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Set $\mathbf{k}=(k_x,0,k_z)\ne 0$. Incompressibility gives $(\mathbf{k}\cdot \tilde{\mathbf{u}})f^{\prime }(\xi )e^{st}=0$. Since $f$ is nonconstant and differentiable, it follows that $\mathbf{k}\cdot \tilde{\mathbf{u}}=0$; thereforeThe background has no advection of a $y$-independent perturbation. Its only remaining cross-advection term is $({\mathbf{u}}^{\prime }\cdot \nabla )\mathbf{U}=(-3\Omega u_x^{\prime }/2+q\Omega u_z^{\prime }){\mathbf{e}}_y$. The velocity/time terms have spatial factor $f$, while $\nabla P^{\prime }=\tilde{P}\mathbf{k}{g}^{\prime }(\xi )e^{st}$. For nonzero pressure amplitude, $g^{\prime }$ must therefore be a constant multiple of $f$; absorbing that multiple into $\tilde{P}$ gives $g^{\prime }=f$. An additive spatial constant in $g$ is a pressure gauge. If $\tilde{P}=0$, there is no such restriction on the unused pressure profile.

Matching coefficients givesBecause the quadratic perturbation term vanishes identically, these amplitudes give an exact single-phase incompressible perturbation of a shear flow, not merely a first-order truncation. Any sufficiently differentiable profile with $g^{\prime }=f$ works locally; boundary conditions may restrict the profiles in a particular domain.