I'm trying to show that the following equation has a nontrivial periodic solution $$x'' + [\ln (2x^2 +(x')^2)]x' + x = 0.$$
I substituted $ y = x $, $ z = x'$ to get $$z' = - \ln(2y^2 + z^2) z - y, \quad y' = z.$$
Using polar coordinate, we obtain $$ yy'+zz'=rr'. $$
Thus, $$ rr' = yz- \ln(2y^2 + z^2)z^2-yz = - \ln(2y^2 + z^2)z^2. $$ Now consider the region $A$ defined by: $$\frac{1}{2}\leq 2y^2+z^2\leq2.$$ $$\frac{1}{2}\leq 2y^2+z^2\leq\ 1\implies\ln(2y^2+z^2)\leq0\implies r'\geq0,$$ $$1\leq 2y^2+z^2\leq 2 \implies\ln(2y^2+z^2)\geq0\implies r'\leq0.$$ Also, since the region $A$ doesn't contain any critical point(which is $(0,0)$) of this system, we can use the Poincaré-Bendixson theorem to conclude that there exists a periodic solution.
My question is as follows : According to my textbook (Elementary differential equations and Boundary value problems by Boyce, Di Prima), we should show that there is a solution of this system that stays in $A$ for all $t\geq{t_0}$ for some ${t_0}$ to use the Poincaré-Bendixson theorem . How can we assure this? I think we can infer this from the information about r', but I'm a bit unclear.