# How do solutions of a PDE depend on parameters?

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

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.

• 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 $\sigma$. – Joonas Ilmavirta May 11 '16 at 21:08
• Also note that this computation makes sense when $\Omega$ has a Lipschitz boundary, so your smoothness assumption can be relaxed considerably. – Graham Cox May 12 '16 at 3:19