Planning a Run Around Recurrence

The task

You have been asked to measure the Landau damping rate at \(k\lambda_D = 0.3\), where \(\gamma = -0.01262\). Choose the grid before running anything.

This is the single most useful arithmetic in Chapter 2, because it takes two minutes and routinely saves a day of wasted compute.

The two competing constraints

You need enough damping times to fit. A clean exponential fit wants the signal to fall by at least two or three \(e\)-foldings, so

\[t_{\rm run} \gtrsim \frac{3}{|\gamma|} = \frac{3}{0.01262} = 238\]

You must finish before recurrence.

\[T_R = \frac{2\pi}{k\,\Delta v} > t_{\rm run}\]

Together:

\[\frac{2\pi}{k\,\Delta v} > \frac{3}{|\gamma|} \qquad\Longrightarrow\qquad \Delta v < \frac{2\pi|\gamma|}{3k}\]

Step 1: required \(\Delta v\)

\[\Delta v < \frac{2\pi \times 0.01262}{3 \times 0.3} = \frac{0.0793}{0.9} = 0.0881\]

Step 2: required \(N_v\)

With \(v_{\max} = 6\) (needed to cover the resonance at \(\omega_r/k = 1.15985/0.3 = 3.87\,v_{th}\) with margin):

\[N_v = \frac{2v_{\max}}{\Delta v} > \frac{12}{0.0881} = 136\]

So \(N_v = 256\) is comfortable; \(N_v = 128\) is marginal; \(N_v = 64\) will fail.

Step 3: verify

\(N_v = 256 \Rightarrow \Delta v = 0.0469 \Rightarrow T_R = 2\pi/(0.3\times0.0469) = 447\). Run to \(t = 240\) and recurrence is far away. ✓

\(N_v = 64 \Rightarrow \Delta v = 0.1875 \Rightarrow T_R = 112\). The wave would have damped by only \(e^{-1.4}\) — a factor of four — before the grid gave the energy back. The fit would be contaminated over most of its window, and the resulting rate would be wrong in a way that looks like weak damping rather than like a bug.

Why \(k\lambda_D = 0.3\) is the hard case

Counter-intuitively, the weakly damped case is harder to measure than the strongly damped one:

| \(k\lambda_D\) | \(|\gamma|\) | damping time | \(t_{\rm run}\) for 3 e-folds | min \(N_v\) at \(v_{\max}=6\) | |---|---|---|---|---| | 0.3 | 0.0126 | 79 | 238 | 136 | | 0.4 | 0.0661 | 15 | 45 | 35 | | 0.5 | 0.1534 | 6.5 | 20 | 13 | | 0.6 | 0.2641 | 3.8 | 11 | 6 |

Weak damping means a long run, which means recurrence must be pushed far away, which means fine velocity resolution. The resolution requirement is set by the physics you are measuring, not by the equation you are solving — a general lesson worth carrying beyond this problem.

There is a second reason the weak case is hard: numerical diffusion from interpolation adds an artificial damping that is a fixed property of the scheme. At \(|\gamma| = 0.0126\) that artefact is a much larger fraction of the answer than at \(|\gamma| = 0.264\), which is why the measured error in the benchmark table is 3.9% at \(k\lambda_D = 0.3\) and 0.3% elsewhere.

Step 4: the cost

Doubling \(N_v\) doubles both memory and time per step. In 1D1V that is cheap. The same reasoning in a 2D2V or 3D3V code is what makes weakly damped kinetic problems genuinely expensive, and it is a large part of why PIC — which has no velocity grid and therefore no recurrence — remains competitive despite its noise.

The general habit

Before any kinetic run, write down:

The timescale of the physics you want to measure (\(1/|\gamma|\), a growth time, a transit time).
The run length needed to measure it — usually a few of those.
Every artefact time in the scheme (\(T_R\), a filling time, a wave-crossing time).
Confirm (3) \(\gg\) (2). If not, change the grid, not the analysis.

Recurrence · Landau damping benchmark · Normalisation & resolution · Semi-Lagrangian solver · Vlasov widget — change \(N_v\) and watch \(T_R\) move