If $\vec{\omega}\cdot\vec{n}=0$ for $t=0$ then it holds for all time I am trying to show that if $\vec{u}$ is a solution of the incompressible Euler equations with zero body force in a bounded domain $D$, and if $\vec{\omega}.\vec{n}=0$ on the boundary of $D$ for $t=0$ then $\vec{\omega}.\vec{n}=0$ also holds for all time $t\geq0$.
I thought that if I can show $\frac{∂\omega}{∂t}=0$ then by time independence of $\omega$ the above is true. Taking the curl of the incompressible Euler equations with $\vec{f}=\vec{0}$ then gives me $\frac{∂\omega}{∂t}=\nabla\times(\vec{u}\times\vec{\omega})$. (Of course after a few lines of vector identities/algebra).
I'm not sure where to go from here.
 A: In inviscid flow, the fluid velocity satisfies the no-flux  condition $\mathbf{u} \cdot \mathbf{n} = 0$ on the boundary $\partial D$, but the tangential component of velocity need not vanish as in viscous flow.
Let $\Sigma(t)$ be a smooth material surface element, bounded by the simple, smooth closed curve $C(t)$. As $\Sigma(t)$  evolves in time,  it contains the same fluid particles present at time $t= 0$.  If $\Sigma(0) \subset \partial D$, then no flux condition ensures that $\Sigma(t) \subset \partial D$ for all $t > 0$.
By Stokes' theorem we have
$$\oint_{C(t)}\mathbf{u} \cdot dl = \int_{\Sigma(t)}(\nabla \times \mathbf{u}) \cdot \mathbf{n} \, dS = \int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS $$
In an inviscid flow with the absence of body forces it follows from Kelvin's circulation theorem that
$$\frac{d}{dt}\oint_{C(t)}\mathbf{u} \cdot dl = \frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS  =0$$
By a generalization of the Leibniz integral rule we can pass the derivative under the integral on the RHS to obtain
$$\frac{d}{dt}\int_{\Sigma(t)} \mathbf{\omega} \cdot \mathbf{n} \, dS  = \int_{\Sigma(t)} \left[\frac{\partial\mathbf{\omega}}{\partial t}+ (\nabla \cdot \mathbf{\omega})\mathbf{u}  \right]\cdot \mathbf{n} \, dS- \oint_{C(t)}(\mathbf{u}\times \mathbf{\omega}) \cdot dl$$
Note that $(\nabla \cdot \mathbf{\omega})\, \mathbf{u} \cdot \mathbf{n} = 0$ and $\mathbf{u}\times \mathbf{\omega} = \mathbf{u}\times (\nabla \times \mathbf{u}) = \mathbf{u} \cdot \nabla\mathbf{u} - \mathbf{u} \cdot \nabla\mathbf{u} = \mathbf{0}$.
Thus,
$$\int_{\Sigma(t)} \frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} \, dS   = 0$$
Under the usual smoothness assumption for the velocity and vorticity fields, since this holds for any $\Sigma(t) \subset \partial D$, we get $\frac{\partial\mathbf{\omega}}{\partial t}\cdot \mathbf{n} = 0$ for all $t > 0$.
