Note that when deriving the more general compressible Navier-Stokes equations,
the diffusion is modeled as the divergence of the stress tensor
@f{eqnarray*}
- \tau = - \mu (2\varepsilon(\textbf{u}) - \frac{2}{3}\nabla \cdot \textbf{u} I),
+ \tau = - \mu \left(2\varepsilon(\textbf{u}) - \frac{2}{3}\nabla \cdot \textbf{u} I\right),
@f}
where $\mu$ is the viscosity of the fluid. With the assumption of $\mu=1$
(assume constant viscosity and non-dimensionalize the equation by dividing out
@f{eqnarray*}
\textrm{div}\; \tau
= -2\textrm{div}\;\varepsilon(\textbf{u})
- = -\triangle \textbf{u} + \nabla \cdot (\nabla\textbf{u})^T
+ = -\triangle \textbf{u} - \nabla \cdot (\nabla\textbf{u})^T
= -\triangle \textbf{u}.
@f}
This is because the $i$th entry of $\nabla \cdot (\nabla\textbf{u})^T$ is given by:
= \sum_j \frac{\partial}{\partial x_j} [(\nabla\textbf{u})]_{j,i}
= \sum_j \frac{\partial}{\partial x_j} \frac{\partial}{\partial x_i} \textbf{u}_j
= \sum_j \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} \textbf{u}_j
-= \frac{\partial}{\partial x_i} \textrm{div}\; \textbf{u}
+= \frac{\partial}{\partial x_i}
+ \underbrace{\textrm{div}\; \textbf{u}}_{=0}
= 0.
@f}
If you can not assume the above mentioned regularity, or if your viscosity is