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?


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.


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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – quarague
    Feb 1 at 9:43
  • $\begingroup$ Thank you very much, I will try your advices. $\endgroup$ Feb 1 at 9:56

1 Answer 1


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 $t\,.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .