Reference request: Is if possible to estimate the local behaviour of the solution of $\nabla \cdot a(x) \nabla f=g$ via constant coefficients?

Question

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.

Answer 1

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.