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If a stretched variable is $x=ϵr$ and an outer expansion contains ${ϵ}^2\log x$, its expression in the fixed-$r$ region is ${ϵ}^2\log r+{ϵ}^2\log ϵ$. The parameter logarithm must therefore be present in the inner expansion, often multiplying a homogeneous solution selected by a boundary condition. An ansatz containing only integer powers with parameter-independent coefficients misses this switchback term. The same reasoning applies to other powers of $ϵ$ and other stretched-coordinate scalings.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

The expansion is singular near the moving lower limit, so expanding $e^{-\epsilon /x}$ throughout the original integral would lose its constant term. Introduce the symmetric exponential integralUnder $x=\epsilon /t$, the discarded piece becomesThis is a uniform Taylor expansion because $t\ge 1$. To derive the logarithmic expansion of a symmetric exponential integral without needing another special function, first exploit the same substitution to writeThe estimate follows from $0\le 1-e^{-\epsilon /x}\le \epsilon /x$. Here $\gamma$ is Euler's constant and the small-argument expansion of the exponential integral fixes the constant as well as the logarithm.

Differentiate under the integral and integrate the derivative of $e^{-x-\epsilon /x}/x$ over $(0,\infty )$; the exponentially small endpoint terms vanish. This gives the ordinary differential equationFor the Frobenius method at its regular singular point, write $J=A\log \epsilon +B+\epsilon (C_1\log \epsilon +D)+\cdots$. Substitution gives $C_1=A$ and $D=B-2A$. The preceding leading calculation sets $A=-1$, $B=-2\gamma$, henceSubtracting the moving-cutoff correction to a symmetric exponential integral yields the requested answer, including the logarithmically enhanced first-order term:Thus the $\epsilon \log \epsilon$ switchback term must be retained when asking for all contributions through order $\epsilon$.

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.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

The solution and its first derivative are continuous at $x=1$: a jump would create a Dirac delta distribution or its derivative, absent from the forcing. Put $r=x-1$, $A=a+1$ and $B=b$. The outer expansions, fixed by the respective endpoint boundary conditions, areBalancing diffusion and advection near $r=0$ gives the interior layer at a simple zero of advection with inner variable $z=r/\epsilon$. The leading inner expansion solves $Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$, so matching to $A$ and $B$ givesFor the next term define the Gaussian drift primitivesWriting $H$ for the Heaviside step function, let $F(z)=zH(-z)+(\sqrt{\pi }/4)\mathrm{erf}|z|$. Its value and first derivative match at zero and $(D^2+2zD)F=2zH(-z)$. Thuswhere the logarithmic overlaps fixHere $C_1,C_2$ are the constants in the supplied large-positive-$z$ limits of $E_0,E_1$. The terms $\epsilon \log \epsilon$ are essential switchback terms; discarding them would not give accuracy through $O(\epsilon )$.

Subtracting the common overlap from the inner expansion and outer expansions gives the additive composite expansionThe standard overlap subtraction leaves $\epsilon rH(-r)$. Replacing $H(-r)$ by $(1-\mathrm{erf}z)/2$ changes the value only by $O({\epsilon }^2)$ uniformly and gives a composite with continuous first derivative. The last term restores the endpoint values through the retained order. When $b=a+1$, $d=0$ and the order-one error-function jump disappears. An $O(\epsilon )$ interior adjustment remains to accommodate the leading outer derivative mismatch, together with logarithmic matching when $A\ne 0$.

Reversing the diffusion sign changes the leading inner equation to $-Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$. Its nonconstant solution grows like the integral of $e^{z^2}$, so bounded matching forces the same leading value $C$ on both sides. The bulk is $C+r$ on the left and $C$ on the right; both endpoint conditions are instead supplied by decaying endpoint layers for reversed diffusion, of width ${\epsilon }^2$. A weaker width-$\epsilon$ interior adjustment matches derivatives and selects $C$. Indeed, its first-order equation has a non-growing solution only ifThe leading structure is consequentlywith the smaller interior correction understood. Unlike the positive-diffusion case, the endpoint values are not transported into two distinct order-one inner limits.

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.

To form an additive composite expansion, subtract the common overlap from the sum of the inner and outer expressions. Their retained common part isThus one composite is $ϵF_0(ϵr)+{ϵ}^2{F}_1(ϵr)-{ϵ}^2(D+1/2)/r$. The last term can be screened by multiplying it by $e^{-ϵr}$ without changing either retained expansion. This gives a useful exponentially decaying version:Its boundary value is $1+o({ϵ}^2)$. If exact satisfaction of the boundary value is desired, use insteadThe small-argument expansion of the exponential integral shows that this normalized composite has the same two retained expansions; it equals one at $r=1$ and tends to zero at infinity.

Factor out the elementary exponential and writeUse a matched asymptotic expansion with $0<\lambda \ll \delta \ll 1$. In the outer part,The given identity for the Euler--Mascheroni constant, integrated by parts, impliesHenceapart from pure higher matching powers in $\delta$.

In the inner part, use $t={\lambda }^2/s$:Since $s={\lambda }^2/t\le {\delta }^2$, expand the last exponential. At the required order,With $a={\lambda }^2/{\delta }^2$, this givesThe overlap terms ${\delta }^2$ and $\log \delta$ cancel, leaving the logarithmic switchback termFinally,so

Use $\rho =ϵr={\delta }^{2}r$ in the supplied small-$\rho$ expansions. The outer approximation re-expanded in inner variables isThis is exactly the large-$r$ re-expansion of the inner result in part 1. In particular, the identityforces the switchback term $ϵ\log ϵ(1-r^{-1/2})$. The coefficient of the homogeneous $E_{3/2}$ contribution at outer order $ϵ$ is fixed to $-2\sqrt{\pi }$, while the remaining inner homogeneous constant is fixed to $\pi +C/4-1$. Thus the two matched asymptotic expansions agree term by term in $1\ll r\ll {ϵ}^{-1}$.

As $x\downarrow 0$, the exponential integral haswhere $\gamma$ is the Euler--Mascheroni constant. Integrating the Taylor series of $E_1^{\prime }(x)=-e^{-x}/x$ gives every nonconstant coefficient; the constant follows from the limiting definition of $\gamma$. A logarithm of a stretched variable can create a switchback term in a matched asymptotic expansion.