Is solution's group property is a characteristic property of autonomous differential dynamic systems?

Question

That means if a one parameter differentiable group $\Phi(x,t):\mathbb{R}^d\times \mathbb{R}\to\mathbb{R}^d$ satisfies $$\Phi(\Phi(x_0,t_1),t_2)=\Phi(x_0,t_1+t_2),\Phi(x_0,0)=x_0$$ holds for $\forall x_0\in\mathbb{R}^d,t_1,t_2\in\mathbb{R}.$ Then is there exist a suitable function $F$ on $\mathbb{R}^d$ s.t. $\dot{\Phi(x(0),t)}=F(\Phi(x(0),t)),x(0)=x$?

Answer 1

The notation may become tricky, that is why we will lighten it by remarking that $x(t) = \Phi(x_0,t)$, where $x_0 = x(0)$ plays the role of an initial condition. Moreover, it allows to eliminate the time-dependence of the flow $\Phi$ with respect to its first variable, since $x_0$ is a constant vector. In consequence${}^1$, we can write $\dot{x}(t) = \dot{\Phi}(x_0,t) = F(\Phi(x_0,t)) = F(x(t))$. Then, the following "trick" permits to "linearize" this differential equation : $\dot{x} = F(x) = (F(x)\cdot\nabla_x)x$, which is solved formally${}^2$ by $x(t) = \left.e^{tF(x)\cdot\nabla_x}x\right|_{x=x_0}$. But, $x(t) = \Phi(x_0,t)$ at the same time, from which one concludes $F(x) = \dot{x}(0) = \left.\dot{\Phi}(x,t)\right|_{t=0}$ in the end.

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${}^1$ N.B. : otherwise, without the aforementioned time-independence in the first variable, we would have $\dot{x}(t) = \frac{\mathrm{d}}{\mathrm{d}t}\Phi(x(t),t) = \nabla_x\Phi(x(t),t)\dot{x}(t) + \partial_t\Phi(x(t),t)$, which would be a more complicated expression to deal with.

${}^2$ N.B. : you may encounter the expression $x(t) = e^{tF(x_0)}x_0$ instead, because the vector field $F(x)\cdot\nabla_x$ might be identified with $F(x)$ alone by abuse of notation, because the tangent space $T_x\Bbb{R}^d$ is isomorphic to $\Bbb{R}^d$.