It looks like you have a 2-d problem, hence there is only one vorticity component omega_z, oriented orthogonal to the x-y plane. It is defined as -du/dy+dv/dx  for a positive rotation sense counter clockwise, and hence it is given by your initial base flow: omega_z is just a constant everywhere (-du/dy). Since you do an inviscid computation, there is no generation of vorticity in the field, i.e. the inlet vorticity integral is kept, unless you impose a source term in the field. By the way, the u initial distribution comes from a viscous case.

Hi Markus,

Thanks for your comment.

I tried equating vorticity to -du/dy but in order to evaluate du/dy I need velocity values at the wall. Since the flow is inviscid , I cannot say u is zero.

Also, I totally agree with your point that there will not be any vorticity generation but I want to see how the initial vorticity patch convects with the velocity field as the time passes. In order to do that I need to specify the initial vorticity field.

There is an excellent paper on the same topic but he used stream vorticity formulation and hence it was easy to specify the boundary conditions.

The value of the velocity itself does not matter. According to your plot the velocity gradient does not alter there initially, meaning that du/dy is constant for the base flow. If the velocity has a jump at the wall then the vorticityhas also one and gets very large (-> infinite) locally. But typically the numerical code smears it a little. So you can start with the constant value of the base vorticity, plus the disturbance vorticity distribution that you want to see convected. Note that the disturbance vorticity should also be reflected by the velocity distribution. Everything else is a result, of physics but also of numerical viscosity.

PS We used extensively the velocity-vorticity formulation, see, e.g., the paper attached.