Bessel Equation and Bessel fuctions

[unscientific]

We first express Bessel's Equation in Sturm-Liouville form through a substitution:

Next, we consider a series solution and replace v by m where m is an integer. We obtain a recurrence relation:

Then, since all these terms must be = 0,

Consider m = 0

First term vanishes and second term =

a₁x = 0

therefore, a₁ = 0. Then by recurrence relation above, a₁ = a₃ = a₅ = ... = a_2n-1 = 0

Then only the even series give non-zero coefficients, so we start with a₀ ≠ 0.

Consider m = 1

Second term vanishes, and first term =

a₀ = 0

By recurrence relation, this implies a_2n = 0

So we start the series with a₁ ≠ 0

Book's Explanation

I don't understand why a₀ = a₁ = ... = a_m-1 = 0

Surely, for any value of m that is ≠ 1 we have:

a_{1}x = \frac{m^2}{1 - m^2} a_0

This doesn't imply anything above? Also, how did they get the resulting Bessel functions? It looks wildly different from mine.

a₀ = a₁ = ... = a_m-1 = 0,

u = \sum_{n=m}^\infty a_{n} x^n = \frac {1}{2^n n!} x^n

 

[AlephZero]

From 8.44 you have $(k^2 - m^2)a_k = -a_{k-2}$

When $k = m$, that is $0a_k = a_{k-2}$.

Using $a_{k-2} = 0$ and the recurrence relation, $a_{k-4} = a_{k-6} = \dots = 0$.

 

[pasmith]

Start from
<br /> -m^2 a_0 + (1 - m^2)a_1 x + \sum_{k= 2}^{\infty} ((k^2 - m^2)a_k + a_{k-2})x^k = 0<br />

We need this equation to hold for all x. Thus looking at the constant term we must have
<br /> m^2 a_0 = 0<br />
Thus if m \neq 0 we must have a_0 = 0. On the other hand if m = 0 then m^2 a_0 = 0 whatever the value of a_0.

Looking at the coefficient of x^1 we have
<br /> (1 - m^2)a_1 = 0<br />
so that if m \neq 1 we are forced to take a_1 = 0. If m = 1 then (1 - m^2)a_1 = 0 whatever the value of a_1.

At order x^2 and higher we find that
(k^2 - m^2)a_k + a_{k-2} = 0
from which it follows that if a_{k-2} = 0 then we must have a_k = 0 unless k = m, in which case we can choose a_m as we see fit. But this only happens if m \geq 2, so that m \neq 0 and m \neq 1, and by the above we must take a_0 = a_1 = 0. The recurrence relation then tells us that a_0 = a_1 = \dots = a_{m - 1} = 0.

We also have
((m + 1)^2 - m^2)a_{m + 1} + a_{m - 1}<br /> = (2m + 1)a_{m+1} + 0 = 0
so that a_k = 0 for k = m + 2n + 1, n \geq 0.

If k = m + 2n, n \geq 0, then
<br /> ((m + 2n)^2 - m^2)a_{m + 2n} + a_{m + 2(n-1)} = 0<br />
so that
<br /> a_{m + 2n} = - \frac{a_{m + 2(n-1)}}{(m^2 + 4mn + 4n^2) - m^2} <br /> = - \frac{a_{m + 2(n-1)}}{4n(m + n)}<br />
which can be solved to obtain
<br /> a_{m + 2n} = \frac{(-1)^n}{4^n n!(m + 1) \times \dots \times (m + n) }a_m = \frac{(-1)^n m!}{2^{2n} n!(m + n)!}a_m<br />
(looking at a_{m + 2}, a_{m + 4}, and a_{m + 6} establishes the pattern which can then be proven by induction on n) from which the given series for J_m(x) follows.