0
$\begingroup$

I have a question about a property of the solutions to the heat equation. Let $u(t,x)$ be a solution to the (one-space dimension) heat equation $$u_t = u_{xx}.$$ Is it true that $\int_{\mathbb{R}}u(t,x) \mathrm{d}x$ is constant with respect to $t$? That is, I want to verify (and find a reference) for the following equation:

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathbb{R}}u(t,x) \mathrm{d}x = 0.$$ Where can I find a reference for this fact? In addition, do we have to assume $u(0,\cdot) \in L^1(\mathbb{R})$ for the above to hold? Or is it true even if the initial condition is not integrable? Many thanks!

Edit: I understand why conservation of mass holds under some conditions specified in an answer below. I still looking for a reference for this statement (I only found it in some private lecture notes, but not in any books or papers). I also want to know whether it is true even if the initial condition satisfies $\int_{\mathbb{R}}u(0,x)\mathrm{d}x = \infty$.

$\endgroup$
1
  • 1
    $\begingroup$ conversation (blah blah...) $\to$ conservation... $\endgroup$
    – Jean Marie
    Commented Jun 11 at 12:24

1 Answer 1

1
$\begingroup$

I would say, to get a reasonable globale notion of total heat $$W(t):=\int_{\mathbb{R}^N} u(t, x) d x\quad W(0)=\int_{\mathbb{R}^N} g(x) d x \quad g\in L^1(\mathbb{R}^n)$$ you want to assume $g\in L^1(\mathbb{R}^n)$. Additionally, since you want to differentiate a parameter integral $W(t)$, having following conditions met will be sufficient to change integration and differentiation:

(i) $u(t, \cdot)$ is $L^1(\mathbb{R}^n)$ for every $t \in (0,\infty)$;

(ii) The $\frac{d}{dt} u(t, x)$ exists in $(0,\infty)$ and for $\lambda$-almost all $x \in \mathbb{R}^n$;

(iii) $\exists h \in L^1(\mathbb{R}^n)$, such that for all $t \in (0,\infty)$ there exist $\lambda$-nullset $N_t$ with $$ \left|\frac{\partial u}{\partial t}(t, x)\right| \leq h(x) \quad \forall x \in \mathbb{R}^n \backslash N_t . $$

So at least in the cases, where your solution $u$ meets these conditions and $u$, $\nabla u$ additionally have sufficient (uniform) decay properties (which depends on your data function), you can simply see for the total heat via divergence theorem

$$ \begin{aligned} \frac{d}{d t} W(t)=\int_{\mathbb{R}^N} u_t(t, x) d x & =\int_{\mathbb{R}^N} \Delta u(t, x) d x=\lim _{R \rightarrow \infty} \int_{B_R(0)} \Delta u(t, x) d x \\ & =\lim _{R \rightarrow \infty} \int_{S_R(0)} u_{\nu_x}(t, x) d x=0 \quad \text { für } t>0 . \end{aligned} $$ Thus $W(t)=W(0)$

New contributor
xajas is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
3
  • $\begingroup$ Thank you! I have 2 questions: 1. Is there a reference of the claim you stated? I've looked through about 4 books on PDEs and couldn't find this fact mentioned, not even as a problem/exercise. 2. I deal with a case where the initial condition is not integrable. I get that your reasoning doesn't work here, but the result fails here as well? Is there a counter example where you start with an infinite mass that becomes in a finite time? $\endgroup$ Commented Jun 11 at 13:23
  • $\begingroup$ 1. Had it in a PDE lecture. 2. So in case you have an initial value problem, but your data function is not integrable, then i think it’s not clear how to get a notion of total heat, which one would naturally conceptualise by simply taking the integral of the solution for fixed time $t$, and in case for $t=0$, the initial data over the given integration domain. Do you have more context?:) $\endgroup$
    – xajas
    Commented Jun 11 at 16:00
  • $\begingroup$ I mean, what happens if $u(t,x)$ is a solution to the heat equation, where $\int_{\mathbb{R}}u(0,x)dx=\infty$ $\endgroup$ Commented Jun 11 at 21:31

You must log in to answer this question.

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