Using a Fourier transform on the wave equation

[Luminous Blob]

Hi, I want to know how to get rid of the time part of the homogeneous wave equation:

$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }<br /> \nabla^2\psi-c^{-2}\pd{\psi}{t}{2} = 0$$

I've read that this can be done using a Fourier transform, with the following given as the "frequency-time Fourier transform pair" that is apparently used for this (according to a document I found on the internet, anyway):

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\omega)e^{-i\omega t}d\omega$$
$$f(\omega)=\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt$$

Now, I've never used Fourier transforms before and all the references I've come across say to use a Fourier transform to get the Helmholtz equation, but don't explain the steps.

Can anyone explain to me step-by-step how one applies a Fourier transform to the above wave equation to get the Helmholtz equation, or provide a good reference for beginners that explains it in reasonable detail?

Also, if there is another way it can be done (without using a Fourier transform) I'd appreciate any explanation. Thanks.

This isn't so much a homework question as me trying to understand some things that I've glossed over in the past (accepting the results without really understanding how they were reached). Should this question be in the maths forum instead, or is it okay to post it here?

 

[hananl]

try:
introduction to Fourier optics. pages 32 (starting on "the helmholtz equation") up to 50.
author: joseph w. goodman.

 

[J77]

To get the Helmholtz, you substitute in:

$$E(r,t)=\Re\{E(r)e^{-i\omega t}\}$$

for the field - this gives you the Hh. eqn. with a propagation constant.

(r is a direction vector)

To tie in with George's very detailed answer, the above treats the field as a single harmonic.

 

[George Jones]

$\psi$ is function of 4 variables, 3 spatial coordinates and time, i.e., $\psi= \psi \left( x,y,z,t \right)$.

Define a new, related function $\tilde{\psi} \left( x,y,z, \omega \right)$ that is also a function of 4 variables, by

$$\tilde{\psi} \left( x,y,z, \omega \right) = \int_{-\infty}^{\infty} \psi \left( x,y,z,t \right) e^{i\omega t}dt.$$

Note that $\tilde{\psi} \left( x,y,z, \omega \right)$ is a function of the same 3 spatial coordinates as $\psi$, but now time is replaced by frequency. By the Fourier inversion theorem,

$$\psi \left( x,y,z, t \right) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t}d \omega.$$

Substitute this into your wave equation, and assume that you can move partial derivatives through the integral signs. This is as as much as you can do for the first term in the wave equation. In the second term, though, the time partial derivatives operate on $e^{-i\omega t}$.

What do you get for the wave equation?

Do things step-by-step, and if at any point things are unclear, just ask more questions.

Edit: I have been correcting mistakes in this post as I see them. I just corrected a serious mistake - the integration variable in the last equation was wrong.

Regards,
George

 

[Luminous Blob]

So, wherever $$\psi$$ appears in the original wave equation, it is replaced by

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t}dt$$

?

Also, what do you mean by "assume that you can move partial derivatives through the integral signs"?

Thanks for your help.

 

[George Jones]

So, wherever $$\psi$$ appears in the original wave equation, it is replaced by

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t}dt?$$

Yes, but note the edit in my previous post that changes the integration variable.

Also, what do you mean by "assume that you can move partial derivatives through the integral signs"?

Thanks for your help.

I just mean that, for example,

$$\frac{\partial^2}{\partial t^2} \int_{-\infty}^{\infty} \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t}d\omega = \int_{-\infty}^{\infty} \frac{\partial^2}{\partial t^2} \left[ \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t} \right] d\omega$$

Similarly for the spatial derivitaves. This is possible if the function are nice enough, and amount to the changing the order of limits.

Regards,
George

 

[Luminous Blob]

So it'd look like this after the substitution?

$$\frac{1}{2 \pi}\int_{-\infty}^{\infty}\nabla^2 \tilde{\psi}\left( x,y,z,\omega \right) e^{-i\omega t}d\omega -c^{-2}\frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{\partial^2}{\partial t^2} \left[ \tilde{\psi} \left( x,y,z,\omega \right) e^{-i\omega t} \right] d\omega=0$$

 

[George Jones]

Right.

There is no need to to anything more to the first term. In the second term, $\tilde{\psi}$ is not a function of time, and so is treated as a constant in terms of the time partials, but $e^{-i \omega t}$ needs to be differentiated twce with respect to time.

Also, collect everything under one integral sign, and 2 pi is missing from the second term.

What results?

Regards,
George

 

[Luminous Blob]

After doing that, you get

$$\frac{1}{2 \pi}\int_{-\infty}^{\infty}e^{-i\omega t} \left( \nabla^2 \tilde{\psi}\left( x,y,z,\omega \right) + c^{-2} \omega^2 \tilde{\psi} \left( x,y,z,\omega \right) \right)d\omega=0$$

?

 

[George Jones]

Yes, and for this to be true in general, what's under the integral must be zero.

After setting the integrand to zero, multiply both sides of the equation by $e^{i\omega t}$.

Regards,
George

 

[Luminous Blob]

Right then, setting the integrand to zero:

$$e^{-i\omega t} \left( \nabla^2 \tilde{\psi}\left( x,y,z,\omega \right) + c^{-2} \omega^2 \tilde{\psi} \left( x,y,z,\omega \right) \right) = 0$$

Which gives

$$\nabla^2\tilde{\psi}+\frac{\omega^2 \tilde{\psi}}{c^2}=0$$

which is...the Helmholtz equation!

HA! You're a legend! Thanks a million for all the help. It all makes a lot more sense now.