= Solution
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 is
$$
Lu+U'v=-p_x,\qquad Lv=-p_y,\qquad Lw=-p_z,\qquad u_x+v_y+w_z=0.
$$
Taking the divergence requires the commutator $[\partial_y,L]=U'\partial_x$. Both this commutator and the divergence of $U'v$ contribute, so $\nabla^2p=-2U'v_x$. Applying the <Laplacian> to $Lv=-p_y$ gives
$$
L\nabla^2v+2U'v_{xy}+U''v_x=-\partial_y\nabla^2p
=2U'v_{xy}+2U''v_x.
$$
Hence the wall-normal velocity equation is
$$
\boxed{L\nabla^2v-U''v_x=0.}
$$
Taking $\partial_z$ of the streamwise momentum equation minus $\partial_x$ of the spanwise equation cancels pressure and yields the inviscid <Squire equation>,
$$
\boxed{L\eta=-U'v_z,\qquad\eta=u_z-w_x.}
$$
This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm1,t)=0$. 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 $q=(D^2-k^2)\widetilde v$. The equations become
$$
(\partial_t+i\alpha U)q-i\alpha U''\widetilde v=0,\qquad
(\partial_t+i\alpha U)\widetilde\eta=-i\beta U'\widetilde v.
$$
For <Couette flow> $U=\lambda y$, $U''=0$, so $q(y,t)=e^{-i\alpha\lambda yt}q_0(y)$ with $q_0=(D^2-k^2)\widetilde v_0$. 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 is
$$
G_k(y,s)=-\frac{\sinh[k(y_<+1)]\sinh[k(1-y_>)]}{k\sinh(2k)}.
$$
It vanishes at both walls and its first derivative has jump one at $y=s$. Thus
$$
\boxed{\widetilde v(y,t)=\int_{-1}^1G_k(y,s)e^{-i\alpha\lambda st}(D_s^2-k^2)\widetilde v_0(s)\,ds.}
$$
At $t=0$, 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 gives
$$
\boxed{\widetilde\eta(y,t)=e^{-i\alpha\lambda yt}\widetilde\eta_0(y)-i\beta\lambda\int_0^t e^{-i\alpha\lambda y(t-s)}\widetilde v(y,s)\,ds.}
$$
For 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 $\alpha=0$, $\beta\ne0$, the Dirichlet operator $D^2-\beta^2$ is invertible and $q_t=0$. Hence $\widetilde v=\widetilde v_0$ and
$$
\boxed{\widetilde\eta(y,t)=\widetilde\eta_0(y)-i\beta\lambda t\widetilde v_0(y).}
$$
Since $\widetilde\eta=i\beta\widetilde u$ in this case, $\widetilde u=\widetilde u_0-\lambda t\widetilde v_0$. 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 $U'(y)$ for a general parallel profile.
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