Brewster H.D. Mathematical Physics

242

Transform

Techniques

in

Physics

The

integral we

have

to

compute

is

f (t) = _1_

f+i

OO

est

ds

21ti

k-iOO

s(s

+

1)

This

integral has poles

at

s = 0

and

s = -1.

We

enclose

the

contour

with a semicircle

to

the

left

of

the

path

in

the

complex

s-plane.

One

has

to

verify

that

the

integral

over

the

semicircle vanishes as

the

radius

goes

to

infinity. Assuming

that

we

have

done

this, then

the

result is simply

obtained

as

2rci

times

the sum

of

the

residues.

The

residues

in

this case are:

Res ;

z = 0 = lim

_e

- = 1

[

ea

1 a

z(z+l)

z~o(z+l)

and

Res ;

z

-1

= lim

~

-=

_e-

t

[

ea

1 a

z(z

+

1)

z~-l

z .

Therefore, we have

fit)

= 21ti(_I_.

(1)

+

_1_.

(-e

-I»)

=

1-

e

-I.

2m 2m

1

Fig. The Contour

used

for Applying the Bromwich Integral to

F(s)

=

---

s(s

+

1)

We

can verify this result using

the

Convolution

Theorem

or

using partial

fraction decomposition.

The

decomposition is simplest.

1 1 1

s(s+l)

= -;-

s+l·

The

first

term

leads

to

an

inverse

transform

of

1

and

the

second

term

gives

an

e-

t

.

Thus,

we

have

verified

the

result

from

doing

a

contour

integration.

Chapter

7

Problems

in

Higher

Dimensions

We will explore several generic examples

of

the solution

of

initial-

boundary value problems involving higher spatial dimensions. These are

described by higher dimensional partial differential equations. The method

of

solution will be

the

method

of

separation

of

variables. This will result

in a system

of

ordinary differential equations for each problem.

Of

these equations, several are boundary value problems, which are

eigenvalue problems. We solve the eigenvalue problems for the eigenvalues

and eigenfunctions, leading to a set

of

product solutions satisfying the partial

differential equation

and

the boundary conditions. The general solution can

be written

as

a linear superposition

of

the product solutions.

As

you

go

through

these examples,

you

will see some

common

features.

For

example, the main equation

that

we

have seen are

the

heat

equation

and

the wave equation.

For

higher dimensional problems these

take the form

u =

kV

2

u

t '

Utt

= c

2

V

2

u.

One can first separate

out

the time dependence. Let

u(r,

t)

=

<I>(r)

T (t).

For

these two equations

we

have

T<I>

= k'JV

2

<1>,

T"<I>

= c

2

'JV2<1>.

Separating

out

the time

and

space dependence, we find

1

T'

V2~

--

=-=-A

k T

~

1 T

W

V2~

--

----A

c

2

T -

~

-

Note

that in each case

we

have a function

of

time equals a function

of

spatial variables. Thus, they must be a constant,

-A

<

O.

The sign

of

A

is

chosen because

we

expect decaying solutions in time for the heat equation

and

oscillations in time for the wave equation.

244

Problems

in

Higher Dimensions

This leads

to

the

respective set

of

equations for

T (t).

T'=

-

'AkT,

T"+

c

2

'AT=

0.

These are easily solved.

We

have

T (t) = T (O)e-

lJa

,

T

(t)

= acos

rot

+ b sin

rot

ro

=

cfi.

.

In

both

cases

the

spatial

equation

becomes

V2<\>

+

'A<\>

=

0.

This

is

called

the

Helmholtz equation.

For

one

dimensional problems, which we

have

already solved,

the

Helmholtz

equation

takes

the

form

'A"

+

'A<\>

=

0.

We

had

to

impose

the

boundary

conditions

and

found

that

there

were a discrete set

of

eigenvalues,

'An'

and

associated eigenfunctions,

<\>n.

In

higher dimensional problems we need

to

further

separate

out

the

space dependence.

We

will again use

the

boundary

conditions

and

find

the

eigenvalues

and

eigenfunctions,

though

they will

be

labelled with more

than

one

index.

The

resulting

boundary

value problems belong

to

a class

of

eigenvalue

problems

called Sturm-Liouville problems.

Also, in

the

solution

of

the

ordinary

differential equations we will

find solutions

other

than

the sines

and

cosines

that

we have seen in previous

problems, such

as

the vibrations

of

a string.

The

special functions also form a

set

of

orthogonal functions, leading

to

general Fourier-type series expansions.

VIBRATIONS

OF

RECTANGULAR

MEMBRANES

Our

first example will

be

the

study

of

the

vibrations

of

a

rectangular

membrane.

You

can

think

of

this as a

drum

with a

rectangular

cross-section.

We

stretch

the

membrane

over

the

drumhead

and

fasten

the

material

to

the

boundaries

of

the rectangle.

The

hight

of

the

vibrating

membrane

is

described by its hight

from

equilibrium, u(x,

y,

t). This

problem

is a

much

simpler example

of

higher

dimensional vibrations than that possed by the oscillating electric

and

magnetic

fields.

The

problem

is

given by a

partial

differential

equation,

u

tt

= c

2

V

2

u = c

2

(u

xx

+ U

yy

)' t > 0, ° < x <

L,

° < y <

H.

a set

of

boundary

conditions,

1l(0,

y, t) = 0, u(L, y, t) = 0, t > 0, ° < y <

H,

u(x,

0,

t) = 0, u(x, H, t) = 0, t > 0, ° < x < L,

and

a

pair

of

initial conditions (since

the

equation

is

second

order

in time),

u(x, y, 0) = f (x,

y),

ut

(x, y, 0) = g(x, y).

Problems

in

Higher

Dimensions

245

The

first step

is

to

separate

the

variables. u(x,

y,

t) = X (x)y

(y)T(t).

Inserting into the wave equation, we

hav~

X(x)y (y)T"(t) = c

2

-X"(x)

Y

(y)T

(t) + X (x)

Y"(y)T

(t».

Dividing by

both

u(x,

y,

t)

and

c

2

,

we obtain

H

IT"

c

2

T

'----v--'

Function

of

t

y

=

X"

Y"

X Y

'---v---'

Function

of

t

and

y

=-A.

--~--------------------~--------.x

L

Fig. The Rectangular Membrane

of

Length L

and

Width H. There are Fixed

Boundary Conditions Along the Edges

We see

that

we

have a function

of

t equals a function

of

x and

y.

Thus,

both expressions are constant. We expect oscillations in time, so

we

chose the

constant to be

A >

O.

(Note. As usual, the primes mean differentiation with

respect to the specific dependent variable. So, there should be no ambiguity.)

and

These lead to two equations.

T"+

C

2

AT=

0

,

X"

Y"

-X+-y

=-A.

The

first equation

is

easily solved. We have

T (t) = a cos

rot

+ b sin

mt,

where

m =

c-JA.

This

is

the angular frequency in terms

of

the separation constant,

or

246 Problems in Higher Dimensions

eigenvalue.

It

leads

to

the

frequency

of

oscillations

for

the

various

harmonics

of

the

vibrating

membrane

as

v=~=~Ji.

2n 2n

Once

we

know

we

can

compute

these frequencies.

Now

we solve

the

spatial equation. Again, we need

to

do

a separation

of

variables.

Rearranging

the

spatial

equation,

we have

X"

Y"

= ---')... = -/l.

Y

~

Function

of

y

X

'-r-'

Function

of

t

Here

we have a

function

of

x equals a function

of

y. So,

the

two

expressions

are

constant,

which we

indicate

with a

second

separation

constant,

-/-1

<

0.

We pick

the

sign in this way because we expect oscillatory

solutions

for

X (x). This leads

to

two equations.

X"

+

r.tX

= 0,

Y"

+

(')...

-

/-1)

Y =

0.

We

now

need

to

use

the

boundary

conditions.

We

have

u(O,

y, t) = °

for

all t > °

and

0<

y < H. This implies

that

X

(o)y

(y)T(t)

= ° for all t

andy

in

the

domain. This

is

only

true

if

X (0) =

0.

Similarly, we find

that

X(L)

= 0,

YeO)

= 0,

and

Y(H)

=

0.

We

note

that

homogeneous

boundary

conditions

are

important

in

carrying

out

this process.

Nonhomogeneous

boundary

conditions could

be

imposed,

but

the

techniques are a

bit

more

complicated

and

we will

not

discuss these techniques here.

The

boundary

values

problems

we need

to

solve are.

X"

+

/-1X

= 0, X (0) = 0, X

(L)

=

0.

Y"+

(')...- /-1)Y= 0, YeO) = 0,

Y(H)

=

0.

We

have seen the first

of

these problems before, except with a

')...

instead

of

a

/-1.

The

solution

is

mrx

(nn)2

X(x)

=1:'

')...

= L ' n =

1,2,3,

...

The

second

equation

is

solved in

the

same way.

The

differences

are

that

our

"eigenvalue"

is

A-

/-1,

the

independent variable

is

y,

and

the

interval

is

[0, H].

Thus,

we

can

quickly write

down

the

solution

as A

instead

of

a

/-1.

The

solution

is

)

.

mrrx

Y(y

=sm

H

,

')...-11=(

n;;

r,

m =

1,2,3,

....

We

have successfully carried

out

the

separation

of

variables for

the

wave

Problems

in

Higher

Dimensions 247

equation

for

the

vibrating

rectangular

membrane.

The

product

solutions

can

be

written as

= (

'"

t + b .

r.,

t)·

rna

.

m7ty

u

nm

a cos ""nm

SIn

""nm

sln-sm--·

L H

Recall

that

(i)

is

given in

terms

of

A.

We

have

that

m7t

(

)

2

Amn-

~n

= H

and

Therefore,

So,

s

The

most general solution

can

now

be written as a linear combination

of

the

product

solutions

and

we

can

solve for the expansion coefficients

that

will

lead

to

a

solution

satisfying

he

initial conditions.

However,

we will first

concentrate

on

the two dimensional harmonics

of

this membrane.

For

the

vibrating string

the

nth

harmonic

corresponded

to

the

function

rna

sin T .

The

various harmonics corresponded

to

the

pure

tones supported

by

the

string. These

then

lead

to

the

corresponding frequencies

that

one

would

hear.

The

actual

shapes

of

the

harmonics

could

be

sketched

by

locating

the

nodes,

or

places

on

the string

that

did

not

move.

In

the

same

way, we

can

explore

the

shapes

of

the

harmonics

of

the

vibrating membrane. These

are

given

by

the

spatial functions

.

rna

.

m7tY

<l>nm(x,y)

=

smTsm/i'

Instead

of

nodes, we will

look

for

the

nodal

curves,

or

nodal

lines.

These

are

the

points

(x,

y)

at

which

<I>(x,

y)

=

O.

Of

course, these

depend

on

the

indices,

nand

m.

248

Y

H

1------,

L

x

Y

H

....

_--_

.....

L

x

Y

H

1-----.

L x

Problems

in

Higher

Dimensions

Y

H

1---....---,

L

Y

H

.....

_

........

·

L

Y

H

1---....---,

·

.......

'-----

·

----0,,----

·

x

L x

Y

H

I--_r_--r--,

L

.Y

H

..

.........

1

......

.

L

Y

H

I--_r_--r--,

· .

..

.........

_I

......

· .

..

_1-

..

_1_

....

· .

x

L x

Fig. The First

few

modes

of

the Vibrating Rectangular Membrane. The Dashed

Lines show the Nodal lines Indicating the Points that do not move for the

Particular Mode

For

example, when n = 1 and m =

1,

we

have

.

moe

.

mny

Slll--Slll--

= °

L H .

These are zero when either

. mtx 0 .

mny

Slll--=

or

Slll--

= °

L'

H·

Of

course, this can only happen for x = 0,

Land

y =

0,

H. Thus,

there are no interior nodal lines.

When

n = 2

and

m =

1,

we

have y = 0,

Hand

. 2nx

Slll--

= °

L '

L L

or, x = 0,

2"'

L.

Thus, there

is

one interior nodal line

at

x

=2".

These

points stay fixed during the oscillation and all other points oscillate

on

either side

of

this line.

A similar solution shape results for the (l,2)-mode, i.e.,

n = 1

and

m=2.

Problems in Higher Dimensions

n=1

nz3

m=1

m=2

m=3

Fig. A three Dimensional view

of

the Vibrating

Rectangular Membrane for the Lowest Modes

249

The nodal lines for several modes for n, m = 1,2, 3 The blocked regions

appear to vibrate independently. The frequencies

of

vibration are easily

computed

usi~g

the formula for

ffi

nm

.

For

completeness,

we

now see how one satisfies the initial conditions.

The

general solution

is

given by a linear superposition

of

the

product

solutions. There are two indices to sum over. Thus, the general solution

is

The

first initial condition

is

u(x, y,

0)

= f (x, y). Setting t = 0 in the

general solution,

we

obtain

00

" " .

mtx

.

m1ty

f(x,

y)

=~

~anmsmLsmH·

n=lm=1

This

is

a double sine series. The goal

is

to find the unknown coefficients

a

nm

. This can be done knowing what

we

already know

about

Fourier sine

series. We can write this as

the

single sum

where

250

Problems

in

Higher

Dimensions

These are two sine series. Recalling that the coefficients

of

sine series

can

be

computed

as

integrals,

we

have

2

rL

mtx

An(y)

=L

.b

J(x,y)sinT

'

21H

.

m1ty

a

=-

A

(y)slO-dy

nm

H n

H·

Inserting the first result into the second,

4

1H

1L

n1ty

m1ty

a = -

J(x,y)sin-sin-dxdy

nm

LH

L H .

y

x

Fig. The Circular Membrane

of

Radius Q A General Point on the Membrane

is

Given by the Distance from the centre,

r,

and the Angle,. There are Fixed Boundary

Conditions Along the Edge at

r = Q .

We

can carry out the same process for satisfying the second initial

condition,

ut

(x, y, 0) = g(x,

y)

for the initial velocity

of

each point.

Inserting this into the general solution, we have

Again, we have a double sine series. But, now we can write down

Fourier coefficients quickly:

4

1H

1L

. ( ).

n1tX

.

m1ty

dxdy

b = g

x,y

SIO--SIO--

nm

wnmLH

L H .

Problems

in

Higher

Dimensions

251

This completes the full solution

of

the vibrating rectangular membrane

problem.

VIBRATIONS OF A

KETTLE

DRUM

In

this section we consider the vibrations

of

a circular membrane

of

radius.

Again we are looking for the harmonics

of

the vibrating membrane, but with

the membrane fixed around the circular boundary given by

x

2

"+

y2

= a

2

.

However, expressing the boundary condition

in

Cartesian coordinates would

be awkward. In this case the boundary condition would be

u = 0 at r =

a.

Before solving the initial-boundary value problem, we have to cast it

in

polar coordinates. This means that we need to rewrite the Laplacian

in

rand

e.

To do so would require that we know how to transform derives

in

x and y

into derivatives with respect to

rand

e.

First recall that transformations

x = r cos

e,

y = r sin e

and

r=~x2+i,

tane=Y.

x

Now, consider a

functionf=

f(x(r,

e),

y(r,

e»

= g(r,

e).

(Technically,

once we transform a given function

of

Cartesian coordinates we obtain a

new function g

of

the polar coordinates. Many texts do not rigourously

distinguish between the two which

is

fine when this point

is

clear.)

Thinking

of

x = x(r,

e)

and y = y(r, 8), we have from the chain rule

for functions

of

two variables.

aj

= ag

ar

+ ag

as

ax

ar

ax

ae

ax

Here we have used

ar

ax

and

Similarly,

ag x ag y

ar

r

x x

_.

,

r

Brewster H.D. Mathematical Physics

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