# Let $\mathrm{div}\,(A\,\mathrm{grad}\,u) + b u = f$. Is $(A\,\mathrm{grad}\,u)$ weakly differentiable?

Let us consider the basic linear elliptic PDE $$\mathrm{div} (A\,\mathrm{grad}\,u) + bu = f,$$ with $f\in L^p,$ $A,b$ uniformly bounded. Do we have, for a weak solution $u\in W^{1,p}(\Omega')$, $$(A\,\mathrm{grad}\,u)\in (W^{1,p})^d(\Omega),\ \mbox{ in }\Omega\subset\subset\Omega'\subset\mathbb{R}^d?$$ or at least do we have $$\mathbf{n}\cdot(A\,\mathrm{grad}\,u)\in L^{p}(\partial\Omega)?$$ This seems to me true at least if $A$ is piecewise Lipschitz uniformly because across an interface where $A$ has a jump, the transmission condition $A\,\mathrm{grad}\,u$ still has matched traces.

My final goal is to know under what least conditions, we have Green's representation formula in the sense of Lebesgue integrals: $$u(\mathbf{y})=\int_{\partial\Omega}\frac{\partial G(\cdot,\mathbf{y})} {\partial \mathbf{n}_A}u- \frac{\partial u}{\partial \mathbf{n}_A}G(\cdot,\mathbf{y})~dS(\mathbf{x}) +\int_{\Omega}fG(\cdot,\mathbf{y})~d\mathbf{x},$$ for $\mathbf{y}\in\Omega$ and $\frac{\partial v}{\partial\mathbf{n}_A}=\mathbf{n}\cdot(A\,\mathrm{grad}\,v).$

Edit: I'm still waiting a counter-example to the above guess when $A$ is bounded only.

As far as I know, there is no such results if you only assume $A\in L^\infty(\Omega)$.

But if you assume that the domain $\Omega$ is partitioned into two subdomains $\Omega_1\cup\Omega_2$ by a smooth interface $\Gamma$ inside $\Omega$, and in each subdomain you assume that $\,A\in W^{1,\infty}(\Omega_1)\cap W^{1,\infty}(\Omega_2)$ ($\,A$ may be discontinuous across the interface), then the solution of $$\left\{\begin{array}{ll} \nabla\cdot(A\nabla u)=f &\mbox{in}~~\Omega,\\ u=0 &\mbox{on}~~\Omega, \end{array}\right.$$ with $f\in L^p(\Omega)$ satisfies $$\|u\|_{ W^{2,p}(\Omega_1)}+\|u\|_{W^{2,p}(\Omega_2)}\leq C\|f\|_{L^p(\Omega)}$$ and the trace of ${\bf n}\cdot A\nabla u$ on the two sides of the interface coincide, with ${\bf n}\cdot A\nabla u\in L^p(\partial\Omega)$.

If $A\in C^{1+\alpha}(\overline\Omega_1)\cap C^{1+\alpha}(\overline\Omega_2)$, then $$\|u\|_{ C^{2+\alpha}(\overline\Omega_1)}+\|u\|_{C^{2+\alpha}(\overline\Omega_2)}\leq C(\|f\|_{C^\alpha(\overline\Omega_1)}+\|f\|_{C^\alpha(\overline\Omega_2)}) .$$

These classical results probably can be found in the following papers

"Gradient estimates for the perfect conductivity problem" by Ellen Shiting Bao, Yanyan Li and Biao Yin

"Gradient Estimates for the Perfect and Insulated Conductivity Problems with Multiple Inclusions" by Ellen Shiting Bao, Yanyan Li and Biao Yin

• Thanks, Buyang! That is what I expected in the beginning. Can you give a counter-example to my guess? Do you have any comments on Green's representation formula? What are the minimal conditions for the representation? May 13 '14 at 15:59
• If $u\in H^2(\Omega_1)\cap H^2(\Omega_2)$, then this Green formula holds. The minimal condition seems to be $u\in H^1(\Omega)$ and $\Delta u\in H^{-1}(\Omega)$. In this case, one can define a cornormal derivative ${\bf n}\cdot A\nabla u$ on the boundary $\partial\Omega$. See the book of "Strongly Elliptic Systems and Boundary Integral Equations", written by William McLean. May 13 '14 at 22:56
• It seems that your PhD supervisor may know my PhD supervisor very well... May 14 '14 at 8:06
• Thanks, Buyang! Yes, your supervisor shares master supervisor with my supervisor and for this reason I also met Weiwei in Xi'an. Could you give me the relevant page number in McLean's book? I have not found it by myself. I also posted another related question. Could you have a look? May 14 '14 at 11:39
• Hi, Page 116 Lemma 4.3 defines the cornormal derivative by $u\in H^1$ and $\Delta u\in H^{-1}$ May 14 '14 at 13:21

Suppose $p<n$. Remark that just from the equation, $$\partial_i (A_{ij} \partial_j u) = A_{ij} \partial_{ij} u + \partial_i A_{ij} \partial_j u,$$ so if the left-hand side is in $L^p$, and $u\in W^{2,p}$, then $\nabla u \in W^{1,p^*}$ with $p^*=np/(n-p)$ from Sobolev, and for the third term to be in $L^p$ it is natural to expect that $$A \in W^{1,q} \cap L^\infty$$ with, using H\"{o}lder's inequality $$\frac{1}{p}= \frac{n-p}{np} + \frac{1}{q},$$ in other words $q = n$. So $A\in W^{1,n}\cap L^\infty$ is the reasonable general assumption (if you don't look at specific geometries as in Bunyang Li's answer).

Next, you can get rid of $f$: assuming that the domain $\Omega$ is sufficiently smooth, solve in $H^{1}_{0}(\Omega)$ $$-\Delta \psi = f$$ and you obtain that $\psi\in W^{2,p}(\Omega)$, and the problem writes $$\textrm{div}(A\nabla u + \nabla \psi)=0 \mbox{ in } \Omega, \quad(\star)$$ do I have that at least locally, $(a\nabla u)\in W^{1,p}(\Omega)$, when $A \in W^{1,n} and bounded and coercive? I think the optimal assumption is VMO and bounded (and elliptic), from Gallouet & Qafsaoui's 2002 paper. But I don't think it is written. An easy case is$d=3$and$A$a scalar (not a matrix). Set$E=\nabla u $. From ($\star$) you obtain$A E +\nabla \psi = \nabla \times H$, with$\phi \in W^{1,p}$, or in other words $$E= A^{-1} \nabla \times H - A^{-1}\nabla \psi$$. The second term is no problem, it is in$W^{1,p}$. Taking the divergence, as$A$is a scalar, you obtain $$\textrm{div} E = \nabla \times H \cdot \nabla A^{-1} + \textrm{div}A^{-1}\nabla \psi$$ Both terms are in$L^p$, so$W^{1,p}$(div). Of course,$\nabla \times E =0$. So you have$E$in$W^{1,p}$(curl). Then you just need$W^{1,p}$(curl)$\cap W^{1,p}$(div)=$W^{1,p}$, which is true (at least locally, you have to be careful on the boundary). • Your arguments sound interesting and quite different from the more direct way: invoking the classical interior regularity result when$A$is$W^{1,n}$to first get the solution in$W^{2,p}$and then$A\nabla u$in$W^{1,p}.\$ Thanks! May 14 '14 at 15:02
• @HuiZhang Which classical regularity result are you thinking of (for my own education)? May 14 '14 at 22:24
• See Gilbarg-Trudinger's book Elliptic Partial Differential Equations of Second Order, section 8.3 Differentiability of Weak Solutions. The proof is based on difference quotients. May 15 '14 at 7:46
• @HuiZhang I just read it, it only deals with Lipschitz coefficients. May 15 '14 at 13:00
• Yes, you are right. My mistake. It seems that proof needs Lipschitz continuity of coefficients. May 15 '14 at 13:20