Proving continuous dependence on initial conditions for flows

I'm looking for assistance in solving a dynamical systems problem.

Considering the flow $$\phi (t;x): \mathbb{R}\times \mathbb{R}^n \rightarrow \mathbb{R}^n$$, that isn't necessarily associated to an ODE, for which I need to prove continuous dependence on initial conditions. I am given the following: let the initial condition $$x_0 \in \mathbb{R}^n$$ be given and show that for all $$T > 0$$ and $$\delta > 0$$ there is an $$\epsilon > 0$$ such that $$||\phi(T;x_0)-\phi(T;\tilde{x_0})|| < \delta$$ for all $$\tilde{x_0}$$ with $$||x_0-\tilde{x_0}|| < \epsilon$$.

I'm not really sure where to start with this problem. I believe it might need to be solved in three steps, proving first Lipschitz Dependence on Initial Conditions, then Smooth Dependence on Initial Conditions and finally Continuous Dependence on Parameters. However, I can't seem to make heads nor tails of it.

Any and all assistance in solving this problem would be greatly appreciated. Thanks in advance!

• If the flow is not associated to an ODE, why do you want to analyze it as a solution to an ODE? What are the properties of the flow? – LutzL Jul 8 at 13:01
• The question says not necessarily associated to an ODE, so I'm supposing it's to take into account flows that are and are not associated, perhaps some specific clause? As for the properties of the flow, no clue, this is all of the information I am given in the question. – Azog4 Jul 8 at 13:03
• What is a flow that is not associated to an ODE? – miraunpajaro Jul 8 at 13:08
• For the ODE case, do you know Grünwald inequality? – miraunpajaro Jul 8 at 13:09
• You need to have some global continuity property, apart from the flow condition $ϕ(t+s;x)=ϕ(t;ϕ(s;x))$, else the task makes no sense. – LutzL Jul 8 at 13:24

Firstofall, for us to guarantee properties of the flow you must have some extra information, I'll address here the case where $$\phi(t,x)$$ is the only solution of the system

$$\begin{cases} x'(t)=f(t,x(t))\\ x(t_0)=x \end{cases}\star$$

where $$f:I\times U{\Bbb R^{n+1}}\rightarrow{\Bbb R^n}$$ is a lipschitz function, that is $$L=\mbox{sup}_{(t,x)\neq(t,y)\in I\times U}\frac{|| f(t,x)-f(t,y) ||} {||x-y||}<\infty$$ is it's Lipschitz constant uniform of time.

We will use the following version of Gronwall's inequality

if $$\psi\leq a +\int_0^t(b\psi(s)+c)ds$$, then $$\psi\leq a e^{bt}+\frac{c}{b}(e^{bt}-1)$$

Then, $$\star$$ is equivalent to the following system $$x(t)=x_0+\int^{t}_{0}f(s,x(s))ds$$,

or $$\phi(t,x)=x+\int^{t}_{0} \frac{\partial\phi(s,x(s))}{\partial t}ds$$=$$x(t)=x_0+\int^{t}_{0}f(s,\phi(s,x))ds$$, $$\star\star$$

then $$||\phi(t,x)-\phi(t,y)||\leq ||x-\tilde{x}||+\int^{t}_{0}||f(s,\phi(s,x))-f(s,\phi(s,y))||ds\leq ||x-\tilde{x}||+\int^{t}_{0}L||\phi(s,x)-\phi(s,y)||ds$$

By the afore stated Gronwall's lemma $$||\phi(t,x)-\phi(t,y)||\leq||x-\tilde{x}||e^{L|t-t_0|}$$

This proofs your question for this case, since given $$T\in I$$ and any $$\epsilon>0$$(in your question it says to show there is some $$\epsilon>0$$ but any choice does there trick and I'm afraid it might be a mistake of the question), then choose $$\delta = \epsilon/e^{L|T-t_0|}>0$$

Then $$||\phi(t,x)-\phi(t,y)||\leq||x-\tilde{x}||e^{L|t-t_0|}<\delta$$ for $$||x-\tilde{x}||<\epsilon$$

For the dynamical systems where the flow is at least differentiable see https://math.stackexchange.com/a/528109/532993

• Thank you for your answer, I guess the nomenclature isn't really important in the end, so long as the working is correct. – Azog4 Jul 8 at 17:57