Finite Element Methods for Maxwell Equations

Transcription

1 CHAPTER 8 Finite Element Methods for Maxwell Equations The Maxwell equations comprise four first-order partial differential equations linking the fundamental electromagnetic quantities, the electric field E, the magnetic induction B, the magnetic field H, the electric flux density D, the electric current density J, and the space charge density ρ: H = J + t D, divd = ρ, E = t B, divb = 0. They are usually supplemented by the following linear constitutive laws D = εe, B = µh, where ε is the dielectric permittivity and µ the magnetic permeability. In the wave form, we have ε 2 tte + ( µ 1 E = t J, div(εe = ρ, µ 2 tth + ( ε 1 H = ( ε 1 J, div(µh = 0. Usually, time harmonic solutions are considered, that is, E(x, t = R(Ê(xe iωt, H(x, t = R(Ĥ(xe iωt, J(x, t = R(Ĵ(xe iωt, where ω > 0 is the angular frequency, then (µ 1 Ê εω2 Ê = iωĵ, div(εê = ρ, (ε 1 Ĥ µω2 Ĥ = (ε 1 Ĵ, div(µĥ = 0. In this chapter we consider adaptive edge element methods for solving the time-harmonic Maxwell equations. We will first introduce the function space H(curl; Ω and its conforming finite element discretization, the lowest order Nédélec edge element method. Then we will derive the a priori and a posteriori error estimate for the edge element method. 107

2 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS 8.1. The function space H(curl; Ω Let Ω be a bounded domain in R 3 with a Lipschitz boundary Γ. we define with the norm H(curl; Ω = {v L 2 (Ω 3 : v L 2 (Ω 3 } v H(curl;Ω = ( v 2 L 2 (Ω + v 2 L 2 (Ω 1/2. We define H 0 (curl; Ω to be the closure of C 0 (Ω3 in H(curl; Ω. H(curl; Ω and H 0 (curl; Ω are Hilbert spaces. Lemma 8.1. Let Ω be a bounded Lipschitz domain. Let v H(curl; R 3 vanish outside Ω. Then v H 0 (curl; Ω. Proof. Suppose for the moment that the domain Ω is strictly starshaped with respect to one of its points. Without loss of generality, we may take the point as the origin. Then Define, for θ (0, 1, θ Ω Ω θ [0, 1 and Ω θω θ > 1. v θ (x = v(x/θ x R 3. It is obvious that v θ has a compact support in Ω for θ (0, 1 and lim v θ = v in H(curl, R 3. θ 1 For any ϵ > 0, let ρ ϵ (x = ϵ d ρ(x/ϵ C 0 (R3 be the mollifier function where ρ(x is defined in (1.4. Recall that ρ ϵ = 0 for x > ϵ. Hence, for ϵ > 0 sufficiently small, ρ ϵ v θ is in C 0 (Ω3 and from Lemma 1.1, lim lim(ρ ϵ v θ = v in H(curl, Ω. ϵ 0 θ 1 In the general case, Ω can be covered by a finite family of open sets Ω 1 i q O i such that each Ω i = Ω O i is Lipschitz, bounded and strictly star-shaped. Let {χ i } 1 i q be a partition of unity subordinate to the family {O i } 1 i q, that is, q χ i C0 (O i, 0 χ i 1, and χ i = 1 in Ω. Then v = q i=1 χ iv in R 3. Clearly χ i v H(curl; Ω with support in Ω i. Therefore, we can finish the proof by using the result for the strictly starshaped domain in the first part of the proof. i=1

3 8.1. THE FUNCTION SPACE H(curl; Ω 109 Theorem 8.1. Let D( Ω be the set of all functions ϕ Ω with ϕ C 0 (R3. Then D( Ω 3 is dense in H(curl; Ω. Proof. Let l belong to H(curl; Ω, the dual space of H(curl; Ω. As H(curl; Ω is a Hilbert space, by Riesz representation theorem, we can associate with l a function u in H(curl; Ω such that where l, v = (u, v + (w, v w = u. v H(curl; Ω, Now assume that l vanishes on D( Ω 3 and let ũ, w be respectively the extension of u, w by zero outside Ω. Then we have R 3 {ũ v + w v} dx = 0 v C 0 (R 3 3. This implies that ũ = w. Therefore w H(curl; R 3, since ũ L 2 (R 3 3. Now by Lemma 8.1 we have w H 0 (curl; Ω. As C 0 (Ω3 is dense in H 0 (curl; Ω, let w ϵ be a sequence of functions in C 0 (Ω3 that tends to w in H(curl; Ω as ϵ 0, then l, v = lim ϵ 0 {( w ϵ, v + (w ϵ, v} = 0 v H(curl; Ω. Therefore, l vanishes on D( Ω 3 implies that l also vanishes on H(curl; Ω. This completes the proof. The following theorem about the tangential trace of the functions in H(curl; Ω is a direct consequence of the above theorem. Theorem 8.2. The mapping γ τ : v v n Γ defined on D( Ω 3 can be extended by continuity to a linear and continuous mapping from H(curl; Ω to H 1/2 (Γ 3. Moreover, the following Green formula holds v n, w Γ = v w dx v w dx w H 1 (Ω 3, v H(curl; Ω. Ω Ω We remark that the trace operator γ τ is not a surjective mapping. The following characterization of the H 0 (curl; Ω follows from the definition of the space H 0 (curl; Ω and Lemma 8.1. Lemma 8.2. We have H 0 (curl; Ω = {v H(curl, Ω : v n = 0 on Γ}.

4 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS The Helmholtz decomposition plays an important role in the analysis and computation of electromagnetic fields. We start with the following generalization of the classical Stokes theorem. Theorem 8.3. Let Ω be a simply connected Lipschitz domain. Then u L 2 (Ω 3 and u = 0 if and only if there exists a function φ H 1 (Ω/R such that u = φ. Proof. We first prove that u = φ for some φ = L 2 loc (Ω. Since φ L 2 (Ω 3, we then easily have φ L 2 (Ω. In fact, using the argument in Lemma 8.1 we may assume Ω is strictly star-shaped. Then we can introduce φ θ as in Lemma 8.1. Since φ L 2 loc (Ω and φ L2 (Ω 3, we know that φ θ φ and D φ θ D φ as θ 1, where D is some compact subset of Ω. Now since φ L 2 loc (Ω we know φ θ L 2 (Ω. By using Poincaré inequality we know that φ θ is a Cauchy sequence in L 2 (Ω as θ 1. Thus there exists a φ 1 L 2 (Ω such that φ θ φ 1 in L 2 (Ω. This implies φ = φ 1 L 2 (Ω. To show u = φ for some φ = L 2 loc (Ω, first we find a sequence of simply connected Lipschitz domain {Ω m } m 1 such that Ω m Ω, Ω m Ω m+1, Ω = m 1 Ω m. In Ω m we can smooth u so that its curl is zero and so we can apply the classical Stokes theorem for C 1 functions. For any ϵ > 0, let ρ ϵ (x = ϵ d ρ(x/ϵ C 0 (R3 be the mollifier function where ρ(x is defined in (1.4. Recall that ρ ϵ = 0 for x > ϵ. Denote by ũ the zero extension of u outside Ω. Then ρ ϵ ũ C 0 (R3 3, and ρ ϵ ũ ũ in L 2 (R 3 3, (ρ ϵ ũ = ρ ϵ ũ. For sufficiently small ϵ, we have x Ωm B(x; ϵ Ω. Thus (ρ ϵ ũ = 0 in Ω m. Now from the classical Stokes theorem, there is a smooth function φ ϵ H 1 (Ω m /R such that ρ ϵ ũ = φ ϵ in Ω m. Let ϵ 0, we know that there is a function φ m H 1 (Ω m such that φ ϵ φ m in H 1 (Ω m /R, and u = φ m in Ω m. But φ m = φ m+1 in Ω m. Thus φ m, φ m+1 differ by only a constant which we can choose as zero. Therefore φ m+1 = φ m in Ω m m 1.

5 8.1. THE FUNCTION SPACE H(curl; Ω 111 This defines a function φ L 2 loc (Ω such that u = φ. Our next goal is to show that a vector field whose divergence vanishes must be a curl filed. We assume Ω has p + 1 connected parts Γ i, 0 i p, and Γ 0 is the exterior boundary. We denote Ω i the domain encompassed by Γ i, 1 i p (see Figure 1. Ω 0 Ω 2 Γ 2 Ω 1 Γ 1 Γ p Ω p Γ 0 O Figure 1. The domain Ω and the ball O. Theorem 8.4. A vector field v L 2 (Ω 3 satisfies div v = 0 in Ω, v n, 1 Γi = 0, 0 i p, (8.1 if and only if there is a vector potential w H 1 (Ω 3 such that v = w. (8.2 Moreover, w may be chosen such that div w = 0 and the following estimate holds w H 1 (Ω C v L 2 (Ω. (8.3 Proof. 1 Let w H 1 (Ω 3 and v = w. Obviously div v = 0. For 0 i p, let χ i C0 (R3 be the cut-off function such that 0 χ i 1, χ i = δ ij in the neighborhood of Γ j. Define v i = (χ i w. Then v n, 1 Γi = v i n, 1 Γ = div v i dx = 0, 0 i p. This shows (8.1. Ω

6 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS 2 Now let us assume (8.1 holds. First we extend v to be a function in R 3 so that its divergence is zero. Let O be a ball containing Ω (see Figure 1. For 0 i p, denote by θ i H 1 (Ω i the solution of the following problem θ 0 = 0 in Ω 0 = O \ ( Ω p i=1 Ω i, n θ 0 = v n on Γ 0, n θ 0 = 0 on O, θ i = 0 in Ω i, n θ i = v n on Γ i, 0 i p. Define v in Ω, ṽ = θ i in Ω i, 1 i p, 0 in R 3 \Ō. Then ṽ L 2 (R 3 and div ṽ = 0. Let ˆv = (ˆv 1, ˆv 2, ˆv 3 T be the Fourier transform of ṽ 3 ˆv(ξ = e 2πi(x,ξ ṽ(x dx, (x, ξ = x i ξ i. R 3 By taking the Fourier transform of (8.1 we obtain Notice that if (8.2 is satisfied we need If div w = 0 we need i=1 ξ 1ˆv 1 + ξ 2ˆv 2 + ξ 3ˆv 3 = 0. (8.4 ˆv = 2πi(ξ 2 ŵ 3 ξ 3 ŵ 2, ξ 3 ŵ 1 ξ 1 ŵ 3, ξ 1 ŵ 2 ξ 2 ŵ 1 T. (8.5 Solving ŵ from (8.4-(8.6 we get ŵ = ξ 1 ŵ 1 + ξ 2 ŵ 2 + ξ 3 ŵ 3 = 0. ( πi ξ 2 (ξ 3ˆv 2 ξ 2ˆv 3, ξ 1ˆv 3 ξ 3ˆv 1, ξ 2ˆv 1 ξ 1ˆv 2 T. We will define w as the inverse Fourier transform of the above function. Obviously w L 2 (Ω 3 3 because ξ j ŵ k 1 2π ( ˆv 1 + ˆv 2 + ˆv 3. (8.7 Now we show w L 2 (Ω 3. Denote by ω C 0 (R3 the function which is 1 in the neighborhood of the origin. Then ŵ(ξ = ω(ξŵ(ξ + (1 ω(ξŵ(ξ.

7 8.1. THE FUNCTION SPACE H(curl; Ω 113 From (8.5 we know that ˆv j (0 = 0. Since ˆv j (ξ is holomorphic, we know that, in the neighborhood of the origin, 3 ˆv j ˆv j (ξ = ξ k + O( ξ 2. ξ k k=1 Thus ŵ is bounded in the neighborhood of the origin. Now ωŵ has the compact support, its inverse Fourier transform is holomorphic and its restriction to Ω belongs to L 2 (Ω 3. On the other hand, (1 ωŵ is zero in the neighborhood of the origin. Hence (1 ωŵ L 2 (R 3 3 and its inverse Fourier transform in L 2 (R 3 3. This proves the inverse Fourier transform of ŵ is in L 2 (Ω 3. Clearly w can be chosen up to an arbitrary constant. Thus (8.3 follows from (8.7, the Parserval identity, and Poincaré inequality. The following Helmholtz decomposition theorem is now a direct consequence of Theorems 8.3 and 8.4. Theorem 8.5. Any vector field v L 2 (Ω 3 has the following orthogonal decomposition v = q + w, where q H 1 (Ω/R is the unique solution of the following problem ( q, φ = (v, φ φ H 1 (Ω, and w H 1 (Ω 3 satisfies div w = 0 in Ω, w n = 0 on Γ. We conclude this section by proving the embedding theorem for function spaces X N (Ω and X T (Ω which will be used in our subsequent analysis X N (Ω = {v L 2 (Ω 3 : v L 2 (Ω 3, divv L 2 (Ω, v n = 0 on Γ}, X T (Ω = {v L 2 (Ω 3 : v L 2 (Ω 3, divv L 2 (Ω, v n = 0 on Γ}. Theorem 8.6. If Ω is a C 1 or convex domain, X N (Ω, X T (Ω are continuously embedded into H 1 (Ω 3. Proof. Without loss of generality, we may assume Ω is also simply connected and has connected boundary. For, otherwise, Ω is the union of finite number of domains Ω k having above properties. We can introduce the partition of unity χ k subordinate to Ω k and apply the result for each χ k v. 1 Let v X T (Ω. By Theorem 8.4, for v, we have vector potential w H 1 (Ω 3 such that w = v, div w = 0 in Ω.

8 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Moreover, w H 1 (Ω C v L 2 (Ω. Thus (v w = 0 and by Theorem 8.3, v w = φ for some function φ H 1 (Ω. Moreover, φ satisfies φ = div v in Ω, n φ = w n on Γ. (8.8 Since w H 1 (Ω 3, w n H 1/2 (Γ. Now if Ω is a C 1 domain, then the regularity theory for elliptic equations implies φ H 2 (Ω. Thus v = w + φ belongs to H 1 (Ω 3. Moreover, ( φ H 2 (Ω C divv L 2 (Ω + w n H 1/2 (Γ Thus C ( divv L 2 (Ω + v L 2 (Ω. v H 1 (Ω C ( divv L 2 (Ω + v L 2 (Ω. 2 Let v X N (Ω. Let O be a ball in R 3 that includes Ω. Denote by ṽ the zero extension to O of v. By Theorem 8.4, there exists a function w H 1 (O 3 such that w = ṽ, div w = 0 in O. Now w is curl free in O\ Ω and by Theorem 8.3, w = ψ for some ψ H 2 (O\ Ω. On the other hand, (ṽ w = 0 and O is simply connected, again by Theorem 8.3, v w = φ for some function φ H 1 (O. Clearly φ satisfies φ = divv in Ω, φ = ψ on Γ. (8.9 If Ω is a C 1 domain, then φ H 2 (Ω and consequently v = w+ φ H 1 (Ω 3. Moreover, v H 1 (Ω C ( divv L 2 (Ω + v L 2 (Ω. 3 If Ω is a convex domain, the regularity theory for elliptic equation ensures that the solution φ of (8.9 is in H 2 (Ω and thus X N (Ω is continuously embedded into H 1 (Ω 3 by using the same argument in 2. For the case of X T (Ω, the regularity of the elliptic equation with Neumann condition in (8.8 is unknown. A different approach is used to prove the embedding theorem. We refer the reader to the monograph [34].

9 8.2. THE CURL CONFORMING FINITE ELEMENT APPROXIMATION The curl conforming finite element approximation In this section we only consider the lowest order Nédélec finite element space. Definition 8.7. The lowest order Nédélec finite element is a triple (K, P, N with the following properties (i K R 3 is a tetrahedron; (ii P = {u = a K + b K x a K, b K R 3 }; (iii N = {M e : M e (u = e (u t dl edge e of K, u P}. M e(u is called the moment of u on the edge e. Note that if u P 1 (K 3 then u is a constant vector, say u = 2b K, which implies (u b K x = 0 in K. We get u = φ + b K x for some φ P 2 (K. When b K = 0, u should approximate the function in L 2 (K 3, the minimum requirement is φ P 1 (K, that is, φ = a K for some constant vector a K in R 3. This motivates the shape functions in P. Lemma 8.3. The nodal basis of the lowest order Nédélec element is {λ i λ j λ j λ i, 1 i < j 4}. Here λ j, j = 1, 2, 3, 4, are barycentric coordinate functions of the element K. Proof. Let K be the tetrahedron with four vertices A 1, A 2, A 3, A 4 corresponding to λ 1, λ 2, λ 3, λ 4, respectively. We first notice that the normal to the face F 123 with vertices A 1, A 2, A 3 is parallel to λ 4. In fact, for any edge e of F 123 with tangential vector t e, λ 4 t e = λ 4 t e = 0. Similarly, we have the normal to the face F 234 is parallel to λ 1. Now we show that the basis function corresponding to the edge e 14 is u 14 = λ 1 λ 4 λ 4 λ 1. In fact, since λ 1 = 0 on the face F 234 and λ 4 = 0 on the face F 123, we have u 14 t e dl = 0 e e 14. It remains to prove e 14 u 14 t 14 dl = 1. Noting that λ 1 + λ 4 = 1 on e 14, we have e u 14 = λ 4 λ 4 (λ 1 + λ 4 = λ 4 + λ 4 (λ 2 + λ 3.

10 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Therefore λ 4 u 14 t 14 dl = λ 4 t 14 dl = dl = 1. e 14 e 14 e 14 t 14 This completes the proof. Let K be a tetrahedron with vertices A i, 1 i 4, and let F K : ˆK K be the affine transform from the reference element ˆK to K x = F K (ˆx = B K ˆx + b K, ˆx ˆK, B K is invertible, so that F K (Âi = A i, 1 i 4. Notice that the normal and tangential vectors n, ˆn and t, ˆt to the faces satisfy n F K = (B 1 K T ˆn / (B 1 K T ˆn, For any scaler function φ defined on K, we associate t F K = B Kˆt / B K t. ˆφ = φ F K, that is, ˆφ(ˆx = φ(b K ˆx + b K. For any vector valued function u defined on K, we associate û = B T Ku F K, that is, û(ˆx = B T Ku(B K ˆx + b K. (8.10 Denote by u = (u 1, u 2, u 3, û = (û 1, û 2, û 3. We introduce ( ui C = u 3 ( j ûi and Ĉ = û j x j x i ˆx j ˆx i Then we have In fact, û i ˆx j = and û i ˆx j û j ˆx i = k,l i,j=1 3. i,j=1 C F K = (B 1 K T ĈB 1 K. (8.11 ( ˆx j k b ki u k x l b lj k,l b ki (u k F K = k,l b kj u k x l b li = k,l b ki u k x l b lj ( uk b ki u l b lj. x l x k This yields Ĉ ij = k,l b ki C kl b lj and hence Ĉ = B T K(C F K B K, that is, (8.11 holds. Lemma 8.4. We have (i u P(K û ˆP( ˆK; (ii u = 0 ˆ û = 0, u P(K;

11 8.2. THE CURL CONFORMING FINITE ELEMENT APPROXIMATION 117 (iii M e (u = 0 Mê(û = 0, u P(K; (iv Let u P(K and F be a face of K. If M e (u = 0 for any edge e F, then u n = 0 on F ; (v If u P(K and M e (u = 0 for any edge e, then u = 0 in K. Proof. (i For u = a K + b K x P(K, we have û(ˆx = B T K(a K + b K (B K ˆx + b K = B T K(a K + b K b K + B T K(b K B K ˆx = B T Ka K + ˆb ˆK ˆx (by B T K(b K B K ˆx ˆx = 0 ˆP( ˆK. (ii From (8.11, it is obvious. (iii By definition, M e (u = e u t dl = e u(f K (ˆx ê ê = e 1 (iv Without loss of generality, we may assume B Kˆt B Kˆt dˆl û ˆtdˆl ê B Kˆt ê = e 1 ê B M ê(û. (8.12 Kˆt F {x = (x 1, x 2, x 3 R 3 : x 3 = 0}. First by Stokes theorem F u n ds = which implies u n = 0 on F, i.e., Let u = (u 1 (x 1, x 2, 0, u 2 (x 1, x 2, 0 and F u t dl = 0 u 1 x 2 u 2 x 1 = 0 on {x 3 = 0}. (8.13 u = a K + b K x = (b 2 x 3 b 3 x 2, b 3 x 1 b 1 x 3, b 1 x 2 b 2 x 1 + a K. (8.13 implies b 3 = 0. Thus u is constant. Note that u t = u t on F. By assumption, we have, M e (u = 0, for any edge e F, which implies u t = 0 on any edge e F. Thus u = 0 in F. This shows u n = (u 2 (x 1, x 2, 0, u 1 (x 1, x 2, 0, 0 = 0 on F.

12 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS (v At first, by (iv we know that u n = 0 on F for any face F. Thus, from Theorem 8.2, u b K dx = (u n b K ds = 0 K which implies b K = 0, that is, u = a K. Now a K n = 0 for any face F yields a K = 0. This lemma induces a natural interpolation operator on K. Definition 8.8. Let K be an arbitrary tetrahedron in R 3 and u W 1,p (K 3 for some p > 2. Its interpolant γ K u is a unique polynomial in P(K that has the same moments as u on K. In other words, M e (γ K u u = 0. Recall that for any bounded Lipschitz domain, the trace theorem says that the trace of any function in W s,p (Ω is in W s 1/p,p ( Ω, where s > 1/p. Thus if u W 1,p (K 3 for some p > 2, u W 1 1/p,p ( F for any face F of K. Again by the trace theorem u W 1 2/p,p ( F. Therefore, M e (u is well-defined for functions u in W 1,p (K, p > 2. The following lemma indicates that the interpolation operator γ K can also be defined for functions with weaker regularity. K Lemma 8.5. For any p > 2, the operator γ K is continuous on the space {v L p (K 3 : v L p (K 3 and v n L p ( K 3 }. Proof. Let p be such that 1/p + 1/p = 1. For an edge e of a face F of K, we let φ be the function which equals to 1 on e and 0 on the other edges of F. Then φ W 1 1/p,p ( F since 1 1/p < 1/2. Denote φ its lifting from W 1 1/p,p ( F to W 1,p (F. Next, we extend φ to be zero on the other faces of K and denote φ its lifting from W 1 1/p,p ( K to W 1,p (K. By Stokes theorem and Green formula M e (v = (v tφ dl F = ( φv n ds F = φ v n ds + φ v n ds F F = v φ dx + φ v n ds. This implies K M e (v C( v L p (K + v n L p (F φ W 1 1/p,p (e F

13 8.2. THE CURL CONFORMING FINITE ELEMENT APPROXIMATION 119 with a constant C depending only on K and p. This completes the proof. Let Ω be a bounded polyhedron and M h be a regular mesh of Ω. We set X h = {u h H(curl; Ω : u h K P(K K M h }. For any function u whose moments are defined on all edges of the mesh M h, we define the interpolation operator γ h by γ h u K = γ K u on K, K M h. Theorem 8.9. Let u H 1 (curl; Ω, that is, u H 1 (Ω 3 and u H 1 (Ω 3. We have u γ h u H(curl;Ω Ch ( u H 1 (Ω + u H 1 (Ω. Proof. First it follows from (8.12 that γ K u = γ ˆKû, i.e., B T K[(γ K u F K ] = γ ˆK[B T K(u F K ]. From (8.10, we have u γ h u L 2 (K det(b K 1/2 B 1 K û γ ˆKû L 2 ( ˆK. Since P 0 ( ˆK 3 P( ˆK, for any ˆp P 0 ( ˆK 3, we have û γ ˆKû L 2 ( ˆK = (I γ ˆK(û + ˆp L 2 ( ˆK. But the degrees of freedom of u may be estimated by using Lemma 8.5 and by using the Sobolev imbedding theorem to get (I γ ˆK(û + ˆp L 2 ( ( û ˆK C + ˆp H 1 ( ˆK + ˆ (û + ˆp H 1 ( ( ˆK = C û + ˆp H 1 ( ˆK + ˆ û H 1 ( ˆK Now, by using Theorem 3.1, ( inf (I γ ˆK(û + ˆp ˆp P 0 ( ˆK L 3 2 ( ˆK C û H 1 ( ˆK + ˆ û H 1 ( ˆK. Mapping back to the original element K and using (8.11 we obtain u γ h u L 2 (K C det(b K 1/2 B 1 K ( û H 1 ( ˆK + ˆ û H 1 ( ˆK C B 1 K B K 2 ( u H 1 (K + B K u H 1 (K Ch K ( u H 1 (K + u H 1 (K. To show the curl estimate, we use the H(div, Ω conforming finite element space W h and the interpolation operator τ h : H(div, Ω W h in Exercise 8.4 to obtain (u γ h u L 2 (Ω (I τ h u L 2 (Ω Ch u H 1 (Ω.

14 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS This proves the theorem Finite element methods for time harmonic Maxwell equations Let Ω be bounded polyhedral domain in R 3. In this section we consider the finite element approximation to the time harmonic Maxwell equation with the boundary condition (α(x E k 2 β(xe = f in Ω, E n = 0 on Ω. Here k > 0 is the wave number. We assume f L 2 (Ω 3, α, β L (Ω such that α α 0 > 0 and β β 0 > 0. The variational formulation is to find E H 0 (curl; Ω such that (α E, v k 2 (βe, v = (f, v v H 0 (curl; Ω. (8.14 The problem (8.14 is not necessarily coercive and thus its uniqueness and existence is not guaranteed. Here we will not elaborate on this issue and simply assume (8.14 has a unique solution E H 0 (curl; Ω for any given f L 2 (Ω 3. Let X 0 h = X h H 0 (curl; Ω. Then the finite element approximation to (8.14 is to find E h X 0 h such that (α E h, v h k 2 (βe h, v h = (f, v h v h X 0 h. (8.15 The discrete problem (8.15 may not have a unique solution. It can be proved that for sufficiently small mesh size h, the problem (8.15 indeed has a unique solution under fairly general conditions on the domain and the coefficients α, β. Here we only consider a special case when the domain is a convex polyhedron and α, β are constants. Theorem Let Ω be a convex polyhedral domain in R 3 and α = 1, β = 1. Then there exists a constant h 0 > 0 such that for h < h 0, the discrete problem (8.15 has a unique solution E h. Moreover, assume that the solution E of (8.14 satisfies E H 1 (Ω 3, E H 1 (Ω 3, then the following error estimate holds E E h H(curl;Ω Ch ( E H 1 (Ω + E H 1 (Ω. Proof. The proof is divided into several steps.

15 8.3. FINITE ELEMENT METHODS FOR TIME HARMONIC MAXWELL EQUATIONS121 by 1 Let a(, : H 0 (curl; Ω H 0 (curl; Ω R be the bilinear form defined a(u, v = ( u, v k 2 (u, v. From (8.14 and (8.15 we know that a(e E h, v h = 0 v h X 0 h. (8.16 Let P h E X 0 h be the projection of E to X0 h defined by ( P h E, v + (P h E, v = ( E, v + (E, v v X 0 h. Thus E E h 2 H(curl;Ω = a(e E h, E E h + (1 + k 2 E E h 2 L 2 (Ω This yields = a(e E h, E P h E + (1 + k 2 E E h 2 L 2 (Ω = ( (E E h, (E P h E + (E E h, E P h E + (1 + k 2 (E E h, P h E E h. E E h H(curl;Ω E P h E H(curl;Ω + (1 + k 2 (E E h, v h sup. ( v h X 0 v h h H(curl;Ω 2 Now we estimate the second term in above estimate. First since E E h H 0 (curl; Ω, there exists a w H 0 (curl; Ω and φ H0 1 (Ω such that E E h = w + φ, divw = 0. (8.18 In fact we can define φ H0 1 (Ω as the solution of the following problem ( φ, v = (E E h, v v H 1 0 (Ω, and let w = E E h φ. Clearly φ L 2 (Ω E E h L 2 (Ω. For any v h X 0 h, we use the following decomposition v h = w h + φ h, w h X 0 h,0, φ h V 0 h, (8.19 where Vh 0 H1 0 (Ω is the conforming linear finite element space having zero trace on the boundary and X 0 h,0 is the subspace of X0 h whose functions are discrete divergence free X 0 h,0 = {u h X 0 h : (u h, v h = 0 v h V 0 h }.

16 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS In fact we can construct φ h Vh 0 problem as the solution of the following discrete ( φ h, v h = (v h, v h v h V 0 h, and let w h = v h φ h. Clearly φ h L 2 (Ω v h L 2 (Ω and thus w h L 2 (Ω C v h L 2 (Ω. Since E E h is discrete divergence free by (8.16, we have by (8.18-(8.19 that (E E h, v h = (E E h, w h = (w, w h + ( φ, w h. ( We will use the duality argument to estimate w L 2 (Ω. Let z H 0 (curl; Ω be the solution of the following problem z k 2 z = w in Ω, z n = 0 on Γ. By the assumption that k 2 is not the eigenvalue of the Dirichlet problem for the Maxwell system, we know that z H 0 (curl; Ω is well-defined and z H(curl;Ω C w L 2 (Ω. Moreover, div z = 0 in Ω as the consequence of div w = 0 in Ω. Thus z X N (Ω. Since Ω is convex, by Theorem 8.5, we have z H 1 (Ω 3 and z H 1 (Ω C w L 2 (Ω. Similarly, noting that z n φ = z φ = n z φ = 0, φ H 1 (Ω, Γ Ω we have z X T (Ω, and hence by Theorem 8.5, z H 1 (Ω 3 and z H 1 (Ω C w L 2 (Ω. Now, since z is divergence free, by (8.16, w 2 L 2 (Ω = a(w, z = a(e E h, z = a(e E h, z γ h z, which implies, by Theorem 8.9, Therefore w 2 L 2 (Ω Ch w L 2 (Ω E E h H(curl;Ω. (w, w h Ch E E h H(curl;Ω w h L 2 (Ω Γ Ch E E h H(curl;Ω v h L 2 (Ω. ( To estimate ( φ, w h, we define v H 0 (curl; Ω such that v = w h, divv = 0. Note that v is the divergence free part in the Helmholtz decomposition of v h. Since Ω is convex, we have v H 1 (Ω 3 and v H 1 (Ω C w h L 2 (Ω. On the other hand, since v = w h L p (Ω for any p > 2, we

17 8.4. A POSTERIORI ERROR ANALYSIS 123 know from Lemma 8.5 that γ h v is well-defined. By using the operator τ h in Exercise 8.3, we have γ h v = τ h v = τ h w h = w h. Thus γ n v w h = ψ h for some ψ h H0 1(Ω. Since γ nv w h P(K, ψ h Vh 0. Now since divv = 0 and w h is discrete divergence free, w h v 2 L 2 (Ω = (w h v, w h γ h v + (w h v, γ h v v which implies = (w h v, γ h v v w h v L 2 (Ω γ h v v L 2 (Ω, w h v L 2 (Ω γ h v v L 2 (Ω. By using Lemma 8.5 we can prove as in Theorem 8.9 that γ h v v L 2 (Ω Ch ( v H 1 (Ω + v L 2 (Ω Ch wh L 2 (Ω. Therefore, since div v = 0, ( φ, w h = ( φ, w h v Ch w h L 2 (Ω φ L 2 (Ω = Ch v h L 2 (Ω φ L 2 (Ω. ( Combining (8.21-(8.22 with (8.20 we obtain (E E h, v h Ch E E h H(curl;Ω v h H(curl;Ω. Substitute it into (8.17 we know that for sufficiently small h, E h is uniquely existent, and by Theorem 8.9 E E h H(curl;Ω Ch ( E H 1 (Ω + E H 1 (Ω. This completes the proof A posteriori error analysis In this section we consider the a posteriori error estimate for the timeharmonic Maxwell equation with homogeneous Dirichlet boundary condition (8.14. We start with the following theorem on the interpolation of nonsmooth functions [8].

18 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Theorem There exists a linear projection Π h : H 1 (Ω 3 H 0 (curl; Ω X 0 h such that for all v H1 (Ω 3 Π h v L 2 (K C( v L 2 ( K + h K v H 1 ( K Π h v L 2 (K C v H 1 ( K K M h, K M h, v Π h v L 2 (K Ch K v H 1 ( K K M h v Π h v L 2 (F Ch1/2 F v H 1 ( F face F F h, where F h is the set of all interior faces of the mesh M h, K and F are the union of the elements in M h having having nonempty intersection with K and F, respectively. Proof. For any edge e E h, let w e X h be the associated canonical basis function of X h, that is, {w e } e Eh be the basis of X h satisfying w e t e dl = 1, e w e t e dl = 0 e e, e E h, e e. On each face F F h with edges {e 1, e 2, e 3 }, we construct a dual basis {q j } of {w i n} as follows (w i n q j ds = δ ij, i, j = 1, 2, 3. (8.23 We claim that F q i L (F Ch 1 F (8.24 which implies that q i L 2 (F C. Without loss of generality, we will prove that (8.24 holds for i = 1. We first find α = (α 1, α 2, α 3 T such that q 1 = α 1 w 1 n+α 2 w 2 n+α 3 w 3 n, (w i n q 1 ds = δ i1, i = 1, 2, 3. It is clear that α is the solution of the linear system ( A F α = (1, 0, 0 T, where A F = (w i n (w j n ds. ( We will show that A F is invertible. Let F be the face F 123 of a tetrahedron K with vertices A i, i = 1, 2, 3, 4 and let e 1, e 2, e 3 be the edges A 2 A 3, A 3 A 1, A 1 A 2, respectively. From Lemma 8.3, w 1 = λ 2 λ 3 λ 3 λ 2, w 2 = λ 3 λ 1 λ 1 λ 3, w 3 = λ 1 λ 2 λ 2 λ 1. F F

19 8.4. A POSTERIORI ERROR ANALYSIS 125 Let b ij = ( λ i n ( λ j n. Since 4 i=1 λ i = 0 and λ 4 is perpendicular to the face F 123, we have 3 b ij = 0 and b ij = b ji. j=1 Therefore, A F can be rewritten as 3b b 33 b 11 3b 33 + b 11 + b 22 3b 22 + b 33 + b 11 A F = F 12 3b 33 + b 11 + b 22 3b b 33 b 22 3b 11 + b 22 + b 33. 3b 22 + b 33 + b 11 3b 11 + b 22 + b 33 3b b 22 b 33 It follows from λ 1 F 234 that which implies that Similarly, λ 1 = 1/the height of K to the face F 234, b 11 = λ 1 n 2 = e F 2. b 22 = e F 2, b 33 = e F 2. Straightforward computation shows that det A F = e e e F c 0, where c 0 is a positive constant that depends only on the minimum angle of the elements in the mesh. Thus A F is invertible. Since A F = O(1, we have A 1 F = O(1 which implies α = A 1 F (1, 0, 0T = O(1, that is, (8.24 holds. Now for each e E h, we assign one of those faces with edge e and call it F e F h. We have to comply with the restriction that for e on the boundary, F e also on the boundary. Then we can define Π h v = ( (v n q F e e ds w e. e E F e h By virtue of (8.23 this defines a projection. Obviously the boundary condition is respected. By (8.24 (v n q Fe e dσ v L 2 (F e q F e e L F 2 (F C v e L 2 (F. e e

20 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Let K e M h be an element with F e as one of its faces. By the scaled trace inequality, we have Therefore, v 2 L 2 (F e C( h 1 e v 2 L 2 (K e + h e v 2 H 1 (K e. Π h v 2 L 2 (K Ch K Ch K e E h e K e E h e K (v nqe Fe F e dσ ( h 1 e v 2 L 2 (K + h e e v 2 H 1 (K e C ( v 2 L 2 ( K + h2 K v 2 H 1 ( K. This proves the first estimate in the theorem. Since Π h is a projection, we know that Π h c K = c K for any constant c K. Thus v Π h v L 2 (K = inf c K (v + c K Π h (v + c K L 2 (K C inf c K ( v + ck L 2 ( K + h K v + c K H 1 ( K Ch K v H 1 ( K, where we have used the scaling argument and Theorem 3.1 in the last inequality. This proves the third inequality. The last inequality can be proved similarly. The proof of the second inequality is left as an exercise. The following regular decomposition theorem is due to Birman-Solomyak. Lemma 8.6. Let Ω be a bounded Lipschitz domain. Then for any v H 0 (curl; Ω, there exists a ψ H 1 0 (Ω and a v s H 1 (Ω 3 H 0 (curl; Ω such that v = ψ + v s in Ω, and ψ H 1 (Ω + v s H 1 (Ω C v H(curl;Ω, where the constant C depends only on Ω. Proof. Let O be a ball containing Ω. We extend v by zero to the exterior of Ω and denote the extension by ṽ. Clearly ṽ H 0 (curl; O with compact support in O. By Theorem 8.4 there exists a w H 1 (O 3 such that and w = ṽ, div w = 0 in O, (8.26 w H 1 (O C ṽ L 2 (O = C v L 2 (Ω. (8.27 2

21 8.4. A POSTERIORI ERROR ANALYSIS 127 Now since O is simply-connected, (w ṽ = 0, by Theorem 8.3, there exists a φ H 1 (O/R such that ṽ = w + φ in O, and, from (8.27, φ H 1 (O C φ H 1 (O C( ṽ L 2 (O + w L 2 (O C v H(curl;Ω, φ H 2 (O\ Ω w H 1 (O C v L 2 (Ω. Since O \ Ω is a Lipschitz domain, by the extension theorem of Nečas, there exists an extension of φ O\ Ω, denoted by φ H 2 (R 3, such that φ = φ in O \ Ω, φ H 2 (R 3 C φ H 2 (O\ Ω C v H(curl;Ω. This completes the proof by letting ψ = φ φ H0 1(Ω and v s = w + φ. Remember that ṽ = v s + ψ in O and v s = ṽ = 0 in O \ Ω. Thus v s H0 1(Ω3. Theorem Let E H 0 (curl; Ω and E h X 0 h be respectively the solutions of (8.14 and (8.15. We have the following a posteriori error estimate E E h H(curl;Ω C 1/2, where ηk 2 =h 2 K f (α Eh + k 2 βe h 2 + F K K M h η 2 K L 2 (K + h2 K ( h F [[n (α E h ]] 2 L 2 (F + h F Here [[ ]] denotes the jump across the interior face F. div (f + k 2 βe h 2 L 2 (K [[(f + k 2 βe h n]] 2 L 2 (F Proof. Let a(, : H 0 (curl; Ω H 0 (curl; Ω R be the bilinear form defined by a(u, v = (α u, v k 2 (βu, v. From (8.14 and (8.15 we know that a(e E h, v h = 0 v h X 0 h. (8.28 The unique existence of the weak solution of the problem (8.14 implies that there exists a constant C > 0 such that a(u, v inf sup C. ( u H 0 (curl;ω u H(curl;Ω v H(curl,Ω 0 v H 0 (curl;ω For any v H 0 (curl; Ω, by Lemma 8.6, there exists a ψ H0 1 (Ω and a v s H 1 (Ω 3 H 0 (curl; Ω such that v = ψ + v s in Ω, and ψ H 1 (Ω + v s H 1 (Ω C v H(curl;Ω. (8.30.

22 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Let r h : H 1 (Ω Vh 0 be the Clément interpolant defined in Chapter 4, Section 4.2.1, and define v h = r h ψ + Π h v s X 0 h. Then by the inf-sup condition (8.29 and (8.28 E E h H(curl;Ω C a(e E h, v sup 0 v H 0 (curl,ω v H(curl,Ω a(e E h, v v h = C sup. 0 v H 0 (curl,ω v H(curl,Ω On the other hand, by integrating by parts, we have a(e E h, v v h = (f, v v h (α E h, (v v h + k 2( βe h, v v h = ( f, ( ψ + v s ( r h ψ + Π h v s ( α E h, (v s Π h v s + k 2( βe h, v s Π h v s + k 2 ( βe h, (ψ r h ψ = ( f (α Eh + k 2 βe h, v s Π h v s K M h + F F h K M h + F F h F [[n (α E h ]] (v s Π h v s ( div (f + k 2 βe h, ψ r h ψ F [[(f + k 2 βe h n]](ψ r h ψ. Now by Theorem 8.11 and (8.30 ( a(e E h, v v h C This completes the proof. K M h η 2 K ( C k M h η 2 K 1/2( vs H 1 (Ω + ψ H 1 (Ω 1/2 v H(curl;Ω. Bibliographic notes. The results in Section 8.1 are taken from Girault and Raviart [34]. Further results on vector potentials on nonsmooth domains can be found in Amrouche et al [2]. The full characterization of the trace for functions in H(curl; Ω can be found in Buffa et al [17]. The Nédélec edge elements are introduced in Nédédec [43, 44]. Lemma 8.5 is taken from [2].

23 8.5. EXERCISES 129 Further properties of edge elements can be found in Hiptmair [36] and Monk [42]. The error analysis in Section 8.3 follows the development in [42] which we refer to for further results. The interpolation operator in Theorem 8.11 is from Beck et al [8] although the proof here is slightly different. In [8] the a posteriori error estimate is derived for smooth or convex domains. Lemma 8.6, which is also known as regular decomposition theorem, is from Birman and Solomyak [10]. Theorem 8.12 is from Chen et al [21] in which the adaptive multilevel edge element method for time-harmonic Maxwell equations based on a posteriori error estimate is also considered Exercises Exercise 8.1. Prove D( Ω 3 is dense in H(div; Ω. Exercise 8.2. The lowest order divergence conforming finite element is a triple (K, D, N with the following properties: (i K R 3 is a tetrahedron; (ii D = {u = a K + b K x a K R 3, b K R}; (iii N = {M F : M F (u = F (u n ds face F of K, u D}. For any vector field u defined on K, let û F K (ˆx = B K u(b K ˆx + b K. Prove that (i u D(K û ˆD( ˆK; (ii div u = 0 div ˆ û = 0 u D(K; (iii M F (u = 0 M ˆF (û = 0 u D(K; (iv If u D(K and M F (u = 0, then u n = 0 on F ; (v If u D(K and M F (u = 0 for any face F, then u = 0 in K. Exercise 8.3. For any function u defined on K such that M F (u is defined on each face F of K. Let τ K u be the unique polynomial in D(K that has the same moments as u on K: M F (τ K u u = 0. Prove that for any p > 2, τ K is continuous on the space {u L p (K 3 : div u L 2 (K}. Exercise 8.4. Let D(K be the finite element space in Exercise 8.2 and W h = {u h H(div; Ω : u h K D(K K M h }. Let τ h be the global interpolation operator τ h u K = τ K u on K, K M h.

24 FINITE ELEMENT METHODS FOR MAXWELL EQUATIONS Prove that X h W h and γ h u = τ h u. Exercise 8.5. Prove that u τ h u L 2 (Ω Ch u H 1 (Ω u H 1 (Ω 3, div (u τ h u L 2 (Ω Ch div u H 1 (Ω u H 1 (Ω 3, div u H 1 (Ω. Exercise 8.6. Prove the second estimate in Theorem 8.11.