Abstract We consider the cubic nonlinear Schrödinger (NLS) equation set on a two-dimensional box of size L with periodic boundary conditions. By taking the large-box limit L → ∞ in the weakly ...

First, we need to find which number when substituted into the equation will give the answer zero. \(f(1) = {(1)^3} + 4{(1)^2} + (1) - 6 = 0\) Therefore \((x - 1)\)is a factor. Factorise the quadratic ...