Does the Laplace's equation $\nabla^2 u=0$ could have smooth compact-supported solutions $\in C_c^\infty$? Any example?
In this answer an user named @NinadMunshi explain that the following function $f:\Bbb{R}^3\to\Bbb{R}$
$$f(x,y,z) = \begin{cases}\exp\left[\frac{-1}{R^2-x^2-y^2-z^2}\right] & x^2+y^2+z^2 < R^2 \\ 0 & x^2+y^2+z^2 \geq R^2\end{cases}$$
Then for $k\in\Bbb{R}^3$ with $|k|=1$, we have that
$$E_i(x,y,z,t) = f(x-ck_xt,y-ck_yt, z-ck_zt)$$
satisfies the wave equation $\frac{1}{c^2}\frac{\partial^2}{\partial t^2}E=\nabla^2 E$.
I would like to know if its possible also for the Laplace's Equation $\nabla^2 u = 0$ to stand smooth bump-like solutions $\in C_c^\infty$, at least in one coordinate.
I believe that if the function is complex-valued $u(\vec{x}) \in \mathbb{C}$ then no harmonic solution could be compact-supported due the Liouville's theorem (complex analysis), at least if is not defined piecewise as the example for the wave equation, but I don't have any intuition of what would happen if the function is real-valued.
Hope you can answer giving a basic example: piecewise solution are allowed, but must be solving the differential equation in the whole domain and not only where the non-zero piecewise section is defined, so it behave as a properly solution to the Laplace's equation which has a smooth bump function behavior (as it does the example for the wave equation).
Motivation
Recently I learned that a scalar 2nd order ordinary differential equation (ODE) require to have a point in time where is locally non-Lipschitz in order to been able of having solution with a finite extinction time (details here, reference on this paper), which brakes uniqueness of solutions.
Since this will apply for scalar smooth bump functions, I would like to know why this singular point is not required on PDEs for having a smooth-bump behavior on any coordinate as is shown for the wave equation example, so I would like to know how the arise for the Laplace's equation, or if instead they aren't possible.
This answer is just for closing this question, since it was given on the comments by @Aruralreader:
Due the Maximum principle apply for the Laplace equation, it implies that the maximum of the function must happen on the "boundaries", so if the solution is a smooth-bump function it will imply that the maximum is on the edges were the function value is zero, so the only possible solution fulfilling both situations is the zero function.