Exercise 0 ().

Additional Exercise - Wave Equation Solution

Exercise 0 ().

Consider the wave equation

(1)

\left\{\begin{array}[]{ll}u_{tt}(x,t)=c^{2}u_{xx}(x,t),&\left(x,t\right)\in% \left(0,2\pi\right)\times\left(0,\infty\right),\\ u(0,t)=u(2\pi,t),&t\in(0,\infty),\\ u(x,0)=f(x),&t\in(0,2\pi),\\ u_{t}\left(x,0\right)=0,&x\in\left(0,2\pi\right),\end{array}\right.

where $f$ is a smooth function on $\left[0,2\pi\right]$ with $f(0)=f(2\pi)=0$ . We will find a solution to the (1) by utilising the Fourier series (like the heat equation). We assume that we can write

u(x,t)=\sum_{n\in\mathbb{Z}}a_{n}(t)e^{inx}

with

a_{n}(t)=\frac{1}{2\pi}\int_{0}^{2\pi}u(x,t)e^{-inx}dx

and that the convergence is such that all interchanging of differentiation and integration is allowed (this will happen, for instance, if we seek a solution that is $C^{4}$ on the periodic domain).

(i)

Show that $a_{n}(t)$ satisfies the equation

$a_{n}^{\prime\prime}(t)+n^{2}c^{2}a_{n}(t)=0.$
(ii)

using he boundary conditions show that

$a_{n}(t)=A_{n}\cos\left(nct\right)$

and find an explicit expression for $\left\{A_{n}\right\}_{n\in\mathbb{Z}}$ which depends on $f$ .
(iii)

Write a solution to (1)

Solution.

(i)

Differentiating the sum and plugging it into our wave equation yields

$\sum_{n\in\mathbb{Z}}a_{n}^{\prime\prime}(t)e^{inx}=c^{2}\sum_{n\in\mathbb{Z}}% a_{n}(t)\left(in\right)^{2}e^{inx}.$

The uniqueness of the Fourier coefficients implies that $a_{n}(t)$ must solve the equation

$a_{n}^{\prime\prime}(t)=-n^{2}c^{2}a_{n}(t),$

which is the desired result.
(ii)

The solution to the above sequence of ODEs is given by

$a_{n}(t)=\begin{cases}A_{0}+B_{0}t,&$n=0$,\\ A_{n}\cos\left(nct\right)+B_{n}\sin\left(nct\right),&n\not=0\end{cases}.$

At this point we will need to use our boundary condition for $u\left(x,0\right)$ and $u_{t}\left(x,0\right)$ . Since

$u_{t}\left(x,t\right)=\sum_{n\in\mathbb{Z}}a_{n}^{\prime}(t)e^{inx}$

we conclude that $u_{t}\left(x,0\right)=0$ implies, together with the uniqueness of the Fourier coefficients, that

$a_{n}^{\prime}(0)=0,\qquad\forall n\in\mathbb{Z}.$

As

$a_{n}^{\prime}(t)=\begin{cases}B_{0},&$n=0$,\\ -ncA_{n}\sin\left(nct\right)+ncB_{n}\cos\left(nct\right),&n\not=0\end{cases}.$

we find that $B_{n}=0$ for all $n\in\mathbb{Z}$ . Thus, we have that

$a_{n}(t)=\begin{cases}A_{0},&n=0,\\ A_{n}\cos\left(nct\right),&n\not=0.\end{cases}=A_{n}\cos\left(nct\right).$

Using the fact that $u(x,0)=f(x)$ we find that

$\sum_{n\in\mathbb{Z}}a_{n}(0)e^{inx}=u\left(x,0\right)=f(x)=\sum_{n\in\mathbb{% Z}}\widehat{f}_{n}e^{inx}$

and, like before, the uniqueness of the Fourier coefficients imply that

$A_{n}=a_{n}(0)=\widehat{f}_{n},\qquad\forall n\in\mathbb{Z}.$
(iii)

Combining the previous results we conclude that our solution is given by

$u(x,t)=\sum_{n\in\mathbb{Z}}\widehat{f}_{n}\cos\left(nct\right)e^{inx}$

∎