\magnification=\magstep1
\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Selected Solutions -- Midterm Problem Set}
\centerline{March 12, 2001}
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\noindent
I. B. As always, the value of a solution of a first order linear
PDE with prescribed values along some curve in the plane 
is only uniquely determined on the subset of the plane consisting
of all points on characteristics {\it that intersect} the 
curve where the values are prescribed.  Here the characteristics
are the curves $y = \arcsin(x) + c$ for all real $c$.  With
the condition $u(x,0) = \sqrt{1-x^2}$, we have to determine
which characteristics intersect the $x$-axis (the given curve).
By considering the graph of the $\arcsin$ function, we see that
only the characteristics with $-\pi/2 \le c \le \pi/2$ intersect
the $x$-axis here.  (If $c < -\pi/2$, the graph is shifted
``too far down'' along the $y$-axis; if $c > \pi/2$ it is
shifted ``too far up''.)  Hence the region where the solution
is uniquely determined is 
$$\{(x,y) : -1\le x \le 1, \arcsin(x)-\pi/2 \le y \le \arcsin(x)+\pi/2\}$$
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\noindent
II. A.  Using D'Alembert's solution of the wave equation:
$$u(x,t) = {1\over 2}(f(x-ct) + f(x+ct)) + {1\over 2c}\int_{x-ct}^{x+ct} g(s)\ ds$$
we assume that if $f(x),g(x) = 0$ for all $|x| > B$.   Then
$|x| > B + ct$ is the same as $x < -(B+ct)$ or $x > B+ct$.  From this 
we get $x + ct < -B$ (which implies $x - ct < -B$ too
since we take $t \ge 0$) or $x - ct > B$ (which implies
$x + ct > B$ too).  Therefore $f(x-ct) = f(x+ct) = 0$.
Similarly the interval $[x-ct,x+ct] \subset (-\infty,-B]$ in the 
first case, and $[x-ct,x+ct] \subset [B,\infty)$ in the second
case.  Therefore the interval of integration is contained
in the range where $g$ is known to be zero.  Hence
$\int_{x-ct}^{x+ct} g(s)\ ds = 0$.  
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\noindent
II. B.  We have
$$KE + PE = \int_{-\infty}^\infty \rho u_t^2 + Tu_x^2\ dx$$
Hence by the hint and the wave equation:
$$\eqalign{
{d\over dt}(KE + PE) &= {d\over dt} \int_{-\infty}^\infty \rho u_t^2 + Tu_x^2\ dx\cr
&=\int_{-\infty}^\infty 2\rho u_t u_{tt} + 2Tu_x u_{xt}\ dx\cr
&=2T\int_{-\infty}^\infty u_t u_{xx} + u_x u_{xt}\ dx\cr}$$
Integrating by parts in the first term ($U = u_t, dV = u_x\ dx$), we 
get 
$$\eqalign{
&=2T u_t u_x|_{-\infty}^\infty +  
\int_{-\infty}^\infty -u_{xt}u_x + u_x u_{xt}\ dx\cr
&=2T u_t u_x|_{-\infty}^\infty + 0\cr}$$
Now, by part A of the problem, if $|x| > B + ct$, we have
$u(x,t) = 0$.  Hence the limit as $x\to \pm\infty$ of 
$u$ or any of its derivatives must be zero.  Hence 
$${d\over dt}(KE + PE) = 0$$
which is what we wanted to show.
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\noindent
IV.  B.  The eigenvalues and eigenfunctions of the boundary value problem 
$$\cases{X''(x) = -\lambda X(x)\cr
         X(0) = 0\cr
         X'(L) = 0\cr}$$
come only from the case $\lambda > 0$, so 
$X = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$.
The condition $X(0) = 0$ implies $A = 0$.  Then the other
boundary condition says 
$B\sqrt{\lambda}\cos(\sqrt{\lambda}L) =0$.
So $\sqrt{\lambda}L = {(2n-1)\pi\over 2}$ for some $n\ge 1$
(that is, $\sqrt{\lambda}L$ has the form (odd integer)$\cdot \pi/2$).

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\leftline{eigenvalues: $\lambda = \left({(2n-1)\pi\over 2L}\right)^2$.}
\leftline{eigenfunctions: $\phi_n(x) = \sin\left({(2n-1)\pi x\over 2L}\right)$.}
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\noindent
IV B.  To derive the orthogonality relation
$$0=\int_0^L\sin\left({(2n-1)\pi x\over 2L}\right)\sin\left({(2m-1)\pi x\over 2L}\right)\ dx$$
when $m \ne n$, use the trig identity
$$\sin(A)\sin(B) = {1\over 2}(\cos(A - B) - \cos(A + B))$$
then integrate.
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\noindent
V.  A.  The Fourier sine coefficients of the given initial
string shape are
$$\eqalign{B_n &= {2\over L}\int_0^L f(x) \sin\left({n\pi x\over L}\right)\ dx\cr
&= {2\over L}\left[\int_0^a {hx\over a} \sin\left({n\pi x\over L}\right)\ dx
+\int_a^L {h(L-x)\over (L-a)} \sin\left({n\pi x\over L}\right)\ dx\right]\cr
&={2hL^2\over a(L-a)}\cdot \sin\left({n\pi a\over L}\right)\cdot {1\over (n\pi)^2}\cr}$$
The solution of the wave equation is
$$u(x,t) = \sum_{n=1}^\infty B_n\sin\left({n\pi x\over L}\right)
\cos\left({n\pi c t\over L}\right)$$
\bye
