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

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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.

Depends on

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)