; solve 
;	s'' = s-s^3 -s'^2
; by leapfrog method (properly ordered Euler method applied to 
; desynchronized s & v variables; we skip de-synch because v=0
; at the beginning, which would not change s in the first half-step)
 
	s = 1d0 + 0.01d0
	v = 0d0
	dt = 1d-5
	for i=1L, 5000000L do begin    ; 1L is long integer type (4 bytes)
	  v_prev = v
	  f = s*(1-s*s) -1000*v^3 ; cases: v^0,  v^2, v^4,  v^1,  v^3   
	  v = v + dt*f
	  s = s + dt*v
	  if(v*v_prev LT 0d0 AND s GT 1d0) then print,i*dt,s,v
	endfor
end


; this script misleadingly looks like Euler method (1st order
; in time) but in reality has accuracy of 2nd order (leapfrog method).
; 
; solutions of this nonlinear oscillator equation are oscillating around 
; s = 1. program catches and prints at the times when v changes sign, 
; that is at the turning points of s(t).
;
; we ran this script five times with different v-dependent term,
; as commented above and observed that for all even powers of velocity
; the fixed point is a circle (amplitude of small oscillatins is 
; strictly constant), and that all odd powers result in the loss
; of amplitude of oscillations 
; An elegant explanation of the reason for this was given in the 
; lecture: the equations are time-reversible if the power is even.
; Reversible systems necessarily have closed trajectories around
; circle fixed points. 
;