Consider the 1D particle-in-a-box problem with time-dependent box length $L(t)>0$ (with $\hbar = m =1$)

\begin{align} i\psi _t &= \textstyle -\frac{1}{2} \psi _{xx} \\ \psi (t,0) &=\psi (t,L(t)) = 0 , \end{align} The Madelung transformation

\[ \psi (t,x) = e^{iS(t,x)} \sqrt{\rho (t,x)} \]

formally gives the fluid-dynamical formulation: \begin{eqnarray*} v_t+v\cdot (\nabla v) &=& \textstyle \frac{1}{2} \nabla \frac{\Delta\sqrt{\rho } }{\sqrt{\rho }} \\ \rho _t +\nabla \cdot (\rho v)&=& 0, \end{eqnarray*}

where $v:=\nabla S$ (and $\nabla $ and $\Delta $ may be read as $\partial /\partial _x $ and $\partial ^2/\partial _x ^2$).

Clearly, we have $\rho (t,0)= \rho (t,L(t)) =0$. But what are the appropriate boundary conditions

for $v$ (if there are any)? Is the (naive?) guess

\[ v(t,0) = 0, \qquad v(t,L(t)) \sim \frac{d}{dt} L(t) \] reasonable?