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unscientific
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Homework Statement
Given the lorentz system for ##\sigma=10, b = \frac{8}{3}, r = 28##, and ##x(t)## from the first lorentz system, show that we can solve for y(t) and z(t) for the modified lorentz system by finding ##\dot E##.[/B]
Homework Equations
The Attempt at a Solution
I have found the 3 fixed points. They are at the origin ##(0,0,0)##, and ##C^{+} = \left( \sqrt{b(r-1)}, \sqrt{b(r-1)}, r-1 \right)## and ##C^{-} = \left(-\sqrt{b(r-1)}, -\sqrt{b(r-1)} , r-1 \right)##. For ##r = 28##, all three points are unstable.
It turns out that the points ##C^{+}, C^{-}## are only stable for ##1 < r < 25##.
For a dynamical system, the Lyapunov exponent ##\lambda## is related to the trajectory in phase space by
[tex]|\delta V(t) | = |\delta V_0| e^{\lambda t} [/tex]
So does this mean that for ##\lambda > 0## these trajectories are replled from one unstable point to another unstable point? I think this is the 'butterfly effect' described somewhere.
Also, I have re-expressed the equations:
[tex]\dot e_x + \dot x = \sigma \left[ (e_y - e_x) + (y-x) \right] [/tex]
[tex]\dot e_y + \dot y = rx - (e_y + y) - x(e_z + z) [/tex]
[tex]\dot e_z + \dot z = x(e_y + y) - b(e_z + z) [/tex]
[tex]\dot E = \frac{2}{\sigma} e_x \dot e_x + 2 e_y \dot e_y + 2 e_z \dot e_z[/tex]
I'm not sure what the question wants..
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