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Mimetic Methods for Diffusion Problems
DANSIS seminar: High order discretisation 21/3/2018
Kennet OlesenPhD
Aarhus University 2013-2016
A diffusion problem is typically constructed out of 2 parts
β’ An equilibrium or balance partβ E.g. mass balance, energy balance etc.
π β π = π
β’ A constitutive partβ E.g. relation between heat flux and a temperature gradient
π = k ππΌ
Merge the 2 parts and you get:
π β k ππΌ = π
k βπΌ = π
Structure of a diffusion problem
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Special interpolation polynomials: Edge polynomials
In [1] interpolation polynomials are derived,which are based on integrated line values
π½β π =
π=1
π
π΅πππ(π) π΅π = ππβ1
ππ
π½ π ππ
ππ π = β
π=0
πβ1πβπ π
ππ ππβ1
ππ
ππ π = 1 πππ π = π0 πππ π β π
β: Lagrange polynomials
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[1]: Gerritsma M.: Edge Functions for Spectral Element MethodsSelected Papers from the ICOSAHOM '09. Springer, 2011
Special interpolation polynomials: Edge polynomials
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π½β π =
π=1
π
π΅πππ(π)
π΅π = ππβ1
ππ
π½ π ππ
πΌβ π =
π=0
π
πΌπβπ(π)
Interpolation in 3D -> tensor products
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πΌβ π1, π2, π3 =
π
π
π
π΄πππ Ξ¨πππ π1, π2, π3
Ξ¨πππ π1, π2, π3 = βπ(π1)βπ(π2)βπ(π3)
Ξ¨πππ π1, π2, π3 = ππ(π1)βπ(π2)βπ(π3)
Ξ¨πππ π1, π2, π3 = βπ(π1)ππ(π2)βπ(π3)
Ξ¨πππ π1, π2, π3 = βπ(π1)βπ(π2)ππ(π3)Ξ¨πππ π1, π2, π3 = βπ(π1)ππ(π2)ππ(π3)
Ξ¨πππ π1, π2, π3 = ππ(π1)βπ(π2)ππ(π3)
Ξ¨πππ π1, π2, π3 = ππ(π1)ππ(π2)βπ(π3)
Ξ¨πππ π1, π2, π3 = ππ(π1)ππ(π2)ππ(π3)
π1 π2 π3
πΉ
Differentiation
β’ Say we have the approximation:
πΌβ π =
π=0
π
πΌπβπ(π)
β’ We choose to approximate the derivative by:
ππΌβ(π)
ππ=
π=1
π
πβπ
πππΌβ(π)ππ
ππ ππ(π)
β’ This reduces to (first fundamental theorem of calculus):
ππΌβ(π)
ππ=
π=1
π
πΌπ β πΌπβ1 ππ(π)
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πΌπ
πΌπβ1
Divergence
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π β πβ =ππ1
β π1, π2, π3ππ1
+ππ2
β π1, π2, π3ππ2
+ππ3
β π1, π2, π3ππ3
π1β π1, π2, π3 =
π=0
π
π=1
π
π=1
π
ππ,π,π1 βπ(π1)ππ(π2)ππ(π3)
π2β π1, π2, π3 =
π=1
π
π=0
π
π=1
π
ππ,π,π2 ππ(π1)βπ(π2)ππ(π3)
π3β π1, π2, π3 =
π=1
π
π=1
π
π=0
π
ππ,π,π3 ππ(π1)ππ(π2)βπ(π3)
π β πβ =
π=1
π
π=1
π
π=1
π
ππ,π,π1 β ππβ1,π,π
1 + ππ,π,π2 β ππ,πβ1,π
2 + ππ,π,π3 β ππ,π,πβ1
3 ππ(π1)ππ(π2)ππ(π3)
πiv
π¬(3,2)
Divergence β Incidence matrix
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The discrete balance equation
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π β πβ = πh
This is written by:
π=1
π
π=1
π
π=1
π
ππ,π,π1 β ππβ1,π,π
1 + ππ,π,π2 β ππ,πβ1,π
2 + ππ,π,π3 β ππ,π,πβ1
3 ππ(π1)ππ(π2)ππ(π3) =
π=1
π
π=1
π
π=1
π
πΉπ,π,π ππ(π1)ππ(π2)ππ(π3)
They share common interpolation polynomials:
π=1
π
π=1
π
π=1
π
ππ,π,π1 β ππβ1,π,π
1 + ππ,π,π2 β ππ,πβ1,π
2 + ππ,π,π3 β ππ,π,πβ1
3 β πΉπ,π,π ππ(π1)ππ(π2)ππ(π3) = 0
This implies:
ππ,π,π1 β ππβ1,π,π
1 + ππ,π,π2 β ππ,πβ1,π
2 + ππ,π,π3 β ππ,π,πβ1
3 β πΉπ,π,π = 0
Exact balance equation for our mesh
Putting together the system of equations
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β’ Problem: More unknowns than equations
β’ I fix this when I discretize the constitutive equationπβ = k ππΌβ
β’ Weigh with πβ and integrate over domain1
ππβ, πβ
Ξ©= πβ, ππΌβ
Ξ©= β π β πβ, πΌβ
Ξ©+ πβ, πΌβ
πΞ©
πΌππ‘πππππ‘πππ ππ¦ ππππ‘π
β’ Choose same amount of discrete points for πΌβ as number of elements
β’ Square system of equations
Transpose of the divergence in thebalance equations
Putting together the system of equations
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β’ Balance equation + constitutive equation β Square system of equations
π π¬(3,2)
π¬(3,2)π½β π 1
ππ―β
π«πΌ
π«π=
π«πΉ
π©βπ«πΌπ΅
Potential boundary conditions weakly enforced
Flux boundary conditions strongly enforced
β’ Multiply balance equation with π½ββ Symmetric system of equations
π π½βπ¬(3,2)
π¬(3,2)π½β π 1
ππ―β
π«πΌ
π«π=
π½βπ«πΉ
π©βπ«πΌπ΅
Complex geometry
Mimetic Methods for Diffusion Problems 12
β’ Mapping will only affect the matrices with expansion polynomials and not the incidence matrices
β’ I.e. mapping will not affect the mimicking property of the balance equation
π¬(3,2)π«π = π«πΉ π¬(3,2)π«π = π«πΉ
Test case
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β’ Choose potential fieldβ’ Calculate π and πβ’ Apply as boundary conditions
β’ Solve on different meshesβ’ Refine mesh
β’ Mesh size βππ β
β’ Evaluate in 100 Γ 100 points in each element
β’ Plot the maximum value in the entire domain
Error of the Potential (πΌ = π)
Residual of balance equation (π = π β πβ β πβ)
Mimicking vector valued fields
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β’ So far we have considered Poissonβs equation of a scalar fieldβ’ But what about Poissonβs equation of a vector field?
β’ For example the equilibrium of forces in Continuum mechanics
π β π = βπ
Mimicking vector valued fields
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β’ The procedure is the same just in 3 directionsπ«π«π = βπ«πΉ
β’ With
π« =
π¬(3,2) π π
π π¬(3,2) π
π π π¬(3,2)
β’ Consider the structural problemβ’ Constitutive equations
β’ πππ =π
ππ₯ππ’π β πππ = πΆππππ πππ
β’ Equilibrium of forces
β’π
ππ₯ππππ + ππ = 0
β’ Symmetry of the stress tensorβ’ πππ β πππ = 0
π―β π« π½β πβ πΉβ π
π½β π« π πβπΉβ π π
π«π
π«π’
π«π
=π©βπ«π’
π΅πΆ
βπ½βπ«πΉ
π
Test case
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Test case
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FEMimetic
Thank you for your time!
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