数理解析研究所講究録 1416 巻 2005 年 30-63

30Blowup in infinite time in the simplified
system of chemotaxis
HIROSHI OHTSUKA
Natural Science Division, Kisarazu National College of Technology

TAKASI SENBA
Department of Applied Mathematics, Faculty of Technology, University of Miyazaki

TAKASHI SUZUKI
Division of Mathematical Science, Department of System Innovation,
Graduate School of Engineering Science, Osaka University

November 19, 2004

1Introduction
The purpose of the present paper is to study the parabolic-elliptic system ofchemotaxis,$u_{t}=\nabla(\nabla u-u\nabla v)$$v(\cdot, t)$$=(G*u)(\cdot, t)$in$u(\cdot, 0)=u0$$\Omega$$\Omega\cross(0, T)$$\Omega\cross(0, T)$on$\frac{\partial u}{\partial\nu}-u\frac{\partial v}{\partial\iota/}=0$

in
in$\partial\Omega\cross$$(0,T)$(1),

where is abounded domain inwith smooth boundaryis anon-negative smooth function defined on , and$\Omega$$\mathrm{R}^{2}$$\partial\Omega$, $u_{0}=u_{0}(x)$$\overline{.\mathrm{Q}}$$(G*u)(x, t)= \int_{\Omega}G(x, x’)u(x’,t)dx’$ ,with $G=G(x, \mathrm{x}’)$ standing for the Green’s function of asecond order linear
elliptic boundary value problem. This system is proposed to describe the31chemotactic aggregation of cellular slime molds $[16, 24]$ , the motion of themean field of many self-gravitating particles $[2, 34]$ , and that of moleculars
under the chemical reaction [11]. Existence of the solution globally in time,
particularly in the context of the threshold of the total mass
, hasbeen studied by several authors [15, 21, 22, 4, 12], while its counter part,the blowup of the solution in finite time, is summarized as the formation ofcollapses with the quantized mass [33].
The asymptotic behavior of the solution globally in time, on the other
hand, has not been clarified so satisfactorily, in spite of several suggestions
obtained ffom the study of stationary solutions [27]. Its counter part is theclassification of the solution blowing-up in infinite time, and [30] conjectured
that this is the case only when the total mass
is so quantized asor times integer, according to the profile of $G(x, x’)$ on the boundary
In more details, each solution, existing globally in time, will converge to aregular stationary solution if is disquantized, while the convergence to a
singular limit of the stationary solution
occur in the other case. This
paper continues the study, and shows, among other things, that if the freeenergy, defined below, is bounded and the total mass is disquantized, thenthe collapses formed in infinite time vanishes almost every moment. Thissuggests that the blowup in infinite time does not occur in this case; thedisquantized total mass and bounded free energy.
To describe the results proven in this paper precisely, we refer to several
fundamental facts on (1). See [30, 29, 32, 33] for the proof of them. First,(1) is written as$\lambda=||u_{0}||_{1}$$\lambda=||u_{0}||_{1}$$8\pi$$4\pi$$\lambda$$\mathrm{v}^{\gamma}\mathrm{i}11$in$u_{t}=\triangle u-f(u)$$\frac{\partial u}{\partial\nu}=g(u)$$u(\cdot, 0)=u_{0}$onin$\Omega\cross(0, T)$$\partial\Omega\cross(0, T)$$\Omega$for$f(u)$ $=\nabla u\cdot\nabla G*u+u\triangle(G*u)$$g(u)=u \frac{\partial G*u}{\partial\nu}|_{\partial\Omega}$

and the elliptic regularity of $G(x, x’)$ combined with the standard fixed point
argument [17] guarantees the unique existence of the solution $u=u(x, t)\in$
with $T>0$ estimated from below by, where$C^{2+\theta,1+\theta/2}(\overline{\Omega}\cross[0, T])$$||u_{0}||_{C^{2+\theta}(\overline{\Omega})}$32
is denoted by, and henceforth the supremum of its existence$T_{\max}\in(0, +\infty]$ . This solution is non-negative, and preserves the total mass;$0<\theta<1$$\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}$$\int_{\Omega}u(x,t)dx=\int_{\Omega}u\mathrm{o}(x)dx(=\lambda)$

Second, the free energy, denoted by $T$
and it holds that$=\mathcal{F}(u)$.

$(^{\underline{q}})$, acts as a Lyapunov function,$\frac{dF}{dt}+\int_{\Omega}u|\nabla(\log u-v)|^{2}dx$ $=0$

(3)

,

where
$F(u)= \int_{\Omega}u(\log u-1)dx-\frac{1}{9_{\sim}}\int\int_{\Omega \mathrm{x}\Omega}G(x, x’)u(x)u(x’)dxdx’$

In the stationary state, in particular, we have logu-v, and therefore, it follows thatby$\frac{\lambda e^{v}}{\int_{\Omega}e^{v}dx}$$=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$, or $u=$$||u||_{1}=\lambda$

and$G*u=v$$u= \frac{\lambda e^{v}}{\int_{\Omega}e^{v}dx}$(4).

Henceforth, we consider the case that $G(x, x’)$ is the Green’s function to
one of the following elliptic problems;$-\triangle v+v=u$in$\Omega$,$\frac{\partial v}{\partial\nu}=0$$- \triangle v=u-\frac{1}{|\Omega|}\int_{\Omega}udx$$-\triangle v=u$in$\Omega$,in$\Omega$

$v=0$

,

on

on

$\partial\Omega$$\frac{\partial v}{\partial\nu}=0$$\partial\Omega$on$\partial\Omega$,$\int_{\Omega}vdx$

$=0$

.

These problems are referred to as the (N), (JL), and (D) fields, respectively.
Then, considering$V=H^{1}(\Omega)$$V= \{v\in H^{1}(\Omega)|\int_{\Omega}vdx$$V=H_{0}^{1}(\Omega)$,$=0\}$

33
provided with the norms$||v||_{V}=(||\nabla v||_{2}^{2}+||v||_{2}^{2})^{1/2}$$||v||_{V}=||\nabla v||_{2}$$||v||_{V}=||\nabla v||_{2}$we obtain the,morphism$\mathrm{i}\mathrm{s}\mathrm{o}$$u\in V’$$\mapsto$,$v= \int_{\Omega}G(\cdot, x’)u(x’)dx’\in V$

and also the Lagrange functional,$\mathcal{W}(u, v)=\int_{\Omega}u(\log u-1)dx+\frac{1}{2}||v||_{V}^{2}-\int_{\Omega}$

defined for$(u, v)\in M_{\lambda}\cross V$

uvdx

(5)

, where $If_{\lambda}=\{u\geq 0|||u||_{1}=\lambda\}$ . This functional

satisfies$\mathcal{W}(u, v)|_{v=G*\mathrm{u}}=F(u)$for$(u, v)\in\Lambda I_{\lambda}\cross V$

and$\mathcal{W}(u, v)|_{u=\lambda e^{v}/\int_{\Omega}e^{v}dx}=J_{\lambda}(v)$, where,$J_{\lambda}(v)= \frac{1}{2}||v||_{V}^{2}-\lambda\log(\int_{\Omega}e^{v}dx)+\lambda\log\lambda-\lambda$

and , respectively, provide equivalentand both $F$ anddefined on
variational structures to the stationary problem (4).
, thenMore precisely, if
is a critical point of $F$ defined on,conversely,
if
is a criticalis a critical point ofonanddefinedpoint ofis a crititical point of $F$
defined on , then. Henceforth,defined on Ma, and in both cases it holds that
denotes the set of stationary solutions of , i.e.,$M_{\lambda}$$J_{\lambda}$$V$$\Lambda l_{\lambda}$$u_{\infty}$$G*u_{\infty}$$J_{\lambda}$$V$$J_{\lambda}$$V$$v_{\infty}$$u_{\infty}= \lambda e^{v_{\infty}}/\int_{\Omega}e^{v_{\infty}}$$F(u_{\infty})=I_{\lambda}(v_{\infty})$$E_{\lambda}$$v$$E_{\lambda}=\{v\in V|v=G*u$ ,$=\{v\in V|\delta J_{\lambda}(v)=0\}$

where$\lambda=||u_{0}||_{1}$.

,$u= \lambda\frac{e^{v}}{\int_{\Omega}e^{v}dx}\}$$v_{\infty}=$

34
As we mentioned, in the case of $T_{\max}<+\infty$ , there is a formation of collapses with the quantized mass [33]. More precisely, if $G(x,x’)$ is associatedwith the (N) or (JL) field, then it holds that$u(x,t)dx$ $arrow\sum_{x_{0}\in \mathrm{S}}m_{*}(x_{0})\delta_{x_{0}}(dx)+f(x)dx$

as

$t$

$\uparrow T_{\max}$$S$ $=${, where A4$x_{0}\in\overline{\Omega}|$$(\overline{\Omega})=C(\overline{\Omega})’$there exists (denotes the blowup set,$x_{k}$,$*$-weakly in$A\Lambda(\overline{\Omega})$denotes the set of measures on$t_{k})arrow(x_{0},$ $T_{\max})$

such that$0\leq f=f(x)\in L^{1}(\Omega)\cap C(\overline{\Omega}\backslash S)$$8\pi$$(x_{0}\in\Omega)$$4\pi$$(x_{0}\in\partial\Omega)$$m_{*}(x\mathrm{o})=\{$$\overline{\Omega}$$u(x_{k},$ $t_{k})$(6),$arrow+\infty$

}

, and

.

Since the total mass is preserved (-), this implies the finiteness of the blowup
points, more precisely,$\underline{\prime y}$$\#(\Omega\cap S)+\#(\partial\Omega\cap S)\leq||u_{0}||_{1}/(4\pi)$.

A similar fact is proven for the case of $T_{\max}=+\infty([30])$ , that is, in thesuch that
admits(N) or (JL) field, any$t_{k}arrow+\infty$$\{t_{k}’\}\subset\{t_{k}\}$$*- \mathfrak{n}^{\gamma}\mathrm{e}\mathrm{a}\mathrm{k}1\mathrm{y}$$u(x, t_{k\prime}’)dx arrow\sum_{x_{0}\in S’}m_{*}(x_{0})\delta_{x_{0}}(dx)+f(x)dx$

where $S’$ denotes the set of ”exhausted)’ blowup points of
$S’=$

{

$x_{0}\in\overline{\Omega}|$

there exists$xlkarrow x_{0}$$\mathrm{s}\mathrm{u}$ch that ($u$$x_{k}’$

in$\Lambda t(\overline{\Omega})$$\{u(\cdot, t_{k}’)\}$

,

(7)

:

, $t_{k}’)arrow+\infty$}

Our conjecture on the blowup in infinite time, therefore, is proven in theinaffirmative in the (N) or (JL) field, if we can deduce $f=0$ from(7), because the total mass of the solution is preserved as (2) and hence it
follows that$S’\neq\emptyset$$\lambda=\sum_{x_{0}\in \mathrm{S}’}m_{*}(x_{0})+||f||_{1}$from (7). More precisely, if we can show $f=0$ by
and$. \lim_{karrow\infty}||u(\cdot, t_{k})||_{\infty}=+\infty$

with$S’\neq\emptyset$, then $T_{\max}=+\infty$$t_{k}arrow+\infty$

35, $m_{*}(x_{0})\in 4\pi N$.
is admitted only when
Taking this approach to the problem, we use the weak solution generated. This fact on the generation of the weak solution is provenduringfor the problem on the flat torus [31], and also for system (1) under the (N)or (JL) field [33].
In more details, any
admitssuch that$\lambda=||u_{0}||_{1}=\sum_{x_{0}\in \mathrm{S}}$$t_{k}’arrow+\infty$$t_{k}arrow+\infty$$\{t_{k}’\}\subset\{t_{k}\}$in$u(x, t_{k}’+t)dxarrow\mu(dx,t)$$C_{*}(-\infty, +\infty;\mathcal{A}4 (\overline{\Omega}))$,(8)where $\mu=\mu(dx, t)$ is a weak solution to (1). This means$\int_{\Omega}\varphi(x)u(x, t_{k}’+t)dxarrow\langle\varphi, \mu(dx, t)\rangle_{C(\Pi),\mathcal{M}(\overline{\Omega})}$locally uniformly in $t\in(-\infty, +\infty)$ for each$\varphi\in C(\overline{\Omega})$on$X= \{\xi\in C^{2}(\overline{\Omega})|\frac{\partial\xi}{\partial\nu}=0$, and if$\partial\Omega\}$$\beta\xi(x, x’)=\nabla\xi(x)|$ $\nabla_{x}G(x, x’)+\nabla\xi(x’)$. $\nabla_{x’}G(x, x’)$$\mathcal{E}_{0}=\{\rho_{\eta}|\eta \in X\}$$\mathcal{E}=\mathcal{E}_{0}\oplus C(\overline{\Omega}\cross\overline{\Omega})\subset L^{\infty}(\Omega\cross \Omega)$then there is $0\leq\nu$ $=\nu(t)$ belonging tothat,$L_{*}^{\infty}(-T, T;\mathcal{E}’)$$\nu(t)|_{C(\overline{\Omega}\mathrm{x}\overline{\Omega})}=\mu\otimes\mu(dxdx’, t)$$\mathrm{a}.\mathrm{e}$for any $T>0$ such

. $t\in(-\infty, +\infty)$ .

Furthermore, the mapping$t\in(-\infty, +\infty)\mapsto\langle\xi, \mu(dx,t)\rangle_{C(\overline{\Omega}),\mathcal{M}(\overline{\Omega})}$is locally absolutely continuous and satisfies$\frac{d}{dt}\langle\xi, \mu(dx_{\dot{l}}t)\rangle_{C(\overline{\Omega}),\mathcal{M}(\overline{\Omega})}=\langle\triangle\xi, \mu(dx, t)\rangle_{C(\overline{\Omega}),\mathcal{M}(\overline{\Omega})}$$+ \frac{1}{2}\langle\rho_{\xi}, \nu(t)\rangle_{\mathcal{E},\mathcal{E}’}$for each $\xi\in X$ .

$\mathrm{a}.\mathrm{e}$.

$t$$\in(-\infty, +\infty)$

36
From (7), the Radon-Nikodym-Lebsesgue decomposition of this $\mu(dx, t)$
has the form$\mu(dx, t)=\mu_{s}(dx, t)$ $+\mu_{a.c}.(dx, t)$$= \sum_{i=1}^{n(t)}m_{*}(x_{i}(t))\delta_{x,(t)}(dx)+f(x, t)dx$

for each $t\in(-\infty, +\infty)$ , where $S_{t}=\{x_{i}(t)|1\leq i\leq n(t)\}$ denotes the set of, and $0\leq f=f(\cdot,t)\in$as
exhausted blowup points of.
The first result proven in this paper is stated as follows.
$\{u(\cdot, t_{k}’+t)\}$$t_{k}’arrow+\infty$$L^{1}(\Omega)\cap C(\overline{\Omega}\backslash S_{t})$Theorem 1 If $G(x, x’)$ is associated with the (N) or (JL)

$ffiel,d_{J}$and$\lambda=||u_{0}||_{1}\not\in 4\tau_{\mathrm{t}}N$$T_{\max}=+\infty$$\lim_{tarrow+\infty}F(u(\cdot, t))>-\infty$

then $\mu_{s}(dx,t)=0a.e$ .

$t$$\in(-\infty, +\infty)$,.

is generally only
Unfortunately, $\in(-\infty, +\infty)\mapsto\mu_{s}(dx, t)$-weakly upper semi-continuous, and the above theorem is not sufficient to
can inferdeduce $\mu_{s}(dx, 0)=0$ , although if this is the case, thenfrom$\in \mathrm{A}6(\overline{\Omega})$

$t$

$*$

$\lambda\in 4\pi N$$1\mathrm{h}’\mathrm{e}$$T_{\max}=+\infty$$\lim_{tarrow+}\sup_{\infty}||u(\cdot, t)||_{\infty}=+\infty$$\lim_{tarrow+\infty}\mathcal{F}(u(\cdot, t))>-\infty$.

If the free energy is unbounded, on the contrary, the solution blows-up in
finite or infinite time; more precisely [29] ,$\Rightarrow$$\lim_{t\uparrow T_{\max}}F(u(\cdot, t))=-\infty$$\lim_{t\uparrow T_{\max}}\int_{\Omega}(u\log u)(x, t)dx=+\infty$.

(9)This means a kind of concentration as the blowup time approaches, and$T_{\max}=$$T_{\max}<+\infty$ may occur always in this case, namely, we suspect that.
implies$+\infty$$\lim_{t\uparrow+\infty}F(u(\cdot, t))>-\infty$

37
The other conjecture of ours is the convergence to a singular limit of
the stationary solution of the total mass quantized non-stationary solution
blowing-up in infinite time. The second theorem of this paper illustrates sucha profile of the solution in a specific case.
Since this theorem is concerned with the (D) field, here we mention somedifferences of this problem from the other cases. Actually, in the study ofthe (D) field, we have not been able to exclude the boundary blowup pointin both cases of blowing-up in finite time and infinite time. Consequently,(6) or (7) holds with $M$,
and replaced byandrespectively, where
denotes the set of continuous functions on with. This difficulty arises because
the value zero on
is not. Similarly, we have (8) withreplaced bydense in
$G(x,
x’)$
when
is associated with the (D) field.
In spite of these obstructions, we can show the following theorem.
$S$$(\overline{\Omega})$$S\cap\Omega$$\Lambda\Lambda(\Omega)=C_{0}(\overline{\Omega})’$$\overline{\Omega}$$C_{0}(\overline{\Omega})$$\partial\Omega$$C^{2}(\overline{\Omega})\cap C_{0}(\overline{\Omega})$$C_{*}(-\infty, +\infty;\mathcal{M}(\overline{\Omega}))$$C(\overline{\Omega})$$C_{*}(-\infty, +\infty;\mathcal{M}(\Omega))$

Theorem 2 If $G(x, x’)$ is associated with the (D) field, $\lambda=||u_{0}||_{1}=8\pi$ ,$T_{\max}=+\infty$ , and, then any
admitssuch that$E_{8\pi}=\emptyset$$t_{k}arrow+\infty$inis absolutety continuous$u(x, t_{k^{\wedge}}’+t)dxarrow 8\pi\delta_{x(t)}(dx)$$t\in(-\infty, +\infty)\mapsto x(t)$$\lim_{tarrow\pm}\inf_{\infty}$

dist(x(t),$\partial\Omega$$\{t_{k}’\}\subset\{t_{k}\}$$L_{*}^{\infty}(-\infty, +\infty;\mathcal{M}(\overline{\Omega}))$$\in\Omega$) $>0$(10)$(-\infty<t<+\infty)$ ,$\frac{dx}{dt}=4\pi\nabla R(x(t))$,
where $R(x)=[G(x, x’)+ \frac{1}{2\pi}\log|x-x’|]_{x=x}$
indicates the Robin function.

The first relation of (10) implies that the local
norm of $u(\cdot, t+t_{k})$becomes arbitrarily small locally uniformly in. Still this isnear
not enough to exclude the boundary blowup point, but we hope that thisconvergence hoIds actually in.
We recall also that
denotes the set of stationary solutions so that
if and only if it is a (regular) solution to$L^{1}$$\partial\Omega$$t\in \mathrm{R}$$C_{*}(-\infty, +\infty;\mathcal{M}(\overline{\Omega}))$$E_{\lambda}$$v_{\infty}\in E_{\lambda}$$- \triangle v_{\infty}=\lambda\frac{e^{v}\infty}{\int_{\Omega}e^{v}\infty}$

in$\Omega$,$v_{\infty}=0$on$\partial\Omega$.

The condition
has been studied in detail [6, 20, 9]. This is actually
the case, if
is simply connected and close to a disc. For such adomain, any solution $u=u(\cdot, t)$ , existing globally in time with,
$E_{8\pi}=\emptyset$$\Omega\subset \mathrm{R}^{2}$$||u_{0}||_{1}=8\pi$38cannot be uniformly bounded, and therefore,holds true. Then, thanks to the concentration lemma [25], we can show thatis subject tothe location of the concentration mass formed during
the ordinary differential equation given by the last relation of (10). We notethat this is a conjugate form of the vortex equation derived from the Euler
equation [19]:$\lim\sup_{t\uparrow+\infty}||u(\cdot, t)||_{\infty}=+\infty$$t_{k^{\wedge}}’arrow+\infty$$\frac{dx}{dt}=4\pi\nabla^{[perp]}R(x(t))$$(-\infty<t<+\infty)$ .

The last result of this paper proves that our conjecture holds in the affirmative if the solution is radially symmetric; more precisely,
Theorem 3 If $\Omega=\{x\in \mathrm{R}^{2}||x|<R\}$ is a disc, $u_{0}=u_{0}(|x|)$ is radiallysymmetric, $G(x, x’)$ is associated with the (N) or (JL) field, andsystem (1), that is,, then the blowup in infinite time does not$\lambda=||u_{0}||_{1}>$$occur’\dot{\iota}n$$8\pi$$1 \mathrm{i}\mathrm{n}1\sup_{t\uparrow+\infty}||u(\cdot, t)||_{\infty}<+\infty$

holds if $T_{\max}=+\infty$,

In this radially symmetric case, if $\lambda\in(0,8\pi)$ then the solution $u=$$u(x, t)$ is uniformly bounded, and the stationary problem admits the unique(constant) solution, denoted by , and furthermore, we have$\underline{u}_{\lambda}$$\lim_{tarrow+\infty}||u(\cdot, t)$$-\underline{u}_{\lambda}||_{\infty}=0$.

See [21, 27, 33] and the dicussion in the next section. On the other hand, theabove theorem guarantees the generic blowup in finite time in this problem; see [30]. Thus, behavior of the solution global in time has beenif.
almost classifed in this case, using
takeThis paper is composed of five sections and two appendices.
preliminaries in the following section, and prove Theorems 1, 2, and 3 in
\S \S 3, 4, and 5, respectively. In the first appendix, we show the proof of (9)appendix is devoted to the proof of aby the method of [29]. The
conentration lemma [25] used in the proof of Theorem 2.
$\lambda>8\pi$$\lambda=||u_{0}||_{1}$$\backslash \lambda^{r}\mathrm{e}$$\mathrm{s}\mathrm{e}\mathrm{c}.\mathrm{o}\mathrm{n}\mathrm{d}$392Preliminaries

In this section, we take several preliminaries and describe the relation between other works and our theorems. See [30, 29, 32, 33] for details of theresult referred to in this section.
First, as is mentioned in the introduction, the stationary problem (4) has
an equivalent varitational structures, $F$ on
andon
These variational structures are regarded as an “unfolding” of the Lagrange functional,
and in particular, it holds that$\mathcal{W}(u, v)\geq\max\{F(_{\backslash }u), J_{\lambda}(v.)\}$$V$$J_{\lambda}$$\Lambda f_{\lambda}$

for $(u, v)\in M_{\lambda}\cross V$

This inequality means$\int_{\Omega}\{u(\log u-1)-uv\}dx+\lambda\log(\int_{\Omega}e^{v}dx)-\lambda\log\lambda+\lambda\geq 0$

for$(u, v)\in \mathbb{J}I_{\lambda}\cross V$(11), and can be proved directly using Jensen’s inequality[22, 4, 12]. In any case, it holds that

$F(u(\cdot, t))\geq J_{\lambda}(v(\cdot, t))$(12)$(t \in[0, T_{\max}))$, because $v=$
for the solution $(u, v)=(u(\cdot, t),$to (1) with$F$
$G*u$ and therefore,holds in this system.
Next, if $u=u(x, t)$ is a solution to (1), then it holds that$v(\cdot, t))$$||u_{0}||_{1}=\lambda$$=\mathcal{W}$$\frac{dJ}{dt}\leq C\lambda^{2}+3|\Omega|\exp(4K^{2}J)$$(t \in[0, T_{\max}))$for$J=J(u)= \int_{\Omega}(u\log u+e^{-1})$

where , $K$ are positive constants determined bycase of$C$$T_{\max}=+\infty$

there are$t_{k}arrow+\infty$

and,$\Omega$, and therefore, in the$\lim_{t\uparrow+}\inf_{\infty}\int_{\Omega}(u\log u)(x, t)dx<+\infty$, $\delta>0$ , and $C>0$ such that$\int_{\Omega}(u\log u)(x, t)dx$ $\leq C$$(t \in[t_{k}, t_{k}+2\delta)$,$k$$=1,2$ ,

$)$

(13)

40
with a constantThen, Moser’s iteration scheme guarantees$t\in[t_{k},tk+\delta]$and $=1,2$ , . . , and therefore,
independent offollows from the parabolic regularity, where$||u(\cdot, t)||_{\infty}\leq C$$\omega(u_{0})\neq\emptyset$

$k$

$C$

$\omega(u_{0})=${$u_{\infty}|$

there exists$t_{k}arrow+\infty$

such that$u(\cdot,$$t_{k})arrow u_{\infty}$in$C^{2+\theta}(\overline{\Omega})$denotes the -limit set of $u=u($ . , obtained from the initial value
argument of iteration is also valid to the other case, i.e., we obtain$t)$$\omega$$u_{0}$}. This$\lim_{t\uparrow T_{\max}}\int_{\Omega}(u\log u)(x, t)dx=+\infty$

if $T_{\max}<+\infty$ .
Since system (1) is provided with the Lyapunov function, the standard
is aargument of the dynamical system [13] guarantees that any, then there. In fact, first, if ,critical point of $F$ defined oninandsuch thatandare, and therefore, it
for $=1,2$ ,. We may assume. This implies $F(u_{1})\geq$
follows that.
and hence $F$ is constant on, on the other hand, the solution to (1) with the initial valueIf, exists globally in time from the local well-posedness, denoted byfor each $t\geq 0$ by the definition. Thisof (1), and it holds that
implies$u_{\infty}\in\omega(u_{0})$$M_{\lambda}$$u_{1}$$t_{k}^{1}<t_{k^{r}}^{2}<t_{k+1}^{1}$$C^{2+\theta}(\overline{\Omega})$$u(\cdot, t_{k}^{2})arrow u_{2}$$u(\cdot, t_{k}^{1}.)arrow u_{1}$$t_{k}^{2}arrow+\infty$$t_{k\wedge}^{1}arrow+\infty$$u_{2}\in\omega(u_{0})$$k$$\cdot$$F(u(\cdot, t_{k+1}^{1}))\geq \mathcal{F}(u(\cdot, t_{k}^{2}))\geq F(u(\cdot, t_{k}^{1}))$$\omega(u\mathrm{o})$$\mathcal{F}(u_{2})\geq \mathcal{F}(u_{1})$$u_{\infty}\in\omega(u_{0})$$T_{t}u_{\infty}$$u_{\infty}$$T_{t}u_{\infty}\in\omega(u_{0})$$F(T_{t}u_{\infty})=F(u_{\infty})$$(t \geq 0)$

and therefore,$\frac{d}{dt}F(T_{t}u_{\infty})|_{t=0}=0$.

Then, we obtain$u_{\infty}= \lambda\frac{e^{v_{\infty}}}{\int_{\Omega}e^{v}\infty dx}$

is a stationary solution to (1).
by (3), and therefore,Thus,
is a critical point ofdefined on for ch. It holds also thatfrom the general theory of dual
variation mentioned in the introduction. From the mass quantization of thefor$v_{\infty}=G*u_{\infty}$$u_{\infty}$$v_{\infty}=G*u_{\infty}$$\omega(u_{0})$$J_{\lambda}$$J_{\lambda}(v_{\infty})=F(u_{\infty})$$V$$\mathrm{e}\mathrm{a}$$u_{\infty}\in$41
non-compact stationary solution sequence $[23, 26]$ , on the other hand, it

follows that$j_{\lambda} \equiv\inf_{v\in E_{\lambda}}J_{\lambda}(v)>-\infty$

forin the cases of the (N) and (JL) fields, and forin thecase of the (D) field. Therefore, obtain the following fact $[14, 29]$ .
$\lambda\not\in 4\mathrm{k}\mathrm{N}$$\lambda\not\in 8\pi N$$\mathrm{w}\mathrm{e}$Theorem 4 If$\mathcal{F}(u_{0})<j_{\lambda_{f}}$then$\lim_{t\uparrow T_{\Phi \mathrm{R}}}\int_{\Omega}(u\log u)(x, t)dx=+\infty$.

Both cases $T_{\max}=+\infty$ and $T_{\max}<+\infty$ are permitted in the above theorem,
but we suspect thatalways implies $T_{\max}<+\infty$ . Actually, if theassumptions of Theorem 1 hold, then we havefrom the conclusion,and this is impossible in the case of. Thus, we obtain the following
theorem.
$F(u_{0})<j_{\lambda}$$\omega(u_{0})\neq\emptyset$$F(u_{0})<j_{\lambda}$

Theorem 5 If $G(x, x’)$ is associated with the (N) or (JL) field, if$F(u_{0})<j_{\lambda}$with, and if $T_{\max}=+\infty$ holds in the previous theorem,
then $\lim_{t\uparrow T_{\max}}F(u(\cdot, t))=-\infty$ .
$\lambda=||u_{0}||_{1}\not\in 4\pi N$$\backslash \lambda^{\gamma}\mathrm{e}$emphasize again what we suspect, that is, $T_{\max}=+\infty$ with
$t\uparrow+\infty\rceil\iota \mathrm{i}\mathrm{m}F(u(\cdot, t))=-\infty$

will not occur, and therefore, $T_{\max}<+\infty$ will hold under the assumption ofTheorem 5. See the descriptions below Theorem 1.

3

Proof of Theorem 1

Given $t_{k}arrow+\infty$ we havesatisfying (8), where $\mu=\mu(dx, t)$
is a weak solution to (1). We shall write for , and furthermore, given$T>0$ , we may assume $t_{k}+2T<t_{k+1}$ , passing to a subsequence. FYom theassumption, then we have}$\{t_{k^{\wedge}}’\}\subset\{t_{k}\}$$t_{k}$$t_{k}’$$\lim_{t\uparrow+\infty}F(u(\cdot, t))>-\infty$$\sum_{k}\int_{t_{k}-T}^{t_{k}+T}dt$$\int_{\Omega}u$$|\nabla(\log u-v)|^{2}(x, t)dx<+\infty$

42
and hence it holds that$. \lim_{karrow\infty}\int_{t_{k}-T}^{t_{k}+T}dt$$\int_{\Omega}u|\nabla(\log u-v)|^{2}(x, t)dx=0$.

have $G(x, x’)\geq-A$ , and therefore, $v(x)\geq-A\lambda$ , where. This implies
stant determined by$1h^{\gamma}\mathrm{e}$$A$

is a con-$\Omega([1])$$u|\nabla(\log u-v)|^{2}=4e^{v}|\nabla(ue^{-v})^{1/2})|^{2}\geq 4e^{-A\lambda}|\nabla(ue^{-v})^{1/2}|^{2}$,and therefore,$f_{k}(x, t)=(ue^{-v})^{1/2}(x, t+t_{k})- \frac{1}{|\Omega_{d}|}\int_{\Omega}(ue^{-v})^{1/2}(x, t +t_{k})dx$

satisfies$\lim_{k^{\wedge}arrow\infty}\int_{-T}^{T}dt$$\int_{\Omega}|\nabla f_{k^{\wedge}}(x, t)|^{2}dx$$=0$ ,$\int_{\Omega}f_{k}.(x_{7}t)dx=0$.

This means.

$f_{k}$$arrow 0$

in $L^{2}(-T, T;H^{1}(\Omega))$ ,and passing to a subsequence (denoted by the same symbol), we obtain$f_{k’}(x, t)arrow 0$$\mathrm{a}.\mathrm{e}$. $(x, t)\in\Omega\cross(-T, T)$ .

On the other hand, we have$\frac{1}{|\Omega|}\int_{\Omega}(ue^{-v})^{1/2}dx$

and therefore, forthat$\mathrm{a}.\mathrm{e}$$\leq\{\frac{1}{|\Omega|}\int_{\Omega}ue^{-v}dx\}^{1/2}\leq(|\Omega|\lambda e^{-A\lambda})^{-1/2}$. $t\in(-T, T)$ , there is $\{t_{k}’\}\subset\{t_{k}\}$ and $C_{0}(t)\geq 0$ such$(ue^{-v})^{1/2}(x, t_{k}’+t)$ $arrow C_{0}(t)$$\mathrm{i}.\mathrm{e}.$$\mathrm{a}.\mathrm{e}$. $x\in\Omega$ ,,$(ue^{-v})(x, t_{k}^{J}.

+t)$ $arrow C_{0}(t)^{2}$$\mathrm{a}.\mathrm{e}$.

$x\in\Omega$

.

{4}

43
Now, relation (8) implies$v(x, t_{k}’+t)$ $arrow\sum_{i=1}^{n(t)}m_{*}(x_{i}(t))G(x, x_{i}(t))+\int_{\Omega}G(x, x’)f(x’, t)dx’$

weakly infor $1<q<2$ by theelliptic estimate [5] applied tothe second equation of (1). This convergence is strong infor $1\leq p<$$x\in\Omega$
by Rellich-Kondrachov’s theorem, and hence.
, passing to a
subsequence. In case $n(t)\geq 1$ and $C_{0}(t)>0$ , this implies$L^{1}$$\mathrm{I}4^{\gamma 1,q}(\Omega)$$U(\Omega)$$\infty$$\mathrm{a}.\mathrm{e}$$e^{-v(x,t_{\acute{k}}+t)}u(x, t_{k}’+t)\}dx=+\infty$$\int_{\Omega}\lim_{karrow\infty}\{e^{v(x,t_{\acute{k}}+t)}$by$m_{*}\geq 4\tau\downarrow$, but the left-hand side is estimated above by$\lim_{karrow}\inf_{\infty}\int_{\Omega}u(x, t_{k}’+t)dx=\lambda$

from Fatou’s lemma. This is impossible, and therefore, $\mu_{s}(dx, t)\neq 0$ implies$C_{0}(t)=0$ ,$\mathrm{i}.\mathrm{e}.$,$(ue^{-v})(x, t_{k}’+t)arrow 0$$\mathrm{a}.\mathrm{e}$.

$x\in\Omega$

.

(15)

On the other hand, $S_{t}=\{x_{i}(t)|1\leq i\leq n(t)\}$ is the set of exhausted
blowup points ofas $tkarrow\infty$ , and therefore,islocally uniformly bounded inby the elliptic regularity. This implies$\{u(\cdot, t_{k}’+t)\}$$\{v(\cdot, t_{k^{\wedge}}’+t)\}$$\overline{\Omega}\backslash S_{t}$$u(x, t_{k}’+t)arrow 0$$\mathrm{a}.\mathrm{e}$

.

(16)

$x\in\Omega$by (15). The parabolic regularity guarantees, on the other hand,$u(\cdot, t_{k}’+t)$$arrow f(\cdot,t)$

locally uniformly in$\overline{\Omega}\backslash B_{t}$in (S), passing to a subsequence, and therefore,.
by(16). This implies the mass quantization,, which contradicts theassumption. Thus, we obtain. $t\in(-T, T)$ , and hence.
$\in(-\infty, +\infty)$ . The proof
is complete$f(x, t)=0\mathrm{a}.\mathrm{e}$$x\in\Omega$$\lambda\in 4\pi N$$\mu_{s}(dx, t)$$t$$=0\mathrm{a}.\mathrm{e}$$\mathrm{a}.\mathrm{e}$44Proof of Theorem 2

4

It is obvious that this theorem follows from the following lemma, where$\mathcal{K}(u)=\frac{1}{9_{\sim}}\iint_{\Omega \mathrm{x}\Omega}G(x,x’)u(x)u(x’)dxdx’$

denotes -1 times the inner (potential) energy. In fact, we have only toconfirm that the first condition of (17), described below, is satisfied for$u^{k}(\cdot, t)=u(\cdot, t_{k}+t)$.

Lemma 6 If $G(x, x’)$ of (1) is associated with the (D) field,, andsequence of the initial values satisfying$\{u_{0}^{k}\}$

is

$a$

$||u_{0}^{k}||_{1}=8\pi$$K_{k}= \inf_{\mathrm{t}\in(0,T)}\mathcal{K}(u^{k}(\cdot, t))arrow+\infty$$F_{k}= \sup_{\mathrm{r}\in(0,T)}\mathcal{F}(u^{k}(\cdot, t))\leq F<+\infty$

then we have$\{u^{k^{A}}’\}\subset\{u^{k^{A}}\}$

(17)

,

such thatin$u^{k’}(x, t)dx$ $arrow 8\pi\delta_{x(t)}(dx)$$L_{*}^{\infty}$( , ; A4
$0$

$T$$(\overline{\Omega})$

(18)

)

as $k’arrow\infty$ , where $u^{k}=u^{k}(x, t)$ denotes the solution to (1) for the initialvalue$u_{0}^{k}(x)$$\omega\subset\subset\Omega$, $t\in(0, T)\mapsto x(t)\in\omega$ is locally absolutely continuous, withdetermined by$F_{f}$and it holds that$\frac{dx}{dt}=4\pi\nabla R(x(t))$$a.e$The show the first condition of (17) for.

$t$$\in(0, T)$$u^{k}(\cdot, t)$.

$=u(\cdot, t_{k}+t)$$\lim_{t\uparrow+\infty}\int_{\Omega}(u\log u)(x, t)dx=+\infty$

.

(18)

, we use(20)

In fact, if this is not the case, then (13) holds, and therefore, there are. This contradicts.
insuch thatand, and we obtain (20).
the assumption,
Now, we have$t_{k}$$arrow+\infty$$v_{\infty}\in E_{\lambda}$$v(\cdot, t_{k})arrow v_{\infty}$$C^{2+\theta}(\overline{\Omega})$$E_{8\pi}=\emptyset$$\lim_{t\uparrow+\infty}\mathcal{K}(u(\cdot, t))=+\infty$

(21)

45
by $F(u(\cdot, t))\leq F(u_{0})$ , and the proof is complete.
Lemma 6 is obtained from its discrete version, the concentration lemma[25] described below. Traditionally, such a kind of lemma is stated in termsof the convergence of the probability measure [7], and we shall adopt thisformulation, putting$P(\Omega)=\{\rho\in L^{1}(\Omega)|\rho\geq 0$,$\int_{\Omega}\rho(x)dx=1\}$$\mathrm{I}(\rho)=\frac{1}{2}\oint\oint_{\Omega \mathrm{x}\Omega}G(x, x’)\rho(x)\rho(x’)$

dxdx’$- \frac{1}{8\pi}\int_{\Omega}(\rho\log\rho)(x)dx$.

First, the dual form of the Trudinger-Moser inequality $[7, 33]$ assures$\sup\{\mathrm{I}(\rho)|\rho\in P(\Omega)\}<+\infty$

(22)

,

and therefore, the value$I_{\Omega}(x)= \sup\{\lim_{karrow+}\sup_{\infty}\mathrm{I}(\rho_{k})|\{\rho_{k}\}\subset P(\Omega)$$\rho_{k}(x)dxarrow\delta_{x}(dx)$is well-defined for each$x$$\in\overline{\Omega}$$*$-weakly in A4,$(\overline{\Omega})\}<+\infty$(23). Next, we have

$I_{\Omega}(x)=I_{B_{1}(0)}(0)+ \frac{1}{2}R(x)$(24)

for $B_{1}(0)=\{x\in \mathrm{R}^{2}||x|<1\}$ (Theorem 3.1 of [7]), and therefore, it holds
that$\Omega_{I_{\infty}}\equiv\{x\in\Omega|I_{\Omega}(x)\geq I_{\infty}\}\subset\subset\Omega$for each$I_{\infty}\in \mathrm{R}$(25), i.e., there is an open set $O$ such that$\Omega_{I_{\infty}}\subset O\subset\overline{O}\subset\Omega$

Given $u\geq 0$ withthat$||u||_{1}=\lambda$

.

(26)

, we have $f=u/\lambda\in P(\Omega)$ . Then, it holds$\mathrm{I}(f)$$=- \frac{1}{8\pi\lambda}\{\int_{\Omega}u(\log u-1)dx$ $- \frac{4\pi}{\lambda}\iint_{\Omega \mathrm{x}\Omega}G(x, x’)u(x)u(x’)dxdx’\}$$- \frac{1}{8\pi\lambda}\{1-\log\lambda\}$,46
and therefore,$\mathrm{w}\mathrm{e}$have$\mathrm{I}(f)=-\frac{1}{64\tau_{1}^{2}}\mathcal{F}(u)+\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$in the case of$\lambda=8\pi$. Thus, Lemma 6 is reduced to the following lemma.

Lemma 7 If $u^{k^{\mathrm{n}}}=8\pi\rho_{k’}(x, t)(k=1,2, \cdot )$ is a solution seqeuence to(1) with $\rho_{k}\in L^{1}(0, T;P(\Omega))$ satisfying$\inf_{t\in(0,T)}\mathcal{K}(\rho_{k}(\cdot, t))arrow+\infty$$\inf_{t\in(0T)}\mathrm{I}(\rho_{k}(\cdot, t))arrow I_{\infty}>-\infty$

then th $ere$ is a subsequence$\{\rho\nu\}\subset\{\rho_{k}\}$$\rho_{k^{\wedge}}’(x, t)dxarrow\delta_{x(t)}(dx)$

where
$s$

$t$

,

such thatin$\in(0, T)\mapsto x_{\infty}(t)\in\Omega_{I_{\infty}}(\subset\subset\Omega)\prime is$$L_{*}^{\infty}(0, T;\mathcal{M}(\overline{\Omega}))$

,

(27)

locally absolutely continuous andatisfies$\frac{dx_{\infty}}{dt}=4\pi\nabla R(x_{\infty}(t))$

$a.e$

.

$t$

$\in(0,T)$

.

(28)

To show the above lemma, we use its discrete version (concentration
lemma), of which proof is given in the second appendix.
Lelluma 8 If$\{\rho_{k}\}\subset P(\Omega)$satisfies$\lim_{karrow\infty}\mathcal{K}(\rho_{k})=+\infty$$\lim_{karrow\infty}\mathrm{I}(\rho_{k})=I_{\infty}>-\infty$$\lim_{karrow\infty}\int_{\Omega}x\rho\kappa.(x)dx=x_{\infty}$

for some$x_{\infty}\in \mathrm{R}^{2}$, then we have$x_{\infty}\in\Omega_{I_{\infty}}$$\rho_{k}(x)dxarrow\delta_{x_{\infty}}(dx)$

$*$

and

-weakly in$\mathcal{M}(\overline{\Omega})$.

47
Now, we give the following.

Proof of Lemma 7:$\backslash \mathrm{V}\mathrm{e}$define$\Omega_{I_{\infty}}$by (25) for$I_{\infty}= \lim_{karrow\infty}\inf_{t\in(0,T)}\mathrm{I}(\rho_{k}(\cdot, t))>-\infty$

and take the open set

$O$

and

$\xi\in C_{0}^{\infty}(\Omega)$satisfying (26) and$\xi|_{\mathit{0}}=1$, respec-

tively.
Rom the assumption, $u^{k}(x, t)$ $=8\pi\rho_{k}(x, t)$ is a solution to (1). We takean arbitrary, and multiply the first equation of (1) by$x=(x_{1},x_{2})$. Then, using the second equation of (1), wefor $i=1,2$ , whereobtain$\eta\in C_{0}^{\infty}(0, T)$$\eta\xi x_{i}$$- \iint_{\Omega \mathrm{x}(0,T)}\eta’(t)x_{\dot{\mathrm{t}}}\xi(x)\rho_{k}(x,t)dxdt$$= \int_{0}^{T}\eta(t)(\int_{\Omega}\triangle(x_{i}\xi(x))\rho_{k}.(x,t)dx)dt$$+4 \pi\int_{0}^{T}\eta(t)(\iint_{\Omega \mathrm{x}\Omega}---i(x, x’)\rho k(x, t)\rho k(x’, t)dxdx’)dt$(25)

by $G(x, d)$ $=G(x’, x)$ , where$—_{i}(x, x’)=\nabla(x_{i}\xi(x))$. $\nabla_{x}G(x, x’)+\nabla(x_{i}’\xi(x’))-\nabla_{x’}G(x, x’)$ .

Here, we have$G(x, x’)= \frac{1}{2\pi}\log\frac{1}{|x-x’|}+K(x, x’)$withholds that

$K\in C^{\theta,2+\theta}(\overline{\Omega}\cross \Omega)\cap C^{2+\theta,\theta}(\Omega \mathrm{x} \overline{\Omega})$for $0<\theta<1$ , and therefore, it$–i-(x, x’)=- \frac{(x-d)(x_{i}\nabla\xi(x)-x_{\dot{\mathrm{t}}}’\nabla\xi(x’))}{2\pi|x-x’|^{2}}-\frac{(x_{i}-x_{i}’)(\xi(x)-\xi(x’))}{2\pi|x-x’|^{2}}$$+\nabla_{x}K(x, x’)\cdot\nabla(x_{i}\xi(x))+\nabla_{x’}K(x, x’)$.

$\nabla(x_{i}’\xi(x’))$

We have also$||_{-\mathrm{i}}^{-}-||_{\infty}\leq C||\xi||_{C^{2}(\overline{\Omega})}$

,

.

(30)

48
and therefore,
is uniformly bounded and cally equi$(0,T)$. Consequently, there isthat admits thecontinuous incontinuous$\{\int_{\Omega}x_{i}\xi(x)\rho_{k}(x, \cdot)dx\}$$1\mathrm{o}$$\{\rho_{k’}\}\subset\{\rho_{k^{\alpha}}\}$$t$$\in(0, T)\mapsto x_{\infty}(t)=.,\mathrm{h}.\mathrm{n}1karrow\infty\int_{\Omega}x\xi(x)\rho_{k^{J}}.(x, t)dx\in \mathrm{R}^{2}$and then, we have$x_{\infty}(t)\in\Omega_{I_{\infty}}$,and$\rho_{k^{J}}(x, t)dxarrow\delta_{x_{\infty}(t)}(dx)$$*$-weakly in$\mathcal{M}(\overline{\Omega})$.

for each $\in(0, T)$ by Lemma 8. This means (27).
have also$t$$\backslash \mathrm{V}\mathrm{e}$$(t \in(\mathrm{O}, T))$$, \lim_{karrow\infty}\int_{\Omega}[\triangle(x\xi(x))]\rho_{k^{d}}(x, t)dx=0$

and$, \lim_{k^{\wedge}arrow\infty}\int\int_{\Omega \mathrm{x}\Omega}---i(x, x’)\rho\nu(x, t)\rho_{k’}.(x’, t)dxdx’=\frac{\partial R}{\partial x_{i}}(x_{\infty}(t))$

by$\xi|_{\mathit{0}}=1$, and therefore, it follow that$- \int_{0}^{T}\eta’(t)x_{\infty}^{i}(t)dt=4\pi$ $\int_{0}^{T}\eta(t)\frac{\partial R}{\partial x_{i}}(x_{\infty}(t))dt$

from (29). Thus,and satisfies (28). The proof is complete.

$t\in(0, T)\mapsto x_{\infty}(t)\in \mathrm{R}^{2}$

5

is locally absolutely continuous,

Proof of Theorem 3

shall descibe the case that $G(x, x’)$ is associated with the (JL) field,
because the proof is similar to the other case of the (N) field. This system isdefined by$\backslash h^{\gamma}\mathrm{e}$in$u_{t}=\nabla(\nabla u-u\nabla v)$

in$0= \triangle v-\frac{\lambda}{|\Omega|}+u$$\frac{\partial u}{\partial\nu}=\frac{\partial v}{\partial\nu}=0$on$\mathrm{Q}\vee$$\Omega\cross$$\cross(0, \infty)$$(0, \infty)$$\partial\Omega\cross(0, \infty)$,

49
with $u=u(r, t)$ , $r=|x|$ , and, and therefore, it holds that$\lambda=||u_{0}||_{1}$(31)$\frac{\partial}{\partial t}\int_{|x|<r}u(x, t)d_{X=}\underline{9}\pi(\frac{\partial u}{\partial r}-u\frac{\partial v}{\partial r})r$$- \underline{9}_{\pi r\frac{\partial v}{\partial r}=}\int_{|x|<r}$

Still we have (7) withandsuch that$\# S$($u(x, t)$$- \frac{\lambda}{|\Omega|}$

)

$dx$

(32)

.

, and therefore, if $T_{\max}=+\infty$ , $u=u(|x|, t)$ ,with some $t_{k}arrow+\infty$ , then there is$<+\infty$$\lim_{karrow\infty}||u(\cdot, tk)||_{\infty}=+\infty$$\{t_{k^{\wedge}}’\}\subset\{t_{k}\}$$u(x, t_{k}^{J})dxarrow 8\pi\delta_{0}(dx)+f(x)$$*$-weakly in(33)$\mathcal{M}(\overline{\Omega})$with $0\leq f=f(|x|)\in L^{1}(\Omega)\cap C(\overline{\Omega}\backslash \{0\})$ . Moreover, by the parabolicand elliptic regularity [21] we obtain the following inequalities, whereisa constant determined by $\in(0, R)$ ,
, and
, and
$\{x\in \mathrm{R}^{2}|\epsilon <|x|<R\}$ :$C$$\lambda=||u_{0}||_{1}$$\epsilon$$||u_{0}||_{\infty}$$\Omega_{\epsilon}=$$\sup_{t\geq 0}||u(\cdot, t)||_{L^{\infty}(\Omega_{\zeta})}\leq C$$\sup_{t\geq 0}||\nabla u(\cdot, t)||_{L^{2}(\Omega_{\epsilon})}\leq C$

(34)

.

$\sup_{t\geq 0}\int_{t}^{t+1}||u_{t}(\cdot, s)||_{L^{2}(\Omega_{\mathrm{E}})}^{2}ds\leq C$

We define$z(r, t) \equiv\frac{1}{2\pi}\int_{|x|<r}(u(x, t)$ $- \frac{\lambda}{|\Omega|})dx$$(0<r<R, t>0)$ ,satisfying$z_{r}=r(u- \frac{\lambda}{|\Omega|})$,$z_{rr}=ru,$ $+u- \frac{\lambda}{|\Omega|}$ ,and then it follows that$\mathcal{L}(z)\equiv z_{t}-z_{rr}+\frac{1}{r}z_{\gamma}-\frac{1}{r}zz_{r}-\frac{\lambda}{|\Omega|}z=0$$z(0, t)$ $=z(R, t)=0$

in$(0, \infty)$.

in $(0, R)$$\cross(0, \infty)$

(32)

50
Lemma 9 We have $W(R, t)$ $<-R^{2}$ for$t$$>0$ ,$W(r, t)$ $= \int_{0}^{r}z(s, t)sds$

where

.

Proof: From the first equation of (1), we have$\int_{\Omega}|x|^{2}u_{t}dx=-\int_{\Omega}’\sim^{2_{X}}(\nabla u-u\nabla v)dx$$=- \underline{9}\int_{\partial\Omega}[(x’ \nu)u]d\sigma+4\lambda+4\pi$$\int_{0}^{R}r^{2}(uv_{r})dr$,

while (32) implies$(0<r<R)$ .

$-rv_{r}(r,t)$ $= \int_{0}^{r}s(u(s, t)$ $- \frac{\lambda}{|\Omega|})ds$

Thus, we obtain$\int_{0}^{R}(uv_{r})(r, t)r^{2}dr=-\int_{0}^{R}ru(r, t)\{\int_{0}^{r}su(s, t)ds-\frac{\lambda r^{2}}{\underline{9}|\Omega|}\}dr$$=- \frac{1}{9_{\sim}}\{\int_{0}^{R}ru(r, t)dr\}^{2}+\frac{\lambda}{2|\Omega|}\int_{0}^{R}r^{3}u(r, t)dr$$=- \frac{\lambda^{2}}{8\pi^{2}}+\frac{\lambda}{4\pi|\Omega|}\int_{\Omega}|x|^{2}udx$,

and therefore,$\frac{dn^{-}\iota}{dt}=4\lambda-\frac{\lambda^{2}}{\underline{0}_{\pi}}+\frac{\lambda}{|\Omega|}77?-4\pi R^{2}u(R, t)$

(36)

for
$m(t)$ $= \int_{\Omega}|x|^{2}u(x, t)dx$$\backslash \mathrm{V}\mathrm{e}$.

shall show
$4 \lambda-\frac{\lambda^{2}}{\underline{9}\pi}+\frac{\lambda}{|\Omega|}r|\tau(t)>0$$(t>0)$ .

(37)

51
In fact, if this is not the case, we have $t_{1}>0$ such that
$4 \lambda-\frac{\lambda^{2}}{2\pi}+\frac{\lambda}{|\Omega|}m(t_{1})\leq 0$.

Then, the standard continuity argument applied to (36) guarantees $m_{t}<0$in, and also $m(t_{2})<0$ for some
. This is a contradiction, andhence we obtain (37).
We have, on the other hand,$(t_{1}, \infty)$$t_{2}>t_{1}$$m(t)=2 \pi\int_{0}^{R}r^{3}u(r, t)dr=\underline{9}\pi\int_{0}^{R}r^{2}[\frac{d}{dr}\int_{0}^{r}u(\rho, t)\rho d\rho]dr$$=2 \pi\{[r^{2}\int_{0}^{r}u(\rho, t)\rho d\rho]_{r=0}^{r=R}-\int_{0}^{R}2r[\int_{0}^{r}u(\rho, t)\rho d\rho]dr\}$$= \frac{\lambda R^{2}}{2}-4\pi W(R, t)$,

and therefore, $W(R, t)$ $<R^{2}$ for$t$$>0$ by (37).

The proof is complete.

Henceforth, we write $t’ k=t_{k}$ in (33) for simplicity. Then, $z^{k}(r, t)=$$z(r, t+t_{k})(k=1,2, \cdots)$ is uniformly bounded and locally equi-continuous
in $(0, R)\cross(-\infty, +\infty)$ by the second inequality of (34), and therefore, there isa subsequence, denoted by the same symbol, converging locally uniformly in$(0, R)\cross$. Prom the parabolic regularity, this limit function, denotedbybelongs to $C^{2,1}((0, R)\cross(-\infty, \infty))$ and satisfies$(-\infty, \infty)$$z^{\infty}=z^{\infty}(r, t)$in $(0, R)\cross$$\mathcal{L}(z^{\infty})=0$$(-\infty, \infty)$

(38)

.

We have, furthermore, $z^{\infty}\in C([0, R]\cross(-\infty, \infty))$ and$z^{\infty}(0t)\}=4$,$z^{\infty}(R, t)=0$$z^{\infty}(r, t) \geq 4-\frac{\lambda r^{2}}{2\pi R^{2}}$for $t\in(-\infty, \infty)$

for $(r, t)$ $\in(0, R)\cross(-\infty, \infty)$(38), we takeandby (33) and (35).

Proof of Theorem 3: Using

then defin$\lambda>8\pi$$\epsilon$$\in(0, \min(\lambda-8\pi, \pi))$$z_{*}(r, t)=4- \frac{\lambda r^{2}}{2\pi R^{2}}+\frac{\epsilon}{2\pi R^{2}}(r-\ell(t))_{+}^{2}$

,

52
where
$\ell(t)=R\exp(-(1-\frac{\epsilon}{\pi})\frac{t}{R^{2}})\in(0, R)$$\backslash \forall \mathrm{e}$.

have$\mathcal{L}(z_{*})=[-\frac{\epsilon}{\pi R^{2}}(r-\ell(t))_{+}\ell’(t)]-[-\frac{\lambda}{\pi R^{2}}+\frac{\epsilon}{\pi R^{2}}\chi[\ell(t),R](r)]$$+ \frac{1}{r}[-\frac{\lambda r}{\pi R^{2}}+\frac{\epsilon}{\pi R^{2}}(r-\ell(t))_{+}]-\frac{1}{r}[-\frac{\lambda r}{7\Gamma R^{2}}+\frac{\epsilon}{\pi R^{2}}(r-\ell(t))_{+}]z_{*}-\frac{\lambda z}{|\Omega|}*$for$\chi_{E}(r)=\{$

1

0

if
if

$r\in E$
$r\not\in E$,

and therefore, if $r\in(0,\ell(t))$ , we have$\mathcal{L}(z_{*})=\frac{\lambda}{\pi R^{2}}-\frac{\lambda}{\pi R^{2}}+\frac{\lambda}{\pi R^{2}}z_{*}-\frac{\lambda}{|\Omega|}z_{*}=0$.

In the other case of $r\in(\ell(t), R)$ , we have$\mathcal{L}(z_{*})=-\frac{\epsilon}{\pi R^{2}}(r-\ell(t))\ell’(t)-\frac{\epsilon}{\pi R^{2}}+\frac{\epsilon}{\pi R^{2}r}(r-\ell(t))$$- \frac{\epsilon}{\pi R^{2}r}(r-\ell(t))[4-\frac{\lambda r^{2}}{\underline{9}\pi R^{2}}+\frac{\epsilon}{\underline{9}\pi R^{2}}(r-\ell(t))^{2}]$$=- \frac{\epsilon}{\pi R^{2}}(r-\ell(t))[\ell’(t)+\frac{\ell(t)}{r(r-\ell(t))}$$+ \frac{1}{r}(4-\frac{\lambda r^{2}}{2\pi R^{2}})+\frac{\epsilon}{2\pi R^{2}r}(r^{2}-2r\ell(t)+l^{2}(t))]$$\leq-\frac{\epsilon}{\pi R^{2}}(r-\ell(t))[\ell’(t)+\frac{\ell(t)}{r(r-\ell(t))}-\frac{\epsilon\ell(t)}{\pi R^{2}}+\frac{1}{r}(4-\frac{\lambda-\epsilon}{\underline{9}_{\pi R^{2}}}r^{2})]$$\leq-\frac{\epsilon}{\pi R^{2}}(r-\ell(t))[\ell’(t)+\frac{\ell(t)}{R^{2}}(1-\frac{\epsilon}{\pi})+\frac{1}{r}(4-\frac{\lambda-\epsilon}{\underline{9}\pi R^{2}}r^{2})]$$\leq-\frac{\epsilon}{\pi R^{2}}(r-\ell(t))[\ell’(t)+(1-\frac{\epsilon}{\pi})\frac{\ell(t)}{R^{2}}]=0$.

Thus, we obtain$\mathcal{L}(z_{*})\leq 0$

in $(0, R)$$\cross$$(-\infty, \infty)$.

53
1Ve have also$z_{*}(R,t)$$=4- \frac{\lambda}{2\pi}+\frac{\epsilon}{2_{\mathcal{T}\mathrm{I}}R^{2}}(R-\ell(t))_{+}^{2}\leq 4-\frac{\lambda-\epsilon}{2\pi}<0$$z_{*}(0, t)=4$$z^{\infty}(r,t-T) \geq 4-\frac{\lambda r^{2}}{2\pi R^{2}}=z_{*}(r, 0)$by (39), and therefore,$z^{\infty}(r, t)\geq z_{*}(r, T+t)$

for $(r, t)\in$. $(0, R)$

for any $T>0$ from the comparison theorem. By making
in $(0, R)$$z^{\infty}(r, t) \geq 4-\frac{\lambda-\epsilon}{\underline{9}_{\mathcal{T}\downarrow R^{2}}}r^{2}$

Since$\epsilon$$\in(0, \min(0, \lambda-8\pi))$$Tarrow\infty$$\cross(0, \infty)$, we obtain.

is arbitrary, this implies
in $(0, R)$$z^{\infty}(r, t) \geq 4-\frac{\lambda_{1}r^{2}}{R^{2}}$where $\lambda_{1}=\min(\lambda-8\pi, \pi)$ . If$(0, \infty)$$\cross$$\lambda\in(8\pi, 9\pi]$In the other case of, we definesame argument it follows that$\lambda>9\pi$$z^{\infty}(r, t) \geq 4-\frac{\lambda_{2}r^{2}}{2\pi R^{2}}$$z_{*}$,(40), then (40) reads;

in $(0, R)$$z^{\infty}(r, t) \geq 4-\frac{4r^{2}}{R^{2}}$$\cross(0, \infty)$$\cross(0, \infty)$, replacingin $(0, R)$

.

(41)

Ai. Then, from the$\lambda \mathrm{b}_{d}\mathrm{v}$$\cross(0, \infty)$,

where $\lambda_{2}=\min(\lambda_{1}-8\pi, \pi)=\min(\lambda-9\pi, \pi)$ . Repeating this, we obtain (41)if$\lambda>8\pi$.

Lemma 9, on the other hand, guarantees$W^{\infty}(R, t) \equiv\int_{0}^{R}z^{\infty}(r, t)rdr\leq R^{2}$,

while (41) implies$W^{\infty}(R, t) \geq\int_{0}^{R}(4-\frac{4r^{2}}{R^{2}})rdr=R^{2}$.

54
This means$W^{\infty}(, Rt)$$=R^{2}$in$(0, \infty)$,

or equivalently,$z^{\infty}(r, t)=4- \frac{4r^{2}}{R^{2}}$in $(0, R)$$\cross(0, \infty)$.

However, this is impossible by$\mathcal{L}(z^{\infty})=\frac{\lambda-8\pi}{|\Omega|}z^{\infty}\neq 0$.

The proof is complete.

A$\backslash \mathrm{V}\mathrm{e}$Proof of (9)use the Lagrange functional defined by (5):$\mathcal{W}(u, v)=\int_{\Omega}u(\log u-1)dx+\frac{1}{2}||v||_{V}^{2}-\int_{\Omega}$

uvdx,

which satisfies$\mathcal{W}(u(\cdot, t)$, $v(\cdot, t))=F(u(\cdot, t))\leq F(u_{0})$for the solution $(u, v)=(u(\cdot, t),$have$v(\cdot, t))$(42)

to (1). Seince $u\log u+e^{-1}\geq 0$ , we$\mathcal{W}(u, v)\geq-|\Omega|e^{-1}-\lambda-\int_{\Omega}$

uvdx

and therefore,$\lim_{t\uparrow T_{\max}}\int_{\Omega}(uv)(x, t)dx=\mathcal{K}(u(\cdot\}t))=+\infty$(43). On the other hand, we canfrom the assumption
elliptic estimate [5] to the second equation of (1) byapply the, and this implies(44)$\sup$$\lim_{t\uparrow T_{\max}}F(u(\cdot, t))=-\infty$$||u(\cdot, t)||_{1}=$$L^{1}$$\lambda$$||v(\cdot, t)||_{W^{1.q}}<+\infty$$t\in[0,T_{\mathrm{m}\mathrm{a}’ \mathrm{c}})$55
for each $q\in[1,2)$ .
By Chang-Yang’s inequality [10], we have a constant $K$ determined bysuch that$\log(\frac{1}{|\Omega|}\int_{\Omega}e^{w}dx)\leq\frac{1}{8\pi}||\nabla w||_{2}^{2}+\frac{1}{|\Omega|}\int_{\Omega}wdx$

for any$w\in H^{1}(\Omega)$$\Omega$$+K$. For each $b>0$ , therefore, we have$\log(\int_{\Omega}e^{bv}dx)\leq\frac{b^{2}}{8\pi}||\nabla v||_{2}^{2}+\frac{b}{|\Omega|}||v||_{1}+K+\log|\Omega|=\frac{b^{2}}{4\pi}W(u, v)$$- \frac{b^{2}}{4\pi}\int_{\Omega}u(\log u-1)dx+\frac{b^{2}}{4\pi}\int_{\Omega}uvdx+\frac{b}{|\Omega|}||v||_{1}+K+\log|\Omega|$

,

while
$b$

$\int_{\Omega}$$uvdx \leq\int_{\Omega}u(\log u-1)dx+\lambda\log(\int_{\Omega}e^{bv}dx)-\lambda\log\lambda+\lambda$

follows from (11). Using these inequalities, we obtain$b(1- \frac{b\lambda}{4\pi})\int_{\Omega}$$uvdx \leq(1-\frac{b^{2}\lambda}{4\pi})\int_{\Omega}u(\log u-1)dx$$+ \lambda\{\frac{b^{2}}{4\pi}\mathcal{W}(u, v)+\frac{b}{|\Omega|}||v||_{1}+K+\log|\Omega|\}-\lambda\log\lambda+\lambda$

Then, taking $0<b$$< \min\{\frac{4\pi}{\lambda}$,$( \frac{4\pi}{\lambda})^{1/2}\}$and $(u, v)=(u(\cdot, t),$.

$v(\cdot, t))$

, we

have
$\lim_{t\uparrow T_{\max}}\int_{\Omega}u(\log u-1)(x, t)dx=+\infty$by (43), (44), and (42). The proof is complete.

B

Proof of Lemma 8

We use several terminologies of the statistical mechanics. First, we have$G(x, x’)>0$ because it is associated with the (D) field, and therefore, (minus)
potential energy is positive for:$\rho\in P(\Omega)$$\mathcal{K}(\rho)=\frac{1}{2}\iint_{\Omega \mathrm{x}\Omega}G(x, x’)\rho(x)\rho(x’)dxdx’>0$.

56
By Proposition 2.1 of [7], we define the entropy functional$\mathcal{E}(\rho)=-\int_{\Omega}\rho(\log\rho-1)dx$,

and obtain$E(s) \equiv\sup_{\rho\in P(\Omega),\mathcal{K}(p)=s}\mathcal{E}(\rho)<+\infty$

for each $s>0$ , and furthermore, this value is attained by some element,, satisfying. Then, Theorem 6.1 of [7] reads;
denoted by$\rho_{s}\in P(\Omega)$$\mathcal{K}(\rho_{s})=s$

Theorem 10 Given .

$s_{k}$$arrow+\infty$, we have$\rho_{s_{\acute{k}}}(x)dxarrow\delta_{x_{\infty}}(dx)$with$x_{\infty}\in\Omega$$*$$\{s_{k}’\}\subset\{s_{k}\}$-weakly insuch that$\Lambda\Lambda(\overline{\Omega})$satisfying $R(x_{\infty})= \sup_{x\in\Omega}R(x)$ .

Our Lemma 8 is an extension, and follows from a similar argument. In
fact, it is easy to see that this lemma is equivalent to the following theorem,
where$\mathcal{E}^{\Delta}(\rho)=E(\mathcal{K}(\rho))-\mathcal{E}(\rho)$

Theorem 11 If.

$\{\rho_{k}\}\subset P(\Omega)satlsf|fies$$. \lim_{karrow\infty}\mathcal{K}(\rho_{k})=+\infty$$\lim_{k^{\wedge}arrow\infty}\mathcal{E}^{\Delta}(\rho_{k})=E_{\infty}^{\Delta}<+\infty$

then we have$\{\rho_{l\hat{\iota}}’\}\subset\{\rho_{k}\}$such that$\rho_{k}’(x)dxarrow\delta_{x_{\infty}}(dx)$with$x_{\infty}\in\Omega$

sa

(45)

,

$*$

-weakly in$\mathcal{M}(\overline{\Omega})$(46)$t^{r}\iota.sfying$$R(x_{\infty}) \geq\sup_{x\in\Omega}R(x)-\frac{1}{4\pi}E_{\infty}^{\Delta}$

.

(45)

57
To prove this theorem, we use a fact obtained in the proof of Lemma 6.2of [7], which is regarded as an improved dual Trudinger-Moser inequality. In
fact, usual dual Trudinger-Moser inequality (22) is represented as$\sup_{\rho\in P(\Omega)}\mathrm{I}_{8\pi}(\rho)<+\infty$

,

where
$\mathrm{I}_{\beta}(\rho)=\mathcal{K}(\rho)+\frac{1}{\beta}\mathcal{E}(\rho)$.

$= \frac{1}{2}\iint_{\Omega \mathrm{x}\Omega}G(x, x’)\rho(x)\rho(x’)dxdx’-\frac{\mathrm{I}}{\beta}\int_{\Omega}\rho(\log\rho-1)dx$

Lemma 12 Each $d>0$ admits $C=C(d)$ such that if $m>0$ , then we
have $=\beta(m)>8\pi$ such thatsatisfying
for any$\int_{A_{1}}\rho dx$

where$A_{1}$$\rho\in P(\Omega)$$\mathrm{I}_{\beta}(\rho)\leq C$$\beta$,$A_{2}\subset\Omega$,$\int_{A_{2}}\rho dx\geq m$,

are measurable sets with dist(Ai,$A_{2}$)$\geq d$.

Now, we give the following.

Proof of Theorem 11: First, we show
(48)$\lim_{karrow\infty}\{1-Q_{k}.(r)\}=0$

for each $0<r<<1$ , where$Q_{k}(r)= \mathrm{s}\mathrm{u}_{\frac{\mathrm{p}}{\Omega}}y\in\int_{\Omega\cap B(y,r)}\rho_{k}(x)dx$

denotes the concentration function of
In fact, definingby$\rho_{k}=\rho_{k}(x)$.

$x_{k}\in\overline{\Omega}$$\int_{\Omega\cap B(x_{k},r/2)}\rho_{k}(x)dx=Q_{k}(r/2)$,

we have$1-Q_{k}(r) \leq 1-\int_{\Omega\ulcorner 1B(x_{k},r)}\rho_{k}(x)dx=\int_{\Omega\backslash B(x_{k},\mathit{7})}\rho_{k}(x)dx$,

58

and therefore,,$\min\{Q_{k}.(r/\underline{9}), 1-Q_{k}(r)\}\leq\min\{\int_{\Omega\cap B(x_{k},r/2)}\rho_{k}(x)dx$

If we apply Lemma 12 forfor each $m>0$ such that$d=r/9_{\sim}$$\int_{\Omega\backslash B(x_{k},r)}\rho_{k}.(x)dx\}$, then we have $C=C(d)$ and $\beta=\beta(m)>$$8\pi$$m \leq\min\{Q_{k}(r/\underline{9}), 1-Q_{k}.(r)\}$$\Rightarrow$$\mathrm{I}_{\beta}(\rho_{k})\leq C$.

Proposition 6.1 of [7], on the other hand, guarantees$(s>>1)$$-8 \pi s-C_{1}\leq E(s)(=\sup_{\rho\in P(\Omega),\mathcal{K}(\rho)=s}\mathcal{E}(\rho))$with a constant$C_{1}$(49), and hence it follows that$\mathrm{I}_{\beta}(\rho_{k})=\mathcal{K}(\rho_{k})+\frac{1}{\beta}\mathcal{E}(\rho_{k})=\mathcal{K}(\rho_{k^{\wedge}})-\frac{1}{\sqrt}\mathcal{E}^{\Delta}(\rho_{k})+\frac{1}{\beta}E(\mathcal{K}(\rho_{k}))$$\geq(1-\frac{8\pi}{\beta})\mathcal{K}(\rho_{k})-\frac{C_{1}}{\beta}-\frac{1}{\beta}\mathcal{E}^{\Delta}(\rho_{k})arrow+\infty$as$karrow\infty$. Prom these relations, we obtain$\lim_{karrow\infty}\min\{Q_{k}(r/-), 1-Q_{k}(r)\}=0$.

and $0<r<<1$ by the standard
Here, we have $Q_{k}.(r)\geq cr^{2}$ for $k=1,2$ ,convering argument, and therefore, (48) follows.
, passing to a subseNext, we show that (46) holds with somequence. In fact, since
is bounded, we have$x_{\infty}\in\overline{\Omega}$$\Omega\subset \mathrm{R}^{2}$$\overline{x_{k}}\equiv\int_{\Omega}x\rho\iota.(x)dxarrow x_{\infty}\in \mathrm{R}^{2}$,

passing to a subsequence. Then, for each $0<r<<1$ , we have
if is large by (48). In this case, it holds that

$1-Q_{k^{n}}(r/\underline{9})\leq r$$k$$| \overline{x_{k}.}-x_{k}|=|\int_{\Omega}(x-x_{k})\rho_{k}(x)dx|\leq\int_{\Omega\cap B(x_{k},r)}.|x-x_{k^{\alpha}}|\rho_{k}.(x)dx$$+ \int_{\Omega\backslash B(x_{k},t)}|x-x_{k}|\rho_{k^{\wedge}}(x)dx\leq r+\dot{\mathrm{d}}\mathrm{i}\mathrm{a}\mathrm{m}\Omega\int_{\Omega\backslash B(x_{k},r/2)}\rho\kappa.(x)dx$,$=r+\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\Omega(1-Q_{k}.(r/9_{\sim}))\leq(1+\mathrm{d}\mathrm{i}\mathrm{a}\ln\Omega)r$

$5\theta$

and therefore,$\lim_{karrow\infty}|\overline{x_{k}}-x_{k}|=0$

In particular, it holds that$x_{\infty}\in\overline{\Omega}$.

. Similarly, we have$| \zeta(x_{k})-\int_{\Omega}\zeta(x)\rho_{k}(x)dx|\leq\int_{\Omega\cap B(ox_{k},r)}|\zeta(x_{k})-\zeta(x)|\rho_{k}(x)dx$$+ \int_{\Omega\backslash B(x_{k},r)}|\zeta(x_{k})-\zeta(x)|\rho_{k}(x)dx=o(1)$for each$\zeta=\zeta(x)\in C(\overline{\Omega})$, and therefore,$\lim_{karrow\infty}|\zeta(x_{k})-\int_{\Omega}\zeta(x)\rho_{k}(x)dx|=0$.

Thus, using$| \zeta(x_{\infty})-\int_{\Omega}\zeta(x)\rho_{k}(x)dx|\leq|\zeta(x_{\infty})-\zeta(\overline{x_{k^{\wedge}}})|$$+| \zeta(\overline{x_{k}})-\zeta(x_{k})|+|\zeta(x_{k})-\int_{\Omega}\zeta(x)\rho_{k}(x)dx|$,

we have$\lim_{karrow\infty}|\zeta(x_{\infty})-\int_{\Omega}\zeta(x)\rho_{k}(x)dx|=0$,

which means (46).
We show (47) and complete the proof. In fact, we have$\mathrm{I}(\rho_{k})=\mathrm{I}_{8\pi}(\rho_{k})=\mathcal{K}(\rho_{k})+\frac{1}{8\pi}\mathcal{E}(\rho_{k})$$= \mathcal{K}(\rho_{k})-\frac{1}{8\pi}\mathcal{E}^{\Delta}(\rho_{k})+\frac{1}{8\pi}E(\mathcal{K}(\rho_{k}))\geq-\frac{1}{8\pi}\mathcal{E}^{\Delta}(\rho_{k})-\frac{C_{1}}{8\pi}$by (49), and therefore,$\lim_{karrow}\inf_{\infty}\mathrm{I}(\rho_{k})>-\infty$

(50)

60
from the assumption. We have$G(x, x’)= \frac{1}{2\pi}\log|x-x’|^{-1}+K(x, x’)$with $R(x)=K(x, x)arrow-\infty$ as$xarrow\partial\Omega$, and also$\mathrm{I}(\rho_{k})=\frac{1}{4\pi}\iint_{\Omega \mathrm{x}\Omega}\log|x-x’|^{-1}\rho_{k}(x)\rho\iota-(x’)dxdx’$

dxdx’$- \frac{1}{8\pi}\int_{\Omega}(\rho_{k}\log\rho_{k})(x)dx+\underline{\frac{1}{9}}\iint_{\Omega \mathrm{x}\Omega}K(x, x’)\rho_{k}(x)\rho_{k}(x’)$$\leq C+\frac{1}{9_{\sim}}\iint_{\Omega \mathrm{x}\Omega}K(x, x’)\rho_{k^{\alpha}}(x)\rho_{k}.(x’)dxdx’$

by the logarithmic Hardy-Littlewood-Sobolev inequality $[8, 3]$ , and therefore,
implies(46) within the case of (50).
Equality (24), on the other hand, implies a sharp form of (49):$x_{\infty}\in\overline{\Omega}$$x_{\infty}\in\Omega$$\lim_{s\uparrow+\infty}\{s+\frac{1}{8\pi}E(s)\}=\sup_{x\in\Omega}I_{\Omega}(x)$forobtain$I_{\Omega}(x)$

,

(51)

defined by (23). (See the proof of Theorem 3.1 of [7].) Then, we$I_{\Omega}(x_{\infty}) \geq\lim_{karrow}\sup_{\infty}\mathrm{I}(\rho_{k}.)=\lim_{karrow}\sup_{\infty}\{\mathcal{K}(\rho_{k}.)+\frac{1}{8\pi}\mathcal{E}(\mathcal{K}(\rho_{k}))-\frac{1}{8\pi}\mathcal{E}^{\Delta}(\rho_{k})\}$$= \lim\sup\{\mathcal{K}(\rho_{k})+\frac{1}{8\pi}\mathcal{E}(\mathcal{K}(\rho_{k}))\}-\frac{1}{8\pi}E_{\infty}^{\Delta}=\sup_{x\in\Omega}I_{\Omega}(x)-\frac{1}{8\pi}E_{\infty}^{\Delta}$by (45). This means (47) and the proof is complete.

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