# lenear dependance on a segment

• Nov 27th 2010, 02:46 AM
transgalactic
lenear dependance on a segment5
$y''+\frac{1}{x}y'+(1-\frac{a^2}{x^2})y=0$
y1(1)=3 y2(1)=-8.1
y1(1)=3 y2'(1)=5.4

i need to deside:
A.
y1 and y2 are lenear dependant on (0,+infinity)
B.
y1 and y2 are not lenear dependant on (0,+infinity)
C.
we cant know if y1 and y2 are dependant only by a single point

i don tknow how to solve this equation
in a similar question i was told to use y=x^a
but here its not working
i am no get a certain a for which the equation is zero
• Nov 27th 2010, 04:05 AM
chisigma
The solutions of the linear second order DE...

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

... are called Bessel functions of order a. The parameter a can be real or integer [or even complex...] and the approach is different in the two cases. Curiously enough the case a 'non integer' is more easy to be 'attached' and the two independent solution are of the type $\displaystyle J(x)= x^{\pm a}\ \sum_{n=0}^{\infty} \alpha_{n}\ x^{n}$...

Kind regards

$\chi$ $\sigma$
• Nov 27th 2010, 04:08 AM
transgalactic
what two cases
how to eproach this one
• Nov 28th 2010, 07:28 AM
chisigma
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

$\chi$ $\sigma$