C0 Interior Penalty Methods
Variational Inequalities
Current Research in Finite Element Methods
CIMPA Summer School
Mumbai, July 2015
References
S.C.Brenner, S., Y. Zhang
Finite element methods for the displacement obstacle problem
of clamped plates
Math. Comp. (2012)
A C0 interior penalty method for an elliptic optimal control problem with state constraints
IMA Vol. Math. Appl. (2014)
Post-processing procedures for an elliptic distributed optimal
control problem with pointwise state constraints
APNUM (2015)
S.C.Brenner, S., H. Zhang and Y. Zhang
A quadratic C0 interior penalty method for the displacement obstacle problem of clamped Kirchhoff plates
SIAM J. Numer. Anal. (2012)
An Obstacle Problem for Clamped Kirchhoff Plates
An Obstacle Problem for Clamped Kirchhoff Plates
Ω = bounded convex polygon (for simplicity)
ψ ∈ C2 (Ω̄),
ψ < 0 on ∂Ω
f ∈ L2 (Ω)
(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)
2
2
D w : D v dx
Ω
fv dx
Ω
2
D w : D v=
2
X
i,j=1
Z
(f , v) =
2
wxi xj vxi xj
An Obstacle Problem for Clamped Kirchhoff Plates
plate
PSfrag replacements u
obstacle
u is the vertical displacement of the midsurface of the thin plate.
f is the vertical load density.
(flexural rigidity =1)
1
a(v, v) − (f , v)
2
is the energy of the plate determined by the displacement v.
An Obstacle Problem for Clamped Kirchhoff Plates
plate
PSfrag replacements u
obstacle
u is the vertical displacement of the midsurface of the thin plate.
f is the vertical load density.
u = argmin
v∈K
(flexural rigidity =1)
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 Optimal Control Problem with Pointwise 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
An Optimal Control Problem with Pointwise 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.
yd is the desired state.
y ≤ ψ is a pointwise constraint on the state.
u ∈ L2 (Ω) is the control.
β > 0 is related to the cost for implementing the control u.
An Optimal Control Problem with Pointwise 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.
An Optimal Control Problem with Pointwise State
Constraint
1
β
ky − yd k2L2 (Ω) + kuk2L2 (Ω)
2
2
minimize
(y, u) ∈ H01 (Ω) × L2 (Ω)
over
subject to
(
−∆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.
Obstacle Problem
Find y = argmin
v∈K
i
1h
ky − yd k2L2 (Ω) + βk∆yk2L2 (Ω)
2
K = {v ∈ H 2 (Ω) ∩ H01 (Ω) : v ≤ ψ
in Ω}
An Abstract Constrained Minimization Problem
V is a (real) Hilbert space with inner product a(·, ·).
K is a nonempty closed convex subset of V.
F : V −→ R is a bounded linear functional on V.
Find
u = argmin
v∈K
h1
2
a(v, v) − F(v)
i
An Abstract Constrained Minimization Problem
V is a (real) Hilbert space with inner product a(·, ·).
K is a nonempty closed convex subset of V.
F : V −→ R is a bounded linear functional on V.
Find
u = argmin
v∈K
h1
2
a(v, v) − F(v)
Riesz Representation Theorem
F(v) = a(uF , v)
for some uF ∈ V
∀v ∈ V
i
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
h1
2
a(v, v) − F(v)
i
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
= argmin
v∈K
h1
2
h1
2
a(v, v) − F(v)
i
a(v, v) − a(uF , v)
i
F(v) = a(uF , v)
∀v ∈ V
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
= argmin
h1
2
h1
a(v, v) − F(v)
i
a(v, v) − a(uF , v)
i
2
i
1h
= argmin a(v − uF , v − uF ) − a(uF , uF )
2
v∈K
v∈K
completing the square
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
= argmin
h1
2
h1
a(v, v) − F(v)
i
a(v, v) − a(uF , v)
i
2
i
1h
= argmin a(v − uF , v − uF ) − a(uF , uF )
2
v∈K
v∈K
= argmin a(v − uF , v − uF )
v∈K
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
= argmin
h1
2
h1
a(v, v) − F(v)
i
a(v, v) − a(uF , v)
i
2
i
1h
= argmin a(v − uF , v − uF ) − a(uF , uF )
2
v∈K
v∈K
= argmin a(v − uF , v − uF )
v∈K
In other words we are looking for thepelement of K closest to
uF with respect to the norm k · ka = a(·, ·) of V.
An Abstract Constrained Minimization Problem
Find
u = argmin
v∈K
= argmin
h1
2
h1
a(v, v) − F(v)
i
a(v, v) − a(uF , v)
i
2
i
1h
= argmin a(v − uF , v − uF ) − a(uF , uF )
2
v∈K
v∈K
= argmin a(v − uF , v − uF )
v∈K
In other words we are looking for thepelement of K closest to
uF with respect to the norm k · ka = a(·, ·) of V.
Since K is a nonempty closed convex subset of V, there is a
unique u ∈ K closest to uF by the Projection Theorem.
A Variational Inequality
A Variational Inequality
Let v ∈ K be arbitrary. The line segment
(1 − t)u + tv for
0≤t≤1
connecting u to v is in the convex set K and hence the function
1
Φ(t) = a (1 − t)u + tv, (1 − t)u + tv) − F (1 − t)u + tv)
2
has a minimum at t = 0.
A Variational Inequality
Let v ∈ K be arbitrary. The line segment
(1 − t)u + tv for
0≤t≤1
connecting u to v is in the convex set K and hence the function
1
Φ(t) = a (1 − t)u + tv, (1 − t)u + tv) − F (1 − t)u + tv)
2
has a minimum at t = 0.
Therefore Φ0 (0) ≥ 0, which is equivalent to the variational inequality
a(u, v − u) ≥ F(v − u)
that holds for all v ∈ K.
A Second Order Variational Inequality
V = H01 (Ω)
K = {v ∈ V : v ≥ ψ a.e.}
(ψ < 0 on ∂Ω)
f ∈ L2 (Ω)
Find
u ∈ argmin
v∈K
h1
2
(∇v, ∇v) − (f , v)
i
obstacle problem for
elastic membranes
A Second Order Variational Inequality
V = H01 (Ω)
K = {v ∈ V : v ≥ ψ a.e.}
(ψ < 0 on ∂Ω)
f ∈ L2 (Ω)
Find
u ∈ argmin
v∈K
h1
2
(∇v, ∇v) − (f , v)
i
Since (∇v, ∇w) is an inner product on H01 (Ω) and K is a
nonempty closed convex subset of H01 (Ω), there is a unique
minimizer u ∈ K characterized by the variational inequality
(∇u, ∇(v − u)) ≥ (f , v − u)
∀v ∈ K
A Second Order Variational Inequality
Brézis-Stampacchia 1968
The solution u of the second order problem belongs to H 2 (Ω)
under appropriate assumptions, i.e., the solutions of second order variational inequalities enjoy the same elliptic regularity as
the solutions of second order boundary value problems.
A Second Order Variational Inequality
Brézis-Stampacchia 1968
The solution u of the second order problem belongs to H 2 (Ω)
under appropriate assumptions, i.e., the solutions of second order variational inequalities enjoy the same elliptic regularity as
the solutions of second order boundary value problems.
(∇u, ∇(v − u)) ≥ (f , v − u)
⇔ (∆u + f , v − u) ≤ 0
∀v ∈ K
∀v ∈ K
which is equivalent to
∆u + f ≤ 0,
u − ψ ≥ 0,
(∆u + f )(u − ψ) = 0
(Complementary Form of the variational inequality)
Finite Element Method
Th is a triangulation of Ω.
Vh ⊂ H01 (Ω) is the P1 finite element space associated with Th .
Πh : H 2 (Ω) ∩ H01 (Ω) −→ Vh is the nodal interpolation operator.
Kh = {v ∈ Vh : v ≥ Πh ψ}
(Πh u ∈ Kh )
Find uh = argmin
v∈Kh
h1
2
i
(∇v, ∇v) − (f , v)
Finite Element Method
Th is a triangulation of Ω.
Vh ⊂ H01 (Ω) is the P1 finite element space associated with Th .
Πh : H 2 (Ω) ∩ H01 (Ω) −→ Vh is the nodal interpolation operator.
Kh = {v ∈ Vh : v ≥ Πh ψ}
(Πh u ∈ Kh )
Find uh = argmin
v∈Kh
h1
2
i
(∇v, ∇v) − (f , v)
There is a unique minimizer uh ∈ Kh characterized by the discrete variational inequality
(∇uh , ∇(v − uh )) ≥ (f , v − uh )
∀ v ∈ Kh
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
Falk (1974)
Brezzi-Hager-Raviart (1977)
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) = (∇(u − uh ), ∇(u − uh ))
= (∇(u − uh ), ∇(u − Πh u)) + (∇(u − uh ), ∇(Πh u − uh ))
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) = (∇(u − uh ), ∇(u − uh ))
= (∇(u − uh ), ∇(u − Πh u)) + (∇(u − uh ), ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (∇uh , ∇(Πh u − uh ))
Cauchy-Schwarz Inequality
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) = (∇(u − uh ), ∇(u − uh ))
= (∇(u − uh ), ∇(u − Πh u)) + (∇(u − uh ), ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (∇uh , ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (f , Πh u − uh )
Discrete Variational Inequality
(∇uh , ∇(v − uh )) ≥ (f , v − uh )
Πh u ∈ Kh
∀ v ∈ Kh
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) = (∇(u − uh ), ∇(u − uh ))
= (∇(u − uh ), ∇(u − Πh u)) + (∇(u − uh ), ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (∇uh , ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (f , Πh u − uh )
= k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
− ((∆u + f ), Πh u − uh )
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) = (∇(u − uh ), ∇(u − uh ))
= (∇(u − uh ), ∇(u − Πh u)) + (∇(u − uh ), ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (∇uh , ∇(Πh u − uh ))
≤ k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
+ (∇u, ∇(Πh u − uh )) − (f , Πh u − uh )
= k∇(u − uh )kL2 (Ω) k∇(u − Πh u)kL2 (Ω)
− ((∆u + f ), Πh u − uh )
≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
− ((∆u + f ), Πh u − uh )
= −((∆u + f ), (Πh u − u) + (u − ψ) + (ψ − Πh ψ) + (Πh ψ − uh ))
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
− ((∆u + f ), Πh u − uh )
= −((∆u + f ), (Πh u − u) + (u − ψ) + (ψ − Πh ψ) + (Πh ψ − uh ))
≤ −((∆u + f ), (Πh u − u) + (ψ − Πh ψ))
− ((∆u + f ), u − ψ) = 0 because (∆u + f )(u − ψ) = 0
− ((∆u + f ), Πh ψ − uh ) ≤ 0
because (∆u + f ) ≤ 0 and
uh − Πh ψ ≥ 0
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
− ((∆u + f ), Πh u − uh )
= −((∆u + f ), (Πh u − u) + (u − ψ) + (ψ − Πh ψ) + (Πh ψ − uh ))
≤ −((∆u + f ), (Πh u − u) + (ψ − Πh ψ))
≤ k∆u + f kL2 (Ω) (kΠh u − ukL2 (Ω) + kψ − Πh ψkL2 (Ω) )
Cauchy-Schwarz Inequality
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
− ((∆u + f ), Πh u − uh )
= −((∆u + f ), (Πh u − u) + (u − ψ) + (ψ − Πh ψ) + (Πh ψ − uh ))
≤ −((∆u + f ), (Πh u − u) + (ψ − Πh ψ))
≤ k∆u + f kL2 (Ω) (kΠh u − ukL2 (Ω) + kψ − Πh ψkL2 (Ω) )
≤ C2 h2
Error Estimate:
k∇(u − uh )kL2 (Ω) ≤ Ch
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) − ((∆u + f ), Πh u − uh )
− ((∆u + f ), Πh u − uh )
= −((∆u + f ), (Πh u − u) + (u − ψ) + (ψ − Πh ψ) + (Πh ψ − uh ))
≤ −((∆u + f ), (Πh u − u) + (ψ − Πh ψ))
≤ k∆u + f kL2 (Ω) (kΠh u − ukL2 (Ω) + kψ − Πh ψkL2 (Ω) )
≤ C2 h2
k∇(u − uh )k2L2 (Ω) ≤ C1 hk∇(u − uh )kL2 (Ω) + C2 h2
1
≤ C3 h2 + k∇(u − uh )k2L2 (Ω) + C2 h2
2
Obstacle Problem for the Clamped Kirchhoff Plate
Find u = argmin
h1
v∈K
Z
a(w, v) =
D2 w : D2 v dx
Ω
fv dx
Ω
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Z
(f , v) =
2
i
a(v, v) − (f , v)
K = {v ∈ H02 (Ω) : v ≥ ψ on Ω}
Obstacle Problem for the Clamped Kirchhoff Plate
Find u = argmin
h1
v∈K
Z
a(w, v) =
D2 w : D2 v dx
Ω
fv dx
Ω
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Z
(f , v) =
2
i
a(v, v) − (f , v)
K = {v ∈ H02 (Ω) : v ≥ ψ on Ω}
Since a(·, ·) is symmetric, bounded and coercive on H02 (Ω),
it defines an inner product on H02 (Ω) and the norm k · ka is
equivalent to the Sobolev norm k · kH 2 (Ω) . Therefore H02 (Ω) is
a Hilbert space under this inner product and K is a non-empty
closed convex subset of this Hilbert space.
Obstacle Problem for the Clamped Kirchhoff Plate
Find u = argmin
h1
v∈K
Z
a(w, v) =
D2 w : D2 v dx
Ω
fv dx
Ω
D2 w : D2 v =
2
X
wxi xj vxi xj
i,j=1
Z
(f , v) =
2
i
a(v, v) − (f , v)
K = {v ∈ H02 (Ω) : v ≥ ψ on Ω}
We can apply the abstract theory to conclude that the obstacle
problem has a unique solution characterized by the variational
inequality
a(u, v − u) ≥ (f , v − u)
∀v ∈ K
Obstacle Problem for the Clamped Kirchhoff Plate
a(u, v − u) ≥ (f , v − u)
∀v ∈ K
Z
a(w, v) =
D2 w : D2 v dx
K = {v ∈
Ω
H02 (Ω)
: v ≥ ψ on Ω}
If u ∈ H 4 (Ω), then u satisfies the complementarity form of the
variational inequality in a strong sense:
∆2 u − f ≥ 0,
u − ψ ≥ 0,
(∆2 u − f )(u − ψ) = 0
and the numerical analysis of the obstacle problem can proceed
as in the second order case.
Regularity Results
3 (Ω)
u ∈ Hloc
under the assumption that ψ ∈ C2 (Ω).
Frehse (1971)
Since u > ψ near ∂Ω and hence ∆2 u = f near ∂Ω, we have
u ∈ H 3 (Ω) by the convexity of Ω.
Kondrat’ev (1967)
Blum and Rannacher (1980)
4 (Ω) even if Ω, f and ψ are
u ∈ C2 (Ω) but in general u 6∈ Hloc
smooth.
Caffarelli and Friedman (1979)
4 regularity means that the complementarity form
The lack of Hloc
of the variational inequality only exists in a weak sense, which
complicates the numerical analysis of the obstacle problem.
A Quadratic C0 Interior Penalty Method
Th = triangulation of Ω
Vh = quadratic finite element space ( ⊂ H01 (Ω))
Kh = {v ∈ Vh : v ≥ ψ at all the vertices of Th }
ah (w, v) =
XZ
T∈Th
+
2
XZ
2
D w : D v dx +
T
e∈Eh
XZ
e∈Eh
e
+σ
e
{{∂ 2 w/∂n2 }}[[∂v/∂n]] ds
{{∂ 2 v/∂n2 }}[[∂w/∂n]] ds
X
e∈Eh
−1
|e|
Z
[[∂w/∂n]][[∂v/∂n]] ds
e
A Quadratic C0 Interior Penalty Method
Th = triangulation of Ω
Vh = quadratic finite element space ( ⊂ H01 (Ω))
Kh = {v ∈ Vh : v ≥ ψ at all the vertices of Th }
Discrete Problem
Find uh = argmin
v∈Kh
h1
2
i
ah (v, v) − (f , v)
Since ah (·, ·) is symmetric and coercive, the discrete obstacle
problem has a unique solution characterized by a discrete variational inequality
ah (uh , v − uh ) ≥ (f , v − uh )
∀ v ∈ Kh
Convergence Analysis
An Auxiliary Obstacle Problem
Find ũh = argmin
h1
v∈K̃h
2
i
a(v, v) − (f , v)
where
K̃h = {v ∈ H02 (Ω) : v(p) ≥ ψ(p) at all the vertices of Th }
⊃ K = {v ∈ H02 (Ω) : v ≥ ψ
on Ω}
Convergence Analysis
An Auxiliary Obstacle Problem
Find ũh = argmin
h1
v∈K̃h
2
i
a(v, v) − (f , v)
where
K̃h = {v ∈ H02 (Ω) : v(p) ≥ ψ(p) at all the vertices of Th }
⊃ K = {v ∈ H02 (Ω) : v ≥ ψ
on Ω}
This is a minimization problem for the functional of the original
obstacle problem but on a larger closed convex set.
Variational Inequality:
a(ũh , v − ũh ) ≥ (f , v − ũh )
∀ v ∈ K̃h
Convergence Analysis
An Auxiliary Obstacle Problem
Find ũh = argmin
h1
v∈K̃h
2
i
a(v, v) − (f , v)
where
K̃h = {v ∈ H02 (Ω) : v(p) ≥ ψ(p) at all the vertices of Th }
⊃ K = {v ∈ H02 (Ω) : v ≥ ψ
on Ω}
The auxiliary obstacle problem is an intermediate obstacle
problem. On one hand it shares the space H02 (Ω) and the funcPSfrag replacements
tional with the original obstacle problem. On the other hand it
shares the same constraint as the discrete obstacle problem.
Kh
Eh
injection
K̃h
K
Error Estimates
By using the auxiliary obstacle problem to connect the continuous and discrete obstacle problems, we can show that
Theorem
ku − uh kh ≤ Ch
kvk2h =
X
T∈Th
|v|2H 2 (T) +
+
X
e∈Eh
X
e∈Eh
|e|k{{∂ 2 v/∂n2 }}k2L2 (e)
|e|−1 k[[∂v/∂n]]k2L2 (e)
Error Estimates
By using the auxiliary obstacle problem to connect the continuous and discrete obstacle problems, we can show that
Theorem
ku − uh kh ≤ Ch
Key Ingredients
I
the three variational inequalities
a(u, v − u) ≥ (f , v − u)
∀v ∈ K
a(ũh , v − ũh ) ≥ (f , v − ũh )
∀ v ∈ K̃h
ah (uh , v − uh ) ≥ (f , v − uh )
∀ v ∈ Kh
I
techniques from the calculus of variations
I
properties of Eh
Error Estimates
By using the auxiliary obstacle problem to connect the continuous and discrete obstacle problems, we can show that
Theorem
ku − uh kh ≤ Ch
This error estimate is optimal since u ∈ H 3 (Ω) and the norm
k · kh behaves like the H 2 (Ω) norm.
Error Estimates
By using the auxiliary obstacle problem to connect the continuous and discrete obstacle problems, we can show that
Theorem
ku − uh kh ≤ Ch
It is not difficult to deduce an L∞ error estimate from the energy
error estimate since k · kh behaves like k · kH 2 (Ω) and
kζkL∞ (Ω) ≤ CΩ kζkH 2 (Ω)
∀ ζ ∈ H 2 (Ω)
by the Sobolev inequality.
Theorem
ku − uh kL∞ (Ω) ≤ Ch
But numerical results indicate that this estimate is not sharp.
General Polygons and General Boundary Conditions
Theorem
ku − uh kh + ku − uh kL∞ (Ω) ≤ Chα
The index of elliptic regularity α is determined by the interior
angles of Ω and the boundary conditions. For clamped plates
we have 12 < α ≤ 1 and we can take α to be 1 if Ω is convex.
For the obstacle problem with the boundary conditions of simply
supported plates, the value of α can be less than 1 even on
convex polygonal domains. But we can improve the order of
convergence using various techniques from the finite element
arsenal.
Theorem
With proper treatment we have
ku − uh kh + ku − uh kL∞ (Ω) ≤ Ch
An Optimal Control Problem with Pointwise State
Constraint
Ω is a bounded convex polygonal domain in R2 .
Find y = argmin
v∈K
i
1h
ky − yd k2L2 (Ω) + βk∆yk2L2 (Ω)
2
K = {v ∈ H 2 (Ω) ∩ H01 (Ω) : v ≤ ψ
in Ω}
An Optimal Control Problem with Pointwise State
Constraint
Ω is a bounded convex polygonal domain in R2 .
Find y = argmin
v∈K
i
1h
ky − yd k2L2 (Ω) + βk∆yk2L2 (Ω)
2
K = {v ∈ H 2 (Ω) ∩ H01 (Ω) : v ≤ ψ
in Ω}
Equivalent Problem
Find y = argmin
v∈K
Z
a(w, v) =
1
2
βa(v, v) + (v, v) − (yd , v)
D2 w : D2 v dx
Ω
K = {v ∈ H 2 (Ω) ∩ H01 (Ω) : v ≤ ψ
in Ω}
A Quadratic C0 Interior Penalty Method
Th = triangulation of Ω
Vh = quadratic finite element space ( ⊂ H01 (Ω))
Kh = {v ∈ Vh : v ≤ ψ at all the vertices of Th }
ah (w, v) =
XZ
T∈Th
+
XZ
D2 w : D2 v dx +
T
e∈Ehi
XZ
e∈Ehi
e
+σ
e
{{∂ 2 w/∂n2 }}[[∂v/∂n]] ds
{{∂ 2 v/∂n2 }}[[∂w/∂n]] ds
X
e∈Ehi
−1
|e|
Z
[[∂w/∂n]][[∂v/∂n]] ds
e
A Quadratic C0 Interior Penalty Method
Th = triangulation of Ω
Vh = quadratic finite element space ( ⊂ H01 (Ω))
Kh = {v ∈ Vh : v ≤ ψ at all the vertices of Th }
Discrete Problem
Find y = argmin
v∈Kh
1
βah (v, v) + (v, v) − (yd , v)
2
Kh = {v ∈ Vh : v ≤ ψ
at the vertices of Th }
Error Estimates
Theorem
Without special treatment we have
ky − yh kh + ky − yh kL∞ (Ω) ≤ Chα
Theorem
With proper treatment we have
ky − yh kh + ky − yh kL∞ (Ω) ≤ Ch
kvk2h = β
X
T∈Th
|v|2H 2 (T) +
X
e∈Ehi
|e|−1 k[[∂v/∂n]]k2L2 (e) + kvk2L2 (Ω)
Error Estimates
Theorem
Without special treatment we have
ky − yh kh + ky − yh kL∞ (Ω) ≤ Chα
Theorem
With proper treatment we have
ky − yh kh + ky − yh kL∞ (Ω) ≤ Ch
We can compute an approximation uh of the optimal control u
from the discrete optimal state yh through post-processing so
that
ku − uh kL2 (Ω) ≈ ky − yh kh
Post-Processing
Procedure 1
Since u = −∆y, we take
uh = −∆h yh
where ∆h is the piecewise defined Laplacian.
Post-Processing
Since u = −∆y, we take
Procedure 1
uh = −∆h yh
where ∆h is the piecewise defined Laplacian.
Procedure 2
Z
Since
Ω
∇y · ∇v dx =
Z
uv dx
Ω
we compute uh ∈ Vh by
Z
Z
∇yh · ∇v dx =
uh v dx
Ω
Ω
∀ v ∈ H01 (Ω)
∀ v ∈ Vh
Post-Processing
Procedure 3 Since
Z
Z
∇u · ∇v dx =
∇(−∆y) · ∇v dx
Ω
ZΩ
=
D2 y : D2 v dx
Ω
we compute uh ∈ Vh by
Z
∇uh · ∇v dx = ah (yh , v)
Ω
∀ v ∈ H01 (Ω) ∩ H 2 (Ω)
∀ v ∈ Vh
where ah (·, ·) is the bilinear form for the quadratic C0 interior
penalty method that approximates the bilinear form
Z
(w, v) −→
D2 w : D2 v dx
Ω
Post-Processing
Procedure 1
numerical differentiation
Procedure 2
numerical differentiation followed by averaging
Procedure 3
numerical differentiation followed by smoothing
Post-Processing
Theorem
For all three procedures we have
kū − ūh kL2 (Ω) ≤ Chα
without special treatments, and
kū − ūh kL2 (Ω) ≤ Ch
with special treatments.
Post-Processing
Theorem
For all three procedures we have
kū − ūh kL2 (Ω) ≤ Chα
without special treatments, and
kū − ūh kL2 (Ω) ≤ Ch
with special treatments.
Procedure 1, which only involves numerical differentiation, is
the cheapest one. Procedure 2, which also requires solving a
problem involving the mass matrix, is more expensive. Procedure 3, which requires solving a problem involving the stiffness
matrix of a second order problem, is the most expensive one.
But in practice Procedure 3 can outperform Procedure 2, which
in turn can outperform Procedure 1.
Comparison
There are finite element methods for this problem that are
based on discrete versions of the optimal control problem (Deckelnick-Hinze 2007, Meyer 2008, Casas-MateosVexler 2014) where
|y − yh |H 1 (Ω) + ku − uh kL2 (Ω) = O(h)
for a smooth Ω or a rectangular Ω.
Comparison
There are finite element methods for this problem that are
based on discrete versions of the optimal control problem (Deckelnick-Hinze 2007, Meyer 2008, Casas-MateosVexler 2014) where
|y − yh |H 1 (Ω) + ku − uh kL2 (Ω) = O(h)
for a smooth Ω or a rectangular Ω.
In our approach we have
ky − yh kh + ku − uh kL2 (Ω) ≤ O(h)
for a smooth Ω or a rectangular Ω, where k · kh is an H 2 like norm. Therefore the discrete state yh in our approach
provides more information for the optimal state y. Moreover the convergence of yh to y in the H 1 norm or the L∞
norm is of higher order in practice.
Comparison
There are finite element methods for this problem that are
based on discrete versions of the optimal control problem (Deckelnick-Hinze 2007, Meyer 2008, Casas-MateosVexler 2014) where
|y − yh |H 1 (Ω) + ku − uh kL2 (Ω) = O(h)
for a smooth Ω or a rectangular Ω.
In our approach we have
ky − yh kh + ku − uh kL2 (Ω) ≤ O(h)
for a smooth Ω or a rectangular Ω, where k · kh is an H 2 like norm.
We also observe high order convergence for the discrete
optimal control uh obtained by the second and third postprocessing procedures.
Numerical Results
The first set of numerical experiments involve the obstacle problem for clamped Kirchhoff plates on the unit square
(−.5, .5) × (−.5, .5)
Numerical Results: Experiment 1
External Force
f =0
Obstacle Function
ψ(x) = 1 − 5|x|2 + |x|4
Numerical Results: Experiment 1
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
2−8
relative error in k · kh order
3.2401 × 10−1
4.5394 × 10−1
−0.48
4.9944 × 10−1
−0.13
−1
3.8333 × 10
0.38
1.9609 × 10−1
0.96
−2
9.2707 × 10
1.08
4.4712 × 10−2
1.05
2.1855 × 10−2
1.03
kvk2h =
X
T∈Th
|v|2H 2 (T) +
+
X
e∈Eh
X
e∈Eh
−1
|e|
L∞ error
1.0000 × 100
3.4417 × 10−1
5.9705 × 10−2
2.6127 × 10−2
3.6557 × 10−3
1.2895 × 10−3
4.1668 × 10−4
1.0245 × 10−4
|e|k{{∂ 2 v/∂n2 }}k2L2 (e)
k[[∂v/∂n]]k2L2 (e)
order
1.53
2.52
1.19
2.83
1.50
1.62
2.02
Numerical Results: Experiment 1
We can also visualize the approximation to the coincidence set
I = {x : u(x) = ψ(x)}
by plotting all the nodes p (vertices and midpoints) that satisfy
the condition
|uh (p) − ψ(p)| ≤ L∞ error
where the L∞ error is estimated by comparing solutions on consecutive levels.
Since
∆2 ψ = ∆2 (1 − 5|x|2 + |x|4 ) > 0
the non-coincidence set is connected.
Caffarelli and Friedman (1979)
Numerical Results: Experiment 1
Coincidence set I
(h = 2−8 )
Numerical Results: Experiment 2
External Force
f =0
Obstacle Function
ψ(x) = 1 − 5|x|2 − |x|4
This is similar to Experiment 1 except that
∆2 ψ = ∆2 (1 − 5|x|2 − |x|4 ) < 0
which implies that the interior of the coincidence set should be
empty.
(Otherwise ∆2 u = ∆2 ψ < 0 in an open set contradicting the fact
that ∆2 u is a nonnegative Borel measure.)
Numerical Results: Experiment 2
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
2−8
relative error in k · kh order
3.4133 × 10−1
4.7596 × 10−1
−0.47
5.1117 × 10−1
−0.10
−1
3.3897 × 10
0.59
1.6913 × 10−1
1.00
−2
7.9146 × 10
1.09
3.8567 × 10−2
1.03
1.8889 × 10−2
1.02
L∞ error
1.0000 × 100
3.3309 × 10−1
7.2578 × 10−2
2.5308 × 10−2
7.6540 × 10−3
1.6226 × 10−3
5.8201 × 10−4
1.0995 × 10−4
order
1.58
2.19
1.51
1.72
2.23
1.47
2.40
Numerical Results: Experiment 2
Coincidence set I
(h = 2−8 )
Numerical Results: Experiment 3
External Force

2

1000e|x|



5000(3 + |x|2 )
f (x) =

−300 sin(πx1 ) sin(πx2 )



−200(1 − |x|3 )
if x1
if x1
if x1
if x1
≤ 0 and x2
> 0 and x2
≤ 0 and x2
> 0 and x2
≥0
≥0
<0
<0
Numerical Results: Experiment 3
A Nonsymmetric Obstacle Function ψ
Numerical Results: Experiment 3
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
relative error in k · kh order
5.8955 × 10−1
4.1882 × 10−1
0.49
5.0682 × 10−1
−0.28
3.6432 × 10−1
0.48
2.0144 × 10−1
0.85
−2
9.8022 × 10
1.04
4.8507 × 10−2
1.02
L∞ error
order
0
5.3265 × 10
9.1903 × 10−1 2.53
9.8753 × 10−1 −0.10
4.4694 × 10−1 1.14
1.8191 × 10−1 1.30
7.9253 × 10−2 1.20
2.4371 × 10−2 1.70
Numerical Results
The second set of numerical experiments involve the optimal
control problem with pointwise state constraint.
Numerical Results: Experiment 4
Ω = (−0.5, 0, 5) × (−0.5, 0.5)
ψ(x) = 0.1
yd = sin 2π(x1 + 0.5)(x2 + 0.5)
β = 10−3
α=1
uniform meshes
Numerical Results: Experiment 4
Coincidence set
(h = 2−8 )
Numerical Results: Experiment 4
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
2−8
relative error of
state in k · kh
2.697 × 100
1.244 × 100
6.764 × 10−1
3.455 × 10−1
1.749 × 10−1
8.643 × 10−2
4.267 × 10−2
2.123 × 10−2
order
1.12
0.88
0.97
0.98
1.01
1.02
1.01
relative error of
state in | · |H 1 (Ω)
3.287 × 100
1.154 × 100
3.394 × 10−1
9.206 × 10−2
2.664 × 10−2
7.282 × 10−3
1.856 × 10−3
4.671 × 10−4
order
1.51
1.77
1.88
1.79
1.87
1.97
1.99
Relative errors of the state in k · kh and | · |H 1 (Ω)
Numerical Results: Experiment 4
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
2−8
proc. 1
1.736 × 100
1.057 × 100
6.797 × 10−1
3.569 × 10−1
1.775 × 10−1
8.578 × 10−2
4.194 × 10−2
2.079 × 10−2
order
0.72
0.64
0.93
1.01
1.05
1.03
1.01
proc. 2
2.675 × 100
1.563 × 100
9.448 × 10−1
3.529 × 10−1
1.217 × 10−1
4.198 × 10−2
1.313 × 10−2
4.442 × 10−3
order
0.78
0.73
1.42
1.54
1.54
1.68
1.56
proc. 3
order
1.729 × 100
1.515 × 100
0.19
7.525 × 10−1 1.01
2.194 × 10−1 1.78
7.167 × 10−2 1.61
2.830 × 10−2 1.34
8.170 × 10−3 1.79
2.945 × 10−3 1.47
Relative errors of the control in k · kL2 (Ω)
Numerical Results: Experiment 4
h
2−1
2−2
2−3
2−4
2−5
2−6
2−7
2−8
proc. 2
order
100
2.148 ×
2.322 × 100
2.395 × 100
1.823 × 100
1.265 × 100
8.536 × 10−1
5.484 × 10−1
3.637 × 10−1
proc. 3
order
100
-0.11
-0.04
0.39
0.53
0.57
0.64
0.59
1.794 ×
1.585 × 100
1.427 × 100
8.604 × 10−1
5.297 × 10−1
3.845 × 10−1
2.191 × 10−1
1.522 × 10−1
Relative errors of the control in | · |H 1 (Ω)
0.18
0.15
0.73
0.70
0.46
0.81
0.53
Numerical Results: Experiment 5
Ω = (−0.5, 0.5) × (−0.5, 0.5)
ψ(x) = sin[2π(x1 + 0.5)(x2 + 0.5)] sin[6π(x1 − 0.5)(x2 − 0.5)] + 0.2


cos[π(x1 − 0.25)]




5(x + 0.25)2 + (x − 0.25)2
1
2
yd =
1

− 2 exp[(x1 + 0.25)2 + (x2 + 0.25)2 ]




 1 + [(x − 0.25)2 + (x + 0.25)2 ]3/2
1
2
2
β = 10−3
α=1
uniform meshes
x1 > 0, x2 > 0
x1 ≤ 0, x2 > 0
x1 ≤ 0, x2 ≤ 0
x1 > 0, x2 ≤ 0
Numerical Results: Experiment 5
Coincidence set
(h = 2−8 )
Numerical Results: Experiment 5
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
relative error of
state in k · kh
4.368 × 10−1
5.684 × 10−1
3.350 × 10−1
1.577 × 10−1
7.552 × 10−2
3.586 × 10−2
1.727 × 10−2
order
-0.38
0.76
1.09
1.06
1.07
1.05
relative error of
state in | · |H 1 (Ω)
2.911 × 10−1
4.403 × 10−1
1.718 × 10−1
4.984 × 10−2
1.908 × 10−2
3.954 × 10−3
1.073 × 10−3
order
-0.60
1.36
1.79
1.39
2.27
1.88
Relative errors of the state in k · kh and | · |H 1 (Ω)
Numerical Results: Experiment 5
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
proc. 1
5.697 × 10−1
5.555 × 10−1
3.855 × 10−1
1.910 × 10−1
8.977 × 10−2
4.522 × 10−2
2.190 × 10−2
order
0.04
0.53
1.01
1.09
0.99
1.05
proc. 2
6.204 × 10−1
8.836 × 10−1
5.062 × 10−1
2.035 × 10−1
9.077 × 10−2
3.558 × 10−2
1.475 × 10−2
order
-0.51
0.80
1.31
1.16
1.35
1.27
proc. 3
5.064 × 10−1
1.049 × 100
5.631 × 10−1
1.992 × 10−1
9.624 × 10−2
3.669 × 10−2
1.549 × 10−2
Relative errors of the control in k · kL2 (Ω)
order
-1.05
0.90
1.50
1.05
1.39
1.24
Numerical Results: Experiment 5
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
proc. 2
order
10−1
9.991 ×
1.760 × 100
1.763 × 100
1.520 × 100
1.226 × 100
8.800 × 10−1
6.804 × 10−1
proc. 3
order
10−1
-0.82
-0.00
0.21
0.31
0.48
0.37
6.770 ×
1.534 × 100
1.382 × 10−1
1.038 × 10−1
8.742 × 10−1
5.743 × 10−1
4.549 × 10−1
-1.18
0.15
0.41
0.25
0.61
0.34
Relative errors of the control in | · |H 1 (Ω)
Numerical Results: Experiment 6
Ω = a pentagon
(−0.5, 0.5)
(0, 0.5)
ψ(x) = 0.1
(0.5, 0)
yd = sin 2π(x1 + 0.5)(x2 +PSfrag
0.5)replacements
β = 10−3
(−0.5, −0.5)
α=
(0.5, −0.5)
1
3
uniform meshes
1
ky − yh kh + ku − uh kL2 (Ω) ≤ Ch 3
Numerical Results: Experiment 6
Coincidence set
(h = 2−9 )
Numerical Results: Experiment 6
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
2−9
relative error of
state in k · kh
1.275 × 100
7.305 × 10−1
3.607 × 10−1
1.958 × 10−1
1.176 × 10−1
7.897 × 10−2
5.772 × 10−2
4.416 × 10−2
order
0.80
1.02
0.88
0.73
0.57
0.45
0.39
relative error of
state in | · |H 1 (Ω)
1.156 × 100
3.484 × 10−1
8.279 × 10−2
2.217 × 10−2
6.526 × 10−3
2.131 × 10−3
8.760 × 10−4
4.570 × 10−4
order
1.73
2.07
1.90
1.76
1.61
1.28
0.94
Relative errors of the state in k · kh and | · |H 1 (Ω)
Numerical Results: Experiment 6
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
2−9
proc. 1
order
2.244 × 100
1.556 × 100
0.53
7.624 × 10−1 1.03
3.944 × 10−1 0.95
2.257 × 10−1 0.81
1.472 × 10−1 0.62
1.061 × 10−1 0.47
8.064 × 10−2 0.40
proc. 2
order
1.729 × 100
1.084 × 100
0.67
3.937 × 10−1 1.46
1.365 × 10−1 1.53
5.331 × 10−2 1.36
2.539 × 10−2 1.07
1.716 × 10−2 0.56
1.312 × 10−2 0.39
proc. 3
order
1.716 × 100
8.585 × 10−1 1.00
2.461 × 10−1 1.80
8.015 × 10−2 1.62
3.234 × 10−2 1.31
9.605 × 10−3 1.75
3.625 × 10−3 1.41
1.416 × 10−3 1.36
Relative errors of the control in k · kL2 (Ω)
Numerical Results: Experiment 6
h
2−2
2−3
2−4
2−5
2−6
2−7
2−8
2−9
proc. 2
order
100
1.074 ×
1.123 × 100
8.438 × 10−1
6.000 × 10−1
4.728 × 10−1
4.8583 × 10−1
6.782 × 10−1
1.047 × 10−1
proc. 3
order
100
-0.06
0.41
0.49
0.34
-0.04
-0.48
-0.63
1.809 ×
1.685 × 100
9.972 × 10−1
1.038 × 10−1
4.498 × 10−1
2.583 × 10−1
1.787 × 10−1
1.254 × 10−1
Relative errors of the control in | · |H 1 (Ω)
0.10
0.76
0.71
0.44
0.80
0.53
0.51
Numerical Results: Experiment 7
Ω = a pentagon
(−0.5, 0.5)
(0, 0.5)
ψ(x) = 0.1
yd = sin 2π(x1 + 0.5)(x2 +PSfrag
0.5)replacements
(0.5, 0)
β = 10−3
(−0.5, −0.5)
α=
1
3
graded meshes
ky − yh kh + ku − uh kL2 (Ω) ≤ Ch
(0.5, −0.5)
Numerical Results: Experiment 7
T0 (left) and T1 (right)
Numerical Results: Experiment 7
h
2−3
2−4
2−5
2−6
2−7
2−8
2−9
relative error of
state in k · kh
5.153 × 10−1
3.433 × 10−1
1.984 × 10−1
1.096 × 10−1
6.013 × 10−2
3.284 × 10−2
1.780 × 10−2
order
0.59
0.79
0.86
0.87
0.87
0.88
relative error of
state in | · |H 1 (Ω)
2.246 × 10−1
1.099 × 10−1
4.501 × 10−2
1.501 × 10−2
5.728 × 10−3
1.949 × 10−4
6.608 × 10−4
order
1.03
1.29
1.58
1.39
1.56
1.56
Relative errors of the state in k · kh and | · |H 1 (Ω)
Numerical Results: Experiment 7
h
2−3
2−4
2−5
2−6
2−7
2−8
2−9
proc. 1
order
5.568 × 10−1
3.776 × 10−1 0.56
2.148 × 10−1 0.81
1.151 × 10−1 0.90
6.221 × 10−2 0.89
3.355 × 10−2 0.89
1.795 × 10−2 0.90
proc. 2
order
4.717 × 10−1
2.953 × 10−1 0.68
1.532 × 10−1 0.95
7.764 × 10−2 0.98
4.084 × 10−2 0.93
2.165 × 10−2 0.92
1.135 × 10−2 0.93
proc. 3
order
1.587 × 100
5.001 × 10−1 1.67
1.428 × 10−1 1.81
5.409 × 10−2 1.40
1.554 × 10−2 1.80
5.675 × 10−3 1.45
1.906 × 10−3 1.57
Relative errors of the control in k · kL2 (Ω)
Numerical Results: Experiment 7
h
2−3
2−4
2−5
2−6
2−7
2−8
2−9
proc. 3
order
100
2.6689 ×
1.5051 × 100
8.0510 × 10−1
5.3368 × 10−1
2.8460 × 10−1
2.0225 × 10−1
1.3918 × 10−1
0.83
0.90
0.59
0.91
0.49
0.54
Relative errors of the control in | · |H 1 (Ω)
Adaptive C0 Interior Penalty Methods
A Residual Based Error Estimator for Clamped Plates
ηh =
X
2
ηe,
1
e∈Eh
+
X
2
(ηe,
2
+
2
ηe,
3)
+
e∈Ehi
X
ηT2
1
2
,
T∈Th
where
ηe,1 =
σ
1
|e| 2
k[[∂uh /∂n]]kL2 (e)
1
ηe,2 = |e| 2 k[[∂ 2 uh /∂n2 ]]kL2 (e)
3
ηe,3 = |e| 2 k[[∂ 3 uh /∂n3 ]]kL2 (e)
ηT = h2T kf − ∆2 uh kL2 (T)
Brenner-Gudi-S (2010)
Adaptive C0 Interior Penalty Methods
Theorem ηh is a reliable and (locally) efficient estimator for
kũh − uh kh .
Since the functions u and ũh have singularities at the same
locations, we can use ηh as an error indicator in adaptive computations for u.
S.C. Brenner, J. Gedicke, S. and Y. Zhang
An a posteriori error analysis for a quadratic C0 interior penalty
method for the obstacle problem of clamped Kirchhoff plates
(in preparation)
Numerical Results for an L-Shaped Domain
Domain
Ω = (−0.5, 0.5)2 \ [0, 0.5]2
External Force
f =0
Obstacle Function
Coincidence Set
ψ(x) = 1 −
h
(x1 +0.25)2
0.22
+
x22
0.352
i
Figure 11. Adaptive mesh with about 3000 nodes for the
Numerical
Results
for an L-Shaped Domain
L-shape
domain example.
η uniform
||u−uh||h uniform
2
10
η adaptive
||u−u || adaptive
h h
1
10
1
0.4
1/2
0
10
1
2
3
10
10
0
4
10
5
10
Quadratic
-IPM on uniform
and adaptive
meshes
Figure
12. C
Convergence
rates for uniform
and adaptive
refinement for the L-shape domain example.
Numerical
Results
for an L-Shaped
Domain
FOURTH
ORDER OBSTACLE
AFEM
January 8, 2014
0.5
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−0.4
−0.5
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
Adaptive mesh with about 3000 nodes
Figure 11. Adaptive mesh with about 3000 nodes for the
L-shape domain example.
8
Figure 13. Adaptive mesh with about 2000 nodes for third
Numerical
for an
L-Shaped
Domain
(!!!) Results
order polynomials
for the
L-shape domain
example.
η uniform
||u−uh||h uniform
2
10
η adaptive
||u−uh||h adaptive
1
10
0
10
1
1
−1
10
3
10
0
4
10
Cubic14.
C -IPM
on uniform
and
adaptive
meshesreFigure
Convergence
rates for
uniform
and adaptive
finement and third (!!!) order polynomials for the L-shape
domain example.
FOURTHResults
ORDER OBSTACLE
AFEM
January 8, 2014
Numerical
for an L-Shaped
Domain
0.5
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−0.4
−0.5
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
Adaptive mesh with about 2000 nodes
Figure 13. Adaptive mesh with about 2000 nodes for third
(!!!) order polynomials for the L-shape domain example.
Open Problems
I
Sharp error estimates in k · kH 1 (Ω) and k · kL∞ (Ω)
I
Convergence in k · kH 1 (Ω) for optimal control generated by
the post-processing procedure 3
I
Convergence and optimality of adaptive methods
I
Fast solvers