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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.

The cubic dispersion symmetry with negative transport supplies two other roots of $w(\nu )=w(k)$:Both lie in the lower half-plane when $k\in D_{+}$. For example this is immediate at large positive imaginary $k$. A root could leave that half-plane only by becoming real, where $\mathrm{Re}w(\nu )=0$, contradicting $\mathrm{Re}w(k)<0$. Connectedness of $D_{+}$ completes the argument; a coalescence merely exchanges the two roots.

Evaluate the global relation at these roots. Because their dispersion values equal $w(k)$, the same boundary transforms occur, andInterpolate this linear polynomial in $\nu$ at ${\nu }_1,{\nu }_2$. Defineand similarly $\mathcal{T}\hat{q}(k,t)$. Substitution givesIndeed, interpolation of ${\nu }^2+1$ yields $({\nu }_1+{\nu }_2)k-{\nu }_1{\nu }_2+1=-2k^2$, which proves the coefficient $-(3k^2+1)$ after subtracting $k^2+1$.

The final-state term has zero contour integral. After cancelling $e^{wt}$ its integrand is $e^{ikx}\mathcal{T}\hat{q}(k,t)$, holomorphic in $D_{+}$ and decaying upon closing the contour upward for $x>0$. The possible square-root singularities are removable: the expression is unchanged by exchanging the roots, and when they coalesce at $\nu$, its limit is $F(\nu )+(k-\nu )F^{\prime }(\nu )$, with the same formula for the final-state transform. Thus the Cauchy integral theorem eliminates this unknown term.

We obtain the fully specified Dirichlet spectral formula for Airy flow with negative transportOnly the initial Fourier transform and the given boundary-value time transform remain.