Chapter 5

Dunes

The muddy colour of many rivers and the milky colour of glacial melt streams are dueto the presence in the water of suspended sediments such as clay and silt. The abilityof rivers to transport sediments in this way, and also (for larger particles) by rollingor saltation as bedload transport, forms an important constituent of the processes bywhich the Earths topography is formed and evolved: the science of geomorphology.

Sediment transport occurs in a variety of different (and violent) natural scenarios.Powder flow avalanches, sandstorms, lahars and pyroclastic flows are all examples ofviolent sediment laden flows, and the kilometres long black sandur beaches of Iceland,laid down by deposition of ash-bearing floods issuing from the front of glaciers, aretestimony to the ability of fluid flows to transport colossal quantities of sediment.In this chapter we will consider some of the landforms which are built through theinteraction of a fluid flow with an erodible substrate; in particular we will focus onthe formation of dunes and anti-dunes in rivers, and aeolian dunes in deserts.

5.1 Patterns in rivers

There are two principal types of patterns which are seen in rivers. The first is apattern of channel form, i. e., the shape taken by the channel as it winds through thelandscape. This pattern is known as a meander, and an example is shown in figure5.1.

The second type of pattern consists of variations in channel profile, and thereare a number of variants which are observed. A distinction arises between profilevariations transverse to the stream flow and those which are in the direction of flow.In the former category are bars; in the latter, dunes and anti-dunes. The formationof lateral bars results in a number of different types of river, in particular the braidedand anastomosing river systems (described below).

All of these patterns are formed through an erosional instability of the uniformstate when water of uniform depth and width flows down a straight channel. Theinstability mechanism is simply that the erosive power of the flowing water increaseswith water speed, which itself increases with water depth. Thus a locally deeperflow will scour its bed more rapidly, forming a positive feedback which generates

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Figure 5.1: A meandering river. Photograph courtesy Gary Parker.

the instability. The different patterns referred to above are associated with differentgeometric ways in which this instability is manifested.

River meandering occurs when the instability acts on the banks. A small oscilla-tory perturbation to the straightness of a river causes a small secondary flow to occurtransverse to the stream flow, purely for geometric reasons. This secondary flow isdirected outwards (away from the centre of curvature) at the surface and inwardsat the bed. As a consequence of this, and also because the stream flow is faster onthe outside of a bend, there is increased erosion there, and this causes the bank tomigrate away from the centre of curvature, thus causing a meander.

high stage

low stage

Figure 5.2: Cross section of a braided river with one lateral bar, which is exposedwhen the river is at low stage (i. e., the river level is low). The instability which causesthe bar is operative in stormflow conditions, when the bar is submerged.

Braided rivers form because of a lateral instability which forms perturbationscalled bars. This is indicated schematically in figure 5.2. A deeper flow at one side ofa river will cause excess erosion of the bed there, and promote the development of a

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lateral bar in stormflow conditions. The counteracting (and thus stabilising) tendencyis for sediments to migrate down the lateral slope thus generated. Bars commonlyform in gravel bed rivers, and usually interact with the meandering tendency to formalternate bars, which form on alternate sides of the channel as the flow progressesdownstream. In wider channels, more than one bar may form across the channel, andthe resulting patterns are called multiple row bars. In this case the stream at lowstage is split up into many winding and connected braids, and the river is referred toas a braided river, as shown in figure 5.3.

Figure 5.3: A braided river. Image from http://www.braidedriver.net.

It is fairly evident that the scouring conditions which produce lateral bars andbraiding only occur during bank full discharge, when the whole channel is submerged.Such erosive events are associated with major floods, and are by their nature occa-sional events. In between such floods, vegetation may begin to colonise the raisedbars, and if there is sufficient time, the vegetative root system can stabilise the sed-iment against further erosion. A further stabilising effect of vegetation is that theplants themselves increase the roughness of the bed, thus diminishing the stress trans-

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mitted to the underlying sediment. If the bars become stably colonised by vegetation,then the braided channels themselves become stabilised in position, and the resultingset of channels is known as an anastomosing river system.

The final type of bedform is associated with waveforms in the direction of flow.Depending on the speed of the flow, these are called dunes or anti-dunes. At highvalues of the Froude number (F > 1), anti-dunes occur, and at low values (F < 1)dunes occur. A related feature is the ripple, which also occurs at low Froude number.Ripples are distinguished from dunes by their much smaller scale. Indeed, ripplesand dunes often co-exist, with ripples forming on the larger dunes. The rest of thischapter focusses on models to describe the formation and evolution of dunes.

5.2 Dunes

Dunes are perhaps best known as the sand dunes of wind-blown deserts. They occurin a variety of shapes, which reflect differences in prevailing wind directions. Wherewind is largely unidirectional, transverse dunes form. These are ridges which form atright angles to the prevailing wind. They have a relatively shallow upslope, a sharpcrest, and a steep downslope which is at the limiting angle of friction for slip. Theair flow over the dune separates at the crest, forming a separation bubble behind thedune. Transverse dunes move at speeds of metres per year in the wind direction.

Linear dunes, or seifs, form parallel to the mean prevailing wind, but are due totwo different prevailing wind directions, which alternatively blow from one or otherside of the dune. Such dunes propagate forward, often in a snakelike manner.

Other types of dunes are the very large star dunes (which resemble starfish), whichform when winds can blow from any direction, and the crescentic barchan dunes,which occur when there is a limited supply of erodible fine sand. They take the shapeof a crab-like crescent, with the arms pointing in the wind direction. Barchan duneshave been observed on Mars. (Indeed, it is easier to find images of dunes on Marsthan on Earth.) Figure 5.4 shows images of the four principal types of dune describedabove.

As already mentioned, dunes also occur extensively in river flow. At very low flowrates, ripples form on the bed, and as the flow rate increases, these are replaced by thelonger wavelength and larger amplitude dunes. These are regular scarped features,whose steep face points downstream, and which migrate slowly downstream. Theyform when the Froude number F < 1 (the lower regime), and are associated withriver surface perturbations which are out of phase, and of smaller amplitude. Thewavelength of dunes is typically comparable to the river depth, the amplitude issomewhat smaller than the depth.

When the Froude number increases further, the plane bed re-forms at F 1, andthen for F > 1, we obtain the upper regime, wherein anti-dunes occur. Whereasdunes are analogous to shock waves, anti-dunes are typically sinusoidal, and are inphase with the surface perturbations, which can be quite large. They may traveleither upstream or (more rarely) downstream. Indeed, for the more rapid flows,backward breaking shocks occur at the surface, and chute and pool sequences form.

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Figure 5.4: Illustrations of four of the most common types of aeolian dunes:transverse (top left), seif (top right), barchan (bottom left), star (bottomright). The satellite view of transverse dunes is in the Namibian desert; source:http://earthasart.gsfc.nasa.gov/images/namib hires.jpg. The seif dunesare from the Grand Erg Oriental, in the Sahara Desert in Algeria. Image fromhttp://www.eosnap.com/public/media/2009/06/algeria/20090614-algeria-full.jpg, courtesy of Chelys. The barchanoid dunes are on Mars, so-calleddark dunes in Herschel Crater. Image courtesy of NASA/JPL/University of Ari-zona, available at http://hirise.lpl.arizona.edu/PSP 002860 1650. Finallythe image of star dunes is from the Rub al Khali, or Empty Quarter, desert atthe borders of Saudi Arabia, Oman, Yemen, and the United Arab Emirates. Image athttp://photography.nationalgeographic.com/staticfiles/NGS/Shared/StaticFiles/Photography/Images/Content/empty-quarter-dunes-765958-xl.jpg, re-produced by permission of National Geographic.

Anti-dunes can be found on rapid outlet streams on beaches; for example I have seenthem on beach streams in Normandy and Ireland, where the velocity is on the orderof a metre per second, and the flow depth may be several centimetres. A common

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Figure 5.5: Antidunes on a beach stream at Spanish Point, Co. Clare, Ireland. Thewaves form, migrate slowly upstream (on a time scale determined by slow sedimenttransport), break and collapse. The process then repeats. Image courtesy of RosieFowler.

observed feature of such flows is their time dependence: anti-dunes form, then migrateupstream as they steepen, leading to hydraulic jumps and collapse of the pattern, onlyfor it to re-form elsewhere. An example of such anti-dunes is shown in figure 5.5. Thesuccession of bedforms as the Froude number increases is illustrated in figure 5.6.Anti-dunes do not form in deserts simply because the Froude number is never highenough.1

Dunes and anti-dunes clearly form through the erosion of the underlying bed, andthus mathematical models to explain them must couple the river flow mechanics withthose of sediment transport. Sediment transport models are described below. Thereare two main classes of bedform models. The most simple and appealing is to combinethe St. Venant equations with an equation for bedform erosion. There are two waysin which sediment transport occurs, as bedload or as suspended load. Each transportmechanism gives a different model, and we shall find that a suspended load transportmodel can predict the instability which forms anti-dunes, but not dunes, which indeedmay occur in the absence of suspended sediment transport.2 On the other hand, theSt. Venant equations coupled with a simple model of bedload transport cannot predictinstability, although such a model can explain the shape and speed of dunes.

The other class of model which has been used describes the variation of streamvelocity with depth explicitly. One version employs potential theory, as is customarily

1The Froude number corresponding to a wind of 20 m s1 = 45 miles per hour over a boundarylayer depth of 1 km is 0.2.

2This also seems to be true of anti-dunes.

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ripples

dunes

increasing

chutepool

flat bed

antidunes

F

lower regime upper regimeF < 1 F > 1

Figure 5.6: The succession of bedforms which are observed as the Froude number isincreased. In the lower regime, where F < 1, we see first ripples and then the largerdune features. Surface perturbations are small. In the upper regime, F > 1; dunesdisappear, giving a flat bed, and then anti-dunes are formed, in phase with surfacewaves. These are often transient features, occurring in flood conditions, and they arelikely to be time dependent also.

done in linearised surface wave theory. At first sight, this appears implausible insofaras the flow is turbulent, and indeed the model can then only explain dunes whenthe bed stress is artificially phase shifted. In order to deal with this properly, it isnecessary to include a more sophisticated description of turbulent flow, and this canbe done using an eddy viscosity model, which is then able to explain dune formation.The issue of analysing the model beyond the linear instability regime is more difficult,and some progress in this direction is described in this chapter. In Appendix B , wediscuss the use of an eddy viscosity in simple models of turbulent shear flows.

5.2.1 Sediment transport

Transport of grains of a cohesionless bed occurs as bedload or in suspension. At agiven flow rate, the larger particles will roll along the bed, while the smaller ones arelifted by turbulent eddies into the flow. Clearly there is a transition between the twomodes of transport: saltating grains essentially bounce along the bed.

Relations to describe sediment transport are ultimately empirical, though theorysuggests the use of appropriate dimensionless groups. The basic quantity is the Shields

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1 10 100 1000 Rep

bedload

1

0.1

suspension

Figure 5.7: The critical Shields stress for the onset of sediment transport, weaklydependent on the particle Reynolds number Rep = uDs/.

stress, defined as the dimensionless quantity

=

gDs. (5.1)

Here is the basal shear stress, = sw is the excess density of solid grains overwater (s is the density of the solid grains, w is the density of water), g is gravity,and Ds is the grain size. In general, grain sizes are distributed, and the Shields stressdepends on the particle size. The shear stress at the bed is usually related to themean flow velocity u by the semi-empirical relation (4.9), i. e.,

= fwu2, (5.2)

where f is a dimensionless friction factor, of typical value 0.010.1. (Larger valuescorrespond to rougher channels.)

Shields found that sediment transport occurred if was greater than a criticalvalue c , which itself depends on flow rate via the particle Reynolds number

Rep =uDs

; (5.3)

(The friction velocity is defined to be

u = (/w)1/2.) (5.4)

Figure 5.7 shows the variation of c with uDs/; except at low flow rates, c 0.06.

5.2.2 Bedload

Various recipes have been given for bedload transport, that due to Meyer-Peter andMuller being popular:

q = K[ c ]3/2+ , (5.5)

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where [x]+ = max(x, 0). Here K = 8, c = 0.047, and q is the dimensionless bedload

transport rate, defined by

q =qb

(gD3s/w)1/2

, (5.6)

qb being the bedload measured as volume per unit stream width per unit time.

5.2.3 Suspended sediment

Suspended sediment transport is effected through a balance between an erosion fluxvE and a deposition flux vD, each having units of velocity. The meaning of these isthat svE is the mass of sediment eroded from the bed per unit area per unit time,while svD is the mass deposited per unit area per unit time.

Erosion

It is convenient to define a dimensionless erosion rate E via

vE = vsE, (5.7)

where vs is the particle settling velocity, given by Stokess formula

vs =gD2s18

, (5.8)

being the dynamic viscosity of water. Various expressions forE have been suggested.They share the feature that E is a concave increasing function of basal stress. Typicalis Van Rijns relationship

E ( c )3/2Re1/5p ; (5.9)typical measured values of E are in the range 103 101.

Deposition

The calculation of deposition flux vD is more complicated, as it is analogous to thecalculation of basal shear stress in terms of mean velocity via an eddy viscosity model,as indicated in Appendix B . We can define the dimensionless deposition flux D bywriting

svD = vscD, (5.10)

where c is the mean column concentration of suspended sediment, measured as massper unit volume of liquid, and D depends on a modified Rouse number R = vs/T u.(Here T is related to the eddy viscosity; specifically

1T is the Reynolds number based

on the eddy viscosity (see (B.9)), so the Rouse number is a Reynolds number basedon particle fall velocity and eddy viscosity.) D increases with R, with D(0) = 1, anda typical form for D is

D =R

1 eR (5.11)(see Appendix B for more detail).

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5.3 The potential model

The first model to explain dune formation dates from 1963, and invoked a potentialflow for the fluid, which was assumed inviscid and irrotational. This is somewhat atodds with the fact that it is the basal stress of the fluid which drives sediment trans-port, but one can rationalise this by supposing that the stress is manifested through abasal turbulent boundary layer. We restrict our attention to two dimensional motionin the (x, z) plane: x is distance downstream, z is vertically upwards. The bed is atz = s(x, t), the free water surface is at z = (x, t), so that the depth h is given by

h = s; (5.12)the geometry is shown in figure 5.8. In the potential flow model, the usual equationsfor the fluid flow potential apply:

2 = 0 in s < z < ,z = t + xx on z = ,

t + g +12 ||2 = constant on z = ,

z = st + xsx on z = s. (5.13)

h

z =

x

z

z = s

Figure 5.8: Geometry of the problem.

The extra equation required to describe the evolution of s is the Exner equation:

(1 n)st

+qbx

= 0, (5.14)

where n is the porosity of the bed; this assumes bedload transport only, and we maytake (see equations (5.5) and (5.2)) qb = qb(u), where qb(u) > 0. Implicitly, we suppose

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a (turbulent) boundary layer at the bed, wherein the basal stress develops througha shear layer; the basal shear stress will then depend on the outer flow velocity. Wedefine

q =qb

1 n, (5.15)so that

s

t+

q

x= 0. (5.16)

In the absence of any dynamic effect of the bed shape on the flow, we would expect u,and thus also q, to increase as s increases, due to the constriction of the flow. If indeedq is an increasing function of the local bed elevation s, then it is easy to see from (5.16)that perturbations to the uniform state s = 0 will persist as forward travelling waves,and if q is convex (q(s) > 0) then the waves will break forwards. We interpret slipfaces as the consequent shocks, so that this is consistent with observations. However,such a simple model does not allow for instability.

A simple way in which instability can be induced in the model is by allowingthe maximum stress to occur upstream of the bed elevation maximum, as is indeedindicated by numerical simulations of the flow. One way to do this is to take

q = q(u|x

), (5.17)

that is to say, the horizontal velocity u = x is evaluated at x and z = s, wherethe phase lag is included to model the notion that in shear flow over a boundary,such a lag is indeed present. Of course (5.17) is a crude and possibly dangerous wayto model this effect.

To examine the linear stability of a uniform steady state we write s = 0, = h,

= Ux+ , q = q(U) +Q, = h+ , (5.18)

and then linearise the equations and boundary conditions (which are applied at theunperturbed boundaries z = 0 and z = h) to obtain

2 = 0 in 0 < z < h;z = t + Ux, t + g + Ux = 0 on z = h;

z = st + Usx, st +Qx = 0 on z = 0, (5.19)

whereQ = q(U)x|x,z=0 . (5.20)

For a mode of wavenumber k, we put

(, s, Q) = (, s, Q) eikx+t, (5.21)and write

= eikx+t[A cosh kz +B sinh kz], (5.22)

280

so that the boundary conditions together with (5.20) become

k[A sinh kh+B cosh kh] = ( + ikU),

( + ikU)[A cosh kh+B sinh kh] + g = 0,

kB = ( + ikU)s,

s+ ikQ = 0,

Q = qikeikA. (5.23)

Some straightforward algebra leads to

[( + ikU)2 + gk tanh kh] + ( + ikU)kqeik[( + ikU)2 tanh kh+ gk] = 0, (5.24)

a cubic for (k).Solution of this is facilitated by the observation that we can expect two modes

to correspond to upstream and downstream water wave propagation, while the thirdcorresponding to erosion of the bed may be much smaller, basically if qb is sufficientlysmall. Specifically, let us assume (realistically) that q & hu. Then we may assumeq & h, and for small q, the roots of (5.24) are approximately the (stable) wavemodes

ik U (gktanh kh

)1/2, (5.25)

and the erosive mode

k2Uq[sin k + i cos k] tanh kh

[F 2 coth kh

kh

][F 2 tanh kh

kh

] , (5.26)where we define the Froude number by

F =Ugh. (5.27)

For the erosive mode, the growth rate is

Re = k2Uq sin k tanh kh

[F 2 coth kh

kh

][F 2 tanh kh

kh

] , (5.28)and the wave speed is

Imk

= kUq cos k tanh kh

F 2 coth khkhF 2 tanh kh

kh

. (5.29)281

00.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3

dunes

anti-dunes

-

+

-

kh

F2

Figure 5.9: Instability diagram for the potential flow model. The regions markedwith a minus sign, above the upper curve and below the lower curve, are regions ofinstability if < 0, more specifically if sin k < 0. The marked distinction betweendunes and anti-dunes is based on the surface/bed phase relation (see (5.30). Wavemotion is downstream if cos k > 0, upstream if cos k < 0.

This gives us the typical instability diagram shown in figure 5.9. For < 0 (morespecifically, sin k < 0) the regions above and below the two curves are unstable,

corresponding to dunes and anti-dunes. The curves are given by F 2 =coth kh

khand

F 2 =tanh kh

kh, respectively.

The phase relation between surface and bed for the erosive bed is given by

s F

2sech kh[F 2 tanh kh

kh

] , (5.30)and this defines wave forms below the lower curve in figure 5.9 as dunes, and thoseabove as antidunes.

Figure 5.9 is promising, at least if sin k < 0, as it will predict both dunes andanti-dunes. To get the wave speed positive, we need in fact to have cos k > 0, thus0 > k > pi/2 (we can take pi < k < pi without loss of generality), whereas wewould generally want k < pi/2 for anti-dunes to migrate backwards.

282

There is a serious problem with this model, beyond the fact that the phase shift is arbitrarily included. The spatial delay is unlikely to provide a feasible modelfor nonlinear studies; indeed, we see that Re k2 at large k, and in the unstableregime this is one of the hallmarks of ill-posedness.

Having said that, it will indeed turn out to be the case that a phase lead ( < 0)really is the cause of instability. A phase lead means that the stress, and thus thebedload transport, takes its maximum value on the upstream face of a bump in thebed. A phase lead will occur because of the effect of the bump on the turbulentvelocity structure above, as we discuss further below. It can also occur through aneffect of bedload inertia (see also question 5.7).

The choice of wave speed in this theory is unclear, since cos k can be positive ornegative. The possibly more likely choice of a positive value implies positive wavespeed.

5.4 St. Venant type models

Since river flow is typically modelled by the St. Venant equations, it is natural to tryusing such a model together with a bed erosion equation to examine the possibilityof instability. This has the added advantage of being more naturally designed forfully nonlinear studies. A St. Venant/Exner model can be written in the form (cf.the footnote following (4.46))

st + qx = 0,

ht + (uh)x = 0,

ut + uux = gS fu2

h gx, (5.31)

where S is the downstream slope, q = q(), = fwu2, and s = h. It is convenientto take advantage of the limit q & hu, just as we did before, and we do so by firstnon-dimensionalising the equations. We choose scales as follows:

s, x, h, h0, u u0, q q0, t h20

q0, (5.32)

and we choose h0, u0 by balancing terms as follows: uh Q0, gS fu2/h; here Q0 isthe (prescribed) volume flow per unit width. We choose q0 as the size of the bedloadtransport equation in (5.5).

With these scales, the dimensionless equations corresponding to (5.31) are

st + qx = 0,

ht + (uh)x = 0,

F 2(ut + uux) = x + (1 u

2

h

),

h = s, (5.33)

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where the parameters are

F =u0gh0

, =q0Q0

, = S. (5.34)

If we now suppose & 1 and & 1, both of them realistic assumptions, then wehave approximately

uh = 1,

12F

2u2 + = 12F2 + 1, (5.35)

supposing that u, h 1 at large distances. Eliminating h and , we have

s = 1 1u+ 12F

2(1 u2), (5.36)

whose form is shown in figure 5.10. In particular, s(1) = (1F 2), so the basic stateu = 1 corresponds to the left hand or right hand root of s(u) depending on whetherthe Froude number F < 1 or F > 1.

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

0 0.5 1 1.5 2 2.5 3 3.5 4

s

u

F = 2

F = 0.5

Figure 5.10: s(u) as given by (5.36) for two typical cases of rapid and tranquil flow.

We also haveds

d=F 2 h3F 2

, (5.37)

so that small perturbations to h = 1 are out of phase (dunes) if F < 1 and in phase(anti-dunes) if F > 1. If we take the dimensionless bedload transport as q 3/2 = u3

284

(the dimensionless basal stress having been scaled with fwu20), so that u = q1/3, then

we see from (5.36) that s = s(q), and s(q) has the same shape as s(u), as shown infigure 5.10.

The whole model reduces to the single first order equation

s(q)qt + qx = 0. (5.38)

Disturbances to the uniform state q = 1 will propagate at speed v(q) = 1/s(q),where v is shown in figure 5.11. For F < 1, v(1) > 0 and v(1) > 0, thus waves inq (and thus s) propagate downstream and form forward facing shocks; this is nicelyconsistent with dunes. For F > 1, v < 0 and v(1) is positive if F < 2, negative ifF > 2 (see question 5.4). Backward facing shocks form, these are elevations in s ifv > 0.

-6

-4

-2

0

2

4

6

0 0.5 1 1.5 2

F = 0.5F = 1.5

v

q

Figure 5.11: The wave speed v(q) = 3q4/3/(1 F 2q) for the tranquil and rapid casesF = 0.5 and F = 1.5.

Unfortunately, the hyperbolic equation does not admit instability. It is straight-forward to insert a lag as before, by writing q(x, t) = q[s(x , t)], or equivalentlys(x, t) = s[q(x+ , t)]. Perturbation of

st + qx = 0,

q = q[s(x , t)], (5.39)via

s = seikx+t, q = 1 + qeikx+t, (5.40)

leads tos+ ikq = 0,

285

q = qeiks, (5.41)

and thus = kq[ sin k i cos k]. (5.42)

This requires sin k < 0 for instability if q(s) > 0 (F < 1) and sin k > 0 if q(s) < 0(F > 1). The long wavelength limit of (5.26) in which kh 0 is precisely (5.42),bearing in mind that (5.26) is dimensional and that q = dq/du there, whereas q =dq/ds in (5.42).

5.5 A suspended sediment model

The shortcoming of both the potential model and the St. Venant/Exner model is thelack of a genuine instability mechanism. We now show that the inclusion of suspendedload can produce instability. Ideally, we would hope to predict anti-dunes, since dunescertainly do not require suspended sediment transport. A St. Venant model includingboth bedload and suspended sediment transport is

ht + (uh)x = 0,

ut + uux = g(S x) fu2

h,

t(hc) +

x(hcu) = s(vE vD),

(1 n)st

+qbx

= (vE vD), (5.43)where c is the column average concentration (mass per unit volume) of suspendedsediment (written as c earlier). The distinction between suspended sediment transportand bedload lies in the source terms due to erosion and deposition, vE and vD, andit is these which will enable instability to occur. We have s = h, and we supposeqb = qb(), = fwu2, whence q = q(u). Additionally (see (5.7) and (5.10)), we write

vE = vsE, svD = vscD, (5.44)

and expect that E = E(u) and D = D(u), with E > 0, D < 0; typically E < 1,D > 1.

We scale (5.43) as before in (5.32), except that we choose the time scale t0, down-stream length scale x0, and concentration scale c0 via

c0 = sE0D0

, t0 =(1 n)h0vsE0

, x0 =Q0vsD0

, (5.45)

where we writeE = E0E

(u/u0), D = D0D(u/u0), (5.46)

286

and choose E0 and D0 so that E and D are O(1), and so that these are consistentwith typical observed suspended loads of 10 g l1. With this choice of scales, weobtain the dimensionless set of equations

s = h,ht + (uh)x = 0,

F 2(ut + uux) =

(1 u

2

h

) x,

h(ct + ucx) = E cD,

st + qx = (E cD), (5.47)where the parameters , F, and are now given by

=E0

(1 n)D0 =c0

s(1 n) , =u0S

vsD0,

F =u0

(gh0)1/2, =

qb0D0Q0E0

=sqb0c0Q0

. (5.48)

Here qb0 is the scale for qb rather than q = qb/(1n). The Froude number is the sameas before, but the parameters and are different: is a measure of the suspendedsediment density relative to the bed density, and is always small; is the ratio of the(small) bed slope to the ratio of settling velocity to stream velocity. For more rapidlyflowing streams, we might expect 1. However, if we suppose that wavelengths ofanti-dunes are comparable to the depth (so x0 h0), then (5.45) implies S & 1.Thus 1 implies x0 h0/S * h0. The parameter is a direct measure of theratio of bedload (sqb0) to suspended load (c0Q0). For * 1, we would revert to ourpreceding bedload model and its scaling, and neglect the suspended load. If we adoptthe Meyer-Peter/Muller relation in (5.5) and (5.6), then (noting that fu20 = gSh0)

qb0 =Kl

(ghS)3/2, (5.49)

and we can write

=

{Kl

(1 n)}S3/2

F; (5.50)

both small or large values are possible.To analyse (5.47), we ignore bedload (put = 0) and take 0. Then

= h+ s, uh = 1, (5.51)

so thatcx = E

(u) cD(u) = st,

x

[12F

2u2 +1

u+ s

]= (1 u3). (5.52)

287

If, in addition, & 1, then, taking s = 0 when h = 1,

s = s(u) = 12F2(1 u2) + 1 1

u, (5.53)

and the entire suspended load model is

s(u)u

t= cD(u) E(u) = c

x. (5.54)

The function s(u) is the same as we derived before in (5.36) and shown in figure5.10. We can in fact write (5.54) as a single equation for u, by eliminating c; thisgives

c =E(u)D(u)

+s(u)D(u)

u

t,

s(u)u

t+

x

[E(u)D(u)

+s(u)D(u)

u

t

]= 0, (5.55)

and the equation for u (or the pair for u, c) is of hyperbolic type. Note that naturalinitial-boundary conditions for (5.54) are to prescribe u at t = 0, x > 0, and c atx = 0, t > 0.

Let us examine the stability of the steady state u = 1, c = 1. We put

u = 1 + Re(Ueikx+t

), c = 1 + Re

(Ceikx+t

), (5.56)

and linearise, to obtain (noting E(1) = D(1) = 1)

ikC = [E(1)D(1)]U C = s(1), (5.57)and thus

=

[E(1)D(1)

s(1)

](k2 ik1 + k2

). (5.58)

If we suppose E > 0, D < 0 as previously suggested, then this model implies insta-bility (Re > 0) for s(1) < 0, i.e. F > 1, and that the wave speed is Im ()/k < 0;thus this theory predicts upstream-migrating anti-dunes.

Two features suggest that the model is not well posed if F > 1. The first isthe instability of arbitrarily small wavelength perturbations; the second is that theunstable waves propagate upstream, although the natural boundary condition for cis prescribed at x = 0.

Numerical solutions of (5.54) are consistent with these observations. In solvingthe nonlinear model (5.54) in 0 < x

u 1 as x, t = 0. (5.60)For F < 1, numerical solutions are smooth and approach the stable solution u = c =1. However, the solutions are numerically unstable for F > 1, and u rapidly blowsup, causing breakdown of the solution.

Some further insight into this is gained by consideration of the solution at x = 0.If c = c0(t) on x = 0 and u = u0(x) on t = 0, then we can obtain u on x = 0 from(5.55), by solving the ordinary differential equation

u

t+E(u)s(u)

=D(u)s(u)

c0(t) (5.61)

with u = u0(0) at t = 0. If we suppose that c = 1 at x = 0, then it is easy to showthat if F < 1 and u(0, 0) < 1/F 2/3, then u(0, t) 1 as t. If on the other hand,F > 1 and u(0, 0) < 1, then u(0, t) 1/F 2/3 in finite time, and the solution breaksdown as u/t ; if u(0, 0) > 1, then u(0, t) , again in finite time if, forexample, E u3. More generally, breakdown of the solution when F > 1 occurs inone of these ways at some positive value of x. Thus this suspended sediment modelshares the same weakness of the phase shift model in not appearing to provide a wellposed nonlinear model.

5.6 Eddy viscosity model

The relative failure of the models above to explain dune and anti-dune formation ledto the consideration of a full fluid flow model, in which, rather than supposing thatthe flow is shear free and that viscous effects were confined to a turbulent boundarylayer, rotational effects were considered, and a model of turbulent shear flow incor-porating an eddy viscosity, together with the Exner equation for bedload transport,was adopted. This allows for a linear stability analysis of the uniform flow over a flatbed via the solution of a suitable Orr-Sommerfeld equation. We shall in fact proceedin somewhat more generality. As an observation, fully-formed dunes have relativelysmall height to length ratios, and thus the fluid flow over them can be approximatelylinearised. Although we use a linear approximation to derive the stress at the bed,we may retain the nonlinear Exner equation for example. In this way we may derivea nonlinear evolution equation for bed elevation.

5.6.1 Orr-Sommerfeld equation

Suppose, therefore, that we have two-dimensional turbulent flow down a slope ofgradient S, governed by the Reynolds equations

ut + uux + wuz = 1

p

x+ T2u+ gS,

wt + uwx + wwz = 1

p

z+ T2w g(1 S2)1/2,

289

ux + wz = 0, (5.62)

where (u,w) are the velocity components and T is an eddy viscosity associated withthe Reynolds stress terms, such as prescribed in (B.9). In the second equation, wecan take g(1 S2)1/2 g since S is small.

We consider perturbations to a basic shear flow u(z) in s < z < which satisfies(5.62) with T taken as constant. (Later, we will study a more realistic eddy viscositymodel.) It is convenient first of all to non-dimensionalise the equations (5.62). In thebasic uniform state, with s = 0 and = h, the shear flow satisfies

Tu

z= gS(h z), (5.63)

whence

u =gS

T

(hz 12z2

), (5.64)

and the column mean flow is

u =1

h

h0

u dz =gS

3Th2. (5.65)

Taking T = T uh, we find that the basal shear stress is

= wTu

z

0

= fwu2, (5.66)

where f = 3T . This gives the relationship between the empirical f and the semi-analytic T . If the bed and hence the flow is perturbed, we would only retain constantT if the volume flux per unit width is the same; this we therefore assume.

We now non-dimensionalise the variables by writing

(u,w) u, (x, z) h, t h/u, p g(h z) wu2. (5.67)The dimensionless equations are

ut + uux + wuz = px + 1R2u+ S

F 2,

wt + uwx + wwz = pz + 1R2w,

ux + wz = 0, (5.68)

and the parameters are a turbulent Reynolds number and the Froude number:

R =uh

T, F =

ugh. (5.69)

The dimensionless basic velocity profile is then

u =gSh2

T u

(z 12z2

), (5.70)

290

and the dimensionless mean velocity is, by definition of u,

1 =gSh2

3T u. (5.71)

SinceT = T uh =

13fuh, (5.72)

this requires

u =

(gSh

f

)1/2. (5.73)

In particular, the dimensionless basic velocity profile is

u = U(z) = 3(z 12z2

). (5.74)

We now suppose that s and are perturbed by small amounts; we may thuslinearise (5.68). We put

(u,w) = (U(z) + z,x), (5.75)whence it follows for small that satisfies the steady state Orr-Sommerfeld equation

U2x U x = R14, (5.76)where we assume stationary solutions in view of the anticipated fact that s evolveson a slower time scale.

The condition of zero pressure at z = is linearised to be

= 1 + F 2p |z=1. (5.77)If F 2 is small, then we may take to be constant, and we do so as we are primarilyinterested in dunes. However, the dimensionless pressure p is only determined up toaddition of an arbitrary constant, which implies that the value of the constant isunconstrained. This represents the vertical translation invariance of the system. If auniform perturbation to s is made, then the response of the (uniform) stream is toraise the surface by the same amount. We can remove the ambiguity by prescribing = 1, with the implication that the mean value of s is required to be zero.

The other boundary conditions on z = s and z = 1 are no slip at the base, noshear stress at the top, and the perturbed volume flux is zero. These imply

= 0, zz = 0 on z = 1, s0

U(z) dz + = 0, U + z = 0 on z = s. (5.78)

Linearisation of this second pair about z = 0 gives

= 0, z = U 0s on z = 0, (5.79)

291

where U 0 = U(0). Our aim is now to solve (5.76) with (5.78) and (5.79) to calculate

the perturbed shear stress. The dimensional basal shear stress is then

= wTu20U

0

[1 + s

U 0U 0

+1

U 0zz|0

], (5.80)

and since f = 3T = TU 0, we may write this as

= fwu2

[1 +

sU 0U 0

+1

U 0zz|0

]. (5.81)

The problem to solve for is linear and inhomogeneous, and so we suppose that

s =

s(k)eikx dk, =

(k)eikx dk. (5.82)

(Note s will evolve slowly in time.) For each wave number k, we obtain

ik[U( k2) U

]=

1

R

[iv 2k2 + k4

], (5.83)

with boundary conditions = = 0 on z = 1,

= 0, = U 0s on z = 0, (5.84)and thus we finally define

= U 0s(z, k), (5.85)where satisfies the canonical problem

ik[U( k2) U ] = 1

R

[iv 2k2 + k4] ,

= = 0 on z = 1,

= 0, = 1 on z = 0. (5.86)

In terms of , the basal (dimensional) shear stress is

= fwu2

[1 s

eikxs(k)(0, k) dk]. (5.87)

Using the convolution theorem, this is

= fwu2

[1 s+

K(x )s() d], (5.88)

where s = s/x, and

K(x) = 12pi

(0, k)ik

eikx dk. (5.89)

292

Depending on K, we can see how may depend on displaced values of s. The formof (5.88) illustrates our previous discussion of the vertical translation invariance ofthe system. For a possible uniform perturbation s = constant, we would obtain amodification to the basic friction law, = fwu2. This is excluded by enforcing thecondition that s has zero mean in x,

limL

1

2L

LL

s(x) dx = 0, (5.90)

which corresponds (for a periodic bed) to prescribing

s(0) = 0. (5.91)

To determine K, we need to know the solution of (5.86) for all k. In general, theproblem requires numerical solution. However, note that R = 1/T , and is reasonablylarge (for a value f = 0.005, R = 3/f = 600). This suggests that a useful meansof solving (5.86) may be asymptotically, in the limit of large R. The fact that wecan obtain analytic expressions for (0, k) means this is useful even when R is notdramatically large, as here.

The solution of the Orr-Sommerfeld equation at large R has a long pedigree, andit is a complicated but mathematically interesting problem. We devote Appendix Cto finding the solution. We find there that, for k > 0,

(0, k) 3(ikRU 0)1/3Ai(0) +O(1), (5.92)where Ai is the Airy function. For k < 0, (0, k) = (0,k), and this leads to

(0, k)ik

{ ceipi/3|k|2/3, k > 0ceipi/3|k|2/3, k < 0, (5.93)

wherec = 3(RU 0)

1/3Ai(0), (5.94)

and c 1.54R1/3 for U 0 = 3, as Ai(0) =1

32/3(23) 0.355. From (5.89), we find

K(x) =c

pi

0

cos[kx pi3 ] dkk2/3

. (5.95)

Evaluating the integral,3 we obtain the simple formula

K(x) =

x1/3, x > 0,

K(x) = 0, x < 0, (5.96)

3How do we do that? The blunt approach is to consult Gradshteyn and Ryzhik (1980), where therelevant formulae are on page 420 and 421 (items 4 and 9 of section 3.761). The quicker way, usingcomplex analysis, is to evaluate

0

1ei d (after a simple rescaling of k, k|x| = ) by rotatingthe contour by pi/2 and using Jordans lemma. Thus

0

1ei d = ()eipi/2.

293

where

=32/3R1/3

{(23)}2 1.13R1/3. (5.97)

For stability purposes, note that

K =

K(k) eikxdk, (5.98)

where

K = (0, k)2piik

=

c exp

[ipi

3sgn k

]2pi|k|2/3 . (5.99)

5.6.2 Orr-Sommerfeld-Exner model

We now reconsider (5.33), which we can write in the form

st + qx = 0,

ht + (uh)x = 0,

F 2(ut + uux) = x + (1

h

),

h = s. (5.100)Here is the local basal stress, scaled with fwu20. We suppose q = q(), so that theExner equation is

s

t+ q()

x= 0. (5.101)

It is tempting to suppose that, writing u = u0u,

= u2[1 s+

K(x )s

(, t) d

]. (5.102)

We would then have, with & 1 and = 1, u 11 s in (5.102). There is a subtle

point here concerning the modified stress. Insofar as we may wish to describe differentatmospheric or fluvial conditions (e. g., the difference between strong and weak windsat different times of day, or rivers in normal or flood stage), we do want to allowdifferent choices of u. However, such conditions also imply different values of h, andthe basis of the solution for the perturbed stress is that the mean depth (and thus

the mean velocity) do not vary. The value of u =1

1 s is a local column average,whereas the u in (5.102) is in addition a horizontal average. Thus, given u0 = u andh0 = h, we define

1 s+

K(x )s

(, t) d, (5.103)

294

and the model consists of the Exner equation (5.101) and the Orr-Sommerfeld stressformula (5.103). Variable u is simply manifested in differing time scales for the Exnerequation.

We linearise by writing = 1 + T , and then

s =

s(k, t)eikx dk, T =

T (k, t)eikx dk, (5.104)

so thatst + ikq

(1)T = 0,

T = s+ 2piKiks, (5.105)and thus, using (5.99), solutions are proportional to et, where

= q(1)[2pik2K + ik ]. (5.106)

When Re K > 0, as for (5.99), the steady state is unstable, with Re k4/3 ask . Specifically, the growth rate is

Re = 12q(1)c|k|4/3, (5.107)

while the wave speed is

Imk

= q(1)(12

3 c|k|1/3 1

); (5.108)

thus waves move downstream (except for very long waves).

5.6.3 Well-posedness

The effect of (5.103) is to cause increased where sx is positive, on the upstreamslopes of bumps. Since u is in phase with s, this implies leads u (i.e., is amaximum before s is); it is this phase lead which causes instability. However, theunbounded growth rate at large wave numbers is a sign of ill-posedness. Without somestabilising mechanism, arbitrarily small disturbances can grow arbitrarily rapidly. Inreality, another effect of bed slope is important, and that is the fact that sedimentwants to roll downslope: in describing the Meyer-Peter/Muller result, no attentionwas paid to the variations of bed slope itself.

For a particle of diameter Ds at the bed, the streamflow exerts a force of approx-imately D2s on it, and it is this force which causes motion. On a slope, there isan additional force due to gravity, approximately gD3ssx. Thus the net stresscausing motion is actually

gDssx. (5.109)In dimensionless terms, we therefore modify the bedload transport formula by writing

q = q(e), e = sx, (5.110)

295

sx

t

Figure 5.12: Development of the dune instability from an initial perturbation near

x = 0 obtained by solving (5.116) using q() = 3/2, K =

x1/3when x > 0, K = 0

otherwise, with parameters = 9.57 and D = 4.3. Separation occurs in this figurewhen t = 0.8, after which the computation is continued as described in the notes atthe end of the chapter. Figure kindly provided by Mark McGuinness.

where

=Dswh0S

. (5.111)

Typical values in water are /w 2, Ds 103 m, h0 2 m, S 103, whence 4; generally we will suppose that O(1).

The effect of this is to replace the definition of in (5.103) by

e = 1 s+

K(x )s

(, t) d sx, (5.112)

(together with (5.101)) and in the stability analysis, T = s[1+2piikKik], whence

= q(1)[ik

{12

3 c|k|1/3 1

}+ 12c|k|4/3 k2

]. (5.113)

This exhibits the classical behaviour of a well-posed model. The system is stable at

high wavenumber, and the maximum growth rate is at k =

(c

3

)3/2. This would be

the expected preferred wavenumber of the instability.Figure 5.12 shows a numerical solution of the nonlinear Exner equation, showing

the growth of dunes from an initially localised disturbance. Because the expressionin (5.112) is only valid for small s, we can equivalently write

q(e) = q( sx) q()Dsx, (5.114)

296

whereD = q() q(1), (5.115)

and the equation has been solved in this form, with the diffusion coefficient D takenas constant, i. e., s satisfies the equation

s

t+

xq [1 s+K sx] = D

2s

x2. (5.116)

As the dunes grow, the model becomes invalid when 1 O(1), and this happenswhen s 1

. This is a representative value for the elevation of both fluvial and aeolian

dunes, and is suggestive of the idea that it is the approach of towards zero whichcontrols eventual dune height. Additionally, when reaches zero, separation occurs,and the model becomes invalid. Possible ways for dealing with this are outlined inthe notes at the end of the chapter. A further issue is that the derivation of (5.112)

becomes invalid when s 1, because then the thickness of the viscous boundary

layer in the OrrSommerfeld equation becomes comparable to the elevation of thedunes. This implies that the OrrSommerfeld equation should now be solved in adomain where the lower boundary can not be linearised about z = 0, and the Fouriermethod of solution can no longer be implemented. It is not clear whether this willfundamentally change the nature of the resultant formula for the stress.

The numerical method used to solve (5.116) is a spectral method. Spectral meth-ods for evolution equations of this sort are convenient, particularly when the integralterm is of convolution type, but they confuse the issue of what appropriate boundaryconditions for such equations should be. In the present case, it is not clear. For ae-olian dunes, it is natural to pose conditions at a boundary representing a shore-line,but it is then less clear how to deal with the integral term. The derivation of this termalready presumes an infinite sand domain, and it seems this is one of those questionsakin to the issue of posing boundary conditions for averaged equations, for examplefor two-phase flow, where a hidden interchange of limits is occurring.

5.7 Mixing-length model for aeolian dunes

Measurements of turbulent fluid flow in pipes, as well as air flow in the atmosphere(and also in wind tunnels), show that the assumption of constant eddy viscosity isnot a good one, and the basic shear velocity profile is not as simple as assumed in thepreceding section. In actual fact, the concept of eddy viscosity introduced by Prandtlwas based on the idea of momentum transport by eddies of different sizes, with thetransport rate (eddy viscosity) being proportional to eddy size. Evidently, this mustgo to zero at a solid boundary, and the simplest description of this is Prandtls mixinglength theory, described in appendix B . In this section, we generalise the previousapproach a little to allow for such a spatially varying eddy viscosity, and we specificallyconsider the case of aeolian dunes, in which a kilometre deep turbulent boundary layerflow is driven by an atmospheric shear flow.

297

5.7.1 Mixing-length theory

The various forms of sand dunes in deserts were discussed earlier; the variety of shapescan be ascribed to varying wind directions, a feature generally absent in rivers. An-other difference from the modelling point of view is that the fluid atmosphere is aboutten kilometres in depth, and the flow in this is essentially unaffected by the under-lying surface, except in the atmospheric boundary layer, of depth about a kilometre,wherein most of the turbulent mixing takes place. Within this boundary layer, thereis a region adjoining the surface in which the velocity profile is approximately loga-rithmic, and this region spans a range of height from about forty metres above thesurface to the roughness height of just a few centimetres or millimetres above thesurface.

Consider the case of a uni-directional mean shear flow u(z) past a rough surfacez = 0, where z measures distance away from the surface. If the shear stress is constant,equal to , then we define the friction velocity u by

u = (/)1/2, (5.117)

where is density. Observations support the existence near the surface of a logarith-mic velocity profile of the form

u =uln

(z

z0

), (5.118)

where the Von Karman constant 0.4, and z0 is known as the roughness length: itrepresents the effect of surface roughness in bringing the average velocity to zero atsome small height above the actual surface.4 Since z0 is a measure of actual roughness,a typical value for a sandy surface might be z0 = 103 m.

Prandtls mixing length theory provides a motivation for (5.118). If we supposethe motion can be represented by an eddy viscosity , so that

= u

z, (5.119)

then Prandtl proposed

= l2uz

, l = z, (5.120)from which, indeed, (5.118) follows. The quantity l = z is called the mixing length.Prandtls theory works well in explaining the logarithmic layer, and in extensionit explains pipe flow characteristics very well; but it has certain drawbacks. Thetwo obvious ones are that it is not frame-invariant; however, this would be easilyrectified by replacing |u/z| by the second invariant 2, where 22 = ij ij, and ijis the strain rate tensor. Also not satisfactory is the rather loosely defined mixinglength, which becomes less appropriate far from the boundary, or in a closed container.

4A better recipe would be u =u

ln(z + z0z0

), which allows no slip at z = 0. See also question

5.11.

298

Despite such misgivings, we will use a version of the mixing length theory to see howit deviates from the constant eddy viscosity assumption.

We want to see how to solve a shear flow problem in dimensionless form. Tothis end, suppose for the moment that we fix u = U on z = d. Then U =(u/) ln(d/z0) determines u (and thus ), and we can define a parameter5 by

=uU

=

ln(d/z0). (5.121)

For d = 103 m, z0 = 103 m, = 0.4, 0.03. Writing u in terms of U rather thanu yields

u = U[1 +

ln(zd

)]. (5.122)

Note also that the basic eddy viscosity is then

= Ud(zd

), (5.123)

and the shear stress is = 2U2. (5.124)

We shall use these observations in scaling the equations. For the atmospheric bound-ary layer, it seems appropriate to assume that U is prescribed from the large scalemodel of atmospheric flow (cf. chapter 3), and that d is the depth of the planetaryboundary layer. Of course, this may be an oversimplified description.

5.7.2 Turbulent flow model

Again we assume a mean two-dimensional flow (u, 0, w) with horizontal coordinatex and vertical coordinate z over a surface topography given by z = s. The basicequations are

ux + wz = 0,

(uux + wuz) = px + 1x + 3z,(uwx + wwz) = pz + 3x 1z, (5.125)

where 1 = 11 and 3 = 13 are the deviatoric Reynolds stresses, and are defined, wesuppose, by

1 = 2ux,

3 = (uz + wx). (5.126)

We ignore gravity here, so that the pressure is really the deviation from the hydrostaticpressure. Our choice of the eddy viscosity will be motivated by the Prandtl mixinglength theory (5.120), but we postpone a precise specification for the moment.

5Note that this definition of is unrelated to its previous definition and use, as for example in(5.100).

299

The basic flow then dictates how we should non-dimensionalise the variables. Wedo so by writing

u = U(1 + u), w U, x, z d,1, 3 2U2, dU, p U2, (5.127)

and then the dimensionless equations are (dropping the asterisk on u)

ux + wz = 0,

ux + px = [1x + 3z {uux + wuz}],wx + pz = [3x 1z {uwx + wwz}],

3 = (uz + wz),

1 = 2ux. (5.128)

5.7.3 Boundary conditions

The depth scale of the flow d is, we suppose, the depth of the atmospheric boundarylayer, of the order of hundreds of metres to a kilometre. Above the boundary layer,there is an atmospheric shear flow, and we suppose that u u0(z), w 0, p 0 asz .6 The choice of u0 is determined for us by the choice of , as is most easilyseen from the case of a uniform flow where u/z = /. The correct boundarycondition to pose at large z is to prescribe the shear stress delivered by the mainatmospheric flow, and this can be taken to be 3 = 1 by our choice of stress scale.Thus we prescribe

3 1, w 0, p 0 as z . (5.129)Next we need to prescribe conditions at the surface. This involves two further

length scales, the length L and amplitude H of the surface topography. Since weobserve dunes often to have lengths in the range 1001000m, and heights in therange 2100m, we can see that there are two obvious distinguished limits, L = d,H = d, and it is most natural to use these in scaling the surface s. In fact sincedunes are self-evolving it seems most likely that they will select length scales alreadypresent in the system. Thus, we suppose that in dimensionless terms the surface isz = s(x), and longer, shorter, taller or smaller dunes can always be introduced asnecessary later, by rescaling s. The surface boundary conditions are then taken to be(recalling the definition of the roughness length)

u = 1, w = 0 on z = s+ z0 , (5.130)

wherez0 =

z0d

= e/. (5.131)

For completeness, we need to specify horizontal boundary conditions, for exampleat x = . We keep these fairly vague, beyond requiring that the variables remainbounded. In particular, we do not allow unbounded growth of velocity or pressure.

6The modelling alternative is to specify velocity conditions on a lid at z = 1.

300

5.7.4 Eddy viscosity

Prandtls mixing length theory in scaled units would imply

= 2(z s)2uz

, (5.132)and we assume this, although other choices are possible. In particular, (5.132) is notframe indifferent, but this is hardly of significance since the eddy viscosity itself isunreliable away from the surface. (We comment further on this in the notes at theend of the chapter.)

To convert to the constant eddy viscosity model of the preceding section, equation(5.68), we would rescale u,w, p 1/, and choose = : thus 2 = 1/R.

5.7.5 Surface roughness layer

The basic shear flow near a flat surface z = 0 is given by (5.122), and in dimensionlessterms is

u =1

ln z; (5.133)

we will require similar behaviour when the flow is perturbed. Suppose, more generally,that as z s,

u a+ b ln(z s) +O(z s), (5.134)which we shall find describes the solution away from the boundary. We put

z = s+ Z, (5.135)

where = e/. (5.136)

Additionally, we write

u = 1+ U, w = sxU + W, 1 = T1, = N. (5.137)

Then we find thatUx +WZ = 0,

3Z

2sxT1Z

+ sxp

Z 0,

p

Z 2

[sx3Z

+T1Z

],

N 2Z2UZ

,

3 N(1 2s2x)U

Z,

T1 = 22sxZ2(U

Z

)2, (5.138)

301

where we have neglected transcendentally small terms proportional to .Correct to O(2), 3 is constant through the roughness layer, and equal to its

surface value , and

2Z2(U

Z

)2, (5.139)

again correct to O(2). The boundary conditions on Z = 1 (i.e., z s = z0 = ) areU = W = 0, thus

U =

lnZ, (5.140)

and this must be matched to the outer solution (5.134). Rewriting (5.140) in termsof u and z, we have

u 1

+

ln(z s), (5.141)

and this is in fact the matching condition that we require from the outer solution.We see immediately that variations of O(1) in u yield small corrections of O() in .

Solving for W , we have

W = ()

[Z lnZ Z], (5.142)

where () =

/x, and in terms of w and z, this is written

w = sx + sxu ()

[ln(z s) 1 +

](z s). (5.143)

Hence the outer solution must satisfy (correct to O(2))

w sx + sxu as z s. (5.144)

5.7.6 Outer solution

We turn now to the solution away from the roughness layer, in the presence of surfacetopography of amplitude O() and length scale O(1). The topography has two effects.The O(1) variation in length scale causes a perturbation on a height scale of O(1),but the vertical displacement of the logarithmic layer by O() causes a shear layerof this thickness to occur. Thus the flow away from the surface consists of an outerlayer of thickness O(1), and an inner shear layer of thickness O(). We begin withthe outer layer.

We expand the variables as

u = u(0) + u(1) + , (5.145)etc., so that to leading order, from (5.128),

u(0)x + w(0)z = 0,

u(0)x + p(0)x = 0,

w(0)x + p(0)z = 0. (5.146)

302

Notice that, at this leading order, the precise form of in (5.132) is irrelevant, as thisouter problem is inviscid. We have

u(0) + p(0) = u0(z), (5.147)

andp(0)x = w

(0)z , p

(0)z = w(0)x , (5.148)

which are the Cauchy-Riemann equations for p(0)+iw(0), which is therefore an analyticfunction, and p(0) and w(0) both satisfy Laplaces equation. The matching conditionsas z s can be linearised about z = 0, and if w(0) = w0 and p(0) = p0 on z = s,then from (5.144) we have

w(0) = sx on z = 0. (5.149)

Assuming also that w(0), p(0) 0 as z , we can write the solutions in the form

w(0) =1

pi

zs d

[(x )2 + z2] , p(0) = 1

pi

(x )s d[(x )2 + z2] , (5.150)

and in particular, p(0) on z = s is given to leading order by p0, where

p0 =1

pi

s d

x = H(sx); (5.151)

the integral takes the principal value, and H denotes the Hilbert transform.The shear velocity profile u0(z) is undetermined at this stage, although we would

like it to be the basic shear flow profile; but to justify this, we need to go to the O()terms. At O(), we have

u(1)x + w(1)z = 0,

u(1)x + p(1)x =

(0)1x +

(0)3z {u(0)u(0)x + w(0)u(0)z },

w(1)x + p(1)z =

(0)3x (0)1z {u(0)w(0)x + w(0)w(0)z },

(0)3 = (0)[u(0)z + w

(0)x ],

(0)1 = 2(0)u(0)x ,

(0) = 2z2u(0)z

. (5.152)We can use the zero-th order solution to write (5.152)2 in the form

u(1)x + p(1)x =

(0)3z

+

x

[ (0)1 12(u(0)2 + w(0)2) + u0(z)(0)

], (5.153)

where (0) is the stream function such that w(0) = (0)x , and specifically, we have

(0) = 12pi

ln[(x )2 + z2]p0() d, (5.154)

303

which can be found (as can the formulae in (5.150)) by using a suitable Greensfunction; (this is explained further below when we find p(1)).

On integrating (5.153), we have to avoid secular terms which grow linearly in x,and we therefore require the integral of the right hand side of (5.153) with respect tox, from to , to be bounded. The integral of the derivative term is certainlybounded; thus the secularity condition requires

(0)3z dx to be bounded, and it is

this condition that determines the function of integration u0(z) in (5.147).For the particular choice of (0) in (5.152), we have

(0) = 2z2(u0 + w

(0)x

)(5.155)

(assumed positive), so that

(0)3 = 2z2

(u0 + w

(0)x

) (u0 + 2w

(0)x

). (5.156)

The condition that (0)3 /z have zero mean is then

z[2z2(u 20 + 2w

(0)2x )] dx = 0, (5.157)

and thus (0)3 /z = 0, where the overbar denotes the horizontal mean. Thus (with

(0)3 = 1 from the condition at z =), u0 is determined via

u 20 + 2w(0)2x =

1

2z2. (5.158)

The non-zero quantity 2w(0)2x represents the form drag due to the surface topography.Note that the logarithmic behaviour of u0 near z = 0 is unaffected by this extra term,and we can take

u0 =1

ln z +O(z2) as z 0. (5.159)

In particular, since p(0) p0+p(0)z |s(zs) as z s, and p(0)z |s = w(0)x |s = sxx,we have

u(0) p0 + 1ln z + sxx(z s) +O(z2) as z s. (5.160)

From (5.156), we now have

(0)3 = 1 + 32z2u0w

(0)x +

x, (5.161)

where we define

=

x

22z2{w(0)2x w(0)2x

}dx. (5.162)

Hence from (5.153),

u(1)+p(1) =

z

[32z2u0w

(0) + ]+ (0)1 12

(u(0)2 + w(0)2

)+u0(z)

(0)+u1(z), (5.163)

304

where u1 must be determined at O(2).

Now u0 1ln z + O(z2), = O(z2), (0)1 = O(z), w

(0) = sx + O(z), (0) =

s zp0 + O(z2) (this last follows from manipulation of (5.154)). Therefore, asz 0,

u(1) = p10+3sx 12[s2x +

{p0 + 1

ln z

}2]+

1

z(s zp0)+u1+O(z), (5.164)

where p10 = p(1)|z=0.

5.7.7 Determination of p10

Define the Greens function

K(x, z; , ) = 14pi

[ln{(x )2 + (z )2}+ ln{(x )2 + (z + )2}] . (5.165)

We then have, for example,

p(0) =

>0

[K2p(0) p(0)2K] d d

=

(Kp(0)

n p(0)K

n

)ds

= 0

Kp(0)

d =

0

Kw(0)

d =

0

w(0)K

d, (5.166)

whence we derive (5.150) for example; the integrals with respect to are taken along = 0.

Next, expanding (5.144) about z = 0, we find

w(1) (su(0))x as z 0. (5.167)Putting

w(1) = sxu0 +W, (5.168)

we deduce the conditionW = (sp0)x on z = 0, (5.169)

and from (5.152)

p(1)x Wz = R,p(1)z +Wx = S,

where

R = (0)1x + (0)3z {u(0)u(0)x + w(0)u(0)z sxu0},

S = (0)3x (0)1z {u(0)w(0)x + w(0)w(0)z + sxxu0}.

305

Also2p(1) = Rx + Sz, (5.170)

and it follows from using the Greens function as in (5.166) that, after some manip-ulation involving Greens theorem,

p(1) =

>0

(RK + SK) d d 0

KW d, (5.171)

and therefore

p10 =1

pi

>0

[(x )R(, ) S(, )](x )2 + 2 d d

1

pi

(sp0) d

x . (5.172)

5.7.8 Matching

Overall, then, the outer solution can be written, as z 0, in the form (using (5.160))

u p0 + 1ln z + sxx(z s) +

[p10 + 3sx 12s2x 12p20 +

p0ln z

122

ln2 z sz p0

+ u1

]. (5.173)

If we define = 1 + A1 +

2A2 + . . . , (5.174)

then (5.141) takes the form

u A1 + 1ln z +

[ sz

+A1

ln z + A2

]+ . . . , (5.175)

and the leading order term can be matched directly to that of (5.173) by choosing

A1 = p0. (5.176)Using (5.176), (5.174) and (5.151), we have

1 + 2pi

s d

x , (5.177)

and this can be compared with (5.103). Whereas in (5.103) K(x) = 0 for x < 0, thekernel K(x) in (5.177) is proportional to 1/x for all x, and thus non-zero for x < 0.Consequently, there is no instability, and to find an instability we need to progress tothe next order term.

Unfortunately, the O() terms do not match because the terms p0ln z in the

two expansions are not equal, and also because of the linear term in (5.173). In orderto match the expansions to O(), we have to consider a further, intermediate layer:this is the shear layer we alluded to earlier.

306

5.7.9 Shear layer

A distinguished limit exists when z = O(), and thus we put

z = s+ , w = sx + [usx +W ],

= N, 1 = T1,

u = p0 + 1ln(z s) + v, (5.178)

and from (5.173) and (5.141) (using (5.174) and (5.176)), we require

v sxx p10 + 3sx 12s2x 12p20 p0

+p0ln +

(u1 1

22ln2

)as ,

v A2 p0ln as 0. (5.179)

It follows from (5.178) that

N = 22u

,

T1 = 2N [ux sxu ],3 = N [u + sxx +O(

2)],

ux +W = 0,

(u+ p)x sxp = 3 [uux +Wu ] +O(2),p = sxx +O(2). (5.180)

Since we have p = p0 + p10 and W = 0 on = 0, then

p = p0 + (p10 sxx) +O(2),W = p0 +O(), (5.181)

and thus v satisfies

vx + p10 sxxx + sxsxx =

[2v + sxx]

[p0

(p0 + 1

ln

)+p0

]+O(),

(5.182)together with (5.179).

The solution of (5.182) is

v p10 12s2x p0 12p20 +

p0ln + sxx + 3sx + V, (5.183)

whereV

x=

[2

V

], (5.184)

and (5.179) impliesV 0 as ,

307

V A2 2p0

ln as 0, (5.185)where

A2 = A2 p10 12s2x

p0 12p20 + 3sx. (5.186)

The solution of (5.184) which tends to zero as is

V =

V (, k)eikx dk, (5.187)

where the Fourier transform V (as thus defined) is given by

V = BK0

[(2ik

)1/2], (5.188)

the square root is chosen so that Re (ik)1/2 > 0,7 and K0 is a modified Bessel functionof order zero. Evidently we require

V A2 2p0

ln() as 0, (5.189)

where the overhat defines the Fourier transform, in analogy to (5.187). Now K0() ln 12 as 0, where 0.5772 is the Euler-Mascheroni constant. Also(

2ik

)1/2=

(2|k|

)1/2exp

[ipi

4sgn k

]; (5.190)

therefore (5.188) implies

V B[ + 12 ln |k| 12 ln 2+ 12 ln +

ipi

4

k

|k|], (5.191)

and matching this to (5.189) implies

B =4p0, (5.192)

whence

A2 =2p0

ln 4p0

[ + 12 ln |k| 12 ln 2+

ipik

4|k|]. (5.193)

We have sx = iks, H(sx) = |k|s, and J sx = |k|s ln |k|, where J sx is theconvolution of J with sx, and J = (i/2pi) ln |k| sgn k. (The convolution theoremhere takes the form f g = 2pif g.) It follows from this that

J(x) = 1pix

[ + ln |x|] . (5.194)7Assuming the principal branch of the square root, this implies we take k = |k|eipi when k is

negative.

308

Thus

A2 =2

(ln 2 2) p0 + pi

sx +

1

piJ sx, (5.195)

and, from (5.186),

A2 =2

(ln 2 2 12

)p0 +

(pi+ 3

)sx

+1

piJ sx p10 12p20 12s2x, (5.196)

where J is given by (5.194), p0 = H(sx) ((5.151)), and p10 is given by (5.172).We can summarise our calculation of the basal shear stress as follows. From

(5.174), (5.176) and (5.151) we have

= 1 + B1 + 2B2 + . . . , (5.197)

whereB1 = 2A1 = 2H (sx) , B2 = 2A2 + A21. (5.198)

Using (5.186) and (5.193), we find after a little algebra that the transform of B2 is

B2 =B1

[2 ln 2+ 2 ln |k|+ ipisgn k + 4 + 1] + C, (5.199)

where C is the transform of

C = 2p10 s2x + 6sx. (5.200)

5.7.10 Linear stability

The Exner equation is, in appropriate dimensionless form,8

st + qx = 0, (5.201)

and since q = q(),

q = q1 2q1p0 + 2[(2A2 + p

20)q

1 + 2p

20q1

]+ . . . , (5.202)

where q1 = q(1), q1 = q(1), q1 = q

(1).Thus s satisfies the nonlinear evolution equation

s

t 2q1

p0x

+

x

[(2A2 + p

20)q

1 + 2p

20q1

] 0. (5.203)This is

s

t p0

x+

x

[q1(2sx + 2J sx 2p10 s2x

)+ 2q1p

20

]= 0,

8Note that the definition of here is that pertaining to the mixing length theory, i. e., (5.121)and not (5.48).

309

p0 = H(sx), (5.204)

where

= 2q1

[1 2

(ln 2 2 12

)],

=pi

+ 3,

=1

pi. (5.205)

We linearise (5.204) for small s by neglecting the terms in s2x and p20. Taking the

Fourier transform (as defined here in (5.187)), we have

st = ikp0 ikq1(2iks+ 4piikJ s 2p10

). (5.206)

From (5.172),

p10 =

0

(a R + b S) d H{(sp0)x}, (5.207)where

a(x, ) =x

pi(x2 + 2), b(x, ) =

pi(x2 + 2). (5.208)

Hence, neglecting the quadratic Hilbert transform term,

p10 = 2pi

0

(aR + bS)d. (5.209)

Calculation of a and b gives

a = i2pi

e|k|sgn k, b = 12pi

e|k| , (5.210)

so that

p10 = 0

[iR sgn k + S]e|k| d. (5.211)

Now

(0)3 = 1 + 3zw(0)x + 2

2z2w(0)2x ,

(0)1 = 2zw(0)z 22z2w(0)x w(0)z ,u(0) = u0(z) p(0),u(0)x = w(0)z ,u(0)z = u

0 + w

(0)x , (5.212)

thus, retaining only the perturbed linear (in s) terms, we have from (5.170)

R ikt1 + t3z + u0wz u0[w iks],S ikt3 t1z iku0[w + iks], (5.213)

310

where w = w(0), andt1 = 2ikzp, t3 = 3ikzw, (5.214)

where p = p(0).Finally, from (5.150),

w(0) = b(x, z) sx, p(0) = a(x, z) sx, (5.215)whence using (5.210),

w = ikse|k|z,p = |k|se|k|z, (5.216)

and we eventually obtain

p10 = s 0

[k2u0(1 + 2e

|k|) |k|u0(1 e|k|) 5ik|k|e|k|]e|k| d. (5.217)

Simplification of this, using the fact that

0

et ln t dt = , where 0.5772 isthe Euler-Mascheroni constant, yields

p10 = s

[2|k|

(ln 2|k|+ ) + 52ik]. (5.218)

Solutions of (5.206) are s = et, where = r ikc, and after some simplification, wefind that the growth rate r is

r = 2k2q1(pi+ 12

), (5.219)

and the wave speed c is

c = 2q1|k|[1 +

2

{(1 1

2pi

)ln |k| ln 4+ + 12

}]. (5.220)

Thus dunes grow, as r > 0, on a time scale of O(1/), while the waveforms movedownstream at a speed c 2q1|k| = O(1).

This apparently more realistic theory for dune-forming instability is less satisfac-tory than the constant eddy viscosity theory, because the growth rate r k2, and thebasic model is again ill-posed. As before, we can stabilise the model by including thedownslope force, thus replacing the stress by the effective stress defined using (5.109).The effect of this is to add a term to the stress definition in (5.174), which can thenbe written as

e = 1 2p0 + 22 (A2 + . . .) sx, (5.221)where the definition of differs from that in (5.111) because of the different scalingused in the aeolian model. Using (5.124), x d and s d, we find

=

Dsd

1

F 2, (5.222)

311

where the Froude number is

F =Ugd. (5.223)

Using values / = 2.6103, Ds/d = 106, 0.03, F 2 0.04 (based on d = 1000m and U = 20 m s1), we find 2.2.

If we consult (5.196), we see that the destabilising term arises from that propor-tional to sx in A2. Effectively we can write

e = 1 + . . .+

[2(2pi

+

)

]s

x+ . . . , (5.224)

where the modification of the coefficient reflects the effect of the terms in J andp10, as indicated by (5.219). We see that the downslope term stabilises the system if > O(2), and thus practically if F 2 < 1. On the Earth, a typical value is F 2 = 0.04,so that the instability is removed, at all wave numbers. This is distinct from theconstant eddy viscosity case, because the stabilising term has the same wave numberdependence as the destabilising one.

If we ignore the stabilising term in , then the situation is somewhat similar tothe rill-forming instability which we will study in chapter 6. There the instabilityis regularised at long wavelength by inclusion of singularly perturbed terms. Themost obvious modification to make here in a similar direction is to allow for a finitethickness of the moving sand layer. It seems likely that this will make a substantialdifference, because the detail of the mixing length model relies ultimately on theexistence of an exponentially small roughness layer through which the wind speeddrops to zero. It is noteworthy that the constant eddy viscosity model does not sharethis facet of the problem.

5.8 Separation at the wave crest

The constant eddy viscosity model can produce a genuine instability, with decay atlarge wave numbers. If pushed to a nonlinear regime, it allows shock formation, al-though it also allows unlimited wavelength growth. The presumably more accuratemixing length theory actually fares somewhat worse. It can produce a very slow in-stability via an effective negative diffusivity, but this is easily stabilised by downslopedrift. It is possibly the case that specific consideration of the mobile sand layer willalleviate this result.

A complication arises at this point. Aeolian sand dunes inevitably form slip faces.There is a jump in slope at the top of the slip face, and the wind flow separates,forming a wake (or cavity, or bubble). One authority is of the opinion that no modelcan be realistic unless it includes a consideration of separation. In this section wewill consider a model which is able to do this. Before doing so, it is instructive toconsider how such separation arises.

If the constant eddy viscosity model has any validity, it suggests that the uniformflat bed is unstable, and that travelling waves grow to form shocks. If the slope withinthe shocks is steep enough to exceed the angle of repose of sand grains (some 34),

312

then a slip face will occur, with the sand resting on the slip face at this angle. Theturbulent flow over the dune inevitably separates at the cusp of the dune, forming aseparation bubble, as indicated in figure 5.13. The formation of a separation bubblemakes the model fundamentally nonlinear, and it provides a possible mechanism forlength scale selection. It is thus an attractive possible way out of the conundrumsconcerning instability alluded to above.

2pi

bubbleair flow

0 ba

Figure 5.13: Separation behind a dune.

It is simplest to treat the separation bubble in the context of the mixing lengththeory, and this we now do, despite our misgivings about its applicability for smallamplitude perturbations. We suppose that there is a periodic sequence of dunes, withperiod chosen to be 2pi. We suppose that there is a slip face, as shown in figure 5.13,and we suppose the corresponding separation bubble occupies the interval (a, b). Wedenote the bubble interval as B, and the corresponding attached flow region as B.

Because our method will use complex variables, it is convenient to rechristen thespace coordinates as x and y, and the corresponding velocity components as u and v.At leading order, the inviscid flow is described by the outer equations (5.146):

ux + vy = 0,

ux + py = 0,

vx + py = 0, (5.225)

and these are valid in y > s. From these it follows that p and v satisfy the Cauchy-Riemann equations, and thus

p+ iv = f(z) (5.226)

is an analytic function, where z = x+ iy.The boundary conditions for p and v are that both tend to zero as y , and

v satisfies the no flow through condition (5.144), v = sx + usx on y = s. Thesecompletely specify the problem in the absence of a separation bubble.

If we suppose that a separation bubble occurs, as shown in figure 5.13, then itsupper boundary is unknown, and must be determined by an extra boundary condition.We let y = s(x) denote this unknown upper boundary, and define the ground surfaceto be y = s0(x); thus s(x) = s0(x) for x B.

313

There are various ways to provide the extra condition. Two such are that thepressure, or alternatively the vorticity, are constant in the bubble. We shall supposethe former, and therefore we prescribe

p = pB for y = s, x B. (5.227)The bubble pressure pB is an unknown constant, and must be determined as part ofthe solution.

Separation occurs because the viscous boundary layer (here, the roughness layer)detaches from the surface, forming a free shear layer at the top of the bubble, whichrapidly thickens to form a more diffuse upper boundary. The assumption of constantpressure in the bubble is essentially a consequence of this shear layer, implying thatmean fluid velocities in the bubble are small.

For small , we can expand the boundary conditions at y = s about y = 0,so that to leading order, the problem becomes that of finding an analytic functionf(z) = p+ iv in the upper half plane Im z > 0, satisfying

f 0 as z ,v = sx on y = 0,

p = pB on y = 0, x B. (5.228)The extra pressure condition should help determine s in B, but the endpoint

locations are not necessarily known. Specification of the behaviour of the solution atthe endpoints is necessary to determine these. Firstly, we expect s to be continuousat the end points:

s(a) = s0(a), s(b) = s0(b). (5.229)

A difference now arises depending on whether a slip face occurs or not. If not, thenthe bed slope is continuous, and at the upstream end point x = a, we might surmisethat boundary layer separation is associated with the skin friction dropping to zero.Now from (5.174) and (5.176), we have the surface stress defined by

= 1 p0, (5.230)

where p0 is the surface pressure. The only apparent interpretation of this which wecan make in our simplified model is to require that

p + on y = 0 as x a B; (5.231)more detailed consideration of the boundary layer structure near the separation pointwould be necessary to be more precise than this. We do not pursue this possibilityhere, mainly because the more relevant situation is when a slip face is present.

If we suppose a slip face is present, then we can presume that separation occursat its top, and this determines the point x = a. In addition, it is natural to supposethat boundary layer detachment occurs smoothly, in the sense that we suppose theslope of s is continuous at a:

s(a+) = s0(a); (5.232)

314

this implies that v is continuous at x = a. If possible, we would like to have smoothreattachment at b, and in addition (and in fact, because of this) continuity of pressurealso:

[p]b+b = [p]a+a = 0, s

(b) = s0(b+). (5.233)We shall in fact find that all these conditions can be satisfied. This is not always

the case in such problems, and sometimes (worse) singularities have to be tolerated.The choice of the behaviour of the solution at the end points actually constitutes themost subtle part of solving Hilbert problems.

5.8.1 Formulation of Hilbert problem

The first thing we do is to analytically continue f(z) into the lower half plane. Specif-ically, we define

G(z) =

12 [f(z) pB] , Im z > 0,

12[f(z) pB

], Im z < 0.

(5.234)

G is analytic in both the upper and lower half planes, and if G+ and G denote thelimiting values of G as z x from above and below, then

G+ +G = is,

G+ G = p pB, (5.235)everywhere on the real axis.

Because of the assumed periodicity in x, we make the following transformations:

= eiz, = eix, G(z) = H(). (5.236)

The geometry of the problem is then illustrated in figure 5.14. The problem to solveis identical to (5.235), replacing G by H, and thus we have the standard Hilbertproblem

H+ H = 0 on B,H+ +H = i0 on B, (5.237)

where 0() = s0(x). We have to solve this subject to the supplementary conditions

H(0) = 12pB, H() = 12pB ; (5.238)the first of these in fact implies the second automatically. We seek to apply theconditions that both 12(p pB) = ReH and 12v = ImH are continuous (thus H iscontinuous) at both endpoints = a = eia and = b = eib. Given H satisfying(5.237), then the separation bubble boundary is given by the solution of

s = 2iH on B, s(a) = s0(a), (5.239)and the pressure on B is given by

p = pB +H+ H on B. (5.240)

315

B

b a

B

Figure 5.14: B and B on the unit circle in the complex plane. B is a branch cutfor the solution of the Hilbert problem (5.237).

Solution

The solution to (5.237), given the location of a and b, is as follows. Define a function() such that

+ + = 0 on B (5.241)

(and is analytic away from B); then(H

)+

(H

)=i0+

, (5.242)

and by the discontinuity theorem, we have

H =()

2pii

B

i0(t) dt

+(t)(t ) + P, (5.243)

where P is an as yet undetermined polynomial. To find P , we must specify , andthis in turn depends on the required singularity structure of the solution.

The smoothness of H is essentially that of , and so we will choose the function

= [( a)( b)]1/2 . (5.244)

The most general choice is = ( a)ma+ 12 ( b)mb+ 12 , where ma and mb are in-tegers, but most of these possibilities can in general be eliminated by requirementseither of continuity or at least integrability of the solution.

We consider the behaviour of the Cauchy integral

() =1

2pii

B

(t) dt

t (5.245)

316

near the end points of integration. Note that in the present case,

(t) =i0(t)

+(t). (5.246)

First suppose that (t) is continuous at an end point.9 Then we have

() = (c)2pii

ln( c) +O(1), (5.247)

where c denotes either end point of B, and the upper and lower signs apply at theright (a) and left (b) hand ends of B, respectively. (5.247) applies as c, with / B.

Similarly, for B,

() = (c)2pii

ln( c) +O(1), (5.248)

where () denotes the principal value of the integral (and () = 12 [+() + ()]).Bearing in mind (5.246), we see that if is unbounded at c, and specifically goesalgebraically to infinity, then the corresponding Cauchy integral is bounded, and thusH will be unbounded (unless the choice of P can be chosen to remove the singularity).Using the definition

H = () [() + P ] , (5.249)

we have from (5.239) and (5.240) that

s = 2iH() = 2i() [() + P ] on B,p pB = 2+() [() + P ] on B. (5.250)

The implication of this is that if is unbounded at an end point, then in generalboth p and s will also be unbounded, unless the choice of P removes the singularity.The worst singularity we can tolerate is an integrable one, thus ( c)1/2.

Now suppose that is bounded at an end point, and specifically ( c)1/2.(Any higher power causes the Cauchy integral to be undefined, because then is notintegrable.) If we define via

(t) (t)(t c)1/2 as t c, (5.251)

then

() =(c)

2( c)1/2 + o(

1

( c)1/2), B,

() = o

(1

( c)1/2), B. (5.252)

9More precisely, should be Holder continuous, that is |(t1) (t2)| < K|t1 t2| , for somepositive .

317

It then follows from (5.250) that s is bounded (and in fact continuous) and p iscontinuous at c. It is because of this that we choose as defined in (5.244), in orderto satisfy the smoothness conditions (5.232) and (5.233).

In this case, the polynomial P must be zero in order to satisfy the condition at =, and we have

H =()

2pii

B

i0(t) dt

+(t)(t ) . (5.253)

We define the integrals

I0 =02pii

B

i0(t) dt

t+(t), I =

1

2pii

B

i0(t) dt

+(t); (5.254)

we thus have H(0) = I0, H() = I, and the conditions in (5.238) correspond toprescribing

I0 = I = 12pB. (5.255)It is a straightforward exercise in contour integration to show that I0 = I, where theoverbar denotes the complex conjugate, therefore (5.255) is tantamount to the singlecondition I0 = 12pB. Because this is a complex-valued integral, (5.255) actuallycomprises two conditions for the two unknown quantities pB and b.

It remains to be seen whether s is continuous at b. Since (5.255) determines b,and s is fully determined by (5.239), it is not obvious that this will be the case. (Ifit were not, we would have to allow for a singularity in the solution at one of the endpoints.)

In fact, it is easy to show that (5.255) automatically implies that s is continuous atb. To show this, it is sufficient to show that s is continuous over the periodic domain[0, 2pi]. Equivalently, we need to show that

I =

2pi0

s dx =B B

i(H+ +H) di

= 0, (5.256)

using (5.239) and (5.237). Denoting contours just inside and outside the unit circleas C+ and C (see figure 5.15), we see that

I = [

C+

H d

+

C

H d

]. (5.257)

H is analytic inside and outside the unit circle. The integral over C+ is thus just2piiH(0) using the residue theorem, while the integral over C can be extended bydeforming the contour out to infinity, whence we obtain the integral 2piiH(). Thus

I = 2pii[I0 I] = 0, (5.258)and continuity of s at b is assured. We have thus obtained a solution in which theseparated streamline leaves and rejoins the surface smoothly, and the pressure iscontinuous at the end points.

318

b aB

C

C

+

B

Figure 5.15: The contours C+ and C lie just inside and outside the unit circle,respectively.

5.8.2 Calculation of the free boundary

In order to solve (5.239) for s, we need to evaluate H on B. There are various waysto do this. One simple one, which may be convenient for subsequent evolution of thebed using spectral methods, is to use a Fourier series representation. Let us supposethat

s(x) =

k=ake

ikx, (5.259)

so that

i0() =

k=dk

k, (5.260)

wheredk = kak. (5.261)

We suppose that the Laurent expansion for i0 extends to the complex plane as ananalytic function with singularities only at 0 and . (This is automatically true forany finite such series.) Then we can write the solution for H as

H = 12 [i0() q()()], (5.262)where q has a Laurent expansion

q =

k=lk

k. (5.263)

Then we obtain s by solvings = s0 + iq (5.264)

319

on [a, b], with s(a) = s0(a). In practice, we would obtain b by shooting.Suppose that

1

()=

r=0

frr+1

, || > 1 (5.265)

(see question 5.10 for one way to calculate the coefficients); then we can write

H12

=

m=dm

mr=0

frr+1

j=lj

j. (5.266)

As , and H 12pB; equating coefficients of j in (5.266) for j 0 yields

lj =r=0

dj+r+1fr, j 0, (5.267)

and for j = 0 we have

pB =r=0

drfr l1. (5.268)

For || < 1, we find1

()=

1

0

r=0

frr, (5.269)

and thusH12

=1

0

r=0

frr

m=

dmm

j=

ljj. (5.270)

As 0, H 12pB; equating powers of j for j 1, we find

lj =1

0

r=0

frdjr, j 1, (5.271)

and for j = 0 we havepB0

=1

0

r=0

frdr l0. (5.272)

Putting these results together, we find that (5.268) and (5.272) together give(bearing in mind that dk = dk)

pB =r=0

drfr + 0

r=0

frdr+1, (5.273)

with the added constraint that pB is real.We can now use the definitions of lj in (5.267) and (5.271) to evaluate iq in

(5.264). Being careful with the arguments, we find that on B,

= 21/21/20 R, (5.274)

320

where

0 = exp[12(a+ b)

], R =

[sin

(x a2

)sin

(b x2

)]1/2, (5.275)

and after some algebra, we have the differential equation for s on B:

s = s0 4R Im[1/20

j=0

r=0

frdj+r+1 exp{i(j + 12

)x}]

, (5.276)

with initial condition s(a) = s0(a). To solve this, guess b; we can then calculate theright hand side. Solving for s, we adjust b by decreasing it if s reaches s0 for x < b,and increase it if s remains > s0 for all x b.

Computational approaches

Complex analysis is all very elegant, but is probably not an efficient way to computea time-evolving interface.