The flow with no solution

The world with no inertia is the collection’s account of creeping flow: drop the inertia, keep the viscous term, and what is left is linear, reversible and solvable. Stokes’ drag on a sphere, $6\pi\mu Ua$, is the standard result and it is right.

Try the same thing for a cylinder and there is no answer. Not an approximate one, not one needing a different method — the problem as posed has no solution.

Four lines #

The general solution of the two-dimensional Stokes equations with the symmetry of a body in a stream is

$$\psi = \left(A r\ln r + B r + \frac{C}{r} + D r^3\right)\sin\theta.$$

Four constants, four conditions: two at the wall and two at infinity.

At infinity the flow must approach $U r\sin\theta$. That forbids $D r^3$, which grows faster than $r$; it forbids $A r\ln r$, which also grows faster than $r$; and it fixes $B = U$. Three of the four constants are gone.

That leaves $C$ and two wall conditions. Setting $\psi(a) = 0$ — no flow through the surface — gives $C = -Ua^2$. And the tangential velocity at the wall is then

$$-\frac{\partial\psi}{\partial r}\bigg|_{r=a} = -\left(U + \frac{Ua^2}{a^2}\right) = -2U,$$

with nothing left to adjust.

The no-slip condition is missed by exactly twice the free-stream velocity, at every radius, every speed and every viscosity. The computation confirms it at nine combinations to $10^{-14}$, and the constancy is the point: this is not a small residual to be improved on, it is a fixed failure.

Why the two thrown-away terms had to go #

It is worth pausing on the two constants the far field killed, because they are the interesting ones and because their absence is the whole difficulty.

$D r^3$ is a straining flow that grows quadratically in velocity. Nothing in an unbounded problem can produce it, and it appears in bounded problems where the outer boundary imposes a strain.

$A r\ln r$ is the Stokeslet — the flow due to a point force. Its velocity grows logarithmically with distance, which is why it cannot be present when the flow must approach a uniform stream, and it is the only term in the family that carries a net force on the body.

So the far-field condition has removed the term that carries the force, and the remaining flow has no force on it. That is the paradox in its most economical form: an unbounded two-dimensional Stokes flow matching a uniform stream exerts no force on the body, and a body with no force on it cannot satisfy no-slip.

Three dimensions escapes because a three-dimensional Stokeslet decays as $1/r$ rather than growing, so it is allowed and the force can be carried.

What the paradox is about, which is not the cylinder #

The natural reading is that something is wrong with two-dimensional flow. It is not. The trouble is the far field, and the sphere has it too.

Dropping inertia is legitimate where viscous stresses dominate, which is near the body. Far away the velocity disturbance decays and the viscous term, which involves second derivatives, decays faster than the inertial term, which involves first derivatives times the velocity. So beyond some radius the neglected term is the larger one.

The crossover is the Oseen length. Comparing $U\,\partial u/\partial x \sim U u'/r$ with $\nu\,\nabla^2 u \sim \nu u'/r^2$ gives $r \sim \nu/U = a/Re$, which is enormous at small Reynolds number and finite at every Reynolds number.

So the Stokes equations are the wrong equations beyond $a/Re$, however small $Re$ is. The limit $Re \to 0$ and the limit $r \to \infty$ do not commute, and Stokes’ paradox is what that non-commutation looks like in two dimensions.

Put a wall somewhere and the problem is fine #

The quickest way to see that the far field is the culprit is to remove it.

With an outer boundary at radius $R$ carrying the uniform stream, there are four conditions for four constants and exactly one solution, computed here by solving the four-by-four directly. All four conditions are met to $10^{-9}$ or better at every box size tried, and the $r\ln r$ term — the one the unbounded problem had to throw away — is non-zero, which is what makes the difference.

Its drag is

$$F = \frac{4\pi\mu U}{\ln(R/a) - \tfrac12 + \cdots},$$

and it agrees with the solve to 1.6 per cent at $R/a = 10^{12}$.

The reciprocal of the drag is a straight line #

The shape of the dependence is the important part, and it is clearest inverted.

$1/F$ against $\ln(R/a)$ is linear, with a slope of 0.079577 — which is $1/4\pi$ to six figures — and it does not approach a horizontal asymptote. A finite unbounded drag would look like a line levelling off. This one keeps going.

So the drag falls to zero as the box grows, logarithmically, and at $R/a = 10^{12}$ it is still 0.47 in units where the closest box gives 9.5. There is no Stokes drag for a cylinder to converge on.

The dependence on the box has a practical consequence that is easy to overlook. A two-dimensional computation of creeping flow past a cylinder must state its domain size, because the drag it computes is a function of it and not merely contaminated by it. A three-dimensional computation of a sphere can report a domain-independent drag once the domain is a few radii across; a two-dimensional one cannot, at any size.

That is the numerical shadow of the paradox, and it is met by anybody who tries the calculation.

Oseen’s resolution, and the logarithm in it #

The fix is to put the inertia back where it matters, which is what Oseen did in 1910: linearise the inertial term about the uniform stream, $\mathbf{u}\cdot\nabla\mathbf{u} \approx U\,\partial\mathbf{u}/\partial x$, which is a good approximation far from the body where the disturbance is small.

The resulting equation is still linear and still solvable, and it behaves correctly at infinity. Matching it to the Stokes solution near the body replaces the artificial outer boundary $R$ by the Oseen length $a/Re$, and Lamb’s result follows:

$$F = \frac{4\pi\mu U}{\tfrac12 - \gamma - \ln(Re/8)}.$$

That is the drag of a cylinder at low Reynolds number, and it is the only case in this subject where the leading term of a small-parameter expansion is a logarithm of the parameter.

Which means it never becomes accurate #

The consequence is the sharpest thing in this essay.

The relative change in the coefficient per decade of Reynolds number is $\ln 10$ divided by the denominator, which is

There is no Reynolds number low enough for the leading term to be right to one per cent. At $10^{-20}$ — a Reynolds number no experiment will ever reach — the coefficient is still moving by nearly five per cent for every further decade.

That is a different kind of asymptotic failure from the ones this collection usually meets. An expansion in powers becomes accurate quickly: halving the parameter quarters a second-order error. An expansion in $1/\ln(1/\epsilon)$ does not become accurate at all in any practical sense, because the logarithm of an enormous number is a modest number.

A limit nothing reaches is the collection’s essay on the general phenomenon — an asymptotic regime whose approach is too slow to be reached — and this is its most extreme instance.

Three dimensions, where the same thing happens one order later #

A sphere has a Stokes solution, so the paradox does not arise at leading order. It arises at the next one but one.

The drag expansion is

$$F = 6\pi\mu Ua\left(1 + \tfrac38 Re + \tfrac{9}{40}Re^2\ln Re + \cdots\right),$$

where the first correction is Oseen’s and is a clean power, and the second carries a logarithm. That is Whitehead’s paradox: the second-order Stokes problem has no solution satisfying the far-field condition, for the same reason the first-order two-dimensional one does not.

The signature is measurable. The ratio of the second correction to the first is not a power of the Reynolds number, and its apparent exponent drifts from 0.835 to 0.410 across two decades — which would be constant if both terms were powers.

So the non-uniformity is the same non-uniformity, displaced by one order. Three dimensions buys one extra term before the far field has to be dealt with.

Spin the cylinder instead, and the paradox is gone #

The diagnosis was that the far field kills the one term carrying a net force. That is a checkable claim, because it predicts that a problem needing no net force should have no paradox at all.

Rotation is the test. Spin the cylinder about its own axis in fluid otherwise at rest, and the answer is one line:

$$u_\theta = \frac{\Omega a^2}{r},$$

which satisfies no-slip at the surface, decays to nothing at infinity, and gives a torque of $4\pi\mu\Omega a^2$ per unit length. No missing constant, no residual, no logarithm, and no dependence on the size of the container.

The difference is the decay rate of the singularity each problem needs. The rotlet — the two-dimensional flow of a point torque — falls as $1/r$, so a torque can be handed out to infinity without disturbing a uniform stream. The Stokeslet — the flow of a point force — grows as $\ln r$, and cannot.

So the paradox is not about two dimensions, not about cylinders and not about the Stokes equations. It is about one singularity’s decay rate, and the sharpest statement of it is that in a plane creeping flow a torque reaches infinity and a force does not.

Why the sphere is luckier than the cylinder #

The difference is worth understanding rather than memorising, and it is about how fast a disturbance decays.

A Stokeslet in three dimensions decays as $1/r$; in two dimensions it decays as $\ln r$, which is to say it grows. So a two-dimensional Stokes solution with a force on the body cannot match a uniform stream at all, and the only solutions that decay are the ones with no net force — which is exactly what the four-line argument found.

That is the same dimensional asymmetry three dimensions are kinder records for ideal flow: a sphere’s disturbance falls as $1/r^3$ and a cylinder’s as $1/r^2$, the added mass differs, and a sphere cannot carry circulation at all. Two dimensions is the harsh case in the exact theory and in the creeping-flow theory both, and for related reasons.

What still works, which is nearly everything #

None of this makes creeping-flow theory unusable, and it is worth saying which of the collection’s low-Reynolds-number results are affected.

A swimmer that cannot go backwards is Taylor’s waving sheet: an infinite sheet, no net force, no far-field matching, and its results — a swimming speed of $\tfrac12 k^2b^2c$ and a work per unit distance with no amplitude in it — are exact.

A force without the flow that makes it is the reciprocal theorem, which is an identity between two Stokes flows and needs no far field at all.

Nothing but the shape of the gap is lubrication theory, where the geometry is thin and the far field never enters.

What is affected is precisely the class of problems in which a net force on a body has to be communicated to infinity in an unbounded fluid — a settling particle, a translating cylinder, a drag measurement. Those are the ones where the neglected inertia dominates the far field, and in two dimensions they are the ones with no answer.

What a measured drag on a fibre actually is #

There is an experimental corollary worth stating, because it explains a familiar difficulty.

Measuring the drag on a long thin fibre at low Reynolds number is a standard experiment and its results have always been awkward: the coefficient depends on the length of the fibre, on the presence of walls, and on how the fibre is held, more strongly than a three-dimensional measurement does.

That is not sloppiness. The drag genuinely depends on whatever is at a distance $a/Re$ — a wall, the fibre’s own ends, the container — because that is where the flow’s far field is being decided. A two-dimensional drag is a property of the whole apparatus, and the only way to get a well-defined number is to say which apparatus.

Slender-body theory for a fibre of finite length handles it by using the length as the outer scale, which puts a $\ln(L/a)$ in the answer rather than a $\ln(1/Re)$. The logarithm does not go away; its argument changes.

A number worth carrying #

The Oseen length $a/Re$ is the thing to remember, because it converts the abstraction into a distance.

A one-millimetre sphere settling in water at a Reynolds number of $10^{-2}$ has an Oseen length of ten centimetres. A one-micron particle at $10^{-6}$ has one of a metre. A bacterium swimming at $10^{-5}$ has one of about a decimetre.

So the region in which Stokes’ equations are the right equations is generously large in every practical case, and the region beyond it — where they are not — is a laboratory rather than a universe. That is why creeping-flow theory works so well for so many things and why its far-field failure is nearly always irrelevant.

It becomes relevant exactly when a net force has to be communicated to infinity, and there the container’s own size is usually smaller than the Oseen length anyway — which means the answer depends on the container, which is the correct answer.

Limits recorded rather than smoothed over #

Lamb’s formula is quoted rather than derived. The matched-asymptotic construction that produces it is several pages and is not attempted here; what is computed is its consequences, and the structure of the failure it repairs.

The bounded solve is a model of the resolution, not the resolution. Replacing the far field by a wall at $R$ and then setting $R \sim a/Re$ reproduces the logarithm and gets the leading behaviour right; it does not get the constant inside the logarithm right, which is what Oseen’s matching supplies.

The sphere’s series is quoted too. The $\tfrac38 Re$ is Oseen’s and the $\tfrac{9}{40}Re^2\ln Re$ is Proudman and Pearson’s, and neither is derived here. What is computed is the drift in the apparent exponent, which is a consequence of the form.

And “no solution” is a statement about the posed problem. The Stokes equations have plenty of two-dimensional solutions; what has no solution is the specific combination of no-slip on a cylinder and a uniform stream at infinity, in an unbounded domain, with inertia entirely absent.

The pattern of a singular perturbation, in general #

It is worth extracting the general shape, because Stokes’ paradox is the textbook case and the shape recurs across this collection.

A regular perturbation problem has a small parameter multiplying a term, and setting the parameter to zero gives a problem of the same character with a solution close to the true one everywhere. Corrections are found by iterating, and they are small everywhere.

A singular perturbation has a small parameter multiplying the highest derivative, or a term that becomes dominant somewhere in the domain. Setting it to zero changes the character of the problem — an order is lost, or a region is misdescribed — and the reduced problem cannot satisfy all the boundary conditions. The failure is confined to a region that shrinks with the parameter, and the resolution is a matched expansion with a different scaling in that region.

Prandtl’s boundary layer is the first example every student meets: the viscous term carries the highest derivative, dropping it loses the no-slip condition, and the layer is the region where it is restored. The thin layer is that story.

Stokes’ paradox is the same structure with the regions reversed. The reduced problem is good near the body and bad far away, so the boundary layer is at infinity — which is why it is harder to see and why it took fifty years between Stokes and Oseen.

And the length the limit invents is a third instance, in which the failure is a short region in the streamwise direction rather than a thin one across the flow.

Three problems, one structure, and in each the resolution is a region with its own scaling that the leading-order theory cannot express.

The residue #

The limit is $Re \to 0$ and the residue is unusual: it is not a quantity, it is the existence of the answer.

Take the limit near the body and everything is fine — the Stokes equations describe the local flow beautifully, and every result in this collection that uses them is safe. Take it uniformly, out to infinity, and in two dimensions the problem has no solution at all, and in three it has one whose second correction does not exist either.

What survives is the far field, where the term that was dropped is always the larger one, however small the parameter multiplying it. That is the definition of a singular perturbation, and Stokes’ paradox is the cleanest example of one in fluid mechanics.