Orthogonal Functions and Fourier Series

CHAPTER 12

Ch12_2

Contents

12.1 Orthogonal Functions12.2 Fourier Series12.3 Fourier Cosine and Sine Series12.4 Complex Fourier Series12.5 Strum-Liouville Problems12.6 Bessel and Legendre Series

Ch12_3

12.1 Orthogonal Functions

The inner product of two functions f1 and f2 on an interval [a, b] is the number

DEFINITION 12.1Inner Product of Function

b

adxxfxfff )()() ,( 2121

Two functions f1 and f2 are said to be orthogonal on an interval [a, b] if

DEFINITION 12.2Orthogonal Function

0)()(),( 2121 b

adxxfxfff

Ch12_4

Example

The function f1(x) = x2, f2(x) = x3 are orthogonal on the interval [−1, 1] since

061

),(1

1

631

1

221

xxdxxff ．

Ch12_5

A set of real-valued functions {0(x), 1(x), 2(x), …}

is said to be orthogonal on an interval [a, b] if

(2)

DEFINITION 12.3Inner Product of Function

nmdxxxb

a nmnm ,0)(),(),(

Ch12_6

Orthonormal Sets

The expression (u, u) = ||u||2 is called the square norm. Thus we can define the square norm of a function as

(3)

If {n(x)} is an orthogonal set on [a, b] with the property that ||n(x)|| = 1 for all n, then it is called an orthonormal set on [a, b].

,)( 22 b

a nn dxx b

a nn dxxx )()( 2

Ch12_7

Example 1

Show that the set {1, cos x, cos 2x, …} is orthogonal on [−, ].

Solution Let 0(x) = 1, n(x) = cos nx, we show that

0 ,0sin1

cos)()() ,( 00

nfornx

n

nxdxdxxx nn

Ch12_8

Example 1 (2)

and

nmnm

xnmnm

xnm

dxxnmxnm

nxdxmxdxxx nmnm

,0)sin()sin(

21

])cos()[cos(21

coscos)()(),(

Ch12_9

Example 2

Find the norms of each functions in Example 1.

Solution

0 ,||||

)2cos1(21

cos||||

,cos

22

n

dxnxnxdx

nx

n

n

n

22 ,1 02

00 dx

Ch12_10

Vector Analogy

Recalling from the vectors in 3-space that

(4)we have

(5)

Thus we can make an analogy between vectors and functions.

,332211 vvvu ccc

3

1232

3

322

2

212

1

1

||||

),(

||||

),(

||||

),(

||||

),(

nn

n

n vv

vuv

v

vuv

v

vuv

v

vuu

Ch12_11

Orthogonal Series Expansion

Suppose {n(x)} is an orthogonal set on [a, b]. If f(x) is defined on [a, b], we first write as

(6)

Then

?)()()()( 1100 xcxcxcxf nn

),(),(),(

)()(

)()()()(

)()(

1100

1100

mnnmm

b

a mnn

b

a m

b

a m

b

a m

ccc

dxxxc

dxxxcdxxxc

dxxxf

Ch12_12

Since {n(x)} is an orthogonal set on [a, b], each term on the right-hand side is zero except m = n. In this case we have

,...2,1,0 ,)(

)()(

)(),()()(

2

2

ndxx

dxxxfc

dxxccdxxxf

b

a n

b

a nn

b

a nnnnn

b

a n

Ch12_13

In other words,

(7)

(8)

Then (7) becomes

(9)

0

),()(n

nn xcxf

2||)(||

)()(

x

dxxxfc

n

b

a n

n

02 )(

||)(||

),()(

nn

n

n xx

fxf

Ch12_14

Under the condition of the above definition, we have

(10)

(11)

A set of real-valued functions {0(x), 1(x), 2(x), …}

is said to be orthogonal with respect to a weight

function w(x) on [a, b], if

DEFINITION 12.4Orthogonal Set/Weight Function

nmdxxxxwb

a nm ,0)()()(

2||)(||

)()()(

x

dxxxwxfc

n

b

a n

n

b

a nn dxxxwx )()(||)(|| 22

Ch12_15

Complete Sets

An orthogonal set is complete if the only continuous function orthogonal to each member of the set is the zero function.

Ch12_16

12.2 Fourier Series

Trigonometric SeriesWe can show that the set

(1)

is orthogonal on [−p, p]. Thus a function f defined on [−p, p] can be written as

(2)

,

3sin,

2sin,sin,,

3cos,

2cos,cos,1 x

px

px

px

px

px

p

1

0 sincos2

)(n

nn xp

nbx

pn

aa

xf

Ch12_17

Now we calculate the coefficients.

(3)

Since cos(nx/p) and sin(nx/p) are orthogonal to 1 on this interval, then (3) becomes

Thus we have

(4)

1

0 sincos2

)(n

p

pn

p

pn

p

p

p

pdxx

pn

bdxxp

nadx

adxxf

000

22)( pax

adx

adxxf

p

p

p

p

p

p

p

pdxxf

pa )(

10

Ch12_18

In addition,

(5)

by orthogonality we have

1

0

sincoscoscos

cos2

cos)(

n

p

pn

p

pn

p

p

p

p

dxxp

nx

pm

bdxxp

mx

pm

a

dxxp

ma

dxxp

mxf

0sincos

0,0cos

p

p

p

p

xdxp

nx

pm

mxdxp

m

Ch12_19

and

Thus (5) reduces to

and so

(6)

nmp

nmxdx

pn

xp

mp

p ,

0,coscos

paxdxp

nxf n

p

p

cos)(

p

pn dxxp

nxf

pa

cos)(

1

Ch12_20

Finally, if we multiply (2) by sin(mx/p) and use

and

we find that

(7)

0sinsin

0 ,0sin

p

p

p

p

xdxp

nx

pm

mxdxp

m

nmp

nmxdx

pn

xp

mp

p ,

0,sinsin

p

p nmp

nmdxx

pn

xp

m,

,0sinsin

Ch12_21

The Fourier series of a function f defined on the

interval (−p, p) is given by

(8)where

(9)

(10)

(11)

DEFINITION 12.5Fourier Series

1

0 )sincos(2

)(n

nn xp

nbx

pn

aa

xf

p

pdxxf

pa )(

10

p

pn dxxp

nxf

pa

cos)(

1

p

pn dxxp

nxf

pb

sin)(

1

Ch12_22

Example 1

Expand (12)

in a Fourier series.

SolutionThe graph of f is shown in Fig 12.1 with p = .

xx

xxf

0,

0,0)(

221

)(01

)(1

0

2

0

0

0

xx

dxxdxdxxfa

Ch12_23

Example 1 (2)

22

0

00

0

0

)1(11cos

cos1

sin1sin

)(1

cos)(01

cos)(1

nn

n

nnx

n

dxnxnn

nxx

dxnxxdx

dxnxxfa

n

n

←cos n = (-1)n

Ch12_24

Example 1 (3)

From (11) we have

Therefore

(13)

nnxdxxbn

1sin)(

10

1

2 sin1

cos)1(1

4)(

n

n

nxn

nxn

xf

Ch12_25

Fig 12.1

Ch12_26

Let f and f’ be piecewise continuous on the interval (−p, p); that

is, let f and f’ be continuous except at a finite number of points

in the interval and have only finite discontinuous at these points.

Then the Fourier series of f on the interval converge to f(x) at a

point of continuity. At a point of discontinuity, the Fourier

series converges to the average

where f(x+) and f(x－ ) denote the limit of f at x from the right

and from the left, respectively.

THEOREM 12.1Criterion for Convergence

2)()( xfxf

Ch12_27

Example 2

Referring to Example 1, function f is continuous on (−, ) except at x = 0. Thus the series (13) will converge to

at x = 0.22

02

)0()0( ff

Ch12_28

Periodic Extension

Fig 12.2 is the periodic extension of the function f in Example 1. Thus the discontinuity at x = 0, 2, 4, …will converge to

and at x = , 3, … will converge to

22)0()0( ff

02

)0()( ff

Ch12_29

Fig 12.2

Ch12_30

Sequence of Partial Sums

Sequence of Partial SumsReferring to (13), we write the partial sums as

See Fig 12.3.

xxxS

xxSS

2sin21

sincos2

4

,sincos2

4 ,

4

3

21

Ch12_31

Fig 12.3

Ch12_32

12.3 Fourier Cosine and Sine Series

Even and Odd Functions

even if f(−x) = f(x)

odd if f(−x) = −f(x)

Ch12_33

Fig 12.4 Even function

Ch12_34

Fig 12.5 Odd function

Ch12_35

(a) The product of two even functions is even.

(b) The product of two odd functions is even.

(c) The product of an even function and an odd function is odd.

(d) The sum (difference) of two even functions is even.

(e) The sum (difference) of two odd functions is odd.

(f) If f is even then

(g) If f is odd then

THEOREM 12.2Properties of Even/Odd Functions

0)(

)(2)(0

a

a

aa

a

dxxf

dxxfdxxf

Ch12_36

Cosine and Sine Series

If f is even on (−p, p) then

Similarly, if f is odd on (−p, p) then

0sin)(1

cos)(2

cos)(1

)(2

)(1

0

00

p

pn

pp

pn

pp

p

xdxp

nxf

pb

xdxp

nxf

pxdx

pn

xfp

a

dxxfp

dxxfp

a

p

nn xdxp

nxf

pbna

0sin)(

2 ,...2,1,0,0

Ch12_37

(i) The Fourier series of an even function f on the interval (−p, p) is the cosine series

(1)where

(2)

(3)

DEFINITION 12.6Fourier Cosine and Sine Series

1

0 cos2

)(n

n xp

na

axf

p

dxxfp

a00 )(

2

p

n dxxp

nxf

pa

0cos)(

2

Ch12_38

(continued)

(ii) The Fourier series of an odd function f on the interval (−p, p) is the sine series

(4)where

(5)

DEFINITION 12.6Fourier Cosine and Sine Series

1

sin)(i

n xp

nbxf

p

n dxxp

nxf

pb

0sin)(

2

Ch12_39

Example 1

Expand f(x) = x, −2 < x < 2 in a Fourier series.

SolutionInspection of Fig 12.6, we find it is an odd function on (−2, 2) and p = 2.

Thus

(6)Fig 12.7 is the periodic extension of the function in Example 1.

ndxx

nxb

n

n

12

0

)1(42

sin

1

1

2sin

)1(4)(

n

n

xn

nxf

Ch12_40

Fig 12.6

Ch12_41

Fig 12.7

Ch12_42

Example 2

The function

shown in Fig 12.8 is odd on (−, ) with p = .From (5),

and so

(7)

x

xxf

0,1

0,1)(

ndxnxb

n

n)1(12

sin)1(2

0

nxn

xfn

n

sin)1(12

)(1

Ch12_43

Fig 12.8

Ch12_44

Gibbs Phenomenon

Fig 12.9 shows the partial sums of (7). We can see there are pronounced spikes near the discontinuities. This overshooting of SN does not smooth out but remains fairly constant even when N is large. This is so-called Gibbs phenomenon.

Ch12_45

Fig 12.9

Ch12_46

Half-Range Expansions

If a function f is defined only on 0 < x < L, we can make arbitrary definition of the function on −L < x < 0.

If y = f(x) is defined on 0 < x < L,

(i) reflect the graph about the y-axis onto −L < x < 0; the function is now even. See Fig 12.10.

(ii) reflect the graph through the origin onto −L < x < 0; the function is now odd. See Fig 12.11.

(iii) define f on −L < x < 0 by f(x) = f(x + L). See Fig 12.12.

Ch12_47

Fig 12.10

Ch12_48

Fig 12.11

Ch12_49

Fig 12.12

Ch12_50

Example 3

Expand f(x) = x2, 0 < x < L, (a) in a cosine series, (b) in a sine series (c) in a Fourier series.

Solution The graph is shown in Fig 12.13.

Ch12_51

Example 3 (2)

(a)

Then

(8)

22

2

0

2

2

0

20

)1(4cos

2

,322

n

Ldxx

Ln

xL

a

LdxxL

a

nL

n

L

1

22

22

cos)1(4

3)(

n

n

xL

n

n

LLxf

Ch12_52

Example 3 (3)

(b)

Hence

(9)

]1)1[(4)1(2

sin2

33

212

0

2

nn

L

n n

Ln

Ldxx

Ln

xL

b

1

23

12

sin]1)1[(2)1(2

)(n

nn

xL

n

nnL

xf

Ch12_53

Example 3 (4)

(c) With p = L/2, n/p = 2n/L, we have

Therefore

(10)The graph of these periodic extension are shown in Fig 12.14.

nL

xdxLn

xL

b

n

Lxdx

Ln

xL

aLdxxL

a

L

n

L

n

L

2

0

2

22

2

0

22

0

20

2sin

2

2cos

2 ,

322

12

22 2sin

12cos

13

)(n

xLn

nx

Ln

n

LLxf

Ch12_54

Fig 12.14

Ch12_55

Periodic Driving Force

Consider the following physical system

(11)

where

(12)

is a half-range sine expansion.

)(2

2

tfxkdt

xdm

1

sin)(n

np tp

nBtx

Ch12_56

Example 4

Referring to (11), m = 1/16 slug, k = 4 lb/ft, the force f(t) with 2-period is shown in Fig 12.15. Though f(t) acts on the system for t > 0, we can extend the graph in a 2-periodic manner to the negative t-axis to obtain an odd function. With p = 1, from (5) we have

From (11) we have

(13)

ntdtntb

n

n

11

0

)1(2sin2

tnn

xdt

xd

n

n

sin)1(2

4161

1

1

2

2

Ch12_57

Example 4 (1)

To find a particular solution xp(t), we substitute (12) into (13). Thus

Therefore

(14)

)64(

)1(32

)1(2)4

161

(

22

1

122

nnB

nBn

n

n

n

n

tnnn

txn

n

p

sin)64(

)1(32)(

122

1

Ch12_58

12.4 Complex Fourier Series

Euler’s formulaeix = cos x + i sin xe-ix = cos x i sin x

(1)

Ch12_59

Complex Fourier Series

From (1), we have

(2)

Using (2) to replace cos(nx/p) and sin(nx/p), then

(3)

,2

cosixix ee

x

iee

xixix

2sin

1

////0

222 n

pxinpxin

n

pxinpxin

nee

bee

aa

1

//0 )(21

)(21

2 n

pxinnn

pxinnn eibaeiba

a

1

/

1

/0

n

pxinn

n

pxinn ececc

Ch12_60

where c0 = a0/2, cn = (an ibn)/2, c-n = (an + ibn)/2. When the function f is real, cn and c-n are complex conjugates.We have

(4)

p

pdxxf

pc )(

121

0 ．

Ch12_61

(5)

p

p

pxin

p

p

p

p

p

p

nnn

dxexf

dxxp

nix

pn

xfp

dxxp

nxf

pidxx

pn

xfp

ibac

/)(21

sincos)(21

sin)(1

cos)(1

21

)(21

Ch12_62

(6)

p

p

pxin

p

p

p

p

p

p

nnn

dxexf

dxp

nix

pn

xfp

dxxp

nxf

pidxx

pn

xfp

ibac

/)(21

sincos)(21

sin)(1

cos)(1

21

)(21

Ch12_63

The Complex Fourier Series of function f defined on

an interval (p, p)is given by

(7)

where (8)

DEFINITION 12.7Complex Fourier Series

n

pxinnecxf /)(

,2,1,0,)(21 /

ndxexfp

cp

p

pxinn

Ch12_64

If f satisfies the hypotheses of Theorem 12.1, a complex Fourier series converges to f(x) at a point of continuity and to the average

at a point of discontinuity.

2)()( xfxf

Ch12_65

Example 1

Expand f(x) = e-x, < x <, in a complex Fourier series.

Solutionwith p = , (8) gives

][)1(2

121

21

)1()1(

)1(

inin

xininxxn

eein

dxedxeec

Ch12_66

Example 1 (2)

Using Euler’s formula

Hence

(9)

eninee

enineenin

nin

)1()sin(cos

)1()sin(cos)1(

)1(

1

1sinh)1(

)1(2)(

)1( 2

n

inin

eec nn

n

Ch12_67

Example 1 (3)

The complex Fourier series is then

(10)

The series (10) converges to the 2-periodic extension of f.

n

inxn en

inxf

1

1)1(

sinh)( 2

Ch12_68

Fundamental Frequency

The fundamental period is T = 2p and then p = T/2. The Fourier series becomes

(11)

where = 2/T is called the fundamental angular frequency.

1

0 )sincos(2 n

nn xnbxnaa

n

xinnec

Ch12_69

Frequency Spectrum

If f is periodic and has fundamental period T, the plot of the points (n, |cn|) is called the frequency spectrum of f.

Ch12_70

Example 2

In Example 1, = 1, so that n takes on the values 0, 1, 2, … Using , we see from (9) that

See Fig 12.17.

1

1sinh ||

2

22

nc

i

n

Ch12_71

Fig 12.17

Ch12_72

Example 3

Find the spectrum of the wave shown in Fig12.18. The wave is the periodic extension of the function f:

21

41

41

41

41

21

,0

,1

,0

)(

x

x

x

xf

Ch12_73

Example 3 (2)

SolutionHere T = 1 = 2p so p = ½. Since f is 0 on (½, ¼) and (¼, ½), (8) becomes

2sin

1

21

41

41

2

)1()(

2/2/2

41

41

221

21

2

nn

c

iee

nine

dxedxexfc

n

ininxin

xinxinn

Ch12_74

Example 3 (3)

It is easy to check that

Fig 12.19 shows the frequency spectrum of f.

2141

410 dxc

Ch12_75

Fig 12.19

Ch12_76

12.5 Sturm-Liouville Problem

Eigenvalue and EigenfunctionsRecall from Example 2, Sec 3.9

(1)

This equation possesses nontrivial solutions only when took on the values n = n22/L2, n = 1, 2, 3,…called eigenvalues. The corresponding nontrivial solutions y = c2 sin(nx/L) or simply y = sin(nx/L) are called the eigenfunctions.

0)(,0)0(,0 Lyyyy

Ch12_77

Example 1

It is left as an exercise to show the three possible cases: = 0, = 2 < 0, = 2 > 0, ( > 0), that the eigenvalues and eigenfunctions for

(2)

are respectively n = n2 = n22/L2, n = 0, 1, 2, …and

y = c1 cos(nx/L), c1 0.

0)(,0)0(,0 Lyyyy

Ch12_78

Regular Sturm-Liouville Problem

Let p, q, r and r be real-valued functions continuous on [a, b], and let r(x) > 0 and p(x) > 0 for every x in the interval. Then

Solve (3)

Subject to(4)(5)

is said to be a regular Sturm-Liouville problem. The coefficients in (4), (5) are assumed to be real and independent of .

0))()((])([ yxpxqyxrdxd

0)()( 11 ayay 0)()( 22 ayay

Ch12_79

(a) There exist an infinite number of real eigenvalues that can be arranged in increasing order 1 < 2 < 3 < … < n < … such that n → as n → .

(b) For each eigenvalue there is only one eigenfunction (except for nonzero constant multiples).

(c) Eigenfunctions corresponding to differenteigenvalues are linearly independent.

(d) The set of eigenfunctions corresponding to the setof eigenvalues is orthogonal with respect to the weight function p(x) on the interval [a, b].

THEOREM12.3 Properties of the Regular Strum-Liouville Problem

Ch12_80

Proof of (d)Let ym and yn be eigenfunctions corresponding to eigenvalues m and n. Then

(6)

(7)

From (6)yn (7)ym we have

0))()((])([ mmm yxpxqyxrdxd

0))()((])([ nnn yxpxqyxrdxd

')(')()()( mnnmnmnm yxrdxd

yyxrdxd

yyyxp

Ch12_81

Integrating the above equation from a to b, then

(8)

Since all solutions must satisfy the boundary condition (4) and (5), from (4) we have

0)(')(

0)(')(

11

11

ayBayA

ayBayA

nn

mm

)]()()()()[(

)]()()()()[(

)()(

ayayayayar

bybybybybr

dxyyxp

mnnm

mnnm

b

a nmnm

Ch12_82

For this system to be satisfied by A1 and B1 not both zero, the determinant must be zero

Similarly from (5), we have

Thus the right-hand side of (8) is zero.Hence we have the orthogonality relation

(9)

0)(')()(')(

0)(')()(')(

bybybyby

ayayayay

mnnm

mnnm

nm

b

a nm dxxyxyxp ,0)()()(

Ch12_83

Example 2

Solve(10)

Solution You should verify that for = 0 and < 0, (10) possesses only trivial solution. For = 2 > 0, > 0, the general solution is y = c1 cos x + c2 sin x. Now the condition y(0) = 0 implies c1 = 0, thus y = c2 sin x. The second condition y(1) + y(1) = 0 implies c2 sin + c2 cos. = 0.

0)1()1(,0)0(,0 yyyyy

Ch12_84

Example 2 (2)

Choosing c2 0, we have(11)

From Fig 12.20, we see there are infinitely many solution for > 0. It is easy to get the values of > 0. Thus the eigenvalues are n = n

2, n = 1, 2, 3, …and the corresponding eigenfunctions are yn = sin nx.

tan

Ch12_85

Fig 12.20

Ch12_86

Singular Sturm-Liouville problem

There are several import conditions of (3) r(a) = and a boundary condition of the type given

in (5) is specified at x = b;(12)

r(b) = 0 and a boundary condition of the type given in (4) is specified at x = a. (13)

r(a) = r(b) = 0 and no boundary condition is specified at either x = a or at x = b; (14)

r(a) = r(b) and boundary conditions y(a) = y(b), y’(a) = y’(b). (15)

Ch12_87

Notes:

Equation (3) satisfies (12) and (13) is said to be a singular BVP.Equation (3) satisfies (15) is said to be a periodic BVP.

Ch12_88

By assuming the solutions of (3) are bounded on [a, b], from (8) we have If r(a) = 0, then the orthogonality relation (9) holds

with no boundary condition at x = a; (16) If r(b) = 0 , then the orthogonality relation (9) hold

s no boundary condition at x = b; (17) If r(a) = r(b) = 0, then the orthogonality relation

(9) holds with no boundary conditions specified at either x = a or x = b;

(18) If r(a) = r(b), then the orthogonality relation (9) ho

lds wuth peroidic boundary conditions y(a) = y(b), y’(a) = y’(b). (19)

Ch12_89

Self-Adjoint Form

In fact (3) is the same as

(20)Thus we can write the Legendre’s differential equation as

(21)Here we find that the coefficient of y is the derivative of the coefficient of y.

0)1('2")1( 2 ynnxyyx

0))()(()()( yxpxqyxryxr

0)1(])1[( 2 ynnyxdxd

Ch12_90

In addition, if the coefficients are continuous and a(x) 0 for all x in some interval, then any second-order differential equation

(22)can be recast into the so-called self-adjoint form (3).

To understand the above fact, we start from a1(x)y + a0(x)y = 0

Let P = a0/a1, = exp( Pdx), = P, theny + Py = 0, y + Py = 0,

Thus d(y)/dx = 0.

0))()(()()( yxdxcyxbyxa

Ch12_91

Now for (22), let Y = y, and the integrating factor be e [b(x)/a(x)] dx. Then (22) becomes

In summary, (22) can become

......)()(

')(/)()(/)()(/)(

Ye

dxd

Yexaxb

Yedxxaxbdxxaxbdxxaxb

(23) 0)()(

)()(

')()(

"

)/()/(

)/()/(

yexaxd

exaxc

yexaxb

ye

dxabdxab

dxabdxab

Ch12_92

In addition, (23) is the same as (3)

dxabdxabdxab

dxabdxabdxab

exaxd

xpexqexr

yexaxd

eyedxd

)/()/()/(

)/()/()/(

)()(

)(,)(,)( where

0)()(

'

Ch12_93

Example 3

In Sec 5.3, we saw that the general solution of the parametric Bessel differential equation

Dividing the Bessel equation by x2 and multiplying the resulting equation by integrating factor e [(1/x)] dx = eln

x = x, we have

)()( is

... ,2 ,1 ,0 ,0)('"

21

2222

xYcxJcy

nynxxyyx

nn

22

22

22

,,/, where

0)('or ,0)('"

xpxnqxr

yx

nxxy

dxd

yx

nxyxy

Ch12_94

Example 3 (2)

Now r(0) = 0, and of the two solutions Jn(x) and Yn(x) only Jn(x) is bounded at x = 0. From (16), the set {Jn

(ix)}, i = 1, 2, 3, …, is orthogonal with respect to the weight function p(x) = x on [0, b]. Thus

(24)provided the i and hence the eigenvalues i = i

2 are defined by a boundary condition at x = b of the type given by (5):

A2Jn(b) + B2Jn(b) = 0 (25)

,,0)()(0 ji

b

jnin dxxJxxJ

Ch12_95

Example 4

From (21), we identify q(x) = 0, p(x) = 1 and = n(n + 1). Recall from Sec 5.3 when n = 0, 1, 2, …, Legendre’s DE possesses polynomial solutions Pn(x). We observer that r(−1) = r(1) = 0 together with that fact that Pn(x) are the only solutions of (21) that are bounded on [−1, 1], to conclude that the set {Pn(x)}, n = 0, 1, 2, …, is orthogonal w.s.t. the weight function p(x) = 1 on [−1, 1]. Thus

nmdxxPxP nm ,0)()(1

1-

Ch12_96

12.6 Bessel and Legendre Series

Fourier-Bessel SeriesWe have shown that {Jn(ix)}, i = 1, 2, 3, …is orthogonal w.s.t. p(x) = x on [0, b] when the i are defined by

(1)This orthogonal series expansion or generalized Fourier series of a function f defined on (0, b) in terms of this orthogonal set is

(2)where

(3)

0)()( 22 bJBbJA nn

1

)()(i

ini xJcxf

20

)(

)()(

xJ

dxxfxJxc

in

b

in

i

Ch12_97

The square norm of the function Jn(ix) is defined by

(4)

The series (2) is called a Fourier-Bessel series.

b

inin dxxxJxJ0

22 )()(

Ch12_98

Differential Recurrence Relations

Recalling from (20) and (21) in Sec 5.3, we have the differential recurrence relations as

(5)

(6)

)()]([ 1 xJxxJxdxd

nn

nn

),()]([ 1 xJxxJxdxd

nn

nn

Ch12_99

Square Norm

The value of (4) is dependent on i = i2. If y = Jn(x)

we have

After we multiply by 2xy’, then

0]['

0'

22222

22

ydxd

nxxydxd

yx

nxxy

dxd

Ch12_100

Integrating by parts on [0, b], then

Since y = Jn(x), the lower limit is 0 for n > 0, because Jn(0) = 0. For n = 0, at x = 0. Thus

(7)

where y = Jn(x).

0

22222

0

22 )('2bb

ynxxydxxy

,)])[()]([

)(2

2222222

0

22

bJnbbJb

dxxxJ

nn

b

n

Ch12_101

Now we consider three cases for the condition (1). Case I: If we choose A2 = 1 and B2 = 0, then (1) is

(8)There are an infinite number of positive roots xi = ib of (8) (see Fig 5.3), which defines i = xi/b. The eigenvalues are positive and then i = i

2 = (xi/b)2. No new eigenvalues result from the negative roots of (8) since Jn(−x) = (−1)nJn(x).

0)( bJn

Ch12_102

The number 0 is not ab eigenvalue for any n since Jn(0) = 0, n= 1, 2, 3, … and J0(0) = 1. When (6) is written as xJn(x) = nJn(x) – xJn+1(x), it follows from (7) and (8)

(9)

).(2

||)(|| 21

22 bJ

bxJ inin

Ch12_103

Case II: If we choose A2 = h 0 and B2 = b, then (1) is

(10)There are an infinite number of positive roots xi = ib for n = 1, 2, 3, …. As before i = i

2 = (xi/b)2. = 0 is not an eigenvalue for n = 1, 2, 3, …. Substituting ibJn(ib) = – hJn(ib) into (7), then

(11)

.0)()( bJbbhJ nn

).(2

||)(|| 22

22222 bJ

hnbxJ in

i

iin

Ch12_104

Case III: If h = 0 and n = 0 in (10), i are defined from the roots of

(12)Though (12) is a special case of (10), it is the only situation fro which = 0 is an eigenvalue. For n = 0, the result in (6) implies J0(b) = 0 is equivalent to J1(b) = 0.

0)(0 bJ

Ch12_105

Since x1 = 1b = 0 is a root of the last equation and because J0(0) = 1 is nontrivial, we conclude that from 1 = 1

2 = (x1/b)2 that 1 is an eigenvalue. But we can not use (11) when 1

= 0, h = 0, n = 0, and n = 0. However from (4) we have

(13)For i

> we can use (11) with h = 0 and n = 0:

(14)

2||1||

2

0

2 bdxx

b

).(2

||)(|| 20

22

0 bJb

xJ ii

Ch12_106

The Fourier-Bessel series of a function f defined on

the interval (0, b) is given by

(i)

(15)

(16)

where the i are defined by Jn(b) = 0.

DEFINITION 12.8Fourier-Bessel Series

1

)()(i

ini xJcxf

b

inin

i dxxfxJxbJb

c02

12 )()(

)(

2

Ch12_107

(continued)

(ii)

(17)

(18)

where the i are defined by hJn(b) + bJ’n(b) = 0.

DEFINITION 12.8Fourier-Bessel Series

1

)()(i

ini xJcxf

b

inini

ii dxxfxJx

bJhnbc

022222

2

)()()()(

2

Ch12_108

(continued)

(iii)

(19)

(20)

where the i are defined by J’0(b) = 0.

DEFINITION 12.8Fourier-Bessel Series

201 )()(

iii xJccxf

,)(2

021 b

dxxfxb

c b

ii

i dxxfxJxbJb

c0 02

02 )()(

)(

2

Ch12_109

If f and f’ are piecewise continuous in the open interval

(0, b), then a Fourier-Bessel expansion of f converges

to f(x) at any point where f is continuous ant to the

converge [f(x+) + f(x-)] / 2 at a point where f is

discontinuous.

THEOREM 12.4Conditions for Convergence

Ch12_110

Example 1

Expand f(x) = x, 0 < x < 3, in a Fourier-Bessel series, using Bessel function of order one satisfying the boundary condition J1(3) = 0.

SolutionWe use (15) where ci is given by (16) with b = 3:

3

0 12

22

2)(

)3(3

2dxxJx

Jc i

ii

Ch12_111

Example 1 (2)

Let t = i x, dx = dt/i, x2 = t2/i

2, and use (5) in the form d[t2J2(t)]/dt = t2J1(t):

)()3(

22)(

isexpansion theTherefore

)3(2

)]([)3(9

2

11 2

2

3

0 22

22

3

xJJ

xf

JdttJt

dtd

Jc

ii ii

iiiii

i

Ch12_112

Example 2

If the i in Example 1 are defined by J1(3) + J1(3) = 0, the only thing changing in the expansion is the value of the square norm. Since 3J1(3) + 3J1(3) = 0 matching (10) when h = 3, b = 3 and n = 1. Thus (18) and (17) yield in turn

)()3()89(

)3(18)(

)3()89(

)3(18

11

21

22

21

22

xJJ

Jxf

J

Jc

ii ii

ii

ii

iii

Ch12_113

The Fourier-Legendre series of a function f defined

on the interval (－ 1, 1) is given by

(i)

(21)

(22)

where the i are defined by Jn(b) = 0.

DEFINITION 12.9Fourier-Legendre Series

0

)()(n

nn xPcxf

1

1)()(

212

dxxPxfn

c nn

Ch12_114

If f and f’ are piecewise continuous in the open interval

(－ 1, 1), then a Fourier-Legendre series (21) converges

to f(x) at a point of continuity ant to the converge

[f(x+) + f(x-) / 2 at a point of discontinuous.

THEOREM 12.5Conditions for Convergence

Ch12_115

Example 3

Write out the first four nonzero terms in the Fourier-Legendre expansion of

Solution From page 269 and (22):

10

01

,1

,0)(

x

xxf

43

123

)()(23

21

1121

)()(21

1

0

1

1 11

1

0

1

1 00

xdxdxxPxfc

dxdxxPxfc

Ch12_116

Example 3 (2)

0)13(21

125

)()(25 1

0

21

1 22 dxxdxxPxfc

12.22. Fig See

...)(3211

)(167

)(43

)(21

)(

Hence3211

)157063(81

12

11)()(

211

0)33035(81

129

)()(29

167

)35(21

127

)()(27

5310

1

0

351

1 55

1

0

241

1 44

1

0

31

1 33

xPxPxPxPxf

dxxxxdxxPxfc

dxxxdxxPxfc

dxxxdxxPxfc

Ch12_117

Fig 12.22

Ch12_118

Alternative Form of Series

If we let x = cos , the x = 1 implies = 0, x = −1 implies = . Since dx = −sin d, (21) and (22) become, respectively,

(23)

(24)

where f(cos ) has been replaced by F().

0

)(cos)(n

nnPcF

,sin)(cos)(2

120

dPFn

c nn