Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 22I / c

Set $y_k(0)=x_0$ and let $M={\sup }_{z\in {\mathbb{R}}^n}\Vert F(z)\Vert <\infty$. On the first interval, $y_k$ is the known constant $x_0$, so the integral equation defines $x_k$. Its endpoint value then defines the constant $y_k$ on the next interval, and induction over the $k$ subintervals defines the pair on all of $[0,1]$. An integral of the bounded, piecewise constant function $F\circ y_k$ is continuous, so every $x_k$ is continuous. In fact,so the approximations are uniformly Lipschitz continuous.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzela-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.