Show that a system of differential equations has a periodic solution

Question

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.

Answer 1

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.

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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.

Answer 2

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: