Let $f:V\rightarrow \mathbb{R}^n$ be locally Lipschitz ($V$ is a subset of $\mathbb{R}\times\mathbb{R}^m\times \mathbb{R}^n$). Suppose we have a function $x:[t_0,\beta[\times W\rightarrow \mathbb{R}^n$ differentiable in the first argument ($W$ is an open subset of $\mathbb{R}^m$, $\beta$ is finite) such that for every $(t,\overrightarrow{\alpha})\in [t_0,\beta[\times W$ we have:
$$(t,\overrightarrow{\alpha},x(t,\overrightarrow{\alpha}))\in V$$
$$x_1(t,\overrightarrow{\alpha})=f(t,\overrightarrow{\alpha},x(t,\overrightarrow{\alpha}))$$
Here $x_1(t,\overrightarrow{\alpha})$ means partial derivative with respect to first argument.
It is also given that the function $g:W\rightarrow \mathbb{R}^n$ given by $g(\overrightarrow{\alpha})=x(t_0,\overrightarrow{\alpha})$ is locally Lipschitz.
Question: Does it follow that the function $x:[t_0,\beta[\times W\rightarrow\mathbb{R}^n$ is continuous ?
I can only prove the conclusion if the hypotheses are strengthened to $f,g$ Lipschitz instead of just merely locally Lipschitz.I would still like to know the answer in the locally Lipschitz case.
Thank you a lot.
Best Answer
The setup may be rewritten in the following form (avoiding special treatment of the parameter):
$\;\;y=(\alpha,x)$, $\; \; y_0=(\alpha,g(\alpha))$, $\; \;v(t,y) = (0,f(t,\alpha,x))$ and the ode: $$ \dot{y} = v(t,y), \ \ y(t_0) = y_0 $$ Now, if $v$ is locally Lipschitz in $y$ then for each $y_0$ there is a maximal solution defined (we apparently only look at $t\geq t_0$) some interval of time $[t_0,\tau)$ where $\tau=\tau(y_0)$ is a lower semi-continuous function of the initial condition $y_0$.
For every $t_0<b<\tau$ there is a neighborhood $U=U_{t_0,b}(y_0)$ (which may be very small) so that a maximal solution $y=\phi^t(t_0,y_1)$ exists for all $y_1\in U$, $t\in [t_0,b]$ and in addition $y_1\in U\mapsto \phi^t(t_0,y_1)$ is $L$-Lipschitz for all $t\in [t_0,b]$ and some $L=L(U)<+\infty$.
[The proof uses the local result which it seems you have understood how to get, and then uses compactness of $[t_0,b]$ to show that the Lipschitz dependency pops out from composing a finite number of local solutions]
Your hypothesis is that given the initial condition $y_0=y_0(\alpha) =(\alpha,g(\alpha))$, with $\alpha \in W$ the maximal solution exists up to, but not including the time $\beta$. Then for any $t_0<b<\beta$ the (composed) flow: $$ (t, \alpha)\in [t_0,b] \times W \mapsto \phi^t(t_0, (\alpha,g(\alpha)))$$ is locally Lipschitz in $\alpha$. Taking the limit $b\rightarrow \beta^-$ you infer that the map is continuous on $[t_0,\beta[\times W$ (but may fail to be locally Lipschitz on this set).