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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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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
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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
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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;.
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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
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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
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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:
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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)