Set $f(z)=\cosh (\pi z/a)$. The hyperbolic cosine addition formula givesFor $x>0$ and $0<y<a$, both factors in the imaginary part are positive. Thus the image lies in the upper half-plane. The three walls map onto the entire real axis:Also $f^{\prime }(z)=(\pi /a)\sinh (\pi z/a)\ne 0$ in the interior.

To verify that this is a one-to-one conformal map onto the whole upper half-plane, factor it through $s=e^{\pi z/a}$. This first map is one-to-one onto $|s|>1$, $\mathrm{Im}s>0$, and $f(z)=(s+s^{-1})/2$. Given any $\mathrm{Im}\zeta >0$, the roots of $s^2-2\zeta s+1=0$ have product one. Neither lies on $|s|=1$, since that would make $\zeta$ real. Exactly one root therefore has $|s|>1$, andforces that root into the upper half of the exterior disk. It has a unique preimage $z=(a/\pi )(\log |s|+i\arg s)$ with $0<\arg s<\pi$. This proves both surjectivity and injectivity. The construction is the hyperbolic-cosine map of a half-strip.

The source location $c$ is complex, with $\mathrm{Re}c>0$ and $0<\mathrm{Im}c<a$. Put $C=f(c)$, so $\mathrm{Im}C>0$. The method of images places an equal line source at $\bar{C}$ below the real axis:For real $\zeta$, the product of the two logarithm arguments is $|\zeta -C{|}^2>0$. Hence the streamfunction is constant along the wall, on locally consistent branches of the complex logarithm, and there is no normal velocity. Pulling back gives the requested complex potential of a two-dimensional potential flow:This is the line source in a closed-end semi-infinite strip, with no extra imposed flow. Since $f^{\prime }(c)\ne 0$, near the physical source $f(z)-f(c)=f^{\prime }(c)(z-c)+O((z-c{)}^2)$. Thus $w=(Q/(2\pi ))\log (z-c)+$ a regular term, verifying that the source strength remains $Q$ under the conformal map. The image source has no preimage inside the physical strip. Finally, as $x\to \infty$, $w(z)\sim Qz/a+$ a constant, so the far-field velocity is $(Q/a,0)$ and the channel carries exactly the injected volume flux $Q$.

Let $\mu =\pi /H$. The end-wall forcing has transverse factor $\cos (\mu y)$, which already satisfies both fixed-wall Neumann boundary conditions. By linearity and separation of variables, seek its periodic response asSubstitution into the wave equation and the moving-wall condition givesThe acoustic waveguide cutoff is thereforeIt is the lowest nonzero transverse-mode frequency. The uniform transverse mode has no cutoff, but is not excited here because ${\int }_0^H\cos (\pi y/H)dy=0$.

For $0<\omega <{\omega }_c$, put $\alpha =\sqrt{{\mu }^2-{\omega }^2/c^2}>0$. The axial equation has solutions $e^{\pm \alpha x}$. Boundedness removes the growing exponential, and ${\Phi }^{\prime }(0)=ϵ\omega$ fixes the coefficient. ThusThis evanescent wave oscillates in time but decays with penetration length $1/\alpha$. Its axial velocity is proportional to $\cos (\omega t)$, while its excess pressure $p^{\prime }=-{\rho }_0{\varphi }_t$ is proportional to $\sin (\omega t)$. They are in phase quadrature, so their mean product, the axial acoustic energy flux, is zero.

For $\omega >{\omega }_c$, put $q=\sqrt{{\omega }^2/c^2-{\mu }^2}>0$. The axial solutions are $e^{\pm iqx}$. The outgoing condition selects $e^{iqx}$ for the chosen time convention, andTaking the real part givesIts axial velocity is $ϵ\omega \cos (\pi y/H)\cos (qx-\omega t)$, so the moving-wall condition is satisfied with the correct sign. This is a traveling acoustic waveguide mode carrying energy toward positive $x$. Its dispersion relation isThe larger axial phase velocity does not imply faster energy transport: the group velocity controls that transport.

At $\omega ={\omega }_c$, the axial equation becomes ${\Phi }^{\prime \prime }=0$. The imposed nonzero derivative forces $\Phi =ϵ{\omega }_{c}x+$ a constant, which is unbounded. Thus there is no bounded periodic response at exact cutoff for this infinite, undamped, velocity-driven model. The divergent coefficients as $q\to 0$ or $\alpha \to 0$ also show why the assumed small perturbations fail sufficiently close to cutoff at fixed forcing. Initial conditions can add free transients; the displayed formulas give the selected periodic forced response.

The fixed upper and lower walls have zero normal velocity. Since $v={\varphi }_y$, their no-penetration boundary conditions areFor the moving end wall, use its level-set equation $F(x,y,t)=x-X(y,t)=0$. The kinematic boundary condition $DF/Dt=0$ gives the exact impermeability relationBoth $v$ and $X_y$ are first order, so their product is second order. A Taylor expansion from $x=X$ to $x=0$ changes the first-order perturbation velocity only at second order as well. ThusAt infinity, select a bounded response with no waves incident from $x=+\infty$. A propagating acoustic waveguide mode must satisfy the outgoing radiation condition for a waveguide mode, while a nonpropagating mode must decay. For a time dependence $e^{-i\omega t}$, a propagating axial factor has the form $e^{iqx}$, $q>0$, and obeys ${\Phi }_x-iq\Phi \to 0$. An evanescent factor is $e^{-\alpha x}$, $\alpha >0$.

Requiring $\varphi \to 0$ for all frequencies would incorrectly exclude the radiated traveling waves. In this lossless channel they retain finite amplitude arbitrarily far downstream.