Part I, Chapter 2
One dimension and tensor products
This chapter presents important examples of finite elements, first in one space
dimension, then in multiple space dimensions using tensor product techniques.
Important computational issues related to the manipulation of high-order
polynomial bases are addressed.
2.1 Legendre and Jacobi polynomials
This section reviews some basic properties of the Legendre and Jacobi polynomials which are often used in the design of high-order finite elements.
Definition 2.1 (Legendre polynomials). The Legendre polynomials are
univariate polynomial functions R → R defined by
L⊥
m (t) =
(−1)m dm
((1 − t)m (1 + t)m ),
2m m! dtm
(2.1)
for all t ∈ R and all m ≥ 0.
The superscript ⊥ refers to the following important orthogonality property.
Proposition 2.2 (L2 -orthogonality). The Legendre polynomials are L2 orthogonal over the interval (−1, 1), and the following holds:
Z 1
2
⊥
L⊥
δmn ,
∀m, n ≥ 0.
(2.2)
m (t)Ln (t) dt =
2m
+1
−1
⊓
⊔
Proof. Use induction and integration by parts.
Legendre polynomials satisfy many other properties. We now list some of
the most important ones. The following Bonnet’s recursion formula holds:
⊥
⊥
(m + 1)L⊥
m+1 (t) = (2m + 1)tLm (t) − mLm−1 (t),
∀m ≥ 1.
♥♥(JLG)
revised
again Feb 9, 2016,
22:33
16
Chapter 2. One dimension and tensor products
⊥
m
⊥
L⊥
m is a polynomial of degree m and Lm (−1) = (−1) , Lm (1) = 1. The m
⊥
roots of Lm are in (−1, 1) and are called Gauss–Legendre nodes; they play an
important role in the design of quadratures; see Chapter 56. The first four
Legendre polynomials are
L⊥
0 (t) = 1,
L⊥
1 (t) = t,
2
1
L⊥
2 (t) = 2 (3t − 1),
3
1
L⊥
3 (t) = 2 (5t − 3t),
and their graphs are shown in Figure 2.1.
1
0.5
0
-0.5
-1
-1
-0.5
0
0.5
1
Fig. 2.1. Legendre polynomials of degree at most 3 in the interval [−1, 1].
Definition 2.3 (Jacobi polynomials). Let α, β ∈ R be such that α > −1
and β > −1. The Jacobi polynomials are univariate polynomial functions
R → R defined by
m
α,β
−α
dm
Jm
(t) = (−1)
(1 + t)−β dt
(1 − t)α+m (1 + t)β+m ,
(2.3)
m
2m m! (1 − t)
for all t ∈ R and all m ≥ 0.
The Jacobi polynomials satisfy the following L2 -orthogonality property on the
interval (−1, 1):
Z 1
α,β
(1 − t)α (1 + t)β Jm
(t)Jnα,β (t) dt = cm,α,β δmn ,
∀m, n ≥ 0,
(2.4)
−1
α+β+1
Γ (m+α+1)Γ (m+β+1)
2
, where Γ is the Gamma function
with cm,α,β = 2m+α+β+1
m!Γ (m+α+β+1)
(recall that Γ (n + 1) = n! for any natural number n). The Jacobi polynomials
satisfy the following recursion formula for all m ≥ 1:
α,β
2(m + 1)(m + α + β + 1)(2m + α + β)Jm+1
(t) =
α,β
(2m + α + β + 1)((2m + α + β + 2)(2m + α + β)t + α2 − β 2 )Jm
(t)
α,β
− 2(m + α)(m + β)(2m + α + β + 2)Jm−1
(t).
Part I. Getting Started
17
α,β
α,β
α,β
Jm
is a polynomial of degree m and Jm
(−1) = (−1)m m+β
m , Jm (1) =
m+α
m . The Legendre polynomials are Jacobi polynomials with parameters
0,0
α = β = 0, i.e., L⊥
m (t) = Jm (t) for all m ≥ 0. The first three Jacobi polynomials corresponding to the parameters α = β = 1 are
J01,1 (t) = 1,
J11,1 (t) = 2t,
J21,1 (t) = 34 (5t2 − 1)
1,1
are related to the integrated Legendre polynoThe Jacobi polynomials Jm
mials as follows (see Exercise 2.4):
Z
t
−1
L⊥
m (s) ds = −
1
1,1
(1 − t2 )Jm−1
(t),
2m
∀m ≥ 1.
(2.5)
We refer to Abramowitz and Stegun [1, Chap. 22] for further results.
2.2 One-dimensional finite elements
In this section, we present important examples of one-dimensional finite elements. Recall that Pk , k ≥ 0, is the real vector space composed of univariate
polynomial functions of degree at most k. For convenience, degrees of freedom
and shape functions of one-dimensional finite elements using the polynomial
space Pk are numbered from 0 to k.
2.2.1 Lagrange (nodal) finite elements
Following Definition 1.8, the degrees of freedom for Lagrange finite elements
are chosen as the values at some set of nodes.
Proposition 2.4 (Lagrange finite element). Let k ≥ 0. Let K be a compact interval with non-empty interior, and let P = Pk . Consider a set of
nsh = k + 1 distinct nodes {al }l∈{0: k} in K. Let Σ = {σl }l∈{0: k} be the linear
forms on P such that σl (p) = p(al ) for all l ∈ {0: k}. Then, (K, P, Σ) is a
Lagrange finite element.
Proof. We verify unisolvence. We observe that dim P = k + 1 = nsh = card Σ
and that a polynomial p ∈ P such that σl (p) = p(al ) = 0 for all l ∈ {0: k} is
of degree at most k and has (k + 1) distinct roots. Hence, p ≡ 0.
⊓
⊔
The shape functions of a one-dimensional Lagrange finite element are the
[a]
Lagrange interpolation polynomials {Ll }i∈{0: k} defined as in (1.2). Following (1.10), the Lagrange interpolation operator acts as follows:
X
[a]
L
IK
(v)(t) =
v(al )Ll (t),
∀t ∈ K,
(2.6)
l∈{0: k}
18
Chapter 2. One dimension and tensor products
L
and possible choices for the domain of IK
are V (K) = C 0 (K) or V (K) =
1,1
W (K), see Exercise 1.7.
The Lagrange interpolation polynomials based on equidistant nodes (including both endpoints) in the interval K = [−1, 1] are henceforth denoted
{Lkl }l∈{0: k} . The graphs of these polynomials are illustrated in Figure 1.1 for
k ∈ {1, 2, 3}, and explicit expressions are as follows:
L10 (t) = 12 (1 − t),
L11 (t) = 12 (1 + t),
L20 (t) = 12 t(t − 1),
L30 (t) =
L22 (t) = 12 (t + 1)t,
L32 (t) =
L21 (t) = (t + 1)(1 − t),
L31 (t) =
L33 (t) =
9
16 (t
27
16 (t
27
16 (t
9
16 (t
+ 13 )(t − 31 )(1 − t),
+ 1)(t − 31 )(t − 1),
+ 1)(t + 31 )(1 − t),
+ 1)(t + 31 )(t − 31 ).
While the choice of equidistant nodes appears somewhat natural, it is appropriate only when working with low-degree polynomials, see §2.2.5.
2.2.2 Modal finite elements
In this section, we illustrate the construction of §1.4.2 in the one-dimensional
setting, say on the interval [−1, 1].
Proposition 2.5 (Legendre finite element). Let k ≥ 0. Let K = [−1, 1]
and let P = Pk . Let Σ = {σl }l∈{0: k} be the nsh = k + 1 linear forms on P
such that
Z
2l + 1 1 ⊥
σl (p) =
Ll (t)p(t) dt, ∀l ∈ {0: k}.
(2.7)
2
−1
Then, (K, P, Σ) is a finite element. The shape functions are the (k + 1) Legendre polynomials of degree ≤ k.
Proof. Use (2.2) and Proposition 1.9.
⊓
⊔
Following (1.13), the
interpolation
operator acts as follows:
Legendre
R
P
2l+1 1
⊥
⊥
=
L (s)v(s) ds Ll (t) for all t ∈ K and all
l∈{0: k}
2
−1 l
m
IK
(v)(t)
v ∈ V (K) = L1 (K).
2.2.3 Hybrid finite elements(♦)
Hybrid finite elements mix nodal and modal degrees of freedom. When constructing so-called H 1 -conforming approximation spaces (see Chapter 13 for a
full treatment in multiple space dimensions), it is important that all the basis
functions but one vanish at each endpoint of the interval, say K = [−1, 1].
This calls for using the values at ±1 as nodal degrees of freedom on Pk . For
k ≥ 2, some or all of the remaining degrees of freedom can be taken to be of
modal type. Taking all of them as moments against polynomials in Pk−2 leads
to the so-called canonical hybrid finite element; a multi-dimensional extension
in a simplex is presented in §3.4.
Part I. Getting Started
19
Proposition 2.6 (Canonical hybrid finite element). Let k ≥ 1. Set K =
[−1, 1] and P = Pk . Define σ0 (p) = p(−1), σk (p) = p(1),
and, if k ≥ 2,
R
let {µl }l∈{1: k−1} be a basis of Pk−2 and define σl (p) = K pµl ds for all l ∈
{1: k − 1}. Set Σ = {σl }l∈{0: k} . Then, the triple (K, P, Σ) is a finite element.
⊓
⊔
Proof. See Exercise 2.5.
g
The corresponding interpolation operator is denoted IK
(the superscript
is consistent with the notation introduced in §10.2 where the letter “g” refers
to the gradient operator). Its action on functions v ∈ W 1,1 (K) is such that
R1 g
g
IK
(v)(±1) = v(±1) and −1 (IK
(v) − v)q ds = 0 for all q ∈ Pk−2 .
g
Proposition 2.7 (Commuting with derivative). Let k ≥ 0. Let IK
be the
interpolation operator built from the canonical hybrid finite element of order
m
(k+1). Let IK
be the interpolation operator built from the modal finite element
g
m ′
of order k. Then, IK
(v)′ = IK
(v ) for all v ∈ W 1,1 (K).
Proof. Integrating
R 1 g by parts and using the properties of theg interpolation operators (i.e., −1 (IK
(v) − v)r ds = 0 for all r ∈ Pk−1 and IK (v)(±1) = v(±1)),
we infer that, for all q ∈ Pk ,
Z 1
Z 1
g
g
g
IK
(v)q ′ dt + [IK
(v)q]1−1
IK
(v)′ q dt = −
−1
=−
This proves that
Z
g
(v)′
IK
−1
1
−1
=
vq ′ dt + [vq]1−1 =
m ′
IK
(v )
Z
1
−1
v ′ q dt =
Z
1
−1
m ′
IK
(v )q dt.
since both functions are in Pk .
⊓
⊔
The shape functions of the canonical hybrid finite element can be computed
explicitly once a choice for the basis functions {µl }l∈{1: k−1} of Pk−2 is made;
an example is proposed in Exercise 2.5 using Jacobi polynomials.
2.2.4 Hierarchical bases(♦)
The notion of hierarchical polynomial bases is important when working with
high-order polynomials. It is particularly convenient in simulations where the
degree k varies from one element to the other.
Definition 2.8 (Hierarchical basis). A sequence of polynomials (Pk )k∈N is
said to be a hierarchical polynomial basis if the set {Pl }l∈{0: k} is a basis of
Pk for all k ∈ N.
The monomial basis (i.e., Pk (t) = tk ) is the simplest example of hierarchical polynomial basis. The Jacobi polynomials introduced in §2.1 form a hierarchical basis which is L2 -orthogonal with respect to the weight (1 − t)α (1 + t)β .
The Lagrange shape functions do not form a hierarchical basis, i.e., increasing k to (k + 1) requires to recompute the whole basis of shape functions.
20
Chapter 2. One dimension and tensor products
Instead, the shape functions of modal elements form, by construction, a hierarchical basis. Finally, in the hybrid case, the shape functions do not form,
in general, a hierarchical basis. One can obtain a hierarchical basis though by
slightly modifying the degrees of freedom. For instance, the following shape
functions form a hierarchical basis:
θ0 (t) = 21 (1 − t),
θl (t) =
θk (t) =
1
4 (1
1
2 (1
− t)(1 +
+ t).
(2.8a)
1,1
t)Jl−1
(t),
∀l ∈ {1: k − 1},
(2.8b)
(2.8c)
Their degrees of freedom can be taken to be σ0 (p) = p(−1), σk (p) = p(1),
R1
1,1
and σl (p) = αl −1 pJl−1
dt + βl− p(−1) + βl+ p(1) for all l ∈ {1: k − 1}, where
R1
1,1
−1
±
αl = 4c−1
l−1,1,1 , βl = −2cl−1,1,1 −1 (1 ± t)Jl−1 dt, and cl−1,1,1 defined in (2.4).
2.2.5 High-order Lagrange elements(♦♦)
The main difficulty comes from the oscillatory nature of the Lagrange polynomials as the number of interpolation nodes grows. This phenomenon is often
referred to as the Runge phenomenon [400] (see also Meray [349]). A classic example illustrating this phenomenon consists of considering the function
f (x) = (1 + x2 )−1 , x ∈ [−5, 5]; the Lagrange interpolation polynomial of f
using n equidistant points over [−5, 5] converges uniformly to f in the interval
(−xc , xc ) with xc ∼ 3.63 and diverges outside this interval.
The approximation quality in the maximum norm of the one-dimensional
Lagrange interpolation operator using nsh = k + 1 distinct nodes is quantified by the Lebesgue constant; see Theorem 1.11. Since the Lebesgue constant is invariant by linear transformation of the interval K, we henceforth
restrict the discussion to K = [−1, 1]. It can be shown that the Lebesgue
constant for equidistant nodes grows exponentially with nsh ; more precisely,
2n
Λnsh ∼ en(log
n+γ) as n → ∞ where γ = .5772 · · · is the Euler constant;
see Trefethen and Weideman [451]. A lower bound for the Lebesgue constant
for any set of points is π2 ln(nsh ) − C for some positive C; see Erdős [208]. If
the Chebyshev nodes al = cos( (2l−1)π
2nsh ), l ∈ N = {1: nsh }, are used instead of
the equidistant nodes, the Lebesgue constant behaves as π2 ln(nsh ) + C + αn ,
with C = π2 (γ + log π8 ) = 0.9625 · · · and αn → 0 as n → ∞, showing that this
choice is asymptotically optimal; see Luttmann and Rivlin [340] and Rivlin
[396, Chap. 4].
One drawback of Chebyshev nodes is that they do not contain the interval
endpoints ±1, which make them unsuitable for constructing H 1 -conforming
finite element spaces; see the discussion at the beginning of §2.2.3. One
set of interpolating nodes realizing this constraint are the Gauss–Lobatto–
Legendre nodes, which are, for k ≥ 1, the (k + 1) roots of the polynomial
′
(1 − t)(1 + t)(L⊥
k ) (t); all these nodes are in [−1, 1] since all the roots of
⊥
Lk are in (−1, 1). The corresponding Lagrange interpolation polynomials for
Part I. Getting Started
21
k = 3 are shown in the left panel of Figure 2.2. The Gauss–Lobatto–Legendre
nodes lead to an asymptotically optimal Lebesgue constant with upper bound
2
π log(nsh ) + C, with C ∼ 0.685; see Hesthaven [280, Conj. 3.2] and Hesthaven et al. [281, p. 106]. Note that the Lagrange polynomial bases using
the Gauss–Lobatto–Legendre nodes are not hierarchical since the set of nsh
Gauss–Lobatto–Legendre nodes is not included in the set of (nsh + 1) nodes.
1
1
0.5
0.5
0
0
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
Fig. 2.2. Left: Lagrange polynomials associated with the Gauss–Lobatto–Legendre
nodes for k = 3. Right: Hybrid nodal/modal shape functions for k = 4.
Another important class of sets of nodes is that consisting of the so-called
Fekete points. Fekete points are defined from a maximization problem. Let
{ai }i∈N be a set of nodes in K = [−1, 1] and let {φi }i∈N be a basis of Pk
(recall that N = {1: nsh }). Consider the (generalized) nsh ×nsh Vandermonde
matrix V with entries Vij = φi (aj ), for all i, j ∈ N . Since the Lagrange polyP
[a]
nomials can be expressed as Li (t) = j∈N (V −1 )ij φj (t) (see Exercise 2.1),
a reasonable criterion for selecting the interpolation nodes is to maximize
the determinant of V with respect to {ai }i∈N (observe that the solution to
this problem does not depend on the chosen basis, since a change of basis
only multiplies the determinant by a factor equal to the determinant of the
change of basis matrix). It is shown in Fejér [229] that the Fekete points and
the Gauss–Lobatto–Legendre nodes coincide on any interval. The notion of
Fekete points extends naturally to any space dimension, but the concept of
Gauss–Lobatto–Legendre nodes can be extended in higher space dimensions
only by invoking tensor product techniques; see §2.3.
22
Chapter 2. One dimension and tensor products
2.3 Tensor product elements
We show in this section that the one-dimensional finite elements presented
in §2.2 can be extended to higher dimensions by using tensor product
techQd
niques when K ⊂ Rd has the Cartesian product structure K = i=1 [zi− , zi+ ]
with real numbers zi± , i ∈ {1: d}, such that zi− < zi+ . A set of this form in Rd
is called a cuboid.
2.3.1 The polynomial space Qk,d
A useful ingredient to build tensor-product finite elements in Rd is the polynomial space
(2.9)
Qk,d = Pk ⊗ . . . ⊗ Pk .
{z
}
|
d times
This space is composed of d-variate polynomial functions q : Rd → R of partial
degree at most k with respect to each variable. Thus,
n
o
Qk,d = span xβ1 1 . . . xβd d , 0 ≤ β1 , . . . , βd ≤ k .
(2.10)
We omit the subscript d and simply write Qk when the context is unambiguous. Let β = (β1 , . . . , βd ) ∈ Nd be a multi-index, define kβkℓ∞ =
maxi∈{1: d} βi , and consider the multi-index set Bk,d = {β ∈ Nd | kβkℓ∞ ≤ k}.
Polynomial functions q ∈ Qk,d can be written in the following generic form:
X
q(x) =
aβ xβ ,
xβ = xβ1 1 . . . xβd d ,
(2.11)
β∈Bk,d
with real numbers aβ . Note that card(Bk,d ) = dim(Qk,d ) = (k + 1)d .
2.3.2 Tensor-product finite elements
We begin with tensor-product Lagrange finite elements with nodes obtained
by invoking the tensor product of nodes along each Cartesian direction. This
leads to the following construction.
Qd
Proposition 2.9 (Tensor-product Lagrange). Let K = i=1 [zi− , zi+ ] be
d
a cuboid in R . Let k ≥ 1 and let P = Qk,d . For all i ∈ {1: d}, consider (k + 1)
distinct nodes {ai,l }l∈{0: k} in [zi− , zi+ ]. For any multi-index β = (β1 , . . . , βd ) ∈
Bk,d , define the node aβ in K with Cartesian components (ai,βi )i∈{1: d} . Let
Σ = {σβ }β∈Bk,d be the linear forms on P such that σβ (p) = p(aβ ) for all
β ∈ Bk,d . Then, (K, P, Σ) is a Lagrange finite element.
Proof. See Exercise 2.6.
⊓
⊔
Part I. Getting Started
23
The shape functions of a tensor-product Lagrange finite element are
products of the one-dimensional Lagrange interpolation polynomials defined
in (1.2):
d
Y
[a ]
∀x ∈ K,
(2.12)
Lβii (xi ),
θβ (x) =
i=1
for all β ∈ Bk,d . The Lagrange interpolation operator acts as follows:
X
L
IK
(v)(x) =
v(aβ )θβ (x),
∀x ∈ K,
(2.13)
β∈Bk,d
and possible choices for its domain are V (K) = C 0 (K) or V (K) = W s,p (K)
L
(v)(x)
with real numbers p ∈ [1, ∞] and s > dp . Note that, in general, IK
cannot be factorized as a product of univariate functions, except when the
+
d
function v has the separated form v(x) = Πi=1
vi (xi ) with vi ∈ C 0 ([x−
i , xi ]),
Q
d
L
L
in which case one can verify that IK
(v)(x) = i=1 I[x
− + (vi )(xi ).
,x ]
i
0110
11
00
Q1
Q2
11
00
11
00
00 00
11
00 11
11
11
00
00
11
00
00 11
11
00
11
1
0
0
1
0
1
0
1
11 00
00
11 00
11
1
0
1
0
0
1
0
1
0
1
i
Q3
1
0
0
1
0
1
0
1
1
0
00
0 11
1
00
11
0
1
00
0 11
1
00
11
0 1
1
0 0
1 00
11
1 0
0
1 1
0 00
11
Table 2.1. Two- and three-dimensional Lagrange finite elements Q1 , Q2 , and Q3 ;
only visible degrees of freedom are shown in three dimensions.
Table 2.1 presents examples for k ∈ {1, 2, 3} in dimensions d ∈ {2, 3}
with equidistant nodes in each Cartesian direction. The bullets conventionally
indicate the location of the nodes. It is useful to use the tensor product of
Gauss–Lobatto–Legendre nodes when k is large, since it can be shown that
the Fekete points in cuboids are the tensor products of the one-dimensional
Gauss–Lobatto–Legendre nodes, see Bos et al. [69].
The tensor-product technique can also be used to build modal and hybrid
nodal/modal finite elements in cuboids; see [69]. Following the early work
of Patera [372], finite element methods based on nodal bases using tensor
products are often referred to as spectral methods.
24
Chapter 2. One dimension and tensor products
2.3.3 Serendipity finite elements(♦♦)
It is possible to remove some nodes inside the cuboid while maintaining the
approximation properties obtained when using full tensor products. This is the
idea of the serendipity finite elements. The corresponding polynomial space,
Sk , is then a proper subspace of Qk . The main motivation is to reduce computational costs without sacrificing the possibility to build H 1 -conforming finite
element spaces and without sacrificing the accuracy of the interpolation operator. Classical two-dimensional examples consist of using the 8 boundary
nodes for k = 2 and the 12 boundary nodes for k = 3 (see Table 2.2). For
k = 4, one uses the 16 boundary nodes plus the barycenter of K. A systematic
construction of the serendipity finite elements for any space dimension and
polynomial degree is devised in Arnold and Awanou [16].
S2
11 00
00
11
11
00
11 00
00
11
S3
00
00
1 1
0
0 00
11 1
0 0
1 11
0011
11
00
11
11
00
1
0
0
1
1 1
0
11
00
0
0
1
00
0
1
0 11
1
00
11
00
11
00
1 0
0
1 00
11 0
1 0
1 11
0011
00
11
S4
0110 0110 1001
01
01 0110
1010 1010 101010
Table 2.2. Two-dimensional serendipity Lagrange finite elements S2 , S3 , and S4 .
Exercises
Exercise 2.1 (Vandermonde matrix). Consider the Lagrange finite ele[a]
ment from Proposition 2.4 with nodes {ai }i∈N in K. Let {Li }i∈N be the
corresponding Lagrange polynomials, see (1.2). Let {φi }i∈N be a basis of Pk .
Show that the Lagrange polynomials can be written in the form
X
[a]
Li (t) =
(V −1 )ij φj (t),
j∈N
where the matrix V has entries Vij = φi (aj ) for all i, j ∈ N . (Note: when the
functions φi are the monomial basis, the matrix V is called a Vandermonde
matrix; otherwise, it is called a generalized Vandermonde matrix.)
Exercise 2.2 (Cross approximation). Let f : X × Y → R be a bivariate
function where X, Y are two nonempty subsets of R. Let nsh ≥ 1, set N =
{1: nsh }, and consider nsh points {xi }i∈N in X and nsh points {yj }j∈N in
Part I. Getting Started
25
Y . Assume that the matrix F ∈ Rnsh ×nsh with entries Fij = f (xi , yj ) is
invertible. Let I CA (f ) : X × Y → R be defined by
X
I CA (f )(x, y) =
(F −T )ij f (x, yj )f (xi , y).
i,j∈N
Prove that I CA (f )(x, yk ) = f (x, yk ) for all x ∈ X and all k ∈ N , and that
I CA (f )(xk , y) = f (xk , y) for all y ∈ Y and all k ∈ N .
Exercise 2.3 (Mass and Vandermonde matrices). Let K = [−1, 1], let
[a]
{a1 , . . . , ansh } be nodes in K, and let {Li }1≤i≤nsh be the corresponding Lansh ×nsh
grange polynomials. Let M ∈ R
be the so-called mass matrix with
R 1 [a]
[a]
entries Mij = −1 Li (t)Lj (t) dt for all i, j ∈ N . Prove that M = (V T V)−1
where V ∈ Rnsh ×nsh is the (generalized) Vandermonde matrix with entries
2
⊥
)1/2 L⊥
Vij = ( 2i−1
i−1 (aj ) and the Li ’s are the Legendre polynomials (see Definition 2.1).
Exercise 2.4 (Integrated Legendre polynomials). Let k ≥ 2 and set
(0)
Pk = {p ∈ Pk | p(±1) = 0}.
(0)
(i) Show that a basis for Pk are the integrated Legendre polynomials
Rt
{ −1 L⊥
l (s) ds}l∈{1: k−1} .
(ii) Prove (2.5). (Hint: consider moments against polynomials in Pm−2 and
the derivative at t = 1.)
Exercise 2.5 (Canonical hybrid element).
(i) Prove Proposition 2.6. (Hint: verify unisolvence.)
1,1
(ii) Compute the shape functions when µl = Jl−1
for all l ∈ {1: k − 1}. (Hint:
1,0
1,1
0,1
consider the polynomials Jk−1
, Jl−1
, ∀l ∈ {1: k − 1}, and Jk−1
.)
Exercise 2.6 (Qk,d Lagrange). Prove Proposition 2.9. (Hint: observe that
Pk
any polynomial p ∈ Qk,d can be written p(x) = id =0 qid (x1 , . . . , xd−1 )xidd
and use induction on d.)
Exercise 2.7 (Bi-cubic Hermite rectangle). Let K be a (non-degenerate)
rectangle with vertices {z1 , z2 , z3 , z4 }. Set P = Q3,2 and
Σ = {p(zi ), ∂x1 p(zi ), ∂x2 p(zi ), ∂x21 x2 p(zi )}1≤i≤4 .
Show that
P (K, P, Σ) is a finite element. (Hint: write p ∈ Q3,2 in the form
p(x) = 1≤i,j≤4 γij θi (x1 )θj (x2 ) where {θ1 , . . . , θ4 } are the shape functions of
the one-dimensional Hermite finite element; see Exercise 1.3.)