Ernest and Virginia Cockrell Chair in. Engineering ..... element method in space with backward Euler or Crank-Nicholson time discretization and Newton or ...
RICE UNIVERSITY
Mixed Finite Element Methods for Variably Saturated Subsurface Flow by
Carol A. San Soucie A Thesis Submitted in Partial Fulfillment of the Requirements for the Degree
Doctor of Philosophy
Approved, Thesis Committee:
Clint N. Dawson, Co-Chairman Associate Professor of Aerospace Engineering and Engineering Mechanics University of Texas at Austin Mary F. Wheeler, Co-Chairman Ernest and Virginia Cockrell Chair in Engineering University of Texas at Austin Dan C. Sorensen Professor of Computational and Applied Mathematics J. Ed Akin Professor of Mechanical Engineering and Materials Science
Houston, Texas April, 1996
Mixed Finite Element Methods for Variably Saturated Subsurface Flow Carol A. San Soucie
Abstract The ow of water through variably saturated subsurface media is commonly modeled by Richards' equation, a nonlinear and possibly degenerate partial dierential equation. Due to the nonlinearities, this equation is dicult to solve analytically and the literature reveals dozens of papers devoted to nding numerical solutions. However, the literature also reveals a lack of two important research topics. First, no a priori error analysis exists for one of the discretization schemes most often used in discretizing Richards' equation, cell-centered nite dierences. The expanded mixed nite element method reduces to cell-centered nite dierences for the case of the lowest-order discrete space and certain quadrature rules. Expanded mixed methods are useful because this simpli cation occurs even for the case of a full coecient tensor. There has been no analysis of expanded mixed methods applied to Richards' equation. Second, no results from parallel computer codes have been published. With parallel computer technology, larger and more computationally intensive problems can be solved. However, in order to get good performance from these machines, programs must be designed speci cally to take advantage of the parallelism. We present an analysis of the mixed nite element applied to Richards' equation accounting for the two types of degeneracies that can arise. We also consider and analyze a two-level method for handling some of the nonlinearities in the equation. Lastly, we present results from a parallel Richards' equation solve code that uses the expanded mixed method for discretization.
Acknowledgments I want to thank the Texaco Graduate Fellowship program, the Center for Subsurface Modeling Industrial Aliates and the United States Department of Energy for supporting this work. I wish to thank my two advisors, Clint Dawson and Mary Wheeler for their advice and help over the years. Clint Dawson has put much time and eort into my instruction and for his encouragement and energy, I am especially grateful. Mary Wheeler has been a strong source of support and inspiration for which I will always be appreciative. I would also like to thank Lawrence Cowsar and Fredrik Saaf who have been the best of \brothers" the last few years. Furthermore, I thank Laurie Feinswog, Cli Nolan and the basement \dwellers" at Rice for their friendship. The time we have all shared has made my years in school memorable. I would never have attended graduate school were it not for the encouragement and preparation given me by high school and college mentors, especially Paul Murrin and Martha Talbert, formerly of the Louisiana School for Math, Science and the Arts, and Elizabeth Swoope and Guillermo Ferreyra of Loiuisiana State University. I owe much gratitude to Leonard Gray who rst introduced me to research and set in motion events which have changed my life. My mother, Dory San Soucie, my father, William San Soucie, and my brother, Paul San Soucie, have always been strong sources of encouragement and love. For that I can never fully express my appreciation. Most importantly, I thank Robert Woodward for his never-ending words of encouragement, for his friendship and love and for the joy and happiness that he brings to life.
Contents Abstract Acknowledgments List of Illustrations List of Tables
ii iii vi vii
1 Introduction
1.1 Introductory Remarks : : : : : : : : : : : : : : : : : : : : : : : : : : 1.2 Previous Work : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1.3 Present Work : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
2 Physical Background 3 Discretization 3.1 3.2 3.3 3.4
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4 An a priori Error Analysis of Richards' Equation
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5 Two-Level Methods for Nonlinear Parabolic Equations 49 5.1 A Two-Level Finite Dierence Scheme : : : : : : : : : : : 5.1.1 A Coarse Grid Nonlinear Finite Dierence Scheme : 5.1.2 Fine Grid Linear Scheme : : : : : : : : : : : : : : : 5.1.3 Extensions to Multiple Levels : : : : : : : : : : : : 5.2 A Two-Level Method for Richards' Equation : : : : : : : :
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6 Implementation and Numerical Results
6.1 Implementation Issues : : : : : : : : : : : : : : : : : : : : : : : 6.2 Numerical Results : : : : : : : : : : : : : : : : : : : : : : : : : : 6.2.1 A Known Solution Test Case : : : : : : : : : : : : : : : : 6.2.2 A One-Dimensional Flow Problem : : : : : : : : : : : : : 6.2.3 A Three-Dimensional Irregular Geometry Flow Problem
7 Conclusions
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7.1 Summary : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7.2 Future Work : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Bibliography
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Illustrations 2.1 Typical van Genuchten curve of water content vs. matric potential. : 2.2 Typical van Genuchten curve of hydraulic conductivity vs. water content. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.3 Typical van Genuchten curve of hydraulic conductivity vs. pressure head. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6.1 The 19-point discretization stencil. : : : : : : : : : : : : : : : : : : : 6.2 Linear plot of convergence data. : : : : : : : : : : : : : : : : : : : : : 6.3 Solutions for the one-dimensional test problem with various time step sizes. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6.4 Solutions for the one-dimensional test problem with various mesh sizes. 6.5 Three-dimensional single permeability layer test case after 45 simulation days. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6.6 Three-dimensional two permeability layer test case after 50 simulation days. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6.7 Three-dimensional irregular geometry test case after 5 simulation days. 6.8 Three-dimensional irregular geometry test case after 20 simulation days. 6.9 Three-dimensional irregular geometry test case after 50 simulation days. 6.10 Three-dimensional irregular geometry test case after 75 simulation days.
11 13 14 75 78 81 81 83 84 85 85 86 86
Tables 6.1 Convergence results for a three dimensional analytic test problem. : : 6.2 Physical data for the one-dimensional ow problem. : : : : : : : : : : 6.3 Physical data for the three-dimensional ow problem. : : : : : : : : :
78 79 82
1
Chapter 1 Introduction 1.1 Introductory Remarks Recent years have seen an increase in attention to modeling the ow of water through variably saturated porous media. This increase arises from heightened interest in nding appropriate sites for waste facilities and in evaluating the impact of current sites on local groundwater systems. One way to gain an understanding of the groundwater systems at these sites is through computer simulations of subsurface ow. A commonly accepted mathematical model of water ow through variably saturated porous media is Richards' equation, a nonlinear parabolic partial dierential equation well known in hydrology and related sciences. Richards' equation is expressed as, @(h) + S @h ? r K (h)rh = f; (1.1) s @t @t where h is the hydraulic head, is the moisture content of the soil, Ss is the speci c storage of the medium, K is the hydraulic conductivity and f is a water source/sink term. The highly nonlinear nature of this equation makes analytical solutions dicult to nd, so this equation is most often solved numerically. In this thesis we formulate ecient discretization schemes based on the mixed nite element method for the solution of Richards' equation, prove a priori error estimates for these methods and show results from a computer program developed for parallel platforms which solves Richards' equation.
1.2 Previous Work Before presenting the results of this thesis, a brief summary of previous work on the analysis and numerical solution of Richards' equation is in order. We rst consider reasons for choosing mixed nite element methods and some results pertaining to these methods. We then mention previous work anaylyzing Richards' equation, and lastly, we discuss the literature pertaining to numerical solutions of the equation.
2 Mixed methods are considered for this work because they conserve mass on a cellby-cell basis. This conservation of mass means that at any time, the ux out of each cell is equal to the ux into the cell plus any source. Galerkin nite elements only guarantee a global conservation of mass, meaning that the ux out of the domain is equal to that into the domain plus any source. Since Richards' equation is actually a conservation equation, mixed methods in some sense require the solution to hold cell-by-cell. Mixed nite element methods for linear elliptic problems have been well studied [17, 56, 9]. Element spaces have been developed for both two and three dimensions and for many dierent element shapes [54, 50, 14, 15, 16]. For these equations analysis has shown that if h is the maximal mesh spacing, then optimal convergence of the lowest order mixed method is O(h) for both the scalar and velocity variables. Moreover, superconvergence of O(h2) has been shown for the pressure and velocity variables at certain points [49, 62, 29, 25]. In the case of linear elliptic equations, Russell and Wheeler [59] have shown that for the lowest order Raviart-Thomas-Nedelec [54, 50] spaces on rectangles and for a diagonal tensor K , the use of certain quadrature rules simpli es the mixed method into a cell-centered nite dierence scheme with a 5 point stencil in two dimensions and a 7 point stencil in three dimensions. Weiser and Wheeler [62] showed that this simpler scheme retains the convergence and superconvergence rates of the original method for both the pressure and velocity. Although much analysis has been done on mixed nite element methods, most of it assumes that K is a diagonal and invertible tensor. However, a full tensor can arise when computing \eective permeabilities" as in upscaling from ne to coarse data [26] or when mapping a rectangular grid into a logically rectangular grid [7]. When the tensor is full it is not possible to derive a nite dierence scheme equivalent to the mixed method. Recently, methods have been developed to handle a full, possibly noninvertible tensor [8, 19, 44]. In particular, Arbogast, Wheeler and Yotov have analyzed the expanded mixed nite element method [8]. This method simultaneously approximates the pressure, its gradient and the ux. Arbogast, Wheeler and Yotov showed that for the lowest order Raviart-Thomas-Nedelec space on parallelepipeds, a cell-centered nite dierence scheme results from this method. In certain discrete norms and for linear elliptic equations, this scheme exhibits superconvergence of O(h2) for the scalar variable and of O(h3=2) for its gradient and ux. However, in the interior of the domain, they show O(h2 ) for the last two of these.
3 There has been relatively little analysis of the mixed method applied to nonlinear equations. Milner [48] developed a mixed method for the solution of two-dimensional second order quasi-linear elliptic equations. He was able to show existence and uniqueness of a solution to his scheme as well as optimal convergence. Dawson and Wheeler [22] in the course of analyzing a two-grid scheme for three-dimensional problems derived optimal order estimates for the expanded mixed method applied to the nonlinear heat equation. Richards' equation is particularly dicult to analyze since it can be degenerate, i.e. the K (h) term can be 0 as can the time derivative of . There has been some recent work on the analysis of degenerate parabolic equations. Rose [57] considered the porous medium equation, which admits solutions lacking the regularity of classical solutions. He developed continuous and discrete time Galerkin nite element approximations and derived estimates based on assumed rates of degeneracy. Nochetto and Verdi [51] also considered degenerate parabolic equations and developed linear Galerkin nite element schemes with error estimates. Arbogast, Wheeler and Zhang [6] made use of the Kirchho transformation in order to develop estimates of the mixed method applied to degenerate parabolic equations but they assume a linear time derivative term. In [4], Arbogast developed error estimates for Galerkin nite elements applied to Richards' equation. He allowed for the time derivative of to be 0 but assumed K > 0. Arbogast, Obeyeskere and Wheeler [5] developed estimates for the Galerkin method applied to Richards' equation in the case that both the time derivative of and the hydraulic conductivity are not 0. When considering previous numerical work on Richards' equation, it is helpful to be familiar with some common formulations of the equation. Dierent formulations of Richards' equation have various advantages and disadvantages depending on the physical situation and the numerical scheme. Various formulations are possible due to a constitutive relationship between pressure head and water content. Probably the most common expression of the equation is formulated in terms of the pressure head only, (Ss + C (h)) @h @t ? r K (h)rh = f; where C (h) = @=@h is the water capacity. While this formulation gives the solution as pressure head, due to the way the time derivative is expressed, numerical schemes based on this form tend to be nonconservative. Another form is based on the water
4 content,
@ ? r D()r = f; @t where D() = K ()=(@=@h). This form is advantageous in that it is in conservative form. However, for saturated media, becomes constant, D approaches in nity, and this form is no longer applicable. Furthermore, is not continuous across interfaces separating layers of two dierent soils. The pressure head is continuous across these discontinuities which makes head-based methods better suited for modeling ow in layered soils. However, researchers have found that schemes for the head-based formulation produce large mass balance errors [18, 39]. The formulation in equation (1.1) is the mixed form. This formulation is also mass conserving and gives the solution in terms of pressure. Many papers have been published discussing numerical solutions to Richards' equation. The most common approaches use a low-order nite dierence or nite element method in space with backward Euler or Crank-Nicholson time discretization and Newton or Picard iteration for the nonlinearities. We now brie y describe some of this work. Allen and Murphy [2] in the context of collocation methods and Celia, Bouloutas and Zarba [18] in the context of nite dierences and nite elements have formulated the modi ed Picard method for handling nonlinearities in the mixed form of Richards' equation. This method applies Picard iteration to the nonlinearities in the hydraulic conductivity, but uses a rst order Taylor expansion of about the previous value for the time derivative term. This expansion results in the same linear system as the standard head-based scheme except that the right hand side also contains the time derivative of at the previous iteration for the given time level. Numerical results show this term helps to preserve mass balance that is lost with the standard head-based schemes. Much work on the solution of the head-based form of the equation has focused on developing mass-conservative schemes for this nonconservative form. In [47], Milly formulated a mass-conservative scheme by using an average value of the water capacity over each time step. This averaging reduced error associated with evaluation of the function at a xed point which may or may not represent the behavior over the entire time step. Kirkland, Hills and Wierenga [43] employ an update for based on computed ux values. This new update removes the nonconservative nature of the head-based scheme and preserves mass balance. Rathfelder and Abriola [53]
5 developed mass-conservative numerical solutions of the head-based form with both nite elements and nite dierences. They make use of a chord-slope approximation of the water capacity term, C . In the nite dierence case, their scheme for the head-based form results in the same discrete system as the scheme of Celia, et.al. for the mixed form of the equation. Hills, Porro, Hudson and Wierenga [39] developed a scheme for the -based form. They modeled the discontinuities of by adding an additional source term expressed as a jump in values across interfaces. Comparisons between their water contentbased and head-based forms show that the -based scheme is far better at conserving mass and is less sensitive to time step size in dry conditions than the pressure-based form. They point out that the main disadvantage of the -based scheme is its inapplicability to saturated ow. Some authors have considered the Kirchho transform to numerically handle degeneracies. Haverkamp and Vauclin [37] compared a nite dierence solution of the Kirchho transformed equation with the head-based form. They found that the transformed equation gave more accurate results but required much more compute time due to the need for integrated values of the transformation. Ross and Bristow [58] have discussed the transformation in the case of discontinuous hydraulic conductivities. They apply the Kichho transformation element-by-element, then couple the elements together through the continuous pressure head at element boundaries. Some authors have looked at variable transformations to switch between saturated and unsaturated conditions thereby using the -based form in unsaturated regions and the head-based form in saturated regions. Kirkland, Hills and Wierenga [43] transform Richards' equation in terms of a single variable which is de ned as water content in dry conditions and pressure head otherwise. With this transformation the scheme generates very little mass balance errors, and is stable over a wide range of conditions. However, they nd degradation in accuracy near the interface between saturated and unsaturated regions. Forsyth, Wu and Pruess [33] developed a similar scheme in that they switch from head-based to -based schemes depending on water saturation values. Their scheme diers from Kirkland et.al. in that the change in variables is performed after the equation is discretized, as opposed to rewriting the original equation in terms of a new, more general variable. Forsyth et.al. switch between pressure and water content by substituting the appropriate variable in the equation for each grid point. Numerical results show a signi cant improvement in computational speed by using this variable substitution method instead of standard
6 head-based methods for dry conditions. The reason for this improvement is that they are able to take larger time steps when the domain is unsaturated. Huyakorn, Thomas and Thompson [40] compared the Newton and Picard methods. For a Galerkin method applied to the head-based form of the equation, they have found that even though the Newton iterations are each slower than the Picard iterations, in general, signi cantly fewer Newton iterations are required for convergence. One diculty with solving Richards' equation numerically is that despite the fact that the equation is parabolic, steep water saturation fronts can occur when modeling
ow of water into very dry media. The saturation fronts can be very dicult to simulate numerically and common methods such as Galerkin nite element methods can produce sharp nonphysical oscillations near these fronts. Forsyth and Kropinski [32] have given monotonicity conditions for a head-based scheme. If these conditions are met, the solution will be non-oscillatory near steep fronts. They further indicate that only upstream weighting [61] for the K term as opposed to central weighting will give nonoscillatory solutions. Abriola and Lang [1] have shown that adding more degrees of freedom by using a higher order method near the front gives more accuracy than if the same number of new unknowns were introduced solely by grid re nement.
1.3 Present Work Until now, there have been no estimates of the mixed nite element method applied to Richards' equation. This thesis will present a number of estimates for the expanded mixed method applied to the equation. We rst present a continuous time analysis for the case where K > 0 and the time derivative of may be zero. This is the case for partially to fully saturated ow. For this situation, bounds of the error in approximating and the negative gradient of hydraulic head are derived in terms of a Holder continuity parameter. Furthermore, an optimal bound for the nonlinear form, 1=2 Z T ; ( ( p ) ? ( P ) ; p ? P ) dt 0 where p is the hydraulic head, is derived. The bound is optimal since it is equal to the order of truncation error for approximation with the same degree polynomial as the approximating space used in the method. In addition, this nonlinear form is bounded below by the error in and above by the error in p. Next, we consider the case where K > 0 and the time derivative of is strictly nonzero. This is the case of strictly partially saturated ow. For this situation,
7 optimal convergence of the hydraulic head and its negative gradient are shown for a fully discrete time scheme. The third case considered occurs when the tensor coecient is positive semide nite, and @=@p 0. This is the case of unsaturated to fully saturated. For this possibly degenerate situation, the Kirchho transform, Zp R(p) = k((}))d}; 0
is used, where p is the hydraulic head and k((p)) is the relative permeability. As seen below, this transformation moves the nonlinearity from the K term to the gradient. In the situation when K = 0, the problem solution lacks enough regularity to formulate a variational problem involving the time derivative of , with trial functions in L2. Thus, we follow the technique of Arbogast, Wheeler and Zhang [6] and formulate an integrated in time scheme. The error estimates for the resulting scheme applied to Richards' equation are optimal in the sense that they reduce to approximation error. Having analyzed the expanded mixed method applied to Richards' equation, we turn to methods of handling the nonlinearities at the level of discretization. The approach used is that of J. Xu [63, 64] and Dawson and Wheeler [22]. In these works, the discretization scheme is applied to the nonlinear equation on a coarse grid, and the equation is then linearized about the coarse grid solution on the ne grid. Xu analyzed this scheme for Galerkin methods applied to nonlinear elliptic equations, and Dawson and Wheeler analyzed the scheme for the expanded mixed method applied to the nonlinear heat equation. As a rst step in applying this scheme to Richards' equation, we analyze the scheme for a superconvergent cell-centered nite dierence method also applied to the nonlinear heat equation. Then the scheme for the expanded mixed method applied to Richards' equation is discussed. Although much computational work has been done in nding ecient ways of solving Richards' equation, to this author's knowledge there have been no published results from a parallel computer code. Results are given from a parallel, threedimensional Richards' equation code, PREQS. This code uses a cell-centered nite difference scheme equivalent to the expanded mixed method with quadrature. One point upstream weighting is used to more accurately model the moving fronts. Parallelism is achieved by spatially decomposing the domain into subdomains and assigning one subdomain to each processor, and extra unknowns are introduced along subdomain interfaces in order to reduce communication requirements.
8 Results are given from a variety of test cases. The rst case is a nonlinear parabolic equation with a source term chosen to guarantee a speci c solution. A three-dimensional convergence analysis is done which indicates a spatial rate of convergence of almost O(h2). The second test case is a one-dimensional Richards' equation problem from Celia, Bouloutas and Zarba [18]. Celia et.al. measure the mass balance ratio which is the total amount of water entering at the boundaries of the domain divided into the time rate of change in water mass. For a mass conserving numerical method, this ratio should always be unity. Celia et.al. report a ratio of 1 for a mixed formulation scheme and ratios signi cantly less than 1 for a head-based scheme. The PREQS code always gives a ratio of 1, indicating conservation of mass. Lastly, results are given for a three-dimensional full tensor Richards' equation case using the general geometry techniques of Arbogast, Wheeler and Yotov [7]. These results indicate that the code predicts reasonable solutions to ow problems on general domains. The rest of this document is organized as follows. In the next chapter an overview of the physical ow problem and assumptions leading to Richards' equation are given. In chapter 3, notation and discretization schemes are introduced. We summarize the mixed and expanded mixed nite element methods as well as discuss the RaviartThomas-Nedelec approximating spaces. In the following chapter an a priori error analysis of the expanded mixed method applied to Richards' equation is presented. Chapter 5 discusses a novel two-level method for handling the nonlinearities in the equation and chapter 6 presents the parallel Richards' equation code and numerical results. Lastly, chapter 7 gives a brief summary of the thesis and indicates directions for future work.
9
Chapter 2 Physical Background In this chapter we give a brief description of the physical laws that lead to Richards' equation for ow of water through variably saturated porous media. For information beyond that presented here, the reader is referred to the books of Bear [11, Chapter 9], Fetter [31, Chapter 4] and Freeze and Cherry [34, Chapter 2]. The physical situation we are modeling is that of water owing into a porous medium lled with air and a small amount of water. Water saturation measures the amount of water in the medium and is de ned as the fraction of total pore space that is lled with water. The term \variably saturated" refers to the possibility that the water saturation, s, can vary between some residual water saturation, sr , and ss , a fully saturated medium. The medium is called unsaturated if water saturation is less than ss and saturated otherwise. Water saturation is closely related to the volumetric water content of the soil, , which is the fraction of total volume that is lled with water. The relationship between and s is, = s; where is the porosity, or amount of pore space per unit volume of the medium, and s is the water saturation. In unsaturated ow, the driving force of the ow is the matric potential, , which has units of Newtons per square meter (N=m2). This potential is caused by surface tension creating a negative pressure on the pore water and is a function of the volumetric water content of the soil, . In unsaturated media, the matric potential is negative and is equal to the negative of capillary pressure, Pc , also having units of N=m2. The capillary pressure is related to the pressures of the other phases by, Pc = pa ? pw ; where pa and pw are the air and water pressures, respectively. For the case of a single water phase owing into a porous medium, it is assumed that the air phase pressure
10 remains constant at atmospheric pressure. Thus, the capillary pressure is no longer a function of pa. In this case, we will consider the water pressure as a gage pressure, i.e. pw = p0w + pa where p0w is the absolute water pressure. Thus, ?Pc = pw . Capillary pressure can also be experimentally measured as a function of water saturation. The resulting curves exhibit hysteresis; they are dierent depending on whether water is
owing into (imbibition) or out of (drainage) the medium. In the work considered here, hysteresis will be neglected. Due to the relationship between capillary pressure and water saturation, we can write s = Pc?1 (pa ? pw ); where pa is constant. This relation shows s as a function of pw . Van Genuchten [38] derived an empirical formula for the water content as a function of matric potential. This relationship is, r ; = r + [1 +s(? ) (2.1) n ]m (2.2) n = 1 ?1 m ; = h1 (21=m ? 1)1?m ; (2.3) b where is the volumetric water content, s is the volumetric water content at s = ss , r is the irreducible minimum water content at s = sr , is the matric potential, m is an experimental parameter based on the soil type and hb is the bubbling pressure (de ned below). For a typical soil-water system, m 0:5; r 0:1; s 0:5 and hb ?355cm. Thus, n 2:0 and 0:005. A typical curve of water content vs. matric potential for these parameters is given in Figure 2.1. The bubbling pressure can be de ned as follows. At atmospheric pressure, the medium is saturated with = s, where s is the highest value can take. Now consider decreasing the matric potential. The medium will remain saturated as the matric potential is decreased until the potential is negative enough that the water will begin to drain. The potential value at which this drainage starts to occur is the bubbling pressure. For saturated ow, the driving force is again a pressure potential. However, the pressure is now positive and the potential > 0. If water is allowed to be slightly compressible, the density is not constant and is related to the water pressure through an equation of state, for example, = 0e (p?p0); (2.4)
11 Water Content vs. Matric Potential
6
10
5
Matric Potential (−cm)
10
4
10
3
10
2
10
1
10
0
0.1
0.2 0.3 0.4 Water Content (dimensionless)
0.5
0.6
Figure 2.1 Typical van Genuchten curve of water content vs. matric potential.
where 0 is water density at atmospheric pressure p0, and is a small constant. The water compressibility constant, , is de ned as the negative of change in water volume per unit volume per change in pressure, or, (2.5) = ? dVwdp=Vw 4:4 10?10m2=N: The total soil-moisture potential, , is the sum of the matric potential and a gravity potential. The gravity potential can be expressed as the product of the water density, , the acceleration of gravity, g, and the height, z, above some reference level. Thus, the total soil-moisture potential is, p = + gz: If this equation is divided by g, the result is the soil moisture potential expressed as energy per unit weight, commonly measured in cm. This potential is, +z h = g = h + z;
12 where h is the matric potential expressed in units of length. The matric potential expressed as a length is often referred to as pressure head, and the soil-moisture potential expressed as a length is referred to as hydraulic head. Darcy's Law for saturated ow and the Buckingham ux law for unsaturated ow relate the ow of water to the gradient of the hydraulic head through the relation,
q = ?K (h)rh;
(2.6)
where q is the soil moisture ux (cm=s) and K (h) is the hydraulic conductivity of the soil. The hydraulic conductivity (cm=s) measures the ability of the soil to transmit water. For a saturated medium, the pore space is lled with water and all the pores participate in the transmission of water. Thus, the hydraulic conductivity is a function of position only. However, for an unsaturated medium, some of the pore space is lled with air. Water will only travel through wetted areas, so for an unsaturated medium, the hydraulic conductivity is a function of the moisture content as well as position. Experiments conducted with ideal, uniform porous media have indicated that the hydraulic conductivity can be written as, K () = kkrw()g ; where k is the intrinsic permeability of the medium (measured in Darcy's where 1 darcy = 10?8 cm2), krw () is the relative permeability of water to air (dimensionless) and is the dynamic viscosity of water (N s=cm2). The relative permeability is the ratio of the unsaturated hydraulic conductivity evaluated at to the saturated hydraulic conductivity, evaluated at s. The value of krw is simply a number between 0 and 1. The hydraulic conductivity can also be expressed as a function of the matric potential. Van Genuchten [38] derived expressions relating the hydraulic conductivity to both the water content and the pressure head. The relationship between K and is expressed as,
K () = Ks Se1=2[1 ? (1 ? Se1=m)m]2; where Se = ( ? r )=(s ? r ) is an eective saturation between 0 and 1, Ks is the saturated hydraulic conductivity and m is the van Genuchten soil parameter. For the typical parameters discussed above, this curve is given in Figure 2.2. The relationship
13 Relative Permeability vs. Water Content 1
Relative Permeability K/K_s (dimensionless)
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
0.1
0.2 0.3 0.4 Water Content (dimensionless)
0.5
0.6
Figure 2.2 Typical van Genuchten curve of hydraulic conductivity vs. water content.
between K and h is, n?1 [1 + (h)n ]?m )2 : K (h) = Ks (1 ? (h[1)+ (h )n ]m=2
(2.7)
Figure 2.3 shows this curve for the above described parameters. Note that the relative permeability is just, K ()=Ks or K (h)=Ks , so these curves also give the relative permeability function. Conservation of mass for ow of water in a porous medium requires that the net rate of water mass ow into a small control volume be equal to the time rate of change of water mass storage within the volume plus any source terms. Combining this statement with equation (2.6) gives, @ (s) ? r (K (h)r ) = f; h @t where f is a source term. This is Richards' equation [55]. The time derivative term can be written as, @ (s) = s @ + s @ + @s : (2.8) @t @t @t @t
14 Relative Permeability vs. Pressure Head
0
Relative Permeability K/K_s (dimensionless)
10
−1
10
−2
10
−3
10
−4
10
−5
10
−6
10
0
10
1
10
2
10 Pressure Head (−cm)
3
10
4
10
Figure 2.3 Typical van Genuchten curve of hydraulic conductivity vs. pressure head.
Assuming an incompressible medium, is constant in time and the rst term is zero. For unsaturated ow, the second term is small relative to the third and the time derivative is just, @ (s) @s = @(h) : @t @t @t For saturated ow, s is a constant equal to ss , and the second term in (2.8) is the only applicable term. Thus, @ (s) = s @ = s @p = S @h ; s @t @t s @t @t where is a the water compressibility constant in (2.4), and Ss is the speci c storage of the aquifer. The speci c storage is de ned as the volume of water that a unit of aquifer releases from storage under a unit decline in the hydraulic head. In the present work, the component of Ss related to the compressibility of the medium will be ignored. The derivation of Ss is as follows. By equation (2.5),
dVw = ? Vwdp:
15 The water volume is just ss VT where VT is the total volume. Assuming VT = 1 and using the relation p = gh,
dVw = ? ssgdh: Taking a unit decline in h, dh = ?1, gives,
dVw = ssg = Ss : For the spatial derivatives, we have r (q). This term can be written as,
r (q) = r q + r q: Since we assume that water is slightly compressible as in (2.4), r is very small and the rst term may be neglected. Canceling the density, we can write Richards' equation as, @(h) + S @ h ? r (K (h)r ) = f; in ; (2.9) s h @t @t where the w subscript has been dropped and is the ow domain. For purposes of analysis, and K are considered functions of h = h + z. Boundary conditions can be stated as, h = D ; on ?D ; ?K (h)rh n = gN ; on ?N ;
(2.10) (2.11)
where ?D [ ?N = @ , ?D 6= ;, and n is an outward pointing, unit, normal vector to . This is the mixed form of Richards' equation. The second term in (2.9) is neglected for unsaturated ow, and the rst term does not apply for saturated ow. Note here that due to the constant (or passive) air phase pressure assumption, Richards' equation ignores the air phase except through its eects on the hydraulic conductivity, K . An initial condition, h = 0(x); t = 0;
(2.12)
completes the speci cation of the problem. Owing to the fact that is a function of h, we can write equation (2.9) as, h ? r (K ( )r ) = f; in ; (2.13) (Ss + C (h)) @@t h h
16 where C (h) = @=@ h denotes the speci c moisture capacity. This form is the head-based form of the equation. Lastly, equation (2.9) may be written as, @ ? r (D()r) = f; in ; (2.14) @t where D() = K ()=C () is the soil-moisture diusivity. This is the -based form of the equation. We have now presented a complete mathematical model of partially saturated subsurface ow for a single water phase. In the remaining chapters, we will analyze and solve this model.
17
Chapter 3 Discretization In this chapter we present a discussion of spatial and temporal discretization techniques employed in this work. We begin by introducing notation, then presenting variational formulations of Richards' equation. Formulations corresponding to both the mixed and expanded mixed nite element methods will be presented. Discrete approximating spaces and approximation schemes will be discussed. Lastly, comments are made on the time discretization method used.
3.1 Notation Let be a domain in IRd with boundary ? = @ , and let ?D be the portion of the boundary where Dirichlet conditions are speci ed and ?N the portion where Neumann conditions are speci ed. We assume that ? = ?D [ ?N . Let L2( ) be the set of square R integrable functions on , i.e., L2( ) = fwj w2d < 1g. Let (L2( ))d denote the space of d-dimensional vectors which have all components in L2( ). Furthermore, let (:; :) denote the L2( ) inner product, scalar and vector, i.e. for f; g 2 L2( ), Z (f; g) = f g d :
Let (:; :)@ denote the inner product and k:k@ its associated norm. Let H ( ; div) be the space of vectors in (L2( ))d which have divergence in L2( ), i.e. H ( ; div) = fv 2 (L2( ))d : r v 2 L2( )g. If f 2 H ( ; div), then, kf kH ( ;div) kf k2L2( ) + kr f k2L2( ) 1=2 : Let Wpk ( ) be the standard Sobolev space [12, p. 27], Wpk ( ) = ff : kf kWpk ( ) < 1g; where, 0 11=p X kf kWpk ( ) = @ kD f kpLp( )A :
L2(@ )
jjk
18 Let Hs ( ) for s a positive integer be the Sobolev space, W2s( ). Denote the inner product for the Hs Sobolev space as, XZ (f; g)s = D f Dg d ; jjs
where f; g 2 Hs( ). Let H?s be the dual space of Hs with norm, (3.1) kgk?s = sup < kg; k > ; s f 2Hs; 6=0g where < :; : > is the duality pairing between Hs and H?s . We will make use of the fractional Sobolev space H 1=2(@ ) with norm [17],
kgk1=2;@ = fv2H 1( ); infvj =gg kvkH 1( ): @
Let V = H ( ; div); V~ = (L2( ))d ; W = L2( ) and = H 1=2(@ ). Let VN and V0 denote subspaces of V with functions whose normal traces on ?N are equal to gN from equation (2.11) and 0, respectively.
3.2 Variational Formulations To avoid confusion, in this and all following chapters, p will denote hydraulic head. Introducing a velocity variable uM = ?K (p)rp and writing equation (2.9) as a system of rst order equations gives, @(p) + S @p + r u = f; (3.2) s M @t @t uM = ?K (p)rp; (3.3) D p = pD ; ? ; (3.4) uM n = gN ; ?N ; (3.5) where n is an outward pointing, unit, normal vector. Multiplying (3.2) by w 2 W then integrating and multiplying (3.3) by K (p)?1 and v 2 V0, then integrating by parts, the problem is formulated as nding (pM ; uM ) 2 (W; VN ) such that, (3.6) ( @(@tpM ) ; w) + (Ss @p@tM ; w) + (r uM ; w) = (f; w); (K (pM )?1uM ; v) ? (pM ; r v) = ?(pD ; v n)?D : (3.7) Equations (3.6)-(3.7) de ne the variational formulation of the mixed form of Richards' equation corresponding to the mixed nite element method.
19 For the expanded mixed nite element method we consider a dierent formulation of the problem, @(p) + S @p + r u = f; (3.8) s @t @t u~ = ?rp; (3.9) u = K (p)u~; (3.10) p = p D ; ?D ; (3.11) u n = g N ; ?N ; (3.12) where two additional unknowns, u~ and u have been introduced. Multiplying (3.8) by w 2 W , multiplying (3.9) by v 2 V0 and multiplying (3.10) by v 2 V~ , then integrating each of the resulting equations, the problem can be formulated as nding (p; u~; u) 2 (W; V~ ; VN ) such that, (3.13) ( @@t(p) ; w) + (Ss @p @t ; w) + (r u; w) = (f; w); (u~; v) ? (p; r v) + (pD ; v n)?D = 0; (3.14) (u; v) = (K (p)u~; v): (3.15) Thus, for the expanded mixed method, a set of three equations in three unknowns is solved.
3.3 Approximating Spaces For mixed nite element methods, the scalar variable p and its velocity are simultaneously approximated. Thus, two approximating spaces are necessary, one for hydraulic head unknowns and one for velocity unknowns. Ideally, these spaces are chosen so that the resulting method has a unique solution. The approximating spaces used in this work are the Raviart-Thomas-Nedelec spaces on rectangles and parallelepipeds, which are now brie y described. Use of these spaces for linear elliptic problems guarantees a unique solution to the mixed method system [13, 45, 54]. Let Th denote a quasi-uniform triangulation of into rectangles with diameter O(h) in two dimensions or parallelepipeds also with diameter O(h) in three dimensions.
20 The Raviart-Thomas-Nedelec (RTN) [54, 50] approximating space of order k on a rectangular element E 2 T is, Vk (E ) = Pk+1;k (E ) Pk;k+1 (E ); d = 2; Vk (E ) = Pk+1;k;k (E ) Pk;k+1;k Pk;k;k+1 (E ); d = 3; where Pr;s;t is the space of polynomials of degree r in the x direction, s in the y direction and t in the z direction. Raviart and Thomas developed these spaces for two dimensions, and Nedelec for three-dimensions. The space L2( ) is approximated by, Wk (E ) = Pk (E ): For the nite dierence scheme presented in Chapter 5, we consider the lowest order RTN space, i.e. k = 0, on parallelepipeds, Vh (E ) = f(1x1 + 1; 2x2 + 2; 3x3 + 3)T : i; i 2 IRg; Wh (E ) = f : 2 IRg; where the last component in Vh should be deleted in two dimensions. We also de ne a hybrid space, Bh L2(@ ), of Lagrange multipliers for the pressure restricted to @ and corresponding to the above RTN spaces [9, 17]. So, on an edge or face e, Bh (e) = f : 2 IRg: The standard nodal basis is used, where for Vh and h the nodes are at the midpoints of edges or faces of the elements, and for Wh the nodes are at the centers of the elements. Denote the grid points by (xi+1=2; yj+1=2); i = 0; : : :; Nx; j = 0; : : : ; Ny ; and de ne xi = 21 (xi+1=2 + xi?1=2); i = 1; : : : ; Nx; yj = 21 (yj+1=2 + yj?1=2); j = 1; : : : ; Ny ; hxi+1=2 = xi+1 ? xi; i = 1; : : :; Nx ? 1; hyj+1=2 = yj+1 ? yj ; j = 1; : : : ; Ny ? 1; hxi = xi+1=2 ? xi?1=2; i = 1; : : : ; Nx; hyj = yj+1=2 ? yj?1=2; j = 1; : : : ; Ny ; h = max (hx; hy ); i;j i j
21 with corresponding notation for a third dimension. De ne discrete inner products corresponding to applications of the midpoint (M), trapezoidal (T) and midpoint by trapezoidal (TM) quadrature rules by Ny Nx X X hxihyj rij sij ; (r; s)M = i=1 j =1 Ny Nx X X
Ny Nx X X y x x x hxihyj+1=2vijy +1=2qijy +1=2; hi+1=2hj vi+1=2j qi+1=2j + (v; q)TM = i=1 j =0 i=0 j =1 N y N x XX x y1 x hi+1=2hj 2 (vi+1=2j?1=2qix+1=2j?1=2 + vix+1=2j+1=2qix+1=2j+1=2) (v; q)T = i=0 j =1 Ny Nx X X hxihyj+1=2 12 (viy?1=2j+1=2qiy?1=2j+1=2 + viy+1=2j+1=2qiy+1=2j+1=2); + i=1 j =0
where a third sum in each is added for the case of three dimensions. We denote the associated norms by k:kR; where R = M, T or TM and by ER(q; r), the error in approximating an integral by the given rule, i.e. ET (q; r) = (q; r) ? (q; r)T . The error in approximating an integral by either the trapezoidal or the trapezoidal by midpoint rule is [20], X X @ (3.16) jEQ(q; v)j C k @ x (q v)kL1(E)h2: E 2Th jj=2 For any 2 L2( ) let ^ denote the L2 projection of onto Wk , i.e. (; w) = (^; w); 8w 2 Wk :
(3.17)
In a similar manner, de ne an L2(?) projection onto k . These two L2 projection operators have the following approximation properties for 2 H k+1( ) and 2 H k+1 (?), k^ ? k C kkrhr ; 0 r k + 1; (3.18) k ^ ? k? C k kr;?hr ; 0 r k + 1; (3.19) k ^ ? k?;M Ch2: (3.20) Associated with the RTN mixed nite element spaces is the projection operator : (H 1( ))d ! Vh , such that for q 2 H k+1 ( ), (r q; w) = (r q; w); 8w 2 Wk ; (q n; )e = (q n; )e ; 8 2 k ;
(3.21) (3.22)
22 where n is an outward pointing, unit, normal vector. The following approximation properties hold for the projection,
kq ? qk C kqkrhr ; 0 r k + 1; (3.23) kr (q ? q)k C kr qkrhr ; 0 r k + 1; (3.24) where e is any element edge or face. Note that q n = q^ n on any boundary edge e.
We will use the following estimate [25] which is true on rectangular or parallelepiped elements. For u and u~ de ned by equations (3.13)-(3.15),
ku ? ukTM + ku~ ? u~kTM Ch2:
(3.25)
The following inverse estimate for discrete polynomial spaces [12, p. 111] will be used extensively in the following analysis, Theorem 3.1 Let Th be a quasi-uniform triangulation of . Let E^ be a reference element. Let P be a space of approximating polynomials and h = fy : yjE 2 PE 8E 2 Th g. Then if P (E^ ) Wpl(E^ ) \ Wqm(E^ ) where 1 p 1; 1 q 1 and 0 m l, then there exists a constant Q such that, 0 11=q 0 11=p X X d d @ kykpWpl (E)A Chm?1+min(0; p ? q ) @ kykqWqm(E)A ; E 2Th
E 2Th
1=s for all y 2 Vh . For s = 1, interpret the expression, PE2Th kyksWsm(E) as maxE2Th kykW1l (E).
Let Wh and Vh be discrete subspaces of W and V, respectively. De ne Vh0 = V 0 \ Vh and VhN = V N \ Vh . Then, the continuous time mixed nite element method is to nd (PM ; UM ) 2 (Wh; VhN ) satisfying, ( @(@tPM ) ; w) + (Ss @P@tM ; w) + (r UM ; w) = (f; w); w 2 Wh ; (3.26) (K (PM )?1UM ; v) ? (PM ; r v) + (pD ; v n)?D = 0; v 2 Vh0 : (3.27) Choosing w in equation (3.26) to be the basis function associated with cell i; j; k, i.e., ( 1; in cell i; j; k, w= (3.28) 0; otherwise,
23 then equation (3.26) implies that mass is conserved within cell i; j; k. For this reason, the mixed method is known to conserve mass on a cell-by-cell basis. For the expanded mixed method, we approximate the scalar variable, its velocity ~ h be disand its ux. Thus, three discrete spaces are needed. Let Wh; Vh and V crete subspaces of W; V and V~ , respectively. Then the expanded mixed method is ~ ; U) 2 (Wh; V~ h ; VhN ) which satisfy, formulated as nding (P; U (P ) ; w) + (S @P ; w) + (r U; w) = (f; w); w 2 W ; (3.29) ( @@t s h @t ~ ; v) ? (P; r v) + (pD ; v n)?D = 0; v 2 Vh0 ; (U (3.30) ~ ); v); v 2 V~ h : (3.31) (U; v) = (K (P )U For this method, we have the freedom to choose V~ not equal to V. However, for this work V~ = V. Note that with w chosen as in (3.28), equation (3.29) again implies conservation of mass over each cell.
3.4 Time Discretization We consider nding the solution to equation (2.9) over a time interval J = (0; T ), where T > 0 is some nal time. Let 0 = t0 < t1 < < tN = T be a given sequence of time steps, tn = tn ? tn?1 and t = maxn tn. Further, assume that there exist constants c and C such that,
ctn tn+1 C tn;
(3.32)
for all n. For = (t; :), let n = (tn; :) and denote the discrete and continuous partial derivatives by, n n?1 dtn = ? tn ; @t = @ @t : In subsequent chapters, the following norms will be used,
kkL1(J ;L2( )) ess sup kk(t); t kkl1(J ;L2( )) 1max kn k; nN
24
! 21 kkL2(J ;L2( )) 0 k(:; t)k2dt ; ! 21 N X n n 2 t k k : kkl2(J ;L2( )) ZT
n=1
This work will use an implicit backward Euler time discretization in order to formulate a discrete time expanded mixed method for Richards' equation. This scheme ~ n ; Un) 2 (Wh; Vh ; VhN ) is to nd for each time step n; n = 1; : : : ; N , functions (P n; U satisfying, (dt(P n ); w) + (SsdtP n ; w) + (r Un; w) = (f n ; w); w 2 Wh; (3.33) n n 0 (3.34) (U~ ; v) ? (P ; r v) + (pD ; v n)?D = 0; v 2 Vh ; ~ n; v); v 2 Vh : (3.35) (Un; v) = (K (P n )U An implicit method is used to prevent the need for taking unnecessarily small time steps. The Discrete Gronwall Inequality [30] will be used in coming chapters. We present it now for the sake of completeness,
Lemma 3.1 Let g(t); f (t) and h(t) be nonnegative functions de ned on [0; T ], t = it; i = 0; : : : ; N ? 1 and g(t) nondecreasing. If, f (t) + h(t) g(t) + C t
tX ?t s=0
f (s);
(3.36)
then,
f (t) + h(t) g(t)eCT :
(3.37)
In conclusion, this chapter has set some notation, introduced the mixed and expanded mixed methods and de ned continuous and discrete time numerical schemes for Richards' equation. The next chapter analyzes these schemes.
25
Chapter 4 An a priori Error Analysis of Richards' Equation In this chapter, we present an error analysis of the expanded mixed nite element method applied to Richards' equation. For simplicity we analyze the form, @(p) + S @p ? r K ((p))rp = f; (4.1) s @t @t where we have made use of the fact that krw is a function of water content, , and have taken K ((p)) = fk(x)krw ((p))gg=. Two degenerate conditions can occur. The rst, K = 0, implies a 0 relative permeability, a condition ocurring in very dry media. The second degeneracy occurs when 0 = 0. This condition happens when the media is fully saturated. We rst discuss the case where K > 0 for all p and allow for the possibility that @(p)=@t = 0. This situation corresponds to partially to fully saturated ow. For clarity, a continuous time estimate is presented. Bounds for k(p) ? (P )kL1(L2) and ku~ ? U~ kL2(L2) are shown for the partially saturated case, i.e. Ss = 0, and a bound on kp ? P kL1(L2) for the saturated case. These proofs closely follow the techniques of Arbogast [4] who derived estimates for the case of Galerkin methods. We have extended his work to account for the expanded mixed nite element method. Next the case where K is bounded above 0 and the derivative of is strictly nonzero is considered. This is the case of strictly unsaturated ow. For this situation, optimal convergence of a discrete time scheme is shown. The next set of estimates are for the case where K 0. This situation corresponds to completely dry to fully saturated ow. The Kirchho transformation [6, 37, 58] is used to analyze this case. A bound for the ux only is presented in the case that @=@p can be 0 and a bound for kp ? P kH?1 is presented for @=@p 6= 0. When K = 0, the solution generally does not have enough regularity to prove optimal bounds. However, the results presented here would be optimal if the solution had the necessary smoothness. These estimates follow the techniques of Arbogast, Wheeler and Zhang [6] for degenerate equations. This work extends their work to the case of the expanded mixed method and to Richards' equation.
26 In the following arguments, C will represent a generic constant independent of mesh and time step sizes, and its value should be assumed dierent at each instance. The arithmetic-geometric inequality, (4.2) ab 2 a2 + 21 b2; a; b; 2 IR; > 0; will be used throughout the analysis.
4.1 Partially to Fully Saturated Flow In this section, the expanded mixed nite element method applied to Richards' equation (4.1) is considered. The following assumptions are made: 1. The tensor K is symmetric and positive de nite. 2. The tensor K is Lipschitz in . Thus, there exists a constant LK independent of two numbers, 1 and 2 such that, kK (1) ? K (2)k LK k1 ? 2k. 3. The function is Lipschitz in p. Thus, there exists a constant L independent of two numbers, p1 and p2 such that, k(p1) ? (p2)k L kp1 ? p2 k. 4. The function (p) is monotone nondecreasing in p. 5. The speci c storage Ss can be 0, i.e. Ss 0. Recall that Ss = 0 in the case of unsaturated ow and Ss > 0 for saturated ow. 6. The derivative @ K is bounded above, j@ K j C . 7. The derivative @t is bounded above, j@tj C . 8. The composition (@p) ?1 is Holder continuous of order ; 0 < 1. Thus, for f; g 2 IR,
j@p(f ) ? @p(g)j C j(f ) ? (g)j :
(4.3)
The following analysis bounds the error in the continuous time expanded mixed method applied to Richards' equation given in equations (3.29)-(3.31). We start with a lemma, proven by Arbogast in [4].
27
Lemma 4.1 Assume (x; p) is monotone and nondecreasing in p 2 IR for each xed x 2 , uniformly Lipschitz in both x and p, and uniformly bounded from above and below. Then, for v and w in IR, Zv ? 1 2 (2 sup j@pj) ((v) ? (w)) ((}) ? (w))d} p w
((v) ? (w))(v ? w):
(4.4)
The following theorem holds for the convergence of the continuous time scheme. Theorem 4.1 Let (P; U~ ; U) 2 (Wh ; Vh; VhN ) satisfy equations (3.29)(3.31). Then, under the assumptions given in 1-8 above and for = 2(k + 1) =(1 + ) with k + 1 > d(1 + )=(2(3 ? 1)) > 0,
k(p) ? (P )kL1(J ;L2( )) + Ss kp^ ? P k2L1(J ;L2 ( )) ~ kL2(J ;L2( )) Ch ; +ku~^ ? U (4.5) Sskp ? P k2L1 (J ;L2( )) C (h + hk+1); (4.6) ku~ ? U~ kL2(J ;L2( )) C (h + hk ); (4.7) where k is the order of the approximating space. Proof The proof will be in two parts. First, a bound for ((p) ? (P ); p ? P ) is found. Then a bound for k(p) ? (P )k in terms of ((p) ? (P ); p ? P ) will be derived. These two pieces are put together in a continuation argument which gives the nal result. Applying the de nitions of the L2 and projections and subtracting (3.29)-(3.31) from (3.13)-(3.15), gives the error equations, (@t((p) ? (P )); w) + (Ss@t(p ? P ); w) = ?(r (u ? U); w); w 2 Wh ; (4.8) ~ ; v) = (^p ? P; r v); v 2 Vh0 ; (u~^ ? U (4.9) ~ ; v) = 0; v 2 Vh : (u ? U; v) ? (K ((p))u~ ? K ((P ))U (4.10) Rewriting (4.10) results in,
~ ); v) (u ? U; v) = (u ? u; v) + (K ((p))(u~ ? u~^ ); v) + (K ((p))(u~^ ? U +((K ((p)) ? K ((P )))u~^ ; v) ?((K ((p)) ? K ((P )))(u~^ ? U~ ); v): (4.11)
28
~ and = u ? U. Let Q1 be a constant whose value Let = p^ ? P; = u~^ ? U R will be determined later. Then, in (4.8) let w = tt e?Q1s (:; s)ds, ! ! Zt Zt ? Q s ? Q s 1 1 @t((p) ? (P )); t e (:; s)ds + Ss@t(p ? P ); t e (:; s)ds ! Zt ? Q s 1 (4.12) + r ; e (:; s)ds = 0: t
Multiplying (4.9) by e?Q1s, integrating from t to t < T holding v xed, then letting v = gives, ! Zt ! Zt ? Q s ? Q s 1 1 e (:; s)ds; = e (:; s)ds; r : (4.13) t
t
R In (4.11), let v = tt e?Q1s(:; s)ds to get, Zt (; e?Q1s(:; s)ds) t ! ! Zt Zt ? Q s ? Q s 1 1 = u ? u; e (:; s)ds + K ((p))(u~ ? u~^ ); e (:; s)ds t t ! ! Zt Zt ? Q s ? Q s + K ((p)) ; e 1 (:; s)ds + (K ((p)) ? K ((P )))u~^ ; e 1 (:; s)ds t t ! Zt ? (K ((p)) ? K ((P ))); e?Q1s (:; s)ds : (4.14) t
Combining (4.12)-(4.14) results in, ! ! Zt Zt ? Q s ? Q s @t((p) ? (P )); e 1 (:; s)ds + Ss@t(p ? P ); e 1 (:; s)ds t t ! ! Zt Zt ? Q s ? Q s + u ? u; e 1 (:; s)ds + K ((p))(u~ ? u~^ ); e 1 (:; s)ds t t ! ! Zt Zt ? Q s ? Q s 1 1 ^ + K ((p)); e (:; s)ds + (K ((p)) ? K ((P )))u~; e (:; s)ds t t ! Zt ? (K ((p)) ? K ((P ))); t e?Q1s (:; s)ds = 0: (4.15) Since K is Lipschitz in ,
j((K ((p)) ? K ((P )))u~^ ; v)j C k(p) ? (P )kku~^ kL1 kvk:
(4.16)
29 Also, by the product rule, !2 Zt Zt 1 ? Q s ? Q s K ((p)) e 1 (:; s)ds = ? 2 @t[K ((p)) e 1 (:; s)ds eQ1t] t t 1 + 2 [Q1K ((p)) + @ K ((p))@t(p)] !2 Zt ? Q s 1 e (:; s)ds eQ1t (4.17) t
Thus, equation (4.15) becomes, ! ! Zt Zt ? Q s ? Q s 1 1 @t((p) ? (P )); e (:; s)ds + Ss @t(p ? P ); e (:; s)ds t t ! 2 Z Zt 1 ? Q s 1 ? 2 @t K ((p)) t e (:; s)ds eQ1tdx !2 Z Zt 1 ? Q s e 1 (:; s)ds eQ1tdx + 2 Q1 K ((p)) t
! ! Zt Zt ? u ? u; t e?Q1s (:; s)ds ? K ((p))(u~ ? u~^ ); t e?Q1s(:; s)ds !2 Z Zt 1 ? Q s ? 2 @ K ((p))@t(p) t e 1 (:; s)ds eQ1tdx ! Zt ? Q s + (K ((p)) ? K ((P ))); e 1 (:; s)ds t Zt + C k(p) ? (P )kk e?Q1s (:; s)dsk: (4.18) t Now consider bounding the rst four right-hand side terms. Rewriting the rst of these gives, Zt Zt ? Q s 1 j(u ? u; t e (:; s)ds)j ku ? ukk t e?Q1s(:; s)dsk ! Zt ? Q s Q t= 2 ? Q t= 2 1 1 1 (ku ? uke ) k e (:; s)dske t
Since K is bounded,
ku ? uk2e?Q1t Zt +C k e?Q1s(:; s)dsk2eQ1t t Zt
j(K ((p))(u~ ? u~^ ); t e?Q1s (:; s)ds)j
(4.19)
30
C ku~ ? u~^ kk
Zt
e?Q1 s(:; s)dsk Zt ku~ ? u~^ k2e?Q1t + C k t e?Q1s(:; s)dsk2eQ1t: t
Since @ K and @t are assumed to be bounded above, Z Zt j 21 @ K ((p))@t(p)( t e?Q1s (:; s)ds)2dxeQ1t)j Zt C k t e?Q1s(:; s)dsk2eQ1t:
Using Theorem 3.1,
(4.20)
(4.21)
Zt
j((K ((p)) ? K ((P ))); t e?Q1s (:; s)ds)j Zt
C kkL1 k(p) ? (P )kk t e?Q1s(:; s)dsk Zt
Ch?d=2kkk(p) ? (P )kk t e?Q1 s(:; s)dsk
Zt ? d 2 2 ? Q t 1 h kk k(p) ? (P )k e + C k t e?Q1s (:; s)dsk2eQ1t:
(4.22)
Combining the above bounds with equation (4.18) results in, ! ! Zt Zt ? Q s ? Q s 1 1 @t((p) ? (P )); e (:; s)ds + Ss@t(p ? P ); e (:; s)ds t t ! 2 Z Zt 1 ? Q s 1 ? 2 @t K ((p)) t e (:; s)ds eQ1tdx !2 Z Zt 1 ? Q s + 2 Q1 K ((p)) e 1 (:; s)ds eQ1tdx
t Zt C k t e?Q1s (:; s)dsk2eQ1t + f1 + h?d kk2gk(p) ? (P )k2e?Q1t + fku ? uk2 + ku~ ? u~^ k2ge?Q1t: (4.23) This equation is integrated in time over (0; t). The rst two left-hand side terms are handled by integration by parts, ! Zt Zt ? Q s 1 @t((p) ? (P )); e (:; s)ds dt 0 t ! Zt Zt ? Q t 0 0 = ? (p ) ? (P ); (:; t)e 1 dt + ((p) ? (P ); p ? P )e?Q1 tdt 0
0
31
Zt
? 0 ((p) ? (P ); p ? p^)e?Q1tdt Zt
Zt
0 ((p) ? (P ); p ? P )e?Q1tdt ? C f 0 kp ? p^k2e?Q1 tdt + k(p0 ) ? (P 0)k2g Zt
?f
Zt 2 ? Q t 1 k(p) ? (P )k e dt + k 0 (:; t)e?Q1tdtk2g: 0
The second term is, Zt Zt Ss (@t(^p ? P ); e?Q1s (:; s)ds)dt 0 t Zt Zt Ss k k2e?Q1tdt ? CSskp^0 ? P 0k2 ? Ssk (:; s)e?Q1s dsk2: 0
0
(4.24)
(4.25)
So, integrating (4.23) over (0; t) gives, Zt Zt ? Q t 1 ((p) ? (P ); p ? P )e dt + Ss kp^ ? P k2e?Q1tdt 0 0 ! 2 Z Z t + 21 K ((p0 )) (:; s)e?Q1 s ds dx 0
!2 Zt Z tZ 1 ? Q s 1 K ((p)) e (:; s)ds eQ1tdxdt + 2 Q1 0
t Zt Zt Zt C 0 k t e?Q1s(:; s)dsk2eQ1tdt + C 0 kp ? p^k2e?Q1 tdt + C k(p0) ? (P 0)k2 Zt Zt 2 ? Q t 0 0 2 1 + CSskp^ ? P k + f k(p) ? (P )k e dt + k (:; t)e?Q1tdtk2g 0 0 Zt + f1 + h?d kk2gk(p) ? (P )k2e?Q1tdt 0 Zt (4.26) + fku ? uk2 + ku~ ? u~^ k2ge?Q1tdt: 0
Since is Lipschitz in p, Zt Zt k(p) ? (P )k2e?Q1tdt kp ? P k2e?Q1tdt 0 0 Zt 0 kp ? p^k2e?Q1tdt Zt + 0 k (:; t)k2e?Q1tdt:
(4.27)
Choosing Q1 to exactly cancel the last left-hand side term in (4.26) with the rst right-hand side term, again using the fact that is Lipschitz and choosing small
32 enough gives, Zt 0
((p) ? (P ); p ? P )e?Q1 tdt + S
s
Zt
0 !2
kp^ ? P k2e?Q1tdt
Z Zt 1 0 + 2 K ((p )) (:; s)e?Q1 s ds dx 0
Zt C 0 kp ? p^k2e?Q1tdt + C k(p0) ? (P 0)k2 + CSskp^0 ? P 0k2 Zt + k 0 (:; t)e?Q1tdtk2 Zt ? d + h 0 kk2k(p) ? (P )k2e?Q1 tdt Zt + fku ? uk2 + ku~ ? u~^ k2ge?Q1tdt 0 Zt (4.28) + 0 k (:; t)k2e?Q1tdt: Before continuing, we present the following lemma. Lemma 4.2 For P and U~ de ned as in equations (3.29)-(3.31) and for ~ , we have,
= p^ ? P and = u~^ ? U k k C k k; (4.29)
Z t Z
t
e?Q1s (:; s)ds
C
e?Q1s(:; s)ds
: (4.30)
t
t
Proof Let satisfy the equation ? = , and let f satisfy, f = ?r; in ; f n = 0; on ?N ; = 0; on ?D ; where we have assumed that ?D 6= ;. Then, recalling (4.9), k k2 = ( ; r f ) = ( ; r f ) = (; f ) kk(kf ? f k + kf k) kk(Chkf k1 + kf k) C kkkk2 C kkk k;
(4.31) (4.32) (4.33)
33 where the last inequality holds by elliptic regularity. In a similar manner (4.30) is shown. Applying approximation results, (3.18) and (3.23), and Lemma 4.2 to equation R (4.28) and noting that 0t e?Q1tdt = Q11 (1 ? eQ11 t ) = C , gives, Zt Zt ? Q t 1 ((p) ? (P ); p ? P )e dt + Ss 0 kp^ ? P k2e?Q1tdt 0 !2 Z Zt 1 0 + (:; s)e?Q1 s ds dx 2 K ((p )) 0 Ch2(k+1) + C k(p0) ? (P 0)k2 + CSskp^0 ? P 0k2 Zt + h?d kk2k(p) ? (P )k2e?Q1tdt: (4.34) 0 This completes the rst part of the proof. We come back to this estimate later. Let w = in (4.8), v = in (4.9) and v = in (4.10) and combine the three resulting equations to get, (@t((p) ? (P )); p^ ? P ) + (Ss @t(^p ? P ); p^ ? P ) + (u ? u; ) ~ ; ) = 0: + (K ((p))u~ ? K ((P ))U (4.35) Now, K ((p))u~ ? K ((P ))U~ = K ((p))(u~ ? u~^ ) + K ((P ))(u~^ ? U~ ) +(K ((p)) ? K ((P )))u~^ : (4.36) Combining equations (4.36) and (4.35), applying approximation properties (3.18) and (3.23), using the assumption that K is Lipschitz and using the arithmeticgeometric inequality, (@t((p) ? (P )); p ? P ) + (Ss@t(^p ? P ); p^ ? P ) + 1 (K ((P )); ) 2 ^ = ?(u ? u; ) ? (K ((p))(u~ ? u~); ) ? ((K ((p)) ? K ((P )))u~^ ; ) +(@t((p) ? (P )); p ? p^) Ch2(k+1) + C k(p) ? (P )k2 + (@t((p) ? (P )); p ? p^): (4.37) Note that, Zp @t ((}) ? (P ))d} = ((p) ? (P ))@tp ? @t(P )(p ? P ) P = ((p) ? (P ))@tp + @t[(p) ? (P )](p ? P ) ?@t(p)(p ? P ): (4.38)
34 So,
@t[(p) ? (P )](p ? P )e?Q2t Zp Zp ? Q t 2 = @t[ ((}) ? (P ))d}e ] + Q2 ((}) ? (P ))d}e?Q2t P
P
+ f@t(p)(p ? P ) ? ((p) ? (P ))@tpge?Q2t:
(4.39)
Consider the last term in this equation and apply the chain rule and Mean Value Theorem, where for some w 2 (P; p),
j@t(p)(p ? P ) ? ((p) ? (P ))@tpj = j(@p(p) ? @p(w))(p ? P )@tpj C j(p) ? (w)j jp ? P j C j(p) ? (P )j jp ? P j C j((p) ? (P ))(p ? P )j2 =(1+ ) + fjp^ ? P j2 + jp ? p^j2g;
(4.40)
where we have used the inequality, [4]
jabj p=q1 p jajp + q jbjq;
(4.41)
for any 1 < p < 1; p1 + q1 = 1, which implies, jaj jbj C jabj2 =(1+ ) + b2. Multiplying (4.37) by e?Q2t, combining with the above bounds and integrating from 0 to t gives, Z tZ Z p Z tZ Z p ? Q 2 @ [ ((}) ? (P )d}]e dxd + Q2 ((}) ? (P ))d}e?Q2 dxd P 0
0 P Zt Zt 1 ? Q 2 + (Ss@ (^p ? P ); p^ ? P )e d + (K ((P )); )e?Q2 d 2 0 0 Zt Zt 2 ? Q 2( k +1) 2 Ch + C k(p) ? (P )k e d + (@t((p) ? (P ); p ? p^)e?Q2 d 0 0 Z Zt + 0 j((p) ? (P ))(p ? P )j2 =(1+ )e?Q2 ddx Z Zt + (jp^ ? P j2 + jp ? p^j2)e?Q2 ddx: (4.42)
0 R Lemma (4.1) implies that, C ((v) ? (w))2 wv (() ? (w))d. So, the above becomes, Zt Zt C (@ k(p) ? (P )k2e?Q2 d + Q2 k(p) ? (P )k2e?Q2 d 0
0
35
Zt Zt + (Ss@ (^p ? P ); p^ ? P )e?Q2 d + 21 (K ((P )); )e?Q2 d 0 0 Zt Zt Ch2(k+1) + C 0 k(p) ? (P )k2e?Q2 d + 0 (@ ((p) ? (P ); p ? p^)e?Q2 d Zt Zt 2 = (1+ ) ? Q 2 e d + 0 kp^ ? P k2e?Q2 d + 0 j((p) ? (P ); p ? P )j Zt (4.43) + kp ? p^k2e?Q2 d 0
Zt
By integration by parts,
(@ ((p) ? (P ); p ? p^)e?Q2 d Zt Zt ? Q 2 = ? ((p) ? (P ); @ (p ? p^))e d + Q2 ((p) ? (P ); p ? p^)e?Q2 d 0 0 ? Q t 0 0 2 +((p) ? (P ); p ? p^)e ? ((p ) ? (P ); p0 ? p^0) Zt Zt 0 k(p) ? (P )k2e?Q2 d + C 0 k@ (p ? p^)k2e?Q2 d Zt Zt +Q2 k(p) ? (P )k2e?Q2 d + Q2 kp ? p^k2e?Q2 d + ^k(p) ? (P )k2e?Q2t 0 0 +C kp ? p^k2e?Q2t + j((p0) ? (P 0); p0 ? p^0 )j: (4.44) 0
Also by integration by parts, Zt Zt 2 ? Q 2 C (@ k(p) ? (P )k )e d = Q2C k(p) ? (P )k2e?Q2 d 0 0 +C k(p) ? (P )k2e?Q2t ?C k(p0) ? (P 0)k2; and Zt
(4.45)
Z t Ss @ kp^ ? P k2e?Q2 d 0 2 0 Zt = Ss2Q2 kp^ ? P k2e?Q2 d + S2s kp^ ? P k2e?Q2 0 ? S2s kp^0 ? P 0k2: (4.46) Combining (4.44) and (4.45) with (4.43) gives, Zt Zt Q2C 0 k(p) ? (P )k2e?Q2 d + C k(p) ? (P )k2e?Q2t + Ss2Q2 0 kp^ ? P k2e?Q2 d Zt + S2s kp^ ? P k2e?Q2 + 41 (K ((P )); )e?Q2 d 0 (Ss@ (^p ? P ); p^ ? P )e?Q2 d =
36
Zt
Zt
Ch2(k+1) + (Q2 + C ) 0 k(p) ? (P )k2e?Q2 d + C 0 k@ (p ? p^)k2e?Q2 d +
Zt 0
j((p) ? (P ); p ? P )j2 =(1+ )e?Q2 d + ^k(p) ? (P )k2e?Q2 t
+ j((p0) ? (P 0); p0 ? p^0 )j + C k(p0) ? (P 0)k2 + S2s kp^0 ? P 0k2:
(4.47)
So, choosing Q2 and to make the rst left-hand side term cancel the second right-hand side term, noting that the third left-hand side term is nonnegative and taking ^ small enough, Zt 1 S s 2 ? Q 2 ? Q t 2 2 C k(p) ? (P )k e + 2 kp^ ? P k e + 4 (K ((P )); )e?Q2 d 0 Zt Ch2(k+1) + 0 j((p) ? (P ); p ? P )j2 =(1+ )e?Q2 d + j((p0) ? (P 0); p0 ? p^0)j +C k(p0) ? (P 0)k2 + S2s kp^0 ? P 0k2: (4.48) Take P 0 = p^0. Then,
k(p0) ? (P 0)k2 C (kp0 ? p^0k2 + kp^0 ? P 0k2) Ch2(k+1): Thus,
Zt 1 (K ((P )); )e?Q2 d + 2 4 0 2 =(1+ ) Z t g; (4.49) C fh2(k+1) + j((p) ? (P ); p ? P )je?Q2(1+ )=2 d
C k(p) ? (P )k2e?Q2t + Ss kp^ ? P k2e?Q2 0
where we have used the Holder inequality with p = (1+ )=2 and q = (1 ? )=(1+ ). This completes the second part of the proof. We now combine the rst two parts to derive the desired estimate. For some xed value C0 independent of h, let T 0 T be the largest value of time for which,
ku~^ ? U~ kL2(J 0;L2( )) C0hd =(3 ?1);
(4.50)
~ = 0 and u~ is where J 0 = (0; T 0) and d is the spatial dimension. Since initially u~^ ? U assumed continuous in time, we must have T 0 > 0. We make use of the following inequality [4]. If 21 < 1, then jabcj jbj + (jaj=(2?1)jbj)(2?1)=jcj:
(4.51)
37 Note that since 0 < 1, k + 1 2(k1++1) . Let t = t be the time when ess supt2(0;J 0) k(p) ? (P )k is attained. Then, combining equations (4.34) and (4.49), k(p) ? (P )k2L1 (J 0;L2( )) + Sskp^ ? P k2L1 (J 0;L2 ( )) + ku~^ ? U~ k2L2(J ;L2( )) !2 =(1+ ) Zt ( k +1)(4 = (1+ )) ? d 2 2 ? Q t 1 ^ ~ Ch + h ku~ ? Uk k(p) ? (P )k e dt
Ch(k+1)(4 =(1+ ))
0
~ k2L2 (J ;L2( ))k(p) ? (P )k2L1 (J 0;L2( )))2 =(1+ ) +(h?dku~^ ? U Ch(k+1)(4 =(1+ )) + 2 =(1+ )fku~^ ? U~ k2L2 (J ;L2) ~ k2L2(J ;L2))(3 ?1)=2 k(p) ? (P )k2L1 (J 0;L2)g; +((h?d2 =(3 ?1)ku~^ ? U
(4.52) where J = (0; t), inequality (4.51) is used to derive the last inequality, and we have assumed 3 > 1 so that = 2 =(1 + ) > 12 . In (4.50), we took the exponent on h to be exactly large enough to cancel the ?d exponent in (4.52). Now, since t t0, we can use (4.50) and take small enough to hide the last two right-hand side terms. Thus, k(p) ? (P )k2L1 (J 0;L2) + Sskp^ ? P k2L1(J 0 ;L2( )) + ku~^ ? U~ k2L2 (J ;L2) Ch(k+1)(4 =(1+ )): (4.53) Without a bound on ku~ ? U~ k or k(p) ? (P )k, we could not have hidden the last right-hand side term. We thus assumed the minimum for one bound and derived the other. We now can improve the rst. Let t = T 0 and again combine equations (4.34) and (4.49), k(p) ? (P )k2L1(J 0 ;L2( )) + Sskp^ ? P k2L1 (J 0;L2( )) + ku~^ ? U~ k2L2(J 0 ;L2( )) Ch(k+1)(4 =(1+ )) + 2 =(1+ )fku~^ ? U~ k2L2(J 0 ;L2) ~ k2L2(J 0;L2 ))3 ?1=2 k(p) ? (P )k2L1(J 0 ;L2)g: (4.54) +((h?2 d=(3 ?1)ku~^ ? U Using the bounds (4.53) and (4.50) and taking small enough gives, ku~^ ? U~ k2L2(J 0;L2) Ch(k+1)(4 =(1+ )): (4.55) We continue by contradiction. Suppose that T 0 < T , that h0 > 0 is xed and that k + 1 > d(1 + )=(2(3 ? 1)) > 0: Then, (4.56) ku~^ ? U~ kL2 (J 0;L2) Ch2(k+1) =(1+ ) 21 C0hd =(3 ?1));
38 for all values of h < h0. Since T 0 < T and T 0 is the maximal value such that (4.50) is true, we have a contradiction. Thus, T 0 = T and,
ku~^ ? U~ k2L2 (J ;L2) Ch(k+1)(4 =(1+ )):
(4.57)
This, together with (4.53) gives the desired result. We have the following corollary,
Corollary 4.1 For the scheme given by (3.29)-(3.31) and = k +1 with k + 1 > d(1 + )=(2(3 ? 1)) > 0, we have, Z T 0
1 ((p) ? (P ); p ? P )dt 2 Ch:
(4.58)
Proof Let t = T in (4.34) to get, ZT
((p) ? (P ); p ? P )e?Q1tdt ZT 2( k +1) ? d ku~^ ? U~ k2k(p) ? (P )k2e?Q1tdt Ch + h 0 Ch2(k+1) + h?dk(p) ? (P )k2L1(J ;L2( ))ku~^ ? U~ k2L2(J ;L2 ( )) Ch2(k+1) + h2(k+1)4 =(1+ )?d Ch2(k+1) + h2(k+1)4 =(1+ )?(k+1)2(3 ?1)=(1+ ) Ch2(k+1): 0
(4.59)
1 As Arbogast [4] points out, the nonlinear form R0T ((p) ? (P ); p ? P )dt 2 tells us something about the error of the scheme, since for two constants c and C , cj(p) ? (P )j (((p) ? (P ))(p ? P )) 12 C jp ? P j: Thus, as the nonlinear form gets smaller, the error in the water content also decreases. Furthermore, the bound on the nonlinear form is optimal, since O(hk+1 ) is the order of truncation error for approximation with polynomials of order k. The estimates in this section show bounds for the case where the equation is degenerate. For this case, Ss can be 0, and we have only a bound on k(p) ? (P )k. In the next section we make a simplifying assumption that allows us to bound the error in the hydraulic head directly.
39
4.2 Strictly Partially Saturated Flow In this section, the case of strictly partially saturated ow is analyzed. It is assumed that the ow never reaches the fully saturated realm and thus that the derivative, @=@p is never zero. The following assumptions are made. 1. The tensor K is symmetric, positive de nite and bounded, i.e. K Kij K for all i and j , where K and K are xed positive constants. 2. The tensor K is Lipschitz in . Thus, there exists a constant LK independent of two numbers, 1 and 2 such that, kK (1) ? K (2)k LK k1 ? 2k. 3. The function is Lipschitz in p. Thus, there exists a constant L independent of two numbers, p1 and p2 such that, k(p1) ? (p2)k L kp1 ? p2 k. 4. The function (p) is monotone increasing in p, 0(p) > 0. 5. The speci c storage Ss = 0. 6. The derivative @pK is bounded above, j@pK j C . In this section, K is considered to be a function of p and not speci cally of (p). The estimates given are simpler than in the previous section. As a consequence, we consider a discrete time scheme. Again consider the variational formulation of Richards' equation given in equations (3.13)-(3.15). The discrete time numerical scheme considered here is that given in equations (3.33)-(3.35). Before analyzing this case, we give without proof a lemma proven in [4] which will be used in the following arguments, Lemma 4.3 Suppose that un, un?1 , vn and vn?1 are real numbers. Suppose also that : IR 7! IR is such that 0 0 Q < 1 and j00j Q for some constant Q. Then, Z u n n n n dt((u) ? (v)) (u ? v ) = dt [() ? (v)]d ? E; (4.60) v where
E C 0f(un ? vn)2 + (un?1 ? vn?1)2 + (tn)2g; for some C 0 depending on Q and jdtuj.
(4.61)
40 The main result of this section is as follows.
Theorem 4.2 Let (P n ; U~ n; Un ) 2 (Wh; Vh ; VhN ) satisfy the equations
(3.33)-(3.35) for each time step n; n = 1; : : : ; N . Then, !1=2 N X n n 2 n N N KkU~ ? u~ k t kP ? p k + C (hk+1 + t): n=1
Proof Subtracting equations (3.13)-(3.15) from equations (3.33)-(3.35) gives the error equations,
(dt((P ) ? (p))n ; w) + (r (Un ? un); w) (pn ) ; w); (4.62) = ?(dt(pn ) ? @@t ~ n ? u~^ n ; v) = (P n ? p^n ; r v); (U (4.63) (Un ? un; v) = (un ? un; v) + (K (pn )(u~^n ? u~n); v) ~ n ? u~^ n ); v) + ((K (P n ) ? K (pn ))u~^ n; v):(4.64) + (K (P n )(U
~ n ? u~^ n and n = Un ? un. Let w = n in (4.62), Let, n = P n ? p^n ; n = U v = n in (4.63) and v = n in (4.64), giving, (dt((P n ) ? (pn )); P n ? pn ) + (r n; n)
(pn ) ; n); = (dt((P ) ? (p))n ; p^n ? pn ) ? (dt(pn ) ? @@t (n ; n ) = ( n; r n ); (4.65) n (n ; n ) = +(un ? un; n) + (K (pn )(u~^ ? u~n); n) + (K (P n )n; n) + ((K (P ) ? K (p))u~^ n; n): (4.66)
Combine these equations to get, (dt((P ) ? (p))n ; P n ? pn ) + (K (P n )n; n)
@ (pn ); n) =(dt((P ) ? (p))n ; p^n ? pn ) ? (dt(pn ) ? @t ? (un ? un; n ) ? (K (pn )(u~^ n ? u~n); n) ? ((K (P n) ? K (pn ))u~^ n ; n ):
(4.67)
41 By the Mean Value Theorem, ((K (P n) ? K (pn ))u~^ n ; n ) = (K 0(z)(P n ? pn )u~^ n; n); where z 2 (P n ; pn). Consider the rst term on the left-hand side of equation (4.67) and apply Lemma 4.3 to give, ! Z Pn Z n n n n (dt((P ) ? (p)) ; P ? p ) dt n (() ? (p ))d dx p ?C fkP n ? pn k2 + kP n?1 ? pn?1 k2
+(tn)2g:
(4.68)
Using the integral form of the remainder in Taylor's Theorem and the Schwarz inequality, the time discretization error term is bounded by, j(dt(pn ) ? @t@ (pn ); n)j C k nk2 + C kdt(pn ) ? @t@ (pn )k2 Z tn 2 C k nk2 + C k 1tn tn?1 @ @t(2p) (t ? tn?1)dtk2 2 (4.69) C k nk2 + C tnk @ @t(2p) k2L2((tn?1;tn);L2): Combining the above bounds with equation (4.67), ! Z Pn Z n dt n (() ? (p ))d dx + (K (P n)n; n )
p
(dt((P ) ? (p))n ; p^n ? pn ) + kn k2 + C fkun ? un k2 + ku~n ? u~^ nk2 + kP n ? pnk2 + kP n?1 ? pn?1 k2 2(p) @ n + C t k 2 k2L2 ((tn?1;tn);L2) + (tn)2g: (4.70) @t
Taking small enough, the kn k2 term can be brought to the left-hand side. Multiply by tn and sum on n = 1; : : : ; N . The rst term on the left-hand side collapses giving, ! Z Z PN Z Pn Z N X n n (() ? (P N ))ddx dt n (() ? (p ))d dx = t N
p p
n=1 Z Z P0 ? 0 (() ? (p0))ddx: (4.71)
p
42 Applying summation by parts, the rst term on the right-hand side becomes, N X (dt ((P n) ? (pn )); p^n ? pn )tn n=1
= ((P N ) ? (pN ); p^N ? pN ) ? ((P 0) ? (p0 ); p^1 ? p1) N n+1 X ? ((P n) ? (pn ); dt(^pn+1 ? pn+1 )) t tn tn n=1 kP N ? pN k2 + C kp^N ? pN k2 + C kP 0 ? p0 k2 + C kp^1 ? p1k2 N n n X + C fkP n ? pn k2 + k @ (^p @t? p ) k2gtn; n=1
(4.72)
where the assumption (3.32) has been used. Since 0 is bounded above and below by positive constants, there exists a constant Q such that, Zv (4.73) Q?1(v ? u)2 (() ? (v))d Q(v ? u)2: u
Combining these bounds gives the following estimate, N X kP N ?pN k2 + Kknk2tn n=1 C kP 0 ? p0k2 + C kp^N
? pN k2 + kp^1 ? p1k2 N n n X + C fkP n ? pn k2 + k @ (^p ? p ) k2 + kun ? un k2 + ku~n ? u~^ nk2gtn @t
n=1 + (tn)2:
(4.74)
Take P 0 = p^0 and apply Gronwall's Lemma 3.1 to equation (4.74) to remove the rst term in the sum on the right-hand side. Taking approximation properties of the L2 and projections results in, N X (4.75) kP N ? pN k2 + K ku~^ ? U~ n k2tn C (h2(k+1) + (t)2); n=1
where k is the order of the approximating space. Thus, in the case of strictly partially saturated ow, convergence for both hydraulic head and velocity is optimal.
43
4.3 Unsaturated to Fully Saturated Flow In this section, the case of ow through unsaturated to fully saturated soil is considered. In this situation, K can be zero, and Richards' equation is degenerate. The general technique of Arbogast, Wheeler and Zhang [6] is followed for this analysis. We write Richards' equation in the following way, @(p) ? r (K (x)k((p))rp) = f; (4.76) @t where in the case of fully saturated ow, we neglect the speci c storage term. We allow for the case that k((p)) = 0, a condition arising when the porous media is at residual saturation. Note here that we assume the relative permeability is only a function of hydraulic head. The results given below can be generalized to the case where relative permeability also depends on position. The following analysis will employ the Kirchho transformation, Zp (4.77) R(p) = k((}))d}; with gradient,
0
rR(p) = k((p))rp: (4.78) De ning u~ = ?rR and u = K (x)u~, equation (4.76) can be written as the follow-
ing equivalent system of equations, @(p) + r u = f; @t u~ = ?rR; u = K (x)u~:
(4.79) (4.80) (4.81)
Alt and Luckhaus [3] state the following regularity results for the above equation,
(p) 2 L1 (J ; L1( )); @t(p) 2 L2(J ; H ?1( )); u~ 2 L2(J ; (L2( ))d ):
(4.82) (4.83) (4.84)
Since @t is only in L2(J ; H ?1( )), a variational formulation of the problem would require trial functions for equation (4.79) to be taken in H 1( ). In order to relax this requirement, an alternate time integrated variational formulation developed by Arbogast, Wheeler and Zhang [6] is considered.
44 For this formulation, we need to integrate in time, but equation (4.82) does not guarantee that (p) exits pointwise everywhere in time. However, we know that physically is de ned at every time and we assume that (p) 2 L1(J ; L1( )) so that (p) exists pointwise for each t. Therefore, (4.79) can be integrated to get, Zt Zt (4.85) (p(:; t)) + r ud = fd + (p0): 0 0 Since physically, the normal components of the ow ux are continuous, r u 2 2 L ( ). Thus, the integral R0t ud is in L2(J ; H ( ; div)), and the following variational formulation can be de ned, Zt Zt (4.86) ((p); w) + (r ud;w) = ( fd; w) + ((p0); w); w 2 W; 0
0
0
0
(u~; v) =(R(p); r v) ? (R(pD ); v n)?D ; v 2 V0; (4.87) (u; v) =(K (x)u~; v); v 2 V: (4.88) ~ ; U) 2 (Wh; Vh ; VhN ) satisThe continuous time numerical scheme is to nd (P; U fying, Zt Zt (4.89) ((P ); w) + (r Ud;w) = ( fd; w) + ((P 0); w); w 2 Wh;
~ ; v) =(R(P ); r v) ? (R(pD ); v n)?D ; v 2 Vh0 ; (4.90) (U (U; v) =(K (x)U~ ; v); v 2 Vh : (4.91) Theorem 4.3 For the numerical scheme given by equations (4.89)(4.91), the following bounds hold, ZT ZT ((P ) ? (p); e?rt(Rd (P ) ? Rd (p))) + kK 1=2 dsk2 0 0 ZT ZT C fe?rT k 0 K 1=2(u~ ? u~^ )d k2 + 0 e?rtkK 1=2(u~ ? u~^ )k2dt ZT ZT + e?rT k (u ? u)d k2 + e?rtku ? uk2dtg: (4.92) 0
0
Proof Making use of the L2 and projections, and then subtracting equations
(4.86)-(4.88) from (4.89)-(4.91) gives the following error equations, Zt ((P ) ? (p); w) + (r (U ? u)d; w) = ((P 0) ? (p0); w); w 2 Wh ; (4.93) 0 ~ ? u~^ ; v) =(Rd (U (P ) ? Rd (p); r v); v 2 Vh0 ; (4.94)
~ ? u~); v) + (u ? u; v); v 2 Vh : (U ? u; v) =(K (x)(U
(4.95)
45 For some constant r de ned later, let w = e?rt(Rd (P ) ? Rd (p)) 2 Wh in (4.93) resulting in, Zt ? rt d d (P ) ? Rd (p))) ((P ) ? (p); e (R(P ) ? R(p)))+(r (U ? u)d; e?rt(Rd 0
=((P 0) ? (p0 ); e?rt(Rd (P ) ? Rd (p))):
(4.96)
In (4.94), let v = e?rt R0t(U ? u)d to give, Zt Zt ? rt ? rt d d ^ ~ (U ? u~; e 0 (U ? u)d ) = (R(P ) ? R(p); r e 0 (U ? u)d ): (4.97)
Lastly, integrate (4.95) from 0 to t holding v xed and multiply by ert. Then, let v = U~ ? u~^ so that, Zt ~ ? u~^ ) (e?rt (U ? u)d; U 0 Zt Zt ~ ? u~^ ): (4.98) =(e?rt K (x)(U~ ? u~)d; U~ ? u~^ ) + (e?rt (u ? u)d; U 0
0
Combining the above three equations results in, Zt ~ ? u~^ ); U~ ? u~^ ((P ) ? (p);e?rt(Rd (P ) ? Rd (p))) + e?rt K (x)(U 0 Zt Zt ? rt ? rt ^ ^ ^ ~ ~ (u ? u); U ? u~ K (x)(u~ ? u~); U ? u~ + e = e 0 0 + ((P 0) ? (p0); e?rt(Rd (P ) ? Rd (p))): (4.99)
~ ? u~^ and integrate (4.99) in time from 0 to T . Then, consider the Let = U second left-hand side term of (4.99), d K (x) Z t ds; Z t ds = 2 K (x); Z t ds : (4.100) dt 0 0 0 R R So, (e?rt 0t K (x); ) = 21 e?rt dtd kK 1=2 0t dsk2: Thus, by integration by parts,
Zt ZT ZT Z t
2 K (x); e?rt ds = 21 re?rt
K 1=2 ds
dt 0 0 0 0
2 Z T 1
1 = 2 ? rT ds
: (4.101) +e 2
K 0 Therefore, (4.99) integrated in time is, Z T 1 Z T ?rt
1=2 Z t
2 ? rt d d (P ) ? (p); e (R(P ) ? R(p)) + 2 0 re K 0 ds dt 0
46
Z T 2
+ e?rT 1
K 1=2 ds
2 0 Z T Z t Zt ZT r ? rt ? rt ^ = 0 e 0 K (x)(u~ ? u~); + 0 e 0 e (u ? u); ZT (P ) ? Rd (p))): (4.102) + ((P 0) ? (p0); e?rt(Rd 0 Now, by integration by parts, ZT Zt e?rt K (x)(u~ ? u~^ ); dt 0 0 Zt ZT Zt ZT Zt ? rt ? rt ^ ^ =r 0 e 0 K (x)(u~ ? u~)d; 0 d dt ? 0 e K (x)(u~ ? u~); 0 d dt ZT ! ZT ? rT ^ + e 0 K (x)(u~ ? u~); 0 d
2 Z t
2 Z T
Zt Z T
1 = 2 ? rt 1 = 2 ? rt
^
Cr 0 e K 0 (u~ ? u~)d dt + r 0 e K 0 d
dt Z t
2 Z T
ZT 1 = 2 ? rt ? rt 1 = 2 2 ^
d
dt + C e kK (u~ ? u~)k dt + e K 0 0 0
2
Z T
Z T
2
1 = 2 ? rT 1 = 2 ? rT + Ce
0 K (u~ ? u~^ )d
+ e
K 0 d
: (4.103) Similarly, ZT Zt ? rt e (u ? u); 0 0
2 Z T
Z t
2 Z T
Z t ? rt
Cr 0 e 0 (u ? u)d dt + ~r 0 e?rt
0 d
dt Z T
Z t
2 ZT ? rt 2 + C e ku ? uk dt + ~ e?rt
d
dt 0 0 0
Z T
2
2
Z T (4.104) + Ce?rT
(u ? u)d
+ ~e?rT
d
: 0 0 Take the initial approximation, P 0, such that, ((P 0) ? (p0); w) = 0; 8w 2 Wh . R This is done by computing P 0 = ?1(d (p0)), where (^p0)jEi = m(1Ei) Ei (p0 ). Thus, combining equations (4.103)-(4.104) with equation (4.102), taking and ~ small enough and choosing r large enough to exactly cancel the second left-hand side term in (4.102) gives,
Z T
2 ZT
? rt 1 = 2 d d ds
((P ) ? (p); e (R(P ) ? R(p))) +
K 0 0
47
0
0
2 Z T
Z T
? rT + e
(u ? u)d
+ e?rtku ? uk2dtg: 0 0
2 Z T
Z T C fe?rT
K 1=2(u~ ? u~^ )d
+ e?rtkK 1=2(u~ ? u~^ )k2dt (4.105)
The theorem just proven bounds the error in the numerical ux by approximation bounds. Thus, once the regularity of the solution is known, the error in the ux will have the same asymptotic behavior as approximation in the discrete space. The next estimate gives a bound in the H?1 norm of the numerical approximation to hydraulic head in the case that @=@p > 0. The above result is used to derive this result. Let 2 H01( ). Then, equation (4.93) and the de nition of the L2 projection imply, ((P ) ? (p); ) =((P ) ? (p); ? ^) + ((P ) ? (p); ^) Zt ^ ^ =((P ) ? (p); ? ) ? r (U ? u)d; Zt 0 ^ =((P ) ? (p); ? ) + (U ? u)d; r ; (4.106) 0
where we have used integration by parts, the de nition of the projection and have again choosen P 0 so that the initial term is 0. Now, assuming is Lipschitz continuous, and again using the de nition of the L2 projection, ((P ) ? (p); ? ^) = ((P ) ? (^p); ? ^) + ((^p) ? (p); ? ^) 0 + Chkp^ ? pkk kH1 : (4.107) Also,
Z t Zt
(U ? u)d; r (U ? u)d
k kH1 : 0 0
(4.108)
Integrating equation (4.95) from 0 to t holding v xed and then taking v = Rt 0 (U ? u)d , results in,
2
Z t Z Z
(U ? u)d
= K (x) t(U~ ? u~)d; t(U ? u)d 0 0 0
48
Zt (u ? u)d; (U ? u)d
02
Z t
0 Z t
2 1 = 2 ~
C K 0 (U ? u~)d + C 0 (u ? u)d
2
Z t
(4.109) + 0 (U ? u)d
: +
Z t
Combining equations (4.106)-(4.109) and recalling the de nition of the H?1 norm, equation (3.1), gives,
k(P ) ? (p)kH?1
Z t
Zt 1 = 2
~
C fhkp^ ? pk + K 0 (U ? u~)d + 0 (u ? u)d
g:
(4.110)
For a given time t, apply the mean value theorem to write,
(P (:; t)) ? (p(:; t)) = 0(w(t))(P (:; t) ? p(:; t)) C (P ? p):
(4.111)
kP ? pkH?1
Z t
Zt 1 = 2
~ C fhkp^ ? pk + K 0 (U ? u~)d + 0 (u ? u)d
g:
(4.112)
Thus,
So, with Theorem 4.3, the error in hydraulic head is bounded in terms of approximation error.
49
Chapter 5 Two-Level Methods for Nonlinear Parabolic Equations The analysis in the previous chapter applies to discretizing the full nonlinear problem on a computational grid of cell diameter h. However, due to the highly nonlinear nature of and K , solving the resulting discrete nonlinear system is computationally very expensive. Thus, alternative schemes which get around solving the full nonlinear mixed formulation of Richards' equation are attractive. One alternative approach is to consider linearizing the equation before discretization. For this approach, one would solve the full nonlinear problem on a coarse grid with a small amount of unknowns, then use that coarse grid solution to linearize the problem on a ne grid. This idea of using a two level scheme for nonlinear problems was rst developed by Xu [63, 64] who applied it to nonlinear elliptic equations with Galerkin nite elements and extended by Dawson and Wheeler [22] to the expanded mixed method applied to nonlinear parabolic equations. Xu showed optimal H1 convergence for both the coarse and ne grids. Dawson and Wheeler showed optimal H1 and L2 estimates for the coarse and ne grids, and for the case of the lowest order Raviart-Thomas-Nedelec space, they showed superconvergence for the coarse grid in both norms. In this work, we will show superconvergence results in certain discrete norms on both grids for a nite dierence scheme applied to the nonlinear heat equation. This is a rst step in trying to apply the two-level technique to Richards' equation. After completing this analysis. We show convergence results for a two-level scheme with the expanded mixed method applied to Richards' equation. For this scheme, the equation is not fully linearized on the ne grid. To fully linearize the equation would require giving up a mass conserving scheme. Previous work has shown that solutions become inaccurate when mass balance is lost. We thus leave the time term nonlinear on the ne grid and just consider linearizing the hydraulic conductivity term.
50 Two quasi-uniform triangulations of are considered, a coarse triangulation with mesh size H denoted by TH , and a ne triangulation with mesh size h denoted by Th . We assume that Th is a re nement of TH . Both these triangulations consist of rectangles in two dimensions and parallelepipeds in three dimensions.
5.1 A Two-Level Finite Dierence Scheme We begin with a nite dierence scheme for the nonlinear heat equation, @p ? r K (p)rp = f; @t ?K (p)rp n = 0:
(5.1) (5.2)
For simplicity we consider homogeneous Neumann boundary conditions. It is straightforward to extend the following results to nonhomogeneous conditions. The following assumptions are made. 1. The tensor K is symmetric and positive de nite. 2. The tensor K is bounded, i.e. there exist positive constants, K and K such that for z 2 IRd,
Kkzk2 ztKz K kzk: 3. Each element of K is twice continuously dierentiable in space and time with derivatives up to second order bounded above by K . 4. The tensor K (p) is Lipschitz continuous in p.
5.1.1 A Coarse Grid Nonlinear Finite Dierence Scheme In this section we develop and give convergence estimates for a nonlinear cell-centered nite dierence scheme on the coarse grid. For simplicity we consider two dimensions and note that extensions to three dimensions are straightforward.
De nition of the Scheme A variational formulation for (5.1)-(5.2) at time tn is to nd (pn ; u~n ; un) 2 (W V V 0) satisfying (@tpn ; w) + (r un ; w) = (f n ; w); 8w 2 W;
(5.3)
51 (u~n ; v) = (pn ; r v); 8v 2 V 0; (un ; v) = (K (pn )u~n ; v); 8v 2 V:
(5.4) (5.5)
~ nH 2 VH and UnH 2 VH0 Cell-centered nite dierence approximations PHn 2 WH ; U to the functions p(tn ; :); u~(tn; :) and u(tn; :), respectively, are chosen for each n = 1; : : : ; N , satisfying (dtPHn ; w) + (r UnH ; w) = (f n ; w); 8w 2 WH ; ~ nH ; v)TM = (PHn ; r v); 8v 2 VH0 ; (U ~ nH ; v)T; 8v 2 VH ; (UnH ; v)TM = (K (PH (PHn ))U
(5.6) (5.7) (5.8)
with PH0 = p^H (t0; :): Recalling that the grid points are denoted by, (xi+1=2; yj+1=2); i = 0; : : :; Nx; j = 0; : : : ; Ny ; and midpoints by,
xi = 12 (xi+1=2 + xi?1=2); i = 1; : : : ; Nx; yj = 21 (yj+1=2 + yj?1=2); j = 1; : : :; Ny ; de ne PH (p) from the values of pij for i = 1; : : : ; N^x and j = 1; : : :; N^y as follows. For points (x; y) such that xi x xi+1; i 2 f1; : : : ; N^xg and yj y yj+1; j 2 f1; : : : ; N^y g, take PH (p)(x; y) to be the bilinear interpolant, PH (p)(x; y) = (pij ( xxi+1 ?? xx ) + pi+1j ( xx ??xix ))( yyj+1 ?? yy ) i+1 i i+1 i j +1 j x ? x x ? x y +(pij+1 ( x i+1 ? x ) + pi+1j+1 ( x ? ix ))( y ??yjy ): i+1
For i = 1; : : : ; N^x ? 1, set
i
i+1
i
j +1
j
y + H y )p ? H y p (2 H 1 2 i1 1 i2 : PH (p)(xi; y1=2) = y y H1 + H 2 This is a two point extrapolation, and by Taylor's theorem j(PH (p) ? p)(xi ; y1=2)j CH 2. For points (x; y) such that xi x xi+1 and y1=2 y y1, de ne PH (p) as the bilinear interpolant between pi;1; pi+1;1; PH (p)(xi; y1=2) and PH (p)(xi+1; y1=2). By interpolation theory jPH (p) ? pj CH 2 for these points. In a similar way de ne
52
PH (p) for (x; y) such that xi x xi+1 and yN^y y yN^y+1=2 as well as for points (x; y) where x1=2 x x1 or xN^x x xN^x+1=2 and yj y yj+1 for j such that 1 j N^y . Lastly, de ne PH (p) at the corners of the domain. Here, three point extrapolation is used,
PH (p)(x1=2; y1=2) = PH (p)1;1=2 + PH (p)1=2;1 ? p1;1 = p1;1=2 + p1=2;1 ? p1;1 + O(H 2 ): By Taylor's theorem, j(PH (p) ? p)(x1=2; y1=2)j CH 2. For points (x; y) such that x1=2 x x1 and y1=2 y y1, de ne PH (p)(x; y) as the bilinear interpolant of PH (p)(x1=2; y1=2); PH (p)(x1=2; y1); PH (p)(x1; y1=2) and p1;1 which is an O(H 2 ) approximation to p(x; y) within this \corner region". Similarly, de ne PH (p) as an O(H 2) approximation to p in the other three \corner" regions. We have just proven the following lemma, Lemma 5.1 If p is twice dierentiable, then for PH (p) de ned above,
kPH (p) ? pkL1 CH 2: If a uniform mesh is used and K is a diagonal tensor, equations (5.6)-(5.8) reduce to a standard nonlinear nite dierence procedure. Denoting PHn by P n , in the interior of , 2 H 2 n ? 1 n fij H + Pij tn = 12 [(K11(PH (P n ))i+1=2j+1=2 + K11(PH (P n ))i+1=2j?1=2)(Pijn ? Pin+1j ) + (K11(PH (P n ))i?1=2j+1=2 + K11(PH (P n ))i?1=2j?1=2)(Pijn ? Pin?1j ) + (K22(PH (P n ))i+1=2j+1=2 + K22(PH (P n ))i?1=2j+1=2)(Pijn ? Pijn+1 ) + (K22(PH (P n ))i+1=2j?1=2 + K22(PH (P n ))i?1=2j?1=2)(Pijn ? Pijn?1 )] 2 (5.9) + Htn Pijn : Existence and uniqueness of a solution to this discrete nonlinear problem is given in the following theorem.
Theorem 5.1 For time tn and t suciently small, there exists a unique solution to equations (5.6)-(5.8).
53
Proof We are seeking a unique solution to the nonlinear equation F (P n) = 0, where
F (P n) = bn + P n + Ht2n A(P n)P n . Here, bn is a vector whose entry corresponding to R grid cell (xi; yj ) is ? Ht2n ij fijn ? Pijn?1 , P is a vector whose ij th entry corresponds to the value of the scalar variable Pijn and A is a matrix function of P n given by the stencil above. By Theorem 5.4.5 of Ortega and Reinbolt [52], if F is continuously dierentiable and uniformly monotone on IRn, then a unique solution to F (P n) = 0 exists. It is easily veri ed that the F de ned above is continuously dierentiable. In order to prove that F is uniformly monotonic we note that uniform monotoncity is equivalent to positive de niteness of the Jacobian, J = F 0, and that a real matrix J is positive de nite if and only if its symmetric part, (J + J T )=2, is positive de nite [10, Lemma 3.1]. Furthermore, we know that if a matrix is stricly diagonal dominant with positive diagonal entries, then the eigenvalues of the matrix have positive real parts [10, Theorem 4.9]. Now, J = I + Ht2n A(P n) + Ht2n A0(P n )P n . Thus, with Ht2n suciently small, we have that the symmetric part of J has positive real eigenvalues and, hence, is positive de nite, making J positive de nite and F uniformly monotonic.
Preliminary Estimates Before we show convergence estimates for this nite dierence scheme, we show convergence for a related linear scheme. The arguments given below closely follow those of Arbogast, Wheeler and Yotov [8] except that we extend their work to time dierenced time dependent problems. Theorem 5.2 For each n = 1; : : : ; N , let (P nH ; U~ nH ; UnH ) 2 (WH VH VH0 ) satisfy (5.10) (r UnH ; w) = (bn; w); 8w 2 WH ; n ~ H ; v)TM = (P nH ; r v); 8v 2 VH0 ; (5.11) (U n (5.12) (UnH ; v)TM = (K (PH (pn ))U~ H ; v)T; 8v 2 VH ; with bn = f n ? @tpn and P 0H = p^0H . We further require the compatibility R condition, R P nH = p^n . Then, (5.13) kUnH ? un kTM + kU~ nH ? u~n kTM CH 2; n n 2 kP H ? p kM CH ; (5.14) (5.15) kdtP nH ? dt pn kM C (H 2 + t):
54
[8].
We will make use of the following lemma proven in Arbogast, Wheeler and Yotov
Lemma 5.2 For the lowest order RTN space on rectangles and for any q = (qx; qy ) 2 H 1( ) and E 2 Tk , @ (q)xk k @q k ; k @x 0;E @x 0;E x
@ (q)y k k @q k : k @y 0;E @y 0;E y
(5.16) (5.17)
In order to prove the above theorem, we will rst prove two preliminary lemmas.
Lemma 5.3 Assume for each n = 1; : : : ; N , that pn ; @tpn 2 W34( ). ~ ;n 2 VH ; P ;n 2 WH ; Z;n 2 VHn; Z~ ;n 2 VH and W ;n 2 Then, there exist U WH such that
~ ;n; v)TM =(P ;n; r v); v 2 VH0 ; (U (5.18) (Z~ ;n; v)TM =(W ;n; r v); v 2 VH0 ; (5.19) ~ ;n; v)T; v 2 VH ;(5.20) (Z;n; v)TM = (K (pn )Z~ ;n; v)T + (@t(K (pn ))U and there exists a constant C independent of H such that for all i; j ,
jPi;j;n ? pni;j j CH 2; jWi;j;n ? @tpni;j j CH 2; ;n ~ny;ij+1=2j CH 2; jU~x;i;n+1=2j ? u~nx;i+1=2j j + jU~y;ij +1=2 ? u ;n ~ny;ij+1=2j CH 2; jZ~x;i;n+1=2j ? @tu~nx;i+1=2j j + jZ~y;ij +1=2 ? @t u ;n n 2 jZx;i;n+1=2j ? @tunx;i+1=2j j + jZy;ij +1=2 ? @tuy;ij +1=2 j CH :
(5.21) (5.22) (5.23) (5.24) (5.25)
Proof Arbogast, Wheeler and Yotov [8] present a lemma which gives the desired ~ ;n above. In order to derive (5.22) and (5.24), we apply a lemma due to P ;n and U
55 Weiser and Wheeler [62] to the pair (@tu~n; @tpn ) which, by de nition, satis es the two equations,
r @tu~n = F n; in ; @tu~n = ?r@tpn; in ; where F n = @tf n + @ttpn . This result gives a W ;n satisfying (5.22) and through (5.19), Z~ ;n satis es (5.24) in the interior of . De ne Z~ on ? by, ;n ~nx;i+1=2j ; Z~x;i +1=2j = @tu ;n ~ny;ij+1=2: Z~y;ij +1=2 = @tu
Then, (5.24) clearly holds on ?. Choosing v in (5.20) to be the basis function associated with node (xi+1=2; yj ), we have for i = 1; : : : ; N^x ? 1, Zx;i;n+1=2j = 21 [K11(pn )i+1=2j+1=2 + K11(pn )i+1=2j?1=2]Z~ x;i;n+1=2j ~ x;i;n+1=2j : + 21 [@t(K11(pn ))i+1=2j+1=2 + @t(K11(pn ))i+1=2j?1=2]U
Since @tun = K (pn )@tu~n + @tK (pn )u~n , Taylor's theorem gives for i = 1; : : : ; N^x ? 1, @tunx;i+1=2j = 21 [K11(pn )i+1=2j+1=2 + K11(pn )i+1=2j?1=2]@tu~nx;i+1=2j + 21 [@t(K11(pn ))i+1=2j+1=2 + @t(K11(pn ))i+1=2j?1=2]u~nx;i+1=2j + O(H 2):
Therefore,
jZx;i;n+1=2j ? @tunx;i+1=2j j C jZ~ x;i;n+1=2j ? @tu~nx;i+1=2j j + O(H 2): ;n n In a similar manner we can bound jZy;ij +1=2 ? @tuy;ij +1=2 j, and (5.25) follows.
We can now extend a corollary from Arbogast, Wheeler and Yotov [8] to arrive at ~ ;n; P ;n; Z;n; Z~ ;n and W ;n in Lemma 5.3, there the following statement. For the U exists a constant C , independent of H , such that
kU~ ;n ? u~n kTM CH 2; kZ~ ;n ? @tu~n kTM CH 2; kZ;n ? @tun kTM CH 2:
56
Lemma 5.4 There exists a constant C independent of H and t such that,
(5.26) kr (dtun ? dt UnH )k CH; kdtu~n ? dtU~ nH kTM + kdtun ? dtUnH kTM C (H 2 + t): (5.27)
Proof To prove this lemma, consider the time dierence of (5.3)-(5.5), (r dtun; w) = (dtbn; w); 8w 2 W; (dt u~n; v) = (dtpn ; r v); 8v 2 V 0; (dt un; v) = (dt(K (pn ))u~n ; v) + (K (pn?1 )dtu~n ; v); 8v 2 V;
(5.28) (5.29) (5.30)
and the time dierence of (5.10)-(5.12), (r dtUnH ; w) = (dtbn; w); 8w 2 WH ; ~ nH ; v)TM = (dtP nH ; r v); 8v 2 VH0 ; (dtU ~ nH ; v)T (dtUnH ; v)TM = (dt(K (PH (pn )))U ~ nH ; v)T; 8v 2 VH : +(K (PH (pn?1 ))dtU
(5.31) (5.32) (5.33)
Subtract (5.31) from (5.28), and subtract (5.32) and (5.33) from (5.19) and (5.20) to give, (r (dtun ? dtUnH ); w) = 0; (5.34) n ~ H ; v)TM = (W ;n ? dtP nH ; r v); (5.35) (Z~ ;n ? dtU n (Z;n ? dtUnH ; v)TM = (K (pn )Z~ ;n ? K (PH (pn?1 ))dtU~ H ; v)T ~ ;n ? dtK (PH (pn ))U~ nH ; v)T: (5.36) +(@t(K (pn )U Using (5.34) and applying the Cauchy-Schwarz inequality we have,
kr (dtun ? dtUnH )k2 = (r (dtun ? dtUnH ); r (dtun ? dtun )) kr (dtun ? dt UnH )kkr (dtun ? dtun )k: Thus, by (3.24) the rst part of the lemma is obtained.
57
~ nH in (5.36), use (5.34) Now, let v = dtun ? dtUnH in (5.35) and v = Z~ ;n ? dtU and combine to get, ~ nH ; Z~ ;n ? dtU~ nH )T (K (pn )Z~ ;n ? K (PH (pn?1 ))dtU ~ nH ; dtun ? dtUnH )TM + (Z;n ? dt UnH ; Z~ ;n ? dtU~ nH )TM = ? (Z~ ;n ? dtU (5.37) ? (@tK (pn )U~ ;n ? dtK (PH (pn))U~ nH ; Z~ ;n ? dtU~ nH )T: Adding (K (PH (pn?1 ))Z~ ;n; Z~ ;n ? dtU~ nH )T to both sides of (5.37), using the boundedness assumption on K , Taylor's Theorem, the Cauchy-Schwarz inequality and (4.2) results in, kZ~ ;n ? dtU~ nH kTM C (kdtun ? Z;nkTM + tkU~ ;nkTM ~ ;nkT +kdt(K (pn ) ? K (PH (pn )))U ~ ;n ? U~ nH )kT + (H 2 + t)kZ~ ;nkT): +kdtK (PH (pn ))(U Taylor's theorem, the estimate (3.25) and Lemma 5.3 imply that kdtun ? Z;nkTM C (H 2 + t). By Lemma 5.3 kU~ ;nkTM and kZ~ ;n kT are bounded. Thus, by Taylor's theorem, the Lipschitz condition on @tK , the boundedness of @tK and the approximation properties of PH , (5.38) kZ~ ;n ? dtU~ nH kTM C (H 2 + t + kU~ ;n ? U~ nH kT ): ~ ;n ? U~ nH kT CH 2. Hence, by By results from Arbogast, Wheeler and Yotov [8], kU the triangle inequality and Lemma 5.3, kdtu~n ? dt U~ nH kTM C (H 2 + t): Now, let v = Z;n ? dtUnH in (5.36) and use the Cauchy-Schwarz inequality to get kZ;n ? dtUnH kTM kK (pn )Z~ ;n ? K (PH (pn?1 ))dtU~ nH kT ~ ;n ? dt (K (PH (pn )))U~ nH kT : +k@tK (pn )U By the Lipschitz assumptions on K and @tK , Taylor's theorem, the approximation properties of PH and the boundedness of @tK , kZ;n ? dt UnH kTM k(K (pn) ? K (PH (pn?1 )))Z~ ;nkT + kK (PH (pn?1 ))(Z~ ;n ? dt U~ nH )kT ~ ;nkT + kdtK (PH (pn ))(U~ ;n ? U~ nH )kT + k(@tK (pn ) ? dtK (PH (pn )))U C (H 2 + t):
58 The triangle inequality and Lemma 5.3 result in,
kdtun ? dt UnH kTM C (H 2 + t):
Remark 5.1.1 By the inverse assumption, the de nition of and (3.25) we have
kU~ nH kL1 kU~ nH ? u~nkL1 + ku~n ? u~n kL1 + ku~n kL1 CH ?d=2kU~ nH ? u~nkTM + ku~n ? u~n kL1 + ku~n kL1 C (H ?d=2H 2 + H + 1); ~ nH kL1 is bounded. where d is the space dimension. Thus, kU
Proof (Of Theorem 5.2) Results (5.13) and (5.14) have been proven by Arbogast,
Wheeler and Yotov [8]. In order to derive (5.15), subtract (5.32)-(5.33) from (5.29)-(5.30) and use the de nition of the L2 projection to give,
~ nH ; v) + ETM (dtU~ nH ; v) = (dt p^nH ? dtP nH ; r v); v 2 VH0 ; (dtu~n ? dtU (5.39) n ~ H ; v) (dtun ? dtUnH ; v) + ETM (dtUnH ; v) = (dt (K (pn ))u~n ? dt(K (PH (pn )))U ~ nH ; v) +(K (pn?1 )dt u~n ? K (PH (pn?1 ))dtU ~ nH ; v) +ET (dt(K (PH (pn )))U +ET (K (PH (pn?1 ))dtU~ nH ; v); v 2 VH : (5.40) Let satisfy the auxilliary problem with n 2 L2( ) de ned later,
where we assume that
?r K (PH (pn?1 ))rn = n; ; K (PH (pn?1 ))rn n = 0; ?;
(5.41) (5.42)
knk2 C knk:
(5.43)
R n = 0. Elliptic regularity implies that
By equations (5.41) and (5.39) and the de nition of , (dtp^nH ? dtP nH ; n )
59 = ?(dtp^nH ? dtP nH ; r K (PH (pn?1 ))rn) ~ nH ; K (PH (pn?1 ))rn) = ?(dtu~n ? dtU ?ETM(dt U~ nH ; K (PH (pn?1 ))rn) ~ nH ; K (PH (pn?1 ))rn ? K (PH (pn?1 ))rn) = ?(dtu~n ? dtU ?(K (PH (pn?1 ))(dtu~n ? dtU~ nH ); rn ? rn) ?(K (PH (pn?1 ))(dtu~n ? dtU~ nH ); rn) ?ETM(dt U~ nH ; K (PH (pn?1 ))rn):
(5.44)
By (5.40), ?(K (PH (pn?1 ))(dtu~n ? dtU~ nH ); rn) =((K (pn?1 ) ? K (PH (pn?1 )))dtu~n; rn) ? (dt un ? dtUnH ; rn) ~ nH ; rn) ? ETM (dtUnH ; rn) + (dt (K (pn ))u~n ? dt(K (PH (pn )))U ~ nH ; rn) + ET (dt(K (PH (pn)))U ~ nH ; rn): (5.45) + ET (K (PH (pn?1 ))dt U We also have by integration by parts, (5.34) and (5.36) ?(dtun ? dtUnH ; rn) = ?(dtun ? dtUnH ; rn ? rn) ? (dtun ? dtUnH ; rn) = ?(dtun ? dtUnH ; rn ? rn) + (r (dtun ? dtUnH ); n ? ^nH ): (5.46) Furthermore, we can write ~ nH ; rn) (dt K (pn )u~n?dtK (PH (pn ))U =((dtK (pn ) ? dtK (PH (pn )))u~n ; rn) ~ nH ); rn): + (dtK (PH (pn ))(u~n ? U (5.47) Using (3.16) and recalling the de nition of the lowest order RTN space gives, X X @ ~ n k @ x (dt UH K (pn )rn)kL1(E)H 2 jETM (dtU~ nH ; K (pn)rn)j C E 2TH jj=2 X @dtU~ nH;x @ (K (pn)rn)xk C (k @x k0;E k @x 0;E E 2TH @dtU~ nH;y @ (K (pn )rn)y k )H 2: (5.48) +k @y k0;E k @y 0;E
60 By Lemma 5.2 and the inverse inequality we have
@dtU~ nH;x @ (d U~ n ? d u~n )k + k @ d u~n k k @x k0;E k @x t x 0;E t H;x @x t x 0;E @ d u~n k C kdtU~ nH;x ? dtu~nxk0;E H ?1 + k @x t x 0;E C kdtU~ nH ? dtu~n k0;E H ?1 + kdtu~n k1;E :
(5.49)
~ nH and K (pn )rn are in VH , by (5.48) and the Cauchy-Schwarz inequality, Since dtU X jETM (dtU~ nH ; K (pn)rn)j C (kdtU~ nH ? dtu~n k0;E + kdtu~n k1;E H )knk2;E H E
C (kdtU~ nH ? dtu~nk0 + kdtu~n k1H )kn k2H:
(5.50)
As done above and noting that K has bounded second derivatives,
jETM (dtUnH ; rn)j C (kdtUnH ? dtunk + kdtun k1H ) knk2H; (5.51) n n jET(dt (K (PH (pn )))U~ H ; rn)j (C + t)(kU~ H ? u~nk + ku~n k1H ) knk2H; (5.52) jET(K (PH (pn?1 ))dtU~ nH ; rn)j C (kdtU~ nH ? dtu~nk + kdtu~n k1H ) knk2H; (5.53) where C in the second and third inequalities depends on K . Combining (5.44) with (5.45)-(5.53), applying approximation properties of the L2 and projections, using Lemma 5.4, and equations (5.13) and (3.25) gives (dtp^nH ? dtP nH ; n ) kdtu~n ? dtU~ nH kkK (pn)rn ? K (pn )rnk ~ nH )kkrn ? rnk +kK (PH (pn?1 ))(dtu~n ? dtU +k(dtK (pn ) ? dtK (PH (pn )))u~nkkrnk ~ n )kkrnk +kdtK (PH (pn ))(u~n ? U +k(K (pn?1 ) ? K (PH (pn?1 )))dt u~nkkrnk +kdtun ? dtUnH kkrn ? rnk +kr (dtun ? dtUnH )kkn ? ^nH k ~ nH kkrnk +kdt(K (pn ))u~n ? dt(K (PH (pn )))U
61 +jETM (dtU~ nH ; K (pn)rn)j + jETM (dtUnH ; rn)j ~ nH ; rn)j + jET(K (PH (pn?1 ))dtU~ nH ; rn)j +jET (dt(K (PH (pn )))U C (H 2 + t)knk2: Taking n = dtp^n ? dtP nH and noting that that R n = R 0 = 0 and applying equation (5.43) we have, kdtp^nH ? dt P nH k C (H 2 + t): (5.54)
Convergence Estimate of the Nonlinear Scheme We now prove the following theorem about the convergence of the above coarse grid nite dierence scheme: Theorem 5.3 Let PHn ; U~ nH and UnH ; n = 1; : : : ; N be de ned as in (5.6)(5.8) with initial values PH0 = p^H (t0; :). Then, there exists a positive constant C , independent of H and t such that N X kPHN ? pN kM + ft KkU~ nH ? u~nk2Tg1=2 C (H 2 + t): (5.55) n=1
Proof Let n = PHn ? P nH ; n = U~ nH ? U~ nH ; n = UnH ? UnH and n = P nH ? pn . Subtracting (dtP nH ; w) + (r UnH ; w) from both sides of (5.6) and using equations (5.10) and (5.3) we have, (dt n ; w) + (r n ; w) = (f n ; w) ? (r UnH ; w) ? (dtP nH ; w) = (@tpn ; w) ? (dtpn ; w) ? (dtn ; w) = (n ; w) ? (dt n; w); w 2 WH ; (5.56) where n is a time truncation term. Subtracting (5.11) from (5.7) results in, (n; v)TM = ( n; r v); v 2 VH0 ; (5.57) and subtracting (5.12) from (5.8) gives, ~ nH ; v)T (n ; v)TM = (K (PH (PHn ))U~ nH ; v)T ? (K (PH (pn ))U ~ nH ; v)T = (K (PH (PHn ))n ; v)T + ((K (PH (PHn )) ? K (PH (P nH )))U ?((K (PH (pn )) ? K (PH (P nH )))U~ nH ; v)T; v 2 VH : (5.58)
62 Letting w = n in (5.56), v = n in (5.57) and v = n in (5.58) gives, (dt n ; n) = ?(r n; n ) + (n; n)M ? (dtn; n ); (5.59) (n; n)TM = ( n; r n); (5.60) n ~ H ; n )T (n; n)TM = (K (PH (PHn ))n; n)T + ((K (PH (PHn )) ? K (PH (P nH )))U ?((K (PH (pn )) ? K (PH (P nH )))U~ nH ; n)T : (5.61) Combining equations (5.59)-(5.61), applying the Cauchy-Schwarz inequality and (4.2) we have, 1 [k nk2 ? k n?1 k2 ] + kK (P (P n ))1=2nk2 H H M M T 2t (dt n ; n)M + kK (PH (PHn ))1=2n k2T (n; n )M ? (dtn ; n)M + ((K (PH (pn )) ? K (PH (P nH )))U~ nH ; n)T ?((K (PH (PHn )) ? K (PH (P nH )))U~ nH ; n )T 21 kn k2M + k nk2M + 21 kdtn k2M + C k(K (PH (pn )) ? K (PH (P nH )))U~ nH k2T +C k(K (PH (PHn )) ? K (PH (P nH )))U~ nH k2T + knk2TM ; where K=2. Now, Z tn jnij j = 1t j tn?1 ptt(xi; yj ; t)(t ? tn)dtj kptt(xi; yj ; :)kL2(tn?1;tn)(t) 21 : So, X (5.62) knk2M t HixHjy kptt(xi; yj ; :)k2L2(tn?1;tn): ij
By the triangle inequality and Theorem 5.2,
kdt nk2M C (H 4 + t2): By the Lipschitz assumption on K , the de nition of PH and Theorem 5.2
k(K (PH (pn )) ? K (PH (P nH )))U~ nH k2T C kpn ? P nH k2M CH 4; k(K (PH (PHn )) ? K (PH (P nH )))U~ nH k2T C k nk2M ; ~ nH kL1 as per Remark 5.1.1. where we have used the boundedness of kU
(5.63) (5.64)
63 Multiplying by 2t, bringing the knk2TM term to the left-hand side, summing on n; n = 1; : : :; N , using (5.62)-(5.64) and applying Gronwall's Lemma 3.1 gives, N X k N k2M ? k 0k2M + t kK (PH (PHn ))1=2n k2T N X
n=1
N X
C t (knk2M + kdtn k2M + kn k2M) + CH 4 + C t k n k2M n=1 n=1 2 4 C (t + H ): The proof is completed by applying the initial conditions on PH0 and P 0H , Theorem 5.2 and the triangle inequality.
5.1.2 Fine Grid Linear Scheme We now consider a linear cell-centered nite dierence scheme on the ne grid where we make use of the nonlinear solution on the coarse grid. Note that since we assume Th is a re nement of TH , we have, WH Wh and VH Vh . ~ nh 2 Vh and Unh 2 VH0 at each time We solve the following problem for Phn 2 Wh; U step n = 1; : : : ; N , (dtPhn; w) = ?(r Unh ; w) + (f n ; w); w 2 Wh; (5.65) ~ nh; v)TM = (Phn ; r v); v 2 Vh0; (5.66) (U ~ nh ; v)T (Unh; v)TM = (K (PH (PHn ))U ~ nH )(Ph(Phn) ? PH (PHn )); v)T; v 2 Vh : (5.67) +(K 0(PH (PHn ))QH (U We de ne QH (u~) as a vector quantity with entries QxH (~ux) and QyH (~uy ). The entry QxH (~ux) is de ned from the values of u~xi+1=2j for i = 0; : : : ; N^x and j = 1; : : : ; N^y as follows. For points (x; y) such that xi?1=2 x xi+1=2; i 2 f1; : : : ; N^xg and yj y yj+1; j 2 f1; : : : ; N^y g, we take QxH (~ux) to be the bilinear interpolant of u~xi?1=2;j ; u~xi+1=2;j ; u~xi?1=2;j+1 and u~xi+1=2;j+1. This leaves a strip half a cell in height along the top and bottom of the domain. We will consider the bottom strip. For i = 0; : : : ; N^x, we set (2H y + H y )~ux ? H y u~x QxH (~ux)(xi+1=2; y1=2) = 1 2 Hiy+1+=2H;1 y 1 i+1=2;2 : 1 2 Now, for points (x; y) such that xi?1=2 x xi+1=2; i 2 f1; : : :; N^xg and y1=2 y y1, we let QxH (~ux)(x; y) be the bilinear interpolant of the two interpolated values
64
QxH (~ux)(xi?1=2; y1=2) and QxH (~ux)(xi+1=2; y1=2) and the two values u~xi?1=2;1 and u~xi+1=2;1. An analogous de ntion is made along the top strip of the domain. The de nition of QyH (~uy ) is similar to the above, except that the strips are along the left and right sides of the domain. The following lemma summarizes the approximation error of QH . Lemma 5.5 If each component of u~ is twice dierentiable, then for QH (u~) de ned above,
kQH (u~) ? u~kL1 CH 2:
Proof By Taylor's theorem we have that the two point extrapolation for the bound-
ary points described above is O(H 2) accurate. Thus, since bilinear interpolation is also O(H 2 ) accurate, the lemma is proven. We turn now to an analysis of the ne grid scheme. Theorem 5.4 Let Phn ; U~ nh and Unh; n = 1; : : : ; N be de ned as in (5.65)(5.67) with initial values Ph0 = p^h (t0; :). Then, there exists a positive constant C , independent of h; H and t such that N X kPhN ? pN kM + ft K kU~ nh ? u~nk2T g1=2 n=1 C (H 4?d=2 + h2 + t):
Proof We can de ne P nh 2 Wh; U~ nh 2 Vh and Unh 2 Vhn at each n = 1; : : : ; N
satisfying equations (5.10)-(5.12) and Theorem 5.2 on the ne grid. ~ nh ? U~ nh; n = Unh ? Unh and n = P nh ? pn . As done in Let n = Phn ? P nh ; n = U Theorem 5.3, we subtract (dtP nh ; w) + (r Unh; w) from both sides of equation (5.65) and combine with equation (5.10) applied to the ne grid. We also subtract (5.11) and (5.12) from (5.66) and (5.67) to give the error equations, (dt n ; w) = ?(r n ; w) + (n; w) ? (dtn ; w); (5.68) (n ; v)TM = ( n ; r v); (5.69) n n n n n ~ h ; v )T ( ; v)TM = (K (PH (PH ))U~ h ; v)T ? (K (Ph(p ))U ~ nH )(Ph(Phn ) ? PH (PHn )); v)T: (5.70) +(K 0(PH (PHn ))QH (U
65 Using Taylor's Theorem, K (Ph(pn )) can be written as
K (Ph (pn)) = K (Ph(PHn )) + K 0(PH (PHn ))(Ph(pn ) ? PH (PHn )) 00 n + K 2( ) (Ph(pn ) ? PH (PHn ))2;
where n is between Ph(pn ) and PH (PHn ). Using this expression in (5.70) and adding and subtracting the derivative term, 0 ~ nH )Ph(pn ); v)T gives, (K (PH (PHn ))QH (U (n; v)TM = (K (PH (PHn ))n; v)T ~ nh )(Ph(pn ) ? PH (PHn )); v)T +(K 0(PH (PHn ))(QH (U~ nH ) ? U ~ nH )(Ph(Phn) ? Ph(pn )); v)T +(K 0(PH (PHn ))QH (U 00 n ~ nh ; v)T: +( K 2( ) (Ph(pn ) ? PH (PHn ))2U
Let w = n ; v = n and v = n in (5.68), (5.69) and (5.71), respectively, and combine to give, 1 [k nk2 ? k n?1k2] + K kn k2 T 2t (dt n ; n) + kK (PH (PHn ))1=2nk2T 21 knk2 + k nk2 + 21 kdtnk2 + kn k2 ~ nh )(Ph(pn ) ? PH (PHn ))k2T +C kK 0(PH (PHn ))(QH (U~ nH ) ? U ~ nH )(Ph(Phn ) ? Ph(pn ))k2T +C kK 0(PH (PHn ))QH (U 00 n ~ nhk2T ; (5.71) +C k K 2( ) (Ph(pn ) ? PH (PHn ))2U
where K=2. Consider now the last three terms of (5.71). The rst of these can be bounded as follows.
kK 0(PH (PHn ))(QH (U~ nH ) ? U~ nh)(Ph(pn ) ? PH (PHn ))k2T C kQH (U~ nH ) ? U~ nhk2TM kPh(pn ) ? PH (PHn )k2L1 ; where,
kQH (U~ nH ) ? U~ nhk2TM kQH (U~ nH ) ? QH (u~n )k2TM + kQH (u~n) ? u~n k2TM ~ nhk2TM : +ku~n ? U
(5.72)
66 Since QH (U~ nH ) is a bilinear interpolant of terms that can be expressed in terms of ~ nH on the coarse grid, it can be shown that, nodal values of U
kQH (U~ nH ) ? QH (u~n )k2TM C kU~ nH ? u~n k2TM;H ;
where k kTM;H denotes the midpoint by trapezoidal norm on the coarse grid. Also, kQH (u~n ) ? u~nk2TM CH 4 by Lemma 5.5. In order to bound the second term in (5.72), write it as,
kPh(pn ) ? PH (PHn )k2L1 kPH (PHn ) ? PH (pn )k2L1 + kPH (pn ) ? pn k2L1 +kpn ? Ph(pn )k2L1 : By the de nition of PH , the quasi-uniformity assumption on the coarse grid and
Theorem 5.3,
kPH (PHn ) ? PH (pn )k2L1 HCd kPHn ? pn k2M;H CH ?d(H 4 + t2); where d is the space dimension. By Lemma 5.1, kPH (pn ) ? pn k2L1 CH 4 and kPh(pn ) ? pn k2L1 Ch4. Thus, kK 0(PH (PHn ))(QH (U~ nH ) ? U~ nh )(Ph(pn ) ? PH (PHn ))k2T C (kU~ nH ? u~n k2TM;H + ku~n ? U~ nhk2TM + H 4)(H 4?d + H ?d t2 + h4): (5.73)
The second to last term in (5.71) can be bounded by,
kK 0(PH (PHn ))QH (U~ nH )(Ph(Phn) ? Ph(pn ))k2T C kU~ nH k2L1 kPh(Phn) ? Ph(pn )k2T C (H ?d kU~ nH ? u~n k2TM + ku~nk2L1 )(kPhn ? P nh k2M + kP nh ? pnk2M ) C (H ?d kU~ nH ? u~n k2TM + C )(k nk2 + h4):
The last term in (5.71) is bounded by, 00 n k K 2( ) (Ph(pn ) ? PH (PHn ))2U~ nhk2T C kPh(pn ) ? PH (PHn )k2L1 kPh(pn ) ? PH (PHn )k2T C (H 4?d + H ?d t2 + h4)(h4 + H 4 + kpn ? PHn k2M;H ) C (H 4?d + H ?d t2 + h4)(h4 + H 4 + t2) C (H 8?d + h4 + t2 + H ?d h4t2 + H ?d t4):
(5.74)
(5.75)
67 Combining equation (5.71) with equations (5.73)-(5.75), taking the kn k2 term to the left side, multiplying by 2t and summing over n; n = 1; : : : ; N where N is the time step at which k nk achieves its maximum value gives, N X 2 N 0 k k ? k k + t Kknk2TM n=1
N X n 2 n 2 t (k k + kdt k ) + C t k nk2 n=1 n=1 N X ~n n 2 ~ kTM;H +C (H 4?d + H ?dt2 + h4)t (kU H ?u n=1 N X 2 4 N ~ nH ? u~nk2 +C (h + k k )t H ?d kU n=1 8 ? d 4 2 +C (H + h + t + H ?d h4t2 + H ?d t4): N X
~ nh k2TM + H 4) + ku~n ? U
Recalling the bound on n, using Theorem 5.2 and recalling the initial conditions on P 0h and Ph0 gives, N X 2 N k k + t Kkn k2TM C (H 8?d + h4 + t2 + H ?d h4t2 + H ?d t4) n=1
+C t
N X
k nk2 + C~ k N k2(H 4?d + H ?d t2):
n=1 H 4?d + H ?d t2 21C~ ,
We can choose H and t such that and the last term can be moved to the left-hand side. Applying Gronwall's Lemma gives N X 2 N k k + t Kkn k2TM C (H 8?d + h4 + t2): n=1
Applying Theorem 5.2 and the triangle inequality gives the desired result.
5.1.3 Extensions to Multiple Levels The above analysis carries through for multiple levels. In this case, the nonlinear problem would still be solved once on the coarsest grid. However, one could have multiple ne grids. On each of these ner and ner grids, the nonlinear term is expanded about the next coarser solution and the resulting linear system is solved. Adding more grids corresponds to adding more Newton-like iterations with each iteration taking place on the next ner grid.
68
5.2 A Two-Level Method for Richards' Equation In this section, we consider applying these two-level ideas to Richards' equation. We discuss only the expanded mixed method and not the superconvergent nite dierence case. In order to get a sucient rate of convergence on the coarse grid to transfer to the ne grid, we make the assumption of strictly partially saturated ow. Under this assumption, the scheme in Section 4.2 can be applied on the coarse grid giving optimal convergence. Thus, we turn to the ne grid. Linearizations of the time derivative term lead to schemes which are nonconservative and give incorrect mass balance. Thus, we consider leaving the time term nonlinear and just linearizing the hydraulic conductivity, K . ~ n ; Un) 2 (Wh; Vh ; Vh ) for each Our discrete time ne grid scheme is to nd (P n; U time step n = 1; : : : ; N satisfying, (dt(P n ); w) + (r Un; w) =(f n ; w); (5.76) ~ n ; v) =(P n ; r v); (U (5.77) ~ n; v) + (K 0(PHn )U~ nH (P n ? PHn ); v): (5.78) (Un ; v) =(K (PHn )U The following theorem gives the convergence behavior of this scheme. Theorem 5.5 For (P n ; U~ n; Un ) 2 (Wh; Vh ; Vh) de ned as in equations (5.76)-(5.78), we have, !1=2 N X n 1 = 2 n n 2 N N kp ? P k+ kK (PH ) (U~ ? u~ )k t n=1 C (hk+1 + t + H 2k+2?d=2 ):
(5.79)
Proof Subtracting the numerical scheme from the variational formulation, equa-
tions (3.13)-(3.15), gives the error equations,
and
(dt((pn ) ? (P n )); w) + (r (un ? Un ); w) =(n; w); ~ n ; v) =(^pn ? P n ; r v); (u~^ n ? U
(5.80) (5.81)
(un ? Un; v) =(un ? un ; v) ? (K (PHn )U~ n ; v) ? (K 0(PHn )U~ nH (P n ? PHn ); v) + (K (pn )u~n ; v):
(5.82)
69
~ n ; and n = un ? Un . Take w = n in (5.80), Let n = p^n ? P n; n = u~^ n ? U v = n in (5.81) and v = n in (5.82) and combine to get, ~ n; n) (dt((pn )?(P n)); n ) ? (K (PHn )U ~ nH (P n ? PHn ); n ) =(n; n) ? (un ? un ; n ) + (K 0(PHn )U ? (K (pn)u~n ; n): (5.83) Expanding K (pn ) gives, 00 n K (pn ) = K (PHn ) + K 0(PHn )(pn ? PHn ) + K (2 ) (pn ? PHn )2: Rewriting (5.83) and using this expansion gives, (dt ((pn )?(P n )); pn ? P n) + (K (PHn )n ; n ) =(n; n ) + (un ? un; n) ? (K (pn )(u~^ n ? u~n); n) ~ nH ; n) + (K (PHn )(pn ? PHn )(u~^ n ? U~ nH ); n) + (( n + (pn ? p^n ))K 0(PHn )U n + 21 ((pn ? PHn )2K 00(n )u~^ ; n ) + (dt((pn ) ? (P n )); pn ? p^n ): (5.84) By Lemma 4.3, we have, (dt ((pn )?(P n )); pn ? P n) Z Z pn dt n ((}) ? (P n ))d}dx
P
? C f(pn ? P n )2 + (pn?1 ? P n?1)2 + (tn)2g:
(5.85)
We bound the time discretization term as in section 4.2, 2 (5.86) knk k @@t2 kL2(tn?1;tn);L2)(tn)1=2: Combining equations (5.84)-(5.86) and applying the Cauchy-Schwarz and the arithmetic-geometric mean inequalities, we have, ! Z Z pn n dt n ((}) ? (P ))d} dx + (K (PHn )n; n)
P
C fkpn ? P n k2 + kpn?1 ? P n?1 k2g + C f(t)2 + k@ttk2L2 (tn?1;tn);L2)tn + k nk2g + (dt((pn ) ? (P n )); pn ? p^n ) + C kun ? unk2 + kn k2 + C ku~^ n ? u~nk2 ~ nH ? u~n kL1 + ku~n kL1 ) + C k(pn ? p^n )U~ nH k2 + C kn kk nk(kU ~ nH )k2 + C k(pn ? PHn )2u~^ nk2: + C k(pn ? PHn )(u~^ ? U (5.87)
70 Let N^ be the time index where maxn kpn ? P n k occurs. Multiply (5.87) by tn and sum on n = 1; : : : ; M , where M N and M N^ . The rst left-hand side term becomes, ! Z Z pn M X n n t dt n ((}) ? (P ))d} dx P
n=1 ! ! Z Z p0 Z Z pM 0 M =? ((}) ? (P ))d} dx + P M ((}) ? (P ))d} dx
P0
? C kp0 ? P 0k2 + C kpM ? P M k2;
(5.88)
where we have used Lemma 4.3. By summation by parts, M X (dt((pn ) ? (P n)); pn ? p^n )tn n=1
=?
MX ?1 n=1
((pn ) ? (P n ); dt(pn+1 ? P n+1 ))tn
? ((p0 ) ? (P 0); p1 ? p^1 ) + ((pM ) ? (P M ); pM ? p^M );
(5.89)
where we have used assumption (3.32) of a quasi-uniform time discretization. The rst right-hand side term of equation (5.87) is bounded by, M X fkpn ? P n k2 + kpn?1 ? P n?1k2gtn n=1
2
So, noting that
M X
kpn ? P n k2tn + kp0 ? P 0k2t:
n=1 k n k kpn ? p^n k + kpn ? P n k,
we have,
M X
kpM ? P M k2+ (K (PHn )n; n)tn n=1 M X
C
n=1
kpn ? P nk2tn + kp0 ? P 0k2 + C (t)2
M X n n 2 n + C kp^ ? p k t + knk2tn n=1 n=1 M X + C k(pn ) ? (P n)kkdt(pn+1 ? p^n+1 )ktn n=1 M X
(5.90)
71 +C
m X n=1
fkun ? unk2 + ku~^ n ? u~k2gtn +
4 X
i=1
Ti:
(5.91)
We bound T1 by. M X T1 =C kn kk nk(kU~ nH ? u~n kL1 + ku~n kL1 )tn
C
n=1 M X
n=1
+C
kn kkpn ? P n k(kU~ nH ? u~n kL1 + ku~n kL1 )tn
M X
n=1
C kpN^
kn kkp^n ? pn k(kU~ nH ? u~nkL1 + ku~nkL1 )tn
? P N^ k
M X n=1
+ C kp^ ? pkL1 (L2) +C
M X n=1
M X n=1
+ C kp^ ? pkL1 (L2) M X n=1
n=1
kpn ? P n k2tn + C
C kpN^ ? P N^ k
+C
!1=2 X !1=2 X M M n n 2 n ~ kUH ? u~ kL1 t + knk2tn n=1 n=1 ! ! 1=2 M M X ~ n n 2 n 1=2 X n 2 n kUH ? u~ kL1 t k k t
knk2tn
knk2tn
M X
n=1
kpn ? p^n k2tn
n=1 !1=2
(H ?d=2(H k+1 + t)) !1=2 M X n 2 n k k t (H ?d=2(H k+1 + t))
n=1
kpn ? P n k2tn + C
M X n=1
kpn ? p^n k2tn;
(5.92)
where we have used the inverse assumption Theorem 3.1 and Theorem 4.2. Now, !1=2 M ^ X ^ n 2 n N N (H ?d=2(H k+1 + t)) kp ? P k k k t n=1 ! M 1 ^ ^ 2 1 X n 2 n N N 2 kp ? P k + 2 k k t (H ?d=2(H k+1 + t))2: (5.93) n=1
We choose H and t such that,
H ?d=2(H k+1 + t) K1=2:
(5.94)
72 Note that in two dimensions, this requires k 0 and in three dimensions, we must have k 1. Then, the second right-hand side term in (5.93) is bounded by, M K X n 2 n (5.95) 2 n=1 k k t : Similarly, !1=2 M X kp^ ? pkL1 (L2) knk2tn (H ?d=2(H k+1 + t)) n=1
M X 1 K 2 2 kp^ ? pkL1 (L2) + 2 knk2tn: n=1
(5.96)
Thus, T1 is bounded by, M M M X X X T1 C kpn ? P n k2tn + C kpn ? p^n k2tn + K2 kn k2tn: (5.97) n=1
n=1
The term T2 is bounded by, M X T2 =C k(pn ? p^n )U~ nH k2tn
C
n=1 M X
n=1
n=1
~ nH ? u~nk2)tn (kpn ? p^n k2ku~n k21 + kpn ? p^n k21kU
Ch2(k+1) + C and T3 by,
T3 =C
C C
M X n=1 M X n=1 M X n=1
M X n=1
h2(k+1)kU~ nH ? u~nk2tn;
k(pn ? PHn )(u~^ n ? U~ nH )k2tn kpn ? PHn k21ku~^ n ? U~ nH k2tn
~ nH k2tn; (H ?d=2(H k+1 + t))2ku~^ n ? U
where we have again used the inverse assumption. Lastly, M X T4 =C k(pn ? PHn )u~^ n k2tn n=1
(5.98)
(5.99)
73
C C
M X n=1 M X n=1
kpn ? PHn k21kpn ? PHn k2ku~^ nk21 tn (H ?d=2(H k+1 + t))2(H k+1 + t)2tn:
(5.100)
Thus, again using Theorem 4.2 we have for these last terms, 4 X jTij C (h2(k+1) + (H ?d=2(H k+1 + t))2(H k+1 + t)2 i=2
+C
M X n=1
kpn ? P n k2tn + C
M X n=1
kpn ? p^n k2tn:
(5.101)
Now, let M = N^ . Applying approximation properties, Gronwall's Lemma 3.1 and taking P 0 = p^0 in (5.91) gives,
kpN^
N^ ^ 2 X N ? P k + (K (PHn )n ; n )tn n=1 C ft2 + H 2(k+1) + h2(k+1) + H ?d H 4(k+1)g:
(5.102)
So, kpN^ ? P N^ k C ft + hk+1 + H ?d=2+2(k+1)g. Combining these equations and taking M = N in (5.91) gives, !1=2 N X n 1 = 2 n 2 N N kp ? P k+ kK (PH ) k t n=1 C (hk+1 + t + H 2k+2?d=2):
The triangle inequality nishes the result.
(5.103)
74
Chapter 6 Implementation and Numerical Results In this chapter the implementation of a C++ three-dimensional Parallel Richards' EQuation Solve code, PREQS, is described. We rst discuss the equation formulation and discretization scheme used. Then, a brief discussion of the nonlinear and linear discrete system solution techniques is given. Lastly, some numerical test cases are presented along with results.
6.1 Implementation Issues In order to achieve parallelism, the parallellepiped domain, , is decomposed into a set of smaller parallelepiped subdomains, denoted by i, so that = [i i. The data corresponding to the cells in each of these subdomains is stored on a single processor. Let ?i = @ i and ?Ii = ?i n ?. Thus, ?Ii is the part of the boundary of subdomain i that is contained in . As shown in Section 5.1, the expanded mixed nite element method with certain quadrature rules simpli es to a cell-centered nite dierence scheme with a 19 point stencil. This stencil is shown in Figure 6.1. In order to implement this nite dierence scheme in parallel, each processor communicates with up to 18 neighbors. To reduce this requirement, we add extra unknowns along the interfaces between subdomains. Adding these unknowns allows for the normal uxes at the interface points to be discontinuous, which is a nonphysical condition. Thus, extra equations which enforce continuity of normal uxes at the interfaces are also introduced. Adding these extra unknowns corresponds to adding a single hydraulic head at the interface points. This value will be \owned" by one processor and communicated to the \non-owner" after updates. Formulating a mixed method with these extra unknowns along interfaces between subdomains was rst done by Glowinski and Wheeler [35] in the context of linear elliptic equations. They used these unknowns to formulate domain decomposition schemes for the mixed method.
75
Figure 6.1 The 19-point discretization stencil. General conditions of the form,
u n + p = g; are considered for the external boundaries. The functions and are assumed to be functions of position only. The data g is a function of both time and position. To de ne the numerical scheme, we de ne a new discrete vector space, V^ h . Let Vi = Vh j i . Then take, V^ h = L Vi. The numerical scheme is de ned as nding ~ n; Un ; Ln) 2 (Wh ; V^ h; V^ h ; h) at each time step n = 1; : : : ; N satisfying, (P n ; U (dt (P n); w) + (r Un; w) = (f n ; w); (U~ n ; v) i;TM = (P n ; r v) i ? (Ln ; v n)?i ; ~ n; v) i;T ; (Un ; v) i;TM = (K (P n )U (Un n; )?i\? = (g + Ln ; )M;?i\?; X n (U n; )?Ii = 0: i
(6.1) (6.2) (6.3) (6.4) (6.5)
The extra unknowns provide a boundary condition for internal interfaces. Thus, the subdomains are coupled only through shared boundaries, and each subdomain will only need to communicate with neighbors sharing interfaces. Hence, subdomains communicate with up to 6 neighbors and not 18.
76 The main disadvantage of this approach is that as the number of subdomains increases, so does the number of extra unknowns. Thus, the algorithm changes as more processors are added, and parallel speedup, in the traditional sense, will be non-optimal. In calculating ow velocities, the technique of one point upstream weighting [46] is employed. For this technique, the relative permeability and its derivative at interfaces are approximated by the value at the cell one point upstream. The upstream point is de ned in the case of a full permeability tensor and cell-centered nite dierences by [21], 8 < k (P ); if k(x)U~ i+1=2 0, (6.6) krw (Pi+1=2) = : rw i+1 krw (Pi ); otherwise. Forsyth and Kropinski [32] and Sammon [61] have shown that upstream weighting is necessary in order to accurately track the uid front. However, by Taylor's Theorem the truncation error associated with upstream weighting is O(h). Thus, we expect this approximation to be only O(h). We use an inexact Newton method [23] with backtracking line search globalization [24, 27] to solve the discrete nonlinear problem. In this method the Jacobian system is solved inexactly, most often with an iterative method. The GMRES Krylov subspace method [60] preconditioned with a Jacobi preconditioner is used in the code. The linear system tolerances are chosen using an algorithm of Eisenstat and Walker [28] which prevents oversolving of the system. Far from the solution, the nonlinear function F and its local linear approximation may disagree signi cantly. In this case, forcing the linear solution to be very accurate may lead to a step which provides little or no progress toward a solution. Furthermore, solving the linear system to a high level of accuracy can be very costly. Eisenstat and Walker give a variety of choices for tolerances [28]. One choice is, m m?1 J (P m?1 )P mk j ; m = j kF (P )k ? kFk(FP(P m)?+ 1)k
where P m is the current Newton iterate. This choice re ects the agreement between the function and its linear model at the previous step. Eisenstat and Walker have shown that for this choice of m, once the iterates are close enough to the solution, the inexact Newton method shows two-step quadratic convergence. This method has been eectively implemented in a two-phase ow code where the compute time decreased signi cantly with this choice for the linear system tolerances [21].
77 The PREQS code uses the kScript scripting language of Keenan [42] in order to provide a exible user interface. The kScript command generation program cmdGen [41] was used to de ne kScript commands and variables. As a result, code input can be set in any units and in any order. Furthermore, Keenan has developed extensive array and vector C++ classes that were used throughout the PREQS code.
6.2 Numerical Results In this section, results are given for the PREQS code applied to various test problems. A three-dimensional nonlinear heat equation with a known solution is considered rst. Then, one-dimensional and three-dimensional partially saturated ow problems are discussed.
6.2.1 A Known Solution Test Case In order to verify the asymptotic rate of convergence, a test problem with a known closed form analytical solution is considered. For this case, and kr are both taken to be p, giving the equation, @p ? r (pkrp) = f; (6.7) @t where f is chosen so that p = xyzt + 1. The problem domain is taken to be the unit cube and, 3 2 1 : 0 0 : 1 0 : 1 7 6 k = 64 0:1 1:0 0:1 75 : 0:1 0:1 1:0 Dirichlet boundary conditions are taken on all sides of the domain and the initial condition is 1 everywhere. The nonlinear iteration tolerance was set to 10?9 , and all problems were solved on a processor mesh of 2 2 1. Time steps of 0:001 were taken, and the discrete L2 error measured at the nal time, 0:1. The time step was taken to ensure that t h2 for all h considered. Table 6.1 gives the discrete L2 error for various mesh sizes. Let Ei be the error after solving on a grid of size hi. Then, Ei = Chri is the discrete L2 error at the last time step. Taking logs results in, (log hi)r + log C = log Ei:
(6.8)
78 Writing this for each data pair (Ei; hi) gives an overdetermined set of equations for the convergence rate r and the log of the constant C . Using Matlab to solve this system, we arrive at the solution r = 1:9; log C = ?6:3411. This line, along with the ve data points, is plotted in Figure 6.2.
Table 6.1 Convergence results for a three dimensional analytic test problem. Grid L2 error 222 4:2947 10?4 444 1:3104 10?4 888 3:4952 10?5 16 16 16 8:8112 10?6 32 32 32 2:1521 10?6
Log−Log Plot of Convergence Data and Best Fit Line
−3
10
−4
10
Log of Error
slope = 1.9
−5
10
−6
10
−2
10
−1
10 Log of Mesh Size
0
10
Figure 6.2 Linear plot of convergence data. The analysis in Chapter 4 indicates that the expanded mixed method with the lowest order space should give at least O(h) spatial convergence. However, the above results indicate that it may be possible to prove better than O(h). Furthermore, the code uses a nite dierence scheme based on superconvergent points similar to the
79 coarse grid scheme proven to be O(h2) for the nonlinear heat equation in Chapter 5. The method used in the PREQS code employs one-point upstream weighting which is O(h) instead of the O(h2) bilinear interpolation. However, for problems where no steep fronts are present, it is not surprising to see better than O(h) convergence even though upstream weighting is used.
6.2.2 A One-Dimensional Flow Problem The next test problem we consider was reported by Celia, Bouloutas and Zarba [18] and is a one-dimensional ow problem. This problem was implemented in order to verify conservation of mass and to examine the eect of lengthening time steps. Physical properties for this test case are given in Table 6.2. The water con-
Table 6.2 Physical data for the one-dimensional ow problem. Domain 1 cm 1 cm 60 cm n m s r kxx; kyy ; kzz kxy ; kxz ; kyz Density Viscosity Porosity
0.0335 2 0.5 0.368 0.102 1:0568 10?4 cm2 0 1gm=cm3 1:124cP 0.368
tent and relative permeability are given by the van Genuchten curves, (2.2) and (2.7). Initial and boundary conditions for pressure head were as follows, h(z; 0) = ?1000cm; h(0; t) = ?1000cm and h(60cm; t) = ?75cm. No ow boundary conditions were taken on the four remaining boundaries. The depth direction was divided into cells of width 2:5cm, and a single processor was used for all results with this test case. Let W denote the time change of water content in the domain over the time of simulation, and let F denote the water mass entering the domain over the time of simulation. Then, N X n ? n?1 X i tnz = X(N ? 0 )z; i W i i n t n=1 i i
80
! N N X X X n n (ui+1=2 ? ui?1=2) ni tn = (un0 n0 ? unB nB )tn; F n=1
i
n=1
where unB n is the ux on the z = 60cm boundary. The mass balance ratio is given by, (6.9) MB W F: If this ratio is unity, the numerical method exactly conserves mass. Celia, et.al. show the mass balance ratio for a head-based form of Richards' equation solved with both a Galerkin nite element method and a nite dierence method. Both schemes show large degradations of the mass balance as the time step increases. For steps of 1 minute, the nite dierence scheme gives a ratio close to 1, but as t goes to 60 minutes, the ratio drops to 0:6. The nite element scheme gives even worse results. However, with the mixed form of Richards' equation, mass should be conserved and the ratio should be close to unity. This test problem was solved with the PREQS code for one simulation day with various time steps ranging from 1 minute to 60 minutes. In all cases, the above mass balance ratio was always unity. Thus, no water was arti cially created or destroyed by the numerical method. Figure 6.3 shows the approximate solutions for four dierent time steps. These solutions are almost identical indicating that the mixed formulation of Richards' equation used in this work prevents degradation in results due to time step increases, unlike h-based forms which degrade quickly with step increases. This degradation can be dramatic and is documented in Celia, et.al. [18]. Figure 6.4 shows approximate solutions after 0.5 simulation days for a xed time step of 15 minutes and varying spatial steps. The solution is converging to a sharper and sharper front indicating that as the grid is re ned, the solution improves.
6.2.3 A Three-Dimensional Irregular Geometry Flow Problem The last problem we consider is a three-dimensional problem over an irregular geometry domain. For this case, the domain can be mapped to a rectangular domain of 100cm 100cm 20cm with a C 2 map, F . The theory of Arbogast, Wheeler and Yotov [7] discusses the transformation of the original problem to one over the rectangular computational domain. Speci cally, if the mapping is C 2, then the problem can
81 Solutions for Varying Time Steps 0 −100 −200 dz = 2.5cm T_f = 1 day
−400 −500 −600 −700
dt = 3min
−800
dt = 10min −900
dt = 30min dt = 60min
−1000 0
10
20
30 Depth (cm)
40
50
60
Figure 6.3 Solutions for the one-dimensional test problem with various time step sizes.
Solutions for Varying Mesh Sizes 0 −100 dt = 15 min
−200
T_f = 0.5 day
−300
Hydraulic Head (cm)
Hydraulic Head (cm)
−300
−400 −500 −600 −700 −800
dz = 0.125cm dz = 0.25cm dz = 0.5cm
−900
dz = 1cm dz = 2cm
−1000 0
10
20
30 Depth (cm)
40
50
Figure 6.4 Solutions for the one-dimensional test problem with various mesh sizes.
60
82 be transformed into an equivalent problem over a rectangular domain which has a convergent solution. The permeability tensor, K, is transformed by,
K = JDF ?1K(DF ?1)T ;
(6.10)
where DF is the Jacobian of the map, F , and J is the determinant of DF . The resulting permeability tensor will be full even in the case that the original is diagonal. Furthermore, the time derivative term is multiplied by J . Applying these transformations allows computation over a regular grid. Physical properties for the original case are given in Table 6.3. The water content
Table 6.3 Physical data for the three-dimensional ow problem. Domain 100 cm 100 cm 20 cm n m s r kxx; kyy ; kzz ; z 10cm kxy ; kxz ; kyz ; z 10cm kxx; kyy ; kzz ; z > 10cm kxy ; kxz ; kyz ; z > 10cm Density Viscosity Porosity
0.0334 2 0.5 0.361 0.102 9:33 10?12 m2 9:33 10?13 m2 9:33 10?10 m2 9:33 10?11 m2 1gm=cm3 1:124cP 0.368
and relative permeability are again given by the van Genuchten curves, (2.2) and (2.7). The computational grid was 20 20 10 divided uniformly over a 2 2 1 processor mesh. As seen in Table 6.3, the domain has two horizontal layers. The top layer has a much higher permeability than the lower. No ow boundary conditions were taken on all boundaries except the top and bottom. On the top face, the x 2 (0; 20cm); y 2 (0; 20cm) section had a hydraulic head boundary condition of -50 cm and no ow conditions everywhere else. On the bottom face, the x 2 (80cm; 100cm); y 2 (0; 20cm) section had a hydraulic head condition of -1000 cm and no ow everywhere else. These conditions eectively placed a source at the top left section of the domain and a sink at the bottom right section.
83 Figure 6.5 shows results for the regular domain in the case that the entire domain has the higher permeabilities given in Table 6.3 after 45 simulation days. The in ltrating water has advanced radially outward from the injection part of the upper boundary and has been pulled downward by gravity. The water has reached the sink boundary condition and has started owing out of the domain. Figure 6.6 shows the regular domain for the two permeability layer case given in Table 6.3 after 50 simulation days. Here, we see that the water does not easily ow into the low permeability region. The water must accumulate enough weight in order to push into this region. Furthermore, the low hydraulic head condition is not felt by the water since it has not yet gotten to that part of the boundary. Hydraulic Head After 45 Days
−729 −661 −593 −525 −457 −389 −322 −254 −186 −118 −50
Figure 6.5 Three-dimensional single permeability layer test case after 45 simulation days.
Lastly, Figures 6.7-6.10 show the hydraulic head after 5, 20, 50 and 75 simulation days for the irregular geometry test case. As can be seen from the gures, water starts owing at the top left of the domain. As time passes, it begins drifting toward the right while gravity pulls it downward. However, when the water reaches the lower permeability region, it must accumulate enough pressure to go further. Instead of going straight down, it pools along the interface between the two regions. As enough pressure builds underneath the injection area, the water is pushed downward and
84 Hydraulic Head After 50 Simulation Days
−772 −699 −627 −555 −483 −411 −338 −266 −194 −122 −50
Figure 6.6 Three-dimensional two permeability layer test case after 50 simulation days.
begins to accumulate at the bottom of the domain. Around 20 days we can see the preference of the water to ow toward the rightmost part of the domain. This is due to the shape of the domain as it is twisted at the oor. Thus, the downward direction is toward the lower front of the domain. By 75 days, most of the upper region has lled with water and the water is starting to ow into the lower region. The eect of the two layers is clearly seen. Furthermore, it is clear that domain shapes have a dramatic eect on the ow and should not be modeled by simple rectangles. In conclusion, we have shown that the PREQS code gives O(h2) accuracy on a model problem with a known solution. Furthermore, the code maintains perfect mass balance and is robust for large time steps. Lastly, the code predicts expected solutions to three-dimensional groundwater problems with full tensor coecients.
85 Hydraulic Head After 5 Simulation Days
−735 −666 −598 −529 −461 −393 −324 −256 −187 −119 −51
Figure 6.7 Three-dimensional irregular
geometry test case after 5 simulation days. Hydraulic Head After 20 Simulation Days
−736 −667 −598 −530 −461 −393 −324 −256 −187 −119 −50
Figure 6.8 Three-dimensional irregular
geometry test case after 20 simulation days.
86 Hydraulic Head After 50 Simulation Days
−736 −667 −598 −530 −461 −393 −323 −256 −187 −119 −50
Figure 6.9 Three-dimensional irregular
geometry test case after 50 simulation days. Hydraulic Head After 75 Simulation Days
−736 −667 −598 −530 −461 −393 −324 −256 −187 −119 −50
Figure 6.10 Three-dimensional irregular
geometry test case after 75 simulation days.
87
Chapter 7 Conclusions 7.1 Summary In this thesis we have analyzed and solved numerical methods for simulating variably saturated subsurface ow. A brief summary of existing literature in this area was presented along with the physical situation that is modeled. We developed and analyzed a number of expanded mixed nite element methods applied to Richards' equation. Optimal convergence was shown for a discrete time scheme applied to the strictly variably saturated case. Optimal convergence was also shown for a nonlinear form in the case of partially to fully saturated ow. The nonlinear form is bounded below by the error in water content and above by the error in hydraulic head. Convergence rates in terms of a Holder continuity rate were shown for the error in water content for this case. Lastly, convergence rates in terms of approximation error were shown for the unsaturated to fully saturated ow case with a continuous time scheme. This scheme was de ned using the Kirchho transform in order to put the equation into a more easily analyzable form. A method for handling nonlinearities was examined where the nonlinear problem is solved on a coarse grid and the problem on the ne grid linearized about the coarse grid solution. This technique is dicult to apply to Richards' equation due to the need to maintain mass balance. Thus, a scheme where the time derivative term on the ne grid was left nonlinear was analyzed. Numerical results were given for a three-dimensional parallel Richards' equation solve code, PREQS. The code exhibited superconvergence on a model problem as well as robustness for large time steps. Mass balance is maintained exactly in the code. Results from a full tensor three-dimensional irregular geometry test case were shown and expected behavior was produced.
88
7.2 Future Work Despite the above results, there is still plenty of work to be done in the area of simulating partially saturated subsurface ow. The PREQS code uses only a Jacobi preconditioner and can be quite slow for long simulations where the linear systems get dicult to solve. Thus, future work will include nding robust preconditioners for the linear nonsymmetric Jacobian systems. Furthermore, the dynamic linear system tolerances described in Chapter 6 can signi cantly slow convergence of the Newton method. Further work will try to identify ways of making the forcing term selection more robust. Richards' equation has been criticized as a simplistic model for expressing the two-phase ow of water and air [36]. However, when applicable, it is much faster to solve Richards' equation than a full two-phase model such as described in [21]. An interesting area for future work would be to closely examine under which numerical and physical conditions the solution of Richards' equation fails to match that of the full two-phase ow problem.
89
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