How to obtain $\xi_{1} = \frac{K}{\lambda A^2}$ in deriving modes centered on a site from the DNLS

Question

The $1$-dimensional discrete nonlinear schrodinger equation with nearest-neighbor interaction is written as $$\begin{align}i\partial_{t} \psi_n + K \left (\psi_{n+1} - 2\psi_{n} + \psi_{n-1} \right ) + \lambda |\psi_n|^2 \psi_n &= 0, \tag{1}\end{align}$$ where $K,\lambda$ are constants.

Following 1, with the ansatz $\psi_n(t) = Af_{n}e^{-i \omega t}$, for real $f_{n}$, we obtain

\begin{align} \omega f_{n} + K(f_{n+1} - 2f_{n} + f_{n-1}) + \lambda A^2 f_{n}^3 &=0 \tag{2}\end{align}

We have the following assumptions:

1. Symmetry $f_{-n} = f_{n}$,
2. Mode centered at site $n = 0$,
3. $f_0 = 1$,
4. $|f_{n}|<< f_{1}$ for $|n|>1$

Questions:

5. What do $A$ and $f_{n}$ represent?
6. How is $\xi_{1} := \frac{K}{\lambda A^2}$ obtained?

Attempt:

Take $n=0$. From eq.$(2)$ we explicitly obtain $f_1$:

\begin{align}\omega + K(2f_{1} - 2) + \lambda A^2 &= 0 \tag{3}\\ f_{1} &= 1 - \frac{\omega}{2K} - \frac{ \lambda A^2}{2K} \tag{4}\end{align}

I see we get something similar to their $\xi_{1}$, my $\boxed{\frac{ \lambda A^2}{2K}}$ term.

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1 - https://doi.org/10.1103/PhysRevE.48.3077

Free access obtained here: https://sci-hub.st/https://doi.org/10.1103/PhysRevE.48.3077

See the material around equation $(5)$ and $(6)$.

Answer 1

The standard DNLS (1-dim) is derived from the continuum NLS $$i\partial_{t}\psi(x,t) = - \partial_{xx}\psi(x,t) - |\psi(x,t)|^2 \psi(x,t)$$ by applying a centered-difference scheme to the spatial operator $$i\partial_{t}\psi_n = \frac{1}{h^2} \left ( \psi_{n+1} - 2\psi_{n} + \psi_{n-1}\right ) + |\psi_n|^2 \psi_n$$

Identify $K := \frac{1}{h^2}$. Consider the ansatz $\psi_n(t) = Af_n e^{-i\omega t}$ to obtain the system

$$\omega f_n + K(f_{n+1} - f_n + f_{n-1}) + A^2 f_n^3 = 0$$

Let $n=0$ and solve for $\frac{\omega}{K}$:

$$\frac{\omega}{K} = 2\left (f_{1} - 1 \right ) - \boxed{\frac{A^2}{K}}$$

The boxed term must be very big because $K$ is very small. This equivalent to saying

"$\frac{K}{A^2}$ is very small"