AN ABSTRACT OF THE DISSERTATION OF
Francis Patricia Medina for the degree of Doctor of Philosophy in Mathematics
presented on May 13, 2014.
Title: Mathematical Treatment and Simulation of Methane Hydrates and Adsorption Models
Abstract approved:
Malgorzata S. Peszynska
In this dissertation we develop mathematical treatment for two important applications: (i) evolution of methane in coalbeds with the associated phenomena of adsorption,
and (ii) formation of methane hydrates in seabed. We use simplified models for (i) and (ii)
since we are more interested in qualitative properties of the solutions rather than direct
applications to engineering.
For methane hydrates we focus on a scalar problem with diffusion only, and we
discuss it as a nonlinear parabolic problem in a single variable with monotone operators.
We show how the problem can be cast in the framework of a free boundary problem. The
particular nonlinearity that we deal with comes from a constraint on one of the variables.
For the simplified model of methane hydrates, we establish well-posedness of the problem
in an abstract weak setting. We also perform simulations with a novel approach based
on semismooth Newton methods. We demonstrate convergence rates of the numerical
approximation which are similar to those for Stefan free boundary value problem.
On the other hand, for adsorption problems, we focus on their structure as systems
of conservation laws, with equilibrium and non-equilibrium type nonlinearities, where
the latter are associated with microscale diffusion. We also work with an unusual type
isotherm called Ideal Adsorbate Solution, which is defined implicitly. For the IAS adsorption system, we show sufficient conditions that render the system hyperbolic. We also
construct numerical approximations for equilibrium and nonequilibrium models.
c
Copyright by Francis Patricia Medina
May 13, 2014
All Rights Reserved
Mathematical Treatment and Simulation of Methane Hydrates and Adsorption Models
by
Francis Patricia Medina
A DISSERTATION
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Presented May 13, 2014
Commencement June 2015
Doctor of Philosophy dissertation of Francis Patricia Medina presented on May 13, 2014
APPROVED:
Major Professor, representing Mathematics
Chair of the Department of Mathematics
Dean of the Graduate School
I understand that my dissertation will become part of the permanent collection of Oregon
State University libraries. My signature below authorizes release of my dissertation to
any reader upon request.
Francis Patricia Medina, Author
ACKNOWLEDGEMENTS
Academic
I am indebted to the National Science Foundation for partially supporting the research done in this dissertation under NSF-DMS 1115827 “Hybrid modeling in porous
media”, P.I.: Malgorzata Peszynska. I want to thank my collaborators Dr. Nathan Gibson, Dr. Malgorzata Peszynska and Dr. Ralph Showalter for sharing their expertise with
me while working in the project related to paper [16].
This dissertation would not be possible without the consistent help , dedication,
mentorship and support of my advisor Malgorzata Peszysnka, who has been extremely
generous with her time and knowledge on topics that I never dreamed in learning.
I want to thank my entire committee for their patience. Specially Dr. Ossiander
and Dr. Parks for their support.
Personal
I wish to thank my the love of my life, my husband, Boris Iskra for believing in me,
for helping in every aspect of my life, for his support. Because, you believed in me more
than anyone else in the darkest times.
I wish to thank my parents Ida and Angel for their love and support, for teaching
me the love for learning, for loving me the way I am. To my sister Angela and her husband
Ryan for supporting me.
I wish to thank my friend Christian Calderon, John Alexopoulus, Edmundo Castillo,
Vivian Klein, Veronika, Theresa Migler, Hieu Do, Sooie-Hoe Loke, Hussain, Patcharee,
Patti Conklin (and the Yert-Conklin family), Margaret Weinberger and the game night
crew Stef, Greg, Stefan and Andrew for their cheerfulness. To my in laws, Zdenka and
Winko for their kindness. Last, but not least, my friend Jeef for always listening and
having a kind word to make me feel better.
I wish to thank Dr. So-Hsiang Chou and Dr. Diomedes Barcenas for their mentorship and friendship. Diomedes, you will be always remembered.
TABLE OF CONTENTS
Page
1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2. BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.1.
Weak and numerical solutions of evolution problems with monotone graphs
5
2.1.1 Function spaces and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Monotone operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.3 Finite element solution to evolution equations . . . . . . . . . . . . . . . . . . .
2.1.3.1 Stationary variational problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.3.2 Finite element for evolution equations . . . . . . . . . . . . . . . . . . . .
5
9
13
13
16
2.2.
Classical Newton Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.3.
Introduction to Nonsmooth Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.3.1 Semismoothness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Piecewise differentiable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 Semismooth Newton Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.
27
29
32
Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.4.1 Weak solutions to conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Linear scalar conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3 Nonlinear scalar conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3.1 Rankine-Hugoniot condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3.2 Entropy solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.3.3 Systems of conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.4 Numerical methods for conservation laws . . . . . . . . . . . . . . . . . . . . . . . .
2.4.5 Godunov method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.6 Numerical methods for systems of conservation laws . . . . . . . . . . . . .
40
41
43
46
48
49
49
50
53
3. MODELING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.1.
Modeling for Methane Hydrates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.2.
Modeling for Adsorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.2.1 Transport with adsorption in ECBM . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Single component adsorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2.1 Shock speed for adsorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
62
64
TABLE OF CONTENTS (Continued)
Page
3.2.3 Diffusion into micropores and transport with memory terms. . . . .
3.2.4 Multicomponent transport with adsorption . . . . . . . . . . . . . . . . . . . . . .
65
67
4. EVOLUTION PROBLEM WITH A MONOTONE GRAPH. ANALYSIS AND
NUMERICAL APPROXIMATIONS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.1.
Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.1.1
4.1.2
4.1.3
4.1.4
4.1.5
4.2.
Literature review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Semidiscrete formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Fully discrete implicit scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Semismooth implicit Newton solver. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Formulation of constraint as an NCC (Nonlinear Complementarity Constraint) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
84
85
89
89
Semismooth Newton for the monotone graph α . . . . . . . . . . . . . . . . . . . . . . . . 91
4.3.1 The case of a singular graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.2 The graph for methane hydrate problem . . . . . . . . . . . . . . . . . . . . . . . .
4.3.3 Stefan problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.
73
73
74
75
79
Numerical approximation for IBVP with a monotone graph . . . . . . . . . . . . . 79
4.2.1
4.2.2
4.2.3
4.2.4
4.2.5
4.3.
Family of monotone graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Construction of a normal convex integrand . . . . . . . . . . . . . . . . . . . . . .
Abstract initial–value problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Comparison principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Summary of analysis and outlook . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
93
94
Numerical results in 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.4.1 One–phase Stefan problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.4.2 Singular graph with an analytical solution . . . . . . . . . . . . . . . . . . . . . . . 100
4.4.3 The methane hydrates problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.5.
Numerical results for methane hydrates in 2D. . . . . . . . . . . . . . . . . . . . . . . . . . . 119
4.5.1 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
4.5.2 Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4.5.3 Remarks about 2D implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
5. ADSORPTION: ANALYSIS AND NUMERICAL APPROXIMATION . . . . . . . 147
TABLE OF CONTENTS (Continued)
Page
5.1.
Scalar conservation law for adsorption with Langmuir isotherm . . . . . . . . . 149
5.2.
Multicomponent system with extended Langmuir isotherm . . . . . . . . . . . . . . 156
5.2.1 Experiments with equilibrium case using Godunov . . . . . . . . . . . . . . . 160
5.3.
Multicomponent system with IAS (Ideal Adsorbate Solution) . . . . . . . . . . . 162
5.3.1
5.3.2
5.3.3
5.3.4
5.3.5
5.3.6
From single component to multicomponent isotherms . . . . . . . . . . . . 162
General case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Multicomponent system with IAS isotherms . . . . . . . . . . . . . . . . . . . . . 164
Eigenvectors and eigenvalues for IAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
CFL condition for Godunov scheme for IAS . . . . . . . . . . . . . . . . . . . . . 167
IAS example for a particular choice of single component isotherms 168
5.3.6.1 IAS calculations when Langmuir volumes are not equal . . . 169
5.3.7 Orders of convergence with IAS and EL . . . . . . . . . . . . . . . . . . . . . . . . . 170
5.3.8 Experiments with IAS vs. EL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
5.3.9 System with nonequilibrium and diffusion . . . . . . . . . . . . . . . . . . . . . . . 174
5.3.10 Numerical experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
6. SUMMARY AND CONCLUSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
LIST OF FIGURES
Figure
Page
Classical Newton’s method with F (x) = x2 + 2.5x + 10−5 . Graph of three
different iterates x0 , x1 and x2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
Semismooth Newton’s method with semismooth function f (x) defined in
(2.37). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.1
Phases and components. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
3.2
Graph of F associated to problem (3.11)-(3.12) . . . . . . . . . . . . . . . . . . . . . . . . .
59
3.3
Phases and components for the adsorption model . . . . . . . . . . . . . . . . . . . . . . .
61
3.4
Illustration of mechanism of adsorption process. Injection of CO2 in the
coalbed for CH4 displacement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
4.1
Selection diagram from [36]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
4.2
Neither αE or α−1 = αW is a function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
4.3
Graph associated to the methane hydrates problem . . . . . . . . . . . . . . . . . . . . .
94
4.4
Graphs associated to the Stefan problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
4.5
Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.0070125. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.6
Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.01465. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.7
Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.090025. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.8
Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.121425. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.9
Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) = 0.25(x+
1) at t = 0.005. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.1
2.2
4.10 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.25(x + 1) at t = 0.075. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
LIST OF FIGURES (Continued)
Figure
Page
4.11 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) = 0.25(x+
1) at t = 0.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
4.12 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.25(x + 1) at t = 0.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
4.13 Numerical solutions for u, v and Sh in for problem (4.68)–(4.69) with
v ∗ (x) = 0.5(x2 + 1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
4.14 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) = 0.5(x2 +
1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
4.15 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) = 0.5(x2 +
1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.16 Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.5(x2 + 1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.17 Numerical solution versus exact solution at time tn for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 121
4.18 Numerical solution versus exact solution at time tn for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 122
4.19 Numerical solution versus exact solution at time tn for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 123
4.20 Numerical solution versus exact solution at time tn for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 124
4.21 Numerical solution versus exact solution at time tn for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 125
4.22 Error orders in L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)) for heat equation
using 2D semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . 126
4.23 Evolution of total amount of CH4 ,u, hydrate saturation Sh and dissolved
CH4 with maximum dissolved methane v ∗ = 0.25. Example 4.5.2.2. . . . . . 128
4.24 Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
LIST OF FIGURES (Continued)
Figure
Page
4.25 Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
4.26 Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
4.27 Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
4.28 Evolution of total amount of CH4 ,u, hydrate saturation Sh and dissolved
CH4 with maximum dissolved methane v ∗ = 0.25(x + 1). Example 4.5.2.3. 133
4.29 Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
4.30 Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
4.31 Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
4.32 Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
4.33 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
4.34 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
4.35 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
4.36 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
4.37 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
4.38 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
LIST OF FIGURES (Continued)
Figure
Page
4.39 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
4.40 Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
5.1
Evolution of numerical numerical solution for problem (2.48a)-(2.48b)
with initial data u(x, 0) = H(x), ∆x = 0.04, ∆t =, T = 1 on the interval
[−1, 2]. Example 5.0.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
5.2
Log-log graph for orders of convergence for Burgers equation numerical solution with initial data H(x) compared with its rarefaction weak
solution. Example 5.0.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
5.3
Evolution of the numerical solution for Example 5.1.0.2 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1. . . . . . . . . . . . . . . . . 152
5.4
Evolution of the numerical solution for Example 5.1.0.3 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1. Initial data
u(x, 0) = H(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
5.5
Orders of convergence for single component adsorption with initial data
H(x). Example 5.1.0.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
5.6
Evolution of the numerical solution for Example 5.1.0.4 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1. Initial data
u(x, 0) = H(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
5.7
Orders of convergence for single component adsorption with initial data
H(x) − H(x − 1). Example 5.1.0.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
5.8
Evolution of numerical solutions for the system (3.32)–(3.33) with initial
data u1 (x, 0) = H(x) and u2 (x, 0) = 1 − H(x). Example 5.2.1.1. . . . . . . . . . 161
5.9
Orders of convergence for EL with u1 (x, 0) = H(x), u2 (x, 0) = 1 −
H(x), M = 1, 000, hmin = 0.001 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
5.10 Orders of convergence for IAS with u1 (x, 0) = H(x), u2 (x, 0) = 1 −
H(x), M = 1, 000, hmin = 0.001 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
LIST OF FIGURES (Continued)
Figure
Page
(0)
(0)
(0)
(0)
5.11 IAS versus extended Langmuir. Initial data u1 = H(x) and u2 =
1 − H(x). VL,1 = VL,2 = b1 = 1 and b2 = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
5.12 IAS versus extended Langmuir. Initial data u1 = H(x) and u2 =
1 − H(x). VL,1 = 3, VL,2 = 6, b1 = 1 and b2 = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . 173
5.13 Extended Langmuir isotherm a1 with parameters VL,1 = 3, VL,2 = 6, b1 =
1 and b2 = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
5.14 IAS isotherm aIAS
with parameters VL,1 = 3, VL,2 = 6, b1 = 1 and b2 = 2. 174
1
5.15 Kinetic versus equilibrium case for extended Langmuir system. Initial
data u1 (x, 0) = H(x) and u2 (x, 0) = 1 − H(x). . . . . . . . . . . . . . . . . . . . . . . . . . . 179
5.16 Kinetic system plots for different values of D1 and D2 . Include the case
D1 = D2 = 0.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
5.17 Extended Langmuir kinetics system solution plots for different values
of τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data:
u1 (x, 0) = H(x), u2 (x, 0) = 1 − H(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
5.18 Extended Langmuir kinetic system solution plots for different values
of τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data
u1 (x, 0) = H(x), u2 (x, 0) = 2(1 − H(x)). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
5.19 Extended Langmuir kinetics system solution plots for different values
of τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data
u1 (x, 0) = 2H(x), u2 (x, 0) = 1 − H(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
5.20 Extended Langmuir kinetic system solution plots for different values
of τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data
u1 (x, 0) = max(0, 1 − |2x − 1|), u2 (x, 0) = 1 − H(x). . . . . . . . . . . . . . . . . . . . . 184
LIST OF TABLES
Table
2.1
2.2
Page
Iterates xn , errors and convergence order for Newton’s method with tolerance 10−6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
Iterates xn , errors, and convergence order for semismooth Newton method
with tolerance 10−6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.1
Convergence and iteration count for (4.59) with b = 0, ϕ = 0, f = λ. . . . . . 100
4.2
Convergence and iteration count for (4.59) with b = 0, ϕ = 0, f = λ. . . . . . 101
4.3
Convergence and iteration count for (4.61)–(4.62) . . . . . . . . . . . . . . . . . . . . . . . 103
4.4
Convergence and iteration count for (4.61)–(4.62) . . . . . . . . . . . . . . . . . . . . . . . 104
4.5
Convergence and iteration count for (4.68)–(4.69). Case vc∗ (x) = 1. . . . . . . 108
4.6
Convergence and iteration count for (4.68)–(4.69). Case vc∗ = 1. . . . . . . . . . 109
4.7
Convergence and iteration count for (4.68)–(4.69). Case va∗ (x) = (1 + x)/2 110
4.8
Convergence and iteration count for (4.68)–(4.69). Case va∗ (x) = (1 + x)/2 111
4.9
Convergence and iteration count for (4.68)–(4.69). Case vn∗ (x) = (1 −
2x + x2 )/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
4.10 Convergence and iteration count for (4.68)–(4.69). Case vn∗ (x) = (1 −
2x + x2 )/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
4.11 L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)) errors for heat equation in 2D
semismooth Newton method. Example 4.5.2.1. . . . . . . . . . . . . . . . . . . . . . . . . . . 126
5.1
Errors and orders of convergence for Burgers equation in grid norms
L1 , L2 and L∞ norms using its rarefaction weak solution. Example 5.0.3.1 149
5.2
Errors and orders of convergence for adsorption problem (5.2) in grid
norms L1 and L2 norm with initial data H(x). Example 5.1.0.3. . . . . . . . . . 154
5.3
Errors and orders of convergence for Burgers equation in grid norms
L1 , L2 and L∞ norm with intial data H(x). Example 5.1.0.4. . . . . . . . . . . . 156
MATHEMATICAL TREATMENT AND SIMULATION OF
METHANE HYDRATES AND ADSORPTION MODELS
1.
INTRODUCTION
In this dissertation we develop mathematical treatments for two important applications: (i) evolution of methane in coalbeds and the associated phenomena of adsorption,
and (ii) formation of methane hydrates in seabed. We use simplified models for (i) and (ii)
since we are more interested in qualitative properties of the solutions rather than direct
applications to engineering.
For methane hydrates we focus on a scalar problem with diffusion only, and we
discuss it as a nonlinear parabolic problem in a single variable with monotone operators.
We show how the problem can be cast in the framework of a free boundary problem. The
particular nonlinearity that we deal with comes from a constraint on one of the variables.
For the simplified model of methane hydrates, we establish well–posedness of the problem
in an abstract weak setting. We also perform simulations with a novel approach based
on semismooth Newton methods. We demonstrate convergence rates of the numerical
approximation which are similar to those for Stefan free boundary value problem.
On the other hand, for adsorption problems, we focus on their structure as systems
of conservation laws, with equilibrium and non–equilibrium type nonlinearities, where the
latter are associated with microscale diffusion. We also work with an unusual type of
isotherm called Ideal Adsorbate Solution, which is defined implicitly. For IAS adsorption
system, we show sufficient conditions that render the system hyperbolic. We also construct
numerical approximations for equilibrium and non–equilibrium models.
2
In this dissertation we develop mathematical and computational aspects related to
two classes of models of methane evolution in subsurface.
Methane is a primary component of natural gas, and thus one of most significant
energy resources. It occurs naturally under the Earth’s surface in several ways. In this
dissertation we discuss methane in coalbeds and methane in methane hydrates. Methane
in coalbeds is associated with phenomena of adsorption, modeled by a system of hyperbolic
conservation laws. On the other hand, evolution of methane in methane hydrates is a free
boundary problem modeled using monotone graphs. The models for these two applications
are known, but not enough rigorous mathematical results are available for their analysis
and numerical approximation.
The mathematical models for methane evolution discussed here, are built with partial differential equations describing the conservation of mass and momentum, with constitutive relationships built from empirical models. The models are transient, and we frame
them as initial boundary value problems. The solutions to these models are not smooth
enough to warrant existence of classical solutions, therefore we define and discuss weak solutions in appropriate function spaces. Nonsmoothness of solutions is also associated with
difficulties in building and analyzing numerical algorithms for their numerical solution.
Our focus is on qualitative properties of the models rather than on quantitative
results. Therefore we use simplified models and are interested in the most significant
first order effects. We do not aim directly to produce simulations that can be used in
engineering.
The outline of this dissertation is as follows. In Chapter 2 we give the basic background to describe the frameworks in adsorption and methane hydrates models. Section
2.1 is devoted to introduce function spaces used in the particular evolution equation that
we study in this dissertation. In the same section, we define monotone operators and
subgradients, and present some results linked with the theoretical development of the
3
analytical theory in Chapter 4. In Section 2.2, we discuss part of the semismooth theory necessary to have a good understanding of the semismooth Newton method used in
the numerical analysis of Chapter 4. Section 2.3 provides the background necessary to
mathematically study our adsorption model. We define weak solutions for conservation
laws, provide examples of such solutions for linear and nonlinear fluxes and discuss the
Godunov method used in our numerical simulations for one single scalar equation and for
systems of conservation laws.
Chapter 3 is devoted to describe the models of evolution of methane hydrates and
adsorption. For the first one, we describe phases and components, the physical constraints
and the simplifications needed to develop the theory in Chapter 4. In Section 3.2, we
describe the single component adsorption model and the multicomponent case. This
serves as preparation for Chapters 4 and 5.
Chapter 4 is totally focused on the evolution problem of methane hydrates. We
pose the problem with a family of monotone graphs and write it as an abstract initial
value problem. Well posedness results are discussed. Section 4.2 deals with the numerical approximation for the evolution equation under study. We specify the fully discrete
scheme to be used and the formulation of the problem using a Nonlinear Complementarity
Constraint which allows us to apply our semismooth Newton solver. In Section 4.3, we
give examples with different graphs like Stefan problem, singular graph and the methane
hydrate problem. Section 4.4 is devoted to present some numerical results in one dimension for examples given in section 4.3. It includes rates of convergence in different norms
and graphs describing the evolution of numerical solutions.
Chapter 5 is dedicated to analysis and numerical approximations of our adsorption
model. In Section 5.1 the single component Langmuir isotherm is described and the
numerical approximation is discussed. Next, in Section 5.2, the multicomponent system
and the extended Langmuir isotherms are introduced. Numerical aspects are discussed
4
and some experiments are shown for different initial data.
Finally, in Section 5.3 we introduce the Ideal Adsorbate Solution for single component and multicomponent systems. We compare IAS approach using single component
Lagmuir isotherms with the extended Langmuir isotherms as a test case. Some experiments are performed after providing CFL conditions for the Godunov scheme applied to
these systems. In addition, we discuss the kinetic model with diffusion and compare with
numerical solutions for the system in equilibrium.
Chapter 6 includes a summary of what we have done and work in progress that was
not included in this dissertation.
5
2.
BACKGROUND
In this section we provide notation and mathematical background needed in the rest
of the dissertation.
We first develop some elements of the abstract theory of variational problems and of
finite element solutions. Next, we provide background needed for special Newton solvers
which we apply to semismooth functions. This material is needed in our discussion of
the methane hydrate model. Finally, we provide an introduction to the treatment of
conservation laws. This material is needed in our discussion of adsorption models.
2.1.
2.1.1
Weak and numerical solutions of evolution problems with monotone
graphs
Function spaces and notation
The purpose of this section is to introduce Sobolev spaces and some of their prop-
erties. We will establish the notation that we will be using through the rest of this
dissertation. Results in this section follow the material in [45].
Throughout, the symbol Ω denotes an open set in the Euclidean space Rn . If
α = (α1 , . . . , αn ) is an n–tuple of nonnegative integers, α is called a multi–index and the
length of α is
|α| =
n
X
αi .
i=1
The partial derivative operators are denoted by Di =
∂
∂xi
for 1 ≤ i ≤ n, and the
higher order derivatives by
Dα = D1α1 · · · Dnαn =
∂xα1 1
∂ |α|
.
· · · ∂xαnn
We denote by C 0 (Ω) the space of continuous functions on Ω.
(2.1)
6
Definition 2.1.1.1. Let k be a nonnegative integer, possible ∞. Then
C k (Ω) = {u| u : Ω → R, Dα u ∈ C 0 (Ω), 0 ≤ |α| ≤ k}
and C0k (Ω) := C k (Ω) ∩ {u : spt(u) compact , spt(u) ⊂ Ω}, where spt(u) is the support of
u which is the set where u does not vanish.
Definition 2.1.1.2. For 1 ≤ p ≤ ∞, Lploc (Ω) denotes the spaces of measurable functions
in Ω that are pth −power integrable on each compact subset of Ω.
Definition 2.1.1.3. Lp (Ω) is the subspace of functions that are pth −power integrable on
Ω. The norm on Lp (Ω) is given by
Z
kukp;Ω =
p
1/p
|u| dx
Ω
and in case p = ∞, it is defined as
kuk∞;Ω = ess sup |u|.
Ω
For p = 2, the space L2 (Ω) is a Hilbert space with the scalar product
Z
f (x)g(x)dx.
(f, g) =
Ω
Remark 2.1.1.1. The notation
R
u(x) dx or sometimes simply
R
udx denotes integration
with respect to Lebesgue measure. The elements of Lp (Ω) are indeed equivalent classes of
functions where two functions are said to be equivalent if they agree in Ω except on a set
of Lebesque measure zero.
Definition 2.1.1.4. Let u ∈ L1loc (Ω). For a given multi-index α, a function v ∈ L1loc (Ω)
is called the αth -weak derivative of u if
Z
Ω
ϕv dx = (−1)|α|
Z
uDα ϕdx
Ω
for all ϕ ∈ C0∞ (Ω). We write this as v = Dα u and call it generalized derivative of u.
7
Dα u is uniquely determined up to sets of Lebesgue measure zero.
Remark 2.1.1.2. Definition 2.1.1.4 extends the notation in (2.1) which originally was
defined for u ∈ C k (Ω).
Now we are ready to define Sobolev spaces.
Definition 2.1.1.5. For p ≥ 1 and k a nonnegative integer, we define the Sobolev space
W k,p (Ω) = Lp (Ω) ∩ {u : Dα u ∈ Lp (Ω), |α| ≤ k}.
The space W k,p (Ω) is equipped with a norm
1/p

Z
kukk,p;Ω = 
X
|Dα u|p dx
.
Ω |α|≤k
On the other hand, for an open set G in Rn we define C m (G) as the linear space of
restrictions to G of functions in C0m (Rn ) with scalar product
(f, g)H k (G) =
X Z
α
α
D f · D g : |α| ≤ k .
G
We denote the corresponding norm by kf kH k (G) .
Definition 2.1.1.6. We define H m (G) to be the completion of the linear space C m (G)
with the norm k · kH m (G) .
From the latter definition arises another important space.
Definition 2.1.1.7. We define H0m (G) to be the closure in H m (G) of C0∞ (G).
In fact, H0m (G) consists of functions in H m (G) which vanish on ∂G together with
their derivatives through order m − 1. For a more detailed treatment see [40].
Next, we need to define the dual of V to understand other norms used in theory of
the numerical treatment of our methane hydrate problem. First, let V and W be linear
spaces over R and consider the set L(V, W ) of linear functions from V to W.
8
Definition 2.1.1.8. Let V be a Banach space. Its dual space V 0 is the linear space of all
continuous linear functionals f : V → R.
It is known (see introduction in [38]) that V 0 is also a Banach space with norm
kf kV 0 ≡ sup {|f (x)| : kxk ≤ 1} . The dual space of H01 (Ω) is denoted by H −1 (Ω).
In what follows, we introduce other spaces involved in the theoretical development
of the methane hydrate problem in Chapter 4. More information about those spaces can
be found in [15].
Definition 2.1.1.9. The space Lp (0, T ; X) consists of all strongly measurable functions
u : [0, T ] → X with
(i) kukLp (0,T ;X) :=
R
T
0
1/p
kukp dt
< ∞ for 1 ≤ p ≤ ∞.
(ii) kukL∞ (0,T ;X) := ess sup0≤t≤T ku(t)k < ∞,
where k · k represents the norm of the Banach space X and u : [0, T ] → X is strongly
measurable if there exist simple functions sk : [0, T ] → X such that sk → f (t) a.e. for
0 ≤ t ≤ T.
Another space appearing in Chapter 4 is given in the following definition.
Definition 2.1.1.10. The space C([0, T ]; X) comprises all the continuous u : [0, T ] → X
with kukC([0,T ];X) := sup0≤t≤T ku(t)k < ∞.
Definition 2.1.1.11. The Sobolev space W 1,p (0, T ; X) consists of all functions u ∈ Lp (0, T ; X)
such that u0 = D1 u belongs to Lp (0, T ; X). Furthermore,

R
1/p

T
p

0 (t)kp dt
kukW (0,T ;X) :=
ku(t)k
+
ku
,
0
1,p


ess sup0≤t≤T (ku(t)k + ku0 (t)k)
(p = ∞).
(1 ≤ p ≤ ∞)
9
2.1.2
Monotone operators
In this section we introduce necessary background to analyse the methane hydrate
problem. Analysis of the evolution equation representing this model and numerics are
done in Chapter 4.
Let B be a set. We denote by hx, yi an element of B × B with x ∈ B, y ∈ B.
Definition 2.1.2.1. Let B be a Banach space. An operator A on B is a relation A ⊂ B×B
with A(x) = {y ∈ B : hx, yi ∈ A} at each x ∈ Dom(A).
Definition 2.1.2.2. Let B be a Banach space. The operator A is accretive on B if for
each hx1 , y1 i, hx2 , y2 i ∈ A and > 0 we have kx1 − x2 k ≤ kx1 + y1 − (x2 + y2 )k.
Let H be a Hilbert space with scalar product (·, ·)H . We say that A is accretive if
(y1 − y2 , x1 − x2 )H ≥ 0 for hx1 , y1 i, hx2 , y2 i ∈ A.
Definition 2.1.2.3. The operator A is m-accretive on B if it is accretive and if
Rg(A + I) = B.
We denote the extended real numbers by R∞ = R ∪ {+∞}.
Definition 2.1.2.4. Let H be a Hilbert space and Ψ : H → R∞ be proper, convex, and
lower-semicontinuous. Then f ∈ H is a subgradient of Ψ at u ∈ H if
(f, w − u)H ≤ Ψ(w) − Ψ(u),
w ∈ H.
We denote by ∂H Ψ(u) any subgradient of Ψ at u.
From now on, we will write ∂Ψ instead of ∂H Ψ to avoid unnecessary subscripts.
It can be shown that, for any Ψ as in Definition 2.1.2.4, the multivalued operator
∂Ψ : H → H is m-accretive. We need the following proposition to prove this assertion, see
[40] Prop. 1.4.
10
Proposition 2.1.2.1. Let K be a closed convex and non-empty set in H, and let Φ : H →
R∞ be convex and lower semi-continuous. If
lim
x∈K, kuk→∞
Φ(u) = ∞,
then there exists a minimum point u0 ∈ K such that
Φ(u0 ) ≤ Φ(z), for all z ∈ K.
Proposition 2.1.2.2. Let Ψ : H → R∞ be given as in Definition 2.1.2.4. Then Ψ is
m-accretive.
Proof. To show that ∂Ψ is accretive, let fj ∈ ∂Ψ(uj ) for j = 1, 2. Then,
(f1 , u2 − u1 )H
≤ Ψ(u2 ) − Ψ(u1 ),
(2.2)
(f2 , u1 − u2 )H
≤ Ψ(u1 ) − Ψ(u2 ).
(2.3)
Adding inequalities (2.2) and (2.3),
(f1 , u2 − u1 )H + (f2 , u1 − u2 )H = (−f1 + f2 , u1 − u2 )H ≤ 0
So (f1 − f2 , u1 − u2 )H ≥ 0 and ∂Ψ is accretive.
It is left to see that Rg(I + ∂Ψ) = H to show that ∂Ψ is m-accretive. We let f0 ∈ H
and define Φ : H → R by
1
Φ(u) = Ψ(u) + kuk2 − (f0 , u)H ,
2
u ∈ H.
Since Ψ is convex, thus Φ has an affine lower bound. Therefore, limkuk→∞ Φ(u) = ∞.
Proposition 2.1.2.1 applies and Φ attains a minimum value at some point u0 ∈ H, i.e. 0 ∈
∂Ψ, which means
(f0 , ũ − u0 )H
1
2
2
≤ Ψ(ũ) − Ψ(u0 ) +
kũk − ku0 k , for all ũ ∈ H.
2
11
Now take ũ = tu + (1 − t)u0 , and use the convexity of Ψ:
(f0 , t(u − u0 ))H
1
kt(u − u0 ) + u0 k2 − ku0 k2
2
t2
≤ t(Ψ(u) − Ψ(u0 )) + t(u0 , u − u0 )H + ku − u0 k2 .
2
≤ t(Ψ(u) − Ψ(u0 )) +
(2.4)
Dividing both sides of inequality of (2.4) by t > 0, we obtain
t
(f0 , u − u0 )H ≤ Ψ(u) − Ψ(u0 ) + (u0 , u − u0 )H + ku − u0 k2 .
2
Letting t → 0+ ,
(f0 , u − u0 )H ≤ Ψ(u) − Ψ(u0 ) + (u0 , u − u0 )H
or,
(f0 − u0 , u − u0 )H ≤ Ψ(u) − Ψ(u0 ).
(2.5)
That is, f0 − u0 ∈ ∂Ψ(u0 ).
We need the following definitions and results when proving solvability in the context
of the methane hydrate problem.
Definition 2.1.2.5. The epigraph of ϕ : V → R∞ is given by
epi(ϕ) ≡ {(u, a) ∈ V × R : ϕ(u) ≤ a}.
Proposition 2.1.2.3. ([40] Prop II.7.7) Let ϕ1 and ϕ2 be convex functions and suppose
there is a point in dom(ϕ1 ) ∩ dom(ϕ2 ) at which ϕ1 is continuous. Then
∂(ϕ1 + ϕ2 ) = ∂ϕ1 + ∂ϕ2 .
Proof. Let f ∈ ∂ϕ1 (u) and g ∈ ∂ϕ2 (u), (so f + g ∈ ∂ϕ1 + ∂ϕ2 ) then
(f, w − u) ≤ ϕ1 (w) − ϕ1 (u),
(2.6)
12
and,
(g, w − u) ≤ ϕ2 (w) − ϕ2 (u),
(2.7)
for all w ∈ V. Adding (2.6) and (2.7) we get
(f + g, w − u) ≤ (ϕ1 + ϕ2 )(w) − (ϕ1 + ϕ2 )(u).
That is f + g ∈ (ϕ1 + ϕ2 )(u). We have just proved the assertion
∂ϕ1 + ∂ϕ2 ⊂ ∂(ϕ1 + ϕ2 ).
Now, let f ∈ ∂(ϕ1 + ϕ2 )(u), which means,
(f, w − u) ≤ (ϕ1 + ϕ2 )(w) − (ϕ1 + ϕ2 )(u),
for all w ∈ V. That is,
ϕ1 (w) − ϕ1 (u) − f (w − u) ≥ ϕ2 (u) − ϕ2 (w),
w ∈ V.
Therefore, the two sets
E := {(w, t) ∈ V × R : ϕ1 (w) − ϕ1 (u) − f (w − u) ≤ t}
and
F := {(w, t) ∈ V × R : ϕ2 (u) − ϕ2 (w) ≥ t}
can only have boundary points in common. By hypothesis, ϕ1 is somewhere continuous
so E is the epigraph of a function somewhere continuous, so E is convex with non-empty
interior. F is convex, since it is the reflection of epi(ϕ2 ). Therefore, there is a closed
non-vertical hyperplane that separates E and F. Hence, there exist g ∈ V 0 and c ∈ R such
that
ϕ1 (w) − ϕ1 (u) − f (w − u) ≥ −g(w) + c ≥ ϕ2 (u) − ϕ2 (w),
w ∈ V.
(2.8)
Taking w = u gives 0 = f (0) ≤ g(w) − c ≤ 0, or c = g(u). Putting this back into (2.8) we
obtain
ϕ1 (w) − ϕ1 (u) ≥ −g(w) + g(u) + f (w) − f (u),
13
so we have
f − g ≡ h ∈ ∂ϕ1 (u).
(2.9)
On the other hand, from the second inequality of (2.8) we get,
g ∈ ∂ϕ2 (u).
(2.10)
From (2.9) and (2.10), we have f = h + g with h ∈ ∂ϕ1 (u) and g ∈ ∂ϕ2 (u), which shows
the inclusion we needed.
2.1.3
Finite element solution to evolution equations
In this section we discuss basic aspects of the finite element method as a preparation
for numerical algorithms for evolution equations with monotone graphs. First, we discuss
a stationary problem and next we discuss a transient problem.
2.1.3.1
Stationary variational problem
We follow [3, 7] and [42] for this exposition.
Consider the following abstract variational problem posed in a Hilbert space
V = H01 (Ω), where Ω is an open, bounded, convex polygonal domain, Ω ⊂ Rd .
Find u ∈ V such that for all v ∈ V, a(u, v) = f (v).
(2.11)
Here, a(·, ·) is a bilinear form and f is a continuous linear functional on V , i.e. f ∈ V 0 .
Recall that a bilinear form, a(·, ·), on a linear space V is a mapping a : V × V → R such
that each of the maps v 7→ a(v, w) and w 7→ a(v, w) is linear on V. If V is a normed linear
space, a(·, ·) is said to be bounded (or continuous) if there is c1 > 0 such that |a(v, w)| ≤
c1 kvkV kwkV for all v, w ∈ V . In addition, a(·, ·) is symmetric if a(v, w) = a(w, v) for all
v, w ∈ V , and is V −elliptic (or coercive) if there is c2 > 0 such that a(v, v) ≥ c2 kvk2V for
all v ∈ V.
The following Theorem establishes the well-posedness of (2.11).
14
Theorem 2.1.3.1 (Lax-Milgram Lemma, [7], Theorem 1.1.3). Let V be a Hilbert space,
let a(·, ·) : V × V → R be a continuous V −elliptic bilinear form, and let f : V → R be
a continuous linear form. Then the abstract variational problem (2.11) has one and only
one solution.
The Galerkin method for the problem (2.11) consists in defining a finite dimensional
counterpart of (2.11) over a finite dimensional subspace Vh of V . We have the discrete
problem:
Find uh ∈ Vh such that for all vh ∈ Vh , a(uh , vh ) = f (vh ).
(2.12)
The construction of a subspace Vh in finite element method is as follows.
1. The triangulation Th , a union of triangular or rectangular elements covering Ω is
established over the set Ω in such a way that the following properties hold:
(a) Ω = ∪K∈Th K.
(b) For each K ∈ Th , the set K is closed and the interior K ◦ is non-empty.
(c) For each distinct K1 , K2 ∈ Th , one has K1◦ ∩ K2◦ = ∅.
(d) For each K ∈ Th , the boundary ∂K is Lipschitz-continuous.
Define the parameter h = maxK∈K hK , where hK = diam(K).
2. The spaces PK , K ∈ Th consists of linear polynomials over K.
3. We define Vh as the space of functions whose restrictions to the elements K are in
PK , so that space Vh contains piecewise polynomials defined on the triangulation
Th . Now in order to have Vh ⊂ H 1 (Ω) we must impose that the functions vh ∈ Vh
are globally continuous over Ω, so that Vh ⊂ C 0 (Ω). See [7] for more details.
We present an example in one dimension for illustration. Let Ω = (0, 1) and 0 =
x0 < x1 < · · · < xM = 1 be a partition of [0, 1]. Set hj = xj − xj−1 , Kj = [xj−1 , xj ], for
j = 1, . . . , M, and h = maxj hj . Let Vh be the linear space of functions v such that
15
(i) v ∈ C 0 ([0, 1]),
(ii) v|[xi−1 ,xi ] is a linear polynomial, i = 1, . . . , M, and
(iii) v(0) = 0, v(1) = 0.
Let {φi : 1 ≤ i ≤ M − 1} be functions with properties (i) − (iii) with the additional
requirement that φi (xj ) = δij , where δij is the Kronecker delta. It can be shown that
{φi : 1 ≤ i ≤ M − 1} is a basis for Vh .
Now consider the problem
−uxx = f in Ω = (0, 1), with u(0) = u(1) = 0,
(2.13)
and f ∈ L2 (Ω). The variational form is (2.11) with
Z
1
ux vx dx,
a(u, v) =
0
and V = H01 (0, 1).
One can easily show that this form is V −continuous and elliptic. Indeed,
|a(u, v)| ≤ kux kL2 (Ω) kvx kL2 (Ω)
≤ kukV kvkV ,
for all u and v on V . On the other hand, a(·, ·) is V −elliptic since
a(v, v) = kvx k2L2 (Ω) = kvk2V ,
is a consequence of the Poincaré–Friedrichs inequality. The coercivity constant is 1.
We seek an approximation uh to the solution u of the problem (2.12) in the finitedimensional space Vh . We can write any vh ∈ Vh as
vh (x) =
M
−1
X
i=1
vi φi (x),
16
with vi = vh (xi ). We pose the finite-dimensional problem (2.12). In terms of the basis
{φi : 1 ≤ i ≤ M − 1} we write
uh (x) =
M
−1
X
Uj φj (x)
j=1
and substitute into (2.12) to get that
M
−1
X
Uj a(φj , φi ) = (f, φi ), for i = 1, . . . , M − 1.
(2.14)
j=1
The system (2.14) of equations can be expressed in matrix form as
Ah U = b,
(2.15)
−1
where U = (Ui )M
i=1 , Ah = (aij )1≤i,j≤M −1 is the stiffness matrix with elements aij =
−1
a(φi , φj ), and b = (bi )M
i=1 the load vector with elements bi = (f, φi ) with i = 1, . . . , M − 1.
The positive definiteness of the matrix Ah is equivalent to the coercivity of the
bilinear form a(·, ·).
It follows from Theorem 2.1.3.1 that (2.15), and therefore (2.12) has a unique solution, the finite element solution of (2.13).
2.1.3.2
Finite element for evolution equations
Now consider a time dependent problem
ut − uxx = f, x ∈ (0, 1), t > 0
(2.16)
u(x, 0) = u0 (x),
(2.17)
u(0, t) = u(1, t) = 0,
(2.18)
where f ∈ L2 (0, 1) and u0 ∈ V. The variational form is
(ut , v) + a(u, v) = (f, v), ∀v ∈ V,
(2.19)
(u(0), v) = (u0 , v).
(2.20)
17
Here u : [0, T ] → V and u0 is the given initial condition.
The fully discrete numerical approximation to (2.19)-(2.20) extends (2.15). Let
0 = t1 < t2 < · · · < tN = T be a uniform partition of [0, T ] and define the uniform time
step as k = tn − tn−1 . We seek U n , n = 1, 2, . . . , N as solutions to
Mh U n + kAh U n = Mh U n−1 + kbn ,
Mh u0 = Mh u0 ,
which is an implicit backward Euler discretization in time of the finite element approximation to (2.19)-(2.20).
It is known that (see [42]) if solutions are smooth, then the numerical approximation
Uhn converges to the analytical solution in L∞ (0, T ; L2 (Ω)) with O(∆t + h2 ) rate. For
example, this rate applies to the solution to the heat equation (2.16)-(2.18).
Next, we consider a nonlinear extension of (2.16)-(2.18)
β(u)t − uxx = f, x ∈ (0, 1), t > 0,
u(x, 0) = u0 (x)
u(0, t) = u(1, t) = 0.
The finite element solution to this problem can be defined analogously to that for the
linear problem and it requires that we solve
Mh β(U n ) + kAh U n = Mh β(U n−1 ) + kbn .
(2.21)
Algebraic solution to (2.21) requires implementation of a nonlinear solver such as Newton
solver discussed in Section 2.3.3.
However, as long as β is a smooth monotone strictly increasing function with a
bounded derivative, the convergence rate remains similar to that for the linear problem
(2.16)-(2.18).
18
For evolution problems with a graph, theory of standard Newton method does not
work. We need its extension to so-called semismooth methods discussed in Section 2.3.
The situation is very different if β is a monotone graph or a degenerate function as
in Porous Medium Equation. In such cases, suboptimal convergence is to be expected.
We comment on those cases later in Section 4.4.
2.2.
Classical Newton Method
In this section we introduce the classical Newton method. Later, in section 2.3., we
introduce a method for a bigger class of functions, the semismooth Newton method.
We follow here the exposition given by [22]. We first introduce basic notation and
then we introduce Newton’s method and its assumptions for such method to converge.
Let F : RN → RN and denote the ith component of F by fi . If the components fi
of F are differentiable at x ∈ RN we define the Jacobian matrix F 0 (x) by
F 0 (x)ij =
∂fi
(x).
∂xj
We seek to solve the problem
F (x) = 0.
(2.22)
We express the fundamental theorem of calculus in a convenient way for us.
Theorem 2.2.0.2 (Theorem 4.0.1,[22]). Let F be differentiable in a open set Ω ⊂ RN
and let x∗ ∈ Ω. Then for all x ∈ Ω sufficiently near x∗
∗
Z
F (x) − F (x ) =
1
F 0 (x∗ + t(x − x∗ ))(x − x∗ )dt.
0
Now we introduce the different types of convergence
Definition 2.2.0.1. Let {xn } ⊂ RN and x∗ ∈ RN . Then
19
• xn → x∗ q-quadratically if xn → x∗ and there is K > 0 such that
kxn+1 − x∗ k ≤ Kkxn − x∗ k2 .
• xn → x∗ q-superlinearly with q-order α > 1, if xn → x∗ and there is a K > 0 such
that
kxn+1 − x∗ k ≤ Kkxn − x∗ kα .
• xn → x∗ q-superlinearly if
kxn+1 − x∗ k
= 0.
n→∞ kxn − x∗ k
lim
• xn → q-linearly with q-factor σ ∈ (0, 1) if
kxn+1 − x∗ k ≤ σkxn − x∗ k
for n sufficiently large.
Definition 2.2.0.2. An iterative method for computing x∗ is said to be locally (q-quadratically,
q-superlinearly, q-linearly, etc.) convergent if iterates converges to x∗ (q-quadratically, qsuperlinearly, q-linearly, etc.) given that the initial data for the iteration is sufficiently
good.
Remark 2.2.0.1. A q-superlinearly convergent sequence is also q-linearly convergent with
q-factor σ for any σ > 0. A q-quadratically convergent sequence is q-superlinearly convergent with q-order 2.
Definition 2.2.0.3. Let Ω ⊂ RN and let G : Ω → RN . We say that G is Lipschitz
continuous on Ω with Lipschitz constant γ if
kG(x) − G(y)k ≤ γkx − yk
for all x, y ∈ Ω.
20
We need to make the following assumptions
Assumptions 2.2.0.1.
1. Equation (2.22) has a solution x∗ .
2. F 0 : Ω → RN ×N is Lipschitz continuous with Lipschitz contant γ.
3. F 0 (x∗ ) is nonsingular.
We denote by B(r) the ball with center x∗ and radius r. The differences e = x − x∗
and en = xn − x∗ . The following lemma is used to show the q-quadratic convergence of
Newton’s method.
Lemma 2.2.0.1 (Lemma 4.3.1, [22], p.69). Assume that Assumptions 2.2.0.1 hold. Then
there is δ > 0 so that for all x ∈ B(δ)
kF 0 (x)k ≤ 2kF 0 (x∗ )k,
kF 0 (x)
−1
k ≤ 2kF 0 (x∗ )
−1
k,
(2.23)
and
kF 0 (x∗ )
−1 −1
k
kek/2 ≤ kF (x)k ≤ 2kF 0 (x∗ )kkek.
Now we are heading to prove the local convergence of Newton’s method. We can
write the Newton’s method as
x+ = xc − F 0 (xc )
−1
F (xc ),
(2.24)
where xc represents the current iterate and x+ is the new iterate. So (2.24) describes the
transition from the xc to x+ .
If we see x+ as the root of the two term Taylor polynomial about xc , we have another
interpretation
Mc (x) = F (xc ) + F 0 (xc )(x − xc ).
Now, we are ready to prove the convergence of Newton’s method. We have completed many of the details in the proof of the Theorem 2.2.0.3.
21
Theorem 2.2.0.3 (Theorem 5.1.1, [22], p. 71). Let Assumptions 2.2.0.1 hold. Then there
are K > 0 and δ > 0 such that if xc ∈ B(δ) the Newton iterate from xc given by (2.24)
satisfies
ke+ k ≤ Kkec k2 .
(2.25)
Proof. First, observe that subtracting x∗ from both sides of equation (2.24) we obtain
e+ = ec − F 0 (xc )
−1
F (xc ).
(2.26)
Choose δ > 0 small enough so Lemma 2.2.0.1 applies. Now, by Theorem 2.2.0.2 we have
Z 1
∗
F (xc ) = F (x ) +
F 0 (x∗ + tec )ec dt.
(2.27)
0
Replacing F (xc ) in equation(2.26) by the expression in (2.27) and taking into account
that F (x∗ ) = 0, we obtain
−1
0
Z
1
e+ = ec − F (xc )
F 0 (x∗ + tec )ec dt
0
Z 1
−1
0
0
0 ∗
= F (xc )
ec F (xc ) −
F (x + tec )ec dt
0
Z 1
−1
0
0
0 ∗
= F (xc )
ec F (xc ) − F (x + tec )ec dt
0
Z 1
−1
0
= F (xc ) ec
F 0 (xc ) − F 0 (x∗ + tec )dt.
(2.28)
0
Therefore,
0
ke+ k ≤ kF (xc )
−1
Z
1
kkec k
kF 0 (xc ) − F 0 (x∗ + tec )kdt
0
0
−1
≤ kF (xc )
−1
Z
1
kkec k
(1 − t)dt γkec k
= kF 0 (xc )
kγkec k /2
≤ (2kF 0 (x∗ )
−1
= γkF 0 (x∗ )
−1
(2.29)
0
2
k)γkec k2 /2
kkec k2 ,
(2.30)
(2.31)
where the inequality in (2.29) is coming from the assumption that F 0 is Lipschitz. The
inequality in (2.30) is due to the relation in (2.23) from Lemma 2.2.0.1.
22
Taking K = γkF 0 (x∗ )−1 k in (2.31) we finally have (2.25).
Now, if we impose Kδ < 1 then we have that xn → x∗ q-quadratically as the
following Theorem summarizes. The short proof of this Theorem can be found in [22].
Theorem 2.2.0.4 (Theorem 5.1.2, [22], p.71). Let Assumptions 2.2.0.1 hold. Then there
is δ such that if x0 ∈ B(δ) the Newton iteration
−1
xn+1 = xn − F 0 (xn )
F (xn )
converges q-quadratically to x∗ .
The Newton algorithm is described in Algorithm 2.2.0.1. We let τ = (τr , τa ) ∈ R2
be a vector of termination tolerances.
Algorithm 2.2.0.1.
Step 1. r0 = kF (x)k
Step 2. Do while kF (x)k > τr r0 + τa
Step 2.1. Compute F 0 (x)
Step 2.2. Solve F 0 (x)s = −F (x)
Step 2.3. x = x + s
Step 2.4. Evaluate F (x)
Example 2.2.0.1. This example illustrates the outcome of an implementation in one
dimension of the Newton’s method. Consider the function F (x) = x2 + 2.5x + 10−5 . One
of the roots of F is x∗ = −0.00000400000640010667. Figure 2.1 shows geometrically how
the classical Newton’s method works.
Outputs from Table 2.1 conforms that Newton has quadratic order of convergence.
23
y
x2 x1
x0
x
FIGURE 2.1: Classical Newton’s method with F (x) = x2 + 2.5x + 10−5 . Graph of three
different iterates x0 , x1 and x2 .
tol= 10−6
n
xn
|x − x∗ |
αn
0
1.5
1.5000040000064
−
1
0.4090890909
0.4090930909154
−
2
0.05043245876
0.0504364587627
1.6111
3
0.000974073232
0.0009780732384
1.8836
4
−0.000003617653448
0.00000038235295
1.9902
TABLE 2.1: Iterates xn , errors and convergence order for Newton’s method with tolerance
10−6
24
2.3.
Introduction to Nonsmooth Analysis
This section is devoted to finite-dimensional semismoothness. We develop the basics
of the theory and illustrate it with several examples. We also introduce semismooth
Newton’s methods. We follow [44] and [8], and provide details of some proofs sketched in
these references.
Let V ⊂ Rn , and f : V → Rm . We denote by Df the set
Df := {x ∈ V : f admits a (Fréchet-)derivative f 0 (x) ∈ Rm×n }.
Suppose that f is Lipschitz continuous near x ∈ V. That is, there exists an open
neighbourhood around x, V (x) ⊂ V on which f is Lipschitz continuous. Recall that by
Radamacher’s theorem, V (x)\Df has Lebesgue measure zero.
Definition 2.3.0.4. Let V ⊂ Rn be open and f : V → Rn be Lipschitz continuous near
x ∈ V. The set
∂B f (x) := {M ∈ Rm×n | there exists (xk ) ⊂ Df : xk → x, f 0 (xk ) → M }
is called the B−subdifferential of f at x.
Definition 2.3.0.5. Let f be as in Definition 2.3.0.4 and x ∈ V. The convex hull ∂f (x) :=
co(∂B f (x)) is the Clarke’s generalized Jacobian of f at x.
In this definition, the convex hull co(X) of a set X is the smallest convex set that
contains X.
The following example is closely related to our application.
Example 2.3.0.2. Consider α(u) = (u − v ∗ )− + v ∗ , where v ∗ is a fixed constant and
(z)− := min(z, 0). Then, ∂B α(0) = {0, 1} and ∂α(0) = [0, 1].
25
Next we define ∂C f (x) := ∂f1 (x)×· · ·×∂fm (x) the Q0i s C− subdifferential (notation
used in [44]). Even though we are not using explicitly ∂C f in this dissertation, we mention
it, since is part of having a better understanding of the semismooth theory.
We develop the following example to illustrate the previous definitions. At the same
time, this simple example is used to illustrate the semismooth Newton’s method.
Example 2.3.0.3. Consider H : R2 → R2 defined by


 f1 (v, S) 
H(v, S) = 
,
f2 (v, S)
where f1 (v, S) = Sv+10(1−S)−c, c is a fixed constant and f2 (v, S) = min(1−v, 1−S). We
find ∂C H(v, S). We have that ∂f1 (v, S) is given by the row vector ∂f1 (v, S) = (S, v − 10)
and ∂B f2 (v, S) is the set the two row vectors {(−1, 0), (0, −1)}. Therefore,
∂f2 (v, S) = co(∂B f2 (v, S))
= {α(−1, 0) + (1 − α)(0, −1) : 0 ≤ α ≤ 1}
= {(−α, α − 1) : 0 ≤ α ≤ 1} .
We obtain for ∂C H(v, S) a set of 2 × 2 matrices.
∂C H(v, S) = {(S, v − 10) × (α, α − 1) : 0 ≤ α ≤ 1}







 S v − 10 
=

:0≤α≤1 .


 −α α − 1

(2.32)
In this example, it is worth observing that ∂B H(v, S) is the set of two 2 × 2 matrices

 





 S v − 10   S v − 10 
∂B H(v, S) = 
, 
 .


 −1

0
0
−1
Hence,
∂H(v, S) = co(∂B H(v, S))
is exactly given by matrix in (2.32).
26
Definition 2.3.0.6. Let X and Y be Banach Spaces and G : X → P(Y ). We say that
G is upper semicontinuous at x if it has the following property: For all > 0, there exist
δ > 0 such that
G(x0 ) ⊂ G(x) + BY
for all x0 ∈ x + δBX ,
where BX and BY represent the unit balls in the spaces X and Y respectively.
The generalized subdifferentials ∂B f, ∂f and ∂C f have the following properties:
Proposition 2.3.0.1. Let V ⊂ Rn be open and f : V → Rm be locally Lipschitz continuous. Then for x ∈ V we have
(a) ∂B f (x) and ∂C f (x) are nonempty, compact, and convex.
(b) ∂B f, ∂f, and ∂C f are locally bounded and semicontinuous.
(c) ∂B f (x) ⊂ ∂f (x) ⊂ ∂C f (x).
(d) If f is continuously differentiable in a neighbourhood of x, then
∂C f (x) = ∂f (x) = ∂B f (x) = {f 0 (x)}.
See proof in [44].
The next proposition shows the chain rule for semismooth functions.
Proposition 2.3.0.2. ([44],Proposition 2.3, p.26) Let V ⊂ Rn and W ⊂ Rl be nonempty
open sets, g : V → W be Lipschitz continuous near x ∈ V, and h : W → Rm be Lipschitz
continuous near g(x). Then, f = h ◦ g is Lipschitz continuous near x and for all v ∈ Rn ,
it holds that
∂f (x)v ⊂ co(∂h(g(x))∂g(x)v) = co{Mh Mg v : Mh ∈ ∂h(g(x)), Mg ∈ ∂g(x)}.
If in addition h(·) is continuously differentiable near the point g(x) then
∂f (x)v = h0 (g(x))∂g(x)v
for all v ∈ Rn .
27
If f (·) is real-valued then the argument v can be omitted.
Corollary 2.3.0.1. Let V ⊂ Rn be open and f : V → Rm be Lipschitz continuous near
x ∈ V. Then
∂fi (x) = eTi ∂f (x) = {Mi : Mi is the ith row of some M ∈ ∂f (x)}.
where ei is the ith unit vector in Rn . That is, eTi = (0, · · · , |{z}
1 , · · · , 0).
i
Proof. Taking h(y) = eTi y = yi and g = f and apply the chain rule in Proposition 2.3.0.2.
Then,
∂fi (x) = h0 (f (x))∂f (x) = eTi ∂f (x) = {Mi : Mi is the ith row of some M ∈ ∂f (x)}.
2.3.1
Semismoothness
Definition 2.3.1.1. Let V ⊂ Rn be nonempty and open. The function f : V → Rm is
semismooth at x ∈ V if it is Lipschitz continuous near x and if
lim
M ∈∂f (x+τ d)
d→s, τ →0+
Md
(2.33)
exists for all s ∈ Rn . If f is semismooth for all x ∈ V, we say that f is semismooth.


 x1 
Example 2.3.1.1. Consider f (x) = min(x), where x is the column vector x = 
.
x2




 s1 
 d1 
Let s and d be the column vectors s = 
 and d = 
 . Then,
s2
d2
∂f (x + τ d) = {(α, 1 − α) : 0 ≤ α ≤ 1} .
Since




 d1 
 d1 
∂f (x + τ d) 
 = {(α, 1 − α) : 0 ≤ α ≤ 1} 

d2
d2
= αd1 + (1 − α)d2 → αs1 + (1 − α)s2
28
as τ → 0 and d → s. So the limit in (2.33) exists for every s ∈ Rn .
We have other characterizations for semismoothess. We will give the following
definition first.
Definition 2.3.1.2 ([44]). Let f : V → Rm be defined on the open set V.
(a) f is directionally differentiable at x in V if the directional derivative
f 0 (x, s) := lim
τ →0+
f (x + τ s) − f (x)
τ
exists for all s ∈ Rn .
(b) f is B−differentiable at x ∈ V if f is directionallly differentiable at x and
f (x + s) − f (x) − f 0 (x, s) = o (ksk) as s → 0
(c) f is α−order B−differentiable at x ∈ V, 0 < α ≤ 1, if f is directionally differentiable
at x and
f (x + s) − f (x) − f 0 (x, s) = O ksk1+α as s → 0.
Now, we give alternative characterizations for semismoothness.
Proposition 2.3.1.1 ([44], p. 29). Let V be an open set in Rn and consider f : V → Rm .
Then for x ∈ V the following statements are equivalent:
(a) f is semismooth at x.
(b) f is Lipschitz continuous near x, f 0 (x, ·) exists, and
sup
M s − f 0 (x, s) = o(ksk) as s → 0.
M ∈∂f (x+s)
(c) f is Lipschitz continuous near x, f 0 (x, ·) exists, and
sup
M ∈∂f (x+s)
kf (x + s) − f (x) − M sk = o(ksk) as s → 0.
(2.34)
29
As a direct consequence of Proposition 2.3.1.1 we have.
Proposition 2.3.1.2. Let V ⊂ Rn be open. If f : V → Rn is continuously differentiable
in a neighborhood of x ∈ V, then f is semismooth at x and ∂f (x) = ∂B f (x) = {f 0 (x)}.
The class of semismooth functions is closed under composition as shown in the next
result.
Proposition 2.3.1.3. Let V ⊂ Rn and W ⊂ Rl be open sets. Let g : V → W be
semismooth at x ∈ V and h : W → Rm be semismooth at g(x) with g(V ) ⊂ W. Then the
composite map f = h ◦ g : V → Rm is semismooth at x and
f 0 (x, ·) = h0 (g(x))g 0 (x, ·)
The following proposition assures that f is semismooth if its component functions
are semismooth. The converse is true as well.
Proposition 2.3.1.4. Let V ⊂ Rn open. The function f : V → Rm is semismooth at
x ∈ V if and only if its component functions are semismooth at x.
Proof. (Proof from [44]). Note first that if f is semismooth at x, then the function fi are
Lipschitz continuous and directionally differentiable at x. By Corollary 2.3.0.1 we have
sup
|fi (x + s) − fi (x) − vs| =
sup
M ∈∂f (x+s)
v∈∂fi (x+s)
T
ei (f (x + s) − f (x) − M s)
= o(ksk) as s → 0,
so fi is semismooth at x. For the converse, note that from Proposition 2.3.0.1 we have
∂f (x) ⊂ ∂C f (x). This completes our proof.
2.3.2
Piecewise differentiable functions
We devote a whole section to piecewise differentiable functions since they are an
important subclass of semismooth functions.
30
Definition 2.3.2.1. A function f : V → Rm defined on the open set V ⊂ Rn is called
P C k -function (1 ≤ k ≤ ∞), if f is continuous and if at every point x0 ∈ V there exists
a neighbourhood W ⊂ V of x0 and a finite collection of C k -functions f i : W → Rm , i =
1, . . . , N, such that
f (x) ∈ {f 1 (x), . . . , f N (x)} for all x ∈ V.
We introduce some notation before describing some properties for the P C k -functions:
• In Definition 2.3.2.1, we say that f is a continuous selection of {f 1 , . . . , f N } in W.
• The set I(x) := {i : f (x) = f i (x)} is an active set index at x ∈ W.
• We say that I e (x) = {i ∈ I(x) : x ∈ int{y ∈ W : f (y) = f i (x)} is the essentially
active set at x.
There are some natural results for P C k -functions”.
Proposition 2.3.2.1. The class of P C k -functions is closed under composition, finite
summation, and multiplication.
As a consequence of Proposition 2.3.2.1 we have a list of examples that we will use
in the next section.
1. The absolute value function: f (t) = |t|, t ∈ R is a P C ∞ −function.
2. The max function f (u, v) = max(u, v), (u, v) ∈ R2 is a P C ∞ −function.
3. From the two previous examples we have that the projection P[a,b] (t) = max{a, min{t, b}}
is in P C ∞ . Therefore, M CP -functions Φ[a,b] are P C k as well. See Section 4.3.3 for
definition of M CP -functions and its application to the Stefan problem.
We don’t provide proofs of the next three propositions, they are proved in [44].
31
Proposition 2.3.2.2. Every P C 1 −function f : V → Rm , on an open set V ⊂ Rn , is
locally Lipschitz continuous.
Proposition 2.3.2.3. Let the P C 1 −function f : V → Rm , V ⊂ Rn open, be a continuous
selection of the C 1 −functions {f 1 , . . . , f N } in a neighbourhood W of x ∈ V. Then f is
B-differentiable at x and, for all y ∈ Rn ,
0
f 0 (x)y ∈ {(f i ) (x)y : i ∈ I e (x)}.
Further, if f is differentiable at x, then
0
f 0 (x) ∈ {(f i ) (x) : i ∈ I e (x)}.
Proposition 2.3.2.4. Let the P C 1 −function f : V → Rm , V ⊂ Rn open, be a continuous
selection of the C 1 −functions {f1 , . . . , f N } in a neighbourhood W of x ∈ V. Then
0
∂B f (x) = {(f i ) (x) : i ∈ I e (x)},
0
∂f (x) = co{(f i ) (x) : i ∈ I e (x)}.
Now, we are ready to establish the semismoothness of P C k -functions.
The general sketch of the proof of the following proposition is in [44]. Extra details
were worked out by us.
Proposition 2.3.2.5. Let f : V → Rm be a P C 1 −function on the open set V ⊂ Rn . Then
f is semismooth. If f is a P C 2 − function, then f is 1-order semismooth. (See definition
of α-order semismooth in Definition 2.3.3.1)
Proof. By Propositions 2.3.2.2 and 2.3.2.3 we have that f is Lipschitz continuous and Bdifferentiable. Let x ∈ V, then, by definition, there is a neighbourhood W ⊂ V of x such
32
that f is a continuous selection of {f 1 , . . . , f N }. We can assume that all f i are active at x,
for all i = 1, . . . , N. Now, by Proposition 2.3.2.4, for all x + s ∈ W and all M ∈ ∂f (x + s)
we have that
M=
0
X
λi (f i ) (x + s),
λi ≥ 0,
X
i∈I e (x+s)
λi = 1.
i
Note that f (x + s) = f i (x + s) for all i ∈ I e (x + s),
X
i
i 0
i
λi (f ) (x + s)s
kf (x + s) − f (x) − M sk = f (x + s) − f (x) −
i∈I e (x+s)
X
0
≤
λi kf i (x + s) − f i (x) − (f i ) (x + s)sk
i∈I e (x+s)
=
X
i∈I e (x+s)
Z
λi x
Z
≤
max
i∈I e (x+s) 0
x+s
(f ) (t)dt − (f ) (x + s)s
i 0
i 0
(2.35)
1
i 0
i 0
(f
)
(x
+
τ
s)s
−
(f
)
(x
+
s)s
dτ
= o(ksk).
Where the equality (2.35) is due to the fundamental theorem of calculus. Hence, f is
semismooth. Now, assume that f i ∈ C 2 for all i. Following similar steps as in the proof of
semismoothness and applying Taylor’s theorem with remainder in the integral form, we
obtain
Z
kf (x + s) − f (x) − M sk ≤ max
i∈I e (xs ) 0
1
00
τ sT (f i ) (x + τ s)s dτ = O ksk2 ,
which proves that f is 1-order semismooth in this case.
2.3.3
Semismooth Newton Method
Let f : V → Rn , where V ⊂ Rn and consider the equation
f (x) = 0,
(2.36)
with f semismooth at V.
We present the algorithm of the semismooth Newton method for the solution of the
equation (2.36) and then develop an example.
33
Algorithm 2.3.3.1.
Step 0. Choose an initial guess x0 and set k = 0.
Step 1. If kf (xk )k ≤ τr r0 + τa , then STOP.
Step 2. Choose Mk ∈ ∂f (xk ) and compute sk from
Mk sk = −f (xk ).
Step 3. Set xk+1 = xk + sk , increment k by one, and go to step 1.
Example 2.3.3.1. Consider the function


 x2 + 2.5x + 10−5 if x ≤ 0,
f (x) =

 x2 + 0.1x + 10−5 if x > 0.
(2.37)
The graph of f is in Figure 2.2. One of the two roots for f is x∗ = −0.00000400000640010667.
y
x
FIGURE 2.2: Semismooth Newton’s method with semismooth function f (x) defined in
(2.37).
From Table 2.2 we can observe that the order of convergence is q-superlinear and
not quadratic.
34
tol= 10−6
n
xn
|x − x∗ |
αn
0
1.5
1.5000040000064
−
1
0.7258032258
0.7258072258128
−
2
0.3395064013
0.3395104012871
1.0466
3
0.1479495538
0.1479535537642
1.0932
4
0.05526425809
0.0552682581009
1.1855
5
0.01445950543
0.0144635054351
1.3614
6
0.001544204349
0.0015482043554
1.6669
7
−0.00007387283424
0.00006987282784
1.3865
TABLE 2.2: Iterates xn , errors, and convergence order for semismooth Newton method
with tolerance 10−6 .
The following proposition, Proposition 2.3.3.1, ensures that under certain assumptions the iterations (xk ) converge locally q-superlinearly. We follow the proof in [44], but
a lot of details were completed by us.
Proposition 2.3.3.1. (See [44], proposition 2.12, page 29). Let f : V → Rn with V an
open subset of Rn . Let x be an isolated solution of (2.36). Assume that
(a) The estimate (2.34) holds at x = x (which, in particular, is satisfied if f is semismooth
at x.)
(b) One of the following conditions holds:
(i) There exists a constant C > 0 such that, for all k, the matrices Mk (see Step 2
in Algorithm 2.3.3.1) are nonsingular with Mk−1 ≤ C.
(ii) There exist constants η > 0 and C > 0 such that, for all x ∈ x + ηB n , every
M ∈ ∂f (x) is nonsingular with M −1 ≤ C.
35
(iii) Every M ∈ ∂f (x) is nonsingular with M −1 ≤ C.
Then there exists δ > 0 such that, for all x0 ∈ x + δB n , (i) holds and Algorithm 2.3.3.1
either terminates with xk = x or generates a sequence (xk ) that converges q-superlinearly
to x.
Proof. We start the proof showing that any of the conditions (ii) or (iii) implies (i). For
that we’ll show that (iii) =⇒ (ii) and (ii) =⇒ (i).
(iii) =⇒ (ii). We followed the proof in [44] but we are completing details extensively. We prove by contradiction. If (ii) does not hold then, there is a sequence xi → x
and Ai ∈ ∂f (xi ) such that for each i, either Ai is singular or k(Ai )−1 k ≥ i. By Proposition
2.3.0.1 we have that ∂f is upper semicontinuos and compact. Therefore, we can extract
a subsequence (Aij ) such that (Aij ) → A for some A ∈ ∂f (x). If an infinite number of
the sequence (Aij ), is such that (Aijm )−1 ≥ ijm , then (Aijm )−1 → ∞ as m → ∞,
contradicting (iii). Otherwise, (Aij ) is eventually singular. That is, there is a positive
integer k such that for all ij ≥ k,
det(Aij ) = 0, so det(Aij ) → det(A) and A is singular,
again, contradicting (iii).
(ii) =⇒ (i): Notice that if (ii) holds, then (i) holds as long as xk ∈ x + ηB n for all
k. Up to now, we can say that if one of the conditions in (b) holds then (i) is true as
long as δ < η. We denote the errors by vk = xk − x. From Step 2 in Algorithm 2.3.3.1,
Mk sk = −f (xk ). Thus, for xk ,
Mk vk+1 = Mk (xk+1 − xk + xk − x)
= Mk (sk + vk )
= M k s k + Mk v k
= −f (xk ) + Mk vk
= −[f (x + vk ) − f (x) − Mk vk ].
(2.38)
36
From (2.34) we have
kMk vk+1 k = kf (x + vk ) − f (x) − Mk vk k = o(kvk k)
as kvk k → 0.
(2.39)
Choosing δ > 0 small enough such that
kMk vk+1 k ≤
1
kvk k,
2C
and using (i),
1
kvk+1 k ≤ kMk−1 kkMk vk+1 k ≤ kvk k,
2
which is
1
kxk+1 − xk ≤ kxk − xk.
2
So, since xk ∈ x + δB n then xk+1 ∈ x + (δ/2)B n and then, xk → x. Adding to this
convergence what we have in (2.39) and (i), we conclude that the rate of convergence is
q-superlinear.
We can have better order of convergence of the semismooth Newton method if we
have what is called a higher-order semismoothness. A classical example among this class
of functions is the Euclidean norm. We carefully develop it in Example 2.3.3.2.
Definition 2.3.3.1. Let V ⊂ Rn be open and 0 ≤ α ≤ 1. A function f : V → Rm is
α−order semismooth at x ∈ V if f is locally Lipschitz continuous near x, f 0 (x, ·) exists,
and
sup
M s − f 0 (x, ·) = O ksk1+α as s → 0.
M ∈∂f (x+s)
If f is α−order semismooth for all x ∈ V, we call f α−order semismooth on V.
We have similar results for α−semismooth functions.
Proposition 2.3.3.2. ([44], Proposition 2.14, p.30) Let f : V → Rm be defined on the
open set V ⊂ Rn . Then for x ∈ V and 0 ≤ α ≤ 1 the following statements are equivalent:
37
(a) f is α−order semismooth at x.
(b) f is Lipschitz continuous near x, α−order B-differentiable at x, and
kf (x + s) − f (x) − M sk = O ksk1+α as s → 0.
sup
(2.40)
M ∈∂f (x+s)
Recall the following definition.
Definition 2.3.3.2. We say that a function f is α−Hölder continuous if there exist
constants K > 0 and 0 < α < 1 such that
kf (x) − f (y)k ≤ Kkx − ykα
for all x, y ∈ V.
Proposition 2.3.3.3. Let V ⊂ Rn be open. If f : V → Rm is differentiable in a neighbourhood of x ∈ V with α−Hölder continuous derivative, 0 ≤ α ≤ 1, then f is α−order
semismooth at x and ∂f (x) = ∂B f (x) = {f 0 (x)}.
Example 2.3.3.2 (The Euclidean norm). Recall that for x ∈ Rn , the Euclidean norm
e : Rn → R is defined by e(x) = kxk2 = (xT x)
Lipschitz on Rn and C ∞
1/2
is a semismooth function. Note that e is
P
on Rn \{0}. Since (e(x))2 = ni=1 x2i , then 2e(x)e0 (x) = 2xT and
e0 (x) =
xT
.
kxk2
By Proposition 2.3.0.1 we have that ∂e(x) = ∂B e(x) = {e0 (x)} for all x ∈ Rn \{0}. For
x = 0, take a vector M ∈ ∂B e(0), then there exists a sequence xk → 0 such that
xTk
→ M ∈ Rn×1 as k → ∞.
kxk k2
T x Note that kxkkk = 1 for all k ≥ 1, and this implies kM k2 = 1. So,
2
2
∂B e(0) ⊂ {v T : v ∈ Rn with kvk2 = 1}.
38
On the other hand, let v ∈ Rn×1 with kvk2 = 1. We need to find a sequence (xk ) such that
xk → 0 and
xk
kxk k2
So xk → 0 and
→ v T . Choose any sequence (yk ) such that yk → 0 and let xk = v T kyk k.
xk
kxk k2
= v T for every k. Hence,
∂B e(0) = {v T : v ∈ Rn with kvk2 = 1} and ∂e(0) = {v T : v ∈ Rn with kvk2 ≤ 1}.
In addition, the hypotheses of Proposition 2.3.3.3 hold for e on Rn \{0} with α = 1.
On the other hand, we can show that e(·) is 1-order semismooth at 0. Indeed, for all
s ∈ Rn \{0} and M ∈ ∂e(s) we have seen that M =
sT
ksk2
and then
e(s) − e(0) − M s = ksk2 −
ssT
ksk2
= ksk2 − ksk2
= 0.
Thus e(·) is also 1−order semismooth at 0.
Since semismooth functions are closed under composition, we expect to have a
similar result for the class of α−order semismooth functions.
Proposition 2.3.3.4. Let V ⊂ Rn and W ⊂ Rl be open sets and 0 ≤ α ≤ 1. Let
g : V → W be α−order semismooth at x ∈ V and h : W → Rm be α−order semismooth at
g(x) with g(V ) ⊂ W. Then the composite map f := h◦g : V → Rm is α−order semismooth
at x. Moreover,
f 0 (x, ·) = h0 (g(x))g 0 (x, ·).
We have a result similar to Corollary 2.3.0.1.
Proposition 2.3.3.5. Let V ⊂ Rn be open. The function f : V → Rm is α−order
semismooth at x ∈ V, 0 ≤ α ≤ 1, if and only if its component functions are α−order
semismooth at x.
The counterpart of Proposition 2.3.3.1 is given by the following result.
39
Proposition 2.3.3.6. Let all assumptions from 2.3.3.1, but assume (2.40) holds at x
with 0 ≤ α ≤ 1 instead of (2.34). Then there exists a δ > 0 such that, for all x0 ∈
x + δB n , Algorithm 2.3.3.1 either terminates with xk = x or generates a sequence (xk )
that converges to x with rate 1 + α.
Proof. The proof is very similar to the one from Proposition 2.3.3.1. It is left to prove the
new rate of convergence. From (2.38), (2.40) and the fact that kMk k is bounded we get
kvk+1 k = O(kvk k)
as kvk k → 0.
In what follows, we give examples of semismooth functions: The Euclidean norm,
the Fischer-Burmeister function and any piecewise continuously differentiable function.
The next example can be found in [44]. Some details were added to this example.
Example 2.3.3.3 (The Fischer-Burmeister function). We define the Fischer-Burmeister
p
function φF B : R2 → R by φF B (x) = x1 + x2 − x21 + x22 . Take φ = φF B to simplify
notation. Note that
φ(x) = f (x) − e(x)
where f (x) := x1 + x2 is linear and e(x) is, as we already saw, 1-order semismooth
and Lipschitz continuous function. Therefore, φ is 1-order semismooth. Moreover, using
Proposition 2.3.3.3 on f we have
∂B φ(x) = f 0 (x) − ∂B e(x),
∂φ(x) = f 0 (x) − ∂e(x).
Therefore, for x 6= 0,
xT
∂φ(x) = ∂B φ(x) = (1, 1) −
.
kxk2
For x = 0,
∂B φ(0) = (1, 1) − y T : kyk2 = 1 ,
∂φ(0) = (1, 1) − y T : kyk2 ≤ 1 .
40
2.4.
Conservation laws
The goal of this section is to introduce the basics of conservation laws. In particular,
we discuss weak solutions and work out some examples. The intention is to provide the
context for our adsorption model, introduced in Section 3.2.3 and discussed in Chapter 5.
We closely follow the exposition in [25].
2.4.1
Weak solutions to conservation laws
Consider the Cauchy problem
(
ut + (f (u))x = 0,
x ∈ R, t > 0,
(2.41a)
u(x, 0) = u0 (x).
(2.41b)
We now introduce the notion of weak solution to (2.41a)-(2.41b). We consider the space
of test functions to be C01 (R × R) (recall that C01 represents the set of functions that are
continuously differentiable with compact support). Next, multiply both sides of the partial
differential equation (2.41a) by φ ∈ C01 and integrate by parts. Assume the integrands are
in L1loc so we can apply Tonelli-Fubini Theorem to interchange the order of integration.
Thus from
∞ Z ∞
Z
0
−∞
φ(x, t)ut + φ(x, t)(f (u))x dx dt = 0,
we obtain
Z
∞
Z
∞
Z
∞Z ∞
φ(x, t)ut dtdx +
−∞
0
0
−∞
φ(x, t)(f (u))x dxdt = 0.
(2.42)
Integrating by parts the first term of (2.42)
Z ∞Z ∞
Z ∞
Z ∞Z ∞
t=a
φ(x, t)ut dtdx =
lim [φ(x, t)u]t=0 dx −
uφt dt
−∞ 0
−∞ a→∞
−∞ 0
Z ∞
Z ∞Z ∞
= −
φ(x, 0)u(x, 0) dx −
uφt dtdx.
−∞
Then the second term is
Z ∞Z ∞
Z
φ(x, t)(f (u))x dxdt =
0
−∞
0
∞
lim [φ(x, t)f (u)]x=a
x=−a dt
a→∞
−∞
Z
(2.43)
0
∞Z ∞
−
f (u)φx dxdt.
0
−∞
(2.44)
41
Now, substitude the right hand sides of (2.43) and (2.44) to obtain
Z
∞Z ∞
Z
∞
[φt u + φx f (u)]dxdt = −
0
φ(x, 0)u(x, 0)dx.
(2.45)
−∞
−∞
After all these calculations we are ready to give the definition of weak solution that will
be used through this chapter.
Definition 2.4.1.1. The function u(x, t) is called the weak solution of the conservation
law if (2.45) holds for all functions φ ∈ C01 (R × R).
2.4.2
Linear scalar conservation law
Consider the linear Cauchy problem (2.41a)-(2.41b) with f (u) = cu and c a positive
constant. The problem we are dealing with now is
ut + cux = 0,
x ∈ R, t > 0,
(2.46a)
u(x, 0) = u0 (x).
(2.46b)
We consider u0 to be Riemman data. That is,



uL if x < 0,
u0 (x) =


uR if x > 0,
(2.47)
where uL and uR are constants. We will show that the travelling wave solution
u(x, t) = u0 (x − ct),
is also a weak solution of the problem (2.46) with (2.47). Obviously since u0 (x) is discontinuous, it is not a classical solution to (2.46) with (2.47).
The travelling wave solution is given by


 uL , x − ct < 0,
u0 (x − ct) =

 uR , x − ct > 0.
42
Now, observe that we are working with continuous functions φ with compact support on
Ω = R × R+ . Suppose that a given function φ in C01 (Ω) has compact support K, then K
is totally contained in the open ball B(0, r) with radius r > 0. So we can assume without
loss of generality that the support is contained in B(0, r). Observe that Ω is the union of
two regions KL and KR , where u = uL in KL and u = uR in KR (determined by the line
x = ct) and ∂B(0, R) ∩ [t > 0] = γ1 (R) ∪ γ2 (R), and R > r. We apply Green’s identity to
the main equation.
Z Z
Z Z
(φt u + φx cu) dxdt +
(φt u + φx cu) dxdt
KL
KR
Z
Z
φ(cuL dt − uL dx) +
=
φ(cuR dt − uR dx)
∂KL
∂KR
Z
Z
φ(cuL dt − uL dx) +
=
φ(cuR dt − uR dx)
[x=ct]
[x=ct]
Z
Z
φ(cuL dt − uL dx) +
+
γ1 (R)
Z 0
Z
φ(cuR dt − uR dx)
γ2 (R)
R
φ(cuL dt − uL dx) +
+
−R
Z
=
φ
[x=ct]
0
cu
L
c
Z
− uL dx +
cu
[x=ct]
Z
−
φ(cuR dt − uR dx)
0
Z
− uR dx
R
c
R
uL φ(x, 0)dx −
uR φ(x, 0)dx.
−R
0
To get the last equality we used the fact that φ = 0 on (B(0, R))c , for all R > r. Integrating
in the real line (t = 0) the integral respect to t disappears, we use the change of variable
t=
x
c
so that dt =
dx
c .
Since the last equality is valid for all R > r, we get that
Z Z
Z Z
Z
φt u + φx cudxdt = −
φt u + φx vudxdt +
KL
0
KR
Z
uL φ(x, 0)dx −
−∞
∞
uR φ(x, 0)dx.
0
Hence, u satisfies (2.45) and it is a weak solution for (2.46a) and (2.46b).
43
2.4.3
Nonlinear scalar conservation law
The simplest and classical example for nonlinear conservation laws is Burger’s equa-
tion. We consider the following Cauchy problem
(
ut + (f (u))x = 0, x ∈ R, t > 0,
u(x, 0) = u0 (x),
where f (u) =
u2
2 .
(2.48a)
(2.48b)
Observe that the graph of f is convex. The Cauchy problem is complete
if we provide initial data u(x, 0) = u0 (x). The characteristics satisfy
x0 (t) = u(x(t), t)
(2.49)
and along each characteristic u(x, t) is constant, since
d
u(x(t), t) =
dt
∂
∂
u(x(t), t)x0 (t) + u(x(t), t)
∂x
∂t
= ut + uux = 0
Observe that x0 (t) is constant by (2.49) and because u is constant. Therefore, the characteristics are straight lines determined by the initial data.
If we have smooth initial data then for t small enough, for each (x, t), we can solve
the equation x = ξ + u(ξ, 0)t for ξ and then, u(x, t) = u(ξ, 0).
For larger t, the characteristics may cross for the first time at a certain time t =
Tf . The function u(x, t) has an infinite slope, then the wave breaks and a shock forms.
Afterwards, we don’t have a classical solution and the weak solution is discontinuous. The
nonconservative form for (2.48a) becomes
ut + uux = 0,
(2.50)
in the case of Burger’s example. Actually, (2.50) should be called the “inviscid Burger’s
equation”, since the original equation that Burger studied included a viscous term and it
was given by ut + uux = εuxx .
44
As a good source of examples we have the Burger’s equation with Riemann data.
Note that the relation between uL and uR is important, since it can determine the form
of the solution. For instance, there are infinitely many weak solutions to inviscid Burger’s
equation when uL < uR . Among them we have,



uL if x < st,
u(x, t) =


uR if x > st,
(2.51)
where
s=
uL + uR
.
2
Another weak solution is the rarefaction wave



uL
if x < uL t,




u(x, t) = x/t if uL t ≤ x ≤ uR t ,






uR
if x > uR t.
(2.52)
(2.53)
It is easy to show that (2.53) is a weak solution and this has been shown in many standard
texts.
We have many other solutions. For example, we show that the following unphysical
shock
u(x, t) =



uL





 um


x/t





 u
R
if x < sm t,
if sm t ≤ x ≤ um t,
(2.54)
if um t ≤ x ≤ uR t,
if x > uR t,
um + uL
. This is a suggested
2
exercise proposed in [25] (Exercise 3.6, p. 30). We prove that u satisfies identity (2.45).
is a weak solution for any um with uL ≤ um ≤ uR and sm =
45
Let I1 =
Z
I1 =
0
R∞ R∞
−∞ 0
Z
φt udtdx and I2 =
∞
−∞ φx f (u)dxdt.
0
uR φt dtdx
0
Z ∞ Z x/sm
+
0
Z
x/uR
0
0
Z
um φ(x, x/um )dx −
Z
0
∞
∞
Z
∞ Z x/um
uR φ(x, x/uR )dx +
0
0
Z
0
x
t2
x/uR
#
φ(x, t)dtdx
∞
um (φ(x, x/sm ) − φ(x, x/um )) dx −
+
x/sm
0
∞
+
uL φt dtdx
0
x/um
uR (φ(x, x/uR ) − φ(x, 0))dx
−∞
"Z
∞Z ∞
∞
Z
uL (−φ(x, 0))dx +
=
Z
um φt dtdx +
(x/t)φt dtdx +
0
On one hand,
∞ Z x/uR
Z
uL φt dtdx
−∞ 0
Z ∞ Z x/um
+
R∞R∞
uL φ(x, x/sm )dx.
0
Continuing we obtain
Z
∞
I1 =
Z
uL (−φ(x, 0))dx +
−∞
Z ∞ Z x/um
+
0
Z
∞
=
−∞
∞ Z x/um
0
Z
=
∞
Z
x
∞
Z
∞
uL (−φ(x, x/sm ))dx
0
0
Z
x
∞
φ(x, t)dtdx +
(um − uL )φ(x, x/sm )dx
t2
x/uR
0
Z ∞
uL (−φ(x, 0))dx +
uR (−φ(x, 0))dx
−∞
∞ Z x/um
Z
+
0
uR (−φ(x, 0))dx
0
um φ(x, x/sm )dx +
φ(x, t)dtdx +
t2
x/uR
0
Z ∞
uL (−φ(x, 0))dx +
uR (−φ(x, 0))dx
Z
+
∞
x/uR
0
x
t2
Z
∞
sm (um − uL )φ(sm t, t)dt.
φ(x, t)dtdx +
0
46
On the other hand, we have
Z
∞ Z sm t
I2 =
∞ Z um t
φx f (uL )dxdt +
0
Z
−∞
∞ Z uR t
φx f (um )dxdt
0
Z
sm t
∞Z ∞
φx f (uR )dxdt
φx f (x/t)dxdt +
+
0
Z
=
Z
um t
∞ 2
uL
Z
0
∞ 2
um
uR t
φ(sm t, t)dt +
[φ(um t, t) − φ(sm t, t)]dt
2
2
0
Z ∞ 2
uR
u2m
φ(uR t, t) −
φ(um t, t) dt
2
2
0
Z ∞ 2
Z ∞ Z uR t
x
uR
φ(x, t) 2 dxdt +
(−φ(uR t, t))dt
t
2
0
0
um t
Z ∞ Z uR t
Z ∞ 2
x
uL u2m
−
φ(sm t, t)dt −
φ(x, t) 2 dxdt
2
2
t
0
um t
0
Z ∞
Z ∞ Z uR t
x
uL + um
(uL − um )
φ(sm t, t)dt −
φ(x, t) 2 dxdt
2
t
0
0
um t
Z ∞ Z uR t
Z ∞
x
(uL − um )sm φ(sm t, t)dt −
φ(x, t) 2 dxdt.
t
0
0
um t
0
+
−
=
=
=
Hence, after cancellations, we obtain
Z
∞
Z
uL (−φ(x, 0))dx +
I1 + I2 =
−∞
Z ∞ Z x/um
+
0
x/uR
|
∞
uR (−φ(x, 0))dx
0
Z
x
∞ Z uR t
φ(x, t)dtdx −
t2
0
{z
} |
a
um t
x
φ(x, t) 2 dxdt.
t
{z
}
b
a and b are the same. Therefore,
Now, notice that using Fubini we have that integrals in Z
∞
I1 + I2 =
Z
uL (−φ(x, 0))dx +
−∞
∞
uR (−φ(x, 0))dx,
0
and u(x, t) given by (2.54) is a weak solution for Burger’s equation.
2.4.3.1
Rankine-Hugoniot condition
In this section we discuss the relationship between the speed s and the states uL
and uR (uL > uR ) from Riemann data. This discussion helps to ensure that we have a
weak solution which is physically meaningful for a conservation law using a general flux
f (u). We start by working with the Burger’s equation example from Section 2.4.3.
47
Recall the weak solution for Burger’s equation given in (2.51) with speed s given
by (2.52). The same discontinuity propagating at a different speed would not be a weak
solution.
The following exposition is inspired by [25].
We can determine the speed of propagation by conservation. That is, integrating
(2.48) over (x1 , x2 ) we get that
d
dt
Z
x2
u(x, t)dx = f (u(x1 , t)) − f (u(x2 , t)), for all x1 , x2 , t.
(2.55)
x1
Then if M is large compared to st, we can apply (2.55) to the weak solution (2.51) for
Burger’s equation,
d
dt
M
Z
u(x, t)dx = f (uL ) − f (uR )
(2.56)
−M
=
=
u2L − u2R
2
1
(uL + uR )(uL − uR ).
2
On the other hand, we can see that integrating (2.51) on [−M, M ] we have
Z
M
u(x, t)dx = (M + st)uL + (M − st)uR .
−M
Therefore,
d
dt
Z
M
u(x, t) dx = s(uL − uR ).
(2.57)
−M
From (2.56) and (2.57) we get
1
(uL + uR )(uL − uR ) = s(uL − uR ),
2
which gives the expression for s in (2.52).
Now, for a general flux function f (u), the same argument gives the following relationship between the speed s and the states uL and uR . That is
f (uL ) − f (uR ) = s(uL − uR ).
(2.58)
48
So in the scalar case
s=
[f ]
f (uL ) − f (uR )
=
,
uL − uR
[u]
(2.59)
where [u] denotes the jump from state uL to the state uR . The relation in (2.58) is called
the Rankine-Hugoniot jump condition.
2.4.3.2
Entropy solutions
Above we have shown that there can be multiple weak solutions to any conservation
law. These include solutions with discontinuities, or discontinuities in the derivative; such
functions cannot be classical solutions to partial differential equations but they make sense
for applications.
The notion of weak solution allows us to enlarge the space of possible solutions to
conservation laws; thus existence of solutions is possible even for problems where singularities develop. It is well known [17, 25] that singularities in conservation laws can be due
to singularities in initial data but also that they can develop from various smooth initial
data. The latter is true when the flux function in nonlinear while for linear problems the
degree of smoothness of solutions is preserved.
Since the weak solutions are sought in a larger space than that of classical solutions,
there arises naturally a question of uniqueness. As we have shown, for example, the
unphysical (2.54) as well as the rarefaction solution (2.53) are both weak solutions to the
same problem.
So there must be a way to limit the class of weak solution to only those that make
sense physically, which would eliminate for example (2.54). This is done with a notion
of entropy solutions which are the limits of diffusive conservation laws with vanishing
diffusion constant (also called vanishing viscosity).
Entropy solutions are formulated to eliminate unphysical shocks and the conditions
to achieve it have been considered by Lax, Oleinik and others [25]. We do not review this
here but mention that for scalar conservation laws with an increasing convex flux function
49
f and Riemann data (2.47) the entropy solutions include a shock such as (2.51) when
uL > uR , and a rarefaction such as (2.54) when uL < uR .
Equivalently, if f (·) is an increasing concave function, the entropy solution is a shock
if uL < uR , and a (forward travelling) rarefaction when uL > uR .
2.4.3.3
Systems of conservation laws
A system of conservation laws can be formulated in the same way as (2.48) except
that now u(x, t) ∈ Rp . The theory for systems of conservation laws is much more complicated than that for the scalar case. Already for a linear system one sees more types
of solutions than shocks and rarefactions and these can propagate in separate directions
depending on the eigenvalues of the Jacobian Df .
A system is called strictly hyperbolic if the eigenvalues of Df are real and distinct.
In the sequel we deal with a practical adsorption system which we transform to
(2.48) and show that it is hyperbolic based on some assumptions on the data.
2.4.4
Numerical methods for conservation laws
In this section, we introduce the numerical methods used to solve conservation laws,
that is, partial differential equations of the form (2.48). In general, we want to stay in the
class of explicit conservative methods. We follow the exposition in [25].
An explicit method is in conservative form if it can be written in the following way
Ujn+1 = Ujn −
k
n
n
n
n
n
n
[F (Uj−p
, Uj−p+1
, . . . , Uj+q
) − F (Uj−p−1
, Uj−p
, . . . , Uj+q−1
)]
h
(2.60)
for some function F of p + q + 1 arguments. F is called the numerical flux function. For
example, if p = 0 and q = 1, then (2.60) becomes
Ujn+1 = Ujn −
k
n
n
) − F (Uj−1
, Ujn )
F (Ujn , Uj+1
h
We explain why the form of (2.61) makes sense. We define the cell average as
Z
1 xj+1/2
n
uj :=
u(x, tn ) dx.
h xj−1/2
(2.61)
(2.62)
50
We can view Ujn as an approximation of the cell average (2.62). We know that since
u(x, t) is a weak solution of (2.41a) and (2.41b), then it satisfies the integral form of the
conservation law,
Z xj+1/2
Z
u(x, tn+1 ) dx =
xj−1/2
xj+1/2
u(x, tn ) dx
xj−1/2
Z
tn+1
−
Z
tn+1
f (u(xj+1/2 , t))dt −
tn
f (u(xj−1/2 , t))dt
tn
Dividing both sides by h and using the definition of unj in (2.62), we obtain
Z tn+1
Z tn+1
1
n+1
n
uj = uj −
f (u(xj+1/2 , t))dt −
f (u(xj−1/2 , t))dt
h tn
tn
(2.63)
Comparing (2.63) with (2.61) we notice that
Z
1 tn+1
f (u(xj+1/2 , t))dt.
F (Uj , Uj+1 ) ≈
k tn
That is, the numerical flux F (Uj , Uj+1/2 ) is the average flux through xj+1/2 over the time
interval [tn , tn+1 ].
2.4.5
Godunov method
In this section, we discuss one of the particular numerical methods that we will be
using for our numerical experiments. This method is the Godunov’s first order method.
In this case, the intercell numerical fluxes are computed using solutions of local Riemann
problems. The first assumption that has to be made is that at a given time tn the data
has a piecewise constant distribution of the same form of the cell average introduced in
(2.62). We can see the data at time tn as a pair of constant states (uni , uni+1 ) separated by
a discontinuity at the intercell boundary xi+1/2 . So, locally, we can define the Riemann
problem
ut + f (u)x = 0,



uni
u0 (x) =


un
i+1
if x < 0,
(2.64)
if x > 0.
51
In this case, we say that at each intercell boundary we have a local Riemann problem
RP (uni , uni+1 ) with initial data (uni , uni+1 ). Over a short time interval, we can solve the
sequence of Riemann problems (2.64) (where the waves of neighbouring cells cannot interact). The exact solution to the global problem is obtained piecing together these Riemann
solutions.
Let ũ(x, t) be the combined solution of RP (uni−1 , uni ) and RP (uni , uni+1 ). This is
the solution over the time interval [tn , tn+1 ]. Moreover, ũ(x, t) is an exact solution of
the original conservation law. We define the approximate solution U n+1 as the following
average
Ujn+1
1
=
h
Z
xj+1/2
ũn (x, tn+1 )dx.
(2.65)
xj−1/2
Using these values, we define new piecewise constant data ũn+1 (x, tn+1 ) and then repeat
the process.
It is useful to show at this point that Godunov’s method is conservative. Indeed,
we assumed that ũn is a weak solution. Therefore,
Z xj+1/2
Z xj+1/2
ũ(x, tn+1 ) dx =
ũ(x, tn ) dx
xj−1/2
xj−1/2
Z
tn+1
Z
tn+1
f (ũ(xj+1/2 , t))dt −
−
f (ũ(xj−1/2 , t))dt .(2.66)
tn
tn
Dividing both sides by h, using (2.65) and noticing that ũn (x, tn ) ≡ Ujn over the cell
(xj−1/2 , xj+1/2 ), we have that equation (2.66) reduces to
Ujn+1 = Ujn −
k
n
n
F (Ujn , Uj+1
) − F (Uj−1
, Ujn )
h
where the numerical flux function is given by
Z
1 tn+1
n
n
F (Uj , Uj+1 ) =
f (ũn (xj+1/2 , t))dt.
k tn
(2.67)
And we are able to write this method in conservative form.
Remark 2.4.5.1. The integral in (2.67) is easy to compute since ũn is constant at the
point xj+1/2 over the time interval (tn , tn+1 ). The values ũn depend only on the data Ujn
52
n
n ) then the
and Uj+1
for this Riemann problem. If we denote this value by u∗ (Ujn , Uj+1
numerical flux in (2.67) becomes
n
n
F (Ujn , Uj+1
) = f (u∗ (Ujn , Uj+1
)),
and Godunov’s method is now given in the form
Ujn+1 = Ujn −
k
n
n
f (u∗ (Ujn , Uj+1
)) − f (u∗ (Uj−1
, Ujn )) .
h
(2.68)
Example 2.4.5.1. In the case of a scalar equation with an increasing and convex flux
n )) = f (U n ) and f (u∗ (U n , U n )) = f (U n ). So
function f (·), we have that f (u∗ (Ujn , Uj+1
j
j−1
j
j−1
(2.68) is the upwind method.
The CFL condition is given by
k
max |f 0 (Ujn )| ≤ 1.
h
For the case of a scalar system of conservation laws, we need to know the signs of
the eigenvalues to properly construct solutions to the Riemann problem.
In the case of a system of equations, we need to avoid interaction between the waves
coming from the points xj−1/2 and xj+1/2 . We need the CFL condition
k
1
max |λk (Ujn )| ≤ , 1 ≤ k ≤ p,
h
2
(2.69)
where λk are eigenvalues for the system.
Since, f is convex increasing, we can numerically solve (5.3) with Godunov’s method,
which is the upwind scheme,
n
Wjn+1 = Wjn − λ f (Wjn ) − f (Wj−1
)
where Wjn ≈ w(xj , tn ), Ujn ≈ u(xj , tn ), λ =
∆t
∆x .
53
2.4.6
Numerical methods for systems of conservation laws
Numerical methods for systems of conservation laws are not just simple extensions
of those for scalar conservation laws. While Godunov method discussed in Section 2.4.5
can be defined for systems, it requires a solution to the local Riemann problem (2.64).
These are quite complicated and depend heavily on the form of the flux function.
Much research was devoted to construction of local Riemann solvers for important
applications such as Euler equations [17], [43] and [25] or shallow water equations. However, for an arbitrary flux function f exact Riemann solvers are not available. Instead,
one can formulate approximate Riemann solvers, and much research was also devoted to
this topic, see e.g. [17], [43] and [25]. In this dissertation our approach to a system of
conservation laws arising in adsorption is to use upwinding combined with fully implicit
handling of nonlinearity in the time derivative term.
We remark that for a nonlinear system of the form (2.48), for which the Jacobian
matrix Df (Ujn ) has nonegative eigenvalues for all Ujn , the upwind method is of the form
(2.61) with F (v, w) = f (v). Now, if Df (Ujn ) has only nonpositive eigenvalues for all
Ujn , then the only information that the upwind method needs is the point to the right.
Therefore, the numerical flux for this case is F (v, w) = f (w).
54
3.
MODELING
In this chapter we describe models for the two applications of interest in this dissertation. Both involve multiple components and multiple phases evolving in subsurface.
Each of these applications involves the evolution of methane either in a flowing phase, or
in a solid phase, or in an adsorbed phase. Mathematical models that we formulate below
describe how the components evolve in time, how they are transported in space, and how
they change from one phase to another.
Below we first explain how such models are built from first principles involving
conservation of mass for each component. Next in Section 3.1. we describe a model for
methane hydrates evolution. In Section 3.2. we describe a model for methane and carbon
dioxide adsorption. We follow [30], [32], [31], [24] and [16].
The equations below are formulated for flow and transport in subsurface. Let Ω ⊂
Rd with 1 ≤ d ≤ 3 be a porous reservoir under the Earth’s surface with depth D(x), for
each x ∈ Ω. We denote by φ a positive coefficient representing the porosity and by K
the permeability coefficient which is a uniformly positive definite tensor. We also denote
by P the pressure, and by T the temperature in the reservoir. The models of methane
evolution take into consideration multiple phases (for example, gas, adsorbed and water)
and multiple components (for example, methane, carbon dioxide, nitrogen and water).
In general, mass conservation equations for each component C can be written
StoC + AdvC + Diff C = qC ,
(3.1)
where the terms StoC , AdvC , Diff C represent storage, advection and diffusion, respectively, and they depend on the component C. Here qC is the source term.
Observe that the term StoC will tell us if the phase transition or adsorption take
place in an equilibrium model or a kinetic model. It tells us how a component C is
55
partitioned between one phase and another. It is given by
!
∂
∂
StoC := (φNC ) :=
∂t
∂t
φ
X
ρp Sp χpC
,
(3.2)
p
where NC is the mass concentration of the component C, Sp is the saturation on phase p
(or volume fraction), ρp density of phase p, χpC is the mass fraction of a component C in
P
a phase p. Naturally, Sp ≥ 0 and p Sp = 1. Also,
X
χpC = 1.
(3.3)
C
Remark 3.0.6.1. In the case of adsorption, the definition (3.2) has to include
∂
∂t ((1
− φ)ρa Sa χaC ).
Now, the term Diff C depends on the application and scale. Usually, the divergence
of diffusive fluxes is determined by Fick’s law, so
!
Diff C := −∇ ·
φ
X
ρp Sp DpC ∇χpC
.
p
The advection term arises from mass fluxes
AdvC := ∇ ·
X
χpC ρp Up ,
(3.4)
p
due to velocities
Up = −K
krp
(∇Pp − ρp G∇D(x)), p = l, g.
µp
(3.5)
In (3.5), the velocities are given via the multiphase extension of Darcy’s law. Here, krp
and µp denote the permeability and viscosity of phase p, respectively, and G the gravity
constant. The phase pressures Pl and Pg (mobile phases) are coupled via the capillary
pressure relation
Pg − Pl = Pc (Sl ).
Here we followed generally accepted notation for components and phases as in [24]. For
the two applications of interest, the models are described in detail in [16, 30, 32].
Below, we make the general model (3.1) specialized for each application.
56
3.1.
Modeling for Methane Hydrates
Methane hydrates are an ice-like substance present in various subsurface reservoirs,
where pressure is high and temperature is low, and where there is enough methane production from microbial activity, or where there are large fluxes from underneath. In
particular, they can be found in the region called Hydrate Ridge off the Pacific Northwest
coast of USA and in Blake Ridge off the Southeast coast of USA.
Methane hydrates are of interest as a possible energy source, as well as a hazard
during drilling of reservoirs in which hydrates are present.
Assumptions 3.1.0.1. Below we assume that the pressure P and temperature T are
known and given by hydrostatic and geothermal gradients (the rates of increasing pressure
and temperature with respect to increasing depth in the Earth’s interior. Both variables
P and T increase linearly with depth). In addition, we assume that the pressure is high
enough and the temperature is low enough in Ω so that both liquid and hydrate phases can
be present, but not gas phase (See [26]).
We assume the distribution of phases and components is as symbolically shown in
Figure 3.1. Let Sh and Sl be the hydrate and liquid saturation, respectively. We assume
that Sl > 0, Sh ≥ 0. In this dissertation, we assume that Sg = 0 but this can be lifted
later. Then,
Sh + Sl = 1,
(3.6)
and each phase has a density ρh and ρl .
Assumptions 3.1.0.2. The components C in the liquid phase are water C = W , salt
C = S, and methane C = M with the corresponding mass fractions denoted by χlW , χlS
and χlM , respectively. On the other hand, the components for the hydrate phase are water
and methane. We denote the mass fractions by χhW and χhM , respectively. Both χhW
57
Phases
Liquid
Methane
Hydrate
Water
Gas
Salt
Components
FIGURE 3.1: Phases and components
and χhM are assumed to be constants since hydrate crystals are typically built from a fixed
proportion of methane and water molecules. In general,
χlW + χlS + χlM = 1.
(3.7)
Assumptions 3.1.0.3. For the moment, we will assume that the salinity χlS = χlS (x)
is a known and fixed quantity. An extension was treated, e.g. in [33].
In a general multicomponent multiphase model for methane hydrates, one has to
solve for P, T, and the solubilities χlM , χlW , χlS and Sl , Sh . In general, one would write
three mass conservation equations, and one energy equation, and use (3.6) and (3.7) to
resolve the remaining equations. Thus there are seven unknowns and six equations. The
remaining equation is a solubility constraint between χlM and Sl discussed later.
Because of Assumption 3.1.0.3, we eliminated χlS and mass equation for C = S.
Assumption 3.1.0.1 lets us eliminate one equation for C = W , thus fixing P as a known
variable. Thus χlW follows from (3.7) and we do not solve for it. Assumption 3.1.0.1 on
T also eliminates the need for the energy equation.
Thus we are left with one mass conservation equation for C = M and with χlM and
Sl and Sh as independent variables. This is equation (3.8) to be complemented by the
solubility constraint,
58
∂
(φ0 Sl ρl χlM + φ0 Sh ρh χhM ) − O · (DM ρl OχlM ) = fM ,
∂t
(3.8)
coming from (3.1). Here fM is the external source of methane, for example, due to bacteria
producing methane from organic matter. Note that the independent variables are χlM and
Sl .
The quantity
NM := Sl ρl χlM + Sh ρh χhM
represents the total amount of methane present in both phases, liquid and hydrate, and
can be used instead of one of χlM , Sl , Sh as an independent variable.
The amount of methane that can be dissolved in the liquid phase is determined by
temperature T, pressure Pl and salinity χlS . Since these are assumed known functions of
depth, this amount, the maximum solubility denoted by χmax
lM , is a function of depth thus
of x. This quantity determines how the total amount of methane, NM is partitioned in
the two phases.
max
We always have χlM (x, t) ≤ χmax
lM (x) for (x, t) ∈ Ω × [0, T ]. If χlM (x, t) < χlM (x),
this means that Sl (x, t) = 1 so only the liquid phase is present and the total amount of
methane is NM = ρl χlM . If χlM and NM > ρl χlM , then χLM is now fixed to χmax
lM , and
hydrate phase forms, and Sh = 1 − Sl , or Sl is the independent variable. The process just
described can be written in the following way.




χlM ≤ χmax

lM



if Sl = 1,
χlM = χmax
if Sl ≤ 1,
lM






(χlM − χmax )(1 − Sl ) = 0.
lM
(3.9)
Usually, χmax
lM increases with depth at least linearly within the hydrate zone. Recent
extensions include its additional dependence on the type of sediment.
Now we develop a simplified version of the model in (3.8) in order to make the
exposition easier to follow.
59
Assumptions 3.1.0.4. The diffusion coefficient DlM is assumed to be constant. We also
assume that the densities ρl , ρh and porosity φ0 are constants as well as the diffusion
coefficient DlM .
Dividing both sides of (3.8) by ρl φ0 , we can rewrite it as
∂
(Sl χlM + R(1 − Sl )) − O · (D0 OχlM ) = f,
∂t
where R :=
ρh
ρl χhM ,
f =
fM
ρ l φ0
and D0 =
DM
φ0 .
(3.10)
Here R is constant due to Assump-
tions 3.1.0.1, 3.1.0.2, 3.1.0.3 and 3.1.0.4.
For the needs of analysis and the setting of our numerical algorithm we further
simplify the notation. We let S = Sl , u =
NM
ρl ,
v = χlM and v ∗ = χmax
lM .
Now, the model (3.10) together with the constraint (3.9) can be written as
∂u
− 4v = f,
∂t
u := Sv + R(1 − S),
hv, Si ∈ F(x, ·),
(3.11)
(3.12)
where F(x, ·) := [0, v ∗ (x)] × {1} ∪ {v ∗ (x)} × (0, 1].
The notation hv, Si ∈ F in (3.12) means that S ∈ F(v), emphasizing that F can be
regarded as a set-valued function. Figure 3.2 shows how the graph associated to problem
(3.11)-(3.12) looks.
S
F(x)
1
v∗
v
FIGURE 3.2: Graph of F associated to problem (3.11)-(3.12)
60
We will assume henceforth that
v∗
:
Ω → R is (at least) piecewise smooth,
min(v ∗ (x)) ≥ v0 > 0,
x∈Ω
R > max(v ∗ (x)).
x∈Ω
(3.13)
All these assumptions are physically grounded. The diffusivity and maximum solubilities
are always nonnegative, while (3.13) follows form thermodynamics.
In Chapter 4 we show well–posedness of an abstract formulation of an initial boundary value problem for (3.11)-(3.12).
3.2.
Modeling for Adsorption
Adsorption is a surface phenomenon in which particles, molecules, or atoms of adsorbate attach to the surface of adsorbent. In this dissertation we are interested in adsorption processes in Enhanced Coalbed Methane (ECBM). These involve, in general multiple
components. In this dissertation we focus on methane C = M , and carbon dioxide C = D.
Figure 3.3 shows all the phases and components that might be present in the adsorption model. In this work we just work with two phase and two components.
3.2.1
Transport with adsorption in ECBM
Below we first describe a single component adsorption model where C = M, where
M denotes the methane component. In this case, methane can be present either in gas
phase p = g, or in the adsorbed phase p = a. Single component adsorption is described in
Sections 3.2.2 and 3.2.3.
Next, we consider the multicomponent case. The components are C = M, D. For
the phases, we have p = g, a, which are gas and adsorbed gas. The mass conservation
61
equation (3.1) gives the evolution of NM and ND . As pointed out in (3.3),
χgM + χgD = 1,
χaM + χaD = 1.
FIGURE 3.3: Phases and components for the adsorption model
FIGURE 3.4: Illustration of mechanism of adsorption process. Injection of CO2 in the
coalbed for CH4 displacement.
In Figure 3.4 the adsorption process for displacement of methane by injecting carbon
62
dioxide is described. The left red arrow signifies that we are injecting carbon dioxide in
the coalbed. The small blue circles on the first rectangle represent the methane molecules
attached to pieces of coal. On the second rectangle, we see small red circles which represent
molecules of carbon dioxide entering to the reservoir where adsorption occurs. The third
rectangle shows how the molecules of methane are detached from the surface of pieces
of coal and going out from the coalbed. The last rectangle together with the blue arrow
means that molecules of methane made their exit from the reservoir.
The two windows to the right of the yellow rectangles show numerical simulation of
the displacement of methane processed just described.
A multicomponent model is described in Section 3.2.4.
3.2.2
Single component adsorption
We write now the single component simplified scalar version of the model (3.1) for
ECBM.
Assumptions 3.2.2.1. Only gas and adsorbed phases are present and Sg + Sa = 1.
From Assumption 3.2.2.1 the storage component is given as
StoM =
∂
[φ0 (ρg Sg χgM ) + (1 − φ0 )ρa Sa χaM ] .
∂t
Also, we have from (3.4) that (since the adsorbed p = a is immobile)
AdvM = ∇ · (Ug χgM ρg ).
We now assume
Assumptions 3.2.2.2.
Diff M = 0, fM = 0.
63
This assumption is reasonable at the time scale of advection. We explain later how
diffusion into micropores is taken into account in the model.
Thus the model (3.1) reads
∂
(φ0 ρg Sg χgM + (1 − φ0 )ρa Sa χaM ) + ∇ · (Ug χgM ρg ) = 0.
∂t
(3.14)
Now we want to look at the nondimensional form of (3.14), thus we simplify further.
Assumptions 3.2.2.3. Let ρg be constant, Ug = 1, and φ0 = 0.5. Also, we combine
(1 − φ0 ) ρρag Sa χaM into a new variable, called v. Also denote by u the value Sg χgM .
The model (3.14) under Assumptions 3.2.2.1, 3.2.2.2 and 3.2.2.3 is now
∂
(u + v) + ux = 0.
∂t
(3.15)
Here u is the concentration of methane component in the mobile phase, and v is the
adsorbed amount. The relation between u and v (and/or their rates) must be provided.
Most commonly, it is assumed that is given by an equilibrium isotherm
v = a(u).
(3.16)
From experimental data, a is a known smooth monotone increasing function which
describes the surface coverage v of adsorbent.
As an example, there is the Langmuir isotherm
a(u) = VL
bu
,
1 + bu
(3.17)
where b and VL are constants. It is derived from the monolayer assumption and equality of
adsorption and desorption rates. Notice that the Langmuir isotherm is concave, increasing,
and Lipschitz.
The kinetic model is given by
∂v
= r(a(u) − v).
∂t
(3.18)
64
Observe that, in this case, v is an exponential follower with rate r of the equilibrium model
a(u). Rate r is also called sometimes the reciprocal of relaxation time τ so r =
1
τ.
(See
[37, 21]). As τ → 0 the kinetic model (3.18) becomes the equilibrium model (3.16).
3.2.2.1
Shock speed for adsorption
Here we repeat calculations introduced in Section 2.4.3.1 for the problem (3.15). We
calculate the shock speed for the Riemann problem from Rankine-Hugoniot condition.
Let w = u + a(u) and f = (a + I)−1 where I is the identity function.
Z
d M
w(x)dx = f (wL ) − f (wR )
dt −M
= uL − uR ,
(3.19)
(3.20)
where the equality in (3.20) is a consequence of f being continuous and increasing. Then,
the equation in (3.19) is equivalent to
Z
d M
(u + a(u))dx = uL − uR .
dt −M
On the other hand,
Z M
(u + a(u))dx = (M + st)(uL + a(uL )) + (M − st)(uR + a(uR )).
(3.21)
(3.22)
−M
Taking derivative respect to t in (3.22) we get
Z
d M
(u + a(u))dx = s(uL + a(uL ) − uR − a(uR )).
dt −M
Comparing (3.21) and (3.23) we obtain
uL − uR = s(uL + a(uL ) − uR − a(uR )),
which is equivalently to
f (wL ) − f (wR ) = s(wL − wR ).
So
s=
[u]
[f ]
=
.
[u + a(u)]
[w]
(3.23)
65
3.2.3
Diffusion into micropores and transport with memory terms.
In many instances there is substantial diffusion from coal seam into the coal matrix.
It is efficient to include this phenomenon in the adsorption model with a kinetic model
(3.18) rather than setting Diff 6= 0 and solving equations in the matrix.
Models with nonequilibrium adsorption can be analysed and approximated either as
a system, or as a single equation. To get the latter, one can solve the differential equation
(3.18) for v in terms of u and substitute back to (3.15). This approach was pursued in
[32], where the resulting model was subsequently truncated.
We now show how this would be done and show the difference between the system
that we consider in this dissertation and the single equation approach from [32].
We calculate v from (3.18) via a convolution. Recall the Volterra convolution
Z
β ∗ u(x, t) :=
t
β(t − s)u(x, s)ds,
0
where β(·) ∈ L1loc . The following scalar linear system is considered
ut + vt + ux
=
0,
(3.24)
vt
=
r(au − v),
a>0
(3.25)
u(·, 0) = u0 , v(·, 0) = v0 .
We rewrite (3.25) as
vt + rv = rau.
(3.26)
Multiply both sides of (3.26) by the integrating factor µ(s) = esr to get
(ers v)s = raers u.
(3.27)
Integrating both sides of (3.27) (and applying the fundamental theorem of calculus) we
obtain
rt
Z
e v(x, t) − v(x, 0) = a
0
t
rers u(x, s)ds.
66
Now, we solve for v(x, t),
−rt
−rt
Z
t
ers u(x, s)ds
v(x, 0) + ae
0
Z t
= e−rt v(x, 0) + a
re−r(t−s) u(x, s)ds
v(x, t) = e
0
=
1
β(t)v0 + a(β ∗ u)(x, t),
r
(3.28)
where
β(t) = re−rt .
Now, notice that
d
[(β ∗ u)(x, t)] = (β ∗ ut )(x, t) + β(t)u0 .
dt
(3.29)
Hence, having found v in terms of u in (3.28), we can obtain an expression for vt using
(3.29). That is
1
vt = β 0 (t)v0 + a(β ∗ ut )(x, t) + aβ(t)u0 .
r
(3.30)
If we substitute (3.30) into (3.24), we get
1
ut + β 0 (t)v0 + β ∗ (au)t + β(t)(au0 ) + ux = 0.
r
Now we reformulate this equation and move the terms dependent on initial condition
to the right hand side
1
ut + β ∗ (au)t + ux = − β 0 (t)v0 − β(t)(au0 ).
r
It is clear that the terms on the right hand side which are associated with the initial
condition on u and v can be understood as source terms to the equation on the left hand
side.
In [32] these terms are dropped and the single equation approach is pursued, and
the problem
ut + β ∗ ut + ux = 0,
x ∈ R, t ∈ (0, T )
(3.31)
67
is analysed. In addition, its nonlinear version as well as that with a weakly singular β are
also considered in [32].
Suitable numerical approximations for the solution u to (3.31) are included in the
class of schemes for scalar conservation laws with memory terms. The author in [32] dealt
with this numerical approximations where the convergence analysis for a Godunov scheme
is done with an approximation for the memory terms. Experiments on that work confirm
that memory terms have a smoothing effect.
In this dissertation we consider the full model
ut + vt + ux = 0,
vt = r(a(u) − v),
without the simplifications.
Remark 3.2.3.1. Our approach to stability analysis in the kinetic case is different from
exposition that was given in [32]. In fact, we will see that we have to change the quantity of
interest (u, v) in order to obtain stability. We will not have a convolution term, therefore,
we will not have the same numerical approximation.
More generally, the model with memory term is similar to double-porosity model
for slightly compressible flow in oil and gas reservoirs with fractures and fissures.
3.2.4
Multicomponent transport with adsorption
Now we consider a multicomponent model for C = M, D. We set u1 = Sg χgM , v1 =
Sa χaM , u2 = Sg χgD , v2 = Sa χaD . The model (3.1) becomes under assumptions similar
to assumptions in Section 3.2.2 a system of two partial differential equations with four
unknowns u1 , v1 , u2 and v2 ,
∂
(u1 + v1 ) +
∂t
∂
(u2 + v2 ) +
∂t
∂u1
∂x
∂u2
∂x
= 0,
(3.32)
= 0.
(3.33)
68
Now we need to specify how vi depends on u1 and u2 for each component i = 1, 2
( i = 1 corresponds to C = M and i = 2 corresponds to C = D).
Such relationships can be formulated as equilibrium or nonequilibrium isotherms
similarly to the single component case. In the equilibrium case v1 = a1 (u1 , u2 ) and
v2 = a2 (u1 , u2 ) and a1 , a2 are known from experiments.
In this dissertation we consider two types of equilibrium isotherms of extended Langmuir case and of Ideal Adsorbate Solution type (IAS). We also consider a nonequilibrium
multicomponent system. The extended Langmuir isotherm is a simple algebraic extension
of the single component Langmuir isotherm (3.17),
a1 (u1 , u2 ) =
a2 (u1 , u2 ) =
b1 VL,1 u1
,
1 + b1 u1 + b2 u2
b2 VL,2 u2
.
1 + b1 u1 + b2 u2
(3.34)
(3.35)
Here bi and VL,i with i = 1, 2 are obtained experimentally. The author in [23] proposed
this extension.
Unfortunately, it has been shown that the extended Langmuir model is not Thermodynamically consistent in the sense that it does not satisfy the Gibbs Isotherm equation
(See [12]).
Therefore, a proper thermodynamic consistent model known as IAS can be proposed.
It is defined implicitly using single component isotherms which are obtained experimentally
in the absence of the second component.
We show how to find IAS isotherms numerically in Section 5.3.1.
For both the extended Langmuir and IAS isotherms we discuss conditions on hyperbolicity of the system (3.34)-(3.35); we propose a numerical approximation and show
numerical simulations.
We also discuss analytical extensions to (3.32)-(3.33) in Section 5.3.9.
69
4.
EVOLUTION PROBLEM WITH A MONOTONE GRAPH.
ANALYSIS AND NUMERICAL APPROXIMATIONS.
In this section we discuss the methane hydrate problem (3.11)–(3.12). We frame it
as an extension to an evolution problem with a monotone graph
∂u
− 4v = f,
∂t
v ∈ α(u) on Ω × (0, T ),
v = 0 on ∂Ω × (0, T ),
u(·, 0) = u0 (·) on Ω.
(4.1)
(4.2)
(4.3)
The problem (4.1)–(4.3) falls in the category of equations known as Porous Medium
Equation which include: the Stefan free boundary value problem, and fast and slow diffusion equations [13, 14]. Solutions to such problems have singularities and therefore only
can be understood in a weak sense.
The methane hydrate problem that we describe in Section 3.1. is similar to the
Stefan problem but differs from it in one very important way.
Recall the simplified model (3.12). As we explained there, the graph α has to be
parametrized by the independent variable x, because of the dependence of the maximum
solubility constraint related to v ∗ on the depth. This is the defining element in the
construction of the graph α. We recall this model for convenience here
∂u
− 4v = f,
∂t
v ∈ α(x; u) on Ω × (0, T ),
v = 0 on ∂Ω × (0, T ),
u(·, 0) = u0 (·) on Ω.
(4.4)
(4.5)
(4.6)
Therefore, we develop new analysis for an extended version (4.4)–(4.6) of (4.1)–
(4.3). This analysis requires a new construction of abstract operators used in the evolution
problem.
70
We also develop a numerical scheme for the methane hydrate problem. The scheme
is similar to the schemes originally developed for the Stefan problem. However, it includes
the extension to α = α(x; ·). Another important result is that we develop a way to treat
the nonlinear graph v ∈ α(x; u). We note that due to α not being a function, in practical implementations most numerical algorithms for (4.1)–(4.3) approximate α with some
function with a steep gradient. In contrast our method for (4.4)–(4.6) does not require
any regularization.
The results on analysis and approximation shown here including convergence in 1D
were published in [16], and we present an expanded version below. This chapter also
includes new 2D results as well as 1D simulation examples not included in [16].
4.1.
Analysis
We will analyse the initial–boundary value problem (4.4)–(4.6) which we first cast
as an abstract initial boundary value problem.
We let H = L2 (Ω), the Lebesgue space with inner product (·, ·), V = H01 (Ω), the
Sobolev space with inner product (u, w)V = (∇v, ∇w) and dual space V 0 = H −1 (Ω). The
Riesz map is given by −∆ : V → V 0 . Note that V ⊂ H ⊂ V 0 and
(f, g)V 0 = ((−4)−1 f, (−4)−1 g)V = f ((−4)−1 g),
f, g ∈ V 0 .
An equation “on Ω” (or “on Ω × (0, T )”) means that it holds in V 0 or L1 (Ω), respectively, for a.e. t ∈ (0, T ), and similarly for ∂Ω and ∂Ω × (0, T ).
We first analyse the initial–boundary–value problem (4.4)–(4.6) with a graph α
without parametrization. That is, we consider (4.1)–(4.3). Next, we discuss (4.4)–(4.6).
For any relation A on H, that is, a subset A of H × H, we define the domain
D(A) = {x : [x, y] ∈ A}.
We consider A = −∆ ◦ α with an appropriate domain D(A), and we formulate
71
(4.1)–(4.3) as
du(t)
+ A(u(t)) 3 f (t) a.e. on (0, T ], u(0) = u0 .
dt
(4.7)
We also have to specify D(A) and the spaces in which (4.7) is considered.
The operator A has been studied extensively. In particular, it has been shown in
[40] that the operator A is m–accretive on (i) L1 (Ω) and on (ii) V 0 .
Thus we can consider two notions of solutions. The first one due to (i) is
u ∈ C([0, T ], L1 (Ω)) with v(t) ∈ W01,1 (Ω),
Ov(t) ∈ L1 (Ω).
(4.8)
The second notion due to (ii) is
u ∈ W 1,1 ((0, T ), V 0 ) with v(t) ∈ V.
(4.9)
Note that it is possible in some particular cases, if the initial data u0 ∈ L1 (Ω) ∩ V 0 , to
have a solution satisfying both notions (4.8) and (4.9).
There are two very well known results on these abstract settings.
Theorem 4.1.0.1 (Theorem 3.1 in [16], p. 821 and Theorem 4.3 in [40], p. 186). Assume
that A is m–accretive on the Hilbert space H. For each u0 ∈ D(A) and f ∈ W 1,1 ((0, T ), H),
there is a unique u ∈ W 1,∞ ((0, T ), H) which satisfies (4.7).
Theorem 4.1.0.2 (Theorem 3.2 in [16], p. 821 and Theorem 3.6 [1]). Assume A = ∂Ψ
is a subgradient on the Hilbert space H, u0 ∈ D(A) and f ∈ L2 ((0, T ), H). Then there is a
√
2
unique u ∈ C([0, T ], H) with t du
dt ∈ L ((0, T ), H) and u(t) ∈ D(A) a.e. which satisfies
(4.7). If u0 ∈ D(A), then
du
dt
∈ L2 ((0, T ), H).
We now describe how to transform the initial–boundary–value problem (4.1)–(4.3)
to the abstract setting (4.7) so we can apply the above theorems.
Take H = V 0 and the operator A = −∆◦α with domain Dom(A) = {u ∈ V 0 ∩L1 (Ω) :
for some v ∈ V, v ∈ α(u)} and −∆v ∈ A(u) for all such v. If α(·) is a maximal monotone
graph onto R, then A is a subgradient in V 0 so we can apply Theorem 4.1.0.2.
72
Additionally, if α or α−1 are Lipschitz functions, then extra regularity results apply.
In order to apply these results just mentioned, we have to show that α associated
to the examples of interest is a maximal monotone graph. This can be done for the
Stefan problem with the graph αST discussed in Section 4.3.3, and for singular graph αE
discussed in Section 4.3.1.
For αM H discussed in Section 4.3.2, we have to do extra work, because it is parametrized
−1
by x, and because the graph βM H = αM
H is only defined for 0 ≤ u ≤ R, thus its is not
maximal.
Remark 4.1.0.1. The Stefan problem α is given by
α = αST = u− + (u − 1)+ .
The Stefan Problem is presented in Section 4.3.3. This graph is a maximal monotone
graph onto R. (See Figure 4.4.) Note that αST is Lipschitz on R (see [39].) Moreover,
it is known that (4.1)–(4.3) has a unique bounded solution with ∇v, vt ∈ L2 (Ω × (0, T )).
See [11, 40]. The continuity of v(x, t) was established in [5]. If not α but α−1 is maximal
monotone onto R the continuity of u(x, t) follows (see [10]).
Remark 4.1.0.2. Now consider the graph α = αE = {0} × (−∞, 1] ∪ [0, ∞) × {1},
presented in Section 4.3.1, Figure 4.2. We see it is monotone. One can make an affine
−1
= αW =
extension resulting in a maximal monotone graph. Note that its inverse αE
(−∞, 1] × {0} ∪ {1} × [0, ∞) and that both αE and αW are not functions. Therefore there
are no results on further regularity. Since we have analytical solutions available in [41],
we use it in [16] to test our numerical scheme in section 4.4.
Remark 4.1.0.3. For our main example,
αM H (u) = (u − v ∗ )− + v ∗ ,
(4.10)
we note that this graph is a translate of the Stefan graph for a fixed x. If the graph does
not depend on x, αM H has the same properties as αST thus the variables involved in the
73
problem share the same properties as the variables in the Stefan problem. This is true as
long as αM H can be extended for all u. See Figure 4.3 in Section 4.3.2.
Observe that for each fixed x both αM H (x; ·) and βM H (x; ·) are monotone graphs.
Moreover, the saturation S = S(x; u) is a function



u ≤ v ∗ (x),
u − R 1,
S=
=
v−R 

 ∗u−R , u > v ∗ (x),
v (x)−R
which by (3.13) is monotone decreasing in u with values in [0, 1] as long as 0 ≤ u ≤ R.
4.1.1
Family of monotone graphs
Now, our problem requires that we deal with a family of maximal monotone graphs
{α(x; ·) : x ∈ Ω}. We will take advantage of the fact that the inverse of α : α−1 (x; ·) =
β(x; ·) is a subgradient on H, for a prescribed convex function φ(x; ·) on R.
Note that in our primary problem, the inverse β(x, ·) is given by
βM H (x, ·) := {(v, v) : v ≤ v ∗ (x)} ∪ {v ∗ (x)} × [v ∗ (x), R).
(4.11)
1
φ(x; v) = v 2 + (R − v ∗ (x))(v − v ∗ (x))+ .
2
(4.12)
We take
Thus β(x, ·) = ∂φ(x; ·), a maximal monotone extension of (4.11).
4.1.2
Construction of a normal convex integrand
In order to tackle the case of a family of monotone graphs, we use part of the theory
developed in [34].
Definition 4.1.2.1. A function ϕ : Ω×R → R∞ is a convex integrand if ϕ(x; ·) : R → R∞
is convex for each x ∈ Ω.
Definition 4.1.2.2. We shall call a convex integrand ϕ normal if ϕ(x, ξ) is proper and
lower semi–continuous in ξ for each x, and if further there exists a countable collection B
of measurable functions w from Ω to R having the following properties:
74
(a) for each w ∈ B, ϕ(x; w(x)) is measurable in x;
(b) B(x) ≡ {w(x) : w ∈ B} satisfies B(x) ∩ Dom(ϕ(x; ·)) is dense in Dom(ϕ(x; ·)) for
each x ∈ Ω, where Dom(ϕ(x; ·)) = {ξ ∈ R : ϕ(x; ξ) < ∞} is the effective domain.
Note that in our primary case, v ∗ is piecewise smooth, so the function ϕ(x, ·) defined
in (4.12) is a normal convex integrand. Moreover, the class B can be constructed from
step functions w with rational values. Now, x 7→ ϕ(x; w(x)) is measurable for each w ∈ H,
so we can define using the normal convex integrand ϕ the function
Z
Φ(w) =
ϕ(x; w(x))dx,
w ∈ H.
(4.13)
Ω
Notice that Φ is proper, lower semicontinuous and convex and u ∈ ∂Φ if and only
if u, v ∈ H, and u(x) ∈ ∂ϕ(x; v(x)) ≡ β(x; v(x)) a.e. on Ω.
4.1.3
Abstract initial–value problem
Our next task is to formulate the initial–boundary–valued problem (4.4)–(4.6) as
an abstract initial value problem as in (4.7).
We define the operator A on V 0 by A(u) = {−4v : v ∈ V, u ∈ H, u ∈ ∂Φ(v)} on
Dom(A) = {u ∈ H : u ∈ ∂Φ(v) for some v ∈ V}. It can be shown that A is m–accretive.
Showing that A is accretive is not difficult. For the m–accretive part, we need to solve
the problem
u ∈ H, v ∈ V : u − 4v = f, u ∈ ∂Φ(v),
for each f ∈ V 0 . This problem can be solved by standard theory of monotone operators
and using the affine bound
|η| ≤ a|ξ| + b for all η ∈ β(x; ξ), x ∈ Ω, ξ ∈ R,
with an a–priori error estimate.
Now we apply Theorem 4.1.0.1 and summarize it in the following theorem.
(4.14)
75
Theorem 4.1.3.1. Assume that Φ : H → R∞ is given by (4.13) as a normal convex integrand, and assume the subgradients β(x; ξ) = ∂ϕ(x; ξ) satisfy (4.14). Let f ∈
W 1,1 ((0, T ), V 0 ) and u0 ∈ ∂Φ(v0 ) for some v0 ∈ V. Then there is a unique pair u ∈
W 1,∞ ((0, T ), V 0 ), v ∈ L∞ ((0, T ), V) which satisfies
4.1.4
du
− 4v(t)
dt
=
f (t) in V 0 for a.e. t ∈ (0, T ),
(4.15)
u(t) ∈ ∂Φ(v(t))
in
H, v(t) ∈ V for all t ∈ (0, T ),
(4.16)
u(·, 0)
=
u0 (·) on Ω.
(4.17)
Comparison principle
We have now a good framework to show a comparison principle for the evolution
problem with a family of graphs. We need to recall the following definition.
Definition 4.1.4.1. Let A be m–accretive on H. Then the Yosida approximation of A is
the operator
Aλ ≡
1
(I − Jλ ),
λ
λ > 0,
where Jλ is its corresponding resolvent Jλ ≡ (I + λA)−1 .
Lemma 4.1.4.1. Assume that u1 and u2 are solutions of the problem
uj − 4vj = fj in L1 (Ω), vj ∈ W01,1 (Ω), vj (x) ∈ α(x; uj (x))a.e. in Ω
(j = 1, 2). (4.18)
Then
k(u1 − u2 )+ kL1 (Ω) ≤ k(f1 − f2 )+ kL1 (Ω) .
(4.19)
Consequently, if f1 ≤ f2 , then u1 ≤ u2 .
Proof. We approximate problem (4.18) by replacing β(x; ·) by its Yosida approximation
βλ (x; ·),
λ > 0, so αλ (x; ξ) = βλ−1 (x; ξ) = α(x; ξ) + λξ is strictly increasing. This
approximated version of problem (4.18) is given by
uλj − 4vjλ = fj in L1 (Ω), vjλ ∈ W01,1 (Ω), vjλ (x) ∈ αjλ (x; uλj (x)) a.e. in Ω
(j = 1, 2).
(4.20)
76
We need an approximation of the Heaviside




0




H (v) = v






1
function.
for v ≤ 0,
for 0 ≤ v ≤ ,
for ≤ v,
and lim→0 H (v) = H0 (v), where
H0 (v) =



0
for v ≤ 0,


1
for v > 0.
The corresponding monotone graph is the extension H(·) with H(0) = [0, 1].
Let σ = H (v1λ − v2λ ), then
σ → H0 (v1λ − v2λ ) ∈ H(uλ1 − uλ2 ) as → 0,
(4.21)
since αλ (x; ·) is strictly increasing. We subtract equations in (4.20), multiply by σ and
integrate over Ω to get
Z
Z
Z
λ
λ
λ
λ
(u1 − u2 )σ dx −
4(v1 − v2 )σ dx = (f1 − f2 )σ dx.
Ω
Ω
(4.22)
Ω
We integrate by parts the second term in (4.22)
Z
Z
2
− 4(v1λ − v2λ )σ dx =
|O(v1λ − v2λ )| H0 (v1λ − v2λ )dx ≥ 0.
Ω
Ω
Dropping this term in (4.22) and letting → 0 results in the inequality
Z
Z
λ
λ
λ
λ
(u1 − u2 )H0 (v1 − v2 )dx ≤
(f1 − f2 )+ dx,
Ω
(4.23)
Ω
where w+ is the positive part of w, w+ = wH0 (w) = max{w, 0}.
Now, we recall from (4.21) that H0 (v1λ − v2λ ) ∈ H(uλ1 − uλ2 ) and use it in the left
hand side of (4.23). We end up with
k(uλ1 − uλ2 )+ kL1 (Ω) ≤ k(f1 − f2 )+ kL1 (Ω) .
(4.24)
It is known that uλj → uj for j = 1, 2 as λ → 0, (see [4]) even in the parametrized case.
With this last remark, we are done with the proof.
77
Corollary 4.1.4.1 ([16], Corollary 3.1, p. 14). Assume the constraint satisfies
−4v ∗ ≥ 0.
(4.25)
Let u be a solution of
u − 4v = f in L1 (Ω),
v = αM H (·; u) in W01,1 .
If 0 ≤ f (x) ≤ R a.e in Ω, then 0 ≤ u(x) ≤ R a.e. in Ω.
Proof. We will follow an argument similar to the one employed in the last Theorem.
λ
λ
Take u2 (x) ≡ R and v2λ (x) = vλ∗ (x) = αM
H (x; R) and note that v2 |∂Ω > 0. Define
f2λ = R − 4vλ∗ = u2 − 4v2λ .
Let u and v satisfy the problem
uλ − 4v λ = f ≤ R in L1 (Ω),
1,1
λ
λ
v λ (x) = αM
H (x; u (x)) in W0 .
Then
uλ − u2 − 4(v λ − v2λ ) = f − f2 ≤ 4vλ∗ ≤ 0.
(4.26)
Multiply (4.26) by H (v λ − v2λ ) and integrate
Z
Z
Z
(uλ − u2 )H (v λ − v2λ )dx − 4(v λ − v2λ )H (v λ − v2λ )dx = (f − f2 )H (v λ − v2λ )dx ≤ 0.
Ω
Ω
Ω
(4.27)
Note that (v λ − v2λ )|∂Ω < 0.
Integrate by parts the second term of (4.27) and following the same reasoning of
last theorem we conclude that k(u − u2 )+ kL1 (Ω) ≤ 0 implies u − u2 ≤ 0. That is
u ≤ R.
We have found that R is an upper bound for u, for the lower bound we let u1 ≡ 0 and
λ
λ
∗
λ
λ
v1λ (x) ≡ vλ∗ (x) = αM
H (x; 0) = 0. Then we define f1 = 0 − 4vλ = u1 − 4v1 . Similarly to
the proof for the upper bound, we let u and v satisfy the problem
uλ − 4v λ = f ≥ 0 in L1 (Ω),
v λ (x) = αM H (x; uλ (x)) in W01,1 .
78
Then
uλ − uλ1 − 4(v λ − v1λ ) ≥ f − f λ = f + 4vλ∗ = f ≥ 0.
(4.28)
Multiplying both sides by H (v λ − v1λ ) and integrating by parts (4.28) we obtain u ≥ uλ1 ≡
0. This finishes our proof for the lower bound.
Now for two solutions uj (t), vj (t) to the evolution problem (4.15)–(4.17) we are back
to solving the implicit in–time problems
unj − un−1
j
tn − tn−1
− 4vjn = fjn ,
vjn ∈ α(x; unj ),
where j = 1, 2, u0j = uj (0), and the solutions for these stationary problems are in L1 (Ω).
The estimate (4.24) carries over to the limit uj (t) of the time–discrete unj and we have
Z
k(u1 (t) − u2 (t))+ kL1 (Ω) ≤ k(u1 (0) − u2 (0))+ kL1 (Ω) +
0
1
k(f1 (s) − f2 (s)+ kL1 (Ω) ds. (4.29)
The reasoning here follows along the usual path (see [40], p. 221).
Corollary 4.1.4.2 (Corollary 3.2, [16]). If u(t), v(t) is a solution of (4.4)–(4.6) with
0 ≤ u(0) ≤ R and 0 ≤ f (t) ≤ −4v ∗ , then we have 0 ≤ u(t) ≤ R.
Proof. As before, for the methane hydrate problem of (4.10) with a subharmonic constraint, we can use the constant solution u2 (t) = R ≥ 0 to bound the solution u2 (t)
of the initial–boundary–value problem (4.4)-(4.6) and taking v2 (t) = v ∗ = α(x; R) and
f2 (t) = −4v ∗ . Indeed, since u(0)−u2 (0) ≤ 0 and f (s)−f2 (s) ≤ 0 then the right hand side
of inequality (4.29) is zero. So k(u(t) − u2 (t))+ kL1 (Ω) ≤ 0 which implies that u(t) ≤ R.
For the lower bound, again, we choose v1 (x) = α(x; 0) = 0 and obtain 0 ≤ u(t).
Remark 4.1.4.1. The comparison principle lets us extend the graph β as before, so the
results for Stefan problem apply. It also lets us formulate uniqueness for the L1 notion of
solutions extending (4.8) for the case α = α(x; ·).
79
4.1.5
Summary of analysis and outlook
In Corollary 4.1.4.1 we required (4.25). When v ∗ ≡ const the graph αM H is, up to
a translation, the same as αST . In this case, the solutions v are continuous. In addition,
(4.25) holds if v ∗ is affine in x.
More generally, (4.25) requires that v ∗ is concave. This might not be always true
in practice, and thus we hope to be able to weaken the sufficient condition (4.25). In
particular, some of the numerical experiments reported in the sequel exhibited similar
convergence properties to those for the affine and constant case, but (4.25) did not hold.
4.2.
Numerical approximation for IBVP with a monotone graph
We start with a literature review. Most of the work on numerical approximation for
free boundary problems was done for Stefan problem (see Section 4.4.1 for a description).
The main difficulties are due to the low regularity of the weak solutions, as well as to the
difficulty of resolving the nonlinear relationship v ∈ α(u). In addition, all the known work
was done for problem (4.1)–(4.3), and did not allow parametrization as in (4.4)–(4.6).
In this section we formulate the numerical model for the generalized version (4.4)–
(4.6) of (4.1)–(4.3). We also show how to implement a fully implicit solver and how to
realise in practice the nonlinear relationship v ∈ α(u). This is done with a semismooth
and Nolinear Complementarity Constraint which was introduced in Section 2.2.
The solver works so well that we extended it to the Stefan problem and the case of
a singular graph, for which neither α nor α−1 is a function.
In what follows we denote by Q = Ω×(0, T ) and L̄2 (0, T ; L2 (Ω)) the space equipped
with the L2 discrete norm as defined in (4.58) with p = 2.
80
4.2.1
Literature review
For Stefan problem and Porous Medium Equation we briefly review known results
on numerical approximation. For Stefan problem, the problem is of the form
∂H(u)
− ∆u + f (u) = 0,
dt
where u represents the temperature and e = H(u) the enthalpy. In our problem the
temperature plays the role of solubility and the enthalpy the role of total amount of
methane. It was shown in [20] first that for the two–phase Stefan problem one can obtain
a rate of convergence of O(h) in L̄2 (0, T ; L2 (Ω)) for temperature and the same rate for
0
enthalpy in L∞ (0, T ; F ), where F = (H 1 (Ω)) .
The authors in [20] develop a regularity theory and make a series of estimates for
both u and H(u). We have to point out that they obtain different rates of convergence
depending on which set of assumptions they use. The regularized version of the enthalpy,
H has to be continuously differentiable and Lipschitz. Let u be a weak solution for
the two-phase Stefan problem, Uhn is the sequence of discrete solutions (see [20], where
Uhn satisfy problem (5.1i)–(5.2i)). Choosing = 0 h4/3 and ∆t ≤ ch4/3 with 0 and c
positive constants, they obtain rate of O(h2/3 ) on L̄(0, T ; L2 (Ω)). With an additional
hypothesis on grid quasiuniformity, i.e., for all h > 0 and triangulation Th we must have
inf T ∈Th
σ(T )
h
≥ γ0 > 0, where σ(T ) is the radius of largest ball contained in T , and
a uniform bound for H (Uhn ) in L2 (Ω), they obtain a rate of convergence for H(u) of
O(h2/3 ) in L∞ (0, T ; F ) ([20], Theorem 5.1, p. 405).
Now, we mention the assumptions needed to get O(h) rate of convergence. Take
the same assumptions as before but now suppose that ∆t ≤ ch2 , = 0 h2 with c, 0 > 0.
Then [20] gives rate of convergence of O(h) in L̄2 (0, T ; L2 (Ω)). However, they have to add
the quasiuniform hypothesis in order to get rate O(h) for H(u) in L∞ (0, T ; F ). With an
extra regularity assumption, we just need 0 ≤ ≤ 0 h2 and ∆t ≤ ch to obtain O(h) for u
in L̄(0, T ; L2 (Ω)).
81
For a more general evolution problem, [36] deals with a fully discrete implicit scheme
for degenerate parabolic problems. The form of the problem for them is
∂e
− ∆u = f, e ∈ β(u) in (0, T ) × Ω,
∂t
where β is a maximal monotone graph, e is energy (or enthalpy) and u is temperature.
Results in this paper apply to both Stefan problem and Porous Medium Equation. They
found general estimates for the fully discrete scheme. They showed that the best rate of
α
α
convergence is O(τ +ln(1/h)1+ 2 h) in L∞ (0, T ; H −1 ), for enthalpy and O(τ +ln(1/h)1+ 2 h)
in L2 (0, T ; L2 (Ω)) for temperature.
In [36] no regularization is needed. They give a rate of convergence for temperature
in L2 (Q). For the energy e, the rate of convergence is given in the norm L∞ (0, T ; H −1 ).
α
Moreover, for u the rate is O(τ + ln(1/h)1+ 2 h) and in the 1D case is O(h4/5 ) in L2 (Q).
Their assumptions are: f ∈ Lip(0, T ; L1 (Ω)) ∩ H 1 (0, T ; H −1 ) ∩ L1 (0, T ; L∞ (Ω)), u0 ∈
L∞ (Ω) ∩ H01 (Ω) ∩ W 2,1 (Ω). Conditions on β are |β(s)| ≤ C(1 + |s|α ) for some α ≥ 0. In
1D we have to choose τ = h4/5 and we need β −1 Lipschitz to get rate of convergence in u.
Moreover, the authors point out how to make a selection out the graph to obtain
a function. In Figure 4.1 we reproduce the diagram shown in [36] and explain how they
make the unique selection. Here B is a primitive for β, i.e., β = ∂B and i is the embedding
Vh
i
∂(B ◦ i)
Vh0
Lp (Ω)
B
R
∂B
i0
0
Lp (Ω)
FIGURE 4.1: Selection diagram from [36].
i : Vh → Lp (Ω). They are working in the framework of the discrete dual space Vh0 . Denote
P0h the L2 projection onto Vh . That is, (P0h v, wh ) = (v, wh ) for all wh in Vh for v ∈ L2 (Ω).
82
The fully discrete scheme is
un+1 ∈ Vh , en+1
∈ P0h β(un+1 ),
(en+1 , vh ) + τ (∇un+1 , ∇vh ) = τ hf n+1/2 , vh i + (en , vh ) ∀vh ∈ Vh .
(4.30)
The authors from [20] show that the solution of the scheme in (4.30) may be found as the
minimum of Φ : Vh → R, given by
Z n
o
τ
Φτ (u) =
|∇u|2 + B(u) − hτ f n+1/2 + en , ui
Ω 2
over Vh .
In lemma 2.4 from [36] it is proved that there is ẽn+1 ∈ β(un+1 ) such that en+1 =
P0h ẽn+1 . This is a consequence of the chain rule for convex functions. For this problem,
that means that the diagram from Figure 4.1 commutes.
Now, B ◦ i is continuous and convex and using the commutativity of the diagram,
∂(B ◦ i) = i0 ◦ ∂B ◦ i. So if en+1 ∈ (i0 ◦ ∂B ◦ i)(un+1 ) then there is ẽn+1 ∈ ∂B(un+1 ) and
i0 (ẽn+1 ) = en+1 which can be shown to be equivalent to Ph0 (ẽn+1 ) = en+1 .
In [27], the problem is written in the form
ut − ∇x · [∇x v + b(r(v))] + f (r(v)) = 0, u ∈ γ(v)
(4.31)
and γ is a maximal monotone graph. They study the case of the Stefan problem (r(v) = v),
with b 6= 0, f 6= 0, u is the enthalpy and v is the temperature in (4.31). For the Stefan
problem, the order of error in u is O(h1/3 ) in the space L2 (Q).
One of the assumptions that they use to obtain this rates of convergence is the
non–degeneracy property, i.e.,


meas {0 < v < s } ∩ 

\
F (t) ≤ C,
(4.32)
0≤t≤T
where F (t) is the phase change (see Section 4.4.1 for more explanation about phase change
in Stefan problem). Choose s = 1 for Stefan problem and s =
Equation.
1
m+1
for Porous Medium
83
The main contribution of [27] is introducing numerical integration to get a scheme
that is easy to implement. They use a smoothing procedure. So regularization is needed,
that means that they replace γ by a smooth function γ . However, their error analysis is
not based in additional regularity assumptions like in [14], [20] and [28].
For Stefan problem, the rates of convergence are in L2 (Q). Temperature is v and
enthalpy is denoted by u. In (4.31), b and f have to be uniformly Lipschitz continuous. In
addition, u0 ∈ L2 (Ω) with v0 = γ −1 (u0 ). They have to assume the non–degeneracy property (4.32). Choosing τ = C1 h4/3 and = C2 h2/3 (C1 and C2 are arbitrary) [[27],Corollary
2, p.806].
For the Porous Medium Equations (γ(s) = s1/m , r(s) = s), assuming the non–
4m
degeneracy property (4.32) and taking τ = C1 h 3m−1 , = C2 h
2(m−1)
3m−1
(where C1 , C2 > 0),
among other assumptions. they get that rate of convergence in u in the norm Lm+1 is
4m
O(h 3m−1 ) [[27], Corollary 4, p. 807].
In [28], the problem is written in the form
γ(u)t − ∇x · [∇x u + b(u)] + f (u) = 0.
(4.33)
Now, u is the temperature and γ represents the enthalpy. Moreover, γ is a maximal
monotone graph in R × R with a singularity at the origin. The author obtained sharp
errors of convergence, O(h) and O(h1/2 ) for u and γ(u) both in L̄2 (0, T ; L2 (Ω)). These
bounds were obtained using regularization. Initially, rates of convergence for the variables
are obtained in suitable Lp spaces, but assuming property in (4.32), results are improved.
Indeed, for the Stefan problem if property in (4.32) is not assumed, taking h4/3 ∼ ∼ τ
then a L2 rate of O(h2/3 ) is proved. On the other hand, if property (4.32) is assumed then
L2 rates of O(h) for temperature and O(h1/2 ) for enthalpy are obtained.
For the Porous Medium Equation, if the non-degeneracy property is assumed and
(m+1)
3m
2
(2m−1)
(2m−1)
∼h
,τ ∼h
then a L rate of O h
is obtained for the density. Now,
4m
if b = f = 0 in (4.33), a Lm+1 rate of convergence of O h (2m−1)(m+1) is obtained. If we
2(m−1)
(2m−1)
84
4(m−1)
4(m−1)
don’t assume the non–degeneracy property (4.32) and ∼ h (3m−1) , τ ∼ h (3m−1) if m ≥ 3
3(m+1)
4
and τ ∼ h (3m−1) if m < 3, they get L2 rate of O(h (3m−1) ) for the density. If b = f = 0 the
same latter rate is obtained, but author indicated that it might not be sharp.
In [35] the author proved the rate of convergence in L2 (Q). The problem is written
in the form
du(t)
+ A(u(t)) 3 0, u(0) = u0 ,
dt
which is a Cauchy problem. In the case of Porous Media or Stefan problem, A(u) =
−∆ ◦ α(u), and α is monotone on H −1 . The author does not use regularization and A
is allowed to be multivalued. For the Stefan problem, α(u) is the temperature and its
rate of convergence is O(h) in L2 (0, ∞; L2 (Ω)), and the gradient of enthalpy u has rate of
convergence O(h1/2 ) in L2 (0, ∞; L2 (Ω)). In order to obtain these rates, the author assumes
that A can be expressed as A = ∂ϕ + B where ∂ϕ is a subgradient and B is monotone. In
addition, it is assumed that there is a constant C > 0 such that (u, w) ≥ −Ckv + wk2 for
all u ∈ D(A), v ∈ ∂ϕ(u) and w ∈ B(u) [[35], Theorem 5, p. 79].
4.2.2
Semidiscrete formulation
In this section we define discretization in time of the problem (4.4)–(4.6) but written
with the graph β = α−1 as discussed in Section 4.1.1, and in a weak sense as discused in
Theorem 4.1.3.1.
We assume uniform time stepping tn = nτ, with a time step τ. We seek finite
difference approximations in time un , v n to u and v. We use fully implicit in time difference
scheme
(un , Φ) + τ (∇v n , ∇Φ)
=
(un−1 , Φ),
un
∈
β(v n ),
(u0 , Φ) := (u0 , Φ).
∀Φ ∈ V
(4.34)
(4.35)
(4.36)
85
Below we formulate a fully discrete scheme by considering a finite dimensional subspace
Vh ⊂ V.
The problem (4.34)–(4.36) could be analysed using Semigroup theory but this will
not be done here. Instead, below we focus on the main challenge which is realising (4.35)
with the β without regularization.
4.2.3
Fully discrete implicit scheme
Now we consider problem (4.34)–(4.36) with a finite element discretization in space.
We follow the notation from Section 2.1.3.
Let Vh ⊂ V be the finite element space given by piecewise linears over a triangulation
of Ω, and associate interpolant Ih with range in Vh . The scheme involves the solution in
vh ∈ Vh at each time step tn , n > 0 of
(unh , Φ) + τ (∇vhn , ∇Φ)
=
(un−1
h , Φ),
unh
∈
βh (vhn ),
(4.38)
(u0h , Φ) := (u0 , Φ),
(4.39)
∀Φ ∈ Vh
(4.37)
where we have to specify what is the choice unh ∈ βh . In general, the selection out of
β(vh (xj )) is not unique unless we specify how to solve (4.37)–(4.39). We present here
a duality argument from [36] (Lemma 2.4) that allows to give a unique meaning to βh .
For graphs such as αE or parameter dependent families such as the one associated to
methane hydrates, αM H (x; u), there are not available results in the literature. We can
extend the scheme (4.37)–(4.39) for this case. We keep the first equation (4.37). However,
we will write in a matrix–vector form as follows. We identify vhn ≈ vn ∈ RM by its degrees
of freedom (v1 , . . . , vm ). Let M be the mass matrix defined by (uh , wh ) = wT Mu and
(∇uh , ∇wh ) = wT Ku for any uh , wh ∈ Vh . So, (4.37) is written in terms of these matrices
as
Mun + τ Kvn = Mun−1 .
(4.40)
86
Instead of using the L2 (Ω) inner product (u, Φ), we use its numerical approximation
(w, Φ)h or mass lumping. In one dimension and in the case of a uniform grid, M is replaced
by hI. Hence, (4.40) can be written as
un + τ Ah vn = un−1 ,
(4.41)
where Ah := M−1 K. Notice that Ah is symmetric positive definite. Now, we deal with
(4.38) and use the pointwise selection
hvjn , unj i ∈ β(xj , ·) := βj (·),
(4.42)
and we complemented with the initial selection in (4.39) to have a complete scheme. We
show that the latter scheme is uniquely solvable for the examples from Section 4.1. Before
proving solvability, we need the following definition and proposition.
Let K be a closed convex set, we define the indicator function as



0
for x ∈ K
Ik (x) =


∞ otherwise .
Lemma 4.2.3.1. For every n > 0 there is a unique solution vn ∈ Rn of the scheme
(4.40),(4.42) and (4.39) for β = βM H ; it is the unique minimizer of the appropriate
functional Ψ(v) for which the latter scheme is the Euler–Lagrange condition.
Proof. We consider the problem solved at every time step u = un and v = vn . The
previous time step f = vn−1 is given.
u + τ Ah v = f ,
uj ∈ βj (vj ), j = 1, 2, . . . , M.
(4.43)
We must define Ψ(v) = ΨM H (v) for β = βM H (x, ·). We consider pointwise–define convex
P
functions ϕj (λ) = 21 λ2 + I(−∞,v∗ (xj )] (λ) and Φ(v) :=
j ϕj (vj ). Now, Ah is defined
on all of RM and it is single valued, which implies that is maximal. We define Ψ as
87
Ψ(v) = 12 τ vAh vT + Φ(v). Proposition 2.1.2.3 applies and we have that the subgradient
of Ψ is given by
∂Ψ(v) = τ Ah v + ∂Φ(v).
(4.44)
We can prove solvability in an analogous way for βE and βW by defining convex functions
ϕE = I(−∞,1) and ϕW (x) = x + I[0,∞) .
Corollary 4.2.3.1. The scheme (4.40),(4.42) and (4.39) is uniquely solvable for each
βM H (x, ·), βST , βE and βW .
Now we prove the comparison principle for the simplest case, d = 1, mass lumping
and uniform grid.
Lemma 4.2.3.2. Consider the problem (4.43) - (4.44) for which Ah is the usual tridiagonal discrete Laplacian and scaled by
1
.
h2
Let solutions u(1) and u(2) with the corresponding
(1)
v(1) and v(2) satisfy (4.43) for f = f (1) , f (2) , respectively. Let also vj
(2)
− vj
= 0 for
boundary indices j. Then the counterpart of (4.19) holds, namely
M
X
(u(1) − u(2) )+ ≤
j=1
M
X
(f (1) − f (2) )+ .
j=1
Proof. We solve (4.43) for vectors u = (u1 , u2 , . . . , uM ) and v = (v1 , v2 , . . . , vM ). We set v0
and vM +1 from boundary conditions. Now denote Dj+1 v = τ
vj+1 −vj
.
h2
With this notation
and (4.43) we have the following identity
uj + (Dj v − Dj+1 v) = fj ,
uj ∈ βj (vj ).
(4.45)
So, writing (4.45) for u(m) and v(m) for m = 1, 2, that is,
(1)
(1)
∈ β j vj
,
(4.46)
(2)
(2)
∈ β j vj
.
(4.47)
(1)
uj
(2)
uj
uj + (Dj v(1) − Dj+1 v(1) ) = fj ,
and
uj + (Dj v(2) − Dj+1 v(2) ) = fj ,
88
We subtract (4.47) from (4.46) and denote w = v(1) − v(2) , then multiply both sides by
H(wj ) and sum over j :
M
X
(1)
(2)
H(wj )(uj − uj )+
j
M
X
H(wj )(Dj w − Dj+1 w) =
j
M
X
(1)
H(wj )(fj
(2)
− fj ) (4.48)
j
≤
M
X
(f (1) − f (2) )+ .
(4.49)
j=1
The last inequality in (4.49) is due to the fact that H(wj ) ∈ [0, 1]. Now we work on the
second term of (4.48) and apply summation by parts. That is,
M
X

H(wj )(Dj w − Dj+1 w) =
j=1
τ 
H(w1 )(w1 − w2 )
h2
+ H(wM )(wM

M
X
− wM +1 ) +
(H(wj ) − H(wj−1 ))(wj − wj−1 )
j=2

=
τ 
H(w1 )w1 + H(wM )wM
h2
− H(w1 )w0 − H(wM )wM +1 +
M
X

(H(wj ) − H(wj−1 ))(wj − wj−1 ) .
j=2
(4.50)
Since H is monotone, the first two terms in (4.50) are nonnegative as well as the last one.
(1)
Now, by hypothesis, wj = vj
(2)
− vj
= 0 for the boundary indices j = 1, M. Hence, the
second term in (4.48) is nonnegative. It remains to show that the first term in (4.48) is
nonnegative. Indeed, if βj is single-valued, we have that pointwise,
(1)
(2)
(1)
(2)
(1)
(2)
(1)
(2)
H(wj )(uj − uj ) = H(vj − vj )(uj − uj ) = (uj − uj )+ ,
and we are done for that case. To treat the multivalued case we use the single-valued
Yosida regularizations for βjλ and define uλ , vλ and wλ . The result is proved for the latter
regularized case, and then we let λ → 0. (see [1, 40]).
89
4.2.4
Semismooth implicit Newton solver.
We want to solve (4.40)–(4.39)–(4.42) with semismooth Newton method. We will
use a double set of degrees of freedom and express (4.42) as Φ(x; u, v) = 0, where Φ is
a semismooth function suitable for α (or β). Recall that in Section 2.3. we showed that
piecewise smooth functions are semismooth. Moreover, in the same Section, it was shown
that the semismooth Newton algorithm converge superlinearly.
For the methane hydrate problem, we can express βM H as a complementarity constraint. Example 2.3.0.3 shows that a similar minimum function is semismooth. We have
seen that for finite dimensional problems it is efficient to use a solver from the class of
semismooth Newton methods. On the other hand, CC–Newton solvers can be applied to
the Stefan problem without the need of regularization, and to αE and αW as well.
4.2.5
Formulation of constraint as an NCC (Nonlinear Complementarity Constraint)
It is clear that in our context




λ ≥ 0,




λ = 0,






µλ
if µ = 0,
if µ ≥ 0,
= 0,
can be written as min(λ, µ) = 0.
Let hu, vi ∈ β be represented as a NCC in the form min(F (u, v), G(u, v)) = 0, where
F and G are semismooth functions. Then (4.39)–(4.40)–(4.42) is solved as
u + τ Ah = b
min(Fj (uj , vj ), Gj (uj , vj )) = 0,
(4.51)
∀j
(4.52)
where b is given by the previous time step. Now, we apply the semismooth Newton method
to solve (4.51)–(4.52). We need to figure out what is the structure of the Jacobian.
90
The following example illustrates how to describe the block of a 8 × 8 Jacobian for
the problem (4.51)–(4.52). It helps us to introduce notation as well.
Example 4.2.5.1. Let u = (0.2, 0.4, 0.3, 0.1) and v = (0.7, 0.8, 0.7, 0.6). Let Fj and Gj
be given by Fj (uj , vj ) = uj and Gj (uj , vj ) = 1 − vj , with 1 ≤ j ≤ 4. Then, J − = {j :
uj −(1−vj ) < 0} = {j : uj +vj −1 < 0}. In this case, J − = {1, 4}, J 0 = {3} and J + = {2}.
So u− = (u1 , u4 ), u0 = u3 and u+ = u2 . On the other hand, v− = (v1 , v4 ), v0 = v3 and
v+ = v2 . In addition, J −,0 = {1, 3, 4}, J +,0 = {2, 3} and v−,0 = (v1 , v3 , v4 ). The Jacobian
is formed by the 4 × 4 blocks J11 = I4×4 , J12 = τ Ah ,

J21
1 0 0 0





 0 0 0 0 


=
,


 0 0 1 0 


0 0 0 1
and

J22
0
0
0 0


 0 −1 0 0

=

 0 0 0 0

0 0 0 0





.



Now, for the more general problem, the Jacobian of (4.51)–(4.52) is given by


 J11 J12 
J =

J21 J22
where Jij ,
1 ≤ i ≤ 2, 1 ≤ j ≤ 2 are blocks of J. Here, J11 = I and J12 = τ Ah are
constants and therefore smooth in u and v. The blocks J21 and J22 are coming from the
equation (4.52) in which we have commented that the left hand side is semismooth.
We now repeat the notation from Example 4.2.5.1 in the more general setting.
Let J − := {j : F (uj , vj ) − G(uj , vj ) < 0}, J + := {j : F (uj , vj ) − G(uj , vj ) > 0}, and
91
J 0 := {j : F (uj , vj ) − G(uj , vj ) = 0}. Note that if j ∈ J − then
(J21 (u, v)jj =
∂
∂F
min(F (uj , vj ), G(uj , vj )) =
,
∂uj
∂uj
(J21 (u, v)jj =
∂
∂G
min(F (uj , vj ), G(uj , vj )) =
.
∂uj
∂uj
and if j ∈ J + then
Observe that in J 0 is precisely where the min(·, ·) is not differentiable, but can be lumped
in implementation with J − .
Similarly, if j ∈ J − then
(J22 (u, v))jj =
∂
∂F
min(F (uj , vj ), G(uj , vj )) =
,
∂vj
∂vj
(J22 (u, v))jj =
∂
∂G
min(F (uj , vj ), G(uj , vj )) =
.
∂vj
∂vj
and if j ∈ J + then
In section 2.3., in Proposition 2.3.2.5 is shown that P C K functions are semismooth. In
our examples, F and G are semismooth so Φ is semismooth.
It has been shown in Section 2.3. that a condition for the Newton’s iterations to
converge superlinearly [condition (i), Proposition 2.3.3.1] is that J −1 (u(k) , v(k) ) can be
bounded for any (u(k) , v(k) ). For each of the following examples we point out why the
corresponding Jacobian is nonsingular.
4.3.
4.3.1
Semismooth Newton for the monotone graph α
The case of a singular graph
We consider here the graph αE and αW to be defined. The graphs are shown in
Figure 4.2. It is clear that neither αE nor αW are functions. We are interested in this
example because (i) there is an analytical solution for the evolution with this graph know
92
from literature [40]. In addition, (ii) the problem has various singular solutions and it
presents an extreme challenge to our numerical algorithm and to the semismooth solver.
We can write the associated problem for the graph αE (·) in the following ways
u ∈ βE (v) ⇐⇒ ΦE (u, v) := min(u, 1 − v) = 0.
(4.53)
The function min was considered in Section 2.3. where it was shown to be a semismooth
function. Thus we obtain the following result
αE
αW
1
1
FIGURE 4.2: Neither αE or α−1 = αW is a function.
Lemma 4.3.1.1. The semismooth Newton algorithm converges superlinearly for the problem (4.53) for any initial guess.
Proof. Since the min function is semismooth, in order to use Proposition 2.3.3.1 we have
to show that the Jacobian of the system is never singular.
In this case, F (u, v) = u and G(u, v) = 1 − v are smooth functions. The Jacobian J
is constant on J + , J 0 and J − . Now, we show that J is nonsingular. Recall the definition
of a characteristic function in a set S,
χS (j) =



1
if j ∈ S


0
if j ∈
/ S.
93
J21 and J22 are diagonal matrices with (J21 )jj = χJ −,0 (j) and (J22 )jj = −χJ + (j) . We check
if the system
J[u, v]T = [0, 0]T
(4.54)
has a nontrivial solution for u and v. First, see the form of J21 and J22 (Example 4.2.5.1 is
useful to visualize the form of the blocks), then it is clear that u−,0 = ~0 at J −,0 . Likewise,
v+,0 = ~0 at J +,0 . The equation
uj + τ M−1 Kvj = 0
is true for every j. If j ∈ J −,0 then τ (M−1 Kv − )j = 0. So (Kv− )j = 0. Since K is positive
definite, then v− = 0. Now, uj = −τ M−1 Kv and we have just proved that v ≡ 0 on all
J. Therefore, uj = 0 for all j ∈ J + . We can conclude that the only solution to (4.54) is
v = u = 0. Finally, consider the iterate u(k−1) , v(k−1) . Suppose that J + in nonempty. For
j ∈ J + we have an equation to define the new iterate u(k) , v(k) . That is,
(k)
(J22 )jj (vj
(k−1)
− vj
(k)
− vj
(k)
− 1)
) = (−1)(vj
= (−1)(vj
(k)
(k−1)
)
(k)
= φ(uj , vj )
= 0.
(k)
We obtain that vj
= 1. These values can be used to eliminate vj from the blocks involving
(k)
J11 and J12 . Similarly, we can see that for any j ∈ J −,0 we obtain uj
4.3.2
= 0.
The graph for methane hydrate problem
Now we discuss the methane hydrate problem. It is not difficult to show that this
problem can be framed as NCC as seen in Section 4.2.5. In the case of the methane
hydrate problem
u ∈ βM H (v) ⇐⇒ ΦM H (u, v) ≡ min(u − v, v ∗ (x) − v) = 0,
(4.55)
94
the original constraint (in terms of the variables u and S) can be written as
min(v ∗ (x) − v, 1 − S) = 0.
v
u
αM H
v∗
(4.56)
βM H
v∗
v∗
R
u
v∗
v
FIGURE 4.3: Graph associated to the methane hydrates problem
Lemma 4.3.2.1. The semismooth Newton algorithm converge superlinearly for the problem associated to (4.56) and is equivalent to switching of variables.
Proof. The proof of the nonsingularity of J is similar to the one from the Lemma 4.3.2.1.
Of course, the definition of the sets J − , J + and J 0 is subject to the constraint min(v ∗ (x)−
v, 1 − S) = 0. To show the equivalence to variable switching, consider the iterates v(k−1)
and S(k−1) . Now, if j ∈ J −,0 := {v ∗ (xj ) − vj ≥ 1 − Sj }, then the entries for S (k) are
set to be Sjk = 1. So, in this row, the independent variable is vj . On the other hand, if
j ∈ J + := {v ∗ (xj ) − vj < 1 − Sj }, the Sj are the independent variables. Hence, the process
just described is equivalent in implementation to variable switching. (See [9])
4.3.3
Stefan problem
In the case of the Stefan problem, we do not need dependence of the graph α on
x. Therefore all the literature results on convergence apply. However, numerical results
associated with the results form literature were only demonstrated for regularized problem
or where extra effort was spent on adaptive gridding and well as on regularization [27].
95
We show here how can we implement Stefan problem without regularization using
semismooth Newton methods. Since, the graph αST cannot be written as a single NCC,
we write each of its pieces as an NCC. Moreover, we can see this problem as a mixed
complementarity problem (MCP).
It is useful to see exactly how we would write each piece of the graph of αST as
an NCC. For example, the top part of the graph can be written as min(1 + v − u, v) =
min(1 − (u − v), v) = 0. The bottom part can be written as min(u − v, −v) = 0.
Now, we want to use the framework for box–constrained variational inequality problems (which is the same as MCP). First, recall the Heaviside graph
H := {hx, yi ∈ R2 : 0 ≤ y ≤ 1, yx ≥ 0, (y − 1)x ≥ 0}.
(4.57)
Let w = u − v, we have
hv, wi ∈ H.
That is, 0 ≤ u ≤ 1 + v, (u − v)v ≥ 0 and (u − v − 1)v ≥ 0. Now, according to [44], (Section
1.21), (4.57) can be expressed in an equivalent way as
y − P[0,1] (y + x) = 0,
where P[0,1] (z) = max{0, min{z, 1}}. But we want to find a function Φ[0,1] : R2 → R with
the property that having Φ[0,1] (u, v) = 0 is equivalent to the MCP. We call a function
like Φ an MCP–function for the interval [0, 1]. To have a more consistent notation, let
Φ[0,1] = ΦST . Hence,
u ∈ βST (v) ⇐⇒ ΦST (u, v) := u − v − PST (u) = 0.
Note that ΦST is semismooth and its Jacobian is nonsingular. We summarize this as
follows.
Lemma 4.3.3.1. The semismooth Newton algorithm converges superlinearly for the Stefan problem.
96
v
u
αST
βST
1
1
1
u
1
v
FIGURE 4.4: Graphs associated to the Stefan problem
The proof is similar to the proof of Lemma 4.3.1.1, omiting the details concerning
the nonsigularity of the Jacobian for the MCP case.
4.4.
Numerical results in 1D
We now perform numerical experiments for the initial boundary value problem (4.4)–
(4.6) in the one–dimensional case. We use the scheme given in (4.41)–(4.42) with a semismooth Newton solver.
We apply the scheme for three different examples of graphs β and α. We use the
singular graph example, Stefan free boundary valued problem example and the methane
hydrate example.
For the first two examples, analytical solutions are available from literature. For
methane hydrates there is not analytical solution available. We study errors for u and
v in Lp (Q) with p = 1, 2. In addition, we consider quasi–norms as in [13]. Moreover,
we observed in all experiments that the average number of Newton’s iterations is mesh–
independent and very small. However, we don’t pretend to do an exhaustive study of the
robustness of our semismooth Newton solver respect to other known solvers.
97
We define the errors for u, v and S that we use in our experiments
!1/p
eu,p :=
X
τ ku − unh kpLp (Ω)
,
(4.58)
n
!1/p
ev,p :=
X
τ kv − vhn kpLp (Ω)
,
n
eq
X Z
|u − unh ||v − vhn |dx.
:=
τ
n
Ω
In all our examples, we take τ = O(h1/2 ) or O(h), M = 1/h. We denote by ru,p the rate
of convergence in eu,p . We exhibit results for eS,p as well.
We test our solver first with the one–phase Stefan problem, and show that we get
the same rates as obtained in the literature; see review in Section 4.2. Since analytical
solutions are available for αE and αW in [40], we test the rates of convergence in these
cases as well.
Next, as we pointed out before, our methane hydrate problem can bee seen as a modified Stefan problem. Thus we expect the rates of convergence to be alike those for Stefan
problem. Whenever we do not have an analytical solution on hand, we find a numerical
solution for a very fine grid. That is, we consider, hmin , much smaller than h for the grids
for which we test convergence. Errors are computed with numerical approximations using
different h0 s against the one obtained with hmin .
4.4.1
One–phase Stefan problem
In this section we use an example of a one phase Stefan free boundary value problem
for which an analytical solution is available and is given in [6].
In what follows DT = {(x, t)|0 < x < s(t), 0 < t ≤ T } where T is a fixed time
and x = s(t) is the location of the interface at time t. BT is the boundary consisting
of the curve {(s(t), t) : 0 ≤ t ≤ T }, and the line segment {(x, 0) : 0 ≤ x ≤ b} on the
characteristic t = 0.
We need to determine two functions v = v(x, t) and the position of free boundary
98
s = s(t), such that the pair satisfies



vxx − vt = 0,







v(0, t) = f (t) ≥ 0,



v(x, 0) = ϕ(x) ≥ 0,







v(s(t), t) = 0,
a < x < s(t),
0 < t ≤ T,
0 < t ≤ T,
(4.59)
0 ≤ x ≤ b ≡ s(0),
0 < t ≤ T,
s0 (t) = −vx (s(t), t),
0 < t ≤ T.
(4.60)
The following two definitions make precise what it means to be a solution of (4.59) for a
given continuous function s(t) and the Stefan problem (4.59)–(4.60).
Definition 4.4.1.1. A solution of (4.59) for a known continuous s(t) is a function v(x, t)
defined in DT ∪ BT such that vxx , ut ∈ C(DT ), u satisfies the condition (4.59) and v ∈
C(DT ∪ BT ) except at points of discontinuity of f and ϕ. Also, u is bounded in DT , which
implies that at a point of discontinuity of f or ϕ, 0 ≤ lim inf v ≤ lim sup u < ∞.
Definition 4.4.1.2. A solution (s, v) of the Stefan problem (4.59)–(4.60) is a pair of
functions s = s(t) and v = v(x, t) such that s(0) = b, s(t) > 0, s ∈ C 1 ((0, T ]) ∩ C 0 ([0, T ]),
u satisfies (4.59) for this s in the sense of Definition 4.4.1.1, vx (s(t), t) exists it is continuous, and s and v satisfy (4.60).
Now, we are interested in the particular case for which an analytical solution can
be found when b = 0, ϕ ≡ 0, and f ≡ λ > 0. In [6], the Lemma 17.3.1 (p. 288), shows
that there exists a unique positive constant c = c(λ) such that the pair
√
s(t) = c(λ) t,
0 < t,
and
(
v λ (x, t) = λ − c(λ) exp
c(λ)2
4
)
v(x, t),
99
where
Z
u(x, t) =
√
x/(2 t)
exp{−ρ2 }dρ,
0
is the unique solution to the Stefan problem (4.59)–(4.60).
Putting this in context for the Stefan problem with constraint as in Section 4.3.3, we
assume initial condition u(x, 0) = v(x, 0) = 0. The parameter λ > 0 defines the boundary
condition v(0, t) = λ, t > 0 at the left end of the domain, and assume the initial condition
u(x, 0) = v(x, 0) = 0. There is only one “phase” in the problem. The free boundary is
x = s(t), with v(s(t), t) = 0. On one hand, u(x, t) = v(x, t) + 1 for x < s(t) ahead of
the free boundary, and on the other hand u(x, t) = v(x, t) for x > s(t) behind the free
boundary. This is dictated by the graph αST , for which the graph of its inverse is given
in Figure 4.4.
The evolution is considered for t ≤ T , when the front of the free boundary reaches
x = 1, so that the right boundary condition v(1, t) = 0, t ≤ T holds. From Table 4.1 we
can observe that the error eu,2 ≈ O h1/2 and ev,2 ≈ O (h), which agrees with the
theory reviewed in Section 4.2. While, ru,1 and rv,1 are generally higher. The computed
L2 rates agree with the theory, and they depend on the scaling between h and τ , while
L1 rates are generally higher.
The Tables 4.1 and 4.2 come from [16]. The values of the tables conform the orders
of convergence obtained in [29].
100
TABLE 4.1: Convergence and iteration count for (4.59) with b = 0, ϕ = 0, f = λ.
4.4.2
1/h
1/τ
Nit
eu,2
ru,2
eu,1
ru,1
32
320
2
2.43e-02
64
640
2
1.69e-02
0.523
1.14e-03
0.983
128
1280
2
1.17e-02
0.535
5.70e-04
1.011
256
2560
2
7.91e-03
0.566
2.76e-04
1.044
512
5120
2
5.22e-03
0.600
1.30e-04
1.082
16
1600
2
3.22e-02
32
3200
2
2.23e-02
0.533
1.67e-03
1.024
64
6400
2
1.55e-02
0.525
8.29e-04
1.015
128
12800
2
1.06e-02
0.548
4.03e-04
1.041
256
25600
2
6.98e-03
0.603
1.88e-04
1.093
16
256
2
3.19e-02
32
1024
2
2.13e-02
0.581
1.69e-03
1.136
64
4096
2
1.40e-02
0.609
7.72e-04
1.133
2.27e-03
3.40e-03
3.72e-03
Singular graph with an analytical solution
The analytical solution for this example is given in [40].
The graph αE = {0} × (−∞, 1] ∪ [0, ∞) × {1} extends




{1}
if x ∈ (0, 1]




α(x) = [0, 1] if x = 0






x
if x ∈
/ [0, 1]
shown in Figure 4.2 and discussed in Section 4.3. Consider the initial boundary value
101
TABLE 4.2: Convergence and iteration count for (4.59) with b = 0, ϕ = 0, f = λ.
1/h
1/τ
Nit
ev,2
rv,2
ev,1
rv,1
32
320
2
5.42e-03
64
640
2
3.45e-03
0.651
3.30e-04
0.936
3.81e-03
0.714
128
1280
2
2.10e-03
0.713
1.67e-04
0.979
2.25e-03
0.758
256
2560
2
1.24e-03
0.760
8.27e-05
1.020
1.31e-03
0.784
512
5120
2
7.06e-04
0.818
3.90e-05
1.083
7.33e-04
0.837
16
1600
2
5.20e-03
32
3200
2
1.92e-03
1.438
1.73e-04
1.482
4.24e-03
1.007
64
6400
2
8.33e-04
1.204
6.99e-05
1.308
2.17e-03
0.963
128
12800
2
4.23e-04
0.978
2.98e-05
1.227
1.10e-03
0.972
256
25600
2
2.27e-04
0.897
1.27e-05
1.225
5.55e-04
0.998
16
256
2
5.65e-03
32
1024
2
2.32e-03
1.282
2.47e-04
1.617
4.13e-03
1.003
64
4096
2
9.62e-04
1.272
8.24e-05
1.588
2.07e-03
1.000
6.32e-04
eq
rq
6.25e-03
4.83e-04
8.51e-03
7.59e-04
8.29e-03
problem
ut − vxx , v ∈ αE (u),
(4.61)
v(·, t) ∈ H01 (0, 1) for t > 0, u(x, 0) = 1.
(4.62)
As shown in [40], the solution for the initial–boundary–valued problem (4.61)–(4.62) is
given by the following pair of functions u and v.

√
√


1 for t ∈ [0, 1/8), x ∈ ( 2t, 1 − 2t),
u(x, t) =


0 otherwise,
102
and v(t) ∈ H01 (0, 1) is defined by

√
√


min{x/ 2t, 1, (1 − x)/ 2t} for x ∈ (0, 1), t ∈ (0, 1/8),
v(x, t) =


0
otherwise.
Notice, they are weak solutions according to the notions (4.8) and (4.9). Neither u nor v
is continuous on (0, 1) × (0, ∞). This example is not covered by the theory exposed in [36]
since α in not a function.
We implement the numerical Algorithm 2.3.3.1 to solve this problem, and use the
representation of the graph αE as in (4.53).
We show the rates of convergence obtained from simulations in Tables 4.3 and 4.4,
extracted from [16].
The Tables 4.3 and 4.4 provide the orders of errors eu,2 ≈ O(h1/2 ), eu,1 ≈ O(h), ev,2 ≈
O(h), ev,1 ≈ O(h) and eq ≈ O(h). This is a good test example, it helps us to once again
check that things are the way we were expecting.
The Figures 4.5–4.8 illustrate the evolution of numerical solutions for u and v and
their exact counter parts.
103
TABLE 4.3: Convergence and iteration count for (4.61)–(4.62)
1/h
1/τ
Nit
eu,2
ru,2
eu,1
ru,1
32
320
2
3.04e-02
64
640
2
2.14e-02
0.502
1.49e-03
0.973
128
1280
2
1.50e-02
0.515
7.42e-04
1.006
256
2560
2
1.03e-02
0.540
3.61e-04
1.038
512
5120
2
6.81e-03
0.601
1.70e-04
1.089
16
1600
2
4.33e-02
32
3200
2
3.06e-02
0.498
2.98e-03
0.969
64
6400
2
2.15e-02
0.513
1.49e-03
0.997
128
12800
2
1.47e-02
0.546
7.28e-04
1.037
256
25600
2
9.69e-03
0.602
3.42e-04
1.089
16
256
2
4.15e-02
32
1024
2
2.90e-02
0.516
2.80e-03
0.953
64
4096
2
1.93e-02
0.591
1.35e-03
1.058
2.92e-03
5.84e-03
5.43e-03
104
TABLE 4.4: Convergence and iteration count for (4.61)–(4.62)
1/h
1/τ
Nit
ev,2
rv,2
ev,1
rv,1
32
320
2
5.88e-03
64
640
2
3.51e-03
0.743
4.82e-04
0.867
2.57e-03
0.893
128
1280
2
2.05e-03
0.771
2.42e-04
0.993
1.34e-03
0.934
256
2560
2
1.19e-03
0.785
1.16e-04
1.057
6.40e-04
1.073
512
5120
2
6.73e-04
0.828
5.72e-05
1.026
3.00e-04
1.094
16
1600
2
9.44e-03
32
3200
2
4.77e-03
0.986
9.40e-04
1.001
5.93e-03
0.971
64
6400
2
2.39e-03
0.993
4.71e-04
0.999
2.99e-03
0.987
128
12800
2
1.23e-03
0.966
2.35e-04
1.001
1.48e-03
1.016
256
25600
2
6.29e-04
0.961
1.17e-04
1.002
7.19e-04
1.040
16
256
2
1.01e-02
32
1024
2
5.25e-03
0.945
9.94e-04
0.936
5.42e-03
0.800
64
4096
2
2.62e-03
1.003
4.88e-04
1.026
2.78e-03
0.964
8.79e-04
eq
rq
4.78e-03
1.88e-03
1.16e-02
1.90e-03
9.44e-03
105
FIGURE 4.5: Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.0070125.
FIGURE 4.6: Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.01465.
106
FIGURE 4.7: Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.090025.
FIGURE 4.8: Numerical and analytical solutions for problem (4.61)–(4.62) with vc∗ ≡ 1
at t = 0.121425.
107
Example 4.4.2.1 (Singular graph with a constant solution). Now, we consider the inverse
−1
of αE , αW = αE
= (−∞, 1] × {0} ∪ {1} × [0, ∞). We extend the graph of αW as we did
with αE , i.e, α(x) = x for x ∈ [0, 1]. The problem now is
ut − vxx = 0, v ∈ αW (u),
(4.63)
v(·, t) ∈ H01 (0, 1) for t > 0, u(x, 0) = 1/2.
(4.64)
It is clear that the constant u ≡ 1/2, v ≡ 0 satisfies (4.63)-(4.64). We found out that with
the semismoooth Newton method, using φW (u, v) = min(1 − u, v), the algorithm converges
just in two iterations to the true solution.
4.4.3
The methane hydrates problem
For this Section, we consider three examples, each associated with a different max-
imum solubility curve.
We consider v ∗ to be either constant or linear or quadratic. More precisely, v ∗ (x)
will be one of the following solubilities
vc∗ (x) = 1,
(4.65)
va∗ (x) = (1 + x)/2,
(4.66)
vn∗ (x) = (1 + 2x − x2 )/2.
(4.67)
For each of the data v ∗ (x) we can now construct an appropriate graph αM H (x; ·) and
equivalently its inverse βM H . We implemented the numerical Algorithm 2.3.3.1 and we
use the representation for αM H given in (4.55).
Notice that it was discussed in Section 4.1., that αM H is a translate of the graph
αST . Therefore we expect the orders of convergence obtained for our schemes to be the
same as for the problem associated to αST even though αM H depends on xj .
108
1/h
1/τ
Nit
eu,2
ru,2
eu,1
ru,1
32
320
2
6.84e-03
64
640
2
4.74e-03
0.530
5.05e-04
0.925
128
1280
2
3.15e-03
0.586
2.57e-04
0.972
256
2560
2
2.14e-03
0.560
1.28e-04
1.005
512
5120
2
1.39e-03
0.623
6.04e-05
1.086
16
1600
2
6.53e-03
32
3200
2
4.28e-03
0.610
3.25e-04
1.138
64
6400
2
2.93e-03
0.547
1.62e-04
1.009
128
12800
2
1.98e-03
0.559
8.04e-05
1.005
256
25600
2
1.31e-03
0.603
3.85e-05
1.063
16
256
2
7.83e-03
32
1024
2
4.39e-03
0.833
4.24e-04
1.436
64
4096
2
2.69e-03
0.705
1.67e-04
1.343
9.58e-04
7.15e-04
1.15e-03
TABLE 4.5: Convergence and iteration count for (4.68)–(4.69). Case vc∗ (x) = 1.
Recall that the problem is
ut − vxx = 0, v ∈ αM H (x; u),
(4.68)
v(·, t) ∈ H01 (0, 1) for t > 0, u(x, 0) ≡ ν = 1.2.
(4.69)
Tables 4.5–4.10 show the rates of convergence for u and v in different norms. We use
different initial data and boundary conditions in these Tables.
In Figures 4.9-4.16, we illustrate the evolution of methane hydrate for the different
solubilities (4.65), (4.66) and (4.67).
From Tables 4.5 and 4.6, we can see that the rates of convergence for vc∗ = 1 are
109
1/h
1/τ
Nit
ev,2
rv,2
ev,1
rv,1
32
320
2
3.71e-03
64
640
2
2.36e-03
0.654
3.36e-04
0.910
2.41e-03
0.672
128
1280
2
1.44e-03
0.713
1.73e-04
0.961
1.47e-03
0.717
256
2560
2
8.50e-04
0.760
8.55e-05
1.014
8.62e-04
0.768
512
5120
2
4.86e-04
0.806
4.05e-05
1.077
4.92e-04
0.810
16
1600
2
2.31e-03
32
3200
2
7.12e-04
1.697
8.15e-05
1.807
1.17e-03
1.239
64
6400
2
3.48e-04
1.029
3.45e-05
1.239
6.03e-04
0.955
128
12800
2
2.17e-04
0.682
1.74e-05
0.986
3.33e-04
0.855
256
25600
2
1.31e-04
0.725
8.52e-06
1.033
1.81e-04
0.883
16
256
2
3.44e-03
32
1024
2
1.31e-03
1.396
1.82e-04
1.811
1.55e-03
1.287
64
4096
2
4.88e-04
1.421
5.10e-05
1.834
6.57e-04
1.239
6.31e-04
eq
rq
3.84e-03
2.85e-04
2.76e-03
6.38e-04
3.79e-03
TABLE 4.6: Convergence and iteration count for (4.68)–(4.69). Case vc∗ = 1.
110
1/h
1/τ
Nit
eu,2
ru,2
eu,1
ru,1
32
320
2
1.58e-02
64
640
2
1.10e-02
0.515
9.46e-04
0.955
128
1280
2
7.64e-03
0.529
4.75e-04
0.994
256
2560
2
5.22e-03
0.551
2.33e-04
1.027
512
5120
2
3.44e-03
0.602
1.09e-04
1.090
16
1600
2
2.19e-02
32
3200
2
1.48e-02
0.564
1.46e-03
1.016
64
6400
2
1.04e-02
0.518
7.27e-04
1.003
128
12800
2
7.11e-03
0.545
3.54e-04
1.039
256
25600
2
4.69e-03
0.601
1.66e-04
1.090
16
256
2
2.19e-02
32
1024
2
1.43e-02
0.622
1.46e-03
1.113
64
4096
2
9.35e-03
0.612
6.72e-04
1.120
1.83e-03
2.95e-03
3.16e-03
TABLE 4.7: Convergence and iteration count for (4.68)–(4.69). Case va∗ (x) = (1 + x)/2
eu,2 ≈ O(h1/2 ), eu,1 ≈ O(h), ev,2 ≈ O(h), ev,1 ≈ O(h) and eq ≈ O(h) as we previously
mentioned.
For the solubilities va∗ (x) = (1 + x)/2 and vn∗ (x) = (1 + 2x − x2 )/2, Tables 4.7 - 4.10
show the same rate of convergence as for vc∗ = 1.
111
1/h
1/τ
Nit
ev,2
rv,2
ev,1
rv,1
32
320
2
2.98e-03
64
640
2
1.88e-03
0.665
2.39e-04
0.928
2.15e-03
0.729
128
1280
2
1.14e-03
0.718
1.22e-04
0.976
1.26e-03
0.768
256
2560
2
6.74e-04
0.762
6.00e-05
1.021
7.27e-04
0.798
512
5120
2
3.84e-04
0.810
2.83e-05
1.083
4.07e-04
0.838
16
1600
2
3.06e-03
32
3200
2
1.17e-03
1.383
1.53e-04
1.488
2.68e-03
0.980
64
6400
2
5.08e-04
1.208
5.91e-05
1.371
1.37e-03
0.964
128
12800
2
2.47e-04
1.039
2.40e-05
1.297
6.94e-04
0.986
256
25600
2
1.29e-04
0.941
1.00e-05
1.263
3.45e-04
1.010
16
256
2
3.12e-03
32
1024
2
1.30e-03
1.262
1.95e-04
1.552
2.57e-03
0.942
64
4096
2
5.52e-04
1.236
6.65e-05
1.552
1.29e-03
0.997
4.56e-04
eq
rq
3.57e-03
4.29e-04
5.29e-03
5.72e-04
4.95e-03
TABLE 4.8: Convergence and iteration count for (4.68)–(4.69). Case va∗ (x) = (1 + x)/2
112
1/h
1/τ
Nit
eu,2
ru,2
eu,1
ru,1
32
320
2
1.07e-02
64
640
2
7.40e-03
0.533
6.33e-04
0.949
128
1280
2
5.09e-03
0.539
3.19e-04
0.988
256
2560
2
3.45e-03
0.561
1.56e-04
1.028
512
5120
2
2.27e-03
0.605
7.39e-05
1.084
16
1600
2
1.39e-02
32
3200
2
9.50e-03
0.555
7.85e-04
1.040
64
6400
2
6.61e-03
0.524
3.88e-04
1.016
128
12800
2
4.50e-03
0.554
1.88e-04
1.043
256
25600
2
2.96e-03
0.604
8.88e-05
1.087
16
256
2
1.43e-02
32
1024
2
9.18e-03
0.636
8.21e-04
1.182
64
4096
2
5.98e-03
0.619
3.67e-04
1.160
1.22e-03
1.61e-03
1.86e-03
TABLE 4.9: Convergence and iteration count for (4.68)–(4.69). Case vn∗ (x) = (1 − 2x +
x2 )/2.
113
1/h
1/τ
Nit
ev,2
rv,2
ev,1
rv,1
32
320
2
2.9e-03
64
640
2
1.89e-03
0.656
2.53e-04
0.910
2.05e-03
0.706
128
1280
2
1.15e-03
0.717
1.29e-04
0.965
1.22e-03
0.751
256
2560
2
6.77e-04
0.763
6.40e-05
1.014
7.07e-04
0.785
512
5120
2
3.86e-04
0.811
3.03e-05
1.080
3.98e-04
0.827
16
1600
2
2.73e-03
32
3200
2
1.00e-03
1.445
1.15e-04
1.635
2.04e-03
1.034
64
6400
2
4.35e-04
1.205
4.36e-05
1.400
1.04e-03
0.969
128
12800
2
2.19e-04
0.990
1.86e-05
1.228
5.33e-04
0.967
256
25600
2
1.19e-04
0.875
8.26e-06
1.172
2.68e-04
0.995
16
256
2
2.97e-03
32
1024
2
1.18e-03
1.330
1.65e-04
1.684
1.98e-03
1.013
64
4096
2
4.86e-04
1.280
5.24e-05
1.655
9.88e-04
1.000
4.75e-04
eq
rq
3.34e-03
3.57e-04
4.18e-03
5.30e-04
3.99e-03
TABLE 4.10: Convergence and iteration count for (4.68)–(4.69). Case vn∗ (x) = (1 −
2x + x2 )/2.
114
FIGURE 4.9:
Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with
v ∗ (x) = 0.25(x + 1) at t = 0.005.
Example 4.4.3.1 (Methane hydrate example with linear constraint). The Figures 4.94.12 represent the evolution of the numerical solutions given an initial data u(x, 0) with a
little perturbation. The source term f = 0, and Dirichlet boundary condition v(x0 , t) = 0
if x0 < 0.5 and v(x0 , t) = v ∗ (x0 , t) if x0 ≥ 0.5.
115
FIGURE 4.10: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.25(x + 1) at t = 0.075.
FIGURE 4.11: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with
v ∗ (x) = 0.25(x + 1) at t = 0.3.
116
FIGURE 4.12: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.25(x + 1) at t = 0.5.
Example 4.4.3.2 (Methane hydrates with quadratic constraint). Now, we consider the
point source
f (x, t) =



15
if t < .05 and |x − .5| < .1,


0
otherwise.
The initial condition u(x, 0) is a linear function, the boundary conditions for v are given by
v(x, 0) = 0.7v ∗ (0) and v(x, 1) = 0.7v ∗ (1). The maximum solubility is given by v ∗ (x) = 0.5(x2 +
1). See Figures 4.13-4.16.
117
FIGURE 4.13: Numerical solutions for u, v and Sh in for problem (4.68)–(4.69) with
v ∗ (x) = 0.5(x2 + 1).
FIGURE 4.14: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with
v ∗ (x) = 0.5(x2 + 1).
118
FIGURE 4.15: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with
v ∗ (x) = 0.5(x2 + 1).
FIGURE 4.16: Numerical solutions for u, v and Sh in problem (4.68)–(4.69) with v ∗ (x) =
0.5(x2 + 1).
119
4.5.
Numerical results for methane hydrates in 2D
Now we extend the one dimensional results to two dimensions just for the case of
methane hydrates. We first give the finite element setting for our problem. For implementation, we made a modification of the codes used in [2].
4.5.1
Algorithm
Here we use the implementation of linear finite elements presented in [2] for station-
ary smooth nonlinear problems and for the heat equation. Our modification works for the
heat equation with a graph.
We consider a 2D implementation of (4.37)–(4.39). One of the main parts is to set
up a general way to solve the nonlinear problem. In the code provided by authors of [2]
the nonlinearity was of semilinear type, i.e., it affected the right hand side source term f .
In our case, nonlinearity is the term associated with the graph β.
More precisely, our time-discrete weak formulation is given by
(β(vhn ), φ) − (β(vhn−1 ), φ) + k(∇vhn , ∇φ) = k(f n+1 , φ), for all φ ∈ Vh .
(4.70)
Then the residual associated to (4.70) is given by
J(vhn , φ)
Z
=
β(vhn )φ dx
Z
∇vhn ∇φ dx
+k
Ω
Z
−k
Ω
n
Z
f φ dx −
Ω
Ω
β(vhn−1 )φ dx,
(4.71)
where the last two terms are known.
In the Newton solver, in order to solve
J(vhn , φ) = 0
we need to make precise the Jacobian ∇v J(vhn , φ). If β is a function (not a graph) then
∇v J(vhn , φ)
Z
=
β
Ω
0
(vhn )φv dx
Z
∇φ · ∇v dx.
+k
Ω
(4.72)
120
The first integral in (4.72) is computed using one numerical integration rule over each
element K and gives rise to nonlinear version of mass matrix M similar to what was
defined in (4.41).
If β is a graph then we cannot use β 0 (v) directly. Instead, we keep the two unknowns
unh and vhn in the same equation as was done in (4.41), and resolve the dependence between
uh and vh with semismooth function apropiate for the methane hydrate example.
4.5.2
Experiments
Example 4.5.2.1 (The heat equation). Let Ω = [0, 1] × [0, 1]. We consider the evolution
equation
ut − 4u = f,
u(x, y, 0) = 0,
u = 0,
(x, y) ∈ Ω,
(x, y) ∈ ∂Ω.
In this example, β is the function β(u) = u.
Here, the exact solution is given by
2 /2
u(x, y, t) = (1 − e−t
) sin(πx) sin(πy).
So the right hand side f is given by
h
i
2
f (x, y, t) = sin(πx) sin(πy) 2π 2 + e−t /2 (t − 2π 2 ) .
The graphs given in Figures 4.17-4.21 are comparisons of the numerical solution
against the exact solution at times tn = 0.000976563, 0.375, 0.625, 0.875 and tn = 1.
In Table 4.11 we show some errors of the numerical solution with respect to the
exact solution in L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)). In Figure 4.22 we show rates
of convergence in L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)) which are O(h2 ) and O(h), respectively. Note that the order of O(h2 ) is obtained since the solution is smooth, which
121
FIGURE 4.17: Numerical solution versus exact solution at time tn for heat equation using
2D semismooth Newton method. Example 4.5.2.1.
122
FIGURE 4.18: Numerical solution versus exact solution at time tn for heat equation using
2D semismooth Newton method. Example 4.5.2.1.
123
FIGURE 4.19: Numerical solution versus exact solution at time tn for heat equation using
2D semismooth Newton method. Example 4.5.2.1.
124
FIGURE 4.20: Numerical solution versus exact solution at time tn for heat equation using
2D semismooth Newton method. Example 4.5.2.1.
125
FIGURE 4.21: Numerical solution versus exact solution at time tn for heat equation using
2D semismooth Newton method. Example 4.5.2.1.
126
h
maxn ku(·, tn ) − uh (·, tn )kL2
maxtn ku(·, tn ) − uh (·, tn )kH 1
1/16
0.00395965
0.0618527
1/32
0.00099498
0.0307794
1/64
0.000248534
0.0152578
0
TABLE 4.11: L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)) errors for heat equation in 2D semismooth Newton method. Example 4.5.2.1.
FIGURE 4.22: Error orders in L∞ ([0, 1], L2 (Ω)) and L∞ ([0, 1], H01 (Ω)) for heat equation
using 2D semismooth Newton method. Example 4.5.2.1.
is not the case in Examples 4.5.2.2, 4.5.2.3 and 4.5.2.4. In the latter examples we don’t
provide convergence rate here, but expect to have the same as in the 1D cases, shown in
[16].
127
Let Ω = [0, 1]×[0, 1] and ∆t = O(∆x2 ). In Examples 4.5.2.2 and 4.5.2.3, we simulate
the evolution of the total amount u of methane present at time tn , the hydrate saturation
Sh = 1 − S, where S is the liquid saturation and the solubility v (dissolved methane). The
graph of the maximum solubility v ∗ is shown (maximum CH4 dissolved) in position (2, 2)
on each figure.
In these examples we do not compute convergence rates since we do not have analytical solutions available. We would have to consider a very fine mesh to compute the
errors. In addition, we already provided rates of convergence for our equivalent examples
in 1D. See Section 4.4.
For Examples 4.5.2.2 and 4.5.2.3 we use initial data u(x, y, 0) = 0.8v ∗ (x, y) + 0.2R +
ξ(x) where ξ ∼ U (0, 0.1) (U (0, 0.1) is the uniform distribution on the interval (0, 0.1)).
The boundary condition is
u(x, y, 0) =



0
if x0 < 0.5,


v ∗ (x0 , y)
if x0 ≥ 0.5,
and the source term is f = 0.
Example 4.5.2.2 (Constant constraint). The initial data and initial conditions are the
counterparts of our one dimensional examples. These examples illustrate how the modification of the code in [2] works. Figures 4.23 to 4.27 show the evolution of the total amount
of CH4 present at five different times tn . Also the evolution for the saturation, Sh and
the dissolved methane, v are shown at these different times.
128
FIGURE 4.23: Evolution of total amount of CH4 ,u, hydrate saturation Sh and dissolved
CH4 with maximum dissolved methane v ∗ = 0.25. Example 4.5.2.2.
129
FIGURE 4.24: Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2.
130
FIGURE 4.25: Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2.
131
FIGURE 4.26: Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2.
132
FIGURE 4.27: Evolution of u, Sh and v with constant maximum solubility v ∗ = 0.25 at
the given time tn . Example 4.5.2.2.
133
FIGURE 4.28: Evolution of total amount of CH4 ,u, hydrate saturation Sh and dissolved
CH4 with maximum dissolved methane v ∗ = 0.25(x + 1). Example 4.5.2.3.
Example 4.5.2.3 (Linear constraint). Now we use the same source term f , boundary
conditions an initial conditions as in Example 4.5.2.2. We also use, linear maximum
solubility v ∗ = 0.5(x + 1). Figures 4.28 to 4.32 show the evolution of the total amount of
CH4 present at five different times tn . Also the evolution for the saturation, Sh and the
dissolved methane, v are shown at these different times.
134
FIGURE 4.29: Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3.
135
FIGURE 4.30: Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3.
136
FIGURE 4.31: Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3.
137
FIGURE 4.32: Evolution of u, Sh and v with linear maximum solubility v ∗ = 0.25(x + 1)
at the given time tn . Example 4.5.2.3.
138
FIGURE 4.33: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
Example 4.5.2.4 (Quadratic constraint). In this case, the maximum solubility is v ∗ (x, y) =
0.5(x2 + 1), and the source term f is the point source



15 if t < .05 and |x − .5| < .1,
f (x, y, t) =


0
otherwise,
and the boundary conditions are given by v(0, y, t) = 0.7v ∗ (0, y) and v(1, y, t) = 0.7v ∗ (1, y).
Figures 4.33 to 4.40 show the evolution of u, v and Sh at eight different times.
139
FIGURE 4.34: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
140
FIGURE 4.35: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
141
FIGURE 4.36: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
142
FIGURE 4.37: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
143
FIGURE 4.38: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
144
FIGURE 4.39: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
145
FIGURE 4.40: Evolution of u, Sh and v with quadratic maximum solubility v ∗ = 0.5(x2 +
1) at the given time tn . Example 4.5.2.4.
146
4.5.3
Remarks about 2D implementation
We can observe from the 2D Examples 4.5.2.3 and 4.5.2.4, that the evolution of
numerical solutions is similar to their 1D counterparts in Examples 4.4.3.1 and 4.4.3.2.
For the implementation, there are many things to take into consideration while
modifying the code in [2]. We have to produce a triangulation, which in this case was
not difficult since we are dealing with a rectangular domain. We have to approximate the
integrals involved in (4.71) and (4.72) using numerical integration. For instance, we might
use the center of gravity of an element T ∈ Th or the midpoint formula to approximate
those integrals. We used the midpoint formula.
We had to modify the functionals in (4.71) and (4.72) which were quite different to
the ones associated to their test problem. We had a time dependent problem and that
required some work to modify their stationary nonlinear code.
147
5.
ADSORPTION: ANALYSIS AND NUMERICAL
APPROXIMATION
In this Chapter we analyse the adsorption model introduced in Section 3.2. We start
by analysing the scalar equation case for single component and Langmuir isotherm. We
then discuss the Godunov numerical scheme for this case and provide CFL stability and
time step conditions for the scheme.
Then we introduce the multicomponent system for adsorption in the IAS case. We
describe the IAS algorithm and show through an example how to pass from single component Langmuir isotherm to multicomponent isotherm using the IAS algorithm. We
provide conditions for the system to be hyperbolic and provide general CFL and time step
conditions for the Godunov scheme.
Later, we implement an implicit scheme and perform some experiments to compare,
the IAS and the extended Langmuir approaches and the equilibrium and nonequilibrium
cases.
To motivate the treatment examples for the adsorption model we revisit the Burger’s
equation introduced in Section 2.4.3.
Example 5.0.3.1. In this example we show the numerical solution versus the rarefaction
weak solution for the Burger’s equation given in (2.53) with uL = 0 and uR = 1. That is,
the Cauchy problem (2.48a)-(2.48b) with initial data H(x). We use the Godunov numerical
method, which in this case is the upwind scheme. We estimate orders of convergence as
well.
We see four snapshots of this experiment in Figure 5.1. Figure 5.2 shows the orders
of convergence using H(x) as initial data. We might observe at first glance that the order
of convergence is slightly better than O(h1/2 ) in L1 and L2 norms. Table 5.1 conforms
this observation.
148
FIGURE 5.1: Evolution of numerical numerical solution for problem (2.48a)-(2.48b) with
initial data u(x, 0) = H(x), ∆x = 0.04, ∆t =, T = 1 on the interval [−1, 2]. Example 5.0.3.1
FIGURE 5.2: Log-log graph for orders of convergence for Burgers equation numerical solution with initial data H(x) compared with its rarefaction weak solution. Example 5.0.3.1
149
h
eL1
eL2
esup
α1
α2
αsup
0.5
0.188125
0.219289
0.302348
-
-
-
0.25
0.109727
0.132401
0.222542
0.777773
0.727912
0.442133
0.125
0.0664485
0.0804833
0.1485
0.723611
0.718158
0.583614
0.0625
0.041317
0.0487384
0.091151
0.685501
0.72363
0.70413
0.03125
0.0251177
0.0292316
0.0530752
0.718032
0.737532
0.78022
0.015625
0.0149695
0.0172963
0.0309132
0.746678
0.757066
0.779815
TABLE 5.1: Errors and orders of convergence for Burgers equation in grid norms L1 , L2
and L∞ norms using its rarefaction weak solution. Example 5.0.3.1
5.1.
Scalar conservation law for adsorption with Langmuir isotherm
The main purpose of this section is to deal with the adsorption model given in (3.15)
with Langmuir isotherm. First, consider the problem
∂
(φu + (1 − φ)a(u)) + ∇ · (vu) = 0
∂t
(5.1)
for the case of an ideal gas. (See [18]). Here, u represents concentration, v velocity and φ
is the porosity. Recall that a is the mass fraction of the adsorbed amount.
We consider a simplified model in which velocity is constant, we take v = 1/2 and
φ = 1/2 in (5.1). Hence, we are now dealing with the problem
(u + a(u))t + ux = 0,
x ∈ R, t > 0,
(5.2)
u(x, 0) = u0 (x),
bu
as in (3.17). Recall that VL is the Langmuir volume capacity and
1 + bu
b is the Langmuir constant and they are both positive.
with a(u) = VL
Model (5.2) is a conservation law. To see it in the canonical form (2.41b), we use
150
the change of variable w = u + a(u) to get
wt + (f (w))x = 0,
(5.3)
where f = (I + a)−1 , and I : R → R is the identity function.
Remark 5.1.0.1. Changing variables in conservation laws does not change the nature of
the problem as long as the system has locally a classical solution. In particular, changing variables does not give an equivalent conservation law across a singularity. In what
follows we only use change of variables to locally analyse behaviour of conservation laws.
Numerically, we always solve the original problem.
Observe that for u ≥ 0 we have that a has a concave increasing graph. Moreover,
a0 (u) =
bVL
> 0,
(1 + VL u)2
1
≤ 1. Both f and a are Lipschitz and f is
1 + a0 (u)
increasing. Since a is concave, f is convex.
and |a0 (u)| ≤ bVL . Also, 0 < f 0 (w) =
Since f is convex increasing, the Godunov method applied to (5.3) becomes the
upwind method as discussed in Section 2.4.5.
n
Wjn+1 = Wjn − λ f (Wjn ) − f (Wj−1
) ,
where Wjn ≈ w(xj , tn ), Ujn ≈ u(xj , tn ), λ =
∆t
∆x .
The CFL condition is given by
λ|f 0 (Wjn )| ≤ 1 for all Wjn .
Equivalently, we are solving
n
Ujn+1 + a(Ujn+1 ) = Ujn + a(Ujn ) − λ Ujn − Uj−1
(5.4)
for Ujn .
Therefore, we have the following condition for the time step
∆t ≤ ∆x(1 + a0 (Ujn )) for all Ujn .
(5.5)
151
Since the constraint on time step depends on the solution, we have to make sure that (5.5)
always holds, or we can simply use ∆t ≤ ∆x remembering that a0 > 0.
Note that in order to apply the scheme in (5.4), we need actually to resolve Wjn+1 =
Ujn+1 + a(Ujn+1 ) to get Ujn+1 . For this we implement a local Newton solver.
In what follows, we present some experiments using the Godunov scheme. Given in
Examples 5.1.0.2, 5.1.0.3 and 5.1.0.4.
Example 5.1.0.2 (Riemann initial data). In Figure 5.3, we show the evolution of the
numerical solutions using the upwind scheme and Riemann initial data for the flux f (w) =
(I + a)−1 (w).
u(x, 0) =



1.5
if x < 0,


0
otherwise.
We denote h = ∆x and the time step by k = ∆t = dt.
152
FIGURE 5.3: Evolution of the numerical solution for Example 5.1.0.2 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1.
Example 5.1.0.2 shows how shocks propagate for the adsorption problem (5.2). In
fact, as predicted above, it should be a shock travelling with speed
s=
[u]
1.5 − 0
=
≈ 0.7143.
[u + a(u)]
1.5 + a(1.5) − 0
From Figure 5.3 we can predict a rate of change of the shock of approximately
0.7291.
0.55 − 0.2
=
0.74 − 0.26
153
Example 5.1.0.3 (Heaviside initial data). This experiment uses the same parameters as
Example 5.1.0.2 with exception of the initial data, which is the Heaviside function H(x).
Figure 5.4 shows the evolution of the numerical solution. See Figure 5.4.
FIGURE 5.4: Evolution of the numerical solution for Example 5.1.0.3 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1. Initial data u(x, 0) = H(x).
Figure 5.4 shows how rarefaction wave forms.
We can observe from Figure 5.5 and Table 5.2 that the orders of convergence for
this case are O(h1 ) in the L1 grid norm and O(h1/2 ) in the L2 grid norm.
Example 5.1.0.4 (“Box” initial data). For this example, we use as initial data “box”
154
FIGURE 5.5: Orders of convergence for single component adsorption with initial data
H(x). Example 5.1.0.3.
h
eL1
eL2
α1
α2
0.5
0.333333
0.471405
-
-
0.25
0.166667
0.333333
1
0.5
0.125
0.0833333
0.235702
1
0.5
0.0625
0.0416667
0.166667
1
0.5
0.03125
0.0208333
0.117851
1
0.5
0.015625
0.0104167
0.0833333
1
0.5
0.0078125
0.00520833
0.0589256
1
0.5
TABLE 5.2: Errors and orders of convergence for adsorption problem (5.2) in grid norms
L1 and L2 norm with initial data H(x). Example 5.1.0.3.
function given by
u(x, 0) = H(x) − H(x − 1).
Figure 5.6 shows the evolution of the numerical solution in this case. This example com-
155
bines a shock and a rarefraction wave.
FIGURE 5.6: Evolution of the numerical solution for Example 5.1.0.4 with T = 1, λ =
0.99. Langmuir constants for the isotherm a: VL = b = 1. Initial data u(x, 0) = H(x).
As we can observe from Figure 5.7 and Table 5.3 the order of convergence are O(h)
in the L1 norm and slightly more than O(h1/2 ) in the L2 norm.
156
FIGURE 5.7: Orders of convergence for single component adsorption with initial data
H(x) − H(x − 1). Example 5.1.0.4.
h
eL1
eL2
esup
α1
α2
αsup
0.5
0.737623
0.609826
0.651842
-
-
-
0.25
0.391396
0.36413
0.514216
0.914256
0.743945
0.342146
0.125
0.207511
0.211991
0.346843
0.915443
0.780446
0.568091
0.0625
0.13156
0.160931
0.406355
0.657467
0.397564
-0.228458
0.03125
0.0807809
0.12334
0.474931
0.703633
0.383803
-0.224975
0.015625
0.0425868
0.076216
0.366074
0.923608
0.694474
0.375583
0.0078125
0.019982
0.0456212
0.332048
1.09171
0.740388
0.140744
TABLE 5.3: Errors and orders of convergence for Burgers equation in grid norms L1 , L2
and L∞ norm with intial data H(x). Example 5.1.0.4.
5.2.
Multicomponent system with extended Langmuir isotherm
Let ~f : Rp → Rp , x ∈ R, t > 0. We will transform
wt + ~f (w)x = 0
(5.6)
157
into
wt + A(w)wx = 0,
(5.7)
where A(w) = D~f (w).
We denote by (λk (w))pk=1 the eigenvalues for A(w).
Definition 5.2.0.1. We say that a system is strictly hyperbolic if for any w ∈ Ω, the
p × p Jacobian matrix
A(w) =
∂fi
(w)
∂wj
1≤i,j≤p
has p distinct eigenvalues λ1 (w) < λ2 (w) < · · · < λp (w).
Recall from Section 2.4.5 that a CFL stability condition for systems was given in
(2.69) for the Godunov scheme as seen in [17, 25, 43]. Recall the CFL condition
1
λ max λk (Wjn ) ≤
1≤k≤p
2
(5.8)
at each Wjn . As a consequence, the time step condition is given by
∆x
1
∆t ≤
min n
2 1≤k≤p λk (Wj ) at each Wjn . We note that in [25] it is mentioned that a less restrictive condition can be
1
used. More precisely, instead of using right hand side in (5.8) we can choose 1. However,
2
we have chosen to use the more restrictive condition in our numerical experiments. For
instance, in Example 5.2.1.1, we work with a particular two component system (p = 2
in (5.6)), the one given in Section 3.2.4 with the two equation system (3.32)–(3.33) with
isotherms (3.34)–(3.35).
We get to the form wt + A(w)wx = 0 by the change
of variables
wi = ui + ai (u)


 a1 (u) 
for i = 1, 2. Here, u = (u1 , u2 ), w = (w1 , w2 ), ~a(u) := 
 and u = ~f (w), where
a2 (u)
−1
2
2
~f (w) = (I + ~a) w = u. Notice that I + ~a : R → R is nonlinear and its components
158
are continuously differentiable in the domain of interest. The system (3.32)–(3.33) can be
expressed in the form (5.6).
In order to find the conditions for the system to be hyperbolic, we need to express
it in the form (5.7). We need to find the Jacobian A(w).
Let B(u) = D(I + ~a)u, then
A(w) = B−1 (w).
(5.9)
More precisely, for each u, B is given by

∂a1
∂a1
(u)
 1 + ∂u1 (u)
∂u
2


B(u) = 

 ∂a2
∂a2
(u)
1+
(u)
∂u1
∂u2




.


We are ready to discuss the system properties. It is not difficult to check that if tr(D~a(u))+
det(D~a(u)) 6= −1, then B(u) is nonsingular. The condition for the system (3.32)–(3.33)
to be strictly hyperbolic is given in the following proposition.
Proposition 5.2.0.1. The system (5.7) is strictly hyperbolic if
∂a2
∂a1
>0
∂u2
∂u1
Proof. Let B11 =
∂a1
∂u1 ,
B12 =
mial for B is given by
1 + B11 − λ
B12
B21
1 + B22 − λ
∂a1
∂u2 ,
B21 =
∂a2
∂u1
and B22 =
∂a2
∂u2 .
(5.10)
The characteristic polyno-
= λ2 − λ(2 + B11 + B22 ) + (1 + B11 )(1 + B22 ) − B12 B21 .
and its eigenvalues are
q
1
2
λ± (B; u) =
2 + B11 + B22 ± (B11 − B22 ) + 4B12 B21 .
2
(5.11)
Since we want the eigenvalues to be real and distinct, we need (B11 − B22 )2 > −4B12 B21 .
From here we can see that a sufficient condition to have an strictly hyperbolic system is
B12 B21 > 0, which is what we wanted to show.
159
Corollary 5.2.0.1. The extended Langmuir system with a1 and a2 given by (3.34) and
(3.35) provided bi , VL,i > 0,
(i = 1, 2) is strictly hyperbolic.
Proof. Here we write the explicit expressions for the partial derivatives
∂a1
∂u1
∂a1
∂u2
∂a2
∂u1
∂a2
∂u2
=
=
=
=
b1 VL,1 (1 + b2 u2 )
(1 + b1 u1 + b2 u2 )2
−b1 b2 VL,1 u1
(1 + b1 u1 + b2 u2 )2
−b1 b2 VL,2 u2
(1 + b1 u1 + b2 u2 )2
b2 VL,2 (1 + b1 u1 )
(1 + b1 u1 + b2 u2 )2
∂a1 ∂a1 ∂a2 ∂a2
,
,
,
:
∂u1 ∂u2 ∂u1 ∂u2
,
(5.12)
,
(5.13)
,
(5.14)
.
(5.15)
We can see that with these particular isotherms condition (5.10) holds. Thus the extended
Langmuir system is strictly hyperbolic.
Proposition 5.2.0.2. The eigenvalues of the extended Langmuir system are positive.
Proof. It follows from (5.11), (5.12) and (5.15) since bi , VL,i > 0,
(i = 1, 2) that λ+ (B; u)
is positive. To show that λ− (B; u) is also positive, it’s enough to show that det (B(u)) > 0
∂a1
∂a2
∂a1 ∂a2
det (B(u)) =
1+
1+
−
∂u1
∂u2
∂u2 ∂u1
∂a1
∂a2
∂a1 ∂a2
∂a1 ∂a2
= 1+
+
+
−
.
∂u1 ∂u2 ∂u1 ∂u2 ∂u2 ∂u1
Since 1 +
∂a1
∂u1
+
∂a2
∂u2
> 0, it suffices to show that
∂a1 ∂a2
∂a1 ∂a2
−
∂u1 ∂u2 ∂u2 ∂u1
=
∂a1 ∂a2
∂u1 ∂u2
b1 VL,1 b2 VL,2
−
∂a1 ∂a2
∂u2 ∂u1
> 0, which follows from
(1 + b1 u1 )(1 + b2 u2 )
(1 + b1 u1 + b2 u2 )4
(b1 b2 )2 VL,1 VL,2 u1 u2
−
.
(1 + b1 u1 + b2 u2 )4
b1 VL,1 b2 VL,2
=
(1 + b1 u1 + b2 u2 ) > 0.
(1 + b1 u1 + b2 u2 )4
160
It is clear that if we find eigenvalues λ(B; u) of B then, we can find eigenvalues
λ(A; w) for A(w). For the Godunov numerical scheme that we present in (2.68), we have
that the CFL stability conditions (2.69) translates for this problem into
1
1
λ max at each Ujn .
≤
n
k=± λk (B; Uj ) 2
This gives the following condition over the time step
∆t ≤
∆x
min λk (B; Ujn )
2 k=±
at each Ujn .
(5.16)
We note that the condition (5.16) does not depend on the type of adsorption isotherm.
For the system (3.32)–(3.33), if the eigenvalues are positive. the Godunov method
becomes the Upwind method. That is,
∆t
n+1
n+1
n+1
n
n
n
n
n
U1,j
+ a1 U1,j
, U2,j
= U1,j
+ a1 U1,j
, U2,j
−
U1,j
− U1,j−1
,
∆x
∆t
n+1
n+1
n+1
n
n
n
n
n
U2,j
+ a2 U1,j
, U2,j
= U2,j
+ a2 U1,j
, U2,j
−
− U2,j−1
,
U2,j
∆x
(5.17)
(5.18)
n ≈ u (x , t ) for k = 1, 2. And the time step ∆t obeys the inequality (5.16).
where Uk,j
k j n
5.2.1
Experiments with equilibrium case using Godunov
Recall the extended Langmuir system (in equilibrium) given in (3.32)–(3.33) with
the extended Langmuir isotherms a1 and a2 defined by (3.34) and (3.35), respectively.
Figure 5.8 shows the evolution of numerical solutions for u1 and u2 with the Godunov
scheme.
Example 5.2.1.1. Consider the system given in (3.32)–(3.33) with initial data u1 (x, 0) =
H(x) and u2 (x, 0) = 1 − H(x). Recall that this system is in equilibrium and that we are
using the extended Langmuir isotherms defined in (3.34)–(3.35). In Figure 5.8 we can see
the evolution of the numerical solutions resulting from the implementation of the Godunov
Scheme.
161
FIGURE 5.8: Evolution of numerical solutions for the system (3.32)–(3.33) with initial
data u1 (x, 0) = H(x) and u2 (x, 0) = 1 − H(x). Example 5.2.1.1.
162
5.3.
Multicomponent system with IAS (Ideal Adsorbate Solution)
In this section we deal with a 2 × 2 system of conservation laws of form (3.32)–
(3.33) as in Section 5.2. However, the isotherms are not given explicitly with an algebraic
expression as in the extended Langmuir case (3.34)–(3.35).
Instead, the isotherms aIAS
(u1 , u2 ) and aIAS
(u1 , u2 ) are formulated implicitly from
1
2
single component isotherms. In practice, this means that a local nonlinear solver has to
be applied to get values of aIAS
(u1 , u2 ), aIAS
(u1 , u2 ) for every u1 , u2 .
1
2
5.3.1
From single component to multicomponent isotherms
In this section, we gain familiarity with the IAS algorithm and show step by step
the difficulties when applied to the particular example of single component Langmuir
isotherm. Let ui be the partial pressure associated to the i component (i = 1, 2), and
a◦µ,i = a◦µ,i (ui ) be the amount of i component adsorbed in the absence of other components.
Here u◦i denotes the hypothetical pressure of the pure component and ai = ai (u1 , u2 ) is
the isotherm when all components are present.
Recall that the ad-hoc expression for Langmuir isotherms are given in (3.34)–(3.35)
and they work well in practice, but they are not thermodynamically consistent. IAS
represents another alternative which is thermodynamically consistent and only assumes
that we already have pure (single) component isotherms a◦µ,i = a◦µ,i (ui ). From this single
component IAS gives an implicit expression that provides the isotherms aIAS
.
i
We write the multicomponent system with isotherms coming from IAS. Then, the
general IAS algorithm is presented, and in Section 5.3.6, we apply this algorithm to single
component Langmuir isotherms.
Algorithm 5.3.1.1.
Step 1. Consider single component isotherms a◦µ,i = a◦µ,i (ui ).
163
Step 2. Get z and ui from
u◦i
Z
z=
0
a◦µ,i (ui )
dui .
ui
Step 3. Set
u1
◦
u1 (z)
+
u2
◦
u2 (z)
=1
(5.19)
and solve for z.
Step 4. Compute
∂z
∂ui
aIAS
(u1 , u2 ) = ui
i
5.3.2
(i = 1, 2).
(5.20)
General case
In this section, we find expressions for
∂z
∂u1
and
∂z
∂u2
from expression (5.19).
Differentiating implicitly (5.19) with respect to u1 , we get
∂
∂u1
u1
◦
u1 (z)
∂
+
∂u1
u2
◦
u2 (z)
= 0.
(5.21)
Now, we work with the left hand side of (5.21),
∂
∂u1
u1
◦
u1 (z)
∂
+
∂u1
u2
◦
u2 (z)
=
=
u◦1 (z)
u1 ∂u◦1 ∂z
−
[u◦1 (z)]2 [u◦1 (z)]2 ∂u1 ∂u1
u2 ∂u◦2 ∂z
− ◦
[u2 (z)]2 ∂u1 ∂u1
#
"
1
u1 ∂u◦1
∂z
u2 ∂u◦2
−
+
u◦1 (z)
∂u1
[u◦1 (z)]2 ∂u1 [u◦2 (z)]2 ∂u1
which gives that
∂z
1
h
=
∂u◦1
u1
∂u1
◦
u1 (z) [u◦ (z)]2 ∂u1 +
1
We obtain an analogous expression for
∂z
u2
i.
(5.22)
i.
(5.23)
:
∂z
1
h
=
◦
∂u
∂u2
2
2
u◦2 (z) [u◦u(z)]
2 ∂u +
2
2
∂u◦2
u2
2 ∂u
◦
1
[u2 (z)]
u1
[u◦1 (z)]2
∂u◦1
∂u2
With (5.22) and (5.23) have now aIAS
(i = 1, 2) from (5.20).
i
164
We write general expressions for
∂aIAS
1
∂u1
=
∂2z
∂z
+ u1 2 ,
∂u1
∂u1
(5.24)
∂aIAS
2
∂u2
=
∂2z
∂z
+ u2 2 ,
∂u2
∂u2
(5.25)
∂aIAS
2
∂u1
∂aIAS
1
∂u2
5.3.3
∂aIAS
i
(i = 1, 2) that we use in the sequel,
∂ui
∂2z
,
∂u1 ∂u2
∂2z
= u1
.
∂u2 ∂u1
= u2
(5.26)
(5.27)
Multicomponent system with IAS isotherms
The purpose of this section is to present some results aiming to give conditions
to solve the multicomponent system involving IAS isotherms. At the beginning of this
section we found conditions for the hyperbolicity, CFL condition and time step condition
for the multicomponet system for the extended Langmuir isotherm. In Section 5.3.1 we
present an example including the single component Langmuir isotherms using the IAS
Algorithm 5.3.1.1. (See [12, 19]).
Now we find conditions for the system to be hyperbolic, for the CFL stability and
for the time step.
We begin by expressing this system as a conservation law.
Recall that the system in question is
∂
(u1 + aIAS
(u)) +
1
∂t
∂
(u2 + aIAS
(u)) +
2
∂t
∂u1
∂x
∂u2
∂x
= 0,
(5.28)
= 0,
(5.29)
with aIAS
given by (5.20). We start by writing the system (5.28)-(5.29) in the form (5.6).
i
Now w1 = u1 + aIAS
(u1 , u2 ), w2 = u2 + aIAS
(u1 , u2 ). Note that we will end up with a
1
2
matrix system AIAS with the same structure as the matrix A in (5.9).
To check hyperbolicity of the IAS system, we investigate the matrix AIAS .
165
Proposition 5.3.3.1. A sufficient condition for the system (5.28)–(5.29) to be strictly
hyperbolic is
u1 u2 > 0.
(5.30)
Proof. Following the calculations as in proof of Proposition 5.2.0.1 we see that we need
∂aIAS
1
∂u2
∂aIAS
2
∂u1
> 0.
But, for the IAS case, this means that
∂aIAS
1
∂u2
∂aIAS
2
∂u1
∂2z
∂2z
∂u1 ∂u2
∂u2 ∂u1
2
∂2z
> 0.
= u1 u2
∂u1 ∂u2
= u1 u2
(5.31)
The first equality comes from expressions (5.25) and (5.26), and (5.31) holds as long as
condition in (5.30) holds, then Proposition 5.3.3.1 applies and system (5.28)–(5.29) is
strictly hyperbolic.
Remark 5.3.3.1. In practice (5.30) is a natural assumption. Recall from Section 3.2. that
u1 and u2 are concentrations of the methane component and carbon dioxide component in
the mobile phase.
The following lemma gives conditions for the eigenvalues of our system to be positive.
Having this information allows us to work easily with the Godunov scheme when working
with single component Langmuir isotherms.
Lemma 5.3.3.1. The eigenvalue λ+ (BIAS ; u) > 0 is positive at every u. On the other
hand, the eigenvalue λ− (BIAS ; u) is positive if and only if
∂aIAS
∂aIAS
1+ 1 + 2 +
∂u1
∂u2
∂aIAS
1
∂u1
∂aIAS
2
∂u2
>
∂aIAS
1
∂u2
∂aIAS
2
∂u1
.
(5.32)
Proof. If aIAS
and aIAS
have the same properties as the Langmuir isotherm, then we can
1
2
say that B11 and B22 are positive numbers and B12 and B21 are negative numbers. On
166
the other hand, notice that λ+ (BIAS ; u) is positive and λ− (BIAS ; u) might be positive as
well. Notice that λ− (BIAS ; u) > 0 is equivalent to
q
2 + B11 + B22 > (B11 − B22 )2 + 4B12 B21 .
(5.33)
IAS for the entries of the matrix BIAS of IAS case, we keep the same
(instead of using Bij
notation of the extended Langmuir matrix, hoping that this does not cause confusion)
Since the left hand side of (5.33) is positive, then we can square both sides. Thus, in order
for (5.33) to hold we need
(2 + B11 + B22 )2 > (B11 − B22 )2 + 4B12 B21
and we get
1 + B11 + B22 + B11 B22 > B12 B21
as condition to have distinct eigenvalues.
After a direct calculation in the case when VL,1 = VL,2 , that is, when IAS isotherms
associated with single component isotherms which are Langmuir, coincide with extended
Langmuir isotherm (see Section 5.3.6 for details), we have that the inequality (5.32) holds
for every u in the domain of interest.
The following remark makes clear what is our domain of interest and opens a discussion about how to define z(u) at u = (0, 0), which is an important issue in implementation.
Remark 5.3.3.2. The variable z(u) defined in IAS Algorithm is twice differentiable in
[0, b] × [0, b] − {(0, 0)}
its domain. That is,
(b > 0). Therefore, the cross derivatives of z are the same on
∂2z
∂u1 ∂u2 (u)
=
∂2z
∂u2 ∂u1 (u),
for all u ∈ [0, b] × [0, b] − {(0, 0)}. When
coding, we need to define z at (0, 0) and we prove in Section 5.3.6 that in the case of
single component Langmuir isotherm that it makes sense to define z = 0 at (0, 0). Hence,
the system is strictly hyperbolic, provided u ∈ (0, b] × (0, b]. In fact, in our adsorption
application, it just makes senses to assume that u1 and u2 are nonnegative since they
represent concentrations. In consequence, z(u) > 0 for all u in this domain.
167
5.3.4
Eigenvectors and eigenvalues for IAS
In passing we also compute expressions which can be useful to study other properties
of the multicomponent system in the future. We start with the characteristic equation
~ = ~0.
(BIAS − λ± (BIAS ; u)I)V
So our two eigenvectors are


~− = 
V

and
−B12
1 + B11 − 12 λ− (BIAS ; u)

~+ = 
V


,

−B12
1 + B11 − 12 λ+ (BIAS ; u)

.
The eigenvalues for the matrix AIAS are given by λ− (AIAS ; w) =
1
λ−
(BIAS ; u)
and
1
. The eigenvectors V− and V+ that we just found are eigenλ+ (BIAS ; u)
vectors for AIAS as well.
λ+ (AIAS ; w) =
Remark 5.3.4.1. Recall that we needed (5.33) in order to have two different eigenvalues
λ− , λ+ . Once we see the form for the eigenvectors, it is more evident that we need these
conditions to have a complete set of eigenvectors {V~− , V~+ }. Clearly, we asserted that the
system was hyperbolic before, since different eigenvalues give a linearly independent set of
eigenvectors.
5.3.5
CFL condition for Godunov scheme for IAS
The Godunov scheme for IAS adsorption system has the same general form as
scheme for extended Langmuir system (5.17)–(5.18) presented in Section 5.2. However,
since the isotherms aIAS
and aIAS
are not given explicitly, an additional element of the
1
2
numerical algorithm is required. Namely, a local nonlinear solver has to be coupled to the
solution of the conservation law.
168
In practice, this can be avoided and the isotherms aIAS
and aIAS
can be precom1
2
puted ahead of the simulation.
In addition, there remains a question of how to choose the time step for IAS system
as we showed, in Godunov method the CFL constraint (5.16) must be satisfied. Since the
isotherms, thus the eigenvalues for IAS system are not known in advance, this may be
quite difficult to find the time step.
Therefore we hope to be able to formulate a simplified version of the process of finding appropriate time step for IAS system depending only on single component isotherms.
However this is not yet available.
5.3.6
IAS example for a particular choice of single component isotherms
To illustrate how IAS algorithm works, we show how it works for the case when
single component isotherms are Langmuir isotherms. This will help us to see clearly the
main differences between the IAS approach and the extended Langmuir isotherm example
that we developed in Section 5.3.1. The single component isotherm is given by
ui
,
1 + bi ui
a◦µ,i (ui ) = VL,i bi
i = 1, 2.
We then apply Step 2 from Algorithm 5.3.1.1:
Z
0
u◦i
a◦µ,i (ui )
dui =
ui
Z
u◦i
VL,i bi
0
Z
u◦i
= VL,i bi
0
ui
1 + bi ui
1
dui
ui
dui
1 + bi ui
= VL,i ln(1 + bi u◦i )
(i = 1, 2).
Therefore, if z = VL,i ln(1+bi u◦i ) then ez/VL,i = 1+bi u◦i and u◦i (z) =
ez/VL,i − 1
bi
(i = 1, 2).
Now, we write the expression (5.19) from Step 3.
b1 u1
z/V
e L,1 −
1
+
b2 u2
z/V
e L,2 −
1
= 1.
(5.34)
169
There is an interesting turn here. If VL,1 = VL,2 = K, equation (5.34) is an explicit
expression. Otherwise, (5.34) is an implicit equation for z. We do the computation for
both cases. If VL,1 = VL,2 then
b1 u1 +b2 u2
z/VL,1
−1
e
= 1 or ez/VL,1 = 1 + b1 u1 + b2 u2 , which gives
z(u) = K ln(b1 u1 + b2 u2 + 1).
In order to apply Step 4 from the Algorithm 5.3.1.1, we need to compute
∂z
∂ui
for i = 1, 2.
In this case,
∂z
bi
=K
∂ui
1 + b1 u1 + b2 u2
(i = 1, 2).
Finally,
Kb1 u1
,
1 + b1 u1 + b2 u2
Kb2 u2
.
1 + b1 u1 + b2 u2
aIAS
(u1 , u2 ) =
1
aIAS
(u1 , u2 ) =
2
We see that these expressions are the same as those for extended Langmuir system
isotherms a1 (u1 , u2 ) and a2 (u1 , u2 ) given in (3.34) and (3.35), respectively.
5.3.6.1
IAS calculations when Langmuir volumes are not equal
Now, if VL,1 6= VL,2 , calculations are a bit more complicated. We now recall implicit
equation for z given by (5.34) in Step 3 and differentiate respect to u1 and u2 in order to
get to Step 4 from Algorithm 5.3.1.1. Differentiating respect to u1
∂
u1
∂
1
b1
+ b2 u2
= 0,
∂u1 ez/VL,1 − 1
∂u1 ez/VL,2 − 1




∂z
1 z/VL,2
∂z
ez/VL,1 − 1 − (u1 /VL,1 )ez/VL,1 ∂u
−
e
∂u1
1
 + b2 u2  k2
 = 0,
b1 
2
2
z/V
z/V
L,1
L,2
(e
− 1)
(e
− 1)
2
b2
∂z
∂z
z/VL,1 − 1)2 ez/VL,2
b1 (ez/VL,2 − 1) ez/VL,1 − 1 − (u1 /VL,1 )ez/VL,1 ∂u
−
u
(e
VL,2 2
∂u1
1
2
2
(ez/VL,1 − 1) (ez/VL,2 − 1)
Solving for
∂z
∂u1 ,
we obtain
∂z
=
∂u1
2
b1 (ez/VL,2 − 1) (ez/VL,1 − 1)
b1 u1 z/VL,2
VL,1 (e
2
− 1) ez/VL,1 +
b2 u2 z/VL,1
VL,2 (e
2
− 1) ez/VL,2
.
= 0.
170
Note that setting VL,1 = VL,2 , we ended up obtaining the Langmuir isotherm aIAS
. Using
1
the symmetry of expression (5.34), the expression for
∂z
=
∂u2
∂z
∂u2
is given by
2
b2 (ez/VL,1 − 1) (ez/VL,2 − 1)
b2 u2 z/VL,1
VL,2 (e
2
− 1) ez/VL,2 +
b1 u1 z/k2
k1 (e
2
− 1) ez/k1
.
Now, we address the point raised in Remark 5.3.3.2 as to how one should define
z(u) at u = (0, 0) when implementing IAS algorithm. In turns out it makes sense to set
z(0) = 0 as shown in the following lemma.
Lemma 5.3.6.1. limu→(0,0) z(u) = 0.
Proof. From Taylor series expansion around zero for ez/VL,i − 1 and z ≥ 0, we get ez/VL,i −
1≥
z
VL,i
which gives us
1
z/VL,i
e
b1 u1
z/V
e L,1 −
1
−1
+
≤
VL,i
z .
b2 P2
z/k
e 2−
So
bi ui
z/VL,i
e
−1
≤
bi VL,i ui
,
z
for i = 1, 2. Therefore,
1
≤ (b1 VL,1 u1 + b2 VL,2 u2 ) · .
z
1
(5.35)
Using equation (5.34), inequality (5.35) is transformed into
1
1 ≤ (b1 VL,1 u1 + b2 VL,2 u2 ) · ,
z
or
0 ≤ z ≤ b1 VL,1 u1 + b2 VL,2 u2 .
(5.36)
Letting u → (0, 0) on right hand side of (5.36) we are done proving the claim.
5.3.7
Orders of convergence with IAS and EL
Figures 5.9 and 5.10 show the resulting order of convergence using Riemman data
as initial condition for extended Langmuir isotherms and IAS with single component
Langmuir isotherms in the L∞ , L1 and L2 norms.
171
[-1,2],h
min
0
10
= 0.001000, M= 1000,T=1, dt=82, u1(x,0)=iemman 1 then 0,u2(x,0)=Heaviside H(x)
Error L2
Error L1
Error Sup
linear
root
-1
10
-2
10
-2
10
-1
10
FIGURE 5.9: Orders of convergence for EL with u1 (x, 0) = H(x), u2 (x, 0) = 1 −
H(x), M = 1, 000, hmin = 0.001
FIGURE 5.10: Orders of convergence for IAS with u1 (x, 0) = H(x), u2 (x, 0) = 1 −
H(x), M = 1, 000, hmin = 0.001
172
We have used the same initial u1 (x, 0) and u2 (x, 0) and parameters in both experiments. The orders of convergence for IAS and extended Langmuir cases are the same.
We got O(h1/2 ) in both the L1 and L2 norms.
5.3.8
Experiments with IAS vs. EL
We compare IAS using two components Langmuir isotherms with Extended Lang-
muir system case in the following experiments.
In Figure 5.11 we can observe that whenever VL,1 = VL,2 we have that the numerical
solutions produced by IAS using single component Langmuir isotherms are pretty similar
to the solutions produced using extended Langmuir isotherms. In Figure 5.12 we have
chosen VL,1 6= VL,2 and numerical solutions corresponding to IAS are far apart from the
numerical solutions using extended Langmuir isotherms. This conform our observations
in Section 5.3.6.1.
(0)
FIGURE 5.11: IAS versus extended Langmuir. Initial data u1
1 − H(x). VL,1 = VL,2 = b1 = 1 and b2 = 2.
(0)
= H(x) and u2
=
173
(0)
FIGURE 5.12: IAS versus extended Langmuir. Initial data u1
(0)
= H(x) and u2
=
1 − H(x). VL,1 = 3, VL,2 = 6, b1 = 1 and b2 = 2.
In Figures 5.13 and 5.14 we show surfaces representing the isotherms a1 and aIAS
.
1
FIGURE 5.13: Extended Langmuir isotherm a1 with parameters VL,1 = 3, VL,2 = 6, b1 =
1 and b2 = 2.
174
with parameters VL,1 = 3, VL,2 = 6, b1 = 1 and b2 = 2.
FIGURE 5.14: IAS isotherm aIAS
1
Comparing Figure 5.13 and Figure 5.14 we can see that the isotherms a1 and aIAS
1
are totally different. This is an example for the case VL,1 6= VL,2 .
5.3.9
System with nonequilibrium and diffusion
We use a fully implicit scheme in the two component kinetic case given by the system
(u1 )t + (v1 )t + (u1 )x − D1 (u1 )xx = 0,
(5.37)
1
(v1 − a1 (u1 , u2 )) = 0,
τ1
(5.38)
(u2 )t + (v2 )t + (u2 )x − D2 (u2 )xx = 0,
(5.39)
1
(v2 − a2 (u1 , u2 )) = 0.
τ2
(5.40)
(v1 )t +
(v2 )t +
Therefore, the fully implicit discretization for (5.37) is written
n+1
n+1
n
n
U1,j
− U1,j
+ V1,j
− V1,j
+k
n+1
n+1
U1,j
− U1,j−1
h
−
k
n+1
n+1
n+1
D
U
−
2U
+
U
1
1,j+1
1,j
1,j−1 = 0,
h2
175
for j = 2, . . . , N − 1, and for j = 1 we set
n+1
U1,1
− u1 (a, 0) = 0.
Remark 5.3.9.1. We cannot use index 0 in MATLAB so the initial data is given by
ui (a, 0), i = 1, 2.
On the other hand, for j = N we set
n+1
n+1
−U1,N
−1 + U1,N = 0.
Now, for (5.38) we have
k
k n+1 n+1 n+1
n
+ 1+
V1,j
− V1,j
= 0,
− a1 U1,j , U2,j
τ1
τ1
(5.41)
for j = 1, . . . , N. We write the corresponding schemes for (5.39) and (5.40) in the same
n and V n .
way as in (5.37) and (5.38) but using U2,j
2,j
We have to assemble the corresponding Newton function by pieces. We obtain an
N by N Jacobian by blocks. We first define residuals Fj,1 , Hj,1 , Fj,2 , Hj,2 in order to find
the Jacobian for Newton’s method implementation.
Fj,1
n+1
n+1
n+1
n+1
U1,j−1
, U1,j
, U1,j+1
, V1,j
D1 λ
2D1 λ
n+1
n+1
= −λ −
U1,j−1 + 1 + λ +
U1,j
h
h
D1 λ n+1
n+1
−
U
+ V1,j
− b1 ,
h 1,j+1
n + V n . For the function corresponding to (5.41) we have
where b1 = U1,j
1,j
k n+1 n+1 k
n+1
n+1
n+1
n+1
Hj,1 U1,j , U2,j , V1,j
= − a1 U1,j , U2,j
+ 1+
V1,j
− b2 ,
τ1
τ1
n . Now, for the discretizations corresponding to the second component we
where b2 = V1,j
obtain
n+1
n+1
n+1
n+1
Fj,2 (U2,j−1
, U2,j
, U2,j+1
, V2,j
)
=
D2 λ
−λ −
h
n+1
U2,j−1
2D2 λ
n+1
+ 1+λ+
U2,j
h
D2 λ n+1
n+1
−
U
+ V2,j
− b3 ,
h 2,j+1
176
n +Vn .
where b3 = U2,j
2,j
k
k
n+1
n+1
n+1
V2,j
− b4 ,
= − a2 (U1,j , U2,j ) + 1 +
τ2
τ2
n+1
n+1
n+1
Hj,2 (U1,j
, U2,j
, V2,j
)
n . For notation use
where b4 = V2,j
n+1
n+1
n+1
n+1
n+1
n+1
Uk = Uk,1
, Uk,2
, . . . , Uk,N
, Vk = Vk,1
, Vk,2
, . . . , Vk,N
,
for k = 1, 2. Let Fk = (F1,k , F2,k , . . . , FN,k ) and Hk = (H1,k , H2,k , . . . , HN,k ) for k = 1, 2.
Here Fj,k and Hj,k depend on Uk and Vk (j = 1, . . . , N, k = 1, 2).
~ : R4N → R4N with
Note that we are considering a function F
~ 1 , V1 , U2 , V2 ) = (F1 , H1 , F2 , H2 ),
F(U

1


 −λ −



0

∂F1
=

..
∂U1 
.



0


0
0
D1 λ
h
2D1 λ
h
1+λ+
−λ −
..
···
0
− Dh1 λ
D1 λ
h
.
1+λ+
..
.
···
−λ −
···
0








∂F1
=
∂V1 






0
0
0
1
0
0 ··· 0
0 0
.. . .
.
.
1
..
0 ··· 0
. . .
. .. .. ..
0
0
..
. 0
1
0
0
0
··· 0
0
0
0
1+λ+
−1
0
0
..
.
− Dh1 λ
..
.
D1 λ
h
∂F1
∂F1
=
= 0.
∂U2
∂V2
Now, for H1 we have
...
2D1 λ
h
··· 0
0








,







2D1 λ
h
0
− Dh1 λ
1








,







177




∂H1

=
∂U1 


n+1
n+1
∂a1
(U1,1
, U2,1
)
− τk1 ∂u
1
0
. ...
.
. . . ..
0
..
.
0
∂H1
=
∂V1



∂H1

=
∂U2 



0
..
0

...
0
0
n+1
n+1
∂a1
. . . − τk1 ∂u
(U1,N
, U2,N
)
1




,



k
1+
I,
τ1
n+1
n+1
∂a1
)
, U2,1
(U1,1
− τk1 ∂u
2
0 ...
..
. ...
.
. . . ..
0
..
.
0
0

0
0
0
n+1
n+1
∂a1
. . . − τk1 ∂u
)
, U2,N
(U1,N
2




,



∂H1
= 0,
∂V2
∂F2
∂F2
=
= 0,
∂U1
∂V1

1


 −λ −



0

∂F2
=

..
∂U2 
.



0


0
0
D2 λ
h
2D2 λ
h
1+λ+
−λ −
..
···
0
D2 λ
h
.
− Dh2 λ
1+λ+
..
.
···
−λ −
···
0
2D2 λ
h
D2 λ
h
0
...
0
..
.
− Dh2 λ
..
.
1+λ+
−1
2D2 λ
h
0
− Dh2 λ
1








,







178








∂F2
=
∂V2 










∂H2

=
∂U1 


0
··· 0
0
0
0
1
0
0 ··· 0
0 0
.. . .
.
.
1
..
0 ··· 0
. . .
. .. .. ..
0
0
..
. 0
1
0
0
0
··· 0
0
0
0
n+1
n+1
∂a2
)
, U2,1
(U1,1
− τk2 ∂u
1
0
...
..
. ...
.
. . . ..
0
..
.
0
0








,







0
0
0
n+1
n+1
∂a2
. . . − τk2 ∂u
)
, U2,N
(U1,N
1





,



∂H2
= 0,
∂V1




∂H2

=
∂U2 


n+1
n+1
∂a2
)
, U2,1
− τk2 ∂u
(U1,1
2
...
..
. ...
.
. . . ..
0
..
.
0
0
∂H2
=
∂V2
5.3.10
0
0
0
0
n+1
n+1
∂a2
. . . − τk2 ∂u
(U1,N
, U2,N
)
2





,



k
1+
I.
τ2
Numerical experiments
In this section we exhibit some experiments in order to compare the kinetic case
presented in Section 5.3.9 and the equilibrium case presented in Section 5.2. for the two
component system. We work on the interval [0, 1], and stop at time T = 0.5. In different
examples we perform experiments with D1 and D2 fixed and τ1 , τ2 → 0 as shown in
179
System EL with λ = 0.99, h = 0.01, τ1 = τ2 = D1 = D2 = 0.01, VL,1 = VL,2 = b1 = b2 .
FIGURE 5.15: Kinetic versus equilibrium case for extended Langmuir system. Initial
data u1 (x, 0) = H(x) and u2 (x, 0) = 1 − H(x).
Figures 5.17, 5.18, 5.19 and 5.20. In one experiment, we fix τ1 and τ2 and use different
coefficients D1 and D2 shown in Figure 5.16.
First, we compare the kinetic case with the equilibrium case.
In Figure 5.15 can be observed that the solutions u1 , u2 in the kinetic case and the
solution in the equilibrium case are close to each other. We perform the same experiment
with different initial condition as shown in Figures 5.16, 5.17, 5.18, 5.19 and 5.20.
In Figure 5.16 we observe that as D1 , D2 → 0, the profiles for u1 , u2 gets sharper.
180
Comparision with different D1 , D2 . kinetic system EL. λ = 0.99, h = 0.01, τ1 = τ2 = 0.01, VL,1 = VL,2 = b1 = b2 .
FIGURE 5.16: Kinetic system plots for different values of D1 and D2 . Include the case
D1 = D2 = 0.
181
System EL Kin vs. Equi with λ = 0.99, h = 0.01, D1 = D2 = 0.01, VL,1 = VL,2 = b1 = b2 and different values for τ1 , τ2 .
Nonequi u1 τ1 = τ2 = 0.1
Nonequi u2 τ1 = τ2 = 0.1
Nonequi u1 τ1 = τ2 = 0.01
Nonequi u2 τ1 = τ2 = 0.01
Nonequi u1 τ1 = τ2 = 0.001
Nonequi u2 τ1 = τ2 = 0.001
Equilibrium u1
Equilibrium u2
Nonequi u1
Nonequi u2
FIGURE 5.17: Extended Langmuir kinetics system solution plots for different values of τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data: u1 (x, 0) =
H(x), u2 (x, 0) = 1 − H(x).
In Figure 5.17, we can see that as τ1 and τ2 go to zero, the solutions for the Kinetic
system are approaching to the solution for the system in equilibrium. We run same
experiment with different initial data u1 (x, 0) and u2 (x, 0), in Figures 5.18 to 5.20.
182
System EL Kin vs. Equi with λ = 0.99, h = 0.01, D1 = D2 = 0.01, VL,1 = VL,2 = b1 = b2 and different values for τ1 , τ2 .
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Equilibrium u1
Equilibrium u2
FIGURE 5.18:
Extended Langmuir kinetic system solution plots for different val-
ues of τ1 (= τ2 ). Includes solution for the equilibrium case.
H(x), u2 (x, 0) = 2(1 − H(x)).
Initial data u1 (x, 0) =
= 0.1
= 0.1
= 0.01
= 0.01
= 0.001
= 0.001
183
System EL Kin vs. Equi with λ = 0.99, h = 0.01, D1 = D2 = 0.01, VL,1 = VL,2 = b1 = b2 and different values for τ1 , τ2 .
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Equilibrium u1
Equilibrium u2
FIGURE 5.19: Extended Langmuir kinetics system solution plots for different values of τ1 (= τ2 ). Includes solution for the equilibrium case.
2H(x), u2 (x, 0) = 1 − H(x).
Initial data u1 (x, 0) =
= 0.1
= 0.1
= 0.01
= 0.01
= 0.001
= 0.001
184
System EL Kin vs. Equi with λ = 0.99, h = 0.01, D1 = D2 = 0.01, VL,1 = VL,2 = b1 = b2 and different values for τ1 , τ2 .
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Nonequi u1 τ1 = τ2
Nonequi u2 τ1 = τ2
Equilibrium u1
Equilibrium u2
FIGURE 5.20: Extended Langmuir kinetic system solution plots for different values of
τ1 (= τ2 ). Includes solution for the equilibrium case. Initial data u1 (x, 0) = max(0, 1 −
|2x − 1|), u2 (x, 0) = 1 − H(x).
We have chosen T = 0.4 in the experiment associated to Figure 5.20. In Figures 5.18,
5.19 and 5.20 we observe a similar tendency as in Figure 5.17. That is, as τi → 0, (i = 1, 2),
the solution u1 , u2 approaches the solution for the equilibrium system.
= 0.1
= 0.1
= 0.01
= 0.01
= 0.001
= 0.001
185
6.
SUMMARY AND CONCLUSIONS
In this dissertation we have shown analysis and simulation results for methane hydrates and adsorption models. With respect to methane hydrates, we were able to express
the model with a monotone graph as an evolution equation with a Nonlinear Complementarity Constraint. We established well posedness of a weak form of that model and
proposed a numerical algorithm to approximate its solutions. The important component
of that algorithm is the use of the semismooth Newton solver which relies on transforming
the constraint to an equivalent form using a nonlinear function from the class of semismooth functions. The solver can be used also for other problems of similar structure and,
in particular, for a singular graph for which neither the graph nor its inverse are functions.
In addition, the solver can also work or be implemented for the Stefan free boundary value
problem. Thus, our results not only extend known theory to cover the methane hydrate
model, but also are applicable to other important models in the literature.
For the adsorption model we have shown how to cast a problem with an implicitly defined multicomponent isotherm in the framework of hyperbolic conservation laws.
We have also defined the conditions upon which the system in strictly hyperbolic. In
addition, we set up a framework for implicit treatment of nonlinearity in the adsorption
problems which let us reuse methods known for hyperbolic systems in a traditional form.
Furthermore, we set up a framework for analysis and approximation of nonequilibrium
problems.
Future work in methane hydrates include adaptive griding time stepping for finite
element simulation. This is nontrivial since none of the adaptive a posteriori error theory
relies on having H 2 solution. In our case we do not even have solutions in H 1 . Another
aspect that can be developed is the role of salinity as a variable in our model.
With respect to the adsorption model, current work includes stability analysis for
186
the kinetic case. It appears that the L2 norm of the two components is not an appropiate
quantity of interest, and we have preliminary work on how to change it. Once the latter
analysis is completed, it would be natural to extend stability analysis to multiple components in the kinetic case. Finally, a different aspect to work on would be the analysis
and numerical simulations when we consider the pressure equation. That means that we
do not assume that the velocity is constant. We also plan to develop more convergence
studies.
187
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