3
$\begingroup$

So I have been given a formula for the spherical Bessel functions in the form of $$ j_\ell(x)=(-x)^\ell \left(\frac{1}{x}\frac{d}{dx}\right)^\ell\frac{\sin(x)}{x} $$ which is Rayleigh's formula. I've been asked to show this satisfies Helmholtz's equation, however, I don't know how to diffentiate the middle part (containing the differential operator). To differentiate this, do I use the product rule as I normally would taking the $-x $, $\frac{1}{x}\frac{d}{dx}$ and $\frac{\sin(x)}{x}$ separately and pulling down the powers?

Many thanks in advance.

$\endgroup$

1 Answer 1

1
$\begingroup$

Quoting Wikipedia, when solving the Helmholtz equation in spherical coordinates by separation of variables, the radial equation has the form: $$ x^2 y'' + 2x y' + \left[x^2-n(n+1)\right] y = 0. \tag{1}$$ For $n=0$, it is straightforward to check that $y_0=\operatorname{sinc}(x)$ is a solution.

Suppose now that a solution of $(1)$ has the form $y=(-x)^n f(x)$. Then $f(x)$ satisfies the ODE:

$$ x\, f + 2(1+n)\,f' +x\, f'' = 0.\tag{2} $$ Can you prove now that a solution of $(2)$ is given by $\left(\frac{1}{x}\frac{d}{dx}\right)^n y_0$?

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .