Math 2.0

Fluid mechanics Definition

Reynolds decomposition

Builds on Turbulence

Definition

Let f(𝐱,t)f(\mathbf{x}, t) and g(𝐱,t)g(\mathbf{x}, t) be flow quantities, such as velocity components, pressure or temperature, of a turbulent flow. Let f↦fˉf \mapsto \bar f be an averaging operation (a time average, a spatial average or an ensemble average) that satisfies the Reynolds rules:

  1. f+g‾=fˉ+gˉ\overline{f + g} = \bar f + \bar g and af‾=afˉ\overline{a f} = a \bar f for every constant aa;
  2. fˉg‾=fˉ gˉ\overline{\bar f g} = \bar f \, \bar g, and in particular fˉˉ=fˉ\bar{\bar f} = \bar f;
  3. ∂f/∂xi‾=∂fˉ/∂xi\overline{\partial f / \partial x_i} = \partial \bar f / \partial x_i and ∂f/∂t‾=∂fˉ/∂t\overline{\partial f / \partial t} = \partial \bar f / \partial t.

The Reynolds decomposition of ff is f=fˉ+f′f = \bar f + f', where fˉ\bar f is the mean and f′=f−fˉf' = f - \bar f is the fluctuation.

Remarks

From the rules, f′‾=fˉ−fˉˉ=0\overline{f'} = \bar f - \bar{\bar f} = 0, and fg‾=fˉ gˉ+f′g′‾.\overline{f g} = \bar f \, \bar g + \overline{f' g'}. The last term, the correlation of fluctuations, is in general not zero; for velocity components it gives the Reynolds stresses.

Example: the ensemble average over a probability space of flow realisations satisfies all three rules, since expectation is linear and commutes with derivatives of smooth fields. Non-example: a spatial filter fˉ=G∗f\bar f = G * f with a smooth kernel GG, as used in large eddy simulation, is linear and commutes with derivatives. But in general fˉˉ≠fˉ\bar{\bar f} \neq \bar f, so it fails rule 2 and f′‾≠0\overline{f'} \neq 0. An average over a finite time window TT satisfies the rules only approximately, when TT is long compared with the fluctuations and short compared with changes of the mean.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Reynolds_decomposition
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Changed3 November 2026

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