Mathematical Methods for Physics and Engineering : A Comprehensive Guide

18.5 BESSEL FUNCTIONS

To determine the required boundary conditions for this result to hold, let us

consider the functionsf(x)=Jν(αx)andg(x)=Jν(βx), which, as will be proved

below, respectively satisfy the equations

x^2 f′′+xf′+(α^2 x^2 −ν^2 )f=0, (18.85)

x^2 g′′+xg′+(β^2 x^2 −ν^2 )g=0. (18.86)

Show thatf(x)=Jν(αx)satisfies (18.85).

Iff(x)=Jν(αx) and we writew=αx,then

df

dx

=α

dJν(w)

dw

and

d^2 f

dx^2

=α^2

d^2 Jν(w)

dw^2

.

When these expressions are substituted into (18.85), its LHS becomes

x^2 α^2

d^2 Jν(w)

dw^2

+xα

dJν(w)

dw

+(α^2 x^2 −ν^2 )Jν(w)

=w^2

d^2 Jν(w)

dw^2

+w

dJν(w)

dw

+(w^2 −ν^2 )Jν(w).

But, from Bessel’s equation itself, this final expression is equal to zero, thus verifying that
f(x) does satisfy (18.85).

Now multiplying (18.86) byf(x) and (18.85) byg(x) and subtracting them gives

d

dx

[x(fg′−gf′)] = (α^2 −β^2 )xfg, (18.87)

wherewehaveusedthefactthat

d

dx

[x(fg′−gf′)] =x(fg′′−gf′′)+(fg′−gf′).

By integrating (18.87) over any given rangex=atox=b,weobtain
∫b

a

xf(x)g(x)dx=

1

α^2 −β^2

[

xf(x)g′(x)−xg(x)f′(x)

]b

a

,

which, on settingf(x)=Jν(αx)andg(x)=Jν(βx), becomes
∫b

a

xJν(αx)Jν(βx)dx=

1

α^2 −β^2

[

βxJν(αx)J′ν(βx)−αxJν(βx)Jν′(αx)

]b

a

.

(18.88)

Ifα=β, and the interval [a, b] is such that the expression on the RHS of (18.88)

equals zero, then we obtain the orthogonality condition (18.84). This happens, for

example, ifJν(αx)andJν(βx) vanish atx=aandx=b,orifJ′ν(αx)andJν′(βx)

vanish atx=aandx=b, or for many more general conditions. It should be

noted that the boundary term is automatically zero at the pointx=0,asone

might expect from the fact that the Sturm–Liouville form of Bessel’s equation

hasp(x)=x.

Ifα=β, the RHS of (18.88) takes the indeterminant form 0/0. This may be