= Solution
Write $\phi'=\Phi(z)e^{ik(x-ct)}$ and let $B_\pm$ be the displacement amplitudes at $z=\pm L$. Use subscripts $I$ for the jet interior and $O$ for the adjoining exterior. Each interface is a material <vortex sheet>, so its linearized <kinematic boundary conditions> are
$$
\Phi'_I(\pm L)=ik(V-c)B_\pm,\qquad
\Phi'_O(\pm L)=-ikcB_\pm.
$$
The equal <mass densities> make the hydrostatic displacement terms cancel. The <dynamic boundary condition for an inviscid interface> therefore gives
$$
\boxed{(V-c)\Phi_I(\pm L)=-c\Phi_O(\pm L),\qquad
\frac{\Phi'_I(\pm L)}{V-c}=\frac{\Phi'_O(\pm L)}{-c}=ikB_\pm.}
$$
The division form assumes $c\ne0,V$; the preceding undivided equations remain the proper conditions in those exceptional cases. Potentials themselves need not be continuous, and their vertical <derivatives> need not be equal: the same material displacement is advected by different tangential base <velocities>. At infinity the exterior potentials decay. These displacement and <pressure> conditions, rather than continuity of $\Phi$ and $\Phi'$, determine the <equal-density top-hat planar-jet dispersion relation>.
Back to article page
