Why are steps not visible in the continuum limit of the Frenkel Kontorova model?

[LagrangeEuler]

http://www.iop.kiev.ua/~obraun/fk-intro.htm#model

Question? Why this model is discreet? As far as I see particles can have any position?

Why then is written in the text then In the continuum-limit approximation we come to the sine-Gordon (SG) equation?

 

[vanhees71]

It's discrete in the sense that it's a model of many point particles. You can take a continuum limit. E.g., as the most simple case you can model a string first as a chain of harmonically coupled point particles. The continuum limit leads to the wave equation for the string.

 

[LagrangeEuler]

Thanks a lot for the answer.

And for example in this paper

http://fmc.unizar.es/people/juanjo/papers/falo93.pdf

steps appear because model is discrete. In continuum limit it would be impossible to see steps in the $v(F)$ (FIG 1). Is it possible to explain why in continuum limit is impossible to see steps?