2
$\begingroup$

The inequality is listed in line 10 on page 29: $\parallel \frac{\partial p}{\partial x_{k}} \parallel_{L^{2}(G_{R^{"}})} \leq c \parallel\nabla \frac{\partial p}{\partial x_{k}} \parallel_{H^{-1}(G_{R})}.$ First I would like to know how $\nabla \partial p / \partial x_{k}$ is defined as a functional in $H^{1}_{0}$. Secondly the proof of the inequality. Thanks in advance.

  • 0
    What is the domain of the $L^2$ norm in the left hand side?2011-12-06

0 Answers 0