How can one find the energy eigenfunctions of a particle in a finite square well via the Klein-Gordon equation? It is said that Klein-Gordon equation is a relativistic version of the Schrodinger equation. In Schrodinger equation, it is straightforward to include potential energy. But for K-G eqn things seem to be more complicated.
To put it specifically, how can one find the energy eigenfunctions of a particle in a finite square well via Klein-Gordon equation?
 A: Schroedinger discovered the so-called Klein-Gordon equation first, before he heard about Heisenberg's work and before he thought of the Schroedinger equation.  The reason he did this is because he knew relativity was true, and so thought that it would be stupid to look for a non-relativistic equation, all he had to do was modify de Broglie's relativistic equation to include a potential.  This is why it is sometimes called the Schroedinger-Klein-Gordon-Fock equation, and sometimes the Schroedinger-Fock equation, and sometimes the Klein-Gordon equation.
He didn't publish because, not knowing about spin, he thought it should describe an electron rotating around the proton, and so he plugged in the mass of the electron, and although the basic energy levels were right, the relativistic fine splitting was wrong, out of the bounds of experimental error, so he gave it up.  Dirac later commented he should have had more confidence in his equation and published anyway.  IN the 50's it was found to more or less correctly describe one kind of meson in the Coulomb field of a proton.
After he heard about Heisenberg's work, noticing it was non-relativistic, he worked out a non-relativistic approximation to his original equation and published that, thinking to himself, oh well at any rate I can salvage something from that idea.  That published equation is now called the Schroedinger equation.  But I suspect he had a guilty conscience about it since he thought he knew that its relativistic version was false...
Now, as to your specific eigenvalue-problem query.  Pauli assigns that exact problem in his lectures on Wave Mechanics.  The method of solution is the same as for the Scrhoedinger eq.  Solve the free equ. in the half-infinite region on the left of the potential well.  Solve it on the right.  In the well, since the potential is constant, the same method (plug in an exponential function and solve for the parameters) works.  Now match boundary conditions at the two interfaces, and see which values of the parameters allow exponential decay in the infinite regions so that the wave function is $L^2$.  Once you choose a parameter in the left region, the parameters in the well and the right region are forced on you by matching the boundary conditions at the interfaces.  
A: K-G is a relativistic equation. So if you want to couple it to external fields you have to use a vector potential $A_\mu$ which will modify the partial-derivative, so that the Klein-Gordon equation:
$$ \left( \partial^\mu \partial_\mu + m^2 \right)\phi = 0 $$
becomes:
$$ \left ( D^\mu D_\mu + m^2 \right) \phi = 0 $$
where $ D_\mu = \partial_\mu + \imath A_\mu $.
Then it remains to figure out what $A_\mu$ will correspond to a given potential - finite square well or otherwise. For this you will likely have to fix a gauge before you can identify the potential as such.
A square well potential will correspond to $ A_\mu = (V(x), 0,0,0)$, where $ V(x) = V_0 \,\forall |x| > \pm a $ and $V(x) = 0, x \in [-a,a] $. The resulting behavior is a little counter-intuitive, yielding something known as the Klein paradox.
For further details see ch. 4 for this excellent thesis on this problem by D. Schaffer.

Edit: This page also has excellent information along these lines.
