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The printed diagram and the word “parallel” describe a Kelvin-Voigt model, not a series Maxwell fluid. In the depicted model both elements have displacement $x$ and their forces add:For $x=a{e}^{i\omega t}$ and the convention $G(\omega )=F_{\mathrm{amplitude}}/x_{\mathrm{amplitude}}$,The storage modulus and loss modulus here are element stiffnesses with units $\mathrm{N}{\mathrm{m}}^{-1}$; conversion to stress/strain moduli requires a geometric factor. If “response” instead denotes displacement per force, it is the compliance $1/(k+i\omega \mu )$. Specifying which ratio is used removes that convention ambiguity. The mean dissipated power is $\mu {\omega }^2|a{|}^2/2$, confirming the positive loss sign for $e^{i\omega t}$.

For completeness, the actual series Maxwell fluid analogue has a common force and additive spring/dashpot extensions. Thus $\dot{x}=\dot{F}/k+F/\mu$. With $\tau =\mu /k$,These different functions cannot both describe the printed parallel arrangement. The distinction also changes the elastic beam's response to a constant load.

Model the polymerization Brownian ratchet by a rigid tip and a Brownian load. Let $a$ be the elementary load advance, ${\zeta }_L$ its drag coefficient, and $D=k_{B}T/{\zeta }_L$ its diffusion coefficient. For a spherical load, ${\zeta }_L=6\pi \mu R_L$, but no load radius is specified, so the rate is naturally expressed in terms of $D$. Assume addition is effectively instantaneous once the gap $h$ reaches $a$, and removal is negligible. The gap reflects at zero, first arrival at $a$ causes insertion, and insertion resets it to zero.

With no opposing force, the mean first-passage time satisfies $D{T}^{\prime \prime }=-1$, $T^{\prime }(0)=0$, $T(a)=0$. Integration gives $T(h)=(a^2-h^2)/(2D)$. One advance per passage givesThe advance-event rate is $2D/a^2$, while $V_0$ is a length per time.

A constant resisting force $F\ge 0$ produces drift $-u_F=-DF/(k_{B}T)$. The diffusion-limited Brownian ratchet with constant load now satisfiesAn integrating factor gives $T^{\prime }(h)=[1-e^{u_{F}h/D}]/u_F$. Integrating again from $h$ to $a$ gives $T(0)=D(e^z-1-z)/u_F^2$, where $z=Fa/(k_{B}T)$. ThereforeAt weak load the ratio is $1-z/3+O(z^2)$; at large load $V\sim (D/a)z^2{e}^{-z}$. Replacing the whole force dependence by $e^{-z}$ would discard the first-passage kinetics of this regime.

For two staggered strands that alternately advance the leading tip by half a monomer, $a=\delta /2$ and $V_0=4D/\delta$. Two aligned strands, as schematically drawn, do not automatically have that half-step: an advance may be $a=\delta$, with the other strand filling the same layer afterward. Use the same geometrical step in the passage problem and in converting event rate to velocity. The limiting formula assumes negligible removal, rigid tips, an overdamped load with constant drag, and fast insertion compared with $a^2/D$. With appreciable removal or additional internal strand states, the renewal reset model must be replaced by a coupled gap/reaction process.

Write $u$ for the effective unloaded addition rate and $w$ for the removal rate of the same elementary tip step $a$. If the supplied on-rate is a bimolecular rate constant, then $u=k_{\mathrm{on}}{c}_m$, with monomer concentration $c_m$; if it already denotes an event rate, use $u=k_{\mathrm{on}}$. Assume gap diffusion is much faster than chemical events, removal is force-independent, and the monomer bath is maintained.

Under a positive constant resisting force, the load equilibrates in the potential $Fh$ above a reflecting tip. The normalized Boltzmann distribution is $p(h)=\beta F{e}^{-\beta Fh}$ for $h\ge 0$, where $\beta =1/(k_{B}T)$. The fraction of time with room to insert isSteric gating reduces addition to $u{e}^{-\beta Fa}$; each addition advances by $a$ and each removal retreats by $a$. The reaction-limited polymerization Brownian ratchet therefore hasFor $u>w>0$, the stall force is $F_s=(k_{B}T/a)\log (u/w)$. Equivalently, the available chemical free energy per effective step is $\Delta \mu =k_{B}T\log (u/w)$, and stall balances $F_{s}a=\Delta \mu$. A half-monomer step requires rates defined for that half-step, not automatically single-strand monomer rates with a gratuitous factor of two.

At exactly zero force a load on an unbounded half-line has no normalizable equilibrium gap density. The unloaded expression is the $F\downarrow 0$ limit, or the usual freely available-gap approximation with a weak positional confinement. For a positive load the equilibration time must remain short compared with reaction times. Load inertia, filament bending, monomer depletion, and force-dependent bond rupture have been excluded. For arbitrary finite reaction and diffusion rates one must solve the joint gap dynamics rather than attach $-aw$ to the diffusion-limited first-passage formula: a removal changes the gap and the distribution from which the next addition starts.

Let $\mathbf{n}$ point from fluid $-$ to fluid $+$, $P_s=I-\mathbf{nn}$, ${\nabla }_s=P_s\nabla$, and $\mathcal{K}={\nabla }_s\cdot \mathbf{n}$. A clean material interface without phase change has continuous velocity, with normal interface speed $V_n={\mathbf{u}}_{\pm }\cdot \mathbf{n}$. Its surface stress is $\gamma P_s$. The surface divergence of that tensor is ${\nabla }_s\gamma -\gamma \mathcal{K}\mathbf{n}$. Balancing forces on a small interfacial patch therefore gives the interfacial stress balance with variable surface tensionThe normal condition is $\mathbf{n}\cdot ({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=\gamma \mathcal{K}$; the tangential condition is $P_s({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=-{\nabla }_s\gamma$. On a stationary sphere with outward normal these reduce to $p_{-}-p_{+}=2\gamma /a$ when viscous normal stresses vanish. This fixes the sign convention for the Marangoni stress rather than leaving the direction ambiguous.

For the bubble take negligible internal dynamic viscosity, no contaminants, and initially the standard effectively insulating gas-bubble approximation. With heat advection neglected, the exterior temperature satisfies Laplace's equation, approaches $T_0+\mathbf{H}\cdot \mathbf{r}$ remotely, and has zero normal heat flux at $r=a$. The temperature dipole around an insulating sphere givesThe zero-flux condition follows by differentiating the radial factor. Let ${\gamma }_T=(d\gamma /dT{)}_{T_0}$. To first order in the weak gradient,

In the laboratory frame the force-free bubble can be described by the harmonic Papkovich–Neuber representation potentialsThey give constant exterior pressure and the decaying source dipole fieldAt the surface $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so the normal kinematic boundary condition holds for translation. In coordinates aligned with $\mathbf{U}$, $u_r=a^{3}U\cos \theta /r^3$ and $u_{\theta }=a^{3}U\sin \theta /(2r^3)$. Direct differentiation givesEquivalently, the surface tangential traction is $(3\mu /a)P_s\mathbf{U}$. The tangential stress condition therefore requiresFor the usual negative ${\gamma }_T$, the bubble moves toward hotter liquid. The normal condition is consistent as well: its viscous dipole is $-6\mu \mathbf{U}\cdot \mathbf{n}/a=3{\gamma }_T\mathbf{H}\cdot \mathbf{n}$, exactly the dipolar part of $2{\gamma }_s/a$. The constant part fixes $p_{\mathrm{gas}}-p_{\infty }=2{\gamma }_0/a$. There is no Stokeslet term, hence no net applied force or drag imbalance.

The thermal assumption matters. The problem gives the exterior thermal diffusivity but no gas thermal conductivity. For an inviscid bubble with conductivity ratio $K=k_{\mathrm{gas}}/k_{\mathrm{liquid}}$, temperature and normal heat-flux continuity instead giveso the same stress calculation givesThe insulating-bubble result uses $K\ll 1$. If instead an undisturbed surface temperature gradient is imposed, corresponding to $K=1$, the coefficient is $1/3$ rather than $1/2$. Exterior thermal diffusivity alone does not select between those thermal boundary conditions. This dependence is explicit rather than silently treating a missing gas property as a prescribed value.

The small parameters can now be stated precisely. With $U_{\ast }=a|{\gamma }_T||\mathbf{H}|/(2\mu )$ and $\nu =\mu /\rho$, requireSmall Reynolds number justifies Stokes flow; small Péclet number allows quasi-static heat conduction without advective distortion; small capillary number and fractional surface-tension variation support a nearly spherical interface and a weak perturbation of its base surface tension. These last two conditions are proportional for this velocity scale but describe different approximations. Linearizing $\gamma (T)$ also needs $a|\mathbf{H}||{\gamma }_{TT}/{\gamma }_T|\ll 1$ if its derivative varies appreciably. Negligible internal viscosity and the insulating specialization require the corresponding gas-to-liquid viscosity and conductivity ratios to be small. Gravity has been excluded; if restored, its buoyancy-induced migration must be small compared with $U_{\ast }$, not merely small enough to avoid deformation.

Finally, surfactants change both the constitutive surface tension and the surface stress dynamics. The surface flow driven by the thermal Marangoni effect carries insoluble surfactant toward the high-tension side, lowering its tension and producing an opposing concentration-driven gradient. When surface diffusion or exchange is slow, this tends to immobilize the interface, suppressing the clean-bubble thermocapillary migration of an insulating bubble, sometimes nearly stopping it. If the surfactant stays uniformly equilibrated, it can instead change ${\gamma }_T$ without a large opposing gradient. Its concentration, surface diffusivity and adsorption kinetics must be supplied to compute a replacement speed; no universal numerical reduction factor follows merely from its presence.

Put $\tau =m/\zeta$. The causal Green function of the displacement equation is $G(t)={\zeta }^{-1}(1-e^{-t/\tau })$ for $t>0$ and zero for $t<0$. Its derivative has the jump needed to give $m{G}^{\prime \prime }+\zeta G^{\prime }=\delta (t)$. For the specified zero position and velocity,For Gaussian white noise these are stochastic convolutions; their first two moments follow from the stated covariance even without assuming Gaussian higher moments. In particular $⟨x⟩=⟨\dot{x}⟩=0$, andThe required equipartition theorem limit therefore fixes the fluctuation-dissipation relation for a Langevin particle:Here $k_B$ denotes the printed Boltzmann constant $k$. Set $D=k_{B}T/\zeta$, the Stokes–Einstein relation. Squaring the displacement kernel givesConsequently ${\lim }_{t\to \infty }⟨x⟩=0$ and ${\lim }_{t\to \infty }⟨x^2⟩=+\infty$, with $⟨x^2⟩=2Dt-3D\tau +o(1)$ for positive temperature. There is no finite equilibrium variance of position in an unconfined fluid. The inertial Brownian displacement with zero initial velocity has short-time variance $2D{t}^3/(3{\tau }^2)+O(t^4)$, rather than the ballistic law associated with an initially thermalized velocity.

For constant $D$, the diffusion equation has fundamental density $P(x,t)=(4\pi Dt{)}^{-1/2}{e}^{-x^2/(4Dt)}$, whose mean is zero and variance is $2Dt$. It describes the large-time, overdamped limit of the Langevin dynamics. Under the usual Gaussian-force assumption, the finite-time position density is Gaussian with the variance above and obeys a time-dependent diffusion equation with coefficient $D(1-e^{-t/\tau }{)}^2$. It does not obey the constant-$D$ equation through the initial inertial transient. The given covariance alone determines the moments, not a Gaussian law; the density statement uses the standard Gaussian thermal-noise model.

Now write $\zeta (x)={\zeta }_0+\epsilon {\zeta }_0^{\prime }x$ and $x=x_0+\epsilon x_1+O({\epsilon }^2)$, with ${\tau }_0=m/{\zeta }_0$ and $D_0=k_{B}T/{\zeta }_0$. For a fixed realization of the additive force, the first-order equation isIf the force amplitude is instead adjusted to local thermal equilibrium, its first-order noise term has zero mean in the inertial Itô formulation and does not change the following first-order mean calculation. The mixed unperturbed moment isLet $M(t)=⟨x_1(t)⟩$. Averaging and solving the scalar forced relaxation equation, with $M(0)=M^{\prime }(0)=0$, givesThus the long-time first-order coefficient and local mean drift areLiterally $\lim M=-\infty$ for ${\zeta }_0^{\prime }>0$, $+\infty$ for ${\zeta }_0^{\prime }<0$, and zero for ${\zeta }_0^{\prime }=0$; the useful finite quantity is $\lim M^{\prime }$. Because the viscosity gradient is only a local expansion, the physical drift result applies in the intermediate window ${\tau }_0\ll t\ll L_{\zeta }^2/D_0$, where $L_{\zeta }={\zeta }_0/(\epsilon |{\zeta }_0^{\prime }|)$. Extrapolating the linear drag law to arbitrarily large displacements could make the drag negative and would invalidate the model.

The isothermal diffusion with position-dependent drag interpretation uses locally matched thermal noise and $D(x)=k_{B}T/\zeta (x)$. Integrating the printed Fickian diffusion equation against $x$, with vanishing boundary fluxes, givesSince $D^{\prime }(x)=-\epsilon k_{B}T{\zeta }_0^{\prime }/{\zeta }_0^2+O({\epsilon }^2)$, this reproduces the same drift after the inertial transient. Equivalently the overdamped Itô diffusion has drift $D^{\prime }$ and noise amplitude $\sqrt{2D}$. Omitting that drift would give $P_t=(DP{)}_{xx}$, which is a different Fokker-Planck equation and not the stated Fickian equation.

Noise conventions cannot be mixed silently. If the force covariance is held at the spatially constant value $F_0=2{\zeta }_0{k}_{B}T$ while drag varies, the local overdamped diffusion coefficient is instead $\tilde{D}=F_0/(2{\zeta }^2)$ and its Itô drift is ${\tilde{D}}^{\prime }/2$. Its density equation is $P_t={\partial }_x[\tilde{D}{P}_x+({\tilde{D}}^{\prime }/2)P]$, not the isothermal Fickian equation. This still gives the same displayed first-order mean drift at $x=0$, but has different spatial diffusion and equilibrium properties. The stated Fickian interpretation therefore requires local thermal-noise matching.

Physically, thermally driven excursions travel farther on the lower-drag side. Their displacement is correlated with velocity, so averaging the position-dependent friction leaves a net drift toward greater diffusivity, without an applied force. This is not a violation of equilibrium: in the isothermal Fickian model the drift and spatially varying noise cancel in the probability flux for a uniform density. It is a local diffusion-induced shift of the centroid of an initially localized particle distribution.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

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.

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.