 Rotationally invariant Neumann reconstruction in an equilateral triangle

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 giveFor $\lambda >0$ 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.