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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.

For $R(\alpha )={\alpha }^{-\gamma }$, the uniform upstream area satisfiesThe condition $\beta >q^{1-\gamma /2}$ is equivalent to ${\alpha }_1<\sqrt{q}$, so this is a supercritical tube flow. The downstream prescribed area is subcritical. The steady equation cannot pass smoothly through $\alpha =\sqrt{q}$: its numerator there is positive under this strict inequality, while its denominator vanishes. A localized, dissipative elastic jump in a quadratic pressure-area tube permits the transition.

Neglect gravity and distributed resistance within the short jump. Area conservation gives equal flux on both sides. Multiply the inviscid momentum equation by $A$ and use the steady flux to put it in conservative form:The elastic contribution follows from ${\rho }^{-1}A{p}_x=c_0^2{A}_0{\alpha }^2{\alpha }_x$. Thus the dimensionless momentum flux $q^2/\alpha +{\alpha }^3/3$ is equal on both sides. If ${\alpha }_2$ is the post-jump area, subtraction and division by ${\alpha }_2-{\alpha }_1\ne 0$ giveThe momentum-flux function decreases up to $\sqrt{q}$ and increases thereafter, so the upstream state has exactly one conjugate state with ${\alpha }_2>\sqrt{q}$. The physically admissible direction is narrow, fast flow into wide, slow flow. The jump conserves momentum and mass, while losing mechanical energy. Indeed the steady Bernoulli equation away from a jump has $B=c_0^2(q^2/{\alpha }^2+{\alpha }^2)/2$. Using the momentum relation to eliminate $q^2$ gives $B_1-B_2=c_0^2({\alpha }_2^2-{\alpha }_1^2)({\alpha }_2-{\alpha }_1{)}^2/(6{\alpha }_1{\alpha }_2)>0$. Imposing conservation of $B$ across the jump would therefore give the wrong relation.

Downstream of the jump the steady equation becomesFor $\alpha >\sqrt{q}>{\alpha }_1$, both numerator and denominator are positive. Thus area grows downstream, and a solution of this type requires ${\alpha }_3\ge {\alpha }_2$. Integration from the jump to the exit givesThe first equation determines ${\alpha }_1$, the cubic determines its subcritical conjugate ${\alpha }_2$, and the integral determines $x_s$. The integral must lie between zero and $GL/c_0^2$ for a jump inside the tube. If ${\alpha }_3<{\alpha }_2$, or the required downstream length exceeds $L$, these boundary data do not admit the described single-jump branch. For admissible data the uniform upstream section can extend to the selected jump location because it is an exact equilibrium of the steady area equation.