# Can Laplace's equation be solved using Fourier transform instead of Fourier series?

Sorry for the long text, but I am unable to make my question more compact.

Any periodic function can be Fourier expanded. Usually, they say in mathematical physics books, if the function is not periodic we use Fourier transform which is more general than Fourier series expansion.

If Fourier transform is more general, cannot we use it to expand a periodic functions as well? Why periodic functions in textbooks are only Fourier expanded but not Fourier transformed?

More specifically, the boundary value problems that we solve in electromagnetism (like in chapter 3 of Griffiths) in which for example some potential is specified on the boundary of some region and we want to find the potential inside that region, this problem is usually solved by separation of variable then eventually applying Fourier series expansion to fit the boundary conditions. Those problems are never solved using Fourier transform, why is that? is it because that in Fourier series expansion one has control on truncating the series to whatever accuracy one wants whereas for Fourier transform one cannot do that? or is it an issue of convergence?

If both are viable there must be some criteria on using one over the other!

If one can point out a reference in which Laplace's equation is solved once with Fourier series and once with Fourier transform that will be greatly appreciated.

The Fourier transform of a periodic function is a delta function at every integer position with coefficient equal to the corresponding Fourier series value. You can show this by multiplying the function by a very wide Gaussian and taking the limit. The mathematical theory is made rigorous in the subject of tempered distributions.

• I see. But why solving Laplace's equation in finding the electric potential over some domain is done via Fourier series and not Fourier transform? – Revo Oct 24 '11 at 21:59
• @Revo: it is traditional to use the smallest usable domain for solving the equation. You can use Fourier transforms to do whatever you do with Fourier series, but then you are computing with the coefficients of the delta-functions. The methods are identical. Since any periodic function is also periodic with twice the period, you can also ask the question of why you use the fundamental period, and not twice the fundamental period, for doing the Fourier series. It's just a natural convention. – Ron Maimon Oct 24 '11 at 22:58
• This is strange, why nobody in books never said that this is a convention? It is spelled out explicitly that there are 2 kinds of electric charges and whatever you call negative or positive is a convention. Double standards ! – Revo Oct 24 '11 at 23:17
• @Revo: This is such a natural convention that nobody even bothers to think of it as a convention--- why would you be so perverse as to use seven times the period as your period? It's ridiculous! When doing mathematics, one has to make such choices all the time, and one has to develop a nose for the natural conventions. For exmaple, some sequences are labelled starting at 0, and some starting at 1, but nobody labels a sequence starting at 11: $A_{11}, A_{12},A_{13},...$ without a good reason. Why not? Some things are better left to the reader to figure out. – Ron Maimon Oct 24 '11 at 23:31

For the Fourier transform of a function to exist, its absolute value must be integrable, $\int\limits_{-\infty}^\infty |f(x)|\mathrm{d}x<\infty$. The absolute value of a periodic function is not integrable on an infinite domain, so no Fourier transform. [To enjoy the full power of Fourier analysis, the function should be square integrable, $\int\limits_{-\infty}^\infty |f(x)|^2\mathrm{d}x<\infty$.]

For the Fourier expansion of a periodic function, the function has to be integrable on the finite domain of one period of the function, $\int\limits_0^L |f(x)|\mathrm{d}x<\infty$, instead of on the whole real line, which many or most of the periodic functions one meets in electromagnetism problems will be.

So, the difference between a Fourier transform and a Fourier expansion of a periodic function is that the integration is on an infinite domain, respectively a finite domain.

Fourier transforms/expansions are well suited to rectilinear coordinate systems, but they are generally less well-suited to problems in which the boundary conditions pick out curved coordinate systems. Fourier analysis is nonetheless often usable as a first approximation, as when electromagnetic fields are directed along a curved wave-guide.

• As Ron points correctly, the Fourier transform exists in a distributional sense. This is not just pedantry - inability or unwillingness to employ distributional methods would make most treatments of physics unintelligible. – BebopButUnsteady Oct 25 '11 at 1:13
• @BebopButUnsteady, 's true. Distributions move us to a significantly higher mathematical level, but it's OK to use them. – Peter Morgan Oct 25 '11 at 3:11
• @PeterMorgan I am just trying to understand what are the cases in which one can use Fourier series and the cases when one can use Fourier transform to solve Laplace's equation, tricks of the trade or best practices say (because so far I have seen only Fourier series being used in standard textbooks. If one can point out an example or a reference where both are used on the same problem that will be greatly appreciated). There must be advantages and disadvantages for each in different situations, it cannot be all merely due to conventions as it was said in other posts by Ron. – Revo Oct 25 '11 at 18:48
• @BebopButUnsteady Sorry for bothering but could you please clarify your last sentence, my English is not perfect :) – Revo Oct 25 '11 at 18:50
• @Revo One trouble with distributions is that novices use them as if they're functions. But we can't, in general, multiply distributions, so they can't be used in nonlinear problems carelessly. en.wikipedia.org/wiki/Generalized_function has something on a few of the ways people have addressed this, but it's not for the faint-hearted. One doesn't have to worry about such matters if one sticks to Fourier series when there are natural or periodic boundary conditions. When the boundary is at infinity, one uses distributions, carefully, as needed. – Peter Morgan Oct 25 '11 at 19:10

You probably can answer this question yourself. You know that any periodic function can be expanded in a Fourier series. If you Fourier transform said series, what do you get?

Hint:

\begin{equation} \int e^{inx}e^{-ipx}dx = 2\pi\delta(p-n) \end{equation}

• Sorry but that does not answer any of my questions – Revo Oct 24 '11 at 21:33