Creating a Deterministic Push-Down Automaton for the Union of two languages Suppose, we have $L_1:=\{w\in\{a,b\}^*\mid \#_a(w) \equiv 0 \mod 4\}$ and $L_2:=\{w\in\{a,b\}^*\mid abaab \text{ is a substring of } w\}$. Now we want to create a Deterministic Push-Down Automaton for $L_1\cap L_2$


I've created PDA's for both languages:

$M'=(Q,\Sigma,\Gamma,\Delta,s,F)$ with $\Sigma=\{a,b\}$, $\Gamma=\{\}$, $Q=\{1,2,3,4\}$, $F=\{0\}$, $s=\{0\}$ and
\begin{align}
\Delta=\{&amp;(0,a,\lambda,1,\lambda),(0,b,\lambda,0,\lambda)(1,a,\lambda,2,\lambda),(1,b,\lambda,1,\lambda),(2,a,\lambda,3,\lambda),(2,b,\lambda,2,\lambda)\\
&amp;(3,a,\lambda,0,\lambda),(3,b,\lambda,3,\lambda)\}
\end{align}
where $\Delta\subseteq Q_{old}\times (\Sigma \cup \{\lambda\})\times(\Gamma\cup \lambda)\times Q_{new}\times\Gamma^*$

and

$M''=(Q,\Sigma,\Gamma,\Delta,s,F)$ with definitions above and $\Sigma=\{a,b\}$, $\Gamma=\{\}$, $Q=\{1,2,3,4,5,6\}$, $F=\{6\}$, $s=\{1\}$ and
\begin{align}
\Delta=\{&amp;(1,a,\lambda,2,\lambda),(1,b,\lambda,1,\lambda),(2,a,\lambda,2,\lambda),(2,b,3,\lambda),(3,a,\lambda,4,\lambda),(3,b,\lambda,1,\lambda),\\&amp;(4,a,\lambda,5,\lambda),(4,b,\lambda,3,\lambda),(5,a,\lambda,2,\lambda),(5,b,\lambda,6,\lambda),(6,a,\lambda,6,\lambda),(6,b,\lambda,6,\lambda)\}
\end{align} 

$M'$ accepts $L_1$ and $M''$ accepts $L_2$, but how to construct $M^{\star}=M'\cap M''$ which accepts $L_1\cap L_2$?

I know, that we could possibly do this with Deterministic Finite Automata, but I want to know how it would work with PDA's.
There is no general algorithm for computing a DPDA $M^*$ as the intersection of two DPDA $M,M'$; that problem is undecidable.  See Undecidable problem intersection of two DCFL languages is DCFL? for a proof.

So, you will have to look for some pattern in this particular problem that makes it easier than the general case.  Hint: what level in the Chomsky hierarchy is $L_1$ at?  what about $L_2$?