lenear dependance on a segment

The second order linear ordinary DE...

$\displaystyle y^{''} + \frac{y^{'}}{x} + (1-\frac{\nu^{2}}{x^{2}})=0$ (1)

... is known as 'Bessel equation'. As I said in the previous post, the case in which $\nu$ is not an integer is easier to manage, so that we start with it. First we observe that (1) is second order linear so that its general solution is of the type...

$\displaystyle y(x)= c_{1}\ \phi(x) + c_{2}\ \psi(x)$ (2)

... where $\phi(*)$ and $\psi(*)$ are two independent solution of (1). We can start with the hypothesis that a solution of (1) can be written as...

$\displaystyle y(x)= x^{\nu}\ \lambda(x)= x^{\nu}\ \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (3)

... where $\lambda(*)$ is an analytic function, so that (1) becomes...

$\displaystyle \sum_{n=0}^{\infty} a_{n}\ x^{n} = - \sum_{n=2}^{\infty} n\ (n-1)\ a_{n}\ x^{n-2} - (1+2\ \nu)\ \sum_{n=1}^{\infty} n\ a_{n}\ x^{n-2}$ (4)

In can be easily verified that all the $a_{n}$ with n odd are nul, so that $\lambda(*)$ is an even function. The coefiicient with n even are computed recursively starting from $a_{0}$ which is arbitrary...

$\displaystyle a_{0}= -4\ (1+\nu)\ a_{2} \implies a_{2} = - \frac{a_{0}}{4\ (1+\nu)}$

$\displaystyle a_{2}= -4\ 2\ \ (2+\nu)\ a_{4} \implies a_{4} = - \frac{a_{2}}{4\ 2\ (2+\nu)}$

...

$\displaystyle a_{2n}= -4\ (n+1)\ (n+1+\nu)\ a_{2n+2} \implies a_{2 (n+1)} = - \frac{a_{2n}}{4\ (n+1)\ (n+1+\nu)}$ (5)

... so that we can write...

$\displaystyle y(x)= a_{0}\ x^{\nu}\ \{1+ \sum_{n=1}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (1+\nu)\ (2+\nu) ... (n+\nu)}\}$ (6)

Some considerations...

a) in (6) there is a term $a_{0}$ and that's justified by the fact that a solution multiplied by a constant is also a solution...

b) in (1) there is $\nu^{2}$ and that means two different solution like (6), one with $\nu >0$ and the other with $\nu<0$ ...

The b) permits us to whrite...

$\displaystyle y(x)= c_{1}\ J_{\nu} (x) + c_{2}\ J_{-\nu} (x)$ (7)

At this point we have to identify $a_{0}$ . At this scope we define the 'factorial function' as ...

$\displaystyle x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt$ (8)

Among the properties of this function the following is very important...

$\displaystyle (1+x)!= (1+x)\ x!$ (9)

... so that we can set $a_{0}= \frac{1}{\nu!}$ for $\nu >0$ and $a_{0}= \frac{1}{(-\nu)!}$ for $\nu <0$ obtaining...

$\displaystyle J_{\nu} = x^{\nu}\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (n+\nu)!}$

$\displaystyle J_{-\nu} = x^{-\nu}\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{x}{2})^{2n}}{n!\ (n-\nu)!}$ (10)

It is important to remember that all that is true only if $\nu$ isn't an integer. Finally an example of the J functions for $\nu=\frac{1}{2}$ ...

http://digilander.libero.it/luposabatini/MHF95.bmp

Kind regards

$\chi$ $\sigma$