continuum models of two phase flow in porous media · pdf filecontinuum models of two phase...

Continuum Models of Two Phase Flow in Porous Media

Michael Shearer

North Carolina State UniversityDepartment of Mathematics

Research supported by NSF grants DMS 0968258 FRG, DMS 0636590 RTG

Shearer (NCSU) Flow in Porous Media 1 / 46

Acknowledgements

Students and former students

Rachel Levy (Harvey Mudd College)Kim Spayd (Gettysburg College)Melissa Strait (NC State University)

Collaborators

Ruben Juanes, Luis Cueto-Felgueroso (MIT Engineering)Zhenzheng Hu (NC State University)


Applications

Oil recovery: water flood

water injection

H O2

Water infiltration under gravity(DiCarlo 2004)

Carbon Sequestration in Deep SalineAquifers


Introduction I

Framework: Conservation of mass; Darcy’s law; incompressibility

Buckley-Leverett flux f (u) (1942) u is saturationnon-convex conservation law ut + f (u)x = 0

Modeling: pressure difference pc across interfaces: interfacial energy

Classical approach: pc = pe(u) equilibrium pressure; decreasing in uut + f (u)x = −(H(u)pe(u)x)x

Dynamic effects: Hassanizadeh and Gray (1991,1993):pc = pe(u)− τut

Netherlands group: undercompressive shocks; sharp shocks

Pop, van Duijn, Peletier, Cuesta, Fan (2006-2012)


Introduction II

2-d stability/instability for Lax shocks: hyperbolic/elliptic system

Phase field modeling: Juanes (2008), Witelski (1998), based onpolymer mixtures: introduce stored energy function with capillaryeffects

Capillary tube gas/liquid finger: degenerate fourth order PDE

Dynamical systems trick for traveling waves

Carbon sequestration: Huppert, Neufeld, Hesse, ... (2008-2011).Discontinuous dependence in conservation law to allow for depositionof CO2 bubbles as plume propagates in brine (deep saline aquifer)

Track interface; non-standard wave interactions


2-D Model

u(x , y , t) : saturation (vol. fraction) of water, (1− u) : oil saturationp(x , y , t) : pressure in water

Conservation of mass with Darcy’s law: velocity v = −λ(u)∇p :

φ∂u

∂t−∇ · (λ(u)∇p) = 0 φ = porosity, λ(u) =

Kk(u)

µ

K = absolute permeability, k(u) = relative permeability, µ = viscosity

Incompressibility: ∇ · vTotal = 0

∇ ·(

Λ(u)∇p + λoil(u)∇pc(u))

= 0

Λ = λwater + λoil , pc = poil − p : capillary pressure;

For simplicity, neglect gravity, set φ = 1.


One-Dimensional Equation for u(x , t) : ut − (λpx)x = 0

∇ · vTotal = 0 =⇒ vTotal = (vT , 0) = constant, after rescaling t

Thus, Λpx + (Λ− λ)pcx = −vT

Eliminate px : px = − vTΛ(u)

−(

1− λ(u)

Λ(u)

)pcx

Thenut + f (u)x = − (H(u)pc

x )x

f (u) = vT λ(u)

Λ(u), H(u) =

λ(u)

Λ(u)(Λ(u)− λ(u))


ut + f (u)x = − (H(u)pcx )x

f (u) = vT λ(u)

Λ(u), H(u) =

λ(u)

Λ(u)(Λ(u)− λ(u))

Fractional flow rate Dissipation coefficient

Recall: Λ(u) is total mobility: Λ(u)− λ(u) = λoil(1− u)λ, λoil are positive, increasing convex functions, with λ(0) = λoil(0) = 0.

0 0.5 10

0.2

0.4

0.6

0.8

1

u

f(u)

0 0.5 10

0.02

0.04

0.06

0.08

0.1

u

H(u)


ut + f (u)x = − (H(u)pcx )x , but what to do about pc?

Buckley-Leverett (1942): ignore it: pc = 0

scalar non-convex conservation law ut + f (u)x = 0

Lax-Oleinik construction: rarefaction-shock interface

Classical approach: equilibrium capillary pressure depends onsaturation: pc = pe(u), a decreasing function: d

dupe(u) < 0.

The PDE is parabolic but degenerate at u = 0, 1.

Max principle: 0 ≤ u ≤ 1 is positively invariant.

Lax-Oleinik solutions replaced by smooth approximation.


Hassanizadeh and Gray (1991, 1993): rate dependence in pc :

pc = pe(u)− τut

Modified Buckley-Leverett equation

ut + f (u)x = −(H(u) ∂

∂x pe(u))x

+ τ (H(u)utx)x

Equation is dissipative-dispersive. No maximum principle, but positiveinvariance of 0 ≤ u ≤ 1 (Yabin Fan thesis 2012)

Analogous equation: modified BBM-Burgers:

ut + (u3)x = αuxx − βuxxt

Traveling waves for undercompressive shocks (Spayd thesis, 2012)

Explicit construction similar to modified KdV-Burgers:

ut + (u3)x = αuxx − βuxxx

(Jacobs, McKinney, Shearer, 1995; LeFloch book, 2002)


Traveling Wave Solutions

u(ξ) = u(x − st)⇒ −su′ + (f (u))′ = [H(u)u′]′ − sτ [H(u)u′′]′

Integrating with boundary conditions u(±∞) = u± leads to secondorder ODE: −s(u − u−) + f (u)− f (u−) = H(u)u′ − sτH(u)u′′

Rescale u = u( x−st√sτ

) and write as first order system of ODEs:

u′ = v

v ′ =1√sτ

v +1

H(u)[s(u − u−)− f (u) + f (u−)]

Singular at u = 0, u = 1, where H(u) = 0


Saddle-Saddle Connections

Phase portraits of ODE system: up to three equilibria(ue , 0) : (u±, 0), (umid, 0), s = (f (ue)− f (u−))/(ue − u−)

Analyze phase plane with separation function R(u−, u+, τ)

0.4 0.6 0.8

−0.5

0

0.5

u

v=uv u

midu u-

+

R>0

0.4 0.6 0.8 1

−0.5

0

0.5

u

v

R<0

0.4 0.6 0.8 1

−0.5

0

0.5

u

v=uv

R = 0

Saddle-saddle connection: R = 0 :heteroclinic orbit from (u−, 0) to (u+, 0) undercompressive shock


Numerical procedure

For τ <∞, fix u− and determine u+ = uΣ(u−) for whichR(u−, u+, τ) = 0Monotonicity of uΣ(u−) from Melnikov integral: derivatives of R

For τ =∞, determine (u−, uΣ) by using exact integral:

12

d

duv 2 =

1

H(u)[s(u − u−)− f (u) + f (u−)] ≡ G (u; u−, u+)

Thus: connection when

∫ u+

u−

G (u; u−, u+) du = 0 : u+ = uΣ(u−)

Asymptotics in corners: e.g., u− → 0, u+ → 1 :

u+ = uΣ(u−) ∼ 1− uM−


Στ Curves

Fix τ > 0, plot (u−, u+) with saddle-saddle connection u− → u+

u−

u+

0 0.2 0.4 0.6 0.8 110

0.2

0.4

0.6

0.8

1

Σ0.1

Σ

Σ

1

∞

u−

u+

0 0.2 0.40.75

0.8

0.85

0.9

0.95

1

0.6

Σ0.1

c: R=0

a: R>0

b: R<0

τ=0.1


Scalar conservation law: ut + f (u)x = 0

Idealization: no capillary pressure; characteristic speed f ′(u)Scale invariant solutions: building blocks for solving initial value problem

Rarefactions

u(x , t) =

u− if x < f ′(u−)t

r( xt ) if f ′(u−)t ≤ x ≤ f ′(u+)t

u+ if x > f ′(u+)t

Shocks

u(x , t) =

{u− if x < st

u+ if x > st

Rankine-Hugoniot condition: shock speed s =f (u+)− f (u−)

u+ − u−


Admissible Shocks

f ′(u+) < s < f ′(u−) Lax shock is admissible if there is a travelingwave from u− = umid (unstable node) to u+ (saddle point)

s > f ′(u±) PLUS: shock has corresponding traveling wave:

undercompressive shock Σ.

Jacobs, McKinney, Shearer (1995), LeFloch Book (2002)

0 10

1

u

f(u)

s

u-u+

Lax

0 10

1

u

f(u)

u+

Σ

u-

Undercompressive


Buckley-Leverett solution 1942

Solve conservation law with initial jump from all water u− = 1 to alloil u+ = 0: water flooding:

ut + f (u)x = 0, u(x , 0) =

{1 if x < 0

0 if x > 0

0 10

1

u

f(u)

u

shock

rarefaction

*

Solution: rarefaction fromu− = 1 to u∗; Lax shock fromu∗ to u+ = 0 :rarefaction-shock.


The Riemann Problem: Classical Solution

Solve conservation law with initial jump from u` to ur

ut + f (u)x = 0, u(x , 0) =

{u` if x < 0

ur if x > 0

ul

ur

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

RS

RS R

R

S

SR: Rarefaction WaveS: Lax ShockRS : Rarefaction - Shock


The Riemann Problem; dynamic capillary pressure

(RP) : ut + f (u)x = 0, u(x , 0) =

{u` if x < 0

ur if x > 0

Solutions of (RP): leading order approximations to solutions of modifiedBuckley-Leverett equation

ul

ur

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

RΣ

RSRS

RΣ

R

R

S

S

SΣ

SΣ

R: Rarefaction WaveS: Admissible Lax ShockΣ : Undercompressive Shock

For every (u`, ur ) ∈ (0, 1), thereis a solution that stays in (0, 1)even though SΣ solutions arenonmonotonic.


PDE Simulations

Full PDE: ut + f (u)x = (H(u)ux)x + τ (H(u)utx)x

u` ∈ (0, ur ) : rarefaction wave

−2 0 2 40

0.2

0.4

0.6

0.8

1

x

u

u

ul

r

u` ∈ (ur , umid) : Lax shock

−2 0 2 40

0.2

0.4

0.6

0.8

1

x

u

u

u

l

r


PDE Simulations - Nonclassical Solutions

u` ∈ (umid , uΣ) :Lax shock u` to uΣ

undercompressive shockuΣ to ur

−2 0 2 40

0.2

0.4

0.6

0.8

1

x

u

u

u

u

l

Σ

r

u` ∈ (uΣ, 1) :rarefaction wave u` to uΣ

undercompressive shockuΣ to ur

−2 0 2 40

0.2

0.4

0.6

0.8

1

x

u

ur

ul

uΣ


‘Sharp’ Traveling waves TW u = u(x − st)

ut + f (u)x = −(

H(u)∂

∂xpe(u)

)x

+ τ (H(u)utx)x

−s(u − u−) + f (u)− f (u−) = −H(u)pe(u)′ − sτH(u)u′′, u(±∞) = u±

after one integration

Singular at u = 0, 1 Can have connections up to u = 1, or down to u = 0depending on relative permeability functions at u = 0, 1

Example: λ(u) = u3/2; λoil(u) = (1− u)3/2

Pop, Cuesta, van Duijn (2011): sharp shock TW with corner at u = 1


0 5 10 15 20 25 30 35

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1u

time = 0

time = 6

time = 3

time = 9

time = 12

x

0 0.2 0.4 0.6 0.8 1

0

1

2u = 1, u+= 0.025

sharp shock

u’

u0.8 0.85 0.9 0.95 1

0

0.5

1u = 0.95, u+= 1

u

u’

Lax shock profile

PDE Simula+on

Phase Portraits: Traveling waves

Rela+ve permeabili+es Integrable case: K(u) = u3/2


Planar Lax shock ut + f (u)x = 0

Interface x = st; velocity v± = −λ±∂xp±; mobility λ± = Kk(u±)/µ

Continuous pressure p = p±(x − st) = − v±λ±

z = − vT

Λ±z , z = x − st;

shock location: z = x − st = 0

u = up-

x = st

x

y

--- -

-u = u+p = (x - st) p+p = (x - st)

Shock speed

s =f (u+)− f (u−)

u+ − u−

Lax condition:

f ′(u+) < s < f ′(u−)

Characteristic speeds f ′(u±)


2-dimensional linearized stability of Lax shocks

2-d equations with pc ≡ 0 variables u, p saturation, pressure:

∂u

∂t−∇ · (λ(u)∇p) = 0

∇ · (Λ(u)∇p) = 0

(1)

Interface x = x(y , t), normal in t, x , y : (−xt , 1,−xy )Jump condition at shock: ([q] = q+ − q−)

−xt [u]− [λ(u)px ] + xy [λ(u)py ] = 0

−[Λ(u)px ] + xy [Λ(u)py ] = 0(2)

Base shock: u = u±, p = p±(x − st), x = st, constants u±, p±,V

s =f (u+)− f (u−)

u+ − u−, f (u) = vT λ(u)

Λ(u), p± = − vT

Λ(u±)


Equations ∂u∂t −∇ · (λ(u)∇p) = 0, ∇ · (Λ(u)∇p) = 0

∂u

∂t− λ′(u)∇u · ∇p − λ(u)∆p = 0

Λ′(u)∇u.∇p + Λ(u)∆p = 0

(1)

Linearize about shock on each side:u = u + U(x , y , t), p = p(x − st) + P(x , y , t)

Ut − λ′(u)pUx − λ(u)∆P = 0

Λ′(u)pUx + Λ(u)∆P = 0(2)

Thus, ∆P = −Λ′(u)

Λ(u)pUx . Subs. into Ut equation:

Ut − p

(λ′(u)− λ(u)

Λ′(u)

Λ(u)

)Ux = 0


CLAIM: Lax shock =⇒ U has to be zero.

Ut − p

(λ′(u)− λ(u)

Λ′(u)

Λ(u)

)Ux = 0, u = u±

Let U(x , y , t) = w(x − st)e iαyeσt . Then, with z = x − st,w ′ = dwdz :

σw − sw ′ − p λ(u)

(λ′(u)

λ(u)− Λ′(u)

Λ(u)

)w ′ = 0

But f (u) = vTλ(u)/Λ(u), p λ(u) = −f (u) so we have

σw = (s − f ′(u)) w ′, with solutions w(z) = a±eβ±z ,

σ = β±(s − f ′(u±))

For a Lax shock, f ′(u+) < s < f ′(u−).Thus, Re σ ≥ 0 implies w(z) does not decay at z = ±∞ unless w ≡ 0.

Consequently, U ≡ 0


2-d stability: perturb pressure in linearized equations

u = u±; p = p±z + P; P(z , y , t) = q±(z)e iαy+σt

z = x − Vt = ae iαy+σt , z = x − x

We seek σ = σ1α, α > 0. Linearized equations: ∆P = 0 :

q′′± − α2q± = 0 (1)

Decaying solutions: q± = b±e∓αz , ±z > 0,

Next: linearize jump conditions: two equations for three coefficients

a, b−, b+


Linearized jump conditions (no perturbation on u)

σa[u] + [λ(u)q′] = 0 (1)

[Λ(u)q′] = 0 (2)

Equation (2):

Λ(u+)b+ = −Λ(u−)b− (3)

Then (1) implies

b−Λ(u−)(f (u+)− f (u−)) = −σα

a(u+ − u−)vT

But (f (u+)− f (u−))/(u+ − u−) = s, the shock speed, so

b−Λ(u−)s = −aσ

αvT (4)


2-dimensional stability: third equation for a, b±

Third equation comes from continuity of pressure (linearized)at z = z(y , t) = ae iαy+σt

p = p±z + q±(z)e iαy+σt , q± = b±e∓αz (±z > 0)

Consequently,p+a + b+ = p−a + b− (5)

Thus, (p+ − p−)a = −σα

vT

sa

(1

Λ(u+)+

1

Λ(u−)

), (from (3,4))

Since p± = − vT

Λ(u±), we obtain, for a 6= 0 :

σ

α= s

Λ(u−)− Λ(u+)

Λ(u−) + Λ(u+)


Interpretation of stability condition: quadratic relativepermeabilities: k(u) = κu2

Lax shocks for u+ < u− ≤ u∗, f ′(u∗) = (f (u+)− f (u∗))/(u+ − u∗)

Stability boundary: u+ = −u− + 2MM+1

0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

u−

u

u

+

US

I

*

M = 0.22M

M + 1=

1

3

Inflection point I of f (u) atuI = 0.2591

S: Stable Lax shocks

U: Unstable Lax shocks


Fingering Instability - Zhengzheng’s Simulations

Full parabolic/elliptic system with capillary pressure pc = −u + τut

Crank-Nicolson time-step; upwind advection; periodic side boundaryconditions, moving frame, plot middle contour u = 1

2 (u− + u+)

∆t = O(10−3), ∆x = ∆y = O(10−2)

Initial condition: u = randomly perturbed hyperbolic tangent

u− = 0.2, u+ = 0,M = 0.05

−3 −1.5 0 1.5 3−3

−1.5

0

1.5

3time = 0

−3 −1.5 0 1.5 3−3

−1.5

0

1.5

3time = 2

−3 −1.5 0 1.5 3−3

−1.5

0

1.5

3time = 4


Numerical Simulations - Stable case

M = 0.2

u− = 0.25, u+ = 0

Oil-water mixture displaces oil

Initial perturbation decays

Zoomed-in contour0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

u−

u+

US

I

time = 0.05

− 0 .05 0 0.05

−12

−6

0

6

12time = 1.25

−12

−6

0

6

12

− 0 .05 0 0.05

time = 20

−12

−6

0

6

12

−0.05 0 0.05


Numerical Simulations: Unstable case: Fingering Instability

M = 0.2

u− = 0.25, u+ = 0.15

weak Lax shock

Initial perturbation grows ⇒fingering instability

0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

u−

u+

(u , u )i i

US

− 0 .05 0 0.05

−12

−6

0

6

12time = 0.05

− 0.05 0 0.05

−12

−6

0

6

12time = 1.5

− 0.05 0 0.05

time = 20

−12

−6

0

6

12


Conclusions: stability/fingering instability in 2-d

Analysis of hyperbolic/elliptic system linearized around 1-d shockLinear dependence of growth rate σ on transverse wave number αdistinguishes stable waves from unstable

Numerical simulations of full parabolic/elliptic system confirm results;(Riaz and Tchelepi (2006) also conducted numerical experiments)

Weak Lax shocks may be stable or unstable

Oil/water mixture displacing oil can be stable.

K. Spayd, M. Shearer and Z. Hu, Applicable Analysis, 2012.


Back to basics: capillary pressure in a capillary tube

x2r(x) 2R

γ = 1/Ca = σ/Uη

Ca =: capillary number

balance between surfacetension σ and viscosity η.

Saturation u : area fraction

u(x , t) = r(x)2/R2.

Write equation in form

ut + f (u)x = −γ∂x [f (u)λ(u)∂xψ]

f (u) = u/(u + (1− u)3)

Fractional flow rate for air/water(Bear 1972)

λ(u) = (1− u)3

ψ = δFδu , F : free energy

F = F 0(u) + κ(u)F 1(ux)

F 0 : bulk free energy

F 1(ux) : interfacial energy

κ(u) : interfacial energy weight


Phase field model: F = F 0(u) + κ(u)F 1(ux)

Cueto-Felgueroso and Juanes, PRL (2008, 2011)

Simple choices:

F 0(u) = −Σ

R(1− u)2u2 + 2

γwg cos θ

R(1− u)2

Double-well potential; tangent construction (like equal area rule)

Σ = γsw − γsg − γwg < 0 spreading parameter

θ : static contact angle Young-Laplace law at contact lineγsw − γsg = γwg cos θ, set γ = γwg

F 1(ux) = Γu2x

Γ = C (θ)γR


Phase field model: ut + f (u)x = −γ∂x [f (u)λ(u)∂xψ]

Variational derivative: ψ = δFδu

ψ = −C1u(1− u)(1− 2u) + C2

√κ(u)∂x(

√κ(u)∂xu), C1 > 0,C2 > 0

Chemical potential for polymer mixture:

de Gennes: J. Chem. Phys. (1980); Witelski, Appl. Math. Lett. (1998)

Benzi, Sbragaglia, Bernachi, Succi: Binary fluid mixtures, PRL (2011)

∂tu + ∂x f (u) = γ∂x [H(u)Q(u)∂xu]− γ∂x[H(u)∂xC2

√κ(u)∂x(

√κ(u)∂xu)

]Q(u) = C1(6u2 − 6u + 1), H(u) = f (u)λ(u)

Cahn-Hilliard type equation


Static solutions: bubbles and compactons

Now choose κ(u) so that a static bubble is an equilibrium solution

air �uid�uid

static air bubble

Ends of bubble are spherical caps (take contact angle θ = 0, or π) :

u(x) = r 2 = 1− x2, 0 ≤ x ≤ 1. Thus u′ = −2x = −2√

1− u, u′′ = −2.

Here, tube radius R = 1, θ = π, ′ = ∂x

∂x [f (u)λ(u)∂xψ] = 0

Consequently,

f (u)λ(u)ψ′ = c0 = 0, since f (0) = 0


ψ(u) = −C1G (u) + C2√κ∂x(√κ∂xu)); G (u) = u(1− u)(1− 2u)

f (u)λ(u)ψ′ = 0 =⇒ ψ = c1 = 0, since κ(0) = 0

Thus,

m(u)(m(u)u′)′ = Ku(1− u)(1− 2u), K = C2/C1,m(u) =√κ(u).

u(x) = 1− x2 =⇒ x =√

1− u : ODE for m(u) :

m(u)

(dm

du4(1− u)− 2m(u)

)= KG (u) = Ku(1− u)(1− 2u).

But κ(u) = m(u)2. Then

dκ

du(1− u)− κ =

K

2u(1− u)(1− 2u).

LHS = ddu ((1− u)κ), so (1− u)κ = K

2 ( 12 u2 − u3 + 1

2 u4). That is,

κ(u) =K

4(1− u)u2


General contact angles θ : Coeff. of interfacial energy κ

Similar result for general contact angle θ, 0 < θ < π :

u = sec2 θ − x2/R2

κ(u) =K (θ)u2(1− u)2

4(sec2 θ − u), 0 ≤ u ≤ 1.

Idea works for more general constitutive laws

ψ(u) = −C1G (u) + C2√κ∂x(√κ∂xu)); G (u) bistable

d

du

(κ(u)(sec2 θ − u)

)=

K (θ)

2G (u)

κ(u) =K (θ)

∫G (u) du

4(sec2 θ − u), 0 ≤ u ≤ 1


PDE simulations Rachel Levy

−10 −5 0 5 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t=0 t=4 t=6

structure: rarefaction and leading (undercompressive) shock/TW fromplateau u = u− to u = 0 at x = st + x0 <∞.


Traveling Waves

∂tu + ∂x f (u) = γ∂x [H(u)G (u)∂xu]− γC∂x [H(u)∂xm(u)∂x(m(u)∂xu)]

u = u(x − st), u(−∞) = u−, u(0) = 0, s = f (u−)/u−

−su + f (u) = γH(u)G (u)u′ − γCH(u)(m(u)(m(u)u′)′)′,

Note: constant of integration is zeroWrite as first order system, β = 1/(Cγ) (multiply 3rd eqn by m(u))

m(u)u′ = v

m(u)v ′ = w

m(u)w ′ = 1C G (u)v + β

m(u)

H(u)(su − f (u))

Eliminate m(u) to remove singularity at origin


ODE system - nondegenerate vector field

Eliminate m(u) : m(u) ddξ = d

dη , u(ξ) = U(η), ...

U ′ = V

V ′ = W

W ′ = 1C G (U)V + β

m(U)

H(U)(sU − f (U))

(U,V ,W ) = (0, 0, 0) now a regular equilibrium

Seek solutions with −∞ < η <∞(U,V ,W )(−∞) = (u−, 0, 0), (u, v ,w)(∞) = (0, 0, 0)

Need real eigenvalues at origin: γ > γ∗ : Ca < Ca∗

Shooting: 1-d unstable manifold from u− to 2-d stable manifold atorigin: u− = u−(β).


Recover original variable u(ξ)→ 0, ξ → 0−

Transform back: m(U(η)) dη = dξU(η) ∼ aeµη as η →∞, (µ < 0 : eigenvalue) implies

ξ = −∫ ∞η

U(y) dy = aeµη/µ

Thus, u(ξ) = U(η) ∼ µξ, as ξ → 0−, slope µ < 0.

0

0.1

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0.5

0.6u

x

Traveling wave MATLAB plot


Summary

Many applications with two-phase porous media flow

Oil recovery context: oil/water interface; Lax shocks may be stable orunstable to 2-D perturbations; hyperbolic analysis gives condition forlinear stability

Yortsos and Hickernell (1989) traveling wave stability; most unstablewave number from dispersion relation tested by Zhengzheng Hu

Phase field analysis for flow in capillary tube; undercompressiveshocks, stationary compactons; degenerate diffusion traveling waves

Hassanizadeh-Gray modeldegenerate diffusion-dispersion; sharp shocks


continuum models of two phase flow in porous media · pdf filecontinuum models of two phase...

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