MATHEMATICAL COMPUTER MODELLING Mathematical
PERGAMON
and Computer
Modelling
30 (1999)
153-178 www.elsevier.nl/locate/n~cn~
Domain Decomposition Methods Applied to Sedimentary Basin Modeling E. ZAKARIAN Institut Frarqais du P&role, 1 et 4, avenue de Bois-P&au 92852 Rueil-Mailmaison Cedex, France R. GLOWINSKI Department of Mathematics, University of Houston Houston, TX 77204-3476, U.S.A. (Received
accepted December
Abstract-A study several domain methods is sedimentary modeling. We a simplified for a basin by a parabolic equation strongly discontinuous gradient interface relaxation are first on a generalized to nonlinear model. 1999 Elsevier Ltd. All
within the of flow in sedimentary Conjugate equation and reserved.
Keywords-Sedimentary methods.
Domain
modeling,
Nonlinear
equations,
INTRODUCTION:
1.
SYNOPSIS
The primary objective of basin modeling is to simulate the evolution of a sedimentary basin 1,~ taking into account the compaction tion.
A sedimentary
layers that have been deposited organic
material
of the porous medium and hydrocarbon formation
basin is a heterogeneous
contained
temperature.
to accumulate
This migration
in reservoirs.
to now, most basin simulators erosion,
and compaction However,
Then,
In order to handle these complex
for short)
have been
chosen.
Indeed,
is realistic
flow through
The authors would like to acknowledge help and suggestions of I. Faille.
the main numerical
the support
@ 1999 Elsevier
is considered
of Institut
Science
although
difficulties Fmn&s
flow. Up
from deposition,
characteristics
along which block
into subregions
This choice,
solid
the porous medium
Dom,ain Decomposition
split basins
model problem
is neglected.
enough to contain
0895.7177/1999/s - see front matter PII: SO895-7177(99)00188-O
geometries,
Over time.
or three-phase
only if their geometrical
are cut by faults,
faults
A simplified
can occur and where compaction models,
hydrocarbons
times.
and migrastratigraphic
into mobile hydrocarbons
takes place as a t.wo-phase
of the porous medium, most real life basins
subdomains.
layers is transformed
have been able to handle those regions resulting
can occur.
computational
of stacked
from the onset of the basin up to present
into some stratigraphic
under the effect of increased
ple enough.
porous medium consisting
that
are sim-
displacements
Methods (DDM, naturally
define
where only one-phase
flow
not too close to usual basin usually encountered
du P&ok
Ltd. All rights reserved.
(I.F.P.)
in basin
and the invaluable
Typeset
by 4n/ls-‘%?
154
E. ZAKARIAN AND R. GLOWINSKI
modeling. The equations governing the physical phenomena are discretized using a Finite Volume Method and the resulting finite-dimensional problem is solved using a nonoverlapping domain decomposition method, where one solves alternatively discrete Dirichlet and Neumann subproblems, as advocated in [l]. The interface problem is first solved by a relaxation-type method and then by a conjugate gradient-like algorithm (as advocated in, e.g., [2,3]). The article is organized as follows. In Section 2, we provide the mathematical model describing the physical phenomena under study. In Section 3, we describe the finite volume discretization of the above model. In Section 4, we discuss the domain decomposition based solution of the discrete problems. Finally, in Section 5, we present and discuss the results of numerical experiments.
2. SEDIMENTARY 2.1.
BASIN
MODELING
Generalities
A well-chosen simplified model is sufficient to reproduce the main numerical difficulties traditionally observed in basin modeling. Such a model can be obtained by modeling the time evolution of the over-pressure (the over-pressure is the difference between the actual water pressure and the hydrostatic pressure) in a very simple basin configuration where compaction is neglected and where stratigraphic layers are not moving with time. The simplified problem can be either linear or nonlinear, depending on the relations between pressure and porosity used in the sedimentary basin model. 2.2. The Mathematical
Model:
(I) Governing
Equations
We consider only incompressible one-phase (the water phase) fluid flow in a two-dimensional cross-section of a sedimentary basin. As already mentioned, we shall model the time evolution of the over-pressure. Starting from standard basin models (see, e.g., [4]), we obtain first the equation modeling the mass conservation of the water, namely
$w) +v. (PWCPVW) = 0,
(2.1)
where, in (2.1), pW,V,, and cp denote the density of the water, the velocity of the water, and the porosity of the medium, respectively. We suppose that ,oWis constant, implying that the fluid is incompressible. We suppose that the Darcy’s law is verified, namely =
uw= cpvw = where, in (2.‘4, U,,
Pw, g,
CL~, and k
-+w -/A&),
(2-2)
denote the Darcy’s velocity, the water pressure, the
gravity, the water viscosity, and the porous medium intrinsic permeability tensor, respectively. The permeability tensor depends strongly on the actual lithology; it can vary by several orders of magnitude-four typically-from one layer to another. The following relations between porosity and pressure are encountered in the specialized literature (see, again, [4]): either cp = cpr + oo(PUJ - PO)
(2.3)
or cp = cpT+ exp(aP,); in (2.3),(2.4), obtain
cp,-,
ao,
Po,
(2.4)
and a are constant quantities.
Combining now (2.1) and (2.3), we
ZE
ap?lJ
croPwdt
-
PWV. E(VPw
- plug) = 0;
(2.5)
Sedimentary
Basin
155
Modeling
similarly, combining (2.1) and (2.4) yields (2.6) Let us define the over-pressure P by P = P,
(3.7)
- pwgz,
with g = Igl. Combining (2.5) and (2.7), we obtain the following linear parabolic equation: =
c+v.~vP=o;
(2.8)
pw
while combining (2.6) and (2.7) yields the nonlinear equation - ‘J. EVP
exp(ap,,g*)g(exp(aP))
= 0,
(2.9)
also of the parabolic type. 2.3.
The
Complete
Mathematical
Model:
(II)
B oundary
Conditions
and Description
of the
Model
It follows from (2.8) and (2.9) that the models that we consider include linear or nonlinear parabolic equations. We suppose that the porous medium region that we consider is of rectangular
shape as the one in Figure 2.1; we denote it by R.
01
2 Figure
3
45
2.1. A domain
I
-i
R and its discretization.
The boundary conditions depend on time and are of the Dirichlet type on the north and south sides and are of the Dirichlet or Neumann type on the east and west sides. Hence, the problem is to find a function P(x, t) : fi x [0, T) 4 R verifying (after simplifying the notation) ;H(P)
- V .l%VP
= 0,
in R x (O,T),
P(G 0) = PO(X),
in 0,
P = PO,
on (rh’ U Ts) x (0, T),
P=PD
or
l?VP.n=FN,
on (Tw U r~) x (0, T),
(2.10)
E. ZAKARIAN AND R. GLOWINSKI
where n is the unit vector of the outward normal at do and where H(P) H(P)
=
ap,
(linear case),
exp(aP),
(nonlinear case),
is defined by
with c~ > 0. The initial value Po is defined as the solution of the following steady-state analogue of (2.10) in R,
-V.EVPo=O, PO(X) = P&,
O),
PO(Z) = PD(z, 0)
or
= K(lc)VPo(z)
3. DERIVATION
on IN U ITS,
.n = FN(z,0),
OnrwurE.
DISCRETE
MODEL
OF THE
(2.11)
3.1. Discretization To discretize problems (2.10),(2.11) we have chosen a finite volume method, like those discussed in [5]. A Cartesian fixed grid (like the one shown in Figure 2.1) is chosen. We suppose that K is constant on each cell and that the boundary between two layers is the union of common interfaces of adjacent cells. The problem is to compute the over-pressure P at the center of each cell or, to be more precise, to find a vector P E RI’ J such that p
=
{pij}l
V ’ I?Vp2 = f,
acp2 (p2 = 0,
in Ri,
= f,
EV(p2 . n2 = -kVp2
af12b,
on y.
. nl,
(4.67b)
Problem (4.65) can be viewed as an optimal control problem in the sense of [13]; since problem (4.27) is well posed it is very easy to see that problem (4.65) has a unique solution. There is no basic difficulty at solving-formally-the
least-squares problem (4.65) by a conju-
gate gradient algorithm operating in L2(r) if we can compute
easily J’(p),
‘dp E L2(y).
Using a
standard perturbation analysis, we obtain, with obvious notation, that Wcl)
=
J’(&WY
=
sY =
(p - lp2)Q
- 972) h
sY
JP
-
cp2N~h
(-1.68)
+
s
(~92 -
~16~2
dy>
sY
where, from (4.67), 6~1 and 6~2 verify
(4.69a) in R2,
a&s - V. I?X76p2 = 0, Q2
EVS$+J. n2 = -EVdpi
on afi2\Y,
= 0,
on 7.
. nl,
(4.69b)
Let, now, pl and p2 be two smooth functions defined over Ri and s22. respectively. and satisfying Pi =O,
Vi = 1,2.
on a%\?,
(4.70)
Multiplying by pi and p:! we obtain from (4.69) that, after summation and integration by parts,
&/ i=l
Oz
(aPi&+EV6Yi. Vpi)
dx = k/p,nV~Yy,
ni dy.
(4.71)
1=1 7
Complete (4.70) by Pl = pa>
on y.
(4.72)
Next, take (4.69b) and (4.72) int o account and transpose I?; it follows then from (4.71) t,hat 2 apihqi CJ’ +I
+
KtVpi . 069,)
dx = 0.
(1.73)
fli (
Integrating by parts over Qi and taking (4.69a) into account, (4.73) yields api - V * R”Vpi)
6qi dx =
J
I?‘Vpl
. ni6p dy +
Y
J Y
ktVp2
. n26y2 dy.
(4.74)
Suppose that pl, p:! verify api - V. E’Vpi RtVp2.
= 0, n2 = 9
in Ri, - p,
vi = 1,2, 011 y.
(4.75) (4.76)
168
E. ZAKARIAN AND R. GLOWINSKI
We have then, from (4.74)-(4.76), 7(v2
-
J’
p)+zdy
=
-
ntVpl
. nlbpddy,
s -I
which, combined with (4.68), implies J’(P) = I-L-
92
IT -
ntVpl
. nl/
(4.77)
v’I_L E L2(Y)?
Y’
where pr is obtained from p, cpr,(~2 via the solution of the following adjoint equations ap2 - V . gtOp2 P2 =o,
am - -V . ntVpl on
Pl = 0,
(4.78a) on y,
n2 = cp2- p,
KtVp2.
on dotz\y,
in 02,
= 0,
=
0, PI =
d%\y,
in CIr, ~2,
(4.78b)
on 7.
We now have the fundamental information that we need to solve (4.65) by a conjugate gradient algorithm; such an algorithm is described as follows: X0 E L2 (y) is given;
(4.79)
solve acpy-V.&q$=O, on
‘p: = 0,
in Rr, ‘py =
%\y,
a& - V. EV& on
cp; = 0,
= 0,
&pi
dfl2\7,
on
P; =o,
= 0,
spy - V. RtVpy = 0, on %\7,
(4.80b)
nr,
on Y,
in 02,
=t K Vp; . n2 = &! - x0,
dfl2\y,
P? =o,
in R2,
. n2 = --EVqy.
api - V . c&p;
(4.80a)
on y,
X0,
(4.81a) on Y,
in Rr,
PY = Pii>
(4.81b)
on Y,
and set
X0- cp!jy - E"opy.
go =
nr
(4.82)
Y’
UI” = g?
(4.83)
For lc 2 0, assuming that A”, g”, wk are known, compute A’+‘, gk+‘, wk+r as follows. Solve a&”
- v. Ev$.Q
cprk = 0, W2”
$Qk = 0,
on
= 0,
&” = wk,
on Q\Y, -
v . kvq2p,” &7$52”
afi2\7,
= ’
0,
on
i-592\7,
8”
aPrk - V . $v& Fr” = 0,
on %\y,
in 02, ’ nr,
(4.8413) on Y,
in Cl2, (4.85a)
Vpz” . n2 = cpzk - wk, = 0,
(4.84a)
on y,
n2 = -EVp-r”
afik - v. k$i2k = 0, @-2k= 0,
in Rr,
in Rr,
PI’” = p2”,
on y.
on Y, (4.85b)
Sedimentary
Basin Modeling
169
set (4.86) and compute (4.87) (4.88) (4.89) 5 5, take
If II!/“+lII~Z(r)/lI.CIoII~“(l)
X = X”+l:
else, compute
(4.90) u,k+l
Do X: = I? + 1 and return The
above algorithm
per iteration combining
higher
least-squares
(p+1
ypP.
+
(4.91)
to (4.84~~). respects
is twice
=
definitely
than
the spirit of the original
the one of the
and conjugate
gradient
Quarteroni
Quarteroni
method,
methodologies
but
method;
I its cost
we can hope
will lead to algorithms
that
which are
at the same time fast and robust,. Conjugate
4.6. 4.6.1.
Generalities:
There
Variants
in Section
(4.8),
at modifying
like those described
situations 4.4.1
of Algorithm
(II)
(4.2)-(4.7):
The
Nonlinear
Case
Synopsis
is no basic difficulty
nonlinear problem
Gradient
that
the
Qunrteroni
algorithms
in Sections DDM
(4.53)-(4.64)
or (4.79). (4.91) t,o handle
2 and 3. After discretization,
is nothing
but a method
wc have shown
to solve t,hc fixed-point
name11 X = G(X),
with operator
G : RN,, 4
IR“‘,J defined in Section
4.4.1.
Operator
G can be made explicit
as
follows: G(P)
= BIPI(CL)+&P,(P.,
wit511pi (CL) E iRNt. i = 1.2, solutions
(-1.92)
+Fw+d.
uf (4.93)
AI (PI (~1) = b1l-l+ clr AZ where,
(Pi)
=
bpt(p)
(-l.94)
+b.w+~2~
in (4.92)-(4.94):
l
B1,B2,B3
l
A1 is a nonlinear
operator from iRN1 into RN1, bl is a N1 x N, matrix
l
A2 is a nonlinear
operator from IRNz into RN2, b2 is it Nz
matrix
are ~“r’,x N1, N, x N2, NO x N, matrices.
x
and d E RN,,; and cl E !RN1:
N1 matrix,
b:s is a _Vz x N,,
and cz E IRNZ.
The various operators
and vectors
and also, for some of them, account,
respectively,
a least-squares
in (4.92)-(4.94)
of the solution
formulation
depend
at the previous
of the fixed-point
on the discretization, time step.
problem
(4.8)
Taking
on tensor
(4.92)-(4.94)
k, into
is given by
(1.95)
170
E. ZAKARIAN
AND R. GLOWINSKI
with j(p)
= f ]](I - B3) P - Blpl
- B2~2
(4.96)
- d112.
where, in (4.96), p1 = p1 (p), p2 = p2(~) are the solutions of equations (4.93)-(4.94), respectively, and where ]] . (1denotes the canonical Euclidian norm of RN-. The least-squares problem (4.95) has therefore the structure of a discrete control problem in the sense of [13]. To solve the leastsquares problem (4.95), it would make sense to apply some of the methods discussed in, e.g., [14] (see also the references therein). In the present article, we shall address the conjugate gradient solution of problem (4.95) (actually, conjugate gradient is one of the methods considered in [14] for the solution of least-squares problems).
The conjugate gradient solution of problem (4.95)
requires the knowledge of the gradient j’(p)
of functional j(.)
at p, V/A E RN-; this important
issue will be discussed in the next paragraph. 4.6.2.
Gradient
Let us define
calculation
R(p) E IWNcL by R(P)
= (~-B~)P--B~PI
(4.97)
v'c1 E IwN-.
-B2~2,
If 6~ is a perturbation of p, we have, from (4.96),(4.97)
b.(p) = (j’ (1-1)) WWiv,=
((I - Bi) R(P) >b&v
