Finite element.

(Finite element) 1. "Finite element" is a triple where the "element domain" $\Omega$ is a closed bounded subset of $\QTR{cal}{R}^{n}$ , the "space of shape functions" $\QTR{cal}{P}$ is a finite dimensional space of functions defined on $\Omega$ and the "nodal variables" are a basis of the dual space .

2. The basis of the space $\QTR{cal}{P}$ with the property

(Dual basis)

is called the "nodal basis" of $\QTR{cal}{P}$ .

3. The set "determines" the space $\QTR{cal}{P}$ if for any $u\in\QTR{cal}{P}$ MATH

Example

(Constant finite element) , $\QTR{cal}{P}$ is the 1-dimensional space of constant functions, MATH

Example

(1-dimensional Lagrange element) , $\QTR{cal}{P}$ is the 2-dimensional space of linear functions, MATH

Example

(2-dimensional Lagrange element) $\Omega$ is a triangle in $\QTR{cal}{R}^{2}$ with vertices , $\QTR{cal}{P}$ is the 3-dimensional space of linear functions, MATH

Proposition

(Identification of nodal variables 1) Let $\QTR{cal}{P}$ 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 $v\in\QTR{cal}{P}$ , MATH

Proof

. Assume the contrary: $\left( 1\right)$ holds but MATH Then we introduce a s.t. MATH Such $\psi_{0}$ 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. MATH Therefore we reduced the number of indexes in the condition (2): MATH Let be a basis of $\QTR{cal}{P}$ . For any $v\in\QTR{cal}{P}$ there are numbers such that MATH Thus $\left( \&\right)$ reads as follows: there are numbers , s.t. for any set of numbers MATH This means that the matrix would be invertible if complemented to an $n\times n$ matrix with an arbitrary row . We obtained a contradiction.

Proposition

(Identification of nodal variables 2) Let $P\left( x\right)$ , be a polynomial of degree $d\geq1$ that vanishes on the hyperplane . Then MATH 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 MATH We have MATH thus the $Q\left( y\right)$ has the form MATH for some polynomial of degree .

Example

(Linear tensor product element) Let be a rectangle with vertices , $\QTR{cal}{P}$ be a set of linear polynomials of coordinates and is given by MATH

We establish that forms a nodal basis by noting that on any edge a function from $\QTR{cal}{P}$ 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 $L_{i},i=1,2$ 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: MATH for any function such that the operation is well defined for .