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Stability of the equilibrium at (0,0)

  1. Aug 1, 2012 #1
    x'=y-x^3 and y'=-x^5

    I've worked the jacobian which is
    [-3x^2 1;-5x^4 0] and the equilibrium is at (0,0)
    so jac = [0 1;0 0]

    and eigenvalues are both 0

    so is the stability non isolated point??? and what i can say about the basin of attraction of the origin?

    Could anyone help me?

    thanks a lot
     
  2. jcsd
  3. Aug 3, 2012 #2

    haruspex

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    Not an area I know much about, but here goes...

    It's fairly clear that the trajectory spirals clockwise around the origin. Question is, does it spiral in or out? Maybe it spirals in within some annulus and out in another.
    For |y| >> |x5|, it approximates the ellipse-like curves 3y2+x6 = c. So set f(x,y) = 3y2+x6 and calculate f'. This pretty much settles the overall behavior, I think.
     
  4. Aug 3, 2012 #3
    but using the λ^2-τ*λ+Δ=0

    if Δ=0 at least one of the eigenvalues is zero. then the origin is not an isolated fixed point. there is either a whole line of fixed points or a plane of fixed points if A=0

    Spirals only satisfy τ^2-4*Δ<0.

    I don't know how to draw it and pplane and XPP just confuse me.

    Thanka a lot
     
  5. Aug 3, 2012 #4
    Google for the terms "stable" or "asymptotically stable" statioary point. First you have to
    understand this terms and then you will find a characterization based on the eigenvalues.
     
  6. Aug 3, 2012 #5

    haruspex

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    As I said, I'm ignorant of the theory - just working from first principles.
    dy/dx = -x5/(y-x3). This is zero on the y-axis and infinite elsewhere along y = x3. For y > x3 > 0, it is negative; for y < x3 > 0 positive, etc. This certainly makes it look like spirals.
    Now consider f = 3y2+x6. f' = 6yy'+6x5x' = -6yx5 + 6x5y - 6x8 = - 6x8
    So f (which is everywhere >= 0) is reducing with time everywhere except on the y-axis. But the trajectory does not stay on the y-axis except at the origin, so if the limit set includes points off the origin it also includes points of the y-axis. Since the value of f constrains those of |x| and |y|, x and y must tend to a limit of 0.
     
  7. Aug 4, 2012 #6

    HallsofIvy

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    The fact that the functions are 0 there mean it is, as you say, an equlibrium- nothing happens there! It is like you were sitting at the top of a hill of bottom of a depression. The fact that the eigenvalues are all 0 means that nothing is happening around the point! It is so flat, in some area around the point, that there no "motion" around the point. Essentially that says that linear approximations (which is what the Jacobean gives you- a linear approximation around the given point) are not enough and you will have to allow higher powers to learn anything about the properties of the system.
     
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