Had to look up non-newtonian fluids, water wasn't listed. But it looks like a non-newtonian fluid just means the shear force is not linearly proportional to the rate of distortion. ie. the 2nd term on the right hand side of the Navier Stokes Equation doesn't work as it should.
(This equation looks complicated but is just F=ma for fluids. The ugly looking left hand side of the equation just allows the mathematics to follow a particular fluid particle. The first term on the right is just the pressure force on the fluid particle of interest, the last is external forces on the particle of fluid we're following, usually gravity.
That 2nd derivative of the velocity in the 2nd term on the right is interesting. The 2nd term again is the viscous force on our particle due to velocity gradients in the surrounding fluid. At first you'd think the first derivative would be OK. But the 2nd is needed for the shear force because if the velocity gradient is linear shear drag from neighboring particles below = shear push from the fluid particles above above and our fluid particle gets no net viscous shear force so doesn't accelerate.
The trouble with the Navier Stokes equation is that even though it makes sense for
looking at one fluid particle swimming amongst all its neighbours, if you don't know what the neighbours are doing you've got a real problem. The Navier Stokes Equation is very difficult to solve.
Sorry got sidetracked by the non-newtonian fluids and tried to demystify the Navier Stokes equation. Took me ages to work out what that famous equation was on about years ago. Do windsurfers need to think about things like this? Maybe maybe not. I've got the red/yellow arrow day off work)
But back to our no.1 issue.

The bubbling reduces the overall density but the biggest bonus is that the bubbles give it some compressibility, which means the above form of the equation is no help at all.