bounded domain problem for the modified · 2013. 1. 21. · bounded domain problem for the modified...
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BOUNDED DOMAIN PROBLEM FOR THE MODIFIEDBUCKLEY-LEVERETT EQUATION
YING WANG †§ AND CHIU-YEN KAO†‡
Abstract. The focus of the present study is the modified Buckley-Leverett (MBL) equation1
describing two-phase flow in porous media. The MBL equation differs from the classical Buckley-2
Leverett (BL) equation by including a balanced diffusive-dispersive combination. The dispersive term3
is a third order mixed derivatives term, which models the dynamic effects in the pressure difference4
between the two phases. The classical BL equation gives a monotone water saturation profile for5
any Riemann problem; on the contrast, when the dispersive parameter is large enough, the MBL6
equation delivers non-monotone water saturation profile for certain Riemann problems as suggested7
by the experimental observations. In this paper, we first show that the solution of the finite interval8
[0, L] boundary value problem converges to that of the half-line [0,+∞) boundary value problem9
for the MBL equation as L → +∞. This result provides a justification for the use of the finite10
interval in numerical studies for the half line problem. Furthermore, we numerically verify that the11
convergence rate is consistent with the theoretical derivation. Numerical results confirm the existence12
of non-monotone water saturation profiles consisting of constant states separated by shocks.13
Key words. conservation laws, dynamic capillarity, two-phase flows, porous media, shock waves,14
pseudo-parabolic equations15
AMS subject classifications. 35L65, 35L67, 35K70, 76S05, 65M0616
1. Introduction. The classical Buckley-Leverett (BL) equation [3] is a simple17
model for two-phase fluid flow in a porous medium. One application is secondary re-18
covery by water-drive in oil reservoir simulation. In one space dimension the equation19
has the standard conservation form20
ut + (f(u))x = 0 in Q = {(x, t) : x > 0, t > 0}u(x, 0) = 0 x ∈ (0,∞) (1.1)u(0, t) = uB t ∈ [0,∞)
with the flux function f(u) being defined as21
f(u) =
0 u < 0,
u2
u2+M(1−u)2 0 ≤ u ≤ 1,1 u > 1.
(1.2)
In this content, u : Q → [0, 1] denotes the water saturation (e.g. u = 1 means pure22
water, and u = 0 means pure oil), uB is a constant which indicates water saturation23
at x = 0, and M > 0 is the water/oil viscosity ratio. The classical BL equation (1.1)24
is a prototype for conservation laws with convex-concave flux functions. The graph25
of f(u) and f ′(u) with M = 2 is given in Figure 1.1.26
Due to the possibility of the existence of shocks in the solution of the hyperbolicconservation laws (1.1), the weak solutions are sought. The function u ∈ L∞(Q) iscalled a weak solution of the conservation laws (1.1) if∫
Q
{u∂φ
∂t+ f(u)
∂φ
∂x
}= 0 for all φ ∈ C∞0 (Q).
†Department of Mathematics, The Ohio State University, Columbus, OH 43210;‡Department of Mathematics, Claremont Mckenna College, CA 91711;
[email protected]; [email protected]§[email protected]
1
2
(a) f(u) = u2
u2+M(1−u)2
0 0.5 10
0.5
1
αu
f(u)
(b) f ′(u) =2Mu(1−u)
(u2+M(1−u)2)2
0 0.5 10
0.5
1
1.5
2
2.5
u
f′ (u)
Fig. 1.1. f(u) and f ′(u) with M = 2.
Notice that the weak solution is not unique. Among the weak solutions, the entropy27
solution is physically relevant and unique. The weak solution that satisfies Oleinik28
entropy condition [11]29
f(u)− f(ul)u− ul
≥ s ≥ f(u)− f(ur)u− ur
for all u between ul and ur (1.3)
is the entropy solution, where ul, ur are the function values to the left and right of the30
shock respectively, and the shock speed s satisfies Rankine-Hugoniot jump condition31
[10, 8]32
s =f(ul)− f(ur)
ul − ur. (1.4)
The classical BL equation (1.1) with flux function f(u) as given in (1.2) has been33
well studied (see [9] for an introduction). Let α be the solution of f ′(u) = f(u)u , i.e.,34
α =
√M
M + 1. (1.5)
The entropy solution of the classical BL equation can be classified into two categories:35
1. If 0 < uB ≤ α, the entropy solution has a single shock at xt = f(uB)
uB.36
2. If α < uB < 1, the entropy solution contains a rarefaction between uB and α37
for f ′(uB) < xt < f ′(α) and a shock at x
t = f(α)α .38
These two types of solutions are shown in Figure 1.2 for M = 2. In either case, the39
entropy solution of the classical BL equation (1.1) is a non-increasing function of x at40
any given time t > 0. However, the experiments of two-phase flow in porous medium41
reveal complex infiltration profiles, which may involve overshoot, i.e., profiles may not42
be monotone [5]. This suggests the need of modification to the classical BL equation43
(1.1).44
To better describe the infiltration profiles, we go back to the origins of (1.1). Let45
Si be the saturation of water/oil (i = w, o) and assume that the medium is completely46
saturated, i.e. Sw + So = 1. The conservation of mass gives47
φ∂Si∂t
+∂qi∂x
= 0 (1.6)
where φ is the porosity of the medium (relative volume occupied by the pores) and qi48
denotes the discharge of water/oil with qw+qo = q, which is assumed to be a constant49
3
(a) uB = 0.7
0 0.5 10
0.5
1
x
t
u
(b) uB = 0.98
0 0.5 10
0.5
1
x
t
u
Fig. 1.2. The entropy solution of the classical BL equation (M = 2, α =q
23≈ 0.8165). (a)
0 < uB = 0.7 ≤ α, the solution consists of one shock at xt
=f(uB)uB
; (b) α < uB = 0.98 < 1,
the solution consists of a rarefaction between uB and α for f ′(uB) < xt< f ′(α) and a shock at
xt
=f(α)α
.
in space due to the complete saturation assumption. Throughout of this work, we50
consider it constant in time as well. By Darcy’s law51
qi = −kkri(Si)µi
∂Pi∂x
, i = w, o (1.7)
where k denotes the absolute permeability, kri is the relative permeability and µi is the52
viscosity of water/oil. Instead of considering constant capillary pressure as adopted53
by the classical BL equation (1.1), Hassanizadeh and Gray [6, 7] have defined the54
dynamic capillary pressure as55
Pc = Po − Pw = pc(Sw)− φτ ∂Sw∂t
(1.8)
where pc(Sw) is the static capillary pressure and τ is a positive constant, and ∂Sw∂t56
is the dynamic effects. Using Corey [4, 12] expressions with exponent 2, krw(Sw) =57
S2w, kro(So) = S2
o , rescaling xφq → x and combining (1.6)-(1.8), the single equation58
for the water saturation u = Sw is59
∂u
∂t+
∂
∂x
[u2
u2 +M(1− u)2
]= − ∂
∂x
[φ2
q2
k(1− u)2u2
µw(1− u)2 + µou2
∂
∂x
(pc(u)φ− τ ∂u
∂t
)](1.9)
where M = µwµo
[14]. Linearizing the right hand side of (1.9) and rescaling the equation60
as in [13, 12], the modified Buckley-Leverett equation (MBL) is derived as61
∂u
∂t+∂f(u)∂x
= ε∂2u
∂x2+ ε2τ
∂3u
∂x2∂t(1.10)
where the water fractional flow function f(u) is given as in (1.2). Notice that, if62
Pc in (1.8) is taken to be constant, then (1.9) gives the classical BL equation; while63
if the dispersive parameter τ is taken to be zero, then (1.10) gives the viscous BL64
equation, which still displays monotone water saturation profile. The third order65
mixed derivative term ε2τuxxt in the MBL equation (1.10) plays an essential role.66
Van Duijn et al. [13] showed that the value τ is critical in determining the type of67
4
the solution profile. In particular, for certain Riemann problems, the solution profile68
of (1.10) is not monotone when τ is larger than the threshold value τ∗, which was69
numerically determined to be 0.61 [13]. The non-monotonicity of the solution profile70
is consistent with the experimental observations [5].71
The classical BL equation (1.1) is hyperbolic, the MBL equation (1.10), however,72
is pseudo-parabolic. Unlike the finite domain of dependence for the classical BL73
equation (1.1), the domain of dependence for the MBL equation (1.10) is infinite. This74
naturally raises the question for the choice of computational domain. To answer this75
question, we will first study the MBL equation equipped with two types of domains76
and corresponding boundary conditions. One is the half line boundary value problem77
ut + (f(u))x = εuxx + ε2τuxxt in Q = {(x, t) : x > 0, t > 0}u(x, 0) = u0(x) x ∈ [0,∞)
u(0, t) = gu(t), limx→∞
u(x, t) = 0 t ∈ [0,∞)
u0(0) = gu(0) compatibility condition
(1.11)
and the other one is finite interval boundary value problem78
vt + (f(v))x = εvxx + ε2τvxxt in Q = {(x, t) : x ∈ (0, L), t > 0}v(x, 0) = v0(x) x ∈ [0, L]
v(0, t) = gv(t), v(L, t) = h(t) t ∈ [0,∞)v0(0) = gv(0), v0(L) = h(0) compatibility condition.
(1.12)
Considering79
u0(x) ={v0(x) for x ∈ [0, L]0 for x ∈ [L,+∞) , gu(t) = gv(t) ≡ g(t), h(t) ≡ 0,
(1.13)we will show the relation between the solutions of problems (1.11) and (1.12). To80
the best knowledge of the authors, there is no such study for MBL equation (1.10).81
Similar questions were answered for BBM equation [1, 2].82
The organization of this paper is as follows. Section 2 will bring forward the83
exact theory comparing the solutions of (1.11) and (1.12). The difference between84
the solutions of these two types of problems decays exponentially with respect to85
the length of the interval L for practically interesting initial profiles. This provides a86
theoretical justification for the choice of the computational domain. Section 3 provides87
the numerical comparison of the solutions of (1.11) and (1.12). The computational88
results show that the difference between the solutions of these two types of problems89
indeed decays exponentially with respect to L, and nearly exponentially with respect90
to 1ε as well. The numerical results also confirm that the water saturation profile91
strongly depends on the dispersive parameter τ value as studied in [13]. For τ > τ∗,92
the MBL equation (1.10) gives non-monotone water saturation profiles for certain93
Riemann problems as suggested by experimental observations [5]. Section 4 gives the94
conclusion of the paper and the possible future directions.95
2. The half line problem versus the finite interval problem. Let u(x, t)96
be the solution to the half line problem (1.11), and let v(x, t) be the solution to the97
finite interval problem (1.12). We consider the natural assumptions (1.13). The goal98
of this section is to develop an estimate of the difference between u and v on the99
spatial interval [0, L] at a given finite time t. The main result of this section is100
5
Theorem 2.1 (The main Theorem). If u0(x) satisfies101
u0(x) ={Cu x ∈ [0, L0]0 x > L0
(2.1)
where L0 < L and Cu, are positive constants, then
‖u(·, t)− v(·, t) ‖H1L,ε,τ
≤ D1;ε,τ (t)e−λLε√τ +D2;ε,τ (t)e−
λ(L−L0)ε√τ
for some 0 < λ < 1, D1;ε,τ (t) > 0 and D2;ε,τ (t) > 0, where
‖Y (·, t) ‖H1L,ε,τ
:=
√∫ L
0
Y (x, t)2 + (ε√τYx(x, t))2 dx
Notice that with the initial condition (2.1) we are considering a Riemann problem.102
Theorem 2.1 shows that the solution to the half line problem (1.11) can be approxi-103
mated as accurately as one wants by the solution to the finite interval problem (1.12)104
in the sense that D1;ε,τ (t), D2;ε,τ (t), λLε√τ
and λ(L−L0)ε√τ
can be controlled.105
To prove theorem 2.1, we first derive the implicit solution formulae for the half106
line problem and the finite interval problem in section 2.1 and section 2.2 respectively.107
The implicit solution formulae are in integral form, which are derived by separating108
the x-derivative from the t-derivative, and formally solving a first order linear ODE109
in t and a second order non-homogeneous ODE in x. In section 2.3, we use Gronwall’s110
inequality multiple times to obtain the desired result in Theorem 2.1.111
2.1. Half line problem. In this section, we derive the implicit solution formula112
for the half line problem (1.11) (with gu(t) = g(t) as given in (1.13)). To solve (1.11),113
we first rewrite (1.11) by separating the x-derivative from the t-derivative,114 (I − ε2τ ∂
2
∂x2
)(ut +
1ετu
)=
1ετu− (f(u))x. (2.2)
By using integrating factor method, we formally integrate (2.2) over [0, t] to obtain115 (I − ε2τ ∂
2
∂x2
)(u− e− t
ετ u0
)=∫ t
0
(1ετu− (f(u))x
)e−
t−sετ ds. (2.3)
Furthermore, we let116
A = u− e− tετ u0, (2.4)
then (2.3) can be written as117
A′′ − 1ε2τ
A =∫ t
0
(− 1ε3τ2
u+1ε2τ
(f(u))x
)e−
t−sετ ds, where ′ =
∂
∂x. (2.5)
Notice that (2.5) is a second-order non-homogeneous ODE in x-variable along with118
the boundary conditions119
A(0, t) = u(0, t)− e− tετ u0(0) = g(t)− e− t
ετ g(0),
A(∞, t) = u(∞, t)− e− tετ u0(∞) = 0.
(2.6)
6
To solve (2.5), we first solve the corresponding linear homogeneous equation with120
the non-zero boundary conditions (2.6). We then find a particular solution for the121
non-homogeneous equation with zero boundary conditions by introducing a Green’s122
function G(x, ξ) and a kernel K(x, ξ) for the non-homogeneous terms u and (f(u))x123
respectively. Combining the solutions for the two non-homogeneous terms and the124
homogeneous part with boundary conditions, we get the solution for equation (2.5)125
satisfying the boundary conditions (2.6):126
A(x, t) = − 1ε3τ2
∫ t
0
∫ +∞
0
G(x, ξ)u(ξ, s)e−t−sετ dξ ds
+1ε2τ
∫ t
0
∫ +∞
0
K(x, ξ)f(u)e−t−sετ dξ ds
+(g(t)− e− t
ετ g(0))e− xε√τ
(2.7)
where the Green’s function G(x, ξ) and the kernel K(x, ξ) are127
G(x, ξ) =ε√τ
2
(e− x+ξε√τ − e−
|x−ξ|ε√τ
), (2.8)
K(x, ξ) = −∂G(x, ξ)∂ξ
=12
(e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
). (2.9)
To recover the solution for the half line problem (1.11), we refer to the definition of128
A in (2.4). Thus, the implicit solution formula for the half line problem (1.11) is129
u(x, t) = − 12ε2τ√τ
∫ t
0
∫ +∞
0
(e− x+ξε√τ − e−
|x−ξ|ε√τ
)u(ξ, s)e−
t−sετ dξ ds
+1
2ε2τ
∫ t
0
∫ +∞
0
(e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
)f(u)e−
t−sετ dξ ds
+(g(t)− e− t
ετ g(0))e− xε√τ + e−
tετ u0(x).
(2.10)
2.2. Finite interval problem. The implicit solution for the finite interval prob-130
lem (1.12) (with gv(t) = g(t) as given in (1.13)) can be solved in a similar way. The131
only difference is that the additional boundary condition h(t) at x = L in (1.12) gives132
different boundary conditions for the non-homogeneous ODE in x-variable. Denote133
AL = v − e− tετ v0, (2.11)
then it satisfies134
(AL)′′ − 1ε2τ
AL =∫ t
0
(− 1ε3τ2
v +1ε2τ
(f(v))x
)e−
t−sετ ds where ′ =
∂
∂x(2.12)
with the boundary conditions
AL(0, t) = v(0, t)− e− tετ v0(0) = g(t)− e− t
ετ g(0),
AL(L, t) = v(L, t)− e− tετ v0(L) = h(t)− e− t
ετ h(0).
These boundary conditions affect both the homogeneous solution and the partic-135
7
ular solution of (2.12) as follows136
AL(x, t) = − 1ε3τ2
∫ t
0
∫ L
0
GL(x, ξ)v(ξ, s)e−t−sετ dξ ds
+1ε2τ
∫ t
0
∫ L
0
KL(x, ξ)f(v)e−t−sετ dξ ds
+ c1(t)φ1(x) + c2(t)φ2(x)
(2.13)
where the Green’s function GL(x, ξ), the kernel KL(x, ξ) and the bases for the homo-137
geneous solutions are138
GL(x, ξ) =ε√τ
2(e2Lε√τ − 1)
(ex+ξε√τ + e
2L−(x+ξ)ε√τ − e
|x−ξ|ε√τ − e
2L−|x−ξ|ε√τ
), (2.14)
139
KL(x, ξ) = − 1
2(e2Lε√τ − 1)
(ex+ξε√τ − e
2L−(x+ξ)ε√τ
+sgn(x− ξ)e|x−ξ|ε√τ − sgn(x− ξ)e
2L−|x−ξ|ε√τ
),
(2.15)
c1(t) = g(t)− e− tετ g(0), c2(t) = h(t)− e− t
ετ h(0), (2.16)
φ1(x) =eL−xε√τ − e
−L+xε√τ
eLε√τ − e−
Lε√τ
, and φ2(x) =e
xε√τ − e−
xε√τ
eLε√τ − e−
Lε√τ
. (2.17)
Thus, the implicit solution formula for the finite interval problem (1.12) is140
v(x, t) =− 1
2ε2τ√τ(e
2Lε√τ − 1)
∫ t
0
∫ L
0
(ex+ξε√τ + e
2L−(x+ξ)ε√τ − e
|x−ξ|ε√τ
−e2L−|x−ξ|ε√τ
)v(ξ, s)e−
t−sετ dξ ds
− 1
2ε2τ(e2Lε√τ − 1)
∫ t
0
∫ L
0
(ex+ξε√τ − e
2L−(x+ξ)ε√τ + sgn(x− ξ)e
|x−ξ|ε√τ
−sgn(x− ξ)e2L−|x−ξ|ε√τ
)f(v)e−
t−sετ dξ ds
+ c1(t)φ1(x) + c2(t)φ2(x) + e−tετ v0(x).
(2.18)
2.3. Comparisons. In this section, we will prove that the solution u(x, t) to141
the half line problem can be approximated as accurately as one wants by the solution142
v(x, t) to the finite interval problem as stated in Theorem 2.1.143
Due to the difference in the integration domains, we do not use (2.10) and (2.18)144
directly for the comparison. Instead, we decompose u(x, t) (v(x, t) respectively) into145
two parts: U(x, t) and uL(x, t) (V (x, t) and vL(x, t) respectively), such that U(x, t)146
(V (x, t) respectively) has zero initial condition and boundary conditions at x = 0147
and x = L. We estimate the difference between u(·, t) and v(·, t) by estimating the148
differences between uL(·, t) and vL(·, t), U(·, t) and V (·, t), then applying the triangle149
inequality.150
151
8
2.3.1. Definitions and lemmas. To assist the proof of Theorem 2.1 in section2.3.3, we introduce some new notations in this section. We first decompose u(x, t) assum of two terms U(x, t) and uL(x, t), such that
u(x, t) = U(x, t) + uL(x, t) x ∈ [0,+∞)
where152
uL = e−tετ u0(x) + c1(t)e−
xε√τ +
(u(L, t)− c1(t)e−
Lε√τ − e− t
ετ u0(L))φ2(x) (2.19)
and c1(t) and φ2(x) are given in (2.16) and (2.17) respectively. With this definition,153
uL takes care of the initial condition u0(x) and boundary conditions g(t) at x = 0 and154
x = L for u(x, t). Then U satisfies an equation slightly different from the equation u155
satisfies in (1.11):156
Ut − εUxx − ε2τUxxt =(ut − εuxx − ε2τuxxt
)−((uL)t − ε(uL)xx − ε2τ(uL)xxt
)= − (f(u))x +
1ετuL(x, t)
(2.20)
In addition, U(x, t) has zero initial condition and boundary conditions at x = 0 and157
x = L, i.e.,158
U(x, 0) = 0, U(0, t) = 0, U(L, t) = 0. (2.21)
Similarly, for v(x, t), let
v(x, t) = V (x, t) + vL(x, t) x ∈ [0, L]
where159
vL = e−tετ v0(x) + c1(t)φ1(x) + c2(t)φ2(x) (2.22)
and c1(t), c2(t) and φ1(x), φ2(x) are given in (2.16) and (2.17) respectively. With160
this definition, vL takes care of the initial condition v0(x) and boundary conditions161
g(t) and h(t) at x = 0 and x = L for v(x, t). Then V satisfies an equation slightly162
different from the equation v satisfies in (1.12):163
Vt − εVxx − ε2τVxxt = − (f(v))x +1ετvL(x, t) (2.23)
with164
V (x, 0) = 0, V (0, t) = 0, V (L, t) = 0. (2.24)
Since, in the end, we want to study the difference between U(x, t) and V (x, t), wedefine
W (x, t) = V (x, t)− U(x, t) for x ∈ [0, L].
Because of (2.20) and (2.23), we have165
Wt − εWxx − ε2τWxxt = − (f(v)− f(u))x +1ετ
(vL − uL). (2.25)
9
In lieu of (2.21) and (2.24), W (x, t) also has zero initial condition and boundary166
conditions at x = 0 and x = L, i.e.,167
W (x, 0) = 0, W (0, t) = 0, W (L, t) = 0. (2.26)
Now, to estimate ‖u− v ‖, we can estimate ‖W ‖ = ‖V − U ‖ and estimate168
‖uL − vL ‖ separately. These estimates are done in section 2.3.3.169
Next, we state the lemmas needed in the proof of Theorem 2.1. The proof of170
the lemmas can be found in the appendix A and [14]. In all the lemmas, we assume171
0 < λ < 1 and u0(x) satisfies172
u0(x) ={Cu x ∈ [0, L0]0 x > L0
(2.27)
where L0 < L and Cu are positive constants. Notice that the constraint λ ∈ (0, 1) is173
crucial in Lemmas 2.3, 2.4.174
Lemma 2.2. f(u) = u2
u2+M(1−u)2 ≤ Du where D = f(α)α and α =
√MM+1 .175
Lemma 2.3.176
(i)∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ dξ ≤ 2ε
√τ
1−λ2 .177
(ii)∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ eλx−ξε√τ dξ ≤ ε
√τ
e(1−λ) .178
(iii)∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ≤ 2Cuε
√τe
λL0ε√τ .179
Lemma 2.4.180
(i)∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ dξ ≤ 2ε
√τ
1−λ2 .181
(ii)∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ eλx−ξε√τ dξ ≤ ε
√τ + ε
√τ
e(1−λ) .182
(iii)∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ≤ 2Cuε
√τe
λL0ε√τ .183
Lemma 2.5.184
(i)∣∣∣φ1(x)− e−
xε√τ
∣∣∣ = e− Lε√τ |φ2(x)| .185
(ii) |φ2(x)| ≤ 1 for x ∈ [0, L] .186
(iii) |φ′2(x)| ≤ 2ε√τ
if ε� 1 for x ∈ [0, L] .187
Last but not least, the norm that we will use in Theorem 2.1 and its proof is188
‖Y (·, t) ‖H1L,ε,τ
:=
√∫ L
0
Y (x, t)2 + (ε√τYx(x, t))2 dx. (2.28)
2.3.2. A proposition. In this section, we will give a critical estimate, which189
is essential in the calculation of maximum difference ‖uL(·, t)− vL(·, t) ‖∞ in section190
2.3.3. By comparing uL(x, t) and vL(x, t) given in (2.19) and (2.22) respectively, it is191
clear that the coefficient u(L, t)−c1(t)e−Lε√τ −e− t
ετ u0(L) for φ2(x) appeared in (2.19)192
needs to be compared with the corresponding coefficient c2(t) for φ2(x) appeared in193
(2.22). We thus define a space-dependent function194
Uc2(x, t) = u(x, t)− c1(t)e−xε√τ − e− t
ετ u0(x) (2.29)
and establish the following proposition195
Proposition 2.6.196
|Uc2(L, t)| ≤ aτ (t)ebτ tετ e− λLε√τ + cτ
t
ετe
(bτ−1)tετ e
−λ(L−L0)ε√τ (2.30)
10
for some parameter-dependent constants aτ , bτ and cτ .197
Proof. Based on the implicit solution formula (2.10) derived in section 2.1, Lemma198
2.2 and the relationship between Uc2 and u given in (2.29), we can get an inequality199
in terms of Uc2200
|Uc2(x, t)| ≤ 12ε2τ√τ
[∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ |Uc2(ξ, s)| e−t−sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ |c1(s)| e−ξε√τ e−
t−sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ |u0(ξ)| e− tετ dξ ds
]+
D
2ε2τ
[∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ |Uc2(ξ, s)| e−t−sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ |c1(s)| e−ξε√τ e−
t−sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ |u0(ξ)| e− tετ dξ ds
].
(2.31)
To show that Uc2(x, t) decays exponentially with respect to x, we pull out an expo-201
nential term by writing Uc2(x, t) = e− λxε√τ e−
tετ U(x, t), where 0 < λ < 1, such that202
U(x, t) = eλxε√τ e
tετ Uc2(x, t), (2.32)
then (2.31) can be rewritten in terms of U(x, t) as follows203
∣∣∣U(x, t)∣∣∣ ≤ 1
2ε2τ√τ
[∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ
∣∣∣U(ξ, s)∣∣∣ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ |c1(s)| eλx−ξε√τ e
sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ds
]+
D
2ε2τ
[∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ
∣∣∣U(ξ, s)∣∣∣ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ |c1(s)| eλx−ξε√τ e
sετ dξ ds
+∫ t
0
∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ds
].
(2.33)
11
Because of Lemmas 2.3–2.4, we can get the following estimate for∣∣∣U(·, t)
∣∣∣∞
based on204
(2.33) :205 ∣∣∣U(·, t)∣∣∣∞≤ 1
2ε2τ√τ
[2ε√τ
1− λ2
∫ t
0
|U(·, s)|∞ ds+ε√τ
e(1− λ)
∫ t
0
|c1(s)|e sετ ds
+2Cuε√τe
λL0ε√τ
∫ t
0
1 ds]
+D
2ε2τ
[2ε√τ
1− λ2
∫ t
0
|U(·, s)|∞ ds+ ε√τ
(1 +
1e(1− λ)
)∫ t
0
|c1(s)|e sετ ds
+2Cuε√τe
λL0ε√τ
∫ t
0
1 ds]
≤∫ t
0
bτετ|U(·, s)|∞ ds+
∫ t
0
aτ (s)ετ
ds
(2.34)
where
bτ =1 +D
√τ
1− λ2, aτ (t) = aτe
tετ + cτe
λL0ε√τ ,
aτ =|c1(·)|∞(1 +D
√τ(e(1− λ) + 1))
2e(1− λ), cτ = Cu(1 +D
√τ).
By Gronwall’s inequality, inequality (2.34) gives that∣∣∣U(·, t)∣∣∣∞≤∫ t
0
aτ (t− s)ετ
ebτ (t−s)ετ ds ≤
(aτe
tετ + cτ
t
ετeλL0ε√τ
)ebτ tετ
Hence |Uc2(x, t)| ≤∣∣∣U(·, t)
∣∣∣∞e−λxε√τ e−
tετ ≤
(aτe
tετ + cτ
tετ e
λL0ε√τ
)ebτ tετ e
−λxε√τ e−
tετ i.e.,206
Uc2(x, t) decays exponentially with respect to x. In particular, when x = L, we207
have208
|Uc2(L, t)| ≤ aτebτ tετ e− λLε√τ + cτ
t
ετe
(bτ−1)tετ e
−λ(L−L0)ε√τ (2.35)
as given in (2.30).209
2.3.3. Proof of Theorem 2.1. In this section, we will first find the maximum210
difference of ‖uL(·, t)− vL(·, t) ‖∞, then we will derive ‖uL(·, t)− vL(·, t) ‖H1L,ε,τ
and211
‖W (·, t) ‖H1L,ε,τ
= ‖U(·, t)− V (·, t) ‖H1L,ε,τ
. Combining these two, we will get an esti-212
mate for ‖u(·, t)− v(·, t) ‖H1L,ε,τ
.213
Proposition 2.7. If u0(x) satisfies (2.27), then
‖uL − vL ‖∞ ≤ E1;ε,τ (t)e−λLε√τ + E2;ε,τ (t)e−
λ(L−L0)ε√τ
where E1;ε,τ (t) = |c1(·)|∞ + aτebτ tετ and E2;ε,τ (t) = cτ
tετ e
(bτ−1)tετ .214
Proof. By the definition of uL and vL given in (2.19) and (2.22) and the assump-tion that u0(x) = v0(x) for x ∈ [0, L], we can get their difference
uL(x, t)− vL(x, t) = c1(t)(e− xε√τ − φ1(x)
)+(Uc2(L, t)− h(t) + e−
tετ h(0)
)φ2(x)
12
Combining Lemmas 2.5(i), 2.5(ii), inequality (2.35), and h(t) ≡ 0, we have215
‖uL(·, t)− vL(·, t) ‖∞ ≤ E1;ε,τ (t)e−λLε√τ + E2;ε,τ (t)e−
λ(L−L0)ε√τ (2.36)
where216
E1;ε,τ (t) = |c1(·)|∞ + aτebτ tετ and E2;ε,τ (t) = cτ
t
ετe
(bτ−1)tετ . (2.37)
217
Proposition 2.8. If u0(x) satisfies (2.27), and E1;ε,τ (t), E2;ε,τ (t) are as inproposition 2.7, then
‖uL(·, t)− vL(·, t) ‖H1L,ε,τ
≤√
5L(E1;ε,τ (t)e−
λLε√τ + E2;ε,τ (t)e−
λ(L−L0)ε√τ
).
218
Proof. Because of the definition of uL and vL given in (2.19) and (2.22), Lemma219
2.5(iii) and inequality (2.35), we have that220
‖ (uL(·, t)− vL(·, t))x ‖∞ ≤ |c1(t)| e−Lε√τ |φ′2(x)|+ |Uc2(L, t)| |φ′2(x)|
≤ 2ε√τ
(E1;ε,τ (t)e−
λLε√τ + E2;ε,τ (t)e−
λ(L−L0)ε√τ
).
(2.38)
Now, combining (2.36) and (2.38), we obtain that221
‖uL(·, t)− vL(·, t) ‖H1L,ε,τ
=
√∫ L
0
|uL − vL|2 +∣∣ε√τ (uL − vL)x
∣∣2 dx≤√
5L(E1;ε,τ (t)e−
λLε√τ + E2;ε,τ (t)e−
λ(L−L0)ε√τ
).
(2.39)
222
Proposition 2.9. If u0(x) satisfies (2.27), then
‖W (·, t) ‖H1L,ε,τ
≤ γ1;ε,τ (t)e−λLε√τ + γ2;ε,τ (t)e−
λ(L−L0)ε√τ
where the coefficients are given by223
γ1;ε,τ (t) = e(M+1)2t2Mε√τ
((M + 1)2
√τ
2M+ 1)√
L
(t
ετ|c1(·)|∞ +
aτbτ
(ebτ tετ − 1)
)γ2;ε,τ (t) = e
(M+1)2t2Mε√τ
((M + 1)2
√τ
2M+ 1)√
Lcτ
·(
t
ετ(bτ − 1)e
(bτ−1)tετ − 1
(bτ − 1)2(e
(bτ−1)tετ − 1)
).
(2.40)
224
Proof. Multiplying the governing equation of W (2.25) by 2W , integrating over225
[0, L], and using integration by parts, we get226
d
dt
∫ L
0
W 2 + (ε√τWx)2 dx
= −ε∫ L
0
2W 2x dx+
∫ L
0
2Wx (f(v)− f(u)) dx+2ετ
∫ L
0
W (vL − uL) dx.
13
Therefore, using the norm we defined earlier in (2.28), and f ′(u) ≤ (M+1)2
2M := C, we227
have228
d
dt‖W (·, t) ‖2H1
L,ε,τ
≤ 2∫ L
0
|Wx||f ′(η)||v − u| dx+2√L
ετ‖ vL − uL ‖∞ ‖W (·, t) ‖H1
L,ε,τ
≤ 2C∫ L
0
|Wx| (|W |+ ‖ vL − uL ‖∞) dx+2√L
ετ‖ vL − uL ‖∞ ‖W (·, t) ‖H1
L,ε,τ
≤ 2Cε√τ
(‖W (·, t) ‖2H1
L,ε,τ+ ‖ vL − uL ‖∞
√L ‖W (·, t) ‖H1
L,ε,τ
)+
2√L
ετ‖ vL − uL ‖∞ ‖W (·, t) ‖H1
L,ε,τ
=2Cε√τ‖W (·, t) ‖2H1
L,ε,τ+(
2Cε√τ
+2ετ
)√L ‖ vL − uL ‖∞ ‖W (·, t) ‖H1
L,ε,τ.
Hence,
d
dt‖W (·, t) ‖H1
L,ε,τ≤ C
ε√τ‖W (·, t) ‖H1
L,ε,τ+(
C
ε√τ
+1ετ
)√L ‖ vL − uL ‖∞ .
By Gronwall’s inequality and (2.36)229
‖W (·, t) ‖H1L,ε,τ
≤∫ t
0
(C
ε√τ
+1ετ
)√L ‖ vL − uL ‖∞ e
C(t−s)ε√τ ds
≤ eCtε√τ
(C
ε√τ
+1ετ
)√L
∫ t
0
E1;ε,τ (s)e−λLε√τ + E2;ε,τ (s)e−
λ(L−L0)ε√τ ds
≤(eCtε√τ
(C
ε√τ
+1ετ
)√L
∫ t
0
E1;ε,τ (s) ds)e− λLε√τ
+(eCtε√τ
(C
ε√τ
+1ετ
)√L
∫ t
0
E2;ε,τ (s) ds)e−λ(L−L0)
ε√τ
≤ eCtε√τ
(C
ε√τ
+1ετ
)√L
(t|c1(·)|∞ +
aτ ετ
bτ(e
bτ tετ − 1)
)e− λLε√τ
+eCtε√τ
(C
ε√τ
+1ετ
)√Lcτετ
(ετ
bτ − 1te
(bτ−1)tετ − (
ετ
bτ − 1)2(e
(bτ−1)tετ − 1)
)e−λ(L−L0)
ε√τ .
Hence
‖W (·, t) ‖H1L,ε,τ
≤ γ1;ε,τ (t)e−λLε√τ + γ2;ε,τ (t)e−
λ(L−L0)ε√τ
where γ1;ε,τ (t) and γ2;ε,τ (t) are given in (2.40).230
Now we are in the position to prove the main theorem of this section.231
Theorem 2.10. If u0(x) satisfies232
u0(x) ={Cu x ∈ [0, L0]0 x > L0
14
where L0 < L and Cu, are positive constants, and E1;ε,τ (t), E2;ε,τ (t), γ1;ε,τ (t), γ2;ε,τ (t)233
are as in (2.37) and (2.40) , then234
‖u(·, t)− v(·, t) ‖H1L,ε,τ
≤ D1;ε,τ (t)e−λLε√τ +D2;ε,τ (t)e−
λ(L−L0)ε√τ (2.41)
for some 0 < λ < 1, and
D1;ε,τ (t) = γ1;ε,τ (t) +√
5LE1;ε,τ (t), D2;ε,τ (t) = γ2;ε,τ (t) +√
5LE2;ε,τ (t).
235
Proof. [Proof of the Main Theorem]
‖u(·, t)− v(·, t) ‖H1L,ε,τ
≤ ‖W (·, t) ‖H1L,ε,τ
+ ‖ vL(·, t)− uL(·, t) ‖H1L,ε,τ
= D1;ε,τ (t)e−λLε√τ +D2;ε,τ (t)e−
λ(L−L0)ε√τ
where
D1;ε,τ (t) =γ1;ε,τ (t) +√
5LE1;ε,τ (t)
=e(M+1)2t2Mε√τ
((M + 1)2
√τ
2M+ 1)√
L
(t
ετ|c1(·)|∞ +
aτbτ
(ebτ tετ − 1)
)+√
5L(|c(·)|∞ + aτebτ tετ ),
D2;ε,τ (t) =γ2;ε,τ (t) +√
5LE2;ε,τ (t)
=e(M+1)2t2Mε√τ
((M + 1)2
√τ
2M+ 1)√
Lcτ ·
·(
t
ετ(bτ − 1)e
(bτ−1)tετ − 1
(bτ − 1)2(e
(bτ−1)tετ − 1)
)+√
5Lcτt
ετe
(bτ−1)tετ .
236
This result gives that ‖u(·, t)− v(·, t) ‖H1L,ε,τ
exponentially decays in L. This237
theorem shows that if λLε√τ
and λ(L−L0)ε√τ
converge to infinity, then the solution v(x, t)238
of the finite interval problem converges to the solution u(x, t) of the half line problem239
in the sense of ‖ · ‖H1L,ε,τ
. This can be achieved either by letting L→∞ or ε→ 0. For240
example, in the extreme case, ε = 0, the half line problem (1.11) becomes hyperbolic241
and the domain of dependence is finite, so, certainly, one only need to consider the242
finite interval problem. This is consistent with the main theorem in the sense that for243
a fixed final time t, if λL > bτ t and λ(L−L0) > (bτ − 1)t, i.e., L > max( bτ tλ ,(bτ−1)t
λ ),244
then ‖u(·, t)− v(·, t) ‖H1L,ε,τ
≤ D1;ε,τ (t)e−λLε√τ + D2;ε,τ (t)e−
λ(L−L0)ε√τ → 0 as ε → 0.245
Theorem 2.10 gives a theoretical justification for using the solution of the finite interval246
problem to approximate the solution of the half line problem with appropriate choice247
of L and ε.248
3. Numerical Results. In this section, we adopt the finite difference schemegiven in [13] to solve the following finite interval initial boundary value problem
vt + (f(v))x = εvxx + ε2τvxxt
v(x, 0) = uBχ{x=0} + 0χ{0<x≤L}v(0, t) = uB , v(L, t) = 0.
(3.1)
15
3.1. Verification of Theorem 2.10. We first verify the theoretical approxi-mation given by Theorem 2.10. This has been done for various choices of parameters,and similar results have been obtained. In this section, we demonstrate how the nu-merical results confirm Theorem 2.10 with the choice τ = 1 and uB = 0.9. Let v(x, t)be the solution to the finite interval boundary value problem (3.1), and u(x, t) bethe solution to the corresponding half-line problem, with the initial and boundaryconditions
u(x, 0) = uBχ{x=0} + 0χ{x>0}
u(0, t) = uB .(3.2)
For the seek of numerical verification to Theorem 2.10, we compare u(x, t) and v(x, t)in the sense of ‖ · ‖∞ norm. Proposition 2.7 and Theorem 2.10, with L0 = 0, give that
‖u(·, t)− v(·, t) ‖∞ ≤ Gε,τ (t)e−λLε√τ , (3.3)
and we numerically verify that ‖u− v ‖∞ decays exponentially with respect to L as249
in (3.3).250
With t = 0.1, ∆x = 0.005, ∆t = 0.001 and M = 2 fixed, we solve (3.1) for two251
different choices of diffusion coefficient, one smaller ε = 0.1 and one larger ε = 1.252
Below we use u and v to denote the numerical solutions for the half line and the253
finiter interval boundary value probelm respetively.254
(a) ε = 0.1We numerically solved (3.1) for L = 0.5, 1, 2, 4, 8, 16, 32, 64, 128, and the obtainednumerical solutions differ by only machine-ε when L ≥ 64. Hence, we use thesolution of (3.1) corresponding to L = 128 as the numerical approximation for thehalf-line problem (3.2) for comparison purpose, i.e., we take u = v{L=128}. Figure
(a) ε = 0.1
0 20 40 60−600
−500
−400
−300
−200
−100
L
ln||u
−v|
| ∞
t = 0.025
t = 0.05
t = 0.075
t = 0.1
(b) t = 0.025
0 2 4 6 80
0.5
1x 10
−9
L
v
L = 0.5L = 1L = 2L = 4
(c) t = 0.1
0 2 4 6 80
0.5
1x 10
−15
L
v
L = 0.5L = 2L = 4L = 8
Fig. 3.1. ε = 0.1, τ = 1, uB = 0.9 and M = 2 are fixed. 3.1(a): ‖u− v ‖∞ decays exponentiallywith respect to L; 3.1(b), 3.1(c): solution profiles with different domain size L’s at t = 0.025 andt = 0.1 respectively.
3.1(a) shows that ‖u− v ‖∞ decays exponentially with respect to the domain sizeL at various time t = 0.025, t = 0.05, t = 0.075 and t = 0.1. Furthermore, itdisplays that ‖u− v ‖∞ increases as time progresses. This numerically shows thatGε,τ (t) in (3.3) is increasing with respect to t. Notice that in Figure 3.1(a), whenL ≤ 8, the ‖u− v ‖∞ decay rate is smaller than that for L ≥ 16. This is especiallysignificant when t > 0.05. This is because when L is too small, at a later time,the domain size is not sufficently large for the wave propagation, which in turnmakes ‖u− v ‖∞ unproportionally large. This can be seen from Figures 3.1(b)
16
and 3.1(c). These two subfigures give the zoom-in view of the solution profilesgotten with different domain size L’s at t = 0.025 (Figure 3.1(b)) and t = 0.1(Figure 3.1(c)). Based on Figure 3.1(b), it is clear that at t = 0.025, L = 2 is toosmall to capture the leading shock; while Figure 3.1(c) shows that at a later timet = 0.1, even L = 4 is too small to capture the leading shock, in fact, even L = 8is too small to capture the correct solution, which corresponds to a significantdrop of decay rate for L ≤ 8 in Figure 3.1(a) when t = 0.1. In addition, the λ in(3.3) is estimated based on
λ = slope×(−ε√τ), (3.4)
where slope is the slope of the line ln ‖u− v ‖∞ versus L. Based on Figure255
3.1(a), λ approximately equals 0.9866, 0.9883, 0.9902, 0.9709 for t equals 0.025,256
0.05, 0.075, 0.1 respectively.257
(b) ε = 1258
The increment of ε enhances the diffusion effect in the solution profiles. It is259
natural to enlarge the domain size to achieve the correct solution. Therefore for260
the larger ε = 1, we numerically solved (3.1) for L = 0.5, 1, 2, 4, 8, 16, 32,261
64, 128, 256, 512 and the obtained numerical solutions differ by only machine-ε262
when L ≥ 256. Hence in this case, we use the solution of (3.1) corresponding to263
L = 512 as the numerical approximation for the half-line problem (3.2), i.e., we264
take u = v{L=512}. Figure 3.2(a) again shows that ‖u− v ‖∞ decays exponentially
(a) ε = 1
0 100 200−250
−200
−150
−100
−50
L
ln||u
−v|
| ∞
t = 0.025
t = 0.05
t = 0.075
t = 0.1
(b) t = 0.025
0 50 1000
0.5
1x 10
−9
L
v
L = 8L = 16L = 32L = 64
(c) t = 0.1
0 50 1000
0.5
1x 10
−15
L
v
L = 16L = 32L = 64L = 128
Fig. 3.2. ε = 1, τ = 1, uB = 0.9 and M = 2 are fixed. 3.2(a): ‖u− v ‖∞ decays exponentiallywith respect to L; 3.2(b), 3.2(c): solution profiles with different domain size L’s at t = 0.025 andt = 0.1 respectively.
265
with respect to the domain size L, and for fixed L, ‖u− v ‖∞ increases as time266
progresses. Figure 3.2(b) shows that when t = 0.025, L = 16 is too small to267
capture the leading shock; whereas Figure 3.2(c) shows that when t = 0.1, even268
L = 32 is not sufficiently large to ensure the accuracy of the solution, which269
corresponds to the significant drop of decay rate for L ≤ 32 in Figure 3.2(a) when270
t = 0.1. In addtion, (3.4) is used to estimate the λ value, and based on Figure271
3.2(a), λ approximately equals 0.9949, 0.9669, 0.9696, 0.9429 for t equals 0.025,272
0.05, 0.075, 0.1 respectively.273
For both smaller and larger diffusion effects, ε = 0.1 and ε = 1 respectively, we haveshown numerically that when ε, τ and uB are fixed, ‖u− v ‖∞ decays exponentiallywith respect to L as given in (3.3). We also numerically showed that Gε,τ (t) in (3.3)is increasing with respect to t. In addition, Figures 3.1(b) and 3.2(b) demonstratethat at a fixed time (t = 0.025), bigger ε = 1 requires larger domain size to ensure
17
the accuracy of the solution. This can be seen in Figures 3.1(c) and 3.2(c) (t = 0.1)too. Furthermore, We numerically estimated the λ value in (3.3) to be
λ ∈ (0.94, 0.99).
This estimate gives the convergence rate for using the solution to the finite interval274
boundary value problem to approximate that to the half-line problem. It provides a275
way to estimate the domain size needed to achieve a desired accuracy.276
3.2. Effect of ε. The theoretical approximation given by Theorem 2.10 does not277
only provide a way to estimate the computational domain size needed to capture the278
solution to the half-line problem, it also shows how the diffusion coefficient ε affects279
‖u(·, t)− v(·, t) ‖. In this section, we verify this numerically based on (3.3).280
With t = 0.05, ∆x = 10−4, ∆t = 10−5 and M = 2 fixed, we numerically solve281
for u and vL=0.1 for four representative ε = 0.001, 0.002, 0.004, 0.008, where again, we282
use u and v to denote the numerical solutions for the half-line and the finite interval283
boundary value problems respectively. The numerical solution u(x, t) of the half-line284
problem was obtained by doubling the domain size starting with L = 0.1, until the285
solutions do not differ more than the tol = 10−10.286
Figure 3.3(a) shows the plot ln ‖u− v ‖∞ versus 1ε at various time t = 0.0125,
t = 0.025, t = 0.0375, t = 0.05. It is shown that ln ‖u− v ‖∞ and 1ε are not linearly
(a) L = 0.1
200 400 600 800 1000−30
−25
−20
−15
−10
−5
1/ε
ln||e
rror
|| ∞
t = 0.0125
t = 0.025
t = 0.0375
t = 0.05
(b) t = 0.0125
0 0.05 0.10
0.2
0.4
0.6
0.8
1
L
v
ε = 0.001ε = 0.002ε = 0.004ε = 0.008
(c) t = 0.05
0 0.05 0.10
0.2
0.4
0.6
0.8
1
L
v
Fig. 3.3. τ = 1, uB = 0.9 and M = 2 are fixed. 3.3(a): ‖u− v ‖∞ decays with respect to 1/ε;3.3(b), 3.3(c): solution profiles with different ε’s at t = 0.0125 and t = 0.05 respectively.
related, this is because if we take logarithm on (3.3),
ln ‖u(·, t)− v(·, t) ‖∞ ≤ lnGε,τ (t)− λL√τ· 1ε, (3.5)
the Gε,τ (t) appeared in the upper bound in (3.5) also depends on ε. Hence the upper287
bound in (3.5) is not a linear function of 1ε . Notice that for a fixed 1
ε , ‖u− v ‖∞288
increases as time progresses, this confirms again that Gε,τ (t) is increasing with respect289
to t. For any fixed time, as 1ε increases, the rate ln ‖u− v ‖∞ decreases slows down.290
Hence, Gε,τ (t) decays slower than exponential decay with respect to 1ε . As a matter of291
fact, Gε,τ (t) could be even increasing with respect to 1ε . Figures 3.3(b), 3.3(c) give the292
solution profiles for all the four different ε’s at t = 0.0125 and t = 0.05 respectively.293
They both show that smaller ε smears out the solution less. Comparing Figures294
3.3(b) and 3.3(c), it shows that the diffusion effects do not change qualitatively as295
time progresses.296
18
3.3. Monotonicity of solution profiles. In this section, we numerically verify297
that the MBL equation (3.1) does deliver non-monotone water saturation profiles as298
observed in experiments [5]. Van Duijn et al [13] provided a numerical uB–τ bifurca-299
tion diagram. In particular, it explicitly spelled out the conditions for the existence300
of non-monotone solutions. Van Duijn et al [13] show that when the dispersive co-301
efficient τ is larger than the threshold value τ∗ ≈ 0.61, it corresponds to an interval302
[u, u], and if the post-shock value uB lies inside this interval, then the solution will303
contain two shocks, one from uB to u, and another one from u to 0, with the shock304
speed f(uB)−f(u)uB−u and f(u)
u respectively.
(a) (τ, uB) = (0.2, 0.9)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(b) (τ, uB) = (1, 0.9)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(c) (τ, uB) = (5, 0.9)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(d) (τ, uB) = (0.2, α)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(e) (τ, uB) = (1, α)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(f) (τ, uB) = (5, α)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(g) (τ, uB) = (0.2, 0.75)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(h) (τ, uB) = (1, 0.75)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
(i) (τ, uB) = (5, 0.75)
0 1 2 3 40
0.2
0.4
0.6
0.8
1
x
v
Fig. 3.4. Numerical solutions to MBL equation (3.1) with different parameter settings. Thecolor coding is for different time: 1
4T (blue), 2
4T (green), 3
4T (magenta) and T (black). In figures
3.4(d) – 3.4(f), α =q
MM+1
=q
23
for M = 2. The red dashed lines indicate the theoretical shock
locations and plateau values.
305
Taking the convergence estimate given in section 3.1 into consideration, we have306
chosen large enough computational domains to seek the solution of (3.1) for three307
representative τ values 0.2, 1, 5. The numerical experiments in Figure 3.4 are carried308
19
out for M = 2, ε = 0.001 and T = 4000 × ε, to get the asymptotic solution profiles,309
and ∆x was chosen to be ε10 and ∆t
∆x was chosen to be 0.1. The solution profiles at T4310
(blue), 2∗T4 (green), 3∗T
4 (magenta) and T (black) are chosen to demonstrate the time311
evolution of the solutions. The red dashed lines are used to denote the theoretical312
shock locations and plateau values for comparison purpose. Since τ = 0.2 is less that313
τ∗ ≈ 0.61, Figures 3.4(a) 3.4(d) 3.4(g) all display monotone solution profiles. However,314
τ = 1 is larger than τ∗ ≈ 0.61, Figures 3.4(b) 3.4(e) 3.4(h) show the solutions for three315
representative uB values 0.9, α, 0.75 respectively. Notice that Figure 3.4(e) is the only316
one that displays non-monotone behavior. This is because among all the three chosen317
uB values, uB = α is the only one that lies inside [uτ=1, uτ=1]. Another choice of318
τ = 5 corresponds to uτ=5 ≈ 0.68 and uτ=5 ≈ 0.98. 3.4(c) 3.4(f) 3.4(i) show that all319
the three representative uB values 0.9, α, 0.75 lie inside [uτ=5, uτ=5], hence all display320
non-monotone solution profiles.321
4. Conclusion. We proved that the solution to the infinite domain problem can322
be approximated by that of the bounded domain problem. This provides a theoret-323
ical justification for using finite domain to calculation the numerical solution of the324
MBL equation (1.10). We also numerically verified the convergence rates are consis-325
tent with the theoretical estimates. The numerical solutions for qualitatively different326
parameter values τ and initial conditions uB show that the jump locations are con-327
sistent with the theoretical calculation and the plateau heights are consistent with328
the numerically obtained values given in [13]. In particular, the numerical solutions329
give non-monotone water saturation profiles, for certain τ and uB values, which is330
consistent with the experimental observations.331
In [14, 12], the two-dimensional space extension of the modified Buckley-Leverett332
equation has been derived. One of the future directions is to study the difference333
between the solution to the quarter-plane problem and finite domain problem to 2D334
MBL equation, and develop numerical schemes to solve it efficiently.335
Appendix A. Proof of the lemmas.336
Proof. [Proof to lemma 2.2] Let g(u) = f(u)u = u
u2+M(1−u)2 , then337
g′(u) =M − (1 +M)u2
(u2 +M(1− u)2)2
> 0 if 0 < u <
√MM+1
= 0 if u =√
MM+1
< 0 if u >√
MM+1
and hence g(u) achieves its maximum at u =√
MM+1 . Therefore, f(u)
u = g(u) ≤ D,338
where D = f(α)α and α =
√MM+1 , and in turn, we have that f(u) ≤ Du for all339
0 ≤ u ≤ 1.340
Proof. [Proof to lemma 2.3 (i)]341 ∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ dξ = ε
√τ−2 + 2e
(λ−1)xε√τ
λ2 − 1≤ 2ε
√τ
1− λ2if λ ∈ (0, 1).
342
Proof. [Proof to lemma 2.3 (ii)]343 ∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ eλx−ξε√τ dξ = xe
(λ−1)xε√τ ≤ ε
√τ
e(1− λ)if λ ∈ (0, 1).
20
344
Proof. [Proof to lemma 2.3 (iii)] Based on the assumption on u0 in (2.27)345 ∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ≤
∫ +∞
0
e− |x−ξ|
ε√τ e
λxε√τ |u0(ξ)| dξ
≤Cueλxε√τ
∫ L0
0
e− |x−ξ|
ε√τ dξ = Cuy1(x)
(A.1)
Calculating y1(x) with the assumption that λ ∈ (0, 1), we get346
y1(x) =
eλxε√τ∫ L0
0e− |x−ξ|
ε√τ dξ ≤ 2ε
√τe
λxε√τ ≤ 2ε
√τe
λL0ε√τ for x ∈ [0, L0]
e(λ−1)xε√τ∫ L0
0e
ξε√τ dξ ≤ ε
√τe
(λ−1)x+L0ε√τ ≤ ε
√τe
λL0ε√τ for x ∈ [L0,+∞)
Therefore, we get the desired inequality347 ∫ +∞
0
∣∣∣e− x+ξε√τ − e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ≤ 2Cuε
√τe
λL0ε√τ .
348
Proof. [Proof to lemma 2.4 (i)]349
350 ∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ eλx−λξε√τ dξ
=ε√τ
λ2 − 1
(−2 + 2λe
(λ−1)xε√τ − 2(λ− 1)e−
2xε√τ
)≤ 2ε
√τ
1− λ2if λ ∈ (0, 1).
351
Proof. [Proof to lemma 2.4 (ii)]352
353 ∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ eλx−ξε√τ dξ
=2e
(λ−3)xε√τ − 2e
(λ−1)xε√τ
−2ε√τ
+ xe(λ−1)xε√τ ≤ ε
√τ +
ε√τ
e(1− λ)if λ ∈ (0, 1).
354
Proof. [Proof to lemma 2.4 (iii)] Based on the assumption on u0 in (2.27)355 ∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ
≤ Cueλxε√τ
∫ L0
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ dξ= Cuy3(x)
(A.2)
Calculating y3(x) with the assumption that λ ∈ (0, 1), we get for x ∈ [0, L0]356
y3(x) ≤ e(λ−1)xε√τ
∫ x
0
(e−ξε√τ + e
ξε√τ ) dξ + e
(λ+1)xε√τ
∫ L0
x
e− ξε√τ dξ ≤ 2ε
√τe
λL0ε√τ
21
and357
y3(x) ≤ e(λ−1)xε√τ
∫ L0
0
(e−ξε√τ + e
ξε√τ ) dξ ≤ ε
√τe
(λ−1)x+L0ε√τ ≤ ε
√τe
λL0ε√τ
for x ∈ [L0,+∞).358
Therefore, we get the desired inequality359 ∫ +∞
0
∣∣∣e− x+ξε√τ + sgn(x− ξ)e−
|x−ξ|ε√τ
∣∣∣ e λxε√τ |u0(ξ)| dξ ≤ 2Cuε
√τe
λL0ε√τ .
Proof. [Proof to lemma 2.5 (i)]360 ∣∣∣φ1(x)− e−xε√τ
∣∣∣ = e− Lε√τ
∣∣∣∣∣e−xε√τ − e
xε√τ
eLε√τ − e−
Lε√τ
∣∣∣∣∣ = e− Lε√τ |φ2(x)| .
361
Proof. [Proof to lemma 2.5 (ii)] Since φ2(x) = exε√τ −e
− xε√τ
eLε√τ −e
− Lε√τ
, we see that φ′2(x) =362
1ε√τexε√τ +e
− xε√τ
eLε√τ −e
− Lε√τ
> 0 and hence φ2(x) ≤ φ2(L) = 1 for x ∈ [0, L].363
Proof. [Proof to lemma 2.5 (iii)] φ′2(x) = 1ε√τexε√τ +e
− xε√τ
eLε√τ −e
− Lε√τ
gives that φ′′2(x) =364
1ε2τ φ2(x) > 0, and hence φ′2(x) ≤ φ′2(L) = 1
ε√τeLε√τ +e
− Lε√τ
eLε√τ −e
− Lε√τ
= 1ε√τe
2Lε√τ +1
e2Lε√τ −1
≤ 2ε√τ
if365
ε� 1 for x ∈ [0, L].366
Acknowledgments. CYK would like to thank Prof. L.A. Peletier for introduc-367
ing MBL equation and Mathematical Biosciences Institute at OSU for the hospitality368
and support.369
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