}$ . Substituting the value of $m_{1}^{r}$ into (4.6) and letting $x_{2}=0$ yields

		$\displaystyle a\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+b\,e^{{\rm i}\,\left(m_{1}^{r}\,x_{1}+\psi(x_{1},0)\right)}+c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+\psi(x_{1},0)\right)}$
		$\displaystyle=a\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+b\,e^{{\rm i}\,\left(m_{1}^{r}\,x_{1}+c_{0}\,x_{1}+h(0)\right)}+c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}+h(0)\right)}$
		$\displaystyle=a\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+b\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}+h(0)\right)}+c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}+h(0)\right)}$
		$\displaystyle=\left(a\,+b\,e^{{\rm i}\,h(0)}\right)\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}+h(0)\right)}=0.$

Letting $A=a\,+b\,e^{{\rm i}\,h(0)}$ and $B=c\,e^{{\rm i}\,h(0)}$ we obtain that

A\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+B\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}\right)}=0

for all $x_{1}$ . Differentiating the last expression with respect to $x_{1}$ and dividing by ${\rm i}$ yields

A\,m_{1}^{i}\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}+B\left(m_{1}^{t}+c_{0}\right)\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}\right)}=0

which written in matrix form is

\left(\begin{matrix}1&1\\ m_{1}^{i}&m_{1}^{t}+c_{0}\end{matrix}\right)\left(\begin{matrix}A\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}\\ B\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}\right)}\end{matrix}\right)=0.

Since $c\neq 0$ , the vector $\left(\begin{matrix}A\,e^{{\rm i}\,\left(m_{1}^{i}\,x_{1}\right)}\\ B\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+c_{0}\,x_{1}\right)}\end{matrix}\right)\neq 0$ , then we obtain $c_{0}=m_{1}^{i}-m_{1}^{t}$ . Since we assumed $c_{0}=m_{1}^{i}-m_{1}^{r}$ , we get that $m_{1}^{r}=m_{1}^{t}$ contradicting the assumption that $m_{1}^{r}\neq m_{1}^{t}$ . If on the other hand, $m_{1}^{i}-m_{1}^{t}=c_{0}$ proceeding in the same way and now using that $b\neq 0$ we get as before $m_{1}^{r}=m_{1}^{t}$ contradicting the initial assumption.

To prove that $m_{2}^{r}=m_{2}^{t}$ , we proceed in the same way but using (7.10).

To complete the proof of (i), since $m_{1}^{t}=m_{1}^{r}$ and $m_{2}^{t}=m_{2}^{r}$ , substituting these into (4.6) yields

(7.12)

a\,e^{{\rm i}\,\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}\right)}+(b+c)\,e^{{\rm i}\,\left(m_{1}^{r}x_{1}+m_{2}^{r}x_{2}+\psi(x_{1},x_{2})\right)}=0.

Differentiating this identity with respect to $x_{1}$ and dividing by ${\rm i}$ yields

a\,m_{1}^{i}\,e^{{\rm i}\,\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}\right)}+(b+c)\,\left(m_{1}^{r}+\psi_{x_{1}}\right)\,e^{{\rm i}\,\left(m_{1}^{r}x_{1}+m_{2}^{r}x_{2}+\psi(x_{1},x_{2})\right)}=0.

That is, we obtain the system of equations

\left(\begin{matrix}1&1\\ m_{1}^{i}&m_{1}^{r}+\psi_{x_{1}}\end{matrix}\right)\left(\begin{matrix}a\,e^{{\rm i}\,\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}\right)}\\ (b+c)\,e^{{\rm i}\,\left(m_{1}^{r}x_{1}+m_{2}^{r}x_{2}+\psi(x_{1},x_{2})\right)}\end{matrix}\right)=0.

Since $a\neq 0$ , the vector $\left(\begin{matrix}a\,e^{{\rm i}\,\left(m_{1}^{i}x_{1}+m_{2}^{i}x_{2}\right)}\\ (b+c)\,e^{{\rm i}\,\left(m_{1}^{r}x_{1}+m_{2}^{r}x_{2}+\psi(x_{1},x_{2})\right)}\end{matrix}\right)$ is a non trivial solution to the system and so $\det\left(\begin{matrix}1&1\\ m_{1}^{i}&m_{1}^{r}+\psi_{x_{1}}\end{matrix}\right)=m_{1}^{r}+\psi_{x_{1}}-m_{1}^{i}=0$ for all $x_{1},x_{2}$ as desired.

Differentiating (7.12) with respect to $x_{2}$ in the same way we obtain that $m_{2}^{r}+\psi_{x_{2}}-m_{2}^{i}=0$ completing the proof of (i).

Proof of (ii). Since $a=0$ , we have simplifying the exponential $e^{{\rm i}\,\psi(X)}$ in (4.6) that

b\,e^{{\rm i}\,{\mathbf{m}}_{r}\cdot X}+c\,e^{{\rm i}\,{\mathbf{m}}_{t}\cdot X}=0.

Differentiating the last equation with respect to $x_{1}$ and letting $x_{2}=0$ yields

b\,m_{1}^{r}\,e^{{\rm i}\,m_{1}^{r}\,x_{1}}+c\,m_{1}^{t}\,e^{{\rm i}\,m_{1}^{t}\,x_{1}}=0.

So we get the system

\left(\begin{matrix}1&1\\ m_{1}^{r}&m_{1}^{t}\end{matrix}\right)\left(\begin{matrix}b\,e^{{\rm i}\,m_{1}^{r}\,x_{1}}\\ c\,e^{{\rm i}\,m_{1}^{t}\,x_{1}}\end{matrix}\right)=0,

and since $c\neq 0$ the determinant of the last matrix must be zero and so $m_{1}^{r}=m_{1}^{t}$ . Similarly, differentiating with respect to $x_{2}$ and letting $x_{1}=0$ we obtain that $m_{2}^{r}=m_{2}^{t}$ .

Proof of (iii). We have

a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0,

and differentiating with respect to $x_{1}$ and dividing by ${\rm i}$ yields

m_{1}^{i}\,a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+\left(m_{1}^{t}+\psi_{x_{1}}\right)c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}=0,

and so

\left(\begin{matrix}1&1\\ m_{1}^{i}&m_{1}^{t}+\psi_{x_{1}}\end{matrix}\right)\left(\begin{matrix}a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}\\ c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}\end{matrix}\right)=0.

Since the vector $\left(\begin{matrix}a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}\\ c\,e^{{\rm i}\,\left({\mathbf{m}}_{t}\cdot X+\psi(X)\right)}\end{matrix}\right)\neq 0$ , the determinant of the matrix equals zero, i.e., $m_{1}^{t}+\psi_{x_{1}}(x_{1},x_{2})=m_{1}^{i}$ for all $X$ and therefore $\psi_{x_{1}}$ is constant, that is, $\psi(x_{1},x_{2})=c_{0}\,x_{1}+g(x_{2})$ . Substituting the value of $\psi$ in the original equation yields

a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+m_{2}^{t}\,x_{2}+c_{0}\,x_{1}+g(x_{2})\right)}=0,

which differentiated with respect to $x_{2}$ gives

m_{2}^{i}\,a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}+\left(m_{2}^{t}+g^{\prime}(x_{2})\right)\,c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+m_{2}^{t}\,x_{2}+c_{0}\,x_{1}+g(x_{2})\right)}=0.

Therefore

\left(\begin{matrix}1&1\\ m_{2}^{i}&m_{2}^{t}+g^{\prime}(x_{2})\end{matrix}\right)\left(\begin{matrix}a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}\\ c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+m_{2}^{t}\,x_{2}+c_{0}\,x_{1}+g(x_{2})\right)}\end{matrix}\right)=0.

Again, since the vector $\left(\begin{matrix}a\,e^{{\rm i}\,{\mathbf{m}}_{i}\cdot X}\\ c\,e^{{\rm i}\,\left(m_{1}^{t}\,x_{1}+m_{2}^{t}\,x_{2}+c_{0}\,x_{1}+g(x_{2})\right)}\end{matrix}\right)\neq 0$ , we obtain $m_{2}^{t}+g^{\prime}(x_{2})=m_{2}^{i}$ , implying that $g^{\prime}(x_{2})$ is constant and so $g(x_{2})=c_{1}\,x_{2}+c_{2}$ and we are done.

This proof of (iv) is the same as the proof of (iii). ∎

8. Conclusion

An analysis of the Maxwell system of electrodynamics in the context of distributions is carried out, leading to the derivation of boundary conditions for the electromagnetic field when the current and charge densities are localized at the interface. Consequently, by representing the electric field as a nonlinear perturbation of a plane wave characterized by a phase discontinuity function, the generalized Snell law is obtained. Furthermore, we derive formulas for the amplitudes of the reflected and transmitted waves in terms of the amplitude of the incident wave.

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