# Does the Continuity Equation Imply it's Zero for Incompressible Flow?

This might be an incredibly simple question, but I haven't been able to figure it out on my own. I'm an engineer by trade, so please forgive my unfamiliarity with vector calculus. I'm interested in the inviscid, incompressible Navier-Stokes equation and how it relates to the Bernoulli equation. The form of the NS equation I'm interested in is as follows:

$$\frac{\partial {\mathbf u} }{\partial t} + ({\mathbf u} \cdot \nabla ){\mathbf u} = -\frac{1}{\rho} \nabla p + {\mathbf g}$$

which can relatively easily be manipulated into the Bernoulli equation if the flow is assumed to be steady-state.

$$\frac{p}{\rho} + \frac{1}{2}\| {\mathbf u} \| ^2 +gh = const.$$

The convective term of NS $$({\mathbf u} \cdot \nabla ){\mathbf u}$$ turns into the velocity term in Bernoulli, $$\frac{1}{2}\| {\mathbf u} \| ^2$$. This is obviously a very important part of the Bernoulli equation. However, if the fluid is incompressible, we have the continuity equation which states that $$\nabla \cdot {\mathbf u}=0$$. Since the dot product is commutative, doesn't this imply that $$({\mathbf u} \cdot \nabla ){\mathbf u} = {\mathbf 0}$$? Obviously this is not the case, I am just wondering where I am going wrong, since vector calc is not my strength.

Since the dot product is commutative, doesn't this imply that $$(\mathbf u\cdot\nabla)\mathbf u=0$$?
No because $$\nabla$$ is an operator that acts on the object to its right. The term in the parenthesis is equal to, $$\mathbf u\cdot\nabla=u_x\partial_x+u_y\partial _y+u_z\partial_z$$ and this is applied to each term in the vector $$\mathbf u$$, $$\left(\mathbf u\cdot\nabla\right)\mathbf u=\left(u_x\partial_x+u_y\partial_y+u_z\partial_x\right)\mathbf u=\left(\begin{array}{c}u_x\partial_xu_x+u_y\partial_yu_x+u_z\partial_xu_x \\ u_x\partial_xu_y+u_y\partial_yu_y+u_z\partial_xu_y \\ u_x\partial_xu_z+u_y\partial_yu_z+u_z\partial_xu_z\end{array}\right)$$