# Limit of time derivative in linear parabolic equation

$M^n$ , $n\ge 1$ , is a compact n-dimensional manifold without boundary. $F_0: M^n\rightarrow \mathbb R^{n+1}$ is smoothly immersion with nonnegative mean curvature. Deform $F(x,t)$ by $$\partial_t F(x,t) = \Delta_g F+ F \\ F(\cdot , 0)=F_0~~~~~~~~$$
$\Delta_g$ is Laplace-Beltrami operator. If the solution exist for all $t\ge0$. Then, how to show $\partial_t F \rightarrow 0$ as $t\rightarrow \infty$ ?

This question is origin from Huisken's Asymptotic behavior for singularities of the mean curvature flow. In fact, Huisken want to prove the limit of $F$ is a sphere. First , he proves the limit manifold $F_\infty(M^n)$ satisfies $$H=\langle F, \nu\rangle$$ $H$ is the mean curvature , $\nu$ is normal vector, $\langle \cdot,\cdot \rangle$ is Euclid inner. For proving this, he needs a monotonicity formula. In my view , the monotonicity formula is not easy to think. And obviously, there is a intuitional way. If we have $\partial _t F \rightarrow 0$, because $\Delta_g F=-H\nu$, we can easily get $H=\langle F, \nu\rangle$ in the limit manifold. Maybe, this is too strong, just $\langle \partial _t F ,\nu \rangle \rightarrow 0$ is enough.