# Showing the integral of the solution to a heat equation is constant

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.

• I think the trick in your first computation is to switch the integration order of x and y and pull out the $v(y)$. I think the second approach could also be made rigorous by separating the domain into inside the ball and outside the ball. Feb 1 at 9:43
• Thank you very much, I will try your advices. Feb 1 at 9:56

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 $$t\,.$$