Every unstable wave is inside one circle

A layer with a kink in it is the collection’s account of Rayleigh’s criterion: an inviscid shear flow cannot be unstable without an inflection point in its velocity profile. That is a statement about whether. This essay is about where and how fast, and the answer to both is bounded by two numbers that can be read off the profile’s two ends.

The theorem #

For an inviscid parallel shear flow, the complex phase speed $c = c_r + i c_i$ of every unstable mode satisfies

$$\left(c_r - \frac{U_{\max} + U_{\min}}{2}\right)^2 + c_i^2 \le \left(\frac{U_{\max} - U_{\min}}{2}\right)^2.$$

Howard proved it in 1961 in half a page. Nothing about the profile enters except its largest and smallest speeds: not its shape, not where it bends, not how many inflection points it has, not the wavenumber, not whether the flow is stratified — Howard’s proof covers the Taylor–Goldstein equation too, which makes it one of very few results in this subject that survives adding physics.

Two consequences are immediate and both are useful before anything is computed. A disturbance cannot travel faster than the fastest part of the flow or slower than the slowest. And the growth rate $kc_i$ cannot exceed $k$ times half the velocity range.

The profiles chosen have as little in common as possible: a shear layer, the same layer rescaled to a different velocity range, a jet, a wake, an asymmetric layer with an off-centre bump, and a double shear layer with two inflection points where the others have one. If the shape mattered, they would disagree.

Checking the instrument first #

The eigenvalues come from a complex shooting solve: integrate Rayleigh’s equation from below with a decaying condition, ask whether the solution decays above, and drive the residual to zero with Newton’s method on $c$. The method works only for growing modes, and that is a feature — for $c_i > 0$ the factor $U - c$ never vanishes, so there is no critical layer and no branch cut, and the class of modes the solver can find is exactly the class the theorem is about.

Two independent checks. Setting $c = 0$ in Rayleigh’s equation for $U = \tanh z$ turns it into

$$\varphi'' + \left(2\,\mathrm{sech}^2 z - k^2\right)\varphi = 0,$$

which is Pöschl and Teller’s equation; its bound state is $\varphi = \mathrm{sech}\,z$ at exactly $k = 1$ and at no other wavenumber, so the neutral point is an identity. Extrapolating the computed branch to zero growth gives $k = 1.000106$, a part in ten thousand, and the residual is the branch’s own curvature rather than the solver’s.

The growth rate below the neutral point is not exact, and there Michalke’s 1964 numbers are the reference: a maximum amplification of 0.1897 at $k = 0.4446$. The solver returns 0.18969 at 0.4442.

One numerical detail earned its place. Newton’s method on a shooting residual will happily “converge” on a point where the derivative is nearly zero, and the residual itself is a number of arbitrary size because the shooting solution is dominated by a growing exponential. Normalising the residual by the size of the two terms it is a difference of — and rejecting anything above $10^{-4}$ — is what distinguishes a root from a place the iteration stopped. Without it the solver reported an unstable mode at a Richardson number of 0.26, where Miles’ theorem forbids one.

What the circle means physically #

The bound has a reading that is worth having before the figures, because it turns an algebraic statement into something a reader can carry.

An unstable mode grows because it extracts energy from the mean shear, and it does that at a critical layer — the height where the flow moves at the wave’s own phase speed. Above that height the wave is being overtaken by the flow; below it, the wave overtakes. The exchange happens where the two match.

If the phase speed were outside the range of $U$, there would be no such height anywhere in the flow. The wave would be moving faster than everything or slower than everything, and there would be nowhere for it to couple to the shear. So the real part of $c$ has to lie between $U_{\min}$ and $U_{\max}$, and that is the horizontal extent of the circle.

The vertical extent — the cap on the growth rate — is the same argument done quantitatively. The wave can only take energy at the rate the shear can supply it across the layer it occupies, and the shear it sees is bounded by the total velocity difference. Half the range appears because the wave has to sit somewhere inside the profile rather than at one end of it.

None of that is a proof, and the proof below is four lines rather than four paragraphs. But it explains why the two end-values are the only things that matter: they are the two limits on where a critical layer can be, and everything the instability does happens there.

Every mode, inside the circle #

Scaling each profile’s eigenvalues on its own centre and radius puts them all in the same picture, and every one of them is inside the unit semicircle. Across all six profiles and every wavenumber on every branch, the closest approach to the boundary is 0.915 of the radius.

The branches themselves have the familiar shape: a band of unstable wavenumbers, a maximum inside it, a neutral point at the top. The theorem holds at every point of every one of them, including the ones near the maximum where the growth is largest and the temptation to think a bound might be violated is strongest.

And it has the invariance it must. Adding a constant to the velocity profile shifts both the circle’s centre and every eigenvalue’s real part by exactly that constant, and leaves the growth rates untouched to fifteen figures. A stability theorem that failed that would be a theorem about a frame rather than about a flow — which is the same requirement the picture belongs to whoever is watching imposes on every other quantity in this collection.

What the bound is worth #

Between 68 and 91 per cent of the radius. That is the useful range for a bound: close enough that it is not vacuous, far enough that it is not an estimate.

Converted into the quantity anybody actually wants — the maximum growth rate — the bound is a factor of two to three loose. The tanh layer’s fastest mode grows at 43 per cent of what the semicircle permits; the wake’s at 32 per cent; the asymmetric layer’s at 55.

So the theorem is not a prediction and was never offered as one. What it is, in practice, is three things.

A search box. A numerical eigenvalue solve needs somewhere to start and somewhere to stop, and the semicircle supplies both from the profile alone. Every eigenvalue in this essay was found from a guess placed inside it.

A sanity check. An eigenvalue outside the circle is a bug, with no further diagnosis needed. That is worth having in a calculation whose answers are otherwise unverifiable — a stability solver is one of the few instruments in this subject with no independent reference to be checked against, so a theorem that can catch it being wrong is not a luxury.

And a statement about what cannot happen. A shear layer between streams differing by ten metres a second cannot grow a disturbance at twenty. Nothing in the profile can arrange it, and no amount of cleverness in the shape can buy it.

Howard’s proof gives a second inequality alongside the circle — the growth rate is also bounded by the maximum of the profile’s own shear — and the two together cut the semicircle down to a wedge. The eigenvalues sit inside both, and inside neither’s edge.

What it does not decide, which is most of what matters #

It is worth being explicit about the gap between what the theorem gives and what a stability calculation is usually for, because the gap is wide.

It does not say whether the flow is unstable. That is Rayleigh’s criterion and Fjørtoft’s, and the kink in the profile is where they live. A perfectly stable flow satisfies Howard’s theorem vacuously, since it has no unstable modes to put anywhere.

It does not say which wavenumber grows fastest. The bound is proportional to $k$, so read naively it says the growth rate rises without limit with wavenumber — and every real profile has a neutral wavenumber above which nothing grows at all. The bound is loosest exactly where the physics is tightest, which is the usual fate of a bound derived without using the profile.

And it does not say what the disturbance looks like. The eigenfunction — where the billow’s amplitude peaks, how far it reaches into the two streams — is what decides how much fluid a Kelvin–Helmholtz event actually mixes, and nothing in the semicircle touches it.

So the theorem belongs to a particular class of results: cheap, exact, general, and about the outside of a problem rather than its inside. The site’s other members of that class are d’Alembert’s zero and Kelvin’s minimum energy, and all three have the same shape — a statement obtained from a conservation law or an inequality, holding for every case, and cheaper than any of the cases.

Why it is true, in a paragraph #

The proof is worth having because it is short and because it explains the indifference to shape.

Write Rayleigh’s equation in terms of $F = \varphi/(U - c)$, multiply by the conjugate and integrate across the layer. The imaginary part gives $c_i \int U' |F|^2 \cdot$ (something) $= 0$ — which for $c_i \ne 0$ forces an integral condition — and the real part then gives a second one. Combining the two and using the elementary inequality $\int (U - U_{\min})(U_{\max} - U) |F|^2 \ge 0$, which holds because the integrand is a product of two non-negative factors, delivers the circle directly.

That last inequality is the whole trick, and it is where the two end-values enter. The profile appears only through the fact that it lies between its own maximum and minimum, which is true of every profile, so no shape can produce a stronger conclusion or evade this one.

Where a bound like this comes from, generally #

The proof’s shape recurs often enough to be worth naming.

Every result of this kind starts by turning the differential equation into an integral statement — by multiplying by a conjugate and integrating, which is the same manoeuvre that gives an energy equation — and then applies an inequality whose only input is that some quantity has a fixed sign. Here the non-negative quantity is $(U - U_{\min})(U_{\max} - U)|F|^2$, and the price of using it is that everything about $U$ between its two ends has been thrown away.

That trade is the general one. A bound is cheap because it discards information, and the information it discards is exactly what would have made it sharp. Sharpening it means putting some of the profile back — which is what Fjørtoft’s refinement of Rayleigh’s criterion does, and what the second inequality above does — and each step back costs generality.

Knowing where on that scale a result sits is most of knowing how to use it. Howard’s theorem is at the cheap end: two numbers in, a region out, no solving. What it buys is a guarantee, and a guarantee that holds for every profile is worth having even when it is a factor of three loose on the profile in hand.

The one profile that nearly touches #

Two of the six get to 0.915 of the radius, and they are the two shear layers — the plain tanh and the same profile rescaled to run from 0.3 to 1.7. That is not a coincidence: the rescaling is a linear map of the velocity, and both the eigenvalue and the circle transform under it in the same way, so the fraction is invariant. Two profiles related by a scaling and a shift are one profile as far as this theorem is concerned.

The Bickley jet and the wake also share a fraction — 0.677 — and for the same reason: the wake used here is $1 - 0.4\,\mathrm{sech}^2 z$, which is the jet’s profile scaled and inverted. So the six profiles are really four, and the theorem’s own invariances are what collapsed them.

That is worth noticing as a matter of method. A demonstration that a bound holds across “six different cases” is worth less than it looks if two pairs of them are related by transformations the bound is invariant under, and the way to find out is to ask what the theorem does not see. Here it does not see a linear map of the velocity, and it does not see a shift in the vertical either — so a profile displaced upwards is also not a new case.

What is a new case is the asymmetric layer and the double one, which are not related to the others by anything, and they come in at 0.848 and 0.837.

The profile that sits exactly on it #

The bound is loose on every profile computed here, and it is not loose in general — it is attained, by the one profile the essay has been treating as a limit rather than a case.

Take the shear layer’s thickness to zero and what is left is a vortex sheet: a discontinuous jump from one stream’s speed to the other’s. Its stability problem is elementary, and its answer is

$$c = \frac{U_1 + U_2}{2} \pm i\,\frac{U_1 - U_2}{2},$$

which is the circle’s centre plus exactly its radius, straight up. The eigenvalue sits on the boundary, at the top of the semicircle, at every wavenumber.

So the theorem is sharp, and the 0.915 the tanh layer reaches is a measure of how far a profile of finite thickness is from a sheet. Every smooth profile falls inside because it has a thickness, and the sheet is the only thing that touches.

Which also explains the bound’s worst feature. Howard’s cap grows as the wavenumber, and on a sheet the growth rate really does — every wavenumber unstable, the shortest waves fastest, and no most-amplified mode at all. A finite thickness is what supplies the neutral wavenumber, and the same finite thickness is what pulls the eigenvalue in off the boundary. One property, both effects.

What it costs to state #

The theorem takes two numbers from a profile and returns a region. That is the cheapest possible computation in stability theory — cheaper than evaluating the profile’s second derivative, which Rayleigh’s criterion needs — and it is worth noticing how little was spent for how strong a conclusion.

The reason it is so cheap is that it discards nearly everything. Two profiles with the same maximum and minimum are the same profile as far as this result is concerned, however differently they behave in between, and the price of the generality is the factor of two to three by which the bound is loose on any particular one.

That trade — everything about the shape given up, a guarantee obtained — is what a bound is, and it is worth having explicitly in mind whenever one is quoted.

What is not claimed #

Inviscid, parallel, two-dimensional. Viscosity can destabilise a flow that is inviscidly stable — which is what makes pipe flow’s transition so strange — and the theorem says nothing about those modes. Three-dimensional disturbances are covered by Squire’s theorem for the unstratified case and are not treated here.

Linear. Every mode here is infinitesimal. The theorem bounds the initial growth of a small disturbance and says nothing about the amplitude a billow reaches or about the secondary instabilities that follow, which is where the street this site cannot draw begins.

Normal modes. Non-modal transient growth — a superposition of decaying modes that grows for a while before it decays — is not bounded by anything here, and in shear flows it can be very large. The theorem is about eigenvalues, and eigenvalues are not the whole story.

Stratification is not treated here, though the theorem covers it. Howard proved the semicircle for the Taylor–Goldstein equation as well as for Rayleigh’s, so a stratified shear layer’s eigenvalues are inside the same circle; that case is sufficient and not necessary, where the interesting bound is a different one.

And the six profiles are a demonstration. They were chosen to differ, not sampled from anything, and nothing here rules out a profile whose eigenvalues approach the boundary more closely. What rules it out is the proof; the six are a check on the proof and on the solver.