= Solution
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> $\widehat\phi(\kappa,y)=\int_{\mathbb R}\phi(x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing <Helmholtz equation> solution is $\widehat\phi=F^-(\kappa)e^{-\gamma y}$ 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 transform
$$
\int_{-\infty}^0(-ib)e^{i(\kappa+a)x}dx=-\frac b{\kappa+a}.
$$
Since the upper normal derivative is $-\gamma F^-$, the <Wiener-Hopf equation> is
$$
\gamma F^-+G^+=\frac b{\kappa+a}.
$$
For 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> gives
$$
\frac b{(\kappa+a)\gamma_+(\kappa)}=
\frac b{(\kappa+a)\gamma_+(-a)}+
\frac b{\kappa+a}\left[\frac1{\gamma_+(\kappa)}-\frac1{\gamma_+(-a)}\right].
$$
The 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>:
$$
\boxed{F^-(\kappa)=\frac{b}{(\kappa+a)\gamma_+(-a)\gamma_-(\kappa)}=
\frac{-i\sqrt{2k_0}\sin(\theta_0/2)}{(\kappa+a)\sqrt{\kappa+k_0}}.}
$$
The scattered field on either side of the screen is therefore
$$
\phi(x,y)=\frac{\operatorname{sgn}y}{2\pi}\int F^-(\kappa)e^{-i\kappa x-\gamma|y|}d\kappa.
$$
Its 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 $\operatorname{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered <geometrical optics> field is
$$
\phi_{\rm GO}=
\begin{cases}
e^{iax-iby},&\pi-\theta_0<\theta<\pi,\\
-e^{iax+iby},&-\pi<\theta<\theta_0-\pi,\\
0,&\text{otherwise}.
\end{cases}
$$
Adding 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 $-ik_0r$, and second derivative $ir/(k_0\sin^2\theta)$. <Stationary phase> therefore supplies $\sqrt{2\pi k_0/r}|\sin\theta|e^{-ik_0r+i\pi/4}$. Substituting $F^-(\kappa_s)$, and using $\operatorname{sgn}(y)|\sin\theta|/\sqrt{1+\cos\theta}=\sqrt2\sin(\theta/2)$, gives
$$
\boxed{\phi_d\sim\sqrt{\frac{2}{\pi k_0r}}\frac{\sin(\theta_0/2)\sin(\theta/2)}{\cos\theta+\cos\theta_0}e^{-ik_0r-i\pi/4}.}
$$
The full asymptotic scattered field is $\phi_{\rm GO}+\phi_d$. 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.
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