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The Finite Element Method in Geodynamics
C. Thieulot
April 19, 2019
Contents
1 Introduction
1.1 Philosophy . . . . . . . . . . . .
1.2 Acknowledgments . . . . . . . . .
1.3 Essential literature . . . . . . . .
1.4 Installation . . . . . . . . . . . .
1.5 What is a fieldstone? . . . . . . .
1.6 Why the Finite Element method?
1.7 Notations . . . . . . . . . . . . .
1.8 Colour maps for visualisation . .
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2 List of tutorials
3 The
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
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physical equations
The heat transport equation - energy conservation equation
The momentum conservation equations . . . . . . . . . . .
The mass conservation equations . . . . . . . . . . . . . . .
The equations in ASPECT manual . . . . . . . . . . . . . .
the Boussinesq approximation: an Incompressible flow . . .
Stokes equation for elastic medium . . . . . . . . . . . . . .
The strain rate tensor in all coordinate systems . . . . . . .
3.7.1 Cartesian coordinates . . . . . . . . . . . . . . . . .
3.7.2 Polar coordinates . . . . . . . . . . . . . . . . . . . .
3.7.3 Cylindrical coordinates . . . . . . . . . . . . . . . .
3.7.4 Sperical coordinates . . . . . . . . . . . . . . . . . .
Boundary conditions . . . . . . . . . . . . . . . . . . . . . .
3.8.1 The Stokes equations . . . . . . . . . . . . . . . . . .
3.8.2 The heat transport equation . . . . . . . . . . . . .
Meaningful physical quantities . . . . . . . . . . . . . . . .
4 The building blocks of the Finite Element Method
4.1 Numerical integration . . . . . . . . . . . . . . . . .
4.1.1 in 1D - theory . . . . . . . . . . . . . . . . .
4.1.2 in 1D - examples . . . . . . . . . . . . . . . .
4.1.3 in 2D/3D - theory . . . . . . . . . . . . . . .
4.2 The mesh . . . . . . . . . . . . . . . . . . . . . . . .
4.3 A bit of FE terminology . . . . . . . . . . . . . . . .
4.4 Elements and basis functions in 1D . . . . . . . . . .
4.4.1 Linear basis functions (Q1 ) . . . . . . . . . .
4.4.2 Quadratic basis functions (Q2 ) . . . . . . . .
4.4.3 Cubic basis functions (Q3 ) . . . . . . . . . .
4.5 Elements and basis functions in 2D . . . . . . . . . .
4.5.1 Bilinear basis functions in 2D (Q1 ) . . . . . .
4.5.2 Biquadratic basis functions in 2D (Q2 ) . . . .
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4.5.3 Eight node serendipity basis functions in 2D (Qs2 )
4.5.4 Bicubic basis functions in 2D (Q3 ) . . . . . . . . .
4.5.5 Linear basis functions for triangles in 2D (P1 ) . . .
4.5.6 Quadratic basis functions for triangles in 2D (P2 ) .
4.5.7 Cubic basis functions for triangles (P3 ) . . . . . .
Elements and basis functions in 3D . . . . . . . . . . . . .
4.6.1 Linear basis functions in tetrahedra (P1 ) . . . . . .
4.6.2 Triquadratic basis functions in 3D (Q2 ) . . . . . .
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5 Solving the flow equations with the FEM
5.1 strong and weak forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Which velocity-pressure pair for Stokes? . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 The compatibility condition (or LBB condition) . . . . . . . . . . . . . . . . . . . .
5.3 Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 The bi/tri-linear velocity - constant pressure element (Q1 × P0 ) . . . . . . . . . . .
5.3.2 The bi/tri-quadratic velocity - discontinuous linear pressure element (Q2 × P−1 ) .
5.3.3 The bi/tri-quadratic velocity - bi/tri-linear pressure element (Q2 × Q1 ) . . . . . .
5.3.4 The stabilised bi/tri-linear velocity - bi/tri-linear pressure element (Q1 × Q1 -stab)
5.3.5 The MINI triangular element (P1+ × P1 ) . . . . . . . . . . . . . . . . . . . . . . . .
5.3.6 The quadratic velocity - linear pressure triangle (P2 × P1 ) . . . . . . . . . . . . . .
5.3.7 The Crouzeix-Raviart triangle (P2+ × P−1 ) . . . . . . . . . . . . . . . . . . . . . .
5.4 Other elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 The penalty approach for viscous flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6 The mixed FEM for viscous flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.1 in three dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.2 Going from 3D to 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 Solving the elastic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.8 Solving the heat transport equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6 Additional techniques and features
6.1 Picard and Newton . . . . . . . . . . . . . . . . . . . .
6.2 The SUPG formulation for the energy equation . . . .
6.3 Tracking materials and/or interfaces . . . . . . . . . .
6.4 Dealing with a free surface . . . . . . . . . . . . . . . .
6.5 Convergence criterion for nonlinear iterations . . . . .
6.6 Static condensation . . . . . . . . . . . . . . . . . . . .
6.7 The method of manufactured solutions . . . . . . . . .
6.7.1 Analytical benchmark I . . . . . . . . . . . . .
6.7.2 Analytical benchmark II . . . . . . . . . . . .
6.7.3 Analytical benchmark III . . . . . . . . . . . .
6.7.4 Analytical benchmark IV . . . . . . . . . . . .
6.8 Assigning values to quadrature points . . . . . . . . .
6.9 Matrix (Sparse) storage . . . . . . . . . . . . . . . . .
6.9.1 2D domain - One degree of freedom per node .
6.9.2 2D domain - Two degrees of freedom per node
6.9.3 in fieldstone . . . . . . . . . . . . . . . . . . . .
6.10 Mesh generation . . . . . . . . . . . . . . . . . . . . .
6.11 Visco-Plasticity . . . . . . . . . . . . . . . . . . . . . .
6.11.1 Tensor invariants . . . . . . . . . . . . . . . . .
6.11.2 Scalar viscoplasticity . . . . . . . . . . . . . . .
6.11.3 about the yield stress value Y . . . . . . . . . .
6.12 Pressure smoothing . . . . . . . . . . . . . . . . . . . .
6.13 Pressure scaling . . . . . . . . . . . . . . . . . . . . . .
6.14 Pressure normalisation . . . . . . . . . . . . . . . . . .
6.15 The choice of solvers . . . . . . . . . . . . . . . . . . .
6.15.1 The Schur complement approach . . . . . . . .
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4.6
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6.16 The GMRES approach . . . . . . . . . . . . . . . . . . . . . . .
6.17 The consistent boundary flux (CBF) . . . . . . . . . . . . . . .
6.17.1 applied to the Stokes equation . . . . . . . . . . . . . .
6.17.2 applied to the heat equation . . . . . . . . . . . . . . . .
6.17.3 implementation - Stokes equation . . . . . . . . . . . . .
6.18 The value of the timestep . . . . . . . . . . . . . . . . . . . . .
6.19 mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.20 Exporting data to vtk format . . . . . . . . . . . . . . . . . . .
6.21 Runge-Kutta methods . . . . . . . . . . . . . . . . . . . . . . .
6.22 Am I in or not? . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.22.1 Two-dimensional space . . . . . . . . . . . . . . . . . . .
6.22.2 Three-dimensional space . . . . . . . . . . . . . . . . . .
6.23 Error measurements and convergence rates . . . . . . . . . . .
6.24 The initial temperature field . . . . . . . . . . . . . . . . . . . .
6.24.1 Single layer with imposed temperature b.c. . . . . . . .
6.25 Single layer with imposed heat flux b.c. . . . . . . . . . . . . .
6.26 Single layer with imposed heat flux and temperature b.c. . . .
6.26.1 Half cooling space . . . . . . . . . . . . . . . . . . . . .
6.26.2 Plate model . . . . . . . . . . . . . . . . . . . . . . . . .
6.26.3 McKenzie slab . . . . . . . . . . . . . . . . . . . . . . .
6.27 Kinematic boundary conditions . . . . . . . . . . . . . . . . . .
6.27.1 In-out flux boundary conditions for lithospheric models
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74
75
75
75
76
79
80
81
83
84
84
84
87
88
88
89
89
89
89
89
90
90
7 fieldstone 01: simple analytical solution
91
8 fieldstone 02: Stokes sphere
93
9 fieldstone 03: Convection in a 2D box
94
10 fieldstone 04: The lid
10.1 the lid driven cavity
10.2 the lid driven cavity
10.3 the lid driven cavity
97
97
97
97
driven cavity
problem (ldc=0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
problem - regularisation I (ldc=1) . . . . . . . . . . . . . . . . . . . .
problem - regularisation II (ldc=2) . . . . . . . . . . . . . . . . . . .
11 fieldstone 05: SolCx benchmark
99
12 fieldstone 06: SolKz benchmark
101
13 fieldstone 07: SolVi benchmark
102
14 fieldstone 08: the indentor benchmark
104
15 fieldstone 09: the annulus benchmark
106
16 fieldstone 10: Stokes sphere (3D) - penalty
108
17 fieldstone 11: stokes sphere (3D) - mixed formulation
109
18 fieldstone 12: consistent pressure recovery
110
19 fieldstone 13: the Particle in Cell technique (1) - the effect of averaging
112
20 fieldstone f14: solving the full saddle point problem
116
21 fieldstone f15: saddle point problem with Schur complement approach - benchmark
118
3
22 fieldstone f16: saddle point problem with Schur complement approach - Stokes
sphere
121
123
23 fieldstone 17: solving the full saddle point problem in 3D
23.0.1 Constant viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
23.0.2 Variable viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
24 fieldstone 18: solving the full saddle point problem with Q2 × Q1 elements
128
25 fieldstone 19: solving the full saddle point problem with Q3 × Q2 elements
130
26 fieldstone 20: the Busse benchmark
132
27 fieldstone 21: The non-conforming Q1 × P0 element
134
28 fieldstone 22: The stabilised Q1 × Q1 element
28.1 The Donea & Huerta benchmark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28.2 The Dohrmann & Bochev benchmark . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28.3 The falling block experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135
136
136
137
29 fieldstone 23: compressible flow (1) - analytical benchmark
138
30 fieldstone 24: compressible flow (2) - convection box
30.1 The physics . . . . . . . . . . . . . . . . . . . . . . . . .
30.2 The numerics . . . . . . . . . . . . . . . . . . . . . . . .
30.3 The experimental setup . . . . . . . . . . . . . . . . . .
30.4 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30.5 Conservation of energy 1 . . . . . . . . . . . . . . . . . .
30.5.1 under BA and EBA approximations . . . . . . .
30.5.2 under no approximation at all . . . . . . . . . . .
30.6 Conservation of energy 2 . . . . . . . . . . . . . . . . . .
30.7 The problem of the onset of convection . . . . . . . . . .
30.8 results - BA - Ra = 104 . . . . . . . . . . . . . . . . . .
30.9 results - BA - Ra = 105 . . . . . . . . . . . . . . . . . .
30.10results - BA - Ra = 106 . . . . . . . . . . . . . . . . . .
30.11results - EBA - Ra = 104 . . . . . . . . . . . . . . . . .
30.12results - EBA - Ra = 105 . . . . . . . . . . . . . . . . .
30.13Onset of convection . . . . . . . . . . . . . . . . . . . . .
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141
141
141
143
143
144
144
145
145
146
148
150
151
152
154
155
31 fieldstone 25: Rayleigh-Taylor instability (1) - instantaneous
157
32 fieldstone 26: Slab detachment benchmark (1) - instantaneous
159
33 fieldstone 27: Consistent Boundary Flux
161
34 fieldstone 28: convection 2D box - Tosi et al, 2015
34.0.1 Case 0: Newtonian case, a la Blankenbach et al., 1989
34.0.2 Case 1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
34.0.3 Case 2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
34.0.4 Case 3 . . . . . . . . . . . . . . . . . . . . . . . . . . .
34.0.5 Case 4 . . . . . . . . . . . . . . . . . . . . . . . . . . .
34.0.6 Case 5 . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 fieldstone 29: open boundary conditions
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165
166
167
169
171
173
175
177
4
36 fieldstone 30: conservative velocity interpolation
180
36.1 Couette flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
36.2 SolCx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
36.3 Streamline flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
37 fieldstone 31: conservative velocity interpolation 3D
181
38 fieldstone 32: 2D analytical sol. from stream function
182
38.1 Background theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
38.2 A simple application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
39 fieldstone 33: Convection in an annulus
186
40 fieldstone 34: the Cartesian geometry elastic aquarium
188
41 fieldstone 35: 2D analytical sol.
41.1 Linking with our paper . . . . .
41.2 No slip boundary conditions . .
41.3 Free slip boundary conditions .
in
. .
. .
. .
annulus from
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
stream function
190
. . . . . . . . . . . . . . . . . . . . . . 193
. . . . . . . . . . . . . . . . . . . . . . 193
. . . . . . . . . . . . . . . . . . . . . . 194
42 fieldstone 36: the annulus geometry elastic aquarium
196
43 fieldstone 37: marker advection and population control
199
44 fieldstone 38: Critical Rayleigh number
200
45 fieldstone: Gravity: buried sphere
202
46 Problems, to do list and projects for students
204
A Three-dimensional applications
206
B Codes in geodynamics
207
C Matrix properties
209
C.1 Symmetric matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
C.2 Schur complement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
5
WARNING: this is work in progress
1
Introduction
1.1
Philosophy
This document was writing with my students in mind, i.e. 3rd and 4th year Geology/Geophysics students at Utrecht University. I have chosen to use jargon as little as possible unless it is a term that is
commonly found in the geodynamics literature (methods paper as well as application papers). There is
no mathematical proof of any theorem or statement I make. These are to be found in generic Numerical
Analysic, Finite Element and Linear Algebra books.
The codes I provide here are by no means optimised as I value code readability over code efficiency. I
have also chosen to avoid resorting to multiple code files or even functions to favour a sequential reading of
the codes. These codes are not designed to form the basis of a real life application: Existing open source
highly optimised codes shoud be preferred, such as ASPECT [297, 253], CITCOM, LAMEM, PTATIN,
PYLITH, ...
All kinds of feedback is welcome on the text (grammar, typos, ...) or on the code(s). You will have
my eternal gratitude if you wish to contribute an example, a benchmark, a cookbook.
All the python scripts and this document are freely available at
https://github.com/cedrict/fieldstone
1.2
Acknowledgments
I have benefitted from many discussions, lectures, tutorials, coffee machine discussions, debugging sessions, conference poster sessions, etc ... over the years. I wish to name these instrumental people in
particular and in alphabetic order: Wolfgang Bangerth, Jean Braun, Rens Elbertsen, Philippe Fullsack,
Menno Fraters, Anne Glerum, Timo Heister, Robert Myhill, John Naliboff, E. Gerry Puckett, Lukas van
de Wiel, Arie van den Berg, Tom Weir, and the whole ASPECT family/team.
1.3
Essential literature
http://www-udc.ig.utexas.edu/external/becker/Geodynamics557.pdf
1.4
Installation
python3.6 -m pip install --user numpy scipy matplotlib
1.5
What is a fieldstone?
Simply put, it is stone collected from the surface of fields where it occurs naturally. It also stands for the
bad acronym: finite element deformation of stones which echoes the primary application of these codes:
geodynamic modelling.
6
1.6
Why the Finite Element method?
The Finite Element Method (FEM) is by no means the only method to solve PDEs in geodynamics, nor
is it necessarily the best one. Other methods are employed very succesfully, such as the Finite Difference
Method (FDM), the Finite Volume Method (FVM), and to a lesser extent the Discrete Element Method
(DEM) [152, 153, 184], or the Element Free Galerkin Method (EFGM) [247].
1.7
Notations
Scalars such as temperature, density, pressure, etc ... are simply obtained in LATEXby using the math
mode: T , ρ, p. Although it is common to lump vectors and matrices/tensors together by using bold
fonts, I have decided in the interest of clarity to distinguish between those: vectors are denoted by an
arrow atop the quantity, e.g. ~ν, ~g , while matrices and tensors are in bold M , σ, etc ...
Also I use the · notation between two vectors to denote a dot product ~u · ~v = ui vi or a matrix-vector
multiplication M · ~a = Mij aj . If there is no · between vectors, it means that the result ~a~b = ai bj is a
~ · ~ν is the velocity divergence while ∇~
~ ν is the velocity gradient tensor.
matrix. Case in point, ∇
1.8
Colour maps for visualisation
In an attempt to homogenise the figures obtained with ParaView, I have decided to use a fixed colour
scale for each field throughout this document. These colour scales were obtained from this link and are
Perceptually Uniform Colour Maps [296].
Field
colour code
Velocity/displacement
CET-D1A
Pressure
CET-L17
Velocity divergence
CET-L1
Density
CET-D3
Strain rate
CET-R2
Viscosity
CET-R3
Temperature
CET-D9
7
8
23
24
full matrix
Schur comp. CG
Schur comp. PCG
full matrix
full matrix
full matrix
full matrix
mixed
mixed
analytical benchmark
convection box
†
†
†
†
†
†
†
†
†
†
†
†
†
elastomechanics
mixed
mixed
mixed
mixed
mixed
mixed
penalty
penalty
mixed
†
numerical benchmark
Q1 × P0
Q1 × P0
Q1 × P0
Q2 × Q1
Q2 × Q1
Q3 × Q2
Q1 × P0
Q1 × P0 R-T
Q1 × Q1 stab
Q1 × P0
Q1 × P0
penalty
analytical benchmark
14
15
16
17
18
19
20
21
22
physical problem
analytical benchmark
Stokes sphere
Blankenbach et al., 1989
Lid driven cavity
SolCx benchmark
SolKz benchmark
SolVi benchmark
Indentor
annulus benchmark
Stokes sphere
Stokes sphere
analytical benchmark
+ consistent press recovery
Stokes sphere
+ markers averaging
analytical benchmark
analytical benchmark
Stokes sphere
Burstedde benchmark
analytical benchmark
analytical benchmark
Busse et al., 1993
analytical benchmark
analytical benchmark
compressible
Q1 × P0
full matrix
formulation
penalty
penalty
penalty
penalty
penalty
penalty
penalty
penalty
penalty
penalty
mixed
penalty
nonlinear
13
outer solver
time stepping
1
2
3
4
5
6
7
8
9
10
11
12
element
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
temperature
List of tutorials
3D
tutorial number
2
elastomechanics
numerical benchmark
analytical benchmark
compressible
nonlinear
physical problem
Rayleigh-Taylor instability
Slab detachment
CBF benchmarks
Tosi et al, 2015
Open Boundary conditions
Cons. Vel. Interp (cvi)
Cons. Vel. Interp (cvi)
analytical benchmark
convection in annulus
elastic Cartesian aquarium
time stepping
formulation
mixed
mixed
mixed
mixed
mixed
X
X
mixed
penalty
temperature
outer solver
full matrix
full matrix
full matrix
full matrix
full matrix
X
X
full matrix
3D
tutorial number
9
element
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 × P0
Q1 , Q2
Q1 , Q2
Q1 × P0
Q1 × P0
Q1
25
26
†
27
†
†
28
† † †
†
29
†
30
†
†
31
†
†
†
32
†
33
† † †
34
† †
†
35
36
Q1
elastic annulus aquarium
† †
†
37
Q1 , Q2
X
X
population control, bmw test
†
†
38
Critical Rayleigh number
XX
Analytical benchmark means that an analytical solution exists while numerical benchmark means that a comparison with other code(s) has been carried
out.
3
The physical equations
Symbol
t
x, y, z
r, θ
r, θ, z
r, θ, φ
~ν
ρ
η
λ
T
~
∇
~
∇·
p
ε̇(~ν)
α
k
Cp
H
βT
τ
σ
3.1
meaning
Time
Cartesian coordinates
Polar coordinates
Cylindrical coordinates
Spherical 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
deviatoric stress tensor
full stress tensor
unit
s
m
m,m,-,m
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
Pa
Pa
The heat transport equation - energy conservation equation
Let us start from the heat transport equation as shown in Schubert, Turcotte and Olson [416]:
ρCp
Dp
DT
~ · k ∇T
~ + Φ + ρH
− αT
=∇
Dt
Dt
with D/Dt being the total derivatives so that
∂T
DT
~
=
+ ~ν · ∇T
Dt
∂t
Dp
∂p
~
=
+ ~ν · ∇p
Dt
∂t
Solving for temperature, this equation is often rewritten as follows:
ρCp
DT
~ · k ∇T
~ = αT Dp + Φ + ρH
−∇
Dt
Dt
~ ν.
A note on the shear heating term Φ: In many publications, Φ is given by Φ = τij ∂j ui = τ : ∇~
10
Φ
= τij ∂j ui
=
=
=
=
=
2η ε̇dij ∂j ui
1
2η ε̇dij ∂j ui + ε̇dji ∂i uj
2
1 d
2η ε̇ij ∂j ui + ε̇dij ∂i uj
2
1
2η ε̇dij (∂j ui + ∂i uj )
2
2η ε̇dij ε̇ij
2η ε̇d : ε̇
1 ~
= 2η ε̇d : ε̇d + (∇
· ~ν)1
3
d
d
~ · ~ν)
= 2η ε̇ : ε̇ + 2η ε̇d : 1(∇
=
=
2η ε̇d : ε̇d
(1)
Finally
~ ν = 2η ε̇d : ε̇d = 2η (ε̇dxx )2 + (ε̇dyy )2 + 2(ε̇dxy )2
Φ = τ : ∇~
3.2
The momentum conservation equations
Because the Prandlt number is virtually zero in Earth science applications the Navier Stokes equations
reduce to the Stokes equation:
~ · σ + ρ~g = 0
∇
Since
σ = −p1 + τ
it also writes
~ +∇
~ · τ + ρ~g = 0
−∇p
Using the relationship τ = 2η ε̇d we arrive at
~ +∇
~ · (2η ε̇d ) + ρ~g = 0
−∇p
3.3
The mass conservation equations
The mass conservation equation is given by
Dρ
~ · ~ν = 0
+ ρ∇
Dt
or,
∂ρ ~
+ ∇ · (ρ~ν) = 0
∂t
~ = 0, i.e. Dρ/Dt = 0 and the remaining
In the case of an incompressible flow, then ∂ρ/∂t = 0 and ∇ρ
equation is simply:
~ · ~ν = 0
∇
11
3.4
The equations in ASPECT manual
The following is lifted off the 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 [416].
Specifically, we consider the following set of equations for velocity u, pressure p and temperature T :
~ · ~ν)1 + ∇p
~ = ρ~g
~ · 2η ε̇(~ν) − 1 (∇
in Ω, (2)
−∇
3
~ · (ρ~v ) = 0
∇
in Ω, (3)
∂T
~
~ · k ∇T
~ = ρH
ρCp
+ ~ν · ∇T
−∇
∂t
1 ~
1 ~
+ 2η ε̇(v) − (∇
· ~ν)1 : ε̇(v) − (∇
· ~ν)1
(4)
3
3
~
+ αT v · ∇p
in Ω,
~ ν + ∇~
~ νT ) is the symmetric gradient of the velocity (often called the strain rate).
where ε̇(~ν) = 21 (∇~
In this set of equations, (253) and (254) 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 (255) 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;
In order to arrive at the set of equations that ASPECT solves, we need to
• neglect the ∂p/∂t. WHY?
• neglect the ∂ρ/∂t . WHY?
from equations above.
—————————————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
12
3.5
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
(253). The primary result of this assumption is that the continuity equation (254) 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
∇ · (ρv) = 0
∂T
+ v · ∇T − ∇ · k∇T = ρH
ρ0 C p
∂t
in Ω,
in Ω,
(5)
(6)
in Ω
(7)
Note that all terms on the rhs of the temperature equations have disappeared, with the exception of the
source term.
13
3.6
Stokes equation for elastic medium
What follows is mostly borrowed from Becker & Kaus lecture notes.
The strong form of the PDE that governs force balance in a medium is given by
∇·σ+f =0
where σ is the stress tensor and f is a body force.
The stress tensor is related to the strain tensor through the generalised Hooke’s law:
X
σij =
Cijkl kl
(8)
kl
where C is the fourth-order elastic tensor. In the case of an isotropic material, this relationship simplifies
to
σij = λkk δij + 2µij
or,
σ = λ(∇ · u)1 + 2µ
(9)
where λ is the Lamé parameter and µ is the shear modulus1 . The term ∇ · u is the isotropic dilation.
The strain tensor is related to the displacement as follows:
=
1
(∇u + ∇uT )
2
The incompressibility (bulk modulus), K, is defined as p = −K∇ · u where p is the pressure with
p
1
= − T r(σ)
3
1
= − [λ(∇ · u)T r[1] + 2µT r[]]
3
1
= − [λ(∇ · u)3 + 2µ(∇ · u)]
3
2
= −[λ + µ](∇ · u)
3
(10)
so that K = λ + 23 µ.
Remark : Eq. (8) and (9) are analogous to the ones that one has to solve in the context of viscous
flow using the penalty method. In this case λ is the penalty coefficient, u is the velocity, and µ is then
the dynamic viscosity.
The Lamé parameter and the shear modulus are also linked to ν the poisson ratio, and E, Young’s
modulus:
2ν
νE
λ=µ
=
with
E = 2µ(1 + ν)
1 − 2ν
(1 + ν)(1 − 2ν)
The shear modulus, expressed often in GPa, describes the material’s response to shear stress. The poisson
ratio describes the response in the direction orthogonal to uniaxial stress. The Young modulus, expressed
in GPa, describes the material’s strain response to uniaxial stress in the direction of this stress.
1 It
is also sometimes written G
14
3.7
The strain rate tensor in all coordinate systems
The strain rate tensor ε̇ is given by
ε̇ =
3.7.1
3.7.2
3.7.3
1 ~
~ νT )
(∇~ν + ∇~
2
(11)
Cartesian coordinates
ε̇xx
=
ε̇yy
=
ε̇zz
=
ε̇yx = ε̇xy
=
ε̇zx = ε̇xz
=
ε̇zy = ε̇yz
=
∂u
∂x
∂v
∂y
∂w
∂z
1 ∂u ∂v
+
2 ∂y
∂x
1 ∂u ∂w
+
2 ∂z
∂x
1 ∂v ∂w
+
2 ∂z
∂y
(12)
(13)
(14)
(15)
(16)
(17)
Polar coordinates
ε̇rr
=
ε̇θθ
=
ε̇θr = ε̇rθ
=
∂vr
∂r
vr
1 ∂vθ
+
r r ∂θ
1 ∂vθ
vθ
1 ∂vr
−
+
2 ∂r
r
r ∂θ
(18)
(19)
(20)
Cylindrical coordinates
http://eml.ou.edu/equation/FLUIDS/STRAIN/STRAIN.HTM
3.7.4
Sperical coordinates
ε̇rr
=
ε̇θθ
=
ε̇φφ
=
ε̇θr = ε̇rθ
=
ε̇φr = ε̇rφ
=
ε̇φθ = ε̇θφ
=
∂vr
∂r
vr
1 ∂vθ
+
r
r ∂θ
1 ∂vφ
r sin θ ∂φ
1
∂ vθ
1 ∂vr
r ( )+
2
∂r r
r ∂θ
1
1 ∂vr
∂ vφ
+r ( )
2 r sin θ ∂φ
∂r r
1 sin θ ∂ vφ
1 ∂vθ
(
)+
2
r ∂θ sin θ
r sin θ ∂φ
15
(21)
(22)
(23)
(24)
(25)
(26)
3.8
Boundary conditions
In mathematics, the Dirichlet (or first-type) boundary condition is a type of boundary condition, named
after Peter Gustav Lejeune Dirichlet. When imposed on an ODE or PDE, it specifies the values that a
solution needs to take on along the boundary of the domain. Note that a Dirichlet boundary condition
may also be referred to as a fixed boundary condition.
The Neumann (or second-type) boundary condition is a type of boundary condition, named after Carl
Neumann. When imposed on an ordinary or a partial differential equation, the condition specifies the
values in which the derivative of a solution is applied within the boundary of the domain.
It is possible to describe the problem using other boundary conditions: a Dirichlet boundary condition
specifies the values of the solution itself (as opposed to its derivative) on the boundary, whereas the Cauchy
boundary condition, mixed boundary condition and Robin boundary condition are all different types of
combinations of the Neumann and Dirichlet boundary conditions.
3.8.1
The Stokes equations
You may find the following terms in the computational geodynamics literature:
• free surface: this means that no force is acting on the surface, i.e. σ · ~n = ~0. It is usually used on
the top boundary of the domain and allows for topography evolution.
• free slip: ~ν · ~n = 0 and (σ · ~n) × ~n = ~0. This condition ensures a frictionless flow parallel to the
boundary where it is prescribed.
• no slip: this means that the velocity (or displacement) is exactly zero on the boundary, i.e. ~ν = ~0.
• prescribed velocity: ~ν = ~νbc
• stress b.c.:
• open .b.c.: see fieldstone 29.
3.8.2
The heat transport equation
There are two types of boundary conditions for this equation: temperature boundary conditions (Dirichlet
boundary conditions) and heat flux boundary conditions (Neumann boundary conditions).
16
3.9
Meaningful physical quantities
• Velocity (m/s):
• Root mean square velocity (m/s):
R
νrms =
|~ν|2 dΩ
dΩ
Ω
ΩR
1/2
=
1
VΩ
Z
1/2
2
|~ν| dΩ
(27)
Ω
In Cartesian coordinates, for a cuboid domain of size Lx × Ly × Lz, the vrms is simply given by:
νrms =
1
Lx Ly Lz
Z
Lx
Ly
Z
!1/2
Lz
Z
2
2
2
(u + v + w )dxdydz
0
(28)
0
0
In the case of an annulus domain, although calculations are carried out in Cartesian coordinates,
it makes sense to look at the radial velocity component vr and the tangential velocity component
vθ , and their respective root mean square averages:
vr |rms =
vθ |rms =
1
VΩ
Z
1
VΩ
Z
vr2
1/2
(29)
dΩ
Ω
vθ2 dΩ
1/2
(30)
Ω
• Pressure (Pa):
• Stress (Pa):
• Strain (X):
• Strain rate (s−1 ):
• Rayleigh number (X):
• Prandtl number (X):
• Nusselt number (X): the Nusselt number (Nu) is the ratio of convective to conductive heat transfer
across (normal to) the boundary. The conductive component is measured under the same conditions
as the heat convection but with a (hypothetically) stagnant (or motionless) fluid.
In practice the Nusselt number Nu of a layer (typically the mantle of a planet) is defined as follows:
Nu =
q
qc
(31)
where q is the heat transferred by convection while qc = k∆T /D is the amount of heat that would
be conducted through a layer of thickness D with a temperature difference ∆T across it with k
being the thermal conductivity.
For 2D Cartesian systems of size (Lx ,Ly ) the Nu is computed [55]
Nu =
1
Lx
R Lx
0
R Lx
1
− Lx 0
k ∂T
∂y (x, y = Ly )dx
kT (x, y = 0)/Ly dx
R Lx
0
∂T
∂y
= −Ly R Lx
0
(x, y = Ly )dx
T (x, y = 0)dx
i.e. it is the mean surface temperature gradient over the mean bottom temperature.
finish. not happy with definition. Look at literature
Note that in the case when no convection takes place then the measured heat flux at the top is the
one obtained from a purely conductive profile which yields Nu=1.
Note that a relationship Ra ∝ Nuα exists between the Rayleigh number Ra and the Nusselt number
Nu in convective systems, see [505] and references therein.
17
Turning now to cylindrical geometries with inner radius R1 and outer radius R2 , we define f =
R1 /R2 . A small value of f corresponds to a high degree of curvature. We assume now that
R2 − R1 = 1, so that R2 = 1/(1 − f ) and R1 = f /(1 − f ). Following [276], the Nusselt number at
the inner and outer boundaries are:
Nuinner =
Nuouter =
f ln f 1
1 − f 2π
Z
ln f 1
1 − f 2π
Z
2π
0
2π
0
∂T
∂r
∂T
∂r
dθ
(32)
dθ
(33)
r=R1
r=R2
Note that a conductive geotherm in such an annulus between temperatures T1 and T2 is given by
Tc (r) =
ln(r/R2 )
ln(r(1 − f ))
=
ln(R1 /R2 )
ln f
so that
∂Tc
1 1
=
∂r
r ln f
We then find:
Nuinner
Nuouter
=
=
f ln f 1
1 − f 2π
Z
ln f 1
1 − f 2π
Z
2π
∂Tc
∂r
∂Tc
∂r
0
0
2π
dθ =
f ln f 1 1
=1
1 − f R1 ln f
(34)
dθ =
ln f 1 1
=1
1 − f R2 ln f
(35)
r=R1
r=R2
As expected, the recovered Nusselt number at both boundaries is exactly 1 when the temperature
field is given by a steady state conductive geotherm.
derive formula for Earth size R1 and R2
• Temperature (K):
• Viscosity (Pa.s):
• Density (kg/m3 ):
• Heat capacity cp (J.K −1 .kg −1 ): It is the measure of the heat energy required to increase the
temperature of a unit quantity of a substance by unit degree.
• Heat conductivity, or thermal conductivity k (W.m−1 .K −1 ). It is the property of a material that
indicates its ability to conduct heat. It appears primarily in Fourier’s Law for heat conduction.
• Heat diffusivity: κ = k/(ρcp ) (m2 .s−1 ). Substances with high thermal diffusivity rapidly adjust
their temperature to that of their surroundings, because they conduct heat quickly in comparison
to their volumetric heat capacity or ’thermal bulk’.
• thermal expansion α (K−1 ): it is the tendency of a matter to change in volume in response to a
change in temperature.
check aspect manual The 2D cylindrical shell benchmarks by Davies et al. 5.4.12
18
4
The building blocks of the Finite Element Method
4.1
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.
4.1.1
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.
b
Z
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 (x)dx ' (b − a)
a
f (a) + f (b)
2
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.
19
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.
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
(36)
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
r=
2
(x − a) − 1
b−a
20
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
x=
2
From this it follows that
b−a
dx =
dr
2
and then
Z
Z b
n
b−a X
b − a +1
f (r)dr '
f (x)dx =
wi f (riq )
2
2 i =1 q
−1
a
q
4.1.2
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
f (x)dx = π
dx = 2π
I=
−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.(36). 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:
Z +1
Z +1
1
I=
f (x)dx =
(mx + p)dx = [ mx2 + px]+1
−1 = 2p
2
−1
−1
Z
+1
Igq =
f (x)dx =
−1
nq
X
wiq f (xiq ) =
iq =1
nq
X
nq
X
wiq (mxiq + p) = m
iq =1
wiq xiq +p
iq =1
|
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
Z +1
Z +1
1
2
I=
f (x)dx =
x2 dx = [ x3 ]+1
−1 =
3
3
−1
−1
and
Z
+1
Igq =
f (x)dx =
−1
nq
X
iq =1
nq
X
wiq f (xiq ) =
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 =
• It also works ∀nq > 2 !
21
2
3
wiq x2iq
4.1.3
in 2D/3D - theory
Let us now turn to a two-dimensional integral of the form
Z +1 Z +1
f (x, y)dxdy
I=
−1
−1
The equivalent Gaussian quadrature writes:
Igq '
nq nq
X
X
f (xiq , yjq )wiq wjq
iq =1 jq
4.2
The mesh
4.3
A bit of FE terminology
We introduce here some terminology for efficient element descriptions [239]:
• For triangles/tetrahedra, the designation Pm × Pn means that each component of the velocity is
approximated by continuous piecewise complete Polynomials of degree m and pressure by continuous
piecewise complete Polynomials of degree n. For example P2 × P1 means
u ∼ a1 + a2 x + a3 y + a4 xy + a5 x2 + a6 y 2
with similar approximations for v, and
p ∼ b1 + b2 x + b3 y
Both velocity and pressure are continuous across element boundaries, and each triangular element
contains 6 velocity nodes and three pressure nodes.
• For the same families, Pm × P−n is as above, except that pressure is approximated via piecewise
discontinuous polynomials of degree n. For instance, P2 × P−1 is the same as P2 P1 except that
pressure is now an independent linear function in each element and therefore discontinuous at
element boundaries.
• For quadrilaterals/hexahedra, the designation Qm × Qn means that each component of the velocity
is approximated by a continuous piecewise polynomial of degree m in each direction on the quadrilateral and likewise for pressure, except that the polynomial is of degree n. For instance, Q2 × Q1
means
u ∼ a1 + a2 x + a3 y + a4 xy + a5 x2 + a6 y 2 + a7 x2 y + a8 xy 2 + a9 x2 y 2
and
p ∼ b1 + b2 x + b3 y + b4 xy
• For these same families, Qm × Q−n is as above, except that the pressure approximation is not
continuous at element boundaries.
• Again for the same families, Qm × P−n indicates the same velocity approximation with a pressure
approximation that is a discontinuous complete piecewise polynomial of degree n (not of degree n
in each direction !)
+
• The designation Pm
or Q+
m means that some sort of bubble function was added to the polynomial
approximation for the velocity. You may also find the term ’enriched element’ in the literature.
• Finally, for n = 0, we have piecewise-constant pressure, and we omit the minus sign for simplicity.
Another point which needs to be clarified is the use of so-called ’conforming elements’ (or ’nonconforming elements’).
Following again [239], conforming velocity elements are those for which the
basis functions for a subset of H 1 for the continuous problem (the first derivatives and their squares are
integrable in Ω). For instance, the rotated Q1 × P0 element of Rannacher and Turek (see section ??) is
such that the velocity is discontinous across element edges, so that the derivative does not exist there.
Another typical example of non-conforming element is the Crouzeix-Raviart element [125].
22
4.4
4.4.1
Elements and basis functions in 1D
Linear basis functions (Q1 )
Let f (r) be a C 1 function on the interval [−1 : 1] with f (−1) = f1 and f (1) = f2 .
Let us assume that the function f (r) is to be approximated on [−1, 1] by the first order polynomial
f (r) = a + br
(37)
Then it must fulfill
f (r = −1)
= a − b = f1
f (r = +1)
= a + b = f2
This leads to
1
1
(f1 + f2 )
b = (−f1 + f2 )
2
2
and then replacing a, b in Eq. (37) by the above values on gets
1
1
f (r) =
(1 − r) f1 + (1 + r) f2
2
2
a=
or
f (r) =
2
X
Ni (r)f1
i=1
with
4.4.2
N1 (r)
=
N2 (r)
=
1
(1 − r)
2
1
(1 + r)
2
(38)
Quadratic basis functions (Q2 )
Let f (r) be a C 1 function on the interval [−1 : 1] with f (−1) = f1 , f (0) = f2 and f (1) = f3 .
Let us assume that the function f (r) is to be approximated on [−1, 1] by the second order polynomial
f (r) = a + br + cr2
(39)
Then it must fulfill
f (r = −1)
f (r = 0)
f (r = +1)
= a − b + c = f1
= a
= f2
= a + b + c = f3
23
This leads to
1
1
(−f 1 + f 3) c = (f1 + f3 − 2f2 )
2
2
and then replacing a, b, c in Eq. (39) by the above values on gets
1
1
2
f (r) =
r(r − 1) f1 + (1 − r )f2 + r(r + 1) f3
2
2
a = f2
b=
or,
f (r) =
3
X
Ni (r)fi
i=1
with
4.4.3
N1 (r)
=
N2 (r)
=
N3 (r)
=
1
r(r − 1)
2
(1 − r2 )
1
r(r + 1)
2
(40)
Cubic basis functions (Q3 )
The 1D basis polynomial is given by
f (r) = a + br + cr2 + dr3
with the nodes at position -1,-1/3, +1/3 and +1.
= a − b + c − d = f1
c
d
b
= f2
f (−1/3) = a − + −
3 9 27
b
c
d
f (+1/3) = a − + −
= f3
3 9 27
f (+1) = a + b + c + d = f4
f (−1)
Adding the first and fourth equation and the second and third, one arrives at
f1 + f4 = 2a + 2c
f2 + f3 = 2a +
2c
9
and finally:
1
(−f1 + 9f2 + 9f3 − f4 )
16
9
c=
(f1 − f2 − f3 + f4 )
16
Combining the original 4 equations in a different way yields
a=
2b 2d
+
= f3 − f2
3
27
2b + 2d = f4 − f1
so that
1
(f1 − 27f2 + 27f3 − f4 )
16
9
d=
(−f1 + 3f2 − 3f3 + f4 )
16
b=
Finally,
24
= a + b + cr2 + dr3
1
=
(−1 + r + 9r2 − 9r3 )f1
16
1
+
(9 − 27r − 9r2 + 27r3 )f2
16
1
+
(9 + 27r − 9r2 − 27r3 )f3
16
1
+
(−1 − r + 9r2 + 9r3 )f4
16
4
X
=
Ni (r)fi
f (r)
i=1
where
N1
=
N2
=
N3
=
N4
=
1
(−1 + r + 9r2 − 9r3 )
16
1
(9 − 27r − 9r2 + 27r3 )
16
1
(9 + 27r − 9r2 − 27r3 )
16
1
(−1 − r + 9r2 + 9r3 )
16
Verification:
• Let us assume f (r) = C, then
fˆ(r) =
X
Ni (r)fi =
X
i
Ni C = C
X
Ni = C
i
so that a constant function is exactly reproduced, as expected.
• Let us assume f (r) = r, then f1 = −1, f2 = −1/3, f3 = 1/3 and f4 = +1. We then have
X
fˆ(r) =
Ni (r)fi
1
1
= −N1 (r) − N2 (r) + N3 (r) + N4 (r)
3
3
= [−(−1 + r + 9r2 − 9r3 )
1
− (9 − 27r − 9r2 − 27r3 )
3
1
+ (9 + 27r − 9r2 + 27r3 )
3
+(−1 − r + 9r2 + 9r3 )]/16
=
[−r + 9r + 9r − r]/16 + ...0...
= r
(41)
The basis functions derivative are given by
25
∂N1
∂r
∂N2
∂r
∂N3
∂r
∂N4
∂r
=
=
=
=
1
(1 + 18r − 27r2 )
16
1
(−27 − 18r + 81r2 )
16
1
(+27 − 18r − 81r2 )
16
1
(−1 + 18r + 27r2 )
16
Verification:
• Let us assume f (r) = C, then
∂ fˆ
∂r
X ∂Ni
=
i
= C
fi
∂r
X ∂Ni
∂r
i
C
[(1 + 18r − 27r2 )
16
+(−27 − 18r + 81r2 )
=
+(+27 − 18r − 81r2 )
+(−1 + 18r + 27r2 )]
=
0
• Let us assume f (r) = r, then f1 = −1, f2 = −1/3, f3 = 1/3 and f4 = +1. We then have
∂ fˆ
∂r
=
X ∂Ni
i
∂r
fi
1
[−(1 + 18r − 27r2 )
16
1
− (−27 − 18r + 81r2 )
3
1
+ (+27 − 18r − 81r2 )
3
+(−1 + 18r + 27r2 )]
1
[−2 + 18 + 54r2 − 54r2 ]
=
16
= 1
=
4.5
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:
26
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:
4
X
u(xM , yM ) =
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
(42)
N3 (x2 , y2 )
=
0
(43)
N3 (x3 , y3 )
=
1
(44)
N3 (x4 , y4 )
=
0
(45)
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
(46)
Ni (x, y) vi
(47)
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
=
1 ∂u 1 ∂v
1 X ∂Ni
1 X ∂Ni
+
=
ui +
vi
2 ∂y
2 ∂x
2 i=1 ∂y
2 i=1 ∂x
(48)
4
(49)
4
˙xy (x, y)
4
(50)
How we actually obtain the exact form of the basis functions is explained in the coming section.
4.5.1
Bilinear basis functions in 2D (Q1 )
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):
27
3===========2
|
|
|
|
|
|
|
|
|
|
0===========1
(r_0,s_0)=(-1,-1)
(r_1,s_1)=(+1,-1)
(r_2,s_2)=(+1,+1)
(r_3,s_3)=(-1,+1)
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) = −0.25(1 − s)
∂r
∂N2
(r, s) = +0.25(1 − s)
∂r
∂N3
(r, s) = +0.25(1 + s)
∂r
∂N4
(r, s) = −0.25(1 + s)
∂r
∂N1
(r, s) = −0.25(1 − r)
∂s
∂N2
(r, s) = −0.25(1 + r)
∂s
∂N3
(r, s) = +0.25(1 + r)
∂s
∂N4
(r, s) = +0.25(1 − r)
∂s
Let us go back to Eq.(47). 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
4
X
Ni (r, s) = C[N1 (r, s) + N2 (r, s) + N3 (r, s) + N4 (r, s)] = C
i=1
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
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
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.
28
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
(51)
Ni (r, s)(ari + bsi )
(52)
i=1
=
4
X
i=1
= a
4
X
Ni (r, s)ri +b
4
X
(53)
Ni (r, s)si
i=1
i=1
{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
(54)
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
(55)
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
=
+
=
(56)
(57)
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.
4.5.2
Biquadratic basis functions in 2D (Q2 )
This element is part of the so-called LAgrange family.
citation needed
Inside an element the local numbering of the nodes is as follows:
3=====6=====2
|
|
|
|
|
|
7=====8=====5
|
|
|
|
|
|
0=====4=====1
(r_0,s_0)=(-1,-1)
(r_1,s_1)=(+1,-1)
(r_2,s_2)=(+1,+1)
(r_3,s_3)=(-1,+1)
(r_4,s_4)=( 0,-1)
(r_5,s_5)=(+1, 0)
(r_6,s_6)=( 0,+1)
(r_7,s_7)=(-1, 0)
(r_8,s_8)=( 0, 0)
The basis polynomial is then
f (r, s) = a + br + cs + drs + er2 + f s2 + gr2 s + hrs2 + ir2 s2
29
The velocity shape functions are given by:
N0 (r, s)
=
N1 (r, s)
=
N2 (r, s)
=
N3 (r, s)
=
N4 (r, s)
=
N5 (r, s)
=
N6 (r, s)
=
N7 (r, s)
=
N8 (r, s)
=
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 )
and their derivatives by:
∂N0
∂r
∂N1
∂r
∂N2
∂r
∂N3
∂r
∂N4
∂r
∂N5
∂r
∂N6
∂r
∂N7
∂r
∂N8
∂r
4.5.3
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
∂N0
∂s
∂N1
∂s
∂N2
∂s
∂N3
∂s
∂N4
∂s
∂N5
∂s
∂N6
∂s
∂N7
∂s
∂N8
∂s
=
= (−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
1
= (1 − r2 ) (2s − 1)
2
1
= r(r + 1)(−2s)
2
1
= (1 − r2 ) (2s + 1)
2
1
= r(r − 1)(−2s)
2
=
= (1 − r2 )(−2s)
Eight node serendipity basis functions in 2D (Qs2 )
Inside an element the local numbering of the nodes is as follows:
3=====6=====2
|
|
|
|
|
|
7=====+=====5
|
|
|
|
|
|
0=====4=====1
(r_0,s_0)=(-1,-1)
(r_1,s_1)=(+1,-1)
(r_2,s_2)=(+1,+1)
(r_3,s_3)=(-1,+1)
(r_4,s_4)=( 0,-1)
(r_5,s_5)=(+1, 0)
(r_6,s_6)=( 0,+1)
(r_7,s_7)=(-1, 0)
The main difference with the Q2 element resides in the fact that there is no node in the middle of the
element The basis polynomial is then
f (r, s) = a + br + cs + drs + er2 + f s2 + gr2 s + hrs2
30
Note that absence of the r2 s2 term which was previously associated to the center node. We find that
N0 (r, s)
=
N1 (r, s)
=
N2 (r, s)
=
N3 (r, s)
=
N4 (r, s)
=
N5 (r, s)
=
N6 (r, s)
=
N7 (r, s)
=
1
(1 − r)(1 − s)(−r − s − 1)
4
1
(1 + r)(1 − s)(r − s − 1)
4
1
(1 + r)(1 + s)(r + s − 1)
4
1
(1 − r)(1 + s)(−r + s − 1)
4
1
(1 − r2 )(1 − s)
2
1
(1 + r)(1 − s2 )
2
1
(1 − r2 )(1 + s)
2
1
(1 − r)(1 − s2 )
2
(58)
(59)
(60)
(61)
(62)
(63)
(64)
(65)
The shape functions at the mid side nodes are products of a second order polynomial parallel to side
and a linear function perpendicular to the side while shape functions for corner nodes are modifications
of the bilinear quadrilateral element:
verify those
4.5.4
Bicubic basis functions in 2D (Q3 )
Inside an element the local numbering of the nodes is as follows:
12===13===14===15
||
||
||
||
08===09===10===11
||
||
||
||
04===05===06===07
||
||
||
||
00===01===02===03
(r,s)_{00}=(-1,-1)
(r,s)_{01}=(-1/3,-1)
(r,s)_{02}=(+1/3,-1)
(r,s)_{03}=(+1,-1)
(r,s)_{04}=(-1,-1/3)
(r,s)_{05}=(-1/3,-1/3)
(r,s)_{06}=(+1/3,-1/3)
(r,s)_{07}=(+1,-1/3)
(r,s)_{08}=(-1,+1/3)
(r,s)_{09}=(-1/3,+1/3)
(r,s)_{10}=(+1/3,+1/3)
(r,s)_{11}=(+1,+1/3)
(r,s)_{12}=(-1,+1)
(r,s)_{13}=(-1/3,+1)
(r,s)_{14}=(+1/3,+1)
(r,s)_{15}=(+1,+1)
The velocity shape functions are given by:
N1 (r) = (−1 + r + 9r2 − 9r3 )/16
N1 (t) = (−1 + t + 9t2 − 9t3 )/16
N2 (r) = (+9 − 27r − 9r2 + 27r3 )/16
N2 (t) = (+9 − 27t − 9t2 + 27t3 )/16
N3 (r) = (+9 + 27r − 9r2 − 27r3 )/16
N3 (t) = (+9 + 27t − 9t2 − 27t3 )/16
N4 (r) = (−1 − r + 9r2 + 9r3 )/16
N4 (t) = (−1 − t + 9t2 + 9t3 )/16
31
4.5.5
N01 (r, s)
= N1 (r)N1 (s) = (−1 + r + 9r2 − 9r3 )/16 ∗ (−1 + t + 9s2 − 9s3 )/16
N02 (r, s)
= N2 (r)N1 (s) = (+9 − 27r − 9r2 + 27r3 )/16 ∗ (−1 + t + 9s2 − 9s3 )/16
N03 (r, s)
=
N3 (r)N1 (s) = (+9 + 27r − 9r2 − 27r3 )/16 ∗ (−1 + t + 9s2 − 9s3 )/16
N04 (r, s)
=
N4 (r)N1 (s) = (−1 − r + 9r2 + 9r3 )/16 ∗ (−1 + t + 9s2 − 9s3 )/16
N05 (r, s)
=
N1 (r)N2 (s) = (−1 + r + 9r2 − 9r3 )/16 ∗ (9 − 27s − 9s2 + 27s3 )/16
N06 (r, s)
=
N2 (r)N2 (s) = (+9 − 27r − 9r2 + 27r3 )/16 ∗ (9 − 27s − 9s2 + 27s3 )/16
N07 (r, s)
= N3 (r)N2 (s) = (+9 + 27r − 9r2 − 27r3 )/16 ∗ (9 − 27s − 9s2 + 27s3 )/16
N08 (r, s)
= N4 (r)N2 (s) = (−1 − r + 9r2 + 9r3 )/16 ∗ (9 − 27s − 9s2 + 27s3 )/16
N09 (r, s)
= N1 (r)N3 (s) =
(66)
N10 (r, s)
= N2 (r)N3 (s) =
(67)
N11 (r, s)
= N3 (r)N3 (s) =
(68)
N12 (r, s)
= N4 (r)N3 (s) =
(69)
N13 (r, s)
= N1 (r)N4 (s) =
(70)
N14 (r, s)
= N2 (r)N4 (s) =
(71)
N15 (r, s)
= N3 (r)N4 (s) =
(72)
N16 (r, s)
= N4 (r)N4 (s) =
(73)
Linear basis functions for triangles in 2D (P1 )
2
|\
| \
|
\
|
\
0=======1
(r_0,s_0)=(0,0)
(r_1,s_1)=(1,0)
(r_2,s_2)=(0,2)
The basis polynomial is then
f (r, s) = a + br + cs
and the shape functions:
4.5.6
1−r−s
N0 (r, s)
=
N1 (r, s)
= r
(75)
N2 (r, s)
= s
(76)
Quadratic basis functions for triangles in 2D (P2 )
2
|\
| \
5
4
|
\
|
\
0===3===1
(r_0,s_0)=(0,0) (r_3,s_3)=(1/2,0)
(r_1,s_1)=(1,0) (r_4,s_4)=(1/2,1/2)
(r_2,s_2)=(0,1) (r_5,s_5)=(0,1/2)
The basis polynomial is then
f (r, s) = c1 + c2 r + c3 s + c4 r2 + c5 rs + c6 s2
32
(74)
We have
f1 = f (r1 , s1 )
= c1
f2 = f (r2 , s2 )
= c1 + c2 + c4
f3 = f (r3 , s3 )
= c1 + c3 + c6
f4 = f (r4 , s4 )
= c1 + c2 /2 + c4 /4
f5 = f (r5 , s5 )
= c1 + c2 /2 + c3 /2
+ c4 /4 + c5 /4 + c6 /4
f6 = f (r6 , s6 )
This can be cast as f = A · c where A
1
1
1
A=
1
1
1
= c1 + c3 /2 + c6 /4
is a 6x6 matrix:
0
0
0
1
0
1
0
1
0
1/2 0 1/4
1/2 1/2 1/4
0 1/2 0
0
0
0
0
1/4
0
0
0
1
0
1/4
1/4
It is rather trivial to compute the inverse of this matrix:
1
0
0
0 0 0
−3 −1 0
4 0 0
−3
0
−1
0
0 4
A−1 =
2
2
0
−4
0
0
4
0
0 −4 4 −4
2
0
2
0 0 −4
In the end, one obtains:
f (r, s)
=
f1 + (−3f1 − f2 + 4f4 )r + (−3f1 − f3 + 4f6 )s
+(2f1 + 2f2 − 4f4 )r2 + (4f1 − 4f4 + 4f5 − 4f6 )rs
+(2f1 + 2f3 − 4f6 )s2
=
6
X
Ni (r, s)fi
(77)
i=1
with
4.5.7
1 − 3r − 3s + 2r2 + 4rs + 2s2
N1 (r, s)
=
N2 (r, s)
= −r + 2r2
N3 (r, s)
= −s + 2s2
N4 (r, s)
=
4r − 4r2 − 4rs
N5 (r, s)
=
4rs
N6 (r, s)
=
4s − 4rs − 4s2
Cubic basis functions for triangles (P3 )
2
|\
| \
7
6
|
\
8 9
5
|
\
0==3==4==1
(r_0,s_0)=(0,0)
(r_1,s_1)=(1,0)
(r_2,s_2)=(0,1)
(r_3,s_3)=(1/3,0)
(r_4,s_4)=(2/3,0)
(r_5,s_5)=(2/3,1/3)
(r_6,s_6)=(1/3,2/3)
(r_7,s_7)=(0,2/3)
(r_8,s_8)=(0,1/3)
(r_9,s_9)=(1/3,1/3)
33
The basis polynomial is then
f (r, s) = c1 + c2 r + c3 s + c4 r2 + c5 rs + c6 s2 + c7 r3 + c8 r2 s + c9 rs2 + c10 s3
N0 (r, s)
N1 (r, s)
N2 (r, s)
N3 (r, s)
N4 (r, s)
N5 (r, s)
N6 (r, s)
N7 (r, s)
N8 (r, s)
N9 (r, s)
9
(1 − r − s)(1/3 − r − s)(2/3 − r − s)
2
9
=
r(r − 1/3)(r − 2/3)
2
9
=
s(s − 1/3)(s − 2/3)
2
27
=
(1 − r − s)r(2/3 − r − s)
2
27
(1 − r − s)r(r − 1/3)
=
2
27
rs(r − 1/3)
=
2
27
=
rs(r − 2/3)
2
27
=
(1 − r − s)s(s − 1/3)
2
27
=
(1 − r − s)s(2/3 − r − s)
2
= 27rs(1 − r − s)
=
(78)
(79)
(80)
(81)
(82)
(83)
(84)
(85)
(86)
(87)
verify those
4.6
4.6.1
Elements and basis functions in 3D
Linear basis functions in tetrahedra (P1 )
(r_0,s_0)
(r_1,s_1)
(r_2,s_2)
(r_3,s_3)
=
=
=
=
(0,0,0)
(1,0,0)
(0,2,0)
(0,0,1)
The basis polynomial is given by
f (r, s, t) = c0 + c1 r + c2 s + c3 t
f1
= f (r1 , s1 , t1 ) = c0
(88)
f2
= f (r2 , s2 , t2 ) = c0 + c1
(89)
f3
= f (r3 , s3 , t3 ) = c0 + c2
(90)
f4
= f (r4 , s4 , t4 ) = c0 + c3
(91)
which yields:
c0 = f1
f (r, s, t)
c1 = f2 − f1
c2 = f3 − f1
c3 = f4 − f1
= c0 + c1 r + c2 s + c3 t
= f1 + (f2 − f1 )r + (f3 − f1 )s + (f4 − f1 )t
= f1 (1 − r − s − t) + f2 r + f3 s + f4 t
X
=
Ni (r, s, t)fi
i
Finally,
34
4.6.2
1−r−s−t
N1 (r, s, t)
=
N2 (r, s, t)
= r
N3 (r, s, t)
= s
N4 (r, s, t)
= t
Triquadratic basis functions in 3D (Q2 )
N1
=
0.5r(r − 1) 0.5s(s − 1) 0.5t(t − 1)
N2
=
0.5r(r + 1) 0.5s(s − 1) 0.5t(t − 1)
N3
=
0.5r(r + 1) 0.5s(s + 1) 0.5t(t − 1)
N4
=
0.5r(r − 1) 0.5s(s + 1) 0.5t(t − 1)
N5
=
0.5r(r − 1) 0.5s(s − 1) 0.5t(t + 1)
N6
=
0.5r(r + 1) 0.5s(s − 1) 0.5t(t + 1)
N7
=
0.5r(r + 1) 0.5s(s + 1) 0.5t(t + 1)
N8
=
0.5r(r − 1) 0.5s(s + 1) 0.5t(t + 1)
N9
=
(1. − r2 ) 0.5s(s − 1) 0.5t(t − 1)
N10
=
0.5r(r + 1) (1 − s2 ) 0.5t(t − 1)
N11
N12
= (1. − r2 ) 0.5s(s + 1) 0.5t(t − 1)
= 0.5r(r − 1) (1 − s2 ) 0.5t(t − 1)
N13
=
(1. − r2 ) 0.5s(s − 1) 0.5t(t + 1)
N14
=
0.5r(r + 1) (1 − s2 ) 0.5t(t + 1)
N15
=
(1. − r2 ) 0.5s(s + 1) 0.5t(t + 1)
N16
=
0.5r(r − 1) (1 − s2 ) 0.5t(t + 1)
N17
=
0.5r(r − 1) 0.5s(s − 1) (1 − t2 )
N18
=
0.5r(r + 1) 0.5s(s − 1) (1 − t2 )
N19
=
0.5r(r + 1) 0.5s(s + 1) (1 − t2 )
N20
=
0.5r(r − 1) 0.5s(s + 1) (1 − t2 )
N21
=
(1 − r2 ) (1 − s2 ) 0.5t(t − 1)
N22
=
(1 − r2 ) 0.5s(s − 1) (1 − t2 )
N23
=
0.5r(r + 1) (1 − s2 ) (1 − t2 )
N24
=
(1 − r2 ) 0.5s(s + 1) (1 − t2 )
N25
=
0.5r(r − 1) (1 − s2 ) (1 − t2 )
N26
=
(1 − r2 ) (1 − s2 ) 0.5t(t + 1)
N27
=
(1 − r2 ) (1 − s2 ) (1 − t2 )
35
5
Solving the flow equations with the FEM
In the case of an incompressible flow, we have seen that the continuity (mass conservation) equation takes
~ · ~v = 0. In other word flow takes place under the constraint that the divergence of its
the simple form ∇
velocity field is exactly zero eveywhere (solenoidal constraint), i.e. it is divergence free.
We see that the pressure in the momentum equation is then a degree of freedom which is needed to
satisfy the incompressibilty constraint (and it is not related to any constitutive equation) [142]. In other
words the pressure is acting as a Lagrange multiplier of the incompressibility constraint.
Various approaches have been proposed in the literature to deal with the incompressibility constraint
but we will only focus on the penalty method (section 5.5) and the so-called mixed finite element method
5.6.
5.1
strong and weak forms
The strong form consists of the governing equation and the boundary conditions, i.e. the mass, momentum
and energy conservation equations supplemented with Dirichlet and/or Neumann boundary conditions
on (parts of) the boundary.
To develop the finite element formulation, the partial differential equations must be restated in an
integral form called the weak form. In essence the PDEs are first multiplied by an arbitrary function and
integrated over the domain.
5.2
Which velocity-pressure pair for Stokes?
The success of a mixed finite element formulation crucially depends on a proper choice of the local
interpolations of the velocity and the pressure.
5.2.1
5.3
The compatibility condition (or LBB condition)
Families
Taylor-Hood
USE [278] section 3.6.2
5.3.1
The bi/tri-linear velocity - constant pressure element (Q1 × P0 )
discussed in example 3.71 of [278]
5.3.2
The bi/tri-quadratic velocity - discontinuous linear pressure element (Q2 × P−1 )
5.3.3
The bi/tri-quadratic velocity - bi/tri-linear pressure element (Q2 × Q1 )
This element, implemented in penalised form, is discussed in [45] and the follow-up paper [46]. CHECK
Biquadratic velocities, bilinear pressure. See Hood and Taylor. The element satisfies the inf-sup
condition [259]p215.
36
5.3.4
The stabilised bi/tri-linear velocity - bi/tri-linear pressure element (Q1 × Q1 -stab)
5.3.5
The MINI triangular element (P1+ × P1 )
discussed in section 3.6.1 of [278]
5.3.6
The quadratic velocity - linear pressure triangle (P2 × P1 )
From [417]. Taylor-Hood elements (Taylor and Hood 1973) are characterized by the fact that the
pressure is continuous in the region Ω. A typical example is the quadratic triangle (P2P1 element).
In this element the velocity is approximated by a quadratic polynomial and the pressure by a linear
polynomial. One can easily verify that both approximations are continuous over the element boundaries.
It can be shown, Segal (1979), that this element is admissible if at least 3 elements are used. The
quadrilateral counterpart of this triangle is the Q2 × Q1 element.
5.3.7
The Crouzeix-Raviart triangle (P2+ × P−1 )
From [130]: seven-node Crouzeix-Raviart triangle with quadratic velocity shape functions enhanced
by a cubic bubble function and discontinuous linear interpolation for the pressure field [e.g., Cuvelier et
al., 1986]. This element is stable and no additional stabilization techniques are required [Elman et al.,
2005].
From [417]. These elements are characterized by a discontinuous pressure; discontinuous on element
boundaries. For output purposes (printing, plotting etc.) these discontinuous pressures are averaged in
vertices for all the adjoining elements.
The most simple Crouzeix-Raviart element is the non-conforming linear triangle with constant pressure
(P1 × P0 ).
p248 of Elman book. satisfies LBB condition.
5.4
Other elements
P1P0 example 3.70 in [278]
37
P1P1
Q2Q2
P2P2
5.5
The penalty approach for viscous flow
In order to impose the incompressibility constraint, two widely used procedures are available, namely the
Lagrange multiplier method and the penalty method [30, 259]. 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 [129], [398] or [243].
The penalty formulation of the mass conservation equation is based on a relaxation of the incompressibility constraint and writes
~ · ~ν + p = 0
(92)
∇
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
~ · ~ν = 0 will be approximately
chooses λ to be a sufficiently large number, the continuity equation ∇
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 [144, 260].
Equation (92) can be used to eliminate the pressure in Eq. (??) so that the mass and momentum
conservation equations fuse to become :
~ · (2η ε̇(~ν)) + λ∇(
~ ∇
~ · ~ν) = ρg = 0
∇
(93)
[330] 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 [43] 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 [225] .
list codes which use this approach
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
1 1 0
2 0 0
∂v
1 1 0 +η 0 2 0 ·
= λ
∂y
0
0
1
0
0
0
|
{z
}
|
{z
}
∂u
+ ∂v
K
Remember that
C
4
4
∂u X ∂Ni
=
ui
∂x
∂x
i=1
∂v X ∂Ni
=
vi
∂y
∂y
i=1
38
∂y
∂x
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
~ · σdΩ +
Ni ∇
Ωe
Ni bdΩ = 0
Ωe
We can integrate by parts and drop the surface term2 :
Z
Z
~ i · σdΩ =
∇N
Ni bdΩ
Ωe
Ωe
or,
Z
∂Ni
∂x
0
∂Ni
∂y
0
∂Ni
∂y
∂Ni
∂x
Ωe
Z
σxx
· σyy dΩ =
Ni bdΩ
Ωe
σxy
Let i = 1, 2, 3, 4 and stack the resulting four equations
∂N1
∂N1
Z
0
σxx
∂x
∂y
· σyy
∂N1
∂N1
Ωe
σxy
0
∂y
∂x
∂N2
∂N2
Z
0
σxx
∂x
∂y
· σyy
∂N2
∂N2
Ωe
σxy
0
∂y
∂x
∂N3
∂N3
Z
0
σxx
∂x
∂y
· σyy
∂N3
∂N3
Ωe
σxy
0
∂y
∂x
∂N4
∂N4
Z
0
σxx
∂x
∂y
· σyy
∂N4
∂N4
Ωe
σxy
0
∂y
∂x
on top of one another.
Z
bx
dΩ =
N1
dΩ
by
Ωe
=
Ni
Ωe
=
N3
Ωe
39
Z
dΩ
=
N4
Ωe
NbT = (N1 bx , N1 by , ...N4 bx , N4 by )
will come back to this at a later stage
Z
dΩ
We easily recognize B T inside the integrals! Let us define
2 We
Z
dΩ
bx
by
bx
by
bx
by
(94)
dΩ
(95)
dΩ
(96)
dΩ
(97)
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
V =
Nb dΩ
B T · [λK + ηC] · BdΩ · |{z}
Ωe
Ωe
{z
} (8x1) | {z }
|
Ael (8×8)
Bel (8×1)
or,
Z
Z
Z
T
T
λB · K · BdΩ +
V =
ηB · C · BdΩ · |{z}
Nb dΩ
Ω
Ω
| Ωe
{z
} | e
{z
} (8x1) | e {z }
Aη
el (8×8)
Aλ
el (8×8)
Bel (8×1)
INTEGRATION - MAPPING
reduced integration [260]
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
5.6
5.6.1
The mixed FEM for viscous flow
in three dimensions
In what follows the flow is assumed to be incompressible, isoviscous and isothermal.
The methodology to derive the discretised equations of the mixed system is quite similar to the one
we have used in the case of the penalty formulation. The big difference comes from the fact that we are
now solving for both velocity and pressure at the same time, and that we therefore must solve the mass
and momentum conservation equations together. As before, velocity inside an element is given by
~νh (~r) =
mv
X
i=1
40
Niν (~r) ~νi
(98)
where Niv are the polynomial basis functions for the velocity, and the summation runs over the mv nodes
composing the element. A similar expression is used for pressure:
ph (~r) =
mp
X
Nip (~r) pi
(99)
i=1
Note that the velocity is a vector of size while pressure (and temperature) is a scalar. There are then ndofv
velocity degrees of freedom per node and ndofp pressure degrees of freedom. It is also very important
to remember that the numbers of velocity nodes and pressure nodes for a given element are more often
than not different and that velocity and pressure nodes need not be colocated. Indeed, unless co-called
’stabilised elements’ are used, we have mv > mp , which means that the polynomial order of the velocity
field is higher than the polynomial order of the pressure field (usually by value 1).
insert here link(s) to manual and literature
Other notations are sometimes used for Eqs.(98) and (99):
~ ν · ~u
uh (~r) = N
~ ν · ~v
v h (~r) = N
~ν ·w
wh (~r) = N
~
~ p · p~
ph (~r) = N
~ ν is the vector containing all basis functions evaluated at location ~r:
where ~ν = (u, v, w) and N
~ v = N ν (~r), N ν (~r), N ν (~r), . . . N ν (~r)
N
1
2
3
mv
p
~p =
N
N1p (~r), N2p (~r), N3p (~r), . . . Nm
(~
r
)
p
(100)
(101)
(102)
and with
~u =
(u1 , u2 , u3 , . . . umv )
(103)
~v
=
(v1 , v2 , v3 , . . . vmv )
(104)
w
~
=
(w1 , w2 , w3 , . . . wmv )
p~ = p1 , p2 , p3 , . . . pmp
(105)
(106)
We will now establish the weak form of the momentum conservation equation. We start again from
~ · σ + ~b
∇
~ · ~v
∇
= ~0
(107)
=
(108)
0
For the Niν ’s and Nip ’regular enough’, we can write:
Z
Z
~ · σdΩ +
Niν ∇
Niν~b dΩ
Ωe
Ωe
Z
~ · ~v dΩ
Nip ∇
= ~0
(109)
=
(110)
0
Ωe
We can integrate by parts and drop the surface term3 :
Z
Z
~ iν · σdΩ =
∇N
Ωe
Z
Niν~bdΩ
(111)
Ωe
~ · ~v dΩ
Nip ∇
=
0
(112)
Ωe
or,
Ωe
0
∂Niν
∂y
0
∂Niν
∂y
0
0
∂Niν
∂x
0
Z
3 We
∂Niν
∂z
0
0
∂Niν
∂x
0
∂Niν
∂z
∂Niν
∂z
0
∂Niν
∂x
∂Niν
∂y
will come back to this at a later stage
41
·
σxx
σyy
σzz
σxy
σxz
σyz
Z
dΩ =
Niν~b dΩ
Ωe
(113)
As before (see section XXX) the above equation can ultimately be written:
σxx
σyy
Z
Z
T σzz
~ b dΩ
N
B ·
dΩ =
Ωe
Ωe
σxy
σxz
σyz
We have previously established that the strain rate vector ~ε̇ is:
P ∂Niν
∂x ui
∂N1ν
i
0
∂u
∂x
∂x
ν
P
∂Ni
∂y vi
∂N1ν
0
∂v
i
∂y
∂y
ν
P
∂Ni
∂w
∂z wi
0
0
i
∂z
~ε̇ =
=
=
P ∂Niν
∂u ∂v
∂Niν
∂N1ν
∂N1ν
∂y + ∂x
(
u
+
v
)
i
i
∂x
∂y
∂x
i ∂y
∂u ∂w
∂N ν
∂z + ∂x
P ∂Niν
1
∂Niν
0
∂z
u
+
w
)
(
i
i
∂z
∂x
i
∂v
∂w
∂N1ν
∂z + ∂y
0
P ∂N ν
ν
∂z
∂Ni
i
( ∂z vi + ∂y wi )
|
0
(114)
···
ν
∂Nm
v
∂x
0
0
0
0
···
0
ν
∂Nm
v
∂y
∂N1ν
∂z
···
0
0
ν
∂Nm
v
∂z
0
···
ν
∂Nm
v
∂x
ν
∂Nm
v
∂x
0
∂N1ν
∂x
···
ν
∂Nm
v
∂z
0
ν
∂Nm
v
∂x
0
ν
∂Nm
v
∂z
ν
∂Nm
v
∂y
∂N1ν
∂y
i
···
{z
B
·
}|
(115)
~ where B is the gradient matrix and V
~ is the vector of all vector degrees of freedom for the
or, ~ε̇ = B · V
~ is mv long. we have
element. The matrix B is then of size 3 × mv and the vector V
σxx
=
−p + 2η ε̇dxx
(116)
σyy
=
−p + 2η ε̇dyy
(117)
σzz
=
−p + 2η ε̇dzz
(118)
σxy
=
2η ε̇dxy
(119)
σxz
=
(120)
σyz
=
2η ε̇dxz
2η ε̇dyz
(121)
Since we here only consider incompressible flow, we have ε̇d = ε̇ so
1
1
1
1
1
1
~
~p ~
~
~σ = −
p + C · ε̇ = − 0 N · P + C · B · V
0
0
0
0
0
with
C = η
2
0
0
0
0
0
0
2
0
0
0
0
0
0
2
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
42
~ε̇ =
ε̇xx
ε̇yy
ε̇zz
2ε̇xy
2ε̇xz
2ε̇yz
(122)
(123)
u1
v1
w1
u2
v2
w2
u3
v3
...
umv
vm v
wmv
{z
~
V
}
Let us define matrix N p of size 6 × mp :
Np =
1
1
1
0
0
0
~p
N
N~ p
N~ p = N~ p
0
0
0
(124)
so that
~
~σ = −N p · P~ + C · B · V
(125)
finally
Z
~ ] dΩ =
B T · [−N p · P~ + C · B · V
Ωe
Z
Nb dΩ
(126)
Ωe
or,
Z
Z
Z
T
p
T
~
~
~ b dΩ
−
B · N dΩ ·P +
B · C · B dΩ ·V =
N
Ωe
Ωe
Ωe
|
{z
}
|
{z
}
| {z }
G
K
(127)
f~
where the matrix K is of size (mv ∗ ndofv × mv ∗ ndofv ), and matrix G is of size (mv ∗ ndofv × mp ∗ ndofp ).
43
Turning now to the mass conservation equation:
Z
~ p∇
~ · ~v dΩ
0 =
N
Ωe
Z
=
~p
N
Ωe
mv
X
∂N ν
∂Niν
∂Niν
ui +
vi +
wi dΩ
∂x
∂y
∂z
i
i=1
N1p
m
Pv
∂Niν
∂x ui
m
Pv
∂Niν
∂y vi
m
Pv
∂Niν
∂z wi
+
+
i=1
i=1
i=1
m
m
m
v
v
v
ν
ν
ν
P
P
P
∂Ni
∂Ni
∂Ni
p
Z N2 i=1 ∂x ui + i=1 ∂y vi + i=1 ∂z wi
m
m
m
v
v
v
dΩ
ν
ν
ν
P
P
P
=
∂Ni
∂Ni
∂Ni
Np
3
Ωe
∂x ui +
∂y vi +
∂z wi
i=1
i=1
i=1
.
.
.
m
m
m
p
Pv ∂Niν
Pv ∂Niν
Pv ∂Niν
Nmp
∂x ui +
∂y vi +
∂z wi
i=1
i=1
i=1
P ∂Niν
∂x ui
i
P ∂Niν
p
p
p
N1 0 0 0
N1
N1
∂y vi
i
N2p
N2p
N2p 0 0 0
P ∂Niν
∂z wi
Z
i
p
p
N3p
N3
N3 0 0 0 ·
=
P ∂Niν
∂N ν
Ωe
( ∂y ui + ∂xi vi )
..
..
..
.. .. ..
i
.
.
.
. . .
P ∂Niν
∂N ν
(
u+ i w)
p
p
p
Nmp Nmp Nmp 0 0 0
i ∂z i ∂x i
P ∂N ν
∂N ν
( ∂zi vi + ∂yi wi )
i
p
p
p
N1
N1
N1 0 0 0
p
p
p
N
N
N
0 0 0
Z
2
2
2
p
p
p
N3
N3
N3 0 0 0
=
·~ε̇ dΩ
..
..
..
.. .. ..
Ωe
.
.
.
. . .
p
Nm
p
p
Nm
p
|
Z
=
p
Nm
p
{z
0
0
dΩ
0
}
Np
~
N · B dΩ · V
p
~
= −GTe · V
(128)
say something about minus sign?
Ultimately we obtain the following system for each element:
~
Ke
Ge
V
f~e
·
=
−GTe
0
0
P~
Such a matrix is then generated for each element and then must me assembled into the global F.E. matrix.
Note that in this case the elemental Stokes matrix is antisymmetric. One can also define the following
symmetric modified Stokes matrix:
~
Ke Ge
V
f~e
·
=
GTe
0
0
P~
This matrix is symmetric, but indefinite. It is non-singular if ker(GT ) = 0, which is the case if t he
compatibility condition holds.
44
CHECK: Matrix K is the viscosity matrix. Its size is (ndofv ∗ Nv ) × (ndofv ∗ Nv ) where ndofv is
the number of velocity degrees of freedom per node (typically 1,2 or 3) and Nv is the number of velocity
nodes. The size of matrix G is (ndofv ∗ Nv ) × (ndofp ∗ Np ) where ndofp (= 1) is the number of velocity
degrees of freedom per node and Np is the number of pressure nodes. Conversely, the size of matrix GT
is (ndofp ∗ Np ) × (ndofv ∗ Nv ). The size of the global FE matrix is N = ndofv ∗ Nv + ndofp ∗ Np Note
that matrix K is analogous to a discrete Laplacian operator, matrix G to a discrete gradient operator,
and matrix GT to a discrete divergence operator.
On the physical dimensions of the Stokes matrix blocks
~ +∇
~ · (2η ε̇) + ρg =
−∇p
~ · ~ν =
∇
We start from the Stokes equations:
0
(129)
0
(130)
The dimensions of the terms in the first equation are: M L−2 T −2 . The blocks K and G stem from
the weak form which obtained by multiplying the strong form equations by the (dimensionless) basis
funstions and integrating over the domain, so that it follows that
~ ] = [G · P~ ] = [f~] = M L−2 T −2 L3 = M LT −2
[K · V
We can then easily deduce:
[K] = M T −1
[G] = L2
On elemental level mass balance. Note that in what is above no assumption has been made about
whether the pressure basis functions are continuous or discontinuous from one element to another.
Indeed, as mentioned in [239], since the weak formulation of the momentum equation involves inte~ the resulting weak form contains no derivatives of pressure. This introduces the
gration by parts of ∇p,
possibility of approximating it by functions (piecewise polynomials, of course) that are not C 0 -continuous,
and indeed this has been done and is quite popular/useful.
It is then worth noting that only discontinuous pressure elements assure an element-level mass balance
[239]: if for instance Nip is piecewise-constant on element e (of value 1), the elemental weak form of the
mass conservervation equation is
Z
Z
Z
~ · ~ν =
~ · ~ν =
~n · ~ν = 0
∇
Nip ∇
Ωe
Γe
Ωe
One potentially unwelcome consequence of using discontinuous pressure elements is that they do not
possess uniquely defined pressure on the element boundaries; they are dual valued there, and often
multi-valued at certain velocity nodes.
On the C matrix
τ
The relationship between deviatoric stress and deviatoric strain rate tensor is
2η ε̇d
1 ~
· ~v )1
= 2η ε̇ − (∇
3
ε̇xx ε̇xy ε̇xz
1 0 0
1
= 2η ε̇yx ε̇yy ε̇yz − (ε̇xx + ε̇yy + ε̇zz ) 0 1 0
3
0 0 1
ε̇zx ε̇zy ε̇zz
2ε̇ − ε̇yy − ε̇zz
3ε̇xy
3ε̇xz
2 xx
3ε̇yx
−ε̇yy + 2ε̇yy − ε̇yy
3ε̇yz
=
η
3
3ε̇zx
3ε̇zy
−ε̇xx − ε̇yy 2ε̇zz
=
45
(131)
(132)
(133)
(134)
so that
2
~τ = η
3
2ε̇xx − ε̇yy − ε̇zz
−ε̇yy + 2ε̇yy − ε̇yy
−ε̇xx − ε̇yy + 2ε̇zz
3ε̇xy
3ε̇xz
3ε̇yz
η
=
3
4
−2
−2
0
0
0
−2
4
−2
0
0
0
−2 0
−2 0
4 0
0 3
0 0
0 0
{z
|
0
0
0
0
3
0
0
0
0
0
0
3
·
C
= C d · ~ε̇
(135)
}
Cd
In the case where we assume incompressible flow
2 0 0 0
0 2 0 0
0 0 2 0
~τ = η
0 0 0 1
0 0 0 0
0 0 0 0
|
{z
ε̇xx
ε̇yy
ε̇zz
2ε̇xy
2ε̇xz
2ε̇yz
from the beginning, i.e. ε = εd , then
0 0
ε̇xx
0 0
ε̇yy
0 0 ε̇zz
= C · ~ε̇
·
0 0
2ε̇xy
1 0
2ε̇xz
0 1
2ε̇yz
}
(136)
On the ’forgotten’ surface terms
5.6.2
Going from 3D to 2D
The world is three-dimensional. However, for many different reasons one may wish to solve problems
which are two-dimensional.
Following ASPECT manual, we will think of two-dimensional models in the following way:
• We assume that the domain we want to solve on is a two-dimensional cross section (in the x − y
plane) that extends infinitely far in both negative and positive z direction.
• We assume that the velocity is zero in the z direction and that all variables have no variation in
the z direction.
As a consequence, two-dimensional models are three-dimensional ones in which the z component of the
velocity is zero and so are all z derivatives. This allows to reduce the momentum conservation equations
from 3 equations to 2 equations. However, contrarily to what is often seen, the 3D definition of the
deviatoric strain rate remains, i.e. in other words:
1 ~
· ~v )1
ε̇d = ε̇ − (∇
3
(137)
and not 1/2. In light of all this, the full strain rate tensor and the deviatoric strain rate tensor in 2D are
given by:
ε
d
ε̇
=
=
ε̇xx
ε̇yx
ε̇zx
1
3
ε̇xy
ε̇yy
ε̇zy
ε̇xz
ε̇yz =
ε̇zz
2 ∂u
− ∂v
∂x ∂y
1 ∂u
∂v
2 ∂y + ∂x
0
1
2
∂u
∂y
− ∂u
∂x
1
2
∂u
∂x
1
2
+
+
∂u
∂y
∂v
∂x
+
0
∂v
∂x
+
0
0
− ∂u
∂x
∂u
∂y
∂v
∂y
0
∂v
2 ∂y
0
−
∂v
∂y
∂v
∂x
0
0
0
(138)
(139)
Although the bottom right term may be surprising, it is of no consequence when this expression of the
deviatoric strain rate is used in the Stokes equation:
~ · 2η ε̇d =
∇
46
FINISH!
In two dimensions the velocity is then ~ν = (u, v) and the FEM building blocks and matrices are
simply:
ε̇xx
~ε̇ = ε̇yy
2ε̇xy
∂u
∂x
=
∂v
∂y
∂u
∂y
+
∂v
∂x
=
∂N1ν
∂x
0
∂N2ν
∂x
0
∂N3ν
∂x
0
...
ν
∂Nm
v
∂x
0
0
∂N1ν
∂y
0
∂N2ν
∂y
0
∂N3ν
∂y
...
0
ν
∂Nm
v
∂x
∂N1ν
∂y
∂N1ν
∂x
∂N2ν
∂y
∂N2ν
∂x
∂N3ν
∂y
∂N3ν
∂x
...
ν
∂Nm
v
∂y
ν
∂Nm
v
∂x
|
{z
B
·
}
|
(140)
we have
so
σxx
=
−p + 2η ε̇xx
(141)
σyy
=
−p + 2η ε̇yy
(142)
σxy
=
+2η ε̇xy
(143)
1
1
~
~σ = − 1 p + C · ~ε̇ = − 1 N~ p · P~ + C · B · V
0
0
with
2
C = η 0
0
0
2
0
0
0
1
or
4
η
−2
C=
3
0
−2
4
0
0
0
3
(144)
(145)
check the right C
Finally the matrix N p is of size 3 × mp :
1
N p = 1 N~ p =
0
5.7
Solving the elastic equations
5.8
Solving the heat transport equation
N~ p
N~ p
0
We start from the ’bare-bones’ heat transport equation (source terms are omitted):
∂T
~
~ · k ∇T
~
ρcp
+ ~ν · ∇T = ∇
∂t
(146)
(147)
In what follows we assume that the velocity vield ~ν is known so that temperature is the only unknwon.
Let N θ be the temperature basis functions so that the temperature inside an element is given by4 :
h
T (~r) =
mT
X
~ θ · T~
N θ (~r)Ti = N
(148)
i=1
where T~ is a vector of length mT The weak form is then
Z
Z
∂T
θ
~
~ · k ∇T
~ dΩ
~
+ ν · ∇T
dΩ =
Niθ ∇
Ni ρcp
∂t
Ω
Ω
4 the
θ superscript has been chosen to denote temperature so as to avoid confusion with the transpose operator
47
(149)
u1
v1
u2
v2
u3
v3
...
umv
vmv
{z
~
V
}
Z
Z
∂T
~ dΩ =
~ · k ∇T
~ dΩ
dΩ +
Niθ ρcp~ν · ∇T
Niθ ∇
Niθ ρcp
∂t
Ω
Ω
Ω
|
{z
} |
{z
} |
{z
}
Z
I
II
i = 1, mT
III
Looking at the first term:
Z
Niθ ρcp
Ω
Z
∂T
dΩ
∂t
=
~ θ · T~˙ dΩ
Niθ ρcp N
(150)
Ω
(151)
so that when we assemble all contributions for i = 1, mT we get:
Z
Z
˙
˙
θ
θ ~˙
θ ~θ
~
~
~
I=
N ρcp N · T dΩ =
ρcp N N dΩ · T~ = M T · T~
Ω
Ω
where M T is the mass matrix of the system of size (mT × mT ) with
Z
T
Mij
=
ρcp Niθ Njθ dΩ
Ω
Turning now to the second term:
Z
~ dΩ
Niθ ρcp~ν · ∇T
Z
∂T
∂T
+v
)dΩ
∂x
∂y
Ω
Z
~θ
~θ
∂N
∂N
=
Niθ ρcp (u
+v
) · T~ dΩ
∂x
∂y
Ω
Niθ ρcp (u
=
Ω
(152)
(153)
(154)
so that when we assemble all contributions for i = 1, mT we get:
!
Z
~θ
~θ
∂N
∂N
θ
~
II =
ρcp N (u
+v
)dΩ · T~ = Ka · T~
∂x
∂y
Ω
where Ka is the advection term matrix of size (mT × mT ) with
Z
ρcp Niθ
(Ka )ij =
Ω
∂Njθ
∂Njθ
+v
u
∂x
∂y
!
dΩ
Now looking at the third term, we carry out an integration by part and neglect the surface term for now,
so that
Z
Z
~ · k ∇T
~ dΩ = −
~ iθ · ∇T
~ dΩ
Niθ ∇
k ∇N
(155)
Ω
Ω
Z
~ iθ · ∇(
~ N
~ θ · T~ )dΩ
= −
k ∇N
(156)
Ω
(157)
with
~N
~θ =
∇
∂x N1θ
∂x N2θ
...
θ
∂x Nm
T
∂y N1θ
∂y N2θ
...
θ
∂ y Nm
T
so that finally:
Z
III = −
θ T ~ ~θ
~
~
k(∇N ) · ∇N dΩ · T~ = −Kd · T~
Ω
where Kd is the diffusion term matrix:
Z
Kd =
~N
~ θ )T · ∇
~N
~ θ dΩ
k(∇
Ω
48
Ultimately terms I, II, III together yield:
˙
M θ · T~ + (Ka + Kd ) · T~ = ~0
What now remains to be done is to address the time derivative on the temperature vector. The most
simple approach would be to use an explicit Euler one, i.e.:
∂ T~
T~ (k) − T~ (k−1)
=
∂t
δt
where T~ (k) is the temperature field at time step k and δt is the time interval between two consecutive
time steps. In this case the discretised heat transport equation is:
M θ + (Ka + Kd )δt · T~ (k) = M θ · T~ (k−1)
other time discr schemes!
49
6
Additional techniques and features
6.1
Picard and Newton
6.2
The SUPG formulation for the energy equation
6.3
Tracking materials and/or interfaces
6.4
Dealing with a free surface
6.5
Convergence criterion for nonlinear iterations
6.6
Static condensation
50
6.7
The method of manufactured solutions
The method of manufactured solutions is a relatively simple way of carrying out code verification. In
essence, one postulates a solution for the PDE at hand (as well as the proper boundary conditions),
inserts it in the PDE and computes the corresponding source term. The same source term and boundary
conditions will then be used in a numerical simulation so that the computed solution can be compared
with the (postulated) true analytical solution.
Examples of this approach are to be found in [142, 96, ?].
. python codes/fieldstone
. python codes/fieldstone saddlepoint
. python codes/fieldstone saddlepoint q2q1
. python codes/fieldstone saddlepoint q3q2
. python codes/fieldstone burstedde
6.7.1
Analytical benchmark I
Taken from [144]. 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
~ν = (u, v) and the pressure p such that
~ + ~b = ~0
−η∆~ν + ∇p
∇·v =0
in Ω
in Ω
v=0
on Γ
where the fluid viscosity is taken as η = 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
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
R
Ω
p dΩ = 0 One can turn to the spatial derivatives of the fields:
∂u
∂x
∂v
∂y
=
(2x − 6x2 + 4x3 )(2y − 6y 2 + 4y 3 )
(158)
=
−(2x − 6x2 + 4x3 )(2y − 6y 2 + 4y 3 )
(159)
~ · ~ν = 0 and
with of course ∇
∂p
∂x
∂p
∂y
=
1 − 2x
(160)
=
0
(161)
The velocity and pressure fields look like:
51
http://ww2.lacan.upc.edu/huerta/exercises/Incompressible/Incompressible Ex1.htm
As shown in [144], If the LBB condition is not satisfied, spurious oscillations spoil the pressure
approximation. Figures below show results obtained with a mesh of 20x20 Q1P0 (left) and P1P1 (right)
elements:
]]
http://ww2.lacan.upc.edu/huerta/exercises/Incompressible/Incompressible Ex1.htm
Taking into account that the proposed problem has got analytical solution, it is easy to analyze
convergence of the different pairs of elements:
http://ww2.lacan.upc.edu/huerta/exercises/Incompressible/Incompressible Ex1.htm
6.7.2
Analytical benchmark II
Taken from [141]. It is for a unit square with ν = µ/ρ = 1 and the smooth exact solution is
u(x, y)
= x + x2 − 2xy + x3 − 3xy 2 + x2 y
v(x, y)
2
p(x, y)
2
3
= −y − 2xy + y − 3x y + y − xy
3 2
= xy + x + y + x y − 4/3
52
(162)
2
(163)
(164)
Note that the pressure obeys
R
Ω
p dΩ = 0
bx
by
6.7.3
−(1 + y − 3x2 y 2 )
=
3
−(1 − 3x − 2x y)
=
(165)
(166)
Analytical benchmark III
This benchmark begins by postulating a polynomial solution to the 3D Stokes equation [141]:
x + x2 + xy + x3 y
y + xy + y 2 + x2 y 2
v=
−2z − 3xz − 3yz − 5x2 yz
(167)
and
p = xyz + x3 y 3 z − 5/32
(168)
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. (229). Note that the pressure fulfills
Z
p(~r)dΩ = 0.
Ω
Constant viscosity In this case, the right hand side writes:
2 + 6xy
f = −∇p + µ 2 + 2x2 + 2y 2
−10yz
2 3
yz + 3x y z
2 + 6xy
= − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
xy + x3 y 3
−10yz
We can compute the components of the strainrate tensor:
ε̇xx
=
1 + 2x + y + 3x2 y
2
(169)
ε̇yy
=
1 + x + 2y + 2x y
(170)
ε̇zz
=
(171)
ε̇xy
=
ε̇xz
=
ε̇yz
=
−2 − 3x − 3y − 5x2 y
1
(x + y + 2xy 2 + x3 )
2
1
(−3z − 10xyz)
2
1
(−3z − 5x2 z)
2
(172)
(173)
(174)
Note that we of course have ε̇xx + ε̇yy + ε̇zz = 0.
Variable viscosity
In this case, the right hand side is obtains through
2 + 6xy
f = −∇p + µ 2 + 2x2 + 2y 2
−10yz
2ε̇xx
2ε̇xy
2ε̇xz
∂µ
∂µ
∂µ
+ 2ε̇xy
+ 2ε̇yy
+ 2ε̇yz
∂x
∂y
∂z
2ε̇xz
2ε̇yz
2ε̇zz
53
(175)
The viscosity can be chosen to be a smooth varying function:
µ = exp(1 − β(x(1 − x) + y(1 − y) + z(1 − z)))
(176)
Choosing β = 0 yields a constant velocity µ = e1 (and greatly simplifies the right-hand side). 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 . In this case
∂µ
∂x
∂µ
∂y
∂µ
∂z
=
−4β(1 − 2x)µ(x, y, z)
(177)
=
−4β(1 − 2y)µ(x, y, z)
(178)
= −4β(1 − 2z)µ(x, y, z)
(179)
[96] has carried out this benchmark for β = 4, i.e.:
µ(x, y, z) = exp(1 − 4(x(1 − x) + y(1 − y) + z(1 − z)))
In a unit cube, this yields a variable viscosity such that 0.1353 < µ < 2.7182, i.e. a ratio of approx. 20
within the domain. We then have:
∂µ
∂x
∂µ
∂y
∂µ
∂z
= −4(1 − 2x)µ(x, y, z)
(180)
= −4(1 − 2y)µ(x, y, z)
(181)
= −4(1 − 2z)µ(x, y, z)
(182)
sort out mess wrt Eq 26 of busa13
6.7.4
Analytical benchmark IV
From [45]. The two-dimensional domain is a unit square. The body forces are:
fx
=
128[x2 (x − 1)2 12(2y − 1) + 2(y − 1)(2y − 1)y(12x2 − 12x + 2)]
fy
=
128[y 2 (y − 1)2 12(2x − 1) + 2(x − 1)(2x − 1)y(12y 2 − 12y + 2)]
(183)
The solution is
u = −256x2 (x − 1)2 y(y − 1)(2y − 1)
v
=
256x2 (y − 1)2 x(x − 1)(2x − 1)
p
=
0
(184)
Another choice:
fx
=
128[x2 (x − 1)2 12(2y − 1) + 2(y − 1)(2y − 1)y(12x2 − 12x + 2)] + y − 1/2
fy
=
128[y 2 (y − 1)2 12(2x − 1) + 2(x − 1)(2x − 1)y(12y 2 − 12y + 2)] + x − 1/2
(185)
The solution is
u = −256x2 (x − 1)2 y(y − 1)(2y − 1)
v
=
256x2 (y − 1)2 x(x − 1)(2x − 1)
p
=
(x − 1/2)(y − 1/2)
54
(186)
6.8
Assigning values to quadrature points
As we have seen in Section ??, the building of the elemental matrix and rhs requires (at least) to assign
a density and viscosity value to each quadrature point inside the element. Depending on the type of
modelling, this task can prove more complex than one might expect and have large consequences on the
solution accuracy.
Here are several options:
• The simplest way (which is often used for benchmarks) consists in computing the ’real’ coordinates
(xq , yq , zq ) of a given quadrature point based on its reduced coordinates (rq , sq , tq ), and passing these
coordinates to a function which returns density and/or viscosity at this location. For instance, for
the Stokes sphere:
def rho(x,y):
if (x-.5)**2+(y-0.5)**2<0.123**2:
val=2.
else:
val=1.
return val
def mu(x,y):
if (x-.5)**2+(y-0.5)**2<0.123**2:
val=1.e2
else:
val=1.
return val
This is very simple, but it has been shown to potentially be problematic. In essence, it can introduce
very large contrasts inside a single element and perturb the quadrature. Please read section 3.3
of [253] and/or have a look at the section titled ”Averaging material properties” in the ASPECT
manual.
• another similar approach consists in assigning a density and viscosity value to the nodes of the FE
mesh first, and then using these nodal values to assign values to the quadrature points. Very often
,and quite logically, the shape functions are used to this effect. Indeed we have seen before that for
any point (r, s, t) inside an element we have
fh (r, s, t) =
m
X
fi Ni (r, s, t)
i
where the fi are the nodal values and the Ni the corresponding basis functions.
In the case of linear elements (Q1 basis functions), this is straightforward. In fact, the basis functions
Ni can be seen as moving weights: the closer the point is to a node, the higher the weight (basis
function value).
However, this is quite another story for quadratic elements (Q2 basis functions). In order to
illustrate the problem, let us consider a 1D problem. The basis functions are
N1 (r) =
1
r(r − 1)
2
N2 (r) = 1 − r2
N3 (r) =
1
r(r + 1)
2
Let us further assign: ρ1 = ρ2 = 0 and ρ3 = 1. Then
ρh (r) =
m
X
ρi Ni (r) = N3 (r)
i
There lies the core of the problem: the N3 (r) basis function is negative for r ∈ [−1, 0]. This means
that the quadrature point in this interval will be assigned a negative density, which is nonsensical
and numerically problematic!
use 2X Q1. write about it !
55
The above methods work fine as long as the domain contains a single material. As soon as there are
multiple fluids in the domain a special technique is needed to track either the fluids themselves or their
interfaces. Let us start with markers. We are then confronted to the infernal trio (a menage a trois?)
which is present for each element, composed of its nodes, its markers and its quadrature points.
Each marker carries the material information (density and viscosity). This information must ultimately be projected onto the quadrature points. Two main options are possible: an algorithm is
designed and projects the marker-based fields onto the quadrature points directly or the marker fields
are first projected onto the FE nodes and then onto the quadrature points using the techniques above.
————————–
At a given time, every element e contains ne markers. During the FE matrix building process, viscosity
and density values are needed at the quadrature points. One therefore needs to project the values carried
by the markers at these locations. Several approaches are currently in use in the community and the
topic has been investigated by [139] and [149] for instance.
elefant adopts a simple approach: viscosity and density are considered to be elemental values, i.e.
all the markers within a given element contribute to assign a unique constant density and viscosity value
to the element by means of an averaging scheme.
While it is common in the literature to treat the so-called arithmetic, geometric and harmonic means
as separate averagings, I hereby wish to introduce the notion of generalised mean, which is a family of
functions for aggregating sets of numbers that include as special cases the arithmetic, geometric and
harmonic means.
If p is a non-zero real number, we can define the generalised mean (or power mean) with exponent p
of the positive real numbers a1 , ... an as:
n
Mp (a1 , ...an ) =
1X p
a
n i=1 i
!1/p
(187)
and it is trivial to verify that we then have the special cases:
M−∞
=
M−1
=
M0
=
lim Mp = min(a1 , ...an )
p→−∞
(minimum)
n
(harm. avrg.)
+
+ · · · + a1n
Y
1/n
n
ai
lim Mp =
(geom. avrg.)
1
a1
1
a2
p→0
(188)
(189)
(190)
i=1
n
M+1
M+2
M+∞
1X
ai
n i=1
v
u n
u1 X
= t
a2
n i=1 i
=
=
(arithm. avrg.)
(191)
(root mean square)
(192)
lim Mp = max(a1 , ...an )
p→+∞
(maximum)
(193)
Note that the proofs of the limit convergence are given in [83].
An interesting property of the generalised mean is as follows: for two real values p and q, if p < q
then Mp ≤ Mq . This property has for instance been illustrated in Fig. 20 of [414].
One can then for instance look at the generalised mean of a randomly generated set of 1000 viscosity
values within 1018 P a.s and 1023 P a.s for −5 ≤ p ≤ 5. Results are shown in the figure hereunder and the
arithmetic, geometric and harmonic values are indicated too. The function Mp assumes an arctangentlike shape: very low values of p will ultimately yield the minimum viscosity in the array while very high
values will yield its maximum. In between, the transition is smooth and occurs essentially for |p| ≤ 5.
56
1e+23
arithm.
M(p)
1e+22
1e+21
geom.
1e+20
harm.
1e+19
1e+18
-4
-2
0
2
p
. python codes/fieldstone markers avrg
57
4
6.9
Matrix (Sparse) storage
The FE matrix is the result of the assembly process of all elemental matrices. Its size can become quite
large when the resolution is being increased (from thousands of lines/columns to tens of millions).
One important property of the matrix is its sparsity. Typically less than 1% of the matrix terms is
not zero and this means that the matrix storage can and should be optimised. Clever storage formats
were designed early on since the amount of RAM memory in computers was the limiting factor 3 or 4
decades ago. [407]
There are several standard formats:
• compressed sparse row (CSR) format
• compressed sparse column format (CSC)
• the Coordinate Format (COO)
• Skyline Storage Format
• ...
I focus on the CSR format in what follows.
6.9.1
2D domain - One degree of freedom per node
Let us consider again the 3 × 2 element grid which counts 12 nodes.
8=======9======10======11
|
|
|
|
| (3) | (4) | (5) |
|
|
|
|
4=======5=======6=======7
|
|
|
|
| (0) | (1) | (2) |
|
|
|
|
0=======1=======2=======3
In the case there is only a
like this:
X
X
X
X
single degree of freedom per node, the assembled FEM matrix will look
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
where the X stand for non-zero terms. This matrix structure stems from the fact that
• node 0 sees nodes 0,1,4,5
• node 1 sees nodes 0,1,2,4,5,6
• node 2 sees nodes 1,2,3,5,6,7
• ...
58
• node 5 sees nodes 0,1,2,4,5,6,8,9,10
• ...
• node 10 sees nodes 5,6,7,9,10,11
• node 11 sees nodes 6,7,10,11
In light thereof, we have
• 4 corner nodes which have 4 neighbours (counting themselves)
• 2(nnx-2) nodes which have 6 neighbours
• 2(nny-2) nodes which have 6 neighbours
• (nnx-2)×(nny-2) nodes which have 9 neighbours
In total, the number of non-zero terms in the matrix is then:
N Z = 4 × 4 + 4 × 6 + 2 × 6 + 2 × 9 = 70
In general, we would then have:
N Z = 4 × 4 + [2(nnx − 2) + 2(nny − 2)] × 6 + (nnx − 2)(nny − 2) × 9
Let us temporarily assume nnx = nny = n. Then the matrix size (total number of unknowns) is
N = n2 and
N Z = 16 + 24(n − 2) + 9(n − 2)2
A full matrix array would contain N 2 = n4 terms. The ratio of N Z (the actual number of reals to store)
to the full matrix size (the number of reals a full matrix contains) is then
R=
16 + 24(n − 2) + 9(n − 2)2
n4
It is then obvious that when n is large enough R ∼ 1/n2 .
CSR stores the nonzeros of the matrix row by row, in a single indexed array A of double precision
numbers. Another array COLIND contains the column index of each corresponding entry in the A array.
A third integer array RWPTR contains pointers to the beginning of each row, which an additional pointer
to the first index following the nonzeros of the matrix A. A and COLIND have length NZ and RWPTR
has length N+1.
In the case of the here-above matrix, the arrays COLIND and RWPTR will look like:
COLIN D = (0, 1, 4, 5, 0, 1, 2, 4, 5, 6, 1, 2, 3, 5, 6, 7, ..., 6, 7, 10, 11)
RW P T R = (0, 4, 10, 16, ...)
6.9.2
2D domain - Two degrees of freedom per node
When there are now two degrees of freedom per node, such as in the case of the Stokes equation in
two-dimensions, the size of the K matrix is given by
NfemV=nnp∗ ndofV
In the case of the small grid above, we have NfemV=24. Elemental matrices are now 8 × 8 in size.
We still have
• 4 corner nodes which have 4 ,neighbours,
• 2(nnx-2) nodes which have 6 neighbours
• 2(nny-2) nodes which have 6 neighbours
59
• (nnx-2)x(nny-2) nodes which have 9 neighbours,
but now each degree of freedom from a node sees the other two degrees of freedom of another node too.
In that case, the number of nonzeros has been multiplied by four and the assembled FEM matrix looks
like:
X X X X
X X X X
X X X X
X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X X X X
X
X
X
X
X X X X
X X X X
X
X
X
X
X X X X
X X X X
X
X
X
X
X X X X
X X X X X X
X
X
X
X
X
X
X X X X X X
X X X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X
X
X X X X
X
X
X
X
X
X
X
X
X X X X
X
X
X
X
X
X
X
X
X X X X
X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X X X X X X
X
X
X
X
X
X
X X X X X X
X X X X
X X X X
X X X X
X X X X
Note that the degrees of freedom are organised as follows:
(u0 , v0 , u1 , v1 , u2 , v2 , ...u11 , v11 )
In general, we would then have:
N Z = 4 [4 × 4 + [2(nnx − 2) + 2(nny − 2)] × 6 + (nnx − 2)(nny − 2) × 9]
and in the case of the small grid, the number of non-zero terms in the matrix is then:
N Z = 4 [4 × 4 + 4 × 6 + 2 × 6 + 2 × 9] = 280
In the case of the here-above matrix, the arrays COLIND and RWPTR will look like:
COLIN D = (0, 1, 2, 3, 8, 9, 10, 11, 0, 1, 2, 3, 8, 9, 10, 11, ...)
RW P T R = (0, 8, 16, 28, ...)
6.9.3
in fieldstone
The majority of the codes have the FE matrix being a full array
a mat = np . z e r o s ( ( Nfem , Nfem ) , dtype=np . f l o a t 6 4 )
and it is converted to CSR format on the fly in the solve phase:
s o l = s p s . l i n a l g . s p s o l v e ( s p s . c s r m a t r i x ( a mat ) , r h s )
Note that linked list storages can be used (lil matrix). Substantial memory savings but much longer
compute times.
60
6.10
Mesh generation
Before basis functions can be defined and PDEs can be discretised and solved we must first tesselate the
domain with polygons, e.g. triangles and quadrilaterals in 2D, tetrahedra, prisms and hexahedra in 3D.
When the domain is itself simple (e.g. a rectangle, a sphere, ...) the mesh (or grid) can be (more or
less) easily produced and the connectivity array filled with straightforward algorithms [452]. However,
real life applications can involve etremely complex geometries (e.g. a bridge, a human spine, a car chassis
and body, etc ...) and dedicated algorithms/softwares must be used (see [456, 180, 507]).
We usually distinguish between two broad classes of grids: structured grids (with a regular connectivity) and unstructured grids (with an irregular connectivity).
On the following figure are shown various triangle- tetrahedron-based meshes. These are obviously
better suited for simulations of complex geometries:
add more examples coming from geo
Let us now focus on the case of a rectangular computational domain of size Lx × Ly with a regular
mesh composed of nelx×nely=nel quadrilaterals. There are then nnx×nny=nnp grid points. The
elements are of size hx×hy with hx=Lx/nelx.
We have no reason to come up with an irregular/ilogical node numbering so we can number nodes
row by row or column by column as shown on the example hereunder of a 3×2 grid:
8=======9======10======11
|
|
|
|
| (3) | (4) | (5) |
|
|
|
|
4=======5=======6=======7
|
|
|
|
| (0) | (1) | (2) |
|
|
|
|
0=======1=======2=======3
2=======5=======8======11
|
|
|
|
| (1) | (3) | (5) |
|
|
|
|
1=======4=======7======10
|
|
|
|
| (0) | (2) | (4) |
|
|
|
|
0=======3=======6=======9
"row by row"
"column by column"
The numbering of the elements themselves could be done in a somewhat chaotic way but we follow
the numbering of the nodes for simplicity. The row by row option is the adopted one in fieldstoneand
the coordinates of the points are computed as follows:
x = np . empty ( nnp , dtype=np . f l o a t 6 4 )
y = np . empty ( nnp , dtype=np . f l o a t 6 4 )
61
counter = 0
f o r j i n r a n g e ( 0 , nny ) :
f o r i i n r a n g e ( 0 , nnx ) :
x [ c o u n t e r ]= i ∗hx
y [ c o u n t e r ]= j ∗hy
c o u n t e r += 1
The inner loop has i ranging from 0 to nnx-1 first for j=0, 1, ... up to nny-1 which indeed corresponds
to the row by row numbering.
We now turn to the connectivity. As mentioned before, this is a structured mesh so that the so-called
connectivity array, named icon in our case, can be filled easily. For each element we need to store the
node identities of its vertices. Since there are nel elements and m=4 corners, this is a m×nel array. The
algorithm goes as follows:
i c o n =np . z e r o s ( (m, n e l ) , dtype=np . i n t 1 6 )
counter = 0
f o r j in range (0 , nely ) :
f o r i in range (0 , nelx ) :
i c o n [ 0 , c o u n t e r ] = i + j ∗ nnx
i c o n [ 1 , c o u n t e r ] = i + 1 + j ∗ nnx
i c o n [ 2 , c o u n t e r ] = i + 1 + ( j + 1 ) ∗ nnx
i c o n [ 3 , c o u n t e r ] = i + ( j + 1 ) ∗ nnx
c o u n t e r += 1
In the case of the 3×2 mesh, the icon is filled as follows:
element id→
node id↓
0
1
2
3
0
1
2
3
4
5
0
1
5
4
1
2
6
5
2
3
7
6
4
5
9
8
5
6
10
9
6
7
11
10
It is to be understood as follows: element #4 is composed of nodes 5, 6, 10 and 9. Note that nodes are
always stored in a counter clockwise manner, starting at the bottom left. This is very important since
the corresponding basis functions and their derivatives will be labelled accordingly.
In three dimensions things are very similar. The mesh now counts nelx×nely×nelz=nel elements
which represent a cuboid of size Lx×Ly×Lz. The position of the nodes is obtained as follows:
x = np . empty ( nnp , dtype=np . f l o a t 6 4 )
y = np . empty ( nnp , dtype=np . f l o a t 6 4 )
z = np . empty ( nnp , dtype=np . f l o a t 6 4 )
c o u n t e r=0
f o r i i n r a n g e ( 0 , nnx ) :
f o r j i n r a n g e ( 0 , nny ) :
f o r k i n r a n g e ( 0 , nnz ) :
x [ c o u n t e r ]= i ∗hx
y [ c o u n t e r ]= j ∗hy
z [ c o u n t e r ]=k∗ hz
c o u n t e r += 1
The connectivity array is now of size m×nel with m=8:
i c o n =np . z e r o s ( (m, n e l ) , dtype=np . i n t 1 6 )
counter = 0
f o r i in range (0 , nelx ) :
f o r j in range (0 , nely ) :
f o r k in range (0 , n e l z ) :
i c o n [ 0 , c o u n t e r ]=nny∗ nnz ∗ ( i
)+nnz ∗ ( j
)+k
i c o n [ 1 , c o u n t e r ]=nny∗ nnz ∗ ( i +1)+nnz ∗ ( j
)+k
i c o n [ 2 , c o u n t e r ]=nny∗ nnz ∗ ( i +1)+nnz ∗ ( j +1)+k
i c o n [ 3 , c o u n t e r ]=nny∗ nnz ∗ ( i
)+nnz ∗ ( j +1)+k
i c o n [ 4 , c o u n t e r ]=nny∗ nnz ∗ ( i
)+nnz ∗ ( j
)+k+1
i c o n [ 5 , c o u n t e r ]=nny∗ nnz ∗ ( i +1)+nnz ∗ ( j
)+k+1
i c o n [ 6 , c o u n t e r ]=nny∗ nnz ∗ ( i +1)+nnz ∗ ( j +1)+k+1
i c o n [ 7 , c o u n t e r ]=nny∗ nnz ∗ ( i
)+nnz ∗ ( j +1)+k+1
c o u n t e r += 1
62
produce drawing of node numbering
63
6.11
Visco-Plasticity
6.11.1
Tensor invariants
Before we dive into the world of nonlinear rheologies it is necessary to introduce the concept of tensor
invariants since they are needed further on. Unfortunately there are many different notations used in the
literature and these can prove to be confusing.
Given a tensor T , one can compute its (moment) invariants as follows:
• first invariant :
TI |2D
3D
TI |
= T r[T ] = Txx + Tyy
= T r[T ] = Txx + Tyy + Tzz
• second invariant :
TII |2D
=
1 2
1X
1
2
2
T r[T 2 ] =
Tij Tji = (Txx
+ Tyy
) + Txy
2
2 ij
2
TII |3D
=
1
1 2
1X
2
2
2
2
2
T r[T 2 ] =
Tij Tji = (Txx
+ Tyy
+ Tyy
) + Txy
+ Txz
+ Tyz
2
2 ij
2
• third invariant :
TIII =
1X
1
T r[T 3 ] =
Tij Tjk Tki
3
3
ijk
The implementation of the plasticity criterions relies essentially on the second invariants of the (deviatoric) stress τ and the (deviatoric) strainrate tensors ε̇:
τII |2D
=
=
=
τII |3D
=
=
=
εII |2D
=
=
=
εII |3D
=
=
1 2
2
2
(τ + τyy
) + τxy
2 xx
1
2
(σxx − σyy )2 + σxy
4
1
(σ1 − σ2 )2
4
1 2
2
2
2
2
2
(τ + τyy
+ τzz
) + τxy
+ τxz
+ τyz
2 xx
1
2
2
2
(σxx − σyy )2 + (σyy − σzz )2 + (σxx − σzz )2 + σxy
+ σxz
+ σyz
6
1
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ1 − σ3 )2
6
1 d 2
(ε̇xx ) + (ε̇dyy )2 + (ε̇dxy )2
2
1 1
1
(ε̇xx − ε̇yy )2 + (ε̇yy − ε̇xx )2 + ε̇2xy
2 4
4
1
(ε̇xx − ε̇yy )2 + ε̇2xy
4
1 d 2
(ε̇xx ) + (ε̇dyy )2 + (ε̇dzz )2 + (ε̇dxy )2 + (ε̇dxz )2 + (ε̇dyz )2
2
1
(˙xx − ˙yy )2 + (˙yy − ˙zz )2 + (˙xx − ˙zz )2 + ˙2xy + ˙2xz + ˙2yz
6
Note that these (second) invariants are almost always used under a square root so we define:
64
τ II =
√
τII
ε̇II =
p
ε̇II
Note that these quantities have the same dimensions as their tensor counterparts, i.e. Pa for stresses and
s−1 for strain rates.
6.11.2
Scalar viscoplasticity
This formulation is quite easy to implement. It is widely used, e.g. [501, 454, 429], and relies on the
assumption that a scalar quantity ηp (the ’effective plastic viscosity’) exists such that the deviatoric stress
tensor
τ = 2ηp ε̇
(194)
is bounded by some yield stress value Y . From Eq. (194) it follows that τ II = 2ηp ε̇II = Y which yields
ηp =
Y
2ε̇II
This approach has also been coined the Viscosity Rescaling Method (VRM) [280].
insert here the rederivation 2.1.1 of spmw16
It is at this stage important to realise that (i) in areas where the strainrate is low, the resulting effective
viscosity will be large, and (ii) in areas where the strainrate is high, the resulting effective viscosity will be
low. This is not without consequences since (effective) viscosity contrasts up to 8-10 orders of magnitude
have been observed/obtained with this formulation and it makes the FE matrix very stiff, leading to
(iterative) solver convergence issues. In order to contain these viscosity contrasts one usually resorts to
viscosity limiters ηmin and ηmax such that
ηmin ≤ ηp ≤ ηmax
Caution must be taken when choosing both values as they may influence the final results.
. python codes/fieldstone indentor
6.11.3
about the yield stress value Y
In geodynamics the yield stress value is often given as a simple function. It can be constant (in space
and time) and in this case we are dealing with a von Mises plasticity yield criterion. . We simply assume
YvM = C where C is a constant cohesion independent of pressure, strainrate, deformation history, etc ...
Another model is often used: the Drucker-Prager plasticity model. A friction angle φ is then introduced and the yield value Y takes the form
YDP = p sin φ + C cos φ
and therefore depends on the pressure p. Because φ is with the range [0◦ , 45◦ ], Y is found to increase
with depth (since the lithostatic pressure often dominates the overpressure).
Note that a slightly modified verion of this plasticity model has been used: the total pressure p is
then replaced by the lithostatic pressure plith .
65
6.12
Pressure smoothing
It has been widely documented that the use of the Q1 × P0 element is not without problems. Aside from
the consequences it has on the FE matrix properties, we will here focus on another unavoidable side
effect: the spurious pressure checkerboard modes.
These modes have been thoroughly analysed [240, 107, 408, 409]. They can be filtered out [107] or
simply smoothed [302].
On the following figure (a,b), pressure fields for the lid driven cavity experiment are presented for
both an even and un-even number of elements. We see that the amplitude of the modes can sometimes
be so large that the ’real’ pressure is not visible and that something as simple as the number of elements
in the domain can trigger those or not at all.
b)
a)
c)
a) element pressure for a 32x32 grid and for a 33x33 grid;
b) image from [142, p307] for a manufactured solution;
c) elemental pressure and smoothed pressure for the punch experiment [454]
The easiest post-processing step that can be used (especially when a regular grid is used) is explained in
[454]: ”The element-to-node interpolation is performed by averaging the elemental values from elements
common to each node; the node-to-element interpolation is performed by averaging the nodal values
element-by-element. This method is not only very efficient but produces a smoothing of the pressure that
is adapted to the local density of the octree. Note that these two steps can be repeated until a satisfying
level of smoothness (and diffusion) of the pres- sure field is attained.”
In the codes which rely on the Q1 × P0 element, the (elemental) pressure is simply defined as
p=np . z e r o s ( n e l , dtype=np . f l o a t 6 4 )
while the nodal pressure is then defined as
q=np . z e r o s ( nnp , dtype=np . f l o a t 6 4 )
The element-to-node algorithm is then simply (in 2D):
count=np . z e r o s ( nnp , dtype=np . i n t 1 6 )
f o r i e l in range (0 , nel ) :
q [ i c o n [ 0 , i e l ]]+=p [ i e l ]
q [ i c o n [ 1 , i e l ]]+=p [ i e l ]
q [ i c o n [ 2 , i e l ]]+=p [ i e l ]
q [ i c o n [ 3 , i e l ]]+=p [ i e l ]
count [ i c o n [ 0 , i e l ]]+=1
count [ i c o n [ 1 , i e l ]]+=1
count [ i c o n [ 2 , i e l ]]+=1
count [ i c o n [ 3 , i e l ]]+=1
q=q/ count
66
Pressure smoothing is further discussed in [260].
produce figure to explain this
link to proto paper
link to least square and nodal derivatives
67
6.13
Pressure scaling
As perfectly explained in the step 32 of deal.ii5 , we often 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. We start from
K G
V
f
·
=
P
h
GT 0
and introduce the scaling coefficient as follows (which in fact does not alter the solution at all):
η
V
K
f
LG
·
=
η
η T
L
0
ηP
Lh
LG
We then end up with the modified Stokes system:
K G
V
f
·
=
P
h
GT 0
where
G=
η
G
L
P=
L
P
η
After the solve phase, we recover the real pressure with P =
5 https://www.dealii.org/9.0.0/doxygen/deal.II/step
32.html
68
h=
η 0
LP .
η
h
L
6.14
Pressure normalisation
When Dirichlet boundary conditions are imposed everywhere on the boundary, pressure is only present
by its gradient in the equations. It is thus determined up to an arbitrary constant. In such a case, one
commonly impose the average of the pressure over the whole domain or on a subsect of the boundary to
be have a zero average, i.e.
Z
pdV = 0
Ω
Another possibility is to impose the pressure value at a single node.
Let us assume that we are using Q1 × P0 elements. Then the pressure is constant inside each element.
The integral above becomes:
Z
XZ
X Z
X
pdV =
pdV =
pe
dV =
p e Ae =
Ω
e
Ωe
e
Ωe
e
where the sum runs over all elements e of area Ae . This can be rewritten
LT P = 0
and it is a constraint on the pressure. As we have seen before ??, we can associate to it a Lagrange
multiplier λ so that we must solve the modified Stokes system:
K
G 0
V
f
GT
0 L · P = h
λ
0
0 LT 0
When higher order spaces are used for pressure (continuous or discontinuous) one must then carry out
the above integration numerically by means of (usually) a Gauss-Legendre quadrature.
. python codes/fieldstone stokes sphere 3D saddlepoint/
. python codes/fieldstone burstedde/
. python codes/fieldstone busse/
. python codes/fieldstone RTinstability1/
. python codes/fieldstone saddlepoint q2q1/
. python codes/fieldstone compressible2/
. python codes/fieldstone compressible1/
. python codes/fieldstone saddlepoint q3q2/
69
6.15
The choice of solvers
Let us first look at the penalty formulation. In this case we are only solving for velocity since pressure is
recovered in a post-processing step. We also know that the penalty factor is many orders of magnitude
higher than the viscosity and in combination with the use of the Q1 × P0 element the resulting matrix
condition number is very high so that the use of iterative solvers in precluded. Indeed codes such as
SOPALE [183], DOUAR [66], or FANTOM [450] relying on the penalty formulation all use direct solvers
(BLKFCT, MUMPS, WSMP).
The main advantage of direct solvers is used in this case: They can solve ill-conditioned matrices.
However memory requirements for the storage of number of nonzeros in the Cholesky matrix grow very fast
as the number of equations/grid size increases, especially in 3D, to the point that even modern computers
with tens of Gb of RAM cannot deal with a 1003 element mesh. This explains why direct solvers are often
used for 2D problems and rarely in 3D with noticeable exceptions [454, 510, 67, 323, 11, 12, 13, 499, 356].
In light of all this, let us start again from the (full) Stokes system:
K G
V
f
·
=
P
h
GT 0
We need to solve this system in order to obtain the solution, i.e. the V and P vectors. But how?
Unfortunately, this question is not simple to answer and the ’right’ depends on many parameters.
How big is the matrix ? what is the condition number of the matrix K ? Is it symmetric ? (i.e. how are
compressible terms handled?).
6.15.1
The Schur complement approach
Let us write the above system as two equations:
KV + GP
= f
(195)
GT V
= h
(196)
The first line can be re-written V = K−1 (f − GP) and can be inserted in the second:
GT V = GT [K−1 (f − GP)] = h
or,
GT K −1 GP = GT K−1 f − h
The matrix S = GT K−1 G is called the Schur complement. It is Symmetric (since K is symmetric) and
Postive-Definite6 (SPD) if Ker(G) = 0. look in donea-huerta book for details Having solved this equation
(we have obtained P), the velocity can be recovered by solving KV = f − GP. For now, let us assume
that we have built the S matrix and the right hand side f˜ = GT K−1 f − h. We must solve SP = f˜.
Since S is SPD, the Conjugate Gradient (CG) method is very appropriate to solve this system. Indeed,
looking at the definition of Wikipedia: In mathematics, the conjugate gradient method is an algorithm for
the numerical solution of particular systems of linear equations, namely those whose matrix is symmetric
and positive-definite. The conjugate gradient method is often implemented as an iterative algorithm,
applicable to sparse systems that are too large to be handled by a direct implementation or other direct
methods such as the Cholesky decomposition. Large sparse systems often arise when numerically solving
partial differential equations or optimization problems.
The Conjugate Gradient algorithm is as follows:
6M
positive definite ⇐⇒ xT M x > 0 ∀ x ∈ Rn \ 0
70
Algorithm obtained from https://en.wikipedia.org/wiki/Conjugate_gradient_method
Let us look at this algorithm up close. The parts which may prove to be somewhat tricky are those
involving the matrix (in our case the Schur complement). We start the iterations with a guess pressure
P0 ( and an initial guess velocity which could be obtained by solving KV0 = f − GP0 ).
r0
= f˜ − SP0
T
= G K
−1
(197)
T
f − h − (G K
−1
G)P0
= GT K−1 (f − GP0 ) − h
T
= G K
−1
KV0 − h
T
= G V0 − h
(198)
(199)
(200)
(201)
(202)
We now turn to the αk coefficient:
αk =
rkT rk
rkT rk
rkT rk
=
=
pk Spk
pk GT K−1 Gpk
(Gpk )T K−1 (Gpk )
We then define p̃k = Gpk , so that αk can be computed as follows:
1. compute p̃k = Gpk
2. solve Kdk = p̃k
3. compute αk = (rkT rk )/(p̃Tk dk )
Then we need to look at the term Spk :
Spk = GT K−1 Gpk = GT K−1 p̃k = GT dk
We can then rewrite the CG algorithm as follows [525]:
• r0 = GT V0 − h
• if r0 is sufficiently small, then return (V0 , P0 ) as the result
• p0 = r0
• k=0
• repeat
– compute p̃k = Gpk
– solve Kdk = p̃k
– compute αk = (rkT rk )/(p̃Tk dk )
71
– Pk+1 = Pk + αk pk
– rk+1 = rk − αk GT dk
– if rk+1 is sufficiently small, then exit loop
T
– βk = (rk+1
rk+1 )/(rkT rk )
– pk+1 = rk+1 + βk pk
– k =k+1
• return Pk+1 as result
We see that we have managed to solve the Schur complement equation with the Conjugate Gradient
method without ever building the matrix S. Having obtained the pressure solution, we can easily recover
the corresponding velocity with KVk+1 = f − GPk+1 . However, this is rather unfortunate because it
requires yet another solve with the K matrix. As it turns out, we can slightly alter the above algorithm
to have it update the velocity as well so that this last solve is unnecessary.
We have
Vk+1 = K−1 (f −GPp+1 ) = K−1 (f −G(Pk +αk pk )) = K−1 (f −GPk )−αk K−1 Gpk = V−αk K−1 p̃k = Vk −αk dk
and we can insert this minor extra calculation inside the algorithm and get the velocity solution nearly
for free. The final CG algorithm is then
solver cg:
• compute V0 = K−1 (f − GP0 )
• r0 = GT V0 − h
• if r0 is sufficiently small, then return (V0 , P0 ) as the result
• p0 = r0
• k=0
• repeat
– compute p̃k = Gpk
– solve Kdk = p̃k
– compute αk = (rkT rk )/(p̃Tk dk )
– Pk+1 = Pk + αk pk
– Vk+1 = Vk − αk dk
– rk+1 = rk − αk GT dk
– if rk+1 is sufficiently small (|rk+1 |2 /|r0 |2 < tol), then exit loop
T
– βk = (rk+1
rk+1 )/(rkT rk )
– pk+1 = rk+1 + βk pk
– k =k+1
• return Pk+1 as result
This iterative algorithm will converge to the solution with a rate which depends on the condition
number of the S matrix, which is not easily obtainable since S is never built. Also, we know that large
viscosity contrasts in the domain will influence this too. Finally, we notice that this algorithm requires
one solve with matrix K per iteration but says nothing about the method employed to do so.
One thing we know improves the convergence of any iterative solver is the use of a preconditioner
matrix and therefore use the Preconditioned Conjugate Gradient method:
72
Algorithm obtained from https://en.wikipedia.org/wiki/Conjugate_gradient_method
In the algorithm above the preconditioner matrix M has to be symmetric positive-definite and fixed,
i.e., cannot change from iteration to iteration.
We see that this algorithm introduces an additional vector z and a solve with the matrix M at each
iteration, which means that M must be such that solving M x = f where f is a given rhs vector must be
cheap. Ultimately, the PCG algorithm applied to the Schur complement equation takes the form:
solver pcg:
• compute V0 = K−1 (f − GP0 )
• r0 = GT V0 − h
• if r0 is sufficiently small, then return (V0 , P0 ) as the result
• z0 = M −1 r0
• p0 = z0
• k=0
• repeat
– compute p̃k = Gpk
– solve Kdk = p̃k
– compute αk = (rkT zk )/(p̃Tk dk )
– Pk+1 = Pk + αk pk
– Vk+1 = Vk − αk dk
– rk+1 = rk − αk GT dk
– if rk+1 is sufficiently small (|rk+1 |2 /|r0 |2 < tol), then exit loop
– zk+1 = M −1 rk+1
T
– βk = (zk+1
rk+1 )/(zkT rk )
– pk+1 = zk+1 + βk pk
– k =k+1
• return Pk+1 as result
73
how to compute M for the Schur complement ?
6.16
The GMRES approach
74
6.17
The consistent boundary flux (CBF)
The Consistent Boundary Flux technique was devised to alleviate the problem of the accuracy of primary
variables derivatives (mainly velocity and temperature) on boundaries, where basis function (nodal)
derivatives do not exist. These derivatives are important since they are needed to compute the heat flux
(and therefore the NUsselt number) or dynamic topography and geoid.
The idea was first introduced in [343] and later used in geodynamics [520]. It was finally implemented
in the CitcomS code [522] and more recently in the ASPECT code (dynamic topography postprocessor).
Note that the CBF should be seen as a post-processor step as it does not alter the primary variables
values.
The CBF method is implemented and used in ??.
6.17.1
applied to the Stokes equation
We start from the strong form:
∇·σ =b
and then write the weak form:
Z
Z
N ∇ · σdV =
Ω
N bdV
Ω
where N is any test function. We then use the two equations:
Z
∇ · (N σ) = N ∇ · σ + ∇N · σ
(chain rule)
Z
(∇ · f ) dV =
f · n dS
(divergence theorem)
Ω
Γ
Integrating the first equation over Ω and using the second, we can write:
Z
Z
Z
N σ · n dS −
∇N · σ dV =
N bdV
Γ
Ω
Ω
On Γ, the traction vector is given by t = σ · n:
Z
Z
Z
N tdS =
∇N · σdV +
N bdV
Γ
Ω
Ω
Considering the traction vector as an unknown living on the nodes on the boundary, we can expand (for
Q1 elements)
2
2
X
X
tx =
tx|i Ni
ty =
ty|i Ni
i=1
i=1
on the boundary so that the left hand term yields a mass matrix M 0 . Finally, using our previous experience
of discretising the weak form, we can write:
M 0 · T = −KV − GP + f
where T is the vector of assembled tractions which we want to compute and V and T are the solutions
of the Stokes problem. Note that the assembly only takes place on the elements along the boundary.
Note that the assembled mass matrix is tri-diagonal can be easily solved with a Conjugate Gradient
method. With a trapezoidal integration rule (i.e. Gauss-Lobatto) the matrix can even be diagonalised
and the resulting matrix is simply diagonal, which results in a very cheap solve [520].
6.17.2
applied to the heat equation
We start from the strong form of the heat transfer equation (without the source terms for simplicity):
∂T
+ v · ∇T = ∇ · k∇T
ρcp
∂t
75
The weak form then writes:
Z
Z
Z
∂T
dV + ρcp
N ρcp
N v · ∇T dV =
N ∇ · k∇T dV
∂t
Ω
Ω
Ω
Using once again integration by parts and divergence theorem:
Z
Z
Z
Z
∂T
dV + ρcp
N v · ∇T dV =
N k∇T · ndΓ −
∇N · k∇T dV
N ρcp
∂t
Ω
Γ
Ω
Ω
On the boundary we are interested in the heat flux q = −k∇T
Z
Z
Z
Z
∂T
dV + ρcp
N v · ∇T dV = − N q · ndΓ −
N ρcp
∇N · k∇T dV
∂t
Ω
Γ
Ω
Ω
or,
Z
Z
N q · ndΓ = −
Γ
N ρcp
Ω
∂T
dV − ρcp
∂t
Z
Z
N v · ∇T dV −
Ω
∇N · k∇T dV
Ω
Considering the normal heat flux qn = q · n as an unknown living on the nodes on the boundary,
qn =
2
X
qn|i Ni
i=1
so that the left hand term becomes a mass matrix for the shape functions living on the boundary. We
have already covered the right hand side terms when building the FE system to solve the heat transport
equation, so that in the end
∂T
− Ka · T − Kd · T
M 0 · Qn = −M ·
∂t
where Qn is the assembled vector of normal heat flux components. Note that in all terms the assembly
only takes place over the elements along the boundary.
Note that the resulting matrix is symmetric.
6.17.3
implementation - Stokes equation
Let us start with a small example, a 3x2 element FE grid:
16 17
8
18 19
9
3
4
89
1213
5
1415
6
0
7
1
01
22 23
11
5
1011
4
0
20 21
10
2
23
45
1
2
67
3
Red color corresponds to the dofs in the x direction, blue color indicates a dof in the y direction.
We have nnp=12, nel=6, NfemV=24. Let us assume that free slip boundary conditions are applied.
The fix bc array is then:
bc fix=[ T T T T T T T T T T T T T T T T T T T T T T T T ]
Note that since corners belong to two edges, we effectively prescribed no-slip boundary conditions on
those.
We wish to compute the tractions on the boundaries, and more precisely for the dofs for which
a Dirichlecht velocity boundary condition has been prescribed. The number of (traction) unknowns
NfemTr is then the number of T in the bc fix array. In our specific case, we wave NfemTr= . This means
that we need for each targeted dof to be able to find its identity/number between 0 and NfemTr-1. We
therefore create the array bc nb which is filled as follows:
76
bc nb=[T T T T T T T T T T T T T T T T T T T T T T T T ]
This translates as follows in the code:
NfemTr=np . sum ( b c f i x )
bc nb=np . z e r o s ( NfemV , dtype=np . i n t 3 2 )
c o u n t e r=0
f o r i i n r a n g e ( 0 , NfemV) :
if ( bc fix [ i ]) :
bc nb [ i ]= c o u n t e r
c o u n t e r+=1
Algorithm 1 dksjfhfjk ghdf
0: kjhg kf
0: gfhfk
The algorithm is then as follows
A Prepare two arrays to store the matrix Mcbf and its right hand side rhscbf
B Loop over all elements
C For each element touching a boundary, compute the residual vector Rel = −fel + Kel Vel + Gel Pel
D Loop over the four edges of the element using the connectivity array
E For each edge loop over the number of degrees of freedom (2 in 2D)
F For each edge assess whether the dofs on both ends are target dofs.
G If so, compute the mass matrix Medge for this edge
H extract the 2 values off the element residual vector and assemble these in rhscbf
I Assemble Medge into NfemTrxNfemTr matrix using bc nb
M cbf = np . z e r o s ( ( NfemTr , NfemTr ) , np . f l o a t 6 4 )
r h s c b f = np . z e r o s ( NfemTr , np . f l o a t 6 4 )
# A
for
# B
i e l in range (0 , nel ) :
. . . compute e l e m e n t a l r e s i d u a l
...
# C
#boundary 0−1
f o r i i n r a n g e ( 0 , ndofV ) :
i d o f 0 =2∗ i c o n [ 0 , i e l ]+ i
i d o f 1 =2∗ i c o n [ 1 , i e l ]+ i
i f ( b c f i x [ i d o f 0 ] and b c f i x [ i d o f 1 ] ) :
i d o f T r 0=bc nb [ i d o f 0 ]
i d o f T r 1=bc nb [ i d o f 1 ]
r h s c b f [ i d o f T r 0 ]+= r e s e l [0+ i ]
r h s c b f [ i d o f T r 1 ]+= r e s e l [2+ i ]
M cbf [ i d o f T r 0 , i d o f T r 0 ]+=M edge [ 0 , 0 ]
M cbf [ i d o f T r 0 , i d o f T r 1 ]+=M edge [ 0 , 1 ]
M cbf [ i d o f T r 1 , i d o f T r 0 ]+=M edge [ 1 , 0 ]
M cbf [ i d o f T r 1 , i d o f T r 1 ]+=M edge [ 1 , 1 ]
# D
# E
#boundary 1−2
#[D]
# F
# H
#
# I
#
#
...
#boundary 2−3
#[D]
...
77
#boundary 3−0
#[D]
...
78
6.18
The value of the timestep
The chosen time step dt used for time integration is chosen to comply with the Courant-Friedrichs-Lewy
condition [14].
h
h2
δt = C min
,
max |v| κ
where h is a measure of the element size, κ = k/ρcp is the thermal diffusivity and C is the so-called CFL
number chosen in [0, 1[.
In essence the CFL condition arises when solving hyperbolic PDEs . It limits the time step in many
explicit time-marching computer simulations so that the simulation does not produce incorrect results.
This condition is not needed when solving the Stokes equation but it is mandatory when solving
the heat transport equation or any kind of advection-diffusion equation. Note that any increase of grid
resolution (i.e. h becomes smaller) yields an automatic decrease of the time step value.
79
6.19
mappings
The name isoparametric derives from the fact that the same (’iso’) functions are used as basis functions
and for the mapping to the reference element.
80
6.20
Exporting data to vtk format
This format seems to be the universally accepted format for 2D and 3D visualisation in Computational
Geodynamics. Such files can be opened with free softwares such as Paraview 7 , MayaVi 8 or Visit 9 .
Unfortunately it is my experience that no simple tutorial exists about how to build such files. There
is an official document which describes the vtk format10 but it delivers the information in a convoluted
way. I therefore describe hereafter how fieldstone builds the vtk files.
I hereunder show vtk file corresponding to the 3x2 grid presented earlier 6.10. In this particular
example there are:
• 12 nodes and 6 elements
• 1 elemental field: the pressure p)
• 2 nodal fields: 1 scalar (the smoothed pressure q), 1 vector (the velocity field u,v,0)
Note that vtk files are inherently 3D so that even in the case of a 2D simulation the z-coordinate of the
points and for instance their z-velocity component must be provided. The file, usually called solution.vtu
starts with a header:
We then proceed to write the node coordinates as follows:
NumberOfComponents= ’ 3 ’ Format= ’ a s c i i ’>
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00
These are followed by the elemental field(s):
−1.333333 e+00
−3.104414 e −10
1 . 3 3 3 3 3 3 e+00
−1.333333 e+00
8 . 2 7 8 4 1 7 e −17
1 . 3 3 3 3 3 3 e+00
Nodal quantities are written next:
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
7 https://www.paraview.org/
8 https://docs.enthought.com/mayavi/mayavi/
9 https://wci.llnl.gov/simulation/computer-codes/visit/
10 https://www.vtk.org/wp-content/uploads/2015/04/file-formats.pdf
81
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
8 . 8 8 8 8 8 5 e −08 −8.278405 e −24 0 . 0 0 0 0 0 0 e+00
8 . 8 8 8 8 8 5 e −08 1 . 6 5 5 6 8 2 e −23 0 . 0 0 0 0 0 0 e+00
0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
1 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
1 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
1 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
1 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00 0 . 0 0 0 0 0 0 e+00
−1.333333 e+00
−6.666664 e −01
6 . 6 6 6 6 6 4 e −01
1 . 3 3 3 3 3 3 e+00
−1.333333 e+00
−6.666664 e −01
6 . 6 6 6 6 6 4 e −01
1 . 3 3 3 3 3 3 e+00
−1.333333 e+00
−6.666664 e −01
6 . 6 6 6 6 6 4 e −01
1 . 3 3 3 3 3 3 e+00
To these informations we must append 3 more datasets. The first one is the connectivity, the second
one is the offsets and the third one is the type. The first one is trivial since said connectivity is needed
for the Finite Elements. The second must be understood as follows: when reading the connectivity
information in a linear manneer the offset values indicate the beginning of each element (omitting the
zero value). The third simply is the type of element as given in the vtk format document (9 corresponds
to a generic quadrilateral with an internal numbering consistent with ours).
0 1 5 4
1 2 6 5
2 3 7 6
4 5 9 8
5 6 10 9
6 7 11 10
4
8
12
16
20
24
9
9
9
9
9
9
The file is then closed with
The solution.vtu file can then be opened with ParaView, MayaVi or Visit and the reader is advised
to find tutorials online on how to install and use these softwares.
82
6.21
Runge-Kutta methods
These methods were developed around 1900 by the German mathematicians Carl Runge and Martin
Kutta. The RK methods are methods for the numerical integration of ODEs. These methods are well
documented in any numerical analysis textbook and the reader is referred to [206, 268].
Any Runge-Kutta method is uniquely identified by its Butcher tableau.
The following method is called the Runge-Kutta-Fehlberg method and is commonly abbreviated
RKF45. Its Butcher tableau is as follows:
0
1/4
1/4
3/8
3/32
9/32
12/13 1932/2197 -7200/2197 7296/2197
439/216
-8
3680/513
-845/4104
1
1/2
-8/27
2
-3544/2565
1859/4104
-11/40
16/135
0
6656/12825 28561/56430 -9/50 2/55
25/216
0
1408/2565
2197/4104
-1/5
0
The first row of coefficients at the bottom of the table gives the fifth-order accurate method, and the
second row gives the fourth-order accurate method.
83
6.22
Am I in or not?
It is quite common that at some point one must answer the question: ”Given a mesh and its connectivity
on the one hand, and the coordinates of a point on the other, how do I accurately and quickly determine
in which element the point resides?”
One typical occurence of such a problem is linked to the use of the Particle-In-Cell technique: particles
are advected and move through the mesh, and need to be localised at every time step. This question could
arise in the context of a benchmark where certain quantities need to be measured at specific locations
inside the domain.
6.22.1
Two-dimensional space
We shall first focus on quadrilaterals. There are many kinds of quadrilaterals as shown hereunder:
I wish to arrive at a single algorithm which is applicable to all quadrilaterals and therefore choose an
irregular quadrilateral. For simplicity, let us consider a Q1 element, with a single node at each corner.
3
2
M
1
0
Several rather simple options exist:
• we could subdivide the quadrilateral into two triangles and check whether point M is inside any of
them (as it turns out, this problem is rather straightforward for triangles. Simply google it.)
• We could check that point M is always on the left side of segments 0 → 1, 1 → 2, 2 → 3, 3 → 0.
• ...
Any of these approaches will work although some might be faster than others. In three-dimensions
all will however become cumbersome to implement and might not even work at all. Fortunately, there is
an elegant way to answer the question, as detailed in the following subsection.
6.22.2
Three-dimensional space
If point M is inside the quadrilateral, there exist a set of reduced coordinates r, s, t ∈ [−1 : 1]3 such that
4
X
i=1
Ni (rM , s, t)xi = xM
4
X
Ni (rM , s, t)yi = yM
i=1
4
X
i=1
84
Ni (rM , s, t)zi = zM
This can be cast as a system of three equations and three unknowns. Unfortunately, each shape function
Ni contains a term rst (as well as rs, rt, and st) so that it is not a linear system and standard techniques
are not applicable. We must then use an iterative technique: the algorithm starts with a guess for values
r, s, t and improves on their value iteration after iteration.
The classical way of solving nonlinear systems of equations is Newton’s method. We can rewrite the
equations above as F (r, s, t) = 0:
8
X
Ni (r, s, t)xi − xM
=
0
Ni (r, s, t)yi − yM
=
0
Ni (r, s, t)zi − zM
=
0
i=1
8
X
i=1
8
X
(203)
i=1
or,
Fr (r, s, t)
=
0
Fs (r, s, t)
=
0
Ft (r, s, t)
=
0
so that we now have to find the zeroes of continuously differentiable functions F : R → R. The
recursion is simply:
rk+1
rk
Fr (rk , sk , tk )
sk+1 = sk − JF (rk , sk , tk )−1 Fs (rk , sk , tk )
tk+1
tk
Ft (rk , sk , tk )
where J the Jacobian matrix:
JF (rk , sk , tk )
=
=
∂Fr
∂r (rk , sk , tk )
∂Fr
∂s (rk , sk , tk )
∂Fr
∂t (rk , sk , tk )
∂Fs
∂r (rk , sk , tk )
∂Fs
∂s (rk , sk , tk )
∂Fs
∂t (rk , sk , tk )
∂Ft
∂Ft
∂Ft
∂r (rk , sk , tk )
∂s (rk , sk , tk )
∂t (rk , sk , tk )
8
8
8
P ∂Ni
P ∂Ni
P
(r
,
s
,
t
)x
(r
,
s
,
t
)x
k
k
k
i
k
k
k
i
∂r
∂s
i=1
i=1
i=1
8
8
8
P
P
P
∂Ni
∂Ni
∂r (rk , sk , tk )yi
∂s (rk , sk , tk )yi
i=1
i=1
i=1
8
8
8
P
P
P
∂Ni
∂Ni
∂r (rk , sk , tk )zi
∂s (rk , sk , tk )zi
i=1
i=1
i=1
∂Ni
∂t (rk , sk , tk )xi
∂Ni
∂t (rk , sk , tk )yi
∂Ni
∂t (rk , sk , tk )zi
In practice, we solve the following system:
rk+1
rk
Fr (rk , sk , tk )
JF (rk , sk , tk ) sk+1 − sk = − Fs (rk , sk , tk )
tk+1
tk
Ft (rk , sk , tk )
Finally, the algorithm goes as follows:
• set guess values for r, s, t (typically 0)
• loop over k=0,...
• Compute rhs= −F (rk , sk , tk )
• Compute matrix JF (rk , sk , tk )
85
• solve system for (drk , dsk , dtk )
• update rk+1 = rk + drk , sk+1 = sk + dsk , tk+1 = tk + dtk
• stop iterations when (drk , dsk , dtk ) is small
• if rk , sk , tk ∈ [−1, 1]3 then M is inside.
This method converges quickly but involves iterations, and multiple solves of 3 × 3 systems which, when
carried out for each marker and at each time step can prove to be expensive. A simple modification can
be added to the above algorithm: iterations should be carried out only when the point M is inside of a
cuboid of size [min xi : max xi ] × [min yi : max yi ] × [min zi : max zi ] where the sums run over the vertices
i
i
i
i
i
i
of the element. In 2D this translates as follows: only carry out Newton iterations when M is inside the
red rectangle!
3
2
M
1
0
Note that the algorithm above extends to high degree elements such as Q2 and higher, even with
curved sides.
86
6.23
Error measurements and convergence rates
What follows is written in the case of a two-dimensional model. Generalisation to 3D is trivial. What
follows is mostly borrowed from [449].
When measuring the order of accuracy of the primitive variables v and p, it is standard to report
errors in both the L1 and the L2 norm. For a scalar quantity Ψ, the L1 and L2 norms are computed as
sZ
Z
||Ψ||1 =
|Ψ|dV
Ψ2 dV
||Ψ||2 =
V
V
For a vector quantity k = (kx , ky ), the L1 and L2 norms are defined as:
sZ
Z
||k||1 =
(|kx | + |ky |)dV
(kx2 + ky2 )dV
||k||2 =
V
V
To compute the respective norms the integrals in the above norms can be approximated by splitting
them into their element-wise contributions. The element volume integral can then be easily computed by
numerical integration using GaussLegendre quadrature.
The respective L1 and L2 norms for the pressure error can be evaluated via
v
u ne nq
nq
ne X
X
uX X
|ep (rq )|2 wq |Jq |
||ep ||1 =
|ep (rq )|wq |Jq |
||ep ||2 = t
i=1 q=1
i=1 q=1
where
ep (rq ) = p(rq ) − pth (rq )
is the pressure error evaluated at the qth quadrature associated with the ith element. ne and nq refer to
the number of elements and the number of quadrature points per element. wq and Jq are the quadrature
weight of the Jacobian associated with point q.
The velocity error ev is evaluated using the following two norms
v
u ne nq
nq
ne X
X
uX X
[|eu (rq )| + |ev (rq )|]wq |Jq |
||ev ||2 = t
||ev ||1 =
[|eu (rq )|2 + ev (rq )|2 ] wq |Jq |
i=1 q=1
i=1 q=1
where
eu (rq ) = u(rq ) − uth (rq )
ev (rq ) = v(rq ) − v th (rq )
We compute the different error norms for ep and ev for a set of numerical experiments with varying
resolution h. We expect the error norms to follow the following relationships:
0
||ev ||1 = Chrv
||ev ||2 = Chrv
||ep ||1 = Chrp
||ep ||2 = Chrp
0
where C is a resolution-independent constant and rp , rp0 and rv , rv0 are the convergence rates for pressure
and velocity, respectively. Using linear regression on the logarithm of the respective error norm and the
resolution, we compute the convergence rates of the numerical solutions.
• For Q1 P0 , the theoretical lower bound for rv0 is 2 and for rp0 it is 1
• For Q2 P−1 , the theoretical lower bound for rv0 is 3 and for rp0 it is 2
We note that when using discontinuous pressure space (e.g., P0 , P−1 ), these bounds remain valid even
when the viscosity is discontinuous provided that the element boundaries conform to the discontinuity.
87
6.24
6.24.1
The initial temperature field
Single layer with imposed temperature b.c.
Let us take a single layer of material characterised by a heat capacity cp , a heat conductivity k and a
heat production term H.
The Heat transport equation writes
ρcp (
∂T
~ )=∇
~ · (k ∇T
~ ) + ρH
+ ~v · ∇T
∂t
At steady state and in the absence of a velocity field, and assuming that the material properties to be
independent of time and space, and that there is no heat production (H = 0), this equation simplifies to
∆T = 0
Assuming the layer to be parallel to the x-axis, this yields to write
T (x, y) = T (y) = αT + β
In order to specify the constants α and β, we need two constraints.
At the bottom of the layer y = yb a temperature Tb is prescribed while a temperature Tt is prescribed
at the top with y = yt . This ultimately yields a temperature field in the layer given by
T (y) =
Tt − Tb
(y − yb ) + Tb
yt − yb
If now the heat production coefficient is not zero, the differential equation reads
k∆T + H = 0
The temperature field is then expected to be of the form
T (y) = −
H 2
y + αy + β
2k
Supplied again with the same boundary conditions, this leads to
β = Tb +
H 2
y − αyb
2k b
ie,
T (y) = −
H 2
(y − yb2 ) + α(y − yb ) + Tb
2k
and finally
α=
Tt − Tb
H
+
(yb + yt )
yt − yb
2k
or,
H
T (y) = − (y 2 − yb2 ) +
2k
Tt − Tb
H
+
(yb + yt ) (y − yb ) + Tb
yt − yb
2k
Taking H = 0 in this equation obviously yields the temperature field obtained previously. Taking
k = 2.25, Tt = 0C, Tb = 550C, yt = 660km, yb = 630km yields the following temperature profiles and
heat fluxes when the heat production H varies:
88
600
0.06
0.05
400
heat flux (W/m2)
Temperature (C)
0.07
H=0.0e-6
H=0.8e-6
H=1.6e-6
500
300
200
100
0
625000
H=0.0e-6
H=0.4e-6
H=0.8e-6
H=1.0e-6
H=1.2e-6
H=1.6e-6
0.04
0.03
0.02
0.01
630000
635000
640000
645000
650000
655000
660000
0
625000
630000
y
635000
640000
645000
650000
655000
660000
y
Looking at the values at the top, which are somewhat estimated to be about 55 − 65mW/m2 [277, table
8.6], one sees that value H = 0.8e − 6 yields a very acceptable heat flux. Looking at the bottom, the heat
flux is then about 0.03W/m2 which is somewhat problematic since the heat flux at the Moho is reported
to be somewhere between 10 and 20 mW/m2 in [277, table 7.1].
6.25
Single layer with imposed heat flux b.c.
Let us now assume that heat fluxes are imposed at the top and bottom of the layer:
We start again from the ODE
k∆T + H = 0
but only integrate it once:
k
dT
+ Hy + α = 0
dy
At the bottom q = k(dT /dy)|y=yb = qb and at the top q = k(dT /dy)|y=yt = qt so that
to finish
6.26
Single layer with imposed heat flux and temperature b.c.
to finish
6.26.1
Half cooling space
6.26.2
Plate model
6.26.3
McKenzie slab
89
6.27
6.27.1
Kinematic boundary conditions
In-out flux boundary conditions for lithospheric models
The velocity on the side is given by
u(y)
= vext
y < L1
vin − vext
=
(y − y1 ) + vext
y2 − y1
= vin
y > y2
u(y)
u(y)
y1 < y < y2
The requirement for volume conservation is:
Z
Ly
Φ=
u(y)dy = 0
0
Having chosen vin (the velocity of the plate), one can then compute vext as a function of y1 and y2 .
Z
Φ
=
y1
Z
y2
u(y)dy +
0
Z
Ly
u(y)dy +
y1
u(y)dy
y2
1
= vext y1 + (vin + vext )(y2 − y1 ) + (Ly − y2 )vin
2
1
1
= vext [y1 + (y2 − y1 )] + vin [ (y2 − y1 ) + (Ly − y2 )]
2
2
1
1
= vext (y1 + y2 ) + vin [Ly − (y1 + y1 )]
2
2
and finally
vext = −vin
Ly − 21 (y1 + y1 )
1
2 (y1 + y2 )
90
7
fieldstone 01: simple analytical solution
This fieldstone was developed in collaboration with Job Mos.
From [142]. 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.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (no-slip)
• direct solver
• isothermal
• isoviscous
• analytical solution
91
vx
1.0
vy
1.0
0.010
0.8
0.8
0.005
y
y
0.4
0.2
x
0.6
0.8
0.01
x
0.6
0.8
1.0
0.000010
0.8
0.000005
0.6
yy
0.8
0.000010
0.000005
0.2
0.4
x
0.6
0.8
1.0
0.04
0.04
0.2
0.4
x
0.6
0.8
p pth
1.0
0.00010
0.00005
0.8
0.00000
0.00005
y
0.000005
0.2
0.0
0.0
0.00010
0.4
0.000010
0.2
0.4
x
0.6
0.8
0.100000
1.0
0.00015
0.2
0.0
0.0
0.00020
0.2
0.4
x
0.6
0.8
velocity
pressure
x2
x1
0.010000
error
0.001000
0.000100
0.000010
0.000001
1.0
0.6
y
y
0.000010
xy
0.000000
0.4
0.000005
0.01
0.1
h
Quadratic convergence for velocity error, linear convergence for pressure error, as expected.
92
0.15
0.00
0.0
0.0
1.0
0.8
1.0
0.02
vy tyth
1.0
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
0.6
0.4
0.0
0.0
0.10
1.0
0.03
0.000000
0.2
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
0.4
0.4
0.005
0.005
0.2
0.2
0.00
0.6
0.000
y
0.6
0.4
0.0
0.0
0.05
0.8
0.005
0.6
0.000
p
1.0
0.010
1.0
0.00025
8
fieldstone 02: Stokes sphere
Viscosity and density directly computed at the quadrature points.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• isothermal
• non-isoviscous
• buoyancy-driven flow
• Stokes sphere
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
0.0
0.0
0.2
0.6
0.2
0.2
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
0.4
x
0.8
0.6
0.000
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
0.8
0.2
0.0
0.4
0.002
0.005
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
93
x
0.6
0.8
1.0
0.2
0.4
x
0.6
0.8
1.0
9
fieldstone 03: 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 [55]:
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
Ra =
αgy ∆T h3 ρ2 cp
αgy ∆T h3
=
κν
kµ
(204)
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)
(205)
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 [55]:
R ∂T
∂y (y = Ly )dx
(206)
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 [55] and [438]. (Note that this benchmark was also carried out and published
in other publications [460, 6, 206, 136, 307] 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 [438]
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.
94
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
0.4
0.6
0.4
20
0.2
0.2
0.4
x
0.6
0.8
40
0.2
0.4
60
xx
1.0
100
0.8
x
0.6
0.8
1.0
yy
1.0
0
100
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
4
0.4
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
95
0.6
0.4
0.4
0.8
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
0.75 0.6
0.2
x
0.6
xy
0.0
0.0
6
0.8
0.6
0.0
0.0
0.4
0.8
7
y
0.50
0.2
0.2
0.00
qx
1.0
0.0
0.0
0.4
0
0.6
200000
0.2
0.2
y
0.4
100
0.4
0.6
y
y
60
0.6
0.4
0
200 1.0
0.8
0.6
0.6
20
0.2
40 0.0
0.0
1.0
0
0.8
2000000.8
Nu
y
0
y
0.6
0.8
20
T
y
0.8
20
1.0
40
400000
1.0
p
y
0.8
0.0
0.0
1.0
40
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
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
96
10
fieldstone 04: The lid driven cavity
The lid driven cavity is a famous Computational Fluid Dynamics test case and has been studied in
countless publications with a wealth of numerical techniques (see [161] for a succinct review) and also in
the laboratory [295].
It models a plane flow of an isothermal isoviscous fluid in a rectangular (usually square) lid-driven
cavity. The boundary conditions are indicated in the Fig. ??a. The gravity is set to zero.
10.1
the lid driven cavity problem (ldc=0)
In the standard case, the upper side of the cavity moves in its own plane at unit speed, while the other
sides are fixed. This thereby introduces a discontinuity in the boundary conditions at the two upper
corners of the cavity and yields an uncertainty as to which boundary (side or top) the corner points
belong to. In this version of the code the top corner nodes are considered to be part of the lid. If these
are excluded the recovered pressure showcases and extremely large checkboard pattern.
This benchmark is usually dicussed in the context of low to very high Reynolds number with the full
Navier-Stokes equations being solved (with the noticeable exception of [408, 409, 107, 154] which focus
on the Stokes equation). In the case of the incompressible Stokes flow, the absence of inertia renders this
problem instantaneous so that only one time step is needed.
10.2
the lid driven cavity problem - regularisation I (ldc=1)
We avoid the top corner nodes issue altogether by prescribing the horizontal velocity of the lid as follows:
u(x) = x2 (1 − x)2 .
(207)
In this case the velocity and its first derivative is continuous at the corners. This is the so-called regularised
lid-driven cavity problem [372].
10.3
the lid driven cavity problem - regularisation II (ldc=2)
Another regularisation was presented in [138]. Here, a regularized lid driven cavity is studied which is
consistent in the sense that ∇·v = 0 holds also at the corners of the domain. There are no-slip conditions
at the boundaries x = 0, x = 1, and y = 0.
The velocity at y = 1 is given by
2
x1 − x
1 − cos(
π)
x ∈ [0, x1 ]
x1
u(x) = 1
x ∈ [x1 , 1 − x1 ]
2
x − (1 − x1 )
1
1 − cos(
π)
x ∈ [1 − x1 , 1]
u(x) = 1 −
4
x1
u(x)
=
1−
1
4
Results are obtained with x1 = 0.1.
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• isothermal
• isoviscous
97
(208)
A 100x100 element grid is used. No-slip boundary conditions are prescribed on sides and bottom. A
zero vertical velocity is prescribed at the top and the exact form of the prescribed horizontal velocity is
controlled by the ldc parameter.
200
ldc0
ldc1
ldc2
150
pressure
100
50
0
-50
-100
-150
-200
0
0.1
0.2
0.3
0.4
0.5
x
98
0.6
0.7
0.8
0.9
1
11
fieldstone 05: 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)
(209)
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) =
(210)
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 [149] (references to earlier uses of the benchmark are
available there) and its analytic solution is given in [519]. It has been carried out in [297] and [210].
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 ([348], 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
99
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
What we learn from this
100
0.6
0.8
1.0
0.015
0.2
0.4
x
0.6
0.8
1.0
0.020
12
fieldstone 06: SolKz benchmark
The SolKz benchmark [399] 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
(211)
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)
(212)
Free slip boundary conditions are imposed on all sides of the domain. This benchmark is presented in
[519] as well and is studied in [149] and [210].
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
y
0.6
0.4
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
vx txth
1.0
x
0.6
0.8
0.2
101
0.4
x
0.6
0.8
1.0
0.6
0.8
0.4
0.2
x
0.6
0.8
1.0
error
0.00100000
0.00010000
0.00001000
0.00000100
0.00000010
0.00000001
0.01
0.6
0.8
1.0
p pth
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
0.8
10
4
0.6
10
5
0.4
10
6
0.2
10
7
0.0
0.0
velocity
pressure
x2
x1
0.01000000
x
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
0.4
0.2
density
0.2
0.2
0.4
1.0
0.8
0.0
0.0
xy
0.6
1.0
x
1.0
0.8
0.5
y
y
102
0.2
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
0.4
0.0
0.0
103
0.0
0.0
0.8
0.6
104
0.2
0.6
vy tyth
1.0
0.4
x
0.8
105
0.6
0.4
0.2
1.0
0.00075
0.00050
0.00025
0.00000
0.00025
0.00050
0.00075
1.0
1.0
y
yy
0.2
1.0
0.8
1.0
0.4
2
4
6
0.2
0.8
y
y
0.4
0.6
0.6
1e 7
0.6
0.4
0.0
0.0
x
0.8
1.0
0.8
0.2
0.2
0.00006
0.0
0.0
6
4
2
0
0.0
0.0
0.00004
0.0
0.0
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
0.8
0.00000
0.4
1.0
0.00075
0.00050
0.00025
0.00000
0.00025
0.00050
0.00075
0.00004
0.00002
0.6
xx
1.0
1.0
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
101
0.2
0.4
x
0.6
0.8
1.0
13
fieldstone 07: 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. [415] derived a simple analytic
solution for the pressure and velocity fields for a circular inclusion under simple shear and it was used in
[139], [436], [149], [297] and [210].
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)
(213)
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.
[139] thoroughly investigated this problem with various numerical methods (FEM, FDM), with and
without tracers, and conclusively showed how various averagings lead to different results. [149] obtained
a first order convergence for both pressure and velocity, while [297] and [210] showed that the use of
adaptive mesh refinement in respectively the FEM and FDM yields convergence rates which depend on
refinement strategies.
102
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
0.2
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
103
1.0
x
0.6
0.8
density
102
y
0.6
y
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
velocity
pressure
x2
x1
1.00000000
0.10000000
0.01000000
0.00100000
0.00010000
0.00001000
0.00000100
0.00000010
0.00000001
0.01
0.1
h
103
0.8
0.0
0.0
1.0
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
0.8
0.6
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
1.0
0.8
error
0.8
vy tyth
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
0.0
0.0
0.4
y
y
0.75
0.2
0.2
0.6
1.00
0.4
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
14
fieldstone 08: the indentor benchmark
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) [280].
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 [454] and are also presented in [200].
The two dimensional punch problem has been extensively studied numerically for the past 40 years
[531, 530, 112, 111, 261, 513, 74, 397] and has been used to draw a parallel with the tectonics of eastern
China in the context of the India-Eurasia collision [443, 344]. 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 [367, 137, 279].
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).
104
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
0.0
0.2
0.4
y
y
0.2
vy
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
0.4
0.5
0.4
0.3
0.2
0.1
0.0
0.0
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
0.2
0.3
0.2
0.1
0.0
0.1
0.2
0.3
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
vx
y
y
y
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
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
0.05
0.0
5
0.2
4
0.00
p
0.2
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
ToDo: smooth punch
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• non-isoviscous
• nonlinear rheology
105
0.4
x
0.6
0.8
1.0
0.0
0.2
0.4
x
0.6
0.8
1.0
15
fieldstone 09: the annulus benchmark
This fieldstone was developed in collaboration with Prof. E.G.P. Puckett.
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),
(214)
vθ (r, θ)
=
f (r) cos(kθ),
(215)
p(r, θ)
=
kh(r) sin(kθ),
(216)
ρ(r, θ)
=
ℵ(r)k sin(kθ),
(217)
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
g
f
f0
g0
= 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
(218)
(219)
(220)
(221)
(222)
(223)
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
106
10
velocity
pressure
x2
x1
1
error
0.1
0.01
0.001
0.0001
0.01
h
107
16
fieldstone 10: 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
108
17
fieldstone 11: stokes sphere (3D) - mixed formulation
This is the same setup as Section 16.
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• 3D
• elemental b.c.
• buoyancy driven
109
18
fieldstone 12: consistent pressure recovery
What follows is presented in [529]. The second part of their paper wishes to establish a simple and
effective numerical method to calculate variables eliminated by the penalisation process. The method
involves an additional finite element solution for the nodal pressures using the same finite element basis
and numerical quadrature as used for the velocity.
Let us start with:
p = −λ∇ · v
which lead to
(q, p) = −λ(q, ∇ · v)
and then
Z
Z
N N dΩ · P = − λ N ∇N dΩ · V
or,
M · P = −D · V
and finally
P = −M −1 · D · V
with M of size (np × np), D of size (np ∗ ndof × np ∗ ndof ) and V of size (np ∗ ndof ). The vector P
contains the np nodal pressure values directly, with no need for a smoothing scheme. The mass matrix
M is to be evaluated at the full integration points, while the constraint part (the right hand side of the
equation) is to be evaluated at the reduced integration point.
As noted by [529], it is interesting to note that when linear elements are used and the lumped matrices
are used for the M the resulting algebraic equation is identical to the smoothing scheme based on the
averaging method only if the uniform square finite element mesh is used. In this respect this method is
expected to yield different results when elements are not square or even rectangular.
——q1 is smoothed pressure obtained with the center-to-node approach.
q2 is recovered pressure obtained with [529].R
All three fulfill the zero average condition: pdΩ = 0.
110
y
0.000
0.4
0.2
0.4
x
0.6
0.8
0.0
0.0
1.0
vx txth
1.0
0.000002
0.2
0.4
x
0.6
0.8
0.05
0.05
0.4
0.00
0.6
0.05
y
y
0.6
0.4
0.10
0.2
0.10
0.2
0.0
0.0
0.15
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
q1 pth
0.8
y
0.6
0.4
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0.000
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
1.0
0.003
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
x1.5
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
1.0
0.15
0.2
0.15
1.0
0.05
0.8
0.00
1.0
p pth
0.0
0.0
1.0
q2
1.0
0.8
x
0.8
0.8
0.2
0.000002
0.6
0.6
0.4
0.000001
0.4
x
0.6
0.000000
0.2
0.4
0.8
0.000001
0.0
0.0
q1
1.0
0.000002
0.2
1.0
0.2
1.0
0.4
0.000002
0.0
0.0
1.0
0.6
0.000001
0.2
0.8
y
y
0.000000
0.4
0.6
0.8
0.000001
0.6
x
vy tyth
1.0
0.8
0.4
0.10
0.2
0.010
0.2
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
111
19
fieldstone 13: the Particle in Cell technique (1) - the effect
of averaging
This fieldstone is being developed in collaboration with BSc student Eric Hoogen.
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:
m
m
X
X
xim =
Ni (rim , sim )xi
yim =
Ni (rim , sim )yi
(224)
i
i
where xi , yi are the coordinates of the vertices of the element.
A third option consists in the use of the so-called Poisson-disc sampling which produces points that
are tightly-packed, but no closer to each other than a specified minimum distance, resulting in a more
natural pattern 11 . Note that the Poisson-disc algorithm fills the whole domain at once, not element after
element.
say smthg about avrg dist
insert here theory and link about Poisson disc
11 https://en.wikipedia.org/wiki/Supersampling
112
Left: regular distribution, middle: random, right: Poisson disc.
16384 markers (32x32 grid, 16 markers per 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
1X
φi
(225)
hφiam =
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
(226)
i
and the harmonic mean:
n
hφihm =
1X 1
n i φi
!−1
(227)
Furthermore, there is a well known inequality for any set of positive numbers,
hφiam
≥
hφigm
≥
hφihm
(228)
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 [414, 139, 449, 381]
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.
113
Let k be the green node of the figures above. Let (r, s) denote the coordinates of a marker inside its
element. For clarity, we define the follow three nodal averaging schemes:
• nodal type A:
fk =
sum of values carried by markers in 4 neighbour elements
number of markers in 4 neighbour elements
• nodal type B:
fk =
sum of values carried by markers inside dashed line
number of markers in area delimited by the dashed line
• nodal type C
fk =
sum of values carried by markers in 4 neighbour elements ∗ Np (r, s)
sum of Np (r, s)
where Np is the Q1 basis function corresponding to node p defined on each element. Since these
functions are 1 on node k and then linearly decrease and become zero on the neighbouring nodes,
this effectively gives more weight to those markers closest to node k.
This strategy is adopted in [1, 341] (although it is used to interpolate onto the nodes of Q2 P−1
elements. It is formulated as follows:
”We assume that an arbitrary material point property f , is discretized via f (x) ' δ(x − xp )fp . We
then utilize an approximate local L2 projection of fp onto a continuous Q1 finite element space.
The corner vertices of each Q2 finite element define the mesh fp is projected onto. The local
reconstruction for a node i is defined by
R
P
Ni (x)f (x)
p Ni (xp )fp
Ωi
ˆ
R
' P
fi =
N
(x)
i
p Ni (xp )
Ωi
where the summation over p includes all material points contained within the support Ωi of the
trilinear interpolant Ni ”.
The setup is identical to the Stokes sphere experiment. The bash script script runall runs the code
for many resolutions, both initial marker distribution and all four 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.
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.
• elemental values yield better results (espcecially at low resolutions)
• harmonic mean yields overal the best results
114
Root mean square velocity results are shown hereunder:
random PD markers, elemental proj
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
regular markers, elemental proj
0.01
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.009
0.008
0.007
0.006
vrms
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.005
0.004
0.003
0.002
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
random markers, elemental proj
0.01
0.009
0.005
0.004
0.003
0.002
0.001
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.02
0.03
0.04
0.05
0.06
h
random markers, nodal proj A
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
0.007
0.006
0.005
0.004
0.003
0.001
0.06
0.07
0.08
0.09
0.1
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
0.03
0.04
0.05
0.06
0.07
0.08
0.09
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
0.007
0.006
0.005
0.004
0.003
0.01
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.008
0.007
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
0.08
0.09
0.1
0.03
0.04
0.05
0.06
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
0.08
0.09
0.06
0.06
0.07
0.08
0.09
0.006
0.005
0.004
0.003
0.1
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
0.1
0.006
0.005
0.004
0.003
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
regular markers, nodal proj C
0.007
0.001
0.09
0.007
0.01
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.008
0.002
0.08
h
0.01
0.001
0.07
0.008
0.1
0.009
0.002
0.05
0.05
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
0.007
0.04
0.07
vrms
0.008
0.03
0.04
0.01
random markers, nodal proj C
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.001
0.02
random markers, nodal proj C
0.01
0.02
0.03
h
0.009
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
0.002
0.01
h
0.01
0.02
0.009
0.001
0.07
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.1
0.008
0.002
0.06
0.09
regular markers, nodal proj B
0.01
0.001
0.08
h
0.009
0.002
0.07
0.01
0.1
vrms
0.008
0.05
0.06
0.009
random markers, nodal proj B
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.04
0.05
0.001
0.02
random markers, nodal proj B
0.01
0.03
0.04
h
0.009
0.02
0.03
0.002
0.01
h
0.01
0.02
regular markers, nodal proj A
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.008
0.002
0.05
0.01
h
0.01
0.001
0.04
0.1
0.009
0.002
0.03
0.09
vrms
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.02
0.08
random PD markers, nodal proj A
0.01
0.01
0.07
h
0.009
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
0.001
0.01
vrms
0.03
vrms
0.02
09
16
25
36
48
64
09
16
25
36
48
64
09
16
25
36
48
64
0.002
0.001
0.01
am,
am,
am,
am,
am,
am,
gm,
gm,
gm,
gm,
gm,
gm,
hm,
hm,
hm,
hm,
hm,
hm,
0.005
0.004
0.003
0.002
0.001
0.01
0.02
0.03
0.04
0.05
0.06
h
0.07
0.08
0.09
0.1
0.01
0.02
0.03
0.04
0.05
0.06
h
0.07
0.08
0.09
0.1
h
Left column: random markers, middle column: Poisson disc, right column: regular markers. First row: elemental
projection, second row: nodal 1 projection, third row: nodal 2 projection, fourth row: nodal 3 projection.
64 markers per element, arithmetic avrg
64 markers per element, geometric avrg
64 markers per element, harmonic avrg
0.01
0.01
0.01
0.009
0.009
0.009
0.008
0.008
0.008
0.007
vrms
0.005
0.004
0.003
0.002
0.001
0.01
0.02
0.03
0.04
0.05
0.06
h
0.07
0.08
0.09
0.1
0.007
reg, el
rand, el
pd, el
reg, nodA
rand, nodA
pd, nodA
reg, nodB
rand, nodB
pd, nodB
reg, nodC
rand, nodC
pd, nodC
0.006
vrms
reg, el
rand, el
pd, el
reg, nodA
rand, nodA
pd, nodA
reg, nodB
rand, nodB
pd, nodB
reg, nodC
rand, nodC
pd, nodC
0.006
0.005
0.004
0.003
0.002
0.001
0.01
0.02
0.03
0.04
0.05
0.06
0.07
h
0.08
0.09
0.1
reg, el
rand, el
pd, el
nodA
nodA
nodA
nodB
nodB
nodB
nodC
nodC
nodC
0.006
vrms
0.007
0.005
reg,
rand,
pd,
reg,
rand,
pd,
reg,
rand,
pd,
0.004
0.003
0.002
0.001
0.01
0.02
0.03
0.04
0.05
0.06
h
Left to right: arithmetic, geometric, harmonic averaging for viscosity.
115
0.07
0.08
0.09
0.1
20
fieldstone f14: solving the full saddle point problem
The details of the numerical setup are presented in Section ??.
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
116
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
0.0
0.010 0.0
1.0
0.00
q
1.0
0.05
0.8
0.10
0.6
y
0.010
vx
1.0
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
0.06
0.01
0.6
0.01
0.6
0.00
y
y
0.00
0.02
0.8
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
0.8
0.0
0.0
0.6
0.0000000
0.4
0.0000005
0.2
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
y
0.8
0.0000005
0.0
0.0
0.02
0.2
0.8
0.0000005
0.2
0.00
0.4
0.8
0.4
0.02
0.6
0.0000010
1.0
vy tyth
0.06
0.04 1.0
y
0.02
0.8
1.0
xy
x
0.6
0.8
1.0
q pth
1.0
0.01
0.00
0.02
0.8
0.04
0.6
0.06
y
1.0
0.03
yy
y
0.03
xx
0.25
1.0
0.10
0.2
0.4
x
0.6
0.8
1.0
0.12
0.4
0.08
0.2
0.0
0.0
0.10
0.2
0.4
x
0.6
0.8
1.0
Unlike the results obtained with the penalty formualtion (see Section ??), the pressure showcases a
very strong checkerboard pattern, similar to the one in [143].
Left: pressure solution as shown in [143]; 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.
117
0.12
21
fieldstone f15: saddle point problem with Schur complement approach - benchmark
The details of the numerical setup are presented in Section ??. The main difference resides in the Schur
complement approach to solve the Stokes system, as presented in Section 6.15 (see solver cg). This
iterative solver is very easy to implement once the blocks K and G, as well as the rhs vectors f and h
have been built.
0.010
vx
1.0
0.010
vy
1.0
p
1.0
0.8
0.005 0.8
0.005 0.8
0.6
0.6
0.6
0.6
0.4
0.05 0.4
0.4
x
0.6
0.8
1.0
0.03
xx
1.0
0.0
0.010 0.0
0.02
0.8
0.01
0.6
0.4
x
0.6
0.8
0.03
yy
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
0.02
0.8
0.01
0.6
0.00
0.4
0.01
0.2
0.0
0.0
0.02
0.2
0.000010
1.0
vx txth
1.0
0.03
0.4
x
0.6
0.8
1.0
0.03
0.10
0.2
0.4
x
0.6
0.8
1.0
0.0
0.0
0.8
0.00
0.4
0.02
0.2
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
1.0
0.0
0.0000100.0
0.2
0.4
x
0.6
0.8
1.0
0.02
0.04 0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.01
0.00010
p pth
0.014
q pth
0.000051.0
0.012
0.010
y
0.008
y
y
y
0.8
0.03
0.2
0.00005
0.6
0.6
0.15
0.4
0.6
0.000000
0.4
x
1.0
0.04
0.6
0.000000
0.4
0.4
0.8
0.6
0.6
0.2
x
0.6
0.8
0.02
0.6
0.000000.8
0.0
0.0
0.4
0.05
0.8
0.000005
0.000005
0.2
0.10
0.2
0.04 1.0
0.8
0.000005
0.2
0.05
0.2
0.8
0.4
0.00
0.15
xy
1.0
0.000010
1.0
vy tyth
0.8
y
0.0
0.010 0.0
1.0
y
y
0.00
0.2
0.05
y
0.2
0.0050.2
y
0.0
0.0
0.4
0.0050.2
0.2
0.00
y
0.000
y
y
0.000
0.4
q
0.05 1.0
0.00010
0.4
0.006
0.000005
0.2
0.00015
0.2
0.004
0.0
0.0000100.0
0.00020
0.0
0.0
0.00025
0.2
0.4
x
0.6
0.8
1.0
0.2
0.4
x
0.6
0.8
1.0
0.002
0.000
Rather interestingly the pressure checkerboard modes are not nearly as present as in Section ?? which
uses a full matrix approach.
Looking at the discretisation errors for velocity and pressure, we of course recover the same rates and
values as in the full matrix case.
118
0.100000
velocity
pressure
x2
x1
0.010000
error
0.001000
0.000100
0.000010
0.000001
0.01
0.1
h
Finally, for each experiment the normalised residual (see solver cg) was recorded. We see that all
things equal the resolution has a strong influence on the number of iterations the solver must perform to
reach the required tolerance. This is one of the manifestations of the fact that the Q1 × P0 element is
not a stable element: the condition number of the matrix increases with resolution. We will see that this
is not the case of stable elements such as Q2 × Q1 .
1.000000000
8x8
12x12
16x16
24x24
32x32
48x48
64x64
128x128
1e-8
0.100000000
normalised residual
0.010000000
0.001000000
0.000100000
0.000010000
0.000001000
0.000000100
0.000000010
0.000000001
0
10
20
30
40
50
# iteration
119
60
70
80
90
100
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Schur complement approach
• isothermal
• isoviscous
• analytical solution
build S and have python compute its smallest and largest eigenvalues as a function of resolution?
120
22
fieldstone f16: saddle point problem with Schur complement approach - Stokes sphere
We are revisiting the 2D Stokes sphere problem, but this time we use the Schur complement approach
to solve the Stokes system, Because there are viscosity contrasts in the domain, it is advisable to use the
Preconditioned Conjugate Gradient as presented in Section 6.15 (see solver pcg).
0.00075
0.00050
1.0
0.6
0.00000
0.4
0.4
0.8
0.0000
0.6
0.00025
0.2
0.2
0.0
0.0
0.0
0.0
0.00050
0.0010
0.0
0.0
0.4
x
0.6
0.8
1.0
0.2
0.4
x
0.6
0.8
1.0
0.0
0.2
0.6
0.0
0.4
0.2
0.2
0.2
0.4
x
0.6
0.8
1.0
0.0015
0.00075
q
1.0
0.8
0.4
0.0005
0.2
0.2
0.2
y
0.6
p
0.0005
1.0
y
0.8
0.00025
y
0.8
vy
0.4
0.4
y
vx
1.0
0.0010
0.0
0.0
0.2
0.2
0.4
x
0.6
0.8
1.0
0.4
0.4
0.0020
0.015
0.015
0.010
0.010
xx
1.0
yy
1.0
0.02
0.020
xy
1.0
0.01 1.0
0.0050.8
0.8
0.6
0.6
0.000
0.4
0.6
0.4
0.6
0.000
0.4
0.2
0.2
0.005
0.2
0.005
0.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.0
0.0
0.010
0.2
0.4
x
0.6
0.8
1.0
0.0
0.0
0.010
0.015
0.015
0.015
y
0.00
y
y
0.0050.8
y
0.8
0.4
0.010
0.2
0.2
0.4
x
0.6
0.8
1.0
0.010.0
0.0
0.2
0.4
x
0.6
0.8
1.0
0.005
0.02
The normalised residual (see solver pcg) was recorded. We see that all things equal the resolution
has a strong influence on the number of iterations the solver must perform to reach the required tolerance.
However, we see that the use of the preconditioner can substantially reduce the number of iterations inside
the Stokes solver. At resolution 128x128, this number is halved.
121
1.000000000
32x32 (prec.)
32x32 (no prec.)
64x64 (prec.)
64x64 (no prec.)
128x128 (prec.)
128x128 (no prec.)
1e-8
0.100000000
normalised residual
0.010000000
0.001000000
0.000100000
0.000010000
0.000001000
0.000000100
0.000000010
0.000000001
0
50
100
150
200
# iteration
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Schur complement approach
• isothermal
• non-isoviscous
• Stokes sphere
122
250
300
350
400
23
fieldstone 17: 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 [141]:
x + x2 + xy + x3 y
y + xy + y 2 + x2 y 2
v=
(229)
2
−2z − 3xz − 3yz − 5x yz
and
p = xyz + x3 y 3 z − 5/32
(230)
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. (229). Following [96], the viscosity is given by the smoothly
varying function
µ = exp(1 − β(x(1 − x) + y(1 − y) + z(1 − z)))
(231)
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
123
(232)
(233)
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
∂p
∂x
∂p
∂y
∂p
∂z
=
2(2 + 6xy)
=
1 + 4xy
= −3 − 10xy
=
1 + 2y 2 + 3x2
=
2(2 + 2x2 )
= −3 − 5x2
= −10yz
=
0
=
2(0)
= yz + 3x2 y 3 z
(234)
= xz + 3x3 y 2 z
(235)
= xy + x3 y 3
(236)
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
(237)
Ω
This is a single constraint associated to a single Lagrange multiplier λ and the global Stokes system takes
the form
K
G 0
V
GT
0 C P
λ
0 CT 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.(237). Its exact expression is as follows:
Z
XZ
XZ X p
X X Z
X
p
p dΩ =
p dΩ =
Ni pi dΩ =
Ni dΩ pi =
Ce · pe
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.
124
23.0.1
Constant viscosity
Choosing β = 0 yields a constant velocity µ(x, y, z) = exp(1) ' 2.718 (and greatly simplifies the righthand 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
2 + 6xy
yz + 3x2 y 3 z
f = − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
−10yz
xy + x3 y 3
Note that there seems to be a sign problem with Eq.(26) in [96].
125
(238)
(239)
(240)
0.01000
velocity
pressure
x3
x2
error
0.00100
0.00010
0.00001
0.1
h
23.0.2
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)
126
and thr right-hand side by
yz + 3x2 y 3 z
2 + 6xy
f = − xz + 3x3 y 2 z + µ 2 + 2x2 + 2y 2
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 − 10xyz
−3z − 5x z
−4 − 6x − 6y − 10x
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 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
127
24
fieldstone 18: solving the full saddle point problem with
Q2 × Q1 elements
The details of the numerical setup are presented in Section ??.
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
features
• Q2 × Q1 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• isothermal
• isoviscous
• analytical solution
128
0.0100000
velocity
pressure
x3
x2
0.0010000
error
0.0001000
0.0000100
0.0000010
0.0000001
0.1
h
129
25
fieldstone 19: solving the full saddle point problem with
Q3 × Q2 elements
The details of the numerical setup are presented in Section ??.
Each element has mV = 16 vertices so in total ndofV × mV = 32 velocity dofs and ndofP ∗ mP = 9
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 .
60===61===62===63===64===65===66===67===68===70
||
||
||
||
50
51
52
53
54
55
56
57
58
59
||
||
||
||
40
41
42
43
44
45
46
47
48
49
||
||
||
||
30===31===32===33===34===35===36===37===38===39
||
||
||
||
20
21
22
23
24
25
26
27
28
29
||
||
||
||
10
11
12
13
14
15
16
17
18
19
||
||
||
||
00===01===02===03===04===05===06===07===08===09
Example of 3x2 mesh. nnx=10, nny=7, nnp=70, nelx=3, nely=2, nel=6
12===13===14===15
||
||
||
||
08===09===10===11
||
||
||
||
04===05===06===07
||
||
||
||
00===01===02===03
06=====07=====08
||
||
||
||
||
||
03=====04=====05
||
||
||
||
||
||
00=====01=====02
Velocity (Q3)
Pressure (Q2)
(r,s)_{00}=(-1,-1)
(r,s)_{01}=(-1/3,-1)
(r,s)_{02}=(+1/3,-1)
(r,s)_{03}=(+1,-1)
(r,s)_{04}=(-1,-1/3)
(r,s)_{05}=(-1/3,-1/3)
(r,s)_{06}=(+1/3,-1/3)
(r,s)_{07}=(+1,-1/3)
(r,s)_{08}=(-1,+1/3)
(r,s)_{09}=(-1/3,+1/3)
(r,s)_{10}=(+1/3,+1/3)
(r,s)_{11}=(+1,+1/3)
(r,s)_{12}=(-1,+1)
(r,s)_{13}=(-1/3,+1)
(r,s)_{14}=(+1/3,+1)
(r,s)_{15}=(+1,+1)
(r,s)_{00}=(-1,-1)
(r,s)_{01}=(0,-1)
(r,s)_{02}=(+1,-1)
(r,s)_{03}=(-1,0)
(r,s)_{04}=(0,0)
(r,s)_{05}=(+1,0)
(r,s)_{06}=(-1,+1)
(r,s)_{07}=(0,+1)
(r,s)_{08}=(+1,+1)
Write about 4 point quadrature.
130
features
• Q3 × Q2 element
• incompressible flow
• mixed formulation
• isothermal
• isoviscous
• analytical solution
0.0001
velocity
pressure
x4
x3
1x10-5
error
1x10-6
1x10-7
1x10-8
1x10-9
1x10-10
0.1
h
velocity error rate is cubic, pressure superconvergent since the pressure field is quadratic and therefore
lies into the Q2 space.
131
26
fieldstone 20: the Busse benchmark
This three-dimensional benchmark was first proposed by [97]. It has been subsequently presented in
[438, 460, 6, 362, 136, 297]. We here focus on Case 1 of [97]: an isoviscous bimodal convection experiment
at Ra = 3 × 105 .
The domain is of size a × b × h with a = 1.0079h, b = 0.6283h with h = 2700km. It is filled
with a Newtonian fluid characterised by ρ0 = 3300kg.m−3 , α = 10−5 K−1 , µ = 8.0198 × 1023 Pa.s,
k = 3.564W.m−1 .K−1 , cp = 1080J.K−1 .kg−1 . The gravity vector is set to g = (0, 0, −10)T . The
temperature is imposed at the bottom (T = 3700◦ C) and at the top (T = 0◦ C).
The various measurements presented in [97] are listed hereafter:
• The Nusselt number N u computed at the top surface following Eq. (206):
RR
∂T
dxdy
z=L ∂y
N u = Lz R R z
T dxdy
z=0
• the root mean square velocity vrms and the temperature mean square velocity Trms
• The vertical velocity w and temperature T at points x1 = (0, 0, Lz /2), x2 = (Lx , 0, Lz /2), x3 =
(0, Ly , Lz /2) and x4 = (Lx , Ly , Lz /2);
• the vertical component of the heat flux Q at the top surface at all four corners.
The values plotted hereunder are adimensionalised by means of a reference temperature (3700K), a
reference lengthscale 2700km, and a reference time L2z /κ.
0.0009
0.0008
0.0007
vrms
0.0006
0.0005
0.0004
0.0003
0.0002
0.0001
0
2x108 4x108 6x108 8x108 1x109 1.2x1091.4x1091.6x109
time/year
2124
2123
2122
2121
2120
2119
2118
2117
2116
2115
0
2x108
4x108
6x108
8x108
time/year
132
1x109 1.2x109 1.4x109 1.6x109
0.002
0.0015
w at Lz/2
0.001
0.0005
0
-0.0005
-0.001
-0.0015
-0.002
0
2x108 4x108 6x108 8x108 1x109 1.2x1091.4x1091.6x109
time/year
0
-0.5
heat flux
-1
-1.5
-2
-2.5
-3
-3.5
1x108
1.5x108
2x108
2.5x108
3x108
3.5x108
time/year
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• 3D
• elemental b.c.
• buoyancy driven
ToDo: look at energy conservation. run to steady state and make sure the expected values are
retrieved.
133
27
fieldstone 21: The non-conforming Q1 × P0 element
features
• Non-conforming Q1 × P0 element
• incompressible flow
• mixed formulation
• isothermal
• non-isoviscous
• analytical solution
• pressure smoothing
try Q1 mapping instead of isoparametric.
134
28
fieldstone 22: The stabilised Q1 × Q1 element
The details of the numerical setup are presented in Section ??.
We will here focus on the Q1 × Q1 element, which, unless stabilised, violates the LBB stability
condition and therefore is unusable. Stabilisation can be of two types: least-squares (see for example
[144]) [447, 285, ?], or by means of an additional term in the weak form as first introduced in [141, 56].
It is further analysed in [359, 310]. Note that an equal-order velocity-pressure formulation that does not
exhibit spurious pressure modes (without stabilisaion) has been presented in [400].
This element corresponds to Bilinear velocities, bilinear pressure (equal order interpolation for both
velocity and pressure) which is very convenient in terms of data structures since all dofs are colocated.
In geodynamics, it is used in the Rhea code [430, 96] and in Gale. It is also used in [307] in its
stabilised form, in conjunction with AMR.
The stabilisation term C enters the Stokes matrix in the (2,2) position:
K G
V
f
·
=
P
h
GT C
Let us take a simple example: a 3x2 element grid.
16,17
8
18,19
9
3
8,9
20,21
10
4
5
10,11
5
4
0
12,13
6
1
0,1
14,15
7
2
2,3
0
22,23
11
4,5
1
2
6,7
3
The K matrix is of size N f emV × N f emV with N f emV = ndof V × nnp = 2 × 12 = 24. The G
matrix is of size N f emV × N f emP with N f emP = ndof P × nnp = 1 × 12 = 12. The C matrix is of
size N f emP × N f emP .
A corner pdof sees 4 vdofs, a side pdof sees 12 vdofs and an inside pdof sees 18 vdofs, so that the
total number of nonzeros in G can be computed as follows:
N ZG = |{z}
4 + 2(nnx − 2) ∗ 12 + 2(nny − 2) ∗ 12 + (nnx − 2)(nny − 2) ∗ 18
|
{z
} |
{z
} |
{z
}
corners
2hor.sides
2vert.sides
Concretely,
• pdof #0 sees vdofs 0,1,2,3,8,9,10,11
• pdof #1 sees vdofs 0,1,2,3,4,5,8,9,10,11,12,13
• pdof #5 sees vdofs 0,1,2,3,4,5,8,9,10,11,12,13,16,17,18,19,20,21
so that the GT matrix non-zero structure then is as follows:
135
insidenodes
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
0
1
2
3
4
5
6
7
8
9
10
11
We will here use the formulation of [141], which is very appealing since it does not require any free
(user-chosen) parameter.
Furthermore the matrix C is rather simple to compute as we will see hereafter.
R
We also impose pdV = 0 which means that the following constraint is added to the Stokes matrix:
K
G 0
V
f
GT C L · P = h
λ
0
0 LT 0
28.1
The Donea & Huerta benchmark
As in [144] we solve the benchmark problem presented in section 6.7.1.
0.10000
error
0.01000
velocity
pressure
x2
x1.5
0.00100
0.00010
0.00001
0.1
h
28.2
The Dohrmann & Bochev benchmark
As in [141] we solve the benchmark problem presented in section 6.7.2.
136
1.00000
0.10000
velocity
pressure
x2
x1
error
0.01000
0.00100
0.00010
0.00001
0.1
h
compare my rates with original paper!
28.3
The falling block experiment
The setup is desscribed in [453].
1x1021
|vz| η1/δρ
1x1020
1x1019
1x1018
1x1017
0.0001
0.01
1
100
10000
η2/η1
137
1x106
29
fieldstone 23: compressible flow (1) - analytical benchmark
This work is part of the MSc thesis of T. Weir (2018).
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,
(242)
in Ω
(243)
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 by12
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
∂v
1 ∂u ∂v
1 ∂u 2 ∂v
−
+
=−
+
∂y 3 ∂x ∂y
3 ∂x 3 ∂y
∂u ∂v
+
∂y
∂x
and then
˙ d (v) =
d
From ~τ = 2η~
τxx
τyy = 2η
τxy
we arrive at:
˙dxx
2/3
˙dyy = 2η −1/3
0
˙dxy
2 ∂u
3 ∂x
1 ∂u
2 ∂y
−1/3
2/3
0
−
1 ∂v
3 ∂y
1 ∂u
2 ∂y
+
1 ∂v
2 ∂x
− 13 ∂u
∂x
0
0 ·
1/2
+
1 ∂v
2 ∂x
+
2 ∂v
3 ∂y
∂u
∂x
∂v
∂y
∂u
∂v
∂y + ∂x
(246)
4/3
= η −2/3
0
~τ = Cη BV
the ASPECT manual for a justification of the 3 value in the denominator in 2D and 3D.
138
(245)
or,
12 See
(244)
−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
(247)
We derive a velocity given by:
vx (x, y) =
With gx (x, y) =
1
x
Cx
Cy
, vy (x, y) =
x
y
(248)
and gy (x, y) = y1 , this leads us to a pressure profile:
4Cy
4Cx
+ 2 + xy + C0
p = −η
3x2
3y
(249)
This gives us a strain rate:
˙xx =
−Cx
x2
˙yy =
−Cy
y2
˙xy = 0
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.
• benchmark #2 (ibench=2): Starting from a density profile of:
ρ = cos(x) cos(y)
(250)
We derive a velocity given by:
Cy
Cx
, vy =
cos(x)
cos(y)
1
and gy = cos(x)
, this leads us to a pressure profile:
!
4Cx sin(x) 4Cy sin(y)
+
+ (sin(x) + sin(y)) + C0
p=η
3 cos2 (x)
3 cos2 (y)
vx =
With gx =
1
cos(y)
(251)
(252)
sin(y)
sin(x)
˙yy = Cy
˙xy = 0
2
cos (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)
˙xx = Cx
(thank you WolframAlpha)
• 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
139
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)
140
30
fieldstone 24: compressible flow (2) - convection box
This work is part of the MSc thesis of T. Weir (2018).
30.1
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
in Ω
∇ · v + v · ∇ρ = 0
ρ
∂T
∂p
ρCp
+ v · ∇T − ∇ · k∇T = ρH + 2η ˙ d : ˙ d + αT
+ v · ∇p
∂t
∂t
(253)
(254)
in Ω,
(255)
Note that this presupposes that the density is not zero anywhere in the domain.
30.2
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.ii13 , 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
13 https://www.dealii.org/9.0.0/doxygen/deal.II/step
32.html
141
K
G0T + Z
G0 + W
0
V
f
·
=
P0
h
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 [306]. This however requires more work since the rhs depends on the
solution and some form of iterations is needed.
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
1
V
Z
1
< T >=
V
Z
vrms =
v 2 dV
V
• The average temperature (Tavrg):
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
Z
1
< Φ >= ΦdV =
2η ε̇ : ε̇dV
V
• The gravitational potential energy (EG)
Z
ρgy (Ly − y)dV
EG =
V
142
• 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...!
30.3
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:
πx
πy
Ly − y
− 0.01 cos( ) sin( ) ∆T + Tsurf
T (x, y) =
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:
Ra =
From the second equation we get α =
Ra =
αg∆T L3 ρ20 Cp
ηk
DiCp
gL ,
Di =
αgL
Cp
which we can insert in the first one:
DiCp2 ∆T L2 ρ20
ηk
or,
η=
DiCp2 ∆T L2 ρ20
Ra k
For instance, for Ra = 104 and Di = 0.75, we obtain α ' 3 · 10−5 and η ' 1025 which are quite reasonable
values.
30.4
Scaling
Following [286], 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
kr
ur =
(refvel)
ρr Cp L
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
143
30.5
Conservation of energy 1
30.5.1
under BA and EBA approximations
Following [306], we take the dot product of the momentum equation with the velocity v and integrate
over the whole volume14 :
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 : ε̇dV =
ΦdV
− (∇ · τ ) · vdV = − τ |v{z
· n} dS +
V
V
V
V
S
=0 (b.c.)
which is the volume integral of the shear heating. Then,
Z
Z
Z
dS
−
∇p · vdV =
p v
·
n
| {z }
V
V
S
=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
(256)
V
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
ρ g · vdV =
ΦdV =
V
Z
0
V
(ρ − ρ0 )g · vdV
(257)
V
Obviously Eqs.(256) and (257) 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 is of course inconsistent.
Since the mass conservation equation
R
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.(256) is not verified by
the code, while (257) is.
In the end:
Z
Z
ΦdV =
(ρ − ρ0 )g · vdV
| V {z } | V
{z
}
visc diss
14 Check:
work grav
this is akin to looking at the power, force*velocity, says Arie
144
(258)
30.5.2
under no approximation at all
Z
Z
Z
∇p · vdV
=
V
p v
· n} dS −
| {z
S
=0 (b.c.)
Z
=
V
∇ · v pdV = 0
(259)
V
1
v · ∇ρ pdV = 0
ρ
(260)
(261)
ToDo:see section 3 of [306] where this is carried out with the Adams-Williamson eos.
30.6
Conservation of energy 2
Also, following the Reynold’s transport theorem [331],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
∂T
∂
∂ρ
d
+
ρCp
ρCp T dV =
(ρCp T )dV +
Aρ v
·
n
dS
=
C
T
dV
dV (262)
p
|
{z
}
dt V
∂t
∂t
V
V ∂t
S
V
|
{z
}
|
{z
}
=0 (b.c.)
I
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
∂ρ
Cp T dV
∂t
Z
II =
ρCp
V
∂T
dV
∂t
Z
Z
= −
Cp T ∇ · (ρv)dV = −
V
V
Z
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
V
Z
=
−ρCp v · ∇T
=
V
Z
Cp T ρ v
· n} dS +
| {z
=0 (b.c.)
ρCp ∇T · vdV
(263)
V
∂p
+ ∇ · k∇T + ρH + Φ + αT
+ v · ∇p dV
(264)
∂t
Z
∂p
+ v · ∇p dV +
∇ · k∇T dV
(265)
+ ρH + Φ + αT
∂t
V
Z
∂p
+ ρH + Φ + αT
+ v · ∇p dV +
k∇T · ndS(266)
∂t
S
Z
∂p
+ v · ∇p dV −
q · ndS (267)
+ ρH + Φ + αT
∂t
S
Finally:
d
I + II =
dt
Z
|V
ρCp T dV
{z
}
Z
Z
∂p
=
ρH + Φ + αT
+ v · ∇p dV −
q · ndS
∂t
V
S
(268)
ET
Z
=
V
Z
Z
Z
∂p
ρHdV +
ΦdV +
αT dV +
αT v · ∇pdV −
∂t
| V {z } | V {z
} |V
{z
}
visc diss
extra
adiab heating
Z
q · ndS
(269)
| S {z }
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
0 =
Cp T ∇ · (ρv)dV = −
Cp T ρ v
·
n
dS
+
ρCp ∇T · vdV
(270)
| {z }
V
V
=0 (b.c.)
V
Z
ρCp ∇T · vdV
=
(271)
V
so that the same term in Eq.(267) vanishes too, and then Eq.(269) 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.
145
30.7
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 [461, 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!
features
• Q1 × P0 element
• compressible flow
• mixed formulation
• Dirichlet boundary conditions (no-slip)
• isoviscous
• analytical solution
• pressure smoothing
Relevant literature: [44, 269, 441, 306, 286, 307, 321, 253]
ToDo:
• heat flux is at the moment elemental, so Nusselt and heat flux on boundaries measurements not as
accurate as could be.
146
• implement steady state detection
• do Ra = 105 and Ra = 106
• velocity average at surface
• non dimensional heat flux at corners [55]
• depth-dependent viscosity (case 2 of [55])
147
results - BA - Ra = 104
30.8
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)
90000
-10000
80000
80000
70000
70000
viscous dissipation
-50000
60000
50000
40000
30000
20000
-70000
10000
-80000
0
(g)
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
1x10-8
vrms (cm/yr)
-2x10-8
-5x10-8
5
0
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
time (Myr)
3
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
0.03
0.01
-4x10-8
4.88
0.04
0.02
-3x10-8
10000 20000 30000 40000 50000 60000 70000 80000 90000100000
6
0.05
0
-1x10-8
0
time (Myr)
3x10-8
Nu
adiabatic heating
-40000
work against gravity
90000
-30000
0 100002000030000400005000060000700008000090000
100000
time (Myr)
0
-60000
heat flux
(f)
time (Myr)
-20000
(j)
-0.0015
0 100002000030000400005000060000700008000090000
100000
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.(269) is verified by (l) and Eq.(258) is verified by (m).
148
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
149
results - BA - Ra = 105
30.9
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
2273.150000000
5000
10000 15000 20000 25000 30000 35000
3000
min(u)
max(u)
min(v)
max(v)
0.005
2500
2000
1500
0
(e)
0
-0.005
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)
work against gravity
900000
-300000
(h)
0.8
2x10-6
0.7
vrms (cm/yr)
1x10-6
5x10-7
0
5000
10000
15000
20000
25000
30000
5000
10000
15000
20000
25000
30000
time (Myr)
1x10-5
35000
(k)
15000
20000
25000
30000
35000
10.53
12
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
35000
time (Myr)
VD
WG
800000
4x10-6
0
10000
14
6x10-6
-8x10-6
5000
16
0
HF
dETdt
8x10-6
0
18
0.2
0
300000
time (Myr)
193.2150
0.3
-1x10-6
400000
(i)
0.4
0.1
500000
0
35000
0.5
-5x10-7
600000
100000
0
0.6
1.5x10-6
10000 15000 20000 25000 30000 35000
200000
time (Myr)
2.5x10-6
5000
time (Myr)
900000
viscous dissipation
adiabatic heating
time (Myr)
0.015
min(T)
max(T)
0
0
10000 15000 20000 25000 30000 35000
-0.01
0
time (Myr)
(g)
5000
500
2273.149999999
0
0
1000
2273.149999999
heat flux
(c)
Nu
2273.150000000
2273.150000000
(l)
-7.73687x1023
10000 15000 20000 25000 30000 35000
0.01
2273.150000000
(d)
5000
3500
2273.150000000
2273.149999999
-7.73687x1023
0
time (Myr)
2273.150000001
(j)
-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.(269) is verified by (l) and Eq.(258) is verified by (m).
150
30.10
results - BA - Ra = 106
151
results - EBA - Ra = 104
30.11
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
250000
time (Myr)
2280
4500
2000
(d)
50000
100000
150000
200000
250000
(e)
0
50000
100000
150000
200000
-0.0004
250000
(f)
time (Myr)
10000
10000
9000
9000
8000
8000
-3000
7000
-5000
-6000
-7000
6000
5000
4000
3000
-8000
2000
-9000
1000
-10000
0
(g)
0
50000
100000
150000
200000
250000
time (Myr)
work against gravity
0
-2000
-4000
(h)
50000
100000
150000
200000
250000
250000
5000
4000
3000
0
50000
100000
150000
200000
250000
time (Myr)
2.4
2.2
0.02
5000
2
4000
3000
1.8
0.015
Nu
vrms (cm/yr)
6000
heat flux
200000
6000
0
7000
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
150000
200000
0.8
(l)
0
50000
100000
150000
200000
250000
time (Myr)
10000
VD
WG
9000
8000
-2000
7000
-3000
6000
-4000
5000
-5000
4000
-6000
3000
-7000
-8000
2000
-9000
1000
-10000
250000
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
viscous dissipation
adiabatic heating
time (Myr)
0
0
-0.0003
500
0
min(u)
max(u)
min(v)
max(v)
-0.0002
1000
2140
250000
-0.0001
1500
2160
200000
0.0001
2500
2180
150000
0.0002
velocity
2200
100000
0.0003
3000
Temperature
2220
50000
0.0004
3500
2240
0
time (Myr)
min(T)
max(T)
4000
2260
-7.76x1023
(c)
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.(269) is verified by (l) and Eq.(258) is verified by (m).
152
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
153
results - EBA - Ra = 105
30.12
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
20000
40000
60000
80000
100000
5000
2260
Temperature
2240
2230
-7.75x1023
(c)
0.003
3500
0.002
2000
20000
40000
60000
80000
100000
120000
time (Myr)
-0.003
-0.004
0
20000
40000
60000
80000
100000
-0.005
120000
(f)
time (Myr)
100000
viscous dissipation
120000
100000
-80000
-120000
(g)
80000
60000
40000
20000
0
20000
40000
60000
80000
100000
120000
time (Myr)
0
(h)
20000
40000
60000
80000
100000
120000
80000
100000 120000
60000
40000
0
0
20000
40000
60000
80000
100000
120000
time (Myr)
0.3
6
5.5
0.25
5
50000
4.5
0.2
30000
20000
10000
4
Nu
vrms (cm/yr)
40000
heat flux
60000
80000
(i)
60000
0.15
3.5
3
2.5
0.1
2
0
1.5
0.05
-10000
-20000
40000
20000
0
time (Myr)
70000
20000
time (Myr)
120000
-100000
(j)
0
0
-60000
120000
min(u)
max(u)
min(v)
max(v)
-20000
-40000
100000
0
-0.002
(e)
80000
0.001
1000
500
0
60000
-0.001
1500
0
40000
0.004
2500
2210
20000
0.005
4000
3000
2220
0
time (Myr)
velocity
2250
120000
min(T)
max(T)
4500
2270
-7.748x1023
0
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
1
0
20000
40000
60000
80000
100000
120000
time (Myr)
20000
0
(k)
0
20000
40000
60000
100000
120000
0.5
(l)
0
20000
40000
60000
80000
100000
120000
time (Myr)
120000
0
AH+VD+HF
dETdt
10000
80000
time (Myr)
VD
WG
100000
0
80000
-10000
-20000
60000
-30000
40000
-40000
20000
-50000
-60000
(l)
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)
154
30.13
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
16x16
32x32
48x48
64x64
6x10-15
vrms trend
4x10-15
2x10-15
0
-2x10-15
-4x10-15
1000
Ra
1.5x10-16
16x16
32x32
48x48
64x64
1x10-16
5x10-17
vrms trend
0
-5x10-17
-1x10-16
-1.5x10-16
-2x10-16
-2.5x10-16
-3x10-16
-3.5x10-16
770
775
780
Ra
155
785
790
Appendix: Looking for the right combination of parameters for the King benchmark.
I run a quadruple do loop over L, ∆T , ρ0 and η0 between plausible values (see code targets.py) and
write in a file only the combination which yields the required Rayleigh and Dissipation number values
(down to 1% accuracy).
a l p h a=3e−5
g=10
hcapa =1250
hcond=3
DTmin=1000
Lmin=1e6
rhomin =3000
etamin=19
;
;
;
;
DTmax=4000
Lmax=3e6
rhomax=3500
etamax=25
;
;
;
;
DTnpts=251
Lnpts =251
r h o n p t s =41
e t a n p t s =100
On the following plots the ’winning’ combinations of these four parameters are shown:
0.26
Di
0.255
0.25
0.245
0.24
9600
9800
10000
10200
10400
Ra
1.05x106
1.045x106
1.04x106
1.035x106
L
1.03x106
1.025x106
1.02x106
1.015x106
1.01x106
1.005x106
1x106
500
1000
1500
2000
2500
3000
3500
4000
4500
DT
9x1023
8x1023
7x1023
eta
6x1023
5x1023
4x1023
3x1023
2x1023
1x1023
2800
3000
3200
3400
3600
3800
4000
4200
rho
We see that:
• the parameter L (being to the 3rd power in the Ra number) cannot vary too much. Although it is
varied between 1000 and 3000km there seems to be a ’right’ value at about 1040 km. (why?)
• viscosities are within 1023 and 1024 which are plausible values (although a bit high?).
• densities can be chosen freely between 3000 and 3500
• ∆T seems to be the most problematic value since it can range from 1000 to 4000K ...
156
31
fieldstone 25: Rayleigh-Taylor instability (1) - instantaneous
This numerical experiment was first presented in [478]. It consists of an isothermal Rayleigh-Taylor
instability in a two-dimensional box of size Lx = 0.9142 and Ly = 1. Two Newtonian fluids are present
in the system: the buoyant layer is
at the bottom of the box and the interface between both fluids
placed
πx
The
bottom fluid is parametrised by its mass density ρ1 and its
is given by y(x) = 0.2 + 0.02 cos L
x
viscosity µ1 , while the layer above is parametrised by ρ2 and µ2 .
No-slip boundary conditions are applied at the bottom and at the top of the box while free-slip
boundary conditions are applied on the sides.
In the original benchmark the system is run over 2000 units of dimensionless time and the timing
and position of various upwellings/downwellings is monitored. In this present experiment only the root
mean square velocity is measured at t = 0: the code is indeed not yet foreseen of any algorithm capable
of tracking deformation.
Another approach than the ones presented in the extensive literature which showcases results of this
benchmark is taken. The mesh is initially fitted to the fluids interface and the resolution is progressively
increased. This results in the following figure:
0.01000
vrms
0.00100
0.00010
0.00001
0.01
0.1
hx
The green line indicates results obtained with my code ELEFANT with grids up to 2000x2000 with the
exact same methodology.
157
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• isothermal
• numerical benchmark
158
32
fieldstone 26: Slab detachment benchmark (1) - instantaneous
As in [413], the computational domain is 1000km × 660km. No-slip boundary conditions are imposed
on the sides of the system while free-slip boundary conditions are imposed at the top and bottom. Two
materials are present in the domain: the lithosphere (mat.1) and the mantle (mat.2). The overriding
plate (mat.1) is 80km thick and is placed at the top of the domain. An already subducted slab (mat.1)
of 250km length hangs vertically under this plate. The mantle occupies the rest of the domain.
The mantle has a constant viscosity η0 = 1021 P a.s and a density ρ = 3150kg/m3 . The slab has a
density ρ = 3300kg/m3 and is characterised by a power-law flow law so that its effective viscosity depends
on the second invariant of the strainrate I2 as follows:
ηef f =
1
1 −1/ns 1/ns −1
1/n −1
1/n −1
1/n −1
(272)
I2
= [(2 × 4.75×1011 )−ns ]−1/ns I2 s = 4.75×1011 I2 s = η0 I2 s
A
2
2
with ns = 4 and A = (2 × 4.75×1011 )−ns , or η0 = 4.75 × 1011 .
Fields at convergence for 151x99 grid.
1x1024
1x1022
1x1021
1x1020
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
1x1023
viscosity (Pa.s)
1x1023
viscosity (Pa.s)
1x1024
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
1x1022
1x1021
0
100
200
300
400
500
600
700
800
900
1x1020
400
1000
450
500
x (km)
1x1026
1x1024
normalised nonlinear residual
viscosity (Pa.s)
1.0000000
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
1x1025
1x1023
1x1022
1x1021
1x1020
0
550
100
200
300
400
500
600
0.0100000
0.0010000
0.0001000
0.0000100
0.0000010
0.0000001
700
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
1e-7
0.1000000
0
5
10
15
20
y (km)
4x10-16
0
25
30
35
40
45
y (km)
1x10-15
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
2x10-16
600
x (km)
1.5x10-15
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
8x10-16
6x10-16
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
1x10-15
5x10-16
exy
eyy
exx
4x10-16
-2x10-16
0
2x10-16
-4x10-16
0
-6x10-16
-8x10-16
-5x10-16
-1x10-15
-2x10-16
0
100
200
300
400
500
x (km)
600
700
800
900
1000
-4x10-16
0
100
200
300
400
500
x (km)
159
600
700
800
900
1000
-1.5x10-15
0
100
200
300
400
500
x (km)
600
700
800
900
1000
Along the horizontal line
4x10-16
1x10-15
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
2x10-16
0
6x10-16
4x10-17
exy
eyy
2x10-17
-4x10-16
2x10-16
0
-6x10-16
0
-2x10-17
-8x10-16
-2x10-16
-4x10-17
-1x10-15
-4x10-16
0
100
200
300
400
500
600
700
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
6x10-17
4x10-16
exx
-2x10-16
8x10-17
nelx=32
nelx=48
nelx=51
nelx=64
nelx=100
nelx=128
nelx=151
8x10-16
0
100
200
300
y (km)
400
500
600
y (km)
Along the vertical line
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• isothermal
• nonlinear rheology
• nonlinear residual
Todo: nonlinear mantle, pressure normalisation
160
700
-6x10-17
0
100
200
300
y (km)
400
500
600
700
33
fieldstone 27: Consistent Boundary Flux
In what follows we will be re-doing the numerical experiments presented in Zhong et al. [520].
The first benchmark showcases a unit square domain with free slip boundary conditions prescribed
on all sides. The resolution is fixed to 64 × 64 Q1 × P0 elements. The flow is isoviscous and the buoyancy
force f is given by
fx
=
0
fy
= ρ0 αT (x, y)
with the temperature field given by
T (x, y) = cos(kx)δ(y − y0 )
where k = 2π/λ and λ is a wavelength, and y0 represents the location of the buoyancy strip. We set
gy = −1 and prescribe ρ(x, y) = ρ0 α cos(kx)δ(y − y0 ) on the nodes of the mesh.
One can prove ([520] and refs. therein) that there is an analytic solution for the surface stress σzz 15
cos(kx)
σyy
[k(1 − y0 ) sinh(k) cosh(ky0 ) − k sinh(k(1 − y0 )) + sinh(k) sinh(ky0 )]
=
ραgh
sinh2 (k)
We choose ρ0 α = 64, η = 1 (note that in this case the normalising coefficient of the stress is exactly
1 (since h = Lx /nelx = 1/64) so it is not implemented in the code). λ = 1 is set to 1 and we explore
62 59
y0 = 63
64 , 64 , 64 and y0 = 32/64. Under these assumptions the density field for y0 = 59/64 is:
We can recover the stress at the boundary by computing the yy component of the stress tensor in the
top row of elements:
σyy = −p + 2η ˙yy
Note that pressure is by definition elemental, and that strain rate components are then also computed in
the middle of each element.
These elemental quantities can be projected onto the nodes (see section ??) by means of the C→N
algorithm or a least square algorithm (LS).
y0=32/64
y0=32/64
0.2
0.001
0.15
0.0008
0.0006
0.1
0.0004
σyy error
σyy
0.05
0
-0.05
-0.15
-0.2
0
0.2
0.4
0.6
0.8
elemental
C-N
LS
cbf
0.
-0.0006
-0.0008
-0.001
1
x
15 Note
0
-0.0002
-0.0004
analytical
elemental
C-N
LS
cbf
-0.1
0.0002
0
0.2
0.4
0.6
x
that in the paper the authors use ραg which does not have the dimensions of a stress
161
0.8
1
y0=59/64
y0=59/64
1
0.003
0.8
0.002
0.6
0.4
0.001
σyy error
σyy
0.2
0
-0.2
-0.4
-0.8
-1
-0.001
analytical
elemental
C-N
LS
cbf
-0.6
0
0.2
0.4
0.6
0.8
0
elemental
C-N
LS
cbf
0.
-0.002
-0.003
1
0
0.2
0.4
x
0.6
0.8
1
x
y0=62
y0=62
1.5
0.1
0.08
1
0.06
0.04
σyy error
σyy
0.5
0
-0.5
-1.5
0
0.2
0.4
0.6
0.8
0
-0.02
-0.04
analytical
elemental
C-N
LS
cbf
-1
0.02
elemental
C-N
LS
cbf
0.
-0.06
-0.08
-0.1
1
0
0.2
0.4
x
0.6
0.8
1
x
y0=63/64
y0=63/64
1.5
0.2
0.15
1
0.1
σyy error
σyy
0.5
0
-0.5
analytical
elemental
C-N
LS
cbf
-1
-1.5
0
0.2
0.4
0.6
0.8
0.05
0
-0.05
elemental
C-N
LS
cbf
0.
-0.1
-0.15
-0.2
1
0
0.2
x
0.4
0.6
0.8
1
x
The consistent boundary flux (CBF) method allows us to compute traction vectors t = σ · n on the
boundary of the domain. On the top boundary, n = (0, 1) so that t = (σxy , σyy )T and ty is the quantity
we need to consider and compare to other results.
In the following table are shown the results presented in [520] alongside the results obtained with
Fieldstone:
Method
Analytic solution
Pressure smoothing [520]
CBF [520]
fieldstone: elemental
fieldstone: nodal (C→N)
fieldstone: LS
fieldstone: CBF
y0 = 63/64
0.995476
1.15974
0.994236
0.824554 (-17.17 %)
0.824554 (-17.17 %)
1.165321 ( 17.06 %)
0.994236 ( -0.13 %)
y0 = 62/64
0.983053
1.06498
0.982116
0.978744 (-0.44%)
0.978744 (-0.44%)
1.070105 ( 8.86%)
0.982116 (-0.10%)
y0 = 59/64 16
0.912506
0.911109
0.912157
0.909574 (-0.32
0.909574 (-0.32
0.915496 ( 0.33
0.912157 (-0.04
%)
%)
%)
%)
y0 = 32/64
0.178136
n.a.
n.a.
0.177771 (-0.20
0.177771 (-0.20
0.178182 ( 0.03
0.177998 (-0.08
We see that we recover the published results with the same exact accuracy, thereby validating our
implmentation.
On the following figures are shown the velocity, pressure and traction fields for two cases y0 = 32/64
and y0 = 63/64.
162
%)
%)
%)
%)
Here lies the superiority of our approach over the one presented in the original article: our code
computes all traction vectors on all boundaries at once.
explain how Medge is arrived at!
compare with ASPECT ??!
gauss-lobatto integration?
pressure average on surface instead of volume ?
163
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• isothermal
• isoviscous
• analytical solution
• pressure smoothing
• consistent boundary flux
164
34
fieldstone 28: convection 2D box - Tosi et al, 2015
This fieldstone was developed in collaboration with Rens Elbertsen.
The viscosity field µ is calculated as the harmonic average between a linear part µlin that depends
on temperature only or on temperature and depth d , and a non-linear, plastic part µplast dependent on
the strain rate:
˙ =2
µ(T, z, )
1
1
+
˙
µlin (T, z) µplast ()
−1
.
(273)
The linear part is given by the linearized Arrhenius law (the so-called Frank-Kamenetskii approximation [?]):
µlin (T, z) = exp(−γT T + γz z),
(274)
where γT = ln(∆µT ) and γz = ln(∆µz ) are parameters controlling the total viscosity contrast due to
temperature (∆µT ) and pressure (∆µz ). The non-linear part is given by [?]:
σY
˙ = µ∗ + √
µplast ()
,
˙ : ˙
(275)
where µ∗ is a constant representing the effective viscosity at high stresses [?] and σY is the yield
stress, also assumed to be constant. In 2-D, the denominator in the second term of equation (275) is
given explicitly by
s
2
2
2
√
p
∂ux
∂uy
1 ∂ux
∂uy
˙ : ˙ = ˙ij ˙ij =
+
+
+
.
(276)
∂x
2 ∂y
∂x
∂y
The viscoplastic flow law (equation 273) leads to linear viscous deformation at low stresses (equation
(274)) and to plastic deformation for stresses that exceed σY (equation (275)), with the decrease in
viscosity limited by the choice of µ∗ [?].
In all cases that we present, the domain is a two-dimensional square box. The mechanical boundary
conditions are for all boundaries free-slip with no flux across, i.e. τxy = τyx = 0 and u · n = 0, where
n denotes the outward normal to the boundary. Concerning the energy equation, the bottom and top
boundaries are isothermal, with the temperature T set to 1 and 0, respectively, while side-walls are
assumed to be insulating, i.e. ∂T /∂x = 0. The initial distribution of the temperature field is prescribed
as follows:
T (x, y) = (1 − y) + A cos(πx) sin(πy),
(277)
where A = 0.01 is the amplitude of the initial perturbation.
In the following Table ??, we list the benchmark cases according to the parameters used.
Case
1
2
3
4
5a
5b
Ra
102
102
102
102
102
102
∆µT
105
105
105
105
105
105
∆µy
1
1
10
10
10
10
µ∗
–
10−3
–
10−3
10−3
10−3
σY
–
1
–
1
4
3–5
Convective regime
Stagnant lid
Mobile lid
Stagnant lid
Mobile lid
Periodic
Mobile lid – Periodic – Stagnant lid
Benchmark cases and corresponding parameters.
In Cases 1 and 3 the viscosity is directly calculated from equation (274), while for Cases 2, 4, 5a, and
5b, we used equation (273). For a given mesh resolution, Case 5b requires running simulations with yield
stress varying between 3 and 5
In all tests, the reference Rayleigh number is set at the surface (y = 1) to 102 , and the viscosity
contrast due to temperature ∆µT is 105 , implying therefore a maximum effective Rayleigh number of
107 for T = 1. Cases 3, 4, 5a, and 5b employ in addition a depth-dependent rheology with viscosity
contrast ∆µz = 10. Cases 1 and 3 assume a linear viscous rheology that leads to a stagnant lid regime.
165
Cases 2 and 4 assume a viscoplastic rheology that leads instead to a mobile lid regime. Case 5a also
assumes a viscoplastic rheology but a higher yield stress, which ultimately causes the emergence of a
strictly periodic regime. The setup of Case 5b is identical to that of Case 5a but the test consists in
running several simulations using different yield stresses. Specifically, we varied σY between 3 and 5 in
increments of 0.1 in order to identify the values of the yield stress corresponding to the transition from
mobile to periodic and from periodic to stagnant lid regime.
34.0.1
Case 0: Newtonian case, a la Blankenbach et al., 1989
90
80
7
7
6
6
5
5
4
4
50
Nu
vrms
60
Nu
70
40
3
3
30
2
2
20
0
1
1
10
0
0.05
0.1
0.15
0.2
time
0.25
0.3
0.35
0
0
0
0.05
0.1
0.15
0.2
time
166
0.25
0.3
0.35
0
10
20
30
40
50
vrms
60
70
80
90
34.0.2
Case 1
In this case µ? = 0 and σY = 0 so that µplast can be discarded. The CFL number is set to 0.5 and the
˙ = µlin (T, z). And since ∆µz = 1 then γz = 0 so that µlin (T, z) = exp(−γT T )
viscosity is given by µ(T, z, )
350
3.5
300
3
2.5
200
Nu
vrms
250
2
150
1.5
100
ELEFANT, 32x32
ELEFANT, 48x48
ELEFANT, 64x64
ELEFANT, 100x100
243.872
50
0
0.001
0.01
0.1
1
0.5
1
3.3987
0.001
0.01
0.1
3.5
0.8
3
0.75
2.5
0.7
2
0.65
1.5
0.6
1
0.55
0.5
0
50
100
150
200
250
300
350
0.5
vrms
0.77368
0.001
0.01
1
1
1
1
0.9
0.8
0.8
0.8
0.7
0.7
0.6
0.6
y
0.5
0.4
y
0.6
y
0.1
time
0.9
0.4
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.1
0
1
time
Nu
time
0.1
0
0.1
0.2
0.3
0.4
0.5
T
0.6
0.7
0.8
0.9
1
0
0.00010
0.00100
0.01000
viscosity
167
0.10000
0
0
50
100
150
200
velocity
250
300
350
400
168
34.0.3
Case 2
600
14
12
500
10
8
Nu
vrms
400
300
6
200
4
100
2
141.518
0
0.001
0.01
0.1
0
1
8.17154
0.001
0.01
0.1
time
1
time
14
0.62
12
0.6
10
0.58
Nu
8
6
0.56
4
0.54
2
0.52
0
0
100
200
300
400
500
600
0.5
vrms
0.60521
0.001
0.01
1
1
1
0.9
0.8
0.8
0.8
0.7
0.7
0.6
0.6
y
0.5
0.4
y
0.6
y
1
time
0.9
0.4
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.1
0
0.1
0.1
0
0.1
0.2
0.3
0.4
0.5
T
0.6
0.7
0.8
0.9
1
0
0.00010
0.00100
0.01000
viscosity
169
0.10000
0
0
50
100
150
200
velocity
250
300
350
400
170
34.0.4
Case 3
110
3.5
100
3
90
80
2.5
60
Nu
vrms
70
50
2
40
1.5
30
20
1
10
100.018
0
0.001
0.01
0.1
0.5
1
3.03472
0.001
0.01
0.1
time
1
time
3.5
0.75
3
0.7
2.5
Nu
0.65
2
0.6
1.5
1
0.55
0.5
0
10
20
30
40
50
60
70
80
90
100
110
0.5
vrms
0.7277
0.001
0.01
1
1
1
0.9
0.8
0.8
0.8
0.7
0.7
0.6
0.6
y
0.5
0.4
y
0.6
y
1
time
0.9
0.4
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.1
0
0.1
0.1
0
0.1
0.2
0.3
0.4
0.5
T
0.6
0.7
0.8
0.9
1
0
0.00010
0.00100
0.01000
viscosity
171
0.10000
0
0
50
100
150
200
velocity
250
300
350
400
172
34.0.5
Case 4
500
16
450
14
400
12
350
10
Nu
vrms
300
250
8
200
6
150
4
100
2
50
79.4746
0
0.001
0.01
0.1
6.40359
0
1
0.001
0.01
0.1
time
1
time
16
0.53
14
0.525
12
0.52
0.515
Nu
10
8
6
0.51
4
0.505
2
0
0.5
0
50
100
150
200
250
300
350
400
450
500
0.495
vrms
0.528634
0.001
0.01
1
1
1
0.9
0.8
0.8
0.8
0.7
0.7
0.6
0.6
y
0.5
0.4
y
0.6
y
1
time
0.9
0.4
0.5
0.4
0.3
0.3
0.2
0.2
0.2
0.1
0
0.1
0.1
0
0.1
0.2
0.3
0.4
0.5
T
0.6
0.7
0.8
0.9
1
0
0.00010
0.00100
0.01000
viscosity
173
0.10000
0
0
50
100
150
200
velocity
250
300
350
400
174
34.0.6
Case 5
8
100
7
6
80
5
Nu
vrms
60
4
3
40
2
20
0
1
41.61
98.183
0
0.5
1
1.5
2
2.5
3
3.5
4
0
4.5
2.6912
7.0763
0
0.5
1
1.5
2
2.5
3
3.5
8
0.68
7
0.66
6
0.64
5
0.62
4
3
4.5
0.6
0.58
0.56
2
0.54
1
0.52
0
4
time
Nu
time
0
10
20
30
40
50
60
70
80
90
100
0.5
vrms
0.66971
0.65206
0
1
2
3
4
5
6
7
time
200
300
150
250
200
100
150
50
v
u
100
0
50
-50
0
-100
-50
-150
-200
-100
min(u)
max(u)
0
1
2
3
4
5
6
-150
7
time
min(v)
max(v)
0
1
2
3
4
time
175
5
6
7
176
35
fieldstone 29: open boundary conditions
In what follows we will investigate the use of the so-called open boundary conditions in the very simple
context of a 2D Stokes sphere experiment.
We start with a domain without the sphere. Essentially, it is what people would call an aquarium.
Free slip boundary conditions are prescribed on the sides and no-slip conditions at the bottom. The top
surface is left free. The fluid has a density ρ0 = 1 and a viscosity η0 = 1. In the absence of any density
difference in the domain there is no active buoyancy force so that we expect a zero velocity field and a
lithostatic pressure field. This is indeed what we recover:
If we now implement a sphere parametrised by its density ρs = ρ0 + 1, its viscosity ηs = 103 and
its radius Rs = 0.123 in the middle of the domain, we see clear velocity field which logically shows the
sphere falling downward and a symmetric return flow of the fluid on each side:
Unfortunately it has been widely documented that the presence of free-slip boundary conditions
affects the evolution of subduction [109], even when these are placed rather far from the subduction zone.
A proposed solution to this problem is the use of ’open boundary conditions’ which are in fact stress
boundary conditions. The main idea is to prescribe a stress on the lateral boundaries (instead of free
slip) so that it balances out exactly the existing lithostatic pressure inside the domain along the side
walls. Only pressure deviations with respect to the lithostatic are responsible for flow and such boundary
conditions allow flow across the boundaries.
We need the lithostatic pressure and compute it before hand (which is trivial in our case but can
prove to be a bit more tedious in real life situations when for instance density varies in the domain as a
function of temperature and/or pressure).
p l i t h = np . z e r o s ( nnp , dtype=np . f l o a t 6 4 )
f o r i i n r a n g e ( 0 , nnp ) :
p l i t h [ i ]=( Ly−y [ i ] ) ∗ rho0 ∗ abs ( gy )
Let us start with a somewhat pathological case: even in the absence of the sphere, what happens when
no boundary conditions are prescribed on the sides? The answer is simple: think about an aquarium
without side walls, or a broken dam. The velocity field indeed shows a complete collapse of the fluid left
and right of the bottom.
177
Let us then continue (still with no sphere) but let us now switch on the open boundary conditions.
Since the side boundary conditions match the lithostatic pressure we expect no flow at all in the absence
of any density perturbation in the system. This is indeed what is recovered:
Finally, let us reintroduce the sphere. This time flow is allowed through the left and right side
boundaries:
Finally, although horizontal velocity Dirichlet boundary conditions and open boundary conditions
are not compatible, the same is not true for the vertical component of the velocity: the open b.c.
implementation acts on the horizontal velocity dofs only, so that one can fix the vertical component to
zero, as is shown hereunder:
We indeed see that the in/outflow on the sides is perpendicular to the boundaries.
Turning now to the actual implementation, we see that it is quite trivial, since all element edges are
vertical, and all have the same vertical dimension hx . Since we use a Q0 approximation for the pressure
178
we need to prescribe a single pressure value in the middle of the element. Finally because of the sign of
the normal vector projection onto the x-axis, we obtain:
if
o p e n b c l e f t and x [ i c o n [ 0 , i e l ]] < e p s : # l e f t s i d e
pmid =0 .5∗( p l i t h [ i c o n [ 0 , i e l ] ] + p l i t h [ i c o n [ 3 , i e l ] ] )
f e l [ 0 ] + = 0 . 5 ∗ hy∗pmid
f e l [ 6 ] + = 0 . 5 ∗ hy∗pmid
i f o p e n b c r i g h t and x [ i c o n [ 1 , i e l ]] > Lx−e p s : # r i g h t s i d e
pmid =0 .5∗( p l i t h [ i c o n [ 1 , i e l ] ] + p l i t h [ i c o n [ 2 , i e l ] ] )
f e l [2] −=0.5∗ hy∗pmid
f e l [4] −=0.5∗ hy∗pmid
These few lines of code are added after the elemental matrices and rhs are built, and before the
application of other Dirichlet boundary conditions, and assembly.
features
• Q1 × P0 element
• incompressible flow
• mixed formulation
• open boundary conditions
• isoviscous
179
36
fieldstone 30: conservative velocity interpolation
In this the Stokes equations are not solved. It is a 2D implementation of the cvi algorithm as introduced
in [490] which deals with the advection of markers. Q1 and Q2 basis functions are used and in both cases
the cvi algorithm can be toggled on/off. Markers can be distributed regularly or randomly at startup.
Three velocity fields are prescribed on the mesh:
• the so-called Couette flow of [490]
• the SolCx solution
• a flow created by means of a stream line function (see fieldstone 32)
36.1
Couette flow
36.2
SolCx
36.3
Streamline flow
6
2.5
Q1, no cvi, RK1
Q1, cvi, RK1
Q2, no cvi, RK1
Q2, cvi, RK1
Q1, no cvi, RK1, **
5
Q1, no
Q1 ,
Q1, no
Q1, no
Q2, no
Q2, no
2
nmarker
nmarker
4
3
cvi,
cvi,
cvi,
cvi,
cvi,
cvi,
RK2
RK2
RK4
RK5
RK2
RK4
1.5
1
2
0.5
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0
time
0.2
0.4
0.6
time
In this case RK order seems to be more important that cvi.
Explore why ?!
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions (free-slip)
• direct solver
• isothermal
• non-isoviscous
• analytical solution
180
0.8
1
37
fieldstone 31: conservative velocity interpolation 3D
181
38
38.1
fieldstone 32: 2D analytical sol. from stream function
Background theory
The stream function is a function of coordinates and time of an inviscid liquid. It allows to determine
the components of velocity by differentiating the stream function with respect to the space coordinates.
A family of curves Ψ = const represent streamlines, i.e. the stream function remains constant along a
streamline. Although also valid in 3D, this approach is mostly used in 2D because of its relative simplicity
REFERENCES.
In two dimensions the velocity is obtained as follows:
∂Ψ ∂Ψ
,−
(278)
v=
∂y
∂x
Provided the function Ψ is a smooth enough function, this automatically insures that the flow is incompressible:
∂u ∂v
∂2Ψ ∂2Ψ
∇·v =
+
=
−
=0
(279)
∂x ∂y
∂xy
∂xy
Assuming constant viscosity, the Stokes equation writes:
− ∇p + µ∆v = ρg
(280)
Let us introduce the vector W for convenience such that in each dimension:
2
∂ u ∂2u
∂p
+µ
+
= ρgx
Wx = −
∂x
∂x2
∂xy
2
∂p
∂ v
∂2v
Wy = −
+µ
+
= ρgy
∂y
∂x2
∂xy
(281)
(282)
Taking the curl of the vector W and only considering the component perpendicular to the xy-plane:
∂Vx
∂ρgy
∂ρgx
∂Vy
−
=
−
∂x
∂y
∂x
∂y
(283)
The advantage of this approach is that the pressure terms cancel out (the curl of a gradient is always
zero), so that:
2
2
∂
∂ v
∂2v
∂ u ∂2u
∂ρgx
∂
∂ρgy
µ ( 2+
µ
+
−
(284)
−
=
∂x
∂x
∂xy
∂y
∂x2
∂xy
∂x
∂y
and then replacing u, v by the their stream function derivatives yields (for a constant viscosity):
4
∂ Ψ ∂4Ψ
∂ρgx
∂4Ψ
∂ρgy
+
+
2
=
µ
−
∂x4
∂y 4
∂x2 y 2
∂x
∂y
(285)
or,
4
∇ Ψ=
∂2
∂2
+
∂x2
∂y 2
∂2
∂2
+
∂x2
∂y 2
Ψ=
∂ρgy
∂ρgx
−
∂x
∂y
(286)
These equations are also to be found in the geodynamics literature, eee Eq. 1.43 of Tackley book, p 70-71
of Gerya book.
38.2
A simple application
I wish to arrive at an analytical formulation for a 2D incompressible flow in the square domain [−1 :
1] × [−1 : 1] The fluid has constant viscosity µ = 1 and is subject to free slip boundary conditions on all
sides. For reasons that will become clear in what follows I postulate the following stream function:
Ψ(x, y) = sin(mπx) sin(nπy)
182
(287)
We have the velocity being defined as:
∂Ψ ∂Ψ
v = (u, v) =
,−
= (nπ sin(mπx) cos(nπy), −mπ cos(mπx) sin(nπy))
∂y
∂x
(288)
The strain rate components are then:
ε̇xx
=
ε̇yy
=
2ε̇xy
=
=
=
=
∂u
= mnπ 2 cos(mπx) cos(nπy)
∂x
∂v
= −mnπ 2 cos(mπx) cos(nπy)
∂y
∂u ∂v
+
∂y
∂x
2
∂ Ψ ∂2Ψ
−
∂y 2
∂x2
2 2
−n π Ψ + m2 π 2 Ψ
(m2 − n2 )π 2 sin(mπx) sin(nπy)
(289)
(290)
(291)
(292)
(293)
(294)
Note that if m = n the last term is identically zero, which is not desirable (flow is too ’simple’) so in
what follows we will prefer m 6= n.
It is also easy to verify that u = 0 on the sides and v = 0 at the top and bottom and that the term
ε̇xy is nul on all four sides, thereby garanteeing free slip.
Our choice of stream function yields:
∇4 Ψ =
∂4Ψ ∂4Ψ
∂2Ψ
+
+ 2 2 2 = π 4 (m4 Ψ + n4 Ψ + 2m2 n2 Ψ) = (m4 + n4 + 2m2 n2 )π 4 Ψ
4
4
∂x
∂y
∂x y
We assume gx = 0 and gy = −1 so that we simply have
(m4 + n4 + 2m2 n2 )π 4 Ψ = −
∂ρ
∂x
(295)
so that (assuming the integration constant to be zero):
ρ(x, y) =
m4 + n4 + 2m2 n2 3
π cos(mπx) sin(nπy)
m
The x-component of the momentum equation is
−
∂p ∂ 2 u ∂ 2 u
∂p
+
+ 2 =−
− m2 nπ 3 sin(mπx) cos(nπy) − n3 π 3 sin(mπx) cos(nπy) = 0
2
∂x ∂x
∂y
∂x
so
∂p
= −(m2 n + n3 )π 3 sin(mπx) cos(nπy)
∂x
and the pressure field is then (once again neglecting the integration constant):
p(x, y) =
m 2 n + n3 2
π cos(πx) cos(πy)
m
R
Note that in this case pdV = 0 so that volume normalisation of the pressure is turned on (when free
slip boundary conditions are prescribed on all sides the pressure is known up to a constant and this
undeterminacy can be lifted by adding an additional constraint to the pressure field).
183
Top to bottom: Velocity components u and v, pressure p, density ρ and strain rate ε̇xy . From left to right:
(m, n) = (1, 1), (m, n) = (1, 2), (m, n) = (2, 1), (m, n) = (2, 2)
184
100
10
error
1
0.1
velocity
pressure
x2
x1
0.01
0.001
0.01
0.1
h
Errors for velocity and pressure for (m, n) = (1, 1), (1, 2), (2, 1), (2, 2)
Velocity arrows for (m, n) = (2, 1)
185
39
fieldstone 33: Convection in an annulus
This fieldstone was developed in collaboration with Rens Elbertsen.
This is based on the community benchmark for viscoplastic thermal convection in a 2D square box
[458] as already carried out in ??.
In this experiment the geometry is an annulus of inner radius R1 = 1.22 and outer radius R2 = 2.22.
The rheology and buoyancy forces are identical to those of the box experiment. The initial temperature
is now given by:
R2 − r
∈ [0, 1]
T (r, θ) = Tc (r) + A s(1 − s) cos(N0 θ)
s=
R2 − R1
where s in the normalised depth, A is the amplitude of the perturbation and N0 the number of lobes. In
this equation Tc (r) stands for the steady state purely conductive temperature solution which is obtained
by solving the Laplace’s equation in polar coordinates (all terms in θ are dropped because of radial
symmetry) supplemented with two boundary conditions:
∂T
1 ∂
r
=0
T (r = R1 ) = T1 = 1
T (r = R2 ) = T2 = 0
∆Tc =
r ∂r
∂r
We obtain
Tc (r) =
log(r/R2 )
log(R1 /R2 )
Note that this profile differs from the straight line that is used in [458] and in section 34.
Examples of initial temperature fields for N0 = 3, 5, 11
Boundary conditions can be either no-slip or free-slip on both inner and outer boundary. However,
when free-slip is used on both a velocity null space exists and must be filtered out. In other words, the
solver may be able to come up with a solution to the Stokes operator, but that solution plus an arbitrary
rotation is also an equally valid solution. This additional velocity field can be problematic since it is used
for advecting temperature (and/or compositions) and it also essentially determines the time step value
for a chosen mesh size (CFL condition).
For these reasons the nullspace must be removed from the obtained solution after every timestep.
There are two types of nullspace removal: removing net angular momentum, and removing net rotations.
We calculate the following output parameters:
• the average temperature < T >
R
Z
T dΩ
1
Ω
R
=
T dΩ
hT i =
VΩ Ω
dΩ
Ω
(296)
• the root mean square velocity vrms as given by equation (27).
• the root mean square of the radial and tangential velocity components as given by equations (29)
and (30).
186
• the heat transfer through both boundaries Q:
Z
q · n dΓ
Qinner,outer =
(297)
Γi,o
• the Nusselt number at both boundaries N u as given by equations (32) and (33).
• the power spectrum of the temperature field:
Z
P Sn (T ) =
Ω
features
• Q1 × P0 element
• incompressible flow
• penalty formulation
• Dirichlet boundary conditions
• non-isothermal
• non-isoviscous
• annulus geometry
187
2
T (r, θ)einθ dΩ .
(298)
40
fieldstone 34: the Cartesian geometry elastic aquarium
This fieldstone was developed in collaboration with Lukas van de Wiel.
The setup is as follows: a 2D square of elastic material of size L is subjected to the following boundary
conditions: free slip on the sides, no slip at the bottom and free at the top. It has a density ρ and is
placed is a gravity field g = −gey . For an isotropic elastic medium the stress tensor is given by:
σ = λ(∇ · u)1 + 2µε
where λ is the Lamé parameter and µ is the shear modulus. The displacement field is u = (0, uy (y))
because of symmetry reasons (we do not expect any of the dynamic quantities to depend on the x
coordinate and also expect the horizontal displacement to be exactly zero). The velocity divergence is
then ∇ · u = ∂uy /∂y and the strain tensor:
!
0
0
ε=
∂u
0 ∂yy
so that the stress tensor is:
σ=
∇ · σ = (∂x
∂y ) ·
λ
∂uy
∂y
0
λ
∂uy
∂y
0
0
∂u
(λ + 2µ) ∂yy
!
0
∂u
(λ + 2µ) ∂yy
=
!
0
∂2u
(λ + 2µ) ∂y2y
!
=
0
ρg
so that the vertical displacement is then given by:
uy (y) =
1 ρg
y 2 + αy + β
2 λ + 2µ
where α and β are two integration constants. We need now to use the two boundary conditions: the
first one states that the displacement is zero at the bottom, i.e. uy (y = 0) = 0 which immediately
implies β = 0. The second states that the stress at the top is zero (free surface), which implies that
∂uy /∂y(y = L) = 0 which allows us to compute α. Finally:
uy (y) =
ρg
y2
( − Ly)
λ + 2µ 2
The pressure is given by
λ + 32 µ
1 + 2µ
2
2
ρg
3λ
p = −(λ + µ)∇ · u = (λ + µ)
(L − y) =
ρg(L − y) =
ρg(L − y)
3
3 λ + 2µ
λ + 2µ
1 + 2µ/λ
In the incompressible limit, the poisson ratio is ν ∼ 0.5. Materials are characterised by a finite Young’s
modulus E, which is related to ν and λ:
λ=
Eν
(1 + ν)(1 − 2ν)
µ=
E
2(1 + ν)
It is then clear that for incompressible parameters λ becomes infinite while µ remains finite. In that case
the pressure then logically converges to the well known formula:
p = ρg(L − y)
188
In what follows we set L = 1000m, ρ = 2800, ν = 0.25, E = 6 · 1010 , g = 9.81.
velocity
pressure
x2
x1
100000000.000000
1000000.000000
error
10000.000000
100.000000
1.000000
0.010000
0.000100
0.000001
0.01
0.1
h
189
41
fieldstone 35: 2D analytical sol. in annulus from stream
function
We seek an exact solution to the incompressible Stokes equations for an isoviscous, isothermal fluid in
an annulus.Given the geometry of the problem, we work in polar coordinates. We denote the orthonormal
basis vectors by er and eθ , the inner radius of the annulus by R1 and the outer radius by R2 . Further,
we assume that the viscosity µ is constant, which we set to µ = 1 we set the gravity vector to g = −gr er
with gr = 1.
Given these assumptions, the incompressible Stokes equations in the annulus are [416]
1 ∂vr
vr
2 ∂uθ
1 ∂ 2 vr
∂p
∂ 2 vr
+
− 2− 2
+
−
∂r2
r ∂r
r2 ∂θ2
r
r ∂θ
∂r
1 ∂vθ
2 ∂vr
1 ∂p
∂ 2 vθ
1 ∂ 2 vθ
vθ
+
+ 2
Aθ =
+ 2
− 2−
∂r2
r ∂r
r ∂θ2
r ∂θ
r
r ∂θ
1 ∂(rvr ) 1 ∂vθ
+
r ∂r
r ∂θ
Ar =
= ρgr
(299)
=
0
(300)
=
0
(301)
Equations (299) and (300) are the momentum equations in polar coordinates while Equation (301) is the
incompressibility constraint. The components of the velocity are obtained from the stream function as
follows:
1 ∂Ψ
∂Ψ
vr =
vθ = −
r ∂θ
∂r
where vr is the radial component and vθ is the tangential component of the velocity vector.
The stream function is defined for incompressible (divergence-free) flows in 2D (as well as in 3D with
axisymmetry). The stream function can be used to plot streamlines, which represent the trajectories of
particles in a steady flow. From calculus it is known that the gradient vector ∇Ψ is normal to the curve
Ψ = C. It can be shown that everywhere u · ∇Ψ = 0 using the formula for u in terms of Ψ which proves
that level curves of Ψ are streamlines:
u · ∇Ψ = vr
1 ∂Ψ ∂Ψ ∂Ψ 1 ∂Ψ
∂Ψ
1 ∂Ψ
+ vθ
=
−
=0
∂r
r ∂θ
r ∂θ ∂r
∂r r ∂θ
In polar coordinates the curl of a vector A is:
1 ∂(rAθ ) ∂Ar
∇×A=
−
r
∂r
∂θ
Taking the curl of vector A yields:
1
r
∂(rAθ ) ∂Ar
−
∂r
∂θ
=
1
r
−
∂(ρgr )
∂θ
Multiplying on each side by r
∂ρgr
∂(rAθ ) ∂Ar
−
=−
∂r
∂θ
∂θ
If we now replace Ar and Aθ by their expressions as a function of velocity and pressure, we see that the
pressure terms cancel out and assuming the viscosity and the gravity vector to be constant we get: Let
us assume the following separation of variables Ψ(r, θ) = φ(r)ξ(θ) . Then
vr =
1 ∂Ψ
φξ 0
=
r ∂θ
r
vθ = −
190
∂Ψ
= −φ0 ξ
∂r
Let us first express Ar and Aθ as functions of Ψ and
Ar
=
=
=
=
∂Ar
∂θ
=
1 ∂vr
vr
2 ∂uθ
∂ 2 vr
1 ∂ 2 vr
+
− 2− 2
+
∂r2
r ∂r
r2 ∂θ2
r
r ∂θ
∂ 2 φξ 0
1 ∂ φξ 0
1 ∂ 2 φξ 0
1 φξ 0
2 ∂
(
(
(
)
+
(
)
+
)
−
) − 2 (−φ0 ξ)
2
2
2
2
∂r
r
r ∂r r
r ∂θ
r
r
r
r ∂θ
φ 0
φ 0
2
φ00
φ0
φ0
φ 000 φξ 0
(
− 2 2 + 2 3 )ξ + ( 2 − 3 )ξ + 3 ξ − 3 + 2 φ0 ξ 0
r
r
r
r
r
r
r
r
φξ 000
φ00 ξ 0
φ0 ξ 0
+ 2 + 3
r
r
r
φξ 0000
φ00 ξ 00
φ0 ξ 00
+ 2 + 3
r
r
r
(302)
(303)
(304)
(305)
(306)
(307)
2
Aθ
∂(rAθ )
∂r
2
∂ vθ
1 ∂vθ
2 ∂vr
1 ∂ vθ
vθ
+
+ 2
+ 2
− 2
∂r2
r ∂r
r ∂θ2
r ∂θ
r
2
1
∂
2 ∂ φξ 0
∂2
∂
1
1
(−φ0 ξ) +
=
(−φ0 ξ) + 2 2 (−φ0 ξ) + 2 (
) − 2 (−φ0 ξ)
2
∂r
r ∂r
r ∂θ
r ∂θ r
r
φ00 ξ
2φξ 00
φ0 ξ
φ0 ξ 00
000
= −φ ξ −
− 2 + 2 + 2
r
r
r
r
=
=
(308)
(309)
(310)
(311)
191
WRONG:
∂(r∆v)
∂r
=
=
=
=
=
∂∆u
∂θ
=
=
=
=
∂
∂
∂v
1 ∂2v
r
+
∂r ∂r
∂r
r ∂θ2
∂2
∂v
∂ 1 ∂2v
r
+
∂r2
∂r
∂r r ∂θ2
∂2
∂
∂Ψ
∂Ψ
∂ 1 ∂2
r
(−
(−
)
+
)
∂r2
∂r
∂r
∂r r ∂θ2
∂r
2
2
3
∂ Ψ
∂ 1 ∂ Ψ
∂
− 2 r 2 −
∂r
∂r
∂r r ∂θ2 ∂r
∂4Ψ
1 ∂3Ψ
∂3Ψ
1 ∂4Ψ
−2 3 − r 4 + 2 2 −
2
2
∂r
∂r r ∂θ 2∂r r ∂θ ∂r
∂ 1 ∂
∂u
1 ∂ u
r
+ 2 2
∂θ r ∂r
∂r
r ∂θ
∂ 1 ∂
∂u
1 ∂3u
r
+ 2 3
∂θ r ∂r
∂r
r ∂θ
∂ 1 ∂
∂ 1 ∂Ψ
1 ∂ 3 1 ∂Ψ
r (
)
+ 2 3(
)
∂θ r ∂r
∂r r ∂θ
r ∂θ r ∂θ
1 ∂3Ψ
1 ∂4Ψ
1 ∂4Ψ
1 ∂2Ψ
− 2
+
+ 3
3
2
2
2
2
r ∂θ
r ∂r∂θ
r ∂r ∂θ
r ∂θ4
(312)
(313)
(314)
(315)
(316)
(317)
(318)
(319)
(320)
Assuming the following separation of variables Ψ(r, θ) = φ(r)ξ(θ) :
∂(r∆v)
∂r
∂∆u
∂θ
=
=
1 0 00
φξ −
r2
1 00
1
1
φξ − 2 φ0 ξ 00 + φ00 ξ 00 +
3
r
r
r
−2φ000 ξ − rφ0000 ξ +
1 00 00
φ ξ
r
1 0000
φξ
r3
(321)
(322)
so that
∂(r∆v) ∂∆u
1
1
1
1
1
1
−
= −2φ000 ξ − rφ0000 ξ + 2 φ0 ξ 00 − φ00 ξ 00 − 3 φξ 00 + 2 φ0 ξ 00 − φ00 ξ 00 − 3 φξ 0000
∂r
∂θ
r
r
r
r
r
r
Further assuming ξ(θ) = cos(kθ) , then ξ 00 = −k 2 ξ and ξ 0000 = k 4 ξ then
∂(r∆v) ∂∆u
1
1
1
1
1
1
−
= −2φ000 ξ − rφ0000 ξ − k 2 2 φ0 ξ + k 2 φ00 ξ + k 2 3 φξ − k 2 2 φ0 ξ + k 2 φ00 ξ − k 4 3 φξ
∂r
∂θ
r
r
r
r
r
r
By choosing ρ such that ρ = λ(r)Υ(θ) and such that ∂θ Υ = ξ = cos(kθ) then we have
−2φ000 ξ − rφ0000 ξ − k 2
1 0
1
1
1
1
1
1
φ ξ + k 2 φ00 ξ + k 2 3 φξ − k 2 2 φ0 ξ + k 2 φ00 ξ − k 4 3 φξ = − λξgr
r2
r
r
r
r
r
η
and then dividing by ξ: (IS THIS OK ?)
−2φ000 − rφ0000 − k 2
1 0
1
1
1
1
1
1
φ + k 2 φ00 + k 2 3 φ − k 2 2 φ0 + k 2 φ00 − k 4 3 φ = − λgr
2
r
r
r
r
r
r
η
−2φ000 − rφ0000 − 2k 2
1 0
1
1
1
φ + 2k 2 φ00 + (k 2 − k 4 ) 3 φ = − λgr
2
r
r
r
η
so
η
λ(r) =
gr
Also not forget Υ =
1
k
000
0000
2 1 0
2 1 00
2
4 1
2φ + rφ + 2k 2 φ − 2k φ − (k − k ) 3 φ
r
r
r
sin(kθ)
192
41.1
Linking with our paper
We have
φ(r)
0
φ (r)
00
φ (r)
000
f (r) =
41.2
η0
g0
= −rg(r)
(323)
0
= −g(r) − rg (r) = −f (r)
(324)
0
(325)
00
= −f (r)
φ (r)
= −f (r)
(326)
φ0000 (r)
= −f 000 (r)
(327)
2f 00 (r) + rf 000 (r) + 2k 2
1
21 0
2
4 1
f
(r)
−
2k
g(r)
f
(r)
+
(k
−
k
)
r2
r
r2
No slip boundary conditions
No-slip boundary conditions inside and outside impose that all components of the velocity must be zero
on both boundaries, i.e.
v(r = R1 ) = v(r = R2 ) = 0
Due to the separation of variables, and since ξ(θ) = cos(kθ) we have
u(r, θ) =
1 ∂Ψ
1
1
= φξ 0 = − φ(r)k sin(kθ)
r ∂θ
r
r
v(r, θ) = −
∂Ψ
= −φ0 (r)ξ = −φ0 (r) cos(kθ)
∂r
It is obvious that the only way to insure no-slip boundary conditions is to have
φ(R1 ) = φ(R2 ) = φ0 (R1 ) = φ0 (R2 ) = 0
We could then choose
φ(r)
0
φ (r)
=
=
(r − R1 )2 (r − R2 )2 F(r)
(328)
2
2
2
2
0
2(r − R1 )(r − R2 ) F(r) + 2(r − R1 ) (r − R2 )F(r) + (r − R1 ) (r − R2 ) F (r)
(329)
which are indeed identically zero on both boundaries. Here F(r) is any (smooth enough) function of r.
We would then have
Ψ(r, θ) = (r − R1 )2 (r − R2 )2 F(r) cos(kθ)
In what follows we will take F(r) = 1 for simplicity.
193
COMPUTE f from φ and then the pressure.
41.3
Free slip boundary conditions
Before postulating the form of φ(r), let us now turn to the boundary conditions that the flow must fulfill,
i.e. free-slip on both boundaries. Two conditions must be met:
• v · n = 0 (no flow through the boundaries) which yields u(r = R1 ) = 0 and u(r = R2 ) = 0, :
1 ∂Ψ
(r = R1 , R2 ) = 0
r ∂θ
194
∀θ
which gives us the first constraint since Ψ(r, θ) = φ(r)ξ(θ):
φ(r = R1 ) = φ(r = R2 ) = 0
• (σ · n) × n = 0 (the tangential stress at the boundary is zero) which imposes: σθr = 0, with
1 ∂v v 1 ∂u
∂
∂Ψ
1 ∂Ψ
1 ∂ 1 ∂Ψ
σθr = 2η ·
− +
=η
(−
) − (−
)+
(
)
2 ∂r
r
r ∂θ
∂r
∂r
r
∂r
r ∂θ r ∂θ
Finally Ψ must fulfill (on the boundaries!):
−
∂ 2 Ψ 1 ∂Ψ
1 ∂2Ψ
+
=0
+
∂r2
r ∂r
r2 ∂θ2
1
1
−φ00 ξ + φ0 ξ + 2 φξ 00 = 0
r
r
or,
1
1
−φ00 + φ0 − k 2 2 φ = 0
r
r
Note that this equation is a so-called Euler Differential Equation17 . Since we are looking for a
solution φ such that φ(R1 ) = φ(R2 ) = 0 then the 3rd term of the equation above is by definition
zero on the boundaries. We have to ensure the following equality on the boundary:
1
−φ00 + φ0 = 0
r
for r = R1 , R2
The solution of this ODE is of the form φ(r) = ar2 + b and it becomes evident that it cannot satisfy
φ(r = R1 ) = φ(r = R2 ) = 0.
Separation of variables leads to solutions which cannot fulfill the free slip boundary conditions
17 http://mathworld.wolfram.com/EulerDifferentialEquation.html
195
42
fieldstone 36: the annulus geometry elastic aquarium
This fieldstone was developed in collaboration with Lukas van de Wiel.
The domain is an annulus with inner radius R1 and outer radius R2 . It is filled with a single elastic
material characterised by a Young’s modulus E and a Poisson ratio ν, a density ρ0 . The gravity g = −g0 er
is pointing towards the center of the domain.
The problem at hand is axisymmetric so that the tangential component of the displacement vector vθ
is assumed to be zero as well as all terms containing ∂θ . The components of the strain tensor are
εrr
=
εθθ
=
εrθ
=
∂vr
∂r
vr
1 ∂vθ
vr
+
=
r r ∂θ
r
1 ∂vθ
vθ
1 ∂vr
−
+
=0
2 ∂r
r
r ∂θ
(330)
(331)
(332)
so that the tensor simply is
ε=
εrr
εrθ
εrθ
εθθ
=
∂vr
∂r
0
0
(333)
vr
r
The pressure is
p = −λ∇ · v = −λ
1 ∂(rvr )
r ∂r
(334)
and finally the stress tensor:
σ = −p1 + 2µε =
r)
r
+ 2µ ∂v
λ 1r ∂(rv
∂r
∂r
0
0
!
r)
λ 1r ∂(rv
+ 2µ vrr
∂r
The divergence of the stress tensor is given by [416]:
∂σrr
σrr −σθθ
θr
+ 1r ∂σ
∂r +
r
∂θ
∇·σ =
∂σrθ
σrθ +σθr
1 σθθ
∂r + r ∂θ +
r
Given the diagonal nature of the stress tensor this simplifies to (also remember that ∂θ = 0):
∂σ
σrr −σθθ
rr
∂r +
r
∇·σ =
0
(335)
(336)
(337)
Focusing on the r-component of the stress divergence:
(∇ · σ)r
=
=
=
=
=
σrr − σθθ
∂σrr
+
∂r
r
∂
1 ∂(rvr )
∂vr
1
1 ∂(rvr )
∂vr
1 ∂(rvr )
vr
λ
+ 2µ
+
λ
+ 2µ
−λ
− 2µ
∂r
r ∂r
∂r
r
r ∂r
∂r
r ∂r
r
∂ 2 vr
1 ∂(rvr ) 2µ ∂vr
1 ∂(rvr ) 2µvr
∂ 1 ∂(rvr )
+ 2µ 2 + λ 2
+
−λ 2
− 2
λ
∂r r ∂r
∂r
r
∂r
r ∂r
r
∂r
r
vr
1 ∂vr
∂ 2 vr
∂ 2 vr
2µ ∂vr
2µvr
λ(− 2 +
+
) + 2µ 2 +
− 2
r
r ∂r
∂r2
∂r
r ∂r
r
∂ 2 vr
1 ∂vr
vr
+ (2µ + λ) 2
−(2µ + λ) 2 + (2µ + λ)
r
r ∂r
∂r
(338)
(339)
(340)
(341)
(342)
So the momentum conservation in the r direction is
(∇ · σ + ρ0 g)r = −(2µ + λ)
vr
1 ∂vr
∂ 2 vr
+ (2µ + λ)
+ (2µ + λ) 2 − ρ0 g0 = 0
2
r
r ∂r
∂r
196
(343)
or,
∂ 2 vr
1 ∂vr
ρ0 g0
vr
+
− 2 =
2
∂r
r ∂r
r
λ + 2µ
(344)
We now look at the boundary conditions. On the inner boundary we prescribe vr (r = R1 ) = 0 while free
surface boundary conditions are prescribed on the outer boundary, i.e. σ · n = 0 (i.e. there is no force
applied on the surface).
The general form of the solution can then be obtained:
vr (r) = C1 r2 + C2 r +
C3
r
(345)
with
C1 =
ρ0 g0
3(λ + 2µ)
C2 = −C1 R1 −
C3
R12
C3 =
k1 + k2
(R12 + R22 )(2µ + λ) + (R22 − R12 )λ
(346)
k2 = λC1 (R12 R23 − R13 R22 )
(347)
and
k1 = (2µ + λ)C1 (2R12 R23 − R13 R22 )
Pressure can then be computed as follows:
1
1 ∂(rvr )
2
= −λ
(3C1 r + 2C2 r) = −λ(3C1 r + 2C2 )
p = −λ∇ · v = −λ
r ∂r
r
(348)
0
1.2x1011
-20000
1x1011
-40000
radial displacement
radial displacement
We choose R1 = 2890km, R2 = 6371km, g0 = 9.81ms−2 , ρ0 = 3300, E = 6 · 1010 , ν = 0.49.
-60000
-80000
-100000
6x1010
4x1010
2x1010
-120000
-140000
8x1010
3x106
3.5x106
4x106
4.5x106
5x106
5.5x106
0
6x106
3x106
3.5x106
4x106
r
4.5x106
r
radial profiles of the displacement and pressure fields
displacement and pressure fields in the domain
197
5x106
5.5x106
6x106
10000.000000
velocity
pressure
x2
x1
100.000000
error
1.000000
0.010000
0.000100
0.000001
0.01
h
198
43
fieldstone 37: marker advection and population control
The domain is a unit square. The Stokes equations are not solved, the velocity is prescribed everywhere
in the domain as follows:
u = −(z − 0.5)
(349)
v
=
0
(350)
w
=
(x − 0.5)
(351)
At the moment, velocity is computed on the marker itself (rk0 algorithm). When markers are advected
outside, they are arbitrarily placed at location (-0.0123,-0.0123).
in construction.
199
44
fieldstone 38: Critical Rayleigh number
This fieldstone was developed in collaboration with Arie van den Berg.
The system is a layer of fluid between y = 0 and y = 1, with boundary conditions T (x, y = 0) = 1
and T (x, y = 1) = 0, characterized by ρ, cp , k, η0 . The Rayleigh number of the system is
Ra =
ρ0 g0 α∆T h3
η0 κ
We have ∆T = 1, h = 1 and choose κ = 1 so that the Rayleigh number simplifies to Ra = ρ0 g0 α/η0 .
~ · σ + ~b = ~0 with ~b = ρ~g . Then the components of the this equation on the
The Stokes equation is ∇
x- and y−axis are:
~ · σ)x
(∇
~ · σ)y
(∇
= −ρ~g · ~ex = 0
(352)
= −ρ~g · ~ey = ρg0
(353)
since ~g and ~ey are in opposite directions (~g = −g0~ey , with g0 > 0). The stream function formulation of
the incompressible isoviscous Stokes equation is then
∇4 Ψ =
g0 ∂ρ
η0 ∂x
Assuming a linearised density field with regards to temperature ρ(T ) = ρ0 (1 − αT ) we have
∂T
∂ρ
= −ρ0 α
∂x
∂x
and then
∇4 Ψ = −
ρ0 g0 α ∂T
∂T
= −Ra
g
η0
∂x
∂x
(354)
For small perturbations of the conductive state T0 (y) = 1 − y we define the temperature perturbation
T1 (x, y) such that
T (x, y) = T0 (y) + T1 (x, y)
The temperature perturbation T1 satisfies the homogeneous boundary conditions T1 (x, y = 0) = 0 and
T1 (x, y = 1) = 0. The temperature equation is
DT
∂T
∂T0 + T1
~
~
ρcp
= ρcp
+ ~ν · ∇T = ρcp
+ ~ν · ∇(T0 + T1 ) = k∆(T0 + T1 )
Dt
∂t
∂t
and can be simplified as follows:
ρcp
∂T1
~ 0
+ ~ν · ∇T
∂t
= k∆T1
~ 1 to be second
since T0 does not depend on time, ∆T0 = 0 and we assume the nonlinear term ~ν · ∇T
order (temperature perturbations and coupled velocity changes are assumed to be small). Using the
relationship between velocity and stream function vy = −∂x Ψ we have v · ∇T0 = −vy = ∂x Ψ and since
κ = k/ρcp = 1 we get
∂Ψ
∂T1
− κ∆T1 = −
(355)
∂t
∂x
Looking at these equations, we immediately think about a separation of variables approach to solve these
equations. Both equations showcase the Laplace operator ∆, and the eigenfunctions of the biharmonic
operator and the Laplace operator are the same. We then pose that Ψ and T1 can be written:
Ψ(x, y, t)
=
AΨ exp(pt) exp(±ikx x) exp(±iky y) = AΨ Eψ (x, y, t)
(356)
T1 (x, y, t)
=
AT exp(pt) exp(±ikx x) exp(±iky y) = AT ET (x, y, t)
(357)
where kx and ky are the horizontal and vertical wave number respectively. Note that we then have
∇2 Ψ = −(kx2 + ky2 )Ψ
∇2 T1 = −(kx2 + ky2 )T1
200
The boundary conditions on T1 , coupled with a choice of a real function for the x dependence yields:
ET (x, y, t) = exp(pt) cos(kx x) sin(nπy).
from here onwards check for minus signs!
The velocity boundary conditions are vy (x, y = 0) = 0 and vy (x, y = 1) = 0 which imposes conditions
on ∂Ψ/∂x and we find that we can use the same y dependence as for T1 . Choosing again for a real
function for the x dependence yields:
EΨ (x, y, t) = exp(pt) sin(kx x) sin(nπz)
We then have
Ψ(x, y, t)
=
AΨ exp(pt) sin(kx x) sin(nπz) = AΨ Eψ (x, y, t)
(358)
T1 (x, y, t)
=
AT exp(pt) cos(kx x) sin(nπz) = AT ET (x, y, t)
(359)
In what follows we simplify notations: k = kx . Then the two PDEs become:
pT1 + κ(k 2 + n2 π 2 ) − kAΨ exp(pt) cos(kx x) sin(nπz) = kAΨ Eθ
(360)
−RaAT cos(kx) sin(nπz) + κ(k 2 + n2 π 2 )2 AΨ = −RaAT EΨ + κ(k 2 + n2 π 2 )2 AΨ = 0
(361)
These equations must then be verified for all ... which leads to write:
p + (k 2 + n2 π 2 )
−k
Aθ
0
=
−Ra k
(k 2 + n2 π 2 )2
AΨ
0
The determinant of such system must be nul otherwise there is only a trivial solution to the problem, i.e.
Aθ = 0 and AΨ = 0 which is not helpful. CHECK/REPHRASE
D = [p + (k 2 + n2 π 2 )](k 2 + n2 π 2 )2 − Ra k 2 = 0
or,
p=
Ra k 2 − (k 2 + n2 π 2 )3
(k 2 + n2 π 2 )2
The coefficient p determines the stability of the system: if it is negative, the system is stable and both
Ψ and T1 will decay to zero (return to conductive state). If p = 0, then the system is meta-stable, and if
p > 0 then the system is unstable and the perturbations will grow. The threshold is then p = 0 and the
solution of the above system is
201
45
fieldstone: Gravity: buried sphere
Before you proceed further, please read :
http://en.wikipedia.org/wiki/Gravity_anomaly
http://en.wikipedia.org/wiki/Gravimeter
Let us consider a vertical domain Lx × Ly where Lx = 1000km and Ly = 500km. This domain
is discretised by means of a grid which counts nnp = nnx × nny nodes. This grid then counts nel =
nelx × nely = (nnx − 1) × (nny − 1) cells. The horizontal spacing between nodes is hx and the vertical
spacing is hy.
Assume that this domain is filled with a rock type which mass density is given by ρmedium =
3000kg/m3 , and that there is a circular inclusion of another rock type (ρsphere = 3200kg/m3 ) at location (xsphere, ysphere) of radius rsphere. The density in the system is then given by
ρsphere inside the circle
ρ(x, y) =
ρmedium outside the circle
Let us now assume that we place nsurf gravimeters at the surface of the model. These are placed
equidistantly between coordinates x = 0 and coordinates x = Lx. We will use the arrays xsurf and ysurf
to store the coordinates of these locations. The spacing between the gravimeters is δx = Lx/(nsurf − 1).
At any given point (xi , yi ) in a 2D space, one can show that the gravity anomaly due to the presence
of a circular inclusion can be computed as follows:
g(xi , yi ) = 2πG(ρsphere − ρ0 )R2
yi − ysphere
(xi − xsphere )2 + (yi − ysphere )2
(362)
where rsphere is the radius of the inclusion, (xsphere , ysphere ) are the coordinates of the center of the
inclusion, and ρ0 is a reference density.
However, the general formula to compute the gravity anomaly at a given point (xi , yi ) in space due
to a density anomaly of any shape is given by:
Z Z
∆ρ(x, y)(y − yi )
g(xi , yi ) = 2G
dxdy
(363)
(x
− xi )2 + (y − yi )2
Ω
where Ω is the area of the domain on which the integration is to be carried out. Furthermore the density
anomaly can be written : ∆ρ(x, y) = ρ(x, y) − ρ0 . We can then carry out the integration for each cell
and sum their contributions:
nel Z Z
X
(ρ(x, y) − ρ0 )(y − yi )
g(xi , yi ) = 2G
dxdy
(364)
2
2
Ωe (x − xi ) + (y − yi )
ic=1
where Ωe is now the area of a single
R R cell. Finally, one can assume the density to be constant within each
cell so that ρ(x, y) → ρ(ic) and
dxdy → hx × hy and then
Ωe
g(xi , yi ) = 2G
nel
X
(ρ(ic) − ρ0 )(y(ic) − yi )
sx sy
(x(ic)
− xi )2 + (y(ic) − yi )2
ic=1
(365)
We will then use the array gsurf to store the value of the gravity anomaly measured at each gravimeter
at the surface.
202
3200
50
3100
0
0.00075
gy
0
50
100
150
3000
0.00050
error(gy)
0.00025
0
200000
400000
0
200000
400000
0.000005
x
600000
800000
1000000
600000
800000
1000000
0.000010
x
To go further
• explore the effect of the size of the inclusion on the gravity profile.
• explore the effect of the ρ0 value.
• explore the effect of the grid resolution.
• measure the time that is required to compute the gravity. How does this time vary with nsurf ?
how does it vary when the grid resolution is doubled ?
• Assume now that ρ2 < ρ1 . What does the gravity profile look like ?
• what happens when the gravimeters are no more at the surface of the Earth but in a satellite ?
• if you feel brave, redo the whole exercise in 3D...
203
46
Problems, to do list and projects for students
• Darcy flow. redo WAFLE (see http://cedricthieulot.net/wafle.html)
• carry out critical Rayleigh experiments for various geometries/aspect ratios. Use Arie’s notes.
• Newton solver
• Corner flow
• elasticity with markers
• Indentor/punch with stress b.c. ?
• chunk grid
• read in crust 1.0 in 2D on chunk
• compute gravity based on tetrahedra
• compare Q2 with Q2-serendipity
• NS a la http://ww2.lacan.upc.edu/huerta/exercises/Incompressible/Incompressible Ex2.htm
• produce example of mckenzie slab temperature
• write about impose bc on el matrix
• constraints
• discontinuous galerkin
• formatting of code style
• navier-stokes ? (LUKAS) use dohu matlab code
• nonlinear poiseuille
• Finish nonlinear cavity case5.
• write about mappings
• write about stream functions
• periodic boundary conditions
• free surface
• zaleski disk advection
• better yet simple matrix storage ?
• write Scott about matching compressible2 setup with his paper
• deal with large matrices.
• compositions, marker chain
• 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
• including phase changes (w. R. Myhill)
• compute strainrate in middle of element or at quad point for punch?
204
• GEO1442 code
• GEO1442 indenter setup in plane ?
• in/out flow on sides for lith modelling
• Fehlberg RK advection
• redo puth17 2 layer experiment
• create stone for layeredflow (see folder one up)
205
A
Three-dimensional applications
In the following table I list many papers which showcase high-resolution models of geodynamical processes
(subduction, rifting, mantle flow, plume transport, ...). Given the yearly output of our community and
the long list of journal in which research can be disseminated, this list can not be exhaustive.
Ref.
[27]
[433]
[388]
[339]
[103]
[157]
[11]
[12]
[378]
[271]
[102]
[72]
[47]
[71]
[73]
[104]
[163]
[411]
[328]
[527]
[301]
[499]
[311]
[382]
[162]
[252]
[346]
[405]
[91]
[455]
[314]
topic
Small-scale sublithospheric convection in the Pacific
Migration and morphology of subducted slabs in the upper mantle
Subduction scissor across the South Island of New Zealand
Influence of a buoyant oceanic plateau on subduction zones
Subduction dynamics, origin of Andean orogeny and the Bolivian orocline
Feedback between rifting and diapirism, ultrahigh-pressure rocks exhumation
Numerical modeling of upper crustal extensional systems
Rift interaction in brittle-ductile coupled systems
Kinematic interpretation of the 3D shapes of metamorphic core complexes
Role of rheology and slab shape on rapid mantle flow: the Alaska slab edge
Complex mantle flow around heterogeneous subducting oceanic plates
Oblique rifting and continental break-up
Influence of mantle plume head on dynamics of a retreating subduction zone
Rift to break-up evolution of the Gulf of Aden
Thermo-mechanical impact of plume arrival on continental break-up
Subduction and slab breakoff controls on Asian indentation tectonics
Modeling of upper mantle deformation and SKS splitting calculations
Backarc extension/shortening, slab induced toroidal/poloidal mantle flow
Sediment transports in the context of oblique subduction modelling
Crustal growth at active continental margins
Dynamics of India-Asia collision
Strain-partitioning in the Himalaya
Collision of continental corner from 3-D numerical modeling
Dependence of mid-ocean ridge morphology on spreading rate
Mid mantle seismic anisotropy around subduction zones
Oblique rifting of the Equatorial Atlantic
Dynamics of continental accretion
Thrust wedges: infl. of decollement strength on transfer zones
Asymmetric three-dimensional topography over mantle plumes
modelled crustal systems undergoing orogeny and subjected to surface processes
Thermo-mechanical modeling ontinental rifting and seafloor spreading
206
resolution
448 × 56 × 64
50 × 50 × 25
17 × 9 × 9
80 × 40 × 80
96 × 96 × 64
100 × 64 × 20
160 × 160 × 12
160 × 160 × 23
67 × 67 × 33
960 × 648 × 160
96 × 96 × 64
150 × 50 × 30
80 × 40 × 80
83 × 83 × 40
100 × 70 × 20
96 × 96 × 64
96 × 64 × 96
352 × 80 × 64
500 × 164 × 100
404 × 164 × 100
257 × 257 × 33
256 × 256 × 40
500 × 340 × 164
196 × 196 × 100
197 × 197 × 53
120 × 80 × 20
256 × 96 × 96
309 × 85 × 149
500 × 500 × 217
96 × 32 × 14
197 × 197 × 197
B
Codes in geodynamics
In what follows I make a quick inventory of the main codes of computational geodynamics, for crust,
lithosphere and/or mantle modelling.
in order to find all CIG-codes citations go to: https://geodynamics.org/cig/news/publications-refbase/
• ABAQUS [198] [298] [334] [355] [369]
• ADELI [249] [487] [57] [58] [196] [221] [493] [106]
• ASPECT
This code is hosted by CIG at https://geodynamics.org/cig/software/aspect/
[29] [297] [22] [458] [133] [188] [518] [251] [131] [253] [403] [404] [21] [451] [69] [361] [439] [515] [132]
[363] [223] [254] [189] [368] [379] [68] [31] [434] [122] [320]
• CitcomS and CITCOMCU
These codes are hosted by CIG at https://geodynamics.org/cig/software/citcomcu/ and
https://geodynamics.org/cig/software/citcoms/.
[428] [349] [521] [476] [524] [245] [50] [442] [475] [119] [51] [426] [52] [38] [373] [440] [41] [120] [53]
[523] [332] [525] [28] [401] [351] [118] [222] [187] [522] [258] [313] [18] [517] [15] [176] [87] [82] [484]
[27] [88] [516] [39] [304] [472] [308] [318] [271] [54] [59] [267] [427] [257] [62] [63] [272] [392] [360] [19]
[121] [175] [81] [281] [20] [492] [419] [5] [327] [528] [61] [60] [422] [134] [473] [494] [491] [490] [248]
[444] [309] [498] [497] [326] [179] [250] [256]
cross check with CIG database
• CONMAN This code is hosted by CIG at https://geodynamics.org/cig/software/conman/
[290] [269] [287] [288] [270] [40] [289]
• DOUAR
[66] [454] [510] [67] [323] [352] [499] [356]
• GAIA
• GALE
This code is hosted by CIG at https://geodynamics.org/cig/software/gale/
[168] [49] [126] [378] [19]
ADD:
Cruz, L.; Malinski, J.; Hernandez, M.; Take, A.; Hilley, G. Erosional control of the kinematics of the
Aconcagua fold-and-thrust belt from numerical simulations and physical experiments 2011 Geology
Goyette, S.; Takatsuka, M.; Clark, S.; Mller, R.D.; Rey, P.; Stegman, D.R. Increasing the usability
and accessibility of geodynamic modelling tools to the geoscience community: UnderworldGUI 2008
Visual Geosciences
Li, Y.; Qi, J. Salt-related Contractional Structure and Its Main Controlling Factors of Kelasu
Structural Zone in Kuqa Depression: Insights from Physical and Numerical Experiments 2012
Procedia Engineering
• GTECTON [230] [231] [77] [78] [232] [177] [322] [24] [25] [335]
• ELEFANT
[458] [325] [80] [299] [451] [375]
• ELLIPSIS
[347] [362] [348] [150] [364] [388] [308] [305]
207
• FANTOM
[450] [11] [12] [13] [160] [455] [158] [159]
• FLUIDITY [136] [186]
• the I2(3)E(L)VIS code
[215][216][214] [217][218][212] [86] [75][204][227][208] [205][226] [414][207][462][164][526] [209] [202][357]
[147][149][312][203][211] [123][146][514] [311][354][328][469][468][527][151][210][337] [148][382][405][489][26][315][435][3
[91][228][26][470] [145][463][406][213][406] [2][329][174]
• LAMEM [414] [282] [300] [340] [301] [117] [170] [169] [380] [171] [116] [283]
• LITMOD [181] [3] [4] [182]
• MILAMIN [130] [511] [193] [324] [194] [448] [275] [336]
• PARAVOZ/FLAMAR [376] [93] [89] [23] [115] [246] [199] [244] [201] [459] [92] [508] [94] [509] [197]
[510] [17] [196] [242] [195] [165] [178] [185] [90] [506] [338] [192] [140]
• PTATIN [370] [1] [341]
• RHEA [95] [430] [9] [96]
• SEPRAN [464] [488] [113] [474] [480] [481] [482] [483] [418] [316] [317] [479] [65] [64] [374] [486] [465]
[437] [477] [466] [42] [109] [16] [110] [350] [467]
• SLIM3D
[377] [396] [72] [73] [71] [70] [252] [294] [114]
• SOPALE
[183]
[156]
[235]
[274]
[502] [32] [183] [155] [36] [500] [385] [37] [34] [264] [386] [263] [485] [503] [390] [35] [387] [384]
[191] [190] [266] [383] [421] [265] [128] [345] [420] [495] [496] [284] [33] [79] [234] [425] [8] [7]
[389] [127] [101] [262] [236] [237] [293] [233] [273] [99] [108] [172] [173] [229] [238] [292] [358]
[224] [10] [100] [255] [319] [98]
• STAGYY [402] [512] [124]
• 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).
[393] [157] [76] [445] [123] [241] [219] [220] [394] [395] [353] [532] [446]
• TERRA [85] [84] [371] [505] [504] [135] [471]
• the YACC code
[458] [457]
• UNDERWORLD 1&2 [432] [348] [410] [303] [365] [105] [339] [433] [431] [167] [103] [102] [47] [411]
[163] [166] [391] [48] [412] [423] [424] [366] [291] [342]
208
C
Matrix properties
C.1
Symmetric matrices
Any symmetric matrix has only real eigenvalues, is always diagonalizable, and has orthogonal eigenvectors.
A symmetric N × N real matrix M is said to be
• positive definite if ~x · M · ~x > 0 for every non-zero vector ~x of n real numbers. All the eigenvalues
of a Symmetric Positive Definite (SPD) matrix are positive. If A and B are positive definite, then
so is A+B. The matrix inverse of a positive definite matrix is also positive definite. An SPD matrix
has a unique Cholesky decomposition. In other words the matrix M is positive definite if and only
if there exists a unique lower triangular matrix L, with real and strictly positive diagonal elements,
such that M = LLT (the product of a lower triangular matrix and its conjugate transpose). This
factorization is called the Cholesky decomposition of M .
• positive semi-definite if ~x · M · ~x ≥ 0
• negative definite if ~x · M · ~x < 0
• negative semi-definite if ~x · M · ~x ≤ 0
The Stokes linear system
K
GT
G
0
v
f
·
=
p
g
is indefinite (i.e. it has positive as well as negative eigenvalues).
A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if
and only if its determinant is 0. Singular matrices are rare in the sense that if you pick a random square
matrix, it will almost surely not be singular.
C.2
Schur complement
From wiki. In linear algebra and the theory of matrices, the Schur complement of a matrix block (i.e.,
a submatrix within a larger matrix) is defined as follows. Suppose A, B, C, D are respectively p × p,
p × q, q × p and q × q matrices, and D is invertible. Let
A B
M=
C D
so that M is a (p + q) × (p + q) matrix. Then the Schur complement of the block D of the matrix M is
the p × p matrix
S = A − B · D −1 · C
Application to solving linear equations: The Schur complement arises naturally in solving a system of
linear equations such as
A · ~x + B · ~y
C · ~x + D · ~y
= f~
= ~g
where ~x, f~ are p-dimensional vectors, ~y , ~g are q-dimensional vectors, and A, B, C, D are as above.
Multiplying the bottom equation by B · D −1 and then subtracting from the top equation one obtains
(A − B · D −1 · C) · ~x = f~ − B · D −1 · ~g
Thus if one can invert D as well as the Schur complement of D, one can solve for ~x, and then by using the
equation C ·~x +D·~y = ~g one can solve for y. This reduces the problem of inverting a (p+q)×(p+q) matrix
to that of inverting a p × p matrix and a q × q matrix. In practice one needs D to be well-conditioned in
order for this algorithm to be numerically accurate.
Considering now the Stokes system:
K
G
~v
f~
·
=
p~
GT −C
~g
209
Factorising for p~ we end up with a velocity-Schur complement. Solving for p~ in the second equation
and inserting the expression for p~ into the first equation we have
Sv · ~v = f~
with
Sv = K + G · C−1 · GT
Factorising for ~v we get a pressure-Schur complement.
Sp · p~ = GT · K−1 · f~
with
210
Sp = GT · K−1 · G + C
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Index
P1 , 32, 34
P2 , 32
P3 , 33
Pm × Pn , 22
Pm × P−n , 22
Q1 × P0 , 91, 97, 99, 105, 112, 116, 120, 122, 139,
146, 158, 160, 164, 179, 180
Q1 , 23, 27
Q2 × Q1 , 22, 123, 128
Q2 , 23, 29, 35
Qs2 , 30
Q3 × Q2 , 131
Q3 , 24, 31
Qm × P−n , 22
Qm × Qn , 22
Qm × Q−n , 22
incompressible flow, 99, 105, 112, 116, 120, 122,
128, 131, 134, 139, 158, 160, 164, 179, 180
isoparametric, 80
isothermal, 91, 97, 99, 105, 112, 116, 120, 122, 128,
131, 134, 139, 158, 160, 164, 180
isoviscous, 91, 97, 116, 120, 128, 131, 139, 146, 164,
179
Lamé parameter, 14
Legendre polynomial, 20
method of manufactured solutions, 51
midpoint rule, 19
mixed formulation, 116, 120, 122, 128, 131, 134,
139, 146, 158, 160, 164, 179
MMS, 51
analytical solution, 91, 99, 116, 120, 128, 131, 134, Neumann boundary condition, 16
Newton’s method, 85
139, 146, 164, 180
Newton-Cotes, 20
arithmetic mean, 113
non-conforming element, 22
non-isoviscous, 99, 105, 112, 122, 134, 180
basis functions, 27
nonconforming Q1 × P0 , 134
Boussinesq, 13
nonlinear, 160
bubble function, 22
nonlinear rheology, 105
bulk modulus, 14, 141
numerical benchmark, 158
bulk viscosity, 12
buoyancy-driven flow, 93
open boundary conditions, 179
CBF, 164
particle-in-cell, 112
CG, 70
penalty formulation, 38, 91, 93, 97, 99, 105, 112,
checkerboard, 66
180
cohesion, 65
piecewise,
22
Compressed Sparse Column, 58
Poisson
ratio,
14
Compressed Sparse Row, 58
preconditioned
conjugate gradient, 72
compressibility, 141
pressure
normalisation,
124
compressible flow, 146
pressure
scaling,
68
conforming element, 22
pressure smoothing, 66, 116, 134, 139, 146, 164
conjugate gradient, 70
connectivity array, 62
quadrature, 20
convex polygon, 61
CSC, 58
rectangle rule, 19
CSR, 58
Rens Elbertsen, 165, 186
Dirichlet boundary condition, 16
divergence free, 36
Drucker-Prager, 65
dynamic viscosity, 12
Gauss quadrature, 20
geometric mean, 113
harmonic mean, 113
hyperbolic PDE, 79
Schur complement, 70
Schur complement approach, 120, 122
second viscosity, 12
shear modulus, 14
SPD, 70
static condensation, 50
Stokes sphere, 93, 122
strain tensor, 14
strong form, 36
structured grid, 61
240
Symmetric Positive Definite, 209
tensor invariant, 64
thermal expansion, 141
trapezoidal rule, 19
unstructured grid, 61
Viscosity Rescaling Method, 65
von Mises, 65
VRM, 65
weak form, 36, 39
work against gravity, 144
Young’s modulus, 14
241
Notes
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finish. not happy with definition. Look at literature . . . . . . . . . . . . . . . . . . . . . . . . 17
derive formula for Earth size R1 and R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
check aspect manual The 2D cylindrical shell benchmarks by Davies et al. 5.4.12 . . . . . . . . 18
insert here figure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
insert here figure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
citation needed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
verify those . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
verify those . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
list codes which use this approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
insert here link(s) to manual and literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
say something about minus sign? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
other time discr schemes! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
sort out mess wrt Eq 26 of busa13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
add more examples coming from geo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
produce drawing of node numbering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
insert here the rederivation 2.1.1 of spmw16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
produce figure to explain this . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
link to proto paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
link to least square and nodal derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
how to compute M for the Schur complement ? . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
to finish . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
to finish . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
build S and have python compute its smallest and largest eigenvalues as a function of resolution?120
explain how Medge is arrived at! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
compare with ASPECT ??! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
gauss-lobatto integration? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
pressure average on surface instead of volume ? . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
cross check with CIG database . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
242
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