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 $T'=-\lambda \,T$.

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 $n=1,\ldots ,n$: $\lambda _{n}=n^{2}$ (eigenvalues); $X_{n}(x)=\sin {\left(n\,x\right)}$ (eigenfunctions)

Hence the separated solutions are of the form

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

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

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

$f(x)=\sum _{n=1}^{\infty }b_{n}\,\sin {\left(n\,x\right)}$

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

$f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {\left(n\,x\right)}+b_{n}\,\sin {\left(n\,x\right)}$

This is the Fourier Series

Fourier found that

$a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)\,\mathrm {d} x$

For $n\geq 1$,

• $a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\,\cos {\left(n\,x\right)}\,\mathrm {d} x$
• $b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\,\sin {\left(n\,x\right)}\,\mathrm {d} x$

Orthogonality Relations

• $\int _{-\pi }^{\pi }\cos {\left(n\,x\right)}\,\cos {\left(m\,x\right)}\,\mathrm {d} x={\begin{cases}2\pi &n=m=0\\\pi &n=m\geq 1\\0&n\neq m\end{cases}}$
• $\int _{-\pi }^{\pi }\sin {\left(n\,x\right)}\,\sin {\left(m\,x\right)}\,\mathrm {d} x={\begin{cases}\pi &n=m\\0&n\neq m\end{cases}}$
• $\int _{-\pi }^{\pi }\cos {\left(n\,x\right)}\,\sin {\left(m\,x\right)}\,\mathrm {d} x=0$. 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 $f(x)$:

${\begin{aligned}f(x)&=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {\left(n\,x\right)}+b_{n}\,\sin {\left(n\,x\right)}\int _{-\pi }^{\pi }f(x)\,\mathrm {d} x&=a_{0}\,\int _{-\pi }^{\pi }\,\mathrm {d} x+\sum _{n=1}^{\infty }a_{n}\int _{-\pi }^{\pi }\cos {\left(n\,x\right)}\,\cos {\left(0\,x\right)}\,\mathrm {d} x+b_{n}\int _{-\pi }^{\pi }\sin {\left(n\,x\right)}\,\sin {\left(0\,x\right)}\,\mathrm {d} x\end{aligned}}$

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

${\begin{aligned}\int _{-\pi }^{\pi }\sin {\left(m\,x\right)}\,\sin {\left(n\,x\right)}\,\mathrm {d} x&={\frac {1}{2}}\int _{-\pi }^{\pi }\left(\cos {\left((m-n)\,x\right)}-{\cancel {\cos {\left((m+n)\,x\right)}}}\right)\,\mathrm {d} x\end{aligned}}$

If $m\neq n$, we get

$\int _{-\pi }^{\pi }\cos {\left((m-n)\,x\right)}\,\mathrm {d} x=\left.{\frac {1}{m-n}}\,\sin {\left((m-n)\,x\right)}\right|_{-\pi }^{\pi }=0$

If $m=n$, then

$\int _{-\pi }^{\pi }\sin ^{2}{\left(n\,x\right)}\,\mathrm {d} x={\frac {1}{2}}\,\int _{-\pi }^{\pi }\cos {\left(0\,x\right)}\,\mathrm {d} x={\frac {1}{2}}\,\int _{-\pi }^{\pi }\,\mathrm {d} x=\pi$