# Show a Result for Pressure Using Navier-Stokes Equation

By taking the divergence of the incompressible Navier-Stokes equation with a zero body force, $\underline{X} = \underline{0}$, show that the pressure satis es $$\frac{\partial^2p}{\partial x_i \partial x_i} = -\rho\frac{\partial u_j}{\partial x_i}\frac{\partial u_i}{\partial x_j}$$

In my notes I have this for an incompressible flow: \begin{align} \underline{\nabla}\cdot\underline{u} &= 0\tag1\\ \frac{\partial\underline{u}}{\partial t} + (\underline{u}\cdot\underline{\nabla})\underline{u} &= -\frac{1}{\rho}\underline{\nabla}p + \underline{X} + \nu\nabla^2\underline{u}\tag2\\ &= -\frac{1}{\rho}\underline{\nabla}p + \nu\nabla^2\underline{u}\quad\text{with \underline{X} = \underline{0}}\end{align}

I'm juse not sure how to take the divergence of all this to get the final result?

For an incompressible fluid it is clear that $\dot\rho=0$. Then via the continuity equation we find $$\nabla\cdot u = 0 .$$ We can now take the divergence of the Navier-Stokes equation and get $$-\nabla^2 P = \rho\nabla_j(u_i\nabla_i u_j).$$ The easiest way to solve this constraint is to convert the NS equation into an equation for the vorticity $\omega=\nabla\times u$. Thus $$\frac{\partial \omega}{\partial t} + u\cdot\nabla {\omega} = \eta {\nabla}^2 {\omega} + \omega\cdot\nabla {u},$$ where $\eta=\eta/\rho$ is the kinetic viscosity.
• @Michael For a rank-2 tensor which corresponds to a matrix, say $A$, its divergence is just the vector $(b_1,b_2,b_3)$, where $b_j$ is the divergence of the $j$th-column or row vector of $A$, depending on the convention. – Chee Han Mar 12 '18 at 20:58