The solution that keeps its nonlinear term

The three exact solutions below this one are solutions of the Navier–Stokes equations, and they are exact for the same reason. The wall that shakes, the layer that stops growing and the started plate are all parallel flows: the velocity points one way and varies across it, so $(\mathbf u\cdot\nabla)\mathbf u$ is identically zero at every point, and what is left is the diffusion equation with a term added. The nonlinearity was not solved. It was arranged out of existence, and the suction layer said so — two exact solutions built out of the same trick, kill the nonlinear term and what remains can be integrated.

There is a fourth, and it does not use the trick. Von Kármán’s rotating disc has a three-dimensional velocity field with all three components non-zero, its convective terms are the same size as its viscous ones everywhere in the layer, and the reduction is exact anyway. What makes it work is not a vanishing but a cancellation of a coordinate: every term in all three momentum equations turns out to scale as $\Omega^2 r$, so the radius divides out and three partial differential equations become three ordinary ones.

The similarity, and what it costs #

Put a plane disc in an unbounded fluid at rest and spin it at $\Omega$. The fluid touching it is dragged round by no-slip, is thrown outward by the swirl it has acquired, and is replaced from above. That much is a description; the similarity is the claim that the description has the same shape at every radius.

Write the height in the only length the problem has — there is no other, since the disc is infinite and the fluid unbounded —

$$\eta = z\sqrt{\Omega/\nu},$$

and write the velocities as

$$u_r = \Omega r\,F(\eta), \qquad u_\theta = \Omega r\,G(\eta), \qquad u_z = \sqrt{\nu\Omega}\,H(\eta).$$

The radial and azimuthal velocities carry a factor $r$, because they are driven by a surface moving at $\Omega r$. The axial one does not, and that asymmetry is the whole of the trick: continuity then reads $H' = -2F$ with no $r$ in it, and substituting into the momentum equations gives

$$H' = -2F,\qquad F'' = F^2 - G^2 + HF',\qquad G'' = 2FG + HG',$$

with $F(0) = H(0) = 0$, $G(0) = 1$, and $F, G \to 0$ far above.

What it costs is the formula. The three solutions below all end in a closed expression — an exponential, an error function, a complex exponential in depth. This one ends in a boundary-value problem with two unknown wall gradients, and the answer is a number found by integration. “Exact” here means the reduction is exact, and the distinction is worth insisting on, because a reduction that is exact and an answer that is closed-form are two different goods and only the first is available past the parallel flows.

Why it is the radius that divides out #

It is worth asking why the similarity exists at all, because the answer is short and it explains which other problems will have one.

There is no length in this problem. The disc is infinite, the fluid is unbounded, and the only quantities available are $\Omega$ (one over a time), $\nu$ (a length squared over a time) and the coordinates. From those, exactly one length can be built — $\sqrt{\nu/\Omega}$ — and it is a property of the fluid and the rotation rather than of any point in the flow. So the vertical structure has no choice: it must be a function of $z$ divided by that length, and nothing else is available for it to depend on.

The radius is different, because $r$ is a length and can be combined with the others. What fixes its role is that the boundary condition itself carries it: the disc’s surface moves at $\Omega r$, so any velocity driven by the surface must be proportional to $r$. The axial velocity is not driven by the surface — it is supplied by continuity, which relates a derivative in $z$ to a derivative in $r$ and removes one power — which is why it alone has no $r$ in it.

That is the whole of the reasoning, and it is a dimensional argument of the kind counting what matters sets out: the number of independent groups a problem can depend on is decided before anything is solved, and here it comes out at one. A problem with a second length in it — a disc of finite radius, a housing above it, a gap to a second disc — has a second group, the similarity is lost, and the equations go back to being partial.

The nonlinear terms, which are not small anywhere #

The claim that this solution keeps its nonlinearity is the essay’s whole point, so it deserves to be measured rather than declared.

At the wall the arithmetic is immediate. $F(0) = 0$ and $H(0) = 0$, so two of the three convective terms vanish there and the third does not: $F''(0) = -G(0)^2 = -1$, exactly. The entire curvature of the radial velocity profile at the surface is the centrifugal term, which is $u_\theta^2/r$ and is as nonlinear as anything in the equations. In every solution below, $F''(0)$ would have been set by a pressure gradient or by a clock.

Above the wall the balance passes to the third term. $HF'$ is the axial convection — fluid being carried down through a varying radial profile — and it is nonlinear because $H$ is not independent: continuity makes it the integral of $F$. From about one penetration depth outward it is the largest term in the equation, and it stays so all the way out.

The middle of the layer shows something else, and it is the reason a numerical solution is needed rather than a perturbation. The three terms there are individually of order a hundredth and their sum is of order a ten-thousandth: computed along the solution, the convective magnitude exceeds the residue they leave by a factor of 96. The equation in that region is a small difference of larger quantities, which is exactly the arithmetic that makes a series expansion in any of them useless.

Three numbers, found rather than quoted #

The solution is two unknowns — $F'(0)$ and $G'(0)$ — shot from the wall until $F$ and $G$ both vanish far above. It is a textbook exercise and its answers are printed in every textbook, and printing them here without the convergence would be quoting an integration rather than reporting a solution.

The far field of this problem has a growing mode, which is what makes shooting delicate: a guess slightly off does not merely give a slightly wrong answer, it diverges. So the march is grown outward by continuation — solved on a short domain, the answer used to seed a longer one — and the honest test is whether the result stops moving. It does. The wall gradients change by $5\times10^{-7}$ between fourteen penetration depths and eighteen, and the axial inflow by $9\times10^{-5}$.

$$F'(0) = 0.5102325, \qquad G'(0) = -0.6159220, \qquad H(\infty) = -0.8844555.$$

The far-field conditions are left at $10^{-14}$, which is the residual of the Newton iteration rather than a statement about the physics.

The contrast with the started plate’s own layer is worth holding beside it. There the fluid is dragged along by a wall that has just begun to move, the nonlinear term is identically zero, and the answer is an error function. Here the fluid is thrown outward by a wall that has been turning forever, the nonlinear term is the whole of the balance at the surface, and the answer is an integration. The two are the same equation with one term restored.

A layer that does not grow along the surface it is on #

The most surprising thing in the solution is what is missing from it, and it takes a moment to notice.

Every boundary layer in the collection so far thickens along the surface. Blasius’ grows as $\sqrt{x}$ and never stops; the suction layer stops but only because fluid is being removed. Here the layer’s thickness is $5.41\sqrt{\nu/\Omega}$, and there is no radius in it.

On a disc a hundred millimetres across turning at a hundred radians a second in water, the layer is 0.541 millimetres at the axis, 0.541 millimetres at the rim and 0.541 millimetres everywhere between. A plate in a stream at the rim speed has a layer growing as the square root of distance, and the two are the same thickness only at one station.

The reason is a cancellation and not a coincidence. A boundary layer’s thickness at distance $x$ under a stream $U$ goes as $\sqrt{\nu x/U}$. On the disc the distance from the axis is $r$ and the local sweeping speed is $\Omega r$, so the estimate is $\sqrt{\nu r/(\Omega r)} = \sqrt{\nu/\Omega}$ — and the radius cancels exactly, not approximately. The Blasius estimate applied locally gives $5.0\sqrt{\nu/\Omega}$ against the solution’s 5.41, which is unusually close agreement for an estimate, and it is close because the similarity the estimate assumes is the similarity the flow actually has.

That is also the answer to why this problem has an exact reduction and the flat plate does not have an exact solution at all. Blasius’ equation is itself a similarity reduction, and it is exact in the same sense; what it is not is exact for the Navier–Stokes equations, since deriving it requires the boundary-layer approximation first. The disc’s reduction requires no approximation. All three momentum equations are kept in full, and the similarity is substituted into them rather than into a simplified version of them.

One consequence of the constant thickness is worth drawing out because it inverts an intuition. On a flat plate the friction coefficient falls along the surface, because the layer thickens and the velocity gradient at the wall slackens. On the disc the layer does not thicken, so the wall stress rises in exact proportion to the local surface speed, and the shear grows linearly with radius while the thickness does not move at all. The outer third of a disc’s radius therefore carries most of the torque — the moment arm grows as $r$ and the stress grows as $r$, so the contribution per unit radius grows as $r^3$ — and the innermost half of a disc contributes one sixteenth of its drag. That is why a disc brake’s friction material is an annulus and not a plate, and it follows from a solution with no fitted constant in it.

It is also worth putting beside the layer that never stops thickening, which is the case every boundary-layer calculation starts from. That layer grows because the only length it has is the distance from the leading edge; this one does not grow because it has a length that is not a distance at all.

The pump with no moving parts except itself #

The axial velocity’s behaviour is the part of the solution that is genuinely odd on first sight and is the part that has made this flow an instrument.

$H$ does not go to zero far from the disc. It goes to $-0.8845$, which means the disc draws fluid down onto itself at $0.8845\sqrt{\nu\Omega}$ — at every radius, at every height, for ever.

The disc is a pump. It throws fluid outward in a thin sheet at its surface and must draw an equal volume down from above, and because the outflow per unit radius grows as $r$ while the area grows as $r^2$, the inflow speed needed is the same everywhere. That uniformity is why a rotating disc electrode is a standard piece of electrochemical apparatus: the rate at which fresh fluid is delivered to its surface does not vary across the surface, so a reaction proceeds at one rate over the whole disc, and the rate is set by one knob. At a hundred radians a second in water the inflow is 8.8 millimetres a second.

The torque comes out of the other wall gradient by an integration and has no fitted constant in it either: $C_M = 2\pi|G'(0)|/\sqrt{Re}$ for a disc wetted on both sides, which is $3.870/\sqrt{Re}$. For the same 0.1-metre disc at a hundred radians a second in water, the torque on one side is 0.0967 newton-metres. Above a rotational Reynolds number of a few hundred thousand the layer goes turbulent and the measured torque exceeds this — a borrowed observation rather than anything computed here, and the shaded band marks where the laminar answer stops being an answer.

The pumping has a second reading that connects it to the rotating flows elsewhere in the collection. A fluid brought into rotation by a boundary reaches its steady state through a secondary circulation rather than by diffusion alone, and that circulation is what makes spin-up far faster than viscosity could manage. The disc’s inflow is the same mechanism seen in isolation, with nothing to spin up.

The same equations with the swirl the other way round #

The reduction does not belong to von Kármán’s problem. It belongs to the geometry, and swapping which side is turning gives a second solution from the same three equations.

Bödewadt’s problem is a disc at rest under fluid rotating at $\Omega$, so $G(0) = 0$ and $G \to 1$. One term has to be added to make it work, and leaving it out is why a first attempt at this diverges: far from the disc the rotating fluid needs a radial pressure gradient to hold it in its circle, $\rho^{-1}\partial p/\partial r = \Omega^2 r$, and that gradient reaches all the way down to the wall where there is no swirl left to balance it. The radial equation gains $+s^2$, with $s$ the far-field swirl ratio.

The result is the flow’s mirror image. $F'(0) = -0.9419709$ and $G'(0) = 0.7728854$: the radial flow near the wall is inward, the axial flow is upward, and the layer is 1.54 times thicker. Fluid spirals in along the floor and is thrown up in the middle, which is the tea-leaf paradox — stir a cup, let it settle, and the leaves gather at the centre rather than being flung to the rim, because the only place they can be carried is where this secondary flow takes them.

Bödewadt’s profiles also overshoot and come back, oscillating about their far-field values as they decay, where von Kármán’s approach monotonically. That is visible in the axial velocity, which is still climbing through its oscillation at nine penetration depths — 1.3353 there, against a limit near 1.35 — and it is the reason the layer edge has to be defined as the last height at which the swirl is still a per cent away rather than the first, which reports a quarter of the real thickness.

What the earlier solutions could not have reached #

The four exact solutions now in this sequence divide cleanly, and the division is worth stating because it is not the one a reader would expect.

The first three are all the same solution wearing different boundary conditions. Stokes’ oscillating wall is the diffusion equation driven at a frequency; the asymptotic suction layer is the diffusion equation with a constant drift; the started plate is the diffusion equation with a step. Each has a closed form, each is obtained by a transform or a substitution, and none of them is a solution of the Navier–Stokes equations in any sense that the nonlinear term participates in. They are exact statements about a linear problem that happens to be what the Navier–Stokes equations reduce to in a parallel flow.

The fourth is a solution of the full equations. Nothing was dropped, nothing cancelled, and all three momentum equations are carried through the substitution intact. What is given up is the closed form.

That trade is the general shape of the subject past the parallel flows, and it is why the list of exact solutions is as short as it is. Almost every one of them is either a parallel flow — where the nonlinearity is absent and a formula survives — or a similarity reduction like this one, where the nonlinearity is present and the answer is an ordinary differential equation. There is very little in between, and the reason is structural rather than a matter of nobody having tried hard enough: a similarity reduction needs a symmetry, and the equations have only so many.

Which is the question the essay above this one asks. If the exact solutions are the reductions the equations’ symmetries permit, then the list is finite, countable, and its length is a fact about a group rather than a measure of how hard people have looked.

One more comparison closes the sequence’s own arc. The wall the fluid is listening to shows that a viscous flow’s response to its boundary is a convolution — the fluid at a height is averaging over a distribution of delays. That statement is available because the problem is linear. Here it is not, and the profile above the disc is not an average of anything: it is a balance struck at every height between three terms that each depend on the answer.

What the similarity does not cover #

The disc is infinite and no disc is. The similarity has no radius in it precisely because there is no radius in the problem, and a real disc has an edge where the outflow leaves and a wake behind it. The solution is a good description in the interior and is not a description of the rim; how far inward the edge’s influence reaches is not something this calculation can say.

The fluid is unbounded above. A disc in a housing has a second wall, and the return flow that wall forces turns the problem into the rotor–stator cavity, whose solutions are not these and whose classification is the subject of the still-open section.

The layer is assumed laminar and it is not, above a few hundred thousand. The measured transition on a rotating disc is itself a landmark — it is one of the cleanest cases of a cross-flow instability, and the spiral vortices that precede it are visible — and none of that is in a solution which assumes the flow depends on $\eta$ alone.

And the shooting is a numerical solution with a numerical error. The convergence figure is what is claimed about it: the wall gradients stop moving at seven figures, the axial inflow at five, and the Bödewadt case’s axial inflow has not converged at all at the furthest the march reaches. Where a number here is quoted to more figures than that, it is being quoted from the integration rather than from the problem.

The controls are the usual ones. The thickness computed at two radii a thousandfold apart agrees exactly, because no radius enters. The nonlinear measure applied to a fluid at rest returns zero, which it must, or it would be measuring the arithmetic rather than the flow. And the calculation refuses a disc that is not turning, a moment at a Reynolds number of nothing, and a guess from which the march cannot converge.

A closing note on where this solution stops being observed rather than where it stops being derived. The cost of going turbulent is the general statement that a laminar friction law is a lower bound once the layer transitions, and on a disc the transition is unusually clean — the spiral vortices that precede it are visible, and they are a cross-flow instability rather than the streamwise one a flat plate has.

Still open: where the same equations stop having one answer #

The two problems above are the ends of a family. Let the fluid far away turn at $s\Omega$ while the disc turns at $\Omega$, and the same three equations carry $s$ as a parameter: $s = 0$ is von Kármán’s, and the Bödewadt case is its counterpart with the roles exchanged. Between and beyond them the family is not as well behaved as either end.

The calculation worth doing is to march $s$ downward through zero into counter-rotation. For $s$ below about $-0.2$ the single solution branch of this problem is known to end, and what replaces it is more than one solution at the same $s$ — a fold in the branch, with two or three profiles satisfying the same boundary conditions, differing in whether the fluid is thrown out near the disc or drawn in. Finding that fold with the shooting solver above, and asking which of the branches a fluid actually adopts, would be the first case in this sequence where an exact reduction does not give an exact answer because it gives several.

Beside it is the case that matters practically, which is two discs rather than one. A rotating disc facing a stationary one at a finite gap admits the same similarity, and the argument about what it produces — whether the two layers stay separate with a core turning between them, or merge into one profile across the gap — is the Batchelor–Stewartson controversy, which ran for twenty years and was settled by finding that both answers occur. That is the next essay’s subject arriving from a different geometry: an exact solution that has stopped being one answer.