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Let $\Omega$ be a bounded and regular open subset $\Omega$ of $\mathbb{R}^N$ and $u:[0,\infty)\times \Omega\to \mathbb{R}$ be a smooth function (for example a smooth solution to a PDE). Thus the function $w=\min(0,u)$ has weak time derivative given by $w_t=u_t.1_{\{u \leq 0 \}}$. Does the following "weak differentiation under the integral sign" holds? $$\frac{d}{dt}\int_\Omega w(t,x)dx=\int_\Omega \frac{\partial}{\partial t}w(t,x)dx,$$ for almost all $t\geq 0$.

We know that differentiation under the integral sign holds for $u$ because it is smooth. But I am wondering if it also holds for a function like $w=\min(0,u)$ which only has a weak derivative.

My guess is that since $w$ has a weak derivative in time (real line), then it must be differentiable for almost all $t\geq 0$ (which is not necessarily true in higher dimension, for example differentiation with respect to space $x$). So $\int_\Omega w(t,x)dx$ must also be differentiable for almost all $t\geq 0$ and so we can use some kind of dominated convergence theorem.

David Lingard
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1 Answers1

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Yes, it does hold, and the only necessary information about $w$ is that it's weakly differentiable. $\newcommand{\pl}{\partial}$

The fact that $w$ has a weak derivative $\pl_t w$ means $$ \int_0^\infty \int_\Omega \pl_t \Phi(t,x) w(t,x) + \Phi(t,x) \pl_t w(t,x) \, dx dt = 0 \quad \text{for } \Phi \in C_c^\infty((0,\infty) \times \Omega). $$ Specializing to $\Phi$ of the form $\Phi(t,x) = \varphi(t)\eta(x)$ with $\varphi \in C_c^\infty(0,\infty)$ and $\eta \in C_c^\infty(\Omega)$, we have $$ \int_0^\infty \int_\Omega (\pl_t \varphi(t,x) w(t,x) + \varphi(t,x) \pl_t w(t,x)) \eta(x) \, dx dt = 0 \quad \text{for all } \varphi, \ \eta. $$ One can choose a sequence of $\eta$'s approximating the function constantly equal $1$ on $\Omega$. If only $w$ and $\pl_t w$ are integrable (in your case, this translates to an integrability condition for $u$ and $\pl_t u$), Lebesgue's dominated convergence theorem implies in the limit that $$ \int_0^\infty \int_\Omega \pl_t \varphi(t,x) w(t,x) + \varphi(t,x) \pl_t w(t,x) \, dx dt = 0 \quad \text{for } \varphi \in C_c^\infty(0,\infty). $$ The above can be rewritten as $$ \int_0^\infty \pl_t \varphi(t,x) \left( \int_\Omega w(t,x) \, dx \right) + \varphi(t,x) \left( \int_\Omega \pl_t w(t,x) \, dx \right) dt = 0 \quad \text{for } \varphi \in C_c^\infty(0,\infty), $$ which is nothing else but the statement that $$ \pl_t \left( \int_\Omega w(t,x) \, dx \right) = \int_\Omega \pl_t w(t,x) \, dx $$ weakly.

If you want to have a.e. pointwise differentiability, just recall that any weakly differentiable function on the line is almost everywhere differentiable (its pointwise derivative coinciding with its weak derivative).

Michał Miśkiewicz
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