1
$\begingroup$

Consider a surface $f:\Omega\to\mathbb{R}^3$ with Gauss map $\nu:\Omega\to\mathbb{S}^2$, and for each $p\in\Omega$ let me denote by $\kappa_1,\kappa_2$ the principal curvatures of $f$ at $p$, that is, the eigenvalues of the shape operator at the point.

In the context of constant mean curvature surfaces, the surface $f$ is called stable for area under a volume constraint if the second variation of area satisfies

\begin{equation} \delta_{u\nu}^2 A_U(f)=-\int_U u\Delta_f u+(\kappa_1^2+\kappa_2^2)u^2\,dS\geq 0 \end{equation} for all normal variations $u\in\mathcal{C}_0^\infty(\Omega,\mathbb{R})$ with $\int_U u\,dS=0$, where $U=\text{supp}\;u$. In this question, $\Delta_f$ represents the standard Laplacian.

I would like to check that a cylinder is not stable. Let me consider a cylinder of height $2\pi$ and of radius $1/(2H)$, with mean curvature $H>0$.

For this surface one can compute the principal curvatures: $\kappa_1=0$ and $\kappa_2=\pm\frac{1}{2H}$ (sign depends on the choice of $\nu$). Then, $\kappa_1^2+\kappa_2^2=\frac{1}{4H^2}$.

Next, I will parametrize the cylinder using

\begin{equation} C(\theta,z)=\left(\frac{1}{2H}\cos\theta,\frac{1}{2H}\sin\theta,z\right),\quad (\theta,z)\in[0,2\pi]\times[0,2\pi]. \end{equation}

Then, the area element $dS$ is $dS=\frac{1}{2H}\,d\theta dz$.

Since the cylinder is not stable, I need to find an admissable variation function $u$ such that $\delta_{u\nu}^2 A_U(f)<0$. For this purpose, let me take the variation function

\begin{equation} u(\theta,z)=\sin\left(\frac{1}{2H}\sin\theta\right). \end{equation}

Although one can check that $\int_U u\,dS=0$, and so $u$ is an admissable function, computing $\delta_{u\nu}^2 A_U(f)<0$ is not doable (not even computer-aided).

Since this example should be example enough to check (even by hand!), I would like to finish my calculations using a different function $u$. Can anyone suggest a variation function $u$ that makes it easier?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

First, since the second variations can be written as \begin{equation} \delta_{u\nu}^2 A_U(f)=\int_S|\nabla u|^2 -(\kappa_1^2+\kappa_2^2)u^2\,dS, \ \ \ \text{ for all } u\in C^\infty_0(S), \end{equation} Using the density of $C^\infty_0(S) \subset W^{1,2}_0(S)$ with the $W^{1,2}$ norm, a CMC surface is stable if \begin{equation} \int_S|\nabla u|^2 -(\kappa_1^2+\kappa_2^2)u^2\,dS \ge 0, \ \ \ \text{ for all } u\in W^{1,2}_0(S), \end{equation}

On the cylinder, for any $\ell >0$, consider $$u_\ell (\theta, z) =\begin{cases} \sin \left( \frac{z}{\ell} \right), & \text{ if } |z|\le \pi \ell, \\ 0, & \text{ otherwise.}\end{cases}$$

Then $u_\ell \in W^{1,2}_0$, $\int_S u_\ell = 0$ and \begin{align} \int_S|\nabla u_\ell|^2 -(\kappa_1^2+\kappa_2^2)u_\ell^2\,dS &= 2\pi \int_{-\pi\ell}^{\pi \ell} \frac{1}{\ell^2} \left| \cos\left( \frac{z}{\ell} \right)\right|^2 - \frac{1}{4H^2} \left| \sin \left( \frac{z}{\ell} \right)\right|^2 \, dz \\ &= 2\pi \left(\frac{1}{\ell^2} - \frac{1}{4H^2}\right) \int_{-\pi\ell}^{\pi \ell} \left| \cos\left( \frac{z}{\ell} \right)\right|^2 \, dz. \end{align}

This term can be negative if $\ell > 2H$. Thus the cylinder is unstable.

$\endgroup$
2
  • $\begingroup$ It might be silly but I don't see what you are doing in the last equality. $\endgroup$
    – Edu
    Commented Aug 21, 2020 at 12:56
  • 1
    $\begingroup$ I used $\int |cos (x)|^2 = \int |\sin (x)|^2$. @Edu $\endgroup$ Commented Aug 21, 2020 at 13:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .