Let $I=(l_0,l_1)$, $y \in H^2(I)$ and $\rho(x) \in L^\infty(I)$ with $\rho(x)>c\geq 0$. Define $D_\rho(x)=\rho(x)\chi_{(a_0,b_0)}$ with $l_0\leq a_0<b_0\leq l_1$. Now, For any $\phi \in C_c^\infty(I)$ I have
$$ \left<(D_\rho y')',\phi\right>=\left<D_\rho'y',\phi\right>+\left<D_\rho y'',\phi\right>$$
this means that $(D_\rho y')'=D_\rho'y'+D_\rho y''$ in distributional sense. Since $D_\rho \in L^\infty$ and $y'' \in L^2(I)$ I get $D_\rho y'' \in L^2(I)$. But I can't say anything about $D_\rho'y'$, so I'm here asking for help in trying to figure out which conditions $\rho(x)$ must satisfy in order to guarantee that $D_\rho' y'$ is in $L^2(I)$.
I know that if I assume that $\rho'(x)$ is bounded almost everywhere, then $D_\rho' y' \in L^2(I)$, but this is a very strong condition and I want weak conditions.
I really appreciate any kind of help!
I don’t know if I am missing something but if $l_0<a_0<b_0<l_1$, then I think that no assumption on $\rho$ can guarantee that the product is in $L^2$. Take $y(x)=x$, then $y’=1$. If $D’_\rho$ were in $L^2$, then the function $D_\rho$ would have to be equal almost everywhere to an absolutely continuous function, see this link ac , but your function has a bad discontinuity at $a_0$ and $b_0$