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Period of limit cycle?

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm given this system:
    [tex]\dot x = Ax^2 y + 1 - (B+1)x[/tex]
    [tex]\dot y = Bx - Ax^2 y [/tex]

    (a) Find the value of B when hopf bifurcation occurs.
    (b) Estimate the period of the limit cycle in terms of ##A## and ##B##.


    2. Relevant equations


    3. The attempt at a solution

    I have found fixed point to be ##(1, \frac{B}{A})##. I have plotted the graphs of ##\dot x = 0## and ##\dot y = 0##. Does bifurcation occur at ##B_H = 0##? Because when that happens the y=intercept of the fixed point approaches ##0##.

    How do I find the period of the limit cycle surrounding the trapping region?

    2008_B1_Q1.png
     
    Last edited: May 24, 2015
  2. jcsd
  3. May 25, 2015 #2

    ehild

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    Linearise the original system by considering small deviations from the critical point, that is, replace x=1+u and y=B/A+v and solve the linearised system of equations.
     
  4. May 25, 2015 #3
    I tried that, and also tried a different approach by finding the eigenvalues of the Jacobian.

    Jacobian Approach
    The jacobian matrix is 2008_B1_Q1_2.png
    Eigenvalue is ##\lambda = 0, -(B+1)##.

    Linearizing approach
    [tex]\delta \dot x = -(B+1)\delta x [/tex]
    [tex]\delta \dot y = B \delta x[/tex]
     
  5. May 25, 2015 #4

    ehild

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    You did not do the linearisation correctly. Why did you ignored y?
     
  6. May 25, 2015 #5
    Ok sorry, here's my proper working.
    [tex]\dot u = A(1+u)^2 (\frac{B}{A}+v) - (B+1)(1+u)[/tex]
    [tex]\dot u = B+Av+ 2Bu - (B+1)(1+u)[/tex]
    [tex]\dot u = Av + u(B - 1) - 1 [/tex]

    Now for ##\dot v##:
    [tex]\dot v = B(1+u) - A(1+2u)(\frac{B}{A} + v)[/tex]
    [tex]\dot v = B(1+u) - A \left( \frac{B}{A} + v + 2u\frac{B}{A} \right) [/tex]
    [tex]\dot v = B(1+u) - B - Av - 2Bu[/tex]
    [tex]\dot v = -(Bu + Av) [/tex]

    Now I try to find solve the equation entirely in terms of ##u, \dot u, \ddot u##:
    [tex]\ddot u = A\dot v + \dot u(B-1)[/tex]
    [tex]\ddot u = -A(Bu+Av) + \dot u(B-1)[/tex]
     
    Last edited: May 25, 2015
  7. May 25, 2015 #6

    ehild

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    Check. That -1 should not be there.
     
  8. May 25, 2015 #7
    =
    Yep that's right. I missed out ##+1## in the first line. The correct equations should now read:
    [tex]\dot u =Av + Bu - u[/tex]
    [tex]\dot v = -(Av + Bu)[/tex]

    Solving for ##v##, I get
    [tex]\ddot v = -B\left( Av + Bu - u \right) - A\dot v [/tex]
    [tex]\ddot v = +B\dot v + Bu - A\dot v[/tex]
    [tex]\ddot v =B\dot v - \left( Av + \dot v \right) - A\dot v[/tex]
    [tex]\ddot v = (B-A-1)\dot v - Av [/tex]

    Does bifurcation occur at ##B_H = 0##? Because when B=0, the trapping region disappears, i.e. no limit cycle.
     
    Last edited: May 25, 2015
  9. May 25, 2015 #8

    ehild

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    Bifurcation is at the fix point if the eigenvalues of the linearised equation are imaginary. So what should be the solutions for v?
    End check the signs in the last equation.
     
    Last edited: May 25, 2015
  10. May 25, 2015 #9
    But from the jacobian I have found the eigenvalue to be ##\lambda = 0## or ##\lambda = -(B+1)##. For positive values of B, the eigenvalue has to be real.
    proxy.php?image=http%3A%2F%2Fs30.postimg.org%2Fop1dxejxd%2F2008_B1_Q1_2.png

    For ##\ddot v = (B-A-1)\dot v + Av ##, I try ##v \propto e^{i\omega t}##. We have
    [tex]-\omega^2 = i\omega(B-A-1) + A[/tex]
     
  11. May 25, 2015 #10

    ehild

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    But it should be periodic. No damping term. What should be B?
    Forget the Jacobian, Go ahead with your method and check the equation for ##\ddot v ## .
     
  12. May 25, 2015 #11
    The first part of this question actually wants us to find the value of ##B = B_H## when bifurcation occurs. How do I find when bifurcation occurs? The eigenvalues of the linearized jacobian matrix are never imaginary.
    rums.com%2Fproxy.php%3Fimage%3Dhttp%253A%252F%252Fs30.postimg.org%252Fop1dxejxd%252F2008_B1_Q1_2.png

    I checked my working again, I got ## \ddot v = (B-A-1)\dot v - Av##.
     
  13. May 25, 2015 #12

    ehild

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    You are right, it is correct. What is the solution for v(t)?

    And how did you get that Jacobian? It looks wrong.
     
  14. May 25, 2015 #13
    Thanks for working with me on this. I got the jacobian by finding the first row using ##\frac{\partial \dot x}{\partial x }= -(B+1)## and ##\frac{\partial \dot x}{\partial y} = 0##, and the second row using ##\frac{\partial \dot y}{\partial x } = B## and ##\frac{\partial \dot y}{\partial y } = 0##. Linearized. I should linearize the jacobian to find the eigenvalues, right? All the questions I have been doing so far are like that.

    Using ANSATZ ##v \propto e^{-\omega t}## we have
    [tex]-\omega^2 = i\omega(B-A-1) - A[/tex]

    So for it to be real, we require ##\omega = \sqrt A = \sqrt {B-1}##?
     
  15. May 25, 2015 #14

    ehild

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    ##\frac{\partial \dot x}{\partial y} = 0## is wrong.

    Working with the Jacobian matrix yields the same result as the method you started to follow. Go ahead with it.
    By the way, you should use the Ansatz
    ##v \propto e^{i \omega t}##
     
  16. May 25, 2015 #15
    For ##\ddot v = (B-A-1)\dot v - Av## and ##v \propto e^{i\omega t}##, we have the same result:
    [tex]-\omega^2 = i\omega (B-A-1) - A [/tex]

    So I shouldn't linearize the jacobian from the start? So we have ##\frac{\partial \dot x}{\partial x} = 2Axy - (B+1)## and ##\frac{\partial \dot x}{\partial y} = Ax^2## and ##\frac{\partial \dot y}{\partial x} = B - 2Axy## and ##\frac{\partial \dot y}{\partial y} = -Ax^2##. This gives the jacobian matrix to be:
    2008_B1_Q1_2.png

    Expanding out to find eigenvalues, we have
    [tex]\lambda^2 + \lambda(A-B+1) + A = 0 [/tex]
    For the eigenvalues to be imaginary, we require
    [tex](A-B+1)^2-4A = 0 [/tex]
    [tex]A - B + 1 = \pm 2\sqrt A[/tex]
    [tex]B_H = A + 1 \pm 2\sqrt A[/tex]
     
  17. May 25, 2015 #16

    ehild

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    Substitute x=1 and y=B/A
    No, it is not true. Use the quadratic formula. When do you get pure imaginary solutions?
     
    Last edited: May 25, 2015
  18. May 25, 2015 #17
    For pure imaginary solutions, we require ##\frac{-b}{2a} = 0## and ##b^2-4ac <0##, so that means ##B=A+1##.

    Therefore the displacement equation becomes ##\dot v = - Av##. ##T = \frac{2\pi}{\sqrt A} =\frac{\pi}{\sqrt{B-1}}##.
     
  19. May 25, 2015 #18

    ehild

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    You meant ##\ddot v = - Av##
    Otherwise, it is correct now.
     
  20. May 25, 2015 #19

    ehild

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    And A should be positive. :smile:
     
  21. May 25, 2015 #20
    Yep, it was stated in the question that both B and A are positive numbers.
     
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