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For a (well-behaved) one-dimensional function $f: [-\pi, \pi] \rightarrow \mathbb{R}$, we can use the Fourier series expansion to write $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n\sin(nx) \right)$$

For a function of two variables, Wikipedia lists the formula

$$f(x,y) = \sum_{j,k \in \mathbb{Z}} c_{j,k} e^{ijx}e^{iky}$$

In this formula, $f$ is complex-valued. Is there a similar series representation for real-valued functions of two variables?

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Substitute $e^{i\omega} = \cos\omega + i\sin\omega$ and $c_{j,k} = a_{j,k} + ib_{j,k}$ in the formula you get from Wikipedia, and look only at the real value of the result. The formula gets a bit unwieldy due to the 4 $\sin\cos$ combinations you get, but it works... –  fgp Oct 12 '12 at 15:06
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3 Answers

up vote 7 down vote accepted

Yes! And these types of expansions occur in a variety of applications, e.g., solving the heat or wave equation on a rectangle with prescribed boundary and initial data.

As a specific example, we can think of the following expansion as a two dimensional Fourier sine series for $f(x,y)$ on $0<x<a$, $0<y<b$: $$ f(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty c_{nm}\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right), \quad 0<x<a,\ 0<y<b, $$ where the coefficients (obtained from the same type of orthogonality argument as in the 1D case) are given by \begin{align} c_{nm}&={\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy\over \int_0^b \int_0^a \sin^2\left({n\pi\, x\over a}\right)\sin^2\left({m\pi\, y\over b}\right)\,dx\,dy}\\ &={4\over a b}\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy, \quad n,m=1,2,3,\dots \end{align}

For example, the picture below shows (left) the surface $$f(x,y)=30x y^2 (1-x)(1-y)\cos(10x)\cos(10y), \quad 0<x<1,\ 0<y<1,$$ and a plot of the two dimensional Fourier sine series (right) of $f(x,y)$ for $n,m,=1,\dots,5$:

Mathematica graphics

Finally, keep in mind that we are not limited just to double sums of the form sine-sine. We could have any combination we like so long as they form a complete orthogonal family on the domain under discussion.

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This representation seems to be valid only if the values of the function on the boundary of the rectangle are zero. –  Beni Bogosel Nov 13 '13 at 11:59
    
Yes, that's why I said, "As a specific example..." The sine functions used there are the eigenfunctions obtained when solving the heat equation on a rectangle where zero boundary conditions are specified. –  JohnD Nov 14 '13 at 0:47
    
What about the case where cosine terms are required? This answer is useless because it does not address the general case. –  Edward Bird Dec 21 '13 at 14:19
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@JohnD only details about the coefficients. The correct formula is:

$$c_{n,m} = \frac{\int_{0}^a \int_0^b f(x,y)sin(\frac{n\pi x}{a})sin(\frac{n\pi y}{b})dxdy}{{\int_{0}^a \int_0^b sin^2(\frac{n\pi x}{a})sin^2(\frac{n\pi y}{b})dxdy}{}}$$

the impression that $c_{n,m}$ is $1$. That's not true. Cheers!

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Yes, it was a typo (I had left off the squares in the denominator). Fixed now. –  JohnD Mar 6 '13 at 4:48
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By Euler's identity $e^{i\varphi}=\cos{\varphi}+i\sin{\varphi}$ trigonometrical Fourier series expansion $$f(x) \sim \dfrac{a_0}{2} + \sum\limits_{n=1}^{\infty} \left( a_n \cos(nx) + b_n\sin(nx) \right)$$ may be easily transformed into exponential form $$f(x) \sim \sum\limits_{n=-\infty}^{\infty}{c_n{e^{inx}}},$$ and vice versa, where $c_k=\dfrac{1}{2\pi}\int\limits_0^{2\pi}f(x){e^{-ikx}} \, dx$ are complex Fourier coefficients. Function $f$ may be real- or complex-valued.

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