# True of false?

1. Apr 25, 2005

### quasar987

Every "spacially periodic" function [i.e. s.t there exist P s.t. f(x+P,t) = f(x,t)] of the form f(x,t) = X(x)cos(wt) is a solution of the wave equation.

Last edited: Apr 25, 2005
2. Apr 25, 2005

### James R

True..........

3. Apr 25, 2005

### jdavel

quasar,

False.

For one thing, if f(x,t) is periodic (and it doesn't have to be periodic), then there's a strict relation between the periodicity in x and in t. In other words, in an example using your format, if f(x,t) = sin(kx)cos(wt), then w/k = v.

It's not too hard to see what the wave equation is saying about f(x,t) if you think about it. Within a multiplicative constant, the two partial derivatives are the same. That means f has to depend on x and t in very similar ways. I think the most general form for f(x,t) that satisfies the W.E. is f(kx-wt). Certainly any function of that form will work. Although that's not really what you were asking.

4. Apr 25, 2005

### jdavel

Uh oh!

James R., I posted before I saw yours. Why do you say it's true?

5. Apr 26, 2005

### James R

Hmm... Here are my thoughts. I think I might have changed my mind!

At first, I thought that any function of the form f(x,t) = X(x)cos wt will describe a standing wave, and so it is necessarily a solution of the wave equation.

But then...

The wave equation, in 1 dimension, is:

$$\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{v^2 \partial t^2} = 0$$

where v is a constant (the wave speed).

Using the function given, we have:

$$\frac{\partial^2 f}{\partial x^2} = \frac{d^2 X}{dx^2}\cos \omega t$$
$$\frac{\partial^2 f}{\partial t^2} = -\omega^2 X(x)\cos \omega t$$

Therefore, we require:

$$\frac{d^2 X}{dx^2} + \frac{X}{v^2} = 0$$

This restricts X(x) to harmonic functions of the form:

$$X(x) = A \sin kx + B \cos kx$$

where A and B are arbitrary constants, but k is restricted:

$$k=\omega / v$$

So, it seems that the most general functions of the given form which satisfy the wave equation are:

$$f(x,t) = [A \sin kx + B \cos kx]\cos \omega t$$

with the above restriction on k.

Does that seem right?

6. Apr 26, 2005

### Galileo

You don't you just work out the simplest case: $f(x,t)=\cos(wt)$.
We clearly have $f(x+P,t)=f(x,t)$ for any t (and any P).
The wave equation clearly doesn't hold in this case (unless $\omega=0$, but I understand $\omega$ can be arbitrary).