# Show that a system of differential equations has a periodic solution

We define a system of differential equations by \begin{align} \frac{dx}{dt} &= x + y − x(x^2 + 3y^2) \\ \frac{dy}{dt} &= −x + y − 2y^3. \end{align} We want to show that there exists a periodic solution to this problem. The first step is to transform the system into polar coordinates; some calculating gives us: \begin{align} \frac{dr}{dt} &= r(1 - r^2) - r^3·\sin^2(θ) \\ \frac{dθ}{dt} &= -\frac1{r^2} + \tan(θ). \end{align} Now how should I proceed? To me the problem seems quite difficult. My thanks.

• I apologise for the sketchy writing of equations. I'm quite new to the website. Commented Dec 18, 2016 at 15:09
• Please see math.stackexchange.com/questions/1007585/… where you can find solution of analogous question with detailed explanation Commented Dec 18, 2016 at 18:21

Using the usual sine bounds one gets $$r-2r^3\le\frac{dr}{dt}\le r-r^3$$ Apart from the origin, the fixed points of the bounds are $$r=\sqrt{\frac12}$$ and $$r=1$$. Take the annulus between these radii as domain of interest. On the boundary cirlces of it, the radial component of the vector field is $$\frac{dr}{dt}\big|_{r=\sqrt{\frac12}}=\sqrt{\frac18}\cos^2θ\ge 0 \text{ and } \frac{dr}{dt}\big|_{r=1}=-\sin^2θ\le 0.$$ In both cases it points inward resp. not outwards (make the annulus slightly larger to get a definitive picture). Now apply the Poincaré-Bendixson theorem.

Note that the angle dynamic is wrong, with $$r^2\dotθ=x\dot y-y\dot x=-r^2+xyr^2$$ one gets $$\dotθ=-1+\frac12r^2\sin(2θ).$$ The angle-averaged dynamic is $$\dot r=r(1-\tfrac32r^2)\\ \dot θ=-1$$ giving a clockwise oriented circle at radius $$\sqrt{\frac23}$$ as first approximation of the limit cycle.

We can attempt to verify boundedness directly. Wlog for unit speed $ds = dt$ we have

$$dr/ds= \cos \psi = r (1-r^2) - ( r \sin \theta)^3$$

$$d\theta/ds= \sin\psi/r = -1/r^2 + \tan \theta$$

where $\psi$ is angle between radius and arc in polar diagram. $\psi= 0,\pi/2$ occur at origin and maximum radii respy. Eliminating $\psi$ we are left with sin and tan which are trig functions with a periodicity $\pi$ halved due to squaring.

$$(-r^3 \sin^3\theta-r^3+r)^2+(r\, \tan\theta-1/r)^2=1,(\pi/2,r,\pi/2),(-4<\theta<8);$$

$r=0$ or the pole occurs for odd multiples of $\theta = (k-1/2) \pi/2 ,$ else $r$ cannot not be finite.

To confirm that the radius is bounded, the eliminant is plotted in rectangular plot:

• This calculation starts from the wrong angle dynamic given in the question. The actual dynamic is much less singular. Commented Apr 18, 2022 at 11:29