How to show $\phi(t,x)$ is a flow of a vector field?

Question

Consider the function $\phi(t,x)$ where $x=(x_1,x_2) \in \mathbb{R}^2$. How can we check if $\phi(t,x)$ is a flow for a vector field? $$\phi(t,x)=(x_1\cos(r^2t)+x_2\sin(r^2t),-x_1\sin(r^2t)+x_2\cos(r^2t))$$ where $r^2=x_1^2+x_2^2$.

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My try:

We need to check if $\phi(0,x)=\phi(t,\phi(-t,x))$. I am wondering if there is a shorter way?

Answer 1

If you write $\zeta(t,z)=ϕ_1(t,x)+iϕ_2(t,x)$ where $z=x_1+ix_2$, you will find that $$ \zeta(t,z)=e^{-i|z|^2t}z $$ and as a consequence $|\zeta(t,z)|=|z|$ constant along the trajectories, so that the group conditions for the flow are easy to prove.

Answer 2

$\phi_t$ is the flow of a vector field if and only $\phi_{t+t'}=\phi_t\circ \phi_{t'}$.

If $\phi_{t+t'}=\phi_t\circ \phi_{t'}$, write $X(x)=lim_{\rightarrow 0}{d\over {dt}}\phi_t(x)$.

${d\over{du}}_{u=t_0}$ $\phi_t(x)={d\over{du}}_{u=0}\phi_{t_0+u}=$

${d\over{du}}_{u=0}\phi_{t_0}\circ\phi_u(x)=\phi_{t_0}^*{d\over{du}}_{u=0}\phi_t(x)=\phi_{t_0}^*X(x)$ which is equivalent to saying that $\phi_t$ is the flow of $X$.

The fact that if $\phi_t$ is the flow of then $\phi_{t+t'}=\phi_t\circ\phi_{t'}$ is a classical argument which is a consequence of the properties of differential equations.