## Existence and Uniqueness of Solutions to the Constant Mean Curvature Equation with Nonzero Neumann Boundary Data in Product Manifold $M^{n}\times{\Bbb R}$

Gao Ya,, Mao Jing,, Song Chunlan

 基金资助: 国家自然科学基金.  11801496国家自然科学基金.  11926352霍英东教育基金会青年教师基金应用数学湖北省重点实验室基金

Received: 2020-01-31

 Fund supported: the NSFC.  11801496the NSFC.  11926352the Fok Ying-Tung Education Foundation (China)the Hubei Key Laboratory of Applied Mathematics (Hubei University)

Abstract

In this paper, we can prove the existence and uniqueness of solutions to the constant mean curvature (CMC for short) equation with nonzero Neumann boundary data in product manifold $M^{n}\times{\Bbb R}$, where $M^{n}$ is an $n$-dimensional $(n\geq2)$ complete Riemannian manifold with nonnegative Ricci curvature, and ${\Bbb R}$ is the Euclidean 1-space. Equivalently, this conclusion gives the existence of CMC graphic hypersurfaces defined over a compact strictly convex domain $\Omega\subset M^{n}$ and with nonzero Neumann boundary data.

Keywords： Constant mean curvature ; Neumann boundary condition ; Convexity ; Ricci curvature ; Product manifold

Gao Ya, Mao Jing, Song Chunlan. Existence and Uniqueness of Solutions to the Constant Mean Curvature Equation with Nonzero Neumann Boundary Data in Product Manifold $M^{n}\times{\Bbb R}$. Acta Mathematica Scientia[J], 2020, 40(6): 1525-1536 doi:

## 1 研究背景及意义

$$$\label{eq:a1} H=\sum\limits_{i=1}^{n}h^i_i=-\frac{\sum\limits_{i, k=1}^{n}g^{ik}D_{i}D_{k}u}{\sqrt{1+|Du|^2}}=-\frac{\sum\limits_{i, k=1}^{n}\left(\sigma^{ik}-\frac{D^{i}uD^{k}u}{1+|Du|^{2}}\right)D_{i}D_{k}u}{\sqrt{1+|Du|^2}}.$$$

我们将利用与文献[3, 定理2.2]的证明类似的思路得到方程($\natural$)的梯度估计.事实上, 这一梯度估计的技巧源自于文献[5].取$a^{ij}:=(1+|Du|^{2})\sigma^{ij}-D^{i}uD^{j}u$, $f=\varepsilon u$, $v=\sqrt{1+|Du|^{2}}$, $\varepsilon>0$.则在方程($\natural$)中的第一个式子可简写为

$\begin{array}{l}0 \ge {\Phi _n} = \frac{{|D\omega |_n^2}}{{|D\omega {|^2}}} - \zeta \\ = \sum\limits_{k = 1}^{n - 1} {\frac{{2{\omega ^k}{D_{{\tau _n}}}{D_{{\tau _k}}}\omega }}{{|D\omega {|^2}}}} - \zeta \\ = \sum\limits_{k = 1}^{n - 1} {\frac{{2{\omega ^k}[{\tau _k}({\tau _n}(\omega )) - ({D_{{\tau _k}}}{\tau _n})\omega ]}}{{|D\omega {|^2}}}} - \zeta \\ = - \sum\limits_{k = 1}^{n - 1} {\frac{{2{\omega ^k}({D_{{\tau _k}}}{\tau _n})(\omega )}}{{|D\omega {|^2}}}} - \zeta \\ = - \sum\limits_{k = 1}^{n - 1} {\frac{{2{\omega ^k}{\omega _j}{{\langle {D_{{\tau _k}}}{\tau _n}, {\tau _j}\rangle }_\sigma }}}{{|D\omega {|^2}}}} - \zeta \\ = \sum\limits_{k = 1}^{n - 1} {\frac{{2{\omega ^k}{\omega _j}{{\langle {D_{{\tau _k}}}{\tau _j}, {\tau _n}\rangle }_\sigma }}}{{|D\omega {|^2}}}} - \zeta \\ = \sum\limits_{k, j = 1}^{n - 1} {\frac{{2{\omega ^k}{\omega _j}h_{kj}^{\partial \Omega }}}{{|D\omega {|^2}}}} - \zeta \\\geq 2\kappa_{1}-\zeta. \end{array}$

$v=\sqrt{1+|Du|^{2}}=\sqrt{1+u^{2}_{1}}$, 且有

$$$\label{e-1} \Phi_{i}=\frac{|D\omega|^{2}_{i}}{|D\omega|^{2}}+\zeta\beta_{i}=0,$$$

$\begin{eqnarray}\label{e-2}0&\geq&\sum\limits_{i, j=1}^{n}a^{ij}\Phi_{ij}\\&=&\sum\limits_{i, j=1}^{n}\frac{a^{ij}|D\omega|^{2}_{ij}}{|D\omega|^{2}}-\zeta^{2}\sum\limits_{i, j=1}^{n}a^{ij}\beta_{i}\beta_{j}+\zeta \sum\limits_{i, j=1}^{n}a^{ij}\beta_{ij}\\&\triangleq &Ⅰ+Ⅱ+Ⅲ.\end{eqnarray}$

$i=1, 2, \cdots, n$时, 通过(2.2)式得到

$$$\label{e-3}\sum\limits_{k=1}^{n}\omega^{k}u_{ki}=\sum\limits_{k=1}^{n}\omega^{k}\omega_{ki}+\sum\limits_{k=1}^{n}\omega^{k}G_{ki}=-\frac{\zeta\beta_{i}v^{2}}{2}+O(v),$$$

$i=2, 3, \cdots, n$时, 有

$$$\label{e-4}u_{1i}=O(1)-\frac{\zeta\beta_{i}v}{2}-\frac{\omega^{i}}{v}u_{ii}.$$$

$\begin{eqnarray} \label{eq:a2}u_{11}&=&O(1)-\frac{\zeta\beta_{1}v}{2}-\sum\limits_{k=2}^{n}\frac{\omega^{k}}{v}\left(O(1)-\frac{\zeta\beta_{k}v}{2}-\frac{\omega^{k}}{v}u_{kk}\right)\nonumber\\&=&O(1)-\frac{\zeta\beta_{1}v}{2}+\sum\limits_{k=2}^{n}\left(\frac{\omega^{k}}{v}\right)^{2}u_{kk}, \end{eqnarray}$

$$$\label{e-5}u_{11}+\sum\limits_{i=2}^{n}\sigma^{ii}u_{ii}=fv+\frac{u_{1}^{2}}{v^{2}}u_{11}=fv+O(1)-\frac{u_{1}^{2}\zeta\beta_{1}}{2v}+\sum\limits_{k=2}^{n}\frac{(u_{1}\omega^{k})^{2}}{v^{4}}u_{kk},$$$

$$$\label{e-6}fv=\frac{u_{11}}{v^{2}}+\sum\limits_{i=2}^{n}\sigma^{ii}u_{ii}=O(\frac{1}{v^{2}})-\frac{\zeta\beta_{1}}{2v}+\sum\limits_{k=2}^{n}\left[\sigma^{kk}+\frac{(\omega^{k})^{2}}{v^{4}}\right]u_{kk},$$$

$$$\label{e-7}II=-\zeta^{2}\sum\limits_{i, j=1}^{n}a^{ij}\beta_{i}\beta_{j}=-\zeta^{2}\left(\beta^{2}_{1}+v^{2}\sum\limits_{i=2}^{n}\sigma^{ii}\beta^{2}_{i}\right),$$$

$$$\label{e-8}III=\zeta\sum\limits_{i, j=1}^{n}a^{ij}\beta_{ij}\geq\zetak_{0}\left(1+(n-1)\eta+(n-1)\eta u^{2}_{1}\right),$$$

$\begin{eqnarray}\label{e-9}\sum\limits_{i, j=1}^{n}a^{ij}|D\omega|^{2}_{ij}&=&2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}u_{kij}-2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}G_{kij}+2\sum\limits_{i, j, k=1}^{n}a^{ij}\sigma^{kk}u_{ki}u_{kj}\\& &4\sum\limits_{i, j, k=1}^{n}a^{ij}\sigma^{kk}u_{ki}G_{kj}+2\sum\limits_{i, j, k=1}^{n}a^{ij}\sigma^{kk}G_{ki}G_{kj}\\&:= &I_{1}+I_{2}+I_{3}+I_{4}+I_{5}.\end{eqnarray}$

$\begin{eqnarray}\label{e-10}I_{1}&=&2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}u_{kij}\\&=&2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}( u_{ijk}+R^{l} _{ikj}u_{l})\\&=&2\sum\limits_{i, j, k=1}^{n}\omega^{k}[(fv^{3})_{k}-(a^{ij})_{k}u_{ij}]+2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}R^{l}_{ikj}u_{l}\\&=&2\sum\limits_{k=1}^{n}(\varepsilon u_{k}\omega^{k}v^{3}+3fv\sum\limits_{l=1}^{n}u^{l}u_{lk}\omega^{k})-4\sum\limits_{i, l, k=1}^{n}\sigma^{ii}u_{ii}u^{l}u_{lk}\omega^{k}\\& &+4\sum\limits_{i, j, k, l=1}^{n}\sigma^{il}\sigma^{1j}u_{1}u_{lk}u_{ij}\omega^{k}+2\sum\limits_{i, j, k, l=1}^{n}a^{ij}\omega^{k}R^{l}_{ikj}u_{l}\\&\geq& \left[6fv-4(u_{11}+\sum\limits_{i=2}^{n}\sigma^{ii}u_{ii})\right]\sum\limits_{k, l=1}^{n}u^{l}u_{lk}\omega^{k}\\ & & +4u_{1}\sum\limits_{i, k, l=1}^{n}\sigma^{il}u_{lk}u_{1i}\omega^{k}+2\sum\limits_{i, j, k=1}^{n}a^{ij}\omega^{k}R^{1}_{ikj}u_{1}\\&\triangleq &I_{11}+I_{12}+I_{13}, \end{eqnarray}$

$$$\label{e-19} \zeta^{2}\beta^{2}_{1}+\sum\limits_{i=2}^{n}u_{1}\sigma^{ii}\zeta^{2}\beta^{2}_{i}v +\frac{3}{8}(1+v^{2})\sum\limits_{i=2}^{n}(\sigma^{ii})^{2}\zeta^{2}\beta^{2}_{i}\geq\zeta^{2}v^{2}\sum\limits_{i=2}^{n}\sigma^{ii}\beta^{2}_{i}$$$

$$$\label{e-20} -\sum\limits_{i=2}^{n}\frac{\left[O(v^{2})-\sigma^{ii}u_{1}\zeta \beta_{1}v^{2}\right]^{2}}{6v^{4}(\sigma^{ii})^{2} +O(v^{2})}\geq-\frac{n-1}{5}\zeta^{2}v^{2}\beta^{2}_{1}+O(1)$$$

$\begin{eqnarray} \label{e-21} I&=&\sum\limits_{i, j=1}^{n}\frac{a^{ij}|D\omega|^{2}_{ij} }{|D\omega|^{2}}=\frac{\sum\limits_{i=1}^{5}I_{i}}{|D\omega|^{2}}\geq O(v)+\alpha^{2}v^{2}\sum\limits_{i=2}^{n}\sigma^{ii}\beta^{2}_{i}-\frac{n-1}{5}\alpha^{2}v^{2}\beta^{2}_{1}. \end{eqnarray}$

## 3 解的存在唯一性

$\varepsilon\rightarrow0$ (必要时提取子序列)可以推断:存在一个常数$\lambda$和一个光滑的函数$u^{\infty}(x)$使得

$\left\{ {\begin{array}{*{20}{l}}{\sum\limits_{k,l = 1}^n {{D_k}} \left[ {{\tilde g_{kl}}(x){D_l}({u^\infty } - {u^\chi })} \right] = \lambda - \chi ,}&{在\Omega {\rm{中}},}\\{{D_{\vec \nu }}({u^\infty } - {u^\chi }) = 0,}&{在\partial \Omega {\rm{上}}}\end{array}} \right.$

$p$在边界$\partial\Omega$上, 故$\Gamma^{n}_{\alpha\beta}=h^{\partial\Omega}_{\alpha\beta}$, $\Gamma^{\beta}_{\alpha n}=-h^{\partial\Omega}_{\alpha\gamma}\sigma^{\gamma\beta}$, $\Gamma^{n}_{\alpha n}=0$, 其中$h^{\partial\Omega}_{ij}$为边界第二基本型.

$p$点处, 有

$g(s)$是光滑凸函数, 且在$s\leq0$时, 有

$\partial\Omega$上, 我们有

$|D\beta|=1$.综上, $\beta=g(m)$为梯度估计中我们所需的辅助函数.

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