Chapter 3 Conforming Finite Element Methods 3.1 Foundations Ritz-Galerkin Method

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1 Chapter 3 Conforming Finite Element Methods 3.1 Foundations Ritz-Galerkin Method Let V be a Hilbert space, a(, ) : V V lr a bounded, V-elliptic bilinear form and l : V lr a bounded linear functional. We want to approximate the variational equation: Find u V such that (3.1) a(u, v) = l(v), v V. We recall that the Lax-Milgram Lemma ensures the existence and uniqueness of a solution of (3.1). Now, given a finite-dimensional subspace V h V, dim V h = n h, the Ritz-Galerkin method is to approximate (3.1) by its restriction to V h : Find u h V h such that (3.2) a(u h, v h ) = l(v h ), v h V h. Again, by the Lax-Milgram Lemma we know that (3.2) has a unique solution u h V h. It is easy to see that the Ritz-Galerkin method gives rise to a linear algebraic system, once we specify a basis of V h. Therefore, let us assume that (ϕ (i) h )n h i=1 is a basis of V h, i.e., (3.3) V h = span(ϕ (1) h,..., ϕ(n h) h ). Then, the solution u h V h of (3.2) can be represented as a linear combination of the basis functions according to (3.4) u h = n h j=1 α j ϕ (j) h. Apparently, u h as given by (3.4) satisfies (3.2) if and only if (3.5) n h j=1 a(ϕ (j) h, ϕ(i) h ) α j = l(ϕ (i) h ), 1 i n h. Obviously, (3.5) represents a linear algebraic system (3.6) A h α h = b h 1

2 2 Ronald H.W. Hoppe in the unknown vector α h = (α 1,..., α nh ) T, where the stiffness matrix A h lr n h n h and the load vector b h lr n h are given by a(ϕ (1) h, ϕ(1) h ) a(ϕ(n h) h, ϕ (1) h ) (3.7) A h := a(ϕ (1) h, ϕ(n h) h ) a(ϕ (n h) h, ϕ (n h) h ) and (3.8) b (i) h := l(ϕ (i) h ), 1 i n h. There are basically two major aspects with regard to the construction of appropriate finite-dimensional subspaces V h : The efficient numerical solution of the linear algebraic system (3.6). The accuracy of the approximation of the solution u V of (3.1) by the solution u h V h of (3.2) Céa s Lemma Céa s Lemma tells us that under the assumptions of the Lax-Milgram Lemma the accuracy of the solution u V of (3.1) by the solution u h V h of (3.2) is as good as the best approximation of u V by a function in V h which reduces this issue to a problem of approximation theory. It is based on the observation that the error u u h is a-orthogonal to V h, i.e., (3.9) a(u u h, v h ) = 0, v h V h, a property which is referred to as Galerkin orthogonality. Definition 3.1 Elliptic projection If the bilinear form a(, ) is symmetric and V-elliptic, it defines an inner product on V, the energy inner product. Then, (3.9) states that the solution u h V h of (3.2) is the projection of the solution u V of (3.1) onto V h which is called the elliptic projection. Lemma 3.1 Céa s Lemma Assume that a(, ) : V V lr is a bounded, V-elliptic bilinear form, i.e., there exist positive constants C and α such that (3.10) (3.11) a(u, v) C u V v V, u, v V, a(u, u) α u 2 V, u V. Assume further that l V is a bounded linear functional and that u V and u h V h are the unique solutions of (3.1) and (3.2), respectively.

3 Finite Element Methods; Chapter 3 3 Then, there holds (3.12) u u h V C α inf v h V h u v h V. Proof. Using the V-ellipticity and boundedness of a(, ) as well as the Galerkin orthogonality (3.9), we find that for any v h V h α u u h 2 V a(u u h, u u h ) = from which we readily deduce (3.12) Triangulations = a(u u h, u v h ) + a(u u h, v h u h ) }{{} = 0 C u u h V u v h V, The construction of conforming finite element spaces is based on a suitable partition of the computational domain. Definition 3.2 Triangulation Let Ω lr d be a bounded domain with a Lipschitz continuous boundary Γ = Ω. A triangulation T h of Ω is a partition of Ω into a finite number of subsets K, called finite elements, such that (T h 1) Ω = K T h K, (T h 2) K = K, K o, K T h, (T h 3) K o 1 K o 2 = for all K 1, K 2 T h, K 1 K 2, (T h 4) K, K T h is Lipschitz continuous. We consider two types of elements: d-simplices and d-rectangles. Definition 3.3 d-simplex A d-simplex K in lr d is the convex hull of d + 1 points a j = (a ij ) d l=1 lr d : d+1 d+1 K = { x = λ j a j 0 λ j 1, λ j = 1 }. j=1 The simplex K is called non degenerate, if any point x lr d can be uniquely represented in the form j=1 j=1 d+1 d+1 x = λ j a j, λ j lr, λ j = 1. j=1

4 4 Ronald H.W. Hoppe Remark 3.1 Non degeneracy of a d-simplex The non degeneracy of a d-simplex is related to the unique solvability of the linear system a 11 a 1,d+1 λ 1 x 1 a d1 a d,d+1 λ d = x d. } 1 {{ 1 } λ d+1 1 =: A Obviously, K is non degenerate if and only if the matrix A is regular. a 4 a 3 a 1 a 2 a 3 a 2 a 1 Figure 3.1: Triangle (d=2) and tetrahedron (d=3) Definition 3.3 Vertices, edges, and faces The points a j, 1 j d + 1, of a d-simplex K are called vertices. An m-dimensional face of a d-simplex K, 0 m d, is an m-simplex whose vertices correspond to vertices of K. A 1-dimensional face is referred to as an edge. For D Ω, we refer to V h (D), F h (D), and E h (D) as the sets of vertices, (d 1)-dimensional faces and edges of T h, respectively. The d-simplex ˆK with vertices â 1 = (0,..., 0) d and â i+1 = e i, 1 i d, is referred to as the unit d-simplex.

5 Finite Element Methods; Chapter 3 5 Definition 3.4 Barycentric coordinates, center of gravity The barycentric coordinates λ j, 1 j d + 1, of a point x lr d with respect to the d + 1 vertices a j of a non degenerate d-simplex K are the components of the unique solution of the linear system (3.13). The center of gravity x S of a non degenerate d-simplex K is the point with 1 λ j (x S ) =, 1 j d + 1. d + 1 Lemma 3.2 Affine transformation Any non degenerate d-simplex K lr d is the image of the unit d- simplex ˆK under an affine transformation F K : lr d lr d ˆx F K (ˆx) = B K ˆx + b K with a nonsingular matrix B K lr d d and b K lr d. Definition 3.5 Simplicial triangulation A triangulation T h of a polyhedral domain Ω lr d is called a simplicial triangulation, if its elements K are d-simplices. Definition 3.6 d-rectangle A d-rectangle K in lr d is the tensor product of d intervals [c i, d i ], c i d i, 1 i d, i.e., d K = [c i, d i ] = { x = (x 1,..., x d ) T c i x i d i, 1 i d }. i=1 A d-rectangle K is said to be non degenerate, if c i < d i, 1 i d. The d-rectangle ˆK := [0, 1] d is called the unit d-rectangle (unit cube) in lr d. a 7 a 8 a 3 a 4 a 5 a 6 a 3 a 4 a 1 a 2 a 1 a 2 Figure 3.2: 2-Rectangle (d=2) and 3-rectangle (d=3)

6 6 Ronald H.W. Hoppe Definition 3.7 Vertices, edges, and faces The points a j, 1 j 2 d, of a d-rectangle K given by a j = (a j1,..., a jd ) T, a ji = c i or a ji = d i, 1 i d, are called vertices. An m-dimensional face of a d-rectangle K, 1 m d 1, is an m-rectangle whose vertices correspond to vertices of K. A 1-dimensional face is referred to as an edge. For D Ω, we refer to V h (D), F h (D), and E h (D) as the sets of vertices, (d 1)-dimensional faces and edges of T h, respectively. Lemma 3.3 Diagonal affine transformation Any non degenerate d-rectangle K lr d is the image of the unit d- rectangle (unit cube) ˆK under a diagonal affine transformation F K : lr d lr d ˆx F K (ˆx) = B K ˆx + b K with a nonsingular diagonal matrix B K = (b K ii ) d i=1 and b K lr d. Definition 3.8 Rectangular triangulation A triangulation T h of a rectangular domain Ω lr d is called a rectangular triangulation, if its elements K are d-rectangles. 3.3 Local specification of finite elements With respect to a triangulation T h of the computational domain Ω lr d, conforming finite element functions are defined locally for the elements K T h and composed in such a way that the resulting globally defined function belongs to the underlying function space V. Definition 3.9 Local trial functions and degrees of freedom Let T h be a triangulation of Ω lr d and K T h. Assume that P K is a linear space of functions p : K lr with P K H 1 (K) and dim P K = n K. the elements of P K are called local trial functions. Let l i P K, 1 i n K be bounded linear functionals l i : P K lr, 1 i n K, and consider (3.14) Σ K := { l i (p) p P K, 1 i n K }. The elements of Σ K are called degrees of freedom.

7 Finite Element Methods; Chapter 3 7 Definition 3.10 Finite elements and unisolvence Let T h be a triangulation of Ω lr d, K T h and P K, Σ K as in Definition 3.9. Then, the triple (K, P K, Σ K ) is called a finite element. A finite element (K, P K, Σ K ) is said to be unisolvent, if any p P K is uniquely determined by its degrees of freedom in Σ K. Definition 3.11 Affine equivalence of finite elements Let T h be a triangulation of Ω lr d, K T h and P K, Σ K as in Definition 3.9. Let further ( ˆK, ˆP ˆK, ˆΣ ˆK) be a reference element. Then, the finite elements (K, P K, Σ K ), K T h, are said to be affine equivalent to the reference element ( ˆK, ˆP ˆK, ˆΣ ˆK), if there exists an invertible affine mapping F K : lr d lr d such that for all K T h (3.15) K = F K ( ˆK), (3.16) P K = {p : K lr p = ˆp F 1 K, ˆp ˆP ˆK}, (3.17) Σ K = {l i : P K lr l i = ˆl i F 1 K, ˆl i ˆΣ ˆK, 1 i n K }. Definition 3.12 Finite element space Let T h be a triangulation of Ω lr d and P K, K T h as in Definition 3.9. Then (3.18) V h := { v h : Ω lr v h K P K, K T h } is called a finite element space. Theorem 3.1 H 1 -conformity of finite elements Let V h be a finite element space and assume that (3.19) (3.20) Then (3.21) P K H 1 (K), K T h, V h C 0 (Ω). V h H 1 (Ω). Proof. Let v h V h. Obviously, v h L 2 (Ω). We have to show that v h admits weak first derivatives wh α L2 (Ω), α = 1, i.e., (3.22) v h D α z dx = ( 1) α wh α z dx, z C0 (Ω). Ω Ω

8 8 Ronald H.W. Hoppe Since v h K H 1 (K), we may apply Green s theorem elementwise and obtain v h D α z dx = (3.23) v h D α z dx = K T h Ω = K T h K D α v h z dx + K T h v h n α D α z dσ = K = K T h K D α v h z dx + K F F h (Ω) F [v h ] n α D α z dσ, where [v h ] denotes the jump [v h ] := v h K1 v h K2 across F = K 1 K 2, K i T h, 1 i 2,. But v h C 0 (Ω), and hence, [v h ] = 0 in (3.23) which proves (3.22) with wh α K := D α v h K, K T h. Corollary 3.1 H 1 0-conformity of finite elements Let V h be a finite element space and assume that (3.24) (3.25) Then (3.26) P K H 1 (K), K T h, V h C 0 0(Ω). V h H 1 0(Ω). Proof. The proof is left as an exercise. 3.4 Lagrangian finite elements of type (k) Definition 3.13 Polynomials of degree k Let K be a d-simplex. For k 0, we define P k (K) as the linear space of all polynomials of degree k on K, i.e., p P k (K), if p(x) = a α x α, a α lr, α k, where x α := d i=1 We note that (3.27) α k x α i i, α := (α 1,..., α d ) ln d 0, α := dim P k (K) = ( k+d d ) = (k + d)! d! k!. d α i. i=1

9 Finite Element Methods; Chapter 3 9 Definition 3.14 Central lattice of order k of K Let K be a d-simplex with vertices a i, 1 i d + 1, and k ln 0. The set d+1 L k (K) := { x = λ i a i λ i {0, 1 k,..., k 1 d+1 k, 1}, λ i = 1 } i=1 is called the central lattice of order k of K. We have card L k (K) = ( k+d d ). i=1 Definition 3.15 Nodal points The elements of the central lattice of order k are called nodal points. For D Ω, we refer to N h (D) as the set of nodal points in D. Definition 3.16 Lagrangian finite element of type k Let T h be a triangulation of Ω lr d. Then, the triple (K, P k (K), L k (K)), K T h is called a Lagrangian finite element of type (k). Lemma 3.4 Unisolvence of Lagrangian FEs of type (k) A Lagrangian finite element of type (k) is unisolvent. Proof. The proof is left as an exercise. Lemma 3.5 Affine equivalence of Lagrangian FEs of type (k) Let e i, 1 i d, be the unit vectors in lr d with respect to the Cartesian coordinate system and (3.28) ˆK := conv(0, e 1,..., e d ), ˆP ˆK := P k ( ˆK), k ln, (3.29) ˆΣ ˆK := {ˆp(ˆx) ˆx L k ( ˆK), ˆp ˆP ˆK}. The Lagrangian finite elements of type k are affine equivalent to the reference element ( ˆK, ˆP ˆK, ˆΣ ˆK. In particular, ˆK is called the reference d-simplex. Proof. Let F K : lr d lr d be the invertible affine mapping with K = F K ( ˆK). Then, P k (K) = {p = ˆp F 1 K ˆp P k( ˆK)}. Moreover, since x L k (K) x = F K (ˆx), ˆx L k ( ˆK), (3.17) is easily verified.

10 10 Ronald H.W. Hoppe Fig. 3.3.a: P K = P 1 (K) (d = 2) Fig. 3.3.b: P K = P 1 (K) (d = 3) Fig. 3.4.a: P K = P 2 (K) (d = 2) Fig. 3.4.b: P K = P 2 (K) (d = 3) Fig. 3.5.a: P K = P 3 (K) (d = 2) Fig. 3.5.b: P K = P 3 (K) (d = 3)

11 Finite Element Methods; Chapter 3 11 Figures 3.3,3.4 and 3.5 illustrate Lagrangian finite elements of type (k) for k = 1, 2, 3 in two and in three dimensions. Note that a Lagrangian finite element of type (1) for d = 2 is called Courant s triangle (cf. Fig. 3.3a). Definition 3.17 Lagrangian finite element space The finite element space V h composed by Lagrangian finite elements of type (k) is called Lagrangian finite element space and denoted by S k (Ω, T h ). In order to ensure conformity of S k (Ω, T h ), we have to require that the simplicial triangulation T h is geometrically conforming. Definition 3.18 Geometrically conforming simplicial triangulation A simplicial triangulation T h of the computational domain Ω lr d is called geometrically conforming, if there holds (T h 5) The intersection of two different elements of the triangulation is either empty, or consists of a common face, or a common edge, or a common vertex. Lemma 3.6 Conformity of Lagrangian finite element spaces Let T h be a geometrically conforming simplicial triangulation of the computational domain Ω lr d. Then, there holds (3.30) S k (Ω, T h ) = {v h C 0 (Ω) v h K P k (K), K T h } H 1 (Ω). Proof. Since P k (K) H 1 (K), K T h, in view of Theorem 3.1 we have only to show that S k (Ω, T h ) C 0 (Ω). We give the proof exemplarily in case d = 2 and k = 2: Let K i T h, 1 i 2, be two adjacent elements such that E = K 1 K 2 E h (Ω) with nodal points a j N h (E), 1 j 3. Let further p i P 2 (K i ), 1 i 2. Then, p i E P 2 (E). Since p 1 (a j ) = p 2 (a j ), 1 j 3, we conclude that p 1 E p 2 E. Corollary 3.2 Conformity of Lagrangian finite element spaces Let S k,0 (Ω, T h ) := {v h S k (Ω, T h ) v h K Ω = 0, K T h, K Ω }. Then, under the same assumptions as in Lemma 3.5 there holds (3.31) S k,0 (Ω, T h ) H 1 0(Ω). Proof. The proof is left as an exercise.

12 12 Ronald H.W. Hoppe Definition 3.19 Nodal basis functions Let N h (Ω) = {x j 1 j n h }. The function ϕ i S k (Ω, T h ) given by (3.32) ϕ i (x j ) = δ ij, 1 i, j n h, is called nodal basis function with supporting point x i N h (Ω). The set of these functions constitutes a basis of S k (Ω, T h ). Lemma 3.7 Representation by barycentric coordinates Let K be a d-simplex with vertices a i, 1 i d + 1 and set (3.33) a ij := 1 2 (a i + a j ), 1 i < j d + 1, a iij := 1 3 (2a i + a j ), 1 i j d + 1, a ijk := 1 3 (a i + a j + a k, 1 i < j < k d + 1. Let λ i, 1 i d, be the barycentric coordinates. Then there holds: (i) For k = 1, the nodal basis functions ϕ i associated with the nodal points a i, 1 i d + 1, admit the representation (3.34) ϕ i = λ i, 1 i d + 1. (ii) For k = 2, the nodal basis functions ϕ i associated with the nodal points a i, 1 i d + 1, admit the representation (3.35) ϕ i = λ i (2λ i 1), 1 i d + 1. ϕ ij associated with the nodal points a ij, 1 i < j d + 1, admit the representation (3.36) ϕ ij = 4 λ i λ j, 1 i < j d + 1. (iii) For k = 3, the nodal basis functions ϕ i associated with the nodal points a i, 1 i d + 1, admit the representation (3.37) ϕ i = 1 2 λ i (3λ i 1) (3λ i 2), 1 i d + 1. ϕ iij associated with the nodal points a iij, 1 i j d + 1, admit the representation (3.38) ϕ iij = λ i λ j (3λ i 1), 1 i d + 1.

13 (3.39) Finite Element Methods; Chapter 3 13 ϕ ijk associated with the nodal points a ijk, 1 i < j < k d+1, admit the representation ϕ ijk = 27 λ i λ j λ k, 1 i < j < k d + 1. Proof. The proof is left as an exercise. a 3 e 2 e 1 a 1 e 3 a 2 Fig. 3.6: Non degenerate triangle with vertices a i and edges e i Lemma 3.8 Geometric characterization Let K be a non degenerate 2-simplex with vertices a i, 1 i 3, and edges e i, 1 i 3 (cf. Fig. 3.6). Moreover, let m ei be the midpoints of the edges and denote by n ei the exterior unit normals to e i. Then, the nodal basis functions ϕ i, 1 i 3, associated with the vertices a i admit the representation (3.40) ϕ i (x) = meas(e i) 2meas(K) n e i (m ei x), x = (x 1, x 2 ) T. Proof. The proof is left as an exercise. 3.5 Hermitian finite elements of type (3) The name Lagrangian finite elements is motivated by polynomial interpolation of Lagrangian type. Indeed, if Π K, K T h, is the associated local interpolation operator, Π K v, v C 0 (K), interpolates the function p in x L k (K). Another type of polynomial interpolation is Hermite interpolation, where both point values and derivatives of a function are interpolated. The counterpart in finite elements are Hermitian finite elements. As an example, we consider Hermitian finite elements of type 3. Definition 3.20 Hermitian finite elements of type (3) Let K be a non degenerate d-simplex with vertices a i, 1 i d + 1, and a ijk := 1 3 (a i + a j + a k ), 1 i < j < k d + 1. Moreover, let (3.41) P K := P 3 (K),

14 14 Ronald H.W. Hoppe (3.42) Σ K := {p(a i ), 1 i d + 1, p(a ijk ), 1 i < j < k d + 1, p x j (a i ), 1 i d + 1, 1 j d}. Then, (K, P K, Σ K ) is called a Hermitian finite element of type (3). Remark 3.1 Equivalent specification of the DOFs We note that in the definition of the degrees of freedom of the Hermitian finite element of type (3) the partial derivatives p x j (a i ), 1 i d + 1, 1 j d, can be replaced by the directional derivatives Dp(a i )(a j a i ), 1 i d + 1, 1 j d, i.e., instead of Σ K we have (3.43) Σ K := {p(a i ), 1 i d + 1, p(a ijk ), 1 i < j < k d + 1, Dp(a i )(a j a i, 1 i d + 1, 1 j d}. Lemma 3.9 Unisolvence of Hermitian finite elements A Hermitian finite element of type (3) is unisolvent. Proof. We give the proof exemplarily in case d = 3. Let p P 3 (K). It suffices to show that (3.44) (3.45) (3.46) p(a i ) = 0, 1 i 4, p(a ijk ) = 0, 1 i < j < k 4, Dp(a i )(a j a i ) = 0, 1 i 4, 1 j 3 implies p 0. Let F ijk, 1 i < j < k 4, be the face with vertices a i, a j, a k and let e ij, e ik, e jk be the corresponding edges. Then p eij P 3 (e ij ), p eik P 3 (e ik ), p ejk P 3 (e jk ). Since the Hermite interpolation polynomial interpolating p(a i ), p(a j ), Dp(a i )(a j a i ), Dp(a j )(a i a j ) is uniquely determined, (3.44) and (3.46) imply p eij 0. Likewise, we deduce p eik 0 and p ejk 0, and hence Consequently, p is of the form p Fijk 0. p = α λ i λ j λ k, α lr.

15 Then, (3.45) implies α = 0, i.e., Finite Element Methods; Chapter 3 15 p Fijk 0, 1 i < j < k 4 = p 0. a 3 a 1 a 2 a 4 a 3 a 2 a 1 Figure 3.6: Hermitian finite element of type (3) Figure 3.6 contains a schematic representation of Hermitian finite elements of type (3). The degrees of freedom, as given by point values, are marked by a dot, whereas the degrees of freedom, associated with partial resp. directional derivatives, are indicated by a circle. Lemma 3.10 Affine equivalence of Hermitian finite elements Let ˆK be the reference d-simplex and ˆP ˆK := P 3 ( ˆK), ˆΣ ˆK := {ˆp(â i ), 1 i d + 1, ˆp(â ijk ), 1 i < j < k d + 1, ˆp ˆx j (â i ), 1 i d + 1, 1 j d}. The Hermitian finite element K, P K, Σ K ) of type (3) is affine equivalent to the reference element ˆK, ˆP ˆK, ˆΣ ˆK) of type (3). Proof. We only have to show how the partial derivatives are transformed. Let F K : lr d lr d be the invertible affine mapping F K (ˆx) = B K ˆx+b K such that K = F K ( ˆK). Then, for ˆp P 3 ( ˆK) and p = ˆp F 1 K

16 16 Ronald H.W. Hoppe we have whence p x i (a j ) = d ˆp ˆx k k=1 (F 1 K (a j)) k x i, Dp(a j ) = B 1 K ˆDˆp(â j ). Definition 3.21 Hermitian finite element space For a simplicial triangulation T h of the polyhedral domain Ω lr d, the finite element space V h composed by Hermitian finite elements of type (3) is called a Hermitian finite element space. It will be denoted by H 3 (Ω; T h ). Lemma 3.11 Conformity of Hermitian finite element spaces Let T h be a geometrically conforming simplicial triangulation of Ω lr d. Then, the the Hermitian finite element space H 3 (Ω; T h ) is conforming, i.e., (3.47) H 3 (Ω; T h ) H 1 (Ω). Proof. The proof is left as an exercise. Definition 3.22 Basis functions of Hermitian finite element spaces We restrict ourselves to the case d = 2 and assume that H 3 (Ω; T h ) is the Hermitian finite element space with respect to a geometrically conforming simplicial triangulation of a polygonal domain Ω lr 2. We denote by b l, 1 l r h, the nodal points that are vertices of triangles T T h and by b l, r h + 1 l s h, those that are centers of gravity of T T h. Then, an appropriate basis (3.48) ϕ k, 1 k s h, ϕ (ν) k, 1 k r h, 1 ν 2, is given by (3.49) ϕ k (b l ) = δ kl, 1 k, l s h, ϕ k (b l ) x ν = 0, 1 k s h, 1 l r h, 1 ν 2, (3.50) ϕ (1) k (b l) = 0, 1 k r h, 1 l s h, ϕ (1) k x 1 (b l ) = δ kl, ϕ (1) k x 2 (b l ) = 0, 1 k, l r h,

17 Finite Element Methods; Chapter 3 17 (3.51) ϕ (2) k (b l) = 0, 1 k r h, 1 l s h, ϕ (2) k x 1 (b l ) = 0, ϕ (2) k x 2 (b l ) = δ kl, 1 k, l r h. 3.6 Lagrangian finite elements of type [k] Let K := d [c i, d i ] lr d be a d-rectangle. For k ln 0 we denote by i=1 Q k (K) the linear space of all polynomials that are of degree k in each of the d variables x i, 1 i d, i.e., p Q k (K) is of the form (3.52) p(x) = α i k a α1...α d x α 1 1 x α x α d d, from which we easily deduce (3.53) dim Q k (K) = (k + 1) d. For the d-rectangle K we consider the point-set (3.54) and define L [k] (K) := {x = (x i1,..., x id ) T x il := c l + i l k (d l c l ), i l {0, 1,..., k}, 1 l d} (3.55) Σ K := {p(x) x L [k] (K)}, p Q k (K). Then there holds: Definition 3.22 Lagrangian finite element of type ([k]) The element (K, Q k (K), Σ K ) is called a Lagrangian finite element of type [k] and will be denoted by S [k] (K). The points x L [k] (K) are refereed to as nodal points. Lemma 3.12 Lagrangian finite element of type ([k]) as a tensor product finite element The Lagrangian finite element of type ([k]) is a tensor product finite element based on tensor product Lagrangian type polynomial interpolation: The polynomial p Q k (K), interpolating in x L [k] (K), has the representation (3.56) p(x) = L(x) p(y). y L [k] (K)

18 18 Ronald H.W. Hoppe Here, the Lagrangian fundamental polynomial L Q k (K) is given by d (3.57) L(x) = L il (x) in terms of the one-dimensional Lagrangian fundamental polynomials k x x i (3.58) L il (x) := l, 1 l d. x il x i l i l i l =0 Proof. Let p Q k (K) and consider an arbitrary edge e j K, j {1, d} as given by d e j := [c j, d j ] {g l }, g l {c l, d l }. j l=1 l=1 Then, p ej P k (e j ) is uniquely determined by its values in x ij := c j + i j k (d j c j ) e j, i j {0, 1,..., k}, and has the Lagrangian representation p ej (x) = k L ij (x) p ej (x ij ), x e j. i j =0 Lemma 3.13 Unisolvence The Lagrangian finite element of type ([k]) is unisolvent. Proof. The proof is an immediate consequence of Lemma Lemma 3.14 Affine equivalence of Lagrangian FEs of type [k] Let ˆK := [0, 1] d be the unit cube in lr d and (3.59) ˆP ˆK := Q k ( ˆK), k ln, (3.60) ˆΣ ˆK := {ˆp(ˆx) ˆx L [k] ( ˆK), ˆp ˆP ˆK}. The Lagrangian finite elements of type [k] are affine equivalent to the reference element ( ˆK, ˆP ˆK, ˆΣ ˆK. In particular, ˆK is called the reference d-rectangle. Proof. The proof is left as an exercise.

19 Finite Element Methods; Chapter 3 19 Fig. 3.7.a: S [1] (K) (d = 2) Fig. 3.7.b: S [1] (K) (d = 3) Fig. 3.8.a: S [2] (K) (d = 2) Fig. 3.8.b: S [2] (K) (d = 3) For k = 1 and k = 2, Figures 3.7 and 3.8 contain illustrations of Lagrangian finite elements of type [k]. Definition 3.23 Lagrangian finite element space Let Ω lr d be given as the union of a finite number of d-rectangles and let T h be a rectangular triangulation of Ω. The finite element space V h composed by Lagrangian finite elements of type [k] is called Lagrangian finite element space and denoted by S [k] (Ω, T h ). Lemma 3.15 Conformity of Lagrangian finite element spaces Let T h be a geometrically conforming rectangular triangulation of the computational domain Ω. Then, there holds (3.61) S [k] (Ω, T h ) = {v h C 0 (Ω) v h K Q k (K), K T h } H 1 (Ω). Proof. The proof is left as an exercise

20 20 Ronald H.W. Hoppe Corollary 3.3 Conformity of Lagrangian finite element spaces Let S [k],0 (Ω, T h ) := {v h S [k] (Ω, T h ) v h K Ω = 0, K T h, K Ω }. Then, under the same assumptions as in Lemma 3.15 there holds (3.62) S [k],0 (Ω, T h ) H 1 0(Ω). Proof. The proof is left as an exercise. Definition 3.24 Nodal basis functions Let N h (Ω) = {x L [k] (K) K T h } and suppose card(n h (Ω)) = n h. The function ϕ i S [k] (Ω, T h ) given by (3.63) ϕ i (x j ) = δ ij, 1 i, j n h, is called nodal basis function with supporting point x i N h (Ω). The set of these functions constitutes a basis of S [k] (Ω, T h ).