I recently tumbled over a statement in a geophysics paper (PDF here). They have a wave equation which they formulate as
$$ \frac{1}{v_0}\frac{\partial^2}{\partial t^2} \begin{pmatrix}p \\ r\end{pmatrix} = \begin{pmatrix} 1+2\epsilon&\sqrt{1+2\delta}\\\sqrt{1+2\delta} &1\end{pmatrix} \begin{pmatrix}G_{\bar x \bar x}+G_{\bar y \bar y}&0\\0&G_{\bar z \bar z} \end{pmatrix} \begin{pmatrix} p\\r \end{pmatrix}\tag{20} $$
and they claim that
To achieve stability, the rotated differential operators $G_{\bar x \bar x}$, $G_{\bar y \bar y}$, and $G_{\bar z \bar z}$ should be self-adjoint and nonpositive definite as are the second-order derivative operators ($\tfrac{\partial^2}{\partial x^2}$, $\tfrac{\partial^2}{\partial y^2}$ and $\tfrac{\partial^2}{\partial z^2}$).
(see eq. 14 and statement under eq. 20). They also claim that self-adjointness and nonpositiveness of the differential operators of the wave equation are necessary to conserve the energy in this system, and that if they were not self-adjoint, numerical instabilities occur.
We have solved the problem by introducing the self-adjointness to the operator matrices in equation 20 to make sure that energy is conserved during the wave propagation to avoid amplitude blowup in the modeling.
In this case the wave equation consists of two coupled elliptical PDEs. What happens in general, when some operators are not bounded and linear?
Unfortunately I don't have the mathematical background to understand this statement.