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\centerline{Mathematics 174 -- Applied Mathematics 2}
\centerline{Discussion/Lab 4 -- Applications of Eigenfunction Expansions}
\centerline{February 29, 2000}
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\noindent
{\it Background}
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Last Thursday, using eigenfunction expansions, we saw that it
was possible to solve the following type of boundary value problem for 
the 1D heat equation with a heat source term:
$$\cases{u_t = ku_{xx} + q(x,t) &\cr
u(0,t) = 0 & \cr
u(L,t) = 0 & \cr
u(x,0) = g(x) & \cr}$$
by expanding $u(x,t)$ and $q(x,t)$ in terms of the eigenfunctions
$\phi_n(x) = \sin\left({n\pi x\over L}\right)$ for the associated
homogeneous problem (i.e. the same type of problem, but with 
$q(x,t)$ replaced by $0$).  If $u(x,t) = \sum_n a_n(t) \phi_n(x)$
and $q(x,t) = \sum_n c_n(t) \phi_n(x)$, then the coefficients 
$a_n(t)$ must satisfy the first order linear ODE's:
$${d a_n(t)\over dt} + \lambda_n k a_n(t) = c_n(t).$$
Hence by the general method for solving equations of this type from
Problem Set 1, 
$$a_n(t) = e^{-\lambda_n k t}\left(a_n(0) + \int_0^t e^{\lambda_n k t} 
c_n(t)\ dt\right)$$
The values $a_n(0)$ are determined by the expansion of the initial
temperature $g(x)$ in terms of the $\phi_n(x)$:
$$g(x) = \sum_{n=1}^\infty a_n(0) \phi_n(x).$$

Today, we want to see that this method is actually very general: it applies
in cases where the eigenfunctions are different from the sines above (and
where there is more than one space dimension), and also
to other equations (e.g. the wave equation).
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\noindent
{\it Discussion/Lab Questions}
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\item{A)}  First, let's consider a BVP for the 2D Heat Equation on 
a circular plate $0 \le r \le a$, with heat sources given by $q(r,\theta,t)$
$$\cases{
u_t = k \nabla^2 u + q(r,\theta,t) & \cr
u(a,\theta,t) = 0 & \cr
u(r,\theta,0) = \alpha(r,\theta) & \cr}\leqno(1)$$
From the associated homogenous problem ($q = 0$), we know the 
``2D eigenfunctions'' (i.e. functions $\phi(r,\theta)$ which are solutions 
of $\nabla^2 \phi = -\lambda \phi$, the equation obtained by {\it separation
of $t$ from the space variables\/}).  Since there are two space dimensions here,
we need two indices to list all of them:  
$$\phi_{mn}(r,\theta) = \cases{J_m\left({z_{mn}r\over a}\right) \cos(m\theta)&\cr
                               J_m\left({z_{mn}r\over a}\right) \sin(m\theta)&\cr}$$
We have $\nabla^2\phi_{mn} = -\lambda_{mn} \phi_{mn}$, where
$\lambda_{mn} = \left({z_{mn}\over a}\right)^2$.
Note: this way of writing things is not really satisfactory, since we are not 
distinguishing between the ``sin'' and ``cos'' forms of $\phi$ in 
the notation. {\it  However, the
exact form of $\phi_{mn}(r,\theta)$ will not be all that relevant in
what we have to do here, so let's not worry about that!}

\item{} {\it The Question:}  Suppose we know the expansions
$$q(r,\theta,t) = \sum_{m,n} c_{mn}(t) \phi_{mn}(r,\theta)$$
and 
$$\alpha(r,\theta) = \sum_{m,n} b_{mn} \phi_{mn}(r,\theta)$$
(the $b_{mn}$ are constants, since $\alpha$ does not depend on $t$ -- 
it's the initial temperature distribution).  Determine 
$a_{mn}(t)$ so that
$$u(r,\theta,t) = \sum_{m,n} a_{mn}(t) \phi_{mn}(r,\theta)$$
solves the problem (1).  (Assume all derivatives can be interchanged
with summations as necessary.)
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\item{B)}  Next we want to consider an inhomogeneous 1D wave equation:
$$u_{tt} = c^2 u_{xx} + q(x,t).$$
Solutions of this type of equation represent the vibrations of strings, in 
the presence of an {\it external driving force} $q(x,t)$ (i.e. a force
different from the tension forces exerted by the string itself).  
Examples would be forces exerted by motions of the air around
the string, forces exerted by another object such as a bow in contact 
with the string, etc.  In particular, suppose we have a string fixed at 
its endpoints, with given initial position and velocity, and
a known driving force $q$.  The displacement $u(x,t)$ satisfies
$$\cases{ u_{tt} = c^2 u_{xx} + q(x,t) &\cr
          u(0,t) = u(L,t) = 0 & \cr
          u(x,0) = f(x) &\cr
          u_t(x,0) = g(x) &\cr}\leqno(2)$$
We write the eigenfunctions of the associated homogeneous
equation as $\phi_n(x)$ as before, and write $\lambda_n$
for the associated eigenvalue.

\bigskip
\item{1)}  Following what we did with the heat equation, suppose
that
$$f(x) = \sum_n b_n \phi_n(x),\qquad g(x) = \sum_n c_n \phi_n(x)$$
(here, $b_n$ and $c_n$ are constants for each $n$) and 
$$q(x,t) = \sum_n d_n(t) \phi_n(x).$$
What differential equations must the coefficients $a_n(t)$ in
the expansion $u(x,t) = \sum_n a_n(t) \phi_n(x)$ satisfy
if $u$ is to be a solution of (2)?

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\item{}  From now on, let's suppose that $q(x,t)$ is of the form
$q(x,t) = \cos(\omega t)$ -- a periodic driving force with frequency
$\omega$.  A solution of the form $a_n(t)\phi_n(x)$ (no summation!)
can be viewed as one of the fundamental modes of vibration
for the string in this situation.  Our next goal is to 
understand the form of these solutions, and how they depend on 
the driving frequency $\omega$.  

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\item{2)}  From part 1, we know that each $a_n$ satisfies
a 2nd order ODE (the equation should depend on $n$ too, since
the corresponding eigenvalue $\lambda_n$ will appear in it).  
We can use a numerical method to generate approximate
solutions of these equations!  Plot $a_n(t)$ versus $t$ with Maple,
following what we did with the {\tt DEplot} command in Lab/Problem 
Set 1.  Take $c = 1$, $L = 3$ for all plotting,
and make the initial conditions $a_n(0) = 1, a_n'(0) = 0$ (these
are chosen abitrarily; in an actual problem, these values
would be determined from the Fourier coefficients of the functions
$f(x),g(x)$ in (2)).
$$\cases{ n = 2 & $\omega = 1,1.5,2,2.094,3,4$\cr
          n = 3 & $\omega = 1,2,3.1,3.1416,4,5$\cr
          n = 4 & your choice -- try to find value(s) of $\omega$\cr
                & where something similar to the behavior at\cr
                & $\omega = 2.094$ for $n = 2$, and $\omega = 3.1416$\cr
                & for $n = 3$ occurs.\cr}$$
Maple technical hint: To minimize the work you need to do, you 
can set up a {\it function\/} that takes $\omega$ as its argument, and
produces the plot you want as the output.

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\item{3)} Discuss your results.  The usual name for the special
behavior at $\omega \doteq 2.094$ for $n = 2$, and $\omega \doteq 3.1416$
for $n = 3$ is {\it resonance}.  Why is that an apt name?  (What 
would be happening to a string vibrating this way??)

\bigskip
\item{4)}  The solutions of the ODE in part 2 can also be
expressed analytically through a method called {\it variation of
parameters}.  The idea is this. Just as in the 1st order linear
case that we saw in Problem Set 1, the general solution of 
$${d^2 a_n \over dt^2} + c^2 \lambda_n a_n = \cos(\omega t)\leqno(3)$$
can be written as the {\it sum of\/} the general solution of the associated
homogeneous equation
$${d^2 a_n \over dt^2} + c^2 \lambda_n a_n = 0,$$
namely
$$\alpha_1 \cos(c \sqrt{\lambda_n} t) + \alpha_2 \sin(c \sqrt{\lambda_n} t)$$
($\alpha_1, \alpha_2$ arbitrary constants), and 
any particular solution of (3).  To find a particular solution of
(3), we can use the fact that if $a(t) = A\cos(\omega t)$, then
the second derivative of $a(t)$ is another constant times the same 
cosine.  What should $A$ be to yield a solution of (3)?  (Note:
``varying the parameter $A$'' yields the solution -- hence the name
of the method.)  

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\item{5)}  Use this method to solve (3) with $n = 2$ and $\omega = 1$:
$${d^2 a_2 \over dt^2} + c^2 \left({2\pi\over 3}\right)^2 a_n = \cos(t)$$
together with the initial conditions $a_2(0) = 1$, $a_2'(0) = 0$.
Plot your solution using the formula you get and compare with the 
corresponding numerical approximation you saw in part 2.

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\item{6)}  In finding the formula for $A$ in part 4, you should have
had to {\it divide} at some point.  That step is not valid if
the quantity you divided by is zero!!  For which $\omega$ does that
happen?  These are called the {\it resonant frequencies of the 
string}.  There is a corresponding sequence of resonant frequencies
for any physical object, determined by the eigenvalues of the
Laplacian on it(!)

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\item{7)}  If $\omega$ is one of the resonant frequencies, there is a
particular solution of (3) that
has a different form.  Show that in this case,
$${1\over 2 \omega} t\sin(\omega t)$$
is a solution of (3).  Again solve (3) with $n = 2$:
$${d^2 a_2 \over dt^2} + c^2 \left({2\pi\over 3}\right)^2 a_n = \cos(\omega t)$$
together with the initial conditions $a_2(0) = 1$, $a_2'(0) = 0$,
in the case that $\omega = c {2\pi\over 3}$.
Plot your solution (taking $c = 1$) and compare with the 
corresponding numerical approximation you saw in part 2.
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\noindent
{\it Assignment}
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Group write-ups due Thursday, March 16.
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