Passing Derivative Under the Integral: Conditions and Applications

  • Thread starter Brian T
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In summary: You could use a smooth cutoff function, but there are other ways to handle this as well.In summary, the conversation discusses the conditions for passing the derivative under the integral in a diffusion equation problem. The solution involves using Green's identity and letting the "boundary" go to infinity. However, the conditions for being able to do this are not fully justified and may require further analysis.
  • #1
Brian T
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Homework Statement


My question involves whether or not we can pass the derivative under the integral in the following question, and what conditions need to be met. My second is question is how to evaluate using Green's identity at the "boundary". See below

A function u satisfies the diffusion equation in [itex]\mathcal{R}^n \times (0,\infty)[/itex] with some initial value.
$$u_t = \Delta u, u(x,0)=v(x) $$
The question is to show that the integral of [itex]u(x,t)[/itex] over the whole domain is constant w.r.t time (physically, this means the total amount of whatever is diffusing remains constant over all space, e.g. mass conservation). We are also given that the value of the solution [itex]u(x,t) \rightarrow 0 \ as \ |x| \rightarrow \infty[/itex], i.e. that it asymptotically approches 0.
To show that it is constant w.r.t time, we show that
$$\frac{d}{dt} \int_{\mathcal{R}^n}u(x,t)d^nx = 0$$
My question is, if we know that the function vanishes at infinity, can we pass the differentiating under the integral. What are the conditions to be able to do this?
2. Relevant eq
3. The Attempt at a Solution

Assuming that we can pass the derivative under the integral,
$$\frac{d}{dt} \int_{\mathcal{R}^n}u(x,t)d^nx = \int_{\mathcal{R}^n} u_t(x,t)d^nx = \int_{\mathcal{R}^n} \Delta u(x,t)d^nx$$
Using Green's identity here, where we take the "boundary" to be a sphere [itex]S^{n-1}_R[/itex] and let the radius R go to infinity.
$$ = lim_{R \rightarrow \infty} \int_{S^{n-1}_R} \frac{\partial u}{\partial n} d^{n-1}s$$
and since the function vanishes asymptotically, it's derivative will be 0 at this "boundary" and so the whole term will be 0,
$$\therefore \frac{d}{dt} \int_{\mathcal{R}^n}u(x,t)d^nx = 0$$

The questions I have with my solution is
(1) in what conditions are we allowed to pass the derivative under the integral as I did in my solution. Was this justified? (Although the vanishing at infinity is weaker than the condition for compact support, I treated them essentially as the same. I'd like more details on this)
(2) Was it justified to use Green's theorem on a "boundary" whose size we let go to infinity?
If anyone has a more rigorous approach at this problem, please clue me in :)
Regards
 
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  • #2
Brian T said:
The questions I have with my solution is
(1) in what conditions are we allowed to pass the derivative under the integral as I did in my solution. Was this justified?
The limit in the definition of the derivative must commute with the integration. As long as it does you will be fine.
Brian T said:
Was it justified to use Green's theorem on a "boundary" whose size we let go to infinity?
Yes, but you are missing that this is not sufficient. It is generally possible for an integral over a growing domain to have a non zero limit even if the integrand goes to zero.
 
  • #3
Orodruin said:
The limit in the definition of the derivative must commute with the integration. As long as it does you will be fine.

Yes, but you are missing that this is not sufficient. It is generally possible for an integral over a growing domain to have a non zero limit even if the integrand goes to zero.

Thanks for the insight. So for your second comment, does this imply that we have to assume that the function has compact support?
 
  • #4
Brian T said:
Thanks for the insight. So for your second comment, does this imply that we have to assume that the function has compact support?
No. Just that you need to argue better. Due to the behaviour of the Green function, it will not have compact support.
 

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