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 elements 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 simplicial Nédélec 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 Nédélec 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.

12.2 The polynomial space Nk,d Let k N and let d 2, 3 (the material of this section extends to any ∈ H ∈ { } d 2). Let Pk,d be the space of the homogeneous polynomials of degree k ≥ H H d d introduced in Definition 11.2. Set Pk,d := [Pk,d] and Pk,d := [Pk,d] . d Definition 12.5 (Nk,d). We define the following real vector space of R - valued polynomials: N := P S , with S := q PH q(x) x =0 . (12.5) k,d k,d ⊕ k+1,d k+1,d { ∈ k+1,d | · } Note that the above sum is direct since P S P PH = 0 . k,d ∩ k+1,d ⊂ k,d ∩ k+1,d { } 2 T 2 T Example 12.6 (Space S2,d). The set ( x2, x1x2) , (x1x2, x1) is a ba- 2 { T− 2 T − } 2 T sis of S2,2, and the set ( x2, x1x2, 0) , ( x3, 0, x1x3) , (x1x2, x1, 0) , 2 T { −2 T 2−T T − (0, x3, x2x3) , (x1x3, 0, x1) , (0, x2x3, x2) , (x2x3, x1x3, 0) , (0, x1x3, x−x )T is a basis of S −. Note that dim(−S ) = 2 and− dim(S ) = 8. − 1 2 } 2,3 2,2 2,3 ⊓⊔ Lemma 12.7 (Dimension of N ). Let k N and d 2. Then, k,d ∈ ≥ (k + d + 1)! dim(N )= . (12.6) k,d k!(d 1)!(k + 2) − 1 Hence, dim(Nk,2) = (k + 1)(k + 3) and dim(Nk,3)= 2 (k + 1)(k + 3)(k + 4). Part III. Finite Element Interpolation 181

H H Proof. (1) Let us ﬁrst prove that the map φ : Pk,d p x p Pk+1,d is ∋ 7→ · ∈ H surjective. By linearity, it suﬃces to prove that for each monomial q Pk+1,d α H ∈ so that q(x)= x with α = k + 1, there is r Pk,d such that q(x)= x r(x). | | d ∈ · Let (ei)i 1: d be the Cartesian basis of R . Since α = k +1 1, there exists ∈{ } α1 | | αi 1 ≥ αd i 1:d so that αi 1. Then setting r(x)= x1 ...xi − ...xd ei, we have r∈{PH }and q(x)=≥x r(x). ∈ k,d · (2) Observing that ker(φ)= Sk,d, the rank Theorem implies that dim(Sk,d)+ H H H k+d 1 dim(im(φ)) = dim P , i.e., dim(S ) = d dim P dim P = d − k k,d k − k+1,d k − k+d k+d 1 k+d (k+d 1)! = − (d )= k − . The sum in (12.5) being direct, we k+1 k − k+1 (k+1)!(d 2)! conclude that −

dim(Nk,d)= d dim(Pk,d) + dim(Sk+1,d) (k + d)! (k + d)! (k + d + 1)! = + (k + 1) = . k!(d 1)! (k + 2)!(d 2)! k!(d 1)!(k + 2) ⊓⊔ − − − d Lemma 12.8 (Trace space). Let H be an aﬃne hyperplane in R , let nH d 1 be a unit normal vector to H, and let TH : R − H be an aﬃne bijective → 2 map with Jacobian matrix JH . Let ΠH = Id nH nH be the ℓ -orthogonal projector onto the tangent space to H (the linear− hyperplane⊗ in Rd parallel to T 1 H). Then, JH ΠH v H Nk,d 1 TH− for all v Nk,d. | ∈ − ◦ ∈ Proof. Identical to the proof of Proposition 12.3. ⊓⊔ Lemma 12.9 (d =2). N = R π (RT ), where R π is the rotation of angle k,2 2 k,2 2 π 2 T 2 T in R , i.e., for x = (x1, x2) R , R π x = ( x2, x1) . 2 ∈ 2 − Proof. See Exercise 12.4. ⊓⊔

Lemma 12.10 (Curl). Assume d 2, 3 . Then v Pk,d for all v N , and if v = 0, there is p P ∈ { such} that v∇×= p∈(that is, v P ).∈ k,d ∇× ∈ k+1,d ∇ ∈ k,d

Proof. That v Pk,d results from Nk,d Pk+1,d. The condition v = 0 together with∇×v ∈N P implies that⊂ there is p P such∇× that ∈ k,d ⊂ k+1,d ∈ k+2,d v = p. The deﬁnition of Nk,d implies that v = p1 + p2 with p1 Pk+1,d ∇ ∇1 ∇ ∈1 and p2 Sk+1,d. We infer that p2(x) p2(0)= 0 ( p2(tx),tx)ℓ2 t− dt = 0, which∇ means∈ that p is constant and that− v = p with∇ p P . 2 ∇R1 1 ∈ k+1,d ⊓⊔

12.3 Simplicial N´ed´elec elements

Let k N and let d 2, 3 . Let K be a simplex in Rd. In this section, ∈ ∈ { } we define degrees of freedom (dof’s) to make the triple (K,Nk,d,Σ) a finite element. The construction can be generalized to any dimension. 182 Chapter 12. H(curl) finite elements

12.3.1 Two-dimensional case Let us orient the three edges E of K by the edge vectors t . Let us ∈ EK E orient K by the two vectors tK,j j 1,2 which are the edge vectors for the two edges of K sharing the vertex{ } with∈{ lowest} enumeration index; note that 2 t is always a basis of R . Let µ e be a basis of P , K,j j 1,2 m m 1: nsh k,1 { } e∈{ } { } ∈{ } with n = dim(P ) = k + 1, and (if k 1) let ψ c be a basis sh k,1 m m 1:nsh c 1 ≥ { } ∈{ } of Pk 1,2, with nsh = dim(Pk 1,2) = k(k + 1). We define the dof’s of the − − 2 two-dimensional Nédélec element (K,Nk,2,Σ) by

e 1 1 e σ (v)= (v t )(µ T − ) dl, E , m 1:n , (12.7a) E,m E · E m ◦ E ∈EK ∈{ sh} | | ZE 1 σc (v)= (v t )ψ dx, j 1:2 , m 1:nc , (12.7b) j,m K · K,j m ∈{ } ∈{ sh} | | ZK 1 where TE is an affine bijective map from the unit simplex S = [0, 1] in R onto E. Since N = R π (RT ) owing to Lemma 12.9 and since the above k,2 2 k,2 degrees of freedom are those of the RTk,2 finite element once theb edges (faces) are oriented by the vectors ν := R π t and K is oriented by the vectors E 2 E ν := R π t , it follows from Proposition 11.15 that the triple (K,N ,Σ) K,j 2 K,j k,2 is a finite element for all k 0. The unit of all the above dof’s is a length. ≥ Remark 12.11 (2D Piola transformations). One can show that AT = 1 1 2 2 det(A)R−π A− R π for all A R × . Hence, the two-dimensional contravari- 2 2 ∈ c d ant and covariant Piola transformations satisfy R π ψ (v)= ψ (R π v). 2 K K 2 ⊓⊔ 12.3.2 Three-dimensional case Let K be a simplex (tetrahedron) in R3. Let collect the six edges of K EK and let K collect the four faces of K. Each edge E K is oriented by the edge vectorF t = z z where z , z are the two vertices∈E of E with p < q E q − p p q (note that t 2 = E ). Each face F is oriented by the two edge k Ekℓ | | ∈ FK vectors tF,j j 1,2 with tF,1 = zq zp, tF,2 = zr zp where zp, zq, zr are { } ∈{ } − − the three vertices of F with pzs zp, where zp, zq, zr, zs are the four − − − 3 vertices of K with p

E TF TE K K S1 S2 F b b Fig. 12.3. Reference edge S1 and reference face S1 with the corresponding maps. b b

1 c (k + 1)k (if k 1), and let ψm m 1: nc be a basis of Pk 2,3 with nsh = 2 ≥ { } ∈{ sh} − 1 (k + 1)k(k 1) (if k 2). The set Σ is deﬁned to be the collection of the 6 − ≥ following linear forms acting on Nk,3:

e 1 1 σ (v)= (v t )(µ T − ) dl, (12.8a) E,m E · E m ◦ E | | ZE f 1 1 σ (v)= (v t )(ζ T − ) ds, (12.8b) F,j,m F · F,j m ◦ F | | ZF 1 σc (v)= (v t )ψ dx, (12.8c) j,m K · K,j m | | ZK e for all E K , m 1:nsh in (12.8a), all F K , j 1, 2 , f∈ E ∈ { } ∈ F c ∈ { } m 1:nsh in (12.8b), and all j 1:3 , m 1:nsh in (12.8c). ∈ { }e e ∈f { } f ∈ { } We denote Σ := σ m 1: ne , Σ := σ f , and E E,m sh F F,j,m j 1,2 ,m 1: nsh c c { } ∈{ } { } ∈{ } ∈{ } Σ = σj,m j 1: 3 ,m 1: nc . { } ∈{ } ∈{ sh} Remark 12.13 (dof’s). The unit of all the dof’s is a length. For the cell c 2 dof’s, we could also have written σ (v)= ℓ− (v e )ψ dx, where ℓ is a j,m K K · j m K length scale associated with K and (ej )j 1: d is the natural Cartesian basis ∈{ } R of Rd. We will see that the definition (12.8c) is more natural when using the covariant Piola transformation to generate Nédélec finite elements. ⊓⊔ Lemma 12.14 (Invariance). Assume that the basis µm m 1: ne (resp., ∈{ sh} { } 1 1 ζm m 1: nf ) is globally invariant by any linear bijective map S : S S { } ∈{ sh} → 2 2 e f (resp., S : S S ). Then the set ΣE (resp., ΣF ) is independent of the affine → b b bijective map TE (resp., TF ). b b Proof. Similar to that of Lemma 11.12; see also Example 11.13 for invariance with respect to vertex permutation. ⊓⊔ The following result is important in view of H(curl)-conformity, Lemma 12.15 (Face unisolvence). Let v N and let F be a face ∈ k,3 ∈FK of K. Let F collect the (three) edges forming the boundary of F , and let nF be a unit normalE to F . The following equivalence holds:

f e [ σ(v)=0, σ ΣF ( E F ΣE)] [ v F nF = 0 ], (12.9) ∀ ∈ ∪ ∪ ∈E ⇐⇒ | ×

Moreover, both assertions in (12.9) imply that ( v) F nF =0. ∇× | · 184 Chapter 12. H(curl) ﬁnite elements

Proof. We only prove the implication in (12.9), since the converse is evident. The proof is an extension of that of Proposition 12.3 accounting for the richer structure of the degrees of freedom. We introduce v := JT Π (v T ); it can F F ◦ F be shown that v N , see Exercise 12.6. The unit simplex S2 is oriented by ∈ k,2 the two edge vectors tj j 1,2 so that JF tj = tF,jb TF , for all j 1, 2 . { } ∈{ } ◦ ∈ { } We obtain for theb face degrees of freedom that b b b 1 1 T (v tj )ζm ds = ((JF (v (v nF )nF ) TF ) tj )ζm ds S2 S2 · S2 S2 − · ◦ · | | Z | | Z b 1 b b b b b b b = b (((v (v nF )nF ) tF,j ) TF )ζm ds S2 S2 − · · ◦ | | Z 1 b = ((v tF,j ) TF )ζm ds b b2 2 · ◦ S ZS | | b 1 1 f = (v t )(ζ T − ) dsb= σ (v)=0. Fb · F,j m ◦ F F,j,m | | ZF One proves similarly that the edge degrees of freedom vanish. This proves T that v = 0 because v Nk,2. Since JF is full rank, we infer that ΠF v F = 0, ∈ | which implies that v F nF = (ΠF v F ) nF = 0. Finally, ( v) F nF = 0 | × | × ∇× | · immediatelyb followsb from v F nF = 0. | × ⊓⊔

Proposition 12.16 (Finite element). (K,Nk,3,Σ) is a ﬁnite element.

Proof. Observe ﬁrst that the cardinality of Σ can be evaluated as follows:

k +1 k +1 card(Σ)=3nc +2 4nf +6ne =3 +8 + 6(k + 1) sh × sh sh 3 2 1 = (k + 1)(k + 3)(k + 4) = dim(N ). 2 k,3 As a result, the statement will be proved once it is established that zero is the only function in Nk,3 that annihilates the degrees of freedom in Σ. Let v N be such that σ(v) = 0 for all σ in Σ. We are going to show that ∈ k,3 v = 0. Owing to Lemma 12.15, v F nF = 0 and ( v) F nF = 0 for any | × ∇× | · face F K. (1) Let∈F us prove that w := v = 0. Since w P RT , it suffices to ∇× ∈ k,3 ⊂ k,3 prove that w annihilates all the dof’s of the RTk,3 element. Since w F nF = 0, w annihilates all the dof’s associated with the faces of K. In addition,| · if k 1, we observe that w q dx = v q dx = v q dx, for all ≥ K · K ∇× · K ·∇× q Pk 1,3, where we have used that v nK = 0 on ∂K. This, in turn, implies ∈ − R ×R R c that K w q dx = 0 since q Pk 2,3 and σ(v) = 0 for all σ Σ if k 2; the statement· is obvious if∇×k =∈ 1. In− conclusion, v = w = 0∈. ≥ (2) UsingR Lemma 12.10, we infer that there is p ∇×P such that v = p. ∈ k+1,3 ∇ The condition v nK = 0 on ∂K implies that p is constant on ∂K. Without loss of generality,× we take this constant equal to zero. This, in turn, implies Part III. Finite Element Interpolation 185 that p = 0 if k 2 (see Exercise 5.4(iii)), while for k 3, we infer that ≤ ≥ p = λ0 ...λ3r where λi, i 0:3 , are the barycentric coordinates in K and ∈ { } α r Pk 3,3. Writing the polynomial r in the form r(x)= aαx , we − α k 3 ∈ 1 | |≤ − α consider the vector-valued function q(x) = aαx1x e1 where α k 3 α1+1P |3|≤ − e1 is the first vector of the canonical basis of R . Since q Pk 2,3, the fact P − that σ(v) = 0 for all σ Σc implies that v q dx =∈ 0. Integration by ∈ K · parts and the fact that p ∂K = 0 then yield 0 = v q dx = p q dx = | R K K λ ...λ r2 dx. In conclusion r = 0, so that v = · p = 0. − ∇· − K 0 3 R ∇ R ⊓⊔ RemarkR 12.17 (Literature). The above finite element has been introduced by Nédélec [186]; see also Weil [245], Whitney [248] for the lowest-order. It is an accepted practice in the literature to call this element either edge element or Nédélec element. See also Bossavit [41, Chap. 3], Hiptmair [149], Monk [180, Chap. 5], and Bonazzoli and Rapetti [35]. ⊓⊔

12.4 Generation of N´ed´elec elements

3 Let K be the reference simplex in R . Let h be an aﬃne simplicial mesh. Let K = T (K) be a mesh cell. Let F beT a face of K; we have F = T (F ) K ∈FK K whereb F is a face of K. Similarly, let E be an edge of K; we have ∈FK ∈EK E = T (Eb )b where E is an edge of K. Using increasing vertex-indexb K ∈ EK enumeration,b Theorem 7.14b shows that it is possible to orient the edges E and E in ab way thatb is compatible with the geometricb map TK ; that is to say, the edge vectors tE and tE are such that tE TK E = JK tE where JK is the ◦ | Jacobianb matrix of TK. Web also orient the faces byb using theb two edge vectors from the vertex with theb lowest index in each face. We ﬁnallyb orient the cells by using the three edge vectors from the vertex with the lowest index in each cell. As a result, we have

tE TK E = JKtE , tF,j TK F = JKtF ,j, tK,j TK = JKtK,j. (12.10) ◦ | ◦ | ◦ b b b b b Recall the covariantb Piola transformationb introduced in (7.11b):b

T ψc (v)= J (v T ), (12.11) K K ◦ K and the pullback by the geometric map such that ψg (q) := q T . K ◦ K Lemma 12.18 (Transformation of dof’s). Let v C0(K) and let q C0(K). The following holds: ∈ ∈ 186 Chapter 12. H(curl) ﬁnite elements

1 1 c g (v tE )q dl = (ψK (v) tE )ψK (q) dl, E K, E E · E E · ∀ ∈E | | Z | | Z b 1 1 b (v t )q ds = (ψc (v) tb )ψg (q)b ds, F , j 1, 2 F,j b K F ,j K K F F · F F · ∀ ∈F ∈{ } | | Z | | Z b 1 1 b (v t )q dx = (ψc (v)bt )ψg (q) dbx, j 1:3 . K,j b K K,j K K K · K K · ∀ ∈{ } Z Z b | | | | b Proof. The first identity is nothing butb (7.18b) (theb fact that TK is affine is not used here). The proofb of the two other identities follows from (12.10) and F K ds = | | ds, dx = | | dx since T is affine. F K K | | | | ⊓⊔ b b Proposition 12.19 (Generation). Let (K, P , Σ) be a simplicial Nédélec el- b b ement. Assume that the map TK is affine and that (12.10) holds. The finite element generated in K using Proposition b6.5bwithb the map (12.11) is a simplicial Nédélec element with the following degrees of freedom:

e 1 1 σ (v)= (v t )(µ T − ) dl, (12.12a) E,m E · E m ◦ K,E | | ZE f 1 1 σ (v)= (v t )(ζ T − ) ds, (12.12b) F,j,m F · F,j m ◦ K,F | | ZF 1 σc (v)= (v t )ψ dx, (12.12c) j,m K · K,j m | | ZK where T = T T : S1 E and T = T T : S2 F are the affine K,E K ◦ E → K,F K ◦ F → bijective maps mappingb vertices with increasing indices.b b b Proof. Let us first prove that P = Nk,3. Let TK (x) = JK x + bK with 3 3 3 c JK R × and bK R . Let v be a member of P , then ψK (v) = p + q, ∈ ∈ T 1 T 1 with p P and q S , yielding v = J− p T − + J− q T − . Since ∈ k,3 ∈ k+1,3 K ◦ Kb K b◦ K each component of q is a homogeneous polynomial of degree (k + 1), web inferb 1 1 1 1 that qb TK− (x) = bq(JK− x JK− bK ) = q(JK− x)+b r(x), wherebr Pk,3, see ◦ − T 1 ∈ Exercise 11.4. As a result,b v = (p + r)+ q, where p := JK− p TK− Pk,3 and T 1 1 ◦ ∈ H q := JbK− q JK− . Noteb that p + r Pk,3band q JK− is a member of Pk+1,3, ◦ ∈H ◦ T 1 which implies that q is also in Pk+1,3. Moreover, (x, JK− bq(JK− x))ℓ2(R3) = 1 1 (J− x, q(Jb− x)) 2 R3 = 0 which, in turn, impliesb that q S . In conclu- K K ℓ ( ) ∈ k+1,3 sion, v Nk,3, meaning that P Nk,3; the converse followsb from a dimen- ∈ ⊂ sion argument.b Finally, the definition of the degrees of freedom results from Lemma 12.18, and the properties of the maps TK,E and TK,F from those of T , T , and T . K E F ⊓⊔ Remarkb 12.20b (Unit). Since all the dof’s scale like a length unit, the shape functions scale like the reciprocal of a length unit. ⊓⊔ Remark 12.21 (Non-affine meshes). Proposition 6.5 together with the map (12.11) can still be used to generate a finite element (K, P ,Σ) if the geometric map TK is non-affine. The function space P and the degrees of freedom in Σ then differ from those of the N element. k,3 ⊓⊔ Part III. Finite Element Interpolation 187 12.5 Other H(curl) finite elements 12.5.1 Nédélec elements of the second kind Nédélec elements of the second kind [187] offer an interesting alternative to those investigated in 12.3 (and often called Nédélec elements of the first kind) § since they allow one to work with finite elements (K, P ,Σ) with P = Pk,d with k 1. The price to pay is that the curl operator maps onto Pk 1,d. This is not a≥ limitation if the functions to be interpolated are curl-free. − Let K be a simplex in R3. The degrees of freedom are attached to the edges of K, its faces (for k 2), and K itself (for k 3). The edge degrees ≥ e ≥ of freedom are the linear forms σE,m defined in (12.8a) for all E K and e e ∈ E m 1:nsh with nsh = dim(Pk,1) (as for the elements of the first kind), the∈ face { degrees} of freedom are moments on each face of K of the tangential component against a set of basis functions of RTk 2,2 up to a contravariant − Piola transformation (instead of basis functions of Pk 1,2 for the elements of the first kind), and the cell degrees of freedom are− moments against a set of basis functions of RTk 3,3 (instead of basis functions of Pk 2,3 for the elements of the first kind).− It is shown in [187] that the triple (K,− P ,Σ) is a finite element.

12.5.2 Cartesian Nédélec elements The Cartesian version of Nédélec elements have been introduced in Nédélec [186, p. 330-333]. Let us briefly review these elements; see Exercise 12.8 for the proofs. We focus on the case d = 3, since two-dimensional Cartesian Nédélec elements can be built by a rotation of the two-dimensional Cartesian Raviart–Thomas elements from 11.5.2. Let k N and define § ∈ N = Q Q Q , (12.13) k,3 k,k+1,k+1× k+1,k,k+1× k+1,k+1,k where the anisotropic polynomial spaces Qα1,α2,α3 are defined in 11.5.2. One 2 § can verify that dimNk,3 = 3(k + 1)(k + 2) . Let K = (0, 1)3 be the unit cube in R3. Let collect the six faces of K; FK let K collect the twelve edges of K. Let TF , F K (resp., TE, E K ) be an affineE geometric map from [0, 1]2 onto F (resp.,∈F [0, 1] onto E). Let∈Ete = 1 e be the canonical basis of R; we orient E K using tE = JEt where JE is f ∈ E 2 the Jacobian matrix of TE. Let tj j 1,2 be the canonical basis of Rb ; we ∈{ } f { } orient F K using tF,j := JF tj , j 1, 2 , where JF is the Jacobianb matrix ∈F ∈{ } 3 of TF . We orient K using the canonical basis tK,j := ej j 1: 3 of R . Let e { } ∈{ } µm m 1:ne be a basis of Pk,b1 with nsh = k +1. Let ζj,m m 1: nf be a ∈{ sh} sh { } { } f ∈{ } basis of the space Qk,k 1 if j = 1 and Qk 1,k if j = 2, with nsh = (k + 1)k − − (if k 1). Let ψ c be a basis of the space Q if j = 1, j,m m 1:nsh k,k 1,k 1 ≥ { } ∈{ } c − −2 Qk 1,k,k 1 if j = 2, and Qk 1,k 1,k if j = 3, with nsh = (k + 1)k (if k 1). Let−Σ be− the set composed− of the− following linear forms: ≥ 188 Chapter 12. H(curl) finite elements

e 1 1 σ (v)= (v t )(µ T − ) dl, (12.14a) E,m E · E m ◦ E | | ZE f 1 1 σ (v)= (v t )(ζ T − ) ds, (12.14b) F,j,m F · F,j j,m ◦ F | | ZF 1 σc (v)= (v t )ψ dx, (12.14c) j,m K · K,j j,m | | ZK e for all E K and m 1:nsh in (12.14a), all F K , j 1, 2 , and m 1:nf∈ Ein (12.14b),∈ and { all }j 1:3 and m 1:∈n Fc in∈ (12.14c). { } ∈{ sh} ∈{ } ∈{ sh} Proposition 12.22 (Finite element). (K,Nk,3,Σ) is a finite element. Using affine geometric maps and the covariant Piola transformation, Carte- sian Nédélec elements can be generated on all the mesh cells of an affine mesh composed of paralleloptopes. Recall, however, that orienting such meshes re- quires some care; see Theorem 7.16.

Exercises

Exercise 12.1 (S ). (i) Prove that for all q S , there is a unique skew- 1,d ∈ 1,d symmetric matrix Q so that q(x)= Qx. (ii) Propose a basis of S1,d. (iii) Show that q S if and only if there is b R3 such that q(x)= b x. ∈ 1,3 ∈ × Exercise 12.2 (Cross product). (i) Prove that (Ab) (Ac) = A(b c) for 3 3 ×3 × any rotation matrix A R × and any vectors b, c R . (Hint: use Exer- cise 7.4.) (ii) Show that∈ (a b) c = (a c)b (b c)a. (Hint:∈ (a b) = ε a b × × · − · × k ikj i j with Levi-Civita tensor εikj , see the proof of Lemma 7.4.) (iii) Prove that (b n) n + (b n)n = b if n is a unit vector. − × × · Exercise 12.3 (N0,3). (i) Prove (12.4). (Hint: verify that tE λq = 1 and t λ = 1.) (ii) Prove that v = v + 1 ( v) (x c ) for·∇ all v N , E ·∇ p − h iK 2 ∇× × − K ∈ 0,3 where v K is the mean-value of v in K and cK is the center of gravity of K. (hHinti : (b x)=2b for b R3.) (iii) Let θe be the shape function ∇× × ∈ E associated with the edge E K . Let F K with unit normal nK F pointing ∈Ee ∈F | outward K. Prove that (θE) F nK F = 0 if E is not an edge of F , and e | × | θE nK F ds = ιE,F (cE cF ) otherwise, where cE is the center of gravity F × | − of E, cF that of F , and ιE,F = 1 if nK F tE points outward F , ιE,F = 1 R − | × otherwise. (Hint: Use Lemma 12.15 and Exercise 11.1(ii).) (iv) Let E collect the two faces sharing E . Prove that F ∈EK e 1 θE dx = ιE,F (cF cK) (cE cF ). K 2 − × − F E Z X∈F (Hint: take the inner product with an arbitrary vector e R3 and introduce the function ψ(x)= 1 e (x c ).) ∈ 2 × − K Part III. Finite Element Interpolation 189

Exercise 12.4 (Rotated RTk,2). Prove Lemma 12.9. (Hint: observe that H R π P = P and S = (R π x)P .) 2 k,2 k,2 k+1,2 2 k,2

Exercise 12.5 (Hodge decomposition). Prove that P = N PH , k,d k,d⊕∇ k+1,d for all k N. (Hint: Compute N PH , dim(P ), dim(N ), and ∈ k,d ∩ ∇ k+1,d k,d k,d dim( PH ).) ∇ k+1,d Exercise 12.6 (Face element). We use the notation from the proof of Lemma 12.15. Let F . Let T : S2 F be an aﬃne bijective ∈ FK F → map. Let JF be the Jacobian matrix of TF . Let v Nk,3 and let v := T ∈ T J (I n n )(v T ). Show that v Nb . (Hint: Compute y v(y) and F 3 − F ⊗ F ◦ F ∈ k,2 apply the result from Exercise 11.4.) b b b b b d Exercise 12.7 (Geometric map TA). Let A be an aﬃne subspace of R of dimension l 1:d 1 , d 2. Let a A and let PA(x) = a + ΠA(x a) ∈ { − } ≥ ∈ d d −d be the orthogonal projection onto A, where ΠA R × . (i) Let n R be such that n (x y) = 0 for all x, y A (we say that∈ n is normal to A∈). Show · − d ∈ that ΠAn = 0. Let t R be such that a + t A (we say that t is tangent ∈ ∈ 1 to A). Show that Π t = t. (ii) Let q P and letq ˜(x) = q(T − P (x)). A ∈ k,l A ◦ A Compute q˜. (iii) Show that there are t1,..., tl tangent vectors and q1,...,ql ∇ l 1 polynomials in P such that q˜(x)= q (T − (x))t for all x A. (iv) k,l ∇ s=1 s A s ∈ Let t be a tangent vector. Show that there is µ Pk,l such that t q˜(x) = 1 P ∈ ·∇ µ(TA− (x)).

Exercise 12.8 (Cartesian Nédélec element). (i) Propose a basis for N0,3 2 and prove that dim Nk,3 = 3(k + 1)(k + 2) . (ii) Prove Proposition 12.22. 190 Chapter 12. H(curl) finite elements Solution to exercises

Exercise 12.1 (S ). (i) Let q S . Since q is homogeneous of degree 1,d ∈ 1,d 1, there is a unique d d matrix Q such q(x) = Qx. Then q S1,d if and only if xTQx = 0 for× all x Rd, which means that the quadratic∈ form 2 ∈ d i 1: d Qiixi + i=j 1: d (Qij + Qji)xixj vanishes for all x R ; hence Q is skew∈{ symmetric.} We6 ∈{ have} established that there is a one-to-one∈ correspon- P P dence between the members of S1,d and the d d skew-symmetric matrices. d(d 1) × ij (ii) Consider the 2− skew-symmetric matrices Q , 1 i < j d deﬁned ij ≤ij ≤ij by Qkl = δkiδlj δkj δli (the only nonzero entries of Q are Qij = 1 and ij − ij ij ij Qji = 1). Then setting q (x) = Q x we have shown that q 1 i

Exercise 12.2 (Cross product). (i) Using Exercise 7.4, we obtain

1 T (Ab) (Ac) = det(A)− A− (b c)= A(b c), × × × T since det(A) = 1 and A− = A. (ii) We have (using summation for repeated indices)

((a b) c) = c ε a ε b × × k − i ikj l ljm m = c a b (δ δ δ δ ) − i l m im kl − il km = (a c)b (b c)a , · k − · k since εikj εljm = δimδkl δilδkm. (iii) We apply the formula− derived in Step (ii) using (n n) = 1. · Exercise 12.3 (N0,3). (i) Since tE = zq zp, (zq zp) λq = λq(zq) − − ·∇ e − λq(zp) = 1, and similarly (zq zp) λp = 1, we infer that E θE tE dl = − e·∇ e − · (λ +λ ) dl = E showing that σ (θ ) = 1. Consider now an edge E′ with E p q | | E E R E′ = E. Then, at least one vertex zp or zq is not in E′, say zp E′. This R 6 e e 6∈ implies that λp 0 in E′ and that tE′ λp = 0; hence, σE′ (θE) = 0. (ii) Let v N ≡.Then v = a + b (x ·∇c ) with a, b R3, and since c is ∈ 0,3 × − K ∈ K the center of gravity of K, we infer that a = v K . Furthermore, using the h i 1 hint yields that v =2b. In conclusion, v = v K + 2 ( v) (x cK ). ∇× e h i ∇× × − (iii) The assertion (θE) F nK F = 0 if E is not an edge of F is a direct | × | e consequence of Lemma 12.15 since the three degrees of freedom of θE attached to F vanish. Assume now that E is an edge of F . Then, observing that e e e π θ n = ι R π (θ (θ n )n ) where R π is the rotation by in E K F E,F 2 E E K F K F 2 2 × | − · | | e e the hyperplane parallel to F , and recalling that R π (θ (θ n )n ) T 2 E E K F K F F − · e| | ◦ is in RT0,2, we can use Exercise 11.1(ii) to infer that θE nK F ds = F × | R Part III. Finite Element Interpolation 191

ιE,F (cE cF ). (iv) Using− the hint, we obtain that

e e e e e θE dx = ( ψ) θE = ψ ( θE) dx (nK F θE) ψ ds. · K K ∇× · K · ∇× − F | × · F K Z Z Z X∈F Z The ﬁrst term on the right-hand side vanishes since ψ has zero mean-value e in K and θE is constant on K. Since the summation in the second term reduces to∇×F owing to item (iii), we infer that ∈FE e e e θE dx = (nK F θE) ψ ds =: T1 + T2, · K − F | × · F E Z X∈F Z with

e T1 = (nK F θE) ψ(cF ) ds, − F | × · F E X∈F Z e T2 = (nK F θE) (ψ ψ(cF )) ds. − F | × · − F E X∈F Z Since ψ(cF ) is constant, we can use item (iii) to evaluate T1, so that 1 T = ι (c c ) (e (c c )) 1 2 E,F E − F · × F − K F E X∈F 1 = ι e ((c c ) (c c )). 2 E,F · F − K × E − F F E X∈F Let us ﬁnally prove that T2 = 0. Since ψ ψ(cF ) has zero mean-value in F , we can write −

e e T2 = (nK F (θE θE(cF ))) (ψ ψ(cF )) ds IF ds. − F | × − · − − F F E F E X∈F Z X∈F Z e e 3 Since θE(x) θE (cF )= b (x cF ) for some b R and since ψ(x) ψ(cF )= 1 e (x c −), we obtain that× − ∈ − 2 × − F 1 IF = (nK F (b (x cF ))) (e (x cF )) 2 | × × − · × − 1 = (nK F b)(x cF ) (e (x cF )) = 0, 2 | · − · × − since nK F (x cF ) = 0. | · − Exercise 12.4 (Rotated RT ). We observe that R π P = P . Moreover, k,2 2 k,2 k,2

k+1 k+1 l+1 k+1 l l k+2 l 0 = (x, q)ℓ2(Rd) = q1,lx1 x2 − + q2,lx1x2 − . Xl=0 Xl=0 192 Chapter 12. H(curl) ﬁnite elements

This implies that q1,k+1 = 0, q2,0 = 0, and q1,l = q2,l+1 for all l 0:k . k −l k l H ∈ { } Hence, q1 = x2r and q2 = x1r with r = l=0 q1,lx1x2− Pk,2. This shows H − ∈ that Sk+1,2 = (R π x)P . We conclude that Nk,2 = R π (RTk,2). 2 k,2 P 2 H Exercise 12.5 (Hodge decomposition). Let v Nk,d Pk+2, then u = H H ∈ ∩ ∇ p where p Pk+2. Observe that p Pk+1,d and x p(x) = (k + 2)p(x). ∇ ∈ ∇ ∈ ·∇ H The assumption v = p Nk,d and the property p Pk+1,d imply that x p(x) = 0, which can∇ be∈ true only if p = 0; hence∇ ∈v = 0. Conclude by ·∇ H using a dimensional argument, i.e., dim(Pk,d) = dim(Nk,d) + dim(Pk+1,d), (k+d+1)! H (k+d+1)! where dim(Nk,d)= k!(d 1)!(k+2) , dim(Pk+1,d)= (k+2)!(d 1)! . − −

Exercise 12.6 (Face element). By definition v = r + q where r Pk,3 and q PH satisfies yTq(y) =0. Let Π := I n n . Let y R∈2, then ∈ k+1,3 F 3 − F ⊗ F ∈ T T T y v(y)= y JF ΠF (v TF )(y) b T T ◦ T = y JF ΠF r(TF (y)) + (JF y) ΠF q(TF (y)) b b b bT T b T = y JF ΠF r(TF (y)) + (JF y) ΠF q(TF (y) TF (0R2 )+ TF (0R2 )) bT T b b T b − = y JF ΠF r(TF (y)) + (JF y) ΠF q(JF (y)+ TF (0R2 )). b b b b We now invoke the result from Exercise 11.4 componentwise; there is t b b b b ∈ Pk,2 such that q(JF (y)+ TF (0R2 )) = q(JF (y)) + t(y). Upon setting s(y) := yTJT Π (r(T (y))+t(y)), where s P , and observing that (J y)Tn b = F F F ∈ k+1,2 F F 0 for all y, we obtainb b b b b b b b T b b b T b y v(y)= s(y) + (J y) q(J (y)) = s(y). b F F Since v P , we have the decomposition v = r + q where r P and k+1,2 b b b b b b b k,2 q PH ∈ . Then ∈ ∈ k+1,2 b T T T b b b b y v(y)= y r(y)+ y q(y)= s(y) P , b ∈ k+1,2 but yTr(y) P and yTq(y) PH . Hence yTq(y) = 0 for all y, which kb+1,b2 b b b b bk+2b,2b b proves that ∈v N . ∈ ∈ k,2 b b b b b b b b b b Exercise 12.7 (Geometric map TA). (i) These are elementary results in linear algebra.b Let t ,..., t be a basis of A a. let n ,..., n be a basis of 1 l − l+1 d span(t1,..., tl)⊥. Let n be a normal vector. Let x = a+n, then PA(x) a = Π (x a) = Π n. Observe that 0 = n (P (x) a) = n Π n, for− every A − A s· A − s· A normal vector ns, l +1 s d. Note also that ts (PA(x) x) = 0, for every tangent vector t , 1 s≤ d≤, i.e., 0 = t (a + Π n· x)=−t ( n + Π n)= s ≤ ≤ s· A − s· − A ts ΠAn. Hence, ΠAn is orthogonal to span(t1,..., tl) span(nl+1,..., nd)= d· ⊕ R , meaning that ΠAn = 0. Let t be a tangent vector and let x = a + t, then x A by definition; hence t = x a = PA(x) a = ΠA(x a)= ΠAt. (ii)∈ Let h Rd. We use the Frechet− derivative− notation and− apply the chain rule: ∈ Part III. Finite Element Interpolation 193

1 1 Dq˜(x)(h)= Dq(T − P (x))(D(T − P (x))(h)) A ◦ A A ◦ A 1 1 = Dq(T − P (x))(DT − (P (x))(DP (x)(h))). A ◦ A A A A 1 1 Note that DPA(x)(h)= ΠAh and DTA− (x′)(h′)= JA− h′ for all x′ A and d ∈ all h′ R . We identify the Frechet derivatives ofq ˜ and q with the gradients, as usual,∈ so that

1 1 q˜(x) h := Dq˜(x)(h)= q(T − P (x)) (J− Π h) ∇ · ∇ A ◦ A · A A T 1 d = Π J− q(T − P (x)) h, h R . A A ∇ A ◦ A · ∀ ∈ Hence T T 1 d q˜(x)= Π J− q(T − P (x)), x R . ∇ A A ∇ A ◦ A ∀ ∈ (iii) Let n any normal vector, then

T T 1 T 1 n q˜(x)= n Π J− q(T − P (x)) = (Π n) J− q(T − P (x))=0, ·∇ · A A ∇ A ◦ A A · A ∇ A ◦ A since we have already established that ΠAn = 0. Hence q˜(x) span(t1,..., tl). Moreover, since P (x)= x for all x A, we have ∇ ∈ A ∈ T T 1 q˜(x)= Π J− q(T − (x)), x A. ∇ A A ∇ A ∀ ∈ Hence, q˜(x) is a l-variate Rd-valued polynomial of degree at most k. The two above∇ arguments show that there exists q ,...,q P such that 1 l ∈ l,d l 1 q˜(x)= q (T − (x))t , x A. ∇ s A s ∀ ∈ s=1 X (iv) Let t be a tangent vector. Then the above arguments show that there is µ P such that ∈ k,l 1 t q˜(x)= µ(T − (x)), x A. ·∇ A ∀ ∈ Exercise 12.8 (Cartesian Nédélec element). (i) to be done (ii) Observe first that card(Σ)=3k2(k + 1)+ 12k(k + 1) + 12(k +1) = 2 3(k + 1)(k + 2) = dim Nk,3.