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First remove the observation time from the spectral amplitude. For any fixed horizon $H>t$, replace $G_0(k,t)$ by $G_0(k,H)$ in the formula of part (d). The difference in the boundary integral is proportional toFor every $s>t$, this integrand is holomorphic inside $D_{+}$, where $\mathrm{Re}w<0$, and decays on its large closing arcs. The Cauchy integral theorem makes the difference zero. This finite-horizon extension of a boundary transform therefore gives the time-independent spectraAll observation-time dependence is now in $e^{ikx-wt}$.

When the boundary history is sufficiently decaying, the natural long-time version isThe integral converges in $\mathrm{Re}w<0$ for bounded boundary data and on its boundary for integrable histories; otherwise one must state an Abel prescription or a suitable temporal growth assumption. Arbitrary smooth $g_0$ need not have such an infinite-time transform. The finite-horizon formula remains exact without that extra assumption, and the source alone does not prescribe a universal long-time limit.

A useful further decomposition, if $g_0(s)=g_{\infty }+r(s)$ with integrable $r$, isThis separates the persistent forcing from the transient spectrum. The pole at $k=i$, where $w=0$, accounts for the decaying stationary profile $g_{\infty }{e}^{-x}$; that profile also follows directly from $q_{xxx}-q_x=0$, spatial decay and its boundary value. For the remaining oscillatory integrals the phase is $i[kx+(k^3+k)t]$. At a fixed ray $x/t=\xi$, its stationary points satisfy $3k^2+1+\xi =0$, or $k=\pm i\sqrt{(1+\xi )/3}$. These time-independent amplitudes expose precisely the poles and saddle points used in the method of steepest descent; allowable deformations must also respect the analyticity strip of the initial transform.

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.

Fix $T>t$. In the representation from (c), the finite-horizon extension of a boundary transform allows $G_0(k,t)$ to be replaced by $G_0(k,T)$. To see this directly, their difference contributesIt has an holomorphic integrand inside $D_{+}$; there $\mathrm{Re}\omega <0$, and the factors $e^{\omega (s-t)}$ and $e^{ikx}$ allow the contour to be closed. Its integral is zero for $x>0$. This replacement puts the observation-time dependence entirely in $e^{ikx-\omega t}$.

Every such exponential satisfies the linear dispersive Stokes equation, becauseConsequently the representation satisfies the partial differential equation. The stipulated smoothness and decay justify the differentiated oscillatory integrals, or equivalently their regularized limits. The fixed horizon is useful here: no artificial source term arises from differentiating the endpoint of a time transform.

At $t=0$, use the original $G_0(k,0)=0$ version. The $F$ integral vanishes after closing inside $D_{+}$, since $F$ is holomorphic there and $x>0$. Fourier inversion of the initial half-line transform therefore givesFor the boundary trace use the fixed-horizon version, with $0<t<T$. First consider the initial-data terms at $x=0$. Differentiating ${\nu }_j^2+k{\nu }_j+k^2-1=0$ givesThus changing variables separately in the two pieces of the $F$ integral giveswhere ${\Gamma }_{-}$ goes from lower-left infinity to $-1/\sqrt{3}$, across to $1/\sqrt{3}$, and then to lower-right infinity along $\mathrm{Re}\omega =0$. For completeness, on the upper-right curved edge one root maps onto the real ray $(-\infty ,-2/\sqrt{3}]$ and the other onto the lower-right curved edge. On the upper-left edge the corresponding images are $[2/\sqrt{3},\infty )$ and the lower-left edge. The middle segment maps to the remaining real intervals $[1/\sqrt{3},2/\sqrt{3}]$ and $[-2/\sqrt{3},-1/\sqrt{3}]$. These real images cancel the outer pieces of the original real-axis integral, leaving exactly ${\Gamma }_{-}$ with the stated orientation.

This contour bounds the lower-central region $\mathrm{Im}k<0$, $\mathrm{Re}\omega >0$. The spatial transform is holomorphic there and $e^{-\omega t}$ decays on its closing arcs for $t>0$. Its integral is therefore zero by the Cauchy integral theorem. The initial-data part of the boundary trace vanishes.

In the remaining boundary integral, use $d\omega =i(1-3k^2)dk$. Along the oriented $\partial D_{+}$, $\omega$ traverses the imaginary axis once from $-i\infty$ to $i\infty$. On its real segment, the imaginary part runs from $-2/(3\sqrt{3})$ to $2/(3\sqrt{3})$; the two curved edges supply the complementary intervals. It follows thatThe last equality is ordinary temporal Fourier inversion, after $\omega =i\xi$. Because $t$ is strictly inside $(0,T)$ it gives the full value, not a truncated-transform endpoint half-value. ThusCompatibility $q_0(0)=g_0(0)$ makes the two traces agree at the corner. Spatial decay follows from the upper-contour exponential and nonstationary oscillatory estimates on the real integral as $x\to \infty$, under the assumed data decay.

The decaying smooth solution is also unique. For a zero-data difference $w$, integration by parts in the linear dispersive Stokes equation givesusing $w(0,t)=0$ and decay at infinity. The initial norm is zero, so $w=0$. This completes verification of the initial value, boundary value and governing equation.