Glassey-Strauss representation of Vlasov-Maxwell systems in a Half Space

Yunbai Cao and Chanwoo Kim [email protected] [email protected]; [email protected]

Abstract.

Following closely the classical works [8]-[12] by Glassey, Strauss, and Schaeffer, we present a version of the Glassey-Strauss representation for the Vlasov-Maxwell systems in a 3D half space when the boundary is the perfect conductor.

Key words and phrases:

Vlasov-Maxwell system, Glassey-Strauss representation, half space, perfect conductor, electromagnetic fields.

1991 Mathematics Subject Classification:

Primary: 35Q61, 35Q83; Secondary: 35Q70.

CK is supported in part by National Science Foundation under Grant No.

1900923

and the Brain Pool program (NRF-2021H1D3A2A01039047) of the Ministry of Science and ICT in Korea.

Yunbai Cao

Department of Mathematics

Rutgers University

Piscataway, NJ 08854, USA

Chanwoo Kim

Department of Mathematics

University of Wisconsin-Madison

Madison, WI 53717, USA

$\&$

Department of Mathematical Sciences

Seoul National University

Seoul 08826, Korea

This paper is dedicated to the memory of the late Bob Glassey.

1. Vlasov-Maxwell systems

Consider the plasma particles of several species with masses $m_{\beta}$ and charges $e_{\beta}$ for $\beta=1,2,\cdots,N,$ which occupy the half space

\Omega=\mathbb{R}^{3}_{+}=\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x_{3}>0\}\ni x.

(1)

The relativistic velocity for each particle is, for the speed of light $c$ ,

\hat{v}_{\beta}=\frac{v}{\sqrt{m_{\beta}^{2}+|v|^{2}/c^{2}}}\ \ \ \text{for}\ \ v\in\mathbb{R}^{3}.

(2)

Denote by $f_{\beta}(t,x,v)$ the particle densities of the species. The total electric charge density (total charge per unit volume) $\rho$ and the total electric current density (total current per unit area) $J$ are given by

	$\displaystyle\rho(t,x)$	$\displaystyle=\int_{\mathbb{R}^{3}}\sum_{\beta}e_{\beta}f_{\beta}(t,x,v)dv,$		(3)
	$\displaystyle J(t,x)$	$\displaystyle=\int_{\mathbb{R}^{3}}\sum_{\beta}\hat{v}_{\beta}e_{\beta}f_{\beta}(t,x,v)dv.$		(4)

The relativistic Vlasov-Maxwell system governs the evolution of $f_{\beta}(t,x,v)$ (see page 140 of the Glassey’s book [8]): for $(t,x,v)\in[0,T]\times\Omega\times\mathbb{R}^{3}$ ,

\partial_{t}f_{\beta}+\hat{v}_{\beta}\cdot\nabla_{x}f_{\beta}+(e_{\beta}E+e_{\beta}\frac{\hat{v}_{\beta}}{c}\times B-m_{\beta}g\mathbf{e}_{3})\cdot\nabla_{v}f_{\beta}=0,

(5)

where $g$ is the gravitational constant (we can easily treat a general given external field). The electromagnetic fields $(E,B)$ is determined by the Maxwell’s equations in a vacuum (in Gaussian units)

$\displaystyle\nabla_{x}\cdot E$	$\displaystyle=4\pi\rho,$	(6)
$\displaystyle\nabla_{x}\times E$	$\displaystyle=-\frac{1}{c}\partial_{t}B,$	(7)
$\displaystyle\nabla_{x}\cdot B$	$\displaystyle=0,$	(8)
$\displaystyle\nabla_{x}\times B$	$\displaystyle=\frac{4\pi}{c}J+\frac{1}{c}\partial_{t}E.$	(9)

2. Boundaries

Plasma particles can face various forms of boundaries in different scales from the astronomic one to the laboratory ([1, 2]). In particular, we are interested in the plasma inside the fusion reactors in this paper. So-called plasma-facing materials, the materials that line the vacuum vessel of the fusion reactors, experience violent conditions as they are subjected to high-speed particle and neutron flux and high heat loads. These require several challenging conditions for the boundary materials, namely high thermal conductivity for efficient heat transport, high cohesive energy for low erosion by particle bombardment, and low atomic number to minimize plasma cooling. Traditionally sturdy metals and alloys such as stainless steel, tungsten, titanium, beryllium, and molybdenum have been used for the boundary material [4]. As these metals have very high electric conductivity, we can regard them as the perfect conductor. This boundary condition is the major interest of the paper (see Section 2.1).

On the other hand, carbon/carbon composites such as refined graphite have excellent thermal and mechanical properties: eroded carbon atoms are fully stripped in the plasma core, thus reducing core radiation. In addition, the surface does not melt but simply sublimes if overheated. For this reason, the majority of the latest machines have expanded graphite coverage tile to include all of the vacuum vessel walls [4]. Graphite is an allotrope of carbon, existing as the collection of thin layers of a giant carbon atoms’ covalent lattice. As there is one delocalized electron per carbon atom, graphite does conduct electricity throughout each layer of the graphite lattice but poorly across different layers. Due to the anisotropic electric conductivity of graphite, one has to employ different boundary conditions from one for the metal boundary.

2.1. Perfect conductor boundary

In this section, we consider the boundary conditions of $(E,B)$ at the boundary $\partial\Omega:=\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x_{3}=0\}$ . For that, actually we consider more general situation: two different media occupy $\mathbb{R}^{3}_{+}$ and $\mathbb{R}^{3}_{-}:=\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x_{3}<0\}$ separately. In case that media are subject to electric and magnetic polarization, it is much more convenient to write the Maxwell’s equations only for the free charges and free currents in terms of SI units (see Chapter 7 in [5]):

$\displaystyle\nabla_{x}\cdot D$	$\displaystyle=\rho_{\text{free}},$	(10)
$\displaystyle\nabla_{x}\times E$	$\displaystyle=-\partial_{t}B,$	(11)
$\displaystyle\nabla_{x}\cdot B$	$\displaystyle=0,$	(12)
$\displaystyle\nabla_{x}\times H$	$\displaystyle=J_{\text{free}}+\partial_{t}D,$	(13)

where $\epsilon_{0}$ is the permittivity of free space and $\mu_{0}$ is the permeability of free space (note that the speed of light $c=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$ ). Here, $D=\epsilon_{0}E+P$ and $H=\frac{1}{\mu_{0}}B-M$ , while an electric polarization $P$ and a magnetic polarization $M$ are determined by appropriate constitutive relations in terms of $E$ and $B$ . For example, a linear medium has

P=\varepsilon_{0}\chi_{e}E,\ \ \ M=\chi_{m}H.

(14)

Here, $\chi_{e}$ and $\chi_{m}$ are called the electric susceptibility and magnetic susceptibility, respectively. In a vacuum, as $\chi_{e}=0=\chi_{m}$ and $\rho=\rho_{\text{free}}$ and $J=J_{\text{free}}$ (all plasma particles/charges are free to move), we recover 6-9.

Denote by $n$ the outward unit normal of $\Omega$ (which is $n=-\mathbf{e}_{3}$ for our case); $[V]$ the jump of $V$ across $\partial\Omega$ : $[V](x_{1},x_{2})=\lim_{x_{3}\downarrow 0}V(x_{1},x_{2},x_{3})-\lim_{x_{3}\uparrow 0}V(x_{1},x_{2},x_{3})$ . Then from 11 and 12 we derive the jump conditions

n\times[E]=0,\ \ \ n\cdot[B]=0.

(15)

In other words, the tangential electric fields $E_{1}$ , $E_{2}$ , and the normal magnetic field $B_{3}$ are continuous across the interface $\partial\Omega$ . We note that 15 hold in general, no matter what constitutive relations hold ([3, 6]). (In special circumstances (e.g. electromagnetic band gap structures), one has to consider a non-zero surface magnetic charge and current, in which 15 should be replaced by discontinuous jump conditions [14]. Such cases are out of our interest in the paper.)

Now we come back to the original situation that the plasma particles stay in a vacuum of the upper half space $\Omega=\mathbb{R}^{3}_{+}$ , while some matter fills the lower half space $\mathbb{R}^{3}_{-}$ . We assume that the current follows the Ohm’s law in the matter:

J_{\text{free}}=\sigma\{\text{Lorentz force}\},

(16)

where Lorentz force equals $e_{\beta}E+e_{\beta}\frac{\hat{v}_{\beta}}{c}\times B$ as the gravitation effect is negligible inside the matter. Here, $\sigma$ is the conductivity of the matter, which equals the reciprocal of the resistivity. The perfect conductors have $\sigma=\infty$ and the dielectrics get $\sigma=0$ , while most of real matter is between them. As the drift speed of electrons/ions in the matter is slow (typical drift speed of electrons is few millimeters per second), we ignore the magnetic effect in the Lorentz force to derive that $\nabla_{x}\cdot J_{\text{free}}=\sigma\nabla_{x}\cdot E$ . We assume that the matter is the linear medium 14 and hence $D=\epsilon_{0}(1+\chi_{e})E$ . We derive that, from the continuity equation and 10,

\partial_{t}\rho_{\text{free}}=-\nabla_{x}\cdot J_{\text{free}}=-\sigma\nabla_{x}\cdot E=-\frac{\sigma}{\epsilon_{0}(1+\chi_{e})}\nabla_{x}\cdot D=-\frac{\sigma}{\epsilon_{0}(1+\chi_{e})}\rho_{\text{free}}.

Hence the charge density $\rho_{\text{free}}$ vanishes in the time scale of $1/\sigma$ , which implies $\rho_{\text{free}}\equiv 0$ inside the perfect conductor $(\sigma=\infty)$ . As a consequence, the charge density and current density accumulate only on the surface/boundary/interface (“Skin effect” [13]). Moreover, 13 and 16 formally imply that $E\equiv 0$ , and then 11 forces $\partial_{t}B=0$ inside the perfect conductor. Therefore by assuming the initial datum of $B$ vanishes in $\mathbb{R}^{3}_{-}$ , we have $B\equiv 0$ in $\mathbb{R}^{3}_{-}$ . On the other hand, the superconductor has $B\equiv 0$ no matter what initial datum of $B$ is (the Meissner effect).

We summarize the above discussion about $E_{1},E_{2},B_{3}$ in 17 and will derive the boundary conditions for $E_{3}$ and $B_{1},B_{2}$ using the equations:

Definition 2.1 (Perfect conductor (or superconductor) boundary condition).

Assume the lower half space $\mathbb{R}^{3}_{-}$ consists of a linear medium 14 of the perfect conductor $\sigma=\infty$ satisfying the Ohm’s law 16. We further assume either the initial magnetic field $B$ totally vanishes or the matter of $\mathbb{R}^{3}_{-}$ is the superconductor. Then $E\equiv 0\equiv B$ in $\mathbb{R}^{3}_{-}$ . Therefore, from 15 we derive boundary conditions of the solutions $(E,B)$ to 6-9:

E_{1}=0=E_{2},\ \ \ B_{3}=0\ \ \ \text{on}\ \ \ \partial\Omega.

(17)

Moreover,

	$\displaystyle\partial_{3}E_{3}=4\pi\rho\ \ \$	$\displaystyle\text{on}\ \ \ \partial\Omega,$		(18)
	$\displaystyle\partial_{3}B_{1}=4\pi J_{2},\ \ \partial_{3}B_{2}=-4\pi J_{1}\ \ \$	$\displaystyle\text{on}\ \ \ \partial\Omega.$		(19)

We only need to derive the boundary conditions for $E_{3}$ , $B_{1}$ , and $B_{2}$ . We achieve them by using the equations 6 and 9. From 6, we have $\partial_{3}E_{3}=-\partial_{1}E_{1}-\partial_{2}E_{2}+4\pi\rho.$ Then 17 formally implies 18.

Now from 9, we have, for $n=-\mathbf{e}_{3}$ ,

\frac{1}{c}n\times\partial_{t}E+\frac{4\pi}{c}n\times J-n\times(\nabla_{x}\times B)=0.

From 17, on $\partial\Omega$ we deduce that $n\times\partial_{t}E=0$ and hence

\begin{split}0=4\pi\begin{bmatrix}J_{2}\\ -J_{1}\\ 0\end{bmatrix}-\begin{bmatrix}0\\ 0\\ -1\end{bmatrix}\times\begin{bmatrix}\partial_{2}B_{3}-\partial_{3}B_{2}\\ -(\partial_{1}B_{3}-\partial_{3}B_{1})\\ \partial_{1}B_{2}-\partial_{2}B_{1}\end{bmatrix}=4\pi\begin{bmatrix}J_{2}\\ -J_{1}\\ 0\end{bmatrix}-\begin{bmatrix}\partial_{3}B_{1}\\ \partial_{3}B_{2}\\ 0\end{bmatrix}.\end{split}

Therefore, we conclude 19.

2.2. Surface charge and surface current

To consider general jump conditions across the interface 20, we need to count a surface charge with density $\sigma_{\text{free}}$ and a surface current with density $K_{\text{free}}$ which are “concentrated” on the interface $\partial\Omega$ (see [13, 5]). Physically, a non-zero surface charge and current exist on the surface of the perfect conductor as the interior electric field is zero (see a survey on the concept of the “perfect” conductor and the surface charge and current in history [16]). Then from 9 and 6, we formally get

n\times[H]=n\times K_{\text{free}},\ \ \ n\cdot[D]=\sigma_{\text{free}}.

(20)

For example, if both media are linear 14 then 20 implies that

n\times(\frac{1}{\mu_{+}}B_{+}-\frac{1}{\mu_{-}}B_{-})=n\times K_{\text{free}},\ \ \ n\cdot(\epsilon_{+}E_{+}-\epsilon_{-}E_{-})=\sigma_{\text{free}},

(21)

where $\epsilon_{\pm}=\epsilon_{0}(1+\chi_{e,\pm})$ and $\mu_{\pm}=\mu_{0}(1+\chi_{m,\pm})$ are the permittivity and permeability for the upper and lower media. In the case of Definition 2.1, the upper half is the vacuum and the lower half is a perfect conductor with $B\equiv 0$ , then 20 implies that

K_{\text{free}\parallel}=\frac{1}{\mu_{0}}B_{\parallel}|_{\partial\Omega},\ \ \ \ \sigma_{\text{free}}=\varepsilon_{0}E_{3}|_{\partial\Omega}.

(22)

We note that 22 is not the boundary condition, but one can measure the surface charge and surface current on the surface of the perfect conductor by evaluating $E$ and $B$ .

On the other hand, unless the dielectric media can be polarized on the interface, both surface charge and current vanish on the interface. This results in the dielectric boundary condition, which is 20 with $K_{\text{free}}=0=\sigma_{\text{free}}$ ([3]). When the media have anisotropic conductivities as graphite, the surface charge and current would not be prescribed simply but determined by PDEs.

3. The Glassey-Strauss representation in $\mathbb{R}^{3}$ ([9])

In the whole space, $E$ and $B$ solve

	$\displaystyle\partial_{t}^{2}E-\Delta_{x}E$	$\displaystyle=-4\pi\nabla_{x}\rho-4\pi\partial_{t}J,$		(23)
	$\displaystyle\partial_{t}^{2}B-\Delta_{x}B$	$\displaystyle=4\pi\nabla_{x}\times J,$		(24)

with the initial data

\begin{split}E|_{t=0}=E_{0},\ \ \partial_{t}E|_{t=0}=\partial_{t}E_{0}:=\nabla_{x}\times B_{0}-4\pi J|_{t=0},\\ B|_{t=0}=B_{0},\ \ \partial_{t}B|_{t=0}=\partial_{t}B_{0}:=-\nabla_{x}\times E_{0}.\end{split}

(25)

Obviously the wave equations suffer from the “loss of derivatives” of $(E,B)$ with respect to the regularity of the source terms $\rho$ and $J$ . As Glassey mentions in his book [8], the key idea of the Glassey-Strauss representation is replacing the derivatives $\partial_{t},\nabla_{x}$ by a geometric operator $T$ in 27 and a kinetic transport operator $S$ in 28:

\begin{split}\partial_{t}=\frac{S-\hat{v}\cdot T}{1+\hat{v}\cdot\omega},\ \ \partial_{i}=\frac{\omega_{i}S}{1+\hat{v}\cdot\omega}+\left(\delta_{ij}-\frac{\omega_{i}\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)T_{j},\end{split}

(26)

while, for $\omega=\omega(x,y)=\frac{y-x}{|y-x|}$ ,

	$\displaystyle T_{i}$	$\displaystyle:=\partial_{i}-\omega_{i}\partial t,$		(27)
	$\displaystyle S$	$\displaystyle:=\partial_{t}+\hat{v}\cdot\nabla_{x}.$		(28)

Note that

T_{j}f(t-|y-x|,y,v)=\partial_{y_{j}}[f(t-|y-x|,y,v)],

(29)

which is a tangential derivative along the surface of a backward light cone [8]. On the other hand, the Vlasov equation 5 implies that

Sf=-\nabla_{v}\cdot[(E+\hat{v}\times B-g\mathbf{e}_{3})f].

(30)

Therefore, in [9, 8], they can take off the derivatives $T_{j}$ , $S$ from $f$ using the integration by parts within the Green’s formula of 23-24 by connecting the source terms to $f$ via 3-4. We refer to [15] for the recent development in the whole space case.

4. Derivation of the Representations in a half space

In this section we review the original Glassey-Strauss representation of $(E,B)$ in a whole space and then generalize the representation to the half space problem when the perfect conductor boundary condition 17-19 of Definition 2.1 holds at the boundary $\partial\Omega$ . For the sake of simplicity, we may assume a single species case $\{\beta\}=\{1\}$ and $m_{\beta}=e_{\beta}=c=1$ by the renormalization.

Consider the perfect conductor boundary condition of Definition 2.1. We derive the representation of $E$ and $B$ satisfying the perfect conductor boundary condition at the boundary $\partial\Omega$ . We adopt convenient notations: $E=(E_{\parallel},E_{3})=(E_{1},E_{2},E_{3})$ , $B=(B_{\parallel},B_{3})=(B_{1},B_{2},B_{3})$ , $\nabla=(\nabla_{\parallel},\partial_{3})=(\partial_{1},\partial_{2},\partial_{3})$ , and

\begin{split}\bar{x}=(x_{\parallel},-x_{3})\ \ \ \text{for}\ \ x=(x_{\parallel},x_{3})=(x_{1},x_{2},x_{3}).\end{split}

(31)

We refere to [7] for previous study on Vlasov equations in half space. We also refer to [11, 12, 17] for the lower dimensional cases.

4.1. Tangential components of the Electronic field in a half space

From 23, 17, and 25, we recall that, in $\Omega$ ,

\begin{split}\partial_{t}^{2}E_{\parallel}-\Delta_{x}E_{\parallel}=G_{\parallel}:=-4\pi\nabla_{\parallel}\rho-4\pi\partial_{t}J_{\parallel},\\ E_{\parallel}|_{t=0}=E_{0\parallel},\ \partial_{t}E_{\parallel}|_{t=0}=\partial_{t}E_{0\parallel},\end{split}

(32)

and

E_{\parallel}=0\ \ \text{ on }\ \partial\Omega.

(33)

To solve the Dirichlet boundary condition, we employ the odd extension of the data: for $i=1,2$ , and $x\in\mathbb{R}^{3}$ ,

\begin{split}G_{i}(t,x_{\parallel},x_{3})=&\mathbf{1}_{x_{3}>0}G_{i}(t,x)-\mathbf{1}_{x_{3}<0}G_{i}(t,\bar{x}),\\ E_{0i}(x_{\parallel},x_{3})=&\mathbf{1}_{x_{3}>0}E_{0i}(x)-\mathbf{1}_{x_{3}<0}E_{0i}(\bar{x}),\\ \partial_{t}E_{0i}(x_{\parallel},x_{3})=&\mathbf{1}_{x_{3}>0}\partial_{t}E_{0i}(x)-\mathbf{1}_{x_{3}<0}\partial_{t}E_{0i}(\bar{x}).\end{split}

(34)

Then the weak solution of $E_{\parallel}(t,x)$ to 32 with data 34 in the whole space $\mathbb{R}^{3}$ takes a form of, for $i=1,2,$

	$\displaystyle E_{i}(t,x)=\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}E_{0i}(y)+E_{0i}(y)+\nabla E_{0i}(y)\cdot(y-x)\right)dS_{y}$
	$\displaystyle\ \ \ \ \ +\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}-t\partial_{t}E_{0i}(\bar{y})-E_{0i}(\bar{y})-\nabla E_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}$
	$\displaystyle\ \ \ \ \ +\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}>0\}}\frac{G_{i}(t-\|y-x\|,y)}{\|y-x\|}dy$		(35)
	$\displaystyle\ \ \ \ \ +\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}<0\}}\frac{-G_{i}(t-\|y-x\|,\bar{y})}{\|y-x\|}dy,$		(36)

where $B(x,t)=\{y\in\mathbb{R}^{3}:|x-y|<t\}$ and $\partial B(x,t)=\{y\in\mathbb{R}^{3}:|x-y|=t\}$ . Clearly the above form satisfies the zero Dirichlet boundary condition 33 at $x_{3}=0$ formally. From now one we regard the above form as the weak solution of 32-33. The rest of task is to express 35 and 36.

Expression of 35: We follow the idea of the Glassey-Strauss (Section 3). From 3-4 and 26,

	$\displaystyle\ref{Eexpanmajor1}=-\int_{B(x;t)\cap\{y_{3}>0\}}\frac{(\partial_{i}\mathbf{\rho}+\partial_{t}{J}_{i})(t-\|y-x\|,y)}{\|y-x\|}\mathrm{d}y$
	$\displaystyle\ \ =-\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}(\partial_{i}f+\hat{v}_{i}\partial_{t}f)(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$
	$\displaystyle\ \ =-\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{\omega_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\omega}(Sf)(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$		(37)
	$\displaystyle\ \ \ \ \ -\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\int_{\mathbb{R}^{3}}\left(\delta_{ij}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)T_{j}f(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}.$		(38)

Here, we followed the Einstein convention (when an index variable appears twice, it implies summation of that term over all the values of the index) and will do throughout the paper.

For 37, we replace $Sf$ with 30 and apply the integration by parts in $v$ to derive that 37 equals

\begin{split}\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}a^{E}_{i}(v,\omega)\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|},\end{split}

(39)

where

a^{E}_{i}(v,\omega):=\nabla_{v}\left(\frac{\omega_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\omega}\right)=\frac{(e_{i}-\hat{v}_{i}\hat{v})(1+\hat{v}\cdot\omega)-(\omega_{i}+\hat{v}_{i})(\omega-(\omega\cdot\hat{v})\hat{v})}{\langle v\rangle(1+\hat{v}\cdot\omega)^{2}}.

(40)

For 38, we replace $T_{j}f$ with 29 and apply the integration by parts to get 38 equals

\begin{split}&-\int_{\partial B(x;t)\cap\{y_{3}>0\}}\omega_{j}\left(\delta_{ij}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)f(0,y,v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(\delta_{i3}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{3}}{1+\hat{v}\cdot\omega}\right)f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y_{\parallel}}{|y-x|}\\ &+\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\int_{\mathbb{R}^{3}}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\omega_{i})}{(1+\hat{v}\cdot\omega)^{2}}f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|^{2}}.\end{split}

(41)

where we have used that, from [9, 8],

\frac{\partial}{\partial y_{j}}\left[\frac{1}{|y-x|}\left(\delta_{ij}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)\right]=\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\omega_{i})}{|y-x|^{2}(1+\hat{v}\cdot\omega)^{2}}.

Expression of 36: In order to study the expression in the lower half space we modify the idea of Glassey-Strauss slightly. Define

\bar{\omega}=\begin{bmatrix}\omega_{1}&\omega_{2}&-\omega_{3}\end{bmatrix}^{T}.

(42)

We use the same $S$ of 28 but

\begin{split}\bar{T}_{3}f&=-\partial_{y_{3}}[f(t-|y-x|,y_{\parallel},-y_{3},v)]=\partial_{y_{3}}f-\bar{\omega}_{3}\partial_{t}f,\\ \bar{T}_{i}f&=\partial_{y_{i}}[f(t-|y-x|,y_{\parallel},-y_{3},v)]=\partial_{y_{i}}f-\bar{\omega}_{i}\partial_{t}f\,\text{ for }\ i=1,2.\end{split}

(43)

Then we get

	$\displaystyle\partial_{t}$	$\displaystyle=\frac{S-\hat{v}\cdot\bar{T}}{1+\hat{v}\cdot\bar{\omega}},$		(44)
	$\displaystyle\partial_{y_{i}}$	$\displaystyle=\bar{T}_{i}+\bar{\omega}_{i}\frac{S-\hat{v}\cdot\bar{T}}{1+\hat{v}\cdot\bar{\omega}}=\frac{\bar{\omega}_{i}S}{1+\hat{v}\cdot\bar{\omega}}+\bar{T}_{i}-\bar{\omega}_{i}\frac{\hat{v}\cdot\bar{T}}{1+\hat{v}\cdot\bar{\omega}}.$		(45)

Therefore, we derive

\begin{split}\partial_{i}+\hat{v}_{i}\partial_{t}=\frac{\bar{\omega}_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\bar{\omega}}S+\left(\delta_{ij}-\frac{\bar{\omega}_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\right)\bar{T}_{j}.\end{split}

(46)

Now we consider 36. From 46,

	$\displaystyle\ref{Eexpanmajor2}=\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}(\partial_{i}f+\hat{v}_{i}\partial_{t}f)(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$
	$\displaystyle\ \ =\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{\bar{\omega}_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\bar{\omega}}(Sf)(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$		(47)
	$\displaystyle\ \ +\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}<0\}\end{subarray}}\int_{\mathbb{R}^{3}}\Big{(}\delta_{ij}-\frac{\bar{\omega}_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}\bar{T}_{j}f(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}.$		(48)

As getting 39, we derive that, with $a_{i}^{E}$ of 40, 47 equals

\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}a^{E}_{i}(v,\bar{\omega})\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}.

(49)

For 48, applying 43 and the integration by parts, we derive that 48 equals

\begin{split}&\int_{\partial B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\bar{\omega}_{j}\Big{(}\delta_{ij}-\frac{\bar{\omega}_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}f(0,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\iota_{3}\Big{(}\delta_{i3}-\frac{\bar{\omega}_{i}\hat{v}_{3}+\hat{v}_{i}\hat{v}_{3}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &-\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\bar{\omega}_{i})}{(1+\hat{v}\cdot\bar{\omega})^{2}}f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|^{2}},\end{split}

(50)

where we have utilized the notation

\iota_{i}=+1\ \ \text{ for }\ i=1,2,\ \ \iota_{3}=-1,

(51)

and the direct computation

\iota_{j}\frac{\partial}{\partial y_{j}}\left[\frac{1}{|y-x|}\Big{(}\delta_{ij}-\frac{\iota_{i}\omega_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}\right]=\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\bar{\omega}_{i})}{|y-x|^{2}(1+\hat{v}\cdot\bar{\omega})^{2}}.

(52)

4.2. Normal components of the Electronic field in a half space

From 23, 18 and 25, we have

\begin{split}\partial_{t}^{2}E_{3}-\Delta_{x}E_{3}=G_{3}:=-4\pi\partial_{3}\rho-4\pi\partial_{t}J_{3},\\ E_{3}|_{t=0}=E_{03},\ \partial_{t}E_{3}|_{t=0}=\partial_{t}E_{03},\end{split}

(53)

and

\partial_{3}E_{3}=4\pi\rho\ \ \text{ on }\ \ \partial\Omega.

(54)

It is convenient to decompose the solution into two parts: one with the Neumann boundary condition of 53 and the zero forcing term and initial data

\begin{split}\partial_{t}^{2}w-\Delta_{x}w=0\ \ &\text{ in }\ \ \Omega,\\ w|_{t=0}=0,\ \partial_{t}w|_{t=0}=0\ \ &\text{ in }\ \ \Omega,\\ \partial_{3}w=4\pi\rho\ \ &\text{ on }\ \ \partial\Omega,\end{split}

(55)

and the other part $\tilde{E}_{3}$ with the initial data of 53 and the zero Neumann boundary condition. We achieve it by the even extension trick. Recall $\bar{x}$ in 31. For $x\in\mathbb{R}^{3}$ , define

\begin{split}G_{3}(t,x)=&\mathbf{1}_{x_{3}>0}G_{3}(t,x)+\mathbf{1}_{x_{3}<0}G_{3}(t,\bar{x}),\\ E_{03}(x)=&\mathbf{1}_{x_{3}>0}E_{03}(x)+\mathbf{1}_{x_{3}<0}E_{03}(\bar{x}),\\ \partial_{t}E_{03}(x)=&\mathbf{1}_{x_{3}>0}\partial_{t}E_{03}(x)+\mathbf{1}_{x_{3}<0}\partial_{t}E_{03}(\bar{x}).\end{split}

(56)

The weak solution $\tilde{E}_{3}$ to 53 with the data 56 in the whole space $\mathbb{R}^{3}$ take a form of

$\displaystyle\tilde{E}_{3}(t,x)$	$\displaystyle=\frac{1}{4\pi t^{2}}\int_{\begin{subarray}{c}\partial B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\left(t\partial_{t}E_{03}(y)+E_{03}(y)+\nabla E_{03}(y)\cdot(y-x)\right)dS_{y}$
	$\displaystyle+\frac{1}{4\pi t^{2}}\int_{\begin{subarray}{c}\partial B(x;t)\\ \cap\{y_{3}<0\}\end{subarray}}\big{(}t\partial_{t}E_{03}(\bar{y})+E_{03}(\bar{y})+\nabla E_{03}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}$
	$\displaystyle+\frac{1}{4\pi}\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\frac{G_{3}(t-\|y-x\|,y)}{\|y-x\|}dy$	(57)
	$\displaystyle+\frac{1}{4\pi}\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}<0\}\end{subarray}}\frac{G_{3}(t-\|y-x\|,\bar{y})}{\|y-x\|}dy.$	(58)

Following the same argument to expand 35 into 37-41 and 36 into 47-50, we derive that

\begin{split}&\ref{tildeE3formula1}\\ &=\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\int_{\mathbb{R}^{3}}a^{E}_{3}(v,\omega)\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{3}+\omega_{3})}{(1+\hat{v}\cdot\omega)^{2}}f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|^{2}}\\ &-\int_{\partial B(x;t)\cap\{y_{3}>0\}}\omega_{j}\left(\delta_{3j}-\frac{(\omega_{3}+\hat{v}_{3})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)f(0,y,v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(1-\frac{(\omega_{3}+\hat{v}_{3})\hat{v}_{3}}{1+\hat{v}\cdot\omega}\right)f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y_{\parallel}}{|y-x|},\end{split}

(59)

\begin{split}&\ref{tildeE3formula2}\\ &=-\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}<0\}\end{subarray}}\int_{\mathbb{R}^{3}}a^{E}_{3}(v,\bar{\omega})\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{3}+\bar{\omega}_{3})}{(1+\hat{v}\cdot\bar{\omega})^{2}}f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|^{2}}\\ &-\int_{\partial B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\bar{\omega}_{j}\Big{(}\delta_{3j}-\frac{\bar{\omega}_{3}\hat{v}_{j}+\hat{v}_{3}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}f(0,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\Big{(}1-\frac{(\bar{\omega}_{3}+\hat{v}_{3})\hat{v}_{3}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}.\end{split}

(60)

Note that the weak derivative $\partial_{3}$ to the form of $\tilde{E}_{3}$ solves the linear wave equation 53 with oddly extended forcing term and the initial data in the sense of distributions. From the argument of Section 4.1, we conclude that

\partial_{3}\tilde{E}_{3}=0\ \ \text{ on }\ \ \partial\Omega.

(61)

4.2.1. Wave equation with the Neumann boundary condition

Now we consider 55. We assume $\rho(t,x)$ for all $t\leq 0$ , which implies $w(t,x)=0$ for all $t\leq 0$ . Define the Laplace transformation:

W(p,x)=\int^{\infty}_{-\infty}e^{-pt}w(t,x)dt,\ \ \ \ R(p,x)=\int^{\infty}_{-\infty}e^{-pt}\rho(t,x)dt.

(62)

Then $W$ solves the Helmholtz equation with the same Neumann boundary condition:

\begin{split}p^{2}W-\Delta_{x}W=0&\ \ \text{ in }\ \Omega,\\ \partial_{n}W=4\pi R&\ \ \text{ on }\ \partial\Omega.\end{split}

(63)

The solution for $(p^{2}-\Delta_{x})\Phi(x)=\delta(x)$ in $\mathbb{R}^{3}$ is known as $\frac{1}{4\pi}\frac{e^{\pm p|x|}}{|x|}$ . We choose

\Phi(x)=\frac{1}{4\pi}\frac{e^{-p|x|}}{|x|}.

(64)

We have the following identities:

Lemma 4.1.

Suppose $u\in C^{2}(\bar{\Omega})$ is an arbitrary function. For a fixed $x\in\Omega$ and $\Phi$ in 64, we have

\begin{split}u(x)&=\int_{\Omega}\Phi(y-x)(p^{2}-\Delta_{x})u(y)dy\\ &+\int_{\partial\Omega}\left[\Phi(y-x)\partial_{n}u(y)-u(y)\partial_{n}\Phi(y-x)\right]dS_{y}.\end{split}

(65)

Proof.

The proof is rather standard. Fix $x\in\Omega$ . Let $0<\varepsilon\ll 1$ , and $B(x,\varepsilon)$ be a ball centered at $x$ with radius $\varepsilon$ such that $B(x,\varepsilon)\subset\Omega$ . Let $V_{\varepsilon}=\Omega-B(x,\varepsilon)$ . Then, by the integration by parts,

\begin{split}&-\int_{V_{\varepsilon}}\Phi(y-x)(\Delta_{y}-p^{2})u(y)dy\\ &=\int_{\partial\Omega}u(y)\partial_{n}\Phi(y-x)dS_{y}+\int_{\partial B(x,\varepsilon)}u(y)\partial_{n}\Phi(y-x)dS_{y}\\ &\ \ -\int_{\partial\Omega}\Phi(y-x)\partial_{n}u(y)dS_{y}-\int_{\partial B(x,\varepsilon)}\Phi(y-x)\partial_{n}u(y)dS_{y}.\end{split}

From 64, $\int_{\partial B(x,\varepsilon)}\Phi(y-x)\partial_{n}u(y)dS_{y}\lesssim 4\pi\varepsilon^{2}\frac{e^{|p|\varepsilon}}{4\pi\varepsilon}\to 0,$ as $\varepsilon\to 0.$ And by direct computation,

\begin{split}&\int_{\partial B(x,\varepsilon)}u(y)\partial_{n}\Phi(y-x)dS_{y}=\int_{\partial B(x,\varepsilon)}u(y)\frac{-(y-x)}{|y-x|}\cdot\nabla\Phi(y-x)dS_{y}\\ =&\frac{1}{4\pi}\int_{\partial B(x,\varepsilon)}u(y)\frac{-(y-x)}{|y-x|}\cdot(-ip|y-x|-1)\frac{e^{-ip|y-x|}(y-x)}{|y-x|^{3}}dS_{y}\\ =&\frac{1}{4\pi}\int_{\partial B(x,\varepsilon)}\left(-(-p|y-x|-1)\frac{e^{-p|y-x|}}{|y-x|^{2}}\right)u(y)dS_{y}\\ =&\left((1-(-p)\varepsilon)e^{-p\varepsilon}\right)\left(\frac{1}{4\pi\varepsilon^{2}}\int_{\partial B(x,\varepsilon)}u(y)dS_{y}\right)\to u(x),\text{ as }\varepsilon\to 0.\end{split}

Combining all together, and letting $\varepsilon\to 0$ we get 65.∎

Next for $x\in\Omega$ , let $\phi^{x}(y)$ be the function such that

\begin{split}(\Delta_{y}-p^{2})\phi_{N}^{x}(y)=0\ \ &\text{ in }\ \ \Omega,\\ \partial_{n}\phi_{N}^{x}(y)=\partial_{n}\Phi(y-x)\ \ &\text{ on }\ \ \partial\Omega.\end{split}

(66)

The integration by parts implies

\begin{split}0=&\int_{\Omega}(\Delta_{y}-p^{2})\phi_{N}^{x}(y)u(y)dy\\ =&\int_{\Omega}(\Delta_{y}-p^{2})u(y)\phi_{N}^{x}(y)dy+\int_{\partial\Omega}[\partial_{n}\phi_{N}^{x}(y)u(y)-\phi_{N}^{x}(y)\partial_{n}u(y)]dS_{y}\\ =&\int_{\Omega}(\Delta_{y}-p^{2})u(y)\phi_{N}^{x}(y)dy+\int_{\partial\Omega}[\partial_{n}\Phi(y-x)u(y)-\phi_{N}^{x}(y)\partial_{n}u(y)]dS_{y}.\end{split}

(67)

By adding 67 to 65, we derive that

\begin{split}u(x)=&-\int_{\Omega}\left(\Phi(y-x)-\phi_{N}^{x}(y)\right)(\Delta_{y}-p^{2})u(y)dy\\ &+\int_{\partial\Omega}(\Phi(y-x)-\phi_{N}^{x}(y))\partial_{n}u(y)dS_{y}.\end{split}

(68)

For the half space $\Omega=\mathbb{R}_{+}^{3}$ , we have, with $\bar{x}$ in 31,

\phi_{N}^{x}(y)=-\Phi(y-\bar{x}).

(69)

Finally, we derive that, from 68 and 69:

Lemma 4.2.

For $\Omega=\mathbb{R}^{3}_{+}$ , and $\Phi$ in 64,

\begin{split}u(x)=&-\int_{\Omega}\left(\Phi(y-x)+\Phi(y-\bar{x})\right)(\Delta_{y}-p^{2})u(y)dy\\ &+\int_{\partial\Omega}(\Phi(y-x)+\Phi(y-\bar{x}))\partial_{n}u(y)dS_{y}.\end{split}

(70)

By applying 69 to 63, we derive that

\begin{split}W(p,x)&=\int_{\partial\Omega}(\Phi(y-x)+\Phi(y-\bar{x}))4\pi R(y)dS_{y}\\ &=2\int_{\mathbb{R}^{2}}\frac{e^{-p((y_{1}-x_{1})^{2}+(y_{2}-x_{2})^{2}+x_{3}^{2})^{1/2}}}{(y_{1}-x_{1})^{2}+((y_{2}-x_{2})^{2}+x_{3}^{2})^{1/2}}R(y_{1},y_{2})dy_{1}dy_{2}.\end{split}

(71)

Using the inverse Laplace transform, we derive that

\begin{split}&w(t,x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{(p_{1}+ip_{2})t}W(p_{1}+ip_{2},x)dp_{2}\\ &\ \ =\frac{1}{\pi}\int_{-\infty}^{\infty}dp_{2}\ e^{(p_{1}+ip_{2})t}\int_{\mathbb{R}^{2}}dy_{1}dy_{2}\frac{e^{-(p_{1}+ip_{2})((y_{1}-x_{1})^{2}+(y_{2}-x_{2})^{2}+x_{3}^{2})^{1/2}}}{(y_{1}-x_{1})^{2}+((y_{2}-x_{2})^{2}+x_{3}^{2})^{1/2}}\\ &\ \ \ \ \ \ \times\int_{-\infty}^{\infty}ds\ e^{-(p_{1}+ip_{2})s}(-\rho(s,y_{1},y_{2})).\end{split}

(72)

Finally, we derive that, using the identity $\int_{-\infty}^{\infty}e^{ip_{2}t}dp_{2}=2\pi\delta(t)$ ,

\begin{split}&w(t,x)=\frac{-1}{\pi}\int_{\mathbb{R}^{2}}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{e^{(p_{1}+ip_{2})(t-s-\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}})}}{\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}}}\rho(s,y_{\parallel})dp_{2}dsdy_{\parallel}\\ &=-2\int_{\mathbb{R}^{2}}\int_{\mathbb{R}}\frac{e^{p_{1}(t-s-\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}})}\delta(t-s-\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}})}{\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}}}\rho(s,y_{\parallel})dsdy_{\parallel}\\ &=-2\int_{\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}}<t}\frac{\rho(t-\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}},y_{\parallel})}{\sqrt{|y_{\parallel}-x_{\parallel}|^{2}+x_{3}^{2}}}dy_{\parallel}.\end{split}

(73)

4.3. Summary

Collecting the terms, we conclude the following formula.

Proposition 1.

\begin{split}&E_{i}(t,x)=\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}E_{0i}(y)+E_{0i}(y)+\nabla E_{0i}(y)\cdot(y-x)\right)dS_{y}\\ &+\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\iota_{i}\big{(}-t\partial_{t}E_{0i}(\bar{y})-E_{0i}(\bar{y})-\nabla E_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\omega_{i})}{|y-x|^{2}(1+\hat{v}\cdot\omega)^{2}}f(t-|y-x|,y,v)\mathrm{d}v\mathrm{d}y\\ &-\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}\frac{(|\hat{v}|^{2}-1)(\hat{v}_{i}+\bar{\omega}_{i})}{|y-x|^{2}(1+\hat{v}\cdot\bar{\omega})^{2}}f(t-|y-x|,\bar{y},v)\mathrm{d}v\mathrm{d}y\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}a^{E}_{i}(v,\omega)\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &-\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}a^{E}_{i}(v,\bar{\omega})\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(\delta_{i3}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{3}}{1+\hat{v}\cdot\omega}\right)f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y_{\parallel}}{|y-x|}\\ &-\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\iota_{i}\left(\delta_{i3}-\frac{\bar{\omega}_{i}\hat{v}_{3}+\hat{v}_{i}\hat{v}_{3}}{1+\hat{v}\cdot\bar{\omega}}\right)f(t-|y-x|,y_{\parallel},0,v)\mathrm{d}v\frac{\mathrm{d}y_{\parallel}}{|y-x|}\\ &-\int_{\partial B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\omega_{j}\left(\delta_{ij}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)f(0,y,v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &+\int_{\partial B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}\bar{\omega}_{j}\Big{(}\delta_{ij}-\frac{\bar{\omega}_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}f(0,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}S_{y}}{|y-x|}\\ &-\delta_{i3}\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\frac{2f(t-|y-x|,y_{\parallel},0,v)}{|y-x|}dvdS_{y}.\end{split}

(74)

4.4. Representation of the Magnetic field in a half space

Next, we solve for $B$ . For $B_{1},B_{2}$ we have, for $i=1,2,$

\begin{split}\partial_{t}^{2}B_{i}-\Delta_{x}B_{i}=&4\pi(\nabla_{x}\times J)_{i}:=H_{i}\ \ \text{ in }\ \ \Omega,\\ \partial_{x_{3}}B_{1}=&4\pi J_{2},\ \partial_{x_{3}}B_{2}=4\pi J_{1}\ \ \text{ on }\ \ \partial\Omega,\\ B_{i}(0,x)=&B_{0i},\partial_{t}B_{i}(0,x)=\partial_{t}B_{0i}\ \ \text{ in }\ \ \Omega.\end{split}

(75)

To solve 75 we write $B_{i}=\tilde{B}_{i}+B_{bi}$ with $\tilde{B}_{i}$ satisfies the wave equation in $(0,\infty)\times\mathbb{R}^{3}$ with even extension in $x_{3}$ :

\begin{split}\partial_{t}^{2}\tilde{B}_{i}-\Delta_{x}\tilde{B}_{i}=&\mathbf{1}_{x_{3}>0}H_{i}(t,x)+\mathbf{1}_{x_{3}<0}H_{i}(t,\bar{x}),\\ \tilde{B}_{i}(0,x)=&\mathbf{1}_{x_{3}>0}B_{0i}(x)+\mathbf{1}_{x_{3}<0}B_{0i}(\bar{x}),\\ \partial_{t}\tilde{B}_{i}(0,x)=&\mathbf{1}_{x_{3}>0}\partial_{t}B_{0i}(x)+\mathbf{1}_{x_{3}<0}\partial_{t}B_{0i}(\bar{x}).\end{split}

(76)

And $B_{bi}$ satisfies

\begin{split}\partial_{t}^{2}B_{bi}-\Delta_{x}B_{bi}=0&\text{ in }\Omega,\\ B_{bi}(0,x)=0,\partial_{t}B_{bi}=0&\text{ in }\Omega,\\ \partial_{x_{3}}B_{b1}=4\pi J_{2},\ \partial_{x_{3}}B_{b2}=-4\pi J_{1}&\text{ on }\Omega.\end{split}

(77)

Then from 76,

\begin{split}&\tilde{B}_{i}(t,x)\\ =&\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}B_{0i}(y)+B_{0i}(y)+\nabla B_{0i}(y)\cdot(y-x)\right)dS_{y}\\ &+\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}t\partial_{t}B_{0i}(\bar{y})+B_{0i}(\bar{y})+\nabla B_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}\\ &+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}>0\}}\frac{H_{i}(t-|y-x|,y)}{|y-x|}dy+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}<0\}}\frac{H_{i}(t-|y-x|,\bar{y})}{|y-x|}dy.\end{split}

Applying 73 to 77,

\begin{split}B_{bi}(t,x)=(-1)^{i}2\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\frac{\hat{v}_{\underline{i}}f(t-|y-x|,y_{\parallel},0,v)}{|y-x|}dvdS_{y},\end{split}

(78)

where we define

\underline{i}=\begin{cases}2,\text{ if }i=1,\\ 1,\text{ if }i=2.\end{cases}

Thus,

\begin{split}&B_{i}(t,x)\\ =&\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}B_{0i}(y)+B_{0i}(y)+\nabla B_{0i}(y)\cdot(y-x)\right)dS_{y}\\ &+\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}t\partial_{t}B_{0i}(\bar{y})+B_{0i}(\bar{y})+\nabla B_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}\\ &+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}>0\}}\frac{H_{i}(t-|y-x|,y)}{|y-x|}dy+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}<0\}}\frac{H_{i}(t-|y-x|,\bar{y})}{|y-x|}dy\\ &+(-1)^{i}2\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\frac{\hat{v}_{\underline{i}}f(t-|y-x|,y_{\parallel},0,v)}{|y-x|}dvdS_{y}.\end{split}

(79)

On the other hand, $B_{3}(t,x)$ satisfies

\begin{split}\partial_{t}^{2}B_{3}-\Delta_{x}B_{3}=&4\pi(\nabla_{x}\times j)_{3}:=H_{3}\text{ in }\Omega,\\ B_{3}(0,x)=&B_{03},\partial_{t}B_{3}(0,x)=\partial_{t}B_{03}\text{ in }\Omega,\\ B_{3}=&0\text{ on }\partial\Omega.\end{split}

Using the odd extension in $x_{3}$ :

\begin{split}H_{3}(t,x)=&\mathbf{1}_{x_{3}>0}H_{3}(t,x)-\mathbf{1}_{x_{3}<0}H_{3}(t,\bar{x}),\\ B_{03}(x)=&\mathbf{1}_{x_{3}>0}B_{03}(x)-\mathbf{1}_{x_{3}<0}B_{03}(\bar{x}),\\ \partial_{t}B_{03}(0,x)=&\mathbf{1}_{x_{3}>0}\partial_{t}B_{03}(x)-\mathbf{1}_{x_{3}<0}\partial_{t}B_{03}(\bar{x}),\end{split}

we get the expression for $B_{3}$ :

\begin{split}&B_{3}(t,x)\\ =&\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}B_{03}(y)+B_{30}(y)+\nabla B_{03}(y)\cdot(y-x)\right)dS_{y}\\ &-\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}t\partial_{t}B_{03}(\bar{y})+B_{03}(\bar{y})+\nabla B_{03}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}\\ &+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}>0\}}\frac{H_{3}(t-|y-x|,y)}{|y-x|}dy-\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}<0\}}\frac{H_{3}(t-|y-x|,\bar{y})}{|y-x|}dy.\end{split}

(80)

Combining 79 and 80, we get for $i=1,2,3$ ,

$\displaystyle B_{i}(t,x)=$	$\displaystyle\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}B_{0i}(y)+B_{0i}(y)+\nabla B_{0i}(y)\cdot(y-x)\right)dS_{y}$
	$\displaystyle+\frac{\iota_{i}}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}t\partial_{t}B_{0i}(\bar{y})+B_{0i}(\bar{y})+\nabla B_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}$
	$\displaystyle+\frac{1}{4\pi}\int_{B(x;t)\cap\{y_{3}>0\}}\frac{H_{i}(t-\|y-x\|,y)}{\|y-x\|}dy$	(81)
	$\displaystyle+\frac{\iota_{i}}{4\pi}\int_{B(x;t)\cap\{y_{3}<0\}}\frac{H_{i}(t-\|y-x\|,\bar{y})}{\|y-x\|}dy$	(82)
	$\displaystyle+(-1)^{i}2(1-\delta_{i3})\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\frac{\hat{v}_{\underline{i}}f(t-\|y-x\|,y_{\parallel},0,v)}{\|y-x\|}dvdS_{y}.$

Using 26, we have

$\displaystyle\ref{Bexpanmajor1}=$	$\displaystyle\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(\nabla_{x}f\times\hat{v})_{i}(t-\|y-x\|,y,v)}{\|y-x\|}dy$
$\displaystyle=$	$\displaystyle\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(\omega\times\hat{v})_{i}}{1+\hat{v}\cdot\omega}Sf(t-\|y-x\|,y,v)dv\frac{dy}{\|y-x\|}$	(83)
	$\displaystyle+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\left((T\times\hat{v})_{i}-\frac{(\omega\times\hat{v})_{i}\hat{v}\cdot T}{1+\hat{v}\cdot\omega}\right)f(t-\|y-x\|,y,v)dv\frac{dy}{\|y-x\|}.$	(84)

For 83, we replace $Sf$ with 30 and apply the integration by parts in $v$ to derive that 83 equals

\begin{split}\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}a^{B}_{i}(v,\omega)\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|},\end{split}

(85)

where

a^{B}_{i}(v,\omega):=\nabla_{v}\left(\frac{(\omega\times\hat{v})_{i}}{1+\hat{v}\cdot\omega}\right)=\frac{\nabla_{v}[(\omega\times v)_{i}]}{\sqrt{1+|v|^{2}}(1+\hat{v}\cdot\omega)}+\frac{(\omega\times v)_{i}(\hat{v}+\omega)}{(\sqrt{1+|v|^{2}}(1+\hat{v}\cdot\omega))^{2}}.

(86)

For 84, we replace $T_{j}f$ with 29 and apply the integration by parts to get 84 equals

\begin{split}&\int_{\partial B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}(\omega\times\hat{v})_{i}\left(1-\frac{\hat{v}\cdot\omega}{1+\hat{v}\cdot\omega}\right)f(0,y,v)dv\frac{dS_{y}}{t}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(-(e_{3}\times\hat{v})_{i}+\frac{(\omega\times\hat{v})_{i}}{1+\hat{v}\cdot\omega}(\hat{v}\cdot e_{3})\right)f(t-|y-x|,y_{\parallel},0,v)dv\frac{dy_{\parallel}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(\omega\times\hat{v})_{i}\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}}f(t-|y-x|,y,v)dvdy.\end{split}

(87)

where we have used that, from [9, 8],

\begin{split}&\partial_{y_{j}}\left(\frac{(\omega\times\hat{v})\hat{v}_{j}}{(1+\hat{v}\cdot\omega)|y-x|}\right)\\ =&\frac{(\omega\times\hat{v})\left(-(\omega\cdot\hat{v})(1+\omega\cdot\hat{v})-(\omega\cdot\hat{v})(1+\omega\cdot\hat{v})-|\hat{v}|^{2}+(\omega\cdot\hat{v})^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}}\\ =&\frac{(\omega\times\hat{v})\left(-2(\omega\cdot\hat{v})-|\hat{v}|^{2}-(\omega\cdot\hat{v})^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}},\end{split}

and

\begin{split}&-\nabla_{y}(\frac{1}{|y-x|})\times\hat{v}+\partial_{y_{j}}\left(\frac{(\omega\times\hat{v})\hat{v}_{j}}{(1+\hat{v}\cdot\omega)|y-x|}\right)\\ &=\frac{(\omega\times\hat{v})\left((1+\hat{v}\cdot\omega)^{2}-2(\omega\cdot\hat{v})-|\hat{v}|^{2}-(\omega\cdot\hat{v})^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}}=\frac{(\omega\times\hat{v})\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}}.\end{split}

Now we consider 82. From 46,

82	$\displaystyle=\iota_{i}\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{(\nabla_{x}f\times\hat{v})_{i}(t-\|y-x\|,\bar{y},v)}{\|y-x\|}dy$
	$\displaystyle=\iota_{i}\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{(\bar{\omega}\times\hat{v})_{i}}{1+\hat{v}\cdot\bar{\omega}}Sf(t-\|y-x\|,\bar{y},v)dv\frac{dy}{\|y-x\|}$	(88)
	$\displaystyle+\iota_{i}\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\left((\bar{T}\times\hat{v})_{i}-\frac{(\hat{v}\cdot\bar{T})(\bar{\omega}\times\hat{v})_{i}}{1+\hat{v}\cdot\bar{\omega}}\right)f(t-\|y-x\|,\bar{y},v)dv\frac{dy}{\|y-x\|}.$	(89)

As getting 85, we derive that, with $a_{i}^{B}$ of 86, 88 equals

\iota_{i}\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}a^{B}_{i}(v,\bar{\omega})\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}.

(90)

For 89, applying 43 and the integration by parts, we derive that 89 equals

\begin{split}&\iota_{i}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}(\bar{\omega}\times\hat{v})_{i}\left(1-\frac{\hat{v}\cdot\bar{\omega}}{1+\hat{v}\cdot\bar{\omega}}\right)f(0,\bar{y},v)dv\frac{dS_{y}}{t}\\ +&\iota_{i}\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(-(e_{3}\times\hat{v})_{i}+\frac{(\bar{\omega}\times\hat{v})_{i}}{1+\hat{v}\cdot\bar{\omega}}(\hat{v}\cdot e_{3})\right)\frac{f(t-|y-x|,y_{\parallel},0,v)}{|y-x|}dvdy_{\parallel}\\ +&\iota_{i}\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\left(\frac{(\bar{\omega}\times\hat{v})_{i}\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}}\right)f(t-|y-x|,\bar{y},v)dvdy,\end{split}

(91)

where we have used the direct computation

\begin{split}\iota_{j}\partial_{y_{j}}\left(\frac{(\bar{\omega}\times\hat{v})\hat{v}_{j}}{(1+\hat{v}\cdot\bar{\omega})|y-x|}\right)=&\frac{(\bar{\omega}\times\hat{v})\left(-(\hat{v}\cdot\bar{\omega})(1+\hat{v}\cdot\bar{\omega})-\hat{v}\cdot\bar{\omega}-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}}\\ =&\frac{(\bar{\omega}\times\hat{v})\left(-2(\hat{v}\cdot\bar{\omega})-|\hat{v}|^{2}-(\hat{v}\cdot\bar{\omega})^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}},\end{split}

and

\begin{split}&-\overline{\nabla_{y}(|y-x|^{-1})}\times\hat{v}+\iota_{j}\partial_{y_{j}}\left(\frac{(\bar{\omega}\times\hat{v})\hat{v}_{j}}{(1+\hat{v}\cdot\bar{\omega})|y-x|}\right)\\ &=\frac{(\bar{\omega}\times\hat{v})\left((1+\hat{v}\cdot\bar{\omega})^{2}-2(\hat{v}\cdot\bar{\omega})-|\hat{v}|^{2}-(\hat{v}\cdot\bar{\omega})^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}}=\frac{(\bar{\omega}\times\hat{v})\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}}.\end{split}

Collecting the terms, we conclude the following formula:

Proposition 2.

\begin{split}&B_{i}(t,x)=\frac{1}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}>0\}}\left(t\partial_{t}B_{0i}(y)+B_{0i}(y)+\nabla B_{0i}(y)\cdot(y-x)\right)dS_{y}\\ &+\frac{\iota_{i}}{4\pi t^{2}}\int_{\partial B(x;t)\cap\{y_{3}<0\}}\big{(}t\partial_{t}B_{0i}(\bar{y})+B_{0i}(\bar{y})+\nabla B_{0i}(\bar{y})\cdot(\bar{y}-\bar{x})\big{)}dS_{y}\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{(\omega\times\hat{v})_{i}\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\omega)^{2}|y-x|^{2}}f(t-|y-x|,y,v)dvdy\\ &+\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}\frac{(\bar{\omega}\times\hat{v})_{i}\left(1-|\hat{v}|^{2}\right)}{(1+\hat{v}\cdot\bar{\omega})^{2}|y-x|^{2}}f(t-|y-x|,\bar{y},v)dvdy\\ &+\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}a^{B}_{i}(v,\omega)\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}a^{B}_{i}(v,\bar{\omega})\cdot(E+\hat{v}\times B-g\mathbf{e}_{3})f(t-|y-x|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\left(-(e_{3}\times\hat{v})_{i}+\frac{(\omega\times\hat{v})_{i}\hat{v}_{3}}{1+\hat{v}\cdot\omega}\right)f(t-|y-x|,y_{\parallel},0,v)dv\frac{dy_{\parallel}}{|y-x|}\\ &+\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\iota_{i}\left(-(e_{3}\times\hat{v})_{i}+\frac{(\bar{\omega}\times\hat{v})_{i}\hat{v}_{3}}{1+\hat{v}\cdot\bar{\omega}}\right)f(t-|y-x|,y_{\parallel},0,v)dv\frac{dy_{\parallel}}{|y-x|}\\ &+\int_{\partial B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\left(\frac{(\omega\times\hat{v})_{i}}{1+\hat{v}\cdot\omega}\right)f(0,y,v)dv\frac{dS_{y}}{t}\\ &+\int_{\partial B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\iota_{i}\left(\frac{(\bar{\omega}\times\hat{v})_{i}}{1+\hat{v}\cdot\bar{\omega}}\right)f(0,\bar{y},v)dv\frac{dS_{y}}{t}\\ &+(-1)^{i}2(1-\delta_{i3})\int_{B(x;t)\cap\{y_{3}=0\}}\int_{\mathbb{R}^{3}}\frac{\hat{v}_{\underline{i}}f(t-|y-x|,y_{\parallel},0,v)}{|y-x|}dvdS_{y}.\end{split}

(92)

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	$\displaystyle\ref{Eexpanmajor1}=-\int_{B(x;t)\cap\{y_{3}>0\}}\frac{(\partial_{i}\mathbf{\rho}+\partial_{t}{J}_{i})(t-\|y-x\|,y)}{\|y-x\|}\mathrm{d}y$
	$\displaystyle\ \ =-\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}(\partial_{i}f+\hat{v}_{i}\partial_{t}f)(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$
	$\displaystyle\ \ =-\int_{B(x;t)\cap\{y_{3}>0\}}\int_{\mathbb{R}^{3}}\frac{\omega_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\omega}(Sf)(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$		(37)
	$\displaystyle\ \ \ \ \ -\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}>0\}\end{subarray}}\int_{\mathbb{R}^{3}}\left(\delta_{ij}-\frac{(\omega_{i}+\hat{v}_{i})\hat{v}_{j}}{1+\hat{v}\cdot\omega}\right)T_{j}f(t-\|y-x\|,y,v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}.$		(38)

	$\displaystyle\ref{Eexpanmajor2}=\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}(\partial_{i}f+\hat{v}_{i}\partial_{t}f)(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$
	$\displaystyle\ \ =\int_{B(x;t)\cap\{y_{3}<0\}}\int_{\mathbb{R}^{3}}\frac{\bar{\omega}_{i}+\hat{v}_{i}}{1+\hat{v}\cdot\bar{\omega}}(Sf)(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}$		(47)
	$\displaystyle\ \ +\int_{\begin{subarray}{c}B(x;t)\\ \cap\{y_{3}<0\}\end{subarray}}\int_{\mathbb{R}^{3}}\Big{(}\delta_{ij}-\frac{\bar{\omega}_{i}\hat{v}_{j}+\hat{v}_{i}\hat{v}_{j}}{1+\hat{v}\cdot\bar{\omega}}\Big{)}\bar{T}_{j}f(t-\|y-x\|,\bar{y},v)\mathrm{d}v\frac{\mathrm{d}y}{\|y-x\|}.$		(48)