MATH 676
–
Finite element methods in
scientific computing
Wolfgang Bangerth, Texas A&M University
http://www.dealii.org/
Wolfgang Bangerth
Lecture 33.25:
Which element to use
Part 2: Saddle point problems
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Wolfgang Bangerth
Stokes
Consider the stationary Stokes equations:
−Δ u+∇ p = f
∇⋅u
= 0
This can equivalently be considered as a minimization
problem:
minu ∈H (Ω)
1
1
2
∥∇ u∥ −(f , u)
2
∇⋅u = 0
d
such that
Let us consider the constraint in variational form:
minu ∈V =H (Ω)
1
such that
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d
1
2
∥∇ u∥ −(f ,u)
2
2
(q , ∇⋅u) = 0 ∀ q ∈Q=L
Wolfgang Bangerth
Stokes
Consider the stationary Stokes equations:
−Δ u+∇ p = f
∇⋅u
= 0
The discrete formulation for this seeks uh ∈V h ⊂V , p h ∈Q h⊂Q :
(∇ v h , ∇ u h )−( ∇⋅v h , ph )−(q h , ∇⋅uh ) = (v h , f )
∀ v h ∈V h , q h ∈Q h
This corresponds to the finite dimensional minimization
problem
minu ∈V
h
h
such that
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1
2
∥∇ uh∥ −( f ,u h )
2
(q h , ∇⋅u h ) = 0 ∀ q h ∈Q h
Wolfgang Bangerth
Stokes
Consider the discrete Stokes equations:
minu ∈V
h
h
such that
1
2
∥∇ uh∥ −( f ,u h )
2
(q h , ∇⋅u h ) = 0 ∀ q h ∈Q h
Here, we have dim Qh constraints on the velocity uh.
Intuitively, if (asymptotically) Qh is “too large” compared to
Vh, then:
●
we have too many constraints on the velocity
●
the velocity does not have enough degrees of freedom.
In this case the discrete solution may not converge.
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Wolfgang Bangerth
Stokes
Consider the discrete Stokes equations:
We only get convergence of discrete solutions of
(∇ v h , ∇ uh )−( ∇⋅v h , ph )−(q h , ∇⋅u h ) = (v h , f )
or equivalently
minu ∈V
h
h
such that
∀ v h ∈V h , qh ∈Qh
1
2
∥∇ uh∥ −( f ,u h )
2
(q h , ∇⋅u h ) = 0 ∀ q h ∈Q h
if the inf-sup/Babuska-Brezzi/LBB condition is satisfied:
There exists a constant c independent of h so that
sup v ∈V
h
h
(∇⋅v h , q h )
≥ c∥qh∥Q
∥v h∥V
∀ q h ∈Q h
Note: V=H1, Q=L2.
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Wolfgang Bangerth
Stokes
The inf-sup condition:
We can write the condition either as:
There exists a constant c independent of h so that
sup v ∈V
h
h
(∇⋅v h , q h )
≥ c∥qh∥Q
∥v h∥V
∀ q h ∈Q h
Or as:
There exists a constant c independent of h so that
inf q ∈Q sup v ∈V
h
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h
h
h
(∇⋅v h , q h )
≥ c
∥v h∥V ∥q h∥Q
Wolfgang Bangerth
Stokes
The inf-sup condition:
The condition...
There exists a constant c independent of h so that
sup v ∈V
h
h
(∇⋅v h , q h )
≥ c∥qh∥Q
∥v h∥V
∀ q h ∈Q h
...can be satisfied by making
●
the velocity space Vh large enough
●
the pressure space Qh small enough
Typical choices (the “Taylor-Hood element”):
●
Vh = Pk+1 , Qh = Pk
on triangles/tetrahedra
●
Vh = Qk+1 , Qh = Qk
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on quadrilaterals/hexahedra
Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk/Pk or Qk/Qk is not stable:
– the constant c goes to zero as h→0
– the matrix has a near-nullspace
– the pressure develops a “checkerboard pattern”:
(using step-22 with equal order elements)
Consequence: We need to make the velocity space
larger or the pressure space smaller!
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Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk+1/Pk or Qk+1/Qk is stable:
– there is a constant c>0
– the matrix remains regular
– the pressure is stable
(using step-22 with non-equal order elements)
Consequence: This works!
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Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk+2/Pk or Qk+2/Qk is stable:
– there is a constant c>0
– the matrix remains regular
– the pressure is stable
(using step-22 with non-equal order elements)
– accuracy is now limited by the low-order pressure
Consequence: This choice is wasteful!
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Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk/Pk or Qk/Qk doesn't work
●
Pk+1/Pk or Qk+1/Qk works
●
Can we find something in between?
Option 1: With a slightly smaller velocity space.
●
●
We can't take away shape functions without either
– violating unisolvency
– making the shape functions discontinuous
(which would make the element non-conforming)
This option doesn't work
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Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk/Pk or Qk/Qk doesn't work
●
Pk+1/Pk or Qk+1/Qk works
●
Can we find something in between?
Option 2a: With a slightly larger pressure space.
●
Recall the bilinear form:
(∇ v h , ∇ u h )−( ∇⋅v h , ph )−(q h , ∇⋅uh ) = (v h , f )
●
●
●
We don't actually need continuity of the pressure
We could try Qk+1/Qk+DGQ0
This actually works!
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Wolfgang Bangerth
Stokes
Why Taylor-Hood (Pk+1/Pk or Qk+1/Qk):
●
Pk/Pk or Qk/Qk doesn't work
●
Pk+1/Pk or Qk+1/Qk works
●
Can we find something in between?
Option 2b: With an even larger pressure space.
●
Recall the bilinear form:
(∇ v h , ∇ u h )−( ∇⋅v h , ph )−(q h , ∇⋅uh ) = (v h , f )
●
●
●
We don't actually need continuity of the pressure
We could try Qk+1/DGQk
This doesn't work, too many pressure degrees of freedom
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Wolfgang Bangerth
Stokes
Choosing a polynomial degree:
We could choose either
●
Pk+1/Pk or Qk+1/Qk
●
Qk+1/Qk+DGQ0
In practice one typically chooses k=1:
●
There are 2*32+22=22 (3*33+23=89) degrees of freedom
per cell in 2d (3d)
●
On a uniform mesh, matrix rows may have up to
– 2*52+32 = 59
– 3*53+33 = 402
entries
●
k>1 yields better accuracy, but matrix starts to get full
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Wolfgang Bangerth
Mixed Laplace
Consider the mixed Laplace equations:
u+ ∇ p = 0
∇⋅u
= f
This can equivalently be considered as a minimization
problem:
minu ∈H (Ω)
1
1
2
∥u∥
2
∇⋅u = f
d
such that
Let us consider the constraint in variational form:
minu ∈V =H (Ω)
1
such that
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d
1
2
∥u∥
2
(q , ∇⋅u) = (q , f )
∀ q ∈Q=L
2
Wolfgang Bangerth
Mixed Laplace
Consider the mixed Laplace equations:
u+ ∇ p = 0
∇⋅u
= f
The discrete formulation for this seeks uh ∈V h ⊂V , p h ∈Q h⊂Q :
(v h ,uh )−( ∇⋅v h , p h )−(q h , ∇⋅u h ) = −(q h , f )
∀ v h ∈V h , qh ∈Qh
This corresponds to the finite dimensional minimization
problem
minu ∈V
h
h
such that
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1
2
∥u h∥
2
(q h , ∇⋅u h ) = (q h , f )
∀ q h ∈Q h
Wolfgang Bangerth
Mixed Laplace
Consider the discrete mixed Laplace equations:
We only get convergence of discrete solutions of
(v h ,uh )−( ∇⋅v h , p h )−(q h , ∇⋅u h ) = −(q h , f )
∀ v h ∈V h , qh ∈Qh
or equivalently
minu ∈V
h
1
2
∥u h∥
2
(q h , ∇⋅u h ) = (q h , f )
h
such that
∀ q h ∈Q h
if the inf-sup/Babuska-Brezzi/LBB condition is satisfied:
There exists a constant c independent of h so that
sup v ∈V
h
h
(∇⋅v h , q h )
≥ c∥qh∥Q
∥v h∥V
∀ q h ∈Q h
Note: V=H(div), Q=L2.
http://www.dealii.org/
Wolfgang Bangerth
Mixed Laplace
We have the same situation as before:
The condition...
There exists a constant c independent of h so that
sup v ∈V
h
h
(∇⋅v h , q h )
≥ c∥qh∥Q
∥v h∥V
∀ q h ∈Q h
...can be satisfied by making
●
the velocity space Vh large enough
●
the pressure space Qh small enough
http://www.dealii.org/
Wolfgang Bangerth
Mixed Laplace
One common choice:
●
Vh = Pk+1 , Qh = Pk
on triangles/tetrahedra
●
Vh = Qk+1 , Qh = Qk
on quadrilaterals/hexahedra
This is again the Taylor-Hood element. It is stable
We can play the same game:
●
●
Can we make the velocity space smaller?
(Less numerical effort with essentially same accuracy.)
Can we make the pressure space larger?
(Better accuracy with only slightly more work.)
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Wolfgang Bangerth
Mixed Laplace
Option 1: Make velocity space smaller.
●
Vh = Qk+1 , Qh = Qk works
●
Vh consists of continuous functions so that we can take
the (weak) gradient, which we needed for Stokes:
(∇ v h , ∇ uh )−( ∇⋅v h , ph )−(q h , ∇⋅u h ) = (v h , f )
●
But we don't need this for mixed Laplace:
(v h ,u h )−( ∇⋅v h , p h )−(q h , ∇⋅uh ) = −(q h , f )
●
∀ v h ∈V h , qh ∈Qh
∀ v h ∈V h , q h ∈Q h
All we need is that the divergence is defined.
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Wolfgang Bangerth
Mixed Laplace
Option 1: Make velocity space smaller.
We need: An element with continuous normal vector
component but possibly discontinuous tangential
component.
This is the Raviart-Thomas element.
This works: Replace
●
Vh = Qk+1 , Qh = Qk
by
●
Vh = Raviart-Thomas(k), Qh = Qk
if k>0
●
Vh = Raviart-Thomas(0), Qh = DGQ0
if k=0
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Wolfgang Bangerth
Mixed Laplace
Option 2: Make pressure space larger.
Recall the bilinear form:
(v h ,u h )−( ∇⋅v h , p h )−(q h , ∇⋅uh ) = −(q h , f )
∀ v h ∈V h , q h ∈Q h
We don't need a continuous pressure.
This works: Replace
●
Vh = Raviart-Thomas(k), Qh = Qk
if k>0
●
Vh = Raviart-Thomas(0), Qh = DGQ0
if k=0
Vh = Raviart-Thomas(k), Qh = DGQk
(see step-20)
by
●
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Wolfgang Bangerth
Mixed Laplace
Option 3: Alternatives
There are any number of alternatives to the Raviart-Thomas
element:
●
Brezzi-Douglas-Marini (BDM)
●
Arnold-Falk-Winther
●
…
●
●
●
●
Most of these use piecewise constant pressures at lowest
order
This leads to very slow convergence (O(h))
For practical applications: use higher orders
Elements are relatively “sparse”, i.e., not too many DoFs
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Wolfgang Bangerth
MATH 676
–
Finite element methods in
scientific computing
Wolfgang Bangerth, Texas A&M University
http://www.dealii.org/
Wolfgang Bangerth