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For a stationary pumped condensate in a closed cross-section, integrate the gain-loss Madelung equations. Vanishing lateral condensate number current and constant integrated axial current imply the displayed balance. It fixes a density-weighted mean, not the pointwise density. If the integrated axial current $J(z)$ varies, the exact relation instead is ${\int }_S{n}^{2}dS=(\gamma /\Gamma )N(z)-[\hslash /(2\Gamma )]J^{\prime }(z)$. Thus lateral no-flux conditions alone are insufficient for a slice-by-slice balance in an axially inhomogeneous state.

Define the physical velocity potential by $\psi =\sqrt{n}\exp (im\varphi /\hslash )$, so the superfluid velocity is $\mathbf{v}=\nabla \varphi$. The condensate number current isMultiplying the pumped Gross–Pitaevskii equation by ${\psi }^{\ast }$ and subtracting its complex conjugate gives the density equation. Equivalently, substituting the amplitude and phase and equating imaginary parts yieldsEquating real parts gives the other gain-loss Madelung equations relation,The last term is the quantum potential, retained wherever $n>0$. Taking a gradient gives the equivalent velocity equationin an irrotational region. In equilibrium $n_t=0$ and $\varphi (\mathbf{x},t)=\Phi (\mathbf{x})-\mu t/m$, soThis convention distinguishes the physical velocity potential from the dimensionless quantum phase $m\varphi /\hslash$; using the latter instead would change the divergence prefactor.

Here equilibrium means a time-independent number density and current, with a common temporal quantum phase. Write $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }\Psi (\mathbf{x})$ with real $\mu$, since an imaginary temporal frequency would change the density. The stationary pumped Gross–Pitaevskii equation isThe chemical potential is the common phase-rotation frequency times $\hslash$; this open system has no conserved particle-number constraint fixing it as in the conservative problem. Local gain and loss need not cancel separately in an inhomogeneous equilibrium, because a stationary condensate number current can transport particles between gain and loss regions.

Let $N={\int }_{S}ndS$ be the particle number per unit axial length, and consider the axially homogeneous equilibrium implicit in a cross-sectional description. Integrating the stationary gain-loss Madelung equations over $S$ and applying the divergence theorem givesThe wall condition $n=0$ means the lateral condensate number current vanishes. This is also directly apparent from $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$ for a regular Dirichlet wavefunction, without assigning a phase at its zeros. Consequently the cylindrical pump-loss balance isIntegrated particle creation equals integrated nonlinear loss; there is no assumption that $n=q$ pointwise. In fact the Cauchy-Schwarz inequality gives $N^2\le |S|{\int }_S{n}^2=|S|qN$, so a nonzero equilibrium obeys $N\le q|S|$.

The axial condition is needed for a slice-by-slice statement. More generally, put $J(z)={\int }_{S}n{\partial }_z\Phi dS$. Lateral no-flux and stationary density giveThus the requested equality also holds whenever the integrated axial current is constant. If axially inhomogeneous equilibria with nonzero axial flux divergence are allowed, the lateral boundary condition alone proves this more general balance, rather than the stated equality at every cross-section.

For a straight singly quantized quantum vortex, use cylindrical distance $r$ from the line and writeThere is no dependence along the vortex line. Substitution into the normalized stationary equation gives, before separating real and imaginary parts,Consequently the wave amplitude and phase-gradient equations areThe second equation can also be written $(r{f}^{2}k{)}^{\prime }=-{\xi }_{I}r{f}^2(1-f^2)$, which remains convenient at the core.

To specify what is meant by radial velocity, use the condensate number current. The time-dependent equation has kinetic operator $-{\nabla }^2$ and therefore number current $\mathbf{j}=2|\Psi {|}^2\nabla S$. Its hydrodynamic superfluid velocity in these units is $\mathbf{u}=2\nabla S$, so $u_r=2k$ and $u_{\theta }=2/r$. In that convention the requested pair isIf instead velocity is defined as the complex argument gradient itself, the preceding $k$ pair is the corresponding convention; the distinction matters because the printed kinetic coefficient differs from that of the normalized conservative equation in Question 1.

Regularity of a unit-charge core gives $f(r)=ar+O(r^3)$ with $a>0$. Integrating the current equation from the origin, with no point source or singular radial flux, givesThe integral is $a^2{r}^4/4+O(r^6)$, so the radial flow near a driven vortex core isFor the phase-gradient convention, $k^{\prime }(0)=-{\xi }_I/4=\alpha /4$. Gain exceeds loss in the depleted core, hence this regular flow is outwards for $\alpha >0$. The real wave amplitude equation also gives $f=ar-{\xi }_{R}a{r}^3/8+O(r^5)$, independently consistent with the linear core behaviour. A nonzero integration constant in the radial-current identity would create a singular $u_r\sim r^{-3}$ and is excluded.

Finally, the literal boundary $\psi \to 1$ from the preceding part cannot apply to a multiplicity-one vortex: its complex argument changes by $2\pi$ around a large circle. Even $f(r)e^{i\theta }$ with $f\to 1$ has different limits along different rays. The usual intended vortex condition concerns the wave amplitude approaching its bulk value, with the winding complex argument retained; it is not a constant complex-field limit. The local equations and core slope derived here do not establish a global stationary vortex satisfying that literal boundary. In a driven system the far-field radial flow and oscillation frequency may also require selection, so no global zero-flow vortex is asserted.