# Nonlinear terms - Phase plane

• February 24th 2011, 12:51 PM
mukmar
Nonlinear terms - Phase plane
Problem: Consider the system $\dot{x} = -y - x^3$, $\dot{y} = x$. Show that the origin is a spiral, although the linearization predicts a center.

I tried converting to polar coordinates by doing $x = r\cos{\theta}$, $y = r\sin{\theta}$. But the equations I got were:
$\dot{r} = -r^3\cos^4{\theta}$ and $\dot{\theta} = 1 + r^2\cos^3{\theta}\sin{\theta}$, which aren't independent equations, so it didn't really help in analyzing the system.

The two identities used were: $x\dot{x} + y\dot{y} = r\dot{r}$ and $\dot{\theta} = \frac{x\dot{y} - y\dot{x}}{r^2}$

Any hints or suggestions would be greatly appreciated!
• February 24th 2011, 02:03 PM
Jester
Well, this is what you have. First

$\dot{r} = -r^3 \cos^3 \theta \le 0$ which means you'll either be on a decaying orbit or a fixed orbit depending on whether $\dot{r} < 0$ or $\dot{r} = 0$. However, if $\dot{r} = 0$ then $\cos \theta = 0$ so either $\theta = \pi/2 \; \text{or}\; 3 \pi/2$ so that $\dot{\theta} = 1$ which means that you'll pass through $\theta = \pi/2 \;\text{or}\; 3 \pi/2$ and then you're back to a decaying orbit thus giving a decaying spiral.
• February 24th 2011, 03:09 PM
mukmar
Thank you for explaining it so succinctly! (Happy)