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\begin{document}
Feb. 15, 1996 \hfill Math 312, Exam 1 
\Sk
\q{40}Consider the following partial differential equation, boundary,
initial value problem:
\begin{eqnarray}
\frac{\p^2u}{\p x^2}+4u &=& \frac{\p u}{\p t},\ 0 < x < \pi,\ 0 < t\\
u(0,t) &=& 0,\ \ 0 < t \\
u(\pi,t) &=& 0,\ \ 0 < t \\
u(x,0) &=& 1,\ 0 < x < \pi
\end{eqnarray}
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Find, using separation of variables, a class of functions which
satisfy (1).
\begin{quote}\item
Look for solutions to (1) of the form $u(x,t)=X(x)T(t)$. This leads to
the equation: $X^{\prime\prime}T + 4XT = XT^\prime$, which leads to
the pair of equations: $X^{\prime\prime}+(4+\lambda)X=0$ and
$T^\prime+\lambda T=0$. The general solutions to these equations are:
$$
X(x) = a\sin\sqrt{4+\lambda\,}x + b\cos\sqrt{4+\lambda\,}x\,,\hbox{
and }T(t) = ce^{-\lambda t}\,.
$$
\end{quote}
\item From your answer to part a. show how to extract a subset of
these functions which also satisfy (2) and (3). 
\begin{quote}\item
Since $X(0)= b$, equation (2) forces $b=0$. The equation $X(\pi)=0$
forces $\sqrt{4+\lambda\,}$ to be an integer. Setting
$n=\sqrt{4+\lambda\,}$, we have the following solutions to the first
three equations:
$$
u_n(x,t) = \sin nx\, e^{-n^2t}e^{4t}
$$
\end{quote}
\item Now show how you would find a solution to the full intitial,
boundary value problem. Be sure to explain how the ``linearity'' of
the problem enables you to do this.
\begin{quote}
We will look for a solution of the form $\ds \sum_{n=1}^\infty b_n\sin
nx\, e^{-n^2t}e^{4t}$. Since the partial differential equation is
linear we know that finite linear combinations of the individual
solutions $u_n$ will satisfy the differential equation, moreover the
boundary conditions (2) and (3) are linear and the equations are
homogeneous. Thus, finite linear combinations of the individual
solutions $u_n$ will also satisfy the boundary conditions. The fact
that an infinite sum will also satisfy these equations will be
addressed later. Thus, the last equation we need to satisfy is (4),
the initial condition. To satisfy (4) we need to be able to pick the
constants $b_n$ so that 
$$
1=\sum_{n=1}^\infty b_n\sin nx\,,\hbox{ for }0 < x < \pi\,.
$$
Thus, we need to calculate the Fourier sine series expansion of $1$ on
the interval $(0,\pi)$. The formulas for the coefficients are: 
$$
b_n = \frac{2}{\pi}\int_0^\pi\sin nx\,dx = \frac{2}{n\pi}(1-(-1)^n)\,.
$$
Thus, only the odd indexed terms will appear and we have as a
tentative solution
$$
u(x,t) = e^{4t}\sum_{k=1}^\infty
\frac{4}{(2k-1)\pi}\sin(2k-1)x\,e^{-(2k-1)^2t}\,.
$$
\end{quote}
\end{list}
\Sk
\q{40}Let $\ds f(x)=\left\{\begin{array}{rc}1+2x,&0\leq x <
1\\(x-2)^2,&1\leq x <3\\-1,&3\leq x < 4\end{array}\right.$.
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Graph the period 4 extension of $f$. Denote this extended
function by $F(x)$. $F(4)=$? $F(6.5)=$?
\begin{quote}
\begin{center}
\epsfig{file=exam1-fig1.eps,height=3in,angle=270} \\
$F(4)=f(0)=1$, and $F(6.5)=f(6.5 - 4) = f(2.5)=1/4$.
\end{center}
\end{quote}
\item At what values of $x$, where $x$ is any real number, will the
Fourier series of $f$ converge? Explain.
\begin{quote}
Since the period 4 extension of $f$ is piecewise smooth on every
interval, the fourier series of $f$ will converge at all real numbers. 
\end{quote}
\item For those values of $x$ for which the Fourier series of $f$
converges, to what value will the series converge? Explain.
\begin{quote}
I will just discuss what happens on the closed interval $[0,4]$. At
$x=0$ the fourier series of $f$ converges to the average of the left
and right hand limits of the extended function $F$, which is zero, for
$x$ in the open interval $(0,1)$, the Fourier series converges to
$f(x)$, since $f$ is continuous on this open interval. At $x=1$ the
Fourier series converges to the average of the left and right hand
limits; this average equals 2. On the open interval $(1,3)$ the 
Fourier series converges to $f$; at $x=3$, the Fourier series
converges to 0. On the open interval $(3,4)$ the series converges to
$f$ and finally at $x=4$ the series converges to zero.
\end{quote}
\end{list}
\newpage
\q{20}Let $P_0(x) =1$, and $\ds P_1(x)=\frac12\frac{d}{dx}(x^2-1)$.
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Show $P_0$ and $P_1$ are orthogonal on $[-1,1]$. Note: the inner
product is $\ds \left<f,g\right>\ =\int_{-1}^1f(x)g(x)\,dx$.
\begin{quote}
$$
\left<P_0,P_1\right> = \int_{-1}^1\frac12\frac{d}{dx}(x^2-1)\,dx =
\frac12(x^2-1)\left|^1_{-1}\right.=0-0=0
$$
\end{quote}
\item Find those scalars $a_0$ and $a_1$ such that
$$
\int_{-1}^1\left[f(x)-\left(a_0P_0(x)+a_1P_1(x)\right)\right]^2\,dx
$$
is minimized. In the above integral 
$f(x)=\left\{\begin{array}{lr}0,&-1<x<0 \\ 1,&0 < x
<1\end{array}\right.$.
\begin{quote}
The scalars which minimize the above integral are the ``Fourier''
coefficients of $f$. Thus, if $f \approx a_0P_0 + a_aP_1$, then we
have 
\begin{eqnarray*}
\left<f,P_0\right> &=&a_0\left<P_0,P_0\right>+a_1\left<P_1,P_0\right>=a_0\left<P_0.P_0\right>\\
\left<f,P_1\right>
&=&a_0\left<P_0,P_1\right>+a_1\left<P_1,P_1\right>=a_1\left<P_1,P_1\right>
\end{eqnarray*}
Solving for the coefficients $a_0$ and $a_1$, we have
\begin{eqnarray*}
a_0 &=& \frac{\left<f,P_0\right>}{\left<P_0,P_0\right>}=\frac12 \\
{}\\
a_1 &=&
\frac{\left<f,P_1\right>}{\left<P_1,P_1\right>}=\frac{1/2}{1/3}=\frac32\,.
\end{eqnarray*}
\end{quote}
\end{list}


\end{document}