12
1,
2
,,,
nn
h ijkhijk
n
hijk
nnn
zh h
zijk
ijk
CC
Ct
hUDD CC
kijk kij



 



11
,,,
nn
h ijkijkijkh
CgX

(10a)
(10b)

where


11,
,
nijk ijk
Ux
c U11
1,1,1/ 2,1,1,1/ 2,
1, 1,
1,
nn n
ijk ijkijk ijk
ijkijkxijk
x
HUDDHUD Dc









2
11
2,2, ,1/2,2,2, ,1/2,
2, 2,2,
,1,
nnn n
ijk ijkijk ijk
ijkijkijkijky ijk
y
Uy
cUHUD DHUDDc









3
11
3, 3,,1/23,3,,1/2
3, 3,3,
,1,
nnn n
ijkij kijkij k
ijkijkijkijkz ijk
z
Uz
c UHUDDHUDDc







and

1,0, .
0, 0.
z
Hz z



12/3
,,
,
1
3
12
,
,,,
nn
hijk hijk
s ijk
nnn
hz hh
zijk
HH
St
KCHH
kijk kij




11
,,.
nn
h ijkijkijkh
HhX

Next, for fluid Equation (1) the fractional steps sinite
difference scheme is given by









,
,,
0
 
1/3
,, 1/ 3
,1
23
1
,, ,,
nn
h ijkh ijknn
sijkhx h
xijk
nn nn
hyh hzh
yz
ijk ijk
n
nn h ijk
h ijkh ijkn
ijk ijk
nn
HH
SKCH
t
KCHKC H
C
CC q
t
KCCijkii jk







31
2hh
zijk
1/3 1
,,,
nn
h ijkijkijkh
HhX


(11a)
(11b)
 

 
2/3 1/3
,,
,
2/3
2
12
,
,,,
nn
h ijkhijk
s ijk
nn n
hy hh
ijk
HH
St
KC HH
kjjik




2/3 1
,,,
nn
h ijkijkijkh
HhX


y
ji
(12a)
(12b)
(13a)
(13b)


00
,0, 0
,,
h ijkijkh ijkijkijkh
CcXHHXX .
,
nn
CH n
t
(14)
The algorithm for a time step is as follows: Assuming
the Approximate solution
,,hi
jkhijk at time is
known, it is needed to find out the approximate solutions
at the next time level. First, compute Darcy’s velocity
n
U, from schemes (8a), (8b), and method of speedup is
used to get the solution of transition sheaf
1/3n
C

2/3n
C
,hijk
along x direction. Secondly, from schemes (9a), (9b) we
obtain the solution of transition sheaf ,hijk . Thirdly,
from schemes (10a), (10b) we obtain solution
1n
C
1/3n
H
,hijk .
Next, from (11a), (11b), by using method of speedup, we
get the solution of transition sheaf
,hijk along x
direction; from (12a), (12b) we obtain the solution of
transition sheaf
2/3n
H
,hijk . Finally, from schemes (13a),
(13b) we obtain solution
1
,
n
h ijk
H. So a complete time
Y. R. YUAN ET AL.
976
step can be taken. At last, it is because of the positive
condition that this finite difference solution exists, being
the sole one.
2.2. The First Order Weighted Upwind Finite
Difference Fractional Steps Scheme
For salt concentration Equation (2), the first order up-
wind finite difference fractional steps scheme is given by






12
1/3
,,
,
23
,,
,,
*,
,,1
,,
nn
nn
h ijkh ijk
n
hijk x
yh z
yZ
ijk
nn
Ux Uy
h ijkh ijk
nnn
ijkh ijkh ijk
CC
t
DCD
CC
qCCi jk

 





3
1/3
1
,
,
2
,,
n
n
x hijk
nn
h
ijk
n
Uz h ijk
CD
C
C
C
i ijk

1,,
ijk h
X
(8a)
1/3
,
nn
h ijkijk
Cg
 (8b)
 

2/3 ,
n
hijk
D
CC

2/3 1,,
nn
ijk h
X


 
2/3 1/3
,,
2
,
12
,,
,
nn
h ijkh ijk
nn
hijkyh
y
CC
C
t
jikj jik



(9a)
,hijk ijk
Cg (9b)
 

12/3
1,
nn
nnn
hh ijk
CC DCC

11
,,,
nn
h ijkijkijkh
CgX


 
,,
3
,
12
,,
,
hijkh ijk
h ijkz
z
C
t
ki
jk kij


(10a)
(10b)
where
 



1,1, 1,1,
1
1
,
nnn n
Ux
ijkijkijkijkx ijk
cUHUHUc
h
 



1, 1, 1,
12
x
nijk i jki jk
Uc c


and 23
,,
,
nn
Uy Uz
ijk ijk
cc

are defined similarly with a con-
stant
0, 1


3
0,1 ,
.
The algorithm is similar to that of Scheme (8)-(13).
3. Convergence Analysis
For brevity we assume 1,hN
T
ijk

,, ,
X
ih,tnt

,nn
WX tWjh khn
,
ijk ijk . Let
hh
π,
H
HcC
 where H and c are the exact
solutions of this problem (1) - (5), and Hh and Ch are the
difference solutions of the schemes (8) - (13). ,, and
1 denote the inner product and the norms on the dis-
crete spaces l2() and h1() corresponding to L2() and
H1() [19,20,29]. First consider the second order scheme.
Theorem I. Suppose that the exact solutions of prob-
lem (1)-(5) satisfy condition:


Adopt the modified method of upwind procedures (8)-
(13). Let the dissectible satisfy relation:
2
tOh .
Then the following error estimates hold:



1, 1,4,
2
,,
,.
Hc WWL W
4, 22 2
,,
H
tctLWHt c


 t LL
 



2
11 2
22
(; )(; );
*2
(;)
hhth
LJhLJh
L
Jl
th
LJl
HHcC dHH
dcCM th


 (15)
where
 
2
2
;;
0
sup ,sup
N
nn
LJS LJS
SS
ntTNtT n
gg gt


g


*
.
H
M
depends on ,
Constant c
1/3n
C2/3n
h
C

and their derivatives.
Proof. First consider the concentration equation. For
Equations (8)-(13), dispel h and , and we
get the following equivalent form:



123
1
11
,, 1
11
,1
1
11
22
2
1
11
33
3
,,,,
,, ,
12
12
12
nnn
nn
h ijkh ijk
nnn
hijkx h
xijk
ijk
nn
yh
yijk
ijk
nn
zh
Zijk
ijk
nnnn
h ijkh ijkhijkijkh
Ux Uy Uz
CC h
CUDDC
t
hUDD C
hUDD C
CCCqC
















 






*,
,
1
1
21
11
1
1
1
22
2
1
11
11
1
1
1
33
3
1
2
2
12
1||
2
12
12
12
nn
ijkh ijk
nn
xh
x
ijk
nn
yt h
y
ijk
nn
xh
x
ijk
nn
zt h
z
ijk
n
ijk
C
h
tUDDC
hUD DdC
hUD D C
hUDDdC
hUD





 
























1
1
1
1
1
33
3
1
1
3
1
1
1
1
1
122
2
1
1
133
3
12
1||
2
12
1,
2
1,,
n
yh
y
nn
zt h
z
ijk
n
x
ijk
nn
xh y
y
nn n
hzth
z
ijk
DC
hUDDdC
h
tUD
h
DCUDD
h
CUDDdC
ijkN


 






 









1,
(16a)
Y. R. YUAN ET AL.
977
11
,
,
ijk ijkh
X


1n
.
(16b)
nn
h ijk
Cg
From Equation (2) (tt) and (16) we have the concentration error equations.


 
1
11 22
11
1
,, 1
11
,1
11
11
11
223 3
23
1
,,
,, ,
12
11
22
nn n
nn
hijk hijk
nnn
hijkx h
xijk
ijk
nnnn
yh zh
yZ
ijk ijk
ijk ijk
nn n
h ijkijkh ijk
Uxu xUyu
h
CUDD
t
hh
UD DUDD
Cc C



 







 

 






11
33
11
,
,,,
11
11
11
111
11
11
11
11
222
22
11
22
11
22
12
nnn
nnn
ijkh ijkijk
yUzuz
nnn
xh
xijk
ijk ijk
nnn
yh
yijk
ijk ijk
cCc
hh
uDUD DC
hh
uDUD DC
h


















 







 







 




11
11
11
333
33
1*, 1*,
,,,
11
11
21111
11 22
12
12
11
22
nnn
zh
zijk
ijk ijk
nnnnnn
ijkh ijkijkijkh ijkh ijk
nnn
xy
xy
ijk
h
uDUD DC
qC cqCC
hh
tuDDcuDDd

 

















 
 

 

 

n
t
c




11
11
1
11 22
12
11
11
111
11 22
12
11
22
11
22
nn
nn
xh yth
xy
ijk ijk
nnn
xy
xy
ijk ijk
hh
UDDCUDDdC
hh
uDDcUD DdC

 











 

 


 




 

 

 


 



ijk
n
th

 


11
11
31111
11 22
12
11
11
133 1
31
1
1
1
12
2
{1( (1(
22
11
22
1||
2
nnn
xy
xy
ijk
nnn
zt
zx
ijk
ijk
nn
xh
hh
tuDDc uDD
hh
uD DdcUD
h
DCUD
 
 
 






11n
c

 
 
 
 
 
 
 
 







1
1
1
23
3
1
31,
1||
2
,1,,1,
nn
yh
y
nn
zt hijk
zijk
h
DC UD
DdC ijkN

 




(17a)
10, ,
n
ijkijk h
X

(17b)
where

12
.
1,
nijk
th

1/3n
h
H2/3n
h
H
Next, consider the fluid equation. For Equations
(11)-(13), dispel and , and we get the
following equivalent form:
Y. R. YUAN ET AL.
978






 


 


 
1
,, 11
,12
1
,
,,
3
0
21
12
1
13
nn
h ijkhijknn nn
sijkhxhhyh
xy
ijk ijk
n
nn h ijk
h ijkh ijknnn
ijkijkh h
zijk
nnn
hxs hyth
xy ijk
nn
hxs hzth
xz
HH
SKCHKCH
t
C
CC qKCC
t
tKCSKCdH
KCSKC dH
 
 
 



 



1
3
nn
hzh
z
ijk
KCH








1
23
311
123 ,1
nn
hys h
yz
ijk
nnnn
hxshys hzth
xyz ijk
KCSKC
tKCSKCSKCdH







, ,1,
nn
zthijk
dH
ijkN
 
11
,.
nn
ijk h
HhX


1n
(18a)
,hijk ijk (18b)
From Equation (1) (tt) and (18) we have the fluid error equations.






 

 




1
1
,
11 1
00
11
33
nn
nn ijkh ijk
ijk ijk
nnnn
hhhijkijk
ijk
nn nn
hh
zz
ijk ijk
cC
Kc KCHq
t
Kc cKCC


 




   







1
111
,1 2 3
ππ
πππ
nn
ijk ijknnnn nn
sijkhxhy hz
xy z
ijkijk ijk
n
ijk
SKCKCKC
t
q
 






 


 


 


 


 


 
2111
12
11
12 1
11
13 2
1
23
nnn
xs yt
xy ijk
nnn n
hxs hythxs
xyx z
ijk
nnn n
hxs hzthys
xzy z
ijk
nn
hys hz
yz
tKcSKcdH
KCSKC dHKcSKc
KCSKC dHKcSKc
KCSKC




 













11
3
11
3
nn
zt ijk
nn
zt ijk
dH
dH


 
The Theory and Application of Upwind Finite Difference Fractional Steps Procedure for Seawater Intrusion