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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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
5.4 Power Frobenius Method(2/20)
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
5.4 Power Frobenius Method(3/20)
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
5.4 Power Frobenius Method(4/20)
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
5.4 Power Frobenius Method(5/20)
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
5.4 Power Frobenius Method(6/20)
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
5.4 Power Frobenius Method(7/20)
(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
5.4 Power Frobenius Method(8/20)
(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
5.4 Power Frobenius Method(9/20)
Frobenius Method. Basis of Solutions. Three Cases(2)
THEOREM 2
Case 2. Double Root r1 = r2 = r. A basis is
(7)
(of the same general form as before) and
(8)
12
5.4 Power Frobenius Method(10/20)
Example (B) : Solve (x2x)y”+(3x1)y’ +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.
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
5.4 Power Frobenius Method(11/20)
(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
5.4 Power Frobenius Method(12/20)
(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
5.4 Power Frobenius Method(13/20)
*(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
5.4 Power Frobenius Method(14/20)
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
5.4 Power Frobenius Method(15/20)
Frobenius Method. Basis of Solutions. Three Cases(3)
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
5.4 Power Frobenius Method(16/20)
Example (C) : Solve (x2x)y”xy’ + 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.
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
5.4 Power Frobenius Method(17/20)
(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
5.4 Power Frobenius Method(18/20)
(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
5.4 Power Frobenius Method(19/20)
*(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
5.4 Power Frobenius Method(20/20)
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
5.5 Bessel’ s Equation. Bessel Functions J
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)
5.5 Bessel’ s Equation. Bessel Functions J
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 .
5.5 Bessel’ s Equation. Bessel Functions J
ν (4/11)
27
5.5 Bessel’ s Equation. Bessel Functions J
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
5.5 Bessel’ s Equation. Bessel Functions J
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
5.5 Bessel’ s Equation. Bessel Functions J
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
5.5 Bessel’ s Equation. Bessel Functions J
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
5.5 Bessel’ s Equation. Bessel Functions J
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
5.6 Bessel Functions of the Second Kind Yν(x) (2/7)
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
5.6 Bessel Functions of the Second Kind Yν(x) (3/7)
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
5.6 Bessel Functions of the Second Kind Yν(x) (4/7)
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
5.6 Bessel Functions of the Second Kind Yν(x) (5/7)
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
5.6 Bessel Functions of the Second Kind Yν(x) (6/7)
40
5.6 Bessel Functions of the Second Kind Yν(x) (7/7)
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
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