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

Then

138

Chapter

5.

Boundary value problems

in

statics

In

working with

differential

operators

and

symmetry, integration

by

parts

is

an

essential

technique,

as the

following

examples

show.

Example

5.5.

Let

LD

be the

operator

defined

above. Then,

if

u,v

€

Cf^O,^],

we

have

Therefore,

LD

is a

symmetric operator.

Example

5.6.

Define

138

Chapter 5. Boundary value problems

in

statics

In

working with differential operators and symmetry, integration by

parts

is

an

essential technique, as the following examples show.

Example

5.5.

Let

LD

be

the operator defined

above.

Then,

if

u, v E

C1[0,

f],

we

have

{l

d2u

(LDu,

v) =

-T

10

dx

2

(x)v(x)

dx

[

du ]

l

{l

du dv

=

-T

dx

(x)v(x)

0 + T

10

dx (x) dx (x) dx

~ ~

r~

~

=

-T

dx

(f)v(f)

+ T dx

(O)v(O)

+ T

10

dx (x) dx (x) dx

{i

du

dv

= T

10

dx (x) dx (x) dx (since

v(O)

=

v(f)

= 0)

[

dV]l

r

~v

=

Tu(x)

dx (x) 0 - T

10

u(x)

dx

2

(x) dx

dv du

(l

~v

=

Tu(f)

dx

(f)

- Tu(O) dx

(0)

- T

10

u(x)

dx

2

(x) dx

(l

~v

=

-T

10

u(x)

dx

2

(x) dx (since

u(O)

=

u(f)

= 0)

= (U,LDV).

Therefore,

LD

is

a symmetric operator.

Example

5.6.

Define

c1[0,ij

=

{u

E C

1

[0,f] :

u(O)

=

u(f)

=

O}

and

MD

:

C1[0,

f) -+

C[O,

f)

by

Then

(l

du

(MDu,v)

=

10

dx(x)v(x)dx

r dv

=

u(x)v(x)l~

-

10

u(x)

dx (x) dx

(l

dv

=

u(f)v(f)

- u(O)v(O) -

10

u(x)

dx (x) dx

(i

dv

= -

10

u(x)

dx (x) dx (since

u(O)

=

u(f)

= 0)

=

-(U,MDV).

5.1.

The

analogy between BVPs

and

linear algebraic systems

139

Therefore,

MD

is not

symmetric.

A

symmetric matrix

A

G

R

nxn

has the

following properties (see Section 3.5):

All

eigenvalues

of A are

real.

Eigenvectors

of A

corresponding

to

distinct eigenvalues

are

orthogonal.

There exists

a

basis

of

R

n

consisting

of

eigenvectors

of A.

Analogous properties exist

for a

symmetric

differential

operator.

In

fact,

the first

two

properties

can be

proved

exactly

as

they were

for

symmetric

matrices.

27

In the

following

discussion,

S is a

subspace

of

C

k

[a,b]

and K

:

S

—>

C[a,b]

is

a

symmetric linear operator.

A

scalar

A is an

eigenvalue

of K if

there exists

a

nonzero function

u

such

that

Just

as in the

case

of

matrices,

we

cannot assume

a

priori

that

A is

real,

or

that

the

eigenfunction

u is

real-valued. When working with complex-valued functions,

the

L

2

inner

product

on [a, 6] is

defined

by

The

properties

hold, just

as for the

complex

dot

product.

We

immediately show

that

there

is no

need

to

allow complex numbers when

working

with symmetric operators. Indeed, suppose

u is a

nonzero function

and A

is

a

scalar

satisfying

Then, since

(Ku,u)

=

(u,Ku)

holds

for

complex-valued functions

as

well

as

real-

valued functions (see Exercise

4), we

have

It is

easy

to

show

that

there must

be a

real-valued

eigenfunction corresponding

to

A

(the proof

is the

same

as for

matrices;

see

Theorem 3.44). Therefore,

we do not

need

to

consider complex numbers

any

longer when

we are

dealing with symmetric

operators.

27

The

third property

is

more

difficult.

We

discuss

an

analogous property

for

symmetric dif-

ferential

operators

in the

subsequent sections

and

delay

the

proof

of

this property until Chapter

9.

5.1.

The

analogy between BVPs and linear algebraic systems 139

Therefore,

MD

is

not

symmetric.

A symmetric matrix A E

Rnxn

has the following properties (see Section 3.5):

• All eigenvalues of A are real.

• Eigenvectors of

A corresponding

to

distinct eigenvalues are orthogonal.

• There exists a basis of R

n consisting of eigenvectors of

A.

Analogous properties exist for a symmetric differential operator. In fact, the first

two properties can be proved exactly as they were for symmetric matrices.

27

In the following discussion, S is a subspace of Ck[a,

b]

and K : S

-t

C[a,

b]

is

a symmetric linear operator. A scalar A

is

an eigenvalue of K if there exists a

nonzero function

u such

that

Ku

=

AU.

Just

as in the case of matrices,

we

cannot assume a priori

that

A

is

real, or

that

the

eigenfunction u

is

real-valued. When working with complex-valued functions, the

L2

inner product on

[a,

b]

is

defined by

(f,g)

=

lb

f(x)g(x)

dx. (5.7)

The properties

(al,g)

=a(f,g),

(f,ag)

=a(f,g)

hold, just as for the complex dot product.

We

immediately show

that

there

is

no need to allow complex numbers when

working with symmetric operators. Indeed, suppose

u

is

a nonzero function and A

is

a scalar satisfying

Ku

=

AU.

Then, since

(Ku,u)

=

(u,Ku)

holds for complex-valued functions as

well

as real-

valued functions (see Exercise 4),

we

have

(Ku,u)

=

(u,Ku)

=?

(AU,U)

=

(U,AU)

=?

A(U,U) =

X(u,u)

=?A=X

=?AER.

It

is

easy

to

show

that

there must be a real-valued eigenfunction corresponding

to

A (the proof

is

the same as for matrices; see Theorem 3.44). Therefore,

we

do not

need to consider complex numbers any longer when

we

are dealing with symmetric

operators.

27The

third

property

is

more

difficult. We discuss

an

analogous

property

for

symmetric

dif-

ferential

operators

in

the

subsequent

sections

and

delay

the

proof

of

this

property

until

Chapter

9.

subject

to

Dirichlet

conditions,

is

symmetric, which means

that

any

eigenvalues

must

be

real

and the

eigenfunctions

will

be

orthogonal.

We

have also seen directly

that

any

eigenvalues must

be

positive. Assuming

we can find the

eigenvalues

and

140

Chapter

5.

Boundary value problems

in

statics

We

now

assume

that

AI,

A2

€ R are

distinct eigenvalues

of the

operator

K,

with

corresponding eigenfunctions

wi,W2

G

S.

We

then have

Since

AI

^

A2,

this

is

only possible

if

(^1,^2)

= 0,

that

is, if

u\

and

u^

are

orthog-

onal. Therefore,

a

symmetric

differential

operator

has

orthogonal eigenfunctions.

The

symmetric

differential

operator

LD has a

special property

not

shared

by

every

symmetric operator. Suppose

A is an

eigenvalue

of LD and u is a

corresponding

eigenfunction,

normalized

so

that

(u,u)

—

1.

Then

Therefore,

every eigenvalue

of LD is

positive. (This explains

why we

prefer

to

work

with

the

negative second derivative

operator—the

eigenvalues

are

then positive.

Actually,

in the

computation above,

it is

only obvious

that

and

therefore

that

A

>

0. How can we

conclude

that,

in

fact,

the

inequality must

be

strict?

See

Exercise

5.)

Summary

We

have

now

seen

that

the

differential

operator

140

Chapter

5.

Boundary value problems

in

statics

We

now assume

that

A1,

A2

E R are distinct eigenvalues of

the

operator

K,

with corresponding eigenfunctions

U1,

U2

E S.

We

then

have

A1(U1,U2)

=

(A1U1,U2)

= (

KU

1,U2)

=

(U1,

Ku

2)

=

(U1'

A2U2)

=

A2(Ut,

U

2).

Since

A1

=I

A2,

this is only possible if

(U1'

U2)

=

0,

that

is, if

U1

and

U2

are orthog-

onal. Therefore, a symmetric differential

operator

has orthogonal eigenfunctions.

The

symmetric differential operator

LD

has a special property

not

shared by

every symmetric operator. Suppose A is

an

eigenvalue of

LD

and

U is a corresponding

eigenfunction, normalized so

that

(u,u) =

1.

Then

A =

A(U,U)

=

(AU,U)

=

(LDU,U)

l

£d2u

=

-T

0

dx

2

(x)u(x)

dx

(5.8)

= T

1£

~~

(x)

~~

(x)

dx

(integration by parts)

= T

1£

(~~(x))

2

dx

>

0.

Therefore, every eigenvalue of

LD

is positive. (This explains why

we

prefer

to

work

with

the

negative second derivative

operator-the

eigenvalues are

then

positive.

Actually, in

the

computation above,

it

is only obvious

that

and therefore

that

A

2:

0.

How can

we

conclude

that,

in fact,

the

inequality

must

be strict? See Exercise 5.)

Summary

We

have now seen

that

the

differential

operator

subject

to

Dirichlet conditions, is symmetric, which means

that

any eigenvalues

must be real

and

the

eigenfunctions will be orthogonal.

We

have also seen directly

that

any eigenvalues must

be

positive. Assuming

we

can find

the

eigenvalues

and

We

develop this method

in the

sections.

5.1.1

A

note

about

direct

integration

The BVP

(5.9)

is

quite elementary; indeed,

we

already showed

in

Chapter

2 how

to

solve this

and

similar problems

by

integrating twice (see Examples

2.2 and

2.3).

There

are two

reasons

why we

spend this chapter developing other methods

for

solv-

ing

(5.9). First

of

all,

the

methods

we

develop, Fourier series

and finite

elements,

form

the

basis

of

methods

for

solving time-dependent PDEs

in one

spatial variable.

Such

PDEs

do not

admit direct solution

by

integration.

Second, both

the

Fourier

se-

ries

method

and the finite

element method generalize

in a

fairly

straightforward

way

to

problems

in

multiple spatial dimensions, whereas

the

direct integration method

does not.

5.1.

The

analogy between BVPs

and

linear algebraic systems

141

eigenfunctions

of

LD,

and

that

there

are

"enough"

eigenmnctions

to

represent

u

and

/, we can

contemplate

a

spectral method

for

solving

Exercises

1. Let

MD

be the

operator

defined

in

Example 5.6.

(a)

Show

that

if /

e

C[0,£]

is

such

that

Mpu

= f has a

solution, then

that

solution

is

unique.

(b)

Show

that

Mpu

— f has a

solution only

if /

e

C[0,^]

satisfies

a

certain

constraint. What

is

that

constraint?

2.

Define

C}[Q,t]

=

{u

€

C

l

[Q,i]

:

u(0)

= 0} and

M

7

:

C)[0,£]

->•

C[0,£]

by

(a)

Show

that

M/w

= / has a

unique solution

for

each

/ 6

C[0,^].

(Hint:

Use

the

fundamental theorem

of

calculus.) This

is

equivalent

to

showing

that

JV(M/)

is

trivial

and ft(M/) =

C[Q,f\.

(b)

Show

that

M/

is not

symmetric.

3.

Consider again

the

differential

operator equation

Lu

—

f

discussed

in

Section

5.1. Suppose

an

equation

LSU

= / is

produced

by

restricting

the

domain

of

L to a

subspace

S of

C

2

[0,£]

(that

is, by

defining

LSU =

Lu

for all u € 5).

(a)

Show

that

LSU

= / has at

most

one

solution

for

each

/

e

(7[0,

t]

provided

N(L)nS

=

{Q}.

5.1. The analogy between BVPs

and

linear algebraic systems 141

eigenfunctions

of

L

D

,

and

that

there

are

"enough" eigenfunctions

to

represent u

and

f,

we

can contemplate a spectral method for solving

tPu

-T

dx

2

=

f(x),

0<

x <

£,

u(o)

= 0, (5.9)

u(£)

=

0.

We develop this method in

the

next two sections.

5.1.1 A note about direct integration

The

BVP

(5.9)

is

quite elementary; indeed,

we

already showed

in

Chapter

2 how

to

solve this

and

similar problems by integrating twice (see Examples 2.2

and

2.3).

There are two reasons why

we

spend this chapter developing

other

methods for solv-

ing (5.9).

First

of all,

the

methods

we

develop, Fourier series

and

finite elements,

form

the

basis of methods for solving time-dependent

PDEs

in one spatial variable.

Such

PDEs

do

not

admit

direct solution by integration. Second,

both

the

Fourier se-

ries

method

and

the

finite element

method

generalize in a fairly straightforward way

to

problems in multiple spatial dimensions, whereas

the

direct integration

method

does not.

Exercises

1. Let

MD

be

the

operator

defined in Example 5.6.

(a) Show

that

if f E

C[O,

£]

is such

that

MDu = f has a solution,

then

that

solution

is

unique.

(b) Show

that

MDu = f has a solution only if f E

C[O,

£]

satisfies a certain

constraint.

What

is

that

constraint?

2. Define C][O,

£]

=

{u

E C

1

[0,

£]

: u(O) =

O}

and

MI

: C][O,

£]

--+

C[0,

£]

by

du

M1u =

dx'

(a) Show

that

M1u

= f has a unique solution for each f E

C[O,£].

(Hint:

Use

the

fundamental theorem

of

calculus.) This is equivalent

to

showing

that

N(M

1

)

is trivial

and

R(M

1

)

=

C[O,£].

(b) Show

that

MI

is not symmetric.

3. Consider again

the

differential

operator

equation

Lu

= f discussed in Section

5.1. Suppose

an

equation

Lsu

= f is produced by restricting

the

domain of

L

to

a subspace S of C

2

[0,£]

(that

is, by defining

Lsu

=

Lu

for all u E S).

(a) Show

that

Lsu

= f has

at

most one solution for each f E

C[O,

£]

provided

N(L)

n S =

{O}.

8.

Repeat Exercise

6

with

K

:

C^[Q,f\

->•

C[Q,f\

defined

by

142

Chapter

5.

Boundary

value

problems

in

statics

(b)

Show

that

L$u

— j has a

unique solution

for

each

/

e

C[0,1]

for

either

of

the

following

choices

of

S:

4.

Suppose

5 is a

subspace

of

C

k

[a,

b]

and K

:

S

—>•

C[a,

6] is a

symmetric

linear operator. Suppose

u,

v are

complex-valued

functions, with

u

= / +

ig,

v

=

w

+

iz,

where

/,

g,

w,

z G 5.

Show

that

holds, where

(•, •) is

defined

by

(5.7).

5.

Explain why,

in the

last

step

of the

calculation (5.8),

the

integral

must

be

positive. (Hint:

If

du/dx

is

zero,

then

u

must

be

constant.

What

constant functions belong

to

C£)[0,^]?)

6.

Define

(a)

Show

that

the

null space

of

L

m

is

trivial.

(b)

Show

that

the

range

of

L

m

is all of

(7[0,^].

(c)

Show

that

L

m

is

symmetric.

(d)

Show

that

all of the

eigenvalues

of

L

m

are

positive.

where

a and b are

positive constants.

142

Chapter

5.

Boundary value problems

in

statics

(b) Show

that

Lsu

= f has a unique solution for each f E C[O,f] for either

of the following choices of

S:

i.

S={UEC

2

[0,f]:

~~(~)=O,

f~u(x)dx=O}.

ii. S =

{u

E C

2

[0,f] :

u(O)

=

~~(O)

=

O}.

4.

Suppose S

is

a subspace of Ck[a,

b]

and

K : S -+

C[a,

b]

is

a symmetric

linear operator. Suppose

u, v are complex-valued functions, with u = f + ig,

v = w +

iz,

where

f,g,w,z

E S. Show

that

(Ku,v)

=

(u,Kv)

holds, where (.,.)

is

defined by (5.7).

5.

Explain

why,

in the last step of the calculation (5.8), the integral

must be positive. (Hint:

If

du/dx is zero, then u must be constant.

What

constant functions belong to C1[0, f]?)

6.

Define

C![O,f]

=

{u

E C

2

[0,f]

u(O)

=

~~(£)

=

O}

and

Lm

:

C~

[0,

f] -+

C[O,

f] by

(a) Show

that

the

null space of

Lm

is

trivial.

(b) Show

that

the range of

Lm

is all of C[o,f].

(c)

Show

that

Lm

is

symmetric.

(d) Show

that

all of the eigenvalues of

Lm

are positive.

7.

Repeat Exercise 6 with

C~[O,f]

and

Lift :

C~[O,f]-+

C[O,f] defined by

C~[O,f]

=

{u

E C

2

[0,f] :

~~(O)

= u(f) =

O},

J2u

Liftu = -

dx

2

'

8.

Repeat Exercise 6 with

K:

C1[0,f]-+

C[O,f] defined by

J2u

Ku=-a

dx2

+bu,

where a and b are positive constants.

5.1.

The

analogy

between

BVPs

and

linear

algebraic

systems

143

9.

Consider

the

differential

operator

M :

C^O,

1]

—>•

C[0,1]

defined

by

Show

that

M is not

symmetric

by

producing

two

functions

u,

v 6

(7^(0,1]

such

that

10.

Define

B

:

C

4

[0,l]

->

C[Q,£\

by

(a)

Determine

the

null

space

of

B.

(b)

Find

a set of

boundary conditions such

that,

if 5 is the

subspace

of

(7

4

[0,1]

consisting

of

functions

satisfying

these boundary conditions, then

B,

restricted

to 5, is

symmetric

and has a

trivial null space.

11. In

this exercise,

we

consider

a new

boundary condition, more complicated

than

Dirichlet

or

Neumann conditions,

that

provides

a

more realistic model

of

certain physical phenomena. This boundary condition

is

called

a

Robin

condition;

we

will

introduce

it in the

context

of

steady-state

heat

flow in a

one-dimensional bar.

Suppose

the

ends

of a bar are

uninsulated

and the

heat

flux

through each

end

is

proportional

to the

difference

between

the

temperature

at the end of the bar

and the

surrounding temperature (assumed constant).

If we let a > 0 be the

constant

of

proportionality

and TO,

TI

be the

temperatures surrounding

the

ends

of the bar at x = 0 and x =

i,

respectively, then

the

resulting boundary

conditions

are

These

can be

rewritten

as

Define

5.1.

The

analogy between BVPs and linear algebraic systems

9.

Consider

the

differential operator M :

C1[0,

1)-+

C[O,

1)

defined by

cPu

du

Mu=

--+

-+5u.

dx

2

dx

143

Show

that

M

is

not symmetric by producing two functions

u,

v E

C1[0,1]

such

that

(Mu,v)

=I

(u,Mv).

10. Define B : C

4

[0,

£)

-+

C[O,

£)

by

d

4

u

Bu

=

dx

4

.

(a) Determine the null space of

B.

(b) Find a set of boundary conditions such

that,

if S

is

the subspace of

C

4

[0,

£]

consisting of functions satisfying these boundary conditions, then

B,

restricted

to

S,

is

symmetric and has a trivial null space.

11. In this exercise,

we

consider a new boundary condition, more complicated

than

Dirichlet or Neumann conditions,

that

provides a more realistic model

of certain physical phenomena. This boundary condition

is

called a Robin

condition;

we

will introduce

it

in the context of steady-state heat

flow

in a

one-dimensional bar.

Suppose the ends of a

bar

are uninsulated

and

the heat flux through each end

is

proportional

to

the

difference between the temperature

at

the

end of the

bar

and the surrounding temperature (assumed constant).

If

we

let a > °

be

the

constant of proportionality and

To,

T£

be

the

temperatures surrounding the

ends of

the

bar

at

x = ° and x =

£,

respectively, then the resulting boundary

conditions are

du

K, dx

(0)

=

-a

(To

-

u(O))

,

du

K, dx

(£)

= a

(T£

- u(£)).

These can be rewritten as

du

-K,

dx

(0) +

£lu(O)

=

£lTD,

du

K, dx

(£)

+

£lu(£)

=

£IT£.

Define

c1[0,

£]

= {u E C

2

[0,

£]

:

-K,

~~

(0)

+

£lu(O)

=

0,

K,

~~(£)

+

£lu(£)

=

O},

and let

LR

:

CMO,

£]

-+

C[O,

£]

be defined by

cPu

LRU

=

-K,

dx

2

·

This

BVP can be

written simply

as

Lpu

= /,

where

LD is the

symmetric linear

differential

operator

defined

in the

last

section.

The

spectral method

for a

symmetric linear system

Ax = b is

based

on

the

fact

that

a

symmetric matrix

A

e

R

nxn

has n

orthonormal eigenvectors,

and

therefore

any

vector

in

R

n

(including

the

right-hand side

b and the

solution

x) can

be

written,

in a

simple way,

as a

linear combination

of

those eigenvectors. (The

reader

may

wish

to

review

Section

3.5 at

this time.)

We

have seen,

at the end of

the

last

section,

that

any

eigenvalues

of LD

must

be

real

and

positive

and

that

eigenfunctions

of LD

corresponding

to

distinct eigenvalues must

be

orthogonal.

We

now

find the

eigenvalue-eigenfunction

pairs

(eigenpairs

for

short)

of LD and

explore

the

following

question:

Can we

represent

the

right-hand side

f(x)

and the

solution

u(x)

in

terms

of the

eigenfunctions?

5.2.1

Eigenpairs

of

—

-j^

under

Dirichlet

conditions

We

will

show

that

the

operator

LD has an

infinite

collection

of

eigenpairs.

To

simplify

the

calculations,

we may as

well

assume

that

T = 1,

since

if

A,w

is an

eigenpair

of

5.2

Introduction

to the

spectral method;

eigenfunctions

In

this

section

and the

next,

we

develop

a

"spectral"

method

for the BVP

144

Chapter

5.

Boundary value problems

in

statics

(As

usual,

we

define

the

operator

in

terms

of the

homogeneous

version

of the

boundary conditions.)

(a)

Prove

that

LR is

symmetric.

(b)

Find

the

null space

of LR.

(subject

to

Dirichlet conditions), then

T\,u

is an

eigenpair

of

under

the

same boundary conditions.

To

find the

eigenpairs,

we

need

to

solve

144

Chapter 5. Boundary value problems

in

statics

(As usual,

we

define

the

operator in terms of the homogeneous version of

the

boundary conditions.)

(a) Prove

that

LR

is

symmetric.

(b) Find the null space of

L

R.

5.2 Introduction

to

the spectral method:

eigenfunctions

In this section and the next,

we

develop a "spectral" method for

the

BVP

J2u

-T

dx2

=/(x),

O<x<£,

u(O)

=

0,

(5.10)

u(£)

=

O.

This BVP can be written simply as

LDu

=

/,

where

LD

is the symmetric linear

differential operator defined in

the

last section.

The spectral method for a symmetric linear system

Ax

= b

is

based on

the fact

that

a symmetric matrix A E

RllXll

has n orthonormal eigenvectors, and

therefore any vector in

Rll

(including the right-hand side b

and

the solution x) can

be written, in a simple

way,

as a linear combination of those eigenvectors. (The

reader may wish to review Section 3.5

at

this time.)

We

have seen,

at

the

end of

the

last section,

that

any eigenvalues of

LD

must be real and positive

and

that

eigenfunctions of

LD

corresponding to distinct eigenvalues must be orthogonal.

We

now find

the

eigenvalue-eigenfunction pairs (eigenpairs for short) of

LD

and explore

the

following question: Can

we

represent the right-hand side

/(x)

and

the

solution

u(x) in terms of

the

eigenfunctions?

5.2.1 Eigenpairs of -

:1:2

under Dirichlet conditions

We

will show

that

the

operator

LD

has an infinite collection of eigenpairs. To

simplify the calculations,

we

may as

well

assume

that

T =

1,

since if

A,

u

is

an

eigenpair of

d

2

dx

2

(subject

to

Dirichlet conditions), then T).., u is

an

eigenpair of

under the same boundary conditions.

J2

-T-

dx

2

To

find the eigenpairs,

we

need

to

solve

d

2

u

-

dx

2

=

AU,

u(O)

= 0,

u(£)

=

O.

(5.11)

We

already know

that,

since

the

eigenvalues

of LD are

distinct,

the

eigenfunc-

tions listed above must

be

orthogonal. This

can

also

be

verified

directly using

the

trigonometric identity

5.2.

Introduction

to the

spectral

method;

eigenfunctions

145

"Solving" this problem means

identifying

those special values

of A

such

that

the

BVP

(5.11)

has a

nonzero solution

u,

and

then determining

u.

(For

any

value

of

A,

the

zero

function

is a

solution

of

(5.11), just

as x

=

0 is a

solution

of Ax = Ax

for

any A. In

both

cases,

the

eigenvalues

are the

scalars

which lead

to a

nonzero

solution.) Since

we

know,

from

the

previous section,

that

all

such

A are

positive,

we

write

A =

O

2

.

We can use the

results

of

Section

4.2 to

write

the

general solution

of

the ODE

and

then

try to

satisfy

the

boundary conditions.

The

characteristic roots

of

(5.12)

are

±0i,

and the

general solution

is

Therefore,

and the first

boundary condition implies

c\

— 0. We

then have

u(x]

—

C2sin(0x),

and the

second

boundarv

condition becomes

This equation only holds

in two

cases:

1. The

value

of

C2

is

zero.

But

then

u(x)

is the

zero function, which cannot

be

an

eigenfunction.

2.

The

value

of A

=

O

2

is

such

that

sin

(Of.)

= 0.

This holds

in the

cases

Solving

for A, we

have

(we

discard

the

case

that

A = 0, for

then

u is the

zero

function).

It

turns

out

that

the

value

of

c^

is

immaterial; these values

of A, and no

others, produce

eigenfunctions

with

the

correct boundary conditions.

We

thus

see

that

the

operator

LD has

infinitely

many

eigenpairs:

5.2. Introduction

to

the

spectral method; eigenfunctions 145

"Solving" this problem means identifying those special values of A such

that

the

BVP

(5.11) has a nonzero solution

u,

and then determining

u.

(For any value of

A,

the zero function

is

a solution

of

(5.11),

just

as x = 0

is

a solution of

Ax

=

AX

for any

A.

In

both

cases, the eigenvalues are the scalars which lead

to

a nonzero

solution.) Since

we

know, from the previous section,

that

all such A are positive,

we

write A =

(J2.

We

can use the results of Section 4.2 to write

the

general solution

of

the

ODE

cf2u

2

dx

2

+ 0 U = 0,

(5.12)

and

then

try

to

satisfy the boundary conditions.

The characteristic roots of

(5.12) are ±Oi, and the general solution

is

u(x) =

Cl

cos

(Ox)

+ C2 sin

(Ox).

Therefore,

u(O)

=

Cl,

and the first boundary condition implies

Cl

=

O.

We

then have u(x) = C2 sin

(Ox),

and the second boundary condition becomes

C2

sin

(Of)

=

O.

This equation only holds in two cases:

1.

The

value of

C2

is

zero.

But

then

u(x) is the zero function, which cannot be

an

eigenfunction.

2.

The

value of A = 0

2

is

such

that

sin

(Of)

=

O.

This holds in the cases

Of

=

±n1f,

n =

1,2,

....

Solving for

A,

we

have

n

2

1f2

A=~,

n=

1,2,

...

(we discard

the

case

that

A =

0,

for

then

u

is

the

zero function).

It

turns out

that

the value of

C2

is

immaterial; these values of

A,

and no others, produce

eigenfunctions with the correct boundary conditions.

We

thus see

that

the

operator

LD

has infinitely many eigenpairs:

(5.13)

We

already know

that,

since the eigenvalues of

LD

are distinct,

the

eigenfunc-

tions listed above must

be

orthogonal. This can also be verified directly using the

trigonometric identity

1

sin Q sin.B =

"2

(cos

(Q

-

.B)

- cos

(Q

+

.B))

146

Chapter

5.

Boundary value problems

in

statics

that

is,

Thus

the

eigenfunctions

are

orthogonal,

and

they

can be

normalized

by

multiplying

each

by

^/2/l.

The first

four

eigenfunctions

V>i,

V>2,

V's,

V>4

a^

6

graphed

in

Figure 5.2.

Figure

5.2.

The first

four eigenfunctions

^15^2,^85^4

ft —

1J-

5.2.2 Representing functions

in

terms

of

eigenfunctions

We

have shown

that

the

operator

—d?/dx

2

,

subject

to

Dirichlet boundary condi-

tions,

has an

infinite number

of

eigenpairs,

and

that

the

eigenfunctions

are

orthog-

onal.

We now

must answer

the

crucial question:

is it

possible

to

represent both

the

solution

and

right-hand side

of

(5.10)

in

terms

of the

eigenfunctions?

(see

Exercise

1).

Also,

for

each

n,

146

Chapter

5.

Boundary value problems

in

statics

(see Exercise 1). Also, for each n,

that

is,

II1/1nll

=

[f,

n =

1,2,

....

Thus the eigenfunctions are orthogonal, and they can be normalized by multiplying

each by

-/27"f.

1

0.5

-0.5

-1

The

first four eigenfunctions

1/11,1/12,1/13,1/14

are graphed in Figure 5.2.

-\""'-'W"'"

-

.........

,,'

,-

",;

" '

:,

,

,:

i '

" ,

,:,'

,

:',

,

:;

,

:'

,

,::.',

:"

jl

o

,.,

\

0.2

\

,

\

,

\

,

\

,

\

'.,'

\

.

\

,

\ ,

\'

: \

0.4

\

"

,:

:"

\

, "

'

......

,,;

x

,

\

\ ,

\t

"

,

-' \

,

0.6

,

'"

,',

I "

'

..

\

, :

, ,:

, :

"

:'

,

---;

0.8

',' :

"

.-

,-

",'

Figure

5.2.

The first four eigenfunctions

1/11,

'1/12,1/13,1/14

(£

=

I),

5.2.2 Representing functions

in

terms of eigenfunctions

1

We

have shown

that

the operator -d? / dx

2

,

subject

to

Dirichlet boundary condi-

tions, has

an

infinite number of eigenpairs, and

that

the eigenfunctions are orthog-

onal.

We

now must answer the crucial question:

is

it

possible

to

represent

both

the

solution and right-hand side of (5.10) in terms of the eigenfunctions?

5.2.

Introduction

to the

spectral

method;

eigenfunctions

147

Let

us

take

any

function

/

e

<7[0,^].

We

know,

from

Section 3.5,

how to find

the

best

approximation

to /

from

a

finite-dimensional

subspace.

We

define

We

now

show several examples

of

this

approximation.

28

In the

following examples,

1

=

1.

Example

5.7.

Let

Then

Figure

5.3

shows

the

errors

in

approximating

/(#)

by

/N(X]

for N =

10,20,40,80.

(The

approximations

are so

good

that,

if

we

were

to

plot

both

f and

/AT

on the

same

graph,

the two

curves would

be

virtually indistinguishable.

Therefore,

we

plot

the

difference

f(x)

—

/N(X)

instead.)

In

this example,

the

approximation error

is

small

and

gets increasingly small

as N

increases.

It is

easy

to

believe

that

28

Many

specific

integrals must

be

computed

to

compute these approximations.

It is

beyond

the

scope

of

this book

to

discuss methods

of

integration; moreover, modern technology

often

allows

us

to

avoid this mechanical

step.

Computer algebra software, such

as

Mathematica,

MATLAB,

Maple,

etc.,

as

well

as the

more

powerful

handheld calculators,

will

compute many integrals symbolically.

The

best

approximation

to /

from

FN

is

Since

(?/>

n

,V>n)

=

^/2

for all n, we

have

Example

5.8.

Let

Then

Figure

5.4

shows

f(x)

together with

the

approximations

/jv(ar)

for N =

10,20,40,80

5.2. Introduction

to

the spectral method; eigenfunctions 147

Let us take any function f E

e[o,

fl.

We

know, from Section 3.5, how to find

the best approximation to

f from a finite-dimensional subspace.

We

define

{

.

(7fX)

.

(27fX)

.

(N7fX)}

FN = span sm e

,sm

-f-

,

...

,SIn

-f-

.

The best approximation

to

f from

FN

is

Since

(1/Jn,1/Jn)

=

f/2

for all

n,

we

have

(1/Jn,

f)

2 r

l

( ) .

(n7fX)

C

n

=

(1/Jn,1/Jn)

="iJo

fx

sm

-f-

dx,

n=1,2,

...

,N.

We

now show several examples of this approximation.

28

In

the following examples,

f =

1.

Example

5.7.

Let

f(x)

=

x(l

-

x).

Then

1

4(1+

(_l)n+1)

c

n

=2

x(1-x)sin(n7fx)dx=

33

.

o n

7f

Figure 5.3 shows the errors

in

approximating

f(x)

by fN(X)

for

N =

10,20,40,80.

(The

approximations are so good that,

if

we were to plot both f

and

fN

on the

same

graph, the two curves would

be

virtually indistinguishable. Therefore, we plot the

difference

f(x)

-

fN(x)

instead.)

In

this example, the approximation error is

small

and

gets increasingly

small

as N increases.

It

is

easy to believe

that

fN(x)

-+

f(x)

as N

-+

00.

Example

5.8.

Let

f(x)

=

I-x.

Then

C

n

= 2 r

1

(1

-

x)

sin (n7fx)

dx

=

~.

J

o

n7f

Figure

5.4

shows

f(x)

together with the approximations fN(X) for N =

10,20,40,80.

28Many specific integrals

must

be

computed

to

compute

these approximations.

It

is beyond

the

scope

of

this

book

to

discuss

methods

of integration; moreover,

modern

technology often allows us

to

avoid this mechanical step.

Computer

algebra software, such as Mathematica,

MATLAB,

Maple,

etc., as well as

the

more powerful

handheld

calculators, will

compute

many

integrals symbolically.

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

Подождите немного. Документ загружается.