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Solving an autonomous system finding critical points

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all critical points and derive the linearised system about each critical point

    [itex]\dot{y}[/itex] = [3y1 + y1y2; y1 + y2 - y22]

    2. Relevant equations



    3. The attempt at a solution

    Setting

    0 = 3y1 + y1y2 so y2 = -3
    0 = y1 + y2 - y22 so y1 = 12

    = [0, 1; 12, -3]
    so p = -3 (<0)
    q = -12 (<0)
    [itex]\Delta[/itex] = 57 (>0)

    Is this right that my only critical point is (12,-3) and is that how I determine the nature of the critical point and whether my critical point is stable or unstable with my values for p, q and [itex]\Delta[/itex]?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2011 #2

    fzero

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    Gold Member

    You have quadratic equations, so there are more solutions, and so more critical points.

    The nature of the critical points can be determined by studying the linearized systems asked for in the problem. Around a critical point [itex](y_1^{(c)},y_2^{(c)})[/itex], expand

    [itex] y_1 = y_1^{(c)} + \epsilon_1,~~y_2=y_2^{(c)}+\epsilon_2,[/itex]

    and keep the terms linear in [itex]\epsilon_{1,2}[/itex] (and derivatives). Solving the linear system tells you how the complete solution must behave in the neighborhood of the critical points.
    Perhaps your [itex]p,q,\Delta[/itex] are somehow relevant to the linear system but you haven't explained how they're defined.
     
  4. Aug 18, 2011 #3

    HallsofIvy

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    Either y2= -3 or y1= 0.

    If y2= -3, then y1= 12, but if y1= 0, then y2- y22= 0.

     
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