# Why consider the Neumann functions $N_\nu$, while having the basis $J_\nu(k\rho)$?

Let's consider a point-charge $$q$$ located in the origin, generating the following potential in cylindrical coordinates $$$$\Phi(\rho,\varphi,z)=\frac{q}{\rho^2 + z^2}$$$$ The problem is how to write it as an expansion of Bessel functions. I thought the Hankel transform could work, as $$\left\lbrace J_{\nu}(k\rho) \right\rbrace_{k\in \mathbb{R}}$$ is a basis for the space of square-integrable functions over the positive real numbers $$\mathbb{R}_{>0}$$. The expansion in terms of this is: $$$$\Phi(\rho,\varphi,z)=\sum_{\nu=-\infty}^{+\infty} \int_{-\infty}^{+\infty} dk A_{\nu}(k) e^{kz}e^{i\nu\varphi}J_\nu(k\rho)$$$$ For each slice at constant $$z$$, we have a basis in the plane $$(\rho,\varphi)$$. But we've assumed the Neumann functions don't appear in the expansion! We did that because they do not form any basis. But in this specific case, they could come in handy, because $$|N_\nu(k\rho)|\rightarrow \infty$$ as $$\rho\rightarrow0$$, and so does the potential.

Then my question is: why do we bother considering $$N_\nu$$? Do they really appear in any problem? Are they useful to express divergent potentials like the one of the point-charge?

Edit: Instead of a point-charge, let's consider now an infinite line charge with uniform density $$\lambda$$, with the potential $$$$\Phi(\rho,\varphi,z) = \frac{-\lambda}{2\pi\rho} \log\left( \frac{\rho}{\rho_0} \right)$$$$ Now the potential diverges for every $$z$$. As it was pointed out, the previous case doesn't present any trouble taking $$z\neq 0$$, but now we have a divergent (possibly non-square-integrable) function in $$(\rho,\varphi)$$ for every $$z\in\mathbb{R}$$ and there is no escape from it. Does $$N_\nu$$ come handy for this new situation?

• Neumann functions do form a basis, but this basis is orthogonal to that consisting of Bessel functions. Together, in the form of Hankel functions, they are useful for representing running waves — e.g. expanding cylindrical waves. – Ruslan May 12 '20 at 21:01

It is not true in general that $$|\Phi|\to\infty$$ as $$\rho\to 0$$. Fixing some nonzero $$z$$, we have that $$\Phi(0,\varphi,z)=\frac{q}{z^2}$$. And, with some nonzero $$z$$, $$\Phi$$ is square-integrable over the positive real numbers, so we can indeed write it as a linear combination of Bessel functions.
If, on the other hand, we take $$z=0$$, then $$\Phi$$ is no longer square-integrable over the positive real numbers, meaning that we are no longer guaranteed to be able to write $$\Phi(\rho,\varphi,0)$$ as a linear combination of Bessel functions.
• That's totally right! Thank you, I haven't realized. But I still have some confusion about the usefulness of $N_\nu$. I'll edit my queston to consider what happens with the infinite line charge, where we can no longer take $z\neq 0$ to make it square-integrable – adiselann May 12 '20 at 21:00