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Math Help - pitchfork bifurcation and sketching full bifurcation diagram

  1. #1
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    Angry pitchfork bifurcation and sketching full bifurcation diagram and more

    given system of ODE

    dx/dt = ax+xy-x^3
    dy/dt = b-x^2-y^2

    first, show that there is saddle-node bifurcation for b=0, without using the theorem

    second, determine the curve in the a-b plane for which there exists a pitchfork bifurcation without using the theorem

    third, sketch the full bifurcation set in the a-b plane (including homoclinic bifurcation)


    Help plz
    Last edited by tk1234; October 12th 2008 at 09:40 PM.
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  2. #2
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    I'm no expert ok, but I find these fascinating so I'd like to try: I'd first solve for the equilibrium points in terms of a and b:

    ax+xy-x^3=0
    b-x^2-y^2=0

    These are:

    e_1=\left\{0,-\sqrt{b}\right\}
    e_2=\left\{0,\sqrt{b}\right\}
    e_3=\left\{-\frac{\sqrt{-1+2 a-\sqrt{1-4 a+4 b}}}{\sqrt{2}},\frac{1}{2} \left(-1-\sqrt{1-4 a+4 b}\right)\right\}
    e_4 \left\{\frac{\sqrt{-1+2 a-\sqrt{1-4 a+4 b}}}{\sqrt{2}},\frac{1}{2} \left(-1-\sqrt{1-4 a+4 b}\right)\right\}
    e_5=\left\{-\sqrt{-\frac{1}{2}+a+\frac{1}{2} \sqrt{1-4 a+4 b}},-\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a+4 b}\right\}
    e_6=\left\{\sqrt{-\frac{1}{2}+a+\frac{1}{2} \sqrt{1-4 a+4 b}},-\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a+4 b}\right\}<br />

    Fixed points are created whenever the equilibrium points are real so the root objects completely control the creation and destruction of fixed points. Immediately we see the creation of e_1 and e_2 whenever b becomes non-negative. I would suspect through further analysis, i.e., linearizing, and calculating eigenvalues, we could determine that these two fixed points are indeed a saddle and node hence the saddle-node bifurcation. Probably other ways too though. In the first plot, I depicted the flow in phase space about these two points (red) when a=4 and b=1 (green section in second plot). The top fixed point looks to be a saddle (points move to then away from it). The bottom point looks a source (node). Points move away from it. That would need to be verified algebraically though.

    The remaining four have real y-component whenever the quantity b>a-1/4 or the region above the line b=a-1/4. These have real x component whenever -1/2+a+1/2\sqrt{1-4a+4b} or -1+2a-\sqrt{1-4a+4b} are non-negative. These regions are depicted in the second plot below with the number of fixed points indicated in each.

    Again, further analysis could determine the types of remaining fixed points as well as further types of bifurations such as Hopf bifurcations. This is just a start.
    Attached Thumbnails Attached Thumbnails pitchfork bifurcation and sketching full bifurcation diagram-phaseportrait.jpg   pitchfork bifurcation and sketching full bifurcation diagram-bifurcation-diagram-b.jpg  
    Last edited by shawsend; October 14th 2008 at 07:14 AM.
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  3. #3
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    I'd like to add a little rigor to my analysis above regarding the fixed points when a=4 and b=1. In this case we have:

    x'=4x+xy-x^3
    y'=1-x^2-y^2

    The fixed points are e_1=(0,1) and e_2=(0,-1). In order to linearize these, I move the points to the origin via change of variables.

    For  e_1 it's u=x and v=y-1.

    For e_2 it's u=x and v=y+1.

    Making these changes results in the new differential equations with fixed points at the origin. In the case of e_1 we have:

    u'=5u+uv-u^3
    v'=-2v-u^2-v^2

    Since the fixed points are now at the origin, the non-linear terms have little impact for u and v values near the origin and the resulting linear system is an approximation to the non-linear system:

    u'=5v+0v
    v'=0u-2v

    The eigen values for this one is -2 and 5 which is a saddle. Likewise, doing the same thing with the other one, I get for the linear system:

    u'=3u+0v
    v'=0u+2v

    The eigenvalues of which are 2 and 3 which means this is a source, i.e,. a node.
    Last edited by shawsend; October 14th 2008 at 11:19 AM.
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  4. #4
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    Thanks !!!
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