Kadomtsev–Petviashvili equation

Remarks

Example: let $v(x, t)$ be a solution of the Korteweg–De Vries equation $v_t + 6 v v_x + v_{xxx} = 0$, and put $u(x, y, t) = v(x, t)$. Then the bracket vanishes and $u_{yy} = 0$, so $u$ solves both KP-I and KP-II.

Non-example: $u(x, y, t) = y^2$ is not a solution, because the first term is $0$ while $3 \sigma u_{yy} = 6 \sigma \ne 0$.

Other normalisations are used, for example $\partial_x(u_t + u u_x + \varepsilon^2 u_{xxx}) + \lambda u_{yy} = 0$ with $\lambda = \pm 1$. They reduce to the form above by scaling the variables, and the sign of the $u_{yy}$ term relative to the dispersive term decides between KP-I and KP-II.