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 $\widetilde{\nabla}^2f=0$ ?)
 A: 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).
A: It is in fact possible to find a smooth $u : D \to \mathbb{R}$ and a smooth $\varphi : \partial D \to \mathbb{R}$ such that
$$
\begin{cases}
L_s u = 0 &\text{in }D \\
u = \varphi &\text{on }\partial D
\end{cases}
$$
for infinitely many elliptic operators $\{L_s\}_s$.  The idea is to build elliptic operators out of elliptic operators in lower dimension.
Here's a very simple example for $D \subset \mathbb{R}^2$, indexed by $s \in (0,\infty)$.  Define $u$ and $\varphi$ via $u(x,y) = ax+b$ and $\varphi = u \vert_{\partial D}$.  For $s \in (0,\infty)$ consider the elliptic operator
$$
L_s = \partial_x^2 + s \partial_y^2.
$$
This is a constant coefficient operator that is elliptic since $s >0$.  A trivial computation reveals that $L_s u =0$ in $D$, and $u = \varphi$ on $\partial D$ by definition.
This idea obviously generalizes to higher dimensions.  Pick a second order elliptic operator $M = \sum_{i,j=1}^n a_{ij} \partial_i \partial_j$ and a smooth, nontrivial $v: \mathbb{R}^n \to \mathbb{R}$ satisfying $Mv =0$ in $\mathbb{R}^n$.  We then consider, for $s \in (0,\infty)$ again,
$$
L_s = M + s \partial_{n+1}^2,
$$
which is an elliptic operator on $\mathbb{R}^{n+1}$. Now suppose we have a bounded domain $D \subseteq \mathbb{R}^{n+1}$.  Then upon setting $u : D \to \mathbb{R}$ and $\varphi : \partial D \to \mathbb{R}$ via $u(x,y) = v(x)$ for $(x,y) \in D$ (now viewing $x \in \mathbb{R}^n$ and $y \in \mathbb{R}$) and $\varphi = u \vert_{\partial D}$, we readily deduce that $L_s u =0$ in $D$ and $u= \varphi$ on $\partial D$ for all $s >0$.
Similar tricks can be used if you want something other than zero on the right side of the equation, say $L_s u = f$.  Just pick a smooth pair $v,f : \mathbb{R}^n \to \mathbb{R}$ satisfying $Mv =f$ and then set $F: D \to \mathbb{R}$ via $F(x,y) =f(x)$ to get $L_s u = F$ in $D$ with $u = \varphi$ on $\partial D$.  These ideas will also work for boundary conditions other than Dirichlet and elliptic operators of higher order.
