this is my first post here. I knocked my head on a differential equation yesterday, this one:
$$
\frac{12 \nu}{x^2} \frac{S(x)''}{S(x)} = -\lambda^2
$$
Where $nu$ is a constant.
The book says the solution is a Bessel function. I have had a little exposure to the Bessel equation and functions, I know how a Bessel equation looks like and worked out the solution using the series method during a small part of a course on quantum mechanics. My problem here is that i don't see how this equations is a Bessel equation and therefore why the solution is a Bessel function of order $1/4$. I was even thinking to figure out the solution with the series method and see if that one looks like that Bessel funcion, but I think that would be stupid since I already did for the general case. It's something that should be automatic, like: this is a Bessel equation $\Rightarrow$ the solution is a Bessel function.
Some help? how can I see that is a Bessel equation? and why the solution is a Bessel function of order $1/4$?
Thank you fo any help in advance.
Best Answer
A transformed form of Bessel differential equation is : $$\frac{d^2y}{dx^2}+(2p+1)x^{-1}\frac{dy}{dx}+(\alpha^2x^{2r-2}+\beta^2x^{-2})y=0$$ which general solution is : $$y(x)=x^{-p}\left(C_1 J_{q/r}(\frac{\alpha}{r}x^r) +C_2 Y_{q/r}(\frac{\alpha}{r}x^r) \right)\quad\text{where}\quad q=\sqrt{p^2-\beta^2}$$ $J$ and $Y$ denote the Bessel functions of first and second kind respectively.
See : http://mathworld.wolfram.com/BesselDifferentialEquation.html (Eqs.3-5).
This formula can be derived from the standard Bessel differential equation thanks to changes of variable and function, but this is a boring calculus. Nevertheless, any relentless one can do it.
Now, comparing the above ODE to the presently considered ODE : $$\frac{d^2S}{dx^2}+\left(\frac{\lambda^2}{12\nu}x^2 \right)S=0$$ we see that they are the same with conditions : $$\begin{cases} y(x)=S(x)\\ 2p+1=0\\ \beta=0\\ 2r-2=2\\ \alpha^2=\frac{\lambda^2}{12\nu} \end{cases} \quad\to\quad \begin{cases} p=-\frac{1}{2}\\ \beta=0\\ r=2\\ \alpha=\frac{\lambda}{\sqrt{12\nu}}\\ q=\sqrt{p^2-\beta^2}=\frac{1}{2}\quad\to\quad \frac{q}{r}=\frac{1}{4} \end{cases} $$ Thus : $$S(x)=\sqrt{x}\left(C_1 J_{1/4}\left(\frac{\lambda}{2\sqrt{12\nu}}x^2\right) +C_2 Y_{1/4}\left(\frac{\lambda}{2\sqrt{12\nu}}x^2\right) \right)$$