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Periodic Poisson residual controls the gathered particle force

Statement

Let $s$ be a zero-mean grid source and let $ϕ$ be the zero-mean solution of $A_{h} ϕ = s$ . Let $\tilde{ϕ}$ be any approximate grid potential, and set $r = s - A_{h} \tilde{ϕ}$ ; then $\overset{ˉ}{r} = 0$ . Define $γ_{N, L} = \max_{\begin{array}{c} k \in K_{N} \\ k \neq 0 \end{array}} \frac{{(\sum_{a = 1}^{d} h^{- 2} \sin^{2} (2 π k_{a} / N))}^{1 / 2}}{4 h^{- 2} \sum_{a = 1}^{d} \sin^{2} (π k_{a} / N)} .$ For the centered field operator $D^{0} = (D_{a}^{0})_{a = 1}^{d}$ , $∥ D^{0} (ϕ - \tilde{ϕ}) ∥_{h} \leq γ_{N, L} ∥ r ∥_{h}, γ_{N, L} \leq \frac{h}{2 \sin (π / N)} \leq \frac{L}{4} .$ For any particle location $x_{p}$ and charge-to-mass ratio $q_{p} / m_{p}$ with $m_{p} > 0$ , the force gather using the weights of Periodic PIC deposition shape and gather operators obeys $∣ \frac{q_{p}}{m_{p}} G_{x_{p}} ⁣ (- D^{0} ϕ + D^{0} \tilde{ϕ}) ∣ \leq \frac{∣ q_{p} ∣}{m_{p}} h^{- d / 2} γ_{N, L} ∥ r ∥_{h} .$ Thus the norm in the force certificate is the actual finite-grid inverse/derivative norm; the explicit $h^{- d / 2}$ conversion is retained when passing from grid $L^{2}$ to a particle gather.

Facts & Assumptions

Given: An odd periodic grid, its centered Poisson operator and field differences, a compatible source, an approximate potential, and nonnegative normalized shape weights.

[F1]

Periodic uniform PIC grid and inner product: $A_{h}$ is the centered periodic stencil and $D_{a}^{0}$ is its centered field difference.

[F2]

Periodic centered-difference Poisson solvability and zero-mean gauge: $A_{h}$ has kernel the constants and range the zero-mean arrays; the Fourier characters form a basis, and on mode $k \neq 0$ , $A_{h}$ has eigenvalue $λ_{h} (k) = 4 h^{- 2} \sum_{a} \sin^{2} (π k_{a} / N) > 0$ . The symbol of $D_{a}^{0}$ is derived from its stencil in [F1].

[F3]

Periodic PIC deposition shape and gather operators: $G_{x_{p}} E = \sum_{j} w_{j} (x_{p}) E_{j}$ with nonnegative weights of total one.

[F4]

The periodic Poisson mean-zero pseudoinverse is linear and self-adjoint: for zero-mean $r$ , $A_{h}^{†} r$ is the unique mean-zero solution.

Proof

1.1F1F2F4givenalgebra

Let $δ ϕ = ϕ - \tilde{ϕ}$ . Since $A_{h} ϕ = s$ , $A_{h} δ ϕ = r$ . The range statement in [F2] gives $\overline{A_{h} \tilde{ϕ}} = 0$ , so $\overset{ˉ}{r} = \overset{ˉ}{s} - \overline{A_{h} \tilde{ϕ}} = 0$ . By [F4], $A_{h}^{†} r$ is the mean-zero solution of $A_{h} u = r$ ; hence $δ ϕ - A_{h}^{†} r$ lies in $\ker A_{h}$ and is constant by [F2]. Expand $r$ in the finite orthogonal Fourier-character basis from [F2]. On each nonzero mode, $A_{h}^{†} r$ has coefficient ${\hat{r}}_{k} / λ_{h} (k)$ by [F2], while the possible constant difference is annihilated by $D^{0}$ .

2.1F1F2F4step 1.1algebra

For a Fourier character $e_{k}$ , direct substitution into the centered difference from [F1] gives $(D_{a}^{0} e_{k})_{j} = \frac{e_{k} (j + e_{a}) - e_{k} (j - e_{a})}{2 h} = \frac{e^{2 π i k_{a} / N} - e^{- 2 π i k_{a} / N}}{2 h} e_{k} (j) = \frac{i \sin (2 π k_{a} / N)}{h} e_{k} (j) .$ By [F2], $A_{h}$ has eigenvalue $λ_{h} (k) > 0$ on every nonzero mode, so [F4] implies that $A_{h}^{†} r$ has coefficient ${\hat{r}}_{k} / λ_{h} (k)$ there. Thus the vector multiplier of $D^{0} A_{h}^{†}$ has Euclidean modulus ${(\sum_{a} h^{- 2} \sin^{2} (2 π k_{a} / N))}^{1 / 2} / λ_{h} (k)$ . Parseval's identity for this finite orthogonal basis therefore gives the first bound with the displayed maximum $γ_{N, L}$ .

3.1F2step 2.1algebra

Put $S = \sum_{a} \sin^{2} (π k_{a} / N)$ . For $k \neq 0$ , some coordinate satisfies $1 \leq ∣ k_{a} ∣ \leq (N - 1) / 2$ because $N$ is odd and $k \in K_{N}$ . Hence $π / N \leq π ∣ k_{a} ∣ / N < π / 2$ ; monotonicity of sine on $[0, π / 2]$ gives $S \geq \sin^{2} (π / N) > 0$ . Since $∣ \sin (2 θ) ∣ \leq 2 ∣ \sin θ ∣$ , the numerator in the definition of $γ$ is at most $2 h^{- 1} \sqrt{S}$ , while its denominator is $4 h^{- 2} S$ . Their ratio is at most $h / (2 \sqrt{S}) \leq h / (2 \sin (π / N))$ . For $N \geq 3$ , $π / N \leq π / 2$ and concavity of sine gives $\sin (π / N) \geq 2 / N$ , so this is at most $h N / 4 = L / 4$ .

4.1F3step 2.1step 3.1algebra∎

Cauchy–Schwarz gives $∣ G_{x_{p}} E ∣ \leq (\sum_{j} w_{j} (x_{p})^{2})^{1 / 2} (\sum_{j} ∣ E_{j} ∣^{2})^{1 / 2}$ . Nonnegativity and $\sum_{j} w_{j} = 1$ imply $\sum_{j} w_{j}^{2} \leq 1$ , while $\sum_{j} ∣ E_{j} ∣^{2} = h^{- d} ∥ E ∥_{h}^{2}$ . Apply this to $E = D^{0} (ϕ - \tilde{ϕ})$ and use steps 2.1–3.1. Multiplying by $∣ q_{p} ∣ / m_{p}$ gives the gathered-force estimate.

Depends on

Used by

Fixed-regularization periodic PIC force consistency on a smooth density Theorem

Dependency tree · one level

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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)