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Quantitative Analysis
Parallel Processing
Numerical Analysis
C++ Multithreading
Python for Excel
Python Utilities
Services
Author
Printable PDF file
I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
IV. Data Analysis.
V. Implementation tools.
VI. Basic Math II.
VII. Implementation tools II.
1. Calculational Linear Algebra.
2. Wavelet Analysis.
3. Finite element method.
4. Construction of approximation spaces.
A. Finite element.
B. Averaged Taylor polynomial.
C. Stable space splittings.
D. Frames.
E. Tensor product splitting.
F. Sparse tensor product. Cure for curse of dimensionality.
5. Time discretization.
6. Variational inequalities.
VIII. Bibliography
Notation. Index. Contents.

Finite element.


efinition

(Finite element) 1. "Finite element" is a triple MATH 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" MATH are a basis of the dual space MATH .

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

MATH (Dual basis)
is called the "nodal basis" of $\QTR{cal}{P}$ .

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

Example

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

Example

(1-dimensional Lagrange element) MATH , $\QTR{cal}{P}$ 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 MATH , $\QTR{cal}{P}$ is the 3-dimensional space of linear functions, MATH

Proposition

(Identification of nodal variables 1) Let $\QTR{cal}{P}$ be an $n$ -dimensional vector space and MATH is a subset of the dual space MATH . Then the following statements are equivalent:

1. MATH is a basis for MATH .

2. For any $v\in\QTR{cal}{P}$ , MATH

Proof

MATH . Assume the contrary: $\left( 1\right) $ holds but MATH Then we introduce a MATH s.t. MATH Such $\psi_{0}$ is linearly independent from MATH . We obtained a contradiction.

MATH . Assume the contrary: MATH is not a basis. Since MATH has $n$ elements it has to be linearly dependent: there is a rearrangement of indexes $k=1,...,n$ and numbers MATH s.t. MATH Therefore we reduced the number of indexes $k$ in the condition (2): MATH Let MATH be a basis of $\QTR{cal}{P}$ . For any $v\in\QTR{cal}{P}$ there are numbers MATH such that MATH Thus $\left( \&\right) $ reads as follows: there are numbers MATH , MATH s.t. for any set of numbers MATH MATH This means that the MATH matrix MATH would be invertible if complemented to an $n\times n$ matrix with an arbitrary row MATH . We obtained a contradiction.

Proposition

(Identification of nodal variables 2) Let $P\left( x\right) $ , MATH be a polynomial of degree $d\geq1$ that vanishes on the hyperplane MATH . Then MATH where the $Q$ is a polynomial of degree $d-1$ .

Proof

We perform an affine change of coordinates from MATH to MATH so that the plane MATH is given by MATH . Let MATH We have MATH thus the $Q\left( y\right) $ has the form MATH for some polynomial MATH of degree $d-1$ .

Example

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

We establish that MATH forms a nodal basis by noting that on any edge MATH a function $u$ from $\QTR{cal}{P}$ is 1-degree polynomial of one variable. Hence, if MATH then such function $u$ vanishes on every edge. But then, by the proposition ( Identification of nodal variables 2 ), MATH where $L_{i},i=1,2$ are 1-degree polynomials and MATH , $i=1,2$ identify two adjacent edges. Thus $c=0$ by examining $u$ on one vertex not on MATH and MATH . Hence MATH forms a nodal basis by the proposition ( Identification of nodal variables 1 ).

Definition

(Interpolant) Let MATH be a finite element and MATH be a corresponding nodal basis. We define the "interpolant" operator: MATH for any function $v$ such that the operation MATH is well defined for $k=1,...,n$ .

Proposition

(Properties of interpolant) The operator $I_{\Omega}$ of the definition ( Interpolant ) has the following properties:

1. $I_{\Omega}$ is a linear operator.

2. MATH , $k=1,...,n$ .

3. MATH . In particular, MATH .

Proof

Direct verification.





Notation. Index. Contents.


















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