Chap6 Series Bessel Legendre

NCCU Wireless Comm. Lab.1

Chapter 6Series Solutions of Linear Equations


Outline Using power series to solve a differential equation. First, we

should decide the point we choose to be the expanding point that is ordinary or not.

If the point is not an ordinary point, decide it a regular or irregular singular point, then use the Frobenius’ series to solve the problem.

Introduce the Bessel equation and the Legendre’s equation.


Introduction In applications, higher order linear equations with variable

coefficients are just as important as, if not more important than, differential equations with constant coefficient.

Considering a equation it does not possess elementary solutions. But we can find two linear independent solutions of

by using the series expansion.

0,y xy

0y xy


6.1 Solutions About Ordinary Points In section 4.7, without understanding that the most higher-order

ordinary equations with variable coefficients cannot be solved in terms of elementary functions.

The usual strategy for solving differential equations of this sort is to assume a solution in the form of an infinite series and proceed in a manner similar to the method of undetermined coefficients.


6.1.1 Review of Power Series Definition

A power series in is an infinite series of the form

Such a series is also said to be a power series centered at a.

For example, the power series is centered at a =1.

ConvergenceA power series is convergent at a specified value of x if its

sequence of partial sums converges- that is,

If the limit does not exist at x, the series is said to be divergent.

x a2

0 1 20

( ) ( ) ( )nn

n

c c x a c x a c x a

0

( 1)nn

n

c x

0

( )nn

n

c x a

( )NS x

0

( ) constant.lim limN

nN n

N N n

S x c x a


Interval of ConvergenceEvery power series has an interval of convergence. The interval of

convergence is the set of all real number x for which the series converges.

Note: We will use the ratio test to see the series is convergence or divergence for

Radius of ConvergenceAs we mentioned that the R is assigned to be an interval boundary to check

the series for its convergence property and R is also called the radius of convergence.

What’s the meaning for the value R?

It means a distance from the point x to the nearest singular point.(see in Theorem 6.1) Bringing a concept, the singular point possess between convergent and divergent region.

,

.

x x the region of R where R will be defined immediately

and R is the radius of convergence


If then a power series converges for and diverges for

For example, if the series converges for x=a or for all x, then R is equal to 0 or

Recall that is equivalent to

Note: A power series may or may not converge at the endpoints a-R or a+R of this interval.

Absolute ConvergenceWithin its interval of convergence a power series converges absolutely.

.

x a R .a R x a R

0,R 0

( )nn

n

c x a

x a R .x a R

0

( ) , .nn

n

c x a converges where a R x a R


Ratio testConvergence of a power series can often be determined by

the ratio test.

Suppose that

If L<1 the series converges absolutely, if L>1 the series diverges, and if L=1 the test is inconclusive.

0

( )nn

n

c x a

11 1

0 ,

( ) .( )lim lim

n

nn n

nn nn n

c n and that

c x a cx a Lc x a c


Example: A power series the ratio test gives

The series converges absolutely for (L<1), we get

The series diverges for L>1, that is

Test for the convergence of the boundary for x=1 or 9.

0

( 5) ,4 ( 1)

n

nn

xn

1

1

( 5)( 1) 14 ( 2) 5 5 ( );

( 5) 4( 2) 44 ( 1)

lim lim

n

n

nn n

n

xnn x x L

x nn

1 5 14

x 1 9.x

9 1.x or x

0 0

1 1

( 4) ( 1) 1, .4 ( 1) ( 1)

(4) 1 9, .4 ( 1) ( 1) int [1,9).

n n

nn n

n

nn n

for x it will convergen n

for x it will divergen n

So the erval of convergence of the series is


A Power Series Defines a FunctionA power series defines a function whose domain is the interval of convergence of the series. If the radius of the convergence is R>0, then f is continuous, differential, and integrable on the interval Thus, and can be found by term-by-term

differentiation and integration.

If is a power series in x, then the first two derivatives are

It will be useful to substitute into the differential equation.

0

( ) ( )nn

n

f x c x a

( , ).a R a R ( )f x ( )f x dx

0

nn

n

y c x

2

0 1 20

1 11 2

1

2 22 3

2

2

2 3 2 ( 1) ( 1) .

n nn n

n

n nn n

n

n nn n

n

Since y c c x c x c x c x

So y c c x nc x nc x

y c c x n n c x n n c x

, , y y and y 2nd


Identity Property

Analytic at a PointA function f is analytic at a point a if it can be represented by a power

series in x-a with a positive or infinite radius of convergence.For example,

0

( ) 0, 0 interval ,

0 .

nn

n

n

If c x a R for all numbers x in the of convergence

then c n

2

0

2 2 2

0

3 2 2 1 2 2 1

0

1 2! ! !

( 1)cos 1 ( 1)2! 2 ! 2 !

( 1) ( 1)sin . 3! (2 1)! (2 1)!

n nx

n

n n nn

n

n n n n

n

x x xe xn n

x x xxn n

x x xx x for xn n

These Taylor series centere

0, ,

, cos , sin 0.x

d at called Maclaurin series

show that e x and x are analytic at x

, , , integrable interval .

For analytic the function is continuous differentialand in the of convergence


Arithmetic of Power SeriesPower series can be combined through the operations of addition,

multiplication, and division.

Example:

2 2 1 2

0 0

3 2 2 1 2 2

3 5 7

Using expansion sin cos

( 1)sin cos(2 1)! 2 !

( 1)( ) (1 )3! (2 1)! 2! 2 !

, .3 20 252

n n m

n m

n n n

the series to describe the x x

x xx xn m

x x x xxn n

x x xx x x

It is obvious that

sin cos interval.the power series x x converges on the same


Shifting the Summation Index It is important to combine two or more summations with different index,

so it may need to shift the summation index. You may see the rule by the following example.

Example 1:Adding Two Power Series 2 1

2 0

0

( 1)

( point 0,

.)

,

n nn n

n n

nn

n

Write n n c x c x

If y c x is a Taylor series at centered the subject

is liking to combine the y xy

From the original subject

1 1

0 1

3 0

start from x start from x

start from x start from x with n with n

2 1 0 2 12

2 0 3 0

( 1) 2 ( 1)

n n n nn n n n

n n n n

n n c x c x c x n n c x c x


2 1

3 0

( 1)

- 3

( 1)

n nn n

n n

nn

Considering the term of n n c x c x

let k n let k n

n n c x

2 1 1 13

3 0 0 0

13

0

2 1 12 3

2 0 0

( 3)( 2)

[( 3)( 2) ]

( 1) 2 [( 3)( 2) ]

: ,

n k kn k k

n n k k

kk k

k

n n kn n k k

n n k

c x k k c x c x

k k c c x

n n c x c x c k k c c x

Note For the shifting summation approachSt

1. minimum , . 2. a

ep You may figure out the order of each component then take the termsout from the summation

Step Substitute new index to the old one for preparing to combin . 3. .

e the summationStep Create a new summation for the original subject


6.1.2 Power Series Solutions Definition 6.1 Ordinary and Singular Points

A point is said to be an ordinary point of the differential equation if both P(x) and Q(x) in the standard form are analytic at A point that is not an ordinary

point is said to be a singular point of the equation.

Considering the differential equation and

–

0x2 1 0( ) ( ) ( ) 0,a x y a x y a x y

( ) ( ) 0y P x y Q x y 0.x

1 02

2 2

( ) ( )For analytic, it means that ( ) 0 in 0.( ) ( )

a x a xa x y y ya x a x

cos 0xy x y e y ln 0.xy x y e y

cos 0 ln 0

cos 0 ln 0

( ) cos , ( )

x x

x x

x

y x y e y y x y e y

For y x y e y For y x y e y

P x x Q x e

( ) ln , ( )0 point, 0 singular point,

( ) ( ) . ( )

xP x x Q x ex is the ordinary x is abecause P x and Q x are analytic because P x is not analyti

.c


Polynomial Coefficients

2 1 0

, points denominator .

if a ( ), a ( ), ,and a ( ) are polynomials with no common factors,

then

A polynomial is analytic at any value x and a rational functionis analytic except at where its is zeroThus x x x

01

2 2

2

0 2 1 0 2 0

0

a ( )a ( ) both rational function P(x)= and Q(x)= are analytic except ( ) ( )

where a ( ) 0.

point ( ) ( ) ( ) 0, ( ) 0, singular point

xxa x a x

x

For x x is an ordinary of a x y a x y a x y if a xwhereas x x is a

2 1 0 2 0 ( ) ( ) ( ) 0, ( ) 0.of a x y a x y a x y if a x


2 2

2

: ( 1)( 3) ( 1) ( 1)( 3) 0 singular points.

( 1) ( 1)(

Example If the equation is x x y x y x x yConsider for the ordinary and

xThe original equation can be rewritten to yx x

2 2 2

2 22 2

( 1)( 3) 0.3) ( 1)( 3)

1. ( ) ( 1)( 3) , ( ) 0 point , 1, -3.

2.

x xy yx x

For a x x x let a x to calculate the possible that make the functionnot analytic so x

P

2 2

2 2

( 1) 1( )( 1)( 3) ( 1)( 3)( 1)( 3) 1 ( )

( 1)( 3) 1 3. 1 2. 1, -3 singular point .

: 1,-3

xxx x x xx xQ x

x x xCombining andSo x are the for this function

Note x

singular point 6.2.

are also called the regularin Definition


Theorem 6.1 Existence of Power Series Solutions If is an ordinary point of the differential equation we can always find two linearly independent

solutions in the form of a power series centered at -that is, A series solution converges at least on some interval defined by whereas R is the distance from to the closest singular point.

A solution of the form is said to be a solution about the ordinary point

0x x

2 1 0( ) ( ) ( ) 0,a x y a x y a x y

0x

00

( ) .nn

n

y c x x

0 ,x x R 0x

00

( )nn

n

y c x x

0.x


Example 2. Power Series SolutionsSolve

0.y xy

-2

0 2

singular points, 6.1 0, .

( -1) into

n nn n

n n

There are no finite Theorem guaranteestwo power series solutions centered at convergent for x

Substituting y c x and y c n n x the

differential equatio

-2 1 0 -2 12

2 0 3 0

0 1 12 3

0 0

2

-3 sec

.

( -1) 2 ( -1)

2 ( 3)( 2)

2 ( 3)( 2)

n n n nn n n n

n n n n

k kk k

k k

let k nfor the ond term

and let k nfor the third term

n

y xy c n n x c x c x c n n x c x

c x c k k x c x

c k k

13

00k

k kk

c c x


12 3

0

2 3

2 3

, 2 ( 3)( 2) 0.

2 0 ( 3)( 2) 0, 0,1,2, .

0 , 0,1,2 .( 3)( 2)

into

kk k

k

k k

kk

x x satisfies c k k c c x

We should let c and k k c c where kcSo we get c and c k

k kSubstitute the value of k the recur

0 03

1 14

25

3 0 06

4 1 17

5 28

,

0, 2 3 6

1, 4 3 12

02, 05 4 20

13, 6 5 30 6 180

14, 7 6 42 12 504

15, 08 7 56 20

sive termc ck c

c ck c

ck c

c c ck c

c c ck c

c ck c


13

0

3 6 4 7

0 1

0

3 6

0

( 3)( 2)

16 180 12 504

into .

, 16 180

kk k

k

nn

n

So we reconstruct the term k k c c x

x x x xto be the form of c c x

We substitue it the origin series y c x

x xSo we get y c

4 7

1

0 1 1 2

3 21

1

3 12

1

.12 504

( ) ( ),

( 1) ( ) 12 3 (3 1)3

( 1) ( )3 4 (3 )(3 1)

nn

n

nn

n

x xc x

Another description of y c y x c y x

we get y x xn n

y x x xn n


Example 3. Power Series SolutionSolve 2( 1) 0.x y xy y

singular points , 0 1, 1 distance

.

There shows at x i and so the power series solutioncentered at will converge at least for x where is the

in the complex plane from the origin to x i

Substituing y c

1

10

-2

2

, ,

( -1) into .

nn

n

nn

n

nn

n

y nc x andx

y c n n x the differential equation


2

1

0 12 0 3 1 1

2

-2

2 2 0

-2

2 4 2

( 1)

(2 ) (6 )

( -1) ( -1)

( -1) ( -1)

nn

n

nn

n

le

let k nfor another summations

n n nn n n

n n n

n n nn n n

n n n

x y xy y nc x

c c x c c c x nc x

c n n x c n n x c x

c n n x c n n x c x

-2

12 0 3

2

12 0 3 2

22 2 2

2

2 6 ( 2)

2 6 ( -1) 1 ( 2)( 1) 0

0 , 1.

( -1) ( 1)t k n

for the fourth termk

kk

k k

k k kk k k

k k k

k

k

c c c x k k kc x

c c c x k k k c k k c

For the equality of the equation to x x

We have

c k x c k x c x

x

2 0 3 2

0 2 3 2

2 0, 6 0, ( -1) 1 ( 2)( 1) 0, k=2,3,4, .

( 1)( 1) -12 , 0, - - , 2.3.4. .( 2)( 1) 2

k k

k k k

c c c and k k k c k k c

k k kc c c c c c kk k k


4 2 0 0

5 3

6 4 0 0

7

2 40 1 0 1

0

, 1 1 1 12, - - -4 4 2 823, - 053 1 1 14, - - 6 2 8 16

5, 0

1 1 (1 ) ( )2 8

nn

n

By iteration of k we get

k c c c c

k c c

k c c c c

k c

So we rewritten the power series y c x to y c x x c x c y x

1 2

2 1 21

2

2

( ).

1 1 3 5 (2 3) ( ) 1 ( 1)2 2 !

( ) , 1.

n nn

n

c y x

nThus y x x xn

y x x where x


Example 4. Three-Term Recurrence Relation If we seek a power series solution for the differential equation

0

nn

n

y c x

(1 ) 0.y x y

2 1

2 0 0

0 2 12 0

3 1 0

(1 ) ( 1)

(2 ) ( 1)

n n nn n n

n n n

n n nn n n

n n n

y x y c n n x c x c x

c c x c n n x c x c x

Let k

2 0 2 11

12 0 2

- 2 1

2 ( 2)( 1) 0

, .1 1.2.3. .2 ( 2)( 1)

kk k k

k

k kk

n k n k n

c c c k k c c x

For the equality x x

c cWe get c c and c kk k


2 0

1 0 013

0 12 1 1

4 0

3 2 015

0

2 30

into 1 ,2

1, 3 2 6 6

1122,

4 3 12 12 24

3, 5 4 120 30

1 1 1(12 6 24

nn

n

Iterating the value k the recurrence relation

with the condition c c

c c cck c

c cc c ck c c

c c cck c

y c x

c x x

4 5 3 4 51

1 1 1 1) ( ).30 6 12 120

x x c x x x x


Nonpolynomial CoefficientsThe next example illustrates how to find a power series solution about the

ordinary point of a differential equation when its coefficients are not polynomials.

Example 5. ODE with Nonpolynomial CoefficientsSolve

0 0x

(cos ) 0.y x y

2

0 2

2

0

0 point , cos 0.

, ( -1) ,

( 1)cos into .2 !

n nn n

n n

n n

n

We see that x is an ordinary of the equationbecause we know x is analytic at x

Substituting y c x y n n c x and

xx the equationn


22

2 0 0

2 3 2 4 6 22 3 4 5 0 1 2

2 32 0 3 1 4 2 0 5 3 1

( 1)(cos ) ( -1)2 !1 1 12 6 12 20 (1 )( )2 24 720

1 12 (6 ) (12 ) (20 ) 0.2 2

, .

k kn n

n nn k n

xy x y n n c x c xk

c c x c x c x x x x c c x c x

c c c c x c c c x c c c x

For the equality x x

2 0 3 1 4 2 0 5 3 1

2 0 3 1 4 0 5 1

2 4 3 50 1

0

1 1 2 0, 6 0, 12 0, 20 0, .2 2

1 1 1 1 , , , , .2 6 12 30

1 1 1 1(1 ) ( ).2 12 6 30

nn

n

We get c c c c c c c c c c

So c c c c c c c c

y c x c x x c x x x


Solution CurvesThe approximate graph of a power series solution can be obtained in several ways. We can always resort to graphing the terms in

the sequence of partial sums of the series—in other words, the graphs of the polynomials

For a large value of N, By this, will give us some information about the behavior of y(x) near the ordinary point.

0

nn

n

y c x

0

.N

nN n

n

S x c x

arg arg 0

lim lim .N

nN nN is l e N is l e n

S x c x y

0

Nn

N nn

S x c x


RemarksEven though we can generate as many terms as desired in series solution either through the use of a recurrence relation or, as in Example 4, by multiplication, it may not be possible to deduce any general term for the coefficients We may have to settle, as we did in Example

4 and 5, for just writing out the first few terms of the series.

0

nn

n

y c x

.nc


6.2 Solutions About Singular Points The two differential equations are similar

only in that they are both examples of simple linear second-order equations with variable coefficients.

We saw in the preceding section that since x=0 is an ordinary point of the first equation, there is no problem in finding two linear independent power series solutions centered at that point.

In the contrast, because x=0 is a singular point(which is defined in Definition 6.1) of the second ODE, finding two infinite series solutions of the equation about that point becomes a more difficult task.

20 and x 0y xy y y


Regular and Irregular Singular Points

Definition 6.2 Regular and Irregular Singular Points

0

2 1 0

singular point ( ) ( ) ( ) 0

.

A at x x of a linear differential equationa x y a x y a x y

is further classified as either regular or irregular The classification againdepends on the

1 0

2 2

standard ( ) ( ) 0.

( ) ( ): ( ) ( ) .( ) ( )

functions P and Q in the formy P x y Q x y

a x a xNote Where P x and Q xa x a x

0

22 1 0 0 0

0

singular point singular point

( ) ( ) ( ) 0, ( ) ( ) ( - ) ( ) ( ) ( - ) . sing

A x is said to be a regular of the differential equation

a x y a x y a x y if the function p x P x x x and q x Q x x xare both analytic at x A

ular point

singular point .that is not regular is said to be an

irregular of the equation


Example 1. Classification of Singular Points

0

20

2 2 2 20 0 0 0 0

singular point ( ) ( ) 0,

( - ) ,

( - ) ( - ) ( ) ( - ) ( ) 0 ( - ) ( - ) ( ) ( ) 0.

If x x is a regular of the differential equation y P x y Q x

then multiplying x x in both side

x x y x x P x y x x Q x y x x y x x p x y q x y

For reg

2

0 0

0

20 0 0

singular point ( ) ( ) ( - ) ( ) ( ) ( - ) .

( - ) ( - ) ( ) ( ) 0, point .

ular p x P x x x and q x Q x x xare both analytic at x

So x x y x x p x y q x y where p and q are analytic at the x x

2 2

2

2 -2 singular points

( - 4) 3( 2) 5 0. ( ) ( ) 0,

3( 2) ( )( - 4

It should be clear that x and x are of

x y x y yComparing with the general form of y P x y Q x y

xit is clearly that P xx

2 2 2 2

20 0

20 0

3 5 ( ) .) ( 2) ( 2) ( - 4)

( - ) ( - ) ( ) ( ) 0,

( ) ( ) ( - ) ( ) ( ) ( - ) .

and Q xx x x

By the equivalent form x x y x x p x y q x y

where p x P x x x and q x Q x x x


22 2

2 -2 singular point.3 5 2, ( ) ( - 2) ( ) ( ) ( - 2) ( ) .

( 2) ( 2) ( ) ( ) 2, 2 singular po

Checking x and x is the regular or irregular

First if x p x x P x and q x x Q xx x

From p x and q x are analytic at x so x is a regular

22

int.

3 5 -2, ( ) ( 2) ( ) ( ) ( 2) ( ) .( 2)( 2) ( 2)

( ) -2, ( ) -2. , -2 singular poi

Second if x p x x P x and q x x Q xx x x

Although q x is analytic at x p x is not analytic at xCombining these conditions x is an irregular

nt.


Theorem 6.2 Frobenius’ Theorem If is a regular singular point of the differential equation then there exists at least one solution of the form

where the number r is a constant to be determined. The series will converge at least on some interval

If we consider a differential equation that has a regular singular point, then we can substitute into the DE like the

approach

we did before by using the power series to solve with the ordinary point.

0x x

2 1 0( ) ( ) ( ) 0,a x y a x y a x y

0 0 00 0

( ) ( ) ( ) ,r n n rn n

n n

y x x c x x c x x

00 .x x R

00

( )n rn

n

y c x x


Example 2. Two series solutions

0

10 1

10 1

0 singular point 3 - 0,

.

,

(

n rn

n

r r n rn

r

Because x is a regular of the differential equationxy y y

we try to find a solution of the form y c x

From y c x c x c x

y c rx c

1 1

0

2 1 2 20 1

0

1 ) ( ) ( )

( 1) (1 ) ( )( 1) ( )( 1) ,

, , into .

3 - 3 ( )( 1

r n r n rn n

n

r r n r n rn n

n

n

r x c n r x n r c x

y c r r x c r rx c n r n r x c n r n r x

now we substitute y y and y the equation

xy y y c n r n r

1 1

0 0 0

1 1 10

1 1 0

-1 2 3

10

4

) ( )

3 ( 1) 3 ( )( 1) ( )

(3 - 2)nd rd

th

n r n r n rn n

n n n

r n r n r n rn n n

n n n

let k nfor and term

r

let k nfor term

x n r c x c x

r r r c x c n r n r x n r c x c x

r r c x

10

( 1)(3 3 1) 0k rk k

k

k r k r c c x


0 1

0

(3 - 2) 0 ( 1)(3 3 1) 0, 0,1, 2, . 0, , ' .

, (3 - 2) 0

k kThis implies r r c and k r k r c c kIf we let c we will see the all terms in recursive term that will equal to zero it s a trivial solution

So we choose r r a

1

1 1

2 1

, 0,1, 2, .( 1)(3 3 1)

0, , 0,1, 2, (1) ( 1)(3 1)

2 , , 0,1,2, . (2)3 ( 1)(3 5)

(1)

kk

kk

kk

cnd c kk r k r

cIf r c kk k

cr c kk k

From we get

01 0 1

0 02 2

03

(2)

5

8 80

168

From we getcc c c

c cc c

cc

03

0 0

2640

! 1 4 (3 2) ! 5n n

cc

c cc cn n n

.

8 11 (3 2)n


0

01

1

23

21

, . - .

1( ) 1! 1 4 (3 2)

1( ) 1! 5 8 11 (3 2)

n

n

n

n

From each set contains the same coefficient c so we omit the termSo we find two independent solutions on the entired x axis

y x x xn n

y x x xn n

1 2

.

( ) ( )

By the ratio test it can be demonstrated that both y x and y x convergefor all finite values of x

1 1 2 2

, .

, superposition , ( ) ( ) . interval ,

that is x

Hence by the principle y c y x c y x is another solutionof this differential equation On any not containing the origin this linearcombinition represents t

.he general solution of the differential equation


Indicial Equation Equation is called the indicial equation of the previous example,

and the value are called the indicial roots, or exponents, of the singularity x=0.

In general, after substituting into the given differential equation and simplifying, the indicial equation is a quadratic

equation in r that results from equating the total coefficient of the lowest power of x to zero.

We solve for the two values of r and substitute these value into a recurrence relation such as

By Theorem 6.2, there is at least one solution of the assumed series form that can be found.

(3 2) 0r r 1 2

20 3

r and r

0

n rn

n

y c x

1 .( 1)(3 3 1)

kk

cck r k r


Example for Indicial Equation

2

2 2

2

0

( ) ( ) 0, ,

( ) ( ) 0.

, ( ) ( )

( ) ( ) .

, ( )

From y P x y Q x y if we multiply x on both side

we get x y x xP x y x Q x y

As the previous concept we know that the p x xP x and

q x x Q x are both analytic

So we let p x a a

2 21 2 0 1 2

0

2 20 1 2 0 1 2

0 0 0

( ) .

, ( ), ( ) into ,

( )( 1) ( ) 0.

n rn

n

n r n r n rn n n

n n n

x a x and q x b b x b x

Substituting y c x p x and q x the differential equation

n r n r c x n r a a x a x c x b b x b x c x

As w

0 0

, , 0, ( -1) 0.

re find the indicial equation we will take from the lowest order of n like xso we let n we get the indicial equation r r ra b


Three Cases Case I:

Case II:

1 2

1

2 2 1 0

0

integer, ( ) ( ) ( ) ( ) ( ) 0

. n rn

n

If r and r are ristinct and do not differ by anthen there exist two linearly independent solutions y xand y x of equation a x y a x y a x y of

the form y c x This is

2.the case of Example

1

1 2

2 1 0

1 00

, integer, ( ) ( ) ( ) 0

( ) , 0

n rn

n

If r r N where N is a positive then there exist two linearlyindependent solutions of equation a x y a x y a x y of the form

y x c x c

2

2

2 1 00

2 10

2

( ) ( ) ln , 0,

constant .

: ( ) ( ) ln ?

( )

n rn

n

n rn

n

y x Cy x x b x b

where C is a that could be zero

Note Why y x is equal to Cy x x b x and how to obtain it

The solution y x can be obtained by y

2 1 21

exp( ( ) )( ) ( ) dx .

( )

P x dxx y x

y x


1 1 1 2 1 1 1

2 2 2 1 2 2 2

1 2 2 1 1 2 2 1

1 2 2 1

( ) ( ) 0,

( ) ( ) 0 ( ) ( ) 0 (1)

( ) ( ) 0 ( ) ( ) 0 (2)

(2) - (1),

( ) 0

( ) (

From y P x y Q x y we get

y P x y Q x y y y P x y Q x y

y P x y Q x y y y P x y Q x y

By

y y y y P x y y y y

y y y y P

1 2 2 1

1 2 2 1

1 2 2 12 21 1

22

1 1

22

1 1

2 1 21

) 0 " "

exp( ( ) )

exp( ( ) )

exp( ( ) )( )

exp( ( ) ) , 1,

exp( ( ) )

x y y y y First order linear differential equation

y y y y c P x dx

c P x dxy y y yy y

c P x dxydy y

c P x dxy let cy y

P x dxso y y

y

2 . ( ) using .So you can calculate y x by the formula


Case III: If then there always exists two linearly independent solutions of of the form

1 2r r

2 1 0( ) ( ) ( ) 0a x y a x y a x y

1

2

1 00

2 11

2 2 1 21

2

( ) , 0

( ) ( ) ln ,

exp( ( ) ) ( ) ( ) ( ) dx .

( )

: ( )

n rn

n

n rn

n

y x c x c

y x Cy x x b x

P x dxThe solution y x can also be obtained by y x y x

y x

Note We will see how the formula of y x works by the following exa

.mple


Example 5. Find the general solution of 0.xy y

1 2 1

2 1 21

2 3 41

1

2 1

( ) 4. ( ) ( ) into

exp( ( ) )( ) ( ) .

( )1 1 1( )2 12 144

: 4. ( ).

exp( 0( ) ( )

You can use y x of Example and find y x by substituting y x

P x dxy x y x dx

y x

y x x x x x

Note There is wrong in Example for y x

dy x y x

22 3 4

1 1 23 4 5

21

22 1 1

)

1 1 12 12 144

1 1 7 19( ) ( )5 7 12 72-

12 121 7 19( ) ln

12 1441 7 19( ) ( ) ln .

12 144 interval (0, )

xdx

x x x x

dxy x y x x dxx xx x x x

y x x x xx

y x y x x y x xx

On the

1 1 2 2, ( ) ( ).the general solution is y c y x c y x


Remark When the difference of indicial roots is a positive integer it sometimes pays to iterate the recurrence relation using

the smaller root first.

Since r is the root of a quadratic equation, it could be complex. Here we do not concern this case.

If x=0 is an irregular singular point, we may not be able to find any solution of the form

1 2r r

1 2( ),r r

2r

0

.n rn

n

y c x


6.3 Two Special Equations The two differential equations

occur frequently in advanced studies in applied mathematics, physics, and engineering.

They are called Bessel’s equation and Legendre’s equation, respectively.

In solving (1) we shall assume whereas in (2) we shall consider only the case when n is a nonnegative integer.

2 2 2

2

( ) 0 (1)

(1- ) - 2 ( 1) 0 (2)

x y xy x y

and x y xy n n y

0,


Solution of Bessel’s Equation Substituting into the Bessel’s equation,

0

n rn

n

y c x

2 2

0 0 0 0

2 2 2 10 1

2 2 22

0

22 2 21 2 2

( )( 1) ( )

( ) ( 1) ( 1)

2 0

0,( , - ) 2 0

n r n r n r n rn n n n

n n n n

r r

k rk k

k

k k

n r n r c x n r c x c x c x

c r r r x c r r r x

k r r c c x

Taking r r r and k r r c c

1 1 2

1

22 2

,

, 0, , 0,1,2, .( 2)( 2 2 )

0, 0, 1,3,5,7, .

2 2 , 1,2,3, , .2 ( )

kk

k

nn

cwe get r r c and c kk k

If c it is easy to say c kcLet k n n so c

n n


02 2

024 2 4

046 2 6

02 2

2 1 (1 )

2 2 (2 ) 2 1 2 (1 )(2 )

2 4 (4 ) 2 1 2 3 (1 )(2 )(3 )

( 1) , 1, 2,3, .2 !(1 )(2 ) ( )

n

n n

cThus c

ccc

ccc

cc nn n

First we defin

02 2

(1 ) ( ),(1 1) (1 ) (1 ) (1 ) ( ) (1 ) ( 1) ( 1)

(1 )!, (1 ) denominator,

( 1) (1 ) , 2 ! (1 )(1 )(2 ) ( )

n

n n

e the gamma function

So we multiply at the numerator and

cc nn n

0 02 2 2

1, 2,3, .

(1 )(1 )(2 ) ( ) !(1 )(2 ) ( ) (1 ),

( 1) (1 ) ( 1) (1 ) .2 ! !(1 )(2 ) ( ) 2 ! (1 )

n n

n n n

From n n n

c ccn n n n


1 2 21 0 2 2

0

220 0

20 0

2

1 0 1 1 10

,

( 1) (1 ) ( 1) (1 )22 ! (1 ) ! (1 ) 2

( 1) (1 )2 , ( ).! (1 ) 2

r nn

n

nn nn

nn n

nn

n

So y x c c x x c x

c c xx xn n n n

xLet a c then y a a J xn n

We denoted th

2

0

2

2

2 2 20

2

0

( 1) ( ) .! (1 ) 2

into ,

( 1) ( ).! (1 ) 2

( 1) ( )! (1 ) 2

nn

n

nn

n

nn

n

xat J xn n

If we take r the equation we will get

xy a a J xn n

xWe denoted that J xn n

Thus y y

1 2 1 2

-

( ) ( ) ( ) ( ).

: ( ) ( ) - , .

x y x a J x a J x

Note J x and J x are called Bessel functions of the first kindof the order and repectively


Example 1. General Solution: Not an Integer

0

, 0,1,2, .

( ) ( ) ( 1) ( ) ( ) ( ) ( 1) ( )0, 0

( ) (0) ( ) lim ( )1, 1

: ( ) ( )

m mm m m m

m mx

m m

First we talk about the properties of Bessel functionsof order m m

i J x J x ii J x J xm

iii J iv Y xm

Note J x and J x are li

, 0,1,2, .near dependent where m

12

2 2

1 2 12

1 1( ) 0 (0, ), .4 2

( ) ( ).

Put out atteneion on this differential equation

x y xy x y on we can see that

Then the general solution is y c J x c J x


Bessel Functions of the Second Kinds -

2 2 21 2 1 2

cos ( ) - ( ) integer, ( ) ,sin

( ) ( )

( - ) 0, ( ) ( ) ( ) ( ).

integer,

J x J xIf by linear mapping Y x

J x and Y x can be the linearly independent solutions of

x y xy x y so y c J x c J x or y c J x c Y x

If

2 2 21 2

' ' lim ( ) .

( ) lim ( ), ( ) ( )

( - ) 0, ( ) ( ).:

m

m m mm

according to L Hospital s rule that Y x exists

By linear mapping Y x Y x Y x and J x are the linearly

independent solutions of x y xy x m y so y c J x c Y xNote W

1 2

2 2 2

( ) ( ), ( ) ( ) .

, ( ) 0 (0, ) model

e drop the solution of y c J x c J xbecause J x and J x are linear dependent from the previous page

So for any value of the general solution of x y xy x yon can be

1 2ed ( ) ( ).

: ( ) second .

by y c J x c Y x

Note Y x is called the Bessel function of the kind of order


Following plot will show us the first and second kind of Bessel function.


Example 2. General Solution: an Integer

Parametric Bessel Equation

2 2

2

1 3 2 3

( - 9) 0 (0, ),

9 3, ( ) ( ).

Considering x y xy x y on

we see and so the general solution isy c J x c Y x

2 2 2 2

1 2

1 2

( - ) 0, ( ) ( )

( ) ( ), integer.

: .

For the parametic equation x y xy x ythe general solution y c J x c Y x ory c J x c J x where

Note We will prove it at the next page


2 2 2 2

2 22

2 2

22 2 2 2 2

2

22 2 2

2

1

( - ) 0, let z= x,dy dythen , y = ,dx

dy d y .

into the DE,

( - ) 0

( - ) 0

,

From x y xy x ydy dzdz dx dz

d y d dzdx dz dz dx dzSubstituting we get

d y dyx x x ydz dz

d y dyz z z ydz dz

So y c J

2 1 2

1 2 1 2

( ) ( ) ( ) ( ). ( ) ( ) ( ) ( ).

z c Y z or y c J z c J zThen y c J x c Y x or y c J x c J x


Example 4. Derivation Using the Series Definition1

2

0

2

0

2 2

0 0

( ) ( ) ( ).

( 1) ( ) ,! (1 ) 2

( 1) (2 )( )! (1 ) 2

( 1) ( 1)2! (1 ) 2 ! (1 ) 2

nn

n

nn

n

n nn n

n n

Derive the formula xJ x J x xJ x

xFrom J xn n

n xxJ xn n

x n xn n n n

2 1

1

2 1

10

( 1)( ) , -1( 1)! (1 ) 2

( 1) ( 1)( ) ( ) ( ).! (2 ) 2

nn

n

kk

k

n xJ x x let k nn n

k xJ x x J x xJ xk k


Spherical Bessel Functions

1

1

-1

- - 1 - - -1

4. ( ) ( ) ( ),

, ( ) ( ) ( ).

( ) ( ) ( ), .

( ) ( ) ( ) ( )

From the example xJ x J x xJ x we divide

x at both side J x J x J xx

J x J x J x multiply x at both sidex

dx J x x J x x J x x J x x Jdx

1

0 1 0 1

( ).

, ( ) ( ) ( ).

x

So J x J x and Y Y x

integer, , 1 3 5, , , , ( )2 2 2

expressed sin ,cos , .

When the order is half an odd that is

the Bessel function of first kind J x

can be in terms of the elementary functions xx and powers of x Such Bessel

.

functions are calledSpherical Bessel functions


Example 5. Spherical Bessel Function with 1 .2

12

122

102

3

expression ( ).

1 ( 1) , ( ) .12 2! (1 )2

1 (1 ) ( ) ( ) , 2

1 10 : (1 )2 23 3 1 3!1: (1 )2 2 2 2

1 (2: (1 )2

nn

n

Find an alternative for J x

xWith we get J xn n

By and we find

n

n

n n n

2 1

122 2 1

10 02

2 1

3 5 2 1

0

12

-1)! .2 !

( 1) 2 ( 1), ( ) .(2 -1)! 2 (2 1)!!2 !

1 1 ( 1) sin - ,3! 5! (2 1)!

2 ( ) sin .

n

nn nn

n nn

nn

n

nn

xSo J x xn x nnn

From x x x x xn

so we get J x xx


Now we bring out an extra concept that is not mentioned in textbook, that is, the Modified Bessel Equation. 2 2 2

2 2 2 2

1 2

1 2

( ) 0 .

( ) 0. ( ) ( ), 0,1,2,3, , ( ) ( ),

u u

u u

x y xy x u y is called the Modified Bessel Equation

It changes from x y xy i x u y We get the generalsolution y c J ix c Y ix where u ory c J ix c J ix where u

2 22

0 0

22

0

22

0

0,1,2, .

( 1) ( 1) ( )! (1 ) 2 ! (1 ) 2

( 1) ( ). ! (1 ) 2

( 1) ( ) ! (1 ) 2

n u n un n n u

un n

n un nu u

un

n un n

un

ix i i xAnd J ixn n u n n u

i xi i I xn n u

i xWhere I x is called then n u

-

" 1 ".

, ( ) ( ).uu u

Modified Bessel function of st kind

Following the similar approach you can get J ix i I x


2 2 2

1 2

1 2

1 2

1

( ) 0, ( ) ( ), , 0,1, 2, .

modify ( ) ( ).

( ) ( ), , 0,1, 2, . modify (

u u

u u

u u

m

From the equation x y xy x u y the solution isy c J ix c J ix where u m mWe it to y A I x A I x

y c J ix c Y ix where u m mWe it to y A I x

2) ( ),( ) ( ) ( ) , ( ) ( ).

2 sin( ) " 2 ".

limm

u uu m u

u m

u

A K xI x I xwhere K x K x K x

uK x is called the Modified Bessel function of nd kind


Example.

Solution of Legendre’s Equation

2 2

2 2 2 2

1 26 6

x ( 4 6) 0.

[(2 ) ( 6) ] 0,

(2 ) (2 ).

Solve y xy x y

x y xy i x ySo the general solutiony c I x c K x

2

0

20 2 1 3

0 point (1- ) - 2 ( 1) 0,

into ,

(1- ) - 2 ( 1) ( 1) 2 ( 1)( 2) 6

(

nn

n

Since x is the ordinary of the equation x y xy y

let y c x substitute the equation and combine the summation

x y xy y c c c c x

n

22

2)( 1) ( )( 1) 0.nn n

n

n c n n c x


0 2

1 3

2

02

13

2

, ( 1) 2 0 ( 1)( 2) 6 0 ( 2)( 1) ( )( 1) 0

( 1) 2!

( 1)( 2) 3!

( )(

n n

n

From the previous equation c cc c

n n c n n ccor c

cc

n nc

2

2 04

3 15

4 06

7

1) , n 2,3,4, .( 2)( 1)

n 2,3,4, into , ( 2)( 3) ( 2) ( 1)( 3)

4 3 4!( 3)( 4) ( 3)( 1)( 2)( 4)

5 4 5!( 4)( 5) ( 4)( 2) ( 1)( 3)( 5)

6 5 6!(

n

n

cn n

Take c we getc cc

c cc

c cc

c

5 15)( 6) ( 5)( 3)( 1)( 2)( 4)( 6)7 6 7!

c c


2 40

3 51

0 1

2 4

3

( 1) ( 2) ( 1)( 3)12! 4!

( 1)( 2) ( 3)( 1)( 2)( 4)3! 5!

( ) ( ), ( 1) ( 2) ( 1)( 3)( ) 1

2! 4!( 1)( 2) ( 3)( 1)( 2)(( )

3!

y c x x

c x x x

c U x c V x where

U x x x

V x x x

54) .5!

( ) , 0,1,2,3, " : integer."( ) ( ) 1, ., , 0,1,2,3, ,

(1

x

i If n n Note Here n means a nonnegativeU x and V x will diverge at x it violate the physic phenomenonSo n n the original equation becomes

2

0 1

- ) - 2 ( 1) 0, 0,1,2,3, , ( ) ( )n n

x y xy n n y n the solution isy c U x cV x


( ) 0, 2,4, , ( ) , ( ) infinit . 1,3,5, , ( ) infinite , ( ) .

, ( ) ( ),

n n

n n

n n

ii When n U x has a finite terms but V x has an e termsn U x has an terms but V x has a finite terms

So we recombine the U x and V x

.: infinite .

( ) , 0, 2, 4,(1)

( ) ,( ) , 1,3,5,(1)

( ) " - '

n

nn

n

n

n

put the same term togetherNote Separate the terms and the finite terms

U x nU

Define P xV x nV

where P x is the first kind of n th order Legendre s e

".

( ) (1), 0, 2,4, ( ) ,

( ) (1), 1,3,5,

( ) "second - ' ".

n nn

n n

n

quation

V x U nQ x

U x V n

where Q x is the kind of n th order Legendre s equation


2 30 1 2 3

3 50

3 51

2 23 5

2

1 1 ( ) 1, ( ) , ( ) (3 1), ( ) (5 3 ).2 2

1 1 1 1( ) ln( )3 5 2 1

1 1 1( ) ( ) ln( ) 13 5 2 1

3 1 1 1 3 3 1 1 3( ) ( ) ln( ) .2 3 5 2 4 1 2

We get P x P x x P x x and P x x x

xQ x x x xx

x xQ x x x x xx

x x xQ x x x x x xx

The general sol

expressed ( ) ( ).1, ( ) , 0.

, 1,3,5,, ( ), ( ) .

, 0,2,4,

( )

n n

n

n n

n

ution can be by y AP x BQ xx Q x will diverge B

odd function nSo the solution is y AP x where P x

even function n

We will show the plot of P x at

.next page


Example from P305. Q19

Properties of

( )nP x

( ) ( ) ( 1) ( ) ( ) (1) 1

( ) ( 1) ( 1) ( ) (0) 0, 0

( ) (0) 0, .

nn n n

nn n

n

i P x P x ii P

iii P iv P n dd

v P n even

2

1 1

1 1

(1- ) - 2 2 0, 1.

1, ( ) ( ),1( ) ( ) ln( ) 1.

2 1

For x y xy y x

We get n so y AP x BQ x wherex xP x x and Q x

x


ConclusionHere we point out two type of these special function, it will be

categorized to represent with a special solution, that we will find out in the mathematics handbook.

Chap6 Series Bessel Legendre

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Transcript of Chap6 Series Bessel Legendre