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 \centerline{\large Math.\ 401, Sec.\
501\hfill    Spring, 2006}
 \bigskip
\centerline{\bfseries\large Homework 13, due May 2}
\medskip
{\narrower {\bf Everybody} should do the first problem, but 
not turn it in.  The rest is an optional assignment for extra 
credit; turn in at most two of the problems.
} \bigskip

 
\begin{enumerate}

\item  {\it [Logan, p.\ 159,  Ex.\ 1.3]\/}\quad
Determine in which regions each equation is hyperbolic,
elliptic, or parabolic.
Subscripts indicate partial differentiation with respect to the
indicated variables.

\begin{enumerate}
\item\quad $t\,u_{tt} + u_{xx} =0$
\item\quad $u_{tt} - u_{xx}=0$
\item\quad $u_{tt} + (1+x^2) u_x -u_t = e^t$
\item\quad $u_{xx} + u_{yy} = f(x,y)$
\end{enumerate}

 
                                                                                
                                                                                
\item  Let $r$ and $\theta$ be polar coordinates, defined in the
usual way:
$$x= r \cos \theta, \quad y = r\sin\theta.$$
Show that
$$\nabla^2u \equiv {\partial^2u\over \partial x^2} +
{\partial ^2u\over \partial y^2} =
{\partial ^2u\over\partial r^2} +\frac1r\,
{\partial  u\over\partial r}
+ \frac1{r^2}\,{\partial ^2u\over \partial \theta^2}\,.$$
{\sl Hint:\/} First show that (acting on any function)
$${\partial \over\partial x} = \cos \theta\, {\partial
\over\partial r} -{\sin \theta\over r}\, {\partial
\over\partial \theta }\,, $$
$${ \partial\over\partial y} = \sin \theta\, {\partial \over
\partial r} +{\cos \theta\over r}\, {\partial\over\partial\theta}
\,.$$
{\sl Hint for hint:\/} The calculations are easier if you use 
some kind
of implicit differentiation.  For example, a quick way to show 
that
$${\partial\theta \over \partial  y} = {\cos \theta \over r}$$
is to differentiate the equation
$$\tan \theta = \frac yx$$
with respect to $y$ (with $x$ fixed) and perform some necessary 
algebra on the
result.  Remember that (for example) a $y$ derivative with $x$ 
fixed is not
the same as a $y$ derivative with $r$ fixed!
 
 
 \item {\it [Logan, p.\ 304, Ex.\ 3.13]\/} \quad
In three dimensions the wave equation is
$${\partial ^2u\over \partial t^2} -c^2 \nabla^2 u =0$$
where $\nabla^2$ is the Laplacian.
For waves with spherical symmetry,
$u=u(r,t)$ and
$$\nabla^2 u = {\partial ^2u\over \partial r^2} + \frac2r \,
{\partial u\over \partial r}\,.$$
By introducing the variable $U = ru$,
show that the general solution for the spherically symmetric
wave equation is
$$ u = \frac1r \, f(r-ct) + \frac1r \, g(r+ct).$$
 






\item  
\begin{enumerate}
 \item Solve the wave equation in $\mathbf R^2$ by Fourier 
 transforms, and rearrange the result into the form
 $$u(x,t) = \int_{-\infty}^\infty dy
 \left[V(x-y, t) f(y) + W(x-y,t)g(y)\right],$$
 where $f$ and $g$ are the initial data (in the notation of the
printed notes), and $W$ and $V$ are certain integral expressions.

 \item What identities for the Fourier transforms of delta
functions and step functions do you need to assume in order to get
your answer to agree with the one on p.\ 137 of the notes?
 Try to boil it down to simple formulas for functions (or
distributions) of {\sl one\/} variable, not three.
\end{enumerate}

 \item  Solve by the d'Alembert method
 (see pp.\ 137--140 of  notes)
  the wave equation for $0<x<\infty$
with Neumann boundary condition (${\partial 
u\over\partial x}(0,t)=0$) and
 initial data
 $$u(x,0) \equiv f(x) = e^{-100(x-10)^2}, \qquad
{\partial u\over \partial t}(x,0) \equiv g(x) =0.$$
 Sketch in space-time the pulses and the paths they follow.
 (The sketch should include at least the time interval
 $0\le t\le 15$.)

\item Same as the previous problem, but with Dirichlet boundary 
condition ($u(0,t)=0$) and initial data
 $$u(x,0) \equiv f(x) =0, \qquad 
{\partial u\over \partial t}(x,0) \equiv g(x)=
 x\,e^{-100(x-10)^2}.$$

                                                                                
\item Solve this problem in a quarter-disk 
($0<\theta<\frac{\pi}2$, $0\le r<1$):
$${\partial ^2u  \over \partial r^2} +\frac1r\,
{\partial u\over \partial r} + \frac1{r^2} {\partial ^2u\over 
\partial\theta^2} ={\partial u\over \partial t}\,,$$
$$u(t,r,0)=0=u(t,r,\pi/2), \qquad
{\partial u\over\partial r}(t,1,\theta) = 0,$$
$$u(0,r,\theta) = g(r,\theta).$$


\end{enumerate}
\end{document}