Let $u$ be a smooth solution to:
$\partial_t u - \Delta u = 0$, $(t, x) \in (0, \infty)$ x $\mathbb{R}^3$
such that $u$ and its derivatives are rapidly decreasing as $|x| \rightarrow \infty$ (say, bounded by $(1 + |x|)^{-4}$ as $|x| \rightarrow \infty$; this is more than necessary). Prove that:
$\int_{\mathbb{R}^3} u(t, x)$ $dx$ is constant in $t$.
I'm honestly completely lost on how to start this question. I know it has to do with the heat equation... I think I have to plug in the fundamental solution into the integral, and then take the derivative with respect to $t$ to show that it's constant. Am I on the right track? Any guidance would be greatly helpful, because I'm very lost with this subject. Thank you.
Let $f(t) = \int_{\mathbb R^3} u(t,x) \, dx$. Then $$ f'(t) = \int_{\mathbb R^3} \partial_t u \, dx = \int_{\mathbb R^3} \Delta u \, dx. $$ Now notice that $\Delta u = \nabla \cdot \nabla u$ (in other words, the Laplacian is the divergence of the gradient) and use the divergence theorem to show that the integral on the right is zero.