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The Finite Element Method in Geodynamics
C. Thieulot
December 31, 2018
Contents
1 Introduction
1.1 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Essential literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Installation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
4
4
2 The physical equations of Fluid Dynamics
2.1 the Boussinesq approximation: an Incompressible flow . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
8
3 The Finite Element Method
3.1 Numerical integration . . . . .
3.1.1 in 1D - theory . . . . .
3.1.2 in 1D - examples . . . .
3.1.3 in 2D/3D - theory . . .
3.2 The mesh . . . . . . . . . . . .
3.3 Elements and basis functions in
3.4 Elements and basis functions in
3.4.1 The Q1 basis in 2D . . .
3.4.2 The Q2 basis in 2D . . .
3.5 The penalty approach . . . . .
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9
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4 Additional techniques
4.1 The method of manufactured solutions . . . . .
4.2 Sparse storage . . . . . . . . . . . . . . . . . . .
4.3 Mesh generation . . . . . . . . . . . . . . . . .
4.4 The value of the timestep . . . . . . . . . . . .
4.5 Tracking materials . . . . . . . . . . . . . . . .
4.6 Visco-Plasticity . . . . . . . . . . . . . . . . . .
4.7 Picard and Newton . . . . . . . . . . . . . . . .
4.8 The choice of solvers . . . . . . . . . . . . . . .
4.9 The SUPG formulation for the energy equation
4.10 Tracking materials and/or interfaces . . . . . .
4.11 Dealing with a free surface . . . . . . . . . . . .
4.12 Pressure normalisation . . . . . . . . . . . . . .
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18
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18
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1D
2D
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5 fieldstone: simple analytical solution
20
6 fieldstone: Stokes sphere
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7 fieldstone: Convection in a 2D box
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8 fieldstone: solcx benchmark
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9 fieldstone: solkz benchmark
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10 fieldstone: solvi benchmark
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11 fieldstone: the indentor benchmark
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12 fieldstone: the annulus benchmark
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1
13 fieldstone: stokes sphere (3D) - penalty
35
14 fieldstone: stokes sphere (3D) - mixed formulation
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15 fieldstone: consistent pressure recovery
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16 fieldstone: the Particle in Cell technique (1) - the effect of averaging
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17 fieldstone: solving the full saddle point problem
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18 fieldstone: solving the full saddle point problem in 3D
18.0.1 Constant viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18.0.2 Variable viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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19 fieldstone: solving the full saddle point problem with Q2 × Q1 elements
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20 fieldstone: The non-conforming Q1 × P0 element
50
21 fieldstone: The stabilised Q1 × Q1 element
51
22 fieldstone: compressible flow (1)
52
23 fieldstone: compressible flow (2)
23.1 The physics . . . . . . . . . . . . . . . . . .
23.2 The numerics . . . . . . . . . . . . . . . . .
23.3 The experimental setup . . . . . . . . . . .
23.4 Scaling . . . . . . . . . . . . . . . . . . . . .
23.5 Conservation of energy 1 . . . . . . . . . . .
23.5.1 under BA and EBA approximations
23.5.2 under no approximation at all . . . .
23.6 Conservation of energy 2 . . . . . . . . . . .
23.7 The problem of the onset of convection . . .
23.8 results - BA - Ra = 104 . . . . . . . . . . .
23.9 results - BA - Ra = 105 . . . . . . . . . . .
23.10results - BA - Ra = 106 . . . . . . . . . . .
23.11results - EBA - Ra = 104 . . . . . . . . . .
23.12results - EBA - Ra = 105 . . . . . . . . . .
23.13Onset of convection . . . . . . . . . . . . . .
A The main codes in computational
A.1 ADELI . . . . . . . . . . . . . . .
A.2 ASPECT . . . . . . . . . . . . .
A.3 CITCOMS and CITCOMCU . .
A.4 DOUAR . . . . . . . . . . . . . .
A.5 GAIA . . . . . . . . . . . . . . .
A.6 GALE . . . . . . . . . . . . . . .
A.7 GTECTON . . . . . . . . . . . .
A.8 ELVIS . . . . . . . . . . . . . . .
A.9 ELEFANT . . . . . . . . . . . . .
A.10 ELLIPSIS . . . . . . . . . . . . .
A.11 FANTOM . . . . . . . . . . . . .
A.12 FLUIDITY . . . . . . . . . . . .
A.13 LAMEM . . . . . . . . . . . . . .
A.14 MILAMIN . . . . . . . . . . . . .
A.15 PARAVOZ/FLAMAR . . . . . .
A.16 PTATIN . . . . . . . . . . . . . .
A.17 RHEA . . . . . . . . . . . . . . .
A.18 SEPRAN . . . . . . . . . . . . .
A.19 SOPALE . . . . . . . . . . . . .
A.20 STAGYY . . . . . . . . . . . . .
A.21 SULEC . . . . . . . . . . . . . .
A.22 TERRA . . . . . . . . . . . . . .
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geodynamics
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2
A.23 UNDERWORLD 1&2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
B fieldstone.py
69
C fieldstone stokes sphere.py
70
D fieldstone convection box.py
70
E fieldstone solcx.py
71
F fieldstone indentor.py
72
G fieldstone saddlepoint.py
73
3
1
Introduction
WARNING: this is work in progress
practical hands-on approach
as little as possible jargon
no mathematical proof
no optimised codes (readability over efficiency). avoiding as much as possible to have to look elsewhere. very
sequential, so unavoidable repetitions (jacobian, shape functions)
FE is one of several methods.
All the python scripts and this document are freely available at https://github.com/cedrict/fieldstone.
1.1
Acknowledgments
Jean Braun, Philippe Fullsack, Arie van den Berg. Lukas van de Wiel. Robert Myhill. Menno, Anne Too many BSc
and MSc students to name indivisually, although Job Mos did produce the very first version of fieldstone as part of
his MSc thesis. The ASPECT team in general and Wolfgang Bangerth and Timo Heister in particular.
1.2
Essential literature
a)
1.3
b)
c)
d)
Installation
python3.6 -m pip install --user numpy scipy matplotlib
4
e)
2
The physical equations of Fluid Dynamics
Symbol
t
x, y, z
v
ρ
η
λ
T
∇
∇·
p
ε̇(v)
α
k
Cp
H
βT
meaning
Time
Cartesian coordinates
velocity vector
mass density
dynamic viscosity
penalty parameter
temperature
gradient operator
divergence operator
pressure
strain rate tensor
thermal expansion coefficient
thermal conductivity
Heat capacity
intrinsic specific heat production
isothermal compressibility
unit
s
m
m· s−1
kg/m3
Pa· s
Pa· s
K
m−1
m−1
Pa
s−1
K−1
W/(m · K)
J/K
W/kg
Pa−1
Let us start from the heat transport equation as shown in Schubert, Turcotte and Olson [47]:
ρCp
DT
Dp
− αT
= ∇ · k∇T + Φ + ρH
Dt
Dt
with
∂T
Dp
∂p
DT
=
+ v · ∇T
=
+ v · ∇p
Dt
∂t
Dt
∂t
In order to arrive at the set of equations that ASPECT solves, we need to neglect the ∂p/∂t. WHY? Also, their
definition of the shear heating term Φ is:
Φ = kB (∇ · v)2 + 2η ε̇d : ε̇d
For many fluids the bulk viscosity kB is very small and is often taken to be zero, an assumption known as the Stokes
assumption: kB = λ + 2η/3 = 0. Note that η is the dynamic viscosity and λ the second viscosity. Also,
τ = 2η ε̇ + λ(∇ · v)1
but since kB = λ + 2η/3 = 0, then λ = −2η/3 so
2
τ = 2η ε̇ − η(∇ · v)1 = 2η ε̇d
3
[from aspect manual] We focus on the system of equations in a d = 2- or d = 3-dimensional domain Ω that describes
the motion of a highly viscous fluid driven by differences in the gravitational force due to a density that depends on
the temperature. In the following, we largely follow the exposition of this material in Schubert, Turcotte and Olson
[47].
Specifically, we consider the following set of equations for velocity u, pressure p and temperature T :
1
in Ω,
(1)
−∇ · 2η ε̇(v) − (∇ · v)1 + ∇p = ρg
3
∇ · (ρv) = 0
ρCp
∂T
+ v · ∇T
∂t
in Ω,
(2)
− ∇ · k∇T = ρH
1
+ 2η ε̇(v) − (∇ · v)1
3
1
: ε̇(v) − (∇ · v)1
3
+ αT (v · ∇p)
∂X
+ ρT ∆S
+ v · ∇X
∂t
where ε̇(u) = 21 (∇u + ∇uT ) is the symmetric gradient of the velocity (often called the strain rate).
5
(3)
in Ω,
In this set of equations, (82) and (83) represent the compressible Stokes equations in which v = v(x, t) is the
velocity field and p = p(x, t) the pressure field. Both fields depend on space x and time t. Fluid flow is driven by the
gravity force that acts on the fluid and that is proportional to both the density of the fluid and the strength of the
gravitational pull.
Coupled to this Stokes system is equation (84) for the temperature field T = T (x, t) that contains heat conduction
terms as well as advection with the flow velocity v. The right hand side terms of this equation correspond to
• internal heat production for example due to radioactive decay;
• friction (shear) heating;
• adiabatic compression of material;
• phase change.
The last term of the temperature equation corresponds to the latent heat generated or consumed in the process of
phase change of material. In what follows we will not assume that no phase change takes place so that we disregard
this term altogether.
6
A note on the shear heating term Φ: In many publications, Φ is given by Φ = τij ∂j ui = τ : ∇v.
Φ
=
τij ∂j ui
(4)
=
2η ε̇dij ∂j ui
(5)
1
= 2η ε̇dij ∂j ui + ε̇dji ∂i uj
2
1
= 2η ε̇dij ∂j ui + ε̇dij ∂i uj
2
1
= 2η ε̇dij (∂j ui + ∂i uj )
2
= 2η ε̇dij ε̇ij
=
(6)
(7)
(8)
(9)
d
2η ε̇ : ε̇
1
= 2η ε̇d : ε̇d + (∇ · v)1
3
=
=
(10)
(11)
2η ε̇d : ε̇d + 2η ε̇d : 1(∇ · v)
d
(12)
d
2η ε̇ : ε̇
(13)
Finally
Φ = τ : ∇v = 2η ε̇d : ε̇d = 2η (ε̇dxx )2 + (ε̇dyy )2 + 2(ε̇dxy )2
7
2.1
the Boussinesq approximation: an Incompressible flow
[from aspect manual] The Boussinesq approximation assumes that the density can be considered constant in all
occurrences in the equations with the exception of the buoyancy term on the right hand side of (82). The primary
result of this assumption is that the continuity equation (83) will now read
∇·v =0
This implies that the strain rate tensor is deviatoric. Under the Boussinesq approximation, the equations are much
simplified:
−∇ · [2η ε̇(v)] + ∇p = ρg
ρ0 C p
∂T
+ v · ∇T
∂t
in Ω,
(14)
∇ · (ρv) = 0
in Ω,
(15)
− ∇ · k∇T = ρH
in Ω
(16)
Note that all terms on the rhs of the temperature equations have disappeared, with the exception of the source term.
8
3
The Finite Element Method
3.1
Numerical integration
As we will see later, using the Finite Element method to solve problems involves computing integrals which are more
often than not too complex to be computed analytically/exactly. We will then need to compute them numerically.
[wiki] In essence, the basic problem in numerical integration is to compute an approximate solution to a definite
integral
Z b
f (x)dx
a
to a given degree of accuracy. This problem has been widely studied [?] and we know that if f (x) is a smooth function,
and the domain of integration is bounded, there are many methods for approximating the integral to the desired
precision.
There are several reasons for carrying out numerical integration.
• The integrand f (x) may be known only at certain points, such as obtained by sampling. Some embedded systems
and other computer applications may need numerical integration for this reason.
• A formula for the integrand may be known, but it may be difficult or impossible to find an antiderivative that is
an elementary function. An example of such an integrand is f (x) = exp(−x2 ), the antiderivative of which (the
error function, times a constant) cannot be written in elementary form.
• It may be possible to find an antiderivative symbolically, but it may be easier to compute a numerical approximation than to compute the antiderivative. That may be the case if the antiderivative is given as an infinite
series or product, or if its evaluation requires a special function that is not available.
3.1.1
in 1D - theory
The simplest method of this type is to let the interpolating function be a constant function (a polynomial of degree
zero) that passes through the point ((a + b)/2, f ((a + b)/2)).
This is called the midpoint rule or rectangle rule.
Z
b
f (x)dx ' (b − a)f (
a
a+b
)
2
insert here figure
The interpolating function may be a straight line (an affine function, i.e. a polynomial of degree 1) passing through
the points (a, f (a)) and (b, f (b)).
This is called the trapezoidal rule.
Z b
f (a) + f (b)
f (x)dx ' (b − a)
2
a
insert here figure
For either one of these rules, we can make a more accurate approximation by breaking up the interval [a, b] into
some number n of subintervals, computing an approximation for each subinterval, then adding up all the results. This
is called a composite rule, extended rule, or iterated rule. For example, the composite trapezoidal rule can be stated
as
!
Z b
n−1
b − a f (a) X
b−a
f (b)
f (x)dx '
+
f (a + k
)+
n
2
n
2
a
k=1
where the subintervals have the form [kh, (k + 1)h], with h = (b − a)/n and k = 0, 1, 2, . . . , n − 1.
a)
b)
The interval [−2, 2] is broken into 16 sub-intervals. The blue lines correspond to the approximation of the red curve
by means of a) the midpoint rule, b) the trapezoidal rule.
9
There are several algorithms for numerical integration (also commonly called ’numerical quadrature’, or simply
’quadrature’) . Interpolation with polynomials evaluated at equally spaced points in [a, b] yields the NewtonCotes
formulas, of which the rectangle rule and the trapezoidal rule are examples. If we allow the intervals between interpolation points to vary, we find another group of quadrature formulas, such as the Gauss(ian) quadrature formulas.
A Gaussian quadrature rule is typically more accurate than a NewtonCotes rule, which requires the same number of
function evaluations, if the integrand is smooth (i.e., if it is sufficiently differentiable).
An n−point Gaussian quadrature rule, named after Carl Friedrich Gauss, is a quadrature rule constructed to
yield an exact result for polynomials of degree 2n − 1 or less by a suitable choice of the points xi and weights wi for
i = 1, . . . , n.
The domain of integration for such a rule is conventionally taken as [−1, 1], so the rule is stated as
Z
+1
f (x)dx =
−1
n
X
wiq f (xiq )
iq =1
In this formula the xiq coordinate is the i-th root of the Legendre polynomial Pn (x).
It is important to note that a Gaussian quadrature will only produce good results if the function f (x) is well
approximated by a polynomial function within the range [−1, 1]. As a consequence, the method is not, for example,
suitable for functions with singularities.
Gauss-Legendre points and their weights.
As shown in the above table, it can be shown that the weight values must fulfill the following condition:
X
wiq = 2
(17)
iq
and it is worth noting that all quadrature point coordinates are symmetrical around the origin.
Since most quadrature formula are only valid on a specific interval, we now must address the problem of their use
outside of such intervals. The solution turns out to be quite simple: one must carry out a change of variables from
the interval [a, b] to [−1, 1].
We then consider the reduced coordinate r ∈ [−1, 1] such that
2
(x − a) − 1
b−a
r=
This relationship can be reversed such that when r is known, its equivalent coordinate x ∈ [a, b] can be computed:
b−a
(1 + r) + a
2
x=
From this it follows that
dx =
and then
Z
b
f (x)dx =
a
b−a
2
Z
b−a
dr
2
+1
f (r)dr '
−1
n
b−a X
wi f (riq )
2 i =1 q
q
10
3.1.2
in 1D - examples
example 1 Since we know how to carry out any required change of variables, we choose for simplicity a = −1,
b = +1. Let us take for example f (x) = π. Then we can compute the integral of this function over the interval [a, b]
exactly:
Z +1
Z +1
I=
f (x)dx = π
dx = 2π
−1
−1
We can now use a Gauss-Legendre formula to compute this same integral:
Z
+1
f (x)dx =
Igq =
−1
nq
X
wiq f (xiq ) =
iq =1
nq
X
wiq π = π
iq =1
nq
X
wiq = 2π
iq =1
| {z }
=2
where we have used the property of the weight values of Eq.(17). Since the actual number of points was never specified,
this result is valid for all quadrature rules.
example 2
Let us now take f (x) = mx + p and repeat the same exercise:
+1
Z
Z
−1
−1
Z
+1
Igq =
f (x)dx =
−1
+1
f (x)dx =
I=
nq
X
wiq f (xiq ) =
1
(mx + p)dx = [ mx2 + px]+1
−1 = 2p
2
nq
X
wiq (mxiq + p) = m
wiq xiq +p
iq =1
iq =1
iq =1
nq
X
|
nq
X
wiq = 2p
iq =1
{z
=0
}
| {z }
=2
since the quadrature points are symmetric w.r.t. to zero on the x-axis. Once again the quadrature is able to compute
the exact value of this integral: this makes sense since an n-point rule exactly integrates a 2n − 1 order polynomial
such that a 1 point quadrature exactly integrates a first order polynomial like the one above.
example 3
Let us now take f (x) = x2 . We have
+1
Z
I=
Z
+1
f (x)dx =
−1
−1
and
Z
+1
Igq =
f (x)dx =
−1
nq
X
1
2
x2 dx = [ x3 ]+1
−1 =
3
3
wiq f (xiq ) =
nq
X
iq =1
iq =1
(1)
• nq = 1: xiq = 0, wiq = 2. Igq = 0
√
√
(1)
(2)
(1)
(2)
• nq = 2: xq = −1/ 3, xq = 1/ 3, wq = wq = 1. Igq =
2
3
• It also works ∀nq > 2 !
3.1.3
in 2D/3D - theory
Let us now turn to a two-dimensional integral of the form
Z +1 Z +1
I=
f (x, y)dxdy
−1
−1
The equivalent Gaussian quadrature writes:
Igq '
nq nq
X
X
f (xiq , yjq )wiq wjq
iq =1 jq
11
wiq x2iq
3.2
The mesh
3.3
Elements and basis functions in 1D
3.4
Elements and basis functions in 2D
Let us for a moment consider a single quadrilateral element in the xy-plane, as shown on the following figure:
Let us assume that we know the values of a given field u at the vertices. For a given point M inside the element in
the plane, what is the value of the field u at this point? It makes sense to postulate that uM = u(xM , yM ) will be
given by
uM = φ(u1 , u2 , u3 , u4 , xM , yM )
where φ is a function to be determined. Although φ is not unique, we can decide to express the value uM as a weighed
sum of the values at the vertices ui . One option could be to assign all four vertices the same weight, say 1/4 so that
uM = (u1 + u2 + u3 + u4 )/4, i.e. uM is simply given by the arithmetic mean of the vertices values. This approach
suffers from a major drawback as it does not use the location of point M inside the element. For instance, when
(xM , yM ) → (x2 , y2 ) we expect uM → u2 .
In light of this, we could now assume that the weights would depend on the position of M in a continuous fashion:
u(xM , yM ) =
4
X
Ni (xM , yM ) ui
i=1
where the Ni are continous (”well behaved”) functions which have the property:
Ni (xj , yj ) = δij
or, in other words:
N3 (x1 , y1 )
=
0
(18)
N3 (x2 , y2 )
=
0
(19)
N3 (x3 , y3 )
=
1
(20)
N3 (x4 , y4 )
=
0
(21)
The functions Ni are commonly called basis functions.
Omitting the M subscripts for any point inside the element, the velocity components u and v are given by:
û(x, y)
=
4
X
Ni (x, y) ui
(22)
Ni (x, y) vi
(23)
i=1
v̂(x, y)
=
4
X
i=1
Rather interestingly, one can now easily compute velocity gradients (and therefore the strain rate tensor) since we
have assumed the basis functions to be ”well behaved” (in this case differentiable):
4
˙xx (x, y)
=
∂u X ∂Ni
=
ui
∂x
∂x
i=1
˙yy (x, y)
=
∂v X ∂Ni
=
vi
∂y
∂y
i=1
˙xy (x, y)
=
1 ∂u 1 ∂v
1 X ∂Ni
1 X ∂Ni
+
=
ui +
vi
2 ∂y
2 ∂x
2 i=1 ∂y
2 i=1 ∂x
(24)
4
(25)
4
12
4
(26)
How we actually obtain the exact form of the basis functions is explained in the coming section.
3.4.1
The Q1 basis in 2D
In this section, we place ourselves in the most favorables case, i.e. the element is a square defined by −1 < r < 1,
−1 < s < 1 in the Cartesian coordinates system (r, s):
add corner numbering
This element is commonly called the reference element. How we go from the (x, y) coordinate system to the (r, s)
once and vice versa will be dealt later on. For now, the basis functions in the above reference element and in the
reduced coordinates system (r, s) are given by:
N1 (r, s)
=
0.25(1 − r)(1 − s)
N2 (r, s)
=
0.25(1 + r)(1 − s)
N3 (r, s)
=
0.25(1 + r)(1 + s)
N4 (r, s)
=
0.25(1 − r)(1 + s)
The partial derivatives of these functions with respect to r ans s automatically follow:
∂N1
(r, s)
∂r
∂N2
(r, s)
∂r
∂N3
(r, s)
∂r
∂N4
(r, s)
∂r
∂N1
(r, s)
∂s
∂N2
(r, s)
∂s
∂N3
(r, s)
∂s
∂N4
(r, s)
∂s
=
−0.25(1 − s)
=
+0.25(1 − s)
=
+0.25(1 + s)
=
−0.25(1 + s)
=
−0.25(1 − r)
=
−0.25(1 + r)
=
+0.25(1 + r)
=
+0.25(1 − r)
Let us go back to Eq.(23). And let us assume that the function v(r, s) = C so that vi = C for i = 1, 2, 3, 4. It then
follows that
v̂(r, s) =
4
X
Ni (r, s) vi = C
i=1
4
X
Ni (r, s) = C[N1 (r, s) + N2 (r, s) + N3 (r, s) + N4 (r, s)] = C
i=1
This is a very important property: if the v function used to assign values at the vertices is constant, then the value of
v̂ anywhere in the element is exactly C. If we now turn to the derivatives of v with respect to r and s:
4
4
X ∂Ni
X ∂Ni
∂v̂
(r, s) =
(r, s) vi = C
(r, s) = C [−0.25(1 − s) + 0.25(1 − s) + 0.25(1 + s) − 0.25(1 + s)] = 0
∂r
∂r
∂r
i=1
i=1
13
4
4
X ∂Ni
X ∂Ni
∂v̂
(r, s) =
(r, s) vi = C
(r, s) = C [−0.25(1 − r) − 0.25(1 + r) + 0.25(1 + r) + 0.25(1 − r)] = 0
∂s
∂s
∂s
i=1
i=1
We reassuringly find that the derivative of a constant field anywhere in the element is exactly zero.
If we now choose v(r, s) = ar + bs with a and b two constant scalars, we find:
v̂(r, s)
=
4
X
Ni (r, s) vi
(27)
Ni (r, s)(ari + bsi )
(28)
i=1
=
4
X
i=1
= a
4
X
Ni (r, s)ri +b
(29)
Ni (r, s)si
i=1
i=1
|
4
X
{z
}
r
{z
|
s
}
= a [0.25(1 − r)(1 − s)(−1) + 0.25(1 + r)(1 − s)(+1) + 0.25(1 + r)(1 + s)(+1) + 0.25(1 − r)(1 + s)(−1)]
+ b [0.25(1 − r)(1 − s)(−1) + 0.25(1 + r)(1 − s)(−1) + 0.25(1 + r)(1 + s)(+1) + 0.25(1 − r)(1 + s)(+1)]
= a [−0.25(1 − r)(1 − s) + 0.25(1 + r)(1 − s) + 0.25(1 + r)(1 + s) − 0.25(1 − r)(1 + s)]
+ b [−0.25(1 − r)(1 − s) − 0.25(1 + r)(1 − s) + 0.25(1 + r)(1 + s) + 0.25(1 − r)(1 + s)]
= ar + bs
(30)
verify above eq. This set of bilinear shape functions is therefore capable of exactly representing a bilinear field. The
derivatives are:
∂v̂
(r, s)
∂r
=
4
X
∂Ni
i=1
=
a
∂r
(r, s) vi
4
X
∂Ni
i=1
∂r
(r, s)ri + b
(31)
4
X
∂Ni
i=1
∂r
(r, s)si
=
a [−0.25(1 − s)(−1) + 0.25(1 − s)(+1) + 0.25(1 + s)(+1) − 0.25(1 + s)(−1)]
+
b [−0.25(1 − s)(−1) + 0.25(1 − s)(−1) + 0.25(1 + s)(+1) − 0.25(1 + s)(+1)]
a
[(1 − s) + (1 − s) + (1 + s) + (1 + s)]
4
b
[(1 − s) − (1 − s) + (1 + s) − (1 + s)]
4
a
=
+
=
(32)
(33)
Here again, we find that the derivative of the bilinear field inside the element is exact: ∂∂rv̂ = ∂v
∂r .
However, following the same methodology as above, one can easily prove that this is no more true for polynomials
of degree strivtly higher than 1. This fact has serious consequences: if the solution to the problem at hand is for
instance a parabola, the Q1 shape functions cannot represent the solution properly, but only by approximating the
parabola in each element by a line. As we will see later, Q2 basis functions can remedy this problem by containing
themselves quadratic terms.
3.4.2
3.5
The Q2 basis in 2D
The penalty approach
In order to impose the incompressibility constraint, two widely used procedures are available, namely the Lagrange
multiplier method and the penalty method [2, 25]. The latter is implemented in elefant, which allows for the
elimination of the pressure variable from the momentum equation (resulting in a reduction of the matrix size).
Mathematical details on the origin and validity of the penalty approach applied to the Stokes problem can for
instance be found in [10], [43] or [23].
The penalty formulation of the mass conservation equation is based on a relaxation of the incompressibility constraint and writes
p
(34)
∇·v+ =0
λ
where λ is the penalty parameter, that can be interpreted (and has the same dimension) as a bulk viscosity. It is
equivalent to say that the material is weakly compressible. It can be shown that if one chooses λ to be a sufficiently
14
large number, the continuity equation ∇ · v = 0 will be approximately satisfied in the finite element solution. The
value of λ is often recommended to be 6 to 7 orders of magnitude larger than the shear viscosity [17, 26].
Equation (34) can be used to eliminate the pressure in Eq. (??) so that the mass and momentum conservation
equations fuse to become :
∇ · (2η ε̇(v)) + λ∇(∇ · v) = ρg = 0
(35)
[36] have established the equivalence for incompressible problems between the reduced integration of the penalty
term and a mixed Finite Element approach if the pressure nodes coincide with the integration points of the reduced
rule.
In the end, the elimination of the pressure unknown in the Stokes equations replaces the original saddle-point
Stokes problem [3] by an elliptical problem, which leads to a symmetric positive definite (SPD) FEM matrix. This is
the major benefit of the penalized approach over the full indefinite solver with the velocity-pressure variables. Indeed,
the SPD character of the matrix lends itself to efficient solving stragegies and is less memory-demanding since it is
sufficient to store only the upper half of the matrix including the diagonal [22] . ToDo: list codes which use this
approach.
15
The stress tensor σ is symmetric (i.e. σij = σji ). For simplicity I will now focus on a Stokes flow in two dimensions.
Since the penalty formulation is only valid for incompressible flows, then ˙ = ˙ d so that the d superscript is
ommitted in what follows. The stress tensor can also be cast in vector format:
σxx
−p
˙xx
σyy = −p + 2η ˙yy
σxy
0
˙xy
˙xx + ˙yy
˙xx
= λ ˙xx + ˙yy + 2η ˙yy
0
˙xy
∂u
∂x
2 0 0
1 1 0
∂v
1 1 0 +η 0 2 0 ·
= λ
∂y
0 0 1
0 0 0
{z
}
{z
}
|
|
∂u
+ ∂v
K
Remember that
∂y
C
4
∂x
4
∂u X ∂Ni
=
ui
∂x
∂x
i=1
∂v X ∂Ni
=
vi
∂y
∂y
i=1
4
4
X
X
∂u ∂v
∂Ni
∂Ni
+
=
ui +
vi
∂y
∂x
∂y
∂x
i=1
i=1
so that
∂u
∂x
∂v
∂y
∂u
∂y
+
∂v
∂x
=
∂N1
∂x
0
∂N2
∂x
0
∂N3
∂x
0
∂N4
∂x
0
0
∂N1
∂y
0
∂N2
∂y
0
∂N3
∂y
0
∂N4
∂y
∂N1
∂y
∂N1
∂x
∂N2
∂y
∂N2
∂x
∂N3
∂y
∂N3
∂x
∂N3
∂y
∂N4
∂x
|
{z
B
·
}
|
u1
v1
u2
v2
u3
v3
u4
v4
{z
V
}
Finally,
σxx
~σ = σyy = (λK + ηC) · B · V
σxy
We will now establish the weak form of the momentum conservation equation. We start again from
∇·σ+b=0
For the Ni ’s ’regular enough’, we can write:
Z
Z
Ni ∇ · σdΩ +
Ωe
Ni bdΩ = 0
Ωe
We can integrate by parts and drop the surface term1 :
Z
Z
∇Ni · σdΩ =
Ωe
Ni bdΩ
Ωe
or,
Z
∂Ni
∂x
0
∂Ni
∂y
0
∂Ni
∂y
∂Ni
∂x
Ωe
1 We
Z
σxx
· σyy dΩ =
Ni bdΩ
Ωe
σxy
will come back to this at a later stage
16
Let i = 1, 2, 3, 4 and stack the resulting four equations on top of one another.
Z
∂N1
∂x
0
∂N1
∂y
∂N2
∂x
0
Ωe
Z
0
∂N2
∂y
∂N3
∂x
0
Ωe
Z
∂N1
∂x
∂N2
∂y
∂N2
∂x
∂N3
∂y
0
∂N3
∂y
∂N4
∂x
0
∂N3
∂x
∂N4
∂y
0
∂N4
∂y
∂N4
∂x
Ωe
Z
∂N1
∂y
0
Ωe
σxx
· σyy dΩ
σxy
σxx
· σyy dΩ
σxy
σxx
· σyy dΩ
σxy
σxx
· σyy dΩ
σxy
Z
=
N1
Ωe
Z
=
Ni
Ωe
Z
=
N3
Ωe
Z
=
N4
Ωe
bx
by
bx
by
bx
by
bx
by
dΩ
(36)
dΩ
(37)
dΩ
(38)
dΩ
(39)
We easily recognize B T inside the integrals! Let us define
NbT = (N1 bx , N1 by , ...N4 bx , N4 by )
then we can write
Z
σxx
B T · σyy dΩ =
Nb dΩ
Ωe
Ωe
σxy
Z
and finally:
Z
Z
T
B · [λK + ηC] · B · V dΩ =
Ωe
Nb dΩ
Ωe
Since V contains the velocities at the corners, it does not depend on the x or y coordinates so it can be taking outside
of the integral:
Z
Z
B T · [λK + ηC] · BdΩ · |{z}
V =
Nb dΩ
Ωe
Ωe
{z
} (8x1) | {z }
|
Ael (8×8)
Bel (8×1)
or,
Z
Z
Z
T
T
λB · K · BdΩ +
V =
Nb dΩ
ηB · C · BdΩ · |{z}
Ω
Ω
| Ωe
{z
} | e
{z
} (8x1) | e {z }
Aλ
el (8×8)
Aη
el (8×8)
INTEGRATION - MAPPING
1. partition domain Ω into elements Ωe , e = 1, ...nel .
2. loop over elements and for each element compute Ael , Bel
3. a node belongs to several elements
→ need to assemble Ael and Bel in A, B
4. apply boundary conditions
5. solve system: x = A−1 · B
6. visualise/analyse x
17
Bel (8×1)
4
Additional techniques
4.1
The method of manufactured solutions
4.2
Sparse storage
4.3
Mesh generation
4.4
The value of the timestep
4.5
Tracking materials
4.6
Visco-Plasticity
4.7
Picard and Newton
4.8
The choice of solvers
4.9
The SUPG formulation for the energy equation
4.10
Tracking materials and/or interfaces
4.11
Dealing with a free surface
4.12
Pressure normalisation
18
so much to do ...
write about impose bc on el matrix
Q3Q2
full compressible
total energy calculations
constraints
compositions, marker chain
van keken initial value with deformed mesh
free-slip bc on annulus and sphere . See for example p540 Gresho and Sani book.
non-linear rheologies (two layer brick spmw16, tosn15)
Picard vs Newton
markers
Schur complement approach
periodic boundary conditions
open boundary conditions
free surface
SUPG
produce fastest version possible for convection box
zaleski disk advection
all kinds of interesting benchmarks
Busse convection pb, compare with aspect
cvi !!!
pure elastic
including phase changes (w. R. Myhill)
derivatives on nodes
Nusselt
discontinuous galerkin
formatting of code style
navier-stokes ? (LUKAS)
pressure smoothing
compute strainrate in middle of element or at quad point for punch?
GEO1442 code
GEO1442 indenter setup in plane ?
in/out flow on sides for lith modelling
Problems to solve:
colorscale
better yet simple matrix storage ?
19
5
fieldstone: simple analytical solution
From [15]. In order to illustrate the behavior of selected mixed finite elements in the solution of stationary Stokes
flow, we consider a two-dimensional problem in the square domain Ω = [0, 1] × [0, 1], which possesses a closed-form
analytical solution. The problem consists of determining the velocity field v = (u, v) and the pressure p such that
−ν∆v + ∇p = b
∇·v =0
v=0
in Ω
in Ω
on Γ
where the fluid viscosity is taken as ν = 1. The components of the body force b are prescribed as
bx
(12 − 24y)x4 + (−24 + 48y)x3 + (−48y + 72y 2 − 48y 3 + 12)x2
=
+(−2 + 24y − 72y 2 + 48y 3 )x + 1 − 4y + 12y 2 − 8y 3
by
(8 − 48y + 48y 2 )x3 + (−12 + 72y − 72y 2 )x2
=
+(4 − 24y + 48y 2 − 48y 3 + 24y 4 )x − 12y 2 + 24y 3 − 12y 4
With this prescribed body force, the exact solution is
Note that the pressure obeys
features
R
Ω
u(x, y)
= x2 (1 − x)2 (2y − 6y 2 + 4y 3 )
v(x, y)
= −y 2 (1 − y)2 (2x − 6x2 + 4x3 )
p(x, y)
= x(1 − x) − 1/6
p dΩ = 0
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (no-slip)
• direct solver
• isothermal
• isoviscous
• analytical solution
20
vx
1.0
vy
1.0
0.010
0.8
0.8
0.005
0.4
0.4
0.2
0.6
0.8
0.01
x
0.6
0.8
1.0
yy
0.8
0.6
0.0000075
0.6
0.8
1.0
0.2
0.4
x
0.6
0.8
1.0
p pth
1.0
0.00010
0.0000050 0.8
0.00005
0.0000025
0.00000
0.6
0.00005
y
y
y
x
0.6
0.04
0.4
0.0000025
0.00010
0.2
0.00000500.2
0.00020
0.0
0.0
0.0000075
0.0
0.0
0.000010
0.4
0.04
0.4
0.000005
0.2
xy
0.0000000
0.000000
0.4
0.15
0.00
0.0
0.0
vy tyth
0.000005
1.0
0.02
1.0
0.8
0.8
0.2
0.03
0.6
0.6
0.4
0.02
x
x
0.6
0.01
0.4
0.4
0.02
0.01
0.2
0.2
0.8
0.02
0.000010
0.2
0.10
1.0
0.03
1.0
0.8
0.0
0.0
0.2
0.0
0.0
1.0
0.00
0.0
0.0
vx txth
1.0
0.8
0.2
0.03
0.4
0.6
0.4
0.02
0.2
x
0.6
0.01
0.2
0.4
y
y
0.00
0.4
0.2
0.8
0.02
0.6
0.010
1.0
0.03
0.8
0.0
0.0
0.0
0.0
1.0
xx
1.0
0.05
y
0.010
x
0.4
0.005
0.005
0.2
0.4
0.00
0.6
0.000
y
y
0.000
y
0.6
0.2
0.05
0.8
0.005
0.6
0.0
0.0
p
1.0
0.010
0.2
0.4
x
0.6
0.8
1.0
0.100000
0.00015
0.00025
0.2
0.4
x
0.6
0.8
1.0
velocity
pressure
x2
x1
0.010000
error
0.001000
0.000100
0.000010
0.000001
0.01
0.1
h
Quadratic convergence for velocity error, linear convergence for pressure error, as expected.
ToDo:
pressure normalisation?
different cmat, a la schmalholz
To go further:
21
1. make your own analytical solution
22
6
fieldstone: Stokes sphere
Viscosity and density directly computed at the quadrature points.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• buoyancy-driven flow
vx
1.0
0.8
0.8
0.0005
0.6
0.4
0.0005
0.0010
0.2
0.4
x
0.6
0.8
1.0
xx
1.0
0.8
0.2
0.6
0.2
0.0
0.0
0.2
0.2
0.002
0.2
0.4
x
0.6
0.8
0.4
0.0
0.0
1.0
yy
0.2
0.010
0.0
0.0
1.0
y
0.0050
0.2
0.010
0.2
0.4
x
0.6
0.8
density
0.4
x
0.6
0.8
0.005
0.010
0.015
0.2
0.4
x
0.6
0.8
1.0
viscosity
2.0
1.0
0.8
1.8
0.8
80
0.6
1.6
0.6
60
0.4
1.4
0.4
40
0.2
1.2
0.2
20
1.0
0.0
0.0
0.0025
0.2
0.000
0.0
0.0
1.0
y
0.0125
0.0075
0.005
100
y
0.0150
0.4
0.010
0.2
1.0
0.0175
0.0100
0.015
0.005
0.2
0.6
xy
0.4
0.005
0.8
1.0
y
0.000
0.4
x
0.8
0.6
y
y
0.4
0.8
x
0.6
0.8
0.6
0.6
0.4
0.005
0.6
0.4
0.2
1.0
0.010
0.005
0.2
0.0
0.4
0.8
0.000
p
0.4
0.001
1.0
0.010
0.8
0.0
0.0
0.001
0.000
0.4
0.2
1.0
y
y
0.0000
0.002
y
0.6
0.0
0.0
vy
1.0
0.0010
0.0
0.0
0.2
0.4
x
0.6
0.8
23
1.0
0.2
0.4
x
0.6
0.8
1.0
7
fieldstone: Convection in a 2D box
This benchmark deals with the 2-D thermal convection of a fluid of infinite Prandtl number in a rectangular closed
cell. In what follows, I carry out the case 1a, 1b, and 1c experiments as shown in [5]: steady convection with constant
viscosity in a square box.
The temperature is fixed to zero on top and to ∆T at the bottom, with reflecting symmetry at the sidewalls (i.e.
∂x T = 0) and there are no internal heat sources. Free-slip conditions are implemented on all boundaries.
The Rayleigh number is given by
αgy ∆T h3 ρ2 cp
αgy ∆T h3
=
(40)
Ra =
κν
kµ
In what follows, I use the following parameter values: Lx = Ly = 1,ρ0 = cP = k = µ = 1, T0 = 0, α = 10−2 ,
g = 102 Ra and I run the model with Ra = 104 , 105 and 106 .
The initial temperature field is given by
T (x, y) = (1 − y) − 0.01 cos(πx) sin(πy)
(41)
The perturbation in the initial temperature fields leads to a perturbation of the density field and sets the fluid in
motion.
Depending on the initial Rayleigh number, the system ultimately reaches a steady state after some time.
The Nusselt number (i.e. the mean surface temperature gradient over mean bottom temperature) is computed as
follows [5]:
R ∂T
∂y (y = Ly )dx
(42)
N u = Ly R
T (y = 0)dx
Note that in our case the denominator is equal to 1 since Lx = 1 and the temperature at the bottom is prescribed to
be 1.
Finally, the steady state root mean square velocity and Nusselt number measurements are indicated in Table ??
alongside those of [5] and [49]. (Note that this benchmark was also carried out and published in other publications
[54, 1, 20, 11, 34] but since they did not provide a complete set of measurement values, they are not included in the
table.)
4
Ra = 10
Ra = 105
Ra = 106
Vrms
Nu
Vrms
Nu
Vrms
Nu
Blankenbach et al
42.864947 ± 0.000020
4.884409 ± 0.000010
193.21454 ± 0.00010
10.534095 ± 0.000010
833.98977 ± 0.00020
21.972465 ± 0.000020
Tackley [49]
42.775
4.878
193.11
10.531
833.55
21.998
Steady state Nusselt number N u and Vrms measurements as reported in the literature.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• Boussinesq approximation
• direct solver
• non-isothermal
• buoyancy-driven flow
• isoviscous
• CFL-condition
24
0.4
20
0.2
0.2
0.4
x
0.6
0.8
40 0.0
0.0
1.0
0.4
100
0.8
x
0.6
0.8
1.0
yy
1.0
0
100
0.4
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.2
200 0.00.0
0.25
0.8
100
0.2
0.4
x
0.6
0.8
1.0
qy
1.0
1.00 0.4
3
0.2
1.25 0.2
2
y
0.4
x
0.6
0.8
1.0
1.50 0.0
0.0
1.75
1
0.2
0.4
x
0.6
0.8
1.0
0
0.6
0.4
0.4
y
0.6
0.0
400000 0.0
1.0
40
x
0.6
0.8
1.0
0.0
rho
0.998
0.996
0.6
0.4
0
0.994
10 0.2
0.2
0.4
x
0.6
0.8
1.0
20 0.0
0.0
30
80
7
70
6
0.2
0.4
x
0.6
0.8
1.0
0.992
5
50
40
30
4
3
20
2
10
0
0.4
0.8
20
10
0.2
0.2
1.0
30
60
5
4
0.4
0.8
xy
0.0
0.0
6
0.8
0.75 0.6
0.2
x
0.6
0.8
7
0.6
0.0
0.0
0.4
0.2
y
0.50
0.2
0.4
0
0.00
qx
1.0
0.0
0.0
0.6
0.4
100
0.8
200000
0.2
0.2
y
0.6
y
y
60
0.4
200 1.0
0.8
0.6
0
y
40
0.2
60
xx
1.0
0.6
20
0.2
T
2000000.8
Nu
0.4
0
y
y
0
0.8
20
0.6
400000
1.0
p
1.0
40
0.8
20
0.6
0.0
0.0
1.0
40
0.8
vrms
1.0
1.0
60
vy
y
60
vx
1
0.00 0.05 0.10 0.15 0.20 0.25 0.30
time
ToDo:
implement steady state criterion
reach steady state
do Ra=1e4, 1e5, 1e6
plot against blankenbach paper and aspect
look at critical Ra number
25
0.00 0.05 0.10 0.15 0.20 0.25 0.30
time
8
fieldstone: solcx benchmark
The SolCx benchmark is intended to test the accuracy of the solution to a problem that has a large jump in the
viscosity along a line through the domain. Such situations are common in geophysics: for example, the viscosity in a
cold, subducting slab is much larger than in the surrounding, relatively hot mantle material.
The SolCx benchmark computes the Stokes flow field of a fluid driven by spatial density variations, subject to a
spatially variable viscosity. Specifically, the domain is Ω = [0, 1]2 , gravity is g = (0, −1)T and the density is given by
ρ(x, y) = sin(πy) cos(πx)
(43)
Boundary conditions are free slip on all of the sides of the domain and the temperature plays no role in this benchmark.
The viscosity is prescribed as follows:
1
f or x < 0.5
µ(x, y) =
(44)
106 f or x > 0.5
Note the strongly discontinuous viscosity field yields a stagnant flow in the right half of the domain and thereby yields
a pressure discontinuity along the interface.
The SolCx benchmark was previously used in [18] (references to earlier uses of the benchmark are available there)
and its analytic solution is given in [57]. It has been carried out in [32] and [21]. Note that the source code which
evaluates the velocity and pressure fields for both SolCx and SolKz is distributed as part of the open source package
Underworld ([39], http://underworldproject.org).
In this particular example, the viscosity is computed analytically at the quadrature points (i.e. tracers are not
used to attribute a viscosity to the element). If the number of elements is even in any direction, all elements (and
their associated quadrature points) have a constant viscosity(1 or 106 ). If it is odd, then the elements situated at the
viscosity jump have half their integration points with µ = 1 and half with µ = 106 (which is a pathological case since
the used quadrature rule inside elements cannot represent accurately such a jump).
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• analytical solution
26
1.0
0.0010
0.8
0.002
0.8
0.6
0.001
0.6
0.6
y
0.0000
0.4
y
0.0005
0.0005
0.0010
0.2
0.0
0.0
0.0015
0.0
0.0
0.4
x
0.6
0.8
1.0
xx
1.0
0.0075
0.8
0.0025
y
0.0000
0.4
0.0
0.0
0.0050
0.0075
0.2
0.4
x
0.6
0.8
1.0
vx txth
1.0
0.4
x
0.6
0.8
0.0100
yy
0.0050
0.0075
0.4
x
0.6
0.8
1.0
vy tyth
0.6
0.8
1.0
0.005
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
p pth
0.010
0.020
0.4
0.00000200.4
0.005
0.0000025
0.2
0.0000030
0.0
0.0
0.010
0.010
0.000
y
0.0
0.0
0.6
0.000
0.2
0.005
0.000003 0.0
0.0
x
0.0100
0.005
0.4
0.6
0.0000015
0.000002 0.2
0.4
0.010
0.6
0.2
0.2
0.020
0.015
y
y
0.000001
0.025
0.0000005
0.8
0.0000010
0.000000
0.4
1.0
xy
0.0000000 1.0
0.000002 0.8
0.000001
0.8
0.6
0.0025
0.2
x
0.6
0.015
0.0025
0.2
0.4
0.8
0.0050
0.4
0.2
1.0
0.0100
0.0000
0.0
0.0
0.2
0.0
0.0
0.0075
0.000003 1.0
0.8
0.1
1.0
0.6
0.0025
0.2
0.2
y
0.6
0.0
0.4
0.002
0.8
0.0050
0.1
0.2
1.0
0.0100
0.2
0.001
0.2
0.2
0.000
0.4
p
1.0
0.003
y
0.8
vy
0.0015
y
vx
1.0
0.2
0.4
x
0.6
0.8
What we learn from this
27
1.0
0.015
0.2
0.4
x
0.6
0.8
1.0
0.020
9
fieldstone: solkz benchmark
The SolKz benchmark [44] is similar to the SolCx benchmark. but the viscosity is now a function of the space
coordinates:
µ(y) = exp(By) with B = 13.8155
(45)
It is however not a discontinuous function but grows exponentially with the vertical coordinate so that its overall
variation is again 106 . The forcing is again chosen by imposing a spatially variable density variation as follows:
ρ(x, y) = sin(2y) cos(3πx)
(46)
Free slip boundary conditions are imposed on all sides of the domain. This benchmark is presented in [57] as well and
is studied in [18] and [21].
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
y
0.6
0.4
0.2
0.2
0.4
x
0.6
0.8
vx txth
1.0
0.4
0.6
0.8
0.0
0.0
0.2
101
x
0.6
0.8
1.0
0.8
0.4
0.2
x
error
y
0.2
0.4
x
0.6
0.8
1.0
0.0008
0.0006
0.0004
0.0002
0.0000
0.0002
0.0004
0.0006
0.0008
0.6
0.8
10
4
10
5
0.4
10
6
0.2
10
7
0.0
0.0
1.0
0.00010000
0.00001000
0.00000100
0.00000010
0.01
p pth
0.6
0.00100000
0.00000001
1.0
0.8
velocity
pressure
x2
x1
0.01000000
0.8
0.0004
0.0003
0.0002
0.0001
0.0000
0.0001
0.0002
0.0003
0.0004
1.0
0.6
0.10000000
0.0
0.0
0.75
0.50
0.25
0.00
0.25
0.50
0.75
0.4
0.6
0.2
density
0.2
x
0.4
1.0
0.8
0.0
0.0
0.4
0.6
1.0
0.6
0.2
0.8
0.5
x
1.0
xy
1.0
0.5
0.4
0.8
0.2
1e 7
0.2
0.6
0.4
0.0
0.0
0.0
x
0.6
1.0
0.2
0.4
0.8
1.0
y
y
102
0.4
0.8
0.4
0.0
0.0
103
0.2
0.6
0.6
104
0.0
0.0
x
vy tyth
1.0
0.2
0.4
0.2
1.0
0.00075
0.00050
0.00025
0.00000
0.00025
0.00050
0.00075
0.8
105
0.4
1.0
yy
1.0
1.0
0.6
0.8
0.2
1.0
0.8
0.6
0.4
2
4
6
0.2
x
0.0
0.0
x
y
y
0.6
0.4
0.00006
0.6
1e 7
0.8
0.2
0.2
0.0
0.0
0.8
1.0
6
4
2
0
0.0
0.0
0.00004
0.4
0.075
0.050
0.025
0.000
0.025
0.050
0.075
0.4
0.00002
0.2
p
0.6
0.2
y
0.8
0.0
0.0
0.00000
0.4
1.0
0.00075
0.00050
0.00025
0.00000
0.00025
0.00050
0.00075
0.8
0.00002
0.6
xx
1.0
1.0
0.00004
y
0.4
0.8
0.00006
y
y
0.6
vy
1.0
y
0.8
0.000100
0.000075
0.000050
0.000025
0.000000
0.000025
0.000050
0.000075
0.000100
y
vx
1.0
0.1
h
28
0.2
0.4
x
0.6
0.8
1.0
10
fieldstone: solvi benchmark
Following SolCx and SolKz, the SolVi inclusion benchmark solves a problem with a discontinuous viscosity field, but
in this case the viscosity field is chosen in such a way that the discontinuity is along a circle. Given the regular nature
of the grid used by a majority of codes and the present one, this ensures that the discontinuity in the viscosity never
aligns to cell boundaries. This in turns leads to almost discontinuous pressures along the interface which are difficult to
represent accurately. [46] derived a simple analytic solution for the pressure and velocity fields for a circular inclusion
under simple shear and it was used in [13], [48], [18], [32] and [21].
Because of the symmetry of the problem, we only have to solve over the top right quarter of the domain (see Fig.
??a).
The analytical solution requires a strain rate boundary condition (e.g., pure shear) to be applied far away from the
inclusion. In order to avoid using very large domains and/or dealing with this type of boundary condition altogether,
the analytical solution is evaluated and imposed on the boundaries of the domain. By doing so, the truncation error
introduced while discretizing the strain rate boundary condition is removed.
A characteristic of the analytic solution is that the pressure is zero inside the inclusion, while outside it follows the
relation
µm (µi − µm ) ri2
cos(2θ)
(47)
pm = 4˙
µi + µm r2
where µi = 103 is the viscosity of the inclusion and µm = 1 is the viscosity of the background media, θ = tan−1 (y/x),
and ˙ = 1 is the applied strain rate.
[13] thoroughly investigated this problem with various numerical methods (FEM, FDM), with and without tracers,
and conclusively showed how various averagings lead to different results. [18] obtained a first order convergence for
both pressure and velocity, while [32] and [21] showed that the use of adaptive mesh refinement in respectively the
FEM and FDM yields convergence rates which depend on refinement strategies.
29
vx
1.0
vy
1.0
0.0
0.2
0.8
0.6
0.6
0.6
0.4
0.6
0.4
0.4
0.4
0.6
0.4
0.2
0.2
0.2
0.8
0.2
0.0
0.0
0.0
0.0
0.0
0.8
1.0
xx
1.0
0.8
1.75
1.0
1.50
0.8
1.25
0.6
0.25
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.00
vx txth
1.0
0.000
0.4
x
0.6
1.50
0.2
0.0
0.0
1.75
0.0
0.0
0.8
0.8
1.0
vy tyth
103
0.6
0.8
0.8
102
y
0.6
y
0.6
0.4
0.4
101
0.2
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
100
0.0
0.0
10.00000000
0.2
0.4
x
0.6
0.8
1.0
velocity
pressure
x2
x1
1.00000000
0.10000000
error
0.01000000
0.00100000
0.00010000
0.00001000
0.00000100
0.00000010
0.00000001
0.01
0.0
0.0
0.1
h
30
1.100
1.075
1.050
1.025
1.000
0.975
0.950
0.925
0.900
0.4
x
0.6
0.8
1.0
p pth
1.00
0.75
0.50
0.25
0.00
0.25
0.50
0.75
1.00
15
10
5
0
5
0.2
1.0
density
1.0
0.8
x
0.2
15
1.0
0.4
0.000
0.4
0.8
0.6
0.005
0.2
0.6
0.8
0.015
0.010
x
xy
1.0
0.020
0.0
0.0
1.0
1.0
0.6
y
y
0.2
0.2
x
0.4
0.4
1.25
0.4
0.2
0.6
0.75
0.2
10
0.8
0.50
0.2
0.020
0.0
0.0
0.25
1.00
5
1.0
0.00
0.4
0.015
0.2
0.0
0.0
1.0
0.6
0.010
0.4
0.8
0.8
0.005
0.6
0.6
yy
1.0
0.8
x
0.4
0.50
0.2
0.4
y
y
0.75
0.4
0.2
0.6
1.00
0
y
0.6
5
y
x
10
10
0.2
0.4
x
0.6
0.8
1.0
1.0
15
100
0.8
0.6
10
1
10
2
10
3
y
0.4
15
y
0.8
y
0.8
y
0.8
0.2
p
1.0
0.4
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
11
fieldstone: the indentor benchmark
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.4
x
0.6
0.8
1.0
1.0
10.0
7.5
5.0
2.5
0.0
2.5
5.0
1.0
20.0
17.5
15.0
12.5
10.0
7.5
5.0
2.5
xx
0.2
0.4
0.2
0.4
x
x
0.6
0.6
0.8
0.8
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.2
vy
0.0
0.2
0.4
y
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.6
0.2
0.4
x
0.6
0.8
1.0
0.4
x
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.6
0.8
1.0
5.0
2.5
0.0
2.5
5.0
7.5
10.0
0.5
0.4
0.3
0.2
0.1
0.0
0.0
6
p
5
4
3
2
0.2
0.4
1.0
yy
0.2
0.8
y
0.2
0.3
0.2
0.1
0.0
0.1
0.2
0.3
x
0.6
0.8
1.0
xy
0.2
0.4
x
0.6
0.8
1.0
103
0.5
0.4
0.3
0.2
0.1
0.0
0.0
102
101
100
0.2
0.4
x
0.6
0.8
1.0
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
vx
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
y
y
y
y
The punch benchmark is one of the few boundary value problems involving plastic solids for which there exists an
exact solution. Such solutions are usually either for highly simplified geometries (spherical or axial symmetry, for
instance) or simplified material models (such as rigid plastic solids) [30].
In this experiment, a rigid punch indents a rigid plastic half space; the slip line field theory gives exact solutions
as shown in Fig. ??a. The plane strain formulation of the equations and the detailed solution to the problem were
derived in the Appendix of [53] and are also presented in [19].
The two dimensional punch problem has been extensively studied numerically for the past 40 years [60, 59, 9, 8, 27,
56, 6, 42] and has been used to draw a parallel with the tectonics of eastern China in the context of the India-Eurasia
collision [51, 38]. It is also worth noting that it has been carried out in one form or another in series of analogue
modelling articles concerning the same region, with a rigid indenter colliding with a rheologically stratified lithosphere
[40, 12, 29].
Numerically, the one-time step punch experiment is performed on a two-dimensional domain of purely plastic
von Mises material. Given that the von Mises rheology yield criterion does not depend on pressure, the density of
the material and/or the gravity vector is set to zero. Sides are set to free slip boundary conditions, the bottom to
no slip, while a vertical velocity (0, −vp ) is prescribed at the top boundary for nodes whose x coordinate is within
[Lx /2 − δ/2, Lx /2 + δ/2].
The following parameters are used: Lx = 1, Ly = 0.5, µmin = 10−3 , µmax = 103 , vp = 1, δ = 0.123456789 and the
yield value of the material is set to k = 1.
The analytical solution predicts that the angle of the shear bands stemming from the sides of the punch is π/4,
that√the pressure right under the punch is 1 + π, and that the velocity of the rigid blocks on each side of the punch is
vp / 2 (this is simply explained by invoking conservation of mass).
10
1
10
2
10
3
0.5
0.4
0.3
0.2
0.1
0.0
0.0
1
20
15
10
5
0
5
10
15
20
5
p (nodal)
4
3
2
0.2
0.4
x
0.6
0.8
1.0
1
6
0.0
5
0.2
4
0.00
p
0.2
0.05
v
u
0.4
0.10
0.4
3
0.6
0.05
2
0.8
0.10
1
1.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.0
0.2
0.4
x
0.6
31
0.8
1.0
0.0
0.2
0.4
x
0.6
0.8
1.0
ToDo: smooth punch
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• non-isoviscous
• nonlinear rheology
32
12
fieldstone: the annulus benchmark
This benchmark is based on Thieulot & Puckett [Subm.] in which an analytical solution to the isoviscous incompressible
Stokes equations is derived in an annulus geometry. The velocity and pressure fields are as follows:
vr (r, θ)
=
g(r)k sin(kθ),
(48)
vθ (r, θ)
=
f (r) cos(kθ),
(49)
p(r, θ)
=
kh(r) sin(kθ),
(50)
ρ(r, θ)
=
ℵ(r)k sin(kθ),
(51)
with
f (r)
g(r)
h(r)
ℵ(r)
A
B
= Ar + B/r,
B
C
A
r + ln r + ,
=
2
r
r
2g(r) − f (r)
=
,
r
g0
f
f0
g
= g 00 − − 2 (k 2 − 1) + 2 + ,
r
r
r
r
2(ln R1 − ln R2 )
,
= −C 2
R2 ln R1 − R12 ln R2
R22 − R12
= −C 2
.
R2 ln R1 − R12 ln R2
(52)
(53)
(54)
(55)
(56)
(57)
The parameters A and B are chosen so that vr (R1 ) = vr (R2 ) = 0, i.e. the velocity is tangential to both inner
and outer surfaces. The gravity vector is radial and of unit length. In the present case, we set R1 = 1, R2 = 2 and
C = −1.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions
• direct solver
• isothermal
• isoviscous
• analytical solution
• annulus geometry
• elemental boundary conditions
33
10
velocity
pressure
x2
x1
1
error
0.1
0.01
0.001
0.0001
0.01
h
34
13
fieldstone: stokes sphere (3D) - penalty
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• 3D
• elemental b.c.
• buoyancy driven
resolution is 24x24x24
35
14
fieldstone: stokes sphere (3D) - mixed formulation
This is the same setup as Section 14.
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• 3D
• elemental b.c.
• buoyancy driven
36
15
fieldstone: consistent pressure recovery
p = −λ∇ · v
q1 is smoothed pressure obtained with the center-to-node approach.
q2 is recovered pressure obtained with [58]. R
All three fulfill the zero average condition: pdΩ = 0.
y
0.000
0.4
0.2
0.4
x
0.6
0.8
vx txth
1.0
0.000002
x
0.6
0.8
0.8
0.4
0.2
0.10
0.0
0.0
0.15
x
0.6
0.8
1.0
0.6
y
0.6
0.4
0.2
0.2
0.4
x
0.6
0.8
1.0
0.10
0.2
0.15
0.2
0.4
x
0.6
0.8
1.0
q2 pth
1.0
0.00002
0.00000
0.00002
0.00004
0.00006
0.2
0.4
0.004
0.8
0.003
1.0
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
1.0
0.8
0.002
0.6
0.4
0.001
0.4
0.2
0.000
0.2
0.2
0.4
x
0.6
0.8
1.0
0.001
0.0
0.0
velocity
pressure (el)
pressure (q1)
pressure (q2)
x2
x1
0.010000
error
0.001000
0.000100
0.000010
0.01
0.8
0.8
0.6
0.0
0.0
0.100000
0.000001
x
0.6
y
0.8
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0.000
0.05
0.4
q1 pth
1.0
0.00
0.0
0.0
0.15
1.0
0.05
y
y
0.05
0.4
q2
1.0
p pth
0.0
0.0
1.0
0.8
0.00
0.6
x
0.8
0.2
0.000002
0.8
0.6
0.4
0.000001
0.6
x
0.6
0.000000
0.4
0.4
0.8
0.000001
0.2
0.2
1.0
0.000002
1.0
0.05
0.2
vy tyth
0.0
0.0
1.0
0.0
0.0
1.0
0.2
q1
1.0
0.8
0.4
0.000002
0.4
0.6
0.6
0.000001
0.2
x
y
y
0.000000
0.4
0.4
0.8
0.000001
0.6
0.2
0.2
0.10
0.2
0.010
1.0
0.8
0.0
0.0
0.0
0.0
1.0
0.05
0.4
0.005
0.2
0.010
0.00
0.6
0.000
0.4
0.005
0.2
0.0
0.0
0.6
0.05
0.8
0.005
y
0.6
0.0
0.0
0.8
0.005
p
1.0
0.010
y
0.8
vy
1.0
0.010
y
vx
1.0
0.1
h
In terms of pressure error, q2 is better than q1 which is better than elemental.
QUESTION: why are the averages exactly zero ?!
TODO:
• add randomness to internal node positions.
• look at elefant algorithms
37
16
fieldstone: the Particle in Cell technique (1) - the effect of averaging
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• non-isoviscous
• particle-in-cell
After the initial setup of the grid, markers can then be generated and placed in the domain. One could simply
randomly generate the marker positions in the whole domain but unless a very large number of markers is used, the
chance that an element does not contain any marker exists and this will prove problematic. In order to get a better
control over the markers spatial distribution, one usually generates the marker per element, so that the total number
of markers in the domain is the product of the number of elements times the user-chosen initial number of markers
per element.
Our next concern is how to actually place the markers inside an element. Two methods come to mind: on a regular
grid, or in a random manner, as shown on the following figure:
In both cases we make use of the basis shape functions: we generate the positions of the markers (random or
regular) in the reference element first (rim , sim ), and then map those out to the real element as follows:
xim =
m
X
Ni (rim , sim )xi
yim =
i
m
X
Ni (rim , sim )yi
i
where xi , yi are the coordinates of the vertices of the element.
When using active markers, one is faced with the problem of transferring the properties they carry to the mesh on
which the PDEs are to be solved. As we have seen, building the FE matrix involves a loop over all elements, so one
simple approach consists of assigning each element a single property computed as the average of the values carried by
the markers in that element. Often in colloquial language ”average” refers to the arithmetic mean:
n
hφiam =
1X
φi
n
k
where < φ >am is the arithmetic average of the n numbers φi . However, in mathematics other means are commonly
used, such as the geometric mean:
!
n
Y
hφigm =
φi
i
and the harmonic mean:
n
hφihm =
1X 1
n i φi
!−1
Furthermore, there is a well known inequality for any set of positive numbers,
hφiam
≥
hφigm
38
≥
hφihm
which will prove to be important later on.
Let us now turn to a simple concrete example: the 2D Stokes sphere. There are two materials in the domain, so
that markers carry the label ”mat=1” or ”mat=2”. For each element an average density and viscosity need to be
computed. The majority of elements contains markers with a single material label so that the choice of averaging does
not matter (it is trivial to verify that if φi = φ0 then hφiam = hφigm = hφihm = φ0 . Remain the elements crossed by
the interface between the two materials: they contain markers of both materials and the average density and viscosity
inside those depends on 1) the total number of markers inside the element, 2) the ratio of markers 1 to markers 2, 3)
the type of averaging.
This averaging problem has been studied and documented in the literature [45, 13, 52, 41]
Nodal projection. Left: all markers inside elements to which the green node belongs to are taken into account.
Right: only the markers closest to the green node count.
The setup is identical as the Stokes sphere experiment. The bash script runall runs the code for many resolutions,
both initial marker distribution and all three averaging types. The viscosity of the sphere has been set to 103 while
the viscosity of the surrounding fluid is 1. The average density is always computed with an arithmetic mean. Root
mean square velocity results are shown hereunder:
random markers, elemental proj
regular markers, elemental proj
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.009
0.008
0.007
vrms
0.006
0.005
0.004
0.003
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.01
0.009
0.008
0.007
0.006
vrms
0.01
0.005
0.004
0.003
0.002
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.002
0.001
0.001
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
h
0.05
0.06
0.07
0.08
0.09
0.1
h
random markers, nodal proj
regular markers, nodal proj
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.009
0.008
0.007
vrms
0.006
0.005
0.004
0.003
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.01
0.009
0.008
0.007
0.006
vrms
0.01
0.005
0.004
0.003
0.002
0.002
0.001
0.001
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
h
0.05
0.06
0.07
0.08
0.09
0.1
h
random markers, nodal proj
regular markers, nodal proj
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.009
0.008
0.007
vrms
0.006
0.005
0.004
0.003
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.01
0.009
0.008
0.007
0.006
vrms
0.01
0.005
0.004
0.003
0.002
0.002
0.001
0.001
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0
h
0.01
0.02
0.03
0.04
0.05
h
39
0.06
0.07
0.08
0.09
0.1
Conclusions:
• With increasing resolution (h → 0) vrms values seem to converge towards a single value, irrespective of the
number of markers.
• At low resolution, say 32x32 (i.e. h=0.03125), vrms values for the three averagings differ by about 10%. At
higher resolution, say 128x128, vrms values are still not converged.
• The number of markers per element plays a role at low resolution, but less and less with increasing resolution.
• Results for random and regular marker distributions are not identical but follow a similar trend and seem to
converge to the same value.
• At low resolutions, elemental values yield better results.
• harmonic mean yields overal the best results
64 markers per element
0.01
0.009
am, reg, el
gm, reg, el
hm, reg, el
am, rand, el
gm, rand, el
hm, rand, el
am, reg, nod1
gm, reg, nod1
hm, reg, nod1
am, rand, nod1
gm, rand, nod1
hm, rand, nod1
am, reg, nod2
gm, reg, nod2
hm, reg, nod2
am, rand, nod2
gm, rand, nod2
hm, rand, nod2
0.008
0.007
vrms
0.006
0.005
0.004
0.003
0.002
0.001
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
h
40
0.08
0.09
0.1
17
fieldstone: solving the full saddle point problem
The details of the numerical setup are presented in Section 5.
The main difference is that we no longer use the penalty formulation and therefore keep both velocity and pressure
as unknowns. Therefore we end up having to solve the following system:
K G
V
f
·
=
or,
A · X = rhs
P
h
GT 0
Each block K, G and vector f , h are built separately in the code and assembled into the matrix A and vector rhs
afterwards. A and rhs are then passed to the solver. We will see later that there are alternatives to solve this approach
which do not require to build the full Stokes matrix A.
Each element has m = 4 vertices so in total ndof V × m = 8 velocity dofs and a single pressure dof, commonly
situated in the center of the element. The total number of velocity dofs is therefore N f emV = nnp × ndof V while
the total number of pressure dofs is N f emP = nel. The total number of dofs is then N f em = N f emV + N f emP .
As a consequence, matrix K has size N f emV, N f emV and matrix G has size N f emV, N f emP . Vector f is of size
N f emV and vector h is of size N f emP .
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• direct solver (?)
• isothermal
• isoviscous
• analytical solution
• pressure smoothing
41
0.010
vx
1.0
0.010
vy
1.0
0.005 0.8
0.005 0.8
0.6
0.6
0.6
0.4
0.0
0.0
0.4
0.0050.2
0.2
0.2
0.4
x
0.6
0.8
1.0
0.00
0.05
0.10
y
y
y
0.000
0.05
0.0
0.010 0.0
0.4
0.15
0.0050.2
0.2
0.4
x
0.6
0.8
1.0
0.00
q
1.0
0.05
0.8
0.10
0.6
y
0.8
0.000
p
1.0
0.0
0.010 0.0
0.20
0.2
0.4
x
0.6
0.8
1.0
0.15
0.4
0.20
0.2
0.25 0.00.0
0.2
0.4
x
0.6
0.8
1.0
0.06
0.8
0.01
0.6
0.02
0.8
0.01
0.6
0.00
y
y
0.00
1.0
0.4
0.01
0.2
0.0
0.0
0.02
0.2
0.4
x
0.6
0.8
1.0
0.4
0.01
0.2
0.0
0.0
0.02
0.2
0.0000010
1.0
vx txth
1.0
0.03
1.0
0.4
x
0.6
0.8
1.0
0.03
0.0
0.0
0.0000005
0.2
0.4
x
0.6
0.8
1.0
0.0
0.0
0.0000010
0.2
0.4
x
0.6
0.8
1.0
0.4
x
0.6
0.8
1.0
0.04
0.6
0.03
0.0
0.0
0.0000010
0.4
0.02
0.2
0.04 0.0
0.0
0.2
0.00
p pth
0.02
0.04
0.06
0.08
0.0000005
0.2
0.2
0.05
0.8
0.4
0.06
y
0.6
0.0000000
0.4
y
0.6
0.0000000
0.4
y
0.6
0.2
0.02
0.2
0.8
0.0000005
0.0
0.0
0.00
0.4
0.8
0.0000005
0.2
0.02
0.6
0.8
0.4
0.04 1.0
0.8
0.0000010
1.0
vy tyth
0.06
y
0.02
xy
x
0.6
0.8
1.0
q pth
1.0
0.10
0.2
0.4
x
0.6
0.8
1.0
0.12
0.02
0.8
0.04
0.6
0.06
0.4
0.08
0.2
0.0
0.0
0.01
0.00
y
1.0
0.03
yy
y
0.03
xx
0.25
0.10
0.2
0.4
x
0.6
0.8
1.0
0.12
Unlike the results obtained with the penalty formualtion (see Section 5), the pressure showcases a very strong
checkerboard pattern, similar to the one in [16].
Left: pressure solution as shown in [16]; Right: pressure solution obtained with fieldstone.
Rather interestingly, the nodal pressure (obtained with a simple center-to-node algorithm) fails to recover a correct
pressure at the four corners.
42
18
fieldstone: solving the full saddle point problem in 3D
When using Q1 ×P0 elements, this benchmark fails because of the Dirichlet b.c. on all 6 sides and all three components.
However, as we will see, it does work well with Q2 × Q1 elements. .
This benchmark begins by postulating a polynomial solution to the 3D Stokes equation [14]:
x + x2 + xy + x3 y
y + xy + y 2 + x2 y 2
v=
(58)
−2z − 3xz − 3yz − 5x2 yz
and
p = xyz + x3 y 3 z − 5/32
(59)
While it is then trivial to verify that this velocity field is divergence-free, the corresponding body force of the Stokes
equation can be computed by inserting this solution into the momentum equation with a given viscosity µ (constant
or position/velocity/strain rate dependent). The domain is a unit cube and velocity boundary conditions simply use
Eq. (58). Following [7], the viscosity is given by the smoothly varying function
µ = exp(1 − β(x(1 − x) + y(1 − y) + z(1 − z)))
(60)
One can easily show that the ratio of viscosities µ? in the system follows µ? = exp(−3β/4) so that choosing β = 10
yields µ? ' 1808 and β = 20 yields µ? ' 3.269 × 106 .
We start from the momentum conservation equation:
˙ =f
−∇p + ∇ · (2µ)
The x-component of this equation writes
fx
∂
∂
∂
∂p
+
(2µ˙xx ) +
(2µ˙xy ) +
(2µ˙xz )
∂x ∂x
∂y
∂z
∂p
∂
∂
∂
∂µ
∂µ
∂µ
= −
+ 2µ ˙xx + 2µ ˙xy + 2µ ˙xz + 2 ˙xx + 2 ˙xy + 2 ˙xz
∂x
∂x
∂y
∂z
∂x
∂y
∂z
= −
Let us compute all the block separately:
˙xx
=
1 + 2x + y + 3x2 y
˙yy
=
1 + x + 2y + 2x2 y
˙zz
= −2 − 3x − 3y − 5x2 y
2˙xy
=
(x + x3 ) + (y + 2xy 2 ) = x + y + 2xy 2 + x3
2˙xz
=
(0) + (−3z − 10xyz) = −3z − 10xyz
2˙yz
=
(0) + (−3z − 5x2 z) = −3z − 5x2 z
In passing, one can verify that ˙xx + ˙yy + ˙zz = 0. We further have
∂
2˙xx
∂x
∂
2˙xy
∂y
∂
2˙xz
∂z
∂
2˙xy
∂x
∂
2˙yy
∂y
∂
2˙yz
∂z
∂
2˙xz
∂x
∂
2˙yz
∂y
∂
2˙zz
∂z
=
2(2 + 6xy)
=
1 + 4xy
= −3 − 10xy
=
1 + 2y 2 + 3x2
=
2(2 + 2x2 )
= −3 − 5x2
= −10yz
=
0
=
2(0)
43
(61)
(62)
∂p
∂x
∂p
∂y
∂p
∂z
= yz + 3x2 y 3 z
(63)
= xz + 3x3 y 2 z
(64)
= xy + x3 y 3
(65)
Pressure normalisation Here again, because Dirichlet boundary conditions are prescribed on all sides the pressure
is known up to an arbitrary constant. This constant can be determined by (arbitrarily) choosing to normalised the
pressure field as follows:
Z
p dΩ = 0
(66)
Ω
This is a single constraint associated to a single Lagrange
K
G
GT
0
0 CT
multiplier λ and the global Stokes system takes the form
0
V
C P
λ
0
In this particular case the constraint matrix C is a vector and it only acts on the pressure degrees of freedom because
of Eq.(66). Its exact expression is as follows:
Z
X
XZ X p
X X Z
XZ
p
Ni dΩ pi =
Ce · pe
p dΩ =
Ni pi dΩ =
p dΩ =
Ω
e
Ωe
e
Ωe
e
i
i
Ωe
e
where pe is the list of pressure dofs of element e. The elemental constraint vector contains the corresponding pressure
basis functions integrated over the element. These elemental constraints are then assembled into the vector C.
18.0.1
Constant viscosity
Choosing β = 0 yields a constant velocity µ(x, y, z) = exp(1) ' 2.718 (and greatly simplifies the right-hand side) so
that
∂
µ(x, y, z) = 0
∂x
∂
µ(x, y, z) = 0
∂y
∂
µ(x, y, z) = 0
∂z
and
fx
=
=
=
fy
=
=
=
fz
=
=
∂p
∂
∂
∂
+ 2µ ˙xx + 2µ ˙xy + 2µ ˙xz
∂x
∂x
∂y
∂z
−(yz + 3x2 y 3 z) + 2(2 + 6xy) + (1 + 4xy) + (−3 − 10xy)
−
−(yz + 3x2 y 3 z) + µ(2 + 6xy)
∂
∂
∂
∂p
+ 2µ ˙xy + 2µ ˙yy + 2µ ˙yz
−
∂y
∂x
∂y
∂z
−(xz + 3x3 y 2 z) + µ(1 + 2y 2 + 3x2 ) + µ2(2 + 2x2 ) + µ(−3 − 5x2 )
−(xz + 3x3 y 2 z) + µ(2 + 2x2 + 2y 2 )
∂p
∂
∂
∂
−
+ 2µ ˙xz + 2µ ˙yz + 2µ ˙zz
∂z
∂x
∂y
∂z
−(xy + x3 y 3 ) + µ(−10yz) + 0 + 0
= −(xy + x3 y 3 ) + µ(−10yz)
Finally
yz + 3x2 y 3 z
2 + 6xy
f = − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
xy + x3 y 3
−10yz
Note that there seems to be a sign problem with Eq.(26) in [7].
44
(67)
(68)
(69)
0.01000
velocity
pressure
x3
x2
error
0.00100
0.00010
0.00001
0.1
h
18.0.2
Variable viscosity
The spatial derivatives of the viscosity are then given by
∂
µ(x, y, z)
∂x
∂
µ(x, y, z)
∂y
∂
µ(x, y, z)
∂z
=
−(1 − 2x)βµ(x, y, z)
=
−(1 − 2y)βµ(x, y, z)
=
−(1 − 2z)βµ(x, y, z)
and thr right-hand side by
yz + 3x2 y 3 z
2 + 6xy
f = − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
(70)
xy + x3 y 3
−10yz
2˙xx
2˙xy
2˙xz
−(1 − 2x)βµ(x, y, z) 2˙xy − (1 − 2y)βµ(x, y, z) 2˙yy − (1 − 2z)βµ(x, y, z) 2˙yz
2˙xz
2˙yz
2˙zz
2 3
yz + 3x y z
2 + 6xy
= − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
xy + x3 y 3
−10yz
2
2 + 4x + 2y + 6x y
x + y + 2xy 2 + x3
−3z − 10xyz
−3z − 5x2 z
− (1 − 2x)βµ x + y + 2xy 2 + x3 − (1 − 2y)βµ 2 + 2x + 4y + 4x2 y − (1 − 2z)βµ
2
−3z − 5x z
−4 − 6x − 6y − 10x2 y
−3z − 10xyz
Note that at (x, y, z) = (0, 0, 0), µ = exp(1), and at (x, y, z) = (0.5, 0.5, 0.5), µ = exp(1 − 3β/4) so that the
45
maximum viscosity ratio is given by
µ? =
exp(1 − 3β/4)
= exp(−3β/4)
exp(1)
By varying β between 1 and 22 we can get up to 7 orders of magnitude viscosity difference.
features
• Q1 × P0 element
• incompressible flow
• saddle point system
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• 3D
• elemental b.c.
• analytical solution
46
19
fieldstone: solving the full saddle point problem with Q2 × Q1 elements
The details of the numerical setup are presented in Section 5.
Each element has mV = 9 vertices so in total ndofV × mV = 18 velocity dofs and ndofP ∗ mP = 4 pressure dofs.
The total number of velocity dofs is therefore N f emV = nnp × ndof V while the total number of pressure dofs is
N f emP = nel. The total number of dofs is then N f em = N f emV + N f emP .
As a consequence, matrix K has size N f emV, N f emV and matrix G has size N f emV, N f emP . Vector f is of size
N f emV and vector h is of size N f emP .
renumber all nodes to start at zero!! Also internal numbering does not work here
The velocity shape functions are given by:
N0V
N1V
N2V
N3V
N4V
N5V
N6V
N7V
N8V
1
1
r(r − 1) s(s − 1)
2
2
1
1
=
r(r + 1) s(s − 1)
2
2
1
1
=
r(r + 1) s(s + 1)
2
2
1
1
=
r(r − 1) s(s + 1)
2
2
1
= (1 − r2 ) s(s − 1)
2
1
=
r(r + 1)(1 − s2 )
2
1
= (1 − r2 ) s(s + 1)
2
1
=
r(r − 1)(1 − s2 )
2
= (1 − r2 )(1 − s2 )
=
47
and their derivatives:
∂N0V
∂r
∂N1V
∂r
∂N2V
∂r
∂N3V
∂r
∂N4V
∂r
∂N5V
∂r
∂N6V
∂r
∂N7V
∂r
∂N8V
∂r
∂N0V
∂s
∂N1V
∂s
∂N2V
∂s
∂N3V
∂s
∂N4V
∂s
∂N5V
∂s
∂N6V
∂s
∂N7V
∂s
∂N8V
∂s
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
1
1
(2r − 1) s(s − 1)
2
2
1
1
(2r + 1) s(s − 1)
2
2
1
1
(2r + 1) s(s + 1)
2
2
1
1
(2r − 1) s(s + 1)
2
2
1
(−2r) s(s − 1)
2
1
(2r + 1)(1 − s2 )
2
1
(−2r) s(s + 1)
2
1
(2r − 1)(1 − s2 )
2
(−2r)(1 − s2 )
1
1
r(r − 1) (2s − 1)
2
2
1
1
r(r + 1) (2s − 1)
2
2
1
1
r(r + 1) (2s + 1)
2
2
1
1
r(r − 1) (2s + 1)
2
2
2 1
(1 − r ) (2s − 1)
2
1
r(r + 1)(−2s)
2
1
(1 − r2 ) (2s + 1)
2
1
r(r − 1)(−2s)
2
(1 − r2 )(−2s)
features
• Q2 × Q1 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• isoviscous
• analytical solution
48
0.0100000
velocity
pressure
x3
x2
0.0010000
error
0.0001000
0.0000100
0.0000010
0.0000001
0.1
h
49
fieldstone: The non-conforming Q1 × P0 element
20
features
• Non-conforming Q1 × P0 element
• incompressible flow
• mixed formulation
• isothermal
• non-isoviscous
• analytical solution
• pressure smoothing
try Q1 mapping instead of isoparametric.
0.005 0.8
0.6
0.4
0.6
0.000
0.0
0.0
0.4
0.0050.2
0.2
0.2
0.4
x
0.6
0.8
1.0
0.05
0.00
0.005 0.8
y
y
0.000
p
1.0
0.05
0.6
0.10
y
0.8
0.010
vy
1.0
0.0
0.010 0.0
0.4
0.15
0.0050.2
0.2
0.4
x
0.6
0.8
1.0
0.00
q
1.0
0.05
0.8
0.10
0.6
y
0.010
vx
1.0
0.0
0.010 0.0
0.20
0.2
0.4
x
0.6
0.8
1.0
0.15
0.4
0.20
0.2
0.25 0.00.0
0.2
0.4
x
0.6
0.8
1.0
0.06
0.01
0.6
0.02
0.8
0.01
0.6
0.00
y
y
0.00
1.0
0.4
0.01
0.2
0.0
0.0
0.02
0.2
0.4
x
0.6
0.8
1.0
0.4
0.01
0.2
0.0
0.0
0.02
0.2
0.0000010
1.0
vx txth
1.0
0.03
1.0
0.4
x
0.6
0.8
1.0
0.03
0.0
0.0
0.0000005
0.2
0.4
x
0.6
0.8
1.0
0.0
0.0
0.0000010
0.2
0.4
x
0.6
0.8
1.0
0.4
x
0.6
0.8
1.0
0.04
0.6
0.03
0.0
0.0
0.0000010
50
0.4
0.02
0.2
0.04 0.0
0.0
0.2
0.00
p pth
0.02
0.04
0.06
0.08
0.0000005
0.2
0.2
0.05
0.8
0.4
0.06
y
0.6
0.0000000
0.4
y
0.6
0.0000000
0.4
y
0.6
0.2
0.02
0.2
0.8
0.0000005
0.0
0.0
0.00
0.4
0.8
0.0000005
0.2
0.02
0.6
0.8
0.4
0.04 1.0
0.8
0.0000010
1.0
vy tyth
0.06
y
0.02
0.8
xy
x
0.6
0.8
1.0
q pth
1.0
0.10
0.2
0.4
x
0.6
0.8
1.0
0.12
0.02
0.8
0.04
0.6
0.06
0.4
0.08
0.2
0.0
0.0
0.01
0.00
y
1.0
0.03
yy
y
0.03
xx
0.25
0.10
0.2
0.4
x
0.6
0.8
1.0
0.12
21
fieldstone: The stabilised Q1 × Q1 element
The details of the numerical setup are presented in Section 5.
51
22
fieldstone: compressible flow (1)
We first start with an isothermal Stokes flow, so that we disregard the heat transport equation and the equations we
wish to solve are simply:
1
−∇ · 2η ε̇(v) − (∇ · v)1 + ∇p = ρg
3
∇ · (ρv) = 0
in Ω,
(71)
in Ω
(72)
The second equation can be rewritten ∇ · (ρv) = ρ∇ · v + v · ∇ρ = 0 or,
1
∇ · v + v · ∇ρ = 0
ρ
Note that this presupposes that the density is not zero anywhere in the domain.
We use a mixed formulation and therefore keep both velocity and pressure as unknowns. We end up having to
solve the following system:
K
G
V
f
·
=
or,
A · X = rhs
P
h
GT + Z 0
Where K is the stiffness matrix, G is the discrete gradient operator, GT is the discrete divergence operator, V the
velocity vector, P the pressure vector. Note that the term ZV derives from term v · ∇ρ in the continuity equation.
Each block K, G , Z and vectors f and h are built separately in the code and assembled into the matrix A and
vector rhs afterwards. A and rhs are then passed to the solver. We will see later that there are alternatives to solve
this approach which do not require to build the full Stokes matrix A.
Remark: the term ZV is often put in the rhs (i.e. added to h) so that the matrix A retains the same structure as
in the incompressible case. This is indeed how it is implemented in ASPECT. This however requires more work since
the rhs depends on the solution and some form of iterations is needed.
In the case of a compressible flow the strain rate tensor and the deviatoric strain rate tensor are no more equal
(since ∇ · v 6= 0). The deviatoric strainrate tensor is given by2
1
1
˙ = (v)
˙
˙
− (∇ · v)1
˙ d (v) = (v)
− T r()1
3
3
In that case:
˙dxx
=
˙dyy
=
2˙dxy
=
∂u 1 ∂u ∂v
2 ∂u 1 ∂v
−
+
−
=
∂x 3 ∂x ∂y
3 ∂x 3 ∂y
1 ∂u ∂v
1 ∂u 2 ∂v
∂v
−
+
=−
+
∂y 3 ∂x ∂y
3 ∂x 3 ∂y
∂u ∂v
+
∂y
∂x
and then
˙ d (v) =
2 ∂u
3 ∂x
−
1 ∂v
3 ∂y
1 ∂u
2 ∂y
+
1 ∂v
2 ∂x
1 ∂u
2 ∂y
+
1 ∂v
2 ∂x
− 13 ∂u
∂x +
2 ∂v
3 ∂y
(73)
(74)
(75)
From ~τ = 2η~d we arrive at:
d
˙xx
τxx
2/3
τyy = 2η ˙dyy = 2η −1/3
τxy
0
˙dxy
−1/3
2/3
0
0
0 ·
1/2
∂u
∂x
∂v
∂y
∂u
∂v
∂y + ∂x
4/3
= η −2/3
0
or,
~τ = Cη BV
2 See
the ASPECT manual for a justification of the 3 value in the denominator in 2D and 3D.
52
−2/3
4/3
0
0
0 ·
1
∂u
∂x
∂v
∂y
∂u
∂v
∂y + ∂x
In order to test our implementation we have created a few manufactured solutions:
• benchmark #1 (ibench=1)): Starting from a density profile of:
ρ(x, y) = xy
(76)
We derive a velocity given by:
vx (x, y) =
With gx (x, y) =
1
x
Cx
Cy
, vy (x, y) =
x
y
and gy (x, y) = y1 , this leads us to a pressure profile:
4Cx
4Cy
p = −η
+ 2 + xy + C0
3x2
3y
(77)
(78)
This gives us a strain rate:
−Cx
−Cy
˙yy = 2
˙xy = 0
x2
y
In what follows, we choose η = 1 and Cx = Cy = 1 and for a unit square domain [1 : 2] × [1 : 2] we compute C0
so that the pressure is normalised to zero over the whole domain and obtain C0 = −1.
˙xx =
• benchmark #2 (ibench=2): Starting from a density profile of:
ρ = cos(x) cos(y)
(79)
We derive a velocity given by:
vx =
With gx =
1
cos(y)
and gy =
Cx
Cy
, vy =
cos(x)
cos(y)
(80)
1
cos(x) ,
this leads us to a pressure profile:
!
4Cx sin(x) 4Cy sin(y)
p=η
+
+ (sin(x) + sin(y)) + C0
3 cos2 (x)
3 cos2 (y)
(81)
sin(y)
sin(x)
˙yy = Cy
˙xy = 0
cos2 (x)
cos2 (y)
We choose η = 1 and Cx = Cy = 1. The domain is the unit square [0 : 1] × [0 : 1] and we obtain C0 as before
and obtain
1
C0 = 2 − 2 cos(1) + 8/3(
− 1) ' 3.18823730
cos(1)
(thank you WolframAlpha)
˙xx = Cx
• benchmark #3 (ibench=3)
• benchmark #4 (ibench=4)
• benchmark #5 (ibench=5)
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• isoviscous
• analytical solution
• pressure smoothing
ToDo:
• pbs with odd vs even number of elements
• q is ’fine’ everywhere except in the corners - revisit pressure smoothing paper?
• redo A v d Berg benchmark (see Tom Weir thesis)
53
23
23.1
fieldstone: compressible flow (2)
The physics
Let us start with some thermodynamics. Every material has an equation of state. The equilibrium thermodynamic
state of any material can be constrained if any two state variables are specified. Examples of state variables include
the pressure p and specific volume ν = 1/ρ, as well as the temperature T .
After linearisation, the density depends on temperature and pressure as follows:
ρ(T, p) = ρ0 ((1 − α(T − T0 ) + βT p)
where α is the coefficient of thermal expansion, also called thermal expansivity:
1 ∂ρ
α=−
ρ ∂T p
α is the percentage increase in volume of a material per degree of temperature increase; the subscript p means that
the pressure is held fixed.
βT is the isothermal compressibility of the fluid, which is given by
1
1 ∂ρ
βT =
=
K
ρ ∂P T
with K the bulk modulus. Values of βT = 10−12 −10−11 Pa−1 are reasonable for Earth’s mantle, with values decreasing
by about a factor of 5 between the shallow lithosphere and core-mantle boundary. This is the percentage increase
in density per unit change in pressure at constant temperature. Both the coefficient of thermal expansion and the
isothermal compressibility can be obtained from the equation of state.
The full set of equations we wish to solve is given by
−∇ · 2η ˙ d (v) + ∇p = ρ0 ((1 − α(T − T0 ) + βT p) g
in Ω
1
∇ · v + v · ∇ρ = 0
in Ω
ρ
∂T
∂p
d
d
ρCp
+ v · ∇T − ∇ · k∇T = ρH + 2η ˙ : ˙ + αT
+ v · ∇p
∂t
∂t
(82)
(83)
in Ω,
(84)
Note that this presupposes that the density is not zero anywhere in the domain.
23.2
The numerics
We use a mixed formulation and therefore keep both velocity and pressure as unknowns. We end up having to solve
the following system:
K
G+W
V
f
·
=
or,
A · X = rhs
P
h
GT + Z
0
Where K is the stiffness matrix, G is the discrete gradient operator, GT is the discrete divergence operator, V the
velocity vector, P the pressure vector. Note that the term ZV derives from term v · ∇ρ in the continuity equation.
As perfectly explained in the step 32 of deal.ii3 , we need to scale the G term since it is many orders of magnitude
smaller than K, which introduces large inaccuracies in the solving process to the point that the solution is nonsensical.
This scaling coefficient is η/L. After building the G block, it is then scaled as follows: G0 = Lη G so that we now solve
K
G0 + W
V
f
·
=
P0
h
G0T + Z
0
After the solve phase, we recover the real pressure with P = Lη P 0 .
adapt notes since I should scale W and Z too. h should be caled too !!!!!!!!!!!!!!!
Each block K, G , Z and vectors f and h are built separately in the code and assembled into the matrix A and
vector rhs afterwards. A and rhs are then passed to the solver. We will see later that there are alternatives to solve
this approach which do not require to build the full Stokes matrix A.
Remark 1: the terms ZV and WP are often put in the rhs (i.e. added to h) so that the matrix A retains the
same structure as in the incompressible case. This is indeed how it is implemented in ASPECT, see also appendix A
of [33]. This however requires more work since the rhs depends on the solution and some form of iterations is needed.
3 https://www.dealii.org/9.0.0/doxygen/deal.II/step
32.html
54
Remark 2: Very often the adiabatic heating term αT (v · ∇p) is simplified as follows: If you assume the vertical
component of the gradient of the dynamic pressure to be small compared to the gradient of the total pressure (in
other words, the gradient is dominated by the gradient of the hydrostatic pressure), then −ρg ' ∇p and then
αT (v · ∇p) ' −αρT v · g. We will however not be using this approximation in what follows.
We have already established that
~τ = Cη BV
The following measurements are carried out:
• The root mean square velocity (vrms):
s
vrms =
1
V
Z
1
V
Z
v 2 dV
V
• The average temperature (Tavrg):
< T >=
T dV
V
• The total mass (mass):
Z
M=
ρdV
V
• The Nusselt number (Nu):
Nu = −
1 1
Lx ∆T
Lx
Z
0
∂T (x, y = Ly )
dx
∂y
• The kinetic energy (EK):
Z
EK =
V
1 2
ρv dV
2
• The work done against gravity
Z
< W >= −
ρgy vy dV
V
• The total viscous dissipation (visc diss)
Z
< Φ >=
1
ΦdV =
V
Z
2η ε̇ : ε̇dV
• The gravitational potential energy (EG)
Z
ρgy (Ly − y)dV
EG =
V
• The internal thermal energy (ET)
Z
ET =
ρ(0) Cp T dV
V
Remark 3: Measuring the total mass can be misleading: indeed because ρ = ρ0 (1 − αT ), then measuring the
total mass amounts to measuring a constant minus the volume-integrated temperature, and there is no reason
why the latter should be zero, so that there is no reason why the total mass should be zero...!
23.3
The experimental setup
The setup is as follows: the domain is Lx = Ly = 3000km. Free slip boundary conditions are imposed on all four
sides. The initial temperature is given by:
Ly − y
πx
πy
T (x, y) =
− 0.01 cos( ) sin( ) ∆T + Tsurf
Ly
Lx
Ly
with ∆T = 4000K, Tsurf = T0 = 273.15K. The temperature is set to ∆T + Tsurf at the bottom and Tsurf at the top.
We also set k = 3, Cp = 1250, |g| = 10, ρ0 = 3000 and we keep the Rayleigh number Ra and dissipation number Di
as input parameters:
αg∆T L3 ρ20 Cp
αgL
Ra =
Di =
ηk
Cp
55
DiCp
gL ,
From the second equation we get α =
Ra =
which we can insert in the first one:
DiCp2 ∆T L2 ρ20
ηk
or,
DiCp2 ∆T L2 ρ20
Ra k
η=
For instance, for Ra = 104 and Di = 0.75, we obtain α ' 3 · 10−5 and η ' 1025 which are quite reasonable values.
23.4
Scaling
Following [31], we non-dimensionalize the equations using the reference values for density ρr , thermal expansivity
αr , temperature contrast ∆Tr (refTemp), thermal conductivity kr , heat capacity Cp , depth of the fluid layer L and
viscosity ηr . The non-dimensionalization for velocity, ur , pressure pr and time, tr become
ur =
kr
ρr Cp L
(refvel)
pr =
ηr kr
ρr Cp L2
(refpress)
tr =
ρr Cp L2
kr
(reftime)
In the case of the setup described hereabove, and when choosing Ra = 104 and Di = 0.5, we get:
alphaT 2.083333e-05
eta 8.437500e+24
reftime 1.125000e+19
refvel 2.666667e-13
refPress 7.500000e+05
23.5
Conservation of energy 1
23.5.1
under BA and EBA approximations
Following [33], we take the dot product of the momentum equation with the velocity v and integrate over the whole
volume4 :
Z
Z
[−∇ · τ + ∇p] · vdV =
ρg · vdV
V
V
or,
Z
−
Z
Z
(∇ · τ ) · vdV +
V
∇p · vdV =
V
ρg · vdV
V
Let us look at each block separately:
Z
Z
Z
Z
Z
− (∇ · τ ) · vdV = − τ |v{z
· n} dS +
τ : ∇vdV =
τ : ε̇dV =
ΦdV
V
S
V
V
V
=0 (b.c.)
which is the volume integral of the shear heating. Then,
Z
Z
Z
∇p · vdV =
p v
· n} dS −
| {z
V
S
V
=0 (b.c.)
∇
| {z· v}
pdV = 0
=0 (incomp.)
which is then zero in the case of an incompressible flow. And finally
Z
ρg · vdV = W
V
which is the work against gravity.
Conclusion for an incompressible fluid: we should have
Z
Z
ΦdV =
ρg · vdV
V
4 Check:
V
this is akin to looking at the power, force*velocity, says Arie
56
(85)
This formula is hugely problematic: indeed, the term ρ in the rhs is the full density. We know that to the value of ρ0
corresponds a lithostatic pressure gradient pL = ρ0 gy. In this case one can write ρ = ρ0 + ρ0 and p = pL + p0 so that
we also have
Z
Z
[−∇ · τ + ∇p0 ] · vdV =
ρ0 g · vdV
V
V
which will ultimately yield
Z
Z
ΦdV =
V
Z
ρ0 g · vdV =
(ρ − ρ0 )g · vdV
V
(86)
V
Obviously Eqs.(85) and (86) cannot be true at the same time. The problem comes from the nature of the (E)BA
approximation: ρ = ρ0 in the mass conservation equation but it is not constant in the momentum conservation
equation, which
R is of course inconsistent. Since the mass conservation equation is ∇ · v = 0 under this approximation
then the term V ∇p · vdV is always zero for any pressure (full pressure p, or overpressure p − pL ), hence the paradox.
This paradox will be lifted when a consistent set of equations will be used (compressible formulation). On a practical
note, Eqs.(85) is not verified by the code, while (86) is.
In the end:
Z
Z
ΦdV =
(ρ − ρ0 )g · vdV
V
{z
}
| {z } |
(87)
V
visc diss
23.5.2
work grav
under no approximation at all
Z
Z
∇p · vdV
=
V
Z
p v
· n} dS −
| {z
S
=0 (b.c.)
Z
=
V
∇ · v pdV = 0
(88)
V
1
v · ∇ρ pdV = 0
ρ
(89)
(90)
ToDo:see section 3 of [33] where this is carried out with the Adams-Williamson eos.
23.6
Conservation of energy 2
Also, following the Reynold’s transport theorem [37],p210, we have for a property A (per unit mass)
Z
Z
Z
d
∂
AρdV =
(Aρ)dV +
Aρv · ndS
dt V
V ∂t
S
Let us apply to this to A = Cp T and compute the time derivative of the internal energy:
Z
Z
Z
Z
Z
d
∂
∂ρ
∂T
ρCp T dV =
(ρCp T )dV +
Aρ v
·
n
dS
=
C
T
dV
+
ρCp
dV
p
{z
}
|
dt V
∂t
∂t
∂t
V
S
V
V
{z
} |
{z
}
|
=0 (b.c.)
I
(91)
II
In order to expand I, the mass conservation equation will be used, while the heat transport equation will be used
for II:
Z
I=
V
Z
II =
V
∂ρ
Cp T dV
∂t
∂T
ρCp
dV
∂t
Z
= −
Z
Cp T ∇ · (ρv)dV = −
V
Z
=
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
V
V
Z
Cp T ρ v
· n} dS +
| {z
=0 (b.c.)
ρCp ∇T · vdV
(92)
V
∂p
+ ∇ · k∇T + ρH + Φ + αT
+ v · ∇p dV
∂t
Z
∂p
+ ρH + Φ + αT
+ v · ∇p dV +
∇ · k∇T dV
∂t
V
Z
∂p
+ ρH + Φ + αT
+ v · ∇p dV +
k∇T · ndS
∂t
S
Z
∂p
+ ρH + Φ + αT
+ v · ∇p dV −
q · ndS
∂t
S
Finally:
57
(93)
(94)
(95)
(96)
d
I + II =
dt
Z
|V
ρCp T dV
{z
}
Z
Z
∂p
=
+ v · ∇p dV −
ρH + Φ + αT
q · ndS
∂t
V
S
(97)
ET
Z
=
Z
Z
∂p
αT v · ∇pdV −
αT dV +
ΦdV +
∂t
V
V
V
{z
} |
{z
}
| {z } |
Z
ρHdV +
V
visc diss
extra
adiab heating
Z
q · ndS
| {z }
(98)
S
heatflux boundary
This was of course needlessly complicated as the term ∂ρ/∂t is always taken to be zero, so that I = 0 automatically.
The mass conservation equation is then simply ∇ · (ρv) = 0. Then it follows that
Z
Z
Z
ρCp ∇T · vdV
(99)
0 =
Cp T ∇ · (ρv)dV = −
Cp T ρ v
· n} dS +
| {z
V
V
=0 (b.c.)
V
Z
ρCp ∇T · vdV
=
(100)
V
so that the same term in Eq.(96) vanishes too, and then Eq.(98) is always valid, although one should be careful when
computing ET in the BA and EBA cases as it should use ρ0 and not ρ.
23.7
The problem of the onset of convection
[wiki] In geophysics, the Rayleigh number is of fundamental importance: it indicates the presence and strength of
convection within a fluid body such as the Earth’s mantle. The mantle is a solid that behaves as a fluid over geological
time scales.
The Rayleigh number essentially is an indicator of the type of heat transport mechanism. At low Rayleigh numbers
conduction processes dominate over convection ones. At high Rayleigh numbers it is the other way around. There is
a so-called critical value of the number with delineates the transition from one regime to the other.
This problem has been studied and approached both theoretically and numerically [55, e.g.] and it was found that
the critical Rayleigh number Rac is
Rac = (27/4)π 4 ' 657.5
in setups similar to ours.
VERY BIG PROBLEM
The temperature setup is built as follows: Tsurf is prescribed at the top, Tsurf + ∆T is prescribed at the bottom.
The initial temperature profile is linear between these two values. In the case of BA, the actual value of Tsurf is of no
consequence. However, for the EBA the full temperature is present in the adiabatic heating term on the rhs of the hte,
and the value of Tsurf will therefore influence the solution greatly. This is very problematic as there is no real way to
arrive at the surface temperature from the King paper. On top of this, the density uses a reference temperature T0
which too will influence the solution without being present in the controlling Ra and Di numbers!!
In light thereof, it will be very difficult to recover the values of King et al for EBA!
58
features
• Q1 × P0 element
• compressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• isoviscous
• analytical solution
• pressure smoothing
Relevant literature: [4, 28, 50, 33, 31, 34, 35, 24]
ToDo:
• heat flux is at the moment elemental, so Nusselt and heat flux on boundaries measurements not as accurate as
could be.
• implement steady state detection
• do Ra = 105 and Ra = 106
• velocity average at surface
• non dimensional heat flux at corners [5]
• depth-dependent viscosity (case 2 of [5])
59
23.8
results - BA - Ra = 104
These results were obtained with a 64x64 resolution, and CFL number of 1. Steady state was reached after about
1250 timesteps.
7x10-6
-7.73687x1023
7.67188x1022
6x10-6
-7.73687x1023
7.67188x1022
5x10-6
-7.73687x1023
7.67188x1022
4x10-6
-7.73687x1023
3x10-6
EG
ET
EK
7.67188x1022
7.67188x1022
-7.73687x1023
7.67188x1022
1x10-6
0
(a)
-7.73687x1023
7.67188x1022
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
7.67188x1022
(b)
-7.73687x1023
0 100002000030000400005000060000700008000090000
100000
time (Myr)
2273.15
-7.73687x1023
-7.73687x1023
7.67188x1022
2x10-6
-7.73687x1023
4500
-7.73687x1023
(c)
time (Myr)
0.0015
min(T)
max(T)
4000
0 100002000030000400005000060000700008000090000
100000
2273.15
min(u)
max(u)
min(v)
max(v)
0.001
3500
2273.15
2273.15
3000
0.0005
velocity
Temperature
2273.15
2500
2000
1500
0
-0.0005
1000
2273.15
-0.001
500
2273.15
(d)
0
100002000030000400005000060000700008000090000100000
time (Myr)
0
(e)
time (Myr)
-10000
80000
80000
70000
70000
-40000
-50000
-60000
60000
50000
40000
30000
20000
-70000
10000
-80000
0
(g)
work against gravity
90000
-30000
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
(h)
60000
50000
40000
30000
20000
10000
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
0
(i)
0.07
8
2x10-8
0.06
7
-2x10-8
(j)
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
8x10-7
0
(k)
2
1
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
90000
0
HF
dETdt
6x10-7
4
3
0.01
-4x10-8
-5x10-8
5
0.03
0.02
-3x10-8
4.88
0.04
Nu
vrms (cm/yr)
-1x10-8
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
6
0.05
0
0
time (Myr)
3x10-8
1x10-8
0 100002000030000400005000060000700008000090000
100000
time (Myr)
90000
viscous dissipation
adiabatic heating
(f)
0
-20000
heat flux
0 100002000030000400005000060000700008000090000
100000
-0.0015
0
(l)
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
VD
WG
80000
70000
4x10-7
60000
2x10-7
50000
0
40000
30000
-2x10-7
20000
-4x10-7
-6x10-7
(l)
10000
0
0 1x1010
2x1010
3x1010
4x1010
5x1010
6x1010
7x1010
8x1010
9x1010
1x1011
0
1x10102x10103x10104x10105x10106x10107x10108x10109x10101x1011
time (Myr)
(m)
AH: adiabatic heating, VD: viscous dissipation, HF: heat flux, WG: work against gravity
time (Myr)
Eq.(98) is verified by (l) and Eq.(87) is verified by (m).
60
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
61
23.9
results - BA - Ra = 105
These results were obtained with a 64x64 resolution, and CFL number of 1. Steady state was reached after about
1250 timesteps.
0.0007
7.67188x1022
0.0006
7.67188x1022
0.0005
-7.73687x1023
-7.73687x1023
-7.73687x1023
7.67188x1022
-7.73687x1023
7.67188x1022
EG
ET
EK
0.0004
0.0003
7.67188x1022
0.0002
0
(a)
5000
10000
15000
20000
25000
30000
35000
time (Myr)
7.67188x1022
(b)
2273.150000001
4500
2273.150000001
4000
Temperature
-7.73687x1023
(c)
2273.150000000
2273.150000000
2273.150000000
2000
1500
1000
500
2273.149999999
0
min(u)
max(u)
min(v)
max(v)
0
-0.005
-0.01
0
5000
10000 15000 20000 25000 30000 35000
time (Myr)
(e)
0
5000
-0.015
10000 15000 20000 25000 30000 35000
(f)
time (Myr)
0
-100000
800000
800000
-200000
700000
700000
-400000
-500000
-600000
600000
500000
400000
300000
-700000
200000
-800000
100000
-900000
0
5000
10000
15000
20000
25000
30000
35000
time (Myr)
(h)
0.8
2x10-6
0.7
vrms (cm/yr)
1x10-6
5x10-7
0
10000
15000
20000
25000
30000
35000
0
5000
10000
15000
20000
25000
30000
35000
1x10-5
(k)
5000
10000
15000
20000
25000
30000
35000
18
10.53
14
12
0.4
0.3
10
8
6
4
2
0
5000
0
10000 15000 20000 25000 30000 35000
(l)
time (Myr)
900000
700000
600000
2x10-6
500000
0
400000
-2x10-6
300000
-4x10-6
200000
-6x10-6
100000
0
5x109 1x1010 1.5x1010 2x1010 2.5x1010 3x1010 3.5x1010
0
5000
10000
15000
20000
25000
30000
time (Myr)
VD
WG
800000
4x10-6
0
0
16
6x10-6
-8x10-6
300000
time (Myr)
193.2150
0
HF
dETdt
8x10-6
400000
0
0.5
0
500000
(i)
0.1
time (Myr)
(l)
5000
0.2
-5x10-7
600000
100000
0
0.6
1.5x10-6
10000 15000 20000 25000 30000 35000
200000
time (Myr)
2.5x10-6
(j)
work against gravity
900000
viscous dissipation
900000
-300000
5000
time (Myr)
0
-1x10-6
10000 15000 20000 25000 30000 35000
0.005
2500
2273.149999999
(g)
5000
time (Myr)
3000
2273.149999999
0
0
0.015
min(T)
max(T)
Nu
10000 15000 20000 25000 30000 35000
3500
2273.150000000
adiabatic heating
5000
0.01
2273.150000000
(d)
-7.73687x1023
0
time (Myr)
2273.150000001
heat flux
-7.73687x1023
velocity
0
-7.73687x1023
-7.73687x1023
7.67188x1022
0.0001
-7.73687x1023
0
5x109
1x1010 1.5x1010 2x1010 2.5x1010 3x1010 3.5x1010
time (Myr)
(m)
AH: adiabatic heating, VD: viscous dissipation, HF: heat flux, WG: work against gravity
time (Myr)
Eq.(98) is verified by (l) and Eq.(87) is verified by (m).
62
35000
23.10
results - BA - Ra = 106
63
results - EBA - Ra = 104
23.11
These results were obtained with a 64x64 resolution, and CFL number of 1. Steady state was reached after about
2500 timesteps
8x10-7
7.7x1022
7.6x1022
6x10-7
-7.74x1023
7.55x1022
5x10-7
7.5x1022
-7.745x1023
7.45x1022
EG
4x10-7
ET
EK
-7.735x1023
7.65x1022
7x10-7
7.4x1022
3x10-7
-7.75x1023
7.35x1022
2x10-7
7.3x1022
1x10-7
-7.755x1023
7.25x1022
0
0
(a)
50000
100000
150000
200000
250000
time (Myr)
7.2x1022
(b)
0
50000
100000
150000
200000
-7.76x1023
250000
(c)
time (Myr)
2280
4500
2200
2180
2140
(d)
50000
100000
150000
200000
250000
2000
(e)
-0.0003
0
50000
100000
150000
200000
-0.0004
250000
(f)
10000
10000
9000
9000
8000
8000
-3000
7000
-5000
-6000
-7000
6000
5000
4000
3000
-8000
2000
-9000
1000
-10000
0
0
50000
100000
150000
200000
250000
time (Myr)
work against gravity
0
-2000
-4000
(h)
50000
100000
150000
200000
250000
200000
250000
6000
5000
4000
3000
0
0
50000
100000
150000
200000
250000
time (Myr)
2.4
2.2
0.02
2
7000
5000
4000
3000
1.8
0.015
Nu
vrms (cm/yr)
6000
0.01
1.2
0.005
1000
1.6
1.4
2000
1
0
0
50000
100000
150000
200000
250000
time (Myr)
0
(k)
1000
0
50000
100000
-1000
200000
0.8
250000
(l)
0
50000
100000
150000
200000
time (Myr)
10000
VD
WG
9000
8000
-2000
7000
-3000
6000
-4000
5000
-5000
4000
-6000
3000
-7000
2000
-8000
1000
-9000
-10000
150000
time (Myr)
0
AH+VD+HF
dETdt
0
(l)
150000
7000
(i)
0.025
8000
(j)
100000
1000
0
9000
-1000
50000
2000
time (Myr)
10000
0
time (Myr)
-1000
(g)
heat flux
0
-0.0002
time (Myr)
viscous dissipation
adiabatic heating
time (Myr)
min(u)
max(u)
min(v)
max(v)
-0.0001
1500
0
250000
0.0001
2500
500
0
200000
0.0002
3000
1000
2160
150000
0.0003
velocity
Temperature
2220
100000
0.0004
3500
2240
50000
time (Myr)
min(T)
max(T)
4000
2260
0
0
50000
100000
150000
200000
0
250000
0
50000
100000
150000
200000
250000
time (Myr)
(m)
AH: adiabatic heating, VD: viscous dissipation, HF: heat flux, WG: work against gravity
time (Myr)
Eq.(98) is verified by (l) and Eq.(87) is verified by (m).
64
250000
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
65
results - EBA - Ra = 105
23.12
These results were obtained with a 64x64 resolution, and CFL number of 1. Simulation was stopped after about 4300
timesteps.
9x10-5
7.7x1022
8x10-5
7x10-5
6x10-5
0
20000
40000
60000
80000
100000
120000
time (Myr)
7.4x1022
(b)
EG
0
20000
40000
60000
80000
100000
5000
Temperature
2240
2230
2220
2210
-7.75x1023
20000
40000
60000
80000
100000
120000
time (Myr)
3500
0.002
3000
2500
2000
-0.002
1000
-0.003
-0.004
0
20000
40000
60000
80000
100000
-0.005
120000
(f)
time (Myr)
0
viscous dissipation
100000
80000
60000
40000
-100000
20000
-120000
0
0
20000
40000
60000
80000
100000
120000
(h)
0
20000
40000
60000
80000
100000
120000
80000
100000 120000
60000
40000
0
0
20000
40000
60000
80000
100000
0.3
6
5.5
5
50000
4.5
0.2
30000
20000
10000
4
Nu
vrms (cm/yr)
40000
0.15
3.5
3
2.5
0.1
2
0
1.5
0.05
-10000
1
0
20000
40000
60000
80000
100000
120000
time (Myr)
20000
0
(k)
0
20000
40000
60000
80000
100000
120000
time (Myr)
0.5
(l)
0
20000
40000
60000
80000
100000
time (Myr)
120000
0
AH+VD+HF
dETdt
10000
VD
WG
100000
0
80000
-10000
-20000
60000
-30000
40000
-40000
20000
-50000
-60000
(l)
120000
time (Myr)
0.25
(j)
60000
80000
(i)
60000
-20000
40000
20000
time (Myr)
70000
20000
time (Myr)
120000
time (Myr)
heat flux
0
1500
100000
(g)
120000
0.001
120000
-80000
100000
-0.001
0
-60000
80000
min(u)
max(u)
min(v)
max(v)
-20000
-40000
60000
0.004
0.003
(e)
40000
0.005
500
0
20000
time (Myr)
4000
0
0
(c)
velocity
2260
2250
120000
min(T)
max(T)
4500
2270
-7.748x1023
time (Myr)
2280
adiabatic heating
-7.746x1023
7.45x1022
1x10-5
work against gravity
EK
ET
7.5x1022
2x10-5
(d)
7.55x1022
-7.744x1023
3x10-5
2200
-7.74x1023
-7.742x1023
4x10-5
(a)
-7.738x1023
7.6x1022
5x10-5
0
-7.736x1023
7.65x1022
0
20000
40000
60000
80000
0
100000 120000
0
20000
40000
60000
80000
100000
120000
time (Myr)
(m)
AH: adiabatic heating, VD: viscous dissipation, HF: heat flux, WG: work against gravity
time (Myr)
66
120000
23.13
Onset of convection
The code can be run for values of Ra between 500 and 1000, at various resolutions for the BA formulation. The value
vrms (t) − vrms (0) is plotted as a function of Ra and for the 10 first timesteps. If the vrms is found to decrease, then
the Rayleigh number is not high enough to allow for convection and the initial temperature perturbation relaxes by
diffusion (and then vrms (t) − vrms (0) < 0. If the vrms is found to increase, then vrms (t) − vrms (0) > 0 and the system
is going to showcase convection. The zero value of vrms (t) − vrms (0) gives us the critical Rayleigh number, which is
found between 775 and 790.
8x10-15
1.5x10-16
16x16
32x32
48x48
64x64
6x10-15
5x10-17
0
vrms trend
4x10-15
vrms trend
16x16
32x32
48x48
64x64
1x10-16
2x10-15
0
-5x10-17
-1x10-16
-1.5x10-16
-2x10-16
-2.5x10-16
-2x10-15
-3x10-16
-4x10-15
-3.5x10-16
770
1000
Ra
775
780
Ra
67
785
790
A
The main codes in computational geodynamics
In what follows I make a quick inventory of the main codes of computational geodynamics, for crust, lithosphere
and/or mantle modelling.
A.1
ADELI
A.2
ASPECT
A.3
CITCOMS and CITCOMCU
A.4
DOUAR
A.5
GAIA
A.6
GALE
A.7
GTECTON
A.8
ELVIS
A.9
ELEFANT
A.10
ELLIPSIS
A.11
FANTOM
A.12
FLUIDITY
A.13
LAMEM
A.14
MILAMIN
A.15
PARAVOZ/FLAMAR
A.16
PTATIN
A.17
RHEA
A.18
SEPRAN
A.19
SOPALE
A.20
STAGYY
A.21
SULEC
SULEC is a finite element code that solves the incompressible Navier-Stokes equations for slow creeping flows. The
code is developed by Susan Ellis (GNS Sciences, NZ) and Susanne Buiter (NGU).
A.22
TERRA
A.23
UNDERWORLD 1&2
68
B
fieldstone.py
69
C
fieldstone stokes sphere.py
D
fieldstone convection box.py
70
E
fieldstone solcx.py
71
F
fieldstone indentor.py
72
G
fieldstone saddlepoint.py
73
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Index
Q1 × P0 , 20, 26, 32, 38, 41, 53, 59
Q2 × Q1 , 43, 48
analytical solution, 20, 26, 41, 48, 50, 53, 59
arithmetic mean, 38
basis functions, 12
Boussinesq, 8
bulk modulus, 54
bulk viscosity, 5
compressibility, 54
compressible flow, 59
dynamic viscosity, 5
Gauss quadrature, 10
geometric mean, 38
harmonic mean, 38
incompressible flow, 26, 32, 38, 41, 48, 50, 53
isothermal, 20, 26, 32, 38, 41, 48, 50, 53
isoviscous, 20, 41, 48, 53, 59
Legendre polynomial, 10
midpoint rule, 9
mixed formulation, 41, 48, 50, 53, 59
Newton-Cotes, 10
non-isoviscous, 26, 32, 38, 50
nonconforming Q1 × P0 , 50
nonlinear rheology, 32
particle-in-cell, 38
penalty formulation, 14, 20, 26, 32, 38
pressure normalisation, 44
pressure smoothing, 41, 50, 53, 59
quadrature, 10
rectangle rule, 9
second viscosity, 5
thermal expansion, 54
trapezoidal rule, 9
weak form, 16
work against gravity, 56
77
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