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Recurrence: The Artefact You Can Predict

Source: PHY653B Ch. 2

Intuition

Run a Vlasov simulation past its damping phase and the "damped" wave comes back — sometimes to a large fraction of its original amplitude. This is neither a bug nor physics. It is the velocity grid, and its arrival time can be predicted before the run starts from nothing but \(\Delta v\).

Landau damping is reversible

Nothing in the collisionless Vlasov equation is dissipative. The wave energy has not been destroyed; it has been converted into fine-scale structure in velocity space. A perturbation initially uniform in \(v\) shears under free streaming as

\[f_1 \propto e^{ik(x - vt)}\]

so after time \(t\) the structure in velocity has wavenumber \(kt\) — the filaments get finer linearly and without limit. The field, which only sees \(\int f\,dv\), averages those filaments to nothing, and reports damping.

In a continuum this goes on forever. On a grid it cannot.

The recurrence time

Once the filament wavelength in \(v\) reaches the grid spacing \(\Delta v\), the discrete velocities can no longer distinguish the structure. They come back into phase when \(k\,\Delta v\,t = 2\pi\):

Recurrence time

\[T_R = \frac{2\pi}{k\,\Delta v}\]

At that moment the filaments reassemble, the velocity integral stops cancelling, and the field springs back to life. The widget marks \(T_R\) and hits it within a few percent; raise \(N_v\) and the revival moves later in exact proportion.

This is a rare and valuable thing: a numerical artefact with a closed-form arrival time. Predicting it and confirming it is a genuine verification exercise — it demonstrates that you understand what your grid is doing, not merely that your output looks plausible.

Worked example

A run uses \(v_{\max} = 6\), \(N_v = 128\), \(k\lambda_D = 0.5\). When does recurrence hit, and is the run long enough to measure \(\gamma = -0.1534\)?

\[\Delta v = \frac{12}{128} = 0.09375, \qquad T_R = \frac{2\pi}{0.5\times0.09375} = 134\]

The damping time is \(1/|\gamma| = 6.5\), so the wave falls by \(e^{-1}\) every 6.5 time units and by \(\sim10^{-9}\) over 134. There is plenty of clean exponential before recurrence — this run is comfortably adequate. Halve \(N_v\) to 64 and \(T_R\) drops to 67, still fine. Drop to \(N_v = 16\) and \(T_R = 17\), barely two damping times: not enough.

Mitigation, and what each cure costs

Cure Effect Price
Increase \(N_v\) \(T_R \propto N_v\) Linear cost; never removes the problem, only postpones it
Velocity-space filter / diffusion Erases filaments below a cutoff You have broken reversibility — and set a floor on measurable damping rates
Add a real collision operator Physically motivated smoothing Correct, but you are now solving a different (collisional) problem
Just stop before \(T_R\) Free Only works if the physics you want happens before then

The honest framing: recurrence is the price of a finite velocity grid, and every cure is a trade. The one thing you must not do is see a recurrence revival and report it as physics — phase-space holes and trapping oscillations are real and can look similar.

Common mistakes

  • Mistaking recurrence for nonlinear saturation. Both interrupt exponential decay. The test: recurrence arrives at \(2\pi/(k\Delta v)\) and moves when you change \(N_v\); trapping arrives at the bounce time and moves when you change amplitude. Change one thing at a time.
  • Assuming a finer grid eventually eliminates it. It never does. \(T_R \to \infty\) only as \(\Delta v\to0\).
  • Adding filtering without saying so. It changes the physics and it changes the smallest damping rate you can honestly claim.

Knowledge graph position

Prerequisites: Vlasov–Poisson, phase mixing. Leads to: filtering strategies, collisional Vlasov solvers, honest run-length planning.

Quiz

Q1 (computational). \(v_{\max} = 5\), \(N_v = 200\), \(k\lambda_D = 0.4\). Find \(T_R\).

Answer

\(\Delta v = 10/200 = 0.05\), so \(T_R = 2\pi/(0.4\times0.05) = 2\pi/0.02 \approx 314\). With \(\gamma = -0.0661\) the damping time is 15, so the run has ~20 damping times before recurrence: ample.

Q2 (conceptual). How do you distinguish recurrence from nonlinear trapping in a run whose damping stops early?

Answer

Change one parameter at a time. Recurrence depends on the grid: double \(N_v\) and the revival time doubles, with the amplitude unchanged. Trapping depends on the amplitude: halve \(A\) and the arrest happens later and weaker, while \(N_v\) makes no difference.

Q3 (MCQ). Recurrence occurs because:

  • (a) energy is injected by the field solver
  • (b) discrete velocities that had dephased come back into phase after \(2\pi/(k\Delta v)\)
  • (c) the time step is too large
  • (d) the boundary conditions are wrong
Answer

(b). Free streaming shears the perturbation at a rate proportional to \(v\); with a discrete set of velocities the relative phases are periodic, and everything realigns after \(2\pi/(k\Delta v)\).