I have a rather naive question.
If I want to interpret the PDE
$\frac{\partial}{\partial t}u=\Delta u$
in the sense of distributions, what does that mean? I do not understand what the time derivative means for the distribution.
If I only had to solve $\Delta u=0$, say, in one space dimension, that is, $\frac{\partial ^2}{\partial x^2}u=0$ in the sense of distributions, then this would mean (in one space dimension)
$$
\langle u'',\varphi\rangle=\langle u,\varphi''\rangle=0 \Leftrightarrow \int_\mathbb{R}u(x)\varphi''(x)\, dx = 0
$$
But now including the partial time derivative, I do not know how to formulate the distributional sense of the PDE. Do I treat the partial time-derivative just as any partial derivative of a distribution, getting
$$
\langle u_t-u_{xx},\varphi\rangle=-\int\int u(x,t)\varphi_t(x,t)\, dx\, dt-\int\int u(x,t)\varphi_{xx}(x,t)\, dx\, dt=0
$$
or do I only consider the spatial partial derivatives of the distribution?
Best Answer
You may get a more authoritative and fully fleshed out answer than mine, but briefly I think it has the potential to mean at least two things. We could interpret the PDE to be understood in what I'd call the "total distributional sense" where we strictly interpret the equation via the usual "integration by parts" type of argument like you outlined in your answer, jointly in the variables $t,x$.
There is also the potential to understand it as an evolution equation for distributions of the space variable that actually have classical differentiability in the $t$-variable. Here we think of a solution $\phi_t$ as a differentiable path $t\mapsto \phi_t \in \mathcal D(\mathbb R^n_x)$ with $\partial_t\phi_t\in\mathcal D(\mathbb R^n)$, where for each $t$ the "point" $\phi_t$ satisfies $\partial_t\phi_t = \Delta\phi_t$ in $\mathcal D(\mathbb R^n_x)$. In this perspective, depending on our assumptions, $\partial_t\phi_t\in\mathcal D(\mathbb R^n)$ may be an axiom or it may be a Proposition.
I think both perspectives get attention in PDE theory.