1
$\begingroup$

Let $\Omega\subseteq\mathbb R^d$ be open.

$v:\Omega\to\mathbb R$ is called weak divergence of $u:\Omega\to\mathbb R^d$ $:\Leftrightarrow$ $$\int_\Omega v\varphi\;{\rm d}\lambda=-\int_\Omega\langle u,\nabla\varphi\rangle\;{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\;.\tag 1$$ In that case, we write $v=\operatorname{div}u$.

Since $$\left\|v\varphi\right\|_{L^1(\Omega)}\le\left\|v\right\|_{L^1(\Omega)}\left\|\varphi\right\|_{L^\infty(\Omega)}$$ by Hölder's inequality and $\varphi\in\mathcal L^\infty(\Omega)$, the integral on the left-hand side of $(1)$ is well-defined, if $v\in\mathcal L^1(\Omega)$. Moreover, since $$\langle u,\nabla\varphi\rangle_{L^2(\Omega;\mathbb R^d)}\le\left\|u\right\|_{L^2(\Omega;\mathbb R^d))}\left\|\nabla\varphi\right\|_{L^2(\Omega;\mathbb R^d))}$$ by the Cauchy–Schwarz inequality and $\nabla\varphi\in\mathcal L^\infty(\Omega;\mathbb R^d)$, the integral on the right-hand side of $(1)$ is well-defined, if $u\in\mathcal L^2(\Omega)$.

Question:

  1. Is $v=\operatorname{div}u$ possible for more general $(u,v)$?
  2. Does $\operatorname{div}u$ exist for each $u\in\mathcal L^2(\Omega;\mathbb R^d)$? If that's the case: How can we prove it (or where can I find a rigorous proof of that) and does the same hold true for more general $u$?
  3. Does $\operatorname{div}u$, if each component of $u$ is weakly differentiable?
$\endgroup$
1
  • $\begingroup$ Why isn't the answer accepted? $\endgroup$ Mar 13, 2023 at 13:25

1 Answer 1

2
$\begingroup$
  1. In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions.
  2. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence.
  3. If each component of $u$ is weakly differentiable with weak derivative $\partial_i u \in L^1_{\text{loc}}(\Omega)$, then $\operatorname{div} u = \sum_{i=1}^d \partial_i u$.
$\endgroup$
2
  • $\begingroup$ How can $(1)$ be defined for distributions? $\endgroup$
    – 0xbadf00d
    Jul 11, 2016 at 10:36
  • $\begingroup$ If $\Phi_i$ are some distributions, you can define $\operatorname{div}(\Phi)(v) = \sum_i \Phi_i(\partial v /\partial x_i)$. $\endgroup$
    – gerw
    Jul 11, 2016 at 17:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .