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?