数理解析研究所講究録
第 1837 巻 2013 年 1-20

1Trudinger-Moser inequality for point vortex mean
field limit with multi-intensities
TAKASHI SUZUKI AND XIAO ZHANG

We study a variational functional associated with point vortex
mean field equation, particularly the extremal case, that is, boundedness of the functional and existence of minimizer.

1

Introduction

In 1949, Onsager [13] used statistical mechanics to describe an ordered structure observed in fluid motion. In the theory of Gibbs, first, the Hamilton
system$\frac{dq_{i}}{dt}=\frac{\partial H}{\partial p_{i}}, \frac{dp_{i}}{dt}=-\frac{\partial H}{\partial q_{i}}, 1\leq i\leq N,$

is introduced in the phase space $x=(q_{1}, \ldots, q_{N},p_{1}, \ldots,p_{N})\in R^{6N}$ . It
induces the micro-canonical measure$d \mu^{H,N}=\frac{1}{\Omega(H)}\cdot\frac{d\Sigma(H)}{|\nabla H|}$

where $d\Sigma(H)$ anddenote the measure on each level set $\{x\in R^{6N}|H(x)=H\}$and the weight factor defined by$\Omega(H)$$dx=dH \cdot\frac{d\Sigma(H)}{|\nabla H|}$

and$\Omega(H)=\int_{H(x)=H\}}\frac{d\Sigma(H)}{|\nabla H|},$

respectively. Then the thermodynamical relation gives the inverse temperature $\beta=1/(k_{B}T)$ by$\beta=\frac{\partial}{\partial H}\log\Omega(H)$2from which emerges the canonical measure$d \mu^{\beta,N}=\frac{e^{-\beta H}dx}{Z(\beta,N)}, Z(\beta, N)=\int_{R^{6N}}e^{-\beta H}dx$

wheredenotes the Boltzmann constant. Then the mean field limt of the, that is, the one point pdf, arises asfactorized density ofunder the principle of equal a priori probabilities.
Onsager [13] used the vorticity equation of Kirchhoff which is derived
from the Euler equation$k_{B}$$N\uparrow+\infty$$d\mu^{\beta,N}$$v_{t}+(v\cdot\nabla)v=-\nabla p,$$\nabla\cdot v=0$

,

in

on

$\nu\cdot v=0$$\Omega\cross(0, T)$$\partial\Omega\cross(0, T)$

,

(1.1)

and
where
is a simply-connected domain with smooth boundaryasIfsoconcentratedis
vector.
denotes the outer unit normal$\Omega\subset R^{2}$$\partial\Omega$$\omega=\nabla\cross v$$\nu$$\omega N(dx, t)=\sum_{i=1}^{N}\alpha_{i}\delta_{x_{i}(t)}(dx)$,equation (1.1) is reduced to$\frac{dx_{i}}{dt}=\nabla_{i}^{\perp}\hat{H}_{N}, 1\leq i\leq N$

for$\hat{H}_{N}(x_{1,\ldots,N}x)=\sum_{i}\frac{\alpha_{i}^{2}}{2}R(x_{j})+\sum_{i<j}\alpha_{i}\alpha_{j}G(x_{i}, x_{j})$

,

where
$\nabla_{i}^{\perp}=(\frac{\partial}{-\partial x}\frac{2\partial}{\partial x_{i1}}) , x_{i}=(x_{i1}, x_{i2})$,$G=G(x, x’)$ denotes the Green’s function, and$R(x)=[G(x, x’)+ \frac{1}{2\pi}\log|x-x’|]_{x=x}$, the equationis the Robin function. For the single intensity caseto which mean field limit of the canonical measure is subject is derived bywith[5, 12]. Namely, it arises in the high energy limit,$\alpha_{i}=\hat{\alpha}$$N\uparrow+\infty$$\hat{\alpha}N=1,\hat{H}_{N}=H,\hat{\alpha}^{2}N\hat{\beta}=\beta,$

3and the one-point pdf takes the limit satisfying.

$\rho=\frac{e^{-\beta\psi}}{\int_{\Omega}e^{-\beta\psi}}, \psi=\int_{\Omega}G(\cdot, x’)\rho(x’)dx’$(1.2)

The rigorous proof [2, 6] for this limit process is valid if $\lambda=-\beta<8\pi$because of the uniqueness of the solution to (1.2) proven by [18]. Equation(1.2) takes the form of the Boltzmann-Poisson equation$- \Delta v=\frac{\lambda e^{v}}{\int_{\Omega}e^{v}}$in$\Omega,$

$v=0$

on

$\partial\Omega$

(1.3)

.

$\alpha\in[-1,1],$
If the distribution of the vortices of the intensity
is subject to the Borel probability measure, then (1.3) is replaced by$N\hat{\alpha}=1,$$\alpha\hat{\alpha},$$P(d\alpha)$$\underline{\alpha e^{\alpha v}}P(d\alpha)$$- \triangle v=\lambda\int_{[-1,1]}\int_{\Omega}e^{\alpha v}$

in$\Omega,$

$v=0$

on

$\partial\Omega.$(1.4)

It is the point vortex mean field equation for the case of multi-intensities.
A formal derivation of this deterministic distribution is done in [16], but
Onsager himself has left a note where (1.4) is shown for the discrete case
(1.5)

$P(d \alpha)=\sum_{i=1}^{\ell}n^{i}\delta_{\alpha_{i}}$(see [3]).
A different model derived by [8] is the stochastic case where relative
intensity $\alpha\in[-1,1]$ is a random variable subject to the distribution function$P(d\alpha)$ . Then it
follows that$- \triangle v=\lambda\frac{\int_{[-1,1]}\alpha e^{\alpha v}P(d\alpha)}{\int_{J-1,1]}\int_{\Omega}e^{\alpha v}P(d\alpha)}, v|_{\partial\Omega}=0$

.

(1.6)

If the intensities are neutral we have$P(d \alpha)=\frac{1}{2}(\delta_{1}+\delta_{-1})$

and then equations (1.4) and (1.6) read$- \triangle v=\frac{\lambda}{2}(\frac{e^{v}}{\int_{\Omega}e^{v}dx}-\frac{e^{-v}}{\int_{\Omega}e^{-v}dx}) , v|_{\partial\Omega}=0$

(1.7)

4
and
$- \Delta v=\frac{\lambda(e^{v}-e^{-v})}{\int_{\Omega}e^{v}+e^{-v}dx}, v|_{\partial\Omega}=0,$respectively.
Equations (1.4) and (1.6) are the Euler-Lagrange equations of the functionals$J_{\lambda}^{d}(v)= \frac{1}{2}\Vert\nabla v\Vert_{2}^{2}-\lambda\int_{1-1,1]}[\log\int_{\Omega}e^{\alpha v}]P(d\alpha)$

and$J_{\lambda}^{s}(v)= \frac{1}{2}\Vert\nabla v\Vert_{2}^{2}-\lambda\log\int_{[-1,1]}[\int_{\Omega}e^{\alpha v}]P(d\alpha)$, respectively. Then the extremal values of for theirdefined forboundedness is a fundamental factor to prescribe the critical state of manystationary point vortices. We study these functionals on$v\in H_{0}^{1}(\Omega)$$\lambda$$E= \{v\in H^{1}(\Omega)|\int_{\Omega}v=0\}$

where is a Riemann surface without boundary.
First, it is obvious that$\Omega$$J_{\lambda}^{d}\geq J_{\lambda}^{s}$by Jensen’s inequality. Next, the Trudinger-Moser-Fontana inequality [4]$\int_{\Omega}e^{4\pi w^{2}}\leq C, \forall w\in E, \Vert\nabla w\Vert_{2}\leq 1$

implies$\inf_{E}J_{8\pi}^{s}>-\infty.$

In fact, we have$\alpha v\leq\frac{1}{16\pi}\Vert\nabla v\Vert_{2}^{2}+4\pi\alpha^{2}\cdot\frac{v^{2}}{\Vert\nabla v\Vert_{2}^{2}}$and hence$\log\int_{[-1,1]}[\int_{\Omega}e^{\alpha v}]P(d\alpha)\leq\frac{\Vert\nabla v\Vert_{2}^{2}}{16\pi}+C, v\in E.$

In fact the value
the following theorem.

$\lambda=8\pi$

is actually the extremal for$J^{s}$

to be bounded by5Theorem 1 ([15]). If$\sup$

it holds that

supp $P=+1$$\inf_{E}J_{\lambda}=-\infty$

for

or

$\inf$

supp $P=-1$(1.8)$\lambda>8\pi.$Proof. We assume sup supp $P=+1$ without loss of generality. If$\alpha>0$

we

have

$ve^{\alpha v}\geq v, \forall v\in R$

and hence$\frac{d}{d\alpha}\int_{\Omega}e^{\alpha v}=\int_{\Omega}ve^{\alpha v}\geq\int_{\Omega}v=0$

Then it holds that$\log\int_{[-1,1]}[\int_{\Omega}e^{\alpha v}]P(d\alpha) \geq \log\int_{[1-\delta,1]}\int_{\Omega}e^{\alpha v}P(d\alpha)$$\geq \log\int_{\Omega}e^{(1-\delta)v}+\log P[1-\delta, 1]$

for $0<\delta<1$ and $v\in E$ . Writing $w=(1-\delta)v$ , then we obtain$J_{\lambda}^{s}(v) \leq \frac{1}{2}\cdot\frac{1}{(1-\delta)^{2}}\Vert\nabla w\Vert_{2}^{2}-\lambda\log\int_{\Omega}e^{w}+C_{\delta}$$= \frac{1}{(1-\delta)^{2}}\{\frac{1}{2}\Vert\nabla w\Vert_{2}^{2}-\lambda(1-\delta)^{2}\log\int_{\Omega}e^{w}\}+C_{\delta}.$Given
it follows that$\lambda>8\pi$, we have $0<\delta\ll 1$ such that$\tilde{\lambda}=\lambda(1-\delta)^{2}>8\pi$. Then$\inf_{E}J_{\lambda}^{s}=-\infty$

from$\inf_{E}J\frac{0}{\lambda}=-\infty$, where$J_{\lambda}^{0}(v)= \frac{1}{2}\Vert\nabla v\Vert_{2}^{2}-\lambda\log\int_{\Omega}e^{v}$

(1.9)
口

Now we turn to the extremal value for$J^{d}$

defined by$\lambda_{*}=\sup\{\lambda|\inf_{E}J_{\lambda}^{d}>-\infty\}.$6
If$\lambda<\lambda_{*}$

then$J_{\lambda}^{d}$takes minimizer on$E$which solves$- \triangle v=\lambda\int_{[-1,1]}\alpha[\frac{e^{\alpha v}}{\int_{\Omega}e^{\alpha v}}-\frac{1}{|\Omega|}]P(d\alpha) , \int_{\Omega}v=0$(1.10).

on does not exist, there will be a formation of
If the minimizer ofsingularity of ground states at the critical level of negative inverse temper. Furthermore, the profile of its singularity is associated withature
the boundedness of the extremal functional indicated by$J_{\lambda_{*}}^{d}$$E$$\lambda=\lambda_{*}$$\inf_{E}J_{\lambda_{*}}^{d}>-\infty$(1.11).

Thus we are addressed by three questions at this moment; prescribing theexact value , boundedness of the extremal functional (1.11), and the exison $E$ . In fact, Ohtsuka-Suzuki
tence or non-existence of the minimizer offor the neutral case (1.7).
[10] showed
In 2010, Ohtsuka-Ricciardi-Suzuki [9] prescribed the profile of singular
limits of the solution to (1.10), and derived a rough estimate,$\lambda_{*}$$J_{\lambda_{*}}^{d}$$\lambda_{*}=16\pi$$\lambda_{*}\geq\inf\{\frac{8\pi}{\int_{[-1,0]}\alpha^{2}P(d\alpha)}, \frac{8\pi}{\int_{[0,1]}\alpha^{2}P(d\alpha)}\}.$The exact value of , however, had been obtained for the discrete case(1.5) by [17], represented in the dual form (see [19]). Taking the limit ofthis inequality, we obtain the following theorem.
$\lambda_{*}$

Theorem 2 ([14]). Under the assumption of (1.8) it holds that$\lambda_{*}=\inf\{\frac{8\pi P(K_{\pm})}{[\int_{K\pm}\alpha P(d\alpha)]^{2}}|K\pm\subset I_{\pm}\cap suppP\}$,(1.12)

where $I+=[0,1]$ and $I_{-}=[-1,0].$ofand the minimizer
To approach (1.11), here we take$\{v_{k}\}\subset
E$
if(1.10)
andtois a solution. Thisis compact, then we have (1.11) with a minimizer. If this is not the case weapply [9] to get the following lemma.
$\lambda_{k}\uparrow\lambda_{*}$$\inf_{E}J_{\lambda_{k}}^{d}$$v_{k}$$(v, \lambda)=(v_{k}, \lambda_{k})$Lemma 1. If the above
subsequence we obtain$\{v_{k}\}\subset E$

is non-compact, then passing to a$\frac{\lambda_{k}e^{\alpha v}k}{\int_{\Omega^{e^{\alpha v}}}k}dxP(d\alpha)arrow\mu(dxP(d\alpha))$$in$ $\mathcal{M}(\Omega\cross[-1,1])$

7
for$\mu(dxP(d\alpha))=[\sum_{x_{0}\in S}m(x_{0}, \alpha)\delta_{x0}(dx)+r(x, \alpha)dx]P(d\alpha)$,(1.13)

where$m(x_{0}, \alpha)\geq 0, 0\leq r=r(x, \alpha)\in L^{1}(\Omega\cross[-1,1], dxP(d\alpha)$

and $S=S+\cup S$ -with$s_{\pm}=\{x_{0}|\exists x_{k}arrow x_{0}s.t.

v_{k}(x_{k})arrow\pm\infty\}$with $\# S<+\infty$ . Furthermore, it holds that
$8 \pi\int_{[-1,1]}m(x_{0}, \alpha)P(d\alpha)=\{\int_{[-1,1]}\alpha m(x_{0}, \alpha)P(d\alpha)\}^{2}$$4\pi\leq n\pm(x_{0})=l_{\pm}|\alpha|m(x_{0}, \alpha)P(d\alpha) , \forall x_{0}\in s_{\pm}$.

(1.14)(1.15)

Henceforth, we assume the non-compactness of the above $\{vk\}\subset E$
although the property described in Lemma 1 is valid to any non-compact
solution sequence to (1.10). If $r=0$ we say that the residual vanishingoccurs to (1.13). Then we obtain the following lemma.

Lemma 2 ([20]). Let $P(d\alpha)$ be non-atomic, supp $P\subset I+,$$\sup\{\alpha\in I+|P([0, \alpha))=0\}>\frac{1}{2}l_{+}\alpha P(d\alpha)$,and$\frac{1}{(\int_{I+}\alpha P(d\alpha))^{2}}<\frac{P(K_{+})}{(\int_{K+}\alpha P(d\alpha))^{2}}$suppP satisfying $K+\neq I+,$ $P(K_{+})<1$ . Then it follows
that (1.11) under the assumption of the residual vanishing of $\{v_{k}\}\subset E$
defined above.

for any$K+\subset I+\cap$

The propery (1.11) is valid for the discrete case (1.5) (see [17]). Hence
there may be the other approach of evaluating its bound uniformly. Here
we note that this approach was successful for the sub-critical case [14], and8also that any counter example to (1.11) has not yet be known. Thus there
may be a chance for (1.11) to be proven by a limit process similar to thesub-critical case. Actually we expect (1.11) for all cases.
In contrast, the argument taken by this paper may have an advantageof picking up the case of the existence of minimizers. More precisely, if we, thenget a contradiction from the non-compactness of the aboveon . So far, the argument employed here
there must be a minimizer to. We have,guarantees (1.11) for both clustered and separated cases offurthermore, the existence of minimizer in the latter case. This property
defined by (1.9), which may bearises even under slight perturbations ofitself does not always take any minimizers on $E.$
surprising because
This paper is composed of three sections. In \S 2 we study the residualvanishing and related properties. Then the notion of partially compact isintroduced and studied in \S 3.
$\{v_{k}\}\subset E$$J_{\lambda_{*}}^{s}$$E$$P(d\alpha)$$J_{8\pi}^{0}$$J_{8\pi}^{0}$

2

Residual Vanishing

The proof of the following fact may be useful to observe the role of residualvanishing for (1.11) to be valid.
Proposition 1. Let
in the previous section withand define the sequence. Then it holds that

$P(d\alpha)=\delta_{1}$$\lambda_{k}\uparrow\lambda_{*}$$J_{\lambda_{k}}^{d}(v_{k})=O(1)$Proof. We have$\lambda_{*}=8\pi$.

and$- \Delta v_{k}=\lambda_{k}(\frac{e^{v_{k}}}{\int_{\Omega}e^{v_{k}}}-\frac{1}{|\Omega|}) , \int_{\Omega}v_{k}=0.$

Since $\# S=1$ we may assume$v_{k}(0) arrow+\infty, \int_{\Omega}e^{v_{k}}arrow+\infty.$Then$\xi_{kk}=v-\log\int_{\Omega}e^{v_{k}}$satisfies$\int_{\Omega}e^{\xi_{k}}=1, e^{\xi_{k}}arrow\delta_{0}.$$\{v_{k}\}\subset E$as9Y.Y. Li’s estimate [7] now guarantees$| \xi_{k}(X)-\log\frac{e^{\xi_{k}(0)}}{(+e^{\xi_{k}(0)}|X|^{2})^{2}}|+|\xi_{k}(0)+\overline{\xi_{k}}|\leq C$

for $|X|\ll 1$ , where $X$ denotes the iso-thermal chart and$\overline{\xi_{k}}=\frac{1}{|\Omega|}\int_{\Omega}\xi_{k}.$Here we have$\log\int_{\Omega}e^{v_{k}}=-\overline{\xi_{k}}$and also$\Vert\nabla v_{k}\Vert_{2}^{2} = \langle-\Delta v_{k}, v_{k}\rangle=-\lambda_{k}(e^{\xi_{k}}-\frac{1}{|\Omega|}, v_{k})=\lambda_{k}\int_{\Omega}e^{\xi_{k}}v_{k}$$= \lambda_{k}(\int_{\Omega}e^{\xi_{k}}\xi_{k}+\log\int_{\Omega}e^{v_{k}})=\lambda_{k}(\int_{\Omega}e^{\xi_{k}}\xi_{k}-\overline{\xi_{k}})$,which implies$\frac{2}{\lambda_{k}}J_{\lambda_{k}}(v_{k}) = \int_{\Omega}\xi_{k}e^{\xi_{k}}+\overline{\xi_{k}}$$= \int_{\Omega}(\xi_{k}-\xi_{k}(0))e^{\xi_{k}}+(\xi_{k}(0)+\overline{\xi_{k}})=O(1)$.
口

The next observation is the following lemma. It shows what is emerged
from the residual vanishing if
is one-sided.
$P(d\alpha)$

Lemma 3. Assume supp $P\subset I+and$ the residual vanishing fordefined in the previous section. Then it holds that $\# S=1$ and (1.12) isattained by $K_{+}=I_{+}$ , that is,$\{v_{k}\}\subset$

$E$

(2.1)

$\lambda_{*}=\frac{8\pi}{(\int_{I_{+}}\alpha P(d\alpha))^{2}}$Proof. By (1.13) with $r=0$ we have$\lambda_{*}=\sum_{xo\in S}m(x_{0}, \alpha)$

,

$P$

-a.e.

$\alpha,$10
while the first equality of (1.14) reads$8\pi l_{+}m(x_{0}, \alpha)P(d\alpha)=\{l_{+}\alpha m(x_{0}, \alpha)P(d\alpha)\}^{2}$$\forall x_{0}\in S$(2.2)

by $suppP\subset I+\cdot$ Then it holds that
$8 \pi\lambda_{*} = \sum_{xo\in S}\{\int_{+}\alpha m(x_{0}, \alpha)P(d\alpha)\}$$\leq \{l\sum_{+xo\in S}\alpha m(x_{0}, \alpha)P(d\alpha)\}^{2}=\{l_{+}\alpha\lambda_{*}P(d\alpha)\}^{2}$(2.3)

and hence$\lambda_{*}\geq\frac{8\pi}{\{\int_{I_{+}}\alpha P(d\alpha)\}^{2}}.$

Therefore, (1.12) is attained for $K+=I+$ and the equality is valid in (2.3)
口which means $\# S=1$ .

The following lemma is useful to ensure the residual vanishing.
Lemma 4. Given a relatively open set denoted by$r=0, dxP(d\alpha)-a.e.

if and only if any$karrow\infty$, we haveon\Omega\cross I_{0}$(2.4)

admits $\{k’\}\subset\{k\}$ such that
$P$ - $a$$\int_{\Omega}e^{\alpha v_{k}’}arrow+\infty,$

Proof. First, assume (2.5), and take$\langle\psi,$$I_{0}\subset I$

. .
$e$

$\alpha\in I_{0}$$\psi\in C(\Omega\backslash S)$$\frac{e^{\alpha v_{k}’}}{\int_{\Omega}e^{\alpha v_{k}’}}\ranglearrow 0,$

$P$

-a.e.

.

(2.5)

. Then it holds that$\alpha\in I_{0}.$Here we have$\frac{1}{|\Omega|}\int_{\Omega}e^{\alpha v_{k}’}\geq\exp(\frac{1}{|\Omega|}\int_{\Omega}\alpha v_{k}’)=1$(2.6)$l \varphi\langle\psi, \frac{e^{\alpha v_{k}’}}{\int_{\Omega}e^{\alpha v_{k}’}}\rangle P(d\alpha)arrow 0$(2.7)

and hence

11
by the dominated convergence theorem, where
it follows that$\varphi\in C_{0}(I_{0})$is arbitrary. Then$l\varphi\langle\psi, r(\cdot, \alpha)\rangle P(d\alpha)=0$

from (1.13), which implies (2.4).
If (2.4) is the case, conversely, it holds that$l \varphi\langle\psi, \frac{e^{\alpha v_{k}}}{\int_{\Omega}e^{\alpha v_{k}}}\rangle P(d\alpha)arrow 0$

for any
we obtain$0\leq\psi\in C(\Omega\backslash S)$$\langle\psi,$

and $0\leq\varphi\in C_{0}(I_{0})$ . Passing to a sub-sequence,$P$$\frac{e^{\alpha v_{k}}}{\int_{\Omega}e^{\alpha v_{k}}}\ranglearrow 0,$-a.e.

$\alpha\in$んby a diagonal argument. Here the elliptic regularity to (1.10) combined with
(2.6) guarantees, whereis an open set. Hence$\Vert v_{k}\Vert_{L\infty(\omega)}=O(1)$$\omega\subset\Omega\backslash S$$\int_{\Omega}e^{\alpha v_{k}}arrow+\infty,$

$P$

-a.e.

$\alpha\in I_{0}$for this subsequence and the proof is complete.
$Now$

口

we show the following theorem.

Theorem 3. If supp $P\subset I+thenr(\cdot, \alpha)=0a.e$ . in for $\alpha>1/2$ . In
particular, the residual vanishing occurs todefined in the previous$P\subset(1/2,1].$
section, provided that supp$\Omega$$\{v_{k}\}\subset E$

Proof. We shall show$\int_{\Omega}e^{\alpha v_{k}}arrow+\infty, \forall\alpha>1/2$

.

(2.8)

In fact, (2.2) implies$l_{+}\alpha m(x_{0}, \alpha)P(d\alpha)\geq 8\pi, \forall x_{0}\in \mathcal{S}$and the right-hand side of (1.10) for$(\lambda, v)=(\lambda_{k}, v_{k})$takes the limit$l_{+} \alpha\mu(dxP(d\alpha))-\frac{\lambda}{|\Omega|}*l_{+}\alpha P(d\alpha)$

(2.9)

12
in

$\mathcal{M}(\Omega)$. Here we have$l_{+} \alpha\mu(dxP(d\alpha))\geq\sum_{xo\in S}l_{+}\alpha m(x_{0}, \alpha)\delta_{x0}(dx)\geq 8\pi\sum_{xo\in S}\delta_{x0}(dx)$by (2.9). Since$\# S\neq\emptyset$, equation (1.10) implies (2.8) by an argument usedin [1].
$E$

口

We conclude this section with the following examples. Henceforth, $vk\in$such that
denotes the minimizer of$\lambda_{k}\uparrow\lambda_{*}.$$J_{\lambda_{k}}^{s}$

Example 1.

$P= \frac{1}{2}(\delta_{1}+\delta_{\gamma}),$$0<\gamma<1.$

Since$\frac{8\pi P(K_{+})}{\{\int_{\kappa_{+}}\alpha P(d\alpha)\}^{2}}=\{\begin{array}{ll}\frac{32\pi}{(1+\gamma)^{2}}, K+=\{1, \gamma\}1^{\cdot}6\pi, K_{+}=\{1\}\frac{1}{\gamma}z, K_{+}=\{\gamma\}\end{array}$it holds that
(2.10)$\lambda_{*}=\inf\{16\pi, \frac{16\pi}{\gamma^{2}}, \frac{32\pi}{(1+\gamma)^{2}}\}=\{\begin{array}{ll}16\pi, \gamma<\sqrt{2}-1\frac{32\pi}{(1+\gamma)^{2}}, \gamma\geq\sqrt{2}-1.\end{array}$

Therefore, the residual vanishing does not occur for $\gamma<\sqrt{2}-1$ by Lemma
3. On the contrary, we have the residual vanishing if $\gamma>1/2$ by Theorem
3. Next, (1.13)imphes$\lambda_{*l_{+}\varphi P(d\alpha)}=l_{+}[\int_{\Omega}r(x, \alpha)dx+\sum_{xo\in S}m(x_{0}, \alpha)]\varphi(\alpha)P(d\alpha)$(2.11). Regarding $suppP=\{1, \gamma\}$ , we put $m_{\alpha}(x_{0})=m(x_{0}, \alpha)$
for any
for $\alpha=1,$ . Then we obtain$\varphi\in C(I_{+})$$\gamma$(2.12)$\lambda_{*}\geq\sum_{xo\in S}m_{1}(x_{0}),\sum_{xo\in S}m_{\gamma}(x_{0})$Equality (1.14), on the other hand, is reduced to (2.2), which means$16\pi(m_{1}(x_{0})+m_{\gamma}(x_{0}))=(m_{1}(x_{0})+\gamma m_{\gamma}(x_{0}))^{2},$$\forall x_{0}\in S$.

(2.13)forBy (2.12)-(2.13) we can conclude $\# S=1$ . We put$\gamma>\sqrt{2}-1$
one
only$\alpha=1,$
is
.
There
, then. If$m_{1}>0$ , satisfying (2.12) and (2.13), that is,withpair of$m_{\alpha}=m_{\alpha}(x_{0})$$x_{0}\in \mathcal{S},$$\lambda_{*}=\frac{32\pi}{(1+\gamma)^{2}}<16\pi$$\gamma$$(m_{\gamma}, m_{1})$$m_{\gamma},$$m_{1}=m_{\gamma}= \frac{32\pi}{(1+\gamma)^{2}}$.

(2.14)

13
Then equalities arise in both inequalities in (2.12), which implies
a.e. for $\alpha=1,$ . Thus we obtain the residual vanishing.
If $\gamma\leq\sqrt{2}-1$ then we have
. If $\gamma=\sqrt{2}-1$ , there arise thecases of $(m_{\gamma}, m_{1})=(16\pi, 16\pi)$ and $(m_{\gamma}, m_{1})=(0,16\pi)$ from (2.12)-(2.13).
In the former case we have the residual vanishing, while in the latter case we
do not havea.e. any more. We may call it mass separation,$\gamma<\sqrt{2}-1$, only $(m_{\gamma}, m_{1})=(0,16\pi)$ satisfies (2.12)regarding. If(2.13). Hence we always have non-residual vanishing and mass separation.
, we take $0<R\ll 1$ such that
Assuming$r(\cdot, \alpha)=0$$\gamma$$\lambda_{*}=16\pi$$r_{\gamma}\equiv r(\cdot, \gamma)=0$$m_{\gamma}=0$$m_{\gamma}=0$$\int_{S_{R}}r_{\gamma}<4\pi, \mathcal{S}_{R}=\bigcup_{x_{0}\in S}B(x_{0}, R)$and define$v_{k}^{\gamma}=v_{k}^{\gamma}(x)$by$- \triangle v_{k}^{\gamma}=\frac{\lambda_{k}}{2}(\frac{e^{\gamma v_{k}}}{\int_{\Omega}e^{\gamma v_{k}}}-\frac{1}{|\Omega|}) , \int_{\Omega}v_{k}^{\gamma}=0.$Then it holds that
1. Now,$\Vert v_{k}^{\gamma}\Vert_{\infty}\leq C$$v_{k}^{1}=v_{k}-v_{k}^{\gamma}$

by Brezis-Merle’s inequality [1] and Lemma

satisfies$- \triangle v_{k}^{1}=\frac{\lambda_{k}}{2}(\frac{V_{k}e^{v_{k}^{1}}}{\int_{\Omega}V_{k}e^{v_{k}^{1}}}\cdot-\frac{1}{|\Omega|}) , \int_{\Omega}v_{k}^{1}=0$,. We have readily shown thatis compact in$0<\alpha<1$ , and. In particular, it holds that
with
for
for

$V_{k}=e^{v_{k}^{\gamma}}>0$$\{v_{k}^{\gamma}\}$$\lambda_{k}\uparrow 16\pi$$C^{2,\alpha}(\Omega)$$\Vert v_{k}^{1}\Vert_{\infty}\uparrow+\infty$$J_{\lambda_{k}}^{d}(vk)=\hat{J}_{k}(v_{k}^{1})+O(1)$$\hat{J}_{k}(v)=\frac{1}{2}\Vert\nabla v\Vert_{2}^{2}-\frac{\lambda_{k}}{2}\log\int_{\Omega}V_{k}e^{v}$

Here Y.Y. Li’s estimate is available even for this case of variable coefficients,
which implies
similarly to Proposition 1. Hence it holds that
(1.11).
Summing up, if $\gamma>\sqrt{2}-1$ then we have the residual vanishing wherea refined version of Lemma 2 is expected to apply to guarantee (1.11). If$\gamma<\sqrt{2}-1$
there arises mass separtion, and then (1.11) by a modification of
Proposition 1. Although the case $\gamma=\sqrt{2}-1$ has not yet been settled down,
the above study may suggest the following. First, if $P(d\alpha)$ is sufficiently
separated aroud $\alpha=1$ and $\alpha=0$ , mass separation and consequently nonresidual vanishig occur. Then the property (1.11) is reduced to the case that$\hat{J}_{k}(v_{k}^{1})=O(1)$14
is clustered near $\alpha=1$ thenis clustered near $\alpha=1$ . Second, if
the residual vanishing occurs, which will make Lemma 2 available. Actually,
the proof of Lemma 2 is based on a kind of Y.Y. Li’s estimate.
We have to note, however, that the weight of two delta functions are
fixed here. Actually, if the positions of two delta functions are sufficiently
clustered and their weights are concentrated at $\alpha=1$ , then we have ais attained.
different phenomena, which guarantees that
$P(d\alpha)$$P(d\alpha)$$J_{\lambda_{*}}^{d}$

Example 2.

$P=\tau\delta_{1}+(1-\tau)\delta_{\gamma},$$0<\gamma<1,0<\tau<1.$

We shall follow the notations used in the previous example. First observation is that$\frac{8\pi}{\frac{b_{\pi}^{\tau}}{\tau}+(1-\tau)\gamma)^{2}}, K_{+}=\{1, \gamma\}$$K_{+}=\{1\}$$\frac{8\pi P(K_{+})}{\{\int_{\kappa_{+}}\alpha P(d\alpha)\}^{2}}=\{$$\frac{8\pi}{(1-\tau)\gamma^{2}}, K+=\{\gamma\}$

implies$\lambda_{*}=\{\frac{\frac{8\pi}{\tau},8\pi}{(\tau+(1-\tau)\gamma)^{2}}, \gamma>\frac{}{}\gamma<\frac{\sqrt{\tau}}{1+,1+\sqrt{\tau}\mathcal{F}_{\tau}^{\mathcal{T}}},$

except for the critical case
(2.13) is replaced by$\gamma=\frac{\sqrt{\tau}}{1+\sqrt{\tau}}$. Inequality (2.12) is still valid, while$8\pi(\tau m_{1}(x_{0})+(1-\tau)m_{\gamma}(x_{0}))=(\tau m_{1}(x_{0})+\gamma(1-\tau)m_{\gamma}(x_{0}))^{2},$Then we can confirm $\# S=1$ similarly.
Treating the separative caseplane crosses the curvein$\gamma<\frac{\sqrt{\tau}}{1+\sqrt{\tau}}$$8\pi/\tau$$\forall x_{0}\in S.$, we observe that the line $m_{1}=$$m_{\gamma}m_{1}$$8\pi(\tau m_{1}+(1-\tau)m_{\gamma})=(\tau m_{1}+\gamma(1-\tau)m_{\gamma})^{2}$

at $m_{\gamma}=0$ and$\gamma<\frac{\sqrt{\tau}}{1+\sqrt{\tau}}$$m_{\gamma}= \frac{1-2\gamma}{\gamma^{2}(1-\tau)}.$$8\pi$(2.15), recalling that $\gamma<1/2$ follows from

. Since$\lambda_{*}\geq m_{1}, m_{\gamma}$, i.e.,
we have mass separation, $m_{\gamma}=0$ , provided that
. In such a case, (1.11) is reduced to the boundedness$\frac{1}{\tau}<\frac{1-2\gamma}{\gamma^{2}(1-\tau)}$$-\sigma+\sqrt{\sigma^{2}+\sigma},$of$\tilde{J}_{k}(v_{k})$$\gamma<$$\sigma=\frac{\tau}{1-\tau}$, where$\tilde{v}_{k}\in E,$$\Vert\tilde{v}_{k}\Vert_{\infty}arrow+\infty,$$\tilde{J}_{k}(v)=\frac{1}{2}\Vert\nabla v\Vert_{2}^{2}-\mu_{k}\log\int_{\Omega}V_{k}e^{v}$(2.16)

15$0<\alpha<1$ . Thiswithcompact inproperty is actually the case by the proof of Proposition 1. Hence (1.11)
arises if
is in the above region.
In the clustered case$\mu_{k}\uparrow\tau\lambda_{*}=8\pi,$$V_{k}=e^{v_{k}^{1}}$$C^{2,\alpha}(\Omega),$$\{v_{k}^{1}\}$$(\gamma, \tau)$$\gamma>\frac{\sqrt{\tau}}{1+\sqrt{\tau}}$

it holds thatcrosses the line
denoted by. If$\lambda_{*}=\frac{8\pi}{(\tau+(1-\tau)\gamma)^{2}}$$m_{1}=\lambda_{*}$(2.17),. In this case the curve (2.15) inonce, with the$m_{\gamma}$

plane-component of the crossing point$m_{\gamma}m_{1}$$m_{\gamma}^{*}$(2.18)$(1-\tau)m_{\gamma}^{*}<4\pi$then we apply Brezis-Merle’s inequality as in Example 1. We obtain$0<\alpha<1$ , and$V_{k}>0$ , compact insatisfying$\mu_{k}\uparrow$$\tau\lambda_{*},$$\{V_{k}\},$$C^{2,\alpha}(\Omega),$$\tilde{v}_{k}\in E$$- \triangle\tilde{v}_{k}=\mu_{k}(\frac{V_{k}e^{\overline{v}_{k}}}{\int_{\Omega}V_{k}e^{\tilde{v}_{k}}}-\frac{1}{|\Omega|}) , \int_{\Omega}\tilde{v}_{k}=0.$

Since, however, thisis compact. Therefore, so is true
for the sequence
defined in the previous section. Henceis attained.
Finally, we shall show that (2.17) with (2.18) actually arises in the caseof $0<1-\tau\ll 1$ and $1/2<\gamma<1$ . First, given $1/2<\gamma<1$ , we have (2.17)
for $0<1-\tau\ll 1$ . Next, plugging
into (2.15), we obtain$\tau\lambda_{*}<8\pi$$\{\tilde{v}_{k}\}\subset E$$\{v_{k}\}\subset E$$\inf_{E}J_{\lambda_{*}}^{d}$$m_{1}=\lambda_{*}$$8 \pi\{\frac{8\pi\tau}{(\tau+(1-\tau)\gamma)^{2}}+(1-\tau)m_{\gamma}^{*}\}$$= \{\frac{8\pi\tau}{(\tau+(1-\tau)\gamma)^{2}}+\gamma(1-\tau)m_{\gamma}^{*}\}^{2}$

which implies$\frac{64\pi^{2_{\mathcal{T}}}}{(\tau+(1-\tau)\gamma)^{4}}\{-\tau+2\gamma+(1-\tau)\gamma^{2}\}$$= \frac{16\pi\gamma m_{\gamma}^{*}}{(\tau+(1-\tau)\gamma)^{2}}+\gamma^{2}(1-\tau)(m_{\gamma}^{*})^{2}$

Then it follows that$\lim_{\tau\uparrow 1}m_{\gamma}^{*}=4\pi\cdot\frac{2\gamma-1}{\gamma}$

for each $1/2<\gamma<1$ and hence (2.18) for $0<1-\tau\ll 1.$

16
3

Partially Compact

If $P$ is divided into two parts, and one of its total collapse mass is lessthanthen (1.11) is reduced to that of the other part. We call such acase the partially compact. It is obvious that mass separation implies both
non-residual vanishing and partially compact. This section is devoted to thegeneral criterion for blowup vamishing to occur. We deal with the cases of, individually.
one-sided and changing-sign
The first theorem is just a generalization of Example 2.
$4\pi$$P(d\alpha)$

Theorem 4. Let $P=\tau P_{\beta}+(1-\tau)P_{\gamma}$ , where $0<\tau<1_{i}0<\gamma<\beta<1,$, respectively.
andandare Borel probability measures on$0<1-\tau\ll
1.$
is attained, provided thatIf $1/2<\gamma<1$ then$andP_{\beta}$$[0, \gamma]$$P_{\gamma}$$[\beta, 1]$$\inf_{E}J_{\lambda_{*}}^{d}$

Proof. Assume the contrary, and let$\{vk\}\subset E$

be the non-compact sequence

defined in \S 1. Then it holds that(31)$\frac{8\pi}{\{\tau\int_{[\beta,1]}\alpha P_{\beta}(d\alpha)+(1-\tau)\int_{[0,\gamma]}\alpha P_{\gamma}(d\alpha)\}^{2}}\geq\lambda_{*}$$\lambda_{*}\geq\sum_{x_{0}\in S}\int_{[\beta,1]}m(x_{0}, \alpha)P_{\beta}(d\alpha),\sum_{x_{0}\in \mathcal{S}}\int_{[0,\gamma]}m(x_{0}, \alpha)P_{\gamma}(d\alpha)$

(32)

and
$8 \pi\{\tau\int_{[\beta,1]}m(x_{0}, \alpha)P_{\beta}(d\alpha)+(1-\tau)\int_{[0,\gamma]}m(x_{0}, \alpha)P_{\gamma}(d\alpha)\}$$= \{\tau\int_{[\beta,1]}\alpha m(x_{0}, \alpha)P_{\beta}(d\alpha)+(1-\tau)\int_{[0,\gamma]}\alpha m(x_{0}, \alpha)P_{\gamma}(d\alpha)\}^{2}(3.3)$

for each $x_{0}\in S$ . Fix $x_{0}\in S$ , and put$X= \int_{[0,\gamma]}m(x_{0}, \alpha)P_{\gamma}(d\alpha) , Y=\int_{[\beta,1]}m(x_{0}, \alpha)P_{\beta}(d\alpha)$

As we have seen, if $X<4\pi$ andis attained.
hence
ifFirst,$\tau\lambda_{*}<8\pi$.

, there is a contradiction, and$\inf_{E}J_{\lambda_{*}}^{d}$$\tau\lambda_{*}<8\pi$$\frac{\tau}{(\tau+(1-\tau)\gamma)^{2}}<1$

by (3.1). Here, (3.4) means (2.17). Next, (3.2) and (3.3)imply$8\pi(\tau Y+(1-\tau)X)\leq(\tau Y+(1-\tau)\gamma X)^{2}, X, Y\leq\lambda_{*}.$

(3.4)

17
Since

$\tau\lambda_{*}<8\pi$

is achieved, $X$ is uniquely determined as$8\pi(\tau Y+(1-\tau)X)=(\tau Y+(1-\tau)\gamma X)^{2}, Y=\lambda_{*}.$

Hence bothand $X<4\pi$ is achieved if $1/2<\gamma<1$ is given and$0<1-\tau\ll 1$ as in Example 2.
口$\tau\lambda_{*}<8\pi$

The next theorem is concerned with the changing-sign case, where (1.15)
is used.

Theorem 5. If$\inf\{\frac{P(K_{\pm})}{\{\int_{K\pm}\alpha P(d\alpha)\}^{2}}|K\pm\subset I\pm\cap$

for $c_{-}=1$ andProof. Fix$c+=1+ \frac{\sqrt{5}}{2}$$x_{0}\in S_{-}$

suppP

then it holds that}$\cdot l_{\pm}|\alpha|P(d\alpha)<c\pm$(3.5)$S_{-}=\emptyset.$, and put$x_{\pm}=l_{\pm}|\alpha|m(x_{0}, \alpha)P(d\alpha)\leq l_{\pm}m(x_{0}, \alpha)P(d\alpha)=Y\pm\cdot$

First, we have$\lambda_{*}\geq m(x_{0}, \alpha)$, $P$-a.e.

$\alpha$, and therefore,$x_{\pm}\leq\lambda_{*l_{+}|\alpha|P(d\alpha)}$$=8 \pi\cdot\inf\{\frac{P(K_{\pm})}{\{\int_{K\pm}\alpha P(d\alpha)\}^{2}}|K\pm\subset I\pm\cap suppP\}$$l_{圭}|\alpha|P(d\alpha)$

.

(3.6)

Next, (1. 14) implies$\{\int_{[-1,1]}\alpha m(x_{0}, \alpha)P(d\alpha)\}^{2}=(X_{+}-X_{-})^{2}$$=8 \pi\int_{[-1,1]}m(x_{0}, \alpha)P(d\alpha)=8\pi(Y_{+}+Y_{-})$$\geq 8\pi(X_{+}+X_{-})^{2}$

(3.7)

18
Here we have

(3.8)

$4\pi\leq X_{-}<8\pi$by (1.15), (3.6), and (3.5) with $c-=1$ . Then (3.7) implies$x_{+}\geq 4(2+\sqrt{5})\pi$(see [11]), which contradicts (3.6) and (3.5) with $c+=4(2+\sqrt{5})\pi.$$\square$

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19[11] H. Ohtsuka and T. Suzuki, blowup analysis to the mean field equation
for arbitrarily-signed vortex, Banach Center Publ. 74 (2006) 185-197$A$[12] Y.B. Pointin and T.S. Lundgren, Statistical mechanics of twodimensional vortices in a bounded container, Phys. Fluids 19 (10)(1976) 1459-1470[13] L. Onsager, Statistical hydrodynamics, Suppl. Nuovo Cimento 6 (1949)

279-287[14] T. Ricciardi and T. Suzuki, Duality and best constant for a TrudingerMoser inequality involving probability measures, Euro J. Appl. Anal.
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Division of Mathematical Science
Department of Systems Innovation
Graduate School of Engineering Scinece
Osaka University

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Machimaneyamacho 1-3
Toyonakashi 560-8531, Japan