Finite volume method for boundary value problemDiscontinuous coefficients and high Peclet number

Praveen. Cpraveen@math.tifrbng.res.in

Tata Institute of Fundamental ResearchCenter for Applicable Mathematics

Bangalore 560065http://math.tifrbng.res.in/~praveen

January 21, 2013

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Discontinuous coefficients

− d

dxa(x)

du

dx= q, x ∈ Ω = (0, 1)

Finite difference scheme

−1

2[D+

x (aD−x ) +D−x (aD+x )]uj = qj , j = 1, 2, . . . , n− 1

or, for j = 1, 2, . . . , n− 1

1

2h2[−(aj−1 + aj)uj−1 + (aj−1 + 2aj + aj+1)uj − (aj + aj+1)uj+1] = qj

Boundary conditionu(0) = 0, u(1) = 1

Discontinuous coefficient

a(x) =

ε 0 ≤ x ≤ x∗1 x∗ < x ≤ 1

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Discontinuous coefficients

Jump condition: au′ is continuous across the discontinuity

ε limx↑x∗

u′(x) = limx↓x∗

u′(x)

Exact solution (q = 0)

u(x) =

αx 0 ≤ x < x∗

εαx+ 1− εα x∗ ≤ x ≤ 1

where

α =1

(1− ε)x∗ + ε

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Discontinuous coefficients

Finite difference solution: Consider the uniform grid xj = jh,j = 0, 1, . . . , n with h = 1

n . Assume xk < x∗ ≤ xk+1. Postulate FDsolution as (u0 = 0 and un = 1 is satisfied)

uj =

αhj 0 ≤ j ≤ kβhj − βhn+ 1 k + 1 ≤ j ≤ n

where (apply FD equation at j = k and j = k + 1)

αh =

[ε1− ε1 + ε

+ ε(n− k) + k

]−1

, βh = εαh

At x = xk = hk, FD solution is

uk =xk

εh1−ε1+ε + (1− ε)xk + ε

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Discontinuous coefficientsCase x∗ = xk+1: Exact solution is

u(xk) =xk

(1− ε)xk+1 + ε

and error in FD solution

uk − u(xk) = O(ε1− ε1 + ε

h

)

Case x∗ = xk + h/2: Exact solution is

u(xk) =xk

(1− ε)xk + ε+ 12h(1− ε)

and error in FD solution

uk − u(xk) = O(

(1− ε)2

ε(1 + ε)h

)5 / 38

Discontinuous coefficients

a(x) error

constant 0continuous O

(h2)

discontinuous O (h)

FD scheme loses its optimal accuracy when coefficients arediscontinuous.

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Finite volume scheme

Non-overlapping finite volumes (vertex-based or vertex-centered)

Ωj = (xj −h

2, xj +

h

2), j = 1, 2, . . . , n− 1

0 1 n− 1 njj − 1 j + 1

j − 1/2 j + 1/2

Ωj

x = 0 x = 1

Integrate over each cell Ωj

−∫

Ωj

(au′)′dx =

∫Ωj

qdx

−[(au′)j+ 12− (au′)j− 1

2] ≈ hqj

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Finite volume scheme

uj+1 − uj =

∫ xj+1

xj

u′dx

=

∫ xj+1

xj

1

a(au′)dx

≈ (au′)j+ 12

∫ xj+1

xj

1

adx

≈ (au′)j+ 12

h

aj+ 12

where1

aj+ 12

=1

2

(1

aj+

1

aj+1

)≈ 1

h

∫ xj+1

xj

1

adx

Flux at x = xj+ 12

(au′)j+ 12

= aj+ 12

uj+1 − ujh

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Finite volume schemeFinite volume scheme

−[aj+ 1

2

uj+1 − ujh

− aj− 12

uj − uj−1

h

]= hqj , j = 1, 2, . . . , n− 1

withu0 = 0, un = 1

Discontinuous coefficient, x∗ = xk + h/2: Discontinuity coinciding withcell face xk+ 1

2

aj+ 12

=

ε, 1 ≤ j < k2ε

1+ε , j = k

1, k < j ≤ n− 1

Postulating FVM solution as before, we get

αh =h

(1− ε)x∗ + ε, βh = εαh

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Finite volume scheme

The FVM solution is exact !!! In general, the error of FV solution isO(h2).

Discontinuous coefficient, x∗ = xk: Discontinuity inside a cell

aj+ 12

is again given as in previous case and the error in uk is O(h2)

(show this).

When dealing with problems with discontinuous coefficients and/orsolutions, finite volume methods are more accurate than finitedifference methods. FVM has better mathematical basis since it isbased on weak formulation.

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If mesh Peclet number is large

µ > 0 and U > 0 are constants

−µu′′ + Uu′ = 0, x ∈ (0, L)

Non-dimensionalize: x→ x/L

− 1

Peu′′ + u′ = 0, x ∈ (0, 1)

Peclet number:

Pe =UL

µ=

convection

diffusion

General solution

u(x) = A+BexPe

Boundary conditions

u(0) = a, u(1) = b

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If mesh Peclet number is largeSolution

u(x) = a+ (b− a)e(x−1)Pe − e−Pe

1− e−Pe

Example

a = 0, b = 1, u(x) =e(x−1)Pe − e−Pe

1− e−Pe

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0Pe=1

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0Pe=50

Boundary layer is present near x = 1 for large value of PeBoundary layer thickness = O

(1

Pe

)for Pe 1

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If mesh Peclet number is largeFinite volume scheme (cell-centered)

1 njj − 1 j + 1

j − 1/2 j + 1/2

Ωjx = 0 x = 1

Integrating over finite volume Ωj

− 1

Pe[u′j+ 1

2

− u′j− 1

2

] + uj+ 12− uj− 1

2= 0

Fluxes: for j = 1, 2, . . . , n− 1

uj+ 12

=1

2(uj + uj+1), u′

j+ 12

=uj+1 − uj

h

Boundary fluxes

u 12

= a, un+ 12

= b, u′12

=u1 − ah/2

, u′n+ 1

2

=b− unh/2

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If mesh Peclet number is large

System of finite volume equations

(1 +6

P)u1 + (1− 2

P)u2 = (2 +

4

P)a

−(1 +2

P)uj−1 +

4

Puj + (1− 2

P)uj+1 = 0, j = 2, 3, . . . , n− 1

−(1 +2

P)un−1 + (−1 +

6

P)un = (−2 +

4

P)b

with the mesh Peclet number P = hPe .

Case P = 0: No convection, only diffusion

uj = A+Bj

Determine A,B from first and last equations.

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If mesh Peclet number is largeCase P = 2:

uj = a, j = 1, 2, . . . , n

Case P 6= 0, 2: Postulate solution of the form

uj = A+Bzj =⇒ z =2 + P

2− PDetermine A,B from first and last equations.

uj = a+b− azn − 1

(−1 +

2− P2

zj)

P z Solution

0 < P < 2 z > 0 monotonicP > 2 z < 0 oscillatory

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If mesh Peclet number is large

Non-oscillatory solution is guaranteed provided h < 2Pe . If Pe 1,

then we need a very fine mesh, h 1.

æ æ æ æ æ æ ææ

æ

æ

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0Pe=10, n=10, P=1

æ æ æ æ æ ææ

æ

æ

æ

0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.5

1.0Pe=40, n=10, P=4

Solution is oscillatory at higher Pe numbers; does not satisfy maximumprinciple

Matlab code: convdiff central ccfvm.m

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If mesh Peclet number is large

Adaptive grid: Boundary layer thickness

∆ =4

Pe

Put 1/3 of total cells in the boundary layer.

Grid refined in the boundary layer

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If mesh Peclet number is large

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Oscillation 6= instability

• Stability refers to influence of perturbations in data (initial andboundary conditions) on the solution

• Consider a perturbation ∆a of the boundary value a. Then

∆uj =zn − zj + zj P2

zn − 1∆a

• Solution is stable in norm ‖·‖ if there is a constant C independentof h such that

‖∆u‖ ≤ C|∆a|

• Case z < −1:

‖∆u‖∞ ≤ C|∆a|, C = 2

(2 +|P |2

)19 / 38

Oscillation 6= instability

• Case −1 ≤ z < 0 and perturbations in b are left as an exercise.

• As h→ 0, the condition h < 2/Pe will be satisfied; the numericalsolution converges to the exact solution.

• In practical computations, we use a finite grid h. It is desirable tohave a scheme free of oscillations for all values of h.

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Upwind scheme

d

dx(Uu)− d

dxD

du

dx= q

Finite volume scheme

(Uu)j+ 12− (Uu)j− 1

2− [(Du′)j+ 1

2− (Du′)j− 1

2] = qjh

Central scheme for convection term

(Uu)j+ 12

=1

2Uj+ 1

2(uj + uj+1)

Upwind scheme for convection term

(Uu)j+ 12

=

Uj+ 1

2uj , Uj+ 1

2≥ 0

Uj+ 12uj+1, Uj+ 1

2< 0

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Upwind scheme

or

(Uu)j+ 12

=1

2(Uj+ 1

2+ |Uj+ 1

2|)uj +

1

2(Uj+ 1

2− |Uj+ 1

2|)uj+1

The flux Uu is biased in the upstream direction. Define

U+ =1

2(U + |U |) ≥ 0, U− =

1

2(U − |U |) ≤ 0

so that

(Uu)j+ 12

= U+j+ 1

2

uj + U−j− 1

2

uj+1

The viscous flux can be discretized as before.

(Du′)j+ 12

= Dj+ 12

uj+1 − ujh

, Dj+ 12

=2DjDj+1

Dj +Dj+1

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Upwind scheme

Scheme can be written as(−U+

j− 12

−Dj− 1

2

h

)︸ ︷︷ ︸

<0

uj−1 +

(U+j+ 1

2

− U−j− 1

2

+Dj− 1

2+Dj+ 1

2

h

)︸ ︷︷ ︸

>0

uj

+

(U−j+ 1

2

−Dj+ 1

2

h

)︸ ︷︷ ︸

<0

uj+1 = qjh

This scheme satisfies maximum principle and hence solutions arenon-oscillatory. There is no condition on the mesh Peclet number.However the scheme is only first order accurate because the convectiveflux is first order accurate

U+j+ 1

2

uj + U−j− 1

2

uj+ 12

= (Uu)(xj+ 12) +O (h)

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Upwind scheme

while centered flux is second order accurate

1

2Uj+ 1

2(uj + uj+1) = (Uu)(xj+ 1

2) +O

(h2)

Upwind flux can also be written as: central flux + dissipation

(Uu)j+ 12

=1

2Uj+ 1

2(uj + uj+1)− 1

2|Uj+ 1

2|(uj+1 − uj)

Formal accuracy is not everything. Each component may be builtaccurately, but the final scheme may still be inaccurate as seen in thecase of central difference scheme at large mesh Peclet numbers.

• Obtain exact FD solution for the problem discussed before.

• Try solving it numerically with an upwind scheme.

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Example of upwind scheme: − 1Peu

′′ + u′ = 0

1 njj − 1 j + 1

j − 1/2 j + 1/2

Ωjx = 0 x = 1

Integrating over finite volume Ωj

− 1

Pe[u′j+ 1

2

− u′j− 1

2

] + uj+ 12− uj− 1

2= 0

Fluxes: for j = 1, 2, . . . , n− 1

uj+ 12

= uj , u′j+ 1

2

=uj+1 − uj

h

Boundary fluxes

u 12

= a, un+ 12

= un, u′12

=u1 − ah/2

, u′n+ 1

2

=b− unh/2

25 / 38

Example of upwind scheme: − 1Peu

′′ + u′ = 0

System of finite volume equations

(1 +2

P)u1 −

1

Pu2 = (1 +

1

P)a

−(1 +1

P)uj−1 + (1 +

2

P)uj −

1

Puj+1 = 0, j = 2, 3, . . . , n− 1

−(1 +1

P)un−1 + (1 +

3

P)un =

2

Pb

with the mesh Peclet number P = hPe .

Matlab code: convdiff upwind ccfvm.m

26 / 38

Schemes of positive type

Lemma (Discrete maximum principle)

Let cn and xn, n = 1, 2, . . . , N be reals satisfying

N∑n=1

cn = 0, c1 > 0 and cn < 0 for n > 1

andN∑n=1

cnxn ≤ 0

Then

xn = x1, n = 1, . . . , N or x1 < maxn>1xn

27 / 38

Schemes of positive type

Proof: We have c1 > 0. Wlog we may take c1 = 1, so that

x1 ≤N∑n=2

(−cn)xn ≤ maxn>1xn

Property of convex combinations:

N∑n=2

(−cn)xn = maxn>1xn iff xn = const = max

n>1xn

If

x1 = maxn>1xn ≤

N∑n=2

(−cn)xn ≤ maxn>1xn

then it implies that x1 = xn, n = 2, 3, . . . , N . Otherwise we have strictinequality.

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Schemes of positive type

FD operator: Let Lh be a linear discrete operator arising from anumerical scheme

Lhψj =∑k∈K

α(j, k)ψj+k

where K is some index set, and α(j, k) are coefficients.

Example: Central difference scheme for Lu = −u′′

Lhψj = − 1

h2ψj−1 +

2

h2ψj −

1

h2ψj+1, K = −1, 0,+1

= c2ψj−1 + c1ψj + c3ψj+1

= α(j,−1)ψj−1 + α(j, 0)ψj + α(j,+1)ψj+1

29 / 38

Definition

The operator Lh is of positive type if∑k∈K

α(j, k) = 0, j = 2, 3, . . . , J − 1

andα(j, k) < 0, k 6= 0, j = 2, 3, . . . , J − 1

The above two conditions imply that

α(j, 0) > 0

Remark: This is related to the characterization of the scheme to bean M-matrix.

30 / 38

Theorem (Discrete maximum principle)

If Lh is of positive type and

Lhψj ≤ 0, j = 2, 3, . . . , J − 1

then local maxima of ψ can occur only for j = 1 or j = J , orψj =constant.

Proof: For any j ∈ 2, 3, . . . , J − 1 let

c1 = α(j, 0) > 0 and ck = α(j, k) ≤ 0, k 6= 0

Then Lemma applies after suitable re-indexing of ck. Considering eachinterior grid point in turn, the theorem follows.

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If the scheme Lh is applied at the boundary point j = J also (e.g.Neumann condition) then

Corollary

If the scheme Lh is of positive type and Lhψj ≤ 0 for j = 2, 3, . . . , J ,then a local maximum of ψ can occur only for j = 1.

Remarks:

• If Lhψj ≥ 0 for j = 2, 3, . . . , J − 1 and Lh is of positive type thenit can be shown that local minima can occur only for j = 1 andj = J .

• Hence, if Lhψj = 0 then local extrema can only occur for j = 1 orj = J .

• We see that if the discretization is of positive type, then it obeys asimilar maximum principle as the differential equation. Hence,spurious wiggles cannot occur.

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Order barrier We would like to have accurate scheme which is free ofwiggles. Since convection term is what causes wiggles in numericalsolution, we would like to have accurate, wiggle-free schemes for thecase of Pe =∞, i.e., for pure convection problem. But

Theorem (Order barrier)

Linear discretization schemes of positive type for the followingdifferential equation

dφ

dx= q(x)

are at most first order accurate.

Proof: For an interior grid point, the scheme is of the form

Lhψj =∑k∈K

α(k)ψj+k, α(k) = O (1/h)

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The local truncation error

τj = Lhφ(xj)− qj =∑k∈K

α(k)φ(xj + kh)− qj

=∑k∈K

α(k)

[φ(xj) + kh

dφ

dx(xj) +

1

2k2h2 d2φ

dx2(xj) +O

(h3)]− dφ

dx(xj)

For τj to be O (h) it is necessary that∑k∈K

α(k) = 0,∑k∈K

kα(k) =1

h

For the scheme to be second order accurate, it is necessary that inaddition ∑

k∈Kk2α(k) = 0

However, since Lh is of positive type, α(k) < 0 for k 6= 0, so that theabove condition cannot be satisfied.

34 / 38

Remark: A similar order barrier theorem holds for time-dependentcase. A way to get around this is to use nonlinear schemes even forlinear problems. In practice, it is preferable to use high order accurateschemes, atleast second order (τj = O

(h2)), in order to get reliable

results at small computational cost.

Remark: The upwind scheme satisfies the conditions of a positivescheme.

Monotone schemes

A scheme for which (ψ = 0 on the boundary)

Lhψ ≥ 0 =⇒ ψ ≥ 0

is called a monotone scheme. For a monotone scheme, the solutiondecreases (or increases) with the right hand side. Schemes of positivetype are monotone, but the converse is not true.

Monotone scheme ⇔ Monotone matrix, Positive scheme ⇔ M-matrixM-matrix ⇒ Monotone matrix

35 / 38

Central FV scheme for −(Du′)′ + (Uu)′ = q

α(j,−1) = −1

2Uj− 1

2−Dj− 1

2

h

α(j,+1) = +1

2Uj+ 1

2−Dj+ 1

2

hα(j, 0) = −α(j,−1)− α(j,+1)

Application of positive scheme condition gives

−2 <Uj+ 1

2h

Dj+ 12

< +2

In terms of a local mesh Peclet number

Pj+ 12

=|Uj+ 1

2|h

Dj+ 12

the condition for positive scheme is

Pj+ 12< 2

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Artificial viscosityConstruct a central scheme for modified PDE, µ = artificial viscosity

d

dx(Uu)− d

dx(D + µ)

du

dx= q

Scheme is of positive type if

µj+ 12≥ 1

2|Uj+ 1

2|h−Dj+ 1

2

We can choose µ = O (h); the scheme is first order accurate. If wechoose

µj+ 12

=1

2|Uj+ 1

2|h

then we obtain the upwind scheme. Thus upwind scheme implicitlyadds numerical dissipation. Upwind scheme = central scheme +dissipation

U+j+ 1

2

uj + U−j− 1

2

uj+1 =1

2Uj+ 1

2(uj + uj+1)− 1

2|Uj+ 1

2|(uj+1 − uj)

37 / 38

Hybrid scheme

For convective flux F = Uu, use blending of upwind and centralscheme; switch locally based on local mesh Peclet number

F cj+ 1

2

=1

2Uj+ 1

2(uj + uj+1), F u

j+ 12

= U+j+ 1

2

uj + U−j− 1

2

uj+1

Switching function

s(P ) =

0, P < 2

1, P ≥ 2

Hybrid scheme

Fj+ 12

= s(Pj+ 12)F u

j+ 12

+ [1− s(Pj+ 12)]F c

j+ 12

Preferable to use a smooth switching function. Hybrid schemesintroduced by Spalding (1972) and Patankar (1980).

38 / 38