# A cylinder is not stable for area under a volume constraint

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

$$$$\delta_{u\nu}^2 A_U(f)=-\int_U u\Delta_f u+(\kappa_1^2+\kappa_2^2)u^2\,dS\geq 0$$$$ for all normal variations $$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

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

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

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

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?

First, since the second variations can be written as $$$$\delta_{u\nu}^2 A_U(f)=\int_S|\nabla u|^2 -(\kappa_1^2+\kappa_2^2)u^2\,dS, \ \ \ \text{ for all } u\in C^\infty_0(S),$$$$ Using the density of $$C^\infty_0(S) \subset W^{1,2}_0(S)$$ with the $$W^{1,2}$$ norm, a CMC surface is stable if $$$$\int_S|\nabla u|^2 -(\kappa_1^2+\kappa_2^2)u^2\,dS \ge 0, \ \ \ \text{ for all } u\in W^{1,2}_0(S),$$$$
On the cylinder, for any $$\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.
• I used $\int |cos (x)|^2 = \int |\sin (x)|^2$. @Edu Commented Aug 21, 2020 at 13:01