Higher Engineering Mathematics, Sixth Edition

508 Higher Engineering Mathematics

which is valid providedvis not a negative integer

and whereAis an arbitrary constant.

Whenc=−v,

a 2 =

a 0

v^2 −(−v+ 2 )^2

=

a 0

v^2 −(v^2 − 4 v+ 4 )

=

−a 0

4 − 4 v

=

−a 0

22 (v− 1 )

a 4 =

a 0

[2^2 (v− 1 )][v^2 −(−v+ 4 )^2 ]

=

a 0

[2^2 (v− 1 )][2^3 (v− 2 )]

=

a 0

24 × 2 (v− 1 )(v− 2 )

Similarly, a 6 =

a 0

26 ×3!(v− 1 )(v− 2 )(v− 3 )

Hence,

y=w=

Bx−v

{

1 +

x^2

22 (v− 1 )

+

x^4

24 ×2!(v− 1 )(v− 2 )

+

x^6

26 ×3!(v− 1 )(v− 2 )(v− 3 )

+···

}

which is valid providedvis not a positiveinteger

and whereBis an arbitrary constant.

The complete solution of Bessel’s equation:

x^2

d^2 y

dx^2

+x

dy

dx

+

(

x^2 −v^2

)

y=0is:

y=u+w=

Axv

{

1 −

x^2

22 (v+1)

+

x^4

24 × 2 !(v+1)(v+2)

−

x^6

26 × 3 !(v+1)(v+2)(v+3)

+···

}

+Bx−v

{

1 +

x^2

22 (v−1)

+

x^4

24 × 2 !(v−1)(v−2)

+

x^6

26 × 3 !(v−1)(v−2)(v−3)

+···

}

(39)

The gamma function

The solution of the Bessel equation of Problem 10 may

be expressed in terms ofgamma functions.is the

upper case Greek letter gamma, and the gamma function

(x)is defined by the integral

(x)=

∫∞

0

tx−^1 e−tdt (40)

and is convergent forx> 0

From equation (40), (x+ 1 )=

∫∞

0

txe−tdt

and by using integration by parts (see page 420):

(x+ 1 )=

[

(

tx

)

(

e−t

− 1

)]∞

0

−

∫∞

0

(

e−t

− 1

)

xtx−^1 dx

=( 0 − 0 )+x

∫∞

0

e−ttx−^1 dt

=x(x) from equation (40)

This is an important recurrence relation for gamma

functions.

Thus, since (x+ 1 )=x(x)

then similarly, (x+ 2 )=(x+ 1 )(x+ 1 )

=(x+ 1 )x(x) (41)

and (x+ 3 )=(x+ 2 )(x+ 2 )

=(x+ 2 )(x+ 1 )x(x),

and so on.

These relationships involvinggamma functionsare used

with Bessel functions.

Bessel functions

The power series solutionof the Bessel equation may be

written in terms of gammafunctions as showninworked

problem 11 below.

Problem 11. Show that the power series solution

of the Bessel equation of worked problem 10 may

be written in terms of the Bessel functionsJv(x)

andJ−v(x)as:

AJv(x)+BJ−v(x)

=

(x

2

)v{ 1

(v+ 1 )

−

x^2

22 (1!)(v+ 2 )

+

x^4

24 (2!)(v+ 4 )

−···

}