TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

Depends on

Used by

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Sources

R. W. Hockney and J. W. Eastwood, Computer Simulation Using Particles, Section 5-5 (standard reference, not scraped)
Nikolaj Nordkvist and Poul G. Hjorth, Classical Mechanics and Symplectic Integration, §3.5 (standard reference, not scraped)