# Flow of an ODE and continuity with respect to initial conditions and parameter - what is the relation between them?

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?

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.
• 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. Nov 2 '21 at 23:21