Consider the bilinear form $$a(u,v)=2\mu\int_\Omega \mathrm{trace}\left(\varepsilon(u)^T\varepsilon(v)\right)dx + \lambda\int_\Omega \mathrm{div}(u)\mathrm{div}(v)dx$$ with $\varepsilon(u)=\frac{1}{2}\left(\nabla u+\nabla u^T\right)$ and $u:\Omega\subset\mathbb R^3\to\mathbb R^3$.
My textbook says that we have $|a(u,v)|\leq d(\lambda+\mu)\|u\|_{H^1(\Omega)}\|v\|_{H^1(\Omega)}$ for all $u,v\in (H^1(\Omega))^3$, but I don't quite know how to get there.
I think by Cauchy-Schwarz we have $$\left(\int_\Omega \mathrm{div}(u)\mathrm{div}(v)dx\right)^2\leq\int_\Omega \mathrm{div}(u)^2dx\cdot\int_\Omega \mathrm{div}(v)^2dx\leq \int_\Omega |\nabla u|^2dx\cdot \int_\Omega |\nabla v|^2dx$$ and hence an estimate $\int_\Omega\mathrm{div}(u)\mathrm{div}(v)dx\leq \|u\|_{H^1(\Omega)}^2\|v\|_{H^1(\Omega)}^2$.
But how can I estimate the first integral? Can anyone give me an input on this? Thanks!
You can as well use Cauchy-Schwarz: $$ \int_\Omega trace(\epsilon(u)^T\epsilon(v)) \le c \|\epsilon(u)\|_{L^2(\Omega)} \|\epsilon(v)\|_{L^2(\Omega)} \le c \| \nabla u\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)} $$