\magnification=\magstep1
\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Discussion 2 -- Separation of Variables and Boundary Value
Problems}
\centerline{February 8, 2001}
\bigskip
\noindent
{\it Background}
\bigskip
Last time in class, we introduced the following ``program'' for finding
solutions of initial-boundary value problems for homogeneous 
linear PDE such as the heat and wave equations:
\bigskip
\item{1)} First, we look for ``completely separated'' solutions
of the relevant PDE:  $u(x,t) = X(x)T(t)$
\item{2)}  Among the separated solutions, we look for those
satisfying the boundary conditions,
\item{3)}  Then, to derive solutions satisfying given initial conditions
we try to form linear combinations of the solutions found in Step 2.

\bigskip
\noindent
For example, for the problem:
$$\cases{u_t = k u_{xx} & (heat equation) \cr
         u(x,0) = f(x) & (given initial temperature) \cr
         u(0,t) = u(L,t) = 0 & all $t$ (ends of wire held at zero temperature)\cr}
         \leqno(1)$$
we saw that $u(x,t) = X(x)T(t)$ solves $u_t = k u_{xx}$ if and only if
$$\cases{T'(t) = -\lambda k T(t)&\cr
         X''(x) = -\lambda X(x).&\cr}$$
This says
$$\cases{T(t) = e^{-\lambda k t} & (up to a constant multiple)\cr
         X(x) = \cases{A e^{\sqrt{-\lambda}x} + B e^{-\sqrt{-\lambda}x} & if $\lambda < 0$\cr
                       A + B x & if $\lambda = 0$\cr
                       A \cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x) & if $\lambda > 0$}\cr}$$
But the boundary conditions $u(0,t) = u(L,t) = 0$ imply $X(0) = X(L) = 0$, 
and this happens with $X(x)$ not identically zero {\it only in the
case that}
$$\sin(\sqrt{\lambda}L) = 0 \Leftrightarrow \lambda = \left({n\pi\over L}\right)^2$$
for some $n \ge 1$.  We get the corresponding separated solutions of
the heat equation and the boundary conditions:
$$u_n(x,t) = e^{-\left({n\pi\over L}\right)^2 k t} 
\sin\left({n\pi x\over L}\right)$$
If coefficients $B_n$ can be found to make
$$\sum_{n=1}^\infty B_n \sin\left({n\pi x\over L}\right) = f(x)\leqno(2)$$
(the initial temperature), then
$$u(x,t) = \sum_{n=1}^\infty B_n u_n(x,t) = 
\sum_{n=1}^\infty B_n e^{-\left({n\pi\over L}\right)^2 k t} 
\sin\left({n\pi x\over L}\right)\leqno(3)$$
is the solution of (1) that we are looking for!  (Substitute $t = 0$
and compare with (2).)
\bigskip
\noindent
Some terminology:
\bigskip
\item{a)} The values $\lambda = \left({n\pi\over L}\right)^2$
are called the {\it eigenvalues} of the 1D BVP $X''(x) = -\lambda X(x)$
with $X(0) = X(L) = 0$.  $X(x)$  {\it  satisfies the boundary
conditions only when $\lambda$ is one of these eigenvalues}.
\item{b)} The corresponding functions 
$X(x) = \sin(\sqrt{\lambda}x) = \sin\left({n\pi x\over L}\right)$
are called the {\it eigenfunctions} of the BVP.
\bigskip
Today, we want to work through several other examples of this 
separation of variables technique and find eigenvalues and
eigenfunctions of the BVP's we get.
\bigskip
\noindent
{\it Discussion Questions}
\bigskip
\item{A)}  In some simple cases, we can find appropriate $B_n$ in (2)
more or less ``by inspection''.  For instance, what is the solution
of the problem (1) if $L = \pi$, and 
$$f(x) = 3\sin(2x) - 4\sin(5x)?$$
Find any one set of coefficients $B_n$ that ``works''; we'll ask
whether there is more than one such set of coefficients later.
\bigskip
\item{B)}  Next, consider the following problem
$$\cases{u_t = k u_{xx} & (heat equation) \cr
         u(x,0) = f(x) & (given initial temperature) \cr
         u_x(0,t) = u_x(L,t) = 0 & all $t$ (ends of wire {\it insulated})\cr}
         \leqno(4)$$
What are the eigenvalues and eigenfunctions for (4)?  (You'll need to 
follow what we did for the boundary conditions $u(0,t) = u(L,t) = 0$;
but there will be differences.)  What does the analog of equation (3)
look like here -- the linear combination giving a solution of (4).
\bigskip
\item{C)}  The same techniques also apply to BVP's for the 
wave equation.  Consider
$$\cases{u_{tt} = k u_{xx} & (wave equation) \cr
         u(x,0) = f(x) & (given initial displacement) \cr
         u_t(x,0) = g(x) & (given initial velocity)\cr
         u(0,t) = u(L,t) = 0 & all $t$ (ends of string fixed at zero displacement)\cr}
         \leqno(6)$$
What are the 2 ODE's we get by separating variables here?  What are
the eigenvalues and eigenfunctions for the equation for $X(x)$,
with $X(0) = X(L) = 0$?  Show that the corresponding solutions
of the wave equation and the boundary conditions can be
written as
$$u_n(x,t) = \left(A_n\cos\left({n\pi ct\over L}\right) + 
B_n\sin\left({n\pi ct\over L}\right)\right)\sin\left({n\pi x\over L}\right)\leqno(7)$$
To get a (possibly infinite) linear combination of these
$u_n(x,t)$ to satisfy the initial conditions, $A_n$ and $B_n$ must
satisfy equations like (2) above.  What do these look like here?
\bigskip
\item{D)}  Note that the solutions (7) of the wave equation represent strings 
that keep oscillating in an exactly periodic fashion forever.  In the real 
world, that never happens because of additional forces that we ignored in 
the derivation of the wave equation -- e.g. air resistance, 
frictional forces at the ends, etc.  In this problem, we will
study a PDE modeling a string moving in the presence of damping 
force(s) proportional to the string's velocity:
$$u_{tt} = c^2 u_{xx} - \beta u_t$$
where $\beta > 0$.  We want to solve boundary value
problems of the form
$$\cases{u_{tt} = c^2 u_{xx} - \beta u_t & \cr
         u(0,t) = 0 & \cr
         u(L,t) = 0 & \cr
         u(x,0) = f(x)\cr
         u_t(x,0) = g(x) & \cr}$$
\medskip
\itemitem{1)} Show that if we apply the separation 
of variables method, looking for separated solutions 
$u(x,t) = X(x)T(t)$, then we get the ordinary differential equations
$$T''(t) + \beta T'(t) + \lambda c^2 T(t) = 0$$
and 
$$X''(x) = -\lambda X(x)$$
\itemitem{2)}  Show that in order to satisfy the boundary conditions
at $x = 0, x = L$, $\lambda = \left({n\pi\over L}\right)^2$ and
$X(x) = \sin\left({n\pi x\over L}\right)$ for some $n \ge 1$.
\itemitem{3)}  Suppose that $\beta$ is relatively small -- 
$$\beta < {2\pi c\over L}.$$
What do the solutions of the equation for $T$ look like in this 
case?
\bigskip
\noindent
{\it Assignment}
\bigskip
Group writeups due Tuesday, February 20.
\bye