Davey-Stewartson Lagrangian formulation

Question

The system is

$i u_t + c_0 u_{xx} + u_{yy} = c_1 |u|^2 u + c_2 u \phi_x,\,$

$\phi_{xx} + c_3 \phi_{yy} = ( |u|^2 )_x.\,$

This is like the NLS but with the extra y-dimension. The NLS has the lagrangian formulation

$\mathcal{L}=\frac{i}{2}(\psi_{t}\bar{\psi}-\psi\bar{\psi}_{t})-|\bigtriangledown\psi|^{2}+|\psi|^{4}.$

For DS since we have two eqns and unknowns, we will have two Euler-Lagrangian equations. But I was wondering if there is a way to express them as one Lagrangian (to make applying Noether's theorem easier).

Answer 1

From an inspection of it, your DS system appears to have the following Lagrangian: $$ \mathcal{L} = \frac{i}{2} (\bar{u} u_t - \bar{u}_t u) - c_0 \bar{u}_x u_x - \bar{u}_y u_y - \frac{c_1}{2} \bar{u}^2 u^2 - c_2 \bar{u} u \phi_x - \frac{1}{2 c_2} (\phi_x)^2 - \frac{c_3}{2 c_2} (\phi_y)^2 $$