Finite Elements

Colin Cotter

January 18, 2019

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The finite element

Given a triangulation T of a domain Ω, finite element spaces aredefined according to

1. the form the functions take (usually polynomial) whenrestricted to each cell,

2. the continuity of the functions between cells.

We also need a mechanism to explicitly build a basis for the finiteelement space.

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Definition 1 (Ciarlet’s finite element)

Let

1. the element domain K ⊂ Rn be some bounded closed set withpiecewise smooth boundary,

2. the space of shape functions P be a finite dimensional spaceof functions on K , and

3. the set of nodal variables N = (N0, . . . ,Nk) be a basis for thedual space P ′.

Then (K ,P,N ) is called a finite element.

P ′ is the dual space to P, defined as the space of linear functionsfrom P to R.

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Examples of dual functions to P include:

1. The evaluation of p ∈ P at a point x ∈ K .

2. The integral of p ∈ P over a line l ∈ K .

3. The integral of p ∈ P over K .

4. The evaluation of a component of the derivative of p ∈ P at apoint x ∈ K .

Exercise 2

Show that the four examples above are all linear functions from Pto R.

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Ciarlet’s finite element provides us with a standard way to define abasis for the P.

Definition 3 (Nodal basis)

Let (K ,P,N ) be a finite element. The nodal basis is the basisφ0, φ2, . . . , φk of P that is dual to N , i.e.

Ni (φj) = δij , 0 ≤ i , j ≤ k . (1)

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Example 4 (The 1-dimensional Lagrange element)

The 1-dimensional Lagrange element (K ,P,N ) of degree k isdefined by

1. K is the interval [a, b] for −∞ < a < b <∞.

2. P is the (k + 1)-dimensional space of degree k polynomials onK ,

3. N = N0, . . . ,Nk with

Ni (v) = v(xi ), xi = a + (b − a)i/k, ∀v ∈ P, i = 0, . . . , k .(2)

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Exercise 5

Show that the nodal basis for P is given by

φi (x) =

∏kj=0,j 6=i (x − xj)∏kj=0,j 6=i (xi − xj)

, i = 0, . . . , k . (3)

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It is useful computationally to write the nodal basis in terms ofanother arbitrary basis ψiki=0.

Definition 6 (Vandermonde matrix)

Given a dual basis N and a basis ψjki=0, the Vandermondematrix is the matrix V with coefficients

Vij = Nj(ψi ). (4)

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Lemma 7

The expansion of the nodal basis φiki=0 in terms of another basisψiki=0 for P,

φi (x) =k∑

j=0

µijψj(x), (5)

has coefficients µij , 0 ≤ i , j ≤ k given by

µ = V−1, (6)

where µ is the corresponding matrix.

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Proof.

The nodal basis definition becomes

δli = Nl(φi ) =k∑

j=0

µijNl(ψj) = (Vµ)ki , (7)

where µ is the matrix with coefficients µij , and V is the matrixwith coefficients Nk(ψj).

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Lemma 8

Let K ,P be as defined above, and let N0,N1, . . . ,Nk ∈ P ′. Letψ0, ψ1, . . . , ψk be a basis for P.Then the following three statements are equivalent.

1. N0,N1, . . . ,Nk is a basis for P ′.2. The Vandermonde matrix with coefficients

Vij = Nj(ψi ), 0 ≤ i , j ≤ k , (8)

is invertible.

3. If v ∈ P satisfies Ni (v) = 0 for i = 0, . . . , k, then v ≡ 0.

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Proof

Let N0,N1, . . . ,Nk be a basis for P ′. This is equivalent to sayingthat given element E of P ′, we can find basis coefficientseiki=0 ∈ R such that

E =k∑

i=0

eiNi . (9)

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Proof (Cont.)

This in turn is equivalent to being able to find a vectore = (e0, e1, . . . , ek)T such that

bi = E (ψi ) =k∑

j=0

ejNj(ψi ) =k∑

j=0

ejVij , (10)

i.e. the equation V e = b is solvable. This means that (1) isequivalent to (2).

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Proof (Cont.)

On the other hand, we may expand any v ∈ P according to

v(x) =k∑

i=0

fiψi (x). (11)

Then

Ni (v) = 0 ⇐⇒k∑

j=0

fjNi (ψj) = 0, i = 0, 1, . . . , k , (12)

by linearity of Ni .

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Proof (Cont.)

So (2) is equivalent to

k∑j=0

fjNi (ψj) = 0, i = 0, 1, . . . , k =⇒ fj = 0, j = 0, 1, . . . , k ,

(13)which is equivalent to V T being invertible, which is equivalent toV being invertible, and so (3) is equivalent to (2).

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Definition 9

We say that N determines P if it satisfies condition 3 of Lemma 8.If this is the case, we say that (K ,P,N ) is unisolvent.

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Corollary 10

The 1D degree k Lagrange element is a finite element.

Proof.

Let (K ,P,N ) be the degree k Lagrange element. We need tocheck that N determines P. Let v ∈ P with Ni (v) = 0 for allNi ∈ N . This means that

v(a + (b − a)i/k) = 0, i =, 0, 1, . . . , k , (14)

which means that v vanishes at k + 1 points in K . Since v is adegree k polynomial, it must be zero by the fundamental theoremof algebra.

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We would like to construct some finite elements with 2D and 3Ddomains K . The following lemma is useful when checking that Ndetermines P in those cases.

Lemma 11

Let p(x) : Rd → R be a polynomial of degree k ≥ 1 that vanisheson a hyperplane ΠL defined by

ΠL = x : L(x) = 0 , (15)

for a non-degenerate affine function L(x) : Rd → R. Thenp(x) = L(x)q(x) where q(x) is a polynomial of degree k − 1.

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Proof

Choose coordinates (by shifting the origin and applying a lineartransformation) such that x = (x1, . . . , xd) with L(x) = xd , so ΠL

is defined by xd = 0. Then the general form for a polynomial is

P(x1, . . . , xd) =k∑

id=0

∑|i1+...+id−1|≤k−id

ci1,...,id−1,id xidd

d−1∏l=1

x ill

,

(16)

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Proof (Cont.)

Then, p(x1, . . . , xd−1, 0) = 0 for all (x1, . . . , xd−1), so

0 =

∑|i1+...+id−1|≤k

ci1,...,id−1,0

d−1∏l=1

x ill

(17)

which means that

ci1,...,id−1,0 = 0, ∀|i1 + . . .+ id−1| ≤ k. (18)

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Proof (Cont.)

This means we may rewrite

P(x) = xd︸︷︷︸L(x)

d∑id=1

∑|i1+...+id−1|≤k−id

ci1,...,id−1,id xid−1d

d−1∏l=1

x ill

︸ ︷︷ ︸

Q(x)

, (19)

with deg(Q) = k − 1.

Equipped with this tool we can consider some finite elements intwo dimensions.

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Definition 12 (Lagrange elements on triangles)

The triangular Lagrange element of degree k (K ,P,N ), denotedPk , is defined as follows.

1. K is a (non-degenerate) triangle with vertices z1, z2, z3.

2. P is the space of degree k polynomials on K .

3. N = Ni ,j : 0 ≤ i ≤ k , 0 ≤ j ≤ i defined by Ni ,j(v) = v(xi ,j)where

xi ,j = z1 + (z2 − z1)i/k + (z3 − z1)j/k . (20)

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Example 13 (P1 elements on triangles)

The nodal basis for P1 elements is point evaluation at the threevertices.

Example 14 (P2 elements on triangles)

The nodal basis for P2 elements is point evaluation at the threevertices, plus point evaluation at the three edge centres.

We now need to check that that the degree k Lagrange element isa finite element, i.e. that N determines P. We will first do this forcases P1 and P2.

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Lemma 15

The degree 1 Lagrange element is a finite element.

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Proof.

Let Π1, Π2, Π3 be the three lines containing the vertices z2 and z3,z1 and z3, and z1 and z3 respectively, and defined by L1 = 0,L2 = 0, and L3 = 0 respectively. Consider a linear polynomial pvanishing at z1, z2, and z3. The restriction p|Π1 of p to Π1 is alinear function vanishing at two points, and therefore p = 0 on Π1,and so p = L1(x)Q(x), where Q(x) is a degree 0 polynomial, i.e. aconstant c . We also have

0 = p(z1) = cL1(z1) =⇒ c = 0, (21)

since L1(z1) 6= 0, and hence p(x) ≡ 0. This means that Ndetermines P.

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Lemma 16

The degree 2 Lagrange element is a finite element.

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Proof.

Let p be a degree 2 polynomial with Ni (p) for all of the degree 2dual basis elements. Let Π1, Π2, Π3, L1, L2 and L3 be defined asfor the proof of Lemma 15. p|Π1 is a degree 2 scalar polynomialvanishing at 3 points, and therefore p = 0 on Π1, and sop(x) = L1(x)Q1(x) with deg(Q1) = 1. We also have0 = p|Π2 = L1Q1|Π2 , so Q1|Π2 = 0 and we conclude thatp(x) = cL1(x)L2(x). Finally, p also vanishes at the midpoint of L3,so we conclude that c = 0 as required.

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Exercise 17

Show that the degree 3 Lagrange element is a finite element.

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Lemma 18

The degree k Lagrange element is a finite element for k > 0.

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Proof

We prove by induction. Assume that the degree k − 3 Lagrangeelement is a finite element. Let p be a degree k polynomial withNi (p) for all of the degree k dual basis elements. Let Π1, Π2, Π3,L1, L2 and L3 be defined as for the proof of Lemma 15. Therestriction p|Π1 is a degree k polynomial in one variable thatvanishes at k + 1 points, and therefore p(x) = L1(x)Q1(x), withdeg(Q1) = k − 1. p and therefore Q also vanishes on Π2, soQ1(x) = L2(x)Q2(x).

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Proof(Cont.)

Repeating the argument again means thatp(x) = L1(x)L2(x)L3(x)Q3(x), with deg(Q3) = k − 3. Q3 mustvanish on the remaining points in the interior of K , which arearranged in a smaller triangle K ′ and correspond to the evaluationpoints for a degree k − 3 Lagrange finite element on K ′. From theinductive hypothesis, and using the results for k = 1, 2, 3, weconclude that Q3 ≡= 0, and therefore p ≡ 0 as required.

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We now consider some finite elements that involve derivativeevaluation.

Example 19 (Cubic Hermite element)

The cubic Hermite element is defined as follows:

1. K is a (nondegenerate) triangle,

2. P is the space of cubic polynomials on K ,

3. N = N1,N2, . . . ,N10 defined as follows:

3.1 (N1, . . . ,N3): evaluation of p at vertices,3.2 (N4, . . . ,N9): evaluation of the gradient of p at the 3 triangle

vertices.3.3 N10: evaluation of p at the centre of the triangle.

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Example 20 (Quintic Argyris element)

The quintic Hermite element is defined as follows:

1. K is a (nondegenerate) triangle,

2. P is the space of quintic polynomials on K ,

3. N defined as follows:

3.1 evaluation of p at 3 vertices,3.2 evaluation of gradient of p at 3 vertices,3.3 evaluation of Hessian of p at 3 vertices,3.4 evaluation of the gradient normal to 3 triangle edges.

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Next we need to know how to glue finite elements together to formspaces defined over a mesh. To do this we need to develop alanguage for specifying connections between finite elementfunctions between element domains.

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Definition 21 (Finite element space)

Let T be a triangulation made of triangles Ki , with finite elements(Ki ,Pi ,Ni ). A space V of functions on T is called a finite elementspace if for each u ∈ V , and for each Ki ∈ T , u|Ki

∈ Pi .

Note that the set of finite elements do not uniquely determine afinite element space, since we also need to specify continuityrequirements between triangles, which we will do in this chapter.

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Definition 22 (Cm finite element space)

A finite element space V is a Cm finite element space if u ∈ Cm

for all u ∈ V .

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Lemma 23

Let T be a triangulation on Ω, and let V be a finite element spacedefined on T . The following two statements are equivalent.

1. V is a Cm finite element space.

2. The following two conditions hold.

2.1 For each vertex z in T , let Kimi=1 be the set of triangles thatcontain z . Then u|K1 (z) = u|K2 (z) = . . . = u|Km(z), for allfunctions u ∈ V , and similarly for all of the partial derivativesof degrees up to m.

2.2 For each edge e in T , let K1, K2 be the two trianglescontaining e. Then u|K1 (z) = u|K2 (z), for all points z on theinterior of e, and similarly for all of the partial derivatives ofdegrees up to m.

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Proof.

V is polynomial on each triangle K , so continuity at points on theinterior of each triangle K is immediate. We just need to checkcontinuity at points on vertices, and points on the interior of edges,which is equivalent to the two parts of the second condition.

This means that we just need to guarantee that the polynomialfunctions and their derivatives agree at vertices and edges (similarideas extend to higher dimensions).

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Definition 24 (Mesh entities)

Let K be a triangle. The (local) mesh entities of K are thevertices, the edges, and K itself. The global mesh entities of atriangulation T are the vertices, edges and triangles comprising T .

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Definition 25 (Geometric decomposition)

Let (K ,P,N ) be a finite element. We say that the finite elementhas a (local) geometric decomposition if each dual basis functionNi can be associated with a single mesh entity w ∈W such thatfor any f ∈ P, Ni (f ) can be calculated from f and derivatives of frestricted to w .

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Definition 26 (Closure of a local mesh entity)

Let w be a local mesh entity for a triangle. The closure of w is theset of local mesh entities contained in w (including w itself).

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Definition 27 (Cm geometric decomposition)

Let (K ,P,N ) be a finite element with geometric decompositionW . We say that W is a Cm geometric decomposition, if for eachlocal mesh entity w , for any f ∈ P, the restriction f |w of f (andthe restriction Dk f |w of the k-th derivative of f to w for k ≤ m)can be obtained from the set of dual basis functions associatedwith entities in the closure of w , applied to f .

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Exercise 28

Show that the Lagrange elements of degree k have C 0 geometricdecompositions.

Exercise 29

Show that the Argyris element has a C 1 geometric decomposition.

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Definition 30 (Discontinuous finite element space)

Let T be a triangulation, with finite elements (Ki ,Pi ,Ni ) for eachtriangle Ki . The corresponding discontinuous finite element spaceV , is defined as

V = u : u|Ki∈ Pi , ∀Ki ∈ T . (22)

This defines families of discontinuous finite element spaces.

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Example 31 (Discontinuous Lagrange finite element space)

Let T be a triangulation, with Lagrange elements of degree k ,(Ki ,Pi ,Ni ), for each triangle Ki ∈ T . The correspondingdiscontinuous finite element space, denoted PkDG, is called thediscontinuous Lagrange finite element space of degree k.

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Definition 32 (Global Cm geometric decomposition)

Let T be a triangulation with finite elements (Ki ,Pi ,Ni ), eachwith a Cm geometric decomposition. Assume that for each globalmesh entity w , the nw triangles containing w have finite elements(Ki ,Pi ,Ni ) each with Mw dual basis functions associated with w .Further, each of these basis functions can be enumeratedNwi ,j ∈ Ni , j = 1, . . . ,Mw , such that

Nw1,j(u|K1) = Nw

2,j(u|K2) = . . . = Nwnw ,j

(u|Kn), , j = 1, . . . ,Mw , forall functions u ∈ Cm(Ω).This combination of finite elements on T together with the aboveenumeration of dual basis functions on global mesh entities iscalled a global Cm geometric decomposition.

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Definition 33 (Finite element space from a global Cm

geometric decomposition)

Let T be a triangulation with finite elements (Ki ,Pi ,Ni ), eachwith a Cm geometric decomposition, and let V be thecorresponding discontinuous finite element space. Then the globalCm geometric decomposition defines a subspace V of V consistingof all functions that u satisfyNw

1,j(u|K1) = Nw2,j(u|K2) = . . . = Nw

nw ,j(u|Knw

), , j = 1, . . . ,Mw forall mesh entities w ∈ T .

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Proposition 34

Let V be a finite element space defined from a global Cm

geometric decomposition. Then V is a Cm finite element space.

Proof.

From the local Cm decomposition, functions and derivatives up todegree m on vertices and edges are uniquely determined from dualbasis elements associated with those vertices and edges, and fromthe global Cm decomposition, the agreement of dual basiselements means that functions and derivatives up to degree magree on vertices and edges, and hence the functions are in Cm

from Lemma 23.

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Example 35

The finite element space built from the C 0 global decompositionbuilt from degree k Lagrange element is called the degree kcontinuous Lagrange finite element space, denoted Pk.

Example 36

The finite element space built from the C 1 global decompositionbuilt from the quintic Argyris element is called the Argyris finiteelement space.

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Next we investigate how continuous functions can be approximatedby finite element functions.

Definition 37 (Local interpolant)

Given a finite element (K ,P,N ), with corresponding nodal basisφiki=0. Let v be a function such that Ni (v) is well-defined for alli . Then the local interpolant IK is an operator mapping v to Psuch that

(IKv)(x) =k∑

i=0

Ni (v)φi (x). (23)

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Lemma 38

The operator IK is linear.

Lemma 39

Ni (IK (v)) = Ni (v), ∀0 ≤ i ≤ k. (24)

Lemma 40

IK is the identity when restricted to P.

Exercise 41

Prove lemmas 38-40.

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