Consider the divergence form uniformly elliptic operator $\nabla \cdot a(x) \nabla$ where the coefficient $a$ are smooth and bounded and $D$ is a bounded and smooth domain of $\mathbb R^d$ $$ \begin{cases} \nabla \cdot a(x) \nabla f (x)=g \text{ in } D \\ f(x) = 0 \text{ in } \partial D, \end{cases} $$ where $g$ for some $g$. Consider now $x_0\in D$ and $\delta < d(x,\partial D)$ and the function $f_{x_0}$ which solves $$ \begin{cases} \nabla \cdot a(x_0) \nabla f^\delta_{x_0} (x)=g \text{ in } B(x_0,\delta)\\ f^\delta_{x_0}(x) = f(x) \text{ in } \partial B(x_0,\delta). \end{cases} $$

I was wondering whether it is possible to bound quantities such as $$ M(x_0,\delta,r,p):=r^{-d}\|f-f^\delta_{x_0}\|_{L^p(B(x_0,r))} $$ for $r < \delta$ and for some $p \in [1,\infty]$. In particular, I was wondering about the case asymptotic behaviour for $r \to 0^+$. That is, can I show that $$ M(p, \gamma):= \sup_{x_0 \in D} \sup_{\delta < d(x,\partial D)\wedge c_a} \sup_{r \le \delta} \frac{M(x_0,\delta,r,p)}{r^\gamma}, $$ is finite for some $\gamma>0$ and some constant $c_a >0$? If so, does that bound depends on the smoothness of $g$?

The idea being that if $\delta$ is sufficiently small, $a(x)\approx a(x_0)$ in the ball $B(x_0,\delta)$ and therefore the two equations should behave similarly. I am not sure if this is indeed enough or if I would need to ask $\delta$ to vanish as well.

I would appreciate any references or even what are the keywords to find such type of estimates.

parametrix method. But I do not know if it is really related to your question. $\endgroup$1more comment