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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.

The rest state is $\rho ={\rho }_0$, $p=p_0$, $\mathbf{u}=0$. At linear order in the perturbations, the continuity equation, Euler momentum equation and the given equation of state becomeFor the acoustic component take $\mathbf{u}=\nabla \varphi$. This is consistent with initially zero vorticity, since taking the curl of the linear momentum equation gives ${\partial }_t(\nabla \times \mathbf{u})=0$. Any pre-existing linear vortical component is time independent and is separate from the potential oscillations being sought.

Spatial integration of ${\rho }_0\nabla {\varphi }_t=-c^2\nabla {\rho }^{\prime }$ gives ${\rho }_0{\varphi }_t+c^2{\rho }^{\prime }=K(t)$. Adding a function of time to the velocity potential removes $K(t)$ without altering the velocity field. HenceSubstitute the first relation into mass conservation:Thus the velocity potential satisfies the wave equationDifferentiating that equation in time and using the same density-potential relation givesAlternatively, differentiate the linear continuity equation in time and take the divergence of the linear momentum equation. The coefficient $c$ is the sound speed, since $dp/d\rho =c^2$.

Let $V(t)$ be any material volume, moving with the velocity field $\mathbf{u}$. The Reynolds transport theorem in the required form isapplied componentwise when $f$ is vector-valued. Mass conservation givesSince the material volume is arbitrary, the local continuity equation follows:For an inviscid fluid with no body force, the balance of linear momentum equates the rate of change of total momentum to the boundary pressure force:using the divergence theorem. The integrand on the left expands asThe bracket vanishes by mass conservation, leaving the Euler momentum equationUnlike the incompressible flow case, the mass density need not be constant along a fluid element.

Since the velocity tends to zero at infinity, the steady Bernoulli equation gives $p+\rho |\mathbf{u}{|}^2/2=p_{\infty }$. On the wall, therefore,Compute the finite source-induced force relative to uniform ambient pressure, or equivalently use the gauge pressure $p-p_{\infty }$. The fluid lies above the wall, so its normal force on the wall points in $-{\mathbf{e}}_y$ for positive pressure. The negative excess pressure found here consequently produces an upward net force, toward the source.

For the symmetric section $-R\le x\le R$, there is no horizontal force, and the vertical excess force isHere an antiderivative of $x^2/(x^2+a^2{)}^2$ is $(1/(2a))\arctan (x/a)-x/[2(x^2+a^2)]$. Taking $R\to \infty$ gives the hydrodynamic attraction of a wall to a line source:Each component is a force per unit length of the wall in the out-of-plane direction. The result is an attraction even though the source sends fluid away from itself: faster tangential motion reduces the wall pressure.

The excess torque about the origin isfor every $R$, because the integrand is odd. Thus the symmetric infinite-wall limit isThe symmetric prescription matters: the separate one-sided moment integrals grow logarithmically, whereas the force integral converges absolutely. This is the appropriate Cauchy principal value for the infinite wall. If one instead asks for the absolute fluid force with no balancing ambient pressure on the other side, its vertical component on a finite section is $-2R{p}_{\infty }+F_y(R)$ and has no finite infinite-wall limit. The finite answer above is the flow-induced load.