Let $f(t,x)$ be piecewise continuous in $t$ and locally Lipschitz in $x$ on $[t_0,t_1]\times{D}$, for some domain $D\subset\mathbb{R}^{n}$. Let $W$ be a compact subset of $D$. Let $x(t)$ be the solution of $\dot{x}=f(t,x)$ tarting at $x(t_0)=x_0\in{W}$. Suppose that $x(t)$ is defined and $x(t)\in{W}$ for all $t\in{[t_0,T)},T<t_1$. Then,
(a) show that $x(t)$ is uniformly continuous on $[t_0,T)$.
(b) Show that $x(T)$ is defined and belongs to $W$ and $x(t)$ is a solution on $[t_0,T]$.
I was able to solve part (a) of the problem which is given as follows.
\begin{align}
\dot{x} & = f(t,x),x_0\in{W},t\in[t_0,T)\\
x(t) & = x(t_0)+\int_{t_0}^{t}f(s,x(s))~ds,\\
x(t) – x(t_0) & = \int_{t_0}^{t}f(s,x(s))~ds,\\
\|x(t)-x(0)\| & \leq \int_{t_0}^{t}\|f(s,x(s))\|~ds\leq{L}(t-t_0),
\end{align}
since $x(t)\in{W},\forall{t}\in[t_0,T)$ and $W$ is a compact set and $f(t,x)$ is Lipschitz on $W$. Since $\|x(t)-x(0)\|\leq{L}(t-t_0)$ over $[t_0,T)$, $x(t)$ is thus uniformly continuous over that time interval.
However, I cannot prove the part (b) of the problem. Any suggestions are greatly appreciated.
Best Answer
Here is an outline of a proof:
Show that $f$ is bounded by some $B$ on $W$ (slight twist since $f$ is only piecewise continuous in $t$).
Then the integral equation for $x$ shows that $x$ is actually Lipschitz with constant $B$ on $[t_0,T)$.
It is straightforward to show that if $x$ is uniformly continuous on $[t_0,T)$ then it can be extended uniquely to a uniformly continuous on $[t_0,T]$ (since a uniformly continuous maps Cauchy sequences into Cauchy sequences). (Note that uniform continuity is required, $x \mapsto \sin {1 \over x}$ is continuous on $(0,1]$ but cannot be extended to a continuous function on $[0,1]$.)
Hence there is some $x_1$ such that $x_1 = \lim_{t \to T, t < T} x(t)$. In particular, continuity of the map $t \mapsto x_0 + \int_{t_0}^t f(x,x(s))ds$ shows that we have $x_1 = x_0 + \int_{t_0}^T f(x,x(s))ds$. The extended $x$ is also Lipschitz continuous with constant $B$.
Since $x(t) \in W$ for all $t \in [t_0,T)$ it follows (since $W$ is closed) that $x_1 \in W$.
It is not entirely clear what the question means when it asks to show that $x$ is a solution on $[t_0,T]$.
I would assume that showing $x(t) = x_0+ \int_{t_0}^t f(s,x(s))ds$ for all $t \in [t_0,T]$ (or equivalently that $x$ is absolutely continuous and $\dot{x} = f(t,x)$ for ae. $t \in [t_0,T]$) is sufficient.