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 \centerline{\large Math.\ 401, Sec.\ 501\hfill    Spring, 2006}
                                                           
\bigskip
                                           
               
\centerline{\bfseries\large Homework 6, due March 3}
\bigskip
                                                                          
\begin{enumerate}

 \item {\it[Bush, p.~156, (i) and (ii)]}\quad
 Find the (lowest-order) composite expansion for the solution of
\begin{enumerate}
 \item\quad $\displaystyle \epsilon y'' + y' +\frac y{x+1} 
=2$.
 \item\quad$\displaystyle \epsilon y'' - y' +\frac y{x+1} 
=2$.
\end{enumerate}
\noindent In both cases [(a) and (b)] consider 
 $$0<\epsilon\ll 1, \qquad 0<x<1, \qquad y(0)=0, \quad y(1)=3. 
$$


 \item  {\it[Bush, p.~156, (iii)--(v)]}\quad
 Find the  composite expansion for the solution of
\begin{enumerate}
 \item \quad$\displaystyle\epsilon y'' + xy' +y =0$,
$$0<\epsilon\ll 1, \qquad 2<x<4, \qquad y(2)=0, \quad y(4)=1. $$
 \item \quad $\displaystyle\epsilon y'' - xy' +y =0$,
$$0<\epsilon\ll 1, \qquad 2<x<4, \qquad y(2)=0, \quad y(4)=1. $$
\item \quad$\displaystyle \epsilon y''+xy'+y=0$,
$$0<\epsilon\ll 1, \qquad -4<x<-2, \qquad y(-4)=1, \quad y(-2)=0. 
$$
\end{enumerate}
\noindent{\bf Do not} work out (c) from the beginning;
 knowing the answers to (a) and (b), you can solve (c) in two or 
three lines by a change of variable.

 \item {\it [Bush, p.~172, (i)]}\quad
$\displaystyle\epsilon y'' -y' +\frac1y=0$,
$$0<\epsilon\ll 1, \qquad 0<x<1, \qquad y(0)=2, \quad y(1)=1. $$

  \item {\it [Bush, p.~172, (ii)]}\quad
$\displaystyle \epsilon y'' +y' +e^y=0$,
$$0<\epsilon\ll 1, \qquad 0<x<1, \qquad y(0)=1, \quad y(1)=-\ln 
2. $$


\item {\it [Logan, p.\ 69,  Ex.\ 3.2(a)]\/} 
\quad Find the lowest-order composite expansion for positive 
$\epsilon$.  Compare the result with the exact solution, either 
numerically or in terms of power series.
 $$\epsilon y'' + 2y' +y=0, \qquad y(0)=0, \quad y(1) =1.$$





\item {\it [Logan, p.\ 69,  Ex.\ 3.4]\/}
\quad Find the lowest-order composite expansion for positive 
$\epsilon$.  Compare the result with the exact solution, either 
numerically or in terms of power series.
$$\epsilon y'' - y' = 2t, 
\qquad y(0) = y(1) = 1.$$

\item {\it [Logan, p.\ 69,  Ex.\ 3.9(a)]\/}
\quad $\epsilon y'' +(t^2 +1) y' -t^3 y =0$,
$$0<\epsilon\ll 1, \qquad y(0) = y(1) = 1.$$

 
  \item {\it [Logan, p.\ 69, Ex.\ 3.9(b)]\/}
\quad $\epsilon y'' +(\cosh t)y' -y=0$,
$$0<\epsilon\ll 1, \qquad
y(0) = y(1) = 1.$$
  {\sl Hint:\/} You may find 
some help with the integral in a handbook under ``Gudermannian
function'' or ``Lobachevsky's angle of 
parallelism''\negthinspace.

\end{enumerate}

 \bigskip\bigskip

\centerline{\bfseries\large Homework 7, due March 8}
\bigskip
                                          
\begin{enumerate}

  \item 
Classify these equations as linear homogeneous, linear 
nonhomogeneous, or nonlinear. 
(Here $u_t \equiv \partial u/\partial t$, etc.)

% old cases from Logan, p.~159:
%\zitem{(a)} $u_t u_{tt} + 3tu =0$
%\zitem{(b)} $e^t u_{tx} -x^2 u = \cos t$
%\zitem{(c)}  $u_{tt} - u_{xx} =0$
%\zitem{(d)}  $u_{tx} +u^2 = \sin x$

\begin{enumerate}
\item\quad  $u_{tt} - u_{xx} = \cos(x-t)$
\item\quad $u_t + u^2 u_x = 0$
\item\quad $u_t + 3t^2u =0$
\item\quad $u_{tt} - u_{xx} = -m^2 u$
\end{enumerate}

\item Why do we not make a distinction between ``homogeneous''
and ``nonhomogeneous'' for {\sl nonlinear\/} equations?
{\sl Hint:\/} Try to classify the equation $y'' + (\cos y)^2 =1$.

%\item  {\it [Logan, p.\ 195,  Ex.\ 3.1(d)]\/}
%Find the eigenvalues and eigenfunctions for the problem
%$$y'' + \lambda y = 0 \quad\mbox{on $0<x<L$}, $$
%$$y'(0) = 0, \qquad y(L) =0.$$

\item {\it [Schaum's, p.\ 46, Ex.\ 2.52]\/}\quad
Find the steady-state temperature in a bar whose ends are located
at $x=0$ and $x=10$, if these ends are kept at 150$^\circ$C
and 100$^\circ$C respectively.

\item Suppose that the boundary conditions (``BC:'') 
on p.~65 of the notes 
(and corresponding lecture on the heat equation with fixed end 
temperatures)
are replaced by 
$$   \frac{\partial u}{\partial x}(t,0) = F_1\,, \qquad 
\frac{\partial u}{\partial x}(t,1) = F_2\,.$$
(That is, the {\sl heat flux\/} through each end of the bars
is held constant.)
What happens when you attempt to find a steady-state solution
as on p.~66?
Distinguish between the two cases $F_1 = F_2$ and $F_1 \ne F_2\,$.
Can you give a physical explanation for your results?

% \item Suppose that the boundary conditions (``BC:'') 
%on p.~68 of the notes are replaced by 
%$$   \frac{\partial w}{\partial x(t,0) = 0, \qquad \
%frac{\partial w}{\partial x}(t,1) = 0.$$
%What changes would be necessary in pp\. 67--72
%(and corresponding lecture on separation of variables in the 
%heat equation)?
%{\sl Warning:\/}  Pay close attention to the possibility 
%that $\lambda$ may not always be positive.

\end{enumerate}
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