The continuous cubic-focusing NLS is $$i \partial_t u(t,x) = -u_{xx} - |u|^2 u$$ A split-step scheme can be performed. Consider the linear split equation: $\partial_t u = -iu_{xx}$
It can be solved using the Fourier transform: \begin{align} \partial_t u + i u_{xx} & = 0\\ \partial_t \hat{u} + i k^2 \hat{u} &=0\\ \int_{\tau_{m}}^{\tau_{m+1}} \frac{d}{dt} \left (e^{ik^2 t}\hat{u} \right )dt &\\ u_{m+1} &= \mathcal{F}^{-1}[e^{-ik^2 \Delta t} \mathcal{F}[u_{m}]] \end{align}
This completes the process for solving the linear part of the split system.
Now I want to do this using the discrete Laplacian.
Question: What is the Fourier transform of the discrete Laplacian? $\delta^2 u = u_{n+1}-2_{n}+u_{n-1}$. That is, instead of $|k|^2$, what is the Fourier symbol for $\delta^2$?
Discrete Fourier Transform:
$\hat{u}(k) = \frac{1}{\sqrt{2\pi}}\sum\limits_{k\in \mathbb{Z}}u(x_m)e^{-imk}$
$u(x_m) = \int\limits_{-\pi}^{\pi}\hat{u}(k)e^{imk}dk$
One has \begin{align} \delta^2u_n&=u_{n+1}-2u_n+u_{n-1}\phantom{\int} \\ &=\int_{-\pi}^{\pi}\hat{u}(k)\left(e^{i(n+1)k}-2e^{ink}+e^{i(n-1)k}\right)dk \\ &=\int_{-\pi}^{\pi}\left(e^{ik}-2+e^{-ik}\right)\hat{u}(k)e^{ink}\,dk \\ &=\int_{-\pi}^{\pi}2(\cos k -1)\,\hat{u}(k)e^{ink}\,dk, \end{align} hence $$ \widehat{\delta^2}=2(\cos k -1). $$