Proving continuous dependence on initial conditions for flows

756 Views Asked by At

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!

1

There are 1 best solutions below

1
On BEST ANSWER

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