Quiz 2

Math 421:03, Fall 2002


Date: 30 October 2002

There are two (2) problems on this quiz. The other one is on the back of this page.


1.Formulate a boundary value problem modeling heat conduction in a thin bar of length $ L$ if the left end is insulated and the right end is kept at a fixed temperate $ T$. The initial temperature distribution in the bar at $ x$ is $ f(x)$. (Just set up the problem and stop!)

Proof. [Solution]The heat equation is $ u_t = k u_{xx}$ for $ 0 < x < L$, $ t > 0$. The boundary conditions are $ u_x(0,t) = 0$ for all $ t > 0$ (because the left side is insulated), and $ u(L,t) = T$ for all $ t > 0$ (because the right side is held to a constant temperature). The initial conditions are $ u(x,0) = f(x)$ for $ 0 \leq x \leq L$. $ \qedsymbol$


2.Solve the boundary-value problem

$\displaystyle u_t$ $\displaystyle = k u_{xx}$ $\displaystyle (0 < x < L, t > 0)$    
$\displaystyle u(0,t)$ $\displaystyle = 0 = u(L,t)$ $\displaystyle (t > 0)$    
$\displaystyle u(x,0)$ $\displaystyle = \begin{cases}0 & 0 \leq x < L/2 \\ 1 & L/2 \leq x \leq L \end{cases}$    

Proof. [Solution]The general solution to this problem is

$\displaystyle u(x,t) = \sum_{n=1}^\infty \sin\left(\frac{\pi n x}{L}\right)
e^{-\pi^2n^2t/L^2},
$

where

$\displaystyle c_n = \frac{2}{L}\int_0^L u(x,0) \sin\left(\frac{\pi n
x}{L}\right) \,dx.
$

In our case we have

\begin{displaymath}
\begin{split}
c_n &= \frac{2}{L}\int_{L/2}^L \sin\left(\fra...
...\left(\frac{\pi n}{2}\right) -
\cos(n\pi)\right).
\end{split}\end{displaymath}

This was enough as far as I was concerned. The simplification is tricker than on the homework, but can be done like this:

$ n$ $ \cos\left(\frac{\pi n}{2}\right)$ $ \cos(n\pi)$ difference
1 0 -1 1
2 -1 1 -2
3 0 -1 1
4 1 1 0
5 0 -1 1
6 -1 1 -2
7 0 -1 1
8 1 1 0

So clearly $ c_n = \frac{2}{\pi n}$ if $ n$ is odd. If $ n$ is even, $ n$ can be written as $ 2m$, but $ c_{2m}$ depends on whether $ m$ is even or odd! The numbers that can be written as 2 times an odd number look like $ 2(2m-1) = 4m-2$. Therefore we have

\begin{multline*}
u(x,t) = \sum_{j=1}^\infty
\frac{2}{\pi (2k-1)}\sin\left(\fr...
...sin\left(\frac{\pi (4m-2) x}{L}\right)
e^{-\pi^2(4m-2)^2t/L^2}.
\end{multline*}

$ \qedsymbol$

\epsfig{file=prob2.eps,height=\textwidth,angle=-90}



Matthew Leingang 2002-11-12