Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 17B / i / Solution

The interface is a material surface, so its defining function $F=z-\zeta (x,y,t)$ obeys $DF/Dt=0$ on each side. Since ${\mathbf{u}}_i=\nabla {\Phi }_i$, the exact kinematic boundary condition isSubstitute ${\Phi }_i=U_{i}x+{\phi }_i$. Products such as ${\phi }_{ix}{\zeta }_x$ are second order, as is the correction from evaluating ${\phi }_{iz}$ at $z=\zeta$ instead of zero. Linearization therefore gives