Discontinuous Galerkin finite volume element methods (DGFVEM) and its

applications to miscible displacement problems

Sarvesh Kumar

sarvesh@iist.ac.in

Department of Mathematics

Indian Institute of Space Science and Technology

Thiruvananthapuram, India

Second International Conference “SCTEMM-2013"

July 08-11, 2013, Yakutsk, Russia

1

Outline of the talk

1. Motivation

2. DGFVEM for Elliptic Problems

• A priori error estimates

• Numerical experiments

3. Applications in oil reservoir studies.

• Problem description

• Discontinuous Finite Volume Element Approximation

• A priori error estimates and numerical experiments

2

Introduction to FVEM and DG methods

1. FVEM are numerical techniques for solving PDEs, like FEM and FDM.

2. FVEM are computaionaly less expensive compared to FEM.

3. Frequently used in conservation laws, computational fluid mechanics

etc .

4. Analysis for FVEM is based on the analysis of FEM.

5. In DG methods no assembly is required.

6. In DG methods no continuity criteria is required.

7. Easy to handle different kinds of boundary conditions.

3

Elliptic Problem

Consider

−∇ · (K∇u) = f in Ω, (2.1)

u = 0 on ∂Ω,

• where Ω is a bounded, convex polygonal domain in R2 with boundary

∂Ω

•K = (kij(x))2×2 is a real valued, symmetric and uniformly positive

definite matrix, i.e., there exists α0 > 0 such that

ξTKξ ≥ α0ξTξ ∀ξ ∈ R2. (2.2)

4

Control volume for standrad FVEM

"!

Figure 1: Primal grid Th and Dual grid V∗h5

Trial and Test Spaces for standard FVEM

Trial Space

Uh =zh ∈ C0(Ω) : zh|T ∈ P1(T ) ∀T ∈ Th

Test Space

Vh =wh ∈ L2(Ω) : wh|V ∗P is a constant ∀V ∗P ∈ V∗h

Remark 1: To obtain optimalL2 error estimate, i.e., o(h2), we need either

f ∈ H1(Ω) and u ∈ H2(Ω) or u ∈ H3(Ω).With only u ∈ H2(Ω)is difficult 1

1 R.E. Ewing et. al., On the accuracy of the finite volume element method based on piecewise linear polynomials, SINUM, 2002

6

Construction of control volume for (DGFVEM)

Figure 2: Primal grid Th and Dual grid T ∗h

7

A control volume in the Dual grid T ∗h

8

Advantages over standard FVEM

• Small support of the control volume compared to standard FVEM.

• Support of the control volumes lies inside the triangle in which they

belong too and does not intersect with other triangles.

• Suitable for parallel computing and implementation of adpative FVEM.

• Easy to implement no assembly process is required.

• Need less regularity assumption on the given data and exact solution

compared to standard FVEM in order to achieve optimal error estimates

in L2 norm. We will see this soon

9

DGFVEM in literature

1. X. Ye, A new discontinuous FVM for elliptic problems, 2004, (SINUM).

2. X. Ye, A discontinuous FVM for the Stokes problems, 2006 (SINUM).

3. S. H. Chou and X. Ye, Unified analysis of finite volume methods

for second order elliptic problems, 2007 (SINUM).

4. Sarvesh Kumar, Neela Nataraj and Amiya K. Pani, DGFVEM for

Second Order Linear Elliptic Problems, 2009 (NMPDE).

5. Sarvesh Kumar, A mixed and DGFVEM for incompressible miscible

displacement problems in porous media, 2012 (NMPDE)

10

Trial and Test Space for DGFVEM

Trial Space:

Vh = vh ∈ L2(Ω) : vh|T ∈ P1(T ) ∀T ∈ Th,

Test Spcae:

Wh = wh ∈ L2(Ω) : wh|T ∗ ∈ P0(T∗) ∀T ∗ ∈ T ∗h ,

Define transfer operator γ : Vh −→ Wh by

γv|T ∗ =1

he

∫e

v|T ∗ds, ∀T ∗ ∈ T ∗h ,

Here e is an edge and he is the length of e.

11

Now, multiplying (2.1) by wh ∈ Wh, integrating T ∗ ∈ T ∗h , applying

Gauss’s divergence theorem and summing up over all the control volumes,

we obtain

−∑T ∗∈T ∗h

∫∂T ∗

K∇u · nwh ds = (f, wh) ∀wh ∈ Wh, (3.1)

where n denotes the outward unit normal vector to ∂T ∗.

Using ac− bd = 12(a + b)(c− d) + 1

2(a− b)(c + d), and

[K∇u = 0], since u ∈ H2(Ω). We have∑T∈Th

∫∂T

K∇u · nwhds =∑e∈Γ

∫e

[wh] · 〈K∇u〉ds. (3.2)

Note: Here Γ denotes all edges and [·] is Jump and< · > is average.

12

Then DGFVEM corresponding to (2.1): Find uh ∈ Vh such that

Ah(uh, vh) = (f, γvh) ∀vh ∈ Vh, (3.3)

where

Ah(φh, ψh) = −∑T∈Th

3∑j=1

∫Aj+1BAj

(K∇φh · n) γψhds

+θ∑e∈Γ

∫e

[γφh] · 〈K∇ψh〉ds−∑e∈Γ

∫e

[γψh] · 〈K∇φh〉ds

+∑e∈Γ

∫e

α

hβe[φh] · [ψh]ds,

here θ ∈ [−1, 1], α and β are penalty parameters.

13

Coercivity and Boundedness

Lemma 4.1 There exists a positive constantC independent of h such that

for α large enough and h small enough

Ah(φh, φh) ≥ C‖|φh‖|2 ∀φh ∈ Vh.

Lemma 4.2 For φh, ψh ∈ Vh, we have

|Ah(φh, ψh)| ≤ C‖|φh‖| ‖|ψh‖|.

Here,

|||v|||2 =∑T∈Th

|∇v|20,T +∑e∈Γ

1

hβe

∫e

[v]2ds

14

Error Estimates

Let uI ∈ Vh be an interpolant of u, which has the following approximation

properties:

|u− uI|s,T ≤ Ch2−sT ‖u‖2,T ∀T ∈ Th, s = 0, 1. (4.1)

• ‖| · ‖| estimates

Theorem 4.1 Let u ∈ H2(Ω) ∩H10(Ω) and uh ∈ Vh be the solutions

of (2.1) and (3.3), respectively. Then there exists a positive constant Cindependent of h such that

‖|u− uh‖| ≤ Ch‖u‖2. (4.2)

15

L2 estimates

Theorem 4.2 2 Letu ∈ H2(Ω)∩H10(Ω) anduh ∈ Vh be the solutions

of (2.1) and (3.3), respectively. Then there exists a positive constant Cindependent of h such that

‖u− uh‖ ≤ Ch2

‖u‖2 +

∑T∈Th

‖f‖21,T

1/2 . (4.3)

Remark: In order to prove above theorem, we needβ = 3, whenθ 6= −1.

2Sarvesh Kumar, Neela Nataraj and Amiya K. Pani, DGFVEM for Second Order Linear Elliptic Problems, 2009 (NMPDE)

16

Numerical Experiments

−∇ · (K∇u) = f in Ω, (4.4)

u = 0 on ∂Ω,

where

• Ω = (0, 1)× (0, 1)

•K =

(1 + x2 0

0 1 + y2

)• f chosen such that u = xy(x− 1)(y − 1).

17

Order of convergence in brokenH1 norm for θ = −1, θ = 0 and θ = 1with β = 1.

2 2.2 2.4 2.6 2.8 3 3.2 3.43.6

3.8

4

4.2

4.4

4.6

4.8

5

−log(h)

−log||

|u−u h|||

θ=−1θ=1θ=0

Slope≈1

18

Order of convergence in L2− norm for θ = 0 and θ = 1 with β = 3.

2 2.2 2.4 2.6 2.8 3 3.2 3.46.5

7

7.5

8

8.5

9

−log(h)

−log||

u−u h||

θ=1θ=0

Slope≈2.3

19

Order of convergence in L2− norm for θ = −1, θ = 0 and θ = 1 with

β = 1

2 2.15 2.3 2.45 2.6 2.75 2.9 3.05 3.2 3.356.5

7

7.5

8

8.5

9

9.5

−log(h)

−log||

u−u h||

θ=−1θ=1θ=0

Slope=2

Slope=2

Slope=2

20

Applications to oil reservoir studies.3

3Sarvesh Kumar, Mixed and DGFEVM for the approximation of incompressible miscible displacement problems in porous media, 2012,NMPDE.

21

Problem description

Russell and Wheeler (1984) J = (0, T ]

u = −κ(x)

µ(c)∇p ∀(x, t) ∈ Ω× J, (5.1)

∇.u = q ∀(x, t) ∈ Ω× J, (5.2)

φ(x)∂c

∂t−∇.(D(u)∇c− uc) = g(x, t, c) ∀(x, t) ∈ Ω× J, (5.3)

with boundary conditions

u.n = 0 ∀(x, t) ∈ ∂Ω× J, (5.4)

(D(u)∇c− uc).n = 0 ∀(x, t) ∈ ∂Ω× J, (5.5)

and initial condition

c(x, 0) = c0(x) ∀x ∈ Ω. (5.6)

22

• µ(c)− the viscosity of the fluid.

µ(c) = µ(0)[1 + (M 1/4 − 1)c

]−4withM =

µ(0)

µ(1)( mobility ratio)

•D(u) is a 2× 2 matrix,

D(u) = φ[dmI + |u|(dlE(u) + dt(I − E(u)))

].

Here, E(u) = uiuj/|u|2, dm is the molecular diffusion and dl, dtare longitudinal and transversal coefficients of dispersion respectively.

• c0(x) is a given initial function, 0 ≤ c0(x) ≤ 1.

23

Finite Volume Element Approximation for the pressure

equation

4 Trial spaces for velocity

Uh =vh ∈ U : vh|T = (a + bx, c + by), ∀T ∈ Th

Trial spaces for pressure

Wh = wh ∈ W : wh|T is a constant ∀T ∈ Th .Test space for velocity

Vh =vh ∈ L2(Ω)2 : vh|T ∗M is a constant vector and vh.n = 0 on ∂Ω

.

HereU =v ∈ L2(Ω)2 : ∇ · v ∈ L2(Ω) and v · n = 0 on ∂Ω

andW = L2(Ω)/R,

4S. H. Chou, D. Y. Kwak and P. Vassilevski , Mixed covolume methods for elliptic problems on triangular grids; SINUM, 1998.

24

Construction of control volume for velocity

!"$#

!"$%

Figure 3: Primal grid Th and Dual grid T ∗h25

Define transfer operator γh : Uh −→ Vh such that

γhvh(x) =

Nm∑j=1

vh(Mj)χ∗j(x) ∀x ∈ Ω, (5.7)

where χ∗j are the scalar characteristic functions corresponding to the

control volume T ∗Mjdefined by

χ∗j(x) =

1, if x ∈ T ∗Mj

0, elsewhere.

andNm is the total number of mid-points of interior edges.

26

Approximation of pressure equation

Multiplying (5.1) by γhvh ∈ Vh, intregarting over T ∗m and using Gauss

divergence thorem, the mixed FVE approximation corresponding to the

pressure equation can be written as: find (uh, ph) ∈ Uh ×Wh such

that

(κ−1µ(ch)uh, γhvh) + b(γhvh, ph) = 0 ∀vh ∈ Uh, (5.8)

(∇.uh, wh) = (q, wh) ∀wh ∈ Wh, (5.9)

where

b(γhvh, wh) = −Nm∑i=1

vh(Mi).

∫∂T ∗Mi

wh nT ∗Mids, ∀vh ∈ Vh, ∀wh ∈ Wh.

Mi is the mid-point of the edgeEi.

27

Completely discrete scheme

Set∂ch∂t|t=tn+1

≈cn+1h − cnh

∆t.

At t = tn: find (unh, p

nh, c

nh) ∈ Uh ×Wh ×Mh such that

(κ−1µ(cnh)unh, γhvh) + (∇ · vh, p

nh) = 0 ∀vh ∈ Uh (5.10)

(∇ · unh, wh) = (qn, wh) ∀wh ∈ Wh. (5.11)

(φcn+1h − cnh

∆t, γzh

)+ Ah(u

nh; cn+1

h , zh) (5.12)

= (g(cn+1h ), γzh) ∀zh ∈Mh.

28

Garding type inequality and boundedness of Ah(uh; ·, ·)

Lemma 5.1 There exist positive constants C and C1 independent of hsuch that for large enough α and small enough h and v ∈ [L∞(Ω)]2

Ah(v;φh, φh) ≥ C‖|φh‖|2 − C1‖φh‖2 ∀φh ∈Mh.(5.13)

Lemma 5.2 For φh, ψh ∈Mh, we have

|Ah(v;φh, ψh)| ≤ C‖|φh‖|‖|ψh‖| (5.14)

Note: Here the bilinear formAh(v; ·, ·) is same as the bilinear form defined

for elliptic problems with θ = −1.

29

Error estimates

Theorem 5.1 Let c and ch be the solutions of (5.3) and (5.12) respectively,

and let ch(0) = c0,h = Rhc(0). Further, assume that ∆t = O(h).

Then, for sufficiently small h there exists a positive constant C(T )independent of h but dependent on the bounds of κ−1 and µ such that

max0≤n≤N

‖c(tn)− cnh‖ ≤ C[h + ∆t

].

Estimates for velocity and pressure

max0≤n≤N

‖u(tn)− unh‖(L2(Ω))2 + ‖p(tn)− pnh‖ ≤ C[h + ∆t]

30

Numerical Experiments

Take q = q+ − q− and g(x, t, c) = cq+ − cq−, where c is the

injection concentration and q+ and q− are the production and injection

rates, respectively.

Test 1(H. Wang et al. SIAM J. Sci. Comput., Vol 22, pp.561-581, 2000)

• Ω = (0, 1000)× (0, 1000) ft2, [0, T ] = [0, 3600] days,

• k = 80, φdm = .1, φdm = 0, φdt = 0, c = 1 andM = 1

• The injection well is located at (1000, 1000) with q− = 30 ft2/day

• The production well is located at (0, 0) with q+ = 30 ft2/day.

• ∆x = ∆y = 50ft in both x and y direction, ∆tc = 120 and

∆tp = 360 days.

31

Test 1 Continued

0

500

1000

0

500

10000

0.2

0.4

0.6

0.8

1

x

(b)

y

c h

x

y

(a)

0 500 10000

100

200

300

400

500

600

700

800

900

1000

Figure 4: Contour plot (a) and surface (b) at t = 3 years

32

0

500

1000

0

500

10000.4

0.5

0.6

0.7

0.8

0.9

1

x

(b)

y

c h

x

y

(a)

0 500 10000

100

200

300

400

500

600

700

800

900

1000

Figure 5: Contour plot (a) and Surface (b) at t = 10 years

33

Test 2

• Data is same as in Test 1.

• We takeκ = 80 md on the sub domain ΩL := (0, 1000)×(0, 500)andκ = 20 md on the sub domain ΩU := (0, 1000)×(500, 1000)

• ∆x = ∆y = 50ft in both x and y direction.

• ∆tc = 120 and ∆tp = 360 days.

34

Test 2

0

500

1000

0

500

10000

0.2

0.4

0.6

0.8

1

x

(b)

y

c h

x

y

(a)

0 500 10000

100

200

300

400

500

600

700

800

900

1000

Figure 6: Contour plot (a) and surface (b) at t = 3 years

35

0

500

1000

0

500

10000.2

0.4

0.6

0.8

1

x

(b)

y

c h

x

y

(a)

0 500 10000

100

200

300

400

500

600

700

800

900

1000

Figure 7: Contour plot (a) and Surface (b) at t = 10 years

36

Order of convergence

2 2.5 3 3.5 4 4.5 5 5.5

5

5.2

5.4

5.6

5.8

6

6.2

6.4

−log(h)

−log||

c−c h||

Slope≈2

Figure 8: Order of convergence in L2− norm

37

THANK YOU.

38