Traces and Duality Lemma
Recall the duality lemma with H 1/2 (∂Ω) := γ0 (H 1 (Ω)) defined as the trace
space of H 1 (Ω) endowed with minimal extension norm; i.e., for w ∈ H 1/2 (∂Ω) ⊂
L2 (∂Ω),
b H 1 (Ω) |w
b ∈ H 1 (Ω), γ0 w
b = w},
kwkH 1/2 (∂Ω) = min{kwk
H −1/2 (∂Ω) := dual to H 1/2 (∂Ω) =: H 1/2 (∂Ω)∗
!
= γν (H(div, Ω)).
Any q ∈ H(div, Ω) (i.e.
H −1/2 (∂Ω) by
q ∈ L2 (Ω, R2 ), div q ∈ L2 (Ω)) defines γν q ∈
(γν q)(w) =: hq · ν, wi∂Ω =
Z
(q · ∇ŵ + ŵ div q)dx
Ω
b ∈ H 1 (Ω) with γ0 w
b = w.
for w ∈ H 1/2 (∂Ω) and w
(Side note:
b ||| +k div qkkwk
b
hq · ν, wi∂Ω ≤ kqk ||| w
b H 1 (Ω)
≤ kqkH(div,Ω) kwk
implies kγν qkH −1/2 (∂Ω) ≤ kqkH(div,Ω) .)
Duality Lemma. (a) There exists exactly one
γν ∈ L(H(div, Ω); H −1/2 (∂Ω))
such that for all q ∈ H 1 (Ω; Rn )
γν q = (γ0 q) · ν a.e. on ∂Ω.
(b) Let h • , • i∂Ω denote the duality brackets of H −1/2 (∂Ω) × H 1/2 (∂Ω). All
q ∈ H(div, Ω) and v ∈ H 1 (Ω) satisfy the formula
Z hγν q, γ0 vi∂Ω =
v div q + q · ∇v dx.
Ω
1
(c) The operator γν is surjective and
k • kH(div)
ker γν = H0 (div, Ω) := D(Ω; Rn )
.
(d) (Duality lemma) For all t ∈ H −1/2 (∂Ω) with t(w) =: ht, wi∂Ω for w ∈
H 1/2 (∂Ω), it holds
ktkH −1/2 (∂Ω) =
ht, wi∂Ω
sup
w∈H 1/2 (∂Ω),
kwk 1/2 =1
H
= sup
v∈H 1 (Ω),
γ0 v6=0
=
inf
ϕ∈H
q∈H(div,Ω),
γν q=t
ht, γ0 vi∂Ω
(Ω) kv − ϕkH 1 (Ω)
inf1
kqkH(div,Ω) .
Proof. Proof of (a). Let q ∈ H(div, Ω). For all v ∈ H 1/2 (∂Ω) v̂ ∈ H 1 (Ω)
denotes the unique weak solution of
−∆v̂ + v̂ = 0 in Ω,
γ0 v̂ = v on ∂Ω.
Then kvkH 1/2 (∂Ω) = kv̂kH 1 (Ω) . Define
Z
Xq (v) := (v̂ div q + q · ∇v̂)dx
Ω
The repeated application of the Cauchy Schwarz inequality shows
Xq (v) ≤ kv̂kL2 (Ω) k div qkL2 (Ω) + kqkL2 (Ω) k∇v̂kL2 (Ω)
≤ kqkH(div) kv̂kH 1 (Ω) = kqkH(div) kvkH −1/2 (∂Ω) .
Hence Xq : H 1/2 (∂Ω) → R is linear and bounded. Thus for any q ∈ H(div, Ω)
there exists g(q) ∈ H −1/2 (∂Ω) with Xq = g(q). Define γν : q 7→ g(q). This operator is linear. The last inequality shows that the operator is also bounded,
more precisely
kγν kL(H(div,Ω);H −1/2 (∂Ω)) ≤ 1.
Moreover, for all functions v ∈ H 1/2 (∂Ω) and q ∈ H 1 (Ω, Rn ), an integration
by parts leads to
Z
h(γ0 q) · ν, vi∂Ω = h(γ0 q) · ν, γ0 v̂i∂Ω = (v̂ div q + q · ∇v̂)dx,
Ω
2
Thus h(γ0 q) · ν, vi∂Ω = hγν q, vi∂Ω . Hence, for all v ∈ H 1/2 (∂Ω) it holds
h(γ0 q) · ν − γν q, vi∂Ω = 0,
i.e., (γ0 q)·ν = γν q ∈ H −1/2 (∂Ω). This implies (a). Moreover, h • , • i∂Ω extends
the scalar product ( • , • )∂Ω in L2 (∂Ω) for smooth functions.
Proof of (b). For all v ∈ H 1 (Ω) and q ∈ H(div, Ω) it holds
Z hγν q, γ0 vi∂Ω =
q · ∇v̂ + v̂ div q dx,
Ω
where v̂ ∈ H 1 (Ω) is such that ∆v̂ + v̂ = 0 and γ0 v = γ0 v̂ in the weak sense.
Since ker γ0 = H01 (Ω) and v − v̂ ∈ H01 (Ω), this equals
Z
Z γν q · γ0 vds =
q · ∇v + v div q dx
∂Ω
Ω
and implies (b).
Proof of (c). For all q ∈ D(Ω; Rn ), γν q = (γ0 q) · ν = 0 a.e. by (a). Hence,
k • kH(div)
H0 (div, Ω) = D(Ω; Rn )
⊆ ker γν .
The proof of ker γν ⊆ H0 (div, Ω) is more technical and can be found in the
literature, i.e., in [Girault, V. and Raviart, P. A., Finite Element Methods
for Navier-Stokes Equations, Springer-Verlag, Berlin, Heidelberg, New York
(1986)]. It remains to show the surjectivity of γν . Given any t ∈ H −1/2 (∂Ω),
the functional
T : H 1 (Ω) → R, v 7→ ht, γ0 vi∂Ω
is linear and bounded, written T ∈ H 1 (Ω)∗ . The Riesz representation z ∈
H 1 (Ω) of T in the Hilbert space H 1 (Ω) satisfies hz, • iH 1 (Ω) = T ( • ). For
ϕ ∈ D(Ω), it follows for γ0 ϕ = 0 that
hz, ϕiH 1 (Ω) = T (ϕ) = ht, γ0 ϕi = 0.
This proves −∆z + z = 0 in the weak sense. In particular, q := ∇z ∈
L2 (Ω; Rn ) and div ∇z = ∆z leads to div q = z ∈ L2 (Ω). Hence, q ∈ H(div, Ω)
and
kqkH(div) = (k div qk2 + kqk2 )1/2 = (kzk2 + k∇zk2 )1/2
= kzkH 1 (Ω) = kT k(H 1 (Ω))∗ .
3
For any v ∈ H 1 (Ω), it follows
Z Z hγν q, γ0 vi∂Ω =
q · ∇v + v div q dx =
∇z · ∇v + vz dx
Ω
Ω
= hz, viH 1 (Ω) = T (v) = ht, γ0 vi∂Ω .
This implies hγν q − t, γ0 vi = 0 for all v ∈ H 1 (Ω), which is γν q − t = 0 in
H −1/2 (∂Ω). Consequently, t = γν q ∈ R(γν ).
Proof of (d). For any t ∈ H −1/2 (∂Ω) let z and q be as above in the proof of
(c). Then
ktkH −1/2 (∂Ω) =
sup
ht, γ0 v̂i
v̂∈H 1 (Ω)
kγ0 v̂k 1/2
=1
H
(∂Ω)
with ht, γ0 v̂i = hγν ∇z, γ0 v̂i = hz, v̂iH 1 (Ω) ≤ kzkH 1 (Ω) kv̂kH 1 (Ω) . Since kv̂kH 1 (Ω) =
1, this implies ktkH −1/2 (∂Ω) ≤ kzkH 1 (Ω) . Conversely, ht, γ0 zi = hz, ziH 1 (Ω) =
kzk2H 1 (Ω) proves ktkH −1/2 (∂Ω) ≥ kzkH 1 (Ω) .
This concludes the proof and characterizes H −1/2 (∂Ω) completely.
Primal PMP with test functions in H 1 (Ω) without (BC) leads to
!
b(u, t; v) = a(u, v) − ht, vi∂Ω = F (v) for all v ∈ H 1 (Ω).
(P)
Theorem. u solves (PMP) ⇐⇒ (u, γν ∇u) solves (P).
Proof. ”⇒” v ∈ H01 (Ω) implies ht, vi∂Ω = 0. Hence u solves (PMP).
1
”⇐” Let u ∈ H0 (Ω) solve (PMP), then p := ∇u ∈ H(div, Ω) leads to t :=
γν p ∈ H −1/2 (∂Ω) so that, for all v ∈ H 1 (Ω), it follows it follows
ht, vi∂Ω = hp · ν, γ0 (v)i∂Ω
Z
= ( p ·∇v + v div p)dx = a(u, v) − F (v).
| {z }
Ω |{z}
=∇u
=−f
Define H0 (div, Ω) := {q ∈ H(div, Ω)|γν q = 0}.
4
Interface trace spaces
For a shape-regular triangulation T of Ω ⊂ Rn into simplices define
Y
H −1/2 (∂K)|∃q ∈ H(div, Ω)∀K ∈ T ,
H −1/2 (∂T ) := {(tK )K∈T ∈
K∈T
γν (q|K ) = tK ∈ H −1/2 (∂K)}
endowed with the norm
k(tK )K∈T kH −1/2 (∂T ) := min{kqkH(div,Ω) |∀K ∈ T , γν (q|K ) = tK }
and
H 1 (T ) := {v ∈ L2 (Ω)|∀K ∈ T , v|K ∈ H 1 (K)}
Y
=
H 1 (K)
K∈T
with
k(vK )K∈T kH 1 (T ) :=
sX
kv|K k2H 1 (K) .
K∈T
Given t ≡ (tK )K∈T ∈ H −1/2 (∂T ) and v ∈ H 1 (T ) define
X
ht, vi∂T :=
htK , v|K i∂K .
K∈T
There exists q ∈ H(div, Ω) such that
tK = γν (q|K ) ∈ H −1/2 (∂K)
for all K ∈ T
and
ht, vi∂T =
XZ
K∈T
(q · ∇v + v div q)dx =
K
Z
(q · ∇N C v + v · div q)dtx
Ω
!
≤ kqkH(div,Ω) kvkH 1 (T ) = ktkH −1/2 (∂T ) kvkH 1 (Ω)
Define
(
b : (H01 (Ω) × H −1/2 (∂T )) × H 1 (T ) → R
b(u, t; v) := ((u, t), v) 7→ aN C (u, v) − ht, vi∂T
5
Theorem. u solves (PMP) and t = (tK )K∈T = (γν (∇u|K ))K∈T if and only
if (u, t) ∈ H01 (Ω) × H −1/2 (∂T ) solves
b(u, t; v) = F (v)
for all v ∈ H 1 (T ).
Proof. The Proof is left as an exercise.
Remark. ht, vi∂T = 0 for v ∈ H01 (Ω).
inf-sup Condition
This section is devoted to some immediate estimation for β > 0. Recall
X := X1 × X2 := H01 (Ω) × H −1/2 (∂T ), Y := H 1 (T ) and the bounded bilinear
form b : X × Y → R with
b(u, t; v) = b((u, t), v) = aNC (u, v) − ht, vi∂T
∀ (u, t) ∈ X, v ∈ Y.
For any (u, t) ∈ S(X) and v ∈ S(Y ) the Cauchy-Schwarz inequality leads
to
b(u, t; v) ≤ ||| u ||| ||| v |||NC +ktkH −1/2 (∂T ) kvkH 1 (T )
q
q
2
2
≤ ||| u ||| +ktkH −1/2 (∂T ) ||| v |||2NC +kvk2H 1 (T ) .
1
With
√ ||| v |||NC ≤ kvkH (T ) the choice of (u, t) and v finally shows, that b(u, t; v)
≤ 2. Given (u, t) ∈ S(X) set M := kb(u, t; • )kH 1 (T )∗ . For u 6= 0 choose
v := u/kukH 1 (T ) to obtain
Z Z
X
ht, ui∂K =
q · ∇u + u div q dx =
u q · νds = 0.
ht, ui∂T =
K∈T
Ω
∂Ω
Hence,
aNC (u, u)
||| u |||2
b(u, t; v) =
=q
.
kukH 1 (T )
2
2
kuk + ||| u |||
The Friedrichs inequality implies kuk ≤ CF (Ω) ||| u ||| with CF ≤ width(Ω)/π.
This leads to
||| u |||
b(u, t; v) ≤ p
≤ M.
1 + CF2 (Ω)
6
Hence
q
||| u ||| ≤ M 1 + CF2 (Ω).
(1)
Given t let q ∈ H(div, Ω) have minimal extension norm in H(div, Ω) with
q · ν = t on ∂K for all K ∈ T . The duality lemma leads to some v ∈ H 1 (T )
with kvkH 1 (T ) = 1 and ktkH −1/2 (∂T ) = ht, vi∂T (i.e. v is the normed Riesz
representation of ht, • i∂T in H 1 (T )). This implies
− ||| u ||| ||| v |||NC +ktkH −1/2 = aNC (u, v) + ktkH −1/2 = b(u, t; v) ≤ M,
whence
ktkH −1/2 (∂T ) − ||| u ||| ≤ M.
(2)
The inequalities (1) and (2) show that
2
1 = ||| u |||
+ktk2H −1/2 (∂T )
q
≤ M (1 + 1 + CF2 (Ω))2 + M 2 (1 + CF2 (Ω)).
2
This leads to
1
2
2
• )k 1
p
≤
M
=
kb(u,
t;
H (T )∗ .
(1 + CF2 (Ω))2 + 1 + CF2 (Ω)
Since this holds for all (u, t) ∈ S(X), it implies
1
0<q
≤ β = inf
sup b(u, t; v).
p
x∈S(X) v∈S(H 1 (T ))
2
2
3 + 2CF (Ω) + 2 1 + CF (Ω)
Splitting Lemmas
Splitting Lemma I. Given real Hilbert spaces X1 , X2 , X := X1 × X2 , {0} =
6
Y1 ⊆ Y and bounded bilinear forms bj : Xj → Y for j = 1, 2, let b : X × Y →
R, (x1 , x2 ; y) 7→ b1 (x1 , y) + b2 (x2 , y). Suppose
(A1) 0 < β1 :=
(A2) 0 < β2 :=
inf
sup b1 (x1 , y1 ),
inf
sup b2 (x2 , y),
x1 ∈S(X1 ) y1 ∈S(Y1 )
x2 ∈S(X2 ) y∈S(Y )
(A3) b2 |X2 ×Y1 = 0.
7
Then b satisfies an inf-sup condition with β > 0 and
0< p
β1 β2
≤ β ≤ β2 .
(β1 + kb1 k)2 + β22
Example (Application to primal dPG for PMP). Let X1 := H01 (Ω), X2 :=
H −1/2 (∂T ) and Y1 := H01 (Ω) ⊆ H 1 (T ) =: Y .
Ad (A1). Show that
β1 =
inf
1
sup
u∈H0 (Ω),||| u |||=1 v∈H 1 (Ω),||| v |||2 +kvk2 =1
a(u, v) = p
0
1
.
1 + CF2 (Ω)
Proof of “≤” is as above. For “≥” utilize the first Dirichlet eigenpair
1/2
−1/2
(λ1 , Φ1 ) with√
kΦ1 k = 1 and ||| Φ1 ||| = λ1 so that CF (Ω) = λ1
and
kΦ1 kH 1 (T ) = 1 + λ1 . Consequently
β1 ≤
a(Φ1 , v)
.
v∈H01 (Ω),kvkH 1 (Ω) =1 ||| Φ1 |||
sup
Since the eigenvectors (Φj )j∈N form an L2 -orthonormal and a-orthogonal
basis of H01 (Ω) the supremum is attained by v := Φ1 /kΦ1 kH 1 ()Ω . This leads
to
√
1
||| Φ1 |||2
||| Φ1 |||
1
λ1
=p
β1 ≤
=
=√
=√
.
||| Φ1 ||| kΦ1 kH 1 (Ω)
kΦ1 kH 1 (Ω)
1 + λ1
1 + λ−1
1 + CF2 (Ω)
Ad (A2). By duality lemma it holds
β2 :=
inf
sup
t∈S(H −1/2 (∂T )) v∈S(H 1 (T ))
−ht, vi∂T =
inf
t∈S(H −1/2 (∂T ))
ktkH −1/2 (∂T ) = 1.
Ad (A3). The Cauchy-Schwarz inequality implies
kb1 k =
sup
sup
aNC (u, v) ≤
u∈S(H01 (Ω)) v∈S(H 1 (T ))
sup
||| u ||| ||| v ||| ≤ 1.
sup
u∈S(H01 (Ω)) v∈S(H 1 (T ))
The proof of kb1 k ≥ 1 is left as an exercise. This leads to the inf-sup estimate
q
1+
CF2 (Ω)
1
p
+( 1+
CF2 (Ω)
+
1)2
=q
8
1
3+
2CF2 (Ω)
≤ β.
p
+2 1+
CF2 (Ω)
Proof of the first
√ splitting lemma. Given (x1 , x2 ) ∈ S(X1 ×X2 ) let s := kx1 kX1
and kx2 kX2 = 1 − s2 for 0 ≤ s ≤ 1. Then (A1) and (A3) imply
β1 s ≤ kb1 (x1 , • )kY1∗ = kb(x1 , x2 ; • )kY1∗ =: M.
Moreover, (A2), the definition of b and triangle inequality show that
p
β2 1 − s2 ≤ kb2 (x2 , • )kY∗ ≤ kb(x1 , x2 ; • )kY∗ + kb1 (x1 , • )kY∗ ≤ M + kb1 ks.
Consequently
f (s) := max{β1 s, β2
p
1 − s2 − kb1 ks} ≤ M.
It remains to compute min f := min0≤s≤1 f (s) ≤ M . Since (x√
1 , x2 ) ∈ S(X) is
arbitrary, this lead to β0 ≤ β. The monotony of β1 s and β2 1 − s2 − kb1 ks
shows that the minimizer s exists in (0, 1) with
p
(kb1 k + β1 )s = β2 1 − s2 .
√
Set κ := β2 /(β1 + kb1 k), so s2 = κ2 (1 − s2 ), whence s = κ/ 1 + κ. Consequently,
β0 = √
β1 κ
1 + κ2
concludes the proof.
Splitting Lemma II. In addition to the notation of the first splitting lemma
with (A1)-(A2), suppose
Y1 := {y ∈ Y | b2 ( • , y) = 0 in X2 }
(then (A3) follows and characterizes maximal Y1 in (A3)) and
N1 := {y1 ∈ Y1 | b1 ( • , y1 ) = 0 in X1 } = {0}.
Then
N := {y ∈ Y | b( • , y) = 0 in X} = 0
and
√
β := q
β12
+
β22
+ kb1
k2
+
p
2β1 β2
(β12
9
+
≤ β.
β22
+ kb1
k2 )2
−
4β12 β22
Example (Application to primal dPG for PMP). Given v ∈ Y1 , then for any
q ∈ H(div, Ω) follows
Z
0 = (v div q + q · ∇NC v) dx.
Ω
Hence, any α = 1, 2 and ϕ ∈ H 1 (Ω) satisfy
Z
0 = (v∂ϕ/∂α + ϕ eα · ∇NC v) dx.
Ω
Hence, ∇NC v is the weak gradient of v ∈ L2 (Ω), i.e. v ∈ H 1 (Ω). Consequently,
Z
v q · ν ds = 0 for all q ∈ H(div, Ω).
∂Ω
This implies v = 0 on ∂Ω, whence v ∈ H01 (Ω). Consequently,
Y1 = {v ∈ H 1 (T )|∀ t ∈ H −1/2 (∂T ), ht, vi∂T = 0} = H01 (Ω).
Moreover,
N1 := {w ∈ H01 (Ω)| aNC ( • , w) = 0 in H01 (Ω)} = {0}.
p
Recall 1 = β2 = kb1 k and 1/ 1 + CF2 (Ω) and compute
√
2
β0 ≤ β = q
p
2(1 + CF2 (Ω)) + 1 + (3 + 2CF2 (Ω))2 − 4(1 + CF2 (Ω))
√
2
=q
.
p
2
2
2
3 + 2CF (Ω) + 5 + 4CF (Ω) + 8CF (Ω)
Proof of the second splitting lemma. Since Y1 is a closed subspace of the Hilbert space Y , there is an orthogonal decomposition Y = Y1 ⊕Y2 with Y1⊥ = Y2 .
Then
0<
inf
sup b2 (x2 , y) =
x2 ∈S(X2 ) y∈S(Y )
inf
sup b2 (x2 , y2 ) = β2 .
x2 ∈S(X2 ) y2 ∈S(Y2 )
Any y2 ∈ Y2 with b2 ( • , y2 ) = 0 in X2 belongs to Y1 , whence y2 ∈ Y1 ∩ Y2 =
{0}. Consequently, b2 |X2 ×Y2 satisfies inf-sup condition with β2 and is nondegenerate. General theory of bilinear forms shows
β2 =
inf
sup b2 (x2 , y2 ) > 0.
y2 ∈S(Y2 ) x2 ∈S(X2 )
10
Given any (x1 , x2 ) ∈ S(X1 × X2 ) there exists a unique solution y2 ∈ Y2 to
b2 ( • , y2 ) = hx2 , • iX2 . From Riesz isomorphism follows kb( • , y2 )kX2∗ = kx2 kX2 .
Then for any y2 ∈ Y2
β2 ky2 kY ≤ kb2 ( • , y2 )kX2∗ = kx2 kX2 .
Since β1 > 0 and N1 = {0}, b1 |X1 ×Y1 satisfies inf-sup conditions and is nondegenerate, whence there exists a unique solution y1 ∈ Y1 to
b1 ( • , y1 ) = h • , x1 iX1 − b1 ( • , y2 ) in X1 .
Consequently,
β1 kY1 kY ≤ kb1 ( • , y1 )kX1∗ ≤ kx1 kX1 + kb1 kky2 kY .
Altogether
b(x, y1 + y2 ) = b1 (x1 , y1 + y2 ) + b2 (x2 , y1 + y2 )
= kx1 k2X1 + b2 (x2 , y2 ) = kx1 k2X1 + kx2 k2Y2 = 1.
On the other hand,
ky1 + y2 k2Y = ky1 k2Y + ky2 k2Y
1
≤ 2 (kx1 kX1 + kb1 kkx2 kX2 /β2 )2 + kx2 k2X2 /β22
β1
β1−2
kb1 kβ1−2 β2−1
kx1 kX1
= (kx1 kX1 , kx2 kX2 )
kx2 kX2
kb1 kβ1−2 β2−1 β2−2 (1 + kb1 k2 /β12 )
is bounded from above by the maximal eigenvalue Λ of the 2 × 2 matrix
!
1
kb
k/β
1
2
2
.
β1−2
1k
kb1 k/β2 β1 +kb
β2
2
This implies
Λ−1/2 ≤
b(x, y1 + y2 )
≤ kb(x, • )kY ∗ .
ky1 + y2 kY
Since x ∈ S(X) is arbitrary, this proves β ≥ Λ−1/2 . The formula follows from
explicit calculations of the above 2 × 2 matrix.
11
Discretization
Define for k ∈ N0
S0k+1 (T ) ⊂ X1 = H01 (Ω)
Pk (E) ⊂ X2 = H −1/2 (∂T )
Xh := S0k+1 × Pk (E)
Yh := Pk+d (T ) ⊂ Y = H 1 (T ).
Suggest d = dimension of domain and all k ∈ N0 . This lecture studies d = 1
for n = 2 space dimensions and k = 0.
Remark ( on P0 (E) ⊂ H −1/2 (∂T )). Given any t0 ∈ P0 (E).
RT0 (T ) ⊂ H(div, Ω) satisfy
Let τRT ∈
∀E ∈ E : t0 = τRT · νE on E.
Then
ht0 , vi∂T =
XZ
(τRT |K · νK )v ds
∂K
K∈T
for all v ∈ H 1 (T ).
Discrete duality lemma. For any t0 ∈ P0 (E) there exists exactly one pRT ∈
RT0 (T ) ⊆ H(div, Ω) such that for all K ∈ T and E ∈ E(K)
(νE · νK |E )t0 = (pRT · νK )|E .
Then
r
kt0 kH −1/2 (∂T ) ≤ kpRT kH(div,Ω) ≤
1+
h2max
kt0 kH −1/2 (∂T ) .
π2
Proof. Recall that kt0 kH −1/2 (∂T ) is the minimum of all kqkH(div,Ω) for any q ∈
H(div, Ω) with
(νE · νK |E )t0 = (q · νK )|E
for all K ∈ T , E ∈ E(K).
(3)
This proves the first inequality. Given any q ∈ H(div, Ω), (pRT − q) · νK = 0
on ∂K defined by the integration-by-parts formula. In particular
Z
Z
0=
(pRT − q) · νK ds =
div(pRT − q) dx for all K ∈ T .
∂K
K
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Consequently,
div pRT = Π0 div q
a.e. in Ω.
An integration by parts shows for any v ∈ H 1 (T ) that
Z
Z
| (pRT − q) · ∇NC v dx| = | (v − Π0 v) div(q − pRT ) dx|
Ω
Ω
≤ hmax /π ||| v |||NC k(1 − Π0 ) div qkL2 (Ω) .
Set v(x) := (Π0 pRT ) · (x − mid(K)) + 1/4(div pRT )|x − mid(K)|2 with
∇NC v = Π0 pRT + 1/2(div pRT ) + 1/2 div pRT ( • − mid(T )) = pRT
in the previous estimate to deduce
Z
Z
2
kpRT kL2 (Ω) =
q · pRT dx + (pRT − q) · ∇NC v dx
Ω
Ω
≤ kqkL2 (Ω) kpRT kL2 (Ω) +
hmax
kpRT kL2 (Ω) k(1 − Π0 ) div qkL2 (Ω) , (4)
π
whence
kpRT kL2 (Ω) ≤ kqkL2 (Ω) +
hmax
k(1 − Π0 ) div qkL2 (Ω) .
π
This and (4) imply with λ = hmax /π
kpRT k2H(div,Ω) ≤ (kqkL2 (Ω) + λk(1 − Π0 ) div qkL2 (Ω) )2 + kΠ0 div qkL2 (Ω)
≤ (1 + λ2 )kqk2L2 (Ω) + (1 + 1/λ2 )λ2 k(1 − Π0 ) div qk2L2 (Ω)
≤ (1 + λ2 )(kqk2L2 (Ω) + k(1 − Π0 ) div qk2L2 (Ω) + kΠ0 div qk).
p
In other words, kpRT kH(div,Ω) / 1 + h2max /π 2 is a lower bound of kqkH(div,Ω)
for all q with (3). By definition of kt0 kH 1/2 (∂T ) as the minimum, this shows
kpRT kH(div,Ω)
p
≤ kt0 kH −1/2 (∂T ) .
1 + h2max /π 2
loc
Annulation property for P := INC
: H 1 (T ) → H 1 (T ) projection onto P1 (T )
defined by
X Z
loc
INC v|K :=
− (v|K ) dsΨE |K ∈ P1 (K) for any v ∈ H 1 (T ), K ∈ T .
E∈E(K) E
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Given any v ∈ H 1 (T ),
q
q
2
loc
loc
2
k(1 − P )vkY = kv − INC vk + ||| v − INC v ||| ≤ 1 + κ2 h2max ||| v |||NC .
Consequently, the Kato lemma implies
kP k = k1 − P k ≤
q
1 + κ2 h2max .
loc
Mean value property of the gradients Π0 ∇NC INC
v for all v ∈ H 1 (T ) leads to
the annulation property
XZ
loc
t0 (v|K − INC
v|K ) ds = ht0 , v − P vi∂T .
K∈T
∂K
Hence, for all xh = (uc , t0 ) ∈ Xh and v ∈ H 1 (T ), it follows
b(xh , v − P v) = aNC (uc , v − P v) − ht0 , v − P vi∂T = 0.
The abstract theory asserts discrete inf-sup condition with β ≤ kP kβh ≤ kbk.
This shows
p
β
≤ βh .
1 + κ2 h2max
The a posteriori analysis involves kF ◦ (1 − P )kY ∗ ≤ κkhT f kL2 (T ) , which is
computable but not of higher order.
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