Mean curvature flow of graphs in generalized Robertson–Walker spacetimes with perpendicular Neumann boundary condition

Abstract

We prove the longtime existence for the mean curvature flow problem with a perpendicular Neumann boundary condition in a generalized Robertson–Walker (GRW) spacetime that obeys the null convergence condition. In addition, we prove that the metric of such a solution is conformal to the one of the leaf of the GRW in asymptotic time. Furthermore, if the initial hypersurface is mean convex, then the evolving hypersurfaces remain mean convex during the flow.

6 Appendix: A Simons’ type identity

proof of Proposition 4.10

Let ${\overline{R}}$ and R be the Riemann tensors of N and $\Sigma _t$ defined by:

$$\begin{aligned} {\overline{R}}(X, Y)Z=\overline{\nabla }_X\overline{\nabla }_YZ-\overline{\nabla }_Y\overline{\nabla }_XZ-\overline{\nabla }_{[X, Y]}Z, \end{aligned}$$

for all $X, Y, Z\in TN$ and

$$\begin{aligned} R(U, V)W=\nabla _U\nabla _VW-\nabla _V\nabla _UW-\nabla _{[U, V]}W, \end{aligned}$$

for all $U, V, W\in T\Sigma _t$.

Set the coordinate vector fields $\{\partial _i\}_{i=1}^{n+1}$ in N. In what follows we use Codazzi’s equation

$$\begin{aligned} \nabla _i a_{jk} -\nabla _j a_{ik}&= \partial _i \langle {{\bar{\nabla }}}_{\partial _j} \nu , \partial _k\rangle - \partial _j \langle {{\bar{\nabla }}}_{\partial _i}\nu , \partial _k\rangle - \langle A(\nabla _{\partial _i}\partial _j), \partial _k\rangle - \langle A\partial _j, \nabla _{\partial _i}\partial _k\rangle \nonumber \\&\,\,\,\,\,\,\,\,+ \langle A(\nabla _{\partial _j}\partial _i), \partial _k\rangle +\langle A\partial _i, \nabla _{\partial _j}\partial _k\rangle =- L_{ijk},. \end{aligned}$$

(61)

being L the (0, 3)-tensor in the hypersurface $\Sigma _t$ given by

$$\begin{aligned} L_{ijk}=\langle {\bar{R}}(\partial _i,\partial _j)\nu , \partial _k\rangle . \end{aligned}$$

(62)

We also need the (Riemannian) Ricci commutation formula

$$\begin{aligned}&\nabla _i \nabla _k a_{j\ell }-\nabla _k\nabla _i a_{j\ell }= -R^s_{ikj} a_{s\ell }-R^s_{ik\ell } a_{js}. \end{aligned}$$

(63)

Using (61) and (63) one obtains

$$\begin{aligned} \nabla _k \nabla _\ell a_{ij}&= \nabla _k \nabla _i a_{\ell j} -\nabla _k L_{\ell ij}\\&= \nabla _i \nabla _k a_{\ell j} - R^s_{ki\ell } a_{sj}- R^s_{kij} a_{\ell s} -\nabla _k L_{\ell ij}\\&= \nabla _i \nabla _j a_{k\ell } -\nabla _i L_{k\ell j}-\nabla _k L_{\ell ij}- R^s_{ki\ell } a_{sj}-R^s_{kij} a_{\ell s}. \end{aligned}$$

Now we should take in account the Gauss equation

$$\begin{aligned} R^s_{ik\ell } = {\bar{R}}_{ik\ell }^s +a_{k\ell } a_i^s - a_{i\ell } a^s_k. \end{aligned}$$

Therefore

$$\begin{aligned} \nabla _k \nabla _\ell a_{ij}&= \nabla _i \nabla _j a_{k\ell } -\nabla _i L_{k\ell j}-\nabla _k L_{\ell ij}\\&\quad - ({\bar{R}}^s_{ki\ell }+a_{i\ell } a^s_k - a_{k\ell } a^s_i) a_{sj}- ({\bar{R}}^s_{kij} +a_{ij}a^s_k -a_{jk}a^s_i) a_{\ell s}. \end{aligned}$$

Taking traces we have

$$\begin{aligned} \Delta a_{ij}&= \nabla _i \nabla _j H -g^{k\ell }(\nabla _i L_{k\ell j}+\nabla _k L_{\ell ij}) - g^{k\ell }({\bar{R}}^s_{ki\ell }a_{sj}+{\bar{R}}^s_{kij}a_{\ell s})\\&\quad - a^k_{i} a^s_k a_{sj}+H a^s_i a_{sj}- a_{ij}|A|^2 +a^\ell _{j}a^s_i a_{\ell s}. \end{aligned}$$

However

$$\begin{aligned} a^\ell _{j}a^s_i a_{\ell s}=a^s_{j}a^\ell _i a_{s \ell }=a^s_{j}a^k_i a_{s k}= a^k_i a_{sj} a^s_{k}. \end{aligned}$$

This implies the following general formula

$$\begin{aligned} \Delta a_{ij}&= \nabla _i \nabla _j H +H a^s_i a_{sj}- a_{ij}|A|^2\\&-g^{k\ell }(\nabla _i L_{k\ell j}+\nabla _k L_{\ell ij}) - g^{k\ell }({\bar{R}}^s_{ki\ell }a_{sj}+{\bar{R}}^s_{kij}a_{\ell s}). \end{aligned}$$

Hence,

$$\begin{aligned} \frac{1}{2} \Delta |A|^2 -|\nabla A|^2= & {} a^{ij}\Delta a_{ij}\\= & {} a^{ij}\nabla _i \nabla _j H +H {\text {tr}} A^3- |A|^4 -g^{k\ell }(\nabla _i L_{k\ell j}+\nabla _k L_{\ell ij}) a^{ij} \\&-g^{k\ell }({\bar{R}}^s_{ki\ell }a_{sj}a^{ij}+{\bar{R}}^s_{kij}a_{\ell s} a^{ij}). \end{aligned}$$

On the other hand

$$\begin{aligned} \partial _t a_{ij}&= \langle {{\bar{\nabla }}}_{\partial _t} {{\bar{\nabla }}}_{\partial _i} \nu , \partial _j\rangle + \langle {{\bar{\nabla }}}_{\partial _i} \nu , {{\bar{\nabla }}}_{\partial _j} \partial _t\rangle \\&= \langle {{\bar{\nabla }}}_{\partial _i} {{\bar{\nabla }}}_{\partial _t} \nu , \partial _j\rangle + \langle {\bar{R}}(\partial _t, \partial _i)\nu , \partial _j\rangle + \langle {{\bar{\nabla }}}_{\partial _i} \nu , {{\bar{\nabla }}}_{\partial _j} \partial _t\rangle \\&= \langle \nabla _{\partial _i} \nabla H, \partial _j\rangle - H\langle {\bar{R}}(\partial _i,\nu )\nu , \partial _j\rangle + \langle {{\bar{\nabla }}}_{\partial _i} \nu , {{\bar{\nabla }}}_{\partial _j} (H\nu )\rangle \\&= \langle \nabla _{\partial _i} \nabla H, \partial _j\rangle - H\langle {\bar{R}}(\partial _i,\nu )\nu , \partial _j\rangle + H a_{i}^k a_{jk}. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{2}\partial _t |A|^2 = a^{ij}\langle \nabla _{\partial _i} \nabla H, \partial _j\rangle - Ha^{ij}\langle \bar{R}(\partial _i,\nu )\nu , \partial _j\rangle + H a_{i}^k a_{jk} a^{ij}. \end{aligned}$$

We conclude that

$$\begin{aligned} \frac{1}{2} Q |A|^2 +|\nabla A|^2&= |A|^4+g^{k\ell }(\nabla _i L_{k\ell j}+\nabla _k L_{\ell ij}) a^{ij} + g^{k\ell }({\bar{R}}^s_{ki\ell }a_{sj}a^{ij}+{\bar{R}}^s_{kij}a_{\ell s} a^{ij})\\&\quad -H\langle {\overline{R}}(\partial _ i, \nu )\nu , \partial _ j\rangle . \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{2} Q|A|^2 +|\nabla A|^2= & {} |A|^4+T_{ij} a^{ij} + g^{k\ell }({\bar{R}}^s_{ki\ell }a_{sj}a^{ij}+{\bar{R}}^s_{kij}a_{\ell s} a^{ij}) -H\langle {\overline{R}}(\partial _ i, \nu )\nu , \partial _ j\rangle , \end{aligned}$$

where $T_{ij}=g^{k\ell }(\nabla _i L_{k\ell j}+\nabla _k L_{\ell ij}) a^{ij}.$ $\square$