# Focusing solutions of Vlasov-Poisson equation on a spherical shell

Kaiwen Tan<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

We study smooth, spherically-symmetric solutions to the Vlasov-Poisson system
and relativistic Vlasov-Poisson system in the plasma physical case. In
particular, we construct solutions that initially possess arbitrarily small
charge densities and electric fields, but attain arbitrarily large value of
these quantities and concentrate on any given spherical shell at some later
time.

††footnotetext: MSC2020: Primary 00A05, Secondary 00A66.

## 1\. Introduction

We consider the single-species Vlasov-Poisson (VP) system:

(1) $\displaystyle\partial_{t}f+v\cdot\triangledown
f+E\cdot\triangledown_{v}f=0,$
$\displaystyle\rho(t,x)=\int_{\mathbb{R}^{3}}f(t,x,v)dv,$ $\displaystyle
E(t,x)=\int_{\mathbb{R}^{3}}\frac{x-y}{|x-y|^{3}}\rho(t,y)dy.$

Here $t>0$ represents time, $x\in\mathbb{R}^{3}$ is the particle position,
$v\in\mathbb{R}^{3}$ is the particle momentum. Also, $f(t,x,v)>0$ is the
density distribution of the particles, $\rho(t,x)$ is the associated charge
density, and $E(t,x)$ is the electric field generated by the charged
particles.

Also, by taking relativistic effects into account, we consider the
relativistic Vlasov-Poisson (RVP) system:

(2) $\displaystyle\partial_{t}f+\hat{v}\cdot\triangledown
f+E\cdot\triangledown_{v}f=0,$
$\displaystyle\rho(t,x)=\int_{\mathbb{R}^{3}}f(t,x,v)dv,$ $\displaystyle
E(t,x)=\int_{\mathbb{R}^{3}}\frac{x-y}{|x-y|^{3}}\rho(t,y)dy,$

while

$\hat{v}=\frac{v}{\sqrt{1+|v|^{2}}}$

is the relativistic velocity.
To complete the description of the systems, we consider the particle
distribution at time zero, denoted as

$f_{0}(x,v)=f(0,x,v)\geq 0.$

Then an important property of both the VP and RVP system, which will be
utilized throughout the paper, is the conservation of total mass [5], namely

$M(t)=\int\int_{\mathbb{R}^{6}}f(t,x,v)dvdx=\int\int_{\mathbb{R}^{6}}f_{0}(x,v)dvdx.$

In this paper, we assume both systems to be spherically symmetric. It is an
established result that for both systems, spherically symmetric initial data
gives rise to global-in-time, spherically symmetric solutions (see [3], [6],
and [7]). Moreover, [3] proves that the charge density and electric field
generated by spherically symmetric solutions of both systems eventually decay
in the $L^{\infty}$ sense. Meanwhile, in the paper [1] and [2] by J. Ben-
Artzi, S. Calogero and S. Pankavich, it is shown that for both systems, one
can construct spherically symmetric solutions with arbitrarily small initial
density and electric field, but over intermediate timescales, the particles
will concentrate to create arbitrarily large density and field. The result is
stated as follows:

######  Theorem 1.

(J. Ben-Artzi, S. Calogero and S. Pankavich) For any constants $C_{1}$ and
$C_{2}$, there exists a smooth, spherically symmetric of the Vlasov-Poisson or
the relativistic Vlasov-Poisson system such that

$\lVert\rho(0)\rVert_{\infty},\quad\lVert E(0)\rVert_{\infty}\leq C_{1}$

but for some time $T>0$,

$\lVert\rho(T)\rVert_{\infty},\quad\lVert E(T)\rVert_{\infty}\geq C_{2}.$

This result is in fact consistent with the decay estimate given by E. Horst in
[3], showing that we can construct cases of focusing solutions in a short time
period, even though they will eventually decay.

However, one may notice that in both papers [1] and [2], the focusing
solutions are constructed in such a way that particles are concentrated
closely around the origin at time $T$. In the case of the VP system, [1]
requires $\mathscr{R}(T)$, the spatial radius of the particles at time $T$, to
satisfy $\mathscr{R}(T)\leq 100\epsilon^{2}$, where $\epsilon>0$ is a
sufficiently small constant. Similarly in [2], the solution of the RVP system
satisfies $\mathscr{R}(T)\leq 20\epsilon$ for small $\epsilon$. This is
because the methodology applied in both papers is to constraint the particles
within a thin shield around the origin at time $T$. Since the supremums of
both charge density and electric field are inversely related to the volume of
the space filled with particles, this method allows the solutions to achieve
arbitrarily large $\lVert\rho(T)\rVert_{\infty}$ and $\lVert
E(T)\rVert_{\infty}$.

Naturally, one may raise the following question: can a similar result be
established by having the particles concentrate within an arbitrary given
shield, but not just close to the origin? Since the shield is randomly away
from the origin, the particles are more scatteringly distributed within a
larger spatial radius, raising the difficulty of whether we can still achieve
the concentrating effect in charge density and electric field. But ideally
this can be achieve by having the particles concentrate within a thin enough
shield. This paper aims at justifying this idea, namely proving that we can in
fact construct focusing solutions for both the VP and RVP systems that have
particles initially distributed in any given range away form the origin and
concentrated within any targeting shield; and such solutions possess a similar
concentrating behavior in charge density and electric field. Specifically,
this paper aims at proving the following result:

######  Theorem 2.

For any constant $C_{1}$, $C_{2}>0$ and $c>b>a>0$, there exists a smooth,
spherically symmetric solution of the 1) Vlasov-Poisson system or 2)
relativistic Vlasov-Poisson system such that for any particle in the initial
data, if we denote its spatial radius at time $t\geq 0$ by $\mathscr{R}(t)$,
then

$\mathscr{R}(0)\geq c,\quad\text{and}\quad a\leq\mathscr{R}(T)\leq b$

for some time $T>0$.
Moreover, we have the estimation that

$\lVert\rho(0)\rVert_{\infty},\quad\lVert E(0)\rVert_{\infty}\leq C_{1}$

but for some time $T>0$,

$\lVert\rho(T)\rVert_{\infty},\quad\lVert E(T)\rVert_{\infty}\geq C_{2}$

Theorem 2 is the result of combining Theorem 6 about the VP system and Theorem
7 about the RVP system, which we will prove in Section 3 and 4 respectively.
In the following Section 2, I will first provide some background information
concerning the spherical symmetry assumption we made and establish several
useful lemmas. Then in Section 3, I will construct a focusing solution for the
VP system, by first displaying our settings for parameters and initial data,
and then prove that the solution satisfies the theorem. Finally in Section 4,
I will construct a focusing solution for the RVP system with similar
methodology but different data and details.

## 2\. Backgrounds and Lemmas

### 2.1. Spherical Symmetry

In this paper, we assume both the VP and RVP systems to be spherical
symmetric. In particular, for both systems, we define the spatial radius $r$,
inward velocity $w$, and square of the angular momentum $l$ of particles at
initial time as follows:

$r=|x|,\quad w=\frac{x\cdot v}{r},\quad l=|x\times v|^{2}.$

Let $f=f(t,r,w,l)$ be the density distribution of the particles under the
spherically symmetric setting. Then using change of variable
$(x,v)\rightarrow(r,w,l)$ (see [1] and [4] for more details), we can reduce
the Vlasov-Poisson system to:

(3)
$\partial_{t}f+w\partial_{r}f+(\frac{l}{r^{3}}+\frac{m(t,r)}{r^{2}})\partial_{w}f=0.$

Similarly (see [2]), the relativistic Vlasov-Poisson system can be reduced to:

(4)
$\partial_{t}f+\frac{w}{\sqrt{1+w^{2}+lr^{-2}}}\partial_{r}f+\left(\frac{l}{r^{3}\sqrt{1+w^{2}+lr^{-2}}}+\frac{m(t,r)}{r^{2}}\right)\partial_{w}f=0.$

In both equations, we know

(5) $m(t,r)=4\pi\int^{r}_{0}s^{2}\rho(t,s)ds,$

and

(6)
$\rho(t,r)=\frac{\pi}{r^{2}}\int^{\infty}_{0}\int^{\infty}_{-\infty}f(t,r,w,l)dw\,dl.$

Also, the electric field is given by

(7) $E(t,x)=\frac{m(t,r)}{r^{2}}\frac{x}{r}$

in both systems.
The total mass of the particles is

(8)
$M=4\pi^{2}\int^{\infty}_{0}\int^{\infty}_{-\infty}\int^{\infty}_{0}f_{0}(r,w,l)dldwdr,$

which, as stated before, is conserved.

Following these descriptions, we can express the characteristics of the
Vlasov-Poisson system in such form:

(9) $\displaystyle\frac{d}{ds}\mathscr{R}(s)=\mathscr{W}(s),$
$\displaystyle\frac{d}{ds}\mathscr{W}(s)=\frac{\mathscr{L}(s)}{\mathscr{R}(s)^{3}}+\frac{m(s,\mathscr{R}(s))}{\mathscr{R}(s)^{2}},$
$\displaystyle\frac{d}{ds}\mathscr{L}(s)=0,$

with the initial conditions

(10) $\mathscr{R}(0)=r,\quad\mathscr{W}(0)=w,\quad\mathscr{L}(0)=l.$

Since the angular momentum of particles is conserved in time on the support of
$f(t)$, we have that $\mathscr{L}(s)=l$ for all $s\geq 0$.

Similarly, we can express the characteristics of the relativistic Vlasovi-
Poisson system as:

(11)
$\displaystyle\frac{d}{ds}\mathscr{R}(s)=\frac{\mathscr{W}(s)}{\sqrt{1+\mathscr{W}(s)^{2}+\mathscr{L}(s)\mathscr{R}(s)^{-2}}}$
$\displaystyle\frac{d}{ds}\mathscr{W}(s)=\frac{\mathscr{L}(s)}{\mathscr{R}(s)^{3}\sqrt{1+\mathscr{W}(s)^{2}+\mathscr{L}(s)\mathscr{R}(s)^{-2}}}+\frac{m(s,\mathscr{R}(s))}{\mathscr{R}(s)^{2}},$
$\displaystyle\frac{d}{ds}\mathscr{L}(s)=0,$

which is also subject to the initial conditions

(12) $\mathscr{R}(0)=r,\quad\mathscr{W}(0)=w,\quad\mathscr{L}(0)=l.$

and conservation of angular momentum as well.

Finally, for both systems, we denote

$S(t)=\\{(r,w,l):f(t,r,w,l)>0\\},$

for $t\geq 0$. In particular, we have

$S(0)=\\{(r,w,l):f_{0}(r,w,l)>0\\}.$

### 2.2. Lemmas

We now give some lemmas that are useful for our proof of the theorem later on.
Most parts of these lemmas are established by Jonathan Ben-Artzi, Simone
Calogero, and Stephen Pankavich in [1] and [2].

The first lemma mainly concerns with the behavior of the characteristics of
the VP system as described by the ODE system in 9 and 10. In particular, we
want to derive an estimate for the particle spatial radius at given time $T>0$
using the initial parameters $r$, $w$, $l$, total mass $M$, and the time $T$.

######  Lemma 3.

Let $l>0$, $M\geq 0$, $r>0$, and $w<0$ be given. Also, let
$(\mathscr{R}(t),\,\mathscr{W}(t),\,\mathscr{L}(t))$ be a solution to 9 and 10
for all $t\geq 0$. Then
(1). There exists a unique $T_{0}>0$ such that $\mathscr{W}(t)<0$ for
$t\in[0,\,T_{0})$, $\mathscr{W}(T_{0})=0$, and $\mathscr{W}(t)>0$ for
$t\in(T_{0},\,\infty)$.
(2). $T_{0}$ satisfies

$T_{0}>\frac{r}{|w|}\left(1-\frac{\sqrt{l+Mr}}{r|w|}\right)=\frac{r}{|w|}-\frac{\sqrt{l+Mr}}{w^{2}}.$

(3). For all $t\in[0,\,T_{0})$, we have

$\mathscr{R}(t)^{2}\leq(r+wt)^{2}+(lr^{-2}+Mr^{-1})t^{2}.$

(4). Let $0<T<T_{0}$. Then $\mathscr{R}(T)$ satisfies

$\mathscr{R}(T)\geq\frac{l}{2}r^{-3}T^{2}+wT+r.$

Proof.  One can find the detailed proof of (1), (2), and (3) from the lemma 3
and 4 in [1]. Thus we only prove (4) here. (1) indicates that for $t\leq
T_{0}$, $\mathscr{W}(t)=\dot{\mathscr{R}}(t)\leq 0$. Therefore, choose time
$T$ such that $0<T<T_{0}$, then we know that
$\mathscr{R}(T)\leq\mathscr{R}(0)=r$. Substituting this result into 9 gives

$\ddot{\mathscr{R}}(T)\geq l\mathscr{R}(T)^{-3}\geq l\mathscr{R}(0)^{-3}.$

Thus we can further deduce that

$\dot{\mathscr{R}}(T)\geq lR(0)^{-3}T+\dot{\mathscr{R}}(0),$
$\mathscr{R}(T)\geq\frac{l}{2}\mathscr{R}(0)^{-3}T^{2}+\dot{\mathscr{R}}(0)T+\mathscr{R}(0),$

which gives the result. $\Box$ The second lemma is an analogue of Lemma 3 for
the ODE system associated with the RVP system, as described in 11 and 12.

######  Lemma 4.

Let $l>0$, $M\geq 0$, $r>0$, and $w<0$ be given. Also, let
$(\mathscr{R}(t),\,\mathscr{W}(t),\,\mathscr{L}(t))$ be a solution to 11 and
12 for all $t\geq 0$, and define

$D=l+Mr\sqrt{1+w^{2}+lr^{-2}}.$

Then
(1). There exists a unique $T_{0}>0$ such that $\mathscr{W}(t)<0$ for
$t\in[0,\,T_{0})$, $\mathscr{W}(T_{0})=0$, and $\mathscr{W}(t)>0$ for
$t\in(T_{0},\,\infty)$.
(2). $T_{0}$ satisfies the bounds

$r\left(1-\sqrt{\frac{D}{r^{2}w^{2}+D}}\right)\leq
T_{0}\leq\frac{-wr^{3}\sqrt{1+w^{2}+lr^{-2}}}{l}.$

(3). For all $t\in[0,\,T_{0})$, we have

$\mathscr{W}(t)^{2}+l\mathscr{R}(t)^{-2}\leq w^{2}+lr^{-2}.$

(4). For all $t\in[0,\,T_{0})$, we have

$\mathscr{R}(t)^{2}\leq\left(r-\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}t\right)^{2}+\frac{D}{r^{2}(1+w^{2}+lr^{-2})}t^{2}.$

(5). Let $0<T<T_{0}$. Then $\mathscr{R}(T)$ satisfies

$\mathscr{R}(T)\geq\frac{l}{2r^{3}(1+w^{2}+lr^{-2})}T^{2}+\frac{w}{\sqrt{1+w^{2}+lr^{-2}}}T+r.$

Proof.  One can find the detailed proof for (1) to (4) from lemma 3 in [2], so
we prove for (5) here. (3) informs us that

$\frac{d}{ds}\mathscr{R}(s)=\frac{\mathscr{W}(s)}{\sqrt{1+\mathscr{W}(s)^{2}+\mathscr{L}(s)\mathscr{R}(s)^{-2}}}\geq\frac{\mathscr{W}(s)}{\sqrt{1+w^{2}+lr^{-2}}},$

and

$\frac{d}{ds}\mathscr{W}(s)\geq\frac{\mathscr{L}(s)}{\mathscr{R}(s)^{3}\sqrt{1+\mathscr{W}(s)^{2}+\mathscr{L}(s)\mathscr{R}(s)^{-2}}}\geq\frac{\mathscr{L}(s)}{r^{3}\sqrt{1+w^{2}+lr^{-2}}}$

Thus

$\mathscr{W}(T)\geq\frac{l}{r^{3}\sqrt{1+w^{2}+lr^{-2}}}T+w.$

Then we have

$\frac{d}{ds}\mathscr{R}(T)\geq\frac{l}{r^{3}(1+w^{2}+lr^{-2})}T+\frac{w}{\sqrt{1+w^{2}+lr^{-2}}}$

and finally

$\mathscr{R}(T)\geq\frac{l}{2r^{3}(1+w^{2}+lr^{-2})}T^{2}+\frac{w}{\sqrt{1+w^{2}+lr^{-2}}}T+r.$

$\Box$ The last lemma provides us with a lower bound for the $L^{\infty}$ norm
of the particle charge density and the electric field at time $t>0$, which can
be applied to both the VP and RVP system.

######  Lemma 5.

Let $f(t,r,w,l)$ be a spherically symmetric solution of the VP or RVP system
with associated charge density $\rho(t,r)$ and electric field $E(t,x)$.
Also, let
$(\mathscr{R}(t,0,r,w,l),\,\mathscr{W}(t,0,r,w,l),\,\mathscr{L}(t,0,r,w,l))$
be a characteristic solution of 9. If at some $T\geq 0$ we have

$a\leq\sup_{(r,w,l)\in S(0)}\mathscr{R}(T,0,r,w,l)\leq b,$

then

$\lVert\rho(T)\rVert_{\infty}\geq\frac{3M}{4\pi(b^{3}-a^{3})},$

and

$\lVert E(T)\rVert_{\infty}\geq\frac{M}{b^{2}}.$

Proof.  Please see lemma 5 in [1] or lemma 4 in [2]. $\Box$

## 3.Focusing Solution of the Vlasov-Poisson System

In this section we attempt to establish the following result:

######  Theorem 6.

For any constant $C_{1}$, $C_{2}>0$ and $c>b>a>0$, there exists a smooth,
spherically symmetric solution of the Vlasov-Poisson system such that for
$(r,w,l)\in S(0)$ we have

$r\geq c,\quad\text{and}\quad a\leq\mathscr{R}(T,0,r,w,l)\leq b$

for some time $T>0$.
Moreover, we have the estimation that

$\lVert\rho(0)\rVert_{\infty},\quad\lVert E(0)\rVert_{\infty}\leq C_{1}$

and

$\lVert\rho(T)\rVert_{\infty},\quad\lVert E(T)\rVert_{\infty}\geq C_{2}$

### 3.1. Parameters and Initial Data Settings

We now begin the construction of the focusing solution. Given constants
$C_{1},\,C_{2}>0$ and $c>b>a>0$, we set a few parameters and choose a smooth,
spherically symmetric solution of the VP system that satisfies the following
conditions:

  1. (i)

Define constant $k=(1+\frac{a^{2}}{b^{2}})\div 2=\frac{a^{2}+b^{2}}{2b^{2}}$,
and
$\epsilon=\frac{\sqrt{k}b-a}{8}=\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{8\sqrt{2}}$.
Also, choose $a_{0}>0$ large enough such that

(13) $\displaystyle a_{0}\geq\text{max}\\{$
$\displaystyle\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{8\sqrt{2}}+c,\,C_{1}^{-1},\,3\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-1},\,$
$\displaystyle
2^{\frac{-11}{6}}\pi^{\frac{1}{3}}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-\frac{1}{3}},\,\frac{16\sqrt{2}(b^{3}-a^{3})C_{2}}{3(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)},\,\frac{4\sqrt{2}b^{2}C_{2}}{\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}\\}.$

  2. (ii)

For $(r,w,l)\in S(0)$, we have

(14) $a_{0}-\epsilon<r<a_{0}+\epsilon.$

Moreover, the initial charge density $\rho_{0}$ satisfies

(15) $\rho_{0}(r)\leq\frac{1}{a_{0}}\quad\text{for all }r>0,$

and

(16) $\rho_{0}(r)=\frac{1}{2a_{0}}\quad\text{for
}r\in[a_{0}-\frac{1}{2}\epsilon,\,a_{0}+\frac{1}{2}\epsilon].$

  3. (iii)

We define time $T>0$ small enough such that

(17) $\displaystyle T\leq\text{min}\\{$
$\displaystyle\frac{a(|\sqrt{k}b-a_{0}-\epsilon|)}{\sqrt{l+(2\pi
a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon)(a_{0}-\epsilon)}},\,$
$\displaystyle\sqrt{\frac{1-k}{l(a_{0}-\epsilon)^{-2}+(8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3})(a_{0}-\epsilon)^{-1}}}b\,\\}.$

  4. (iv)

Finally, for $(r,w,l)\in S(0)$, we let

(18) $\frac{a-a_{0}+\epsilon}{T}\leq w\leq\frac{\sqrt{k}b-a_{0}-\epsilon}{T},$

and we require $l>0$.

######  Remark.

In order for the inequality in 18 to be valid, we need to check
$a-a_{0}+\epsilon\leq\sqrt{k}b-a_{0}-\epsilon$, namely,

$\epsilon<\frac{\sqrt{k}b-a}{2}.$

Our choice of constant $\epsilon$ can verify this point.

######  Remark.

To see that these choices of parameters are feasible, we now provide an
example of a smooth function that satisfy all the above conditions.

First of all, based on the given constants $C_{1}$, $C_{2}$, $a$, $b$, and
$c$, we first fix constants $k$, $\epsilon$, and $a_{0}$ that satisfy the
requirements in (i), and choose constant $T>0$ that it satisfies (iii).
Now we choose $H:\,[0,\infty)\rightarrow[0,\infty)$ to be a smooth function
that satisfies

$\int_{\mathbf{R}^{3}}H(|u|^{2})du=\frac{T^{3}}{2a_{0}}$

with $supp(H)\subset[0,1]$. We then re-scale this function with a scalar
$\delta>0$ by defining

$H_{\delta}(|u|^{2})=\frac{1}{\delta^{3}}H(\frac{|u|^{2}}{\delta^{2}}).$

Then we will find that

$\int_{\mathbf{R}^{3}}H_{\delta}(|u|^{2})du=\frac{T^{3}}{2a_{0}}$

and $supp(H_{\delta})\subset[0,\delta^{2}]$.

Next, we choose a smooth cut-off function $\chi\in
C^{\infty}((0,\infty),[0,1])$ such that

$\begin{cases}\chi(r)=1&\text{for }\quad a_{0}-\frac{1}{2}\epsilon\leq|r|\leq
a_{0}+\frac{1}{2}\epsilon\\\ \chi(r)=0&\text{for }\quad|r|\leq
a_{0}-\epsilon\text{ or }|r|\geq a_{0}+\epsilon\\\ \end{cases}$

We then fix constants $d=\frac{a+\sqrt{k}b}{2}$,
$\delta=\frac{\sqrt{k}b-a-4\epsilon}{2}=\frac{\sqrt{k}b-a}{4}>0$, and define
function

$f_{0}(x,v)=H_{\delta}(|\left(1-\frac{d}{|x|}\right)x+Tv|^{2})\chi(|x|).$

Clearly $f_{0}$ is smooth, by claiming that it satisfies the requirements in
(ii) and (iv) we finish our construction of the solution.

To verify this, we first notice that $r=|x|$, so that we can obtain from the
definition of $\chi$ that

$a_{0}-\epsilon\leq r\leq a_{0}+\epsilon.$

Then an immediate observation is that $d=\frac{a+\sqrt{k}b}{2}\leq
a_{0}-\epsilon\leq r$ by our choice of parameters in (i). Thus we know
$0<1-\frac{d}{|x|}<1$.
Next, the support of $h_{\delta}$ indicates us that

$|\left(1-\frac{d}{|x|}\right)x+Tv|^{2}\leq\delta^{2}.$

Using change of variables and the relation $|v|^{2}=w^{2}+lr^{-2}$, we can
obtain that

$(r-d+Tw)^{2}+T^{2}lr^{-2}\leq\delta^{2},$

and thus $|r-d+Tw|\leq\delta$. Expanding this inequality informs us that

$\frac{d-\delta-r}{T}\leq w\leq\frac{d+\delta-r}{T}.$

We’ve set $d$ and $\delta$ such that $d-\delta=a+2\epsilon$ and
$d+\delta=\sqrt{k}b-2\epsilon$. Moreover, we know that $a_{0}-\epsilon\leq
r\leq a_{0}+\epsilon$. Therefore we can deduce that

$\frac{a+2\epsilon-(a_{0}+\epsilon)}{T}\leq\frac{d-\delta-r}{T}\leq
w\leq\frac{d+\delta-r}{T}\leq\frac{\sqrt{k}b-2\epsilon-(a_{0}-\epsilon)}{T}.$

which satisfies the requirement in (iv).

Finally, to check that the solution satisfies (ii), we notice that

$\int_{\mathbf{R}^{3}}H_{\delta}(|\left(1-\frac{d}{|x|}\right)x+Tv|^{2})dv=\frac{1}{T^{3}}\int_{\mathbf{R}^{3}}H_{\delta}(|u|^{2})du=\frac{1}{2a_{0}}.$

and thus we can calculate

$\rho_{0}(x)=\int_{\mathbf{R}^{3}}f_{0}(x,v)=\left(\int_{\mathbf{R}^{3}}H_{\delta}(|\left(1-\frac{d}{|x|}\right)x+Tv|^{2})dv\right)\chi(|x|)=\frac{1}{2a_{0}}\chi(|x|),$

which satisfies the requirements in (ii).
Therefore, we’ve verified $f_{0}(x,v)$ to be a smooth function that satisfies
all the settings for the initial data.

######  Remark.

The choice of $a_{0}$ might seem complicated, but using the expression of
orders we can see how it’s connected to other given constants.

  * •

$a_{0}\sim O(C_{1}^{-1})$. $C_{1}$ is the required upper bound for
$\lVert\rho(0)\rVert_{\infty}$ and $\lVert E(0)\rVert_{\infty}$. $a_{0}$ being
inversely related to $C_{1}$ indicates that at time zero, if we want the
charge density and electric field to be small, then the particles should be
placed further from the origin, so that they’re distributed more scatteringly.

  * •

$a_{0}\sim O(C_{2})$. $C_{2}$ is the required lower bound for
$\lVert\rho(T)\rVert_{\infty}$ and $\lVert E(T)\rVert_{\infty}$. $a_{0}$ being
positively relating to $C_{2}$ indicate that larger concentrating density and
electric field requires larger spatial radius of particle distribution at time
zero. This is generally because such distribution allows a larger total mass
of particles, which makes a stronger concentration effect possible.

  * •

$a_{0}\geq\frac{8\sqrt{2}(b^{3}-a^{3})C_{2}}{3(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}\geq\frac{2(b^{3}-a^{3})C_{2}}{3\epsilon}$,
showing that $a_{0}$ is also positively related to the thickness and volume of
the targeting shield of concentration, which is $\sim O(b^{3}-a^{3})$.

Moreover, we can use the conditions on $a_{0}$ to derive an estimation on the
scale and order of other parameters, such as the total particle mass $M$ and
the initial charge density $\rho_{0}$. In fact, these estimates are the
strongest motivation behind our complicated choice of $a_{0}$, as we will see
how they help with our calculation and proof later on.

  * •

16 and 15 indicates that $\rho_{0}\sim O(a_{0}^{-1})$. The inverse
relationship between $\rho_{0}$ and $a_{0}$ not only helps us meet the upper
bound requirement on $\lVert\rho(0)\rVert_{\infty}$, but also help us control
the order of the total particle mass.

  * •

An immediate result following the relationship between $\rho_{0}$ and $a_{0}$,
and the inequailties 14, 15, and 16, is that we can derive a range for the
total particle mass $M$:

(19) $\displaystyle M$
$\displaystyle=\int_{\mathbb{R}^{3}}\rho_{0}(x)dx=4\pi\int^{a_{0}+\epsilon}_{a_{0}-\epsilon}\rho_{0}(r)r^{2}dr\leq\frac{4\pi}{a_{0}}\int^{a_{0}+\epsilon}_{a_{0}-\epsilon}\rho_{0}(r)r^{2}dr$
$\displaystyle\leq 8\pi a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3},$

and

(20) $\displaystyle M$ $\displaystyle\geq
4\pi\int^{a_{0}+\frac{1}{2}\epsilon}_{a_{0}-\frac{1}{2}\epsilon}\rho_{0}(r)r^{2}dr=\frac{2\pi}{a_{0}}\int^{a_{0}+\frac{1}{2}\epsilon}_{a_{0}-\frac{1}{2}\epsilon}r^{2}dr$
$\displaystyle=2\pi a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon^{3}.$

  * •

Finally, 18 tells us that

$a-a_{0}+\epsilon\leq a-r\leq wT\leq\sqrt{k}b-r\leq\sqrt{kb}-a_{0}-\epsilon.$

$r+wT$ is a rough estimation of the particle position at time $T$. Thus our
choice of $T$ in 17 and $w$ in 18 is meant to control the particle position at
time of concentration.

### 3.2. Proof of Theorem 6

Proof.  Our strategy is to set the particles far away from the targeting
shield at time zero, and endow them with large enough inward velocity. Then by
balancing between the initial inward velocity $w$ and the time $T$, we can
make the particles concentrate within the targeting shield in a short time
period while they’re still moving inwards. Since we set the initial particle
spatial radius $r$ to be large enough, the particles are scattering at time
zero and have low charge density and electric field. Meanwhile, the total
particle mass $M$ is controlled to be large enough, so that both the charge
density and electric field will rise drastically once the particles
concentrate at time $T$.

We first check the particles’ distributing region at time zero. For
$(r,w,l)\in S(0)$, we have

$r>a_{0}-\epsilon=a_{0}-\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{8\sqrt{2}}>c$

by our choice of $a_{0}$. Thus the particles are initially distributed in the
given region.
Next, we check that the initial charge density and electric field are small
enough. At time zero, we notice that by 15,

$\lVert\rho(0)\rVert_{\infty}\leq\frac{1}{a_{0}}\leq C_{1}.$

Moreover, using the conditions in 7 and 19, we can derive an upper bound for
the initial electric field in each region.

  1. (1)

For $|x|<a_{0}-\epsilon$, since the particles have initial spatial radius
$r\geq a_{0}-\epsilon$, this region is empty and we know that $|E(0,x)|=0$.

  2. (2)

For $|x|>a_{0}+\epsilon$, we know from 7 that

$|E(0,x)|\leq\frac{M}{r^{2}}\leq\frac{8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3}}{(a_{0}+\epsilon)^{2}}\leq
8\pi a_{0}^{-1}\epsilon+\frac{8\pi}{3}a_{0}^{-3}\epsilon^{3}.$

  3. (3)

Lastly, for $a_{0}-\epsilon\leq|x|\leq a_{0}+\epsilon$, we have

$|E(0,x)|\leq\frac{M}{r^{2}}\leq\frac{8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3}}{(a_{0}-\epsilon)^{2}}\leq
32\pi a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}.$

Therefore, an uniform upper bound for the initial electric field is

$\lVert E(0)\rVert_{\infty}\leq 32\pi
a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3},$

and we want to show that this value is bounded by $C_{1}$. Since this value is
inversely related to $a_{0}$, we should be able to do so by choosing $a_{0}$
large enough. Specifically, we’ve chosen
$a_{0}\geq\text{max}\\{3\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-1},\,2^{\frac{-11}{6}}\pi^{1\over
3}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-\frac{1}{3}}$. Then we can calculate
an upper bound for the two terms separately that

$32\pi
a_{0}^{-1}\epsilon=2\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)a_{0}^{-1}\leq\frac{2}{3}C_{1}$

and

$\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}=\frac{\sqrt{2}\pi}{192}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)^{3}a_{0}^{-3}\leq\frac{1}{3}C_{1}.$

Thus by adding these results together, we can get

$\lVert E(0)\rVert_{\infty}\leq 32\pi
a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}\leq C_{1}.$

Therefore, we’ve proven that our solution satisfies the conditions on the
initial particle spatial radius, charge density, and electric field.

We now look at time $T$, which is the presumptive time for the particles to
concentrate. Lemma 3 informs us that particles with inward initial velocity
will first move inwards until a time $T_{0}$ and then move outwards. To have a
more accurate control over the particle location,we want the concentration to
happen in a short time period during which the particles are still moving
inwards. This is to say, we want $T<T_{0}$. To meet this requirement, Lemma 3
implies that we will need

(21) $T\leq\frac{r}{|w|}-\frac{\sqrt{l+Mr}}{w^{2}}\quad\text{or }\quad
wT\geq-r+\frac{\sqrt{l+Mr}}{|w|}.$

Now, in 17 we’ve asked
$T\leq\frac{a(|\sqrt{k}b-a_{0}-\epsilon|)}{\sqrt{l+(2\pi
a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon)(a_{0}-\epsilon)}}$. Since for
$(r,w,l)\in S(0)$, we’ve let $\frac{a-a_{0}+\epsilon}{T}\leq
w\leq\frac{\sqrt{k}b-a_{0}-\epsilon}{T}$, we can bound $|w|$ by

$|w|\geq\left|\frac{\sqrt{k}b-a_{0}-\epsilon}{T}\right|\geq\frac{\sqrt{l+(2\pi
a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon)(a_{0}-\epsilon)}}{a}\geq\frac{\sqrt{l+Mr}}{a}.$

Therefore,

$\frac{\sqrt{l+Mr}}{|w|}\leq a=(a-a_{0}+\epsilon)+(a_{0}-\epsilon)\leq wT+r,$

so that the inequality in 21 is satisfied, and we’ve established $T\leq T_{0}$
as we want.

Now that the concentration is proven to happen in a short time period while
the particles are moving inwards, we want to further verify that the particles
are located within the targeting shield at time $T$. Specifically, we want
that for all $(r,w,l)\in S(0)$,

$a\leq\mathscr{R}(T,0,r,w,l)\leq b.$

To do so, we can utilize the lower and upper bound for $\mathscr{R}(T)$ we’ve
developed in Lemma 3. Specifically, we will need

(22) $\frac{l}{2}r^{-3}T^{2}+wT+r\geq a$

and

(23) $(r+wT)^{2}+(lr^{-2}+Mr^{-1})T^{2}\leq b^{2}.$

Notice that the term $wT$ appears in both inequalities. A physical
interpretation is that this term, the product of initial inward velocity and
time of concentration, is a rough estimation of the distance the particles
travelled inward in a short time period $T$. Thus controlling this term will
help us satisfy both inequalities 22 and 23. The range we choose for $w$ in 18
tells us that

$a-a_{0}+\epsilon\leq wT\leq\sqrt{k}b-a_{0}-\epsilon.$

Then immediately we have

$\frac{l}{2}r^{-3}T^{2}+wT+r\geq
wT+r\geq(a-a_{0}+\epsilon)+(a_{0}-\epsilon)=a,$

so that 22 is satisfied.
Similarly, since $wT\leq\sqrt{k}b-a_{0}-\epsilon$, we have
$r+wT\leq\sqrt{k}b$, so that

$(r+wT)^{2}\leq kb^{2}.$

Then in order to verify 23, we only need to show that the remaining term on
the left side, $(r^{-2}+Mr^{-1})T^{2}$, is smaller than $b^{2}-kb^{2}$. Since
we have estimated $r$ and $M$ using constants $a_{0}$ and $\epsilon$, this
inequality can be achieved simply by setting $T$ small enough. Specifically,
in 17 we’ve set $T\leq\sqrt{\frac{1-k}{l(a_{0}-\epsilon)^{-2}+(8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3})(a_{0}-\epsilon)^{-1}}}b$.
Plug this value into the term $(r^{-2}+Mr^{-1})T^{2}$ gives

$(r^{-2}+Mr^{-1})T^{2}\leq\left(l(a_{0}-\epsilon)^{-2}+(8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3})(a_{0}-\epsilon)^{-1}\right)T^{2}\leq(1-k)b^{2}$

Therefore, combining these two result, we have

$(r+wT)^{2}+(lr^{-2}+Mr^{-1})T^{2}\leq kb^{2}+(1-k)b^{2}=b^{2},$

so that 23 is confirmed. As we’ve verified both inequalities 22 and 23, we can
use Lemma 3 to confirm that for all $(r,w,l)\in S(0)$, we have
$a\leq\mathscr{R}(T)\leq b$.

Finally, we’re left to verify the charge density and electric field at time
$T$ is large enough. Since we’ve established that the particles will be
concentrated within $a\leq\mathscr{R}(T)\leq b$ at time $T$, we can use Lemma
5 to infer that both $\lVert\rho(T)\rVert_{\infty}$ and $\lVert
E(T)\rVert_{\infty}$ are inversely related to the total particle mass. More
specifically, Lemma 5 tells us that in order for
$\lVert\rho(T)\rVert_{\infty}$, $\lVert E(T)\rVert_{\infty}\geq C_{2}$, we
will need

$M\geq max\\{\frac{4\pi}{3}(b^{3}-a^{3})C_{2},\,b^{2}C_{2}\\}.$

Since we can estimate $M$ with

$2\pi a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon^{3}\geq 2\pi
a_{0}\epsilon,$

we know that $M$ is positively related with $a_{0}$. Therefore, we can satisfy
both conditions by setting $a_{0}$ large enough. In 24, we’ve set
$a_{0}\geq\frac{16\sqrt{2}(b^{3}-a^{3})C_{2}}{3(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}$.
Plug this value into our estimation of $M$ gives

$M\geq 2\pi
a_{0}\epsilon\geq\frac{(\pi\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}{4\sqrt{2}}a_{0}\geq\frac{4\pi}{3}(b^{3}-a^{3}).$

Similarly, we’ve asked
$a_{0}\geq\frac{4\sqrt{2}b^{2}C_{2}}{\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}$,
which provides us with

$M\geq 2\pi
a_{0}\epsilon\geq\frac{(\pi\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}{4\sqrt{2}}a_{0}\geq
b^{2}C_{2}.$

Thus by Lemma 5, $\lVert\rho(T)\rVert_{\infty}\geq C_{2}$ and $\lVert
E(T)\rVert_{\infty}\geq C_{2}$ is verified.

Till now, we’ve confirmed that the solution of the VP system we choose is
initially located at region $|x|\geq c$, and concentration within the
targeting shield $a\leq|x|\leq b$ at some time $T$. Moreover, we’ve verified
that the solution satisfies the conditions on charge density and electric
field at time zero and time $T$, namely $\lVert\rho(0)\rVert_{\infty}\leq
C_{1}$, $\lVert E(0)\rVert_{\infty}\leq C_{1}$; and
$\lVert\rho(T)\rVert_{\infty}\geq C_{2}$, $\lVert E(T)\rVert_{\infty}\geq
C_{2}$. Therefore, Theorem 6 is proven. $\Box$

## 4\. Focusing Solution of the Relativistic Vlasov-Poisson System

In this section, we want to prove a similar result for the relativistic
Vlasov-Poisson System, namely

######  Theorem 7.

For any constant $C_{1}$, $C_{2}>0$ and $c>b>a>0$, there exists a smooth,
spherically symmetric solution of the relativistic Vlasov-Poisson system such
that for $(r,w,l)\in S(0)$ we have

$r\geq c,\quad\text{and}\quad a\leq\mathscr{R}(T,0,r,w,l)\leq b,$

For some $T>0$.
Moreover, we have the estimation that

$\lVert\rho(0)\rVert_{\infty},\quad\lVert E(0)\rVert_{\infty}\leq C_{1},$

but

$\lVert\rho(T)\rVert_{\infty},\quad\lVert E(T)\rVert_{\infty}\geq C_{2}.$

### 4.1. Parameters and Initial Data Settings

We now begin constructing the focusing solution of the RVP system, which
shares certain degrees of similarity with the solution of the VP system in the
previous section, but is in fact more complicated in term of choice of
parameters and settings of initial data.

Given constants $C_{1},\,C_{2}>0$ and $c>b>a>0$, we introduce the following
parameters and choose a smooth, spherically symmetric solution of the RVP
system that satisfies the following conditions:

  1. (i)

Define constant $k=(1+\frac{a^{2}}{b^{2}})\div 2=\frac{a^{2}+b^{2}}{2b^{2}}$,
and
$\epsilon=\frac{\sqrt{k}b-a}{4}=\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{4\sqrt{2}}$.
Also, choose constant $a_{0}>0$ large enough such that

(24) $\displaystyle a_{0}\geq\text{max}\\{$
$\displaystyle\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{4\sqrt{2}}+c,\,C_{1}^{-1},\,6\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-1},\,$
$\displaystyle
2^{\frac{5}{6}}\pi^{\frac{1}{3}}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-\frac{1}{3}},\,\epsilon^{3},\,\left(\frac{\sqrt{86\pi}}{a}\right)^{3},\,2\epsilon,\,\left[\frac{344\pi}{(1-k)b^{2}}\right]^{\frac{3}{2}},\,$
$\displaystyle\frac{8\sqrt{2}(b^{3}-a^{3})C_{2}}{3(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)},\,\frac{2\sqrt{2}b^{2}C_{2}}{\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}\\}.$

  2. (ii)

For $(r,w,l)\in S(0)$, we want

(25) $a_{0}-\epsilon<r<a_{0}+\epsilon.$

Moreover, the initial charge density $\rho_{0}$ satisfies

(26) $\rho_{0}(r)\leq\frac{1}{a_{0}}\quad\text{for all }r>0,$

and

(27) $\rho_{0}(r)=\frac{1}{2a_{0}}\quad\text{for
}r\in[a_{0}-\frac{1}{2}\epsilon,\,a_{0}+\frac{1}{2}\epsilon].$

  3. (iii)

Denote

$N=\frac{a_{0}-\epsilon-a}{a_{0}+\epsilon-\sqrt{k}b}.$

Our definition of $\epsilon$ and choice of $a_{0}$ informs us that $0\leq
a_{0}+\epsilon-\sqrt{k}b<a_{0}-\epsilon-a$. Therefore $N>1$.
We then pick constant $h>0$ such that

$h\leq\\{\frac{N-1}{2(N+1)}a_{0}^{2},\,\frac{\sqrt{N}-1}{1+\sqrt{N}}a_{0}^{2}\\}$

We then demand that for $(r,w,l)$,

(28) $-a_{0}^{3}-ha_{0}<w<-a_{0}^{3}+ha_{0}.$

  4. (iv)

For $(r,w,l)\in S(0)$, we want

(29) $0\leq l<(ha_{0}-\epsilon)(a_{0}-\epsilon)^{2}$

  5. (v)

Finally, we choose time $T>0$ such that

(30)
$(a_{0}+\epsilon-\sqrt{k}b)\frac{\sqrt{2+(a_{0}^{3}+ha_{0})}}{a_{0}^{3}-ha_{0}}\leq
T\leq(a_{0}-\epsilon-a)\frac{\sqrt{1+(a_{0}^{3}-ha_{0})}}{a_{0}^{3}+ha_{0}}.$

######  Remark.

We first verify that the inequality in (v) is in fact valid. To do so, we will
need to show that

$(a_{0}+\epsilon-\sqrt{k}b)\frac{\sqrt{2+(a_{0}^{3}+ha_{0})}}{a_{0}^{3}-ha_{0}}\leq(a_{0}-\epsilon-a)\frac{\sqrt{1+(a_{0}^{3}-ha_{0})}}{a_{0}^{3}+ha_{0}},$

that is,

$\frac{a_{0}^{3}+ha_{0}}{a_{0}^{3}-ha_{0}}\frac{\sqrt{2+(a_{0}^{3}+ha_{0}^{3})^{2}}}{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}\leq\frac{a_{0}-\epsilon-a}{a_{0}+\epsilon-\sqrt{k}b}=N.$

This is where our choice of $h$ in (iii) comes into play. We can see that
since $h<\frac{\sqrt{N}-1}{1+\sqrt{N}}a_{0}^{2}$, we have
$(1+\sqrt{N})ha_{0}<(\sqrt{N}-1)a_{0}^{3}$. Thus
$2ha_{0}<(\sqrt{N}-1)(a_{0}^{3}-ha_{0})$, so

$1+\frac{ha_{0}}{a_{0}^{3}-ha_{0}}=\frac{a_{0}^{3}+ha_{0}}{a_{0}^{3}-ha_{0}}\leq\sqrt{N}.$

Moreover, since $h<\frac{N-1}{2(N+1)}a_{0}^{2}$, we have
$2ha_{0}^{4}(N+1)\leq(N-1)a_{0}^{6}$. Therefore
$0\leq(N-1)a_{0}^{6}-2ha_{0}^{4}(N+1)$. So

$0\leq(N-1)+(N-1)a_{0}^{6}-2ha_{0}^{4}(N+1)+(N-1)h^{2}a_{0}.$

Therefore

$1+(a_{0}^{3}+ha_{0})^{2}\leq 2N+N(a_{0}^{3}-ha_{0})^{2},$

so

$\frac{2+(a_{0}^{3}+ha_{0}^{3})^{2}}{1+(a_{0}^{3}-ha_{0})^{2}}\leq N.$

Thus

$\frac{a_{0}^{3}+ha_{0}}{a_{0}^{3}-ha_{0}}\frac{\sqrt{2+(a_{0}^{3}+ha_{0}^{3})^{2}}}{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}\leq\sqrt{N}\times\sqrt{N}=\frac{a_{0}-\epsilon-a}{a_{0}+\epsilon-\sqrt{k}b},$

and the claim we make in (v) about the choice of $T$ is valid.

######  Remark.

To see that these choices of parameters are feasible, we now, similar to the
VP case, provide an example of a smooth function that satisfy all the above
conditions.

First of all, based on the given constants $C_{1}$, $C_{2}$, $a$, $b$, and
$c$, we first fix constants $k$, $\epsilon$, and $a_{0}$ that satisfies the
requirements in (i), and choose constant $T>0$ that satisfies (iii).
Next, we choose $H:\,[0,\infty)\rightarrow[0,\infty)$ to be a smooth function
that satisfies

$\int_{\mathbf{R}^{3}}H(|u|^{2})du=\frac{1}{2a_{0}}$

with $supp(H)\subset[0,1]$. We then re-scale this function with a scalar
$\delta>0$ by defining

$H_{\delta}(|u|^{2})=\frac{1}{\delta^{3}}H(\frac{|u|^{2}}{\delta^{2}}).$

Then we will find that

$\int_{\mathbf{R}^{3}}H_{\delta}(|u|^{2})du=\frac{1}{2a_{0}}$

and $supp(H_{\delta})\subset[0,\delta^{2}]$.

Next, we choose a smooth cut-off function $\chi\in
C^{\infty}((0,\infty),[0,1])$ such that

$\begin{cases}\chi(r)=1&\text{for }\quad a_{0}-\frac{1}{2}\epsilon\leq|r|\leq
a_{0}+\frac{1}{2}\epsilon\\\ \chi(r)=0&\text{for }\quad|r|\leq
a_{0}-\epsilon\text{ or }|r|\geq a_{0}+\epsilon\\\ \end{cases}$

We then fix constants $d=a_{0}^{3}-a_{0}$, and $\delta=ha_{0}-\epsilon>0$, and
define function

$f_{0}(x,v)=H_{\delta}(|\left(1+\frac{d}{|x|}\right)x+v|^{2})\chi(|x|).$

Clearly $f_{0}$ is smooth. We then move on to claim that it satisfies (ii),
(iii), and (iv)

To verify this, we first notice that since $r=|x|$, we can obtain from the
definition of $\chi$ that

$a_{0}-\epsilon\leq r\leq a_{0}+\epsilon.$

Next, the support of $h_{\delta}$ indicates us that

$|\left(1+\frac{d}{|x|}\right)x+v|^{2}\leq\delta^{2}.$

Using change of variables and the relation $|v|^{2}=w^{2}+lr^{-2}$, we can
obtain that

$(r+d+w)^{2}+lr^{-2}\leq\delta^{2}.$

This inequality indicates two things. Firstly, we have $|r+d+w|\leq\delta$,
that is,

$-a_{0}^{3}-ha_{0}\leq-(a_{0}^{3}-a_{0})-(a_{0}+\epsilon)-(ha_{0}-\epsilon)\leq-
d-r-\delta\leq w,$

and

$w\geq-
d-r+\delta\geq-(a_{0}^{3}-a_{0})-(a_{0}-\epsilon)+(ha_{0}-\epsilon)\geq-
a_{0}^{3}+ha_{0}.$

Therefore the conditions in (iii) are satisfied.
Secondly, it also informs us that $lr^{-2}\leq\delta$. Therefore

$l\leq(ha_{0}-\epsilon)(a_{0}-\epsilon)^{2}$

and (iv) is verified.

Finally, to check the solution satisfies (ii), we notice that

$\int_{\mathbf{R}^{3}}H_{\delta}(|\left(1+\frac{d}{|x|}\right)x+v|^{2})dv=\int_{\mathbf{R}^{3}}H_{\delta}(|u|^{2})du=\frac{1}{2a_{0}}.$

and thus we can calculate that

$\rho_{0}(x)=\int_{\mathbf{R}^{3}}f_{0}(x,v)=\left(\int_{\mathbf{R}^{3}}H_{\delta}(|\left(1+\frac{d}{|x|}\right)x+v|^{2})dv\right)\chi(|x|)=\frac{1}{2a_{0}}\chi(|x|),$

which satisfies (ii) as we want.
Therefore, we’ve verified $f_{0}(x,v)$ to be a smooth function that satisfies
all the settings for the initial data.

######  Remark.

Since our choices of initial spatial radius $r$ and initial charge density
$\rho_{0}$ in (ii) are identical to those in the VP case, we can derive the
same estimation on the total particle mass $M$ using $a_{0}$ and $\epsilon$.
Specifically, we have

(31) $2\pi a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon^{3}\leq M\leq 8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3}.$

### 4.2. Proof of Theorem 7

Proof.  The proof of Theorem 7 shares a similar structure with that of Theorem
6, i.e. the VP case. We start with examining the spatial radius, charge
density, and electric field at time zero. Next we show that our choice of $T$
enable the concentration to happen while the particles are moving inwards.
After that, we use the estimations of $\mathscr{R}(T)$ in Lemma 4 to show that
the particles are concentrated within the targeting shield. Finally we use
Lemma 5 to verify the conditions of charge density and electric field at time
$T$.

To begin with, we first verify that the particle distribution, charge density,
and electric field at time zero satisfies the requirements in Theorem 7. For
$(r,w,l)\in S(0)$, we know

$r>a_{0}-\epsilon=a_{0}-\frac{\sqrt{a^{2}+b^{2}}-\sqrt{2}a}{4\sqrt{2}}>c$

by our choice of $a_{0}$.
Moreover, since $a_{0}\geq C_{1}^{-1}$, our setting of the initial charge
density in 27 informs us that

$\lVert\rho(0)\rVert_{\infty}\leq\frac{1}{a_{0}}\leq C_{1}.$

Next we check that the electric field is small enough at time zero. Since our
choice of the initial spatial radius and initial charge density in 25, 26, and
27 is exactly the same as that in the VP case 14, 15, and 16, we can simply
apply an identical argument about the electric field in each region at time
zero, namely that

$\begin{cases}|E(0,x)|=0\quad\quad\text{for }|x|\leq a_{0}-\epsilon,\\\
|E(0,x)|\leq 32\pi
a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}\quad\quad\text{for
}a_{0}-\epsilon\leq|x|\leq a_{0}+\epsilon,\\\ |E(0,x)|\leq 8\pi
a_{0}^{-1}\epsilon+\frac{8\pi}{3}a_{0}^{-3}\epsilon^{3}\quad\quad\text{for
}|x|\geq a_{0}+\epsilon.\end{cases}$

Again, an uniform bound then becomes

$\lVert E(0)\rVert_{\infty}\leq 32\pi
a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}.$

Then by choosing $a_{0}$ large enough we ensures that this value is smaller
than $C_{1}$. Since we choose
$a_{0}\geq\text{max}\\{6\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-1},\,2^{\frac{5}{6}}\pi^{1\over
3}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)C_{1}^{-\frac{1}{3}}$ in 24 and set
$\epsilon=\frac{(\sqrt{a^{2}+b^{2}}-\sqrt{a})}{4\sqrt{2}}$ in (i), we can
estimate the two terms in the right side separately that

$32\pi
a_{0}^{-1}\epsilon=4\sqrt{2}\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)a_{0}^{-1}\leq\frac{2}{3}C_{1}$

and

$\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}=\frac{\sqrt{2}\pi}{24}(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)^{3}a_{0}^{-3}\leq\frac{1}{3}C_{1}.$

Therefore, we can deduce that

$\lVert E(0)\rVert_{\infty}\leq 32\pi
a_{0}^{-1}\epsilon+\frac{32\pi}{3}a_{0}^{-3}\epsilon^{3}\leq C_{1}.$

Thus we’ve proven that our solution satisfies the conditions on the initial
particle spatial radius, charge density, and electric field.

We now look at the particle distribution at time $T$. Similar to the case of
the VP system, we want to verify that the concentration happens during the
period which the particles move inwards, namely $T<T_{0}$. Now Lemma 4 informs
us that

$T_{0}\geq r\left(1-\frac{D}{r^{2}w^{2}+D}\right)\geq r-\frac{\sqrt{D}}{|w|},$

where

$D=l+Mr\sqrt{1+w^{2}+lr^{-2}}.$

Therefore our first step is to derive a reasonable estimation for $D$. Our
choice of initial data in 28 and 29 tells us that

(32) $\displaystyle D$ $\displaystyle=l+Mr\sqrt{1+w^{2}+lr^{-2}}$
$\displaystyle\leq(ha_{0}-\epsilon)(a_{0}-\epsilon)^{2}+(8\pi
a_{0}\epsilon+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{3})(a_{0}+\epsilon)\sqrt{2+(a_{0}^{3}+ha_{0})^{2}}$
$\displaystyle\leq(ha_{0}-\epsilon)(a_{0}-\epsilon)^{2}+(8\pi
a_{0}^{2}\epsilon+\frac{8\pi}{3}\epsilon^{3}+8\pi
a_{0}\epsilon^{2}+\frac{8\pi}{3}a_{0}^{-1}\epsilon^{4})\sqrt{2+(a_{0}^{3}+ha_{0})^{2}}$

by 25 and 31. Moreover, since we’ve set $a_{0}\geq 1$ and
$a_{0}\geq\epsilon^{3}$ (that is, $\epsilon\leq a_{0}^{\frac{1}{3}}$), we can
simplify the inequality that

(33) $\displaystyle D$ $\displaystyle\leq(ha_{0})^{2}+(8\pi
a_{0}^{\frac{7}{3}}+\frac{8\pi}{3}a_{0}+8\pi
a_{0}^{\frac{5}{3}}+\frac{8\pi}{3}a_{0}^{\frac{1}{3}})\sqrt{2+(a_{0}^{3}+ha_{0})^{2}}$
$\displaystyle\leq(2a_{0}^{2})^{2}+(8\pi
a_{0}^{\frac{7}{3}}+\frac{8\pi}{3}a_{0}^{\frac{7}{3}}+8\pi
a_{0}^{\frac{7}{3}}+\frac{8\pi}{3}a_{0}^{\frac{7}{3}})\sqrt{2+(a_{0}^{3}+2a_{0}^{2})^{2}}$
$\displaystyle\leq
4a_{0}^{4}+\frac{64\pi}{3}a_{0}^{\frac{7}{3}}\sqrt{2+9a_{0}^{6}}$
$\displaystyle\leq
4a_{0}^{4}+\frac{64\pi}{3}a_{0}^{\frac{7}{3}}\sqrt{16a_{0}^{6}}$
$\displaystyle\leq 4a_{0}^{2}+\frac{256\pi}{3}a_{0}^{\frac{16}{3}}$
$\displaystyle\leq 86\pi a_{0}^{\frac{16}{3}}.$

With this estimation in hand, we can now verify our choice of focusing time
$T$. We first see that if we have
$\frac{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}{a_{0}^{3}+ha_{0}}\leq 1$, then

$T\leq(a_{0}-\epsilon-a)\frac{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}{a_{0}^{3}+ha_{0}}\leq
a_{0}-\epsilon-a.$

Then since we’ve set $a_{0}\geq\left(\frac{\sqrt{86\pi}}{a}\right)^{3}$, that
is, $\sqrt{86\pi}a_{0}^{-\frac{1}{3}}\leq a$, we have

$\frac{\sqrt{D}}{|w|}\leq\frac{\sqrt{86\pi
a_{0}^{\frac{16}{3}}}}{a_{0}^{3}}\leq\frac{\sqrt{86\pi}a^{\frac{8}{3}}}{a_{0}^{3}}\leq\sqrt{86\pi}a_{0}^{-\frac{1}{3}}\leq
a.$

Therefore we can deduce that

$T\leq a_{0}-\epsilon-a\leq(a_{0}-\epsilon)-\frac{\sqrt{D}}{|w|}\leq
r-\frac{\sqrt{D}}{|w|}\leq T_{0}.$

To make this happen, we will need
$1+(a_{0}^{3}-ha_{0})^{2}\leq(a_{0}^{3}+ha_{0})$, that is, $4ha_{0}^{4}\geq
1$. Since we’ve set $a_{0}$ large enough, this condition can be satisfied.
Therefore, $T\leq T_{0}$ is verified, and the concentration in fact happens
while the particles are moving inwards.

We now turn to the location of particles at time $T$. For all $(r,w,l)\in
S(0)$, we want $a\leq\mathscr{R}(T,0,r,w,l)\leq b$, so that the particles are
distributed in the targeting range. To do so, we will need the estimations on
$\mathscr{R}(T)$ we derived in Lemma 4. Our choice of parameters in 25, 28,
and 29 informs us that

$\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}T\leq\frac{a_{0}^{3}+ha_{0}}{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}\times(a_{0}-\epsilon-a)\frac{\sqrt{1+(a_{0}^{3}-ha_{0})^{2}}}{a_{0}^{3}+ha_{0}}\leq
a_{0}-\epsilon-a.$

Therefore by Lemma 4,

(34) $\displaystyle\mathscr{R}(T)$
$\displaystyle\geq\frac{l}{2r^{3}(1+w^{2}+lr^{-2})}T^{2}+\frac{w}{\sqrt{1+w^{2}+lr^{-2}}}T+r$
$\displaystyle\geq r-\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}T$
$\displaystyle\geq(a_{0}-\epsilon)-(a_{0}-\epsilon-a)$ $\displaystyle\geq a.$

Similarly, we know

$\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}T\geq\frac{a_{0}^{3}-ha_{0}}{\sqrt{2+(a_{0}^{3}+ha_{0})}}\times(a_{0}+\epsilon-\sqrt{k}b)\frac{\sqrt{2+(a_{0}^{3}+ha_{0})}}{a_{0}^{3}-ha_{0}}\geq
a_{0}+\epsilon-\sqrt{k}b.$

Therefore

(35) $\displaystyle\left(r-\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}T\right)^{2}$
$\displaystyle\leq\left(a_{0}+\epsilon-(a_{0}+\epsilon-\sqrt{k}b)\right)^{2}$
$\displaystyle\leq kb^{2}.$

Moreover, since we know $T\leq a_{0}$, and we’ve set $a_{0}\geq 2\epsilon$ (so
that $a_{0}-\epsilon\geq\frac{1}{2}a_{0})$, we can also calculate

(36) $\displaystyle\frac{D}{r^{2}(1+w^{2}+lr^{-2})}T^{2}$
$\displaystyle\leq\frac{D}{(a_{0}-\epsilon)^{2}(1+(a_{0}^{3}+ha_{0})^{2})}a_{0}^{2}$
$\displaystyle\leq\frac{D}{\frac{1}{4}a_{0}^{2}\times a_{0}^{6}}a_{0}$
$\displaystyle\leq\frac{86\pi
a_{0}^{\frac{16}{3}}}{\frac{1}{4}a_{0}^{8}}a_{0}^{2}$ $\displaystyle\leq
344\pi a_{0}^{-\frac{2}{3}}.$

We’ve chosen $a_{0}$ large enough such that
$a_{0}\geq\left[\frac{344\pi}{(1-k)b^{2}}\right]^{\frac{3}{2}}$, which is
equivalent to say $344\pi a_{0}^{-\frac{2}{3}}\leq(1-k)b^{2}$. Then we can
deduce that

$\frac{D}{r^{2}(1+w^{2}+lr^{-2})}T^{2}\leq(1-k)b^{2}.$

Finally using Lemma 4, we can reach the conclusion that since $T\in[0,T_{0})$,

(37) $\displaystyle\mathscr{R}(T)$
$\displaystyle\leq\left(r-\frac{|w|}{\sqrt{1+w^{2}+lr^{-2}}}T\right)^{2}+\frac{D}{r^{2}(1+w^{2}+lr^{-2})}T^{2}$
$\displaystyle\leq kb^{2}+(1-k)b^{2}$ $\displaystyle\leq b^{2},$

so that $\mathscr{R}(T)\leq b$. Therefore, for $(r,w,l)\in S(0)$, we’ve
verified that

$a\leq\mathscr{R}(T,0,r,w,l)\leq b.$

Finally, we’re left to verify that the charge density and electric field are
large enough at time $T$. Here again we apply an identical argument as in the
VP case. We use Lemma 5 to deduce that the conditions
$\lVert\rho(T)\rVert_{\infty}$, $\lVert E(T)\rVert_{\infty}\geq C_{2}$ are
satisfied if we have

$M\geq max\\{\frac{4\pi}{3}(b^{3}-a^{3})C_{2},\,b^{2}C_{2}\\}.$

We know that $M\geq 2\pi a_{0}\epsilon+\frac{\pi}{6}a_{0}^{-1}\epsilon^{3}$.
Then since we’ve set $a_{0}$ large enough, namely

$a_{0}\geq
max\\{\frac{8\sqrt{2}(b^{3}-a^{3})C_{2}}{3(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)},\,\frac{2\sqrt{2}b^{2}C_{2}}{\pi(\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}\\}$

we can easily verify that

$M\geq 2\pi
a_{0}\epsilon\geq\frac{(\pi\sqrt{a^{2}+b^{2}}-\sqrt{2}a)}{2\sqrt{2}}a_{0}\geq
max\\{\frac{4\pi}{3}(b^{3}-a^{3}),\,b^{2}C_{2}\\}.$

Thus by Lemma 5, $\lVert\rho(T)\rVert_{\infty}\geq C_{2}$ and $\lVert
E(T)\rVert_{\infty}\geq C_{2}$ is verified. Therefore, Theorem 7 is proven.
$\Box$

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