Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 81 (Physiological Fluid Dynamics)

Take positive volumetric flow rate in the direction of increasing $x$. For an arterial pressure wave, the characteristic admittance of an arterial wave givesThe backward wave contributes the opposite flow sign. Write $p_j$ for each junction's complex pressure amplitude, suppressing the common $e^{i\omega t}$ factor, and put $r=RY$, $C_{\beta }=\cos \beta$, $S_{\beta }=\sin \beta$. The topology is the two paths $K\to L\to N$ and $K\to M\to N$; their intermediate and common distal junctions each have a hydraulic resistance. The coordinate convention in the source puts the upstream end at $-\ell$ and the downstream end at zero in each segment. Consequently the wave transfer matrix of an arterial segment isEquivalently $q_u=Y(C_{\beta }{p}_u-p_d)/(i{S}_{\beta })$ and $q_d=Y(p_u-C_{\beta }{p}_d)/(i{S}_{\beta })$, with their continuous limits understood when $S_{\beta }=0$.

Symmetry gives $p_L=p_M=p_s$. Distal mass conservation and the resistive relation $Q_N=p_N/R$ giveAt $L$, the incoming flow supplies both its bed and segment $D$, soChoose the incident amplitude $P_A$ at $K$. If $P_{AR}$ is the reflected amplitude in vessel $A$, pressure continuity equation gives $p_K=P_A+P_{AR}$, while inlet flow is $2Y(P_A-P_{AR})=2Y(2P_A-p_K)$. Equating this to the two outgoing segment flows yieldsEliminating $p_s,p_K$ from these three relations gives the wave transmission through a resistively loaded arterial loop resultFor large $r$ at fixed phase, $p_N=2P_A{e}^{-2i\beta }+O(r^{-1})$ and $p_M=C_{\beta }{p}_N+O(r^{-1})$. HenceThese leading terms must not be used as nonzero denominators near a phase at which the normal intermediate flow vanishes.

Once $C$ is removed, the remaining connected segments form the chain $K\to L\to N\to M$, of total length $3\ell$. For $RY\gg 1$, each hydraulic resistance draws a small flow, so to leading order the far end at $M$ is reflecting, with zero segment flow there. Successive wave transfer matrices of an arterial segment giveThe incident-plus-reflected inlet relation remains $q_K=2Y(2P_A-p_K)$. Therefore, defining $D=3e^{3i\beta }+e^{-3i\beta }$,The modulus of $D$ equals $\gamma =|3e^{4i\beta }+e^{-2i\beta }|$, because the expressions differ by a unit-modulus factor. Comparison with the normal flows gives the amplitude ratiosFor small phases the cosines are positive and the absolute values can be removed. The original PDF prints an additional $\cos (\beta /2)$ in the middle numerator. That factor does not follow from the displayed equal-length network: the transfer relation from $M$ to $K$ above fixes $p_M=8P_A/D$, with no such factor. The corrected middle ratio is shown here.

Since ${\gamma }^2=10+6\cos 6\beta$, one has $\gamma =4[1-27{\beta }^2/8+O({\beta }^4)]$. Thus the three corrected ratios, in the order $N,M,L$, areAll three exceed one for sufficiently small nonzero $\beta$; they equal one at zero frequency. This is an oscillatory arterial wave reflection effect, not an increase in steady conductance. Removing a branch changes the standing wave pattern and the total inlet pressure associated with a fixed incident wave. Reflected waves can enhance junction pressures and hence peripheral flow amplitudes. Neither total inlet pressure nor total inlet flow was held fixed, and no extra energy source is implied.

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.

Use the pressure scale $\rho {\hat{U}}^2$, and denote the scaled transverse velocity by $v$. The steady incompressible flow equations under the prescribed coordinate scaling areThe no-slip boundary condition is $u=v=0$ at $y=1$ and at the lower wall $y=ϵF(x)$, taking $F=0$ upstream. Far upstream, $u\to U_0=6y(1-y)$, $v\to 0$, and $p_x\to -12\lambda /R$. These are the dimensional Navier-Stokes equations divided by the streamwise and transverse inertial scales, respectively; the transverse equation's ${\lambda }^{-2}$ is essential.

A consistent nonlinear wall-layer regime can be chosen asNear a wall the oncoming plane Poiseuille flow is linear in distance, so a layer of thickness $ϵ$ has $u=O(ϵ)$ and $v=O({ϵ}^2)$. Its inertia is $O({ϵ}^2)$; its transverse viscous term is $(\lambda /R)O({ϵ}^{-1})=O({ϵ}^2)$, explaining the distinguished balance. The resulting pressure perturbation is $O({ϵ}^2)$. In the core the induced transverse pressure variation is $O(ϵ/{\lambda }^2)$, negligible compared with that pressure perturbation when ${\lambda }^2ϵ\gg 1$, equivalently $\lambda \gg R^{1/7}$. Also $\lambda \ll R$ makes both the indentation and the viscous boundary layer thin. Streamwise viscous derivatives are smaller still. Thus the core perturbation is inviscid at first order and pressure is common to the two wall layers at their leading order.

Write $u=U_0+ϵu_1+\cdots$ and $v=ϵv_1+\cdots$. The core pressure gradient only enters at order ${ϵ}^2$, so first-order continuity equation and streamwise momentum giveEliminating $u_{1x}$ shows $(v_1/U_0{)}_y=0$. Let that ratio be $-A^{\prime }(x)$; integration in $x$, with the unperturbed upstream condition, then givesThis displacement form automatically has zero first-order change in total core flux, because ${\int }_0^1{U}_0^{\prime }dy=0$.

For the lower boundary layer, usewhere $p_0^{\prime }=-12{ϵ}^3$ is the smaller upstream pressure gradient. The extra term in $v$ removes the velocity induced by the moving coordinate surface. The chain rule gives ${\partial }_x{|}_y={\partial }_x{|}_z-F^{\prime }{\partial }_z$ and ${\partial }_y={ϵ}^{-1}{\partial }_z$; the wall-slope terms cancel from both continuity equation and convection. At leading order the Prandtl boundary-layer equation isMatching to the core near the lower wall gives $U\sim 6[z+F+A]$, so $H_{\mathrm{lower}}=F+A$.

For the upper boundary layer, set $y=1-ϵz$, $u=ϵU$, $v=-{ϵ}^{2}V$. The same equations, wall data and common $P(x)$ result, but the core matching gives $U\sim 6[z-A]$, so $H_{\mathrm{upper}}=-A$. The two pressure-driven wall-layer problems have the same upstream shear, wall conditions and pressure forcing. Under the question's uniqueness assumption their solutions coincide, and their displacement limits must coincide. Thus the symmetric viscous interaction in an indented Poiseuille channel gives

For $F=b{x}^{1/3}$, one has $A=-(b/2)x^{1/3}$ and $H=(b/2)x^{1/3}$. Matching $U\sim 6[z+H(x)]$ in a similarity solution $U=x^{\beta }{G}^{\prime }(z{x}^{-\gamma })$ requires $\beta =\gamma =1/3$. Equating the powers in convection $x^{2\beta -1}$, viscosity $x^{\beta -2\gamma }$ and pressure gradient $x^{\sigma -1}$ gives $\sigma =2/3$. HenceUse the stream function $\psi =x^{2/3}G(\eta )$, with $\eta =z{x}^{-1/3}$. ThenDirect differentiation gives $U{U}_x+V{U}_z=x^{-1/3}[(G^{\prime }{)}^2/3-2G{G}^{\prime \prime }/3]$, whereas $-P^{\prime }+U_{zz}=x^{-1/3}[-2\bar{p}/3+G^{\prime \prime \prime }]$. The one-third-power similarity of an indented channel boundary layer is therefore governed byThe original PDF has $(G^{\prime }{)}^2$ in this equation; the converted TeX loses the derivative and gives $G^2$. Existence remains conditional on a value of $\bar{p}$ for which this boundary value problem has a solution, as stipulated in the question. The similarity describes the stated thin-indentation regime on $x>0$.

Measure $z$ from the rigid tube wall towards the air core, so $0<z<h$. To leading order in $h_0/a$ and small interface slope, the curvature entering the Young–Laplace equation isThe air pressure is spatially uniform at the order retained. Thus the linear liquid pressure perturbation associated with thickness amplitude $H$ is $p_1=-{\sigma }_0(a^{-2}-k^2)H$. The circumferential curvature favors growth at long wavelength; the curvature opposes it at short wavelength.

The lubrication theory momentum equation is $\mu u_{zz}=p_x$. With the wall no-slip boundary condition $u(0)=0$ and zero surface shear stress, $u_z(h)=0$, its solution and flux areLeading thin-film mass conservation is $h_t+q_x=0$. For a mode $e^{\beta t+ikx}$, this gives the capillary instability of an annular liquid liningTherefore the unstable band is $\boxed{0<k^2<a^{-2}}$. Maximization with respect to $k^2$ gives $k_{\ast }^2=1/(2a^2)$ andThese curvature coefficients use $a$ instead of $a-h_0$ consistently at leading order in the thin-lining approximation.

Let the positive undisturbed surface tension be $\bar{\sigma }={\sigma }_0-A{\Gamma }_0>0$. An insoluble surfactant perturbation $g$ produces ${\sigma }_1=-Ag$. The tangential Marangoni stress changes the surface condition to $\mu u_z(h)={\sigma }_x$. Solving the lubrication theory momentum equation now givesAt leading order, conservation of insoluble surfactant on a moving interface reduces to ${\Gamma }_t+(\Gamma u_s{)}_x=0$. Corrections from the changing circumference are smaller by $h_0/a$. Likewise the normal-pressure contribution $-{\sigma }_1/a$ gives a coupling smaller by $h_0/a$ than the retained tangential Marangoni stress, so it is omitted at the same order.

For thickness and concentration amplitudes $H,g$ proportional to $e^{\beta t+ikx}$, put $D=a^{-2}-k^2$ and use $p_1=-\bar{\sigma }DH$. The two conservation equations becomeSet $s=k^2{a}^2$, $\lambda =h_0^3\bar{\sigma }/(3\mu a^4)$ and $\alpha =h_{0}A{\Gamma }_0/(\mu a^2)$. The uncoupled capillary growth and elastic relaxation rates are $L=\lambda s(1-s)$ and $S=\alpha s$. The off-diagonal product is $3LS/4$, so the determinant is $(\beta -L)(\beta +S)+3LS/4$. Thus the surfactant stabilization of an annular liquid lining relation isThis proves the dispersion relation before its lettered deductions. The physical assumption $\bar{\sigma }>0$ is needed for the claimed capillary instability band.

For positive $\lambda ,\alpha$ and $0<s<1$, the constant term of the quadratic equation is negative. Its roots are therefore real with opposite signs, so one has positive growth. At $s>1$, both the coefficient of $\beta$ and the constant term are positive. If the roots are real they are both negative; if complex they have negative real parts. Thus the growing band remains $\boxed{0<k^2{a}^2<1}$. At $s=0$ or $1$ there is a neutral root. If $\alpha =0$, the result reduces continuously to the clean-film capillary branch with a passive zero root.

For $\alpha /\lambda \ll 1$, follow the unstable root near the clean-film value $L=\lambda s(1-s)$. Write $\beta =L+\Delta$ in $(\beta -L)(\beta +S)+3LS/4=0$. At first order in $S$, $L\Delta +3LS/4=0$, givingThis expansion is uniform near the maximizing wavenumber. Differentiation gives $s_{\ast }=1/2-3\alpha /(8\lambda )+O((\alpha /\lambda {)}^2)$. Evaluating the growth there yieldsThe change in maximizing wavenumber only affects the maximum at second order. The first-order decrease therefore has exactly the requested coefficient, relative to the clean-film rate with its surface tension replaced by $\bar{\sigma }$.

The quarter-rate limit is the strong-elasticity limit $\lambda /\alpha \ll 1$, not the weak-elasticity limit. The original PDF repeats $\alpha /\lambda \ll 1$ here; that is inconsistent with the preceding small-$\alpha$ result. In fact at $\alpha =0$ the maximum is $\lambda /4$, directly contradicting a quarter of that value in the same limit. The converted TeX repairs this particular ratio.

For $\alpha \gg \lambda$ and wavenumbers near the maximum, $S=\alpha s\gg L$. In the quadratic equation, the growing root is $O(\lambda )$, so its square and the $-L\beta$ term are small compared with $S\beta$. The leading balance is $S\beta -LS/4=0$. ConsequentlyThe second root is a rapidly damped surfactant-relaxation mode. Thus the stated quarter-rate result is correct after reversing the printed small-parameter ratio.

Without surfactant, capillary pressure drives a surface flow towards a growing bulge. Advection compresses the insoluble surfactant in converging parts of that motion, increasing concentration and lowering surface tension. The resulting gradient supplies a tangential Marangoni stress that opposes the surface motion. This additional resistance slows capillary-driven redistribution of liquid even after the base-tension decrease has already been accounted for.

In the strong-elasticity limit, concentration gradients suppress tangential interface motion. The surface behaves as effectively immobile at leading order. Setting $u_s\approx 0$ in the velocity formula gives ${\sigma }_x\approx h{p}_x/2$, henceone-quarter of the clean shear-free mobility $h^3/(3\mu )$. This mobility reduction explains the quarter-rate limit; it slows the unstable capillary mode without removing the long-wave unstable band.