Part III, Chapter 12 H(Curl) Finite Elements 12.1 the Lowest-Order Case

Part III, Chapter 12 H(Curl) Finite Elements 12.1 the Lowest-Order Case

Part III, Chapter 12 H(curl) finite elements The goal of this chapter is to construct Rd-valued finite elements (K, P ,Σ) with d 2, 3 such that (i) Pk,d P for some k 0 and (ii) the degrees of freedom∈{ in Σ }fully determine the⊂ tangential components≥ of the polynomials in P on all the faces of K. The first requirement is key for proving conver- gence rates on the interpolation error; the second one is key for constructing H(curl)-conforming finite element spaces (see Chapter 17). The finite ele- ments introduced in this chapter are used, e.g., in Chapter 36 to approx- imate (simplified forms of) Maxwell’s equations which constitute a funda- mental model in electromagnetism. The focus here is on defining a reference element and generating finite elements on the mesh cells. The interpolation error analysis is done in Chapter 13. We detail the construction for the sim- plicial N´ed´elec finite elements of the first kind; some alternative elements are outlined at the end of the chapter. 12.1 The lowest-order case Let us consider the lowest-order N´ed´elec finite element. Let d 2, 3 be the space dimension, and define the polynomial space ∈{ } N = P S , (12.1) 0,d 0,d ⊕ 1,d where S = q PH q(x) x =0 , i.e., 1,d { ∈ 1,d | · } 0 x3 x2 x2 x3 0 −x1 S1,2 = span −x1 , S1,3 = span , , . (12.2) −x2 x1 0 − n o d(d+1) The sum in (12.1) is indeed direct, so that dim(N0,d) = 2 =: d′ (i.e., d′ = 3 if d = 2 and d′ = 6 if d = 3); note that d′ is the number of edges of a d simplex in R . The space N0,d has several further interesting properties. (a) One has P N , in agreement with the first requirement stated above. 0,d ⊂ 0,d 178 Chapter 12. H(curl) finite elements (b) The gradient of v N0,d is skew-symmetric; indeed, only the component q S contributes to∈ the gradient, and differentiating q(x) x with respect ∈ 1,d · to xi and xj , i = j, yields ∂iqj + ∂j qi = 0. (c) If v N0,d is curl-free, then v is constant; indeed,6 v being curl-free means that v∈is symmetric, and hence ∇ v = 0 owing to (b). (d) The tangential component of v N0,d along an affine∇ line in Rd is constant along that line; indeed, let x, y∈be two distinct points on the line, say L, with tangent vector tL, then there is λ R such that t = λ(x y) and setting v = r + q with r P and q S ∈, we infer L − ∈ 0,d ∈ 1,d that v(x) tL v(y) tL = (q(x) q(y)) tL = λq(x y) (x y) = 0. Let K·be− a simplex· in Rd and− let ·collect the edges− · of −K. Any edge E EK ∈ K is oriented by fixing an edge vector tE so that tE ℓ2 = E ; conventionally, weE set t = z z where z , z are the two endpointsk k of| E| with p < q. We E q − p p q denote by Σ the collection of the following linear forms acting on N0,d: 1 σe (v)= (v t ) dl, E . (12.3) E E · E ∀ ∈EK | | ZE e Note that the unit of σE(v) is a length. A graphic representation of the degrees of freedom is shown in Figure 12.1. The arrow indicates the edge orientation. 4 3 1 3 1 2 2 Fig. 12.1. N0,d finite element in two (left) and three (right) dimensions. Proposition 12.1 (Face (edge) unisolvence, d = 2). Let v N0,2. Let e ∈ E K be an edge of K. Then σE(v)=0 implies that v E tE =0. ∈E | · Proof. Immediate since we have seen above that v E tE is constant. | · ⊓⊔ Proposition 12.2 (Finite element, d =2). (K,N0,2,Σ) is a finite element. Proof. Since dim(N0,2) = card(Σ) = 3, we just need to verify that the only function v N0,2 that annihilates the three degrees of freedom in Σ is zero. ∈ 2 This follows from Proposition 12.1 since span tE E K = R . { } ∈E ⊓⊔ The same results hold for d = 3, but the proofs are more intricate since the tangential component on an affine hyperplane of a function in N0,3 is not necessarily constant. Let F be a face of K and let us fix a unit normal ∈FK vector nF to F . There are two ways to define the tangential component of a function v on F : one can define it either as v n or as Π v = v × F F − Part III. Finite Element Interpolation 179 I 2 (v, nF )ℓ2 nF where ΠF = 3 nF nF is the matrix associated with the ℓ - orthogonal projection onto the− linear⊗ hyperplane parallel to F . We will use both definitions. The second one is more geometric, and the first one is more convenient when working with the operator; the two definitions produce 2 ∇× ℓ -orthogonal vectors since (v n ,Π v) 2 = 0, see Figure 12.2. × F F ℓ nF v ΠF (v) v nF F × Fig. 12.2. Two possible definitions of the tangential component. Proposition 12.3 (Face unisolvence, d = 3). Let v N . Let F ∈ 0,3 ∈ FK be a face of K and let F collect the three edges of K forming the boundary e E of F . Then σE (v)=0 for all E F implies that v F nF = 0. ∈E | × Proof. Let S2 be the unit simplex in R2. Let T : S2 F be defined by F → TF (0, 0) = zp, TF (1, 0) = zq, TF (0, 1) = zr, where zp, zq, zr are the three vertices of Fb enumerated by increasing index. Let J b be the 3 2 Jacobian F × matrix of TF . Note that by definition, the vector JF y is parallel to F and 2 TF (y) zp = JF y for any y R . Let v = r + q with r P0,3 and q S1,3. − T ∈ ∈ ∈ Let us set v := JF ΠF (v TF ). Let us show that v b N0,2. Indeed, for all y R2, we have ◦ ∈ ∈ b b b b T b (y, v(y)) 2 2 = (y, J Π (v(T (y)))) 2 2 b ℓ (R ) F F F ℓ (R ) T = (y, JF ΠF (r + q(TF (y))))ℓ2(R2) b b b b T b = (y, JF ΠF (r + q(zp)+ q(JF y)))ℓ2(R2) b T b = (y, JF ΠF (r + q(zp)))ℓ2(R2) + (JF y, q(JF y))ℓ2(R2). b b Setting c := JT Π (r + q(z )) R2 and using that q S , we infer that F F b p ∈ b∈ 1,3 b (y, v(y))ℓ2(R2) = (y, c)ℓ2(R2). Since v P1,2, v = r + q where r P0,2 and H ∈ ∈ 2 q P ; then, (y, r) 2 R2 + (y, q(y)) 2 R2 = (y, c) 2 R2 , for all y R . This ∈ 1,2 b ℓ ( ) ℓ ( ) ℓ ( ) ∈ impliesb b b that the quadraticb b form (y,bq(y))ℓ2(R2b) is zero;b b hence vb N0,2. Let 2 ∈ nowb E be any ofb theb three edgesb b ofb S . Then Eb =b TF (E) is oneb of the three edges of F . We obtain that b b b b b b b T (v t ) dl = (J Π (v T ) t ) dl · E F F ◦ F · E ZE b ZE b b b b Eb b = ((v T ) t ) dl = | | (v t ) dl = E σe (v)=0. ◦ F · E E · E | | E ZE ZE b | b| b b 180 Chapter 12. H(curl) finite elements 2 Since v N0,2 annihilates the three edge degrees of freedom in S , Propo- ∈ T sition 12.2 implies that v = 0. Since im(ΠF ) is orthogonal to ker(JF ), we conclude that the tangential component of v on F is zero. b b ⊓⊔ b Proposition 12.4 (Finite element, d =3). (K,N0,3,Σ) is a finite element. Proof. Since dim(N0,3) = card(Σ) = 6, we just need to verify that the only function v N that annihilates the six degrees of freedom in Σ is zero. ∈ 0,3 Face unisolvence implies that v F nF = 0, for all F K. Let (e1, e2, e3) be the canonical basis of R3. Using| × (2.21), we infer that∈ F ( v) e dx = K ∇× · i ∂K (v nK ) ei ds = 0. Since v is actually constant on K, we have − v = ×0, and· we have seen that∇× this implies that v PR , i.e., v = p ∇×R ∈ 0,3 ∇ for some p P1,3. Integrating p along the edges of K, we infer that p takes the same value∈ at all the vertices∇ of K; hence p is constant, which in turn implies that v is zero. ⊓⊔ One can verify that the shape functions are given by θe (x)= λ (x) λ λ (x) λ , E , (12.4) E p ∇ q − q ∇ p ∀ ∈EK for all x K, where t = z z . The component of θe along any edge vector ∈ E q − p E t ′ , E′ , is constant; it is equal to one for E′ = E and to zero for E′ = E. E ∈EK 6 See Exercise 12.3 for additional properties of the N0,3 shape functions.

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