% LaTeX file -- PM 450 Winter 2017, Assignment 1

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\begin{document}
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%%%%%%%   TITLE   %%%%%%
{\noindent\textbf{\large PM 450 \hfill Assignment 1 \hfill Due Wednesday January 18.}} 
\vskip 2em 

\begin{Ex}


\item Consider the DE:  $y' = 1+xy$ and $y(0)=0$ for $x \in [-b,b]$, where $b>0$. 
\begin{parts}
\item Reduce this to finding the fixed point of a mapping $T$.
 Show that when $b=1$, the map $T$ is a contraction mapping.
\item Prove that the DE  has a unique solution on $[-b, b]$ for any $b>0$.
Hence deduce that there is a unique solution on the whole line $\bR$.
\item Start with $f_0(x) = 1$ and compute $f_n(x) = T^n f_0$ by induction.
Prove directly (rather than by quoting a theorem) that the sequence $f_n$ converges uniformly on $[-b,b]$.
\end{parts}
 

\item Consider the DE $y'' - x^{-1}y'+ x^{-2}y = 0$ for $x\in [1,3]$.
\begin{parts}
\item Check that $y = x$ is a solution. Look for a solution of the form $f(x) = xg(x)$ by showing that $g'$ satisfies a 1st order DE, and solving it.
\item Show that the set of solutions that you obtain is a 2-dimensional vector space.
\item Show that the initial value conditions $y(1)=a_0$ and $y'(1) = a_1$ determine a unique solution from this set. 
\end{parts}


\item  Consider $f'(x) + f(x)^2 = 4xf(x) - 4x^2 + 2$ for $x \in \bR$ and $f(0)=2$.
\begin{parts}
\item Show that this DE satisfies a  local Lipschitz condition on some smaller region around $x=0$; 
and hence deduce that there is a local solution.
\item Solve this DE explicitly.
\hint Find the DE satisfied by $g(x) = f(x) - 2x$ and solve it first.
\item Hence find the maximal continuation of the solution.
\end{parts}


\item Consider $y' = \sin\bigg( \dfrac{x^5+3x^2-1}{\sqrt{219 - 2y^2}} \bigg)$ and $y(2) = 3$. 
Prove that there is a unique
solution on $[-5,9]$.\\[.3ex]
\hint Show that a solution must satisfy $|y| \le 10$.  Obtain a Lipschitz condition valid in this range.


\item Consider the DE: $y'' = (1+(y')^2)^{3/2}$ and $y(0) = 0$, $y'(0) =  0$.
\begin{parts}
\item Show that $f(x) = 1-\sqrt{1-x^2}$ is the unique solution on $[-1,1]$.
\item This solution does not continue further, yet $|f(x)| \le 1$. Why does this not contradict the Continuation Theorem?
\end{parts}


\item Consider the DE: $xyy'=y^2-1$ and $y(1)=a > 0$.
\begin{parts}
\item Show that this DE satisfies a local Lipschitz condition in $y$ as long as $x,y$ are both positive.
\item Solve the DE and find the largest interval on which a solution exists.
\item Observe that all solutions pass through $(0,1)$ with the same slope. 
What happens when the solution is continued through this point into the second quadrant?
Is this a problem for the theory? 
\end{parts}




\end{Ex}
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