Axisymmetry of the pressure #

For a classical solution of the forced Navier–Stokes equations eq:nse:forced on the unit cylinder whose velocity and force are both axisymmetric, the pressure is axisymmetric too: p (Q_φ x, t) = p (x, t) for every angle φ (pressure_rotZ_invariant). Rotation invariance of the pressure is therefore a consequence of the momentum equation and not an extra standing assumption of Section sec:aniso:equations.

The argument is the one of design note R8. The system is covariant under the rigid rotations (isClassicalSolutionOn_rotField), so the rotated triple (ℛ_φ u, p ∘ Q_φ⁻¹, ℛ_φ f) solves it as well. Axisymmetry identifies ℛ_φ u with u and ℛ_φ f with f on the cylinder, and the cylinder is open, so the two systems have the same velocity, the same force and the same derivatives of the velocity there. Subtracting the two momentum equations leaves only the two pressure gradients, whence

∂_i (p ∘ Q_φ⁻¹) = ∂_i p on the cylinder.

On a fixed time slice the difference y ↦ p (Q_φ⁻¹ y, t) - p (y, t) therefore has vanishing spatial derivative on the unit ball, which is convex and open, so it is constant there. Its value at the origin is zero because every rotation fixes the axis, and the constant is therefore zero.

Classical derivatives depend only on the values on the cylinder #

The pressure gradient is unchanged by a rotation of the point #

The rotated pressure slice differs from the pressure slice by a constant #