2. Proof of the main theorem

In this section, we prove Theorem 1.2. Let $S^2_+$ be the two-dimensional hemisphere given by

\begin{equation*} S^2_+=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2=1, y\geq 0\}, \end{equation*}

which is a two-dimensional manifold with boundary

\begin{equation*} \partial S^2_+=\{(x,0,z)\in\mathbb{R}^3: x^2+z^2=1\}. \end{equation*}

Then $S^2_+$ has a local parametrization given by

(2.1)\begin{equation} \overrightarrow{x}(r,\theta)=(\sin r\cos \theta,\sin r\sin\theta, \cos r), \end{equation}

where $0\leq r\leq \pi$ and $0\leq \theta\leq \pi$. Note that the boundary $\partial S^2_+$ corresponds to the points when θ = 0 or $\theta=\pi$. The induced metric on $S^2_+$ by $\mathbb{R}^3$ is given by

\begin{equation*} dr^2+\sin^2 rd\theta^2. \end{equation*}

Then the product space $S^2_+\times S^1$ is a three-dimensional manifold with boundary $\partial (S^2_+\times S^1)=\partial S^2_+\times S^1$. Moreover,

(2.2)\begin{equation} g=dr^2+\sin^2 rd\theta^2+dt^2 \end{equation}

is a Riemannian metric on $S^2_+\times S^1$. Equipped with the metric g in (2.2), the boundary $\partial (S^2_+\times S^1)$ is a totally geodesic and hence is not strictly convex. It follows from (2.2) that

(2.3)\begin{equation} \begin{split} \nabla_{\frac{\partial}{\partial\theta}}\frac{\partial}{\partial r}=\cot r\frac{\partial}{\partial\theta}, ~~\nabla_{\frac{\partial}{\partial\theta}}\frac{\partial}{\partial t}=0,\\ \nabla_{\frac{\partial}{\partial r}}\frac{\partial}{\partial r}=0,~~ \nabla_{\frac{\partial}{\partial r}}\frac{\partial}{\partial t}=\nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial r}=0,~~ \nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial t}=0. \end{split} \end{equation}

Suppose r(t) is a smooth function on S ¹ such that

(2.4)\begin{equation} r(t+2\pi)=r(t). \end{equation}

Define $f:[0,\pi]\times S^1\to S^2_+\times S^1$ given by

(2.5)\begin{equation} f(\theta,t)=(\overrightarrow{x}(r(t),\theta),t), \end{equation}

where $\overrightarrow{x}$ is defined as in (2.1). Then

\begin{equation*} \vec{e}_1:=\frac{1}{\sin r}\frac{\partial}{\partial\theta} ~\text{and}~~ \vec{e}_2:=\frac{1}{\sqrt{1+r'^2}}\left(r'\frac{\partial}{\partial r}+\frac{\partial}{\partial t}\right) \end{equation*}

are unit vectors with respect to g, and $\{\vec{e}_1,\vec{e}_2\}$ spans the tangent space of $[0,\pi]\times S^1$. Hence the unit normal $\vec{n}$ is given by

\begin{equation*} \vec{n}=\frac{1}{\sqrt{1+r'^2}}\left(\frac{\partial}{\partial r}-r'\frac{\partial}{\partial t}\right). \end{equation*}

By (2.3), we see that

(2.6)\begin{equation} (\nabla_{\vec{e}_1}\vec{n})^T=\nabla_{\vec{e}_1}\vec{n}=\left(\frac{\cot r}{\sqrt{1+r'^2}}\right) \vec{e}_1~~\mbox{and }~~ (\nabla_{\vec{e}_2}\vec{n})^T=-\frac{r''}{(1+r'^2)^{\frac{3}{2}}} \vec{e}_2, \end{equation}

where $(\nabla_{\vec{e}_2}\vec{n})^T$ denotes the tangential component of $\nabla_{\vec{e}_2}\vec{n}$. It follows from (2.6) that the principal curvatures of $[0,\pi]\times S^1$ are $\frac{\cot r}{\sqrt{1+r'^2}}$ and $-\frac{r''}{(1+r'^2)^{\frac{3}{2}}}$, which implies that the mean curvature is

\begin{equation*} H=\frac{1}{2\sqrt{1+r'^2}}\left(\cot r-\frac{r''}{1+r'^2}\right). \end{equation*}

Hence, $[0,\pi]\times S^1$ is minimal if

(2.7)\begin{equation} \frac{r''}{1+r'^2}=\cot r. \end{equation}

On the other hand, the outward conormal vector $\vec{\nu}$ is the unique unit vector in $T_p([0,\pi]\times S^1)$ which is orthogonal to $T_p\big(\partial ([0,\pi]\times S^1)\big)$. Thus, we have

(2.8)\begin{equation} \vec{\nu}=\pm\frac{1}{\sin r}\frac{\partial}{\partial\theta}. \end{equation}

Note that the tangent plane of $\partial(S^2_+\times S^1)$ is spanned by $\displaystyle\left\{\frac{\partial}{\partial r}, \frac{\partial}{\partial t}\right\}$. Hence, it follows from (2.8) that $\vec{\nu}$ is orthogonal to $\partial(S^2_+\times S^1)$, i.e.,

(2.9)\begin{equation} \vec{\nu}\perp \partial(S^2_+\times S^1). \end{equation}

In view of (2.9), we conclude that if r(t) is a solution to (2.4) and (2.7), then the embedding f of $[0,\pi]\times S^1$ defined in (2.5) is a free boundary minimal surface in $S^2_+\times S^1$. As pointed out in [Reference Hsieh and Wang3], the general solutions to (2.4) and (2.7) are elliptic functions for a sequence of initial values $r(t=0)=c_1 \gt c_2 \gt c_3 \gt \cdots\to 0$. The corresponding sequence of embedded free boundary minimal surfaces certainly cannot have a convergent subsequence. This proves our main theorem.