(Finite element) 1. "Finite element" is a triple
where the "element domain"
is a closed bounded subset of
,
the "space of shape functions"
is a finite dimensional space of functions defined on
and the "nodal variables"
are a basis of the dual space
.

2. The basis
of the space
with the
property

(Dual basis)

is called the "nodal basis" of
.

3. The set
"determines" the space
if for any

Example

(Constant finite element)
,
is the 1-dimensional space of constant functions,

Example

(1-dimensional Lagrange element)
,
is the 2-dimensional space of linear
functions,

Example

(2-dimensional Lagrange element)
is a triangle in
with vertices
,
is the 3-dimensional space of linear
functions,

Proposition

(Identification of nodal
variables 1) Let
be an
-dimensional
vector space and
is a subset of the dual space
.
Then the following statements are equivalent:

1.
is a basis for
.

2. For any
,

Proof

.
Assume the contrary:
holds but
Then we introduce a
s.t.
Such
is linearly independent from
.
We obtained a contradiction.

.
Assume the contrary:
is not a basis. Since
has
elements it has to be linearly dependent: there is a rearrangement of indexes
and numbers
s.t.
Therefore we reduced the number of indexes
in the condition
(2):
Let
be a basis of
.
For any
there are numbers
such
that
Thus
reads as follows: there are numbers
,
s.t. for any set of numbers
This means that the
matrix
would be invertible if complemented to an
matrix with an arbitrary row
.
We obtained a contradiction.

Proposition

(Identification of nodal
variables 2) Let
,
be a polynomial of degree
that vanishes on the hyperplane
.
Then
where the
is a polynomial of degree
.

Proof

We perform an affine change of coordinates from
to
so that the plane
is given by
.
Let
We
have
thus the
has the
form
for some polynomial
of degree
.

Example

(Linear tensor product element) Let
be a rectangle with vertices
,
be a set of linear polynomials of coordinates
and
is given
by

We establish that
forms a nodal basis by noting that on any edge
a function
from
is 1-degree polynomial of one variable. Hence, if
then such function
vanishes on every edge. But then, by the proposition
(
Identification of nodal
variables 2
),
where
are 1-degree polynomials and
,
identify two adjacent edges. Thus
by examining
on one vertex not on
and
.
Hence
forms a nodal basis by the proposition
(
Identification of nodal
variables 1
).

Definition

(Interpolant) Let
be a finite element and
be a corresponding nodal basis. We define the "interpolant"
operator:
for any function
such that the operation
is well defined for
.

Proposition

(Properties of interpolant) The operator
of the definition (
Interpolant
) has the following
properties: