It's a question that has puzzled me for a long time.

Every PDE textbook I've ever seen tells me that, Fourier transform can be used to solve linear constant-coefficient differential equations on an infinite domain, but none of them includes an explanation about what boundary condition is actually used when Fourier transform "kills" the derivative. Some materials, for example this seems to suggest that, the boundary condition is $0$ at $\pm\infty$, but it's not true. A counter example is

$$y'(x)+y(x)=\sin (x)$$

The general solution of this equation is

(* Here's the corresponding Mathematica code *)
DSolve[y'[x] + y[x] == Sin[x], y[x], x]

$$y(x)= c_1 e^{-x}+\frac{1}{2} (\sin (x)-\cos (x))$$

while the solution given by Fourier transform and inverse Fourier transform is

(* Here's the corresponding Mathematica code *)
fou = FourierTransform[#, x, w] &;
fou[y'[x]] + fou@y[x] == fou@Sin[x] /. HoldPattern@FourierTransform[__] :> Y[w]
Solve[%, Y[w]][[1, 1, -1]]
InverseFourierTransform[%, w, x]

$$y(x)=\frac{1}{2} (\sin (x)-\cos (x))$$

Clearly the boundary condition isn't $y(\pm\infty)=0$ or $y'(\pm\infty)=0$.

What boundary condition / restriction is imposed when Fourier transform is used for solving differential equations?


When a given PDE can be solved using the Fourier transform in $L^2(\mathbb{R})$, then the cited material is morally right: If a function in $L^2(\mathbb{R})$ has a limit at $\pm \infty$, then this limit equals $0$. However, as noted by user TrialAndError below, there are functions in $L^2(\mathbb{R})$ which do not converge to $0$ at infinity. Such examples do almost converge to $0$ though, e.g. we have for any $\epsilon>0$ that $|\{x\in \mathbb{R}\setminus B_r(0): f(x)>\epsilon\}|\to 0$ as $r\to\infty$.

As you noted, the solution of your example is not at all converging to zero. This is because arriving at this solution requires the calculus of Fourier transforms of tempered distributions! For example, $\sin(x)$ is not in $L^2(\mathbb{R})$ and its distributional Fourier transform involves Dirac deltas, which are also not in $L^2(\mathbb{R})$. While tempered distributions are a larger space than $L^2(\mathbb{R})$, they still exclude functions that diverge exponentially, which is why Mathematica gives you the particular solution you mention.

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  • $\begingroup$ Thanks for the answer. Er… do you have any recommended reference if one (who is poor at math) wants to learn more about the issue? I tried to read the wiki page of tempered distributions but it's somewhat over my head. $\endgroup$ – xzczd Jan 5 '17 at 14:25
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    $\begingroup$ I don't know if it is worth the effor, at the end the message is easy: the usual Fourier transform of the 19th century is defined for $ L^2$ functions and these trivially vanish at infinity. Tempered distributions are objects that in particular include functions that do not vanish (like he constant 1 function) for which a Fourier transform can be defined with some more effort as well. If it turns out that you cannot solve a PDE with the old Fourier transform, for example because the right hand side is not in $ L^2$, then you always know that the solution is not necessarily decaying at infty $\endgroup$ – Bananach Jan 5 '17 at 16:22
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    $\begingroup$ @Bananach : There are $L^2(\mathbb{R})$ functions that don't vanish at $\infty$. You can construct an example with unit height step functions of decreasing widths placed at the integers. $\endgroup$ – Disintegrating By Parts Jan 8 '17 at 2:07
  • $\begingroup$ You are right. I was only thinking of functions that actually do have a limit. Once again, to my embarassment, the word "trivially" served as an alert of an incorrect statement ;) $\endgroup$ – Bananach Jan 8 '17 at 10:32

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