Orthogonal Functions and Fourier Series

CHAPTER 12

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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

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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

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

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Example

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

xxdxxff ．

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

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

DEFINITION 12.3Inner Product of Function

nmdxxxb

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

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Orthonormal Sets

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

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

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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

nxdxdxxx nn

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Example 1 (2)

dxxnmxnm

nxdxmxdxxx nmnm

,0)sin()sin(

])cos()[cos(21

coscos)()(),(

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Example 2

Find the norms of each functions in Example 1.

Solution

0 ,||||

)2cos1(21

cos||||

dxnxnxdx

22 ,1 02

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Vector Analogy

Recalling from the vectors in 3-space that

(4)we have

Thus we can make an analogy between vectors and functions.

,332211 vvvu ccc

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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

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

),(),(),(

)()()()(

dxxxcdxxxc

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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 ,)(

)(),()()(

dxxxfc

dxxccdxxxf

a nnnnn

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In other words,

Then (7) becomes

),()(n

nn xcxf

2||)(||

dxxxfc

||)(||

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Under the condition of the above definition, we have

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||)(||

)()()(

dxxxwxfc

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

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Complete Sets

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

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12.2 Fourier Series

Trigonometric SeriesWe can show that the set

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

2sin,sin,,

2cos,cos,1 x

0 sincos2

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Now we calculate the coefficients.

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

Thus we have

0 sincos2

22)( pax

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In addition,

by orthogonality we have

sincoscoscos

0sincos

0,0cos

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Thus (5) reduces to

and so

0,coscos

paxdxp

pn dxxp

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

we find that

0sinsin

0 ,0sin

0,sinsin

,0sinsin

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The Fourier series of a function f defined on the

interval (−p, p) is given by

(8)where

DEFINITION 12.5Fourier Series

0 )sincos(2

pn dxxp

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Example 1

Expand (12)

in a Fourier series.

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

dxxdxdxxfa

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Example 1 (2)

)1(11cos

sin1sin

cos)(01

cos)(1

dxnxnn

dxnxxdx

dxnxxfa

←cos n = (-1)n

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Example 1 (3)

From (11) we have

Therefore

nnxdxxbn

1sin)(

2 sin1

cos)1(1

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Fig 12.1

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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

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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

)0()0( ff

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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

)0()( ff

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Fig 12.2

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Sequence of Partial Sums

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

See Fig 12.3.

2sin21

sincos2

,sincos2

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Fig 12.3

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12.3 Fourier Cosine and Sine Series

Even and Odd Functions

even if f(−x) = f(x)

odd if f(−x) = −f(x)

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Fig 12.4 Even function

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Fig 12.5 Odd function

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(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

)(2)(0

dxxfdxxf

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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

nn xdxp

0sin)(

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

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

(1)where

DEFINITION 12.6Fourier Cosine and Sine Series

0 cos2

a00 )(

n dxxp

0cos)(

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(continued)

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

(4)where

DEFINITION 12.6Fourier Cosine and Sine Series

sin)(i

n dxxp

0sin)(

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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.

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

)1(4)(

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Fig 12.6

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Fig 12.7

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Example 2

The function

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

and so

ndxnxb

n)1(12

sin)1(2

sin)1(12

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Fig 12.8

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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.

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Fig 12.9

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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.

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Fig 12.10

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Fig 12.11

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Fig 12.12

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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.

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Example 3 (2)

)1(4cos

cos)1(4

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Example 3 (3)

]1)1[(4)1(2

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

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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.

aLdxxL

22 2sin

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Fig 12.14

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Periodic Driving Force

Consider the following physical system

is a half-range sine expansion.

tfxkdt

sin)(n

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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

ntdtntb

)1(2sin2

sin)1(2

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Example 4 (1)

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

Therefore

)1(2)4

sin)64(

)1(32)(

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12.4 Complex Fourier Series

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

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Complex Fourier Series

From (1), we have

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

cosixix ee

pxinpxin

//0 )(21

pxinnn

pxinnn eibaeiba

pxinn ececc

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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

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sincos)(21

sin)(1

cos)(1

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sincos)(21

sin)(1

cos)(1

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The Complex Fourier Series of function f defined on

an interval (p, p)is given by

where (8)

DEFINITION 12.7Complex Fourier Series

pxinnecxf /)(

,2,1,0,)(21 /

ndxexfp

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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

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Example 1

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

Solutionwith p = , (8) gives

][)1(2

)1()1(

xininxxn

dxedxeec

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Example 1 (2)

Using Euler’s formula

eninee

enineenin

)1()sin(cos

)1()sin(cos)1(

1sinh)1(

)1(2)(

eec nn

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Example 1 (3)

The complex Fourier series is then

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

inxn en

sinh)( 2

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Fundamental Frequency

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

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

0 )sincos(2 n

nn xnbxnaa

xinnec

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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.

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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.

1sinh ||

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Fig 12.17

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Example 3

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

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Example 3 (2)

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

dxedxexfc

ininxin

xinxinn

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Example 3 (3)

It is easy to check that

Fig 12.19 shows the frequency spectrum of f.

410 dxc

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Fig 12.19

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12.5 Sturm-Liouville Problem

Eigenvalue and EigenfunctionsRecall from Example 2, Sec 3.9

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

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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

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

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

0)(,0)0(,0 Lyyyy

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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

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(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

From (6)yn (7)ym we have

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

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

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

yyxrdxd

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Integrating the above equation from a to b, then

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

0)(')(

ayBayA

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

ayayayayar

bybybybybr

dxyyxp

a nmnm

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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

0)(')()(')(

bybybyby

ayayayay

a nm dxxyxyxp ,0)()()(

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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

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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.

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Fig 12.20

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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)

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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.

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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)

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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

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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

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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

......)()(

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

Yexaxb

Yedxxaxbdxxaxbdxxaxb

(23) 0)()(

)/()/(

yexaxd

yexaxb

dxabdxab

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

dxabdxabdxab

xpexqexr

yexaxd

eyedxd

)/()/()/(

)(,)(,)( where

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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)('"

xYcxJcy

nynxxyyx

,,/, where

0)('or ,0)('"

xpxnqxr

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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

jnin dxxJxxJ

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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

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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

0)()( 22 bJBbJA nn

ini xJcxf

dxxfxJxc

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

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

inin dxxxJxJ0

22 )()(

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Differential Recurrence Relations

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

)()]([ 1 xJxxJxdxd

),()]([ 1 xJxxJxdxd

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Square Norm

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

we have

After we multiply by 2xy’, then

nxxydxd

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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

where y = Jn(x).

22 )('2bb

ynxxydxxy

,)])[()]([

2222222

bJnbbJb

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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

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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)

||)(|| 21

bxJ inin

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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

.0)()( bJbbhJ nn

||)(|| 22

22222 bJ

hnbxJ in

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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

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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:

2||1||

2 bdxx

||)(|| 20

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The Fourier-Bessel series of a function f defined on

the interval (0, b) is given by

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

DEFINITION 12.8Fourier-Bessel Series

ini xJcxf

i dxxfxJxbJb

12 )()(

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(continued)

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

ini xJcxf

ii dxxfxJx

bJhnbc

022222

)()()()(

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(continued)

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

201 )()(

iii xJccxf

dxxfxb

i dxxfxJxbJb

02 )()(

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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

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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:

2dxxJx

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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):

isexpansion theTherefore

)]([)3(9

JdttJt

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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(

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The Fourier-Legendre series of a function f defined

on the interval (－ 1, 1) is given by

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

DEFINITION 12.9Fourier-Legendre Series

nn xPcxf

dxxPxfn

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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):

)()(23

)()(21

xdxdxxPxfc

dxdxxPxfc

Ch12_116

Example 3 (2)

0)13(21

)()(25 1

1 22 dxxdxxPxfc

12.22. Fig See

...)(3211

Hence3211

)157063(81

11)()(

0)33035(81

)()(29

)35(21

)()(27

xPxPxPxPxf

dxxxxdxxPxfc

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,

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

)(cos)(n

,sin)(cos)(2

Orthogonal Functions and Fourier Series

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