Past exam of the fluids and waves course of the University of Oxford / 2017 / 1 / a

For a scalar fluid property $f(\mathbf{x},t)$, the Eulerian time derivative ${\partial }_{t}f$ measures change at a fixed Eulerian coordinate. The material derivative, also called the Lagrangian time derivative, measures change while following one fluid element. If its Lagrangian trajectory is $\mathbf{X}(\mathbf{a},t)$, where $\mathbf{a}$ is its fixed Lagrangian coordinate, then ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$.

The chain rule therefore givesThe last term accounts for entering a region where the property differs. For example, a spatially varying but time-independent velocity field has zero Eulerian time derivative, while a moving fluid element can have nonzero material acceleration. The same formula applies componentwise to a vector field.

Let $V(t)$ be a smoothly moving material volume, carried by a smooth velocity field $\mathbf{u}$, and let $f$ be continuously differentiable. The material-volume form of the Reynolds transport theorem statesEquivalently, the divergence theorem giveswhere $\mathbf{n}$ is the outward unit normal vector. The second integral measures the flux carried across the instantaneous boundary by its motion.

For the proof, use the particle map $\mathbf{x}=\mathbf{X}(\mathbf{a},t)$ from $V(0)$, with positive Jacobian determinant $J=\det (\partial \mathbf{X}/\partial \mathbf{a})$. Euler's volume identity isIt is the Jacobian evolution of a smooth flow: if $F=\partial \mathbf{X}/\partial \mathbf{a}$, differentiation gives $F_t=(\nabla \mathbf{u})F$, and differentiation of the determinant gives $J_t=J\mathrm{tr}(F^{-1}{F}_t)=J\nabla \cdot \mathbf{u}$. The change of variables formula now moves the integral to the fixed reference volume:Finally, $Df/Dt+f\nabla \cdot \mathbf{u}={\partial }_{t}f+\nabla \cdot (f\mathbf{u})$, giving the equivalent boundary form.

Write $\rho$ for the mass density. A material volume contains the same fluid elements at all times, so conservation of mass requiresby the Reynolds transport theorem. Since the initial material volume can be chosen arbitrarily, continuity of the integrand forces it to vanish pointwise. Thus the continuity equation isFor positive mass density, the incompressibility condition $D\rho /Dt=0$ reduces this to $\nabla \cdot \mathbf{u}=0$. Mass conservation itself does not require incompressible flow.