= Rotationally invariant Neumann reconstruction in an equilateral triangle
{title2=$q(s)=l^{-1}\sum_{n\in\mathbb Z}Q_ne^{-2\pi ins/l}$}
For the <modified Helmholtz equation> $\Delta q-4\lambda q=0$ with identical oriented outward <Neumann boundary data> on all sides of an <equilateral triangle>, uniqueness forces rotation invariance. Put $\chi(k)=l(k+\lambda/k)/2$, $\Psi(k)=\int_{-l/2}^{l/2}e^{(k+\lambda/k)s}f(s)ds$, and $a=e^{2\pi i/3}$. At $\chi(k_n)=i\pi n$, paired <global relations> give
$$
Q_n=\frac{i[\cosh\chi(\bar a k_n)\Psi(k_n)+(-1)^n\Psi(\bar a k_n)+\Psi(a k_n)]}{(k_n-\lambda/k_n)\sinh\chi(\bar a k_n)}.
$$
For $\lambda>0$ one may choose $k_n=i[\pi n/l+\sqrt{(\pi n/l)^2+\lambda}]$; both denominator factors are nonzero. The derivation retains the tangential-derivative term in each side spectrum. Its endpoint contribution vanishes at these samples because rotation invariance makes all vertex values equal. For $\lambda=0$ the data must have zero total flux and an additive constant remains; negative parameters at <Neumann Laplacian> eigenvalues require the <Fredholm solvability condition> and allow homogeneous additions.
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