Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 331 / 1 / b / iii

For an even velocity potential, take ${\Phi }_I=C\cosh (kz)$ and the decaying exterior amplitudes equal at the upper and lower interfaces. At $z=L$, kinematics gives $A=ic{B}_{+}$ and $C\sinh (kL)=i(V-c)B_{+}$. The dynamic condition becomes $-cA=(V-c)C\cosh (kL)$. ThereforeWriting $s=\tanh (kL)$, its two roots areThe interior vertical velocity is odd, so $B_{-}=-B_{+}$; this is a varicose mode of a planar jet. As with the odd potential, $k>0$, finite $L>0$ and $V\ne 0$ give a growing root. The parity of the potential and the parity of the interface displacements are opposite because the vertical velocity is its derivative.