Proof. The function $(x,\,y)\mapsto W(x-y)+\frac 12 V(x)+\frac 12 V(y)$ is lower semicontinuous and blows up at infinity by (2.4), hence it is bounded from below by a constant $-c_1$ with $c_1>0$. Therefore, the energy $I$ in (2.6) is well defined (possibly equal to $+\infty$) and $\inf I>-\infty$. Note also that the two representations (1.1) and (2.4) coincide whenever the interaction energy is well defined and not equal to $-\infty$.

Assumption (2.5) guarantees that $\inf I<+\infty$. Indeed, writing $\{V<+\infty \}$ as the union over $n\in \mathbb {N}$ of the compact sets $\{V\leq n\}$, we must have $\operatorname {cap}(\{V\leq n_0\})>0$ for some $n_0\in \mathbb {N}$, that is, there exists a probability measure $\mu _0$ with support in $\{V\leq n_0\}$ such that

\[ \iint_{\mathbb{R}^2\times\mathbb{R}^2}W_0(x-y)\,{\rm d}\mu_0(x)\,{\rm d}\mu_0(y)<{+}\infty. \]

By (2.2) this implies that

\[ \iint_{\mathbb{R}^2\times\mathbb{R}^2}W(x-y)\,{\rm d}\mu_0(x)\,{\rm d}\mu_0(y)<{+}\infty. \]

On the other hand,

\[ \int_{\mathbb{R}^2}V(x)\,{\rm d}\mu_0(x)\leq n_0, \]

hence $I(\mu _0)<+\infty$.

Existence of a minimizer follows by the Direct Method of the Calculus of Variations. Let $(\mu _n)_n$ be a minimizing sequence. Since $\inf I<+\infty$, there exists a constant $C>0$ such that $I(\mu _n)\leq C$ for every $n\in \mathbb {N}$. By (2.4) for every $M>0$ there exists a compact set $K\subset \mathbb {R}^2$ such that

\[ W(x-y)+\frac12 V(x)+\frac12 V(y)\geq M \quad \text{ for } (x,y)\not\in K\times K. \]

Therefore, for every $n\in N$:

\begin{align*} C \ \geq\ I(\mu_n) & \geq M(\mu_n\otimes\mu_n)\big((K\times K)^c\big) -c_1\big(\mu_n(K)\big)^2 \\ & \geq M\big(1- \big(\mu_n(K)\big)^2\big) -c_1 \\ & \geq M \mu_n(K^c) -c_1. \end{align*}

This inequality implies that the sequence $(\mu _n)_n$ is tight, hence, up to subsequences, $(\mu _n)_n$ converges narrowly to some $\mu \in \mathcal {P}(\mathbb {R}^2)$ (we refer to [Reference Ambrosio, Gigli and Savaré1, Chapter 5] for the definition of tightness, narrow convergence, and their properties). Since the integrand in $I$ is lower semicontinuous and bounded from below, we have:

\[ I(\mu)\leq\liminf_{n\to\infty} I(\mu_n), \]

hence $\mu$ is a minimizer.

Let now $\mu$ be a minimizer. In particular, $I(\mu )<+\infty$. By (2.4) there exists a compact set $K\subset \mathbb {R}^2$ such that

(2.9)\begin{equation} W(x-y)+\frac12 V(x)+\frac12 V(y)\geq I(\mu)+1 \quad \text{ for } (x,y)\not\in K\times K. \end{equation}

By taking $K$ larger if needed, we can assume that $\mu (K)>0$. We claim that ${\operatorname {supp}\mu \subset K}$. Assume by contradiction that $\mu (K)<1$ and define:

\[ \tilde\mu:=\frac{1}{\mu(K)}\mu{\unicode{x1D5AB}} K \in\mathcal{P}(\mathbb{R}^2). \]

By (2.9) we deduce that

\begin{align*} I(\tilde\mu) & = \frac{1}{\big(\mu(K)\big)^2}\left(I(\mu) - \iint_{(K\times K)^c}\Bigg(W(x-y)+\frac12 V(x)+\frac12 V(y)\Bigg)\,{\rm d}\mu(y)\,{\rm d}\mu(x) \right) \\ & \leq \frac{1}{\big(\mu(K)\big)^2} \big( I(\mu) -(I(\mu)+1)(1-\big(\mu(K)\big)^2) \big) \\ & = I(\mu) +1-\frac1{\big(\mu(K)\big)^2} < I(\mu), \end{align*}

where the last inequality follows from the fact that $\mu (K)<1$. This contradicts the minimality of $\mu$.

We conclude by showing that $\mu$ satisfies (2.7)–(2.8). Let $\nu \in \mathcal {P}(\mathbb {R}^2)$ be a competitor such that its support is compact and $I(\nu )<+\infty$. For $\varepsilon \in (0,\,1)$ we have that $(1-\varepsilon )\mu +\varepsilon \nu \in \mathcal {P}(\mathbb {R}^2)$. Therefore, by minimality:

\[ I(\mu) \leq I((1-\varepsilon)\mu+\varepsilon\nu). \]

Expanding the energy on the right-hand side yields:

\[ 0\leq{-}2\varepsilon\int_{\mathbb{R}^2} W\ast\mu\,{\rm d}\mu +2\varepsilon\int_{\mathbb{R}^2} W\ast\mu\,{\rm d}\nu + \varepsilon\int_{\mathbb{R}^2}V\,d(\nu-\mu)+O(\varepsilon^2). \]

Dividing by $2\varepsilon$ and sending $\varepsilon \to 0^+$ lead to the following inequality:

(2.10)\begin{equation} \int_{\mathbb{R}^2} \Bigg(W\ast\mu+\frac12 V\Bigg)\,{\rm d}\nu\geq \int_{\mathbb{R}^2} \Bigg(W\ast\mu+\frac12V\Bigg)\,{\rm d}\mu =:c \end{equation}

for every $\nu \in \mathcal {P}(\mathbb {R}^2)$ with compact support and $I(\nu )<+\infty$.

Set $P:=W\ast \mu +\frac 12 V$ and assume by contradiction that

\[ \operatorname{cap}\big(\big\{x\in\mathbb{R}^2: \ P(x)< c\big\}\big)>0. \]

Note that by Fatou's lemma $P$ is lower semicontinuous, thus $\{P< c\}$ is a Borel set. By definition of capacity there exists a compact set $K\subset \mathbb {R}^2$ of positive capacity such that $P(x)< c$ for every $x\in K$. Therefore, there exists $\nu \in \mathcal {P}(K)$ such that

\[ \iint_{\mathbb{R}^2\times\mathbb{R}^2}W(x-y)\,{\rm d}\nu(x)\,{\rm d}\nu(y)<{+}\infty. \]

Moreover, the confinement energy of $\nu$ is also finite, since

(2.11)\begin{equation} \int_{\mathbb{R}^2}\frac12V\,{\rm d}\nu = \int_K \frac12V\,{\rm d}\nu < \int_K (c-W\ast\mu)\,{\rm d}\nu = c-\int_K W\ast\mu\,{\rm d}\nu \end{equation}

and the right-hand side is finite by the bound from below of $W$ on compact sets. Having finite energy and compact support, the measure $\nu$ has to satisfy condition (2.10). However, (2.11) contradicts (2.10). This proves (2.8).

To prove (2.7) we note that by (2.8) we have $W\ast \mu +\frac 12V\geq c$ q.e., hence $\mu$-a.e. On the other hand, $c$ is by definition the integral of $W\ast \mu +\frac 12 V$ with respect to $\mu$, so necessarily (2.7) holds true.

Proof of proposition 3.2 The heuristic idea to prove (3.7) is to apply Plancherel theorem, as we did formally in (1.3), and write:

\[ \int_{\mathbb{R}^2}W\ast\nu\,{\rm d}\nu =2\pi\int_{\mathbb{R}^2}\widehat W(\xi)|\widehat\nu(\xi)|^2\,{\rm d}\xi. \]

Taking into account (3.5) and the fact that

\[ \widehat\nu(0)=\frac1{2\pi}\int_{\mathbb{R}^2}\,{\rm d}\nu=0, \]

the right-hand side reduces to

\[ 2\pi\int_{\mathbb{R}^2}\widehat W(\xi)|\widehat\nu(\xi)|^2\,{\rm d}\xi = 2\pi\int_{\mathbb{R}^2}\frac{\widehat\Psi(\xi)}{|\xi|^2}|\widehat\nu(\xi)|^2\,{\rm d}\xi. \]

However, this argument is only formal, since we do not have enough regularity to apply Plancherel theorem. Our strategy is to prove (3.7) by approximation. This is a rather delicate argument, since both sides of (3.7) may a priori be not finite and neither $W$ nor $\nu$ have a sign.

For $\varepsilon >0$, let $\varphi _\varepsilon$ be a radial mollifier supported on $B_\varepsilon (0)$. Let $\nu _\varepsilon :=\nu \ast \varphi _\varepsilon \in C^\infty _c(\mathbb {R}^2)\subset \mathcal {S}$. Thus, $\widehat {\nu _\varepsilon }$ belongs to $\mathcal {S}$ and, in particular, to $L^\infty (\mathbb {R}^2)$. We note that $W\ast \nu _\varepsilon \in C^\infty (\mathbb {R}^2)$ and, since $\widehat {\nu _\varepsilon }$ is smooth and $\widehat {\nu _\varepsilon }(0)=0$, we have that

\[ \widehat{W\ast\nu_\varepsilon}=2\pi\widehat W\widehat{\nu_\varepsilon}\in L^1(\mathbb{R}^2). \]

Using these properties one can show that Plancherel theorem holds for $W\ast \nu _\varepsilon$ and $\nu _\varepsilon$, that is,

(3.8)\begin{equation} \int_{\mathbb{R}^2}(W\ast\nu_\varepsilon)(x)\nu_\varepsilon(x)\,{\rm d}x =2\pi\int_{\mathbb{R}^2}\widehat W(\xi)|\widehat{\nu_\varepsilon}(\xi)|^2\,{\rm d}\xi = 2\pi\int_{\mathbb{R}^2}\frac{\widehat\Psi(\xi)}{|\xi|^2}|\widehat{\nu_\varepsilon}(\xi)|^2\,{\rm d}\xi \end{equation}

for every $\varepsilon >0$.

We now want to pass to the limit in (3.8), as $\varepsilon \to 0^+$. For every $\xi \in \mathbb {R}^2$ we have

\[ \widehat{\nu_\varepsilon}(\xi)=2\pi\widehat\nu(\xi)\widehat\varphi(\varepsilon\xi) \ \to 2\pi\widehat\nu(\xi)\widehat\varphi(0)=\widehat\nu(\xi), \]

as $\varepsilon \to 0^+$. Therefore, $\widehat \Psi (\xi )|\widehat {\nu _\varepsilon }(\xi )|^2/|\xi |^2$ converges to $\widehat \Psi (\xi )|\widehat \nu (\xi )|^2/|\xi |^2$ for a.e. $\xi \in \mathbb {R}^2$, as $\varepsilon \to 0^+$. Moreover, $|\widehat \varphi (\varepsilon \cdot )|\leq C\|\varphi \|_{L^1}$. Either by dominated convergence or by Fatou's lemma we can thus pass to the limit in the last integral in (3.8).

To pass to the limit on the left-hand side of (3.8), we observe that

\[ \int_{\mathbb{R}^2}(W\ast\nu_\varepsilon)(x)\nu_\varepsilon(x)\,{\rm d}x =\int_{\mathbb{R}^2}(W\ast\varphi_\varepsilon\ast\varphi_\varepsilon)\ast\nu\,{\rm d}\nu, \]

where we used that $\varphi _\varepsilon$ is a radial, hence even, function. Set $\psi _\varepsilon :=\varphi _\varepsilon \ast \varphi _\varepsilon$ and note that $\psi _\varepsilon$ has the same properties as $\varphi _\varepsilon$ (it is radial, compactly supported, non-negative, and integrates to one). Since $W$ is continuous as a function with values in $\mathbb {R}\cup \{+\infty \}$, we have that $(W\ast \psi _\varepsilon )(x)\to W(x)$, as $\varepsilon \to 0^+$, for every $x\in \mathbb {R}^2$. Moreover, by (2.2):

(3.9)\begin{equation} (W\ast\psi_\varepsilon)(x) \leq (W_0\ast\psi_\varepsilon)(x)+C_2 \quad \text{ for every } x\in\mathbb{R}^2. \end{equation}

Since $W_0$ is superharmonic and $\psi _\varepsilon$ is radial, we have

(3.10)\begin{equation} (W_0\ast\psi_\varepsilon)(x) \leq W_0(x) \leq W(x)- C_1 \quad \text{ for every } x\in\mathbb{R}^2, \end{equation}

where the last inequality follows again from (2.2).

Let $M>0$ be such that $\operatorname {supp}\nu \subset B_M(0)$ and let $c_M$ be the minimum of $W$ on $B_{4M}(0)$. Combining (3.9) and (3.10), we deduce that

(3.11)\begin{equation} 0\leq (W\ast\psi_\varepsilon)(x) -c_M\leq W(x)+C_2- C_1 -c_M \quad \text{ for every } x\in B_{2M}(0) \end{equation}

and for every $\varepsilon >0$ small enough.

We now write:

\begin{align*} \int_{\mathbb{R}^2}(W\ast\psi_\varepsilon)\ast\nu\,{\rm d}\nu & = \iint_{\mathbb{R}^2\times\mathbb{R}^2}(W\ast\psi_\varepsilon)(x-y)\,{\rm d}\mu_0(x)\,{\rm d}\mu_0(y) \\ & \quad + \iint_{\mathbb{R}^2\times\mathbb{R}^2}(W\ast\psi_\varepsilon)(x-y)\,{\rm d}\mu_1(x)\,{\rm d}\mu_1(y) \\ & \quad -2 \iint_{\mathbb{R}^2\times\mathbb{R}^2}(W\ast\psi_\varepsilon)(x-y)\,{\rm d}\mu_0(x)\,{\rm d}\mu_1(y). \end{align*}

Since $\mu _0$ and $\mu _1$ have finite interaction energy, we can pass to the limit in the first two integrals on the right-hand side by (3.11) and dominated convergence. As for the last integral, it goes to the limit either by dominated convergence or by Fatou's lemma. This completes the proof of (3.7).

To conclude, assume that

(3.12)\begin{equation} 0=\int_{\mathbb{R}^2}W\ast\nu\,{\rm d}\nu =2\pi\int_{\mathbb{R}^2}\frac{\widehat\Psi(\xi)}{|\xi|^2}|\widehat\nu(\xi)|^2\,{\rm d}\xi. \end{equation}

Since $\widehat \Psi$ cannot be identically zero (otherwise, $W$ would be constant) and is continuous outside $0$, there exist $\delta >0$, $\eta >0$, and $\xi _0\neq 0$ such that $\widehat \Psi (\xi )>\eta$ for all $\xi \in B_\delta (\xi _0)$. Thus, (3.12) implies that $\widehat \nu =0$ on $B_\delta (\xi _0)$. On the other hand, $\nu$ is a distribution with compact support, so by Paley–Wiener theorem $\widehat \nu$ is the restriction to $\mathbb {R}^2$ of an entire function. Therefore, $\widehat \nu$ has to be identically zero on $\mathbb {R}^2$, that is, $\nu =0$.

Proof of theorem 4.1 By applying the Gauss averaging principle we can explicitly compute the Coulomb potential of $\mu _0$: indeed, using polar coordinates we have:

\[ (W_0\ast\mu_0)(x) ={-}\frac1\pi \int_{B_1(0)}\log|x-y|\,{\rm d}y ={-}\frac1\pi \int_0^1\int_{-\pi}^\pi \log|x-r {\rm e}^{i\theta}|\,{\rm d}\theta\, r\,{\rm d}r. \]

By (4.2) we deduce that

\[ (W_0\ast\mu_0)(x) = \begin{cases} \displaystyle\frac12-\frac12|x|^2 & \text{ for } |x|\leq 1,\\ - \log|x| & \text{ for } |x|>1. \end{cases} \]

Using this formula, one can immediately check that $(W_0\ast \mu _0)(x)+\frac 12 |x|^2=\frac 12$ for $x\in B_1(0)$ and $(W_0\ast \mu _0)(x)+\frac 12 |x|^2\geq \frac 12$ for $x\not \in B_1(0)$, that is, $\mu _0$ satisfies the Euler–Lagrange conditions. By the results of the previous section this is enough to conclude that $\mu _0$ is the unique minimizer of $I_0$.

Proof of theorem 4.2 We write a general domain enclosed by an ellipse as $E=RE_0$, where $R\in SO(2)$ and

\[ E_0=\Bigg\{ x\in\mathbb{R}^2: \ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}\leq 1\Bigg\} \]

and we set $\chi :=\chi _E/|E|$. The theorem is proved if we show that there exist $a_1,\,a_2>0$ and $R\in SO(2)$ such that

(4.4)\begin{equation} (W\ast\chi)(x)+\frac12 |x|^2=c \quad \text{for every } x\in E, \end{equation}

and

(4.5)\begin{equation} (W\ast\chi)(x)+\frac12 |x|^2\geq c \quad \text{for every } x\in \mathbb{R}^2\setminus E \end{equation}

for some constant $c\in \mathbb {R}$.

Note that $W\ast \chi$ is a $C^1$ function in $\mathbb {R}^2$. To compute its expression we would like to make use of the inversion formula, that is, write $W\ast \chi$ as an integral of its Fourier transform. However, $\widehat W$ is not a function, it is only a tempered distribution. To circumvent this difficulty we apply this strategy to the gradient of $W\ast \chi$. In fact, the Fourier transform of $\nabla (W\ast \chi )$ is given by

\[ i\xi\widehat{W\ast\chi}(\xi)=2\pi i\xi\widehat W(\xi)\widehat\chi(\xi), \]

so the presence of the factor $\xi$ annihilates the singular part of $\widehat W$. Note that ${\nabla (W\ast \chi)}$ is a continuous function that behaves as $1/|x|$ at infinity, so it is a tempered distribution.

Step 1: The inversion formula for $\nabla (W\ast \chi )$. We recall that

\[ \widehat{\chi_{B_1(0)}}(\xi)=\frac{J_1(|\xi|)}{|\xi|}, \]

where $J_1$ denotes the Bessel function of the first kind of order $1$. There is no explicit formula for $J_1$; however, it is well known that $J_1$ has the following behaviour:

(4.6)\begin{align} & J_1(|\xi|)\simeq \frac12|\xi| \quad \text{ as } \xi\to0, \end{align}

(4.7)\begin{align} & |J_1(|\xi|)|\leq \frac{C}{|\xi|^{1/2}} \quad \text{ as } |\xi|\to\infty, \end{align}

see, e.g. [Reference Lebedev13, Section 5.16].

For $a=(a_1,\,a_2)\in \mathbb {R}^2$ we denote the diagonal matrix ${\rm diag}(a_1,\,a_2)$ by $D(a)$. By writing $x\in E$ as $x=RD(a)y$ with $y\in B_1(0)$, we obtain:

\[ \widehat\chi(\xi)=\frac1\pi\widehat{\chi_{B_1(0)}}\big(D(a)R^T\xi\big)=\frac1\pi \frac{J_1(|D(a)R^T\xi|)}{|D(a)R^T\xi|}. \]

By the bounds (4.6)–(4.7) we deduce that the function $2\pi i\xi \widehat W\widehat \chi$ belongs to $L^1(\mathbb {R}^2)$. Therefore, since $\nabla (W\ast \chi )$ is a continuous function, the inversion formula holds, that is:

\[ \nabla(W\ast\chi)(x)=\int_{\mathbb{R}^2} i\xi\frac{\widehat\Psi(\xi)}{|\xi|^2}\widehat\chi(\xi){\rm e}^{ix\cdot\xi}\,{\rm d}\xi ={-}\int_{\mathbb{R}^2}\xi\frac{\widehat\Psi(\xi)}{|\xi|^2}\widehat\chi(\xi)\sin(x\cdot\xi)\,{\rm d}\xi \]

for every $x\in \mathbb {R}^2$, where we used that $\widehat \Psi$ and $\widehat \chi$ are even functions. Passing to polar coordinates we obtain:

\begin{align*} \nabla(W\ast\chi)(x) & ={-}\int_{\mathbb{S}^1}\int_0^\infty y\widehat\Psi(y)\widehat\chi(\rho y)\sin(\rho x\cdot y)\,{\rm d}\rho\,{\rm d}{\mathcal{H}}^1(y) \\ & ={-}\frac1\pi \int_{\mathbb{S}^1}\int_0^\infty y \widehat\Psi(y) \frac{J_1(\rho |D(a)R^Ty|)}{\rho |D(a)R^Ty|} \sin(\rho x\cdot y)\,{\rm d}\rho \,{\rm d}{\mathcal{H}}^1(y). \end{align*}

Setting $r:=\rho |D(a)R^Ty|$, we deduce that

(4.8)\begin{equation} \nabla(W\ast\chi)(x)={-}\frac1\pi \int_{\mathbb{S}^1}\frac{y \widehat\Psi(y)}{|D(a)R^Ty|} \int_0^\infty \frac{J_1(r)}{r} \sin(r \alpha(x,y))\,{\rm d}r\,{\rm d}{\mathcal{H}}^1(y) \end{equation}

for every $x\in \mathbb {R}^2$, where we set:

(4.9)\begin{equation} \alpha(x,y):=\frac{x\cdot y}{|D(a)R^Ty|}. \end{equation}

Using formula (5), p. 99, in [Reference Erdélyi, Magnus, Oberhettinger and Tricomi5], one can compute the improper integral:

(4.10)\begin{equation} \int_0^\infty \frac{J_1(r)}{r} \sin(r \alpha)\,{\rm d}r=\begin{cases} \alpha & \text{ if } 0\leq\alpha\leq 1, \\ \dfrac1{\alpha+\sqrt{\alpha^2-1}} & \text{ if } \alpha>1. \end{cases} \end{equation}

Moreover, we have that, if $x\in E$, then

\[ |x\cdot y|=|D(a)^{{-}1}R^Tx\cdot D(a)R^Ty|\leq |D(a)R^Ty|, \]

that is, $|\alpha (x,\,y)|\leq 1$ for every $x\in E$. We conclude that

(4.11)\begin{align} \nabla(W\ast\chi)(x)& ={-}\frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(y)}{|D(a)R^Ty|} \alpha(x,y)y \,{\rm d}{\mathcal{H}}^1(y)\nonumber\\ & \quad ={-}\frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(y)}{|D(a)R^Ty|^2}(x\cdot y)y \,{\rm d}{\mathcal{H}}^1(y) \end{align}

for every $x\in E$. Formula (4.11) shows that $\nabla (W\ast \chi )$ is a homogeneous polynomial of degree $1$ inside $E$. In other words, up to an additive constant, $W\ast \chi$ is a homogeneous polynomial of degree $2$ inside $E$.

Step 2: Solving the first Euler–Lagrange condition. Solving (4.4) corresponds to finding $a_1,\,a_2>0$ and $R\in SO(2)$ such that

\[\nabla{(W\ast\chi)}(x) + x = 0 \quad \text{for every } x \in E.\]

By formula (4.11) this translates into the following system of three equations:

(4.12)\begin{equation} \frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(y)}{|D(a)R^Ty|^2} y_j y_k\,{\rm d}{\mathcal{H}}^1(y)=\delta_{jk} \quad \text{ for every } j,k=1,2, \end{equation}

where $\delta _{jk}$ is the Kronecker delta.

Let us denote by $\mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ the set of positive definite $2\times 2$ symmetric matrices. Note that $M:=RD(a)^2R^T\in \mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ and $|D(a)R^Ty|^2=My\cdot y$. Solving system (4.12) is therefore equivalent to finding $M\in \mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ such that

(4.13)\begin{equation} \frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(y)}{My\cdot y} y_j y_k \,{\rm d}{\mathcal{H}}^1(y)=\delta_{jk} \quad \text{ for every } j,k=1,2. \end{equation}

Let us denote the entries of $M$ by $M_{jk}$ for $j,\,k=1,\,2$. Note that, if we multiply by $M_{jk}$ the $jk$ equation in (4.13) and we sum over $j,\,k$, we obtain:

\[ \operatorname{tr}(M)=M_{11}+M_{22}=\frac1\pi \int_{\mathbb{S}^1} \widehat\Psi(y) \,{\rm d}{\mathcal{H}}^1(y)=2, \]

where we used (3.6). Since $\operatorname {tr}(M)=a_1^2+a_2^2$, where $a_1,\,a_2$ are the semi-axes of the candidate ellipse, we deduce that, if a minimizing ellipse exists, then necessarily its semi-axes satisfy the relation:

\[ a_1^2+a_2^2=2. \]

To solve (4.13) it is convenient to introduce the function:

\[ f(M):={-}\frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(y)\log(My\cdot y) \,{\rm d}{\mathcal{H}}^1(y) +\operatorname{tr}(M) \]

defined for every $M\in \mathbb {M}^{2\times 2}_{\operatorname {sym},+}$. It is immediate to see that conditions (4.13) are satisfied if we find $M_0\in \mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ such that $\nabla _{\!M} f(M_0)=0$, where with a slight abuse of notation we denoted by $\nabla _{\!M}$ the operator:

\[ \nabla_{\!M}=\Bigg(\frac{\partial}{\partial M_{11}}, \frac{\partial}{\partial M_{22}}, \ \frac{\partial}{\partial M_{12}}\Bigg). \]

We claim that $f$ has a minimizer in $\mathbb {M}^{2\times 2}_{\operatorname {sym},+}$. Since $\mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ is an open set, the claim will imply the existence of a critical point of $f$ in this set and thus, of a solution to (4.13).

Given $M\in \mathbb {M}^{2\times 2}_{\operatorname {sym},+}$, we first look at the behaviour of the function $f$ along the line $tM$ for $t>0$, that is, we consider the function:

\[ g(t):=f(tM)={-}\frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(y)\log(My\cdot y) \,{\rm d}{\mathcal{H}}^1(y) -2\log t+t \operatorname{tr}(M) \]

for $t>0$, where the last equality follows from (3.6). It is immediate to see that $g$ is minimized at $t_\ast =2/\operatorname {tr}(M)$. Therefore, minimizing $f$ over $\mathbb {M}^{2\times 2}_{\operatorname {sym},+}$ is equivalent to minimizing $f$ on the set:

\[ \mathcal{M}:=\big\{M\in\mathbb{M}^{2\times2}_{\operatorname{sym},+}: \ \operatorname{tr}(M)=2\big\}. \]

By diagonalization, any $M\in \mathcal {M}$ can be written as $M=QD(b)Q^T$ with $Q\in SO(2)$ and $b=(\beta,\,2-\beta )$, $\beta \in (0,\,2)$. Using this representation and a change of variables we have that

\[ f(M)=f(QD(b)Q^T)={-} \frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(Qy)\log(\beta y_1^2+(2-\beta)y_2^2) \,{\rm d}{\mathcal{H}}^1(y) +2. \]

Therefore, setting

\[ \gamma(\beta, Q):={-} \frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(Qy)\log(\beta y_1^2+(2-\beta)y_2^2) \,{\rm d}{\mathcal{H}}^1(y) \]

for every $\beta \in [0,\,2]$ and $Q\in SO(2)$, it is enough to show that $\gamma$ has a minimizer in $(0,\,2)\times SO(2)$ to conclude that $f$ has a minimizer in $\mathcal {M}$ and thus in $\mathbb {M}^{2\times 2}_{\operatorname {sym},+}$.

The function $\gamma$ is finite and continuous on the compact set $[0,\,2]\times SO(2)$. Therefore, it has a minimizer $(\beta _0,\,Q_0)$ in this set. It remains to prove that $\beta _0$ is neither $0$ nor $2$.

By assumption there exists a constant $C_0>0$ such that

\[ \widehat\Psi(\xi)\geq C_0 \quad \text{ for every } \xi\in\mathbb{S}^1. \]

Using this inequality and (3.6), we obtain that

\begin{align*} \frac{\partial}{\partial\beta}\gamma(\beta, Q_0) & = \frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(Q_0y)\frac{2y_2^2-1}{\beta y_1^2+(2-\beta)y_2^2} \,{\rm d}{\mathcal{H}}^1(y) \\ & \leq \frac4{2-\beta} - \frac1\pi \int_{\mathbb{S}^1}\frac{C_0}{\beta y_1^2+(2-\beta)y_2^2} \,{\rm d}{\mathcal{H}}^1(y) \end{align*}

for every $\beta \in (0,\,2)$. Since the right-hand side goes to $-\infty$ as $\beta \to 0^+$, we deduce that there exists some $\delta >0$ such that

\[ \frac{\partial}{\partial\beta}\gamma(\beta, Q_0) <0 \]

for every $\beta \in (0,\,\delta )$. Hence, $\beta _0$ cannot be equal to zero. One can show in a similar way that $\beta _0$ cannot be $2$ either. This concludes the proof of the step.

Step 3: The first Euler–Lagrange condition implies the second one. To complete the proof, we show that, if $\chi =\chi _E/|E|$ satisfies (4.4), then it satisfies (4.5), as well. Assume $\chi$ satisfies (4.4). By (4.4) it is enough to prove that the potential $(W\ast \chi )(x)+\frac 12 |x|^2$ grows in the radial direction, that is:

\[ \nabla(W\ast\chi)(x)\cdot x + |x|^2\geq0 \quad \text{for every } x\in \mathbb{R}^2\setminus E. \]

Let $x\in \mathbb {R}^2\setminus E$. We can write $x=tx^0$ with $x^0\in E$. By (4.4) we have that

\[ \nabla(W\ast\chi)(x^0)\cdot x^0 + |x^0|^2 = 0, \]

which can be written by (4.11) as

\[ -\frac1\pi \int_{\mathbb{S}^1} \widehat\Psi(y)\alpha^2(x^0,y)\,{\rm d}{\mathcal{H}}^1(y) + |x^0|^2 = 0, \]

where $\alpha$ is defined in (4.9). Multiplying this equation by $t^2$ yields:

(4.14)\begin{equation} -\frac1\pi \int_{\mathbb{S}^1} \widehat\Psi(y)\alpha^2(x,y)\,{\rm d}{\mathcal{H}}^1(y) + |x|^2 = 0. \end{equation}

By the inversion formula (4.8) and (4.10) we can write:

\begin{align*} & \nabla(W\ast\chi)(x)\cdot x +|x|^2 = |x|^2-\frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(y)\alpha^2(x,y)\chi_{[{-}1,1]}(\alpha(x,y))\,{\rm d}{\mathcal{H}}^1(y) \\ & \quad -\frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(y)\frac{|\alpha(x,y)|}{|\alpha(x,y)|+\sqrt{\alpha^2(x,y)-1}}\chi_{\mathbb{R}\setminus[{-}1,1]}(\alpha(x,y))\,{\rm d}{\mathcal{H}}^1(y). \end{align*}

Owing to (4.14), the expression above reduces to

\begin{align*} & \nabla(W\ast\chi)(x)\cdot x +|x|^2\\ & \quad = \frac1\pi \int_{\mathbb{S}^1}\widehat\Psi(y)|\alpha(x,y)|\sqrt{\alpha^2(x,y)-1}\,\chi_{\mathbb{R}\setminus[{-}1,1]}(\alpha(x,y))\,{\rm d}{\mathcal{H}}^1(y) \geq 0. \end{align*}

This concludes the proof.

Proof. For $\varepsilon >0$ we consider the kernels:

\[ W_\varepsilon(x)={-}(1+\varepsilon)\log|x|+\kappa(x) \]

and we denote by $I_\varepsilon$ the corresponding energies with confinement $|x|^2$. Arguing as in (3.4), it is immediate to see that

\[ \widehat{W_\varepsilon}(\xi)=c_\varepsilon\delta_0+\frac{\widehat\Psi(\xi)+\varepsilon}{|\xi|^2} \]

with $\widehat \Psi +\varepsilon >0$ on $\mathbb {S}^1$. By theorem 4.2 for every $\varepsilon >0$ the unique minimizer $\mu _\varepsilon$ of $I_\varepsilon$ is the normalized characteristic function of a set $E_\varepsilon$ of the form:

\[ R_\varepsilon^TE_\varepsilon= \Bigg\{ x\in\mathbb{R}^2: \ \frac{x_1^2}{(a_{1,\varepsilon})^2}+\frac{x_2^2}{(a_{2,\varepsilon})^2}\leq 1\Bigg\} \]

with $R_\varepsilon \in SO(2)$ and $(a_{1,\varepsilon })^2+(a_{2,\varepsilon })^2=2$. The sequence $(\mu _\varepsilon )_\varepsilon$ is tight, since the support of $\mu _\varepsilon$ is contained in the closed ball of centre $0$ and radius $\sqrt 2$ for every $\varepsilon >0$. Therefore, up to subsequences, $(\mu _\varepsilon )_{\varepsilon }$ converges narrowly to a measure $\mu _0\in \mathcal {P}(\mathbb {R}^2)$. We claim that $\mu _0$ is the minimizer of $I$. Indeed, since the supports of $\mu _\varepsilon$ and $\mu _0$ are uniformly bounded, on these sets $W_0$ is bounded from below by some constant $-c_0$, with $c_0>0$. Therefore:

\[ I_\varepsilon(\mu_\varepsilon)\geq I(\mu_\varepsilon)- c_0\varepsilon \]

and by lower semicontinuity:

\[ \liminf_{\varepsilon\to0^+} I_\varepsilon(\mu_\varepsilon)\geq \liminf_{\varepsilon\to0^+} I(\mu_\varepsilon) \geq I(\mu_0). \]

On the other hand, if $\mu$ is any measure in $\mathcal {P}(\mathbb {R}^2)$ with compact support and such that

(4.15)\begin{equation} \iint_{\mathbb{R}^2\times\mathbb{R}^2}\big(-\log|x-y|+\kappa(x-y)\big)\,{\rm d}\mu(y)\,{\rm d}\mu(x)<{+}\infty, \end{equation}

then by minimality:

\[ \limsup_{\varepsilon\to0^+} I_\varepsilon(\mu_\varepsilon)\leq \lim_{\varepsilon\to0^+} I_\varepsilon(\mu) = I(\mu). \]

Therefore, $\mu _0$ minimizes $I$ over all measures with compact support and satisfying (4.15). By theorem 2.1 and proposition 3.2 we conclude that $\mu _0$ is the minimizer of $I$ on the whole $\mathcal {P}(\mathbb {R}^2)$.

Up to subsequences, we can assume that the rotations and the semi-axes converge, that is, $R_\varepsilon \to R$, $a_{1,\varepsilon } \to a_1$, and $a_{2,\varepsilon }\to a_2$, as $\varepsilon \to 0^+$, with $R\in SO(2)$, $a_1,\,a_2\geq 0$, and $a_1^2+a_2^2=2$. Therefore, we have two cases: either both $a_1$ and $a_2$ are strictly positive, or one of them is $0$ and the other one is $\sqrt 2$. In the first case $\mu _0$ is the normalized characteristic function of the set $E$ given by

\[ R^TE= \Bigg\{ x\in\mathbb{R}^2: \ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}\leq 1\Bigg\}. \]

In the second case, $\mu _0$ is the push-forward of measure (1.7) through the rotation map $g(x)=Rx$ for $x\in \mathbb {R}^2$ if $a_1=0$, or $g(x)=RJx$ for $x\in \mathbb {R}^2$ if $a_2=0$. Here, $J$ denotes a $\pi /2$-rotation.

To conclude the proof assume that $a_1=0$ and $a_2=\sqrt 2$. We want to prove that $\widehat \Psi (R e_1)=0$. By (4.12) and the change of variable $z=R_\varepsilon ^Ty$ we have that

\[ \frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(R_\varepsilon z)+\varepsilon}{(a_{1,\varepsilon})^2z_1^2+(a_{2,\varepsilon})^2z_2^2} (R_\varepsilon z\cdot e_j)^2 \,{\rm d}{\mathcal{H}}^1(z)=1 \quad \text{ for } j=1,2 \]

for every $\varepsilon >0$. Summing over $j$ and applying Fatou lemma yield:

\[ \frac1{2\pi} \int_{\mathbb{S}^1} \frac{\widehat\Psi(R z)}{z_2^2} \,{\rm d}{\mathcal{H}}^1(z)\leq \liminf_{\varepsilon\to0} \frac1\pi \int_{\mathbb{S}^1} \frac{\widehat\Psi(R_\varepsilon z)+\varepsilon}{(a_{1,\varepsilon})^2z_1^2+(a_{2,\varepsilon})^2z_2^2} \,{\rm d}{\mathcal{H}}^1(z) = 2. \]

This implies that the function $z\mapsto z_2^{-2}\widehat \Psi (R z)$ is integrable on $\mathbb {S}^1$, hence $\widehat \Psi (R e_1)=0$.

The case $a_1=\sqrt 2$ and $a_2=0$ can be treated analogously.

Proof. We set $W_\alpha (x)=-\log |x|+\kappa _\alpha (x)$, where

\[ \kappa_\alpha(x):=\alpha \frac{x_1^2}{|x|^2}. \]

Since we can write:

\[ \kappa_\alpha (x)=\frac\alpha2 \frac{x_1^2-x_2^2}{|x|^2}+\frac\alpha2, \]

lemma 3.1 ensures that

(5.2)\begin{equation} \widehat{W_\alpha}(\xi)= c_\alpha\delta_0+\frac{\widehat\Psi_\alpha(\xi)}{|\xi|^2} \end{equation}

with

\[ \widehat\Psi_\alpha(\xi)=(1-\alpha)\frac{\xi_1^2}{|\xi|^2}+(1+\alpha)\frac{\xi_2^2}{|\xi|^2}, \quad \xi\neq0. \]

Assume $|\alpha |<1$. Since $\widehat \Psi _\alpha >0$ on $\mathbb {S}^1$ in this case, by theorem 4.2 the unique minimizer is the normalized characteristic function of the domain enclosed by an ellipse of semiaxes $a_{1,\alpha },\, a_{2,\alpha }$. We note that $W_\alpha$ is symmetric with respect to the coordinate axes, that is:

\[ W_\alpha({-}x_1,x_2)=W_\alpha(x_1,-x_2)=W_\alpha(x) \quad \text{ for every } x\in\mathbb{R}^2. \]

By uniqueness the minimizer must have the same symmetry, that is, the ellipse is symmetric with respect to the coordinate axes. Therefore, in system (4.12) the rotation $R$ is necessarily the identity matrix and the equation for $j=1$, $k=2$ is trivially satisfied by symmetry. In other words, using the expression of $\widehat \Psi _\alpha$, the semiaxes $a_{1,\alpha },\, a_{2,\alpha }$ satisfy:

\[ \frac1\pi \int_{\mathbb{S}^1} \frac{(1-\alpha)y_1^2+(1+\alpha)y_2^2}{(a_{1,\alpha})^2 y_1^2 + (a_{2,\alpha})^2 y_2^2} y_j^2\,{\rm d}{\mathcal{H}}^1(y)=1 \quad \text{ for every } j=1,2. \]

It is immediate to see that $a_{1,\alpha }=\sqrt {1-\alpha }$, $a_{2,\alpha }=\sqrt {1+\alpha }$ is a solution of this system. It is indeed the unique solution, since we proved in step 3 of the proof of theorem 4.2 that any solution of (4.12) is automatically a minimizer and the minimizer is unique. We conclude that for $|\alpha |<1$ the minimizer is given by the measure (1.9).

Let now $\alpha =1$. Arguing as in the proof of corollary 4.3, one can show that the minimizer of $I_1$ has to be the limit of the minimizer of $I_\alpha$, as $\alpha \to 1^-$. Therefore, the minimizer is the semicircle law (1.7). The same argument applies to $\alpha =-1$.

If $\alpha >1$, let us denote by $\mu _{\rm sc}$ the semicircle law in (1.7) and let $\mu$ be any other measure in $\mathcal {P}(\mathbb {R}^2)$. Then

\[ I_{\alpha}(\mu)\geq I_{1}(\mu)>I_1(\mu_{\rm sc})=I_\alpha(\mu_{\rm sc}), \]

where the last equality follows from the fact that the anisotropic interaction is $0$ on the semicircle law. A similar argument shows that (5.1) is the unique minimizer of $I_\alpha$ for $\alpha <-1$.

Proof of theorem 5.3 Up to rotating the axes and replacing $\kappa (x)$ by $\kappa (Rx)$, we can assume without loss of generality that $R$ is the identity matrix and $E=E_0$. Theorem 2.1 and proposition 3.2 guarantee that the minimizer exists and is unique. Moreover, it is characterized by Euler–Lagrange conditions (2.7)–(2.8), which take the following form:

(5.3)\begin{align} & (W\ast\mu)(x)=c \quad \text{ for } \mu\text{-a.e. } x\in\operatorname{supp}\mu, \end{align}

(5.4)\begin{align} & (W\ast\mu)(x)\geq c \quad \text{ for q.e. } x\in E_0. \end{align}

To prove the theorem it is enough to show that the measure $\mu _{\partial E_0}$ satisfies (5.3)–(5.4).

For $t>0$, we consider the set:

\[ E_t=\Bigg\{ x\in\mathbb{R}^2: \ \frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}\leq 1+t\Bigg\}. \]

By formula (4.11) for the potential of a general ellipse we have that

(5.5)\begin{equation} \nabla(W\ast\chi_{E_t})(x)={-} a_1a_2 \int_{\mathbb{S}^1} \frac{\widehat\Psi(y)}{|D(a)y|^2}(x\cdot y)y \,{\rm d}{\mathcal{H}}^1(y) \end{equation}

for every $x\in E_t$ and every $t\geq 0$.

Let now $x$ be in the interior of $E_0$. Since $x\in E_t$ for every $t\geq 0$, equation (5.5) holds for every $t\geq 0$. Differentiating both sides of (5.5) with respect to $t$ and applying the coarea formula yield:

\[ 0= \frac{{\rm d}}{{\rm d}t}\Bigg( \int_{E_t} \nabla W(x-y)\,{\rm d}y\Bigg) = \frac12 \int_{\partial E_t} \nabla W(x-y)\Bigg(\frac{y_1^2}{a_1^4}+\frac{y_2^2}{a_2^4}\Bigg)^{{-}1/2}\,{\rm d}{\mathcal{H}}^1(y) \]

for a.e. $t\geq 0$. For $x$ in the interior of $E_0$ the right-hand side is a continuous function of $t$, therefore we deduce that

\[ 0= \frac12 \int_{\partial E_0} \nabla W(x-y)\Bigg(\frac{y_1^2}{a_1^4}+\frac{y_2^2}{a_2^4}\Bigg)^{{-}1/2}\,{\rm d}{\mathcal{H}}^1(y)=\pi a_1a_2\nabla(W\ast\mu_{\partial E_0})(x) \]

for every $x$ in the interior of $E_0$. Since $W\ast \mu _{\partial E_0}$ is a continuous function in $\mathbb {R}^2$, this implies that $W\ast \mu _{\partial E_0}$ is constant in $E_0$, that is, (5.3)–(5.4) are satisfied for $\mu =\mu _{\partial E_0}$.