Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 75 (Physiological Fluid Dynamics)

Use a linear, lossless arterial pressure wave model with an incompressible fluid of mass density $\rho$, small wall displacement and negligible mean-flow advection compared with the wave speed. Let the parent vessel have equilibrium area $A_1$ and speed $c_1$. Linearizing mass conservation and axial momentum gives $a_t+A_1{u}_x=0$ and $\rho u_t=-p_x$. With wall compliance $p=(dp/dA)a$, these imply $c_1^2=(A_1/\rho )dp/dA$. A stiffer wall therefore has larger wave speed. A downstream travelling pressure wave carries flux $Q_{+}=Y_1{p}_{+}$, where $Y_1=A_1/(\rho c_1)$ is the characteristic admittance of an arterial wave; a returning wave has $Q_{-}=-Y_1{p}_{-}$.

Place the reflecting junction at $x=L$. Treat the daughters as outgoing-wave loads, with combined admittance $Y_d={\sum }_j{A}_j/(\rho c_j)$ and no further returning waves. At the junction, pressure continuity and flux conservation give $p_I+p_R=p_T$ and $Y_1(p_I-p_R)=Y_d{p}_T$. Thus the bifurcation reflection of an arterial pressure wave hasA decrease in total area gives $0<r<1$ if the parent and daughter wave speeds are equal. More generally this requires $Y_d<Y_1$, an assumption about both area and stiffness. To isolate the effect of ageing, assume all relevant wave speeds increase by the same factor, so their ratios and hence $r$ stay fixed. Unchanged geometry alone does not ensure unchanged $r$ if the vessels stiffen by different factors.

At a station $x<L$, put $\varphi =\omega (t-x/c_1)$ and $\vartheta =2\omega (L-x)/c_1$, the round-trip phase to the reflector. The incident pressure harmonic and its reflection giveFor time convention $e^{i\varphi }$, the two phasors are $P=1+r{e}^{-i\vartheta }$ and $V=1-r{e}^{-i\vartheta }$. Their maximum times are shifted by $-\arg P/\omega$ and $-\arg V/\omega$, respectively. Hence the pressure-flow phase lag from one arterial reflection isIndeed $V/P$ has the argument of $1-r^2+2ir\sin \vartheta$. On the branch $0<\vartheta <\pi$, this phase lag is positive: the pressure maximum follows the flow maximum. There is no cycle-wrap ambiguity on this branch because $1-r^2>0$.

The age trend need not have the same sign at every station. Differentiating the phase givesIncreasing $c_1$ decreases $\vartheta$. At a distal station with $0<\vartheta <\pi /2$, the lag therefore decreases. At a more proximal, ascending-aortic station with $\pi /2<\vartheta <\pi$, it increases. Thus one sufficient set of assumptions reproducing both trends isAs long as ageing keeps the stations in these ranges, the distal lag falls while the proximal lag rises. The effect is interference between pressure waves that add and flow waves that subtract, not merely the shortening of one travel time. The lag is largest at a round-trip phase of $\pi /2$.

This is a conditional explanation, with fixed representative harmonic frequency, fixed reflector location, negligible attenuation, negligible additional reflections and common stiffness scaling. The given stiffness and geometry facts alone do not specify the phase ranges, so they do not force either age trend without these extra assumptions. Here the two stations lie on the incident-plus-reflected segment. A station beyond the sole junction in an ideal outgoing daughter has only a transmitted wave and zero pressure-flow lag in this model. Actual peripheral measurements beyond that junction require returning waves from further junctions or terminal loads; a single-reflector calculation cannot establish their lag. Non-sinusoidal waveforms, distributed viscosity, mean flow and differently evolving reflection coefficients can also shift peak times. These limitations matter when applying the harmonic explanation to a real arterial network.

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.

For $\gamma =2$, ${\alpha }_1=(q/\beta {)}^{1/2}\to 0$. The jump cubic is dominated by ${\alpha }_2^3\sim 3q^2/{\alpha }_1$, since the other two polynomial terms contain powers of the much smaller ${\alpha }_1$. HenceIn the downstream integral, $\alpha \ge {\alpha }_2\to \infty$ and $q=O(1)$, soIts relative corrections are uniformly small over this interval. Integration therefore givesThe stated sufficient condition $0<{\alpha }_3^2-3^{2/3}q{\beta }^{1/3}<GL/c_0^2$ places the jump in the downstream half of the tube, $L/2<x_s<L$, to this order. The full condition for $0<x_s<L$ is the weaker upper bound $2GL/c_0^2$, because the displayed length relation contains a factor two. The stricter printed bound is sufficient rather than necessary. This large-$\beta$ branch also requires the downstream prescribed area to scale at least as ${\beta }^{1/6}$; keeping ${\alpha }_3$ fixed while sending $\beta$ to infinity would make its right-hand side negative.

Scale axial distance with $\lambda \hat{h}$ and normal velocity with $\hat{U}/\lambda$, while the dimensionless axial velocity has scale $\hat{U}$. Let $r=R+y$; keep $R>0$ fixed on the asymptotic scales. In the core, the ratio of transverse viscous terms to axial convective terms is of order $\lambda /Re$. A small indentation and the associated displacement create tangential slip of order $ϵ$ at the edges of the wall layers.

For a layer of relative thickness $\delta \ll ϵ$, that slip makes axial convection of order ${ϵ}^2/\lambda$, while transverse viscous diffusion is of order $ϵ/(Re{\delta }^2)$. Balancing them givesUnsteady diffusion gives the equivalent estimate $\delta \sim (ReSt{)}^{-1/2}$ when $\lambda St=O(ϵ)$. Thus a consistent regime supporting thin viscous boundary layers and an inviscid core isAt a location where both imposed unsteadiness and slip vanish, a shear-dominated entrance balance can instead have thickness $(\lambda /Re{)}^{1/3}$; the same strict Reynolds-number condition keeps even that scale below $ϵ$. The reduction below assumes that no wall-layer thickening invalidates the stipulated small thickness. The normal flux displaced by a slip layer is of order $ϵ\delta$, smaller than the retained ${ϵ}^2$ core flux. Tangential no slip is imposed within these layers, not directly on the leading inviscid core.

The weakly nonlinear, unsteady and dispersive core terms must enter at the same order. Axial nonlinearity first contributes at $O({ϵ}^2)$, while unsteadiness contributes $O(ϵ\lambda St)$ and transverse-inertia pressure variation contributes $O(ϵ{\lambda }^{-2})$. Thus take the distinguished limitEquivalently, $\lambda =O({ϵ}^{-1/2})$, $St=O({ϵ}^{3/2})$ and the wall-layer condition becomes $Re\gg {ϵ}^{-7/2}$. These orderings imply the initially stipulated long wavelength, low Strouhal number and large Reynolds number. In the core $\lambda /Re=o({ϵ}^3)$, so perturbation viscous terms are beyond the retained order.

Subtract the undisturbed fully developed state and scale pressure by $\rho {\hat{U}}^2$. The inviscid core equations arePut $u=u_0+ϵu_1+{ϵ}^2{u}_2+\cdots$ and $v=ϵv_1+{ϵ}^2{v}_2+\cdots$. At first order $p_{1y}=0$ andThe wall's normal movement first enters at second order in this scaling, so $v_1=0$ at both walls. Taking the regular wall limits, where $u_0=0$ and $u_0^{\prime }\ne 0$, gives $p_{1x}=0$. The arbitrary constant pressure perturbation can be set to zero.

Define the first-order axisymmetric stream function by $u_1={\psi }_{1y}/r$ and $v_1=-{\psi }_{1x}/r$. The axial equation becomes $u_0{\psi }_{1yx}-u_0^{\prime }{\psi }_{1x}=0$, or $({\psi }_{1x}/u_0{)}_y=0$. Consequently ${\psi }_1=A(x,t)u_0(y)$ after absorbing an $x$-independent profile into the background. This proves the annular inviscid-core displacementIt gives exactly the stated core expansion without requiring an explicit calculation of $u_0(y)$. The factors $1/(R+y)$ arise from cylindrical mass conservation.

Continue the expansion established in part (b), with $p={ϵ}^2{p}_2+\cdots$ after removal of the background pressure. The radial momentum equation at order ${ϵ}^2$ givesIntegrating across the core, then differentiating in $x$, givesThis is the dispersive contribution from transverse inertia.

At second order the axial momentum equation isFor the displacement fields just obtained,Use $u_0=0$ on each wall. The lower and upper regular core limits respectively givewhere $v_2(1)=0$ at the fixed outer wall.

At the moving inner wall $y=ϵF$, no penetration requires $v=ϵ\lambda St{F}_t+ϵu{F}_x$. Expanding about $y=0$, the axial edge velocity is $ϵ{\gamma }_0(F+A/R)+O({ϵ}^2)$ and $v_{1y}(0)=-{\gamma }_0{A}_x/R$. ThusorThe thin wall layers do not change this normal-flux condition at the retained order, because their flux correction is $o({ϵ}^2)$.

Subtract the two axial pressure-gradient limits and use the radial-pressure difference. With ${\alpha }_1={\gamma }_0/R+{\gamma }_1/(R+1)$ and ${\alpha }_2={\gamma }_0^2/R^2-{\gamma }_1^2/(R+1{)}^2$, the result is the forced annular displacement equationEvery coefficient follows from the given profile and wall slopes; no explicit base-flow solution has been used. The derivation is conditional on the regular core limits and wall layers remaining asymptotically thin.

Away from forcing, set $F=0$. For a linear sinusoidal disturbance $A=\mathrm{Re}[\hat{A}{e}^{i(kx-\omega t)}]$, the reduced equation givesSince $\sigma ,\beta ,{\alpha }_1>0$, the phase velocity $c_p=\omega /k=\sigma k^2/(\beta {\alpha }_1)$ is positive for every nonzero wavenumber. The group velocity isThus both the linear phase and wave packet travel downstream; the sign of $k$ does not introduce an upstream branch.

Now seek a permanent nonlinear wave with $\xi =x+ct$, for which the propagation velocity is $-c$. Substitution gives $\sigma A^{(3)}-\beta {\alpha }_{1}c{A}^{\prime }-{\alpha }_{2}A{A}^{\prime }=0$. Integration with vanishing far-field data givesMultiplying by $A^{\prime }$ and integrating again, with zero integration constant, givesHere ${\alpha }_2>0$, because ${\gamma }_0>{\gamma }_1>0$ and $R<R+1$. A smooth nonzero wave approaching zero requires $c>0$: for $c<0$ the right-hand side is negative at sufficiently small nonzero $A$. At $c=0$ the allowed positive branch has no nonzero turning point, so it cannot form a smooth localized pulse.

When $c>0$, the nonzero turning point is $A_m=-3\beta {\alpha }_{1}c/{\alpha }_2$. The energy curve is positive between $A_m$ and zero and negative below $A_m$, giving a homoclinic negative pulse. The positive branch has no turning point and cannot return to zero. ThereforeAn explicit solution, also confirming smooth exponential tails, isThis establishes the opposite propagation directions in the annular displacement equation: linear waves move downstream while its localized nonlinear depression waves move upstream.

Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

The unstressed shape has equal radii at $x=\pm L$, so its contributions cancel when subtracting the elastic end pressures. The pressure difference is thereforeThe multiplier is a pressure scale: $\alpha$ has dimensions pressure per length and $2Q/U$ is a length.

In the narrowest part of the film, $H=O(1)$, so $h\sim s=2Q/U$. Thereforeis exactly the thin-annulus requirement needed to replace the local circumference by $2\pi a$ and neglect curvature across the film. The axial lubrication scale is $\ell =\sqrt{s/\kappa }$, so the small-slope assumption additionally requires $s/\ell =\sqrt{\kappa s}\ll 1$. The assumed negligible inertia imposes its corresponding small inertial-to-viscous ratio, of order $\rho U{s}^2/(\mu \ell )\ll 1$. Thus small $C$ is necessary, while slope and inertia conditions are independent parts of self-consistency. If the lubrication approximation is required all the way to large scaled endpoints, the leading parabolic gap further requires $C{\tilde{L}}^2\ll 1$ and $\sqrt{\kappa s}\tilde{L}\ll 1$. Alternatively the thin neck can be matched to a separate outer solution; its pressure contribution must then be assessed.

Take the small-$\lambda$ and large-$\tilde{L}$ limits at fixed $C$ first, then expand in small $C$. Seek $H=H_0+\lambda H_1+\cdots$. The leading equation is $H_{0X}=X$, soAt the next order,For large $\tilde{L}$, extend the integrals determining the localized film resistance to the whole line. The even leading profile has equal end values, so the force condition at order $\lambda$ becomesThe left side follows from $H_1(-\tilde{L})-H_1(\tilde{L})=-\int H_{1X}dX$. Using the supplied integrals gives $I_1=2\pi /b$, $I_2=2\pi /b^3$, $I_3=3\pi /b^5$. HenceThis fixes the integration constant in $H_0$. In terms of $z=b^{-2}$ it reads $2z-3z^2=C(8/3-2z)$. The root near $z=2/3$ givesThe correction has a minus sign. The printed $1+C$ is inconsistent with the supplied differential equation and force condition: substituting $z=(2/3)(1+C)$ gives $2z-3z^2=-(4/3)C+O(C^2)$, whereas $C(8/3-2z)=(4/3)C+O(C^2)$. The discrepancy is not a choice of flux sign, because the positive-gap scaling here requires $s=2Q/U>0$. The leading value $b^{-2}=2/3$ is unaffected by this error.

Use the force integral to calculate the leading pressure drop, retaining its first nonzero order in both $C$ and $\lambda$:At $b^2=3/2$, the bracket is $4\pi /(3b)$. With the definitions of $C$ and $\lambda$ this gives the small-speed compliant-plug lubrication asymptoticsThus the requested leading pressure-drop expression remains valid despite the sign error in its subsidiary correction.

Finally, the absolute downstream pressure selects $s$ and hence $Q/U$. At the downstream matching region the leading shape givesbecause $s{X}^2/2$ cancels the unstressed parabolic term. For example, if the downstream excess pressure is fixed andthen $s\simeq 2d/b^2\simeq 4d/3$, andThis is a linear pressure-speed law on the specified small-$\lambda$ branch. If $p_d-p_0$ depends on speed or cell state, its relation must instead be substituted to obtain the corresponding law. If $d\le 0$, the stated leading positive-gap branch does not exist. Without the absolute-pressure relation, the problem gives a family parameterized by $Q/U$, rather than a unique $\Delta p(U)$. Errors from finite $\tilde{L}$, higher powers of $\lambda$ and higher powers of $C$ have been omitted consistently in this leading approximation. Resolving the displayed $O(C)$ correction to $b^{-2}$ requires the omitted finite-domain and higher-$\lambda$ corrections to be smaller than $C$; otherwise only its leading value is justified.