1. ## Bessel differencial equation

please please can someone hep me on this as i dont have a clue where to start. I know its a ,ong question but if someone coul do itthen would really give me a bse to understand the topic

Edgar

Show that Jn(px) satisfies

x^2y'' + xy' + (p^2x^2 - n^2)y = 0;

and deduce

[x d/dx Jn(px)] (Subscript)x + ((p^2)x -n^2/x) Jn(px) = 0:

Show that the integral between l and 0 of xJn(px)Jn(qx) dx =

l/(q^2 - p^2) [pJn(ql)J'n(pl) - qJn(pl)J'n(ql)] ;

and, using l'Hopital's rule,

the integral between l and 0 of
xJn^2(px) dx =l^2/2 Jn'^2(pl) + 1 - (n^2/p^2l^2)Jn^2(pl)

Deduce that, if l is such that Jn(pl), Jn(ql) are both zero (i.e. pl and ql are both
roots of Jn(x)), then
the integarl between l and 0 of
xJn(px)Jn(qx) dx = 0; p doesnt equal q

2. Originally Posted by edgar davids
please please can someone hep me on this as i dont have a clue where to start. I know its a ,ong question but if someone coul do itthen would really give me a bse to understand the topic

Edgar

Show that Jn(px) satisfies

x^2y'' + xy' + (p^2x^2 - n^2)y = 0;
What is "Jn(px)"?

The equation,
$x^2y''+xy'+(p^2x^2-n^2)y=0$
Look like an analogue of the Bessel equation.
I can show that there exists a solution satisfiying the Frobenius method by shows that these functions are Legendre polynomials of orders 1 and 2. If that is what you are asking.

3. Jn(x) is a series solution to bessels equation

4. Originally Posted by edgar davids
Jn(x) is a series solution to bessels equation
Okay.

Given, the Bessel equation,
$x^2y''+xy'+(x^2-n^2)y=0$
The Bessel function of order $n\,$,
$J_n(x)$ satisfies the differencial equation.

We need to show that,
$J_n(px)$ satisfies the differencial equation.
$x^2y''+xy'+(x^2p^2-n^2)y=0$ (*)

Since,
$x^2J''_n(x)+xJ'_n(x)+(x^2-n^2)J_n(x)=0$
We have that,
$(px)^2J''_n(px)+(px)J'_n(px)+((px)^2-n^2)J_n(px)=0$
$\boxed{p^2x^2J''_n(px)+pxJ'_n(px)+(p^2x^2-n^2)J_n(px)=0}$

With the boxed equation we can show that $y=J_n(px)$ satisfies the differencial equation (*).

Note that,
$y=J_n(px)$
$y'=pJ'_n(px)$
$y''=p^2J''_n(px)$

Substitute that into (*) to get,
$p^2x^2J''_n(px)+pxJ'_n(px)+(x^2p^2-n^2)J_n(px)$
And this expression is equal to zero by the boxed equation.
Hence $J_n(px)$ satisfies (*).