Let $\Omega\subset\mathbb{R}^3$ be a bounded Lipschitz domain. Define the Hilbert space $$ H(div;\Omega):=\{u\in (L^2(\Omega))^3:\nabla\cdot u\in L^2(\Omega)\} $$ equipped with the graph norm $$ \|u\|_{H(div;\Omega)}:=\big(\|u\|^2_{(L^2(\Omega))^3}+\|\nabla\cdot u\|^2_{L^2(\Omega)}\big)^{1/2} $$ I have to prove that $$ \big|\int_{\partial\Omega}(u\cdot\nu)\phi\mathrm{d}A\big|\le\|u\|_{H(div;\Omega)}\|\phi\|_{H^1(\Omega)} $$ where $u\in(C^{\infty}(\bar{\Omega}))^3$,$\phi\in H^1(\Omega)$ and $\nu$ is the outward unitary normal vector on $\partial\Omega$.

The book I'm using says that the equality above is the Cauchy-Schwarz inequality, but I don't think so. I'm trying to prove it, but I can't.

Any help?

  • 1
    $\begingroup$ Are you sure you should have a constant of $1$ rather than some $C$? Cauchy-Schwarz gives $\left | \int_{\partial \Omega} (u \cdot \nu) \phi dA \right | \leq \| u \cdot \nu \|_{L^2(\partial \Omega)} \| \phi \|_{L^2(\partial \Omega)}$, which I would then bound in your form using the divergence and trace theorems. But the trace theorem gives a constant factor, which is at least the surface area divided by the volume (which could be a lot larger than $1$). $\endgroup$ – Ian Nov 19 '15 at 16:30
  • $\begingroup$ Exactly!!! I thought the same! Yes, the constant they use is $C=1$. The book is Peter Monk's "Finite Element Methods for Maxwell's Equations" $\endgroup$ – avati91 Nov 19 '15 at 16:34

By divergence theorem $$ \int_{\partial \Omega }u\cdot n \ \phi=\int_\Omega div(u\phi) = \int_\Omega \phi div(u) + \nabla \phi \cdot u. $$ This gives $$ \left|\int_{\partial \Omega }u\cdot n \ \phi\right|\le \|\phi\|_{L^2(\Omega)}\|div(u)\|_{L^2(\Omega)} + \|\nabla \phi\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \le \|\phi\|_{H^1(\Omega)}\|u\|_{H(div,\Omega)}. $$


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