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For $q=e^{i\pi /6}$ define $I_0(C)={\int }_0^{\infty }\mathrm{Ai}(q(s-C))ds$ and $I_1(C)={\int }_0^{\infty }s\mathrm{Ai}(q(s-C))ds$. Both converge by Airy function decay. Integrating its differential equation gives $I_1-C{I}_0=iq\mathrm{Ai}\prime (-qC)$. These moments convert a decaying vorticity profile into its velocity and displacement for three-layer long-wave shear-flow matching.

The Airy equation is the linear dispersive partial differential equation $q_t+q_{xxx}=0$. Under the spatial Fourier transform, each mode evolves by multiplication by $e^{i{k}^{3}t}$.

For positive time its fundamental solution is $(3t{)}^{-1/3}\mathrm{Ai}(x/(3t{)}^{1/3})$, connecting this evolution equation to the Airy function without identifying the two concepts.

Named after George Biddell Airy, the standard Airy functions solve the Airy ordinary differential equation. Their Wronskian is $\mathrm{Ai}(Z){\mathrm{Bi}}^{\prime }(Z)-{\mathrm{Ai}}^{\prime }(Z)\mathrm{Bi}(Z)=1/\pi$, so they form a fundamental solution pair, including for complex $Z$.

Digital Library of Mathematical Functions, section 9.7 gives the Airy function asymptotic expansions. Equations 9.7.5 and 9.7.7 give the positive-argument exponential forms; equations 9.7.9 and 9.7.11 give the negative-argument oscillatory forms. Their leading terms determine the factor two and phase shift in the Airy turning-point connection formula.

This second-order ordinary differential equation has the independent entire Airy functions $\mathrm{Ai}$ and $\mathrm{Bi}$ as a standard basis. It is distinct from the dispersive PDE in Airy equation. Its general solution is $a_1\mathrm{Ai}(Z)+a_2\mathrm{Bi}(Z)$.

Let $y$ solve the Airy ordinary differential equation $y^{\prime \prime }(s)=sy(s)$. For $t>0$, set $s=x/(3t{)}^{1/3}$ and define the displayed similarity solution. Differentiating the profile equation gives $y^{\prime \prime \prime }=y+s{y}^{\prime }$, so the chain rule yieldsTheir sum is zero, proving the Airy equation $u_t+u_{xxx}=0$. Every real linear combination of the two standard Airy functions gives such a solution. This is a solution of the linear dispersive equation; the nonlinear Korteweg-De Vries equation includes an extra term and does not follow from the same arbitrary profile construction.

At equal argument and large integer order, the Bessel function of the first kind has phase $n(\sin t-t)=-n{t}^3/6+\cdots$ near the endpoint. The cubic stationary endpoint yields the displayed positive constant. It is also the zero-argument value of the Airy function in the uniform Bessel turning-point approximation.

The local inner solution is a combination of Airy functions $\mathrm{Ai},\mathrm{Bi}$. Decay in the forbidden region removes $\mathrm{Bi}$. The remaining Airy turning-point connection formula is oscillatory on the allowed side and represents an incident/reflected pair, not a single upward branch. Without the forbidden-side condition the local differential equation alone leaves both coefficients free.

An astronomer whose name is attached to the Airy functions.

Portraits of George Biddell Airy held in the MacTutor collection.

A Dirichlet boundary condition at a hard endpoint and a single smooth classical turning point give the leading WKB quantization condition ${\int }_0^{x_{\mathrm{t}}}k(x)dx=(n+3/4)\pi$, with $n=0,1,\dots$. The phase differs from the two-soft-turning-point condition because the hard wall imposes an exact wavefunction zero rather than an Airy function connection phase. The WKB validity condition is still checked in the interior.

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.

The WKB approximation, also called JWKB, describes a short local de Broglie wavelength in a slowly varying potential energy. For $p(x)=\sqrt{2m[E-V(x)]}$ in a classically allowed region, its two local waves areThe phase follows the classical momentum; its amplitude varies to conserve probability current. In a forbidden region, putting $\kappa (x)=\sqrt{2m[V(x)-E]}$ gives $\psi \simeq {\kappa }^{-1/2}[C{e}^{{\int }^x\kappa /\hslash }+D{e}^{-{\int }^x\kappa /\hslash }]$, with boundary conditions selecting the physical combination.

The WKB validity condition is $\hslash |p^{\prime }|/p^2\ll 1$, with the analogous condition on $\kappa$. It fails at a turning point where $E=V$. For a simple turning point, locally linearizing V yields an Airy function; matching fixes the relative amplitudes and a $\pi /4$ phase. Two such turning points give the two-turning-point WKB quantization conditionFor an opaque barrier, the WKB tunnelling exponent instead givesThe factor two distinguishes probability from amplitude. Matching determines the prefactor; the exponential alone is not accurate near the barrier top. These formulas connect bound state energy eigenvalues and quantum tunnelling to classical actions. ^[ref]

The real Airy function asymptotics used for the Airy turning-point connection formula are, for $z,s\to +\infty$,These real-axis asymptotics fix both normalization and phase.^[ref] Near $a$, $I\sim 2z^{3/2}/3$, $\Phi \sim 2(-z{)}^{3/2}/3$, and ${\kappa }^{-1/2}\sim {\ell }^{1/2}{z}^{-1/4}$. Matching to the constants of c gives $C_A=2\sqrt{\pi }{\ell }^{1/2}D$ and $C_B=\sqrt{\pi }{\ell }^{1/2}G$. ThereforeIn particular, the decaying component matches the cosine with the requested $-\pi /4$ shift, with a factor two in its oscillatory coefficient. The growing component matches a sine with the indicated minus sign. Expanding into travelling waves givesThe factor two is essential when converting the real oscillatory components to incident and reflected travelling waves.

In the central classically forbidden region, define the positive decay momentum $\chi (x)=\sqrt{2m[V(x)-E]}$ and the dimensionless action integral $K(x)={\hslash }^{-1}{\int }_x^{x_1}\chi (y)dy$. The general leading WKB approximation isBoth coefficients are allowed: this interval has finite length, so the exponential that grows towards the barrier centre cannot be discarded by a condition at infinity. The WKB validity condition holds away from the classical turning point $x_1$; a local Airy function solution supplies its continuation there.

The WKB approximation fails at a simple classical turning point $x_t$, because $p(x_t)=0$ makes its amplitude singular and violates the WKB validity condition. Locally replace $V(x)-E$ by $V^{\prime }(x_t)(x-x_t)$. For a forbidden region on the right, $V^{\prime }(x_t)>0$ and the rescaled coordinate $z=[2m{V}^{\prime }(x_t)/{\hslash }^2{]}^{1/3}(x-x_t)$ changes the time-independent Schrödinger equation into the Airy ordinary differential equation ${\psi }_{zz}-z\psi =0$. Reversing the coordinate handles a forbidden region on the left.

The two independent Airy functions have asymptotic forms, for $z,w>0$,Match these to the WKB approximation in an overlap region where the linearization of the potential energy and the large-argument asymptotic expansions both apply. If $K_t={\hslash }^{-1}|{\int }_{x_t}^x\chi dx|$ on the forbidden side and $s_t={\hslash }^{-1}|{\int }_{x_t}^{x}pdx|$ on the allowed side, the Airy turning-point connection formula isThus a decaying exponential acquires a factor two in its allowed-side amplitude, whereas a growing exponential does not. The phase convention uses positive integrals measured away from the classical turning point, so it applies at both ends of the well without an extra orientation sign. These are leading semiclassical matching rules for simple, separated classical turning points.

Use the amplitudes and positive action integrals of part (a), and define the right-well phaseMatching at the inner classical turning point gives $C=A$, $D=2B$. Matching the exterior decaying wavefunction at the outer classical turning point givesThe trigonometric identitytherefore yieldsAll three pieces on $x>0$ are now fixed up to $E$ and $F$:The Airy function descriptions replace the apparent singularities at the classical turning points. For the permitted energy eigenvalues, part (d) extends this right-hand wavefunction to the left by even or odd parity; $F$ is then determined by wavefunction normalization. At arbitrary $E$, the decaying right-hand solution instead has a unique continuation by the Schrödinger equation, which need not decay at negative infinity.

For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.

By Stirling's approximation, for fixed $a$,Apply this with $a=1/6,5/6$ and use $\Gamma (r+1)=r\Gamma (r)$. Their exponents add to one, and henceMore explicitly, the ratio is $1-5/(36r)+O(r^{-2})$. The exact ratio of neighboring terms isThus the Airy function expansion is factorially divergent, and optimal truncation retains about $|\sigma |$ terms.

The printed summation starts at $r=1$, but its leading $r=0$ term is missing. The Gamma reflection formula gives $\Gamma (1/6)\Gamma (5/6)=2\pi$, so $Y_0=1$. The correctly normalized dominant Airy function expansion starts at zero. This correction is essential for the exponentially small remainder in the next part; if the printed series is used literally, its remainder also contains the entire missing leading term.

Let $P(z)=1/(2\sqrt{\pi }{z}^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma |+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):Here the divergent sum means its lateral Borel summation, not ordinary summation. To see why the pole approximation controls this remainder, write $a_r={\sigma }^r{Y}_r$ and insert $\Gamma (r)={\int }_0^{\infty }{e}^{-v}{v}^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma )$; $v=\sigma t$ produces $I(\sigma ,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the Borel transform of the coefficient series isThe simple pole supplies the leading Stokes phenomenon; its weaker logarithmic singularity supplies the smaller corrections. This is the integral justification for retaining the factorial tail.

For $\arg \sigma =\varphi /\sqrt{|\sigma |}$, expand $\sigma =n+i\varphi \sqrt{n}+O(1)$. Part (i), with $\mu =\varphi$, now givesThis formula uses the upper-pole prescription and the upper Airy function switching ray: near $\arg z=2\pi /3$, take $\arg \sigma =3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma {|}^{-1/2})=O(|z{|}^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is error-function smoothing of a Stokes multiplier, not a discontinuity of the analytic Airy function. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma /2}$ would invalidate this remainder formula.

The outer solution satisfies $y(x)[x+y(x)]=0$, so on a connected interior interval the nonoscillatory branches are $y=0$ and $y=-x$. Both happen to solve the differential equation exactly away from any endpoint mismatch. For a fixed positive endpoint coordinate $q$ ($q=a$ or $q=1$), introduce the inward distance $\xi =(x-a)/\sqrt{ϵ}$ on the left or $\xi =(1-x)/\sqrt{ϵ}$ on the right. The leading nonlinear endpoint layer with a quadratic reaction obeysMatching to zero fixes $E=0$, so $(Y^{\prime }{)}^2=-q{Y}^2-2Y^3/3$. It is negative for every sufficiently small nonzero $Y$. Thus no real nonconstant layer can approach zero: the zero outer branch requires $\alpha =\beta =0$, and then $y=0$ is an exact solution.

Matching to $-q$ fixes $E=q^3/6$. The first integral factors asConsequently an endpoint value $\eta$ can match this branch only if $\eta \le q/2$. For $-q<\eta <q/2$, there are two choices: a direct approach to $-q$, and an initial excursion to $q/2$ followed by a return to $-q$. Both are translations ofThe negative translation gives the direct layer; the positive translation gives the excursion. At $\eta =q/2$ the two coincide, with the maximum at the wall. For $\eta <-q$, the unique admissible direction is a monotone approach from below:where ${\xi }_0$ is fixed by $Y(0)=\eta$. At $\eta =-q$ only the constant leading layer matches at a finite inward coordinate; a pulse translated infinitely far from the wall is outside this endpoint-layer classification.

Thus the leading admissibility condition for the negative outer branch is $\alpha \le a/2$ and $\beta \le 1/2$. Under strict inequalities, each endpoint has two choices when its datum lies between $-q$ and $q/2$, and one choice below $-q$. Their independent combinations give one, two or four formal matched profiles. In particular, for $\alpha =\beta =0$ there are four negative-bulk profiles in addition to the exact zero solution, so uniqueness certainly cannot be inferred from the outer equation. The endpoint equalities and the constant-layer cases are degenerate; higher-order endpoint matching is needed to decide exact finite-$ϵ$ counts at these cutoffs. These conditions and counts concern leading families under the stated exclusion of interior layers and rapid oscillations.

A matched asymptotic expansion for the negative branch isThe following sketch shows the four choices for zero endpoint data, together with the exact zero solution. It plots leading composite profiles, not numerical claims of exact finite-$ϵ$ multiplicity.

Additional interior layers cannot be excluded merely by counting the outer roots. Freezing $x=q$ gives a homoclinic orbit to $-q$ with the same hyperbolic-secant pulse. Its location would require a further matching or solvability condition; it is not an arbitrary parameter in the original variable-coefficient problem. Indeed, put $z=y+x$. The exact equation and an energy identity areAn isolated order-one pulse with exponentially small tails on both sides would decrease this energy by an algebraic amount, whereas its two tail energies would be exponentially small. This obstructs a freely standing isolated pulse well away from the walls. It does not by itself classify oscillatory solutions or pulse interactions with endpoint layers, so a complete global uniqueness claim would exceed this leading construction.

When $a=\alpha =0$, the two outer roots meet at the left endpoint. Let $x={ϵ}^{p}X$, $y={ϵ}^{q}Y$. Balancing $xy$ with $y^2$ gives $p=q$, and balancing $ϵy^{\prime \prime }$ with them gives $1-q=2q$. The one-third-power scaling at a nonlinear turning endpoint is thereforeIts left condition is $Y(0)=0$. Matching to the negative branch requires $Y(X)+X\to 0$ and $Y^{\prime }(X)+1\to 0$ as $X\to \infty$; the right endpoint is then handled by the earlier $q=1$ layer. Write $Y=-X+u$. In the intermediate matching region a small correction satisfies $u^{\prime \prime }-Xu=0$, so $u=C\mathrm{Ai}(X)+D\mathrm{Bi}(X)$. The growing Airy function excludes $D$; the decaying Airy function supplies a permissible stable matching direction. In fact $Y=-X$ is an exact inner solution satisfying $Y(0)=0$, establishing existence of an inner match without solving a nonlinear shooting problem. Since $\mathrm{Ai}(0)\ne 0$, the wall condition excludes an infinitesimal nonzero decaying correction about this solution; it does not prove absence of other finite-amplitude inner solutions. The right wall still requires $\beta \le 1/2$ at leading order.

Matching instead to zero requires $Y\to 0$. Its linearized equation is $Y^{\prime \prime }+XY=0$, with solutions $\mathrm{Ai}(-X)$ and $\mathrm{Bi}(-X)$. Their large-$X$ forms have envelope $X^{-1/4}$ and phase $2X^{3/2}/3$; although their values decay, they generate rapid spatial oscillations in the overlap region. The imposed nonoscillatory assumption removes these nonzero tails. The exact $Y=0$ remains, and a globally zero outer branch again requires $\beta =0$. The PDF's integration hint uses an inverse hyperbolic tangent; the converted TeX's inverse tangent would give a different and incorrect primitive.