Given a quiver, how do you get the indecomposable injective modules from indecomposable projective modules? Given a quiver, we know that it is easy to get the indecomposable projective modules, but the  indecomposable injective modules are not easy to get. 

How do you get the indecomposable injective modules from indecomposable projective modules? 

For example, $Q = (Q_0,Q_1)$ is the quiver
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
4 & \ra{\alpha_{43}} & 3  \\
\da{\alpha_{42}} & & \da{\alpha_{31}}  \\
2 & \ras{\alpha_{21}} & 1  \\
\end{array}$$
and $\mathcal{I} = \langle\alpha_{42}\alpha_{21}-\alpha_{43}\alpha_{31}\rangle$ the admissible ideal of$ KQ$. 
 A: I don't think projectives are easier to calculate for a bound quiver algebra. Whereas the projectives are the modules spanned by $e_i$, so $Ae_i$ if you are working with left modules (or $e_iA$ if you prefer right modules), the injectives are given by the $k$-duals of the projectives of the opposite algebra, so $D(e_iA)$ for left modules (or $D(Ae_i)$ for right modules). So if projectives are given by just taking all paths starting in one vertex, the injectives can be calculated by taking all paths ending in one arrow (the opposite algebra is given by the quiver with all arrows reversed, the relations accordingly).
So for your example of the "commutative square" you'll get as projectives (I'm now working with left modules)
$P_1=S_1$, $P_2$ is of length $2$ with socle $S_1$, $P_3$ is of length $2$ with socle $S_1$, $P_4$ is of length $4$ with socle $S_1$ and socle squared modulo socle $S_2$ and $S_3$.
Similarly for the injectives: $I_1=P_4$, $I_2$ is of length two with top $S_4$, $I_3$ is of length $2$ with top $S_4$ and $I_4=S_4$. 
