If I have a bilinear form
$$a : (H^1(\Omega))^3 \times (H^1(\Omega))^3 \mapsto \mathbb{R} \hspace{0.9in} a(\vec{u}, \vec{v}) = \int_{\Omega}{(D\vec{u})^T C (D\vec{v})\, d\Omega} $$
I would like to find the kernel of this bilinear form when $u \in V$ such that, $$V = \{ (u_x, u_y, u_z) \in (H^1(\Omega))^3 | u_x(0,0,0) = u_y(0,0,0) = u_z(0,0,0) = 0 \}$$
I would like to comment what are C and D operators. D is Symmetric derivative operator. $$D\vec{u} = \begin{bmatrix} \frac{\partial u_x}{\partial x} \\ \frac{\partial u_y}{\partial y} \\ \frac{\partial u_z}{\partial z} \\ \frac{1}{2}( \frac{\partial u_x}{\partial y} + \frac{\partial u_y}{\partial x}) \\ \frac{1}{2}( \frac{\partial u_y}{\partial z} + \frac{\partial u_z}{\partial y}) \\ \frac{1}{2}( \frac{\partial u_x}{\partial z} + \frac{\partial u_z}{\partial x})\end{bmatrix} : \Omega \mapsto \mathbb{R}^6$$ C is fourth order isotropic elasticity tensor in voigt notation. http://web.mit.edu/16.20/homepage/3_Constitutive/Constitutive_files/module_3_with_solutions.pdf pg no - 18