Thread: 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?

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 it is easier directly study the direction of in .


Fernando Revilla

Thanks for answering.

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.

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.

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 and and you'll have partitioned:





...

This will provide you information about the existence of closed orbits around .


Fernando Revilla

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?

Since then so that is decreasing which means that if you're on a circle of some radius say , you're entering the circle.

Argh. Forgot about the minus sign. Thanks to both of you for your help.