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Dynamical System

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{dU}{dz} = V, \frac{dV}{dz}=k+cU-6U^{2}[/itex] c [itex]\in[/itex] ℝ

    Find the fixed points of the system (these are solutions U=U*, V=V* where U*,V* [itex]\in[/itex] ℝ) and determine the value of k so that the origin is a fixed point of the system

    2. Relevant equations



    3. The attempt at a solution

    I'm not too familiar with dynamical systems but I believe that we want

    [itex]\frac{dU}{dz}|_{V=V^{*}} = V^{*} = 0[/itex]

    and

    [itex]\frac{dV}{dz}| _{U=U^{*}} = k+cU-6U^{2} = k+cU^{*}-6(U^{*})^{2} = 0[/itex]

    So we want a value of k so that the origin is a fixed point of the system. I'm not quite sure what this means.
     
  2. jcsd
  3. Apr 27, 2013 #2

    tiny-tim

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    Hi BrainHurts! :smile:
    It means dU/dz and dV/dz are both 0 at (U,V) = (0,0) :wink:
     
  4. Apr 27, 2013 #3
    So the origin is a fixed point of the system when k=6(U*)2 + cU*?
     
  5. Apr 27, 2013 #4

    tiny-tim

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    (you mean " - cU* " :wink:)

    that's a strange way of putting it, but yes …

    so what is k ? :smile:
     
  6. Apr 27, 2013 #5
    yes haha sorry, k is some arbitrary real number, but for (U,V)=(0,0) being a fixed point implies that

    V*=0

    and 6U*2-cU*-k=0

    this gives us

    U*= [itex]\frac{c\pm\sqrt{c^{2}+24k}}{12}[/itex], but we want U*=0 which means that means that k=0 right?

    [itex]0=\frac{c\pm\sqrt{c^{2}-24k}}{12} \Rightarrow -c=\pm\sqrt{c^{2}-24k} \Rightarrow c^{2}=c^{2} - 24k[/itex] and we get k = 0
     
  7. Apr 27, 2013 #6

    Dick

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    Sure, k=0. You didn't really need the quadratic equation to draw that conclusion, but ok.
     
  8. Apr 27, 2013 #7
    Ok going on to the next part of this question

    c) Find the equations for the orbits of the system, these are the trajectories in the UV-plane determined by functions of the form V=V(U) and show that a first integral of the orbit equation is

    V2=2kU + cU2-4U3+h

    So in doing so I got

    [itex]\frac{dV}{dU}=\frac{\frac{dV}{dz}}{\frac{dU}{dz}}=\frac{k+cU-6U^{2}}{V} \Rightarrow VdV = (k+cU-6U^{2})dU [/itex] and this gives us the desired result.

    So one place where I read what orbits are are objects like V(U), V(V(U))... so I'm not really sure when to stop.
     
  9. Apr 28, 2013 #8
    nm found some good reading on this

    http://math.colorado.edu/~jkeller/math4430fall10/Orbits.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
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