I am not quite sure if I can fully follow and/or agree what you say in points (i) and (ii), but I do think the expression you suggest is correct (within the appropriate context) and is analogous to what is used in, e.g., the Ginzburg-Landau and Gross-Pitaevskii mean-field theories for superconductors and superfluids, respectively.
However, I would be cautious about associating any strict physical meaning to the energy density but rather think it as being defined only formally, in the sense that its integral over the system's spatial domain gives the expectation value of the Hamiltonian in the state in question.
If psi is real, both expressions should coincide. Now, when it comes to liking things, I prefer the second expression---just because it is the sándwich of psi with a known operator (the total quantum energy).
I think the question is not correctly formulated. If you discuss stationary states one must consider the virial theorem that imparts unique relations between expectation values of the kinetic energy and the potential. This is trivial for bound states, but for the continuum one must find a well-defined spectral density that permits a reasonable decomposition of the initial wave packet.
Hence if you discuss an evolving wave packet, you can Fourier transform the equation of motion. This does not lead to an eigenvalue equation, but rather to an inhomogeneous equation from which one might analyse the evolution in more detail.
As I come back to this thread, two more comments occur to me.
1. Sincé the first term in the hamiltonian is usually writen as a constant times del-squared psi, couldn't you define the first term in u with that expression, instead of (del psi)squared?
2. Shouldn't you have a plus or minus (not a product) sign between @u/@t and div j? Besides, how do you define j?
Maybe these are stupid questions, but I haven't read QM for a long time.
On the other hand, I know that here one is supposed to answer questions, not to ask more of them.
In quantum mechanics, a particle must be considered as a whole, not as a local perturbation. Therefore, the energy density may be locally negative, only the probability density must be everywhere positive.