Define: $$ \mathcal{L} = -\sum_{i,j} a_{ij} \partial_{x_i} \partial_{x_j}, $$ where $a_{i,j}=a_{j,i}$. If $a_{i,j}=\delta_{i,j}$ the operator is the Laplacian which is known to be self-adjoint in the $L^2$ norm. For other choices of coefficients $a_{i,j}$ - is the operator $\mathcal{L}$ self adjoint in the $L^2 $ norm?
If we restrict our attention to the $\mathbb{R}^3 $ case, we have that $$ \int_\Omega v \mathcal{L}u \, d \vec{x} = \int_{\partial \Omega} v(A\nabla u) \cdot \hat{n} dS - \int_\Omega (A\nabla u)\cdot \nabla v d \vec{x}, $$ $$ \int_\Omega u \mathcal{L}v \, d \vec{x} = \int_{\partial \Omega} u(A\nabla v) \cdot \hat{n} dS - \int_\Omega (A\nabla v)\cdot \nabla u d \vec{x}, $$ where $v \in C_c^{\infty} (\Omega)$, $A = (a_{i,j})$. From the symmetricity of $A$, the second term in both equalities is the same, but is there any reason for the first term to also be equal in the two equalities?