# Physical accuracy of Hankel function solution to cylindrical voltage wave propagation?

The well-known solution to an outward-travelling wave in cylindrical coordinates (in an unbounded medium) is the Hankel function of the first kind:

$$H^{\left(1\right)}_n (\rho,t) = \left(J_n (\rho,t) + iY_n (\rho,t)\right) e^{i\omega t}$$

I am interested in the electrical case, specifically a TEM ($TM_{00}$) mode guided by two parallel plates embedded in a vacuum, where a voltage is excited at the origin, say

$$V(0,t) = e^{i\omega \left(t + \pi/2\right)}$$

According to the well-known solution, the voltage of the travelling wave solution at $(\rho = 0, t = 0)$ should then diverge instead of being equal to unity. Even close to the origin, the solution easily exceeds unity. However, there is no physical mechanism present by which the voltage magnitude may exceed that of the excitation voltage. Furthermore, applying this single boundary condition to the expression

$$V(\rho,t) = \left(A J_n (\rho,t) + iBY_n (\rho,t)\right) e^{i\omega t}$$

yields B = 0, resulting in a standing wave expression (which is obviously not correct).

It therefore appears that the Hankel function is not an appropriate solution in this case. Is there a more appropriate expression? Otherwise, how can this apparent problem be solved?

Your problem stems from the fact that you're trying to define a boundary condition on the "boundary" $$\rho = 0$$, which is not, properly speaking, a surface. This same problem arises when you try to solve Laplace's equation $$\nabla^2 V = 0$$ assuming cylindrical symmetry (i.e., $$V$$ is a function of $$\rho$$ only.) The general solution in this case is $$V(\rho) = A \ln \rho + B,$$ and demanding that $$V(0) = C \neq 0$$ is impossible, as there are no values of $$A$$ and $$B$$ that satisfy these constraints. We can fix this problem by demanding that $$V(R_0) = C$$ for some radius $$R_0$$; in this case, a solution for the constants $$A$$ and $$B$$ does exist, with $$B = C - A \ln R_0$$: $$V(\rho) = C + A \ln \left( \frac{\rho}{R_0} \right).$$ But, of course, this solution is not well-behaved as $$R_0 \to 0$$. Simply put, a solution to $$\nabla^2 V = 0$$ with $$V(0) = C \neq 0$$ does not exist.
The same thing would happen if we try to solve Laplace's equation in spherical coordinates with $$V(\infty) = 0$$ while demanding that $$V(0) = C \neq 0$$. Again, the issue is that the general solution does not allow for $$\lim_{r \to 0} V(r)$$ to be any finite number other than zero.