The complete solution for the time dependent equation with an infinite potential step is found by the method of images. Given any initial wavefunction
$$ \psi_0(x) $$
for x<0, you write down the antisymmetric extension of the wavefunction
$$ \psi_0(x) = \psi_0(x) - \psi_0(-x) $$
And you solve the free Schrodinger equation. So any solution of the free Schrodinger equation gives a solution for the infinite potential step. This is not completely trivial to make, because the solutions do not vanish in any region. But, for example, the spreading delta-function
$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } $$
Turns into the spreading, reflecting, delta function
$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } - {1\over \sqrt{2\pi it}} e^{-(x+x_0)^2\over it } $$
You can do the same thing with the spreading Gaussian wavepacket, just subtract the solution translated to +x from the solution translated to -x. In this case, normalizing the wavefunction is hard when the wavefunction start out close to the reflection wall.
Time independent infinite potential wall
The solution to the time independent problem of the infinite potential wall are all wavefunctions of the form
$$ \sin(kx) $$
for all k>0. Superposing these solutions gives all antisymmetric functions on the real line.
To find this solution, note that the time independent problem (eigenvalue problem) for the Schrodinger equation is solved by sinusoidal waves of the form $e^{ikx}$, and you need to superposes these so that they are zero at the origin, to obey the reflection condition. This requires that you add two k-waves up with opposite signs of k and opposite sign coefficients.
The opposite sign of k just means that the wave bounces off the wall (so that k changes sign), while the opposite sign of the coefficient means that the phase is opposite upon reflection, so that the wave at the wall cancels.
General solution
The time dependent problem for a time independent potnetial is just the sum of the solutions to the time independent problem with coefficients that vary in time sinusoidally.
If the eigenfunctions $\psi_n$ are known, and their energies $E_n$ are known, and the potential doesn't change in time, then the,
$$ \psi(t) = \sum_n C_n e^{-iE_n t} \psi_n(x) $$
is the general solution of the time dependent problem. This is so well known that generally people don't bother saying they solved the time-dependent problem once they have solved the eigenvalue problem.
The general solution of the time-dependent Schrodinger equation for time dependent potentials doesn't reduce to an eigenvalue problem, so it is a different sort of thing. this is generally what people understand when you say solving the time-dependent equation, and this reflects the other answers you are getting. I don't think this was the intent of your question, you just wanted to know how to solve the time dependent equation for a time independent potential, in particular, for an infinite reflecting potential wall. This is just the bouncing solution described above.
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.
Best Answer
Of course there are. That's why we solve the time-independent version of the Schrödinger equation to begin with: because given any eigenfunction $\psi_0$ of the hamiltonian with eigenvalue $E$, the phase-evolved combination $$ \psi(t) = e^{-iEt/\hbar}\psi_0 $$ is a solution of the time-dependent Schrödinger equation, and, moreover, any linear combination of such solutions is still a solution.
There is, of course, an understandable prejudice against taking a stationary state as an initial condition for the TDSE (but it is just a human prejudice with no real meat to back it up). If that really bothers you, then you can just take a nontrivial linear combination, like, say, $$ \psi = \frac{\psi_{100}+\psi_{210}}{\sqrt{2}}, $$ and it will then show oscillations in both the position-space and momentum-space probability distributions. To borrow from my answer to Is there oscillating charge in a hydrogen atom?, the explicit wavefunction is given by
This combination gives you an explicit analytical solution of the time-dependent Schrödinger equation. Now, again, it is understandable to dismiss this as somehow "not being a real wavepacket", partly because from some perspectives it might feel "too easy", but all of those are human prejudices with very little support on well-defined and truly meaningful mathematical criteria on the initial conditions or the corresponding solution. This is an honest-to-goodness electronic wavepacket moving under the influence of a point-charge nucleus.