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The setting is on evolving hypersurfaces. So for each time $t$, $\Gamma(t)$ is a hypersurface given by the zero level set of the function $\phi(x,t)$. Consider a ball, then the hypersurface has $\phi(x,t) = x_1^2 + ... + x_n^2 - R(t)^2$ as the level set. So for each time $t, \Gamma(t)$ is a ball of radius $R(t)$.

Associated with each surface $\Gamma(t)$ is a normal velocity $V(x,t)$ and a mean curvature $H(x,t)$. Suppose the surface evolves by the rule V = -H. After some calculations, let's say that $V = \dot{R}$ and $H = -\sqrt{x_1^2 + ... + x_n^2}$.

  • One can then solve this ODE $V=-H$ to get R(t).

  • But because the mean curvature $H$ is defined only on points on the surface, can't we rewrite $H = -\sqrt{x_1^2 + ... + x_n^2}$ to be $H = -R$? Because if $x_1^2 + ... + x_n^2 - R^2(t) \neq 0$, then then $H(x,t)$ is meaningless.

But if I make this substituion and then solve the ODE, i get a different answer obviously. What's the right thing to do?

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    So if I worked out $H = \frac{1}{x_1^2 + ... + x_n^2}$ I can just rewrite that to be $H = \frac{1}{R}$? I am not sure what to do with time dependent quantities like these.2012-04-16

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Considering that

  • $H$ is a function defined on a time-dependent manifold, and
  • $R$ is a function of time

it is only possible to express $H$ as a function of $R$ if everymanifold $\Gamma(t)$ has constant mean curvature. But this is exactly the case for the spheres. The conclusion is yes, it is possible to write $H(t)=1/R(t)$. One should keep in mind that this single-variable function $H$ is not the same as $H(x,t)$, but they are related by
$H(x,t)=H(t) \quad \text{when } \ x\in \Gamma(t)$