Stable Capillary Minimal Surfaces

We set the following notations. Suppose $(M^{n+1},\partial M)$ is the ambient manifold we’re interested in. We say $\Sigma$ is a capillary minimal surface with contact ange $\theta$ if $\Sigma$ is a minimal surface with boundary $\partial \Sigma$ and it has contact angle $\theta$ with $\partial M$ along $\partial \Sigma$.


First Variation Formula

Suppose $\partial \Sigma$ bounds a region $W$ in $\partial M$ with outer unit normal $\overline{\nu}$, then $\Sigma$ is a capillary minimal surface if and only if it is a critical point of the following functional, $$ \mathcal{A}(\Sigma)=\left|\Sigma\right|-\cos \theta \left|W\right|. $$ This can be seen from the following argument.

Let $X$ be the variational vector field. Then, $$ \left.\frac{d}{dt}\right|_{t=0}\mathcal{A}(\Sigma_t)=\int_{\Sigma}\mathrm{div}_\Sigma XdA-\cos\theta\int_{ M}\mathrm{div}_MXdA. $$ Now if we suppose $X|_{\partial M} \in \mathfrak{X}(\partial M)$, then $$ \int_{ M} \mathrm{div}_MXdA= \int_{ \partial \Sigma} X\cdot \overline{\nu} ds. $$

So $$ \left. \frac{d}{dt}\right|_{t=0}\mathcal{A}(\Sigma_t)={} \int_{ \Sigma} X \cdot \nu H dA+\int_{ \partial \Sigma} (X \cdot \eta-\cos \theta X\cdot \overline{\nu})ds. $$ Suppose $\eta$ makes angle $\varphi$ with $T(\partial M)$, (i.e., $\eta=\cos \varphi \overline{\nu}+\sin \varphi \overline{\eta}$), then $X\cdot \eta = \cos \theta X \cdot \overline{\nu}$. Hence, $\Sigma$ is a critical point if and only if $H\equiv 0$ and $\varphi=\theta$ along $\partial \Sigma$.

Second variation formula

We will show that if $\Sigma$ is a capillary minimal surface with contact angle $\theta$, then $$ \left. \frac{d^2}{dt^2}\right|_{t=0}\mathcal{A}(\Sigma_t)=\int_{ \Sigma} \left|\nabla \phi\right|^2-\left|A\right|^2\phi^2+ \int_{ \partial \Sigma} \left( \frac{A _{\partial M}(\overline{\nu},\overline{\nu})}{\sin \theta}+ \cot \theta A(\eta,\eta) \right)\phi^2. $$

We choose the special vector fields as $X = \left< X, \nu \right>\nu +\left< X, \eta \right> \eta= \phi \nu + \cot \theta \phi \eta=\frac{1}{\sin \theta} \phi \overline{\nu}$.

We still write $\phi= X\cdot \nu$. We take derivative with respect to $t$ of the first variation formula and we have $$ \left. \frac{d^2}{dt^2}\right|_{t=0}\mathcal{A}(\Sigma_t)={} \left. \frac{d}{dt}\right|_{t=0}\int_{ \Sigma} X \cdot \nu H dA+\left. \frac{d}{dt}\right|_{t=0}\int_{ \partial \Sigma} (X \cdot \eta-\cos \theta X\cdot \overline{\nu})ds. $$ For the first term, we have \begin{align*} \left. \frac{d}{dt}\right|_{t=0}\int_{ \Sigma}X\cdot \nu_t H_t dA_t= {} & \int_{ \Sigma} X\cdot \nu H' dA + \int_{ \Sigma} H \left. \frac{d}{dt}\right|_{t=0}(X\cdot \nu_t)dA_t=\int_{ \Sigma} X\cdot \nu H' dA \end{align*} since $H=0$. We can use $H' = \Delta_{\Sigma}\phi+\left|A\right|^2\phi$ to get the interior term for the second variation formula. Here, we use the notation $(\cdot)’=\frac{d}{dt}|_{t=0}(\cdot)$.

For the boundary term, we need to compute \begin{align*} \left. \frac{d}{dt}\right|_{t=0} \int_{ \partial \Sigma} (X\cdot \eta_t - \cos \theta X \cdot \overline{\nu}_t)ds_t ={} & \int_{ \partial \Sigma} \left< X, \eta-\cos \theta \overline{\nu} \right>' ds. \end{align*}

So we need to compute $\eta', \overline{\nu}'$.

First, let’s compute $\nu'$. Let $\left\{ e_i \right\}$ be the orthonormal basis of $T_p \Sigma$. Then \begin{align*} \left< \nu', e_i \right> ={} & -\left< \nu, e_i' \right> = -\left< \nu ,D _{e_i}X \right> =-\left< \nu, D _{e_i}(\phi \nu + \left< X, e_j \right> e_j) \right> \\ ={} & -e_i(\phi)-\left< X, e_j \right> h_{ij}. \end{align*}

Hence $\nu' = - \nabla_{\Sigma}\phi-A(X^\top ,e_i )e_i$.

Now $ \left< \eta', \nu \right> =- \left< \eta,\nu' \right> =\frac{\partial }{\partial \eta}\psi + A(X^T, \eta)$.

We also have \begin{align*}\left< \eta', \tau_i \right> ={}& -\left< \eta, \tau_i' \right> = -\left< \eta, D _{\tau_i}X \right> = \phi A(\eta, \tau_i)+\cot \theta \tau_i(\phi) \end{align*}

So $$ \eta' = \left( \frac{\partial \phi}{\partial \eta}+A(X^T, \eta) \right)\nu + \phi A(\eta,\tau_i)\tau_i+\cot \theta \nabla_{\partial \Sigma} \phi. $$ Hence \begin{align*} \left< X, \eta'-\cos \theta \overline{\nu}' \right> ={} & \frac{\phi}{\sin \theta}\left< \overline{\nu}, \eta'-\cos \theta \overline{\nu}' \right> =\frac{\phi}{\sin \theta} \left< \overline{\nu}, \eta' \right> \\ ={} & \phi \left( \frac{\partial \phi}{\partial \eta} + \cot \theta\phi A(\eta,\eta)\right) \end{align*}

Moreover, $$ \left< X', \eta - \cos \theta \overline{\nu} \right> = \left< \nabla_{X}X, \sin \theta \overline{\eta} \right> = \frac{\phi^2}{\sin\theta} A _{\partial M}(\overline{\nu},\overline{\nu}). $$

Generalized Bernstein Theorem

We know the curvature estimate for stable capillary surfaces will be related to the generalized Bernstein Theorem in half space $H = \left\{ x_1\ge 0 \right\}$.

Here, we will prove the generalized Bernstein Theorem for capillary surfaces in $H^3$.

If $(\Sigma,\partial \Sigma) \hookrightarrow (H^3, \partial H)$ is a stable capillary minimal surface with quadratic area growth condition. Then $\Sigma$ is flat.

The stability inequality becomes $$ \int_{ \Sigma} \left|A\right|^2 \phi ^2dA\le \int_{ \Sigma} \left|\nabla \phi\right|^2dA+ \int_{ \partial \Sigma} \cot \theta A(\eta,\eta)\phi^2ds. $$ Dealing with boundary term

Let’s choose $w = 1+\left< \nu, \overline{\eta} \right>\cos \theta $. Replace $\phi$ by $\phi w$ in stability inequality. We note \begin{align*} \int_{ \Sigma} \left|\nabla (\phi w)\right|^2={} & \int_{ \Sigma} w^2 \left|\nabla \phi\right|^2+ \phi^2 \left|\nabla w\right|^2+\frac{1}{2}\int_{ \Sigma} \left< \nabla \phi^2, \nabla w^2 \right> \\ ={} & \int_{ \Sigma} w^2 \left|\nabla \phi\right|^2-\phi^2 w \Delta w + \int_{ \partial \Sigma} \phi^2 w \frac{\partial w}{\partial \eta}. \end{align*}

Note that we have $$ \Delta w = \cos \theta \Delta \left< \nu ,\overline{\eta} \right> = -\cos \theta \left< h_{ij}e_j, \overline{\eta}\right> _i= -\cos \theta\left|A\right|^2 \left< \nu, \overline{\eta} \right> $$ and $$ \frac{\partial w}{\partial \eta}=\cos \theta \left< \frac{\partial \nu}{\partial \eta}, \overline{\eta} \right> =-\cos \theta \sin \theta A(\eta,\eta)=-\cot \theta A(\eta,\eta)w. $$ Here, we’ve used $w=1-\cos ^2\theta=\sin ^2\theta$ along $\Gamma$.

So in summary, we will get $$ \int_{ \Sigma} \left|A\right|^2 w\phi^2 dA\le \int_{ \Sigma} \left|\nabla \phi\right|^2w^2dA. $$

So we will get the following inequality $$ \int_{ \Sigma} \left|A\right|^2\phi^2dA\le C_\theta \int_{ \Sigma} \left|\nabla \phi\right|^2dA. $$ Choose special test functions (logarithmic cutoff function)

We will control the curvature $\left|A\right|$ in the half ball $B_1 \cap H$.

Choose $\phi$ as follows $$ \phi= \begin{cases} 1, & \left|x\right|\le 1, \\ 1- \frac{\log \left|x\right|}{n}, & 1\le \left|x\right|\le e^n,\\ 0, & \left|x\right|\ge e^n. \end{cases} $$

Then, we can find \begin{align*} \int_{ \Sigma \cap B_1} \left|A\right|^2dA\le {} & C \int_{ \Sigma \cap B _{e^n}\backslash B_1} \frac{1}{\left|x\right|^2 n^2} dA\le C \sum_{i =1}^{n} \int_{ \Sigma \cap B _{e^i}\backslash B_{e^{i-1}}}\frac{1}{e^{2i-2} n^2}dA \\ \le {} & C \sum_{i =1}^{n} \frac{C e^2}{n^2}= \frac{C}{n}. \end{align*} Now we may choose $n\rightarrow 0$ to get $\left|A\right|\equiv 0$ in $B_1$. The unique continuation of minimal surfaces tells us $\Sigma$ is flat.

Gaoming Wang
Gaoming Wang

My research interests include Geometric Analysis and Partial Differential Equations.