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If ${\beta }_1,{\beta }_2,{\beta }_3$ are the affine barycentric coordinates of an equilateral triangle of side $\ell$, then $h={\prod }_j\sin (\pi {\beta }_j)=\frac{1}{4}{\sum }_j\sin (2\pi {\beta }_j)$ vanishes on the boundary and satisfies $\Delta h=-16{\pi }^{2}h/(3{\ell }^2)$. Thus the problem $\Delta q-4\lambda q=0$ with zero Dirichlet boundary data has distinct solutions zero and $h$ at the displayed negative parameter. Their normal traces differ, demonstrating failure of a uniquely determined Dirichlet-to-Neumann map at resonance.

Consider the Helmholtz equation $(\Delta +k^2)u=0$ in $z>0$, with $k>0$, boundary value $u(x,0)=g(x)$, and upward outgoing or decaying propagation. Taking the tangential Fourier transform, with frequency $\xi$, givesThus the upward normal derivative is the Fourier multiplier $i\beta (\xi )$. The outward unit normal of the upper half-space points downward, so its Dirichlet-to-Neumann map has multiplier $-i\beta (\xi )$ instead. The normal orientation and the square-root branch determine both signs.

For $|\xi |<k$ the modes propagate; for $|\xi |>k$ they are evanescent waves decaying with height. At high frequency $\beta (\xi )\sim i|\xi |$, so the map grows in magnitude like one tangential derivative. Small-amplitude boundary roughness can therefore have a large derivative effect when it varies on a short spatial scale.

Interpret the printed coordinate notation as the square corners $1+i$, $1-i$, $-1-i$, $-1+i$. This listed order is clockwise, contrary to the counterclockwise convention in (ii). Keep the printed first side directed from top to bottom. Then $m_1=1$, $h_1=-i$, and $z(s)=1-is$. Put $g_1(s)=u(1,-s)$ and $q_1(s)=u_x(1,-s)$, where $q_1$ is the outward normal derivative on the right side. The pullback formula, without any orientation shortcut, givesIndeed $h_1{u}_z-{\bar{h}}_1{u}_{\bar{z}}=-i(u_z+u_{\bar{z}})=-i{u}_x$, and $i\beta (\lambda h_1+{\bar{h}}_1/\lambda )=\beta (\lambda -{\lambda }^{-1})$. If one reverses the side to match the counterclockwise convention, its parameter is $1+is$ and its integrand is $-W_1(-s,\lambda )$. Reversing every side multiplies the global relation by minus one, leaving its zero value unchanged.

The unheaded numerical reconstruction request. The four unknown Neumann boundary conditions are four functions on the sides, not four scalar values. The polygonal global relation is linear in their outward normal derivatives, and its remaining terms depend only on the prescribed Dirichlet boundary data.

Choose a finite approximation on each side, for example an expansion of $q_j(s)$ in Legendre polynomials or piecewise polynomials. Substitute those expansions into the consistently oriented global relation. At chosen nonzero spectral parameters for a linear boundary value problem, integrate the exponential kernels against each basis function to assemble a complex linear system; the known right-hand side is obtained by integrating the given Dirichlet boundary data. Use enough independent samples to resolve all side coefficients, and preferably oversample. The conjugate global relations for the modified Helmholtz equation provide useful companion tests; for complex data use the corresponding independent adjoint relation rather than assuming the traces real.

Solve the scaled system by least-squares solution using a stable factorization such as a singular value decomposition. Sampling directions should probe all sides, and exponential row scaling avoids overflow and poor conditioning. Refine the side approximation and spectral samples until the recovered traces and unused global relation residuals stabilize. Corner incompatibilities or limited corner regularity call for mesh refinement or enriched endpoint basis functions. This realizes a numerical Dirichlet-to-Neumann map without first discretizing the whole interior.

The underlying Dirichlet problem is uniquely solvable in the usual trace class for $\Delta u-4{\beta }^{2}u=0$ with $\beta >0$: the homogeneous problem has ${\int }_{\Omega }(|\nabla u{|}^2+4{\beta }^2|u{|}^2)dA=0$. This supports the boundary reconstruction, although uniqueness of the continuous problem alone does not guarantee that an arbitrary finite set of spectral samples is well conditioned.

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