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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.

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: