The <kinetic helicity conservation law> implies
$$
\frac{Dq}{Dt}=\boldsymbol\omega\cdot\nabla\left(\frac{u^2}{2}-w\right)-q\nabla\cdot\mathbf u.
$$
Therefore $q$ is materially constant exactly when the right side vanishes. <Incompressible flow> together with constancy of $w-u^2/2$ along vortex lines is sufficient. In compressible flow, the density-normalized quantity satisfies $D(q/\rho)/Dt=-(\boldsymbol\omega/\rho)\cdot\nabla(w-u^2/2)$.

Material conservation of kinetic helicity density

Write $q=\mathbf u\cdot\boldsymbol\omega$ for the <kinetic helicity density>, and retain $\mathcal B=w+u^2/2$. Since $\boldsymbol\omega$ is a <curl>, $\nabla\cdot\boldsymbol\omega=0$. Differentiating the density and using the equations from the preceding part gives
$$
\partial_tq=-\boldsymbol\omega\cdot\nabla\mathcal B+\mathbf u\cdot\nabla\times(\mathbf u\times\boldsymbol\omega).
$$
The <divergence of a cross product> gives
$$
\nabla\cdot\{\mathbf u\times(\mathbf u\times\boldsymbol\omega)\}=(\mathbf u\times\boldsymbol\omega)\cdot\boldsymbol\omega-\mathbf u\cdot\nabla\times(\mathbf u\times\boldsymbol\omega).
$$
The first term on the right is zero, while $\boldsymbol\omega\cdot\nabla\mathcal B=\nabla\cdot(\mathcal B\boldsymbol\omega)$. Thus the <kinetic helicity conservation law> has flux
$$
\boxed{\mathbf F_{H_k}=\mathcal B\boldsymbol\omega+\mathbf u\times(\mathbf u\times\boldsymbol\omega)=q\mathbf u+\left(w-\frac{u^2}{2}\right)\boldsymbol\omega,\qquad\partial_tq+\nabla\cdot\mathbf F_{H_k}=0.}
$$
The last equality for the flux uses the <vector triple product identity>. Integrating and applying the <divergence theorem> gives $d\int_Vq\,dV/dt=-\int_{\partial V}\mathbf F_{H_k}\cdot\mathbf n\,dS$ for a fixed volume. Hence its total <kinetic helicity> is conserved when the boundary flux vanishes; for example, tangency of both <velocity field> and <vorticity> to the boundary is sufficient. The local <conservation law> does not by itself imply that its density follows each fluid particle unchanged.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 1 / b / Solution

Expanding the <kinetic helicity conservation law> separates <advection> from the other flux:
$$
\boxed{\frac{Dq}{Dt}=\boldsymbol\omega\cdot\nabla\left(\frac{u^2}{2}-w\right)-q\nabla\cdot\mathbf u.}
$$
Thus <material conservation of kinetic helicity density> holds precisely when
$$
\boxed{\boldsymbol\omega\cdot\nabla\left(\frac{u^2}{2}-w\right)=q\nabla\cdot\mathbf u.}
$$
An especially useful sufficient pair of conditions is <incompressible flow>, $\nabla\cdot\mathbf u=0$, and constancy of $w-u^2/2$ along integral curves of the <vorticity>, $\boldsymbol\omega\cdot\nabla(w-u^2/2)=0$. Together these conditions make both terms vanish without cancellation. They are not necessary individually, because the two terms in the displayed balance can cancel. An <irrotational flow> is a trivial conserved case with $q=0$.

For contrast, combining the balance with <mass conservation> gives
$$
\frac D{Dt}\left(\frac q\rho\right)=-\frac{\boldsymbol\omega}{\rho}\cdot\nabla\left(w-\frac{u^2}{2}\right).
$$
Consequently constancy of $w-u^2/2$ along integral curves of the <vorticity> makes $q/\rho$ an advected scalar even in <compressible flow>; it does not generally make $q$ itself constant.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 1 / c / Solution

For smooth inviscid <barotropic fluid> motion without gravity or magnetic forces, let $w$ be the <specific enthalpy> with $\nabla w=\rho^{-1}\nabla p$, and set $q=\mathbf u\cdot\boldsymbol\omega$. Then
$$
\partial_tq+\nabla\cdot\left[q\mathbf u+\left(w-\frac{u^2}{2}\right)\boldsymbol\omega\right]=0.
$$
This follows by dotting the <barotropic vorticity transport> equation with $\mathbf u$ and using the <divergence of a cross product>. A volume integral is constant when the resulting boundary flux vanishes.

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 Incoming links: Kinetic helicity conservation law