I'm working on an unassessed course problem,
Suppose that the velocity field in an incompressible fluid is given by $$\vec{u}=(\alpha x-\Omega y, -y+\Omega x,0),$$ where $\alpha$ and $\Omega$ may be functions of time. Find $\alpha$ and the vorticity $\vec{\omega}$.
If the density and kinematic viscosity are constant, show that if $\Omega$ is not constant in time then the fluid must be viscous.
I get $$\alpha=1,\hspace{1em}\vec{\omega}=(0,0,2\Omega),$$ agreeing with the solution booklet.
Setting up to use the vorticity equation $$\frac{\partial\vec{\omega}}{\partial t}+(\vec{u}\cdot\nabla)\vec{\omega}=(\vec{\omega}\cdot\nabla)\vec{u}+\nu\nabla^2\vec{\omega},$$ I get \begin{align} & (\vec{u}\cdot\nabla)\vec{\omega}=\vec{0}, \hspace{1em} (\vec{\omega}\cdot\nabla)\vec{u}=\vec{0}, \end{align} also agreeing with the solution booklet.
Then the solution booklet goes on to reason \begin{align} \therefore \; & \frac{\partial\vec{\omega}}{\partial t}=\nu\nabla^2\vec{\omega} \\ \therefore \; & \frac{\partial\vec{\omega}}{\partial t} \neq 0 \implies \nu \neq 0 \implies \text{ the fluid is viscous, as required.} \end{align} But doesn't $$\nabla^2\vec{\omega}=\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)(0,0,2\Omega)=\vec{0}?$$ What am I missing?
The question itself is ill-posed.
Your velocity field is $$\boldsymbol u(t,x,y)= \begin{bmatrix}x-\Omega(t) y \\ -y+\Omega(t)x\end{bmatrix}\tag{1}$$ I claim that $(1)$ satisfies the momentum equation only if $\Omega$ is constant in time. The momentum equation is $$\partial_t\boldsymbol u+(\boldsymbol u\cdot \nabla)\boldsymbol u-\nu\Delta\boldsymbol u=-\nabla p\tag{2}$$ If we calculate the LHS of $(2)$ using $(1)$, we get (NOTE $\Delta \boldsymbol u=0$) $$\frac{\partial}{\partial t}\begin{bmatrix}x-\Omega(t) y \\ -y+\Omega(t)x\end{bmatrix}+(x-\Omega(t)y)\frac{\partial}{\partial x}\begin{bmatrix}x-\Omega(t) y \\ -y+\Omega(t)x\end{bmatrix}+(-y+\Omega(t)x)\frac{\partial}{\partial y}\begin{bmatrix}x-\Omega(t) y \\ -y+\Omega(t)x\end{bmatrix}=-\nabla p(t,x,y)$$ Carrying out the computations, $$\begin{bmatrix}-y~\dot{\Omega}(t) \\ x~\dot{\Omega}(t)\end{bmatrix}+(x-\Omega(t)y)\begin{bmatrix}1 \\\Omega(t) \end{bmatrix}+(-y+\Omega(t)x)\begin{bmatrix}-\Omega(t) \\ -1 \\ \end{bmatrix}=-\nabla p(t,x,y)$$
This simplifies to $$\begin{bmatrix}-y~\dot{\Omega}(t)+x-x~{\Omega(t)}^2 \\ x~\dot{\Omega}(t)+y-y~{\Omega(t)}^2\end{bmatrix}=-\nabla p(t,x,y)$$
Now, if you take the curl of the right hand side, you of course get $$\nabla\times(-\nabla p)=0\tag{*}$$ However, if you take the curl of the left hand side, you get $$\nabla\times\begin{bmatrix}-y~\dot{\Omega}(t)+x-x~{\Omega(t)}^2 \\ x~\dot{\Omega}(t)+y-y~{\Omega(t)}^2 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 2\dot{\Omega}(t)\end{bmatrix}\tag{**}$$ The only way for $(*)$ and $(**)$ to be compatible is if $\dot{\Omega}(t)=0$, i.e, $\Omega$ is constant.
Edit: Implications.
The implications of this is that the 2D vorticity equation, $\partial_t\omega=\nu\Delta\omega$, being derived from the momentum equation, is completely invalid for this $\boldsymbol u$ field unless $\dot{\Omega}=0$.