1. ## 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?

2. 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 $(0,0)$ it is easier directly study the direction of $v(x,y)=(-y-x^3,x)$ in $\mathbb{R}^2$ .

Fernando Revilla

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.

4. 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 $y=-x^3$ and $x=0$ and you'll have $\mathbb{R}^2$ partitioned:

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

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

...

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

Fernando Revilla

5. 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?

6. Since $r'=-r^3\cos^4 \theta$ then $r' \le 0$ so that $r'$ is decreasing which means that if you're on a circle of some radius say $r = r_0$, you're entering the circle.

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