advanced series
TRANSCRIPT
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
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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
between x = 7 and x = 10; y = 0; f (x) = 0, 7 < x <10
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
between x = 7 and x = 9; y = 0; f (x) = 0, 7 < x <9
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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:
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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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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
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Ex. Find the Fourier Series expansion of
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t2 -1 < t > 1f(t) = f (t + 2)
-1 1
f(t)
t
t2
(Even function)
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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).
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=
=
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)
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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)
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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
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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,
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f(t) = +
S = S =
Let
= A
Harmonic Cosine Series
Or
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A
b
S =
S =
f(t) = A +
= tan
S =
S = Harmonic Sine Series
EXPONENTIAL FORM OF THE FOURIER SERIES:
Adding,
Subtracting,
Substituting,
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= 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,
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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.
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C
=
C
f(t)
A……. …….
2
2
Solution:
f(t) =
T = 2 , Pulse width =
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f(t)
2
2
A
<
< t < ,
C
=
=
=
=
Magnitude Spectrum
AVERAGE POWER OF PERIODIC SIGNALS (FUNCTIONS)
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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 =
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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
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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,
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4. Find the Fourier series for the function f(t) defined by
and
Solution: Since
When n=1,
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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
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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:
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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
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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:
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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)
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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)
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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
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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
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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:
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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.
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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
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(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
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(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
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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
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(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
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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
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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
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10. y” + 2xy’ + 5y =0
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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
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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.
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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.
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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
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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
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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
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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
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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
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Supplementary Problems
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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.
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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.
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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
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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,
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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:
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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
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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)
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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.
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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
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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.
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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.
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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 .
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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
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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.
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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.
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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
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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
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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
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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).
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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
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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
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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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