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✓ 17 results · all verified · 15 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Particle in cell and particle mesh kinetic methods

1 · Prerequisites

2 · Summary

This page fixes a periodic electrostatic particle/grid model on $T_{L}^{d}$ , $d \in {1, 2, 3}$ , with an odd uniform grid, a nonnegative partition-of-unity deposition shape, a uniform neutralizing background, centered differences, and a zero-mean periodic Poisson gauge. The finite Fourier solver is exact-arithmetic; floating-point FFT residuals require a separate certificate. Deposition and potential gathering are adjoints under the cell-volume grid pairing. A centered-field residual estimate uses the actual inverse/derivative norm and retains the grid-to-particle $h^{- d / 2}$ factor.

Two force branches are kept distinct. The ordinary particle-mesh interaction uses a fixed- $ε$ smooth periodic assignment kernel, which makes the finite particle ODE globally smooth and Lipschitz. A separate variational grid energy defines the canonical force used for the leapfrog geometry theorem. Adjoint gathering alone does not identify the ordinary interpolated force with that variational force, and neither symplecticity nor reversibility gives exact discrete energy conservation.

For fixed $ε$ , the smooth-density force error is bounded by $L^{d / 2} h^{2 - d / 2} (γ_{N, L} C_{s r c} + C_{3}) + C_{i n t} h^{2}$ , with $γ_{N, L} \leq L / 4$ and source/field derivative constants stated in the theorem. Vector quadrature uses Euclidean norms and a supremum over all shifted kernel/field products, preserving the scalar $1 / 12$ coefficient without a dimension factor. The regularized continuum force is globally Lipschitz, so the XD-08 Dobrushin estimate propagates a prescribed initial $W_{1}$ error and a uniform force defect. Classical RK4 contributes a conditional $P^{- 1 / 2} C_{R K} Δ t^{4}$ term on uniform grid times $t_{n} = n Δ t \leq T$ with $0 < Δ t \leq h_{0}$ , when the $C^{4}$ field and all paired stages stay in the declared compact tube and satisfy its local-defect bounds. The grid need not end at $T$ . The resulting estimate is at fixed $ε$ ; it asserts no singular $ε \to 0$ limit and no universal empirical $P^{- 1 / 2}$ rate.

Grid crossing, spectral aliasing, and stochastic loading are separate effects. The CIC example shows that a piecewise-linear assignment weight has a derivative jump at a node, so the smooth-tube RK4 theorem is not supplied by CIC regularity. The Fourier alias example concerns spectral collocation sampling, not the centered finite-difference solver. The chosen loading enters through the actual initial discrepancy $δ_{0}$ ; no sampling law is assumed. A local periodic solvability proof is included because the inspected elliptic supplier page states a rectangular Dirichlet solver and does not provide the periodic Poisson interface planned in the design. Owner reconciliation is recorded in batch evidence.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Periodic uniform PIC grid and inner product

Definition

Fix a spatial dimension $d \in {1, 2, 3}$ , side length $L > 0$ , and odd integer $N \geq 3$ . Put $h = L / N$ and $T_{L}^{d} = (R / L Z)^{d}$ . The periodic grid has index set $J_{N} = (Z / N Z)^{d}$ and nodes $x_{j} = h j$ modulo $L$ . A grid scalar is a real array $u = (u_{j})_{j \in J_{N}}$ ; a grid vector has $d$ real components at each node. Define $⟨ u, v ⟩_{h} = h^{d} \sum_{j \in J_{N}} u_{j} v_{j}, ∥ u ∥_{h} = ⟨ u, u ⟩_{h}^{1 / 2}, \overset{ˉ}{u} = L^{- d} h^{d} \sum_{j \in J_{N}} u_{j} .$ The same pairing is componentwise for grid vectors. For the coordinate vector $e_{a}$ , define the centered field difference and positive periodic Poisson stencil by $(D_{a}^{0} u)_{j} = \frac{u_{j + e_{a}} - u_{j - e_{a}}}{2 h}, (A_{h} u)_{j} = \sum_{a = 1}^{d} \frac{2 u_{j} - u_{j + e_{a}} - u_{j - e_{a}}}{h^{2}} .$ All indices wrap modulo $N$ . The discrete Fourier index set is $K_{N} = {- \frac{N - 1}{2}, \dots, \frac{N - 1}{2}}^{d}$ ; the character at $k \in K_{N}$ is $j \mapsto e^{2 π i k \cdot j / N}$ . This odd-grid convention has no Nyquist mode.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Composite periodic trapezoidal error in a uniform tensor grid

Statement

Let $L > 0$ , odd $N \geq 3$ , $h = L / N$ , and $f \in C^{2} (T_{L}^{d})$ for $d \in {1, 2, 3}$ . With the uniform tensor grid $x_{j} = h j$ and indices taken modulo $N$ , $∣ h^{d} \sum_{j \in (Z / N Z)^{d}} f (x_{j}) - \int_{[0, L]^{d}} f (x) d x ∣ \leq \frac{L^{d} h^{2}}{12} \sum_{a = 1}^{d} ∥ \partial_{a a} f ∥_{\infty} .$

Facts & Assumptions

Given: A periodic $C^{2}$ function and a uniform periodic tensor grid with $N$ cells in each coordinate.

[F1]

Periodic uniform PIC grid and inner product: the periodic grid uses $h = L / N$ , wraps indices modulo $N$ , and its tensor nodes are $x_{j} = h j$ .

Proof

1.1F1algebra

In one coordinate, on a cell $[a, a + h]$ , put $K (x) = (x - a) (a + h - x) / 2$ . Two integrations by parts give $\frac{h}{2} (f (a) + f (a + h)) - \int_{a}^{a + h} f (x) d x = \int_{a}^{a + h} K (x) f^{''} (x) d x .$ Since $K \geq 0$ and $\int_{a}^{a + h} K (x) d x = h^{3} / 12$ , the absolute cell error is at most $h^{3} ∥ f^{''} ∥_{\infty} / 12$ . Sum over the $N$ periodic cells, whose endpoint contributions assemble into the periodic trapezoid sum $h \sum_{j} f (j h)$ . This gives the one-dimensional error bound $L h^{2} ∥ f^{''} ∥_{\infty} / 12$ .

2.1F1step 1.1algebra∎

For each coordinate $a$ , let $Q_{a}$ denote the periodic trapezoid operator in that coordinate and $I_{a}$ integration over that coordinate. Both have sup-norm operator bound at most $L$ , since their weights are nonnegative and sum to $L$ . The telescoping identity $Q_{1} \dots Q_{d} - I_{1} \dots I_{d} = \sum_{a = 1}^{d} Q_{1} \dots Q_{a - 1} (Q_{a} - I_{a}) I_{a + 1} \dots I_{d}$ holds by cancellation. In the $a$ th summand, the one-dimensional bound from step 1.1 applies after the later-coordinate integrations; differentiation in coordinate $a$ commutes with those integrations. The other $d - 1$ operators contribute at most $L^{d - 1}$ in sup norm. Thus that summand is bounded by $L^{d} h^{2} ∥ \partial_{a a} f ∥_{\infty} / 12$ . Summing these bounds gives the displayed estimate, because $Q_{1} \dots Q_{d} f = h^{d} \sum_{j} f (x_{j})$ and $I_{1} \dots I_{d} f = \int_{[0, L]^{d}} f$ .

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Periodic PIC deposition shape and gather operators

Definition

On the torus and grid of Periodic uniform PIC grid and inner product, an algebraic assignment shape is any family of functions $w_{j} : T_{L}^{d} \to [0, \infty)$ such that $\sum_{j \in J_{N}} w_{j} (x) = 1$ for every $x$ . For the smooth fixed- $ε$ branch used in the force-error and ODE results, fix $ε > 0$ , put $α_{ε} = e^{- ε / L}$ and choose the explicit even periodic kernel $η_{ε} (x) = L^{- d} \prod_{a = 1}^{d} \frac{1 - α_{ε}^{2}}{1 - 2 α_{ε} \cos (2 π x_{a} / L) + α_{ε}^{2}} .$ For this smooth branch and each $x \in T_{L}^{d}$ , set $Z_{ε, h} (x) = h^{d} \sum_{j \in J_{N}} η_{ε} (x_{j} - x), w_{j}^{ε, h} (x) = \frac{h^{d} η_{ε} (x_{j} - x)}{Z_{ε, h} (x)} .$ Each factor in the kernel denominator is positive, so $Z_{ε, h} (x) > 0$ and the smooth-branch weights are well-defined. For any algebraic shape family above and finitely many particle positions $X = (X_{1}, \dots, X_{P})$ with charges $q_{1}, \dots, q_{P} \in R$ , write $Q = \sum_{p = 1}^{P} q_{p}$ and define deposited charge density $ρ_{j} (X) = h^{- d} \sum_{p = 1}^{P} q_{p} w_{j} (X_{p}) - Q / L^{d} .$ For any grid scalar $g$ , its potential gather is $(G_{X} g)_{p} = \sum_{j} w_{j} (X_{p}) g_{j}$ . For any grid vector $E$ , its field gather is defined componentwise by the same weights, $(G_{X} E)_{p} = \sum_{j} w_{j} (X_{p}) E_{j}$ . On the smooth fixed- $ε$ branch, $w_{j}$ means the explicit family $w_{j}^{ε, h}$ above.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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 $k_{a} = 0$ .

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The periodic Poisson mean-zero pseudoinverse is linear and self-adjoint

Statement

For any real grid array $s$ , define its mean-zero periodic Poisson pseudoinverse by $A_{h}^{†} s : = the unique mean-zero grid array u such that A_{h} u = s - \overset{ˉ}{s} 1,$ where $\overset{ˉ}{s} = L^{- d} ⟨ s, 1 ⟩_{h}$ . This definition is well-defined by Periodic centered-difference Poisson solvability and zero-mean gauge. It gives $A_{h} A_{h}^{†} s = s - \overset{ˉ}{s} 1$ , $A_{h}^{†} 1 = 0$ , and a linear self-adjoint map on the full finite grid space.

Facts & Assumptions

Given: The real grid space, its weighted inner product and mean, and the periodic centered Poisson operator.

[F1]

Periodic uniform PIC grid and inner product: $⟨ u, v ⟩_{h} = h^{d} \sum_{j} u_{j} v_{j}$ and $\overset{ˉ}{s} = L^{- d} ⟨ s, 1 ⟩_{h}$ .

[F2]

Periodic centered-difference Poisson solvability and zero-mean gauge: $A_{h}$ is self-adjoint, its range is the zero-mean subspace, and every zero-mean source has a unique mean-zero solution.

Proof

1.1F1F2given

For any $s$ , the source $s - \overset{ˉ}{s} 1$ has mean zero by [F1]. By [F2] it has a unique mean-zero solution; this defines $A_{h}^{†} s$ , proves $A_{h} A_{h}^{†} s = s - \overset{ˉ}{s} 1$ , and gives $A_{h}^{†} 1 = 0$ .

2.1F1F2step 1.1algebra

Let $a, b \in R$ and grid arrays $s, t$ . The array $a A_{h}^{†} s + b A_{h}^{†} t$ has mean zero, and applying $A_{h}$ gives $a s + b t - (a \overset{ˉ}{s} + b \overset{ˉ}{t}) 1 = a s + b t - \overline{a s + b t} 1$ . Uniqueness in [F2] therefore yields $A_{h}^{†} (a s + b t) = a A_{h}^{†} s + b A_{h}^{†} t$ .

3.1F1F2step 1.1step 2.1algebra∎

Both $A_{h}^{†} s$ and $A_{h}^{†} t$ have mean zero. Hence, using [F2] and step 1.1, $⟨ A_{h}^{†} s, t ⟩_{h} = ⟨ A_{h}^{†} s, t - \overset{ˉ}{t} 1 ⟩_{h} = ⟨ A_{h}^{†} s, A_{h} A_{h}^{†} t ⟩_{h} = ⟨ A_{h} A_{h}^{†} s, A_{h}^{†} t ⟩_{h} = ⟨ s - \overset{ˉ}{s} 1, A_{h}^{†} t ⟩_{h} = ⟨ s, A_{h}^{†} t ⟩_{h} .$ The middle equality uses self-adjointness of $A_{h}$ . This proves that the full-space linear map $A_{h}^{†}$ is self-adjoint.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Fourier diagonal solver for the periodic centered-difference Poisson equation

Definition

Input an odd $N \geq 3$ periodic grid as in Periodic uniform PIC grid and inner product and a real grid source $s$ . If $\overset{ˉ}{s} \neq 0$ , return the status incompatible-source. Otherwise compute the finite discrete Fourier coefficients ${\hat{s}}_{k} = N^{- d} \sum_{j} s_{j} e^{- 2 π i k \cdot j / N}$ . Set ${\hat{ϕ}}_{0} = 0$ and, for each $k \neq 0$ , set ${\hat{ϕ}}_{k} = {\hat{s}}_{k} / λ_{h} (k)$ with $λ_{h} (k)$ from Periodic centered-difference Poisson solvability and zero-mean gauge. Return the inverse finite Fourier transform $ϕ_{j} = \sum_{k \in K_{N}} {\hat{ϕ}}_{k} e^{2 π i k \cdot j / N}$ and status success. All arithmetic here is exact real or complex arithmetic; an implementation using floating-point FFTs requires a separate perturbation analysis.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Periodic PIC deposition conserves charge and satisfies Poisson compatibility

Statement

For positions $X_{1}, \dots, X_{P} \in T_{L}^{d}$ and real charges $q_{1}, \dots, q_{P}$ , define $Q = \sum_{p} q_{p}$ and an admissible algebraic assignment shape $w_{j}$ as in Periodic PIC deposition shape and gather operators. Then $h^{d} \sum_{j \in J_{N}} h^{- d} \sum_{p = 1}^{P} q_{p} w_{j} (X_{p}) = Q .$ Consequently the neutralized source $ρ_{j} (X) = h^{- d} \sum_{p} q_{p} w_{j} (X_{p}) - Q / L^{d}$ has mean zero and is in the range of the periodic Poisson operator; it therefore has a unique mean-zero potential for the gauge $\overset{ˉ}{ϕ} = 0$ by Periodic centered-difference Poisson solvability and zero-mean gauge.

Facts & Assumptions

Given: The finite periodic grid, particle positions and charges, the normalized assignment shape, and the periodic Poisson operator.

[F1]

Periodic PIC deposition shape and gather operators: $ρ_{j} = h^{- d} \sum_{p} q_{p} w_{j} (X_{p}) - Q / L^{d}$ , and an algebraic assignment has nonnegative weights.

[F2]

Periodic PIC deposition shape and gather operators: every algebraic assignment shape satisfies $\sum_{j} w_{j} (x) = 1$ for every particle location.

[F3]

Periodic centered-difference Poisson solvability and zero-mean gauge: the range of $A_{h}$ is exactly the zero-mean grid subspace, and every zero-mean source has a unique mean-zero solution.

Proof

1.1givenF1F2algebra

Before subtracting the background, sum the assigned density with cell volume. Since the sum is finite, interchange its particle and grid sums; [F2] gives $h^{d} \sum_{j} h^{- d} \sum_{p} q_{p} w_{j} (X_{p}) = \sum_{p} q_{p} \sum_{j} w_{j} (X_{p}) = \sum_{p} q_{p} = Q .$

2.1F1step 1.1algebra

The background has total grid charge $h^{d} ∣ J_{N} ∣ Q / L^{d} = L^{d} Q / L^{d} = Q$ because $h^{d} ∣ J_{N} ∣ = L^{d}$ . Hence $⟨ ρ, 1 ⟩_{h} = 0$ , equivalently $\overset{ˉ}{ρ} = 0$ .

3.1F3step 2.1∎

By [F3], this zero-mean source has a solution and fixing its mean to zero selects exactly one potential.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Periodic PIC regularization gives a smooth compatible Poisson solution

Statement

Let $η_{ε}$ be the explicit product kernel in Periodic PIC deposition shape and gather operators, with $α_{ε} = e^{- ε / L} \in (0, 1)$ . Let $μ$ be a finite signed Borel measure on $T_{L}^{d}$ with mass $Q$ and total variation $∥ μ ∥_{T V}$ . Put $\hat{μ} (k) = \int_{T_{L}^{d}} e^{- 2 π i k \cdot y / L} d μ (y), κ_{k} = 2 π k / L .$ Then $η_{ε}$ is strictly positive, even, smooth, and has integral one. The regularized neutralized source $ρ_{ε}^{μ} (x) = \int_{T_{L}^{d}} η_{ε} (x - y) d μ (y) - Q / L^{d}$ is $C^{\infty}$ with zero mean. Its unique zero-mean periodic solution of $- Δ ϕ_{ε}^{μ} = ρ_{ε}^{μ}$ is the absolutely and uniformly differentiable Fourier series $ϕ_{ε}^{μ} (x) = L^{- d} \sum_{k \in Z^{d} ∖ {0}} \frac{(\prod_{a = 1}^{d} α_{ε}^{∣ k_{a} ∣}) \hat{μ} (k)}{∣ κ_{k} ∣^{2}} e^{i κ_{k} \cdot x} .$ For every integer $r \geq 0$ , all spatial derivative series of order at most $r$ converge absolutely and uniformly; their bounds may depend on $r, ε, L, ∥ μ ∥_{T V}$ .

Facts & Assumptions

Given: $d \in {1, 2, 3}$ , $L, ε > 0$ , the product periodic kernel from the shape definition, and a finite real signed measure $μ$ on the torus.

[F1]

Periodic PIC deposition shape and gather operators: $η_{ε} (x) = L^{- d} \prod_{a} (1 - α_{ε}^{2}) / (1 - 2 α_{ε} \cos (2 π x_{a} / L) + α_{ε}^{2})$ with $α_{ε} = e^{- ε / L}$ .

Proof

1.1F1algebra

For $0 < α < 1$ , the two geometric series $\sum_{n \geq 0} α^{n} e^{i n θ}$ and $\sum_{n \geq 1} α^{n} e^{- i n θ}$ converge absolutely and uniformly. Adding them and simplifying the denominator gives $\sum_{n \in Z} α^{∣ n ∣} e^{i n θ} = 1 + \frac{α e^{i θ}}{1 - α e^{i θ}} + \frac{α e^{- i θ}}{1 - α e^{- i θ}} = \frac{1 - α^{2}}{1 - 2 α \cos θ + α^{2}} .$ The denominator is $(1 - α)^{2} + 2 α (1 - \cos θ) > 0$ . The constant Fourier coefficient is $1$ , so integrating the absolutely convergent series over one period gives mean one. Applying this in every coordinate proves the positivity, evenness, smoothness, and unit integral of $η_{ε}$ .

2.1F1step 1.1algebra

Multiplying the coordinate Fourier series from step 1.1 gives $η_{ε} (x) = L^{- d} \sum_{k \in Z^{d}} (\prod_{a} α_{ε}^{∣ k_{a} ∣}) e^{i κ_{k} \cdot x} .$ The product coefficients are absolutely summable, so uniform convergence permits termwise integration against the finite measure $μ$ . The $k = 0$ term is $Q / L^{d}$ and is removed by neutralization. For each $k \neq 0$ , $∣ \hat{μ} (k) ∣ \leq ∥ μ ∥_{T V}$ . For every integer $m \geq 0$ , $\sum_{n \in Z} (1 + ∣ n ∣)^{m} α_{ε}^{∣ n ∣} < \infty$ by the ratio test; taking products bounds the corresponding $d$ -dimensional weighted sums. Hence the source and every formally differentiated Fourier series converge absolutely and uniformly, proving that $ρ_{ε}^{μ} \in C^{\infty}$ and has zero mean.

3.1step 2.1algebra∎

For $k \neq 0$ , $- Δ e^{i κ_{k} \cdot x} = ∣ κ_{k} ∣^{2} e^{i κ_{k} \cdot x}$ . Dividing the nonzero source coefficients in step 2.1 by this positive eigenvalue gives the displayed $ϕ_{ε}^{μ}$ ; the weighted geometric summability in step 2.1 justifies termwise differentiation and the equation. Its zero Fourier coefficient is set to zero, so it has mean zero. If two mean-zero $C^{2}$ potentials solve the equation, their difference $u$ satisfies $Δ u = 0$ . Periodic integration by parts gives $\int_{T_{L}^{d}} ∣ \nabla u ∣^{2} = - \int u Δ u = 0$ , so $\nabla u = 0$ , $u$ is constant, and its zero mean forces $u = 0$ . This proves uniqueness.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Periodic PIC deposition and potential gather are adjoint

Statement

Let $ρ_{j}^{+} = h^{- d} \sum_{p} q_{p} w_{j} (X_{p})$ be the deposited density before background subtraction for an admissible algebraic shape $w_{j}$ of Periodic PIC deposition shape and gather operators, and let $G_{X}$ be its scalar potential gather. For every grid scalar $ϕ$ , $⟨ ρ^{+}, ϕ ⟩_{h} = \sum_{p = 1}^{P} q_{p} (G_{X} ϕ)_{p} .$ For the neutralized density $ρ = ρ^{+} - Q / L^{d}$ the corresponding identity is $⟨ ρ, ϕ ⟩_{h} = \sum_{p} q_{p} (G_{X} ϕ)_{p} - Q \overset{ˉ}{ϕ} .$ Thus the background term vanishes in the zero-mean potential gauge.

Facts & Assumptions

Given: The grid inner product, particle positions and charges, and the deposition and gather formulas.

[F1]

Periodic uniform PIC grid and inner product: $⟨ u, v ⟩_{h} = h^{d} \sum_{j} u_{j} v_{j}$ and $\overset{ˉ}{ϕ} = L^{- d} ⟨ ϕ, 1 ⟩_{h}$ .

[F2]

Periodic PIC deposition shape and gather operators: $ρ_{j}^{+} = h^{- d} \sum_{p} q_{p} w_{j} (X_{p})$ and $(G_{X} ϕ)_{p} = \sum_{j} w_{j} (X_{p}) ϕ_{j}$ .

Proof

1.1givenF1F2algebra

Substitute [F2] into the left pairing and cancel $h^{d} h^{- d}$ : $⟨ ρ^{+}, ϕ ⟩_{h} = h^{d} \sum_{j} h^{- d} \sum_{p} q_{p} w_{j} (X_{p}) ϕ_{j} = \sum_{p} q_{p} \sum_{j} w_{j} (X_{p}) ϕ_{j} .$ The finite inner sum is $(G_{X} ϕ)_{p}$ by [F2], proving the first identity.

2.1F1step 1.1algebra∎

The background contributes ${⟨ Q / L^{d}, ϕ ⟩}_{h} = Q L^{- d} h^{d} \sum_{j} ϕ_{j} = Q \overset{ˉ}{ϕ}$ by [F1]. Subtract this term from step 1.1 to obtain the second identity.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Smooth periodic PIC assignment weights have bounded derivatives

Statement

For fixed $ε > 0$ and periodic grid, the smooth-branch weights $w_{j}^{ε, h}$ of Periodic PIC deposition shape and gather operators are $C^{\infty}$ on $T_{L}^{d}$ . For every multi-index $β$ , $\max_{j} ∥ \partial^{β} w_{j}^{ε, h} ∥_{\infty} < \infty$ . They are nonnegative and sum to one at every point.

Facts & Assumptions

Given: The explicit smooth periodic kernel $η_{ε}$ , its sampled normalizer $Z_{ε, h}$ , and the displayed weights.

[F1]

Periodic PIC deposition shape and gather operators: $Z_{ε, h} (x) = h^{d} \sum_{j} η_{ε} (x_{j} - x)$ and $w_{j}^{ε, h} (x) = h^{d} η_{ε} (x_{j} - x) / Z_{ε, h} (x)$ , where $Z_{ε, h} (x) > 0$ .

[F2]

Periodic PIC regularization gives a smooth compatible Poisson solution: $η_{ε}$ is strictly positive and $C^{\infty}$ on the compact torus.

Proof

1.1F1F2algebra

By [F2], each numerator in [F1] is a smooth periodic function. The finite sum defining $Z_{ε, h}$ is smooth and strictly positive; compactness gives a positive minimum. The reciprocal of a positive smooth function is smooth, so every ratio $w_{j}^{ε, h}$ is $C^{\infty}$ . Each derivative is continuous on the compact torus and hence bounded; there are finitely many indices $j$ , so their maximum is finite.

2.1F1step 1.1algebra∎

Each weight is nonnegative by [F1] and positivity of $η_{ε}$ . Summing [F1] over $j$ gives $\sum_{j} w_{j}^{ε, h} (x) = Z_{ε, h} (x) / Z_{ε, h} (x) = 1$ .

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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.

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Fixed-regularization periodic PIC pair kernel

Definition

Fix $ε > 0$ , $d \in {1, 2, 3}$ , $L > 0$ , and a regularized one-species model with total charge $Q$ and total mass $M > 0$ . Let $c = Q / M$ . For each position $y \in T_{L}^{d}$ , define the continuous unit-source field by taking the zero-mean solution $ϕ_{ε}^{y}$ from Periodic PIC regularization gives a smooth compatible Poisson solution for $μ = δ_{y}$ , and setting $E_{ε} (x, y) = \int_{T_{L}^{d}} η_{ε} (z - x) (- \nabla ϕ_{ε}^{y} (z)) d z = \frac{1}{L^{d}} \sum_{k \in Z^{d} ∖ {0}} \frac{- i κ_{k}}{∣ κ_{k} ∣^{2}} (\prod_{a = 1}^{d} α_{ε}^{2 ∣ k_{a} ∣}) e^{i κ_{k} \cdot (x - y)} .$ The series converges absolutely with all derivatives by geometric summability. For the grid branch, let $ϕ_{h}^{y}$ be the mean-zero solution of $A_{h} ϕ_{h}^{y} = h^{- d} w^{ε, h} (y) - L^{- d} 1$ whose compatibility follows from Periodic PIC deposition conserves charge and satisfies Poisson compatibility, and put $E_{ε, h} (x, y) = \sum_{j \in J_{N}} w_{j}^{ε, h} (x) (- D^{0} ϕ_{h}^{y})_{j} .$ If $z = (x, v)$ and $z^{'} = (y, w)$ are lifted phase-space states in $R^{2 d}$ , the continuum and grid mean-field kernels are $K_{ε} (z, z^{'}) = (v, c Q E_{ε} (x, y)), K_{ε, h} (z, z^{'}) = (v, c Q E_{ε, h} (x, y)) .$ The spatial components are periodic in their position variables. A finite equal-weight system ${\dot{z}}_{p} = P^{- 1} \sum_{r = 1}^{P} K_{ε, h} (z_{p}, z_{r})$ uses particle charge $Q / P$ and mass $M / P$ . No claim about the limit $ε \to 0$ is part of this definition.

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Partition and quadrature bounds for a periodic PIC shape

Statement

Let $d, L, N, h, η_{ε}, Z_{ε, h}$ , and $w_{j}^{ε, h}$ have the meanings in Periodic uniform PIC grid and inner product and Periodic PIC deposition shape and gather operators. For a finite signed Borel measure $μ$ on the torus, write $∥ μ ∥_{T V} = ∣ μ ∣ (T_{L}^{d})$ and define $(P_{h}^{ε} μ)_{j} = h^{- d} \int_{T_{L}^{d}} w_{j}^{ε, h} (x) d μ (x) .$ Put $C_{η} = \frac{L^{d}}{12} \sum_{a = 1}^{d} ∥ \partial_{a a} η_{ε} ∥_{\infty}, Z_{\min} = L^{d} \min_{x \in T_{L}^{d}} η_{ε} (x) > 0.$ Then $\sum_{j} w_{j}^{ε, h} (x) = 1$ for every $x$ , and $Z_{ε, h} (x) \geq Z_{\min}$ . The measure assignment satisfies $∣ (P_{h}^{ε} μ)_{j} - (η_{ε} * μ) (x_{j}) ∣ \leq \frac{C_{η} ∥ η_{ε} ∥_{\infty}}{Z_{\min}} ∥ μ ∥_{T V} h^{2} .$ For $g \in C^{2} (T_{L}^{d}; R^{m})$ with finite $m \geq 1$ (including scalar $m = 1$ ), use the Euclidean norm $∣ \cdot ∣_{2}$ and $∥ g ∥_{\infty} = \sup_{y} ∣ g (y) ∣_{2}$ . Define the translation-uniform constant $C_{η, g} = \frac{L^{d}}{12} \sum_{a = 1}^{d} \sup_{x, y \in T_{L}^{d}} {∣ \partial_{y_{a} y_{a}} [η_{ε} (y - x) g (y)] ∣}_{2} .$ The gathered nodal samples satisfy the explicit bound $∣ \sum_{j} w_{j}^{ε, h} (x) g (x_{j}) - (η_{ε} * g) (x) ∣ \leq \frac{C_{η, g} + C_{η} ∥ g ∥_{\infty}}{Z_{\min}} h^{2} (x \in T_{L}^{d}) .$

Facts & Assumptions

Given: The periodic grid and pairing, the strictly positive normalized smooth kernel, its sampled normalizer, and the weights defined in Periodic uniform PIC grid and inner product and Periodic PIC deposition shape and gather operators.

[F1]

Periodic uniform PIC grid and inner product: $h = L / N$ , the grid is the uniform periodic tensor grid, and $h^{d} \sum_{j}$ is its volume-weighted sum.

[F2]

Periodic PIC deposition shape and gather operators: $Z_{ε, h} (x) = h^{d} \sum_{j} η_{ε} (x_{j} - x)$ and $w_{j}^{ε, h} (x) = h^{d} η_{ε} (x_{j} - x) / Z_{ε, h} (x)$ .

[F3]

Composite periodic trapezoidal error in a uniform tensor grid: for $f \in C^{2} (T_{L}^{d})$ , the uniform periodic tensor-grid trapezoid error is at most $(L^{d} h^{2} / 12) \sum_{a} ∥ \partial_{a a} f ∥_{\infty}$ .

[F4]

Periodic PIC regularization gives a smooth compatible Poisson solution: $η_{ε}$ has unit integral.

Proof

1.1F1F2algebra

Strict positivity of $η_{ε}$ implies $Z_{ε, h} (x) > 0$ . Summing the weights in [F2] gives $\sum_{j} w_{j}^{ε, h} (x) = Z_{ε, h} (x) / Z_{ε, h} (x) = 1$ , and each weight is nonnegative. Since every summand in $Z$ is at least $h^{d} \min η_{ε}$ , and $h^{d} ∣ J_{N} ∣ = L^{d}$ , $Z_{ε, h} (x) \geq Z_{\min}$ .

1.2F1F3F4algebra

For fixed $x$ , put $f_{x} (y) = η_{ε} (y - x) g (y)$ and let $q_{x} = h^{d} \sum_{j} f_{x} (x_{j}) - \int f_{x} (y) d y \in R^{m}$ . For every Euclidean unit vector $v$ , apply the scalar estimate [F3] to $v \cdot f_{x}$ . Since $∣ v \cdot \partial_{a a} f_{x} ∣ \leq ∣ \partial_{a a} f_{x} ∣_{2}$ , this gives $∣ v \cdot q_{x} ∣ \leq C_{η, g} h^{2}$ . If $q_{x} \neq 0$ , take $v = q_{x} / ∣ q_{x} ∣_{2}$ ; if $q_{x} = 0$ the bound is immediate. Thus $∣ q_{x} ∣_{2} \leq C_{η, g} h^{2}$ , with no dimension factor. The integral is $(η_{ε} * g) (x)$ because the kernel is even. Projecting that integral onto its unit direction in the same way, nonnegativity and the unit integral [F4] give $∣ (η_{ε} * g) (x) ∣_{2} \leq ∥ g ∥_{\infty}$ . The defining supremum is over both $x$ and $y$ : translation of the kernel alone does not translate $g$ .

2.1F1F2F3F4step 1.1algebra

Apply [F3] to $y \mapsto η_{ε} (y - x)$ ; translation does not change its second-derivative suprema and its integral is $1$ by [F4]. Thus $∣ Z_{ε, h} (x) - 1 ∣ \leq C_{η} h^{2}$ . By [F2], $(P_{h}^{ε} μ)_{j} = \int_{T_{L}^{d}} \frac{η_{ε} (x_{j} - x)}{Z_{ε, h} (x)} d μ (x) .$ Subtract $(η_{ε} * μ) (x_{j})$ , factor $η_{ε} (x_{j} - x)$ , and use $∣ Z^{- 1} - 1 ∣ = ∣ Z - 1 ∣ / Z \leq C_{η} h^{2} / Z_{\min}$ . Integrating against the total variation measure gives the first bound.

3.1F2step 1.1step 1.2algebra∎

The gathered sum in the Statement is $h^{d} \sum_{j} f_{x} (x_{j}) / Z_{ε, h} (x)$ by [F2]. Subtract the integral, use the quadrature error from step 1.2 divided by $Z \geq Z_{\min}$ , and bound the Euclidean norm of the normalization correction by $∣ Z^{- 1} - 1 ∣ ∥ g ∥_{\infty} \leq C_{η} h^{2} ∥ g ∥_{\infty} / Z_{\min}$ . This is the second bound.

Remarks

The shifted-product constant is finite and can be bounded entirely by separate derivative suprema: $C_{η, g} \leq \frac{L^{d}}{12} \sum_{a = 1}^{d} (∥ \partial_{a a} η_{ε} ∥_{\infty} ∥ g ∥_{\infty} + 2 ∥ \partial_{a} η_{ε} ∥_{\infty} ∥ \partial_{a} g ∥_{\infty} + ∥ η_{ε} ∥_{\infty} ∥ \partial_{a a} g ∥_{\infty}) .$ This follows by differentiating the product twice and using the Euclidean triangle inequality. The unshifted product $η_{ε} (y) g (y)$ alone does not supply a uniform bound for all gather locations $x$ .

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The periodic PIC Poisson algorithm returns the unique gauged solution

Statement

For every real zero-mean source $s$ on the grid of Periodic uniform PIC grid and inner product, Fourier diagonal solver for the periodic centered-difference Poisson equation terminates and returns the unique $ϕ$ satisfying $A_{h} ϕ = s$ and $\overset{ˉ}{ϕ} = 0$ . Its incompatible-source branch is taken exactly when the equation has no solution.

Facts & Assumptions

Given: A real grid source and the exact-arithmetic finite Fourier procedure.

[F1]

Periodic centered-difference Poisson solvability and zero-mean gauge: the Fourier characters diagonalize $A_{h}$ , $λ_{h} (k) > 0$ for $k \neq 0$ , and a zero-mean source has a unique zero-mean solution.

[F2]

Fourier diagonal solver for the periodic centered-difference Poisson equation: the procedure checks $\overset{ˉ}{s}$ , sets ${\hat{ϕ}}_{0} = 0$ , divides each other coefficient by $λ_{h} (k)$ , and applies the inverse finite Fourier transform.

Proof

1.1F1F2algebra

The source and grid contain finitely many real numbers, so computing its mean and finite Fourier transform terminates. If $\overset{ˉ}{s} \neq 0$ , then the constant Fourier coefficient of $s$ is nonzero while $A_{h}$ annihilates constants and has zero constant coefficient on every output, so $A_{h} ϕ = s$ is impossible.

2.1F1F2step 1.1algebra

If $\overset{ˉ}{s} = 0$ , its zero Fourier coefficient is zero. By [F1], every other eigenvalue is positive, so all divisions in [F2] are defined. For real input, ${\hat{s}}_{- k} = \overline{{\hat{s}}_{k}}$ and $λ_{h} (- k) = λ_{h} (k)$ ; therefore the returned coefficients have the same conjugate symmetry, and pairing the $k$ and $- k$ terms in the inverse transform shows every $ϕ_{j}$ is real. For every nonzero mode, the returned coefficient satisfies $λ_{h} (k) {\hat{ϕ}}_{k} = {\hat{s}}_{k}$ , and the zero mode satisfies $0 = {\hat{s}}_{0}$ . Inverse Fourier transformation therefore gives $A_{h} ϕ = s$ and ${\hat{ϕ}}_{0} = 0$ gives $\overset{ˉ}{ϕ} = 0$ .

3.1F1step 2.1∎

The coefficient list is finite, hence inverse transformation terminates. Uniqueness of the returned gauged solution follows from [F1].

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Classical RK4 step for the fixed-grid regularized PIC particle ODE

Definition

Input an odd periodic grid with $N \geq 3$ , a fixed $ε > 0$ , particle count $P \geq 1$ , total charge $Q$ , total mass $M > 0$ , a state $Z = (z_{1}, \dots, z_{P})$ with $z_{p} = (X_{p}, V_{p}) \in R^{2 d}$ , and step $Δ t > 0$ . The grid, mass, and shape must satisfy the domains in Periodic uniform PIC grid and inner product, Fixed-regularization periodic PIC pair kernel, and Periodic PIC deposition shape and gather operators; otherwise return status invalid-input. Use the exact-evaluation phase field of Fixed-regularization periodic PIC pair kernel: $F_{p} (Z) = \frac{1}{P} \sum_{r = 1}^{P} K_{ε, h} (z_{p}, z_{r}) .$ At each of four classical RK stages, evaluate this complete field at that stage's particle state: deposit every charge $Q / P$ , subtract the neutralizing background, solve its zero-mean Poisson system with Fourier diagonal solver for the periodic centered-difference Poisson equation, compute $E_{h} = - D^{0} ϕ_{h}$ , and gather $E_{h}$ at each stage position $X_{p}$ . Multiply the gathered field by $c = Q / M$ and return $F_{p} = (V_{p}, c (G_{X} E_{h})_{p})$ , using that stage's velocity $V_{p}$ . With $k_{1} = F (Z)$ , $k_{2} = F (Z + \frac{Δ t}{2} k_{1})$ , $k_{3} = F (Z + \frac{Δ t}{2} k_{2})$ , and $k_{4} = F (Z + Δ t k_{3})$ , return $Z^{+} = Z + \frac{Δ t}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) .$ Every stage source is compatible by Periodic PIC deposition conserves charge and satisfies Poisson compatibility, and The periodic PIC Poisson algorithm returns the unique gauged solution supplies each exact finite field solve. Thus no field-solve-failure branch occurs for admissible input. All operations use exact real or complex arithmetic; no floating-point stability or error bound is asserted.

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Matched variational Hamiltonian branch for periodic PIC

Definition

Use the smooth assignment branch of Periodic PIC deposition shape and gather operators, real particle charges $q_{p}$ , positive masses $m_{p}$ , and its neutralized grid density $ρ (X)$ . Let $ϕ (X) = A_{h}^{†} ρ (X)$ denote the unique mean-zero solution from Periodic centered-difference Poisson solvability and zero-mean gauge. Define the grid potential energy $U_{h} (X) = \frac{1}{2} ⟨ ρ (X), A_{h}^{†} ρ (X) ⟩_{h} = \frac{1}{2} ⟨ ρ (X), ϕ (X) ⟩_{h}$ and the finite-dimensional separable Hamiltonian $H_{h} (X, p) = \frac{1}{2} p^{T} M^{- 1} p + U_{h} (X), M = diag (m_{1} I_{d}, \dots, m_{P} I_{d}) .$ The corresponding canonical ODE is $\dot{X} = M^{- 1} p$ , $\dot{p} = - \nabla_{X} U_{h} (X)$ . Its force is the variational force obtained by differentiating the deposited shape in $U_{h}$ ; this definition does not identify that force with ordinary field interpolation $G_{X} E$ .

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Fixed-regularization periodic PIC kernels are globally Lipschitz

Statement

For fixed $ε > 0$ , odd grid size $N$ , and $Q, M$ as in Fixed-regularization periodic PIC pair kernel, define $M_{x}^{ε, h} = \sup_{x, y \in T_{L}^{d}} ∥ D_{x} (c Q E_{ε, h}) (x, y) ∥_{o p}, M_{y}^{ε, h} = \sup_{x, y \in T_{L}^{d}} ∥ D_{y} (c Q E_{ε, h}) (x, y) ∥_{o p}$ and $L_{ε, h} = \max {1, M_{x}^{ε, h}, M_{y}^{ε, h}}$ . These are finite mesh- and regularization-dependent constants. The phase kernel $K_{ε, h}$ is globally Lipschitz in the two-variable convention $∣ K_{ε, h} (z, z^{'}) - K_{ε, h} (\tilde{z}, {\tilde{z}}^{'}) ∣ \leq L_{ε, h} (∣ z - \tilde{z} ∣ + ∣ z^{'} - {\tilde{z}}^{'} ∣) .$ The continuum kernel $K_{ε}$ is also globally Lipschitz with the explicit admissible constant $L_{ε} = \max {1, \frac{∣ c Q ∣}{L^{d}} \sum_{k \in Z^{d} ∖ {0}} \prod_{a = 1}^{d} α_{ε}^{2 ∣ k_{a} ∣}} .$

Facts & Assumptions

Given: The fixed periodic grid, smooth normalized weights, the finite grid Poisson inverse, the continuum regularized Fourier kernel, and the phase-state definition.

[F1]

Fixed-regularization periodic PIC pair kernel: $E_{ε, h} (x, y)$ is a finite weighted sum of entries of the fixed matrix $- D^{0} A_{h}^{†}$ ; $E_{ε}$ is the displayed absolutely convergent Fourier series.

[F2]

Smooth periodic PIC assignment weights have bounded derivatives: every fixed-grid weight is $C^{\infty}$ with all derivatives bounded on the compact torus.

[F3]

The periodic Poisson mean-zero pseudoinverse is linear and self-adjoint: $A_{h}^{†}$ is linear on grid sources, so the pair field is a finite sum of products of its source and gather weights with fixed entries of $- D^{0} A_{h}^{†}$ .

Proof

1.1F1F2F3givenalgebra

By [F2], each grid weight is smooth and has bounded derivatives of every finite order. By [F1], $E_{ε, h}$ is a finite sum of products of such weights and fixed matrix entries, so it is $C^{\infty}$ and all its derivatives are bounded. In particular, both derivative suprema $M_{x}^{ε, h}, M_{y}^{ε, h}$ are finite. More explicitly, if $B_{j k}$ are the vector entries of $- D^{0} A_{h}^{†}$ and $C_{w} = \max_{j} ∥ \nabla w_{j} ∥_{\infty}$ , then $M_{x}^{ε, h} \leq ∣ c Q ∣ h^{- d} C_{w} \sum_{j, k} ∣ B_{j k} ∣ ∥ w_{k} ∥_{\infty}, M_{y}^{ε, h} \leq ∣ c Q ∣ h^{- d} C_{w} \sum_{j, k} ∣ B_{j k} ∣ ∥ w_{j} ∥_{\infty},$ which are finite explicit bounds for the defining suprema.

1.2F1algebra

Any finite-order derivative of the continuum series multiplies each coefficient by finitely many components of $κ_{k}$ . The coefficient already has one factor $∣ κ_{k} ∣^{- 1}$ ; after $r$ additional derivatives, its magnitude is bounded by a constant times $(1 + ∣ k ∣)^{r} \prod_{a} α_{ε}^{2 ∣ k_{a} ∣}$ . These bounds are summable because each one-dimensional geometric series with a polynomial factor converges. Thus every differentiated series converges uniformly, and all derivatives are bounded. In particular, for a first derivative the factor $∣ κ_{k} ∣$ cancels the denominator $∣ κ_{k} ∣^{2}$ , so its norm is bounded by the displayed sum times $∣ c Q ∣ / L^{d}$ . Periodicity permits the same bounds on Euclidean lifts.

2.1givenstep 1.1step 1.2algebra∎

First change $z = (x, v)$ while holding $z^{'} = (y, w)$ fixed. The output difference has components $(v - \tilde{v}, c Q (E_{ε, h} (x, y) - E_{ε, h} (\tilde{x}, y)))$ . Its squared Euclidean norm is at most $∣ v - \tilde{v} ∣^{2} + (M_{x}^{ε, h})^{2} ∣ x - \tilde{x} ∣^{2}$ , hence its norm is at most $\max {1, M_{x}^{ε, h}} ∣ z - \tilde{z} ∣$ . Next change $z^{'}$ while holding $\tilde{z}$ fixed; this changes only the field component, by at most $M_{y}^{ε, h} ∣ z^{'} - {\tilde{z}}^{'} ∣$ . Adding these bounds gives the stated two-variable estimate with $L_{ε, h}$ . The same block-norm calculation using the continuum derivative bounds from step 1.2 gives the estimate with $L_{ε}$ .

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Fixed-regularization periodic PIC force consistency on a smooth density

Statement

Fix $ε > 0$ and a finite signed Borel measure $μ$ on $T_{L}^{d}$ with total mass $Q$ . Let $ρ_{ε} = η_{ε} * μ - Q / L^{d}$ and let $ϕ_{ε}$ be its unique zero-mean periodic Poisson potential from Periodic PIC regularization gives a smooth compatible Poisson solution. Put $E_{ε} = - \nabla ϕ_{ε}$ . Use the shape and periodic centered-difference Poisson operator of Periodic PIC deposition shape and gather operators and Periodic uniform PIC grid and inner product. Let $s_{h} = P_{h}^{ε} μ - Q / L^{d}$ , let $A_{h} ϕ_{h} = s_{h}$ with ${\overset{ˉ}{ϕ}}_{h} = 0$ , set $E_{h, j} = - D^{0} ϕ_{h}$ , and gather $G_{h} E_{h}$ using the same shape. Define $C_{d e p} = \frac{C_{η} ∥ η_{ε} ∥_{\infty}}{Z_{\min}} ∥ μ ∥_{T V}, C_{F D} = \frac{1}{12} \sum_{a = 1}^{d} ∥ \partial_{a}^{4} ϕ_{ε} ∥_{\infty}, C_{s r c} = C_{d e p} + C_{F D},$ where $C_{η}, Z_{\min}$ are from Partition and quadrature bounds for a periodic PIC shape. Put $C_{3} = \frac{1}{6} {(\sum_{a = 1}^{d} ∥ \partial_{a}^{3} ϕ_{ε} ∥_{\infty}^{2})}^{1 / 2}, C_{i n t} = \frac{C_{η, E} + ∥ E_{ε} ∥_{\infty} C_{η}}{Z_{\min}}, C_{η, E} = \frac{L^{d}}{12} \sum_{a = 1}^{d} \sup_{x, y \in T_{L}^{d}} {∣ \partial_{y_{a} y_{a}} [η_{ε} (y - x) E_{ε} (y)] ∣}_{2} .$ For vector fields, $∥ V ∥_{\infty} = \sup_{x} ∣ V (x) ∣_{2}$ (the maximum of Euclidean lengths for grid fields); the derivatives in $C_{η, E}$ use this same Euclidean norm. Then the source $s_{h}$ is compatible and $∥ G_{h} E_{h} - η_{ε} * E_{ε} ∥_{\infty} \leq L^{d / 2} h^{2 - d / 2} (γ_{N, L} C_{s r c} + C_{3}) + C_{i n t} h^{2} .$ The displayed $γ_{N, L}$ is the actual finite-grid inverse/derivative norm of Periodic Poisson residual controls the gathered particle force and satisfies $γ_{N, L} \leq L / 4$ . Thus for $d \in {1, 2, 3}$ this fixed- $ε$ force error tends to zero as $h \to 0$ . Every constant depends on the stated smooth regularized solution and kernel; this is not an $ε \to 0$ estimate.

Facts & Assumptions

Given: A zero-mean periodic density, its stated smooth potential and field, a fixed $ε > 0$ , and odd periodic grids with the specified shape and Poisson operators.

[F1]

Partition and quadrature bounds for a periodic PIC shape: the smooth weights are nonnegative and normalized; assigned samples differ from $(η_{ε} * μ) (x_{j})$ by at most $C_{η} ∥ η_{ε} ∥_{\infty} ∥ μ ∥_{T V} h^{2} / Z_{\min}$ , and gathered samples of a $C^{2}$ real vector field differ from its $η_{ε}$ convolution by the stated Euclidean trapezoidal bound with translation-uniform shifted-product derivatives.

[F2]

Periodic centered-difference Poisson solvability and zero-mean gauge: $A_{h}$ has range exactly the zero-mean arrays and a unique mean-zero solution for every such source.

[F3]

Periodic Poisson residual controls the gathered particle force: if $A_{h} ϕ = s$ is the zero-mean solution and $r = s - A_{h} \tilde{ϕ}$ , then $\overset{ˉ}{r} = 0$ and $∥ D^{0} (ϕ - \tilde{ϕ}) ∥_{h} \leq γ_{N, L} ∥ r ∥_{h}$ , with $γ_{N, L} \leq L / 4$ .

[F4]

Periodic PIC regularization gives a smooth compatible Poisson solution: $η_{ε} * μ - Q / L^{d}$ has a unique zero-mean $C^{\infty}$ periodic potential obtained by dividing each nonzero Fourier coefficient by its positive Laplacian eigenvalue.

Proof

1.1F1F2F4givenalgebra

The assigned source has mean zero: by the partition identity in [F1], $h^{d} \sum_{j} (P_{h}^{ε} μ)_{j} = \int \sum_{j} w_{j} (x) d μ (x) = Q$ , so subtracting the background gives zero mean. By [F4], the continuum potential and field are smooth; [F2] gives the unique grid potential. The assignment and centered-stencil Taylor bounds give the residual $r_{h} = s_{h} - A_{h} (ϕ_{ε} ∣_{h}), ∥ r_{h} ∥_{\infty} \leq C_{s r c} h^{2} .$ Here [F1] bounds the measure-assignment error by $C_{d e p} h^{2}$ , and for each coordinate the centered second difference has remainder at most $h^{2} ∥ \partial_{a}^{4} ϕ_{ε} ∥_{\infty} / 12$ , whose sum is $C_{F D} h^{2}$ .

2.1F1F3step 1.1algebra

The weighted grid norm satisfies $∥ r_{h} ∥_{h} \leq L^{d / 2} ∥ r_{h} ∥_{\infty}$ because the grid has $N^{d}$ points and $h^{d} N^{d} = L^{d}$ . In [F3], take $s = s_{h}$ , $ϕ = ϕ_{h}$ , and $\tilde{ϕ} = ϕ_{ε} ∣_{h}$ ; its residual is exactly $r_{h}$ from step 1.1, so $∥ D^{0} (ϕ_{h} - ϕ_{ε} ∣_{h}) ∥_{h} \leq γ_{N, L} L^{d / 2} C_{s r c} h^{2}$ . Taylor's theorem for the centered first difference gives $∥ D^{0} (ϕ_{ε} ∣_{h}) - (\nabla ϕ_{ε}) ∣_{h} ∥_{\infty} \leq C_{3} h^{2}$ , hence its grid $L^{2}$ norm is at most $L^{d / 2} C_{3} h^{2}$ .

3.1F1step 2.1algebra

For any nonnegative normalized weights, $\sum_{j} w_{j} (x)^{2} \leq 1$ , so Cauchy–Schwarz gives $∣ G_{x} V ∣ \leq h^{- d / 2} ∥ V ∥_{h}$ . Apply this to the two grid-field errors in step 2.1. Their gathered contribution is at most $L^{d / 2} h^{2 - d / 2} (γ_{N, L} C_{s r c} + C_{3})$ .

4.1F1step 3.1algebra∎

The remaining difference is gathering the exact nodal samples $E_{ε} (x_{j})$ instead of the continuous field. Apply the Euclidean-vector bound of [F1] with $m = d$ and $g = E_{ε}$ . Its translation-uniform constant is exactly $C_{η, E}$ in the Statement, so this difference is at most $C_{i n t} h^{2}$ . The unit-vector projection proof in that supplier introduces no $\sqrt{d}$ factor. Add this to step 3.1 to get the displayed estimate. Since $γ_{N, L} \leq L / 4$ and $2 - d / 2 > 0$ for $d \leq 3$ , both terms tend to zero.

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The matched periodic PIC grid force is a gradient and its ODE conserves energy

Statement

For the Hamiltonian grid model of Matched variational Hamiltonian branch for periodic PIC, let $ϕ = A_{h}^{†} ρ (X)$ . Its position gradient is $\nabla_{X_{p}} U_{h} (X) = q_{p} \sum_{j \in J_{N}} \nabla w_{j}^{ε, h} (X_{p}) ϕ_{j} .$ Along every differentiable solution of $\dot{X} = M^{- 1} p$ , $\dot{p} = - \nabla_{X} U_{h} (X)$ , the continuous semidiscrete energy $H_{h} (X (t), p (t))$ is constant. This is an invariant of the exact finite-grid ODE; no time-discrete energy conservation is asserted.

Facts & Assumptions

Given: Smooth periodic shape weights, the neutralized deposition, the mean-zero inverse $A_{h}^{†}$ , positive masses, and a differentiable solution of the stated Hamiltonian ODE.

[F1]

Matched variational Hamiltonian branch for periodic PIC: $U_{h} = \frac{1}{2} ⟨ ρ, A_{h}^{†} ρ ⟩_{h}$ , $H_{h} = \frac{1}{2} p^{T} M^{- 1} p + U_{h}$ , and the ODE is $\dot{X} = M^{- 1} p$ , $\dot{p} = - \nabla U_{h}$ .

[F2]

The periodic Poisson mean-zero pseudoinverse is linear and self-adjoint: $A_{h}^{†}$ is linear, maps a source to the unique mean-zero solution of its mean-zero part, and is self-adjoint.

[F3]

Periodic PIC deposition conserves charge and satisfies Poisson compatibility: the neutralized density $ρ (X)$ has zero mean.

[F4]

Smooth periodic PIC assignment weights have bounded derivatives: the smooth weights are differentiable and satisfy $\sum_{j} w_{j}^{ε, h} (x) = 1$ for every $x$ .

Proof

1.1F1F2F3F4givenalgebra

For a variation $δ X$ , differentiating the partition identity in [F4] gives $\sum_{j} \nabla w_{j}^{ε, h} (X_{p}) = 0$ for each $p$ . Thus the finite variation $(D ρ (X) [δ X])_{j} = h^{- d} \sum_{p} q_{p} \nabla w_{j}^{ε, h} (X_{p}) \cdot δ X_{p}$ has zero mean; $ρ (X)$ itself has zero mean by [F3]. Since $A_{h}^{†}$ is self-adjoint on the zero-mean subspace by [F2], differentiating $U_{h}$ gives $D U_{h} (X) [δ X] = \frac{1}{2} ⟨ D ρ, A_{h}^{†} ρ ⟩_{h} + \frac{1}{2} ⟨ ρ, A_{h}^{†} D ρ ⟩_{h} = ⟨ D ρ, ϕ ⟩_{h} = \sum_{p} q_{p} (\sum_{j} \nabla w_{j} (X_{p}) ϕ_{j}) \cdot δ X_{p} .$ Identifying coefficients of arbitrary $δ X_{p}$ yields the displayed gradient.

2.1F1step 1.1givenalgebra∎

Differentiate $H_{h}$ along a solution. Symmetry of $M^{- 1}$ gives $\frac{d}{d t} H_{h} = (\nabla U_{h}) \cdot \dot{X} + (M^{- 1} p) \cdot \dot{p} = (\nabla U_{h}) \cdot M^{- 1} p - (M^{- 1} p) \cdot \nabla U_{h} = 0.$ Therefore $H_{h}$ is constant in time on the solution interval.

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Periodic PIC particle and force errors under a prescribed initial coupling

Statement

Fix $ε > 0$ and grid width $h$ . Let $K_{ε}$ and $K_{ε, h}$ be the phase kernels of Fixed-regularization periodic PIC pair kernel, and let $L_{ε}$ be the two-variable global Lipschitz constant of Fixed-regularization periodic PIC kernels are globally Lipschitz. Suppose a probability characteristic solution $μ_{t}$ for $K_{ε}$ and an equal-weight finite-particle grid solution with empirical law $ν_{t}^{h}$ are defined on $[0, T]$ , with finite first moments and integrable path suprema. Assume the uniform grid-force defect $δ_{ε, h} = \sup_{z, z^{'} \in R^{2 d}} ∣ K_{ε, h} (z, z^{'}) - K_{ε} (z, z^{'}) ∣ < \infty .$ If $δ_{0} = W_{1} (ν_{0}^{h}, μ_{0})$ is the actual initial empirical-measure error, then for every $0 \leq t \leq T$ , $W_{1} (ν_{t}^{h}, μ_{t}) \leq e^{2 L_{ε} t} (δ_{0} + t δ_{ε, h}) .$ No rate for $δ_{0}$ is implied: in particular, no dimension-free $P^{- 1 / 2}$ Wasserstein rate is assumed.

Facts & Assumptions

Given: A fixed regularization and grid, a finite equal-weight particle path, a continuum characteristic solution, finite first moments, a prescribed initial coupling, and the uniform pair-kernel force defect.

[F1]

Fixed-regularization periodic PIC pair kernel: the equal-weight particle system is ${\dot{z}}_{i} = P^{- 1} \sum_{j} K_{ε, h} (z_{i}, z_{j})$ and the continuum kernel is $K_{ε}$ .

[F2]

Fixed-regularization periodic PIC kernels are globally Lipschitz: $K_{ε, h}$ and $K_{ε}$ have two-variable Lipschitz constants $L_{ε, h}$ and $L_{ε}$ , respectively, in the stated sum convention.

[F3]

The coupling definition of the 1-Wasserstein distance: $W_{1}$ is the infimum of $\int ∣ z - z^{'} ∣ d π$ over Borel probability couplings of finite-first-moment laws; no minimizing coupling is part of the definition.

[F4]

Characteristic solution of a probability mean-field equation: a characteristic solution is the pushforward of its initial probability by an absolutely continuous label flow satisfying its integral equation.

[F5]

Gronwall's integral inequality with variable and constant coefficients: if $u (t) \leq b \int_{0}^{t} u (s) d s$ for a nonnegative continuous $u$ and $b \geq 0$ , then $u (t) = 0$ .

[F6]

Probability W1 stability under bounded force defects: for two characteristic flows of the same kernel with defects bounded by $η_{μ}, η_{ν}$ , every initial coupling of cost $D_{0}$ gives transported cost at most $e^{2 L_{ε} t} (D_{0} + (η_{μ} + η_{ν}) t)$ .

Proof

1.1F1F2F5givenalgebra

If particles $i$ and $j$ start at the same phase state, subtract their grid equations. The terms with common second argument satisfy $∣ z_{i} (t) - z_{j} (t) ∣ \leq L_{ε, h} \int_{0}^{t} ∣ z_{i} (s) - z_{j} (s) ∣ d s,$ because the two-variable Lipschitz bound in [F2] applies with only the first argument changed. By [F5], the two paths agree. Thus the finite particle flow defines a single path $X_{t} (a)$ for each distinct atom $a$ of $ν_{0}^{h}$ , independent of the label representing $a$ .

2.1F1F2F4givenstep 1.1algebra

Extend $X_{t}$ by the identity off the finite support of $ν_{0}^{h}$ . This is a jointly Borel map, its paths are absolutely continuous on that support, and their path suprema are integrable because the support is finite and each path is continuous on $[0, T]$ . Also $ν_{t}^{h} = (X_{t})_{#} ν_{0}^{h}$ . For a support point $a$ , the finite system equation can be written as $r_{h} (t, a) = \int (K_{ε, h} (X_{t} (a), X_{t} (b)) - K_{ε} (X_{t} (a), X_{t} (b))) d ν_{0}^{h} (b) .$ Thus ${\dot{X}}_{t} (a) = \int K_{ε} (X_{t} (a), X_{t} (b)) d ν_{0}^{h} (b) + r_{h} (t, a)$ . This measurable residual has norm at most $δ_{ε, h}$ by the uniform defect hypothesis. The continuum characteristic flow has residual zero.

3.1F2F3F4F6step 1.1step 2.1

Fix any prescribed initial coupling $π_{0}$ of $ν_{0}^{h}$ and $μ_{0}$ with cost $D_{0}$ . Apply [F6] to the two characteristic flows in steps 1.1 and 2.1, with $η_{μ} = 0$ and $η_{ν} = δ_{ε, h}$ . The transported coupling has cost at most $e^{2 L_{ε} t} (D_{0} + t δ_{ε, h})$ , so by [F3] the same expression bounds $W_{1} (ν_{t}^{h}, μ_{t})$ .

4.1F3step 3.1∎

Step 3.1 holds for every initial coupling. Taking the infimum over its initial cost $D_{0}$ gives $δ_{0}$ by [F3], yielding the claimed inequality. No coupling attaining the infimum is selected.

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Fourth-order RK4 trajectory error for the fixed-grid periodic PIC ODE

Statement

Fix $ε > 0$ , grid width $h$ , particle count $P \geq 1$ , and final time $T > 0$ . Apply classical RK4 from Classical RK4 step for the fixed-grid regularized PIC particle ODE to the autonomous finite-dimensional PIC system $\dot{Z} = F_{ε, h} (Z), (F_{ε, h} (Z))_{p} = \frac{1}{P} \sum_{r = 1}^{P} K_{ε, h} (z_{p}, z_{r}), z_{p} = (X_{p}, V_{p}) .$ Let $Z (t)$ be an exact solution on $[0, T]$ . Assume $F_{ε, h}$ is $C^{4}$ on an open set containing a common compact convex tube $T$ ; the exact path, all numerical states, and every RK4 stage generated from both the exact and numerical starting states lie in $T$ ; and all derivatives of $F_{ε, h}$ through order four are bounded there. Choose a uniform step $0 < h_{t} \leq h_{0}$ with $h_{0} > 0$ , initialize exactly, and retain every time $t_{k} = k h_{t} \leq T$ for integers $k \geq 0$ ; $T / h_{t}$ need not be an integer. The stage hypotheses concern each step ending by $T$ . Put $L_{F} = 2 L_{ε, h}, L_{*} = L_{F} e^{L_{F} h_{0}},$ and choose $C_{4}$ so that the exact-start defect on this tube is at most $C_{4} s^{5}$ for admitted steps $0 < s \leq h_{0}$ ending by $T$ . The local-order supplier gives such constants for a sufficiently small positive $h_{0}$ under the stated smoothness and tube assumptions. Then, for every grid time $t_{k} = k h_{t} \leq T$ , $∥ Z_{k} - Z (t_{k}) ∥_{2} \leq {\begin{cases} C_{4} \frac{e^{L_{*} T} - 1}{L_{*}} h_{t}^{4}, & L_{*} > 0, \\ C_{4} T h_{t}^{4}, & L_{*} = 0. \end{cases}$ The tube, exact-solution, and stage-admissibility hypotheses are conditional; this is a fixed- $ε$ , fixed-grid finite-particle time error, not a continuum or singular-force estimate.

Facts & Assumptions

Given: The exact smooth finite-grid ODE solution, exact initialization, the common compact convex tube, bounded derivatives through order four, admissible RK4 stages from exact and numerical starts, and a uniform partial grid with $0 < h_{t} \leq h_{0}$ and the stated uniform local-defect bound.

[F1]

Fixed-regularization periodic PIC kernels are globally Lipschitz: the two-variable phase kernel has constant $L_{ε, h}$ in the sum norm convention.

[F2]

Orders and scalar stability functions of Euler, midpoint, and classical RK4: classical RK4 has uniform local order at least four for a $C^{4}$ autonomous field with bounded derivatives and its exact-start stages in a common compact convex tube.

[F3]

One-step Lipschitz bound for classical RK4 on an admitted tube: if the field is $L_{F}$ -Lipschitz on the stages from two admitted starts, the one-step RK4 map has bound $(1 + L_{*} h_{t}) ∥ u - v ∥_{2}$ with $L_{*} = L_{F} e^{L_{F} h_{0}}$ .

Proof

1.1F1F2givenalgebra

The state field is a mean over $P$ pair interactions. For two stacked states $Z, \tilde{Z}$ , [F1] gives $∣ (F_{ε, h} (Z))_{p} - (F_{ε, h} (\tilde{Z}))_{p} ∣ \leq L_{ε, h} ∣ z_{p} - {\tilde{z}}_{p} ∣ + \frac{L_{ε, h}}{P} \sum_{r} ∣ z_{r} - {\tilde{z}}_{r} ∣ .$ Taking the stacked Euclidean norm and applying finite Cauchy–Schwarz to the average term shows $∥ F_{ε, h} (Z) - F_{ε, h} (\tilde{Z}) ∥_{2} \leq 2 L_{ε, h} ∥ Z - \tilde{Z} ∥_{2} = L_{F} ∥ Z - \tilde{Z} ∥_{2}$ . The $C^{4}$ tube hypotheses therefore permit [F2], which gives a uniform exact-start local defect at most $C_{4} h_{t}^{5}$ .

2.1F2F3givenstep 1.1algebra

Let $E_{k} = ∥ Z_{k} - Z (t_{k}) ∥_{2}$ . The RK4 stages from the paired exact and numerical starts are admitted by assumption, so [F3] bounds the difference of their one-step outputs by $(1 + L_{*} h_{t}) E_{k}$ . Adding the exact-start defect from step 1.1 by the triangle inequality gives $E_{k + 1} \leq (1 + L_{*} h_{t}) E_{k} + C_{4} h_{t}^{5}$ , with $E_{0} = 0$ .

3.1step 2.1algebra∎

Unroll step 2.1 over the first $k$ steps to get $E_{k} \leq C_{4} h_{t}^{5} \sum_{j = 0}^{k - 1} (1 + L_{*} h_{t})^{j}$ , including the empty sum at $k = 0$ . If $L_{*} > 0$ , the finite geometric sum gives $E_{k} \leq C_{4} h_{t}^{4} [(1 + L_{*} h_{t})^{k} - 1] / L_{*}$ . Since $1 + u \leq e^{u}$ for $u \geq 0$ and $k h_{t} = t_{k} \leq T$ , this is at most $C_{4} h_{t}^{4} (e^{L_{*} T} - 1) / L_{*}$ . If $L_{*} = 0$ , the sum gives $E_{k} \leq C_{4} k h_{t}^{5} = C_{4} t_{k} h_{t}^{4} \leq C_{4} T h_{t}^{4}$ . These are the claimed bounds on every retained partial-grid time, without requiring a step at $T$ .

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Leapfrog geometry and finite-time order for the matched periodic PIC Hamiltonian

Statement

Let $H_{h} (X, p) = \frac{1}{2} p^{T} M^{- 1} p + U_{h} (X)$ be Matched variational Hamiltonian branch for periodic PIC, with $M$ symmetric positive definite and $U_{h} \in C^{3} (R^{d P})$ periodic. Let $Φ_{Δ t} = K_{Δ t / 2} \circ D_{Δ t} \circ K_{Δ t / 2}$ be the leapfrog map with the kick and drift conventions of One-step leapfrog map as force and drift shears. Then:

For every signed step $Δ t$ , $Φ_{Δ t}^{- 1} = Φ_{- Δ t}$ .
Every step is symplectic.
Assuming Countable Choice, every step preserves Lebesgue volume on phase space.
On a finite interval $[0, T]$ , if the exact and numerical paths, all kick/drift stages, and all line segments in the comparison lie in a common compact convex tube and the Hamiltonian vector field has bounded derivatives through order two there, the leapfrog trajectory error is at most $C_{V} T e^{L T} Δ t^{2}$ for exact initialization and sufficiently small $Δ t = T / n$ .

The exact continuous flow conserves $H_{h}$ by The matched periodic PIC grid force is a gradient and its ODE conserves energy. No claim says the leapfrog map preserves $H_{h}$ exactly or that modified-energy estimates hold outside the analytic tube hypotheses of the geometric-integration suppliers.

Facts & Assumptions

Given: A smooth periodic potential, symmetric positive-definite mass matrix, the kick/drift convention, an admitted step, and (for the finite-time error) the stated compact convex tube and exact initialization.

[A1]

Countable Choice is assumed only in claim 3 (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

One-step leapfrog map as force and drift shears: $K_{a} (q, p) = (q, p - a \nabla U_{h} (q))$ , $D_{b} (q, p) = (q + b M^{- 1} p, p)$ , and $Φ_{Δ t} = K_{Δ t / 2} \circ D_{Δ t} \circ K_{Δ t / 2}$ .

[F2]

Canonical kick and drift shears satisfy the symplectic Jacobian identity: for $U_{h} \in C^{2}$ and $M$ symmetric positive definite, each kick and drift Jacobian satisfies $A^{T} J A = J$ .

[F3]

Local leapfrog and kick-then-drift symplectic Euler are symplectic: the admitted local leapfrog composition is $C^{1}$ and symplectic under those hypotheses.

[F4]

Local and finite-time trajectory error for symplectic Euler and leapfrog: on a common compact convex tube with $U_{h} \in C^{3}$ , bounded derivatives through order two of the Hamiltonian vector field, exact initialization, and admitted stages, leapfrog has finite-time global error at most $C_{V} T e^{L T} Δ t^{2}$ .

[F5]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: under Countable Choice a $C^{1}$ diffeomorphism changes Lebesgue measure by the absolute value of its Jacobian determinant.

[F6]

The matched periodic PIC grid force is a gradient and its ODE conserves energy: the continuous semidiscrete Hamiltonian flow conserves the exact $H_{h}$ .

Proof

1.1F1algebra

The kicks and drifts in [F1] have inverses $K_{- a}$ and $D_{- b}$ . Reversing the composition gives $Φ_{Δ t}^{- 1} = K_{- Δ t / 2} \circ D_{- Δ t} \circ K_{- Δ t / 2} = Φ_{- Δ t}$ . All maps are global on the periodic lift, so no intermediate-domain restriction is lost.

1.2F1F2F3given

The potential is smooth, hence $C^{2}$ , and $M$ is symmetric positive definite. Each admitted kick and drift satisfies the symplectic Jacobian identity [F2]; composing the three maps preserves that identity by the chain rule, as stated in [F3].

1.3A1F1F5givenalgebra

The Jacobian of a kick is $(\begin{array}{cc} I & 0 \\ - \frac{Δ t}{2} \nabla^{2} U_{h} & I \end{array})$ and the Jacobian of a drift is $(\begin{array}{cc} I & Δ t M^{- 1} \\ 0 & I \end{array})$ . Each is block triangular with identity diagonal blocks, so its determinant is $1$ ; multiplicativity gives $\det D Φ_{Δ t} = 1$ . The global kick/drift maps are $C^{1}$ diffeomorphisms, so [A1] permits applying [F5] to the map and its inverse, proving preservation of Lebesgue volume.

2.1F4F6given∎

Under the tube and smoothness hypotheses, [F4] gives the stated global order-two error bound. The continuous invariant is the separate identity [F6]; it is not an assertion about the discrete map.

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Staged fixed-regularization PIC error budget

Statement

Fix $ε > 0$ , grid width $h$ , particle count $P \geq 1$ , and $T > 0$ . Use uniform RK4 times $t_{n} = n Δ t$ for integers $n \geq 0$ , with $0 < Δ t \leq h_{0}$ , where $h_{0}$ is the positive admitted step-width bound in the RK4 tube/defect hypotheses below. Retain all $t_{n} \leq T$ ; no integer value of $T / Δ t$ is required. Assume the smooth fixed-regularization hypotheses of Fixed-regularization periodic PIC force consistency on a smooth density, and that the translated unit-source regularized densities $η_{ε} (\cdot - y) - L^{- d}$ satisfy its derivative bounds uniformly in $y$ . Let $δ_{ε, h}$ be the uniform phase-kernel defect from Periodic PIC particle and force errors under a prescribed initial coupling. The consistency theorem gives the explicit estimate $δ_{ε, h} \leq ∣ c Q ∣ [L^{d / 2} h^{2 - d / 2} (γ_{N, L} C_{s r c}^{(1)} + C_{3}^{(1)}) + C_{i n t}^{(1)} h^{2}],$ where $C_{s r c}^{(1)}, C_{3}^{(1)}, C_{i n t}^{(1)}$ are the constants of that theorem evaluated at a unit point source; translation of the continuum source leaves their derivative suprema unchanged, while the shape quadrature constants are independent of the translate. Define the RK4 empirical measure by $ν_{P, h}^{Δ t} (t_{n}) = P^{- 1} \sum_{p = 1}^{P} δ_{z_{p}^{Δ t} (t_{n})}$ . Put $δ_{0} = W_{1} (ν_{0}^{h}, μ_{0}), C_{R K} = {\begin{cases} C_{4} (e^{L_{*} T} - 1) / L_{*}, & L_{*} > 0, \\ C_{4} T, & L_{*} = 0, \end{cases}$ using the constants of Fourth-order RK4 trajectory error for the fixed-grid periodic PIC ODE. For exact initialization, a common admissible initial coupling, and the same compact-tube hypotheses, the RK4 empirical measure $ν_{P, h}^{Δ t} (t_{n})$ and regularized characteristic solution $μ_{ε} (t_{n})$ satisfy $W_{1} ⁣ (ν_{P, h}^{Δ t} (t_{n}), μ_{ε} (t_{n})) \leq \frac{C_{R K}}{\sqrt{P}} Δ t^{4} + e^{2 L_{ε} T} (δ_{0} + T δ_{ε, h}), t_{n} \leq T .$ The terms are time error, prescribed initial empirical error, and fixed- $ε$ mesh-force error. No universal rate for $δ_{0}$ and no $ε \to 0$ singular Vlasov–Poisson limit is asserted.

Facts & Assumptions

Given: The fixed- $ε$ smooth regularized characteristic solution, the finite grid particle solution, a prescribed initial coupling of finite cost, exact initialization, and the common RK4 tube and stage assumptions.

[F1]

Fixed-regularization periodic PIC force consistency on a smooth density: the gathered grid field error is bounded by $L^{d / 2} h^{2 - d / 2} (γ C_{s r c} + C_{3}) + C_{i n t} h^{2}$ with all constants explicit in the source/field derivatives.

[F2]

Periodic PIC particle and force errors under a prescribed initial coupling: its proof transports every prescribed initial coupling of cost $D_{0}$ to an exact-grid/continuum coupling of cost at most $e^{2 L_{ε} t} (D_{0} + t δ_{ε, h})$ .

[F3]

Fourth-order RK4 trajectory error for the fixed-grid periodic PIC ODE: with $0 < Δ t \leq h_{0}$ , exact initialization gives stacked state error at most $C_{R K} Δ t^{4}$ at every retained uniform time $t_{n} = n Δ t \leq T$ , even when the grid does not end at $T$ .

[F4]

The coupling definition of the 1-Wasserstein distance: $W_{1}$ is the infimum of coupling costs for probability measures with finite first moment.

Proof

1.1F1givenalgebra

Apply [F1] to each unit source $μ = δ_{y}$ . The explicit constants in [F1] depend on the source through $∥ μ ∥_{T V}$ and suprema of derivatives of its continuum potential and field. Here $∥ δ_{y} ∥_{T V} = 1$ , and those continuum functions for $δ_{y}$ are translates of the functions for $δ_{0}$ ; their derivative suprema are therefore independent of $y$ . For the corrected shifted-product quadrature constant, translating the unit-source field by $y$ is absorbed by translating both its spatial variable and the free gather location in the joint supremum. That constant and the other shape quadrature constants therefore do not depend on the source translate. One set of constants $C_{s r c}^{(1)}, C_{3}^{(1)}, C_{i n t}^{(1)}$ gives the stated uniform phase-kernel defect bound, without asserting translation invariance of the finite grid kernel.

1.2F2F3givenalgebra

Work at a retained time $t = t_{n} = n Δ t \leq T$ on the stated grid, whose step obeys $0 < Δ t \leq h_{0}$ . Let $A$ be the finite support of $ν_{0}^{h}$ and $I_{a} = {p : z_{p}^{h} (0) = a}$ , $m_{a} = ∣ I_{a} ∣$ for $a \in A$ . For any prescribed coupling $π_{0}$ of $ν_{0}^{h}$ and $μ_{0}$ , define a measure on the second coordinate for each label $p \in I_{a}$ by $λ_{p} (B) = π_{0} ({a} \times B) / m_{a}$ . Since $ν_{0}^{h} ({a}) = m_{a} / P$ , each $λ_{p}$ has mass $1 / P$ , and for each $a$ the sum over $p \in I_{a}$ is $π_{0} ({a} \times \cdot)$ . By the duplicate-path argument in Periodic PIC particle and force errors under a prescribed initial coupling, all labels in $I_{a}$ share the same exact path $X_{t} (a)$ . Thus summing these label measures gives exactly the pushforward $Π_{t} = (X_{t}, Y_{t})_{#} π_{0}$ , the Dobrushin-transported coupling, where $Y_{t}$ is the continuum characteristic flow. In particular, $Π_{t}$ has the exact grid and continuum marginals. Its cost is at most $e^{2 L_{ε} T} (D_{0} + T δ_{ε, h})$ by [F2]. Replace $z_{p}^{h} (t)$ in each summand by the RK4 state $z_{p}^{Δ t} (t)$ . The resulting measure is a coupling of $ν_{P, h}^{Δ t} (t)$ and $μ_{ε} (t)$ , and pointwise triangle inequality increases its cost by at most $P^{- 1} \sum_{p} ∣ z_{p}^{Δ t} - z_{p}^{h} ∣$ . By Cauchy–Schwarz this is at most $P^{- 1 / 2} ∥ Z^{Δ t} - Z^{h} ∥_{2} \leq P^{- 1 / 2} C_{R K} Δ t^{4}$ using [F3]. This gives the claimed bound with $D_{0}$ in place of $δ_{0}$ .

2.1F4step 1.2∎

The bound of step 1.2 holds for every prescribed initial coupling. Taking the infimum of $D_{0}$ gives $δ_{0}$ by [F4], without selecting a minimizing plan. This proves the stated staged error budget.

5 · Examples, counterexamples and false statements

None yet.