If $\frac{{\rm d}X^{s,\:x}}{{\rm d}t}(t)=v(t,X^{s,\:x}(t))$ and $X^{s,\:x}_s=x$, can we show that $X^{s,\:X^{r,\:x}(s)}(t)=X^{r,\:x}(s+t)$?

Question

Let $\tau>0$, $d\in\mathbb N$, $v:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ and $X^{s,\:x}:[0,\tau]\to\mathbb R^d$ denote the solution of \begin{align}\frac{{\rm d}X}{{\rm d}t}(t)&=v(t,X(t))\;\;\;\text{for all }t\in(s,\tau)\tag1\\ X(s)&=x\tag2\end{align} for $(s,x)\in[0,\tau]\times\mathbb R^d$. Now let $$T_{s,\:t}(x):=X^{s,\:x}(t)\;\;\;\text{for }0\le s\le t\le\tau\text{ and }x\in\mathbb R^d.$$

Let $x\in\mathbb R^d$. Are we able to show that $$T_{s,\:t}\left(T_{r,\:s}(x)\right)=T_{r,\:s+t}(x)\tag3$$ with $s+t\le\tau$?

Answer 1

Let as usual $\phi(s,(t,x))$ or $\phi(s;t,x)$ denote the solution of the IVP with initial condition $\phi(t,(t,x))=x$.

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It is certainly true if you define the propagators $\Pi_{s,t}(x)=\phi(s;(t,x))$ to be the solution of the IVP through the point $(t,x)$ at time $s$ that then $$ \Pi_{s,r}=\Pi_{s,t}\circ\Pi_{t,r} $$

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If you want to also have the additive structure represented you would need to define $$ X_s(t,x)=\phi(t+s;(t,x)) $$ so that then $X_{s+r}=X_s\circ X_r$.

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Your version still mixes these two ideas.