Preprint-Series of the Institute of Applied Mathematics
The invertibility of the isoparametric mapping for pyramidal and prismatic nite elements Peter Knabner and Gerhard Summ Institute of Applied Mathematics, Martensstrae 3, 91058 Erlangen, Germany. e-mail: knabner,
[email protected]
April 15, 1999
Summary We consider the isoparametric transformation, which
maps a given reference element onto a global element given by its vertices, for multi-linear nite elements on pyramids and prisms. We present easily computable conditions on the position of the vertices, which ensure that the isoparametric transformation is bijective. Mathematics Subject Classi cation (1991): 65N30, 65N50
1 Introduction The partition of a complex three-dimensional domain into a mesh consisting of hexahedral nite elements is a dicult task, especially if one wants to obtain a locally re ned grid, whereas the partition into a tetrahedral mesh is much easier to perform. On the other hand hexahedral meshes provide a better approximation than corresponding tetrahedral meshes with the same number of degrees of freedom. Therefore it is desirable to combine tetrahedrons and hexahedrons into one mesh. Unfortunately the faces of tetrahedrons are triangles, and the faces of hexahedrons are quadrilaterals. Thus, to get a conforming grid consisting of both tetrahedrons and hexahedrons, it is necessary to use in addition nite elements, which have both triangular and quadrangular faces. Those elements can be pyramids (one quadrangular face and four triangular faces) or prisms (three quadrangular and two triangular faces).
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P. Knabner and G. Summ
Furthermore, the use of prismatic elements instead of tetrahedrons may be desirable to resolve thin layers typical for geological applications. For a mesh consisting of tetrahedrons and prisms, in addition pyramidal elements are necessary to make it consistent (see e.g. [3]). The common convergence proof for the nite element method employs an estimation of the interpolation error on a single nite element, which is obtained by a transformation to a given reference element, the application of the Bramble-Hilbert lemma on the reference element, and a back-transformation to the global element (cf. [1]). For ane families of nite elements this transformation is performed by an ane-linear mapping, so the question of the invertibility of this transformation reduces to the evaluation of the constant value of the Jacobian determinant. To gain more exibility also isoparametric nite elements are used. In these cases the Jacobian determinant is not constant any more, and it is not evident, for which choices of the vertices the transformation remains invertible. In this paper we investigate the invertibility of the isoparametric transformation for multi-linear pyramidal and prismatic nite elements. We present easy to check relations for the vertices, such that the corresponding isoparametric mapping is invertible if and only if these conditions are satis ed. In Section 2 below we recall some de nitions (e.g. of isoparametric equivalence between nite elements) and introduce a decomposition of the isoparametric transformation to simplify the following considerations. The case of pyramidal nite elements is considered in Section 3, and the case of prismatic nite elements in Section 4. In Section 5 we nally illustrate the obtained conditions with some examples of degenerated elements, which violate these conditions.
2 General considerations We recall the de nition of a nite element, as it can be found in many textbooks(e.g. [1], Section 10). De nition 2.1 A nite element is a triple (K; P; ) consisting of 1. a closed and connected set K Rn with nonempty interior and Lipschitz-continuous boundary, 2. a nite-dimensional ansatz space P of real-valued functions, 3. a P -unisolvent set of linear forms = fi : P ! R : i = 1; : : : ; Ng, the degrees of freedom. Here P -unisolvent means, that the mapping := (i )i=1;:::;N : P ! RN is bijective. In particular this requires N = dim(P ) and
Invertibility of the isoparametric mapping
3
implies the existence of N functions 'i 2 P , i = 1; : : : ; N , ful lling j ('i ) = ij , j = 1; : : : ; N , which form a basis of P . In this paper we will only consider Lagrange nite elements, i.e., the degrees of freedom are given by the Dirac functionals ai , such that i (p) = ai (p) = p(ai ) for all p 2 P and some node ai 2 K . The following de nition of isoparametric equivalence makes use of this restriction. De nition 2.2 A Lagrange nite element (K; P; ^ )^ is^called isoparametrically equivalent to a reference element K; P; , if there exists a bijective mapping FK : K^ ! Rn , such that FK 2 (P^ )n ; (2.1) ^ (2.2) n K = FK?(K1 ) ; ^ o (2.3) P = p = p^ FK p^ 2 P ; o n (2.4) = ai ai = FK (^ai ); a^i 2 ^ : In practical applications neither the nite element (K; P; ), nor the transformation FK is given. Only the nodes ai 2 K are given (and therefore the set of degrees of freedom ). Then the following choice for FK ful lls the conditions (2.1) and (2.4):
FK : K^
! Rn
; FK (^x) =
X
dim(P^ )
i=1
ai '^i (^x) for all x^ 2 K^ ; (2.5)
where '^i , i = 1; : : : ; dim(P^ ) are the basis functions of P^ ful lling '^i (^aj ) = ij for j = 1; : : : ; dim(P^ ). K and P are then de ned by (2.2) and (2.3). Only one condition in the de nition of isoparametric equivalence remains open: the question whether the transformation FK de ned by (2.5) is bijective. In the following sections we answer this question for multi-linear ansatz spaces on pyramidal and prismatic nite elements. In both cases the nodes are given by the vertices of the geometric element K . By the implicit function theorem we can only deduce, that the mapping FK is locally invertible, if the determinant of the Jacobian matrix does not vanish. But in the case of multi-linear ansatz functions the second derivative of FK is constant, such that we can apply the ideas used by Ciarlet ([1], proof of Theorem 37.2) to conclude that FK is globally invertible, if and only if for all x^ 2 K^ the Jacobian determinant det DFK (^x) does not vanish.
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Before we address the question, under which conditions on the nodes ai the transformation FK is invertible, we introduce a decomposition of FK into an ane-linear and a reduced nonlinear part to simplify the following considerations. Girault [2] employed this decomposition in the case of bilinear quadrilateral nite elements. Assume that we are given the nodes a^i , i = 1; : : : ; N , of the reference element K^ and ai , i = 1; : : : ; N , of the global element K . Then we can de ne an ane-linear transformation FT , which maps the nsimplex T^ spanned by n + 1 vertices a^i , i 2 IT f1; : : : ; N g onto the n-simplex T spanned by the vertices ai , i 2 IT . Of course the indices i 2 IT should be chosen suitably such that measn(T^) > 0. (In the case n = 3, this means, that the four nodes a^i , i 2 IT , must not lie in one plane.) If FT is invertible, the nodes a~i := FT?1 (ai ), i = 1; : : : ; N , can be used to de ne a nonlinear mapping FK~ : K^ ! Rn by replacing ai by a~i in (2.5). The following lemma shows, that the composition FT FK~ equals FK , such that we can compute the Jacobian determinant det DFK (^x) = det DFT det DFK~ (^x) by separate computations of det DFT and det DFK~ (^x).
Lemma 2.1 With the de nitions above and under the additional condition 1 2 P^ it holds
?F F (^x) = F (^x) K T K~
for all x^ 2 Rn :
(2.6)
Proof First we note P that the basis functions '^i of P^ form a parP N N tition of unity, i.e., i=1 '^i 1, because i=1 '^i (^aj ) = 1 for all j 2 f1; : : : ; N g and ^ is P^ -unisolvent. Furthermore the ane-linear mapping FT can be written as FT (x) = BT x + bT for all x 2 Rn , where BT 2 Rn;n and bT 2 Rn . Therefore we can conclude
! X N N ?F F (^x) = F X a~i '^i (^x) = (BT a~i ) '^i (^x) + bT T T K~ i=1 i=1 N N N X X X =
=
i=1
N X i=1
(BT a~i ) '^i (^x) +
FT (~ai) '^i (^x) =
i=1
N X i=1
'^i (^x) bT =
i=1
(BT a~i + bT ) '^i (^x)
ai '^i (^x) = FK (^x) ;
where x^ 2 Rn can be chosen arbitrarily. ut
Invertibility of the isoparametric mapping
5
3 Pyramidal nite elements The following de nition of the pyramidal reference element has been proposed by Wieners [4]. In contrast to Zgainski et al. [5] the ansatz space consists of piecewise multi-linear polynomials.
^ P^ ; ^ 3.1 The pyramidal reference element K;
The reference element is given by K^ = conv fa^i ; i = 1; : : : ; 5g, which is spanned by the vertices 001 001 011 011 001 a^1 = @ 0 A ; a^2 = @ 0 A ; a^3 = @ 1 A ; a^4 = @ 1 A ; a^5 = @ 0 A : 1 0 0 0 0 Considering Lagrange nite elements the degrees of freedom are given by ^ = fa^i ; i = 1; : : : ; 5g. The ansatz space P^ is given in terms of its basis functions '^i , which ful ll the condition '^i (^aj ) = ij :
'^1 := 1 ? ? ? + + '^2 :=
? ?
'^3 := '^4 := '^5 :=
; >
; ;>
;
; ; ;> + ; ; ; > ? ? ; ; :
(3.1)
Obviously we have P 1 P^ , where P 1 is the set of linear polynomials in (; ; )T 2 R3 , and the basis functions '^i , i = 1; : : : ; 5, are not continuously di erentiable. The following lemma, which can be found in a similar form in [4], gives the reason for this astonishing fact.
Lemma 3.1 There exists no continuously di erentiable ansatz function for the reference pyramid K^ , which is linear along the edges. Proof We consider the basis function ', that satis es the conditions '(^ai ) = i1 for i = 1; : : : ; 5. By assumption ' is linear along the edges of K^ , such that we have 'j[^a1 ;a^5 ] = 1 ? and 'j[^ai ;a^5 ] = 0 for i = 2; 3; 4, where [^ai ; a^j ] denotes the edge with vertices a^i and a^j .
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P. Knabner and G. Summ
If we assume ' to be continuously di erentiable, we can conclude r'(^a5 )T (^a5 ? a^i ) = 0 for i = 2; 3; 4 (3.2) and r'(^a5 )T (^a5 ? a^1 ) = 1 : (3.3) But as a^1 = a^2 ? a^3 + a^4 we get from (3.2) r'(^a5 )T (^a5 ? a^1 ) = r'(^a5 )T (^a5 ? (^a2 ? a^3 + a^4 )) = r'(^a5 )T (^a5 ? a^2 ) ? r'(^a5 )T (^a5 ? a^3) + r'(^a5 )T (^a5 ? a^4 ) = 0 ; which is a contradiction to (3.3). ut At rst sight this lack of regularity leads to problems when applying the common convergence proof for the nite element method. The back-transformation to the global element requires an upper bound for the norm of the second derivative kD2 FK kL2 (see [1], Theorem 37.1), which is not de ned at any point x^ = (; ; )T with = and 6= 0. However, we can circumvent this diculty if we do not consider the whole pyramidal element K = FK (K^ ), but split it into two \tetrahedrons" Ti := FK (T^i ), i = 1; 2, where we de ne T^1 := conv fa^1 ; a^2 ; a^3 ; a^5 g and T^2 := conv fa^1 ; a^3 ; a^4 ; a^5 g. This end we introduce some notation, which we adopt from [1]. For Rn and a function v : ! R we denote by kvk0; the L2 ( ){norm, and by kvk0;1; the L1 ( ){norm of v on . For each integer m 0 let H m ( ) be the Sobolev{space of those functions v 2 L2 ( ), for which the partial derivatives up to order m (in the distributional sense) belong to L2 ( ). We equip H m ( ) with the usual norm k km; and seminorms j jk; for k m. Furthermore we denote by Lk (Rn ; Rn ) the space of all k-linear mappings A : (Rn )k ! Rn , that is equipped with the operator norm kAkLk := sup fkA(v1 ; : : : ; vn)kRn : kvikRn 1; i = 1; : : : ; kg, where k kRn is an arbitrary norm on Rn . Note that DFK (^x) 2 L1(R3 ; R3 ) and D2 FK (^x) 2 L2 (R3 ; R3 ). Finally we introduce the interpolation operator K : Given a nite element (K; P; ) and a function v : K ! Rn suciently smooth, such that i (v), i = 1; : : :P ; N , are well de ned, the P -interpolant K v is de ned by K v := Ni=1 i (v)'i , where 'i, i = 1; : : : ; N , are the basis functions of P . We make the following assumptions on the isoparametric mapping FK :
(3.4) kdet DFK k0;1;K^ det DFK?1 0;1;K C1 ;
Invertibility of the isoparametric mapping
kDF k
C2 h ;
2 K L
0;1;K^
D FK L 0;1;T^ C3 h2 ;
?1
DFK L 0;1;K C4 h?1 ; 1
2
1
i
7
(3.5) (3.6) (3.7)
where (3.6) holds for i = 1; 2. Here Ci , i = 1; : : : ; 4, are constants independent of the mesh size parameter h > 0 (typically the diameter of K ), which is supposed to tend to zero, when the mesh is re ned. Note that (3.4) is equivalent to the invertibility of FK , and (3.5){(3.7) require additional restrictions on the position of the nodes ai , which are not discussed here. For the case of quadratic ansatz functions on n-simplices Ciarlet ([1], Theorem 37.2) gives sucient conditions for the validity of (3.5){(3.7). Theorem 3.1 Assume that the isoparametric mapping FK satis es the conditions (3.4){(3.7), the nite element (K; P; ) is de ned by ^ P; ^ ^ ), and let (2.2){(2.4) using the pyramidal reference element (K; v 2 H 2 (K ). Then the interpolation error on K satis es for suciently small h( C4 ) kv ? K vk1;K Ch (jvj1;K + jvj2;K ) : (3.8) Proof First we note that K v is well de ned, because H 2 (K ) is continuously embedded in C 0 (K ). In addition we have K v 2 H 1 (K ), since for i = 1; 2 we have K vjTi 2 C 1(Ti ) and K v 2 C 0 (K ). Therefore kv ? K vk1;K is well de ned and we can split kv ? K vk21;K = kv ? K vk21;T1 + kv ? K vk21;T2 : (3.9) An application of the transformation formula ([1], Theorem 37.1) and of (3.7) yields for i = 1; 2
kv ? K vk21;Ti kdet DFK k0;1;T^i C42 h?2 v^ ? K^ v^ 21;T^i ; (3.10) where we have used v^ := v FK 2 H 2 (T^i ), and 1 C42 h?2 . The last term can be estimated by the Bramble{Hilbert lemma ([1], Theorem 28.1), to get
v^ ? v^
2 C 2 jv^j2 : (3.11) 5 2;T^i K^ 1;T^i The back-transformation to the global element yields by another application of the transformation formula and of (3.5) { (3.6)
jv^j22;T^i det DFK?1 0;1;Ti max C24; C32 h4 (jvj1;Ti + jvj2;Ti )2 : (3.12) Since the combination of (3.10) { (3.12) gives (3.8) where K is substituted by Ti , joining the contributions of T1 and T2 according to (3.9) yields the assertion. ut
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P. Knabner and G. Summ
3.2 Conditions for the invertibility of the isoparametric mapping
The following considerations address the question, under which conditions on the nodes ai , i = 1; : : : ; 5, the isoparametric mapping FK de ned by (2.5) using the basis functions '^i given in (3.1) is invertible on K^ . Assume that we are given ve nodes ai , i = 1; : : : ; 5, (the vertices of the global element K ). We will use the decomposition (2.6) of FK into an ane-linear and a nonlinear part, that was introduced at the end of Section 2. The index set IT is chosen to be IT = f1; 2; 4; 5g, such that T^ = conv fa^1 ; a^2 ; a^4 ; a^5 g is the unit tetrahedron, for which measn (T^) = 1=6. Hence FT is given by FT (^x) = DFK (^a1 )^x + a1 , such that FT is not invertible if and only if det DFK (^a1 ) = 0. Since DFK (^a1 ) = (a2 ? a1 ; a4 ? a1 ; a5 ? a1 ) the vertices ai, i 2 IT , must not lie in one and the same plane to ensure that FT is invertible. If the above condition on the invertibility of FT is ful lled, we can compute a~3 = FT?1 (a3 ), and de ne the nonlinear transformation FK~ using a~i = a^i for i 2 IT , to get
01 01 ; > @ A @ A FK~ = + (~a3 ? a^3 ) + ; :
(3.13)
Using the notation a~3 = (~x3 ; y~3 ; z~3 )T a straightforward computation
yields
01 + y~ ; > 3 det DFK~ @ A = 1 ? ? ? + y~3 + x~3 + zz~~3 + x~ ; :
3 3 Thus det DFK~ is a linear polynomial in each of the two tetrahedra T^1 = conv fa^1 ; a^2 ; a^3 ; a^5 g and T^2 = conv fa^1; a^3 ; a^4 ; a^5 g and therefore attains its extremum values in K^ at one of the vertices. The values at the vertices a^i , i = 1; : : : ; 4 are given by 8 9 1 ; i = 1 > > < = y ~ ; i = 2 3 det DFK~ (^ai ) = > x~ + y~ + z~ ? 1 ; i = 3 > ; : 3 3x~3 3 ; i = 4 ; whereas det DFK~ (^a5 ) is not de ned uniquely, as FK~ is not continuously di erentiable at any point x^ = (; ; )T with = and 6= 0.
Thus we have to consider the limits det DFK~ (^x) = x~3 : lim det DFK~ (^x) = y~3 and x^lim ! a^ 5 x^!a^ 5
Invertibility of the isoparametric mapping
9
Combining these results together yields the following theorem. Theorem 3.2 Let FK~ be the isoparametric transformation for pyramidal elements de ned by (3.13). Then the following equivalence holds 8 9 x~3 > 0 < = y~3 > 0 det DFK~ (^x) > 0 for all x^ 2 K^ , : ^ : (3.14) ^ x~3 + y~3 + z~3 > 1 ;
4 Prismatic nite elements The following de nition of the reference element for prismatic nite elements is standard and can be found in many textbooks, e.g. [1], Section 12.
^ P; ^ ^ 4.1 The prismatic reference element K;
In the prismatic case we have K^ = conv fa^i ; i = 1; : : : ; 6g, where the vertices are given by 001 011 001 a^1 = @ 0 A ; a^2 = @ 0 A ; a^3 = @ 1 A ; 0 0 0 001 011 001 a^4 = @ 0 A ; a^5 = @ 0 A ; a^6 = @ 1 A : 1 1 1 Again, the degrees of freedom are given by ^ = fa^i ; i = 1; : : : ; 6g, and the ansatz space P^ is given in terms of the following basis functions '^i , which meet the conditions '^i (^aj ) = ij : '^1 := 1 ? ? ? + + ; '^2 := ? ; '^3 := ? ; (4.1) '^4 := ? ? ; '^5 := ; '^6 := : Obviously the ansatz space P^ spanned by these basis functions ful lls P 1 P^ .
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P. Knabner and G. Summ
4.2 Conditions for the invertibility of the isoparametric mapping
The question, under which conditions on the nodes ai , i = 1; : : : ; 6, the isoparametric mapping FK de ned by (2.5) using the basis functions '^i given in (4.1) is invertible on K^ , is a little bit more dicult to answer than in the pyramidal case. Using the decomposition (2.6) of FK into an ane-linear and a nonlinear part, that was introduced at the end of Section 2, we can reduce this problem mainly to the case, that only two vertices a~i di er from a^i . Here we choose IT = f1; 2; 3; 4g, such that again T^ = conv fa^1 ; a^2 ; a^3; a^4 g is the unit tetrahedron and the ane-linear transformation FT is given by FT (^x) = DFK (^a1 )^x + a1 , where DFK (^a1 ) = (a2 ? a1; a3 ? a1 ; a4 ? a1 ). If the vertices ai , i 2 IT , do not lie in one and the same plane, FT is invertible and we can compute a~i = FT?1 (ai ) for i = 5; 6. Using a~i = a^i for i 2 IT the nonlinear transformation FK~ is given by
01 01 FK~ @ A = @ A + (~a5 ? a^5) + (~a6 ? a^6 ) ;
(4.2)
such that the computation of the Jacobian matrix yields
?
DFK~ (; ; )T =
a^2 + (~a5 ? a^5 ) ; a^3 + (~a6 ? a^6 ) ; a^4 + (~a5 ? a^5 ) + (~a6 ? a^6 ) 0 1 + (~x5 ? 1) 1 x~6 (~x5 ? 1) + x~6 A; y~5 1 + (~y6 ? 1) y~5 + (~y6 ? 1) =@ (~z5 ? 1) (~z6 ? 1) 1 + (~z5 ? 1) + (~z6 ? 1) where we have used again a~i = (~xi ; y~i ; z~i )T for i = 5; 6. Thus the Jacobian determinant is given by
?
J (; ; ) := det DFK~ (; ; )T = 1 + (~z5 ? 1) + (~z6 ? 1) + ((~x5 ? 1) + (~y6 ? 1)) + ((~y6 ? 1)(~z5 ? 1) ? y~5 (~z6 ? 1)) (4.3) + ((~x5 ? 1)(~z6 ? 1) ? x~6 (~z5 ? 1)) + ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5 ) 2 : Since J is linear in and for xed, a substantial simpli cation applies.
Invertibility of the isoparametric mapping
11
Lemma 4.1 Let J , de ned in (4.3), be the Jacobian determinant
of the isoparametric transformation FK~ for prismatic nite elements de ned in (4.2). Then the following equivalence holds
8 J (0; 0; ) > 0 9 < = J (; ; ) > 0 8 (; ; )T 2 K^ , : ^ J (1; 0; ) > 0 ; 8 2 [0; 1] : ^ J (0; 1; ) > 0 ? ? ? Proof For (; ; )T 2 K^ we have (; )T 2 T^ = conv 0 ; 1 ; 0 2
0
0
1
(T^2 is the unit triangle). If we choose = 0 xed, J (; ; 0 ) is a linear polynomial in and and therefore attains its extremum values in ^T2 at the corners ?00; ?10; ?01. This proves the assertion. ut In the remainder of this section we study the quadratic polynomials J (0; 0; ), J (1; 0; ) and J (0; 1; ). The following lemma will be useful.
Lemma 4.2 Assume that we are given a quadratic polynomial p de ned by p( ) = c0 + c1 + c2 2 ful lling p(0) > 0 and p(1) > 0. Then there exists a root 2 (0; 1) if and only if c0 < c2 and c1 ?2pc0 c2 : Proof \)": Since p(0) and p(1) are both positive and p( ) = 0, there exists a second root 2 (0; 1) (maybe = ) such that p( ) = c2 ( ? )( ? ) = c2 2 ? c2 ( + ) + c2 : A comparison of coecients yields c0 = c2 and c1 = ?c2 ( + ). As c0 = p(0) > 0 and 2 (0; 1), we can conclude c2 = 1 > 1 ) c < c : 0 2 c0 Furthermore we have 0
p
p 2
p
p
? = ? 2 + , ? ( + ) ?2
and nally p p c1 = ?c2 ( + ) ?2c2 = ?2c2 c0 =c2 = ?2pc0 c2 : \(": The well known solution formula for quadratic equations yields
p
2 = (c ) = ?c1 + c1 ? 4c0 c2 1
2c2
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P. Knabner and G. Summ
is a root of p. As c1 ?2pc0 c2 the discriminant c21 ? 4c0 c2 is non-negative and therefore 2 R. derivative of with respect If c1 < ?2pc0 c2 , we can consider the to c1 . As c2 > c0 > 0 and c1 < ?2pc0 c2 < 0 we obtain
!
d 1 2c1 dc1 = 2c2 pc21 ? 4c0 c2 ? 1 < 0 ; and therefore depends monotonically on c1 . As p(1) = c0 + c1 + c2 > 0 we have ?(c0 + c2 ) < c1 < ?2pc0 c2 . Considering the upper and the lower bound for c1 , we get p (?(c0 + c2 )) = 1 and (?2pc0 c2 ) = c0 =c2 2 (0; 1) ; such that the monotonicity of yields the assertion. ut 4.2.1 Characterization of J (0; 0; ) > 0 From (4.3) we get by setting = = 0:
J (0; 0; ) = 1 + ((~x5 ? 1) + (~y6 ? 1)) + ((~x5 ? 1)(~y6 ? 1) ? x~6y~5 ) 2 : Seeking for conditions under which J (0; 0; ) has a root in [0; 1], we get the following result: Lemma 4.3 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric transformation FK~ for prismatic nite elements de ned in (4.2). Then there exists 2 [0; 1], such that J (0; 0; ) 0 if and only if one of the following conditions is satis ed i) x~5 y~6 x~6y~5 ; (4.4) ( ) x~5 + y~6 < x~5 y~p 6 ? x~6 y~5 ii) ^ x~5 + y~6 ? 2 ?2 (~x5 ? 1)(~y6 ? 1) ? x~6 y~5 : (4.5) Proof As J (0; 0; 0)=1 we can conclude from the continuity of J (0; 0; )
9 2 [0; 1] : J (0; 0; ) 0 , 9 2 [0; 1] : J (0; 0; ) = 0 : Obviously J (0; 0; ) has a root in [0; 1], if J (0; 0; 1) = x~5y~6 ? x~6 y~5 0 , x~5y~6 x~6y~5 : This yields (4.4). As J (0; 0; ) is a quadratic polynomial, there might be the case, that there is a minimum such that J (0; 0; ) 0, but J (0; 0; 1) > 0.
Invertibility of the isoparametric mapping
13
Then we can apply Lemma 4.2 with c0 = 1, c1 = x~5 + y~6 ? 2 and c2 = (~x5 ? 1)(~y6 ? 1) ? x~6 y~5 to conclude that in this case we have c0 < c2 , x~5 + y~6 < x~5 y~6 ? x~6y~5 ; p p c1 ?2 c0 c2 , x~5 + y~6 ? 2 ?2 (~x5 ? 1)(~y6 ? 1) ? x~6 y~5 ; which results in condition (4.5). ut 4.2.2 Characterization of J (1; 0; ) > 0 From (4.3) we get by setting = 1 and = 0: J (1; 0; ) = z~5 + (~x5 ? 1 + z~5(~y6 ? 1) ? y~5 (~z6 ? 1)) + ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5 ) 2 : Lemma 4.4 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric transformation FK~ for prismatic nite elements de ned in (4.2). Then there exists 2 [0; 1], such that J (1; 0; ) 0 if and only if one of the following conditions is satis ed i) z~5 0 ; (4.6) ii) y~6 (~x5 + z~5 ? 1) y~5 (~x6 + z~6 ? 1) ; (4.7) 8 9 z~5 < (~x5 ? 1)(~y6 ? 1) ? x~6 y~5 > > < = iii) : (4.8) ^ x ~ ? 1 + z ~ (~ y ? 1) ? y ~ (~ z ? 1) 5 5 6 5 6 > > p : ; ?2 z~5 ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5) Proof Condition (4.6) results directly from J (1; 0; 0) = z~5 . If J (1; 0; 0) = z~5 > 0 we can again conclude by continuity that 9 2 [0; 1] : J (1; 0; ) 0 , 9 2 [0; 1] : J (1; 0; ) = 0 : Therefore we look for conditions under which J (1; 0; ) has a root in [0; 1]. Obviously there exists such a root, if J (1; 0; 1) 0 , y~6(~x5 + z~5 ? 1) y~5(~x6 + z~6 ? 1) ; which is exactly (4.7). Finally we turn to the case that J (1; 0; ) has two roots (maybe one double root) in (0; 1). Applying Lemma 4.2 using c0 = z~5 , c1 = x~5 ? 1 + z~5 (~y6 ? 1) ? y~5 (~z6 ? 1) and c2 = (~x5 ? 1)(~y6 ? 1) ? x~6y~5 we obtain in this case c0 < c 2 , z~5 < (~x5 ? 1)(~y6 ? 1) ? x~6y~5 ; p y6 ? 1) ? y~5 (~z6 ? 1) c1 ?2 c0 c2 , x~5 ? 1 + z~p 5 (~ ?2 z~5 ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5) ; which results in (4.8). ut
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4.2.3 Characterization of J (0; 1; ) > 0 Lemma 4.5 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric transformation FK~ for prismatic nite elements de ned in (4.2). Then there exists 2 [0; 1], such that J (0; 1; ) 0 if and only if one of the following conditions is satis ed i) z~6 0 ; (4.9) ii) x~5 (~y6 + z~6 ? 1) x~6 (~y5 + z~5 ? 1) ; (4.10) 8 9 z~6 < (~x5 ? 1)(~y6 ? 1) ? x~6 y~5 > > < = iii) : (4.11) ^ y ~ ? 1 + z ~ (~ x ? 1) ? x ~ (~ z ? 1) 6 6 5 6 5 > > p : ; ?2 z~6 ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5) Proof Since J (0; 1; ) is given by J (0; 1; ) = z~6 + (~y6 ? 1 + z~6 (~x5 ? 1) ? x~6 (~z5 ? 1)) + ((~x5 ? 1)(~y6 ? 1) ? x~6 y~5 ) 2 ; interchanging the roles of z~5 and z~6 , x~5 and y~6 , and y~5 and x~6 respectively in the proof of Lemma 4.4 yields the desired result. ut Summarizing these results yields Theorem 4.1 Let FK~ be the isoparametric transformation for prismatic elements de ned in (4.2). Then det DFK~ (^x) > 0 for all x^ 2 K^ if and only if no one of the conditions (4.4) { (4.11) is satis ed. Proof Due to Lemma 4.1 it suces to check the positivity of J = det DFK~ on the edges [^ai ; a^i+3 ], i = 1; 2; 3. Thus the negation of the conditions obtained in Lemmas 4.3 { 4.5 yields the assertion. ut 4.3 Geometric interpretation ? The determinant of a 3 3-matrix M = m(1) ; m(2) ; m(3) can be represented as triple scalar product of its columns m(i) : E D det(M ) = m(1) ; m(2) ; m(3) := m(1) m(2) m(3) ; and its absolute value jdet M j can be interpreted as the volume of the parallelepiped spanned by the vectors m(i) . Therefore we can compute the Jacobian determinant of the isoparametric mapping FK~ de ned in (4.2) as triple scalar product, and obtain ? (4.12) J (; ; ) = DFK~ (; ; )T = D E a^2 + (~a5 ? a^5 ) ; a^3 + (~a6 ? a^6) ; a^4 + (~a5 ? a^5 ) + (~a6 ? a^6 ) :
Invertibility of the isoparametric mapping
15
4.3.1 Characterization of J (0; 0; ) = 0 From (4.12) we get by setting = = 0
J (0; 0; ) = h(^a2 + (~a5 ? a^5 ) ) ; (^a3 + (~a6 ? a^6) ) ; a^4i = hp2 ( ) ? p1 ( ) ; p3 ( ) ? p1 ( ) ; p1 (1 + ) ? p1 ( )i ;
?
?
where we have used p1( ) := FK~ (0; 0; )T , p2 ( ) := FK~ (1; 0; )T ? T and p3 ( ) := FK~ (0; 1; ) . Seeking for roots of J (0; 0; ) we can conclude that J (0; 0; ) = 0, if and only if p1 ( ), p2 ( ), p3 ( ) and p1 (1 + ) lie in one and the same plane. Thus we have
J (0; 0; ) = 0 , f1 ( ) k g1 _ f1( ) \ g1 6= ; ; where f1 ( ) denotes the straight line passing through p2 ( ) and p3 ( ), and g1 the straight line passing through p1 ( ) and p1 (1 + ), (resp. passing through a^1 and a^4 ). As f1 ( ) and g1 can be represented by
8 0 1 9 < = f1 ( ) = :FK~ @ 1 ? A ; 2 R;
8 001 9 < = and g1 = :FK~ @ 0 A ; 2 R; ;
we can formulate this in the following lemma: Lemma 4.6 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric mapping FK~ de ned in (4.2). Then J (0; 0; ) = 0 if and only if one of the following conditions is satis ed
011 001 i) 9 2 R : FK~ @ 0 A ? FK~ @ 1 A = (^a4 ? a^1 ) ; 001 0 1 ii) 9 ; 2 R : FK~ @ 1 ? A = FK~ @ 0 A :
By de nition, the edge [^a1 ; a^4 ] of K~ is a part of the straight line face opposite to [^a1 ; a^4 ], which can be represented by ?(;the g1F, and T ; ; 2 [0; 1] , is a subset of the bundle of straight 1 ? ; ) ~ K lines ff1 ( ) ; 2 [0; 1]g. In this sense Lemma 4.6 yields relations between an edge of K~ and the opposite face. Examples for prismatic elements, where one edge of K~ and the opposite face intersect, are given in section 5.2, see Fig. 5.2 and Fig. 5.3.
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4.3.2 Characterization of J (1; 0; ) = 0 From (4.12) we get by setting = 1 and = 0
J (1; 0; ) = h(^a2 + (~a5 ? a^5 ) ) ; (^a3 + (~a6 ? a^6 ) ) ; (^a4 + (~a5 ? a^5))i = h? (p1 ( ) ? p2 ( )) ; p4 ( ) ? p2 ( ) ; p2 (1 + ) ? p2 ( )i ;
?
where we have used p4 ( ) := FK~ (1; 1; )T in addition to the above de ned points p1 ( ) and p2 ( ). Again, we de ne straight lines h( ) passing through p1 ( ) and p4 ( ), and g2 passing through p2 ( ) and p2 (1 + ) (resp. passing through a^2 and a~5 ), and obtain
J (1; 0; ) = 0 , h( ) k g2 _ h( ) \ g2 6= ; : We note that h( ) and g2 can be represented by
8 01 9 < = h( ) = :FK~ @ A ; 2 R;
8 011 9 < = and g2 = :FK~ @ 0 A ; 2 R; :
As we will see below, we can substitute h( ) by f2 ( ), the straight line passing through p1 ( ) and p3 ( ), which can be represented by
8 001 9 < = f2 ( ) = :FK~ @ A ; 2 R; :
This way we get analogously to section 4.3.1 a relation between the edge [^a2 ; a~5 ] of K~ and the face opposite to this edge.
Lemma 4.7 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric mapping FK~ de ned in (4.2). Then J (1; 0; ) = 0 if and only if one of the following conditions is satis ed
001 001 i) 9 2 R : FK~ @ 1 A ? FK~ @ 0 A = (~a5 ? a^2 ) ; 011 001 ii) 9 ; 2 R : FK~ @ A = FK~ @ 0 A :
Invertibility of the isoparametric mapping
17
Proof The above considerations yield (cf. the derivation of Lemma 4.6), that J (1; 0; ) = 0 if and only if
001 011 9 2 R : FK~ @ 1 A ? FK~ @ 0 A = (~a5 ? a^2 ) ; 01 011 _ 9 ; 2 R : FK~ @ A = FK~ @ 0 A :
The proof of the equivalence of these relations to the assertion of the lemma consists of three parts, which can be shown easily.
01 011 1: 9 2 R n f1g; 2 R : FK~ @ A = FK~ @ 0 A 001 011 , 9 2 R n f?1g; 2 R : FK~ @ A = FK~ @ 0 A ;
, = + . and = where we have = 1 ? , = 1?? 1+ 1+
011 011 2: 9 2R : FK~ @ 1 A = FK~ @ 0 A 001 001 , 9 2 R : FK~ @ 1 A ? FK~ @ 0 A = (~a5 ? a^2 ) ;
where we have = ? .
011 001 3: 9 2 R : FK~ @ 1 A ? FK~ @ 0 A = (~a5 ? a^2) 001 011 , 9 2R : FK~ @ ?1 A = FK~ @ 0 A ;
where we have = ? . ut
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4.3.3 Characterization of J (0; 1; ) = 0 From (4.12) we get by setting = 0 and = 1 J (0; 1; ) = h(^a2 + (~a5 ? a^5 ) ) ; (^a3 + (~a6 ? a^6 ) ) ; (^a4 + (~a6 ? a^6))i = hp4 ( ) ? p3 ( ) ; ? (p1 ( ) ? p3 ( )) ; p3 (1 + ) ? p3 ( )i : Again, we consider the straight lines h( ) passing through p1 ( ) and p4 ( ), and g3 passing through p3 ( ) and p3 (1 + ) (resp. passing through a^3 and a~6 ), and obtain J (0; 1; ) = 0 , h( ) k g3 _ h( ) \ g3 6= ; : Again we can substitute h( ), in this case by f3 ( ), the straight line passing through p1 ( ) and p2 ( ). We note that f3 ( ) and g3 can be represented by 8 01 9 8 001 9 < = < = f3 ( ) = :FK~ @ 0 A ; 2 R; and g3 = :FK~ @ 1 A ; 2 R; :
Hence we obtain in analogy to Lemma 4.7: Lemma 4.8 Let J , de ned in (4.3), be the Jacobian determinant of the isoparametric mapping FK~ de ned in (4.2). Then J (0; 1; ) = 0 if and only if one of the following conditions is satis ed 001 011 i) 9 2 R : FK~ @ 0 A ? FK~ @ 0 A = (~a6 ? a^3 ) ;
ii) 9 ; 2 R :
01 001 FK~ @ 0 A = FK~ @ 1 A :
5 Examples The Theorems 3.2 and 4.1 summarize the conditions, which ensure the positivity of the Jacobian determinant of the simpli ed isoparametric transformation FK~ . In combination with a test for the invertibility of the ane-linear mapping FT , an algorithm for grid generation can use these conditions to check automatically the invertibility of the isoparametric transformation FK to a global element K , whose nodes ai are given. First one has to check the invertibility of the ane-linear mapping FT , which is of the form FT (x) = BT x + bT for x 2 Rn , where BT 2
Invertibility of the isoparametric mapping
19
Rn;n and bT 2 Rn , by computing the Jacobian determinant det BT . If det BT does not vanish, FT is invertible and one can compute a~i = F ?1 (ai) = B ?1 (ai ? bT ) for i 2= IT . Since in the cases considered here T
T
n = 3, the computation of the Jacobian determinant det BT and the inversion of BT can be done explicitly, and are not computationally too expensive. Finally the conditions in (3.14) and (4.4) { (4.11), respectively, can by checked easily. We illustrate this approach with some examples. 5.1 Veri cation of the invertibility of the isoparametric transformation for pyramidal nite elements
Assume that the nodes ai , i = 1; : : : ; 5, of the global pyramidal element K are given. As the index set is de ned by IT = f1; 2; 4; 5g, we have BT = (a2 ? a1 ; a4 ? a1 ; a5 ? a1 ) and bT = a1 . If FT is invertible, we can compute a~3 = FT?1 (a3 ) = BT?1 (a3 ? a1 ) and check nally, if the conditions in (3.14) are ful lled. Example 5.1 For simplicity we assume a1 = (0; 0; 0)T (this can be achieved by a simple translation and therefore is no restriction). Furthermore let a2 = h(1; 0; 1)T , a3 = h(3; 1; 3)T , a4 = h(2; 1; 0)T and a5 = h(0; 0; 2)T , where h > 0 is the mesh size parameter. We immediately obtain 01 2 01 BT = h @ 0 1 0 A : 102 As det BT = 2h3 > 0, BT is invertible and a short computation yields a~3 = (1; 1; 1)T , such that obviously (3.14) is satis ed . However, if we have a3 = h(0; ?1; 4)T , while the other nodes ai , i = 1; 2; 4; 5, remain unchanged, we get a~3 = (2; ?1; 1)T , such that (3.14) is violated. Fig. 5.1 shows K~ for this case. It is obvious that the exterior angle between the faces convfa1 ; a2 ; a5 g and convfa2 ; a3 ; a5 g is less than 180 . 5.2 Veri cation of the invertibility of the isoparametric transformation for prismatic nite elements
If the nodes ai , i = 1; : : : ; 6, of a prismatic nite element K are given, we can verify the invertibility of the corresponding isoparametric transformation the same way as in the case of a pyramidal element considered above. First, we check the invertibility of the
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a5
a3 a4
Fig. 5.1.
a1
a2
A pyramidal element that violates condition (3.14)
ane-linear mapping FT by computing the Jacobian determinant of BT = (a2 ? a1 ; a3 ? a1 ; a4 ? a1 ) (here the index set IT is de ned by IT = f1; 2; 3; 4g, cf. Section 4). If det BT 6= 0, we can compute a~5 = FT?1 (a5 ) and a~6 = FT?1 (a6 ) and nally check the conditions (4.4) { (4.11). If no one of these conditions is satis ed, the isoparametric transformation FK is globally invertible on K^ . Example 5.2 As the veri cation of the invertibility of FT does not di er from the pyramidal case, we restrict the following example to the consideration of the nonlinear part FK~ by assuming ai = a^i for i = 1; 2; 3; 4, such that FT = id and therefore a~i = ai for i = 5; 6. If a5 = (1; 1; z5 )T and a6 = (2; 1; z6 )T , (4.4) is satis ed, no matter of the values of z5 and z6 . Choosing z5 = z6 = 1 the conditions (4.7) and (4.10) coincide with (4.4), and accordingly (4.8) and (4.11) coincide with (4.5). Hence (4.7) and (4.10) are ful lled, too. Fig. 5.2 shows the resulting element K for z5 = z6 = 1. Apparently, the straight line passing through a2 and a5 penetrates the face with the vertices a1 , a3 , a4 and a6 , which corresponds to the second condition in Lemma 4.7. A straightforward computation shows, thatpthis conp dition is satis ed for = = = 1=2, and that = 1=2 is a root of J (; ; ) = 1 ? 2 .
Invertibility of the isoparametric mapping
21
a5
a6
a4
a3
a1
Fig. 5.2.
a2
A prismatic element that satis es condition (4.4)
From an inspection of Fig. 5.2 and, more mathematically, from condition (4.4) we can conclude easily, that FK would be invertible, if a5 and a6 were switched. This remedy does not work, if we consider a5 = (1; ?1; 1)T and a6 = (?2; 1; 1)T . Again, (4.4) (and accordingly (4.7) and (4.10)) is satis ed in this case, but if we switch a5 and a6 , (4.5) is satis ed instead of (4.4). Fig. 5.3 shows the resulting element K for a5 = (?2; 1; 1)T and a6 = (1; ?1; 1)T . The kind of degeneration, when (4.6) or (4.9) are ful lled, is evident. The validity of (4.7) or (4.8) and accordingly of (4.10) or (4.11) lead to similar shapes of the global prismatic element K as in Fig. 5.2 and Fig. 5.3.
6 Conclusion We considered the isoparametric mapping for multi-linear pyramidal and prismatic nite elements, and developed conditions for the position of the vertices which ensure, that the isoparametric mapping is bijective. To our knowledge, for trilinear hexahedral nite elements this problem is unsolved up to now. Maybe the techniques used in this paper can be used also to develop conditions for the invertibility of the hexahedral isoparametric mapping.
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a5 a4
a6
a3
a1 a2
Fig. 5.3.
A prismatic element that satis es condition (4.5)
References 1. P.G. Ciarlet: Basic Error Estimates for Elliptic Problems, in: P.G. Ciarlet and J.L. Lions, eds., Handbook of numerical analysis, Vol. II: Finite Element Methods (Part 1), North{Holland, Amsterdam, 1991. 2. V. Girault: A local projection operator for quadrilateral nite elements, Math. Comput. 64, 212 (1995) 1421{1431. 3. C. Tapp: Anisotrope Gitter | Generierung und Verfeinerung (in German), Ph.D. Thesis, University of Erlangen, 1999, to appear. 4. C. Wieners: Conforming discretizations on tetrahedrons, pyramids, prisms and hexahedrons, Preprint, University of Stuttgart. 5. F.-X. Zgainski et al.: A New Family of Finite Elements: The Pyramidal Elements, IEEE Transactions on Magnetics 32, 3 (1996) 1393{1396.
