Symmetry and M-matrix issues for the O-method on an Unstructured Grid

Transcription

1 Symmetry and M-matrix issues for the O-method on an Unstructured Grid G. T. Eigestad Department of Mathematics, University of Bergen, Norway I.Aavatsmark Norsk Hydro Research Centre, Bergen, Norway M.Espedal Department of Mathematics, University of Bergen, Norway Abstract. More sophisticated discretization methods than the traditional controlvolume finite-difference methods, have been proposed by Aavatsmark et al. in recent papers for solving the mass balance equations for porous media flow. These methods are based on a local representation of fluxes across cell-edges of control volumes (CVs). This paper will focus on mathematical properties of the discrete operator that arises when an elliptic term of the form - (K p) is discretized based on these discretization principles. 1. Introduction As mathematical modeling of fluid flow in reservoirs has become more sophisticated, the need for discretization methods to handle flexible grids has arisen. One class of discretization methods, the so called multi point flux approximation methods (MPFA) [1]-[7], [8], [19], seems to be suitable when the flow equations are to be discretized on unstructured grids [4],[5],[7], [19]. These methods are finite volume based. Common for MPFA methods is that they yield local representations of fluxes across edges of control volumes, and allow for heterogeneity, anisotropy and general geometry. It is a well known fact that the traditional Galerkin finite element method is not locally mass conservative for media with discontinuous conductivity. This has lead to formulations where fluxes also are considered. Mixed finite element (MFE) methods [13], and versions of these [14], [15] solve for both pressures and fluxes. Adjoint divergence methods [18] are similar approaches. These methods do not have the property that fluxes have a local representation. Similarities between MPFA-methods and certain (control volume) mixed finite element methods have been investigated in [15]. A common approach in commercial reservoir simulation is to apply finite difference discretization, and assume diagonal permeability representation [17]. For general geometries, heterogeneities and/or anc 2002 Kluwer Academic Publishers. Printed in the Netherlands. main.tex; 12/06/2002; 12:41; p.1

2 2 isotropies, alternative discretization methods are needed for accurate modeling of the fluid flow. Convergence of MPFA methods has not been proven analytically, but some numerical tests have been performed by Jeannin et. al [16] and in [22]. The tests indicate that the convergence rates of pressure and flux are of the same order as for MFE methods when the grid refinement is uniform. Numerical tests have also been made in [8], where the convergence rates are similar to MFE. This paper extends the discussion of the O-method in [4] and [5], where the methods for discretization on unstructured grids were derived for general inhomogeneous, anisotropic media. We will examine properties such as symmetry, diagonal dominance and M-matrix [21] property for the discrete elliptic operator. It is a well known fact that symmetric M-matrices are positive definite (and have non-negative inverses). The M-matrix property has also a nice physical impact, as the scheme in this case satisfies the maximum principle. Although this paper does not discuss multi phase flow, this is important for stability when gas might be present in the oil phase. If nonphysical maxima or minima may exist, gas may come out from the oil phase where it should not be present. This may give rise to numerical instabilities. When the O-method is applied on a structured grid, and curvilinear coordinates are used, the coefficient matrix will be symmetric [3]. It will in general not be an M-matrix, though, but stable solutions will be obtained for a wider range of problems. In contrast, the coefficient matrix is in general not symmetric when the O-method is applied on unstructured grids. It is therefore of interest to examine how a grid should be constructed in order for symmetry and M-matrix property to be satisfied. Note that the starting point will not be a specific medium; we are merely interested in finding grids that yield symmetric coefficient matrices. When this is known, the results can be used in a practical gridding algorithm. A brief summary of the governing equations and the discretization principles is given in Section 2. We further illustrate how the discrete operator is obtained and formulate the symmetry problem. The homogeneous case is investigated, and it will be proven that symmetry always is satisfied for certain polygonal control volumes. Simple conditions can further be posed on a dual triangular grid in order to obtain M-matrix property. In Section 3, a symmetry algorithm is applied to construct synthetic grids for which symmetry of the discrete operator is satisfied. Section 4 investigates cases of heterogeneity and anisotropy that are associated with layered porous media. The element wise contributions to the coefficient matrix are studied. This is used to prove symmetry under certain conditions, and analyse local M-matrix property. main.tex; 12/06/2002; 12:41; p.2

3 3 2. Discretization principles The PDEs that describe fluid flow in porous media are originally derived on integral form. Discretization of the integral form is referred to as the Control Volume (CV) formulation. A fully implicit CV formulation for a general conservation law can be expressed as m n m n 1 + t n f n j = Qn. (1) Here t denotes the time step, n the time level, m is accumulated mass, and Q is a source term. The discussion below will regard one phase flow in porous media, where special interest will be given to the flux term f. Across interface j of a control volume, the flux at time level n will be given by fj n = (K u) n nds, (2) where u is the fluid potential (pressure) and n is the outer unit normal vector of interface j of the control volume. The permeability of the medium, K, is approximated to be constant on each CV, and is assumed to be a symmetric, positive definite tensor. The principal directions of the permeability are allowed to vary from control volume to control volume. In [1]-[3] and [6] the discretization principles for certain MPFAmethods on structured grids are outlined. Advantages of the methods are shown numerically. The nodes for the pressures are located within the control volumes. Pressures will be the only unknowns, but the discretization methods are derived based on flux constraints across cell edges. Hence, fluxes are embedded in the discrete set of pressure equations Interaction regions A brief summary of how the discrete pressure equations are derived for the O-method on polygonal grids is given here. The polygonal grid cells will in this case be control volumes. It is useful to use a triangular grid for constructing the polygonal grid, and this grid will be termed the dual triangular grid. A polygonal grid constructed from a triangular grid is depicted in Figure 1a. Each triangular grid cell will now be an interaction region. By our grid construction algorithm [4] three grid cells meet in the cluster centre, denoted x 0, which is also referred to as the triangle point. The edges that separate the grid cells within an interaction region, will be termed sub interfaces or half edges. A whole cell edge that separates main.tex; 12/06/2002; 12:41; p.3

4 4 x 3 x _ x n x 2 n _ x n a. Dual grids b. Interaction region Figure 1. a: Polygonal control volumes (bold lines) and dual triangular grid. b: Three polygonal grid cells meet in one triangular interaction region. Vertices of triangle, x i, i = 1, 3, are nodes of polygonal CVs. x i, i = 1,3, are dividing points. 3 0 _ x 2 x 2 two grid cells, will be fully enclosed in two neighbouring interaction regions. Dividing points are points for which half edges are connected across two neighbouring interaction regions. Details are found in [4]. Eq. (2) is the flux expression for the continuum model. In order to obtain discrete equations, the fluxes across the 3 sub interfaces of each interaction region must be approximated. A looping over all the interaction regions to construct the coefficient matrix can then be performed. This is further illustrated in sections 2.5 and 2.6. Each vertex of an interaction region is a node of a CV, and the idea behind the O-method is that the fluxes across sub interfaces are determined by the local interaction between the three nodes in this restricted area. An interaction region with relevant entities for a triangular interaction region is depicted in Figure 1b. For each sub interface, the flux based on (2) will be assumed to be a weighted sum of the potential values of the 3 vertices of the triangular interaction region: f i = t ij u j. (3) These weights are referred to as transmissibilities, and will be determined based on flux and potential constraints, see [1], [2], [4], and the brief discussion below. Index i refers to the local numbering of sub interfaces (corresponding to normal vectors in Figure 1b), whereas j is the local number of the node (vertex of the interaction region). Both i and j run from 1 to 3 in the counterclockwise direction. Since there are 3 sub interfaces contained in a triangular interaction region, 9 transmissibilities have to be determined for each interaction region. In matrix notation the 3 fluxes across the half edges can be expressed as f = Tu, (4) where f = [f 1, f 2,f 3 ] T and u = [u 1,u 2,u 3 ] T. The matrix T is a 3 3 matrix whose rows are the transmissibilities for each of the 3 half main.tex; 12/06/2002; 12:41; p.4

5 edges. Row number i of the transmissibility matrix corresponds to sub interface i. Since there is no flow across sub interfaces if the potentials have equal values in the interaction region, the row sums of T are zero: 3 t ij = 0, i = 1,2,3. (5) j= Derivation of transmissibilities A detailed derivation of the transmissibilities for sub interfaces for the O-method is given in [4]. The following briefly illustrates the principles of the O-method for triangular interaction regions. On each CV the fluid potential is assumed to vary linearly, and in a triangular interaction region 9 degrees of freedom are available for hinging together the potentials. The transmissibilities of (4) are derived from a physical point of view, and fluxes are required to be continuous across sub interfaces. The flux across a sub interface of a control volume is given by Eq. (2), and is a constant since the fluid potential is assumed to vary linearly. Hence, 3 degrees of freedom will be needed in order for flux continuity to be satisfied. Since the node values (vertices of interaction region) must be honoured, one is then only left with 3 degrees of freedom for potential continuity. In the setting of the O-method, relaxed potential continuity conditions are proposed, and potential is only assumed to be continuous at the dividing points. This closes the system of equations for determining the transmissibilities of sub interfaces of an interaction region. Note that in the discrete system of equations, only potential values at the grid nodes are used. Requiring potential continuity at dividing points have merely been done to evaluate discrete fluxes. For this grid type, the O-method is also referred to as the triangle method, since the local interaction between nodes in a triangle determines fluxes across each half edge Mathematical properties of the discrete operator The transmissibilities derived in [4],[5] depend only on geometry and permeability. Vectors involved in the transmissibilities are depicted in Fig. 2, and are used in the scalar coefficients, which we define: ω ± ik = nt i K kν ± k 2F k. (6) The vectors ν ± k and n i are normal vectors of half triangle edges and sub interfaces respectively. The numbering of the permeability tensors K k main.tex; 12/06/2002; 12:41; p.5

6 6 ν + 3 n 2 ν 3 ν 1 n 3 n 1 ν + 2 ν + 1 ν 2 Figure 2. Interaction region and vectors used in scalar-coefficients corresponds to the vertices of the interaction region, and F k is the area of the variation triangle x k x k x m(k). The quantity m(k) is a backward integer function given by m(k) = { k 1, for k > 1, 3, for k = 1. (7) Likewise, a forward integer function will be used: p(k) = { k + 1, for k < 3, 1, for k = 3. (8) The following 4 matrices will be used for analytical expressions of the transmissibilities. ω11 0 ω+ 11 A = ω 22 + ω 22 0, (9) 0 ω 33 + ω 33 B = ω 11 + ω ω 22 + ω 22 0, (10) 0 0 ω33 ω+ 33 ω C = 0 ω 23 + ω 23, ω31 0 ω+ 31 (11) 0 ω 12 + ω 12 0 D = 0 0 ω 23 + ω 23. ω 31 + ω (12) Using the flux continuity and relaxed potential continuity constraints, it was shown in [4] that one arrives with the following expressions for T in Eq. (4) main.tex; 12/06/2002; 12:41; p.6

7 T = A (A C) 1 (D B) + B, (13) or alternatively T = C (A C) 1 (D B) + D. (14) Analytical expressions for each of the elements of the transmissibility matrix will be complicated for the general case, but may easily be implemented for numerical purposes Boundaries For CVs touching upon the boundary, artificial grid cells will be included in the interaction regions. There will then be a no-flow boundary between the artificial cell and two active grid cells. This can be illustrated by assuming that for instance grid cell 3 in Figure 1b is inactive. The contour x 2 x 0 x 3 is then a no-flow boundary, and only transmissibilities associated with flux interface x 1 x 0 need to be found. Since cell 3 is inactive, this implies that t 13 = 0. The flux is then trivially given by a two-point flux molecule, and t 12 = t 11. In [4] it is shown that the transmissibilities obtained have the same character as a harmonic average The discrete operator The construction of the discrete elliptic operator by the O-method resembles similarities with the procedure of assembly of inner products for the finite element method [20], and is illustrated by the following discussion. The resulting coefficient-matrix arises when the integrated form of the flow term, Ω (K u)dω, is discretized. This term will be approximated as (K u )dω = Au. Ω The coefficient matrix A is n n, u = [u 1,u 2,...,u n ] T, and n is the number of control volumes the physical domain Ω has been divided into. Since the integrated form of the flow term is to be discussed, element i of the vector on the right hand side will represent the integral (K u )dω i = Ω i K u n ds i, S i where Ω i is CV number i, and S i is the (closed) contour of Ω i. The non-zero elements of row i of A multiplied by their associated potential values, represent the total outflux from control volume i. 7 main.tex; 12/06/2002; 12:41; p.7

8 8 In the following it is assumed that cell i is not located at the boundary. For one phase, incompressible flow, where no sources or sinks are present, the total outflow from cell i is equal to zero. Since this also has to be valid for the case of equal potentials in all grid cells, the line sum of row i is equal to zero: n a ij = 0. (15) j= Symmetry equations Our first task is to examine symmetry of the discretized operator. The continuous operator (K ) is a self adjoint operator, and the discrete analogue should ideally be symmetric, [18]. If symmetry is satisfied, it is often easier to show stability of a scheme. For notational purposes, a symmetry-function is defined for each pair of non-zero elements of the coefficient matrix: g ij = a ij a ji. (16) One way to construct a polygonal grid from a triangular grid (without necessarily honouring the media) is to use the geometric centre of the triangle as triangle point, and use midpoints of the triangle-edges as dividing points (see Sec 2.1). Numerical tests show that the coefficient matrix is in general not symmetric by this approach. If symmetry is not satisfied, a symmetry equation can be defined; g ij = 0, and a solution to this equation might be sought. The transmissibilities of (4) are functions of both the vertices, dividing points and the triangle point of an interaction region: t ij = t ij ( x 1, x 2, x 3,x 1,x 2,x 3,x 0 ). In addition, they are also dependent on the permeabilities of the grid cells, but since our starting point is to investigate synthetic grids, permeability values will be kept fixed even if the shape of the control volume changes. For each whole cell edge that is enclosed by two neighbouring interaction regions (see Figure 3), there will be one symmetry function/equation associated with it. The transmissibilities involved will then origin from these two interaction regions. In this discussion, the nonlinear symmetry equation will be solved with respect to the triangle point x 0, while the other points are fixed. This is further motivated by the proof of Theorem 1 below. main.tex; 12/06/2002; 12:41; p.8

9 9 Interaction region 1 xj xk F A B C D E xl xi Interaction region 2 Figure 3. The whole cell-edge that separates cell i and cell j, is fully enclosed by two neighbouring interaction regions. Positive directions for half edges indicated by the arrows. Figure 3 shows two neighbouring interaction regions in which the cell edge between cell i and cell j is fully enclosed. Cell numbers i,j,k and l are global numbers of the grid cells involved, and A,B,C,D,E and F are half edges that separate control volumes. The fluxes across each half edge are described by flux molecules for which all potential values in their respective interaction regions contribute. To examine how the elements a ij and a ji in the coefficient-matrix arise, we will look at how the potential values u i and u j contribute in the flow terms for cell j and cell i respectively. The flow term for cell i is approximated as follows: (K u )dv i f A + f B f C + f D +... V i Fluxes not listed in the above expression (fluxes from other interaction regions) do not involve the potential value of cell j. It then follows that a ij is some combination of four transmissibilities; two from each interaction region. The element a ji arises in the same manner by studying the flow term (K u )dv j f F f B + f C f E +... V j The symmetry function is a non linear function of x 0 which is a combination of up to 8 different transmissibilities. The non linearity is seen through the dependence of the scalar coefficients Homogeneous medium Permeabilities are defined to be constant on each polygonal control volume. For a homogeneous medium, permeabilities will be constant on both CVs and also within the interaction regions. Simple expressions main.tex; 12/06/2002; 12:41; p.9

10 10 for the transmissibilities can now be given. In [4] it is shown that the transmissibilities (for the three half-edges) are given by t ij = 1 2F nt i Kν p(j). (17) The index i is the local edge number, and j is the local node number. Both are numbered in the counter-clockwise direction as in Figure 1b. The quantity p(j) is the forward integer function introduced in (8), n i is the unit normal vectors of half edge i, and ν p(j) denotes outward normal vectors of the whole triangle edges. F is the area of the whole interaction region. The following can now be proved: Theorem 1 Suppose that the polygonal grid is constructed from a triangular mesh, and the dividing points are defined as x i = 1 2 (x i+x p(i) ). For a homogeneous medium, the coefficient matrix will then be symmetric. Proof: The symmetry equation for one cell edge arises from some combination of transmissibilities from two neighbouring interaction regions. Assume that we split the analysis, and look at each of the two interaction regions (that together contain the whole cell edge) separately. This means that we look at the contribution to elements a ij and a ji from one specific interaction region, and form a local symmetry function/equation for this region. We investigate the edge that separates cell i and cell j in Figure 4, and assume without loss of generality that they have local numbering 2 and 3 respectively. To distinguish the analysis from the case where the whole edge is analyzed, the local contribution to matrix elements a ij will be denoted m ij. The indexing ij now refers to the local numbering of the vertices in the interaction region; index i is the cell for which outflow/inflow is considered, and j is the cell that cell i interacts with. The local symmetry equation for the pair of elements will then be m 23 m 32 = (t 23 t 13 ) (t 32 t 22 ) = 1 2F (nt 2 nt 1 )Kν 1 1 2F (nt 3 nt 2 )Kν 3 = 1 2F [ηt 1 Kν 1 η T 2 Kν 3 ]. (18) The vectors η 1 = n 2 n 1 and η 2 = n 3 n 2 are sketched in Fig. 4a, and it is easily seen that η 1 = 1 2 ν 3, η 2 = 1 2 ν 1. (19) main.tex; 12/06/2002; 12:41; p.10

11 11 Hence, it follows that m 23 m 32 = 1 4F [νt 3 Kν 1 ν T 1 Kν 3] = 0, (20) because K is symmetric. The contribution from this interaction region to this specific pair of matrix elements of a global coefficient matrix, yields symmetry. The same will hold for the neighbouring interaction region, and symmetry is therefore achieved for this specific pair of elements. Since the same argument will hold for all cell-edges of a homogeneous medium, symmetry will be satisfied globally. x 3 x 1 ν 3 η 1 ν 2 η 2 C h31 h23 B ν 1 x 2 A h 12 a. Interaction region b. Decoupling of equations Figure 4. a: Interaction region with relevant vectors. Dividing points defined as midpoint of triangle edges. b: Interaction region covering three half edges. We will further examine the contribution from this interaction region to the off-diagonal element m 23 of the global coefficient matrix, given from Eq. 18: t 23 t 13 = 1 2F nt 2 Kν 1 1 2F nt 1 Kν 1 = 1 4F νt 3 Kν 1. (21) This expression is independent of the position of the triangle point, such that the question regarding M-matrix property is independent of the positioning of x 0. If the medium is isotropic, it is then sufficient to consider the scalar product (ν 3,ν 1 ), which is negative if the angle associated with the corner x 1 is less than π/2. All other local contributions to the global coefficient matrix will arise in the same manner, and it is then seen that if all angles are less than π/2, all contributions to off-diagonal terms are non-positive. The matrix is then irreducibly diagonally dominant and symmetric, and it will follow that the matrix is an M-matrix, see [21]. It is a well known fact that symmetric M-matrices are positive definite. The M-matrix property of the coefficient matrix is the property that insures that the discrete set of equations satisfies a monotonicity principle, while the positive definiteness implies stability of the methods. The requirement main.tex; 12/06/2002; 12:41; p.11

12 12 that all angles of the interaction regions are less than π/2 can, however, be relaxed. A sufficient condition for the matrix to be an M-matrix is that the dual grid (the grid consisting of the triangular interaction regions) is a Delaunay grid [4] since contributions from neighbouring interaction regions can compensate for wrong signs. A Delaunay grid is a grid for which the sum of two opposite angles of a triangle edge is less than or equal to π. The same constraints apply for the control volume finite element method [11], [12] since this method gives the same coefficient matrix as for the O-method for the homogeneous case. If the physical domain has a uniform anisotropy (still homogeneous medium), this will correspond to a uniform stretching of the medium. The above restrictions on the dual grid must then apply to the grid in a reference system for which the medium is isotropic. From the proof of symmetry for the homogeneous case, it was seen that a decoupling of symmetry equations could be done. In the following subsection, the general case is discussed, and it is shown that a complete decoupling of symmetry equations always is possible Decoupling of symmetry equations A whole cell edge consists of two half edges from two neighbouring interaction regions, as seen from Fig. 3. The symmetry function (16) is built up by transmissibilities for half edges from two neighbouring interaction regions. The symmetry function for a whole cell edge can be split into two half edge symmetry functions, denoted h ij,1 and h ij,2. The relationship between the symmetry function for the whole edge, g ij, and the half edge functions h ij is formally given by g ij = h ij,1 + h ij,2. (22) Indices 1 and 2 refer to the two neighbouring interaction regions. Each h ij is given by h ij = m ij m ji, where m ij and m ji denote contributions from this interaction region to matrix elements a ij and a ji respectively. Each half edge symmetry equation, h ij = 0, can then be solved. When both equations are satisfied, the symmetry equation for the whole edge is satisfied. For every interaction region where three active cells interact, three half edges are contained herein. When the symmetry equation for one of the half edges is solved, it could be the case that the symmetry equations for the remaining two half edges in the same interaction region were not satisfied. The next theorem shows that if the local symmetry equation is satisfied for one half edge, then all the local symmetry equations are satisfied: main.tex; 12/06/2002; 12:41; p.12

13 Theorem 2 The sum of half-edge symmetry functions, h ij, for each cell in an interaction region is equal to zero: Cell 1: h 12 + h 13 = 0. (23) Cell 2: h 21 + h 23 = 0. (24) Cell 3: h 31 + h 32 = 0. (25) Moreover, if the associated symmetry equation is solved for one of the half edges, the symmetry equations for the remaining two half edges are automatically satisfied. Proof: Half edge symmetry functions will be denoted by h ij, i,j = 1,3. The indices correspond to the local numbering of vertices in the interaction region; i is the cell for which outflow/inflow is considered, and j is the cell with which it interacts. These functions have positive directions as indicated by Fig. 4b. It is trivial that h ij = h ji, so that there will only be one equation to solve for each half edge. We only proof the first identity above, as the proof of the other two are identical. Consider the expressions for the net out-flux for the parts of the cells that are contained in the interaction region depicted in Fig. 4b. The flux edges have been labeled A, B and C, and fluxes are positive in the counter-clockwise direction: Cell 1: (K 1 u)dv 1 = f A f C. V 1 Cell 2: (K 2 u)dv 2 = f B f A. V 2 Cell 3: (K 3 u)dv 3 = f C f B. V 3 The definition of the symmetry functions yields h 12 + h 13 = (m 12 m 21 ) + (m 13 m 31 ). (26) 13 main.tex; 12/06/2002; 12:41; p.13

14 14 The local contributions are m 12 = (t A2 t C2 ), m 21 = (t B1 t A1 ), m 13 = (t A3 t C3 ), m 31 = (t C1 t B1 ). Eq. (26) then becomes: h 12 + h 13 = (t A1 + t A2 + t A3 ) (t C1 + t C2 + t C3 ) + (t B1 t B1 ).(27) Using (5), the desired result is obtained. To prove the rest of the theorem, we use the fact that h ij = h ji, and rewrite (23), (24) and (25) as: h 12 + h 13 = 0, h 12 + h 23 = 0, h 13 h 23 = 0. From these three equations it is readily verified that if one of the half edge symmetry equations h 12 = 0, h 13 = 0 or h 23 = 0 is satisfied, then all of them are satisfied. Hence, it is sufficient to solve the symmetry equation for one of the three half edges in order for all three equations to be satisfied. Having obtained this result, a looping over all interaction regions can be done, and the symmetry equation associated with one of the edges in each of the regions can be solved with respect to the triangle point. This solution procedure may be applied to any polygonal grid, and synthetic test cases are examined in the next section. The symmetry equation to solve then only depends on the triangle point contained in the interaction region h i = h i (x 0i ). (28) For every interaction region there is only one equation to solve, whereas two degrees of freedom are available. This indicates flexibility for obtaining a symmetric operator. In the numerical examples below, the x coordinate of the triangle points will be fixed, and the half edge symmetry equation for each interaction region is solved with respect to the y coordinate. Because there are more degrees of freedom than equations to solve, this allows for the formulation of an optimization problem instead. One could for instance solve the symmetry problem with respect to the triangle point with the constraint that the triangle point deviates from some chosen point or line in an optimal way. This issue has not been pursued in this paper. main.tex; 12/06/2002; 12:41; p.14

15 15 NS Figure 5. Possible path from half-edge NS to a boundary half-edge may be paved. The question whether the solutions of the local symmetry problem are the only solutions for symmetry of the discrete operator have been omitted so far. As it turns out, local symmetry solutions are the only solutions: Lemma If the local symmetry equations are not satisfied, global symmetry cannot be achieved. Proof: We first show an intuitive property for an interaction region in which only two active grid cells interact. As was shown in Sec. 2.4, the flux across the only non-zero sub interface in such an interaction region will be given by a two point flux molecule. Assume that this edge is the edge that separates active cells i and j, and that the local numbering is such that j = p(i), as defined by Eq. (8). The local contribution to the elements of the global coefficient matrix with (global) numbering corresponding to the cells in this interaction region, will be denoted m ij and m ji respectively. Then m i,p(i) = t i,p(i), and m p(i),i = t ii. Since t ii = t i,p(i) when this edge is the only non-zero sub interface, local symmetry is satisfied for this interaction region. We now proceed to considering the gridding of a domain as in Fig. 5. Assume that local symmetry is not satisfied for the inner half edge labelled NS, and that the local deviation from symmetry is d NS. From Eq. (22) it is seen that global symmetry for the whole cell edge that NS is part of, could be satisfied if the location of the triangle point in the neighbouring interaction region compensated for the deviation from non symmetry for NS. Different paths of half edges can then be paved such that it connects NS with an interaction region in which only two active grid cells interact, and where local symmetry is satisfied. For all the half edges which the connection path consists of, the local deviation from symmetry is ±d NS. Hence, the second to last half edge of the path will yield non symmetry for a pair of elements in the global coefficient matrix. Global symmetry of the coefficient matrix can therefore not be satisfied unless all the local deviations from symmetry for each half edge are zero. This concludes the proof. main.tex; 12/06/2002; 12:41; p.15

16 16 3. Synthetic symmetry examples Based on the local symmetry equations, an algorithm for obtaining a symmetric coefficient matrix has been implemented in a triangular grid generator. Our polygonal control volumes are further constructed from the triangular grids, where the triangular grid cells correspond to interaction regions of the polygonal grid. Midpoints of triangle edges are used as dividing points (see Sec. 2.1), whereas the triangle points of interaction regions are chosen as solutions of local symmetry equations. As was shown in Sec. 2.7, triangle points may have an arbitrary position in interaction regions where all interacting cells have the same permeability. For the general heterogeneous case, a local symmetry equation needs to be solved in each interaction region where local heterogeneity occurs. Different cases will be categorized according to the local heterogeneity of the interaction region. For the anisotropic test cases we have limited the numerical experiments to cases where the principal directions of the permeability tensors are the same for all cells that interact in an interaction region. As mentioned above, there are two degrees of freedom when solving the symmetry equation with respect to the triangle point. In our synthetic numerical experiments we have hence chosen to keep either the x- or y-coordinate fixed when seeking a position of the triangle point for which symmetry is obtained. The examples below are used to illustrate where the triangle points may be located for different cases of heterogeneity. A general rule for obtaining symmetry has not been found, but for layered media, rules can be stated. Example 1: K 1 K 2 = K 3 This example will deal with an isotropic case where one grid cell in the interaction regions has different permeability from the other two. The permeability of this cell will be denoted K 1. The baricentre was used as the initial position for the triangle points, and Newton s method was applied to solve the symmetry problem with respect to the x-coordinate of the triangle points. The result for a specific case is depicted in Fig. 6b, where two interaction regions are affected. For the rightmost interaction regions the triangle point is located such that the local layer edge (edge that separates layers with different permeability) becomes a straight line. The location of the triangle point in the other interaction region seems more arbitrary, but it deviates only slightly from the initial position. If, for this interaction region, the initial guess of x 0 is chosen closer to a point that coincides with the straight line that connects the main.tex; 12/06/2002; 12:41; p.16

17 17 K 1 K 1 a. Heterogeneous and isotropic b. Heterogeneous and isotropic Figure 6. a: Initial positions of triangle points. b: Positions of triangle points that yield symmetry. dividing points, the symmetry solution will be such that the local layer edge is a straight line. K 1 K 1 a. Fully heterogeneous case b. Anisotropic case Figure 7. a: Position of triangle points for fully heterogeneous case. b: Position of triangle points for anisotropic case Example 2: Fully heterogeneous medium Let now all three grid cells that interact in one interaction region have different permeability. The positions of the triangle points that yield symmetry for such a region and one of its neighbouring interaction regions are visualized in Fig. 7a for some specific permeability values. For the rightmost interaction region the same symmetry solution as in Example 1 is obtained. The position of the triangle point in the fully heterogeneous interaction region seems arbitrary, and it s significance is not known. If K 1 is allowed to increase further, the triangle point seems to converge to some limit. Example 3: Anisotropy Let now anisotropy be introduced for K 1 for the grid/medium studied in Example 1, whereas the permeability is isotropic for the other cells. The triangle point that yields symmetry for the rightmost interaction regions is only altered slightly, and changes only slightly if the ratio k 1y /k 1x increases. For the second interaction region, however, our Newton solver for symmetry does not converge even for small anisotropy main.tex; 12/06/2002; 12:41; p.17

18 18 ratios. The reason for this is seen when plotting the (local) symmetry function for anisotropy ratio k 1y /k 1x = 1/2, and fixed y-coordinate. The symmetry function does not have any zeros in the region where it previously did, and this is also the case for larger anisotropy ratios. We have performed tests where the y-coordinates have been set to different values, and solutions have been sought in the x-direction. No zeros are found for these cases either. If we instead study a slightly different grid, see Figure 7b, symmetry solutions are obtained. The anisotropy ratio k 1y /k 1x was gradually increased, and the symmetry solution for the extreme case k 1y /k 1x = 500 is depicted in Fig. 7b. For the uppermost interaction region, the triangle point seems to be located near a straight line that separates two dividing points. Contrary to the isotropic case, it is not the local layer edge that becomes a straight line. For larger anisotropy ratios the x- coordinate seems to converge to some limit close to this line. Example 4: Layered medium Layered media naturally fall into the scope of the test cases in example 1 and 3. For an isotropic example, a possible set of triangle points that yield symmetry is depicted in Fig. 8. These points are located such that the local layer edges are straight lines as in Example 1. For this case the triangle points that are solutions to the symmetry problem seem more realistic than when baricentres of the triangles are used because of the zig-zag pattern the edge then would have. The whole layer edge is only piecewise straight. This is due to the arbitrary positioning of the triangles, and could be fixed if a suitable restriction was put on the triangular grid. If the medium is anisotropic and all permeabilities have the same principal directions and the same anisotropy ratio on both sides of the layer edge, the numerical tests again show that the triangle points should be located such that the local layer edges are straight lines. However, if the anisotropy ratios are not the same on both sides, the solutions will in general not be located at straight lines between the dividing points. Further tests show that if the local layer edge is chosen to be a straight line, and the principal directions of the permeability tensors are orthogonal and parallel to this edge respectively, symmetry is satisfied regardless of the anisotropy ratios Practical gridding issues The impact of the results above is that certain cases will be ideal for the triangle method. For isotropic, but heterogeneous media, a case like main.tex; 12/06/2002; 12:41; p.18

19 19 k 1y k 2y = k 3y k 1x 1 x k 2x = k 3x a. Extracted area b. Local layer edge Figure 8. a: To obtain symmetry, triangle points may be located on straight lines that connect dividing points. b: Two of the interacting cells have the same permeability. For anisotropic media the eigenvectors will be assumed parallel and orthogonal to the local layer edge K 1 K 2 K 2 K 1 a. Isotropic medium b. Ansotropic medium Figure 9. Triangularisation/honouring of physical layering near permeability contrasts. Control volumes (bold lines) drawn on outer side of discontinuity line. the one depicted in Fig. 9a can easily be triangularised such that the edges of the polygonal control volumes honour the physical layering. For anisotropic media, where the principal directions of the permeability tensors are aligned with the layer edges, a situation like the one depicted in Fig. 9b is easy to triangularise to assure that symmetry of the discrete operator is satisfied. 4. Local contributions to coefficient matrix, layered medium This section analyses the local contributions from an interaction region to the total coefficient matrix for the case of a layered medium. It will be assumed that the local layer edge of the interaction region is a straight line. The cell with different permeability from the other two cells will have local numbering 1, and the two remaining cells 2 and 3. Hence, K 1 K 2 = K 3, see Fig. 8b. Angles associated with corners x i of an interaction region will be denoted α i. For cases with anisotropy, it will be assumed that the eigenvectors of K i,i = 1,2,3 are orthogonal and parallel respectively to the local layer main.tex; 12/06/2002; 12:41; p.19

20 20 edge. The eigenvalues of the permeability tensors will be denoted k ix and k iy respectively. For isotropic media it is irrelevant to talk about directions of the eigenvectors of K, and the local layer edge can have an arbitrary position in any reference system. With the above assumptions we will be able to prove symmetry for this case. Secondly, we will be able to analyse M-matrix property locally for an interaction region Isotropic medium We start out by combining and rewriting eqs. (13) and (14). θ 1 ζ 1 γ 1 T = A θ 2 ζ 2 γ 2 + B = C θ 3 ζ 3 γ 3 θ 1 ζ 1 γ 1 θ 2 ζ 2 γ 2 + D, (29) θ 3 ζ 3 γ 3 where the matrix consisting of γ, θ and ζ parameters is the matrix product (A C) 1 (D B). One can perform some tedious, but straight forward calculations to obtain the following relations between the parameters (when K 2 = K 3 ). γ 2 = 1 2, γ 3 γ 1 = 1 2. (30) θ 2 = 0, θ 1 = θ 3. (31) ζ 2 = 1 2, ζ 1 ζ 3 = 1 2. (32) From (29), (30), (31) and (32), the following expressions follow: γ 1 = 1 2 (ω 12 ω+ 11 ) ω11 + ω+ 11. (33) ω+ 12 θ 1 = ζ 1 = ω ω 11 ω11 + ω+ 11. (34) ω (ω ω 12 ) (ω ω 12 ) ω 11 + ω+ 11 ω+ 12. (35) Using the definition of the scalar coefficients (6), it follows that the above parameters are constant when x 0 varies along the local layer edge. The contributions from one interaction region to off-diagonal terms of the total coefficient matrix will now be analyzed. In total, there will main.tex; 12/06/2002; 12:41; p.20

21 be 3 pairs of contributions to examine from one interaction region: m 13 and m 31, m 12 and m 21, m 23 and m 32 (same notation as in Sec. 2.8). The question of non positive off-diagonal elements of the coefficient matrix will in general depend on two neighbouring interaction regions for each pair of elements, since contributions from one interaction can compensate for positive contributions from its neighbouring interaction region. Only contributions to off-diagonal terms from one interaction region are discussed here. The local contributions m 31 = t 31 t 21 and m 13 = t 13 t 33 are first considered. The transmissibilities involved in m 31 can be expressed based on (13) and (29): t 21 = θ 1 ω θ 2ω 22, (36) t 31 = θ 2 ω θ 3ω 33. (37) From (34) it follows that θ 1 = θ 2 range between 0 and 1 for different heterogeneities and different triangles. Combining (36) and (37) yields. m 31 = t 31 t 21 = θ 1 (ω33 ω+ 22 ). (38) By the same analysis as in (18) and (19), it further follows that (ω 33 ω+ 22 ) = 1 2F (ν 2 )T K 2 ν + 2, (39) where F is 1/4 of the area of the whole interaction region. This term is negative for α 2 < π/2. Hence, m 31 < 0 with this angle constraint. For this element it is therefore both a necessary and sufficient condition for it to be negative that α 2 < π/2. The element m 13 = t 13 t 33 can be studied in a similar way. It is straightforward to show that it can be written as 21 m 13 = γ 1 (ω11 + ω+ 11 ω 33 ) (ω 11 + ω 33 + ω+ 33 ). (40) Now m 31 and m 13 can be compared, and it will be shown that they are equal: Since the dividing points of the triangles are midpoints of the triangle edges, n 1 and ν + 2 are parallel. A direct calculation shows that θ 1 = k 1x /(k 1x + k 2x ), so that (38) becomes m 31 = k 1x 2F(k 1x + k 2x ) ν 2x k 2xν + 2x, (41) main.tex; 12/06/2002; 12:41; p.21

22 22 in the local coordinate system where the y-axis is parallel to the local layer edge. Furthermore, (33) can be rewritten as γ 1 = 1 2 (k 1x k 2x ) ν+ 1x. (42) k 1x + k 2x ν + 2x After some manipulation, Eq. (40) reads m 13 = γ 1(n T 1 K 1ν nt 3 K 3ν + 2 ) (nt 1 K 1ν + 1 nt 3 K 3ν + 1 ). 2F Inserting (42) and noting that n 1 n 3 = ν + 2, and that n 1y = n 3y = 0 in the specific coordinate system we are using, the following is obtained. m 13 = k 1x 2F(k 1x + k 2x ) k 2xν 2x ν+ 2x, (43) which is the same as (41). Hence, symmetry is satisfied for layered media when the local layer edge is a straight line. Note also that this result generalizes to anisotropic cases with different anisotropy ratios on opposite sides of the local layer edge (provided that the principal directions of the permeability tensors are orthogonal and parallel to the local layer edge). The next contribution to analyse is m 23 = t 23 t 13. Using (29) it follows that the transmissibilities t 23 and t 13 can be written as t 23 = γ 1 ω γ 2ω 22, (44) t 13 = γ 1 ω 11 + γ 3ω + 11, (45) From the definition of the scalar coefficients, (6), it is seen that they vary linearly with the y-coordinate when the triangle point is moved along the (straight) local layer edge. Since γ 1,γ 2,γ 3 then are constant, it follows that t 23 and t 13 vary linearly with x 0 along the local layer edge. Using (45) and (44), m 23 can be written: m 23 = t 23 t 13 = γ 1 ω ω 22 γ 1ω 11 γ 3ω (46) From (6), (30) and (33) it can be shown that m 32 is constant when x 0 varies along the local layer edge. Since symmetry is satisfied, we know that m 32 = m 23. Similar to the discussion of m 13, the sign of m 23 can now be investigated. Since m 23 is constant along the local layer edge, it is sufficient to examine one position of x 0 in order to analyze the sign of the term. main.tex; 12/06/2002; 12:41; p.22

23 23 ν 1 α 1 ν 1 x 0 α 3 α 2 ν + 2 α 1 ν + 1 l α 3 α 2 ν + 2 ν 2 ν 2 a. Relevant vectors for m 23 b. α 2 and α 3 equal. Figure 10. Interaction regions and relevant vectors. Triangle point may be located arbitrarily along local layer edge Without loss of generality, assume that x 0 is located at x 1, such that n 2 = ν 1 and ω± 11 vanishes. Eq. (46) then yields m 23 = 1 4F ((ν 1 )T K 2 ν 2 + 2γ 1(ν 1 )T K 2 ν + 2 ). (47) From (33) it follows that γ 1 ranges between 0 and 0.5 when k 2x > k 1x and α 2 < π/2. If, in addition, the angles α 1 and α 3 each are smaller than π/2, then ω 22 ± is always negative. The following inequality is then obtained for isotropic K 2 : m 23 < 1 4F (ν 1 )T K 2 ν 2 < 0. (48) Hence, the contributions to the global stiffness matrix, m 23 and m 32, are negative with this angle condition and permeability restriction. The case when k 1x > k 2x is further investigated for a specific example in Sec. 4.2, where anisotropy also is allowed. The last contribution to investigate is m 12 = t 12 t 32. From (13) and (29), we can write the transmissibilities as t 12 = ζ 1 ω 11 + ζ 3ω (49) t 32 = ζ 2 ω ζ 3ω 33. (50) From (32), (49) and (50) m 12 will be given by m 12 = ζ 1 ω 11 + ζ 3(ω + 11 ω 33 ) 1 2 ω+ 33. (51) It can also be shown that this contribution is constant when x 0 varies along the local layer edge. To investigate the sign of it, we may, without loss of generality, assume that x 0 is located such that n 3 = 0. Then main.tex; 12/06/2002; 12:41; p.23

24 24 n 1 = ν + 2, and from (6) it follows that ω± 33 = 0. From (51) the following is then obtained. m 12 = 1 2F (ζ 1(ν + 2 )T K 1 ν 1 + ζ 3(ν + 2 )T K 1 ν + 1 ). (52) Similar to the analysis of m 23 and m 31 it can be shown that m 12 < 0 provided that α 2,α 3 < π/2. Note that these angle conditions are only sufficient, but not necessary conditions Anisotropy As it turns out, anisotropy will only affect the local contribution m 23. The effect of anisotropy is eliminated for m 12 and m 31 as follows: It is easily verified that the parameters γ i,θ i,ζ i,i = 1,3 (30-32) are independent of anisotropy when the principal directions of the permeability tensors are orthogonal and parallel to the local layer edge. The contribution m 31 is given by (38). The parameters θ 1 and θ 2 range between 0 and 1. Left multiplication on ν + 2 by K 2 is independent of k 2y since ν + 2 is assumed to be parallel to the eigenvector which corresponds to k 2x. The sign of m 13 is therefore unchanged if anisotropy is introduced, and it is again a (necessary and) sufficient condition for m 13 to be negative that α 2 < π/2. For m 12, much the same argument will apply. Since the permeability tensors K i are symmetric, the vector multiplication may be interchanged, and (52) can be rewritten: m 12 = 1 2F (ζ 1(ν 1 )T K 1 ν ζ 3(ν + 1 )T K 1 ν + 2 ). Since left multiplication on ν + 2 by K i is independent of anisotropy, m 12 is also unaffected. A more thorough analysis will be needed for m 23. When k 1x < k 2x, the first inequality of (48) is still valid if anisotropy is allowed. For the specific position of x 0 we are looking at, ω22 = 1 2F (ν 1 )T K 2 ν 2, and this term may be positive if k 2x > k 2y. How much larger k 2x needs to be will depend on the triangle as well. The two terms of Eq. (47) may now have different signs, and the first inequality of (48) is not sharp enough for further analysis. It is also straightforward to see that m 23 can become positive even for angles smaller than π/2. If k 1x > k 2x, γ 1 becomes negative. In addition, ω 22 + is always non-positive for α 3 < π/2, so that the last term of Eq. (47) may be positive as well. Hence, m 23 may in the worst case consist of two positive terms. Next, a specific case is analyzed for obtaining a non-positive m 23. main.tex; 12/06/2002; 12:41; p.24

25 25 Example of geometry, heterogeneity and anisotropy Assume that an interaction region as depicted in Fig. 10b is given, where the angles α 2 and α 3 are equal. We have earlier commented on the fact that if k 1x > k 2x, one of the terms in Eq. (47) may be positive. We will now investigate what angular condition should be posed for α 1 in order to make m 23 non-positive for given anisotropy and permeability difference across the local layer edge. A direct calculation of γ 1, Eq. (33), yields: γ 1 = 1 (k 2x k 1x )ν 1x + 2 (k 2x + k 1x )ν 2x + = 1 (k 2x k 1x ) 4 (k 2x + k 1x ). It is now clear that the fraction becomes non-positive if k 1x > k 2x. For this case, the last element of Eq. (47) will be positive when α 2 < π 2, and could potentially yield a positive m 23. Assume that the height from x 1 to the the edge x 2 x 3 is some distance l, see Fig. 10b. Since x 0, without loss of generality, can be assumed to be located such that n 2 = ν 1, ω 22 and ω+ 22 can be expressed by α 1 and l. Simple geometry yields ω 22 = l2 8F (tan2 ( α 1 2 )k 2x k 2y ), ω + 22 = l2 k 2x 4F tan2 ( α 1 2 ), so that Eq. (46) yields: m 23 = l2 16F (tan2 ( α 1 2 )k 2x tan 2 ( α 1 2 )k 2x k 1x k 2x + k 1x k 2x k 2y ). (53) For a non-positive contribution, the bracket sum must be non-positive, and upon division by k 2x, we obtain the following angular condition: tan( α 1 2 ) < k 2y /k 2x 1 (1 k 1x /k 2x )/(1 + k 1x /k 2x ). (54) When the medium is homogeneous and isotropic, the restriction on the angle is then that α 1 2 < π 4, which agrees with results obtained in Sec Summary This paper has examined how the coefficient matrix is built up when an elliptic term is discretized by the O-method on polygonal control volumes with triangular dual grid. For the homogeneous case, symmetry is always satisfied when dividing points are midpoints of triangle main.tex; 12/06/2002; 12:41; p.25