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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.

An inner expansion resolves a region in which a stretched independent variable makes otherwise small effects enter the leading balance.

An inner variable is a stretched coordinate such as $\xi =(x-x_0)/\delta (ϵ)$ that remains order one inside a thin region of width $\delta (ϵ)$. Its scale is chosen by a dominant balance, rather than by the size of a coefficient alone.

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$ with $y(0)=0$, $y(1)=1$, set $X=x/ϵ$ and $Y_{ϵ}(X)=y_{ϵ}(ϵX)$. The chain rule givesThus both derivative terms survive in the inner differential equation $Y_{ϵ}^{\prime \prime }+Y_{ϵ}^{\prime }=0$. The exact domain is $0\le X\le 1/ϵ$, withFor bounded $X$, $Y_{ϵ}\to Y_0$ as $ϵ\to 0^{+}$: the inner expansion has a nonconstant limiting profile even though the outer solution is constant. The finite-$ϵ$ profiles nearly collapse onto $Y_0$ after rescaling, with their remaining difference coming from the normalization at the distant right endpoint.

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.

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.

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.

Write $y(x)=Y(X)$ with $X=(1-x)/\sqrt{ϵ}$. The transformed equation isSince the rescaled forcing contains $\sqrt{ϵ}$, an inner expansion naturally uses powers of $\sqrt{ϵ}$. Its leading term satisfiesThe asymptotic matching condition with the leading outer solution isThe general solution is $Y_0=1+A{e}^X+B{e}^{-X}$. Matching eliminates $A{e}^X$, and the boundary condition gives $B=-1$. HenceThis tends to the boundary value zero at $x=1$ and to the matching value one on the interior side of the layer. The negative sign in the exponential follows from measuring the inner variable inward from the right endpoint.

For an optional uniform leading approximation, combine the leading inner and outer terms and subtract their common overlap value one. The additive composite expansion isIt satisfies the right boundary condition exactly; its left-end discrepancy is the exponentially small value $-e^{-1/\sqrt{ϵ}}$. The requested leading inner expansion is $Y_0$ above.

The common value in the overlap region is $e$. The additive composite expansion therefore givesThis leading approximation has $y_{\mathrm{comp},0}(2)=0$ and $y_{\mathrm{comp},0}(1)=1-e^{1-2/ϵ}$; the left endpoint error is exponentially small. It rises from near one following the outer solution $e^{x-1}$, approaches a value close to $e$, then drops rapidly to zero in the right boundary layer. Its first derivative isFor small $ϵ$ the largest value lies slightly inside the interval. For this additive composite expansion, the first derivative vanishes atThis locates the rounded top of the sketch; the rapid final drop has the $O(ϵ)$ scale found above. The additive composite expansion is a leading approximation, so its exponentially small endpoint error does not imply an exponentially small error throughout the interval: neglected algebraic corrections to the outer expansion and inner expansion remain.

The reduced equation and the left boundary condition give the leading outer solutionIntroduce the right inner variable $X=(2-x)/ϵ\ge 0$ and write $Y(X)=y(2-ϵX)$. The chain rule gives $y^{\prime }=-Y_X/ϵ$ and $y^{\prime \prime }=Y_{XX}/{ϵ}^2$. Multiplying the rescaled differential equation by $ϵ$ gives the exact equationAt leading order, $Y_{0,XX}+2Y_{0,X}=0$, with general solution $Y_0=A+B{e}^{-2X}$. The right boundary condition gives $Y_0(0)=0$. The asymptotic matching condition in the overlap region $ϵ\ll 2-x\ll 1$ givesThus $A=e$ and $B=-e$, yielding the leading inner expansionEquivalently, the inner variable $\xi =2(2-x)/ϵ$ gives $Y_0=e(1-e^{-\xi })$. Both normalizations describe the same boundary layer.

With the negative coefficient of $y^{\prime \prime }$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left boundary condition to the outer solution, givingThe boundary layer still has width $O(ϵ)$. Put $X=(1-x)/ϵ$ and $Y(X)=y(1-ϵX)$. The rescaled equation isIts leading inner expansion has $Y_0=A+B{e}^{-X}$. The asymptotic matching condition is $Y_0(\infty )=e^{-1/2}$, while $Y_0(0)=0$. ThusSubtracting the overlap $e^{-1/2}$ in the additive composite expansion givesIts graph starts exponentially close to one, follows the gently decreasing outer solution, and drops from near $e^{-1/2}$ to zero in a right boundary layer of width $O(ϵ)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/ϵ}$. Its right endpoint slope is $-e^{-1/2}(1+1/ϵ)$, showing the rapid final descent.

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

For an assay begun by mixing free enzyme and substrate, take $\boxed{[S](0)=S_0>0,[E](0)=E_0>0,[C](0)=[P](0)=0}$. The complex has not yet formed and no product has been made. The later small-enzyme limit assumes $E_0\ll S_0$. Other preincubation protocols could require nonzero initial complex, but these initial conditions are the ones used by the stated inner solution.

The fast scale resolves the rapid complex buildup that is compressed into a thin initial layer on the substrate-depletion scale. The outer scale follows depletion over a much longer time and cannot satisfy $c(0)=0$ by itself.

In the overlap $1\ll {\tau }_f\ll {ϵ}^{-1}$, the inner solution tends to $(s,c)=(1,1/(1+K))$. The outer solution as ${\tau }_s\downarrow 0$ has the same limit. Subtracting this common part produces a leading composite approximation,It satisfies the leading initial values and tends to the outer quasi-steady solution after the initial layer. Corrections of order $ϵ$ describe the small substrate loss during complex buildup.

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.

Assume $ϵ\downarrow 0$ and fixed $A>0$. Away from the degenerate endpoint $x=0$, the reduced equation is $A{y}_0^{\prime }+y_0=0$. The positive drift makes the rapidly varying homogeneous mode decay as one moves away from the left boundary, so the outer solution is selected by the right boundary condition:It does not satisfy the left value: its limit there is $L=e^{1/A}$.

If the layer width is $\delta$, diffusion is $O(ϵ/{\delta }^2)$, while $Ax{y}^{\prime }$ is $O(A)$ because the drift vanishes linearly. Their dominant balance gives the square-root boundary layer at a vanishing driftWriting $y=Y(\xi )$ gives $Y^{\prime \prime }+\xi Y^{\prime }+[\sqrt{ϵ}/A^{3/2}]\xi Y=0$. The first nonzero inner expansion satisfies $Y_0^{\prime \prime }+\xi Y_0^{\prime }=0$, hence $Y_0^{\prime }=B{e}^{-{\xi }^2/2}$. Imposing $Y_0(0)=1$ and matching $Y_0(\infty )=L$ givesThe overlap region is $\sqrt{ϵ/A}\ll x\ll 1$ for fixed $A$. The additive composite expansioncombines the two leading approximations and satisfies the left value exactly and the right value up to an exponentially small term. Only the first nonzero terms were requested; higher matching terms need not be added.

For fixed $A<0$, the decaying fast mode is at the other endpoint. The outer solution is now fixed by $y(0)=1$, giving $e^{-x/A}$. At $x=1$ the drift does not vanish, so the right layer has width $O(ϵ/|A|)$, with leading transient proportional to $e^{-|A|(1-x)/ϵ}$. In particular its leading inner profile is $e^{-1/A}+(1-e^{-1/A})e^{-\eta }$, where $\eta =|A|(1-x)/ϵ$. The vanishing drift at the left endpoint does not require an order-one mismatch layer for these data.

For $A=0$, the reduced equation is $x{y}_0=0$, which cannot supply the nonzero endpoint values. The equation becomes an Airy ordinary differential equation: $x={ϵ}^{1/3}\xi$ gives $Y^{\prime \prime }+\xi Y=0$. The exact solutions are combinations of $\mathrm{Ai}(-x/{ϵ}^{1/3})$ and $\mathrm{Bi}(-x/{ϵ}^{1/3})$. For $x>0$ they oscillate, with local wavelength of order $\sqrt{ϵ/x}$, rather than forming a decaying exponential layer matched to a smooth nonzero outer profile. The $O({ϵ}^{1/3})$ region resolves the turning endpoint, but oscillations extend throughout the interval. Boundary-value resonances can occur at isolated parameter values. Thus setting $A=0$ is not a continuous substitution into either fixed-sign layer approximation.

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.