Continuous dependence of solutions to ODEs on parameters

Question

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.

Answer 1

I will argue that yes, if that $x:[t_0,\beta[\times W\to\mathbb R^n$ exists then it is continuous.

Let's prove sequential continuity: fix $(t_n,\alpha_n)$ converging to $(t^*,\alpha_0)$ (all in $[t_0,\beta[\times W$). We need to show $x(t_n,\alpha_n)\to x(t^*,\alpha_0).$ We will construct a neighbourhood of $(t^*,\alpha_0)$ on which $x$ is continuous, and since some tail of the sequence $(t_n,\alpha_n)$ is in this neighbourhood, this will give the necessary convergence. Set $T=\tfrac12(t^*+\beta),$ so $t_0\leq t^*<T<\beta.$

By Lemma 1 below there exist $A,X,L>0$ such that whenever $t,\alpha$ satisfy:

• $0\leq t\leq T$ and
• $\|\alpha-\alpha_0\|\leq A$ and
• $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$

then

$$\|x_1(t,\alpha)-x_1(t,\alpha_0)\|\leq L(\|\alpha-\alpha_0\|+\|x(t,\alpha)-x(t,\alpha_0)\|).\tag{*}$$

If necessary, shrink $A$ further to ensure that $g$ is Lipschitz on the region $\|\alpha-\alpha_0\|\leq A.$ This is possible by the local Lipschitz property of $g.$

By continuity of $g$, we can shrink $A$ if necessary to also ensure that whenever $\|\alpha-\alpha_0\|\leq A$ we have

$$\|\alpha-\alpha_0\|, \|g(\alpha)-g(\alpha_0)\|\leq \tfrac 1 2 X e^{-LT}.$$

We will now show:

$$\|x(t,\alpha)-x(t,\alpha_0)\|\leq(\|\alpha-\alpha_0\|+\|g(\alpha)-g(\alpha_0)\|)e^{Lt}\tag{G}$$ for all $0\leq t\leq T$ and $\|\alpha-\alpha_0\|\leq A.$ Note that the right-hand-side is at most $X.$

Suppose not. Fix some $\alpha$ for which (G) fails. There are two cases:

• (G) fails very badly: $\|x(t,\alpha)-x(t,\alpha_0)\|\geq X$ for some $0\leq t<T$ and $\|\alpha-\alpha_0\|\leq A$
• (G) fails mildly: $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$ for some $0\leq t\leq T$ and $\|\alpha-\alpha_0\|\leq A$

For the very bad case, we can take $t$ to be infimal such that $\|x(t',\alpha)-x(t',\alpha_0)\|\geq X,$ for some fixed $\alpha.$ In the mild case, just pick any $(t,\alpha)$ such that (G) fails. We now apply Gronwall's inequality to (*); for both the very bad and mild cases, we have arranged that (*) applies for all smaller $t.$ Gronwall gives:

\begin{align*} \|x(t,\alpha)-x(t,\alpha_0)\| &\leq \|\alpha-\alpha_0\|+\|x(t,\alpha)-x(t,\alpha_0)\|\\ &\leq (\|\alpha-\alpha_0\|+\|g(\alpha)-g(\alpha_0)\|)e^{Lt} \end{align*} as required. This proves (G).

This means that solutions starting close to $\alpha_0$ stay in the set $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$. This means they obey an ODE with a global Lipschitz condition on $[0,T]$. We can therefore apply Lemma 2 below to get a neighbourhood of $(t^*,\alpha_0)$ on which $x$ is continuous.

Lemma 1

For this argument we need that there is a neighbourhood of $C=\{(t,\alpha_0,x(t,\alpha_0))\mid 0\leq t\leq T\}$ on which $f$ is Lipschitz. Note that $C$ is compact - it's the graph of the continuous function $x|_{[0,T]\times \{\alpha_0\}}.$ So this is a special case of the more general statement:

For all metric spaces $X,Y$, compact subsets $C\subseteq X$, and locally Lipschitz functions $f:X\to Y$, there exists a neighbourhood of $C$ on which $f$ is Lipschitz.

The proof is to suppose otherwise. Then for each $n\geq 1$ there are distinct points $x_n,x'_n$ in the $1/n$-neighbourhood $B_X(C,1/n)$ with $d_Y(f(x_n),f(x'_n))\geq n d_X(x_n,x'_n) $. By compactness of $C$ we can restrict to a subsequence $n_k\to\infty$ such that $x_{n_k}$ and $x'_{n_k}$ converge to some $x\in C$. Since $f$ is locally Lipschitz, there exist $L,\epsilon>0$ such that $f$ is $L$-Lipschitz on $B(x,\epsilon)$. But for sufficiently large $k$ we have $x_{n_k},x'_{n_k}\in B(x,\epsilon)$ and $n_k>L$, which contradicts the choice of $x_{n_k},x'_{n_k}.$

Lemma 2

If $f$ is Lipschitz on $V'\subseteq V$, and $g$ is Lipschitz on $W'\subseteq W$, and $(t,\alpha,x(t,\alpha))\in V'$ for all $(t,\alpha)\in[0,T]\times W'$, then $x$ is continuous on $[t_0,T)\times W'.$

This is stated as known in the original question.

Answer 2

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).

Answer 3

Let's fix $t_0 = 0$. We will consider the case when $f$ doesn't depend on $t$. I find it easier to prove the more general (well-known) statement

Let $f : V \rightarrow \mathbb{R}^n$ be locally Lipschitz ($V$ is open) . Define $\displaystyle D := \bigcup_{x \in V} I_x \ \times \{x\} \subseteq \mathbb{R} \times V $, where $I_x$ is the unique maximal interval of the unique solution $y(\cdot,x):I_x \rightarrow \mathbb{R}^n$ of $$ y' = f(y) $$ Then $D$ is open and $y:D \rightarrow \mathbb{R}^n$ is continuous, we call this the local flow of $f$

Your claim follows by letting $Y = (\alpha,x):D \subseteq V \rightarrow \mathbb{R}^m \times \mathbb{R}^n$ be the local flow of $F=(0,f)$. In other-words, for all $(t,\alpha_0,x_0) \in D$.

$$ \alpha'(t,\alpha_0,x_0) = 0 \quad ; \quad x'(t,\alpha_0,x_0) = f(\alpha_0,x(t,\alpha_0,x_0))$$

with $\alpha(0,\alpha_0,x_0) = \alpha_0$ and $x(0,\alpha_0,x_0)=x_0$.

Now your assumption:

" Suppose we have a function $x:[0=t_0,\beta) \times W\rightarrow \mathbb{R}^n$ differentiable .... "

Can be restated by saying that $[0,\beta) \times \mathrm{graph}(g) \subseteq D$. By the theorem above, we conclude that the following composition is continuous;

$$[0,\beta) \times W \hookrightarrow [0,\beta) \times \mathrm{graph}(g) \subseteq D \xrightarrow{Y} \mathbb{R}^m \times \mathbb{R}^n \xrightarrow{\pi_2} \mathbb{R}^n $$

where the first map is the obvious one, and $\pi_2$ denote the projection on the "second" argument. This is clearly "your" map.