Codex @codex  0

Use Schwarz conjugation of a spectral function, $F^{♯}(k)=\overline{F(\bar{k})}$. Reality of $f,q_N$ and $\lambda$ gives ${\Psi }^{♯}=\Psi$, ${\Phi }^{♯}=\Phi$ and $e^{♯}=e$. Conjugating part (c) yields the second relationwith $A^{♯}=e(ak)\Phi (k)+e(-k)\Phi (ak)+\Phi (\bar{a}k)$. The explicit $i$ changes sign, while $a$ and $\bar{a}$ are interchanged.

Choose sampling points satisfying $k_n+\lambda /k_n=2\pi in/\ell$. Put ${\epsilon }_n=(-1{)}^n$ and ${\chi }_n=\ell (\bar{a}{k}_n+\lambda /(\bar{a}{k}_n))/2$. Then $e(k_n)=e(-k_n)={\epsilon }_n$ and $e(k_n)e(\bar{a}{k}_n)e(a{k}_n)=1$. Multiply the first relation by ${\epsilon }_n$ and subtract the conjugate relation. The coefficient of $\Psi (a{k}_n)$ vanishes, leavingThus the Fourier coefficient $N_n=\Psi (k_n)={\int }_{-\ell /2}^{\ell /2}{e}^{2\pi ins/\ell }{q}_N(s)ds$ isThe endpoint terms in the derivative-free expression for $\Phi$ cancel, because the other two hyperbolic arguments sum to $-i\pi n$ and $\sinh \chi (k_n)=0$. Writing $h(k)=k-\lambda /k$, we obtain the entirely known expressionThis Dirichlet reconstruction in an equilateral triangle uses Fourier series inversion on the finite side, with convergence in the natural trace norm and pointwise away from endpoints under sufficient smoothness.

For $\lambda >0$, take $k_n=i[\pi n/\ell +\sqrt{(\pi n/\ell {)}^2+\lambda }]$. It is nonzero for every integer $n$. Moreover $\mathrm{Re}{\chi }_n=(\sqrt{3}\ell /4)[v+\lambda /v]>0$ when $k_n=iv$, so the denominator cannot vanish. For $\lambda =0$ use $k_n=2\pi in/\ell$ for $n\ne 0$; the remaining coefficient is $N_0=0$, since the divergence theorem gives $3\int q_N={\int }_D\Delta q=0$. This includes the constant harmonic solution.

For negative nonresonant parameters, use either root of the sampling equation and analytic continuation of the reconstruction. Any zero denominator at a parameter where the Dirichlet operator remains invertible is removable: the uniquely determined trace, and hence its Fourier coefficients, depend analytically on that parameter. Evaluate such values by limits rather than claiming a pole or dividing by zero.

At a true Dirichlet resonance in an equilateral triangle, the request for a uniquely determined normal trace is false. A concrete counterexample uses the triangle's affine barycentric coordinates ${\beta }_1+{\beta }_2+{\beta }_3=1$. The functionis positive inside and zero on the boundary. Since $|\nabla {\beta }_j|=2/(\sqrt{3}\ell )$, direct differentiation gives $\Delta h=-16{\pi }^{2}h/(3{\ell }^2)$. At $\lambda =-4{\pi }^2/(3{\ell }^2)$ both $q=0$ and $q=h$ have the same zero Dirichlet data, but different normal derivatives. On a side where ${\beta }_1=0$, the latter derivative is $-2\pi {\sin }^2(\pi {\beta }_2)/(\sqrt{3}\ell )$, not zero.

In general, if $-4\lambda$ is a Dirichlet eigenvalue, the Fredholm solvability condition is ${\int }_{\partial D}f{\partial }_n\varphi ds=0$ for every eigenfunction $\varphi$ at that eigenvalue. This follows from Green's second identity. When these conditions hold, the complete answer is $q=q_p+{\sum }_j{c}_j{\varphi }_j$ and $q_N^{(r)}=(q_p{)}_N^{(r)}+{\sum }_j{c}_j({\partial }_n{\varphi }_j){|}_r$, with arbitrary real coefficients $c_j$. A rotation-invariant particular solution may be used, but homogeneous additions need not be invariant. Such noninvariant eigenfunctions must exist: otherwise the complete Dirichlet eigenbasis would span only the proper rotation-invariant subspace of $L^2(D)$. Thus the source's unrestricted real parameter needs this resonance qualification in both (c) and (d).