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

28

Chapter

2.

Models

in one

dimension

reference

configuration moves

to (x,

u(x,

t))

at

time

t. We are

thus postulating

that

the

motion

of the

string

is

entirely

in the

transverse direction (this

is one of the

severe

assumptions

that

we

mentioned

in the

previous paragraph). Granted this

assumption,

we now

derive

the

differential

equation satisfied

by the

displacement

u.

Since,

by

assumption,

a

string does

not

resist bending,

the

internal restoring

force

of the

string under tension

is

tangent

to the

string itself

at

every point.

We

will

denote

the

magnitude

of

this restoring

force

by

T(x,

t). In

Figure 2.5,

we

display

a

part

of the

deformed

string, corresponding

to the

part

of the

string between

x

and x + Ax in the

reference

configuration, together with

the

internal

forces

at the

ends

of

this

part,

and

their magnitudes.

In the

absence

of any

external

forces,

the

sum

of

these internal

forces

must balance

the

mass times acceleration

of

this

part

of

the

string.

To

write down

these

equations,

we

must decompose

the

internal

force

into

its

horizontal

and

vertical components.

Figure

2.5.

A

part

of

the

deformed

string.

We

write

n =

n(x,

t) for the

force

at the

left

endpoint

and 9 for the

angle this

force

vector makes with

the

horizontal.

We

then have

a

reasonable approximation

is

with

Assuming

that

\du/dx

<C

1 at

every point,

and

noting

that

28

Chapter

2.

Models

in

one dimension

reference configuration moves

to

(x,

u(x,

t))

at

time

t.

We

are thus postulating

that

the motion of the string

is

entirely in

the

transverse direction (this

is

one of

the

severe assumptions

that

we

mentioned in

the

previous paragraph). Granted this

assumption,

we

now derive the differential equation satisfied by the displacement u.

Since, by assumption, a string does not resist bending, the internal restoring

force of the string under tension

is

tangent

to

the string itself

at

every point.

We

will denote the magnitude

ofthis

restoring force by

T(x,

t). In Figure 2.5,

we

display

a

part

of the deformed string, corresponding

to

the

part

of the string between x

and x +

6x

in the reference configuration, together with

the

internal forces

at

the

ends of this part, and their magnitudes. In the absence of any external forces, the

sum of these internal forces must balance the mass times acceleration of this

part

of

the

string.

To

write down these equations,

we

must decompose the internal force

into its horizontal and vertical components.

Figure

2.5.

A part

of

the deformed string.

We

write n = n(x, t) for

the

force

at

the left endpoint

and

0 for

the

angle this

force vector makes with the horizontal.

We

then have

ni

+

n~

=

T(x,

t)2,

with

nl

=

-T(x,

t) cos

(0),

n2

=

-T(x,

t) sin

(0).

Assuming

that

lou/oxl

« 1

at

every point, and noting

that

au

tan

(0)

=

8x(x,t),

a reasonable approximation

is

cos

(0)

==

1,

sin

(0)

==

tan

(0)

=

~~

(x,

t).

where

p is the

density

of the

string

(in

units

of

mass

per

length).

Since this holds

for

all x and

Ax

sufficiently

small,

we

obtain

the

differential

equation

2.3.

The

wave

equation

for a

vibrating string

29

We

then

obtain

Similarly,

the

force

at the

right

end of the

part

of the

string under consideration

is

We

then

see

that

is

(an

approximation

to) the

horizontal component

of the

total

force

on the

part

of

the

string,

and

is

(an

approximation

to) the

vertical component.

By

assumption,

the

horizontal component

of the

acceleration

of the

string

is

zero,

and so

Newton's second

law

yields

thus

the

tension

is

constant throughout

the

string

at

each point

in

time.

We

will

assume

that

the

length

of the

string never changes much

from

its

length

in the

reference

configuration

(which

is

£),

and

therefore

that

it

makes sense

to

assume

that

T is

independent

of t as

well,

that

is,

that

T is a

constant.

Applying Newton's

second

law to the

vertical component yields

We

recognize this

as the

homogeneous wave equation.

It is

usual

to

write this

in

the

form

where

c

2

=

T/p.

The

significance

of the

parameter

c

will

become clear

in

Chapter

7.

In

the

case

that

an

external body

force

is

applied

to the

string

(in the

vertical

direction),

the

equation becomes

2.3. The

wave

equation for a vibrating string

29

We

then obtain

n(x,

t)

===

(

-T(x,

t),

-T(x,

t)

~~

(x,

t))

.

Similarly, the force

at

the right end of the

part

of the string under consideration

is

n(x

+ Llx, t)

===

(T(X

+ Llx, t),

T(x

+ Llx, t)

~~

(x + Llx,

t))

.

We

then see

that

T(x

+ Llx, t) -

T(x,

t)

is

(an approximation to) the horizontal component of

the

total

force on the

part

of

the

string, and

au au

T(x

+ Llx, t)

ax

(x + Llx, t) -

T(x,

t)

ax

(x,

t)

is

(an approximation to) the vertical component.

By assumption,

the

horizontal component of the acceleration of the string

is

zero, and

so

Newton's second law yields

T(x

+ Llx, t) -

T(x,

t) = 0;

thus the tension

is

constant throughout the string

at

each point in time.

We

will

assume

that

the length of the string never changes much from its length in the

reference configuration (which is

f), and therefore

that

it makes sense to assume

that

T

is

independent of t as well,

that

is,

that

T

is

a constant. Applying Newton's

second law

to

the vertical component yields

t+1':!..x

a

2

u

au au

ix

p

at

2

(s, t) ds = T

ax

(x + Llx, t) - T

ax

(x, t)

{X

+1':!..

x a2u

=

ix

T

ax

2

(s, t) ds,

where p

is

the density of

the

string (in units of mass per length). Since this holds

for all x and Llx sufficiently small,

we

obtain the differential equation

a

2

u a

2

u

p

at

2

= T

ax

2

'

0 < x <

£,

t >

O.

(2.18)

We

recognize this as the homogeneous wave equation.

It

is

usual

to

write this in

the form

a

2

u a

2

u

at

2

-

c

2

ax

2

= 0, ° < x <

£,

t > 0,

where

c2

= T /

p.

The significance of the parameter c will become clear in Chapter

7.

In the case

that

an external body force

is

applied to the string (in the vertical

direction), the equation becomes

a

2

u a

2

u

at

2

-

c

2

ax

2

=

I(x,

t), 0 < x <

£,

t>

O.

(2.19)

Exercises

1.

What

units must

f(x,t)

have

in

(2.19)?

2.

What

are the

units

of the

tension

T in the

derivation

of the

wave equation

for

the

string?

3.

What

are the

units

of the

parameter

c in

(2.18)?

4.

Suppose

the

only external

force

applied

to the

string

is the

force

due to

gravity.

What

form

does (2.19) take

in

this case? (Let

g be the

acceleration

due to

gravity,

and

take

g to be

constant.)

5.

Explain

why a

homogeneous Neumann condition models

an end of the

string

that

is

allowed

to

move

freely

in the

vertical direction.

6.

Suppose

that

an

elastic string

is fixed at

both ends,

as in

this section,

and it

sags under

the

influence

of an

external

force

f(x)

(f is

constant with respect

to

time).

What

differential

equation

and

side conditions does

the

equilibrium

displacement

of the

string satisfy? Assume

that

/ is

given

in

units

of

force

per

length.

2.4

Suggestions

for

further

reading

If

the

reader wishes

to

learn more about

the use of

mathematical models

in the

sciences,

an

excellent place

to

start

is the

text

by Lin and

Segel

[36],

which com-

bines modeling with

the

analysis

of the

models

by a

number

of

different

analytical

techniques.

Lin and

Segel cover

the

models

that

form

the

basis

for

this book,

as

well

as

many others.

A

more advanced

text,

which

focuses

almost entirely

on the

derivation

of

differential

equations

from

the

basic principles

of

continuum mechanics,

is

Gurtin

[21].

30

Chapter

2.

Models

in one

dimension

Exercise

1

asks

the

reader

to

determine

the

units

of /.

The

natural boundary conditions

for the

vibrating string,

as

suggested above,

are

homogeneous Dirichlet conditions:

One

can

also imagine

that

one or

both ends

of the

string

are

allowed

to

move

freely

in

the

vertical direction (perhaps

an end of the

string slides along

a

frictionless

pole).

In

this case,

the

appropriate boundary condition

is a

homogeneous Neumann

condition (see Exercise

5).

30

Chapter

2.

Models in

one

dimension

Exercise 1 asks the reader to determine the units of f.

The

natural boundary conditions for the vibrating string, as suggested above,

are homogeneous Dirichlet conditions:

u(O,

t) = u(£, t) = 0, t >

0.

One can also imagine

that

one or

both

ends of the string are allowed

to

move freely

in the vertical direction (perhaps

an

end of the string slides along a frictionless

pole).

In

this case,

the

appropriate boundary condition

is

a homogeneous Neumann

condition (see Exercise 5).

Exercises

1.

What

units must

f(x,

t) have in (2.19)?

2.

What

are the units of

the

tension T in

the

derivation of the wave equation for

the string?

3.

What

are

the

units of the parameter c in (2.18)?

4. Suppose

the

only external force applied

to

the string

is

the force due

to

gravity.

What

form does (2.19) take in this case? (Let g be the acceleration due

to

gravity,

and

take g

to

be constant.)

5.

Explain why a homogeneous Neumann condition models

an

end of the string

that

is

allowed to move freely in

the

vertical direction.

6.

Suppose

that

an

elastic string is fixed

at

both

ends, as in this section,

and

it

sags under the influence of

an

external force

f(x)

(f

is constant with respect

to

time).

What

differential equation

and

side conditions does the equilibrium

displacement of the string satisfy? Assume

that

f

is

given in units of force

per length.

2.4

Suggestions for further reading

If

the reader wishes

to

learn more

about

the use of mathematical models in the

sciences,

an

excellent place

to

start

is

the

text

by Lin and Segel

[36],

which com-

bines modeling with the analysis of the models by a number of different analytical

techniques. Lin and Segel cover the models

that

form

the

basis for this book, as

well as many others.

A more advanced text, which focuses almost entirely on

the

derivation of

differential equations from the basic principles of continuum mechanics,

is

Gurtin

[21].

Chapter

3

linear algebra

The

solution techniques presented

in

this book

can be

described

by

analogy

to

techniques

for

solving

where

A is an n x n

matrix

(A 6

R

nxn

)

and x and b are

n-vectors

(x,b

€

R

n

).

Recall

that

such

a

matrix-vector

equation represents

the

following

system

of n

linear

equations

in the n

unknowns

x\,

#2,

• •

•,

x

n

:

Before

we

discuss methods

for

solving

differential

equations,

we

review

the

funda-

mental facts about systems

of

linear (algebraic) equations.

3.1

Linear

systems

as

linear operator equations

To

fully

appreciate

the

point

of

view

taken

in

this book,

it is

necessary

to

understand

the

equation

Ax = b not

just

as a

system

of

linear equations,

but as a finite-

dimensional linear operator equation.

In

other words,

we

must

view

the

matrix

A

as

defining

an

operator

(or

mapping,

or

simply function)

from

R

n

to

R

n

via

matrix

multiplication:

A

maps

x 6

R

n

to y = Ax

e

R

n

.

(More generally,

if A is not

square,

say A

e

R

mxn

,

then

A

defines

a

mapping

from

R

n

to

R

m

,

since

Ax

G

R

m

for

each

x

e

R

n

.)

The

following

language

is

useful

in

discussing operator equations.

Definition

3.1.

Let X and Y be

sets.

A

function

(^operator,

mapping,)

/

from

X

to Y is a

rule

for

associating with

each

x

e

X a

unique

y £

Y,

denoted

y =

f(x).

The set X is

called

the

domain

of

f,

and the

range

of f is the set

31

Eseential

Chapter 3

ial linear algebra

The

solution techniques presented in this book can be described by analogy

to

techniques for solving

Ax=b,

where A

is

an

n x n

matrix

(A

E RllXll)

and

x

and

bare

n-vectors (x, b E

Rll).

Recall

that

such a matrix-vector equation represents

the

following system of n linear

equations in

the

n unknowns

Xl,

X2,

...

,

Xn:

anXl

+

al2X2

+ ... +

alnX

n

b

l

,

a2lXl

+

a22X2

+ ... +

a2nXn

b

2

,

Before

we

discuss methods for solving differential equations,

we

review

the

funda-

mental facts

about

systems of linear (algebraic) equations.

3.1 Linear systems

as

linear operator equations

To fully appreciate

the

point of view taken in this book,

it

is

necessary

to

understand

the

equation

Ax

= b not

just

as a system of linear equations,

but

as a finite-

dimensional linear operator equation. In

other

words,

we

must view

the

matrix A

as defining

an

operator (or mapping, or simply function) from

Rll

to

Rll

via matrix

multiplication: A maps x

E

Rll

to

Y =

Ax

E

Rll.

(More generally, if A is not

square, say A

E RffiXll,

then

A defines a mapping from

Rll

to

Rffi, since

Ax

E Rffi

for each x E Rll.)

The

following language is useful in discussing operator equations.

Definition

3.1.

Let X and Y

be

sets. A function

(operator,

mapping) f from X

to Y

is

a rule for associating with

each

X E X a unique y E

Y,

denoted y = f (x) .

The

set X

is

called

the domain

of

f,

and the range

of

f

is

the set

R(A)

=

{f(x)

E Y : x E

X}.

31

32

Chapter

3.

Essential linear algebra

We

write

f

:

X

->

Y

("f

maps

X

into

Y")

to

indicate

that

f is a

function

from

X

toY.

The

reader should recognize

the

difference

between

the

range

of a

function

/ : X

—)•

Y and the set Y

(which

is

sometimes called

the

co-domain

of /). The set

Y

merely identifies

the

type

of the

output values

/(#);

for

example,

if Y = R,

then

every

/(#)

is a

real number.

On the

other hand,

the

range

of / is the set of

elements

of

Y

that

are

actually

attained

by /. As a

simple example, consider

/ : R

—>

R

defined

by

f(x)

=

x

2

.

The

co-domain

of / is R, but the

range

of /

consists

of the

set of

nonnegative numbers:

In

many cases,

it is

quite

difficult

to

determine

the

range

of a

function.

The co-

domain,

on the

other hand, must

be

specified

as

part

of the

definition

of the

function.

A

set is

just

a

collection

of

objects (elements); most

useful

sets have operations

defined

on

their elements.

The

most important sets used

in

this book

sue

vector

spaces.

Definition

3.2.

A

vector space

V is a set on

which

two

operations

are

defined,

addition

(if

u,

v

6

V,

then

u

+ v

e

V) and

scalar multiplication

(if

u € V and a

is

a

scalar, then

cm €

V).

The

elements

of the

vector

space

are

called

vectors.

(In

this

book,

the

scalars

are

usually

real

numbers,

and we

assume this unless otherwise

stated.

Occasionally

we use the set of

complex

numbers

as the

scalars.

Vectors

will

always

be

denoted

by

lower

case

boldface

letters.)

The

two

operations must

satisfy

the

following

algebraic

properties:

For

every vector space considered

in

this book,

the

verification

of

these vector

space properties

is

straightforward

and

will

be

taken

for

granted.

Example

3.3.

The

most common example

of a

vector

space

is

(real)

Euclidean

n-space:

32

Chapter

3.

Essential linear algebra

We write f : X

--+

Y

(lif

maps X into

Y")

to indicate that f is a function from X

to

Y.

The reader should recognize the difference between the range of a function

f : X

--+

Y and the set Y (which

is

sometimes called the co-domain of

f).

The set

Y merely identifies

the

type of the

output

values

f(x);

for example, if Y =

R,

then

every

f (x)

is

a real number. On

the

other hand, the range of f

is

the

set of elements

of

Y

that

are actually attained by f.

As

a simple example, consider f : R

--+

R

defined by

f(x)

= x

2

•

The co-domain of f

is

R,

but

the range of f consists of the

set of nonnegative numbers:

R(f)

= [0,00).

In many cases, it

is

quite difficult

to

determine the range of a function. The co-

domain, on the other hand, must be specified as

part

of the definition of

the

function.

A set

is

just a collection of objects (elements); most useful sets have operations

defined on their elements. The most important sets used in this book are

vector

spaces.

Definition

3.2.

A vector space V is a

set

on which two operations are defined,

addition

(if

u,

v E

V,

then u + v E

V)

and scalar multiplication

(if

u E V and a

is a scalar, then

au

E

V).

The elements

of

the vector space

are

called vectors. (In

this

book,

the scalars are usually real numbers, and

we

assume this unless otherwise

stated. Occasionally

we

use the

set

of

complex numbers as the scalars. Vectors will

always

be

denoted

by

lower case boldface letters.}

The two operations

must

satisfy the following algebraic properties:

1.

u + v = v + u for all

u,

v E

V.

2.

(u

+

v)

+ w = u +

(v

+

w)

for all

u,

v,

wE

V.

3.

There is a zero vector 0

in

V with the property that u + 0 = u for all u E

v.

4.

For each u E

V,

there is a vector

-u

E V such that u +

(-u)

=

o.

5.

a(u

+

v)

=

au

+

av

for all

u,

v E V and for all scalars

a.

6.

(a

+

P)u

=

au

+

pu

for all u E V and for all scalars

a,

p.

7.

a(pu)

=

(ap)u

for all u E V and for all scalars

a,

p.

8.

1 u = u for all u E

v.

For every vector space considered in this book,

the

verification of these vector

space properties

is

straightforward and will be taken for granted.

Example

3.3.

The

most

common example

of

a vector space is (real) Euclidean

n-space:

3.1.

Linear

systems

as

linear

operator

equations

33

Vectors

in

R

n

are

usually written

in

column form,

as

it is

convenient

at

times

to

think

of

u

G

R

n

as an n x 1

matrix.

Addition

and

scalar

multiplication

are

defined

componentwise:

Example

3.4.

Apart from Euclidean

n-space,

the

most common vector

spaces

are

function

spaces

—vector

spaces

in

which

the

vectors

are

functions. Functions (with

common domains)

can be

added

together

and

multiplied

by

scalars,

and the

algebraic

properties

of a

vector

space

are

easily

verified.

Therefore,

when

defining

a

function

space,

one

must

only

check

that

any

desired

properties

of

the

functions

are

preserved

by

addition

and

scalar multiplication. Here

are

some important examples:

1.

C[a,

b]

is

defined

to be the set of all

continuous, real-valued functions

defined

on the

interval

[a,

b].

The sum

of

two

continuous functions

is

also

continuous,

as is any

scalar multiple

of a

continuous function.

Therefore,

C[a,

b]

is a

vector

space.

2.

C

l

[a,

b]

is

defined

to be the set of all

real-valued, continuously

differentiate

functions

defined

on the

interval

[a,b].

(A

function

is

continuously

differen-

tiate

if

its

derivative exists

and is

continuous.)

The sum

of

two

continuously

differentiate

functions

is

also

continuously

differentiate,

and the

same

is

true

for a

scalar multiple

of a

continuously

differentiate

function.

Therefore,

C

l

[a,

b]

is a

vector

space.

3. For any

positive

integer

k,

C

k

[a,

b]

is the

space

of

real-valued

functions

defined

on [a,

b]

that have

k

continuous derivatives.

Many vector spaces

that

are

encountered

in

practice

are

subspaces

of

othe

vector

spaces.

Definition

3.5.

Let V be a

vector

space,

and

suppose

W is a

subset

ofV

with

the

following

properties:

1.

The

zero vector

belongs

to

W.

2.

Every linear combination

of

vectors

in W is

also

in

W.

That

is,

if

x, y G W

and

a,

/?

G

R,

then

3.1. Linear systems

as

linear operator equations

33

Vectors

in

R

n

are usually written

in

column form,

as

it

is convenient at times to think

of

u

ERn

as

an n x 1 matrix. Addition and

scalar multiplication

are

defined componentwise:

u + v =

(Ul,

U2,

...

,

un)

+

(VI,

V2,

...

,V

n

)

=

(Ul

+

VI,

U2

+

V2,

•

..

,Un

+ V

n

),

au

=

a(Ul,U2,

...

,Un)

=

(aUl,aU2,

...

,aUn).

Example

3.4.

Apart

from Euclidean n-space, the

most

common vector spaces are

function spaces

-vector

spaces

in

which the vectors are functions. Functions (with

common domains) can

be

added together and multiplied

by

scalars, and the algebraic

properties

of

a vector space are easily verified. Therefore, when defining a function

space, one

must

only check that any desired properties

of

the functions are preserved

by

addition and scalar multiplication. Here

are

some important examples:

1.

C[a,

b]

is defined to

be

the set

of

all continuous, real-valued functions defined

on the interval

[a,

b].

The

sum

of

two continuous functions is also continuous,

as

is any scalar multiple

of

a continuous function. Therefore, C[a,

b]

is a

vector space.

2.

C

1

[a,

b]

is defined to

be

the set

of

all real-valued, continuously differentiable

functions defined on the interval

[a,

b].

(A function is continuously differen-

tiable

if

its derivative exists and is continuous.) The

sum

of

two continuously

differentiable functions

is also continuously differentiable, and the same is

true for a scalar multiple

of

a continuously differentiable function. Therefore,

C

1

[a,

b]

is a vector space.

3.

For any positive integer k, Ck[a,

b]

is the space

of

real-valued functions defined

on

[a,

b]

that have k continuous derivatives.

Many

vector spaces

that

are encountered in practice

are

subspaces

of

other

vector spaces.

Definition

3.5.

Let

V

be

a vector space, and suppose W is a subset

of

V with the

following properties:

1.

The zero vector belongs to

W.

2.

Every linear combination

of

vectors

in

W is also

in

W.

That is,

if

x,

yEW

and

a,

j3

E

R,

then

ax

+

j3y

E

W.

34

Chapter

3.

Essential

linear

algebra

Then

we

call

W a

subspace

ofV.

A

subspace

of a

vector space

is a

vector space

in its own

right,

as the

reader

can

verify

by

checking

that

all the

properties

of a

vector space

are

satisfied

for a

subspace.

Example

3.6.

We

define

The set

C^ja,

b]

is a

subset

of

C

2

[a,b],

and

this subset contains

the

zero function

(hopefully

this

is

obvious

to the

reader).

Also,

if

u,v

e

6%[a,

b],

a,j3

G

R,

and

w

=

au

+

f3v,

then

where

7 and 8 are

nonzero

real

numbers. Then, although

W is a

subset

of

C

2

[a,b],

it is not a

subspace.

For

example,

the

zero function

does

not

belong

to

W,

since

it

does

not

satisfy

the

boundary

conditions. Also,

if

u,

v

e

W and a,

J3

€ R,

then,

with

w =

au

+ flv, we

have

Example

3.8.

We

define

and

so w

e

C^[a,b].

This

shows that

C^[a,b]

is a

subspace

of

C

2

[a,b].

The

examples

will

be

used throughout

this

book.

The

letters

"D"

and

"N"

stand

for

Dirichlet

and

Neumann, respectively (see

Section

2.1,

for

example).

The

following

provides

an

important

nonexample

of a

subspace.

Similarly,

Example

3.7.

We

define

w

e

C

2

[a,

b]

(since

C

2

[a,b]

is a

vector

space);

and

w(a)

—

au(a)

+

/3v(a)

=

a-Q

+

0-Q

=

Q,

and

similarly

w(b)

= 0.

Therefore,

w € Cp[a,b],

which shows that

C^a,b]

is a

subspace

of

C

2

[a,b].

The

set

C^[a,b]

is

also

a

subset

of

C

2

[a,b],

and it can be

shown

to be a

subspace.

Clearly

the

zero function

belongs

to

Cj^[a,b].

If

u,v

€

C^[a,b],

a,/?

£ R, and

w

= au + fiv,

then

34

Chapter

3. Essential linear algebra

Then

we call W a subspace

of

V.

A subspace of a vector space

is

a vector space in its own right, as the reader

can verify by checking

that

all the properties of a vector space are satisfied for a

subspace.

Example

3.6.

We

define

C1[a,

b]

=

{u

E C

2

[a,

b]

:

u(a)

= u(b) = o}.

The

set

Cb[a,

bj

is a subset

of

C

2

[a,

b],

and

this subset contains the zero

function

(hopefully this is obvious to the reader). Also,

if

u,v

E Cb[a,b],

a,f3

E

R,

and

w =

au

+

f3v,

then

•

wE

C

2

[a,

bj

(since C

2

[a,

bj

is a vector space); and

•

w(a)

=

au

(a) + f3v(a) =

0.·0

+

f3

. 0 = 0,

and

similarly w(b) =

o.

Therefore,

wE

Cb[a,

b],

which shows

that

Cb[a,

bj

is a subspace

of

C

2

[a,

bj.

Example

3.7.

We

define

C~,[a,bj={uEC2[a,bj:

~~(a)=~~(b)=O}.

The

set

CF,-[a,

bj

is also a subset

of

C

2

[a,

bj,

and

it

can

be

shown

to

be

a subspace.

Clearly the zero

function

belongs to

C;'

[a,

bj.

If

u,

v E

C;'

[a,

b],

a,

f3

E R, and

w =

au

+

f3v,

then

dw

du

dw

dx

(a) = a

dx

(a) +

f3

dx

(a) = a

·0+

f3

. 0 =

o.

Similarly,

~:

(b)

= 0,

and

so w E C;'[a,bj. This shows

that

C;'[a,bj

is

a subspace

of

C

2

[a,

bj.

The previous two examples will be used throughout this book. The letters

"D"

and "N"

stand

for Dirichlet

and

Neumann, respectively (see Section 2.1, for

example).

The following provides an important

nonexample

of a subspace.

Example

3.8.

We

define

W

=

{u

E C

2

[a,bj :

u(a)

=

'Y,

u(b) = 8},

where'Y

and

8 are nonzero real numbers. Then, although W

is

a subset

of

C

2

[a,

bj,

it

is

not

a subspace. For example, the zero

function

does

not

belong to

W,

since

it

does

not

satisfy the boundary conditions. Also,

if

u,

v E

Wand

a,

f3

E

R,

then,

with w =

au

+

f3v,

we have

w(a) =

au(a)

+ f3v(a) =

a'Y

+

f3'Y

=

(a

+

(3)"(.

7

The

word

"operator"

is

preferred over "function"

in

this context, because

the

vector spaces

themselves

are

often

spaces

of

functions.

3.1.

Linear systems

as

linear

operator

equations

35

Thus

w(a)

does

not

equal

7,

except

in the

special

case

that

a +

(3

=

I.

Similarly,

w(b)

does

not

satisfy

the

boundary

condition

at the

right endpoint.

The

concept

of a

vector space allows

us to

define

linearity, which describes

many simple processes

and is

indispensable

in

modeling

and

analysis.

Definition

3.9.

Suppose

X and Y are

vector

spaces,

and f : X

—>•

Y is an

operator

(or

function,

7

or

mapping)

with domain

X and

range

Y.

Then

f is

linear

if and

only

if

This

condition

can be

expressed

as the

following

two

conditions, which together

are

equivalent

to

(3.1):

A

linear

operator

is

thus

a

particularly simple kind

of

operator;

its

simplicity

can be

appreciated

by

comparing

the

property

of

linearity

(e.g.

f(x

+

y}

=

f(x)

+

f(y)}

with common nonlinear operators:

^/x

+ y

^

^-\-^Jy->

sin (x + y)

^

sin

(x)

+

sin(?/),

etc.

f

(x)

= Ax,

Example

3.10.

The

operator

defined

by a

matrix

A G

R

mxn

via

matrix-vectoi

multiplication,

is

linear;

the

reader

should

verify

this

if

necessary (see Exercise

7).

Moreover,

it

can

be

shown that

every

linear operator

mapping

R

n

into

R

m

can be

represented

by

a

matrix

A €

R

mxn

in

this

way

(see Exercise

8).

This

explains

why the

study

of

(finite-dimensional)

linear

algebra

is

largely

the

study

of

matrices.

In

this

book,

matrices

will

be

denoted

by

upper

case

boldface

letters.

Example

3.11.

To

show that

the

sine function

is not

linear,

we

observe that

while

Example

3.12.

Differentiation

defines

an

operator

3.1. Linear systems

as

linear operator equations

35

Thus w(a) does

not

equal

,,(,

except

in

the special case that a +

j3

= 1. Similarly,

w(b) does

not

satisfy the boundary condition at the right endpoint.

The

concept

of

a vector space allows

us

to

define linearity, which describes

many

simple processes

and

is indispensable

in

modeling

and

analysis.

Definition

3.9.

Suppose X and

Yare

vector spaces, and f : X -+ Y is an operator

(or function,

7

or mapping) with domain X and range

Y.

Then f is linear

if

and

only

if

f(ax

+

j3z)

=

af(x)

+ j3f(z) for all

a,

j3

E

R,

x, z E

X.

(3.1)

This condition can

be

expressed

as

the following two conditions, which together

are

equivalent to (3.1):

1.

f(ax)

=

af(x)

for all x E X and all a E

R;

2.

f(x

+ z) = f(x) + f(z) for all

x,z

E

X.

A linear

operator

is

thus

a

particularly

simple

kind

of

operator;

its

simplicity

can

be

appreciated

by

comparing

the

property

of

linearity (e.g.

f(x

+ y) =

f(x)

+

f(y))

with

common nonlinear

operators:

Jx

+ y

-:j:.

-/X

+

v:y,

sin

(x

+ y)

-:j:.

sin (x) +

sin (y), etc.

Example

3.10.

The operator defined by a matrix A E R

ffixn

via matrix-vector

multiplication,

f(x) =

Ax,

is linear; the reader should verify this

if

necessary (see Exercise 7). Moreover,

it

can

be

shown that every linear operator mapping Rn into Rffi can

be

represented

by

a

matrix

A E Rffixn

in

this way (see Exercise

8).

This explains why the study

of

(finite-dimensional) linear algebra is largely the study

of

matrices.

In

this

book,

matrices will

be

denoted by upper case boldface letters.

Example

3.11.

To

show that the sine function is

not

linear, we observe that

sin

(2i)

= sin

(7f)

= 0,

while

2sin(~)

=2·1=2.

Example

3.12.

Differentiation defines an operator

d~

: C

1

[a,

b]

-+

C[a,

b],

7The word "operator" is preferred over "function" in this context, because

the

vector spaces

themselves are often spaces

of

functions.

36

Chapter

3.

Essential

linear

algebra

and

this

operator

is

well

known

to be

linear.

For

example,

since

In

general,

the

kth

derivative

operator

defines

a

linear

operator

mapping

C

k

[a,

b

into

C[a,b].

This

is why

linearity

is so

important

in the

study

of

differential

equa-

tions.

Since

a

matrix

A

e

R

nxn

defines

a

linear operator

from

R

n

to

R

n

,

the

linear

system

Ax = b can be

regarded

as a

linear

operator

equation. From this point

of

view,

the

questions posed

by the

system are:

Is

there

a

vector

x G

R

n

whose image

under

A is the

given vector

b? If so, is

there only

one

such vector

x? In the

we

will

explore these questions.

The

point

of

view

of a

linear operator equation

is

also

useful

in

discussing

differential

equations.

For

example, consider

the

steady-state heat

flow

problem

from

Section 2.1.

We

define

a

differential

operator

L

D

:

C|,[0,^]

->

C[Q,l]

by

Then

the BVP

(3.2)

is

equivalent

to the

operator equation

(the reader should notice

how the

Dirichlet boundary conditions

are

enforced

by

the

definition

of the

domain

of

LD).

This

and

similar examples

will

be

discussed

throughout this chapter

and in

detail

in

Section

5.1.

Exercises

1.

In

elementary algebra

and

calculus courses,

it is

often

said

that

/ : R

->•

R

is

linear

if and

only

if it has the

form

f(x)

— ax +

&,

where

a and b are

constants. Does this agree with

Definition

3.9?

If

not, what

is the

form

of a

linear

function

/ : R

—^

R?

2.

Show explicitly

that

/ : R

->

R

defined

by

f(x)

=

^/x

is not

linear.

3. For

each

of the

following

sets

of

functions, determine whether

or not it is a

vector space.

(Define

addition

and

scalar multiplication

in the

obvious way.)

If

it is

not,

state

what property

of a

vector space

fails

to

hold.

36

Chapter

3. Essential linear algebra

and this operator is well known to

be

linear. For example,

d~

[2

sin (x) - 3e

X

]

= 2 cos (x) - 3e

x

,

since

d~

[sin

(x)] = cos (x),

d~

[eX]

=

eX.

In

general, the

kth

derivative operator defines a linear operator mapping

Ck

[a,

b]

into C[a,

b].

This is why linearity is so important

in

the study

01

differential equa-

tions.

Since a matrix A E Rnxn defines a linear operator from

Rn

to

Rn,

the linear

system

Ax

= b can be regarded as a linear operator equation. From this point of

view, the questions posed by the system are: Is there a vector x

E Rn whose image

under A

is

the

given vector

b?

If

so,

is

there only one such vector x? In

the

we

will explore these questions.

The point of view of a linear operator equation

is

also useful in discussing

differential equations. For example, consider

the

steady-state heat

flow

problem

(Pu

-/1,

ox

2

=

I(x),

0 < x <

C,

u(O)

= 0,

u(e)

= 0

from Section 2.1.

We

define a differential operator

LD

: C1[0,C]-+

C[O,C]

by

~u

LDu

=

-/1,

dx

2

'

Then

the

BVP (3.2)

is

equivalent

to

the operator equation

LDU=I

(3.2)

(the reader should notice how the Dirichlet boundary conditions are enforced by

the definition of the domain of

LD)'

This and similar examples will be discussed

throughout this chapter and in detail in Section 5.1.

Exercises

1.

In elementary algebra and calculus courses,

it

is

often said

that

1 : R

-+

R

is

linear if and only if

it

has the form

I(x)

=

ax

+

b,

where a and b are

constants. Does this agree with Definition 3.9?

If

not, what

is

the form of a

linear function

1 : R

-+

R?

2.

Show explicitly

that

1 : R

-+

R defined by

I(x)

=

v'x

is

not linear.

3.

For each of the following sets of functions, determine whether or not it

is

a

vector space. (Define addition and scalar multiplication in the obvious way.)

If

it

is

not,

state

what property of a vector space fails to hold.

3.1. Linear

systems

as

linear

operator equations

37

(d)

P

n

,

the set of all

polynomials

of

degree

n or

less.

(e)

The set of all

polynomials

of

degree exactly

n.

4.

Prove

that

the

differential

operator

L

:

C

l

[a,

b]

—>

C[a,

b]

defined

by

is

not a

linear operator.

5.

Prove

that

the

differential

operator

L

:

C

l

[a,b]

—>•

C[a,b]

defined

by

is

not a

linear operator.

6.

Prove

that

the

differential

operator

M

:

C

2

[a,

b]

—)•

<7[a,

b]

defined

by

is

a

linear operator.

7.

(a) Let A e

R

2x2

.

Prove

that

(Write

and

write

out

A(ax

+

/3y)

explicitly.)

(b)

Now

repeat

part

7a for A

e

R

nxn

.

The

proof

is not

difficult

when

one

uses summation notation.

For

example,

if the

(i,

j)-entry

of A is

denoted

aij,

then

Write

(A(ax

+

/?y))

i

and

(o:Ax

+

/5Ay)

i

in

summation notation,

and

show

that

the first can be

rewritten

as the

second using

the

elementary

properties

of

arithmetic.

3.1. Linear systems

as

linear operator equations

(a)

{J

E

C[O,

1]

: 1(0) =

O}

(b)

{J

E

C[O,

1]

: 1(0) = I}

(c)

{1

E

C[O,

1]

:

10

1

1(x)

dx

=

O}

(d) P

n

,

the set of all polynomials of degree n or less.

(e) The set of all polynomials of degree exactly

n.

4.

Prove

that

the

differential operator L : C

1

[a,

b]

-+

C[a,

b]

defined by

du

Lu=u-

dx

is

not a linear operator.

5.

Prove

that

the differential operator L : C

1

[a,

b]

-+

C[a,

b]

defined by

is

not a linear operator.

du 3

Lu

= -

+u

dx

6.

Prove

that

the differential operator M : C

2

[a,

b]

-+

C[a,

b]

defined by

is

a linear operator.

d

2

u du

Mu=

--2-+3u

dx

2

dx

7.

(a) Let A E R2x2. Prove

that

A(ax

+

f3y)

=

aAx

+ f3Ay for all a,f3 E

R,

X,y E R2.

(Write

and write out

A(ax

+

f3y)

explicitly.)

37

(b)

Now

repeat

part

7a for A E Rnxn. The proof

is

not difficult when one

uses summation notation. For example, if the

(i,

j)-entry

of A

is

denoted

aij,

then

n

(Ax)i = L

aijXj.

j=1

Write

(A

(ax +

f3Y))i

and

(aAx

+ f3AY)i in summation notation, and

show

that

the first can be rewritten as

the

second using the elementary

properties of arithmetic.

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

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