Show that the following system of ordinary differential equations $$\frac{dx}{dt}=0.5x+2.5y-x(x^2+y^2),$$ $$\frac{dy}{dt}=-0.5x+1.5y-y(x^2+y^2).$$ has at least one periodic solution.
I tried to find a conserved function to the system, but it seems not easy and I cannot come up with other ideas. Are there any other useful ideas to deal with such problem proving the existence of periodic solutions? Thanks!
The system has the form $\newcommand{\vv}{{\vec v}}$ $$ \dot{\vv}=A\vv-r^2\vv,~~ \vv=\pmatrix{x\\y},~r=|\vv|,~A=\frac12\pmatrix{1&5\\-1&3} $$ The stationary points of it are $\vv=0$ and the vectors of eigenpairs $(|\vv|^2,\vv)$, if the matrix had real eigenvalues.
Claim: There are no real eigenvalues and the origin is an outward spiral.
If you set $u=r^2=x^2+y^2$ you get $$ \dot u=\vv^TB\vv-2u^2 \text{ with } B=A+A^T=\pmatrix{1&2\\2&3}, $$ so that $$ \lambda_\min u-2u^2\le \dot u\le\lambda_\max u-2u^2 $$ where the $λ_{\min,\max}$ are the eigenvalues of $B$. The left side is not helpful, however, the right side is negative at $u=\lambda_\max$.
Claim: This makes $0<|\vv|<\sqrt{\lambda_\max}$ a trapping region with no equilibrium points.