When does the time independent Schrödinger equation have a physical solution? In some cases, such as finite and infinite square wells, the Hamiltonian has energy eigenstates which correspond to physical wavefunctions.
In other cases, such as a one dimensional universe with constant potential, it doesn't. It has the plane wave, but that's not normalizable. (Also, for all Hamiltonians, the constant zero wavefunction is an eigenstate, but that's equally non normalizable.)
Is there any physical meaning to whether the Hamiltonian has energy eigenstates? Under what sort of situations does it have energy eigenstates?
 A: The time-independent Schrodinger equation is mainly useful for describing standing waves. It has serious shortcomings when used to describing traveling waves. If you have an example like a constant potential, then there are only traveling-wave solutions, and the time-independent Schrodinger equation may be the wrong tool for the job.
Physically, the constant-potential example has realistic solutions that are wave packets. These packets have to be constructed from a superposition of different energies. The time-independent equation restricts you to describing a single energy eigenstate, so the physically realistic packets aren't solutions. The packets spread out over time, and this is physically necessary behavior (caused ultimately by the Heisenberg uncertainty principle) that the time-independent Schrodinger equation can't describe.
A: In my opinion,if a potential function is authentic, the solution of the time independent Schrodinger equation  will have physics meaning. You say a one dimensional universe with constant potential,in fact ,this kind of potential is not existed. I think the plane wave cannot be normalized is a reflection of this.
A: Picking up on your comment that plane waves are not renormalisable:
Only infinite plane waves are not renormalisable, and an infinite plane wave is not physically realistic simply because we can't make, or indeed even observe) infinite objects. Any plane wave we can observe will be finite and therefore normalisable.
An infinite plane wave represents an object with zero uncertainty in the momentum, and therefore infinite uncertainty in position. The converse of this is that it's also physically unrealistic to have a solution representing a particle with a perfectly known position, i.e. a delta function, because then it has infinitely uncertain momentum.
These limitations aren't really to do with special properties of the potential. It's perfectly possible to take a superposition of plane wave solutions and construct perfectly realistic wavepackets.
