I'm currently brushing up on my ODE theory by reading through some texts.

I was told that the system $$x'_1=x_2\hspace{5mm}x'_2=-x_1+(1-x_1^2-x_2^2)x_2\hspace{5mm}\Big('=\frac{d}{dt}\Big)$$ has a rather interesting property:
Apparently, all periodic solutions of this system are of the form $\varphi=(\varphi_1,\varphi_2)$, where $\varphi_1(t)=\sin(t+c)$, $\varphi_2(t)=\cos(t+c)$, and $c$ is an arbitrary constant. (This is excluding the trivial periodic solution $\varphi=0$ of course.)

Is there an easy proof to see why this is true? It looks like a really interesting result that comes out of nowhere (or in my opinion at least), and I'm hoping that the reasoning behind it could help me understand periodic solutions better.

  • $\begingroup$ Can I know where did you find this exercise? $\endgroup$ – Felice Iandoli Sep 6 '14 at 17:48

Yes, put the DE into polar coordinates. Re-stated the de appears to be $\theta' = -1$, $r' = 1-r^2$. Here $(x_1,x_2) = r(\cos \theta, \sin\theta)$.

  • $\begingroup$ That certainly makes a lot more sense now. Is there a way from that new system to get our periodic solutions? I'm probably missing something easy here, but I'm a bit forgetful on how to solve it. $\endgroup$ – Dustin Tran Nov 30 '11 at 22:04
  • $\begingroup$ The variables $\theta$ and $r$ are independent, so $\theta = -t + c$ is a solution, the only $r'=0$ solution is $r=1$ (constant). Re-write that in terms of $x_1,x_2$ and you have your answer. If $r \neq 1$ you can check that $r(t)$ converges to $1$ -- you can solve for it explicitly since it's a separable DE. It's an exponential decay, so it's not a periodic solution. $\endgroup$ – Ryan Budney Nov 30 '11 at 22:18

You can also consider the function $\rho(x,y)=\frac{1}{2}(x^2)+(y)^2$. (In my answer x'=y). Taking the derivative in time of this function you can see that is zero. So you have that the norm of the solution in unchanging in time and the solution must be a circle!


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