# Compute $\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx$ in the sense of distributions where $u$ solves a PDE

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?

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

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.
• Maybe I got it: do you mean that $S(t)$ is the "interface" (i.e. the boundary of the support)? – Jay May 11 at 12:36
• 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). – Jay May 11 at 12:43
• if the derivative vanishes there is no finite answer; for the same reason that $\int \delta(x^2)dx$ diverges. – Carlo Beenakker May 11 at 13:24