Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 82 / 2

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

Take a cut-on mode with $k>0$, put $\lambda =n\pi /h$ and $\alpha ={\tan }^{-1}(\lambda /k)$, so $k=k_0\cos \alpha$, $\lambda =k_0\sin \alpha$, and initially take $n>0$. The acoustic waveguide mode splits into two acoustic plane waves:At the upper edge $y=h$, the upward-going constituent is incident from below. Use the reflected local coordinate $Y_{+}=h-y$; its incident potential becomes $(-1{)}^n{e}^{-ikx+i\lambda Y_{+}}/2$, with local incidence angle ${\theta }_0=\pi -\alpha$. At the lower edge use $Y_{-}=y+h$ and the downward-going constituent; it has the same local incident amplitude and incidence angle. At each wall the other constituent is the geometrical optics reflection of this incident half-amplitude, so it must not be counted again as an independent incident wave at the same edge.

Let $D(r)=\sqrt{2/(\pi k_{0}r)}{e}^{-i{k}_{0}r-i\pi /4}$. For a far observer, the upper and lower local observation angles are $-\theta$ and $\theta$, and their distances are $r_{+}=r-h\sin \theta +O(h^2/r)$ and $r_{-}=r+h\sin \theta +O(h^2/r)$. Applying the Wiener-Hopf solution of rigid half-plane diffraction separately givesThe sign difference is the reversal of the upper-edge transverse coordinate. Their sum is the leading independent-edge radiation from an open planar duct:Restore $e^{i\omega t}$ if a time-dependent potential is desired. This is the leading large-$k_{0}h$ approximation, neglecting further diffraction between edges, and assumes a far-field distance $r\gg k_0{h}^2$ as well as $r\gg h$. It is not a uniform approximation for a mode approaching cutoff. The zero mode follows as the grazing-incidence limit $\alpha =0$; its beam-direction limit is treated separately below.

For $n>0$, $k_{0}h\sin \alpha =n\pi$. Hence the numerator and denominator of the combined independent-edge radiation from an open planar duct both vanish at $\theta =\pm \alpha$. At the positive angle, the ratio has the removable limitAt the negative angle the ratio changes sign, as does $\sin (\theta /2)$, so both directions give the same total velocity potential. Since $2\sin (\alpha /2)\cos (\alpha /2)=\sin \alpha$, the result isFor the zero mode, use $\alpha =0$ directly: $\sin (\theta /2)\sin (k_{0}h\sin \theta )/(\cos \theta -1)\to -k_{0}h$. Its two directions coincide andThe two-edge field is finite in the constituent propagation directions. Its magnitude there is enhanced by a factor of order $kh$ over an ordinary isolated-edge diffracted amplitude away from a shadow boundary. An isolated-edge angular formula has a pole in these directions, because its geometrical optics and edge-wave split is nonuniform; the exact single-edge field is not infinite. In the combined field the equal and opposite pole residues cancel, and the difference of endpoint propagation phases produces the finite limit above. Taking this combined limit describes the leading far-field beam; it does not justify evaluating either nonuniform single-edge expression separately at its pole.