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I have a question regarding some things I've seen in previous exams and couldn't find any justification for it.

Let us a look at the following problem: $$\left\{ \matrix{ {u_t}\left( {x,t} \right) = {u_{xx}}\left( {x,t} \right) \hfill \cr u\left( {x,0} \right) = f\left( x \right) \hfill \cr {u_x}\left( {0,t} \right) = g\left( t \right) \hfill \cr {u_x}\left( {L,t} \right) = h\left( t \right) \hfill \cr 0 < x < L{\rm{ }},{\rm{ }}t > 0 \hfill \cr} \right.$$

For the sake of simplicity let us assume the functions $f,g,h$ are infinitely differentiable on the real line.

In many exam exercises I've encountered, the following line boggled me, since no justification for it was provided: $$\mathop {\lim }\limits_{t \to \infty } {d \over {dx}}u\left( {x,t} \right) = {d \over {dx}}\mathop {\lim }\limits_{t \to \infty } u\left( {x,t} \right)$$ It is rather well known that interchanging limit and derivative is not always allowed.

But since I've encountered this numerous times and every time it was treated as something "trivial" in the official solutions provided by my professor.

So am I missing some theorem regarding the solutions of the heat equation and the ability to interchange limit and derivative?

zokomoko
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  • As there is no function $h$ in your PDE or associated data, I'm assuming that is a typo for $u(L,t)$, yes? Or is it actually supposed to be $g$? Also, what do you know about the form of the solution $u$? – Matthew Cassell Jun 03 '19 at 14:50
  • Thanks, typo corrected! I just wish to know if under the conditions mentioned, is it allowed to interchange the derivative and the limit.. and if so, under what theorem – zokomoko Jun 03 '19 at 14:59
  • It might be worth looking at https://math.stackexchange.com/questions/409178/can-i-exchange-limit-and-differentiation-for-a-sequence-of-smooth-functions, since the question is very similar, and the answer given could answer your question about the conditions for interchanging limits and differentiation operators. – bluestool Jun 03 '19 at 18:19
  • Yes I am familiar with thar theorem. But since no type of convergence was proved in the solutions I've seen leads me to believe some other theorem was used, probably using that fact that the function is a solution to the heat equation. – zokomoko Jun 03 '19 at 18:47
  • @bluestoolbl you don't have to assume uniform convergence, all you need is uniform integrability. – ibnAbu Jun 03 '19 at 19:11

1 Answers1

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Set $ u_n (x) = u (x, n) $, where $ n \in \mathbb {N} $

You are essentially asking whether you could differentiate the limit $\lim_{n \to \infty} u_n(x)$ and what is it equal to.

This might help :

This is a stronger version (uniformly integrability condition) of what you need:

If a sequence of absolutely continuous functions {$f_n$} converges pointwise to some $f$ and if the sequence of derivatives {$f_n’$} converges almost everywhere to some $g$ and if {$f_n’$} is uniformly integrable then $\lim\limits_{n\mapsto \infty} f_n’ = g= f’$ almost everywhere. Where the derivative of $f$ is $f’$. If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.

Proof : by FTC $f_n(x) – f_n(a) = \int_a^x f_n’ dx$

By Vitali convergence theorem : $\lim\limits_{n\mapsto \infty}\int_a^x f_n’ dx = \int_a^x g dx$

Therefore $\lim\limits_{n\mapsto \infty}( f_n(x) – f_n(a))= \int_a^x g dx$

$f(x)-f(a) = \int_a^x g dx$

$f(x)’=g$ almost everywhere

If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.

ibnAbu
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