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The periodic Poisson circulant has one compatible zero mode

Statement

For each $N \geq 3$ , let $C_{N} = C (c)$ be the periodic cycle Laplacian circulant generated by $c_{0} = 2$ , $c_{1} = - 1$ , $c_{N - 1} = - 1$ , and $c_{j} = 0$ otherwise. With $ω = e^{- 2 π i / N}$ , its Fourier eigenvalues are $λ_{k} = 2 - ω^{k} - ω^{- k}$ . Exactly $λ_{0}$ vanishes, so $C_{N} x = f$ is solvable exactly when $\sum_{j = 0}^{N - 1} f_{j} = 0$ ; on the zero-mean subspace the solution is unique. For $N = 4$ and $f = (1, - 1, 1, - 1)^{T}$ , the solution is $x = f / 4$ .

Facts & Assumptions

Given: The circulant convention, canonical DFT, and complex exponential used in the Statement.

[F1]

The circulant generated by $c$ has entries $c_{(i - j) m o d N}$ , and the canonical DFT has negative-sign entries $N^{- 1 / 2} e^{- 2 π i k j / N}$ . Toeplitz, circulant, and finite-index conventions for structured solvers The unitary discrete Fourier transform on $Z / N Z$

[F2]

Circulants are diagonalized by the canonical DFT with eigenvalues $\sum_{j} c_{j} e^{- 2 π i k j / N}$ ; compatibility is equivalent to vanishing of the transformed right-hand side at each zero eigenvalue. The canonical finite DFT diagonalizes every circulant

[F3]

The complex exponential satisfies $\exp (z + w) = \exp (z) \exp (w)$ , $∣ \exp (i y) ∣ = 1$ for real $y$ , and $\exp z = 1$ exactly when $z \in 2 π i Z$ . $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$

[F4]

The complex numbers form a field, so a square of a complex number vanishes only when the number vanishes. $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$

Proof

1.1givenF1F2F3algebra

By [F1]–[F2], the $k$ th eigenvalue of this generator is $λ_{k} = 2 - ω^{k} - ω^{k (N - 1)} .$ Since $ω^{N} = \exp (- 2 π i) = 1$ by [F3], $ω^{k (N - 1)} = ω^{- k}$ , giving the stated symbol.

2.1givenF3F4step 1.1algebra

Put $z = ω^{k}$ . It is nonzero by [F3], and $λ_{k} = 0 ⟺ z + z^{- 1} = 2 ⟺ (z - 1)^{2} = 0 ⟺ z = 1,$ where the middle equivalence follows after multiplying by $z$ , and the last uses [F4]. By the exponential kernel characterization in [F3], $ω^{k} = 1$ exactly when $- 2 π i k / N \in 2 π i Z$ , equivalently $k / N$ is an integer. For $0 \leq k < N$ this means $k = 0$ . Thus only the constant Fourier mode is a zero mode.

3.1F1F2step 2.1algebra

At $k = 0$ , $(F_{N} f)_{0} = N^{- 1 / 2} \sum_{j} f_{j}$ . Hence [F2] gives solvability precisely when $\sum_{j} f_{j} = 0$ . All nonzero modes then determine $F_{N} x$ uniquely, while the zero-mode coordinate is free; imposing $\sum_{j} x_{j} = 0$ fixes that coordinate to zero, proving uniqueness on the zero-mean subspace.

3.2givenF1F3step 1.1step 2.1algebra

When $N = 4$ , $ω = \exp (- π i / 2)$ , so the addition law gives $ω^{2} = \exp (- π i)$ and $\exp (- π i) \exp (π i) = 1$ . Euler's identity gives $\exp (π i) = - 1$ , hence $ω^{2} = - 1$ . Since $ω \neq 0$ , this implies $ω^{- 1} = - ω$ , and the symbol from step 1.1 gives $(λ_{0}, λ_{1}, λ_{2}, λ_{3}) = (0, 2, 4, 2)$ . The vector $f = (1, - 1, 1, - 1)^{T}$ has sum zero. Directly, each cyclic neighbor of $f_{j}$ is $- f_{j}$ , so $(C_{4} f)_{j} = 2 f_{j} - (- f_{j}) - (- f_{j}) = 4 f_{j}$ . Therefore $x = f / 4$ solves $C_{4} x = f$ , and its zero mean gives the unique zero-mean solution.

4.1step 3.1step 3.2algebra∎

The compatibility condition is exactly orthogonality to the constant null mode, and division is only performed on the nonzero Fourier modes. The concrete $N = 4$ vector verifies the normalization directly.

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Sources

Ralf Hiptmair, Numerical Methods for Computational Science and Engineering, Lecture Document, §§4.2.1 and 4.5.1 (standard reference, not scraped)