Let $\Omega$ be an open and connected subset of $\mathbb R \times \mathbb R^n$ and let $f$ be a function from $\Omega$ to $\mathbb R^n$ such that the IVP $$\begin{cases} \dot u = f(\cdot,u) \\ u(t_0)=x_0 \end{cases}$$ admits a solution (and only one) for every $(t_0,x_0) \in \Omega$. Fix $(t_0,x_0) \in \Omega$ and let $u$ be the unique (maximal) solution to the IVP. Let $I$ be the domain of $u$. Then, define $\psi$ such that:
- $\psi(x,t,t)=x$
- $\displaystyle{\frac{\partial\psi(x,t,s)}{\partial s}\bigg|_{s=t} = f(t,x)}$
- $\psi(\psi(x,t_1,t_2),t_2,t_3)=\psi(x,t_1,t_3)$
It should be possible to prove that $\psi(x_0,t_0,\cdot)=u$, right? However I’m having trouble doing so. Obviously $\psi(x_0,t_0,t_0)=x_0$, but how can I prove that $$\frac{\partial\psi(x_0,t_0,s)}{\partial s}\bigg|_{s=t} = f(t,\psi(x_0,t_0,t)) \quad \text{for every } t \in I$$ to show that $\psi(x_0,t_0,\cdot)$ solves the IVP?
If you define $x=\psi(x_0,t_0,t)$, then from $$ \frac{\partial\psi(x,t,s)}{\partial s}\bigg|_{s=t} = f(t,x) $$ we get $$ \frac{\partial\psi(\psi(x_0,t_0,t),t,s)}{\partial s}\bigg|_{s=t} = f(t,\psi(x_0,t_0,t)) $$ and since $\psi(\psi(x_0,t_0,t),t,s)=\psi(x_0,t_0,s)$, also $$ \frac{\partial\psi(x_0,t_0,s)}{\partial s}\bigg|_{s=t} = f(t,\psi(x_0,t_0,t)). $$