Fluid-loaded membrane edge scattering

A pinned semi-infinite membrane scatters an incoming guided mode into reflected guided modes and outgoing sound. For unit incident displacement the scattered endpoint value is $-1$. Its Half-range Fourier transform therefore has nonzero endpoint terms even though the total displacement vanishes. The Wiener-Hopf equation kernel is $2/(m{\omega }^2-T{k}^2)+\gamma /({\rho }_0{\omega }^2)$ with the outgoing branch ${\gamma }^2=k^2-{\omega }^2/c_0^2$.

For identical unit-density, unit-sound-speed fluids above and below a sheet on $x<0$, use time $e^{i\omega t}$ and transform $\int e^{ikx}\varphi dx$. The outgoing acoustic square-root branch obeys ${\gamma }^2=k^2-{\omega }^2$. If $J$ is the pressure jump upper minus lower, $s={\eta }^{\prime }(0)$ and $d^{+}$ the right Half-range Fourier transform of the scattered normal derivative, the Wiener-Hopf equation isThe total displacement is pinned, so its transformed second derivative is $s-k^2{\eta }^{-}$. The bare-sheet pole $b=\omega \sqrt{m/T}$ in reconstructed displacement must cancel: $J(b)=Ts$. This is an incoming-pole cancellation at a pinned membrane edge condition, distinct from the outgoing coupled-mode pole at the zero of $K$.

Reconstructing the left scattered displacement in fluid-loaded membrane edge scattering divides by $m{\omega }^2-T{k}^2$. The pole at the lower bare-membrane wavenumber $b=\omega \sqrt{m/T}$ must cancel to preserve outgoing half-plane analyticity. This gives a linear condition determining the otherwise unknown endpoint slope $A$. The prescribed incident pole in the total field is a different singularity.