I'm trying to prove the fowllowing problem:
Let $u(x, t)$ be the solution to the heat equation
$$
\begin{aligned}
u_t – \Delta u &= 0\;\;\;\; \text{in} \;\mathbb{R}^d\times\mathbb{R}_+\,,\\
u(x, 0) &= v\;\;\;\; \text{in} \;\mathbb{R}^d\times\{0\}\,.
\end{aligned}
$$
Show that if $\int_{\mathbb{R}^d}|v(x)|\, \mathrm{d}x < \infty$, then there holds
$$
\int_{\mathbb{R}^d}u(x, t)\, \mathrm{d}x = \text{constant} = \int_{\mathbb{R}^d}v(x)\, \mathrm{d}x\,.
$$
What I tried to do is, since we know the solution is given by the representation
$$
u(x, t) = (H_t*v)(x) = (4\pi t)^{-d/2}\int_{\mathbb{R}^d} v(y) \exp(-|x-y|^2/4t) \,\mathrm{d}y,
$$
where $H_t$ is the heat kernel, we can differentiate the integral of $u$ with respect to $t$ and show the result is $0$. But the calculation turns out to be
$$
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathbb{R}^2} u\,\mathrm{d}x &= \frac{\mathrm{d}}{\mathrm{d}t}[ (4\pi t)^{-d/2}\int_{\mathbb{R}^2}\int_{\mathbb{R}^d} v(y) \exp(-|x-y|^2/4t) \,\mathrm{d}y\mathrm{d}x]\\
&= \int_{\mathbb{R}^2}\int_{\mathbb{R}^d}(|x-y|^2/4t^2 – d/2t) (4\pi t)^{-d/2} v(y) \exp(-|x-y|^2/4t) \,\mathrm{d}y\mathrm{d}x,
\end{aligned}
$$
and I am not able to show this equals to $0$. Can this problem be solved by this way?
P.S.
Anathor way I came up with is the following: Since $u_t = \Delta u$, we can write
$$
\frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathbb{R}^d} u = \int_{\mathbb{R}^d} u_t = \int_{\mathbb{R}^d} \Delta u = \lim_{R\to\infty}\int_{B_R(0)}\Delta u \\
= \lim_{R\to\infty} \int_{\partial B_R(0)}\frac{\partial u}{\partial n},
$$
and the last integral shoud equals $0$ from the regularity of $u$ provided by its represntation $u(x, t) = (H_t * v)(x)$. But I don't know if this is correct or rigorous.
EDIT.
As quarague and Kurt G. suggested, since $\int |v| < \infty$ and $\exp(-|x-y|^2/4t)<1$, the use of Fubini's theorem could be justfied and we can write
$$
\begin{aligned}
&\int(4\pi t)^{-d/2}\int v(y) \exp(-|x-y|^2/4t) \,\mathrm{d}y\,\mathrm{d}x \\
=\, &(4\pi)^{-d/2}\int v(y)\,\mathrm{d}y \int t^{-d/2}\exp(-|x-y|^2/4t) \,\mathrm{d}x.
\end{aligned}
$$
Now it suffices to show that
$$
\begin{aligned}
&\frac{\mathrm{d}}{\mathrm{d}t}\int t^{-d/2}\exp(-|x-y|^2/4t) \,\mathrm{d}x\\
=&\, t^{-d/2-1}\int (\frac{|x|^2}{4t}-\frac{d}{2}) \exp(-|x|^2/4t)\,\mathrm{d}x\\
=&\, 0,
\end{aligned}
$$
which can be done with integrating by parts.
Best Answer
Use Fubini to show that \begin{align} & (4\pi t)^{-d/2}\int_{\mathbb{R}^2}\int_{\mathbb{R}^d} v(y) \exp(-|x-y|^2/4t) \,\mathrm{d}y\,\mathrm{d}x\\ &=\int_{\mathbb{R}^2}v(y)\underbrace{\int_{\mathbb{R}^d}(4\pi t)^{-d/2} \exp(-|x-y|^2/4t) \,\mathrm{d}x}_{1}\,\mathrm{d}y \end{align} is independent of $t\,.$