Finite Element Method - Advanced Concepts 
In this page we will discuss several advanced concepts that appear in the development of Finite Element Method. We start with some basical concepts that we learned in our Calculus, such as contiuous functions, derivative of a function, etc.

Continuous Function: f(x) is continuous at x0 if f(x) is defined at x0 and the limit of f(x) at x0 as x goes to x0 is f(x0). Mathematically, we have 


lim
x® x0
f(x) = f(x0)

For example, the following functions are continuous functions in their domain (all x values, where the function is defined). 

All Polynomial Functions: f(x) = 2x + 3; g(x) = 5x6 + 4x5 - 2x3 + 1 

All Trignometric Functions: f(x) = sin(x); g(x) = cos(x); h(x) = tan(x) 

All Exponential Functions and Logarithmic Functions: f(x) = ex; g(x) = 2x; h(x) = ln(x) 

The following function is not continuous at x = 2 as shown in the following image, where the curve has as jump at x = 2: 

f(x) =  ì
í
î
3x + 1     if    x > 2 
x2 -2     otherwise 

Differentiable Function: f(x) is differentiable at x0 if f(x) is continuous at x0 and the following limit exits, which is called the derivative of f(x) at x0 and is denoted by f¢(x0) or 

d f(x0)
dx
lim
x® x0
f(x) - f(x0)
x - x0

Partial Derivatives are defined for multi-variable functions. A multi-variable function is a function that has two or more parameters. For example f(x, y) = x2 + y2 is a two-variable function, it represents a parabolic surface in three dimension space. When we take a derivative respect to one variable, we have a so called partial derivative. So for f(x, y) = x2 + y2 we can have partial derivative with respect to  x or y variables. They are denoted as 

f(x, y)
x
    and  f(x, y)
y
Or, fx(x, y) and fy(x, y) respectively. 

Frequently, we also use fx(x, y) and fy(x, y) to denote the partial derivative against x and y variables. 

Example Let f(x, y) = x2 + 3xy + 2y3 + 5, find fx(x, y) and fy(x, y) 

Solution: To find fx(x, y), we treat y as constant, just like the constant term 5 in f(x, y). So fx(x, y) = 2x + 3y. 

Similarly, to find fy(x, y), we treat x as constant. So fy(x, y) = 3x + 6y. 

f(x) is said to be continuous over a closed interval [a, b] if f(x) is continuous at each point in the interval. 

f(x) is said to be differentiable over open an interval (a, b) if f(x) is differentiable at each point in the interval. 

C[a, b] is the set of all continuous over interval [a, b] 

C1[a, b] is the set of all differential functions whose first derivative is continuous over interval [a, b]. 

Cn [a, b] is the set of all differential functions whose nth derivative is continuous over interval [a, b], here n is an postive integer. 

Cn0 [a, b] is the set of all differential functions u whose nth derivative is continuous over interval [a, b] and u(a) = u(b) = 0. 

Linear Space V: A linear space V is a set of elements with the addition of elements and multiplication by a scalar defined. And the addition and scalar multiplication satisfies a set of specifical properties. 

For example, C[a, b] , C1[a, b] and Cn[a, b] are all linear spaces. Since from Algebra and Calculus, we know we can add two functions and mutiply a function by a real number. Also all polynomial functions forms a linear space with the usual function addition and scalar multiplication. 

Linearly Independent Elements: Let {v1, v2, v3 , ¼, vn} be a set of elements of linear space V, we say v1, v2, v3 , ¼, vn are linearly independent if 

c1v1+ c2v2+ c3v3+ ¼+ cnvn = 0

if and only if c1 = c2 = c3 = ¼ = cn = 0. For example x1, x2, x3, ¼, xn is a set of linearly independent vectors in the linear space of all polynomials with degreen less or equal to n (denoted by Pn (x)). 

Basis of Linear Space: {v1, v2, v3 , ¼, vn} Ì V is a basis of a linear space V if all element of V are of form 

v = c1v1+ c2v2+ c3v3+ ¼+ cnvn
here v Î V is any element. 

In the finite element method, we will construct piecewise polynomial functions as a basis for the approximation to the solutions of our problem. 

An Ordinary Differential Equation(ODE) is an equation that involves only one independent variable and an unknown one variable function and its derivatives. 

If an ODE contains upto kth order derivatives of the unknown function y(x), we say the ODE is of order k, or kth order ODE. Generally, for unknown function y(x), we can write the first order ODE equation in the form 

F(x, y, y ¢) = 0

and the second order ODE as 

F(x, y, y ¢, y ¢¢) = 0

Example: Here are example of ODEs 

(a) First order ODE: 2y¢+ sin(x) y + 5x3 = 0
(b) Second order ODE: exy¢¢+ 3x(y¢)2 + sin(y) = 0
 In Boundary Value Problems  we want to find an unknown function y(x) which satisfies a second order ODE over an interval [a, b] and with fixed conditions about y(x) at x = a and x = b. 

Example: A boundary value problem of ODE used in beam stress analysis, here y(x) describes the deflection of a beam of length 1 with variable cross section represented by q(x). The deflection is due to the added stresses p(x) and f(x). 

d
dx
(p(x) dy
dx
) + q(x)y = f(x),     for 0 < x < 1

with boundary conditions y(0) = y(1) = 0 

Functional: A functional I defined on a linear space V (Especially linear spaces of functions, like C[a, b] , C1[a, b] , Cn[a, b] and Cn0[a, b]) is a function whose argument is element of V and whose value is real number. We normally write as I : V ® R, where R is the set of all real numbers. 

Example: For the above second order ODE we can associate a functional 

I[u]: C10[a, b] ® R
as below: 
I[u] =  ó
õ
1

0

{p(x) [u¢(x)]2+ q(x)[u(x)]2 - 2f(x)u(x)} dx

One can show that one way to solve this ODE is to find u0(x) Î C10[a, b], so that I reaches its minimum at u0(x). That is 

I[u0(x)] £ I[u(x)] for all u(x) in C10[a, b]. We call u0(x) a minimizer of I[u]. If the ODE has an unique solution, then the minimizer is also unique if it exists. 

The Finite Element Method helps us to solve the minizing problem approximately. Instead of finding  minimizer over the infinite space C10[a, b] , we search for the minimizer over a finite dimensional subspace V, usually a linear space of all polynomials with certain specific properties. This way, as we will see, will transform a problem of solving differential equation into a problem of solving an Algebraic system of linear equations as we have done in PreCalculus, Algebra, and Linear Algebra.