lecture 03

Boundary Value Problems Finite Element Method

Introduction to the Finite Element MethodLecture 02 & 03

P.S. Koutsourelakis

[email protected] Hollister Hall

August 30 - September 13 2010Last Updated: September 13, 2010

[email protected] Cornell University

Lecture 02 & 03


Boundary Value Problem

Example

Consider a bar of length L, cross-sectional area A which is held fixed on theleft end and pulled with a force F on the right end and stretched with adistributed force b(x) along its length. Whats will be the deformation of thebar u(x) at each point x?

A

b(x)

Fu(x)

LBoundary Value Problem (BVP)

EAd2 u(x)

dx2+ b(x) = 0 x (0,L)

with boundary conditions: u(0) = 0, E A dudx |x=L = F


Lecture 02 & 03



Boundary Value Problem (BVP)

EAd2 u(x)

dx2+ b(x) = 0 x (0,L)

with boundary conditions: u(0) = 0, E A dudx |x=L = FAlthough 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 witharbitrary boundary shapes/conditions.

we need a general computational method that is able to produceefficiently, accurate solutions of BVPs.


Lecture 02 & 03




EAd2 u(x)

dx2+ b(x) = 0 x (0,L)

Approximate derivatives with finite differences, i.e.:

dudx = limh0

u(x+h/2)u(xh/2)h

u(x+h/2)u(xh/2)h for 0 < h




EAd2 u(x)

dx2+ b(x) = 0 x (0,L) (1)

define N grid points xi = i h where h = LN and let ui = u(xi)

if h is small enough then I can approximate:

d2 u(x)dx2

|x=xi ui+1 2ui + ui1

h2

substitute in Equation (1) for x = xi , i to obtain N algebraicequations w.r.t N unknowns ui .

this is the the Finite Difference Method


Lecture 02 & 03


Finite Difference Method (FDM)

Ed2 u(x)

dx2+ b(x) = 0 x (0, L)

(discretization)

EAui+1 2ui + ui1

h2+ b(xi) = 0 xi , 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 intime-dependent problems to discretize time-derivatives.

In the Finite Element Method (FEM) we approximate the solutionof an equivalent form of the PDE.


Lecture 02 & 03


Finite Element Method

Figure: The FREU(E)D roadmap to Finite Elements


Lecture 02 & 03



Roadmap to FEM approximations:1 Function Spaces: We are going to define where we are going to to be

looking for solutions, i.e. which function space2 Reformulate: We are going to reformulate the original problem, i.e.

the PDE and Boundary Conditions - Weak form.3 Equivalence: 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 ofthe latter and vice versa.

4 Unique: We are going to show that the solution is unique.5 Equivalence 2: We are going to look at some equivalent forms which

can be considered as special cases.6 Discretization: We are going to propose ways to discretize all these

equivalent [email protected] Cornell UniversityLecture 02 & 03


Function Spaces

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 theirproperties. If is an open subset of R (or Rn) in general, then:C() contains all functions defined on which are continuous.Ck () contains all functions defined on which have continuousderivatives up to order k .Ckb () same as C

k () plus the function is boundedL2() contains all functions defined on which are square integrablei.e.:

u2(x) dx < +

H1() contains all functions in L2 whose derivatives are also squareintegrable i.e.:

|du/dx |2(x) dx < +


Lecture 02 & 03


Function Spaces


EAd2 u(x)

dx2+ b(x) = 0 x (0, L)

with boundary conditions: u(0) = u0, E A dudx |x=L = F

We are going to look for solutions in the trial or candidate solution spaceS:

S = {u(x) : (0, L) R|u(0) = u0,

(0,L)E A

du(x)dx

2

dx < +}

Observe that:

u S satisfy exactly only one of the two boundary conditions. ThisBC is called essential.u S have finite strain energy!u S are continuous and bounded.


Lecture 02 & 03


Function Spaces


EAd2 u(x)

dx2+ b(x) = 0 x (0, L)

with boundary conditions: u(0) = u0, E A dudx |x=L = FWe are going to look for solutions in the space S:

S = {u(x) : (0, L) R|u(0) = u0,

(0,L)E A

du(x)dx

2

dx < +}

Observe that:The space S is much larger than what the PDE and BC wouldimply.Even though a 2nd order derivative of appears in the BVP, we arelooking for solutions that are guaranteed to have a 1st orderderivative.Even though a force BC must be satisfied, we are looking forsolutions that are not a priori guaranteed to satisfy it.

we have RELAXED already the original [email protected] Cornell UniversityLecture 02 & 03



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 theoriginal, i.e. any solution of the former is a solution of the latter and viceversa.

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

5 If that wasnt enough, we are going to look at some equivalent formswhich can be considered as special cases.

6 We are going to propose ways to discretize all these equivalent forms.


Lecture 02 & 03


Weak FormsBoundary Value Problem (BVP)

EAd2 u(x)

dx2+ b(x) = 0 x (0, L) (2)

with boundary conditions: u(0) = u0, E A dudx |x=L = F

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

S = {u(x) : (0, L) R|u(0) = u0,

(0,L)E A|

dudx

|2(x) dx < +}

An arbitrary u S will not satisfy Equation (2) exactly (unless it is thesolution) and in general there will be a residual R(x)

R(x) = EAd2 u(x)

dx2+ b(x) 6= 0

There will also be a residual R(L) = E A dudx |x=L F because u S donot a priori satisfy this BC


Lecture 02 & 03


Weak Forms

How can we make those residuals zero?

R(x) = EAd2 u(x)

dx2+ b(x) R(L) = F EA

dudx

|x=L

(Bubnov)-Galerkin approach

Figure: Boris Galerkin (1871-1945)


Lecture 02 & 03


Weak Forms

How can we make those residuals zero?

R(x) = EAd2 u(x)

dx2+ b(x) R(L) = F EA

dudx

|x=L

(Bubnov)-Galerkin or Weighted Residual approach: define another setof functions called weight functions w W:

W = {w(x) : (0, L) R|w(0) = 0,

(0,L)E A|

dw(x)dx

|2 dx < +}

Find u S such that for all w W:

L0 w(x)R(x) dx = 0

w(L)R(L) = w(L)(F EA dudx |x=L

)= 0

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.


Lecture 02 & 03


Weak Forms

After some mainipulation (see Integration by parts)

STRONG form:

EA d2 u(x)dx2

+ b(x) = 0u(0) = u0F = EA dudx |x=L

WEAK form:{ L

0 EAdudx

dwdx dx =

L0 b(x)w(x)dx + w(L)F w W

u S

Does the weak form remind you of something?

Principle of Virtual WorkThe necessary and sufficient condition for a system in equilibrium is thatthe work done by internal forces should be equal to the work done byexternals loads for any kinematically acceptable virtual displacement


Lecture 02 & 03


Weak Forms

WEAK form:{ L

0 EAdudx

dwdx dx =

L0 b(x)w(x)dx + w(L) F w W

u S

Principle of Virtual Work:a virtual kinematically acceptable displacement w(x) is one thatdoes not violate displacement boundary conditions u(0) = u0, i.e.w(0) = 0.Work of internal forces:

Wint(w) = A

u(x)

stress from u

w (x)

strain from w

dx = A

Edudx

dwdx

dx

Work of external forces:

Wext(w) =

b(x) w(x) dx + F w(L)


Lecture 02 & 03











Lecture 02 & 03


Equivalence of Strong and Weak forms


A) Strong Weak:

EA u,xx + b(x) = 0

(EA u,xx + b(x))w(x)dx = 0 w W

[EA u,x v ]

L0

EA u,x v,x dx

+

b(x)w(x)dx = 0

EAu,x(L)w(L)

EA u,x v,x dx

+

b(x)w(x)dx = 0 (since w(0) = 0)

Fw(L)

EA u,x v,x dx

+

b(x)w(x)dx = 0 (since EAu,x(L) = F )


Lecture 02 & 03




B) Weak Strong:

L

0EA

dudx

dwdx

dx = L

0b(x)w(x)dx + w(L)F w W

w(L)(F EAu,x(L)) + [EA u,x w ]

L0

EA u,x w,x dx

+

b(x)w(x)dx = 0 (since w(0) = 0)

w(L)(F EAu,x(L)) +

(EA u,xx + b(x))w(x)dx = 0 w W

Take w(x) = (EA u,xx + b(x)) x(L x)


Lecture 02 & 03












Lecture 02 & 03


Uniqueness

Uniqueness

Suppose u1, u2 S that satisfy the weak form: L

0 EAdujdx

dwdx dx =

L0 b(x)w(x)dx + w(L) F v W, j = 1, 2

L

0 EA(du1dx

du2dx )

dwdx = 0

Let w = u1 u2 (is this a legitimate member of W?) L

0EA(

du1dx

du2dx

)2dx = 0

du1dx

du2dx

= 0

u1(x) u2(x) = C

u1(x) = u2(x) (since u1(0) = u2(0) = u0)

Note what happens if only force BC are [email protected] Cornell University

Lecture 02 & 03


Uniqueness










Lecture 02 & 03


Alternative/Special Forms

Alternative/Special FormsLets rewrite the weak form (i.e principle of virtual work in mechanics):

L0 EA

dudx

dwdx dx =

L0 b(x)w(x)dx + w(L) F v W

u Sw W

more compactly as:

a(u,w) = f (w) w W

where:- a(u,w) =

L0 EA

dudx

dwdx dx

- f (w) = L

0 b(x)w(x)dx + w(L) FDefine the functional (i.e. a function of functions) (u):

(u) =12

a(u, u) f (u)

=12

EA|dudx

|2dx

b(x)u(x)dx u(L) F


Lecture 02 & 03



Alternative/Special FormsPrinciple of Minimum Potential Energy:The equilibrium configuration u minimizes the potential energy of thesystemDoes the solution of the weak form minimize the potential energy?Consider configurations u = u + w where u S is the solution of theweak form and w W (note that u S). Then

(u + w) =12

a(u + w , u + v) f (u + w)

=

(12

a(u, u) f (u))

+ (a(u,w) f (w))

+ a(w ,w)

= (u) + 0 + a(w ,w)

(u)

can you show the minimizer u is [email protected] Cornell University

Lecture 02 & 03




STRONG form WEAK form Min. Pot. Energy

the usefulness of the Min. Potential Energy Principle is (at least)twofold:

By formulating the problem as an optimization problem we canmake use of all these methods/algorithms that have beendevelopped to solve them.Since (u) (u) u S we have an error bound. No matterwhat approximation u of the solution we come up with, its potentialenergy will always be greater than the potential energy of the truesolution u.


Lecture 02 & 03












Lecture 02 & 03


Discretization

Discretization

We started with the PDE and BCs (STRONG form):

EA d2 u(x)dx2

+ b(x) = 0u(0) = u0F = EA dudx |x=L

and derived an equivalent WEAK form (Principle of Virtual Work):

L0 EA

dudx

dwdx dx =

L0 b(x)w(x)dx + w(L) F v W

u Sw W

and another one (Principle of Minimum Potential Energy)

u = argminuS(u) =12

EA|dudx

|2dx

A b(x)u(x)dx u(L) F


Lecture 02 & 03


Discretization

Discretization

We now have to work with integrals (WEAK form) instead of differential(STRONG form) operators which are more stable in numericaloperations.

In FEM we discretize the solution in order to find approximations, i.e.discretize the functions appearing in the weak forms.

Let S S and W W finite dimensional subsets (approximations) tothe function spaces of interest.


Lecture 02 & 03


Discretization

Discretization

Let S S and W W finite dimensional subsets (approximations) tothe function spaces of interest.

Example:W = span{Nj(x), j = 1, . . . , n}

i.e. any w(x) W can be written as:

w(x) = c1N1(x) + c2N2(x) + . . .+ cnNn(x)W

WN1(x)

N2(x)

Can we select any Nj(x)? No we have to make sure that Nj(0) = 0 and|dNj(x)/dx |2dx < + since:

W = {w(x) : (0, L) R|w(0) = 0,

(0,L)E A|

dwdx

|2(x) dx < +}


Lecture 02 & 03


Discretization

Discretization

What about S S?

S = {u(x) : (0, L) R|u(0) = u0,

(0,L)E A|

dudx

|2(x) dx < +}

We can define a similar approximation using different basis functions, orwe can simply add a N0(x) such that N0(0) = u0 and considerapproximations:

u(x) = N0(x) + d1N1(x) + d2N2(x) + . . .+ dnNn(x)

such that:N0(0) = u0

Note that this way: u(0) = u0Note that when u0 = 0 trivially N0(x) = 0


Lecture 02 & 03


Discretization

DiscretizationDiscretize Principle of Minimum Potential Energy:

(u) =12

EA|dudx

|2dx L

0A b(x)u(x)dx u(L) F

or more succintly:(u) =

12

a(u, u) f (u)

Substituting for u:

(u) =12

a(N0(x) +n

i=1

diNi(x), N0(x) +n

j=1

djNj(x)

f (N0(x) +n

j=1

djNj(x)) =12

n

i=1

n

j=1

didj a(Ni(x),Nj(x))

+n

j=1

dj a(N0(x),Nj(x))n

j=1

dj f (Nj(x)) + f (N0(x))


Lecture 02 & 03


Discretization

DiscretizationDiscretized Principle of Minimum Potential Energy:

(u) =12

n

i=1

n

j=1

didj a(Ni(x),Nj(x)) +n

j=1

dj a(N0(x),Nj(x))

n

j=1

dj f (Nj(x)) + f (N0(x))

To minimize w.r.t. u:

dj= 0 j = 1, 2, . . . , n

which leads to n equations:

n

i=1

di a(Ni(x),Nj(x)) + a(N0(x),Nj(x)) f (Nj(x)) j


Lecture 02 & 03


Discretization

DiscretizationDiscretized Principle of Minimum Potential Energy:

n

i=1

di a(Ni(x),Nj(x)) + a(N0(x),Nj(x)) f (Nj(x)) j

or:K d = F

where:

K =

a(N1(x),N1(x)) . an(N1(x),Nn(x)a(N2(x),N1(x)) . an(N2(x),Nn(x)

. . .a(Nn(x),N1(x)) . an(Nn(x),Nn(x)

F T = [f (N1(x)) a(N0(x),N1(x)), .., f (Nn(x)) a(N0(x),Nn(x))]

This is also called the Ritz-Galerkin method


Lecture 02 & 03


Discretization

DiscretizationExample:

EAd2 u(x)

dx2+ b(x) = 0 u(0) = u0, E A

dudx

|x=L = F

with: E = 1, A = 1, L = 1, u0 = 0, F = 1/2, b(x) = x and solution:

u(x) = x3/6

For n = 2 and:

N1(x) = x N2(x) = x2 u(x) = d1 x + d2 x

2

Since u0, trivially N0(x) = 0

K d = F[ 1

0 1 1dx 1

0 1 2xdx 10 1 2xdx

10 2x 2xdx

] [d1d2

]

=

[ 10 (x) xdx + 1/2 1

0 (x) x2dx + 1/2

]


Lecture 02 & 03


Discretization

Discretization

Example:

EAd2 u(x)

dx2+ b(x) = 0 u(0) = u0, E A

dudx

|x=L = F

with: E = 1, A = 1, L = 1, u0 = 0, F = 1/2, b(x) = x and solution:

u(x) = x3/6

[1 11 4/3

] [d1d2

]

=

[1/61/4

]

[d1d2

]

=

[1/12

1/4

]

u(x) = 1

12x +

14

x2


Lecture 02 & 03


Discretization

Discretization

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x

-0.05

0

0.05

0.1

0.15

0.2u(

x) -

disp

lace

men

t

exactapproximation

u(1/2) = u(1/2) and u(1) = u(1)

E A dudx |x=L =5

12 6= F =12

(u) = 7288 > (u) = 1

40


Lecture 02 & 03


Discretization

Discretization

Discretized WEAK form (Principle of Virtual Work): L

0EA

dudx

dwdx

dx = L

0b(x)w(x)dx + w(L) F w W

or more succintly:a(u, w) = f (w) w W

Substituting for u and w :

a(N0 +n

i=1

diNi(x),n

j=1

cjNj(x)) = f (n

j=1

cjNj(x))

i

j

di a(Ni(x),Nj(x))cj +

j

a(N0(x),Nj(x))

=

j

cj f (Nj(x))


Lecture 02 & 03


Discretization

Discretization

n

j=1

cj

(n

i=1

di a(Ni(x),Nj(x)) + a(N0(x),Nj(x)) f (Nj(x))

)

= 0

If this is to hold w(x) =

j cjNj(x), all the coefficient of cj aboveshould be 0:

n

i=1

di a(Ni(x),Nj(x)) + a(N0(x),Nj(x)) f (Nj(x)) = 0 j

K d = F

where:

K is a n n matrix such that: Ki,j = a(Ni (x), Nj (x)) = EA L0

dNidx

dNjdx dx

d is a n-dimensional vector of djF is a n-dimensional vector such that Fj = f (Nj (x)) a(N0(x), Nj (x))


Lecture 02 & 03


Discretization

Discretization

Observe that if the same basis functions are used for S and Wthen the resulting system from the Principle of Virtual Work:

is symmetriccoincides with the system arising from the Principle of MinimumPotential Energy.

This is also called the Bubnov-Galerkin method.Note however that the WEAK form does not require that thesame basis functions are used for S and W .For example in the previous problem, one could use:

u(x) = d1x + d2x2 and w(x) = c1x2 + c2x3

The resulting system will still be valid but non-symmetricThis is also called the Petrov-Galerkin method


Lecture 02 & 03

Boundary Value ProblemsFinite Element MethodEquivalence of Strong and Weak formsUniquenessAlternative/Special FormsDiscretization

lecture 03

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