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Waves travelling in opposite directions relative to their local background currents can share a laboratory phase velocity. Their interaction can then lock their phases and extract energy from the mean shear, producing growing disturbances. In a layered flow, the coupling is often exponentially weak when the interface separation exceeds the waves' decay length. A dispersion relation for two density interfaces in uniform shear gives a simple explicit example.

For stable density jumps, $J>0$ and $q=J/(2\alpha )>0$. Let $y=C^2$. The two roots of the dispersion relation for two density interfaces in uniform shear areThey are real, and $y_{+}>0$. An exponentially growing normal mode exists precisely when $y_{-}<0$, giving a pair of imaginary phase speeds; the member with positive imaginary part grows in $e^{ik(x-ct)}$ for $k>0$.

Since $y_{+}{y}_{-}=(1-q{)}^2-q^2{E}^2$, this condition is $|1-q|<qE$. Solving both inequalities yieldsAt either endpoint $y_{-}=0$, so the exponential growth rate vanishes. Outside the open band both squared phase speeds are nonnegative and the interfacial modes have real frequencies. This is a modal instability test for the stably stratified configuration; unstable density inversions would require a separate analysis. Within the band the dimensional growth rate is $k\Delta U\sqrt{-y_{-}}/2$.

Take $k>0$ without loss of generality, so $\alpha =kh/2>0$, and require decay as $Z\to \pm \infty$. The base velocity is $\tilde{U}=Z$ everywhere; there is no jump in its derivative. Let $C=\tilde{c}$ and $E=e^{-2\alpha }$. A decaying representation that already solves the layer equations isIt is continuous at both interfaces. At $Z=-1$ the value is $A+BE$ and the derivative jump is $-2\alpha A$; at $Z=1$ they are $B+AE$ and $-2\alpha B$. The density anomaly drops by one across each interface. The jump conditions for stratified inviscid shear flow therefore reduce toFor nonreal $C$ the displacement denominators do not vanish. Neutral limiting values are obtained by continuation, rather than by dividing by zero at an interface.

Set $q=J/(2\alpha )$. A nonzero mode requires the determinant to vanish:Expansion gives the dispersion relation for two density interfaces in uniform shear,Uniform shear throughout the exterior is important: replacing it with constant exterior velocities would introduce vorticity jumps and give a different dispersion relation.

Put $y={\tilde{c}}^2$ and $s=J/(2\alpha )>0$. The dispersion relation for two density interfaces in uniform shear isIts discriminant is $16s+4s^2{e}^{-4\alpha }>0$, so both roots $y_{\pm }$ are real. Their sum is positive; therefore one is negative precisely when their product is negative. That condition isSolving for $J$ givesIn this interval $y_{-}<0$, so $\tilde{c}=i\sqrt{-y_{-}}$ is the growing normal mode. At either endpoint the corresponding root is zero and the mode is neutral.

Write $V_0=\Delta U/2$, $z_{\pm }=\pm h/2$, $B=g\Delta \rho /(2{\rho }_0)$ and $d=B/(2k)$. The velocity is linear through all three layers, so $U(z_{\pm })=\pm V_0$ and $U^{\prime }$ has no jump. Each interface has density drop $\Delta \rho /2$. The jump conditions for stratified inviscid shear flow becomeBetween interfaces, the Taylor–Goldstein equation is ${\hat{w}}^{\prime \prime }-k^2\hat{w}=0$. A decaying solution can therefore be writtenIts derivative jumps are $-2k{b}_{\pm }$, while its values at the interfaces are $b_{\pm }+e^{-kh}{b}_{\mp }$. Hence the coefficients satisfyThe determinant must vanish. With $\tilde{c}=c/V_0$, $\alpha =kh/2$ and $J=g\Delta \rho h/({\rho }_0\Delta U^2)$, one has $d/V_0^2=J/(2\alpha )$. Expanding gives the dispersion relation for two density interfaces in uniform shear: