just found a new time stepping mode. now we've got three to choose from (not counting the delayed time and the acceleration modes, which I consider very experimental, and which are not included in the source code).

mode 0 (discrete)

f(x,t+1) = S[snm] f(x,t)

mode 1 (smooth)

f(x,t+dt) = f(x,t) + dt S[snm'] f(x,t) with snm' = 2*snm-1

this becomes in the limit dt->0 in the Euler integration sense

d_t f(x,t) = S[snm'] f(x,t)

mode 2 (smooth 2)

f(x,t+c) = (1-c) f(x,t) + c S[snm] f(x,t) = f(x,t) + c (S[snm] f(x,t) - f(x,t))

which becomes in the limit c->0 (Euler integration)

d_t f(x,t) = S[snm] f(x,t) - f(x,t) = (S[snm] - 1) f(x,t)

but becomes in the limit c->1

f(x,t+1) = S[snm] f(x,t)

so what we've done here is, we have invented a linear combination of the two cases c=1 and c=0 that we want to have, but don't know what is in between. the simplest solution is a linear combination. for infinitesimal c it generates an equation of the form we want to have, but not with an operator we are satisfied with. why? because it doesn't generate the same generic gliders as in the mode 0 case. we want to be able to switch between discrete and smooth time stepping and get the exact same glider form. only one time hopping in discrete steps and the other time gliding along smoothly. neither of the two smooth modes achieves that unfortunately. but both lead to interesting forms of life with gliders, oscillators and stable structures, so they have a reason to exist in their own right.

or in other words: in the mode 1 case we get the right form of the equation, but with a different operator. that this operator has to be different from the mode 0 operator is clear, but we haven't yet found the right one. in the mode 2 case we get a familiar operator (at least including the mode 0 one), but a slightly different form of equation. so what we seek is a sort of mixture. always obeying the limit c->1 where we have to get the discrete time stepping.

the search for the real smooth time stepping mode will go on, this is just the beginning.

probably the real solution to the problem involves derivatives in x on the right hand side somehow.