ADI scheme for partially dimension reduced heat conduction ...techmat.vgtu.lt/seminarai/ciegisReducedDimension20200908.pdf · e-mail: [email protected] MMK seminaras, Vilnius, 2020.09.08

ModelsADI schemes

Computational experimentsConclusions and Future Work

ADI scheme for partially dimensionreduced heat conduction models

R. ČIEGIS

Vilnius Gediminas Technical University, Vilnius, Lithuaniae-mail: [email protected]

MMK seminaras, Vilnius, 2020.09.08

R. ČIEGIS ADI scheme for dimension reduced model

ModelsADI schemes


This is a joint work with G. Panasenko (Institute Camille JordanUMR CNRS 5208, University of Lyon, Saint-Etienne, France)K. Pileckas, V. Šumskas (Vilnius University)

Publication:

R. Čiegis, G. Panasenko, K. Pileckas, V. Šumskas, ADI scheme forpartially dimension reduced heat conduction models. Computers &Mathematics with Applications. Volume 80, Issue 5, 1 September2020, Pages 1275-1286https://doi.org/10.1016/j.camwa.2020.06.01


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Outline

Models

ADI schemes

Computational experiments

Conclusions and Future Work


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Mathematical Model in 3D

Let us assume that initial and boundary conditions and allcoefficients satisfy the radial symmetry condition, thus we get thefollowing problem

∂u

∂t=

1

r

∂

∂r

(r∂u

∂r

)+∂2u

∂z2+f (r , z , t), (r , z , t) ∈ QT=Ω×(0,T ], (1)

u(r , 0, t)=g1(r , t), u(r , l , t)=g2(r , t), (r , t) ∈ (0,R]× (0,T ], (2)

r∂u

∂r= 0, 0 < z < l , r = 0 and r = R, 0 < t ≤ T , (3)

u(r , z , 0) = u0(r , z), (r , z) ∈ Ω. (4)


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Let S(u)

S(u) =2

R2

∫ R0

ru(r , z , t)dr

denote the averaging operator.

We assume that the initial condition u0 and source function fsatisfy the relations

u0(r , z) = S(u0), f (r , z , t) = S(f ), (z , t) ∈ (0, l)× (0,T ].

It means that u0 and f do not depend on r within the tube T

Denote a reduced tube Tδ = D × (δ, l − δ) andΩδ = {(r , z) ∈ (0,R)× (δ, l − δ)}.


ModelsADI schemes


Function U is called an approximate solution to problem (1) – (4)if it satisfies the following problem

∂U

∂t=

1

r

∂

∂r

(r∂U

∂r

)+∂2U

∂z2+f (z , t), (r , z , t)∈(Ω \ Ωδ)×(0,T ], (5)

∂U

∂t=∂2U

∂z2+ f (z , t), (r , z , t) ∈ Ωδ × (0,T ], (6)

U(r , 0, t)=g1(r , t), U(r , l , t)=g2(r , t), (r , t)∈(0,R]×(0,T ], (7)

r∂U

∂r=0, z∈(0, δ) ∪ (l − δ, l), r=0, r=R, 0 < t≤T , (8)

U(r , z , 0) = u0(r , z), (r , z) ∈ Ω. (9)

In Ωδ × (0,T ] the solution U do not depend on r , i.e.

U(r , z , t) = S(U), (r , z , t) ∈ Ωδ × (0,T ].


ModelsADI schemes


From the weak form of the heat equation, it follows that theconjugation conditions are valid at the truncations of the tube

U∣∣z=δ−0 = U

∣∣z=δ+0

, U∣∣z=l−δ−0 = U

∣∣z=l−δ+0, (10)

∂S(U)

∂z

∣∣∣z=δ−0

=∂U

∂z

∣∣∣z=δ+0

,∂U

∂z

∣∣∣z=l−δ−0

=∂S(U)

∂z

∣∣∣z=l−δ+0

. (11)

The conditions (10) are classical and mean that U is continuousat the truncation points.The remaining two conditions (11) are nonlocal and they definethe conservation of full fluxes along the separation lines.


ModelsADI schemes


For functions defined on the grid Ωh × ωt we introduce the discreteoperators with respect to z and r :

∂zUnjk :=

Unjk − Unj ,k−1H

, Ah2Unjk := −

1

H

(∂zU

nj ,k+1 − ∂zUnj ,k

).

∂rUnjk :=

Unjk−Unj−1,kh

, Ah1Unjk :=−

1

r̃jh

(rj+ 1

2∂rU

nj+1,k−rj− 1

2∂rU

nj ,k

),

where

r̃0 =1

8h, r̃j = rj , 1 ≤ j < J, r̃J =

1

2

(R − h

4

), r− 1

2= 0, rJ+ 1

2= 0.


ModelsADI schemes


Then the heat conduction problem (1)-(4) is approximated by thefollowing Alternating Direction Implicit (ADI) scheme

Un+ 1

2jk − U

njk

τ/2+Ah1U

n+ 12

jk +Ah2U

njk = f

n+ 12

jk , (rj , zk) ∈ ω̄r × ωz , (12)

Un+1jk − Un+ 1

2jk

τ/2+ Ah1U

n+ 12

jk +Ah2U

n+1jk = f

n+ 12

jk , (rj , zk) ∈ ω̄r×ωz .


ModelsADI schemes


LemmaIf a solution of the problem (1)-(4) is sufficiently smooth, then theapproximation error of ADI scheme (12) is O(τ2 + h2 + H2).

Proof.The solution of ADI scheme (12) satisfies the scheme

Un+1jk − Unjk

τ+ Ah1

(Un+1jk + Unjk2

)+ Ah2

(Un+1jk + Unjk2

)+τ2

4Ah1A

h2

(Un+1jk − Unjkτ

)= f

n+ 12

jk .

which is equivalent to the classical symmetrical finite differencescheme.


ModelsADI schemes


LemmaThe discrete operators Ah1 and A

h2 are symmetric and non-negative

and positive definite operators, respectively.

Proof.Applying the summation by parts formula and taking into accountthe boundary conditions for vectors u, v we get

(Ah2u, v) =K−1∑k=1

(Ah2u)kvkH = (∂zu, ∂zv ].

It follows that Ah2 is symmetric. Since Ah2φl = λlφl has a

complete set of eigenvectors φl , l = 1, . . . ,K − 1, and alleigenvalues are positive λl > 0, A

h2 is a positive-definite

operator.


ModelsADI schemes


Proof.Now we consider the operator Ah1. Applying the summation byparts formula we get

[Ah1u, v ]r = (∂ru, ∂rv ]r .

It follows from the obtained equality, that Ah1 is symmetricoperator.The eigenvalue problem Ah1ψl = µlψl has a complete set ofeigenvectors ψl , l = 0, . . . , J, one eigenvalue µ0 = 0 and theremaining eigenvalues are positive µl > 0.


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LemmaADI scheme (12) is unconditionally stable.

Proof.The Fourier stability analysis is used. Let us consider the solution of ADIscheme (12) in the case when boundary conditions gj = 0, j = 1, 2. Sinceoperators Ah1 and A

h2 commute, the solution of (12) can be written as

Un+1jk =J∑

l=0

K−1∑r=0

cn+1lr ψl(rj)φr (zk).

Substituting this formula into equations (12) we obtain the stabilityequations for each mode

cn+1lr = qlrcnlr , qlr =

(1− 0.5τλr )(1− 0.5τµl)(1 + 0.5τλr )(1 + 0.5τµl)

.

Since eigenvalues λr > 0, µl ≥ 0, then the ADI scheme (12) isunconditionally stable in the L2 norm.


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The backward Euler scheme

The heat conduction problem (5)-(11) is approximated by thebackward Euler scheme

Un+1jk −Unjk

τ+Ah1U

n+1jk +A

h2U

n+1jk =f

n+1jk , (rj , zk)∈ω̄r×(ωz1 ∪ ωz3),

(13)

Un+10k − Un0k

τ+ Ah2U

n+10k = f

n+10k , zk ∈ ωz2,

Un+10K1 −Un0K1

τ+

1

H2

(−Sh(Un+1K1−1)+2U

n+10K1−Un+10,K1+1

)=f n+10K1 , (14)

Un+10K2 − Un0K2

τ+

1

H2

(− Sh(Un+1K2+1) + 2U

n+10K2− Un+10,K2−1

)= f n+10K2 ,

Un+1j0 = g1(rj , tn+1), Un+1jK = g2(rj , t

n+1). (15)


ModelsADI schemes


Here Sh denotes the discrete averaging operator

Sh(Unk ) =

2

R2

J∑j=0

r̃jUnjkh.

Note that equations (14) approximate the nonlocal fluxconjugation conditions:

J∑j=0

r̃j

(Un+10,K1+1 − Un+10,K1H

− H2

(Un+10K1 − Un0K1τ

− f n+10K1))

h (16)

=J∑

j=0

r̃j

(Un+10,K1 − Un+1j ,K1−1H

+H

2

(Un+10K1 − Un0K1τ

− f n+10K1))

h.


ModelsADI schemes


For U,V ∈ Dh, such that Uj0 = UjK = 0, Vj0 = VjK = 0, rj ∈ ω̄rthe formulas

(U,V )=J∑

j=0

r̃j

( K1−1∑k=1

UjkVjkH+K−1∑

k=K2+1

UjkVjkH)

h+R2

2

K2∑k=K1

U0kV0kh,

‖U‖ = (U,U)1/2

define a scalar product and a norm in this vector space (in 2Ddiscrete space !!!).


ModelsADI schemes


Let us define two operators for U ∈ Dh:

Ah1U =

{Ah1U.,k , (rj , zk) ∈ ω̄r × (ωz1 ∪ ωz3),0, zk ∈ ω̄z2,

Ah2U =

Ah2Ujk , (rj , zk) ∈ ω̄r × (ωz1 ∪ ωz3),Ah2U0k , zk ∈ ωz2,

1H2

(− Sh(U.,K1−1) + 2U0K1 − U0,K1+1

), k = K1,

1H2

(− Sh(U.,K2+1) + 2U0K2 − U0,K2−1

), k = K2.


ModelsADI schemes


Then we can write the backward Euler scheme as

Un+1 − Un

τ+ (Ah1 +Ah2)Un+1 = F n+1, (rj , zk) ∈ Ωh,RD .

LemmaThe discrete operators Ah1 and Ah2 are symmetric and non-negativeand positive definite operators, respectively.


ModelsADI schemes


Proof.Applying the summation by part formula we get

(Ah1U,V ) =J∑

j=1

rj− 12

( K1−1∑k=1

∂rUjk∂rVjkH +K−1∑

k=K2+1

∂rUjk∂rVjkH)

h

= (U,Ah1V ).

Thus, Ah1 is symmetric and non-negative definite operator.

(Ah2U,V ) =J∑

j=0

r̃j( K1∑

k=1

∂zUjk ∂zVjkH +K∑

k=K2+1

∂zUjk ∂zVjkH)

h

+R2

2

K2∑k=K1+1

∂zU0k ∂zV0kH.

The nonlocal conjugation conditions (16) are used to derive these

equalities.R. ČIEGIS ADI scheme for dimension reduced model

ModelsADI schemes


ADI scheme

Then ADI scheme is written as

Un+12 − Un

τ/2+Ah1Un+

12 +Ah2Un=F n+1/2, (rj , zk) ∈ Ωh,RD , (17)

Un+1 − Un+12

τ/2+Ah1Un+

12 +Ah2Un+1 = F n+1/2. (18)

LemmaIf Un is the solution of ADI scheme (17)-(18), when f n ≡ 0 andgn1 = g

n2 ≡ 0, then the following stability estimate is valid

‖(I + τ2Ah2)Un‖ ≤ ‖(I +

τ

2Ah2)U0‖. (19)


ModelsADI schemes


Proof.The ADI scheme (17)-(18) gives Un+1 = RUn with

R =(

I +τ

2Ah2)−1(

I − τ2Ah1)(

I +τ

2Ah1)−1(

I − τ2Ah2).

We rewrite this relations as(I +

τ

2Ah2)

Un+1 = R̃(

I +τ

2Ah2)

Un.

By induction we prove that(I +

τ

2Ah2)

Un = R̃n(

I +τ

2Ah2)

U0.

It follows from Lemma 2 that ‖R̃‖ ≤ 1, thus

‖R̃n‖ ≤ ‖R̃‖n ≤ 1.


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Due to nonlocal conjugation conditions the classical factorizationalgorithm should be modified in order to solve 1D subproblems(18).

LemmaThe unique solution of the linear system of equations (18) existsand it can be computed by using the efficient factorizationalgorithm.

We get a linear system of two equations to find Un+10K1 , Un+10K2

:{A11U

n+10K1

+ A12Un+10K2

= B1

A21Un+10K1

+ A22Un+10K2

= B2,


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The accuracy of the reduced dimension model

We investigate the accuracy of the reduced dimension model(5)-(11). In the first example a domain with two nodes and oneedge is used.For the space discretization a uniform grid Ωh with J = 100,K = 1600 is used, and integration in time is done with τ = 0.0005.Table 1 gives for a sequence of reduction parameter δ errors e(δ)

e(δ) = max(rj ,zk )∈Ωhh

∣∣∣UNjk − UNjk (δ)∣∣∣of the reduced dimension model


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Table : Errors e(δ) of the discrete solution of the reduced dimensionmodel (5)-(11) for a sequence of truncation parameter δ.

δ = 0.05 δ = 0.1 δ = 0.15 δ = 0.2 δ = 0.25

e(δ) 0.2471 0.0377 0.0056 0.00083 0.00013

CPU time for computing the full model solution is 11.4 seconds,while for the reduced dimension model and δ = 0.25 the time isreduced to 5.9 seconds, for δ = 0.1 the CPU time is reduced to2.5 seconds.


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Example 2. In order to show the robustness of the proposeddiscrete scheme a domain with three full dimension nodes and twoedges is considered. One additional node takes into account theinfluence of the source function.

For this test problem we get a linear system of four equations tofind Un+10K1 , U

n+10K2

, Un+10K3 , Un+10K4

. A matrix of this system istridiagonal, thus the standard factorization algorithm is used tosolve it.


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Table : CPU time for computing the reduced dimension model (5)-(11)solution for a sequence of truncation parameter δ. The column δ∗ givesCPU time for computing the full model solution.

δ = δ∗ δ = 0.25 δ = 0.2 δ = 0.15 δ = 0.1

CPU time (δ) 24.9 17.0 14.6 12.2 9.8


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Figure : Full model and reduced dimension model with δ = 0.1


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Conclusions and Future Work

1.A finite volume method is used to approximate space differentialoperators with nonclassical conjugation conditions between the 3Dand 2D parts.2. The ADI scheme leads to non-iterative implementationalgorithm and a set of one dimensional linear systems are solved byusing the factorization algorithm. An efficient modification of thebasic factorization algorithm is developed to resolve non-localconjugation conditions.3. It is proved that the proposed discrete scheme is unconditionallystable.4. Future work: more complicated geometry, parallel algorithms,Navier-Stokes model.


ModelsADI schemesComputational experimentsConclusions and Future Work

Documents

ADI scheme for partially dimension reduced heat conduction ...techmat.vgtu.lt/seminarai/ciegisReducedDimension20200908.pdf · e-mail: [email protected] MMK seminaras, Vilnius, 2020.09.08