# Linearization

• Feb 12th 2011, 09:26 AM
veronicak5678
Linearization
x'=-y-x^3
y'=x

Using linearization, I see that the only fixed pt, (0,0) is a center. But the real parts of the eigenvalues are 0, so maybe linearization hasn't worked. I tried changing to polar to analyze, and I now have

theta' = 1+rcos^3(theta)
r'=-r^3cos^4(theta)

What do these tell me about the vector field? Have I made a mistake?
• Feb 12th 2011, 10:20 AM
FernandoRevilla
Quote:

Originally Posted by veronicak5678
What do these tell me about the vector field? Have I made a mistake?

If you want to study the stability of the equilibrium point $\displaystyle (0,0)$ it is easier directly study the direction of $\displaystyle v(x,y)=(-y-x^3,x)$ in $\displaystyle \mathbb{R}^2$ .

Fernando Revilla
• Feb 12th 2011, 10:42 AM
veronicak5678

The prompt to my question tells me to analyze using polar coordinates to see whether or not the origin is a center.
I think I did make a mistake in my first post. I now have
theta'=1+r^2cos^3(theta)sin(theta)
but I still don't know what that tells me.
Sorry the post is hard to read.
• Feb 13th 2011, 12:39 AM
FernandoRevilla
Quote:

Originally Posted by veronicak5678
I think I did make a mistake in my first post. I now have theta'=1+r^2cos^3(theta)sin(theta)

Right.

Quote:

but I still don't know what that tells me.

I would insist that is better in this case to directly study de direction of the vector field. Draw $\displaystyle y=-x^3$ and $\displaystyle x=0$ and you'll have $\displaystyle \mathbb{R}^2$ partitioned:

$\displaystyle (i)\quad (x>0)\;\wedge (-y-x^3>0) \Rightarrow v(x,y)\equiv (+,+)$

$\displaystyle (ii)\quad (x=0)\;\wedge (-y-x^3>0)\Rightarrow v(x,y)\equiv (+,0)$

...

This will provide you information about the existence of closed orbits around $\displaystyle (0,0)$ .

Fernando Revilla
• Feb 13th 2011, 03:16 PM
veronicak5678
I know that your way is easier, but my professor says I should be able to use the values I have of r' and theta' to determine the direction of the non-linear spiral by looking at the speed in the radial direction. I think the origin is a growing spiral because r' will be greater than 0. Does this make sense, or am I not seeing this correctly?
• Feb 14th 2011, 05:35 AM
Jester
Since $\displaystyle r'=-r^3\cos^4 \theta$ then $\displaystyle r' \le 0$ so that $\displaystyle r'$ is decreasing which means that if you're on a circle of some radius say $\displaystyle r = r_0$, you're entering the circle.
• Feb 14th 2011, 06:46 AM
veronicak5678
Argh. Forgot about the minus sign. Thanks to both of you for your help.