1 chapter 5 (5.4~5.6) ◎ series solutions of odes. 微分方程式之級數解 ◎ special...

1

Chapter 5(5.4~5.6)

◎Series Solutions of ODEs. 微分方程式之級數解 ◎Special Functions 特殊方程式

Kreyszig, E. “Advanced Engineering Mathematics” (9th ed. )

2

5.4 Frobenius Method 5.5 Bessel’ s Equation. Bessel Funct

ions Jν(x)

5.6 Bessel Functions of the Second Kind Yν(x)

Contents

3

5.4 Power Frobenius Method(1/20)

Frobenius Method ( 傅洛貝尼斯法 )

THEOREM 1

Let b(x) and c(x) be any functions that are analytic at x = 0. Then the ODE

(1)

has at least one solution that can be represented in the form

(2)

where the exponent r may be any (real or complex) number (and r is chosen so that a0 ≠ 0).

4


We shall now explain the Frobenius method for solving (1). Multiplication of (1) by x2 gives the more convenient form

(1') x2y" + xb(x)y' + c(x)y = 0.

We first expand b(x) and c(x) in power series,

b(x) = b0 + b1x + b2x2 + , ‥‥ c(x) = c0 + c1x + c2x2 + ‥‥or we do nothing if b(x) and c(x) are polynomials.

5


Then we differentiate (2) term by term, finding

By inserting all these series into (1') we readily obtain xr[r(r – 1)a0 + ] + (‥‥ b0 + b1x + )‥‥ xr(ra0 + )‥‥ + (c0 + c1x + )‥‥ xr(a0 + a1x + ) = 0‥‥

6


We now equate the sum of the coefficients of each power xr, xr+1, xr+2, to zero. The equatio‥‥n corresponding to the power xr is

[r(r – 1) + b0r + c0]a0 = 0

Since by assumption a0 ≠ 0, the expression in the brackets [ ] must be zero. This gives‥‥ (4) r(r – 1) + b0r + c0 = 0

This important quadratic equation is called the indicial equation ( 指示方程式 ).

7


Frobenius Method. Basis of Solutions. Three Cases(1)

THEOREM 2

Suppose that the ODE (1) satisfies the assumptions in Theorem 1. Let r1 and r2 be the roots of the indicial equation (4). Then we have the following three cases.Case 1. Distinct Roots Not Differing by an Integer. A basis is

(5)

and

(6)

8


Example (A) : Solve 4xy”+2y’ +y= 0 by Frobenius method with coefficients represented up to and including x3 ？

Since & 代入 ODE

** 取最小冪次項 xr1 ( i.e. m=0) 之係數為零 indicial eq.

4r(r1)+2r=0 r(2r1) =0 r =0 or r=1/2 (case 1.)

0

1)('m

rmm xarmy

0

2)1)(("m

rmm xarmrmy

00

1

0

2 0)(2)1)((4m

rmm

m

rmm

m

rmm xaxarmxarmrmx

00

1

0

1 0)(2)1)((4m

rmm

m

rmm

m

rmm xaxarmxarmrm

9


(when r=0)

級數式

註標變換 ( 第 1,2 項 m1=s, m=s+1 ；第 3 項 m=s)

[4(s+1)s+2(s+1)]as+1 + as=0 recurrence

a1= a0/2! ; a2= a1/12=a0/4! ; a3= a2/30= a0/6! …

取 a0=1 ＃

00

1

0

1 02)1(4m

mm

m

mm

m

mm xaxamxamm

001

01 0)1(2)1(4

s

ss

s

ss

s

ss xaxasxsas

ss ass

a)1)(24(

11

xxxxxa

xa

xa

axy cos...!6

1

!4

1

!2

11...)

!6!4!2( 3230200

00

1

10


(when r=1/2)

級數式

註標變換 ( 第 1,2 項 m1/2=s, m=s+1/2 ；第 3 項 m+1/2=s, m=s1/2)

[4(s+1)s+2(s+1)]as+1 + as=0 recurrence

a1= a0/3! (s=1/2); a2= a1/20=a0/5! (s=3/2); a3= a2/42=a0/7! (s=5/2); …

取 a0=1

＃

0

2/1

0

2/1

0

2/1 0)2

1(2)

2

1)(

2

1(4

m

mm

m

mm

m

mm xaxamxamm

02/1

02/1

02/1 0)1(2)1(4

m

ss

m

ss

m

ss xaxasxsas

2/12/1 )1)(24(

1

ss ass

a

xxxxxxa

xa

xa

axy sin...!7

1

!5

1

!3

1...)

!7!5!3( 2/72/52/32/130200

02/1

2

11



THEOREM 2

Case 2. Double Root r1 = r2 = r. A basis is

(7)

(of the same general form as before) and

(8)

12


Example (B) : Solve (x2x)y”+(3x1)y’ +y= 0 by Frobenius method with coefficients represented up to and including x3 ？

Since & 代入 ODE


r(r1)r=0 r2=0 r = 0 (case 2.)

0

1)('m

rmm xarmy

0

2)1)(("m

rmm xarmrmy

00

1

0

22 0)()13()1)(()(m

rmm

m

rmm

m

rmm xaxarmxxarmrmxx

0 0

1

0

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

rmm

m

rmm

m

rmm xarmxarmrmxarmrm

00

1 0)(m

rmm

m

rmm xaxarm

13


(when r=0)

級數式

註標變換 ( 第 2,4 項 m1=s, m=s+1 ；其它項 m=s)

(s+1)2 as+1 + (s+1)2as=0 recurrence as+1 = as

a0= a1 = a2= a3 = a4= …

取 a0=1

03)1()1(00

1

0 0

1

0

m

mm

m

mm

m

mm

m

mm

m

mm xaxmaxamxammxamm

0)1(3)1()1(00

10 0

10

s

ms

m

ss

m

ss

m

ss

m

ss xaxasxasxassxass

xxxxxaxaxaay

1

1...1...)( 323

02

0001

14


(second solution) By the method of reduction of order

(x2x)y”+(3x1)y’ +y= 0 y”+(3x1)/(x2x) y’ + 1/(x2x) y= 0

y2 = uy1 where u = Udx &

&

＃

pdx

ey

U21

1

xxdxxx

xpdx ln)1ln(2

132

xe

yU

pdx 1121

xUdxu ln x

xuyy

1

ln12

15


*(second solution) By the method of undetermined coefficients

&

代回原式

&

整理得

其中

註標變換

1

11

12 lnln 2

m

mm

m

mm

r xAxyxAxxyy 1

1

112 ln''

m

mm xmA

x

yxyy

2

2211

12 )1('2

ln""

m

mm xAmm

x

y

x

yxyy

]ln')[13(])1(

'2ln")[( 1

1

11

2

2211

12 m

mm

m

mm xmA

x

yxyxxAmm

x

y

x

yxyxx

0]ln[1

1

m

mm xAxy 0}')13("){(ln 111

2 yyxyxxx

m

mm

m

mm

m

mm xmAxAmmxAmmyyxy

1

1

22111 3)1()1([2'2'2

0]1

1

1

m

mm

m

mm xAxmA 0

)1(

2

)1(

2

)1(

22'2'2

22111

xxx

xyyxy

0)1(3)1()1(10

111

12

s

ss

s

ss

s

ss

s

ss

s

ss xAxAsxsAxsAsxAss

16


S=0 –A1 =0 A1 =0

s=1 (-2A2+3A1-2A2+A1)=0 (-4A2+4A1)=0 A2 =0

s≧2 As+1=As

A1 =A2 =A3 =….= 0 &

＃

0])1(3)1()1([ 112

sssss

s AAssAsAsAss

01

m

mm xA

x

xxyy

1

lnln12

17



THEOREM 2

Case 3. Roots Differing by an Integer. A basis is

(9)

(of the same general form as before) and

(10)

where the roots are so denoted that r1 – r2 > 0 and k may turn out to be zero.

18


Example (C) : Solve (x2x)y”xy’ + y = 0 by Frobenius method with coefficients represented up to and including x3 ？

Since & 代入 ODE


r(r1) =0 r = 1 or r = 0 (case 3.)

0

1)('m

rmm xarmy

0

2)1)(("m

rmm xarmrmy

00

1

0

22 0)()1)(()(m

rmm

m

rmm

m

rmm xaxarmxxarmrmxx

0 00

1

0

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

rmm

m

rmm

m

rmm

m

rmm xaxarmxarmrmxarmrm

19


(when r1=1)

級數式

註標變換 ( 第 1,3,4 項 m+1=s, m=s 1 ；其它項 m=s)

[s(s 1) s +1] as-1 (s+1)sas = 0 recurrence

a1 = 0‧a0 = 0 a1 = 0 & also a2 = a3 = a4 = … = 0

取 a0=1 y1 = x1a0 = x

0)1()1(1

11

110

1

s

ss

s

ss

s

ss

s

ss xaxasxassxass

0

1

0

1

00

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

mm

m

mm

m

mm

m

mm xaxamxammxmam

1

2

)1(

)1(

ss a

ss

sa

20


(second solution) By the method of reduction of order

(x2x)y” xy’ +y= 0 y” x/(x2x) y’ + 1/(x2x) y= 0

y2 = uy1 where u = Udx &

&

＃

pdx

ey

U21

1

)1ln(2

xdx

xx

xpdx

1ln12 xxuyy

2221

111

xxx

x

y

eU

pdx

x

xUdxu1

ln

21


*(second solution) r2=0 By the method of undetermined coefficients

&

&

代回原式

&

整理得

其中

註標變換

0

10

012 lnln

m

mm

m

mm xAxkyxAxxkyy 1

1

112 ln''

m

mm xmA

x

kyxkyy

2

2211

12 )1('2

ln""

m

mm xAmm

x

ky

x

kyxkyy

]ln'[])1(

'2ln")[( 1

1

11

2

2211

12 m

mm

m

mm xmA

x

kyxkyxxAmm

x

ky

x

kyxkyxx

0]ln[0

1

m

mm xAxky 0}'"){(ln 111

2 yxyyxxxk

01

1

22

1111 0])1()1(['2'2

m

mm

m

mm

m

mm

m

mm xAxmAxAmmxAmm

x

kykykyxky

kx

kykykyxky 1

111 '2'2

0)1()1(011

12

s

ss

s

ss

s

ss

s

ss xAxsAxsAsxAssk

22


s=0 常數 -k + A0 = 0 A0 = k

s=1 –2A2 – A1 + A1 =0 A2 =0

s≧2 A3 = A2/6 = 0, A4 = A3/3 = 0 ...

A2 =A3 =A4 =….= 0 & A0 = k , A1 = unknown

取 A0=1 及去除 y1=x 部分仍為基底＃

ss Ass

sA

)1(

)1( 2

1

11001012 ln)(ln yAAxxAxAAxkyy

1ln2 xxy

23

5.5 Bessel’ s Equation. Bessel Functions J

ν (1/11)One of the most important ODEs in applied mathematics is

Bessel’s equation,

(1) x2y" + xy' + (x2 –ν2)y = 0The parameter ν in (1) is a given number. We assume that

ν is real and nonnegative.

＊ The solutions of Bessel’s equation are Bessel’s functions.

＊ By Frobenius method we can derive the indical equation as (r +ν)(r –ν) = 0 & The roots are r1 = ν (≧ 0)

and r2 = –ν

24


ν (2/11)One of the most important ODEs in applied mathematics is Bessel’s equation,

(1) x2y" + xy' + (x2 –ν2)y = 0The parameter ν in (1) is a given number. We assume that ν is real and nonnegative.

＊ The solutions of Bessel’s equation are Bessel’s functions.

25

If the integer values of ν are denoted by n, For ν= n the rela

tion (7) becomes

(8)

With particular and n!(n + 1) (‥‥ n + m) = (m + n)!

so that (8) simply becomes

(10)


ν (3/11)

26

Then a particular solution of (1), denoted by Jn(x) and give

n by

(11)

Jn(x) is called the Bessel function of the first kind of orde

r n. （ n 階第一類貝索函數） The series (11) converges for all x .


ν (4/11)

27


ν (5/11)For n = 0 we obtain the Bessel function of order 0

(12)

which looks similar to a cosine. For n = 1 we obtain the Bessel function of order 1

(13)

.

28


ν (6/11)Bessel Functions J ν(x) for any ν≥ 0. Then

(19)

With these coefficients and r = r1 =νwe get a particular soluti

on, denoted by J ν(x) and given by

(20)

J ν(x) is called the Bessel function of the first kind of order ν . The series converges for all x

29


ν (7/11)

General Solution of Bessel’s Equation

THEOREM 1

If ν is not an integer, a general solution of Bessel’s equation for all x ≠ 0 is(22) y(x) = c1Jν(x) + c2J –ν(x)

Linear Dependence of Bessel Functions Jn and J–n

THEOREM 2

For integer ν = n the Bessel functions Jn(x) and J–n(x) are linearly dependent, because(23) J–n(x) = (–1)nJn(x) (n = 1, 2, ).‥‥

30


ν (8/11)Derivatives, Recursions

THEOREM 3

The derivative of Jν(x) with respect to x can be expressed by Jν-1(x) or Jν+1(x) by the formulas

(24)

Furthermore, Jν(x) and its derivative satisfy the recurrence relations

(24)

31


ν (9/11)

Elementary Jν for Half-Integer Order ν

THEOREM 4

Bessel functions Jν of orders ±1/2, ±3/2, ±5/2 ‥‥are elementary; they can be expressed by finitely many cosines and sines and powers of x. In particular,

(25)

32

5.5 Bessel’s Equation. Bessel Functions Jν (10/11)

From (24c) with ν= 1/2 and ν= –1/2 and (25) we obtain

respectively, and so on.

33

5.5 Bessel’s Equation. Bessel Functions Jν (11/11)

利用參數變換使非標準 Bessel’s equation 轉換成標準形式

Example (A) : x2y" + xy' + (2x2 ν2)y = 0

令 x = z x = z/ & dz/dx =

So: &

代回原式

＃

Where if is not an integer Y (z) = c1J(z) + c2 J(z)

Or y(x) = c1J(x) + c2 J(x) ＃

'' Ydx

dz

dz

dy

dx

dyy 2"

''" Y

dx

dz

dz

dy

dx

dyy

0)()')(()"()( 2222 YzYz

Yz

0)('" 222 YzzYYz

34

5.6 Bessel Functions of the Second Kind Yν(x) (1/7)

n = 0: Bessel Function of the Second Kind Y0(x)

When n = 0, Bessel’s equation can be written

(1) xy" + y' + xy = 0

Then the indicial equation has a double root r = 0. This is Case 2.

In this case we first have only one solution, J0(x). We see that the desir

ed second solution must be of the form

35


n = 0: Bessel Function of the Second Kind Y0(x)

By comparing the coefficient, we find A1 = A3 = = 0 ‥‥

Then, we obtain the result

36


Since J0 and y2 are linearly independent functions, they form a ba

sis of (1) for x > 0.

Of course, another basis is obtained if we replace y2 by an indepe

ndent particular solution of the form a(y2 + bJ0), where a (≠ 0) and

b are constants. It is customary to choose a = 2π and b = γ – ln 2,

where the number γ = 0.577 215 664 90 is the so-called ‥‥ Eule

r constant.

The standard particular solution thus obtained is called the Bessel

function of the second kind of order zero is denoted by Y0(x)

37


For this reason we introduce a standard second solution Y (x) de

fined for all ν by the formula which is called the Bessel function o

f the second kind of order ν denoted by Yν(x).

38


General Solution of Bessel’s Equation

THEOREM 1

A general solution of Bessel’s equation for all values of ν (and x > 0) is

(9) y(x) = C1Jν(x) + C2Yν(x)

39

利用參數變換使非標準 Bessel’s equation 轉換成標準形式

Example (B) : xy" y' + 4xy = 0

令 xu = y & 2x=z x = z/2 & dz/dx = 2

So: & Also, &

代回原式

＃

Where =1 U(z) = c1J1(z) + c2Y1(z)

Or y(x) = x [c1J1(2x) + c2Y1(2x)] ＃

'' xuuy

0)2

(2

4)'2

()"'2)(2

( uzz

uz

uxuuz

0)1('" 222 UzzUUz

"'2" xuuy 2'' Udx

dz

dz

duu 4"

'" U

dx

dz

dz

duu

0)'22

()"42

'22)(2

( 2 UzUz

UUz

Uz


40


For the modified Bessel’s equation,

x2y" + xy' –(x2 + ν 2)y = 0

＊ The solutions of modified Bessel’s equation are modified Bessel’s functions. i.e.

y(x) = C1Iν(x) + C2Kν(x)

Where & )]()([sin2

)( xIxIxK

02

2

)1(!2)(

mm

m

mm

xxI

1 chapter 5 (5.4~5.6) ◎ series solutions of odes. 微分方程式之級數解 ◎ special...

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