Variational Formulation & the Galerkin Method

Institute of Structural Engineering

Method of Finite Elements I

Variational Formulation & the Galerkin Method

Dr. Konstantinos Agathos

Institute of Structural Engineering ETH Zürich

8 April 2020



Today’s Lecture Contents:

• Introduction

• Strong form

– Strong form of a 1D bar

– Strong form solution for a 1D bar

• Weak form

– Potential minimization

– Principle of Virtual Work

– Galerkin Method

– Weak form solution



Learning goals:

• Understanding strong and weak forms through a simple example

• Demonstrating the equivalence between the two as well as the differences

• Demonstrating the equivalence between alternative weak formulations

• Understanding the advantages of weak formulations in the development of approximate solution methods



FEM applications:

• Simple structural mechanics



FEM applications:

• Not so simple structural mechanics



FEM applications:

• Solid mechanics – Geometrical nonlinearities

Experimental Test – Bertoldi (2010) Simulation (Aguzzi & Zaccherini)



FEM applications:

• Solid mechanics - Fracture



FEM applications:

• Fluid mechanics

Openengineering.com



FEM applications:

• Multiphysics - fluid structure interaction

Boncoraglio et al. 2020



FEM for PDEsThe wide range of applicability is attributed to the fact that FEM is essentially a tool

for solving partial differential equations (PDEs), for instance:

Bernoulli-Euler beam equilibrium equations:

2D elasticity (Navier equations):

Laplace equation:

4

4z y

d wEI f x

dx

0

0

xyxxx

xy yy

y

Fx y

Fx y

2 2

2 2

, ,0

x y x y

x y



©Carlos Felippa

FEM across dimensions

http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.Slides.d/IFEM.Ch06.Slides.pdf



FEM for PDEsA series of steps is typically followed by all FE methods:

Strong form

Weak form

Discretization

Solution

Elasticity, heat conduction,

Navier stokes, etc.

Potential minimization,

principle of virtual work,

Galerkin method, etc.

Isoparametric formulation,

Lagrange polynomials,

Hermite polynomials, etc.

Linear solvers, Newton

Raphson, arc length, etc.

Today’s main

focus!

Only a few simple

examples



Strong formGeneral form of 2D second order partial differential equations (PDEs)

2 2 2

2 2, 2 ( , ) , , , , ,

u u u u uA x y B x y C x y x y u

x x y y x y

Dirichlet or essential boundary conditions

2

0 elliptic

0 parabolic

0 hyperbolic

B AC

Categorization:

0 0 0,u x y u

,m n

m n

uu x u

x y

Boundary conditions (BCs):

Neumann of natural boundary conditions



Strong form – 1D bar

Assumptions:

• Constant cross section

• Linear elastic material

• Loads applied at the centroid of the cross section

• Arbitrary distributed load

Illustrative example – 1D bar with a distributed and an end load



Equilibrium of forces along directionx

0

0

A x F x dx A x dx

d xA x F x dx A x A dx

dx

A x

F x dx A x d x

A dxdx

0

d xA F x

dx

Strong form



Constitutive Equation (Hooke’s law)

duE E

dx

Kinematic equation (strain definition)

du

dx

Equilibrium Equation

2

2

d xA dx F x

dx

d uEA F x

dx

Strong form



Neumann (natural) boundary conditions

( )x L x L

du du RA x L R AE R

dx dx EA

Dirichlet (essential) boundary conditions

0 0u x

Equilibrium Equation

2

2

d uEA F x

dx

Strong form



Strong form solution

Assuming no distributed load:

0F x

2

20

d uEA

dx

x L

du R

dx EA

0 0u x




Assuming no distributed load:

0F x

2

20

d uEA

dx

x L

du R

dx EA

0 0u x

The solution should be of the form: 0 1u x c c x

Equilibrium is satisfied: 2

0 1

20

d c c xEA

dx

From the essential B.C.: 0 0c

From the natural B.C.:1

Rc

EA

Final form of the solution: R

u x xEA




Similarly, assuming a linear

distributed load and no end load:

0

F x ax

R

2

2

d uEA ax

dx

0x L

du

dx

0 0u x






0

F x ax

R

2

2

d uEA ax

dx

0x L

du

dx

0 0u x

The solution should be of the form: 2 3

0 1 2 3u x c c x c x c x

From the equilibrium equation:

2 2 3

0 1 2 3

2 32

2 3

2 6

0,6

d c c x c x c x axEA ax c c x

dx EA

ac c

EA



2 3

0 1 2 3u x c c x c x c x




0

F x ax

R

2

2

d uEA ax

dx

0x L

du

dx

0 0u x

The solution should be of the form:

From the equilibrium equation: 3

0 16

au x c c x x

EA

From the essential B.C.: 0 0c

From the natural B.C.:2 2

1 102 2x L

du aL aLc c

dx EA EA

Final form of the solution: 2

3

2 6

aL au x x x

EA EA




• Analytical solutions satisfy the PDE at every point of the domain, thus

the PDE is called the “strong” form of the problem

• It is not possible to derive such solutions for complex combinations of

PDEs, geometries and BCs

• Typically, approximate, numerical solutions are sought for

• In what follows, some tools to systematically derive such solutions will

be presented



Potential minimization

Principle of stationary potential energy

“Among all admissible configurations of a conservative system, those

that satisfy the equations of equilibrium, make the potential energy

stationary with respect to small admissible variations of the

displacements.”

where:

Admissible are all configurations that satisfy compatibility and

essential boundary conditions




For the problem of the bar, potential energy has two components:

• Internal strain energy

• Work produced by the distributed and concentrated loads:

2 2

0 0

1 1 1 1

2 2 2 2

L L

V V A

A

du du du duU dV E dV E dAdx EA dx

dx dx dx dx

0

( )

L

W F x u x dx Ru L




Total potential energy:

In order to obtain the equilibrium configuration, the minimum of the above

expression should be sought

2

0 0

1( )

2

L Ldu

U W EA dx F x u x dx Ru Ldx




The expression derived for the total potential energy is a functional, i.e. a mapping

from a function space to the real numbers.

In simpler words:

• Functions take numbers as input and return numbers as output, i.e. they map

numbers to numbers

• Functionals take functions as input and return numbers as output, i.e. they map

functions to numbers

To minimize a functional, we need some additional definitions



For the derivatives of the variation, the following applies:

The derivative of the variation is equal to the variation of the derivative of a function!


Variation of a function

The variation of a function is defined as an arbitrary and

sufficiently smooth function that vanishes at the points where

boundary conditions are applied:

n n n

n n n

d d u d u

dx dx dx

u

u x

u x




Variation of a functional

The variation of a functional of a function and its derivatives

is defined as:

0

, , , , , , , ,lim

n nF u u u u F u u u uF

F u

, , , nu u u

• Similar to functions, the variations of functionals vanish at stationary points.

• The variational operator has several common properties to differentiation, for

instance:

• Also:

,F Q F Q FQ F Q Q F

F x dx F x dx




Total potential energy:

Variation of the total potential energy:

2

0 0

1( )

2

L Ldu

EA dx F x u x dx Ru Ldx

In order to minimize the potential

energy functional, its variation

should be computed and set to zero

0 0

( )

L Ldu du

EA dx F x u x dx R u Ldx dx




Stationarity condition:

In order to minimize the potential

energy functional, its variation

should be computed and set to zero

0 0

0 0

0

( ) 0

( )

L L

L L

du duEA dx F x u x dx R u L

dx dx

du duEA dx u x F x dx u L R

dx dx



Principle of Virtual Work

Principle of stationary potential energy:

0 0

0 0

( )

or

( )

L L

L L

du duEA dx u x F x dx u L R

dx dx

du d uEA dx u x F x dx u L R

dx dx

The above is equivalent to the Principle of Virtual Work if are considered as

virtual displacements u x



b b

a a

b b

x b x a

a a

b b

x b x a

a a

f g f g f g

f g dx f g f g dx

f g f g f g dx f g dx

f g dx f g f g f g dx

Reminder: Integration by parts


The principle of stationary potential energy is equivalent to the equilibrium equations and

natural BCs, to show that integration by parts is required.



Potential minimizationPrinciple of stationary potential energy:

0 0

( ) 0

L Ldu du

EA dx u x F x dx u L Rdx dx

Integration by parts for the first term:

2

2

0 0

0

0

0

L L

g g g

gf f f fx L x

x L x

du du du du d uEA dx EA u EA u EA u dx

dx dx dx dx dx

du duEA u L EA u

dx dx

2

2

0

Ld u

EA u dxdx

why?




2

2

0 0

2

2

0

( ) 0

0

L L

x L

L

x L

du d uEA u L EA u dx u x F x dx u L R

dx dx

d u duEA F x u dx EA R u L

dx dx

Combining the two parts:

Since both and are arbitrary, the equation only holds if:u u L

2

20

0x L

d uEA F x

dx

duEA R

dx

Equilibrium equation

Natural BC

The principle of minimum potential

energy is equivalent to the equilibrium

equations and natural BCs!



The Galerkin Method

2

2

d uEA F x

dx Differential equation →

Residual →

Weighted Residual →

2

20

d ur x EA F x

dx

0

0

L

w x r x dx

w xWhere is an arbitrary weight function that should vanish at the points where

essential BCs are applied, for the 1D bar case considered: 0 0w



2 2

2 2

0 0 0

0 0

L L Ld u d u

w EA F dx wEA dx wFdxdx dx

The Galerkin Method

Weighted Residual

To further process this expression, integration by parts is required



b b

a a

b b

x b x a

a a

b b

x b x a

a a

f g f g f g

f g dx f g f g dx

f g f g f g dx f g dx

f g dx f g f g f g dx

Reminder (again!): Integration by parts

The Galerkin Method



2

2

0

0

L

f f f

g g gx L x

d u du duwEA dx wEA wEA

dx dx dx

0

0

L

gf

L

dw duEA dx

dx dx

dw duw L R EA dx

dx dx

The Galerkin Method

Integration by parts of the first term of the weighted residual

0 0w

x L

duEA R

dx

Essential BC:

Weight:



The Galerkin Method

Substituting in the initial expression:

2

2

0 0

0 0

0 0

0

0

L L

L L

L L

d uwEA dx wFdx

dx

dw duw L R EA dx wFdx

dx dx

dw duEA dx wFdx w L R

dx dx



Weak form comparison

0 0

L Ldw du

EA dx wFdx w L Rdx dx

0 0

( )

L Ldu du

EA dx u F dx u L Rdx dx

0 0

( )

L Ldu d u

EA dx u F dx u L Rdx dx

Principle of stationary potential energy:

Principle of Virtual Work:

Galerkin method:

All weak formulations presented are equivalent:

u

u

w

displacement variation

virtual displacement

weight



Weak vs Strong form

0 0

0 0

L Ldw du

EA dx wFdx w L Rdx dx

u

Weak form:

2

2

0 0

x L

d uEA F

dx

du R

dx EA

u x

Strong form:

Natural BCs are part of the weak form

Highest derivative is of order 1

Degree of continuity required:

Equilibrium is satisfied in an integral,

“weak” sense

Natural BCs are explicitly imposed

Highest derivative is of order 2

Degree of continuity required:

Equilibrium is satisfied everywhere in

a “strong” sense

0C 1C



Weak form solution

A general process:

Assume a general form for the solution

Plug into weak form

Obtain unknown coefficients

In this process:

The problem formulation poses restrictions with respect to possible forms of the

solution

The weak form is much less restrictive than the corresponding strong form



Weak form solution - ExactLinear distributed load and no

end load:

, 0F x ax R

2

3

2 6

aL au x x x

EA EA

Analytical solution:

0 0

0 0, 0 0

L Ldw du

EA dx wFdxdx dx

u w

to be determined as part of the solution

Polynomial displacements and weights of cubic order are assumed:

2 3

0 1 2 3

2 3

0 1 2 3

i

i

u x a a x a x a x a

w x b b x b x b x b

Galerkin weak form:

arbitrary




Weak form solution - Exact

2

3

2 6

aL au x x x

EA EA

From weak form:

From essential BCs: 0 00 0, 0 0 0u w a b

2 3 2 3

1 2 3 1 2 3 2 3

1 2 3

0 0

L Ld b x b x b x d a x a x a xEA dx b x b x b x Fdx

dx dx




After a few rearrangements…

with:

1 1 2 2 3 3 0b I b I b I

2 32 2 3

1 1 2 3 1 2 3

0

3 3 4 42 3 2

2 1 2 3 1 2 3

0

4 4 5 52 3 5 3

3 1 2 3 1 2 3

0

2 33

4 32 4 6

3 2 4

3 93 6 9

2 5 5

L

L

L

ax aLI a a x a x dx La L a L a

EA EA

ax L L aLI a x a x a x dx L a a a

EA EA

ax L L aLI a x a x a x dx L a a a

EA EA


2

3

2 6

aL au x x x

EA EA




Since are arbitrary, the equation can only hold if1 1 2 2 3 3 0b I b I b I ib 1 2 3 0I I I

This results in a linear system of equations, whose solution yields the exact coefficients:

3

22 3

1 13 4 42

2 2

54 5 3 33

32

4 30

3 2 4

3 96

52 5

aLaLL L L EA

a a EAL L aL

L a aEA

a a aaLL L

L EAEA


2

3

2 6

aL au x x x

EA EA

why?





2

3

2 6

aL au x x x

EA EA

The weak form is equivalent to the differential equation

If the assumed form of the solution can represent the exact one, then the exact solution

will be obtained

What happens if the assumed form solution cannot represent exact one?



Weak form solution - Approximate


2

3

2 6

aL au x x x

EA EA

Linear displacements and weights are assumed:

0 1

0 1

u x a a x

w x b b x

From weak form:

From essential BCs: 0 00 0, 0 0 0u w a b

1 1

1 1

0 0

L Ld b x d a xEA dx b x Fdx b

dx dx 1 1EALa b

3 2

13 3

aL aLa

EA



Weak form solution - ApproximateExact solution:

2

3

aLu x x

EA

23

2 6

aL au x x x

EA EA

Approximate solution:

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2

Length (m)

Strong Form Galerkin-Cubic order Galerkin-Linear

Variational Formulation & the Galerkin Method

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