One channel, one flux, two flows

The solution that keeps its nonlinear term closed by saying that past the parallel flows the exact solutions are similarity reductions: the nonlinearity survives, the closed form does not, and the answer is an ordinary differential equation. It also said the disc’s reduction ends with an answer. This one does not.

Take two plane walls meeting along a line, at $\pm\alpha$, and look for a flow that is purely radial. That assumption alone does almost all the work. Continuity forces the speed to fall as $1/r$, and writing $u_r = \nu F(\theta)/r$ turns the Navier–Stokes equations into

$$F''' + 2FF' + 4F' = 0, \qquad F(\pm\alpha) = 0.$$

One equation, one variable, the nonlinear term intact. And past a threshold the problem has two solutions carrying the same flux through the same geometry, with nothing to choose between them.

Solved by inverting an integral rather than by shooting #

Where the one-over-r comes from #

The step that makes the whole thing work is worth pausing on, because it is the only place the Navier-Stokes equations are used before the reduction and it is not a similarity argument.

Continuity in plane polar coordinates, with no $\theta$-component at all, reads $\partial(r u_r)/\partial r = 0$. So $r u_r$ cannot depend on the radius, and the speed must fall as $1/r$ exactly — not approximately, not to leading order, and with no assumption about the profile’s shape. The whole radial dependence of the solution is fixed by mass conservation before any momentum equation is written down.

What remains is a function of $\theta$ alone, and substituting it into the two momentum equations gives two equations for one unknown. That should over-determine the problem and does not: the $\theta$-momentum equation says only that $\partial p/\partial\theta$ is a particular expression, which integrates to give the pressure, and the radial one — after eliminating the pressure by cross-differentiation — is the third-order equation above. So the count works out, and it works out because the flow was assumed to have one component rather than two.

That is the difference between this reduction and the rotating disc’s. There, all three components are non-zero and the similarity is a genuine guess about how each scales with radius, which the three momentum equations then have to be consistent with. Here the radial dependence is a theorem and only the profile is guessed. It makes the wedge flow the cheaper of the two to derive and, as it turns out, the stranger of the two to solve.

The equation integrates twice. Once gives $F'' + F^2 + 4F = A$; multiplying by $F'$ and integrating again gives

$$(F')^2 = P(F) = 2A(F-a) - \tfrac23(F^3 - a^3) - 4(F^2 - a^2),$$

with $a = F(0)$ the centreline value and $F'(0) = 0$ by symmetry. So the profile is an inverse function, obtained from one quadrature,

$$\theta(F) = \int_F^{a}\frac{\mathrm dF}{\sqrt{P(F)}},$$

and the wall condition $\theta(0) = \alpha$ is a single equation for $A$.

That is worth insisting on as a method rather than a convenience. Marching the third-order equation outward from the centreline — the obvious thing, and what the disc needed — diverges before reaching the wall for every trial curvature once the profile has thin wall layers, and the divergence reads as the solution branch ending when it is the integrator giving up. Three versions of this calculation reported a converging channel as having no solution above a modest flux for exactly that reason. The quadrature has no such failure: $P$ is a cubic whose roots are available in closed form, and the only difficulty is a square-root singularity at each end of the range, which a squared substitution removes exactly.

The controls say the method works. Every solved profile fills the wedge it was asked for to $3\times10^{-13}$, and in a narrow wedge the flux comes back to a parabola’s $\tfrac43 F(0)\alpha$ to $4\times10^{-5}$, with the difference falling as the square of the angle.

The narrow-wedge limit is worth recognising before the wide one is examined. What it returns is the parabola a pipe or a channel carries, which is the exact solution every treatment of viscous flow begins with — so the wedge problem contains the textbook case as the first term of its own expansion, and the corrections to it are second order in the angle.

Where a diverging channel runs out #

Put a flux through the wedge outward — the apex upstream, the channel widening — and raise it.

The profile flattens in the middle and steepens at the walls, which is what a channel does when inertia starts to matter. And then it stops. The wall shear is $-\sqrt{P(0)}$, and

$$P(0) = -2Aa + \tfrac23 a^3 + 4a^2$$

reaches zero at $A = a^2/3 + 2a$ — a closed condition, not a numerical one. At that value the wall shear vanishes: the flow has separated. Beyond it $P(0)$ is negative, the square root has no real value, and there is no purely outward profile of this form at all.

That is a separation with nothing approximated anywhere near it. There is no boundary-layer assumption, no thin-layer expansion, no neglect of any term: it is the exact solution of the full equations running out of real numbers. Separation is usually introduced as a phenomenon of boundary layers, described by a boundary-layer equation, and diagnosed by a criterion. Here it is the discriminant of a cubic.

Separation is normally a boundary-layer statement, and it is worth saying what is different here. How much uphill a layer can take prices it the usual way, with a thin-layer approximation and a criterion applied to a profile; and when the flow lets go is the account of what happens afterwards. Neither is available to this calculation and neither is needed: the wall shear here is the square root of a cubic, and the cubic runs out.

A threshold that is a gamma function #

The threshold depends on the wedge angle, and it depends on it in a way with a limit.

Written as $\alpha^2 F(0)$ the threshold is 9.6155 at a half-angle of 0.4 radians, 10.1381 at 0.2, 10.2691 at 0.1 and 10.3019 at 0.05. It is approaching something, and the something has a closed form.

In the narrow-wedge limit the whole problem collapses. Scaling $f = \alpha^2 F$ on $\zeta = \theta/\alpha$ removes the angle from the equation entirely, leaving $f'' + f^2 = \text{const}$, so every threshold in the problem becomes a pure number in $f$. At separation the cubic factors — $P = \tfrac23 f(a-f)(a+f)$ — and the quadrature becomes elementary:

$$\alpha^2 F(0) \Big|_{\text{sep}} = \tfrac32\left[\int_0^1\frac{\mathrm dt}{\sqrt{t-t^3}}\right]^2 = \tfrac38\left[\frac{\Gamma(1/4)\,\Gamma(1/2)}{\Gamma(3/4)}\right]^2 = 10.312779.$$

Computing that integral by quadrature gives 2.6220609 against the gamma functions’ 2.6220576, and solving the narrow-wedge problem with the same solver used for every other angle returns 10.312779 — agreement to a part in $10^9$ between three routes that share no arithmetic.

10.31 is the number quoted throughout the literature as the separation criterion for this flow, and it is a ratio of gamma functions in exactly the way the 4.02 in the two-thirds law is. Both are elementary integrals over a finite range with a power-law integrand, and in both cases what looks like a measured constant is a value of the beta function.

What the threshold is in a workshop #

The criterion is written in a combination of variables chosen to have a limit, and it is worth converting it into something a diffuser designer would recognise.

$\alpha^2 F(0) = 10.31$ with $F(0) = u_{\max} r/\nu$ becomes $\alpha \cdot (u_{\max}\, \alpha r/\nu) = 10.31$, and $\alpha r$ is the channel’s half-width $h$ at radius $r$. So the criterion is

$$\alpha \cdot \frac{u_{\max} h}{\nu} = 10.31,$$

a wedge angle times a Reynolds number built on the local half-width. Everything about the shape is in the angle and everything about the flow is in the Reynolds number, and their product is a constant.

Read that way it says something sharp. Double the angle and the Reynolds number at which the flow separates halves. A diffuser at a half-angle of 3.5 degrees — 0.061 radians, which is the seven-degree total the handbooks recommend — separates in this model at a half-width Reynolds number of 169. A diffuser at 10 degrees total separates at 118. Neither is a large number; both are far below anything a real duct runs at.

Which is the honest reading of the classical rule rather than a vindication of it. A laminar diverging channel separates at almost any Reynolds number a real device operates at, and the reason a seven-degree diffuser works is that its flow is turbulent, and a turbulent layer carries far more momentum near the wall and survives a far steeper pressure rise. The exact solution says what happens without that help, and what happens is that the flow cannot stay attached. The rule of thumb is about turbulence; this number is what the rule is a repair for.

The constant’s provenance is worth one more sentence, because this shape has been met before. The 4.02 at the bottom of the two-thirds law is $-\Gamma(-2/3)$, arrived at the same way: an elementary integral over a finite range with a power-law integrand, whose value is a beta function and is therefore a ratio of gammas. A number that looks measured and is not is a recurring shape rather than a coincidence.

The converging channel, which has no such wall #

Run the flux the other way — apex downstream, channel narrowing — and none of this happens.

The converging profile develops a flat core and thin layers against each wall, and the wall shear grows with the flux instead of falling, so there is nothing to reach. The asymmetry sits in the sign of one cubic term: $P(0)$ carries $\tfrac23a^3$, which is positive when $a$ is positive and has to be beaten by $-2Aa$, and negative-to-positive when $a$ is negative so that nothing competes with it.

That is the whole of why a nozzle is easy and a diffuser is hard, written without a boundary layer in it. A converging channel accelerates its fluid and the pressure falls along the flow; a diverging one decelerates it and the pressure rises, and a fluid slowed by a rising pressure eventually has none left near the wall to slow. The exact solution knows that, and the number at which it happens is the one above.

One further feature of the converging solution is worth noting because it is the same object seen elsewhere. At high flux the profile is a uniform core with thin layers against each wall, which is the structure a duct develops after it has forgotten its inlet — except that here it is reached by raising the flux rather than by going further along, and the core is accelerating rather than uniform.

Two flows, one flux #

Past separation the purely outward branch has ended. The equation has not.

$P$ is a cubic with a root at $F = a$ by construction, so dividing that out leaves a quadratic whose two roots are available exactly. When one of them lies below zero, there is a second kind of profile: $F$ falls from the centreline value, crosses zero inside the channel, turns at that root, and climbs back to zero at the wall. The wall is the second zero rather than the first, and the fluid next to each wall is running the other way.

The two branches overlap in flux, and that is the point. In a wedge of 0.2 radians a flux of 24.71 is carried by an outward profile with a centreline value of 105.1, and by a reversed profile with a centreline value of 979.9 whose backflow reaches $-333.9$. The two fluxes agree to nine parts in a hundred thousand, which is the resolution of the search rather than a difference.

Both satisfy the full Navier–Stokes equations. Both satisfy no-slip at both walls. Both are steady, both are symmetric, both are exact. The problem as posed — this geometry, this flux — has two answers, and the equations do not contain the information that would choose between them.

And 105.1 against 979.9 is not a technicality about a branch touching. They are different flows: one delivers its flux with a gentle profile and a modest peak, the other with a jet in the middle and recirculation against the walls, and a dye trace in the two would look nothing alike. The reversed branch goes on past what is drawn, and further branches exist above it with two reversals, three, and so on — the same construction with the profile turning round more times.

What non-uniqueness does to a calculation #

There is a practical consequence, and it is the one that makes this more than a curiosity.

A numerical solver handed this geometry and this flux will converge to a solution. Which one depends on where it started — a solver initialised from rest, or from a parabola, finds the outward branch; one initialised from a previous run at a higher flux may find the reversed one. Neither converged run carries any indication that the other exists, and both will satisfy every residual the solver checks, to whatever tolerance was asked for.

That is the general shape of the problem and it is not confined to wedges. A steady solver looks for a zero of a residual; where the residual has several zeros it finds the one in whose basin it started, and “converged” means only that. The safeguards that exist are continuation — walk the parameter slowly and watch for the branch to fold — and perturbing a converged answer to see whether it returns. Neither is automatic and neither is usually done.

The wedge is a good place to learn it because the multiplicity is computable in closed form: the cubic’s roots say exactly where the second branch begins, so a solver’s behaviour can be checked against an answer rather than against another solver.

The wedge also has a corner, and the corner has its own exact answer. Moffatt’s eddies are the infinite sequence of counter-rotating vortices a viscous fluid forms in a sharp corner at vanishing Reynolds number, in the same geometry as this one. They are a creeping-flow solution of the same wedge — the nonlinear term dropped rather than kept — and they occupy the region near the apex where the solution here is singular.

What decides it, and it is not on this page #

The honest position is that this calculation establishes non-uniqueness and says nothing whatever about selection.

What is missing is time. Both solutions are steady, and steadiness is the assumption that was made at the start; a real flow arrives at one of them from some history, and which one it arrives at is a question about the unsteady equations. It may also be a question about stability — one branch may be unstable and never observed — and stability is not decided by the steady equations either. Both questions need calculations this essay does not have.

What can be said is that the existence of the second branch is not an artefact of the assumption of purely radial flow. That assumption is strong and is examined below, but it is symmetric between the two branches: they are both solutions of the same reduced problem, and if the reduction is legitimate for one it is legitimate for the other.

What the reduction assumes, and where it fails #

Purely radial flow is an assumption about the answer. Nothing derives it. It is a guess at the shape of the solution which turns out to be consistent, and consistency is not uniqueness — the same equations admit flows with a $\theta$-component that this ansatz cannot represent, and a real diffuser at high flux is full of them.

Symmetry is assumed too. Every profile here is even in $\theta$. The equation admits asymmetric solutions — reversed against one wall and not the other — and they are not computed. In a real diverging channel the asymmetric separation is the one that is actually observed, which makes this a real omission rather than a tidy one.

The apex is singular and the model does not describe it. The speed goes as $1/r$, so it is unbounded at the line where the walls meet: three decades closer to the apex is a thousand times faster. The flux through every arc is the same, which is what makes the singularity a genuine feature of the solution rather than a failure of it — the same flux squeezed through an ever-narrower gap — and it also means the solution is a description of the region far from the apex and of nothing else. A real wedge has a throat, an inlet, or a source of finite size.

The solution is steady and laminar. Diverging flow at the fluxes where the second branch exists is, in practice, unsteady and usually turbulent, and none of that is here.

And this is not the wedge flow that appears in boundary-layer theory. Falkner and Skan’s similarity solutions are also called wedge flows, they are also indexed by an angle, and they are a different problem entirely: there the wedge is a body with an external stream running past it, the solution is a boundary layer on its surface with an outer flow imposed, and the reduction is of the boundary-layer equations rather than of the full ones. The Falkner–Skan family also separates at a computable angle — a favourable exponent below about $-0.0904$ — and that number and the 10.31 here are unrelated, arising from different equations about different geometries. The two are worth keeping apart deliberately, because a reader who has met one will recognise the words of the other and the arithmetic does not transfer in either direction.

The arithmetic is checked in the usual way. The narrow wedge returns a parabola’s flux; every profile fills the wedge it was asked for; the separation constant agrees with its own gamma-function closed form to a part in a million and with an independent solve of the limiting problem to a part in $10^9$; and the two branches’ fluxes agree at the crossing. The calculation refuses a wedge with no angle, a wedge opened past a half-plane, a profile with no centreline value, and a reversed profile asked of a converging channel — where, the arithmetic being what it is, there is nothing to reverse.

Still open: how many solutions there are, and what counts them #

Two branches have been found here and a third and fourth are visible in the construction — each one turning round once more on its way to the wall. That suggests the count is not two but a sequence, and the question the next calculation would answer is what indexes it.

The natural conjecture is that the number of reversals is the index, and that the branch with $k$ reversals appears at its own threshold, each threshold a different pure number in $\alpha^2 F(0)$. Computing the first half-dozen of those thresholds, and asking whether they are also gamma functions or whether only the first is elementary, is a bounded piece of work with the solver already here — the roots of the cubic are exact and the only new thing needed is a profile that walks between them more than twice.

Beside it is the question the rotating disc left. A rotating disc facing a stationary one admits the same kind of similarity, and the argument about what it produces — one profile across the gap, or two layers with a core turning between them — ran for twenty years and ended in the discovery that both occur. That is this essay’s phenomenon in a different geometry, and computing the two branches of the two-disc problem with the disc solver would put the same statement on both: an exact reduction is a statement about what solutions look like, and not about how many there are.