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Let $a(t)=a_0+a_1(t)$ and use $s=|y|$ for the source radius, reserving $r$ for observation distance. Incompressible spherical potential flow satisfies $s^2{u}_s=a^2\dot{a}$. Its velocity potential, tending to zero at infinity, and velocity areThe unsteady Bernoulli equation givesThe surface pressure is independent of angle. Hence the integrated surface force is zero, $f=p(a,t){\int }_{S}ndS=0$, and the leading compact surface dipole contribution vanishes. This also makes $\mathcal{F}=0$ in the notation of the preceding part.

The volume flux is $\mathcal{V}=4\pi a^2\dot{a}$, so the acoustic thickness noise contribution isThe leading part is linear in the radial oscillation. Keeping the quadratic terms explicitly gives ${\rho }_0[a_0^2{\ddot{a}}_1+a_0\frac{d^2}{d{t}^2}(a_1^2)]/(c_0^{2}r)$.

For the acoustic quadrupole, use $T_{ij}\simeq {\rho }_0{u}_i{u}_j$. To leading quadratic order,ThusKeeping the instantaneous lower limit $s=a(t)$ instead gives $S_{ij}=4\pi {\rho }_0{a}^3{\dot{a}}^2{\delta }_{ij}/3$, whose leading small-oscillation term is the expression just obtained. Although the stress moment is isotropic, its contracted second derivative is not zero: $n_i{n}_j{\delta }_{ij}=1$.

For a sinusoidal radius perturbation $a_1=A\cos \omega t$, the leading compact radiation moments of a pulsating spherical bubble areWriting $\delta =A/a_0$ and $\alpha =\omega a_0/c_0$, the quadrupole amplitude relative to the linear monopole is $2\delta {\alpha }^2/3$. The monopole also has a quadratic correction, $-2{\rho }_0{a}_0{A}^2{\omega }^2\cos 2\omega \tau /(c_0^{2}r)$; the volume-stress quadrupole is smaller than this quadratic monopole correction by ${\alpha }^2/3$.

In the small-amplitude compact regime, the linear monopole dominates, the leading dipole is zero, and the volume quadrupole is weaker and begins at twice the oscillation frequency. The compact formulas require $\alpha \ll 1$ as well as a far observer. The stated small surface Mach number $|{\dot{a}}_1|/c_0\sim \delta \alpha \ll 1$ alone does not imply compactness. The incompressible near field has a decaying tail, but its stress moment is dominated by distances of order $a_0$; an acoustic matching region is still needed outside it. These are leading compact contributions, not a claim to retain every finite-wavelength correction at the smaller quadrupole order.

Interpret the surface delta as ${\delta }_s=|\nabla F|\delta (F)$, so integration against it gives a surface integral. If $F$ is signed distance, this equals the notation $\delta (F)$ directly. This convention is needed because an arbitrary rescaling of the level-set function must not change the physical source.

Define the volume flux $\mathcal{V}(t)={\int }_S{V}_{n}dS$ and total pressure force $f(t)={\int }_{S}pndS$. Here $\mathcal{V}$ is the rate of change of body volume, not the volume itself. The surface acoustic monopole contributes, in the acoustic compact-source approximation,The surface acoustic dipole contributesThe second term is a nonradiating near-field term and is smaller in the acoustic far field. Writing $\mathcal{F}(t)=x\cdot f(t)$ with the observation point fixed, the leading augmentation is thereforeThe printed dipole coefficient is missing a factor $1/c_0$. The extra inverse speed follows from differentiating $\tau =t-r/c_0$, and is required dimensionally: the printed expression with $c_0^2$ has the units of density times speed. This corrected acoustic loading noise term also agrees with the far-field acoustic force and stress moments. The monopole coefficient is unchanged.

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.

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.

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.

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.