Math 5520 Lecture 28 Jeffrey Connors April 6, 2016 University of Connecticut

Math 5520 Lecture 28
Jeffrey Connors
University of Connecticut
April 6, 2016
Non-conforming FE spaces can also be considered
Consider the model problem
−∇ · A∇u = f , on Ω,
u = 0, on ∂Ω.
For conveniece, we will let A be constant and SPD. The weak
problem is to find u ∈ H01 (Ω) such that
a(u, v ) = L(v ), ∀v ∈ H01 (Ω).
Now, consider using the following FE space:
Vh =
v : v |E ∈ P1 (E ), ∀E ∈ τh
and v is continuous at all edge midpoints .
Then Vh * H01 (Ω) (non-conforming).
Break up the variational form by element
We define a “discrete” variational form
XZ
ah (u, v ) =
A∇u · ∇v dx, ∀u, v ∈ Vh + H01 (Ω).
E ∈τh
E
Note that a(u, v ) = ah (u, v ) if u, v ∈ H01 (Ω). The FE solution
uh ∈ Vh satisfies
ah (uh , vh ) = L(vh ), ∀vh ∈ Vh .
Now define a norm by
X
kw k2h =
|∇w |2L2 (E ) , ∀w ∈ Vh + H01 (Ω).
E ∈τh
We will have coercivity (constant α1 ) and continuity (constant α2 )
of ah (·, ·) over Vh + H01 (Ω), since A is constant and SPD.
The following result is a generalization of Ceá’s lemma
Lemma (“Second Strang Lemma”)
ku−uh kh ≤ C
|ah (u, wh ) − L(wh )|
inf ku − vh kh + sup
vh ∈Vh
kwh kh
wh ∈Vh \{0}
!
.
Proof. Apply the triangle inequality:
ku − uh kh ≤ ku − vh kh + kvh − uh kh .
Then we focus on bounding the term kvh − uh kh . Apply coercivity:
α1 kuh − vh k2h ≤ ah (uh − vh , uh − vh ).
Then add and subtract u in the first argument;
α1 kuh − vh k2h ≤ ah (u − vh , uh − vh )
+ ah (uh , uh − vh ) − ah (u, uh − vh ).
Proof of lemma...
It follows that
α1 kuh − vh k2h ≤ α2 ku − vh kh kuh − vh kh
+ |ah (u, uh − vh ) − L(uh − vh )|
|ah (u, uh − vh ) − L(uh − vh )|
⇒ kuh − vh kh ≤ C ku − vh kh +
kuh − vh kh
!
|ah (u, wh ) − L(wh )|
.
≤ C ku − vh kh + sup
kwh kh
wh ∈Vh \{0}
Now use the bound with the first inequality in the proof:
ku − uh kh ≤ C
|ah (u, wh ) − L(wh )|
ku − vh kh + sup
kwh kh
wh ∈Vh \{0}
!
The desired result follows by taking the infimum over vh ∈ Vh .
.
Bounding the approximation error
Assume:
I Ω is convex and polygonal,
I f ∈ L2 (Ω).
Then H 2 -ellipticity holds for our problem; u ∈ H 2 (Ω). Set
vh = Ih u in the lemma. We note next that
ku − Ih ukh ≤ C h|u|H 2 (Ω) .
This actually follows from our earlier interpolation error arguments;
the steps bounded the error on each element and then summed
over the elements. It remains to bound the consistency error term.
Since u ∈ H 2 ,
XZ
ah (u, wh ) − L(wh ) =
A∇u · ∇wh − fwh dx
E ∈τh
=
XZ
E ∈τh
E
E
− (∇ · A∇u + f ) wh dx +
XZ
E ∈τh
∂E
(A∇u · nE ) wh dσ.
Use the fact that we have a strong solution
It follows that
ah (u, wh ) − L(wh ) =
XZ
(A∇u · nE ) wh dσ.
∂E
E ∈τh
Denote the set of three edges of each E by SE , so e ∈ SE means
that e is an edge of E . Define
Z
1
w eh =
wh dσ.
|e| e
Then
ah (u, wh ) − L(wh ) =
X XZ
E ∈τh e∈SE
since
R
e∈SE1
A∇u · nE1 dσ +
R
e∈SE2
(A∇u · nE ) (wh − w eh ) dσ,
e
A∇u · nE2 dσ = 0.
Get into the form of error estimates over the edges
Note that since
Z
(wh − w eh ) dσ = 0,
e
we can add in the constant ∇(Ih u) · nE :
X XZ
ah (u, wh ) − L(wh ) =
A∇ (u − Ih u) · nE (wh − w eh ) dσ.
E ∈τh e∈SE
e
Apply Cauchy-Schwarz...
|ah (u, wh ) − L(wh )|
X X
≤
kA∇ (u − Ih u) · nE kL2 (e) kwh − w eh kL2 (e)
E ∈τh e∈SE
The terms with wh − w eh are easily bounded since wh is linear. The
interpolation errors are handled using previous results.
Apply the estimates below and the result follows
It holds that
1/2
kA∇ (u − Ih u) · nE kL2 (e) ≤ C hE |u|H 2 (E )
1/2
kwh − w eh kL2 (e) ≤ C hE |wh |H 1 (E ) .
Therefore,
|ah (u, wh ) − L(wh )| ≤ C
X
hE |u|H 2 (E ) |wh |H 1 (E )
E ∈τh
≤ C h |u|H 2 (Ω) kwh kh .
Then dividing through by kwh kh and taking the supremum over
wh ∈ Vh yields the desired result.