Math 6410 • ODE

Let \(f\in C^1(U,\R^n)\) for \(U\subset\R^n\) and \(x_0\in U\). Given the Banach space
\(X=C([0,T],\R^n)\) with norm \(\norm{x}=\max_{0\le t\le T}\abs{x(t)}\), let
\[
  K(x)(t)=x_0+\int_0^t f(x(s))\dd s
\]
for \(x\in X\). Define
\(V=\{x\in X\mid\norm{x-x_0}\le\epsilon\}\) for fixed \(\epsilon>0\) and suppose
\(K(x)\in V\) (which holds for sufficiently small \(T\)), so that \(K:V\to V\) with
\(V\) a closed subset of \(X\).
\begin{enumerate}
\item[(a)] Give the definition of a locally Lipschitz function in an open set \(U\) of a
normed vector space.
\item[(b)] Using the fact that \(f\) is locally Lipschitz in \(U\) with Lipschitz constant
\(L_0\), and taking \(x,y\in V\) show that
\[
  \abs{K(x(t))-K(y(t))}\le L_0t\norm{x-y}.
\]
Hence, show that
\[
  \norm{K(x)-K(y)}\le L_0T\norm{x-y},\qquad x,y\in V.
\]
\item[(c)] State the contraction mapping principle on a Banach space.
\item[(d)] Choosing \(T<1/L_0\), apply the contraction mapping principle to show that
the integral equation has a unique continuous solution \(x(t)\) for all
\(t\in[0,T]\) and sufficiently small \(T\). Hence establish existence and uniqueness
of the initial value problem
\[
  \frac{dx}{dt}=f(x),\qquad x(0)=x_0.
\]
\end{enumerate}