Let $\Omega$ be an open and bounded subset of $\mathbb{R}^n$ with a $C^1$ boundary. Let $u=u(t, x)$ be the solution to the following diffusion problem: $$\left\{\begin{matrix} \frac{\partial u}{\partial t}(t, x) = \Delta_x u(t, x), & x \in \Omega, t > 0\\ \frac{\partial u}{\partial \mathbf{n}} (t, x) = 0, & x \in \partial \Omega, t > 0\\ u(0, x) = g(x), & x \in \Omega \end{matrix}\right.$$ Show that, for any $t > 0$: $$\int_\Omega u(t,x) dx = \int_\Omega g(x) dx$$
I have problems understanding the following reasoning:
Consider an arbitrary $t>0$. Then $$\int_\Omega \Delta_x u(t, x) dx = \int_\Omega \frac{\partial u}{\partial t}(t, x) dx = \frac{\partial}{\partial t}\int_\Omega u(t, x) dx = E'(t)$$
So if I define the energy functional as $$E(t) = \int_\Omega u(t, x) dx$$
then $$E(0) = \int_\Omega u(0, x) dx = \int_\Omega g(x) dx$$
How do I follow from this that $E$ is constant and therefore those two integrals are equal?
I feel like I'm missing the whole point of energy methods.
Edit:
I suppose that since the integrand $\int_\Omega \Delta_x u(t, x) dx$ depends only on $x$ for a given $t$, then I could the Green's formula and thus $\int_\Omega \Delta_x u(t, x) dx = \int_{\partial \Omega} \frac{\partial u}{\partial \mathbf{n}} (t, x) = 0$, meaning that $E'(t) = 0$ for all $t>0$?