Question. Let $f: \mathbf{R} \to \mathbf{R}$ be an analytic function. Is there a rotationally symmetric harmonic function $u$ on the circular cylinder $D \times \mathbf{R}$ so that along the axis of rotation $\{ (0,0) \} \times \mathbf{R}$, $u = f$ and $\frac{\partial u}{\partial x} = 0 = \frac{\partial u}{\partial y}?$

• If necessary, feel free to restrict the behavior of $f$ for $\lvert z \rvert$ large, for example to impose some kind of decay.
• In cylindrical polar coordinates $(\rho,\theta,z)$, the function satisfies the PDE $ \frac{\partial^2 u}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial u}{\partial \rho} + \frac{\partial^2 u}{\partial z^2} = 0$. From this point of view the question looks like an initial value problem, with initial values $u(\rho = 0,z) = f(z)$ and $\frac{\partial u}{\partial \rho}(\rho = 0,z) = 0$. The question would then become: is the Cauchy–Kovalevskaya solution defined up to $\rho = 1$? However, the PDE is singular at the boundary $\rho = 0$.