An isoperimetric problem with a competing nonlocal singular term

Abstract

In this paper, we investigate the minimization of a functional in which the usual perimeter is competing with a nonlocal singular term comparable (but not necessarily equal to) a fractional perimeter. The motivation for this problem is a cell motility model introduced in some previous work by the first author. We establish several facts about global minimizers with a volume constraint. In particular we prove that minimizers exist and are radially symmetric for small mass, while minimizers cannot be radially symmetric for large mass. For large mass, we prove that the minimizing sequences either split into smaller sets that drift to infinity or must develop intricate non-symmetrical shape. Finally, we connect these two alternatives to a related minimization problem for the optimal constant in a classical interpolation inequality (a Gagliardo–Nirenberg type inequality for fractional perimeter).

Appendices

A Proof of Proposition 1.1

First, we write

$$\begin{aligned} P_K(E)&= \int _{|z|\ge R} K(z) \int _{\mathbb {R}^n} |\chi _E(x)-\chi _E(x+z)|dx dz\\&\quad +\int _{|z|\le R} K(z) \int _{\mathbb {R}^n} |\chi _E(x)-\chi _E(x+z)|dx\, dz. \end{aligned}$$

Using the bound $\int _{\mathbb {R}^n} |\chi _E(x)-\chi _E(x+z)| dx \le 2|E|$ in the first integral and $\int _{\mathbb {R}^n} |\chi _E(x)-\chi _E(x+z)|dx \le |z| \int _{\mathbb {R}^n} |D\chi _E|$ in the second integral, we get:

$$\begin{aligned} P_K(E) \le 2|E| \int _{|z|\ge R} K(z) dz+ P(E) \int _{|z|\le R} |z| K(z) \, dz \le n \omega _n \left( 2|E| \frac{R^{-s}}{s}+ P(E) \frac{R^{1-s}}{1-s} \right) . \end{aligned}$$

Optimizing with respect to R by taking $R=\frac{2|E|}{P(E)}$ yields the result.

B Proof of Lemma 5.4

It is equivalent to prove

$$\begin{aligned} \left| \int _{\partial B_1}{(\psi (1+\lambda )-\psi (1))d{\mathcal {H}}^{n-1}(x)}-P(B_1)\beta \lambda \right| \le \frac{C}{s(1-s)}P(B_1)r_0^{n-s}\lambda ^2. \end{aligned}$$

(B.1)

Using (5.9), we write

$$\begin{aligned} \int _{\partial B_1}{\psi (1+\lambda )d{\mathcal {H}}^{n-1}(x)}&=\frac{1}{2}\int _{\mathbb {R}^n}{\int _{\mathbb {R}^n}{\left| \chi _{B_{r_0}}(x)-\chi _{B_{r_0}}(x+\frac{z}{1+\lambda })\right| (1+\lambda )^{n}K(z)dz}dx}\\&=\frac{1}{2}(1+\lambda )^{n}\int _{\mathbb {R}^n}{\phi \left( \frac{z}{1+\lambda }\right) K(z)dz}, \end{aligned}$$

where we defined a radially symmetric function $\phi (|z|):=\phi (z)=\int _{\mathbb {R}^n}|\chi _{B_{r_0}}(x)-\chi _{B_{r_0}}(x+z)|dx.$ Setting

$$\begin{aligned} F(\lambda ):=(1+\lambda )^{n}\int _{\mathbb {R}^n}{\phi \left( \frac{z}{1+\lambda }\right) K(z)dz}, \end{aligned}$$

which satisfies in particular, $F(0)=P_K(B_{r_0})=\int _{\partial B_1}{\psi (1)d{\mathcal {H}}^{n-1}(x)},$ we see that (B.1) is equivalent to showing that there exists $\tilde{\beta }=P(B_1)\beta \in \mathbb {R}$ such that

$$\begin{aligned} |F(\lambda )-F(0)-\tilde{\beta }\lambda |<\frac{C}{s(1-s)}r_0^{n-s}\lambda ^2 \end{aligned}$$

for all $|\lambda |<1/2$. Clearly, this holds if we take $\tilde{\beta }=F'(0)$ and prove

$$\begin{aligned} |F'(0)|\le \frac{C}{s(1-s)}r_0^{n-s},\quad \sup _{|\lambda |<1/2}{|F''(\lambda )|}\le \frac{C}{s(1-s)}r_0^{n-s}. \end{aligned}$$

First, we note that for all $0\le t\le 2r_0$, we have:

$$\begin{aligned} \phi (t)= & {} 2|B_{r_0}|-4\int _{t/2}^{r_0}{\omega _{n-1}(r_0^2-\tau ^2)^{\frac{n-1}{2}}d\tau }, \phi '(t) =4\omega _{n-1}\left( r_0^2-\frac{t^2}{4}\right) ^{\frac{n-1}{2}},\ \phi ''(t)\\= & {} -(n-1)\omega _{n-1}t\left( r_0^2-\frac{t^2}{4}\right) ^{\frac{n-3}{2}}, \end{aligned}$$

while when $t>2r_0$, we find $\phi (t)=2|B_{r_0}|$, $\phi '(t)=0$ and $\phi ''(t)=0$. Next, we compute:

$$\begin{aligned} F'(\lambda )&=n(1+\lambda )^{n-1}\int _{\mathbb {R}^n}{\phi \left( \frac{z}{1+\lambda }\right) K(z)dz} -(1+\lambda )^{n-2}\int _{\mathbb {R}^n}{\phi '\left( \frac{|z|}{1+\lambda }\right) |z|K(z)dz} \end{aligned}$$

which implies

$$\begin{aligned} \tilde{\beta }:=F'(0)&=n\int _{\mathbb {R}^n}{\phi (z)K(z)dz}-\int _{\mathbb {R}^n}{\phi '(|z|)|z|K(z)dz}\\&=nP_K(B_{r_0})-4P(B_1)\omega _{n-1}r_0^{n-s}\int _0^{2}{\left( 1-\frac{t^2}{4}\right) ^{\frac{n-1}{2}}t^n K(t)dt}. \end{aligned}$$

We deduce

$$\begin{aligned} |\beta |=\left| \frac{\tilde{\beta }}{P(B_1)}\right| \le \frac{C}{s(1-s)}r_0^{n-s}. \end{aligned}$$

(B.2)

Finally, we have

$$\begin{aligned} F''(\lambda )&=n(n-1)(1+\lambda )^{n-2}\int _{\mathbb {R}^n}{\phi \left( \frac{z}{1+\lambda }\right) K(z)dz} -(2n+2)(1+\lambda )^{n-3}\\&\qquad \int _{\mathbb {R}^n}{\phi '\left( \frac{|z|}{1+\lambda }\right) |z|K(z)dz}\\&\quad +(1+\lambda )^4\int _{\mathbb {R}^n}{\phi ''\left( \frac{|z|}{1+\lambda }\right) |z|^2K(z)dz}=J_1+J_2+J_3. \end{aligned}$$

When $\lambda <1/2$, the first integral satisfies

$$\begin{aligned} |J_1|\le \frac{C}{(1+\lambda )^2}P_K(B_{(1+\lambda )r_0})\le CP_s(B_{(1+\lambda )r_0})\le \frac{C}{s(1-s)}r_0^{n-s}. \end{aligned}$$

For the second integral, we find

$$\begin{aligned} |J_2|&\le C\int _{0}^{2r_0(1+\lambda )}{t^{n-1}\phi '\left( \frac{t}{1+\lambda }\right) t K(t)dt}\le Cr_0^{n-s}\int _{0}^{2(1+\lambda )}{t^{-s}\left( 1-\frac{t^2}{4(1+\lambda )^2}\right) ^{\frac{n-1}{2}}dt}\\&\le \frac{C}{1-s}r_0^{n-s}, \end{aligned}$$

and the last integral is bounded by

$$\begin{aligned} |J_3|&\le C\int _{0}^{2r_0(1+\lambda )}{t\left( r_0^2-\frac{t^2}{4(1+\lambda )^2}\right) ^{\frac{n-3}{2}}t^2t^{-n-s}t^{n-1}dt}\\&=Cr_0^{n-s}\int _0^{2(1+\lambda )}{\left( 1-\frac{t}{2(1+\lambda )}\right) ^{\frac{n-3}{2}}\left( 1+\frac{t}{2(1+\lambda )}\right) ^{\frac{n-3}{2}}t^{2-s}dt}\le Cr_0^{n-s} \end{aligned}$$

We deduce $\sup _{|\lambda |<1/2}{|F''(\lambda )|}\le \frac{C}{s(1-s)}r_0^{n-s}$ which complete the proof.

C Proof of Lemma 1.2

The result is likely known but since we could not find a reference for it, we provide here a short proof following the original argument of Polya [28] (see also Blumenthal-Getoor [3]).

Since K(x) is the solution of (1.6), we have

$$\begin{aligned} {\widehat{K}}(\xi ) = \int _{\mathbb {R}^n} e^{ix\cdot \xi } K(x)\, dx = \frac{1}{1+|\xi |^s} \end{aligned}$$

and we introduce the function k(x) such that ${\widehat{k}}(\xi ) = \frac{e^{-|\xi |^{2 p}}}{1+|\xi |^s}$ for $p\in {\mathbb {N}}$ with $2p>n+s$. Since the function $\displaystyle \xi \mapsto {\widehat{K}}(\xi ) - {\widehat{k}}(\xi )= \frac{1-e^{-|\xi |^{2p}}}{1+|\xi |^s}$ is $C^{2p}$, its Fourier transform decays faster than $|x|^{2p}$. Hence we only need to show that k(x) has the appropriate behavior as $|x|\rightarrow \infty$.

Since ${\widehat{k}}(\xi )$ is radially symmetric, we can write (see for instance [29]):

$$\begin{aligned} k(x)&= \frac{1}{(2\pi )^{\frac{n}{2}}} \int _0^\infty {\widehat{k}}\left( t\right) \left( \frac{t}{|x|}\right) ^{\frac{n}{2} -1}J_{\frac{n}{2}-1}( |x| t)\, t \, dt\\&= \frac{|x|^{-n-s}}{(2\pi )^{\frac{n}{2}}} \int _0^\infty {\widehat{k}}\left( \frac{t}{ |x|}\right) |x|^{s} t^{\frac{n}{2}} J_{\frac{n}{2}-1}( t)\, dt \end{aligned}$$

where $J_\nu$ denotes the Bessel function (of the first kind) of order $\nu$ for which we recall that

$$\begin{aligned} \frac{d}{dt}\left( t^{\frac{n}{2}}J_{\frac{n}{2}}(t)\right) = t^{\frac{n}{2}}J_{\frac{n}{2}-1}(t). \end{aligned}$$

We can thus write

$$\begin{aligned} |x|^{n+s} k(x)&=\frac{1}{(2\pi )^{\frac{n}{2}}} \int _0^\infty {\widehat{k}}\left( \frac{t}{|x|}\right) |x|^{s} t^{\frac{n}{2}} J_{\frac{n}{2}-1}( t)\, dt\\&=\frac{1}{(2\pi )^{\frac{n}{2}}} \int _0^\infty {\widehat{k}}\left( \frac{t}{ |x|}\right) |x|^{s} \frac{d}{dt}(t^{\frac{n}{2}}J_{\frac{n}{2}}(t)) \, dt\\&=\frac{-1}{(2\pi )^{\frac{n}{2}}} \int _0^\infty {\widehat{k}} '\left( \frac{t}{ |x|}\right) |x|^{s-1} t^{\frac{n}{2}}J_{\frac{n}{2}}(t) \, dt\\&=\frac{-1}{(2\pi )^{\frac{n}{2}}} {\mathrm {Re}}\int _0^\infty {\widehat{k}} '\left( \frac{t}{|x|}\right) |x|^{s-1} t^{\frac{n}{2}}H^{(1)}_{\frac{n}{2}}(t) \, dt, \end{aligned}$$

where the Hankel functions of the first kind is defined by

$$\begin{aligned} H^{(1)}_\nu (x) = J_\nu (x)+iY_\nu (x). \end{aligned}$$

(where $Y_\nu (x)$ is the Bessel functions of the second kind). Following [28], we now change the contour of integration so that it is a straight line $L_\eta$ from 0 to $\infty$ which makes a small positive angle $\eta$ with the real axis. We recall the following asymptotic formula:

$$\begin{aligned} H^{(1)}_{\frac{n}{2}} \sim \sqrt{\frac{2}{\pi z}} e^{i\left( z-\frac{(n+1)\pi }{4}\right) }, \text{ for } |z|\rightarrow \infty , \qquad -\pi<\arg z<\pi . \end{aligned}$$

It follows that along the line $L_\eta$, we have (for large |z|)

$$\begin{aligned} | H^{(1)}_{\frac{n}{2}}(z)| \le C \exp (-(\sin \eta ) t),\quad |{\widehat{k}} ' (z) |\le C \exp (-\sin (2p\eta )t^{2p}), \end{aligned}$$

so as long as $2p\eta <\pi$ we can write (all integrals are convergent):

$$\begin{aligned} |x|^{n+s} k(x) = \frac{-1}{(2\pi )^{\frac{n}{2}}} {\mathrm {Re}}\int _{L_\eta } {\widehat{k}} '\left( \frac{z}{|x|}\right) |x|^{s-1} z^{\frac{n}{2}}H^{(1)}_{\frac{n}{2}}(z) \, dz, \end{aligned}$$

and since ${\widehat{k}}(r)\sim -sr^{s-1}$ when $r\rightarrow 0^+$, we obtain:

$$\begin{aligned} \lim _{|x|\rightarrow \infty } |x|^{n+s} k(x) = \frac{s}{(2\pi )^\frac{n}{2}}\mathrm {Re}\int _{L_\eta } z^{\frac{n}{2}+s-1}H^{(1)}_{\frac{n}{2}}(z) \, dz. \end{aligned}$$

Finally, since the integral converge as long as $\sin \eta >0$, we can rotate $L_\eta$ until $\eta = \pi /2$ to get:

$$\begin{aligned} \lim _{|x|\rightarrow \infty } |x|^{n+s} k(x)&= \frac{s}{(2\pi )^\frac{n}{2}}\mathrm {Re}\int _0^\infty i (it)^{\frac{n}{2}+s-1}H^{(1)}_{\frac{n}{2}}(it) \, dt \\&= \frac{s}{(2\pi )^\frac{n}{2}}\frac{2}{\pi }\mathrm {Re}\int _0^\infty i^{s-1} t^{\frac{n}{2}+s-1} K_{\frac{n}{2}}(t) \, dt, \end{aligned}$$

where the modified Bessel function of second kind $K_{\frac{n}{2}}(t) = \frac{\pi }{2} i^{\frac{n}{2}+1}H^{(1)}_{\frac{n}{2}}(it)$ satisfies the formula (see [13])

$$\begin{aligned} \int _0^\infty t^{\frac{n}{2}+s-1} K_{\frac{n}{2}}(t) \, dt = 2^{\frac{n}{2} +s-2} \Gamma \left( \frac{n+s}{2}\right) \Gamma \left( \frac{s}{2}\right) . \end{aligned}$$

We deduce

$$\begin{aligned} \lim _{|x|\rightarrow \infty } |x|^{n+s} k(x)&= \frac{s 2^{s-1}}{\pi ^{\frac{n}{2}+1}} \sin \left( s\frac{\pi }{2} \right) \Gamma \left( \frac{n+s}{2}\right) \Gamma \left( \frac{s}{2}\right) \\&= \frac{s 2^{s-1}}{\pi ^{\frac{n}{2}}} \frac{\Gamma \left( \frac{n+s}{2}\right) \Gamma \left( \frac{s}{2}\right) }{ \Gamma \left( 1+\frac{s}{2}\right) |\Gamma \left( -\frac{s}{2}\right) |} = \frac{2^{s}}{\pi ^{\frac{n}{2}}} \frac{\Gamma \left( \frac{n+s}{2}\right) }{ |\Gamma \left( -\frac{s}{2}\right) |} \end{aligned}$$

where we used the fact that $\Gamma (1+\frac{s}{2}) \Gamma (-\frac{s}{2}) = \frac{\pi }{-\sin (\frac{\pi }{2} s)}$ (Euler’s reflection formula).