# Why does including the pressure in this FVM for Stokes 2nd Problem lead to wrong solutions?

I'm trying to learn how to use finite volume methods and I want to solve a more general case of Stokes' second problem i.e. an infinite half-plane oscillating harmonically with no-slip boundary conditions.

I've been following the code from this lecture and changed the boundary conditions to:

u[1:-1, 1]  = u[1:-1, 2]                #left
u[1:-1, -1] = u[1:-1, -2]               #right
u[-1,1:]    = 2*np.cos(n*dt) - u[-1,1:] #top
u[0, 1:]    = u[1,1:]                   #bottom

v[1:,0] = v[1:,1]      #left
v[1:,-1] = v[1:,-2]    #right
v[-1,1:-1] = 0.0       #top
v[1,1:-1] = 0.0        #bottom


(My thinking was that this would imply no gradients at the left, right, and bottom boundary so the flow would just continue out of/into the volume and to oscillate the top by setting the value through a ghost cell.)

This leads to vortex formations however

But we know by the analytical solution $$u(y,t) = e^{-y\sqrt{\frac{\omega}{2\nu}} } \cos\left(\omega t - y\sqrt{\frac{\omega}{2\nu} } \right)$$ that there will be no velocity component in the y-direction. And it's not like this vortex is just a small deviation.

However if I simply exclude the projection step/set the pressure to 0 everywhere I recover the analytical solution. Why does this happen? From what I understand the pressure is essentially used to enforce that the flow field is divergenceless, however $$u(y)$$ obviously also has 0 divergence so why does the projection method fail here?