For $x\in \mathbb{R}$, let $u(x,t) = (\Gamma_{D} (\cdot , t ) \ast g)(x)$ with $g(x) \ge 0,$ and $\int_{-\infty}^{\infty}g(x) dx < \infty$. Prove that for all $t>0$,
$$ \int_{-\infty}^{\infty} u(x,t) dx = \int_{-\infty}^{\infty} g(x) dx,$$
the meaning of this statement is that the total heat is conserved.
Consider $$E(t) := \int_{-\infty}^{\infty} u(x,t)dx$$ and show that $E'(t) =0$. You may assume that for any $t>0$, $\lim\limits_{x\to -\infty} u_{x}(x,t) = \lim\limits_{x\to \infty} u_{x}(x,t) =0$.
Attempt at solution
\begin{align}
E(t) &:=\int_{-\infty}^\infty u(x,t) dx\\
E(t) &= \int_{-\infty}^{\infty} \int_{-\infty}^\infty \frac{1}{\sqrt{4\pi k t}} \exp{\left(\frac{-(x-y)^2}{4kt}\right)}g(y) \ dy\ dx\\
\frac{\partial}{\partial t}E(t) &= \int_{-\infty}^\infty \int_{-\infty}^\infty g(y) \frac{\partial}{\partial t} \left(\frac{1}{\sqrt{4\pi k t}} \exp{\left(\frac{(x-y)^2}{4kt}\right)}\right) \ dy\ dx\ \\
\frac{\partial}{\partial t}E(t) &= \int_{-\infty}^\infty \int_{-\infty}^\infty g(y) \left( \frac{-1}{4 \sqrt{\pi d t^{3}}} + \frac{(x-y)^{2}}{8\sqrt{\pi d^{3} t^{5}}}\right)\exp{\left(\frac{(x-y)^2}{4kt}\right)} \ dy\ dx \
\end{align}
I honestly do not know how to proceed from here. I feel like I'm off on the wrong track and that I'm missing something else for this proof.
Best Answer
Plugging in the kernel is not the way to go, just plug in the heat equation itself:
$$\frac{d}{dt}\int_{-\infty}^\infty u\:dx = \int_{-\infty}^\infty \partial_tu\:dx = \int_{-\infty}^\infty \partial_x^2u\:dx =0$$
because the integral of a derivative is always zero.