Boundary conditions, physical wave-functions and domain of Hamiltonian Context :
In quantum mechanics, time evolution is described by a one-parameter unitary group, acting on the Hilbert space of states. Under Stone's theorem (with the right hypotheses), this group has a self-adjoint generator, which is called the Hamiltonian.
For a particle living in a 1D box, the Hilbert space is $L^2([0,\ell])$ (with $\ell>0$ the length of the box) and the Hamiltonian operator is : $$ H = -\frac{\hbar^2}{2m}\frac{d}{d x^2} + V(x) \tag{1}$$
We also need to specify the domain on which $H$ is defined and self-adjoint. Now, it is pointed out here and in this P.SE post, that this is equivalent to fixing boundary conditions.
Then, the (unitary) time evolution operator is $U(t) = e^{-iHt/\hbar}$ and is well defined on the whole Hilbert space.
Questions:

*

*Concretely, how does changing the domain for $H$ change the time evolution operator $U(t)$ ?

*Presumably, $\psi = 1/\sqrt{\ell} \in L^2([0,\ell])$ is a valid wave-function, whichever boundary condition we choose, since it lives in the Hilbert space. How do boundary conditions determine its time evolution?

 A: Here is the deal: canonical quantization forces your $(1)$ to be a quantum Hamiltonian. This operator drives (generates) time evolution iff it is self-adjoint. An operator $A$, if unbounded and defined on a (complex) separable Hilbert space $\mathcal{H}$, can be correctly defined only on a proper subset of $\mathcal{H}$ (called the maximal domain of $A$), if we require the operator to be closed. Self-adjoint operators are always closed, therefore their maximal domain is always proper: $D(A)\subsetneq \mathcal{H}$. This is basic stuff.
Come back to $(1)$. We want $H$ to be self-adjoint when seen as operator $H:D(H)\rightarrow L^2 ([0,l])$. For any well behaved real function $V(x)$, $D(H) = D\left(\frac{d^2}{d x^2}\right)\subsetneq L^2 ([0,l])$.
We can choose only one boundary condition for non-constant elements in $L^2 ([0,l])$, which renders $H$ self-adjoint on its maximal domain.
So your first question is almost trivial: $U(t)$ exists only when $H$ is self-adjoint, which happens only in the presence of a particular boundary condition: $$\psi\in L^2 ([0,l]),~ \psi (0) = \psi (l)=0 \tag{2}$$ or  the trivial (normalized) wavefunction: $\psi$ is constant, i.e. $$\psi (x) = \frac{\exp{i\alpha}}{\sqrt{l}},\alpha\in\mathbb{R} \tag{3}$$ As soon as you consider functions from the Hilbert space outside these two possible boundary conditions, then the Hamiltonian is no longer self-adjoint (not even symmetric, because you cannot make the boundary term vanish), so that $U(t)$ does not exist. Existence of constant wavefunctions is an artifact of using an unphysical restraining of motion in a box with "hard" walls. Usually PIB involve no dynamics at all ($V(x) \equiv 0$), case in which a constant wavefunction is trivially the $0$ vector, so that only option $(2)$ remains.
The second question gets an answer immediately: Since the constant function $(3)$ is a trivial solution to the spectral equation (energy spectrum = numeric value of $V(x), x\in [0,1]$), then there is no impact of choosing particular boundary conditions on the exact form of time evolution: $U(t) = \exp (-i H(x)t /\hbar)$ which acts trivially on a constant wavefunction, as a phase-factor containing the numeric value of the potential.
