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Lorentz Equations - Chaos and Stability

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    The figure below shows the path of a particle governed by the Lorenz equations with r = 28, σ = 10, b = 8/3. The x'es and boxes show points where the path crosses the plane z = r − 2σ > 0.


    (a) Which indicator shows a decreasing z and which shows an increasing z?

    (b) Show the length of element ## | \delta x | ## between the two paths either grows or decays exponentially if aligned with one of the eigenfunctions of jacobian ##\frac{J + J^T}{2}##.

    (c) Find ##\frac{J + J^T}{2}## and its eigenvalues at (0, 0, r-2σ). Hence deduce that of ##\delta x ## grows is in the x-y plane while it decays is along the direction of z-axis.

    (d) Show the volume decreases exponentially with ##\delta V = \delta V_0 e^{−(σ+1+b)t}##
    Since ##\frac{(J + J^T)}{2}## is symmetric, its eigenfunctions are orthogonal. Show that for a cubic element where the three displacement directions are along the eigenfunctions in section (c) decays at the same rate.

    2. Relevant equations
    Lorentz equations are given by:

    [tex]\dot x = \sigma(y-x)[/tex]
    [tex]\dot y = rx - y - xz [/tex]
    [tex]\dot z = xy - bz [/tex]

    3. The attempt at a solution

    Part (a)

    For ## \dot z < 0##, ##xy < bz \approx 21##.

    So the boxes represent decreasing z, the x'es represent increasing z.

    Part (b)

    How do I show it either grows or decays exponentially? Do I put in z = r − 2σ and find the eigenvalues? Wouldn't it be part (c)? I think this part is simpler than it seems.

    Part (c)

    The matrix ##\frac{J + J^T}{2}## becomes:


    I found the eigenvalues to be:

    [tex]\lambda_{1,2} = -\frac{ -(\sigma + 1) \pm \sqrt{ (\sigma+1)^2 - 4(\sigma - \frac{9}{4} \sigma^2) } }{2} [/tex]
    [tex]\lambda_3 = -b[/tex]

    Part (d)

    [tex]\nabla \cdot \vec u = -(\sigma + 1 + b) [/tex]

    Then the result follows.

    This is all that I have managed to do so far, would appreciate any input, thank you!
  2. jcsd
  3. Nov 28, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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