Many boundary value problems with non-homogeneous boundary data may be cast in the abstract variational form: find u ∈ X such that γu = g ∈ M and
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kvk 2 Hk+s
|v| 2 Hk+s
is the Slobodetskij seminorm and k · k Hk
kvk H−1/2
(b) kσ h k H(div;Ω) . kg h k H−1/2
(b) kw h k H(curl;Ω) . kr h k H−1/2
(b) kΠv − vk L2
h F kg h k 2 L2
h F kg h k 2 L2
h 2 F |g i,F | 2 kϕ i,F k 2 L2
kg i,F ϕ i,F k 2 H−1/2
(3.2) kuk H3/2+s
Theorem 3.3. Suppose that f ≡ 0 in (3.1) and let σ = ∇u. Then for each K ∈ T h we have kσk H1/2+s
kσ h k H(div;Ω) =kσ h k L2
.kg h k H−1/2
.kg h k H−1/2
h 2(1/2+s) K kσk 2 H1/2+s
kσk 2 L2
kσk 2 L2
h 2(1/2+s) K kg h k 2 Hs
h 2(1/2+s) F kg h k 2 Hs
h F kg h k 2 L2
h F kg h k 2 L2
kσ − σ h k H (div;Ω) + ku − u h k L2
ku − wk L2
kφ h k H1/2
(4.2) kφ h k 2 H1/2
α 2 j kφ j k 2 H1/2
kψ ` k 2 H−1/2
By equations (4.2) and (4.3) it follows that kφ h k 2 H1/2
α 2 j kcurl Γ φ j k 2 H−1/2
and the result is then a consequence of [1, Formula (5.13)]. Lemma 4.3. For all v h ∈ V h with div v h = 0, there exists w h ∈ N h such that ∇ × w h = v h and kw h k L2
kv h k H(div,Ω) . kdiv r h k H−1/2
km h k H−1/2
(4.5) kφ h k H1/2
(4.6) ku h k H1
This can be accomplished by first taking u ∈ H 1 (Ω) whose trace is φ h and satisfying kuk H1
kw h + ∇u h k H(curl,Ω) ≤ kw h k H(curl,Ω) + k∇u h k L2
ku − vk H(curl;Ω) + kg − g h || H−1/2
kD s ωk L∞
(A.2) kuk H3/2+s
∇(ωu) · n ≡ 0 on ∂Ω. To bound the first term we let v ∈ H 1/2−s (Ω), and define m(v) = |D 1K
4(ωu)v ≤ (k∇ω · ∇uk L2
. (h −1/2−s T k∇uk L2
k − 4(ωu)k H−1/2+s
kuk H3/2+s
ku − m(u)k L2
k∇uk H1/2+s
≤ (2k∇ω · ∇uk L2
. (h −1/2−s K k∇uk L2
(A.3) kvk L2
k − 4(ωu)k H−1/2+s
k∇(ωu) · nk Hs
. h −1/2−s K ku∇ωk L2
. h −3/2−s K kuk L2
kωgk Hs
kuk H3/2+s
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