Metric tensor from hyperbolic PDE It is clear that when a differential equation is composed of the second partial derivatives only, it could be written in the form
$$
g^{\mu\nu} \frac{\partial^2 \psi}{\partial x^\mu \partial x^\nu} = 0
$$
with $g^{\mu\nu}$ denoting the metric tensor. Is there any general way of obtaining coefficients of $g^{\mu\nu}$ from hyperbolic PDE containing derivatives of lower orders? Probably by means of changing variables.
My naive intuition suggests that the dispersion brought by the lower order terms should be "transferable" to the metric tensor (possibly nonconstant in time and space).
 A: Suppose that we have an equation of the form
$$
g^{\rho \sigma} \frac{\partial^2 \psi}{\partial x^\rho \partial x^\sigma} + A^\tau \frac{\partial \psi}{\partial x^\tau} = 0 \tag{1}
$$
and we suspect that under a coordinate transformation it is secretly the wave equation.  In general, the scalar d'Alembertian in terms of coordinate derivatives is
$$
\Box \psi=g^{\rho \sigma}(\partial_\rho \partial_\sigma \psi -\Gamma^\tau {}_{\rho\sigma}\partial_\tau \psi)
$$
which means that if equation (1) is truly the wave equation, we must have $A^\tau = - g^{\rho \sigma} \Gamma^{\tau} {}_{\rho \sigma}$ for this to work.  But it is also not too hard to show that
$$
-g^{\rho \sigma} \Gamma^{\tau} {}_{\rho \sigma} = \frac{1}{\sqrt{|g|}} \partial_\lambda (\sqrt{|g|} g^{\tau\lambda}) 
$$
and since the metric in terms of the coordinates $x^\mu$ is known (it can be "read off" of the higher-derivative term) then it should be easy to simply calculate this quantity and see if it is equal to $A^\tau$.  It is also easy to envision situations where we have $A^\tau \neq - \frac{1}{\sqrt{|g|}} \partial_\lambda (\sqrt{|g|} g^{\tau\lambda})
$, and so the equation (1) is not actually equivalent to the wave equation in some set of coordinates.
The one wrinkle to this argument is that we could allow for a conformal rescaling of equation (1), so that we instead have
$$
\tilde{g}^{\rho \sigma} \frac{\partial^2 \psi}{\partial x^\rho \partial x^\sigma} + \tilde{A}^\tau \frac{\partial \psi}{\partial x^\tau} = 0 \tag{2}
$$
where $\tilde{g}^{\rho \sigma} = \Omega^2 g^{\rho \sigma}$ and $\tilde{A}^\tau = \Omega^2 A^\tau$ for some scalar function $\Omega > 0$.  This is entirely equivalent to (1), but allows us a bit more freedom to find a set of coordinates.  Going through the same logic as above,  equation (2) will be equivalent to the wave equation if we have
$$
\tilde{A}^\tau = - \frac{1}{\sqrt{|\tilde{g}|}} \partial_\lambda \left( \sqrt{|\tilde{g}|} \tilde{g}^{\tau\lambda} \right) \\
\Omega^2 A^\tau = - \frac{\Omega^{D}}{\sqrt{|g|}} \partial_\lambda \left( \Omega^{-D+2} \sqrt{|g|} g^{\tau\lambda} \right) \\
A^\tau = - \frac{1}{\sqrt{|g|}} \partial_\lambda (\sqrt{|g|} g^{\tau\lambda}) + \left( D-2 \right) g^{\tau \lambda} \partial_\lambda (\ln \Omega)
$$
and so equation (1) is equivalent to the wave equation if there exists a scalar function $\Omega$ satisfying
$$
- \left( D-2 \right) \partial_\lambda (\ln \Omega) = A_\lambda + \frac{1}{\sqrt{|g|}} g_{\lambda \tau} \partial_\sigma (\sqrt{|g|} g^{\tau\sigma}).
$$
It is possible to imagine situations where we can solve this for $\Omega$, but it is also possible to imagine situations in which no such $\Omega$ exists.  In particular, the "curl" of the left-hand side must be zero, since $\partial_{[\mu} \partial_{\nu]} (\ln \Omega)$ must vanish; but it would not be hard to find instances where the curl of the right-hand side is not zero.  And in the case of a two-dimensional manifold, this whole rigamarole doesn't get us anywhere because the wave equation is conformally invariant;  if it doesn't work for the equation as written, it won't work at all.
