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Write $\alpha =A/A_0$ and $G=g\sin \theta >0$. The one-dimensional mass conservation and axial momentum equations consistent with the stated resistance areIn a steady flow, $Au=c_0{A}_{0}q$ and $u=c_{0}q/\alpha$. Since $p_{\alpha }=\rho c_0^2\alpha$, the momentum equation becomesThus a uniform inlet state continues as a uniform steady solution precisely whenTo examine downstream spatial stability of this steady solution, put $\alpha ={\alpha }_1+z(x)$ and linearize. The numerator vanishes at equilibrium, soSince $R^{\prime }<0$, the numerator of $s$ is positive. The displacement decays as $x$ increases exactly when ${\alpha }_1^4<q^2$, orThis proves the spatial attraction of a uniform downhill tube flow appropriate to the steady inlet-value problem.

The tube wave speed is $c^2=(A/\rho )dp/dA=c_0^2{\alpha }^2$, while $u=c_{0}q/\alpha$. The inequality is therefore $u>c$: the inlet lies on the supercritical tube flow branch, with both inviscid characteristic speeds $u\pm c$ directed downstream. A subcritical state has a characteristic carrying downstream boundary information upstream; the critical state makes the steady area equation singular unless its forcing also vanishes.

Spatial attraction must not be confused with temporal stability. The stated hypotheses do not guarantee decay of all time-dependent perturbations. For example, take $R={\alpha }^{-2}$ and a uniform supercritical state. The damping coefficient multiplying $u$ is $f=R_0{A}_0/\alpha$, and for perturbations $e^{i(kx-\omega t)}$ about area $\bar{A}$ and speed $\bar{u}$, set $s_t=-i(\omega -k\bar{u})$. Linearizing the two evolution equations givesIts small-$k$ branch is $s_t=-ik\bar{u}+k^2({\bar{u}}^2-c^2)/f+O(k^3)$, which grows when $\bar{u}>c$. This is a friction-induced instability of a collapsible-tube flow. Consequently the advertised stability is valid as downstream stability of the steady solution; an unrestricted temporal-stability claim would require further assumptions.