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Let $G_0(k),G_l(k)$ denote the Half-range Fourier transforms of $g_0,g_l$, and let $N_0(k),N_l(k)$ denote those of the upward derivatives $n_0,n_l$. Also setIntegrate the tangential derivatives in the three spectral functions by parts. For example, ${\int }_0^{\infty }{e}^{-ikx}{g}_0^{\prime }=-h(0)+ik{G}_0$, and ${\int }_0^l{e}^{ky}{h}^{\prime }=e^{kl}h(l)-h(0)-kH$. All corner values cancel in ${\rho }_L+{\rho }_B+{\rho }_T=0$. The result is the semistrip Laplace spectral global relation

For $\kappa >0$, use the sine transformsReality of $q$ makes $W(\kappa )$ and $W(-\kappa )$ real, so taking imaginary parts removes the unknown left-side normal derivative. At $k=\kappa$ and $k=-\kappa$ the two consequences of the global relation areSubtracting these equations and then substituting back gives the semistrip Dirichlet-to-Neumann sine transforms entirely in terms of the prescribed Dirichlet boundary data:These are the requested transforms of $q_y$ itself. The bottom outward Neumann boundary data have the opposite sign. The apparent division at $\kappa =0$ is handled by a continuous limit whenever the corresponding moments exist; there is no singularity for any $\kappa >0$.

In the semistrip Laplace spectral global relation, reality makes $W(\pm \kappa )$ real for $\kappa >0$. Its imaginary parts at those two arguments remove the unknown vertical-side normal derivative. With $G_j^s(\kappa )={\int }_0^{\infty }\sin (\kappa x)g_j(x)dx$ and $S_j(\kappa )={\int }_0^{\infty }\sin (\kappa x)q_y(x,j)dx$, the result isThe lower outward normal derivative is $-q_y$, so its Dirichlet-to-Neumann map coefficient is $-S_0$, whereas the upper coefficient is $S_l$.

// Target: analysis.bigb