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A high Fourier mode aliases to a different periodic PIC grid force

Statement

On $T_{2 π}$ , take the smooth source $ρ (x) = \cos (6 x)$ and solve $- ϕ^{''} = ρ$ with zero mean. Its exact electric field $E = - ϕ^{'}$ is $E (x) = \frac{1}{6} \sin (6 x)$ . On the five-node Fourier-collocation grid $x_{j} = 2 π j / 5$ , the sampled source is exactly $\cos (x_{j})$ . The mean-zero collocation solution is $ϕ_{N} (x) = \cos x$ , and the computed field is $E_{N} (x) = \sin x$ . At every grid node, $E_{N} (x_{j}) = 6 E (x_{j})$ . This is a spectral-sampling aliasing example; it is not the centered finite-difference operator defined on the main PIC page.

Facts & Assumptions

Given: The period $2 π$ , the mode $ρ (x) = \cos (6 x)$ , and the five-node periodic Fourier-collocation convention.

[F1]

Periodic Fourier collocation grid and even-grid Nyquist convention: the nodes are $x_{j} = 2 π j / N$ and the odd- $N$ interpolant contains modes $∣ k ∣ \leq (N - 1) / 2$ .

[F2]

Periodic Fourier modes alias modulo the grid size: if $k \equiv r (m o d N)$ , then $e^{i k x_{j}} = e^{i r x_{j}}$ at every node.

[F3]

Periodic Fourier collocation differentiation matrix and Periodic Fourier collocation differentiates its interpolant exactly: the differentiation matrix differentiates the unique trigonometric interpolant exactly at its nodes.

Proof

1.1F1F2givenalgebra

Since $6 \equiv 1 (m o d 5)$ , [F2] gives $e^{i 6 x_{j}} = e^{i x_{j}}$ for each $j$ . Taking real parts yields $ρ (x_{j}) = \cos (x_{j})$ , whose unique odd-grid interpolant is $\cos x$ by [F1]. On the five-mode space $∣ k ∣ \leq 2$ , the collocation operator $- \partial_{x}^{2}$ multiplies mode $k$ by $k^{2}$ , so its only zero mode is the constant. Therefore $ϕ_{N} = \cos x$ is the unique zero-mean collocation solution to these sampled data.

2.1F1F2F3step 1.1algebra∎

Direct differentiation gives $- \frac{d^{2}}{d x^{2}} (\cos (6 x) / 36) = \cos (6 x)$ and $- \frac{d}{d x} (\cos (6 x) / 36) = \sin (6 x) / 6$ . Since $ϕ_{N} = \cos x$ , exact collocation differentiation [F3], applied twice to its interpolant, gives computed field $E_{N} = \sin x$ . At each node, $\sin (6 x_{j}) = \sin (x_{j} + 2 π j) = \sin x_{j}$ , so $E_{N} (x_{j}) = 6 E (x_{j})$ .

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Sources

Jie Shen and Tao Tang, Spectral and High-Order Methods with Applications, Chapter 2 (standard reference, not scraped)