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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 $a_1\mathrm{Ai}(Z)+a_2\mathrm{Bi}(Z)$.

For $\mathrm{Re}p>0$, put $\lambda =p^{1/3}$ using the principal cube root and $r_{1,2}=\lambda e^{\pm i\pi /3}$. The decaying Green function for ${\partial }_s^3+p$ on the real line isThe identities among the three characteristic roots show that $R_p$ and $R_p^{\prime }$ are continuous, while $R_p^{\prime \prime }$ has jump one. Its distributional derivative therefore satisfies $({\partial }_s^3+p)R_p={\delta }_0$. This is the kernel of the resolvent operator $(p+{\partial }_s^3{)}^{-1}$, obtained by the time Laplace transform of the Airy equation.

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.

For the Airy equation with opposite dispersive sign, $q_t-q_{xxx}=0$ on $x>0$, let $Q$ be the initial Half-range Fourier transform and $F_j(k,t)={\int }_0^t{e}^{i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$. Three integrations by parts give the displayed global relation. The spatial transform is holomorphic for $\mathrm{Im}k<0$, whereas the finite-time spectral boundary transforms are entire functions. The symmetry $k\mapsto \alpha k$, ${\alpha }^3=1$, preserves their time exponent.

// Target: analysis.bigb

For the Airy equation $q_t+q_{xxx}=0$ on $0<x<L$, define $F_j={\int }_0^t{e}^{-i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$ and $G_j$ analogously at $L$. Set $\mathcal{F}=k^2{F}_0-ik{F}_1-F_2$ and $\mathcal{G}=k^2{G}_0-ik{G}_1-G_2$. Integrating the local relation with flux $-q_{xx}-ik{q}_x+k^{2}q$ proves the displayed global relation. The dispersion relation is invariant under multiplication of $k$ by a cube root of unity; the resulting three equations determine the three missing boundary transforms for data $q(0,t),q(L,t),q_x(L,t)$.

The scalar linear partial differential equation $u_t+u_x+u_{xxx}=0$ is an Airy equation with a first-order transport term. It is a linearization of the Korteweg-De Vries equation with an advective term retained. It is distinct from the viscous incompressible-flow Stokes equation. On the whole line its spatial Fourier transform evolves by $e^{i(k^3-k)t}$.

For the Airy equation, let $P=e^{-ikx+\omega t}q$. Its local relation follows by arranging the flux so that the unwanted spatial derivatives cancel:ConsequentlyThus the dispersion relation and the required flux are given byThe calculation is polynomial in the complex spectral parameter, so the local relation holds for every $k\in \mathbb{C}$.

The given Riccati ansatz is $v^{\prime }=-i{v}^2+iz/6$. Differentiating it and substituting for $v^{\prime }$ showsIt therefore selects $C=i/6$ in the preceding scaled Painleve equation. Write $v=-i{\Psi }^{\prime }/\Psi$. The quadratic terms cancel in the ansatz, leavingThus every nonzero solution of the indicated Airy equation gives the local mKdV solutionon intervals where $\Psi \ne 0$. For example $\Psi (z)=A\mathrm{Ai}(-z/6^{1/3})+B\mathrm{Bi}(-z/6^{1/3})$, with constants not both zero, has the required equation. This branch is generally complex and meromorphic; the printed ansatz itself includes $i$. A real $v$ on a real interval would require simultaneously $v^{\prime }=0$ and $v^2=z/6$, which is impossible on an interval. Hence this particular ansatz does not furnish a nontrivial real-valued mKdV profile.

Take $L=-d^2/d{x}^2+x$. The Airy equation is $Lu=0$. To discuss positivity of this linear operator, use its homogeneous variation domain $D(L)=\{v\in H^2(0,1):v(0)=0,v^{\prime }(1)=0\}$. Integration by parts givesThe boundary term is zero at both ends. The same calculation with two different functions proves that the associated form is a symmetric bilinear form. Positivity here concerns the homogeneous domain; the condition $u(0)=1$ instead defines an affine set of admissible solutions.

For the variational problem defineand minimizeThe Sobolev trace at zero is meaningful in $H^1$. No derivative boundary value is imposed on this trial space: the Neumann boundary condition will emerge naturally. For $v\in V$, Cauchy-Schwarz inequality and integration giveso $a(v,v)\ge \frac{2}{3}\Vert v{\Vert }_{H^1}^2$. Thus $a$ is a bounded, symmetric, coercive bilinear form on the Hilbert space $V$. The Lax-Milgram theorem says that a bounded coercive bilinear form and a bounded linear functional determine a unique weak solution. Apply it toand set $u=1+v$. This is precisely the first variation condition $a(u,w)=0$ for $J$.

Compactly supported test functions first yield $u^{\prime \prime }=xu$ in the sense of distributional derivatives. Since $xu\in L^2$, the solution is in $H^2$. Integration by parts then leaves $u^{\prime }(1)w(1)=0$ for every $w\in V$, hence $u^{\prime }(1)=0$. The other boundary condition is built into $1+V$. Conversely a solution of this boundary value problem satisfies the first variation condition. Finally,This proves existence, uniqueness and the global minimum characterization, including the natural boundary condition. It is the mixed-boundary Airy energy principle.

Define the initial Half-range Fourier transform $Q(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$, and, for each available boundary trace, its finite-time spectral boundary transformTemporarily let $f_j={\partial }_x^{j}q(0,t)$ also denote the missing trace. Three integrations by parts in $x$ giveConsequently the backward-sign Airy half-line global relation isThe spatial transform is holomorphic below the real axis, while each finite-time $F_j$ is an entire function. Moreover, $F_j(\alpha k,t)=F_j({\alpha }^{2}k,t)=F_j(k,t)$ because ${\alpha }^3=1$.

Suppose the missing derivative has order $j$, and put $r=2-j$. Its contribution to the right-hand side is $a_j{k}^r{F}_j$, where $a_0=1$, $a_1=-i$, $a_2=-1$. Apply the inversion from the preceding part at time $t$ and selectThe rotated coefficients of that missing trace now equal minus its real-line coefficient. Its entire contribution is thereforeThe real-axis pieces cancel. What remains is the boundary of the middle upper sector, outward on $\arg k=\pi /3$ and inward on $\arg k=2\pi /3$. In this sector $\mathrm{Im}{k}^3\le 0$, andis bounded by ${\int }_0^t|f_j(s)|ds$. The factor $e^{ikx}$ decays exponentially throughout the closing arc, overcoming $k^r$. There are no poles. Thus the Cauchy integral theorem proves the cubic-dispersion elimination of one missing boundary trace and removes every occurrence of the unknown $f_j$.

For a compact explicit answer, use only the two prescribed traces to form $S(k,t)=Q(k)+H(k,t)$. The three choices areFor each row the solution isAll quantities in this formula are transforms of the given data. On the contour rays $k^3$ is real, so the time exponential is oscillatory; the integrals use the usual limiting interpretation of Fourier inversion. The decay estimate used for elimination belongs to the middle sector, not to $D$ or $E$, where $e^{-i{k}^3(t-s)}$ can grow.

At $t=0$, $F_j=0$ and the preceding inversion recovers $q_0$. The spectral exponential satisfies the Airy equation with the required minus sign. Differentiating the finite-time amplitudes produces only derivatives of the Dirac delta supported at $x=0$ and zero rotated closed-contour contributions, so the interior equation also holds. Finally, two solutions with the same data have $Q=H=0$ for their difference, so this representation gives zero: each of the three prescribed pairs determines the solution uniquely in the smooth decaying class.

Put $W=(D^2-1)\hat{v}$ and ${\delta }^3=R{e}^{-1}$. The constant-shear Orr-Sommerfeld equation givesLet $a=e^{i\pi /6}$ and $Z=a(y-c-i{\delta }^3)/\delta$. Since $d/dy=(a/\delta )d/dZ$ and $a^3=i$, the equation becomesThis is the Airy ordinary differential equation, yielding the Airy reduction of the Orr-Sommerfeld equation in constant shear. It is distinct from the evolution PDE represented by the existing Airy equation article.

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

Differentiating the first auxiliary equation in $t$, the second in $x$, and equating ${\psi }_{xt}$ and ${\psi }_{tx}$ leaves preciselyso the Airy equation is their compatibility condition.

Multiplication of ${\psi }_x-ik\psi =q$ by $e^{-ikx}$ givesHenceFor $\mathrm{Im}k\ge 0$, the kernel $e^{ik(x-y)}$ is bounded when $y\le x$; dominated differentiation makes ${\psi }_{+}$ analytic for $\mathrm{Im}k>0$. Rapid decrease gives

For real $k$, multiply the time equation by $e^{-ikx}$ and let $x\to +\infty$. The rapidly decreasing terms on its right vanish, leavingThereforeand the Fourier inversion theorem gives

Differentiate the Airy ordinary differential equation $y^{\prime \prime }=sy$ to obtain $y^{(3)}=y+s{y}^{\prime }$. Combining part (a)'s derivatives yieldsTo make this vanish for an arbitrary Airy profile, the time powers must agree: $a-1=4a$, giving $a=-1/3$. Their common coefficient must then vanish, giving $-1/3+b^{-1}=0$, hence $b=3$. ThereforeFor $t>0$ the two terms cancel exactly:This is an Airy similarity solution of the linear dispersive equation. The Airy equation here is the linear partial differential equation $u_t+u_{xxx}=0$, the reduced equation specified in the question; it is distinct from the profile's Airy ordinary differential equation. The construction works for every linear combination of its two independent profiles, including zero. No condition at $t=0$ is supplied, and this singular time is excluded. With real cube roots the final formula can also be used separately on $t<0$.

A quantum particle above a hard reflecting wall in a linearly increasing potential $V(x)=Fx$ for $x>0$. The Schrödinger equation reduces to the Airy equation. Decay at infinity selects $\mathrm{Ai}$, while the wall requires $\mathrm{Ai}(-z_n)=0$. Thus ${\psi }_n\propto \mathrm{Ai}(x/\ell -z_n)$ and the energies are the displayed values. The $n$th state has $n-1$ interior nodes and a decaying tail beyond its classical turning point. A gravitational realization has $F=mg$; gravitational neutron bound-state transmission tests such confinement experimentally.

The Scorer functions are particular solutions of inhomogeneous Airy equations: ${\mathrm{Hi}}^{\prime \prime }(x)-x\mathrm{Hi}(x)=1/\pi$ and ${\mathrm{Gi}}^{\prime \prime }(x)-x\mathrm{Gi}(x)=-1/\pi$. They occur when an endpoint participates in an Airy-type transition. The NIST Digital Library of Mathematical Functions gives their standard normalizations.