Stability of Near Cylindrical Stationary Solutions to Weighted-Volume Preserving Curvature Flows

Abstract

In this paper we consider a class of weighted-volume preserving curvature flows acting on hypersurfaces that are trapped within two parallel hyperplanes and satisfy an orthogonal boundary condition. In previous work the stability of cylinders under the flows was considered; it was found that they are stable provided the radius satisfies a certain condition. Here we consider flows where there is a critical radius such that the cylinders are only stable on one side of the critical value, and the weight function is a linear combination of the elementary symmetric functions, the reason for such a choice is made clear in the appendix. We find that in such instances bifurcation from the cylindrical stationary solutions occurs at the critical radius and we determine a condition on the speed and weight functions such that the nearby non-cylindrical stationary solutions are stable under the flow. In particular, we find that high dimensional half-period unduoids close to the cylinder are stable stationary solutions to the volume preserving mean curvature flow under axially symmetric, volume preserving perturbations. We will also highlight the specific cases of homogeneous speed functions and the mixed-volume preserving mean curvature flows.

Appendix: A Flow Invariant

In this appendix we aim to determine under what conditions the flow (1) has an invariant weighted-volume type quantity. For simplicity we consider the case where the hypersurfaces are axially symmetric, i.e., Eq. (6), in which case the question becomes when does there exist a second order operator Q such that $\int Q(\rho )\,dz$ is independent of time. To perform the analysis we again consider the equivalent flow on the circle given in (8). To simplify notation we will define $l(u'):=\sqrt{1+u'^2}$.

Lemma 6.1

There exists $Q(u)=q(u,u',u'')$ such that

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz=n\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}v\Xi \left( \varvec{\kappa }_{u}\right) u^{n-1}\,dz \end{aligned}$$

(49)

if and only if $\Xi \left( \varvec{\kappa }\right) =\sum _{a=0}^n c_aE_a\left( \varvec{\kappa }\right)$ for some constants $c_a\in \mathbb {R}$. In this case (49) is satisfied by the Q in (9).

Proof

We proceed with the “only if” case by calculating the linearization of Q inside an integral and use integration by parts

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz=&\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\frac{\partial q}{\partial x_1}\left( u,u',u''\right) v+\frac{\partial q}{\partial x_2}\left( u,u',u''\right) \frac{\partial v}{\partial z}\\&+\frac{\partial q}{\partial x_3}\left( u,u',u''\right) \frac{\partial ^2 v}{\partial z^2}\,dz\\ =&\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}v\left( \frac{\partial q}{\partial x_1}(u,u',u'')-\frac{\partial }{\partial z}\left( \frac{\partial q}{\partial x_2}\left( u,u',u''\right) \right. \right. \\&-\left. \left. \frac{\partial }{\partial z}\left( \frac{\partial q}{\partial x_3}\left( u,u',u''\right) \right) \right) \right) \,dz. \end{aligned}$$

By now expanding the derivatives we obtain

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz&= \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\left( \frac{\partial q}{\partial x_1}(u,u',u'')-\frac{\partial ^2q}{\partial x_1\partial x_2}\left( u,u',u''\right) u'\right. \\&\quad \left. -\,\frac{\partial ^2q}{\partial x_2^2}\left( u,u',u''\right) u''-\frac{\partial ^2q}{\partial x_3\partial x_2}\left( u,u',u''\right) u''' \right. \\&\quad +\,\frac{\partial ^3 q}{\partial x_1^2\partial x_3}\left( u,u',u''\right) u'^2\left. +\,2\frac{\partial ^3q}{\partial x_1\partial x_2\partial x_3}\left( u,u',u''\right) u'u'' \right. \\&\quad +2\frac{\partial ^3q}{\partial x_1\partial x_3^2}\left( u,u',u''\right) u'u'''\\&\quad \left. +\,\frac{\partial ^2q}{\partial x_1\partial x_3}\left( u,u',u''\right) u'' +\frac{\partial ^3q}{\partial x_2^2\partial x_3}\left( u,u',u''\right) u''^2\right. \\&\quad \left. +\,2\frac{\partial ^3 q}{\partial x_2\partial x_3^2}\left( u,u',u''\right) u''u'''+\frac{\partial ^2q}{\partial x_2\partial x_3}\left( u,u',u''\right) u'''\right. \\&\quad \left. +\,\frac{\partial ^3q}{\partial x_3^3}\left( u,u',u''\right) u'''^2 +\frac{\partial ^2q}{\partial x_3^2}\left( u,u',u''\right) u''''\right) v\,dz. \end{aligned}$$

However since $\Xi \left( \varvec{\kappa }_{u}\right) u^{n-1}$ does not depend on $u''''$ we see that we require $\frac{\partial ^2q}{\partial x_3^2}=0$ and hence q is linear in $x_3$. Let $q\left( x_1,x_2,x_3\right) =\alpha \left( x_1,x_2\right) x_3+\beta \left( x_1,x_2\right)$, then:

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz&= \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\left( 2\frac{\partial \alpha }{\partial x_1}\left( u,u'\right) u'' +\frac{\partial \beta }{\partial x_1}\left( u,u'\right) +\frac{\partial ^2\alpha }{\partial x_1\partial x_2}\left( u,u'\right) u'u''\right. \\&\quad \left. -\,\frac{\partial ^2\beta }{\partial x_1\partial x_2}\left( u,u'\right) u' -\frac{\partial ^2\beta }{\partial x_2^2}\left( u,u'\right) u'' +\frac{\partial ^2\alpha }{\partial x_1^2}\left( u,u'\right) u'^2\right) v\,dz. \end{aligned}$$

This is now linear in $u''$, hence $\Xi \left( \varvec{\kappa }_{u}\right) u^{n-1}$ must also be linear in $u''$ for (49) to hold. This in turn means that $\Xi \left( \varvec{\kappa }\right)$ must be linear in $\kappa _n$, however by symmetry it is therefore linear in all $\kappa _a$. Thus it must be a linear combination of the $E_a$s.

To see that if $\Xi$ is a linear combination of the $E_a$s we do obtain a Q such that (49) holds we linearize the Q given in (9) in parts. Firstly consider $Q_0(u)=\frac{1}{n}u^n$:

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ_0(u)[v]\,dz=\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}u^{n-1}v\,dz=\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}vE_0u^{n-1}\,dz. \end{aligned}$$

Now consider $Q_a(u)=\frac{1}{a}E_{a-1}\left( \varvec{\kappa }_{u}\right) u^{n-1}L(u)$ for $1\le a\le n+1$ and use the notation $\left( {\begin{array}{c}b\\ k\end{array}}\right) =0$ for $k>b$. This means that

$$\begin{aligned} q\left( x_1,x_2,x_3\right) =\frac{1}{a}\left( \left( {\begin{array}{c}n-1\\ a-1\end{array}}\right) x_1^{n-a}l(x_2)^{2-a} -\left( {\begin{array}{c}n-1\\ a-2\end{array}}\right) x_1^{n+1-a}l(x_2)^{-a}x_3\right) , \end{aligned}$$

with $\frac{dl}{dx_2}=x_2l(x_2)^{-1}$. A standard computation then gives

$$\begin{aligned}&\frac{\partial q}{\partial x_1}\left( u,u',u''\right) -\frac{\partial }{\partial z} \left( \frac{\partial q}{\partial x_2}\left( u,u',u''\right) -\frac{\partial }{\partial z} \left( \frac{\partial q}{\partial x_3}\left( u,u',u''\right) \right) \right) \\&\quad =\left( {\begin{array}{c}n-1\\ a\end{array}}\right) u^{n-a-1}l(u')^{-a}-\left( {\begin{array}{c}n-1\\ a-1\end{array}}\right) u^{n-a}l(u')^{-(a+2)}u''\\&\quad =E_{a}\left( \varvec{\kappa }_{u}\right) u^{n-1}, \end{aligned}$$

where we set $E_{n+1}=0$. Using the formula for $\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz$ from the start of this proof we have:

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ_a(u)[v]\,dz=\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}vE_{a}\left( \varvec{\kappa }_{u}\right) u^{n-1}\,dz, \end{aligned}$$

Therefore if $\Xi \left( \varvec{\kappa }\right) =\sum _{a=0}^n c_aE_a\left( \varvec{\kappa }\right)$ and we set $Q(u)=n\sum _{a=0}^{n+1} c_aQ_a(u)$ for some $c_{n+1}\in \mathbb {R}$ (as in Eq. (9)) we have

$$\begin{aligned} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)[v]\,dz&=n\sum _{a=0}^{n+1}c_a\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ_a(u)[v]\,dz\\&=n\sum _{a=0}^{n+1} c_a\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}vE_{a}\left( \varvec{\kappa }_{u}\right) u^{n-1}\,dz\\&=n\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}v\Xi \left( \varvec{\kappa }_{u}\right) u^{n-1}\,dz. \end{aligned}$$

$\square$

The conditions for an invariant of the flow follow easily.

Corollary 6.2

There exists a non-zero invariant of the flow (8) of the form $\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}Q(u)\,dz$, where Q is a second order operator satisfying $Q(0)=0$, if $\Xi \left( \varvec{\kappa }\right) =\sum _{a=0}^n c_aE_a\left( \varvec{\kappa }\right)$ for some $c_a\in \mathbb {R}$, $0\le a\le n$.

Proof

Set Q as in (9). Then

$$\begin{aligned} \frac{d}{dt}\left( \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}Q(u)\,dz\right)&= \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}DQ(u)\left[ \frac{\partial u}{\partial t}\right] \,dz\\&=n\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\frac{\partial u}{\partial t}\Xi \left( \varvec{\kappa }_{u}\right) u^{n-1}\,dz\\&=n\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\left( \frac{1}{\int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}\Xi \left( \varvec{\kappa }_{u}\right) \,d\mu _{u}} \int _{{\fancyscript{S}}_{\frac{d}{\pi }}^{1}}F\left( \varvec{\kappa }_{u}\right) \Xi \left( \varvec{\kappa }_{u}\right) \,d\mu _{u} -F\left( \varvec{\kappa }_{u}\right) \right) \\&\quad \times \Xi \left( \varvec{\kappa }_{u}\right) \,d\mu _{u}\\&=0. \end{aligned}$$

$\square$

By using a theorem by Hadwiger, see [8] for the original proof and [14] for a simplified proof, we are able to reformulate this corollary in terms of valuations. A valuation, $\mu$, is a function over the space of convex sets such that $\mu (\varnothing )=0$ and for any convex sets, K and L, such that $K\cup L$ is also convex, it holds that

$$\begin{aligned} \mu \left( K\cup L\right) +\mu \left( K\cap L\right) =\mu \left( K\right) +\mu \left( L\right) . \end{aligned}$$

Hadwiger’s theorem states that the continuous, rigid motion invariant valuations are precisely the linear combinations of mixed volumes. Therefore we obtain the following:

Corollary 6.3

Any continuous, rigid motion invariant valuation is an invariant for a flow of the form (1).