Let $\Omega \subseteq \mathbb R^n$ be a bounded region. Furthermore, let $\Omega_T = \{(x, t) \in \Omega \times (0, T]\}$ denote the parabolic cylinder and $\partial_* \Omega_T$ the parabolic boundary, i.e. the set $\overline{\Omega_T} \backslash \Omega_T = \{(x, 0) : x \in \Omega\}$. Furthermore, let $u \in C^{2; 1}(\overline{\Omega_T})$ (i.e. $u$ is $C^2$ on $\Omega_T$ and $C^1$ on $\partial_* \Omega_T$) be a solution of the heat equation
$$u_t = \Delta_x u \quad \text{ on } \Omega_T, \qquad u \mid_{\partial_* \Omega_T} = 0$$
(where $\Delta_x$ is the Laplacian with respect to $x$). Furthermore, we set $E(t) := \int_\Omega u(x, t)^2 d x$ for $t \in [0, T]$. I now want to show that $E \in C^1(0, T)$ with $E'(t) \leq 0$.
Now I think that $E$ is a $C^1$-function follows immediately from the Leibniz-rule? (i.e. being able to pull the $\frac d{dt}$ into the integral over $x$.) So I would have only left to show that $E'(t) \leq 0$.
For that, I tried to concretely calculate the integral. Using the aforementioned Leibniz rule, I've calculated
$$E'(t) = \frac d{dt} \int_\Omega u(x, t)^2 d x = \int_\Omega \frac d{dt} u(x, t)^2 d x = \int_\Omega 2 u(x, t) u_t(x, t) d x $$
And using the preliminary $u_t = \Delta_x u$, this should become
$$= \int_\Omega 2 u(x, t) \Delta_x u d x$$
but from here on I'm a bit lost... if I have done everything correctly up to this point, then how can I conclude that this integral is $\leq 0$? I assume I have to somehow use the 2nd preliminary (that $u = 0$ on the parabolic boundary) since I haven't used that yet, but I'm not sure how.