= Solution
Write $\psi=a e^{iS}$ with $a=\sqrt\rho$ and $v=S_x$. Then
$$
\psi_{xx}=e^{iS}
\left(a_{xx}-av^2+i(2a_xv+av_x)\right),
$$
while
$$
-2i\psi_t=e^{iS}(2aS_t-2ia_t).
$$
Equating imaginary parts gives
$$
\rho_t=-(\rho v)_x,
$$
and equating real parts and differentiating in $x$ gives
$$
v_t=
\partial_x\left[
\frac{(\sqrt\rho)_{xx}}{2\sqrt\rho}
+\frac{1-\rho}{2}\right]
+\partial_x\left(-\frac{v^2}{2}\right).
$$
Thus
$$
\boxed{A=-\rho v},
\qquad
\boxed{B=\frac{(\sqrt\rho)_{xx}}{2\sqrt\rho}
+\frac{1-\rho}{2}},
\qquad
\boxed{C=-\frac{v^2}{2}}.
$$
This is the hydrodynamic form of the <defocusing nonlinear Schrödinger equation>.
Back to article page
