= Solution
Assume incompressible inviscid layers, gravity $-g\mathbf e_y$, no <surface tension>, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla\cdot\mathbf u_j=0$ and $\partial_t\mathbf u_j+\mathbf u_j\cdot\nabla\mathbf u_j=-\rho_j^{-1}\nabla p_j-g\mathbf e_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta(x,t)$ each layer obeys $u_{jy}=\eta_t+u_{jx}\eta_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this <vortex sheet>.
Write $\mathbf u_j=(U_j,0)+\nabla\phi_j$, with $U_1=U$, $U_2=-U$. Then $\nabla^2\phi_j=0$, and the wall conditions are $\partial_y\phi_1(h)=\partial_y\phi_2(-h)=0$. Linearizing the interface conditions at zero gives
$$
\partial_y\phi_j(0)=(\partial_t+U_j\partial_x)\eta,
$$
$$
\rho_1(\partial_t+U\partial_x)\phi_1-\rho_2(\partial_t-U\partial_x)\phi_2=(\rho_2-\rho_1)g\eta.
$$
The second follows from linear Bernoulli <pressure> $p_j'=-\rho_j(\partial_t+U_j\partial_x)\phi_j$ and the <hydrostatic pressure> shift $-\rho_jg\eta$.
For $k>0$, use amplitudes $\phi_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and $\phi_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma+ikU)A/[k\sinh kh]$ and $B_2=(\sigma-ikU)A/[k\sinh kh]$. The dynamic condition becomes
$$
\rho_1(\sigma+ikU)^2+\rho_2(\sigma-ikU)^2=-(\rho_2-\rho_1)gk\tanh kh.
$$
<Completing the square> yields
$$
\sigma=ikU\frac{\rho_2-\rho_1}{\rho_1+\rho_2}\pm\sqrt{\frac{4\rho_1\rho_2}{(\rho_1+\rho_2)^2}k^2U^2-\frac{\rho_2-\rho_1}{\rho_1+\rho_2}gk\tanh kh}.
$$
A positive <real part> occurs precisely when
$$
\boxed{U^2>\frac{\rho_2^2-\rho_1^2}{4\rho_1\rho_2}\frac gk\tanh kh.}
$$
The function $\tanh(kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its <derivative> has numerator $kh\operatorname{sech}^2(kh)-\tanh(kh)<0$, since $\sinh(2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero <wavenumber> is
$$
\boxed{U_{\rm crit}=\sqrt{\frac{(\rho_2^2-\rho_1^2)gh}{4\rho_1\rho_2}}.}
$$
At equality every $k>0$ still satisfies the strict growth inequality, since $\tanh(kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\rm crit}$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4-shear-threshold.png]
{title=Normalized finite-depth shear-instability threshold}
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