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A stipulated cold two-stream periodic PIC experiment

Statement

Here is a fully specified cold two-stream input for the fixed- $ε$ periodic grid model. Let $d = 1$ , $L = 2 π$ , $N = 63$ , $P = 126$ , $Q = M = 1$ , $ε = 1 / 5$ , and $Δ t = 10^{- 3}$ with $4000$ steps to $T = 4$ . For $σ \in {- 1, + 1}$ and $j \in Z / 63 Z$ , assign charge and mass $1 / 126$ and set $X_{σ, j} (0) = (\frac{2 π j}{63} + σ \frac{1}{100} \cos \frac{2 π j}{63}) (m o d 2 π), V_{σ, j} (0) = \frac{σ}{2} .$ Use the smooth assignment shape, neutralizing background, centered periodic Poisson solve, and exact-arithmetic update of Classical RK4 step for the fixed-grid regularized PIC particle ODE. The two beams are cold because each beam has one velocity value. This is a stipulated numerical experiment only: it asserts no growth rate, nonlinear saturation, or physical validation.

Facts & Assumptions

Given: The finite parameter tuple above, the two beam labels, and the fixed-regularization periodic PIC definitions.

[F1]

Fixed-regularization periodic PIC pair kernel: the equal-weight particle system uses total charge $Q$ , total mass $M$ , and the fixed smooth grid interaction kernel.

[F2]

Periodic PIC deposition conserves charge and satisfies Poisson compatibility: normalized assignment deposits total charge $Q$ , and subtracting $Q / L$ gives a compatible zero-mean source.

[F3]

Classical RK4 step for the fixed-grid regularized PIC particle ODE: the exact-arithmetic RK4 step evaluates the periodic PIC field at each stage and returns the updated particle state.

Proof

1.1givenalgebra

There are $2 \cdot 63 = 126$ labels. Their positive charges sum to $126 / 126 = 1 = Q$ and their positive masses sum to $126 / 126 = 1 = M$ . Within each fixed $σ$ , every velocity equals $σ / 2$ , so the two populations are cold counter-streaming beams.

2.1F1F2F3givenalgebra∎

The stated positions are points of $T_{2 π}$ after reduction modulo $2 π$ , and $N = 63$ is odd as required by the grid convention. At each of the finitely many RK4 stages, the smooth-shape deposition conserves charge $Q = 1$ ; subtracting the uniform background $Q / L = 1 / (2 π)$ makes the source compatible by [F2]. By [F3], the exact finite Fourier Poisson solve therefore returns a potential at every stage, so this exact-arithmetic experiment is defined. No physical conclusion follows from this specification.

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Sources

R. W. Hockney and J. W. Eastwood, Computer Simulation Using Particles, Chapters 2 and 4 (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 (standard reference, not scraped)