Gockenbach M.S. Partial Differential Equations. Analytical and Numerical Methods

168

Chapter

5.

Boundary value problems

in

statics

Example 5.16. (Homogeneous Dirichlet conditions)

We

will solve

the

fol-

lowing

BVP,

applying

the

procedure given

above:

1.

We

have

already

seen that

the

eigenfunctions

of

the

negative second derivative

operator,

on the

interval

[0,2]

and

subject

to

Dirichlet conditions,

are

The

effect

of

multiplying

the

operator

by 3 is to

multiply

the

eigenvalues

by 3,

so the

eigenvalues

are

2.

We

write

the

unknown

solution

u as

The

problem

now

reduces

to

computing

0,1,0,3,0,3,

—

3.

The

Fourier sine

coefficients

of

f(x]

= x

3

on the

interval

[0,2]

are

4-

The

Fourier sine

coefficients

of

—3d?u/dx

2

are

just

Thus

5.

Finally,

the

differential

equation

can be

written

as

168 Chapter

5.

Boundary value problems

in

statics

Example

5.16.

(Homogeneous

Dirichlet

conditions)

We

will solve the fol-

lowing

BVP,

applying the procedure given above:

3

d2U

_ 3

-

dx

2

- X ,

u(O)

= 0,

u(2) =

O.

0<

x < 2,

(5.30)

1.

We

have already seen that the eigenfunctions

of

the negative second derivative

operator, on the interval

[0,2]

and subject to Dirichlet conditions, are

sin

(n;x),

n =

1,2,3,

....

The effect

of

multiplying the operator by 3 is to multiply the eigenvalues by 3,

so the eigenvalues are

3n

2

7f2

An

=

-4-'

n = 1,2,3,

....

2.

We

write the unknown solution u

as

00

u(x)

= L an sin

(n;x).

n=l

The problem now reduces to computing

al,

a2, a3,

....

3.

The Fourier sine coefficients

of

f(x)

= x

3

on the interval

[0,21

are

and

so

4.

The Fourier sine coefficients

of

-3~u/dx2

are

just

Thus

5.

Finally, the differential equation can

be

written

as

5.3.

Solving

the BVP

using Fourier

series

169

so we

have

This

yields

and

so

is

the

solution

to

(5.30)

Example 5.17.

(Inhomogeneous

Dirichlet conditions)

solve

a BVP

with

inhomoqeneous

boundary conditions:

We

must

first

shift

the

data

to

transform

the

problem

to one

with homogeneous

boundary

conditions.

The

function

p(x]

— 3

—

Sx

satisfies

the

boundary

conditions,

so

we

define

v = u

—

p and

solve

the BVP

1. The

eigenfunctions

are now

and

the

eigenvalues

are

2.

We

write

the

solution

v in the

form

5.3.

Solving

the

BVP

using

Fourier

series

169

so

we

have

This yields

and so

is the solution to (5.30).

Example

5.17.

(Inhomogeneous

Dirichlet

conditions)

BVP

with inhomogeneous boundary conditions:

d

2

u

--

=

I-x

dx

2

'

u(O)

= 3,

u

(~)

=

-1.

1

0<

x <

2'

(5.31)

We

must

first shift the data to transform the problem to one with homogeneous

boundary conditions. The function p(x) = 3 -

8x

satisfies the boundary conditions,

so

we

define v = u - p and solve the B

VP

d

2

v

--=l-x

dx

2

'

v(O)

= 0,

v

(~)

=

O.

1.

The eigenfunctions

are

now

1

0<

x < 2'

sin (2mfx) , n =

1,2,3,

...

,

and the eigenvalues

are

2.

We write the solution v

in

the form

00

v(x)

= L

an

sin (2mfx).

n=l

170

Chapter

5.

Boundary

value problems

in

statics

Exercises

In

Exercises 1-4, solve

the

BVPs using

the

method

of

Fourier series,

shifting

the

data

if

necessary.

If

possible,

30

produce

a

graph

of the

computed solution

by

plotting

a

partial

Fourier series with enough terms

to

give

a

qualitatively

correct

graph.

(The

number

of

terms

can be

determined

by

trial

and

error;

if the

plot

no

longer

changes,

qualitatively, when more terms

are

included,

it can be

assumed

that

the

partial

series

contains enough terms.)

30

That

is, if you

have

the

necessary technology.

3. The

Fourier sine

coefficients

of

f(x)

= 1

—

x are

and

so

4-

The

Fourier sine

coefficients

of

—d?v/dx

2

are

and

so

5.

We

thus obtain

The

solution

v is

then

or

which

implies

that

This

yields

170

Chapter

5.

Boundary value problems

in

statics

3.

The Fourier sine coefficients

of

f(x)

= 1 - x

are

1

1/2

2-(-1)n

4

(1

- x) sin (2mrx) dx = , n =

1,2,3,

...

,

o

nn

and

so

00

2 -

(_1)n

1-

x = L sin (2nnx).

nn

n=l

4.

The Fourier sine coefficients

of

_~V/dX2

are

and

so

-

~~

(x) =

f:

4n

2

n

2

a

n

sin(2nnx).

n=l

5.

We

thus obtain

00

002_(_1)n

L 4n

2

n

2

a

n

sin(2nnx) = L sin (2nnx),

n=l

nn

which implies that

or

4

22

2-(-1)n

n n an = , n =

1,2,3,

...

,

nn

2-(-1)n

an = 4 3 3

,n

=

1,2,3,

....

nn

The solution v is then

This yields

Exercises

002_(_1)n.

v(x) = L 4 3 3 sm (2nnx).

n=l

n n

002_(_1)n

u(x)=3-8x+L

4

33

sin (2nnx).

n=l

n n

In Exercises 1-4, solve

the

BVPs using the method of Fourier series, shifting the

data

if

necessary.

If

possible,3o produce a graph of the computed solution by plotting a

partial Fourier series with enough terms

to

give a qualitatively correct graph. (The

number of terms can be determined by trial

and

error; if

the

plot no longer changes,

qualitatively, when more terms are included,

it

can be assumed

that

the

partial

series contains enough terms.)

30That

is,

if

you have

the

necessary technology.

5.

Solve (5.21)

to get

(5.22).

6.

Suppose

u

satisfies

5.3. Solving

the BVP

using Fourier

series

171

3. The

results

of

Exercises 5.2.2

and

5.2.3

will

be

useful

for

these problems:

4. The

results

of

Exercises 5.2.2

and

5.2.3

will

be

useful

for

these problems:

and the

Fourier sine series

of u is

Show

that

it is not

possible

to

represent

the

Fourier sine

coefficients

of

—d?u/dx

2

in

terms

of

61,6

2

,

&3,

—

This

shows

that

it is not

possible

to use the

"wrong"

eigenfunctions

(that

is,

eigenfunctions corresponding

to

different

boundary

conditions)

to

solve

a

BVP.

7.

Suppose

u

satisfies

and has

Fourier

quarter-wave

sine series

5.3. Solving the BVP using Fourier

series

1.

(a) -

~:~

= 1,

u(O)

=

u(l)

= 0

(b) -

~:~

= 1,

u(O)

= 0,

u(l)

= 1

2.

(a) -~ = eX,

u(O)

=

u(l)

= 0

(b)

-~:~

= eX,

u(O)

=

u(l)

= 1

3.

The results of Exercises 5.2.2 and 5.2.3 will be useful for these problems:

(

a) -

d

2

u = X

u(O)

= 0

du

(1)

= 0

dx2'

,

dx

(b)

-~

= x,

~~(O)

=

0,

u(l)

= 1

(c)

-~+2u=l,

u(O)=u(l)=O

(

d) -

d

2

u + u - 0

u(O)

- 0

u(l)

- 1

dX2

-, -,

-

4.

The

results of Exercises 5.2.2 and 5.2.3 will be useful for these problems:

(a)

-~:~

= x

2

,

u(O)

=

u(l)

= 0

(b)

-~

+

2u

=

~

-

x,

u(O)

= u(2) = 0

(c)

-~:~

=x+sin(1fx),

u(O)

=

~~(1)

=0

(d)

-~:~

=

f(x),

~~(O)

=

u(l)

=

0,

where

5.

Solve (5.21) to get (5.22).

6.

Suppose u satisfies

f(x)

=

{I,

0 < x <

~,

0,

~

< x <

1.

du

u(O)

= 0,

dx

(f) = 0,

and the Fourier sine series of u

is

171

Show

that

it

is not possible to represent the Fourier sine coefficients of

-~u/

dx

2

in terms of b

1

,

b

2

,

b

3

, • .

..

This shows

that

it

is

not possible

to

use

the

"wrong"

eigenfunctions

(that

is, eigenfunctions corresponding

to

different boundary

conditions)

to

solve a BVP.

7.

Suppose u satisfies

du

u(O)

= 0,

dx

(£)

= 0,

and has Fourier quarter-wave sine series

~

.

((2n

-1)1fX)

u(x)

=

~ansm

2£

.

(Hint:

The

computation

of the

Fourier

coefficients

of

—Td?u/dx

2

is

similar

to

(5.23)).

8.

Consider

an

aluminum

bar of

length

1 m and

radius

1 cm.

Suppose

that

the

side

of the bar is

perfectly insulated,

the

ends

of the bar are

placed

in ice

baths,

and

heat

energy

is

added throughout

the

interior

of the bar at a

constant

rate

of

0.001

W/cm

3

.

The

thermal conductivity

of the

aluminum alloy

is

1.5W/(cmK).

Find

and

graph

the

steady-state

temperature

of the

bar.

Use

the

Fourier series method.

9.

Repeat

the

previous exercise, assuming

that

the

right

end of the bar is

per-

fectly

insulated

and the

left

is

placed

in an ice

bath.

10.

Consider

the

string

of

Example 5.12. Suppose

that

the

right

end of the

string

is

free

to

move vertically (along

a

frictionless pole,

for

example),

and a

pressure

of

2200

dynes

per

centimeter,

in the

upward direction,

is

applied along

the

string

(recall

that

a

dyne

is a

unit

of

force—one

dyne equals

one

gram-centimeter

per

square

second).

What

is the

equilibrium displacement

of the

string?

How

does

the

answer change

if the

force

due to

gravity

is

taken into account?

5.4

Finite

element methods

for

BVPs

As

we

mentioned near

the end of the

last section,

the

primary utility

of the

Fourier

series

method

is for

problems with constant

coefficients,

in

which case

the

eigenpairs

can be

found

explicitly. (For problems

in two or

three dimensions,

in

order

for the

eigenfunctions

to be

explicitly computable,

it is

also necessary

that

the

domain

on

which

the

equation

is to be

solved

be

geometrically simple.

We

discuss this

further

in

Chapter

8.) For

problems with nonconstant

coefficients,

it is

possible

to

perform

analysis

to

show

that

the

Fourier series method applies

in

principle—the

eigenfunctions

exist, they

are

orthogonal,

and so

forth.

The

reader

can

consult

Chapter

5 of

Haberman [22]

for an

elementary introduction

to

this kind

of

analysis,

which

we

also discuss

further

in

Section 9.7. However, without explicit formulas

for

the

eigenfunctions,

it is not

easy

to

apply

the

Fourier series method.

These remarks apply,

for

example,

to the BVP

(where

k is a

positive

function).

The

operator

K

:

C?JO,£]

—>

C[Q,l]

defined

by

172

Chapter

5.

Boundary

value

problems

in

statics

Show

that

the

Fourier series

of

—Td?u/dx

2

is

172

Chapter

5.

Boundary value problems in statics

Show

that

the

Fourier series of

-Td

2

ufdx

2

is

_Td

2

u(

) _

~

T(2n

-

1)27f2

.

((2n

-l)7fX)

dx2

x -

L..J

4£2

an

sm

2f

.

n=l

(Hint:

The

computation

ofthe

Fourier coefficients of

-Td

2

ufdx

2

is

similar

to

(5.23)).

8. Consider

an

aluminum

bar

of length 1 m

and

radius 1 cm. Suppose

that

the

side of the

bar

is

perfectly insulated,

the

ends of

the

bar

are placed in ice baths,

and

heat

energy is added throughout

the

interior of

the

bar

at

a constant

rate

of 0.001 W

fcm

3

.

The

thermal conductivity of the aluminum alloy is

1.5W/(cmK).

Find

and graph

the

steady-state

temperature

of

the

bar. Use

the

Fourier series method.

9. Repeat

the

previous exercise, assuming

that

the

right end of

the

bar

is

per-

fectly insulated

and

the

left is placed in

an

ice

bath.

10. Consider

the

string of Example 5.12. Suppose

that

the right end of

the

string is

free

to

move vertically (along a frictionless pole, for example),

and

a pressure of

2200 dynes per centimeter, in

the

upward direction, is applied along

the

string

(recall

that

a dyne is a unit of

force-one

dyne equals one gram-centimeter

per

square second).

What

is

the

equilibrium displacement of

the

string? How

does

the

answer change

if

the

force due

to

gravity is taken into account?

5.4 Finite element methods

for

BVPs

As

we

mentioned near the end of

the

last section,

the

primary utility of the Fourier

series method

is

for problems with constant coefficients, in which case

the

eigenpairs

can

be

found explicitly. (For problems in two or three dimensions, in order for

the

eigenfunctions to

be

explicitly computable,

it

is also necessary

that

the

domain

on which

the

equation

is

to

be solved be geometrically simple.

We

discuss this

further in

Chapter

8.) For problems with nonconstant coefficients,

it

is

possible

to

perform analysis

to

show

that

the

Fourier series method applies in

principle-the

eigenfunctions exist, they are orthogonal,

and

so forth. The reader can consult

Chapter

5 of Haberman

[22]

for

an

elementary introduction

to

this kind of analysis,

which

we

also discuss further in Section 9.7. However, without explicit formulas for

the

eigenfunctions, it is

not

easy

to

apply

the

Fourier series method.

These remarks apply, for example,

to

the

BVP

-

d~

(k(X)~:)

=

f(x),

0 < x <

£,

u(O)

=

0,

(5.32)

u(£) = 0

(where k is a positive function).

The

operator

K:

Cb[O,£]

-t

C[O,£]

defined by

Ku

=

--

k(x)-

d (

dU)

dx dx

(5.33)

5.4.

Finite

element

methods

for

BVPs

173

is

symmetric (see Exercise

1),

and

there exists

an

orthogonal sequence

of

eigenfunc-

tions. However, when

the

coefficient

k(x]

is not a

constant,

there

is no

simple

way

to find

these eigenfunctions. Indeed, computing

the

eigenfunctions requires much

more work

than

solving

the

original BVP.

Because

of the

limitations

of the

Fourier series approach,

we now

introduce

the

finite

element method,

one of the

most

powerful

methods

for

approximating solutions

to

PDEs.

The finite

element method

can

handle both variable

coefficients

and,

in

multiple

spatial

dimensions, irregular geometries.

We

will

still restrict ourselves

to

symmetric operators, although

it is

possible

to

apply

the finite

element method

to

nonsymmetric problems.

The finite

element method

is

based

on

three ideas:

1. The BVP is

rewritten

in its

weak

or

variational

form,

which expresses

the

problem

as

infinitely

many scalar equations.

In

this

form,

the

boundary con-

ditions

are

implicit

in the

definition

of the

underlying vector space.

2.

The

Galerkin

method

is

applied

to

"solve

the

equation

on a finite-dimensional

subspace."

This results

in an

ordinary linear system (matrix-vector equation)

that

must

be

solved.

3. A

basis

of

piecewise

polynomials

is

chosen

for the finite-dimensional

subspace

so

that

the

matrix

of the

linear system

is

sparse

(that

is, has

mostly zero

entries).

We

describe each

of

these ideas

in the

following

sections, using

the BVP

(5.32)

as

our

model problem.

We

always assume

that

the

coefficient

k(x]

is

positive, since

it

represents

a

positive physical parameter

(stiffness

or

thermal conductivity,

for

example).

5.4.1

The

principle

of

virtual

work

and the

weak

form

of a BVP

When

an

elastic material

is

deformed,

it

stores potential energy

due to

internal

elastic

forces.

It is not

obvious

from

first

principles

how to

measure (quantitatively)

this elastic potential

energy;

however,

we can

deduce

the

correct definition

from

the

equations

of

motion

and the

principle

of

conservation

of

energy.

Suppose

an

elastic bar, with

its

ends

fixed, is in

motion,

and its

displacement

function

u(x,

t)

satisfies

the

homogeneous wave equation:

We

now

perform

the

following

calculation

(a

trick—multiply

both sides

of the

wave

equation

by

du/dt

and

integrate):

5.4. Finite element methods for BVPs

173

is symmetric (see Exercise 1),

and

there exists

an

orthogonal sequence of eigenfunc-

tions. However, when

the

coefficient k(x)

is

not a constant, there is no simple way

to

find these eigenfunctions. Indeed, computing

the

eigenfunctions requires much

more work

than

solving

the

original BVP.

Because of

the

limitations of

the

Fourier series approach,

we

now introduce

the

finite element method, one of

the

most powerful methods for approximating solutions

to

PDEs.

The

finite element method can handle

both

variable coefficients and, in

multiple spatial dimensions, irregular geometries.

We

will still restrict ourselves

to

symmetric operators, although it

is

possible

to

apply

the

finite element method

to

nonsymmetric problems.

The

finite element method is based on three ideas:

1.

The

BVP

is rewritten in its weak or variational

form,

which expresses

the

problem as infinitely many scalar equations.

In

this form,

the

boundary con-

ditions are implicit in

the

definition of

the

underlying vector space.

2.

The

Galerkin method is applied

to

"solve

the

equation on a finite-dimensional

subspace." This results in

an

ordinary linear system (matrix-vector equation)

that

must be solved.

3. A basis of

piecewise polynomials is chosen for

the

finite-dimensional subspace

so

that

the matrix of

the

linear system is sparse

(that

is, has mostly zero

entries).

We describe each of these ideas in

the

following sections, using the BVP (5.32) as

our model problem.

We

always assume

that

the

coefficient k(x) is positive, since

it

represents a positive physical parameter (stiffness or thermal conductivity, for

example).

5.4.1 The principle of virtual

work

and

the

weak

form of a BVP

When

an

elastic material is deformed, it stores potential energy due

to

internal

elastic forces.

It

is not obvious from first principles how

to

measure (quantitatively)

this

elastic potential energy; however,

we

can deduce

the

correct definition from

the

equations of motion

and

the

principle of conservation of energy.

Suppose

an

elastic bar, with its ends fixed,

is

in motion,

and

its displacement

function

u(x, t) satisfies

the

homogeneous wave equation:

a

2

u a (

au)

Ap(x)at

2

-Aax

k(x)ax

=0,

O<x<f,

t>o,

u(O,

t) =

0,

t > 0,

(5.34)

u(f, t) =

0,

t >

0.

We

now perform

the

following calculation

(a

trick-multiply

both

sides of

the

wave

equation by

au/at

and integrate):

174

Chapter

5.

Boundary value problems

in

statics

This

equation implies

that

the sum of the two

integrals

is

constant with respect

to

time,

that

is,

that

is

conserved. Obviously,

the first

integral represents

the

kinetic energy (one-half

mass times velocity squared),

and we

define

the

second integral

to be the

elastic

potential energy.

The

elastic potential energy

is the

internal energy arising

from

the

strain

du/dx.

We

now

return

to the bar in

equilibrium,

that

is, to the bar

satisfying

the

BVP

(5.32).

We

have

an

expression

for the

elastic potential energy

of the

bar:

The

bar

possesses another

form

of

potential energy,

due to the

external

force

/

acting upon

it. The

proper name

of

this energy depends

on the

nature

of the

force;

if

it

were

the

force

due to

gravity,

for

example,

it

would

be

called

the

gravitational

Also,

Applying

integration

by

parts,

Therefore,

we

obtain

Applying

Theorem

2.1 to

(5.35) yields

174

Chapter

5.

Boundary value problems

in

statics

Applying integration by parts,

1

£ a (

au)

au

[auau]£

1£

au a

2

u

- -

k(x)-

-dx

= -

k(x)--

+

k(x)---dx

o ax ax at ax at 0 0 ax axat

r£

au a

2

u ( . au au )

=

10

k(x) ax atax dx smce at (0, t) = at (l,

t)

= 0 .

Also,

Therefore,

we

obtain

Ap(x)-

- - dx +

Ak(x)-

- -

1

£ a

[1

(au)2]

1£

a

[1

(au)2]

o at 2 at 0 at 2 ax

dx

=0.

(5.35)

Applying Theorem

2.1

to (5.35) yields

a [ 1

r£

( a ) 2 1

rl

( a ) 2 ]

at

"210

Ap(x)

a~

dx

+

"210

Ak(x)

a~

dx

= 0, t >

o.

(5.36)

This equation implies

that

the sum of the two integrals

is

constant with respect

to

time,

that

is,

that

1

r£

(a

) 2 1

r£

(a

) 2

"210

Ap(x)

a~

dx

+

"210

Ak(x)

a:

dx

is

conserved. Obviously, the first integral represents the kinetic energy (one-half

mass times velocity squared), and

we

define the second integral

to

be the elastic

potential energy. The elastic potential energy

is

the internal energy arising from

the

strain au/ax.

We

now return

to

the

bar

in equilibrium,

that

is,

to

the

bar

satisfying the

BVP

(5.32).

We

have an expression for the elastic potential energy of

the

bar:

1

r£

(d)2

"210

Ak(x)

d~

(x)

dx.

The

bar

possesses another form of potential energy, due

to

the external force f

acting upon it. The proper name of this energy depends on the nature of

the

force;

if it were the force due to gravity, for example, it would be called the gravitational

5.4.

Finite

element

methods

for

BVPs

175

potential energy.

We

will

call this energy

the

external potential

energy.

This energy

is

different

in a

fundamental

way

from

the

elastic potential energy:

it

depends

on the

absolute position

of the

bar,

not on the

relative position

of

parts

of the bar to

other

parts.

For

this reason,

we do not

know

the

external potential energy

of the bar in its

"reference" (zero displacement)

position

(whereas

the bar has

zero

elastic

potential

energy

in the

reference

position).

We

define

£Q

to be the

external

potential

energy

of

the bar in the

reference position,

and we now

compute

the

change

of

energy between

the bar in its

reference position

and the bar in its

equilibrium position.

The

change

in

potential energy

due to the

external

force

/ is

equal

to the

work done

by /

when

the bar is

deformed

from

its

reference position

to its

equilibrium position.

The

work

done

by /

acting

on the

part

P of the bar

originally between

x and

x

+

Ax

is

approximately

(recall

that

u(x)

is the

displacement

of the

cross-section

of the bar

originally

at

x—thus

u(x)

is the

distance traveled

by

that

cross-section).

To add up the

work

due

to the

external

force

acting

on all

little

parts

of the

bar,

we

integrate:

We

now

have

an

expression

for the

total

potential energy

of the bar in its

equilibrium position:

If

/ and

u

are

both positive,

for

instance, then

the bar has

less potential energy

in

its

equilibrium position (think

of

gravity).

This

accounts

for the

sign

on the

second

integral.

wish

to

prove

the

following

fact:

the

equilibrium displacement

u of

the

bar is the

displacement with

the

least

potential

energy.

That

is, if

w

^

u is any

other displacement satisfying

the

Dirichlet conditions, then

We

can

write

any

displacement

w as w = u + v

(just take

v = w

—

w), and

the

conditions

u(Q)

— 0,

w;(0)

= 0

imply

that

v(0)

— 0, and

similarly

at x = t.

Thus

v

must belong

to

C|)[0,

^].

For

brevity,

we

shall write

V =

Cf^O,^],

the set of

allowable displacements

v.

Therefore,

we

wish

to

prove

that

We

have

5.4. Finite element methods for BVPs

175

potential energy.

We

will call this energy

the

external potential energy. This energy

is different in a fundamental way from

the

elastic potential energy:

it

depends on

the

absolute position of the bar,

not

on

the

relative position of

parts

of

the

bar

to

other

parts. For this reason,

we

do not know

the

external potential energy of

the

bar

in its

"reference" (zero displacement) position (whereas

the

bar

has zero elastic potential

energy in

the

reference position).

We

define

Co

to

be

the

external potential energy of

the

bar

in

the

reference position,

and

we

now compute

the

change of energy between

the

bar

in its reference position

and

the

bar

in its equilibrium position.

The

change

in potential energy due

to

the

external force f is equal

to

the

work done by f when

the

bar

is deformed from its reference position

to

its equilibrium position.

The

work done by f acting on

the

part

P of

the

bar

originally between x

and

x +

~x

is approximately

f(x)

~

force per unit volume

volume

,.,.-"-..

A~x

u(x)

~

distance

(recall

that

u(x)

is

the

displacement of

the

cross-section of

the

bar

originally

at

x-thus

u(x) is

the

distance traveled by

that

cross-section). To

add

up

the

work

due

to

the

external force acting on all little

parts

of the bar,

we

integrate:

l'

Af(x)u(x)

dx.

(5.37)

We

now have

an

expression for

the

total

potential energy of

the

bar

in its

equilibrium position:

Cpot(u)

=

~

l'

Ak(x)

(:~

(X)) 2 dx

-1'

Af(x)u(x)

dx +

co.

If

f

and

u are

both

positive, for instance,

then

the

bar

has less potential energy in

its equilibrium position (think of gravity). This accounts for

the

sign on

the

second

integral.

wish

to

prove

the

following fact: the equilibrium displacement u

of

the bar is the displacement with the least potential energy.

That

is, if w

¥-

u is any

other displacement satisfying

the

Dirichlet conditions,

then

Cpot(w)

>

Cpot(u).

We

can write any displacement w as w = u + v (just take v = w - u),

and

the

conditions

u(O)

=

0,

w(O)

= 0 imply

that

v(O)

=

0,

and

similarly

at

x =

£.

Thus v must belong

to

c1

[0,

fl.

For brevity,

we

shall write V =

C1

[0,

f],

the

set of

allowable displacements

v. Therefore,

we

wish

to

prove

that

Cpot(u

+ v) >

Cpot(u)

for all v E

V.

We

have

1

r'

(dU

dV) 2 (

Cpot(u

+ v) =

"210

k(x) dx (x) + dx (x) dx -

10

f(x)

(u(x) + vex)) dx +

Co

that

is,

176

Chapter

5.

Boundary

value

problems

in

statics

Using

integration

by

parts,

we

have

(The boundary term disappears

in the

above calculation since

v(0]

=

v(t]

= 0.)

The

equilibrium displacement

u

satisfies

the

differential

equation

from

(5.32),

as we

derived

in

Section 2.2. Therefore,

the

integral above vanishes,

and we

have

If

v is not

identically zero, then neither

is

dv/dx

(only

a

constant function

has

a

zero derivative,

and v

cannot

be

constant unless

it is

identically zero, because

v(Q)

=v(l)

=

G).

Since

whenever

v

^

0

(the integral

of a

nonzero, nonnegative

function

is

positive),

we

have obtained

the

desired result:

176

Chapter

5.

Boundary value problems

in

statics

III

((dU)2

du

dv

(dV

)2)

=2 0 k(x) dx(X)

+2

dx

(X)dx(X)+ dx(X) dx

-

fot

f(x)u(x) dx -

fot

f(x)v(x)

dx +

[0

1

It

(dU)

2

It

du

dv

="2

0 k(x) dx (x) dx + 0 k(x) dx (x) dx (x) dx

1

It

(

dv

) 2

lf

+"2

0 k(x) dx

(x)

dx - 0 f(x)u(x) dx - 0 f(x)v(x) dx +

[0;

that

is,

[pot(U

+

v)

= [pot(u) +

fof

k(x)

~~

(x)

~:

(x) dx -

fol

f(x)v(x) dx

11t

(d

)2

+"2

0 k(x)

~(x)

dx.

(5.38)

Using integration by parts,

we

have

fof

k(x)

~~

(x)

~:

(x) dx -

fot

f(x)v(x) dx

=

[k(X)~~(X)V(X)]:

-

fof

(d~

(k(x)~~(x)))V(X)dX-

fof

f(x)v(x) dx

= -

fot

(d~

(k(X)~~(X))

+

f(X))

v(x)dx.

(The boundary

term

disappears in

the

above calculation since

v(O)

=

v(£)

= 0.)

The

equilibrium displacement U satisfies

d (

du

)

dx k(x) dx (x) +

f(x)

= 0,

the

differential equation from (5.32), as

we

derived in Section 2.2. Therefore, the

integral above vanishes,

and

we

have

11£

(dv)2

[pot(U

+

v)

=

[pot(U)

+ 2 0 k(x) dx (x) dx.

If

v is

not

identically zero,

then

neither is

dv

/ dx (only

<l:

constant function has

a zero derivative,

and

v cannot be constant unless

it

is identically zero, because

v(O)

=

v(£)

= 0). Since

11t

(d

)2

2 0 k(x)

d:(x)

dx>O

whenever

v:/;O

(the integral of a nonzero, nonnegative function is positive),

we

have obtained

the

desired result:

5.4.

Finite

element methods

for

BVPs

177

This result gives

us a

different

understanding

of the

equilibrium

state

of the

bar—the

bar

assumes

the

state

of

minimal potential energy. This complements

our

earlier understanding

that

the bar

assumes

a

state

in

which

all

forces

balance.

The

potential

energy

£

poi

defines

a

function mapping

the

space

of

physically

meaningful

displacements into

R, and the

equilibrium displacement

u is its

mini-

mizer.

A

standard

result

from

calculus

is

that

the

derivative

of a

real-valued

function

must

be

zero

at a

minimizer. Above

we

computed

the

strong form.

We can

show directly

that

the two

forms

of the BVP are

equivalent,

in

the

sense

that,

for / 6

C[0,

I],

u

satisfies

the

strong

form

if and

only

if it

satisfies

the

weak

form.

5.4.2

The

equivalence

of the

strong

and

weak forms

of the BVP

The

precise

statement

of the

weak

form

of the BVP is:

We

have already seen

the

proof

of the

following

fact:

If u

satisfies (5.39),

the

strong

form

of the

BVP, then

u

also satisfies (5.40),

the

weak form. This

was

proved

(see

(5.38)).

The

linear term

in

this expression must

be

D£

pot

(u)v,

the

directional

derivative

of

£

pot

at u in the

direction

of v. The

minimality

of the

potential energy

then implies

that

D£

poi

(u}

=

0, or,

equivalently,

This yields

(as we saw

above),

which

is

called

the

principle

of

virtual work. (The term linear

in u

dominates

the

work

£

po

t

(u +

v}

—

£

po

t

(u)

when

the

displacement

v is

arbitrarily

small

("virtual"),

and is

called

the

virtual work.

The

principle says

that

the

virtual

work

is

zero when

the bar is at

equilibrium.)

The

principle

of

virtual work

is

also called

the

weak

form

of the

BVP;

it is

the

form

that

the BVP

would take

if our

basic principle were

the

minimality

of

potential

energy

rather

than

the

balance

of

forces

at

equilibrium.

By

contrast,

we

will

call

the

original BVP,

5.4. Finite element methods for BVPs

177

This

result gives us a different

understanding

of

the

equilibrium

state

of

the

bar-the

bar

assumes

the

state

of minimal

potential

energy.

This

complements

our

earlier

understanding

that

the

bar

assumes a

state

in which all forces balance.

The

potential

energy

[pot

defines a function

mapping

the

space of physically

meaningful displacements

into

R,

and

the

equilibrium displacement u is

its

mini-

mizer. A

standard

result from calculus is

that

the

derivative

of

a real-valued function

must

be

zero

at

a minimizer. Above we

computed

[pot

(u

+ v) =

[pot

(u) + a

term

linear in v + a

term

quadratic

in v

(see (5.38)).

The

linear

term

in

this

expression

must

be

D[pot(u)v,

the

directional

derivative of

[pot

at

u in

the

direction of v.

The

minimality

of

the

potential

energy

then

implies

that

D[pot(u) = 0, or, equivalently,

D[pot(u)v = ° for all v E

V.

This

yields

i

f du

dv

if

o k(x) dx (x) dx (x) dx - 0

f(x)v(x)

dx = ° for all v E V

(as we saw above), which is called

the

principle

of

virtual work.

(The

term

linear

in

u dominates

the

work

[pot

(u

+

v)

-

[pot

(u)

when

the

displacement v

is

arbitrarily

small

("virtual"),

and

is called

the

virtual work.

The

principle says

that

the

virtual

work is zero when

the

bar

is

at

equilibrium.)

The

principle of

virtual

work

is

also called

the

weak form of

the

BVP;

it

is

the

form

that

the

BVP

would

take

if

our

basic principle were

the

minimality of

potential

energy

rather

than

the

balance of forces

at

equilibrium.

By

contrast,

we

will call

the

original BVP,

d ( du )

- dx k(x) dx (x)

=f(x),O<x<f,

u(o) = 0, (5.39)

u(f) = 0,

the

strong form. We

can

show directly

that

the

two forms of

the

BVP

are

equivalent,

in

the

sense

that,

for f E

C[O,

£],

u satisfies

the

strong

form

if

and

only if

it

satisfies

the

weak form.

5.4.2

The

equivalence

of

the strong

and

weak forms

of

the

BVP

The

precise

statement

of

the

weak form of

the

BVP

is:

i

f du

dv

if

find u E V such

that

0 k(x) dx (x) dx (x) dx = 0

f(x)v(x)

dx for all v E

V.

(5.40~

We have already seen

the

proof of

the

following fact:

If

u satisfies (5.39),

the

strong

form of

the

BVP,

then

u also satisfies (5.40),

the

weak form.

This

was proved

Gockenbach M.S. Partial Differential Equations. Analytical and Numerical Methods

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