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I'm studying physics, so I'm sorry if I'll write some inexact things in this post. I wish you can understand me.

If we have 1D wave equation:

$$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$

we say that it's always possible (for physics situations...) to decompose the generic solution $f(x+ct)+g(x-ct)$ using Fourier Transforms in $e^{i\omega(x/c\mp t)}.$

But the FT is

$$\frac{1}{2\pi} \int_R \psi(\omega)e^{i\omega t}d\omega$$

and we have two variables, $x$ and $t$, how can we use the FT with

$$ \int_R \psi(\omega)e^{i(k(\omega)x\mp\omega t)}d\omega \; ?$$

We have two different types of temporal evolution, $+ct$ and $-ct$... I have seen situations with no dispersion law $\omega=kc$, where we consider only a wave $\psi=f(x+ct)$ and use the point $x=0$ to find the spectrum of waves $f(\omega)$, : $$ f(\omega)=\frac{1}{2\pi}\int_R f(0,t)e^{-i\omega t} $$


$$ f(x,t)=\int_R f(\omega) e^{i\omega(x/c+t)} $$

but can someone explain me the general method?

In 2D or 3D I haven't any idea about how I can do all this. Someone can explain me this too? Can you link me a .pdf online where this is explained? Thank you very much!

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up vote 6 down vote accepted

There are actually two separate questions in your post.

Fourier transform of the travelling waves

The first question is about Fourier decomposition of the travelling wave solutions $\psi(x,t) = f(x + ct) + g(x-ct)$. It suffices to notice the following change of variables: let $u = x + ct$ and $v = x - ct$. Then inverting the change of variables you get $x = \frac12 (u+v)$ and $t = \frac1{2c} (u - v)$.

Then we can re-write $\psi$ in this new coordinate system as $f(u) + g(v)$. Notice that $f$ is constant along $v$ direction, and $g$ along the $u$ direction.

Since now $f$ is a function of one variable only, you can take the Fourier transform in the $u$ variable:

$$ f(u) = \int \hat{f}(\omega_u) \exp ( i u \omega_u ) d\omega_u $$

So with a further change of variables, you have that

$$ f(x,t) = f(x + ct) = \int \hat{f}(\omega) \exp (i\omega(x + ct)) d\omega $$

as desired (notice that I normalized the exponential differently. You can rescale coordinates once more to get it to be the form with $x/c + t$ if you so wish). By similarly playing with the $v$ variable you get a Fourier representation of $g$.

Now, to find $\hat{f}(\omega_u)$, we take the Fourier transform in the $u$ variable

$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u) \exp (i u \omega_u) du $$

Now remember that $f$ is independent of $v$! So writing $f = f(u,v) = f(u)$, for any curve $v = v(u)$ the integral is the same, that is, we can re-write

$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(u,v(u)) \exp(i u \omega_u) du $$

Now, choose the curve to be $v = -u$, which translates to $x = 0$ by our change of variable formula before. And do a change of variable back to $t,x$, we have that along the curve $v = -u$, $u = ct$. So the integral becomes

$$ \hat{f}(\omega_u) = \frac{1}{2\pi} \int f(0,t) \exp (i ct \omega_u) c dt$$

again, the extra factor of $c$ can be removed by a further change of variables on $\omega_u$.

Two and higher dimensions

In higher dimensions, the situation is analogous, but we do not have the method of characteristics for us to start the first step. So instead, we start with the general high dimensional Fourier transform. (What I am about to write is mathematically non-rigorous, but morally correct.) Consider space-time coordinates $(x_1,\ldots,x_n,t)$ and space-time frequencies $(\xi_1,\ldots,\xi_n, \omega)$. We take the full Fourier transform of a solution to the wave equation in the following way. Suppose

$$ \partial_t^2 \psi = c\triangle \psi $$

where $\triangle$ is the Laplacian in $n$ dimensions. Consider the Fourier transform

$$ \hat{\psi}(\xi_1,\ldots,\xi_n,\omega) = \int_{\mathbb{R}^{n+1}} f(x_1,\ldots,x_n,t) \exp [ -2\pi i (x_1\xi_1 + \cdots + x_n\xi_n + ct\omega) ] c dx_1\cdots dx_n dt $$

(this is the slightly questionable step. In general this integral will not converge; but it captures the idea of the thing.) Using the properties of the Fourier transform to replace derivatives by multiplication of the frequencies, you get that $\hat{\psi}$ solves the algebraic equation

$$ - \omega^2 \hat{\psi} = -(\xi_1^2 + \cdots + \xi_n^2) \hat{\psi} $$


$$ (\xi_1^2 + \cdots + \xi_n^2 - \omega^2) \hat{\psi} = 0 $$

As an algebraic equation, this means that if the number inside the first parenthesis is not zero, we must have $\hat{\psi} = 0$. And that $\hat{\psi}$ is only non-zero where the term in the parenthesis vanishes. (In mathematical jargon, this indicates that $\hat{\psi}$ is a distribution/measure, and not strictly speaking a function.) And in particular, we have the "dispersion law" that the only space-time frequencies contributing to $\psi$ are those where

$$ \omega = \pm \sqrt{ \xi_1^2 + \cdots + \xi_n^2 } $$

So we can write $\hat{\psi} = \hat{\psi}_+ + \hat{\psi}_-$, the forward and backward waves, with $\hat{\psi}_\pm$ only functions of $\xi_1,\ldots, \xi_n$. This gives us the plane wave decomposition of a solution to the wave equation by taking the inverse Fourier transform (to save space, I write $x$ for the vector $(x_1,\ldots,x_n)$ and ditto $\xi$, using $\cdot$ to denote the scalar product)

$$ \psi(x,t) = \int_{\mathbb{R}^n} \hat{\psi}_+(\xi) e^{2\pi i(\xi\cdot x + c|\xi|t)} + \hat{\psi}_-(\xi) e^{2\pi i(\xi\cdot x - c|\xi| t)} d\xi $$

Remark this is only one of the very many ways of solving the linear wave equation. (In a recent PDE course where I work, the lecturer allegedly gave 6 different methods...) Maths textbook I know tend not to want to go into the above Fourier transform derivation, because it is not very rigorous (it can be made rigorous with a lot of additional preparations). The above idea should be treated more as a morally correct guiding principle.

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One further remark: if you look at the final decomposition in higher dimensions, you'd see that it is much, much more natural to take the spectrum of your wave at $t=0$ rather than at $x = 0$; you should think that the frequency is determined by the wave-number via the dispersion relation, rather than the other way around. – Willie Wong Dec 12 '10 at 13:53
Thank you for very interesting and relevant answer. – Boy Simone Feb 15 '11 at 9:24

The things that you wanted to know is on pages 184--189 of Stein-Shakarchi 's Fourier Analysis. In particular, you will see how the solution is written as $$f(x+ct)+g(x-ct)$$ on p. 188. (Substitute $\frac{1}{2}(f(x+t)+G(x+t))$ and $\frac{1}{2}(f(x-t)-G(x-t))$ in the book for your $f(x+ct)$ and $g(x-ct)$ respectively), where $G$ denotes the integral (primitive) of $g$.

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+1 for Stein-Shakarchi. It is a nice introductory book that also presents some very classical points-of-view hard to find in other intro. texts. – Willie Wong Dec 12 '10 at 14:50

The general idea behind transform methods for solving PDEs is that they turn the differential equation into an ODE. In the case of the wave equation, the Fourier transform acts on the space variable $x$ only, not $t$. Since derivatives with respect to $x$ get transformed into multipliation by factors of $2\pi i \xi$ under the action of the transform, the result is a second order ODE in the variable $t$.

Note that there are multiple conventions about the definition of the Fourier transform, having to do with where the factor of $2\pi$ goes, but for instance, one can define it by: $$ F(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i \xi x}\, dx $$ In particular, notice that the transform is a function of $\xi$ (usually thought of as frequency). Compare this with the definition you stated, in which the independent variable in the transformed function is $t$. I think this may be the source of your confusion. The method is basically the same in higher dimensions, letting $\xi$ be an $n$-vector.

A good reference (although not online) is Strauss's text on Partial Differential Equations. I also feel compelled to point out that one does not need Fourier transforms to derive the general solution of the wave equation.

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