1

Chapter 4 Power Series Solutions

4.1 Introduction

0dy

ydx

This chapter will focus on the discussion of linear equation with nonconstant-coefficient equations

Solving the differential equation listed in the following with the series method

(1)

To solve by the series method, we seek a solution in the form of a power series expansion about any desired point x=x0,

00

( ) ( )nnn

y x a x x

2

20 1 2

0

( ) nn

n

y x a x a a x a x

We choose x0=0 for simplicity, then

(2a)

and

2 1 20 1 2 1 2 3( ) 2 3

dy da a x a x a a x a x

dx dx

(2b)

Putting Eq. (2a,b) into (1) gives

1 21 0 2 1 3 2( ) (2 ) (3 ) 0a a a a x a a x (4)

Because the right hand side of Eq.( 4) is 0, it implies that

1 0 1 0

2 1 2 1 0

3 2 3 2 0

0

2 0 / 2 / 2

3 0 / 3 / 6

a a a a

a a a a a

a a a a a

3

Where a0 remains arbitrary. Thus, we have

2 30

1 1( ) (1 )

2 6xy x a x x x Ce

In this chapter, we will discuss the following topics.

4.2 Power series solutions

4.3 The method of Frobenius

4.4 Legendre functions

4.5 Singular integrals

4.6 Bessel functions

( ) ( )

( ) ( )

( )

n nn n

n n nn

n nn n n n

d di a x a xdx dx

ii A B A B

iii if A x B x A B

(6)

Eq. (6) us derived based on the following operations

4

4.2 Power series solutions

Finite sum of a series

Infinite sum of a series

1 21

N

n Nk

a a a a

1 2 31

kk

a a a a

The sequence of partial sums of the series (2) as

1 1 2 1 2 3 1 2 3, , s a s a a s a a a

(1)

(2)

(3)

and so on

1

n

n kk

s a

(4)

5

If the limit of the sequence Sn exists, as n→∞, and equals some number s, then we say that the series (2) is convergent, and that it converges to s; otherwise it is divergent. That is the limit of an infinite series is defined as the limit (if that limit exists) of its sequence of partial sums.

1 1

lim limn

k k nn n

k k

a a s s

By definition , it means that to each number ε>0, no matter how small, there exists an integer N such that s-s∣ n <εfor all n>N.∣

lim nns s

Theorem 4.2.1 Cauchy Convergence Theorem

An infinite series is convergent if and only if its sequence of partial sums sn is a Cauchy sequence-that is, if to each >o there corresponds an integer N() such that s∣ m-sn <∣ for all m and n greater than N.

6

A series of the form

20 0 1 0 2 0

0

( ) ( ) ( ) ,nn

n

a x x a a x x a x x

Where the an’s are numbers called the coefficients of the series, x is a variable, and x0 is a fixed point called the centerof the series. The series may converge at some points on the x axis and diverge at others.

(6)

Theorem 4.2.2 Interval of Convergence of Power Series

The power series (6) converges at x=x0. If it converges at other points as well, then those points necessarily comprise an interval x-x∣ 0 <R centered at x∣ 0 and, possibly, one or both endpoints of that interval (Fig. 1), where R can be determined from either of the formulas if the limits in the denominators exist and are nonzero.

7

1

1 1

limlimn

n nn

nn

R or Ra aa

If the limits in Eq. (7a,b) are zero, then Eq. (6) converges for all x (i.e. for every finite x, no matter how large), and we say that “R = ∞.“ If the limits fail to exist by virtue of being infinite, then R = 0 and Eq. (6) converges only at x0.

We call x-x∣ 0 <R the interval of convergence, and R the ∣radius of convergence. If a power series converges to a function f on some interval, we say that it represents f on that interval, and we call f its sum function.

Fig. 1 Interval of convergence of power series

(7a,b)

8

Example 1 Find R of the series0

! nn x

Example 2 Find R of the series0

( 1) [( 5) / 2]n nx

Example 3 Find R of the series4

( 1)

( 1)

n

nn

x

9

Theorem 4.2.3 Manipulation of power series

(a) Termwise differentiation (or integration) permissible A power series may be differentiated (or integrated) termwise within its interval of convergence I. The series that results has the same interval of convergence I and presents the derivative (or integral) of the sum function of the original series.

n 00

-10 0 0

0 0 1

f(x)= a ( ) ,

( ) ( - ) [ ( - ) ] ( - )

n

n n nn n n

If x x within I then

d df x a x x a x x na x x

dx dx

1 1

0 00

0 0

( ) ( )( ) ( - )

1

n nb b n

n na a

b x a xf x dx a x x dx a

n

(12)

(11)

10

Two power series (about the same point xo) may be added (or subtracted or multiplied) termwise within their common interval of convergence I. The series that results has the same interval of convergence I and presents the sum (or difference or product) of their two sum functions.

(b) Termwise addition (or subtraction or multiplication) permissible.

0 00 0

If ( ) ( ) and ( ) ( ) n nn nf x a x x g x b x x

00

( ) ( ) ( )( - )nn nf x g x a b x x

Letting z = x – x0

0 0

0 n 1 n-1 n 00

( ) ( ) ( ( )( )

= (a b +a b + +a b )

n nn n

n

f x g x a z b z

z

(14)

(13)

11

to hold in some common interval of convergence, it must be true that an=bn for each n. In particular, if

00

( ) 0nn

n

a x x

in some interval, then each an must be zero.

(c) If two power series are equal, then their corresponding coefficients must be equal. That is, for

0 00 0

( ) ( )n nn n

n n

a x x b x x

12

Taylor Series: Taylor series of a given function f(x) about a chosen point x0, which we denote here as , is defined as the infinite series

0 xTS f

0

20 00 0 0

( )0

00

( ) ( ) ( ) ( ) ( )

1! 2!

( )( )

!

x

nn

f x f xTS f f x x x x x

f xx x

n

( )0

00

( )( ) ( )

!

nnf x

f x x xn

The purpose of Taylor series is to represent the given function.

Three conditions need to be met.

1. f have a Taylor series about that point.2. The series converge in some interval x-x∣ 0 <R, for R>0.∣3. The sum of the Taylor series is equal to f in the interval.

13

If a function is represented in some nonzero interval∣x-x0 <R by its Taylor series (i.e. exists, and ∣converges to f(x)), then f is said to be analytic at x0. If a function is not analytic at x0, then it is singular there and X0 is called singular point of f.

0 xTS f

4.2.2 Power series solution of differential equations

Theorem 4.2.4 Power series solution

If p and q are analytic at x0, then every solution of

( ) ( ) 0y p x y q x y

is too, and can be found in the form

00

( ) ( )nnn

y x a x x

14

Example 4

Solve by the power series method. 0y y

15

Example 5

Solve the initial-value problem

by the power series method in the interval 4 x≦ ＜∞ .

( 1) 2( 1) 0 (4) 5, (4) 0x y y x y y y

16

4.3 The Method of Frobenius

4.3.1 Singular points

( ) ( ) 0y p x y q x y

For the differential equation (1), we can find two LI solutions as power series expansions about any point xo at which both p and q are analytic. We call such a point xo an ordinary point of the equation (1)

(1)

Definition 4.3.1 Regular and Irregular Singular Point of (1)

Let x0 be a singular point of p and/or q. Xo is a regular singular point of Eq. (1) if (x-x0)p(x) and (x-x0)2q(x) are analytic at x0. Otherwise, X0 is an irregular singular point of Eq.(1) if it is not a regular singular point.

17

Example 1

Example 2

18

4.3.2 Method of Frobenius

( ) ( ) 0y p x y q x y

Consider the equation

to have a regular singular point at the origin. There is no loss of generality in assuming it to be at the origin since, if it is at x=x0≠0, we can always make a change of variableξ= x-x0 to move it to the origin in terms of the new variable ξ .

(5)

Multiplying Eq. (5) by x2 and rearranging terms as

2 2[ ( )] [ ( )] 0x y x xp x y x q x y (6)

Since x = 0 is a regular singular point, it follows that xp(x) and x2q(x) can be expanded about the origin in convergent Taylor series,

19

20 1 0 1( ) ( ) 0x y x p p x y q q x y (7)

Locally, in the neighborhood of x = 0, we can approximate (7) as

20 0 0x y p xy q y (8)

which is Cauchy-Euler equation. As such, Eq. (8) has at least one solution in the form xr, for some constant r.

Returning to Eq. (7), it is reasonable to expect that equation, likewise, to have at least one solution that behaves like xr (for some value of r) in the neighborhood of x = 0. That is, we expect it to have at least one solution of the form

20 1 2( ) ( )ry x x a a x a x (9)

20

where the power series factor is needed to account for the deviation of y(x), away from x= 0, from its asymptotic behavior y(x)~a0xr as x → 0. That is, in place of the power series expansion

0

( ) nn

n

y x a x

(10)

that is guaranteed to work when x = 0 is an ordinary point of Eq.(5), it appears that we should seek y(x) in the more general form

0 0

( ) r n n rn n

n n

y x x a x a x

(11)

If x=0 is a regular singular point.

21

Example 32 26 7 (1 ) 0x y xy x y (0 < x <∞) (12)

22

Theorem 4.3.1 Regular Singular Point; Frobenius Solution

Let x = 0 be a singular point of the differential equation

( ) ( ) 0y p x y q x y (x > 0) (37)

20 1 0 1with ( ) and ( )xp x p p x x q x q q x

Having radii of convergence R1, R2, respectively. Let r1, r2 be the roots of the indicial equation

20 0( 1) 0r p r q (38)

where r1 r≧ 2 if the roots are real.

Seeking y(x) in the form

0 0

( ) r n n rn n

n n

y x x a x a x

a0≠ 0 (39)

23

with r =r1 inevitably leads to a solution

11 0

0

( ) (a 0)r nn

n

y x x a x

(40)

where a1, a2, … are known multiples of a0, which remains arbitrary. For definiteness, we choose a0=1 in Eq. (40). The form of the second LI solution, y2(x), depends on r1 and r2 as follows:(i) r1 and r2 distinct and not differing by an integer. Then with r = r2, Eq. (39) yields

22 0

0

( ) ( 0)r nny x x b x b

(41a)

(ii) Repeated roots, r1 = r2 ≡r. Then y2(x) can be found in the form

2 11

( ) ( ) ln r nny x y x x x c x

(41b)

24

(iii) r1 - r2 equal to an integer. Then y2(x) can be found in the form

22 1

0

( ) ( ) ln r nny x y x x x d x

(41c)

Example 4

2 2( ) 0 (0 < x < )x y x x y y (49)

25

2 11

( ) ( ) ln r nny x y x x x c x

Derive in case of r1 = r2 = r

10 0

( ) r n n rn n

n n

y x x a x a x

Assume is known, let us

assume y2(x)=A(x)y1(x).

Substituting y2 into Eq.(37) gives

1 1 1 1(2 ) ( ) 0A y A y py y py qy (42)

The last term in Eq. (42) is zero, so Eq. (42) becomes

1 1(2 ) 0A y A y py

Replacing , multiplying through by dx, and dividing by y1 and gives

(43)

(44)1

1

2 0dydA

pdxA y

by dA /dxA A

26

Setting C = 1, we have

(45)

(47)

1ln 2 ln ( ) ln for 0A y p x dx c c Integrating

and 2

1ln ( )A y

p x dxC

so0

1 20 1

( )( ) ln ( )

22 21 11

( )( ) (1 2 )(1 )

pp p x dxp x dx p x p xx

rr

e e e eA x C C C

y x x a xx a x

L L

LL

0 1 12

1( ) (1 ) ( ) (ln )

r pA x C k x A x C x k x

x L L

2 1 1 1

1 1 1

11

( ) ( ) ( ) (ln ) ( )

( ) ln ( ) (1 )

( ) ln

r

r nn

y x A x y x x k x y x

y x x k x x a x

y x x x c x

L

L L(48)

27

Example 5

0 (0 < x < )xy y

28

4.4 Legendre Functions

4.4.1 Legendre Polynomials

The differential equation2(1 ) 2 0x y xy y

where l is a constant, is known as Legendre’s equation

The x interval is -1<x<1, and Eq. (1) has regular singular points at each endpoints, x =1 and -1.

Find the power series solutions about the ordinary point x = 0.

0

( ) kk

k

y x a x

(1)

(2)

Putting Eq. (2) into Eq. (1) leads to the recursion formula

29

2

( 1) ( 0,1, 2, )

( 1)( 2)k k

k ka a k

k k

(3)

2 0 3 1 4 0

2 (6 ), ,

2 6 24a a a a a a

The general solution is

2 4 60

3 51

0 1 1 2

(6 ) (20 )(6 )( ) 1

2 24 720

2 (12 )(2 )

6 120

( ) ( )

y x a x x x

a x x x

a y x a y x

(4)

30

Radii of convergence

1

1 1 = 1

( 1)limlim

( 1)( 2)n

kkn

Rk kak ka

So each series converges in -1<x<1. However, for arbitrary values of the parameter the functions y1(x) and y2(x) given in Eq. (4) grow unboundly as x→±1, as illustrated in Fig. 1 for = 1.

Fig. 1 y1(x) and y2(x) in Eq. 4 for = 1.

Nonetheless, for certain specific values of l one series of the other, in Eq. (4), will terminate and thereby be bounded on the interval since it is then a finite degree polynomial!

31

If happens to be such that

( 1)n n (6)

For any integer n = 0, 1, 2, …, then we can see from Eq. (3) that one of the two series terminates at k = n: if n is odd, then the second series contains only a finite number of terms; if n is even, then the first series contains only a finite number of terms. In either of these cases the series which reduces to a finite sum is known as a Legendre polynomial.

2 40

3 51

0 1 1 2

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

2! 4!

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

3! 5!

( ) ( )

n n n n n ny x a x x

n n n n n na x x x

a y x a y x

(6’)

Then Eq. (4) becomes

32

In fact, it can be shown that they are given explicitly by the formula:

21( ) ( 1)

2 !

nn

n n n

dP x x

n dx (8)

which result is known as Rodrigues’s formula

Fig. 2 Graphs of the first five Legendre polynomials

33

4.5 Singular Integrals; Gamma Function

4.5.1 Singular integrals

An integral is said to be singular (or improper) if one or both integration limits are infinite and/or if the integrand is unbounded on the interval; otherwise, it is regular (or proper).

For example, if

521 23 0

2 100

3 41 0

1

xx

x

eI xe dx I dx

xdx

I I xe dxx

(1)

Considering the function of ( )af x dx

Analogous to the definition

0 0

limN

n nNn n

a a

(2)

34

Define ( ) lim ( )

X

a aXI f x dx f x dx

If the limit exists, we say that is convergent; if not, it is divergent.

Comparison tests: If are series of finite positive terms, and an~Kbn as n→∞ for some finite constant K, then S1 and S2 both converge or both diverge.

1 20 0

n nS a and S b

In calculus, the p-series converges if p > 1 and diverges if p 1.≦ 1

1pn

Considering the series , we observe thatas n→∞.

41

2 3

5

n

n

4 3

2 3 2

5

n

n n

31

1 is convergent because it is a - series with 3.p p

n

35

Analogous to the p-series, we study the horizontal p-integral

1 (a>0)

paI dx

x

(5)

where p is a constant.

Xa1-p

lim ln ( 1)1 1

lim =1

lim ( 1)1-p

XaX

X x

p pa aX

X

x p

I dx dxx x x p

1-

lim ln is infinite and hence does not exist, and similar for

lim 1, whereas the latter does exist 1.

X

p

X

X

X if p if p

(6)

36

Theorem 4.5.2 Comparison Tests

1 2( ) and I ( ) , where f(x) and g(x) are a a

I f x dx g x dx

Let

positive and bounded on a x (a) If there exist constants K and X such that f(x) Kg(x)≦ (for all x X), then the convergence of I≧ 2 implies the convergence of I1, and divergence of I1 implies the divergence of I2.

(b) If f(x) ～ C g(x) as x→∞, for some finite constant C, then I1 and I2 both converge or both diverge.

Theorem 4.5.1 Horizontal p-Integral

The horizontal p-integral, converges if p > 1 and diverges if p 1.≦

1padx

x

37

Theorem 4.5.3 Absolute Convergence

If converges, then so does ,

and we say that the latter converges absolutely.

( )af x dx

( )af x dx

Example 2 Consider 22

sin

3 1

xI dx

x

Example 1 Consider 49

2 3

5

xI dx

x

38

Example 3 Consider 100 0.01

0

xI x e dx

Example 4 Consider 3 ln

dxI

x x

39

All the aforementioned cases are that the upper limit is ∞, let us consider the left endpoint x = a. If the intergrand f(x) blows up as x→ a. We define

0( ) lim ( )

b b

a aI f x dx f x dx

where ε→ 0 through positive values.

Considering vertical p-integral

(10)

(11)0

1b

pI dx

x

b

0

0 0 1-p

0

lim ln ( 1)1 1

lim =1

lim ( 1)1-p

b

b b

p p

x p

I dx dxx x x p

According to Eq. (10)

(12)

40

Theorem 4.5.4 Vertical p-Integral

The vertical p-integral, converges if p < 1 and diverges if p > 1.

Theorem 4.5.5 Comparison Test

If , where 0 < b <∞. If f(x)~K/xp as x→ 0 for

some constant K and p, and f(x) is continuous on 0 < x b, ≦

then I converges if p < 1 and diverges if p 1. ≧

0( )

bI f x dx

Example 5 Test the integral for convergence/divergence.

4

302

sin 2xdx

x

0

1 ( 0)

b

pdx b

x

41

4.5.2 Gamma function

1

0( ) x tx t e dt

The integral is singular for two reasons: first, the upper limit is ∞ and, second, the integrand blows up as t→0 if the exponent (x-1) is negative.

0 0 0

00

( ) lim ( ) lim ( ) ( )

= lim ( ) lim ( ) ( ) ( )

T T

T T

T

T

f t dt f t dt f t dt f t dt

f t dt f t dt f t dt f t dt

1 2

0( ) ( 1)x t x tx t e x t e dt

(15)

(16)( ) ( 1) ( 1)x x x

42

( 1) ( ) ( 1) ( 1)

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

n n x n n n

n n n n

since

0(1) 1te dt

Eq. (18) becomes

( 1) !n n

(18)

(19)

(20)

1( )2

and

(21)

For x < 0, let us define

( 1)( ) for 0

xx x

x

(22)

Example 6 Evaluate2

3

0( ) tx t e dt

Fig. Gamma function

43

4.6 Bessel Functions

The differential equation

2 2 2( ) 0x y xy x y (1)

where νis a nonnegative real number, is known as Bessel’s equation of order ν. The equation appears in a wideRange of application such as steady and unsteady diffusion in cylindrical region.

Putting the equation into a standard form by dividing x2 through, we see

2 2

2

1 ( )( ) , ( )

xp x and q x

x x

There is one singular point, x = 0, and that it is a regular singular point because xp(x)=1, and x2q(x) = x2 – ν2 are analytic at x = 0.

44

4.6.1 ν≠integer

Seeking a Frobenius solution for the Bessel’s equation about the regular singular point x = 0 by letting

00

( ) (a 0)k rn

k

y x a x

(2)

2 22

0

=0 k rk k

k

k r a a x

(3)

where a0 ≠ 0 and a-2=a-1≡0. Equating to zero the coefficientof each power of x in Eq. (3) gives

2 20

2 21

2 22

0 : ( ) 0

1: (( 1) ) 0

2 : (( ) ) 0k k

k r a

k r a

k r k a a

(4a)

(4b)

(4c)

45

Since a0 ≠ 0, Eq. (4a) gives the indicial equation r2-2 = 0.

(a) Considering r=+, then Eq. (4b) gives a1=0 and Eq. (4c) gives the recursion relation

2

1 ( 2)

( 2 )k ka a kk k

2 02

02

( 1)

2 !( )( 1) ( 1)

( 1) ( 1) =

2 ! ( 1)

k

k k

k

k

a ak k k

ak k

L

(5)

(6)

From Eq. (5), together with the fact that a1 = 0, it follows that a1 = a3 = a5 =…= 0 and that

(7)

46

20

0

( 1)( ) 2 ( 1) ( )

! ( 1) 2

kk

K

xy x a

k k

we have the solution

(8)

Dropping the a02νΓ(ν+1) scale factor, we call the resulting solution the Bessel function of the first kind, of order :

2

0

( 1)( ) ( ) ( )

2 ! ( 1) 2

kk

K

x xJ x

k k

(9)

(b) Considering r=-. Changing to – everywhere on the right side of Eq. (9). Denoting the result as J(x), the Bessel function of the first kind, of order –, we have

2

0

( 1)( ) ( ) ( )

2 ! ( 1) 2

kk

K

x xJ x

k k

47

22

1 1( )

( 1)2 ( 2)2J x x x

The general solution of Eq. (1) is

( ) ( ) ( )y x AJ x BJ x Writing Eqs. (9) and (10)

(11)

(12)

(13)2

2

1 1( )

(1 )2 (2 )2J x x x

Fig. 1 J1/2 (x) and J-1/2(x)

1

2

1

2

2( ) sin

2( ) cos

J x xx

J x xx

(14a)

(14b)

48

4.6.2 ν= integer

If is a positive integer n, then

( 1) ( 1) ( )!k n k n k

And the solution expressed by Eq. (9) gives

2

0

( 1)( ) ( )

!( )! 2

kk n

nk

xJ x

k k n

(15)

We need to be careful with Eq. (10) because if = -n, then the Γ(k-n+1) in Eq. (10) is undefined when its argument is zero or a negative integer, namely, for k = 0, 1, 2, …, n-1, and it equals 1/(k-n)! For k = n, n+1, n+2,…, in which case Eq. (10)becomes

2( 1)( ) ( )

!( )! 2

kk n

nk n

xJ x

k k n

(17)

49

2

0

( 1)( ) ( )

( )! ! 2

m nm n

nm

xJ x

m n m

Since Γ(k-n+1) is undefined at k=0,1,2,…,(n-1), rather than “∞”. Replacing the dummy summation index k by m according to k-n=m,

If (-1)n is factored out, the series that remains is the same as that given in Eq. (15), so that

( ) ( 1) ( )nn nJ x J x (18)

J ( ) ( ) are linearly dependent. Thus

whereas ( ) ( ) are LI if is not an integer. n nJ x and J x

J x and J x

We have only one linearly independent solution thus far for the case where = n. namely, y1(x) = Jn(x).

50

Let us begin with n=0, i.e. y1(x)=J0(x), for repeated indicial roots, r=±n=0, which corresponds to case (ii) of that theorem4.3.1, to obtain a second LI solution y2(x) we rely on Theorem 4.3.1.Accordingly, we seek y2(x) in the form

2 01

( ) ( ) ln kky x J x x c x

(19)

and obtain2 4

2 0 2

1 1( ) ( ) ln ( ) (1 ) ( )

2 2 (2!) 2

x xy x J x x (20)

which is called Y0(x), the Neumann function of order zero.

Thus, we have two LI solutions, y1(x) = J0(x), and y2(x)=Y0(x). However, following Weber, it proves to be convenient and standard to use, in place of Y0(x), a linear combination of J0(x) and Y0(x), namely,

2 0 0 0

2( ) Y ( ) ( ln 2) ( ) ( )y x x J x Y x

(21)

51

where2 4

0 2 4 2

0 6

6 2

1(ln ) ( ) (1 )

2 2 2 2 (2!)2( )

1 1(1 )

2 3 2 (3!)

x x xJ x

Y xx

(22)

is Weber’s Bessel function of the second kind, of order zero.γ= 0.5772157 is known as Euler’s constant. The graphs of J0(x) and Y0(x) are shown in Fig. 2.

0 0

2( ) ~ 1, Y ( ) ~ lnJ x x x

as x→0

and

0 0

2 2( ) ~ cos( ), Y ( ) ~ sin( )

4 4J x x x x

x x

as x→0

52

For n = 1, 2, 3,…the indicial roots r =± n differ by an integer, which corresponds to case (iii) of Theorem 4.3.1. Using the theorem, and the ideas given above for Y0 we obtain Weber’s Bessel function of the second kind, of order n.

12

0

2

0

1 ( 1)!(ln ) ( ) ( )

2 2 ! 22( )

( 1) ( ) ( )1( )

2 !( )! 2

nk n

nk

n kk n

k

x n k xJ x

kY x

k k n x

k k n

(25)

4.6.3 General solution of Bessel equation

2 2 2( ) 0x y xy x y The solution of Bessel equation

If is not an integer

is an integer and = n = 0, 1, 2, …

( ) ( ) ( )y x AJ x BJ x

( ) ( ) ( )n ny x AJ x BY x

Eq.(9) Eq.(10)

Eq.(15) Eq.(25)

53

for all value of .

( ) ( ) ( )y x AJ x BY x

If we define(cos ) ( ) ( )

( )sin

x J x J xY x

x

for noninteger n, then the limit of Y(x) as →n (n = 0,1, 2,..)gives the same result as Eq. (25). Furthermore, J(x) and Y(x)are LI so we can express the general solution of Eq. (1) as

(27)

4.6.5 Modified Bessel equation

2 2 2( ) 0x y xy x y (33)

(26)

is an integer

54

Let t = ix (or x = it), we can convert Eq. (33) to the Bessel equation

2 2 2( ) 0t Y tY t n Y The general solution is

( ) ( ) ( )n ny x AJ ix BY ix

(34)

(35)

2

0

2

0

( 1)( ) ( )

!( )! 2

1 ( )

!( )! 2

kk n

nk

n k n

k

ixJ ix

k k n

xi

k k n

(36)

and

Define ( ) ( )nn nI x i J ix

2

0

1( ) ( )

!( )! 2k n

nk

xI x

k k n

(37)

known as the modified Bessel function of the first kind, and order n.

55

In place of Yn(ix) it is standard to introduce, as a second real-valued solution, the modified Bessel function of the second kind, and order n.

1( ) ( ) ( )2

nn n nK x i J ix iY ix

(38)

and the graphs of these functions are plotted in Fig. 4.

Fig. 4 Graphs of I0(x) and K0(x).

As a general solution of Eq. (33) we have

( ) ( ) ( )n ny x AI x BK x

Whereas the Bessel functions are oscillatory, the modified Bessel functions are not.

(40)

56

Example 1 Solve 2 0xy y xy

More generally, the equation

( ) 0a cd dyx bx y

dx dx

1

and y

t bx u x

(46)

(47)

22 2 2

2( ) 0

d u dut t t udt dt

(48)

57

If we choose 2 1

2 2

aand

c a c a

(49)

1

( ) ( )y x x Z b x

(50)

where

( ) ( ) ( )Z x AJ x BY x

Example 2

58

Problems for Chapter 4Exercise 4.2 3.(b) 、 (h) 、 (k) 6. ； 7.(b) 、 (f) 11.(b) 、 (c) 、 (e)

Exercise 4.3 1.(a)-(g) 、 (l) 、 (p) 2.; 3.(a) 、 (b) 6.(b) 、 (g) 、 (m) 、 (s) 10.

Exercise 4.4 1.; 5. 6.(a) ； 7. (a) 10.(a) 11.(a) 、 (b)

Exercise 4.6 4.(a) ； 5.(a) ； 6.(a) ； 7.(a) 8.(b) ； 12.(a) 、 (e) 、 (i)

Exercise 4.5 1.; 2. 3.(a) 、 (c) 、 (d) 、 (f) 、 (g) 、 (h) 4. 6.(a) 、 (b) 7.(a) 、 (b) 、 (f) 9. 10.(a) 、 (b) 、 (d) 14. 16.(b) 17. 19.(a) 、 (b) 、 (f) 、 (g)