# On minimizers in the liquid drop model

Otis Chodosh  OC: Department of Mathematics, Bldg. 380, Stanford University,
Stanford, CA 94305, USA<EMAIL_ADDRESS>and  Ian Ruohoniemi  IR:
Department of Mathematics, University of Minnesota-Twin Cities, Minneapolis,
MN, 55455, USA<EMAIL_ADDRESS>
###### Abstract.

We prove that round balls of volume $\leq 1$ uniquely minimize in Gamow’s
liquid drop model.

##  1\. Introduction

For $\Omega\subset\mathbb{R}^{3}$ measurable we define the liquid drop
functional

(1) $\mathcal{E}(\Omega):=P(\Omega)+D(\Omega).$

where

(2) $D(\Omega):=\frac{1}{2}\iint_{\Omega\times\Omega}\frac{dxdy}{|x-y|}$

is the Coulomb energy and
$P(\Omega)=\sup\left\\{\int_{\Omega}\operatorname{div}X:X\in
C^{1}_{c}(\mathbb{R}^{3};\mathbb{R}^{3}),|X|\leq 1\right\\}$ is the perimeter
of $\Omega$.

The functional (1) was introduced by Gamow in 1928 as a model for the nucleus
of an atom [Gam30]. The minimization problem

(3) $E(V):=\inf_{|\Omega|=V}\mathcal{E}(\Omega).$

has recently received considerable attention for its connections with various
problems in mathematical physics. We refer to the recent survey [CMT17] for
further discussion.

One basic feature of (1) is the fact that the perimeter $P(\Omega)$ is
minimized at a round ball (by the isoperimetric inequality) while the Coulomb
term $D(\Omega)$ is maximized at a round ball (by Riesz rearrangement
[Rie30]). Thus, the two terms are in competition in the minimization problem
(3). A well-known conjecture suggests that the perimeter completely dominates
in (3) until the Coulomb energy becomes strong enough to prevent the existence
of minimizers.

To state the conjecture precisely, write $B_{1}\subset\mathbb{R}^{3}$ for the
unit ball and set

(4) $V_{*}:=\frac{2-2^{\frac{2}{3}}}{2^{\frac{2}{3}}-1}\frac{|B_{1}|P(\partial
B_{1})}{\frac{1}{2}\iint_{B_{1}\times
B_{1}}|x-y|^{-1}dxdy}=5\frac{2-2^{\frac{2}{3}}}{2^{\frac{2}{3}}-1}\approx
3.51.$

The significance of (4) is that for $V>V_{*}$, the configuration of two balls
of volume $V/2$ placed infinitely far apart has less energy than a single ball
of volume $V$. We then have (cf. [CP11]):

######  Conjecture 1.

  1. (a)

For $V\leq V_{*}$, the round ball of volume $V$ uniquely minimizes
$\mathcal{E}(\cdot)$ among measurable sets $\Omega\subset\mathbb{R}^{3}$ with
$|\Omega|=V$.

  2. (b)

For $V>V_{*}$ no minimizer exists.

Our main result confirms (a) for the range $V\leq 0.28V_{*}$:

######  Theorem 2.

Round balls $B_{R}$ with $|B_{R}|\leq 1$ uniquely minimize
$\mathcal{E}(\cdot)$ among measurable sets $\Omega\subset\mathbb{R}^{3}$ with
$|\Omega|=|B_{R}|$.

Previously Knüpfer–Muratov proved [KM13, KM14] (see also [BC14, Jul14, MZ15,
FFM+15, CNT22]) proved that balls uniquely minimize $\mathcal{E}(\cdot)$ among
sets of sufficiently small volume $V$. It seems that Theorem 2 is the first
result concerning (a) in Conjecture 1 to go beyond the the “isoperimetric
dominated” (small volume) regime where $D(\Omega)\ll P(\Omega)$ in (1) allows
the repulsive term $D(\Omega)$ to be treated as a lower-order term (cf. [KM13,
p. 1132]). Along these lines, it appears to be the first quantitative estimate
(cf. [MZ15]).

###  1.1. Other results concerning Conjecture 1

  1. (1)

Knüpfer–Muratov [KM14] as well as Lu–Otto [LO14] proved that there is
$V_{N}>0$ large so that no minimizer exists for sets of volume $V>V_{N}$.
Frank–Killip–Nam later obtained [FKN16] the quantitative estimate $V_{N}\leq
8\approx 2.38V_{*}$.

  2. (2)

Frank–Nam have proven [FN21] that a minimizer exists $\mathcal{E}(\Omega)$
among sets with $|\Omega|=V$ for $V\leq V_{*}$ (this covers precisely the
conjecturally sharp range of volumes).

See also [ACO09, CP10, CP11, RW11, CS13, RW14, KMN16, Fra19, EFK20].

###  1.2. Organization

Section 2 recalls some facts about the Newtonian potential. Section 3 contains
a conditional outer-minimizing result for minimizers of $\mathcal{E}(\cdot)$.
In Section 4 we use integral geometry to prove new isoperimetric-type
inequalities involving the Coulomb energy. We compute the first variation of
energy in Section 5 and show how the isoperimetric-type inequalities, the
Minkowski inequality, and the first variation combine in Section 6. Finally,
in Section 7 we prove Theorem 2 using a continuity argument and the results
from Section 6. Appendix A contains a proof of Santaló’s formula. Section B
contains a brief discussion of the maximal binding energy per particle.

###  1.3. Acknowledgements

O.C. was supported by Terman Fellowship and an NSF grant (DMS-2304432). We are
grateful to Rupert Frank for his interest and suggestions.

##  2\. The Newtonian potential

For $\Omega\subset\mathbb{R}^{3}$ measurable, we recall that

(5) $v_{\Omega}(x)=\int_{\Omega}\frac{dy}{|x-y|}$

is the _Newtonian potential_ of $\Omega$.

######  Lemma 3.

The Newtonian potential of a round ball satisfies

$v_{B_{R}}(x)=\begin{cases}\frac{2\pi}{3}(3R^{2}-|x|^{2})&|x|\leq R\\\
\frac{4\pi R^{3}}{3|x|}&|x|\geq R.\end{cases}$

In particular, the Coulomb energy satisfies
$D(B_{R})=\frac{16\pi^{2}}{15}R^{5}$.

###### Proof.

Note that $v_{B_{R}}(0)=4\pi\int_{0}^{R}rdr=2\pi R^{2}$. Moreover, $v_{B_{R}}$
is rotationally invariant and solves $\Delta v_{B_{R}}=-4\pi\chi_{B_{R}}$.
Thus, $\frac{\partial^{2}v_{B_{R}}}{\partial r^{2}}+\frac{2}{r}\frac{\partial
v_{B_{R}}}{\partial r}=-4\pi$ for $r\in(0,R)$. This implies that
$v_{B_{R}}(r)=\frac{2\pi}{3}(3R^{2}-r^{2})$ for $r\leq R$. Arguing similarly
for $r>R$ (and using $\lim_{r\to\infty}v_{B_{R}}(r)=0$) we find
$v_{B_{R}}(r)=\frac{4\pi R^{3}}{3r}$ for $r\geq R$. This proves the first
assertion and the second follows from

$D(B_{R})=2\pi\int_{0}^{R}\frac{2\pi}{3}(3R^{2}-r^{2})r^{2}dr=\frac{16\pi^{2}}{15}R^{5}.$

This completes the proof. ∎

######  Lemma 4.

For any measurable set $\Omega\subset\mathbb{R}^{3}$ with $|\Omega|=|B_{R}|$,
the Newtonian potential satisfies $v_{\Omega}(x)\leq v_{B_{R}}(0)=2\pi R^{2}$.

###### Proof.

Apply the “bathtub principle” [LL01, Theorem 1.14] with
$f(y)=-\frac{1}{|x-y|}$ and $G=|\Omega|$. ∎

##  3\. The outer-minimizing property

We say that a measurable Caccioppoli set $\Omega\subset\mathbb{R}^{3}$ is
_outer-minimizing_ if any $\Omega\subset\hat{\Omega}$ measurable has
$P(\Omega)\leq P(\hat{\Omega})$. We say that $\Omega\subset\mathbb{R}^{3}$
measurable is _strictly outer-minimizing_ if any $\Omega\subset\hat{\Omega}$
measurable with $P(\hat{\Omega})\leq P(\Omega)$ must have
$|\hat{\Omega}\setminus\Omega|=0$. In this section we establish the following
(conditional) outer-minimizing property for minimizers of $\mathcal{E}$.

######  Proposition 5.

Fix $V=\frac{4}{3}\pi R^{3}\leq 1$ and assume that $E(V)=\mathcal{E}(B_{R})$.
If $\Omega\subset\mathbb{R}^{3}$ has $|\Omega|=V$ and
$\mathcal{E}(\Omega)=E(V)$ then $\Omega$ is strictly outer-minimizing.

###### Proof.

Consider $\Omega\subset\hat{\Omega}\subset\mathbb{R}^{3}$ measurable with
$P(\hat{\Omega})\leq P(\Omega)$. Write $|\hat{\Omega}|=\frac{4}{3}\pi
R^{3}(1+\alpha^{3})$ for some $\alpha\geq 0$. Our goal is to prove that
$\alpha=0$.

Write $\Gamma:=\hat{\Omega}\setminus\Omega$ and
$\tilde{\Omega}=(1+\alpha^{3})^{-\frac{1}{3}}\hat{\Omega}$. Note that
$|\tilde{\Omega}|=|\Omega|$ and $|\Gamma|=\frac{4}{3}\pi R^{3}\alpha^{3}$. The
assumption $P(\hat{\Omega})\leq P(\Omega)$ implies

$\displaystyle P(\tilde{\Omega})$
$\displaystyle=(1+\alpha^{3})^{-\frac{2}{3}}P(\hat{\Omega})\leq(1+\alpha^{3})^{-\frac{2}{3}}P(\Omega).$

Recalling the definition of the Newtonian potential $v_{\Omega}$ in (5) we
have

$\displaystyle D(\tilde{\Omega})$
$\displaystyle=(1+\alpha^{3})^{-\frac{5}{3}}D(\hat{\Omega})$
$\displaystyle=(1+\alpha^{3})^{-\frac{5}{3}}D(\Omega)+(1+\alpha^{3})^{-\frac{5}{3}}\int_{\Gamma}v_{\Omega}(x)dx+(1+\alpha^{3})^{-\frac{5}{3}}D(\Gamma)$
$\displaystyle\leq(1+\alpha^{3})^{-\frac{2}{3}}D(\Omega)+(1+\alpha^{3})^{-\frac{5}{3}}2\pi
R^{2}|\Gamma|+(1+\alpha^{3})^{-\frac{5}{3}}D(B_{\alpha R})$
$\displaystyle=(1+\alpha^{3})^{-\frac{2}{3}}D(\Omega)+(1+\alpha^{3})^{-\frac{5}{3}}\frac{8\pi^{2}}{3}R^{5}\alpha^{3}+(1+\alpha^{3})^{-\frac{5}{3}}\frac{16\pi^{2}}{15}R^{5}\alpha^{5}.$

In the third line we used the estimate
$(1+\alpha^{3})^{-\frac{5}{3}}\leq(1+\alpha^{3})^{-\frac{2}{3}}$, Lemma 4, and
the fact that balls maximize $D(\cdot)$ for fixed volumes. Since
$\mathcal{E}(\Omega)=E(V)\leq\mathcal{E}(\tilde{\Omega})$ we find

$\displaystyle\mathcal{E}(\Omega)\leq(1+\alpha^{3})^{-\frac{2}{3}}\mathcal{E}(\Omega)+(1+\alpha^{3})^{-\frac{5}{3}}\frac{8\pi^{2}}{3}R^{5}\alpha^{3}+(1+\alpha^{3})^{-\frac{5}{3}}\frac{16\pi^{2}}{15}R^{5}\alpha^{5}.$

Rearranging and using $\mathcal{E}(\Omega)=E(V)=\mathcal{E}(B_{R})$ we find

$\displaystyle
0\leq\left(1+\alpha^{3}-(1+\alpha^{3})^{\frac{5}{3}}\right)\mathcal{E}(B_{R})+\frac{8\pi^{2}}{3}R^{5}\alpha^{3}+\frac{16\pi^{2}}{15}R^{5}\alpha^{5}.$

Using $\mathcal{E}(B_{R})=4\pi R^{2}+\frac{16\pi^{2}}{15}R^{5}$ this can be
simplified to read

(6)
$0\leq\left(1+\alpha^{3}-(1+\alpha^{3})^{\frac{5}{3}}\right)\left(\frac{15}{\pi}R^{-3}+4\right)+10\alpha^{3}+4\alpha^{5}.$

Since $\frac{15}{\pi}R^{-3}\geq 20$ and
$1+\alpha^{3}-(1+\alpha^{3})^{\frac{5}{3}}<0$ we find

(7) $0\leq
12\left(1+\alpha^{3}-(1+\alpha^{3})^{\frac{5}{3}}\right)+5\alpha^{3}+2\alpha^{5}.$

We now use (7) to prove that $\alpha=0$. Since
$(1+\alpha^{3})^{\frac{5}{3}}\geq 1+\frac{5}{3}\alpha^{3}$, we obtain

$0\leq-8\alpha^{3}+5\alpha^{3}+2\alpha^{5}$

which gives

(8) $\alpha>0\quad\Rightarrow\quad\alpha^{2}\geq\frac{3}{2}.$

On the other hand we have that

$(1+\alpha^{3})^{\frac{5}{3}}=\alpha^{5}(1+\alpha^{-3})^{\frac{5}{3}}\geq\alpha^{5}+\frac{5}{3}\alpha^{2}.$

Using this instead in (7) we find

$0\leq 12-20\alpha^{2}+17\alpha^{3}-10\alpha^{5}:=f(\alpha).$

We note that $f(1)=-1$. Moreover,

$f^{\prime}(\alpha)=-\alpha(40-51\alpha+50\alpha^{3}):=-\alpha g(\alpha).$

It’s easy to check that $g(1)=39$. Since

$g^{\prime}(\alpha)=150\alpha^{2}-51\geq 0$

for $\alpha\geq 1$ we find that $g(\alpha)>0$ for $\alpha\geq 1$. Thus
$f^{\prime}(\alpha)<0$ for $\alpha\geq 1$. Finally, this implies that
$f(\alpha)<0$ for $\alpha\geq 1$, so (7) implies that $\alpha<1$. Combined
with (8) this gives $\alpha=0$ as claimed. ∎

######  Remark 6.

It appears that the above proof could be extended to cover the range
$V\leq\frac{20}{11}$.

##  4\. Santaló’s formula and isoperimetric-type estimates

One of the ingredient in the proof of Theorem 2 are new isoperimetric-type
inequalities relating Coulomb energy terms to perimeter and volume. To state
these inequalities, we recall (2) and also define111It’s important to note the
factor of $\frac{1}{2}$ that appears in (2) does not appear in (9).

(9)
$D^{\partial}(\Omega)=\iint_{\Omega\times\partial\Omega}\frac{dxdy}{|x-y|}=\int_{\partial\Omega}v_{\Omega}(x)dx,$

where the Newtonian potential $v_{\Omega}(x)$ is defined in (5). In this
section we prove:

######  Theorem 7.

For a non-empty compact set $\Omega\subset\mathbb{R}^{3}$ with smooth boundary
we have

(10) $|\Omega|^{3}<\frac{3}{16\pi}P(\Omega)^{2}D(\Omega)$

and

(11) $|\Omega|^{2}\leq\frac{1}{12\pi}P(\Omega)D^{\partial}(\Omega)$

Equality holds in (11) if and only if $\Omega$ is a round ball.

The proof of Theorem 7 relies on an integro-geometric formula due to Santaló
[San04]. We refer to e.g., [BP72, Cro80, Cro84, Cro08, KK19, HM22] for related
results.

######  Remark 8.

Knüpfer–Muratov proved that (10) holds with some abstract constant via an
interpolation argument [KM14, Lemma 7.1] and more recently,
Knüpfer–Muratov–Novaga [KMN20, Theorem 2.2] proved222The inequality is not
stated explicitly in this form in [KMN20] but follows from their work in a
straightforward manner; cf. Lemma 19. a version of (10) with an explicit
constant (they obtain $\frac{4}{9\pi}$ in place of $\frac{3}{16\pi}$). Our
proof of (10) is completely different from theirs and yields a better (but
still not sharp) constant. See also Appendix B some further discussion.

###  4.1. Santaló’s formula

Suppose that $\Omega\subset\mathbb{R}^{N}$ is a compact set with smooth
boundary. Let $U:=\Omega\times\SS^{N-1}\subset T\Omega$ denote the unit sphere
bundle over $\Omega$. We write $U_{x}:=\\{x\\}\times\SS^{N-1}$ for the fiber
over $x$. For $\nu(x)$ the _inwards_ pointing unit normal to $\partial\Omega$,
we set

$U^{+}:=\\{(x,\theta)\in\partial\Omega\times\SS^{N-1}:\left\langle\theta,\nu(x)\right\rangle>0\\}$

and $U_{x}^{+}:=U^{+}\cap U_{x}$. Below we write $\cos
u=\left\langle\theta,\nu(x)\right\rangle$ for $u=(x,\theta)\in U^{+}$. Observe
that $U$ (resp. $U^{+}$) is a smooth submanifold of
$\mathbb{R}^{N}\times\mathbb{R}^{N}$ and thus inherits a Riemannian metric. We
will abuse notation and write $du$ for the induced volume measure on both
manifolds as well as on a fixed fiber $U_{x},U_{x}^{+}$. We define a function
$\ell:U\to[0,\infty)$ defined by

$\ell(x,\theta):=\sup\\{t\geq 0:x+s\theta\in\Omega\textrm{ for all }0\leq
s\leq t\\}.$

The following result is the main tool used in the proof of Theorem 7.

######  Theorem 9 (Santaló).

Set $\Phi((x,\theta),t)=(x+t\theta,\theta)$ for $t<\ell(x,\theta)$. Then,

$\int_{U}f(u)du=\int_{U^{+}}\int_{0}^{\ell(u)}f(\Phi(u,t))\cos u\,dtdu$

for any $f\in L^{1}(U)$.

This is proven in [San04, pp. 336–338] (see also [Ber65, p. 286]) (in the more
general setting of the unit sphere bundle on a Riemannian manifold). We sketch
the proof of Theorem 9 in Appendix A. In the sequel will only ever use the
following special case.

######  Proposition 10.

For $\alpha>-1$ we have

$\int_{U}\ell(u)^{\alpha}du=\frac{1}{\alpha+1}\int_{U^{+}}\ell(u)^{\alpha+1}\cos
u\,du.$

###### Proof.

We take $f(u)=\ell(u)^{\alpha}$ in Theorem 9. Note that for $u\in U^{+}$ and
$t\in(0,\ell(u))$, $\ell(\Phi(u,t))=\ell(u)-t$. Thus

$\int_{U}\ell(u)^{\alpha}du=\int_{U^{+}}\int_{0}^{\ell(u)}(\ell(u)-t)^{\alpha}\,\cos
u\,dtdu=\frac{1}{\alpha+1}\int_{U^{+}}\ell(u)^{\alpha+1}\cos u\,du.$

This completes the proof. ∎

Taking $\alpha=0$ yields the following commonly used expression:

######  Corollary 11.

We have

$\omega_{N}|\Omega|=\int_{U^{+}}\ell(u)\cos u\,du.$

We have used $\omega_{N}$ to denote the area of the unit-sphere in
$\mathbb{R}^{N}$.

###  4.2. Integral estimates for Coulomb energies

We now show how Santaló’s formula can be used to prove Theorem 7.

######  Lemma 12.

For $\Omega\subset\mathbb{R}^{3}$ compact with smooth boundary we have

$\int_{U^{+}}\ell(u)^{3}\cos u\,du\leq 12D(\Omega).$

###### Proof.

For $x\in\Omega$, we can use spherical coordinates centered at $x$ to write

$\int_{\Omega}\frac{dy}{|x-y|}\geq\int_{U_{x}}\int_{0}^{\ell(u)}r^{-1}r^{2}drdu=\frac{1}{2}\int_{U_{x}}\ell(u)^{2}du.$

Integrating with respect to $x$, we obtain

$\displaystyle D(\Omega)$
$\displaystyle\geq\frac{1}{4}\int_{x\in\Omega}\int_{U_{x}}\ell(u)^{2}dudx$
$\displaystyle=\frac{1}{4}\int_{U}\ell(u)^{2}du$
$\displaystyle=\frac{1}{12}\int_{U^{+}}\ell(u)^{3}\cos udu.$

We used Santaló’s formula (in the form of Proposition 10) in the final
equality above. ∎

Using this we can now prove the first inequality in Theorem 7.

###### Proof of (10).

Corollary 11 gives

$4\pi|\Omega|=\int_{U^{+}}\ell(u)\cos\,u\,du.$

Thus, using Hölder’s inequality we find

$\displaystyle(4\pi)^{3}|\Omega|^{3}$ $\displaystyle\leq\left(\int_{U^{+}}\cos
u\,du\right)^{2}\left(\int_{U^{+}}\ell(u)^{3}\cos\,u\,du\right)$
$\displaystyle=P(\Omega)^{2}\left(\int_{S^{2}_{+}}\cos
u\,du\right)^{2}\int_{U^{+}}\ell(u)^{3}\cos u\,du$
$\displaystyle=\pi^{2}P(\Omega)^{2}\int_{U^{+}}\ell(u)^{3}\cos u\,du$
$\displaystyle\leq 12\pi^{2}P(\Omega)^{2}D(\Omega)$

where we used Lemma 12 in the final step. To complete the proof, we note that
if equality held in Hölder’s inequality then $\ell(u)=K$ for a.e. $u\in U^{+}$
and $K$ some fixed constant. This cannot occur, so the inequality at that step
must be strict. This completes the proof. ∎

######  Lemma 13.

For $\Omega\subset\mathbb{R}^{3}$ compact with smooth boundary

$\int_{U^{+}}\ell(u)^{2}du\leq 2D^{\partial}(\Omega).$

###### Proof.

For $x\in\partial\Omega$ we can use spherical coordinates centered at $x$ to
write

$\int_{\Omega}\frac{dy}{|x-y|}\geq\int_{U_{x}^{+}}\int_{0}^{\ell(u)}r^{-1}r^{2}drdu=\frac{1}{2}\int_{U_{x}^{+}}\ell(u)^{2}du.$

Integrating with respect to $x\in\partial\Omega$ completes the proof. ∎

Using this we can now prove the second inequality in Theorem 7.

###### Proof of (11).

Corollary 11 gives

$4\pi|\Omega|=\int_{U^{+}}\ell(u)\cos\,u\,du.$

Thus, using Hölder’s inequality we find

$\displaystyle 16\pi^{2}|\Omega|^{2}$
$\displaystyle\leq\left(\int_{U^{+}}\cos^{2}u\,du\right)\left(\int_{U^{+}}\ell(u)^{2}du\right)$
$\displaystyle=P(\Omega)\left(\int_{S^{2}_{+}}\cos^{2}u\,du\right)\left(\int_{U^{+}}\ell(u)^{2}du\right)$
$\displaystyle=\frac{2\pi}{3}P(\Omega)\left(\int_{U^{+}}\ell(u)^{2}du\right)$
$\displaystyle=\frac{4\pi}{3}P(\Omega)D^{\partial}(\Omega)$

Rearranging this proves the inequality. It is easy to see that equality holds
when $\Omega$ is a ball. Conversely, if equality holds in the application of
Hölder’s inequality we find that $\ell(u)=K\cos u$ for a.e. $u\in U^{+}$ so
$\Omega$ must be a round ball. This completes the proof. ∎

##  5\. The first variation of energy

In this section we compute the first variation of $\mathcal{E}(\cdot)$ and
particular deduce the value of the Lagrange multiplier at a volume constrained
critical point following the scaling method of [BdC84]. Such a calculation has
appeared previously in several places in the literature including [BC14,
Theorem 3.3], [Fra19, Lemma 12], and [Jul17, Lemma 2].

######  Lemma 14.

If $\Omega$ is a compact set with smooth boundary that’s stationary with
respect to $\mathcal{E}(\cdot)$ for volume preserving variations then

$H_{\partial\Omega}+v_{\Omega}=\frac{2P(\Omega)}{3|\Omega|}+\frac{5D(\Omega)}{3|\Omega|}$

along $\partial\Omega$.

###### Proof.

The first variation of area implies that
$H_{\partial\Omega}+v_{\Omega}=\lambda$ is constant along $\partial\Omega$. It
thus remains to determine $\lambda$. For $t\in(-\delta,\delta)$ (for some
$\delta>0$ small) we denote $\tilde{\Omega}_{t}$ the family of regions
obtained by flowing $\partial\Omega$ with unit normal speed (to the outside).
Then, we choose $\eta_{t}$ so that $t\mapsto\eta_{t}\tilde{\Omega}_{t}$ has
constant volume. Because
$\frac{d}{dt}\big{|}_{t=0}|\tilde{\Omega}_{t}|=|P(\Omega)$ we find

$\frac{d}{dt}\Big{|}_{t=0}\eta_{t}=-\frac{P(\Omega)}{3|\Omega|}$

(using the first variation of volume). Thus

$\mathcal{E}(\eta_{t}\tilde{\Omega}_{t})=\eta_{t}^{2}P(\tilde{\Omega}_{t})+\eta_{t}^{5}\frac{1}{2}\iint_{\tilde{\Omega}_{t}\times\tilde{\Omega}_{t}}\frac{dxdy}{|x-y|}.$

Differentiating this at $t=0$ and using the first variation formulae from the
proof of [BC14, Theorem 3.3] we find

$0=-\frac{2P(\Omega)^{2}}{3|\Omega|}-\frac{5P(\Omega)}{6|\Omega|}\iint_{\Omega\times\Omega}\frac{dxdy}{|x-y|}+\int_{\partial\Omega}(H_{\partial\Omega}(x)+v_{\Omega}(x))dx.$

The assertion thus follows from the fact that
$H_{\partial\Omega}(x)+v_{\Omega}(x)=\lambda$ is constant. ∎

The following result is not used in the proof of Theorem 2 but may be of some
independent interest.

######  Proposition 15.

Suppose that $\Omega\subset\mathbb{R}^{3}$ is a compact set with smooth
boundary stationary with respect to $\mathcal{E}(\cdot)$ for variations fixing
$|\Omega|=V\leq\frac{4}{3}\left(\frac{10}{3}\right)^{\frac{1}{2}}\approx
2.43$. Then $\partial\Omega$ is connected and has positive mean curvature
$H_{\partial\Omega}>0$.

###### Proof.

Write $|\Omega|=V=|B_{R}|$. Combining Lemmas 4 and 14 with (10) we find

$\displaystyle H_{\partial\Omega}+2\pi R^{2}$ $\displaystyle\geq
H_{\partial\Omega}+v_{\Omega}$
$\displaystyle=\frac{2P(\Omega)}{3|\Omega|}+\frac{5D(\Omega)}{3|\Omega|}$
$\displaystyle>\frac{2P(\Omega)}{3|\Omega|}+\frac{80\pi|\Omega|^{2}}{9P(\Omega)^{2}}$
$\displaystyle\geq 2\left(\frac{10\pi}{3}\right)^{\frac{1}{3}},$

where we used AM-GM in the last line. Thus, $H_{\partial\Omega}>0$ for
$R\leq\left(\frac{10}{3\pi^{2}}\right)^{\frac{1}{6}}$. This completes the
proof of mean-convexity.

Finally, if $\partial\Omega$ has more than one component, then there will be
some component $\Sigma$ so that $\Sigma=\partial\Gamma$ for a bounded open
component of $\mathbb{R}^{3}\setminus\Omega$. By touching $\Sigma$ from the
outside by a sphere, we see that the mean curvature of $\Sigma$ is positive at
some $p\in\Sigma$, when measured with respect to $\Gamma$. However, this means
that the mean curvature at $p$ when measured with respect to $\Omega$ is
negative. This contradicts the above estimate. ∎

##  6\. Roundness via the Minkowski inequality

Suppose that $\Omega\subset\mathbb{R}^{3}$ is a compact set with smooth
boundary. We say that $\Omega$ satisfies the _Minkowski inequality_ if

(12) $(16\pi
P(\Omega))^{\frac{1}{2}}\leq\int_{\partial\Omega}H_{\partial\Omega}(x)dx.$

Minkowski proved [Min03] that if $\Omega$ is convex then (12) holds. We will
need the following result of Huisken (see [AFM22, Corollary 1.3] for a proof):

######  Theorem 16.

If $\Omega\subset\mathbb{R}^{3}$ is a compact outer-minimizing set with smooth
boundary then $\Omega$ satisfies the Minkowski inequality (12). If equality
holds then $\Omega$ is a round ball.

It’s conjectured that “outer-minimizing” can be relaxed to “mean-convex.” See
[GL09, DHMT16, Gla22, CEK23, Bre21] for related results.

######  Proposition 17.

Suppose that a compact set $\Omega\subset\mathbb{R}^{3}$ has smooth boundary
and minimizes $\mathcal{E}(\cdot)$ among measurable regions with fixed volume
$|\Omega|=V\leq 1$. If $\Omega$ satisfies the Minkowski inequality (12) then
$\Omega$ agrees with a round ball.

###### Proof.

Write $|\Omega|=V=\frac{4}{3}\pi R^{3}$. Lemma 14 yields

$\int_{\partial\Omega}H_{\partial\Omega}(x)dx+D^{\partial}(\Omega)=\frac{2P(\Omega)^{2}}{3|\Omega|}+\frac{5P(\Omega)}{3|\Omega|}D(\Omega)$

Combining this with the Minkowski inequality (12) (assumed to hold) and the
isoperimetric-type inequality from (11), we find

$(16\pi
P(\Omega))^{\frac{1}{2}}+\frac{12\pi|\Omega|^{2}}{P(\Omega)}\leq\frac{2P(\Omega)^{2}}{3|\Omega|}+\frac{5P(\Omega)}{3|\Omega|}D(\Omega)$

Because $\Omega$ minimizes $\mathcal{E}(\cdot)$ among sets of the same volume,
we have $D(\Omega)\leq\mathcal{E}(B_{R})-P(\Omega)$. Thus, we arrive at the
expression

(13) $(16\pi
P(\Omega))^{\frac{1}{2}}+\frac{12\pi|\Omega|^{2}}{P(\Omega)}\leq\frac{5P(\Omega)}{3|\Omega|}\mathcal{E}(B_{R})-\frac{P(\Omega)^{2}}{|\Omega|}.$

Note that $x:=P(\Omega)\geq 4\pi R^{2}$ by the isoperimetric inequality with
equality only for the round ball. We will show that (13) implies that $x=4\pi
R^{2}$. Using

$\mathcal{E}(B_{R})=4\pi R^{2}+\frac{16\pi^{2}}{15}R^{5}$

the inequality (13) can be written as

$f_{R}(x):=-\frac{3}{4\pi}R^{-3}x^{2}+\left(5R^{-1}+\frac{4}{3}\pi
R^{2}\right)x-4\pi^{\frac{1}{2}}x^{\frac{1}{2}}-\frac{64\pi^{3}}{3}R^{6}x^{-1}\geq
0.$

We will check that $f_{R}(4\pi R^{2})=0$, $f^{\prime}(4\pi R^{2})\leq 0$ (as
long as $V\leq 1$, and $f^{\prime\prime}_{R}(x)<0$ for $x\geq 4\pi R^{2}$.
From this, we see that $x=4\pi R^{2}$ as desired.

Since all inequalities used above hold as equalities at the round ball, we
deduce that $f_{R}(4\pi R^{2})=0$. Next, we compute

$f^{\prime}_{R}(4\pi R^{2})=-2R^{-1}+\frac{8\pi}{3}R^{2}$

so $f^{\prime}_{R}(4\pi R^{2})\leq 0$ for $V=\frac{4}{3}\pi R^{3}\leq 1$.
Finally, we compute

$\displaystyle f^{\prime\prime}_{R}(x)$
$\displaystyle=-\frac{3}{2\pi}R^{-3}+\pi^{\frac{1}{2}}x^{-\frac{3}{2}}-\frac{128\pi^{3}}{3}R^{6}x^{-3}.$

Using $x^{-\frac{3}{2}}\leq(4\pi R^{2})^{-\frac{3}{2}}$, we can estimate

$-\frac{3}{2\pi}R^{-3}+\pi^{\frac{1}{2}}x^{-\frac{3}{2}}\leq-\frac{11}{8\pi}R^{-3},$

implying that $f^{\prime\prime}_{R}(x)<0$ for $x\geq 4\pi R^{2}$. This
completes the proof. ∎

##  7\. Proof of Theorem 2

Recall the definition of $E(V)$ in (3). The map $V\mapsto E(V)$ is continuous
(cf. [FL15, p. 4444]). Let
$E_{B}(V):=\mathcal{E}(B_{(\frac{3V}{4\pi})^{\frac{1}{3}}})$ denote the
profile corresponding to the energy of the round ball of volume $V$ (clearly
$E_{B}(V)$ is also continuous). Let

(14) $V_{B}:=\sup\\{V>0:E(V^{\prime})=E_{B}(V^{\prime})\textrm{ for all
$V^{\prime}\leq V$}\\}.$

By [KM14, Theorem 3.2] (cf. [BC14, Jul14, FFM+15]) we have that $V_{B}>0$.
Moreover, by [BC14, Theorem 2.10] if $V<V_{B}$ then the ball uniquely
minimizes $\mathcal{E}(\cdot)$ for fixed volume $V$ (and minimizes for
$V=V_{B}$).

######  Lemma 18.

If $V_{B}<V_{*}$ there is $\Omega\subset\mathbb{R}^{3}$ compact with smooth
boundary so that $|\Omega|=V_{B}$, $\mathcal{E}(\Omega)=E(V_{B})$ but $\Omega$
is not a round ball.

###### Proof.

By [FN21, Theorem 1] and the definition of $V_{B}$ there are
$\Omega_{j}\subset\mathbb{R}^{3}$ that minimize $\mathcal{E}(\Omega_{j})$
among sets of fixed volume so that $|\Omega_{j}|\searrow V_{B}$ but
$\Omega_{j}$ are not round balls. By [KM14, Lemma 7.2] we may assume that the
sets $\Omega_{j}$ are uniformly essentially bounded. Thus, by standard
compactness results for sets of finite perimeter (cf. [Sim83, §2.6]) we can
pass to a subsequence with $\Omega_{j}\to\Omega$ in $L^{1}$ and
$P(\Omega)\leq\limsup_{j\to\infty}P(\Omega_{j})$. As in [FL15, Lemma 2.3] we
have $D(\Omega_{j})\to D(\Omega)$. Thus, by continuity of $E(\cdot)$ we find

$\mathcal{E}(\Omega)\leq E(V_{B}),$

so $\Omega$ is a minimizer of $\mathcal{E}(\cdot)$ among sets of volume
$V_{B}$. By [BC14, Theorem 2.7], we can assume that $\Omega$ is compact with
smooth boundary.

Suppose that $\Omega$ was a round ball (up to a set of measure zero). Then,
the strict stability of the ball (in the range of volumes under consideration)
implies that $\Omega_{j}$ must also be a round ball for $j$ sufficiently large
(see the proof of [BC14, Theorem 2.10]). This contradicts our assumption that
the $\Omega_{j}$ were not round balls. This completes the proof. ∎

With these preparations we can now finish the proof of our main theorem. Note
that the proof below actually proves that there is $\varepsilon>0$ so that
round balls of volume $\leq 1+\varepsilon$ uniquely minimize.

###### Proof of Theorem 2.

As observed above, it suffices to prove that $V_{B}$ (defined in (14))
satisfies $V_{B}>1$. Thus, for the sake of contradiction we assume that
$V_{B}\leq 1<V_{*}$. By Lemma 18 there exists $\Omega\subset\mathbb{R}^{3}$
compact with smooth boundary so that $|\Omega|=V_{B}$,
$\mathcal{E}(\Omega)=E(V_{B})$, but $\Omega$ is not a round ball.

Let $R=\left(\frac{3V_{B}}{4\pi}\right)^{\frac{1}{3}}$ so that
$|\Omega|=|B_{R}|$ and $\mathcal{E}(\Omega)=\mathcal{E}(B_{R})$. The
assumption $V_{B}\leq 1$ and condition
$\mathcal{E}(\Omega)=\mathcal{E}(B_{R})=E(V_{B})$ allows us to apply
Proposition 5 to conclude that $\Omega$ is outer-minimizing. Thus, by
Huisken’s Theorem 16, $\Omega$ satisfies the Minkowski inequality (12). Since
$V_{B}\leq 1$ we can then apply Proposition 17 to conclude that $\Omega$ is a
round ball. This is a contradiction. ∎

##  Appendix A Proof of Theorem 9

Write $U^{\circ}$ for the interior of $U$ and

$\tilde{U}:=\\{(u,t)\in U^{+}\times(0,\infty):t<\ell(u)\\}.$

Then, we define

$\Phi:\tilde{U}\to
U^{\circ},\qquad\Phi:((x,\theta),t)\mapsto(x+t\theta,\theta).$

It is clear that $\Phi$ is a diffeomorphism. We now compute the Jacobian of
$\Phi$ at $p:=((x,\theta),t)\in T\tilde{U}$. We can assume that
$T_{x}\partial\Omega=\mathbb{R}^{N-1}\times\\{0\\}$. Note that (using the
splitting $T_{p}\tilde{U}=T_{x}\partial\Omega\times
T_{\theta}\SS^{N-1}\times\mathbb{R}$ and
$T_{\Phi(p)}U=T_{x+t\theta}\mathbb{R}^{N}\times T_{\theta}\SS^{n-1}$):

$\displaystyle d\Phi_{p}(v,0,0)$ $\displaystyle=(v,0)$ $\displaystyle
d\Phi_{p}(0,w,0)$ $\displaystyle=(tw,w)$ $\displaystyle d\Phi_{p}(0,0,1)$
$\displaystyle=(\theta,0).$

In particular, if we choose an orthonormal basis $w_{1},\dots,w_{N-1}$ for
$T_{\theta}\SS^{N-1}$ and use $e_{1},\dots,e_{N}$ for the standard basis on
$\mathbb{R}^{N}$ we find that the matrix of $d\Phi_{p}$ becomes

$\left(\begin{matrix}\operatorname{Id}_{(N-1)\times(N-1)}&*&*\\\
0_{1\times(n-1)}&t&\left\langle\theta,e_{N}\right\rangle\\\
0_{(N-1)\times(N-1)}&\operatorname{Id}_{(N-1)\times(N-1)}&0_{(N-1)\times
1}\end{matrix}\right)$

which has determinant $\pm\left\langle\theta,e_{N}\right\rangle=\pm\cos u$.
The assertion thus follows from the change of variables formula.

##  Appendix B Binding energy per particle

We note that the sharp constant in (10) is closely related to a question posed
by Frank–Lieb [FL15] concerning the quantity

(15)
$e_{*}:=\min_{\Omega\subset\mathbb{R}^{3}}\frac{\mathcal{E}(\Omega)}{|\Omega|}$

where the minimum is taken over all measurable sets with $0<|\Omega|<\infty$.
By [FL15, Remark 3.6] the sharp constant in (10) is closely related to
$e_{*}$:

######  Lemma 19.

For $\Omega\subset\mathbb{R}^{3}$ measurable we have

$|\Omega|^{3}\leq\frac{27}{4e_{*}^{3}}P(\Omega)^{2}D(\Omega).$

Equality holds if and only if $\mathcal{E}(\Omega)=e_{*}|\Omega|$.

As such, the inequality (10) yields the following estimate.

######  Corollary 20.

$e_{*}>(36\pi)^{\frac{1}{3}}\approx 3.836$.

######  Remark 21.

Note that in [KMN20, Theorem 2.2] Knüpfer–Muratov–Novaga derived the lower
bound $e_{*}\geq\left(\frac{243\pi}{16}\right)^{\frac{1}{3}}\approx 3.627$.

One can compare these estimates to the upper bound

$e_{*}\leq\min_{R}\frac{\mathcal{E}(B_{R})}{|B_{R}|}=3\left(\frac{9\pi}{5}\right)^{\frac{1}{3}}\approx
5.345$

obtained by optimizing the energy density of round balls.

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