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

How can I compute the following distributional derivative?

$$ \partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx.$$

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First, a simple case that $\Omega\subset\mathbb{R}$ and that $u(t,x)$ vanishes at a countable set of values $x_n(t)$, $n=1,2,\ldots$, with $\partial_x u(t,x)\neq 0$ at each $x_n(t)$.

Then $$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx=\partial_t\int_\Omega\theta[u(t,x)]\, dx$$ $$= ∫_\Omega \delta[u(t,x)]\partial_t u(t,x)\,dx = \sum_n\lim_{x\rightarrow x_n(t)}\frac{\partial_t u(t,x)}{|\partial_x u(t,x)|}.$$ In the first equality I introduced the unit step function $\theta(s)$, in the second equality I used that the derivative of the step function is a Dirac delta function, with the chain rule, in the third equality I used that $\delta[f(x)]=\sum_n \delta(x-x_n)/|f'(x_n)|$ if $f(x_n)=0$ with a nonvanishing first derivative $f'$. (See for example, Wikipedia.)


Now more generally, if $\Omega\subset\mathbb{R}^N$, let $S(t)$ be the $N-1$ dimensional surface on which $u(t,x)=0$, and assume that the gradient $\nabla_x u(t,x)$ does not vanish on $S(t)$. Then the desired result is a surface integral $\int_Sd\sigma(x)$, $$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx= ∫_\Omega \delta[u(t,x)]\partial_t u(t,x)\,dx=\int_{S(t)}\frac{\partial_t u(t,x)}{|\nabla_x u(t,x)|}d\sigma(x).$$ For the last identity, see for example Wikipedia.

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  • $\begingroup$ Maybe I got it: do you mean that $S(t)$ is the "interface" (i.e. the boundary of the support)? $\endgroup$ – Jay May 11 at 12:36
  • $\begingroup$ Also: (1) is there anything we can do if the gradient does vanish on $S(t)$? (2) what is the relationship between the last formula and the distributional derivative of the function, i.e. $\langle \int_{\{u(t,\cdot) >0\}} 1 dx, \partial_t \phi \rangle$? (since $\partial_t u$ is to be intended in the distributional sense). $\endgroup$ – Jay May 11 at 12:43
  • $\begingroup$ if the derivative vanishes there is no finite answer; for the same reason that $\int \delta(x^2)dx$ diverges. $\endgroup$ – Carlo Beenakker May 11 at 13:24
  • $\begingroup$ Do you mean that in that case the derivative blows up? $\endgroup$ – Jay May 11 at 13:35
  • $\begingroup$ yes, the derivative diverges; I'm not sure what you mean by the "test function"; all of these manipulations are in distributional sense $\endgroup$ – Carlo Beenakker May 11 at 13:38

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