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I have a simple question on a Lipschitz flow. Recall that a flow on a non-empty set $X$ is a map $\Phi: \mathbb R \times X \to X$ satisfying

$$\Phi(0,x) = x \text{ and } \Phi(s, \Phi(t, x)) = \Phi(s + t, x)$$

Now let $X$ be a Banach space and $F$ a Lipschitz-function on $X$. Now it is well known that there exists for every $x \in X$ a unique function $u_x:\mathbb R \to X$ which is continuously strongly differentiable and satisfies:

$$u_x'(t) = F(u_x(t)) \text{ where $x \in \mathbb R$ and } u_x(0) = x.$$ Where $u'$ is the strong derivative in the Banach space.

Now we can set $\Phi(t,x) = u_x(t)$. This should be a flow, why is that? The first property is easy to verify, but the second one requires us to show that $u_{u_x(t)}(s) = u_x(s + t)$. Can someone give a hint how I obtain this? I'm probably missing something simple.

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1 Answer 1

up vote 2 down vote accepted

This follows from uniqueness of solutions to the initial value problem

$$v_x'(t) = F(v_x(t)),\qquad\qquad v_x(0) = u_x(s).$$

For every $s\in \mathbb R$, $u_{u_x(t)}(s)$ and $u_x(s + t)$ are both solutions with the same initial data at $t=0$.

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