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.



Reference request: Is if possible to estimate the local behaviour of the solution of $\nabla \cdot a(x) \nabla f=g$ via constant coefficients?
As Willie Wong pointed out, there are issues with the quantity you are looking at. But you can fine-tune your question in the one dimensional case, as you cannot expect better regularity than in $1$-d case. Not trying to insult anyone's intelligence, but the solution is in that case an explicit function of $a$ and $g$.


Take $\Omega=(-\ell,\ell)\subset\mathbb{R},$ and $g=1$. The solution
in $H\_{0}^{1}\left(\Omega\right)$ of
$$
\textrm{div}\left(a\nabla f\right)=1\text{ in }\left(-\ell,\ell\right)
$$
is
$$
f=\int\_{-\ell}^{x}\frac{t-z\_{\ell}}{a(t)}dt,
$$
and
$$
f^{\prime}\left(x\right)=\frac{1}{a(x)}\left(x-z\_{\ell}\right),
$$
with
$$
z\_{\ell}=\frac{\int\_{-\ell}^{\ell}\frac{t}{a(t)}dt}{\int\_{-\ell}^{\ell}\frac{1}{a(t)}dt}
$$
The solution $f-f\_{x\_{0}}^{\delta}$ satisfy
\begin{align\*}
\textrm{div}\left(a\left(x\_{0}\right)\nabla\left(f-f\_{x\_{0}}^{\delta}\right)\right) & =\textrm{div}\left(\left(a\left(x\_{0}\right)-a\right)\nabla f\right)\text{when }\left|x-x\_{0}\right|<\delta.\\
f-f\_{x\_{0}} & =0\text{ when }\left|x-x\_{0}\right|=\delta
\end{align\*}
in other words $f-f\_{x\_{0}}^{\delta}\in H\_{0}^{1}\left(x\_{0}-\delta,x\_{0}+\delta\right)$
satisfies
$$
\left(f-f\_{x\_{0}}^{\delta}\right)^{\prime\prime}=\left(\left(\frac{1}{a\left(x\right)}-\frac{1}{a\left(x\_{0}\right)}\right)\left(x-z\_{\ell}\right)\right)^{\prime},
$$
that is,
$$
f\left(x\right)-f\_{x\_{0}}^{\delta}\left(x\right)=\int\_{x\_0-\delta}^{x}\left(\left(\frac{1}{a\left(t\right)}-\frac{1}{a\left(x\_{0}\right)}\right)\left(t-z\_{\ell}\right)-\kappa\right)dt,
$$
with
$$
\kappa=\frac{1}{2\delta}\int\_{x\_0-\delta}^{x\_0+\delta}\left(\frac{1}{a\left(t\right)}-\frac{1}{a\left(x\_{0}\right)}\right)\left(t-z\_{\ell}\right)dt.
$$


If for example $a=\frac{1}{2\ell+x}$, $z\_\ell=\frac16 \ell$, $\kappa=\frac13 \delta^2$,
$$
f\left(x\right)-f\_{x\_{0}}^{\delta}\left(x\right)=\frac{(2x\_0-\ell+4x)(x\_0+\delta-x)(x\_0-\delta-x)}{24\delta}.
$$
At $x=x\_0$, $f\left(x\_0\right)-f\_{x\_{0}}^{\delta}\left(x\_0\right)=\frac{\delta}{24}(\ell-6x\_0)$, which is not null anywhere on the interval.


