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5.3 One of Two Special Equations - Bessel’s Equation
1. Bessel’s equation and solution:
Bessel’s differential equation: x 2 y UU xy U Ÿx 2 " 7 2 y 0, where 7 (read as nu) is a real number.
Since a 2 Ÿx x 2 0 when x 0, the equation has singular points: x 0. Since
Ÿx 2 " 7 2 xPŸx x 1x 1 Ÿa 0 1 ,
x 2 QŸx x 2
x 2 " 7 2 Ÿb 0 "v 2 x2
are analytic, x 0 is a regular singular point. By Frobenius’ Theorem, there exists at least one solutions of
.
the form: y ! m0 c m x mr .
a. The indicial equation:
rŸr " 1 a 0 r b 0 rŸr " 1 r " v 2 r 2 " r r " v 2 r 2 " v 2 0.
Indicial roots: r o7.
.
yU .
! c m Ÿm r x mr"1
and y UU m0
! c m Ÿm r Ÿm r " 1 x mr"2 .
m0
b. Substitute y, y U and y UU into the equation x 2 y UU xy U x 2 y " 7 2 y 0.
.
! c m Ÿm r Ÿm r " 1 x
m0
.
mr
! c m Ÿm r x
.
mr
m0
! cmx
.
mr2
m0
" ! 7 2 c m x mr 0.
m0
In the 3rd summation, replace m 2 by k : m k " 2, k m 2, when m 0, k 2.
In the 1st, 2nd and the last summation replace m by k .
.
.
.
.
k0
k0
k2
k0
! c k Ÿk r Ÿk r " 1 x kr ! c k Ÿk r x kr ! c k"2 x kr " ! 7 2 c k x kr 0.
¡ c 0 Ÿr Ÿr " 1 c 0 Ÿr " 7 2 c 0 ¢x r .
! k2
c 1 Ÿr 1 Ÿr c 1 Ÿr " 7 2 c 1
c k Ÿk r Ÿk r " 1 c k Ÿk r c k"2 " 7 2 c k
x kr 0
c 0 Ÿr Ÿr " 1 c 0 Ÿr " 7 2 c 0 c 0 Ÿr 2 " v 2 0
c 1 Ÿr 1 Ÿr c 1 Ÿr " 7 2 c 1 c 1 ¡Ÿr 1 2 " v 2 ¢ 0
c k Ÿk r Ÿk r " 1 c k Ÿk r c k"2 " 7 2 c k c k ¡Ÿr k 2 " v 2 ¢ c k"2 0, k 2, 3, . . .
From the first equation, since c 0 p 0, r ov. When r ov, Ÿr 1 2 " v 2 p 0 in the 2nd equation.
So, c 1 0. Let r v. Then we have the following recurrence relation:
c 0 is arbitrary, c 1 0
ck "1
"1
c k"2 c k"2 , for k 2, 3, . . . .
kŸk 2v Ÿv k 2 " v 2
c. Since c 1 0, c 3 0, . . . , c 2m"1 0, for m 1, . . . .
Now consider k 2m, where m 1, . . . .
"1
c 2m"2 2 "1
c 2m"2 , m 1, . . . .
c 2m 2mŸ2m 2v 2 mŸm v 1
m 1, c 2 "1
c0
2 2 Ÿ1 Ÿ1 v m 2, c 4 Ÿ"1 2
"1
c
c0
2
2 2 Ÿ2 Ÿ2 v 2 2Ÿ2 2!Ÿ2 v Ÿ1 v :
Ÿ"1 m
c0
2 2m m!Ÿm v Ÿm " 1 v . . . Ÿ1 v m m, c 2m .
! c 2m x
yv .
c0 !
2mv
m0
.
m0
Ÿ"1 m
x 2mv
2 2m m!Ÿm v Ÿm " 1 v . . . Ÿ1 v .
Ÿ"1 m Ÿv 1 2 v
Ÿ"1 m
x 2mv c 0 Ÿv 1 2 v ! 2mv
x 2mv
2mv
2
m!Ÿm v 1 2
m!Ÿm
v
1 m0
c0 !
m0
.
c0 !
2
m0
2mv
Ÿ"1 m
x 2mv
m!Ÿm v 1 .
Gamma function: Ÿx ; t x"1 e "t dt
0
Properties: Ÿi Ÿx 1 xŸx ; Ÿii Ÿn 1 n!
It is also called the generalized factorial.
TI-89: ;Ÿt^Ÿx " 1 ' e^Ÿ"t , t, 0, . Example Evaluate 1
2
1
2
5
2
and .
=
TI-89: ;Ÿt^Ÿ"1/2 ' e^Ÿ"t , t, 0, . 5
2
3
2
1
3
2
3
2
(use (-) for –)
3
2
1
1
2
3
2
1
2
1
2
3
4
= 1. 329 34
2. The General Solution of Bessel’s Differential Equation:
Let
.
J v Ÿx !
m0
.
Ÿ"1 m
Ÿ"1 m
x 2mv x v ! 2mv
x 2m for x in I.
2mv
2
m!Ÿm v 1 2
m!Ÿm
v
1 m0
Since x 0 is the only singular point of Bessel’s equation, the radius of convergence R is . and therefore
the interval of convergence I is either "., 0 or 0, . . J v Ÿx is called the Bessel function of the first
kind of order v.
Case 1. If v is not an integer, a general solution of Bessel’s equation for all x p 0 is
y C 1 J v Ÿx C 2 J "v Ÿx .
Case 2. If v is a nonnegative integer, say v n, then
S
.
J n Ÿx !
m0
2
Ÿ"1 m
x 2mn X
2 2mn m!Ÿm n !
n
2
1
=x cosŸx " =Ÿ 2 4 2
1
0
10
20
30
x
40
50
60
-1
-2
J n Ÿx red n 0, green n 1, blue n 2
J Ÿx 0.
lim
xv. n
S Note that
J "n Ÿx Ÿ"1 n J n Ÿx which can be shown as follows.
.
J "n Ÿx !
m0
Ÿ"1 m
x 2m"n 2m"n
2
m!Ÿm " n !
.
!
mn
2
2m"n
Ÿ"1 m
x 2m"n
m!Ÿm " n !
Let m s n Ÿs m " n . Then
.
J "n Ÿx !
mn
.
!
s0
Ÿ"1 m
x 2m"n 2 2m"n m!Ÿm " n !
.
!
s0
Ÿ"1 sn
x 2Ÿsn "n
2 2Ÿsn "n Ÿs n !Ÿs n " n !
.
Ÿ"1 sn
Ÿ"1 m
2sn
n
x
x 2mn Ÿ"1 n J n Ÿx .
Ÿ"1 !
2mn
2 2sn Ÿs n !s!
2
Ÿm
n !m!
m0
Since J "n Ÿx Ÿ"1 n J n Ÿx , J n and J "n are linearly dependent. Define
1
Y v Ÿx ¡J v Ÿx cosŸv= " J "v Ÿx ¢, v may not be an integer.
sinŸv= Y v Ÿx is called the Bessel function of the second kind of order v. The another form of the general
solution:
y C 1 J v Ÿx C 2 Y v Ÿx .
2 J 0 Ÿx + ln x
Y 0 Ÿx =
2
3
2
" =
.
!
k1
Ÿ"1 k
1 1 . . . 1 x 2k ,
2
k
2 2k Ÿk! 2
+ 0. 57721566
n
" x=
.
!
k0
n"1
"n
" x=
2 J n Ÿx + ln x
Y n Ÿx =
2
Ÿ"1 k
2 2kn k!Ÿk n !
!
k0
Ÿn " k " 1 ! 2k
x
2 2k"n k!
1 1 . . . 1
2
k
1 1 . . . 1
2
kn
x 2k , + 0. 57721566.
Since
Y n Ÿx lim
Y Ÿx ,
v¯n v
when v n,
y C 1 J n Ÿx C 2 Y n Ÿx .
3. Properties of Bessel Functions:
a. J "n Ÿx Ÿ"1 n J n Ÿx b. J n Ÿ"x Ÿ"1 n J n Ÿx (when n is even, J m is even, and when n is odd, J m is odd. )
U
U
U
c. xJ v Ÿx vJ v Ÿx " xJ v1 Ÿx , J 0 Ÿx "J 1 Ÿx ,
Y 0 Ÿx "Y 1 Ÿx 9x 2 y UU 9xy U 9x 2 " 1 y 0.
4
It is not in a Bessel equation form yet. Divide both sides of the equation by 9 :
9x 2 y UU 9xy U 9x 2 " 1 y 0 ® x 2 y UU xy U x 2 " 1 y 0
4
36
This is a Bessel equation with v o 1 . Since v is not an integer,
6
y C 1 J 1 Ÿx C 2 J " 1 Ÿx 6
6
Example Find the general solution of the equation:
.
C 1 x 1/6 !
m0
C 1 x 1/6
1
2 1/6 C 2 x "1/6
C 1 x 1/6
Ÿ"1 m
1
2 2m 6 m!Ÿm 7
6
1
2 "1/6 "
5
6
.
7
6
1
2 13/6 "
x 2m C 2 x "1/6 !
m0
13
6
1
2 11/6 x2 11
6
2
1
2 25/6 x2 2m"
1
6
19
6
1
2 23/6 Ÿ"1 m
m!Ÿm 5
6
x 2m
x 4 ". . .
17
6
x 4 ". . .
0. 960 311 " 0. 205 781 x 2 0. 02 37x 4 " C
C 2 x "1/6
0. 994 397 " 0. 298 319 x 2 0. 0407 x 4 . . .
Example Consider the differential equation x 2 y UU xy U 2x 2 " 1 y 0.
9
dy
d2y
, and 2 .
dz
dz
ii Find the general solution in z with expressing the first 3 nonzero terms for each solution.
iii Find the general solution in x with expressing the first 3 nonzero terms for each solution.
i Set z 2 x and rewrite the differential equation as a Bessel equation in z,
i The equation x 2 y UU xy U 2x 2 " 1 y 0 is not in a Bessel equation form.
9
Let z 2 x. Then we have the following substitution:
4
z , and dz dx
2
x
dy
dy dz
dx
dz dx
and the equation becomes:
yU 2
dy
,
dz
y UU 2
z 2 Ÿ2 d y 2
dz 2
which is a Bessel equation in z:
2
dy U
dy U dz
dy
d2y
2 2 2
d Ÿ 2
dz
dx
dz dx
dz
dz
z Ÿ 2 dy Ÿz 2 " 1 y 0,
9
dz
2
z 2 y UU zy U Ÿz 2 " 1 y 0, v o 1
3
9
1
ii Since 7 o 3 , the general solution of the differential equation is:
y C 1 J 1 Ÿ 2 x c 2 J " 1 Ÿ 2 x 3
.
C1 !
m0
3
.
1
1
Ÿ"1 Ÿ 2 x 2m 3
Ÿ"1 m Ÿ 2 x 2m" 3
C
2!
1
2m" 13
2 2m 3 m!Ÿm 1 13 m!Ÿm 1 " 13 m0 2
m
.
C 1 x 1/3 !
m0
.
C 1 x 1/3 !
m0
.
1
Ÿ"1 m 2 m 6
1
2 2m 3 m!Ÿm 2 m
1
6
Ÿ"1 m
m!Ÿm 4
3
4
3
1
Ÿ"1 m 2 m" 6
1
2 2m" 3 m!Ÿm x 2m C 2 x "1/3 !
m0
.
x 2m C 2 x "1/3 !
m0
2 m"
1
6
Ÿ"1 m
m!Ÿm 2
3
2
3
x 2m
x 2m
iii
y C 1 x 1/3
1
" 7/6 1 7 x 2 13/6 1 10 x 4 " C
2 1!Ÿ 3 2 2!Ÿ 3 2 1/6 0!Ÿ 43 C 2 x "1/3
C 1 x 1/3
C 2 x "1/3
1
" 5/6 1 5 x 2 11/6 1 8 x 4 " C
2 1!Ÿ 3 2 2!Ÿ 3 2 "1/6 0!Ÿ 23 0. 9958 " 0. 3734x 2 0. 0400x 4 " C
0. 8299 " 0. 6224x 2 0. 0934x 4 " C
Example Consider the differential equation 4x 2 y UU 4xy U x " 1 y 0.
36
dy
d2y
, and 2 .
dz
dz
ii Find the general solution in z with expressing the first 3 nonzero terms for each solution.
iii Find the general solution in x with expressing the first 3 nonzero terms for each solution.
i Set z x and reduce the differential equation to a Bessel equation in z,
i The equation 4x 2 y UU 4xy U x " 1 y 0 is not a Bessel equation.
36
Let z 2 x. Then we have the following substitution:
dy 1
Ÿ z x , or x z 2 , and dz 1 dz 2z
dx
2 x
yU y UU 5
dy U
dy U dz
d
dx
dz dx
dz
dy
dy dz
dy
dy
Ÿ 1 1
2z dz
dx
dz dx
dz 2 x
1 dy
2z dz
dz dx
2
dy "1
d2y 1
1 1 d y " 1 dy
Ÿ
Ÿ
Ÿ
2z
dz 2z 2
4z 2 dz 2
4z 3 dz
dz 2 2z
and the equation becomes:
dy
d2y
4z 4 Ÿ 1 2 2 " 1 3
4z dz
4z dz
which is a Bessel equation in z:
4z 2
dy 1
Ÿ Ÿz 2 " 1 y 0
36
dz 2z
v o1
z 2 y UU zy U Ÿz 2 " 1 y 0,
6
36
1
ii Since 7 o 6 , the general solution of the differential equation is:
y C1J 1 Ÿ x C2J" 1 Ÿ x 6
6
.
C1 !
m0
.
C1 !
m0
.
1
Ÿ"1 m Ÿ x 2m 6
1
2 2m 6 m!Ÿm 1 2
2m 16
Ÿ"1 m
m!Ÿm .
C 1 x 1/12 !
m0
2 2m
1
6
1
6
Ÿ"1 m
m!Ÿm m0 2
.
x m 12 C 2 !
1
7
6
1
Ÿ"1 m Ÿ x 2m" 6
C2 !
m0
2m" 16
2
2m" 16
Ÿ"1 m
m!Ÿm .
7
6
1
6
m!Ÿm 1 "
x m C 2 x "1/12 !
m0
2 2m"
1
6
1
5
6
x m" 12
Ÿ"1 m
m!Ÿm 5
6
xm
iii
y C 1 x 1/12
C 2 x "1/2
c 1 x 1/12
c 2 x "1/12
1
" 13/6 1 13 x 25/6 1 19 x 2 " C
7
2 2!Ÿ 6 2 1!Ÿ 6 2 0!Ÿ 6 1/6
2
"1/6
1
" 11/6 1 11 x 23/6 1 17 x 2 " C
5
2 2!Ÿ 6 2 1!Ÿ 6 0!Ÿ 6 0. 9006 " 0. 1930x 0. 0111x 2 " C
0. 9984 " 0. 2995x 0. 0204x 2 " C .
Example Verify that y x n J n Ÿx is a particular solution of the equation: xy UU Ÿ1 " 2n y U xy 0. Find
a particular solution of the equation: xy UU " y U xy 0.
Check:
U
U
U
UU
y U nx n"1 J n Ÿx x n J n Ÿx , y UU nŸn " 1 x n"2 J n Ÿx nx n"1 J n Ÿx nx n"1 J n Ÿx x n J n Ÿx xy UU Ÿ1 " 2n y U xy
xŸnŸn " 1 x n"2 J n x nx n"1 J Un x nx n"1 J Un x x n J UUn x Ÿ1 " 2n Ÿnx n"1 J n x x n J Un x xŸx n J n Ÿx "n 2 x n J n x 2n J UUn x n1 J Un x 2n J n
x n Ÿx 2 J UUn Ÿx xJ Un Ÿx Ÿx 2 " n 2 J n Ÿx 0
xy UU " y U xy xy UU Ÿ1 " 2Ÿ1 y U xy 0,
n1
A particular solution is:
.
y x J 1 Ÿx x !
1
m0
Ÿ"1 m
x 2m1 2 2m1 m!Ÿm 1 !
.
!
m0
Ÿ"1 m
x 2m2
2 2m1 m!Ÿm 1 !
1 x 2 " 1 x 4 1 x 6 ". . .
16
2
384
Example One mathematical model for the free undamped motion of a mass on an aging spring is given by
my UU ke ")t y 0, where ) 0.
6
2 k e ")t/2 . Reduce the differential equation to a Bessel equation in s.
i Let s )
m
ii Find the general solution of the differential equation in a series form.
iii Solve the initial-value problem: 4y UU e "0.1t y 0, yŸ0 1, y U Ÿ0 " 12 .
i The equation my UU ke ")t y 0 is not a Bessel equation.
ds 2
)
dt
k
m
" ) e ")t/2 "
2
k ")t/2 " ) s,
m e
2
k e ")t/2 ) 2 s.
d 2 s " k " ) e ")t/2 )
m
m
2
2
2
dt 2
dy ds
dy
dy
,
")s
2
ds dt
dt
ds
d2y
dy
dy
dy
ds d
")s
")s
d
d
2
2
dt dt
ds dt
ds
dt
ds
dt 2
2
2
2
2 dy
dy
d y ds
d y
")
")s ")s
) s
2
2
2
2
2
2
ds
ds
dt
ds
ds
d2y
2
kŸe ")t/2 y
2
dt
2
) 2 s 2 d y ) 2 s dy k s ) m
m
2
2
2 k
ds
ds 2
2
2
d y
dy
s2 2 s
s2y 0 ®
m )
2
ds
ds
d2y
d2y
dy
dy
s2 2 s
s2y 0 ® s2 2 s
Ÿs 2 " 0 y 0, v 0.
ds
ds
ds
ds
ii The general solution is:
my UU ke ")t y m
y C 1 J 0 Ÿs C 2 Y 0 Ÿs C 1 J 0
iii 4y UU e "0.1t y y UU 1 e "0.1t y 0,
4
2
)
k ")t/2
m e
C2Y0
2
)
2
y
k ")t/2
m e
y U Ÿ0 " 12 .
yŸ0 1,
2
) 0. 1, m 4, k 1. s )
k ")t/2 2
m e
0. 1
1 e "0.1t/2 10e "0.05t
4
y C 1 J 0 Ÿ10e "0.05t C 2 Y 0 Ÿ10e "0.05t ,
U
U
y U C 1 Ÿ"0. 5e "0.05t J 0 Ÿ10e "0.05t C 2 Ÿ"0. 5e "0.05t Y 0 Ÿ10e "0.05t yŸ0 C 1 J 0 Ÿ10 C 2 Y 0 Ÿ10 C 1 Ÿ"0. 2459 C 2 Ÿ0. 0557 1
y U Ÿ0 "0. 5C 1 Ÿ"J 1 Ÿ10 " 0. 5C 2 Ÿ"Y 1 Ÿ10 0. 5C 1 Ÿ0. 0435 0. 5C 2 Ÿ0. 2490 " 1
2
C 1 Ÿ"0. 2459 C 2 Ÿ0. 0557 1
0. 5C 1 Ÿ0. 0435 0. 5C 2 Ÿ0. 2490 " 12
2 J 0 Ÿx + ln x
Y 0 Ÿx =
2
2
" =
.
!
k1
,
C 1 "4. 786 962
C 2 "3. 179 788
Ÿ"1 k
1 1 . . . 1 x 2k , + 0. 57721566
2
k
2 2k Ÿk! 2
y "4. 786 962J 0 Ÿ10e "0.05t " 3. 179 788Y 0 Ÿ10e "0.05t Let
7
.
J 0 Ÿx !
m0
Ÿ"1 m
x 2m X 1 " 1 x 2 1 x 4
4
64
2 2m m!Ÿm !
2 1 " 1 x 2 1 x 4 + ln
Y 0 Ÿx X =
4
64
2 " 1 Ÿ1 x 2 1
1
" =
4
16Ÿ4 2 1 " 1 x 2 1 x 4 + ln
=
4
64
1
2
y X "4. 786 962 = 1 " Ÿ10e "0.05t 2 4
" 3. 179 788
x
2
1 x4 1
1 1 1 x6
2
2
3
64Ÿ36 2 " 1 x 2 3 x 4 11 x 6
x
" =
4
128
13 824
2
1 Ÿ10e "0.05t 4
64
"0.05t
2 1 " 1 Ÿ10e "0.05t 2 1 Ÿ10e "0.05t 4 + ln Ÿ10e
=
4
2
64
2 " 1 Ÿ10e "0.05t 2 3 Ÿ10e "0.05t 4 11 Ÿ10e "0.05t 6
3. 179 788 =
4
128
13 824
625
2
"0.1t
"0.2t
e
"4. 786 962 = 1 " 25e
4
2 1 " 25e "0.1t 625 e "0.2t ¡+ lnŸ5e "0.05t ¢
" 3. 179 788 =
4
1875
2
"0.1t
e "0.2t 171 875 e "0.3t
3. 179 788 = "25e
8
216
1
0.8
0.6
0.4
0.2
0
8
2
4
6
8
10
t
12
14
16
18
20
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