c 2012 Institute for Scientific
Computing and Information
INTERNATIONAL JOURNAL OF
NUMERICAL ANALYSIS AND MODELING
Volume 9, Number 2, Pages 247–256
NUMERICAL SOLUTIONS FOR NONEQUILIBRIUM SOLUTE
TRANSPORT WITH FIRST-ORDER DECAY AND
ZERO-ORDER PRODUCTION
MING CUI∗ AND YAZHU DENG
Abstract. Solute transport in the subsurface is often considered to be a nonequilibrium process. Nonequilibrium during transport of solutes in porous medium has been categorized as either
transport-related or sorption-related. For steady state flow in a homogeneous soil and assuming a
linear sorption process, we will consider advection-diffusion adsorption equations. In this paper,
numerical methods are considered for the mathematical model for steady state flow in a homogeneous soil with a linear sorption process. The modified upwind finite difference method is adopted
to approximate the concentration in mobile regions and immobile regions. Optimal order l2 - error
estimate is derived. Numerical results are supplied to justify the theoretical work.
Key words. Solute transport, error estimate, modified upwind finite difference.
1. Introduction
Solute transport in the subsurface is often considered to be a nonequilibrium process. Nonequilibrium during transport of solutes in porous medium has been categorized as either transport-related or sorption-related. Transport nonequilibrium
(also called physical nonequilibrium) is caused by slow diffusion between mobile
and immobile water regions. These regions are commonly observed in aggregated
soils [8, 12] or under unsaturated flow conditions [2, 13, 14, 15], or in layered or
otherwise heterogeneous groundwater systems. Sorption-related nonequilibrium results from either slow intrasorbent diffusion [1] or slow chemical interaction [7]. In
most of these models, the soil matrix is conceptually divided into two types of sites;
sorption is assumed to be instantaneous for one type and rate-limited for the other
type.
Solute transfer between immobile/mobile water regions or instantaneous/ ratelimited sorption sites is commonly described by a first-order rate expression or by
Fica’s law if the geometry of the porous matrix can be specified. Models that
are based on well-defined geometry are difficult to apply to actual field situations,
they require information about the geometry of the structural units that are rarely
available [6]. Hence, the first-order rate formulation has been extensively used to
model underground contaminant transport. We start with a brief outline of two-site
nonequilibrium models as well as the two-region physical nonequilibrium models
which were given in [16]. General solutions are derived for the volume-averaged
solute concentration using Laplace transforms in [16].
Model
(1) Two-site Nonequilibrium Transport Model
Received by the editors October 30, 2010 and, in revised form, March 3, 2010.
2000 Mathematics Subject Classification. 35R35, 49J40, 60G40.
This research was supported by National Nature Science Foundation under Grant 10801092
and Independent Innovation Foundation of Shandong University, IIFSDU (No.2010TS006) and
Excellent Young Science Foundation of Shandong Province (No. 2007BS01021) and LASG State
Key Laboratory Special fund.
∗ Corresponding author Ming Cui: [email protected].
247
248
M. CUI AND Y. DENG
The two-site sorption model makes a distinction between type-1 (equilibrium)
and type-2 (first-order kinetic) sorption sites [9] and is given by
(1)
(1 +
f ρk ∂c
)
θ ∂t
∂2c
∂c
αρ
−v
−
[(1 − f )kc − sk ] − µl c
2
∂x
∂x
θ
f ρµs,e c
f ργs,e (x)
+ γl (x) +
,
θ
θ
= D
−
∂sk
= α[(1 − f )kc − sk ] − µs,k sk + (1 − f )γs,k (x).
∂t
where c is the volume-averaged concentration of the liquid phase; s is the concentration of the sorbed phase; D is the dispersion coefficient; θ is the volumetric
water content; v = q/θ is the average pore water velocity in which q is the volumetric water flux density; ρ is the bulk density; µl and µs are first-order decay
coefficients for degradation in the liquid and sorbed phases, respectively; γl and γs
are zero-order production terms for the liquid and sorbed phase, respectively; k is a
distribution coefficient for linear sorption; α is a first-order kinetic rate coefficient;
f is the fraction of exchange sites assumed to be at equilibrium; x is distance; t
is time; and the subscripts e and k refer to equilibrium and kinetic sorption sites,
respectively.
(2)
(2) Two-Region Nonequilibrium Transport Model
The two-region transport model assumes that the liquid phase can be partitioned
into mobile (flowing) and immobile (stagnant) regions and that solute exchange
between the two liquid regions can be modeled as a first-order process. The model
is given by
(3)
(θm + f ρk)
∂cm
∂t
=
−
(4) (θim + (1 − f )ρk)
∂cim
∂t
∂ 2 cm
∂cm
−q
− α(cm − cim )
∂x2
∂x
(θm µl,m + f ρkµs,m )cm + θm γl,m (x) + f ργs,m (x),
θm D m
= α(cm − cim ) − (θim µl,im + (1 − f )ρkµs,im )cim
+θim γl,im (x) + (1 − f )ργs,im (x).
where the subscripts m and im refer to mobile and immobile liquid regions, respectively; the subscripts l and s refer to the liquid and sorbed phases, respectively;
f represents the fraction of sorption sites that equilibrates with the mobile liquid
phase and α is a first-order mass transfer coefficient governing the rate of solute
exchange between mobile and immobile liquid regions. Note that θ is equal to
θm + θim .
If we employ dimensionless parameters listed in Table 1, equations (1), (2) and
(3), (4) reduce to the same dimensionless form. Dimensionless equations of the
nonequilibrium model for the case of linear sorption are given by
(5)
(6)
βR
∂C1
1 ∂ 2 C1
∂C1
=
−
− ω(C1 − C2 ) − µ1 C1 + γ1 (X),
2
∂T
P ∂X
∂X
(1 − β)R
∂C2
= ω(C1 − C2 ) − µ2 C2 + γ2 (X).
∂T
NUMERICAL SOLUTIONS FOR NONEQUILIBRIUM SOLUTE TRANSPORT
249
Table 1. Dimensionless Parameters for the Two-Site and TwoRegion Transport Models
parameter
two-site
two-region
T
vt/L
vt/L
X
x/L
x/L
P
vL/D
vm L/Dm
R
1+ρk/θ
1+ρk/θ
β
θ+f ρk
θ+ρk
θm +f ρk
θ+kρ
ω
α(1−β)RL
v
αL
θv
C1
c
c0
cm
c0
C2
sk
(1−f )kc0
cim
c0
µ1
(θµl +f ρkµs,e )L
θv
(θm µl,m +f ρkµs,m )L
θv
µ2
(1−f )ρkµs,k L
θv
(θim µl,im +(1−f )ρkµs,im )L
θv
γ1
L(θγl +f ργs,e )
θvc0
L(θm γl,m +f ργs,m )
θvc0
γ2
L(1−f )ργs,k
θvc0
L(θim γl,im +(1−f )ργs,im )
θvc0
Here c0 and L represent characteristic concentrations and lengths, respectively.
where C1 is the reduced volume-averaged solute concentration, C2 is the reduced
kinetically adsorbed concentration, µ is a first-order decay coefficient, γ is a zeroorder production coefficient, X and T are space and time variables, respectively; β,
R, P and ω are adjusted model parameters and subscripts 1 and 2 on µ and γ refer
to equilibrium and nonequilibrium sites, respectively. We shall assume that ω and
µ cannot be negative. Notice that zero-order production terms in (5) are functions
of position X, but that first-order rate coefficients are assumed to be constant.
Subscripts 1 and 2 refer to mobile and immobile regions if dimensionless transport
equations are interpreted in terms of the two-region model.
These dimensionless transport equations given by (5) (6) will be solved for the
following general initial and boundary conditions:
(7)
C1 (X, 0) = C2 (X, 0) = Ci (X),
δ ∂C1
+ C1 )|X=0 = C0 (T ),
P ∂X
with δ = 0 for a first-type and δ = 1 for a third-type boundary condition and
(8)
(−
∂C1
(∞, T ) = 0,
∂X
where Ci is the initial concentration and C0 is the boundary concentration.
The organization of the paper is as follows. In section 2, the modified upwind
difference methods is proposed to the dimensionless mathematical model for steady
(9)
250
M. CUI AND Y. DENG
state flow in a homogeneous soil with a linear sorption process. Optimal order
estimate in l2 - norm is derived. In section 3, numerical results were presented for
the nonequilubrium transport model which was given in [16].
2. The modified upwind finite difference method
There have been many numerical methods for miscible displacement in porous
media. In [5] the finite difference method is proposed combining with the method
of characteristic method for convection-dominated problems in porous media. An
alternating direction method combined with a modified method of characteristic is
presented in [4] for miscible displacement influenced by mobile water and immobile
water. Yi-rang Yuan [17] formulated a modified upwind finite difference procedure
for compressible two-phase displacement problems. The modified upwind finite
difference method is efficient to treat the equation with significant convection and
has the second-order accuracy in space variable.
We assume that coefficients satisfy:
0 < a∗ ≤ a(x) ≤ a∗ ,
0 < b∗ ≤ b(x) ≤ b∗ ,
0 < D∗ ≤ D(x) ≤ D∗ ,
where a∗ , a∗ , b∗ , b∗ , D∗ , D∗ are positive constants.
We also assume that the solution c1 , c2 of (5) and (6) satisfy:
(10)
c1 , c2 ∈ W 1,∞
\
L∞ (W 4,∞ ),
∂c1 ∂c2
∂ 2 c1 ∂ 2 c2
,
∈ L∞ (W 1,∞ ),
,
∈ L∞ (W 1,∞ ).
∂t ∂t
∂t2 ∂t2
For convenience, we introduce new parameters:
u = 1, D = 1/P, a = βR b = (1 − β)R.
For the sake of simplicity, denote the bounded domain Ω = [0, 1], the space step
h = N1 , code xi = ih, i = 0, 1...N , the time step tn = n∆t, and cni = c(xi , tn ),
n
wij
= w(xij , tn ).
Let δx , δy and δx̄ , δȳ stand for standard backward difference respectively.
Define
Di+1,j + Di,j
Di−1,j + Di,j
Di+ 12 ,j =
, Di− 12 ,j =
.
2
2
Similarly, we can define Di,j+ 21 , Di,j− 12 .
n
n
Now, assume {C1,ij
, C2,ij
} are known. Then the modified upwind finite difference
n+1
n+1
method for (5) (6) is given by: finding that {C1,ij
, C2,ij
}, for 1 ≤ i, j ≤ N − 1
satisfying
(11)
n+1
n+1
n
n
C2,ij
− C2,ij
C1,ij
− C1,ij
n+1
n+1
+ bij
+ δU n+1 ,x C1,ij
+ δW n+1 ,y C1,ij
aij
∆t
∆t n+1 −1
−1
n+1
n+1
n+1
h |Uij |
h |Wij |
− 1 + 2 Dij
δx̄ (Dδx Cij ) + 1 + 2 Dij
δȳ (Dδy Cij )
n+1
n+1
+µ1,ij C1,ij
+ µ2,ij C2,ij
= γ1,ij + γ2,ij ,
(12)
bij
n+1
n
C2,ij
− C2,ij
n+1
n+1
n+1
= ω(C1,ij
− C2,ij
) − µ2,ij C2,ij
+ γ2,ij ,
∆t
where
−1
−1
n
n
n
δU n ,x C1,ij
= Uijn H(Uijn )Dij
Di− 21 ,j δx̄ Cij
+ 1 − H(Uijn ) Dij
Di+ 12 ,j δx Cij
,
NUMERICAL SOLUTIONS FOR NONEQUILIBRIUM SOLUTE TRANSPORT
n
δW n ,y C1,ij
251
−1
−1
n
n
n
n
n
= Wij H(Wij )Dij Di,j− 12 δȳ Cij + 1 − H(Wij ) Dij Di,j+ 12 δy Cij ,
1, z ≥ 0,
0, z < 0.
and initial values are given by:
H(z) =
0
0
C1,ij
= c1,0 (xij ), C2,ij
= c2,0 (xij ), 0 ≤ i, j ≤ N.
3. Convergence Analysis
Let ζ = c1 − C1 and ξ = c2 − C2 , where c1 , c2 are the solutions of (5) (6) and
C1 , C2 are numerical solutions of difference scheme (11) (12) respectively. Then we
define the inner product and norm in discrete space l2 (Ω).
(f n , g n ) =
N
X
n 2
fijn gij
h , kf n k2 = (f n , f n ).
i,j=1
By introduce the induction hypothesis
sup {kξ n k, kζ n k, kδxζ n k + kδy ζ n k} → 0, (h, ∆t) → 0.
(13)
0≤n≤L
and some techniques and applying discrete Gronwall Lemma, the following theorem
can be obtained.
Theorem 3.1. Suppose that the solution of the problem (5) (6) satisfies the condition (10) and the time and space discretization satisfy the relationship ∆t = O(h2 ),
then the following error estimate holds for the modified upwind difference scheme
(11) (12):
(14)
where
kc1 − C1 kL∞ ([0,T ];h1 ) + kc2 − C2 kL∞ ([0,T ];l2 ) + kdt (c1 − C1 )kL2 ([0,T ];l2 )
+ kdt (c2 − C2 )kL2 ([0,T ];l2 ) ≤ M ∗ ∆t + h2 ,
n
kgkL∞ (J;X) = sup kg kX ,
n∆t≤T
kgkL2(J;X) = sup
n∆t≤T
(
L
X
kg n k2X
∆t
n=0
) 21
,
2 2 ∂ c1 ∂ c2 M =M 2 ,
, kc1 kL∞ (W 4,∞ ) , kc1 kW 1,∞ .
∂t L∞ ∂t2 L∞
∗
∗
Proof. (6) and (12) can be combined to get:
n+1
n
ξij
− ξij
n+1
n+1
n+1
= ω ζij
− ξij
− µ2,ij ξij
+ ε2 (xij , tn+1 )
∆t
2 ∂ c2 where εn+1
≤
M
2,ij
∂t2 ∞ ∆t, then ε2 = O (∆t) .
L
Suppose the time step and the space step satisfy the relationship ∆t = O h2 .
and introduce the hypothesis (13)
(15)
bij
sup {kξ n k, kζ n k, kδx ζ n k + kδy ζ n k} → 0,
(h, ∆t) → 0.
0≤n≤L
n
Take the inner product of the relation (15) against the test function δt ξij
=
n+1
n
n
ξij − ξij = dt ξij ∆t, then
n
(bdt ξ n , dt ξ n ) ∆t = ω ζ n+1 − ξ n+1 , dt ξ n ∆t− µ2 ξ n+1 , dt ξ n ∆t+ εn+1
∆t.
2,ij , dt ξ
252
M. CUI AND Y. DENG
Thus, we get:
2 2 2
2
kdt ξ n k ∆t ≤ M ζ n+1 + ξ n+1 ∆t + M (∆t) ∆t.
(16)
From (5) and (11), we get:
(17)
n+1
n+1
n
n
ζij
− ζij
ξij
− ξij
n+1
n+1
aij
+ bij
+ δU n+1 ,x ζij
+ δW n+1 ,y ζij
∆t
∆t
n+1
n+1
+ µ1,ij ζij
+ µ2,ij ξij
=
−1
n+1
−1
n+1
)δx̄ (Dδx ζij
) + (1 + h2 |Wijn+1 |Dij
)δȳ (Dδy ζij
)}
{(1 + h2 |Uijn+1 |Dij
+
ε1 (xij , tn+1 ), 1 ≤ i, j ≤ N − 1
n+1
ζij
= 0,
(18)
xij ∈ ∂Ω1 .
where, we have used the fact that
n+1
n+1 ∂c
δU n+1 ,x cn+1
−
U
ij
∂x
ij
n+1 n+1
n+1 n+1
n+1
Uij + Uij
Uij
− Uij
−1
−1
n+1
n+1 ∂c
=
Dij
Di− 1 ,j δx̄ cn+1
+
D
D
δ
c
−
U
1
ij
ij
i+ 2 ,j x ij
2
2
2
∂x
ij
!
3 h n+1 −1
∂
c
2
=−
U
Dij δx̄ Dδx cn+1 ij + O ∂x3 ∞ ∞ h ,
2 ij
L (L )
n+1
n+1 ∂c
δW n+1 ,y cn+1
−
W
ij
∂y
ij
n+1 n+1
n+1
Wijn+1 − Wijn+1 −1
Wij + Wij
n+1
n+1 ∂c
−1
δ
c
−
W
=
Dij Di,j− 1 δy cn+1
+
D
D
y ij
ij
ij
i,j+ 1
2
2
2
2
∂y
ij
!
3 ∂
h n+1 −1
c
2
=−
Wij
Dij δy Dδy cn+1 ij + O ∂y 3 ∞ ∞ h ,
2
L (L )
and the facts that
−1
∂cn+1
h n+1 −1
∂
D
− 1+
Uij
Dij
δx̄ Dδx cn+1 ij
∂x
∂x
2
ij
!
4 c
h n+1 −1
∂
2
=
U
Dij δx̄ Dδx cn+1 ij + O ∂x4 ∞ ∞ h .
2 ij
L (L )
=
−1
∂
∂cn+1
h
−1
D
− 1 + |Wijn+1 |Dij
δȳ Dδy cn+1 ij
∂y
∂y
2
ij
!
4 h n+1 −1
∂
c
2
Wij
Dij δȳ Dδy cn+1 ij + O ∂y 4 ∞ ∞ h .
2
L (L )
Then, we get the following estimates:
(19)
n
n
cn+1
cn+1
1,ij − c1,ij
2,ij − c2,ij
n+1
aij
+ bij
+ δU n+1 ,x cn+1
1,ij + δW n+1 ,y c1,ij
∆t
∆t
−1
−1
n+1
|U n+1 |
n+1
h |Wij |
− 1 + h2 Dijij
δx̄ (Dδx cn+1
)
+
1
+
δ
(Dδ
c
)
ȳ
y
ij
ij
2 Dij
n+1
n+1
+µ1,ij cn+1
1,ij + µ2,ij c2,ij − γ1,ij − γ2,ij = ε1,ij .
NUMERICAL SOLUTIONS FOR NONEQUILIBRIUM SOLUTE TRANSPORT
253
where
εn+1
1,ij
(
)
2 ∂ 2 c1 ∂ c2 2
≤M ∂t2 ∞ ∞ , ∂t2 ∞ ∞ , kc1 kL∞ (W 4,∞ ) (∆t + h .)
L (L )
L (L )
n
Take the inner product of the relation (17) against the test function δt ζij
=
n
n
− ζij = dt ζij ∆t, we get
n+1
ζij
+
+
(20)
+
=
(adt ζ n , dt ζ n ) ∆t + (bdt ξ n , dt ζ n ) ∆t
δU n+1 ,x ζ n+1 , dt ζ n ∆t + δW n+1 ,y ζ n+1 , dt ζ n ∆t
−1 n+1
Dδx ζ n+1 , δx [ 1 + h2 U n+1 D−1
ζ
− ζn ]
−1 n+1
Dδy ζ n+1 , δy [ 1 + h2 W n+1 D−1
ζ
− ζn ]
n
−µ1 ζ n+1 , dt ζ n ∆t + −µ2 ξ n+1 , dt ζ n ∆t + εn+1
∆t.
1,ij , dt ζ
The term on the left-hand side of (20) can be estimated by
Dδx ζ n+1 , δx [(1 + h2 U n+1 D−1 )−1 (ζ n+1 − ζ n )]
= Dδx ζ n+1 , δx (1 + h2 |U n+1 |D−1 )−1 (ζ n+1 − ζ n )
+ Dδx ζ n+1 , (1 + h2 |U n+1 |D−1 )−1 δx (ζ n+1 − ζ n ) .
Note that
−1
h n+1 −1
|D
=
δx 1 + |U
2
1+
n+1 n+1 −1
U
− U
i,j
i+1,j Dij
h n+1 −1
U
D
1 + h U n+1 D−1
h
2
2
i+1,j
ij
2
i,j
ij
n+1 −1
h Dij
2 δx Uij
n+1 −1 .
≤
h n+1 −1
D
1 + 2 Ui+1,j Dij
1 + h2 Ui,j
ij
then
≥
Dδx ζ n+1 , δx [(1 + h2 U n+1 D−1 )−1 (ζ n+1 − ζ n )]
−1
1
δx ζ n+1
Dδx ζ n+1 , 1 + h2 U n+1 D−1
2
2
−1
2
− Dδx ζ n , 1 + h2 U n+1 D−1
δx ζ n − δx ζ n+1 ∆t − ǫ kdt ζ n k ∆t.
Similarly, we have
h
−1 n+1
i
Dδy ζ n+1 , δy 1 + h2 W n+1 D−1
ζ
− ζn
−1
1
Dδy ζ n+1 , 1 + h2 W n+1 D−1
≥
δy ζ n+1
2
2
−1
2
− Dδy ζ n , 1 + h2 W n+1 D−1
δy ζ n − δy ζ n+1 ∆t − ǫ kdt ζ n k ∆t.
Using the fact that
2
δU n+1 ,x ζ n+1 , dt ζ n ∆t ≤ M δx ζ n+1 ∆t + ǫ kdt ζ n k2 ∆t,
2
2
δW n+1 ,y ζ n+1 , dt ζ n ∆t ≤ M δy ζ n+1 ∆t + ǫ kdt ζ n k ∆t.
254
M. CUI AND Y. DENG
Then, we have
kdt ζ n k2 ∆t + kδx ζ n+1 k2 − kδx ζ n k2 + kδy ζ n+1 k2 − kδy ζ n k2
(21)
≤ M {kdt ξ n k2 + k∇h ζ n+1 k2 + k∇h ζ n k2 + kζ n+1 k2 + kξ n+1 k2 }∆t
+M {(∆t)2 + h4 }∆t.
If (21), is summed in time for 0 ≤ n ≤ L, we have
L
P
2
2
kdt ζ n k ∆t + ∇h ζ L+1 n=0
o
L n
P
≤ M
kdt ξ n k2 + k∇h ζ n+1 k2 + kζ n+1 k2 + kξ n+1 k2 ∆t
(22)
+ M
n=0
L n
P
n=0
o
2
(∆t) + h4 ∆t.
Similarly, for (16), we sum in time for 0 ≤ n ≤ L and get
L
X
(23)
2
kdt ξ n k ∆t ≤ M
n=0
L n
o
X
n+1 2 n+1 2
ζ
+ ξ
+ (∆t)2 ∆t.
n=0
0
0
Noting that ζ = ξ = 0, we have
L
L
L+1 2
P
P
≤M
ζ
kζ n k2 ∆t + ǫ
kdt ζ n k2 ∆t,
n=0
n=0
(24)
L
L
L+1 2
P
P
2
2
n
≤M
ξ
kξ k ∆t + ǫ
kdt ξ n k ∆t.
n=0
n=0
then
≤
L P
2
2 2
2
kdt ζ n k ∆t + kdt ξ n k ∆t + ζ L+1 1 + ξ L+1 n=0
o
n
o
L n
P
2
2
2
2
M
k∇h ζ n k + kζ n k + kξ n k ∆t + M (∆t) + h4 .
n=0
An application of the Gronwall lemma shows that
n
o
L 2
2 P
2
2
2
(25)
kdt ζ n k + kdt ξ n k ∆t + ζ L+1 1 + ξ L+1 ≤ M (∆t) + h4 .
n=0
It is easy to know that the induction hypothesis holds. Thus, we get the proof
of the theorem 3.1.
4. Numerical Examples
For the nonequilubrium transport model in one dimension [7, 8, 16],
∂c1
∂ 2 c1
∂c1
=D 2 −
− ω(c1 − c2 ) − µ1 c1 + γ1 (x),
∂t
∂x
∂x
(26)
∂c2
(1 − β)R
= ω(c1 − c2 ) − µ2 c2 + γ2 (x).
∂t
Van Genuchten[7, 8, 16] presented the analytical solutions for different initial and
boundary conditions. In Fig.1 and Fig. 2, the analytical solution and the solution
of modified upwind finite difference method are compared. It is shown that the
modified upwind finite difference method(MUDM) is more accurate that upwind
difference method(UDM).
βR
NUMERICAL SOLUTIONS FOR NONEQUILIBRIUM SOLUTE TRANSPORT
255
Example 1. Initial Value Problem with Stepwise Initial Distribution.We choose
D = 0.1, β = 0.5, R = 2, ω = 1, ν1 = ν2 = 0.2, γ1 = γ2 = 0, ∆t = 0.0025 and
h = 0.05. Let


 0.3, 0 ≤ x < 0.5,
1, 0.5 ≤ x < 1.
c1,0 (x) = c2,0 (x) =

 0.1, 1 ≤ x
0.45
1 exact
2 modified upwind
3 upwind difference
0.4
concentration−−C
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
x
2
2.5
3
Figure 1. Calculated resident equilibrium C1 distribution versus
distance x at T = 1
Example 2. Boundary Value Problem with Stepwise Boundary Condition. We
choose D = 0.25, β = 0.5, R = 2, ν1 = ν2 = 0, γ1 = γ2 = 0. and ∆t = 0.006, h =
0.08. Assume that the initial concentration is zero. The boundary condition is given
by
1, 0 < t ≤ 3,
c0 (t) =
0, t > 3.
0.4
data 1
data 2
data 3
0.35
concentration−C
0.3
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
x
8
10
12
Figure 2. Calculated resident equilibrium C1 distribution versus
distance x at T = 3
In Fig.2, data 2 is the analytical solution c1 at T = 3. data 1 is obtained by
modified upwind finite difference method and data 3 is obtained by upwind finite
difference method.
256
M. CUI AND Y. DENG
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School of Mathematics, Shandong University, Jinan, China 250100; LASG, Institute of Atmospheric Physics, Chinese Academy of Sciences, Beijing 100029, China.
E-mail : [email protected]; [email protected]
School of Mathematics, Shandong University, Jinan, China 250100
E-mail : [email protected]