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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.