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Periodic centered-difference Poisson solvability and zero-mean gauge

Statement

Use the grid, inner product, and operator $A_{h}$ of Periodic uniform PIC grid and inner product. Then $A_{h}$ is self-adjoint and nonnegative, $\ker A_{h}$ consists exactly of constant arrays, and $ran A_{h} = {s : \overset{ˉ}{s} = 0}$ . Consequently, for every compatible source $\overset{ˉ}{s} = 0$ , there is a unique solution $ϕ$ of $A_{h} ϕ = s$ with the gauge $\overset{ˉ}{ϕ} = 0$ .

For $k \in K_{N}$ , the Fourier character $e_{k} (j) = e^{2 π i k \cdot j / N}$ is an eigenvector with $λ_{h} (k) = \frac{4}{h^{2}} \sum_{a = 1}^{d} \sin^{2} ⁣ (\frac{π k_{a}}{N}) .$ Only $k = 0$ has eigenvalue zero.

Facts & Assumptions

Given: $d \in {1, 2, 3}$ , odd $N \geq 3$ , $h = L / N$ , the periodic grid, its weighted inner product, and $A_{h}$ .

[F1]

Periodic uniform PIC grid and inner product: $A_{h} u$ is the periodic negative centered Laplacian, and $⟨ u, v ⟩_{h} = h^{d} \sum_{j} u_{j} v_{j}$ .

Proof

1.1givenF1algebra

Reindex the finite periodic sums in each coordinate. For any real arrays $u, v$ , $⟨ u, A_{h} v ⟩_{h} = h^{d} \sum_{a = 1}^{d} \sum_{j} \frac{(u_{j + e_{a}} - u_{j}) (v_{j + e_{a}} - v_{j})}{h^{2}} .$ The right side is symmetric in $u, v$ and is nonnegative when $u = v$ . Thus $A_{h}$ is self-adjoint and nonnegative.

2.1F1step 1.1algebra

If $⟨ u, A_{h} u ⟩_{h} = 0$ , every nonnegative squared difference in step 1.1 is zero, so $u_{j + e_{a}} = u_{j}$ for every $j, a$ . Coordinate steps connect every pair of indices in $J_{N}$ , hence $u$ is constant. Conversely a constant is annihilated by $A_{h}$ .

3.1step 2.1algebra

For a symmetric finite-dimensional linear map, $ran A_{h} = (\ker A_{h})^{⊥}$ : symmetry gives $ran A_{h} \subseteq (\ker A_{h})^{⊥}$ , and rank-nullity makes the dimensions equal. By step 2.1, $(\ker A_{h})^{⊥}$ is precisely the zero-mean subspace. Therefore each zero-mean $s$ is in the range, and any two solutions differ by a constant; imposing $\overset{ˉ}{ϕ} = 0$ selects exactly one.

4.1F1algebra∎

For $e_{k} (j) = e^{2 π i k \cdot j / N}$ , substitution in the stencil from [F1] gives $(A_{h} e_{k})_{j} = \frac{1}{h^{2}} \sum_{a} (2 - e^{2 π i k_{a} / N} - e^{- 2 π i k_{a} / N}) e_{k} (j) = λ_{h} (k) e_{k} (j) .$ The characters form a basis because their finite geometric sums are orthogonal: in one coordinate $\sum_{j = 0}^{N - 1} e^{2 π i (k - ℓ) j / N}$ is $N$ if $k = ℓ$ modulo $N$ and $0$ otherwise; taking products gives $N^{d}$ independent characters. Since every $k_{a} \in {- (N - 1) / 2, \dots, (N - 1) / 2}$ , $λ_{h} (k) = 0$ exactly when every $k_{a} = 0$ .

Depends on

Periodic uniform PIC grid and inner product

Used by

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Sources

Benjamin Seibold, MIT 18.336 Lecture 5, Finite Difference Approximation (standard reference, not scraped)
A. Myers, P. Colella, and B. Van Straalen, A 4th-Order Particle-in-Cell Method with Phase-Space Remapping for the Vlasov–Poisson Equation, Section 2.2 (standard reference, not scraped)