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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)
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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
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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.
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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
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(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,
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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)
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-
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)
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-
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)
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-
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)
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-
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}.
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-
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.
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-
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).
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-
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.
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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)
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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)
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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)
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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 .
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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.
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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)
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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)
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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.