= Solution
Write the common density as $\rho$ and take positive bending stiffness $\beta$. With $q=|k|>0$, use decaying <velocity potentials>
$$
\phi_+=A_+e^{-qy}e^{ikx+\sigma t},\qquad\phi_-=A_-e^{qy}e^{ikx+\sigma t},
$$
added to the upper base potential $Ux$ and the lower zero base potential. They solve the <Laplace equation>, as required by incompressibility and irrotational flow. The two linear <kinematic boundary conditions> on the membrane are
$$
-qA_+=(\sigma+ikU)C,\qquad qA_-=\sigma C.
$$
The linearized <Bernoulli equation> gives perturbation pressures $p'_+=-\rho(\sigma+ikU)A_+$ and $p'_-=-\rho\sigma A_-$. Substituting the kinematic conditions and imposing the stipulated pressure difference gives
$$
p'_- -p'_+=-\frac\rho q[\sigma^2+(\sigma+ikU)^2]C=\beta k^4C.
$$
Completing the square yields the <bending-membrane Kelvin-Helmholtz dispersion relation>
$$
\boxed{\left(\sigma+\frac{ikU}{2}\right)^2=\frac{U^2k^2}{4}-\frac{\beta|k|^5}{2\rho}.}
$$
The imaginary shift convects the disturbance at mean fluid speed $U/2$. For $U\ne0$ there is a growing root exactly when
$$
\boxed{0<|k|<k_{\max},\qquad k_{\max}=\left(\frac{\rho U^2}{2\beta}\right)^{1/3}.}
$$
At $k=0$ the uniform displacement is neutral; at $|k|=k_{\max}$ the square-root growth vanishes. Above this cutoff both roots are imaginary. To maximize the growth, maximize its square $U^2q^2/4-\beta q^5/(2\rho)$ over the unstable interval. Its derivative is $U^2q/2-5\beta q^4/(2\rho)$, so its unique positive maximum is
$$
\boxed{q_* =\left(\frac{\rho U^2}{5\beta}\right)^{1/3}=\left(\frac25\right)^{1/3}k_{\max},\qquad
\operatorname{Re}\sigma_{\max}=|U|q_*\sqrt{\frac3{20}}.}
$$
For $U=0$ there is no <Kelvin-Helmholtz instability>. Positivity of $\beta$ is the physical stiffness assumption behind the finite cutoff: $\beta=0$ gives no upper unstable cutoff, and $\beta<0$ destabilizes arbitrarily short waves.
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