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\begin{document} \thispagestyle{empty} %%%%%%% 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} \end{document}