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Let $F(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$ and $G_0(k,t)={\int }_0^t{e}^{w(k)s}{g}_0(s)ds$. Using the roots of cubic dispersion symmetry with negative transport, put $\mathcal{T}F=[(k-{\nu }_2)F({\nu }_1)+({\nu }_1-k)F({\nu }_2)]/({\nu }_1-{\nu }_2)$. The Fokas method formula is an initial real-line integral minus an integral over $\partial D_{+}$ of $e^{ikx-wt}[\mathcal{T}F+(3k^2+1)G_0]$, both divided by $2\pi$. The boundary is oriented with $D_{+}$ on the left. Its unknown final-state transform has zero integral by analyticity and upper-half-plane decay.

Let $A_j$ be the weights of the cubic Stokes dispersion symmetry and set $F(k)={\sum }_j{A}_j(k){\hat{q}}_0({\nu }_j(k))$. Orient $\partial D_{+}$ with $D_{+}$ on its left. Evaluating the global relation at its two lower-half-plane roots and interpolating the linear expression $ik{G}_1+G_2$ removes both unknown boundary traces. The resulting Fokas method formula isThe discarded transform of the solution is holomorphic in $D_{+}$ and has zero integral after closing its contour for $x>0$. Initial compatibility and sufficient smoothness and spatial decay are required. At the boundary, use a fixed time horizon later than $t$ before Fourier inversion, to avoid an endpoint half-value.

In a Fokas method boundary integral over $\partial D_{+}$, where $\mathrm{Re}\omega <0$ inside, the contribution of replacing $G(k,t)$ by $G(k,T)$ is an integral of $e^{ikx}{\int }_t^T{e}^{\omega (k)(s-t)}g(s)ds$ times an holomorphic coefficient. For $x>0$ it vanishes by the Cauchy integral theorem and the exponential decay in the upper domain. The transformed amplitude is then independent of the observation time, which simplifies differentiation of the partial differential equation. Temporal inversion at $0<t<T$ recovers the full boundary value rather than the half-value of a truncated transform at its endpoint.

Here $\alpha =e^{2\pi i/3}$. The missing-trace matrix has determinant $-\sqrt{3}k\Delta (k)$. Its nonzero zeros are the eigenvalues in the parametrization $Av=i{k}^{3}v$ of $A=-d^3/d{x}^3$ with $v(0)=v(L)=v^{\prime }(L)=0$. Integration by parts gives $\mathrm{Re}⟨v,Av⟩=-|v^{\prime }(0){|}^2/2$. No nonzero zero lies in $\mathrm{Re}(-i{k}^3)\le 0$: on its boundary the determinant equation reduces, after rotation, to an equality between a unit-modulus exponential and $\cosh u-i\sqrt{3}\sinh u$, whose modulus exceeds one unless $u=0$. The zero at the origin is removable in the original boundary problem. There are nevertheless infinitely many genuine negative eigenvalues of $A$, obtained from $e^{-3s/2}=2\cos (\sqrt{3}s/2+\pi /3)$, $s>0$, with $k=-is/L$. Absence of contour poles in the Fokas method does not imply absence of discrete spectrum of the operator.

For $\alpha >0$, let $L$ be the upper branch $\mathrm{Im}k=(\alpha +\sqrt{{\alpha }^2+4(\mathrm{Re}k{)}^2})/2$, oriented left to right. Above $L$, $i\alpha -k$ lies in the lower half-plane, where a decaying half-line Fourier transform is analytic. For $x>0$, closing upward proves that the displayed integral vanishes. An additional multiplier analytic above $L$ with at most polynomial growth is allowed, using exponential decay and, if needed, regularized contours. This supplies a null term in a Fokas method inversion formula without changing the function in the interior.

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 dispersion symmetry elimination of a boundary trace uses the symmetry of the dispersion relationFor $k\in \overline{D^{+}}$, $\mathrm{Im}\nu (k)\le 0$, so the global relation is valid at $\nu (k)$. Since $i\nu +\alpha =-ik$, it givesSubstitution into the contour integral representation produces an unwanted integral ${\int }_{\partial D^{+}}{e}^{ikx}\hat{u}(\nu (k),t)dk$. Its integrand is analytic in $D^{+}$ and decays on closing the contour upwards; Jordan lemma makes this integral zero for $x>0$. Hence the Fokas method eliminates the unknown normal derivative:All quantities here are determined by the prescribed initial and Dirichlet boundary data up to time $t$.

For verification and for numerical evaluation it is useful to evaluate the spectral contour integrals, giving a half-line drift reflection kernel. Putand define the half-line drift boundary kernelFubini's theorem, the Gaussian Fourier transform and contour deformation give the equivalent causal formulaFor the reflected initial term, ${\hat{u}}_0(i\alpha -k)={\int }_0^{\infty }{e}^{\alpha y+iky}{u}_0(y)dy$ on the contour; the evaluated Gaussian supplies the necessary large-$y$ decay. If $e^{\alpha y}{u}_0(y)$ is not integrable, truncate the initial conditions first, evaluate, and pass to the limit using Gaussian bounds. No extra exponential-decay assumption on the original data is needed for this kernel formula.

The boundary kernel follows particularly simply fromThis calculation independently checks both the sign and the coefficient of the boundary forcing.

The unknown Neumann boundary condition can be removed by the Fokas method. Evaluate the global relation for the half-line free Schrodinger equation at $-k$:Consequently $k{G}_0-i{G}_1=2k{G}_0-{\hat{q}}_0(-k)+e^{i{k}^{2}t}\hat{q}(-k,t)$. In the representation from part (a), the last term has integral ${\int }_{\partial D_{+}}{e}^{ikx}\hat{q}(-k,t)dk=0$, by the Cauchy integral theorem and Jordan lemma. It is analytic in the upper half-plane, and the exponential decays on the closing first-quadrant arc. Hence an expression involving only the given data isOne may replace $t$ by $T$ in the upper limit defining $G_0$: the contribution of boundary times $s>t$ closes to zero in $D_{+}$, since $e^{-i{k}^2(t-s)}$ then decays there. Using $t$ makes causality transparent.

For an explicit proof of uniform convergence, it is useful to apply boundary lifting before inversion. Set $h(x)=e^{-x}$, $c=g_0(0)$, $w_0(x)=q_0(x)-ch(x)$, andThe compatibility condition gives $w_0(0)=0$. The Fourier sine transform of $w=q-g_{0}h$ satisfies $S_t+i{k}^{2}S=Hf$, with initial value $S_0$. The integrating factor therefore gives the equivalent representationThis last integral is absolutely and uniformly convergent for $x\ge 0$, $0\le t\le T$ under concrete sufficient hypotheses $w_0,w_0^{\prime },w_0^{\prime \prime }\in L^1$, decay of the boundary terms, and $g_0\in C^2[0,T]$. Indeed, two integrations by parts give $|S_0(k)|\le \Vert w_0^{\prime \prime }{\Vert }_1/k^2$ for $k\ge 1$. Another integration by parts, this time in $s$, givesThus the initial term is uniformly $O(k^{-2})$ and the forcing term uniformly $O(k^{-3})$. For $0<k<1$, use $|S_0(k)|\le \Vert w_0{\Vert }_1$ and the bounded time integral. An integrable majorant proves the claimed uniform convergence and permits evaluation at both boundaries.

At $t=0$, Fourier sine inversion gives $q(x,0)=c{e}^{-x}+w_0(x)=q_0(x)$. At $x=0$, the integral vanishes, giving $q(0,t)=g_0(t)$, including the compatible corner. To verify the equation, note that $h^{\prime \prime }=h$ and the transformed equation impliesAdding the lifted part gives $i{q}_t+q_{xx}=0$. Under the stated smoothness, this holds classically in the interior; differentiated spectral integrals can first be Gaussian-regularized, or read in the sine-transform $L^2$ sense and then identified with the smooth solution. Uniform convergence of $q$ itself does not require claiming uniform convergence of every differentiated integral at the corner.