Conjugate global relations for the modified Helmholtz equation ($\tilde{\mathcal{G}}(\lambda )=\overline{\mathcal{G}(\overline{\lambda })}$)

The two modified Helmholtz adjoint plane waves $v_{\lambda }$ and ${\tilde{v}}_{\lambda }=\overline{v_{\overline{\lambda }}}$ give boundary identities $\mathcal{G}(\lambda )=0$ and $\tilde{\mathcal{G}}(\lambda )=0$. For real solution traces, $\tilde{\mathcal{G}}(\lambda )=\overline{\mathcal{G}(\overline{\lambda })}$. Reality is needed for this conjugation argument; the parametrization separately yields the reciprocal identity $\tilde{\mathcal{G}}(\lambda )=\mathcal{G}(1/\lambda )$.

On the square $[-1,1{]}^2$, write normal traces in ${\varphi }_n(s)=\sin [n\pi (s+1)/2]$. Set $p_n=n\pi /2$, ${\omega }_n=\sqrt{k^2+p_n^2}$, $r_n=({\omega }_n+p_n)/k$. Paired spectral samples $\{r_n,r_n^{-1}\}$, $\{-r_n,-r_n^{-1}\}$, $\{i{r}_n,i/r_n\}$, $\{-i{r}_n,-i/r_n\}$ give adjoint tests $e^{\mp {\omega }_{n}y}{\varphi }_n(x)$ and $e^{\mp {\omega }_{n}x}{\varphi }_n(y)$. Each test vanishes on the adjacent sides, removing their unknown normal derivatives.

Let $R_{jn}={\int }_{\partial \Omega }f{\partial }_n{v}_{jn}ds$ for a square sine adjoint test, and put ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$, ${\delta }_n=e^{-2{\omega }_n}$. The opposite-side normal-trace coefficients obeywith the same formulas for $L,R$. When the known Dirichlet boundary condition is integrated exactly, these are exact coefficients, since orthogonality eliminates all other unknown modes.

For sine collocation of square modified Helmholtz global relations, each opposite-side mode has a scaled matrixThe assembled system has strict diagonal dominance, and its spectral condition number of a positive-definite matrix is below $(1+e^{-\pi })/(1-e^{-\pi })<1.091$. The conclusion concerns the explicit paired sine rows, not arbitrary uncombined complex rows.