C0 Interior Penalty Methods
Introduction
Current Research in Finite Element Methods
CIMPA Summer School
Mumbai, July 2015
Overview
Overview
I
Introduction
Motivations
Derivations
Overview
I
Introduction
Motivations
Derivations
I
Convergence Analysis
Standard A Priori Error Analysis
Medius Error Analysis
A Posteriori Error Analysis
Overview
I
Introduction
Motivations
Derivations
I
Convergence Analysis
Standard A Priori Error Analysis
Medius Error Analysis
A Posteriori Error Analysis
I
Fast Solver
Multigrid
Overview
I
Introduction
Motivations
Derivations
I
Convergence Analysis
Standard A Priori Error Analysis
Medius Error Analysis
A Posteriori Error Analysis
I
Fast Solver
Multigrid
I
Variational Inequalities
General References
• B. and Scott, The Mathematical Theory of Finite Element
Methods (Third Edition), Springer-Verlag, 2008.
(Springer International Edition 2012)
• B., C0 Interior Penalty Methods, Frontiers in Numerical
Analysis-Durham 2010, Springer-Verlag, 2012.
Outline of Lecture
I
Examples of Fourth Order Problems
I
Classical Finite Element Methods
I
C0 Interior Penalty Methods
I
Enriching Operators
I
Extensions
Examples of Fourth Order Problems
Bernoulli Beam (clamped)
d2 d2 u EI 2 = f
dx2
dx
a<x<b
u(a) = u0 (a) = 0 = u(b) = u0 (b)
u = vertical displacement,
f = vertical load density
E = Young’s modulus,
I = moment of inertia
1111
0000
u
f
111
000
Kirchhoff Plate (clamped)
∆
2Et3
∆u = f
3(1 − σ 2 )
u = ∂u/∂n = 0
in Ω
on ∂Ω
u = vertical displacement,
f = vertical load density
E = Young’s modulus,
σ = Poisson ratio
t = thickness of plate
11
00
00
11
00
11
111
000
000
111
u
f
11
00
00
11
00
11
111
000
000
111
Kirchhoff Plate (simply supported)
2Et3
∆
∆u = f
3(1 − σ 2 )
u = ∆u= 0
in Ω
on ∂Ω
Clamped Plate
u=
∂u
=0
∂n
on ∂Ω
Kirchhoff Plate (simply supported)
2Et3
∆
∆u = f
3(1 − σ 2 )
u = ∆u= 0
in Ω
on ∂Ω
Clamped Plate
u=
I
∂u
=0
∂n
on ∂Ω
Both boundary conditions for the clamped plate are essential boundary conditions, i.e., they are imposed on the
space for the variational (or weak) formulation of the problem.
Kirchhoff Plate (simply supported)
2Et3
∆
∆u = f
3(1 − σ 2 )
u = ∆u= 0
in Ω
on ∂Ω
Clamped Plate
u=
∂u
=0
∂n
on ∂Ω
I
Both boundary conditions for the clamped plate are essential boundary conditions, i.e., they are imposed on the
space for the variational (or weak) formulation of the problem.
I
The condition ∆u = 0 for the simply supported plate is a
natural boundary condition, i.e., it is implied by the variational (or weak) formulation of the problem (for sufficiently
smooth solutions).
Stokes Problem (2D)
−ν∆u + ∇p = f
in Ω
∇·u=0
in Ω
u=0
Ω = a simply connected domain
u = fluid velocity,
f = force density,
p = pressure
ν = viscosity
on ∂Ω
Stokes Problem (2D)
−ν∆u + ∇p = f
in Ω
∇·u=0
in Ω
u=0
Ω = a simply connected domain
Stream Function ψ
∇×ψ =u
on ∂Ω
Stokes Problem (2D)
−ν∆u + ∇p = f
in Ω
∇·u=0
in Ω
u=0
on ∂Ω
Ω = a simply connected domain
Stream Function ψ
∇×ψ =u
Boundary value problem for ψ
ν∆2 ψ = ∇ × f
ψ = ∂ψ/∂n = 0
(after simplification)
in
Ω
on ∂Ω
Strain Gradient Elasticity
−div σ = f
in Ω ⊂ Rd (d = 2, 3)
σ = 2µ ∗ (u) + λ [tr ∗ (u)]I
∗ (u) = (1 − γ 2 ∆)(u)
1
(u) = (∇ + ∇T )u
2
u = displacement of an elastic material
σ = stress, f = force density
µ and λ are the Lamé constants.
(u) = standard strain, ∗ (u) = modified strain
γ is a parameter.
Strain Gradient Elasticity
−div σ = f
in Ω ⊂ Rd (d = 2, 3)
σ = 2µ ∗ (u) + λ [tr ∗ (u)]I
∗ (u) = (1 − γ 2 ∆)(u)
1
(u) = (∇ + ∇T )u
2
I
γ=0:
standard elasticity system
I
γ 6= 0 :
fourth order system
(∗ (u) = (u))
Strain Gradient Elasticity
−div σ = f
in Ω ⊂ Rd (d = 2, 3)
σ = 2µ ∗ (u) + λ [tr ∗ (u)]I
∗ (u) = (1 − γ 2 ∆)(u)
1
(u) = (∇ + ∇T )u
2
I
γ=0:
standard elasticity system
(∗ (u) = (u))
I
γ 6= 0 :
fourth order system
I
Strain gradient elasticity is a phenomenological theory
for capturing the scale effect and localization due to
non-homogeneity at the microscopic level.
Cahn-Hilliard Equation
∂c
= ∇ · β(c)∇(µ(c) − ∆c)
∂t
in
Ω × (0, T)
plus initial and boundary conditions
Ω ⊂ Rd (d = 1, 2, 3)
c(x, t) = concentration of one of the two substances being
tracked (0 ≤ c ≤ 1)
β(c) = mobility
µ(c) = derivative of the free energy
Cahn-Hilliard Equation
∂c
= ∇ · β(c)∇(µ(c) − ∆c)
∂t
I
phase segregation of binary alloys
I
two-phase fluid flow
I
image processing
I
self-assembly of nanovoids
I
planet formation
in
Ω × (0, T)
Cahn-Hilliard Equation
∂c
= ∇ · β(c)∇(µ(c) − ∆c)
∂t
in
Ω × (0, T)
This can be solved numerically by an implicit time discretization
combined with the Newton-Raphson scheme for a fourth order
nonlinear problem at each time step.
An Obstacle Problem
Ω = bounded polygon
ψ ∈ C2 (Ω̄),
f ∈ L2 (Ω)
ψ < 0 on ∂Ω
(external force)
(obstacle function)
K = {v ∈ H02 (Ω) : v ≥ ψ on Ω}
(v ∈ H02 (Ω) if and only if ∂ α u/∂xα ∈ L2 (Ω) for |α| ≤ 2 and
v = ∂v/∂n = 0 on ∂Ω.)
An Obstacle Problem
Ω = bounded polygon
ψ ∈ C2 (Ω̄),
f ∈ L2 (Ω)
ψ < 0 on ∂Ω
(external force)
(obstacle function)
K = {v ∈ H02 (Ω) : v ≥ ψ on Ω}
Find u = argmin
v∈K
Z
a(w, v) =
h1
2
i
a(v, v) − (f , v)
D2 w : D2 v dx
Ω
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Z
(f , v) =
f v dx
Ω
(bending of a clamped Kirchhoff plate over an obstacle)
An Obstacle Problem
plate
PSfrag replacements u
obstacle
u is the vertical displacement of the midsurface of the thin plate.
1
a(v, v) − (f , v)
2
is the energy of the plate determined by the displacement v.
An Obstacle Problem
plate
PSfrag replacements u
obstacle
u is the vertical displacement of the midsurface of the thin plate.
u = argmin
v∈K
h1
2
a(v, v) − (f , v)
i
The obstacle problem is to find the plate that has minimum energy among all admissible plates.
An Obstacle Problem
plate
PSfrag replacements u
obstacle
u is the vertical displacement of the midsurface of the thin plate.
Variational Inequality
a(u, v − u) ≥ (f , v − u)
∀v ∈ K
An Optimal Control Problem with State Constraint
minimize
over
subject to
1
β
ky − yd k2L2 (Ω) + kuk2L2 (Ω)
2
2
(y, u) ∈ H01 (Ω) × L2 (Ω)
(
−∆y
y
= u in Ω
≤ ψ a.e. in Ω
y ∈ H01 (Ω) is the state (temperature distribution).
yd is the desired state.
y ≤ ψ is a pointwise constraint on the state.
u ∈ L2 (Ω) is the control (heat source).
β > 0 is related to the cost for implementing the control u.
An Optimal Control Problem with State Constraint
minimize
over
subject to
1
β
ky − yd k2L2 (Ω) + kuk2L2 (Ω)
2
2
(y, u) ∈ H01 (Ω) × L2 (Ω)
(
−∆y
y
= u in Ω
≤ ψ a.e. in Ω
If Ω is convex or smooth, then y belongs to H 2 (Ω) by elliptic
regularity and we can rewrite the problem as
i
1h
Find y = argmin ky − yd k2L2 (Ω) + βk∆yk2L2 (Ω)
2
v∈K
K = {v ∈ H 2 (Ω) ∩ H01 (Ω) : v ≤ ψ
in Ω}
(a fourth order obstacle problem with the boundary conditions
of a simply supported plate)
Numerical Solutions for 4th Order Problems
Two main difficulties
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We need to discretize the fourth order problems so that the
solutions of the discrete problems are good approximations
of the solutions of the continuous problem.
Such schemes are usually complicated.
I
We want to solve the discrete problem accurately and efficiently.
This is difficult because the discrete system is very illconditioned.
Classical Finite Element Methods
Conforming Methods
Conforming Methods
Since the continuous problems are posed on the Sobolev space
H 2 (Ω), the finite element spaces of conforming methods must
be subspaces of H 2 (Ω), i.e., they are C1 finite element spaces.
The advantage of conforming methods is that they are always
convergent by Galerkin orthogonality. The disadvantage is that
they are complicated in 2D (and more so in 3D).
Conforming Methods
Since the continuous problems are posed on the Sobolev space
H 2 (Ω), the finite element spaces of conforming methods must
be subspaces of H 2 (Ω), i.e., they are C1 finite element spaces.
I
C1 continuity imposes many conditions on the vertices and
the edges of an element.
I
Need many dofs in order to satisfy all these conditions.
Conforming Methods
Argyris (TUBA) triangular elements
P5 = hx1m x2n : m + n ≤ 5i
u
(21 dof)
6 value of the normal derivative
value of the function
i values of the first order derivatives
k values of the second order derivatives
Conforming Methods
Bogner-Fox-Schmit rectangular elements
Q3 = hx1m xjn : m, n ≤ 3i
u
h
(16 dof)
value of the function
values of the first order derivatives
value of the mixed second order derivative
Conforming Methods
Macro elements
C1 piecewise cubic polynomials
(12 dof)
Nonconforming Methods
Nonconforming Methods
Nonconforming finite element methods were invented because
C1 finite element methods are too complicated. Nonconforming
methods are simpler since we only require the finite element
functions and their derivatives to satisfy some weak continuity
conditions.
Nonconforming Methods
Morley element
P2
(6 dof)
Nonconforming Methods
Incomplete biquadratic element
P2 + hx12 x2 , x1 x22 i
(8 dof)
Nonconforming Methods
I
It takes a lot of ingenuity to construct nonconforming finite
elements that work (especially for more complicated fourth
order problems).
I
They are only low order elements (no natural hierarchy),
which are not efficient for smooth solutions.
I
Very little is known about 3D nonconforming elements for
fourth order problems.
Mixed Finite Element Methods
Mixed Finite Element Methods
Find (ω, u) ∈ H 1 (Ω) × H01 (Ω) such that
Z
Z
ωµ dx −
∇µ · ∇u dx = 0
Ω
ZΩ
Z
∇ω · ∇v dx
=
f v dx
Ω
Ω
∀ µ ∈ H 1 (Ω)
∀ v ∈ H01 (Ω)
In this mixed formulation, the biharmonic problem with the
boundary conditions of clamped plates is split into two second
order problems, we only need to use finite element spaces that
are subspaces of H 1 (Ω), i.e., C0 elements.
Mixed Finite Element Methods
I
It is not easy to come up with correct mixed formulations
for more complicated fourth order problems.
Mixed Finite Element Methods
I
It is not easy to come up with correct mixed formulations
for more complicated fourth order problems.
I
In the mixed formulation we use a finite element space for
the unknown ω and a finite element space for the unknown
u. The mixed method only works if the finite element pair
satisfies the Ladyzhenskaya-Babuška-Brezzi condition.
Mixed Finite Element Methods
I
It is not easy to come up with correct mixed formulations
for more complicated fourth order problems.
I
In the mixed formulation we use a finite element space for
the unknown ω and a finite element space for the unknown
u. The mixed method only works if the finite element pair
satisfies the Ladyzhenskaya-Babuška-Brezzi condition.
It is not easy to find such finite element pairs!
Mixed Finite Element Methods
I
It is not easy to come up with correct mixed formulations
for more complicated fourth order problems.
I
In the mixed formulation we use a finite element space for
the unknown ω and a finite element space for the unknown
u. The mixed method only works if the finite element pair
satisfies the Ladyzhenskaya-Babuška-Brezzi condition.
I
For the biharmonic problem with boundary conditions of
simply supported plates, some mixed methods can miss
the leading singularity and produce a wrong solution
(Sapondzhyan paradox).
Mixed Finite Element Methods
I
It is not easy to come up with correct mixed formulations
for more complicated fourth order problems.
I
In the mixed formulation we use a finite element space for
the unknown ω and a finite element space for the unknown
u. The mixed method only works if the finite element pair
satisfies the Ladyzhenskaya-Babuška-Brezzi condition.
I
For the biharmonic problem with boundary conditions of
simply supported plates, some mixed methods can miss
the leading singularity and produce a wrong solution
(Sapondzhyan paradox).
I
At the end one still needs to solve a saddle point problem,
which is more complicated than solving an SPD problem.
C0 Interior Penalty Methods
C0 Interior Penalty Methods
The lowest order elements in this family are as simple as
the classical nonconforming finite elements.
C0 Interior Penalty Methods
The lowest order elements in this family are as simple as
the classical nonconforming finite elements.
We can compute smooth solutions using higher order elements in this family which are as efficient as higher order
C1 elements and much simpler (in 2D and 3D).
C0 Interior Penalty Methods
The lowest order elements in this family are as simple as
the classical nonconforming finite elements.
We can compute smooth solutions using higher order elements in this family which are as efficient as higher order
C1 elements and much simpler (in 2D and 3D).
Unlike mixed methods, this approach can be extended in
a straight-forward way to more complicated fourth order
problems (such as the fourth order elliptic systems that appear in strain-gradient elasticity problems).
C0 Interior Penalty Methods
The lowest order elements in this family are as simple as
the classical nonconforming finite elements.
We can compute smooth solutions using higher order elements in this family which are as efficient as higher order
C1 elements and much simpler (in 2D and 3D).
Unlike mixed methods, this approach can be extended in
a straight-forward way to more complicated fourth order
problems (such as the fourth order elliptic systems that appear in strain-gradient elasticity problems).
The C0 interior penalty methods belong to the class of discontinuous Galerkin methods (where the discontinuity involves the first order derivatives).
C0 Interior Penalty Methods
Biharmonic Equation
∆2 u = f
in Ω
with different boundary conditions
Ω = bounded polygonal domain in R2
∆=
∂2
∂2
+
∂x12 ∂x22
Biharmonic Equation
∆2 u = f
in Ω
with different boundary conditions
Ω = bounded polygonal domain in R2
∆=
∂2
∂2
+
∂x12 ∂x22
Boundary Conditions of Clamped Plates
u=0
∂u
=0
∂n
on ∂Ω
essential
on ∂Ω
essential
Biharmonic Equation
∆2 u = f
in Ω
with different boundary conditions
Ω = bounded polygonal domain in R2
∆=
∂2
∂2
+
∂x12 ∂x22
Boundary Conditions of Simply Supported Plates
u=0
on ∂Ω
essential
∆u = 0
on ∂Ω
natural
Biharmonic Equation
∆2 u = f
in Ω
with different boundary conditions
Ω = bounded polygonal domain in R2
∆=
∂2
∂2
+
∂x12 ∂x22
Boundary Conditions of the Cahn-Hilliard Type
∂u
=0
∂n
∂(∆u)
=0
∂n
on ∂Ω
essential
on ∂Ω
natural
Boundary Conditions of Clamped Plates
Ω = bounded polygonal domain
∆2 u = f
∂u
=0
u=
∂n
f ∈ L2 (Ω)
in Ω
on ∂Ω
Boundary Conditions of Clamped Plates
Ω = bounded polygonal domain
∆2 u = f
∂u
=0
u=
∂n
f ∈ L2 (Ω)
in Ω
on ∂Ω
Variational/Weak Problem
Find u ∈ H02 (Ω) such that
Z
a(u, v) =
f v dx
∀ v ∈ H02 (Ω)
Ω
Z
a(w, v) =
Ω
D2 w : D2 v dx
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
(u = 0 and ∂u/∂n = 0 are both essential boundary conditions
that are imposed on the solution space H02 (Ω).)
Boundary Conditions of Clamped Plates
Ω = bounded polygonal domain
∆2 u = f
∂u
=0
u=
∂n
f ∈ L2 (Ω)
in Ω
on ∂Ω
Variational/Weak Problem
Find u ∈ H02 (Ω) such that
Z
a(u, v) =
f v dx
∀ v ∈ H02 (Ω)
Ω
Z
a(w, v) =
Ω
D2 w : D2 v dx
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Since a(·, ·) is bounded and coercive on H02 (Ω), the variaitonal/weak problem has a unique solution.
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh (⊂ H01 (Ω)) = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh (⊂ H01 (Ω)) = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
We can also use Qk Lagrange finite element spaces.
k=2
k=3
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh (⊂ H01 (Ω)) = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
These are standard C0 finite element spaces for second order
problems.
Discrete Problem
Discrete Problem
Key Observation
The solution u of the continuous problem
Z
a(u, v) =
f v dx
∀ v ∈ H02 (Ω)
Ω
Z
a(w, v) =
D2 w : D2 v dx
Ω
f ∈ L2 (Ω)
also satisfies a mesh-dependent problem
Z
ah (u, v) =
f v dx
∀ v ∈ Vh
Ω
obtained by
I
integration by parts
I
symmetrization
I
stabilization
An Integration by Parts Formula
Z
Z
(∆w)z dx =
T
∂T
∂w
z ds −
∂n
Z
∇w · ∇z dx
T
An Integration by Parts Formula
Z
Z
(∆w)z dx =
∂T
T
Z
2
Z
(∆ u)v dx =
T
∂T
∂w
z ds −
∂n
∂(∆u)
v ds −
∂n
w = ∆u
Z
∇w · ∇z dx
T
Z
∇(∆u) · ∇v dx
T
and z = v
An Integration by Parts Formula
Z
Z
(∆w)z dx =
∂T
T
Z
Z
2
(∆ u)v dx =
Z∂T
T
=
∂T
∂w
z ds −
∂n
Z
∇w · ∇z dx
T
Z
∂(∆u)
v ds − ∇(∆u) · ∇v dx
∂n
ZT
∂(∆u)
v ds −
∆(∇u) · ∇v dx
∂n
T
An Integration by Parts Formula
Z
Z
(∆w)z dx =
∂T
T
∂w
z ds −
∂n
Z
∇w · ∇z dx
T
Z
∂(∆u)
v ds − ∇(∆u) · ∇v dx
(∆ u)v dx =
∂n
T
∂T
Z
ZT
∂(∆u)
=
v ds −
∆(∇u) · ∇v dx
∂n
∂T
T
Z Z
∂
∂(∆u)
v ds −
∇u · ∇v ds
=
∂n
∂T ∂n
∂T
Z
+ ∇(∇u) : ∇(∇v) dx
Z
2
Z
T
w = ∇u and z = ∇v
An Integration by Parts Formula
Z
Z
(∆w)z dx =
∂T
T
∂w
z ds −
∂n
Z
∇w · ∇z dx
T
Z
∂(∆u)
v ds − ∇(∆u) · ∇v dx
(∆ u)v dx =
∂n
T
∂T
Z
ZT
∂(∆u)
=
v ds −
∆(∇u) · ∇v dx
∂n
∂T
T
Z Z
∂
∂(∆u)
v ds −
∇u · ∇v ds
=
∂n
∂T ∂n
∂T
Z
+ ∇(∇u) : ∇(∇v) dx
Z
2
Z
T
∆2 u = f
∇(∇u) = D2 u
∇(∇v) = D2 v
An Integration by Parts Formula
Z
Z
(∆w)z dx =
∂T
T
∂w
z ds −
∂n
Z
∇w · ∇z dx
T
Z
∂(∆u)
v ds − ∇(∆u) · ∇v dx
(∆ u)v dx =
∂n
T
∂T
Z
ZT
∂(∆u)
=
v ds −
∆(∇u) · ∇v dx
∂n
∂T
T
Z Z
∂
∂(∆u)
v ds −
∇u · ∇v ds
=
∂n
∂T ∂n
∂T
Z
+ ∇(∇u) : ∇(∇v) dx
Z
2
Z
T
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds + D2 u : D2 v dx
∂T ∂n
T
Z
Notation
Let v be a piecewise H 2 function.
e = interior edge shared by T±
hh ∂ v ii
∂n
=
∂v +
∂n
−
T+
∂v −
n
∂n
The unit normal n points from T− to T+ .
v± = v
T_
e
T±
The definition of [[∂v/∂n]] is independent of the choice of T± .
Notation
Let v be a piecewise H 2 function.
e = interior edge shared by T±
hh ∂ v ii
∂n
=
∂v +
∂n
−
T+
∂v −
n
∂n
The unit normal n points from T− to T+ .
v± = v
e
T_
T±
e = boundary edge, e ⊂ ∂T
hh ∂v ii
∂n
=−
∂ vT
∂n
The unit normal n points outside Ω.
vT = vT
T
e
n
Notation
Let v be a piecewise H s function for some s > 25 .
e = interior edge shared by T±
nn ∂ 2 v oo
∂n2
T+
1 h ∂ 2 v+ ∂ 2 v− i
=
+
2
∂n2
∂n2
The unit normal n points from T− to T+ .
v± = v
n
T_
e
T±
The definition of {{∂ 2 v/∂n2 }} is independent of the choice of
T± .
Notation
Let v be a piecewise H s function for some s > 25 .
e = interior edge shared by T±
nn ∂ 2 v oo
∂n2
T+
1 h ∂ 2 v+ ∂ 2 v− i
=
+
2
∂n2
∂n2
The unit normal n points from T− to T+ .
v± = v
n
e
T_
T±
e = boundary edge, e ⊂ ∂T
nn ∂ 2 v oo
∂n2
=
∂ 2 vT
∂n2
The unit normal n points outside Ω.
vT = v
T
T
e
n
A Mesh-Dependent Problem for u
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂T ∂n
T
Z
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂T ∂n
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
∂T
T∈Th
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂T ∂n
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
∂T
T∈Th
Each interior edge is shared by two triangles. Since v and the
derivatives of u are continuous and the normals from the two
triangles are pointing at opposite directions, the contributions
from the two triangles cancel.
There is also no contribution from the boundary edges because
v = 0 on ∂Ω.
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂T ∂n
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
Eh = the set of all the edges in Th
e∈Eh
A Mesh-Dependent Problem for u
Z
Z
f v dx =
∂T
T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂T ∂n
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
Z
f v dx =
Ω
e∈Eh
X Z ∂ 2 u hh ∂v ii
ds
D u : D v dx +
2
∂n
T
e ∂n
XZ
T∈Th
2
2
e∈Eh
A Mesh-Dependent Problem for u
Z
f v dx =
Ω
X Z ∂ 2 u hh ∂v ii
D u : D v dx +
ds
2
∂n
T
e ∂n
XZ
T∈Th
2
2
e∈Eh
A Mesh-Dependent Problem for u
X Z ∂ 2 u hh ∂v ii
f v dx =
D u : D v dx +
ds
∂n2 ∂n
Ω
T∈Th T
e∈Eh e
n ∂ 2 u oohh ∂v ii
XZ
XZ n
2
2
ds
=
D u : D v dx +
∂n2
∂n
T
e
Z
XZ
T∈Th
2
2
e∈Eh
Assuming that u is sufficiently smooth the trace of ∂ 2 u/∂n2
from the two sides of e are identical.
A Mesh-Dependent Problem for u
X Z ∂ 2 u hh ∂v ii
f v dx =
D u : D v dx +
ds
∂n2 ∂n
Ω
T∈Th T
e∈Eh e
n ∂ 2 u oohh ∂v ii
XZ
XZ n
2
2
ds
=
D u : D v dx +
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
o hh ∂u ii
XZ n
ds
+
∂n2 | ∂n
{z }
e∈Eh e
Z
XZ
2
2
=0
symmetrization
A Mesh-Dependent Problem for u
X Z ∂ 2 u hh ∂v ii
f v dx =
D u : D v dx +
ds
∂n2 ∂n
Ω
T∈Th T
e∈Eh e
n ∂ 2 u oohh ∂v ii
XZ
XZ n
2
2
ds
=
D u : D v dx +
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂u ii
XZ n
ds
+
∂n2
∂n
e∈Eh e
X 1 Z hh ∂u ii hh ∂v ii
+σ
ds
|e| e | ∂n
∂n
{z
}
e∈Eh
Z
XZ
2
2
=0
stabilization
|e| is the length of e.
σ > 0 is a penalty parameter.
A Mesh-Dependent Problem for u
The solution u of the continuous problem satisfies
Z
ah (u, v) =
f v dx
∀ v ∈ Vh
Ω
n ∂ 2 w oohh ∂v ii
XZ n
ah (w, v) =
D w : D v dx +
ds
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
+
ds
∂n2
∂n
e
e∈Eh
X 1 Z hh ∂w iihh ∂v ii
ds
+σ
|e| e ∂n
∂n
XZ
2
2
e∈Eh
Eh = set of edges
|e| = length of e
{{·}} = average
[[·]] = jump
σ = penalty parameter
Discrete Problem
Find uh ∈ Vh such that
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
n ∂ 2 w oohh ∂v ii
XZ n
ds
ah (w, v) =
D w : D v dx +
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
+
ds
∂n2
∂n
e
e∈Eh
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
XZ
2
2
e∈Eh
Eh = set of edges
|e| = length of e
{{·}} = average
[[·]] = jump
σ = penalty parameter
Discrete Problem
Find uh ∈ Vh such that
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
Galerkin Orthogonality
Z
ah (u, v) =
fv dx
∀ v ∈ Vh
fv dx
∀ v ∈ Vh
Ω
Z
ah (uh , v) =
Ω
ah (u − uh , v) = 0
∀ v ∈ Vh
Discrete Problem
This is an interior penalty method obtained through integration
by parts, symmetrization and stabilization.
Discrete Problem
This is an interior penalty method obtained through integration
by parts, symmetrization and stabilization.
ah (w, v) =
XZ
T∈Th
D2 w : D2 v dx
T
|
{z
}
piecewise version of continuous variational form
(ibp)
ohh ∂v ii
X Z nn ∂ 2 w o
+
ds
∂n2
∂n
e
e∈Eh
|
{z
}
consistency
(ibp)
ohh ∂w ii
X Z nn ∂ 2 v o
X 1 Z hh ∂w ii hh ∂v ii
+
ds + σ
ds
∂n2
∂n
|e| e ∂n
∂n
e∈Eh e
e∈Eh
|
{z
} |
{z
}
symmetrization
stabilization
Discrete Problem
Since the finite element functions are globally continuous, this
is a C0 interior penalty method. It is a discontinuous Galerkin
method for fourth order problems, where the discontinuity is in
the normal derivative across element boundaries. The discrete
problem is a SPD problem when the penalty parameter is sufficiently large. Therefore it preserves the SPD property of the
continuous problem.
Discrete Problem
Since the finite element functions are globally continuous, this
is a C0 interior penalty method. It is a discontinuous Galerkin
method for fourth order problems, where the discontinuity is in
the normal derivative across element boundaries. The discrete
problem is a SPD problem when the penalty parameter is sufficiently large. Therefore it preserves the SPD property of the
continuous problem.
Since Vh is not a subspace of H 2 (Ω), this is a nonconforming
method. The second essential boundary condition (∂u/∂n) = 0
is also not imposed on the finite element space Vh .
Discrete Problem
Since the finite element functions are globally continuous, this
is a C0 interior penalty method. It is a discontinuous Galerkin
method for fourth order problems, where the discontinuity is in
the normal derivative across element boundaries. The discrete
problem is a SPD problem when the penalty parameter is sufficiently large. Therefore it preserves the SPD property of the
continuous problem.
Since Vh is not a subspace of H 2 (Ω), this is a nonconforming
method. The second essential boundary condition (∂u/∂n) = 0
is also not imposed on the finite element space Vh .
These conditions are enforced by the penalty term
X 1 Z hh ∂w ii hh ∂v ii
ds
σ
|e| e ∂n
∂n
e∈Eh
as h ↓ 0.
Discrete Problem
Since the finite element functions are globally continuous, this
is a C0 interior penalty method. It is a discontinuous Galerkin
method for fourth order problems, where the discontinuity is in
the normal derivative across element boundaries. The discrete
problem is a SPD problem when the penalty parameter is sufficiently large. Therefore it preserves the SPD property of the
continuous problem.
The study of discontinuous Galerkin methods for higher order
problems was initiated in a paper by Baker.
Reference
Baker
Finite element methods for elliptic equations using nonconforming elements
Math. Comp. 1977
Discrete Problem
Since the finite element functions are globally continuous, this
is a C0 interior penalty method. It is a discontinuous Galerkin
method for fourth order problems, where the discontinuity is in
the normal derivative across element boundaries. The discrete
problem is a SPD problem when the penalty parameter is sufficiently large. Therefore it preserves the SPD property of the
continuous problem.
Reference
Engel, Garikipati, Hughes, Larson, Mazzei, and Taylor
Continuous/discontinuous finite element approximations of fourth order elliptic problems in structural and continuum mechanics with applications to
thin beams and plates, and strain gradient elasticity
Comput. Methods Appl. Mech. Engrg. 2002
Coercivity of ah (·, ·)
n ∂ 2 w oohh ∂v ii
XZ n
ah (w, v) =
D w : D v dx +
ds
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
ds
+
∂n2
∂n
e∈Eh e
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
XZ
2
2
e∈Eh
Coercivity of ah (·, ·)
n ∂ 2 w oohh ∂v ii
XZ n
ah (w, v) =
D w : D v dx +
ds
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
ds
+
∂n2
∂n
e∈Eh e
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
e∈Eh
n ∂ 2 v oohh ∂v ii
XZ
XZ n
2 2
ds
ah (v, v) =
|D v| dx + 2
∂n2
∂n
T∈Th T
e∈Eh e
hh ii
X 1
∂v 2
+σ
|e| ∂n L2 (e)
XZ
2
2
e∈Eh
Coercivity of ah (·, ·)
n ∂ 2 w oohh ∂v ii
XZ n
ah (w, v) =
D w : D v dx +
ds
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
ds
+
∂n2
∂n
e∈Eh e
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
e∈Eh
n ∂ 2 v oohh ∂v ii
XZ
XZ n
2 2
ds
ah (v, v) =
|D v| dx + 2
∂n2
∂n
T∈Th T
e∈Eh e
hh ii
X 1
∂v 2
+σ
|e| ∂n L2 (e)
XZ
2
2
e∈Eh
A Mesh-Dependent Norm
hh ii
X
X 1
∂v 2
kvk2h =
|v|2H 2 (T) + σ
|e| ∂n L2 (e)
T∈Th
e∈Eh
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
∀ v ∈ Vh
Consequently ah (·, ·) is SPD on Vh and hence the discrete
problem is well-posed.
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
Proof.
∀ v ∈ Vh
n ∂ 2 v oohh ∂v ii
XZ n
ah (v, v) =
+2
ds
∂n2
∂n
T∈Th
e∈Eh e
hh ii
X 1
∂v 2
+σ
|e| ∂n L2 (e)
e∈Eh
n ∂2v o
ohh ∂v ii
XZ n
= kvk2h + 2
ds
∂n2
∂n
e
X
|v|2H 2 (T)
e∈Eh
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
Proof.
∀ v ∈ Vh
n ∂ 2 v oohh ∂v ii
XZ n
ah (v, v) =
+2
ds
∂n2
∂n
T∈Th
e∈Eh e
hh ii
X 1
∂v 2
+σ
|e| ∂n L2 (e)
e∈Eh
n ∂2v o
ohh ∂v ii
XZ n
= kvk2h + 2
ds
∂n2
∂n
e
X
|v|2H 2 (T)
e∈Eh
It suffices to show that
XZ n
ohh ∂v ii 1
n ∂2v o
ds ≤ kvk2h
2
2
∂n
∂n
2
e
e∈Eh
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
Z Z
1
1 Z
2
2
g2
f2
fg ≤
X
X 1 X 1
2
2
b2n
an bn ≤
a2n
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
X
1 X
1
2
2
|v|2H 2 (T)
|e|−1 k[[∂v/∂n]]k2L2 (e)
≤2 C
T∈Th
e∈Eh
The constant C depends only on the shape regularity of Th
and the degree of the polynomials in Vh .
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
X
1 X
1
2
2
|v|2H 2 (T)
|e|−1 k[[∂v/∂n]]k2L2 (e)
≤2 C
T∈Th
≤ δC
X
T∈Th
e∈Eh
1 X −1
|v|2H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e)
δ
e∈Eh
1
2ab ≤ δa2 + b2
δ
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
X
1 X
1
2
2
|v|2H 2 (T)
|e|−1 k[[∂v/∂n]]k2L2 (e)
≤2 C
T∈Th
≤ δC
X
T∈Th
e∈Eh
1 X −1
|v|2H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e)
δ
e∈Eh
X
1 X 2
=
|e|−1 k[[∂v/∂n]]k2L2 (e)
|v|H 2 (T) + 2C
2
T∈Th
e∈Eh
δ=
1
2C
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
X
1 X
1
2
2
|v|2H 2 (T)
|e|−1 k[[∂v/∂n]]k2L2 (e)
≤2 C
T∈Th
≤ δC
X
T∈Th
e∈Eh
1 X −1
|v|2H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e)
δ
e∈Eh
X
1 X 2
=
|e|−1 k[[∂v/∂n]]k2L2 (e)
|v|H 2 (T) + 2C
2
T∈Th
e∈Eh
1 X 2
σ X −1
≤
|v|H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e)
2
2
T∈Th
e∈Eh
σ ≥ 4C
Coercivity of ah (·, ·)
X Z nn ∂ 2 v oohh ∂v ii 2
ds
2
∂n
∂n
e
e∈Eh
X
1 X
1
2
2
2
2
2
−1
2
≤2
|e|k{{∂ v/∂n }}kL2 (e)
|e| k[[∂v/∂n]]kL2 (e)
e∈Eh
e∈Eh
X
1 X
1
2
2
|v|2H 2 (T)
|e|−1 k[[∂v/∂n]]k2L2 (e)
≤2 C
T∈Th
≤ δC
X
T∈Th
e∈Eh
1 X −1
|v|2H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e)
δ
e∈Eh
X
1 X 2
=
|e|−1 k[[∂v/∂n]]k2L2 (e)
|v|H 2 (T) + 2C
2
T∈Th
e∈Eh
1 X 2
σ X −1
1
≤
|v|H 2 (T) +
|e| k[[∂v/∂n]]k2L2 (e) = kvk2h
2
2
2
T∈Th
e∈Eh
Boundary Conditions of Simply Supported Plates
Ω = bounded polygonal domain
∆2 u = f
u = ∆u = 0
f ∈ L2 (Ω)
in Ω
on ∂Ω
Boundary Conditions of Simply Supported Plates
Ω = bounded polygonal domain
∆2 u = f
u = ∆u = 0
f ∈ L2 (Ω)
in Ω
on ∂Ω
Variational/Weak Problem Find u ∈ H 2 (Ω) ∩ H01 (Ω) such that
Z
a(u, v) =
f v dx
∀ v ∈ H 2 (Ω) ∩ H01 (Ω)
Ω
Z
a(w, v) =
Ω
D2 w : D2 v dx
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
(u = 0 is an essential boundary condition that is imposed on
the solution space H 2 (Ω)∩H01 (Ω), whereas ∆u = 0 is a natural
boundary condition.)
Boundary Conditions of Simply Supported Plates
Ω = bounded polygonal domain
∆2 u = f
u = ∆u = 0
f ∈ L2 (Ω)
in Ω
on ∂Ω
Variational/Weak Problem Find u ∈ H 2 (Ω) ∩ H01 (Ω) such that
Z
a(u, v) =
f v dx
∀ v ∈ H 2 (Ω) ∩ H01 (Ω)
Ω
Z
a(w, v) =
Ω
D2 w : D2 v dx
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Since a(·, ·) is bounded and coercive on H 2 (Ω) ∩ H01 (Ω), the
variational/weak problem has a unique solution.
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh (⊂ H01 (Ω)) = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
We can also use Qk Lagrange finite element spaces.
k=2
k=3
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh (⊂ H01 (Ω)) = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
These are the same finite element spaces for the clamped
plates.
A Mesh-Dependent Problem for u
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
∂T
T∈Th
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
(Eh = the set of all the edges in Th )
e∈Eh
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
e∈Eh
On any edge e ⊂ Ω
0 = ∆u
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
e∈Eh
On any edge e ⊂ Ω
0 = ∆u =
∂2u ∂2u
+ 2
∂n2
∂t
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
T∈Th
e∈Eh
On any edge e ⊂ Ω
0 = ∆u =
∂2u ∂2u
∂2u
+
=
∂n2 |{z}
∂t2
∂n2
=0
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
∂n
∂n2 ∂n
T∈Th ∂T
e∈Eh e
X Z ∂ 2 u hh ∂v ii
=
ds
2
∂n
e ∂n
i
e∈Eh
Ehi is the set of the edges of Th interior to Ω.
A Mesh-Dependent Problem for u
Z
Z
f v dx =
∂T
T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
X Z ∂(∆u)
v ds = 0
∂n
T∈Th ∂T
XZ ∂
X Z ∂ 2 u hh ∂v ii
−
∇u · ∇v ds =
ds
∂n
∂n2 ∂n
T∈Th ∂T
e∈Eh e
X Z ∂ 2 u hh ∂v ii
=
ds
2
∂n
e ∂n
i
e∈Eh
Z
f v dx =
Ω
XZ
T∈Th
T
D2 u : D2 v dx +
X Z ∂ 2 u hh ∂v ii
ds
2
∂n
e ∂n
i
e∈Eh
A Mesh-Dependent Problem for u
Z
f v dx =
Ω
XZ
T∈Th
T
D2 u : D2 v dx +
X Z ∂ 2 u hh ∂v ii
ds
2
∂n
e ∂n
i
e∈Eh
A Mesh-Dependent Problem for u
Z
f v dx =
Ω
XZ
T∈Th
=
T
X Z ∂ 2 u hh ∂v ii
ds
2
∂n
e ∂n
i
e∈Eh
n ∂ 2 u oohh ∂v ii
XZ n
ds
D u : D v dx +
∂n2
∂n
e
T
i
XZ
T∈Th
D2 u : D2 v dx +
2
2
e∈Eh
n
n ∂2u o
o
∂2u
=
∂n2
∂n2
A Mesh-Dependent Problem for u
Z
f v dx =
Ω
XZ
T∈Th
=
T
X Z ∂ 2 u hh ∂v ii
ds
2
∂n
e ∂n
i
e∈Eh
n ∂ 2 u oohh ∂v ii
XZ n
ds
D u : D v dx +
∂n2
∂n
e
T
i
XZ
T∈Th
D2 u : D2 v dx +
2
2
e∈Eh
n ∂2v o
o hh ∂u ii
XZ n
+
ds
∂n2 | ∂n
e
i
{z }
e∈Eh
=0
symmetrization
A Mesh-Dependent Problem for u
Z
f v dx =
Ω
XZ
T∈Th
=
T
X Z ∂ 2 u hh ∂v ii
ds
2
∂n
e ∂n
i
e∈Eh
n ∂ 2 u oohh ∂v ii
XZ n
ds
D u : D v dx +
∂n2
∂n
e
T
i
XZ
T∈Th
D2 u : D2 v dx +
2
2
e∈Eh
n ∂2v o
ohh ∂u ii
XZ n
+
ds
∂n2
∂n
e
e∈Ehi
X 1 Z hh ∂u ii hh ∂v ii
+σ
ds
|e| e | ∂n
∂n
i
{z
}
e∈Eh
=0
stabilization
A Mesh-Dependent Problem for u
The solution u of the continuous problem satisfies
Z
ah (u, v) =
f v dx
∀ v ∈ Vh
Ω
ah (w, v) =
n ∂ 2 w oohh ∂v ii
XZ n
D w : D v dx +
ds
∂n2
∂n
e
T
i
XZ
T∈Th
2
2
e∈Eh
n ∂2v o
ohh ∂w ii
XZ n
+
ds
∂n2
∂n
e
i
e∈Eh
X 1 Z hh ∂w iihh ∂v ii
ds
+σ
|e| e ∂n
∂n
i
e∈Eh
Ehi = set of interior edges
|e| = length of e
{{·}} = average
[[·]] = jump
σ = penalty parameter
Discrete Problem
Find uh ∈ Vh such that
Z
ah (u, v) =
f v dx
∀ v ∈ Vh
Ω
ah (w, v) =
n ∂ 2 w oohh ∂v ii
XZ n
ds
D w : D v dx +
∂n2
∂n
T
e
i
XZ
T∈Th
2
2
e∈Eh
n ∂2v o
ohh ∂w ii
XZ n
+
ds
∂n2
∂n
e
e∈Ehi
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
i
e∈Eh
Galerkin Orthogonality
ah (u − uh , v) = 0
∀ v ∈ Vh
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
∀ v ∈ Vh
where
kvk2h =
X
T∈Th
|v|2H 2 (T) + σ
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
i
e∈Eh
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
∀ v ∈ Vh
where
kvk2h =
X
T∈Th
|v|2H 2 (T) + σ
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
i
e∈Eh
Consequently ah (·, ·) is SPD on Vh and hence the discrete
problem is well-posed.
Boundary Conditions of the Cahn-Hilliard Type
Ω = bounded polygonal domain
f ∈ L2 (Ω)
∆2 u = f
∂u/∂n = ∂(∆u)/∂n = 0
in Ω
on ∂Ω
Boundary Conditions of the Cahn-Hilliard Type
Ω = bounded polygonal domain
f ∈ L2 (Ω)
∆2 u = f
∂u/∂n = ∂(∆u)/∂n = 0
in Ω
on ∂Ω
Variational/Weak Problem I Find u ∈ V such that
Z
a(u, v) =
f v dx
∀v ∈ V
Ω
Z
a(w, v) =
2
2
D w : D v dx
Ω
2
2
D w:D v=
2
X
wxi xj vxi xj
i,j=1
where
V = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω}
(∂u/∂n = 0 is an essential boundary condition imposed on the
solution space V.)
Boundary Conditions of the Cahn-Hilliard Type
The bilinear form a(·, ·) is not coercive on V because the constant function 1 belongs to V and a(1, 1) = 0.
V = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω}
Z
a(w, v) =
D2 w : D2 v dx
Ω
Boundary Conditions of the Cahn-Hilliard Type
The bilinear form a(·, ·) is not coercive on V because the constant function 1 belongs to V and a(1, 1) = 0.
The variational/weak problem I is solvable if and only if f satisfies
Z
Z
(†)
0=
f · 1 dx =
f dx
Ω
Ω
Boundary Conditions of the Cahn-Hilliard Type
The bilinear form a(·, ·) is not coercive on V because the constant function 1 belongs to V and a(1, 1) = 0.
The variational/weak problem I is solvable if and only if f satisfies
Z
Z
(†)
0=
f · 1 dx =
f dx
Ω
Ω
Under condition (†) the solution of the variational/weak problem
I is unique up to an additive constant. In particular it has a
unique solution in the subspace
V∗ = {v ∈ V : v(p∗ ) = 0}
where p∗ is a corner of Ω.
Boundary Conditions of the Cahn-Hilliard Type
Variational/Weak Problem II
Assume that f ∈ L2 (Ω) satisfies the constraint
Z
(†)
f dx = 0
Ω
Find u ∈ V∗ such that
Z
a(u, v) =
f v dx
∀ v ∈ V∗
Ω
where V∗ = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω and v(p∗ ) = 0}.
Boundary Conditions of the Cahn-Hilliard Type
Variational/Weak Problem II
Assume that f ∈ L2 (Ω) satisfies the constraint
Z
(†)
f dx = 0
Ω
Find u ∈ V∗ such that
Z
a(u, v) =
f v dx
∀ v ∈ V∗
Ω
where V∗ = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω and v(p∗ ) = 0}.
Since the bilinear form a(·, ·) is bounded and coercive on V∗ ,
the variational/weak problem II has a unique solution, which
is also a solution of the variational/weak problem I due to the
condition (†).
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
We can also use Qk Lagrange finite element spaces.
k=2
k=3
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
Note that no boundary condition is imposed on the finite element space because, as in the case of the clamped plates, the
essential boundary condition ∂u/∂n = 0 can be enforced by
the penalty term as h ↓ 0.
Finite Element Spaces
Th = a simplicial triangulation of Ω
Vh = Pk (k ≥ 2) Lagrange finite element space
k=2
k=3
Subspace Vh∗
Vh∗ = {v ∈ Vh : v(p∗ ) = 0}
where p∗ is a corner of Ω.
A Mesh-Dependent Problem for u
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
A Mesh-Dependent Problem for u
Z
Z
f v dx =
T
∂T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
XZ
T∈Th
∂T
∂(∆u)
v ds = 0
∂n
Each interior edge is shared by two triangles. Since v and the
derivatives of u are continuous and the normals from the two
triangles are pointing at opposite directions, the contributions
from the two triangles cancel.
There is also no contribution from the boundary edges because
∂(∆u)/∂n = 0 on ∂Ω.
A Mesh-Dependent Problem for u
Z
Z
f v dx =
∂T
T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
XZ
T∈Th
−
∂T
∂(∆u)
v ds = 0
∂n
∂
X Z ∂ 2 u hh ∂v ii
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
XZ
T∈Th
[[∂v/∂t]] = 0 over interior edges.
∂ ∂u = 0 on boundary edges.
∂t ∂n
e∈Eh
A Mesh-Dependent Problem for u
Z
Z
f v dx =
∂T
T
∂(∆u)
v ds −
∂n
Z
∂
∇u · ∇v ds+ D2 u : D2 v dx
∂n
∂T
T
Z
Summing up over all the triangles in Th ,
XZ
T∈Th
−
∂
X Z ∂ 2 u hh ∂v ii
∇u · ∇v ds =
ds
2
∂n
∂T ∂n
e ∂n
XZ
T∈Th
Z
f v dx =
Ω
∂T
∂(∆u)
v ds = 0
∂n
e∈Eh
X Z ∂ 2 u hh ∂v ii
D u : D v dx +
ds
2
∂n
T
e ∂n
XZ
T∈Th
2
2
e∈Eh
A Mesh-Dependent Problem for u
The solution u of the continuous problem satisfies
Z
ah (u, v) =
f v dx
∀ v ∈ Vh
Ω
n ∂ 2 w oohh ∂v ii
XZ n
ah (w, v) =
D w : D v dx +
ds
∂n2
∂n
T∈Th T
e∈Eh e
n ∂2v o
ohh ∂w ii
XZ n
+
ds
∂n2
∂n
e
e∈Eh
X 1 Z hh ∂w iihh ∂v ii
ds
+σ
|e| e ∂n
∂n
XZ
2
2
e∈Eh
Eh = set of edges
|e| = length of e
{{·}} = average
[[·]] = jump
σ = penalty parameter
Discrete Problem
Find uh ∈ Vh∗ such that
(Vh∗ = {v ∈ Vh : v(p∗ ) = 0})
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh∗
Ω
n ∂ 2 w oohh ∂v ii
XZ n
D w : D v dx +
ah (w, v) =
ds
∂n2
∂n
e∈Eh e
T∈Th T
ohh ∂w ii
n ∂2v o
XZ n
ds
+
∂n2
∂n
e
e∈Eh
X 1 Z hh ∂w iihh ∂v ii
+σ
ds
|e| e ∂n
∂n
XZ
2
2
e∈Eh
Galerkin Orthogonality
ah (u − uh , v) = 0
∀ v ∈ Vh∗
Coercivity of ah (·, ·)
Lemma
For σ sufficiently large, we have
1
ah (v, v) ≥ kvk2h
2
∀ v ∈ Vh∗
where
kvk2h =
X
T∈Th
|v|2H 2 (T) + σ
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
e∈Eh
Consequently ah (·, ·) is SPD on Vh∗ and hence the discrete
problem is well-posed.
Summary
I
The C0 interior penalty method for the biharmonic equation
with the boundary conditions of simply supported plates
share the same finite element spaces with the C0 interior penalty method for the biharmonic equation with the
boundary conditions of clamped plates, but uses a slightly
different bilinear form (Ehi instead of Eh ).
I
The C0 interior penalty method for the biharmonic equation with the boundary conditions of the Cahn-Hilliard type
shares the same bilinear form with the C0 interior penalty
method for the biharmonic equation with the boundary conditions of clamped plates, but uses a slightly different finite
element space (no boundary condition + vanishing at a
corner).
Enriching Operators
Enriching Operators
Enriching Operators
Clamped Plates
Solution space for the continuous problem is H02 (Ω).
Solution space for the discrete problem is a Pk (or Qk ) Lagrange
finite element space Vh ⊂ H01 (Ω).
Enriching Operators
Clamped Plates
Solution space for the continuous problem is H02 (Ω).
Solution space for the discrete problem is a Pk (or Qk ) Lagrange
finite element space Vh ⊂ H01 (Ω).
The goal is to construct an embedding (i.e., one-to-one map)
Eh from Vh into H02 (Ω) so that
|v−Eh v|H 2 (Ω;Th ) =
X
T∈Th
|v−Eh v|2H 2 (T) +
hh
ii
1
X 1
2
∂(v − Eh v) 2
|e|
∂n
L2 (e)
e∈Eh
provides a measure for the distance between v ∈ Vh and
H02 (Ω) with respect to | · |H 2 (Ω;Th ) .
(The piecewise H 2 (semi-) norm | · |H 2 (Ω;Th ) is identical to the
mesh-dependent norm k · kh when σ = 1.)
C1 Relatives
C1 Relatives
We say that a C1 element is a (rich) relative of a Lagrange
element if
they have the same element domain,
the shape functions of the Lagrange finite element are also
shape functions for the C1 element,
the dofs for the Lagrange element are also dofs for the C1
element.
C1 Relatives
We say that a C1 element is a (rich) relative of a Lagrange
element if
they have the same element domain,
the shape functions of the Lagrange finite element are also
shape functions for the C1 element,
the dofs for the Lagrange element are also dofs for the C1
element.
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000000000000000000000
C1 Relatives
We say that a C1 element is a (rich) relative of a Lagrange
element if
they have the same element domain,
the shape functions of the Lagrange finite element are also
shape functions for the C1 element,
the dofs for the Lagrange element are also dofs for the C1
element.
Averaging
Averaging
Vh = Q2 Lagrange Finite Element Space
Eh : Vh −→ Wh ⊂ H02 (Ω)
where Wh is the Q4 BFS finite element space.
Averaging
Vh = Q2 Lagrange Finite Element Space
Eh : Vh −→ Wh ⊂ H02 (Ω)
where Wh is the Q4 BFS finite element space.
Let v ∈ Vh . Since the dofs of Eh v along the boundary of Ω
must vanish by the condition that Eh v ∈ H02 (Ω), we only need to
specify the dofs at the nodes interior to Ω.
Averaging
Let N be a dof for the Q4 BFS finite element space associated
with the interior node p. We define
N(Eh v) =
1 X
N(vT )
|Tp |
T∈Tp
where Tp is the set of elements that share the node p and vT
is the restriction of v to T.
Averaging
Let N be a dof for the Q4 BFS finite element space associated
with the interior node p. We define
N(Eh v) =
1 X
N(vT )
|Tp |
T∈Tp
where Tp is the set of elements that share the node p and vT
is the restriction of v to T.
Eh v (p) = v(p)
Averaging
Let N be a dof for the Q4 BFS finite element space associated
with the interior node p. We define
N(Eh v) =
1 X
N(vT )
|Tp |
T∈Tp
where Tp is the set of elements that share the node p and vT
is the restriction of v to T.
4
1X
∇(Eh v) (p) =
∇vi (p)
4
i=1
Averaging
Let N be a dof for the Q4 BFS finite element space associated
with the interior node p. We define
N(Eh v) =
1 X
N(vT )
|Tp |
T∈Tp
where Tp is the set of elements that share the node p and vT
is the restriction of v to T.
2
∂(E v) 1 X ∂vi h
(p) =
(p)
∂n
2
∂n
i=1
Averaging
Let N be a dof for the Q4 BFS finite element space associated
with the interior node p. We define
N(Eh v) =
1 X
N(vT )
|Tp |
T∈Tp
where Tp is the set of elements that share the node p and vT
is the restriction of v to T.
4
∂ 2 (E v) 1 X ∂ 2 vi h
(p) =
(p)
∂x1 ∂x2
4
∂x1 ∂x2
i=1
Properties of Eh
Properties of Eh
Eh is one-to-one.
This is due to the fact that Eh v equals v at all the nodes for the
Q2 Lagrange finite element space.
Properties of Eh
Eh is one-to-one.
This is due to the fact that Eh v equals v at all the nodes for the
Q2 Lagrange finite element space.
Let Πh : H02 (Ω) −→ Vh be the nodal interpolation operator defined by
(Πh w)(p) = w(p)
where p is any node for the Q2 Lagrange finite element space.
Then Πh is a left inverse for Eh , i.e.
Πh ◦ Eh = IdVh
[Πh (Eh v)](p) = (Eh v)(p) = v(p)
∀ v ∈ Vh
Properties of Eh
Estimates for Eh v
|v − Eh v|2H 2 (Ω;Th ) +
≤
i
Xh
2
−2
2
h−4
T kv − Eh vkL2 (T) + hT |v − Eh v|H 1 (T)
T∈Th
2
C|v|H 2 (Ω;Th )
∀ v ∈ Vh
Properties of Eh
Estimates for Eh v
|v − Eh v|2H 2 (Ω;Th ) +
≤
i
Xh
2
−2
2
h−4
T kv − Eh vkL2 (T) + hT |v − Eh v|H 1 (T)
T∈Th
2
C|v|H 2 (Ω;Th )
∀ v ∈ Vh
The derivation of these estimates uses the fact that all norms on
a finite dimensional vector space are equivalent together with
scaling arguments.
Properties of Eh
Estimates for Eh v
|v − Eh v|2H 2 (Ω;Th ) +
≤
i
Xh
2
−2
2
h−4
T kv − Eh vkL2 (T) + hT |v − Eh v|H 1 (T)
T∈Th
2
C|v|H 2 (Ω;Th )
∀ v ∈ Vh
The derivation of these estimates uses the fact that all norms on
a finite dimensional vector space are equivalent together with
scaling arguments.
Corollary
|Eh v|H 2 (Ω) = |Eh v|H 2 (Ω;Th )
≤ |Eh v − v|H 2 (Ω;Th ) + |v|H 2 (Ω;Th )
≤ C|v|H 2 (Ω;Th )
∀ v ∈ Vh
Properties of Eh
Estimates for Eh ◦ Πh : H02 (Ω) −→ H02 (Ω)
(Πh ◦ Eh = IdVh )
Xh
2
hT−4 kζ − Eh Πh ζk2L2 (T) + h−2
T |ζ − Eh Πh ζ|H 1 (T)
T∈Th
i
+ |ζ − Eh Πh ζ|2H 2 (T) ≤ C|ζ|2H 2 (Ω)
∀ ζ ∈ H02 (Ω)
Xh
T∈Th
2
hT−6 kζ − Eh Πh ζk2L2 (T) + h−4
T |ζ − Eh Πh ζ|H 1 (T)
i
2
2
+ h−2
|ζ
−
E
Π
ζ|
∀ ζ ∈ H 3 (Ω) ∩ H02 (Ω)
2
h
h
T
H (T) ≤ C|ζ|H 3 (Ω)
Properties of Eh
Estimates for Eh ◦ Πh : H02 (Ω) −→ H02 (Ω)
(Πh ◦ Eh = IdVh )
Xh
2
hT−4 kζ − Eh Πh ζk2L2 (T) + h−2
T |ζ − Eh Πh ζ|H 1 (T)
T∈Th
i
+ |ζ − Eh Πh ζ|2H 2 (T) ≤ C|ζ|2H 2 (Ω)
∀ ζ ∈ H02 (Ω)
Xh
T∈Th
2
hT−6 kζ − Eh Πh ζk2L2 (T) + h−4
T |ζ − Eh Πh ζ|H 1 (T)
i
2
2
+ h−2
|ζ
−
E
Π
ζ|
∀ ζ ∈ H 3 (Ω) ∩ H02 (Ω)
2
h
h
T
H (T) ≤ C|ζ|H 3 (Ω)
The derivation of the first set of estimates uses standard interpolation estimates for Πh and the estimates for Eh .
The derivation of the second set of estimates uses the BrambleHilbert lemma.
Properties of Eh
Estimates for Eh ◦ Πh : H02 (Ω) −→ H02 (Ω)
(Πh ◦ Eh = IdVh )
Xh
2
hT−4 kζ − Eh Πh ζk2L2 (T) + h−2
T |ζ − Eh Πh ζ|H 1 (T)
T∈Th
i
+ |ζ − Eh Πh ζ|2H 2 (T) ≤ C|ζ|2H 2 (Ω)
∀ ζ ∈ H02 (Ω)
Xh
T∈Th
2
hT−6 kζ − Eh Πh ζk2L2 (T) + h−4
T |ζ − Eh Πh ζ|H 1 (T)
i
2
2
+ h−2
|ζ
−
E
Π
ζ|
∀ ζ ∈ H 3 (Ω) ∩ H02 (Ω)
2
h
h
T
H (T) ≤ C|ζ|H 3 (Ω)
These estimates indicate that Eh ◦ Πh is a quasi-local interpolation operator from H02 (Ω) into a C1 finite element space that
satisfies the correct discretization error estimates.
Properties of Eh
Estimates for Eh ◦ Πh : H02 (Ω) −→ H02 (Ω)
(Πh ◦ Eh = IdVh )
Xh
2
hT−4 kζ − Eh Πh ζk2L2 (T) + h−2
T |ζ − Eh Πh ζ|H 1 (T)
T∈Th
i
+ |ζ − Eh Πh ζ|2H 2 (T) ≤ C|ζ|2H 2 (Ω)
∀ ζ ∈ H02 (Ω)
Xh
T∈Th
2
hT−6 kζ − Eh Πh ζk2L2 (T) + h−4
T |ζ − Eh Πh ζ|H 1 (T)
i
2
2
+ h−2
|ζ
−
E
Π
ζ|
∀ ζ ∈ H 3 (Ω) ∩ H02 (Ω)
2
h
h
T
H (T) ≤ C|ζ|H 3 (Ω)
If v = Πh ζ for some ζ ∈ H 3 (Ω), then
|v−Eh v|H 2 (Ω;Th ) ≤ |Πh ζ−ζ|H 2 (Ω;Th ) +|ζ−Eh Πh ζ|H 2 (Ω;Th ) ≤ Ch|ζ|H 3 (Ω)
which indicates that |v − Eh v|H 2 (Ω;Th ) provides a measure for the
distance between v and H02 (Ω) with respect to | · |H 2 (Ω;Th ) .
Other Boundary Conditions
Other Boundary Conditions
Boundary Conditions of Simply Supported Plates
We can construct an enriching operator from the Lagrange finite
element space Vh (⊂ H01 (Ω)) (same finite element space for
clamped plates) into H 2 (Ω) ∩ H01 (Ω) with similar properties. But
the piecewise H 2 norm | · |H 2 (Ω;Th ) is defined by
|v|2H 2 (Ω;Th ) =
X
T∈Th
|v|2H 2 (T) +
X 1
k[[v]]k2L2 (e)
|e|
i
e∈Eh
Other Boundary Conditions
Boundary Conditions of Simply Supported Plates
We can construct an enriching operator from the Lagrange finite
element space Vh (⊂ H01 (Ω)) (same finite element space for
clamped plates) into H 2 (Ω) ∩ H01 (Ω) with similar properties. But
the piecewise H 2 norm | · |H 2 (Ω;Th ) is defined by
|v|2H 2 (Ω;Th ) =
X
T∈Th
|v|2H 2 (T) +
X 1
k[[v]]k2L2 (e)
|e|
i
e∈Eh
Boundary Conditions of the Cahn-Hilliard Type
We can construct an enriching operator from the Lagrange finite
element space Vh∗ (different from the finite element space for
clamped plates) into
V∗ = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω and v(p∗ ) = 0}
with similar properties. The piecewise H 2 norm | · |H 2 (Ω;Th ) is
identical to the one for clamped plates.
Post-Processing by Eh
Post-Processing by Eh
Let uh ∈ Vh be the solution of a C0 interior penalty method
for the biharmonic equation with the boundary condition of the
clamped plates (respectively simply supported plates or CahnHilliard type) and Eh be an enriching operator from Eh into
H02 (Ω) (respectively H 2 (Ω) ∩ H01 (Ω) or V∗ ).
Then Eh uh is a conforming approximation of the boundary
value problem obtained from uh by post-processing.
Post-Processing by Eh
Let uh ∈ Vh be the solution of a C0 interior penalty method
for the biharmonic equation with the boundary condition of the
clamped plates (respectively simply supported plates or CahnHilliard type) and Eh be an enriching operator from Eh into
H02 (Ω) (respectively H 2 (Ω) ∩ H01 (Ω) or V∗ ).
Then Eh uh is a conforming approximation of the boundary
value problem obtained from uh by post-processing.
It can be shown, by using the error estimates of the C0 interior
penalty methods (to be discussed) and properties of Eh , that
Eh uh is a C1 approximate solution with correct estimates.
Therefore C0 interior penalty methods are relevant even if we
only want C1 approximate solutions.
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
∀ v ∈ Vh
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
∀ v ∈ Vh
The piecewise H 2 norm is the one for simply supported plates:
|v|2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
i
e∈Eh
Therefore it also holds for the stronger norm for clamped plates:
|v|2H 2 (Ω;Th ) =
X
T∈Th
|v|2H 2 (T) +
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
e∈Eh
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
Proof.
∀ v ∈ Vh
kvkL2 (Ω) + |v|H 1 (Ω) ≤ kv − Eh vkL2 (Ω) + |v − Eh v|H 1 (Ω)
+ kEh vkL2 (Ω) + |Eh v|H 1 (Ω)
Eh is an enriching operator for simply supported plates that
maps Vh into H 2 (Ω) ∩ H01 (Ω).
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
Proof.
∀ v ∈ Vh
kvkL2 (Ω) + |v|H 1 (Ω) ≤ kv − Eh vkL2 (Ω) + |v − Eh v|H 1 (Ω)
+ kEh vkL2 (Ω) + |Eh v|H 1 (Ω)
kv − Eh vkL2 (Ω) ≤ Ch2 |v|H 2 (Ω;Th )
|v − Eh v|H 1 (Ω) ≤ Ch|v|H 2 (Ω;Th )
Xh
T∈Th
i
2
−2
2
2
h−4
kv
−
E
vk
+
h
|v
−
E
v|
h L2 (T)
h H 1 (T) ≤ C|v|H 2 (Ω;Th )
T
T
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
Proof.
∀ v ∈ Vh
kvkL2 (Ω) + |v|H 1 (Ω) ≤ kv − Eh vkL2 (Ω) + |v − Eh v|H 1 (Ω)
+ kEh vkL2 (Ω) + |Eh v|H 1 (Ω)
kv − Eh vkL2 (Ω) ≤ Ch2 |v|H 2 (Ω;Th )
|v − Eh v|H 1 (Ω) ≤ Ch|v|H 2 (Ω;Th )
kEh vkL2 (Ω) + |Eh v|H 1 (Ω) ≤ CΩ |Eh v|H 2 (Ω)
Eh v ∈ H 2 (Ω) ∩ H01 (Ω)
classical Poincaré-Friedrichs inequality
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
Proof.
∀ v ∈ Vh
kvkL2 (Ω) + |v|H 1 (Ω) ≤ kv − Eh vkL2 (Ω) + |v − Eh v|H 1 (Ω)
+ kEh vkL2 (Ω) + |Eh v|H 1 (Ω)
kv − Eh vkL2 (Ω) ≤ Ch2 |v|H 2 (Ω;Th )
|v − Eh v|H 1 (Ω) ≤ Ch|v|H 2 (Ω;Th )
kEh vkL2 (Ω) + |Eh v|H 1 (Ω) ≤ CΩ |Eh v|H 2 (Ω)
|Eh v|H 2 (Ω) ≤ C|v|H 2 (Ω;Th )
A Poincaré-Friedrichs Inequality
Let Vh (⊂ H01 (Ω)) be a Pk (or Qk ) Lagrange finite element
space. We have
kvkL2 (Ω) + |v|H 1 (Ω) ≤ C|v|H 2 (Ω;Th )
∀ v ∈ Vh
This inequality also holds for v ∈ Vh∗ (finite element space
for boundary conditions of the Cahn-Hilliard type), where the
piecewise H 2 norm is given by
|v|2H 2 (Ω;Th ) =
X
T∈Th
|v|2H 2 (T) +
X 1
k[[∂v/∂n]]k2L2 (e)
|e|
e∈Eh
An Application of the Poincaré-Friedrichs Inequality
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
T∈Th
|v|2H 2 (T) +
X
e∈Eh
|e|−1 k[[∂uh /∂n]]k2L2 (e)
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
T∈Th
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
+σ
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
σ≥1
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
T∈Th
=
X
kuh k2h
+σ
X
e∈Eh
|e|−1 k[[∂uh /∂n]]k2L2 (e)
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
T∈Th
=
X
+σ
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
kuh k2h
≤ 2ah (uh , uh )
ah (v, v) ≥ 21 kvk2h
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
kuh k2h
≤ 2ah (uh , uh )
Z
= 2 f uh dx
Ω
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
T∈Th
=
X
+σ
X
e∈Eh
|e|−1 k[[∂uh /∂n]]k2L2 (e)
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
T∈Th
=
X
+σ
X
e∈Eh
kuh k2h
≤ 2ah (uh , uh )
Z
= 2 f uh dx
Ω
≤ 2kf kL2 (Ω) kuh kL2 (Ω)
|e|−1 k[[∂uh /∂n]]k2L2 (e)
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
|uh |2H 2 (Ω;Th ) =
X
|v|2H 2 (T) +
T∈Th
≤
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
|v|2H 2 (T)
T∈Th
=
X
+σ
X
|e|−1 k[[∂uh /∂n]]k2L2 (e)
e∈Eh
kuh k2h
≤ 2ah (uh , uh )
Z
= 2 f uh dx
Ω
≤ 2kf kL2 (Ω) kuh kL2 (Ω) ≤ Ckf kL2 (Ω) |uh |H 2 (Ω;Th )
Poincaré-Friedrichs Inequality
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
After cancellation we have |uh |H 2 (Ω;Th ) ≤ Ckf kL2 (Ω) .
Equivalently
X
T∈Th
|uh |2H 2 (T) + σ
X 1
k[[∂uh /∂n]]k2L2 (e) ≤ Ckf k2L2 (Ω)
|e|
e∈Eh
An Application of the Poincaré-Friedrichs Inequality
C0 Interior Penalty Method for Clamped Plates
Z
ah (uh , v) =
f v dx
∀ v ∈ Vh
Ω
After cancellation we have |uh |H 2 (Ω;Th ) ≤ Ckf kL2 (Ω) .
Equivalently
X
T∈Th
|uh |2H 2 (T) + σ
X 1
k[[∂uh /∂n]]k2L2 (e) ≤ Ckf k2L2 (Ω)
|e|
e∈Eh
Since 1/|e| ↑ ∞ as h ↓ 0,
k[[∂uh /∂n]]kL2 (e) → 0 as h ↓ 0
In other words, C1 continuity and the essential boundary condition ∂u/∂n = 0 are enforced by the penalty term as h ↓ 0.
Reference
• B., K. Wang and J. Zhao
Poincaré-Friedrichs inequalities for piecewise H 2 functions
Numer. Funct. Anal. Optim., 2004
Extensions
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
∀v ∈ V
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
∀v ∈ V
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
V = H02 (Ω) for the boundary conditions of clamped plates,
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
∀v ∈ V
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
V = H02 (Ω) for the boundary conditions of clamped plates,
V = H 2 (Ω) ∩ H01 (Ω) for the boundary conditions of simply
supported plates,
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
∀v ∈ V
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
V = H02 (Ω) for the boundary conditions of clamped plates,
V = H 2 (Ω) ∩ H01 (Ω) for the boundary conditions of simply
supported plates,
Z
∗
V = V if γ 6= 0 and V = V if γ = 0 (and
f dx = 0) for
Ω
the boundary conditions of the Cahn-Hilliard type.
V = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω}
V∗ = {v ∈ H 2 (Ω) : ∂v/∂n = 0 on ∂Ω and v(p∗ ) = 0}
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
Modifications
Include
Z h
i
β(x)∇w · ∇v + γ(x)wv dx
Ω
in the definition of the bilinear form ah (w, v).
∀v ∈ V
Problems with Lower Order Terms
Find u ∈ V such that
Z h
Z
i
2
2
(D u : D v) + β(x)∇u · ∇v + γ(x)uv dx =
f v dx
Ω
∀v ∈ V
Ω
where β, γ ∈ L∞ (Ω) are nonnegative functions and
Modifications
Include
Z h
i
β(x)∇w · ∇v + γ(x)wv dx
Ω
in the definition of the bilinear form ah (w, v).
For boundary conditions of the Cahn-Hilliard type, remove
the condition v(p∗ ) = 0 in the definition of the finite element
space if γ 6= 0.
Problems with Less Regular Data
Problems with Less Regular Data
Given any F ∈ V 0 , we have a well-posed continuous problem:
Find u ∈ V such that
Z h
i
(D2 u : D2 v) + β(x)∇u · ∇v + γ(x)uv dx = F(v)
Ω
∀v ∈ V
Problems with Less Regular Data
Given any F ∈ V 0 , we have a well-posed continuous problem:
Find u ∈ V such that
Z h
i
(D2 u : D2 v) + β(x)∇u · ∇v + γ(x)uv dx = F(v)
Ω
Discrete Problem
Find uh ∈ Vh such that
ah (u, v) = F(v)
∀ v ∈ Vh
does not make sense in general, since Vh 6⊂ V.
∀v ∈ V
Problems with Less Regular Data
Given any F ∈ V 0 , we have a well-posed continuous problem:
Find u ∈ V such that
Z h
i
(D2 u : D2 v) + β(x)∇u · ∇v + γ(x)uv dx = F(v)
Ω
Discrete Problem
Find uh ∈ Vh such that
ah (u, v) = F(v)
∀ v ∈ Vh
does not make sense in general, since Vh 6⊂ V.
Modification
Find uh ∈ Vh such that
ah (u, v) = F(Eh v)
∀ v ∈ Vh
where Eh is the enriching operator from Vh into V.
∀v ∈ V
Problems on Smooth Domains
We can also combine the C0 interior penalty approach with the
isoparametric approach to solve fourth order boundary value
problems on smooth domains.
Reference
• B., M. Neilan and L.-Y. Sung
Isoparametric C0 interior penalty methods
Calcolo, 2013
Other C0 Interior Penalty Methods
• G.N. Wells and N.T. Dung
A C0 discontinuous Galerkin formulation for Kirchhoff
plates
Methods Appl. Mech. Engrg., 2007
• J. Huang, X. Huang and W. Han
A new C0 discontinuous Galerkin method for Kirchhoff
plates,
Methods Appl. Mech. Engrg., 2010
• T. Gudi, H.S. Gupta and N. Nataraj
Analysis of an interior penalty method for fourth order problems on polygonal domains
J. Sci. Comp., 2013