MATHEMATICAL PHYSICS

V. A. BOROVIKOV

ON THE TWO-DIMENSIONAL DIFFRACTION PROBLEM FOR A POLYGON

(Presented by Academician V. A. Fock on 3 I 1962)

1. Let \(S\) be a convex polygon. Denote by \(\Gamma(x, y, \xi, \eta, t)\) the Green’s function for the wave equation in the exterior domain and with zero boundary conditions on \(S\), i.e. the solution of the equation

\[ \frac{\partial^2 \Gamma}{\partial x^2} + \frac{\partial^2 \Gamma}{\partial y^2} - \frac{\partial^2 \Gamma}{\partial t^2} =0 \]

with boundary conditions \(\Gamma|_S=0\) and initial conditions

\[ \Gamma(x, y, \xi, \eta, 0)=0, \]

\[ \Gamma'_t(x, y, \xi, \eta, 0)=\delta(x-\xi, y-\eta). \]

A method is proposed that makes it possible to determine the form and intensity of the singularities of the function \(\Gamma(x, y, \xi, \eta, t)\).

The Fourier transform in \(t\) of the function

\[ \Gamma_1(x, y, \xi, \eta, k) = \int_{-\infty}^{\infty} e^{ikt}\Gamma(x, y, \xi, \eta, t)\,dt \tag{1} \]

satisfies the equation

\[ \frac{\partial^2 \Gamma_1}{\partial x^2} + \frac{\partial^2 \Gamma_1}{\partial y^2} + k^2\Gamma_1 = \delta(x-\xi, y-\eta), \]

the boundary conditions \(\Gamma_1|_S=0\), is an analytic function of the parameter \(k\), and, for \(\operatorname{Im} k>0\), is bounded at infinity, i.e. is the Green’s function for the stationary diffraction problem*. The asymptotics of \(\Gamma_1\) as \(k\to\infty\), as is evident from formula (1), is determined by the singularities of the function \(\Gamma(x, y, \xi, \eta, t)\) as a function of \(t\) for fixed \(x, y, \xi, \eta\). Therefore it proves possible, for arbitrary \(x, y, \xi, \eta\), to indicate the principal term of the asymptotics of \(\Gamma_1\) as \(k\to\infty\); for the second term of the asymptotics one obtains recurrent formulas (in the number of vertices of the polygon), which makes it possible to determine the form of the second term of the asymptotics for the simplest types of \(S\).

The method developed in principle makes it possible to find any term of the asymptotics.

1. We shall show how the function \(\Gamma(x, y, \xi, \eta, t)\) is constructed. Suppose first that, in the simplest case, when \(S\) is an angle, the Green’s function is known to us. Let us construct the Green’s function for an arbitrary polygon \(S\) with the aid of this Green’s function.

Fix some instant of time \(t\). Then \(\Gamma(x, y, \xi, \eta, t)\), obviously, will be equal to zero when \((x-\xi)^2+(y-\eta)^2>t^2\), i.e. outside the circle \(H_t\) with center at the point \(\xi,\eta\) and radius \(t\), and will depend only on the form of \(S\)

* It is easy to show that the integral (1) converges at infinity for \(\operatorname{Im} k>0\). At present we do not have a proof, not raising doubts from physical considerations, of the convergence of the integral (1) for \(\operatorname{Im} k=0\).

inside this circle. Let now \(t<l/2\), where \(l\) is the length of the smallest side of \(S\). Then, whatever \(\xi\) and \(\eta\) may be, the intersection of \(H_t\) and \(S\), if it is nonempty, will have the form either of a chord of \(H_t\), or, if the point \(\{\xi,\eta\}\) is close to some vertex of \(S\), the form of an angle inside \(H_t\). Therefore, if we can construct the Green’s function for an angle, we can also construct it for an arbitrary polygon, but only for \(t<l/2\), where \(l\) is the length of the smallest side of \(S\).

We have constructed the Green’s function \(\Gamma(x,y,\xi,\eta,t)\) for small, but finite, times \(t<l/2\). Let us show how to determine \(\Gamma(x,y,\xi,\eta,t)\) for an arbitrary time \(t\). We use the obvious formula:

\[ \Gamma(x,y,\xi,\eta,T) = \iint \left[ \Gamma(x,y,\alpha,\beta,t)\, \Gamma'_{T-t}(\alpha,\beta,\xi,\eta,T-t) + \Gamma'_t(x,y,\alpha,\beta,t)\, \Gamma(\alpha,\beta,\xi,\eta,T-t) \right]\,d\alpha\,d\beta, \]

where the integration is carried out over the domain exterior to \(S\). If \(t\) and \(T-t<l/2\), then the integrand is known to us, so that we obtain an expression for the Green’s function \(\Gamma(x,y,\xi,\eta,t)\) for \(t<l\). Although the integral, obviously, cannot be evaluated in elementary functions, knowing the singularities of the functions standing on the right-hand side under the integral sign, it is easy to determine the singularities of the left-hand side. By analogous steps one can proceed further in \(t\).

Fig. 1

1. Let us now construct the Green’s function for an angle. This problem was solved by G. I. Petrashen et al. \({}^{(1)}\); however, the solution obtained by them seems to us too complicated. Thus, from their formulas it is not apparent that the magnitude of the jump of the Green’s function on the front of the wave scattered by the vertex of the angle, as well as the magnitude of the jump of any derivative, are expressed in terms of trigonometric functions. We shall therefore give another formula for the required Green’s function.

Let the vertex of the angle be at the origin, and let its sides in the polar coordinate system \(\rho,\varphi\) lie along the rays \(\varphi=0\), \(\varphi=\Phi\) \((\pi<\Phi<2\pi)\). Let the point \(\xi,\eta\) have polar coordinates \(R,\varphi_0\) \((0<\varphi_0<\Phi)\). Then \(\tau=t-R\) is the time elapsed after the cylindrical wave excited at the point \(R,\varphi_0\) reaches the vertex of the angle, and the front of the wave scattered by the vertex \(S\) will have the equation \(\rho=\tau\). For \(\rho>\tau\), i.e. ahead of the front of this wave, the solution is determined from simple geometrical considerations and consists of the sum of the incident and reflected waves; the incident wave has the form
\(\frac{1}{2\pi}\bigl(\tau^2-\rho^2+2R(\tau+\rho\cos(\varphi-\varphi_0))\bigr)^{-1/2}\) where the radical expression is positive, and is equal to zero where it is negative; the reflected waves have an analogous form. For \(\rho<\tau\), i.e. behind the front of the wave scattered by the vertex \(S\), the solution is given by the formula

\[ \Gamma = \frac{1}{2\pi\cdot 4\pi i} \int_{\gamma} \frac{1}{\sqrt{\tau^2-\rho^2+2R[\tau+\rho\cos\alpha]}}\, H(\alpha+\varphi,\varphi_0,\Phi)\,d\alpha, \tag{2} \]

where

\[ H(\alpha+\varphi,\varphi_0,\Phi) = \frac{\pi}{2\Phi} \left[ \operatorname{ctg}\frac{\pi}{2\Phi}(\alpha+\varphi-\varphi_0) - \operatorname{ctg}\frac{\pi}{2\Phi}(\alpha+\varphi+\varphi_0) \right]. \]

We specify the form of the contour of integration \(\gamma\). The function
\[ \tau-\rho^2+2R[\tau+\rho\cos\alpha] \]
for \(\rho<\tau\) has zeros at the points \(\pm\pi\pm i\omega\), where
\[ \operatorname{ch}\omega=\frac{\tau^2+\rho^2+2R\tau}{2R\rho}. \]
The contour \(\gamma\) goes around the branch points of the function
\[ (\tau^2-\rho^2+2R(\tau+\rho\cos\alpha))^{-1/2} \]
as shown in Fig. 1.

Using formula (2), it is easy to specify the magnitude of the jump of the Green’s function for \(\tau=\rho\). Namely, at the point \(\tau,\tau,\varphi\) the Green’s function has a discontinuity of the first kind, whose magnitude is equal to
\[ \frac{1}{4\pi\sqrt{R\tau}}\,[H(-\pi+\varphi,\varphi_0,\Phi)-H(\pi+\varphi,\varphi_0,\Phi)]. \]

1. Let us give an example of the asymptotics of the Green’s function for the stationary diffraction problem, obtained by the method indicated above. Let \(S\) have the form shown in Fig. 2. Here \(a\) is the source, and \(x\) is the point of observation. The meaning of the quantities \(r_a,\varphi_a,\Phi_1,\Phi_2,\Phi_3,T_{12},T_{23},r_x,\varphi_x\) is clear from Fig. 2. We assume that \(\Phi_1-\varphi_a>\pi\), \(\Phi_3-\varphi_x>\pi\), so that the shortest optical path from the point \(a\) to the point \(x\) is the broken line \(aABCx\). Let the length of this broken line be \(\psi\). Put
   \[ u(\Phi,\varphi)=\left(\frac{\pi}{2\Phi}\right)^2 \left[ \frac{1}{\cos^2\frac{\pi}{2\Phi}(\pi+\varphi)} - \frac{1}{\cos^2\frac{\pi}{2\Phi}(\pi-\varphi)} \right]. \]

Fig. 2

Then, as \(k\to\infty\), the Green’s function \(\Gamma_1(a,x,k)\) of the stationary diffraction problem on \(S\) has the following asymptotics:
\[ \Gamma_1(a,x,k)=e^{ik\psi}\left[\left(\frac{i}{k}\right)^4F_1+\left(\frac{i}{k}\right)^5F_2+O(k^{-6})\right], \]
where \(F_1\) and \(F_2\) are expressed in the following way in terms of the quantities introduced above, \(r_a,\varphi_a,\Phi_1,\Phi_2,\Phi_3,T_{12},T_{23},r_x,\varphi_x\), and the function \(u(\Phi,\varphi)\):
\[ F_1= \frac{1}{(T_{12}T_{23})^{3/2}\sqrt{r_ar_x}(2\pi)^2}\, u'_\varphi(\Phi_2,0)\cdot u(\Phi_3,\varphi_x)\cdot u(\Phi_1,\varphi_a), \]
\[ \begin{aligned} F_2={}& -\frac{u_\varphi(\Phi_2,0)} {8(2\pi)^2(T_{12}T_{23})^{3/2}\sqrt{r_ar_x}} \left\{ \frac{u(\Phi_1,\varphi_a)\,[4u(\Phi_3,\varphi_x)+u''_{\varphi\varphi}(\Phi_3,\varphi_x)]}{r_x} \right.\\ &\left. \qquad\qquad\qquad\qquad\qquad +\frac{u(\Phi_3,\varphi_x)\,[4u(\Phi_1,\varphi_a)+u''_{\varphi\varphi}(\Phi_2,\varphi_a)]}{r_a} \right\}\\ &-\frac{u'_\varphi(\Phi_2,0)} {\sqrt{r_ar_x}(2\pi)^2(T_{12}T_{23})^{3/2}} \left\{ \frac{[\,{}^5/_8\,u(\Phi_1,\varphi_a)+{}^1/_2\,u''_{\varphi\varphi}(\Phi_1,\varphi_a)]\,u(\Phi_3,\varphi_x)}{T_{12}} \right.\\ &\left. \qquad\qquad\qquad\qquad\qquad +\frac{[\,{}^5/_8\,u(\Phi_3,\varphi_x)+{}^1/_2\,u''_{\varphi\varphi}(\Phi_3,\varphi_x)]\,u(\Phi_1,\varphi_a)}{T_{23}} \right\}. \end{aligned} \]

1. We now describe how the asymptotics of the Green’s function of the stationary diffraction problem is structured in the case of an arbitrary polygon \(S\). Suppose first that the source \(a\) and the observation point \(x\) do not lie in the zone of direct visibility or on the extension of any side of \(S\) (on the boundary of shadow and light). We construct all possible optical paths connecting \(a\) and \(x\). It turns out that to each optical path there corresponds its own contribution to the asymptotics of the Green’s function, beginning with terms of order \(k^{1/2-3n/2}\), where \(n\) is the number

internal vertices of an optical path. We call an optical path a broken line whose first segment joins the source \(a\) with one of the vertices of \(S\), having no more common points with \(S\); whose last segment joins the observation point \(x\) with one of the vertices of \(S\), also having no more common points with \(S\); and whose internal segments coincide with sides of \(S\). For example, in the situation shown in Fig. 2, there are no optical paths with fewer than three internal vertices; there is a unique optical path \(aABCx\) with three internal vertices, and two optical paths \(aABABCx\) and \(aABCBCx\) with five internal vertices.

If the source and the observation point are in the zone of direct visibility, then to the asymptotics obtained in this way one must add \(H_0^{(2)}(kr)\), where \(r\) is the distance between the points \(a\) and \(x\), i.e., the Green’s function for free space.

Finally, if the source or the observation point is near the extension of one of the sides of \(S\) (i.e., on the boundary between shadow and light for one of the optical paths), then for the contribution to the asymptotics corresponding to this optical path one can write an expression analogous to the Fresnel integral.

Let us note that when we are interested in the asymptotics of the Green’s function up to some fixed power of \(k\), we have to deal only with a finite number of optical paths.

An analogous result holds if the boundary condition on \(S\) has the form \(\partial u/\partial n = 0\). In this case the function \(H(\alpha+\varphi,\varphi_0,\Phi)\) (see (2)) has the form not of a difference, but of a sum of cotangents. In this case as well each optical path has its own contribution to the asymptotics, but it already begins with terms of the form \(k^{-1/2-n/2}\), where \(n\) is the number of internal vertices of the optical path.

References

1. G. I. Petrashen’, V. G. Nikolaev, D. P. Kouzov, Uch. zap. LGU, No. 246, 5 (1958).