# Bessel's equation and second order DE

• Apr 19th 2010, 05:48 AM
Silverflow
Bessel's equation and second order DE
Hi all!
I've hit a wall on the question I'm working on. Can someone look over my work and see where I may have possible gone wrong?

The question is:
Show that if $\displaystyle y(x)$ is a solution of Bessel's equation of order p, then, $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution of $\displaystyle z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0$.

Here is my working.
Considering that $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution, then:
$\displaystyle w'(z)=-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c}$ and
$\displaystyle w''(z)=(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$

Subing into $\displaystyle L[w]$,
$\displaystyle L[w]=z^2(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$+$\displaystyle (2c+1)z(-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c})+[a^2b^2z^{2b}+(c^2-p^2b^2)]z^{-c}y(az^b)$.

I simplify to find:
$\displaystyle bz^{-c}(a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)).$
As$\displaystyle L[w]=0$, that means $\displaystyle a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)=0$.
Making the substitution of $\displaystyle x=az^b$
I get the following:
$\displaystyle x^2y''(x)+bxy'(x)+b(x^2-p^2)y(x)=0$.
That equation is soo close to Bessel's equation. I go through my working and I can't find why I have those b's are there.

Thank you very much for your time.
• Apr 19th 2010, 06:40 AM
Jester
Quote:

Originally Posted by Silverflow
Hi all!
I've hit a wall on the question I'm working on. Can someone look over my work and see where I may have possible gone wrong?

The question is:
Show that if $\displaystyle y(x)$ is a solution of Bessel's equation of order p, then, $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution of $\displaystyle z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0$.

Here is my working.
Considering that $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution, then:
$\displaystyle w'(z)=-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c}$ and
$\displaystyle w''(z)=(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$

Subing into $\displaystyle L[w]$,
$\displaystyle L[w]=z^2(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$+$\displaystyle (2c+1)z(-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c})+[a^2b^2z^{2b}+(c^2-p^2b^2)]z^{-c}y(az^b)$.

I simplify to find:
$\displaystyle bz^{-c}(a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)).$
As$\displaystyle L[w]=0$, that means $\displaystyle a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)=0$.
Making the substitution of $\displaystyle x=az^b$
I get the following:
$\displaystyle x^2y''(x)+bxy'(x)+b(x^2-p^2)y(x)=0$.
That equation is soo close to Bessel's equation. I go through my working and I can't find why I have those b's are there.

Thank you very much for your time.

You're missing a term in the second derivative. There should be two y' terms - you only have one. I got (when combined)

$\displaystyle ab(b-2c-1)z^{b-c-2} y'\left(a z^b\right)$
• Apr 19th 2010, 04:00 PM
Silverflow
It works!! Thank you very much!
• Apr 20th 2010, 02:44 AM
Silverflow
The next part of the question is to use what it given above, and find a general solution for

$\displaystyle x^2y''+(1/8+x^4)y=0$

I have no idea how to do this. I can try to relate coefficients of this DE to the one I mentioned in my first thread, but that doesn't work. Substituting $\displaystyle x=az^b$ doesn't seem to work either...
• Apr 20th 2010, 04:28 AM
Jester
Quote:

Originally Posted by Silverflow
The next part of the question is to use what it given above, and find a general solution for

$\displaystyle x^2y''+(1/8+x^4)y=0$

I have no idea how to do this. I can try to relate coefficients of this DE to the one I mentioned in my first thread, but that doesn't work. Substituting $\displaystyle x=az^b$ doesn't seem to work either...

Choose $\displaystyle a, b, \; \text{and}\; c$ in

$\displaystyle z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0$

so that you obtain

$\displaystyle x^2y''+(1/8+x^4)y=0$.

This then will give you

$\displaystyle w = z^{-c} y(az^b).$
• Apr 20th 2010, 04:31 AM
Silverflow
Thanks very much.