MATH 414 Lecture 6

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Fourier Series

Emerged from Fourier's theory of heat.

Suppose we have heat distribution over a bar of length $\pi$ such that $u(0,t)=0$ to $u(\pi ,t)=0$, where $u$ is a function of position $x$ and time $t$: $u(x,t)$.

Heat flow equation: $\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2}$

Our boundary conditions are that $u(0,t)=u(\pi ,t)=0$

Separation of Variables

Keep the homogeneous equations

Then look for separation solutions (normal modes, esp. in vibrations problems)

Assume $u(x,t)=X(x)T(x)$ (hence the "separation" in the name)

Get equations of the form: $X^{″}+\lambda X=0$, $X(0)=X(\pi )=0$, and $T^{\prime }=-\lambda T$.

1. the last is easy to solve: $T(t)=A{\mathrm{e}}^{-\lambda t}$
2. The first two set up an eigenvalue problem. There are solutions for only certain values of $\lambda$ for $n=1,\dots ,n$: ${\lambda }_n=n^2$ (eigenvalues); $X_n(x)=\sin \left(nx\right)$ (eigenfunctions)

Hence the separated solutions are of the form

${\mathrm{e}}^{-n^{2}t}\sin \left(nx\right)$

We expect $u(x,t)=\sum _{n=1}^{\mathrm{\infty }}{b}_n{\mathrm{e}}^{-n^{2}t}\sin \left(nx\right)$

Initial conditions $u(x,0)=f(x)$ implies

$f(x)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \left(nx\right)$

This series shall henceforth be called the sine series

Problems

1. How do we get the $b_n$'s?
2. Does the series represent $f(x)$ in any sense?
3. Does the series $u(x,t)=\sum _{n=1}^{\mathrm{\infty }}{b}_n{\mathrm{e}}^{-n^{2}t}\sin \left(nx\right)$ solve the heat flow problem?

The Fourier Series

Let's look at heat flow in a ring of radius $1$. We get the same sort of equations.

Given $f(x)$, we need a series of the form

$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \left(nx\right)+b_n\sin \left(nx\right)$

This is the Fourier Series

Fourier found that

$a_0=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\mathrm{d}x$

For $n\ge 1$,

• $a_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos \left(nx\right)\mathrm{d}x$
• $b_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin \left(nx\right)\mathrm{d}x$

Orthogonality Relations

• $\underset{-\pi }{\overset{\pi }{\int }}\cos \left(nx\right)\cos \left(mx\right)\mathrm{d}x=\{\begin{array}{ll}2\pi & n=m=0\\ \pi & n=m\ge 1\\ 0& n\ne m\end{array}$
• $\underset{-\pi }{\overset{\pi }{\int }}\sin \left(nx\right)\sin \left(mx\right)\mathrm{d}x=\{\begin{array}{ll}\pi & n=m\\ 0& n\ne m\end{array}$
• $\underset{-\pi }{\overset{\pi }{\int }}\cos \left(nx\right)\sin \left(mx\right)\mathrm{d}x=0$. This is because $\cos x$ is even, $\sin x$ is odd, and the product of an even function and an odd function is odd. So by symmetry the integral is $0$.

Using these relations, we can integrate the Fourier series definition of $f(x)$:

$\begin{array}{rlr}f(x)& =a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \left(nx\right)+b_n\sin \left(nx\right)\underset{-\pi }{\overset{\pi }{\int }}f(x)\mathrm{d}x& =a_0\underset{-\pi }{\overset{\pi }{\int }}\mathrm{d}x+\sum _{n=1}^{\mathrm{\infty }}{a}_n\underset{-\pi }{\overset{\pi }{\int }}\cos \left(nx\right)\cos \left(0x\right)\mathrm{d}x+b_n\underset{-\pi }{\overset{\pi }{\int }}\sin \left(nx\right)\sin \left(0x\right)\mathrm{d}x\end{array}$

We want to show that $\underset{-\pi }{\overset{\pi }{\int }}\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x=\pi {\delta }_{m,n}$

$\begin{array}{rl}\underset{-\pi }{\overset{\pi }{\int }}\sin \left(mx\right)\sin \left(nx\right)\mathrm{d}x& =\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}(\cos ((m-n)x)-\cancel{\cos ((m+n)x)})\mathrm{d}x\end{array}$

If $m\ne n$, we get

$\underset{-\pi }{\overset{\pi }{\int }}\cos ((m-n)x)\mathrm{d}x={\frac{1}{m-n}\sin ((m-n)x)|}_{-\pi }^{\pi }=0$

If $m=n$, then

$\underset{-\pi }{\overset{\pi }{\int }}{\sin }^2\left(nx\right)\mathrm{d}x=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}\cos \left(0x\right)\mathrm{d}x=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}\mathrm{d}x=\pi$