Stability for a Second Type Partitioning Problem

Abstract

In this paper, we study stability and instability problem for type-II partitioning problem. First, we make a complete classification of stable type-II stationary hypersurfaces in a ball in a space form as totally geodesic n-balls. Second, for general ambient spaces and convex domains, we give some topological restriction for type-II stable stationary immersed surfaces in two dimension. Third, we give a lower bound for the Morse index for type-II stationary hypersurfaces in terms of their topology.

Appendix A: Second Variational Formula

In this appendix, we prove Proposition 2.2, namely, the second variation formula of area functional (2.7) under admissible wetting-area-preserving variations. For simplicity, we use $\langle \cdot , \cdot \rangle$ to denote all the inner products in the following computation. The computation is very close to the one by Ros–Souam [26].

Firstly, applying the admissible condition, (2.4) and (2.5), we get

$$\begin{aligned} 0=\langle Y,{\bar{N}}\rangle =\langle Y, \sin \theta \,\mu -\cos \theta \,\,\nu \rangle =\sin \theta \,\langle Y,\mu \rangle -\varphi \cos \theta \,\quad \text {on}\,\,\partial M, \end{aligned}$$

Therefore,

$$\begin{aligned} \langle Y,\mu \rangle =\varphi \, \cot \theta \, \quad \text {on}\,\,\partial M. \end{aligned}$$

(A.1)

So we can define

$$\begin{aligned} Y=Y_{0}+\varphi \nu \triangleq Y_{1}+\cot \theta \, \varphi \mu +\varphi \nu \end{aligned}$$

(A.2)

where $Y_{1}$ denotes the tangent part of Y to $\partial M$.

On the other hand, from

$$\begin{aligned} {\bar{\nu }}=\cos \theta \, \mu + \sin \theta \,\nu , \end{aligned}$$

we see Y can be also expressed as follows

$$\begin{aligned} Y=Y_{1}+\frac{\varphi }{\sin \theta }(\cos \theta \,\mu +\sin \theta \nu )=Y_{1}+\frac{\varphi }{\sin \theta }{\bar{\nu }}. \end{aligned}$$

(A.3)

We use a prime to denote the time derivative at $t=0$ in the following.

Proposition A.1

[26] Let ${\tilde{\nabla }}$ denote the gradient on $\partial M$ for the metric induced by x and $Y_{0}$ (resp. $Y_{1}$) the tangent part of Y to M (resp. to $\partial M$). Let also $S_{0}$, $S_{1}$, and $S_{2}$ denote, respectively, the shape operator of M in ${\overline{M}}$ with respect to $\nu$, of $\partial M$ in M with respect to $\mu$ and of $\partial M$ in $\partial B$ with respect to ${\bar{\nu }}$. Then

• (1) $\nu '=S_{0}(Y_{0})-\nabla \varphi .$

• (2) $\mu '=-(h(Y_{0},\mu )+\nabla _{\mu }\varphi )\nu -\varphi S_{0}(\mu )+\varphi h(\mu , \mu )\mu +S_{1}(Y_{1})-\cot \theta \,{\tilde{\nabla }}\varphi .$

• (3) ${\bar{\nu }}'=-h^{\partial B}(Y,{\bar{\nu }}){\bar{N}}+S_{2}(Y_{1})-\frac{1}{\sin \theta }{\tilde{\nabla }}\varphi .$

Proof

to prove (1), let $\{e_{i}\}_{i=1}^{n}$ be an orthonormal basis of $T_{p}M$ for some $p\in M$. Put $e_{i}(t)=(x(t,\cdot ))_{*}(e_{i})$, then using the fact $\langle e_{i}(t),\nu (t)\rangle =0$ and $[e_{i}(t), Y(t)]=0$, we have

$$\begin{aligned} \nu '&=\sum _{i=1}^{n}\langle \nu ', e_{i}\rangle e_{i}=-\sum _{i=1}^{n}\langle \nu , e_{i}'\rangle e_{i} \\&=-\sum _{i=1}^{n}\langle \nu , {{\bar{\nabla }}}_{e_{i}}Y\rangle e_{i}=-\sum _{i=1}^{n}\langle \nu ,{{\bar{\nabla }}}_{e_{i}}(Y_{0}+\varphi \nu ) \rangle e_{i} \\&=\sum _{i=1}^{n}\langle S_{0}(Y_{0}), e_{i}\rangle e_{i}-\sum _{i=1}^{n}d\varphi (e_{i})e_{i} \\&=S_{0}(Y_{0})-\nabla \varphi \end{aligned}$$

As a consequence of (1), we get

$$\begin{aligned} \langle \mu ',\nu \rangle =-\langle \mu ,\nu '\rangle =-h(Y_{0}, \mu )+\nabla _{\mu }\varphi . \end{aligned}$$

(A.4)

Let now $\{T_\alpha \}_{\alpha =1}^{n-1}$ be an orthonormal basis of $T_{p}(\partial M)$ for some $p\in \partial M$. As before, put $T_\alpha (t)=(x(t,\cdot ))_{*}(T_\alpha )$, then we can use (A.2) and $[T_\alpha (t), Y(t)]=0$

$$\begin{aligned} \langle \mu ', T_\alpha \rangle&=-\langle \mu ,T'_{\alpha }\rangle =-\langle \mu ,{{\bar{\nabla }}}_{T_\alpha }Y\rangle =-\langle \mu , {{\bar{\nabla }}}_{T_\alpha }(Y_{1}+\cot \theta \, \varphi \mu +\varphi \nu )\rangle \nonumber \\&=-\langle \mu , {{\bar{\nabla }}}_{T_\alpha }Y_{1}\rangle -\cot \theta \, d\varphi (T_\alpha )-\varphi \langle \mu , {{\bar{\nabla }}}_{T_\alpha }\nu \rangle \end{aligned}$$

(A.5)

Thanks to (A.4) and (A.5), we have

$$\begin{aligned} \mu '&=\langle \mu ',\mu \rangle \mu +\langle \mu ', \nu \rangle \nu +\sum _{\alpha =1}^{n-1}\langle \mu ', T_\alpha \rangle T_\alpha \nonumber \\&=(-h(Y_{0}, \mu )+\nabla _{\mu }\varphi )\nu -\sum _{\alpha =1}^{n-1}\langle \mu , {{\bar{\nabla }}}_{T_\alpha }Y_{1}\rangle T_\alpha -\cot \theta \,\sum _{\alpha =1}^{n-1}d\varphi (T_\alpha )T_\alpha \nonumber \\&\quad -\varphi \sum _{\alpha =1}^{n-1}\langle S_{0}(\mu ), T_\alpha \rangle T_\alpha \nonumber \\&=(-h(Y_{0}, \mu )+\nabla _{\mu }\varphi )\nu +S_{1}(Y_{1})-\cot \theta \,{\tilde{\nabla }}\varphi -\varphi (S_{0}(\mu )-h(\mu ,\mu )\mu ) \end{aligned}$$

(A.6)

The formula (2) follows from (A.6).

To prove (3), we use $[T_\alpha (t), Y(t)]=0$ again and (A.3)

$$\begin{aligned} \langle {\bar{\nu }}',T_\alpha \rangle&=-\langle {\bar{\nu }},T'_{\alpha }\rangle =-\langle {\bar{\nu }},{{\bar{\nabla }}}_{T_\alpha }Y\rangle \nonumber \\&=-\langle {\bar{\nu }},{{\bar{\nabla }}}_{T_\alpha }Y_{1}\rangle -\frac{1}{\sin \theta }\,d\varphi (T_\alpha )\nonumber \\&=\langle S_{2}(Y_{1}),T_\alpha \rangle -\frac{1}{\sin \theta }\,d\varphi (T_\alpha ) \end{aligned}$$

(A.7)

Therefore, the formula (3) now follows from (A.7) and the fact $\langle {\bar{\nu }}',{\bar{N}}\rangle =-h^{\partial B}(Y,{\bar{\nu }}).$ $\square$

Proposition A.2

$$\begin{aligned} \langle S_{1}(Y_{1}),Y_{1}\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi ,Y_{1}\rangle =\cos \theta \,\,\langle Y, {\bar{\nu }}'\rangle +\sin \theta \,h^{\partial B}(Y_{1},Y_{1})\quad \text {along} \quad \partial M.\quad \end{aligned}$$

Proof

By Proposition A.1 (3) and the fact $\langle Y,{\bar{N}}\rangle =0$, we have

$$\begin{aligned} \cos \theta \,\,\langle Y,{\bar{\nu }}'\rangle&=\cos \theta \,\langle Y,-h^{\partial B}(Y,{\bar{\nu }}){\bar{N}}+S_{2}(Y_{1})-\frac{1}{\sin \theta }{\tilde{\nabla }}\varphi \rangle \nonumber \\&=\langle D_{Y_{1}}(\cos \theta \,{\bar{\nu }}), Y_{1}\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi ,Y_{1}\rangle \nonumber \\&=\langle D_{Y_{1}}(\mu -\sin \theta {\bar{N}}),Y_{1}\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi ,Y_{1}\rangle \nonumber \\&=\langle S_{1}(Y_{1}),Y_{1}\rangle -\sin \theta \,h^{\partial B}(Y_{1},Y_{1})-\cot \theta \,\langle {\tilde{\nabla }}\varphi ,Y_{1}\rangle . \end{aligned}$$

(A.8)

$\square$

Now we are ready to prove Proposition 2.2.

Proof of Proposition 2.2

Firstly, from (2.1), we have

$$\begin{aligned} A''(0)=\int \nolimits _{M}&H'\varphi +H\varphi '\, dA+\int \nolimits _{\partial M}\langle Y', \mu \rangle +\langle Y, \mu '\rangle ds+\int \nolimits _{\partial M}\langle Y,\mu \rangle \frac{d}{dt}\Big |_{t=0}ds_{t}. \end{aligned}$$

Notice that $A'(0)=0$ for any wetting-area-preserving variations if and only if $H=0$ in M and $\theta$ is constant. Moreover, we have the well-known formula (see [28])

$$\begin{aligned} H'=-(\Delta \varphi +|h|^{2}\varphi +\overline{\mathrm{Ric}}(\nu ,\nu )\varphi ). \end{aligned}$$

It follows that

$$\begin{aligned} A''(0)&=-\int \nolimits _{M}\varphi (\Delta \varphi +|h|^{2}\varphi +\overline{\mathrm{Ric}}(\nu ,\nu )\varphi )\, dA+\int \nolimits _{\partial M}\langle Y', \mu \rangle +\langle Y, \mu '\rangle ds\nonumber \\&\quad +\int \nolimits _{\partial M}\langle Y,\mu \rangle \frac{d}{dt}\Big |_{t=0}ds_{t}. \end{aligned}$$

(A.9)

So to prove the formula for $A''(0)$, we need to compute

$$\begin{aligned} \int \nolimits _{\partial M}\langle Y', \mu \rangle +\langle Y, \mu '\rangle ds+\int \nolimits _{\partial M}\langle Y,\mu \rangle \frac{d}{dt}\Big |_{t=0}ds_{t}. \end{aligned}$$

(A.10)

Firstly, we compute the first term of (A.10). By using (2.4), we have

$$\begin{aligned} \langle Y',\mu \rangle&=\langle Y',\sin \theta \, {\bar{N}}+\cos \theta \, {\bar{\nu }}\rangle \nonumber \\&=\sin \theta \, \langle D_{Y}Y,{\bar{N}}\rangle +\cos \theta \,\langle Y',{\bar{\nu }}\rangle \nonumber \\&=-\sin \theta \, h^{\partial B}(Y,Y)+\cos \theta \,\langle Y',{\bar{\nu }}\rangle . \end{aligned}$$

(A.11)

By (A.3), we obtain that

$$\begin{aligned} h^{\partial B}(Y,Y)= h^{\partial B}(Y_{1},Y_{1})+\frac{2\varphi }{\sin \theta }\, h^{\partial B}(Y_{1},{\bar{\nu }})+\frac{\varphi ^{2}}{\sin ^{2}\theta }\, h^{\partial B}({\bar{\nu }},{\bar{\nu }}). \end{aligned}$$

(A.12)

Applying (A.11) and (A.12), we have

$$\begin{aligned} \int \nolimits _{\partial M}\langle Y', \mu \rangle ds&=-\sin \theta \int \nolimits _{\partial M}h^{\partial B}(Y,Y)ds+\cos \theta \,\int \nolimits _{\partial M}\langle Y',{\bar{\nu }}\rangle ds\nonumber \\&=-\sin \theta \, \int \nolimits _{\partial M}h^{\partial B}(Y_{1},Y_{1})+\frac{2\varphi }{\sin \theta } \, h^{\partial B}(Y_{1},{\bar{\nu }})\nonumber \\&\quad +\frac{\varphi ^{2}}{\sin ^{2}\theta }\, h^{\partial B}({\bar{\nu }},{\bar{\nu }})ds+\cos \theta \,\int \nolimits _{\partial M}\langle Y',{\bar{\nu }}\rangle ds. \end{aligned}$$

(A.13)

On the other hand, using (2.4) and (2.5), we get

$$\begin{aligned} h(Y_{1},\mu )=\langle \mu ,{{\bar{\nabla }}}_{Y_{1}}\nu \rangle&=\langle \sin \theta {\bar{N}}+\cos \theta \,{\bar{\nu }}, {{\bar{\nabla }}}_{Y_{1}}(-\cos \theta \,{\bar{N}}+\sin \theta {\bar{\nu }})\rangle \nonumber \\&=-\langle {\bar{\nu }}, {{\bar{\nabla }}}_{Y_{1}}{\bar{N}}\rangle =-h^{\partial B}(Y_{1}, {\bar{\nu }}) \end{aligned}$$

(A.14)

Therefore, inserting (A.14) into (A.13), we get the first term of (A.10)

$$\begin{aligned} \int \nolimits _{\partial M}\langle Y', \mu \rangle ds&=-\sin \theta \int \nolimits _{\partial M}h^{\partial B}(Y,Y)\,ds+\cos \theta \,\int \nolimits _{\partial M}\langle Y',{\bar{\nu }}\rangle \,ds\nonumber \\&=-\sin \theta \int \nolimits _{\partial M}h^{\partial B}(Y_{1},Y_{1})-\frac{2\varphi }{\sin \theta } \,h(Y_{1},\mu )\nonumber \\&\quad +\frac{\varphi ^{2}}{\sin ^{2}\theta } \,h^{\partial B}({\bar{\nu }},{\bar{\nu }})ds+\cos \theta \int \nolimits _{\partial M}\langle Y',{\bar{\nu }}\rangle \,ds \end{aligned}$$

(A.15)

By Proposition A.1 (2), Proposition A.2 and (A.2), we can compute the second term of (A.10) as follow

$$\begin{aligned}&\int \nolimits _{\partial M}\langle Y, \mu '\rangle \, ds\nonumber \\&\quad =\int \nolimits _{\partial M}(-h(Y_{0},\mu )+\nabla _{\mu }\varphi )\langle Y,\nu \rangle -\varphi \langle S_{0}(\mu ), Y\rangle +\varphi h(\mu ,\mu )\langle Y,\mu \rangle \nonumber \\&\qquad +\langle S_{1}(Y_{1}),Y\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi , Y\rangle \,ds\nonumber \\&=\int \nolimits _{\partial M}\varphi \nabla _{\mu }\varphi -2h(Y_{0},\mu )\varphi +\varphi h(\mu ,\mu )\langle Y_{1}+\cot \theta \, \varphi \mu +\varphi \nu ,\mu \rangle \nonumber \\&\qquad +\langle S_{1}(Y_{1}),Y\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi , Y\rangle \, ds\nonumber \\&=\int \nolimits _{\partial M}\varphi \nabla _{\mu }\varphi -2h(Y_{0},\mu )\varphi +\cot \theta \, \varphi ^{2}h(\mu ,\mu )\nonumber \\&\qquad +\langle S_{1}(Y_{1}),Y\rangle -\cot \theta \,\langle {\tilde{\nabla }}\varphi , Y\rangle \,ds\nonumber \\&=\int \nolimits _{\partial M}\varphi \nabla _{\mu }\varphi -2h(Y_{1}+\cot \theta \, \varphi \mu ,\mu )\varphi +\cot \theta \, \varphi ^{2}h(\mu ,\mu )\nonumber \\&\qquad +\cos \theta \,\langle Y,{\bar{\nu }}'\rangle +\sin \theta \,h^{\partial B}(Y_{1},Y_{1})\,ds\nonumber \\&=\int \nolimits _{\partial M}\varphi \nabla _{\mu }\varphi -2h(Y_{1}, \mu )\varphi -\cot \theta \, \varphi ^{2}h(\mu ,\mu )\nonumber \\&\qquad +\cos \theta \,\langle Y,{\bar{\nu }}'\rangle +\sin \theta \,h^{\partial B}(Y_{1},Y_{1})\,ds. \end{aligned}$$

(A.16)

Next, we compute the third boundary term of (A.10) by using (2.4)

$$\begin{aligned} \int \nolimits _{\partial M}\langle Y,\mu \rangle \frac{d}{dt}\Big |_{t=0}ds_{t}= & {} \int \nolimits _{\partial M}\langle Y, \sin \theta {\bar{N}}+\cos \theta \,{\bar{\nu }}\rangle \frac{d}{dt}\Big |_{t=0}ds_{t}\nonumber \\= & {} \cos \theta \,\int \nolimits _{\partial M}\langle Y,{\bar{\nu }}\rangle \frac{d}{dt}\Big |_{t=0}ds_{t}. \end{aligned}$$

(A.17)

Therefore, putting (A.15), (A.16), (A.17) into (A.9), we get

$$\begin{aligned} A''(0)&=-\int \nolimits _{M}\varphi (\Delta \varphi +(|h|^{2}+\overline{\mathrm{Ric}}(\nu ,\nu ))\varphi )dA \\&\quad +\int \nolimits _{\partial M}\varphi (\nabla _{\mu }\varphi -q\varphi )ds+\cos \theta \,\frac{d}{dt}\Big |_{t=0}\left( \int \nolimits _{\partial M}\langle Y, {\bar{\nu }}\rangle ds_{t}\right) \\&=-\int \nolimits _{M}\varphi (\Delta \varphi +(|h|^{2}+\overline{\mathrm{Ric}}(\nu ,\nu ))\varphi )dA+\int \nolimits _{\partial M}\varphi (\nabla _{\mu }\varphi -q\varphi )ds \end{aligned}$$

where in the last equality we have used the wetting-area-preserving condition

$$\begin{aligned} A_W'(0)=\frac{d}{dt}\Big |_{t=0}\left( \int \nolimits _{\partial M}\langle Y, {\bar{\nu }}\rangle ds_{t}\right) =0 \end{aligned}$$

and the expression (2.8) for q. The proof is completed. $\square$