Wronskian of complex second order linear differential equation I have asked this question in Physics stack exchange earlier and am reposting here. While studying analogue models of gravity I have come across a differential equation of the form: \begin{align} \frac{d^2y}{dz^2} + \omega^2 (z)~ y(z) = 0 \end{align} where $z$ is a complex variable and $\omega(z), y(z)$ are functions of the complex variable $z$. * Am I right in saying that this equation models a dissipative system whenever $Im(\omega) \neq 0$ ? Secondly, if the above equation contained only real variables and their real valued functions, we could have defined a Wronskian. * Can I construct a Wronskian similarly here? If yes, then that Wronskian should be a constant function of $z$, like a 'conserved quantity'. * Does this contradict with the fact that the above differential equation is dissipative?

To add to @Yiorgos S. Smyrlis's comment: The Wronskian relates two linearly independent solutions, so it's not a conserved quantity in the usual sense, i.e. $F(y,y_z) = c$ which should hold for _every_ solution $y$.

As for the dissipativity: this is somewhat open for interpretation, but if you would define a Hamiltonian in the 'usual' way, i.e. \begin{equation} H(y',y,z) = \frac{1}{2} y'^2 + \frac{1}{2}\omega(z)^2 y^2, \end{equation} then \begin{equation} \frac{\text{d}}{\text{d} z} H = y'\left(y'' + \omega(z)^2 y\right) + \frac{1}{2} \omega \omega' y^2 = \frac{1}{2} \omega \omega' y^2. \end{equation} The only way you could call this 'dissipative' if the right hand side is negative. However, as you're dealing with complex valued functions, the concept of 'negative' doesn't really exist anymore, since the complex numbers don't have a natural ordering.