If $(D^2+2\zeta \cdot D + q) r= f$ has solution $r$, then what does it mean?
$D^2$ is the Laplacian, $D=\nabla$, $\zeta \in \mathbb{C}^n$, $q \in L^{\infty}(\Omega)$.
I wonder if it's the same as in DEs, that plugging in $r$ will make the equation have equal sides?
Would it be wrong to then say that $r=\frac{f}{(D^2+2\zeta \cdot D +q)}$?