TISE asymmetric infinite potential well boundary conditions and normalisation I am attempting to solve the time-independent Schrodinger equation as a numerical analysis exercise, but my QM is a bit weak. I have the following potential and I want the energy/eigenvalue.  \begin{equation*} V(x) =  \begin{cases} 
      \infty & (- \infty , 0)\cup (2l, \infty)  \\
      0 & x \in [0,l]\\
      V_0 & x \in [l,2l] 
   \end{cases}
 \end{equation*}
I was wondering if this was a correct way of attacking it.
I have found solutions $\psi_1(x)$ and $\psi_2(x)$ for  $x \in [0,l]$ and $x \in [l,2l]$ by making initial guesses of $\psi_1'(0)$, $\psi_2'(2l)$ and $E$ which I want to use to extract the true solution by the shooting method. For the $x \in [l,2l]$ case I used a negative step size to traverse backwards, I was unsure if this was correct but I dont think starting at $\psi_2'(l)$ would suffice because it's not at the boundary where potential is infinite and the wave function is 0, so there's no good information for an initial/boundary value .
My main question is when it is time to "clean up" my guesses for the true values of the constants should I normalise with $\displaystyle \int_{0}^{l} |\psi_1|^2$ and $\displaystyle \int_{l}^{2l} |\psi_2|^2$ or $\displaystyle \int_{-\infty}^{\infty} |\psi_1|^2$ and $\displaystyle \int_{-\infty}^{\infty} |\psi_2|^2$ or even $\displaystyle \int_{-\infty}^{\infty} |\psi|^2$ where $\psi$ takes the appropriate values depending on the region. I am also unsure if I am looking for a single eigenvalue that works for both $\psi_1(x)$ and $\psi_2(x)$ or for $E_i$ s.t. $\hat{H}\psi_i(x) = E_i\psi_i(x)$ 
Apologies if this should be in scicomp.stackexchange or is a little basic, thanks.
 A: I) First, I am doing it analytically,
Solutions of Schrodinger Eqution:
$\psi_I(x; 0<x\le l)=A \sin k_1 x+B \cos k_1 x  \Rightarrow  \psi_I(x)=A \sin k_1 x$ (after using the BC: $\psi(0)=0$)
$\psi_{II}(x; l<x\le 2l)=C \sin k_2 x+D \cos k_2 x  \Rightarrow \psi_{II}(x)=C (\sin k_2 x - $
$\tan k_22l \cos k_2x)$ (after using the BC: $\psi_{II}(2l)=0)$.
where $k_1=\sqrt(e);  k_2=\sqrt(e-V_0)$ (I have taken $\hbar^2=1=2m$)
Now matching the solutions at $l$: $\psi_{I}(l)=\psi_{II}(l),$ and $\psi_{I}'(l)=\psi_{II}'(l),$ This eliminates the coefficients $A$ and $C$.
We get the following transcendental Eq.,
$\frac{\tan k_1l}{k_1l}=\frac{\sin k_2l-\tan k_2 2l \cos k_1 l}{k_2l(\cos k_2l+\tan k_2 2l \sin k_2 l)} \Rightarrow \frac{\tan (k_1(e)l)}{k_1(e)l}-\frac{\sin (k_2(e)l)-\tan (k_2(e) 2l) \cos (k_1(e) l)}{k_2(e)l [\cos (k_2(e)l)+\tan (k_2(e) 2l) \sin (k_2(e) l)]} {=}0$
Example: Let's take $l=1$ and $V_0=2.$
First two roots of this equation are: $\textbf {3.367, 10.944} $; These are the first two eigenvalues.
II) Numerically, There is a nice way to do it in Mathematica,
Here, I am integrating the Schrodinger equation from ( with BCs:$   \psi(0)=0,\psi'(0)=1)$ $0$ to $2l$, then looking for zeros of $\psi(2l,e)$.
k[e_] := Sqrt[e];
v0 = 2;
l = 1;
pot[x_] := Which[0 < x <= l, 0, a < x <= 2*l, v0];
sol1 = ParametricNDSolve[{sy1''[x] + sy1[x]*(k[e]^2 - pot[x]) == 0,sy1[0] == 0, sy1'[0] == 1}, {sy1}, {x, 0, 2*l}, {e}];
sy2[x_, e_] := sy1[e][x] /. sol1;
Plot[sy2[2*l, e], {e, 0, 20}]
FindRoot[sy2[2*l, e] == 0, {e, 4}]
FindRoot[sy2[2*l, e] ==0, {e,10}]


Here I have taken $l=1$ and $V_0=2.$
The above plot is for "sy2[2 l, e]" ( i.e. $\psi(2l,e)$). The first two roots are  {e $\rightarrow$ $\textbf {3.36726, 10.9443} $}; These are first two eigenvalues.
A: Because you are only interested in the energy eigenvalues, you don't need to normalize $\psi$. Observe that if $\psi$ is a solution to the Schrodinger equation
$$-\frac{\hbar^{2}}{2m}\frac{{\rm d}^{2}\psi}{{\rm d}x^{2}}+V\left(x\right)\psi=E\psi$$
then so as $A\psi$ for every $A\neq 0$ with the same $E$. Also, since this is a second order differential equation, you only need to set two conditions on $\psi$. Because you are trying to solve the problem numerically, you need to set the conditions at $x=0$ (I assume you solve forwards and not backwards). I argue that you must require
$$\begin{cases}\psi\left(x=0\right)=0\\\psi^{\prime}\left(x=0\right)=1\end{cases}$$
The first condition is the consequence of the infinite potential at the boundaries. The second condition is equivalent to $\psi^{\prime}\left(x=0\right)\neq 0$ - because remember! We don't care about the normalization! Now you use the shooting method. You guess $E$, solve the ODE and get $\psi\left(x=2L\right)$. Keep guessing till you get $\psi\left(x=2L\right)=0$. In other words, use a root finder to solve the equation
$$\psi_{E}\left(x=2L\right)=0$$
for $E$. Start the iterations at the guess $E_{0}=0$ for the ground state. For higher states you'll need to use trial and error. I'd recommend you to plot $\psi_{E}\left(x=2L\right)$ as a function of $E$. The zeros of this graph are the eigenvalues.
