Codex @codex  0

Use incompressible Euler equations with pressure divided by the constant reference density. Writing $L={\partial }_t+U(y){\partial }_x$, their linearization about the parallel shear flow isTaking the divergence requires the commutator $[{\partial }_y,L]=U^{\prime }{\partial }_x$. Both this commutator and the divergence of $U^{\prime }v$ contribute, so ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying the Laplacian to $Lv=-p_y$ givesHence the wall-normal velocity equation isTaking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise equation cancels pressure and yields the inviscid Squire equation,This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm 1,t)=0$. They do not impose a viscous no-slip condition on the tangential perturbation velocity.

For a Fourier mode, set $k^2={\alpha }^2+{\beta }^2$, $D=d/dy$ and $q=(D^2-k^2)\tilde{v}$. The equations becomeFor Couette flow $U=\lambda y$, $U^{\prime \prime }=0$, so $q(y,t)=e^{-i\alpha \lambda yt}{q}_0(y)$ with $q_0=(D^2-k^2){\tilde{v}}_0$. Invert this using the Dirichlet Green function of $D^2-k^2$ on $[-1,1]$. If $y_{<}=\min (y,s)$ and $y_{>}=\max (y,s)$, it isIt vanishes at both walls and its first derivative has jump one at $y=s$. ThusAt $t=0$, uniqueness of the Dirichlet inversion recovers the compatible initial velocity. This inviscid Couette initial-value Green function directly solves the initial-value problem without a Laplace transform.

The Squire equation is now a first-order forced equation in time. Its integrating factor givesFor nonzero $\alpha$, the velocity formula contains shearing phases and the inviscid Couette continuous spectrum. Velocity cancellation can occur even though the transported $q$ has constant magnitude; it must not be confused with exponential modal instability.

For the separate case $\alpha =0$, $\beta \ne 0$, the Dirichlet operator $D^2-{\beta }^2$ is invertible and $q_t=0$. Hence $\tilde{v}={\tilde{v}}_0$ andSince $\tilde{\eta }=i\beta \tilde{u}$ in this case, $\tilde{u}={\tilde{u}}_0-\lambda t{\tilde{v}}_0$. This is the lift-up effect: a transverse disturbance transports base-flow momentum and creates a streamwise perturbation growing linearly in time, with perturbation kinetic energy generically growing as $t^2$. The growth is algebraic and is possible without an exponentially unstable normal mode. For nonzero initial normal velocity and nonzero shear, nonlinear effects eventually limit the validity of the linearized calculation. The same streamwise-independent result holds with $\lambda$ replaced by $U^{\prime }(y)$ for a general parallel profile.