MATH 676
–
Finite element methods in
scientific computing
Wolfgang Bangerth, Texas A&M University
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Wolfgang Bangerth
Lecture 21.5:
Boundary conditions
Part 1: Theory
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Wolfgang Bangerth
General considerations
There are many kinds of boundary conditions:
●
Dirichlet conditions
●
Neumann conditions
●
Robin conditions
●
●
Strong conditions
Natural conditions
Force conditions
●
Tractions conditions
●
…
Question: What do they all mean, where do they enter
the picture, and how do we deal with them?
●
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Wolfgang Bangerth
General considerations
There are many kinds of boundary conditions:
●
u=g
on ∂ Ω
Dirichlet conditions
●
Neumann conditions
a ∂ u /∂ n=g
on ∂ Ω
●
Robin conditions
bu±a ∂ u/∂ n=g
on ∂ Ω
●
●
Strong conditions
Natural conditions
Force conditions
●
Tractions conditions
●
…
Question: What do they all mean, where do they enter
the picture, and how do we deal with them?
●
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Wolfgang Bangerth
Laplace equation
Laplace equation with zero boundary values:
●
Strong form:
−Δ u=f
u=0
●
Weak form: find u∊V0=H10 so that
(∇ v , ∇ u)=(v , f )
●
in Ω
on ∂ Ω
∀ v ∈V 0
This is equivalent to finding the minimizer of
1
2
minu∈V J (u)= ∥∇ u∥ −(f ,u)
2
0
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Wolfgang Bangerth
Laplace equation
Laplace equation with non-zero boundary values:
●
Strong form:
−Δ u=f
u=g
●
in Ω
on ∂ Ω
Weak form: find u∊Vg={v∊H1: v=g on the boundary}
(∇ v , ∇ u)=(v , f )
∀ v∈? ?
Question: What is the appropriate test space here?
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Wolfgang Bangerth
Laplace equation
Laplace equation with non-zero boundary values:
●
Strong form:
−Δ u=f
u=g
●
Equivalent to finding the minimizer of
minu∈V
●
g
1
2
J (u)= ∥∇ u∥ −(f ,u)
2
The minimizer needs to satisfy the optimality condition:
J (u)≤J (u+ε v)
●
in Ω
on ∂ Ω
∀ u+ε v ∈V g
Since u=g on ∂ Ω we need v=0 on ∂ Ω so that u+ε v∈V g
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Wolfgang Bangerth
Laplace equation
Laplace equation with non-zero boundary values:
●
Strong form:
−Δ u=f
u=g
●
Equivalent to finding the minimizer of
minu∈V
●
g
1
2
J (u)= ∥∇ u∥ −(f ,u)
2
For differentiable J the optimality condition is:
J ' (u)(v)=lim ε→0
●
in Ω
on ∂ Ω
J (u+ε v)−J (u)
=(∇ v , ∇ u)−(v , f )=0
ε
∀v
The appropriate test space is then the tangent space,
v ∈V 0 ={φ∈H 1 : φ∣∂Ω =0}
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Wolfgang Bangerth
Laplace equation
Laplace equation with non-zero boundary values on
parts of the boundary:
●
Strong form:
−Δ u=f
u=g
∂ u/ ∂ n=0
●
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
The corresponding weak form is then
J ' (u)(v)=( ∇ v , ∇ u)−(v , f )=0
●
∀v
The appropriate test space is again the tangent space,
v∈V 0 ={φ∈H 1 : φ∣∂ Γ =0}
1
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Wolfgang Bangerth
Laplace equation
From strong to weak form:
●
Strong form:
−Δ u=f
u=g
∂ u/∂ n=h
●
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
Take equation and multiply by test function:
(v ,−Δ u)Ω =(v , f )Ω
●
∀v
Integrate by parts, get boundary terms:
(∇ v , ∇ u)Ω−(v ,n⋅∇ u)∂ Ω=(v , f )Ω
●
∀ v∈V 0
Split boundary term:
(∇ v , ∇ u)Ω−(v ,n⋅∇ u)Γ −( v , n⋅∇ u)Γ =(v , f )Ω
1
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2
∀ v ∈V 0
Wolfgang Bangerth
Laplace equation
From strong to weak form:
●
Strong form:
−Δ u=f
u=g
∂ u/∂ n=h
●
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
Consider the boundary terms separately:
(∇ v , ∇ u)Ω− v⏟ , n⋅∇ u − v , n⋅∇
⏟u =(v , f )Ω
(
●
=0
) (
Γ1
=h
)
∀ v∈V 0
Γ2
This leads to the final form:
(∇ v , ∇ u)Ω=(v , f )Ω + ( v , h ) Γ
2
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∀ v∈V 0
Wolfgang Bangerth
Laplace equation
From strong to weak form:
●
Strong form:
−Δ u=f
u=g
∂ u/∂ n=h
●
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
Consider the boundary terms separately:
(∇ v , ∇ u)Ω − v⏟ ,n⋅∇ u − v , n⋅∇
⏟u =(v , f )Ω
(
●
●
=0
) (
Γ1
=h
)
∀ v ∈V 0
Γ2
We call boundary conditions
– strong, if a term disappears because of the test space
– natural, if a term gets replaced and goes to the r.h.s.
For Laplace: Dirichlet=strong, Neumann=natural
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Wolfgang Bangerth
Laplace equation
What about Robin conditions:
●
Strong form:
−Δ u=f
u=g
b u+∂ u/∂ n=h
●
in Ω
on Γ1 ⊂∂Ω
on Γ2=∂ Ω \ Γ1
This yields:
(∇ v , ∇ u)Ω − v⏟ , n⋅∇ u − v , n⋅∇
⏟u =(v , f )Ω
)
∀ v ∈V 0
(∇ v , ∇ u)Ω + ( v ,b u )Γ =(v , f )Ω +(v , h)Γ
∀ v ∈V 0
(
●
) (
=0
Γ1
=h−bu Γ
2
And in final form:
2
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2
Wolfgang Bangerth
Laplace equation
What about equations with coefficients:
●
Strong form:
−∇⋅A ∇ u=f
u=g
? ?=h
●
in Ω
on Γ1 ⊂∂Ω
on Γ2 =∂ Ω \ Γ1
This yields after integration by parts:
(∇ v , ∇ u)Ω− v⏟ , n⋅A ∇ u − v , n⋅A
∇ u =(v , f )Ω
⏟
(
●
=0
) (
)
Γ2
To replace the third term, we need that the natural
(Neumann) boundary condition reads
n⋅A ∇ u=h
●
??
Γ1
∀ v ∈V 0
on Γ2
This corresponds with the physical definition of flux.
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Wolfgang Bangerth
More general equations
Question: Is it generally true that
●
Dirichlet
= strong boundary condition
●
Neumann
= natural boundary condition
or does this only apply to the Laplace equation?
Answer:
No! It all depends on the spaces and bilinear forms.
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Wolfgang Bangerth
Example: Mixed Laplace
Consider the mixed Laplace equation:
A −1 u+ ∇ p=0
∇⋅u=f
p=g
n⋅u=0
in Ω
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
Multiplying by test functions, integrating by parts yields
−1
(v , A u)Ω−( ∇⋅v , p )Ω + ( v , n p ) ∂Ω− ( q , ∇⋅u )Ω=−(q , f )Ω
∀ v∈U , q ∈P
with spaces (“sort of”)
U ={v∈H (div ): n⋅v∣Γ =0}
P=L2
2
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Wolfgang Bangerth
Example: Mixed Laplace
Consider the mixed Laplace equation:
A −1 u+ ∇ p=0
∇⋅u=f
p=g
n⋅u=0
in Ω
in Ω
on Γ1⊂∂ Ω
on Γ2 =∂ Ω \ Γ1
Considering boundary terms separately:
−1
(v , A u)Ω− ( ∇⋅v , p ) Ω+ v , n ⏟p
(
+ n⋅v
⏟ , p − ( q , ∇⋅u )Ω=−(q , f )Ω
) (
=g Γ
1
=0
)
Γ2
∀ v ∈U , q∈P
This yields:
−1
(v , A u)Ω− ( ∇⋅v , p ) Ω−( q , ∇⋅u )Ω=−(q , f )Ω−( v , n g ) Γ
1
Here:
p=g
n.u=0
∀ v ∈U , q∈ P
→ Dirichlet → natural boundary condition
→ Dirichlet → strong boundary condition
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Wolfgang Bangerth
Example: Mixed Laplace
What happens with conflicting boundary conditions:
A −1 u+ ∇ p=0
∇⋅u=f
p=g
n⋅u=0
in Ω
in Ω
on ∂Ω
on ∂ Ω
After integration by parts we then have
−1
(v , A u)Ω− ( ∇⋅v , p ) Ω+ n⋅v
⏟ , ⏟p
(
=0
)
−( q , ∇⋅u ) Ω=−(q , f )Ω
∀ v ∈U , q∈P
= g ∂Ω
This yields:
−1
(v , A u)Ω− ( ∇⋅v , p ) Ω−( q , ∇⋅u )Ω=−(q , f )Ω
∀ v∈U , q∈P
Here: p=g does not appear at all. It will be ignored.
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Wolfgang Bangerth
Example: Stokes
Consider the Stokes equations:
−ηΔ u+ ∇ p=f
∇⋅u=0
in Ω
in Ω
Question: What boundary conditions can we pose?
Answer: Form the bilinear form:
(η ∇ v , ∇ u)Ω−( v ,n⋅η ∇ u ) ∂Ω−( ∇⋅v , p )Ω +( n⋅v , p )∂Ω−( q , ∇⋅u )Ω=(v , f )Ω
∀v ,q
Combine boundary terms:
(η ∇ v , ∇ u)Ω−( ∇⋅v , p )Ω−( v ,n⋅[η ∇ u− p I ] )∂ Ω−( q , ∇⋅u )Ω=(v , f )Ω
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∀ v ,q
Wolfgang Bangerth
Example: Stokes
Consider the Stokes equations:
−ηΔ u+ ∇ p=f
∇⋅u=0
in Ω
in Ω
Question: What boundary conditions can we pose?
Answer: Consider the boundary term:
( v ,n⋅[η ∇ u− p I ]) ∂Ω
We can then consider the following boundary conditions:
●
Prescribed velocity:
●
Prescribed traction force: n⋅[η ∇ u− p I ]∣∂Ω =h
u∣∂Ω =g
→
v∣∂Ω =0
Here: Dirichlet=strong, Neumann=natural
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Wolfgang Bangerth
Example: Stokes
Stokes equations with Dirichlet conditions:
−ηΔ u+∇ p=f
∇⋅u=0
u=g
in Ω
in Ω
on ∂ Ω
The complete formulation is then:
Find u∈U g ={v ∈ H 1 (Ω)d : v∣∂Ω =g}, p∈ L2 (Ω) so that
(η ∇ v , ∇ u)Ω−( ∇⋅v , p )Ω−( q , ∇⋅u )Ω=(v , f )Ω
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∀ v ∈U 0, q∈ P
Wolfgang Bangerth
Example: Stokes
Stokes equations with Neumann conditions:
−η Δ u+∇ p=f
∇⋅u=0
n⋅[η ∇ u− pI ]=h
in Ω
in Ω
on ∂Ω
The complete formulation is then:
Find u∈U =H 1 (Ω)d , p∈ L2 (Ω) so that
(η ∇ v , ∇ u)Ω −( ∇⋅v , p ) Ω−( q , ∇⋅u )Ω=(v , f )Ω +( v , h)∂Ω
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∀ v ∈U , q∈P
Wolfgang Bangerth
Example: Stokes
Question: What if we prescribe u.n=g?
Answer: Then v.n=0. Consider again the boundary term:
( v , n⋅[ η ∇ u− p I ] )∂Ω = (n⊗n)
⏟v +( I −n⊗n) v , n⋅[η ∇ u− p I ]
(
=( n⋅v )n=0
)
∂Ω
=( v ,( I −n⊗n) ( n⋅[η ∇ u− p I ] ) )∂Ω
Consequence: If we prescribe only the normal component
of the velocity, we also need to prescribe the tangential
component of the traction:
u⋅n∣∂Ω =g
( I −n⊗n) ( n⋅[η∇ u− pI ]) =h
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Wolfgang Bangerth
Example: Stokes
Question: What if we prescribe the physical traction force:
n⋅[2 ηε(u)− p I ]∣∂Ω =h
1
T
where ε(u)= ( ∇ u+∇ u )
2
Answer: Consider again the boundary term:
( v , n⋅[η∇ u− p I ]) ∂Ω
Consequence: You can't do this.
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Wolfgang Bangerth
Example: Stokes
Question: What if we prescribe the physical traction force:
n⋅[2 ηε(u)− p I ]∣∂Ω =h
1
T
where ε(u)= ( ∇ u+∇ u )
2
Answer: You can't do this. However, if we formulate the
Stokes equations as
−2 η ∇⋅ε(u)+ ∇ p=f
∇⋅u=0
in Ω
in Ω
then the weak form is
(2 ηε(v ),ε(u))Ω −( ∇⋅v , p )Ω −( v ,n⋅[2 ηε(u)− p I ] )∂ Ω
−( q , ∇⋅u )Ω=(v , f )Ω
∀ v ,q
and we can impose this boundary condition.
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Wolfgang Bangerth
Example: Stokes
Note: The two formulations of the Stokes equations,
−ηΔ u+ ∇ p=f
∇⋅u=0
and
−2 η ∇⋅ε(u)+ ∇ p=f
∇⋅u=0
in Ω
in Ω
in Ω
in Ω
are equivalent because
2 η ∇⋅ε(u)=η∇⋅( ∇ u)+η ∇⋅( ∇ uT )=ηΔ u+ η ∇ ( ∇⋅u
⏟ )=η Δ u
=0
They only differ in whether we can impose:
●
The unphysical traction
n⋅[η ∇ u− p I ]∣∂Ω =h
●
The physical traction
n⋅[2 ηε(u)− p I ]∣∂Ω =h
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Wolfgang Bangerth
Summary
●
●
●
●
Boundary conditions are complicated
They can be resolved by looking at the boundary terms
after integrating by parts
You can impose strong conditions:
– relating to the test function in these terms
– appear as part of the function space for the solution
– are zero in the function space for the test function
You can impose natural conditions:
– relating to the solution in these terms
– are replaced by their boundary value
– are brought to the right hand side of the weak form
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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