Show a Result for Pressure Using Navier-Stokes Equation

Question

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

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?

Answer 1

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.

Answer 2

To show how this elliptic equation for the pressure emerges from the traditional momentum equation, take the divergence

$$\partial_t \operatorname{div}u + \operatorname{div}(u\cdot \nabla u)+\operatorname{div}\nabla p = \Delta(\operatorname{div}u) + \operatorname{div}f$$

and use that $u$ has vanishing divergence $$-\Delta p = \operatorname{div}(u\cdot \nabla u)-\operatorname{div}f$$

You can rewrite this using the "divergence of a dyad rule" $$\operatorname{div}(u\otimes u) = u\underbrace{\operatorname{div}u}_{=0} + u\cdot \nabla u$$

which yields $$-\Delta p = \operatorname{div}\operatorname{div}(u\otimes u) -\operatorname{div}f$$

In case you require this in Einstein index notation with implicit double-index summations, you can write $$[(u\cdot \nabla)u]_j \equiv u_i \partial_i u_j.$$

This quantity has 1 free index (j), and is therefore a 1-tensor (or vector). Now we apply the divergence to it and get $$\partial_j u_i \partial_i u_j$$

Which is exactly what you have in the first line of your question.