Dispersion symmetry elimination of a boundary trace ($\omega (\nu (k))=\omega (k)$)

To eliminate an unknown boundary transform by a dispersion relation symmetry, evaluate the global relation at $\nu (k)$ only where the spatial transform remains analytic. Substitute the resulting trace identity into the contour representation. An unwanted transform vanishes only if its transformed argument and exponential admit a valid contour deformation and decay estimate. For constant drift, $\nu (k)=i\alpha -k$ maps the upper domain $\mathrm{Re}(k^2-i\alpha k)<0$ into the lower transform half-plane.

For the backward-sign Airy half-line global relation, suppose exactly one trace of derivative order $j\in \{0,1,2\}$ is unknown. Its spectral multiplier is a constant times $k^r$, $r=2-j$. Use the rotated null contours for half-range Fourier inversion with the displayed coefficients. The unknown contribution becomes an integral over the boundary of the middle upper sector $\pi /3<\arg k<2\pi /3$. There $\mathrm{Im}{k}^3<0$, sois bounded, while $e^{ikx}$ decays exponentially. The Cauchy integral theorem therefore removes that entire unknown contribution. The resulting solution depends on the initial transform and the two prescribed boundary transforms only.

// Target: analysis.bigb

For the linear dispersive Stokes equation, $\omega (\nu )=\omega (k)$ factors as $(\nu -k)({\nu }^2+k\nu +k^2-1)=0$. In $D_{+}=\{\mathrm{Im}k>0,\mathrm{Re}\omega (k)<0\}$ the two displayed other roots lie in the lower half-plane, so both can be used in the global relation. To check the root locations, at $k=ib$ they have imaginary part $-b/2$. A root could reach the real axis only where $\mathrm{Re}\omega (k)=0$, so continuity preserves their signs throughout this connected domain. The branch points $\pm 2/\sqrt{3}$ lie outside its closure. The weights $A_1=(k-{\nu }_2)/({\nu }_1-{\nu }_2)$ and $A_2=({\nu }_1-k)/({\nu }_1-{\nu }_2)$ obey $A_1+A_2=1$ and ${\nu }_j^{\prime }=-A_j$.