Intuition behind focusing vs defocusing in integrable systems like NLS, KdV, mKdV

Question

The following are examples of integrable systems arising from the AKNS system (check out AKNS paper here and a short Wikipedia description)

1. Non-Linear Schrodinger equation
2. Korteweg-de Vries equation
3. modified Korteweg-de Vries equation

I am paying attention the last one (mKdV) that has the form $$u_t\pm3u^2u_x+u_{xxx} = 0$$ I know that the $\pm$ represent two cases that change fundamentally change the dynamics of the system. I would like intuition: 1) on how to recognize whether this is the focusing or defocusing case, and 2) the effect of the non-linearity.

I know there are focusing and defocusing cases for the first two equations too, but how does the sign of the non-linearity show this? and what is the intuitive effect of focusing/defocusing? Is there a stability result associated to each case?

Answer 1

The focusing and defocusing are determined by the signs of the nonlinear and spatial derivative terms, $\pm 3u^2u_x$, and $+u_{xxx}$. Note in the simple dispersion relation $u_t+u_{xxx} = 0$, known as linearized KDV, dispersion flows to the left. This can be solved using Fourier Transform.

In the nonlinear equation $u_t\pm 3u^2u_x = 0$, one can think of this as a modified version of the Hopf equation $u_t+uu_x = 0$ giving rise to waves whose heights determine their speed. This gives a basic mathematical model of wave collapse (solutions exist for finite time because they attain infinite gradient at some point). We gain even further intuition from the simplest pde (advection equation) $u_t+cu_x = 0$ whose solutions are left traveling waves with constant speed $c>0$ and right traveling waves with speed $c<0$. Thus, Hopf-equation says that points of $u$ will travel left or right depending on the sign of $u$.

Thus if $c = \pm 3u^2$, we're guaranteed that solutions only travel in one direction (given the initial condition is of one sign). If $c = +3u^2u_x$, solutions will travel to the left (matching linearized KDV) - this gives focusing MKDV. If $c=-3u^2$, then solutions will travel to the right (the opposite direction of the dispersion term), giving rise to defocusing MKDV.