Compute the derivative $\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx$ in the sense of distributions

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Let $u:\Omega\subset \mathbb R^N \to \mathbb R$ be bounded function that solves (in the sense of distributions) an evolution PDE $\partial_t u(t,x)= L(u(t,\cdot))(x)$, where $L$ is some elliptic operator in divergence form.

In a post on MathOverflow, a calculation was made regarding the time derivative $$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx$$ in the following way:

$$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx= ∫_\Omega \delta[u(t,x)]\partial_t u(t,x)\,dx=\int_{S(t)}\frac{\partial_t u(t,x)}{|\nabla_x u(t,x)|}d\sigma(x).$$

  • How can one rewrite this calculation emphasizing the fact that we are doing a derivative in the sense of distributions, i.e. $$\langle \partial_t T, \phi\rangle = -\langle T, \partial_t\phi \rangle $$ where $T$ is the distribution associated with $\int_{\{u(t,\cdot) >0\} } 1\, dx$?

  • And how can we use the fact that $u$ solves the PDE above in the final formula?