Solve the system $$ x'=ax-by $$ $$ y'=bx+ay $$ and show that at any point $(x_0, 0)$ on the $x$-axis, the Poincare map for the focus at the origin is given by $P(x_0)=x_0\exp(2\pi a/|b|)$. Is it stable or unstable?
I try to rewrite it into polar coordinate: let $x=r\cos \theta, y=r\sin \theta$ $$ r'=ar $$ $$ \theta'=b $$
Then the flow is given by $$ \phi_t(r_0,\theta_0)=(r_0e^{at}, bt+\theta_0) $$
So it seems that the Poincare map is given by $$ P(r_0)=r(2\pi)=r_0e^{2\pi a} $$
But there is not $|b|$? Where do I miss something?
To obtain the Poincaré map, we must solve the initial value problem for the initial value $(x_0,0)$ and wait until the solution intersects the $x$-axis. Let the initial point $(x_0,0)$ correspond to the angle $\theta(0)= \theta_0$. The state moves around the origin with a constant angular velocity $b$ counterclockwise, if $b>0$, and clockwise, if $b<0$. In the first case, the intersection with the $x$-axis will occur when $\theta(t)= \theta_0+2\pi$, and in the second case, when $\theta(t)= \theta_0-2\pi$. Since $\theta(t)= \theta_0+bt$, we have for the intersection point $bt= 2\pi$ and $bt= -2\pi$ respectively; $t>0$, so the intersection occurs when $t=2\pi/b$ for $b>0$ and when $t=-2\pi/b$ for $b<0$. Both cases can be combined into the formula $t=2\pi/|b|$. It remains to substitute $r(2\pi/|b|)=r_0 e^{2\pi a/|b|}$.