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.


1 Answer 1


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.

  • $\begingroup$ Many thanks! This estimate is just the kind of thing I imagined there should be. It was a surprise to me that the solution depends Lipschitz continuously on $\sigma$. $\endgroup$ May 11, 2016 at 21:08
  • 1
    $\begingroup$ Also note that this computation makes sense when $\Omega$ has a Lipschitz boundary, so your smoothness assumption can be relaxed considerably. $\endgroup$
    – Graham Cox
    May 12, 2016 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.