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Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.

Write the known Dirichlet boundary condition on side $j$ as $f_j(s)$ and the unknown outward normal derivative as $h_j(s)$, with $-1<s<1$. For any linear combination $v$ of the adjoint waves, the global relation isExpanding the $h_j$ in a finite basis and enforcing these identities at selected collocation points for a global relation gives a linear system with a known right side. The square permits especially simple paired tests.

Let $p_n=n\pi /2$, ${\eta }_n=\sqrt{m^2+p_n^2}$, and ${\phi }_n(s)=\sin [p_n(s+1)]$ for $n\ge 1$. Define four real adjoint solutionsThey satisfy the modified Helmholtz equation because ${\eta }_n^2-p_n^2=m^2$. Each is a linear combination of two adjoint exponentials, hence is obtained from the two spectral global relations. More explicitly, put ${\rho }_n^{\pm }=({\eta }_n\pm p_n)/2$, so ${\rho }_n^{+}{\rho }_n^{-}=c_m$. For the $E_{+}$ convention above, the required spectral pairs areAppropriate phase-weighted differences produce $\sin [p_n(s+1)]$. For example the top test is $e^{-{\eta }_n}[e^{i{p}_n}{E}_{+}({\rho }_n^{-})-e^{-i{p}_n}{E}_{+}({\rho }_n^{+})]/(2i)$. The conjugate relation ensures a real system for real data. This is sine collocation of square modified Helmholtz global relations.

The functions ${\phi }_n$ are an orthonormal Fourier sine basis on $[-1,1]$. Set $h_{j,n}={\int }_{-1}^1{h}_j{\phi }_{n}ds$ and compute the known quantity $R_{j,n}={\int }_{\partial \Omega }f{\partial }_n{v}_{j,n}ds$ by numerical integration. Since ${\phi }_n(\pm 1)=0$, each adjoint test vanishes on both adjacent sides. On its own side it equals ${\phi }_n$, and on the opposite side it equals ${\delta }_n{\phi }_n$, where ${\delta }_n=e^{-2{\eta }_n}$. Thus the square modified Helmholtz Dirichlet-to-Neumann coefficients obeyEach block is inverted explicitly: for opposite sides $j,j^{\prime }$,Compute these coefficients for $1\le n\le N$ and reconstruct $h_j^{(N)}={\sum }_{n=1}^N{h}_{j,n}{\phi }_n$. This is a semi-analytical scheme: the spectral basis integrals and two-by-two inverses are explicit, while the known boundary integrals are evaluated numerically. Increase $N$ and the quadrature resolution until the desired convergence is observed; compatible smooth side data have convergent normal-trace expansions, while corner singularities require the usual weaker trace interpretation and more careful quadrature.

The requested weak interaction between sides is particularly clear: the adjacent-side unknown traces contribute exactly zero, and the opposite-side coefficient is $e^{-2\sqrt{m^2+(n\pi /2{)}^2}}\to 0$. The own-side coefficient stays equal to one. Thus diagonal dominance of paired square global-relation collocation is genuine after pairing; individual uncombined exponential samples need not have the same conditioning. The eigenvalues of each block are $1\pm {\delta }_n$, so its condition number is $(1+{\delta }_n)/(1-{\delta }_n)$.

There is also a direct localization check for an individual adjoint wave. On side $j$ with outward unit normal $n_j$ and tangent $t_j$, the normalized test $e^{\eta (n_j\cdot r-1)+ip{t}_j\cdot r}$ has $\eta =\sqrt{m^2+p^2}$. Its opposite-side magnitude is $e^{-2\eta }$, and its magnitude integrates to at most $1/\eta$ along either adjacent side. This side localization of modified Helmholtz plane waves explains why suitable large spectral parameters suppress remote-side effects even before the exact sine cancellation.

Finally the interior solution can be evaluated from the computed traces using the two-dimensional modified Helmholtz fundamental solutionHere $K_0$ is the Modified Bessel function of the second kind, and $(-\Delta +m^2){\Gamma }_m=\delta$. Replace $h$ by its computed finite Fourier sine series and use numerical integration; at interior points the boundary kernels are smooth. The sign follows from Green second identity with this fundamental-solution convention. This completes the numerical integration of the boundary value problem with a Dirichlet boundary condition, rather than stopping at an equation for its unknown boundary derivatives.

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.