The profile Rayleigh cleared and viscosity did not

A layer with a kink in it gave Rayleigh’s criterion of 1880: a parallel shear flow without viscosity cannot grow a wave unless its velocity profile has an inflection point. Every wavelength at once took the profile with the most extreme inflection, a vortex sheet, and every unstable wave is inside one circle bounded how fast any inviscid wave can grow. All three are statements about a fluid with no viscosity, and all three are exact.

The flow between two parallel plates driven by a pressure gradient is the simplest flow there is that they clear. Its profile is a parabola, curving the same way at every height, with no inflection point anywhere. By Rayleigh’s criterion no wave on it can grow. Heisenberg argued in his 1924 dissertation that it is unstable anyway, and nobody could check until Lin computed a critical Reynolds number in 1945 and Orszag pinned it to 5772.22 in 1971. The wave that grows is there because of viscosity, which everywhere else in the subject damps disturbances. The argument here is that this is not a correction to Rayleigh’s verdict but a separate mechanism, and that it has a signature: remove the viscosity and the instability goes away again.

Rayleigh’s verdict on a channel, and the term it leaves out #

The conventions are these. The walls are at $y = \pm 1$, the profile is $U = 1 - y^2$, and the Reynolds number is built on the centreline speed and the half-width. A disturbance is a travelling wave with stream function $\psi = \varphi(y)\,e^{i\alpha(x - ct)}$: α is its wavenumber in half-widths, $c = c_r + ic_i$ its complex phase speed, and its amplitude grows as $e^{\alpha c_i t}$ in units of half-width over centreline speed. Linearising the Navier–Stokes equations about the parabola gives the Orr–Sommerfeld equation,

$$(U - c)(\varphi'' - \alpha^2\varphi) - U''\varphi = \frac{1}{i\alpha\mathrm{Re}}\left(\varphi'''' - 2\alpha^2\varphi'' + \alpha^4\varphi\right),$$

with $\varphi = \varphi' = 0$ at both walls, so that both velocity components vanish there. Setting the right-hand side to zero leaves Rayleigh’s equation, of second order, and with only the normal velocity required to vanish at the walls. For $U'' = -2$ it has no growing solution. The viscous term does two things at once: it raises the order of the equation to four, and it lets the no-slip condition be imposed on the streamwise velocity. Both matter, and they matter in thin layers the inviscid solution does not have.

The equation is solved here by Chebyshev collocation with 79 interior points, the clamped wall conditions built into the fourth-derivative operator, every eigenvalue found by a QR iteration, and single eigenvalues then followed through wavenumber and Reynolds number by Rayleigh-quotient iteration.

Seventy-nine waves, and one of them growing #

At a Reynolds number of ten thousand and a wavenumber of one, all but one of the seventy-nine waves decay. They lie in three branches: slow waves whose motion is concentrated near the walls, fast waves near the centre travelling at almost the centreline speed, and a column of strongly damped ones at a phase speed of about two-thirds. The next least stable after the growing wave is a centre wave at $c = 0.9646 - 0.0352i$, decaying ten times faster than the one above the axis grows.

That one is a wall wave, with $c = 0.23752649 + 0.00373967i$. Orszag published it to eight figures in 1971 as the test of his own Chebyshev method, and the solver here reproduces it to 1.3 × 10⁻⁹, which is the rounding in the eighth figure. It travels at under a quarter of the centreline speed, so it moves with the fluid at a height where $U = c_r$, 0.873 of the way from the centreline to the wall. That height, the critical layer, is where the argument goes next.

The band of growing waves opens at 5772 and narrows as viscosity goes #

Following the growing wave down in Reynolds number, its growth rate reaches zero, and minimising that neutral Reynolds number over wavenumber gives the critical point: Re = 5772.2218 at α = 1.020548, against the literature’s 5772.22 and 1.02056. The first wave to grow has a wavelength of 6.16 half-widths and travels at 0.264 of the centreline speed.

Above it the band of growing wavenumbers opens into a tongue. At Re = 6,000 it runs from 0.971 to 1.060. At ten thousand it is 0.797 to 1.095, and the upper edge reaches its highest wavenumber, about 1.10, near 8,500. Beyond that both edges fall: 0.479 to 0.871 at 10⁵ and 0.321 to 0.630 at 10⁶. The band never closes at any finite Reynolds number, but it moves steadily towards long waves, and Lin’s asymptotic theory has its two edges falling as Re^(−1/7) and Re^(−1/11). Between half a million and a million the computed edges fall at local rates of −0.166 and −0.145, still steeper than either — a million is nowhere near the asymptote, and neither branch has reached it.

What the tongue’s shape says is the argument. The inviscid problem is the limit of infinite Reynolds number, and in that limit the band of growing waves is squeezed towards zero wavenumber, which is where Rayleigh’s verdict is recovered. The instability lives at finite viscosity and disappears in both directions: below 5772 viscosity damps every wave, and far above it there is too little viscosity to sustain the mechanism.

This is also not where channels become turbulent. In the laboratory they do so at Reynolds numbers of order a thousand on the same scales, a factor of five below the critical point, through disturbances of finite amplitude and through growth the eigenvalues do not describe. The critical number is the earliest a linear wave can grow, which a transition Reynolds number is not.

The wave feeds at the critical layer and pays at the wall #

A disturbance draws energy from a mean shear through its Reynolds stress: the product of its streamwise and normal velocities, averaged over a wavelength, which is non-zero only if the two are not a quarter of a period out of phase — the same quantity whose appearance makes averaged equations unclosed. The rate of production is that stress times the shear. The rate of loss is the viscosity times the wave’s own vorticity squared, the price of a gradient paid by the disturbance rather than the mean flow.

For the growing wave at Re = 10⁴ the two are in quite different places. The production is concentrated in a band around the critical layer at $y = 0.873$, peaking at 0.89: 79 per cent of it comes from within a tenth of a half-width of that height, and none of the channel produces negative energy. The dissipation piles up against the wall, inside a layer whose thickness is of order $(\alpha\mathrm{Re})^{-1/3}$, 0.046 here. That layer is where the wave’s own oscillating pressure gradient along the wall puts vorticity into the fluid, at a rate with no viscosity in it; viscosity decides only how far that vorticity spreads before the gradient reverses. The amplitude itself is largest on the centreline and says nothing about where the action is.

The stress profile explains why viscosity is needed. Across the channel the Reynolds stress climbs gently from zero on the centreline, steps sharply across the critical layer, and is brought back to zero at the wall inside the wall layer. In the inviscid theory of a neutral wave the stress is constant everywhere except at the critical layer, where it jumps by an amount proportional to $U''/U'$ there, and without viscosity there is no wall layer to take it back to zero: an inviscid neutral wave on a parabola has nowhere to put the stress it would need. Viscosity supplies both the smoothed jump at the critical layer and the phase shift at the wall, and the two together make a stress that feeds the wave.

Less viscosity means less loss, and past a point less feeding #

The energy identity makes the balance exact: for any mode, twice its growth rate times its energy equals the production minus the dissipation. The eigenvalue solver never uses that identity, and the computed modes satisfy it to 3.5 × 10⁻¹¹.

Along the fastest-growing wave, at the critical point, production and dissipation are equal, which is what a neutral wave means. Just above it, at Re = 6,000, production is 7.62 × 10⁻³ per unit of wave energy against dissipation of 7.25 × 10⁻³. Dissipation then falls all the way, to 3.2 × 10⁻⁴ by 10⁶, as a smaller viscosity would suggest. Production does not. It rises to 1.05 × 10⁻² near Re 2 × 10⁴ and then falls, to 4.6 × 10⁻³ by 10⁶, faster than the dissipation it is outrunning, because the stress that produces it is itself created by viscosity at the critical and wall layers. By a Reynolds number of a million the wave makes fourteen times more energy than it loses and still grows more slowly than it did at fifty thousand, where it made four and a half times more.

The fastest growth peaks at a finite Reynolds number #

The most amplified wave grows fastest at a Reynolds number near 5 × 10⁴, at a rate of 7.69 × 10⁻³, with wavenumber 0.775 and phase speed 0.158. That is an e-folding time of 130 half-widths over the centreline speed. A packet of such waves travels not at the phase speed but at the group velocity, 0.242 of the centreline speed, so it grows by a factor of e for every 31.5 half-widths it moves downstream: even the fastest instability of a channel is slow when measured in distance. By Re = 10⁶ the best rate has fallen to 4.27 × 10⁻³. A single wave fares worse: the wave of wavenumber one grows only between Re = 5,815 and about 32,000, fastest at 4.15 × 10⁻³ near fifteen thousand, and decays again once the narrowing band has moved below it.

The comparison that makes the mechanism concrete is plane Couette flow, the linear profile $U = y$ between plates sliding past each other. It has no inflection point either, and no curvature at all. Its least stable wave at α = 1 decays with $c_i = -0.119$ at Re = 10³, −0.052 at 10⁴ and −0.023 at 10⁵ — approaching neutral from below as the viscosity is reduced and never crossing, as Romanov proved in 1973 for every Reynolds number. The two channels differ in one thing, the curvature $U''$, and it is exactly the thing the inviscid critical-layer jump is proportional to. The quantity Rayleigh’s criterion examines for an inflection is the quantity the viscous mechanism needs to be non-zero — the same second derivative, doing opposite jobs in the two theories.

The flat-plate boundary layer is the other profile the mechanism matters for. It has no inflection point in zero pressure gradient either, and it grows the same kind of wave — the Tollmien–Schlichting wave — above a Reynolds number on displacement thickness of about 520, a value Jordinson computed in 1970 with the same equation.

What the numbers mean in a channel of water #

Put the waves into a real channel. Water at 20 °C has a kinematic viscosity of about $1.0 \times 10^{-6}\,\mathrm{m^2/s}$, so a channel with a half-width of one centimetre reaches the critical Reynolds number at a centreline speed of 0.577 m/s. The first wave to grow there has a wavelength of 6.16 cm and a frequency of 2.47 Hz, slow enough to follow by eye in dye, and its packet moves at 0.383 of the centreline speed.

At Re = 5 × 10⁴ the same channel runs at 5.0 m/s. The fastest wave there has a wavelength of 8.11 cm and a frequency of 9.7 Hz, grows by a factor of e every 0.26 seconds, and its packet does so every 31.5 cm of channel. Growing by $e^9$ — the factor of about 8,100 that the e^N method takes as typical of an ordinary wind tunnel — needs 2.8 m of channel, 280 half-widths, over which the flow must stay quiet enough that nothing else happens first.

That arithmetic is why the instability was disbelieved for a generation. The wave is linear only while it is small, it grows slowly in distance, and any disturbance large enough to see easily is already large enough to set off a different route to turbulence, which in a channel operates at about a fifth of the critical Reynolds number. The instability is real and exact, and in most apparatus it never gets the length it needs — the gap between a critical Reynolds number and a transition Reynolds number, written in metres.

An oblique wave is a slower copy of a straight one #

Every result so far is for waves with crests perpendicular to the flow. A three-dimensional wave with streamwise wavenumber α and spanwise wavenumber β obeys the same equation with $\alpha^2 + \beta^2$ in the operators and α alone in the viscous factor, and Squire noticed in 1933 that this makes it a two-dimensional wave of wavenumber $k = \sqrt{\alpha^2 + \beta^2}$ at the reduced Reynolds number $\alpha\mathrm{Re}/k$. So a wave at angle θ to the flow first grows at 5772.22 divided by cos θ: 6,665 at 30°, 8,163 at 45°, 11,544 at 60° and 22,302 at 75°. Solving the oblique neutral points with the three-dimensional operators, without using the rule, reproduces those to 3 × 10⁻⁹. The first wave to grow exponentially is two-dimensional.

That is a statement about exponential growth only. The transient amplification that the non-normal essay describes is largest for disturbances with no streamwise variation at all — streamwise vortices lifting slow fluid off the walls — and Squire’s rule has nothing to say about it. The straight-crested wave wins the race to be unstable and loses the race to cause turbulence.

The solver against the numbers it was not given #

Orszag’s eigenvalue is reproduced to 1.3 × 10⁻⁹, and the whole-spectrum QR and the single-mode iteration agree on it to 1.9 × 10⁻¹³, so the two eigen-solvers check each other. The critical Reynolds number is 5772.2218 against 5772.22 and the wavenumber 1.020548 against 1.02056, the last differing in the sixth figure because the neutral Reynolds number is flat at its minimum. The energy identity holds to 3.5 × 10⁻¹¹. An oblique wave and its Squire equivalent agree to 1.2 × 10⁻¹³, which is rounding, since the two matrices are the same matrix. Plane Couette flow’s least stable wave at Re = 5 × 10⁴ decays, with $c_i = -0.0298$. The calculation refuses a wave of zero wavenumber, a negative Reynolds number, a grid of eight intervals, a profile given without its curvature and a neutral curve asked for below the critical point.

What the linear picture cannot show #

Transition. Channels become turbulent at Reynolds numbers well below 5772, through finite-amplitude disturbances and non-modal growth, and above it through waves that have grown large enough to break down. Neither is in a linear eigenvalue problem.

What happens to a wave once it is large. A Tollmien–Schlichting wave of finite amplitude develops a secondary, three-dimensional instability of its own, and it is that instability rather than the primary wave that produces turbulent spots.

Other profiles. A boundary layer grows along the plate, so treating its profile as parallel is an approximation whose error is largest exactly near its critical point; the channel is one of the few flows for which the parallel assumption is exact.

The asymptotic band. The rates at which the band’s edges fall at infinite Reynolds number are known from theory and are not reached by 10⁶. What happens between the computed range and the limit is a matter of asymptotic analysis, not of the numbers here.

Still open: where a channel stops being able to stay turbulent #

The linear critical point is exact and is not where anything practical happens. A channel made turbulent and then slowed does not return to laminar flow at 5772, nor at the thousand or so where it first became turbulent, but at a lower Reynolds number still, where turbulence survives only as isolated oblique bands that split and decay. Whether that lower threshold is a sharp critical point with the statistics of a phase transition, as experiments and simulations on pipes and channels have increasingly suggested, and what number it sits at for a channel on these scales, is not settled. It is the question that replaces “when does a wave first grow” once the eigenvalue has answered it: a question about whether a disturbance that already exists can keep itself going, with the energy budget computed here taken to amplitudes where the wave feeds on itself rather than on the mean flow.