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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

Use the Half-range Fourier transform and the finite-time spectral boundary transformswhere $g_j={\partial }_x^{j}q(0,\cdot )$ for $j=0,1,2$. The spatial transforms are holomorphic in $\mathrm{Im}k<0$, while these finite-time transforms are entire functions. For now $g_2$ is an unknown boundary trace. Integrating the local relation over the half-strip, using spatial decay at infinity, gives the global relationThe boundary sign follows from integrating the flux derivative: its contribution is minus the flux at zero.

Fourier inversion first gives the initial real-line integral plus a real-line integral of $F$. To put the boundary contribution on its useful complex contour, defineOrient its full boundary with $D_{+}$ on the left. It consists of the real axis from left to right and the upper curve $C:k=u+i\sqrt{1+3u^2}$ from right to left. The real-axis piece must not be discarded for this dispersion. The contour integration representation isTo justify the replacement, the integral of $e^{ikx-wt}F$ along $C$ is zero by closing above $C$. In that region $\mathrm{Re}w\ge 0$, and each boundary term is an integral of $e^{-w(t-s)}$ over $0\le s\le t$ times a polynomial in $k$. This has no growing temporal exponential there; $e^{ikx}$ suppresses the upper closing arc for $x>0$. The Cauchy integral theorem then applies. The figure shows the full contour and, separately, the sectors used in the nonlinear problem.

One may replace $F(k,t)$ by $F(k,T)$ for any finite $T\ge t$. Their difference is a future-time integral over $t<s<T$; its factor $e^{w(s-t)}$ decays inside $D_{+}$, so its integral around $\partial D_{+}$ vanishes. All contour formulas are understood as limits of truncated contours, with standard small displacements if a boundary transform needs an Abel limit. Compatibility of the initial and boundary data controls the corner behavior. This is the two-trace half-line cubic dispersion representation.

Use the negative-exponential Half-range Fourier transformFor the boundary traces $g_j(t)={\partial }_x^{j}q(0,t)$, define finite-time spectral boundary transforms $G_j(k,t)={\int }_0^t{e}^{w(k)s}{g}_j(s)ds$. Integrating the conservation law in $x$ and then time gives the global relationThe plus sign of the boundary contribution follows because the integrated flux is $-E(0,t)X(0,t,k)$.

Fourier inversion consequently gives a real-line initial integral plus a boundary integral. To put the latter in the complex plane, defineOrient $\partial D_{+}$ from the left upper infinity, through $i$, to the right upper infinity, with $D_{+}$ on the left. Each $G_j$ is entire, and in the region between the real axis and this contour, $\mathrm{Re}w\ge 0$. The causal factors $e^{-w(t-s)}$ are bounded there and $e^{ikx}$ decays for $x>0$. The Cauchy integral theorem, with the usual large-arc limiting argument, deforms the boundary integral to giveInitial smoothness, spatial decay and corner compatibility justify inversion; if necessary the integrals are first regularized before contour deformation. The unknown Neumann-type traces $g_1,g_2$ will be removed by the global relation.

Use negative-exponential spatial Fourier transforms and finite-time spectral boundary transforms:Write $\mathcal{F}=k^2{F}_0-ik{F}_1-F_2$ and $\mathcal{G}=k^2{G}_0-ik{G}_1-G_2$. Integrate the local relation first over $0<x<L$ and then over time. The left and right endpoint fluxes have opposite signs, giving the finite-interval Airy global relationAll these spatial and temporal integral transforms are entire functions of $k$ for sufficiently smooth data on the finite intervals. Put $\alpha =e^{2\pi i/3}$ and $k_j={\alpha }^{j}k$, $j=0,1,2$. The dispersion relation obeys $\omega (k_j)=\omega (k)$; hence the same global relation holds at all three $k_j$. In particular, the missing traces satisfy the concrete linear systemwhere, with a common upper time limit $t$,The finite-time spectral boundary transforms $F_j,G_j$ are invariant under these rotations because their time kernel depends only on $k^3$. The coefficients multiplying them are not invariant; these differences make the three equations useful for eliminating the missing boundary traces.

LetThe upper domain is the sector $\pi /3<\arg k<2\pi /3$; the lower domain consists of the sectors $\pi <\arg k<4\pi /3$ and $5\pi /3<\arg k<2\pi$. Orient their boundaries with the corresponding domain to the left. For $0<x<L$ and $0<t<T$, the Fokas method gives the integral representationHere $\mathcal{F},\mathcal{G}$ are the combinations of all six finite-time spectral boundary transforms defined above. Initially, Fourier inversion of the global relation gives the first integral together with $-{\int }_{\mathbb{R}}{e}^{ikx-\omega t}\mathcal{F}/(2\pi )$ and $+{\int }_{\mathbb{R}}{e}^{ik(x-L)-\omega t}\mathcal{G}/(2\pi )$. For the left endpoint, deform through the upper complementary sectors, where the spatial exponential decays andis bounded. This replaces the real line by $\partial D_{+}$. For the right endpoint, deform in the lower half-plane; the real portions of $\partial D_{-}$ run from right to left. Thus its integral is the negative of the real-line integral, explaining the minus sign in the last displayed term. The arc contributions vanish by spatial exponential decay and integration by parts in the transforms. Oscillatory real-line integrals can equivalently be defined by a decaying cutoff and its limit.

Finally use the finite-horizon extension of a boundary transform. The difference between upper limits $t$ and $T$ contains ${\int }_t^T{e}^{\omega (s-t)}{f}_j(s)ds$, bounded in $D$, while the appropriate spatial factor decays inside $D_{+}$ or $D_{-}$. The Cauchy integral theorem makes this difference vanish. Keeping $T>t$ also avoids an artificial endpoint half-value in temporal inversion. The integral representation at this stage deliberately includes the unknown traces; the rotated global relations eliminate them in the next step.

The rotated global relations form a square system for the three missing boundary traces. Solve this system and substitute the resulting finite-time spectral boundary transforms into the integral representation. The pieces containing $Q(k_j,T)$ have zero contour contribution by analyticity, rotation of the relevant sectors and the Cauchy integral theorem. The remaining expression involves only $Q_0,F_0,G_0,G_1$. The finite-interval Airy spectral determinant has no nonremovable zeros in the contour domains, as established below, so this elimination introduces no extra spectral data there. Differentiating the representation, and using temporal inversion and compatibility at the corners, verifies the partial differential equation, initial value and the three prescribed boundary values.

For a direct Hadamard well-posedness check, the homogeneous difference satisfiesThis proves uniqueness and continuous dependence for homogeneous boundary data. More generally subtract a smooth quadratic boundary lift $b(x,t)$ satisfying the three prescribed endpoint conditions. The remaining problem has homogeneous boundary conditions and forcing $-b_t-b_{xxx}$. Its semigroup generator is $A=-d^3/d{x}^3$ on $H^3(0,L)$ with $v(0)=v(L)=v^{\prime }(L)=0$. It is a dissipative operator; for every real $\zeta >0$ the equation $(\zeta -A)v=h$ has a unique solution, since the homogeneous third-order boundary system is injective by the same energy identity and therefore invertible. Thus $A$ is maximal dissipative and generates a contraction semigroup. This gives existence and continuous dependence on the lifted data and forcing. Smooth compatible data therefore define a well-posed problem. This operator argument and the spectral elimination describe the same evolution.

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.

Define the negative-exponential Half-range Fourier transformswhich are holomorphic for $\mathrm{Im}k<0$. Write $g_j(s)={\partial }_x^{j}q(0,s)$ and define the finite-time spectral boundary transformsIntegrate the local relation over $0<x<\infty$, $0<s<t$. The spatial boundary term at infinity vanishes by decay, and the term at $x=0$ is subtracted. Consequently the half-line linear dispersive Stokes global relation isFourier inversion on the real axis gives an initial-data integral plus an integral of $B$. To put the boundary term on a complex contour, setFor $k=u+iv$, $\mathrm{Re}\omega =v(3u^2-v^2-1)$, so $D_{+}$ is the connected upper domain $v>0$, $v^2>3u^2-1$. Orient $\partial D_{+}$ with this domain on the left: from upper-left infinity down to $-1/\sqrt{3}$, along the real segment to $1/\sqrt{3}$, and then to upper-right infinity.

The finite-time spectral boundary transforms are entire functions. In the upper regions outside $D_{+}$, where $\mathrm{Re}\omega \ge 0$, their time-integrated factors satisfy $|e^{-\omega t}{e}^{\omega s}|\le 1$ for $0\le s\le t$. Together with $e^{ikx}$, this permits contour deformation of the two remaining real-axis pieces to the curved boundary of $D_{+}$. The resulting representation isThe contour orientation accounts for the plus sign in this convention. This formula still contains the two unknown derivative traces through $G_1,G_2$.

Apply Fourier inversion to the global relation on the real axis, extending $u$ by zero to negative $x$. At an interior point $x>0$ this givesThe finite-time spectral boundary transforms are entire functions of $k$. Define the upper decay domainOrient $\partial D^{+}$ from its left infinite end through $i\alpha$ to its right infinite end, so that $D^{+}$ lies to the left. Its finite point is $i\alpha$, not zero. On the contour, $\mathrm{Re}\omega =0$; the factor $e^{ikx}$ decays because $x>0$.

For the boundary term, write $e^{-\omega t}{G}_j={\int }_0^t{e}^{-\omega (t-s)}{g}_j(s)ds$. Between the real axis and $\partial D^{+}$, $\mathrm{Re}\omega \ge 0$, so this factor has no exponential growth. Contour deformation and Jordan lemma, with the usual cutoffs for oscillatory integrals, therefore giveThis is the requested complex-plane representation with the unknown finite-time spectral boundary transform still present. The decay domain must agree with the sign of the drift in the dispersion relation.

Use the spectral parameter for a linear boundary value problem $k\in \mathbb{C}$ and define the dispersion relation $\omega (k)=k^2-i\alpha k$. Direct differentiation gives the divergence formIndeed the coefficient of $u$ left over after expansion is $\omega -k^2+i\alpha k=0$, and the remaining factor is $u_t-u_{xx}-\alpha u_x$. Thus this is equivalent to the advection-diffusion equation for every $k$.

Introduce the Half-range Fourier transforms and finite-time spectral boundary transformswhere $g_1(t)=u_x(0,t)$ is the unknown normal derivative with the positive-$x$ convention. The outward normal at zero instead gives $-g_1$. The spatial transforms are analytic for $\mathrm{Im}k<0$ and continuous on the real axis under the stated decay assumptions. Integrating the divergence form on $(0,\infty )\times (0,t)$ gives the global relationThe sign follows from the lower spatial endpoint: the integrated spatial derivative is minus its value at zero. This sign will determine the boundary-forcing term in the solution.