- #1

- 222

- 23

## Homework Statement

## Homework Equations

## The Attempt at a Solution

First write ##\phi(x,t)## as its transform

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx} \widetilde{\phi}(p,t) \, \mathrm{d}p##

which I then plug into the PDE in the question to get:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4} \, \mathrm{d}p = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}(p^4-p^2) \widetilde{\phi}(p,t) \, \mathrm{d}p##

Which neatens to:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\Big[\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}-(p^4-p^2) \widetilde{\phi}(p,t)\Big] \, \mathrm{d}p = 0##

which can only hold if the part in the square bracket obeys

##\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}=(p^4-p^2) \widetilde{\phi}(p,t)##

So this is where I am now. I've found a differential equation obeyed by ##\widetilde{\phi}(p,t)##, however it's a PDE, not an ODE. So I'm either mistaken with my understanding of what the precise definition of an ODE is, or I've gone wrong with my logic in the working out.

To be honest the wording of the question itself seem a touch ambiguous. Is it even possible for a multi-variable function to be part of an ODE?

Any help would be appreciated, thanks.