Vorticity constant along a streamline In the case of a 2D incompressible flow  $\omega  = - \nabla^2 \psi$ where $\psi$ is the stream function.
From the vorticity equation of motion I am able to find the following:
$$
\frac{\mathrm{d}\omega}{\mathrm{d}t} = \frac{\partial\omega}{\partial t} + (u\cdot\nabla) \omega = 0 
$$
Is this meant to show that in the case of steady flow $\nabla^2 \psi$ is constant on a streamline?
How does this show this? I am very confused.
 A: Presumably, this is inviscid flow since the viscous terms are absent in your vorticity equation. Since the flow is two-dimensional, there is only one non-zero vorticity component $\omega$ aligned in the direction normal to the plane of flow. Hence, there is no vortex-stretching term $\bf{\omega} \cdot \nabla\mathbf{u}$.
In steady flow, the vorticity equation reduces to $\mathbf{u}\cdot \nabla \omega = 0$.
The gradient $\nabla \omega$ is always oriented in a direction normal to a level curve $\omega = \text{constant}.$ The condition $\mathbf{u} \cdot \nabla \omega = 0$ implies that the velocity at a point is tangential to such a curve. Hence, level curves of $\omega$ correspond to streamlines.
Velocity field is tangential to streamlines (level curves of streamfunction)
In two-dimensional, incompressible flow we can define a stream function $\psi$ by a line integral along any path $C$ joining $(0,0)$ to $(x,y)$
$$\psi(x,y) = \int_C \mathbf{u} \cdot \mathbf{n} dl = \int_C u \, dy - v \, dx.$$
Using the incompressiblity condition $\nabla \cdot \mathbf{u} =  \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$ and Green's theorem it follows that the line integral is independent of path and $\psi$ is a well-defined function which is related to the velocity field by
$$u  = \frac{\partial \psi}{\partial y}, \,\,\,\ v  = -\frac{\partial \psi}{\partial x}. $$
Hence,
$$\mathbb{u} \cdot \nabla \psi = u \frac{\partial \psi}{\partial x} + v \frac{\partial \psi}{\partial y}= u(-v) + v(u) = 0.$$
