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http://pvamu.edu/aam
Appl. Appl. Math.
ISSN: 1932-9466
Applications and Applied
Mathematics:
An International Journal
(AAM)
Vol. 05, Issue 2 (December 2010), pp. 585 – 604
(Previously, Vol. 05, Issue 10, pp. 1682 – 1701)
Finite Element Analysis in Porous Media for Incompressible
Flow of Contamination from Nuclear Waste
Abbas Al-Bayati, Saad A. Manaa and Ekhlass S. Ahmed
Department of Mathematics
University of Mosul
Mosul, Iraq
profabbasalbayati@yahoo.com
Received: March 19, 2010; Accepted: July 28, 2010
Abstract
A non-linear parabolic system is used to describe incompressible nuclear waste disposal
contamination in porous media, in which both molecular diffusion and dispersion are
considered. The Galerkin method is applied for the pressure equation. For the brine,
radionuclide and heat, a kind of partial upwind finite element scheme is constructed.
Examples are included to demonstrate certain aspects of the theory and illustrate the
capabilities of the kind of partial upwind finite element approach.
Keywords: Finite element method; Galerkin method; incompressible flow; nuclear waste
MSC (2000) No.: 65L60, 74S05, 82D75
1. Introduction
The proposed disposal of high-level nuclear waste in underground repositories is an
important environmental topic for many countries. Decisions on the feasibility and safety
of the various sites and disposal methods is based, in part, on numerical models for
describing the flow of contaminated brines and groundwater through porous or fractured
media under severe thermal regimes caused by the radioactive contaminants. A fully
discrete formulation is given in some detail to present key ideas that are essential in code
development. The non-linear couplings between the unknowns are important in modeling
the correct physics of flow.
In this model one obtains a convection-diffusion equations which represent a mathematical
model for a case of diffusion phenomena in which underlying flow is present;
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w and bw correspond to the transport of w through the diffusion process and the
convection effects, respectively, where  and  denoted respectively the gradient operator
and the Laplacian operator in the spatial coordinates.
The prediction of an accurate numerical solution for the convection dominated flow
problem is one of the more difficult tasks in computational fluid dynamics. The central
difference method and the conventional Galerkin method have consistently produced
unphysical oscillatory solutions. One successful technique for solving such a problem is
known as the upwind algorithm, which was originally devised for the finite difference
method. In the finite element method, a popular technique known as the streamline upwind
Petrov-Galerkin (SUPG) method is used. This method modifies the weighting functions by
using the local velocity to dictate an upstream direction of these functions. Such
modification eliminates the oscillatory behavior for some convection problems. For more
details, see Wansophark and Pramote (2008).
In this paper, we have considered the fluid flow in porous media using a Galerkin method
for the pressure equation and a kind of partial upwind finite element scheme is constructed
for the convection dominated saturation (or concentration) equation, the trace
concentration of ith radionuclide and the heat equation. For more details of this subject see
Douglas (2001, 2002), Huang (2000) and Chen et al. (2009).
2. Model Equations
The model for incompressible flow and transport of contaminated brine in porous media
can be described by a differential system that can be put into the following form, Gaohong
and Cheng (1999).
Fluid:
(a) . u   q  Rs' ,

k ( x)

(b) u    (c) P  a (c)P.

(1)
Brine:

c
 u.c  .( Ec c)  g (c).
t
(2)
Heat:
d2
T
 c p u.T  .( EH T )  Q(u, T , c, p ).
t
Radionuclide:
c
 K i i  u.ci  .( Ec ci )  fi (c, c1 ,..., cN ),
t
(3)
(4)
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with the boundary conditions

( a )

(b)

 (c )
( d )


 (e)

u.n  0,
( Ec c  cu ).n  0,
on 
on 
( Ec ci  ci u ).n  0,
on 
( EH T  c pTu ).n  0, on 
p
 0,
n
(5)
( x, t )    (0, T ]
and the initial conditions
( a )

(b)

( c )
(d )
p( x, 0)  p0 ( x);
x  ,
c( x, 0)  c0 ( x);
x  ,
ci ( x, 0)  c0i ( x);
x  ,
T ( x, 0)  T0 ;
x  ,
(6)
where n is the unit outer normal to  , x    R 2 , t  (0, T ] ; u is the Darcy velocity; P is
1  cw , q  q( x, t )
is
the
production
term;
the
pressure;
Rs'  Rs' (c)  [c sK s f s /(1  cs )](1  c) is the salt dissolution term; k(x) is the permeability of
the rock;  (c) is the viscosity of the fluid, c dependent upon c , the concentration of the
brine in the fluid and T is the temperature of the fluid,
d 2   c p  (1   )  R c pR ,
EH  Dc pw  K m' I , K m'  km / 0 , D  ( Dij )
 (T u  ij  ( L  T )ui u j / u ),
and
Q(u , T , c, p )  {[U 0  c p T0 ].u  [U 0  c p (T  T0 ))  ( p /  )][ q  Rs' ]}  qL  qH  qH .
Ec  D  Dm I , and g(c)  -c{[csK s f s /(1  c s )](1  c)}  qc  Rs' ,
c i is the trace concentration of the i-th radionuclide, and
f i (c, c1 , c2 ,..., c N )  ci {q  [csK s f s /(1  cs )](1  c)}  qci  qci  qoi
N
  k ij  j K jc j  i K ici .
j 1
The reservoir  will be taken to be of unit thickness and will be identified with a bounded
domain in R 2 . We shall omit gravitational terms for simplicity of exposition, no significant
mathematical questions arises the lower order terms are included.
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A. Al-Bayati et al.
We assume that:
(A1)
a (c), Rs' (c), g (c), f i (c, c1 ,....., c N ), Q(u , T , c, p )  C01 ( R ) ,
 ( x), K i , d 2  H ' (), q  L (0, T ; H 1 ())
0  c0  a (c) , Rs' (c), g (c), ( x), f i (c, c1 ,..., cn ), K i , d 2 , Q(u , T , c, p )  c1 , c  R, x  
Dm  0 ,  l   t  0
(A2) The solution of the problem (1-6) are regular:
C ( x, t )  L2 (0, T ; H 2 ())  L (0, T ; w' ())
Ci ( x, t )  L2 (0, T ; H 2 ())  L (0, T ; w' ())
P( x, t )  L (0, T ; H r 1 ()) , (r  2)
ct , ctt , cttt  L (0, T ; H 1 ()) ; pt , ptt  L (0, T ; L ())
(A3)
For any   L () , the boundary values problem:
     , x  

x  ,
 0,
n
2
there exists unique solution   H 2 () and a positive constant M such that 
2
M 
[Manaa (2000)].
3. Finite Element Spaces
Consider a regular family {Th } of triangulation defined over  , where h is the longest
diameter of a triangular element with the triangular Th , we have a set of close triangles
{ei }(1  i  N e ) and a set of nodes {Pi }(1  i  N P  M P ) where Pi (1  i  N P ) are interior
nodes in  and Pj ( N P 1  j  N P  M p ) are boundary nodes on  . We put hs to be the
maximum side length of triangles and k to be minimum perpendicular length of triangles
for all e  Th .
Definition 3.1. A family Th
constant
of triangulations is of weakly acute type, if there exists a
 0  0 independent of h such that, the internal angle  of any triangle
ei  Th
satisfies  0     2 .
Definition 3.1.1. Let i ( p), (1  i  M ), be the continuous function in  s.t. i ( p), is
linear on each e  Th and i ( p j )   ij for any nodal point pj.. We denote Mh be the linear
span of i , (1  i  M ), i.e., a finite dimensional subspace of H 1 () and defined by:
M h  {z h | C (); z h is a linear function on e , e  Th } . Also, define a subspace of
H 01 () be: M 0 h  {zh | zh  M h ; zh ( Pk )  0, k  M  1,..., K } .
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We associate the index set   { j  i : Pj is adjacent to Pi } . Let Pi Pj Pk , be three vertices of
triangular element e and i ,  j , k be barycentric coordinates. We have the following
definitions [see Hu and Tian (1992)].
Definition 3.2. With each vertex Pi belonging to triangle e, the barycentric subdivision  ie
is given by:  ie  {P | P  e ; i ( P)   j ( P) , i ( P)  k ( P), Pj  e} , and the barycentric
domain  i associated with vertex Pi in  is given by  i   ie , e  Th .
Definition 3.3. With the characteristic function  i ( x) of barycentric domain  i , the mass
lumping operator  : w  C ()  wˆ  L () is defined by wˆ ( p ) 
NP  M P

w( pi ) i ( p ) .
i
Using interpolation theory in Sobolev space [Ciarlet (1978)] and inverse inequality, with
step-length hc , we have the relation between ŵ and w from the following lemma:
Lemma 3.1. There exists a constant C such that:
w  wˆ o , p  Chc w 1, p ,
wh 1  Mhc1 wh ,
w  M h , p  1
(7)
 wh  M 0 h .
(8)
Lemma 3.2. There exist constants C1 ,C2  0 such that:
C1 w 0, p  wˆ 0, p  C2 w 0, p , w  M h .
(9)
Definition 3.4. Let {M h } be a family of finite dimensional subspaces of C () , which is
piecewise polynomial space of degree less or equal to r with step length hP and the
P  [1, ], r  2 , there exists a constant M such that for
following property: for
r 1
0  q  2 and   wP () :
inf   x
x{ M h }
q, p
 Mh r 1q  r 1, p .
Similarly, we define {N h } be a family of finite-dimensional subspace of C ()  C () ,
which is piecewise polynomial space of degree less or equal to r-1 with the similar
property as Mh and 0  q  r  1. We also assume the families {M h } and {N h } satisfy inverse
inequalities:

L
 MhP1  ,
[Manaa (2000)].

L
 MhP1  ,
  M h
(10)
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4. Error Estimates
Let   0 is time step and N  T  . We use a Galerkin finite element method for the
pressure and velocity and partial upwind finite element scheme for brine, radionuclide, and
heat.
Let C 0  M h be a L2 () - projection of c 0 in M h :
(c 0  C 0 , z h )  0
z h  M h .
We can get P 0 Vh such that
k ( x)
P 0 , v)  (q 0 , v)  ( Rs' , v) , v Vh and U 0 Wh from
0
)
 P dx  0 , (  (c
0

U0 
k ( x)
P 0  a (C 0 )P 0
0
 (C )
If the approximate solution {P m ,U m , C m , Cim (i  1,2,..., N ), T m } Vh  Wh  M h  M hN  Rh is
known, we want to find {P m1 ,U m1 , C m1 , Cim1 (i  1,2,..., N ),T m1} Vh  Wh  M h  M hN  Rh
at t  t
m 1
, with five steps . Let (.,.) denote the inner product in L ()
2
Step 1.
Find Cm+1for m  0 ,1, ,N   1, such that
(ˆD Cˆ m , zˆh )  ( Ec C m1 / 2 , z )  R (U m , C m1 / 2 , z h )  ( gˆ (C m1 / 2 ), zˆ h ) z h  M h ,
(11)
where
D C m  (C m1  C m ) /  , C m1 / 2  (C m1  C m ) / 2 and with zi  z h ( Pi ) ,
Cim 1 / 2  C m 1 / 2 ( Pi ) ,
and
 ijm   U m .nij d , here nij is the unit outer normal to ij .
ij
The partial upwind coefficients should be required that [Hu and Tian (1992)].
(a )  ijm   mji  1
(b) max{1/ 2,1  ij1}   ij  1, if  ij  0,
max{1/ 2,1  ij1}   ji  1, if  ij  0.
Step 2.
Find Cim1 , for m  0 ,1, ,N  1 , such that:
(12)
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(ˆ Kˆ i D Cˆ im , zˆh )  ( Ec Cim 1 , z )  R(U m , Cim 1 , zh )
 ( fˆi (C m 1/ 2 , C1m 1 ,..., CNm 1 ), zˆh ) zh  M hN ,
where
591
(13)
M
R (U m , Cim 1 , zh )   zi  ijm ( ijmCim 1/2   mji C mj 1/2 ) .
i 1
ji
Step 3.
Find Tm+1 such that:
(d 2 D Tˆ m , rˆh )  ( E H T m 1/2 , rh )  R (c pU m , T m 1/2 , rh )
 (Qˆ (U m , T m , C m , P m ), rˆh )
(14)
rh  Rh ,
where
M
R(c pU m , T m1 / 2 , rh )   ri   ijm ( ijmTi m1/ 2   mjiT jm1/ 2 ) .
i 1
ji
Step 4.
Find Pm+1 such that:
P
m 1
dx  0 , (

k ( x)
 (C m1 )
P m1 , v)  ( q m1 , v)  ( Rs' , v) , v Vh
.
(15)
Step 5.
Find Um+1 as:
U m 1  
k ( x)
P m 1  a (C m 1 )P m 1 .
m 1
 (C )
(16)
Lemma 4.1. Let p Vh be the elliptic projection of p  H 1 () into Vh defined by
(a(c)p, v)  (a(c)p, v) , v Vh .
Then, there exists a constant k1 such that
r 1
p  p  h p p  p  k1 p r 1 h p .
Proof:
See Quarteroni and Valli (1997).
Some Important Remarks:
1. U m1.n  0 in  .
2. If U l  u l  ChP (0  l  m) , then U m1  u m1  ChP .
(17)
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and if U m  u m  ChP , then U m  C1 .
(18)
3. We will make the inductive assumption that if U l
U m1
L
L
 k * (0  l  m) , then
k*
(19)
4. From Manaa (2000), if Th is regular triangulation of weakly acute type we have
wh 1  6 / k wˆ h
, wh  M h
(20)
Theorem 4.1. C satisfies discrete mass conservation law
 ˆ D Cˆ

m
d    gˆ m 1/2 (C )d  , m  1, 2, , N .
(21)

Proof:
In (10), let z h  1 , then ( EcC m1 / 2 , 1)  0 and
M
M
R(U m , C m1 / 2 ,1)  1  ijmCijm1 / 2  
i 1
j i
eTh
 (
m
ij
Cijm1 / 2   ijmCijm1 / 2 )  0 .
pi , p je ,i  ji
Then (21) holds.
Theorem 4.2. Ci satisfies discrete mass conservation law
 ˆKˆ D Cˆ
i

m
i
d   f i m1 / 2 d , i  1,2,...., N

Proof:
Same as Theorem 4.1.
Lemma 4.2. Let
c :[0, t ]  M h and ci :[0, t ]  M hN , i  1,..., N
such that
( Ecc  c , z )   (c  c , z )  0, z  M h ,
( Ec(ci  ci ), z )   ((ci  ci ), z )  0, z  M hN , t  J
(22)
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and let
c  c   , ci  ci   i ,
then,
 1  Mhc ,


 Mhc2 ,
t
t
 i 1  Mhc ,
 i

 Mhc2 , i
t
t
 Mhc
1
and
 Mhc .
1
Proof:
See Ciarlet (1978).
Lemma 4.3. For all zh  M h and   c  c ,   c  C ,
2
((u m .c m 1/2 ), zh )  R(U m , C m 1/2 , zh )  M (hc2   m 1/2   m 1/ 2
2
2
  m 1/ 2  u m  U m  zh
2
2
2
)   z h ,
where   0 is arbitrary small constant.
Proof:
((u m .c m 1/2 ), zh )  R(U m , C m 1/ 2 , zh )  ((u m  U m ).c m 1/2 , zh )
 (U m(c m 1/2  C m 1/2 ), zh )  (U m .C m 1/2 , zh  zˆh )
 [(U m .C m 1/2 , zˆh )  R (U m , C m 1/2 , zh )]  J 1  J 2  J 3  J 4.
With (A1) and (A2) we have:
J 1  ((u m  U m ).c m 1/2 , zh )  M u m  U m . zh
2
2
 M ( u m  U m  zh ).
Using (A1), (19), (20) and (9) we have:
(23)
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J 2  (U m .(c m 1/2  C m 1/2 ), zh )  M [ (c m 1/2  c m 1/2 )  (c m 1/2  C m 1/2 ) ]. zh
 M [  m 1/2   m 1/ 2 ]. zh
 M[
6 m 1/2 2
6 m 1/2 2

]  M zh
ˆ

k
k
2
2
2
2
 M (  m 1/ 2   m 1/2  zh ).
From (A2), (19), (20), (7) and (8) we have
J 3  (U mC m 1/2 , zh  zˆh )  M U m C m 1/2 . zh  zˆh
 M  m 1/2 zh  zˆh  M  m 1/2 zh  zˆh
 M c m 1/ 2 zh  zˆh  k1  k2  k3
2
2
k1  M [hc2   m 1/2 ]   zh
2
2
k2  M [hc2   m 1/2  zh ]
2
k3  M [hc2  zh ]
2
2
2
2
J 3  M [hc2   m 1/2   m 1/2  zh ]   zh .
Using (11), (9), (18) and (19) implies
J4
M
 z   (C
m 1 / 2
i
i 1
 Cijm1/ 2 )U m .nij d  C m1 / 2 .U m zˆh .
ji ij
Since Cijm1/ 2   ijm Cim1/ 2   mji C mj1/ 2 ,
J4 
 
eTh
P Pje ,i  j
( zi  z j )  (C m1 / 2  Cijm1 / 2 )U m .nij d 
ije
M C m1 / 2  c m1 / 2  c m1 / 2  c m1 / 2  c m1 / 2 zˆh
M
h
e
( z h e )
eTh

 (C

m 1 / 2
 Cijm1 / 2 )d
PiPj e ,i  j  e
ij
 M [  m1 / 2   m1 / 2  c m1 / 2 ] z h
and in element e we have C m1/ 2  Cijm1/ 2  2 (C m1/ 2 e) he , then
J 4  3M 2  he3 ( zh e) . (C m 1/2 e)  M [  m 1/2   m 1/2  c m 1/2 ] zh
eTh
AAM: Intern. J., Vol. 05, Issue 2 (December 2010) [Previously, Vol. 05, Issue 10, pp. 1682 – 1701]
 M (hc2   m1 / 2
2
  m 1 / 2
2
  m 1 / 2
2
2
595
2
 z h )   z h /
Hence, J1+J2+J3+J4 Implies (23).
Lemma 4.4. For all z h  M h and  i  ci  ci ,  i  ci  Ci ,
2
((u m .ci m 1/2 ), zh )  R(U m , Ci m 1/ 2 , zh )  M (hc2  i m 1/2   i m 1/2
2
2
  i m 1/2  u m  U m  zh
2
2
2
)   z h ,
(24)
where   0 is arbitrary small constant.
Proof:
((u m .ci m 1/2 ), zh )  R (U m , Ci m 1/2 , zh )  ((u m  U m ).ci m 1/ 2 , zh )
 (U m(ci m 1/2  Ci m 1/2 ), zh )  (U m .Ci m 1/2 , zh  zˆh )
 [(U m .Ci m 1/2 , zˆh )  R(U m , Ci m 1/2 , zh )]  J 1  J 2  J 3  J 4.
Same as Lemma 4.3.
Lemma 4.5. There exists a positive constant k2 such that:
P m1  P m1  k 2 C m1  c m1
(25)
Proof:
We have
(a (C m 1 )P m 1 ,  )  ( q m 1/ 2 , v)  ( Rs' (C m 1 ), v),
(26)
(a(c m 1 )p m 1 ,  )  (q m1/2 , v)  ( Rs' (c m1 ), v).
(27)
subtracting (27) from (26), we get
(a (C m 1 )( P m 1  P m 1 ),  )  ([a(C m 1 )  a(c m 1 )]P m 1 ,  )  ( Rs' (C m 1 )  ( Rs' (c m 1 ), v)
Let   P m1  P m1  Vh , then
2
( P m 1  P m 1 )  (a (C m 1 )( P m 1  P m 1 ), ( P m 1  P m 1 ))
 ([a (c m 1 )  a (C m 1 )]P m 1 , ( P m 1  P m 1 ))  ( Rs' (C m 1 )  ( Rs (c m 1 ), P m 1  P m 1 ) .
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A. Al-Bayati et al.
Using (A1), we have
 M P m 1
C m 1  c m 1
( P m 1  P m 1 )  C m 1  c m 1
P m 1  P m 1
with lemma 4.1 and (A2) if hP >0 is sufficiently small
P m1  P m1  k1 P
r 1
hPr  M , so we have (25).
Lemma 4.6. There exists a positive constant k3 such that
U m1  u m1  k3 ( C m1  c m1  hPr )
(28)
Proof:
U m1  u m1  a (C m1 )P m1  a (c m1 )P m1
 a (C m1 )( P m1  p m1 )  a (C m1 )  a (c m1 )
p m1 .
From (A1) and (A2) , we have
 const. P m1  p m1  C m1  c m1
p m1
L
we have P m1  p m1  P m1  p m1  P m1  p m1 .
Using Lemma 4.1 and Lemma 4.5, we get
P m1  p m1  k 2 C m1  c m1  k1 P m1
r 1
h pr , from (A2), (28) holds.
Theorem 4.3. For all m  l  N if    0 , then
c l 1  C l 1  M (  hc  hPr ) , where M independent of  and hc .
Proof:
Multiply (2) by zh and integrating by parts to obtain
t  (m  1 / 2) . Let wm1/ 2  w(., (m  1 / 2) ) and wm1/ 2  ( wm1  wm ) / 2 .
Then,
AAM: Intern. J., Vol. 05, Issue 2 (December 2010) [Previously, Vol. 05, Issue 10, pp. 1682 – 1701]
597
( D c m , zh )  ( Ecc m 1/2 , zh )  (u m c m 1/2 , zh )
c
m 1/ 2 ), z h )
t
 (( Ecc m 1/2  Eccm 1/2 ), zh )  ( (u m c m 1/2  u m1/2 cm 1/ 2 ), zh ).
 ( g (c m 1/2 ), zh )  ( ( D c m 
(29)
Let e  c  C  (c  c )  (C  c )     . Subtract (10) from (29), to obtain:
(ˆ D eˆ m , zˆh )  ( Ecem 1/2 , zh )  ( R (U m , C m 1/2 , zh )
 (u mc m 1/2 , zh ))  ((ˆ D cˆ m , zˆh )  ( D c m , zh ))
c
m 1/2 ) , zh )
t
 (( Ecc m 1/2  Eccm 1/2 ), zh )  ( (u mc m 1/2  u m1/2 cm 1/2 ), zh ).
 (( g (c m 1/2 ), zh )  ( gˆ (C m 1/2 ), zˆh ))  ( ( D c m 
Hence,
(ˆ D ˆ m , zˆh )  ( Ec m 1/ 2 , zh )  (ˆ D ˆ m , zˆh )  ( Ec m 1/2 , zh )
((u mc m 1/2 , zh )  R (U m , C m 1/2 , zh ))  (( g (c m 1/ 2 ), zh )  ( gˆ (C m 1/2 ), zˆh ))
( ( D c m , zh )  (ˆ D cˆ m , zˆh ))  (( ( D c m 
c
t
m 1/ 2
) , zh )
 (( Ecc m 1/2  Eccm 1/2 ), zh )  ( (u mc m 1/2  u m1/2 cm 1/2 ), zh )
 I1  I 2  I 3  I 4  I 5  I 6.
In (30) , let zh   m1 / 2  M h , and using (A1) the Left Hand Side (LHS) is
2

ˆ ˆ m1  ˆ m
2
2
2
 c0  m1 / 2 .
From (A1), we have:
I1  (ˆD ˆ m , zˆh )  M ( D ˆ m
2
2
 ˆ m1 / 2 .
Using (7) and (8), we have
D ˆ m  D  m  D  m  D ˆ m  D  m  Mhc D  m .
1
From (9) we have
ˆ m1 / 2  M  m1 / 2  M (  m1   m )
2
I1  M (  m1   m
2
 D  m
2
2
 hc2 D  m ) .
1
(30)
598
A. Al-Bayati et al.
2
2
Using (A1), we have I 2  M  m 1 / 2    m 1 / 2 .
From Lemma 4.3,
2
I 3  M (hc2   m 1/2   m 1/2
2
2
2
2
  m 1/ 2  u m  U m )    m 1/2 .
Let   
c m1  c m

Girault (1979))]
 (
c
t
m1/ 2
 1 / 24
 3c
t 3
2
) . Using (A1), (A2) we get [Raviart and
I 4  Mhc (1   2 ) z h 1  Mhc (1   2 ) z h
 Mhc z h  Mhc z h 1  Mhc  2 z h  Mhc 2 z h 1
2
 M (hc2  hc2 4 )    m1/ 2
2
2
2
I 5  M (hc2   m1 / 2   m1 / 2 )    m1 / 2 .
Let I6=K1+K2+K3. So,
2
2
K1  M ( 4   m 1   m ),
2
K 2  M  2    m 1/2 ,
2
2
K 3  M ( 2   m 1   m ).
Then, we get
I 6  M ( 2   m1 2   m 2 )    m1 / 2 .
Relation (30) can be written now as
2
ˆ ˆ m 1  ˆ m
2
2
2
2
2
2
 c0  m 1/2  M (  m 1   m  D  m  hc2 D  m   m 1
1
2
2
2
2
  m 1/2   m 1/2  u m  U m  hc2   2 )    m 1/2 .
Take the summation from 0 to l, where m  l  N , and C 0  c 0
so  0  0
2
AAM: Intern. J., Vol. 05, Issue 2 (December 2010) [Previously, Vol. 05, Issue 10, pp. 1682 – 1701]
l 1
2
ˆ ˆ l 1  M 1   m   M 2 D 
2
2
L2 ( L2 )
 M 2 hc2 D 
2
m 0
l
599
l
L2 ( H 1 )
 M 2   1   M 2 (hc2   2 )  M 3  u m  U m 
2
m 0
2
m 0
Where
w
2
L2  ( x )
N
  wm  .
2
m 0
x
Using (7) and (A1), we get
6 ˆ l 1
 l 1  ˆ l 1  Mhc  l 1  ˆ l 1  Mhc
  M ˆ l 1
k
1
ˆ ˆ l 1  M 0 ˆ l 1 ,
so
l 1
2
l
 l 1  M 1   m   M 3  u m  U m   M 2 ( D 
2
2
m0
m 0
2
L2 ( L2 )
 
 hc2 D 
2
L2 ( H 1)
2
L2 ( H 1 )
(31)
 h   ).
2
c
2
From Lemma 4.6, we have
l

m 0
2
l
u m  U m   M (  m   
2
2
L2 ( L2 )
 h p2 r )
m0
From (31) and (32), we have
l 1
 l 1  M 1   m   M 2 (hc2   2  h p2 r ) .
2
2
m 0
From Gronwall inequality, we get:
 l 1  M (hc    hPr ) .
It is c l 1  C l 1  M (hc    hPr ) .
Theorem 4.4. For all m  l  N if    0 , then
ci l 1  Ci l 1  M (  hc  hPr ) ,
where M independent of  and hc .
(32)
600
A. Al-Bayati et al.
Proof:
Multiply (4) by zh and integrate by parts to obtain
t  (m  1 / 2) . Let wm1/ 2  w(., (m  1 / 2) ) and w m1/ 2  ( w m1  w m ) / 2 .
Then
( K i D ci m , zh )  ( Ecci m 1/2 , zh )  (u m ci m 1/2 , zh )
 ( fi (c m 1/2 , c1m 1/2 ,..., cNm 1/ 2 ), zh )  ( Ki ( D ci m 
ci
t
m 1/ 2
(33)
), zh )
(( Ecci m 1/2  Ecci ,m 1/2 ), zh )  ( (u m ci m 1/ 2  u m1/2 ci ,m 1/2 ), zh ).
Let e  ci  Ci  (ci  ci )  (ci  Ci )   i   i . Subtract (10) from (33) to obtain:
(ˆ Ki D eˆi m , zˆh )  ( Ecei m 1/2 , zh )  ( R(U m , Ci m 1/2 , zh )
 (u m ci m 1/ 2 , zh ))  ((ˆ D cˆi m , zˆh )  ( D ci m , zh ))

 ( ( fi (c m 1/2 , c1m 1/2 ,..., cNm 1/2 ), zh )  ( f i (C m 1/ 2 , C1m 1/2 ,..., C Nm 1/2 ), zˆh ))  ( Ki ( D ci m 
ci
t
 (( Ecci m 1/ 2  Ecci ,m 1/2 ), zh )  ( (u mci m 1/2  u m1/2 ci ,m 1/2 ), zh ).
Hence,
(ˆ Kˆ i D ˆi m , zˆh )  ( Eci m 1/2 , zh )  (ˆ Kˆ i D ˆi m , zˆh )  ( Ec i m 1/2 , zh )
((u mc m 1/2 , z )  R (U m , C m 1/2 , z ))  (( fˆ (C m 1/2 , C m 1/2 ,..., C m 1/2 ), zˆ )
i
h
i
h
i
1
N
h

 ( f i (c m 1/2 , c1m 1/2 ,..., cNm 1/2 ), zh ))  ( K i D ci m , zh )  (ˆ K i D cˆi m , zˆh ))
ci
m 1/2
 Ecci ,m 1/2 ), zh )
m 1/2 ) , z h )  (( Ecci
t
( (u m ci m 1/ 2  u m1/2 ci ,m 1/2 ), zh )  I1  I 2  I 3  I 4  I 5  I 6.
(( ( D ci m 
The rest of the proof is the same as in Theorem 4.3.
Theorem 4.5. With the assumption (A1)~ (A3), if hc  O(hP ),  O(hP ) , then
cC
L ( l 2 )
 u U
L ( l 2 )
 M (  hc  hPr ) .
Proof:
We will prove, first the inductive assumption (19). We have
(34)
m 1/2
), zh )
AAM: Intern. J., Vol. 05, Issue 2 (December 2010) [Previously, Vol. 05, Issue 10, pp. 1682 – 1701]
U m1
L
 a (C m )
L
P m1
L
 k 4 P m1
L
601
.
But,
P m 1
L
 P m 1  P m 1
L
 P m 1
When hc  o(hP ) ,   o(hP ) (r  2) ,
L
 MhP1 P m 1  P m 1  k5  2k5 .
P m1
L
 k *,
U m1  u m1  k 3 ( C m1  c m1  hPr ) , m  0,1,..., N
(35)
and from Theorem 4.3,
c m1  C m1  M (hc    hPr ) , m  0,1,..., N .
(36)
Equations (35) and (36) implies
c m1  C m1  u m1  U m1  M (  hPr  hc ) , m  0,1,..., N .
This will complete the proof.
5. Numerical Applications
Example 5.1.
A two dimensional diffusion-convection problem of interest is concerned with a
description of sediment transport in channels. The governing differential equation is [Smith
et al. (1973)]:
dx
c
c c
 2c
 2c

d
 ux
 uy

y
2
2

x

y t
x
y
,
where c is the sediment concentration, dx and dy are the sediment diffusion
coefficients ,ux and uy are the fluid velocity component in the x and y directions. The
boundary conditions of the problem are
c u y
 c  M 1c,
y d y
c u y

 M2
y d y
(M 1 is constant),
at y  0,
(M 2 is constant), at y  -56.0.
We use a Kind of upwind finite element method to solve this problem with the mesh as
triangular element with 82 nodes and take the values ux=1. E-6, uy=0.49 , dx=0, dy=0.0135,
602
A. Al-Bayati et al.
we take the time step τ=300 and the number of time steps Nτ =30 for θ =1 ,the time step
τ=100 and the number of steps Nτ =90 for θ = 1/2 , and also the time step τ=300 and the
number of steps Nτ =30 for θ = 2/3 . In figure (5.1) we show that we can solve this
problem by decreasing the time step τ and increasing the number of time steps Nτ for a
kind of partial upwind finite element method.
Example 5.2.
In this example, we solve a purely convective problem in one dimension [Smith et al.
(1973)]:
u y
c c
 ,
y t
where c is the concentration in the region  2  y  0 , subject to the boundary conditions
c  1,
y  0,
0  t  0.2,
c  0,
c
 0,
y
y  0,
t  0.2,
y  2,
for all t.
We discretized this region into 200 triangular elements with 202 nodes and with the same
distance between any two nodes is 0.02 in both directions x and y, and take uy =1.0, the
time step τ=0.04 and the number of time steps Nτ =25 for θ =0.5. Figure (5.2) shows the
computed solution after one second and it draws between the concentration and coordinate
y, we can see that while y convergence to zero the value of concentration convergence
oscillation to solution.
1.00
Y=-56
=1,
0.80
=0.5,
Concentration
=2/3,
=300,
=30
=100,
=90
=300,
=30
0.60
=0.5,
=300,
=30
Y=-28
0.40
0.20
Y 0
0.00
0.00
40.00
80.00
120.00
160.00
Time (minutes)
Figure 5.1. A kind of partial upwind F.E.M. for triangular element at Different values of θ in Example 5.1.
AAM: Intern. J., Vol. 05, Issue 2 (December 2010) [Previously, Vol. 05, Issue 10, pp. 1682 – 1701]
1.20
Exact solution at
t=1 second
603
A kind of partial upwind FEM
Galerkin method for triangular element
0.80
Concentration
Galerkin method for quadrilateral element
0.40
0.00
-0.40
-2.00
-1.60
-1.20
-0.80
-0.40
0.00
Coordinate Y
Figure 5.2. Solutions in Example 5.2 after one second for θ =0.5, τ =0.04 , Nτ =25
6. Conclusions
We used the system with large coupled of strongly non-linear partial differential equations
arising from the contamination of nuclear waste in a porous media. For the incompressible
case the method satisfied the discrete mass conservation law for approximate C , Ci and
~
derived the error estimates in L (0, T , L2 ()) . The two examples included clearly
demonstrate certain aspects of the theory and illustrate the capabilities of a kind of partial
upwind F.E.M.
Acknowledgement
The author would like to thank the editor and, especially, the reviewers for their valuable
comments.
604
A. Al-Bayati et al.
REFERENCES
Ciarlet, P.G. (1978). The finite Element Method for Elliptic Problems. North-Holland,
Amsterdam, New York.
Chen, M.-C.; Hsieh, P. W.; Li, C.T.; Wang, Y.T. and Suh-Yuh Y. (2009). On two iterative leastsquares finite element schemes for the incompressible Navier-Stokes problem. Numerical
Functional Analysis and Optimization, 30, 436-461.
Gaohong, W. and Cheng A. (1999). Convergence of finite Element Method for Compressible
flow of contamination from nuclear waste. (Chinese Series), 3, 21-31.
Huang, C. (2000). Implementation of a locally conservative Eulerian Lagrangian method applied
to nuclear contaminant transport, "Numerical Treatment of Multiphase Flow in Porous
Media, State of Art", Lecture Notes in Physics. Springer-Verlag, Heidelberg, 552, 179-189.
Hu J. W. and Tian C. S. (1992). A Galerkin Partial Upwind Finite Element Method for the
Convection Diffusion Equations. (Chinese Series), 446-459.
Jim Douglas, Jr. (2001). The transport of nuclear contamination in fractured porous media.
Journal of the Korean Mathematical Society, 38, 723-761.
Jim Douglas, Jr. (2002). Parameter Estimates for a Limit Model for High-Level Nuclear Waste
Transport in Porous Media. Mathematical and Numerical Treatment, Contemporary
Mathematics, 295.
Manaa, S. A. (2000). A kind of Upwind F.E.M. for Numerical Reservoir Simulation. Ph.D
Thesis , Nankai University, China.
Quarteroni, A. and Valli, A. (1997). Numerical Approximation of partial differential Equations.
Springer-Verlag.
Raviart, P. A. and Girault, V. (1979). Finite Element Approximation of the Navier-Stokes
Equations. Lecture notes in mathematics, Springer-Verlag.
Smith, I. M., Farraday , R. V. and O’Connor, B. A. (1973), Rayleigh-Ritz and Galerkin Finite
elements for diffusion-convection water problems. Resources Research, 9, No.3, 593-606.
Wansophark, N. and Pramote D. (2008). Streamline upwind finite element method using 6-node
triangular element with adaptive remeshing technique for convective-diffusion problems.
Appl. Math. Mech. -Engl. Ed., 29, 439–1450.