The wall the fluid is listening to

The wall that shakes solves Stokes’ second problem exactly: a wall sliding back and forth in its own plane sends a wave into the fluid that dies within one wavelength, with a penetration depth of the square root of twice the viscosity over the frequency.

That is one wall motion. This essay asks the general question — for any wall motion, what is the fluid at a given height responding to? — and the answer is a single kernel that contains both classical solutions and a good deal more.

The kernel #

For a wall moving in its own plane with velocity $U(t)$, the fluid at height $y$ has

$$u(y,t) = \int_0^\infty K(y,r)\,U(t-r)\,dr, \qquad K(y,r) = \frac{y}{2\sqrt{\pi\nu}\,r^{3/2}}\,e^{-y^2/4\nu r}.$$

That is a weighted average of the wall’s past, and the weight is a genuine probability density: it is positive everywhere and it integrates to exactly one over all delays. Over the finite window computed here the quadrature reads 0.98242 against the closed form’s 0.99748, a gap of 1.5 per cent that is the window rather than the kernel. So the fluid at a height is literally the wall’s velocity, averaged over its own history with a fixed weighting.

The weighting has a shape worth reading rather than quoting. It is zero at zero delay — nothing has had time to arrive — rises to a peak, and then falls off as $r^{-3/2}$, which is a heavy tail.

Three summaries, and two of them are misleading #

The mode is at $y^2/6\nu$ — two thirds of a second at two millimetres in water. That is the delay the fluid is most likely to be responding to, and it is the number a diffusion argument would give.

The median is at $1.099\,y^2/\nu$, which is 4.40 seconds: 6.6 times later than the mode. Half of what the fluid is doing is a response to something older than that.

And the mean does not exist. The tail falls as $r^{-3/2}$, so the integral of the delay against the kernel diverges. Computing it over a window gives a number that grows as the square root of the window and never settles: at two millimetres the median is 4.396 seconds and the mean over a window of two hundred thousand seconds is 507.8 — a hundred times the median, and still rising with the window.

That last is not a pathology to be apologised for. It is the statement that a diffusive memory has no characteristic age, and it is the same fact that gives the Basset force in the drag that integrates a whole history its missing time constant. A distribution with a mode and a median and no mean is exactly what a fluid at a distance from a wall is responding to.

How the delay grows with height #

Both the mode and the median go as the square of the height, with a measured exponent of 2.00000. That is the diffusion law, and its practical form is that doubling the distance from a wall quadruples how far back the fluid there is listening.

Six heights above a wall in water, in seconds:

Every column goes as the height squared: a factor of sixteen in height gives 256 in delay, exactly. A millimetre above a wall the median delay is 1.10 seconds and the ninetieth percentile is 31.7 — a spread of 29 within one height. A centimetre above, the median is nearly two minutes. Ten centimetres above, three hours — which is why a container of water spun up by its own walls takes so long to notice, and is how long a fluid takes to forget it was not rotating without the rotation to help.

Why a diffusive memory has a heavy tail #

The exponent in the tail is where the whole of the strangeness lives, and it is worth deriving because the derivation is three lines and says which problems share it.

Vorticity generated at the wall at some instant spreads outwards, and after a delay $r$ it occupies a region of thickness $\sqrt{\nu r}$ with its strength spread over that thickness. The amount of it that has reached height $y$ is therefore proportional to $1/\sqrt{\nu r}$, provided $r$ is large enough that the spreading has passed $y$ at all.

That gives the tail directly. Once $\sqrt{\nu r}$ exceeds $y$ the exponential in the kernel is close to one and what is left is $r^{-3/2}$ — the $r^{-1/2}$ of the spreading, times another $r^{-1}$ from the fact that a diffusing quantity of fixed total is being spread thinner.

So a heavy tail is what conservation plus spreading produces. The old contributions are individually small and there are a great many of them, and the two effects nearly cancel — which is exactly the condition for an integral to converge slowly or, in the case of the mean, not at all.

Both classical solutions, out of one kernel #

The kernel is worth trusting only if it reproduces what is already known, and it does.

The impulsively started wall. Convolve the kernel with a step and the result is the complementary error function of $y/2\sqrt{\nu t}$ — Rayleigh’s solution — to within 3·10⁻⁵.

The oscillating wall. Convolve the same kernel with a cosine and the result is the exponentially decaying wave with its one-radian-per-penetration-depth phase lag — Stokes’ solution — to within 9·10⁻⁷.

Those are not two problems. They are one memory, given two inputs, and the phase lag that the wall that shakes computes is the kernel’s delay expressed as an angle at one frequency.

What the phase lag looks like from here #

The connection is worth spelling out because it explains a number that is otherwise a coincidence.

Stokes’ solution has the fluid at depth $y$ lagging the wall by $y/\delta$ radians, where $\delta$ is the penetration depth. Written as a time, that lag is $y/(\delta\omega)$, which at one penetration depth is $1/\omega$ — a radian’s worth of period.

The kernel says the same thing differently: the fluid is averaging over a distribution of delays whose mode is $y^2/6\nu$. At $y = \delta = \sqrt{2\nu/\omega}$ that is $1/(3\omega)$, which is the same order and not the same number — because a lag measured from a sinusoid is a property of the whole distribution rather than of its mode, and the tail pulls it later.

So the phase lag of an oscillating wall is not the delay; it is what the delay distribution does to one frequency. A wall driven at two frequencies at once produces two different lags from one kernel, which is exactly what a memory means and what a single “diffusion time” cannot express.

What the solver computed, and how it was checked #

The kernel is written down and its cumulative distribution has a closed form — the fraction of the response that has arrived by delay $r$ is $\mathrm{erfc}(y/2\sqrt{\nu r})$ — so the quantiles are exact and the numerical integration is only used to check them.

Four checks. That the numerical integral matches the closed-form total to three per cent over the window, which is what says the sampling is adequate on a heavy-tailed function. That the mean diverges, by requiring the partial mean to grow by more than a factor of three when the window grows by a thousand — it grows by 34, which is the square root. That the median is at least five times the mode, since a distribution whose median and mode nearly coincided would not support the essay. And that the delay goes as the square of the height, to three per cent.

Then the two convolutions, against their closed forms. Those are the checks that matter most, because they test the kernel itself rather than statements about it.

Why this is the shape of every diffusive memory #

The quantity a wall cannot take away, however long it is listened to, is the subject of what viscosity cannot take away; and the model that assumes the wall has finished listening is the theory with no memory in it.

The kernel above is specific to a plane wall, and its shape is not.

Any quantity governed by a diffusion equation, responding at a distance to something imposed at a boundary, has a response with the same three properties: zero at zero delay, because nothing arrives instantly; a peak at a distance squared over a diffusivity, because that is the only time the problem contains; and a heavy algebraic tail, because diffusion never finishes.

That is why the same structure appears in the Basset force, in the transient heat flux from a suddenly heated surface, in the concentration downstream of a released tracer, and in the pressure response of a well in a porous formation. The exponents differ with the dimensionality; the absence of a characteristic age does not.

The contrast worth keeping is with a relaxation memory, which the fluid that has not finished its last deformation computes: there the kernel is an exponential, it has a time constant, and there is a definite age past which the fluid has genuinely forgotten. Diffusion has no such age, and models that give it one have put it in by hand.

What the wall is really doing #

It is worth naming the physical object the kernel describes, because “the fluid is listening to the wall” is a metaphor and the mechanism is not.

A wall that moves generates vorticity: there is no other way for it to affect the fluid, and where vorticity comes from is this collection’s account of the wall as the only source. Once generated, that vorticity diffuses outwards and is carried nowhere else, because in this problem there is no convection at all.

So the kernel is a record of arrivals. The velocity at a height is the accumulated effect of every sheet of vorticity the wall has ever emitted, each one having spread by an amount set by how long ago it left — and the weighting is simply how much of each sheet has reached that height by now.

Read that way, the missing mean stops being surprising. The wall has been emitting vorticity for as long as it has been moving, none of it is destroyed, and all of it is still somewhere. What arrives at a given height is dominated by recent emissions in number and by old ones in extent, and the two balance almost exactly.

What this says about a boundary condition #

There is a modelling consequence worth stating, because it is the reason unsteady wall problems are awkward.

A no-slip wall is usually treated as a boundary condition: the fluid velocity there equals the wall’s, now. That is exactly true and it is not the whole of the coupling, because the fluid away from the wall is coupled to the wall’s past rather than its present.

So a computation that starts an unsteady wall problem from a uniform initial condition has, in effect, asserted that the wall has been at rest for ever — and the transient it computes is the fluid learning that this was false. The time that takes is the kernel’s, and at a distance from the wall it is long.

A short unsteady run near a wall is measuring its own initial condition for longer than most people expect, and the diagnostic is exactly the height-squared law above: the region that has forgotten the start extends to $\sqrt{6\nu t}$ and no further.

How much history a computation has to keep #

The practical form of all this is a question anybody solving an unsteady wall problem has to answer, and the kernel answers it.

To get the response at height $y$ right to a per cent, the window of wall history kept must extend to roughly the ninety-ninth percentile of the delay distribution — which, because the tail falls as $r^{-3/2}$, is about $10^4$ times the mode. That is a demanding requirement and it is why unsteady wall problems are solved by marching the diffusion equation rather than by evaluating the convolution.

Marching is the cheap way of keeping the whole history, and that is worth seeing as what it is: a finite-difference solution of the diffusion equation carries the memory implicitly, in the current profile, which is a complete summary of everything the wall has done. The profile is the state, and the kernel is what has to be evaluated when the profile is not available.

That is the general trade in any problem with a memory. Either carry a state that summarises the history — which needs a field — or carry the history itself, which needs a convolution. Viscoelastic codes make the same choice, and the fluid that has not finished its last deformation is where the exponential kernel makes the first option cheap.

What the picture cannot show #

The kernel is drawn on logarithmic axes, which is the only way to show a distribution spanning five decades of delay, and logarithmic axes flatter a heavy tail: the region beyond the median occupies a third of the plot and half the probability, which is not how it looks on linear ones.

Nothing here draws the fluid. The kernel is a weighting, not a velocity field, and the figures showing profiles are borrowed from the essay that computes them at one frequency.

Where the distribution is measurable #

A weighting is an abstraction, and it is worth saying how anybody would see it, because two of the three routes are practical.

Drive the wall with a step and watch a profile develop. The velocity at a height against time is the kernel’s own cumulative distribution, directly — it is the complementary error function, and its quantiles are the delays quoted here. That is a laboratory measurement and it has been made many times.

Drive the wall at several frequencies and read the phase lags. The lag at each frequency is a different moment of the same distribution, so a sweep is a way of sampling it. This is exactly what a rheometer does to a fluid rather than to a layer, and the inversion from lags back to a kernel is the standard one.

Or drive it with noise and cross-correlate. The cross-correlation between the wall’s velocity and the fluid’s at a height is the kernel itself, up to the input’s own spectrum. That is the cleanest route in principle and the hardest in practice, because the response at a height is small and the tail that carries the interesting behaviour is smaller still.

The same kernel acting on a particle rather than on a wall is the essay before this one.

Why the kernel has no mean, and what to use instead #

The absence of a mean delay is the strangest property here and it is worth saying what to do about it.

The mean diverges because the tail is heavy. The kernel falls as a power rather than an exponential, and the integral of time against that power does not converge. Any measured mean is therefore a statement about the length of the record rather than about the fluid, and doubling the record changes the answer.

The median does not diverge, and neither does any other quantile, because a quantile is a statement about where the mass is rather than about how far the tail reaches. So the honest summary of a diffusive memory is a quantile pair — a median and a ninetieth percentile — and not a mean with a spread.

And the practical consequence is a warning about time constants. Fitting an exponential to a diffusive memory gives a time constant that depends on the window fitted over, which is exactly the symptom that says the wrong model is being fitted.

Who found it, and when #

The kernel is the Green’s function of the diffusion equation on a half-line with a Dirichlet boundary, which is Fourier’s problem and Fourier’s answer, from 1822. Stokes’ and Rayleigh’s wall solutions are particular convolutions of it and were obtained directly rather than through it.

Reading it as a distribution over delays is a much later habit and belongs to the theory of linear systems rather than to fluid mechanics — the language of impulse responses, kernels and heavy tails arrived from control and signal processing, and applying it here makes the classical solutions look like two data points on one curve.

Limits recorded rather than smoothed over #

One dimension, one wall, no pressure gradient. The kernel is for a flow driven only by the wall, with nothing happening in the other directions. Adding a pressure gradient adds a second forcing with its own kernel; adding a second wall makes the domain finite and gives the problem a longest time after all.

No convection. Everything here is a wall in a fluid otherwise at rest. In a boundary layer the vorticity is convected downstream as well as diffusing outwards, so the memory becomes a memory of what happened upstream rather than earlier — which is a layer that is an integral of everything upstream, and is the same kernel with a coordinate change.

Laminar. In a turbulent flow the effective diffusivity is not a constant and is not molecular, so the same shape holds with a different and position-dependent coefficient — which changes every number here and none of the structure.

The quantiles are exact and the integrals are not. The closed-form cumulative distribution is what the quantiles come from; the numerical integrations are checks on the sampling and are quoted to three per cent rather than better.

And the missing mean is a property of the idealisation. A real fluid is bounded, so the delay distribution is cut off at the time the far boundary matters, and the mean exists. It is very long, and the point stands: there is no intrinsic time in the problem, and any time quoted has come from a geometry.