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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.

The real roots of $\sin x=x+ϵ(1-x^2)$, with $ϵ>0$, can be counted by solving for the parameter: $ϵ=E(x)=(x-\sin x)/(x^2-1)$. Positive values occur only on $(-1,0)$ and $(1,\infty )$. Direct differentiation and the half-angle identities give the derivative in the title. It is negative except at isolated points, so $E$ is strictly decreasing on both intervals, with limits $+\infty$ and zero. There are exactly two real roots for each positive parameter. As $ϵ\to 0^{+}$, dominant balance gives $x_{-}\sim -(6ϵ{)}^{1/3}$ and $x_{+}\sim {ϵ}^{-1}$. The exact root count supplements the local asymptotic expansions: a small-root expansion alone would omit the root escaping to infinity.

In $ϵy^{\prime \prime }+xy+y^2=0$, balancing all three terms where the outer branches $0$ and $-x$ meet gives the displayed dominant balance and the inner equation $Y^{\prime \prime }+XY+Y^2=0$. Matching to $-X$ and writing $Y=-X+u$ gives $u^{\prime \prime }-Xu+u^2=0$. Its linearized decaying correction is an Airy function; the growing correction is excluded. Matching to zero instead gives $Y^{\prime \prime }+XY=0$, whose nonzero Airy function combinations oscillate. Thus the acceptable matching branches differ qualitatively at the turning endpoint.

Write the equation as $ϵx^3-(x-1{)}^2=0$. For $ϵ\to 0^{+}$, two roots of a polynomial approach the double root one. Their splitting follows a fractional-power dominant balance, rather than an ordinary expansion in integer powers.

Put $\delta =\sqrt{ϵ}$ and $x=1+a\delta +b{\delta }^2+\cdots$. Substitution givesThus $a=\pm 1$ and $b=3/2$. The two bounded roots have the Puiseux series expansionsThe first two nonzero terms are $1\pm \sqrt{ϵ}$; the displayed correction also resolves their common order-$ϵ$ shift.

The third root escapes to infinity because the leading polynomial degree drops when $ϵ=0$. Balance $ϵx^3$ against $x^2$, giving $x=O({ϵ}^{-1})$. Set $X=ϵx$; the equation becomesFor $X=1+Aϵ+B{ϵ}^2+\cdots$, the first two coefficient equations are $A+2=0$ and $B+2A^2+2A-1=0$. HenceIts first two nonzero terms are ${ϵ}^{-1}-2$. These three distinct branches account for all roots of the cubic for sufficiently small positive $ϵ$. The Vieta formulas provide a consistency check: their sum is ${ϵ}^{-1}$, with the constant and order-$ϵ$ terms cancelling.

If $ϵ$ approaches zero through negative values, the bounded pair is complex: $x_{\pm }=1\pm i\sqrt{|ϵ|}+\frac{3}{2}ϵ+O(|ϵ{|}^{3/2})$. The large-root expansion is unchanged as a real asymptotic formula. The positive-parameter regime is used for the boundary-layer problem below.

Suppose the right-end boundary layer has width $\ell (ϵ)$, and let $X=(1-x)/\ell$. An order-one change of $y$ over this width gives $ϵy^{\prime \prime }=O(ϵ/{\ell }^2)$. Near $x=1$, the leading terms $y-x^2$ are order one. The dominant balance is thereforeThusThe inner variable satisfies $X\ge 0$ inside the domain and resolves the reaction boundary layer with square-root thickness.

Take $ϵ\to 0^{+}$. A root on the smaller scale has $x=ϵu$, giving $u+2+ϵu^3=0$. Put $u=u_0+ϵu_1+O({ϵ}^2)$. Successive coefficients give $u_0=-2$ and $u_1+u_0^3=0$, hence $u_1=8$. The real root is thereforeThe other dominant balance is between the cubic and linear terms, requiring $x={ϵ}^{1/2}z$. Set $t={ϵ}^{1/2}$; the equation becomes $z^3+z+2t=0$. At $t=0$ the two nonzero roots are $z_0=\pm i$. With $z=z_0+t{z}_1+O(t^2)$, the first correction satisfies $(3z_0^2+1)z_1+2=0$, so $-2z_1+2=0$ and $z_1=1$. ThusThe limiting roots in both rescalings are simple, so the implicit function theorem justifies these regular expansions in their respective small parameters. The fractional powers in the original variable are a fractional-power root expansion. The real derivative $3x^2+ϵ$ is strictly positive; the cubic therefore has exactly one real root. Its other roots are the displayed complex conjugate pair, accounting for all three roots. Their leading real corrections sum to $-2ϵ+ϵ+ϵ=0$, as required by the absence of a quadratic coefficient.

Setting $ϵ=0$ gives the reduced first-order linear differential equation $y^{\prime }=y$. Its nonzero solutions cannot satisfy the zero value at $x=2$, so one endpoint needs a boundary layer. To determine which endpoint, freeze the coefficient at $x=x_b>0$ in the rapidly varying part: $ϵy^{\prime \prime }-x_b{y}^{\prime }=0$ has fast mode $e^{x_b(x-x_b)/ϵ}$. It decays into the interval from the right endpoint but grows into the interval from the left endpoint. HenceThe dominant balance $ϵ/{\delta }^2\sim x_b/\delta$ gives $\delta \sim ϵ/x_b$, so a convenient local thickness is $ϵ/2$. The left boundary condition must select the outer solution. Equivalently, a putative left inner variable $X=(x-1)/ϵ$ would give $Y_{XX}-Y_X=0$, whose nonconstant solution grows like $e^X$ and cannot satisfy the asymptotic matching condition as $X\to \infty$.

Take $ϵ\to 0^{+}$. There are two real branches, on different scales. For a root tending to zero, the Taylor series givesThe dominant balance is $x^3\sim -6ϵ$, so a fractional-power root expansion yieldsMore precisely $x_{-}=-(6ϵ{)}^{1/3}+O(ϵ)$. An ordinary power series in integer powers of $ϵ$ would miss this scale.

There is also a root escaping to positive infinity. Boundedness of the sine term makes the leading dominant balance $x-ϵx^2=0$. The nonzero branch isIndeed the equation rearranges to $ϵx_{+}^2-x_{+}=ϵ-\sin x_{+}$, which yields $x_{+}={ϵ}^{-1}+O(1)$ on this branch. The bounded oscillatory correction need not have a constant limiting value.

To check that these are all the real solutions, solve for the parameter:Positive values occur only on $(-1,0)$ and $(1,\infty )$; $x=\pm 1$ are not roots. A useful exact derivative isIt vanishes only at isolated points, so $E$ is strictly decreasing on each of these intervals. Its endpoint limits are $+\infty$ and zero on each interval. This monotone parametrization of sine perturbation roots gives exactly one root in each interval for every $ϵ>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,ϵ)\mapsto (-x,-ϵ)$ reflects these two branches and gives all real solutions for that sign as well.

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 the neo-Hookean solid, the axial Cauchy stress tensor under the unloaded lateral boundary conditions isWith $K,\mu >0$, the spring balance is equivalent, after multiplication by $\alpha >0$, to the cubic equationFor a fixed $\gamma \ge 1$, write its left side as $h(\alpha )$. It obeys $h(0)=-\mu <0$ and tends to $+\infty$ as $\alpha \to \infty$. MoreoverThus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha >0$. For $\gamma >1$,so the unique equilibrium has $1/\gamma <\alpha <1$. The tissue is elastically compressed, while its actual length $\alpha \gamma$ exceeds one and compresses the spring.

To determine the large-growth limit without presuming a power law, solve for $\gamma$:This is a strictly decreasing function of $\alpha >0$, since its derivative isIt tends to $+\infty$ only as $\alpha \to 0^{+}$. Consequently unlimited growth forces $\alpha \to 0^{+}$, andMore precisely, the dominant balance is $K\gamma {\alpha }^3\sim \mu$, givingThe surrounding spring softens the divergence relative to rigid confinement but does not bound the compressive stress. Both axial extension and lateral expansion help accommodate growth:whose product is the total volume $\gamma$, as required by elastic incompressibility and the imposed growth tensor.

In the outer solution, derivative terms are small and $\sqrt{y}\sim \epsilon k$, so $y_{\mathrm{out}}\sim {\epsilon }^2{k}^2$. To resolve its mismatch with the zero left boundary, set $x=\epsilon \xi$, $y={\epsilon }^{2}Y$. The original equation becomes exactly $Y^{\prime \prime }-\sqrt{Y}+k=0$. The left inner variable is therefore $\xi =x/\epsilon$, with $Y(0)=0$ and $Y\to k^2$ as $\xi \to +\infty$.

Multiplying by $Y^{\prime }$ gives the first integral of a square-root reaction layer. Matching $Y=k^2$, $Y^{\prime }=0$ fixes its constant:The positive derivative branch below $k^2$ begins with $Y^{\prime }(0)=\sqrt{2k^3/3}$ and approaches the plateau exponentially. Thus the left layer has thickness $O(\epsilon )$, not $O(\sqrt{\epsilon })$ despite the coefficient on the original second derivative. Its small amplitude is essential in the dominant balance.

The square-root reaction problem with nested boundary layers has a small positive plateau $y\sim {\epsilon }^2{k}^2$, an $O(\epsilon )$ left-end layer, and two distinct scales on the right. The broad right layer has width $O(\sqrt{\epsilon })$; its inner edge needs a thinner $O(\epsilon )$ transition to the plateau. A single right-end scale cannot describe both matches.

Here is a uniformly matched leading solution. Let $L(\xi )$ solveusing the increasing branch below $k^2$. DefineThen $\Phi (L(\xi ))=\Phi (0)-\xi$. For the right-hand branch, define $R(\xi )>k^2$ by $\Phi (R(\xi ))=\xi$; it tends to $k^2$ as $\xi \to -\infty$. These formulas follow from the first integral of a square-root reaction layer and are expanded in solution and solution.

For sufficiently small $\epsilon$, let $x_{\ast }=1-\epsilon \Phi ({\epsilon }^{-2})$. Then $R((1-x_{\ast })/\epsilon )={\epsilon }^{-2}$, so the right profile meets the prescribed order-one endpoint. The composite asymptotic expansion isThe subtraction removes the common plateau counted twice. Its boundary residuals are exponentially small: the right profile has almost reached $k^2$ at $x=0$, and the left profile has almost reached $k^2$ at $x=1$. In each active layer the opposite profile is exponentially close to its plateau, so the composite reproduces that layer's governing equation and the shared outer solution to the retained order.

The right transition location satisfies $x_{\ast }=1-2\sqrt{3\epsilon }+o(\sqrt{\epsilon })$. Outside its thin transition, the broad right layer iswhere the bracket is positive; it matches through the thin layer to ${\epsilon }^2{k}^2$, rather than continuing as a fourth power on the wrong side. The dominant balances and the end-region matches are detailed below.

The limiting polynomial has a double root at $x=-1$, so two roots require a Puiseux series. For positive $ϵ$, put $\delta =\sqrt{ϵ}$ and $x=-1+a\delta +b{\delta }^2+\cdots$. Substitution and dominant balance giveThus $a=\pm 1$ and $b=-3/2$. The two finite roots have the three-term asymptotic expansionsThis is square-root splitting of a double polynomial root. For negative $ϵ$, the same formulas apply with a chosen complex square root and describe a complex conjugate pair.

The third root is a divergent algebraic root in a singular perturbation. Set $x=X/ϵ$: the equation for $X$ is $X^3+X^2+2ϵX+{ϵ}^2=0$. Its nonzero limiting root is $X=-1$. Substituting $X=-1+Aϵ+B{ϵ}^2+\cdots$ yields $A=2$ and $B=3$, henceAs an independent check, Vieta formulas give the sum of roots as $-1/ϵ$: the two finite-root expansions sum to $-2-3ϵ+O({ϵ}^2)$, giving the same large-root expansion.

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.

Set $x=\delta X$ and write $y(x)=Y(X)$. The equation becomesA dominant balance must compare all three coefficients, rather than just the two derivative terms.

For $-1<\alpha <1$, balancing diffusion and drift gives $\delta ={ϵ}^{1/(\alpha +1)}$. Their common coefficient tends to infinity, so the reaction term is smaller and the leading inner equation isThis derivative is integrable at infinity, allowing an ordinary localized endpoint adjustment. Thus the thickness is ${ϵ}^{1/(\alpha +1)}$ in this range.

For $\alpha =1$, $\delta =\sqrt{ϵ}$ makes all three terms comparable. Keep all of them:One decaying solution is $e^{-X^2/2}$, and reduction of order gives the other. This is a square-root-thickness layer with all terms retained.

For $\alpha >1$, the first variation scale is $\delta =\sqrt{ϵ}$. Keep diffusion and reaction, and drop drift at this scale:This leading motion is oscillatory, not a monotone decaying boundary-layer profile. To describe localization one must also resolve the slower damping envelope. Removing the first derivative in the original equation bygivesNear the endpoint the wavelength is $O(\sqrt{ϵ})$, while the envelope changes on $x=O({ϵ}^{1/(\alpha +1)})$. Hence an oscillatory endpoint region is possible, but it has distinct wavelength and envelope scales; calling the diffusion-drift balance alone its leading equation would incorrectly discard the reaction term. This generalizes the oscillatory endpoint with a quadratically vanishing drift.

For $\alpha \le -1$, the diffusion-drift balance would require ${\delta }^{\alpha +1}=ϵ$, which gives no shrinking positive power scale. At every shrinking scale the singular drift dominates diffusion, so no regular finite-amplitude diffusion boundary layer of the preceding kind occurs. The coefficient itself is singular at zero; singular endpoint solutions are a different issue from a bounded boundary-layer correction.

In a layer of thickness $\ell \ll 1$, the third-derivative and first-derivative terms have sizes $ϵ{\ell }^{-3}y$ and ${\ell }^{-1}y$. Their dominant balance requires ${\ell }^2\sim ϵ$, soThe local fast modes are exponentials with rates $\pm 1/\sqrt{ϵ}$. A mode decaying into the interval can occur at either endpoint: $e^{-x/\sqrt{ϵ}}$ at zero and $e^{-(1-x)/\sqrt{ϵ}}$ at one. The zero fast exponent represents the matching outer component. The term $xy$ is lower order on either fast scale, and even smaller near zero; it creates no further independent layer thickness.

In a thin boundary layer, take streamwise velocity and distance to be order one, set $z=\delta \zeta$, and use incompressibility to set $w=\delta W$. The streamwise convective acceleration is order one, whereas its dominant viscous contribution after division by the Reynolds number is ${\mathrm{Re}}^{-1}{\delta }^{-2}{u}_{\zeta \zeta }$. Their dominant balance selectsWith this choice, the exact rescaled Navier-Stokes equations areThus the leading pressure is independent of $\zeta$. Matching it to the outer solution from part (a), evaluated near $z=0$, gives $p=p_{\ast }-x^2/2$, hence $-p_x=x$. The leading boundary-layer equation is thereforeThe exact surface shear stress condition rescales to $u_{\zeta }+{\delta }^2{W}_x=\delta \beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential derivative also vanishes there. The leading surface boundary conditions areMatching to the outer solution requiresThe second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta \to 0$ would impose an extra condition not supplied by the leading outer solution. This is the stress-driven stagnation-point boundary layer.

For $ϵy^{\prime \prime }+x^{\alpha }{y}^{\prime }+y=0$ near zero, dominant balance gives a diffusion-drift layer of thickness ${ϵ}^{1/(\alpha +1)}$ for $-1<\alpha <1$, with the reaction term negligible. At $\alpha =1$, all three terms survive on the scale $\sqrt{ϵ}$. For $\alpha >1$, diffusion and reaction first balance on wavelength $\sqrt{ϵ}$, while the drift damps the oscillation on the larger envelope scale ${ϵ}^{1/(\alpha +1)}$. Removing the first derivative gives the envelope factor $\exp [-x^{\alpha +1}/(2ϵ(\alpha +1))]$. For $\alpha \le -1$, no shrinking algebraic diffusion layer of this regular bounded type is obtained.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

For fixed $k>0$ and endpoint values $y(0)=0$, $y(1)=1$, the outer solution is $y\sim {\epsilon }^2{k}^2$. The left boundary layer has width $\epsilon$. At the right, a layer of width $\sqrt{\epsilon }$ carries the order-one rise, and a translated layer of width $\epsilon$ connects its inner edge to the small plateau. The dominant balance depends on both amplitude and derivative scale; using the same scale at both ends loses a necessary match.