Higher order variations of constant mean curvature surfaces

Abstract

We study the third and fourth variation of area for a compact domain in a constant mean curvature surface when there is a Killing field on $\mathbf{R}^3$ whose normal component vanishes on the boundary. Examples are given to show that, in the presence of a zero eigenvalue, the non negativity of the second variation has no implications for the local area minimization of the surface.

A Proof of Theorem 2.1

(I) and (IV) are the same as (I) and (IV) in Theorem 1.3 in [4].

Assume that ${\lambda }_1 < 0$. Let ${\psi }_i$ be the eigenfunction corresponding to the i-th eigenvalue ${\lambda }_i$ of the problem

$$\begin{aligned} L\psi =-\lambda \psi , \quad \psi \in H^1_0-\{0\}, \end{aligned}$$

(74)

where

$$\begin{aligned} L=\Delta +||d\nu ||^2. \end{aligned}$$

We choose $\{{\psi }_i\}$ so that they form an orthonormal basis for $L^2(\Sigma )$. Since ${\psi }_1$ does not change sign,

$$\begin{aligned} \int _\Sigma {\psi }_1 \;d\Sigma \ne 0. \end{aligned}$$

(75)

For a function $u \in C^{2+\alpha }_0(\Sigma )$, set

$$\begin{aligned} v = - \frac{\int _\Sigma u \;d\Sigma }{\int _\Sigma {\psi }_1 \;d\Sigma } {\psi }_1 + u. \end{aligned}$$

(76)

Then

$$\begin{aligned} \int _\Sigma v \;d\Sigma = 0. \end{aligned}$$

(77)

By setting

$$\begin{aligned} a = - \frac{\int _\Sigma u \;d\Sigma }{\int _\Sigma {\psi }_1 \;d\Sigma }, \end{aligned}$$

v is represented as

$$\begin{aligned} v = a{\psi }_1 + u. \end{aligned}$$

When (74) has no zero eigenvalue, by the Fredholm alternative, there exists a unique function $u \in C^{2+\alpha }_0(\Sigma )$ satisfying $Lu = 1$. In this case, by using the Green’s formula and (77), we see that

$$\begin{aligned} I[v]= & {} - \int _\Sigma vLv \;d\Sigma \nonumber \\= & {} - \int _\Sigma (a^2{\psi }_1L{\psi }_1 + a{\psi }_1Lu + auL{\psi }_1 + uLu)\;d\Sigma \nonumber \\= & {} a^2{\lambda }_1\int _\Sigma {\psi }_1^2\;d\Sigma - 2a\int _\Sigma {\psi }_1 Lu \;d\Sigma - \int _\Sigma u \;d\Sigma \nonumber \\= & {} a^2 {\lambda }_1 + \int _\Sigma u \;d\Sigma . \end{aligned}$$

(78)

Since ${\lambda }_1 < 0$, $I[v] < 0$ holds if $\int _\Sigma u \;d\Sigma < 0$ holds. In this case, in view of Lemma 2.1, X is unstable, which proves (II-3).

Next, when $u = {\psi }_2$, from (78) and the orthonormality of $\{{\psi }_i\}$, we see

$$\begin{aligned} I[v]= & {} a^2{\lambda }_1\int _\Sigma {\psi }_1^2\;d\Sigma - 2a \int _\Sigma {\psi }_1 L{\psi }_2 \;d\Sigma - \int _\Sigma {\psi }_2 L{\psi }_2 \;d\Sigma \\= & {} a^2{\lambda }_1\int _\Sigma {\psi }_1^2\;d\Sigma + 2a{\lambda }_2 \int _\Sigma {\psi }_1{\psi }_2 \;d\Sigma + {\lambda }_2 \int _\Sigma {\psi }_2^2 \;d\Sigma \\= & {} {\lambda }_1 \biggl (\int _\Sigma {\psi }_2 \;d\Sigma \biggr )^2 \biggl (\int _\Sigma {\psi }_1 \;d\Sigma \biggr )^{-2} + {\lambda }_2. \end{aligned}$$

Therefore, if ${\lambda }_2 \le 0$ and $\int _M{\psi }_2 \ne 0$, then $I[v] < 0$ and so X is unstable, which proves (III-A).

Next we prove (II-1). Set

$$\begin{aligned} E_1 = \{a{\psi }_1 |a \in \mathbf{R}\}, \qquad E_1^{\perp } = \biggl \{u \in C^{2+\alpha }_0(\Sigma ) \bigg | \int _\Sigma {\psi }_1u \;d\Sigma = 0\biggr \}. \end{aligned}$$

Again by the Fredholm alternative, there exists a unique function $u \in C^{2+\alpha }_0(\Sigma )$ satisfying $Lu = 1$. If

$$\begin{aligned} \int _\Sigma u \;d\Sigma \ge 0, \end{aligned}$$

then

$$\begin{aligned} I[u] = - \int _\Sigma u Lu \;d\Sigma = - \int _\Sigma u \;d\Sigma \le 0. \end{aligned}$$

(79)

Note

$$\begin{aligned} \lambda _1 = I[\psi _1]= & {} \mathrm{min}\biggl \{I[u] \; \bigg | \; u \in H^1_0(\Sigma ) \ \mathrm{and} \ \int _\Sigma u^2 \;d\Sigma = 1\biggr \}, \nonumber \\ \lambda _i = I[\psi _i]= & {} \mathrm{min}\biggl \{I[u] \; \bigg | \; u \in H^1_0(\Sigma ), \ \int _\Sigma u^2 \;d\Sigma = 1 \nonumber \\&\hbox { and } \displaystyle {\int _\Sigma u\psi _j \;d\Sigma = 0} \hbox { for } j \in \{1, \ldots , i - 1\}\biggr \}, \quad i = 2, 3, \ldots . \end{aligned}$$

(80)

(79), (80) and ${\lambda }_2 > 0$ imply $u \notin E_1^{\perp }$. Therefore, any $v \in F_0$ is represented as follows:

$$\begin{aligned} v = w + bu, \quad b \in \mathbf{R}, \ \ w \in E_1^{\perp }. \end{aligned}$$

Then

$$\begin{aligned} I[v]= & {} - \int _\Sigma (b^2uLu + buLw + bwLu + wLw) \;d\Sigma \\= & {} - b^2\int _\Sigma u \;d\Sigma - 2b \int _\Sigma w \;d\Sigma + I[w]\\= & {} b^2\int _\Sigma u \;d\Sigma - 2b \int _\Sigma (w + bu) \;d\Sigma + I[w]\\= & {} - b^2 I[u] + I[w]. \end{aligned}$$

Note $I[w]>0$ if $w\ne 0$. Hence, $I[v]\ge 0$ holds, which implies that X is stable. If $\int _\Sigma u \;d\Sigma >0$, then X is strictly stable, which proves (II-1). If $\int _\Sigma u \;d\Sigma =0$, then $I[v]=0$ if and only if $v=bu$ ($b \in \mathbf{R}$), which proves (II-2).

Lastly, we prove (III-B). Again by the Fredholm alternative, there exists a unique function $u \in E^{\perp } \cap C^{2+\alpha }_0(\Sigma )$ that satisfies $Lu = 1$.

When $\int _\Sigma u \;d\Sigma < 0$, we can prove that X is unstable by the same way as in the proof of (II-2). (III-B1) is proved by the same way as the proof of (II-1). When $\int _\Sigma u \;d\Sigma = 0$,

$$\begin{aligned} I[u] = - \int _\Sigma uLu \;d\Sigma = - \int _\Sigma u \;d\Sigma = 0 = {\lambda }_2. \end{aligned}$$

Assume that $u \in E_1^{\perp }$. By (80), u is an eigenfunction corresponding to ${\lambda }_2 = 0$, that is, $Lu = 0$, which is a contradiction. Therefore, $u \notin E_1^{\perp }$, and hence, we can prove the stability of X by the same way as in the proof of (II-2).