Can a function satisfy two different elliptic PDEs? Let $D\subset \mathbb{R}^2$ be a simply connected finite domain and consider two different 2nd order elliptic operators $\nabla^2$ and $\widetilde{\nabla}^2$. We set some continuous arbitrary function $\Omega_{\mathsf{B}}$ defined on the boundary of $D$.
Apart from the case of $\widetilde{\nabla}^2$ and ${\nabla}^2$ being linearly dependent, is there an $\Omega$ such that
$$
\begin{align}
\nabla^2\Omega=0\:\:\:\text{with}\:\:\:\Omega\Big|_{\partial D}=\Omega_{\mathsf{B}}\\
\widetilde{\nabla}^2\Omega=0\:\:\:\text{with}\:\:\:\Omega\Big|_{\partial D}=\Omega_{\mathsf{B}}
\end{align} 
$$
By different I mean, for instance, that writing $\nabla^2=a_{ij}(x,y)\:\partial^2_{ij}$ and $\widetilde{\nabla}^2=\tilde{a}_{ij}(x,y)\:\partial^2_{ij}$  we have that $a_{ij}\neq\tilde{a}_{ij}$ for at least one point in the interior of $D$.
(As a special case I wish to take $\nabla^2$ to be the Laplacian, so the question would be rephrased as: given an elliptic $\widetilde{\nabla}^2$, can we find a harmonic function $f$ with fixed boundary value such that $\widetilde{\nabla}^2f=0$ ?)

Consider a linear function $f.$ Notice that it is in the kernel of any second order operator.
Otherwise, consider the operator $L=\nabla^2 - \widetilde{\nabla}^2.$  You are looking for a solution with vanishing boundary condtions.
If this difference is elliptic, then existence and uniqueness hold, and there is no such function. However, unless I am confused, if the difference of the operators is not elliptic, there may be a solution (the wave equation comes to mind).
