Mean Curvature Flow equation, where does it come from?

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We consider a compact, uniformally convex, $n$-dimensional surface $M=M_0$ without boundary imbedded in $\mathbb{R}^n$.

We want to find a family of maps satisfying the evolution equation

$\frac{\partial}{\partial t}F(\cdot,t)=\Delta_t F(x,t)$.

$F(\cdot,0)=F_0$.

Where we have $\Delta_t F(x,t)=-H(x,t)\cdot \nu(x,t)$.

I'm having trouble with the last step. I understand that the mean curvature is just the trace of the Hessian which is the Laplacian. But is there an explicit way to do this computation?

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To explain the notation, $F$ is a vector valued function $F = (F^1, \cdots F^{n+1})$ and

$$\Delta_t = (\Delta_t F^1 , \cdots , \Delta_t F^{n+1}\ ).$$

Now if $(x^1, \cdots, x^n)$ is a local coordinate on $M$ with induced metric $g_{ij}$ (depending on $t$), then for any function $f: M \to \mathbb R$,

$$\Delta f = g^{ij} \nabla^2_{ij} f =g^{ij} \left(\frac{\partial f}{\partial x^i \partial x^j} - \Gamma_{ij}^k \frac{\partial f}{\partial x^k}\right).$$

Note that

$$\nabla_i \frac{\partial F}{\partial x^j} = \Gamma_{ij}^k \frac{\partial F}{\partial x^k}$$

Thus

\begin{split} \Delta F &= g^{ij} \left(\frac{\partial ^2F}{\partial x^i \partial x^j} - \Gamma_{ij}^k \frac{\partial F}{\partial x^k}\right) \\ &= g^{ij} \left(\frac{\partial ^2F}{\partial x^i \partial x^j} - \nabla_i \frac{\partial F}{\partial x^j} \right) \\ &= g^{ij}\left( \frac{\partial ^2F}{\partial x^i \partial x^j}\right)^\perp \\ &= \vec H = -H \vec v. \end{split}