Time-dependent Schrödinger equation in 1D infinite potential well with central potential barrier Note: this is the first time I've attempted to solve a differential equation, I have no experience with ODEs, PDEs, or QM outside of what I've read in the past two days. If I'm missing something extremely obvious, it's probably because I never learned it.

The problem is basically the classic one-dimensional particle in a box set up, but with an infinite potential added at $0$.
Solve the time-dependent Schrödinger equation in position basis...
$$\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)+i\hbar\frac{\partial}{\partial t}\psi(x,t)-V(x)\psi(x,t)=0$$
...where the potential, $V$, is given piecewise by...
$$V(x)=\begin{cases}0&x\in(-1,0)\cup(0,1)\\\infty&\text{otherwise}\end{cases}$$
Because I am primarily interested in the behavior of the wavefunction for the given boundary condition, and not the actual values, I will simplify by setting $\hbar=1$ and $2m=1$ (ignoring units), so that...
$$\frac{\partial^2}{\partial x^2}\psi(x,t)+i\frac{\partial}{\partial t}\psi(x,t)-V(x)\psi(x,t)=0$$
As per convention, assume that $\int_\Bbb{R}|\psi(x,t)|^2\ dx=1$ and that $\psi(x,t)=0$ wherever $V(x)=\infty$. For convenience sake, I will also define the function...
$$u(x)=\begin{cases}1&x\in(-1,0)\cup(0,1)\\0&\text{otherwise}\end{cases}$$
...so that a solution can be easily written as $\psi(x,t)=f(x,t)\cdot u(x)$.
Trivial Solutions
There are constant solutions of the form $\psi(x,t)=z\cdot u(x)$, where...
$$z=a\pm i\sqrt{\frac{1}{2}-a^2}\qquad:\qquad a\in\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$$
Nontrivial Solutions
The obvious nontrivial solutions are sums of stationary states...
$$\psi_n(x,t)=A_n\sin(n\pi x)e^{-i n^2\pi^2 t}\cdot u(x)$$
...which can be derived from the general solution to the one-dimensional particle in a box problem (see any QM textbook). Similar solutions exist for the left and right side of the well with...
$$\psi_n(x,t)=A_n\sqrt{2}\sin(n\pi x)e^{-in^2\pi^2 t}\cdot w(x)$$
...with $w$ being another piecewise function confining all nonzero values of $\psi$ to the desired interval (similar to $u$).
Question(s)
Are these the only solutions, or are there other [nontrivial] solutions which can be obtained analytically? In particular, are there solutions such that:


*

*$\psi(x_0,0)=c$ for some $x\in (-1,0)\cup(0,1)$ and arbitrary $c\in \Bbb{C}$?

*$|\psi(x,0)|^2=\delta(x-x_0)$ for some $x_0\in (-1,0)\cup(0,1)$? $\quad(\delta$  is the Dirac delta function$)$
Additionally, I would like to know if the following is true:
For all solutions $\psi$, if there exists a value $t_0\in[0,\infty)$ such that $\psi(x,t_0)=0$ for all $x\in(-1,0)$, then $\psi(x,t)=0$ for all $x\in (-1,0)$ and $t\in[0,\infty)$.
(In other words, if the probability of finding a particle in the left side of the box is ever $0$, then it is always $0$).

An alternative set up for this question could be to set...
$$V_\alpha(x)=\begin{cases}0&x\in(-1,0)\cup(0,1)\\\alpha & x=0\\\infty &\text{otherwise}\end{cases}$$
...and examine the behaviour of solutions to...
$$\frac{\partial^2}{\partial x^2}\psi(x,t)+i\frac{\partial}{\partial t}\psi(x,t)-V_\alpha(x)\psi(x,t)=0$$
...for increasingly large $\alpha$. If there is a formula $\psi(x,t)=f(x,t,\alpha)$ which gives solutions for a particular set of starting/boundary conditions - say $|\psi(x,0)|^2=\delta(x-1/2)$, $\psi(-1,t)=0$, and $\psi(1,t)=0$ - then we can use $\lim_{\alpha\to\infty}f(x,t,\alpha)$ to answer the above questions.
This seems like the most appropriate method, but it also sounds really difficult, especially since each initial condition would require a separate formula and solutions might become chaotic in response to small changes near $x=0$.
 A: Here is a sketched answer to OP's questions.


*

*Values of the potential $V(x)$ within a set on the $x$-axis of Lebesgue measure zero are irrelevant. 

*Let us instead assume that the infinite square well potential is modified with a Dirac delta distribution
$$ V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0, \tag{1}$$
where the half-length of the well is $d=1$ in OP's case. 

*The solutions $\psi_n(x)$, $n\in \mathbb{N}$, to the TISE were derived in my Phys.SE answer here. Briefly, there are 


*

*(i) an infinite sequence of odd real solutions$^1$
$$ \psi(x)~=~A\sin(k x), \qquad |x|~\leq~d,  \tag{2}$$
who do not feel the Dirac delta distribution, with quantization condition
$$\frac{kd}{\pi}~\in~ \mathbb{N};\tag{3}$$ 

*(ii) an infinite sequence of even real solutions 
$$ \psi(x)~=~A\sin (k (d-|x|)), \qquad |x|~\leq~d,\tag{4}$$
who do, with quantization condition
$$  \frac{2mV_0}{\hbar^2}\tan(kd)~=~-2k.\tag{5}$$


*The complete solution to the TDSE is then a linear combination
$$ \Psi(x,t)~=~\sum_{n\in \mathbb{N}} c_n \psi_n(x) \exp\left(-\frac{i}{\hbar} E_n t\right), \qquad E_n~=~\frac{\hbar^2k_n^2}{2m} .\tag{6} $$

*Given an initial wave function profile $\Psi(x,0)$ the $c_n$-coefficients can be determined from the formula
$$ c_n~:=~ \int _{[-d,d]} \!\mathrm{d}x~\psi_n(x) \Psi(x,0). \tag{7}$$
--
$^1$ The normalization constant $A>0$ is assumed to normalize the TISE solution 
$$\int_{[-d,d]} \!\mathrm{d}x~ |\psi(x)|^2~=~1 .\tag{8}$$
