At lesson, the teacher considers a flow $\Phi$ given by the solutions of the ode system for $t\in[0, T]$ and $x\in\mathbb R^d$,
$$
\begin{cases}
y'(s)=b(y(s), s),&s\leq T\\
y(t)=x
\end{cases},\label{1}\tag{\*}
$$
that is $\Phi(x, t, s)=y(s)$ solving \eqref{1}. He said that we will be mostly concerned with $\Phi(\cdot, 0, \cdot)$. The field $b$ is assumed to be Lipschitz continuous in both variables and bounded.


Then, he intoduces the inverse $\Psi$ of the above flow as follows: $\Psi(x, 0, s)=y(s)$ satisfying
$$
\begin{cases}
y'(s)=-b(y(s), t-s),&s<t\leq T\\
y(0)=x
\end{cases},
$$
and he said that $\Psi$ is such that
$$
\Phi(\Psi(x, 0, s), 0, s)=x,\quad \Psi(\Phi(x, 0, s), 0, s)=x.\label{2}\tag{\*\*}
$$
I do not understand \eqref{2}. Can someone help me? Maybe is the definition of the inverse wrong?


Thank you



Definition and properties of the inverse of the flow of an ODE
I think this question is better suited for math.stackexchange. But here's a short answer anyway:


The flow is a map $\Phi : \mathbb{R}\_+ \times \mathbb{R}^d \to \mathbb{R}^d$ and if you take an initial value $y\_0$ at time $t\_0 = 0$, then $t \mapsto \Phi(t, y\_0)$ is the trajectory of the solution to your ODE with initial value $y\_0$ forward in time (non-negative t).


Now you may also be interested in the behaviour backwards in time and that's how $\Psi$ comes into play. Your equation $(\*\*)$ simply says, that $\Psi$ and $\Phi$ are compatible in the following sense. If you start at $x$ and go $s$ units of time backwards via $\Psi$, you'll be in $\Psi(x,0,s)$ and if you then start from that point and go $s$ units in time forward via $\Phi$, you'll end up in $\Phi(\Psi(x,0,s),0,s)$ and $(\*\*)$ simply says that that's the same position as if you switched up the role of $\Phi$ and $\Psi$, i.e. first went forward in time via $\Phi$ and then backward via $\Psi$.


