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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.

The dispersion relation has two additional arguments with the same value. Indeed,Thus use the cubic Stokes dispersion symmetryBoth roots are in the lower half-plane for $k\in D_{+}$. At $k=iv$, $v>0$, their imaginary parts are $-v/2$. A root cannot cross the real axis in $D_{+}$, since a real root would make $\mathrm{Re}\omega (k)=\mathrm{Re}\omega (\nu )=0$. Continuity therefore proves the claim throughout this connected domain. The branch points $\pm 2/\sqrt{3}$ are outside $\overline{D_{+}}$; interchanging the two root labels does not change the formula below.

Evaluate the global relation at each ${\nu }_j$, where its analyticity requirement holds:The left side is linear in ${\nu }_j$. Interpolate it to the argument $k$ with weightsThey satisfy $A_1+A_2=1$ and $A_1{\nu }_1+A_2{\nu }_2=k$. HenceFor the last coefficient, useadding $1-k^2$ and subtracting the interpolated $1-{\nu }_j^2$ gives $1-3k^2$.

The remaining solution-transform contribution is ${\int }_{\partial D_{+}}{e}^{ikx}{\sum }_j{A}_j\hat{q}({\nu }_j,t)dk$. Its integrand is holomorphic in $D_{+}$: the transformed arguments remain in the lower half-plane. The weights are bounded at infinity, and $e^{ikx}$ supplies decay on the upper closing arc for $x>0$. The Cauchy integral theorem therefore makes this integral zero. PutThe Dirichlet spectral representation for the linear dispersive Stokes equation is nowAll quantities in this expression are transforms of the prescribed initial and boundary values. Neither derivative trace remains.