A shear layer and a row of vortices are one formula, and not one flow

The vorticity nothing decides asked what fixes the inside of a closed streamline, and showed that Euler’s equations cannot say: a steady two-dimensional inviscid flow only needs vorticity to be some function of the streamfunction, and which function is information the equations were never given. Any choice of that function gives a steady flow. Few of them can be solved by hand, and the handful that can has an outsized place in the subject because a closed form is something a reader can inspect.

One of them is the most photogenic exact solution in fluid mechanics. J. T. Stuart wrote it down in 1967:

$$\psi = \ln\left(\cosh y + A\cos x\right), \qquad 0 \le A \le 1 .$$

It has one parameter. Set $A=0$ and $\psi=\ln\cosh y$, whose velocity is $u=\tanh y$, which is the hyperbolic-tangent shear layer: the smooth profile that nearly every calculation of the Kelvin–Helmholtz instability begins with. Let $A$ grow and the streamlines open into a row of closed eyes joined by thin braids. Let $A$ reach one and all the vorticity lands on a row of points. The whole of a shear layer’s supposed life appears to be there in a single line of algebra, and it is an irresistible picture to hang next to a photograph of a mixing layer. This essay says what it does and does not show.

What the family holds fixed #

The velocity follows by differentiating,

$$u = \frac{\sinh y}{\cosh y + A\cos x},\qquad v = \frac{A\sin x}{\cosh y + A\cos x},$$

and as $y$ goes to plus or minus infinity, $u$ goes to $\pm1$ whatever $A$ is. So every member is a layer between the same two streams, and by Stokes’s theorem every member carries the same circulation in each period, $4\pi$. The numerical integral, done over the plane for each $A$, gives $4\pi$ to the four figures it was asked for at both $A=0.5$ and $A=0.9$.

That is the first thing that makes the family look like a history. The vorticity is simply being gathered, and the total is not changing. It is

$$\omega = -\frac{1-A^2}{(\cosh y + A\cos x)^2} = -(1-A^2)\,e^{-2\psi},$$

a function of $\psi$ alone, as a steady inviscid flow needs. At the centre of an eye it is $(1+A)/(1-A)$ times the layer’s peak, and on the braid between eyes it is $(1-A)/(1+A)$ times it. At $A=0.5$ that is 3 and a third; at $A=0.9$, 19 and a nineteenth. Vorticity piles up in the eye and the braid is stripped bare, and the separatrix’s half-height is $\operatorname{arccosh}(1+2A)$.

The limit is exact: at $A=1$ the denominator vanishes at $x=\pi$, $y=0$, and the vorticity becomes a string of delta functions, each of strength $4\pi$, whose flow is $\ln(\cosh y+\cos x)$. That is the velocity of a row of equal point vortices spaced $2\pi$ apart, summed by Lamb in closed form. The row is singular and is reached only in the limit, and it is the one member of the family whose peak vorticity is infinite. So a parameter with a range of one takes a smooth flow to a singular one, through steady flows, and every intermediate state is an exact solution.

A mean profile that goes the wrong way #

The first symptom that it is not a history comes from averaging along the layer. A known integral gives the mean velocity in closed form,

$$U(y) = \frac{\sinh y}{\sqrt{\cosh^2 y - A^2}},$$

and that is a surprise. It says that the mean profile gets sharper as the eyes grow.

The explanation is straightforward once seen. A row of point vortices has a velocity that decays exponentially away from the row, so a short distance above it the flow is uniform, and its mean is the discontinuous one. The eye contains closed streamlines going both ways, and averaging along $x$ cancels them. So as the vorticity is gathered, the mean profile loses its width. Momentum thickness, the usual measure of how far a layer has developed, is

$$\theta = \frac14\int\left(1-U^2\right)dy ,$$

and it does the following.

In a real mixing layer the momentum thickness grows as it rolls up, and this is the fact that first ought to worry anyone reading the family as a time history. Winant and Browand’s measurements of 1974, and every computation since, show $\theta$ rising through the saturation of the instability, and rising in steps through pairings. A mixing layer gains momentum thickness because it is taking fluid in from both sides. Stuart’s family has it falling, slowly at first and then very fast. Rise against fall is not a matter of degree. It says that as the family’s parameter increases, it does something opposite to what the flow does.

The opening of the eye is itself a useful piece of information. Near $A=0$, the eye’s half-height is $2\sqrt{A}$, which is the square-root law of any nonlinear critical layer: a wave of small amplitude $A$ has closed streamlines around its critical level, of a width proportional to the square root of its amplitude. That is the same structure as the critical level that pays a water wave, seen from the inside, and it is why a linear calculation that ignores the eye is never quite justified near the critical level, however small the wave.

The area of fluid at each vorticity #

The decisive argument takes one line. In a two-dimensional inviscid flow, each parcel keeps its vorticity and its area. It follows that the amount of fluid with vorticity beyond any stated level is a constant of the motion, for every level at once. That is an infinite family of conservation laws, the Casimirs of two-dimensional Euler, and it is why a smooth 2D flow can never raise its peak vorticity. Whatever else the flow does, however the parcels move, the graph of area against vorticity is fixed from the initial condition onward.

Now read that graph off the family.

The tanh layer’s vorticity never exceeds one, and at $A=0.5$ the area of fluid above one is finite and the peak is 3. Nothing inviscid can do that. The vorticity at the centre of a rolled-up inviscid tanh layer cannot exceed the peak it started with, however tightly it is wound. The eye may hold the fluid that started near $y=0$ in a tight spiral, but it holds it at the original vorticity, mixed with low-vorticity fluid in fine filaments. If those are averaged, a coarse picture has a lower peak than the layer began with, the very opposite of Stuart’s concentration. That is exactly why a spiral is a legible record: the layers of the rolled-up sheet keep their order and their strength.

To go along Stuart’s family in the direction it seems to be pointing, then, the vorticity would have to be gathered from the braids and moved into a smaller area at a greater value: precisely what the transport of vorticity forbids. A viscous flow is permitted to diffuse it, but diffusion spreads vorticity and lowers peaks, so viscosity is the wrong direction too. No unforced flow, viscous or not, can start from a tanh layer and arrive at any other member of this family with a higher peak.

Each member lives on a different sheet #

There is a tidy way to say all of this, due to Arnold in 1965. Think of the space of all flows of a given period, and in it, of the set of flows that can be reached from one another by moving parcels about without changing them: rearrangements of a given vorticity. Those sets are sheets, and they fill the space as the pages of a book fill the book. An inviscid flow never leaves its page, since each parcel keeps its vorticity. Every steady flow is a point on some page where the energy does not change to first order under any rearrangement; a maximum or minimum of energy on its page is a stable flow, and that is the whole of Arnold’s stability theorem.

Stuart’s parameter moves from page to page. Each $A$ has a different area curve, so it is on a different sheet, and the family’s curve runs across the book rather than along any page. This is what the eye misreads when looking at the dial. Along a page the fluid moves. Across pages the fluid is replaced. A sequence of pictures in which each one is a valid flow can therefore be a sequence in which no fluid at all has gone from one picture to the next, and what survives being wound up is exactly the property that distinguishes the two. The same reading is why two eddies can stand but not for long: a pair of vortices can be fixed in place by arithmetic without that saying anything about how they would have got there, and a steady flow carries no account of its own arrival. The same is true of the street behind a cylinder, whose picture is steady only in a moving frame and whose existence is not explained by being a solution.

The energy is not available either #

There is a second conserved quantity, and it agrees. Take the kinetic energy per period relative to a step of the same far speeds, a convergent integral.

The tanh layer sits below the step, at $-2\pi$ per period, and each increase of $A$ raises the energy. To gather vorticity in a mixing layer’s frame takes energy, and the energy the family asks for has no source in an unforced flow. A row of point vortices needs an infinite amount. A rolling-up layer, in contrast, moves along a line of constant energy. So there is a clean geometric statement: if the family is a curve in the space of steady flows, the Kelvin–Helmholtz roll-up is a trajectory that crosses that curve at most at one point, and it cannot follow it.

What the family is good for #

None of this makes the family useless. Each member is a real steady Euler flow, and the question “is this flow stable” is a precise and answerable one. Pierrehumbert and Widnall in 1982 took Stuart’s family as a base state and found the instabilities that matter to a mixing layer, the two-dimensional subharmonic that pairs neighbouring eyes and a three-dimensional mode of the kind that makes streamwise rib vortices. That work is why the formula is so widely known: not as a history but as a usable instantaneous base state, with an adjustable amount of concentration, in a problem where no real base state could be written down. As a local fit of what a rolled-up layer’s streamlines look like, it does well, as long as the fit is for the shape and not for the vorticity. The eyes in a photograph are the right shape. The vorticity inside them is a function of $\psi$ only if something has homogenised it, as the Prandtl–Batchelor argument says viscosity eventually will, and then it is uniform rather than exponential.

A shear layer is a map of a sphere #

The form $\omega=-(1-A^2)e^{-2\psi}$ is the surprising thing here, and it connects this flow to subjects with no fluid in them. It is Liouville’s equation of 1853,

$$\nabla^2\psi = (1-A^2)\,e^{-2\psi},$$

and Liouville was not interested in flows. He was solving for a metric. If $e^{-2\psi}$ multiplies the ordinary $dx^2+dy^2$ to make a new geometry, the condition that the new geometry has a constant curvature is this equation. Every solution of it is a map from the plane to a constant-curvature surface. Stuart’s family gives surfaces of curvature $1-A^2$, a sphere of radius $1/\sqrt{1-A^2}$, with each period of the layer wrapped around the sphere once. Checked at scattered points by central differences, $e^{2\psi}\nabla^2\psi-(1-A^2)$ comes out at about $10^{-7}$, the size of the finite difference’s own error. And it explains a number already met. The area of that sphere is $4\pi/(1-A^2)$ and its curvature $1-A^2$, so their product is $4\pi$ whatever $A$ is: the circulation per period is the sphere’s total curvature, which Gauss and Bonnet say is the same for every member. The sphere swells as $A$ approaches one, and nearly all its area comes from inside the eye, which is the point vortex. It is not an analogy in any dynamical sense, and it does not help compute anything here; but the existence of a closed form is explained by it. The equation is integrable because of what it means geometrically.

The same equation has a whole second life as a balance of forces. Vorticity that depends exponentially on the streamfunction is a Boltzmann distribution. Joyce and Montgomery found in 1973 that the most probable arrangement of many like-signed point vortices has exactly that form, with the streamfunction playing the part of an energy; so the family is also a set of statistical equilibria of a vortex gas. An isothermal, self-gravitating gas sheet has the density profile $\operatorname{sech}^2 y$, found by Spitzer in 1942, which is the vorticity of the tanh layer. A current sheet in a plasma has the magnetic field $\tanh y$, Harris’s sheet of 1962, and its streamfunction with islands in it, Fadeev’s equilibrium of 1965, is the same formula as Stuart’s. Space physicists draw magnetic cat’s eyes in the Earth’s magnetotail with exactly this expression. Each of those is a balance between an exponential response and a Poisson equation, and nothing in it is fluid mechanics until a parcel starts to move.

What the picture cannot show #

A time. This is the essential omission and the reason for this essay. The family has a parameter, and the dial on the figure looks like a clock, but nothing is moving. The figure’s streamlines belong to a steady flow in the frame of the eyes, and sliding $A$ does not correspond to any physical process. A real roll-up must be computed, as a sheet or a smooth layer, and its picture is different in the most visible way: its eyes have spirals in them.

Stability. Each flow in the figure is steady and the figure says nothing about whether it would stay that way. It does not; pairing is the rule, and the row of point vortices itself is unstable.

Three dimensions and viscosity. The closed form belongs to two inviscid dimensions. A real layer develops ribs and breaks down, and it is not periodic: real eyes come in a distribution of sizes.

A unique way of adding more eyes. The formula spaces its eyes at a stated period and treats one period as all there is. The freedom of $\omega=F(\psi)$ also means other families exist. Stuart’s is the simplest in closed form and not the only one that interpolates: Mallier and Maslowe’s array of alternating vortices of 1993 shares its algebra, as does a family with eyes of mixed strength.

Conventions and checks #

Lengths are in units of the wavelength over $2\pi$; speeds in units of half the velocity difference; the layer’s vorticity sign is written as negative, as $\omega=\partial v/\partial x-\partial u/\partial y$ with $u$ rising through $y$, and the areas are plotted against its magnitude. Momentum thickness uses $\frac14(1-U^2)$, which makes the tanh layer’s $\theta$ equal one half. Energies are per period and per unit depth, relative to the step’s $\frac12$ per unit area, integrated out to $y=\pm14$ where the integrand is negligible.

Besides those, the model refuses $A=1$ and $A<0$: the first is the singular limit, the second a phase shift of another member. The area curve is by counting grid cells and is good to about the third figure; its use here is qualitative and very robust, since the curves differ by factors.

Who wrote it, and when #

Liouville’s equation is from 1853. Stuart’s solution was published in the Journal of Fluid Mechanics in 1967, in a paper on finite-amplitude oscillations in laminar mixing layers, the nonlinear partner of Michalke’s linear problem for $\tanh y$. The same expression had already been written for a plasma by Fadeev and his colleagues in 1965. Pierrehumbert and Widnall’s stability calculation is from 1982, and the measurements of the stepwise thickening of a mixing layer, by Winant and Browand, from 1974.

Still open: where a real layer crosses the family #

A real inviscid roll-up never moves along this family, but at some moment it may be close to a member. The next calculation would march a tanh layer through saturation at high resolution, coarse-grain the vorticity over a filter that hides the filaments, and find which $A$ minimises the distance between the coarse field and Stuart’s, frame by frame. It would ask how close the best fit ever gets, whether the fit’s $A$ rises and then saturates, and, since the coarse field’s own peak is below the layer’s, what a fit with a lower peak than $A=0$ even means: whether the right family for a real saturated layer is a different function $F(\psi)$, the one mixing selects, rather than an exponential that nobody chose.