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An additive composite expansion combines an inner expansion and an outer expansion by subtracting their common behavior in the overlap region:The subtraction removes the part counted twice. Near the boundary layer, the outer expansion cancels the overlap contribution; away from the boundary layer, the inner expansion cancels it. An error estimate is needed to establish accuracy across the whole domain.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the outer solution is one, the leading inner expansion is $1-e^{-x/ϵ}$, and their overlap is one. Thus $y_{\mathrm{comp}}=1-e^{-x/ϵ}$. With $q=e^{-1/ϵ}$, comparison with the exact normalized solution givesThe supremum norm error is therefore exponentially small on the entire interval, even though the outer solution alone has error one. This is an error bound for function values; errors in derivatives have additional powers of ${ϵ}^{-1}$.

The composite satisfies the differential equation exactly but has right boundary condition $y_{\mathrm{comp}}(1)=1-q$. Dividing it by $1-q$ restores that boundary condition and recovers the exact solution. This example makes both the strength and the limit of matching explicit: it gives a global approximation, while finite-$ϵ$ normalization can still carry information beyond every algebraic order.

An asymptotic matching condition requires the inner expansion and outer expansion, re-expressed in a common overlap region, to agree to the retained order. Both function values and slopes may constrain the free constants. In the square-root reaction problem with nested boundary layers, the shared small plateau sets the first integral of a square-root reaction layer, while its large-field fourth-power tail fixes the neighboring broad layer.

If an outer expansion contains $-(\epsilon /2)\log x$ and the layer coordinate is $x=\sqrt{\epsilon }z$, its overlap contains $-(\epsilon /2)\log z-(\epsilon /4)\log \epsilon$. Matching forces an inner expansion term $-(\epsilon /4)\log \epsilon \mathrm{erf}z$ when that homogeneous profile tends to one in the overlap and vanishes at the boundary. The logarithm multiplies the matching profile, rather than an arbitrary constant.

For $ϵy^{\prime \prime }+a(x)y^{\prime }+b(x)y=0$ with smooth $a$ bounded positively away from zero, the first-order equation for the outer expansion is selected by the right boundary condition. The left mismatch is repaired on the scale $\xi =x/ϵ$. If $a(0)=a_0>0$, the leading inner transient is $B{e}^{-a_0\xi }$. Its next equation is $Y_1^{\prime \prime }+a_0{Y}_1^{\prime }=-a^{\prime }(0)\xi Y_0^{\prime }-b(0)Y_0$. Matching its constant and linear large-$\xi$ terms to the outer expansion permits an order-$ϵ$ composite approximation. A sign reversal of the drift places the transient at the opposite endpoint.

The equation $ϵy^{\prime \prime }+x{y}^{\prime }+y=0$ integrates to $ϵy^{\prime }+xy=C$. Equal endpoint data $y(-1)=y(1)=A$ force $C=0$ and give the unique Gaussian solution above. Its central width is $\sqrt{ϵ}$, with amplitude $A{e}^{1/(2ϵ)}$. The exponentially large central mode is missed by an assumed bounded algebraic outer expansion $C/x$. Changes of order one relative to either endpoint value occur over distance $O(ϵ)$ from that endpoint.

If a leading outer expansion diverges as $\log x$ and a derivative coefficient contains $x-\epsilon z$, balancing that nonlinear correction gives a layer larger than $O(\epsilon )$. Write $x=\epsilon LX$ with $L=\log (1/\epsilon )$ and subtract the large negative constant from $z$. Matching often introduces a $\log L$ correction and yields an endpoint slope of order $1/(\epsilon L)$. The balance must be derived from the actual equation; logarithmic divergence alone does not fix a universal layer width.

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.

If $h$ has a simple interior zero $u_0$ and $a\to 0^{+}$, the inner variable $v=h^{\prime }(u_0)(u-u_0)/a$ turns the local integral into $(1/|h^{\prime }(u_0)|){\int }_{\mathbb{R}}(1+v^2{)}^{-1}dv$. Thus a peak with vanishing width can retain finite area. Slowly varying positive regularization factors cancel between peak height and width when the numerator has the matching factor. An ordinary outer expansion can miss this contribution even though the peak is exponentially narrow.

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.

An outer expansion describes the solution away from a thin or distant region resolved by an inner expansion. It is matched to that inner expansion in their common overlap region. Its range of validity must be specified: an approximation at each fixed positive $x$ need not have uniform convergence down to an endpoint.

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, try the asymptotic expansion $y\sim y_0+ϵy_1+{ϵ}^2{y}_2+\cdots$ at fixed $x>0$. Substitution gives $y_0^{\prime }=0$ and $y_n^{\prime }=-y_{n-1}^{\prime \prime }$ for $n\ge 1$. Expanding the right boundary condition gives $y_0(1)=1$ and $y_n(1)=0$ for $n\ge 1$. Therefore $y_0=1$ and every higher coefficient vanishes. The missing boundary layer is exponentially small at each fixed positive $x$, invisible to all these powers of $ϵ$.

The exact solution supplies a quantitative limit. For every fixed $a\in (0,1)$,This is uniform convergence on $[a,1]$. On $[0,1]$ the same supremum norm of the error is exactly one. The inner expansion supplies the endpoint information that this outer expansion cannot encode.

For the positive-drift equation, the outer expansion obeys $(1+x)y_0^{\prime }+y_0=0$. Select the boundary condition at the right endpoint, obtaining $y_0=2/(1+x)$. The first correction satisfiesIntegrating $[(1+x)y_1{]}^{\prime }=-4/(1+x{)}^3$ givesThe boundary condition mismatch is at $x=0$. Set $X=x/ϵ$ and use the inner expansion $Y=Y_0+ϵY_1+\cdots$. The inner equation is $Y_{XX}+(1+ϵX)Y_X+ϵY=0$. At leading order, matching to $2$ and imposing $Y(0)=1$ give $Y_0=2-e^{-X}$. At first order,The outer expansion in the overlap is $2+ϵ(3/2-2X)$. Matching its constant and linear terms givesThe additive composite expansion adds inner and outer approximations and subtracts the common overlap:It meets the left condition exactly and the right condition up to exponentially small terms, with an $O({ϵ}^2)$ uniform asymptotic error. This is the first-order composite expansion for a positive drift specialized to the present coefficients.

Reversing the drift moves the decaying boundary layer to $x=1$; the outer solution is selected at $x=0$. For the second equation, $y_{\mathrm{out}}=1+x$ is in fact an exact solution, but predicts $2$ at the right boundary instead of $1+ϵ$. Put $X=(1-x)/ϵ$. The inner equation becomesMatching to $2-ϵX$ givesIndeed, $Y_1^{\prime \prime }+2Y_1^{\prime }=-2+(1+2X)e^{-2X}$ and $Y_1(0)=1$, as required by the order-$ϵ$ boundary value. Since $d/dx=-{ϵ}^{-1}d/dX$, $Y_0^{\prime }(0)=2$ and $Y_1^{\prime }(0)=-4$ give the endpoint derivative in a negative-drift boundary layer:The order-one term cannot be obtained by differentiating the leading inner solution alone.

Multiply the equation by $x^2$. Its left side is an exact derivative, yielding the first integralFor a solution continuous at the singular endpoint, the limiting value is $C=-2{ϵ}^3\gamma$. This can also be established from the exact integrating-factor formula below, without assuming uniform bounds on derivatives near zero.

Away from zero, the outer expansion has leading term $y\sim C/x^2$. The endpoint value at one requires $C\sim {ϵ}^3$ if an exponentially amplified homogeneous term is to be avoided. It follows already thatTo exhibit the matching and explain the required boundedness, first set $x=ϵX$, $y=ϵY(X)$ for a representative without an order-one homogeneous contribution. The leading inner expansion obeysIntegration giveswhere $\mathrm{Ei}$ is the exponential integral. For $X\to \infty$, $Y\sim X^{-2}$, matching the outer term ${ϵ}^3/x^2$. For $X\to 0$, $Y\sim -1/X$, so the intermediate solution behaves as $-{ϵ}^2/x$ and reaches order one on a second scale $x={ϵ}^2\xi$.

On this second scale, the first integral becomes, at leading order, ${\xi }^2{Y}_0^{\prime }-2Y_0=1$. Equivalently its differentiated equation is $Y_0^{\prime \prime }+(2/\xi -2/{\xi }^2)Y_0^{\prime }=0$, as in the hint. The matching solution that tends to zero as $\xi \to \infty$ isIt has $Y_0\sim -1/\xi$ at large $\xi$, matching the intermediate region, and $Y_0(0)=-1/2$. Thus the nested layers at an essential singular endpoint connect an order-${ϵ}^3$ outer solution to an order-one endpoint value without an unbounded overshoot.

The issue of uniqueness is clarified by exact integration. PutFor every fixed $ϵ>0$ and every prescribed exact value of $\gamma$, the solution satisfying the right endpoint isAs $x\to 0$, the integral is asymptotic to $\mu (x)/(2{ϵ}^2)$, proving $y(0)=\gamma$. Since $h(1)>0$, the two exact endpoint values fix the two constants: for a prescribed exact $\gamma$, this finite-$ϵ$ boundary-value problem is unique.

However, specifying only the leading value $\gamma \sim -1/2$ and uniform order-one size does not select a unique asymptotic family. To see this constructively, choose $x_m=\sqrt{2}{ϵ}^{3/2}$ and an arbitrary bounded real constant $A$. WriteThe right endpoint fixesEndpoint Laplace's method gives $I=ϵ\mu (1)[1+2ϵ+O({ϵ}^2)]$. Therefore every bounded choice of $A$ haswith changes in $A$ affecting $\gamma$ only by $O({ϵ}^{-3}{e}^{-1/ϵ})$. These solutions are uniformly order one: near zero their particular part has the finite limit $\gamma$, near one it follows the small outer solution, and the freely chosen value at $x_m$ is $Ah(x_m)=O(1)$. The inner-layer homogeneous contribution is also bounded. Indeed $h$ vanishes at zero, is order one near $x_m$, and is exponentially small at one.

Consequently the order-one selected value is $-1/2$, but algebraic asymptotic endpoint data leave an order-one interior freedom. This is beyond-all-orders boundary-value sensitivity. If “constant” were interpreted as fixing $\gamma =-1/2$ exactly and independently of $ϵ$, the omitted algebraic corrections would excite an exponentially large interior homogeneous term; a genuinely uniformly bounded exact family requires the higher-order tuning of $\gamma$. Constancy in $x$ does not preclude that parameter dependence.

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.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

Set $s=x+1\in [0,2]$. The drift now vanishes quadratically at the left endpoint, and the outer expansion selected from $s=2$ is proportional to $e^{1/s}$, which becomes singular as $s\downarrow 0$. The oscillatory endpoint with a quadratically vanishing drift is not an ordinary monotone layer of width ${ϵ}^{1/3}$: balancing just diffusion and drift on that scale neglects the zeroth-order $y$ term.

The formal local exponential rates obey $ϵ{\lambda }^2+s^2\lambda +1=0$. For $s\gg {ϵ}^{1/4}$ their slow and fast branches are approximately $-s^{-2}$ and $-s^2/ϵ$. They coalesce when $s^4\sim 4ϵ$. This locates the main transition a distance $O({ϵ}^{1/4})$ from the left endpoint.

A reliable procedure removes the first derivative exactly:Use WKB approximation for this transformed equation on each side of its unique positive zero $s_t=\sqrt{2}{ϵ}^{1/4}+O({ϵ}^{1/2})$. For $s>s_t$ the transformed modes are exponential; for $0<s<s_t$ they are oscillatory. Their physical amplitudes include the displayed exponential prefactor.

At $s_t$, $P^{\prime }(s_t)=O({ϵ}^{3/4})$, so the Airy turning-point connection formula usesThis layer connects the oscillatory and exponential WKB approximation branches. On the transition scale $s={ϵ}^{1/4}S$, the exact equation is ${ϵ}^{1/2}{v}_{SS}=(S^4/4-1+{ϵ}^{1/4}S)v$, displaying a small effective WKB approximation parameter ${ϵ}^{1/4}$.

Close to $s=0$, the local wavelength is $O(\sqrt{ϵ})$: setting $s=\sqrt{ϵ}\xi$ gives $Y_{\xi \xi }+Y+\sqrt{ϵ}{\xi }^2{Y}_{\xi }=0$, whose leading solutions are sine and cosine. This endpoint form imposes $y(-1)=1$. Within the oscillatory region, the prefactor changes by order one over the additional envelope scale $s=O({ϵ}^{1/3})$. The $\sqrt{ϵ}$ wavelength and ${ϵ}^{1/3}$ envelope are subscales of the same oscillatory region, not additional omitted ordinary layers.

Use the outer exponential/regular branches for $s\gg {ϵ}^{1/4}$, oscillatory WKB approximation inside the $O({ϵ}^{1/4})$ transition region, an $O({ϵ}^{5/12})$ Airy neighborhood of $s_t$, and the $O(\sqrt{ϵ})$ local endpoint form to apply the left boundary data. Match their constants and enforce the right boundary data at $s=2$; there is no separate right endpoint layer required. Matching amplitudes need not remain $O(1)$, and parameter values admitting a homogeneous Dirichlet solution require a separate solvability check rather than an assumed uniform algebraic expansion.

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 $O({ϵ}^2)$.