Why does the Schrödinger equation work so well for the hydrogen atom despite the relativistic boundary at the nucleus? I have been taught that the boundary conditions are just as important as the differential equation itself when solving real, physical problems.
When the Schrödinger equation is applied to the idealized hydrogen atom it is separable and boundary conditions are applied to the radial component. I am worried about the $r=0$ boundary near the nucleus. Near the proton, the electron's kinetic energy will be relativistic and looking at the Schrödinger equation itself for how this boundary should behave seems dangerous because its kinetic energy term is only a non-relativistic approximation.
Is there any physical intuition, or any math, that I can look at that should make me comfortable with the boundary condition in this region?
 A: The boundary conditions that pick up the hydrogen wave functions are the "constraints" placed on the wave function solutions. Remember that the observable is the probability distribution from the $Ψ^*Ψ$, not a particular location. Please read the link. The solutions are within the quantum mechanic postulates after all.
There isn't any relativistic boundary condition, because there are no orbits, only probability distributions.
Thus the solutions do not have a singularity at r=0, and in general there is a small probability of finding the electron at the origin, if the quantum numbers allow for an interaction, as with electron capture in nuclei. For the hydrogen atom there is not enough energy for a neutron to appear. 
A: The boundary condition at r=0 is that the wave function should be finite. The Schrödinger equation for hydrogen UC atoms and likely all atoms has solutions with negative $\cal l$, which are rejected  because they diverge at r=0. See for example Schiff's textbook on quantum mechanics.
As for relativistic effects, you may want to compare the hydrogen energy expressions for Dirac, better, Klein-Gordon - no spin, and Schrödinger. Check out another great text, Itzykson and Zuber, for these . 
A: In solving the Schroedinger radial equation there is no boundary condition applied at $r=0$.  At $r=\infty$ yes, $R(r)$ must tend to zero - so we reject the positive exponential solution;  any change in that would have massive consequences. But there is no constraint laid on $R(r)$ or indeed $R'(r)$ as $r \to 0$. 
So there's not a change in the boundary condition. There is a change in the kinetic and potential energies due to relativistic effects and the fact that the proton is not a point charge. These do have an effect - but very small, as the volume concerned is about $10^{-15}$ of the volume of the atom.  (Actually atomic physicists experiments can detect these effects, at least for large $Z$ atoms, thanks to some very clever and precise optical experiments.) But this is a small effect, not the game changer that a new boundary condition could give.
A: Good question. Your claim that

near the proton the electron's kinetic energy will be relativistic

is not as straightforward as it might seem. The electron's kinetic energy $\langle \hat{T} \rangle = \langle \hat{p}^2 \rangle / (2m)$ is a nonlocal quantity that can be equivalently expressed as either of the two integrals
$$\langle \hat{T} \rangle = \frac{1}{2m} \int d^3x\ \psi^*(x) \left(-\hbar^2 \nabla^2 \right) \psi(x) = \frac{1}{2m} \int d^3x\ |\hbar\, {\bf \nabla} \psi(x)|^ 2.$$
So the electron's kinetic energy "at" a particular location is not well-defined; it could be the value of either of the two integrands above at that point (or, indeed, of any other integrand that integrates to the same value over all of space).
The latter expression is the more natural one to use, though, because at least it's positive-semidefinite. We still have the problem that $\hbar^2 |\nabla \psi(0)|^2/(2m)$ is a "kinetic energy density" (whatever that is) rather than an actual kinetic energy, so we can't speak of how relativistic the electron is "at" the nucleus. (We could integrate over the empirical size of the nucleus, but I don't think that's really what your question is getting at - you're not asking about when the electron is literally inside the nucleus, but when it's close enough to the potential center that it's intuitively moving very quickly.)
But none of this really matters - the point is that since the integrand is positive-definite, the contribution to the kinetic energy over any particular region is always less than (or equal to) the total kinetic energy over every region. So to meaningfully check whether relativistic effects need to be taken into account, you need to calculate the total kinetic energy over all of space. This turns out to be $\hbar^2/(2m a^2) = m e^4/(2 \hbar^2) = (\alpha^2 /2) m c^2$, where $\alpha$ is the fine-structure constant. Relativistic effects are negligible if the kinetic energy is much less than the electron's rest energy, which corresponds to the condition that $\alpha^2/2 = 1/37538 \ll 1$, which, reassuringly, is true.
