# 2. Finite element spaces: local to global¶

In this section, we discuss the construction of general finite element spaces. Given a triangulation $$\mathcal{T}$$ of a domain $$\Omega$$, finite element spaces are defined according to

1. the form the functions take (usually polynomial) when restricted to each cell (a triangle, in the case considered so far),

2. the continuity of the functions between cells.

We also need a mechanism to explicitly build a basis for the finite element space. We first do this by looking at a single cell, which we call the local perspective. Later we will take the global perspective, seeing how function continuity is enforced between cells.

## 2.1. Ciarlet’s finite element¶

The first part of the definition is formalised by Ciarlet’s definition of a finite element.

Definition 2.1 (Ciarlet’s finite element)

Let

1. the element domain $$K\subset \mathbb{R}^n$$ be some bounded closed set with piecewise smooth boundary,

2. the space of shape functions $$\mathcal{P}$$ be a finite dimensional space of functions on $$K$$, and

3. the set of nodal variables $$\mathcal{N}=(N_0,\ldots,N_k)$$ be a basis for the dual space $$P'$$.

Then $$(K,\mathcal{P},\mathcal{N})$$ is called a finite element.

For the cases considered in this course, $$K$$ will be a polygon such as a triangle, square, tetrahedron or cube, and $$P$$ will be a space of polynomials. Here, $$P'$$ is the dual space to $$P$$, defined as the space of linear functions from $$P$$ to $$\mathbb{R}$$. Examples of dual functions to $$P$$ include:

1. The evaluation of $$p\in P$$ at a point $$x\in K$$.

2. The integral of $$p\in P$$ over a line $$l\in K$$.

3. The integral of $$p\in P$$ over $$K$$.

4. The evaluation of a component of the derivative of $$p\in P$$ at a point $$x\in K$$.

Exercise 2.2

Show that the four examples above are all linear functions from $$P$$ to $$\mathbb{R}$$.

Exercise 2.3

For a domain $$K$$ and shape space $$P$$, is the following functional a nodal variable? Explain your answer.

$N_0(p) = \int_K p^2 \,d x.$

Ciarlet’s finite element provides us with a standard way to define a basis for the $$P$$, called the nodal basis.

Definition 2.4 ((local) nodal basis)

Let $$(K,\mathcal{P},\mathcal{N})$$ be a finite element. The nodal basis is the basis $$\{\phi_0,\phi_2,\ldots,\phi_k\}$$ of $$\mathcal{P}$$ that is dual to $$\mathcal{N}$$, i.e.

$N_i(\phi_j) = \delta_{ij}, \quad 0\leq i,j \leq k.$

We now introduce our first example of a Ciarlet element.

Definition 2.5 (The 1-dimensional Lagrange element)

The 1-dimensional Lagrange element $$(K,\mathcal{P},\mathcal{N})$$ of degree $$k$$ is defined by

1. $$K$$ is the interval $$[a,b]$$ for $$-\infty<a<b<\infty$$.

2. $$\mathcal{P}$$ is the ($$k+1$$)-dimensional space of degree $$k$$ polynomials on $$K$$,

3. $$\mathcal{N}=\{N_0,\ldots,N_k\}$$ with

$N_i(v) = v(x_i), \, x_i = a + (b-a)i/k, \quad \forall v\in \mathcal{P},\, i=0,\ldots,k.$
Exercise 2.6

Show that the nodal basis for $$\mathcal{P}$$ is given by

$\phi_i(x) = \frac{\prod_{j=0,j\ne i}^k (x-x_j)}{\prod_{j=0,j\ne i}^k (x_i-x_j)}, \quad i=0,\ldots,k.$

## 2.2. Vandermonde matrix and unisolvence¶

More generally, It is useful computationally to write the nodal basis in terms of another arbitrary basis $$\{\psi_i\}_{i=0}^k$$. This transformation is represented by the Vandermonde matrix.

Definition 2.7 (Vandermonde matrix)

Given a dual basis $$\mathcal{N}$$ and a basis $$\{\psi_i\}_{i=0}^k$$, the Vandermonde matrix is the matrix $$V$$ with coefficients

$V_{ij} = N_j(\psi_i).$

This relationship is made clear by the following lemma.

Lemma 2.8

The expansion of the nodal basis $$\{\phi_i\}_{i=0}^k$$ in terms of another basis $$\{\psi_i\}_{i=0}^k$$ for $$\mathcal{P}$$,

$\phi_i(x) = \sum_{j=0}^k \mu_{ij}\psi_j(x),$

has coefficients $$\mu_{ij}$$, $$0\leq i,j\leq k$$ given by

$\mu = V^{-1},$

where $$\mu$$ is the corresponding matrix.

Proof

The nodal basis definition becomes

$\delta_{ij} = N_j(\phi_i) = \sum_{l=0}^k\mu_{il}N_j(\psi_l) = \sum_{l=0}^k \mu_{il}V_{lj} = (\mu V)_{ij},$

where $$\mu$$ is the matrix with coefficients $$\mu_{ij}$$, and $$V$$ is the matrix with coefficients $$N_j(\psi_i)$$.

Exercise 2.9

Consider the following finite element.

• $$K$$ is the interval $$[0,1]$$.

• $$P$$ is the quadratic polynomials on $$K$$.

• The nodal variables are:

$N_0[p] = p(0), \, N_1[p]=p(1), \, N_2=\int_0^1 p(x) \,d x.$

Find the corresponding nodal basis.

Given a triple $$(K,\mathcal{P},\mathcal{N})$$, it is necessary to verify that $$\mathcal{N}$$ is indeed a basis for $$\mathcal{P}'$$, i.e. that the Ciarlet element is well-defined. Then the nodal basis is indeed a basis for $$\mathcal{P}$$ by construction. The following lemma provides a useful tool for checking this.

Lemma 2.10 (dual condition)

Let $$K,\mathcal{P}$$ be as defined above, and let $$\{N_0,N_1,\ldots,N_k\}\in \mathcal{P}'$$. Let $$\{\psi_0,\psi_1,\ldots,\psi_k\}$$ be a basis for $$\mathcal{P}$$.

Then the following three statements are equivalent.

1. $$\{N_0,N_1,\ldots,N_k\}$$ is a basis for $$\mathcal{P}'$$.

2. The Vandermonde matrix with coefficients

$V_{ij} = N_j(\psi_i), \, 0\leq i,j\leq k,$

is invertible.

3. If $$v\in\mathcal{P}$$ satisfies $$N_i(v)=0$$ for $$i=0,\ldots,k$$, then $$v\equiv 0$$.

Proof

Let $$\{N_0,N_1,\ldots,N_k\}$$ be a basis for $$\mathcal{P}'$$. This is equivalent to saying that given element $$E$$ of $$\mathcal{P}'$$, we can find basis coefficients $$\{e_i\}_{i=0}^k\in \mathbb{R}$$ such that

$E = \sum_{i=0}^k e_iN_i.$

This in turn is equivalent to being able to find a vector $$e=(e_0,e_1,\ldots,e_k)^T$$ such that

$b_i = E(\psi_i) = \sum_{j=0}^k e_j N_j(\psi_i) = \sum_{j=0}^k e_jV_{ij},$

i.e. the equation $$V{e}={b}$$ is solvable. This means that (1) is equivalent to (2).

On the other hand, we may expand any $$v\in \mathcal{P}$$ according to

$v(x) = \sum_{i=0}^k f_i \psi_i(x).$

Then

$N_i(v)=0 \iff \sum_{j=0}^k f_jN_i(\psi_j) = 0, \quad i=0,1,\ldots,k,$

by linearity of $$N_i$$. So (2) is equivalent to

$\sum_{j=0}^k f_jN_i(\psi_j) = 0, \quad i=0,1,\ldots,k \implies f_j=0, \, j=0,1,\ldots,k,$

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

This result leads us to introducing the following terminology.

Definition 2.11 (Unisolvence.)

We say that $$\mathcal{N}$$ determines $$\mathcal{P}$$ if it satisfies condition 3 of Lemma 2.10. If this is the case, we say that $$(K,\mathcal{P},\mathcal{N})$$ is unisolvent.

We can now go and directly apply this lemma to the 1D Lagrange elements.

Corollary 2.12

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

Proof

Let $$(K,\mathcal{P},\mathcal{N})$$ be the degree $$k$$ Lagrange element. We need to check that $$\mathcal{N}$$ determines $$\mathcal{P}$$. Let $$v\in \mathcal{P}$$ with $$N_i(v)=0$$ for all $$N_i\in \mathcal{N}$$. This means that

$v(a+(b-a)i/k)=0, \, i=,0,1,\ldots,k,$

which means that $$v$$ vanishes at $$k+1$$ points in $$K$$. Since $$v$$ is a degree $$k$$ polynomial, it must be zero by the fundamental theorem of algebra.

Exercise 2.13

Consider the following proposed finite element.

• $$K$$ is the interval $$[0,1]$$.

• $$P$$ is the linear polynomials on $$K$$.

• The nodal variables are:

$N_0[p] = p(0.5), N_1=\int_0^1 p(x) \,d x.$

Is this finite element unisolvent? Explain your answer.

## 2.3. 2D and 3D finite elements¶

We would like to construct some finite elements with 2D and 3D domains $$K$$. The fundamental theorem of algebra does not directly help us there, but the following lemma is useful when checking that $$\mathcal{N}$$ determines $$\mathcal{P}$$ in those cases.

Lemma 2.14

Let $$p(x):\mathbb{R}^d\to\mathbb{R}$$ be a polynomial of degree $$k\geq 1$$ that vanishes on a hyperplane $$\Pi_L$$ defined by

$\Pi_L = \left\{ x: L(x)=0\right\},$

for a non-degenerate affine function $$L(x):\mathbb{R}^d\to \mathbb{R}$$. Then $$p(x)=L(x)q(x)$$ where $$q(x)$$ is a polynomial of degree $$k-1$$.

Proof

Choose coordinates (by shifting the origin and applying a linear transformation) such that $$x=(x_1,\ldots,x_d)$$ with $$L(x)=x_d$$, so $$\Pi_L$$ is defined by $$x_d=0$$. Then the general form for a polynomial is

$P(x_1,\ldots,x_d) = \sum_{i_d=0}^k \left(\sum_{|i_1+\ldots+i_{d-1}|\leq k-i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d}\prod_{l=1}^{d-1} x_{l}^{i_l}\right),$

Then, $$p(x_1,\ldots,x_{d-1},0)=0$$ for all $$(x_1,\ldots,x_{d-1})$$, so

$0 = \left(\sum_{|i_1+\ldots+i_{d-1}|\leq k}c_{i_1,\ldots,i_{d-1},0} \prod_{l=1}^{d-1}x_{l}^{i_l}\right)$

which means that

$c_{i_1,\ldots,i_{d-1},0} = 0, \quad \forall |i_1+\ldots+i_{d-1}|\leq k.$

This means we may rewrite

\begin{align}\begin{aligned}P(x) = {L(x)}\underbrace{\left(\sum_{i_d=1}^k\sum_{|i_1+\ldots+i_{d-1}|\leq k - i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d-1}\prod_{l=1}^{d-1} x_{l}^{i_l}\right)},\\P(x) = \underbrace{x_d}_{L(x)}\underbrace{\left(\sum_{i_d=0}^{k-1}\sum_{|i_1+\ldots+i_{d-1}|\leq k - i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d-1}\prod_{l=1}^{d-1} x_{l}^{i_l}\right)}_{Q(x)},\end{aligned}\end{align}

with $$\deg(Q)=k-1$$.

Exercise 2.15

The following polynomial vanishes on the line $$y=1-x$$. Show that it satisfies the result of the previous theorem.

$x^{5} + 5 x^{4} y - x^{4} + 6 x^{3} y^{2} - 4 x^{3} y - 2 x^{2} y^{3} - 2 x^{2} y^{2} - 3 x y^{4} + 4 x y^{3} + y^{5} - y^{4}$

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

Definition 2.16 (Lagrange elements on triangles)

The triangular Lagrange element of degree $$k$$ $$(K,\mathcal{P},\mathcal{N})$$, denoted $$Pk$$, is defined as follows.

1. $$K$$ is a (non-degenerate) triangle with vertices $$z_1$$, $$z_2$$, $$z_3$$.

2. $$\mathcal{P}$$ is the space of degree $$k$$ polynomials on $$K$$.

3. $$\mathcal{N}=\left\{N_{i,j}:0\leq i \leq k, \, 0\leq j \leq i\right\}$$ defined by $$N_{i,j}(v)=v(x_{i,j})$$ where

$x_{i,j} = z_1 + (z_2-z_1)\frac{i}{k} + (z_3-z_1)\frac{j}{k}.$

We illustrate this for the cases $$k=1,2$$.

Example 2.17 (P1 elements on triangles)

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

Example 2.18 (P2 elements on triangles)

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

We now need to check that that the degree $$k$$ Lagrange element is a finite element, i.e. that $$\mathcal{N}$$ determines $$\mathcal{P}$$. We will first do this for $$P1$$.

Lemma 2.19

The degree $$1$$ Lagrange element on a triangle $$K$$ is a finite element.

Proof

Let $$\Pi_1$$, $$\Pi_2$$, $$\Pi_3$$ be the three lines containing the vertices $$z_2$$ and $$z_3$$, $$z_1$$ and $$z_3$$, and $$z_1$$ and $$z_3$$ respectively, and defined by $$L_1=0$$, $$L_2=0$$, and $$L_3=0$$ respectively. Consider a linear polynomial $$p$$ vanishing at $$z_1$$, $$z_2$$, and $$z_3$$. The restriction $$p|_{\Pi_1}$$ of $$p$$ to $$\Pi_1$$ is a linear function vanishing at two points, and therefore $$p=0$$ on $$\Pi_1$$, and so $$p=L_1(x)Q(x)$$, where $$Q(x)$$ is a degree 0 polynomial, i.e. a constant $$c$$. We also have

$0 = p(z_1) = cL_1(z_1) \implies c=0,$

since $$L_1(z_1)\neq 0$$, and hence $$p(x)\equiv 0$$. This means that $$\mathcal{N}$$ determines $$\mathcal{P}$$.

Exercise 2.20

Let $$K$$ be a rectangle, $$P$$ be the polynomial space spanned by $$\{1, x, y, xy\}$$, let $$\mathcal{N}$$ be the set of dual elements corresponding to point evaluation at each vertex of the rectangle. Show that $$\mathcal{N}$$ determines the finite element.

Exercise 2.21

Let $$K$$ be a triangle, and $$P$$ be the space of quadratic polynomials. Let $$N$$ be the set of nodal variables given by point evaluation at each edge midpoint together with the nodal variables given by integral of the function along each edge. Show that $$N$$ determines $$P$$.

This technique can then be extended to degree 2.

Lemma 2.22

The degree $$2$$ Lagrange element is a finite element.

Proof

Let $$p$$ be a degree $$2$$ polynomial with $$N_i(p)$$ for all of the degree $$2$$ dual basis elements. Let $$\Pi_1$$, $$\Pi_2$$, $$\Pi_3$$, $$L_1$$, $$L_2$$ and $$L_3$$ be defined as for the proof of Lemma . $$p|_{\Pi_1}$$ is a degree 2 scalar polynomial vanishing at 3 points, and therefore $$p=0$$ on $$\Pi_1$$, and so $$p(x)=L_1(x)Q_1(x)$$ with $$\deg(Q_1)=1$$. We also have $$0=p|_{\Pi_2} =L_1Q_1|_{\Pi_2}$$, so $$Q_1|_{\Pi_2}=0$$ and we conclude that $$p(x)=cL_1(x)L_2(x)$$. Finally, $$p$$ also vanishes at the midpoint of $$L_3$$, so we conclude that $$c=0$$ as required.

The technique extends further to degree 3.

Exercise 2.23

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

Going beyond degree 3, we have more than 1 nodal variable taking point evaluation inside the triangle. To deal with this, we use the nested triangular structure of the Lagrange triangle.

Lemma 2.24

The degree $$k$$ Lagrange element is a finite element for $$k>3$$.

Proof

We prove by induction. Assume that the degree $$k-3$$ Lagrange element is a finite element. Let $$p$$ be a degree $$k$$ polynomial with $$N_i(p)$$ for all of the degree $$k$$ dual basis elements. Let $$\Pi_1$$, $$\Pi_2$$, $$\Pi_3$$, $$L_1$$, $$L_2$$ and $$L_3$$ be defined as for the proof of lemma 2.19. The restriction $$p|_{\Pi_1}$$ is a degree $$k$$ polynomial in one variable that vanishes at $$k+1$$ points, and therefore $$p(x)=L_1(x)Q_1(x)$$, with $$\deg(Q_1)=k-1$$. $$p$$ and therefore $$Q$$ also vanishes on $$\Pi_2$$, so $$Q_1(x)=L_2(x)Q_2(x)$$.

Repeating the argument again means that $$p(x)=L_1(x)L_2(x)L_3(x)Q_3(x)$$, with $$\deg(Q_3)=k-3$$. $$Q_3$$ must vanish on the remaining points in the interior of $$K$$, which are arranged in a smaller triangle $$K'$$ and correspond to the evaluation points for a degree $$k-3$$ Lagrange finite element on $$K'$$. From the inductive hypothesis, and using the results for $$k=1,2,3$$, we conclude that $$Q_3\equiv=0$$, and therefore $$p\equiv0$$ as required.

## 2.4. Some more exotic elements¶

We now consider some finite elements that involve derivative evaluation. The Hermite elements involve evaluation of first derivatives as well as point evaluations.

Definition 2.25 (Cubic Hermite elements on triangles)

The cubic Hermite element is defined as follows:

1. $$K$$ is a (nondegenerate) triangle,

2. $$\mathcal{P}$$ is the space of cubic polynomials on $$K$$,

3. $$\mathcal{N}=\{N_1,N_2,\ldots,N_{10}\}$$ defined as follows:

• $$(N_1,\ldots,N_3)$$: evaluation of $$p$$ at vertices,

• $$(N_4,\ldots,N_9)$$: evaluation of the gradient of $$p$$ at the 3 triangle vertices.

• $$N_{10}$$: evaluation of $$p$$ at the centre of the triangle.

It turns out that the Hermite element is insufficient to guarantee functions with continuous derivatives between triangles. This problem is solved by the Argyris element.

Definition 2.26 (Quintic Argyris elements on triangles)

The quintic Argyris element is defined as follows:

1. $$K$$ is a (nondegenerate) triangle,

2. $$\mathcal{P}$$ is the space of quintic polynomials on $$K$$,

3. $$\mathcal{N}$$ defined as follows:

• evaluation of $$p$$ at 3 vertices,

• evaluation of gradient of $$p$$ at 3 vertices,

• evaluation of Hessian of $$p$$ at 3 vertices,

• evaluation of the gradient normal to 3 triangle edges.

## 2.5. Global continuity¶

Next we need to know how to glue finite elements together to form spaces defined over a triangulation (mesh). To do this we need to develop a language for specifying connections between finite element functions between element domains.

Definition 2.27 (Finite element space)

Let $$\mathcal{T}$$ be a triangulation made of triangles $$K_i$$, with finite elements $$(K_i,\mathcal{P}_i,\mathcal{N}_i)$$. A space $$V$$ of functions on $$\mathcal{T}$$ is called a finite element space if for each $$u\in V$$, and for each $$K_i\in\mathcal{T}$$, $$u|_{K_i}\in \mathcal{P}_i$$.

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

Definition 2.28 (Finite element space)

A finite element space $$V$$ is a $$C^m$$ finite element space if $$u\in C^m$$ for all $$u\in V$$.

The following lemma guides use in how to inspect the continuity of finite element functions.

Lemma 2.29 (Continuity lemma)

Let $$\mathcal{T}$$ be a triangulation on $$\Omega$$, and let $$V$$ be a finite element space defined on $$\mathcal{T}$$. The following two statements are equivalent.

1. $$V$$ is a $$C^m$$ finite element space.

2. The following two conditions hold.

• For each vertex $$z$$ in $$\mathcal{T}$$, let $$\{K_i\}_{i=1}^m$$ be the set of triangles that contain $$z$$. Then $$u|_{K_1}(z)=u|_{K_2}(z)=\ldots = u|_{K_m}(z)$$, for all functions $$u\in V$$, and similarly for all of the partial derivatives of degrees up to $$m$$.

• For each edge $$e$$ in $$\mathcal{T}$$, let $$K_1$$, $$K_2$$ be the two triangles containing $$e$$. Then $$u|_{K_1}(z) = u|_{K_2}(z)$$, for all points $$z$$ on the interior of $$e$$, and similarly for all of the partial derivatives of degrees up to $$m$$.

Proof

$$V$$ is polynomial on each triangle $$K$$, so continuity at points on the interior of each triangle $$K$$ is immediate. We just need to check continuity 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 polynomial functions and their derivatives agree at vertices and edges (similar ideas extend to higher dimensions). We achieve this by assigning nodal variables (and their associated nodal basis functions) appropriately to vertices, edges etc. of each triangle $$K$$. First we need to introduce this terminology.

Definition 2.30 (local and global mesh entities)

Let $$K$$ be a triangle. The local mesh entities of $$K$$ are the vertices, the edges, and $$K$$ itself. The global mesh entities of a triangulation $$\mathcal{T}$$ are the vertices, edges and triangles comprising $$\mathcal{T}$$.

Having made this definition, we can now talk about how nodal variables can be assigned to local mesh entities in a geometric decomposition.

Definition 2.31 (local geometric decomposition)

Let $$(K,\mathcal{P},\mathcal{N})$$ be a finite element. We say that the finite element has a (local) geometric decomposition if each dual basis function $$N_i$$ can be associated with a single mesh entity $$w\in W$$ such that for any $$f\in\mathcal{P}$$, $$N_i(f)$$ can be calculated from $$f$$ and derivatives of $$f$$ evaluated on $$w$$.

Exercise 2.32

Consider the finite element defined by:

1. $$K$$ is the unit interval $$[0,1]$$

2. $$P$$ is the space of quadratic polynomials on $$K$$,

3. The nodal variables are:

$N_0[v] = v(0), N_1[v] = v(1), N_2[v] = \int_0^1 v(x)\,d x.$

Find the corresponding nodal basis for $$P$$ in terms of the monomial basis $$\{1, x, x^2\}$$. Provide the $$C^0$$ geometric decomposition for the finite element (demonstrating that it is indeed $$C^0$$).

To discuss $$C^m$$ continuity, we need to introduce some further vocabulary about the topology of $$K$$.

Definition 2.33 (closure of a local mesh entity)

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

This allows us to define the degree of continuity of the local geometric decomposition.

Definition 2.34 ((C^m) geometric decomposition)

Let $$(K,\mathcal{P},\mathcal{N})$$ be a finite element with geometric decomposition $$W$$. We say that $$W$$ is a $$C^m$$ geometric decomposition, if for each local mesh entity $$w$$, for any $$f\in \mathcal{P}$$, the restriction $$f|_w$$ of $$f$$ (and the restriction $$D^kf|_w$$ of the $$k$$-th derivative of $$f$$ to $$w$$ for $$k\leq m$$) can be obtained from the set of dual basis functions associated with entities in the closure of $$w$$, applied to $$f$$.

The idea behind this definition is that if two triangles $$K_1$$ and $$K_2$$ are joined at a vertex, with finite elements $$(K_1,\mathcal{P}_1, \mathcal{N}_1)$$ and $$(K_2, \mathcal{P}_2, \mathcal{N}_2)$$, then if the $$\mathcal{N}_1$$ variables associated with the vertex applied to a function $$u$$ agree with the corresponding $$\mathcal{N}_2$$ variables also associated with that vertex also applied to $$u$$, then the function $$u$$ will be $$C^m$$ continuous through the vertex. Similarly, if $$K_1$$ and $$K_2$$ are joined at an edge, then if the corresponding $$\mathcal{N}_1$$ and $$\mathcal{N}_2$$ nodal variables associated with that edge agree when applied to $$u$$, then $$u$$ will be $$C^m$$ continuous through that edge. We just need to define these correspondences.

We explore this definition through a couple of exercises.

Exercise 2.35

Show that the Lagrange elements of degree $$k$$ have $$C^0$$ geometric decompositions.

Exercise 2.36

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

We now use the geometric decomposition to construct global finite element spaces over the whole triangulation (mesh). We just need to define what it means for elements of the nodal variables from the finite elements of two neighbouring triangles to “correspond”.

We start by considering spaces of functions that are discontinuous between triangles, before defining $$C^m$$ continuous subspaces.

Definition 2.37 (Discontinuous finite element space)

Let $$\mathcal{T}$$ be a triangulation, with finite elements $$(K_i,P_i,\mathcal{N}_i)$$ for each triangle $$K_i$$. The associated discontinuous finite element space $$V$$, is defined as

$V = \left\{u: u|_{K_i} \in P_i, \, \forall K_i \in \mathcal{T}\right\}.$

This defines families of discontinuous finite element spaces.

Example 2.38 (Discontinuous Lagrange finite element space)

Let $$\mathcal{T}$$ be a triangulation, with Lagrange elements of degree $$k$$, $$(K_i,P_i,\mathcal{N}_i)$$, for each triangle $$K_i\in \mathcal{T}$$. The corresponding discontinuous finite element space, denoted $$Pk$$ DG, is called the discontinuous Lagrange finite element space of degree $$k$$.

Next we need to associate each nodal variable in each element to a vertex, edge or triangle of the triangulation $$\mathcal{T}_h$$, i.e. the global mesh entitles. The following definition explains how to choose this association.

Definition 2.39 (Global (C^m) geometric decomposition)

Let $$\mathcal{T}$$ be a triangulation with finite elements $$(K_i,\mathcal{P}_i,\mathcal{N}_i)$$, each with a $$C^m$$ geometric decomposition. Assume that for each global mesh entity $$w$$, the $$n_w$$ triangles containing $$w$$ have finite elements $$(K_i,\mathcal{P}_i,\mathcal{N}_i)$$ each with $$M_w$$ dual basis functions associated with $$w$$. Further, each of these basis functions can be enumerated $$N^w_{i,j}\in\mathcal{N}_i$$, $$j=1,\ldots,M_w$$, such that $$N^w_{1,j}(u|_{K_1})=N^w_{2,j}(u|_{K_2})=\ldots = N^w_{n_w,j}(u|_{K_n}), \quad, j=1,\ldots,M_w$$, for all functions $$u\in C^m(\Omega)$$.

This combination of finite elements on $$\mathcal{T}$$ together with the above enumeration of dual basis functions on global mesh entities is called a global $$C^m$$ geometric decomposition.

Now we use this global $$C^m$$ geometric decomposition to build a finite element space on the triangulation.

Definition 2.40 (Finite element space from a global (C^m) geometric decomposition)

Let $$\mathcal{T}$$ be a triangulation with finite elements $$(K_i,\mathcal{P}_i,\mathcal{N}_i)$$, each with a $$C^m$$ geometric decomposition, and let $$\hat{V}$$ be the corresponding discontinuous finite element space. Then the global $$C^m$$ geometric decomposition defines a subspace $$V$$ of $$\hat{V}$$ consisting of all functions that $$u$$ satisfy $$N^w_{1,j}(u|_{K_1})=N^w_{2,j}(u|_{K_2})=\ldots = N^w_{n_w,j}(u|_{K_{n_w}}), \quad j=1,\ldots,M_w$$ for all mesh entities $$w\in\mathcal{T}$$.

The following result shows that the global $$C^m$$ geometric decomposition is a useful definition.

Lemma 2.41

Let $$V$$ be a finite element space defined from a global $$C^m$$ geometric decomposition. Then $$V$$ is a $$C^m$$ finite element space.

Proof

From the local $$C^m$$ decomposition, functions and derivatives up to degree $$m$$ on vertices and edges are uniquely determined from dual basis elements associated with those vertices and edges, and from the global $$C^m$$ decomposition, the agreement of dual basis elements means that functions and derivatives up to degree $$m$$ agree on vertices and edges, and hence the functions are in $$C^m$$ from Lemma 2.29.

We now apply this to a few examples, which can be proved as exercises.

Example 2.42

The finite element space built from the $$C^0$$ global decomposition built from degree $$k$$ Lagrange element is called the degree $$k$$ continuous Lagrange finite element space, denoted $$Pk$$.

Example 2.43

The finite element space built from the $$C^1$$ global decomposition built from the quintic Argyris element is called the Argyris finite element space.

In this section, we have built a theoretical toolbox for the construction of finite element spaces. In the next section, we move on to studying how well we can approximate continuous functions as finite element functions.