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\begin{document}


\section{Ma 221}

\section{\protect\Large BOUNDARY VALUE PROBLEMS}

\vspace{1pt}

\subsection{Homogeneous Boundary Value Problems}

\vspace{1pt}

Consider the following problem:

\vspace{1pt}

\ 

\vspace{1pt}%
\begin{equation*}
\left. 
\begin{array}{c}
\text{\textbf{D.E}}\mathbf{.}L\left[ y\right] =a_{0}\left( x\right)
y^{\prime \prime }+a_{1}\left( x\right) y^{\prime }+a_{2}\left( x\right)
y=0\qquad a\leq x\leq b \\ 
\qquad \text{B.C. \ }\alpha _{1}y\left( a\right) +\beta _{1}y^{\prime
}\left( a\right) =0\qquad \alpha _{1}^{2}+\beta _{1}^{2}\neq 0 \\ 
\qquad \text{B.C. \ }\alpha _{2}y\left( b\right) +\beta _{2}y^{\prime
}\left( b\right) =0\qquad \alpha _{2}^{2}+\beta _{2}^{2}\neq 0%
\end{array}%
\right\} \qquad \qquad \left( 1\right)
\end{equation*}

Here $\alpha _{1},\alpha _{2},\beta _{1},$ and $\beta _{2}$ are constants.

\vspace{1pt}

\begin{example}
\begin{equation*}
y^{\prime \prime }=0\qquad y^{\prime }(0)=y^{\prime }(1)=0
\end{equation*}
\end{example}

$\left( \text{Here }\alpha _{1}=\alpha _{2}=0\right) $

\qquad \qquad $\Longrightarrow y=Ax+b\qquad y^{\prime }(x)=A\qquad y^{\prime
}(0)=y^{\prime }(1)=A=0$

\qquad \qquad $\Longrightarrow y(x)=b\qquad b$ any constant.

The Boundary Value Problem $(1)$ is called linear and homogeneous since if $%
u_{1}(x)$\ and $u_{2}(x)$ satisfy it, $\Longrightarrow
c_{1}u_{1}(x)+c_{2}u_{2}\left( x\right) $\ also does.

\vspace{1pt}

\begin{example}
\begin{equation*}
y^{\prime \prime }-6y^{\prime }+5y=0\text{ \ \ \ \ }y\left( 0\right) =1\text{
\ \ \ }y\left( 2\right) =1
\end{equation*}
\end{example}

Solution: \ The characteristic equation is%
\begin{equation*}
r^{2}-6r+5=\left( r-5\right) \left( r-1\right) =0
\end{equation*}

so $r=1,5$

\vspace{1pt}

Thus%
\begin{equation*}
y\left( x\right) =c_{1}e^{x}+c_{2}e^{5x}
\end{equation*}

\begin{eqnarray*}
y\left( 0\right) &=&c_{1}+c_{2}=1 \\
y\left( 2\right) &=&c_{1}e^{2}+c_{2}e^{10}=1
\end{eqnarray*}

Thus from the first equation $c_{2}=1-c_{1}$ and the second equation becomes%
\begin{equation*}
c_{1}e^{2}+\left( 1-c_{1}\right) e^{10}=1
\end{equation*}

\begin{equation*}
c_{1}\left( e^{2}-e^{10}\right) =1-e^{10}
\end{equation*}

\begin{equation*}
c_{1}=\frac{1-e^{10}}{e^{2}-e^{10}}
\end{equation*}

\begin{equation*}
c_{2}=1-\frac{1-e^{10}}{e^{2}-e^{10}}=\frac{1}{e^{2}-e^{10}}\left(
e^{2}-1\right)
\end{equation*}

\begin{equation*}
y=\frac{1-e^{10}}{e^{2}-e^{10}}e^{x}+\frac{e^{2}-1}{e^{2}-e^{10}}e^{5x}
\end{equation*}

SNB check%
\begin{eqnarray*}
y^{\prime \prime }-6y^{\prime }+5y &=&0 \\
y\left( 0\right) &=&1 \\
y\left( 2\right) &=&1
\end{eqnarray*}%
, Exact solution is: $\left\{ \frac{e^{5x}}{e^{2}-e^{10}}\left(
e^{2}-1\right) -\frac{e^{x}}{e^{2}-e^{10}}\left( e^{10}-1\right) \right\}
\allowbreak $ \ \ \ \ 

\vspace{1pt}

Remark. The homogeneous Boundary Value Problem (B.V.P.) always possesses the
solution $y\left( x\right) =0$.

\vspace{1pt}

Question. When does there exist a nonzero solution to $(1)$?

\qquad Let $y_{1}\left( x\right) $ and $y_{2}\left( x\right) $ be two
linearly independent solutions of $L\left[ y\right] =0$. $\ \
\Longrightarrow $ $y\left( x\right) =c_{1}y_{1}+c_{2}y_{2}$ is the general
solution of the DE.

B.C. \ $\Longrightarrow \left. 
\begin{array}{c}
\alpha _{1}y\left( a\right) +\beta _{1}y^{\prime }\left( a\right) =0 \\ 
\alpha _{2}y\left( b\right) +\beta _{2}y^{\prime }\left( b\right) =0%
\end{array}%
\right\} \qquad $ and $y\left( x\right)
=c_{1}y_{1}+c_{2}y_{2}\Longrightarrow $

\vspace{1pt}

\qquad \qquad $c_{1}\left[ \alpha _{1}y_{1}\left( a\right) +\beta
_{1}y_{1}^{\prime }\left( a\right) \right] +c_{2}\left[ \alpha
_{1}y_{2}\left( a\right) +\beta _{1}y_{2}^{\prime }\left( a\right) \right]
=0 $

\qquad \qquad $c_{1}\left[ \alpha _{2}y_{1}\left( b\right) +\beta
_{2}y_{1}^{\prime }\left( b\right) \right] +c_{2}\left[ \alpha
_{2}y_{2}\left( b\right) +\beta _{2}y_{2}^{\prime }\left( b\right) \right]
=0 $.

\qquad The above are two equations for $c_{1}$ and $c_{2}$. We want a
nontrivial solution. Let $B_{a}\left( u\right) =\alpha _{1}u\left( a\right)
+\beta _{1}u^{\prime }\left( a\right) $ and $B_{b}\left( u\right) =\alpha
_{2}u\left( b\right) +\beta _{2}u^{\prime }\left( b\right) $. Then the
determinant of the coefficients of the above system must equal zero. Thus we
require

\vspace{1pt}%
\begin{equation*}
\left| 
\begin{array}{ll}
B_{a}\left( y_{1}\right) & B_{a}\left( y_{2}\right) \\ 
B_{b}\left( y_{1}\right) & B_{b}\left( y_{2}\right)%
\end{array}%
\right| =0\qquad \left( 2\right)
\end{equation*}

\vspace{1pt}

Theorem 1. The homogeneous linear B.V.P. $(1)$ has a nontrivial solution if
and if $(2)$ holds.

\vspace{1pt}

Theorem 2. If $u\left( x\right) $ is a particular nontrivial solution of the
B.V.P. $(1)$, then all solutions are given by $y=cu\left( x\right) $ where $%
c $ is an arbitrary constant.

\vspace{1pt}

Proof. Let $v\left( x\right) $ be any solution, $u\left( x\right) $ a
particular solution of the B.V.P. $(1)$ $\ \Longrightarrow \alpha
_{1}u\left( a\right) +\beta _{1}u^{\prime }\left( a\right) =0$ and

$\alpha _{1}v\left( a\right) +\beta _{1}v^{\prime }\left( a\right) =0$ since 
$u$ and $v$ both satisfy the first B.C. These equations may be regarded as
equations for $\alpha _{1},\beta _{1}$. However, since by assumption $\alpha
_{1}$ and $\beta _{1}$ are not both zero $\Longrightarrow $

\vspace{1pt}

\qquad \qquad $\left| 
\begin{array}{ll}
u\left( a\right) & u^{\prime }\left( a\right) \\ 
v\left( a\right) & v^{\prime }\left( a\right)%
\end{array}
\right| =0=W\left[ u,v\right] _{x=a}\Longrightarrow W\left[ u\left( x\right)
,v\left( x\right) \right] =0$ for $a\leq x\leq b$

\vspace{1pt}

$\Longrightarrow u$ and $v$ are two LD solutions of the D.E. $%
\Longrightarrow $ there exist constants $c_{1},c_{2}\neq 0$ such that $%
c_{1}u\left( x\right) +c_{2}v\left( x\right) =0$ for $a\leq x\leq
b\Longrightarrow v\left( x\right) =-\dfrac{c_{1}}{c_{2}}u\left( x\right)
=cu\left( x\right) $.

\vspace{1pt}

\begin{example}
\begin{equation*}
y^{\prime \prime }-\lambda ^{2}y=0\qquad \lambda \neq 0\qquad y\left(
0\right) =y\left( 1\right) =0
\end{equation*}
\end{example}

\vspace{1pt}

The general solution is $y=c_{1}e^{\lambda x}+c_{2}e^{-\lambda x}.$ The B.C
\ $y\left( 0\right) =0$. $\Longrightarrow c_{1}+c_{2}=0,$ whereas the
condition $y\left( 1\right) =0$ leads to

$c_{1}e^{\lambda }+c_{2}e^{-\lambda }=0.$ The two equations for $c_{1}$ and $%
c_{2}$ are

\begin{eqnarray*}
c_{1}+c_{2} &=&0 \\
c_{1}e^{\lambda }+c_{2}e^{-\lambda } &=&0
\end{eqnarray*}

The determinant of the coefficients is $\left\vert 
\begin{array}{ll}
1 & 1 \\ 
e^{\lambda } & e^{-\lambda }%
\end{array}%
\right\vert \neq 0$. $\Longrightarrow c_{1}=c_{2}=0\Longrightarrow $ the
only solution is $y\equiv 0$.

\vspace{1pt}

\subsection{Eigenvalue Problems}

\vspace{1pt}

The following special kind of B.V.P. is called an eigenvalue problem.

\begin{equation*}
\left. 
\begin{array}{lll}
L\left[ y\right] +\lambda y=0 &  & a\leq x\leq b \\ 
\text{B.C. \ }\alpha _{1}y\left( a\right) +\beta _{1}y^{\prime }\left(
a\right) =0\qquad &  & \alpha _{1}^{2}+\beta _{1}^{2}\neq 0 \\ 
\text{B.C. \ }\alpha _{2}y\left( b\right) +\beta _{1}y^{\prime }\left(
b\right) =0\qquad &  & \alpha _{2}^{2}+\beta _{2}^{2}\neq 0%
\end{array}%
\right\} (\ast )
\end{equation*}

\vspace{1pt}

Here $L\left[ y\right] =a_{0}\left( x\right) y^{\prime \prime }+a_{1}\left(
x\right) y^{\prime }+a_{2}\left( x\right) y,$ and $\lambda $ is a parameter.

\vspace{1pt}

Again $y\equiv 0$ is a solution \emph{for all} $\lambda $. However, we are
interested in nontrivial (nonzero) solutions.

\vspace{1pt}

Definition. If a nontrivial solution of the B.V.P. $(\ast )$ exists for a
value $\lambda =\lambda _{i}$, then $\lambda _{i}$ is called an \emph{%
eigenvalue} of $L$ (relevant to the B.Cs.). The corresponding nontrivial
solution $y_{i}\left( x\right) $ is called an \emph{eigenfunction}.

\begin{example}
Find the eigenvalues and eigenfunctions for
\end{example}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y^{\prime \prime }+\lambda y=0,\qquad y^{\prime }\left( 0\right) =0,\qquad
y\left( 1\right) =0
\end{equation*}

We must consider three cases; $\lambda <0,\lambda =0,$, and $\lambda >0$ .

\vspace{1pt}

I. $\lambda <0$. Let $\lambda =-\alpha ^{2}$ where $\alpha \neq 0$. Then the
differential equation becomes 
\begin{equation*}
y^{\prime \prime }-\alpha ^{2}y=0
\end{equation*}
and has the general solution 
\begin{equation*}
y=c_{1}e^{\alpha x}+c_{2}e^{-\alpha x}.
\end{equation*}%
The boundary conditions $\Longrightarrow $

$y^{\prime }\left( 0\right) =c_{1}\alpha -c_{2}\alpha =0$ or $c_{1}=c_{2}$ ,
and $y\left( 1\right) =c_{1}e^{\alpha }+c_{2}e^{-\alpha }=0$ .$%
\Longrightarrow c_{1}=c_{2}=0$.

Thus for $\lambda <0$, the only solution is $y=0$.

\vspace{1pt}

II. $\lambda =0$. The solution is $y=c_{1}x+c_{2}$. The BCs imply $%
c_{1}=c_{2}=0$. Again the only solution is $y=0$.

III. $\lambda >0$. Let $\lambda =\beta ^{2}$ where $\beta \neq 0$. The DE
becomes 
\begin{equation*}
y^{\prime \prime }+\beta ^{2}y=0
\end{equation*}
and has the general solution 
\begin{equation*}
y=c_{1}\sin \beta x+c_{2}\cos \beta x.
\end{equation*}
The BCs imply

\vspace{1pt}

$y^{\prime }\left( 0\right) =c_{1}\beta \cos 0-c_{2}\beta \sin 0=c_{1}\beta
=0$. \qquad Hence\qquad $c_{1}=0,$ , since $\beta \neq 0$.\qquad Thus 
\begin{equation*}
y=c_{2}\cos \beta x.
\end{equation*}

Now $y\left( 1\right) =c_{2}\cos \beta =0$. Since we want a nontrivial
solution we cannot have $c_{2}=0$.

Hence 
\begin{equation*}
\cos \beta =0\Longrightarrow \beta =\frac{2n+1}{2}\pi ,n=0,\pm 1,\pm 2,...
\end{equation*}%
We therefore have the eigenvalues 
\begin{equation*}
\lambda _{n}=\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2},
\end{equation*}%
and eigenfunctions 
\begin{equation*}
y_{n}\left( x\right) =C_{n}\cos \left( \frac{2n+1}{2}\right) x,
\end{equation*}%
for $n=0,1,2,...$ Note the negative values of $n$ do not give additional
eigenfunctions since $\cos \left( -t\right) =\cos t$.

\begin{example}
\vspace{1pt}Find the eigenvalues and eigenfunctions for%
\begin{equation*}
y^{\prime \prime }-12y^{\prime }+4\left( 7+\lambda \right) y=0\text{ \ \ \ }%
y\left( 0\right) =y\left( 5\right) =0
\end{equation*}
\end{example}

Solution: The characteristic equation is 
\begin{equation*}
r^{2}-12r+4\left( 7+\lambda \right) =0
\end{equation*}

so%
\begin{equation*}
r=\frac{+12\pm \sqrt{144-4\left( 4\right) \left( 7+\lambda \right) }}{2}%
=6\pm 2\sqrt{2-\lambda }
\end{equation*}
Thus we have 3 cases to deal with, $2-\lambda <0,2-\lambda =0,$ and $%
2-\lambda >0.$

Case I: $2-\lambda >0.$ Let $2-\lambda =\alpha ^{2}$ where $\alpha \neq 0.$
The the general homogeneous solution is%
\begin{equation*}
y\left( x\right) =C_{1}e^{\left( 6+2\alpha \right) x}+C_{2}e^{\left(
6-2\alpha \right) x}
\end{equation*}

The BCs imply%
\begin{eqnarray*}
C_{1}+C_{2} &=&0 \\
C_{1}e^{\left( 6+2\alpha \right) 5}+C_{2}e^{\left( 6-2\alpha \right) 5} &=&0
\end{eqnarray*}%
, Solution is: $\left\{ C_{2}=0,C_{1}=0\right\} .$ Thus $y=0$ and there are
no eigenvalues for this case.

\vspace{1pt}

Case II: $\lambda =2.$ Then 
\begin{equation*}
y\left( x\right) =C_{1}e^{6x}+C_{2}xe^{6x}
\end{equation*}

The BCs imply%
\begin{eqnarray*}
C_{1} &=&0 \\
C_{2}\left( 5\right) e^{30} &=&0\Rightarrow C_{2}=0
\end{eqnarray*}

Therefore $\lambda =2$ is not an eigenvalue.

\vspace{1pt}

Case III: $2-\lambda <0.$ Let $2-\lambda =-\beta ^{2}$ where $\beta \neq 0.$
Then $r=6\pm 2\beta i.$ The solution to the DE is 
\begin{equation*}
y\left( x\right) =C_{1}e^{6x}\sin 2\beta x+C_{2}e^{6x}\cos 2\beta x
\end{equation*}

The BCs imply%
\begin{eqnarray*}
y\left( 0\right) &=&C_{2}=0 \\
y\left( 5\right) &=&C_{1}e^{30}\sin 10\beta =0
\end{eqnarray*}

Thus 
\begin{equation*}
10\beta =n\pi ,\text{ \ \ }n=1,2,\ldots
\end{equation*}

or%
\begin{equation*}
\beta =\frac{n\pi }{10}\text{ \ \ }n=1,2,\ldots \text{\ }
\end{equation*}

and the eigenvalues are 
\begin{equation*}
\lambda =2+\beta ^{2}=2+\frac{n^{2}\pi ^{2}}{100}\text{ \ \ }n=1,2,\ldots
\end{equation*}

\vspace{1pt}

The eigenfunctions are 
\begin{equation*}
y_{n}\left( x\right) =A_{n}e^{6x}\sin \left( \frac{n\pi }{5}\right) x
\end{equation*}

\begin{example}
\begin{equation*}
\ y^{\prime \prime }+\lambda y=0\qquad y\left( \pi \right) =y\left( 2\pi
\right) =0
\end{equation*}
\end{example}

Solution: \ There are 3 cases to consider. \ $\lambda <0,\lambda =0,$, and $%
\lambda >0$ .

\vspace{1pt}

I. $\lambda <0$. Let $\lambda =-\alpha ^{2}$ where $\alpha \neq 0$. Then the
differential equation becomes 
\begin{equation*}
y^{\prime \prime }-\alpha ^{2}y=0
\end{equation*}%
and has the general solution 
\begin{equation*}
y\left( x\right) =c_{1}e^{\alpha x}+c_{2}e^{-\alpha x}.
\end{equation*}

Then 
\begin{eqnarray*}
y\left( \pi \right) &=&c_{1}e^{\alpha \pi }+c_{2}e^{-\alpha \pi }=0 \\
y\left( 2\pi \right) &=&c_{1}e^{2\alpha \pi }+c_{2}e^{-2\alpha \pi }=0
\end{eqnarray*}

Thus from the first equation 
\begin{equation*}
c_{2}=-c_{1}e^{2\alpha \pi }
\end{equation*}%
and the second equation implies%
\begin{equation*}
c_{1}\left( e^{2\alpha \pi }-1\right) =0
\end{equation*}

Hence $c_{1}=0$ and thus $c_{2}=0,$ so $y=0$ is the only solution. \ There
are no negative eigenvalues.

\vspace{1pt}

II. $\lambda =0.$ \ Then we have $y^{\prime \prime }=0$ so 
\begin{equation*}
y\left( x\right) =c_{1}x+c_{2}
\end{equation*}

\begin{eqnarray*}
y\left( \pi \right) &=&c_{1}\pi +c_{2}=0 \\
y\left( 2\pi \right) &=&2c_{1}\pi +c_{2}=0
\end{eqnarray*}

Therefore $c_{1}=c_{2}=0$ and $y=0,$ so $0$ is not an eigenvalue.

\vspace{1pt}

III. \ $\lambda >0.$ Let $\lambda =\beta ^{2}$ The DE becomes%
\begin{equation*}
y^{\prime \prime }+\beta ^{2}y=0
\end{equation*}%
so%
\begin{equation*}
y\left( x\right) =c_{1}\sin \beta x+c_{2}\cos \beta x
\end{equation*}%
The initial conditions yield 
\begin{eqnarray*}
y\left( \pi \right) &=&c_{1}\sin \beta \pi +c_{2}\cos \beta \pi =0 \\
y\left( 2\pi \right) &=&c_{1}\sin 2\beta \pi +c_{2}\cos 2\beta \pi =0
\end{eqnarray*}

This system will have a non-trivial solution if and only if%
\begin{equation*}
\left\vert 
\begin{array}{cc}
\sin \beta \pi & \cos \beta \pi \\ 
\sin 2\beta \pi & \cos 2\beta \pi%
\end{array}%
\right\vert =0
\end{equation*}%
That is if and only if%
\begin{equation*}
\sin \beta \pi \cos 2\beta \pi -\cos \beta \pi \sin 2\beta \pi =\sin \left(
\beta \pi -2\beta \pi \right) =-\sin \beta \pi =0
\end{equation*}

Thus we must have 
\begin{equation*}
\beta \pi =n\pi \text{ \ \ \ \ }n=1,2,3,\ldots
\end{equation*}

\vspace{1pt}or

\begin{equation*}
\beta =n\text{ \ \ \ \ }n=1,2,3,\ldots
\end{equation*}

Hence the eigenvalues are 
\begin{equation*}
\lambda =\beta ^{2}=n^{2}\text{ \ }n=1,2,3,\ldots
\end{equation*}

The two equations above for $c_{1}$ and $c_{2}$ become

\begin{eqnarray*}
c_{1}\sin n\pi +c_{2}\cos n\pi &=&0 \\
c_{1}\sin 2n\pi +c_{2}\cos 2n\pi &=&0
\end{eqnarray*}%
Thus $c_{2}=0$ and $c_{1}$ is arbitrary. The eigenfunctions are 
\begin{equation*}
y_{n}\left( x\right) =a_{n}\sin nx
\end{equation*}

\vspace{1pt}

Remark. If $\overrightarrow{u}$ and $\overrightarrow{v}$ are $2$ vectors,
then $\overrightarrow{u}\perp \overrightarrow{v}\Leftrightarrow 
\overrightarrow{u}\cdot \overrightarrow{v}=0$

$\overrightarrow{u}=\left( x_{1},...,x_{n}\right) \qquad \overrightarrow{v}%
\left( y_{1},...,y_{n}\right) $ \ As $n\rightarrow \infty \qquad 
\overrightarrow{u}\cdot \overrightarrow{v}\rightarrow \int x_{i}y_{i}$.

\vspace{1pt}

Definition. Let $f\left( x\right) ,g\left( x\right) $ be two continuous
functions on $\left[ a,b\right] $. We define the \emph{inner product} of $f$
and $g$ in an interval $a\leq x\leq b$, denoted by $<f,g>,$ by

\qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
<f,g>=\int_{a}^{b}f\left( x\right) g\left( x\right) dx.
\end{equation*}

Definition. Two functions $f$ and $g$ are said to be \emph{orthogonal} on $%
\left[ a,b\right] $ if

\begin{equation*}
<f,g>=0.
\end{equation*}

\vspace{1pt}

Example. $\int_{0}^{\pi }\sin x\cos xdx=\frac{\sin ^{2}x}{2}|_{0}^{\pi }=0$
Therefore $\sin x$ and $\cos x$ are orthogonal on$\left[ 0,\pi \right] $.

\vspace{1pt}

Definition. The set of functions $\left\{ f_{1},f_{2},...\right\} $ is
called an \emph{orthogonal} set $<f_{i},f_{j}>=0\qquad i\neq j$.

\vspace{1pt}

Example. $\left\{ 1,\cos \dfrac{\pi x}{L},\cos \dfrac{2\pi x}{L},\,...,\cos 
\dfrac{n\pi x}{L},\text{ }...\right\} $ is an orthogonal set on $\left[ 0,L%
\right] $

Remark. For vectors we have the following: if $\overrightarrow{u}=\left(
u_{1},...,u_{n}\right) $ then the length of $\overrightarrow{u}=\left\| 
\overrightarrow{u}\right\| =\left( \sum u_{i}^{2}\right) ^{\frac{1}{2}}=%
\sqrt{\overrightarrow{u}\cdot \overrightarrow{u}}$. Motivated by this we
have the following definition.

\vspace{1pt}

Definition. Let $f\left( x\right) $ be a continuous function on a$\leq x\leq
b$. Then the \emph{norm} of $f$ is defined by

\begin{center}
\begin{equation*}
\left\| f\right\| =\sqrt{<f,f>}=\sqrt{\int_{a}^{b}f^{2}\left( x\right) dx}.
\end{equation*}
\end{center}

\vspace{1pt}

Example. $0\leq x\leq 1\qquad \left\| x^{2}\right\|
^{2}=<x^{2},x^{2}>=\int_{0}^{1}x^{4}dx=\dfrac{x^{5}}{5}|_{0}^{1}=\frac{1}{5}$

$\Longrightarrow \left\| x^{2}\right\| =\frac{1}{\sqrt{5}}$.

Remark. Let $y=\frac{x^{2}}{\left\| x^{2}\right\| }=\frac{x^{2}}{\sqrt{5}}%
\Longrightarrow \left\| y\right\| =\frac{\left\| x^{2}\right\| }{\sqrt{5}}=1$%
.

Definition. If $\left\| f\right\| =1$, then $f$ \ is said to be \emph{%
normalized}.

\vspace{1pt}

Definition. A set of functions $\left\{ \phi _{1},\phi _{2},...\right\} $ is
called \emph{orthonormal} if

(1) the set is orthogonal, and

(2) each has norm 1. Therefore $\left\{ \phi _{1},\phi _{2},...\right\} $ is
an orthonormal set $\Leftrightarrow $

\vspace{1pt}

\begin{center}
\begin{equation*}
<\phi _{i},\phi _{j}>=\delta _{ij}=\left\{ 
\begin{array}{c}
0\qquad i\neq j \\ 
1\qquad i=j%
\end{array}%
\right.
\end{equation*}
\end{center}

\vspace{1pt}

\begin{example}
$\left\{ \sin \left( nx\right) \right\} =\left\{ \sin x,\sin 2x,\sin
3x,\,...\right\} $ on$\left[ 0,\pi \right] $ \ is an orthogonal set since
\end{example}

\begin{eqnarray*}
&<&\sin \left( mx\right) ,\sin \left( nx\right) >=\int_{0}^{\pi }\sin mx\sin
nx\,dx=\frac{1}{2}\int_{0}^{\pi }\left[ \cos \left( m-n\right) x-\cos \left(
m+n\right) x\right] dx\qquad m\neq n \\
&=&\frac{1}{2}\left[ \dfrac{\sin \left( m-n\right) x}{m-n}-\dfrac{\sin
\left( m+n\right) x}{m-n}\right] _{0}^{\pi } \\
&=&\frac{1}{2}\left[ \dfrac{\sin \left( m-n\right) \pi }{m-n}-\dfrac{\sin
\left( m+n\right) \pi }{m+n}\right] =0\text{ \ \ \ }m\neq n
\end{eqnarray*}

\vspace{1pt}

since $m$ and $n$ are integers.

Now 
\begin{eqnarray*}
&<&\sin nx,\sin nx>=\int_{0}^{\pi }\sin ^{2}nxdx \\
&=&\frac{1}{2}\int_{0}^{\pi }\left( 1-\cos 2nx\right) dx \\
&=&\frac{1}{2}\left( x-\frac{\sin 2nx}{2n}\right) |_{0}^{\pi }=\frac{\pi }{2}%
.
\end{eqnarray*}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad

Therefore 
\begin{equation*}
\left\| \sin nx\right\| =<\sin nx,\sin nx>^{\frac{1}{2}}=\sqrt{\frac{\pi }{2}%
}
\end{equation*}%
$\Longrightarrow $ this set is not orthonormal. We can make an orthonormal
set from these functions by dividing each element in the original by $\sqrt{%
\frac{\pi }{2}}\Longrightarrow \left\{ \sqrt{\frac{2}{\pi }}\sin nx\right\} $
is orthonormal set $(n=1,2,$ $...)$.

\vspace{1pt}

\paragraph{Properties of the inner product.}

\begin{equation*}
1.<f,g>=<g,f>\text{since}\qquad \int_{a}^{b}f\left( x\right) g\left(
x\right) dx=\int_{a}^{b}g\left( x\right) f\left( x\right) dx
\end{equation*}

\begin{equation*}
2.<\alpha f+\beta g,h>=\alpha <f,h>+\beta <g,h>\text{ since }\int \left(
\alpha f+\beta g\right) dx=\alpha \int fdx+\beta \int gdx\text{ }
\end{equation*}

\qquad \qquad

\begin{eqnarray*}
3.a. &<&f,f>=0iff=0 \\
b. &<&f,f>>0iff\neq 0
\end{eqnarray*}

\ \ 

Remarks. (1) It will be necessary when dealing with partial differential
equations to \textquotedblleft expand\textquotedblright\ an arbitrary
function $f\left( x\right) $ in terms of an orthogonal set of functions $%
\left\{ \psi _{n}\right\} $.

\vspace{1pt}

(2) Recall that in 3 space, if $\overrightarrow{u}_{1}=\left( 1,0,0\right) ,%
\overrightarrow{u}_{2}=\left( 0,1,0\right) ,$ and $\overrightarrow{u}%
_{3}=\left( 0,0,1\right) $ then $\overrightarrow{v}=\left( \alpha
_{1},\alpha _{2},\alpha _{3}\right) =\alpha _{1}u_{1}+\alpha
_{2}u_{2}+\alpha _{3}u_{3}$.

Note that

$<\overrightarrow{u}_{1},\overrightarrow{v}>=\overrightarrow{u}_{1}\cdot 
\overrightarrow{v}=<\overrightarrow{u}_{1},\alpha _{1}\overrightarrow{u}%
_{1}+\alpha _{2}\overrightarrow{u}_{2}+\alpha _{3}\overrightarrow{u}_{3}>=<%
\overrightarrow{u}_{1},\alpha _{1}\overrightarrow{u}_{1}>+<\overrightarrow{u}%
_{1},\alpha _{2}\overrightarrow{u}_{2}>+<\overrightarrow{u}_{1},\alpha _{3}%
\overrightarrow{u}_{3}>$

\qquad \qquad \qquad $\qquad \qquad \qquad \qquad \qquad \qquad \qquad
\qquad \qquad =\alpha _{1}<\overrightarrow{u}_{1},\overrightarrow{u}%
_{1}>+\alpha _{2}<\overrightarrow{u}_{1},\overrightarrow{u}_{2}>+\alpha _{3}<%
\overrightarrow{u}_{1},\overrightarrow{u}_{3}>=\alpha _{1}$

Also $<\overrightarrow{u}_{2},\overrightarrow{v}>=\alpha _{2}$ and $<%
\overrightarrow{u}_{3},\overrightarrow{v}>=\alpha _{3}$.

\vspace{1pt}

Suppose we are given a set of orthogonal \ functions $\left\{ \psi
_{n}\right\} $ on $\left[ 0,L\right] ,$ and we desire to expand a function $%
f\left( x\right) $ given on $\left[ 0,L\right] $ in terms of them. Then we
want

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
f\left( x\right) =\sum\limits_{n=1}^{\infty }\alpha _{n}\psi _{n}\left(
x\right) .
\end{equation*}

\vspace{1pt}

Question. What does $\alpha _{k}=$?

\vspace{1pt}

Consider 
\begin{eqnarray*}
&<&\psi _{k},f\left( x\right) >=<\psi _{k},\sum\limits_{1}^{\infty }\alpha
_{n}\psi _{n}> \\
&=&<\psi _{k},\alpha _{1}\psi _{1}+\alpha _{2}\psi _{2}+\cdots > \\
&=&\alpha _{1}<\psi _{k},\psi _{1}>+\cdots +\alpha _{k}<\psi _{k},\psi _{k}>+%
\vspace{1pt}\alpha _{k+1}<\psi _{k},\psi _{k+1}>+\cdots
\end{eqnarray*}

\qquad \qquad

But $<\psi _{k},\psi _{j}>=0$ if $j\neq k$ since the set $\left\{ \psi
_{k}\right\} $ is orthogonal.

\vspace{1pt}

$\Longrightarrow $%
\begin{equation*}
<\psi _{k},f\left( x\right) >=\alpha _{k}<\psi _{k},\psi _{k}>=\alpha
_{k}\left\| \psi _{k}\right\| ^{2}
\end{equation*}

\vspace{1pt}

Therefore 
\begin{equation*}
\alpha _{k}=\dfrac{\int_{0}^{L}f\left( x\right) \psi _{k}\left( x\right) dx}{%
\left\| \psi _{k}\right\| ^{2}}=\dfrac{\int_{0}^{L}f\left( x\right) \psi
_{k}\left( x\right) dx}{\int_{0}^{L}\left[ \psi _{k}\left( x\right) \right]
^{2}dx}\qquad k=1,2,...\qquad \qquad \qquad \left( \ast \right)
\end{equation*}

$\left( \ast \right) $ is the formula for the coefficients in the expansion
of a function $f\left( x\right) $\ in terms of a set of orthogonal functions.

\vspace{1pt}

\subsubsection{Ordinary Fourier Series}

\vspace{1pt}

\paragraph{Fourier Sine Series}

\vspace{1pt}

Consider the eigenvalue problem

\begin{equation*}
D.E.y^{\prime \prime }+\lambda y=0\qquad 0\leq x\leq L\qquad B.C.y\left(
0\right) =y\left( L\right) =0\qquad
\end{equation*}

\vspace{1pt}

We shall first solve this problem. There are 3 cases to consider - \ $%
\lambda \,<0,\lambda =0,\lambda >0.$

\vspace{1pt}

I. $\lambda \,<0.$ \ Let $\lambda \,=-\alpha ^{2}$ where $\alpha \neq 0.$ \
The DE becomes\qquad

\vspace{1pt}

\begin{equation*}
y^{\prime \prime }-\alpha ^{2}y=0
\end{equation*}%
so 
\begin{equation*}
y\left( x\right) =c_{1}e^{\alpha x}+c_{2}e^{-\alpha x}
\end{equation*}%
Then $y\left( 0\right) =0$ implies

\vspace{1pt}%
\begin{equation*}
c_{1}+c_{2}=0
\end{equation*}%
so $c_{2}=-c_{1}$ and 
\begin{equation*}
y\left( x\right) =c_{1}\left[ e^{\alpha x}-e^{-\alpha x}\right]
\end{equation*}

But then 
\begin{equation*}
y\left( L\right) =c_{1}\left[ e^{\alpha L}-e^{-\alpha L}\right] =0
\end{equation*}
So $c_{1}=0$ and hence $c_{2}=0$ \ and thus $y\left( x\right) =0$ and there
are no negative eigenvalues.

\vspace{1pt}

II. $\lambda \,=0$ The the equation becomes $y^{\prime \prime }=0$ and $%
y=c_{1}x+c_{2}$ \ and the BCs imply $y=0.$

\vspace{1pt}

III. \ $\lambda \,>0.$ \ Let $\lambda =\beta ^{2}$ where $\beta \neq 0$ \
The DE becomes%
\begin{equation*}
y^{\prime \prime }+\beta ^{2}y=0
\end{equation*}

Thus%
\begin{equation*}
y=c_{1}\sin \beta x+c_{2}\cos \beta x
\end{equation*}%
$y\left( 0\right) =c_{2}=0$. Also%
\begin{equation*}
y\left( L\right) =c_{1}\sin \beta L=0
\end{equation*}%
so%
\begin{equation*}
\beta =\frac{n\pi }{L}\text{ \ \ \ }n=1,2,3,\ldots
\end{equation*}

$\Longrightarrow $%
\begin{equation*}
\lambda _{n}=\dfrac{n^{2}\pi ^{2}}{L^{2}}\text{ \ }n=1,2,3,\ldots
\end{equation*}%
are the eigenvalues, whereas the eigenfunctions are

\begin{equation*}
\sin \sqrt{\lambda _{n}}x=\sin \dfrac{n\pi }{L}x=\psi _{n}\text{ \ }%
n=1,2,3,\ldots
\end{equation*}%
These functions form an orthogonal set.

\vspace{1pt}

Hence if 
\begin{equation*}
f\left( x\right) =\sum\limits_{1}^{\infty }\alpha _{k}\sin \dfrac{k\pi x}{L}
\end{equation*}
then from $\left( \ast \right) $ above 
\begin{equation*}
\alpha _{k}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \sin \dfrac{k\pi x}{L}dx,
\end{equation*}%
since%
\begin{equation*}
\int_{0}^{L}\left[ \psi _{k}\left( x\right) \right] ^{2}dx=\frac{L}{2}.
\end{equation*}

\vspace{1pt}

These formulas are for the Fourier \textit{sine} series for $f\left(
x\right) $ on $0<x<L$.

Remarks. 1. At $x=0$ and $x=L\qquad \sum \alpha _{k}\sin \dfrac{k\pi x}{L}$
gives $0$ for $f\left( x\right) $. Therefore unless $f\left( 0\right)
=f\left( L\right) =0$ the Fourier series is not good at the end points.

\qquad \qquad 2. Since $\sin \dfrac{k\pi }{L}\left( x+2L\right) =\sin \left( 
\dfrac{k\pi }{L}x+2k\pi \right) =\sin \dfrac{k\pi x}{L}$, we see that the
Fourier series yields $f\left( x+2L\right) =f\left( x\right) \Longrightarrow 
$ Fourier series has period $2L$. For $-L<x<0$

\vspace{1pt}

we have $\sum\limits_{1}^{\infty }\alpha _{k}\sin \dfrac{k\pi x}{L}%
=\sum\limits_{1}^{\infty }\alpha _{k}\sin \left( \dfrac{-k\pi (-x)}{L}%
\right) $

\qquad \qquad \qquad \qquad \qquad \qquad $=-\sum\limits_{1}^{\infty }\alpha
_{k}\sin \dfrac{k\pi (-x)}{L}\qquad -L<x<0\Longrightarrow L>-x>0$

\qquad \qquad \qquad \qquad \qquad \qquad $=-f\left( -x\right) ,$ where $%
f\left( x\right) $ is value of series in $0<x<L$.

\vspace{1pt}

Therefore the Fourier sine series converges to function $F\left( x\right) $
where

\vspace{1pt}

\vspace{1pt}\qquad \qquad 
\begin{equation*}
F\left( x\right) =\left\{ 
\begin{array}{c}
f\left( x\right) \qquad 0<x<L \\ 
-f\left( -x\right) \qquad -L<x<0%
\end{array}%
\right. \qquad \qquad F\left( x+2L\right) =F\left( x\right)
\end{equation*}

This is the odd periodic extension of $f\left( x\right) $ with period $2L$.
Unless $f\left( \pm kL\right) =0\qquad F\left( x\right) $ will be
discontinuous at $\pm L,\pm 2L,...$ Note that the function $f\left( x\right) 
$ is given on $\left[ 0,L\right] $ only, where the Fourier Sine series
extends it to a function $F\left( x\right) $ which is define on $-\infty
<x<\infty $.

\qquad

Suppose that the graph of the function $f\left( x\right) $ is given by the
figure below.

\vspace{1pt}

\vspace{1pt}\FRAME{dtbpF}{3.0978in}{2.4699in}{0pt}{}{}{image1.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 3.0978in;height 2.4699in;depth
0pt;original-width 194.25pt;original-height 154.3125pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'/document/graphics/Image10.gif';file-properties "XNPEU";}}

Then the Fourier sine series generates a function $F\left( x\right) $
defined on -$\infty <x<\infty $ whose graph is given below.

\vspace{1pt}

\vspace{1pt}\FRAME{dtbpF}{5.738in}{2.8971in}{0pt}{}{}{image2.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 5.738in;height 2.8971in;depth
0pt;original-width 458.4375pt;original-height 229.625pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'/document/graphics/Image20.gif';file-properties "XNPEU";}}

\vspace{1pt}

\begin{example}
Find the Fourier sine series of
\end{example}

\vspace{1pt}

\begin{center}
\begin{equation*}
f\left( x\right) =\left\{ 
\begin{array}{c}
1\qquad 0<x<\frac{\pi }{2} \\ 
0\qquad \frac{\pi }{2}<x<\pi%
\end{array}%
\right.
\end{equation*}%
$\qquad $

$\vspace{1pt}$
\end{center}

\vspace{1pt}\FRAME{dtbpF}{2.4803in}{1.708in}{0pt}{}{}{image3.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 2.4803in;height 1.708in;depth
0pt;original-width 225.0625pt;original-height 154.3125pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'/document/graphics/Image30.gif';file-properties "XNPEU";}}

Now 
\begin{equation*}
f\left( x\right) =\sum \alpha _{n}\sin \dfrac{n\pi x}{L}=\sum\limits_{1}^{%
\infty }\alpha _{n}\sin nx,
\end{equation*}
since $2L=2\pi \Longrightarrow L=\pi $.

The formula above for the coefficients in the Fourier sine series implies

\vspace{1pt}

\begin{equation*}
\alpha _{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \sin \dfrac{n\pi x}{L}dx=%
\dfrac{2}{\pi }\int_{0}^{\pi }f\left( x\right) \sin nxdx
\end{equation*}

\begin{eqnarray*}
\alpha _{n} &=&\dfrac{2}{\pi }\int_{0}^{\frac{\pi }{2}}1\cdot \sin nxdx+%
\dfrac{2}{\pi }\dint_{\dfrac{\pi }{2}}^{\pi }0\cdot \sin nxdx=-\dfrac{2}{\pi 
}\dfrac{\cos nx}{n}|_{0}^{\dfrac{\pi }{2}} \\
&=&-\dfrac{2}{\pi n}\left[ \cos \dfrac{n\pi }{2}-1\right]
\end{eqnarray*}

\vspace{1pt}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
\alpha _{n}=\left\{ 
\begin{array}{c}
\dfrac{2}{\pi n}\qquad \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\qquad n\text{ }odd
\\ 
\left( \dfrac{-2}{\pi n}\right) \left[ \left( -1\right) ^{\frac{n}{2}}-1%
\right] \text{ \ \ \ }n\text{ }even%
\end{array}%
\right.
\end{equation*}

Therefore

\vspace{1pt}

\begin{eqnarray*}
f\left( x\right) &=&\sum\limits_{1}^{\infty }\alpha _{n}\sin nx \\
&=&\dfrac{2}{\pi }\left[ \sin x+\frac{2}{2}\sin 2x+\frac{1}{3}\sin 3x+0\cdot
\sin 4x+\frac{1}{5}\sin 5x+\frac{2}{6}\sin 6x+\cdots \right]
\end{eqnarray*}

Note that our function $f\left( x\right) $ on $0\leq x\leq \pi $ is extended
to the following on $-\infty <x<\infty $.

\vspace{1pt}

\vspace{1pt}\FRAME{dtbpF}{6.333in}{3.5622in}{0pt}{}{}{image4.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 6.333in;height 3.5622in;depth
0pt;original-width 435.125pt;original-height 243.9375pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'/document/graphics/Image40.gif';file-properties "XNPEU";}}

\vspace{1pt}

What we have done with \textit{sine} functions can be done with \textit{%
cosine} functions.

\vspace{1pt}

\paragraph{Fourier Cosine Series.}

\vspace{1pt}

This comes from eigenvalue problem

\vspace{1pt}

\begin{equation*}
\text{D.E. \ \ }y^{\prime \prime }+\lambda y=0\qquad \text{B.C. \ \ \ \ }%
y^{\prime }\left( 0\right) =y^{\prime }\left( L\right) =0
\end{equation*}

$\vspace{1pt}$

\begin{equation*}
\lambda _{n}=\dfrac{n^{2}\pi ^{2}}{L^{2}}
\end{equation*}%
are the eigenvalues and 
\begin{equation*}
\psi _{n}=\cos \dfrac{n\pi x}{L}
\end{equation*}%
are the eigenfunctions, $n=0,1,2,...$.

\vspace{1pt}

Note $\lambda _{0}=0\Longrightarrow \psi _{0}=1$ which is an eigenfunction.
Now we want to write

\vspace{1pt}

\qquad 
\begin{equation*}
f\left( x\right) =\beta _{0}+\sum\limits_{1}^{\infty }\beta _{n}\cos \dfrac{%
n\pi x}{L}
\end{equation*}

Proceeding as above in our derivation of the constants in the Fourier Sine
series, we get for the constants in the Fourier Cosine series

\vspace{1pt}

\vspace{1pt}\qquad \qquad \qquad 
\begin{equation*}
\beta _{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \dfrac{n\pi x}{L}dx%
\text{ \ }n=1,2,3,\ldots \qquad \beta _{0}=\frac{1}{L}\int_{0}^{L}f\left(
x\right) dx
\end{equation*}

To see where the formula for $\beta _{0}$ comes from note

\qquad \qquad $<\psi _{0},f\left( x\right) >=\psi _{0},\beta _{0}\psi
_{0}>=<1,1>\beta _{0}$

\vspace{1pt}

\qquad $\Longrightarrow \beta _{0}=\dfrac{\int_{0}^{L}1\cdot f\left(
x\right) dx}{\int_{0}^{L}1^{2}dx}=\frac{1}{L}\int_{0}^{L}f\left( x\right) dx$%
.

Note the book writes%
\begin{equation*}
f\left( x\right) \symbol{126}\frac{a_{0}}{2}+\sum\limits_{1}^{\infty
}a_{n}\cos \dfrac{n\pi x}{L}
\end{equation*}%
and%
\begin{equation*}
a_{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \dfrac{n\pi x}{L}dx\text{
\ }n-0,1,2,\ldots
\end{equation*}

\vspace{1pt}

Thus 
\begin{equation*}
\beta _{0}=\frac{a_{0}}{2}
\end{equation*}

\vspace{1pt}

Again the Fourier series is periodic with period $2L$. However, now $f\left(
-x\right) =f\left( x\right) $ since \textit{cosine} is an even function.
Here the Fourier Cosine series extends $f\left( x\right) $ which is given on 
$\left[ 0,L\right] $ to a function $F\left( x\right) $ which is defined on $%
-\infty <x<\infty $ as

\qquad 
\begin{equation*}
F\left( x\right) =\left\{ 
\begin{array}{c}
f\left( x\right) \qquad 0<x<L \\ 
f\left( -x\right) \text{ }-L<x<0%
\end{array}%
\right. \qquad F\left( x+2L\right) =F\left( x\right) .
\end{equation*}

\vspace{1pt}

\vspace{1pt}If the graph of $f\left( x\right) $ looked as below\FRAME{dtbpF}{%
2.6377in}{2.1024in}{0pt}{}{}{image5.gif}{\special{language "Scientific
Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file
"F";width 2.6377in;height 2.1024in;depth 0pt;original-width
194.25pt;original-height 154.3125pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename
'/document/graphics/Image50.gif';file-properties "XNPEU";}}

then $F\left( x\right) ,$ the \textit{even }extension of $f\left( x\right) $%
, would look like

\vspace{1pt}\FRAME{dtbpF}{5.3558in}{2.7043in}{0pt}{}{}{image6.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 5.3558in;height 2.7043in;depth
0pt;original-width 458.4375pt;original-height 229.625pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'/document/graphics/Image60.gif';file-properties "XNPEU";}}

\vspace{1pt}

Example. Find the Fourier Cosine series for $f\left( x\right) =1,\qquad
0<x<4 $

\qquad \qquad $L=4$

\qquad \qquad $f\left( x\right) =\beta _{0}+\sum\limits_{1}^{\infty }\beta
_{n}\cos \dfrac{n\pi x}{4}\qquad \beta _{0}=\frac{1}{4}\int_{0}^{4}f\left(
x\right) dx=\frac{1}{4}\int_{0}^{4}1\cdot dx=1$

\vspace{1pt}

\qquad \qquad $\beta _{k}=\frac{2}{4}\int_{0}^{4}1\cdot \cos \dfrac{n\pi x}{4%
}dx=\frac{1}{2}\left[ \frac{\sin \dfrac{n\pi x}{4}}{\dfrac{n\pi }{4}}\right]
_{0}^{4}=\ \dfrac{2}{4\pi n}\left[ \sin 0\right] =0$

Therefore $f\left( x\right) =1$ is its own Fourier Cosine series. The
function is simply extended.

\vspace{1pt}

\vspace{1pt}\FRAME{dtbpF}{6.1479in}{1.8827in}{0pt}{}{}{image7.gif}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 6.1479in;height 1.8827in;depth
0pt;original-width 422.3125pt;original-height 128pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'/document/graphics/Image70.gif';file-properties "XNPEU";}}

\begin{example}
Find the Fourier cosine series of
\end{example}

\vspace{1pt}

\begin{center}
\begin{equation*}
f\left( x\right) =\left\{ 
\begin{array}{c}
1\qquad 0<x<\frac{\pi }{2} \\ 
0\qquad \frac{\pi }{2}<x<\pi%
\end{array}%
\right.
\end{equation*}%
$\qquad $
\end{center}

$\vspace{1pt}$The graph of $f\left( x\right) $ is given below.

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Note that this is the same function as in the previous example.

Now 
\begin{equation*}
f\left( x\right) =b_{0}+\sum_{n=1}^{\infty }b_{n}\cos \left( \dfrac{n\pi x}{L%
}\right) =b_{0}+\sum\limits_{1}^{\infty }b_{n}\cos nx,
\end{equation*}%
since the function is given on $\left[ 0,L\right] \Longrightarrow L=\pi $.

\begin{eqnarray*}
b_{0} &=&\frac{1}{L}\int_{0}^{L}f\left( x\right) dx=\frac{1}{\pi }\int_{0}^{%
\frac{\pi }{2}}1dx=\frac{1}{2} \\
b_{n} &=&\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \left( \frac{n\pi }{L}%
x\right) dx=\frac{2}{\pi }\int_{0}^{\frac{\pi }{2}}1\cdot \cos nxdx \\
&=&\frac{2}{n\pi }\left[ \sin nx\right] _{0}^{\frac{\pi }{2}}=\frac{2}{n\pi }%
\sin \left( \frac{n\pi }{2}\right)
\end{eqnarray*}

If $n$ is even, then $\sin \left( \frac{n\pi }{2}\right) =0.$ When $n$ is
odd, say $n=2k+1,k=0,1,2,\ldots $ then $\sin \left( \frac{n\pi }{2}\right)
=\pm 1,$ depending on whether $k$ is even or odd. Thus%
\begin{equation*}
b_{n}=\left\{ 
\begin{array}{c}
0\text{ \ }n\text{ even} \\ 
\frac{2}{n\pi }\left( -1\right) ^{k}\text{ \ }n\text{ odd, \ }%
n=2k+1,k=0,1,2,\ldots%
\end{array}%
\right.
\end{equation*}

Thus

\vspace{1pt}%
\begin{eqnarray*}
f\left( x\right) &=&b_{0}+\sum\limits_{1}^{\infty }b_{n}\cos
nx=b_{0}+b_{1}\cos x+b_{2}\cos 2x+\cdots \\
&=&\frac{1}{2}+\frac{2}{\pi }\cos x+0\cos 2x-\frac{2}{2\pi }\cos 3x+0\cos 4x+%
\frac{2}{5\pi }\cos 5x+\cdots
\end{eqnarray*}

The graph of the even extension of the given function is

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\vspace{1pt}

\begin{example}
\vspace{1pt}(a) Find the first four nonzero terms of the Fourier \textit{%
cosine} series for the function 
\begin{equation*}
f(x)=x\text{ \ on }0<x<1
\end{equation*}
\end{example}

Solution: $\qquad $%
\begin{equation*}
f\left( x\right) =\beta _{0}+\sum\limits_{1}^{\infty }\beta _{n}\cos \dfrac{%
n\pi x}{L}
\end{equation*}

where 
\begin{equation*}
\beta _{0}=\frac{1}{L}\int_{0}^{L}f\left( x\right) dx\text{ and }\beta _{n}=%
\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \dfrac{n\pi x}{L}dx\text{ \ }%
n=1,2,3,\ldots
\end{equation*}

Here $L=1$ so

\vspace{1pt}%
\begin{equation*}
f\left( x\right) =\beta _{0}+\sum\limits_{1}^{\infty }\beta _{n}\cos n\pi x
\end{equation*}

\begin{eqnarray*}
\beta _{0} &=&\frac{1}{1}\int_{0}^{1}xdx=\frac{1}{2} \\
\beta _{n} &=&\frac{2}{1}\int_{0}^{1}x\cos n\pi xdx=\frac{2}{n^{2}\pi ^{2}}%
\left. \left( \cos n\pi x+n\pi x\sin n\pi x\right) \right\vert _{0}^{1} \\
&=&\frac{2}{n^{2}\pi ^{2}}\left( \cos n\pi -1\right) =\frac{2}{n^{2}\pi ^{2}}%
\left( \left( -1\right) ^{n}-1\right) \text{ \ }n=1,2,3,\ldots
\end{eqnarray*}

Hence $\beta _{1}=-\frac{4}{\pi ^{2}},$ \ $\beta _{2}=0,$ \ $\beta _{3}=-%
\frac{4}{9\pi ^{2}},$ \ $\beta _{4}=0,$ \ $\beta _{5}=-\frac{4}{25\pi ^{2}}$

Therefore%
\begin{equation*}
f\left( x\right) =\frac{1}{2}-\frac{4}{\pi ^{2}}\cos \pi x-\frac{4}{9\pi ^{2}%
}\cos 3\pi x-\frac{4}{25\pi ^{2}}\cos 5\pi x
\end{equation*}

Note: The book gives the formulas%
\begin{equation*}
f\left( x\right) =\frac{\beta _{0}}{2}+\sum\limits_{1}^{\infty }\beta
_{n}\cos \dfrac{n\pi x}{L}
\end{equation*}

where 
\begin{equation*}
\beta _{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \dfrac{n\pi x}{L}dx%
\text{ \ }n=0,1,2,3,\ldots
\end{equation*}%
Using this formula we get 
\begin{equation*}
\beta _{0}=\frac{2}{1}\int_{0}^{1}xdx=1
\end{equation*}%
Therefore, the first term in the book's formula for the Fourier cosine
series is $\frac{\beta _{0}}{2}=\frac{1}{2}$ as before.

(b) Sketch the graph of the function represented by the Fourier cosine
series in (a) on $-3<x<3.$

\vspace{1pt}\vspace{1pt}\vspace{1pt}\vspace{1pt}

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\vspace{1pt}

\begin{example}
(a) Find the Fourier \textit{sine} series for the function 
\begin{equation*}
f(x)=x\text{ \ on }0<x<1
\end{equation*}
\end{example}

Solution:%
\begin{equation*}
f\left( x\right) =\sum\limits_{1}^{\infty }\alpha _{k}\sin \dfrac{k\pi x}{L}
\end{equation*}%
where 
\begin{equation*}
\alpha _{k}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \sin \dfrac{k\pi x}{L}dx,%
\text{ \ \ }k=1,2,3,\ldots
\end{equation*}

Here $L=1$ so 
\begin{equation*}
f\left( x\right) =\sum\limits_{1}^{\infty }\alpha _{k}\sin \left( k\pi
x\right)
\end{equation*}%
where 
\begin{equation*}
\alpha _{k}=2\int_{0}^{1}f\left( x\right) \sin \left( k\pi x\right) dx,\text{
\ \ }k=1,2,3,\ldots
\end{equation*}

Thus%
\begin{eqnarray*}
\alpha _{k} &=&2\int_{0}^{1}x\sin \left( k\pi x\right) dx=2\left[ \frac{1}{%
\left( k\pi \right) ^{2}}\left( \sin k\pi x-k\pi x\cos k\pi x\right) \right]
_{0}^{1}= \\
&=&-2\left[ \frac{1}{k\pi }\cos k\pi \right] =\frac{2}{k\pi }\left(
-1\right) ^{k+1}\text{ \ \ \ }k=1,2,3,\ldots
\end{eqnarray*}

Thus 
\begin{equation*}
f\left( x\right) =\frac{2}{\pi }\sum\limits_{1}^{\infty }\frac{\left(
-1\right) ^{k+1}}{k}\sin \left( k\pi x\right)
\end{equation*}

\vspace{1pt}

(b) Sketch the graph of the function represented by the Fourier \textit{sine}
series in 5 (a) on $-3<x<3.$

Solution: \ 

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\subsubsection{Full Fourier Series (Omit)}

\vspace{1pt}

\qquad This comes from the eigenvalue problem

\begin{equation*}
D.E.y^{\prime \prime }+\lambda y=0\qquad B.C.y\left( 0\right) =y\left(
2L\right) \qquad y^{\prime }\left( 0\right) =y^{\prime }\left( 2L\right)
\qquad 0\leq x\leq 2L
\end{equation*}

\vspace{1pt}

The eigenvalues are 
\begin{equation*}
\lambda _{n}=\dfrac{n^{2}\pi ^{2}}{L^{2}}
\end{equation*}%
$n=0,1,2,...,$, whereas the eigenfunctions are

\vspace{1pt}

\qquad \qquad 
\begin{equation*}
\psi _{n}=a_{n}\cos \dfrac{n\pi x}{L}+b_{n}\sin \dfrac{2\pi x}{L}\qquad
n=0,1,2,..
\end{equation*}%
$.$

Note that for this problem the function $f\left( x\right) $ is given on $%
\left[ 0,2L\right] $ since the eigenvalue problem is given on this interval.
This is a different interval than that for Fourier Sine and Fourier Cosine
series.

\vspace{1pt}

$\Longrightarrow $%
\begin{equation*}
f\left( x\right) =a_{0}+\sum\limits_{n=1}^{\infty }\left( a_{n}\cos \dfrac{%
n\pi x}{L}+b_{n}\sin \dfrac{n\pi x}{L}\right)
\end{equation*}

where 
\begin{equation*}
a_{0}=\frac{1}{2L}\int_{0}^{2L}f\left( x\right) dx,\qquad a_{n}=\frac{1}{L}%
\int_{0}^{2L}f\left( x\right) \cos \dfrac{n\pi x}{L}dx\qquad b_{n}=\frac{1}{L%
}\int_{0}^{2L}f\left( x\right) \sin \dfrac{n\pi x}{L}dx
\end{equation*}

\vspace{1pt}

\begin{example}
Find full Fourier series for 
\begin{equation*}
f\left( x\right) =\left\{ 
\begin{array}{c}
1\qquad 0<x<\frac{\pi }{2} \\ 
0\text{ \ \ \ \ \ }\frac{\pi }{2}<x<\pi%
\end{array}%
\right.
\end{equation*}%
$\qquad 2L=\pi \Longrightarrow L=\dfrac{\pi }{2}$
\end{example}

\vspace{1pt}

$a_{0}=\dfrac{1}{\pi }\int_{0}^{\pi }f\left( x\right) dx=\ \dfrac{1}{\pi }%
\int_{0}^{\dfrac{\pi }{2}}1\cdot dx=\frac{1}{2}$

\vspace{1pt}

$a_{n}=\frac{1}{\dfrac{\pi }{2}}\int_{0}^{\pi }f\left( x\right) \cos \dfrac{%
n\pi x}{\frac{\pi }{2}}dx=\frac{2}{\pi }\int_{0}^{\dfrac{\pi }{2}}1\cdot
\cos 2nx$ $dx=\frac{2}{\pi }\int_{0}^{\dfrac{\pi }{2}}\cos 2nx$ $dx=\frac{2}{%
\pi }\dfrac{\sin 2nx}{2n}|_{0}^{\dfrac{\pi }{2}}=0$

\vspace{1pt}

$b_{n}=\frac{2}{\pi }\int_{0}^{\dfrac{\pi }{2}}\sin 2nx\,dx=-\frac{2}{\pi }%
\frac{\cos 2nx}{2n}|_{0}^{\dfrac{\pi }{2}}=\ \dfrac{1}{\pi n}\left[ \cos
n\pi -\cos 0\right] \qquad n=1,2,...$

\vspace{1pt}

\qquad \qquad $b_{n}=-\dfrac{1}{\pi n}\left[ \left( -1\right) ^{n}-1\right]
=\left\{ 
\begin{array}{c}
+\dfrac{2}{\pi n}\qquad n\text{ odd} \\ 
0\text{ \ \ \ \ \ \ \ \ \ \ }n\text{ even}%
\end{array}
\right. $

\vspace{1pt}

\qquad \qquad 
\begin{equation*}
f\left( x\right) =\frac{1}{2}+\frac{2}{\pi }\left[ \sin 2x+\frac{1}{3}\sin
6x+\frac{1}{5}\sin 10x+.\text{ }.\text{ }.\right]
\end{equation*}

\vspace{1pt}

\subsubsection{\protect\Large The Vibrating String}

\vspace{1pt}

It may be shown that the equation governing a string of length $L$ vibrating
is

\qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y_{xx}\left( x,t\right) =\dfrac{\partial ^{2}y}{\partial x^{2}}=\dfrac{1}{%
\alpha ^{2}}y_{tt}\left( x,t\right) \qquad \qquad (1)
\end{equation*}

\vspace{1pt}

Equation $(1)$ is called the wave equation. Suppose string is held fixed at
the ends $x=0$ and $x=L$

\vspace{1pt}

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\vspace{1pt}

$\Longrightarrow \qquad $%
\begin{equation*}
(2a)\ \text{\ \ }y\left( 0,t\right) =0\qquad t\geq 0\qquad B.C.
\end{equation*}

\vspace{1pt}

\qquad \qquad 
\begin{equation*}
(2b)\ \text{\ \ }y\left( L,t\right) =0\qquad t\geq 0\qquad B.C.
\end{equation*}

\vspace{1pt}

Also suppose at $t=0$ the string has displacement $y=f\left( x\right) $ and
is released from rest

$\vspace{1pt}$

$\Longrightarrow \qquad \ $%
\begin{equation*}
(3a)\text{ \ \ }\ y\left( x,0\right) =f\left( x\right) \qquad 0\leq x\leq
L\qquad I.C.
\end{equation*}

\qquad \qquad 
\begin{equation*}
(3b)\ \text{\ \ \ }y_{t}\left( x,0\right) =0\qquad 0\leq x\leq L\qquad \ \
I.C.
\end{equation*}

\vspace{1pt}

In order to solve the above problem we shall assume $y\left( x,t\right)
=X\left( x\right) T\left( t\right) $ separation of variables $%
\Longrightarrow y_{x}=X^{\prime }T\qquad y_{xx}=X^{\prime \prime }T\qquad
y_{tt}=XT^{\prime \prime }$. Note that $X^{\prime },T^{\prime },...$ are 
\emph{ordinary }derivatives of $X$ with respect to $x$ and $T$ with respect
to $t.$ Now the P.D.E. $\left( 1\right) $

$\vspace{1pt}$

$\Longrightarrow $%
\begin{equation*}
X^{\prime \prime }T=\frac{1}{\alpha ^{2}}XT^{\prime \prime }
\end{equation*}%
$\Longrightarrow \qquad $%
\begin{equation*}
\dfrac{X^{\prime \prime }}{X}=\dfrac{1}{\alpha ^{2}}\dfrac{T^{\prime \prime }%
}{T}
\end{equation*}%
.

\vspace{1pt}

Note that the left hand side is a function of $x$ only, whereas the right
hand side is a function of $t$ only. This implies that each side must equal
the same constant. Therefore

$\vspace{1pt}$

\begin{center}
\begin{equation*}
\dfrac{X^{\prime \prime }}{X}=\dfrac{1}{\alpha ^{2}}\dfrac{T^{\prime \prime }%
}{T}=k
\end{equation*}
\end{center}

Hence we get the two \emph{ordinary} differential equations\qquad

$\vspace{1pt}$

\begin{center}
\begin{equation*}
X^{\prime \prime }-kX=0\qquad and\ \ \ \ \ T^{\prime \prime }-\alpha ^{2}kT=0
\end{equation*}
\end{center}

\vspace{1pt}

Now $y\left( 0,t\right) =X\left( 0\right) T\left( t\right) =0\Longrightarrow
X\left( 0\right) =0,$ whereas $y\left( L,t\right) =X\left( L\right) T\left(
t\right) =0\Longrightarrow X\left( L\right) =0$. Therefore we must solve the
problem

$\vspace{1pt}$

\begin{center}
\begin{equation*}
X^{\prime \prime }-kX=0\qquad X\left( 0\right) =X\left( L\right) =0.
\end{equation*}
\end{center}

\vspace{1pt}

There are three cases. If $k=0\Longrightarrow X\equiv 0$. If $%
k>0\Longrightarrow X=c_{1}e^{\sqrt{k}x}+c_{2}e^{-\sqrt{k}x}$.

and the boundary conditions $\Longrightarrow c_{1}=c_{2}=0$.

\vspace{1pt}

For the case $k<0,$ let $k=-\lambda ^{2}$

$\vspace{1pt}\Longrightarrow $%
\begin{equation*}
X^{\prime \prime }+\lambda ^{2}X=0\qquad X\left( 0\right) =X\left( L\right)
=0
\end{equation*}

\vspace{1pt}

This is an eigenvalue problem. The solution to the DE is

\begin{equation*}
X=c_{1}\sin \lambda x+c_{2}\cos \lambda x
\end{equation*}

$X\left( 0\right) =0\Longrightarrow c_{2}=0$ whereas $X\left( L\right) =0$ $%
\ \Longrightarrow c_{1}\sin \lambda =0$ $\ \Longrightarrow \lambda =\frac{%
n\pi }{L}$ for $n=\pm 1,\pm 2,\pm 3,...$.

Since $\sin \left( -x\right) =-\sin x$ we may disregard the negative values
of $n$.

Therefore

\vspace{1pt}

\begin{center}
\begin{equation*}
X_{n}\left( x\right) =c_{n}\sin \dfrac{n\pi }{L}x\qquad n=1,2,3,...
\end{equation*}
\end{center}

\vspace{1pt}

For $T\left( t\right) $ we have the equation 
\begin{equation*}
T^{\prime \prime }+\alpha ^{2}\lambda ^{2}T=0,
\end{equation*}%
since $k=-\lambda ^{2}$. Thus

$\vspace{1pt}$

\begin{center}
\begin{equation*}
\NEG{T}_{n}\left( t\right) =c\sin \alpha \lambda t+d\cos \alpha \lambda
t=a_{n}\sin \dfrac{n\pi \alpha }{L}t+b_{n}\cos \dfrac{n\pi \alpha t}{L}.
\end{equation*}
\end{center}

\vspace{1pt}

But $y_{t}\left( x,0\right) =0\Longrightarrow T^{\prime }\left( 0\right) =0.$
Now $T^{\prime }\left( t\right) =a_{n}\left( \alpha \dfrac{n\pi }{L}\right)
\cos \alpha \dfrac{n\pi }{L}t-b_{n}\left( \alpha \dfrac{n\pi }{L}\right)
\sin \alpha \dfrac{n\pi t}{L}$, so $T^{\prime }\left( 0\right) =0$ $%
\Rightarrow a_{n}=0$ for all $n$ .

Therefore

$\vspace{1pt}$

\begin{center}
\begin{equation*}
T_{n}\left( t\right) =b_{n}\cos \dfrac{n\pi \alpha t}{L},
\end{equation*}
\end{center}

\vspace{1pt}

and we have finally that

\vspace{1pt}

\qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y_{n}\left( x,t\right) =X_{n}\left( x\right) T_{n}\left( t\right) =c_{n}\sin 
\dfrac{n\pi x}{L}\times b_{n}\cos \dfrac{n\pi \alpha t}{L}
\end{equation*}

\vspace{1pt}

Let $c_{n}\times b_{n}=d_{n}.$

We note that 
\begin{equation*}
y_{n}\left( x,t\right) =d_{n}\sin \dfrac{n\pi x}{L}\cos \dfrac{n\pi \alpha t%
}{L}\text{ \ \ }n=1,2,3,\ldots
\end{equation*}

\vspace{1pt}

satisfies the P.D.E. $y_{xx}=\dfrac{1}{\alpha ^{2}}y_{tt}$ $\ \left(
1\right) \qquad $and the boundary conditions $y\left( 0,t\right) =y\left(
L,t\right) =0$ $\left( 2a,2b\right) ,$ as well as the initial condition $%
y_{t}\left( 0\right) =0$ $\left( 3b\right) $.

\vspace{1pt}

What about the condition $y\left( x,0\right) =f\left( x\right) $? Notice that

\begin{equation*}
y\left( x,t\right) =\sum\limits_{1}^{\infty }d_{n}\sin \dfrac{n\pi x}{L}\cos 
\dfrac{n\pi \alpha t}{L}
\end{equation*}

is also a solution since of $\left( 1\right) ,\left( 2a,b\right) $ and $%
\left( 3b\right) $. Thus $y\left( x,t\right) $ is solution of everything
except condition $\left( 3a\right) ,$ namely, $y\left( x,0\right) =f\left(
x\right) $.

But

\vspace{1pt}

\qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y\left( x,0\right) =\sum\limits_{1}^{\infty }d_{n}\sin \dfrac{n\pi x}{L}%
=f\left( x\right) .
\end{equation*}

\vspace{1pt}

Therefore if $f$ has a Fourier sine series expansion we let

\begin{center}
$\vspace{1pt}$
\end{center}

$\Longrightarrow $%
\begin{equation*}
d_{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \sin \dfrac{n\pi x}{L}dx.
\end{equation*}

\vspace{1pt}

Now with these coefficients $d_{n}$

\vspace{1pt}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y\left( x,t\right) =\sum\limits_{1}^{\infty }d_{n}\sin \dfrac{n\pi x}{L}\cos 
\dfrac{n\pi \alpha t}{L}
\end{equation*}

is a solution to entire problem $\left( 1\right) ,\left( 2a,2b\right)
,\left( 3a,3b\right) $.

\vspace{1pt}

\begin{example}
\begin{eqnarray*}
y_{xx} &=&y_{tt}\qquad y\left( 0,t\right) =y\left( L,t\right) =0 \\
y_{t}\left( x,0\right) &=&0 \\
y\left( x,0\right) &=&2\sin \dfrac{\pi x}{L}
\end{eqnarray*}
\end{example}

\qquad \qquad \qquad \qquad

\qquad Here $\alpha =1$ and $f\left( x\right) =2\sin \dfrac{\pi x}{L}$\qquad
\qquad

Now 
\begin{equation*}
y\left( x,t\right) =\sum\limits_{1}^{\infty }d_{n}\sin \dfrac{n\pi x}{L}\cos 
\dfrac{n\pi t}{L}
\end{equation*}

\vspace{1pt}

\begin{equation*}
d_{n}=\frac{2}{L}\int_{0}^{L}\sin \dfrac{\pi x}{L}\sin \dfrac{n\pi x}{L}dx=%
\frac{2}{L}\int_{0}^{L}2\sin \dfrac{\pi x}{L}\sin \dfrac{n\pi x}{L}%
dx=0\qquad n=2,3,...
\end{equation*}

\vspace{1pt}

\begin{equation*}
d_{1}=\frac{2}{L}\int_{0}^{L}\left( 2\right) \sin ^{2}\dfrac{n\pi x}{L}dx=%
\frac{4}{L}\left[ \int_{0}^{L}\left( \frac{1-\cos \tfrac{2n\pi x}{L}}{2}%
\right) \right] dx=\frac{4}{L}\left[ \frac{x}{2}-\left( \tfrac{\sin 2\tfrac{%
n\pi x}{L}}{\dfrac{2n\pi }{L}}\right) \right] _{0}^{L}=2
\end{equation*}

\vspace{1pt}

$\Longrightarrow $ solution is

\vspace{1pt}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
y\left( x,t\right) =2\sin \dfrac{n\pi }{L}\cos \dfrac{\pi t}{L}.
\end{equation*}

\begin{example}
Solve:\qquad
\end{example}

\vspace{1pt}

\begin{eqnarray*}
\text{P.D.E.} &\text{:}&\ u_{xx}-16u_{tt}=0 \\
\text{B.C.'s} &\text{:}&\ \ u(0,t)=0\qquad u_{x}(1,t)=0 \\
\text{I.C.} &\text{:}&\ \ u(x,0)=-3\sin \frac{5\pi x}{2}+23\sin \frac{11\pi x%
}{2};\text{ \ \ \ }u_{t}(x,0)=0
\end{eqnarray*}

Solution: We assume 
\begin{equation*}
u\left( x,t\right) =X\left( x\right) T\left( t\right)
\end{equation*}

The PDE implies%
\begin{equation*}
\frac{X^{\prime \prime }}{X}=16\frac{T^{\prime \prime }}{T}=k\text{ \ }k%
\text{ a constant}
\end{equation*}

Then we have the two ordinary DEs%
\begin{eqnarray*}
X^{\prime \prime }-kX &=&0 \\
T^{\prime \prime }-\frac{1}{16}kT &=&0
\end{eqnarray*}

The boundary conditions for $X\left( x\right) $ are%
\begin{equation*}
X\left( 0\right) =X^{\prime }\left( 1\right) =0
\end{equation*}

so that the eigenvalue problem for $X$ is%
\begin{equation*}
X^{\prime \prime }-kX=0\text{ \ \ }X\left( 0\right) =X^{\prime }\left(
1\right) =0
\end{equation*}

For nontrivial solutions we let $k=-\beta ^{2},\beta \neq 0$ and get%
\begin{equation*}
X^{\prime \prime }+\beta ^{2}X=0
\end{equation*}%
so%
\begin{equation*}
X\left( x\right) =C_{1}\sin \beta x+C_{2}\cos \beta x
\end{equation*}

\begin{equation*}
X\left( 0\right) =0\Rightarrow C_{2}=0
\end{equation*}

Thus%
\begin{equation*}
X^{\prime }\left( x\right) =C_{1}\beta \cos \beta x
\end{equation*}

and $X^{\prime }\left( 1\right) =0\Rightarrow $%
\begin{equation*}
\beta =\left( \frac{2n+1}{2}\right) \pi \text{ \ \ \ }n=0,1,2,\ldots
\end{equation*}

Therefore%
\begin{equation*}
X_{n}\left( x\right) =a_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\text{ \ \ 
}n=0,1,2,\ldots
\end{equation*}

Since%
\begin{equation*}
k=-\beta ^{2}=\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}
\end{equation*}

The equation for $T\left( t\right) $ becomes%
\begin{equation*}
T^{\prime \prime }+\frac{1}{16}\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}T=0
\end{equation*}

so%
\begin{equation*}
T_{n}\left( t\right) =b_{n}\sin \left( \frac{2n+1}{8}\right) \pi t+c_{n}\cos
\left( \frac{2n+1}{8}\right) \pi t\text{ \ \ \ \ }n=0,1,2,\ldots \text{\ }
\end{equation*}

The BC $u_{t}(x,0)=0\Rightarrow T^{\prime }\left( 0\right) =0.$ Since

\begin{equation*}
T_{n}^{\prime }\left( t\right) =b_{n}\left( \frac{2n+1}{8}\right) \pi \cos
\left( \frac{2n+1}{8}\right) \pi t-c_{n}\left( \frac{2n+1}{8}\right) \pi
\sin \left( \frac{2n+1}{8}\right) \pi t
\end{equation*}

we see that $b_{n}=0$ so that 
\begin{equation*}
T_{n}\left( t\right) =c_{n}\cos \left( \frac{2n+1}{8}\right) \pi t\text{ \ \
\ \ }n=0,1,2,\ldots \text{\ }
\end{equation*}

Thus%
\begin{equation*}
u_{n}\left( x,t\right) =X_{n}\left( x\right) T_{n}\left( t\right) =D_{n}\sin
\left( \frac{2n+1}{2}\right) \pi x\cos \left( \frac{2n+1}{8}\right) \pi t%
\text{ \ \ }n=0,1,2,\ldots
\end{equation*}

Let 
\begin{equation*}
u\left( x,t\right) =\sum_{n=0}^{\infty }u_{n}\left( x,t\right)
=\sum_{n=0}^{\infty }D_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\cos \left( 
\frac{2n+1}{8}\right) \pi t\text{ }
\end{equation*}

Then 
\begin{equation*}
u(x,0)=\sum_{n=0}^{\infty }D_{n}\sin \left( \frac{2n+1}{2}\right) \pi
x=-3\sin \frac{5\pi x}{2}+23\sin \frac{11\pi x}{2}
\end{equation*}

Therefore%
\begin{equation*}
D_{2}=-3\text{ \ \ }D_{5}=23\text{ \ \ }D_{n}=0\text{ \ }n\neq 2,5
\end{equation*}

The final solution is then%
\begin{equation*}
u\left( x,t\right) =-3\sin \left( \frac{5\pi x}{2}\right) \cos \left( \frac{%
5\pi t}{8}\right) +23\sin \left( \frac{11\pi x}{2}\right) \cos \frac{11\pi t%
}{8}
\end{equation*}

$u\left( x,0\right) =\allowbreak -3\sin \frac{5}{2}\pi x+23\sin \frac{11}{2}%
\pi x$\FRAME{dtbpFX}{3.0007in}{1.9993in}{0pt}{}{}{Plot}{\special{language
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$u\left( x,.1\right) =\allowbreak -3\sin \frac{5}{2}\pi x\cos 0.062\,5\pi
+23\sin \frac{11}{2}\pi x\cos 0.137\,5\pi \allowbreak $

$u\left( x,.4\right) =\allowbreak -3\sin \frac{5}{2}\pi x\cos 0.25\pi
+23\sin \frac{11}{2}\pi x\cos 0.55\pi \allowbreak $

$u\left( x,.6\right) =\allowbreak -3\sin \frac{5}{2}\pi x\cos 0.375\,\pi
+23\sin \frac{11}{2}\pi x\cos 0.825\,\pi \allowbreak $

$u\left( x,.8\right) =\allowbreak 23\sin \frac{11}{2}\pi x\cos
1.\,\allowbreak 1\pi $

\vspace{1pt}

\begin{example}
Solve
\end{example}

\ \ 
\begin{eqnarray*}
\text{PDE \ \ \ \ \ \ \ }u_{xx}-16u_{tt} &=&0 \\
\text{BCs \ \ }u(0,t) &=&0\qquad u_{x}(1,t)=0 \\
\text{IC \ \ \ \ }u(x,0) &=&-6\sin \left( \frac{3\pi x}{2}\right) +13\sin
\left( \frac{11\pi x}{2}\right) \\
\text{IC \ \ }u_{t}(x,0) &=&0
\end{eqnarray*}

You must derive the solution. \ Your solution should not have any arbitrary
constants in it. Show \textbf{all }steps.

Solution: \ Let $u\left( x,t\right) =X\left( x\right) T\left( t\right) .$
Then the PDE implies

\vspace{1pt}%
\begin{equation*}
X^{\prime \prime }T=16XT^{\prime \prime }
\end{equation*}%
or%
\begin{equation*}
\frac{X^{\prime \prime }}{X}=16\frac{T^{\prime \prime }}{T}=-\lambda ^{2}
\end{equation*}%
since we will need sines and cosines in the $X$ part of the solution.

Thus%
\begin{eqnarray*}
X^{\prime \prime }+\lambda ^{2}X &=&0 \\
T^{\prime \prime }+\frac{\lambda ^{2}}{16}T &=&0
\end{eqnarray*}%
The BCs are 
\begin{equation*}
X\left( 0\right) =X^{\prime }\left( 1\right) =0
\end{equation*}%
\begin{equation*}
X\left( x\right) =a_{n}\sin \lambda x+b_{n}\cos \lambda x
\end{equation*}

$X\left( 0\right) =0$ implies that $b_{n}=0,$ so 
\begin{equation*}
X\left( x\right) =a_{n}\sin \lambda x
\end{equation*}%
\begin{equation*}
X^{\prime }\left( x\right) =a_{n}\lambda \cos \lambda x
\end{equation*}%
so%
\begin{equation*}
X^{\prime }\left( 1\right) =a_{n}\lambda \cos \lambda =0
\end{equation*}%
Hence $\lambda =\frac{2n+1}{2}\pi ,$ \ $n=0,1,2,\ldots $ and%
\begin{equation*}
X_{n}\left( x\right) =A_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\text{ \ \ 
}\ n=0,1,2,\ldots
\end{equation*}%
Also%
\begin{equation*}
T^{\prime \prime }+\frac{\lambda ^{2}}{16}T=T^{\prime \prime }+\frac{\left(
2n+1\right) ^{2}\pi ^{2}}{64}T=0
\end{equation*}

\begin{equation*}
T_{n}\left( t\right) =c_{n}\sin \left( \frac{2n+1}{8}\right) \pi t+d_{n}\cos
\left( \frac{2n+1}{8}\right) \pi t
\end{equation*}

$u_{t}(x,0)=0$ implies that $c_{n}=0$ and 
\begin{equation*}
T_{n}\left( t\right) =d_{n}\cos \left( \frac{2n+1}{8}\right) \pi t
\end{equation*}%
Thus%
\begin{equation*}
u_{n}\left( x.t\right) =B_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\cos
\left( \frac{2n+1}{8}\right) \pi t\text{ \ \ }n=0,1,2,\ldots
\end{equation*}%
Let 
\begin{equation*}
u\left( x,t\right) =\sum_{n=0}^{\infty }u_{n}\left( x.t\right)
=\sum_{n=0}^{\infty }B_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\cos \left( 
\frac{2n+1}{8}\right) \pi t
\end{equation*}

\begin{equation*}
u\left( x,0\right) =\sum_{n=0}^{\infty }B_{n}\sin \left( \frac{2n+1}{2}%
\right) \pi x=-6\sin \left( \frac{3\pi x}{2}\right) +13\sin \left( \frac{%
11\pi x}{2}\right)
\end{equation*}%
Therefore $B_{1}=-6,B_{5}=13$ and $B_{n}=0$ for $n\neq 1,5$ so

\begin{equation*}
u\left( x,t\right) =-6\sin \left( \frac{3\pi x}{2}\right) \cos \left( \frac{%
3\pi }{8}\right) t+13\sin \left( \frac{11\pi x}{2}\right) \cos \left( \frac{%
11\pi }{8}\right) t
\end{equation*}

\vspace{1pt}

\begin{example}
Solve:\qquad
\end{example}

\vspace{1pt}

\begin{eqnarray*}
\text{P.D.E.} &\text{:}&\ u_{xx}-16u_{tt}=0 \\
\text{B.C.'s} &\text{:}&\ \ u(0,t)=0\qquad u_{x}(1,t)=0 \\
\text{I.C.} &\text{:}&\ \ u(x,0)=-3\sin \frac{5\pi x}{2}+23\sin \frac{11\pi x%
}{2};\text{ \ \ \ }u_{t}(x,0)=2\pi \sin \frac{3\pi x}{2}
\end{eqnarray*}

Solution: We assume 
\begin{equation*}
u\left( x,t\right) =X\left( x\right) T\left( t\right)
\end{equation*}

The PDE implies%
\begin{equation*}
\frac{X^{\prime \prime }}{X}=16\frac{T^{\prime \prime }}{T}=k\text{ \ }k%
\text{ a constant}
\end{equation*}

Then we have the two ordinary DEs%
\begin{eqnarray*}
X^{\prime \prime }-kX &=&0 \\
T^{\prime \prime }-\frac{1}{16}kT &=&0
\end{eqnarray*}

The boundary conditions for $X\left( x\right) $ are%
\begin{equation*}
X\left( 0\right) =X^{\prime }\left( 1\right) =0
\end{equation*}

so that the eigenvalue problem for $X$ is%
\begin{equation*}
X^{\prime \prime }-kX=0\text{ \ \ }X\left( 0\right) =X^{\prime }\left(
1\right) =0
\end{equation*}

For nontrivial solutions we let $k=-\beta ^{2},\beta \neq 0$ and get%
\begin{equation*}
X^{\prime \prime }+\beta ^{2}X=0
\end{equation*}%
so%
\begin{equation*}
X\left( x\right) =C_{1}\sin \beta x+C_{2}\cos \beta x
\end{equation*}

\begin{equation*}
X\left( 0\right) =0\Rightarrow C_{2}=0
\end{equation*}

Thus%
\begin{equation*}
X^{\prime }\left( x\right) =C_{1}\beta \cos \beta x
\end{equation*}

and $X^{\prime }\left( 1\right) =0\Rightarrow $%
\begin{equation*}
\beta =\left( \frac{2n+1}{2}\right) \pi \text{ \ \ \ }n=0,1,2,\ldots
\end{equation*}

Therefore%
\begin{equation*}
X_{n}\left( x\right) =a_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\text{ \ \ 
}n=0,1,2,\ldots
\end{equation*}

Since%
\begin{equation*}
k=-\beta ^{2}=\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}
\end{equation*}

The equation for $T\left( t\right) $ becomes%
\begin{equation*}
T^{\prime \prime }+\frac{1}{16}\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}T=0
\end{equation*}

so%
\begin{equation*}
T_{n}\left( t\right) =b_{n}\sin \left( \frac{2n+1}{8}\right) \pi t+c_{n}\cos
\left( \frac{2n+1}{8}\right) \pi t\text{ \ \ \ \ }n=0,1,2,\ldots \text{\ }
\end{equation*}

Thus%
\begin{equation*}
u_{n}\left( x,t\right) =X_{n}\left( x\right) T_{n}\left( t\right) =D_{n}\sin
\left( \frac{2n+1}{2}\right) \pi x\sin \left( \frac{2n+1}{8}\right) \pi
t+E_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\cos \left( \frac{2n+1}{8}%
\right) \pi t\text{ \ \ }n=0,1,2,\ldots
\end{equation*}

Let 
\begin{equation*}
u\left( x,t\right) =\sum_{n=0}^{\infty }u_{n}\left( x,t\right)
=\sum_{n=0}^{\infty }\left[ D_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\sin
\left( \frac{2n+1}{8}\right) \pi t\text{ }+E_{n}\sin \left( \frac{2n+1}{2}%
\right) \pi x\cos \left( \frac{2n+1}{8}\right) \pi t\right]
\end{equation*}

\begin{eqnarray*}
&&u_{t}\left( x,t\right) \\
&=&\sum_{n=0}^{\infty }\left[ D_{n}\left( \frac{2n+1}{8}\right) \pi \sin
\left( \frac{2n+1}{2}\right) \pi x\cos \left( \frac{2n+1}{8}\right) \pi t%
\text{ }-E_{n}\left( \frac{2n+1}{8}\right) \pi \sin \left( \frac{2n+1}{2}%
\right) \pi x\sin \left( \frac{2n+1}{8}\right) \pi t\right]
\end{eqnarray*}

Then 
\begin{equation*}
u\left( x,0\right) =\sum_{n=0}^{\infty }E_{n}\sin \left( \frac{2n+1}{2}%
\right) \pi x=-3\sin \frac{5\pi x}{2}+23\sin \frac{11\pi x}{2}
\end{equation*}

Therefore%
\begin{equation*}
E_{2}=-3\text{ \ \ }E_{5}=23\text{ \ \ }E_{n}=0\text{ \ }n\neq 2,5
\end{equation*}

\begin{equation*}
u_{t}\left( x.0\right) =\sum_{n=0}^{\infty }D_{n}\left( \frac{2n+1}{8}%
\right) \pi \sin \left( \frac{2n+1}{2}\right) \pi x=2\pi \sin \frac{3\pi x}{2%
}
\end{equation*}

Thus $D_{1}\left( \frac{3}{8}\right) \pi =2\pi $ so $D_{1}=\frac{16}{3}$ and 
$D_{n}=0$ \ $n\neq 1$

The final solution is then%
\begin{equation*}
u\left( x,t\right) =\frac{16}{3}\sin \left( \frac{3\pi x}{2}\right) \cos 
\frac{3\pi t}{8}-3\sin \left( \frac{5\pi x}{2}\right) \cos \left( \frac{5\pi
t}{8}\right) +23\sin \left( \frac{11\pi x}{2}\right) \cos \frac{11\pi t}{8}
\end{equation*}

\subsection{\protect\Large The Heat Equation}

\vspace{1pt}

Consider a cylinder parallel to $x-$axis

\vspace{1pt}

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'/document/graphics/Image90.gif';file-properties "XNPEU";}}

Let $u$ denote the temperature in the cylinder. Suppose the ends $x=0$ and $%
x=L$ are kept at zero temperature whereas at $t=0$ the initial temperature
distribution is $u=f\left( x\right) $. It may be shown that $u=u\left(
x,t\right) $ satisfies the P.D.E.

\vspace{1pt}

$\qquad $%
\begin{equation*}
u_{xx}=\dfrac{1}{k}u_{t}\qquad 0<x<L,\qquad t>0,\qquad (1)
\end{equation*}

\vspace{1pt}

where $k$\ is a constant and $k>0$

.

Equation $(1)$ is called the heat equation. The physical conditions of the
problem imply

\vspace{1pt}

\qquad 
\begin{equation*}
\ B.C.\ u\left( 0,t\right) =0=u\left( L,t\right) \qquad t\geq 0\qquad \ \ \
\ (2)
\end{equation*}

\qquad 
\begin{equation*}
\ I.C.\ \ u\left( x,0\right) =f\left( x\right) \qquad 0\leq x\leq L\qquad
\qquad (3)
\end{equation*}

\vspace{1pt}

We want to determine $u\left( x,t\right) $, i.e. the temperature in the
cylinder at any point $x$ at any time $t$. Again we use separation of
variables. The assumption $u\left( x,t\right) =X\left( x\right) T\left(
t\right) $ leads to

\vspace{1pt}

\qquad \qquad \qquad \qquad 
\begin{equation*}
\dfrac{X^{\prime \prime }\left( x\right) }{X\left( x\right) }=\dfrac{1}{k}%
\dfrac{T^{\prime }\left( t\right) }{T\left( t\right) }=-\lambda ^{2}
\end{equation*}

$\vspace{1pt}$

$\Longrightarrow X^{\prime \prime }+\lambda ^{2}X=0\qquad X\left( 0\right)
=X\left( L\right) =0$ and $T^{\prime }+k\lambda ^{2}T=0$.

\begin{center}
\vspace{1pt}
\end{center}

$\Longrightarrow X_{n}=c_{n}\sin \dfrac{n\pi x}{L}\qquad n=1,2,...\qquad
\lambda _{n}=\dfrac{n\pi }{L}\Longrightarrow $

$\vspace{1pt}$

\begin{center}
\begin{equation*}
T^{\prime }+k\dfrac{n^{2}\pi ^{2}}{L^{2}}T=0
\end{equation*}
\end{center}

$\vspace{1pt}$

$\Longrightarrow $%
\begin{equation*}
T\left( t\right) =d_{n}e^{-\left( \frac{n\pi }{L}\right) ^{2}kt}
\end{equation*}%
$\Longrightarrow $

$\vspace{1pt}$

\begin{center}
\begin{equation*}
u_{n}\left( x,t\right) =a_{n}e^{-\left( \frac{n\pi }{L}\right) ^{2}kt}\sin 
\dfrac{n\pi x}{L}
\end{equation*}
\end{center}

\vspace{1pt}

satisfies $(1)$ and $\left( 2\right) $ $\ \Longrightarrow $%
\begin{equation*}
u_{n}\left( x,t\right) =\sum_{1}^{\infty }a_{n}e^{-\left( \frac{n\pi }{L}%
\right) ^{2}kt}\sin \dfrac{n\pi x}{L}
\end{equation*}

\vspace{1pt}

also satisfies $\left( 1\right) $ and \ $\left( 2\right) $.

\vspace{1pt}

We need to satisfy $\left( 3\right) $ namely, $u\left( x,0\right) =f\left(
x\right) $ However,

$\vspace{1pt}$

\begin{center}
\begin{equation*}
u\left( x,0\right) =\sum\limits_{1}^{\infty }a_{n}\sin \dfrac{n\pi x}{L}
\end{equation*}
\end{center}

Thus we take $a_{n}$ to be the Fourier sine coefficients of $f\left(
x\right) .$ Hence

$\vspace{1pt}$

\begin{center}
\begin{equation*}
a_{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \sin \dfrac{n\pi x}{L}dx.
\end{equation*}
\end{center}

Remark. The factor $e^{-\left( \frac{n\pi }{L}\right) ^{2}kt}\longrightarrow
0$ as $t\longrightarrow \infty \Longrightarrow \underset{t\rightarrow \infty 
}{\lim }u\left( x,t\right) =0$ as expected from the physical problem.

\begin{example}
Solve the problem:
\end{example}

\qquad

\qquad 
\begin{eqnarray*}
\text{P.D.E.}\text{: \ \ \ } &&\ u_{xx}-8u_{t}=0 \\
\text{B.C.} &\text{:}&\ \ u(0,t)=0\qquad u_{x}(1,t)=0 \\
\text{I.C.} &\text{:}&u(x,0)=-2\sin \frac{3\pi }{2}x+10\sin \frac{9\pi }{2}x
\end{eqnarray*}

Solution: Let $u\left( x,t\right) =X\left( x\right) T\left( t\right) .$ Then
the PDE implies%
\begin{equation*}
\frac{X^{\prime \prime }}{X}=8\frac{T^{\prime }}{T}=k\text{ \ }k\text{ a
constant}
\end{equation*}

Then we have the two ODEs%
\begin{eqnarray*}
X^{\prime \prime }-kX &=&0 \\
T^{\prime }-\frac{1}{8}kT &=&0
\end{eqnarray*}

The BCs for $X\left( x\right) $ are%
\begin{equation*}
X\left( 0\right) =X^{\prime }\left( 1\right) =0
\end{equation*}

The boundary conditions for $X\left( x\right) $ are%
\begin{equation*}
X\left( 0\right) =X^{\prime }\left( 1\right) =0
\end{equation*}

so that the eigenvalue problem for $X$ is%
\begin{equation*}
X^{\prime \prime }-kX=0\text{ \ \ }X\left( 0\right) =X^{\prime }\left(
1\right) =0
\end{equation*}

For nontrivial solutions we let $k=-\beta ^{2},\beta \neq 0$ and get%
\begin{equation*}
X^{\prime \prime }+\beta ^{2}X=0
\end{equation*}%
so%
\begin{equation*}
X\left( x\right) =C_{1}\sin \beta x+C_{2}\cos \beta x
\end{equation*}

\begin{equation*}
X\left( 0\right) =0\Rightarrow C_{2}=0
\end{equation*}

Thus%
\begin{equation*}
X^{\prime }\left( x\right) =C_{1}\beta \cos \beta x
\end{equation*}

and $X^{\prime }\left( 1\right) =0\Rightarrow $%
\begin{equation*}
\beta =\left( \frac{2n+1}{2}\right) \pi \text{ \ \ \ }n=0,1,2,\ldots
\end{equation*}

Therefore%
\begin{equation*}
X_{n}\left( x\right) =a_{n}\sin \left( \frac{2n+1}{2}\right) \pi x\text{ \ \ 
}n=0,1,2,\ldots
\end{equation*}

The equation for $T\left( t\right) $ with $k=-\beta ^{2}=\left( \frac{2n+1}{2%
}\right) ^{2}\pi ^{2}$ is%
\begin{equation*}
T^{\prime }+\frac{1}{8}\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}T=0
\end{equation*}

Thus%
\begin{equation*}
T_{n}\left( t\right) =b_{n}e^{-\frac{1}{8}\left( \frac{2n+1}{2}\right)
^{2}\pi ^{2}t}\text{ \ \ }n=0,1,2,\ldots
\end{equation*}

Therefore we have%
\begin{equation*}
u_{n}\left( x,t\right) =D_{n}\sin \left( \frac{2n+1}{2}\right) \pi xe^{-%
\frac{1}{8}\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}t}\text{ \ \ }%
n=0,1,2,\ldots
\end{equation*}

To satisfy the initial condition we let%
\begin{equation*}
u\left( x,t\right) =\sum_{n=0}^{\infty }D_{n}\sin \left( \frac{2n+1}{2}%
\right) \pi xe^{-\frac{1}{8}\left( \frac{2n+1}{2}\right) ^{2}\pi ^{2}t}
\end{equation*}

Now%
\begin{equation*}
u\left( x,0\right) =\sum_{n=0}^{\infty }D_{n}\sin \left( \frac{2n+1}{2}%
\right) \pi x=-2\sin \frac{3\pi }{2}x+10\sin \frac{9\pi }{2}x
\end{equation*}

This means%
\begin{equation*}
D_{1}=-2\text{ \ \ }D_{4}=10\text{ \ and \ }D_{n}=0,\text{ \ }n\neq 1,2
\end{equation*}

The solution to the problem is then%
\begin{equation*}
u\left( x,t\right) =-2\sin \left( \frac{3}{2}\right) \pi xe^{-\frac{1}{8}%
\left( \frac{3}{2}\right) ^{2}\pi ^{2}t}+10\sin \left( \frac{9}{2}\right)
\pi xe^{-\frac{1}{8}\left( \frac{9}{2}\right) ^{2}\pi ^{2}t}
\end{equation*}

\vspace{1pt}\pagebreak 

\subsubsection{Additional Examples}

\vspace{1pt}

\begin{example}
Wave Equation Example
\end{example}

\vspace{1pt}

Problem 1 Section 10.6

\vspace{1pt}

Find a formal solution to the vibrating string problem governed by the given
initial-boundary value problem.

\vspace{1pt}

\begin{eqnarray*}
u_{tt} &=&u_{xx},\text{ \ \ \ \ }0<x<1,\text{ \ \ }t>0 \\
u\left( 0,t\right) &=&u\left( 1,t\right) =0,\text{ \ \ }t>0 \\
u\left( x,0\right) &=&x\left( 1-x\right) ,\text{ \ \ }0<x\,<1 \\
u_{t}\left( x,0\right) &=&\sin 7\pi x,\text{ \ }0<x<1
\end{eqnarray*}

\vspace{1pt}

Let $u\left( x,t\right) =X\left( x\right) T\left( t\right) .$ \ Then the PDE
leads to 
\begin{equation*}
\frac{X^{\prime \prime }}{X}=\frac{T^{\prime \prime }}{T}=-\lambda ^{2}
\end{equation*}

\vspace{1pt}We then have two ODEs%
\begin{eqnarray*}
X^{\prime \prime }+\lambda ^{2}X &=&0 \\
T^{\prime \prime }+\lambda ^{2}T &=&0
\end{eqnarray*}

Therefore 
\begin{equation*}
X\left( x\right) =a\cos \lambda x+b\sin \lambda x
\end{equation*}

The BCs for $X\left( x\right) $ are $X\left( 0\right) =X\left( 1\right) =0.$
Thus, $a=0$ and $\lambda =n\pi ,$ $n=1,2,\ldots $ and 
\begin{equation*}
X_{n}\left( x\right) =c_{n}\sin n\pi x\text{ \ }n=1,2,\ldots
\end{equation*}%
Also%
\begin{equation*}
T_{n}\left( t\right) =d_{n}\cos n\pi t+e_{n}\sin n\pi t\text{ \ \ }%
n=1,2,\ldots
\end{equation*}%
so%
\begin{equation*}
u_{n}\left( x,t\right) =\left[ a_{n}\cos n\pi t+b_{n}\sin n\pi t\right] \sin
n\pi x\text{ \ }n=1,2,\ldots \text{\ }
\end{equation*}

Thus we let%
\begin{equation*}
u\left( x,t\right) =\sum_{n=1}^{\infty }u_{n}\left( x,t\right)
=\sum_{n=1}^{\infty }\left[ a_{n}\cos n\pi t+b_{n}\sin n\pi t\right] \sin
n\pi x
\end{equation*}%
We want%
\begin{equation*}
u\left( x,0\right) =x\left( 1-x\right) =\sum_{n=1}^{\infty }a_{n}\sin n\pi x
\end{equation*}%
Therefore the constants $a_{n}$ are given by the formula for the Fourier
sine series coefficients with $L=1$ so%
\begin{eqnarray*}
a_{n} &=&\frac{2}{1}\int_{0}^{1}x\left( 1-x\right) \sin n\pi xdx \\
&=&2\left( \int_{0}^{1}x\sin n\pi xdx+\int_{0}^{1}x^{2}\sin n\pi xdx\right)
\end{eqnarray*}

Integration by parts yields%
\begin{equation*}
\int_{0}^{1}x\sin n\pi xdx=-\frac{1}{n\pi }\cos n\pi =-\frac{\left(
-1\right) ^{n}}{n\pi }
\end{equation*}%
and%
\begin{eqnarray*}
\int_{0}^{1}x^{2}\sin n\pi xdx &=&-\frac{1}{n\pi }\cos n\pi -\frac{2}{%
n^{2}\pi ^{2}}\left( -\frac{1}{n\pi }\cos n\pi +\frac{1}{n\pi }\right) \\
&=&-\frac{\left( -1\right) ^{n}}{n\pi }+\frac{2\left[ \left( -1\right) ^{n}-1%
\right] }{n^{3}\pi ^{3}}
\end{eqnarray*}%
Therefore for $n=1,2,\ldots $%
\begin{equation*}
a_{n}=2\left\{ -\frac{\left( -1\right) ^{n}}{n\pi }+\frac{\left( -1\right)
^{n}}{n\pi }-\frac{2\left[ \left( -1\right) ^{n}-1\right] }{n^{3}\pi ^{3}}%
\right\} =-\frac{4\left[ \left( -1\right) ^{n}-1\right] }{n^{3}\pi ^{3}}
\end{equation*}%
Note that 
\begin{equation*}
a_{n}=\left\{ 
\begin{array}{c}
0\text{ \ \ if }n\text{ is even} \\ 
\frac{8}{n^{3}\pi ^{3}}\text{ \ if \ }n\text{ \ is odd}%
\end{array}%
\right.
\end{equation*}%
Since 
\begin{equation*}
u_{t}\left( x,t\right) =\sum_{n=1}^{\infty }\left[ -a_{n}\left( n\pi \right)
\sin n\pi t+b_{n}\left( n\pi \right) \cos n\pi t\right] \sin n\pi x
\end{equation*}%
then%
\begin{equation*}
u_{t}\left( x,0\right) =\sum_{n=1}^{\infty }b_{n}\left( n\pi \right) \sin
n\pi x=\sin 7\pi x
\end{equation*}%
Hence $7\pi b_{7}=1$ so $b_{7}=\frac{1}{7\pi }$ and $b_{n}=0$ for $n\neq 7.$

\vspace{1pt}

Substituting these constants into the expression 
\begin{equation*}
u\left( x,t\right) =\sum_{n=1}^{\infty }u_{n}\left( x,t\right)
=\sum_{n=1}^{\infty }\left[ a_{n}\cos n\pi t+b_{n}\sin n\pi t\right] \sin
n\pi x
\end{equation*}
above and letting $n=2k+1,$ \ $k=0,1,2,\ldots $ since $n$ is odd yields

\begin{equation*}
u\left( x,t\right) =\frac{1}{7\pi }\sin 7\pi t\sin 7\pi x+\sum_{k=0}^{\infty
}\frac{8}{\left[ \left( 2k+1\right) \pi \right] ^{3}}\cos \left( 2k+1\right)
\pi t\sin \left( 2k+1\right) \pi x
\end{equation*}

\vspace{1pt}

\begin{example}
Use separation of variables, \ $u(x,t)=X(x)T(t)$, \ to find two ordinary
differential equations which $X(x)$ \ and \ $T(t)$ \ \ must satisfy to be a
solution of
\end{example}

\qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
-3x^{2}t^{4}\dfrac{\partial ^{2}u}{\partial x^{2}}+(x-2)^{4}(t+6)^{3}\dfrac{%
\partial ^{2}u}{\partial t^{2}}=0.
\end{equation*}

Note: \ Do \textbf{not }solve these ordinary differential equations.

Solution: \ \ $u_{x}\left( x,t\right) =X^{^{\prime }}\left( x\right) T\left(
t\right) ,u_{xx}=X^{\prime \prime }T$

\textbf{\ \qquad \qquad \qquad \qquad }%
\begin{equation*}
-3x^{2}t^{4}X^{\prime \prime }(x)T(t)+(x-2)^{4}(t+6)^{3}X(x)T^{\prime \prime
}(t)=0
\end{equation*}

\qquad \qquad $\Longrightarrow $%
\begin{equation*}
\dfrac{-3x^{2}X^{\prime \prime }}{(x-2)^{4}X}=-\dfrac{(t+6)^{3}T^{\prime
\prime }}{t^{4}T}=k.
\end{equation*}

\qquad \qquad $\Longrightarrow $%
\begin{equation*}
3x^{2}X^{\prime \prime }+k(x-2)^{4}X=0\ \text{and}\ (t+6)^{3}T^{\prime
\prime }+kt^{4}T=0.
\end{equation*}

\vspace{1pt}\vspace{1pt}\pagebreak 

\begin{example}
Solve
\end{example}

\vspace{1pt}

\begin{eqnarray*}
\text{PDE \ \ \ \ \ \ \ \ }u_{xx} &=&4u_{tt} \\
\text{BCS \ \ \ }u_{x}(0,t) &=&0\qquad u_{x}(\pi ,t)=0 \\
\text{ICs \ \ \ \ }u(x,0) &=&0\text{ \ \ \ \ \ }u_{t}(x,0)=-9\cos
(4x)+16\cos (8x)
\end{eqnarray*}

You must derive the solution. \ Your solution should not have any arbitrary
constants in it.

Solution:

$\vspace{1pt}$Let $u\left( x,t\right) =X\left( x\right) T\left( t\right) .$
Then differentiating and substituting in the PDE yields

\QTP{Body Math}
$\vspace{1pt}$

\begin{center}
\begin{equation*}
X^{\prime \prime }T=4XT^{\prime \prime }
\end{equation*}

$\Rightarrow \qquad $%
\begin{equation*}
\dfrac{X^{\prime \prime }}{X}=4\dfrac{T^{\prime \prime }}{T}
\end{equation*}
\end{center}

Using the argument that the left hand side is purely a function of $x$ and
the right hand side is purely a function of $t$, and the only way that they
can be equal is if they are equal to a constant, we get

$\vspace{1pt}$

\begin{equation*}
\dfrac{X^{\prime \prime }}{X}=4\dfrac{T^{\prime \prime }}{T}=k\qquad k\text{
\ a constant}
\end{equation*}

\vspace{1pt}

This yields the two \textit{ordinary differential equations }

$\vspace{1pt}$

\begin{equation*}
X^{\prime \prime }-kX=0\qquad \text{and}\qquad T^{\prime \prime }-\dfrac{1}{4%
}kT=0
\end{equation*}

\vspace{1pt}

The boundary condition $u_{x}(0,t)=0$ implies, since $u_{x}\left( x,t\right)
=X^{\prime }\left( x\right) T\left( t\right) $ that

$X^{\prime }\left( 0\right) $\vspace{1pt}$T\left( t\right) =0.$ We cannot
have $T\left( t\right) =0,$ since this would imply that $u\left( x,t\right)
=0$. Thus $X^{\prime }\left( 0\right) =0.$ Similarly, the boundary condition 
$u_{x}\left( \pi ,t\right) =0$ leads to $X^{\prime }\left( \pi \right) =0.$

$\vspace{1pt}$

We now have the following boundary value problem for $X\left( x\right) :$

$\vspace{1pt}$

\begin{equation*}
X^{\prime \prime }-kX=0\qquad X^{\prime }\left( 0\right) =X^{\prime }\left(
\pi \right) =0
\end{equation*}

\vspace{1pt}

For $k>0,$the only solution is $X=0.$ For $k=0$ we have $X=Ax+B.$ $X^{\prime
}\left( x\right) =A,$ so the BCs imply that $X^{\prime }\left( 0\right)
=X^{\prime }\left( \pi \right) =A=0.$ 
\begin{equation*}
X\left( x\right) =B,\text{ \ \ }B\neq 0
\end{equation*}

\vspace{1pt}is a nontrivial solution corresponding to the eigenvalue $k=0.$

For $k<0,$ let $-k=\alpha ^{2},$ where $\alpha \neq 0.$ Then we have the
equation%
\begin{equation*}
X^{\prime \prime }+\alpha ^{2}X=0
\end{equation*}

and 
\begin{eqnarray*}
X\left( x\right) &=&c_{1}\sin \alpha x+c_{2}\cos \alpha x \\
X^{\prime }\left( x\right) &=&c_{1}\alpha \cos \alpha x-c_{2}\alpha \sin
\alpha x
\end{eqnarray*}

\begin{equation*}
X^{\prime }\left( 0\right) =c_{1}\alpha =0
\end{equation*}

so $c_{1}=0.$%
\begin{equation*}
X^{\prime }\left( \pi \right) =-c_{2}\alpha \sin \alpha \pi =0
\end{equation*}

Therefore $\alpha =n,$ $n=1,2,\ldots $ and the solution is

\QTP{Body Math}
$\vspace{1pt}$

\begin{center}
\begin{equation*}
k=-n^{2}\qquad X_{n}\left( x\right) =a_{n}\cos nx\qquad n=1,2,3,\ldots
\end{equation*}
\end{center}

\vspace{1pt}The case $k=0$ implies that the equation for $T$ becomes $%
T^{\prime \prime }=0,$ so $T=At+B.$ The initial condition $u\left(
x,0\right) =0$ implies $X\left( x\right) T\left( 0\right) =0$ so that $%
T\left( 0\right) =0.$ Thus $B=0$ and $T=At$ for $k=0.$

Substituting the values of $k=-n^{2}$ into the equation for $T\left(
t\right) $ leads to

$\vspace{1pt}$

\begin{equation*}
T^{\prime \prime }+\dfrac{n^{2}}{4}T=0
\end{equation*}

\vspace{1pt}

which has the solution 
\begin{equation*}
T_{n}\left( t\right) =B_{n}\sin \dfrac{nt}{2}+C_{n}\cos \dfrac{nt}{2},\text{
\ \ }n=1,2,3,...
\end{equation*}

The initial condition $u\left( x,0\right) =0$ implies $X\left( x\right)
T\left( 0\right) =0$ so that $T\left( 0\right) =0.$ Thus $c_{n}=0.$ For $n=0$
the equation for $T$ becomes $T^{\prime \prime }=0,$ and has the solution $%
T\left( t\right) =B_{0}t+C_{0}.$ The condition $T\left( 0\right) =0$ implies
that $C_{0}=0,$ so $T_{0}\left( t\right) =B_{0}t$

We now have the solutions

$\vspace{1pt}$

\begin{eqnarray*}
u_{n}\left( x,t\right) &=&X_{n}\left( x\right) T_{n}\left( t\right)
=A_{n}\cos nx\sin \dfrac{nt}{2}\qquad n=1,2,3,... \\
u_{0}\left( x,t\right) &=&A_{0}t
\end{eqnarray*}

\vspace{1pt}

Since the boundary conditions and the equation are linear and homogeneous,
it follows that

$\vspace{1pt}$

\begin{equation*}
u\left( x,t\right) =\sum_{n=0}^{\infty }u_{n}\left( x,t\right)
=A_{0}t+\sum_{n=1}^{\infty }A_{n}\cos nx\sin \dfrac{nt}{2}
\end{equation*}

\vspace{1pt}

satisfies the PDE, the boundary conditions, and the first initial condition.
Since

\QTP{Body Math}
$\vspace{1pt}$

\begin{center}
$\vspace{1pt}$%
\begin{equation*}
u_{t}\left( x,t\right) =A_{0}+\sum_{n=1}^{\infty }A_{n}\left( \dfrac{n}{2}%
\right) \cos nx\cos \dfrac{nt}{2}
\end{equation*}
\end{center}

the last initial condition leads to

$\vspace{1pt}$

\begin{equation*}
u_{t}(x,0)=-9\cos (4x)+16\cos (8x)=A_{0}+\sum_{n=1}^{\infty }A_{n}\left( 
\dfrac{n}{2}\right) \cos nx.
\end{equation*}

\vspace{1pt}

Matching the cosine terms on both sides of this equation leads to

$\vspace{1pt}$

$A_{4}\left( \dfrac{4}{2}\right) =-9$ \ so that $A_{4}=-\frac{9}{2}$ \ and $%
A_{8}\left( \dfrac{8}{2}\right) =16$ so that $A_{8}=4.$ All of the other
constants must be zero, since there are no cosine terms or constant terms on
the left to match with. Thus

$\vspace{1pt}$

\begin{equation*}
u\left( x,t\right) =-\frac{9}{2}\cos 4x\sin 2t+4\cos 8x\sin 4t
\end{equation*}

\vspace{1pt}

\begin{example}
Consider the non-homogeneous problem\qquad
\end{example}

\vspace{1pt}

\begin{eqnarray*}
\text{P.D.E.} &:&\ u_{xx}=9u_{t} \\
\text{B.C.}\prime \text{s} &:&\ \ u_{x}(0,t)=0\qquad u(1,t)=2 \\
\text{I.C.} &:&\ \ u(x,0)=-3\cos \frac{7\pi }{2}x+2
\end{eqnarray*}

\subparagraph{i) \ }

Let 
\begin{equation*}
v(x,t)=u(x,t)-2
\end{equation*}
and show that $v(x,t)$ satisfies the\qquad \qquad \qquad \qquad\ 

homogeneous problem

\qquad \qquad \qquad\ 
\begin{eqnarray*}
\text{P.D.E}. &:&v_{xx}=9v_{t} \\
\text{B.C.} &:&\ \ \ v_{x}(0,t)=0\qquad v(1,t)=0 \\
\text{\ I.C.} &:&\ \ \ \ v(x,0)=-3\cos \frac{7\pi }{2}x
\end{eqnarray*}

\qquad \qquad \qquad\ 

\vspace{1pt} Solution to i)$\qquad $%
\begin{equation*}
u_{xx}\left( x,t\right) =v_{xx}\left( x,t\right) \qquad u_{x}\left(
x,t\right) =v_{x}\left( x,t\right)
\end{equation*}

\qquad \qquad \qquad 
\begin{equation*}
u_{tt}\left( x,t\right) =v_{tt}\left( x,t\right) \qquad u_{t}\left(
x,t\right) =v_{t}\left( x,t\right)
\end{equation*}

\qquad \qquad \qquad 
\begin{equation*}
u\left( 1,t\right) =2\text{ and }u\left( x,t\right) -2=v\left( x,t\right)
\Longrightarrow v\left( 1,t\right) =0
\end{equation*}

\qquad \qquad \qquad 
\begin{equation*}
u_{x}\left( 0,t\right) =0\Longrightarrow v_{x}\left( 0,t\right) =0
\end{equation*}%
$\qquad \qquad $

\qquad \qquad 
\begin{equation*}
u\left( x,0\right) =-3\cos \frac{7\pi }{2}+2\ \text{and}\ u\left( x,t\right)
-2=v\left( x,t\right) \Longrightarrow v\left( x,0\right) =-3\cos \frac{7\pi 
}{2}
\end{equation*}

\vspace{1pt}

ii) \ 

Solve the above problem for $v(x,t)$.\qquad

Solution to ii)$\qquad $Let \ $v\left( x,t\right) =X\left( x\right) T\left(
t\right) $

\qquad \qquad \qquad then \ 
\begin{equation*}
X^{\prime \prime }T=9XT^{\prime }\Longrightarrow \frac{X^{\prime \prime }}{X}%
=9\frac{T^{\prime }}{T}=k
\end{equation*}

resulting in the ordinary differential equations:

\qquad 
\begin{equation*}
X^{\prime \prime }-kX=0\ \ \text{and}\ \ T^{\prime }-\frac{k}{9}T=0
\end{equation*}

Boundary Conditions become: \ \ 
\begin{eqnarray*}
X^{\prime }\left( 0\right) T\left( t\right) &=&0\ \text{and}\ X\left(
1\right) T\left( t\right) =0 \\
&\Longrightarrow &X^{\prime }\left( 0\right) =0\ \text{and}\ X\left(
1\right) =1
\end{eqnarray*}

\qquad \qquad \qquad \qquad \qquad \qquad

Solving the differential equation $X^{\prime \prime }-kX=0$ consider all
values of $k$

$k<0$ \ let \ $k=-u^{2};\quad u>0$

$\qquad $%
\begin{equation*}
X^{\prime \prime }+u^{2}X=0
\end{equation*}
\ has the solution: \ 
\begin{equation*}
X\left( x\right) =c_{1}\cos ux+c_{2}\sin ux
\end{equation*}

\qquad and \ 
\begin{equation*}
X^{\prime }\left( x\right) =-c_{1}u\sin ux+c_{2}u\cos ux
\end{equation*}

\qquad B.C. $\Longrightarrow X\left( 1\right) =c_{1}\cos u+c_{2}\sin u=0$ \
and \ $X^{\prime }\left( 0\right) =c_{2}u=0$

$\qquad \Longrightarrow c_{2}=0$ \ thus \ $c_{1}\cos u=0$%
\begin{equation*}
\Longrightarrow u_{n}=\frac{\left( 2n-1\right) \pi }{2}\quad n=1,2,...
\end{equation*}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad 
\begin{equation*}
\Longrightarrow k_{n}=-\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{4}\quad
n=1,2,...
\end{equation*}

\qquad \qquad \qquad \qquad \qquad \qquad so \ \ 
\begin{equation*}
X_{n}\left( x\right) =c_{n}\cos \frac{\left( 2n-1\right) \pi }{2}x\text{ \ \ 
}n=1,2,...
\end{equation*}

\vspace{1pt}

The other cases for $k,$ namely $k=0$ and $k>0$ yield only the trivial
solution since

$k=0\Longrightarrow X^{\prime \prime }=0\quad $which has the solution: \ $%
X\left( x\right) =c_{1}x+c_{2}$ and $X^{\prime }\left( x\right) =c_{1}$

\qquad B.C. $\Longrightarrow X\left( 1\right) =c_{1}+c_{2}=0$ \ and \ $%
X^{\prime }\left( 0\right) =c_{1}=0\Longrightarrow c_{2}=0$

\qquad thus \ $X\left( x\right) \equiv 0$ \ is the trivial solution.

$k>0$ \ let \ $k=u^{2};\quad u>0$

\qquad $X^{\prime \prime }-u^{2}X=0$ \ has the solution: \ $X\left( x\right)
=c_{1}e^{ux}+c_{2}e^{-ux}$

\qquad and \ $X^{\prime }\left( x\right) =c_{1}ue^{ux}-c_{2}ue^{-ux}$

\qquad B.C. $\Longrightarrow $ \ $X^{\prime }\left( 0\right)
=c_{1}u-c_{2}u=0\Longrightarrow c_{1}=c_{2}$

\qquad and \ \ $X\left( 1\right) =c_{1}e^{u}+c_{2}e^{-u}=0$ \ $%
\Longrightarrow c_{1}e^{u}+c_{1}e^{-u}=0\Longrightarrow c_{1}\left(
e^{u}+e^{-u}\right) =0$

\qquad $\Longrightarrow c_{1}=c_{2}=0\quad $thus \ $X\left( x\right) \equiv
0 $ \ is the trivial solution.

Using the non-trivial solution \ 
\begin{equation*}
k_{n}=-\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{4}\quad \ \ \ \ X_{n}\left(
x\right) =c_{n}\cos \frac{\left( 2n-1\right) \pi }{2}x,\text{ \ }n=1,2,...
\end{equation*}

the equation $\ $%
\begin{equation*}
T^{\prime }-\frac{k}{9}T=0
\end{equation*}
\ becomes 
\begin{equation*}
T^{\prime }+\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{36}T=0
\end{equation*}

\qquad solving by separating \ \ 
\begin{equation*}
\frac{T^{\prime }}{T}=-\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{36}%
\Longrightarrow \int \frac{T^{\prime }}{T}=-\int \frac{\left( 2n-1\right)
^{2}\pi ^{2}}{36}
\end{equation*}

$\qquad $%
\begin{equation*}
\Longrightarrow \ln T=-\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{36}%
t+c\Longrightarrow T_{n}\left( t\right) =c_{n}e^{-\frac{\left( 2n-1\right)
^{2}\pi ^{2}}{36}t}
\end{equation*}

Therefore \ \ 
\begin{eqnarray*}
v_{n}\left( x,t\right) &=&X_{n}\left( x\right) T_{n}\left( t\right) \\
&=&c_{n}\cos \frac{\left( 2n-1\right) \pi x}{2}\,e^{-\frac{\left(
2n-1\right) ^{2}\pi ^{2}}{36}t}
\end{eqnarray*}

$\qquad $

\qquad so we let 
\begin{equation*}
\ v\left( x,t\right) =\sum\limits_{n=1}^{\infty }c_{n}\cos \frac{\left(
2n-1\right) \pi x}{2}\,e^{-\frac{\left( 2n-1\right) ^{2}\pi ^{2}}{36}t}
\end{equation*}

Using\ \ I.C. to compute coefficients we have

\qquad\ 
\begin{equation*}
v\left( x,0\right) =\sum\limits_{n=1}^{\infty }c_{n}\cos \frac{\left(
2n-1\right) \pi x}{2}\,=-3\cos \frac{7\pi x}{2}
\end{equation*}

by equating coefficients: $\ c_{1}=0,c_{2}=0,c_{3}=0,c_{4}=-3,c_{4}=0,...$

\begin{equation*}
v\left( x,t\right) =-3\cos \frac{7\pi x}{2}\,e^{-\frac{49\pi ^{2}}{36}t}
\end{equation*}
\ is the solution.

\vspace{1pt}

iii) \ \ Now use the results of b) i) and ii) to find $u(x,t)$.

Solution to iii)

\begin{equation*}
u\left( x,t\right) =v\left( x.t\right) +2
\end{equation*}

so$\qquad $%
\begin{equation*}
u\left( x,t\right) =-3\cos \frac{7\pi x}{2}\,e^{-\frac{49\pi ^{2}}{36}t}+2
\end{equation*}

\vspace{1pt}

\qquad $\qquad $

\end{document}

%%%%%%%%%%%%%%%%%%%%%%% End /document/Chap11.tex %%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%% Start /document/webmath.cst %%%%%%%%%%%%%%%%%%%%%

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BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_RIGHT=36
PARAGRAPH_INDENT_LEFT=36
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAG_BEHAVIOR=LISTSTART
FONT_SIZE=10
TAGBAR_DELETE=0
LEADIN_ALIGNLEFT=1
FONT_SIZE=10

[remark]
ALIAS=Remark
TAG_TYPE=THEOREMENV
TAG_LEADIN=Remark
BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_RIGHT=36
PARAGRAPH_INDENT_LEFT=36
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAG_BEHAVIOR=LISTSTART
FONT_SIZE=10
TAGBAR_DELETE=0

[exercise]
ALIAS=Exercise
TAG_TYPE=THEOREMENV
TAG_LEADIN=Exercise
BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_RIGHT=36
PARAGRAPH_INDENT_LEFT=36
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAG_BEHAVIOR=LISTSTART
FONT_SIZE=10
TAGBAR_DELETE=0

[axiom]
ALIAS=Axiom
TAG_TYPE=THEOREMENV
TAG_LEADIN=Axiom
BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_RIGHT=36
PARAGRAPH_INDENT_LEFT=36
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAG_BEHAVIOR=LISTSTART
FONT_SIZE=10
TAGBAR_DELETE=0

[newenvironment]
TAG_TYPE=TLTEMPLATE
PARAGRAPH_INDENT_LEFT=36
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=18
TAG_BEHAVIOR=LISTSTART

[tiny_big]
TAG_TYPE=TEXT
FONT_SIZE=7
TEXT_COLOR=R255,G0,B255
TAGBAR_DELETE=1

[tiny]
TAG_TYPE=TEXT
FONT_SIZE=6
TEXT_COLOR=R255,G0,B255
TAGBAR_DELETE=1

[scriptsize]
TAG_TYPE=TEXT
FONT_SIZE=8
TAGBAR_DELETE=1

[footnotesize]
TAG_TYPE=TEXT
FONT_SIZE=9
TAGBAR_DELETE=1

[small]
TAG_TYPE=TEXT
ALIAS=Smaller
FONT_SIZE=9
TAGBAR_DELETE=0

[normalsize]
TAG_TYPE=TEXT
FONT_SIZE=10
TEXT_COLOR=R:0 G:128 B:128 
TAGBAR_DELETE=1

[large]
TAG_TYPE=TEXT
FONT_SIZE=14
TEXT_COLOR=R255,G0,B255
TAGBAR_DELETE=1

[Large]
TAG_TYPE=TEXT
ALIAS=Bigger
FONT_SIZE=16

[LARGE]
TAG_TYPE=TEXT
FONT_SIZE=16
TAGBAR_DELETE=1

[huge]
TAG_TYPE=TEXT
FONT_SIZE=20
TEXT_COLOR=R255,G0,B255
TAGBAR_DELETE=1

[Huge]
TAG_TYPE=TEXT
FONT_SIZE=22
TEXT_COLOR=R255,G0,B255
TAGBAR_DELETE=1

[thebibliography]
ALIAS=Bibliography item
TAG_TYPE=BIBENV
TAG_LEADIN=bibitem
TAG_BEHAVIOR=NOSTRUCTURETAGS|NOMULTIPARA
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=18
TAGBAR_DELETE=1
FONT_SIZE=10
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1

[quotation]
TAG_TYPE=PARA
ALIAS=Block Quote
PARAGRAPH_INDENT_RIGHT=24
PARAGRAPH_INDENT_LEFT=24
BKGROUND_COLOR=R:240 G:240 B:240 
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
FONT_FACE=Times New Roman
NEXT_TAG=Body Text
FONT_SIZE=10
TAGBAR_DELETE=0

[quote]
TAG_TYPE=PARA
ALIAS=Body Quote
PARAGRAPH_INDENT_LEFT=0
NEXT_TAG=Body Quote
BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_FIRST=-11
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=7
PARAGRAPH_LEADING_AFTER=4
LINE_LEADING=0
LINE_JUSTIFICATION=1
FONT_FACE=Times New Roman
FONT_WEIGHT=700
FONT_SLANT=1
TAGBAR_DELETE=0
TEXT_COLOR=R:0 G:128 B:128 
FONT_SIZE=10
PARAGRAPH_INDENT_RIGHT=0

[center]
TAG_TYPE=PARA
ALIAS=Body Center
LINE_JUSTIFICATION=2
FONT_FACE=Times New Roman
NEXT_TAG=Body Center
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
TAGBAR_DELETE=0
BKGROUND_COLOR=R:255 G:255 B:255 

[verbatim]
TAG_TYPE=PARA
ALIAS=Preformatted
PARAGRAPH_INDENT_RIGHT=36
PARAGRAPH_INDENT_LEFT=36
TAG_BEHAVIOR=FORCESTEXT|NOLINEBREAK|PRESERVESPACES
FONT_FACE=Courier New
FONT_SIZE=10
NEXT_TAG=Body Verbatim
BKGROUND_COLOR=R:255 G:255 B:255 
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1

[newfield]
TAG_TYPE=TLTEMPLATE
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[f_0]
ALIAS=Title
TAG_TYPE=FIELD
TAG_LEADIN=Title: 
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_1]
ALIAS=Author
TAG_TYPE=FIELD
TAG_LEADIN=Author: 
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
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[f_2]
ALIAS=Address
TAG_TYPE=FIELD
TAG_LEADIN=Address: 
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_7]
ALIAS=Date
TAG_TYPE=FIELD
TAG_LEADIN=Date: 
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[abstract]
TAG_TYPE=FIELD
TAG_LEADIN=Abstract:
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_11]
ALIAS=Make Title
TAG_TYPE=FIELD
TAG_LEADIN=Make Title
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_12]
ALIAS=Make LOF
TAG_TYPE=FIELD
TAG_LEADIN=Make LOF
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_13]
ALIAS=Make LOT
TAG_TYPE=FIELD
TAG_LEADIN=Make LOT
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[f_14]
ALIAS=Make TOC
TAG_TYPE=FIELD
TAG_LEADIN=Make TOC
TAG_BEHAVIOR=FRONTMATTER|NOSTRUCTURETAGS|NOENVTAGS
PARAGRAPH_INDENT_LEFT=35
PARAGRAPH_INDENT_FIRST=0

[frak]
ALIAS=Fraktur
TAG_TYPE=TEXT
TAG_BEHAVIOR=FORCESMATH
FONT_SLANT=1
TEXT_COLOR=R:192 G:192 B:192 
MATH_COLOR=R:255 G:0 B:255 
TAGBAR_DELETE=1
FONT_SIZE=10
FONT_WEIGHT=400

[Bbb]
ALIAS=Blackboard bold
TAG_TYPE=TEXT
TAG_BEHAVIOR=FORCESMATH
TEXT_COLOR=R:192 G:192 B:192 
TAGBAR_DELETE=1
FONT_SIZE=10

[newlist]
TAG_TYPE=TLTEMPLATE
PARAGRAPH_INDENT_LEFT=70
PARAGRAPH_INDENT_FIRST=0

[Hyperlink]
TAG_TYPE=INTERNAL
TEXT_COLOR=R0,G128,B0
MATH_COLOR=R0,G128,B0

[ProgramCall]
TAG_TYPE=INTERNAL
TEXT_COLOR=R128,G0,B0
MATH_COLOR=R128,G0,B0


[ProgramCall]
TAG_TYPE=INTERNAL
TEXT_COLOR=R128,G0,B0
MATH_COLOR=R128,G0,B0

[Margin_Hint]
ALIAS=Margin Hint
TAG_TYPE=NOTE
ICON_NAME=MarginHint
ICON_SIZE=0.49in,0.53in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0

[Solution_Note]
ALIAS=Solution Note
TAG_TYPE=NOTE
ICON_NAME=Solution
ICON_SIZE=0.23in,0.26in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0

[Proof_Theorem]
ALIAS=Proof
TAG_TYPE=PROOF
ICON_NAME=Proof
ICON_SIZE=0.23in,0.26in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0

[Prob_Solv_Hint]
ALIAS=Problem Solving Hint
TAG_TYPE=NOTE
ICON_NAME=ProbSolvHint
ICON_SIZE=0.51in,0.64in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0

[Note]
TAG_TYPE=NOTE
ICON_NAME=Note
ICON_SIZE=0.50in,0.51in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0

[Answer]
ALIAS=AnswerNote
TAG_TYPE=NOTE
ICON_NAME=Answer
ICON_SIZE=0.26in,0.26in,0.00in
FONT_SIZE=10
PARAGRAPH_INDENT_LEFT=18
PARAGRAPH_INDENT_FIRST=0
PARAGRAPH_INDENT_REST=0
PARAGRAPH_LEADING_BEFORE=0
PARAGRAPH_LEADING_AFTER=0
LINE_LEADING=0
LINE_JUSTIFICATION=1
TAGBAR_DELETE=0


[FKeys]
0x0001=(Remove Item Tag)
0x0002=Body Text
0x0003=(Normal)
0x0004=bf
0x0005=em
0x0006=enumerate
0x0007=itemize
0x0008=cal
0x000a=section
0x000b=subsection

%%%%%%%%%%%%%%%%%%%%%% End /document/webmath.cst %%%%%%%%%%%%%%%%%%%%%%

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%%%%%%%%%%%%%%%%%%%%% Start /document/N4QRHX0F.xvz %%%%%%%%%%%%%%%%%%%%

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