Weak divergence implies weak differentiability of components? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:05:01Z http://mathoverflow.net/feeds/question/110448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110448/weak-divergence-implies-weak-differentiability-of-components Weak divergence implies weak differentiability of components? Beni Bogosel 2012-10-23T17:01:15Z 2012-10-28T21:22:08Z <p>Suppose $\Omega$ is an open set in $\Bbb{R}^N$ and $\sigma : \Omega \to \Bbb{R}^N$ is a field with all components belonging to $L^2(\Omega)$.</p> <p>We say that $\sigma$ has <em>weak divergence</em> if there exists a function $w \in L^2(\Omega)$ such that for all $\varphi \in C_c^\infty (\Omega)$ we have</p> <p>$$\int_\Omega \sigma \cdot \nabla \varphi=-\int_\Omega w \varphi.$$</p> <p>My question is:</p> <blockquote> <p>Can we establish a result of the form: if $\sigma$ has weak divergence then each component of $\sigma$ is weakly differentiable?</p> </blockquote> <p>The idea is that I've seen this technique in proving that if a function $u$ is $H^1(\Omega)$ and it satisfies some convenient weak condition then $\nabla u$ has a weak divergence and therefore $u \in H^2(\Omega)$. The book where I've seen this technique is aimed for engineers, and therefore it is not very rigorous. That is why I've asked this question.</p> http://mathoverflow.net/questions/110448/weak-divergence-implies-weak-differentiability-of-components/110940#110940 Answer by Bazin for Weak divergence implies weak differentiability of components? Bazin 2012-10-28T21:22:08Z 2012-10-28T21:22:08Z <p>So $\sigma=\sum_{1\le j\le n}\sigma_j(x)\frac{\partial}{\partial x_j}$ is a vector field with distributions coefficients $\sigma_j$ and divergence in $L^2$: $$\sum_{1\le j\le n}\frac{\partial \sigma_j}{\partial x_j}\in L^2.$$ If I understand your question correctly, you ask if this implies that each $\sigma_j$ belongs to $L^2$. Of course not since you can choose all $\sigma_j=0$ for $j\ge 2$ and $\sigma_1$ to be any distribution in the variables $(x_2,\dots, x_n)$.</p> <p>Note nevertheless that when $n=2$ and $\text{div} \sigma=0$, then there exists a function $\psi(x_1,x_2)$ such that $$\sigma_1=\frac{\partial \psi}{\partial x_2},\quad \sigma_2=-\frac{\partial \psi}{\partial x_1}.$$ If $\sigma$ happens to be continuous (resp. locally $L^2$), then $\psi$ is locally Lipschitz continuous (resp. locally $H^1)$.</p> <p>For the general statement that you mention, note that the gradient operator is elliptic, which is not the case of the divergence operator.</p>