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Let $$\frac{dx}{dt}=f(t, x, \lambda)$$ be a parametric differential equation, where $f:D\subset \mathbb{R}\times \mathbb{R}^n\times \mathbb{R}^k \to \mathbb{R}^n$ ($\lambda$ is the parameter).

Suppose that for every $\lambda$ the ODE admits uniqueness of solutions. My lecturer defined the parameterized flow of this equation as follows ($(t_0, x_0, \lambda_0)\in D$ is fixed) : it is the function $\alpha:I_1 \times I_0 \times G_0 \times \Lambda_0\in \mathcal{N}(t_0, t_0, x_0, \lambda_0)\to \mathbb{R}^n$ defined as follows: for every $(\tau, \xi, \lambda)\in I_0\times G_0\times \Lambda_0$ the function $\alpha(., \tau, \xi, \lambda):I_1\to \mathbb{R}^n$ is the unique solution of the Cauchy problem with the initial condition $f(\tau)=\xi$.

He then went on to show that if $f$ is continuous and Lipschitz in the second variable then this $\alpha$ is a continuous function and he told us that this shows the continuous dependence of the ODE with respect to the initial conditions and the parameter.

But here I am a bit puzzled. I knew that if there is no parameter then the continuous dependence with respect to the initial conditions for an equation of the form $\frac{dx}{dt}=f(t, x)$is something like this: for every $\epsilon>0$, there is some $\delta>0$ (which depends on $\epsilon$ and the point where I am given the initial condition) such that if $\phi$ and $\psi$ are two solutions of the ODE and $||\phi(t_0)-\psi(t_0)||\le \delta$, then $||\phi(t)-\psi(t)||\le \epsilon$ for every $t$ for which these solutions are defined $(*)$. Basically, if the initial values are sufficiently close, then the solutions will be arbitrarily close.

However, even though intuitively I can see that these notions are the same, I can't see exactly how the fact that my flow $\alpha$ is continuous implies something like $(*)$. I think that I somethow need to write the definition of continuity, but I get confused with all those arguments of $\alpha$. Could you please explain this to me?

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The continuous dependence on the initial conditions pretty much means, in your notation, that $$ \|\phi(t)-\psi(t)\|\leq Ce^{Lt}\|\phi(t_0)-\psi(t_0)\|. $$ That is for any finite time moment $t$ (for which both solutions must be defined) you can find sufficiently close initial conditions such that both solutions are close enough. It does not mean by any reason that the two solutions will stay close for any $t$.

Take, e.g., $\dot x=x$, $\phi(t_0)=\epsilon/2,\psi(t_0)=-\epsilon/2$. Clearly you can find time moment $t$ such that your two solutions will be as far from each other as you wish.

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  • $\begingroup$ yes, I do realize now that I wrote something stupid, even though the math part is correct. How can I see the connection between the continuity of that flow and this continuous dependence on the initial conditions? This is what puzzles me because they should define the same concept. $\endgroup$
    – MathIsCool
    Nov 2 '21 at 21:58
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    $\begingroup$ This fact is not obvious, but it can be proven. $\endgroup$
    – Artem
    Nov 2 '21 at 21:59
  • $\begingroup$ I am happier now that the connection is not obvious :) (I was afraid that I am missing some trivial restatement). Could you guide me how to prove that the continuity of the flow implies continuous dependence on the initial conditions and the parameter in the sense that you wrote in your post? Or do you know any good reference on this? I tried to find a book that discusses this in the spirit I presented the things, but I couldn't. $\endgroup$
    – MathIsCool
    Nov 2 '21 at 22:04
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    $\begingroup$ Or, much better, see Arnold's book, ODE, it is paragraph 31 in the third edition. It gives all the details. $\endgroup$
    – Artem
    Nov 2 '21 at 22:25
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    $\begingroup$ I am not sure that I exactly follow your last question; take, say, $\dot x=f(t,x),x(\tau)=\xi$ and assume that you have a solution $\phi(t,\tau,\xi)$. It is a function not only $t$ but also a function of $\tau$ and $\xi$. Being continuous with respect to $\tau$ and $\xi$ exactly means continuous dependence on the initial conditions. $\endgroup$
    – Artem
    Nov 2 '21 at 23:21

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