Consider the IVP
$$x' = f(t, x)$$
$$x(0) = x_0$$
with complete flow $\phi(t, x_0)$. If $f$ is smooth, then the flow has the group property, namely
$$\forall s, t \in \mathbb{R}: \phi(t + s, x) = \phi(t, \phi(s, x)).$$
Interpretation: Consider a particle in (say) $\mathbb{R}^2$ whose motion is governed by the differential equation. Let $x_2 = \phi(t + s, x_0).$ In both of these scenarios, you get up at the same point $x_2$:
Scenario 1: Start at point $x_0$ and time $ = 0$, and let time flow until $t + s$.
Scenario 2: Start at point $x_1 := \phi(s, x_0):= x_1$, reset time $ = 0$, and then let time flow until $t$.
Question: The fact that the group property holds is completely magical to me. Is there any intuition for why it's true?
What's even more troubling is this: let's compute the derivative of motion for the particle at point $x_1$ in both scenarios.
Scenario 1: $\phi'(s, x_0) = f(s, \phi(s, x_0)) = f(s, x_1).$
Scenario 2: $\phi'(0, x_1) = f(0, \phi(0, x_1)) = f(0, x_1).$
In fact, we can say something stronger. The points $x$ and $\phi(s, x)$ have the same orbit until $\phi$.
I really don't see why why $f(s, x_1)$ and $f(0, x_1)$ have to be in the same direction for all $s$. In fact they really don't, so there must be something wrong with my logic above.
Best Answer
The group property holds in that form for an autonomous equation. For a non-autonomous equation, the flow mapping needs the start point, the flow duration, and the flow start time (or something equivalent, like the start and end points). This means you can write a generalized group property in a form like $\phi(t+s;x,0)=\phi(t;\phi(s;x,0),s)$, where my $\phi$ arguments are flow duration, start point, flow start time.
$f(t,x)=t$ gives a simple explicit example where the group property from the question does not hold.
Incidentally, the need for regularity here is just so there's uniqueness of solution so that $\phi$ is well defined without further specification of selection of solutions.