# Real-valued 2D Fourier series?

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?

-
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

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 $n,m,=1,\dots,5$:

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.

-
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. – user3728501 Dec 21 '13 at 14:19

The full real-valued 2D Fourier series is: \begin{align} f(x, y) & = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\alpha_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\beta_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\gamma_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\delta_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ \end{align}

The coefficients are found with: $$\alpha_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \beta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \gamma_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \delta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy$$ \begin{align} \text{Where } \kappa & = 1 \text{ if } n = 0 \text{ and } m = 0 \\ & = 2 \text{ if } n = 0 \text{ or } m = 0\\ & = 4 \text{ if } n> 0 \text{ and } m > 0 \end{align} Example plot

-

@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!

-
Yes, it was a typo (I had left off the squares in the denominator). Fixed now. – JohnD Mar 6 '13 at 4:48