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An intermediate asymptotic region has a scale between two other scales and provides their connecting matched asymptotic expansion. It can be needed because a coefficient changes its dominant form before the narrowest region's highest-derivative balance is reached. For example, a coefficient $x+ϵ$ changes near $x=O(ϵ)$ even when the fast inner expansion has width $O({ϵ}^2)$. The intermediate solution must be matched at both ends rather than assigned an independent boundary value.

For a travelling wave of large positive speed in the cylindrical film equation, the leading core is proportional to $1-\cos x$ over an axial interval of length $2\pi$. Edge coordinates $X=(3c{)}^{1/3}(x-x_e)$ reduce the transition equation to $H^3{H}_{XXX}=H-1$. Its real positive exponent gives a monotone trailing approach and its complex stable exponents give damped capillary waves ahead, with physical wavelength $4\pi a/(3^{5/6}{c}^{1/3})$. Matching the two quadratic edge expansions selects the uniform film thickness; this is a matched asymptotic expansion, valid while the pulse itself remains thin relative to the cylinder.

A matched asymptotic expansion constructs approximations in regions with different distinguished limits and fixes their free constants by requiring agreement in an overlap region. The inner expansion resolves rapid variation; the outer expansion describes the surrounding region. Matching compares their common limiting behavior, rather than imposing distant boundary conditions directly on a local approximation.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the leading outer solution is $y_0=1$. The inner variable $X=x/ϵ$ gives $Y_0^{\prime \prime }+Y_0^{\prime }=0$ and $Y_0(0)=0$, hence $Y_0=A(1-e^{-X})$. In the overlap region $ϵ\ll x\ll 1$, the inner variable tends to infinity while the outer coordinate tends to zero. The asymptotic matching condition isso $A=1$. One concrete path through the overlap region is $x=\sqrt{ϵ}$: then $x\to 0$ and $X={ϵ}^{-1/2}\to \infty$. Matching selects the free constant that the left boundary condition alone could not determine.

Why choose the right boundary condition for the outer solution? If instead one selects $y_0=0$ from the left, a proposed right boundary layer with $Z=(1-x)/ϵ$ satisfies $U_0^{\prime \prime }-U_0^{\prime }=0$. Its general solution is $A+B{e}^Z$. Matching to zero as $Z\to \infty$ forces $A=B=0$, contradicting $U_0(0)=1$. The direction in which the exponential function decays determines which endpoint can carry the boundary layer; endpoint selection is part of the calculation.

The outer solution satisfies $y(x)[x+y(x)]=0$, so on a connected interior interval the nonoscillatory branches are $y=0$ and $y=-x$. Both happen to solve the differential equation exactly away from any endpoint mismatch. For a fixed positive endpoint coordinate $q$ ($q=a$ or $q=1$), introduce the inward distance $\xi =(x-a)/\sqrt{ϵ}$ on the left or $\xi =(1-x)/\sqrt{ϵ}$ on the right. The leading nonlinear endpoint layer with a quadratic reaction obeysMatching to zero fixes $E=0$, so $(Y^{\prime }{)}^2=-q{Y}^2-2Y^3/3$. It is negative for every sufficiently small nonzero $Y$. Thus no real nonconstant layer can approach zero: the zero outer branch requires $\alpha =\beta =0$, and then $y=0$ is an exact solution.

Matching to $-q$ fixes $E=q^3/6$. The first integral factors asConsequently an endpoint value $\eta$ can match this branch only if $\eta \le q/2$. For $-q<\eta <q/2$, there are two choices: a direct approach to $-q$, and an initial excursion to $q/2$ followed by a return to $-q$. Both are translations ofThe negative translation gives the direct layer; the positive translation gives the excursion. At $\eta =q/2$ the two coincide, with the maximum at the wall. For $\eta <-q$, the unique admissible direction is a monotone approach from below:where ${\xi }_0$ is fixed by $Y(0)=\eta$. At $\eta =-q$ only the constant leading layer matches at a finite inward coordinate; a pulse translated infinitely far from the wall is outside this endpoint-layer classification.

Thus the leading admissibility condition for the negative outer branch is $\alpha \le a/2$ and $\beta \le 1/2$. Under strict inequalities, each endpoint has two choices when its datum lies between $-q$ and $q/2$, and one choice below $-q$. Their independent combinations give one, two or four formal matched profiles. In particular, for $\alpha =\beta =0$ there are four negative-bulk profiles in addition to the exact zero solution, so uniqueness certainly cannot be inferred from the outer equation. The endpoint equalities and the constant-layer cases are degenerate; higher-order endpoint matching is needed to decide exact finite-$ϵ$ counts at these cutoffs. These conditions and counts concern leading families under the stated exclusion of interior layers and rapid oscillations.

A matched asymptotic expansion for the negative branch isThe following sketch shows the four choices for zero endpoint data, together with the exact zero solution. It plots leading composite profiles, not numerical claims of exact finite-$ϵ$ multiplicity.

Additional interior layers cannot be excluded merely by counting the outer roots. Freezing $x=q$ gives a homoclinic orbit to $-q$ with the same hyperbolic-secant pulse. Its location would require a further matching or solvability condition; it is not an arbitrary parameter in the original variable-coefficient problem. Indeed, put $z=y+x$. The exact equation and an energy identity areAn isolated order-one pulse with exponentially small tails on both sides would decrease this energy by an algebraic amount, whereas its two tail energies would be exponentially small. This obstructs a freely standing isolated pulse well away from the walls. It does not by itself classify oscillatory solutions or pulse interactions with endpoint layers, so a complete global uniqueness claim would exceed this leading construction.

When $a=\alpha =0$, the two outer roots meet at the left endpoint. Let $x={ϵ}^{p}X$, $y={ϵ}^{q}Y$. Balancing $xy$ with $y^2$ gives $p=q$, and balancing $ϵy^{\prime \prime }$ with them gives $1-q=2q$. The one-third-power scaling at a nonlinear turning endpoint is thereforeIts left condition is $Y(0)=0$. Matching to the negative branch requires $Y(X)+X\to 0$ and $Y^{\prime }(X)+1\to 0$ as $X\to \infty$; the right endpoint is then handled by the earlier $q=1$ layer. Write $Y=-X+u$. In the intermediate matching region a small correction satisfies $u^{\prime \prime }-Xu=0$, so $u=C\mathrm{Ai}(X)+D\mathrm{Bi}(X)$. The growing Airy function excludes $D$; the decaying Airy function supplies a permissible stable matching direction. In fact $Y=-X$ is an exact inner solution satisfying $Y(0)=0$, establishing existence of an inner match without solving a nonlinear shooting problem. Since $\mathrm{Ai}(0)\ne 0$, the wall condition excludes an infinitesimal nonzero decaying correction about this solution; it does not prove absence of other finite-amplitude inner solutions. The right wall still requires $\beta \le 1/2$ at leading order.

Matching instead to zero requires $Y\to 0$. Its linearized equation is $Y^{\prime \prime }+XY=0$, with solutions $\mathrm{Ai}(-X)$ and $\mathrm{Bi}(-X)$. Their large-$X$ forms have envelope $X^{-1/4}$ and phase $2X^{3/2}/3$; although their values decay, they generate rapid spatial oscillations in the overlap region. The imposed nonoscillatory assumption removes these nonzero tails. The exact $Y=0$ remains, and a globally zero outer branch again requires $\beta =0$. The PDF's integration hint uses an inverse hyperbolic tangent; the converted TeX's inverse tangent would give a different and incorrect primitive.

Let $h=\sqrt{1-m^2}$ be the complementary modulus of the complete elliptic integral of the first kind. With $\phi =\pi /2-\theta$, write the integrand as $({\sin }^2\phi +h^2{\cos }^2\phi {)}^{-1/2}$. Its logarithmic endpoint asymptotic of the complete elliptic integral comes from $\phi =O(h)$; an expansion at fixed $\phi$ alone misses the constant accompanying the logarithm.

Choose an intermediate cutoff $h\ll d\ll 1$. In the endpoint region, the integral isOutside this region, the leading integral isThe arbitrary cutoff cancels in this matched asymptotic expansion, leavingThese are the divergent logarithm and the finite constant, the requested first two terms. If terms are instead grouped by powers of the complementary modulus, the next refinement is $K(m)=\log (4/h)+(h^2/4)[\log (4/h)-1]+O(h^4\log (1/h))$; its coefficients agree with the DLMF expansion.

Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.

Reversing the drift reverses the endpoint carrying the boundary layer. The outer equation is $-(1+x{)}^2{y}_0^{\prime }+y_0=0$, and now its condition comes from $x=0$:Its value at $x=1$ is $e^{1/2}$, so it cannot by itself satisfy the right boundary value. Put $\xi =(1-x)/ϵ$ there. The leading inner equation is $Y^{\prime \prime }+4Y^{\prime }=0$, with a correction proportional to $e^{-4\xi }$ that decays into the domain. Thus the outer region has size $O(1)$ and the right endpoint layer has width $O(ϵ/4)=O(ϵ)$. One would expand the coefficient near $x=1$, solve the successive equations for the inner expansion, match as $\xi \to \infty$, and form a matched asymptotic expansion by subtracting the overlap. No turning region occurs because the drift does not vanish on this interval.

Assume $ϵ\to 0^{+}$. Since the drift $(1+x{)}^2$ is positive, the fast mode decays to the right; the matched asymptotic expansion has a left boundary layer at $x=0$ of width $O(ϵ)$, and the outer expansion uses the condition at $x=1$.

Write $y\sim y_0+ϵy_1$. The first two equations for the outer expansion areEnforcing $y_0(1)=1$ and $y_1(1)=0$, defineThen the outer asymptotic expansion isTo derive the correction, set $y_1=Eq$. Cancellation of the homogeneous terms leaves $q^{\prime }=-2/(1+x{)}^5-1/(1+x{)}^6$; integration and $q(1)=0$ give the expression above.

In the left boundary layer, set $\xi =x/ϵ$ and $y=Y_0(\xi )+ϵY_1(\xi )+\cdots$. The equation becomes $Y^{\prime \prime }+(1+ϵ\xi {)}^2{Y}^{\prime }+ϵY=0$. Put $A=e^{1/2}$ and $B=1-A$. The leading inner equation and matching give $Y_0^{\prime \prime }+Y_0^{\prime }=0$, $Y_0(0)=1$, $Y_0(\infty )=A$, hence $Y_0=A+B{e}^{-\xi }$. At the next order,Since $q(0)=53/80$, put $C=53A/80$. The matched inner correction satisfying $Y_1(0)=0$ isFor verification, the differential operator $d^2/d{\xi }^2+d/d\xi$ maps $e^{-\xi }P(\xi )$ to $e^{-\xi }(P^{\prime \prime }-P^{\prime })$. The large-$\xi$ common part is $A+ϵ(C-A\xi )$, precisely the small-$x$ outer expansion in the overlap $ϵ\ll x\ll 1$.

Adding the inner expansions and outer expansions and subtracting their common part gives the first-order composite expansion for a positive driftThis is uniformly correct through $O(ϵ)$ on $0\le x\le 1$. It satisfies the left boundary condition exactly; at the right boundary the remaining error is exponentially small. Polynomial growth in the inner correction is harmless because it is multiplied by the decaying exponential. For the smooth nonvanishing drift, the next matched terms are uniformly $O({ϵ}^2)$.

Let the actual dimensional thickness be $d=ϵah$, and let $\bar{z}$ increase downward. The leading curvature of the outer free surface isThe Young–Laplace equation gives $p_{\bar{z}}=-\gamma (d_{\bar{z}}/a^2+d_{\bar{z}\bar{z}\bar{z}})$. In the lubrication approximation, the downward velocity at distance $y$ from the wall is $(\rho g-p_{\bar{z}})(dy-y^2/2)/\mu$, from no slip and zero free-surface shear. Integrating this profile givesCircumferential corrections are higher order in $d/a$, so volume conservation is $d_{\bar{t}}+q_{\bar{z}}=0$ at leading order. DefineThen the thin liquid film on a vertical cylinder satisfiesThe circumferential-curvature term $h_z$ is retained along with axial curvature.

The perturbation calculation givesLong waves with $0<|k|<1$ grow, while $|k|>1$ decay; the fastest growth is $1/12$ at $|k|=1/\sqrt{2}$. Disturbances drift downward at speed $G$. The dimensional unstable wavelengths exceed $2\pi a$, with most unstable wavelength $2\pi \sqrt{2}a$. This is the thin-film version of the Rayleigh–Plateau instability driven by circumferential curvature and opposed by short-wave axial curvature.

For a travelling wave, one integration fixed by $H\to 1$ givesor $H^{\prime \prime \prime }+H^{\prime }=3c(H-1)/H^3-G(1-H^{-3})$. In the large positive-speed core, substitute the expansion using the supplied assumptions. The left side at order $c^{2/3}$ is $H_2^{\prime \prime \prime }+H_2^{\prime }$, while the gravity term is order one and $3c/H^2$ is only order $c^{-1/3}$. Thus $H_2^{\prime \prime \prime }+H_2^{\prime }=0$. Its general solution is a constant plus a sine and cosine. The vanishing leading height and slope at both endpoints selectThe core length is $2\pi$ in the axial coordinate; its amplitude is fixed by transition matching.

At either edge set $X=(3c{)}^{1/3}(x-x_e)$, with $x_e=0$ at the trailing edge and $x_e=2\pi$ at the leading edge. Keeping the height $H$ of order one, the exact wave equation becomesThus the transition equation is $H^3{H}_{XXX}=H-1$. Its linearization about $H=1$ has exponents $1$ and $-1/2\pm i\sqrt{3}/2$. Behind the pulse, approaching the uniform film as $X\to -\infty$ permits only the positive real exponent. The positive-growing, nontrivial branch has one amplitude that is removed by translation, giving the relevant unique $H_{-}$. In front, decay as $X\to +\infty$ permits the two real oscillatory amplitudes; translation removes one parameter and leaves a one-parameter family $H_{+}$. These are the unstable manifold and stable manifold dimensions for the transition system. The literal boundary condition also admits the constant solution and an opposite departing branch, so uniqueness here is specifically for the branch matching a raised positive pulse.

Match the supplied trailing quadratic asymptotic to the core near zero. It gives $K_2=\kappa 3^{2/3}$, because $\frac{1}{2}{c}^{2/3}{K}_2{x}^2=\frac{1}{2}\kappa (3c{)}^{2/3}{x}^2$. The leading maximum occurs at $x=\pi$, soThe front tail has the form $H-1\sim C{e}^{-X/2}\cos (\sqrt{3}X/2+\theta )$. Its capillary waves occur ahead of the descending pulse, not behind it; the trailing tail is monotone to leading order. Converting its frequency to physical distance gives

This large solitary pulse on a cylindrical film sketch shows the leading core and the front-tail eigenmodes; the edge transitions are schematic and the ringing is enlarged for visibility. The thin-film asymptotics require both $c\gg 1$ and $ϵc^{2/3}\ll 1$, rather than taking speed to infinity at fixed film slenderness.

At the next core order, $H_0^{\prime \prime \prime }+H_0^{\prime }=-G$. Integrating and imposing zero $H_0^{\prime }$ at both endpoints givesThe constants in the two quadratic transition asymptotics match the core endpoint heights: $A_0+B_0={\beta }_{-}$ and $A_0+B_0-2\pi G={\beta }_{+}$. Hence $2\pi G={\beta }_{-}-{\beta }_{+}$, and the dimensional uniform thickness selected by matching isWith the supplied constants this fixes $G=3.75/(2\pi )$; $B_0$ is not separately fixed by these height conditions. The thickness selection and capillary wavelength follow from matched asymptotic expansion, not from the linear instability calculation.

Let $\delta =1-m\downarrow 0$. This is the complete elliptic integral of the first kind in parameter notation; the modulus used in some definitions is $\sqrt{m}$. Near the endpoint, $s=\pi /2-\vartheta$ givesUse matched asymptotic expansion with an intermediate cutoff $b$ satisfying $\sqrt{\delta }\ll b\ll 1$. Away from the endpoint the leading integral iswhile the endpoint integral isAdding them removes the arbitrary cutoff and gives the logarithmic endpoint asymptotic of the complete elliptic integralThe order of the next term is therefore $(1-m)\log (1/(1-m))$, rather than merely $1-m$. More explicitly, if $L=\log (4/\sqrt{\delta })$,The logarithmic correction arises from the next terms integrated through the overlap. Its coefficient can also be found by substituting $L+\delta (aL+b)$ into the Gauss hypergeometric equation satisfied here, $m(1-m)K_{mm}+(1-2m)K_m-K/4=0$, giving $a=1/4$, $b=-1/4$.

For fixed $x>0$, set $y=y_0+\epsilon y_1+{\epsilon }^2{y}_2+\cdots$ and impose the data at one at each order. The leading equation gives $y_0^{\prime }=0$, hence $y_0=1$. At the next two orders,The boundary conditions $y_1(1)=y_2(1)=0$ give the three-term outer expansionIts ordering fails when $\epsilon /x=O(1)$, locating a boundary layer of thickness $O(\epsilon )$ at zero. Put $x=\epsilon \xi$ and $y=Y_0(\xi )+\epsilon Y_1(\xi )+\cdots$. The rescaled equation isIts leading equation is ${\xi }^2{Y}_0^{\prime }=Y_0^3$. Matching to the positive outer value one fixes the sign and integration constant:At the next order the bracket $\xi (Y_0^2-1)+2Y_0^2$ vanishes identically, leavingThe general solution is $Y_1=(1+k\xi ){\xi }^{1/2}/(\xi +2{)}^{3/2}$. In the overlap its expansion isThe outer solution re-expressed on this scale has the corresponding terms $1-1/\xi +3/(2{\xi }^2)+\epsilon (1-2/\xi )+\cdots$. Thus matched asymptotic expansion determines $k=1$, consistently in both displayed orders, and the two-term inner expansion isThis positive branch tends continuously to zero at $x=0$, with a square-root cusp. An infinite endpoint derivative is compatible with the degeneracy of the original equation's $x^2{y}^{\prime }$ coefficient.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

For the matched asymptotic expansion, first determine the outer expansion. Setting $\epsilon =0$ away from $x=0$ gives $y_0^{\prime }=y_0$, and the right boundary condition sets $y_0=e^x$. At first order,soThere is no outer $\sqrt{\epsilon }$ term: its homogeneous coefficient would be forced to zero by the right boundary condition. The logarithmic overlap forces a switchback term in the inner expansion.

Near zero, balance $\epsilon y^{\prime \prime }$ against $2x{y}^{\prime }$ to get the square-root boundary layer at a vanishing drift:For fixed $z$ expand $Y=Y_0+s{Y}_1+s^2{Y}_2+\cdots$, allowing logarithmic dependence on $s$ in $Y_2$. With ${\mathcal{L}}_0={\partial }_z^2+2z{\partial }_z$, the leading equation is ${\mathcal{L}}_0{Y}_0=0$. Its bounded solutions are a constant plus an error function. The left boundary condition and the overlap value one fixThe following two solution sections complete the half-order and first-order matching; their headings retain the scaffold's organization of the supplied hints.

At fixed $r$, the leading equation is $(r^2{f}_0^{\prime }{)}^{\prime }=0$. Its solution compatible with the eventual decaying far field and the boundary value at one is $f_0=1/r$. This decay implies $f=O(ϵ)$ at distances $r=O({ϵ}^{-1})$. Balancing radial derivatives against the linear screening term identifiesThe distant region is governed by the modified Helmholtz equation; its decaying homogeneous profile is $e^{-x}/x$.

For a matched asymptotic expansion, write the fixed-$r$ approximation as $f=f_0+ϵf_1+{ϵ}^2{f}_2+\cdots$, allowing logarithms of $ϵ$ in the coefficients. Successive equations areThe boundary condition imposes $f_0(1)=1$ and $f_1(1)=f_2(1)=0$. Before matching, their integrated forms can be writtenIn particular a pure power series with parameter-independent coefficients will be insufficient: the logarithmic overlap creates a switchback term.

In the distant region set $f=ϵF_0(x)+{ϵ}^2{F}_1(x)+\cdots$. The scaled equation is $f_{xx}+2f_x/x-f=x{f}^3/ϵ$, soDecay and leading matching give $F_0=e^{-x}/x$. The radial modified Helmholtz equation gives the supplied particular integral in terms of the exponential integral:As $x\downarrow 0$, the small-argument expansion of the exponential integral yieldswhere $\gamma$ is the Euler--Mascheroni constant. Substitute $x=ϵr$ to compare the two expansions in $1\ll r\ll {ϵ}^{-1}$:The constant at order $ϵ$ fixes $C_1=-1$. Its $1/r$ coefficient then fixes $\alpha +\frac{1}{2}\log 2=1$, and the constant at order ${ϵ}^2$ fixes $D$. HenceThe required inner expansion at fixed $r$ isAt fixed positive $x=ϵr$, the outer expansion isBoth display every term through the requested order, including the ${ϵ}^2\log ϵ$ switchback term in the fixed-$r$ region.