Partial Differential Equations - Modelling and ... - ResearchGate
Partial Differential Equations - Modelling and ... - ResearchGate Partial Differential Equations - Modelling and ... - ResearchGate
172 B. Maury Proof. Let us show that the assumption of Proposition 3 is met. It is sufficient to prove that any ξ ∈ V ′ which vanishes over K identifies through b(·, ·) to some ˜ξ ∈ V , i.e. there exists ˜ξ ∈ V such that 〈ξ,v〉 = b(˜ξ,v) ∀v ∈ V. Note that, as ξ vanishes over K, it can be seen as a linear functional defined on K ⊥ , so that it is equivalent to establish that T : V −→ (K ⊥ ) ′ defined by ˜ξ ↦−→ ζ : 〈ζ,v〉 = b(˜ξ,v) ∀v ∈ K ⊥ is surjective. We denote by T ⋆ ∈L(K ⊥ ,V) the adjoint of T . For all w ∈ K ⊥ , |T ⋆ (T ⋆ w, v) b(w, v) (Bw,Bv) w| =sup =sup =sup v≠0 |v| v≠0 |v| v≠0 |v| ≥ |Bw|2 . |w| As B has closed range, |Bw| ≥C |w| for all w in (ker B) ⊥ = K ⊥ ,sothat |T ⋆ w|≥C 2 |w| ∀w ∈ K ⊥ , from which we conclude that T is surjective. Remark 1. Note that Proposition 3 is strictly stronger than its corollary. Consider the standard situation V = H 1 (Ω) where Ω is a smooth, bounded domain, a(u, v) = ∫ ∇u ·∇v, and 〈ϕ, v〉 = ∫ fv, where f is L 2 , and b(v, v) = ∫ ∂Ω v2 . In this situation the corollary cannot be used, because the trace operator from H 1 (Ω) ontoL 2 (∂Ω) does not have a close range. On the other hand, one can establish that ∫ ∂u 〈ξ,v〉 = ∂Ω ∂n v and, as the solution u is regular (u ∈ H 2 (Ω)), its normal derivative (in H 1/2 (∂Ω)) can be built as the trace of a function ˜ξ in H 1 (Ω), so that Proposition 3 holds true. We conclude this section by some considerations concerning the saddlepoint formulation of the constrained problem. We consider again the closed situation: Proposition 4. Under the assumptions of Corollary 1, there exists λ ∈ Λ such that a(u, v)+(λ, Bv) =〈ϕ, v〉 ∀v ∈ V. (5) The solution is unique in B(V ) (which identifies to Λ/ ker B ⋆ ). Proof. The proof of this standard property can be found in [BF91]. In fact, it has just been established in the proof of Corollary 1: λ is simply B ˜ξ. Asfor uniqueness in B(V ), consider two solutions λ 1 , λ 2 . The equation (5) implies that λ 2 − λ 1 is in ker B ⋆ = B(V ) ⊥ .
Numerical Analysis of a Finite Element/Volume Penalty Method 173 Proposition 5. Under the assumptions of Proposition 4 (the assumptions (1) and B(V ) is closed), we introduce λ ε = 1 ε Buε . Then |λ ε − λ| = O(ε), where λ is the unique solution of (5) in B(V ). Proof. Subtracting the variational formulations for u and u ε , we get (λ ε − λ, Bv) =a(u ε − u, v) ∀v ∈ V. Now, as the range of B is closed, and λ ε − λ ∈ B(V )=(kerB ⋆ ) ⊥ ,wehave the inf-sup condition (see, e.g., [BF91]) so that sup v∈V (λ ε − λ, Bv) |v| β |λ ε − λ| ≤sup (λε − λ, Bv) |v| ≥ β |λ ε − λ| , =sup a(uε − u, v) |v| ≤‖a‖|u ε − u| , which ensures the first order convergence thanks to Corollary 1. 2.2 Discretized Problem We consider now a family (V h ) h of inner approximation spaces (V h ⊂ V )and the associated penalized/discretized problems ⎧ ⎪⎨ Find u ε h ∈ V h such that Jh(u ε ε h) = inf J h(v), ε v∈V ⎪⎩ Jh(v ε h )= 1 2 a(v h,v h )+ 1 (6) 2ε b(v h,v h ) −〈ϕ, v h 〉. As far as we know, there does not exist any general theory which would give an upper bound for the error |u − u ε h | as the sum of a discretization error (typically h of h 1/2 for volume penalty, depending on whether the mesh is boundary-fitted or not), and a penalty error (typically ε for closed-range penalty terms). We propose here two general properties which are direct consequences of standard arguments. They are suboptimal in the sense that neither of them is optimal from both standpoints (discretization and penalty), but, at least, they make it possible to recover the behavior in extreme situations (when ε goes to 0 much quicker than h, and the opposite situation). We shall need the following lemma: Lemma 1. Under the assumptions (1), there exists C>0 such that b(u ε ,u ε ) ≤ Cε|u − u ε | .
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172 B. Maury<br />
Proof. Let us show that the assumption of Proposition 3 is met. It is sufficient<br />
to prove that any ξ ∈ V ′ which vanishes over K identifies through b(·, ·) to<br />
some ˜ξ ∈ V , i.e. there exists ˜ξ ∈ V such that<br />
〈ξ,v〉 = b(˜ξ,v) ∀v ∈ V.<br />
Note that, as ξ vanishes over K, it can be seen as a linear functional defined<br />
on K ⊥ , so that it is equivalent to establish that T : V −→ (K ⊥ ) ′ defined by<br />
˜ξ ↦−→ ζ : 〈ζ,v〉 = b(˜ξ,v)<br />
∀v ∈ K ⊥<br />
is surjective. We denote by T ⋆ ∈L(K ⊥ ,V) the adjoint of T . For all w ∈ K ⊥ ,<br />
|T ⋆ (T ⋆ w, v) b(w, v) (Bw,Bv)<br />
w| =sup =sup =sup<br />
v≠0 |v| v≠0 |v| v≠0 |v|<br />
≥ |Bw|2 .<br />
|w|<br />
As B has closed range, |Bw| ≥C |w| for all w in (ker B) ⊥ = K ⊥ ,sothat<br />
|T ⋆ w|≥C 2 |w| ∀w ∈ K ⊥ ,<br />
from which we conclude that T is surjective.<br />
Remark 1. Note that Proposition 3 is strictly stronger than its corollary. Consider<br />
the st<strong>and</strong>ard situation V = H 1 (Ω) where Ω is a smooth, bounded<br />
domain, a(u, v) = ∫ ∇u ·∇v, <strong>and</strong> 〈ϕ, v〉 = ∫ fv, where f is L 2 , <strong>and</strong><br />
b(v, v) = ∫ ∂Ω v2 . In this situation the corollary cannot be used, because the<br />
trace operator from H 1 (Ω) ontoL 2 (∂Ω) does not have a close range. On the<br />
other h<strong>and</strong>, one can establish that<br />
∫<br />
∂u<br />
〈ξ,v〉 =<br />
∂Ω ∂n v<br />
<strong>and</strong>, as the solution u is regular (u ∈ H 2 (Ω)), its normal derivative (in<br />
H 1/2 (∂Ω)) can be built as the trace of a function ˜ξ in H 1 (Ω), so that Proposition<br />
3 holds true.<br />
We conclude this section by some considerations concerning the saddlepoint<br />
formulation of the constrained problem. We consider again the closed<br />
situation:<br />
Proposition 4. Under the assumptions of Corollary 1, there exists λ ∈ Λ<br />
such that<br />
a(u, v)+(λ, Bv) =〈ϕ, v〉 ∀v ∈ V. (5)<br />
The solution is unique in B(V ) (which identifies to Λ/ ker B ⋆ ).<br />
Proof. The proof of this st<strong>and</strong>ard property can be found in [BF91]. In fact, it<br />
has just been established in the proof of Corollary 1: λ is simply B ˜ξ. Asfor<br />
uniqueness in B(V ), consider two solutions λ 1 , λ 2 . The equation (5) implies<br />
that λ 2 − λ 1 is in ker B ⋆ = B(V ) ⊥ .