1 Introduction

Let us introduce the geometric evolution which we consider in this paper. Given an initial set \(E\subset \mathbb {R}^n\), we define its evolution \(E_t\) according to fractional mean curvature flow as follows: the velocity at a point \(x\in \partial E_t\) is given by

(1.1)

where \(s\in (0,1)\) is a fixed parameter and \(\nu \) is the outer normal at \(\partial E_t\) in x. The fractional mean curvature of a set has been introduced in [5] as the first variation of the fractional perimeter functional, and it has been proved in [1] that for sufficiently smooth sets E the rescaled fractional mean curvature \((1-s)H_s(x, E)\) converges as \(s\rightarrow 1\) to the classical mean curvature of E at x. The evolution law (1.1) can be interpreted as the \(L^2\)-gradient flow of the fractional perimeter.

Existence and uniqueness of viscosity solutions to a level set formulation of (1.1) has been provided in [8, 15], and qualitative properties of smooth solutions have been studied in [19]. However, we point out that the short-time existence of smooth solutions has not yet been proved. In [6], the convergence to the fractional mean curvature flow of a threshold dynamics scheme is proved; this result was adapted to the anisotropic case, even in presence of a driving force in [9], where it is also shown that the flow preserves convexity. It has also been observed that the geometric law (1.1) presents some different behavior with respect to the classical mean curvature flow: we refer for instance to the paper [10] about the formation of neck-pinch singularities, and to the paper [7] about fattening and non-fattening phenomena.

In this paper, we are interested in the homothetically shrinking solutions for the flow (1.1). A homothetic solution to (1.1) is a self-similar solution to (1.1): substituting \(E_t=\lambda (t) E\) in (1.1), it is easy to see, using scale invariance of the fractional mean curvature, that this is equivalent to \(\lambda ' (t) x\cdot \nu =-\frac{1}{\lambda (t)^s}H_s(x,E)\) for all \(x\in \partial E\). So homothetically shrinking solutions to (1.1) are given by the solutions to (1.1) with initial datum every set \(E\subseteq \mathbb {R}^n\) of class \(C^{1,1}\) which satisfies

$$\begin{aligned} x\cdot \nu = c\, H_s(x, E)\qquad \text {for some constant}\ c>0. \end{aligned}$$
(1.2)

Homothetically shrinking solutions are particularly relevant in the analysis of the classical mean curvature flow, as they are canonical examples of singularities, in the sense that any solution converges to a self-shrinker, if properly rescaled around a singular point. This result follows from an important monotonicity formula established by Huisken in [14] for the mean curvature flow. The analog of such formula in the fractional setting is still an open problem. We recall moreover that, at the moment, the existence theorem for local in time regular solutions of (1.1), even if expected, has not been proved.

It is well-known that the only embedded planar curve which is homothetically shrinking under curvature flow is the circle [2], whereas in higher dimensions there exist other smooth embedded surfaces which are self-shrinkers for the mean curvature flow, starting from the rotationally symmetric torus discovered by Angenent [3], and then going to more complex configurations as punctured compact surfaces or non-compact asymptotically conical surfaces, see [4, 16]. However, it is easy to show that the ball is the only self-shrinker which is also radially symmetric.

In the fractional setting, the classification of self-shrinkers is still at a very early stage. As far as we know, we provide here the first examples of fractional self-shrinkers which are different from balls and cylinders. More precisely, in Sect. 2, we show the existence of homothetic solutions to the flow (1.1) which are radially symmetric, and have a prescribed number of boundary spheres (see Theorem 2.3). Moreover, in the case of a single annulus, we show uniqueness of the ratio R / r for which the flow starting from the annulus \(B_R\setminus B_r\) self-similarly shrinks to a point. The existence of such radially symmetric self-shrinkers, different from balls, is a new feature compared with the local case, and it is due to the nonlocal nature of the fractional mean curvature.

A natural question arising about self-similar shrinkers is the issue of their dynamic stability. In the case of the classical mean curvature flow, the study of the dynamic stability of self-shrinkers was initiated in [11], and later developed by other authors. From the convergence results in [12, 13] it follows that the balls is dynamically stable under mean curvature flow (see also [17, 18, 20] for a discussion of the stability of the Wulff-Shape as homothetic solution of the anisotropic and crystalline curvature flow). Moreover, in [11] it is shown that balls and cylinders are the only stable self-shrinkers.

In the fractional case, none of such results is currently available, in particular it is not known whether the ball is dynamically stable, and if convex sets shrink to a round point at the singular time. We discuss in this paper the stability issue for the class of solutions that we construct in Theorem 2.3. In particular, in Sect. 3 we show that the radial self-shrinkers different from the ball are all dynamically unstable (see Theorem 3.1).

2 Existence of Symmetric Self-Shrinkers

We start with a technical result which will be useful in the sequel. We denote by \(B_r\) the ball of center 0 and radius \(r>0\), and we let \(B_r(x)=x+B_r\). Moreover, we recall that, by the scale invariance of fractional mean curvature, for all \(x_r\in \partial B_r\) there holds (see [19, Lemma 2])

$$\begin{aligned} H_s(x_r, B_r)=\frac{k(n)}{r^s}\quad \text {where } k(n):=H_s(x_1, B_1). \end{aligned}$$

Lemma 2.1

Let \(x\in \mathbb {R}^n\setminus \{0\}\) and \(\delta \ne 0\). Then, as \(\delta \rightarrow 0\), the following estimate holds:

$$\begin{aligned} \int _{\partial B_{|x|+\delta }} \frac{1}{|y-x|^{n+s}}\,\mathrm{{d}}y =\frac{1}{|\delta |^{1+s}} (c+ o(1)) \,, \end{aligned}$$

for a constant \(c>0\) depending only on n and s.

Proof

Up to a rotation of the reference system, we can assume that \(x=-|x|e_1\). By the change of coordinates \(y' = (y-x) \), we get,

$$\begin{aligned} \int _{\partial B_{|x|+\delta }} \frac{1}{|y-x|^{n+s}}\,\mathrm{{d}}y =\int _{\partial B_{|x|+\delta }(|x|e_1)} \frac{1}{|y'|^{n+s}}\,\mathrm{{d}}y' \end{aligned}$$

Note that

$$\begin{aligned} \int _{\partial B_{|x|+\delta }(|x|e_1)\cap \{y'\cdot e_1>|x|\}} \frac{1}{|y'|^{n+s}}\,\mathrm{{d}}y'\leqslant \frac{n\omega _{n} \,(|x|+\delta )^{n-1}}{2\,|x|^{n+s}}\leqslant C. \end{aligned}$$
(2.1)

Moreover, we write \(\partial B_{|x|+\delta }(|x|e_1)\cap \{y'\cdot e_1<|x|\}=\{(f(z),z)\, |\, z\in \mathbb {R}^{n-1}, z\in B'_{|x|+\delta }\}\), where \(B'_r\subseteq \mathbb {R}^{n-1}\) denotes the ball of center 0 and radius r in \(\mathbb {R}^{n-1}\) and \(f(z)=|x|-\sqrt{(|x|+\delta )^2 -|z|^2}\). Therefore, denoting \(R_\delta :=\frac{|x|+\delta }{|\delta |}\), we get

$$\begin{aligned}&\int _{\partial B_{|x|+\delta }(-x)\cap \{y'\cdot e_1<|x|\}} \frac{1}{|y'|^{n+s}}\,\mathrm{{d}}y' = \int _{B'_{|x|+\delta }} \frac{|x|+\delta }{\sqrt{(|x|+\delta )^2-|z|^2}} \frac{1}{\left( f(z)^2+|z|^2 \right) ^{\frac{n+s}{2}}}\mathrm{{d}}z\\&\quad = \frac{ (n-1)\omega _{n-1} }{|\delta |^{s+1}} \int _0^{R_\delta } \frac{R_\delta }{\sqrt{R_\delta ^2-\rho ^2}} \rho ^{n-2} \left[ \left( \frac{\delta }{|\delta |}-\frac{\rho ^2}{R_\delta +\sqrt{R_\delta ^2-\rho ^2}}\right) ^2+\rho ^2 \right] ^{-\frac{n+s}{2}}\mathrm{{d}}\rho . \end{aligned}$$

Let

$$\begin{aligned} g_\delta (\rho ):= \frac{R_\delta }{\sqrt{R_\delta ^2-\rho ^2}} \rho ^{n-2}\left[ \left( \frac{\delta }{|\delta |} -\frac{\rho ^2}{R_\delta +\sqrt{R_\delta ^2-\rho ^2}}\right) ^2 +\rho ^2 \right] ^{-\frac{n+s}{2}}. \end{aligned}$$

Now we observe that there exists \(C=C(n,s)>0\), such that

$$\begin{aligned} \int _{R_\delta /2}^{R_\delta } g_\delta (\rho )\mathrm{{d}}\rho \leqslant \frac{C}{R_\delta ^{s+1}} =C\frac{|\delta |^{s+1}}{(|x|+\delta )^{s+1}}. \end{aligned}$$
(2.2)

Moreover, taking \(|\delta |\) sufficiently small such that \(R_\delta >4\), we get that there exists a dimensional constant \(C=C(n,s)>0\) such that

$$\begin{aligned} g_\delta (\rho )\chi _{(0,R_\delta /2)}(\rho ) \leqslant C \chi _{(0,1)}(\rho )+C \frac{1}{\rho ^{s+2}} \chi _{(1, +\infty )}(\rho )\in L^1(0, +\infty ), \end{aligned}$$
(2.3)

where \(\chi _{(a,b)}\) is the characteristic function of the interval (a, b). Using (2.3) and observing that

$$\begin{aligned} g_\delta (\rho ) \rightarrow \frac{\rho ^{n-2}}{(1+\rho ^2)^\frac{n+s}{2}} \qquad \text {as}\ \delta \rightarrow 0, \end{aligned}$$

we conclude by (2.2) and by Lebesgue dominated convergence theorem that

$$\begin{aligned}&\int _{\partial B_{|x|+\delta }(-x)\cap \{y_1'<|x|e_1\}} \frac{1}{|y'|^{n+s}}\,\mathrm{{d}}y'\\&\quad = \frac{ (n-1)\omega _{n-1} }{|\delta |^{s+1}}\left[ \int _0^{\frac{R_\delta }{2}}g_\delta (\rho )\mathrm{{d}}\rho +\int _{\frac{R_\delta }{2}}^{R_\delta }g_\delta (\rho )\mathrm{{d}}\rho \right] =\frac{c+o(1)}{|\delta |^{s+1}} \end{aligned}$$

where

$$\begin{aligned} c := (n-1)\,\omega _{n-1} \int _0^{+\infty }\frac{\rho ^{n-2}}{(1+\rho ^2)^\frac{n+s}{2}} \,\mathrm{{d}}\rho \,. \end{aligned}$$

The conclusion follows by this estimate and (2.1). \(\square \)

First of all, we look to the simplest example of rotationally symmetric set different from a ball. We show that there exists a unique value of the ratio \(\frac{R}{r}\) which depends on the dimension n and on the fractional power \(s\in (0,1)\) such that the annulus \(B_R\setminus B_r\) is a self-shrinker.

Proposition 2.2

Let \(n\geqslant 1\). Then, for all \(R>0\) fixed, there exists a unique \(r=r(n,s)\in (0,R) \) depending only on R, \(s\in (0,1)\) and n, such that the flow (1.1) with initial datum the annulus

$$\begin{aligned} A:= B_R\setminus B_r \end{aligned}$$

is a homothetically shrinking solution of the flow.

Proof

Up to rescaling the set we fix \(R=1\). We observe that A is a solution to (1.2) if and only if for some \(c>0\),

$$\begin{aligned} 1= c H_s(x_1, A)\text { for all}\ x_1\ \text {with}\ |x_1|=1\quad \text { and } \quad r=-cH_s(x_r, A) \text { for all} \ x_r\ \text {with}\ |x_r|=r \end{aligned}$$

and so if and only if

$$\begin{aligned} H_s(x_r, A)=-r H_s(x_1, A). \end{aligned}$$
(2.4)

By rotational invariance, we get that \(H_s(x_r, A), H_s(x_1, A)\) do not depend on the points \(x_r, x_1\), but only on \(0<r<1\). Moreover, they are both continuous functions with respect to r, due to the continuity of the fractional mean curvature with respect to \(C^2\)-convergence of sets (see [8, Sect. 5.2]). We consider the following function defined for \(r\in (0, 1)\)

$$\begin{aligned} f_s(r)= H_s(x_r, A)+r H_s(x_1, A). \end{aligned}$$
(2.5)

Note that the function \(f_s\) is continuous on (0, 1). To prove the statement it is sufficient to show that there exists a unique \(r=r(n,s)\) such that \(f_s(r(n,s))=0\).

Let \(r, r'\) such that \( 0<r<r'<1\). By the inclusions \(A_{1,r'}:= B_1\setminus B_{r'}\subseteq A\subseteq B_1\), we get, by the monotonicity of the fractional mean curvature (see [8, Sect. 5.2]), that

$$\begin{aligned} H_s(x_1, A_{1,r'})>H_s(x_1, A)\geqslant H_s(x_1, B_1)= k(n) >0. \end{aligned}$$
(2.6)

This implies that

$$\begin{aligned} r\in (0, 1)\mapsto H_s(x_1, A) \text { is monotone increasing and positive}. \end{aligned}$$
(2.7)

Moreover, we observe, recalling the definitions, that

$$\begin{aligned} H_s(x_r, A)=-H_s(x_r, B_r)+2\int _{\mathbb {R}^n\setminus B_1} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y. \end{aligned}$$

Note that if \(r'>r\) then \(H_s(x_{r'}, B_{r'})= \frac{k(n)}{(r')^{s}}< \frac{k(n)}{r^s}= H_s(x_r, B_r)\), whereas for \(1>r'>r\), \(|x_r-y|\geqslant |\frac{r'}{r} x_{r}-y|\) for all \(y\in \mathbb {R}^n\setminus B_1\), and \(x_ r\) with \(|x_r|=r\). Therefore, by symmetry of the kernel we have that

$$\begin{aligned} \int _{\mathbb {R}^n\setminus B_1} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y< \int _{\mathbb {R}^n\setminus B_1} \frac{1}{|x_{r'}-y|^{n+s}}\mathrm{{d}}y \end{aligned}$$

for all \(x_r, x_{r'}\) with \(|x_r|=r, |x_{r'}|=r'\). Using these facts we conclude that

$$\begin{aligned} r\in (0, 1)\mapsto H_s(x_r, A) \text { is monotone increasing}. \end{aligned}$$
(2.8)

Due to (2.7), (2.8), we notice that the function \(f_s(r)\) defined in (2.5) is monotone increasing. Now, we claim that \(\lim _{r\rightarrow 0} f_s(r)=-\infty \) and that \(\lim _{r\rightarrow 1}f_s(r)=+\infty \). If the claim is true, then the proof is concluded.

First of all we observe that

$$\begin{aligned} H_s(x_r, A)= & {} \int _{\mathbb {R}^n\setminus B_{1}} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\\&+\lim _{\varepsilon \rightarrow 0} \left( \int _{B_{r-\varepsilon }} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y -\int _{B_1\setminus B_{r+\varepsilon } }\frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\right) . \end{aligned}$$

This implies that \(\lim _{r\rightarrow 0} H_s(x_r, A)=-\infty \), and so also \(\lim _{r\rightarrow 0} f_s(r)=-\infty \).

Moreover, recalling Lemma 2.1 we get that

$$\begin{aligned}&H_s(x_r, A) \\&\quad =H_s(x_r, B_r)+\lim _{\varepsilon \rightarrow 0}\left( 2\int _{B_{r-\varepsilon }} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y-2\int _{B_1\setminus B_{r+\varepsilon } }\frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\right) \\&\quad =\frac{k(n)}{r^s}+\lim _{\varepsilon \rightarrow 0}\left( 2\int _\varepsilon ^{r} \int _{\partial B_{r-\delta }}\frac{1}{|x_r-y|^{n+s}} \mathrm{{d}}y\mathrm{{d}}\delta -2\int _{\varepsilon }^{1-r}\int _{\partial B_{r+\delta }} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\mathrm{{d}}\delta \right) \\&\quad =\frac{k(n)}{r^s}+ 2\int _{1-r}^{r}\left[ \int _{\partial B_{r-\delta }} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y- \int _{\partial B_{r+\delta }}\frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\right] \mathrm{{d}}\delta \\&\quad =\frac{k(n)}{r^s}+ 2\big (c + o(1)\big ) \left( \frac{1}{(1-r)^s}-\frac{1}{r^s}\right) . \end{aligned}$$

So, \(\lim _{r\rightarrow 1} H_s(x_r, A)=+\infty \), which permits to conclude that \(\lim _{r\rightarrow 1} f_s(r)=+\infty \).

\(\square \)

We now look for more general symmetric self-shrinkers, given by the union of a finite number of annuli.

Theorem 2.3

For all \(N\geqslant 1\) and all \(R>0\) there exists an increasing sequence \(0< r_1< \ldots<r_{2N-1}<r_{2N}=R\), depending only on n, s and N, such that such that the flow (1.1) with initial datum

$$\begin{aligned} E:= \bigcup _{k=1}^N \left( B_{r_{2k}}\setminus B_{r_{2k-1}}\right) \end{aligned}$$

is a homothetically shrinking solution of (1.1).

Similarly, for all \(N\geqslant 1\) and \(R>0\) there exists an increasing sequence \(0< \tilde{r}_0<\tilde{r}_1<\ldots<\tilde{r}_{2N-1}< r_{2N}=R\), depending only on n, s and N, such that such that the flow (1.1) with initial datum

$$\begin{aligned} \widetilde{E}:= B_{\tilde{r}_0}\cup \, \bigcup _{k=1}^N \left( B_{\tilde{r}_{2k}}\setminus B_{\tilde{r}_{2k-1}}\right) \end{aligned}$$

is a homothetically shrinking solution of (1.1).

Proof

The argument is similar to that in the proof of Proposition 2.2. As before, up to rescaling the sets E, \(\widetilde{E}\), we can assume \(r_{2N}=1\). Then, we want to find radii \(r_i\) in such a way that, letting \(x_{r_i}\in \partial B_{r_i}\), there hold

$$\begin{aligned} f_i(r_1,\ldots ,r_{2N-1}):= r_i H_s(x_1,E) + (-1)^{i-1} H_s(x_{r_i},E)=0 \qquad \forall \ i\in \{1,\ldots ,2N-1\} \end{aligned}$$
(2.9)

and

$$\begin{aligned} f_i(r_0,\ldots ,r_{2N-1}):= r_i H_s(x_1,\widetilde{E}) + (-1)^{i-1} H_s(x_{r_i},\widetilde{E})=0 \quad \forall \ i\in \{0,\ldots ,2N-1\}. \end{aligned}$$
(2.10)

Notice that the functions \(f_i\) are all continuous in their domain of definition.

We divide the proof into four steps. In the first step, we deal with the case \(N=1\) , and in step 2, 3 and 4 we consider the case \(N>1\). For \(N>1\) we provide the proof just of (2.9) for the existence of the set E, since the analogous assertion (2.10) for \(\widetilde{E}\) follows similarly.

Step 1. The case \(N=1\) for E has been proved in Propositions 2.2. Thus, we consider the set \(\tilde{E}\).

First of all, we fix \(r_1\in (0,1)\) and we prove that there exists \(r_0=r_0(r_1)\in (0, r_1)\) such that \(f_0(r_0(r_1), r_1)=0\) for all \(r_1\in (0,1)\). Due to the monotonicity properties of the fractional mean curvature, fixed \(r_1\in (0,1)\) we get

$$\begin{aligned} \lim _{r_0\rightarrow 0} H_s(x_{1},\tilde{E})=H_s(x_1, A_{1,r_1})>0\qquad \lim _{r_0\rightarrow r_1} H_s(x_{1},\tilde{E})=H_s(x_{1}, B_1)=k(n)>0. \end{aligned}$$

Moreover, by definition we get that,

$$\begin{aligned} H_s(x_{r_0},\tilde{E})= & {} -2\int _{B_{1}\setminus B_{r_1}} \frac{1}{|x_{r_0}-y|^{n+s}}\,\mathrm{{d}}y +H_s(x_{r_0}, B_{r_0})\\= & {} -2\int _{B_{1}\setminus B_{r_1}} \frac{1}{|x_{r_0}-y|^{n+s}} \,\mathrm{{d}}y+\frac{k(n)}{r_0^s}, \end{aligned}$$

from which we conclude that

$$\begin{aligned} \lim _{r_0\rightarrow 0}H_s(x_{r_0},\tilde{E})=+\infty \qquad \lim _{r_0\rightarrow r_1}H_s(x_{r_0},\tilde{E})=-\infty . \end{aligned}$$

Therefore, we obtain that

$$\begin{aligned} \lim _{r\rightarrow 0} f_0(r, r_1)=-\infty \qquad \lim _{r\rightarrow r_1} f_0(r, r_1)=+\infty . \end{aligned}$$

By continuity of the function \(f_0\), we deduce that for all \(r_1 \in (0,1)\) there exists at least one \(r=r(r_1)\in (0, r_1)\) such that

$$\begin{aligned} f_0(r(r_1), r_1)=0\,. \end{aligned}$$
(2.11)

We choose as \(r_0(r_1)\) to be the smallest among all possible \(r(r_1)\in (0, r_1)\) which solve (2.11). Observe that due to this choice, the function \(r\rightarrow f_1(r_0(r), r)\) is continuous. To conclude, it is sufficient to prove that that there exists \(r_1\in (0,1)\) such that \( f_1(r_0(r_1), r_1)=0\). Indeed, this would imply that \((B_1\setminus B_{r_1})\cup B_{r_0(r_1)}\) is a solution to (1.2).

Observe that \(\lim _{r\rightarrow 0}r_0(r)=0\), and therefore, we get

$$\begin{aligned} \lim _{r\rightarrow 0} f_1(r_0(r), r)= -\infty . \end{aligned}$$
(2.12)

We now claim that

$$\begin{aligned} \lim _{r\rightarrow 1} f_1(r_0(r), r)= +\infty . \end{aligned}$$
(2.13)

Recalling Lemma 2.1, we observe that as \(r\rightarrow 1\),

$$\begin{aligned} H_s(x_{1}, \tilde{E})= & {} 2\int _{B_{r}\setminus B_{r_0(r)}} \frac{1}{|x_1-y|^{n+s}}\,\mathrm{{d}}y + H_s(x_1, B_1)\nonumber \\= & {} 2\int _{r_0(r)}^{r}\int _{\partial B_t} \frac{1}{|x_1-y|^{n+s}}\,\mathrm{{d}}y\,\mathrm{{d}}t+k(n)\nonumber \\= & {} 2\big (c + o(1)\big ) \left( \frac{1}{(1-r)^s} -\frac{1}{(1-r_0(r))^s}\right) +k(n) \end{aligned}$$
(2.14)

where the constant \(c=c(n,s)>0\) is given by Lemma 2.1. Similarly, we have that

$$\begin{aligned}&H_s(x_{r},\tilde{E}) \nonumber \\&\quad =\lim _{\varepsilon \rightarrow 0} \left( 2\int _{B_{r-\varepsilon }\setminus B_{r_0(r)}} \frac{1}{|x_{r}-y|^{n+s}}\,\mathrm{{d}}y -2\int _{B_{1}\setminus B_{r+\varepsilon }} \frac{1}{|x_{r}-y|^{n+s}}\,\mathrm{{d}}y\right) +H_s(x_{r}, B_{r})\nonumber \\&\quad =\lim _{\varepsilon \rightarrow 0}\left( 2 \int _{r_0(r)}^{r-\varepsilon }\int _{\partial B_t} \frac{1}{|x_{r}-y|^{n+s}}\,\mathrm{{d}}y-2\int _{r+\varepsilon }^1 \int _{\partial B_t} \frac{1}{|x_{r}-y|^{n+s}}\,\mathrm{{d}}y\right) +\frac{k(n)}{r^s}\nonumber \\&\quad =2 \big (c + o(1)\big ) \left( - \frac{1}{(r-r_0(r))^s} +\frac{1}{(1-r)^s}\right) +\frac{k(n)}{r^s} \end{aligned}$$
(2.15)

and

$$\begin{aligned} H_s(x_{r_0(r)},\tilde{E})= & {} - 2\int _{r_n}^ {1}\int _{\partial B_t} \frac{1}{|x_{r_0(r)}-y|^{n+s}}\,\mathrm{{d}}y+ \frac{k(n)}{(r_0(r))^s}\nonumber \\= & {} - 2\big (c + o(1)\big ) \left( \frac{1}{(r-r_0(r))^s} -\frac{1}{(1-r_0(r))^s} \right) + \frac{k(n)}{(r_0(r))^s}.\nonumber \\ \end{aligned}$$
(2.16)

Therefore as \(r\rightarrow 1\) by (2.14) and (2.15)

$$\begin{aligned} f_1(r_0(r), r)=2\big (c + o(1)\big ) \left( \frac{1+r}{(1-r)^s} -\frac{1}{(r-r_0(r))^s}\ - \frac{r }{(1-r_0(r))^s}\right) + O(1). \end{aligned}$$
(2.17)

We claim that

$$\begin{aligned} \lim _{r\rightarrow 1}\frac{r-r_0(r) }{1-r_0(r)}=1. \end{aligned}$$
(2.18)

Note that the claim is equivalent to

$$\begin{aligned} \lim _{r\rightarrow 1} \frac{1-r}{1-r_0(r)}=0 =\lim _{r\rightarrow 1}\frac{1-r}{r-r_0(r)} \end{aligned}$$

and this implies immediately, recalling (2.17), that \(\lim _{r\rightarrow 1} f_1(r_0(r),r)=+\infty \).

To prove (2.18) we recall that \(f_0(r_0(r),r)=0\) and using (2.14) and (2.16) we get

$$\begin{aligned} 2\big (c + o(1)\big ) \left( \frac{r_0(r)}{(1-r)^s} -\frac{r_0(r)+1}{(1-r_0(r))^s}+ \frac{1}{(r-r_0(r))^s} \right) +r_0(r)k(n)+ \frac{k(n)}{(r_0(r))^s}=0 \end{aligned}$$

from which we deduce that

$$\begin{aligned} \frac{r_0(r)}{(1-r)^s}+\frac{1}{(r-r_0(r))^s} =\frac{1+r_0(r)}{(1-r_0(r))^s} + O(1). \end{aligned}$$
(2.19)

Recalling that

$$\begin{aligned} \frac{1}{(1-r)^s}\geqslant \frac{1}{(1-r_0(r))^s} \,\quad \text {and}\quad \frac{1}{(r-r_0(r))^s} \geqslant \frac{1}{(1-r_0(r))^s}\, \end{aligned}$$

from (2.19) we get that

$$\begin{aligned} \frac{1}{(1-r_0(r))^s}\leqslant \frac{1}{(r-r_0(r))^s} \leqslant \frac{1}{(1-r_0(r))^s} + O(1)\,, \end{aligned}$$

which gives the claim (2.18).

By continuity of \(f_1\), from (2.12) and (2.13), it follows that there exists \(r_1\in (0,1)\) such that \( f_1(r_0(r_1), r_1)=0\), which gives the thesis.

Step 2 We pass now to consider the case \(N>1\). We provide a proof of the existence of a sequence of radii \(r_i\) which solves (2.9). We shall determine \(r_i\) by induction on i.

For \(i=1\) we observe that, given a choice of \(0<r_2<\ldots< r_{2N-1}<1\), we have

$$\begin{aligned} \lim _{r_1\rightarrow 0}H_s(x_{r_1},E) = -\infty \quad \text {and} \quad \lim _{r_1\rightarrow r_2}H_s(x_{r_1},E) = +\infty . \end{aligned}$$

By continuity of the function \(f_1\) it follows that there exists \(\bar{r}_1=\bar{r}_1(r_2, \ldots r_{2N-1}) \in (0,r_2)\) such that \(f_1(\bar{r}_1,\ldots ,r_{2N-1})=0\). As before, in case of multiple solutions we choose the smallest one. Notice that \(\bar{r}_1\) is continuous as a function of \(r_2,\ldots , r_{2N-1}\). Notice also that, if we fix \(r_3,\ldots ,r_{2N-1}\) and let \(r_2\rightarrow r_3\), letting \(F:= B_{\bar{r}_1} \cup A_{r_3,r_2}\) and proceeding as in Step 1, we get

$$\begin{aligned}&H_s(x_{\bar{r}_1},E) \\&\quad = - H_s(x_{\bar{r}_1},F) + O(1) =2 \big (c+o(1)\big ) \left( \frac{1}{|r_2-\bar{r}_1|^{n+s}} -\frac{1}{|r_3-\bar{r}_1|^{n+s}}\right) + O(1) \end{aligned}$$

Since \(f_1(\bar{r}_1,\ldots ,r_{2N-1})=0\), we also have \(H_s(x_{\bar{r}_1},E)=-\bar{r}_1 H_s(x_{\bar{r}_1},E) = O(1)\), whence

$$\begin{aligned} \lim _{r_2\rightarrow r_3} \frac{|r_2-\bar{r}_1|}{|r_3-\bar{r}_1|} = 1. \end{aligned}$$
(2.20)

Step 3 Let now \(2\leqslant i<2N-1\). By induction assumption, for all \(j<i\) there exist continuous functions \(\bar{r}_j(r_i,\ldots ,r_{2N-1}))\) such that \(f_j(\bar{r}_1,\ldots , \bar{r}_{i-1}, r_i, \ldots , r_{2N})=0\). In view of (2.20), we shall also assume that

$$\begin{aligned} \lim _{r_i\rightarrow r_{i+1}} \frac{|r_i-\bar{r}_{i-1}|}{|r_{i+1} -\bar{r}_{i-1}|} = 1, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \lim _{r_i\rightarrow r_{i+1}} \frac{|r_{i+1}-r_{i}|}{|r_{i} -\bar{r}_{i-1}|} = 0. \end{aligned}$$
(2.21)

Given a choice of \(r_j\) for \(j>i\), we want to find \(\bar{r}_i\) such that

$$\begin{aligned} f_i(\bar{r}_1,\ldots , \bar{r}_{i}, r_{i+1}, \ldots , r_{2N})=0 \end{aligned}$$
(2.22)

and

$$\begin{aligned} \lim _{r_{i+1}\rightarrow r_{i+2}} \frac{|r_{i+1}-\bar{r}_{i}|}{|r_{i+2}-\bar{r}_{i}|} = 1. \end{aligned}$$
(2.23)

We first notice that

$$\begin{aligned} \lim _{r_i\rightarrow 0} f_i(\bar{r}_1,\ldots , \bar{r}_{i-1}, r_i, \ldots , r_{2N-1}) = \lim _{r_i\rightarrow 0} (-1)^{i-1} H_s(x_i,E) = -\infty . \end{aligned}$$

We now consider the limit \(r_i\rightarrow r_{i+1}\). Reasoning as in Step 1, we get

$$\begin{aligned} (-1)^{i-1} H_s(x_{r_i},E)= & {} 2\big ( c+o(1)\big ) \left( \frac{1}{|r_{i+1}-r_i|^s} +\sum _{j=k}^{i-1} \frac{(-1)^{i-k}}{|r_i-\bar{r}_k|^s} \right) + O(1), \end{aligned}$$

and therefore, recalling (2.21),

$$\begin{aligned}&\lim _{r_i\rightarrow r_{i+1}} f_i(\bar{r}_1,\ldots , \bar{r}_{i-1}, r_i, \ldots , r_{2N-1}) \\&\quad = \lim _{r_i\rightarrow r_{i+1}} (-1)^{i-1} H_s(x_{r_i},E) +O(1)\\&\quad = \lim _{r_i\rightarrow r_{i+1}} \left( \frac{1}{|r_{i+1}-r_i|^s} +\sum _{j=k}^{i-1} \frac{ (-1)^{i-k}}{|r_i-\bar{r}_k|^s}\right) = +\infty . \end{aligned}$$

By continuity of \(f_i\) it follows that there exists \(\bar{r}_i\) such that \(f_i(\bar{r}_1,\ldots , \bar{r}_{i}, r_{i+1}, \ldots , r_{2N})=0\). As before, in case of multiple solutions, we choose the smallest one.

We now show (2.23). If we fix \(r_{i+2},\ldots ,r_{2N-1}\) and let \(r_{i+1}\rightarrow r_{i+2}\), from (2.22) we get \(H_s(x_{\bar{r}_i},E) = O(1)\), which implies

$$\begin{aligned} -\frac{1}{|r_{i+2}-\bar{r}_{i}|^s} + \frac{1}{|r_{i+1}-\bar{r}_i|^s} +\sum _{j=k}^{i-1} \frac{ (-1)^{i-k}}{|\bar{r}_i-\bar{r}_k|^s} = O(1). \end{aligned}$$

Multiplying by \(|r_{i+1}-\bar{r}_i|^s\) and recalling (2.21) we then get

$$\begin{aligned} \lim _{r_{i+1}\rightarrow r_{i+2}} \frac{|r_{i+1}-\bar{r}_i|^s}{|r_{i+2} -\bar{r}_{i}|^s} - \sum _{j=k}^{i-1} \frac{ (-1)^{i-k}|r_{i+1} -\bar{r}_i|^s}{|\bar{r}_i-\bar{r}_k|^s} = \lim _{r_{i+1}\rightarrow r_{i+2}} \frac{|r_{i+1}-\bar{r}_i|^s}{|r_{i+2}-\bar{r}_{i}|^s} = 1, \end{aligned}$$

which gives (2.23).

Step 4. Finally, for \(i=2N-1\) we still have

$$\begin{aligned} \lim _{r_{2N-1}\rightarrow 0} f_{2N-1}(\bar{r}_1,\ldots , \bar{r}_{2N-2}, r_{2N-1}) = -\infty . \end{aligned}$$

We now consider the limit \(r_{2N-1}\rightarrow 1\). Recalling (2.23) with \(i=2N-2\), as in Step 1 we get

$$\begin{aligned} H_s(x_{r_{2N-1}},E)= & {} 2 \big ( c+o(1)\big ) \left( \frac{1}{(1-r_{2N-1})^s} - \frac{1}{(r_{2N-1}-\bar{r}_{2N-2})^s}\right) + O(1) \\ H_s(x_{1},E)= & {} 2\big ( c+o(1)\big ) \left( \frac{1}{(1-r_{2N-1})^s} - \frac{1}{(1-\bar{r}_{2N-2})^s}\right) + O(1)\\= & {} 2\big ( c+o(1)\big ) \, \frac{1}{(1-r_{2N-1})^s} + O(1), \\= & {} 2\big ( c+o(1)\big ) \, \frac{1}{(1-r_{2N-1})^s} + O(1). \end{aligned}$$

Therefore, we have

$$\begin{aligned} 0= & {} \lim _{r_{2N-1}\rightarrow 1} f_{2N-1}(\bar{r}_1,\ldots , \bar{r}_{2N-2}, r_{2N-1})\\= & {} \lim _{r_{2N-1}\rightarrow 1} \left( r_{2N-1} H_s(x_{1},E) +H_s(x_{r_{2N-1}},E)\right) \\= & {} \lim _{r_{2N-1}\rightarrow 1} 2\big ( c+o(1)\big ) , \frac{1+r_{2N-1}}{|1-r_{2N-1}|^s} = +\infty . \end{aligned}$$

As before, it follows that there exists \(\bar{r}_{2N-1}\) such that \(f_{2N-1}(\bar{r}_1,\ldots , \bar{r}_{2N-1}) =0\). \(\square \)

Remark 2.4

An interesting question which is left open by the previous result is the issue of uniqueness for self-shrinkers with a prescribed number of boundary spheres. In the simplest case, that is the annulus, in Proposition 2.2 we prove uniqueness of the ratio \(\frac{R}{r}\) for which the annulus \(B_R\setminus B_r\) is a self-similar shrinker.

From Theorem 2.3 we readily obtain the existence of cylindrical self-shrinkers.

Corollary 2.5

Let \(k<n\). For all \(N\geqslant 1\) and \(R>0\) there exists an increasing sequence \(0< r_1<\ldots<r_{2N-1}<r_{2N}=R\), depending only on k, s and N, such that such that the flow (1.1) with initial datum

$$\begin{aligned} C:= \mathbb {R}^{n-k}\times \bigcup _{j=1}^N \left( B^k_{r_{2j}} \setminus B^k_{r_{2j-1}}\right) \end{aligned}$$

is a homothetically shrinking solution of (1.1), where \(B^k_r\) denotes the ball of radius r in \(\mathbb {R}^k\).

Similarly, for all \(N\geqslant 1\) and \(R>0\) there exists an increasing sequence \(0< \tilde{r}_0<\tilde{r}_1<\ldots<\tilde{r}_{2N-1} <\tilde{r}_{2N}=R\), depending only on k, s and N, such that such that the flow (1.1) with initial datum

$$\begin{aligned} \widetilde{C}:=\mathbb {R}^{n-k}\times B^k_{\tilde{r}_0}\cup \, \bigcup _{j=1}^N \left( B^k_{\tilde{r}_{2j}}\setminus B^k_{\tilde{r}_{2j-1}}\right) \end{aligned}$$

is a homothetically shrinking solution of (1.1).

Remark 2.6

We observe that the radii r(n, s) in Proposition 2.2, \(r_i(n,s), \tilde{r}_i(n,s)\) in Theorem 2.3 and Corollary 2.5 all satisfy \(\lim _{s\rightarrow 1} r(n,s)=\lim _{s\rightarrow 1} r_i(n,s)=\lim _{s\rightarrow 1} \tilde{r}_i(n,s)=R\).

We give a brief justification of this fact just for the simplest case, that is the case of r(n, s) in Proposition 2.2, the others being completely analogous. We recall that if \(E\subset \mathbb {R}^n\) is a compact set with \(C^2\) boundary then \((1-s)H_s(x,E)\) converges uniformly as \(s\rightarrow 1\) to the classical mean curvature \(H(x,\partial E)\) (see [1]). Under the same notation as in the proof of Proposition 2.2, we note that for \(s=1\) the function \(f_1(r)\) defined in (2.5) is given by \(r-\frac{1}{r}\) (this is also true for the functions \(f_i\) defined in the proof of Theorem 2.3, that is \(f_i= r_i-\frac{1}{r_i}\)). So, by uniform convergence of the curvatures, we get that if \((r_k)_k\) is a sequence with \(r_k\in (0,1)\) and \(r_k\rightarrow 1\), there exists \((s_k)_k\) with \(0<s_k<1\) such that \(f_s(t)<0\) for \(t\in (0, r_k]\) and \(s\geqslant s_k\). This implies that \(r(n, s)>r_k\) for all \(s\geqslant s_k\) and that \(s_k\rightarrow 1\), since \(\lim _{r\rightarrow 1}f_s(r)=+\infty \) for all \(s<1\).

3 Stability

We now discuss the dynamic stability of the symmetric self-shrinkers constructed in the previous section. By definition, self-shrinkers are stationary solutions to the flow

$$\begin{aligned} \partial _t x\cdot \nu =-H_s(x,E)+x\cdot \nu . \end{aligned}$$
(3.1)

If the initial datum is rotationally symmetric as in Theorem 2.3 then (3.1) becomes a system of ODE’s in the radii \(r_i\), and Theorem 2.3 guarantees the existence of a stationary point for every number of radii. We are interested in the stability of such critical points, with respect to perturbations which are orthogonal to the vector \((r_1,\ldots r_{2N})\) (or resp. \((r_0,\ldots r_{2N})\)) given by the radii. Indeed this vector corresponds to a rescaling of the initial datum, and therefore, gives a direction of instability for the system which is not geometrically significant.

In the symmetric situation, we can rewrite (3.1) as the system of ODE’s

$$\begin{aligned} \dot{r}_i= (-1)^{i-1}H_s(x_i,E)+r_i \qquad i\leqslant 2N. \end{aligned}$$
(3.2)

Theorem 3.1

Fix \(N\geqslant 1\), and let E (resp. \(\tilde{E}\)) be the symmetric shrinker given by Theorem (2.3), corresponding to the stationary point \((\bar{r}_1,\ldots \bar{r}_{2N})\) (resp. \((\bar{r}_0,\ldots \bar{r}_{2N})\)) for the system (3.2). Then, the Morse index of such point is at least 2, in particular the corresponding homothetic solution is dynamically unstable.

Proof

We shall prove the assertion for the shrinker E, since the proof for \(\tilde{E}\) is analogous.

For the reader convenience, we first present in detail the case \(N=1\), corresponding to an annulus \(A=B_{\bar{r}_2}\setminus B_{\bar{r}_1}\). The system (3.2) then becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{r_1}= H_s(x_{r_1}, A)+r_1= \frac{k(n)}{r_1^s}\\ \qquad +2\lim _{\varepsilon \rightarrow 0}\left( \int _{B_{r_1-\varepsilon }} \frac{1}{|x_{r_1}-y|^{n+s}}\mathrm{{d}}y -\int _{B_{r_2}\setminus B_{r_1+\varepsilon }} \frac{1}{|x_{r_1} -y|^{n+s}}\mathrm{{d}}y\right) +r_1\\ \dot{r}_2= -H_s(x_{r_2}, A)+r_2=-\frac{k(n)}{r_2^s}-2 \int _{B_{r_1}} \frac{1}{|x_{r_2}-y|^{n+s}}\mathrm{{d}}y+r_2. \end{array}\right. } \end{aligned}$$
(3.3)

We define the function \(g(r_1, r_2)= \left( g_1(r_1, r_2), g_2(r_1, r_2)\right) \) as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} g_1(r_1, r_2):= \frac{k(n)}{r_1^s}+2\lim _{\varepsilon \rightarrow 0} \left( \int _{B_{r_1-\varepsilon }} \frac{1}{|x_{r_1}-y|^{n+s}}\mathrm{{d}}y -\int _{B_{r_2}\setminus B_{r_1+\varepsilon }} \frac{1}{|x_{r_1} -y|^{n+s}}\mathrm{{d}}y\right) +r_1\\ g_2(r_1, r_2):=-\frac{k(n)}{r_2^s}-2\int _{B_{r_1}} \frac{1}{|x_{r_2}-y|^{n+s}}\mathrm{{d}}y+r_2. \end{array}\right. } \end{aligned}$$

We now compute the Jacobian matrix Dg at the point \((\bar{r}_1, \bar{r}_2)\) which is a stationary point for (3.3), that is \(g(\bar{r}_1, \bar{r}_2)=0\).

We observe the following fact: for \(\delta \ne 0\), \(\varepsilon >0\), \(R>r>|\delta |\), there hold

$$\begin{aligned} \int _{B_{r+\delta -\varepsilon }} \frac{1}{|x_{r+\delta }-y|^{n+s}}\mathrm{{d}}y= & {} \left( \frac{r}{r+\delta }\right) ^s\int _{B_{r-\varepsilon \frac{r}{r+\delta }}} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y\\ \int _{B_R\setminus B_{r+\delta +\varepsilon }} \frac{1}{|x_{r+\delta }-y|^{n+s}}\mathrm{{d}}y= & {} \left( \frac{r}{r+\delta }\right) ^s\int _{B_{\frac{Rr}{r+\delta }} \setminus B_{r+\varepsilon \frac{r}{r+\delta }}} \frac{1}{|x_r-y|^{n+s}}\mathrm{{d}}y. \end{aligned}$$

Thus, using these equalities we get that the derivative of \(g_1\) at \((\bar{r}_1, \bar{r}_2)\) are given by

$$\begin{aligned} \partial _{r_1} g_1(\bar{r}_1, \bar{r}_2)= & {} -\frac{sk(n)}{\bar{r}_1^{s+1}}\nonumber \\&-\frac{2s}{\bar{r}_1}\lim _{\varepsilon \rightarrow 0}\left( \int _{B_{\bar{r}_1-\varepsilon }} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}dy-\int _{B_{\bar{r}_2} \setminus B_{\bar{r}_1+\varepsilon }} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}\mathrm{{d}}y\right) \nonumber \\&+\frac{2\bar{r}_2}{\bar{r}_1} \int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}\mathrm{{d}}y+1 \nonumber \\= & {} -\frac{s}{r}g_1(\bar{r}_1, \bar{r}_2)+s+1+\frac{2\bar{r}_2}{\bar{r}_1} \int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}\mathrm{{d}}y \end{aligned}$$
(3.4)
$$\begin{aligned}= & {} s+1+\frac{2\bar{r}_2}{\bar{r}_1} \int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}\mathrm{{d}}y \nonumber \\ \partial _{r_2} g_1(\bar{r}_1, \bar{r}_2)= & {} -2 \int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_1}-y|^{n+s}}\mathrm{{d}}y. \end{aligned}$$
(3.5)

Analogously, we observe that for \(\delta \ne 0\), \(R>r>|\delta |\), there holds

$$\begin{aligned} \int _{B_r} \frac{1}{|x_{R+\delta }-y|^{n+s}}\mathrm{{d}}y =\left( \frac{R}{R+\delta }\right) ^s\int _{B_{\frac{Rr}{R+\delta }}} \frac{1}{|x_R-y|^{n+s}}\mathrm{{d}}y. \end{aligned}$$

Using this equality, we compute the derivative of \(g_2\) at \((\bar{r}_1, \bar{r}_2)\):

$$\begin{aligned} \partial _{r_1} g_2(\bar{r}_1, \bar{r}_2)= & {} -2\int _{\partial B_{\bar{r}_1}} \frac{1}{|x_{\bar{r}_2}-y|^{n+s}}\mathrm{{d}}y\nonumber \\ \partial _{r_2} g_2(\bar{r}_1, \bar{r}_2)= & {} \frac{sk(n)}{\bar{r}_2^{s+1}} +\frac{2s}{\bar{r}_2}\int _{B_{\bar{r}_1}} \frac{1}{|x_{\bar{r}_2}-y|^{n+s}}\mathrm{{d}}y +\frac{2\bar{r}_1}{\bar{r}_2}\int _{\partial B_{\bar{r}_1}} \frac{1}{|x_{\bar{r}_2}-y|^{n+s}}\mathrm{{d}}y+1\nonumber \\= & {} -\frac{s}{\bar{r}_2}g_2(\bar{r}_1,\bar{r}_2)+s+1 +\frac{2\bar{r}_1}{\bar{r}_2}\int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_2}-y|^{n+s}}\mathrm{{d}}y\nonumber \\= & {} s+1+\frac{2\bar{r}_1}{\bar{r}_2}\int _{\partial B_{\bar{r}_2}} \frac{1}{|x_{\bar{r}_2}-y|^{n+s}}\mathrm{{d}}y. \end{aligned}$$
(3.6)

Note that, using (3.5) and (3.6),

$$\begin{aligned} Dg(\bar{r}_1,\bar{r}_2)(\bar{r}_1,\bar{r}_2)^t= (s+1)(\bar{r}_1,\bar{r}_2)^t \end{aligned}$$

so that \((\bar{r}_1,\bar{r}_2)\) is an eigenvector with eigenvalue \(s+1>0\). Moreover, by (3.6), we observe that \(\partial _{r_2} g_2(\bar{r}_1, \bar{r}_2)>s+1\). This implies that

$$\begin{aligned} \max _{v:\,|v|=1} v Dg(\bar{r}_1, r_2) v^t\geqslant (0,1) Dg(\bar{r}_1, \bar{r}_2)(0,1)^t > s+1, \end{aligned}$$
(3.7)

which gives that \(Dg(\bar{r}_1, r_2) \) has a second eigenvalue bigger than \(s+1\), and then in particular positive.

We now consider the general case of a self-shrinker

$$\begin{aligned} E:= \bigcup _{k=1}^N \left( B_{\bar{r}_{2k}} \setminus B_{\bar{r}_{2k-1}}\right) . \end{aligned}$$
(3.8)

We also let \(\bar{r}=(\bar{r}_1,\ldots ,\bar{r}_{2N}),\,g (\bar{r})=(g_1(\bar{r}),\ldots ,g_{2N}(\bar{r})) \in \mathbb {R}^2N\), where

$$\begin{aligned} g_{i}(\bar{r}) :=\,&-H_s(x_{i},E)+\bar{r}_{i}= -\frac{k(n)}{\bar{r}_{i}^s} + \bar{r}_{i} +2\sum _{j< i} (-1)^{i-j} \int _{B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y\\&-2\sum _{j> i} (-1)^{i-j} \int _{\mathbb {R}^n\setminus B_{\bar{r}_j}} \frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y, \end{aligned}$$

if the index i is even, and

$$\begin{aligned} g_i(\bar{r}) :=\,&H_s(x_i,E)+r_i =-H_s(x_i,\mathbb {R}^n\setminus E)+r_i\\ =\,&- \frac{k(n)}{\bar{r}_{i}^s} + \bar{r}_{i}+2\sum _{j< i} (-1)^{i-j} \int _{B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y\\&-2\sum _{j> i} (-1)^{i-j} \int _{\mathbb {R}^n\setminus B_{\bar{r}_j}} \frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y, \end{aligned}$$

if i is odd. Notice that, since \(\bar{r}\) is a stationary solution to (3.2), we have \(g(\bar{r})=0\).

We compute, for \(j\ne i\),

$$\begin{aligned} \frac{\partial g_i}{\partial r_j}(\bar{r}) = 2 (-1)^{i-j} \int _{\partial B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial g_i}{\partial r_i}(\bar{r})= & {} - \frac{s k(n)}{\bar{r}_{i}^{s+1}} + 1 -2\frac{s}{\bar{r}_i}\sum _{j< i} (-1)^{i-j} \int _{B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y\\&+\,2\frac{s}{\bar{r}_i}\sum _{j> i} (-1)^{i-j} \int _{\mathbb {R}^n\setminus B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y \\&-\,\frac{2}{\bar{r}_i}\sum _{j\ne i} (-1)^{i-j} \bar{r}_j \int _{\partial B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y\\= & {} -\,\frac{2}{\bar{r}_i} g_i(\bar{r}) + s+ 1 -\frac{2}{\bar{r}_i} \sum _{j\ne i} (-1)^{i-j} \bar{r}_j \int _{\partial B_{\bar{r}_j}} \frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y \\= & {} s+ 1 +2 \sum _{j\ne i} \frac{\bar{r}_j}{\bar{r}_i}\, (-1)^{i-j+1} \int _{\partial B_{\bar{r}_j}}\frac{1}{|x_{\bar{r}_i}-y|^{n+s}}\,\mathrm{{d}}y. \end{aligned}$$

Notice that

$$\begin{aligned} Dg(\bar{r}) \bar{r}^t = \sum _j \frac{\partial g_i}{\partial r_j} (\bar{r}) r_j = (s+1) \bar{r}^t, \end{aligned}$$

so that \(\bar{r}\) is an eigenvector with eigenvalue \(s+1>0\).

Now, we claim that

$$\begin{aligned} \frac{\partial g_{2N}}{\partial r_{2N}}(\bar{r}) > s+1. \end{aligned}$$
(3.9)

If the claim is true, then reasoning as in (3.7), we conclude that there exists an eigenvalue of \(Dg(\bar{r})\) which is strictly greater than \(s+1\) (and then positive), so that the Morse index of \((\bar{r}_1,\ldots \bar{r}_{2N})\) is at least 2.

Since

$$\begin{aligned} \frac{\partial g_{2N}}{\partial r_{2N}}(\bar{r}) = s+1 +\frac{2}{\bar{r}_{2N}} \sum _{j=1}^{2N-1} (-1)^{j-1} \,\bar{r}_{2N-j} \int _{\partial B_{\bar{r}_{2N-j}}} \frac{1}{|x_{\bar{r}_{2N}}-y|^{n+s}}\,\mathrm{{d}}y, \end{aligned}$$

to get the claim (3.9) it is sufficient to prove that for all \(1\leqslant i<j<2N\) there holds

$$\begin{aligned} \bar{r}_i \int _{\partial B_{\bar{r}_i}}\frac{1}{|x_{\bar{r}_{2N}} -y|^{n+s}}\,\mathrm{{d}}y < \bar{r}_j \int _{\partial B_{\bar{r}_j}} \frac{1}{|x_{\bar{r}_{2N}}-y|^{n+s}}\,\mathrm{{d}}y. \end{aligned}$$
(3.10)

We shall prove a slightly stronger statement, namely that

$$\begin{aligned} r\mapsto h(r) := \int _{\partial B_{r}}\frac{1}{|x_{\bar{r}_{2N}} -y|^{n+s}}\,\mathrm{{d}}y\quad \text { is strictly increasing on}\ (0,\bar{r}_{2N}). \end{aligned}$$
(3.11)

Indeed, we compute

$$\begin{aligned} h'(r)= & {} \int _{\partial B_{r}} \nabla \left( \frac{1}{|x_{\bar{r}_{2N}} -y|^{n+s}}\right) \cdot \nu (y)\,\mathrm{{d}}y= \int _{B_{r}} \Delta \left( \frac{1}{|x_{\bar{r}_{2N}}-x|^{n+s}}\right) \,\mathrm{{d}}x\\= & {} (n+s) (s+2) \int _{B_{r}} \frac{1}{|x_{\bar{r}_{2N}} -x|^{n+s+2}}\,\mathrm{{d}}x >0, \end{aligned}$$

which shows (3.11), and so proves (3.10). \(\square \)

Remark 3.2

It would be interesting to determine exactly the Morse index of the stationary points \((\bar{r}_1,\bar{r}_2, \dots , \bar{r}_{2N})\) (resp. \((\bar{r}_0,\bar{r}_1, \dots , \bar{r}_{2N})\)) of the flow (3.2). In the simplest case \(N=1\), we proved in Theorem 3.1 that the index of \((\bar{r}_1,\bar{r}_2)\) is equal to 2.

It would also be interesting to understand if the ball is dynamically stable for any perturbation, not necessarily radial, as it happens for the standard mean curvature flow [11, 13].