Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have some function $V(x,y)$ which is periodic in x with period L. I wish to expand $V(x,y)$ in terms of a fourier sine (for simplicity) series in $x$, is it always the case that I may write the following?

$$V(x,y) = \sum_{n = 1}^{\infty} a_n \sin(\frac{n\pi x}{L})f_n(y)$$

Where the $f_n(y)$ are whatver functions of $y$ needed to satisfy the equality -- do they need to have a particular form or any properties other than continuity?

It seems to me that we might require a certain type of function $V(x,y)$ to do this, does anyone know what the neccessary and sufficient conditions on $V(x,y)$ are to be able to expand in this way?

share|improve this question
add comment

1 Answer

Not really. Your expansion will, rather, take the form

$$V(x,y) = \sum_{n=1}^{\infty} \: \sum_{m=1}^{\infty} a_{nm} \sin{\left ( \frac{n \pi x}{L}\right)} f_m(y)$$

$f_m$ will be defined by boundary conditions on the $y$ boundaries.

share|improve this answer
1  
You could take $\phi_n(y) = \sum_{m=1}^\infty a_{nm} f_m(y)$, then $V(x,y) = \sum_{n=1}^\infty \sin(\frac{n \pi x} {L}) \phi_n(y)$. –  copper.hat Apr 11 '13 at 20:07
1  
Yes, but you still have to find the $a_{nm}$ and the $f_m$ from the BC's. –  Ron Gordon Apr 11 '13 at 20:20
    
@RonGordon. So I am always going to be able to separate the variables in the sum this way? Is it true that if we couldn't write $V$ as you/I have written it then $V$ would simply not have a Fourier Series in $x$? –  user27182 Apr 11 '13 at 20:41
    
You can separate like this when you have rectangular geometry and separable boundary conditions. (If you have BC's that have mixes of values over both directions, then you may not be able to have this representation.) On your 2nd question you are right, but it doesn't mean no FS period. You could have a FS in, say, $xy$ in some bizarre scenario. –  Ron Gordon Apr 11 '13 at 20:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.