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\centerline{Mathematics 174 -- Applied Mathematics 2}
\centerline{Discussion 2 -- The Rayleigh Quotient and Applications}
\centerline{February 3, 2000}
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\noindent
{\it Background}
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We consider a general regular Sturm-Liouville boundary value problem
$$\cases{
L(\phi) = {d \over dx}\left(p(x)\phi\right) + q(x)\phi = -\lambda\sigma(x)\phi&\cr
\beta_1\phi(a) + \beta_2\phi'(a) = 0&\cr
\beta_3\phi(b) + \beta_4\phi'(b) = 0&\cr}\leqno(1)$$
Let $V$ denote the vector space of $C^2$ functions on $[a,b]$ satisfying
the boundary conditions in (1).  For any $u \in V$ (any {\it test function})
we can define the {\it Rayleigh Quotient}:
$$RQ(u) =  {- \langle L(u),u\rangle_1 \over \langle u,u\rangle_\sigma}$$
At the end of class on Tuesday, we stated 
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\proclaim Theorem. Let $\lambda_1$ be the minimal eigenvalue of (1).  Then
$$\lambda_1 = \min_{u \in V} RQ(u)$$

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Today we want to prove this result and use it to study several boundary
value problems for the 1-dimensional heat equation.
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\noindent
{\it Discussion Questions}
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\item{A)}  To deduce the theorem, proceed as follows:
\itemitem{1.}  First, if $u \in C^2([a,b])$ then the last portion of the 
Sturm-Liouville theorem implies that the eigenfunction expansion
$$u(x) \sim \sum_{n=1}^\infty a_n\phi_n(x)$$
($a_n$ are the generalized Fourier coefficients of $u$) converges
to $u(x)$ for all $x \in (a,b)$.  Substitute this expression for $u$
into the Rayleigh quotient and work with the result to show that 
$$RQ(u) = {\sum_{n=1}^\infty a_n^2\lambda_n \langle \phi_n,\phi_n\rangle_\sigma\over 
           \sum_{n=1}^\infty a_n^2\langle \phi_n,\phi_n\rangle_\sigma}$$
\itemitem{2.}  Recall that $\lambda_1 < \lambda_n$ for all $n > 1$.
Use this to show that $RQ(u) \ge \lambda_1$ for all $u$.  
\itemitem{3.}  For which $u$ does $RQ(u)$ achieve its minimum value of 
$\lambda_1$?  

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\item{B)}  Use the Rayleigh Quotient to show that the minimal eigenvalue
$\lambda_1$ of 
$$\cases{{d^2\phi\over dx^2} = -\lambda (x^4 + x^2 + 1)\phi &\cr
          \phi(0) = 0 &\cr  
          \phi'(1) = 0 & \cr}$$
is strictly positive.  (You will need to rule out $\lambda_1  = 0$ with
a separate argument.)

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\item{C)}  In our study of boundary value problems for the 1-D heat equation 
last semester, we concentrated almost exclusively on the cases where
the material of the rod had constant thermal properties (that is $c,\rho,K_0$
were assumed constant).  One nice aspect of the Sturm-Liouville theory
is that it lets us remove that restriction.  Namely, let's now consider
a rod extending along the interval $[0,L]$ in $x$, with no 
heat sources.  Then the temperature
function $u(x,t)$ satisfies
$$c(x)\rho(x) u_t = (K_0(x) u_x)_x\leqno(2)$$
\itemitem{1.}  In all cases of physical interest, the functions $c(x), \rho(x),
K_0(x)$ have {\it strictly positive\/} values on $[0,L]$.  Show that
separation of variables $u(x,t) = h(t)\phi(x)$ leads to a regular Sturm-Liouville
problem for $\phi(x)$ in the case that the ends of the rod are maintained
at zero temperature for all times, and in the case that the ends of the 
rod are insulated.  (What functions play the roles
of the $p, q, \sigma$ in the general form (1) here?)
\itemitem{2.}  In the case that the ends of the rod are insulated, show
that the minimal eigenvalue is $\lambda_1 = 0$.  (Hint: the Rayleigh Quotient
can be used to rule out the possibility $\lambda_1 < 0$.) 
\itemitem{3.}  Still in the case that the ends of the rod are insulated,
let $u(x,0) = f(x)$ be the initial temperature distribution.
By writing the solution of (2) and the boundary conditions $u_x(0,t) = 0,
u_x(L,t) = 0$ using the Sturm-Liouville eigenfunctions and taking a limit
as $t \to \infty$, determine the corresponding equilibrium temperature.
(Note: you don't have specific $\phi_n(x)$ to work with here because
$c(x),\rho(x),K_0(x)$ are not given explicitly.  Your answer should be 
completely general.)
\itemitem{4.}  Now assume $u(0,t) = u(L,t) = 0$ for all $t$.
Using the Rayleigh Quotient, show that if
$$\cases{ 0 < m \le K_0(x) \le M & for all $x \in [0,L]$ and\cr
          0 < r \le c(x)\rho(x) \le R & for all $x \in [0,L]$ and\cr}$$
then
$$ {m\over R}\cdot \left(\pi\over L\right)^2 \le \lambda_1 \le
   {M\over r}\cdot \left(\pi\over L\right)^2$$
In particular, since $m, R > 0$, we have $\lambda_1 > 0$.
\itemitem{5.}  Still assuming $u(0,t) = u(L,t) = 0$, suppose $u(x,0) = f(x)$.  
The solution of the boundary value problem will look like
$$u(x,t) = \sum_{n=1}^\infty a_n e^{-\lambda_n t} \phi_n(x)$$
where 
$$a_n = {\langle f, \phi_n\rangle_\sigma \over
         \langle \phi_n, \phi_n\rangle_\sigma}$$
for the appropriate weight function $\sigma$.  
Which term here decays to zero {\it most slowly}?  (As $t\to \infty$, 
this will be the ``most visible'' term in the solution.)
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\noindent
{\it Assignment}
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Group write-ups due Thursday, February 10.
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