Introduction to the Finite Element Method - Lecture 02

Lecture 02 

pk285@ 

Boundary 

Value 

Problems 

Finite Element 

Method 

Function Spaces 

Weak Forms 

Introduction to the Finite Element Method 


P.S. Koutsourelakis 

pk285@cornell.edu 

369 Hollister Hall 

September 3 2008

Boundary Value Problem 


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Boundary 

Value 

Problems 




Weak Forms 

Example 

Consider a bar of length L, cross-sectional area A which is held 

fixed on the left end and pulled with a force F on the right end and 

stretched with a distributed force b(x) along its length. Whats will 

be the deformation of the bar u(x) at each point x? 

b(x) 

A 

u(x) 

L 

F 

Boundary Value Problem (BVP) 

EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (1) 

with boundary conditions: u(0) = 0, 

E A du 

dx | x=l = F



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Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (2) 

with boundary conditions: u(0) = 0, E A du 

dx | x=l = F 

Although it is straightforward to derive a closed-form 

solution (right?) things are not necessarily so if: 

elastic modulus varies E(x) 

cross-sectional area varies A(x) 

if we are considering two or three dimensional versions 

with arbitrary boundary shapes/conditions. 

we need a general computational method that is able to 

produce efficiently, accurate solutions of BVPs.



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Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (3) 

Approximate derivatives with finite differences, i.e.: 

du 

dx = lim u(x + h) − u(x) u(x + h) − u(x) 

≈ for 0 < h



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Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (4) 

define N grid points x i = i h where h = L N 

and let 

u i = u(x i ) 

if h is small enough then I can approximate: 

d 2 u(x) 

dx 2 | x=xi ≈ u i+1 − 2u i + u i−1 

h 2 (5) 

substitute in Equation (4) for x = x i , ∀i to obtain N 

algebraic equations w.r.t N unknowns u i . 

this is the the Finite Difference Method

Finite Difference Method (FDM) 


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Boundary 

Value 

Problems 




Weak Forms 

E d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) 

⇓ 

(discretization) 

EA u i+1 − 2u i + u i−1 

h 2 + b(x i ) = 0 ∀x i , i = 1, 2,...,N 

Observe that in FDM we approximate the PDE itself 

FDM is still used in a wide range of problems and we 

will use it in time-dependent problems to discretize 

time derivatives. 

In the Finite Element Method (FEM) we approximate 

the solution of the PDE.

Finite Element Method 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

Roadmap to FEM approximations: 

1 We are going to define where we are going to to be looking 

for solutions, i.e. which function space 

2 We are going to reformulate the original problem, i.e. PDE 

and BC. 

3 We are going to show that this new form is actually 

equivalent to the original, i.e. any solution of the former is a 

solution of the latter and vice versa. 

4 We are going to show that the solution is unique. 

5 If that wasn’t enough, we are going to look at some 

equivalent forms which can be considered as special cases. 

6 We are going to propose ways to discretize these alternate 

forms.



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Boundary 

Value 

Problems 




Weak Forms 

Since we are going to be approximating solutions of PDEs 

i.e. functions, it makes sense to recap some of the basic 

function spaces and their properties. If Ω is an open subset 

of R (or R n ) in general, then: 

C(Ω) contains all functions defined on Ω which are 

continuous. 

C k (Ω) contains all functions defined on Ω which have 

continuous derivatives up to order k. 

Cb k(Ω) same as Ck (Ω) plus the function is bounded 

L 2 (Ω) contains all functions defined on Ω which are square 

integrable i.e.: ∫ 

u 2 (x) dx < +∞ (6) 

Ω 

H 1 (Ω) contains all functions in L 2 whose derivatives are also 

square integrable i.e.: 

∫ 

|du/dx| 2 (x) dx < +∞ (7) 

Ω



pk285@ 

Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (8) 

with boundary conditions: u(0) = u 0 , 

E A du 

dx | x=l = F 

We are going to look for solutions in the space S: 

∫ 

S = {u(x) : (0, L) → R|u(0) = u 0 , E A| du 

dx |2 (x) dx < +∞} 

(0,L) 

Observe that: 

u ∈ S satisfy exactly only one of the two boundary 

conditions. This BC is called essential. 

u ∈ S have finite strain energy! 

u ∈ S are continuous and bounded.



pk285@ 

Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (9) 

with boundary conditions: u(0) = u 0 , E A du 

dx | x=l = F 


∫ 

S = {u(x) : (0, L) → R|u(0) = u 0 , 

(0,L) 

E A| du 

dx |2 (x) dx < +∞} 

Observe that: 

The space S is much larger than what the PDE and BC 

would imply. 

Even though a 2nd order derivative of appears in 

Equation (9), we are looking for solutions that are 

guaranteed to have a 1st order derivative 

Even though a “force” BC must be satisfied, we are 

looking for solutions that are not a priori guaranteed to 

satisfy it.

Finite Element Method 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

Roadmap to FEM approximations: 

1 We are going to define where we are going to to be looking 

for solutions, i.e. which function space 

2 We are going to reformulate the original problem, i.e. PDE 

and BC. 

3 We are going to show that this new form is actually 

equivalent to the original, i.e. any solution of the former is a 

solution of the latter and vice versa. 

4 We are going to show that the solution is unique. 

5 If that wasn’t enough, we are going to look at some 

equivalent forms which can be considered as special cases. 

6 We are going to propose ways to discretize these alternate 

forms.

Weak Forms 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 


EA d 2 u(x) 

dx 2 + b(x) = 0 ∀x ∈ (0, L) (10) 

with boundary conditions: u(0) = u 0 , E A du 

dx | x=l = F 


∫ 

S = {u(x) : (0, L) → R|u(0) = u 0 , 

(0,L) 

E A| du 

dx |2 (x) dx < +∞} 

A u ∈ S will not satisfy Equation (10) exactly (unless it is the 

solution) and in general there will be a residual R(x) 

R(x) = EA d 2 u(x) 

dx 2 + b(x) (11) 

There will also be a residual R(L) = E A du 

dx | x=l − F because 

u ∈ S do not a priori satisfy this BC

Weak Forms 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

How can we make those residuals zero? 

R(x) = EA d 2 u(x) 

dx 2 + b(x) R(L) = F − EA du 

dx | x=l 

(Bubnov)-Galerkin approach 

Figure: Boris Galerkin (1871-1945)

Weak Forms 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

How can we make those residuals zero? 

R(x) = EA d 2 u(x) 

dx 2 + b(x) R(L) = F − EA du 

dx | x=l 

(Bubnov)-Galerkin or Weighted Residual approach: define 

another set of functions called weighting functions v ∈ V: 

∫ 

V = {v(x) : (0, L) → R|v(0) = 0, E A| dv 

dx |2 (x) dx < +∞} 

(0,L) 

Find u ∈ S such that for all v ∈ V: 

∫ L 

0 

v(x)R(x) dx + v(L)R(L) = 0 (12) 

Note that the residual is not zero in the STRONG sense i.e. 

R(x) = 0 ∀x but the condition is enforced WEAKLY as above.

Weak Forms 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

STRONG form: 

WEAK form: 

⎧ ∫ L 

⎨ 

0 

u ∈ S 

⎩ 

v ∈ V 

du dv 

EA 

dx 

⎧ 

⎨ 

⎩ 

EA d 2 u(x) 

dx 2 + b(x) = 0 

u(0) = u 0 

F = EA du 

dx | x=l 

dx dx = ∫ L 

A b(x)v(x)dx + v(L)F ∀v ∈ V 

0 

Does the weak form remind you of something? 

Principle of Virtual Work 

The necessary and sufficient condition for a system in 

equilibrium is that the work done by internal forces should be 

equal to the work done by externals loads for any 

kinematically acceptable virtual displacement

Weak Forms 


pk285@ 

Boundary 

Value 

Problems 




Weak Forms 

WEAK form: 

⎧ ∫ L 

⎨ 

0 

u ∈ S 

⎩ 

v ∈ V 

du dv 

EA 

dx 

Principle of Virtual Work: 

dx dx = ∫ L 

A b(x)v(x)dx + v(L) F ∀v ∈ V 

0 

a virtual kinematically acceptable displacement v(x) is 

one that does not violate displacement boundary 

conditions u(0) = u 0 , i.e. v(0) = 0. 

Work of internal forces: 

∫ 

W int (v) = A 

σ u (x) 

} {{ } 

stress from u 

∫ 

ǫ v (x) dx = A 

} {{ } 

strain from v 

Work of external forces: 

∫ 

W ext (v) = A b(x) v(x) dx + F v(l) 

E du 

dx 

dv 

dx dx

Introduction to the Finite Element Method - Lecture 02

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