Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 75 / 3

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.