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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?

Jay
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  • I learned this kind of computations in the notes of Sergiu Klainerman: https://web.math.princeton.edu/~seri/homepage/courses/Analysis2011.pdf (See how he proves the divergence theorem, pag. 73) – Giuseppe Negro May 14 '19 at 19:59
  • @GiuseppeNegro Thank you for the reference. In this specific problem, it is the integral that confuses me. – Jay May 14 '19 at 20:00
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    This post is very much related. – Giuseppe Negro May 14 '19 at 20:05
  • The difficulty is to define $ d\sigma(x)$, it is quite obvious for a given $u$ it exists, but obtaining a simple expression for any $u$ isn't – reuns May 14 '19 at 20:28
  • @GiuseppeNegro Thank you. But unfortunately, I'm not able to derive an answer to this question from it. How can it be done? – Jay May 17 '19 at 21:19
  • $v(x) = u(t,x)$ then $\delta(v(x)) = \lim_{h \to 0} \frac{1_{v(x) \in [-h,+h]}}{2h}$.

    Let $S = { x, v(x)=0}$. If $v$ is smooth and $\nabla v(x_0)$ never vanishes for $x_0\in S$ then $S$ is a hypersurface, locally it is an hyperplane and $\nabla v(x)$ is constant, so it suffices to check the case $v(x) = \nabla v(x_0) . (x-x_0)$ for which $\int_{\Bbb{R}^N} \delta(v(x)) \phi(x)d^N x = \int_S\frac{\phi(x)}{|\nabla v(x)|_2}d\sigma_S(x)$ where $\sigma_S$ is the Minkowski content measure of $S$. From there it suffices to extend from test functions $\phi$ to distributions.

     – reuns May 17 '19 at 22:06
  • @reuns Could you expand your comment in an answer, please? – Jay May 17 '19 at 22:14
  • A thought, if supp $\phi=U$, $$-(\partial_tT,\phi)=\int_U\partial_t \phi(t)\int_{u(x,t)>0}dxdt =\int_{{(x,t) : u(x,t)>0, \ t\in U }}\nabla_{x_1,\dots,x_N,t}\cdot\begin{bmatrix}0\ \vdots\0\ \phi(t)\end{bmatrix} dxdt $$ lets you use Gauss's theorem. I believe $$\partial {(x,t) : u(x,t)>0, \ t\in U } \= {(x,t) : u(x,t)=0, \ t\in U }\cup {(x,t) : u(x,t)>0, \ t\in \partial U }.$$ On the set $ {(x,t) : u(x,t)>0, \ t\in \partial U }$, there is no contribution since $U$ is the support of $\phi$. So someone who can integrate on the level set ${u(x,t)=0}$ (not me) can finish. – Calvin Khor May 17 '19 at 22:37

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