# Solutions of the 2D radially symmetric Vlasov-Maxwell system with an initial
focusing phase

Katherine Zhiyuan Zhang  Courant Institute of Mathematical Sciences, New York
University

###### Abstract.

We study radially symmetric solutions to the 2D Vlasov-Maxwell system and
construct solutions that initially possess arbitrarily small $C^{k}$ norms
($k\geq 1$) for the charge densities and the electric fields, but attain
arbitrarily large $L^{\infty}$ norms of them at some later time.

##  1\. Introduction

The behavior of solutions for Vlasov models describing plasma has been an
important topic that caught wide attention. An important example is the
Vlasov-Poisson equation (VP) (in which the magnetic field is absent)

(1.1) $\partial_{t}f+v\cdot\nabla_{x}f+E\cdot\nabla_{v}f=0,$ (1.2)
$E=\nabla_{x}\phi,\quad\Delta_{x}\phi=\int_{\mathbb{R}^{3}}fdv.$

In Ben-Artzi-Calogero-Pankavich [2], [3] and Zhang [27], it is shown that
there exist spherically symmetric solutions of 3D VP such that the particle
density and the electric field are initially as small as desired, but become
as large as desired at some later time (called ”focusing solutions”). Namely,
for any positive constants $\eta$, $N$, $b$, $\epsilon_{0}$ and integer
$\beta\geq 1$, there exists a smooth, spherically symmetric solution of VP,
such that

(1.3)
$\operatorname{supp}_{x}f(0,x,v)\subset\\{\frac{1}{2}b\leq|x|\leq\frac{3}{2}b\\}$

and

(1.4) $\|\rho(0)\|_{C^{\beta}},\ \|E(0)\|_{C^{\beta}}\leq\eta,$

while for some $T>0$,

(1.5) $\|\rho(T)\|_{L^{\infty}_{|x|\leq\epsilon_{0}}},\
\|E(T)\|_{L^{\infty}}\geq N.$

These results were inspired by a similar study by G. Rein and L. Teagert [25],
which proves the existence of focusing solutions for the gravitational Vlasov-
Poisson system, in which the electric force provides an attractive effect
instead of a repulsive effect on the particles. It is striking that in
contrast to the results in [2], [3], [27], a recent result by S. Pankavich
[23] in 2021 shows that any spherically symmetric solution (with initial data
in $C^{1}(\mathbb{R}^{3}\times\mathbb{R}^{3})$) must decay for $t$
sufficiently large. Namely, there exists $C>0$, such that
$\|\rho(t)\|_{L^{\infty}}\leq C(1+t)^{-3}\ ,\ \|E(t)\|_{L^{\infty}}\leq
C(1+t)^{-2}$ for all $t\geq 0$. There is no contradiction between the
conclusions in [2], [3], [27] and [23].

However, for a system that involves a magnetic field, the study of focusing
solutions is absent. In this paper, we consider the Vlasov-Maxwell system (VM)
on the 2D plane:

(1.6)
$\partial_{t}f+v\cdot\nabla_{x}f+(E_{1}+v_{2}B,E_{2}-v_{1}B)\cdot\nabla_{v}f=0,$
(1.7) $\begin{split}&\partial_{x_{1}}E_{1}+\partial_{x_{2}}E_{2}=\rho,\\\
&\partial_{t}E_{1}=\partial_{x_{2}}B-j_{1},\\\
&\partial_{t}E_{2}=-\partial_{x_{1}}B-j_{2},\\\
&\partial_{t}B=\partial_{x_{2}}E_{1}-\partial_{x_{1}}E_{2}.\\\ \end{split}$

Here, $f(t,x,v)\geq 0$ is the density distribution of the particles. In the
equation, $x\in\mathbb{R}^{2}$ is the particle position, $v\in\mathbb{R}^{2}$
is the particle momentum, and $E=(E_{1},E_{2})$ is the electric field, $B$ is
the magnetic field. The macroscopic charge density
$\rho:=\int_{\mathbb{R}^{2}}fdv$, and the $i$-th component of the current
density is $j_{i}:=\int_{\mathbb{R}^{2}}v_{i}fdv$ for $j=1$, $2$. The system
VM enjoys the conservation of the total mass

(1.8)
$M(t)=\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}f(t,x,v)dvdx=\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}f_{0}(x,v)dvdx:=M,\
\forall t.$

In this paper, we consider the VM equation (1.6) – (1.7) with initial data
supported on an arbitrary annulus that has small $L^{\infty}$ norm, and obtain
radially symmetric solutions that become large and are concentrated near the
origin at some later time. Specifically, we prove the following result:

######  Theorem 1.1.

For any integer $\beta\geq 1$ and any positive constants $\eta$, $N$,
$\epsilon_{0}$, there exists a smooth, radially symmetric solution of the VM
equation (1.6) – (1.7), such that

(1.9) $\operatorname{supp}_{x}f(0,x,v)\subset\\{\frac{1}{2}\leq|x|\leq 1\\}$

and

(1.10) $\|\rho(0)\|_{C^{\beta}}\leq\eta,\ \|E(0)\|_{C^{\beta}}\leq\eta,\
\|B(0)\|_{C^{\beta}}\leq\eta,$

while for some $T>0$,

(1.11) $\|\rho(T)\|_{L^{\infty}_{|x|\leq\epsilon_{0}}}\geq
N,\quad\|E_{r}(T)\|_{L^{\infty}_{|x|\leq\epsilon_{0}}}\geq N$

provided that a $C^{1}$ solution $(f,E,B)$ exists until the time $T$. Here
$E_{r}$ is the $r$-component of the electric field $E$ under the polar
coordinates.

######  Remark 1.2.

The solution constructed in Theorem 1.1 has nontrivial magnetic field. On the
other hand, it is more tricky to see if $\|E_{\varphi}\|_{L^{\infty}}$ or
$\|B\|_{L^{\infty}}$ grows large in $[0,T]$ ($E_{\varphi}$ is the
$\varphi$-component of the electric field $E$ under the polar coordinates),
since the equations of $E_{\varphi}$ and $B$ are intertwined, see Lemma 2.1.

######  Remark 1.3.

The assumption
$\operatorname{supp}_{x}f(0,x,v)\subset\\{\frac{1}{2}\leq|x|\leq 1\\}$ can be
replaced by
$\operatorname{supp}_{x}f(0,x,v)\subset\\{\frac{1}{2}b\leq|x|\leq\frac{3}{2}b\\}$
with a minor change in the proof.

######  Remark 1.4.

Theorem 1.1 is _conditional_ since there is no known global existence result
for classical solutions of (1.6) – (1.7) which contains a nonzero magnetic
field. In fact, the existence of global classical solutions to the non-
relativistic VM system remains an unresolved challenging problem in the field.

The Vlasov-Poisson and the Vlasov-Maxwell systems are among the most important
models for dilute plasmas (see [4], [5]). The issue of the existence of
global-in-time solutions for Vlasov-Poisson is addressed in [1], [11], [14],
[21], [24], etc.. For the Vlasov-Maxwell equation, this question is much
harder and more intriguing. Various progress has been made throughout the last
several decades, see, for example, [12], [13], [15], [20], [22], [26]. The
reader can also refer to Glassey [6] for a thorough introduction of the basic
Cauchy theory of the Vlasov-Poisson and the Vlasov-Maxwell systems. Moreover,
for Vlasov-Poisson, a series of study on its detailed large time behavior has
been carried out, see, for instance, [7], [8], [9], [10], [16], [17], [18],
[23].

The proof of our main result Theorem 1.1 relies upon an analysis of the
particle trajectory and the mass conservation property of the equation. With a
magnetic field, the analysis of the particle trajectory is more complicated.
We make use of the dimension reduction brought in by the radial symmetry
setting, and carry out an estimate on the electromagnetic field, so as to
obtain enough information of the particle trajectory during the time interval
during which the focusing behavior happens.

The contents of the paper are arranged as follows. In Section 2, we describe
the basic settings on the problem, in particular we discuss some
representation formulas for the electromagnetic field. In Section 3, we give
key lemmas that provide estimates for the electromagnetic field and
description of the particle trajectories, which allow us to observe the
focusing phenomena. Section 4 is devoted to the proof of Theorem 1.1, which
involves a careful selection of parameters and computation of the norms of
$\rho(t,x)$ and $E_{r}(t,x)$.

##  2\. Settings

We use the polar coordinates $(r,\varphi)$. We assume radial symmetry in the
initial data, so $f(0)=f(0,r)$, $(E,B)(0)=(E,B)(0,r)$. We write (1.6) – (1.7)
in the polar coordinates:

(2.1)
$\partial_{t}f+v_{r}\partial_{r}f+v_{\varphi}\frac{1}{r}\partial_{\varphi}f+(E_{r}+v_{\varphi}B+\frac{v_{\varphi}^{2}}{r})\partial_{v_{r}}f+(E_{\varphi}-v_{r}B-\frac{v_{r}v_{\varphi}}{r})\partial_{v_{\varphi}}f=0,$
(2.2)
$\begin{split}&\frac{1}{r}\partial_{r}(rE_{r})+\frac{1}{r}\partial_{\varphi}E_{\varphi}=\rho,\\\
&\partial_{t}E_{r}=\frac{1}{r}\partial_{\varphi}B-j_{r},\\\
&\partial_{t}E_{\varphi}=-\partial_{r}B-j_{\varphi},\\\
&\partial_{t}B=\frac{1}{r}\partial_{\varphi}E_{r}-\frac{1}{r}\partial_{r}(rE_{\varphi}),\\\
\end{split}$

and assign the following radially symmetric initial and boundary conditions:

(2.3) $\begin{split}&f(0,r,v_{r},v_{\varphi})=f_{0}(r,v_{r},v_{\varphi})\geq
0,\\\ &E_{\varphi}(0,r)=E_{\varphi,0}(r),\\\ &B(0,r)=B_{0}(r),\\\
&E_{r}(0,0)=0.\\\ \end{split}$

where $E_{\varphi,0}$ and $B_{0}$ are given $C^{1}$ functions satisfying

(2.4) $\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}\leq\eta.$

Writing the equations

(2.5) $\dot{X}=V,\ \dot{V}=E(t,X)+B(V_{2},-V_{1})$

under the polar coordinates, we obtain that the forward characteristics
$(r(s),\varphi(s))$ of the Vlasov equation are described by the following ODE
system:

(2.6)
$\begin{split}&\frac{d^{2}r}{ds^{2}}=r\big{(}\frac{d\varphi}{ds}\big{)}^{2}+E_{r}+r\dot{\varphi}B,\\\
&\frac{d}{ds}\big{(}r^{2}\frac{d\varphi}{ds}\big{)}=rE_{\varphi}-r\dot{r}B.\\\
\end{split}$

for $s\geq 0$, with the initial conditions

(2.7) $r(0)=r_{0},\ \varphi(0)=\varphi_{0},\ \dot{r}(0)=\dot{r}_{0},\
\dot{\varphi}(0)=\dot{\varphi}_{0}.$

We denote

(2.8)
$S(t):=\\{(r,\varphi,\dot{r},\dot{\varphi}):f(t,r,\varphi,\dot{r},\dot{\varphi})>0\\}.$

In particular,

(2.9)
$S(0):=\\{(r,\varphi,\dot{r},\dot{\varphi}):f_{0}(r,\varphi,\dot{r},\dot{\varphi})>0\\}.$

(Notice that for the radial symmetry of the solution to (1.6) – (1.7) can be
propagated, see [19], so we have
$f_{0}(r,\varphi,\dot{r},\dot{\varphi})=f_{0}(r,\dot{r},\dot{\varphi})$,
$f(t,r,\varphi,\dot{r},\dot{\varphi})=f(t,r,\dot{r},\dot{\varphi})$.)

We introduce the following

######  Lemma 2.1.

Let $(f,E_{r},E_{\varphi},B)$ be a solution to the 2D radially symmetric VM.
Let $P_{\pm}=rE_{\varphi}\pm rB$, $t_{1}(t,r)=\max\\{0,t-r\\}$. (Notice that
$t_{1}(t,r)=\max\\{0,t-r\\}=0$ when $t<r$.) We have

(2.10) $rE_{r}(t,r)=\int_{0}^{r}s\rho(t,s)ds,$ (2.11)
$\begin{split}&rE_{\varphi}(t,r)=\frac{1}{2}[P_{+}(t_{1}(t,r),r-t+t_{1}(t,r))+P_{-}(0,r+t)]\\\
&\quad+\frac{1}{2}\int_{t_{1}(t,r)}^{t}(B-rj_{\varphi})(\tau,r-t+\tau)d\tau+\frac{1}{2}\int_{0}^{t}(B-rj_{\varphi})(\tau,r+t-\tau)d\tau,\\\
\end{split}$ (2.12)
$\begin{split}&rB(t,r)=\frac{1}{2}[P_{+}(t_{1}(t,r),r-t+t_{1}(t,r))-P_{-}(0,r+t)]\\\
&\quad+\frac{1}{2}\int_{t_{1}(t,r)}^{t}(B-rj_{\varphi})(\tau,r-t+\tau)d\tau-\frac{1}{2}\int_{0}^{t}(B-rj_{\varphi})(\tau,r+t-\tau)d\tau.\\\
\end{split}$

###### Proof.

The proof is similar to the argument in Section 2.2 in [19] so we omit it. ∎

##  3\. Estimate on the Field and the Particle Trajectories

We introduce the following lemma estimating the electromagnetic field.

######  Lemma 3.1.

Let $(f,E_{r},E_{\varphi},B)$ be a solution to the 2D radially symmetric VM.
Assume

(3.1) $\begin{split}M&=\iint f(t,x,v)dvdx=\iint f_{0}(x,v)dvdx\\\
&=2\pi\int_{0}^{+\infty}\int_{\mathbb{R}^{2}}rf_{0}(r,v)dvdr=2\pi\int_{0}^{+\infty}r\rho_{0}(r)dr<+\infty.\end{split}$

Then we have, for any $T_{1}>0$,

(3.2) $|E_{r}(T_{1},R)|\leq\frac{M}{2\pi R}.$

Moreover, let $(r,\varphi)(t,r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})$
solve the ODE (2.6) – (2.7). If for all $t\in[0,T_{1}]$,

(3.3) $r(t,r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\leq
r_{0},\quad\inf_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}r(t,r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\geq 6t,\quad
r(t)^{2}\dot{\varphi}(t)\leq\frac{4}{3}r_{0}^{2}\dot{\varphi}_{0},$

and for some positive numbers $K$, $K_{1}$,

(3.4) $\begin{split}&\rho(t,r)\leq K,\\\
&\sup_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}|\dot{\varphi}_{0}|\leq K_{1},\\\
&\sup_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in S(0)}|r_{0}|\leq
1,\\\ \end{split}$

with $\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}\leq\frac{2KK_{1}T_{1}}{R}$, then
for any $R\geq 6T_{1}$,

(3.5)
$|(E_{\varphi},B)(T_{1},R)|\leq\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+\frac{6KK_{1}T_{1}}{R}.$

###### Proof.

(3.2) follows from (2.10) – (2.12), due to that
$|\int_{0}^{r}s\rho(T_{1},s)ds|\leq\frac{M}{2\pi}$ for all $T_{1}>0$.

For the proof of (3.5), we first observe

(3.6) $\int_{0}^{T_{1}}(Rj_{\varphi})(\tau,R+T_{1}-\tau)d\tau\leq
2KK_{1}T_{1}.$

This is because on the support of the integral we have
$Rj_{\varphi}(\tau,R+T_{1}-\tau)=\int
R(R+T_{1}-\tau)\dot{\varphi}f(\tau,R+T_{1}-\tau,v)dv\leq\int\frac{3}{2}R^{2}\dot{\varphi}f(\tau,R+T_{1}-\tau,v)dv\leq\int
2r_{0}^{2}\dot{\varphi}_{0}f(\tau,R+T_{1}-\tau,v)dv\leq 2K_{1}\int
f(\tau,R+T_{1}-\tau,v)dv=2K_{1}\rho(\tau,R+T_{1}-\tau,v)$. Here we have used
$R^{2}\dot{\varphi}\leq\frac{4}{3}r_{0}^{2}\dot{\varphi}_{0}$, which follows
by the assumption of the lemma.

Assume that for some $t\in[0,T_{1}]$, $\sup_{r>0}|rB(t,r)|\geq
r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+6KK_{1}T_{1}$. Let
$t_{0}:=\inf\\{t>0:\sup_{r>0}|rB(t,r)|\geq
r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+6KK_{1}T_{1}\\}$. We have, for some
$r>0$,

(3.7) $\begin{split}&r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+6KK_{1}T_{1}\leq
rB(t_{0},r)\\\ &\leq
r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+\frac{1}{2}\int_{0}^{t_{0}}B(\tau,r-t_{0}+\tau)d\tau-\frac{1}{2}\int_{0}^{t_{0}}B(\tau,r+t_{0}-\tau)d\tau+2KK_{1}T_{1},\\\
\end{split}$

which implies

(3.8)
$\frac{1}{2}\int_{0}^{t_{0}}B(\tau,r-t_{0}+\tau)d\tau-\frac{1}{2}\int_{0}^{t_{0}}B(\tau,r+t_{0}-\tau)d\tau\geq
4KK_{1}T_{1}.$

Therefore there must exists some $t^{\prime}\in[0,t_{0}]$,
$r^{\prime}\in[r-t_{0},r+t_{0}]$, such that
$B(t^{\prime},r^{\prime})\geq\frac{4KK_{1}T_{1}}{t_{0}}\geq\frac{20KK_{1}T_{1}}{r-t_{0}}\geq\frac{20KK_{1}T_{1}}{r}$
(here we used that $t_{0}<\frac{r-t_{0}}{5}$, which follows from (3.3)). Since
$r^{\prime}\in[r-t_{0},r+t_{0}]$, we have $r^{\prime}\geq\frac{r}{2}$. Hence
$B(t^{\prime},r^{\prime})\geq\frac{20KK_{1}T_{1}}{r}\geq\frac{10KK_{1}T_{1}}{r^{\prime}}>\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+\frac{6KK_{1}T_{1}}{r^{\prime}}$.
A contradiction to the definition of $t_{0}$. Using (2.11), we obtain the
estimate for $E_{\varphi}$.

This completes the proof of the lemma. ∎

We also give the following lemma, which describes the behavior of the particle
trajectories of VM when the particle trajectories satisfy $0<r_{0}\leq 1$,
$\dot{r}_{0}<0$, $\dot{\varphi}_{0}>0$ at time $0$.

######  Lemma 3.2.

Let $0<r_{0}\leq 1$, $\dot{r}_{0}<0$, $\dot{\varphi}_{0}>0$, and let
$(r(t),\varphi(t))$ be a solution to (2.6) and (2.7) for all $t\geq 0$. Assume
that $|(E_{r},E_{\varphi},B)(t)|\leq\frac{m}{3r}$ for some constant $m>0$
(independent of $r$, $t$) on the support of $f(t)$. Define:

$\mathcal{A}:=\dot{r}_{0}^{2}+r_{0}^{-2}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2mr_{0}^{-1}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})-2m\ln
r_{0},$
$\mathcal{B}:=2mr_{0}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})+(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2m.$

Assume that

(3.9) $\begin{split}&-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}>0,\\\
&\mathcal{A}>0,\quad\mathcal{B}>0,\quad\mathcal{A}r_{0}^{2}-\mathcal{B}>0.\\\
\end{split}$

Also, we assume that $T_{0}\leq\frac{1}{100}r_{0}$, and for $s\in[0,T_{0}]$,
$\dot{r}(s)<0$. Then for all $s\in[0,T_{0}]$,

$r(s)^{2}\leq(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}s)^{2}+\mathcal{B}r_{0}^{-2}s^{2}.$

###### Proof.

Integrating the second line of (2.6) over $[0,s]$ and using
$|(E_{r},E_{\varphi},B)(t)|\leq\frac{m}{3r}$, we learn that

$r^{-2}(-ms+\frac{1}{3}mr-\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})\leq\dot{\varphi}\leq
r^{-2}(ms-\frac{1}{3}mr+\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})$

provided $r>0$. Hence by the first line of (2.6) we obtain

$\ddot{r}(s)\leq\frac{1}{r^{3}}\max\\{(ms-\frac{1}{3}mr+\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2},(\frac{1}{3}mr-\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}\\}+\frac{m}{r}+m\max\\{\dot{\varphi},0\\},$
$\ddot{r}(s)\geq\frac{1}{r^{3}}\min\\{(-\frac{1}{3}mr+\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2},(-ms+\frac{1}{3}mr-\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}\\}-\frac{m}{r}-m|\dot{\varphi}|.$

Since $-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}>0$, and for
$s\in[0,T_{0}]$, $s\leq T_{0}\leq\frac{1}{100}r_{0}$, we deduce from the
inequalities above that

$0<r^{-2}(-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})\leq\dot{\varphi}\leq
r^{-2}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}),$ $\ddot{r}(s)\leq
r^{-3}\max\\{(-\frac{1}{2}mr+\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2},(\frac{1}{2}mr-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}\\}+mr^{-1}+m\max\\{\dot{\varphi},0\\},$
$\ddot{r}(s)\geq
r^{-3}\min\\{(-\frac{1}{2}mr+\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2},(\frac{1}{2}mr-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}\\}-mr^{-1}-m|\dot{\varphi}|.$

Let us denote $\mathcal{C}=\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}$.
Using $r(s)\leq r_{0}$ for $s\in[0,T_{0}]$, from the inequality $\ddot{r}\leq
r^{-3}\max\\{(-\frac{1}{2}mr+\mathcal{C})^{2},(\frac{1}{2}mr-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}\\}+mr^{-1}+m\max\\{\dot{\varphi},0\\}$
and
$0<r^{-2}(-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})\leq\dot{\varphi}\leq
r^{-2}\mathcal{C}$, we have

$\begin{split}\ddot{r}&\leq
r^{-3}(-\frac{1}{2}mr+\mathcal{C})^{2}+mr^{-1}+mr^{-2}\mathcal{C}\\\ &\leq
r^{-3}\mathcal{C}^{2}+mr^{-1}+mr^{-2}\mathcal{C}.\\\ \end{split}$

Multiplying this inequality by $\dot{r}$, we obtain

$\frac{1}{2}\big{(}\frac{d(\dot{r}^{2})}{ds}\big{)}\geq
r^{-3}\dot{r}\mathcal{C}^{2}+mr^{-1}\dot{r}+mr^{-2}\dot{r}\mathcal{C}.$

Integrating yields

(3.10)
$\dot{r}^{2}\geq\dot{r}_{0}^{2}+(r_{0}^{-2}-r^{-2})\mathcal{C}^{2}+2m\ln(\frac{r}{r_{0}})+2m(r_{0}^{-1}-r^{-1})\mathcal{C}.$

For $s\in[0,T_{0}]$, multiplying the inequality (3.10) by $r^{2}$ and using
$r(s)\leq r_{0}$, we obtain

(3.11)
$\begin{split}\big{(}\frac{1}{2}\frac{d}{ds}(r^{2})\big{)}^{2}&\geq\dot{r}_{0}^{2}r^{2}+(r_{0}^{-2}r^{2}-1)\mathcal{C}^{2}+2m\ln(\frac{r}{r_{0}})r^{2}+2m(r_{0}^{-1}r^{2}-r)\mathcal{C}\\\
&=[\dot{r}_{0}^{2}+r_{0}^{-2}\mathcal{C}^{2}+2mr_{0}^{-1}\mathcal{C}]r^{2}+2m\ln(\frac{r}{r_{0}})r^{2}-2mr\mathcal{C}-\mathcal{C}^{2}.\\\
\end{split}$

By the property of the function $y=x^{2}\ln x$, we have
$\ln(\frac{r}{r_{0}})r^{2}=\ln(r)r^{2}-\ln(r_{0})r^{2}\geq-\frac{1}{2e}-r^{2}\ln
r_{0}\geq-1-r^{2}\ln r_{0}$. Hence we have

(3.12)
$\begin{split}\big{(}\frac{1}{2}\frac{d}{ds}(r^{2})\big{)}^{2}&\geq[\dot{r}_{0}^{2}+r_{0}^{-2}\mathcal{C}^{2}+2mr_{0}^{-1}\mathcal{C}]r^{2}+2m(-1-r^{2}\ln
r_{0})-2mr_{0}\mathcal{C}-\mathcal{C}^{2}\\\
&=[\dot{r}_{0}^{2}+r_{0}^{-2}\mathcal{C}^{2}+2mr_{0}^{-1}\mathcal{C}-2m\ln
r_{0}]r^{2}-2mr_{0}\mathcal{C}-\mathcal{C}^{2}-2m\\\
&=\mathcal{A}r^{2}-\mathcal{B},\\\ \end{split}$

where

(3.13)
$\begin{split}\mathcal{A}&:=\dot{r}_{0}^{2}+r_{0}^{-2}\mathcal{C}^{2}+2mr_{0}^{-1}\mathcal{C}-2m\ln
r_{0}\\\
&=\dot{r}_{0}^{2}+r_{0}^{-2}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2mr_{0}^{-1}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})-2m\ln
r_{0},\\\ \end{split}$ (3.14)
$\begin{split}\mathcal{B}&:=2mr_{0}\mathcal{C}+\mathcal{C}^{2}+2m\\\
&=2mr_{0}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})+(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2m.\\\
\end{split}$

With the assumptions we have $\mathcal{A}>0$ and $\mathcal{B}>0$.

If $r(s)>\sqrt{\frac{\mathcal{B}}{\mathcal{A}}}$, then
$r_{0}>\sqrt{\frac{\mathcal{B}}{\mathcal{A}}}$. Moreover,
$\mathcal{A}r_{0}^{2}-\mathcal{B}>0$ due to the assumptions. We have

$-\frac{1}{2}\frac{d}{ds}(r(s)^{2})\big{/}\sqrt{\mathcal{A}r(s)^{2}-\mathcal{B}}\geq
1\ ,$

which implies

$\frac{d}{ds}\sqrt{\mathcal{A}r(s)^{2}-\mathcal{B}}\leq-\mathcal{A}\ .$

Integrating gives

$\sqrt{\mathcal{A}r(s)^{2}-\mathcal{B}}\leq\sqrt{\mathcal{A}r(0)^{2}-\mathcal{B}}-\mathcal{A}s\
.$

Therefore

(3.15)
$\begin{split}r(s)^{2}&\leq\mathcal{A}s^{2}+r_{0}^{2}-2s\sqrt{\mathcal{A}r_{0}^{2}-\mathcal{B}}\\\
&=(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}s)^{2}+\mathcal{B}r_{0}^{-2}s^{2}\
.\\\ \end{split}$

Hence in general

$r(s)^{2}\leq\max\\{\frac{\mathcal{B}}{\mathcal{A}},(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}s)^{2}+\mathcal{B}r_{0}^{-2}s^{2}\\}\
.$

Noticing that $\frac{\mathcal{B}}{\mathcal{A}}$ is actually the minimum of the
parabola
$(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}s)^{2}+\mathcal{B}r_{0}^{-2}s^{2}$
(with respect to $s$), attained at
$s=s_{m}:=\frac{\sqrt{\mathcal{A}r_{0}^{2}-\mathcal{B}}}{\mathcal{A}}$.
Therefore we conclude, for all $s\in[0,T_{0}]$,

(3.16)
$\begin{split}r(s)^{2}&\leq(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}s)^{2}+\mathcal{B}r_{0}^{-2}s^{2}.\\\
\end{split}$

This completes the proof of the lemma.

∎

##  4\. Proof of Theorem 1.1

Recall that

(4.1)
$S(t):=\\{(r,\varphi,\dot{r},\dot{\varphi}):f(t,r,\varphi,\dot{r},\dot{\varphi})>0\\},$
(4.2)
$S(0):=\\{(r,\varphi,\dot{r},\dot{\varphi}):f_{0}(r,\varphi,\dot{r},\dot{\varphi})>0\\}.$

We now give the proof of Theorem 1.1.

###### Proof.

We only prove the case $\beta=1$ here since the case $\beta>1$ is proved in a
similar way.

Without loss of generality, we assume $0<\eta<1$ and $N>1$. Let
$\epsilon\in(0,1)$ be a small positive constant to be determined. Let $k$, $l$
and $\alpha$ be positive constants satisfying $k<\frac{1}{3}l$, $\alpha<l-k$,
$\alpha>4k$, $l>10\alpha+10k$.

Let $H:[0,+\infty)\times[0,+\infty)\rightarrow[0,+\infty)$ be any function
satisfying $\int_{\mathbb{R}^{2}}H(|u_{1}|^{2},|u_{2}|^{2})du_{1}du_{2}=1$
with $\operatorname{supp}(H)\subset[0,1)\times(0,1)$, and rescale it for any
$\epsilon\in(0,1)$:

$H_{\epsilon}(|u_{1}|^{2},|u_{2}|^{2})=\frac{1}{\epsilon^{4k}}H\big{(}\frac{|u_{1}|^{2}}{\epsilon^{4k}},\frac{|u_{2}|^{2}}{\epsilon^{4k}}\big{)}\
.$

so that
$\int_{\mathbb{R}^{2}}H_{\epsilon}(|u_{1}|^{2},|u_{2}|^{2})du_{1}du_{2}=1$ and
$\operatorname{supp}(H_{\epsilon})\subset[0,\epsilon^{2k})\times(0,\epsilon^{2k})$.
For any $\epsilon>0$, $x$, $v\in\mathbb{R}^{2}$, define

$h_{\epsilon}(x,v)=H_{\epsilon}\big{(}|r-|\dot{r}|\epsilon^{2l-k}|^{2},|\dot{\varphi}-\epsilon^{-l}|^{2}\big{)}\
.$

We choose the cut-off function $\chi_{0,1}\in C^{\infty}((0,\infty);[0,1])$
satisfying $|\chi_{0,1}(x)|\leq 1$, $\chi_{0,1}(x)=1$ for $|x|<\frac{1}{2}$
or, and $\chi_{0,1}(x)=0$ for $|x|>1$. Let
$\chi_{m,n}(x):=\chi_{0,1}(\frac{1}{n}(x-m))$, and take
$\chi(r)=\chi_{\frac{3}{4},\frac{1}{4}}(r)$, then $\chi(r)$ is a smooth
function supported on $[\frac{1}{2},1]$, and $\chi(r)=1$ for
$r\in[\frac{5}{8},\frac{7}{8}]$. For $r>0$, we define

(4.3) $\|\frac{d^{k}}{dr^{k}}(\chi(r))\|_{L^{\infty}}\leq c_{k}.$

In particular $c_{0}=1$. Denote $d_{k}:=\sum_{j=0}^{k}c_{j}$.

Let $\epsilon\in(0,1)$ be a small positive constant to be determined later. We
choose the initial data on the density distributions of the ions and the
electrons to be

(4.4)
$f_{0}(x,v)=\epsilon^{\alpha-k+2l}h_{\epsilon}(x,v)\chi(|x|)\Gamma(\dot{r})\
,$

where $\Gamma(\dot{r})=0$ for $\dot{r}\geq 0$, and $\Gamma(\dot{r})=1$ for
$\dot{r}<0$. Notice that $\Gamma$ does not have any effect on the smoothness
of $f_{0}(x,v)$. Note that $\operatorname{supp}f_{0}$ is a region near the set
$\\{r-|\dot{r}|\epsilon^{2l-k}=0,\dot{\varphi}=\epsilon^{-l}\\}$. Moreover,
the support of $f_{0}$ with respect to $x$ is $\\{\frac{1}{2}\leq|x|\leq
1\\}$. Let $M=\iint f(t,x,v)dvdx=\iint
f_{0}(x,v)dvdx=2\pi\int_{0}^{+\infty}\int_{\mathbb{R}^{2}}rf_{0}(r,v)dvdr$.
Using $dv=dv_{1}dv_{2}=rd\dot{r}d\dot{\varphi}$, we compute,

(4.5) $\frac{\pi}{16}\epsilon^{\alpha}\leq M\leq 2\pi\epsilon^{\alpha}.$

For any $(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in S(0)$, we select
$\epsilon$ small enough, so the following holds:

(4.6) $\frac{1}{2}\leq r_{0}\leq 1,\
|r_{0}-|\dot{r}_{0}|\epsilon^{2l-k}|^{2}<\epsilon^{4k},\
|\dot{\varphi}_{0}-\epsilon^{-l}|^{2}<\epsilon^{4k}.$

Hence we have

(4.7) $\begin{split}&\frac{1}{2}\leq r_{0}\leq 1,\\\
&\frac{1}{2}\epsilon^{k-2l}<(r_{0}-\epsilon^{2k})\epsilon^{k-2l}<|\dot{r}_{0}|<(r_{0}+\epsilon^{2k})\epsilon^{k-2l}<2\epsilon^{k-2l},\\\
&\frac{1}{2}\epsilon^{-l}<\epsilon^{-l}-\epsilon^{2k}<\dot{\varphi}_{0}<\epsilon^{-l}+\epsilon^{2k}<2\epsilon^{-l},\\\
\end{split}$

with $\dot{r}_{0}<0$.

From (4.4), we have

(4.8) $\rho(0,r)=\frac{1}{2}\epsilon^{\alpha}\chi(r).$

We let $\epsilon$ be small enough such that

(4.9) $\|\rho(0)\|_{L^{\infty}}\leq
2\pi\epsilon^{\alpha}\leq\frac{1}{10}\eta,$

Noticing that $\rho(t,0)$ is only a function of $r$, we choose $\epsilon$ to
be small enough such that

(4.10) $\|\rho(0)\|_{C^{1}_{x}}\leq\|\frac{\partial}{\partial
r}\rho(0)\|_{L^{\infty}}\leq\frac{1}{2}\epsilon^{\alpha}\|\frac{\partial}{\partial
r}(\chi(r))\|_{L^{\infty}}\leq\frac{1}{2}\epsilon^{\alpha}d_{1}\leq\eta.$

For the electric field, we let $\epsilon$ be small enough such that for all
$r\in[1/2,+\infty)$,

(4.11)
$\begin{split}|E_{r}(0,r)|&=\frac{1}{r}\int_{0}^{r}\int_{\mathbb{R}^{2}}sf(0,s,v)dsdv\\\
&\leq\frac{1}{r}\int_{0}^{+\infty}\int_{\mathbb{R}^{2}}sf(0,s,v)dsdv\\\ &\leq
2\cdot\frac{\pi}{16}\epsilon^{\alpha}\cdot\frac{1}{2\pi}\\\ &\leq\eta,\\\
\end{split}$

and moreover,

(4.12)
$\begin{split}|\partial_{r}E_{r}(0,r)|&=\big{|}\partial_{r}\big{(}\frac{1}{r}\int_{0}^{r}\int_{\mathbb{R}^{2}}sf(0,s,v)dsdv\big{)}\big{|}\\\
&\leq|\frac{1}{r^{2}}\int_{0}^{r}\int_{\mathbb{R}^{2}}sf(0,s,v)dsdv|+|\frac{1}{r}\int_{\mathbb{R}^{2}}rf(0,r,v)dv|\\\
&\leq|\frac{1}{r^{2}}\int_{0}^{+\infty}\int_{\mathbb{R}^{2}}sf(0,s,v)dsdv|+|\int_{\mathbb{R}^{2}}f(0,r,v)dv|\\\
&\leq
4\cdot\frac{\pi}{16}\epsilon^{\alpha}\cdot\frac{1}{2\pi}+\|\rho(0)\|_{L^{\infty}}\\\
&\leq\eta.\\\ \end{split}$

Notice that for all $r\in[0,1/2)$ we have $|E_{r}(0,r)|=0$,
$|\partial_{r}E_{r}(0,r)|=0$.

Furthermore, we set the initial data (with $\epsilon>0$ sufficiently small so
that $\epsilon^{\alpha}<\eta$) such that

(4.13)
$|(E_{\varphi},B)(0,r)|\leq\frac{\epsilon^{\alpha}}{200r}\leq\frac{12\epsilon^{\alpha-4k-l}}{r}\quad\text{and}\quad\|(E_{\varphi},B)(0)\|_{C^{1}}\leq\frac{\epsilon^{\alpha}}{200}.$

Also, we have

(4.14) $|E_{r}(t,r)|\leq\frac{M}{2\pi r}\leq\frac{\epsilon^{\alpha}}{r}$

for all $t>0$. Now, let $\epsilon$ be small enough such that the following
conditions hold with $m=100\epsilon^{\alpha-4k-l}$. These conditions
correspond to the assumptions in Lemma 3.2:

(4.15)
$\begin{split}&|E_{r}(t,r)|\leq\frac{\epsilon^{\alpha}}{r}\leq\frac{m}{3r},\\\
&-\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}>0,\\\
&\mathcal{A}:=\dot{r}_{0}^{2}+r_{0}^{-2}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2mr_{0}^{-1}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})-2m\ln
r_{0}\\\
&\quad\in[r_{0}^{2}(\epsilon^{2k-4l}-100\epsilon^{-2l}),r_{0}^{2}(\epsilon^{2k-4l}+100\epsilon^{-2l})]\subset[\frac{1}{2}r_{0}^{2}\epsilon^{2k-4l},2r_{0}^{2}\epsilon^{2k-4l}],\\\
&\mathcal{B}:=2mr_{0}(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})+(\frac{1}{2}mr_{0}+r_{0}^{2}\dot{\varphi}_{0})^{2}+2m\\\
&\quad\in[r_{0}^{4}(\epsilon^{-2l}-100\epsilon^{\alpha-4k-2l}),r_{0}^{4}(\epsilon^{-2l}+100\epsilon^{\alpha-4k-2l})]\subset[\frac{1}{2}r_{0}^{4}\epsilon^{-2l},2r_{0}^{4}\epsilon^{-2l}],\\\
&\mathcal{A}r_{0}^{2}-\mathcal{B}>0.\\\ \end{split}$

Let
$T:=\epsilon^{-k+2l}-300\epsilon^{\alpha-5k+3l}-300\epsilon^{k+2l}-300\epsilon^{3l-2k}<\epsilon^{-k+2l}\ll\frac{1}{100}r_{0}$.

Claim. For $t\in[0,T]$, $r\leq 10$, we have

(4.16)
$\begin{split}|(E_{\varphi},B)(t,r)|&\leq\min\\{\frac{\epsilon^{\alpha}}{200r},\|(E_{\varphi},B)(0)\|_{L^{\infty}}\\}+\frac{12\epsilon^{\alpha-4k-l}}{r}\\\
&\leq\min\\{\frac{\epsilon^{\alpha}}{200r},\frac{\epsilon^{\alpha}}{200}\\}+\frac{12\epsilon^{\alpha-4k-l}}{r}\leq\frac{24\epsilon^{\alpha-4k-l}}{r}\leq\frac{m}{3r}.\\\
\end{split}$

and

(4.17) $\inf_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}r(t,r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\geq 10t,$ (4.18)
$\begin{split}&\dot{r}<0.\\\ \end{split}$

Proof of Claim. We carry out a bootstrapping argument. If (4.16), (4.17) and
(4.18) hold on $[0,\tilde{T}]$ with $\tilde{T}<T$, we can show that on
$[0,\tilde{T}]$, (4.16) holds with $\frac{24\epsilon^{\alpha-4k-l}}{r}$
replaced by $\frac{20\epsilon^{\alpha-4k-l}}{r}$ on the right hand side using
the argument in Lemma 3.1 (that is,
$|(E_{\varphi},B)(t,r)|\leq\frac{20\epsilon^{\alpha-4k-l}}{r}$). Indeed, for
$t\in[0,\tilde{T}]$, we have

(4.19)
$\begin{split}\rho(t,r)&=\int_{\operatorname{supp}f}f(s,r,\dot{r},\dot{\varphi})rd\dot{r}d\dot{\varphi}\leq\int_{\operatorname{supp}f}\|f_{0}\|_{L^{\infty}}rd\dot{r}d\dot{\varphi}\\\
&\leq\epsilon^{\alpha-k+2l}\epsilon^{-4k}\epsilon^{k-2l}\epsilon^{-l}\epsilon^{k-l}=\epsilon^{\alpha-3k-2l}\\\
\end{split}$

and

(4.20)
$\begin{split}&\dot{r}(t)<0,\quad\sup_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}|\dot{\varphi}_{0}|\leq
2\epsilon^{-l},\quad\sup_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}|r_{0}|\leq 1.\\\ \end{split}$

Since $\dot{r}(t)<0$ on $t\in[0,\tilde{T}]$, we have $r\leq 1<10$ on
$t\in[0,\tilde{T}]$ for all particle trajectories. Integrating the second line
of (2.6) over $[0,t]$ and using $|(E_{r},E_{\varphi},B)(t)|\leq\frac{m}{3r}$,
we learn that for all $t\in[0,\tilde{T}]$,

$-mt+\frac{1}{3}mr(t)-\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0}\leq
r(t)^{2}\dot{\varphi}(t)\leq
mt-\frac{1}{3}mr(t)+\frac{1}{3}mr_{0}+r_{0}^{2}\dot{\varphi}_{0},$

from which we deduce
$R^{2}\dot{\varphi}\leq\frac{4}{3}r_{0}^{2}\dot{\varphi}_{0}$ using the
smallness of $\epsilon$. Repeat the proof of Lemma 3.1, we can show that the
result in Lemma 3.1 holds on $[0,\tilde{T}]$ with $\frac{6KK_{1}T}{R}$
replaced by $\frac{4KK_{1}T}{R}$ 111In the proof of Lemma 3.1, replace
$6KK_{1}T$ by $4KK_{1}T$ in the hypothesis ”for some $t\in[0,T]$
$\sup_{r>0}|rB(t,r)|\geq r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+6KK_{1}T$”,
in the definition $t_{0}:=\inf\\{t>0:\sup_{r>0}|rB(t,r)|\geq
r\|(E_{\varphi,0},B_{0})\|_{L^{\infty}}+6KK_{1}T\\}$, and in the left hand
side of (3.7). Then continue with the proof to show an adapted version of
Lemma 3.1 with $\frac{6KK_{1}T}{R}$ replaced by $\frac{4KK_{1}T}{R}$ in
(3.5)., we have

(4.21)
$\begin{split}|(E_{\varphi},B)(t,r)|&\leq\|(E_{\varphi},B)(0)\|_{L^{\infty}}+\frac{4\epsilon^{\alpha-3k-2l}\cdot
2\epsilon^{-l}\cdot\epsilon^{-k+2l}}{r}\\\
&\leq\frac{\epsilon^{\alpha}}{200}+\frac{8\epsilon^{\alpha-4k-l}}{r}\leq\frac{20\epsilon^{\alpha-4k-l}}{r}\\\
\end{split}$

for $t\in[0,\tilde{T}]$, $r\leq 10$.

Now, with (4.15), we apply the ODE analysis in Lemma 3.2. We deduce that for
all $(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in S(0)$ and
$t\in[0,\tilde{T}]$,

$r(t)^{2}\leq(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}t)^{2}+\mathcal{B}r_{0}^{-2}t^{2}.$

We conclude that on $[0,\tilde{T}]$ (4.17) and (4.18) hold with $10t$ replaced
by $100t$ in (4.17) and $\dot{r}<0$ on $[0,\tilde{T}]$ (provided that
$\epsilon$ is sufficiently small). Therefore the interval $[0,\tilde{T}]$ on
which (4.16), (4.17) and (4.18) hold can be prolonged.

By Lemma 3.2, we have the following estimate from the argument above for
sufficiently small $\epsilon$:

(4.22)
$\begin{split}r(T)^{2}&\leq(r_{0}-\sqrt{\mathcal{A}-\mathcal{B}r_{0}^{-2}}T)^{2}+\mathcal{B}r_{0}^{-2}T^{2}\\\
&\leq[r_{0}-\sqrt{r_{0}^{2}(\epsilon^{2k-4l}-100\epsilon^{-2l})-r_{0}^{4}(\epsilon^{-2l}+100\epsilon^{\alpha-4k-2l})r_{0}^{-2}}\\\
&\quad\cdot(\epsilon^{-k+2l}-300\epsilon^{\alpha-5k+3l}-300\epsilon^{k+2l}-300\epsilon^{3l-2k})]^{2}+2r_{0}^{2}\epsilon^{-2l-2k+4l}\\\
&\leq
10000r_{0}^{2}\epsilon^{2l-2k}+100000r_{0}^{2}\epsilon^{2\alpha-8k+2l}+2r_{0}^{2}\epsilon^{2l-2k}\\\
&\leq 40000r_{0}^{2}\epsilon^{2l-2k}.\\\ \end{split}$

Therefore

(4.23) $r(T)\leq 200r_{0}^{2}\epsilon^{l-k}.$

Now we have

(4.24)
$\begin{split}R_{+}=R_{+}(T)&:=\sup_{(r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in
S(0)}r(T,r_{0},\varphi_{0},\dot{r}_{0},\dot{\varphi}_{0})\in[\frac{1}{2}r_{0}\epsilon^{l-k},200r_{0}\epsilon^{l-k}]\\\
&\subset[\frac{1}{4}\epsilon^{l-k},200\epsilon^{l-k}].\\\ \end{split}$

By the conservation of mass, we have

(4.25) $\|\rho(T)\|_{L^{\infty}_{|x|\leq R_{+}}}\geq\frac{M}{\pi
R_{+}^{2}}\geq\frac{\pi}{16}\frac{\epsilon^{\alpha}}{\pi(200\epsilon^{l-k})^{2}}=\frac{1}{640000}\epsilon^{\alpha-2l+2k}.$

By (2.10),

(4.26)
$\begin{split}|E_{r}(T,R_{+})|&=\frac{1}{R_{+}}\int_{0}^{R_{+}}\int_{\mathbb{R}^{2}}sf(T,s,v)dsdv\\\
&=\frac{1}{R_{+}}\int_{0}^{+\infty}\int_{\mathbb{R}^{2}}sf(T,s,v)dsdv\\\
&\geq\frac{1}{200\epsilon^{l-k}}\frac{\pi}{16}\epsilon^{\alpha}\frac{1}{2\pi}\\\
&=\frac{1}{6400}\epsilon^{\alpha-l+k},\\\ \end{split}$

Selecting $\epsilon$ small enough, we have

(4.27) $\|\rho(T)\|_{L^{\infty}_{|x|\leq
R_{+}}}\geq\frac{1}{640000}\epsilon^{\alpha-2l+2k}>N,\quad\|E_{r}(T)\|_{L^{\infty}_{|x|\leq
R_{+}}}\geq\frac{1}{6400}\epsilon^{\alpha-l+k}>N$

with $R_{+}<\epsilon_{0}$.

This completes the proof of Theorem 1.1.

∎

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