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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.

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.