+ \mu \delsq \vvec$$ If the fluid is incompressible in the sense that $\rho$ is a constant in both time and space, we have the condition: $$\del.\vvec = 0$$ To analyze matters further, we introduce the vorticity vector $$\omvec = \del \times \vvec$$ In the following discussion we demand incompressibility but allow for non-zero vorticity. Using simple identities from vector calculus the Navier-Stokes equations may then be recast in the form $$\rho(\frac{\partial \vvec}{\partial t} - \vvec \times \omvec) + \del (p + \frac{1}{2} \rho \vvec^2
)= -\mu \del \times \omega .$$ Taking the curl of this, we arrive at the vorticity equation $$\frac{\partial \omvec}{\partial t} + \vvec.\del \omvec - \omvec.\del \vvec = \nu \delsq \omvec$$ where the kinematic viscosity $\nu = \mu/\rho$.

The stream function
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Since the velocity field is divergence-free, we may introduce a vector potential $\Psivec$ such that $$\vvec = \del \times \Psivec$$ and furthermore we may choose it so that it is divergence free: $$\del.\Psivec = 0$$ The vector potential can be reduced to a single function when there is an appropriate symmetry. The resulting object is a stream function. For example, planar 2D flow is obtained by setting (and note that this automatically satisfies satisfies 