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Let's consider the heat equation $u_t-u_{xx}=0$. A simple method to solve this PDE is to do a Fourier transform of the equation and get the ODE $\hat{u_t} + x^2 \hat u_{xx}=0$. We can solve this, do a back transformation and obtain a solution to our original PDE.

Most people don't care why $\widehat{u_t} = \frac{d}{dt} \hat u$ is justified. When one writes it out, we get into the situation that we need to interchange limit with integration. This is not trivially justified.

I would like to know in which situations we are justified in doing this. Someone gave me a hint to consider the question from another viewpoint, namely to consider the spatial Fourier transform as a Banach space values function and consider generalized derivatives, but this idea wasn't made more precise.

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It's reasonable to question this interchange of operations. You're looking at trying to justify $$ \frac{d}{dt}\int_{-\infty}^{\infty}e^{-isx}u(x,t)dt = \int_{-\infty}^{\infty}e^{-isx}u_{t}(x,t)dt. $$ The correct norm for the heat equation is the $L^1(\mathbb{R})$ norm because one would typically have positive termperatures, and the total heat (or energy) would be proportional to the integral of $u$ over all $x\in\mathbb{R}$. The natural conditions to justify the above are a continuity condition $$ \lim_{h\downarrow 0}\int_{-\infty}^{\infty}|u(t+h,x)-u(t,x)|dx =0 $$ and a derivative condition $$ \lim_{h\downarrow 0}\int_{-\infty}^{\infty}\left|\frac{1}{h}\{u(t+h,x)-u(t,x)\}-u_t(t,x)\right|dx=0. $$ These two vector conditions combine to allow one to justify the interchange, provided you assume $u_{t}(t,x)\in L^1(\mathbb{R})$ in the spatial variable $x$. Anything less makes it almost impossible to justify the interchange, which is why the interchange is often ignored in order to arrive at a solution.

You can show that these conditions do hold for the final solution obtained by the Fourier transform method. The second condition may not hold at $t=0$, though the first will; and both hold for all other $t > 0$. The second condition at $t=0$ requires spatial smoothness of the initial data distribution $u(0,x)$. It turns out these are natural conditions in the setting of $C^0$ semigroup theory, at least for $t > 0$. The first condition must hold in $C^0$ semigroup theory for $t=0$, while the second cannot hold unless $u(0,x)$ is in the domain of the generator of the semigroup, which is $\frac{d^2}{dx^2}$ in this case. $C_0$ semigroup theory is a vector theory devised to deal with time evolution systems.

It's reasonable to impose these additional stability conditions, and to put restrictions on the functions to be in $L^1$. Why? Because solutions are not necessarily unique unless you do impose such conditions. Being in $L^1$ and imposing stability conditions of this type lead to unique solutions that can be obtained using the Fourier transform method, because the interchange of operations is permitted.

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  • $\begingroup$ Thans a lot, this answer was very helpful for me! I think textbook authors should be more precise in this kind of things... $\endgroup$ – Sebastian Bechtel Sep 16 '16 at 17:57
  • $\begingroup$ @SebastianBechtel : You're welcome. I had questioned the same things until I learned about $C^0$ semigroup theory. Pazy's book on the subject is a nice one. Naturally it requires a Functional Analysis background. $\endgroup$ – DisintegratingByParts Sep 16 '16 at 18:32
  • $\begingroup$ @TrialAndError: I think the "derivative condition" should equal 0, as well. Why do you call them the "vector conditions"? $\endgroup$ – el_tenedor Nov 13 '16 at 10:24
  • $\begingroup$ @el_tenedor : It is a vector derivative, meaning that $\| \frac{1}{t}\{h(t)-h(0)\}-k\|\rightarrow 0$ as $t\downarrow 0$. That such a derivative exists is a stronger requirement than pointwise derivatives only. The norm here is the $L^2(\mathbb{R})$ norm. $\endgroup$ – DisintegratingByParts Nov 13 '16 at 14:29
  • $\begingroup$ @TrialAndError: Thanks. Your post is still missing the $=0$ part, i.e. $$\lim_{h \to 0} \left\|\frac{1}{h}(u(t + h, \cdot) - u(t, \cdot)) - u'(t, \cdot)\right\|_1 = 0$$. $\endgroup$ – el_tenedor Nov 13 '16 at 14:42

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