 Solution

For constant density, divide the incompressible Euler equations by $\rho$. The vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ converts them into ${\partial }_t\mathbf{u}=\mathbf{u}\times \mathbf{\omega }-\nabla (p/\rho +|\mathbf{u}{|}^2/2)$. Taking the curl eliminates the gradient. Expand the curl of the cross product, use $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, and obtainThis is the vorticity equation, including vortex stretching. Without constant density there would in general be an additional baroclinic term; the displayed source equations use the constant-density interpretation.

For the specified velocity, direct differentiation gives $\mathbf{\omega }=2\Omega (t)\mathbf{k}$. It is spatially uniform, while $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=2\beta \mathbf{\omega }$. Thus ${\Omega }^{\prime }=2\beta \Omega$ andIf the positive $z$ direction agrees with the initial vorticity, $2\Omega (0)={\omega }_0$ as written in the question. For opposite rotation the initial component is signed and the vector has the corresponding minus sign.

For a material particle, $d(x^2+y^2)/dt=2x(-\beta x-\Omega y)+2y(-\beta y+\Omega x)=-2\beta (x^2+y^2)$, while $dz/dt=2\beta z$. The initial material circle therefore evolves intoThe azimuthal rotation changes the labels of points on the circle but not the circle as a set. Its counterclockwise circulation, viewed from positive $z$, isfor the positive-orientation convention. The strain contributes no tangential velocity. Thus circulation stays constant although vorticity grows and the enclosed material area shrinks, illustrating Kelvin's circulation theorem for a material loop in constant-density inviscid flow with no nonconservative body force.