Series Solutions And Frobenius

8/3/2019 Series Solutions And Frobenius

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COLLEGE OF ENGINEERING DESIGN, ART ANDTECHNOLOGY

School of Engineering

ENGINEERING MATHEMATICS III (EMT 2101)

LECTURE NOTES

Department of Mechanical Engineering (2011/2012)

May 12

MAKERERE UNIVERSITY


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COURSE OUTLINE

1. POWER SERIES SOLUTIONS TO O.D.ES1.1Power series Solutions1.2 Frobenius method

2. GAMMA AND BETA FUNCTIONS2.1 Integral definition of Gamma and Beta Functions2.2Properties of Gamma and Beta Functions2.3Relationships between Gamma and Beta Functions2.4Definitions of Gamma Functions for Negative Values of Argument.2.5Generalization of the Laplace Transform by means of Gamma Function2.6Other applications of the Gamma Function

3. BESSEL FUNCTIONS3.1Bessels Equation and its solutions.3.2Familiarization with characteristics and Graphs of Bessel Functions3.3 Integrals Involving Bessel Functions3.4Orthogonality of Bessel Functions3.5Bessel series3.6Modified (hyperbolic) Bessel Functions3.7Applications

4. LEGENDRE FUCTIONS4.1Legendres Equation and its Solutions4.2Legendres Polynomials4.3The generating function for Legendres Polynomials


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4.4Orthogonality of Legendres Polynomials4.5Legendre Series4.6Relationship between Legendres Polynomials and their Derivatives4.7Legendres Function of the Second kind4.8Associated Legendre Equation and Its Solutions4.9Orthogonality Relations for the Associated Legendre Functions4.10 Familiarization with characteristics and Graphs of Legendres Polynomials and

Associated Legendre Functions

4.11 Applications5. FOURIER AND LAPLACE TRANSFORMS

5.1Direct and Inverse Fourier Transforms and their Applications5.2Properties of Fourier and Laplace Transforms5.3Solutions to Ordinary Differential Equations by Transform Techniques5.4Transforms of Partial Fractions5.5Derivatives and Products of Functions5.6Transforms of Quadratic Factors5.7The Unit Step Function5.8The Impulse Function5.9Translation and Periodic Functions5.10 Solutions of Simultaneous Ordinary Differential Equations5.11 Applications of transform Methods to solutions of Engineering Problems-oscillatory

motion and Plane motion, Electric circuits.etc

5.12 Differentiation and Integration of Transforms.


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6. APPLIED STATISTICS AND PROBABILITY6.1 Introduction to applied Statistics :Data description and Representation, Descriptive statistics

and inferential statistics, Numerical Characterization and Summarization of data

6.2 Introduction to probability .Set Theory, Conditional Probability, Total Probability, BayesTheorem, Random Variables, Classification of Random Variables, Functions of Random

Variables, Expectations of functions of Random Variables, Moment Generating Functions,

6.3 Examples of Discrete and Continuous Random DistributionsASSESMENT

2 CATS @ 20%

REFERENCES:

1. Advanced Engineering Mathematics by Erwin Kreyszig2. Concepts of Engineering Mathematics by Dr. V.P Mishra3. Mathematical Methods in Physical Sciences by Mary L. Boas4. Advanced Engineering Mathematics by Wylie and Barrett5. Advanced Engineering Mathematics By Dean G. Guffy6. Advanced Engineering Mathematics by Allan Jeffrey


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INTRODUCTION:

Differential Equations are of fundamental importance in Engineering because many physical

laws and relations appear mathematically in the form of such Equations. In maths II, differential

equations were derived from physical or other problems (modeling). These were obtained as

ODEs of the First or Second order i.e. involving only the first or second derivative of the

dependent variable. Methods applied for solving included separation, use of integrating factors,

some Laplace transforms, Fourier Series etc. In Math III, we look at other advanced techniques

of solving some special ODEs.

ODEs may be divided into two large classes namely Linear and non-linear Equations. Whereas

non-Linear equations are difficult in general, Linear equations are much simpler because their

solution have general properties that facilitate working with them and there are standard

methods for solving many practically important linear differential equations.

A linear ODE is one which is linear in the dependent variable (unknown) and its derivatives. A

linear equation is an algebraic equation in which each item is either a constant or the product of

a constant and the first power of a single variable. Linear equations can have one or more

variables, example y = mx+b. the origin of the name linear comes from the fact that the set of

solutions of such an equation forms a straight line.

Example of a linear ODEs ; all ODEs which can be written in the form )()( xryxpy ,

)()()( xryxqyxpy


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A non-linear ODE is one which is not linear. In the dependent variable eg )()()(2

xryxpy ,

02)(2

yyyyyx . Less technically a non-linear system is any problem where the

variables to be solved cannot be written as a linear combination of independent components.

An ODE can be homogeneous or non-homogeneous. A homogeneous differential equation in

simple terms is one of the formn

n

dx

yd )(=0 non-homogeneous is of the form

0)( n

n

dxyd .


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POWER SERIES SOLUTIONS TO ODEs:

ODEs by nature of their solution fall into two broad categories, in the first category are ODEs

that can be solved exactly (in closed forms) in terms of functions usually studied in elementary

calculus such as rational algebraic, trigonometric and inverse trigonometric, exponential and

logarithmic and hyperbolic functions.

In the second category on the other hand, we have ODEs that cannot be solved exactly in terms

of elementary functions by application of any known methods. To this latter category, belong

certain important for Engineering applications, linear, homogeneous, second order ODEs.

0r(x)yq(x)y''p(x)y' Where r(x)andq(x),p(x), are real numbers, polynomials or analytic

functions (Function that can be represented in form of Taylor series) which are of interest to us.

There are essentially two methods today for solving such equations, either by means of infinite

power series or by means of numerical techniques. In this module, we consider techniques

based on the use of infinite power series.

If a homogeneous linear differential equation has constant coefficients, it can be solved by

algebraic methods, and its solutions are elementary functions known from calculus (ex, xcos

etc) however if such an equation has variable coefficients (functions of x), it must be usually

solved by other means. One typical method is the power series method which yields solutions in

the form of power series. An extension to this method is called the Frobenious method.


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Overview of Power Series:

A power series is an infinite series of the form

k

ok

k xa xy )(0

Rx xo = ...)()(2

21 xaxaa ooo xx

Where ,...,,21 aaao are constants called the coefficients of the series. xo is a constant called

the centre of expansion of the series. k is a summation index and real number, particularly a

non- negative integer. x the independent variable is a real number bounded on the interval

Rx x 0 R being the radius of convergence of the series.

In mathematics, the radius of convergence of a power series is a quantity/either anon-negative

real number or , that represents a domain (within the radius) in which the series will

converge. Within the radius of convergence, power series converges absolutely.

Examples of Series:

1. Taylor Series:This is a representation of a function as an infinite sum of terms calculated from the values of its

derivatives at a single point (Introduced by Brook Taylor in 1715)

2. Maclaurin Series:This is technically a Taylor series centered at zero (origin) (Introduced by Collin Maclaurin).

Taylor Series is of the form:


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k

k

n

axk

af

0 !=

...!3!2!1

3'''

2'''

axaf

axaf

axaf

af

...!3!2

1!

...!5!3!12

1sin

...!4!2

1!2

1cos

...11

1

e.g

series.maclaurinthegivesIt;0t whencoefficientheis!

where

32

0

53

0

12

42

0

2

2

0

xxx

ke

xk

xx

x

k

xx

xx

ak

af

k

k

x

k

kk

k

kk

k

k

k

x

xx

x

xx

The Series Solution to ODEs:

To motivate the discussion of the power series method of solution, let us start with a simple

example of solving the first order ODE. Subject to initial conditions .

ODE subject to initial conditions

ProblemValueInitial-IVP10,0' yyy

1. Intuitive approach;


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We know the solution to this as being an exponential function

1Athen

'

xx

x

x

AeAe

Aey

Aey

Solution; ,x

ey

On the other hand however, we recall that virtually all elementary functions encountered in

calculus that feature in exact solutions of ODEs can be represented as infinite power series.

...2210

xaaaxa xxy ook

k

Where ak are coefficients that have to be determined from auxiliary equations known as initial

boundary equations . On this basis, we could therefore legitimately seek a solution for the

equation in the form (see next page)

xfyydx

dy

yy

where

0'


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0

:1kthen:01-kkIf

seriestheshiftingi.eharmoniseto1-kksetting

0

0

that;impliesthis

.. .32'

.. .,

100'solving

expansion.ofcentertheisorigin

1

1k

1

2

1

1

1

1

0

1

1

12

321

1

1

3

3

2

210

0

0

k

kk

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

kk

xaka

xaxka

xaxka

xkaxaxaaxkay

xaxaxaaRxxay

yyy

xy a


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12

then,0solutions;lnon triviaFor x

1,

10

1

1

kk

aak

kaka

k

kk

arbitraryisthatassumewe 0a

!3233;3

22;2

;1For

002

3

012

01

aaaak

aaak

aak

!4!344;4 0034

aaaak

Generally;

0!

0 kk

aak

equationoriginalourinbackFeeding

x

k

kk

k

eak

xax

k

ay

0000

0

!!


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conditionsInitialourinFeeding

.is;solutionThe

1

10

0

xey

a

y

Example: 2

Solve 0'' yy by the power series method..

Postulate a solution of the form.

22

2

1

1

4

4

3

3

2

210

''

'

.. .

kk

k

k

k

k

k

ok

k

xakky

xkay

xaxaxaxaaxay

But 0'' yy .

Substituting the derivatives into the ODE

2set

02

2

2

kk

xaxakkk

ok

k

k

k

k


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2 2

2

2

22 0)(k k

k

k

k

k xaxakk

2

2

2

2 0)(k

k

kk xaakk

0solutions;trivial-nonFor x

2

1

0

2

2

2

kkk

aa

aakk

k

k

kk

.andofin termssother termexpresswearbitrary,areandcase,For this 1010 aaaa

.. .!5!3

...!4!2

1

...!5!4!3!2

!545;5

!434;4

!3;3

;!2

;For

53

1

42

0

5

1

4

0

3

1

2

0

10

13

5

02

4

1

3

0

2

xxxa

xxa

xaxaxaxaxaay

aaak

aaak

aak

aak


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seriesoddsin...!5!3

seriesevencos...!4!2

1But

53

42

xxx

x

xxx

Remember the coefficients

seriesoddforis!12

1and

seriesevenforis!2

1

112

02

k

aa

k

aa

k

k

k

k

Therefore our expression can be simplified to give

0

12

1

0

2

0

12

0

12

2

0

2

0

!121

!2

1

k

kk

k

kk

k

k

k

k

k

k

k

k

k

kx

ak

xa

xaxaxay

xaxay sincos 10


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Example: 3

Solve the Initial Value Problem 1)0(',0)0(0'2'' yyyxyy using the power series

method.

Solution

We start our search for the power series solution by postulating a series of the form.

k

k

kxay

0

ODEesatisfy thshouldsolutionpostulatedthe,tcoefficienthedetermineTo ka

0221

.ofpowersequalobtainshift towe;2set

021

021

2

2

2

2

3

2

2

2k

01

2

2k

0

1

1

2

2k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

k

xaxakxakk

xkk

xaxkaxakk

xaxkaxxakk

termslikecollecting

02212

0221

2

2

3

2

3

202

2

2

3

2

0

2

k

k

k

k

k

kk

k

k

k

k

k

kk

xakxaakkaa

xakxaakk


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05212

012212

02212

2

3

202

2

3

202

2

3

2202

k

k

kk

k

k

kk

k

k

kkk

xakakkaa

xakakkaa

xakaakkaa

Obtaining the re;ationships

31

522.

21;

From

305212.

and02.1

2

0

2

2

02

kkk

aka

aa

kakakk

aa

k

k

kk

!4

3

;4

!3

;3:For

24

13

aa

k

aa

k


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!7

59

67

9

!6

37

56

7

!5

5

45

5

15

7

04

6

13

5

aaa

aaa

aaa

arbitraryareandThus 10 aa

Therefore

110'

000From

...!7

95

!5

5

!3...

!6

73

!4

3

!21

!7

59

!6

73

!5

5

!4

3

!32!

solution;heRemember t

1

0

753

1

642

0

7

1

6

0

5

1

4

0

3

1

2

0

1

ay

ay

xxxxa

xxxay

xaxaxaxaxaxaxaay o

.!7

95

!5

5

!3

753

xxx

xy


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Basic Theories and concepts of Power series:

Theorem 1:

Given the ODE of the generic form

0''' yxryxqyxp

If the coefficient terms xrxqxp and, are

a) Real numbersb) Polynomialsc) Analytic functions at a point 0x , then 0x is an ordinary point of the ODE.The undetermined coefficients ka can be determined by requiring that the postulated solution

satisfies the differential equation in question and the radius of convergence R can be determined

using the ratio test given R .

a) R kk

alimk

1

b)k

k

a

a 1

klim

1R

Provided these limits exist and are not zero.

R is at least as great as the distance from 0x to the nearest singular point of the ODE.

0x is said to be an ordinary point of the ODE if the leading coefficient xp isnt equal to zero

at 0x (point of expansion) or infinity; otherwise 0x is said to be a singular point of the ODE

i.e. xp =0; at 0x , then 0x = singular point.


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Radius of convergence:

The series

...202010

0

xxaxxaaxxa ok

k

k

Always converges at 0xx , because then all its terms except for the first oa , are zero. In

exceptional cases, this may be the only x for which the series converges. Such a series is of no

practical interest.

If there are further values of x for which the series converges, these values form the interval,

called the convergence interval. If this interval is finite, it has the midpoint 0x

, so that it is of the

form.

R0 xx And the series converges for all x .

Such that R0 xx and Diverges for all x such that R0 xx .

Example:

Determine the radius of convergence for the Maclaurins type series solution of the Legendre

equation of the order n.

1.Radius

01or10R

1;0For

1For

1tcoefficienleading

solution;

001'2''-1

2

2

2

xxp

xxp

xxp

nynnxyyx


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Analytic Functions:

A function is said to be analytic at a point 0x if and only if it has a Taylor series in powers of

0xx which represent it in some open interval containing 0x with a radius of convergence

0R i.e.

0finitelim0

xfxfxx

Properties

1. A function which is analytic at a point has a derivative of all orders on an open intervalcontaining that point.

2. On an open interval where they are analytic, sums, differences and products of analyticfunctions and the quotients of two analytic functions i.e.

xgxf

Are both analytic at 0x , then 0xg

Theorem:

For every power series of the form

0

0

k

k

k xxay , One of the following statements is true.

a) The series only converges at 0xx therefore 0R , such series are of no Engineeringrelevance.

b) There is a finite number R greater than zero such that for all values of x , if R0 xx ,the series converges and otherwise Diverges.


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Illustration

Find the 5th

degree polynomial approximation to the series solution of the Initial Value Problem

of;

10',200'''1 yyyxyyx

Step one:

Postulating a solution;

The ODE in question belongs to the family of ODEs in the form:

0''' yxryxqyxp

In comparison, 1,,1 xrxxqxxp are all Coefficient functions being polynomials, or

real numbers, Therefore the ODE in question can be solved using the power series method.

Observe that the origin (point about which the initial conditions are prescribed) like all other

real numbers 1x , is an ordinary point of the ODE. Hence the ODE has at least one

Maclaurins type series solution about the origin of the form

Where R is the radius of convergence and ka is the coefficient of the kth term of the series.

Step two:

Determining R and ka .

By definition R , is the distance between 0x and the nearest singular point of the ODE. In this

case 00 x , and the only singularity due to 01 xxp is 1x thus 1R

R,0

xxayk

k

k


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The coefficients can be determined to within an arbitrary constant by requiring that the

postulated solution satisfies the ODE in question. Hence substituting the postulated solution and

its derivatives yields the expression.

1 0

2

2

011k k

k

k

k

k

k

k

k

xaxkaxakkx

It is our intention to combine all the terms on the left hand side into one series term under one

summation. For this to be possible, same power ofx must appear in the general term of the

general series terms. This can be achieved by shifting the summation indices in all series to

bring down the exponent of the independent variablex to 2k i.e. the least available.

In so doing, we achieve the following,

a)The exponent of the independent variable x is the same in all the series.b)k is the largest subscript on a in any series.

2...01221

02121

yeildsThis

2-kk2-kk1

011

2

2

2

2

3

21

3 2

2

2

2

2

2

2

2

3

1

1 0

2

2

1

2

kxaakkxakakk

xaxakxakkxakk

kk

xaxkaxakkxakk

k

k

kk

k

k

kk

k k

k

k

k

k

k

k

k

k

k

k

k k

k

k

k

k

k

k

k

k

k

k


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Extracting the terms that have 0x yields.

0221123

21220

k

kkkk akakkakkaaa

For non-trivial solutions; 0x

6120

19

5

3

624

5

4

2

!3

2

3

2

211

01211

2From

2

1From

01211.2

02.1

1034

5

10234

10123

21

21

21

0

2

21

20

aaaaa

aaaaa

aaaaa

k

aaka

aakkakk

akakkakk

a

a

akakkakk

aa

kkk

kkk

kkk

kkk


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We discover that our solution is in the form

2110 yayay

...120

20

24

4

6

2

...120

19

24

5

621

543

1

5432

0

xxx

xa

xxxx

ay

Invoking the initial conditions

1,2

10',20

10

aa

yy

On substituting the conditions to get the necessary powers ofx , [particular solution]. We have

20

3

42

54

2xx

xxy

Exercise:

Using the power series method, solve the initial Value problems

1. 60y,10y0241 2 yyxyx 2. 00y,10y02 yyxy


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METHOD OF FROBENIOUS(Extension of the of the Power series Method):

The solution of an ODE of the form 0''' yxryxqyxp about a point 0x depends on the

nature of 0x . If 0x is an ordinary point of the ODE i.. the leading coefficient. 0xp , at 0x

then the ODE has at least one power series solution of the form.

R, 00

0

xxxxayk

kAbout 0x If however 0xp at 0x , it is termed as singular

point of the ODE and is denoted as 0x .

This implies that the quotient s xpxq

and/or xpxr

have discontinuities at 0x and are not analytic

there.

We will continue to note that the manifested discontinuities may be removed by ;

0'''

is;ODEThe

)

)

2

0

0

yxp

xry

xp

xqy

xp

xrxxb

xpxqxxa

Singular Points (Regular and irregular singular Points)

A regular point of an ODE 0''' yxqyxpy is a point 0x at which the coefficient s p and q

are analytic. If a point is regular, then the power series method can be applied. If 0x is not

regular, then it is called a singular point.


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A singular point 0x of the ODE

0''' yxryxqyxp

Can be regular if the product coefficients

xpxq

xx 0 and

xpxr

xx

2

0 are analytic at 0x .

Mathematically;

finitelim

andfinitelim

2

0

0

0

0

xp

xrxx

xp

xqxx

xx

xx

Otherwise x is said to be an irregular singularity of the ODE.

Remember;

0x is a regular point of the ODE

0''' yxryxqyxp If xqxp , and xr are analytic and 0xp and as seen previously

can be solved by the power series method.

Example:

Find and classify the singularities of the ODE

03'25''2 23 yxyxyxx

Solution:

The given ODE belongs to the family of ODEs of the form

0''' yxryxqyxp


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In comparison 23 xxxp , 25 xxq , 23xxr , 2,0;02 03 xxxxp ,

ifregularbetosaidis0x

Since the first product coefficient is not analytic at the origin, the origin is an irregular

singularity of the ODE in question.

question.inODEtheofysingularitregularais

2,atanalyticaretscoefficienproductbothSince

023

lim2

32lim

5.225

lim2

252lim

;2

23

22

2

3232

0

o

o

xx

xx

x

x

x

x

xx

xx

x

x

xx

xx

x

Theorem:

In case0x is an irregular singularity of the singularity of the ODE 0''' yxryxqyxp ,

there are no power series solutions to that ODE. The solution techniques applicable to problems

in that domain are beyond the scope of our discussion. We will again note that if0x a regular

)2(

25lim

2

25lim

0For

finitelim

finitelim

2030

2

00

0

xx

x

xx

xx

x

xxxp

xr

xxxpxq

xx

xx

oxx


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singularity of the ODE 0''' yxryxqyxp ; we postulate an extended solution of the

form;

0; 00

0

axxay

ck

k

k

0n the deleted neighborhood.

ccR 000

D

xxxx

Where c are the roots of the individual equation (indicating equation)

This is known as the method of Frobenius and the associated series is known as the Frobenius

type series. The discussion will be limited to those cases where the origin ( 00 x ) is a regular

singularity of the ODE. 0''' yxryxqyxp .

If the ODE 0''' yxryxqyxp has analytic coefficient functions xrxqxp ,, and a

regular singularity at the origin 00 x so that the product coefficients xpxq

x and xpxr

x2 have

valid Maclaurin series.

20

2

1

0

R

R

xxrxpxrx

xxqxp

xqx

k

k

k

k

k

k

And if the roots 1c and 2c of the indicial equation 0rc1-qc 002 are identified such that

the real part of c1 is at least equal to the real part of c2 . If 21 ReRe cc then the ODE has a


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complete solution 21 byayy where a , b are parameters on the deleted neighborhood of the

origin. R0 x . R Is not less than the smallest of 1R and 2R

21 RandRminR 1y And 2y are linearly independent solutions with the following

mathematical structure;

The ODE in question is 0''' yxryxqyxp the postulated Frobenious solution is of the

form.

solutionstindependenlinearlyandofcomposedis 21

0

000

yyy

xaxy

xaxxay

k

k

k

c

k

ck

kk

ck

k

Case I:

Distinct roots not differing by an integer then

20

0

2

0

10

1

R00,

R0,0,

2

1

xbxbxy

xaxaxy

k

k

k

c

k

k

kc

i.e.

...

...

2

2102

2

2101

2

1

xbxbbxy

xaxaaxy

c

c

Case II:


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31

Roots differing by an integer . If thenInteger,positive21 cc

0

R0,0,

01

0

2

0

101

2

1

bxInxkyxbxy

xaxaxy

k

k

k

c

k

k

k

c

e.g.

...22101 xaxaaxy c

...221012 2 xbxbbxxInkyy c

If the roots 021 cc , and kmay turn out to be zero.

Case III:

Double roots If ccc 21

.. .

...

..

0,

2

2111

2

2101

1

12

0

01

xbxbxxInyy

xaxaaxy

ge

xbxxInyy

axaxy

c

c

k

k

k

c

k

k

k

c

Technically, the Frobenius method is similar to the power series method once the roots of the

indicial equation has been determined.

Observations


32/49

32

1.In all cases, the ODE has at least one Frobenius type series solution 1y , associated with thealgebraically bigger indicial root.

2.In case I, both 1y and 2y are Frobenius type series.3.In case II, the parameter kis unrestricted i.e .can take on any values.4.In case III, 2y is never a Frobenius type series. In the special case where 0kb , for 2,1 yk is

said to be a purely logarithmic case (1yxIn )

Example 1:

Find the Frobenius type series solution of the O.D.E 02 xyyyx near the orign. Give the

general solution and discuss its convergence.

Solution

Step 1:Postulating a solution

The ODE 02 xyyyx belongs to a family of ODEs of the form

0 yxryxqyxp .In comparison xxp , 2xq and xxr

All coefficient functions being real numbers or polynomials and hence analytic for all x .

Therefore the O.D.E in question can be solved using the power series method. Observe that the

origin is a singular point of the ODE the only singularity. Since both product coefficients are

analytic at the origin,

0limlim

,22limlim

2

0

2

00

000

xxp

xrxr

xp

xxqq

xx

xx


33/49

33

The origin is a regular singularity of the ODE in question , therefore there exists at least one

Frobenius type series solution of the form

expansion.seriesin thetermtheoftcoefficienthe-

equation,indicialtheofrootsthe-,econvergencofradiustheisRwhere0

origin.about theoodneighbourhdeletedin the0, 00

th

k

ck

k

k

ka

cRx

axay

Determining R, C and the coeffcients

By definition R is as great as the distance between the centre of expansion of the postulated

series solution and the nearest singularity

0R

00

x

x

The coefficients ka can be to within an arbitrary constant by requiring that the postulated

solution satisfies the ODE in question.

0

21

21

1

0

1

2

2

1

10

0

1...21'

...

k

ck

k

cccck

k

ck

k

ccc

k

ck

k

xackckxacxacxcaxacky

xaxaxaxaxay

Substituting the postulated solution and its derivatives into the ODE in question;

0210

1

00

2

k

ck

k

ck

k

k

k

ck

k xaxxackxackckx

Algebraic manipulation to make each term on the LHS a Power Series yields;.


34/49

34

0210

11

00

1

k

ck

k

ck

k

k

k

ck

k xaxackxackck

Factoring out cx

02)1(0 0 0

111

c

k k k

k

k

k

k

k

k xxaxackxackck

The only way the product can be equal to zero is when;

010 0

11

k k

k

k

k

k xaxackck Since 0c

x For non-trivial solutions

We seek to combine all terms on the left hand side of the above expression under one

summation. This can be achieved by first shifting the summation indices in all series such that

the power of variable x is 1k . Setting 2 kk in the second series yields;

0 2

1

2

1 01k k

k

k

k

k xaxackck

To combine the terms in the above expression under one summation, we then extract all terms

in the first series that are not in the second series.

2

1

2

1

2

01

1

001211

k

k

k

k

k

k

xaxackckxaccxacc

2

1

2

0

1

1

0 01211k

k

kk xaackckxaccxacc

Since the above expression is identically equal to zero, then

2011.3

0)2(1.2

01.1

2

1

0

kaackc

acc

acc

kk

These results imply that


35/49

35

1. 001 0 acc

Indicial roots 1,0 21 cc

Check with

0r,2-1cwhen

01

00

00

2

q

rqc

equation.indicialanis010

012

2

0

2

cc

cc

rcc

2. arbitraryis;01c,02 1ac 3.

2arbitraryis

11

2

ka

ckck

aa kk

The general solution

In such a case i.e. integerve21 cc ,1,0 21 cc It saves time to commence the search for

an infinite series solution by considering the algebraically smaller indicial root. And if a general

solution is not possible this way, there will be only one Frobenius type series with the

algebraically larger indicial root and the second linearly independent solution for the formation

of the desired general solution (method) of variation of parameters etc.

Therefore considering 1c

,2

1

12

ka

ckck

a kk 0a And 1a are arbitrary.

2

02

aa

;

!6

06

aa


36/49

36

Even coefficients;

3,2,1,0,!2

1 02

m

m

aa

m

m

!31

3 aa ; !51

5 aa

!12

1 112

m

aa m

122

0

;

mmkk

ck

k aaaxay

0

2

12

0

12

2

m

m

m

m

m

m xaxay

Factoring out x

x

m

xa

m

xa

y

x

xa

x

xay

m m

mmmm

m m

m

m

m

m

0 0

12

1

2

0

0 0

12

12

2

2

!12

1

!2

1

x

xaxay

sincos 10

Since the solution contains 2 arbitrary constants and two linearly independent solutions

x

xy

cos1 and

x

xy

sin2 , It is clearly the solution of the ODE in question and there is no need

to consider the other identical root.

Convergence of the general solution:


37/49

37

In the seriesx

xaxay

sincos 10 , It is well known that the cosine and sine functions converge

for all x . Observe however that y is a rational function with a denominator x and hence y is

not analytic at the origin which is the only singularity of the ODE in question, therefore

x

xaxay

sincos 10 is analytic for all x except for 0x


38/49

Example 2:

Find the Frobenius type series solution of the ODE 0'2''4 yyxy .

Solution

The ODE in question belongs to the family ODEs of the form 0''' yxryxqyxp . In

comparison, xp = x4 , xq =2, xr =1; all coefficient functions being polynomials and/or real

numbers hence analytic for all x . Therefore the ODE can be solved using the power series

method. Observe that the origin ( x =0) is a singular point, the only singularity of the ODE.

04

limlim

2

1

4

2limlim

2

0

2

00

000

x

x

xp

xrxr

x

x

xp

xqxq

xx

xx

Since both product coefficients are analytic at the origin, the origin is the only singularity of the

ODE, there exists at least one Frobenius type series solution of the form

0, 0 axay ckk in the deleted neighborhood R0 x

econvergencofradiusR , radiusindicialc , seriesintermtheoftcoefficien thk ka

DeterminingR , ka , c :

By definition R is as great as the distance between the center of expansion of the postulated

series solution and the nearest singularity. In this case, the origin which is the center of

expansion of the postulated series solution. ie the only singularity therefore R

The coefficients ka can be determined to within an arbitrary constant by requiring that the

postulated solution satisfies the ODE in question. Therefore substituting the postulated solution

and its derivatives into the ODE yields the expression


39/49

39

0122

intoormedlly transfalgebraicabecanexpressionabovethe,0iesolutionstrivial-nonFor

0214

yieldsLHSon thetermstheofonmanipulatiAlgebraic

0214

0

1

0

0

1

0

1

0

1

0

12

0

k

k

k

ek

k

k

ck

k

k

ck

k

k

ck

k

ok

k

ck

k

k

ck

k

ck

k

k

xaxackck

x

xaxackxackck

xaxackxackck

We seek to combine the terms on the LHS of the expression under one summation. This can be

achieved by first shifting the summation we set kto 1k and the result will be

0122 11

1

1

0

kk

k

k

k

k

xaxackck

To combine the series in the above expression under one summation we extract the first term in

the 1st series and combining the rest under one summation yields

0122122 11

1

1

0

kkkk

xaackckxacc

Since the above expression is identically equal to zero, for non- trivial solutions

10122)2

01221)

1

0

kaackck

acc

kk

0,02(1)from 0 a , 012 cc ,2

11

c , 02 c


40/49

40

Check : The indicial equation

0)1( 02

rc ocq , 0212

cc , 0,2

121 cc

From (2) 122

1

ckck

aa kk

Since2

121 cc is not an integer, we will associate 1c with ka and 2c with kb ie kac 1 ,

kbc 2 obtaining two linearly independent solutions 1y and 2y both being Frobenius type

series

0,0

21

0

1

axayk

k

k

0,0

0

2

bxbyk

k

k

!5

!3

constantarbitraryanis

1,122

,2

1gconsiderin

02

01

0

1

aa

aa

a

kkk

aa

c

kk


41/49

41

0gconsiderin

!121

!7

0

03

c

kaa

aa

k

k

1,

122

11

kb

kkb kk

Making the substitution

xBxAy

cy

yxy

k

xby

xy

k

xay

k

kk

k

kk

sincos

cos

!2

1

sin

!12

1

2

1

2

0

2

02

1

0

12

01

!6

!4

2!

constantarbitraryanis

0

3

0

2

0

1

0

bb

bb

bb

b


42/49

42

Convergence

Testing for convergence of the composite function

xBxAy sincos

existnotdoeslim

0lim

0

0

x

x

x

x

The general solution xBxAy sincos , x0

Exercise

Find a series solution to the ODE 0'2'' xyyxy near the origin.

Solution 0,sinhcosh 10

x

x

xaxaxy

xxrxqxxp ,2,

When ;00 x 0xp ; origin is a singular point of the ODE, the only singularity.

0limlim

22

limlim

2

0

2

00

00

0

x

xx

xp

xrxr

x

x

xp

xqxq

xx

xx

;00 x is a regular singularity of the ODE; There exists at least one Frobenius type series

solution of the form

0k

ck

kxay R0,00 xa Where R is the radius of convergence.

c -roots of the indicial equation , ka -coefficient of the kth

term

DeterminingR , ka and c

00 x 0x R is as great as the distance between the centre of expansion of the postulated

series solution and the nearest singularity 0R


43/49

43

Coefficient ka can be determined to within an arbitrary constant by requiring that the postulated

solution satisfies the ODE in question.

ck

k

kxay

0

10

'

ckk

kxacky

20

1''

ckk

kxackcky

Hence substituting the postulated solution and its derivatives into the ODE in question;

0210

1

0

2

0

ckk

k

ck

k

k

ck

k

k xaxxackxackck

Algebraic manipulation to make each term on the LHS a series yields

021 10

1

0

1

0

ckk

k

ck

k

k

ck

k

k xaxackxackck

Factoring out cx

0221 10

1

0

1

0

kk

k

k

k

k

k

k

k

c xaxackxackckx

021 10

1

0

kk

k

k

k

k

c xaxackckx

The product above is identically equal to zero but 0cx , therefore

01 10

1

0

kk

k

k

k

k xaxackck

We seek to combine all terms of the LHS of the above expressions under the summation. This

can be achieved by first shifting the summation indices in all series such that the variable x is

1k setting kto 2k on the second series yields.

01 12

2

1

0

kk

k

k

k

k xaxackck


44/49

44

To combine the terms in the above expression under one summation, we then extract all terms

in the first series that doesnt exist in the second series.

012112

1

2

1

2

0

1

1

0

k

k

k

k

k

k

xaxackckxaccxacc

01211 12

2

0

1

1

0

kk

kk xaackckxaccxacc

Since the above expression is identically equal to zero then.

constantarbitraryanis0,1,0

;01

checking

1,01.From

2;01.3

021.2

01.1

001

00

2

1

2

1

0

aacc

rcqc

cc

kaackck

acc

acc

kk

constantarbitraryanis,02,1

0212.

1

1

acc

acc

constantsarbitraryareand21

1

201

102

2

2

aakkk

aa

ckck

aa

kaackck

kk

kk

kk

integer,since 21 vecc we find the infinite series solution by considering the algebraically

smaller indicial roots and if this is not possible, this way, there will be one Frobenius type series

solution with the algebraically larger indicial root.

Therefore considering 1c


45/49

45

,...3,2,1,0!12

1

,...3,2,1,0!2

1

!9

!8

!7

!6

!5543245

!443234

!332

2

1

12

0

2

1

9

0

8

17

0

6

113

5

0024

11

3

0

2

mm

aa

mm

aa

aa

aa

aa

aa

aaaa

aaaa

aaa

aa

m

m

m

m

From the postulated series solution

m

mm

m

mm

mmk

k

ck

k

xaxay

aaacxay

2

012

12

02

122

0

,1,

Factoring out x :

012

0

12

2

0

2

mm

m

m

m

m xaxay

x

xaxay

x

m

xa

m

xa

y om m

mmmm

sinhcosh

!12

1

!2

1

10

0

12

1

2

0


46/49

46

Exercise

Find the general solution of the ODE 05'''2 2 yxxyyx near the origin. (20 marks)

Postulating a solution

The ODE 05'''2 2 yxxyyx belongs to the family of ODEs of the form

0''' yxryxqyxp In comparison 2xp , xxq , and 5 xxr , all coefficientfunctions being real numbers or polynomials and hence analytic for all x .

Therefore, the ODE 05'''2 2 yxxyyx can be solved using the power series method.Observe that x =0, is a singular point of the ODE and the only singularity.

Since

2

1

2limlim

200

x

xx

xp

xqx

xxand

2

5

2

5limlim

2

2

0

2

0

x

xx

xp

xrx

xx

It follows thatx =0 is a regular singular point of the ODE. Thence, there exists a Frobenius type

series solution of the form

0, 00

axay

k

ck

k

On the deleted neighborhood R0 x of the origin, where R is the radius of convergence

and ka is the coefficient of theth

k term of the series.

Determining c and ka :

The coefficients ka can be determined to within an arbitrary constant by requiring that the

postulated solution satisfies the ODE in question.

Hence substituting the postulated solution and its derivatives into the ODE yields the expression

05120

1

0

2

0

2

ckk

k

ck

k

k

ck

k

k xaxxackxxackckx

Algebraic manipulation to make each term on the LHS a power series yields

05120

1

000

ckk

k

ck

k

k

ck

k

k

ck

k

k xaxaxackxackck


47/49

47

Factoring out cx and combining 1st

, 2nd

and 4th

series under one summation

0 0

1 0512k k

k

k

k

k xaxackckck

To combine the series in the above expression under one summation, we then extract all termsin the 1

stseries that are not in the 2

ndseries 1.e. the 1

stterm in this case and combining the rest

under one summation yields:

1

10 0512512k

k

kk xaackckckaccc

Since the above expression is identically equal to zero, for non-trivial values ofx :

,...5,4,3,2,105122.

equationindicialthe,0512.1

1

0

kaackckck

accc

kk

These results imply that

1. 0512 0 accc 00 a ,the indicial roots are2

51 c and 12 c

2. 151

1

kk a

ckckcka for ,....4,3,2,1k

The mathematical structure of solution:

In this case i.e. 2721 cc is not an integer, the two linearly independent solutions of the ODE

have the following mathematical structure.

0; 00

1

1

axaxyk

k

k

con the interval

1R0 x

Where ka is the Fourier coefficient associated with the indicial root 251 c , hence

0; 00

1

2

bxbxyk

k

k

con the interval 2R0 x where kb is the Fourier coefficient

associated with the indicial root 12 c , hence it is important to note here that both 1y and 2y

Are Frobenius type series solutions.

The Fourier coefficients ka and kb

721

kk

aa kk and 72

1

kk

bb kk 1k more specifically.


48/49

48

0a and 0b are arbitrary constants which can be determined by imposing the initial conditions to

the resultant general solution if given.

72

5311355432156

311355432135

1135432114

13532113

352132

51

!72!

105!321

1917151311954321195

17151311954321175

151311954321154

1311931133

11921112

91

1

05

6

04

5

03

4

02

3

01

2

0

1

0

3

05

6

04

5

03

4

02

3

012

0

1

kk

bb

bbb

bbb

bbb

bbb

bbb

bb

kk

aka

aaa

aaa

aaa

aaa

aaa

aa

k

k

kk

k


49/49

Convergence of the linearly independent solutions:

Consistent with the fact are polynomials, the ratio test gives

921limlim1

kka

a

kk

k

kand

521limlim1

kkb

bk

k

k

kwhich implies.

That ,RR 21 hence the Frobenius type series in the expression for 1y and 2y converge for all x

on the deleted neighborhood ofx =0, and 1y is a solution there. Its important to note however

that 1y is not defined on the interval 0, but is finite at 0x where 2y is not defined.

The general solution:

The general solution to the ODE in question is hence given by

0

1

0

25

k

kk

k

kk xbBxxaxAy

Where A andB are parameters of the ODE on the deleted neighborhood x0 of the origin

0x

Exercises:

1. Find the Frobenius type series of the ODE 0'2'' xyyxy near the origin. Give the generalsolution and discuss its convergence.

Find the general solution to the ODE 0'''1 yxyyxx near the origin and determine the

values of x for which each of the two independent solutions 1y and 2y and the general solution

21 ByAyy is valid.

Hint: Having evaluated the indicial roots 21 and cc and the recurrence relationship for ka , using

the algebraically greater indicial root 1c , obtain the 1st

solution 1y to the ODE of the general

form 0''' yxryxqyxp The other solution can be obtained using the method ofvariation of parameters, specifically from the expression

dxy

eyy

dxxp

xq

2

1

12)(

And the general solution

dxy

eByAyy

dxxp

xq

2

1

11)(

(25 marks)

Date post:	06-Apr-2018
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Series Solutions And Frobenius

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