Proof. Since $G(\cdot,\,\cdot )\in L^1(D\times D)$, we know that $E$ is bounded from above on the set $\mathcal {M}_{\varepsilon }(D)$. Now we choose a maximization sequence $\{\omega ^n\}\subset \mathcal {M}_{\varepsilon }(D)$ of $E$, that is,

\[ \lim_{k\to +\infty}E(\omega^n)=\sup_{\omega \in \mathcal{M}_{\varepsilon}(D)}E(\omega). \]

By direct computations we can prove that $\mathcal {M}_{\varepsilon }$ is a sequentially compact subset of $L^2(D)$ in the weak topology. So we may assume that, up to a subsequence, $\omega ^n\to \omega ^{\varepsilon }$ weakly in $L^2(D)$ as $n\to +\infty$ for some $\omega ^{\varepsilon }\in \mathcal {M}_{\varepsilon }$. So

\[ k(x)\omega^n\to k(x)\omega^{\varepsilon } \quad \text{in}\ L^2(D) \text{weak topology}. \]

By elliptic regularity theory,

\[ \mathcal{G}(k\omega^{n})=\int_{D}G(x,y)\omega^{n}\,{\rm d}\theta(x)\to \int_{D}G(x,y)\omega^{\varepsilon}\,{\rm d}\theta(x)=\mathcal{G}(k \omega^{\varepsilon })\quad \text{in}\ W^{1,p}(D) \]

for any $p>1$, from which we deduce that

\[ \lim_{n\to+\infty}E(\omega^n) =E(\omega^\varepsilon). \]

So $\omega ^{\varepsilon }$ is a maximizer of $E$ over $\mathcal {M}_{\varepsilon }$.

Proof. Let $\omega ^{\varepsilon }$ be a maximizer. For any ${\omega }\in \mathcal {M}_{\varepsilon }$, we set

\[ \omega_{(s)}=\omega^{\varepsilon }+s({\omega}-\omega^{\varepsilon }),\quad \text{for} \ s\in[0,1]. \]

Since $\mathcal {M}_{\varepsilon }$ is a convex set, $\omega _{(s)}\in \mathcal {M}_{\varepsilon }$ for any $s\in [0,\,1]$. So $E(\omega _{(s)})\leq E(\omega ^\varepsilon )$, which implies that

\[ 0 \ge \left.\frac{{\rm d} E(\omega_{(s)})}{{\rm d}s}\right|_{s=0^+} =\int_{D}({\omega}-\omega^{\varepsilon}) \mathcal{G}(k \omega^{\varepsilon}) \,{\rm d}\theta(x), \]

that is,

\[ \int_{D}\omega \mathcal{G}(k \omega^{\varepsilon}) \,{\rm d}\theta(x) \leq \int_{D}\omega^{\varepsilon } \mathcal{G}(k \omega^{\varepsilon}) \,{\rm d}\theta(x) \]

for any ${\omega }\in \mathcal {M}_{\varepsilon }(D).$ By the definition of $\mathcal {M}_{\varepsilon }(D)$ and the bathtub principle (see [Reference Lieb and Loss26]), we get for any $i=1,\,\ldots,\,l$

(3.5)\begin{equation} \begin{aligned} sgn(d_i) \mathcal{G}(k \omega^{\varepsilon}) & \ge \mu_i^{\varepsilon} \quad \mbox{on} \left\{sgn(d_i)\omega_i^{\varepsilon}=\frac{1}{\varepsilon^2}\right\}\cap B_i, \\ sgn(d_i) \mathcal{G}(k \omega^{\varepsilon}) & = \mu_i^{\varepsilon} \quad \mbox{on}\ \left\{0< sgn(d_i)\omega_i^{\varepsilon}<\frac{1}{\varepsilon^2}\right\}\cap B_i, \\ sgn(d_i) \mathcal{G}(k \omega^{\varepsilon}) & \le \mu_i^{\varepsilon} \quad \mbox{on} \{sgn(d_i)\omega_i^{\varepsilon}=0\}\cap B_i, \end{aligned} \end{equation}

where $\mu _i^{\varepsilon }$ is a constant satisfying

(3.6)\begin{equation} \mu_i^{\varepsilon}=\inf\{s\in\mathbb{R}\mid|\{x\in B_i\mid sgn(d_i) \mathcal{G}\left(k \omega^{\varepsilon} \right) >s\}|_{\theta}\le |d_i|\varepsilon^2\}. \end{equation}

Notice that $|\{x\in B_i\mid sgn(d_i) \mathcal {G}(k \omega ^{\varepsilon } ) >s\}|_{\theta }$ means that the $\,{\rm d}\theta (x)-$measure of the set $\{x\in B_i\mid sgn(d_i) \mathcal {G}(k \omega ^{\varepsilon } ) >s\}$. Thus using (3.5), we get that

\[ \omega_i^{\varepsilon}= sgn(d_i)\frac{1}{ \varepsilon^2}\chi_{\{ \psi_i^{\varepsilon}>0 \}\cap B_i }, \]

where $\psi _i^{\varepsilon }=sgn(d_i)\mathcal {G}(k \omega ^{\varepsilon } )-\mu ^{\varepsilon }_i$. So (3.3) is proved.

It remains to prove that $\mu _i^{\varepsilon }\geq C_0$ for some $C_0$. In fact for any $x\in B_i$,

(3.7)\begin{align} sgn(d_i)\mathcal{G}\left( k \omega^{\varepsilon } \right)(x) & =\mathcal{G}(k |\omega_i^{\varepsilon }|)(x)+sgn(d_i)\sum_{j\neq i}\mathcal{G}(k \omega_j^{\varepsilon })(x) \nonumber\\ & \geq{-}\sum_{j\neq i}k_1|d_j|\max_{x\in B_i, y\in B_j}|G(x,y)|. \end{align}

Combining (3.6) with (3.7), we can get (3.4) by letting $C_0=-\sum _{j\neq i}k_1|d_j| \max _{x\in B_i, y\in B_j}|G(x,\,y)|$.

Proof. We choose

\[ \tilde{\omega}^{\varepsilon}=\frac{sgn(d_i)}{\varepsilon^2}\chi_{B_{t(\varepsilon)\varepsilon}(x_{0,i})}+\sum_{k=1,k\neq i}^{l}\omega_k^{\varepsilon}=\sum_{k=1}^{l}\tilde{\omega}_k^{\varepsilon}, \]

where $t(\varepsilon )\in (({1}/{2})\sqrt {{|d_i|}/{\pi k(x_{0,i})}},\, 2\sqrt {{|d_i|}/{\pi k(x_{0,i})}})$ satisfies $\int _{B_i}\tilde {\omega }_i^{\varepsilon }\,{\rm d}\theta (x)=d_i$. Direct calculations show that $t(\varepsilon )$ exists for $\varepsilon$ sufficiently small and $\lim _{\varepsilon \to 0^+}t(\varepsilon )=\sqrt {{|d_i|}/{\pi k(x_{0,i})}}$. Then $\tilde {\omega }^{\varepsilon }\in \mathcal {M}_{\varepsilon }(D)$ and $E(\omega ^\varepsilon )\geq E(\tilde {\omega }^{\varepsilon })$, which implies that

(3.11)\begin{align} E(\omega^\varepsilon)& \geq{-}\frac{1}{4\pi}\int_{D}\int_{D}\ln|{x}-{y}|\tilde{\omega}_i^\varepsilon({x})\tilde{\omega}_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\notag\\ & \quad - \frac{1}{2}\int_{D}\int_{D}h({x},{y})\tilde{\omega}_i^\varepsilon({x})\tilde{\omega}_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y) \nonumber\\ & \quad +E\left(\sum_{j\neq i}\omega_j^{\varepsilon}\right)+O(1). \end{align}

Since the diameter of $supp(\tilde {\omega }_i^\varepsilon )$ is $\sqrt {({|d_i|}/{\pi k(x_{0,i})})}\varepsilon +o(\varepsilon )$, we have

(3.12)\begin{equation} -\frac{1}{4\pi}\int_{D}\int_{D}\ln|{x}-{y}|\tilde{\omega}_i^\varepsilon({x})\tilde{\omega}_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y) \geq \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+O(1). \end{equation}

By the choice of $B_i$ we obtain

(3.13)\begin{equation} \left|\frac{1}{2}\int_{D}\int_{D}h({x},{y})\tilde{\omega}_i^\varepsilon({x})\tilde{\omega}_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y)\right| = O(1), \end{equation}

Taking (3.12), (3.13) into (3.11) and using (3.10), we get the desired result.

Proof. Let us first prove that

(3.15)\begin{equation} sgn(d_i)\int_{D}(sgn(d_i)\mathcal{G}(k\omega^\varepsilon)-\mu_i^{\varepsilon})\omega_i^\varepsilon \,{\rm d}\theta(x)=O(1). \end{equation}

Using the definition of $B_i$ we get

(3.16)\begin{align} & sgn(d_i)\int_{D}(sgn(d_i)\mathcal{G}(k\omega^\varepsilon)-\mu_i^{\varepsilon})\omega_i^\varepsilon \,{\rm d}\theta(x)\notag\\ & \quad =\int_{D}(sgn(d_i)\mathcal{G}(k\omega^\varepsilon)-\mu_i^{\varepsilon}+C_0)|\omega_i^\varepsilon| \,{\rm d}\theta(x)+|C_0d_i| \nonumber\\ & \quad \leq \int_{D}(sgn(d_i)\mathcal{G}(k\omega_i^\varepsilon)-\mu_i^{\varepsilon}+C_0)_+|\omega_i^\varepsilon| \,{\rm d}\theta(x)+ O(1). \end{align}

Define $P^\varepsilon _i=(sgn(d_i)\mathcal {G}(k\omega _i^\varepsilon )-\mu _i^{\varepsilon }+C_0)_+$ and $\bar {P}^\varepsilon _i=(sgn(d_i)\mathcal {G}(k\omega ^\varepsilon )-\mu _i^{\varepsilon }+C_0)_+$. So by (3.4), $P^\varepsilon _i,\,\bar {P}^\varepsilon _i\in H^1_0(D)$. On the one hand, we get

(3.17)\begin{equation} \int_{D}P^\varepsilon_i \omega_i^\varepsilon \,{\rm d}\theta(x)= \int_{D}|\nabla P^\varepsilon_i|^2 \,{\rm d}{x}. \end{equation}

On the other hand, by the choice of $C_0$ we have $supp(\bar {P}^\varepsilon _i)\cap B_i\subseteq supp(\omega _i^\varepsilon ),\,$ which implies that

\begin{align*} \int_{D}P^\varepsilon_i |\omega_i^\varepsilon| \,{\rm d}\theta(x) & \leq \int_{D}\bar{P}^\varepsilon_i |\omega_i^\varepsilon| \,{\rm d}\theta(x)+O(1)\\ & \leq\frac{1}{ \varepsilon^2}\int_{supp(\omega_i^\varepsilon)}\bar{P}^\varepsilon_i \,{\rm d}\theta(x)+O(1)\\ & \leq \frac{k_1}{ \varepsilon^2}|supp(\omega_i^\varepsilon)|^{{1}/{2}}\left(\int_{supp(\omega_i^\varepsilon)}(\bar{P}^\varepsilon_i)^2\,{\rm d}x\right)^{{1}/{2}}+O(1). \end{align*}

Using the continuous embedding $W^{1,1}(B_i)\subset L^2(B_i)$, we have

\begin{align*} & \int_{D}P^\varepsilon_i |\omega_i^\varepsilon| \,{\rm d}\theta(x)\\ & \quad \leq\frac{Ck_1}{ \varepsilon^2}|supp(\omega_i^\varepsilon)|^{{1}/{2}}\left(\int_{B_i}\bar{P}^\varepsilon_i\,{\rm d}{x}+\int_{B_i}|\nabla \bar{P}^\varepsilon_i| \,{\rm d}{x}\right)+O(1)\\ & \quad \leq \frac{Ck_1}{ \varepsilon^2}|supp(\omega_i^\varepsilon)|^{{1}/{2}}\left(\int_{supp(\omega_i^\varepsilon)}P^\varepsilon_i\,{\rm d}{x}+\int_{supp(\omega_i^\varepsilon)}|\nabla P^\varepsilon_i| \,{\rm d}{x}\right)+O(1)\\ & \quad \leq \frac{Ck_1}{k_0}|supp(\omega_i^\varepsilon)|^{{1}/{2}}\int_{D}P^\varepsilon_i\omega_i^\varepsilon \,{\rm d}\theta(x)\notag\\ & \qquad + \frac{Ck_1 }{ \varepsilon^2}|supp(\omega_i^\varepsilon)|^{{1}/{2}} \int_{supp(\omega_i^\varepsilon)}|\nabla P^\varepsilon_i| \,{\rm d}{x}+O(1). \end{align*}

So for $\varepsilon$ sufficiently small, we get $\int _{D}P^\varepsilon _i |\omega _i^\varepsilon | \,{\rm d}\theta (x)\leq ({Ck_1 }/{ \varepsilon ^2})|supp(\omega _i^\varepsilon )|^{{1}/{2}} \int _{supp(\omega _i^\varepsilon )} |\nabla P^\varepsilon _i| \,{\rm d}{x}+O(1).$ By Hölder's inequality,

(3.18)\begin{align} \int_{D}P^\varepsilon_i |\omega_i^\varepsilon| \,{\rm d}\theta(x)& \leq\frac{Ck_1 }{ \varepsilon^2}|supp(\omega_i^\varepsilon)| \left(\int_{supp(\omega_i^\varepsilon)}|\nabla P^\varepsilon_i|^2 \,{\rm d}{x}\right)^{{1}/{2}}+O(1) \nonumber\\ & \leq C k_1 \left(\int_{D}|\nabla P^\varepsilon_i|^2 \,{\rm d}{x}\right)^{{1}/{2}}+O(1). \end{align}

Combining (3.17), (3.18) and (3.16), we get (3.15).

Notice that

\[ 2E_i(\omega^\varepsilon)=sgn(d_i)\int_{D}(sgn(d_i)\mathcal{G}(k\omega^\varepsilon)-\mu_i^{\varepsilon})\omega_i^\varepsilon \,{\rm d}\theta(x)+|d_i|\mu^\varepsilon_i. \]

So using lemma 3.5, we get (3.14).

Proof. By the choice of $\mathcal {M}_\varepsilon (D)$, we know that $k_1|supp(\omega _i^\varepsilon )|\geq |d_i| \varepsilon ^2.$ This implies that

\[ diam\left(supp( \omega_i^\varepsilon)\right)\geq r_1\varepsilon \]

for some $r_1> 0$.

On the other hand, for any $x\in supp( \omega _i^\varepsilon )$, using (3.3) we have $\psi _i^{\varepsilon }(x)\geq 0$, which shows that

\begin{align*} \mu_i^{\varepsilon}& \leq{-}\frac{1}{2\pi}\int_{D}\ln|x-y ||\omega_i^{\varepsilon}|(y)\,{\rm d}\theta(y)-\int_{D}h(x,y)|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)\notag\\ & \quad +sgn(d_i)\sum_{j=1,j\neq i}^l\mathcal{G}(k\omega_j^{\varepsilon})(x)\\ & \leq{-}\frac{1}{2\pi}\int_{D}\ln|x-y||\omega_i^{\varepsilon}|(y)\,{\rm d}\theta(y)+O(1). \end{align*}

Thus by (3.14), we have

\[ \frac{|d_i|}{2\pi}\ln\frac{1}{\varepsilon}\leq{-}\frac{1}{2\pi}\int_{D}\ln|x-y||\omega_i^{\varepsilon}|(y)\,{\rm d}\theta(y)+O(1). \]

From the classical estimates in [Reference Turkington37], we get $diam(supp( \omega _i^\varepsilon ))\le R_1\varepsilon$ for some $R_1>1$.

Proof. By proposition 3.2, we know that $|\omega ^\varepsilon _i|$ is a vortex patch with height ${1}/{\varepsilon ^2}$, so $\zeta ^\varepsilon _i$ is a vortex patch with height 1. So the limiting function $\zeta ^*_i$ is also a vortex patch with height 1, that is, $\zeta ^*_i=\chi _{U^*}$ for some set $U^*\subseteq B_{R_1}(0)$.

Since $k$ is a $C^0$ function and $\lim _{\varepsilon \to 0^+}\bar {X}_i^\varepsilon =\mathbf {x^*_i}$, we have

\[ k(\varepsilon x+\bar{X}_i^\varepsilon)\to k(\mathbf{x^*_i})\quad \text{uniformly in}\ B_{R_1}(0), \]

so by (3.19)

(3.23)\begin{equation} |d_i|=\lim_{\varepsilon \to 0^+}\int_{B_{R_1}(0)}k(\varepsilon x+\bar{X}_i^\varepsilon)\zeta^\varepsilon_i(x)\,{\rm d}x= k(\mathbf{x^*_i})|U^*|. \end{equation}

On the one hand, by the definition of $E(\omega ^\varepsilon )$ and lemma 3.6, we get

\begin{align*} E(\omega^\varepsilon)& ={-}\frac{1}{4\pi}\sum_{i=1}^l\iint\ln|{x}-{y}| \omega_i^\varepsilon({x}) \omega_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\\ & \quad - \frac{1}{2}\sum_{i=1}^l\iint h({x},{y})\omega_i^\varepsilon({x})\omega_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y) \\ & \quad +\frac{1}{2}\sum_{1\leq i\neq j\leq l}\iint G(x,y)\omega_i^\varepsilon({x})\omega_j^\varepsilon({y})\theta(x)\,{\rm d}\theta(y)\\ & = \sum_{i=1}^l\frac{1}{4\pi}\iint\ln\frac{1}{\varepsilon|{x}-{y}|} k(\varepsilon x+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({x}) k(\varepsilon y+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({y})\,{\rm d}x\,{\rm d}y\\ & \quad -\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})+o(1)\\ & = \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l \frac{1}{4\pi}\int_{B_{R_1}(0)}\int_{B_{R_1}(0)}\\ & \quad \times \ln\frac{1}{|{x}-{y}|} k(\varepsilon x+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({x}) k(\varepsilon y+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({y})\,{\rm d}x\,{\rm d}y\\ & \quad -\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})+o(1). \end{align*}

Since $\zeta ^\varepsilon _i\to \zeta ^*_i$ in $L^p$ weak topology and $k(\varepsilon x+\bar {X}_i^\varepsilon )\to k(\mathbf {x^*_i})$ uniformly as $\varepsilon \to 0$, we have

\begin{align*} & \iint\ln\frac{1}{|{x}-{y}|} k(\varepsilon x+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({x}) k(\varepsilon y+\bar{X}_i^\varepsilon)\zeta_i^\varepsilon({y})\,{\rm d}x\,{\rm d}y \\ & \quad =k(\mathbf{x^*_i})^2\iint\ln\frac{1}{|x-y|}\zeta^*_i(x)\zeta^*_i(y)\,{\rm d}x\,{\rm d}y+o(1), \end{align*}

where we have used the $L^p$ theory in elliptic equations and the compact embedding theorem. Thus

(3.24)\begin{align} E(\omega^\varepsilon)& =\sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l \frac{k(\mathbf{x^*_i})^2}{4\pi}\int_{B_{R_1}(0)}\int_{B_{R_1}(0)}\ln\frac{1}{|x-y|}\zeta^*_i(x)\zeta^*_i(y)\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad -\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})+o(1) \nonumber\\ & \leq \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l \frac{k(\mathbf{x^*_i})^2}{4\pi}\int_{B_{R_1}(0)}\int_{B_{R_1}(0)}\notag\\ & \quad \times \ln\frac{1}{|x-y|} \chi_{B_{\sqrt{{|d_i|}/{\pi k(\mathbf{x^*_i})}}}(0)}(x)\chi_{B_{\sqrt{{|d_i|}/{\pi k(\mathbf{x^*_i})}}}(0)}(y)\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad -\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})+o(1) \nonumber\\ & =\sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^lQ_i(k(\mathbf{x_i^*}))-\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})+o(1) \end{align}

where the second inequality we have used (3.23) and the Riesz's rearrangement inequality.

On the other hand, for any $z_i\in B_i (i=1,\,\ldots,\,l)$, we choose a function $\hat {\omega }^\varepsilon =\sum _{i=1}^l\hat {\omega }^\varepsilon _i,\,$ where $\hat {\omega }^\varepsilon _i$ is defined by

\[ \hat{\omega}^\varepsilon_i=\frac{sgn(d_i)}{\varepsilon^2}\chi_{B_{\tau_i(\varepsilon)\varepsilon}(z_i)}. \]

Here $\tau _i(\varepsilon )\in (({1}/{2})\sqrt { {|d_i|}/{\pi k(z_i)}},\, 2\sqrt {{|d_i|}/{\pi k(z_i)}})$ is chosen to satisfy $\int _{B_i}\hat {\omega }^\varepsilon _i(x) {\rm d}\theta (x)=d_i$. Then direct calculation shows that such $\tau _i(\varepsilon )$ exists for $\varepsilon$ sufficiently small and $\lim _{\varepsilon \to 0^+}\tau _i(\varepsilon )=\sqrt {{|d_i|}/{\pi k(z_i)}}$. By the definition of $\hat {\omega }^\varepsilon$, we obtain $\hat {\omega }^\varepsilon \in \mathcal {M}_\varepsilon (D)$.

For $E(\hat {\omega }^\varepsilon )$, similarly as calculations in (3.24), we obtain

(3.25)\begin{align} E(\hat{\omega}^\varepsilon) & ={-}\frac{1}{4\pi}\sum_{i=1}^l\iint\ln|{x}-{y}| \hat{\omega}_i^\varepsilon({x}) \hat{\omega}_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\notag\\ & \quad - \frac{1}{2}\sum_{i=1}^l\iint h({x},{y})\hat{\omega}_i^\varepsilon({x})\hat{\omega}_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y) \nonumber\\ & \quad +\frac{1}{2}\sum_{1\leq i\neq j\leq l}\iint G(x,y)\hat{\omega}_i^\varepsilon({x})\hat{\omega}_j^\varepsilon({y})\theta(x)\,{\rm d}\theta(y) \nonumber\\ & = \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^lQ_i(k(z_i))-\mathcal{H}_l(z_1,\ldots,z_l)+o(1). \end{align}

By $E(\omega ^\varepsilon )\geq E(\hat {\omega }^\varepsilon )$, (3.24) and (3.25), we get

\[ \sum_{i=1}^lQ_i(k(\mathbf{x^*_i}))-\mathcal{H}_l(\mathbf{x^*_1},\mathbf{x^*_2},\ldots,\mathbf{x^*_l})\geq \sum_{i=1}^lQ_i(k(z_i))-\mathcal{H}_l(z_1,\ldots,z_l)\quad \forall z_i\in B_i. \]

Thus we get (3.21). Notice that $\mathcal {H}_l(x_1,\,x_2,\,\ldots,\,x_l)-\sum _{i=1}^lQ_i(k(x_i))=\Gamma _l(x_1,\,x_2, \ldots,\,x_l)+C_0^*$, where $C_0^*$ is a universal constant. Then by the assumption that $(x_{0,1},\,\ldots,\,x_{0,l})$ is a strict local minimizer of $\Gamma _l$, we get $\mathbf {x^*_i}=x_{0,i}.$

It suffices to prove (3.22). Indeed by (3.24) we have

\begin{align*} & \iint\ln\frac{1}{|x-y|}\zeta^*_i(x)\zeta^*_i(y)\,{\rm d}x\,{\rm d}y\notag\\ & \quad = \iint\ln\frac{1}{|x-y|}\chi_{B_{\sqrt{{|d_i|}/{\pi k(\mathbf{x^*_i})}}}(0)}(x)\chi_{B_{\sqrt{{|d_i|}/{\pi k(\mathbf{x^*_i})}}}(0)}(y)\,{\rm d}x\,{\rm d}y. \end{align*}

Using strict Rearrangement inequality (see theorem 3.9, [Reference Lieb and Loss26]), there exists a translation $\bar {\mathcal {T}}$ such that $\mathcal {T}(\zeta ^*_i)=\chi _{B_{\sqrt {{|d_i|}/{\pi k(\mathbf {x^*_i})}}}(0)}$. Notice that both the centre of $\zeta ^*_i$ and the centre of $B_{\sqrt {{|d_i|}/{\pi k(\mathbf {x^*_i})}}}(0)$ are the origin, we get $\bar {\mathcal {T}}=id$, namely, $\zeta ^*_i=\chi _{B_{\sqrt {{|d_i|}/{\pi k(\mathbf {x^*_i})}}}(0)}$.

Finally, by (3.19) we have $\int _{B_{R_1}(0)}k(\varepsilon x+\bar {X}_i^\varepsilon )(\zeta ^\varepsilon _i(x))^p\,{\rm d}x=|d_i|,\,$ which implies that

\[ \lim_{\varepsilon \to 0^+}\|\zeta^\varepsilon_i\|_{L^p}= \left(\frac{|d_i|}{k(\mathbf{x^*_i})}\right)^{{1}/{p}}=\|\zeta^*_i\|_{L^p}. \]

Using the strict convexity of $L^p$ norm, we finish the proof.

Proof. Using (3.24), one can immediately get (3.26). For (3.27), notice that for any $x\in \partial supp(\omega _i^\varepsilon )$, $\mu _i^\varepsilon =sgn(d_i)\int _{D}G(x,\,y) \omega ^\varepsilon (y) \,{\rm d}\theta (y)$. So by lemma 3.6 and proposition 3.7,

(3.28)\begin{align} \mu_i^\varepsilon& =sgn(d_i)\int_{D}G(x,y) \omega^\varepsilon(y)\,{\rm d}\theta(y) \nonumber\\ & = \frac{1}{2\pi}\int_{B_i}\ln\frac{1}{|x-y|}|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)-\int_{B_i}h(x,y)|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)\notag\\ & \quad +sgn(d_i)\sum_{j\neq i}d_jG(x_{0,i},x_{0,j})+o(1)\nonumber\\ & = \frac{1}{2\pi}\int_{B_i}\ln\frac{1}{|x-y|}|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)-|d_i|h(x_{0,i},x_{0,i})\notag\\ & \quad +sgn(d_i)\sum_{j\neq i}d_jG(x_{0,i},x_{0,j})+o(1). \end{align}

Let $x=\varepsilon x'+\bar {X}^\varepsilon _i$, then $x'\in supp(\zeta _i^\varepsilon )$. By the definition of $\zeta _i^\varepsilon$, we get

(3.29)\begin{align} & \frac{1}{2\pi}\int_{B_i}\ln\frac{1}{|x-y|}|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)\notag\\ & \quad =\frac{1}{2\pi}\int_{B_{R_1}(0)}\ln\frac{1}{\varepsilon|x'-y'|}k(\varepsilon y'+\bar{X}^\varepsilon_i)\zeta_i^\varepsilon(y')\,{\rm d}y' \nonumber\\ & \quad = \frac{|d_i|}{2\pi}\ln\frac{1}{\varepsilon}+\frac{1}{2\pi}\int_{B_{R_1}(0)}\ln\frac{1}{|x'-y'|}k(\varepsilon y'+\bar{X}^\varepsilon_i)\zeta_i^\varepsilon(y')\,{\rm d}y'. \end{align}

By proposition 3.7 and the continuity of $k$, we have

(3.30)\begin{align} & \frac{1}{2\pi}\int_{B_{R_1}(0)}\ln\frac{1}{|x'-y'|}k(\varepsilon y'+\bar{X}^\varepsilon_i)\zeta_i^\varepsilon(y')\,{\rm d}y'\notag\\ & \quad =\frac{k(x_{0,i})}{2\pi} \int_{B_{\sqrt{{|d_i|}/{\pi k(x_{0,i})}}}(0)}\ln\frac{1}{|x^*-y'|}\,{\rm d}y'+o(1), \end{align}

where $x^*$ is any point of $\partial B_{{\sqrt {{|d_i|}/{\pi k(x_{0,i})}}}}(0)$. Taking (3.29), (3.30) into (3.28), we get (3.27).

Proof. By proposition 3.2, we know that $\omega ^\varepsilon$ has the form

\[ \omega^\varepsilon= \sum_{i=1}^l\frac{sgn(d_i)}{ \varepsilon^2} \chi_{\left\{\mathcal{G}(k\omega^\varepsilon)-\mu_i^\varepsilon>0\right\}\cap B_i}. \]

By lemma 3.6, we have $diam(supp(\omega _i^\varepsilon ))=O(\varepsilon )$. Moreover, by lemmas 3.6, 3.7 and the assumption that $(x_{0,1},\,\ldots,\,x_{0,l})$ is the strict local minimizer of $\Gamma _l$, the support set of $\omega _i^\varepsilon$ tends to $x_{0,i}$ as $\varepsilon \to 0^+$, namely,

\[ \lim_{\varepsilon \to 0^+}\sup_{x\in supp(\omega_i^\varepsilon)}|x- x_{0,i}|=0. \]

By proposition 3.7 and lemma 3.10, we get (3)(4) in theorem 2.3. The proof of theorem 2.3 is thus complete.

Proof. Let $l=1$ and $d_1=I$ in theorem 2.3, we get theorem 2.1. Let $\lambda ={1}/{\varepsilon ^2}$ and $u^\lambda =u^\varepsilon,\, c^\lambda =-\mu ^\varepsilon$, we get theorem 1.1.

Proof. The proof is similar to that of proposition 3.1. So we omit it here.

Proof. For each ${\omega }\in \mathcal {N}_{\varepsilon,T}$, we choose test functions

\[ \omega_{(s)}=\omega^{\varepsilon,T}+s({\omega}-\omega^{\varepsilon,T}),\quad \text{for} \ s\in[0,1]. \]

Since $\omega ^{\varepsilon,T}$ is a maximizer, we get $\mathcal {E}(\omega _{(s)})\leq \mathcal {E}(\omega _{(0)})$, which implies that $({{\rm d}\mathcal {E}(\omega _{(s)})})/{{\rm d}s}|_{s=0^+}\leq 0$, that is,

\begin{align*} & \int_{D}\omega \left( \mathcal{G}(k\omega^{\varepsilon,T})-\sum_{i=1}^lf_i^{{-}1}(sgn(d_i)\varepsilon^2\omega_i^{\varepsilon,T})sgn(d_i)\right)\,{\rm d}\theta(x)\\ & \quad \leq \int_{D}\omega^{\varepsilon,T}\left( \mathcal{G}(k\omega^{\varepsilon,T})-\sum_{i=1}^lf_i^{{-}1}(sgn(d_i)\varepsilon^2\omega_i^{\varepsilon,T})sgn(d_i)\right)\,{\rm d}\theta(x) \end{align*}

for all ${\omega }\in \mathcal {N}_{\varepsilon,T}.$ Using the bathtub principle, we obtain

(4.5)\begin{align} \begin{aligned} sgn(d_i) \mathcal{G}(k\omega^{\varepsilon,T}) -f_i^{{-}1}(sgn(d_i)\varepsilon^2 \omega_i^{\varepsilon,T}) & \ge \sigma_i^{\varepsilon,T} \quad \mbox{on} \left\{sgn(d_i)\omega_i^{\varepsilon,T}=\frac{T}{\varepsilon^2}\right\}\cap B_i, \\ sgn(d_i) \mathcal{G}(k\omega^{\varepsilon,T}) -f_i^{{-}1}(sgn(d_i)\varepsilon^2 \omega_i^{\varepsilon,T}) & = \sigma_i^{\varepsilon,T} \quad \mbox{on} \left\{0< sgn(d_i)\omega_i^{\varepsilon,T}<\frac{T}{\varepsilon^2}\right\}\cap B_i, \\ sgn(d_i) \mathcal{G}(k\omega^{\varepsilon,T}) -f_i^{{-}1}(sgn(d_i)\varepsilon^2 \omega_i^{\varepsilon,T}) & \le \sigma_i^{\varepsilon,T} \quad \mbox{on} \{sgn(d_i)\omega_i^{\varepsilon,T}=0\}\cap B_i, \end{aligned} \end{align}

where $\sigma _i^{\varepsilon,T}$ is a constant satisfying

(4.6)\begin{align} \sigma_i^{\varepsilon,T}=\inf\left\{s\in\mathbb{R}\mid|\{x\in B_i\mid sgn(d_i) \mathcal{G}(k\omega^{\varepsilon,T}) -f_i^{{-}1}(sgn(d_i)\varepsilon^2 \omega_i^{\varepsilon,T}) >s\}|_{\theta}\le \frac{|d_i|\varepsilon^2}{T}\right\}. \end{align}

Define $\psi _i^{\varepsilon,T}=sgn(d_i)\mathcal {G}(k\omega ^{\varepsilon,T})-\sigma ^{\varepsilon,T}_i$, then by (4.5) one has

(4.7)\begin{equation} \omega_i^{\varepsilon, T}= sgn(d_i)\left( \frac{1}{ \varepsilon^2}f_i(\psi_i^{\varepsilon,T})\chi_{\{ 0<\psi_i^{\varepsilon,T}< f_i^{{-}1}(T) \}\cap B_i }+\frac{T}{\varepsilon^2}\chi_{\{ \psi_i^{\varepsilon,T}\geq f_i^{{-}1}(T) \}\cap B_i}\right). \end{equation}

So we get (4.3).

It remains to prove that $\sigma _i^{\varepsilon, T}\geq -f_i^{-1}(T)-C_0.$ For any $x\in B_i$,

(4.8)\begin{align} & sgn(d_i) \mathcal{G}(k\omega^{\varepsilon,T}) (x) -f_i^{{-}1}(sgn(d_i)\varepsilon^2 \omega_i^{\varepsilon,T})(x)\notag\\ & \quad \geq{-}\sum_{j\neq i}|d_j|\max_{x\in B_i, y\in B_j}G(x,y) -f_i^{{-}1}(T). \end{align}

Choose $C_0=\sum _{j\neq i}|d_j|\max _{x\in B_i, y\in B_j}G(x,\,y)$. Combining (4.6) and (4.8), we get (4.4).

Proof. We choose a function $\tilde {\omega }^{\varepsilon, T}=({sgn(d_i)}/{\varepsilon ^2})\chi _{B_{t(\varepsilon )\varepsilon }(x_{0,i})}+ \sum _{j\neq i}\omega _j^{\varepsilon,T}= \sum _{k=1}^l\tilde {\omega }^{\varepsilon, T}_k$. Here $t(\varepsilon )\in (({1}/{2})\sqrt { {|d_i|}/{\pi k(x_{0,i})}},\, 2\sqrt {{|d_i|}/{\pi k(x_{0,i})}})$ satisfies $\int _{B_i}\tilde {\omega }_i^{\varepsilon,T} {\rm d}\theta (x)=d_i$. Direct calculations shows that $t(\varepsilon )$ exists for $\varepsilon$ sufficiently small and $\lim _{\varepsilon \to 0^+}t(\varepsilon )=\sqrt {{|d_i|}/{\pi k(x_{0,i})}}$. Then $\tilde {\omega }^{\varepsilon, T}\in \mathcal {N}_{\varepsilon, T}(D)$. Notice that

\begin{align*} & \frac{1}{\varepsilon^2} \int_{D}F_i(sgn(d_i)\varepsilon^2\tilde{\omega}^{\varepsilon, T}_i)\,{\rm d}\theta(x)\leq \frac{1}{\varepsilon^2}F_i(1)\int_{\chi_{B_{t(\varepsilon)\varepsilon }(x_{0,i})}}\,{\rm d}\theta(x)=F_i(1)|d_i|\\ & \frac{1}{2}\int_{D}G(x,y)\tilde{\omega}^{\varepsilon, T}_i(x)\tilde{\omega}^{\varepsilon, T}_i(y) \,{\rm d}\theta(x)\,{\rm d}\theta(y)\geq \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon} +O(1), \end{align*}

so we get $\mathcal {E}(\tilde {\omega }^{\varepsilon, T})\geq ({d_i^2}/{4\pi })\ln ({1}/{\varepsilon })+\sum _{j\neq i}\mathcal {E}_j(\omega ^{\varepsilon,T})+O(1)$. By $\mathcal {E}(\omega ^{\varepsilon, T})\geq \mathcal {E}(\tilde {\omega }^{\varepsilon, T})$ and (4.9), we get (4.10).

Proof. By the definition of $\mathcal {E}_i$ and assumption $(f_2)'$, we get

(4.12)\begin{align} 2\mathcal{E}_i(\omega^{\varepsilon,T})& =\int_{D}\omega_i^{\varepsilon,T} \mathcal{G}(k\omega_i^{\varepsilon,T}) \,{\rm d}\theta(x) -\frac{2}{\varepsilon^2}\int_{D}F_i(sgn(d_i)\varepsilon^2\omega_i^{\varepsilon, T})\,{\rm d}\theta(x) \nonumber\\ & \leq \int_{D}\omega_i^{\varepsilon,T}( \mathcal{G}(k\omega^{\varepsilon, T})-sgn(d_i)\sigma_i^{\varepsilon,T} ) \,{\rm d}\theta(x)\nonumber\\ & \quad -2\delta_1 \int_{D}|\omega_i^{\varepsilon, T}|f_i^{{-}1}(sgn(d_i)\varepsilon^2\omega_i^{\varepsilon, T})\,{\rm d}\theta(x) +|d_i|\sigma_i^{\varepsilon, T}\nonumber\\ & = \int_{\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\}}|\omega_i^{\varepsilon, T}|\left(sgn(d_i)\mathcal{G}(k\omega^{\varepsilon, T})-\sigma_i^{\varepsilon, T}-f_i^{{-}1}(T) \right) \,{\rm d}\theta(x)\nonumber\\ & \quad +(1-2\delta_1)\int_{D}|\omega_i^{\varepsilon, T}|f_i^{{-}1}( \varepsilon^2|\omega_i^{\varepsilon, T}|)\,{\rm d}\theta(x)+|d_i|\sigma_i^{\varepsilon, T} \nonumber\\ & \leq\int_{\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\}}|\omega_i^{\varepsilon, T}|\left(\psi_i^{\varepsilon,T}-f_i^{{-}1}(T)-C_0 \right)_+ \,{\rm d}\theta(x)\nonumber\\ & \quad +|d_i||1-2\delta_1|f_i^{{-}1}(T)+|d_i|\sigma_i^{\varepsilon, T}+C_0|d_i|, \end{align}

where $\psi _i^{\varepsilon,T}=sgn(d_i)\mathcal {G}(k\omega ^{\varepsilon, T})-\sigma _i^{\varepsilon, T}$.

To estimate the first term of the last line of (4.12), we define $W_i^{\varepsilon,T}=(\psi _i^{\varepsilon, T}-f_i^{-1}(T) -C_0 )_+$ and $\bar {W}_i^{\varepsilon,T}=(sgn(d_i)\mathcal {G}(k\omega _i^{\varepsilon, T})-\sigma _i^{\varepsilon,T}-f_i^{-1}(T)-C_0 )_+ .$ Using (4.4), we have $\bar {W}_i^{\varepsilon,T},\, W_i^{\varepsilon,T}\in H^1_0(D)$.

On the one hand,

(4.13)\begin{equation} \int_{D} |\omega_i^{\varepsilon,T}|\bar{W}_i^{\varepsilon, T}\,{\rm d}\theta(x)=\int_{D}|\nabla\bar{W}_i^{\varepsilon,T}|^2\,{\rm d}x. \end{equation}

On the other hand, by (4.3), we get $supp(W_i^{\varepsilon, T})\cap B_i\subseteq \{|\omega _i^{\varepsilon,T}|={T}/{\varepsilon ^2}\}$, which implies that

\begin{align*} \int_{D}|\omega_i^{\varepsilon,T}|\bar{W}_i^{\varepsilon,T}\,{\rm d}\theta(x) & \leq \int_{D}|\omega_i^{\varepsilon,T}|W_i^{\varepsilon,T}\,{\rm d}\theta(x)+O(1) \nonumber\\ & =\int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\omega_i^{\varepsilon,T}|W_i^{\varepsilon, T}\,{\rm d}\theta(x)+O(1). \end{align*}

Using the Sobolev imbedding $W^{1,1}(B_i)\subset L^2(B_i)$, we get

\begin{align*} & \int_{D}|\omega_i^{\varepsilon, T}|\bar{W}_i^{\varepsilon,T}\,{\rm d}\theta(x) \nonumber\\ & \quad \leq\frac{T}{\varepsilon^2}\int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}W_i^{\varepsilon,T}\,{\rm d}\theta(x)+O(1)\\ & \quad \leq \frac{CT}{\varepsilon^2}\left|\left\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\right\}\right|^{{1}/{2}}\left( \int_{B_i}W_i^{\varepsilon, T}\,{\rm d}x+\int_{B_i}|\nabla W_i^{\varepsilon,T}|\,{\rm d}x\right)+ O(1)\\ & \quad \leq C \left|\left\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\right\}\right|^{{1}/{2}} \int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}sgn(d_i)\omega_i^{\varepsilon, T}\bar{W}_i^{\varepsilon, T}\,{\rm d}x \\ & \quad \quad +\frac{CT}{\varepsilon^2}\left|\left\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\right\}\right|^{{1}/{2}}\int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\nabla\bar{W}_i^{\varepsilon,T}|\,{\rm d}x+O(1). \end{align*}

So by the fact that $|\{|\omega _i^{\varepsilon,T}|={\Lambda }/{\varepsilon ^2}\}|=O(\varepsilon ^2)$ and Hölder's inequality, for $\varepsilon$ sufficiently small we get

(4.14)\begin{align} & \int_{D}|\omega_i^{\varepsilon, T}|\bar{W}_i^{\varepsilon,T}\,{\rm d}\theta(x) \nonumber\\ & \quad \leq\frac{CT}{\varepsilon^2}\left|\left\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\right\}\right|^{{1}/{2}}\int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\nabla\bar{W}_i^{\varepsilon,T}|\,{\rm d}x+O(1) \nonumber\\ & \quad \leq\frac{CT}{\varepsilon^2}\left|\left\{|\omega_i^{\varepsilon, T}|=\frac{T}{\varepsilon^2}\right\}\right|\left( \int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\nabla\bar{W}_i^{\varepsilon, T}|^2\,{\rm d}x\right)^{{1}/{2}} +O(1) \nonumber\\ & \quad \leq C|d_i|\left( \int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\nabla\bar{W}_i^{\varepsilon, T}|^2\,{\rm d}x\right)^{{1}/{2}} +O(1). \end{align}

Combining (4.13) and (4.14), we get $\int _{D}|\omega _i^{\varepsilon, T}|\bar {W}_i^{\varepsilon,T}\,{\rm d}\theta (x)=O(1),\,$ which implies that

(4.15)\begin{equation} \int_{\{|\omega_i^{\varepsilon, T}|={T}/{\varepsilon^2}\}}|\omega_i^{\varepsilon, T}|W_i^{\varepsilon,T}\,{\rm d}\theta(x)=O(1). \end{equation}

Taking (4.15) into (4.12) and using lemma 4.3, we get (4.11).

Proof. For any $x\in B_i$, using the definition of $\psi _i^{\varepsilon, T}$ and the rearrangement inequality

\begin{align*} \psi_i^{\varepsilon, T}(x)& =sgn(d_i)\mathcal{G}(k\omega^{\varepsilon,T})(x)-\sigma^{\varepsilon,T}_i \\ & \leq\frac{1}{2\pi}\int_{D}\ln\frac{1}{|x-y|}|\omega_i^{\varepsilon, T}|(y)\,{\rm d}y\theta(y)-\sigma_i^{\varepsilon, T}+O(1)\\ & \leq \frac{k_1T}{2\pi\varepsilon^2}\int_{B_{\sqrt{{|d_i|}/{(\pi k_1T)} }\varepsilon}(0)}\ln\frac{1}{|y|}\,{\rm d}y-\sigma_i^{\varepsilon,T}+O(1)\\ & = \frac{|d_i|}{2\pi}\ln\frac{1}{\varepsilon}+\frac{|d_i|}{4\pi}\ln T-\sigma_i^{\varepsilon,T}+O(1). \end{align*}

Thus using lemma 4.4, we obtain

\[ \psi_i^{\varepsilon, \Lambda}(x)\leq |1-2\delta_1|f_i^{{-}1}(T)+\frac{|d_i|}{4\pi}\ln T+O(1). \]

It follows from assumption $(f_3)$ that for each $a_0 > 0$, $\lim _{s\to +\infty }f_i(s)e^{-a_0 s}=0,\,$ which implies that $\lim _{s\to +\infty } \tau f^{-1}(s)-\ln s=+\infty.$ Thus we can choose $T=T_0$ sufficiently large such that

\[ (1-|1-2\delta_1|)f_i^{{-}1}(T_0)>\frac{|d_i|}{4\pi}\ln T_0+O(1), \]

that is, $|1-2\delta _1|f_i^{-1}(T_0)+({|d_i|}/{4\pi })\ln T_0+O(1)< f_i^{-1}(T_0)$. Thus we have $|\{\psi _i^{\varepsilon,T}\geq f_i^{-1}(T) \}\cap B_i|=0.$ Using lemma 4.2, we get (4.17).

Proof. Since $|\omega _i^\varepsilon |\leq {T_0}/{\varepsilon ^2}$ and $\int _{B_i}\omega _i^\varepsilon \,{\rm d}\theta (x)=d_i$, we get $|supp( \omega _i^\varepsilon )|\geq C\varepsilon ^2$, which implies that $diam(supp( \omega _i^\varepsilon ))\geq \bar {r}_1\varepsilon$ for some $\bar {r}_1>0$.

Similarly as the proof of lemma 3.6, one can get the existence of $\bar {R}_1>1$ such that

\[ diam\left(supp( \omega_i^\varepsilon)\right)\leq \bar{R}_1\varepsilon.\]

□

Proof. On the one hand, using lemma 4.6 and the definition of $x^*_i$, we get

\begin{align*} & \frac{1}{2}\int_{D}\int_{D}h({x},{y})\omega_i^\varepsilon({x})\omega_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y)=\frac{d_i^2}{2}h(x_i^*,x_i^*)+o(1),\\ & \sum_{1\leq i\neq j\leq l}\iint G(x,y)\omega_i^\varepsilon({x})\omega_j^\varepsilon({y})\theta(x)\,{\rm d}\theta(y)=\sum_{1\leq i\neq j\leq l}d_id_jG(x_i^*,x_j^*)+o(1). \end{align*}

By the definition of $\xi _i^\varepsilon$, we have

\begin{align*} & -\frac{1}{4\pi}\iint\ln|{x}-{y}| \omega_i^\varepsilon({x}) \omega_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\\ & \quad = \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\frac{1}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi_i^\varepsilon({x})\xi_i^\varepsilon({y})k(\varepsilon x+\hat{X}_i^\varepsilon)k(\varepsilon y+\hat{X}_i^\varepsilon)\,{\rm d}x\,{\rm d}y. \end{align*}

Since $supp(\xi _i^\varepsilon )\subseteq B_{\bar { R}_1}(0)$ and $||\xi ^\varepsilon _i||_{L^\infty }\leq T_0$, we obtain

(4.23)\begin{align} & -\frac{1}{4\pi}\iint\ln|{x}-{y}| \omega_i^\varepsilon({x}) \omega_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\nonumber\\ & \quad =\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\frac{k(x_i^*)^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y+o(1). \end{align}

For the term involving $F_i$, we have

\begin{align*} \frac{1}{\varepsilon^2}\int_{D}F_i(sgn(d_i)\varepsilon^2\omega_i^\varepsilon)\,{\rm d}\theta(x)& =\int_{B_{\bar{ R}_1}(0)}F_i(\xi_i^\varepsilon)(x)k(\varepsilon x+\hat{X}_i^\varepsilon)\,{\rm d}x \nonumber\\ & =k(x_i^*)\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x+o(1). \end{align*}

Taking those into the definition of $\mathcal {E}$, we get

(4.24)\begin{align} \mathcal{E}(\omega^\varepsilon) & = \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l\frac{k(x_i^*)^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\mathcal{H}_l(x_1^*,\ldots,x_l^*) -\sum_{i=1}^lk(x_i^*)\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x+o(1). \end{align}

On the other hand, for any $z_i\in B_i (i=1,\,\ldots,\,l)$, we choose test functions $\hat {\omega }^\varepsilon =\sum _{i=1}^l\hat {\omega }^\varepsilon _i,\,$ where $\hat {\omega }^\varepsilon _i$ is defined by

\[ \hat{\omega}^\varepsilon_i=\frac{sgn(d_i)}{\varepsilon^2}\xi_i^\varepsilon\left(\frac{\cdot{-}z_i}{\bar{\tau}_i(\varepsilon)\varepsilon}\right). \]

$\bar {\tau }_i(\varepsilon )\in (({1}/{2})\sqrt {{k(x_i^*)}/{k(z_i)}} ,\, 2\sqrt {{k(x_i^*)}/{k(z_i)}} )$ is chosen to satisfy $\int _{B_i}\hat {\omega }^\varepsilon _i(x)\,{\rm d}\theta (x)=d_i$. By (4.20) and (4.21), one can prove that such $\bar {\tau }_i(\varepsilon )$ exists for $\varepsilon$ sufficiently small and $\bar {\tau }_i(\varepsilon )=\sqrt {({k(x_i^*)}/{k(z_i)})}+o(1)$. Now we calculate the energy expansion of $\hat {\omega }^\varepsilon$. It is not hard to prove that

\begin{align*} & - \frac{1}{2}\sum_{i=1}^l\iint h({x},{y})\hat{\omega}_i^\varepsilon({x})\hat{\omega}_i^\varepsilon({y})\theta(x)\,{\rm d}\theta(y) \\ & \quad +\frac{1}{2}\sum_{1\leq i\neq j\leq l}\iint G(x,y)\hat{\omega}_i^\varepsilon({x})\hat{\omega}_j^\varepsilon({y})\theta(x)\,{\rm d}\theta(y)\\ & \qquad ={-}\mathcal{H}_l(z_1,\ldots,z_l)+o(1). \end{align*}

Similar as (4.23), we get

(4.25)\begin{align} & -\frac{1}{4\pi}\iint\ln|{x}-{y}| \hat{\omega}_i^\varepsilon({x}) \hat{\omega}_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y) \nonumber\\ & \quad = \frac{d_i^2}{4\pi}\ln\frac{1}{\bar{\tau}_i(\varepsilon)\varepsilon}+\frac{k(z_i)^2\bar{\tau}_i(\varepsilon)^4}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y+o(1). \end{align}

Taking $\bar {\tau }_i(\varepsilon )=\sqrt {({k(x_i^*)}/{k(z_i)})}+o(1)$ into (4.25) we obtain

\begin{align*} & -\frac{1}{4\pi}\iint\ln|{x}-{y}| \hat{\omega}_i^\varepsilon({x}) \hat{\omega}_i^\varepsilon({y})\,{\rm d}\theta(x)\,{\rm d}\theta(y)\\ & \quad = \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\frac{d_i^2}{8\pi}\ln k(z_i)-\frac{d_i^2}{8\pi}\ln k(x_i^*)\\ & \qquad +\frac{k(x_i^*)^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y+o(1). \end{align*}

For the term involving $F_i$, we get

\begin{align*} \frac{1}{\varepsilon^2}\int_{D}F_i(sgn(d_i)\varepsilon^2\hat{\omega}_i^\varepsilon)\,{\rm d}\theta(x)& =\int_{B_{\bar{ R}_1}(0)}F_i(\xi_i^\varepsilon)(x')k(\bar{\tau}_i(\varepsilon)\varepsilon x'+z_i)\bar{\tau}_i(\varepsilon)^2\,{\rm d}x'\\ & = k(x_i^*)\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x+o(1). \end{align*}

Taking those into the definition of $\mathcal {E}(\hat {\omega }^\varepsilon )$, we get

(4.26)\begin{align} \mathcal{E}(\hat{\omega}^\varepsilon) & =\sum_{i=1}^l\left( \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\frac{d_i^2}{8\pi}\ln k(z_i)-\frac{d_i^2}{8\pi}\ln k(x_i^*)\right.\nonumber\\ & \quad \left.+\frac{k(x_i^*)^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y\right) \nonumber\\ & \quad -\mathcal{H}_l(z_1,\ldots,z_l)-\sum_{i=1}^lk(x_i^*)\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x+o(1). \end{align}

Since $\mathcal {E}(\omega ^\varepsilon )\geq \mathcal {E}(\hat {\omega }^\varepsilon )$, by (4.24) and (4.26) we get

\[ \sum_{i=1}^l\frac{d_i^2}{8\pi}\ln k(x^*_i)-\mathcal{H}_l(x^*_1,\ldots,x^*_l)\geq \sum_{i=1}^l\frac{d_i^2}{8\pi}\ln k(z_i)-\mathcal{H}_l(z_1,\ldots,z_l)\quad \forall\ z_i\in B_i. \]

Thus we get (4.22). By the assumption that $(x_{0,1},\,\ldots,\,x_{0,l})$ is a strict local minimizer of $\Gamma _l$, we have $x^*_i=x_{0,i}.$

Proof. On the one hand, by the Riesz's rearrangement inequality, we have

\begin{align*} & \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\xi_i^\varepsilon(x)\xi_i^\varepsilon(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad \le \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\tilde{\xi}^\varepsilon(x)\tilde{\xi}^\varepsilon(y)\,{\rm d}x\,{\rm d}y, \end{align*}

which implies that

(4.32)\begin{align} & \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\xi_i^*(x)\xi_i^*(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad \le \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\tilde{\xi}_i^*(x)\tilde{\xi}_i^*(y)\,{\rm d}x\,{\rm d}y. \end{align}

On the other hand, let $\tilde {\omega }^\varepsilon =\tilde {\omega }^\varepsilon _i+\sum _{j=1,j\neq i}^l\omega ^\varepsilon _j\in \mathcal {N}_\varepsilon (D)$ satisfying

\[ \tilde{\omega}^\varepsilon_i(x)=\left\{ \begin{array}{ll} \displaystyle\dfrac{sgn(d_i)}{\varepsilon^2}\tilde{\xi}_i^\varepsilon \left(\dfrac{x-\hat{X}_i^\varepsilon}{c(\varepsilon)\varepsilon}\right) & \text{if} \ x\in B_{\bar{ R}_1}(\hat{X}_i^\varepsilon), \\ 0 & \text{if} \ x\in D\backslash B_{\bar{ R}_1}(\hat{X}_i^\varepsilon), \end{array} \right. \]

where $c(\varepsilon )$ is a constant such that $\int _{B_i}\tilde {\omega }^\varepsilon _i\,{\rm d}\theta (x)=d_i$. Then $c(\varepsilon )=1+o(1)$. Similarly as the proof of proposition 4.7, we get

\begin{align*} \mathcal{E}({\omega}^\varepsilon) & = \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+ \frac{d_i^2}{4\pi}\int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\xi_i^\varepsilon(x)\xi_i^\varepsilon(y)\,{\rm d}x\,{\rm d}y\\ & \quad -\frac{d_i^2}{2}h(x^*_i,x^*_i)+\sum_{j=1,j\neq i}^ld_id_jG(x^*_i,x^*_j)\\ & \quad -\frac{1}{\varepsilon^2}\int_{B_i}F_i(sgn(d_i)\varepsilon^2\omega^\varepsilon_i)\,{\rm d}\theta(x)+\mathcal{E}\left(\sum_{j=1,j\neq i}^l\omega_j^\varepsilon\right)+o(1),\\ \mathcal{E}(\tilde{\omega}^\varepsilon) & = \frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+ \frac{d_i^2}{4\pi}\int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\tilde{\xi}_i^\varepsilon(x)\tilde{\xi}_i^\varepsilon(y)\,{\rm d}x\,{\rm d}y\\ & \quad -\frac{d_i^2}{2}h(x^*_i,x^*_i)+\sum_{j=1,j\neq i}^ld_id_jG(x^*_i,x^*_j)\\ & \quad -\frac{1}{\varepsilon^2}\int_{B_i}F_i(sgn(d_i)\varepsilon^2\tilde{\omega}^\varepsilon_i)\,{\rm d}\theta(x)+\mathcal{E} \left(\sum_{j=1,j\neq i}^l\omega_j^\varepsilon\right)+o(1), \end{align*}

and

\[ \frac{1}{\varepsilon^2}\int_{B_i}F_i(sgn(d_i)\varepsilon^2\omega^\varepsilon_i)\,{\rm d}\theta(x)=\frac{1}{\varepsilon^2}\int_{B_i}F_i(sgn(d_i)\varepsilon^2\tilde{\omega}^\varepsilon_i)\,{\rm d}\theta(x)+o(1). \]

Since $\mathcal {E}(\tilde {\omega }^\varepsilon )\le \mathcal {E}({\omega }^\varepsilon )$, we conclude that

\begin{align*} & \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\xi_i^\varepsilon(x)\xi_i^\varepsilon(y)\,{\rm d}x\,{\rm d}y\\ & \quad \ge \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\tilde{\xi}^\varepsilon(x)\tilde{\xi}^\varepsilon(y)\,{\rm d}x\,{\rm d}y+o(1), \end{align*}

which implies that

\begin{align*} & \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\xi_i^*(x)\xi_i^*(y)\,{\rm d}x\,{\rm d}y\\ & \quad \geq \int_{B_{\bar{ R}_1}(0)}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x-y|}\tilde{\xi}_i^*(x)\tilde{\xi}_i^*(y)\,{\rm d}x\,{\rm d}y. \end{align*}

Thus the equality holds in (4.32). By the strict Riesz's rearrangement inequality (see theorem 3.9, [Reference Lieb and Loss26]), there exists a translation $\mathcal {T}$ such that $\mathcal {T}\xi _i^*=\tilde {\xi }_i^*$. Since

\[ \int_{B_{\bar{ R}_1}(0)}x\xi_i^*(x)\,{\rm d}x=\int_{B_{\bar{ R}_1}(0)}x\tilde{\xi}_i^*(x)\,{\rm d}x=0, \]

we get $\xi _i^*=\tilde {\xi }^*_i$.

Proof. For any $R>\bar { R}_1$, notice that $\xi _i^\varepsilon$ is uniformly bounded in $L^\infty (B_{2R}(0))$. Thus, by (4.28) and classical elliptic estimates, $\Psi ^\varepsilon _i$ is uniformly bounded in $W^{2,p}(B_{R}(0))$ for every $1\le p<+\infty$. By the Sobolev embedding theorem, we may conclude that $\Psi ^\varepsilon _i$ is compact in $C^{1,\alpha }(B_{R}(0))$ for every $0<\alpha <1$. Then up to a subsequence we may assume $\Psi ^\varepsilon \to \Psi$ in $C^{1,\alpha }(B_{R}(0))$. By (4.28) and (4.29), $\Psi$ satisfies

\[ \begin{cases} -\Delta \Psi =k(x^*_i)\xi_i^*=k(x^*_i)f_i(\Psi)\quad x\in\mathbb{R}^2,\\ \displaystyle\int_{\mathbb{R}^2}f_i(\Psi) \,{\rm d}x=\frac{|d_i|}{k(x^*_i)}. \end{cases} \]

By lemma 4.9, $\xi _i^*$ is a radial function. Using the Green's function representation, $\Psi$ is also radial. By the uniqueness of radial solutions of (4.30), we have $\Psi =\Psi ^*_i$.

By the strong maximum principle, one can show that for any $x\in \{x\in B_{\bar { R}_1}(0)\mid \Psi _i^*(x)=0 \}$, $|\nabla \Psi _i^*(x)|\neq 0$. Thus by the implicit function theorem, we get that for $\varepsilon$ sufficiently small, $\{x\in B_{\bar { R}_1}(0)\mid \Psi _i^\varepsilon (x)=0 \}$ is a $C^1$ curve and converges to $\{x\in B_{\bar { R}_1}(0)\mid \Psi _i^*(x)=0 \}$ in $C^1$ sense as $\varepsilon \to 0$.

Proof. Using (4.27) and lemma 4.10, one can immediately get the result.

Proof. The proof is based on the proof of proposition 4.7. By (4.24), we have

(4.34)\begin{align} \mathcal{E}(\omega^\varepsilon) & = \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l\frac{k(x_{0,i})^2}{4\pi}\iint\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\mathcal{H}_l(x_{0,1},\ldots,x_{0,l}) -\sum_{i=1}^lk(x_{0,i})\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x+o(1). \end{align}

By lemma 4.10 and corollary 4.11, we obtain

\begin{align*} & \frac{k(x_{0,i})^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}\xi^\varepsilon_i(x)\xi^\varepsilon_i(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad =\frac{k(x_{0,i})^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}f_i\circ\Psi_i^*(x)f_i\circ\Psi_i^*(y)\,{\rm d}x\,{\rm d}y+o(1),\\ & k(x_{0,i})\int_{\mathbb{R}^2}F_i(\xi_i^\varepsilon)(x)\,{\rm d}x=k(x_{0,i})\int_{\mathbb{R}^2}F_i\circ f_i\circ\Psi_i^*(x)\,{\rm d}x+o(1). \end{align*}

Taking those into (4.34), we have

(4.35)\begin{align} \mathcal{E}(\omega^\varepsilon) & = \sum_{i=1}^l\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\sum_{i=1}^l\frac{k(x_{0,i})^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}f_i\circ\Psi_i^*(x)f_i\circ\Psi_i^*(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad-\mathcal{H}_l(x_{0,1},\ldots,x_{0,l}) -\sum_{i=1}^lk(x_{0,i})\int_{\mathbb{R}^2}F_i\circ f_i\circ\Psi_i^*(x)\,{\rm d}x+o(1). \end{align}

However, by the definition of $\Psi _i^*$ and (4.31), we get

(4.36)\begin{align} & \sum_{i=1}^l\frac{k(x_{0,i})^2}{4\pi}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{1}{|x-y|}f_i\circ\Psi_i^*(x)f_i\circ\Psi_i^*(y)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad-\sum_{i=1}^lk(x_{0,i})\int_{\mathbb{R}^2}F_i\circ f_i\circ\Psi_i^*(x)\,{\rm d}x \nonumber\\ & \quad = \frac{1}{4\pi}\sum_{i=1}^l\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\ln\frac{k(x_{0,i})^\frac{1}{2}}{|x-y|}f_i\circ U_i(x)f_i\circ U_i(y)\,{\rm d}x\,{\rm d}y-\sum_{i=1}^l\int_{\mathbb{R}^2}F_i\circ f_i\circ U_i(x)\,{\rm d}x \nonumber\\ & \quad = \sum_{i=1}^l\frac{d_i^2}{8\pi}\ln k(x_{0,i})+C_0, \end{align}

where $C_0=({1}/{4\pi })\sum _{i=1}^l\int _{\mathbb {R}^2}\int _{\mathbb {R}^2}\ln ({1}/{|x-y|})f_i\circ U_i(x)f_i\circ U_i(y)\,{\rm d}x\,{\rm d}y-\sum _{i=1}^l\int _{\mathbb {R}^2}F_i\circ f_i\circ U_i(x)\,{\rm d}x$ is a constant dependent of $f_i$. Taking (4.36) into (4.35), we get the result.

Proof. Similarly as the proof of lemma 4.12, one can immediately get

\begin{align*} \mathcal{E}_i(\omega^\varepsilon) & =\frac{d_i^2}{4\pi}\ln\frac{1}{\varepsilon}+\frac{d_i^2}{8\pi}\ln k(x_{0,i})-\frac{d_i^2}{2}h(x_{0,i},x_{0,i})\\ & \quad +\frac{1}{4\pi} \iint_{\mathbb{R}^2\times \mathbb{R}^2}\ln\frac{1}{|x-y|}f_i\circ U_i(x)f_i\circ U_i(y)\,{\rm d}x\,{\rm d}y\\ & \quad - \int_{\mathbb{R}^2}F_i\circ f_i\circ U_i(x)\,{\rm d}x+o(1). \end{align*}

For (4.38), notice that for any $x\in \partial (supp(\omega _i^\varepsilon ))$, $\sigma _i^\varepsilon =sgn(d_i)\int _{D}G(x,\,y) \omega ^\varepsilon (y) \,{\rm d}\theta (y)$, which implies that

(4.39)\begin{align} \sigma_i^\varepsilon & =\frac{1}{2\pi}\int_{B_i}\ln\frac{1}{|x-y|}|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y)-|d_i|h(x_{0,i},x_{0,i})\nonumber\\ & \quad +sgn(d_i)\sum_{j=1,j\neq i}^ld_jG(x_{0,i},x_{0,j})+o(1). \end{align}

Let $x=\varepsilon x'+\hat {X}^\varepsilon _i$, then $x'\in \partial (supp(\xi _i^\varepsilon ))$. So

(4.40)\begin{align} & \frac{1}{2\pi}\int_{B_i}\ln\frac{1}{|x-y|}|\omega_i^\varepsilon(y)|\,{\rm d}\theta(y) \nonumber\\ & \quad =\frac{|d_i|}{2\pi}\ln\frac{1}{\varepsilon}+\frac{1}{2\pi}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x'-y'|}k(\varepsilon y'+\hat{X}^\varepsilon_i)\xi_i^\varepsilon(y')\,{\rm d}y'. \end{align}

By lemma 4.10 and the continuity of $k$, we have

(4.41)\begin{align} & \frac{1}{2\pi}\int_{B_{\bar{ R}_1}(0)}\ln\frac{1}{|x'-y'|}k(\varepsilon y'+\hat{X}^\varepsilon_i)\xi_i^\varepsilon(y')\,{\rm d}y' \nonumber\\ & \quad =\frac{k(x_{0,i})}{2\pi}\int_{\mathbb{R}^2}\ln\frac{1}{|x'-y'|}f_i\circ\Psi_i^*(y')\,{\rm d}y'+o(1) \nonumber\\ & \quad =\frac{1}{2\pi}\int_{\mathbb{R}^2}\ln\frac{k(x_{0,i})^\frac{1}{2}}{|x^*-y'|}f_i\circ U_i(y')\,{\rm d}y'+o(1) \nonumber\\ & \quad =\frac{|d_i|}{4\pi}\ln k(x_{0,i})+\frac{1}{2\pi}\int_{\mathbb{R}^2}\ln\frac{1}{|x^*-y'|}f_i\circ U_i(y')\,{\rm d}y'+o(1), \end{align}

where $x^*$ is any point of $\partial \{x\in \mathbb {R}^2\mid f_i\circ U_i(x)=0\}$. Taking (4.40), (4.41) into (4.39), we get (4.38).

Proof. The proof is similar to the proof of theorem 2.3. By lemma 4.5, we know that $\omega ^\varepsilon$ has the form

\[ \omega^\varepsilon= \sum_{i=1}^lsgn(d_i)\frac{1}{ \varepsilon^2}f_i(\psi_i^{\varepsilon})\chi_{\{ \psi_i^{\varepsilon}>0 \}\cap B_i }, \]

where $\psi _i^{\varepsilon }=sgn(d_i)\mathcal {G}(k\omega ^{\varepsilon })-\sigma ^{\varepsilon }_i$ for some $\sigma ^{\varepsilon }_i\in \mathbb {R}.$ By lemma 4.6, $diam(supp(\omega _i^\varepsilon ))=O(\varepsilon )$.

Moreover, by proposition 4.7, the support set of $\omega _i^\varepsilon$ tends to $x_{0,i}$ as $\varepsilon \to 0^+$, namely

\[ \lim_{\varepsilon \to 0^+}\sup_{x\in supp(\omega_i^\varepsilon)}|x- x_{0,i}|=0. \]

By lemmas 4.10 and 4.13, we get (3)(4) in theorem 2.4. The proof of theorem 2.4 is thus complete.

Proof. Let $l=1$ and $d_1=I$ in theorem 2.4, we get theorem 2.2. Let $\lambda ={1}/{\varepsilon ^2}$ and $u^\lambda =u^\varepsilon,\, c^\lambda =-\mu ^\varepsilon$, we get theorem 1.3.