Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 81 / 2

The first equation is mass conservation for an incompressible flow whose area may change: local area storage balances the divergence of the volume flux $\alpha u$. The two left-hand momentum terms are local acceleration and convective acceleration, while $-p_x$ is the driving pressure gradient and $-R(\alpha )u$ models viscous drag per unit mass. The pressure-area law of a collapsible tube expresses transverse wall balance using internal minus prescribed external pressure.

For example, choose a reference area ${\mathcal{A}}_{\ast }$, length $L$ and velocity $U_{\ast }$. If dimensional variables are $\mathcal{A},s,t_d,v,P$, takeThe dimensionless external pressure uses the same scale. A dimensional friction rate $r_f(\mathcal{A})$ becomes $R(\alpha )=L{r}_f({\mathcal{A}}_{\ast }\alpha )/U_{\ast }$, and the dimensional wall pressure law is divided by $\rho U_{\ast }^2$. These choices make both inertia and the pressure-gradient coefficient unity; other equivalent reference scales are possible. On the mechanically stable wall branch, increasing area requires increased transmural pressure, whereas an enlarged tube reduces resistance. Thus $\boxed{{\tilde{P}}^{\prime }(\alpha )>0,R^{\prime }(\alpha )<0}$.

Steady mass conservation makes $\alpha u=Q$. For uniform area, $u=U=Q/{\alpha }_0$ is constant and the pressure-area law of a collapsible tube gives $p_x=p_e^{\prime }(x)$. The acceleration vanishes, so momentum balance requiresThe external pressure must therefore decrease linearly along a positive-flow tube. Its additive constant is free; the constant transmural pressure is $\tilde{P}({\alpha }_0)$.

Put $U=Q/{\alpha }_0$, $R_0=R({\alpha }_0)$, $c_0^2={\alpha }_0{\tilde{P}}^{\prime }({\alpha }_0)$ and $\nu =\omega -kU$. Let the area and velocity perturbations have amplitudes $a$ and $w$. Linearizing mass conservation and momentum, and using $R_0^{\prime }=-n{R}_0/{\alpha }_0$, givesFor nonzero $k$, eliminate $w=\nu a/(k{\alpha }_0)$ to obtain the friction-induced instability of a collapsible-tube flow relationAt the stated threshold $nU=c_0$, it factors asTherefore one mode has real $\boxed{\omega =k(U+c_0)}$, while the other has $\omega =k(U-c_0)-i{R}_0$ and decays. The PDF's real-frequency statement is valid for the neutral branch, not for both roots. The Fourier convention $e^{i(kx-\omega t)}$ makes positive imaginary frequency mean growth.

Complete the square in the dispersion relation:If the imaginary part of the square root has magnitude $b$, thenOne root grows exactly when $b>R_0/2$. For $k\ne 0$, squaring the equivalent positive inequality gives $n^2{U}^2>c_0^2$. Thus positive flow is unstable whenever $\boxed{U>c_0/n}$. This proves the entire stated range, not just a small neighborhood of threshold. Zero wavenumber has only a neutral uniform-area change and a damped velocity mode.

Near threshold write $nU=c_0(1+\delta )$ and follow the neutral branch as $\nu =k{c}_0+\delta {\nu }_1+O({\delta }^2)$. Substitution yieldsTaking its imaginary part gives the growth rateArea increases reduce friction, causing a velocity increase that feeds the area disturbance through mass conservation. The pressure-driven base flow supplies the energy; the positive friction coefficient does not by itself prevent instability of that maintained flow.

In the travelling coordinate $X=x-ct$, mass conservation becomes $[\alpha (u-c){]}^{\prime }=0$. Thus the conserved relative flux is $Q$ andFor a periodic peristaltic pumping solution, the neglected-inertia momentum equation isTake ${\alpha }_0$ to be the spatial mean area, so $⟨{\alpha }_1⟩=⟨{\alpha }_2⟩=0$. This normalization, and zero net pressure drop over a period, are implicit in the specified periodic forcing. The constant leading state has no pressure gradient, giving $Q_0=-{\alpha }_{0}c$. Put $B={\tilde{P}}_0^{\prime }$ and $d=R_{0}c/{\alpha }_0$. At first order,Period averaging gives $Q_1=0$, and the zero-mean solution isFor the second-order mean balance, expand the velocity and friction:Averaging the exact pressure derivative gives zero, so $⟨R_0{u}_2+R_1{u}_1⟩=0$. Hence the weak peristaltic pumping of a collapsible tube result isThe mean laboratory volumetric flow rate is $c⟨\alpha ⟩+Q={ϵ}^2{Q}_2+O({ϵ}^3)$, forward with the wave when $c>0$. The negative leading relative flux is therefore consistent with positive mean pumping.