Let \(f \in C^1(U,\mathbb {R}^n)\) for \(U \subset \mathbb {R}^n\) and \(x_0 \in U\). Given the Banach space \(X=C([0,T],\mathbb {R}^n)\) with norm \(\|x\|=\max _{0\le t\le T}|x(t)|\), let \[ K(x)(t)=x_0+\int _0^t f(x(s))\,ds \] for \(x \in X\). Define \(V=\{x\in X\mid \|x-x_0\|\le \epsilon \}\) 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 \(X\).
- Using the fact that \(f\) is locally Lipschitz in \(U\) with Lipschitz constant \(L_0\), and taking \(x,y\in V\), show that \[ |K(x(t))-K(y(t))|\le L_0t\|x-y\|. \] Hence, show that \[ \|K(x)-K(y)\|\le L_0T\|x-y\|,\quad x,y\in V. \]
- 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 \[ \frac {dx}{dt}=f(x),\quad x(0)=x_0. \]
- Show that the solution of the equation \[ \dot {x}=x^2,\quad x(0)=x_0 \] blows up in finite time. Does this contradict the result of part (b)?