advanced series

CHAPTER IV

FOURIER SERIES

Some functions can be expressed in the form of infinite series of sines and

cosines. Problems involving various forms of oscillations are common in fields of

modern technology and Fourier series, enable us to represent periodic functions

as an infinite trigonometrical series in sine and cosine terms. One important

advantage of Fourier series is that it can represent a function discontinuities

whereas Maclaurin’s and Taylor’s series require the function to be continuous

throughout.

PERIODIC FUNCTIONS

A function f(x) is said to be periodic if the function values repeat at regular

intervals of the independent variable. The regular interval between repetitions is

the period of oscillation.

y

f(x)

Period

a. y = sin x – this is the obvious example of periodic function which goes

through its complete range of values while x increases from 00 to 3600.

The period therefore is 3600 or 2 radians and the amplitude, the

maximum displacement from the position of rest is 1

Engr. Tirso A. Ronquillo Instructor

112

0Xl + p

xl

1

0 2

-1

Period = 2

b. y = 5 sin 2x

5

amplitude

0 2

Period is 180 or and there are thus 2 complete cycles in 360 or 2.

Amplitude is 5.

In y = A sin (nx),

A = amplitude

Period = 2/n =3600/n

Graph of y = A cos x have the same characteristics

Exercise: In each of the following, state:

a. the amplitude


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-5

b. the period

1. y = 3 sin 5x 5. y = 5 cos 4x

2. y = 2 cos 2x 6. y = 2 sin x

3. y = sin x/2 7. y = 3 cos 6x

4. y = 4 sin 2x 8. y = 6 sin (2x/3)

NON – SINUSOIDAL PERIODIC FUNCTIONS

Although we introduced the concept of periodic function via sine curve, a

function can be periodic without being obviously sinusoidal in appearance.

Example: In the following cases, the x – axis carries a scale of t in milliseconds

f(t) period = 8ms

0 6 8 14 16


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f(t)

2 5 6 80

Period 6ms

ANALYTICAL DESCRIPTION OF A PERIODIC FUNCTION

A periodic function can be defined analytically in many cases.

Example:

a. between x = 0 and x = 4; y = 3 i.e. f(x) = 3, 0 < x <4

b. between x = 4 and x = 6; y = 0 i.e. f(x) = 0, 4 < x < 6

So we could define the function by:

f (x) =3 0 <x <4

f (x) = 0 4 < x <6

f( x) = f (x + 6)

the last is indicating that the function is periodic with a period of 6.

Example:

Between x = 0 and x = 2; y = x : f (x) = x, 0 < x < 2

Between x = 2 and x = 6; y = (-x/2) + 3 : f (x) = 3 – (x/2), 2 <x < 6


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2 f(t)

-4 0 3 5 10Period = 5ms

0 4 6 10 12

Period = 6ms

0 2 6 12

2

f(x) = f(x+6)period = 6 ms

Define analytically the periodic function shown.

between x = 0 and x = 3, y = -x + 2, f (x) = 2 – x, 0 < x < 3

between x = 3 and x = 5; y = -1 , f (x) = -1, 3 < x <5

f (x) = f (x + 5); Period = -5

between x = 0 and x = 4; y = 3; f (x) = 3, 0 < x <4

between x = 4 and x = 7; y = 5; f (x) = 5, 4 < x < 7


Period = 10; f (x) = f (x + 10)

between x = 0 and x = 4; y = x; f (x) = 0, 0 < x <4

between x = 4 and x = 7; y = 4; f (x) = 4, 4 < x < 7



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1. 2

0-1 2 5 7

2. 5 3

0 4 7 10

3. 4

0

4 7 9

Period = 9; f (x) = f (x + 9)

4.

Periodic function – f (t) is said to be periodic of period T if there exists a number

T such that f (t) = f (t + T) for all t. T must be the smallest non-zero number for

which f(t) = f(t + T).

Example 1:

F(t) = sint is periodic of 2 since for any t, sint = sin (t + 2). While it is also

true that f(t) = f(t + 4), it is not correct to say that the period T = 4 since T must

the smallest number for which f(t) = f(t + T).

Example 2:

The function whose graph appears above is periodic and its period is T = 4 since

f(t) = f(t + 4).


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0 4 7

10

15 19 23

f(t)-1 0 2 3 4 6 7 8

t

TAYLOR’S SERIES FOURIER SERIES

1. Gives close fit of the curve f(t)

for Small neighborhood of t

interval.

2. Uses the variable t as its

fundamental element.

1. Gives the best approximation of

the curve in the fairly wide

interval.

2. Uses sine and cosine as its

fundamental element.

A function f(t) can be expressed in terms of the Fourier Series Fn( t ) of

the form.

If it satisfies the following conditions:

1. It is periodic over a period T (usually 2)

f (t) = f (t + T) = f (t + 2)

2. It is a single valued function.

3. It contains finite number of finite discontinuities over an interval.

4. It has finite maxima and minima over the period.

5. In the expression of f (t) in terms of the Fourier series nt, the value of a0,

a1, a2 , …, an,b1, b2, b3, …, bn can be determined if the following conditions

are met:

a.) the value of the integral and is the same

over


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the period.

b)If the function is multiplied by or and the result is

integrated over a period of T then the result is equal to the integral of

each term of the series multiplied by or .

To solve for an, multiply f(t) by and thus integrate

From the orthogonality relations of sines and cosines


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it can be noted that some terms in the right side in eqn (2) will vanish except the

terms

Now,

Solving for an,

To solve for bn, multiply f(t) by and then integrate over period T.

For the orthogonality relations between sines and cosines note that all terms in

the right side of equation (3) vanish except the term .

Now,


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=

Solving for bn,

To solve for the coefficient a0, set n=0 in the formula for an:


121

Johann Peter Gustav Lejeune Dirichlet

Born: 13 Feb 1805 in Düren, French Empire (now Germany)Died: 5 May 1859 in Göttingen, Hanover (now Germany)

Lejeune Dirichlet's family came from the Belgian town of Richelet where Dirichlet's grandfather lived. In Paris by May 1822, Dirichlet soon contracted smallpox. In return Dirichlet taught German to General Foy's wife and children. Dirichlet proved case 1 and presented his paper to the Paris Academy in July 1825. General Foy died and Dirichlet decided to return to Germany. From 1827 Dirichlet taught at Breslau but Dirichlet encountered the same problem which made him choose Paris for his own education, namely that the standards at the university were low. The quieter life in Göttingen seemed to suit Dirichlet. We should now look at Dirichlet's remarkable contributions to mathematics.

Shortly after publishing this paper Dirichlet published two further papers on analytic number theory, one in 1838 with the next in the following year. This work led him to the Dirichlet problem concerning harmonic functions with given boundary conditions. Dirichlet's work is published in Crelle's Journal in 1828. . Cauchy's work itself was shown to be in error by Dirichlet who wrote of Cauchy's paper:-

... important parts of mathematics were influenced by Dirichlet. With Dirichlet began the golden age of mathematics in Berlin.


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http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cauchy.html

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cauchy.html

Dirichlet Conditions (named after Peter Gustav Lejuene Dirichlet, German

Mathematician)

Set of conditions that the function f(t) must satisfy so that it can be expanded into

a Fourier series

1. f(t) must be periodic and single-valued in each period

2. f(t) must have finite number of finite discontinuities during any

period

3. f(t) must have finite number of maxima and minima during any

period

4. f(t) must be absolutely integrable in any period, that is

If the function f(t) satisfies the Dirichlet conditions, then, on the Fourier

series of f(t) converges to f(t) at all points where f(t) is continuos

at each point of discontinuity of f(t), the Fourier series converges to the

average of the values approached by f(t) from the right and from the left of

the point of discontinuity t=to, that is,

Examples:

1.Determine the Fourier series expansion of the function

Solution:

The function given above is a square wave

1


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--π 0 π 2π

-1

Using Euler-Fourier Formulas


124

Ex. Find the Fourier Series expansion of


125

t2 -1 < t > 1f(t) = f (t + 2)

-1 1

f(t)

t

t2

(Even function)


126

A0 = T = 2 t0 = -1

=

=

=

A0 =

An =

=

= = 2 - u = t2 dv = cos

du = 2t2-1dt v = = 2

= u = t dv = sin

du = dt v =

= 2 -

= 2 - 2

Now, f(t) = t

SYMMETRICAL PROPERTIES

I. EVEN FUNCTIONS

A function f(x) is considered even function if f(x) = f(-x), that is, replacing x

by –x won’t change the value of f(x).


127

=

=

An =

bn = 0

bn =

=

=

=

f(t) =

Even function has a graph which is symmetric to vertical axis.

Examples are functions like x2n, cos nx a)x4, b) cos x

f(x)

x

The Fourier series of an even function f(x) is given by

f(x) = a0/2 + a0 cos nx an = 2/p p f(x) cos n x dx

0 p n = 0, 1, 2,3,….

Note that an even function involves only cosine terms (that is bn = 0)

II. ODD FUNCTIONS

A function f(x) is considered odd function if f(x) = -f(-x)


128

That is replacing x by –x will just change the sign of the function. Examples are

the functions such as, x2n + 1, sin nx , (sin x)2n +1 , a)x3 , b) sin x. Consider the odd

function below

For odd function, folding the right half of the curve about the vertical axis

and rotating it about x-axis, will make it coincide with the left half.

Any line joining the point (x, f(x) ) and the origin will pass through (-x, f(-

x)) thus, an odd function is symmetric with respect to the origin.

The Fourier series of an odd functions involves only sine terms(that is, an = 0)

f(x) = bn sin nx

bn = 2/p p f(x) sin n x

0 p

n = 0, 1, 2, 3, 4,….

PROPERTIES:

1. The product of an odd and even function is an odd function.

PROOF:

Let g(x) an odd function

h(x) an even function

and let f(x) = g(x)h(x) 1

f(-x) = g(-x)h(-x)

= -g(x)h(x)


129

f(x)

x

= -g(x)h(x)

f(x) = f(-x) from 1, thus f(x) is an odd function

2. The product of the two odd functions is an even function.

PROOF: let g(x) and h(x) be the two odd functions and

Let f(x) = g(x)h(x) 1

f(-x) = g(-x)h(-x)

= -g(x)-h(x)

= g(x)h(x)

f(-x) = f(x) from 1

thus, f(x) is an even function

3. The product of the two even function is an even function

PROOF: let g(x) and h(x) be the two even functions and

Let f(x) = g(x)h(x) 1

f(-x) = g(-x)h(-x)

f(-x) = g(x)h(x)

f(-x) = f(x) from 1

thus, f(x) is an even function

4. n f(x) dx = 2n f(x) dx

if f(x) is an even function

5. n f(x) dx = 0

if f(x) is an odd function

FOURIER COSINE SERIES – If f(x) is an even function of period 2p which

satisfies the Dirichlet conditions, then FS of f(x) is


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f(x) = a0/2 + a0 cos nx/p

where an = 2/p p f(x) cos n x dx 0 p

We can create an old periodic function F1(x) identical to f(x) in 0<x<6 by

duplicating the curve for f(x) over the rest of the x-axis. Do this and see that the

function F1(x) that you created is periodic of period 2p = 6 and the curve is

symmetrical about the origin, and hence, is odd. Furthermore, F1(x) satisfies the

Dirichlet conditions and hence, can be expanded into a Fourier series which will

be a Fourier sine series F1(x) is odd. This series is also the Fourier series

representation of f(x) in the interval 0<x<6. It is known right at the start that ao = 0

and hence, we need only to solve for bn using the simplified formula (12).

F1(x) is not the only periodic function that can be created which is identical

to f(x) in 0<x<6. An even function can be created by drawing from x = 6 to x = 0 a

curve which is the mirror image of the curve of f(x) from x = 0 to x = 6. Do this

and afterwards, create a periodic function and call this function F2(x). What is the

periodic of F2(x)? F2(x) is an even periodic function which satisfies the Dirichlet

conditions and hence, it can be expanded into a Fourier series that contains no

sine terms, i.e., a Fourier cosine series. Only ao need to be computed using the

simplified formula (11).

Can you think of other periodic functions that are identical to f(x) in 0<x<6?

USE OF FOURIER SERIES IN ORDINARY DIFFERENTIAL EQUATIONS:

Example : Obtain the general solution of the differential equation


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F(t) = (D2 + 4D + 85)y

Where f(t) is the periodic function in example 2 and 4.

Solution: The inhomogeneous term f(t) is expressed as a Fourier series (see

answer to example 4) so that the equation can be written as

(D2 + 4D + 85)y = bn sin nt / 2 (a)

where bn = 8 sin n (n)2 2

General Solution: Y = Yc + Yp

The general solution of (a)is y = yc + yp where yc is the complementary function

and Yp is the particular solution. The complementary function is the general

solution of the associated homogeneous equation, i.e., the general solution of

(D2 + 4D + 85)yc = 0 (b)

From your study of differential equation, you know that the characteristic

equation of (b) is m2 + 4m + 85 = 0 and the roots of the characteristic equation

are m = -2 9i. Hence, the complementary function is 0

Yc = e2t (c1cos 9t + c2sin 9t)

Denote the differential operator of the (D2 + 4D +85) by F(D). the particular

solution of (a) is the function of yp not containing arbitrary constants which

satisfies (a). Equation (a) can be split up into an infinite number of differential

equations three of which are listed below.

F(D)y1 = b1sint/2 f(D)y2 = b2 sin 2t/2 f(D)yn = bn sin nt/2 (c)

Summing the above equation, we get

F(D)(y1 + y2 + ……..+ yn + ……..) = bn sin nt/2

Thus, yp = y1 + y2 + …… + yn +….. = yn, i.e., yp is the sum of the particular

solutions of the set of the equations (c). For convenience, let wn = nt/2. Then,

the last equation in set (c) can be written as

f(D) = yn = bn sinwnt (d)

using the methods of undetermined coefficients, we write the particular

solution of (d) as

yn = An coswnt + Bn sinwntEngr. Tirso A. Ronquillo Instructor

132

Where An and Bn are to be chosen so that the above yn satisfies (d). Substituting

yn into (d) we get, after some simplifications,

(-wn2An + 4wn Bn + 85An) cos wnt + (-wn

2Bn - 4wn An + 85Bn) sinwnt = bn sinwnt

Equating the coefficients of cos wnt,

(-wn2An + 4wn Bn + 85An) = 0. (e)

Equating the coefficient of sinwnt, we get

(-wn2Bn - 4wn An + 85Bn) = bn (f)

Solve (e) and (f) simultaneously to obtain

An = - 4wnbn

[16wn2 + ( 85 –wn

2)2]

Bn = ( 85 – wn2)bn

[16wn2 + (85-wn

2)2]

The particular solution yn of (d) is now known. Since the particular solution of (a)

is yp = yn, then

Yp = _ 4wnbn cos wnt + ( 85 – wn ) bn sin wnt

. [16 wn2 + ( 85 – wn

2 )2] 16 wn2 + ( 85 – wn

2 )2]

The problem is now solved and the general solution is y = yc + yp.

ALTERNATE FORMS OF THE FOURIER SERIES

Given the standard trigonometric form of Fourier series:

For the right triangle,


133

f(t) = +

S = S =

Let

= A

Harmonic Cosine Series

Or


134

A

b

S =

S =

f(t) = A +

= tan

S =

S = Harmonic Sine Series

EXPONENTIAL FORM OF THE FOURIER SERIES:

Adding,

Subtracting,

Substituting,


135

= Phase angle

= tan

f(t) =

e = cos

e = cos

cos

f(t) = nn

T

ntj

T

ntj

n bee

AA

1

22

0

22

2

22

T

ntj

T

ntjee

sin

f(t) = nn

T

ntj

T

ntj

n bee

AA

1

22

0

22

2

22

T

ntj

T

ntjee

=

Let

2,

2,

2 00 nn

nnn

n

jbAC

jbACC

A

Shifting index for –n in C -n n

C

Thus,


136

f(t) = C

f(t) = C

f(t) = C

Cn

C0

-1 1

= C

= C

f(t) = C

Where C

Simplifying:

Example :Determine the magnitude spectrum of the pulse train below using

exponential Fourier series.


137

C

=

C

f(t)

A……. …….

2

2

Solution:

f(t) =

T = 2 , Pulse width =


138

f(t)

2

2

A

<

< t < ,

C

=

=

=

=

Magnitude Spectrum

AVERAGE POWER OF PERIODIC SIGNALS (FUNCTIONS)


139

Let f(t) be a periodic signal/function the average power (P) is given by,

Exercises

1. Obtain a Fourier series expansion of sin x.

a. determine the period

2p = 2

p =


140

b. determine ao, a1, a2, … an, b1, b2, …bn

f(x) = sin x

ao = 1/ 2 sin x dx

= 1/ [cos 2 - cos 0]

= 0

f(x) = sin x = ao / 2 + a1 cos x / + a2 cos 2x / + … b1sin x / + b2sin 2x /

an cos nx / + bn sin nx /

Solving for a1,

a1 = 1/ sin x cos x / dx

= 1/ sin x cos x dx

=1/ [ sin2 x /2]

= 1/ 2 [ sin2 2 - sin2 0]

a1 = 0

2. Find the Fourier series corresponding to the function f(x) = 1 – x in the interval

-1 x 1.

Period 2p = 2; p = 1

d = -1

an = 1/p 2pd f(x) cos nx/p dx

bn = 1/p f(x) sin nx/p dx

ao =

ao =

ao = [1-1/2] – [-1-1/2]

ao = 2

ao = = (1-x)(1/n )sin n x + 1/n [cosnx n]

ao = 0


141

bn =

Using integration by parts

= (1-x)(-1/n cos nx) -

= -(1-x)(1/ncos nx) + sin n - sin(-n)= -1/n cosn - 1/ncosn

bn = 2 cos n

n

3. Find the Fourier series for the function whose waveform is shown in Figure 1.

f(t)

1

0 t

Figure 1

Solution: The function f(t) can be expressed analytically as

Since the average value of f(t) over a period is zero,


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The first integral on the right-hand side equals zero.Letting t= - in the second integral, we get

Now integrating by parts,

Hence,

Since cos n = (-1)n

Similarly, from(1.10)

Hence,


143

4. Find the Fourier series for the function f(t) defined by

and

Solution: Since

When n=1,


144

f t( )

A

-T - 0 T

t

When

When

When

Hence

SERIES SOLUTION OF DIFFERENTIAL EQUATION

Finding the general solution of a linear differential equation rests on

determining a fundamental set of solutions of the homogenous equation. To deal

with the much larger class of equations having variable coefficients it is

necessary to extend our search for solutions beyond the familiar elementary

functions of calculus. The principal tool that we need is the representation of a

given function by a power series. The basic idea is similar to that in the method


145

of undetermined coefficients. We assume that the solutions of a given differential

equation have power series expansions, and then we attempt to determine the

coefficients so as to satisfy the differential equation

Use of Summation Notation

According to the summation notation, a series such as

uo + u1 + u2 + u3 + . . . + un (1)

with a finite number of terms is represented by

(2)

which is read the sum of all terms having the form u j, where j goes from 0 to n.

The sign is a Greek capital letter sigma, j is called the summation index or

briefly index, and we can read (2) briefly as sigma, or sum, of uj from j=0 to n. As

for definite integrals j=0 is referred to as the lower limit, while n is referred to as

the upper limit.

In the case where we have an infinite series

uo + u1 + u2 + u3 + . . . (3)

we represent it by

(4)

where the upper limit n is replaced by .

The following are some examples of the use of the summation index.

Example 1.

Example 2:


146

Example 3:

taking 0! = 1.

We can of course use other limits besides o and n or . For instance,

i.e., the sum of uj where j goes from 2 to 6. We can also use other indexes. Thus

the sum just given can be represented in any of the forms

, ,

The following are two important properties of summation notation whose

proofs are easily verified by writing out the terms on each side.

Z1.

2. , for any independent of j.

The results are also valid if other limits besides 0 and n are used provided

they are the same in each sum. However, if any limit is infinite, the series must

be convergent.

For purposes of finding power series solutions of differential equations, it

is convenient to use the notation

(5)

where we shall agree that aj = 0 for all negative integer values of j, i.e., j=-1, -2,

-3, . . .

SHIFT OF INDEX SUMMATION

The index of summation in an infinite series is a dummy

parameter just as the integration variable in a definite integral is a dummy


147

variable. Thus it is immaterial which letter is used for the index of summation.

For example,

Just as we make changes of the variable of integration in a definite

integral, we find it convenient to make changes of summation indices in

calculating series solution of differential equations.

Examples:

1. Write as a series whose first term corresponds to n = 0 rather

than n=2.

Solution:

Let m = n – 2; then n = m + 2 and n = 2 corresponds to m = 0.

Hence

(1)

By writing out the first few terms of each of these series, you can verify

that they contain precisely the same terms. Finally, in the series on the

right side of Eq. (1), we can replace the dummy index m by n, obtaining

(2)

In effect, we have shifted the index upward by 2, and compensated by

starting to count at a level 2 lower than originally.

2, Write the series

as a series whose generic term involves (x - xo)n rather than (x - xo)n-2.

Solution:


148

Again, we shift the index by 2 so that n is replaced by n + 2 and start

counting 2 lower. We obtain

3. Write the expression

as a series whose generic term involves xn+r.

Solution:

First take the x2 inside the summation, obtaining

.

Next, shift the index down by 1 and start counting 1 higher. Thus

.

LINEAR EQUATION AND POWER SERIES

Consider the homogenous linear equation of the second order,

b0(x)y’’ + b1(x)y’ + b2(x)y = 0 (1)

with polynomial coefficients. If b0(x) does not vanish at x = 0, then in some

interval about x = 0, staying away from the nearest point where b0(x) does

vanish, it is safe to divide throughout by b0(x). Thus we replace equation (1) by

y’’ + p(x)y’ + q(x)y = 0, (2)

in which the coefficients p(x), q(x) are rational functions of x with denominators

that do not vanish at x = 0.

We shall now show that it is reasonable to expect a solution of (2)

that it is a power series in x and that contains two arbitrary constants. Let y =

y(x) be a solution of equation (2). We assign arbitrarily the values of y and y’ at x

= 0; y(0)=A, y’(0) = B.

Equation (2) yields

y’’(x) = - p(x)y’(x) – q(x)y(x), (3)


149

so y’’(0) may be computed directly, because p(x) and q(x) are well behaved at

x=0. From equation (3) we get

y’’’(x) = - p(x)y’’(x) – p’(x)y’(x) – q(x)y’(x) – q’(x)y(x), (4)

so y’’’(0) can be computed once y’’(0) is known.

The above process can be continued as long as we wish; therefore, we

can determine successively yn(0) for as many integral values of n as maybe

desired. Now, by Maclaurin’s formula in calculus,

y(x) = y(0) + (5)

that is, the right member of (5) will converge to the value y(x) throughout some

interval about x = 0 if y(x) is sufficiently well behaved at and near x = 0. Thus we

can determine the function y(x) and are led to a solution in power series form.

ORDINARY POINTS AND SINGULAR POINTS

For a linear differential equation

b0(x)yn + b1(x)y(n-1) + …… + bn(x)y = R(x) (1)

with polynomial coefficients, the point x = x0 is called an ordinary point of the

equation if bo(xo) 0. A singular point of the linear equation (1) is any point x = x1

for which b0(x1) = 0.

Examples:

1. Determine the singular points and ordinary points of the Bessel equation

of order v.

x2y’’ + xy’ + (x2 – v2)y = 0 (2)

Solution:

The point x = 0 is a singular point since P(x) = x2 is zero there. All

other points are ordinary points of equation (2).

2. Determine the singular points and ordinary points of the Legendre

equation

(1 – x2)y’’ – 2xy’ + ( + 1)y = 0, (3)


150

where is a constant.

Solution:

The singular points are the zeros of P(x) = 1 – x2, namely the points

x = 1. All other points are ordinary points.

Supplementary Problems

1. (x2 + 4)y’’ – 6xy’ + 3y = 0

2. x(3 – x)y’’ – (3 – x)y’ + 4xy = 0

3. 4y’’ + 3xy’ + 2y = 0

4. x(x – 1)2y’’ + 3xy’ + (x – 1)y = 0

5. x2y’’ + xy’ + (1 – x2)y = 0

REGULAR SINGULAR POINTS

Consider the equation

P(x)y’’ + Q(x)y’ + R(x)y = 0 (1)

in the neighborhood of a singular point x0. Recall that if the functions P, Q, and R

are polynomials having no common factors, the singular points of equation (1)

are the points for which P(x) = 0.

If the limits

is finite


151

and

is finite,

then x = x0 is a regular singular point (RSP) of equation (1).

Any singular point of equation (1) that is not regular singular point is called

irregular singular point (ISP) of (1).

Examples:

1. In example 2 we observe that the singular points of the Legendre

equation

(1 – x2)y’’ – 2xy’ + ( + 1)y = 0

are x = 1. Determine whether these singular points are regular or

irregular singular points.

Solution:

We consider the point x = 1 first and also observed that on dividing

by 1 – x2 the coefficients of y’ and y are - 2x/ (1 – x2) and ( + 1)/(1 – x2),

respectively. Thus we calculate

and


152

Since these limits are finite, the point x = 1 is a regular singular point. It

can be shown in a similar manner that x = -1 is also a regular singular

point.

2. Determine the singular points of the differential equation

, and classify

them as regular

or irregular.

Solution:

Dividing the differential equation by

we have

so and The

singular points are x=o and x=2. Consider x=0. We have

Since these limits are finite, x = 0 is a regular point.

For x = 2 we have

so

the limit does not exist;

hence x =2 is an irregular singular point.

Supplementary Problems:


153

Find all singular points of the given equation and

determine whether

each one is regular or irregular.

1.

2.

3.

4.

5.

6.

Bessel

function

7.

8.

9.

10.


154

SOLUTIONS OF DIFFERENTIAL EQUATION NEAR AN ORDINARY POINT

Suppose that x = 0 is an ordinary point of the linear equation

(1)

It is proved in more that there is a solution

(2)

that contains two arbitrary constants, namely, a0 and a1 and

converges inside a circle with center at x = 0 and extending out to the singular

point (or points) nearest x = 0. If the differential equation has no singular points in

the finite plane, then the solution (2) is valid for all finite x. There remains, of

course, the job of finding an for n 2.

Example:

Solve the equation

(1)

near the ordinary point x = 0.

Solution:

The equation we are considering has no singular points in the finite

plane. Hence we may expect to find a solution

(2)

valid for all real x and with a0 and a1 arbitrary. Substituting the series into (1)

gives us


155

(3)

We now change the indexing of the terms in the second sum in (3),

so that the series will involve xn-2 in its general term. Thus (3) becomes

(4)

We can now add the two series to obtain

(5)

because the first two terms of the first sum in (4) are zero.

We now use the fact that, for a power series to vanish identically

over any interval, each coefficient in the series must be zero. Thus, for (5) to

be valid in some interval, it must be true that

for n ≤2

This relation may be written (since n ≤2).

Relation (6) may be used to obtain an for n ≤2 in terms of a0 and a1

which are left arbitrary. We have


156

(6)

In writing out these particular cases of equation (6), we have taken pains to

keep the a’s with even or odd subscripts in separate columns. If we now

multiply the corresponding members of the equations of the first column,

We obtain

which simplifies to

A similar argument applied to the right column in the foregoing array gives us

We now wish to substitute the expressions for the a’s back into the

assumed series for y,

Since we have different forms for a2k and a2k+1, we first rewrite (2) in the form

and then we use (7) and (8) to obtain


157

for k1 (7)

for k1 (8)

(9)

It is possible to rewrite equation (9) in the form

The two series in (10) are the Maclaurin series for the functions cos 2x and sin

2x, so that finally we may write

Thus we have shown that the solution of equation (1) is a linear combination of

cos 2x and sin 2x.

Series Solution of Differential Equation Near an Ordinary Point

Example:

(1-x2)y’’ – 6xy’ – 4y = 0

Solution:

Singular points: 1 – x2 = 0

x1 = -1 , x2= 1

thus, there’s a solution of the equation for |x| < 1

-1 < x < 1

Assume the series solution:

n

from y = n

y’ = n-1

y’’ = n-2

(1-x2) n-2 - 6x n-1 - 4 n = 0


158

(10)

n-2 - n - 6 n - 4 n = 0

n-2 - n = 0

n-2 - n

n-2 - n-2 x n-2 =0

[ - n-2 ] x n-2 = 0

- n-2 = 0

n-2 ] = 0

nAn – (n+2)A n-2 = 0

An= n+2 An-2 = 0 recurrence relation

n

A0 & A 1 are arbitrary

n Even n Odd

2 A2 = 4/2 A0 3 A3 = 5/3 A1


159

4 A4 = 6/4 A2 5 A5 = 7/5 A 3

6 A6 = 8/6 A4 7 A7 = 9/7 A 5

. .

. .

. .

2k A2k = 2k+2 A2k-2 2k+1 A2k +1 = 2k+3 A2k-1

2 2k+1

multiplying A’s ( in “even group”)

A2A4A6 …..A2k = 4 6 8 ……(2k+2) A2A4A6 …..A2k-2

2 4 6 …..2k

A2k = (k+1)A0 for k > 1

multiplying A’s ( in “odd group”)

A3A5A7 …..A2k+1 = 3 7 9 ……(2k+3) A1A3A5 …..A2k-1

3 5 7 …(2k+1)

A2k+1 = 2k+3 A1 for k>1

3

Now, the solution is

n = k

= A0+A1x+ 2kx2k + 2k+1x2k+1

y = A0+A1x+ 0x2k + (2k+3)A1x2k+1

3


160

y = [1 + x2k]A0 + [x + (2k+3)x2k+1] A1

3

y = A0 x2k + A1 (2k+3)x2k+1

3

Supplementary Problems:

Solve the given differential equation by means of a power series about the

given point x0. Find the recurrence relation; also find the first four terms in each of

two linearly independent solutions (unless the series terminates sooner). If

possible, find the general term in each solution.

1. y’’ – y =0, xo = 0

2. y” – xy’ – y =0, xo = 0

3. y’’ – xy’ – y = 0, xo = 1

4. y’’ + k2x2y = 0, xo = 0, k is a constant

5. (1 – x)y’’ + y = 0, x0 = 0

For problems 6-10, find the general solution valid near the origin, unless it is

otherwise requested.

6. (1+x2)y” – 4xy’ +6y = 0

7. (1+x2)y’ + 10xy’ + 20y = 0

8. (x2 + 4)y” + 2xy’ – 12y = 0

9. (x2-s9)y” + 3xy’ –3y = 0


161

10. y” + 2xy’ + 5y =0


162

Ferdinand Georg Frobenius

Born: 26 Oct 1849 in Berlin-Charlottenburg, Prussia (now Germany)Died: 3 Aug 1917 in Berlin, Germany

Georg Frobenius's father was Christian Ferdinand Frobenius, a Protestant parson, and his mother was Christine Elizabeth Friedrich.

Back at the University of Berlin he attended lectures by Kronecker, Kummer and Weierstrass. In 1874, after having taught at secondary school level first at the Joachimsthal Gymnasium then at the Sophienrealschule, he was appointed to the University of Berlin as an extraordinary professor of mathematics.

Weierstrass, strongly believing that Frobenius was the right person to keep Berlin in the forefront of mathematics, used his considerable influence to have Frobenius appointed.

Frobenius was the leading figure, on whom the fortunes of mathematics at Berlin university rested for 25 years. We should not be too hard on Frobenius for, as Haubrich explains in [5], Frobenius's attitude was one which was typical of all professors of mathematics at Berlin at this time:-

Applied mathematics, in his opinion, belonged to the technical colleges.

The view of mathematics at the University of Göttingen was, however, very different. This was a time when there was competition between mathematians in the University of Berlin and in the University of Göttingen, but it was a


163

competition that Göttingen won, for there mathematics flourished under Klein, much to Frobenius's annoyance. Frobenius hated the style of mathematics which Göttingen represented.

To gain an impression of the quality of Frobenius's work before the time of his appointment to Berlin in 1892 we can do no better than to examine the recommendations of Weierstrass and Fuchs when Frobenius was elected to the Prussian Academy of Science in 1892.

On the development of analytic functions in series.

The theory of linear differential equations.

On Pfaff's problem.

On adjoint linear differential operators...

The theory of elliptic and Jacobi functions...

On Sylow's theorem.

The theory of biquadratic forms.

On the theory of surfaces with a differential parameter.

In his work in group theory, Frobenius combined results from the theory of algebraic equations, geometry, and number theory, which led him to the study of abstract groups. This paper also gives a proof of the structure theorem for finitely generated abelian groups. In 1884 he published his next paper on finite groups in which he proved Sylow's theorems for abstract groups (Sylow had proved theorem as a result about permutation groups in his original paper).

In his next paper in 1887 Frobenius continued his investigation of conjugacy classes in groups which would prove important in his later work on characters. It was in the year 1896, however, when Frobenius was professor at Berlin that his really important work on groups began to appear. In that year he published five papers on group theory and one of them über die Gruppencharactere on group characters is of fundamental importance. He wrote in this paper:-

This paper on group characters was presented to the Berlin Academy on July 16 1896 and it contains work which Frobenius had undertaken in the preceding few months. It is worth noting, however, that although we think today of Frobenius's paper on group characters as a fundamental work on representations of groups, Frobenius in fact introduced group characters in this work without any reference to representations.


164

Over the years 1897-1899 Frobenius published two papers on group representations, one on induced characters, and one on tensor product of characters. This work was published in 1897. Frobenius's character theory was used with great effect by Burnside and was beautifully written up in Burnside's 1911 edition of his Theory of Groups of Finite Order.

Frobenius had a number of doctoral students who made important contributions to mathematics. Frobenius collaborated with Schur in representation theory of groups and character theory of groups. Frobenius's representation theory for finite groups was later to find important applications in quantum mechanics and theoretical physics which may not have entirely pleased the man who had such "pure" views about mathematics.

In [5] Haubrich gives the following overview of Frobenius's work:-

Even his analytical work was guided by algebraic and linear algebraic methods.

John von Neumann (born Johann von Neumann) was a child prodigy, born into a banking family in Budapest, Hungary. About 20 of von Neumann's 150 papers are in physics; the rest are distributed more or less evenly among pure mathematics (mainly set theory, logic, topological group, measure theory, ergodic theory, operator theory, and continuous geometry) and applied mathematics (statistics, numerical analysis, shock waves, flow problems, hydrodynamics, aerodynamics, ballistics, problems of detonation, meteorology, and two nonclassical aspects of applied mathematics, games and computers). At the instigation and sponsorship of Oskar Morganstern, von Neumann and Kurt Gödel became US citizens in time for their clearance for wartime work. During the war, von Neumann's expertise in hydrodynamics, ballistics, meteorology, game theory, and statistics, was put to good use in several projects.

Postwar von Neumann concentrated on the development of the Institute for Advanced Studies (IAS) computer and its copies around the world.


165

SOLUTION OF DIFFERENTIAL EQUATION NEAR REGULAR SINGULAR

POINT

The Method of Frobenius (named after Ferdinand Georg Frobenius)

Let x = a be a regular singular point of the differential equation

p(x)y” + q(x)y’ + r(x)y = 0 (1)

where p(x), q(x), r(x) are polynomials, and suppose that x = a is a regular

singular point, i.e.,

Then (1) has a Frobenius-type solution of the form

where the series apart from the factor(x – a)c converges for all x such that Ix – aI

< R and where R is the distance from x = a to the nearest singularity (other than

a). the series may or may not converge for Ix – aI = R but definitely diverges for

Ix – aI > R.

Series Solution of Differential Equation Near the Regular Singular Point (RSP)

Consider the differential Equation,

p(x) y + q(x) y + r(x) y = 0

if x = a js a RSP of then a frobenius type solution

if the form


166

andboth exist

cj

j

c axaaxaaxaaaxy )( )()()( 2

210

1

1

Cj (x-c1) j+c = (x- c1)c Cj (x-c1) j

j=0 j=0

= (x- c1)c C0 + C1 (x-c1) + C2 (x-c1)2 + … + Cj (x-c1)j exists.

Normalizing

y + q(x) y + r(x) p(x) p(x)

x = C1 is RSP of

iff.

The solution exist at /x-c1/ < R and diverges at /x-c1/ > r where R = radius of convergence

Examples:

1. Find the Frobenius solution of 4xy” + 2y’ + y = 0.

Solution:

We have p(x) = 4x, q(x) = 2, r(x) = 1, and x =0 is a singular point.

Since

both exist, we see that x = 0 is a regular singular point, so that there exists a

Frobenius-type solution of the form


167

and

1

1

lim (x-a) q(x) and x-a p(x)

lim (x-a) r(x) both exist x-a p(x)

where the summation limits - and for j are omitted, and where aj = 0 for

negative integers j. Differentiation of (1) yields

Substituting (1) and (2) in he given differential equation, we have

Since aj =0 for j negative, the first value of j for which we get any information

from (3) is j = -1 ( j = -2 yields 0 = 0). For j = -1(3) becomes

from which c = 0,1/2. The equation (4) for determining c is often called the

indicial equation, and the values of c are called the indicial roots, in this case

c=0,1/2. There are two cases which must be considered, corresponding to the two

values of c.

Case 1, c = 0. In this case (3) becomes

Putting j = 0, 1, 2 . . . , we find

from which


168

from whichor, since a0 o, (2c)(2c-1) = 0 (4)

or

Case 2, c = ½, in this case (3) becomes

Putting j = 0, 1, 2, ……yields

From which

From the solutions (5) and (6) we obtain the general solution

The observant student will note that the two series in (7) represent,

3. Find Frobenius-type solutions for x2y” + xy’ + (x2 – 1) = 0.

Solution: Here p(x) = x2, q(x) = x, r(x) = x2 – 1 and x = 0 is a singular point.

Since

It follows that x = is a regular singular point so that, by the theorem above,

there is a Frobenius-type solution of the form

Substituting this in the given differential equation yields


169

Which can be written as the single summation

Setting the coefficient of xj+c equal to zero and simplifying we get

(j + c + 1)(j + c - 1))aj +aj – 2 = 0 (9)

putting j = 0 in (9) yields (c+1)(c-1)ao = 0, which since c o yields the indicial

equation.

(c+1)(c – 1) = 0 so that c = 1, -1

are the indicial roots. There are two cases:

Case 1, c = -1. Putting c = -1 in (9) we have

putting j =1 leads to a1 =0, but putting j = 2 does not lead to a meaningful value

for a2 since we assumed ao 0. Thus we cannot obtain any solution in this case.

Case2, c = 1. Putting c = 1 in (9) yields

Putting j = 1, 2, 3, ………leads to

Note that all aj with odd subscripts are zero because they are all multiples of

a1=0. The required solution is thus given by


170

Supplementary Problems


171

Friedrich Wilhelm Bessel

Wilhelm Bessel's father was a civil servant in Minden.

Bessel's brilliant work was quickly recognized and both Leipzig and Greifswald offered him posts. A doctorate was awarded by the University of Göttingen on the recommendation of Gauss, who had met Bessel in Bremen in 1807 and recognized his talents.

Also during this period, in 1812, he was elected to the Berlin Academy.

The Königsberg Observatory was completed in 1813 and Bessel began observing there.

From 1840 on, Bessel's health deteriorated. Let us examine Bessel's work in more detail. Bessel's work in determining the constants of precession, nutation and aberration won him further honours, such as a prize from the Berlin Academy in 1815. Bessel published Bradley's stellar positions in 1818 in a work which gives the proper motion of stars.

In 1830 Bessel published the mean and apparent positions of 38 stars over the 100 year period 1750-1850. Bessel also worked out a method of mathematical analysis involving what is now known as the Bessel function.


172

It was not the first time that special cases of the functions had appeared, Jacob Bernoulli, Daniel Bernoulli, Euler and Lagrange having studied special cases of them earlier. In fact it was probably Lagrange's work on elliptical orbits that first suggested to Bessel to work on the Bessel functions.

This remarkable man who left formal education at the age of 14 made contributions beyond astronomy and mathematics.

Bessel also had a very significant impact on university teaching despite the fact that he never had a university education.


173

SERIES SOLUTIONS OF SOME IMPORTANT DIFFERENTIAL

EQUATIONS

In the process of formulating various applied problems, several important

differential equations have been found leading to special functions named after

their discoverers. One such equation is called Bessel’s differential equation,

after German astronomer Wilhelm Friedrich Bessel, who discovered it in

formulating a problem in planetary motion. Another is called Legendre’s

differential equation, after the French mathematician and astronomer Adrien

Marie Legendre.

I. Bessel’s Differential Equation (named after Friedrich Wilhelm Bessel –

German mathematician)

The equation x2y” + xy’ +(x2 – n2)y = 0 (1)

Where n may have any value but is usually taken to be an integer, is known as

Besssel’s equation of order n. By methods of the previous section it is seen that

the question has a regular singular point at x = 0. Thus there exist a Frobenius-

type solution having the form

If we substitute (2) into given equation, we obtain

Or

This can be written

or


174

putting j = 0 in (3) and noting that a-2 = 0 while ao 0, this becomes

(c2 – n2)ao = 0 or c2 – n2 = 0

which yields the required indicial roots c = n. Let us consider the two case,

putting j = 1, (4) shows that a1 = 0 since a-1 = 0. Furthermore, (4) also shows that

a3 = 0, a5 = 0, a7 = 0 ………, or all a’s with odd subscript is zero.

Putting j = 2, 4, 6, ……. In succession yields

where the rule of formation is apparent. The required series solution is thus

given by

which can also be written in either of the forms

on looking at the terms in the last series we note that the denominators contain

factorials, i.e., 1!, 2!, 3!, ……Also present in these denominators are n +1,


175

becomes (3) case this In .0 , nncCase 1,

(n+1)(n+ 2), (n + 1)(n + 2)(n + 3)……, which would become factorials, I.e., (n+1)!,

(n + 2)!, (n + 3)!,….. if we multiplied each of them by n!. Another thing that we

notice is that the terms of the series all involve x/2, while the multiplying factor in

front involves only x. A particular solution supplying these factorials and

introducing x/2 instead of x in the factor is obtained by choosing

in which case (5) becomes

This is a solution for n = 0, 1, 2, 3, ……, where we define 0! = 1 so as to make (6)

agree with (5) for n = 0.

Expressing Jn (x) in terms of r(n)

Jn (x) = (x/2)n

r(1/2) = c = n n 0 c = - n n 0

Jn (x) = (x/2)-n

For n greater than zero but not an integer the solution of the Bessel Equation is given by:

Y = C1 Jn (x) + C2 J- n (x) for n 0,1,2,3 …

For integer values of n,

J- n (x) = (-1)n Jn(x)

Graph:


176

1 (x/2)2 (x/2)4 (x/2)6 …r(n+1) r(n+2) 2! r(n+3) 3! r(n+4)

1 (x/2)2 (x/2)4 … r(-n+1) r(-n+2) 2! r(-n+3)

In order to allow for the possibility that n is any positive number, we can

use the generalization of the factorial called the gamma function. The function is

defined by

then we have the recurrence formula

Thus if n is a positive integer, say 1, 2, 3, ……, then

and in general (n + 1) = n!, n = 1, 2, 3, …….. (9)

we have the special value

so that using (8)

To determine the gamma function for any positive number, it suffices

because of (8)b to know its value for numbers between 0 and 1. These are

available on tables.

With this use of the gamma function to generalize factorials we can write

(6) as


177

Which is a solution of Bessel’s equation for all n 0. Now just as we give names

to people so we can talk about them , so we give names to important

functions. We call (11) by the name Jn(x),. Thus

To treat this case, we could go back to the result (3) and rework all the

coefficients. However, this is not necessary since all we have to do is replace n

by –n in (11) wherever it occurs. This lead us to

A question immediately arises as to how we interpret the gamma function for a,

negative number. Thus for example, if n = 5/2, the first two denominators in (13)

are (-3/2) and (-1/2). The definition (7) cannot be used since it is applicable

only for n 0 where the integral converges,. We can however agree to extend

the range of definition to n 0 by use of (8), i.e.,

(n + 1) = n(n) for all n. (14)


178

Thus, for example, on putting n = -1/2 and n = -3/2 in (14) we find

If we put n = 0 in (14), we have normally 1 = 0(0) which lead us to define

1/(0)=0 or (0) as infinite. Similarly, if we put n= -1 in (14), we have (0) = -

1(-1) or 1/(-1) = 0. The process can b e continue and we find

Using these interpretations, (13) becomes meaningful for all n 0 and represents

a solution of (1).

Now if n is positive but is not an integer, the solution (13) is not bounded

at x = 0 while (11) is bounded (and in fact is zero) for x = 0. This implies that the

two solutions are linearly independent so that the general solution of (1) is

Where we have also included n 0 in (16), since for n = 0, Jn(x) and J-n(x) both

reduce to Jo(x), and (16) clearly does not give the general solution.

If n is an integer, Jn(x) and J-n(x) are linearly dependent. In fact we have

For example if n = 3, (13) becomes on using (15)

Also from (12) we have if n = 3

Comparing (18) and (19), we see that J-3(x)n = -J3(x) agreeing with (17) for n = 3.

Similarly (17) can be established for all integers n.


179

To obtain a solution which is linearly independent of Jn(x) in case n is an

integer. We find for the general solution

This is the general solution regardless of whether n is an integer or not and thus

includes (16). A disadvantages of the second solution in (20) is that we do not

have it in explicit form so that we cannot graph it., etc.

To obtain a second solution in explicit form, we can resort to the following

interesting device. If n is not an integer, it follows from the fact that the Jn(x) and

J-n(x) are both solutions of (1) that a solution is also given by

And that this solution is linearly independent of Jn(x). If on the other hand n is an

integer, (21) assumes an indeterminate form 0/0 because of (17) since cosn =

(-1)n. We are thus led to consider as a possible solution for n equal to an integer

This limits can be found using L’Hopital’s rule as in calculus and yields a solution.

We call this second solution, which is linearly independent of Jn(x) whether n is or

not an integer, the Bessel function of the second kind of order n, reserving the

name Bessel function of the first kind of order Jn(x). Using this we can thus write

the general solution of (1) for all n as

Let us now attempt to gain some familiarity with these Bessels functions.

We can start out with the simplest case where n = 0. The Bessel function of the

first kind in this case is given by


180

Now we can do just as much with (24) as we can with ex. For example, after a

laborious calculations we can tabulate the results:

Jo(0) = 1 , Jo(1) = 0.77, J0(2) = - 0.22, Jo(3) = - 0.26, Jo(4) = - 0.40

We may graph Jo(x) for x 0 and obtain that shown in figure 1 below. The

graph for x 0 is easily obtained since it is symmetric to 0 is easily obtained

since it is symmetric to y –axis.

It will be seen that the graph is oscillatory in character, resembling the

graphs of the damped vibrations. The graph also reveals that there are roots of

the equation Jo(x) = 0, also called zero’s of Jo(x), obtained as points of

intersection of the graph with the x – axis. Investigations reveals that there are

infinitely many roots which are all real and positive.

In a similar manner for JI(x) we have

the graph of which is also shown in the figure 1. It should be noted that the roots

of J1(x) = 0 lie between those of Jo(x) = 0.


181

The graphs of Jn(x) for other values of n are similar to those in figure 1. It

is possible to show that the roots of Jn(x) = 0 real, and that between any two

successive positive roots of Jn(x) = 0 there is one and only root of Jn+1(x) = 0. This

is illustrated in table 1 in which we have listed for purposes of reference the first

few positive zeros of Jo(x), J1(x), J2(x), J3(x).

It is of interest that if we take the differences between successive zeros of

Jo(x) we obtain 3.1153, 3.1336, 3.1378, 3.1394,………, suggesting the conjecture

that these differences approach = 3.14159…. A similar observation on the

differences of successive zeros of J1(x), J2(x),…also leads to such a conjecture.


182

2 4 6 8 10 12 14

1.0

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1.0

Yo(x)

Y1(x)

x

y

Fig. 1

The conjecture can actually be proved and since the successive zeros of sin x or

cos x differ by , the approximate description of he Bessel function Jn(x) as a

“damped sine wave” is further warranted.

The wave characteristics of Bessel function seem very much like the

shapes of water waves which could be generated for example by dropping stone

into the middle of a large puddle of water. The water waves thus generated

would resemble the surface revolution generated by revolving the curves of Fig.1

about the y – axis. Indeed it does turn out that in the theory of hydro dynamics

such waves having the shape of Bessel functions do arise.

In like manner the Bessel function of the second kind Yo(x), Y1(x) are

shown in the Fig. 2. Note that these are not bounded as x 0 as we have

already indicated above. On putting n = ½ and n = -1/2 in (11), we find

which shows that J1/2(x), J-1/2(x) are in fact elementary functions obtained in term

of sine x and cos x. it is interesting to note that in such case the differences

between successive zeros are exactly equal to .


183

There are many identities which connect the various Bessel functions. A

few of the important ones are as follows:

This is a recurrence formula, which enables us to obtain Jn+1(x) when we are

given Jn(x) and Jn-1(x).

This is often called the generating function for Bessel functions.

SUPPLEMENTARY PROBLEMS


184

x

2 4 6 8

1.0 0.8 0.6 0.4 0.2 0-0.2-0.4-0.6-0.8-1.0

y

Fig. 2

Yo(x) Y1(x)

1. Solve Bessel’s equation x2y” + xy’ +(x2 –1/9)y = 0 directly using the method

of Frobenius.

2. Write the general solution of

(a) x2y” +xy’ + (x2 - 9)y = 0.

(b) (b) x2y” +xy’ +(x2 –8)y = 0.

3. Find the solution of x2y” +xy’ (4x2 – 1)y = 0, which is bounded at x = 0 and

satisfies the condition y(2) =5.


185

Adrien-Marie Legendre

Born: 18 Sept 1752 in Paris, FranceDied: 10 Jan 1833 in Paris, France

With no need for employment to support himself, Legendre lived in Paris and concentrated on research.

In 1782 Lagrange was Director of Mathematics at the Academy in Berlin and this brought Legendre to his attention. He wrote to Laplace asking for more information about the prize winning young mathematician.

Legendre next studied the attraction of ellipsoids. Legendre submitted his results to the Académie des Sciences in Paris in January 1783 and these were highly praised by Laplace in his report delivered to the Académie in March. Over the next few years Legendre published work in a number of areas. In particular he published on celestial mechanics with papers such as Recherches sur la figure des planètes in 1784 which contains the Legendre polynomials; number theory with, for example, Recherches d'analyse indéterminée in 1785; and the theory of elliptic functions with papers on integrations by elliptic arcs in 1786.

Legendre's career in the Académie des Sciences progressed in a satisfactory manner.


186

Following the work of the committee on the decimal system on which Legendre had served, de Prony in 1792 began a major task of producing logarithmic and trigonometric tables, the Cadastre. Legendre and de Prony headed the mathematical section of this project along with Carnot and other mathematicians.

Each section of the Institut contained six places, and Legendre was one of the six in the mathematics section.

Legendre published a book on determining the orbits of comets in 1806. For example Gauss had proved the law of quadratic reciprocity in 1801 after making critical remarks about Legendre's proof of 1785 and Legendre's much improved proof of 1798 in the first edition of Théorie des nombres.

To his credit Legendre used Gauss's proof of quadratic reciprocity in the 1808 edition of Théorie des nombres giving proper credit to Gauss.

Legendre's major work on elliptic functions in Exercises du Calcul Intégral appeared in three volumes in 1811, 1817, and 1819. In the first volume Legendre introduced basic properties of elliptic integrals and also of beta and gamma functions.

LEGENDRE’S DIFFERENTIAL EQUATION

The equation ( 1 – x2 ) y’’ – 2xy’ + n ( n+ 1 ) y = 0 (1)

is known as Legendre’s equation of order n. * It is seen that x = 0 is an ordinary

point of the equation, so that we can obtain solutions of the form

( 2 )

We could also use a Frobenius-type solution but would come out with c = 0,

which is equivalent to ( 2 ). Since x = ±1 are singular points of ( 1 ), the series

( 2 ) which is obtained should at least converge in the interval –1 < x < 1.

Substituting ( 2 ) to ( 1 ) we have


187

Putting j = -2 in ( 3 ) shows that a0 is arbitrary. Putting j = -1 in ( 3 ) shows that a1

is arbitrary. Thus we can expect to get two arbitrary constants and therefore the

general solution.

From (3),

Putting j= 0,1,2,3,… in succession we find

Then (2) becomes

If n is not an integer, both of these series converge for –1 < x < 1, but they can

be shown to diverge for x = ± 1. If n is a positive integer or zero, one of the series

terminates, i.e. is a polynomial, while the other series converges for –1 < x < 1

but diverges for x = ±1.

Let us find only the polynomial solutions. For n = 0,1,2,3,…,we obtain


188

Which are polynomials of degree 0,1,2,3,…It is convenient to multiply each of

these by a constant so chosen that the resulting polynomial has the value 1 when

x = 1. The resulting polynomials are called Legendre polynomials and are

denoted by Pn(x). The first few are given by

For any Legendre polynomial we have

It should be noted that the Legendre polynomials are the only solutions of

Legendre’s equation which are bounded in the interval –1 ≤ x ≤ 1, since the

series giving all other solutions diverge for x = ±1.

Since we know that Pn(x) is a solution of Legendre’s equation, we can use

Theorems 3 or 8, to find the general solution. We find

This second solution is related to the non – terminating series in (4). If we denote

the second solution by Qn(x), the general solution can be written

y = APn(x) + BQn(x) (6)

These functions are often known collectively as Legedre functions

Some important results involving Legendre polynomials are as follows:

1. (7)Engr. Tirso A. Ronquillo Instructor

189

This is a recurrence formula for the Legendre polynomials

2. (8)

This is known as Rodrigues’ formula for Legendre polynomials

3. (9)

This is called the generating function for Legendre polynomials.

If we make the defintion Pn(x) = 0 for n= -1, -2,…,(9) can be written in the

convenient form

(10)

where the summation is taken from .

In terms of Rodrigue’s Formula

P

Generating Function of Legendre Polynomials

Recurrence Formulas

1. P


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2. P

Legendre Functions of the Second Kind (Q if is given by

Q Q

Orthogonal of Legendre Polynomial

1. if mn

2.

Series of Legendre Polynomial

If f(x) satisfies Dirichlet conditions then at every point of continuity of f(x0

in the interval –1<x<1 there will exist a Legendre series expansion having the

form

f(x) = A

where: A

At any point of discontinuity, the given series

converges to

which can be used to replace f(x).


191

Exercices

Find a solution of ( 1 – x2 )y’’ – 2 xy’ +2y = 0 in the interval such that y(0) = 3, y’(0) = 4.

Ans. Y = 4x + 3 -

Pierre-Simon Laplace

Born: 23 March 1749 in Beaumont-en-Auge, Normandy, FranceDied: 5 March 1827 in Paris, France


192

Pierre-Simon Laplace's father, Pierre Laplace, was comfortably well off in the cider trade. At the age of 16 Laplace entered Caen University. Although Laplace was only 19 years old when he arrived in Paris he quickly impressed d'Alembert. This paper contained equations which Laplace stated were important in mechanics and physical astronomy.

The year 1771 marks Laplace's first attempt to gain election to the Académie des Sciences but Vandermonde was preferred. We have already mentioned some of Laplace's early work. Laplace's reputation steadily increased during the 1770s. Laplace was promoted to a senior position in the Académie des Sciences in 1785. Two years later Lagrange left Berlin to join Laplace as a member of the Académie des Sciences in Paris. Laplace married on 15 May 1788. Delambre also wrote concerning Laplace's leadership of the Bureau des Longitudes:-

Laplace had already discovered the invariability of planetary mean motions. In the Mécanique Céleste Laplace's equation appears but although we now name this equation after Laplace, it was in fact known before the time of Laplace. The first edition of Laplace's Théorie Analytique des Probabilités was published in 1812.. The group strongly advocated a mathematical approach to science with Laplace playing the leading role. Many of Laplace's other physical theories were attacked, for instance his caloric theory of heat was at odds with the work of Petit


193

and of Fourier. At the time that his influence was decreasing, personal tragedy struck Laplace. Laplace's son, Charles-Emile, lived to the age of 85 but had no children.

On the morning of Monday 5 March 1827 Laplace died.


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