 Solution

Take the curl of the linear momentum equation. For constant $\mathbf{\Omega }$ and zero divergence, $\nabla \times (\mathbf{\Omega }\times \mathbf{u})=-(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. With $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, this givesTake another curl and use part (a): $-{\partial }_t{\nabla }^2\mathbf{u}=2\Omega {\partial }_z\mathbf{\omega }$. Differentiate in time and substitute the previous equation:ConsequentlyThe two curl operations remove the pressure gradient while retaining the rotational restoring force.