32.6. THE BESSEL EQUATIONS 629

and multiplying by a constant as above, you can obtain

Jm (x) =∞

∑k=0

(−1)k

k!Γ(k+m+1)

( x2

)2k+m=

∞

∑k=0

(−1)k

k!(k+m)!

( x2

)2k+m

as one solution. In fact, you could consider simply replacing m with −m in the above, butthis will not work out. It won’t work out roughly because Γ(k−m+1)=±∞ for k+1≤m.Thus the sum will reduce to

∞

∑k=m

(−1)k

k!Γ(k−m+1)

( x2

)2k−m

Changing the variable of summation to k = j+m, this becomes

(−1)m∞

∑j=0

(−1) j

( j−m)!Γ( j+1)

( x2

)2 j+m= (−1)m

∞

∑j=0

(−1) j

Γ( j−m+1) j!

( x2

)2 j+m

= (−1)m Jm (x)

so the new solution obtained by replacing m with −m is nothing more than (−1)m timesthe old solution. It follows that there is no way to obtain the general solution as a linearcombination of these two. The second solution must involve a logarithmic term and willtherefore, be unbounded near 0. However, it is convenient to define

J−m (x)≡ (−1)m Jm (x)

The way this is dealt with is to define the second solution as

Ym (x)≡ limν→m

Yν (x)

because the limit does exist for all x > 0.One other important consideration is easy to get which is that the solutions to Bessel’s

equation must oscillate about 0 like sines and cosines.

Proposition 32.6.1 Let y be a solution of the Bessel equation

x2y′′+ xy′+(x2−ν

2)y = 0

Then y has infinitely many zeros.

Proof: Change the independent variable to s where x = es. Thus, letting y(s) = y(x) ,it will suffice to show that s→ y(s) has infinitely many zeros. Doing the transformationyields the following differential equation for s→ y(s)

y′′ (s)+(e2s−ν

2)y(s) = 0

Obviously for all s large enough, e2s−ν2 > 1. Consider now the equation

z′′+ z = 0

The idea is to show that if a,b are successive zeros of z for a large enough that for s > a,e2s−ν2 > 1 it follows that y must have a zero in [a,b]. Since z has infinitely many zeros, itfollows that so does y.