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Write the common density as $\rho$ and take positive bending stiffness $\beta$. With $q=|k|>0$, use decaying velocity potentialsadded 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 areThe linearized Bernoulli equation gives perturbation pressures $p_{+}^{\prime }=-\rho (\sigma +ikU)A_{+}$ and $p_{-}^{\prime }=-\rho \sigma A_{-}$. Substituting the kinematic conditions and imposing the stipulated pressure difference givesCompleting the square yields the bending-membrane Kelvin-Helmholtz dispersion relationThe imaginary shift convects the disturbance at mean fluid speed $U/2$. For $U\ne 0$ there is a growing root exactly whenAt $k=0$ 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 $U^2{q}^2/4-\beta q^5/(2\rho )$ over the unstable interval. Its derivative is $U^{2}q/2-5\beta q^4/(2\rho )$, so its unique positive maximum isFor $U=0$ 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.