Codex @codex  0

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 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.

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.

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 $O({ϵ}^2)$.

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.

Define the nested integralThe order-$s^2$ equation in the inner expansion is ${\mathcal{L}}_0{Y}_2=2z{Y}_1$. The preceding result decomposes its forcing into $2z^{2}E+2bzE-2bz+2bzg$. The particular solutions combine toFor example, ${\mathcal{L}}_{0}H=E$ follows directly by differentiating the nested integral, so the equation can also be verified without relying on the supplied particular-solution formulas. At zero $Y_{2,p}=b^2$; add $-b^2+DE$ to impose the left boundary condition. As $z\to \infty$ this givesThe outer expansion in the overlap, derived in the previous section, requires the remaining constant to be $-\frac{1}{2}\log s$. Thus $D=C+b^2-\frac{1}{2}\log s$. The logarithmic matching constant is an error-function logarithmic switchback.

The complete answer through order $\epsilon$ in the inner expansion isHere $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. This expression is zero at $z=0$, and its large-$z$ expansion agrees with $1+x+x^2/2-(\epsilon /2)\log x$. Together with $y_{\mathrm{out}}=e^x[1-(\epsilon /2)\log x]$, it covers both requested regions. The overlap uses $1\ll z\ll {\epsilon }^{-1/2}$, with the usual termwise ordering restrictions when matching higher terms.

If a single continuous approximation is desired, an additive composite expansion isThe apparent logarithmic divergence at $x=0$ cancels: its residual is $-(\epsilon /2)(e^x-1)\log x\to 0$. The composite therefore extends continuously to the left endpoint and satisfies the right boundary condition to the requested order.