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Lorenz+Shaw

Lorenz+Shaw - 1(10 is appropriate for cold water too high...

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Van der Pol Oscillator Equations du dt = v dv dt =( ± - u 2 ) v - u with initial conditions u (0) = 1, v (0) = 0. Take for example ± =4and t f =40 . Shaw Oscillator Equations In the van der Pol equations, set u v and v →- u ,andth enaddth e sinusoidal forcing term to the (wrong!) new dv/dt equation to obtain (up to constants) the Shaw oscillator equations: du dt =0 . 7 v +10 u (0 . 1 - v 2 ) dv dt = - u +0 . 25 sin(1 . 57 w ) dw dt =1 with initial conditions u (0) = - 0 . 73, v (0) = 0, w (0) = 0 ( w is time t ). Take t f =100 . Lorenz Equations dx dt = σ ( y - x ) dy dt = x ( r - z ) - y dz dt = xy - bz The Lorenz equations model Rayleigh-B´ enard convection. For a counter- rotating vortex, x ( t ) the angular velocity, y ( t →∞ ) T at the middle right edge, and z ( t →∞ )
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Unformatted text preview: 1 (10 is appropriate for cold water; too high for air), r is the Rayleigh number 2 , and b is the aspect ratio of the vortex cell. For σ = 10, r = 28, and b = 8 / 3, the solution is a strange attractor with fractal dimension ≈ 2 . 06. Take the initial conditions x (0) = 0, y (0) = 1, and z (0) = 0, with t f = 30. 1 Dimensionless ratio ν/κ of momentum diFusivity (kinematic viscosity) to thermal diFusivity. 2 Dimensionless number describing heat ±ow—if the Rayleigh number is below (above) a critical value, heat transfer is dominated by conduction (convection)....
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