Linda Fahlberg-Stojanovska

1

Manipulations with Bessel Functions

1. Express the integral 3J dx with the Bessel functions Jo(x) and J1(x)

2. Express the second derivative of J1(x) with the Bessel functions Jo(x) and J1(x)

3. Do a change of variable y=u2 and solve the DE: 2

22yyy y

x y

Solutions:

1. 1 04 ( ) ( )J x J x Cx

2. 1 022 1( 1)J J

xx

3. 2

oy J x

Proofs: 4. Show: 1 0( )J x dx J x C

5. Show: 3 1 22( )J x dx J x J x Cx

http://www.wolframalpha.com/input/?i=BesselJ[0%2Cx] http://www.wolframalpha.com/input/?i=BesselJ[1%2Cx] http://www.wolframalpha.com/input/?i=BesselJ[2%2Cx] http://www.wolframalpha.com/input/?i=BesselJ[3%2Cx]

Basic formula: 2

20

( 1) 2 ! ( 1)

m m

mm

xJ xm m

Gamma Function: ( 1) !n n , ( 1) ( )n n n

[1] 1 12

n n nnJ J Jx

[2] 1 12

n nn

J JJ , 0 1J J

12

12

2 sin( )J x x x , 1

21

2

2 cos( )J x x x

Linda Fahlberg-Stojanovska

2

Expansions

2

2 00

2

2 20

2 4 6 8

2 4 6 8

( 1) 2 ! ( 0 1)

( 1)2 ( !)

12 1! 1! 2 2! 2! 2 3! 3! 2 4! 4!

m m

o mm

m m

mm

xJ xm m

xm

x x x x

21

1 2 10

2 1

2 1 20

3 5 7 9

3 5 7 9

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

2 ( !) ( 1)

[ ...]2 2 1! 1! 2 2 2! 2! 3 2 3! 3! 4 2 4! 4! 5

m m

mm

m m

mm

xJ x xm m

xm m

x x x x x

2 2

2 2 20

2 2

2 2 20

2 4 6 8 10

2 4 6 8 10

( 1) 2 ! ( 2 1)

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

[ ...]2 1 2 1! 1! 3 2 2 2! 2! 4 3 2 3! 3! 5 4 2 4! 4! 6 5

m m

mm

m m

mm

xJ xm m

xm m m

x x x x x

2 3

3 2 30

2 32 3

2 3 2 30 0

3 5 7 9 11

3 5 7 9 11

( 1) 2 ! ( 3 1)

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

[ ...]2 0!3! 2 1! 4! 2 2! 5! 2 3! 6! 2 4! 7!

m m

mm

m m mm

m mm m

xJ xm m

x xm m m m

x x x x x

3 5 7

3 3 5 2 7 2 ...]2 3! 2 (1!) 4 3 2 2 (2!) 5 4 3x x xJ x

Linda Fahlberg-Stojanovska

3

1. Express the integral 3J dx with the Bessel functions Jo(x) and J1(x)

We have: 3 1 22( ) ( ) ( )J x dx J x J x Cx

(see proof below #5).

We have: 1 12

n n nnJ J Jx

(no proof provided)

So 2 1 02 1( ) ( ) ( )J x J x J xx

and 3 1 04( ) ( ) ( )J x dx J x J x Cx

.

2. Express the second derivative of J1(x) with the Bessel functions Jo(x) and J1(x)

We have: 1 12

n nn

J JJ , 0 1J J

0 21 2

J JJ

The trick here is NOT to change 2J with formula [1], but to take the derivative and then change 2J

with formula [2] and then change 3J with formula [1].

0 2 1 31 1 1

1 3 1 2 1

1 2 1 1 0

1 02

1( )2 2 2

3 1 3 1 2 24 4 4 4

1 1 2( )

2 1( 1)

J J J JJ J J

J J J J Jx

J J J J Jx x x

J Jxx

Linda Fahlberg-Stojanovska

4

3. Do a change of variable y=u2 and solve the DE: 2

22yyy y

x y

Solution: 2

2

2

( ) ( ) 2 ( ) 2

( ) (2 ) 2 ( )

2( ) 2

y ud d dy y u u u u udx dx dxd d du d du d duy y u u udx dx dx dx dx dx dx

u u u

Substituting these into the DE: 2

2 22

2 2 2

2

22 2( ) 2 22

22( ) 2 2 2( )

0

uuuuu u u ux u

uuu u u u ux

uuu u ux

So: Either 0u or

0uu ux

Multiply through by x2 to put in “standard” Bessel form. 2 2 0x u xu x u

If you are allowed, immediately write: 1 0( ) ( )u x C J x so 2 20( ) ( )y x u C J and you are done.

Otherwise see below for details:

Linda Fahlberg-Stojanovska

5

Our mega formula is: 2 2 2 2( 2 ) ( ( 1) ) 0p q p px y xy a bx y c dx b a p x b x

With solution:

1 2( ) ( ) ( )px q qy x x e C J x C Y x

Where1

2a

, bp

, 1

2| |dq

, 1

221 (1 ) 42

a cq

(a) if d<0,….

(b) if n , Y can be replaced by J (for d 0) and …

Where

2

20

( 1) 2 ! ( 1)

m m

mm

xJ x xm m

and cos( ) ( ) ( )( )

sin( )J x J xY x

Notice that 0 ( ) 0Y x

Anyway doing all the calculations: 2 2 0x u xu x u . In our mega formula, this means

1. 2 1pa bx which implies a=1 and b=0 or p=0 … 2. which (together with 1) implies

2 2qc dx x so c=0, d=1 and q= 1 So we get:

0 , 0 ,

12|1| 1

1 , 1 0 0

2 1

So 0 0

1 0 2 1 0( ) ( ) 0u x x e C J x C C J 2 2

0( ) ( )y x u C J

References:

1. http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Methodes-

Maths/white/math/s8/s8eqbess/s8eqbess.html

2. http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Methodes-

Maths/white/math/s8/ex8_1/ex8_1.html

Linda Fahlberg-Stojanovska

6

4. Show: 1 0( )J x dx J x C

2 0

2 00

2

2 20

2 4 6 8

2 4 6 8

( 1) 2 ! ( 0 1)

( 1)2 ( !)

12 1! 1! 2 2! 2! 2 3! 3! 2 4! 4!

m m

o mm

m m

mm

xJ xm m

xm

x x x x

2 1

1 2 10

2 4 6 8

2 4 6 8

( 1) 2 ! ( 1 1)

[1 ...]2 2 1! 1! 2 2 2! 2! 3 2 3! 3! 4 2 4! 4! 5

m m

mm

xJ xm m

x x x x x

2 11 2 1

0

2 2

2 10

2 4 6 8 10

3 5 7 9

2 4

4

( 1)d 2 ! ( 1 1)

( 1)2 ! ( 1 1) 2 2

1 1 1 1 1 ...2 2 4 2 1! 1! 2 6 2 2! 2! 3 8 2 3! 3! 4 10 2 4! 4! 5

1 12 1 2 2 2 2 1! 1

mm

mm

m m

mm

J x x C x dxm m

xCm m m

x x x x xC

x xC6 8 10

5 7 9

2 4 6 8 10

4 6 8 10

0

1 1 1 ...! 2 2 3 2 2! 2! 3 2 4 2 3! 3! 4 2 5 2 4! 4! 5

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

x x x

x x x x xC

J x

Linda Fahlberg-Stojanovska

7

5. Show: 3 1 22( )J x J x J x Cx

4 6 8

3 4 2 6 2 8 2

4 4 6 8

4 2 6 8 2

dx ...]2 2 (0!) 3 2 1 3 2 (1!) 4 3 2 4 2 (2!) 5 4 3

...2 2 (0!) 3 2 1 3 2 4 3 2 4 2 (2!) 5 4 3

x x xJ x C

x x x xC

2 4 6 8

1 2 4 6 8 [1 ...]2 2 1! 1! 2 2 2! 2! 3 2 3! 3! 4 2 4! 4! 5x x x x xJ x

So: 2 4 6 8

1 2 4 6 82 [ 1 ...]

2 1! 1! 2 2 2! 2! 3 2 3! 3! 4 2 4! 4! 5x x x xJ x

x

2 2 4 6 8

2 2 2 4 6 8

1[ ...]2 2 1 2 1! 1! 3 2 2 2! 2! 4 3 2 3! 3! 5 4 2 4! 4! 6 5x x x x xJ x

So: 2 4 6 8

2 3 4 6 8 ...2 1 2 1! 1! 3 2 2 2! 2! 4 3 2 3! 3! 5 4

x x x xJ x

Thus:

1 2

2 2 4 4

2 3 4 4

6 6 8 8

6 6 8 8

4 6 82

4 6 8

4 6

3

2

12 1! 1! 2 2 1 2 2! 2! 3 2 1! 1! 3 2

...2 3! 3! 4 2 2! 2! 4 3 2 4! 4! 5 2 3! 3! 5 4

1 1 10 ...2 2 0! 2! 3 3 2 1! 3! 4 4 2 2! 4! 51

4 2 0! 3!

J x J xx

x x x x

x x x x

x x xC x

x xC8

5 7

232 3

0

1 1 ...]6 2 1! 4! 8 2 2! 5!

( 1)2 ! ( 3 1)

mm

mm

x

C x dx J x dxm m