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\begin{document}

{\bf Question}

The wave equation with constant speed $c>0$ is given by

$$c^2u_{xx}-u_{tt}=0.$$

\begin{description}
\item[(a)]
Classify this equation and identify whether it is elliptic,
parabolic or hyperbolic in the $(x,y)$ plane. Hence \emph{state}
the standard from of the equation in characteristic coordinates
$\xi,\eta$ and show that the general solution is

$$u(x,t)=F(x-ct)+G(x+ct)$$

for arbitrary functions $F$ and $G$.

\item[(b)]
Show that, in general, the following initial value system

\begin{eqnarray*} c^2u_{xx}-u_{tt} & = & 0\\ u(x,0) & = & f(x)\\
u_{t}(x,0) & = & g(x) \end{eqnarray*}

$$0<x<\pi,t>0$$

may be solved to give
$$u(x,t)=\ds\frac{1}{2}\{f(x+ct)+f(x-ct)\}+\ds\frac{1}{2c}\ds\int_{x-ct}^{x+ct}
ds g(s).$$

You must show clear working and carefully define any parameters
you use.

\item[(c)]
Write down the solution of the wave equation in the following
specific cases

$$\begin{array}{cll} i) & f(x)=0 & g(x)=\sin x,\\ ii) &
f(x)=\ds\frac{1}{(x+1)} & g(x)=0,\\ iii) & f(x)=0 & g(x)=1,\\ iv)
& f(x)=0 & g(x)=\ds\frac{1}{(x^2+1)}. \end{array}$$

Identify any singularities of $u(x,t)$ in the region $t>0$ and
hence comment on the validity of each solution.
\end{description}

\medskip

{\bf Answer}

$$c^2u_{xx}-u_{tt}=0$$ 2nd order linear homogeneous constant
coefficient PDE.

Discriminant: $a=c^2,\ b=0,\ c=-1$

$$b^2-a\lq\lq c "=+c^2>0\ \ \rm{everywhere}$$

Therefore hyperbolic equation everywhere

Characteristic coordinates given by:

$cdt^2-dx^2=0$

$\Rightarrow (cdt-dx)(cdt+dx)=0$

$\Rightarrow cdt-dx=0$ or $cdt+dx=0$

$\Rightarrow \ds\frac{dx}{dt}=+c\ \ \ \ds\frac{dx}{dt}=-c$

$\Rightarrow x=ct+\xi\ \ \ x=ct+\eta$

Therefore $\xi=(x-ct);\ \eta=(x+ct)$

and hyperbolic wave equation transforms to

$$u_{\xi\eta}=0$$

$$\Rightarrow u=F(\xi)+G(\eta)$$

Hence

$$u(x,t)=F(x-ct)+G(x+ct)\ \ \ F,\ G\ \rm{arbitrary}$$

\begin{description}
\item[(i)]
$f(x)=0,\ g(x)=\sin x$

\begin{eqnarray*} \Rightarrow u(x,t) & = &
\ds\frac{1}{2c}\ds\int_{x-ct}^{x+ct}\sin s\, ds\\ & = &
\ds\frac{1}{2c} [-\cos s]_{x-ct}^{x+ct}\\ & = &
\ds\frac{1}{2c}[-\cos(x+ct)+\cos(x-ct)]\\ & = &
\ds\frac{1}{2c}[-\cos x\cos ct+\sin x\sin ct\\ & & +\cos x\cos
ct+\sin x\sin ct]\\ \rm{Therefore}\ u(x,t) & = &
\un{+\ds\frac{1}{c}\sin x\sin ct} \end{eqnarray*}

Solution valid for all $t$, for all finite $x$.

\item[(ii)]
$f(x)=\ds\frac{1}{x+1},\ g(x)=0$

\begin{eqnarray*} \Rightarrow u(x,t) & = &
\ds\frac{1}{2}\left[\ds\frac{1}{(x+ct+1)}+\ds\frac{1}{(x-ct+1)}\right]\\
& = & \ds\frac{1}{2}\ds\frac{2x+2}{(1+x)^2-c^2t^2}\\ & = &
\un{\ds\frac{x+1}{(1-x)^2-c^2t^2}} \end{eqnarray*}

Solution has a singularity when $1+x=\pm ct$ so not a sensible
result.

\item[(iii)]
$f(x)=0,\ g(x)=1$

Therefore \begin{eqnarray*} u(x,t) & = &
\ds\frac{1}{2}(0+0)+\ds\frac{1}{2c}\ds\int_{x-ct}^{x+ct}ds\\ & = &
\ds\frac{1}{2c}(x+ct-x+ct)\\ & = & \un{t} \end{eqnarray*}

Solution grows with $t$. Not sensible as $t \to \infty$.


\item[(iv)]
$f(x)=0,\ g(x)=\ds\frac{1}{1+x^2}$

\begin{eqnarray*} u(x,t) & = &
\ds\frac{1}{2}(0+0)+\ds\frac{1}{2c}\ds\int_{x-ct}^{x+ct}\ds\frac{ds}{1+s^2}\\
& = & \ds\frac{1}{2c}[\arctan s]_{x-ct}^{x+ct}\\ & = &
\ds\frac{1}{2c}[\arctan(x+ct)-\arctan(x-ct)] \end{eqnarray*}

Assuming arctan defined on specified range, solution is valid flor
all $x$, for all $t$ (as $t \to \infty$) and $x$ finite.

$$u(x,t) \to
\ds\frac{1}{2c}\left[\ds\frac{\pi}{2}+\ds\frac{\pi}{2}\right]=\ds\frac{\pi}{2c}$$

(NB energy considerations if $x$ larger $\cdots$)
\end{description}

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