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!
Best Answer
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.