# Keller-Segel model with Logarithmic Interaction and nonlocal reaction term

Shen Bian111Corresponding author. Email<EMAIL_ADDRESS> Quan
Wang222Email<EMAIL_ADDRESS>

Beijing University of Chemical Technology, 100029, Beijing

###### Abstract

We investigate the global existence and blow-up of solutions to the Keller-
Segel model with nonlocal reaction term
$u\left(M_{0}-\int_{{\mathbb{R}}^{2}}udx\right)$ in dimension two. By
introducing a transformation in terms of the total mass of the populations to
deal with the lack of mass conservation, we exhibit that the qualitative
behavior of solutions is decided by a critical value $8\pi$ for the growth
parameter $M_{0}$ and the initial mass $m_{0}$. For general solutions, if both
$m_{0}$ and $M_{0}$ are less than $8\pi$, solutions exist globally in time
using the energy inequality, whereas there are finite time blow-up solutions
for $M_{0}>8\pi$ (It involves the case $m_{0}<8\pi$) with any initial data and
$M_{0}<8\pi<m_{0}$ with small initial second moment. We also show the infinite
time blow-up for the critical case $M_{0}=8\pi.$ Moreover, in the radial
context, we show that if the initial data
$u_{0}(r)<\frac{m_{0}}{M_{0}}\frac{8\lambda}{(r^{2}+\lambda)^{2}}$ for some
$\lambda>0$, then all the radially symmetric solutions are vanishing in
$L_{loc}^{1}({\mathbb{R}}^{2})$ as $t\to\infty$. If the initial data
$u_{0}(r)>\frac{m_{0}}{M_{0}}\frac{8\lambda}{(r^{2}+\lambda)^{2}}$ for some
$\lambda>0$, then there could exist a radially symmetric solution satisfying a
mass concentration at the origin as $t\to\infty.$

##  1 Introduction

In this work, we study the Keller-Segel model with logistic sources in
dimension two

$\displaystyle\left\\{\begin{array}[]{ll}u_{t}=\Delta
u-\nabla\cdot\left(u\nabla
c\right)+u\left(M_{0}-\int_{{\mathbb{R}}^{2}}udx\right),{}{}&x\in{\mathbb{R}}^{2},~{}t\geq
0,\\\ -\Delta c=u,{}{}&x\in{\mathbb{R}}^{2},~{}t\geq 0,\\\ u(x,0)=u_{0}(x)\geq
0,{}{}&x\in{\mathbb{R}}^{2}.\end{array}\right.$ (1.4)

Here the initial mass is defined as
$m_{0}:=\int_{{\mathbb{R}}^{2}}u_{0}(x)dx.$ This model is developed to
describe the biological phenomenon chemotaxis [8, 10] by introducing nonlocal
terms in the logistic growth factor. The first equation states the random
(Brownian) diffusion of the cells with a bias directed by the chemoattractant
concentration. The chemoattractant $c$ is directly released by the cells,
diffuses on the substrate and can be expressed

$\displaystyle c(x,t)=-\frac{1}{2\pi}\int_{\mathbb{R}^{2}}u(y,t)\log|x-y|dy.$
(1.5)

The model at hand supposes that the cells production with logistic sources.
Here $M_{0}$ can be viewed as the mass capacity [12] and sometimes also called
Malthusian parameter, induces an exponential growth for low density
populations [9]. The nonlocal term $\int_{{\mathbb{R}}^{2}}udx$ describes the
influence of the total mass of the species in the growth of the population,
which is a competitive term limiting such growth [9].

As pointed out in [6, 12], in the absence of the reaction term
$u\left(M_{0}-\int_{{\mathbb{R}}^{2}}udx\right)$, (1.4) has the property that
the total mass is conserved
$\int_{{\mathbb{R}}^{2}}u_{0}(x)dx=\int_{{\mathbb{R}}^{2}}u(x,t)dx$ and the
solution of

$\displaystyle\left\\{\begin{array}[]{ll}u_{t}=\Delta
u-\nabla\cdot\left(u\nabla c\right),{}{}&x\in{\mathbb{R}}^{2},~{}t\geq 0,\\\
-\Delta c=u,{}{}&x\in{\mathbb{R}}^{2},~{}t\geq 0,\\\ u(x,0)=u_{0}(x)\geq
0,{}{}&x\in{\mathbb{R}}^{2}.\end{array}\right.$ (1.9)

exists globally when the initial mass $m_{0}<8\pi,$ while the solution becomes
a singular measure for $m_{0}>8\pi.$ A natural problem is that does the
nonlocal reaction change this effect?

To analyze this issue, we consider the total mass

$\displaystyle m(t)=\int_{{\mathbb{R}}^{n}}u(x,t)dx$ (1.10)

which satisfies

$\displaystyle\frac{dm(t)}{dt}=m(t)\left(M_{0}-m(t)\right)$ (1.11)

and we obtain

$\displaystyle m(t)=\frac{M_{0}}{1+\overline{C}e^{-M_{0}t}}.$ (1.12)

where

$\displaystyle\overline{C}=\frac{M_{0}-m_{0}}{m_{0}}.$ (1.13)

We find that the mass increases in time when $m_{0}<M_{0}$ and versus when
$m_{0}>M_{0},$ in both cases it converges to $M_{0}$ as $t\to\infty.$ The aim
of this paper is to explore the influence of $M_{0},m_{0}$ on the dynamics of
solutions when the nonlocal reaction is present.

In order to put our system in perspective, we recall some related cases.
Actually, chemotaxis models with local sources (where
$u(M_{0}-\int_{{\mathbb{R}}^{2}}udx)$ is replaced by
$a_{0}-u(a_{1}-a_{2}u^{\gamma})$ have been studied recently, which describes
the situation where the influence of the nonlocal terms is neglected, and the
global existence results have been obtained for different parameters
$a_{0},a_{1},a_{2}$, see [7, 9, 11, 15] and the references therein. Logistic
growth described by nonlocal terms has been used in the context of chemotaxis,
which suggests a growth coefficient rate in a competitive system modelling
cancer cells behavior. For instance, the nonlocal term in [13] is given in the
form $\mu_{1}u\left(1-\int_{\Omega}k_{1}(x,y)u(y)dy\right)$ where $u$ denotes
the cancer cells density. We refer the reader to [9, 13] for more details.

To illustrate our results, we observe that the striking feature of (1.4) is
the absence of mass conservation which makes a difficult task to work with
(1.4) directly. To overcome such difficulty, we employ a transformation

$\displaystyle\rho(x,t)=\frac{u(x,t)}{m(t)}$ (1.14)

where $m(t)$ is defined as (1.12), and it transforms system (1.4) into a
diffusion-aggregation equation for $\rho(x,t)$ as follows

$\displaystyle\left\\{\begin{array}[]{ll}\rho_{t}=\Delta\rho-m(t)\nabla\cdot\left(\rho\nabla
w\right),{}{}&x\in{\mathbb{R}}^{2},~{}t\geq 0,\\\ -\Delta
w=\rho,{}{}&x\in{\mathbb{R}}^{2},~{}t\geq 0,\\\
\rho(x,0)=\frac{u_{0}(x)}{m_{0}}\geq
0,{}{}&x\in{\mathbb{R}}^{2}.\end{array}\right.$ (1.18)

Here we should mention that (1.18) admits mass conservation
$\int_{{\mathbb{R}}^{2}}\rho(x,t)dx=\int_{{\mathbb{R}}^{2}}\rho(x,0)dx=1$ via
the transformation, the price we pay is that the coefficient in front of the
concentration is $m(t)$ instead of a constant which also brings barriers to
mathematical analysis. We remark that for the case that the concentration with
constant sensitivity $\chi$ (without loss of generality, we impose $\chi=1$),
there have been a large amount of results [2, 3, 12] and there is a threshold
$8\pi$ on the initial mass separating global existence and finite time blow-up
as we mentioned before. Although we can’t take advantage of the initial mass
to determine the dynamical behaviors of (1.18) due to
$\int_{{\mathbb{R}}^{2}}\rho(x,0)dx=1,$ this fact still stimulates us to
employ the mass conservation of (1.18) to establish global existence and blow-
up to system (1.4). Precisely, the main results of the general solutions are
stated as follows in connection with $M_{0},m_{0}$ and $8\pi.$

  * •

Global existence: for $m_{0}<M_{0}<8\pi,$ there exists a weak solution
globally in time with bounded initial second moment. Furthermore, the weak
solution satisfies the energy inequality (2.3), see Theorem 2.3. In addition,
for $M_{0}<m_{0}<8\pi,$ all solutions of (1.4) exist globally by the
comparison principle.

  * •

$m_{0}<M_{0}=8\pi:$ solutions exist globally in time, see Theorem 2.3. For
$M_{0}=8\pi,$ every stationary solution uniquely assumes a radially symmetric
form in ${\mathbb{R}}^{2}$ up to translation and has infinite second moment.
This result provides that the solution blows up as a delta dirac at the center
of mass as $t\to\infty$ with finite second moment at any time. We will comment
further on these issues in Section 3.

  * •

Finite time blow-up: for $M_{0}>8\pi,$ the weak solution blows up at finite
time, see Theorem 2.6. For $M_{0}<8\pi<m_{0},$ if the initial second moment is
less than a constant depending on $M_{0},m_{0},$ then there exist solutions
blow up at finite time, see Theorem 2.7.

  * •

For $M_{0}<m_{0}=8\pi,$ we can infer from [2] that solutions might have
infinite time blow-up by the comparison principle. However, we can’t exclude
the possibility that solutions may be global in time. For $M_{0}=8\pi<m_{0}$,
both global existence and finite time blow-up can occur, although this is
still an open question.

Let’s emphasise that in the radial context, there are steady states to (1.4)
only for $M_{0}=8\pi$ given by a one-parameter family
$U_{s}(r)=\frac{8\lambda}{(r^{2}+\lambda)^{2}}$ with $\lambda>0.$ The
stationary solutions play a critical role on the initial data separating
global existence and blow-up. If the initial data
$u_{0}(r)<\frac{m_{0}}{M_{0}}\frac{8\lambda}{(r^{2}+\lambda)^{2}}$, then all
the radially symmetric solutions are vanishing in
$L_{loc}^{1}({\mathbb{R}}^{2})$ as $t\to\infty$. If the initial data
$u_{0}(r)>\frac{m_{0}}{M_{0}}\frac{8\lambda}{(r^{2}+\lambda)^{2}}$, then there
could exist a radially symmetric solution satisfying a mass concentration at
the origin as $t\to\infty.$ See Section 4.2.1 and Section 4.2.2 for more
details.

The results are organized as follows. Section 2 shows the global existence and
the finite time blow-up of the weak solution to (1.4). Section 2.1 detects the
global existence of solutions with the help of the energy inequality for
$M_{0}<8\pi$. The finite time blow-up is considered in Section 2.2 with
bounded initial second moment for $M_{0}>8\pi$. Section 3 explores the
infinite time blow-up when $M_{0}=8\pi.$ Section 4 is devoted to the global
existence and the mass concentration at the origin for the radially symmetric
solutions.

##  2 Global existence and finite time blow-up for general solutions

In this section, we prove the global well-posedness for solutions with small
$M_{0}$ and blow-up for large $M_{0}.$ Here an energy inequality (which is
derived to prove global existence) introduces a threshold on $M_{0}$, below
which the solution exists globally and above which the finite time blow-up
occurs with the aid of the second moment. In order to analyse more precisely,
we use the usual definition of solutions in distribution sense as in [12].

######  Definition 2.1.

Let $u_{0}(x)\geq 0$ be the initial data satisfying
$\int_{{\mathbb{R}}^{2}}u_{0}(x)|\log
u_{0}(x)|dx+\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx<\infty$ and
$T\in(0,\infty).$ $u$ is a weak solution to system (1.4) if it satisfies:

  * (i)

Regularity:

$\displaystyle u\in L^{\infty}\left(0,T;L_{+}^{1}\cap
L^{p}({\mathbb{R}}^{2})\right),~{}~{}\mbox{for~{}any}~{}~{}1\leq p\leq\infty,$
$\displaystyle\partial_{t}u\in
L^{2}\left(0,T;H^{-1}({\mathbb{R}}^{2})\right),$ $\displaystyle u^{r}\in
L^{2}\left(0,T;H^{1}({\mathbb{R}}^{2})\right)~{}~{}\mbox{for~{}any}~{}~{}r\geq
1/2.$

  * (ii)

For any $\psi\in C_{0}^{\infty}({\mathbb{R}}^{2})$ and any $0<t<\infty$,

$\displaystyle\int_{{\mathbb{R}}^{2}}\psi
u(\cdot,t)dx-\int_{{\mathbb{R}}^{2}}\psi u_{0}(x)dx$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi
udxdt-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)}{|x-y|^{2}}u(x,t)u(y,t)dxdydt$
$\displaystyle+\int_{0}^{T}\int_{{\mathbb{R}}^{2}}u(\cdot,t)\psi(x)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u(\cdot,t)dx\right)dt.$
(2.1)

###  2.1 Global existence for $M_{0}\leq 8\pi$

In this subsection, we will show the global existence of a weak solution for
both cases $M_{0}=8\pi$ and $M_{0}<8\pi$. Firstly we recall the logarithmic
HLS inequality which is prepared to the estimates of the weak solution.

######  Lemma 2.2 ([12]).

Let $f\geq 0\in L^{1}({\mathbb{R}}^{2})$ such that $f\log f\in
L^{1}({\mathbb{R}}^{2})$. Set $\int_{{\mathbb{R}}^{2}}fdx=M,$ then

$\displaystyle\int_{{\mathbb{R}}^{2}}f\log
fdx+\frac{2}{M}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}f(x)f(y)\log|x-y|dxdy\geq
M\left(\log M-1-\log\pi\right).$ (2.2)

Now we show the main results for $M_{0}\leq 8\pi$.

######  Theorem 2.3 (Global existence for $M_{0}\leq 8\pi$).

Assume $\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx<\infty$ and
$\int_{{\mathbb{R}}^{2}}u_{0}(x)\left|\log u_{0}(x)\right|dx<\infty.$ When the
initial data satisfies $u_{0}\in L_{+}^{1}\cap L^{\infty}({\mathbb{R}}^{2})$
and

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{0}(x)dx<M_{0}\leq 8\pi,$ (2.3)

then there exists a global weak solution to (1.4) satisfying that for any
$0<t<\infty$

$\displaystyle\int_{{\mathbb{R}}^{2}}u(\cdot,t)\left|\log
u(\cdot,t)\right|dx+\int_{{\mathbb{R}}^{2}}u(\cdot,t)|x|^{2}dx$
$\displaystyle\leq C\left(e^{t},\int_{{\mathbb{R}}^{2}}u_{0}(x)|\log
u_{0}(x)|dx,\int_{{\mathbb{R}}^{2}}u_{0}(x)|x|^{2}dx\right),$ (2.4)
$\displaystyle\|u(\cdot,t)\|_{L^{p}({\mathbb{R}}^{2})}$ $\displaystyle\leq
C\left(e^{t},\|u_{0}\|_{L^{p}({\mathbb{R}}^{2})}\right),\quad 1\leq
p\leq\infty.$ (2.5)

Furthermore, the weak solution satisfies the entropy dissipation

$\displaystyle
F[u(\cdot,t)]+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\frac{u(x,s)}{m(s)}\left|\nabla\log
u(x,s)-\nabla c(x,s)\right|^{2}dxds$
$\displaystyle+\int_{0}^{t}\frac{M_{0}-m(s)}{2m(s)}\int_{{\mathbb{R}}^{2}}u(x,s)c(x,s)dxds\leq
F[u_{0}(x)],$ (2.6)

where the free energy

$\displaystyle F(u)=\frac{1}{m(t)}\int_{{\mathbb{R}}^{2}}u\log
udx-\frac{1}{2m(t)}\int_{{\mathbb{R}}^{2}}ucdx-\log m(t)$ (2.7)

and $m(t)=\frac{M_{0}}{1+\frac{M_{0}-m_{0}}{m_{0}}e^{-M_{0}t}}$.

Proof.  We decompose the proof into two parts. In Steps 1-8, a regularized
equation is constructed and a priori estimates are established to obtain the
global existence of a weak solution to (1.4). Furthermore, Steps 9-10 give the
second moment and the energy inequality of the weak solution.

Following the method of [14], we take a cut-off function $0\leq\psi_{1}(x)\leq
1$

$\displaystyle\psi_{1}(x)=\left\\{\begin{array}[]{ll}1&\mbox{ if }|x|\leq
1,\\\ 0&\mbox{ if }|x|\geq 2,\end{array}\right.$

where $\psi_{1}(x)\in C_{0}^{\infty}({\mathbb{R}}^{d})$. Define

$\displaystyle\psi_{R}(x):=\psi_{1}(x/R),$ (2.8)

as $R\to\infty,~{}\psi_{R}\to 1$, then there exist constants $C_{1},C_{2}$
such that
$|\nabla\psi_{R}(x)|\leq\frac{C_{1}}{R},~{}~{}|\Delta\psi_{R}(x)|\leq\frac{C_{2}}{R^{2}}.$
This cut-off function will be used to derive the existence of the weak
solution.

Step 1 (Approximated problem) In order to show the existence of a weak
solution, we firstly consider the regularized problem for $\varepsilon>0$

$\displaystyle\left\\{\begin{array}[]{ll}\partial_{t}u_{\varepsilon}=\Delta
u_{\varepsilon}-\nabla\cdot(u_{\varepsilon}c_{\varepsilon})+u_{\varepsilon}\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right),&x\in{\mathbb{R}}^{2},t\geq
0,\\\ -\Delta c_{\varepsilon}=J_{\varepsilon}\ast
u_{\varepsilon},&x\in{\mathbb{R}}^{2},t\geq 0,\\\
u_{\varepsilon}(x,0)=u_{0\varepsilon}(x),&x\in{\mathbb{R}}^{2}.\end{array}\right.$
(2.12)

Here $J_{\varepsilon}$ is the regularizing kernel

$\displaystyle
J_{\varepsilon}(x)=\frac{1}{\pi}\frac{\varepsilon^{2}}{\left(|x|^{2}+\varepsilon^{2}\right)^{2}}$
(2.13)

with $\int_{{\mathbb{R}}^{2}}J_{\varepsilon}dx=1.$ Simple computations show
that

$\displaystyle
c_{\varepsilon}(x,t)=-\frac{1}{4\pi}\int_{{\mathbb{R}}^{2}}\log\left(|x-y|^{2}+\varepsilon^{2}\right)u_{\varepsilon}(y,t)dy.$
(2.14)

The regularized initial data $u_{0\varepsilon}\in
C^{\infty}({\mathbb{R}}^{2})$ is a sequence of approximation for $u_{0}(x)$
and satisfies

$\displaystyle\left\\{\begin{array}[]{ll}\|u_{0\varepsilon}\|_{L^{p}({\mathbb{R}}^{2})}\leq\|u_{0}\|_{L^{p}({\mathbb{R}}^{2})}~{}~{}\mbox{for~{}any}~{}~{}1\leq
p\leq\infty,\\\\[2.84526pt] u_{0\varepsilon}(x)\to
u_{0}(x)~{}~{}\mbox{in}~{}~{}L^{q}({\mathbb{R}}^{2})~{}~{}\mbox{for}~{}~{}1\leq
q<\infty,\\\\[2.84526pt]
\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0\varepsilon}(x)dx\to\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx~{}~{}\mbox{as}~{}~{}\varepsilon\to
0,\\\\[2.84526pt] \int_{{\mathbb{R}}^{2}}u_{0\varepsilon}|\log
u_{0\varepsilon}|dx\to\int_{{\mathbb{R}}^{2}}u_{0}|\log
u_{0}|dx~{}~{}\mbox{as}~{}~{}\varepsilon\to 0.\\\ \end{array}\right.$ (2.19)

Then following the standard parabolic theory, the system on
$(u_{\varepsilon},c_{\varepsilon})$ admits a unique smooth solution with fast
decay in space, when the initial data is regularized and truncated. Moreover,
there exists $T>0$ such that for $0<t<T$ and any $1\leq r\leq\infty$

$\displaystyle u_{\varepsilon}\in L^{\infty}(0,T;L^{r}({\mathbb{R}}^{2})).$

Step 2 (Global $L^{1}$ norm of $u_{\varepsilon}$ and the boundedness of the
second moment) In this step, we will provide some a priori estimates for
$u_{\varepsilon}$. Firstly, multiplying (2.12) with any test function
$\psi_{R}(x)$ gives

$\displaystyle\frac{d}{dt}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(\cdot,t)\psi_{R}(x)dx$
$\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\Delta\psi_{R}(x)dx-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi_{R}(x)-\nabla\psi_{R}(y)]\cdot(x-y)}{|x-y|^{2}+\varepsilon^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)dxdy$
$\displaystyle+\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\psi_{R}(x)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right).$
(2.20)

Letting $\psi_{R}(x)$ be defined as in (2.8), then

$\displaystyle\left|\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\Delta\psi_{R}(x)dx\right|\leq\frac{C}{R^{2}}\int_{R^{2}}u_{\varepsilon}dx,$
$\displaystyle\left|\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi_{R}(x)-\nabla\psi_{R}(y)]\cdot(x-y)}{|x-y|^{2}+\varepsilon^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)dxdy\right|\leq\frac{C}{R^{2}}\left(\int_{R^{2}}u_{\varepsilon}dx\right)^{2}.$

Passing to the limit $R\to\infty$ arrives at

$\displaystyle\frac{d}{dt}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx=\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right)$
(2.21)

and we compute that for any $t>0$

$\displaystyle
m(t):=\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx=\frac{M_{0}}{1+\frac{M_{0}-m_{0}}{m_{0}}e^{-M_{0}t}}.$
(2.22)

Now we will prove that the second moment is bounded in time provided the
bounded initial second moment $\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x,0)dx$.
Consider a test function $\psi_{R}(x)\in C_{0}^{\infty}({\mathbb{R}}^{2})$
that grows nicely to $|x|^{2}$ as $R\to\infty$. We can infer from (2.1) that
$\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(\cdot,t)\psi_{R}(x)dx$ is uniformly
bounded, thus we may pass to the limit using the Lebesgue monotone convergence
theorem to obtain

$\displaystyle\frac{d}{dt}\int_{R^{2}}|x|^{2}u_{\varepsilon}dx=$
$\displaystyle
4\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx-\frac{1}{2\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{|x-y|^{2}}{|x-y|^{2}+\varepsilon^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)dxdy$
$\displaystyle+\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right)$
$\displaystyle\leq$ $\displaystyle
4\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx+M_{0}\int_{R^{2}}|x|^{2}u_{\varepsilon}dx.$

Then one gets by integrating from $0$ to $t$ in time

$\displaystyle\int_{R^{2}}|x|^{2}u_{\varepsilon}dx\leq
C\left(e^{t},\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}(x,0)dx\right).$
(2.23)

Step 3 (Boundedness of $\int_{{\mathbb{R}}^{2}}u_{\varepsilon}|\log
u_{\varepsilon}|dx$ for $M_{0}\leq 8\pi$) Recalling (2.22) we shall use the
notation

$\displaystyle
F(u_{\varepsilon})=\int_{{\mathbb{R}}^{2}}\frac{u_{\varepsilon}}{m(t)}\log\frac{u_{\varepsilon}}{m(t)}dx-\frac{1}{2}\int_{{\mathbb{R}}^{2}}\frac{u_{\varepsilon}}{m(t)}c_{\varepsilon}dx.$
(2.24)

Then a straightforward computation shows that

$\displaystyle\frac{d}{dt}F(u_{\varepsilon})=-\int_{{\mathbb{R}}^{2}}\frac{u_{\varepsilon}}{m(t)}\left|\nabla\left(\log\frac{u_{\varepsilon}}{m(t)}-c_{\varepsilon}\right)\right|^{2}dx-\frac{M_{0}-m(t)}{2m(t)}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}c_{\varepsilon}dx.$
(2.25)

Thus for $m_{0}<M_{0},$ differentiating it with respect to $t$ yields

$\displaystyle F(u_{\varepsilon})\leq F(u_{0\varepsilon}).$ (2.26)

It can be checked that

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx-\frac{1}{2}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}c_{\varepsilon}dx\leq
m(t)F(u_{0\varepsilon})+m(t)\log m(t).$ (2.27)

Hence from (2.2) we obtain the inequality

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log u_{\varepsilon}dx$
$\displaystyle\leq$ $\displaystyle m(t)F(u_{0\varepsilon})+m(t)\log
m(t)-\frac{1}{8\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log(|x-y|^{2}+\varepsilon^{2})dxdy$
$\displaystyle\leq$ $\displaystyle m(t)F(u_{0\varepsilon})+m(t)\log
m(t)-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log|x-y|dxdy$
$\displaystyle\leq$ $\displaystyle m(t)F(u_{0\varepsilon})+m(t)\log
m(t)-\frac{m^{2}(t)}{8\pi}\left(\log
m(t)-1-\log\pi\right)+\frac{m(t)}{8\pi}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx,$ (2.28)

For $m(t)<M_{0}\leq 8\pi,$ it follows from (2.22) that
$\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log u_{\varepsilon}dx$ can be
estimated respectively

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx\leq\left\\{\begin{array}[]{ll}\frac{8\pi M_{0}}{m_{0}(8\pi-
M_{0})}~{}C\left(F(u_{0\varepsilon}),M_{0},m_{0}\right),&\mbox{for}~{}M_{0}<8\pi,\\\\[5.69054pt]
\frac{8\pi\left(m_{0}e^{M_{0}t}+M_{0}-m_{0}\right)}{8\pi(M_{0}-m_{0})}~{}C\left(F(u_{0\varepsilon}),M_{0},m_{0}\right),&\mbox{for}~{}M_{0}=8\pi.\end{array}\right.$
(2.31)

On the other hand, the following holds

$\displaystyle 1_{\\{\varphi(x)\leq
1\\}}\varphi(x)\left|\log\varphi(x)\right|$ (2.32) $\displaystyle=$
$\displaystyle
1_{\\{\varphi(x)<e^{-2|x|^{2}}\\}}\varphi(x)\log\frac{1}{\varphi(x)}+1_{\\{e^{-2|x|^{2}}\leq\varphi(x)\leq
1\\}}\varphi(x)\log\frac{1}{\varphi(x)}$ $\displaystyle\leq$ $\displaystyle
1_{\\{\varphi(x)<e^{-2|x|^{2}}\\}}\sqrt{\varphi(x)}+1_{\\{e^{-2|x|^{2}}\leq\varphi(x)\leq
1\\}}2\varphi(x)|x|^{2}$ $\displaystyle\leq$ $\displaystyle
e^{-|x|^{2}}+2\varphi(x)|x|^{2}.$ (2.33)

Hence combining (2.31) with (2.23) it gives our result for this step

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\left|\log
u_{\varepsilon}\right|dx$
$\displaystyle=\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx+2\int_{{\mathbb{R}}^{2}}1_{\\{u_{\varepsilon}\leq
1\\}}u_{\varepsilon}\left|\log u_{\varepsilon}\right|dx$
$\displaystyle\leq\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx+2\int_{{\mathbb{R}}^{2}}e^{-|x|^{2}}dx+4\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}dx$
$\displaystyle\leq C_{1}+C_{2}(e^{t}).$ (2.34)

Step 4 (Equi-integrabiliy in $L^{p}$ norms for $1<p<\infty$) Testing (2.12)
with $p(u_{\varepsilon}-K)_{+}^{p-1}$ we obtain

$\displaystyle\frac{d}{dt}\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx+\frac{4(p-1)}{p}\int_{{\mathbb{R}}^{2}}\left|\nabla(u_{\varepsilon}-K)_{+}^{p/2}\right|^{2}dx$
$\displaystyle=$ $\displaystyle(p-1)\int_{{\mathbb{R}}^{2}}J_{\varepsilon}\ast
u_{\varepsilon}(u_{\varepsilon}-K)_{+}^{p}dx+Kp\int_{{\mathbb{R}}^{2}}J_{\varepsilon}\ast
u_{\varepsilon}(u_{\varepsilon}-K)_{+}^{p-1}dx+p\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}u_{\varepsilon}dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right)$
$\displaystyle=$
$\displaystyle(p-1)\int_{{\mathbb{R}}^{2}}J_{\varepsilon}\ast(u_{\varepsilon}-K+K)(u_{\varepsilon}-K)_{+}^{p}dx+Kp\int_{{\mathbb{R}}^{2}}J_{\varepsilon}\ast(u_{\varepsilon}-K+K)(u_{\varepsilon}-K)_{+}^{p-1}dx$
$\displaystyle+p\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}(u_{\varepsilon}-K+K)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx\right)$
$\displaystyle\leq$
$\displaystyle(p-1)\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p+1}dx+K(2p-1)\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx+K^{2}p\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}dx$
$\displaystyle+M_{0}p\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx+KpM_{0}\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}dx.$
(2.35)

We use GNS inequality to get

$\displaystyle\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p+1}dx\leq
C(p)\int_{{\mathbb{R}}^{2}}\left|\nabla(u_{\varepsilon}-K)_{+}^{p/2}\right|^{2}dx\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}dx.$
(2.36)

By (2.1) one has

$\displaystyle\int_{u_{\varepsilon}>K}(u_{\varepsilon}-K)_{+}dx\leq\int_{u_{\varepsilon}>K}u_{\varepsilon}dx\leq\frac{\int_{u_{\varepsilon}>K}u_{\varepsilon}|\log
u_{\varepsilon}|dx}{\log
K}\leq\frac{\int_{{\mathbb{R}}^{2}}u_{\varepsilon}|\log
u_{\varepsilon}|dx}{\log K}\leq\frac{C(t)}{\log K}.$ (2.37)

Moreover, $\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}dx$ can be
estimated as

$\displaystyle\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p-1}dx$
$\displaystyle=\int_{K\leq u_{\varepsilon}\leq
K+1}(u_{\varepsilon}-K)_{+}^{p-1}dx+\int_{u_{\varepsilon}>K+1}(u_{\varepsilon}-K)_{+}^{p-1}dx$
$\displaystyle\leq\int_{K\leq u_{\varepsilon}\leq
K+1}1dx+\int_{u_{\varepsilon}>K+1}(u_{\varepsilon}-K)_{+}^{p}dx$
$\displaystyle\leq\frac{\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx}{K}+\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx.$
(2.38)

Collecting (2.36), (2.37) and (2.1) we can go further from (2.1) that

$\displaystyle\frac{d}{dt}\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx+\frac{4(p-1)}{p}\int_{{\mathbb{R}}^{2}}\left|\nabla(u_{\varepsilon}-K)_{+}^{p/2}\right|^{2}dx$
$\displaystyle\leq$ $\displaystyle C(p)\frac{C(t)}{\log
K}\int_{{\mathbb{R}}^{2}}\left|\nabla(u_{\varepsilon}-K)_{+}^{p/2}\right|^{2}dx+C(K,p)\frac{\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx}{K}+C(K,p)\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx.$
(2.39)

Choosing $K$ large enough such that $\frac{C(p)C(t)}{\log
K}<\frac{4(p-1)}{p}$, we obtain

$\displaystyle\frac{d}{dt}\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx+C(p,K)\int_{{\mathbb{R}}^{2}}\left|\nabla(u_{\varepsilon}-K)_{+}^{p/2}\right|^{2}dx\leq
C_{1}+C_{2}\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx$ (2.40)

and subsequently for any $0<t<T$

$\displaystyle\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)_{+}^{p}dx\leq
C\left(e^{t},\int_{{\mathbb{R}}^{2}}u_{0\varepsilon}^{p}dx\right).$ (2.41)

Step 5 (Boundedness of $L^{p}$ norm for $1<p\leq\infty$) Now we can go further
to prove that $\int_{{\mathbb{R}}^{2}}u_{\varepsilon}^{p}dx$ is bounded as
follows.

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}^{p}dx=$
$\displaystyle\int_{u_{\varepsilon}\leq
K}u_{\varepsilon}^{p}dx+\int_{u_{\varepsilon}>K}u_{\varepsilon}^{p}dx$
$\displaystyle\leq$ $\displaystyle
K^{p-1}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx+\int_{u_{\varepsilon}>K}(u_{\varepsilon}-K+K)^{p-1}u_{\varepsilon}dx$
$\displaystyle\leq$ $\displaystyle
K^{p-1}M_{0}+\max(2^{p-2},1)\left(\int_{u_{\varepsilon}>K}(u_{\varepsilon}-K)^{p-1}u_{\varepsilon}dx+K^{p-1}M_{0}\right)$
$\displaystyle\leq$ $\displaystyle
K^{p-1}M_{0}+\max(2^{p-2},1)\left(K\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)^{p-1}dx+\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)^{p}dx+K^{p-1}M_{0}\right)$
$\displaystyle\leq$ $\displaystyle
K^{p-1}M_{0}+\max(2^{p-2},1)\left(\int_{{\mathbb{R}}^{2}}u_{\varepsilon}dx+K\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)^{p}dx+\int_{{\mathbb{R}}^{2}}(u_{\varepsilon}-K)^{p}dx+K^{p-1}M_{0}\right),$

where the last line is given by (2.1). Therefore, (2.41) guarantees that for
any $0<t<T$

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}^{p}dx\leq
C(M_{0},e^{t},K,p)$ (2.42)

for any $1<p<\infty.$ Furthermore, integrating (2.40) from $0$ to $T$ we have
that for any $T>0$

$\displaystyle\nabla u_{\varepsilon}^{\frac{p}{2}}\in
L^{2}\left(0,T;L^{2}({\mathbb{R}}^{2})\right)~{}~{}\mbox{for~{}any}~{}~{}1<p<\infty.$
(2.43)

Finally, making use of the iterative method in the spirit of [1] we have the
uniformly boundedness

$\displaystyle u_{\varepsilon}\in
L^{\infty}\left(0,T;L^{\infty}({\mathbb{R}}^{2})\right).$ (2.44)

Step 6 (Time regularity) It directly follows from (2.42), (2.43) and (2.44)
that

$\displaystyle\|\nabla
u_{\varepsilon}^{r}\|_{L^{2}(0,T;L^{2}({\mathbb{R}}^{2}))}\leq
C,\quad\mbox{for~{}any}~{}~{}r>1/2,$ (2.45)
$\displaystyle\|u_{\varepsilon}\nabla
c_{\varepsilon}\|_{L^{\infty}(0,T;L^{\infty}({\mathbb{R}}^{2}))}\leq C,$
(2.46)
$\displaystyle\|\partial_{t}u_{\varepsilon}\|_{L^{2}(0,T;H^{-1}({\mathbb{R}}^{2}))}\leq
C.$ (2.47)

Then by Lemma 4.23 in [4] one has that for any bounded domain $\Omega$, there
exists a subsequence $u_{\varepsilon}$ without relabeling such that

$\displaystyle u_{\varepsilon}\to
u~{}~{}\mbox{in}~{}~{}L^{2}\left(0,T;L^{\bar{p}}(\Omega)\right),\quad\mbox{for~{}any}~{}1\leq\bar{p}<\infty.$
(2.48)

By a standard diagonal argument, the following uniform strong convergence
holds true that for any $R>0$

$\displaystyle u_{\varepsilon}\to
u~{}~{}\mbox{in}~{}~{}L^{2}\left(0,T;L^{\bar{p}}(B_{R})\right).$ (2.49)

Let us turn back to sketch the proof of (2.46) and (2.47). Actually, the term

$\displaystyle\nabla
c_{\varepsilon}=-\frac{1}{2\pi}\int_{{\mathbb{R}}^{2}}\frac{x-y}{|x-y|^{2}+\varepsilon^{2}}u_{\varepsilon}(y,t)dy$
(2.50)

can be estimated from the Young inequality that

$\displaystyle\|\nabla c_{\varepsilon}\|_{L^{\infty}({\mathbb{R}}^{2})}$
$\displaystyle\leq\frac{1}{2\pi}\left\|\int_{0<|x-y|\leq
1}\frac{u_{\varepsilon}(y,t)}{|x-y|}dy+\int_{|x-y|>1}\frac{u_{\varepsilon}(y,t)}{|x-y|}dy\right\|_{L^{\infty}({\mathbb{R}}^{2})}$
$\displaystyle\leq\frac{1}{2\pi}\left(\|u_{\varepsilon}\|_{L^{\infty}({\mathbb{R}}^{2})}\left\|\frac{1}{|x|}\right\|_{L^{1}(0<|x|\leq
1)}+\|u_{\varepsilon}\|_{L^{1}({\mathbb{R}}^{2})}\right)$ $\displaystyle\leq
C\left(\|u_{\varepsilon}\|_{L^{\infty}({\mathbb{R}}^{2})}+\|u_{\varepsilon}\|_{L^{1}({\mathbb{R}}^{2})}\right).$
(2.51)

The bound of $u_{\varepsilon}\nabla c_{\varepsilon}\in
L^{\infty}(0,T;L^{\infty}({\mathbb{R}}^{2}))$ follows by using
$u_{\varepsilon}\in L^{\infty}(0,T;L^{1}({\mathbb{R}}^{2}))$.

Moreover, (2.47) can be deduced by using the previous estimates that for any
test function $\psi\in L^{2}\left(0,T;H^{1}({\mathbb{R}}^{2})\right)$

$\displaystyle\left|\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\partial_{t}u_{\varepsilon}~{}\psi
dxdt\right|$ $\displaystyle\leq\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\left|\nabla
u_{\varepsilon}\cdot\nabla\psi\right|dxdt+\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\left|u_{\varepsilon}\nabla
c_{\varepsilon}\cdot\nabla\psi\right|dxdt$ $\displaystyle\leq\left(\|\nabla
u_{\varepsilon}\|_{L^{2}(0,T;{\mathbb{R}}^{2})}+\|\sqrt{u_{\varepsilon}}\nabla
c_{\varepsilon}\|_{L^{\infty}(0,T;L^{\infty}({\mathbb{R}}^{2}))}\sqrt{T\|u_{\varepsilon}\|_{L^{1}({\mathbb{R}}^{2})}}~{}\right)\|\nabla\psi\|_{L^{2}(0,T;L^{2}({\mathbb{R}}^{2}))}$
$\displaystyle\leq C\|\nabla\psi\|_{L^{2}(0,T;L^{2}({\mathbb{R}}^{2}))}.$
(2.52)

Step 7 (Strong convergence of $u_{\varepsilon}$) Furthermore, we will take
advantage of the second moment estimate (2.23) to establish the strong
convergence of $u_{\varepsilon}$ in $L^{2}(0,T;L^{q}({\mathbb{R}}^{2}))$ and
extend (2.49) to the whole space. We compute that for any $1\leq q<\infty$

$\displaystyle\int_{0}^{T}\|u_{\varepsilon}\|_{L^{q}(|x|>R)}^{2}dt$
$\displaystyle\leq\int_{0}^{T}\|u_{\varepsilon}\|_{L^{\infty}(|x|>R)}^{2(q-1)/q}\|u_{\varepsilon}\|_{L^{1}(|x|>R)}^{2/q}dt$
$\displaystyle\leq\int_{0}^{T}\|u_{\varepsilon}\|_{L^{\infty}(|x|>R)}^{2(q-1)/q}\frac{\left(\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}dx\right)^{2/q}}{R^{4/q}}dt\to
0~{}~{}\mbox{as}~{}~{}R\to\infty,$ (2.53)

and the weak semi-continuity of
$L^{2}\left(0,T;L^{q}({\mathbb{R}}^{2})\right)$ implies

$\displaystyle\int_{0}^{T}\|u\|_{L^{q}(|x|>R)}^{2}dt\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{0}^{T}\|u_{\varepsilon}\|_{L^{q}(|x|>R)}^{2}dt\to
0~{}~{}\mbox{as}~{}~{}R\to\infty.$ (2.54)

Therefore, the following inequality is derived that for any $1\leq q<\infty,$
as $R\to\infty,\varepsilon\to 0,$

$\displaystyle\int_{0}^{T}\|u_{\varepsilon}-u\|_{L^{q}({\mathbb{R}}^{2})}^{2}dt=\int_{0}^{T}\left(\|u_{\varepsilon}-u\|_{L^{q}(|x|>R)}+\|u_{\varepsilon}-u\|_{L^{q}(|x|\leq
R)}\right)^{2}dt$ $\displaystyle\leq$
$\displaystyle~{}C(q)\left(\int_{0}^{T}\|u_{\varepsilon}\|_{L^{q}(|x|>R)}^{2}+\int_{0}^{T}\|u\|_{L^{q}(|x|>R)}^{2}dt+\int_{0}^{T}\|u_{\varepsilon}-u\|_{L^{q}(|x|\leq
R)}^{2}dt\right)\to 0.$ (2.55)

In the last inequality, the first term goes to zero due to (2.1), the second
term is given by (2.54) and (2.49) provides the third term. Hence one has

$\displaystyle u_{\varepsilon}\to
u~{}~{}\mbox{in}~{}~{}L^{2}\left(0,T;L^{r}({\mathbb{R}}^{2})\right),\quad\mbox{for~{}any}~{}1\leq
r<\infty.$ (2.56)

Step 8 (Existence of the weak solution) Now multiplying (2.12) by $\psi\in
C_{0}^{\infty}({\mathbb{R}}^{2})$ and integrating it with respect to $x$ and
$t$, we get the weak formulation for $u_{\varepsilon}$

$\displaystyle\int_{{\mathbb{R}}^{2}}\psi
u_{\varepsilon}(\cdot,t)dx-\int_{{\mathbb{R}}^{2}}\psi u_{0\varepsilon}dx$
$\displaystyle=$ $\displaystyle\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi
u_{\varepsilon}dxdt-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)}{|x-y|^{2}+\varepsilon^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)dxdydt$
$\displaystyle+\int_{0}^{T}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(\cdot,t)\psi(x)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(\cdot,t)dx\right)dt.$
(2.57)

It’s to be noticed that (2.49) directly yields

$\displaystyle\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi
u_{\varepsilon}dxdt\to\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi
udxdt,\quad\varepsilon\to 0.$ (2.58)

As to the second term of the right side of (2.1), note that

$\displaystyle\left|\int_{0}^{T}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)\left(\frac{1}{|x-y|^{2}}-\frac{1}{|x-y|^{2}+\varepsilon^{2}}\right)u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)dxdydt\right|$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\int_{0}^{T}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)}{|x-y|}dxdydt\leq
C\varepsilon\int_{0}^{T}\|u_{\varepsilon}\|_{L^{4/3}({\mathbb{R}}^{2})}^{2}dt\leq
C\varepsilon.$ (2.59)

In addition, by Cauchy inequality we have

$\displaystyle\left|\int_{0}^{T}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)\left(\frac{u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)}{|x-y|^{2}}-\frac{u(x,t)u(y,t)}{|x-y|^{2}}\right)dxdydt\right|$
$\displaystyle\leq$ $\displaystyle
C\left(\int_{0}^{T}\iint_{\Omega\times\Omega}|u_{\varepsilon}(x)-u(x)|u_{\varepsilon}(y)dxdy+\iint_{\Omega\times\Omega}|u_{\varepsilon}(y)-u(y)|u(x)dxdydt\right)$
$\displaystyle\leq$ $\displaystyle
C\int_{0}^{T}\|u_{\varepsilon}-u\|_{L^{2}(\Omega)}^{2}dt\int_{0}^{T}\|u_{\varepsilon}\|_{L^{2}(\Omega)}^{2}dt,$
(2.60)

where the last line is given by the semi-continuity of
$\|u\|_{L^{2}({\mathbb{R}}^{2})}\leq\displaystyle\liminf_{\varepsilon\to
0}\|u_{\varepsilon}\|_{L^{2}({\mathbb{R}}^{2})}$. Thus taking the limit
$\varepsilon\to 0$ and combining (2.59) and (2.60) we conclude

$\displaystyle\int_{0}^{T}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)\left(\frac{u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)}{|x-y|^{2}+\varepsilon^{2}}-\frac{u(x,t)u(y,t)}{|x-y|^{2}}\right)dxdydt\to
0.$ (2.61)

Thanks to (2.58), (2.61) and (2.56), passing to the limit $\varepsilon\to 0$
in (2.1) one has that for any $0<t<T$

$\displaystyle\int_{{\mathbb{R}}^{2}}\psi
u(\cdot,t)dx-\int_{{\mathbb{R}}^{2}}\psi u_{0}(x)dx$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi
udxdt-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi(x)-\nabla\psi(y)]\cdot(x-y)}{|x-y|^{2}}u(x,t)u(y,t)dxdydt$
$\displaystyle+\int_{0}^{T}\int_{{\mathbb{R}}^{2}}u(\cdot,t)\psi(x)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u(\cdot,t)dx\right)dt.$
(2.62)

This gives the existence of a global weak solution.

Step 9 (The second moment of the weak solution) Consider a test function
$\psi_{R}(x)\in C_{0}^{\infty}({\mathbb{R}}^{2})$ and $\psi_{R}(x)=|x|^{2}$
for $|x|<R,\psi_{R}(x)=0$ for $|x|\geq 2R,$ letting $\psi=\psi_{R}$ in (2.1)
arrives at

$\displaystyle\int_{{\mathbb{R}}^{2}}\psi_{R}u(\cdot,t)dx-\int_{{\mathbb{R}}^{2}}\psi_{R}u_{0}(x)dx$
$\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{{\mathbb{R}}^{2}}\Delta\psi_{R}udxdt-\frac{1}{4\pi}\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\frac{[\nabla\psi_{R}(x)-\nabla\psi_{R}(y)]\cdot(x-y)}{|x-y|^{2}}u(x,t)u(y,t)dxdydt$
$\displaystyle+\int_{0}^{T}\int_{{\mathbb{R}}^{2}}u(\cdot,t)\psi_{R}(x)dx\left(M_{0}-\int_{{\mathbb{R}}^{2}}u(\cdot,t)dx\right)dt.$
(2.63)

Moreover, we follow the same lines as (2.1) to (2.22) to claim that

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x,t)dx=m(t).$ (2.64)

As before, since $\Delta\psi_{R}(x)$ and
$\frac{[\nabla\psi_{R}(x)-\nabla\psi_{R}(y)]\cdot(x-y)}{|x-y|^{2}}$ are
bounded, thus the first two terms in the right-hand side of (2.1) are bounded.
As a consequence, as $R\to\infty$, we may pass to the limit using the Lebesgue
monotone convergence theorem with $u\in L^{1}({\mathbb{R}}^{2})$ and obtain
that for any $t>0$

$\displaystyle\int_{{\mathbb{R}}^{2}}|x|^{2}u(x,t)dx=$
$\displaystyle\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx+4\int_{0}^{t}m(s)ds-\frac{1}{2\pi}\int_{0}^{t}m^{2}(s)ds$
$\displaystyle+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}|x|^{2}u(x,s)dx(M_{0}-m(s))ds.$
(2.65)

Then by Gronwall’s inequality we have

$\displaystyle\int_{{\mathbb{R}}^{2}}|x|^{2}u(x)dx\leq C.$ (2.66)

Step 10 (The energy inequality of the weak solution) Integrating (2.25) in
time from $0$ to $t$ follows

$\displaystyle\frac{1}{m(t)}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}\log
u_{\varepsilon}dx-\frac{1}{2m(t)}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}c_{\varepsilon}dx-\log
m(t)+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\frac{u_{\varepsilon}}{m(s)}\left|\nabla\log
u_{\varepsilon}-\nabla c_{\varepsilon}\right|^{2}dxds$ $\displaystyle+$
$\displaystyle\int_{0}^{t}\frac{M_{0}-m(s)}{2m(s)}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}c_{\varepsilon}dxds=\frac{1}{m_{0}}\int_{{\mathbb{R}}^{2}}u_{0\varepsilon}\log
u_{0\varepsilon}dx-\frac{1}{2m_{0}}\int_{{\mathbb{R}}^{2}}u_{0\varepsilon}c_{0\varepsilon}dx-\log
m_{0}.$ (2.67)

The aim of the final step is taking limit $\varepsilon\to 0$ in (2.1) to
deduce the energy inequality (2.3).

For the sake of passing limit in (2.1), we split it into three parts. Firstly
it is derived from Lemma 4.5 in the ”Appendix” that

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x,t)\log
u(x,t)dx\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)\log u_{\varepsilon}(x,t)dx.$
(2.68)

Secondly, the lower semi-continuity of the energy dissipation is followed from
Lemma 4.9 of [4] that for any $t>0$

$\displaystyle\int_{0}^{t}\frac{1}{m(s)}\int_{{\mathbb{R}}^{2}}\left|2\nabla\sqrt{u}-\sqrt{u}\nabla
c\right|^{2}dxds\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{0}^{t}\frac{1}{m(s)}\int_{{\mathbb{R}}^{2}}\left|2\nabla\sqrt{u_{\varepsilon}}-\sqrt{u_{\varepsilon}}\nabla
c_{\varepsilon}\right|^{2}dxds.$ (2.69)

It remains to verify the strong convergence of
$\int_{{\mathbb{R}}^{2}}u_{\varepsilon}c_{\varepsilon}dx$ that

$\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)c_{\varepsilon}(x,t)dx\to\int_{{\mathbb{R}}^{2}}u(x,t)c(x,t)dx~{}~{}\mbox{a.e.~{}in}~{}~{}(0,T).$
(2.70)

We write

$\displaystyle-4\pi\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)c_{\varepsilon}(x,t)-u(x,t)c(x,t)dx$
$\displaystyle=$
$\displaystyle\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log\frac{|x-y|^{2}+\varepsilon^{2}}{|x-y|^{2}}dxdy+2\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}u_{\varepsilon}(x,t)(u_{\varepsilon}(y,t)-u(y,t))\log|x-y|dxdy$
$\displaystyle+2\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}(u_{\varepsilon}(x,t)-u(x,t))u(y,t)\log|x-y|dxdy=:I_{1}+I_{2}+I_{3}.$
(2.71)

For $I_{1},$ we compute that for any $0<t<T$

$\displaystyle I_{1}=$ $\displaystyle\iint_{\\{|x-y|\geq
1\\}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log\frac{|x-y|^{2}+\varepsilon^{2}}{|x-y|^{2}}dxdy+\iint_{\\{|x-y|<1\\}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log\frac{|x-y|^{2}+\varepsilon^{2}}{|x-y|^{2}}dxdy$
$\displaystyle\leq$ $\displaystyle\iint_{\\{|x-y|\geq
1\\}}u_{\varepsilon}(x,t)u_{\varepsilon}(y,t)\log(1+\varepsilon^{2})dxdy$
$\displaystyle+C\|u_{\varepsilon}\|_{L^{\infty}({\mathbb{R}}^{2})}\|u_{\varepsilon}\|_{L^{1}({\mathbb{R}}^{2})}\int_{0}^{1}r\log\frac{r^{2}+\varepsilon^{2}}{r^{2}}dr\to
0,~{}~{}\mbox{as}~{}~{}\varepsilon\to 0.$ (2.72)

Moreover, we deal with $I_{2}$ or $I_{3}$ as follows.

$\displaystyle\frac{1}{2}|I_{2}|\leq$ $\displaystyle\iint_{|x-y|\geq
1}u_{\varepsilon}(x,\cdot)|u_{\varepsilon}(y,\cdot)-u(y,\cdot)|\log|x-y|dxdy$
$\displaystyle+\iint_{|x-y|<1}u_{\varepsilon}(x,\cdot)|u_{\varepsilon}(y,\cdot)-u(y,\cdot)|\left|\log|x-y|\right|dxdy=:II_{1}+II_{2}.$

Here it can be estimated that as $\varepsilon\to 0$

$\displaystyle II_{1}=$ $\displaystyle\iint_{|x-y|\geq
1}u_{\varepsilon}(x,\cdot)\left|u_{\varepsilon}(y,\cdot)-u(y,\cdot)\right|^{1/2}\left|u_{\varepsilon}(y,\cdot)-u(y,\cdot)\right|^{1/2}\log|x-y|dxdy$
$\displaystyle\leq$ $\displaystyle\iint_{|x-y|\geq
1}u_{\varepsilon}(x,\cdot)^{1/2}u_{\varepsilon}(x,\cdot)^{1/2}\left|u_{\varepsilon}(y,\cdot)-u(y,\cdot)\right|^{1/2}\left|u_{\varepsilon}(y,\cdot)-u(y,\cdot)\right|^{1/2}|x-y|dxdy$
$\displaystyle\leq$ $\displaystyle
C\left(\|u_{\varepsilon}\|_{L^{1}({\mathbb{R}}^{2})}\right)\|u_{\varepsilon}-u\|_{L^{1}({\mathbb{R}}^{2})}^{1/2}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x)|x|^{2}dx\to
0$ (2.73)

and

$\displaystyle II_{2}\leq$
$\displaystyle\iint_{|x-y|<1}u_{\varepsilon}(x,\cdot)\left|u_{\varepsilon}(y,\cdot)-u(y,\cdot)\right|\log\frac{1}{|x-y|}dxdy$
$\displaystyle\leq$ $\displaystyle
C\|u_{\varepsilon}-u\|_{L^{1}({\mathbb{R}}^{2})}\|u_{\varepsilon}\|_{L^{\infty}({\mathbb{R}}^{2})}\int_{0}^{1}r\log\frac{1}{r}dr\to
0.$ (2.74)

Getting (2.73) and (2.1) together and handling $I_{3}$ similarly yield that as
$\varepsilon\to 0$,

$\displaystyle I_{2}\to 0,\quad I_{3}\to 0.$ (2.75)

Thus (2.70) can be obtained from (2.71).

Hence combining (2.68), (2.69) and (2.70), letting $\varepsilon\to 0$ in (2.1)
leads to

$\displaystyle\int_{{\mathbb{R}}^{2}}\frac{u(x,t)}{m(t)}\log\frac{u(x,t)}{m(t)}dx-\frac{1}{2m(t)}\int_{{\mathbb{R}}^{2}}u(x,t)c(x,t)dx+\int_{0}^{t}\int_{{\mathbb{R}}^{2}}\frac{u(x,s)}{m(s)}\left|\nabla\log
u(x,s)-\nabla c(x,s)\right|^{2}dxds$ $\displaystyle+$
$\displaystyle\int_{0}^{t}\frac{M_{0}-m(s)}{2m(s)}\int_{{\mathbb{R}}^{2}}u(x,s)c(x,s)dxds\leq\frac{1}{m_{0}}\int_{{\mathbb{R}}^{2}}u_{0}(x)\log
u_{0}(x)dx-\frac{1}{2m_{0}}\int_{{\mathbb{R}}^{2}}u_{0}(x)c_{0}dx-\log m_{0},$

which is the desired inequality (2.3) and thus completes the proof. $\Box$

######  Remark 2.4.

For $M_{0}<m_{0}<8\pi,$ then $M_{0}-m(t)<0$ for all $t>0.$ Therefore, $u(x,t)$
is a sub-solution of the keller-segel system $v_{t}=\Delta
v-\nabla\cdot(v\nabla w),~{}-\Delta w=v$ which admits a global solution when
$\int_{{\mathbb{R}}^{2}}v(x,0)dx<8\pi$. By the comparison principle, all
solutions of (1.4) exist globally.

###  2.2 Finite time blow-up for $M_{0}>8\pi$

In this subsection, we will show the blow-up statement with finite initial
second moment. We start with the case $M_{0}>8\pi$ in which we use the
standard argument relying on the evolution of the second moment of solutions
as done in [3, 12]. By passing to the limit from Steps 1-8 and adapting the
argument (2.1) in Step 9 of Theorem 2.3 without further computation we obtain

######  Lemma 2.5 (The Second Moment).

Assume $\int_{R^{2}}|x|^{2}u_{0}(x)dx<\infty$ and $u_{0}(x)\in L_{+}^{1}\cap
L^{\infty}({\mathbb{R}}^{2})$. Let $u(x,t)$ be the weak solution to system
(1.4), then the the second moment satisfies

$\displaystyle
m_{2}(t):=\int_{R^{2}}|x|^{2}u(x,t)dx=m(t)\left[4t-\frac{1}{2\pi}\ln\left(\frac{m_{0}}{M_{0}}e^{M_{0}t}+\frac{M_{0}-m_{0}}{M_{0}}\right)+\frac{\int_{R^{2}}|x|^{2}u_{0}(x)dx}{m_{0}}\right]$
(2.76)

where
$m(t)=\left(\frac{1}{M_{0}}-\left(\frac{m_{0}-M_{0}}{M_{0}m_{0}}\right)e^{-M_{0}t}\right)^{-1}.$

We notice that the second moment evolution is more complicated than in the
classical system corresponding to mass conservation where the time derivative
of the second moment is a constant. An easy consequence of the previous lemma
is the following blow-up result.

######  Theorem 2.6 (Finite time blow-up for $M_{0}>8\pi$).

Assume $\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx<\infty$. If $M_{0}>8\pi$,
then the solution of (1.4) blows up in finite time and there exists a
$0<T<\infty$ such that

$\displaystyle\displaystyle\limsup_{t\to
T}\|u(\cdot,t)\|_{p}=\infty,~{}~{}\mbox{for any}~{}~{}p>1.$ (2.77)

Proof.  Our proof hinges on the second moment evolution. Firstly for
$M_{0}>8\pi,$ the second moment $\int_{{\mathbb{R}}^{2}}|x|^{2}u(x,t)dx$ can
be estimated that

$\displaystyle\int_{{\mathbb{R}}^{2}}|x|^{2}u(x,t)dx\leq\left\\{\begin{array}[]{ll}\left(4-\frac{M_{0}}{2\pi}\right)M_{0}t+\frac{M_{0}}{m_{0}}\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx-\frac{M_{0}}{2\pi}\ln\frac{m_{0}}{M_{0}}&\mbox{for}~{}m_{0}\leq
M_{0},\\\\[5.69054pt]
\left(4-\frac{M_{0}}{2\pi}\right)m_{0}t+\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx,&\mbox{for}~{}m_{0}>M_{0}.\end{array}\right.$
(2.80)

Consider $M_{0}>8\pi$, it enables us to get that the second moment of the weak
solution will become negative after some time and contradicts the non-
negativity of $u$. Therefore, there is a $T^{*}>0$ such that
$\displaystyle\lim_{t\to T^{*}}m_{2}(t)=0$, and using Hölder’s inequality one
has

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x,t)dx=\int_{|x|\leq
R}u(x,t)dx+\int_{|x|>R}u(x,t)dx\leq
CR^{2(p-1)/p}||u||_{L^{p}}+\frac{1}{R^{2}}m_{2}(t),~{}for~{}all~{}p>1.$ (2.81)

Choosing $R=\left(\frac{Cm_{2}(t)}{||u||_{L^{p}}}\right)^{1/(a+2)}$ with
$a=2(p-1)/p$, we obtain

$\displaystyle||u||_{L^{1}}\leq
C||u||_{L^{p}}^{\frac{2}{a+2}}m_{2}(t)^{\frac{a}{a+2}},$ (2.82)

which induces that

$\displaystyle\limsup_{t\to T^{*}}||u||_{L^{p}}\geq\lim_{t\to
T^{*}}\frac{||u(x,t)||_{L^{1}}^{\frac{a+2}{2}}}{Cm_{2}(t)^{\frac{a}{2}}}=\infty.$
(2.83)

Thus the proof is completed. $\Box$

In addition, we know that the solution exists globally for $m_{0}<M_{0}<8\pi$.
On the contrary, we could obtain the finite time blow-up result for
$M_{0}<8\pi<m_{0}$.

######  Theorem 2.7 (Finite time blow-up for $M_{0}<8\pi<m_{0}$).

For $M_{0}<8\pi<m_{0}$, assume
$\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx<C(M_{0},m_{0})$, where
$C(M_{0},m_{0})$ is given by

$\displaystyle C(M_{0},m_{0}):=-\frac{(8\pi-
M_{0})m_{0}}{2M_{0}\pi}\ln{\frac{m_{0}-M_{0}}{8\pi-
M_{0}}}+\frac{4m_{0}}{M_{0}}\ln{\frac{m_{0}}{8\pi}},$

then the solution of (1.4) blows up in finite time.

Proof.  Denote $h(t)=\frac{1}{m(t)}\int_{R^{2}}|x|^{2}u(x,t)dx$, from (2.76)
we find,

$\displaystyle\frac{dh(t)}{dt}=4-\frac{m(t)}{2\pi},~{}~{}\frac{d^{2}h(t)}{dt^{2}}=-\frac{m(t)\left(M_{0}-m(t)\right)}{2\pi}.$

For $M_{0}<8\pi<m_{0}$, we are able to find that $h(t)$ firstly decreases from
the initial data and reaches the global minimum
$\frac{\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx-C(M_{0},m_{0})}{8\pi}$. Thus
we conclude that $m_{2}(t)$ should become negative in finite time when the
initial second moment is small, namley
$\int_{{\mathbb{R}}^{2}}|x|^{2}u_{0}(x)dx<C(M_{0},m_{0})$. This suggests that
the solution blows up in finite time as proved in Theorem 2.6. Therefore the
blow-up statement for the case $M_{0}<8\pi<m_{0}$ is completed without further
comment. $\Box$

##  3 Infinite time blow-up for $m_{0}<M_{0}=8\pi$

As mentioned earlier, the solution exists globally for $0<m_{0}<M_{0}=8\pi$.
In this section, we will firstly show that nontrivial steady state solutions
can exist only in the case $M_{0}=8\pi$. The infinite second moment of the
steady states could serve to give a hint on the further proof of infinite time
blow-up.

###  3.1 Steady states

This subsection is primarily devoted to the analysis on the steady solutions
of (1.4). Keeping (1.12) in mind we say that $m(t)\to M_{0}$ as $t\to\infty$,
and the stationary equation is followed in the sense of distribution

$\displaystyle\left\\{\begin{array}[]{ll}\Delta
U_{s}(x)-\nabla\cdot(U_{s}(x)\nabla C_{s}(x))=0,~{}~{}x\in{\mathbb{R}}^{2},\\\
-\Delta C_{s}(x)=U_{s}(x),~{}~{}x\in{\mathbb{R}}^{2}\end{array}\right.$ (3.3)

with

$\displaystyle\int_{{\mathbb{R}}^{2}}U_{s}(x)dx=M_{0}.$ (3.4)

Now three equivalent statements for the stationary solutions are shown and one
has the constant chemical potential inside the support of the steady
solutions.

######  Proposition 3.1 (Three equivalent statements for the steady states).

Let $\Omega\in{\mathbb{R}}^{2}$ be a connected open set. Assuming that
$U_{s}\log U_{s}\in L^{1}({\mathbb{R}}^{2})$ and $U_{s}\in
L_{+}^{1}({\mathbb{R}}^{2})$ with $\int_{{\mathbb{R}}^{2}}U_{s}dx=M_{0}$,
$U_{s}\in C(\bar{\Omega})$ and
$U_{s}>0~{}~{}\mbox{in}~{}~{}\Omega,~{}~{}U_{s}=0~{}~{}\mbox{in}~{}~{}{\mathbb{R}}^{2}\setminus\Omega$.
Moreover, if $\Omega$ is unbounded, assume that $U_{s}$ decays at infinity.

Assume also $C_{s}\in C^{2}({\mathbb{R}}^{2})$ is the Newtonian potential
satisfying

$\displaystyle\Delta U_{s}-\nabla\cdot(U_{s}\nabla
C_{s})=0,~{}~{}\mbox{in}~{}~{}{\mathbb{R}}^{2},$ (3.5)
$\displaystyle\mu_{s}=\log U_{s}-C_{s},\mbox{ in }{\mathbb{R}}^{2}$ (3.6)

in the sense of distribution. Then the following three statements are
equivalent:
(i) No dissipation: $\int_{\Omega}U_{s}\big{|}\nabla\mu_{s}\big{|}^{2}dx=0$.
(ii) $U_{s}$ is the minimizer of the total interaction energy
$F(u)=\frac{1}{m(t)}\int_{{\mathbb{R}}^{2}}u\log
udx-\frac{1}{2m(t)}\int_{{\mathbb{R}}^{2}}ucdx-\log m(t)$.
(iii) The chemical potential satisfies

$\displaystyle\mu_{s}(x)=\bar{C},$ $\displaystyle\qquad\forall
x\in\mbox{Supp}(U_{s})$ (3.7)

where $\mbox{Supp}(U_{s})={\mathbb{R}}^{2}$ and

$\displaystyle\bar{C}=\frac{1}{M_{0}}\left(\int_{{\mathbb{R}}^{2}}U_{s}\log
U_{s}-U_{s}c_{s}dx\right).$ (3.8)

Proof.  Firstly we prove (i) $\Leftrightarrow$ (iii): (i) is directly from
(iii). (i)$\Rightarrow$ (iii): Suppose
$\int_{\Omega}U_{s}|\nabla\mu_{s}|^{2}dx=0$. It follows from $U_{s}>0$ at any
point $x_{0}\in\Omega$ that $\nabla\mu_{s}=0$ in a neighborhood of $x_{0}$ and
thus $\mu_{s}$ is a constant in this neighborhood. By the connectedness of
$\Omega$ we deduce that $\mu_{s}\equiv\bar{C}$ in $\Omega$.

In order to prove (ii) $\Leftrightarrow$ (iii), we define the minimizer of
$F(u)$: for any $\varphi\in C_{0}^{\infty}(\Omega),$ let
$\overline{\Omega}_{0}=\mbox{ supp }\varphi$ with
$\int_{\Omega}\varphi(x)dx=0,~{}\Omega_{0}\subset\subset\Omega$. There exists

$\varepsilon_{0}:=\frac{\displaystyle\min_{y\in\overline{\Omega}_{0}}U_{s}(y)}{\displaystyle\max_{y\in\overline{\Omega}_{0}}\left|\varphi(y)\right|}>0,$

such that $U_{s}+\varepsilon\varphi\geq 0$ in $\Omega$ for
$0<\varepsilon<\varepsilon_{0}$. Now $U_{s}$ is a critical point of $F(u)$ in
$\Omega$ if and only if

$\displaystyle\frac{d}{d\varepsilon}\Big{|}_{\varepsilon=0}F\left(U_{s}+\varepsilon\varphi\right)=0,\quad\forall\varphi\in
C_{0}^{\infty}(\Omega).$ (3.9)

The above definition yields that for any $\varphi\in C_{0}^{\infty}(\Omega)$
with $\int_{\Omega}\varphi(x)dx=0$

$\displaystyle\int_{\Omega}\left(\log U_{s}-C_{s}\right)\varphi dx=0.$ (3.10)

For any $\psi\in C_{0}^{\infty}(\Omega),$ denoting

$\displaystyle\varphi=\psi-\frac{U_{s}}{M_{0}}\int_{\Omega}\psi dx,$ (3.11)

(3.10) becomes

$\displaystyle\int_{\Omega}\left(\mu_{s}-\int_{\Omega}\frac{U_{s}}{M_{0}}\mu_{s}dx\right)\psi
dx=0,\quad\mbox{for~{}any}~{}~{}\psi\in C_{0}^{\infty}(\Omega).$ (3.12)

This implies

$\displaystyle\mu_{s}=\log U_{s}-C_{s}=\bar{C},~{}~{}a.e.\mbox{ in }\Omega,$
(3.13)

where

$\displaystyle\bar{C}=\frac{1}{M_{0}}\int_{\Omega}U_{s}\log
U_{s}-U_{s}C_{s}dx.$ (3.14)

In addition, it follows from (2.2) that

$\displaystyle 2\int_{{\mathbb{R}}^{2}}U_{s}\log U_{s}dx+2M_{0}(1+\log\pi-\log
M_{0})\geq\int_{{\mathbb{R}}^{2}}U_{s}C_{s}dx\geq\int_{\Omega}U_{s}C_{s}dx.$
(3.15)

Recalling $U_{s}\log U_{s}\in L^{1}({\mathbb{R}}^{2})$ we can further deduce
that

$\displaystyle\bar{C}\geq\frac{1}{M_{0}}\left(\int_{\Omega}U_{s}\log
U_{s}dx-2\int_{{\mathbb{R}}^{2}}U_{s}\log U_{s}dx-2M_{0}(1+\log\pi-\log
M_{0})\right).$ (3.16)

Now we claim $\Omega={\mathbb{R}}^{2}.$ If $\Omega$ is bounded, then (3.13)
implies $\bar{C}=-\infty$ at the boundary of $\Omega.$ This contradiction with
(3.16) implies $\Omega$ is unbounded and the connected unbounded open set is
${\mathbb{R}}^{2}.$ Hence we complete the proof for (i) $\Leftrightarrow$
(iii) and (ii) $\Leftrightarrow$ (iii). $\Box$

From Proposition 3.1, we can obtain that the stationary equation is

$\displaystyle\left\\{\begin{array}[]{ll}\log
U_{s}-C_{s}=\bar{C},\quad\mbox{in}~{}~{}{\mathbb{R}}^{2},\\\ -\Delta
C_{s}=U_{s},\quad\mbox{in}~{}~{}{\mathbb{R}}^{2}.\end{array}\right.$ (3.19)

Letting $\phi=\log U_{s}$ in (3.19), the steady equation reduces to

$\displaystyle-\Delta\phi=e^{\phi},\quad\mbox{in}~{}~{}{\mathbb{R}}^{2}.$
(3.20)

It has been proved in [5, 16] that all the solutions $\phi(x)\in
C^{2}({\mathbb{R}}^{2})$ of (3.20) uniquely assume the radial form up to
translation

$\displaystyle\phi(x)=\log\frac{32\lambda^{2}}{(4+\lambda^{2}|x|^{2})^{2}},~{}~{}\lambda>0.$
(3.21)

It provides us with the explicit expression of the stationary solutions

$\displaystyle
U_{s}(x)=e^{\phi(x)}=\frac{32\lambda^{2}}{(4+\lambda^{2}|x|^{2})^{2}},~{}~{}\lambda>0,$
(3.22)

with $\int_{{\mathbb{R}}^{2}}U_{s}(x)dx=8\pi.$ One readily check that there
are nontrivial solutions to (3.3) only in the case of a special value of the
total mass. Combining with (3.4) it gives our final estimates.

######  Lemma 3.2.

There are nontrivial steady states to (1.4) only for

$\displaystyle M_{0}=8\pi,$ (3.23)

given by a one-parameter family with $\lambda>0$

$\displaystyle U_{s}(x)=\frac{32\lambda^{2}}{(4+\lambda^{2}|x|^{2})^{2}}$
(3.24)

with infinite second moment $\int_{{\mathbb{R}}^{2}}|x|^{2}U_{s}(x)dx=\infty.$

###  3.2 Infinite time blow-up

For $m_{0}<M_{0}=8\pi,$ combining (1.14) and Theorem 2.3 we deduce that
solutions to system (1.18) exist globally for any $0<t<\infty$ and satisfies

$\displaystyle\int_{{\mathbb{R}}^{2}}\rho(\cdot,t)|\log\rho(\cdot,t)|dx+\int_{{\mathbb{R}}^{2}}\rho(\cdot,t)|x|^{2}dx<\infty.$
(3.25)

In this subsection, we will further show that any solution of (1.4) will
converge towards a delta dirac at the center of mass via system (1.18). The
main tool are the free energy functional

$\displaystyle
F[\rho](t)+\int_{0}^{t}\left(\int_{{\mathbb{R}}^{2}}\rho|\nabla\left(\log\rho-m(s)w\right)|^{2}dx+\frac{m^{\prime}(s)}{2}\int_{{\mathbb{R}}^{2}}\rho
wdx\right)ds\leq F[\rho_{0}].$ (3.26)

with $F[\rho]:=\int_{{\mathbb{R}}^{2}}\rho\log\rho
dx-\frac{m(t)}{2}\int_{{\mathbb{R}}^{2}}\rho wdx$. To prove this result,
applying the mass conservation of $\rho$ we follow a procedure analogous to
Lemma 3.6 and Corollary 3.2 in [2] to obtain

######  Lemma 3.3.

Assume
$\int_{{\mathbb{R}}^{2}}\left(|x|^{2}+\left|\log\rho_{0}(x)\right|\right)\rho_{0}(x)dx<\infty$.
If $\rho_{0}\in L_{+}^{1}\cap L^{\infty}({\mathbb{R}}^{2})$ and
$m_{0}<M_{0}=8\pi$, given any solution $\rho$ to (1.18), we have

$\displaystyle\lim_{t\to\infty}\int_{{\mathbb{R}}^{2}}\rho(x,t)\log\rho(x,t)dx=+\infty$
(3.27)

and

$\displaystyle\lim_{t\to\infty}\rho(x,t)=\delta_{M_{1}}~{}~{}~{}weakly-^{*}~{}as~{}measures.$

where $M_{1}$ is the center of mass in infinite time.

Proof.  Without loss of generality, let $\int_{R^{2}}x\rho(x,t)dx=0$. Now we
assume by contradiction that there exists an increasing sequence of time
$\\{t_{k}\\}\to\infty$ such that

$\displaystyle\int_{{\mathbb{R}}^{2}}\rho(x,t_{k})\log\rho(x,t_{k})dx<\infty.$

Combining with the boundedness of the second moment of $\rho$, we can find a
subsequence of $\rho(x,t_{k})$ (without relabelling for simplicity) that
converges weakly to $\rho_{\infty}(x)$ in $L^{1}({\mathbb{R}}^{2})$ by
Dunford-Pettis theorem. In addition, by $\rho=\frac{u(x,t)}{m(t)}$ and (2.76)
one has that the second moment of the limiting density $\rho_{\infty}(x)$
satisfies

$\displaystyle 0<\int_{{\mathbb{R}}^{2}}|x|^{2}\rho_{\infty}(x)dx<\infty.$
(3.28)

On the other hand, it follows from the definition (1.12) of $m(t)$ and the
logarithmic H-L-S inequality (2.2) that $F(\rho)$ is bounded from below:

$\displaystyle\begin{array}[]{ll}F(\rho)&=\int_{{\mathbb{R}}^{2}}\rho(x,t)\log\rho(x,t)dx-\frac{m(t)}{2}\int_{{\mathbb{R}}^{2}}\rho(x,t)w(x,t)dx\\\
\\\
&\geq\int_{{\mathbb{R}}^{2}}\rho(x,t)\log\rho(x,t)dx-4\pi\int_{{\mathbb{R}}^{2}}\rho(x,t)w(x,t)dx\\\
\\\
&\geq\int_{{\mathbb{R}}^{2}}\rho(x,t)\log\rho(x,t)dx+2\iint_{{\mathbb{R}}^{2}\times{\mathbb{R}}^{2}}\rho(x)\rho(y)\log|x-y|dxdy\\\
\\\ &\geq C,\end{array}$

where the last line holds due to $\int_{{\mathbb{R}}^{2}}\rho(\cdot,t)dx=1$
for any $0<t<\infty$. So from (3.26) we have

$\displaystyle
0\leq\lim_{t\to\infty}\int_{0}^{t}\left(\int_{{\mathbb{R}}^{2}}\rho|\nabla\log\rho-m(s)\nabla
w|^{2}dx+\frac{m^{\prime}(s)}{2}\int_{{\mathbb{R}}^{2}}\rho wdx\right)ds\leq
F[\rho_{0}]-\liminf_{t\to\infty}F[\rho](t).$

As a consequence,

$\displaystyle\lim_{t\to\infty}\int_{t}^{\infty}\left(\int_{{\mathbb{R}}^{2}}\rho|\nabla\log\rho-m(s)\nabla
w|^{2}dx+\frac{m^{\prime}(s)}{2}\int_{{\mathbb{R}}^{2}}\rho wdx\right)ds=0.$

Noting that
$\displaystyle\lim_{t\to\infty}m(t)=8\pi,\displaystyle\lim_{t\to\infty}m^{\prime}(t)=0$,
we have, up to the extraction of subsequences , that the limit
$\varrho_{\infty}(s,x)$ of $(s,x)\mapsto\varrho_{\infty}(t+s,x)$ when t goes
to infinity satisfies

$\displaystyle\nabla\log\varrho_{\infty}-8\pi\nabla
w_{\infty}=0,~{}w_{\infty}=-\frac{1}{2\pi}\log|\cdot|*\varrho_{\infty}.$
(3.29)

Now reviewing Section 3.1 we find that $\varrho_{\infty}$ are the family of
stationary solutions to (1.18) with infinite second momentum by virtue of the
transformation (1.14) and Lemma 3.2. This fact contradicts with (3.28) and
thus the first result holds true.

Moreover, by making use of the mass conservation of $\rho$, a careful reading
of the proof of Lemma 3.1 and Corollary 3.2 in [2] gives the latter desired
result. $\Box$

Applying (1.14) we are able to show our main result of system (1.4) in this
section.

######  Theorem 3.4 (Infinite Time Blow-up).

Assume $\int_{{\mathbb{R}}^{2}}\left(\left|\log
u_{0}(x)\right|+|x|^{2}\right)u_{0}(x)dx<\infty$ and $u_{0}\in L_{+}^{1}\cap
L^{\infty}({\mathbb{R}}^{2})$. If
$\int_{{\mathbb{R}}^{2}}u_{0}(x)dx<M_{0}=8\pi$, the the weak solution $u$ to
(1.4) exists globally in time and converges to a delta dirac of mass $M_{0}$
concentrated at the center of mass as $t\to\infty$.

##  4 Radially symmetric solutions

In this section, we firstly study the radially symmetric solutions of (1.18).
The single equation we derive from (1.18) with radial symmetry satisfies a
comparison principle. We can reach the steady state of the reduced system and
have a new intuition of the relationship among $M_{0},m_{0}$ and $8\pi$
besides the second moment. Subsequently, we divide it into $M_{0}<8\pi$ and
$M_{0}>8\pi$. In these cases, the comparison principle plays an important role
to capture the full structures of the reduced system. Finally, we transfer the
system (1.18) back to the original system (1.4) and summarize the dynamical
behaviors of radial symmetric solutions in different cases.

###  4.1 Reduced system and steady states

Radially symmetric solutions of the system (1.18) on $\rho(t,r),w(t,r)$ is
equivalent to

$\displaystyle\left\\{\begin{array}[]{ll}\frac{\partial}{\partial
t}(r\rho)=(r\rho^{\prime})^{\prime}-m(t)(r\rho w^{\prime})^{\prime},{}{}&t\geq
0,~{}r>0,\\\ -(rw^{\prime})^{\prime}=r\rho,{}{}&t\geq 0,~{}r>0,\\\
\rho^{\prime}(t,r=0)=0,{}{}&t\geq 0,\\\ \rho(t=0,r)=\rho_{0}(r)\geq
0,{}{}&r\geq 0.\end{array}\right.$ (4.5)

This can be reduced to a single equation on $M(t,r)$ which is defined as

$\displaystyle M(t,r)=2\pi\int_{0}^{r}\sigma\rho(t,\sigma)d\sigma=-2\pi
rw^{\prime}(t,r).$ (4.6)

Integrating (4.5) we arrive at

$\displaystyle\left\\{\begin{array}[]{ll}\frac{\partial}{\partial
t}M(t,r)=r\left(\frac{M^{\prime}}{r}\right)^{\prime}+\frac{m(t)}{2\pi
r}M^{\prime}M,{}{}&t\geq 0,~{}r>0,\\\ M(t,0)=0,~{}M(t,\infty)=1,{}{}&t\geq
0,\\\ M(0,r)=2\pi\int_{0}^{r}\sigma\rho_{0}(\sigma)d\sigma,{}{}&r\geq
0.\end{array}\right.$ (4.10)

Next we write for time independent solutions

$\displaystyle\left\\{\begin{array}[]{ll}r^{2}\left(\frac{\bar{M}^{\prime}}{r}\right)^{\prime}+\frac{M_{0}}{2\pi}\bar{M}^{\prime}\bar{M}=r\bar{M}^{\prime\prime}-\bar{M}^{\prime}+\frac{M_{0}}{4\pi}\left(\bar{M}^{2}\right)^{\prime},~{}~{}r>0,\\\\[2.84526pt]
\bar{M}(0)=0,~{}~{}\bar{M}^{\prime}(r)>0.\end{array}\right.$ (4.13)

A simple manipulation leads to

$\displaystyle\bar{M}_{\lambda}(r)=\frac{8\pi}{M_{0}\left(1+\lambda
r^{-2}\right)}\quad\mbox{for some}~{}~{}\lambda>0.$ (4.14)

Coming back to the density we obtain

$\displaystyle\bar{\rho}_{\lambda}(r)=\frac{8\lambda}{M_{0}}\frac{1}{(r^{2}+\lambda)^{2}}.$
(4.15)

It is worthwhile to note that (4.15) is consistent with (3.24) up to a
parameter. Recalling $\bar{M}(\infty)=1$, we need to emphasize that there are
radial steady states to (4.10) only for $8\pi=M_{0}$, given by
$\bar{M}_{\lambda}(r)=\frac{1}{1+\lambda r^{-2}}$.

###  4.2 Refined existence and blow-up

In terms of (4.5) and (4.14), we can use a comparison argument to construct
sub-or super-solutions in the spirit of [12]. The analysis for the case
$M_{0}>8\pi$ and $M_{0}<8\pi$ are quite different in view of the expression of
(4.14). Hence, in the following we shall separate these two cases to discuss.

####  4.2.1 Global existence for $M_{0}<8\pi$

######  Lemma 4.1 (Global solutions).

Let $m_{0}<M_{0}$ and $\bar{M}_{\lambda_{0}}(r)$ is defined as (4.14) for some
$\lambda_{0}>0$. We assume that

$\displaystyle 1<\frac{8\pi}{M_{0}},\quad M(0,r)<\bar{M}_{\lambda_{0}}(r)$
(4.16)

for any $r>0$, then the solution of (4.5) vanishes in
$L^{1}({\mathbb{R}}^{2})$ locally. That’s

$\displaystyle M(t,r)\to 0~{}~{}as~{}~{}t\to\infty$ (4.17)

uniformly in interval $0\leq r<R$ for any $R>0.$

Proof.  Due to $M_{0}<8\pi$ and $M(0,r)<\bar{M}_{\lambda_{0}}(r)$, we may
choose $0<\mu<1$ such that

$\displaystyle\mu\frac{8\pi}{M_{0}}>1~{}~{}~{}\mbox{and}~{}~{}~{}M(0,r)<\mu\bar{M}_{\lambda_{0}}(r),~{}\forall
r>0.$ (4.18)

In view of (1.12) and $m_{0}<M_{0}$, we also have

$\displaystyle m_{0}<m(t)<M_{0}$ (4.19)

for any $0<t<\infty$. Then we look for a supersolution to (4.5) as follows:

$\displaystyle\bar{N}(t,r)=\min\left(1,\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}\right)$
(4.20)

where

$\displaystyle\lambda(t)=\lambda_{0}+\frac{M_{0}}{\mu\pi}\left(1-\mu\right)\left(\mu\frac{8\pi}{M_{0}}-1\right)t.$
(4.21)

Furthermore, we may define $R(t)$ as

$\displaystyle 1=\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)R(t)^{-2}\right)}$
(4.22)

which gives rise to

$\displaystyle\left\\{\begin{array}[]{ll}r\geq R(t),&\bar{N}=1,\\\ r\leq
R(t),&\bar{N}=\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}.\end{array}\right.$
(4.25)

Now we claim that $\bar{N}(t,r)$ is a supersolution to (4.5). Denote the
operator $L$

$\displaystyle L(\bar{N}):=\frac{\partial}{\partial
t}\bar{N}-r\left(\frac{\bar{N}^{\prime}}{r}\right)^{\prime}-\frac{m(t)}{2\pi
r}\bar{N}^{\prime}\bar{N}.$ (4.26)

For $r\leq R(t)$, with the fact (4.19) we derive that

$\displaystyle
L(\bar{N})=L\left(\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}\right)$
$\displaystyle=\bar{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\lambda(t)r^{-2}}{1+\lambda(t)r^{-2}}\left(1-\mu\frac{m(t)}{M_{0}}\right)\right)$
$\displaystyle\geq\bar{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\lambda(t)R(t)^{-2}}{1+\lambda(t)R(t)^{-2}}\left(1-\mu\frac{m(t)}{M_{0}}\right)\right)$
$\displaystyle=\bar{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{M_{0}}{\mu\pi}\left(1-\mu\frac{m(t)}{M_{0}}\right)\left(\mu\frac{8\pi}{M_{0}}-1\right)\right)$
$\displaystyle\geq\bar{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{M_{0}}{\mu\pi}\left(1-\mu\right)\left(\mu\frac{8\pi}{M_{0}}-1\right)\right)$
$\displaystyle=0.$ (4.27)

Here the last line holds by the choice (4.21) of $\lambda(t)$. We also find
that for $r>R(t)$, $L(\bar{N})=L(1)=0$. Hence recalling $M(0,r)\leq
M(0,\infty)=1$ and (4.18) respectively, the comparison principle implies that

$\displaystyle
M(t,r)\leq\min\left(1,\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}\right)=\bar{N}(t,r)$
(4.28)

for all $0\leq t<\infty$ and $0\leq r<\infty.$ Moreover, by (4.21) we find

$\displaystyle\lambda(t)\to\infty,$ (4.29)

then $R(t)\to\infty$ for $t$ large enough. Therefore, for any given interval
$r\in(0,R)$ we have

$\displaystyle
M(t,r)\leq\mu\frac{8\pi}{M_{0}\left(1+\lambda(t)R^{-2}\right)}\to
0~{}~{}\mbox{as}~{}~{}t\to\infty.$ (4.30)

Thus completes the proof. $\Box$

####  4.2.2 Infinite time blow-up for $M_{0}>8\pi$

Next we prove a refined blow-up result (with a weaker condition than the
second moment condition) which is close enough to exhibit the shape of the
chemotactic collapse solution.

######  Lemma 4.2 (Blow-up).

Assume $M_{0}<m_{0}$ and $\bar{M}_{\lambda_{0}}(r)$ is defined as (4.14) for
some $\lambda_{0}>0$. If

$\displaystyle\frac{8\pi}{M_{0}}<1,\quad M(0,r)>\bar{M}_{\lambda_{0}}(r)$
(4.31)

for any $r>0$ and there is no finite time blow-up, then the solution to (4.5)
blows up in infinite time. Furthermore, the solution has a mass concentration
at the origin, that’s

$\displaystyle M(t,r(t))>\frac{8\pi}{M_{0}}$ (4.32)

for $r(t)\to 0$ as $t\to\infty$.

Proof.  Since $\frac{8\pi}{M_{0}}<1$, We firstly choose $1<\mu_{0}<\mu_{1}$
such that

$\displaystyle\mu_{0}\frac{8\pi}{M_{0}}<\mu_{1}\frac{8\pi}{M_{0}}<1~{}~{}~{}\mbox{and}~{}~{}~{}M(0,r)>\mu_{1}\bar{M}_{\lambda_{0}}(r)>\mu_{0}\bar{M}_{\lambda_{0}}(r),~{}\forall
r>0.$ (4.33)

Combining $M_{0}<m_{0}$ with (1.12) also leads to

$\displaystyle M_{0}<m(t)<m_{0}.$ (4.34)

We consider the function

$\displaystyle\underline{N}(t,r)=\max\left(\mu_{1}\frac{8\pi}{M_{0}\left(1+\lambda_{0}r^{-2}\right)},\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}\right),$
(4.35)

where $\lambda(t)=\lambda_{0}e^{At}$ for some $A<0,$ we will prove that
$\underline{N}$ is a subsolution as it is the maximum of two subsolutions.

Firstly we denote
$N_{1}:=\mu_{1}\frac{8\pi}{M_{0}\left(1+\lambda_{0}r^{-2}\right)}$ and compute

$\displaystyle\frac{\partial}{\partial
t}N_{1}-r\left(\frac{N_{1}^{\prime}}{r}\right)^{\prime}-\frac{m(t)}{2\pi
r}N_{1}^{\prime}N_{1}$
$\displaystyle=N_{1}\frac{r^{-2}}{1+\lambda_{0}r^{-2}}\frac{8\lambda_{0}}{\lambda_{0}+r^{2}}\left(1-\mu_{1}\frac{m(t)}{M_{0}}\right)$
$\displaystyle\leq
N_{1}\frac{r^{-2}}{1+\lambda_{0}r^{-2}}\frac{8\lambda_{0}}{\lambda_{0}+r^{2}}\left(1-\mu_{1}\right)$
$\displaystyle<0.$ (4.36)

Then it follows from (4.33) that $N_{1}$ is a subsolution to (4.5). Secondly,
we have to prove that
$\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}$ is a subsolution in
the interval $0<r\leq R(t)$ where it achieves the maximum, i.e.

$\displaystyle\mu_{1}\frac{8\pi}{M_{0}\left(1+\lambda_{0}R(t)^{-2}\right)}=\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda(t)R(t)^{-2}\right)}.$
(4.37)

By the definition of $R(t)$, there exists a constant $R_{0}$ such that $r\leq
R(t)\leq R_{0}$ with
$\mu_{1}\frac{8\pi}{M_{0}\left(1+\lambda_{0}R_{0}^{-2}\right)}=\mu_{0}\frac{8\pi}{M_{0}},$
then using (4.34) and a direct calculation provide

$\displaystyle\frac{\partial}{\partial
t}\underline{N}-r\left(\frac{\underline{N}^{\prime}}{r}\right)^{\prime}-\frac{m(t)}{2\pi
r}\underline{N}^{\prime}\underline{N}$
$\displaystyle=\underline{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\lambda(t)}{\lambda(t)+r^{2}}\left(1-\mu_{0}\frac{m(t)}{M_{0}}\right)\right)$
$\displaystyle\leq\underline{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\lambda(t)}{\lambda(t)+r^{2}}\left(1-\mu_{0}\right)\right)$
$\displaystyle\leq\underline{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\lambda(t)R_{0}^{-2}}{\lambda_{0}R_{0}^{-2}+1}\left(1-\mu_{0}\right)\right)$
$\displaystyle=\underline{N}\frac{r^{-2}}{1+\lambda(t)r^{-2}}\left(-\lambda^{\prime}(t)+\frac{8\mu_{0}R_{0}^{-2}\lambda(t)}{\mu_{1}}\left(1-\mu_{0}\right)\right)$
$\displaystyle=0$ (4.38)

by choosing

$\displaystyle\lambda(t)=\lambda_{0}e^{-\frac{8\mu_{0}R_{0}^{-2}}{\mu_{1}}(\mu_{0}-1)t}=\lambda_{0}e^{At}.$
(4.39)

We notice that
$\underline{N}=\mu_{1}\frac{8\pi}{M_{0}\left(1+\lambda_{0}r^{-2}\right)}$ for
$r>R(t)$. Now keeping (4.33) in mind, the comparison principle gives

$\displaystyle
M(t,r)\geq\underline{N}(t,r)\geq\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda(t)r^{-2}\right)}=\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda_{0}e^{At}r^{-2}\right)}.$
(4.40)

Therefore, at $r=e^{\frac{A}{3}t}$ we find

$\displaystyle
M(t,r)\geq\mu_{0}\frac{8\pi}{M_{0}\left(1+\lambda_{0}e^{\frac{A}{3}t}\right)}$
(4.41)

and the subsequent mass concentration at $r=0$

$\displaystyle M(t,r(t))>\frac{8\pi}{M_{0}}$ (4.42)

for $r(t)\to 0$ as $t\to\infty.$ This is our desired estimate. $\Box$

###  4.3 Main results for the radially symmetric solutions of (1.4)

Now we are ready to summarize the main results of (4.5) as follows.

######  Theorem 4.3.

Assume that the initial data $\rho(0,r)$ is radially symmetric,
$\bar{\rho}_{\lambda_{0}}(r)$ is defined as (4.15) for some $\lambda_{0}>0.$

  * (1)

If $M_{0}<8\pi$ and

$\displaystyle\rho(0,r)<\bar{\rho}_{\lambda_{0}}(r)$ (4.43)

for any $r>0,$ then the radially symmetric solution of (4.5) vanishes in
$L^{1}({\mathbb{R}}^{2})$ locally as $t\to\infty.$

  * (2)

If $M_{0}>8\pi$ and

$\displaystyle\rho(0,r)>\bar{\rho}_{\lambda_{0}}(r)$ (4.44)

for any $r>0,$ then there could exist one solution to (4.5) which blows up in
finite time or has a mass concentration at the origin such that

$\displaystyle 2\pi\int_{0}^{r(t)}\rho(t,\sigma)\sigma
d\sigma>\frac{8\pi}{M_{0}}$ (4.45)

for $r(t)\to 0$ as $t\to\infty.$

By using (1.14), we transfer Theorem 4.3 to the original system (1.4).

######  Theorem 4.4.

Assume that $u_{0}(r)$ is the radially symmetric initial data to (1.4).

  * (1)

If $m_{0}<M_{0}<8\pi$ and

$\displaystyle
u_{0}(r)<\frac{8m_{0}}{M_{0}}\frac{\lambda_{0}}{\left(\lambda_{0}+r^{2}\right)^{2}},~{}~{}\forall
r>0$ (4.46)

for some $\lambda_{0}>0,$ then any radially symmetric solution of (1.4)
vanishes in $L^{1}({\mathbb{R}}^{2})$ locally as $t\to\infty.$

  * (2)

If $m_{0}>M_{0}>8\pi$ and

$\displaystyle
u_{0}(r)>\frac{8m_{0}}{M_{0}}\frac{\lambda_{0}}{\left(\lambda_{0}+r^{2}\right)^{2}},~{}~{}\forall
r>0$ (4.47)

for some $\lambda_{0}>0,$ then any radially symmetric solution of (1.4) has
finite time blow-up or has a mass concentration at the origin such that

$\displaystyle 2\pi\int_{0}^{r(t)}u(t,\sigma)\sigma d\sigma>8\pi$ (4.48)

for $r(t)\to 0$ as $t\to\infty.$

we comment that, in the blow-up result, the limitation on the second moment
are in fact useless in the radially symmetric case and they can be relaxed by
the comparison condition with the steady state (4.14) which corresponds to an
infinite second moment. In contrast to Section 2.2, infinite initial second
moment can produce infinite time blow-up.

## Appendix

######  Lemma 4.5.

Let $u_{\varepsilon},u\in L_{+}^{1}({\mathbb{R}}^{2})$ satisfy

$\displaystyle\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}dx<\infty,\quad\int_{{\mathbb{R}}^{2}}|x|^{2}udx<\infty,$
(4.49) $\displaystyle\int_{{\mathbb{R}}^{2}}u_{\varepsilon}|\log
u_{\varepsilon}|dx<\infty,$ (4.50) $\displaystyle
u_{\varepsilon}\rightharpoonup u~{}~{}(\varepsilon\to
0)~{}~{}\mbox{weakly~{}~{}in~{}~{}}L^{1}({\mathbb{R}}^{2}).$ (4.51)

Then

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x)|\log u(x)|dx<\infty,$ (4.52)

and

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x)\log
u(x)dx\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x)\log u_{\varepsilon}(x)dx.$ (4.53)

Proof.  In the following, $C,C_{1},C_{2}$ stand for finite positive constants
that are independent of $\varepsilon$. We will give the proof step by step.

  * (i)

We firstly show that

$\displaystyle\displaystyle\lim_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}(1+|x|^{\beta})u_{\varepsilon}(x)dx=\int_{{\mathbb{R}}^{2}}(1+|x|^{\beta})u(x)dx,\quad\mbox{for~{}~{}any}~{}~{}0<\beta<2.$
(4.54)

In fact, for any $R>1,$

$\displaystyle\left|\int_{{\mathbb{R}}^{2}}(1+|x|^{\beta})(u_{\varepsilon}(x)-u(x))dx\right|$
$\displaystyle\leq$
$\displaystyle\left|\int_{{\mathbb{R}}^{2}}(1+|x|^{\beta})1_{\\{|x|\leq
R\\}}(u_{\varepsilon}(x)-u(x))dx+\int_{|x|>R}(1+|x|^{\beta})u_{\varepsilon}(x)dx+\int_{|x|>R}(1+|x|^{\beta})u(x)dx\right|.$

We deal with the last two terms as follows.

$\displaystyle\int_{|x|>R}(1+|x|^{\beta})u_{\varepsilon}(x)dx+\int_{|x|>R}(1+|x|^{\beta})u(x)dx$
$\displaystyle\leq$ $\displaystyle
2\left(\int_{|x|>R}|x|^{\beta}u_{\varepsilon}(x)dx+\int_{|x|>R}|x|^{\beta}u(x)dx\right)$
$\displaystyle\leq$
$\displaystyle\frac{2}{R^{2-\beta}}\left(\int_{{\mathbb{R}}^{2}}|x|^{2}u_{\varepsilon}(x)dx+\int_{{\mathbb{R}}^{2}}|x|^{2}u(x)dx\right)\leq\frac{C}{R^{2-\beta}}.$

Since, by weak convergence,

$\displaystyle\left|\int_{{\mathbb{R}}^{2}}(1+|x|^{\beta})1_{\\{|x|\leq
R\\}}(u_{\varepsilon}(x)-u(x))dx\right|\to 0,\quad\varepsilon\to 0.$

It follows that

$\displaystyle\displaystyle\limsup_{\varepsilon\to
0}\left|(1+|x|^{\beta})(u_{\varepsilon}(x)-u(x))\right|\leq\frac{C}{R^{2-\beta}},\quad\forall
R>1.$

Letting $R\to\infty$ leads to (4.54).

  * (ii)

Let’s prove

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x)|\log u(x)|dx<\infty.$ (4.55)

Actually, by convexity of $y\mapsto y\log y,y\geq 0,$ one has

$\displaystyle u_{\varepsilon}\log u_{\varepsilon}\geq u\log u+(1+\log
u)(u_{\varepsilon}-u).$

From this we have, for any $m\in\mathbb{N}$,

$\displaystyle 1_{\\{1\leq u(x)\leq m\\}}u_{\varepsilon}\log u_{\varepsilon}$
$\displaystyle\geq$ $\displaystyle 1_{\\{1\leq u(x)\leq m\\}}u\log
u+1_{\\{1\leq u(x)\leq m\\}}(1+\log u)(u_{\varepsilon}-u).$

Hence

$\displaystyle\int_{{\mathbb{R}}^{2}}1_{\\{1\leq u(x)\leq m\\}}u\log
udx+\int_{{\mathbb{R}}^{2}}1_{\\{1\leq u(x)\leq m\\}}(1+\log
u)(u_{\varepsilon}-u)dx$ $\displaystyle\leq\int_{{\mathbb{R}}^{2}}1_{\\{1\leq
u(x)\leq m\\}}u_{\varepsilon}\log u_{\varepsilon}dx$
$\displaystyle\leq\displaystyle\sup_{\varepsilon>0}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}|\log
u_{\varepsilon}|dx=:C_{1}(<\infty).$

Since, by weak convergence,

$\displaystyle\displaystyle\lim_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}1_{\\{1\leq u(x)\leq m\\}}(1+\log
u)(u_{\varepsilon}-u)dx=0,$

it follows by passing to the limit $\varepsilon\to 0$ that

$\displaystyle\int_{{\mathbb{R}}^{2}}1_{\\{1\leq u(x)\leq m\\}}u\log udx\leq
C_{1}.$

Then letting $m\to\infty$ gives by Levi monotone convergence

$\displaystyle\int_{{\mathbb{R}}^{2}}1_{\\{u(x)\geq 1\\}}u\log udx\leq C_{1}.$

This together with (2.33) and the boundedness of
$\int_{{\mathbb{R}}^{2}}|x|^{2}u(x)dx$ yields

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x)|\log u(x)|dx$
$\displaystyle=\int_{{\mathbb{R}}^{2}}1_{\\{u(x)<1\\}}u(x)|\log
u(x)|dx+\int_{{\mathbb{R}}^{2}}1_{\\{u(x)\geq 1\\}}u(x)|\log u(x)|dx$
$\displaystyle\leq\int_{{\mathbb{R}}^{2}}e^{-|x|^{2}}dx+2\int_{{\mathbb{R}}^{2}}u(x)|x|^{2}dx+C_{1}<\infty.$

Hence we close the proof of (4.55).

  * (iii)

Next, to prove (4.53), we consider the truncation

$\displaystyle g_{m}(x)=u(x)\wedge m+\frac{1}{m}e^{-|x|}.$

By convexity we have

$\displaystyle u_{\varepsilon}\log u_{\varepsilon}\geq g_{m}(x)\log
g_{m}(x)+(1+\log g_{m}(x))(u_{\varepsilon}-g_{m}),$

i.e.

$\displaystyle g_{m}(x)\log g_{m}(x)\leq u_{\varepsilon}\log
u_{\varepsilon}+(1+\log g_{m}(x))(g_{m}-u_{\varepsilon}).$ (4.56)

We firstly use (4.55) to check the condition of $g_{m}\log g_{m}$ for using
Lebesgue dominated convergence, i.e.

$\displaystyle\int_{{\mathbb{R}}^{2}}g_{m}|\log g_{m}|dx\leq C_{2}.$ (4.57)

By convexity of $y\mapsto y\log y$ on $y\geq 0$ one has

$\displaystyle(a+b)\log(a+b)$
$\displaystyle=\left(\frac{1}{2}2a+\frac{1}{2}2b\right)\log\left(\frac{1}{2}2a+\frac{1}{2}2b\right)$
$\displaystyle\leq\frac{1}{2}2a\log 2a+\frac{1}{2}2b\log 2b,~{}~{}a,b\geq 0.$
(4.58)

We also have

$\displaystyle(a+b)\log\frac{1}{a+b}\leq
a\log\frac{1}{a}+b\log\frac{1}{b},~{}~{}a,b\geq 0.$ (4.59)

Now we use (Appendix) to see that if $g_{m}\geq 1,$ then, because $m\geq 4,$
we must have $u(x)>1/2,$ so that

$\displaystyle g_{m}(x)|\log g_{m}(x)|=g_{m}(x)\log g_{m}(x)$
$\displaystyle\leq$ $\displaystyle(u(x)\wedge m)\log(2u(x)\wedge
m)+\frac{1}{m}e^{-|x|}\log\left(\frac{2}{m}e^{-|x|}\right)$
$\displaystyle\leq$ $\displaystyle
u(x)\log(2u(x))+\frac{1}{m}e^{-|x|}\frac{2}{m}e^{-|x|}$ $\displaystyle\leq$
$\displaystyle u(x)+u(x)|\log u(x)|+e^{-2|x|}.$

While if $g_{m}(x)<1,$ then $u(x)<1,$ then by (4.59) we have

$\displaystyle g_{m}(x)|\log
g_{m}(x)|=\left(u(x)+\frac{1}{m}e^{-|x|}\right)\log\left(\frac{1}{u(x)+\frac{1}{m}e^{-|x|}}\right)$
$\displaystyle\leq$ $\displaystyle u(x)|\log
u(x)|+\frac{1}{m}e^{-|x|}\log(me^{|x|})$ $\displaystyle\leq$ $\displaystyle
u(x)|\log u(x)|+\frac{2}{\sqrt{m}}e^{-|x|/2},$

where we used $\log y=2\log\sqrt{y}\leq 2\sqrt{y}$ for $y\geq 0.$ So we have
proved an $L^{1}$ control

$\displaystyle g_{m}(x)|\log g_{m}(x)|\leq u(x)+u(x)|\log
u(x)|+e^{-|x|/2}+e^{-2|x|}.$ (4.60)

The aim next is to deal with the second term in the right hand side of (4.56).
Using

$\displaystyle u(x)\wedge m=u(x)-(u(x)-m)_{+},$

we handle

$\displaystyle(1+\log g_{m}(x))(g_{m}(x)-u_{\varepsilon}(x))$ $\displaystyle=$
$\displaystyle(1+\log g_{m}(x))(u(x)-u_{\varepsilon}(x))-(1+\log
g_{m}(x))(u(x)-m)_{+}+(1+\log g_{m}(x))\frac{1}{m}e^{-|x|}.$ (4.61)

By considering $g_{m}(x)\geq 1$ and $g_{m}(x)<1$ we have

$\displaystyle|1+\log g_{m}(x)|\leq 1+\left(\log(m+1)\vee(|x|+\log
m)\right)\leq 2(1+|x|)\log(m+1).$

Therefore we have

$\displaystyle|1+\log g_{m}(x)|\frac{1}{m}e^{-|x|}\leq
2\frac{\log(m+1)}{m}(1+|x|)e^{-|x|},$ (4.62)

and by (4.54),

$\displaystyle\displaystyle\lim_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}(1+\log g_{m}(x))(u(x)-u_{\varepsilon}(x))=0.$ (4.63)

Moreover we have

$\displaystyle-(1+\log g_{m}(x))(u(x)-m)_{+}\leq 0.$ (4.64)

In fact the case $u(x)\leq m$ is obvious. For the case $u(x)>m$ we have
$g_{m}(x)>m>1,$ so that $\log g_{m}(x)>0.$ So (4.64) holds true.

From (4.56), collecting (Appendix)-(4.64) we have by taking lower limit
$\varepsilon\to 0$ that

$\displaystyle\int_{{\mathbb{R}}^{2}}g_{m}(x)\log
g_{m}(x)dx\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x)\log
u_{\varepsilon}(x)dx+2\frac{\log(m+1)}{m}\int_{{\mathbb{R}}^{2}}(1+|x|)e^{-|x|}dx,\quad\forall
m\geq 4.$ (4.65)

Finally by continuity one has

$\displaystyle g_{m}(x)\log g_{m}(x)\to u(x)\log u(x)\quad(m\to\infty).$
(4.66)

Then it follows (4.60) and Lebesgue dominated convergence and (4.65) that

$\displaystyle\int_{{\mathbb{R}}^{2}}u(x)\log
u(x)dx=\displaystyle\lim_{m\to\infty}\int_{{\mathbb{R}}^{2}}g_{m}(x)\log
g_{m}(x)dx\leq\displaystyle\liminf_{\varepsilon\to
0}\int_{{\mathbb{R}}^{2}}u_{\varepsilon}(x)\log u_{\varepsilon}(x)dx.$ (4.67)

Thus we close the proof. $\Box$

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