Let $\Omega\subset\mathbb R^n$ be a bounded smooth domain and $\sigma\_1,\sigma\_2:\Omega\to(c^{-1},c)$ measurable (for some constant $1<c<\infty$).
Let $f\in H^{1/2}(\partial\Omega)=H^1(\Omega)/H^1\_0(\Omega)$.
Let $u\_i\in H^1(\Omega)$, $i=1,2$, be the solution of
$$
\begin{cases}
\operatorname{div}(\sigma\_i\nabla u\_i)=0 & \text{in }\Omega\\
u\_i=f & \text{on }\partial\Omega.
\end{cases}
$$
Is there an estimate like
$$
\|\nabla u\_1-\nabla u\_2\|\_{L^p}
\leq
\|f\|\_{H^{1/2}(\partial\Omega)}
C(c,\Omega,n,p)\omega(\|\sigma\_1-\sigma\_2\|\_{?})
$$
where $C$ is a constant depending on the given parameters, $\omega$ is a modulus of continuity (depending on the same parameters) and the question mark norm is some norm?
What I'm most interested in is the choice of the norm and shape of $\omega$ (linear or worse).
I would like to have $p$ as large as possible ($p=\infty$ would be great), but $p=2$ would also be good.


If an estimate like this is not known (or is known not to exist), are there results in the same spirit for the stability of a solution in perturbations of the equation?
It is no problem if you need to assume more regularity of $\sigma$; smoothness is ok, but analyticity would be too much.



How do solutions of a PDE depend on parameters?
If $p=2$ and $\sigma\_1, \sigma\_2 \in L^\infty(\Omega)$ one has the estimate
$$\|\nabla u\_1 - \nabla u\_2\|\_{L^2(\Omega)} \leq C \|f\|\_{H^{1/2}(\partial\Omega)} \|\sigma\_1 -\sigma\_2\|\_{L^\infty(\Omega)}$$ where $C = C(\Omega,c)$.


To see this, write $0 = (u\_1-u\_2) \operatorname{div}(\sigma\_1 \nabla u\_1 - \sigma\_2 \nabla u\_2)$ and integrate by parts, using the fact that $u\_1 - u\_2$ vanishes on the boundary to obtain
\begin{align\*}
0 &= \int \nabla(u\_1-u\_2)\cdot (\sigma\_1 \nabla u\_1 - \sigma\_2 \nabla u\_2) \\
&= \int \nabla(u\_1-u\_2)\cdot [\sigma\_1 (\nabla u\_1 - \nabla u\_2) + (\sigma\_1 - \sigma\_2) \nabla u\_2]
\end{align\*}
hence
\begin{align\*}
\int \sigma\_1 |\nabla(u\_1-u\_2)|^2 &= \int (\sigma\_2 - \sigma\_1)(\nabla u\_1 - \nabla u\_2)\cdot \nabla u\_2 \\
&\leq \|\sigma\_1 - \sigma\_2\|\_{L^\infty} \|\nabla u\_1 - \nabla u\_2\|\_{L^2} \| \nabla u\_2 \|\_{L^2}.
\end{align\*}


Then the estimate follows from the arithmetic–geometric mean inequality and the fact that $\|\nabla u\_2\|\_{L^2} \leq C \|f\|\_{H^{1/2}}$ for some constant $C = C(\Omega,c)$.


EDIT: As Joonas pointed out, the left-hand side of the above inequality is bounded below by $c\|\nabla u\_1 - \nabla u\_2\|^2\_{L^2}$, so one simply divides by $\|\nabla u\_1 - \nabla u\_2\|\_{L^2}$ to complete the proof; the AGM inequality is not needed.


A more general argument can be found in Appendix C of [this paper](http://dx.doi.org/10.1080/03605302.2015.1025979).


