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For the given harmonic potentials, ${\nabla }^2(\mathbf{\Phi }\cdot \mathbf{x})=2\nabla \cdot \mathbf{\Phi }$. Taking the divergence of the velocity representation therefore gives $\nabla \cdot \mathbf{u}=0$. Taking its Laplacian gives ${\nabla }^2\mathbf{u}=-2\nabla (\nabla \cdot \mathbf{\Phi })$, so the Stokes equations hold withThis is the Papkovich–Neuber representation in the sign and normalization of this problem. The potentials used in other conventions can differ by an overall factor and sign; the velocity and pressure must be scaled together.

(a) Point force. Put the singularity at the origin, $r=|\mathbf{x}|$, and choose $\mathbf{\Phi }=\mathbf{F}/(8\pi \mu r)$, $\chi =0$. The resulting Stokeslet isThe harmonic calculation applies away from the force. Distributionally this is the Green function for $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}{\delta }^{(3)}(\mathbf{x})=0$, with zero divergence.

(b) Point source. Set $\mathbf{\Phi }=0$, $\chi =-Q/(4\pi r)$. The three-dimensional point source field isIts outward volume flux across every enclosing sphere is $Q$. The divergence is zero away from the source and equals $Q{\delta }^{(3)}(\mathbf{x})$ distributionally, so the singular point represents volume injection rather than an incompressible material point.

(c) Stresslet. Let $G_{ij}(\mathbf{x})=({\delta }_{ij}/r+x_i{x}_j/r^3)/(8\pi \mu )$ and use a symmetric trace-free stresslet tensor $S$ with the convention $u_i=-S_{jk}{\partial }_k{G}_{ij}$. Differentiating the Stokeslet and using symmetry and zero trace givesFor an axial force-dipole flow with axis $\mathbf{n}$, write $S=s(\mathbf{nn}-I/3)$. Then the radial factor is proportional to $s[3(\mathbf{n}\cdot \hat{\mathbf{x}}{)}^2-1]$. Specifying the derivative convention is essential for interpreting the signs of image strengths.

(d) Source dipole. A positive source and negative source separated by a vector $\mathbf{\ell }$ have moment $\mathbf{M}=Q\mathbf{\ell }$. Their dipole limit is minus the directional derivative of the unit source field. Hence the source dipole hasIt carries no net volume flux, and unlike a stresslet it is an irrotational potential flow.

Now set $\mathbf{r}=\mathbf{x}-d\mathbf{n}$ and $\mathbf{R}=\mathbf{x}+d\mathbf{n}$, with $d>0$. Define the unscaled kernels $K(\mathbf{R})=\mathbf{n}/R+(\mathbf{n}\cdot \mathbf{R})\mathbf{R}/R^3$, $H(\mathbf{R})=\mathbf{R}/R^3$, and let ${\partial }_n=\mathbf{n}\cdot {\nabla }_{\mathbf{R}}$. The no-slip image system of a normal Stokeslet isThe reflected Stokeslet has force $-F\mathbf{n}$. With the conventions above, the remaining image strengths areThe isotropic part of $S$ produces zero velocity, because ${\partial }_j{G}_{ij}=0$. Its inclusion simply puts the tensor in standard trace-free form. The pressure follows by superposition:

To verify the no-slip boundary condition explicitly, write $\mathbf{x}=\mathbf{\rho }$ on the plane, with $\mathbf{\rho }\cdot \mathbf{n}=0$, and let $b^2={\rho }^2+d^2$. ThenThe $b^{-5}$ terms cancel, while the remaining image contribution is $2d(\mathbf{R}-d\mathbf{n})/b^3=2d\mathbf{\rho }/b^3$. Thus both normal and tangential velocity vanish. All image singularities lie outside the fluid, so the physical half-space contains only the prescribed force.

For the far field, expand about the wall origin. The pair of opposite forces starts with $-2d{\partial }_{n}K$; the image stresslet cancels this entire $r^{-2}$ term. The first surviving field isWriting $z=\mathbf{x}\cdot \mathbf{n}$ and now $r=|\mathbf{x}|$, this isTherefore the generic fixed-angle far-field velocity decays as $r^{-3}$, rather than the $r^{-1}$ free-space Stokeslet or an uncancelled $r^{-2}$ stresslet. Along the wall it is exactly zero; at fixed height and large horizontal distance the leading tangential field instead scales as ${\rho }^{-4}$, consistent with the angular zeros of the displayed far field.

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.

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.

Write the perturbation Stokes streamfunction as $\psi =\varphi (r)e^{ikz-i\omega t}$, with $\omega =kc$ and nonzero $k$. Its radial and axial velocity amplitudes are $u=-ik\varphi /r$ and $w={\varphi }^{\prime }/r$. These automatically obey incompressibility. Linearizing the radial and axial Euler equations, with pressure amplitude $P$, givesThusDifferentiating the first expression, the terms involving $U^{\prime }{\varphi }^{\prime }$ cancel. Equating the two expressions for $P^{\prime }$ proves the axisymmetric inviscid pipe stability equation:At the axis, finite smooth axisymmetric velocity requires $\varphi =O(r^2)$, after choosing the irrelevant constant in the streamfunction. At the wall, impermeability gives $\varphi (a)=0$. There is no tangential no-slip condition in this inviscid problem. A decoupled azimuthal disturbance satisfies $ik(U-c)u_{\theta }=0$ and does not introduce a growing discrete mode when $U\ne c$.

For real $k$, suppose $c=c_r+i{c}_i$ with $c_i\ne 0$, so division by $U-c$ is legitimate. Multiply the ordinary differential equation by ${\varphi }^{\ast }/[r(U-c)]$, integrate, and apply integration by parts. The endpoint terms vanish by the preceding regularity and wall conditions, leavingTaking its imaginary part yieldsA growing temporal mode therefore requires a sign change of $(U^{\prime }/r{)}^{\prime }$, rather than a sign change of $U^{\prime \prime }$ alone. This is the cylindrical counterpart of Rayleigh's inflection-point theorem. If this derivative vanishes identically, the positive first integral rules out a nontrivial growing real-wavenumber mode as well. Isolated zeros without a sign change do not make the second integral vanish for a nonzero regular mode.

For the parabolic Poiseuille flow, $U^{\prime }/r=-2U_0/a^2$ is constant. Under the specified exclusion of critical levels, division givesSetting $\varphi =rf$, or equivalently choosing $n=1$ in $r^{-n}\varphi$, turns this intoIt is the Modified Bessel differential equation of order one. Hence $\varphi =r[A{I}_1(kr)+B{K}_1(kr)]$. Since $r{K}_1(kr)$ tends to a nonzero constant at the axis, it gives a singular radial velocity and must be excluded. The admissible mode shape is thereforeFor $k=0$, direct solution gives $\varphi =C{r}^2$ after axis regularity, and the wall condition again forces the trivial solution. For nonzero $k$, the nonzero roots of $I_1$ are imaginary. Writing $j_{1,l}>0$ for the zeros of the ordinary Bessel function $J_1$, the evanescent potential modes of inviscid Poiseuille flow areTheir perturbation vorticity vanishes: ${\partial }_{z}u-{\partial }_{r}w=k^2\varphi /r-({\varphi }^{\prime }/r{)}^{\prime }=0$. These modes grow exponentially in one axial direction and decay in the other. They can describe end-forced or spatially localized potential fields, but neither sign gives a bounded normal mode on a whole infinite pipe. Moreover, the reduced radial equation contains no $c$: it supplies no temporal eigenvalue condition. An arbitrary imaginary part assigned to $\omega =kc$ here is consequently not evidence for a physical temporal instability.

The real-wavenumber calculation does exclude growing discrete axisymmetric modes of this inviscid profile. It does not establish every form of stability: singular neutral disturbances with $c$ in the range of $U$, transient dynamics, non-axisymmetric disturbances, and finite-amplitude transition require separate treatment. The exclusion $U\ne c$ specifically removes the inviscid continuous spectrum.

With viscosity, linearize the Navier-Stokes equations instead and eliminate pressure to obtain a fourth-order cylindrical counterpart of the Orr-Sommerfeld equation. Impose both $\varphi (a)=0$ and ${\varphi }^{\prime }(a)=0$, with appropriate regularity at the axis. The resulting eigenvalue problem involves the Reynolds number and determines complex frequencies, rather than leaving $c$ arbitrary. Viscosity supplies diffusion, no-slip boundary layers and critical-layer regularization, so the inviscid sign argument alone no longer decides its spectrum. That observation does not imply that viscous pipe flow must have a growing linear eigenmode; finite-amplitude pipe transition is a different question.

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.

Introduce the slow time $T=ϵt$ and expand $y=y_0(t,T)+ϵy_1(t,T)+\cdots$. At leading order writeThe first-order equation isIts resonant sine and cosine components must vanish by the solvability condition in the method of multiple scales. Let angle brackets denote a full-period average. The cosine projection of the nonlinear term is zero, since $⟨{\sin }^n\psi \cos \psi ⟩=0$. The sine projection givesThis is the averaged oscillator damping by a power of velocity.

If $n$ is odd, defineThen $R_T=-C_n{R}^n$, givingFor every fixed $n$, these leading multiple-scale expansions are uniform on bounded slow-time intervals $0\le ϵt\le T_0$. The unforced phase remains $t$ to this order. A bounded first-order correction supplies the small adjustment to the initial derivative and initial displacement; it does not change this leading result.

If $n$ is even, the average of the odd power ${\sin }^{n+1}\psi$ vanishes. Thus $R_T=0$ andThe first-order forcing has only a constant and even harmonics, so it produces no fundamental resonance. Its bounded particular solution can be supplemented by homogeneous sine and cosine terms to satisfy the initial conditions. There can be an $O({ϵ}^2)$ frequency correction, but its phase accumulation is only $O(ϵ)$ on the present time scale. In contrast to the odd case, the even-power force is invariant under velocity reversal and the equation is reversible; calling it positive damping for both signs of velocity would be wrong. The mechanical energy identity makes the distinction clear:This is nonpositive for odd $n$ and changes sign for even $n$; its cycle average reproduces the amplitude equation above.

For large odd $n$, the central binomial coefficient givesThis form reveals several time regimes. For $L_nϵt\ll 1$, $R=1-C_nϵt+\cdots$; at $t=O(1/(ϵ\sqrt{n}))$, the amplitude change is only $O(1/n)$. For $L_nϵt\gg 1$ but $\log (L_nϵt)\ll n$, the leading change is $R\simeq 1-\log (L_nϵt)/n$. In particular, at $t=O(1/ϵ)$ it is only $O(\log n/n)$, despite the long elapsed time. High powers damp mainly near velocity maxima, and even a small amplitude reduction strongly suppresses subsequent damping.

An order-one amplitude reduction in the averaged law requires $\log (L_nϵt)=O(n)$. Formally, $t=e^{(n-1)\sigma }/(ϵL_n)$ gives $R\sim e^{-\sigma }$, and for fixed odd $n>1$ its late envelope is an algebraic power $t^{-1/(n-1)}$. These exponential-in-$n$ time scales are predictions of extending the averaged law; the fixed-$n$, bounded-$ϵt$ multiple-scale error estimate alone does not establish uniform accuracy that far. A joint large-$n$, small-$ϵ$ approximation also needs control of the increasingly sharp nonlinear peaks, rather than assuming fixed-$n$ error constants are uniform. Taking $n$ fixed first and then examining the large-$n$ envelope avoids that unsupported interchange of limits.

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.

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.