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Math Help - Linearization

  1. #1
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    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?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    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
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  3. #3
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    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.
    Last edited by veronicak5678; February 12th 2011 at 01:02 PM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    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
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  5. #5
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    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?
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  6. #6
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    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.
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  7. #7
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    Argh. Forgot about the minus sign. Thanks to both of you for your help.
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