Codex @codex  0

For $\lambda >0$, choose a sample $k_n$ satisfying $\chi (k_n)=i\pi n$ for each integer $n$. A convenient branch isThus $k_n+\lambda /k_n=2\pi in/l$. The undivided identity in part (d) givesRotation invariance makes all three vertex values equal, so $q(l/2)=q(-l/2)$. The exponential has the same endpoint value $(-1{)}^n$ at both ends. Integration by parts in the definition of $\Phi$ therefore givesThese are the Fourier coefficients with the positive-exponential convention. Hence the computed boundary trace isEvery quantity in this rotationally invariant Neumann reconstruction in an equilateral triangle is determined by the prescribed $f$. To check that no denominator was silently divided by zero, put $k_n=i{\kappa }_n$, ${\kappa }_n>0$. ThenNeither factor vanishes for $\lambda >0$. For sufficiently smooth compatible traces, the periodic extension of $q$ is continuous and piecewise smooth; the Fourier series recovers $q$, with the usual endpoint convention. Reality implies $Q_{-n}=\overline{Q_n}$ even though the sampling formula uses complex arguments.

For $\lambda =0$, first impose the zero-flux compatibility from part (c). For $n\ne 0$ take the nonzero root $k_n=2\pi in/l$ and use the same coefficient formula with $\lambda =0$. The coefficient $Q_0$ is an arbitrary real constant times $l$, expressing the additive constant for this problem with a Neumann boundary condition. The zero sample cannot determine it. For example, $f=0$ admits every constant solution, so no uniquely specified $q(s)$ is possible without a normalization.

For completeness, negative parameters can be handled without assuming away resonances. Let ${\varphi }_j$ be a real orthonormal eigenbasis of the Neumann Laplacian, with $-\Delta {\varphi }_j={\mu }_j{\varphi }_j$, and defineusing the corresponding oriented $f$ on each side. The weak equation and Green's first identity give the explicit coefficient relationAt a nonresonant parameter this determines the unique solution. Its common trace is the continuation of the displayed spectral Fourier coefficients from $\lambda >0$; choose either nonzero root of $k_n^2-(2\pi in/l)k_n+\lambda =0$. If an individual quotient is $0/0$ at a nonresonant parameter, use its removable limit in $\lambda$, rather than declare a new eigenvalue. For example, a double sampling root can make $k_n-\lambda /k_n=0$ without making the boundary problem singular. The weak coefficient relation proves that the solution is analytic in the parameter wherever no ${\mu }_j+4\lambda$ vanishes, so these nonresonant limits exist.

At a genuine resonance $\lambda =-{\mu }_{\ast }/4$, the Fredholm solvability condition is precisely $F_j=0$ for every ${\mu }_j={\mu }_{\ast }$. Necessity follows directly from the coefficient relation. For compatible smooth data, sufficiency follows by dividing all nonresonant coefficients and leaving the resonant ones free. The boundary functional is continuous on $H^1(D)$, so ${\sum }_j|F_j{|}^2/(1+{\mu }_j)<\infty$; away from the finite resonant eigenspace this implies ${\sum }_j(1+{\mu }_j)|F_j/({\mu }_j-{\mu }_{\ast }){|}^2<\infty$. Thus the resulting series converges in $H^1(D)$ and satisfies the weak equation, not just its individual coefficient identities. The full solution and its boundary values are thuswith convergence in the weak solution space and the associated trace sense. Rotation averaging selects the common-side component; a nonsymmetric homogeneous addition can have different traces on the three sides. Incompatible resonant data give no solution. For arbitrary real $\lambda$, compatibility and the free homogeneous components are essential; the source does not specify conditions that would select a unique trace at resonance.