@Carl Love First of all thanks for the reply. Energy transfers between these two lines of oscillators, one oscillator at a time. For example, at t=0 oscillator x1 has the total energy of the system. Then energy tranfers thgrouh propagating breather to oscillator y2 then x3 etc. That can be seen at the two red-blue graphs. "An impulsive excitation is applied at t = 0, to the leading oscillator of one of the lattices – the ‘‘excited lattice’’ – while the other, ‘‘absorbing lattice’’, is not directly forced. This is equivalent to an initial velocity of the excited oscillator at t = 0 with all other oscillations assumed at rest. We are interested only in primary wave transmission, and not be concerned by reflections at the boundaries."
1. It shows which oscillator "has energy" at a specific time.
2. Yes it is the normalised energy.
3. - This is the only thing I am not sure myself. I try to figure this out but I tihnk is the kinetic energy after some tranformation.
4. The equations are right. For example if we have 6 oscillators then we have 6 equations - mx''1 equation, my''1, mx''2, my''2, mx''3, my''3.
5. Initial positions x1(0) = y1(0) = 0 (for the rest I think we can use some small step h). Initial velocities x'1(0) = v0 = 0.06 m/s , the rest is zero.
6.It is the initial velocity of x1. At t=0 everything is at rest except x1.
8. n=50 (50 oscilltors per line)
Thanks again. Ask me anything more you want.