Fourier series

A well-known example of orthonormal functions is the set of sine and cosine functions $\sqrt{\frac{1}{a}} \sin( n \pi x/a)$ and $\sqrt{\frac{1}{a}} \cos(n \pi x/a)$ with $1 \le n$ for sine functions and $0 \le n$ for cosine functions ⁸ defined in interval $[0,a]$. The expansion

$$F(x) = A_0 + \sum_n ( A_n \cos( n \pi x/a) + B_n \sin( n \pi x/n)$$

is the expansion of the function $F(x)$ in Fourier series and the coefficients $A_n$'s and $B_n$'s are the Fourier coefficients. A generalization of the Fourier series to more than one dimension is obvious.

As a concrete example of one of the application of orthogonal functions, consider the solution of a problem in which the potential $\phi(\vec r)$ for a charge distribution $\rho(\vec r)$is to be determined. The charge distribution is placed in a conducting cubic box of unit length and the sides of the box are grounded. That is, the potential on the surface of the box is zero. The charge distribution is confined in the box so $\rho(\vec r)$ also vanishes on the sides of the box. We need to solve the Poisson's equation

$$\nabla^2 \phi(\vec r) = - 4 \pi \rho(\vec r)$$ (62)

We expand the potential and the charge density in Fourier series. Here $a=1$ in all dimensions⁹ and because the potential and the charge density vanishes at the boundaries, only sine functions are required. We therefore have

$$\rho (\vec r) = \sum_{i,j,k} \rho_{i.j.k} \sin(i \pi x) \sin(j \pi y) \sin(k \pi z)$$ (63)

$$\mathrm {and}$$

$$\phi(\vec r) = \sum_{i,j,k} \phi_{i.j.k} \sin(i \pi x) \sin(j \pi y) \sin(k \pi z)$$ (64)

The Fourier coefficients $\rho_{i,j,k}$ are determined from the known charge distribution. Substituting these Fourier expansions in the Poisson's equation, we get an algebraic equation for the Fourier coefficients of the potential

$$\phi_{i,j,k} = \frac{ 4 \pi \rho_{i,j,k}}{\pi^2 ( i^2 + j^2 + k^2)}$$ (65)

and the potential is

$$\phi(\vec r) = \sum_{i,j,k} \frac{ 4 \rho_{i,j,k}}{\pi ( i^2 + j^2 + k^2)} \sin(i \pi x) \sin(j \pi y) \sin(k \pi z)$$ (66)

For a point charge placed at the center of the box, $\rho(\vec r) = \delta(x-0.5) \delta(y-0.5) \delta(z-0.5)$ and

$$\rho_{i,j,k} = \sin(i \pi /2) \sin(j \pi /2) \sin(k \pi /2)$$

$$= (-1)^{(i+j+k-3)/2} \;\;\;\;\; \mathrm {for} \;\;\; i, j, k \;\;\; \mathrm {odd} \;\;\; \mathrm {integer}$$ (67)

With this the expression for the potential is

$$\phi(\vec r) = \sum_{i,j,k \; \mathrm {odd} \; \mathrm {integer}}... ...(i+j+k-3)/2}}{\pi ( i^2 + j^2 + k^2)} \sin(i \pi x) \sin(j \pi y) \sin(k \pi z)$$ (68)

Note that the potential has correct boundary condition by construction. The computed potential for $i,j,k \le n$ with $n$ of 11, 31, 101 is shown in Fig(2.6.1). One finds that the convergence is not very bad for the distance away from the position of the charge. This is because the potential has a singularity at the position of the charge and it is not easy to reasonably reproduce that with small number of terms.

Figure 9: Calculated potential for a point charge at the center of grounded conducting cubic box of unit length. The result is shown for different number of terms. The left panel shows the graph for $y=0.25$ and $z=0.4$ and the right panel is for $y=0.5$ and $z=0.5$. The result for $n=11, 31, 101$ are shown with red, green and blue lines.In the left panel the blue and green lines overlap.