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For a top-hat planar jet in stationary fluid of the same mass density, odd and even velocity potentials have the same temporal growth rate. Put $s=\tanh (|k|L)$. Their dispersion relations use $s$ and $1/s$, but $\sqrt{s}/(1+s)=\sqrt{1/s}/(1+1/s)$. Their real phase velocities are respectively $Vs/(1+s)$ and $V/(1+s)$. The common growth rate is strictly below $|kV|/2$ for finite $|k|L>0$, approaching the isolated vortex sheet value when the sheets decouple.

For an odd velocity potential, chooseAt $z=L$ the two kinematic conditions give $A=ic{B}_{+}$ and $C\cosh (kL)=i(V-c)B_{+}$. The dynamic condition is $-cA=(V-c)C\sinh (kL)$. Eliminating $A,C,B_{+}$ gives, with $s=\tanh (kL)$,The lower interface gives the same equation by odd parity of the velocity potential. The interior vertical velocity is even, so $B_{-}=B_{+}$: the displacement is a sinuous mode of a planar jet even though the potential is odd.

For $L>0$, $V\ne 0$ and $k>0$, $0<s<1$ and one root has $c_i>0$. Thus every nonzero wavenumber is unstable in this ideal top-hat planar jet. The phrase “all choices” is understood with these assumptions: $k=0$ has no exponential wave growth, and $V=0$ has no shear.