How does the substitution x = 1/z transform Bessel's Equation?

[Philip Land]

Hi!

When we want to look at different singular points for e.g Bessel's eq. $$u´´(x) + \frac{u'(x)}{x} + (1- \frac{n^2}{x^2})u(x)$$.

We usually evaluate the equation letting x= 1/z. But I don't algebraically see how such a substitution ends up with $$w´´(z) +( \frac{2}{z}- \frac{1}{z^2})z*w'(z) + \frac{1}{z^4}(1- n^2 z^2)w(z)$$.

Letting x= 1/z, and derive both sides gives $1/z^2 z'$ but I simply don't know how to go from u(x) to w(z) which is very central and should be very basic and just one microstep in long calculations lol.

 

[Orodruin]

It is assumed that $w(z) = u(1/z)$. Differentiate both sides with respect to $z$ (two times) and insert into Bessel's differential equation and you should end up with something that looks like your last expression. However, note that your second term cannot be correct as the prefactor needs to have only a single power of $z$. Also note that the derivatives of $w$ are with respect to $z$, not with respect to $x$. (Or rather, $w'$ denotes the derivative of $w$ as a function, i.e., the derivative with respect to the argument of the function.)