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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 $q(0,t),q(L,t),q_x(L,t)$.

For the linear dispersive Stokes equation $q_t+q_x+q_{xxx}=0$, let $\omega (k)=i(k-k^3)$. Its local relation is obtained with $X=q_{xx}+ik{q}_x+(1-k^2)q$. Integrating ${\partial }_t(e^{-ikx+\omega t}q)+{\partial }_x(e^{-ikx+\omega t}X)=0$ over the half-line gives the displayed global relation, with negative-exponential Half-range Fourier transforms and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. The spatial transforms are holomorphic for $\mathrm{Im}k<0$, whereas the finite-time boundary transforms are entire functions.

Let $E=e^{-ikx+w(k)t}$. Expanding the proposed local relation gives $E[q_t+wq+X_x-ikX]$. Choose the derivative terms in $X$ to cancel the mixed spectral terms:Indeed,so the resulting divergence is exactly $E(q_t+q_x-q_{xxx})$. This establishes the identity for every complex spectral parameter $k$, with the given differentiability. The cubic dispersion relation has the reverse sign from the usual $q_t+q_{xxx}=0$ convention; that sign affects both the spectral sectors and the number of required boundary conditions.

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.

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 $k\in \mathbb{C}$.

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.

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the modified Helmholtz equation, including its Laplace equation case $\lambda =0$. Equivalently,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.

Use the linear dispersive Stokes equation with time variable $t$. Put $E_k=e^{-ikx+\omega (k)t}$. Expanding the required local relation givesChoose $X=q_{xx}+ik{q}_x+(1-k^2)q$. Direct differentiation then yieldsThe unwanted coefficient of $q$ cancels precisely when $\omega (k)=ik(1-k^2)$. HenceWith these choices, the local relation is exactly $E_k(q_t+q_x+q_{xxx})=0$ for every complex spectral parameter for a linear boundary value problem.

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 $G_1,G_2$.

Take the negative-exponential Half-range Fourier transform and setThe advection-diffusion equation has the local relation with flux $X=-q_x-(ik+\alpha )q$. Integration over the half-line gives the half-line drift global relationFor the upper decay domainwrite $k=u+iv$. It is the region above $v=(\alpha +\sqrt{{\alpha }^2+4u^2})/2$, in particular above $v=\alpha$. Its boundary is oriented from upper-left infinity through $i\alpha$ to upper-right infinity, with the domain on the left. Fourier inversion and deformation of the boundary terms giveUse the dispersion symmetry elimination of a boundary trace $\nu (k)=i\alpha -k$. It preserves $\omega$ and maps $D_{+}$ into the lower half-plane. Because $i\nu +\alpha =-ik$, the global relation at $\nu$ impliesThe last term contributes an holomorphic upper-domain integral multiplied by $e^{ikx}$, so it vanishes upon closing the contour for $x>0$. The solution representation with only known data is consequentlyFor the boundary value $\cos t$, direct integration givesThis quotient is an entire function of $\omega$ when its apparent singularities at $\omega =\pm i$ are filled by their limits. Those points are not genuine forcing poles of the finite-time transform.

To obtain an exponentially decaying spectral integrand, choose $0<\theta <\pi /4$ and the exponential-decay contour for half-line drift diffusionDeform $\partial D_{+}$ to $\Gamma$. No pole is crossed: the initial-data denominator in that boundary integral has its only pole at $k=i(\alpha -a)$, which lies below $i\alpha$ since $a>0$. The apparent singularities of the forcing transform are removable. In the swept outer sectors, $e^{-\omega t}{G}_0={\int }_0^t{e}^{-\omega (t-s)}\cos sds$ has the appropriate bounded growth for the deformation. The contour formula becomesEvaluate the second quotient as a removable expression at $\omega =\pm i$, rather than splitting it into artificial residue terms. On either ray,Since $\cos 2\theta >0$, the transient factors have Gaussian decay as $r\to \infty$. The nontransient forcing term is $O(r^{-1})e^{-rx\sin \theta }$ and thus also decays exponentially for every $x>0$. The real-axis integral has the Gaussian factor $e^{-k^{2}t}$. These estimates provide the requested improvement over integration along $\mathrm{Re}\omega =0$.

The first panel shows the cubic domain from Question 1. The second uses $\alpha =1$ and $\theta =\pi /8$; its marked initial-data pole corresponds to $a=0.6$ and lies below both boundary contours.

An independent real-kernel form also makes the initial and boundary traces explicit. Apply the Dirichlet gauge transform for constant drift, $u=e^{\alpha x/2}q$, and then multiply by $e^{{\alpha }^{2}t/4}$ to obtain the heat equation. The odd-reflected heat kernel for initial data and its Dirichlet boundary-forcing heat-kernel formula yieldThe first kernel tends to the initial delta kernel on $x>0$ as $t\downarrow 0$ and vanishes at $x=0$. For the second kernel, the substitution $v=x/(2\sqrt{\tau })$ gives the unit-mass approximate boundary delta as $x\downarrow 0$, so its trace is $\cos t$; it vanishes initially for fixed $x>0$. Differentiating the kernels for $x,t>0$ verifies the advection-diffusion equation. The integrals converge for every positive $a,\alpha$, even when the exponentially weighted initial datum grows, because the spatial Gaussian controls that growth. Thus this gives the same decaying solution without introducing an additional restriction on $a$.

For a general decaying initial datum, an unweighted Fourier sine transform does not directly diagonalize the drift operator. DefineTwo integrations by parts for $q_{xx}$ and one for $q_x$ giveThe cosine transform remains unknown, so this is not a closed sine-transform evolution. In contrast, the weighted sine transform for half-line drift diffusion of $u=e^{\alpha x/2}q$ is closed:When the weighted initial transform exists, Fourier sine inversion gives $q(x,t)=e^{-\alpha x/2}(2/\pi ){\int }_0^{\infty }\sin (kx)\tilde{S}(k,t)dk$. This method requires suitable decay of $e^{\alpha x/2}{q}_0$, not merely decay of $q_0$. For the exponential datum it is an ordinary integrable sine transform only when $a>\alpha /2$. At $a\le \alpha /2$ one would need a generalized or regularized transform instead; the contour and real-kernel formulas above already work without that restriction. Therefore the ordinary sine transform of $q$ is not a closed solution method; a weighted sine transform works under the stated extra weighted-decay condition.

Set $\Theta =e^{-ikx+i{k}^{2}t}$. Expanding the proposed local relation givesThus its vanishing is exactly $i{q}_t+q_{xx}=0$. Integrating over $x>0$, using decay at infinity, yieldsIntegrate in time and use the initial data to obtain the global relationHere ${\tilde{g}}_j(z,t)={\int }_0^t{e}^{iz\tau }{g}_j(\tau )d\tau$, with $g_0=q(0,\cdot )$ and $g_1=q_x(0,\cdot )$. In particular these finite-time transforms are entire in $z$.

Write $F(k)=e^{ikx-i{k}^{2}t}$ and $B_{\pm }(k)=k{\tilde{g}}_0(k^2,t)\pm i{\tilde{g}}_1(k^2,t)$. Substitution into part (i), including replacing $k$ by $-k$ in the reflected transform, givesFor the boundary integrands, rotate the negative real ray to the positive imaginary ray through the second quadrant. There are no singularities, and for $k=a+ib$ with $a\le 0,b\ge 0$,After writing $F{B}_{+}$ as its finite time integral, this estimate and the usual smooth-data arc bounds justify the rotation. The positive real ray stays fixed and the new imaginary ray is directed downwards, givingThe same argument applies separately to the ${\tilde{g}}_0$ and ${\tilde{g}}_1$ terms. Thus the net boundary integral in the inversion formula isFor Dirichlet boundary conditions choose $c=-1$, eliminating the unknown normal derivative. For Neumann boundary conditions choose $c=+1$, eliminating the unknown boundary value. The resulting unified transform representations areEach contains only the initial transform and the prescribed boundary transform. The signs depend on the PDF's orientation of $L$, checked above.

For $q_t+q_x-q_{xxx}=0$ on $x>0$, its local relation has flux $(1+k^2)q-ik{q}_x-q_{xx}$. With $Q_0(k)={\int }_0^{\infty }{e}^{-ikx}q(x,0)dx$ and $G_j(k,T)={\int }_0^T{e}^{w(k)s}{\partial }_x^{j}q(0,s)ds$, its global relation is $e^{wT}\hat{q}=Q_0+(1+k^2)G_0-ik{G}_1-G_2$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}w(k)<0\}$, choose the other cubic root $\nu (k)$ in the lower half-plane. Then the known spectral density is $-Q_0(\nu )+(k^2-{\nu }^2)G_0-i(k-\nu )G_1$. The remaining final-state transform has zero integral around $\partial D_{+}$ by the Cauchy integral theorem and exponential decay. The full boundary includes the real axis and the hyperbola $\mathrm{Im}k=\sqrt{1+3(\mathrm{Re}k{)}^2}$, oriented with the domain on the left.