Is this a valid way for deriving the ODE for a lattice vibration of a one dimensional crystal? Consider the following lattice:

I want to derive a differential equation that describes the forces acting on the $n$-th atom in the lattice. Each atom is coupled to its neighbour ($n+1,n-1)$ by a spring with constant $K$ and to its neighbours neighbour ($n+2,n-2$)by a spring with constant $K'$. From Ashcrof/Mermin I know that the equation of motion for three atoms without interaction between $n$ and the $(n+2,n-2)$ atoms is:
$$M\ddot{u}=-\frac{\partial U^{harm}}{\partial u(na)}=-K \left [ 2u(na)-u([n+1]a)-u([n-1]a)\right ] \tag{1}\label{1}$$
where $u(na)$ is the displacement along the line from it's equilibrium position, of the ion that oscilltes about $na$. 
Now I want to expand this and consider the interaction with the $(n-2,n+2)$ atoms. This is what I have tried:
$$F_{total}=M \ddot{u}=-K \left [ 2u(na)-u([n+1]a)-u([n-1]a)\right ]-K' \left  [2u(na)-u([n+2]a)-u([n-2]a)\right ] \tag{2}\label{2}$$
I have two questions:


*

*How is equation $(1)$ derived? I know that according to Hooke's law the force exerted by a spring is give by $F=-kx$ where $k$ is the spring constant and $x$ the displacement from equilibrium. However, here the displacement from equilibrium is not only dependant on $u(na)$ but also on all the other atoms $u([n+1]a), u([n-1]a)$...etc. How is that factored into the equation?

*Does the differential equation $(2)$ fully describe the motion of the $n$-th atom or do I also need to factor in the interaction between the $n\pm2$ and $n\pm1$ atoms? Ashcroft/Mermin seem to omit some interactions between atoms for equation $(1)$ but I don't understand when and why I am allowed to omit interactions.

 A: The force directly acting on particle $s$ in the harmonic approximation is proportional to the displacement from equilibrium, $u$:
$$F = - k (x-x_{eq})=-k u$$
If we take into account only nearest-neighbours interactions we will have the total force acting on particle $s$ will depend on the total displacement from equilibrium on the right side ($u_r$) and the same quantity on the left side ($u_l$)
$$F_s= -k u_l +k u_r = - k(u_{s+1} - u_s)+k(u_{s-1}-u_s) = k(u_{s+1}+u_{s-1}-2u_s)$$
So far so good. But as Aschroft-Mermin correctly points out, particle $s$ is actually coupled to the second-, third-, foruth-... nearest neighbours. So we should actually write, as you correctly concluded in the case of second nearest-neighbours,
$$F_s = \sum_{i=1}^{\infty} k_i (u_{s+i}+u_{s-i}-2u_s)$$
It turns out, however, that $k_{i+1} \ll k_i$. Why is that? This is because the coupling between two atoms is dependent on how much their orbitals are overlapping. Since orbitals are pretty much localized in the vicinity of the nucleus, the overlap of orbital $s$ and orbital $s+1$ will be much more than the one between orbital $s$ and orbital $s+2$. This can be seen explicitly by computing the overlap integral between the orbitals of the two atoms:
$$\int \psi_l^* (\vec r) \psi_m (\vec r - \vec R) d \vec r$$
