Talk:Separatrix (mathematics)

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Some help

Have almost zero experience with wikipedia, but think I can help with this stub. May need someone to come in after me and clear up the format.

Going to expand the example and explain it a bit better. Bob The Tough 14:27, 12 June 2007 (UTC)[reply]

Hmm - my style is scruffy at best, sorry about that. Would be easier to explain the phase space with the aid of a diagram, but don't have the time to add this (and don't know how). If no one does this within a week I may try to do the picture myself. --Bob The Tough 15:15, 12 June 2007 (UTC)[reply]

Well, been longer that I thought, but finally had some time to add the phase space diagram to the page. Hopefully this will make the example far easier to understand. The article is still in need of cleaning up (the use of maths in sentences looks messy) and probably some more mathematically accurate description of what a separatrix is. Anyway, hope I helped.

Sorry - forgot to sign previous talk. Infact ALL the talking done here is myself, Bob. --Bob The Tough 11:30, 13 August 2007 (UTC)[reply]

Comment

Moved from Separatrix (dynamical systems)

Surely - better to show derivation of the constant of integration of the motion, H

using d(dx/dt)/dt = ( dx/dt )* d(dx/dt)/dx i.e. v * dv/dt [ v = dx/dt ] i.e. ( df/dt = (df/dx)*(dx/dt) and let f = dx/dt

so integrating dv/dt w.r.t. x gives (1/2) (v^2) = (1/2) [( dx/dt) ^2 ] [ briefly, dv/dt = dv/dx * dx/dt = v * dv/dx and integrating dv/dt w.r.t. x gives (1/2) v^2 ]

and to choose a datum (ZERO) for H corresponding to the pendulum at rest at the lowest point:

H' = H + (m)g/l

Then the circular (elliptical) orbits at small x are directly obtained from noting that since cos(x) ~ 1 - (1/2) x^2

then 2* H' (small) = (m)(dx/dt)^ 2 + ((m)g/l)* x^2