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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.

Joako
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    The answer is no: solutions of Laplace’s equation are analytic, and no nonzero, smooth, compactly supported function $u$ can be analytic at the boundary of its support. This is because all derivatives must be zero outside the support of $u$, and thus by continuity, all derivatives of $u$ are zero on the boundary of the support, which means the Taylor expansion of $u$ on the boundary of its support is identically zero. If $u$ was analytic there, then $u$ would be zero on an open ball containing some piece of its support, which is impossible. – User8128 Jul 04 '22 at 21:03
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    Have a look at the maximum principle for Laplace’s equation. – A rural reader Jul 05 '22 at 00:24
  • @Aruralreader Thanks for commenting. I have reviewed in Wikipedia the Maximum Principle for the Laplace's equation, but how is described there, I don't understand Why it not apply also for the wave equation?.. maybe is that the case and the example I give is mistaken? (I made it work in $\mathbb{R}^{1+1}$, but I cannot make it work for $\mathbb{R}^{2+1}$). Hope you can elaborate into this on an answer. – Joako Jul 07 '22 at 03:21

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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.

Joako
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