Analysis of Single and Multi Phase Flows in Porous
Media
Luan T. Hoang
Joint with Akif Ibragimov, Thinh Kieu, Tuoc Phan.
Department of Mathematics and Statistics, Texas Tech University
http://www.math.ttu.edu/∼lhoang/
luan.hoang@ttu.edu
Applied Mathematics Seminar
Department of Mathematics and Statistics
Texas tech University
September 3, 10 and 17, 2014
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Outline
1
Introduction
2
Single-phase Forchheimer flows
Uniform Gronwall estimates
Interior L∞ -estimates for pressure
Interior Ls -estimates for pressure gradient
Interior L∞ -estimates for pressure gradient
Interior L∞ -estimates for time derivative of pressure
Interior L2 -estimates for pressure’s Hessian
3
Two-phase incompressible Forchheimer flows
One-dimensional problem
Multi-dimensional problem
Steady states
Linearized problem
In bounded domains
In unbounded domains
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Table of Contents
1
Introduction
2
Single-phase Forchheimer flows
3
Two-phase incompressible Forchheimer flows
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Introduction: Darcy’s and Forchheimer’s flows
Fluid flows in porous media with velocity u and pressure p:
Darcy’s Law:
αu = −∇p,
Forchheimer’s “two term” law
αu + β|u| u = −∇p,
Forchheimer’s “three term” law
Au + B |u| u + C|u|2 u = −∇p.
Forchheimer’s “power” law
au + c n |u|n−1 u = −∇p,
Here α, β, a, c, n, A, B, and C are empirical positive constants.
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Generalized Forchheimer equations
[Aulisa-Bloshanskaya-H.-Ibragimov 2009]
Generalizing the above equations as follows
g (|u|)u = −∇p.
Let G (s) = sg (s). Then G (|u|) = |∇p| ⇒ |u| = G −1 (|∇p|). Hence
u=−
∇p
g (G −1 (|∇p|))
K (ξ) = Kg (ξ) =
⇒ u = −K (|∇p|)∇p,
1
1
=
,
−1
g (s)
g (G (ξ))
sg (s) = ξ.
Class FP(N, α
~ ). Let N > 0, 0 = α0 < α1 < α2 < . . . < αN ,
n
o
FP(N, α
~ ) = g (s) = a0 s α0 + a1 s α1 + a2 s α2 + . . . + aN s αN ,
where a0 , aN > 0, a1 , . . . , aN−1 ≥ 0. Notation: αN = deg(g ),
~a = (a0 , a1 , . . . , aN ), a = ααN N+1 ∈ (0, 1), b = ααN N+2 ∈ (0, 1).
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Historical remarks
Darcy-Dupuit: 1865
Forchheimer: 1901
Other nonlinear models: 1940s–1960s
Incompressible fluids: Payne, Straughan and collaborators since
1990’s, Celebi-Kalantarov-Ugurlu since 2005 (Brinkman-Forchheimer)
Derivation of non-Darcy, non-Forchheimer flows: Marusic-Paloka and
Mikelic 2009 (homogenization for Navier–Stokes equations), Balhoff
et. al. 2009 (computational)
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Works on generalized Forchheimer flows
A. Single-phase flows.
1990’s Numerical study
L2 -theory (for slightly compressible flows):
Aulisa-Bloshanskaya-H.-Ibragimov (2009), H.-Ibragimov: Dirichlet
B.C. (2011), H.-Ibragimov Flux B.C. (2012),
Aulisa-Bloshanskaya-Ibragimov total flux, productivity index (2011,
2012), Inhomogeneous media Celik-H.(in preparation).
Lα -theory: H.-Ibragimov-Kieu-Sobol (2012-preprint)
L∞ , W 1,p -theory: H.-Kieu-Phan (2014), Celik-H.(in preparation).
W 1,∞ -theory: interior H.-Kieu (2014-preprint), global Celik-H.-Kieu
(in preparation).
B. Multi-phase flows.
One-dimensional case: H.-Ibragimov-Kieu (2013).
Multi-dimensional case: H.-Ibragimov-Kieu (preprint).
Note: there are more works on Forchheimer flows (2-terms or 3 terms).
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Table of Contents
1
Introduction
2
Single-phase Forchheimer flows
Uniform Gronwall estimates
Interior L∞ -estimates for pressure
Interior Ls -estimates for pressure gradient
Interior L∞ -estimates for pressure gradient
Interior L∞ -estimates for time derivative of pressure
Interior L2 -estimates for pressure’s Hessian
3
Two-phase incompressible Forchheimer flows
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Single-phase Forchheimer flows
Let ρ be the density. Continuity equation
dρ
+ ∇ · (ρu) = 0.
dt
For slightly compressible fluid:
dρ
1
= ρ,
dp
κ
where κ 1. Then
dp
= κ∇ · K (|∇p|)∇p + K (|∇p|)|∇p|2 .
dt
Since κ 1, we neglect the last terms, after scaling the time variable:
dp
= ∇ · K (|∇p|)∇p .
dt
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Degeneracy - The Degree Condition
Lemma
Let g (s,~a) be in class FP(N, α
~ ). One has for any ξ ≥ 0 that
C1 (~a)
C2 (~a)
≤ K (ξ,~a) ≤
,
(1 + ξ)a
(1 + ξ)a
C3 (~a)(ξ 2−a − 1) ≤ K (ξ,~a)ξ 2 ≤ C2 (~a)ξ 2−a .
Degree Condition (DC)
deg(g ) ≤
4
n(2 − a)
⇐⇒ 2 ≤ (2 − a)∗ =
.
n−2
n − (2 − a)
Under the (DC), W 1,2−a (U) ⊂ L2 (U).
This talk: No conditions on the degree.
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IBVP
We study the resulting parabolic equation for the pressure p = p(x, t):
∂p
= ∇ · (K (|∇p|)∇p),
∂t
x ∈ U, t > 0.
Dirichlet boundary data:
p = ψ(x, t),
x ∈ Γ, t > 0,
where ψ(x, t) is known.
The initial data
p(x, 0) = p0 (x) is given.
Extension Ψ(x, t) of ψ(x, t) to x ∈ Ū, t ≥ 0.
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For α ≥ 1, we define
A(α, t) =
hZ
|∇Ψ(x, t)|
α(2−a)
2
dx
i 2(α−a)
α(2−a)
+
hZ
U
|Ψt (x, t)|α dx
i
α−a
α(1−a)
U
for t ≥ 0, and
A(α) = lim sup A(α, t)
and β(α) = lim sup[A0 (α, t)]− .
t→∞
Also, define α∗ =
an
2−a ,
t→∞
and for α > 0 define
α
b = max α, 2, α∗ .
Whenever β(α) is in use, it is understood that the function t → A(α, t)
belongs to C 1 ((0, ∞)).
For a function f : [0, ∞) → R, we denote by Envf a continuous and
increasing function F : [0, ∞) → R such that F (t) ≥ f (t) for all t ≥ 0.
Denote p = p − Ψ .
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Theorem (H.-Ibragimov-Kieu-Sobol)
Let α > 0.
(i) For all t ≥ 0,
Z
Z
α
b
α
b
|p̄(x, t)| dx ≤ C 1 +
|p̄(x, 0)|αb dx + [EnvA(b
α, t)] α−a
.
U
U
(ii) If A(b
α) < ∞ then
Z
lim sup
t→∞
α
b b
|p̄(x, t)|α dx ≤ C 1 + A(b
α) α−a
.
U
(iii) If β(b
α) < ∞ then there is T > 0 such that
Z
α
b
α
b b
b
|p̄(x, t)|α dx ≤ C 1 + β(b
α) α−2a
+ A(b
α, t) α−a
for all t ≥ T .
U
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For gradient and time derivative estimates, we denote
Z
hZ
i 2−a
r (1−a)
2
G1 (t) =
|∇Ψ(x, t)| dx +
|Ψt (x, t)|r0 dx 0
U
U
hZ
i1
r
+
|Ψt (x, t)|r0 dx 0 ,
U
Z
Z
2
G2 (t) =
|∇Ψt (x, t)| dx +
|Ψt (x, t)|2 dx,
U
U
Z
G3 (t) = G1 (t) + G2 (t), G4 (t) = G3 (t) +
|Ψtt |2 dx.
U
with r0 =
Define
n(2−a)
(2−a)(n+1)−n .
Z
H(ξ) =
ξ2
√
K ( s)ds
for ξ ≥ 0.
0
Then
H(ξ) ∼ K (ξ)ξ 2 ∼ ξ 2−a .
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For t ≥ 0, recall from H.-Ibragimov (2011) that
Z tZ
Z
Z t
H(|∇p|)dxdτ ≤ C
p̄ 2 (x, 0)dx + C
G1 (τ )dτ,
0
U
U
0
and
Z tZ
Z
H(|∇p|)(x, t)dx +
|p̄t (x, τ )|2 dxdτ
U
0
U
Z
Z t
2
≤
H(|∇p(x, 0)|) + p̄ (x, 0) dx + C G3 (τ )dτ.
U
0
Let α ≥ b
2. For t > 0, recall from H.-Ibragimov-Kieu-Sobol that
Z
2
|p̄t (x, t)| dx ≤ C (1 + t
U
−1
Z
Z
α
) 1 + |p̄(x, 0)| dx + H(|∇p(x, 0)|)dx
U
U
Z t
α
G4 (τ )dτ .
+ [EnvA(α, t)] α−a +
0
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Uniform Gronwall estimates
Lemma
For t ≥ 1,
Z t Z
Z
Z t
H(|∇p|)dxdτ ≤ C
p̄ 2 (x, t − 1)dx + C
G1 (τ )dτ,
t−1 U
U
t−1
Z
Z
Z
1 t
H(|∇p|)(x, t)dx +
p̄t2 (x, τ )dxdτ
2
U
t−1/2 U
Z
Z t
2
≤C
p̄ (x, t − 1)dx + C
G3 (τ )dτ,
U
t−1
Z
Z
Z t
2
2
p̄t (x, t)dx ≤ C
p̄ (x, t − 1)dx + C
G4 (τ )dτ.
U
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U
t−1
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Interior L∞ -estimates for pressure
Theorem
Let U 0 b U. If T0 ≥ 0, T > 0 and θ ∈ (0, 1) then
sup
κ1
κ1 α−a
kp(t)kL∞ (U 0 ) ≤ C (1 + T ) κ0 1 + (θT )−1
[T0 +θT ,T0 +T ]
· 1 + kpkLα (U×(T0 ,T0 +T ))
κ2
.
Proof by De Giorgi’s technique with cut-off functions in both spatial and
time variables.
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Parabolic Poincaré-Sobolev inequality
For each T > 0, denote QT = U × (0, T ). Recall α∗ =
an
2−a .
Lemma
Assume α ≥ 2 and α > α∗ . Let p = α 1 +
2−a
n
− a. Then
kukLp (QT ) ≤ C (1 + δT )1/p [[u]],
where δ = 1 in general, δ = 0 in case u vanishes on the boundary ∂U, and
[[u]] = ess sup ku(t)kLα (U) +
Z
T
0
[0,T ]
Z
|u|α−2 |∇u|2−a dxdt
1
α−a
.
U
In case U = BR , the inequality holds with [[u]] defined by
R
1
n( α
− p1 )
ess sup ku(t)kLα (BR ) +R
[0,T ]
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n+2−a
− pn
α−a
Z
T
0
Single and Multi Phase Flows
Z
|u|α−2 |∇u|2−a dxdt
1
α−a
.
BR
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Fast decaying geometry sequences with multiple rates
Lemma
Let {Yi }∞
i=0 be a sequence of non-negative numbers satisfying
Yi+1 ≤
m
X
Ak Bki Yi1+µk ,
i = 0, 1, 2, · · · ,
k=1
where Ak > 0, Bk > 1 and µk > 0 for k = 1, 2, . . . , m. Let
B = max{Bk : 1 ≤ k ≤ m} and µ = min{µk : 1 ≤ k ≤ m}.
If
m
X
Ak Y0µk ≤ B −1/µ
then lim Yi = 0.
k=1
In particular, if Y0 ≤ min{(m−1 A−1
k B
limi→∞ Yi = 0.
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− µ1 1/µk
)
Single and Multi Phase Flows
i→∞
: 1 ≤ k ≤ m} then
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Theorem
Let U 0 b U.
(i) If t ∈ (0, 1) then
κ2
κ1 1
kp(t)kL∞ (U 0 ) ≤ Ct − α−a 1 + kp̄0 kLα + [EnvA(α, t)] α−a + kΨkLα (U×(0,t)) .
If t ≥ 1 then
κ2
1
kp(t)kL∞ (U 0 ) ≤ C 1 + kp̄0 kLα + [EnvA(α, t)] α−a + kΨkLα (U×(t−1,t)) .
(ii) If A(α) < ∞ then
1
lim sup kp(t)kL∞ (U 0 ) ≤ C 1 + A(α) α−a + lim sup kΨkLα (U×(t−1,t))
t→∞
κ2
.
t→∞
(iii) If β(α) < ∞ then there is T > 0 such that for all t ≥ T ,
1
1
kp(t)kL∞ (U 0 ) ≤ C 1 + β(α) α−2a + A(α, t) α−a + kΨkLα (U×(t−1,t))
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κ2
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.
20
Interior Ls -estimates for pressure gradient
Ladyzhenskaya-Uraltseva type embedding:
Lemma
For each s ≥ 1, there exists a constant C > 0 depending on s such that for
each smooth cut-off function ζ(x) ∈ Cc∞ (U), the following inequality holds
Z
hZ
K (|∇p|)|∇p|2s+2 ζ 2 dx ≤ C max |p|2
K (|∇p|)|∇p|2s−2 |∇2 p|2 ζ 2 dx
suppζ
U
U
Z
i
+
K (|∇p|)|∇p|2s |∇ζ|2 dx ,
U
for every sufficiently regular function p(x) such that the right hand side is
well-defined.
Key property.
−aK (ξ) ≤ ξK 0 (ξ) ≤ 0.
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We establish the basic step for the Ladyzhenskaya-Uraltseva iteration.
Lemma
For each s ≥ 0, if T0 ≥ 0, T > 0, and ζ(x, t) is a smooth cut-off function
then
Z
Z
ζ dx+
2s+2 2
|∇p(x, t)|
sup
[T0 ,T0 +T ] U
Z
T0 +T
T0
Z
≤C
K (|∇p|)|∇p|
T0
T0 +T
2s+2
Z
K (|∇p|)|∇2 p|2 |∇p|2s ζ 2 dxdt
U
Z
2
T0 +T
Z
|∇p|2s+2 ζ|ζt |dx.
|∇ζ| dxdt+C
U
T0
U
As a consequence, for 0 < θ0 < θ < 1,
Z
T0 +T
T0
Z
K (|∇p|)|∇p|2s+4 ζ 2 dxdt ≤ C
sup
[T0 +θ0 T ,T0 +T ]
U
Z
T0 +T
Z
·
T0
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kpk2L∞ (V )
(1 + K (|∇p|)|∇p|2s+2+a )(|∇ζ|2 + ζ|ζt |)dxdt.
U
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Proposition
Let U 0 b V b U, T0 ≥ 0, T > 0 and θ ∈ (0, 1). If s ≥ 2 then
Z
T0 +T
Z
T0 +θT
s−2
K (|∇p|)|∇p|s dxdt ≤ C 1 + (θT )−1
U0
· 1+
sup
[T0 +θT /2,T0 +T ]
Z
kpk2L∞ (V )
T0 +T
Z
·
T0
Z
s−2
(1 + K (|∇p|)|∇p|2 )dxdt,
U
s+a−1
|∇p(x, t)|s dxdt ≤ C 1 + (θT )−1
sup
t∈[T0 +θT ,T0 +T ] U 0
· 1+
sup
[T0 +θT /2,T0 +T ]
Z
kpk2L∞ (V )
T0 +T
Z
·
T0
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s−2+a
(1 + K (|∇p|)|∇p|2 )dxdt.
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Theorem
Let U 0 b U and s ≥ 2. If t ∈ (0, 2) then
Z
U0
|∇p(x, t)|s dx ≤ Ct −µ1 (1 + kp̄0 kLα )µ2 +2
µ2
1
· 1 + [EnvA(α, t)] α−a + kΨkLα (U×(0,t))
Z t
· 1+
G1 (τ )dτ ,
0
If t ≥ 2 then
Z
U0
|∇p(x, t)|s dx ≤ C (1 + kp̄0 kLα )µ2 +α
µ2 +α
1
· 1 + [EnvA(α, t)] α−a + kΨkLα (U×(t−2,t))
Z t
· 1+
G1 (τ )dτ .
t−1
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Theorem
Let U 0 b U and s ≥ 2.
(i) If A(α) < ∞ then
Z
lim sup
t→∞
µ2 +α
1
|∇p(x, t)|s dx ≤ C 1+A(α) α−a +lim sup kΨkLα (U×(t−1,t))
t→∞
U0
Z t
· 1 + lim sup
G1 (τ )dτ .
t→∞
t−1
(ii) If β(α) < ∞ then there is T > 0 such that for all t > T ,
Z
|∇p(x, t)|s dx
µ2 +α
1
1
≤ C 1 + β(α) α−2a + sup A(α, ·) α−a + kΨkLα (U×(t−2,t))
U0
[t−1,t]
Z
t
· 1+
G1 (τ )dτ .
t−1
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Interior L∞ -estimates for pressure gradient
For each m = 1, 2, . . . , n, denote um = pxm and u = (u1 , u2 , . . . , un ) = ∇p.
We have
P
h
ui ∂m ui i
∂um
0
= ∂m (∇ · (K (|u|)u)) = ∇ · (K (|u|)∂m u) + ∇ · K (|u|) i
u .
∂t
|u|
Since ∂i um = ∂m ui , we have
∂
· · · , ∂m un ) = (∂1 um , · · · , ∂n um ) = ∇um , and
m u = (∂m u1 ,P
P
i ui ∂m ui =
i ui ∂i um = u · ∇um . Therefore, we rewrite (1) as
h
u · ∇um i
∂um
= ∇ · (K (|u|)∇um ) + ∇ · K 0 (|u|)
u .
∂t
|u|
Then use De Giorgi’s iteration.
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Weighted parabolic Sobolev-Poincaré inequality
Lemma
Given W (x, t) > 0 on QT . Let r be a number that satisfies
Set
def
% = %(r ) == 4(1 − 1/r ∗ ).
2n
n+2
< r < 2.
Then
n 1
Z
2−r o
r
%r
kukL% (QT ) ≤ C [[u]]2,W ;T δT % +ess sup
W (x, t)− 2−r χsupp u (x, t)dx
U
t∈[0,T ]
where δ = 1 in general, δ = 0 in case u vanishes on the boundary ∂U, and
[[u]]2,W ;T = ess sup ku(t)kL2 (U) +
[0,T ]
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Z
T
Z
0
Single and Multi Phase Flows
W (x, t)|∇u|2 dxdt
1
2
.
U
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Theorem
Let U 0 b V b U. For any T0 ≥ 0, T > 0, and θ ∈ (0, 1), if
t ∈ [T0 + θT , T0 + T ] then
k∇p(t)kL∞ (U 0 ) ≤ C (1 + (θT )−1 )
s1 +1
2
s1
λ 2 k∇pkL2 (V ×(T0 +θT /2,T0 +T )) ,
where
λ = λ(T0 , T , θ; V ) =
Z
T0 +T
T0 +θT /2
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Z
(1 + |∇p|)
as0
2−s0
dxdt
2−s0
s0
.
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Theorem
If t ∈ (0, 2) then
k∇p(t)kL∞ (U 0 ) ≤ Ct −κ4 /2 (1 + kp̄0 kLα )s3 (κ5 +1)
s3 κ5
1
· 1 + EnvA(α, t) α−a + kΨkLα (U×(0,t))
Z t
s3 /2
· 1+
.
G1 (τ )dτ
0
If t ≥ 2 then
k∇p(t)kL∞ (U 0 ) ≤ C (1 + kp̄0 kLα )s3 (κ5 +α/2)
s3 (κ5 +α/2)
1
· 1 + [EnvA(α, t)] α−a + kΨkLα (U×(t−2,t))
Z t
s3 /2
G1 (τ )dτ
.
· 1+
t−1
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Theorem
(i) If A(α) < ∞ then
lim sup k∇p(t)kL∞ (U 0 )
t→∞
s3 (κ5 +α/2)
1
≤ C 1 + A(α) α−a + lim sup kΨkLα (U×(t−1,t))
t→∞
Z t
s3 /2
· 1 + lim sup
.
G1 (τ )dτ
t→∞
t−1
(ii) If β(α) < ∞ then there is T > 0 such that when t > T we have
k∇p(t)kL∞ (U 0 )
s3 (κ5 +α/2)
1
1
≤ C 1 + β(α) α−2a + sup A(α, ·) α−a + kΨkLα (U×(t−2,t))
[t−1,t]
Z
t
· 1+
G1 (τ )dτ
s3 /2
.
t−1
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Interior L∞ -estimates for time derivative of pressure
Let q = pt . Then
∂q
= ∇ · K (|∇p|)∇p t .
∂t
Using De Giorgi’s iterations and weighted parabolic Sobolev-Poincaré
inequality, we obtain
Proposition
Let U 0 b V b U. If T0 ≥ 0, T > 0 and θ ∈ (0, 1), then
sup
s1
kpt kL∞ (U 0 ) ≤ C λ 2 (1 + (θT )−1 )
[T0 +θT ,T0 +T ]
s1 +1
2
kpt kL2 (U×(T0 ,T0 +T )) ,
where
λ = λ(T0 , T , θ; V ) =
Z
T0 +T
T0 +θT /2
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Z
as0
(1 + |∇p|) 2−s0 dxdt
2−s0
s0
.
V
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Theorem
Let U 0 b U. For t ∈ (0, 2),
kpt (t)kL∞ (U 0 ) ≤ Ct
−κ6 /2
1
· 1 + [EnvA(α, t)] α−a
κ7 Z
1/2
.
H(|∇p(x, 0)|)dx
U
Z t
κ8 s3 /2
.
+ kΨkLα (U×(0,t))
1+
G3 (τ )dτ )
1 + kp̄0 kLα
1+
0
For t ≥ 2,
kpt (t)kL∞ (U 0 ) ≤ C (1 + kp̄0 kLα )κ9
κ9
1
· 1 + [EnvA(α, t)] α−a + kΨkLα (U×(t−2,t))
Z t
s3 /2
· 1+
G3 (τ )dτ
.
t−1
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Theorem
Let U 0 b U.
(i) If A(α) < ∞ then
κ9
1
lim sup kpt (t)kL∞ (U 0 ) ≤ C 1 + A(α) α−a + lim sup kΨkLα (U×(t−1,t))
t→∞
t→∞
Z t
s3 /2
.
G3 (τ )dτ
· 1 + lim sup
t→∞
t−1
(ii) If β(α) < ∞ then there is T > 0 such that for all t > T ,
κ9
1
1
kpt (t)kL∞ (U 0 ) ≤ C 1 + β(α) α−2a + sup A(α, ·) α−a + kΨkLα (U×(t−2,t))
[t−1,t]
Z
t
· 1+
G3 (τ )dτ
s3 /2
.
t−1
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33
Interior L2 -estimates for pressure’s Hessian
We estimate the L2 -norm of the Hessian ∇2 p = (pxi xj )i,j=1,2,...,n .
Lemma
Let U 0 b V b U. For t > 0,
a Z
1/2
k∇2 p(t)kL2 (U 0 ) ≤ C 1 + k∇p(t)kL∞ (V )
[|∇p|2−a + |pt |2 ]dx
.
U
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Theorem
Let U 0 b U.
(i) For t ∈ (0, 2), one has
2
k∇ p(t)kL2 (U 0 ) ≤ Ct
1
· 1 + EnvA(α, t) α−a
Z
1/2
H(|∇p(x, 0)|)dx
U
Z t
κ12 s4 /2
+ kΨkLα (U×(0,t))
1+
.
G4 (τ )dτ
−κ10 /2
(1 + kp̄0 kLα )
κ11
1+
0
(ii) If t ≥ 2 then one has
k∇2 p(t)kL2 (U 0 ) ≤ C (1 + kp̄0 kLα )κ11
κ11
1
· 1 + [EnvA(α, t)] α−a + kΨkLα (U×(t−2,t))
Z t
s4 /2
· 1+
G4 (τ )dτ
.
t−1
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Theorem (Continued)
(iii) If A(α) < ∞ then
κ11
1
lim sup k∇2 pkL2 (U 0 ) ≤ C 1 + A(α) α−a + lim sup kΨkLα (U×(t−1,t))
t→∞
t→∞
Z t
s4 /2
.
· 1 + lim sup
G4 (τ )dτ
t→∞
t−1
(iv) If β(α) < ∞ then there is T > 0 such that for t > T ,
κ11
1
1
k∇2 p(t)kL2 (U 0 ) ≤ C 1+β(α) α−2a + sup A(α, ·) α−a +kΨkLα (U×(t−2,t))
[t−1,t]
Z
· 1+
t
G4 (τ )dτ
s4 /2
.
t−1
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Table of Contents
1
Introduction
2
Single-phase Forchheimer flows
3
Two-phase incompressible Forchheimer flows
One-dimensional problem
Multi-dimensional problem
Steady states
Linearized problem
In bounded domains
In unbounded domains
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Two-phase incompressible Forchheimer flows
For each ith-phase (i = 1, 2), saturation Si ∈ [0, 1], density ρi ≥ 0,
velocity ui ∈ Rn , and , and pressure pi ∈ R. The saturations satisfy
S1 + S2 = 1.
Each phase’s velocity obeys the generalized Forchheimer equation.
Conservation of mass holds for each of the phases:
∂t (φρi Si ) + div(ρi ui ) = 0,
i = 1, 2.
Due to incompressibility of the phases, i.e. ρi = const. > 0, it is reduced to
φ∂t Si + div ui = 0,
i = 1, 2.
Let pc be the capillary pressure between two phases, more specifically,
p1 − p2 = pc .
Darcy’s flows. Kruzkov, Sukorjanski, Alt, DiBenedetto, Cances, Mikelic,
Galusinski, Saad, Chemetov, Neves ...
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Denote S = S1 and pc = pc (S). Then
gi (|ui |)ui = −fi (S)∇pi ,
i = 1, 2,
∇p1 − ∇p2 = pc0 (S)∇S.
Hence
F2 (S)g2 (|u2 |)u2 − F1 (S)g1 (|u1 |)u1 = ∇S,
where
Fi (S) =
1
pc0 (S)fi (S)
,
i = 1, 2.
In summary,
0 ≤ S = S(x, t) ≤ 1,
St = −div u1 ,
St = div u2 ,
∇S = F2 (S)G2 (u2 ) − F1 (S)G1 (u1 ).
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One-dimensional problem
Assumption A.
f1 , f2 ∈ C ([0, 1]) ∩ C 1 ((0, 1)),
f1 (0) = 0,
f10 (S) > 0,
f2 (1) = 0,
f20 (S) < 0 on (0, 1).
Assumption B.
pc0 ∈ C 1 ((0, 1)),
pc0 (S) > 0 on (0, 1).
Theorem (H.-Kieu-Ibragimov 2013)
• There are 16 types of non-constant steady states (based on their
monotonicity and asymptotic behavior as x → ±∞).
• The steady states which are never zero nor one are linearly stable.
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40
Multi-dimensional problem
In Rn , steady states:
div u1 = div u2 = 0,
∇S = F2 (S)G2 (u2 ) − F1 (S)G1 (u1 ).
Steady states with geometric constraints:
u∗1 (x) = c1 |x|−n x,
u∗2 (x) = c2 |x|−n x,
S∗ (x) = S(|x|),
where c1 , c2 are constants and S(r ) is a solution of the following ODE:
S 0 = F (r , S(r ))
for r > r0 ,
S(r0 ) = s0 ,
0 < S(r ) < 1.
where s0 is always a number in (0, 1) and
F (r , S(r )) = G2 (c2 r 1−n )F2 (S) − G1 (c1 r 1−n )F1 (S).
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Theorem
There exists a maximal interval of existence [r0 , Rmax ), where
Rmax ∈ (r0 , ∞], and a unique solution S ∈ C 1 ([r0 , Rmax ); (0, 1)).
Moreover, if Rmax is finite then either
lim S(r ) = 0
−
r →Rmax
or lim S(r ) = 1.
−
r →Rmax
Theorem
If solution S(r ) exists in [r0 , ∞), then it eventually becomes monotone
and, consequently, s∞ = limr →∞ S(r ) exists.
In case n = 2 and c12 + c22 > 0, let s ∗ = (f1 /f2 )−1
(i) If c1 ≤ 0 and c2 ≥ 0 then s∞ = 1.
(ii) If c1 ≥ 0 and c2 ≤ 0 then s∞ = 0.
(iii) If c1 , c2 < 0 then s∞ = s ∗ .
(iv) If c1 , c2 > 0 then s∞ ∈ {0, 1, s ∗ }.
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c1 a10
c2 a20
.
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Linearized problem
The formal linearized system at the steady state (u∗1 (x), u∗2 (x), S∗ (x)) is
σt = −div v1 ,
∇σ =
σt = div v2 ,
0
F2 (S∗ )G2 (u∗2 )v2 + F20 (S∗ )σG2 (u∗2 )
− F1 (S∗ )G01 (u∗1 )v1 + F10 (S∗ )σG1 (u∗1 ) .
Let v = v1 + v2 . Then div v = 0. Assume v = V(x, t) is given . Let
B = B(x) = F2 (S∗ )G02 (u∗2 ) + F1 (S∗ )G01 (u∗1 ),
A = A(x) = B(x)−1
b = b(x) = F20 (S∗ )G2 (u∗2 ) − F10 (S∗ )G1 (u∗1 ),
c = c(x, t) = F1 (S∗ )G01 (u∗1 )V(x, t).
Decoupling the linearized system:
h
i
σt = ∇ · A(∇σ − σb) + ∇ · (Ac),
v2 = A(∇σ − σb) + Ac,
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v1 = V − v2 .
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Lemma
For any c12 + c22 > 0 and x 6= 0, matrices B(x) and A(x) are symmetric,
invertible and positive definite.
Also, matrix B has the following special property:
B(x)x =
2 n
X
o
Fi (Ŝ(|x|)) gi (|ci ||x|1−n ) + gi0 (|ci ||x|1−n )|ci ||x|1−n x
i=1
= φ(|x|)x,
where
φ(r ) =
2
X
Fi (Ŝ(r )) gi (|ci |r 1−n ) + gi0 (|ci |r 1−n )|ci |r 1−n .
i=1
Now consider “good” steady states.
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In Bounded domains
def
Let R > r0 > 0, U ⊂ U == BR \ B̄r0 . Denote Γ = ∂U, D = U × (0, ∞)
and D = U × (0, ∞).
Initial-boundary value problem (IBVP):

h
i

σ
=
∇
·
A(∇σ
−
σb)
+ ∇ · (Ac) on U × (0, ∞),

 t
σ = g (x, t)
on Γ × (0, ∞),


σ = σ (x)
on U × {t = 0}.
0
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Condition (E1). F1 , F2 ∈ C 7 ((0, 1)) and V ∈ Cx6 (D̄); Vt ∈ Cx3 (D̄).
Theorem
def
Assume (E1) and ∆4 == supD (|V(x, t)| + |∇V(x, t)|) + supΓ×[0,∞) |g (x, t)|
is finite. Then the solution σ(x, t) of the linearized equation satisfies
h
i
sup |σ(x, t)| ≤ C e −η1 t sup |σ0 (x)| + ∆4
for all t > 0.
x∈U
U
Moreover,
lim sup sup |σ(x, t)| ≤ C ∆5 ,
t→∞
x∈U
where
h
i
∆5 = lim supt→∞ supx∈U (|V(x, t)| + |∇V(x, t)|) + supx∈Γ |g (x, t)| .
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Theorem
def
Assume (E1), and ∆6 == supD (|V(x, t)| + |∇V(x, t)| + |∇2 V(x, t)|) and
def
∆7 == supΓ×[0,∞) |g (x, t)| are finite. Then for any U 0 b U, there is M̃ > 0
such that for i = 1, 2, x ∈ U 0 and t > 0,
i
√
1 h −η1 t
sup |vi (x, t)| ≤ M̃ 1 + √
e
sup |σ0 (x)| + ∆6 + ∆6 + ∆7 .
t
U
x∈U 0
Consequently, if
n
o
lim sup(|V(x, t)| + |∇V(x, t)| + |∇2 V(x, t)|) + sup |g (x, t)| = 0,
t→∞
x∈U
x∈Γ
then for any x ∈ U,
lim v1 (x, t) = lim v2 (x, t) = 0.
t→∞
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t→∞
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Structure and Transformation
Rewrite vector function b(x) explicitly as
|c2 | c2
|c1 | c1 b(x) = F20 (S∗ (x))g2 ( n−1 ) n − F10 (S∗ (x))g1 ( n−1 ) n x = λ(|x|)x,
|x|
|x|
|x|
|x|
where
|c2 | c2
|c1 | c1
λ(r ) = F20 (Ŝ(r ))g2 ( n−1 ) n − F10 (Ŝ(r ))g1 ( n−1 ) n .
r
r
r
r
By defining
Z 2 p
Z |x|
1 |x|
λ( ξ)dξ =
r λ(r )dr ,
Λ(x) =
2 r02
r0
we have for x 6= 0 that
b(x) = ∇Λ(x).
Let
w (x, t) = e −Λ(x) σ(x, t).
Then w satisfies
wt − ∇ · A∇w − ∇Λ · A∇w = e −Λ ∇ · (Ac).
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New system
Define the differential operator
Lw = ∂t w − ∇ · (A∇w ) − b · A∇w .
Corresponding IBVP for w (x, t) is


Lw = f0
w (x, 0) = w0 (x)


w (x, t) = G (x, t)
in U × (0, ∞),
in U,
on Γ × (0, ∞),
where w0 (x) and G (x, t) are given initial data and boundary data,
respectively, and f0 (x, t) is a known function.
• For the velocities, we have
v2 = A ∇(e Λ w ) − e Λ w b + Ac = A e Λ ∇w + we Λ ∇Λ − e Λ w b + Ac.
Thus,
v2 = e Λ A∇w + Ac,
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v1 = V − v2 .
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Lemma of growth in time of Landis type
Barrier function. Define
(
ϕ(x)
t −s e − t
W (x, t) =
0
if t > 0,
if t ≤ 0,
where the number s > 0 and the function ϕ(x) > 0 will be decided later.
Then
n
o
ϕ
LW = t −s−2 e − t t − s + ∇ · (A∇ϕ) + b · A∇ϕ + ϕ − (A∇ϕ) · ∇ϕ .
Thus, LW ≤ 0 if
s ≥ ∇ · (A∇ϕ) + b · A∇ϕ
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and ϕ ≤ (A∇ϕ) · ∇ϕ.
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We will choose ϕ to satisfy
A∇ϕ = κ0 x,
where κ0 is a positive constant selected later. Equivalently,
∇ϕ = κ0 A−1 x = κ0 Bx = κ0 φ(|x|)x.
Define for x ∈ Ū the function
Z |x|
ϕ(x) = κ0 ϕ0 +
r φ(r )dr ,
where ϕ0 =
r0
Select
C0
C0 r02
and κ0 =
.
2
2C1
def
s = sR == κ0 (n + C2 R).
Lemma
2,1
The function W (x, t) belongs to Cx,t
(D) ∩ C (D̄) and satisfies LW ≤ 0 in
D.
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Lemma of growth in time
We fix s = sR and also the following two parameters
q=
κ0 C0
2s
and η0 =
r 2s
0
R
,
and denote D1 = U × (0, qR 2 ].
Lemma (Lemma of growth in time)
2,1
Assume w (x, t) ∈ Cx,t
(D1 ) ∩ C (D̄1 ). If
Lw ≤ 0 on D1
and
then
max{0, sup w (x, qR 2 )} ≤
U
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w ≤ 0 on Γ × (0, qR 2 ),
1
max{0, sup w (x, 0)}.
1 + η0
U
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Proof.
Let M = max{0, supŪ w (x, 0)}, W̃ = M[1 − ηW ], η > 0 selected later,
t1 = qR 2 .
Applying maximum principle for W̃ gives
w (x, t1 ) ≤ W̃ (x, t1 ) ≤ M(1 − ηC (s, R)) = M(1 − η0 ) ≤ M/(1 + η0 ).
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Proposition (Homogeneous problem)
2,1
Assume w (x, t) ∈ Cx,t
(D) ∩ C (D̄) satisfies
Lw = 0 in D
Let η1 =
ln(1+η0 )
.
qR 2
and
w = 0 on Γ × (0, ∞).
Then
− e −η1 t inf |w (x, 0)| ≤ w (x, t) ≤ (1 + η0 )e −η1 t sup |w (x, 0)|
U
∀(x, t) ∈ D.
U
Proposition (Non-homogeneous problem)
Assume f0 ∈ C (D̄) and
def
∆1 == supU×(0,∞) |f0 (x, t)| + supΓ×(0,∞) |G (x, t)| < ∞ The solution
2,1
w (x, t) ∈ Cx,t
(D) ∩ C (D̄) satisfies
|w (x, t)| ≤ C e −η1 t sup |w0 (x)| + ∆1 ∀(x, t) ∈ D.
U
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Bernstein’s estimates
Proposition
Assume f0 ∈ C (D̄), ∇f0 ∈ C (D), ∆1 < ∞ and
def
∆3 == sup |∇f0 | < ∞.
D
2,1
For any U 0 b U there is M̃ > 0 such that if w (x, t) ∈ Cx,t
(D) ∩ C (D̄) is a
3
1
solution of (16) that also satisfies w ∈ Cx (D) and wt ∈ Cx (D), then
h
√ i
1 ih
|∇w (x, t)| ≤ M̃ 1+ √ e −η1 t sup |w (x, 0)|+∆1 + ∆3
t
U
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∀(x, t) ∈ U 0 ×(0, ∞
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In unbounded domains
Outer domain U = Rn \ B̄r0 .
Notation. For R > r > 0, denote Or = Rn \ B̄r , Or ,R = BR \ B̄r .
Let Γ = ∂U = {x : |x| = r0 } and D = U × (0, ∞).
Similar IBVP for σ(x, t):

h
i

σ
=
∇
·
A(∇σ
−
σb)
+ ∇ · (Ac) on U × (0, ∞),

t

σ = g (x, t)
on Γ × (0, ∞),


σ = σ (x)
on U × {t = 0}.
0
Define the differential operator
Lw = ∂t w − ∇ · (A∇w ) − b · A∇w .
Corresponding IBVP for w (x, t) is


Lw = f0
w (x, 0) = w0 (x)


w (x, t) = G (x, t)
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in U × (0, ∞),
in U,
on Γ × (0, ∞).
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Maximum principle for unbounded domain
Theorem
2,1
Let T > 0 and w (x, t) be a bounded function in Cx,t
(UT ) ∩ C (ŪT ) that
solves Lw = f0 in UT , where f0 ∈ C (ŪT ). Then
sup |w (x, t)| ≤ sup |w (x, t)| + (T + 1) sup |f0 |.
∂p UT
ŪT
ŪT
Barrier function:
def
ϕ(x)
W (x, t) == (T − t)−s e T −t
for (x, t) ∈ Or0 ,R × (0, T ),
where constant s > 0 and function ϕ(x) > 0 will be decided later.
Elementary calculations give
n
o
ϕ
LW = (T −t)−s−2 e T −t (T −t) s−∇·(A∇ϕ)−b·A∇ϕ +ϕ−(A∇ϕ)·∇ϕ .
Then LW ≥ 0 if
s ≥ ∇ · (A∇ϕ) + b · A∇ϕ and ϕ ≥ (A∇ϕ) · ∇ϕ.
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Choose
Z
|x|
ϕ(x) = κ1 ϕ1 +
r φ(r )dr ,
r0
where ϕ1 =
C1 r02
2
> 0 and κ1 =
C1
2C0 ,
and
def
s = sR == C3 (1 + R).
• With this barrier function, we can prove the maximum principle in the
outter domain, and hence the uniqueness of bounded solution w .
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Lemma of growth in spatial variables
Let R > 0 and ` ≥ R + r0 . Denote
OR (`) = O`−R,`+R = {x ∈ Rn : ||x|−`| < R}
and S` = {x ∈ Rn : |x| = `}.
Define the barrier function of Landis type
ψ(x)
1
W(x, t) =
e − t+1 for |x| ≥ r0 , t ≥ 0,
s
(t + 1)
where parameter s > 0 and function ψ > 0. Then LW ≤ 0 if
s ≥ ∇ · (A∇ψ) + b · A∇ψ
We can choose s = C3 (1 + R) and
Z
ψ(x, t) = κ2
(*)
and ψ ≤ (A∇ψ) · ∇ψ.
|x|
(r − `)φ(r )dr .
`
Lemma
Given any R > 0 and ` ≥ R + r0 . Then the function W(x, t) in (*)
2,1
belongs to Cx,t
(D) ∩ C (D̄) and satisfies LW ≤ 0 on OR (`) × (0, ∞).
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Lemma (Lemma of growth in spatial variables)
Given T > 0, let
R = R(T ) = C4 (1 + T ),
1
1
η0 = η0 (T ) = 1 − C (T +1)
,
5
2
(T + 1)2C5 (T +1)
3
where C4 = max{1, κ8C
} and C5 = C3 C4 . Suppose
2 eC0
2,1
w (x, t) ∈ Cx,t
(UT ) ∩ C (ŪT ) satisfies Lw ≤ 0 on UT and w (x, 0) ≤ 0 on
Ū. Let ` be any number such that ` ≥ R + r0 , then
max 0,
sup
S` ×[0,T ]
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w (x, t) ≤
1
max 0,
sup
w (x, t) .
1 + η0
ŌR (`)×[0,T ]
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Dichotomies
Lemma
Let T > 0 and R, η0 and w (x, t) be as in Lemma 36. For i ≥ 1, let
m̄i = max 0,
sup
w (x, t) .
Sr0 +iR ×[0,T ]
Part A (Dichotomy for one cylinder). Then for any i ≥ 1, we have either
of the following cases.
(a) If m̄i+1 ≥ m̄i−1 , then m̄i+1 ≥ (1 + η0 )m̄i .
(b) If m̄i−1 ≥ m̄i+1 , then m̄i−1 ≥ (1 + η0 )m̄i .
Part B (Dichotomy for many cylinders). For any k ≥ 0, we have the
following two possibilities:
(i) There is i0 ≥ k + 1 such that m̄i0 +j ≥ (1 + η0 )j m̄i0 for all j ≥ 0.
(ii) For all j ≥ 0, m̄k+j ≤ (1 + η0 )−j m̄k .
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Theorem
2,1
Let w ∈ Cx,t
(UT ) ∩ C (ŪT ) be a bounded solution of the IBVP on UT
with f0 ∈ C (ŪT ). If
lim w0 (x) = 0,
|x|→∞
sup |f0 (x, t))| = 0,
lim
|x|→∞ 0≤t≤T
then
lim
r →∞
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sup |w (x, t)| = 0.
Sr ×[0,T ]
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Corollary
2,1
Let w (x, t) ∈ Cx,t
(D) ∩ C (D̄) be a bounded solution of (16) on D with
f0 ∈ C (D̄). Assume w0 ∈ C (Ū), G ∈ C (Γ × [0, ∞)) are bounded, satisfy
same conditions as above for each T > 0. Then there exists an increasing,
continuous function r (t) > 0 satisfying limt→∞ r (t) = ∞ such that
lim
sup |w (x, t)| = 0.
t→∞
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x∈Ōr (t)
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Dealing with weight e Λ(x)
From w (x, t), we return to σ(x, t) = we Λ(x) .
• In the case n ≥ 3,
0 < C7−1 ≤ e Λ(x) ≤ C7
∀|x| ≥ r0 .
• In the case n = 2,
e Λ(x) ≤ C8
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∀|x| ≥ r0 .
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Theorem
Let n ≥ 3. Assume (E1) and
def
∆10 == max{sup |σ0 (x)|, sup |g (x, t)|} < ∞,
U
Γ×[0,∞)
def
∆11 == sup |∇ · (A(x)c(x, t))| < ∞.
D
Then,
2,1
(i) There exists a solution σ(x, t) ∈ Cx,t
(D) ∩ C (D̄). This solution is
unique in class of solutions σ(x, t) that satisfy
sup |σ(x, t)| < ∞ for any T > 0.
U×[0,T ]
(ii) There is C > 0 such that for (x, t) ∈ D,
|σ(x, t)| ≤ C ∆10 + ∆11 (t + 1) .
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Theorem (continued)
(iii) In addition, if
lim σ0 (x) = 0
|x|→∞
and
then
lim
r →∞
lim
sup |∇·(A(x)c(x, t))| = 0 for each T > 0,
|x|→∞ 0≤t≤T
sup |σ(x, t)| = 0 for any T > 0,
Sr ×[0,T ]
and furthermore, there is a continuous, increasing function r (t) > 0 with
limt→∞ r (t) = ∞ such that
lim
sup |σ(x, t)| = 0.
t→∞
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x∈Ōr (t)
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Theorem
Let n = 2 and Ŝ(r ) be a solution (for the steady state) with c1 , c2 < 0.
Assume (E1) and
def
∆12 == max{sup e −Λ(x) |σ0 (x)|, sup |g (x, t)|} < ∞,
U
Γ×[0,∞)
def
∆13 == sup e −Λ(x) |∇ · (A(x)c(x, t))| < ∞.
D
Then the following statements hold true.
2,1
(D) ∩ C (D̄). This solution is
(i) There exists a solution σ(x, t) ∈ Cx,t
unique in class of solutions σ(x, t) that satisfy
sup e −Λ(x) |σ(x, t)| < ∞ for any T > 0.
U×[0,T ]
(ii) There is C > 0 such that for (x, t) ∈ D,
|σ(x, t)| ≤ C ∆12 + ∆13 (t + 1) .
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Theorem (continued)
(iii) In addition, if
lim e −Λ(x) σ0 (x) = 0
|x|→∞
and
lim
sup e −Λ(x) |∇ · (A(x)c(x, t))| = 0
|x|→∞ 0≤t≤T
for each T > 0, then
lim
sup |σ(x, t)| = 0 for any T > 0,
r →∞
Sr ×[0,T ]
and furthermore, there is a continuous, increasing function r (t) > 0 with
limt→∞ r (t) = ∞ such that
lim
sup |σ(x, t)| = 0.
t→∞
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x∈Ōr (t)
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THANK YOU FOR YOUR ATTENTION!
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