Solving the free Schroedinger equation in Madelung variables Can explicit solutions to the (unforced) Linear Schrodinger equation (LSE) be found via the Madelung transformation?
(A note on motivation. I am trying to constrain the behavior of the phase of $A$ for a more complicated higher order nonlinear Schrodinger equation, to see if shocks form in $|A|$. To do this, I want to see if I understand how the phase is related to the amplitude in the much simpler LSE). 
To motivate this, consider
$$iA_t+A_{xx} = 0$$
for $A$ a complex valued function $t\in(0,\infty)$ and $x\in (-\infty,\infty)$. One may solve this using Fourier Transforms. That is, letting
$$A=\frac{1}{2\pi}\int_{-\infty}^{\infty} a (k,t) e^{-ikx}\ dk,$$
we have $a(k,t)=a(k,0)e^{ik^2 x}$, where the initial condition $a(k,0)$ may be found via
$$a(k,0) = \int_{-\infty}^{\infty} A(x,0)e^{ikx} \ dx.$$
For particular initial conditions, i.e. a Gaussian envelope, explicit solutions may be found. 
Next, let $A= \sqrt{\rho} e^{i\theta}$ for $\rho, \theta$ real valued functions of $(x,t)$. The LSE becomes two coupled equations:
$$ \rho_t +( u\rho )_x=0,$$
$$ u_t+uu_x -2\frac{\partial}{\partial x} \left(\frac{1}{\sqrt{\rho}}\frac{\partial^2 \sqrt{\rho}}{\partial x^2}\right)=0,$$ 
where $u =2\theta_x$. The analogy with hydrodynamics (the first equation is the statement of mass conservation, while the second is momentum conservation) is now obvious. This helps aiding the physical interpretation of the LSE.
However, is it possible to then solve explicitly for $\rho$ and $u$?
It seems like the method of characteristics will yield some progress, especially for simple examples but I cannot seem to generalize my results in any kind of insightful way. 
 A: Madelung himself, 1927 discussed the structure of the solutions, and so did Barna et al 2017, Buyukasik & Pashaev 2010, etc...
I am not expert in such systems of equations, but for the free Schroedinger system one should at least review how a dispersive 1-d free Gaussian wavepacket presents, to find one's bearings. Since I did not find it anywhere, I might as well put it down here for the benefit of students who might be curious about the polar representation picture.
I'll take the "stationary" packet with its maximum always hovering over the x origin, and nondimensionalize $\hbar=1, ~ m=1$ out for simplicity, but I will keep the 1/2 in front of the kinetic term, egregious as it might well be in your field. Quantum physicists might well lynch me if I took $m=1/2$ instead. So Schroedinger's free equation is
$$
i\partial_t \psi = -\frac{1}{2}\partial_x^2 \psi,
$$
solved by normalized Gaussian wavepackets 
$$\bbox[yellow]{\Large
\psi= \sqrt[4]{\frac {2}{\pi (1+2it)^2} } ~e^{-\frac{x^2}{1+2it}}=\sqrt{\rho} ~e^{iS}, \\   
\rho=\psi^* \psi = \sqrt{\frac {2}{\pi (1+4t^2)} } ~~ \Large e ^{-\frac{2x^2}{1+4t^2}}, \\
S=\frac{2tx^2}{1+4t^2} -\frac{i}{4}\ln \frac{1-2it}{1+2it} }~~.
$$
Note the velocity 
$$
v\equiv j/\rho= \frac{1}{2i\rho}(\psi^* \partial_x \psi -\psi \partial_x \psi^*)=\partial_x S= \left ( \frac{4tx}{1+4t^2} \right )
$$
to be used in the continuity equation
$$
0=\partial_t \rho + \partial_x j= \partial_t \rho + \partial_x (v \rho)=\partial_t \rho + \partial_x \left (\rho ~\frac{4tx}{1+4t^2}\right ),
$$
which you may readily check.  The farther away you are from the origin the faster you spread. Further check the large t limit.
The real part of the polar form of the Schroedinger equation that Madelung explored  is the QHJ equation for S,
$$\bbox[yellow]{
0=\partial_t S  +\frac{1}{2}  (\partial_x S)^2 +Q \\
Q\equiv - \frac{1}{2} \frac{ \partial_x^2   \sqrt{\rho}}{\sqrt{\rho}}=\frac {1}{1+4t^2}\left (1-\frac{2x^2}{1+4t^2}\right) }  ,
$$
the Q being the quantum potential—the curvature of the wf amplitude. Note this is of lower order than the one you have: it has effectively integrated it once!
It is possible this explicit expression, and the WP translating it with group velocity $k_0$, instead,
$$    \begin{align}
 \psi &= \frac{ \sqrt[4]{2/\pi}}{\sqrt{1 + 2it}} e^{-\frac{1}{4}k_0^2} ~ e^{-\frac{1}{1 + 2it}\left(x - \frac{ik_0}{2}\right)^2}\\
               &= \frac{ \sqrt[4]{2/\pi}}{\sqrt{1 + 2it}} e^{-\frac{1}{1 + 4t^2}(x - k_0t)^2}~ e^{i \frac{1}{1 + 4t^2}\left((k_0 + 2tx)x - \frac{1}{2}tk_0^2\right)} ~\Longrightarrow
\end{align}  \\   \Large
\rho= \frac{ \sqrt{2/\pi}}{\sqrt{1+4t^2}}~e^{-\frac{2(x-k_0t)^2}{1+4t^2}} , $$ 
might help you with bottom-rung checking of your methods... 
