 Solution

Let the perturbation phase be $\vartheta =\mathbf{k}(t)\cdot \mathbf{r}$ and write $\psi ={\psi }_0+\mathrm{Re}[\Pi (t)e^{i\vartheta }]$. Incompressible flow imposes $\mathbf{k}\cdot \mathbf{v}=0$. Applying the background advective derivative to the phase givesCancel this position-dependent phase by choosingThe perturbation acting on the basic flow gives $({\mathbf{u}}^{\prime }\cdot \nabla ){\mathbf{u}}_0=-2A{u}_x^{\prime }{\mathbf{e}}_y$. Its self-advection vanishes exactly: the physical velocity ${\mathbf{u}}^{\prime }$ is perpendicular to $\mathbf{k}$, while every spatial derivative of ${\mathbf{u}}^{\prime }$ is proportional to a component of $\mathbf{k}$. Thus ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }=0$, including the products between the complex mode and its conjugate. Substituting into the full Navier-Stokes equation therefore gives an exact incompressible shearing wave, with no small-amplitude approximation:In components these equations areDifferentiating the transversality constraint closes the pressure:The coefficient $4A$ includes both the shear term in the momentum equation and $\dot{\mathbf{k}}\cdot \mathbf{v}$; treating the wavevector as fixed would miss one contribution. Initial data with $\mathbf{k}(t_0)\cdot \mathbf{v}(t_0)=0$ remain transverse under this closed evolution. These shearing waves are exact in an unbounded local shear flow or with compatible shearing-periodic boundaries; arbitrary rigid boundary conditions need not admit a single such wave.