Simple analytical models of glacier-climate interactions - by Prof. J ...

SIMPLE ANALYTICAL MODELS OF GLACIER - CLIMATE
INTERACTIONS
J. Oerlemans
IMAU, Utrecht University
j.oerlemans@phys.uu.nl
Introduction
1. A mass-balance model
2. Ice deformation: perfect plasticity
3. A simple glacier model
4. A glacier model with a more complicated geometry
5. A volume time scale for valley glaciers
6. Including feedback between glacier length and ice thickness
7. Steady state ice-sheet profiles
1

Introduction
In these lectures I present simple analytical models of glaciers and the interaction of
glaciers with climate. Do such models make sense in a time that computer power is almost
unlimited?
May be it is a matter of taste.
Glaciers shape their own surface climate. The height-mass balance feedback (HBMfeedback
in the following) is the dominating mechanism. It is a direct consequence of the
fact that the specific balance increases with altitude (and this is because air temperature
decreases with altitude and precipitation increases with altitude). It is the HMB-feedback
that may turn small glaciers into huge ice sheets. Without the HMB-feedback the glacial
cycles of the Pleistocene would not have occurred.
Therefore, if we want to study how glaciers and ice sheets grow and shrink, we need to
know about their geometry. In many cases the details are not important. For instance, if
we prescribe climatic conditions in such a way that the specific balance above the ablation
line is constant, the precise form of the ice sheet-profile above this line does not matter!
For valley glaciers we will see that it is the relation between mean surface elevation and
glacier size that matters.
These lectures do not form a systematic treatment, but have a kaleidoscopic nature. There
is a general idea, however. The philosophy is that glacier-climate interactions should first
of all be considered as a continuity problem, in which the glacier mechanics can be
strongly parameterised and most of the attention goes to the total mass budget.
The material presented in these lectures can also be found in:
• C.J. van der Veen (1999): Fundamentals of Glacier Dynamics. Balkema, 406 pp.
• J. Oerlemans (2001): Glaciers and Climate Change. Balkema, 148 pp.
2

1. A mass-balance model
The processes governing the exchange of heat and mass between glacier surface and
atmosphere contain many nonlinear features, and one may well wonder whether an
analytical approach to glacier mass-balance modelling makes any sense. Let us have a
look.
We start by ignoring the daily cycle and assume that the daily loss of mass due to melting
is determined by a potential surface energy flux ψ. Moreover, we assume that this flux
has a sinusoidal shape over the year (period T):
ψ = A 0 + A 1 cos 2π t/T . (1.1)
The coefficients A 0 and A 1 contain both a temperature effect and a solar radiation effect.
We do not include a phase shift in time (which is irrelevant) and take A 1 > 0. In Fig. 1.1
eq. (1.1) is compared to the surface energy flux (daily mean values) as measured by an
automatic weather station (located on the snout of the Morteratschgletscher, Switzerland).
It is clear that eq. (1.1) does not work for the wintertime. However, this is not a problem
because in this period little melting takes place.
Ablation occurs when Ψ is positive, at a rate ψ/L m (L m is the latent heat of melting). Note
that there is never any ablation when A 0 < -A 1 , and there is ablation all year round when
A 0 > A 1 . To keep the model tractable accumulation has to be prescribed in a simple way.
We assume that the accumulation rate is constant when ψ < 0 and zero otherwise. Fig.
1.2 depicts the case when -A 1 < A 0 < A 1 . There is ablation from t = 0 onwards and it
will stop at time te, given by:
t e = T 2π arccos - A 0
A 1
. (1.2)
Fig. 1.1
300
200
total energy flux
sine fit
Morteratschgletscher
year 2000
100
(W m 2 )
0
-100
-200
0 50 100 150 200 250 300 350 day
3

Fig. 1.2
300
200
Surface energy flux (W m -2 )
100
0
-100
ablation
ablation
t e
t e
accumulation
ablation
ablation
-200
accumulation
-300
0 50 100 150 200 250 300 350
Time (days)
We anticipate that the variation of the surface energy flux with altitude is first of all
governed by the parameter A 0 . The upper curve in Fig. 1.2 would correspond to a lower
location on the glacier.
Because the ψ-curve is symmetrical with respect to t = T/2, the total ablation M and
accumulation A can be written as:
M = 2
Lm
0
t e
A 0 + A 1 cos 2π t
T
dt
= 2 A 0 t e
L m
+ A 1 T sin 2π t e
π L m T
(1.3)
A = 2
T/2
P dt
t e
= 2 (T/2 - t e ) P (1.4)
Here P is the precipitation rate, assumed to be constant throughout the year. The mass
balance model can now be formulated as:
(i) A 0 < -A 1 (no ablation): (1.5)
A = T P ,
M = 0 .
(ii) -A 1 < A 0 < A 1 (ablation during part of year): (1.6)
A = 2 (T/2 - t e ) P ,
M = - 2 A 0 t e
L m
- A 1 T sin 2π t e
π L m T
.
4

(iii) A 0 > A 1 (continuous ablation): (1.7)
A = 0 .
M = - A 0 T
L m
,
where te is given by eq. (1.2). The specific balance b is:
b = A + M . (1.8)
To evaluate a mass balance profile we have to specify how A 0 , A 1 and P depend on
altitude. We start by assuming that A 1 is constant, and that A 0 and P vary linearly with
altitude h according to:
A 0 = - β E (h - h ref ) (β E > 0) , (1.9)
P = P sl + γ P h . (1.10)
Note that h ref can be interpreted as the altitude at which the melt season lasts for six
months. For precipitation we take zero altitude (sea level) as a reference point (P sl ). For
given β E , the amplitude of the potential energy flux (A 1 ) determines the altitudinal range
where snow is a fraction of the total precipitation. Below this range there is no snow;
above this range all precipitation falls as snow. At low altitudes where ablation is
continuous the balance gradient is determined entirely by β E . Altogether, to calculate a
balance profile we need to specify five parameters. It is not difficult to fit the model to
observed balance profiles that are reasonably smooth.
Fig. 1.3
2000
Nigardsbreen
ablation
1600
Altitude (m)
1200
observed
calculated
balance
snow
precip
800
400
-10 -5 0 5
[mwe]
5

Fig. 1.3 shows that the model is able to reproduce the balance profile of Nigardsbreen
(model parameter values: A 1 = 110 W m -2 , β E = 0.115 W m -2 , P sl = 2 m, γ P = 0.0012,
h ref = 1325 m). With this choice of parameter values the model not only simulates the
balance profile accurately, but it also reproduces realistic features such as a significant
amount of melt in the highest mass balance interval (1900-2000 m) and little snow
accumulation on the lowest part of the glacier.
The simple mass balance model can be used for climate sensitivity tests. The precipitation
rate can be changed and the new resulting balance profile can easily be calculated. For a
temperature change the situation is more complicated, because now the change in A 0 and
A 1 has to be known. Detailed energy balance studies suggest that the first-order effect of
a 1 K temperature rise is to increase A 0 by typically 10 W/m 2 . The implications for the
mass balance of Nigardsbreen, calculated with the model described above, are shown in
Fig. 1.4.
Fig. 1.4
2000
Nigardsbreen
1600
Altitude (m)
1200
800
400
reference
+2 K
+2 K ; +20 %
-12 -8 -4 0 4
Specific balance (mwe)
Problems
• What are the most important processes that contribute to the altitudinal gradient in the
surface energy flux β E ?
• Estimate from Fig. 1.1 how much ice can be melted on a hot summer day on the snout
of the Morteratschgletscher.
6

2. Ice deformation: perfect plasticity
Perfect plasticity provides a description of ice deformation that is very tractable in simple
analytical models of ice sheets, ice shelves and glaciers.
In the following we use a Cartesian coordinate system with the z-axis pointing upwards.
The ice velocity is in the x-direction and is denoted by u. In terms of simple shearing
flow, perfect plasticity can be considered as an asymptotic case of Glen's law when the
exponent n goes to infinity. It implies the existence of a yield stress τ 0 , such that
τ xz > τ 0 : d u
dz → ∞ , (2.1)
τ xz ≤ τ 0 : d u
dz = 0 .
Eq. (2.1) means that, as long as stress is being built up by gravity, the ice will deform in
such a way that the yield stress is approached but never exceeded. Therefore the ice
velocity is constant and all the shear is concentrated at the base where
τ xz = ρ g H
d h
dx
= τ 0 . (2.2)
Here ρ is ice density, g is the gravitational acceleration, H is the ie thickness and h the
surface elevation. Values for the yield stress used in the literature are typically in the
0.5 . 10 5 to 3 . 10 5 Pa range, where the smaller values apply to ice caps with low mass
turnover and the high values to active valley glaciers. For a given value of τ 0 , the product
of surface slope and ice thickness is constant. Eq. (2.2) can thus be used to estimate ice
thickness from the surface slope.
The simplest possible model for a symmetric ice cap resting on a flat bed is based on the
theory of perfect plasticity. In this case H=h, so
d h 2
d x
= 2 τ 0
ρ g . (2.2)
This can be integrated to give:
h 2 (x) - h 2 (0) = 2 τ 0
ρ g (x-x 0 ) . (2.3)
To construct a solution for an ice cap we have to prescribe its size (L) and the ice
thickness at the boundaries (we take it to be zero). The result is (Weertman, 1961):
7

h(x) =
2 τ 0
ρ g x for 0 ≤ x ≤ L 2 , (2.4)
h(x) =
2 τ 0
ρ g (L-x) for L 2 ≤ x ≤ L .
The profile is shown in Fig. 2.1 (in the example shown L = 200 km and
2 τ 0 / ρ g = 10 m). Although it provides a reasonable first approximation to the shape of
an ice cap, there are some deficiencies. First of all the solution is not valid close to the ice
divide, where the surface slope is very small and longitudinal stresses dominate;
therefore, here the ice flow cannot be approximated by plane shear. Secondly, a property
of the perfectly plastic ice cap is that, given the size, its thickness does not depend on the
mass balance. Although we will see later that this dependence is generally weak, in some
applications it is of importance. The ice velocity or mass flux is not directly related to the
profile, but can only be obtained from the conservation of mass. For instance, for one
half of the ice sheet:
H u =
L
L/2
b n dx
. (2.5)
In this equation u is the vertically averaged horizontal ice velocity and b is the specific
balance rate.
Next we consider the isostatic depression of the bed. A balance is present only when the
following condition is met (ρ i and ρ m are the ice and mantle density, respectively):
ρ i H + ρ m b = ρ i (h-b) + ρ m b = 0 . (2.6)
The height of the bed b then is (b < 0):
Fig. 2.1
1000
h (m)
750
500
250
0
0 50 100 150 200
x (km)
8

=
ρ i h
ρ i - ρ m
= -δ h . (2.7)
The parameter δ is of the order of 1/3. The ice thickness is (1+δ)h and we can substitute
this in eq. (2.2):
ρ g (1+δ) h d h
d x = τ 0 . (2.8)
Therefore the solution now becomes
h(x) =
2 τ 0
ρ g (1+δ) x , (2.9)
and we conclude that the profile remains parabolic with a slightly modified 'plasticity
parameter' µ isos
µ isos =
2 τ 0
ρ g (1+δ)
. (2.10)
The perfectly plastic ice-sheet model can be used to demonstrate how the HMB-feedback
leads to hysteresis when we consider the response of a bounded ice-sheet to climate
change. Suppose that the specific balance rate is a linear function of height:
b n = β (h - E) , (2.11)
where E is the equilibrium-line altitude and β the balance gradient (assumed to be
constant). Now the total mass budget of an ice cap will be positive when
h > E (2.12)
So for h < E an ice sheet cannot exist. However, in the case that h > E > 0 a situation
without an ice sheet also represenst an equilibrium state. For E < 0 there has to be an ice
sheet.
Altogether this can be visualised in a solution diagram (Fig. 2.2), showing the
equilibrium states as a function of the climatic conditions. In this simple model the ice
sheet can have only two states (for given L). Climate change is represented by a shift of
the equilibrium line. The critical points are shown by black dots.
9

Fig. 2.2
ice volume
•
0
(cold)
•
E
0 (warm)
h mean
Problems
• The mean ice thickness for a perfectly plastic ice sheet can be written as H = c L 1/2 .
Find the constant c.
• Suppose that a perfectly plastic ice sheet is axi-symmetrical rather than 'onedimensional'.
How would this affect the profile?
• An axi-symmetric perfectly plastic ice sheet has radius R. Find an expression for the
vertical mean ice velocity u(r). Assume that the specific balance rate b is constant (and
positive). There is a problem at r=R. Can you solve it?
• For the Greenland ice sheet the mean value of E is about 1200 m, for the Antarctic ice
sheet the equilibrium line is 'below sea level'. Can you locate these ice sheets on the
solution diagram of Fig. 2.2?
10

3. A simple glacier model
We consider a glacier that has a uniform width, rests on a bed with a constant slope s,
and behaves perfectly plastically in 'a global sense' (Fig. 3.1).
Fig. 3.1
Altitude
h = b + H
b = b
0
- sx
equilibrium line
L
x
The specific balance is written as
b = β (h - E) , (3.1)
where E is the equilibrium-line altitude and β the balance gradient (assumed to be
constant). The glacier is in balance when the total mass budget is zero:
L
b n dx = β
0
0
L
(H + b 0 - s x - E) dx = 0
. (3.2)
Solving for glacier length L yields:
L = 2 (H m + b 0 - E)
s
. (3.3)
In this expression H m is the mean ice thickness. Note that the solution does not depend
on the balance gradient! The next step is to find an equation for H m . We use again the
concept of perfect plasticity:
H m =
τ 0
ρ g s
. (3.4)
Substituting this in eq. (3.3) yields:
11

L = 2 s
τ 0
ρ g s + b 0 - E . (3.5)
The solution is shown in Fig. 3.2 (parameter values b 0 -E = 500 m, τ 0 /ρ g = 10 m). For
reference the solution for zero ice thickness is also plotted. In this case the intercept of
the equilibrium line and bed is simply at x = L/2. For the full solution ice thickness
increases with decreasing slope, which implies an upward shift of the equilibrium point
(intercept of equilibrium line and glacier surface). There is no solution for a flat bed,
unless the equilibrium line is allowed to slope upwards.
Fig. 3.2
80
60
L (km)
40
20
full solution
0
H m
= 0
-20
0 0.05 0.1 0.15 0.2 0.25
Slope of bed
The simple model can be used to make an order-of-magnitude estimate of climate
sensitivity. Differentiating Eq. (3.5) with respect to E yields:
d L
dE
= - 2 s
. (3.6)
So glaciers on a bed with a smaller slope are more sensitive in an absolute sense. It is
tempting to use eq. (3.6) to make a first-order estimate of the response of glacier length
to a change in free atmospheric temperature Ta. We assume that the equilibrium-line is
linked to this temperature, which implies that d E/d T a = -γ -1 . Here γ is the temperature
lapse rate in the atmosphere, typically -0.007 K km -1 ). It follows that
12

d L
d T a
= ∂ L
∂ E
d E
d T a
= 2
γ s . (3.7)
So we have arrived at the remarkable result that, for the given simple model, only two
parameters are needed to estimate the sensitivity of glacier length to atmospheric
temperature, namely, the characteristic bed slope and the temperature lapse rate! Fig. 3.3
shows d L/d T a in dependence of s for the above mentioned valueof the lapse rate. Larger
valley glaciers typically have mean slopes between 0.1 and 0.2, implying that a 1 K
temperature rise would lead to a 1 to 3 km decrease in glacier length. These figures
appear reasonable, and we can conclude that the simple glacier model provides an
interesting first-order description of the relation between climate change and glacier
response.
Fig. 3.3
0
dL/dT a
(km K -1 )
-2
-4
-6
-8
large valley glaciers
0.05 0.1 0.15 0.2 0.25
Slope of bed
Problems:
• The absolute change in glacier length for a given change in E is larger when the slope
is smaller. However, does this also apply to the relative change in L?
• What are, in your judgement, the largest deficiencies of the model presented above?
13

4. A glacier model with a more complicated geometry
Glaciers having a constant width are very rare. Larger valley glaciers often show a pattern
of a relatively wide accumulation basin and a narrower snout that flows into the valley.
Examples are shown in Fig. 4.1.
Fig. 4.1
Nigardsbreen
Abramov Glacier
N
N
1800
400
600
800
1000
1200
1600
1400
1400
2200
1800
1000
800
1800
2000
1952 m
600
2200
2600
400
2400
2987 m
2 km
2901 m
2 km
To handle such a situation we use a very schematic geometry in which the glacier consists
of two parts having a different width. Again, L is the glacier length and the upper basin
has length L ub (Fig. 4.2).
Fig. 4.2
L ub
L
0
x
14

For L < L ub the solution is given by eq. (3.3):
L = 2 (H + b 0 - E)
s
. (4.1)
When the mass budget of the upper basin is positive L will be larger than Lub. The length
of the glacier is then determined by
L ub L
(H + b 0 - s x - E) dx + ξ (H + b 0 - s x - E) dx = 0
0
L ub
. (4.2)
Here ξ is the width of the glacier tongue divided by the width of the upper basin.
Evaluating the integrals yields:
- 1 2 ξ s L2 + ξ (H + b 0 - E) L + (1 - ξ) (H + b 0 - E) L ub - 1 2 s (1 - ξ) L ub
2 = 0 (4.3)
This is a quadratic equation for L which is easily solved:
L = - 1 s
E' + E' 2 + 2 s 1-ξ
ξ
1/2
E' L ub + 1 2 s L ub
2
(4.4)
In this expression E' = E - b 0 - H (note that E' < 0).
The solution is illustrated in Fig. 4.3. In this example the upper basin has a length of
5 km. Other parameter values are b 0 =2000 m, s=0.1, H=100 m. L is plotted as a
function of E for three different values of ξ. For ξ=0.25, the width of the upper basin is
four times that of the glacier tongue.
Fig. 4.3
20
L (km)
15
10
ξ =4
ξ=1
ξ =0.25
largest sensitivity
5
0
1400 1500 1600 1700 1800 1900 2000
E (m)
15

In this case the glacier length increases rapidly when the equilibrium line sinks below
1850 m. A clear maximum of dL/dE exists when the glacier starts to form the snout.
When the equilibrium line sinks still further, in the end dL/dE will approach the value of
the original model glacier of uniform width. When the lower part of the glacier is wider
than the upper part the opposite is seen. The sensitivity is at a minimum when the budget
of the upper basin is just positive.
In conclusion we can state that glaciers with a narrow tongue are the ones that are most
sensitive to climate change, especially when the glacier front is just below the upper
basin.
16

5. A volume time scale for valley glaciers
A time scale for the adjustment of glaciers to climate change can be derived from the
requirement of mass continuity (Jóhanesson et al, 1989; Haeberli and Hoelzle, 1995).
We perform a linear perturbation analysis. Conservation of ice volume V can be
expressed as
d V
d t
= d A H m
d t
= H m d A
d t
+ A dH m
d t
. (5.1)
Here A is the glacier area and H m the mean ice thickness. Now we write
V = V 0 + V', A = A 0 +A' and H m = H m,0 + H m ', where the reference state is defined
as (V 0 , A 0 , H m,0 ). By substitution in eq. (5.1) and by neglecting higher-order terms we
then obtain the perturbation equation
d V'
d t
= H m,0 d A'
d t
+ A 0 dH m '
d t
. (5.2)
Next we assume (see Fig. 7.1):
A' = w f L' , (5.3)
Here w f is the characteristic width of the glacier tongue.
Fig. 6.1.
L'
total glacier
area A
w f
flowline
If the mean thickness of the glacier does not change we have for the change of ice volume
with time
17

d V'
d t
= w f H m,0 d L'
d t
= amount of mass added . (5.4)
In a perturbation analysis we have two contributions to the mass added to or removed
from the glacier:
change of volume = A 0 B' + w f B f L' . (5.5)
B' is the perturbation of the balance rate (constant over the glacier), and B f the
characteristic balance rate at the glacier front. The first term is simply the amount of ice
added or removed by the balance perturbation on the reference glacier area. The second
term is the amount of ice lost or gained because the length of the glacier deviates from that
of the reference state. Combining eqs. (5.4)-(5.6) yields
d L'
d t
=
A 0
w f H m,0
B' +
B f
H m,0
L' . (5.7)
The equilibrium state is obtained by setting the time derivative to zero. This gives:
L' = - A 0
w f B f
B' . (5.8)
The response time for glacier length according to this model is apparently
t L = -
H m,0
B f
. (5.9)
Some examples on response times are given in the table below. The quantities marked
with • are the input data. Eq. (6.1) has been used to calculate H o . The response time is
then obtained from eq. (5.9).
•Ao
•Lo
•s
•β
Ho
•W f
•B f
t L
(km 2 )
(km)
(m 2 yr -1 )
(m)
(m)
(m yr -1 )
(yr)
Nigardsbreen 48.8 10.4 0.13 0.0078 125 500 -10 13
Abramov Glac. 25.9 9.1 0.09 0.0055 136 700 -4 34
Rhonegletscher 18.5 9.8 0.13 0.0066 122 900 -5 24
Franz-Josef Gl. 36 11.4 0.21 0.0125 107 600 -22 5
Aletschgletscher 86.8 24.7 0.1 0.0066 215 1200 -8 27
Ob. Grindelw.gl. 10.1 5.0 0.25 0.0066 65 500 -7 9
18

Problem:
• The time scale derived above is not very accurate, because it ignores the HMBfeedback.
Repeat the analysis and include the HMB-feedback by assuming that
H m ' = η L'. Find η by linearising eq. (6.1). Calculate the newly derived time scale
for the glaciers in the table. Conclusion?
19

6. Including feedback between glacier length and ice thickness
In the analysis of section 3 we made thickness a function of the bed slope, but not of the
glacier length. This is an obvious shortcoming. A more advanced analysis can be made
by using the relation:
H m =
1/2
µ L
, (6.1)
1+ ν s
where µ and ν are positive constants. Actually, eq. (6.1) fits rather well results form a
numerical glacier model in which s and L are systematically varied (Oerlemans, 2001).
Note that for s = 0 eq. (6.1) reduces to the relation between ice thickness and glaciers
length for a perfectly plastic ice sheet on a flat bed (section 2). In the simple model, the
expression for L was:
L = 2 (H m + b 0 - E)
s
. (6.2)
By substituting eq. (6.1) we obtain (E ' = E -b 0 ):
L = 2 s
1/2
µ L
- E ' . (6.3)
1 + ν s
This quadratic equation is most conveniently solved by setting N = L 1/2 . We then have
N 2 -
2 µ 1/2
s (1 + ν s) 1/2
N + 2 E '
s
= 0 . (6.4)
The determinant is
Det =
4 µ
s 2 (1 + ν s)
- 8 E '
s
. (6.5)
Real solutions exist only when Det ≥ 0. The first term is always positive. Therefore a real
solution exists even for small positive values of E ', that is, when the equilibrium line is
higher than the highest part of the bed at x = 0, but below the glacier surface. This
nonlinearity, of course, reflects the HMB-feedback. The solution for L reads
L =
µ 1/2
s (1 + ν s) 1/2 ± µ
s 2 (1 + ν s)
- 2 E '
s
1/2
2
. (6.6)
20

Note that values of L for which N < 0 are spurious and should not be considered.
An example is shown in Fig. 6.1. Parameter values are: µ = 9 m, ν = 20, s = 0.06 .
Because L = 0 is a stable solution for E ' > 0 (equilibrium line above the bed
everywhere), there are two stable branches and one unstable branch (dotted). In terms of
catastrophe theory this model represents a fold, but because we add the condition that L
should be positive it appears as a distorted cusp. The branching of the steady-state
solutions that shows up here was in fact found numerically a long time ago (Oerlemans,
1981). It should also be mentioned that the dynamics of the present glacier model are
similar to those of a perfectly plastic ice sheet on a flat bed with a sloping equilibrium line
Weertman (1961).
The range of E '-values for which two stable solutions exist is
0 ≤ E ' < E ' crit , (6.7)
where the critical point E' crit is found by setting Det = 0:
E ' crit =
µ
2 s (1 + ν s)
. (6.8)
Therefore, for increasing slope of the bed, the HMB-feedback becomes weaker and the
critical point approaches the origin. The result for the linear analytical model is also
shown. In this caseτ 0 /ρg was set to 9 m, because this is consistent with eq. (6.1) for
s = 0. The squares in Fig. 6.1 shows the dependence of glacier length on E' as found by
a numerical plane-shear model for a glacier of constant width. Clearly, the nonlinear
analytical model matches the numerical result very well.
Fig. 6.1.
14
12
10
L (km)
8
6
4
linear model
2
0
nonlinear model
•
•
-200 -150 -100 -50 0 50
E' (m)
21

From the foregoing analysis it is clear that the bed slope is a very important parameter.
For smaller slopes (ice sheets / ice caps) nonlinear effects are more important than for
larger slopes (valley glaciers). To classify glacier beds with a characteristic slope only
provides not mroe than a schematic picture, of course. Nevertheless, it is instructive to
compae curves for different slopes and to speculate how existing glaciers and ice caps
might fit in (see Fig. 6.2).
Fig. 6.2
1 10 6
s=0.006
8 10 5
L (m)
6 10 5
4 10 5
• Greenland
2 10 5
0
s=0.02
s=0.1 Aletschgl.
Vatnajökull
•
-1000 -750 -500 -250 0 250 500 750
E' (m)
22

7. Steady state ice-sheet profiles
The perfectly plastic ice-sheet model can be improved by calculating the profile for simple
plane shear (Vialov, 1958). In this case the deviatoric stress tensor S ij has only one
nonzero component:
S xz (z) = (H-z) ρ g sin α . (7.1)
Here H is the ice thickness, z the height above the bed, ρ ice density and α the surface
n
slope. This can be combined with Glen's law for simple shear (du/dz = 2 A S xz ) to give
du
dz
= 2 A (H-z) ρ g sin α n . (7.2)
Integrating this equation twice with respect to z yields the vertical mean 'horizontal' ice
velocity U
U = A* H n+1 sin α n + U s (7.3)
In this equation U s is the sliding velocity. A* is the effective flow parameter, which can
be expressed in A, ρ, g and n (see Problems).
Next we consider an axi-symmetric configuration (Fig. 7.1). The vertically-integrated
continuity equation then takes the form:
∂ H
∂ t
= -∇⋅(H U) + b → ∂ H
∂ t
= - ∂(r H U r )
r ∂ r
+ b (7.4)
therefore for the steady state we find:
d (r H U r ) = b r d r → U r = b 2 H-1 r (7.5)
Fig. 7.1
H
φ
R
r
23

We assume that the ice-sheet base is flat and replace sin α by dH/dx. With zero sliding
velocity we then have:
U r = - A* H n+1 d H
d r
n
. (7.6)
Now the velocity can be eliminated by combining eqs. (7.5) and (7.6):
1+2/n
H
dH
dr
1/n
= - b r
1/n . (7.7)
2 A*
Integration with respect to r yields
n
2n+2 H 2+2/n - H 0
2+2/n = - n
n+1
1/n
b
2 A*
r 1+1/n , (7.8)
or
H 2+2/n - H 0
2+2/n = - 2
1/n
b
2 A*
r 1+1/n . (7.9)
Next we have to apply a boundary condition, namely H = 0 at r = R. This leads to a
relation between the ice thickness in the centre and the ice-sheet radius:
H 0
2+2/n = 2
1/n
b
2 A*
R 1+1/n . (7.10)
The final solution then becomes
H(r) = 2 n/(2n+2) b/(2A*) 1/(2n+2) R 1+1/n - r 1+1/n n/(2n+2) . (7.11)
The most commonly used exponent in Glen's law is 3. In this case we have
H(r) = 2 3/8 b/(2A*) 1/8 R 4/3 - r 4/3 3/8 . (7.12)
The solution is shown in Fig. 7.2, together with the profile of a pefectly plastic ice sheet
as derived in section 2. The plane shear solution is shown for two values of A*: one for
which the height at the centre is equal to that of the pefectly plastic solution, and one in
which the mean ice thickness is approximately equal to the mean ice thockness for the
perfectly plastic ice sheet.
24

Fig. 7.2
3500
3000
2500
h (m)
2000
1500
1000
500
perfectly plastic
plane shear 1
plane shear 2
0
0 200 400 600 800 1000 1200 1400 1600
r (km)
A few interesting conclusions can be draw from the plane-shear solution. For n=3 eq.
(7.10) reads
H 0 = 2
1/8
b
2 A*
R 1/2 (7.13)
This shows that the dependence of the ice thickness on the accumulation rate is quite
weak. Halving the accumulation rate reduces the ice thickness by only 8%. The same
applies to the relation between ice thickness and flow parameter. On the other hand, the
dependence on ice-sheet radius is larger. Halving the radius reduces the ice thickness by
30%.
It is hard to judge which model performs best. It is clear that the perfectly plastic model
is not very accurate in the central part of an ice sheet. However, in many applications this
does not matter at all. One can try to check the validity of the profiles by comparison with
observations. This does not give definite answers as to which profile performs best (Van
der Veen, 1999). The less steep slope of the parabolic profile closer to the ice edge seems
more realistic in many cases, albeit for the wrong reason (in reality softer ice and sliding
over deformable beds leads to reduced ice thickness).
It is noteworthy that the dependence of ice thickness on the radius is the same for the
plane-shear and perfectly-plastic models. Altogether, when interest is in the large-scale
response of ice sheets to environmental change, and when changes in ice volume are
expected to be mainly the result of changes in R, then the perfectly plastic model is not a
bad choice.
25

Problems:
• Find an expression for A* in eq. (7.3).
• Suppose that the accumulation rate is not constant but increases with r: b = B R r/R.
Find a new expression for the plane-shear solution and compare with the case of
constant b.
• The flow parameter A depends on ice temperature, approximately following the
relation A = A 0 exp(-Q/RT). Find out for the plane-shear solution by how much the
maximum ice thickness changes when the characteristic ice temperature drops from
275 K to 285 K. Parameter values: A 0 = 1.14⋅10 10 Pa -3 yr -1 , Q = 100 kJ mol -1 .
• We design a schematic mass continuity model for a marine ice sheet. Suppose that the
ice sheet rests on a bed that slopes downwards at a constant rate: b = b 0 - s r and that
the accumulation rate is constant (a). The ice-sheet edge is in the sea, and the
azimuthally averaged ice velocity is proportional to the water depth d (so U R = f * d).
Find an expression for the total mass budget of the ice sheet and solve for the icesheet
radius R.
Apply the model to the West Antarctic ice sheet (set b 0 = 0 for simplicity) and try to
fimd out the sensitivity of R to changes in accumulation rate (use
R = 600 km, f = 1 yr -1 , a = 0.25 m yr -1 ).
REFS
Haeberli W. and M. Hoelzle (1995): Application of inventory data for estimating
characteristics of and regional climate-change effects on mountain glaciers: a pilot
study with the European Alps. Annals of Glaciology 21, 206-212.
Jóhannesson T., C.F. Raymond and E.D. Waddington (1989): Time-scale for
adjustment of glaciers to changes in mass balance. Journal of Glaciology 35, 355-
369.
Oerlemans J. (1981): Some basic experiments with a vertically-integrated ice-sheet
model. Tellus 33, 1-11.
Vialov S.S. (1958): Regularities of glacial shields movement and the theory of plastic
viscous flow. International Association of Hydrology, Scientific Publication 47, 266-
275.
Weertman J. (1961): Stability of ice-age ice sheets. Journal of Geophysical Research
66, 3783-3792.
26

Problem HMB-feedback and response time (P 1)
Conservation of ice volume V can be expressed as
d V
d t
= d A H m
d t
= H m d A
d t
+ A dH m
d t
. (P 1.1)
Here A is the glacier area and H m the mean ice thickness. Now we write
V = V 0 + V', A = A 0 +A' and H m = H m,0 + H m ', where the reference state is defined
as (V 0 , A 0 , H m,0 ). By neglecting higher-order terms we then obtain the perturbation
equation
d V'
d t
= H m,0 d A'
d t
+ A 0 dH m '
d t
. (P 1.2)
Again we assume:
A' = w f L' , (P 1.3)
Here w f is the characteristic width of the glacier tongue. Now we include the height-mass
balance feedback by assuming that the change in mean thickness is proportional to the
change in glacier length:
H m ' = η L' . (P 1.4)
So for the change of ice volume with time we find
d V'
d t
= η A 0 + w f H m,0 d L'
d t
= amount of mass added . (P 1.5)
We now have three contributions to the mass added to or removed from the glacier:
change of volume = A 0 B' + A 0 β H m ' + w f B f L' = A 0 B' + A 0 β η' + w f B f L' .
(P 1.6)
B' is the perturbation of the balance rate (constant over the glacier), β the balance gradient
and B f the characteristic balance rate at the glacier front. The first term is simply the
amount of ice added or removed by the balance perturbation on the reference glacier area.
The second term represents the feedback between balance rate and surface elevation. The
third term is the amount of ice lost or gained because the length of the glacier deviates
from that of the reference state. Combining the equations yields
27

d L'
d t
=
A 0
η A 0 + w f H m,0
B' + η β A 0 + w f B f
η A 0 + w f H m,0
L' . (P 1.7)
The equilibrium state is obtained by setting the time derivative to zero. This gives:
A
L' = - 0 B'
. (P 1.8)
η β A 0 + w f B f
The response time for glacier length according to this model is apparently
t L = - η A 0 + w f H m,0
η β A 0 + w f B f
. (P 1.9)
Some examples on response times are given in the table below. The quantities marked
with • are the input data. Eq. (6.1) has been used to calculate H o and η. The response
time is then obtained from eq. (P 1.9). The last column shows the response time in the
absence of the HMB-feedback. Apparently for most glaciers the HMB-feedback is
important.
•Ao
•Lo
•s
•β
Ho
η
•W f
•B f
t L
t L (η=0)
(km 2 )
(km)
(m 2 yr -1 )
(m)
(m)
(m yr -1 )
(yr)
(yr)
Nigardsbreen 48.8 10.4 0.13 0.0078 125 0.0060 500 -10 129 13
Abramov Glac. 25.9 9.1 0.09 0.0055 136 0.0075 700 -4 167 34
Rhonegletscher 18.5 9.8 0.13 0.0066 122 0.0062 900 -5 60 24
Franz-Josef Gl. 36 11.4 0.21 0.0125 107 0.0047 600 -22 21 5
Aletschgletscher 86.8 24.7 0.1 0.0066 215 0.0044 1200 -8 90 27
Ob. Grindelw.gl. 10.1 5.0 0.25 0.0066 65 0.0065 500 -7 32 9
28

Problem ice-sheet profile (P 2)
For the steady-state radial velocity we now have:
d (r H U r ) = b R r2
R
d r → U r =
b R
3 R H-1 r 2 (P 2.1)
This can be combined again with the expression for the vertical mean ice velcocity in the
case of simple shearing flow:
U r = - A* H n+1 d H
d r
n
(P 2.2)
We get
1+2/n
H
dH
dr
= -
1/n
b R
3 R A*
r 2/n (P 2.3)
Integration with respect to r yields
n
2n+2 H 2+2/n - H 0
2+2/n = - n
n+2
1/n
b R
3 R A*
r 1+2/n , (P 2.4)
or
H 2+2/n - H 0
2+2/n = - 2n+2
n+2
1/n
b R
3 R A*
r 1+2/n . (P 2.5)
Next we have to apply a boundary condition, namely H = 0 at r = R. This leads to a
relation between the ice thickness in the centre and the ice-sheet radius:
H 0
2+2/n = 2n+2
n+2
1/n
b R
3 R A*
R 1+2/n . (P 2.6)
The final solution then becomes
H(r) = 2n+2
n+2
n/(2n+2)
b R /(3 R A*) 1/(2n+2) R 1+2/n - r 1+2/n n/(2n+2) . (P 2.7)
For n=3 we have:
29

H(r) = 8/5 3/8 b R /(3 R A*) 1/8 R 5/3 - r 5/3 3/8 (P 2.8)
3500
3000
2500
b=const
b=b R
r/R
2000
H (m)
1500
1000
500
0
-500
0 500 1000 1500 2000
r (km)
30

Problem West Antarcic ice sheet (P 3)
The mass budget equation is obtained by setting the total accumulation equal to the total
flux across the grounding line. This flux equals the ice velocity times the outlet cross
section. Therefore
π a R = 2 π R f * ρ w
ρ i
d 2 = 2 π R f d 2 (P 3.1)
The water depth at the grounding line equals b 0 - s R, so we have
d 2 = b 0
2 + s
2 R
2 - 2 b0 s R (P 3.2)
Combining yields
2 f s 2 R 2 - (4 b 0 f s + a) R + 2 f b 0
2 = 0 (P 3.3)
Special case: b 0 = 0 (a purely marine ice sheet). So the highest point of the continent is
just at sea level. Eq. (P 2.3) reduces to
2 f s 2 R 2 - a R = 0 (P 3.4)
→ R = a
2 f s 2 (P 3.5)
Application to the WAIS (R = 600 km, f = 1 yr -1 , a = 0.25 m yr -1 ):
1/2
→ s = a = 0.25
1/2 = 0.00046 (P 3.6)
2 f R 2 x 1 x 600,000
Sensitivity to changes in accumulation rate:
∂R
∂a = 1
2 f s 2 = 2.37x106 = 23.7 km/% (P 3.7)
31