# How to get $\nabla p=\nabla (-\Delta)^{-1} \nabla \cdot (u\cdot \nabla u)$ from the Navier-Stokes equations?

The incompressible Navier-Stokes equations are given by:

$$u_t + u \cdot \nabla u -\Delta u + \nabla p=0,\qquad \text{div } u=0,$$ where $u$ is the velocity field and $p$ is the pressure.

Here is a fact that I don't know why it's true from the above equations: $$\nabla p=\nabla (-\Delta)^{-1} \nabla \cdot (u\cdot \nabla u).$$

Thanks!

• Do you know what is $(-\Delta)^{-1}$? – Jack Nov 17 '16 at 0:22
• @Jack I just know it's the inverse of Laplacian operator. Maybe it's not enough, though. – Feipi Wen Nov 17 '16 at 0:25

Formally, you have the so called pressure poisson equation: $$-\Delta p=\nabla\cdot((u\cdot\nabla)u)\tag{1}$$
Now applying the inverse of $-\Delta$ on both sides, you get $$p=(-\Delta)^{-1}\nabla\cdot((u\cdot\nabla)u)$$ and thus $$\nabla p=\nabla(-\Delta)^{-1}\nabla\cdot((u\cdot\nabla)u)$$
To see how to get (1), consider $$\nabla\cdot(u_t+(u\cdot\nabla)u)=\nabla\cdot(\Delta u-\nabla p)$$ and use the divergence-free condition for $u$. On the LHS, you have $$\partial_t(\nabla\cdot u)+\nabla\cdot((u\cdot\nabla)u)=0+\nabla\cdot((u\cdot\nabla)u)$$ On the RHS, you have $$\Delta(\nabla\cdot u)-\Delta p=-\Delta p.$$