Intuition behind focusing vs defocusing in integrable systems like NLS, KdV, mKdV The following are examples of integrable systems arising from the AKNS system (check out AKNS paper here and a short Wikipedia description)


*

*Non-Linear Schrodinger equation

*Korteweg-de Vries equation

*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? 
 A: 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.
