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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≥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 Z˙=Fε,h(Z),(Fε,h(Z))p=1P∑r=1PKε,h(zp,zr),zp=(Xp,Vp). Let Z(t) be an exact solution on [0,T]. Assume Fε,h is C4 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 Fε,h through order four are bounded there. Choose a uniform step 0<ht≤h0 with h0>0, initialize exactly, and retain every time tk=kht≤T for integers k≥0; T/ht need not be an integer. The stage hypotheses concern each step ending by T. Put LF=2Lε,h,L∗=LFeLFh0, and choose C4 so that the exact-start defect on this tube is at most C4s5 for admitted steps 0<s≤h0 ending by T. The local-order supplier gives such constants for a sufficiently small positive h0 under the stated smoothness and tube assumptions. Then, for every grid time tk=kht≤T, ∥Zk−Z(tk)∥2≤{C4eL∗T−1L∗ht4,L∗>0,C4Tht4,L∗=0. 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<ht≤h0 and the stated uniform local-defect bound.

[F1]

Fixed-regularization periodic PIC kernels are globally Lipschitz: the two-variable phase kernel has constant Lε,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 C4 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 LF-Lipschitz on the stages from two admitted starts, the one-step RK4 map has bound (1+L∗ht)∥u−v∥2 with L∗=LFeLFh0.

Proof

1.1F1F2givenalgebra

The state field is a mean over P pair interactions. For two stacked states Z,Z~, [F1] gives ∣(Fε,h(Z))p−(Fε,h(Z~))p∣≤Lε,h∣zp−z~p∣+Lε,hP∑r∣zr−z~r∣. Taking the stacked Euclidean norm and applying finite Cauchy–Schwarz to the average term shows ∥Fε,h(Z)−Fε,h(Z~)∥2≤2Lε,h∥Z−Z~∥2=LF∥Z−Z~∥2. The C4 tube hypotheses therefore permit [F2], which gives a uniform exact-start local defect at most C4ht5.

2.1F2F3givenstep 1.1algebra

Let Ek=∥Zk−Z(tk)∥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∗ht)Ek. Adding the exact-start defect from step 1.1 by the triangle inequality gives Ek+1≤(1+L∗ht)Ek+C4ht5, with E0=0.

3.1step 2.1algebra∎

Unroll step 2.1 over the first k steps to get Ek≤C4ht5∑j=0k−1(1+L∗ht)j, including the empty sum at k=0. If L∗>0, the finite geometric sum gives Ek≤C4ht4[(1+L∗ht)k−1]/L∗. Since 1+u≤eu for u≥0 and kht=tk≤T, this is at most C4ht4(eL∗T−1)/L∗. If L∗=0, the sum gives Ek≤C4kht5=C4tkht4≤C4Tht4. These are the claimed bounds on every retained partial-grid time, without requiring a step at T.

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