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# Sommerfeld type integrals for discrete diffraction problems

Three discrete problems for Helmholtz equation is studied analytically using Sommerfeld integral approach. They are the problem with point source on a whole plane, the problem of diffraction by a half-plane, and the problem of diffraction by a right-angled wedge. It is showed that total field is represented as an integral from an algebraic function on a manifold. The latter is torus. For the problem with a point source a recursive relation is introduced. For half-plane and wedge problems solutions are obtained in terms of Sommerfeld integral.

• 2 publications
• 1 publication
05/22/2016

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## 1 Introduction

In the beginning of 20th century Sommerfeld introduced closed integral solution for the problem of diffraction by a half-plane [Sommerfeld1954]. It was done in a very elegant way with the help of reflection method. Namely, he reduced the half-plane problem to the problem of plane wave propagation on two-sheeted surface. Then, using plane wave decomposition integral he solved the problem. This integral with particular contour of integration was named after Sommerfeld. Later Sommerfeld integral approach was applied to a number of problems such as problem of diffraction by a strip [Shanin2003a], by a wedge [Babich2008] and some others [Luneburg1997, Hannay2003].

Nowadays with the growth of computational power problems on discrete grids draw more attention. Recently, several discrete diffraction problems were solved rigorously using Wiener-Hopf approach [Sharma2015a, Sharma2015b, Sharma2015c]. In the current work we want to apply Sommerfeld integral approach to some of them. We show that in the discrete case Sommerfeld integral is essentially an integral on torus from algebraic function (elliptic integral) and we derive such integrals for the following problems:

• The problem for Green’s function on a plane,

• The problem of diffraction by a half-plane,

• The problem of diffraction by a right-angled wedge.

## 2 Discrete Green’s function on a plane

### 2.1 Problem formulation

Consider the Green’s function for a simplest stencil discrete 2D Helmholtz equation. Namely, let function , , obey the equation

 u(m+1,n)+u(m−1,n)+u(m,n−1)+u(m,n+1)+(K2−4)u(m,n)=δm,0δn,0 (1)

The wavenumber parameter is close to positive real, but has a small positive imaginary part mimicking attenuation in the medium. The radiation condition imposed on is that it should decay exponentially as .

### 2.2 Preliminary step. Reducing the number of computations of the integral

Our aim is to tabulate function for some set of values . It is clear that

 u(m,n)=u(−m,n)=u(m,−n)=u(−m,−n).

Thus, one should tabulate only for non-negative .

Let it be necessary to tabulate all with

 |m|+|n|≤N.

A naive approach requires computations of the integral. However, here we show that one can compute integrals. Namely, we will compute the integrals only for , and all other values find by using “cheap” recursive relations.

Compute the values of row by row. Each row is a set of values with , , , i. e. the rows are in fact diagonals.

Let all values with be already computed, and it is necessary to compute the values with . Find by integration. Then use (1) rewritten as a recursive relation:

 u(M+1−n,n)+u(M+2−n,n−1)= (2)
 −u(M+1−n,n−2)−u(M−n,n−1)+(4−K2)u(M−n,n−1).

Note that all values in the right have the sum of indices , thus they are computed on the previous steps. The left-hand side is a recursive relation for .

### 2.3 Double integral representation

Apply a double Fourier transform to (

1). This transform and its inverse are as follows:

 U(ξ1,ξ2)=∞∑m,n=−∞u(m,n)exp{−i(mξ1+nξ2)}, (3)
 u(m,n)=14π2∫∫π−πU(ξ1,ξ2)exp{i(mξ1+nξ2)}dξ1dξ2. (4)

The result is the following representation of the field :

 u(m,n)=14π2∫∫π−πexp{i(mξ1+nξ2)}D(ξ1,ξ2)dξ1dξ2, (5)

where

 D(ξ1,ξ2)≡2cosξ1+2cosξ2−4+K2. (6)

Introduce the variables

 x=eiξ1,y=eiξ2. (7)

Also introduce the function

 ^D(x,y)≡x+x−1+y+y−1−4+K2. (8)

The integral (5) can be rewritten as

 u(m,n)=−14π2∫σ∫σxmyn^D(x,y)dxxdyy (9)

where contour is the unit circle in the -plane passed in the positive direction anti-clockwise.

### 2.4 Single integral representation

The integral (9) can be taken with respect to one of the variables by the residue integration. There are four cases, possibly intersecting.

Case :

Consider the integral (9). Fix and study the integral with respect to . The -plane is shown in Fig. 1. One can see that there are four possible singular points in this plane. Two of them are the roots of the dispersion equation

 ^D(x,y)=0 (10)

considered with respect to . The roots are

 y(x)=Ξ(x),y(x)=Ξ−1(x) (11)

where

 Ξ(x)=−K2−4+x+x−12+√(K2−4+x+x−1)2−42 (12)

The value of the square root is chosen in such a way that . Note that cannot be equal to 1 if since is not real. The integrand of (9) has poles at the points (11).

Beside (11), there maybe singularities at two other points: and (note that is a certain point of ). However, the presence of singularities at these points depends on the value of . For example, if then the integrand is regular at .

Thus, the only singularity of the integrand inside the circle is . Apply the residue method. The result is

 u(m,n)=12πi∫σxmΞn(x)Ξ(x)∂y^D(x,Ξ(x))dxx. (13)

Obviously,

 y∂y^D(x,y)=y−y−1. (14)

Thus, (13) can be rewritten as

 u(m,n)=12πi∫σxmyndxx(y−y−1),y=Ξ(x). (15)

Case :

The same analysis can be made for the singular points in the -plane for a fixed . This anayisis shows that there may exist a singularity at , but the behavior at is regular. This means that the integrand has no branching at , and the integrand decays not slower than . For such an integrand one can apply the residue theorem to the domain . The result is

 u(m,n)=−12πi∫σxmyndxx(y−y−1),y=Ξ−1(x). (16)

Case :

This case can be considered similarly to . The representation for the field is

 u(m,n)=12πi∫σxmyndyy(x−x−1),x=Ξ(y). (17)

Case :

The field is

 u(m,n)=−12πi∫σxmyndyy(x−x−1),x=Ξ−1(y). (18)

Let us demonstrate that, for example, for a homogeneous version of (1) is valid:

 u(m+1,n)+u(m−1,n)+u(m,n−1)+u(m,n+1)+(K2−4)u(m,n)=0 (19)

For this, apply the representation (15) to the points of the stencil:

 (m,n),(m,n+1),(m,n−1),(m+1,n),(m−1,n).

As the result, get

 u(m+1,n)+u(m−1,n)+u(m,n−1)+u(m,n+1)+(K2−4)u(m,n)=
 12πi∫σ[x+x−1+y+y−1+(K2−4)]xmyndxx(y−y−1).

Since , the expression in the brackets is equal to zero.

### 2.5 A recursive relation for u(m,0)

Let be and . Rewrite the representation (15) in the form

 u(m,0)=12πi∫σxmdxz(x),z(x)=√(x2+(K2−4)x+1)2−4x2. (20)

Using the proof of Legendre’s theorem for the Abelian integrals [Bateman1955], derive a recursive formula for . Introduce the constants as follows:

 a0=1,a1=2(K2−4),a2=(K2−4)2−2,a3=2(K2−4). (21)

Using these constants one can write

 z2(x)=x4+a3x3+a2x2+a1x+a0. (22)

Then note that

 ddx(xmz)=(m+2)xm+3z(x)+(m+3/2)a3xm+2z(x)+
 (m+1)a2xm+1z(x)+(m+1/2)a1xmz(x)+ma0xm−1z(x). (23)

Substituting this identity into (20) and taking into account that contour of integration is closed, get

 −(m+2)u(m+3,0)=(m+3/2)a3u(m+2,0)+
 (m+1)a2u(m+1,0)+(m+1/2)a1u(m,0)+ma0u(m−1,0). (24)

### 2.6 Field representation by integration on a manifold. Plane wave decomposition

Consider and being complex variables. Let be , , where is a compactified complex plane, that is a Riemann sphere.

Each point thus belongs to . Let us describe the set of points such that equation (10) is valid. Obviously, this is an analytic manifold of complex dimension 1 or of real dimension 2. This manifold will be referred to as .

Consider defined by (12). Now consider it as a double-valued function, thanks to the presence of the square root in it. Let us study the Riemann surface of this function. Topologically, there is no difference between and the Riemann surface of .

Function has four branch points. They are the points where the argument of the square root in (12) is equal to zero, i. e.

 η1,1=−d2+√d2−42,d=K2−2, (25)
 η1,2=−d2+√d2−42,d=K2−6, (26)
 η2,1=−d2−√d2−42,d=K2−2, (27)
 η2,2=−d2−√d2−42,d=K2−6. (28)

The values , , , possess the following property that can be checked directly. For

 Υ(x)≡x(y−y−1)=√(x−η1,1)(x−η1,2)(x−η2,1)(x−η2,2). (29)

Obviously, the left-hand side of (29) is the denominator of the integrand of (15).

Exactly two of these branch points are located inside the circle . One can check that the branch points are the points at which .

The scheme of the Riemann surface is shown in Fig. 2. The branch points are connected by cuts shown by bold curves. For definiteness, the the branch cuts are conducted along the lines at which

 y=Ξ(x)=eiθ,θ∈R.

The sides of the cuts labeled by equal Roman number should be connected with each other.

Topologically, is a torus (i. e. it has genus equal to 1). This can be eacily understood, since is obtained by taking two spheres, making two cuts, and connecting their shores.

One of the sheets drawn in Fig. 2 is called physical, and the other is unphysical (the naming is meaningless) . The physical sheet is the one on which for . Respectively, on the unphysical sheet for . Note that and cannot be equal to 1 on , since is not real.

We find useful to mark four “infinity points” belonging to :

 Inf 1 : x=0,y=0, Inf 2 : x=0,y=∞, Inf 3 : x=∞,y=∞, Inf 4 : x=∞,y=0.

Note that belongs to . Points Inf 1 and Inf 4 belong to the physical sheet, while points Inf 2 and Inf 3 belong to the unphysical sheet.

The notations of infinity points and branch points on are shown in Fig. 3.

The statement that is an analytic manifold means that in each (small enough) proximity of any point of one can introduce a complex local variable , such that all transhormation matrices between the neighboring local variables are biholomorphic. It is clear that such local variables can be: for all points except four branch points and two infinities Inf 3 and Inf 4; for the branch points; for the infinities Inf 3 and Inf 4.

An analytic 1-form can be defined in the manifold [Gurvitz1968] by introducing a formal expression , where is a local variable (discussed above) in some proximity, and is an analytic function in this proximity. In neighboring proximities the representations can be different (say, and ), but they should match in an obvious way:

 f2=f1dα1dα2.

The 1-form can be analytic/meromorphic if the functions are analytic/holomorphic. In the same sense the form can have zero or a pole of some order.

Analyticity of a 1-form is an important property since one

One can see that the form

 Ψ=dxx(y−y−1) (30)

is analytic everywhere on . Let us prove this. The statement is trivial everywhere except the branch points and the infinities. Consider the infinities. At the points Inf 1 and Inf 2 it is easy to show that as , and the denominator is non-zero. At the points Inf 3 and Inf 4 one can show that as , thus . A change to the variable shows that the form is regular.

Finally, consider the branch points (25)–(28). As it has been mentioned, one can take as a local variable at these points. An important observation is that due to the theorem about an implicit function,

 dydx=−∂x^D∂y^D (31)

everywhere on . Thus,

 dxx(y−y−1)=−dyy(x−x−1). (32)

At the branch points the denominator of the right-hand side of (32) is not zero, so the form is regular.

The representation (15) can be considered as a contour integral of the form

 ψm,n=i2πxmynΨ (33)

along some contour drawn directly on . The contour is, indeed, shown in Fig. 2. This statement is quite trivial.

What is less trivial, is that three other representations, (16), (17), (18) can be represented as the contour integrals of the same form on , but taken along some other contours. Namely, the contours of integration for the representations (16), (17), (18), are shown in Fig. 4. They are denoted by , , , respectively. The contour for (15) is denoted by for uniformity.

Note that the form is, generally, not analytic on . Depending on and , it can have poles at the infinity points. The list of conditions of regularity for the infinity points is as follows:

 Inf 1 : m+n≥0, Inf 2 : m−n≥0, Inf 3 : −m−n≥0, Inf 4 : −m+n≥0.

The domains of regularities at infinities are shown in Fig. 5.

The representations (15)–(18) can be written in the common form

 u(m,n)=∫σjwm,nΨ, (34)

where

 wm,n=wm,n(x,y)=xmyn (35)

is the “plane wave” form. Note that the integration is held on , so , and any such obey the homogeneous stencil equation (19). Representation (34) can be considered as a plane wave decomposition.

Note that the contours , , , can be deformed into each other. As we mentioned, is a torus. Topologically, the relative positions of the contours and the infinity points are shown in Fig. 6

One can see that carrying the contours in the direction labeled by the red arrows corresponds to moving the observation point in the -plane in the clockwork direction. The representations are converted into each other, and every time there is a region where at least two representations are valid simultaneously.

### 2.7 Sommerfeld integral for Green’s function problem

Sommerfeld integral for this problem is formally a plane wave integral (34) with contour of integration that does not cross the line of propagating waves (locus of points ). After a simple analysis of one can obtain the result shown in figure 7.

One can notice that this curve is topologically equivalent to the one of the canonical sections of . To prove it let us show that domains and are simply connected. For simplicity consider the case (obviously, the topology of domains should be the same for any ). In this case domain covers physical sheet and domain covers unphysical sheet. Thus, resulting domains are linearly connected, and any closed contour lying in any of domains can be collapsed through infinity point.

Contour for Sommerfeld integral consists of two closed non-trivial contours lying at different sides of curve of propagating waves.An example of such a contour is shown in figure 8.

In figure 9 we plot these contours on torus .

Finally, Sommerfeld integral takes form:

 u(m,n)=∫γ1+γ2wm,nΨ. (36)

Obviously, contour is equivalent to the since there is no poles lying on the curve of propagating waves. Nevertheless, representation (36) seems to be more convenient when the problem for an incident plane wave is considered.

## 3 Diffraction by a Dirichlet half-plane

### 3.1 Problem formulation

Let the discrete Helmholtz equation

 uscm+1,n+uscm−1,n+uscm,n+1+uscm,n−1+(K2−4)uscm,n=0 (37)

be satisfied everywhere except line . On this line the following boundary condition should be satisfied:

 uscm,0=−uinm,0,m>=0, (38)

where is an incident plane wave:

 uinm,n=exp{iKmcosθin+iKnsinθin}. (39)

Here an angle of incidence. In order to satisfy the discrete Helmholtz equation, incident wave should satisfy the dispersion equation:

 D(xin,yin)=0, (40)

where we introduced a notation:

 xin=exp(ikcosθin),yin=exp(iksinθin). (41)

Introduce total field as a sum of incident and scattered field:

 um,n=uscm,n+uinm,n. (42)

Also the scattered field should satisfy the radiation condition.

### 3.2 Formulation on a branched surface

Consider a branched surface of continuous variables . For this, parametrize the points by the relations

 m=rcosϕ,n=sinϕ,r>0,0<ϕ≤4π. (43)

Thus, the points become defined on a surface with two sheets reminding the Riemann surface of the function .

Define an integer lattice on the branched surface. There are two points having coordinates for any pair except . Denote these points by , where as an index labelling the sheet somehow (say, by separating the surface into sheets by making an appropriate cut). The pair will be called an affix of the point.

We assume that there is a wave field defined on the points of the branched surface.

Each point except has exactly four stencil neighbors having affixes , , , . We say that equation (37) is valid on the branched discrete plane at some point if it is valid for the value of at and at four its neighbours.

### 3.3 Sommerfeld integral for half-plane problem

First let us first construct Sommerefeld integral for a plane wave on a plane. We search for

 um,n(xin,yin)=∫γ1+γ2A(x,y)wm,nΨ, (44)

where is some algebraic function on the torus . It is well known from the theory of elliptic functions [Bateman1955] that non-trivial function on the torus should have at least two poles of the first order or one pole of the second order. For definiteness, let us suppose that function has one simple pole corresponding to the incident wave , and the other simple pole corresponding to an arbitrary point . It can be checked directly that such function has the following form

 A(x,y,xin,yin)=cx+ay+b. (45)

Here constants and satisfy the following system of linear equations:

 xin+ayin+b=0, (46)
 xar+ayar+b=0. (47)

Let us choose constant in a way that in point the residue of integral (44) will be equal to . We obtain

 c=xin(yin−y−1in)(ay′(xin)+1)2πi, (48)

where is the first derivative of in the point . Thus, the integral (44) with (45) is Sommerfeld integral for a plane wave on plane.

Let us construct a plane wave solution on a branched surface. Following Sommerfeld ideas we need to be build a function that covers torus twice and has a unity pole corresponding to a plane wave. There are several obvious candidates that cover twice, such as

 X1(x,y)=√(x−η1,1)(x−η1,2),X2(x,y)=√(x−η2,1)(x−η2,2), (49)
 X3(x,y)=√(x−η1,1)(x−η2,2),X4(x,y)=√(x−η1,2)(x−η2,1), (50)
 X5(x,y)=√(x−η1,1)(x−η2,1),X6(x,y)=√(x−η1,2)(x−η2,2). (51)

Then, multiplying with function (45) we obtain the function with desired properties. Thus on a branched surface Sommerfeld integral has form:

 uzm,n(xin,yin)=∫γ1+γ2Xi(x,y)A(x,y,xin,y)wm,nΨ, (52)

To choose which function should be used one need to check the validity of radiation conditions. It can be showed after simple computation that only the integral with

 X2(x,y) (53)

satisfies them.

Finally, solution for the half-plane problem can be obtained using reflection principle:

 um,n=uzm,n(xin,yin)−uzm,n(xin,y−1in). (54)

### 3.4 Wiener-Hopf solution

Let us find the solution of the half-plane diffraction problem using the Wiener-Hopf approach. First, let us symmetrize the problem. Namely, represent the incident field (39) as a sum:

 uinm,n=12(uinm,n+urefm,n)+12(uinm,n−urefm,n)≡us,inm,n+ua,inm,n, (55)

where

 urefm,n=exp{iKmcosθin−iKnsinθin}. (56)

Then, study the equation (37) separately for the symmetrical and anti-symmetrical part of the field. One can check directly that anti-symmetrical problem is trivial, i.e.:

 ua,scm,n=0. (57)

Thus the solution of symmetrical problem coincide with the solution of (37), i.e.

 us,scm,n≡uscm,n.

Without loss of generality we can suppose that .

Introduce direct and inverse bilateral -transform as follows:

 F(z)=Z{fn}=n=+∞∑n=−∞fnz−n, (58)
 fn=Z{F(z)}=12πi∫δF(z)zn−1dz, (59)

where is a unit circle passing in a counterclockwise direction. To obtain functional equation let us apply -transform to boundary condition (38). We have

 Usc(z)=−Uin+(z)+G−(z), (60)

where is a unilateral -transform of :

 Uin+(z)=n=+∞∑n=0uinn,0z−n≡=zz−xin, (61)

and is some unknown function analytical inside the unit circle. Function is analytical outside the unit circle [Sharma2015b]. Equation (60) cannot have unique solution, since it also involve unknown function that is analytical in some ring. To introduce a second functional equation let us study a combination

 wm,n=12(uscm+1,n+uscm+1,n)+uscm,n+1+(K2−2)uscm,n. (62)

One can check directly that

 wm,0=0,m<0, (63)

and

 W+(z)=Υ(z)Usc(z), (64)

where were introduced in (29), and

 W+(z)=n=+∞∑n=0wn,0z−n (65)

is analytical outside the unit circle. Combining (60) and (64) we obtain the following Wiener-Hopf equation:

 W+(z)Υ(z)=−zz−xin+G−(z). (66)

The equation can be easily factorized. The solution is as follows:

 W+(z)=−z√(z−η2,1)(z−η2,2)√(xin−η1,1)(xin−η1,2)z−xin. (67)

The scattered field is given by the following integral

 uscm,n=−12πi∫δzmy|n|√(xin−η1,1)(xin−η1,2)√(z−η1,1)(z−η1,2)(z−xin)dz. (68)

It can be showed directly after some algebra that (68) is equivalent to (54).

## 4 Diffracton by a right-angled wedge

### 4.1 Problem formulation

Let the discrete Helmholtz equation (37) be satisfied everywhere except the domain (see figure 10).

On the boundary of this domain the following conditions should be satisfied:

 uscm,0=−uinm,0,m≥0,usc0,n=−uin0,n,n≥0. (69)

where is an incident plane wave (39). Also the scattered field should satisfy the radiation condition.

Using reflection method this problem can be reduced to the problem of wave propagation on three-sheeted surface [Sommerfeld1954]. It can be checked directly that total field on three sheeted surface is related to the total field of original problem by the following formula:

 um,n(xin,yin)≡uscm,n+uinm,n=uzm,n(xin,yin)−uzm,n(xin,y−1in)−uzm,n(x−1in,yin). (70)

### 4.2 Sommerfeld integral on three-sheeted surface

We will search integral in the form (44) as in previous sections. Here we need to construct function covering torus three times and having a pole corresponding to the incident wave with zero residue. Unfortunately, there are no obvious candidates like it is for two-branched surface. Let us study from the topological point of view. The Riemann diagram of torus is shown in figure 11.

It can be noticed that function which covers three times should have Riemann diagram that is shown in figure 12.

It can be easily proved that function having Riemann surface has the following structure:

 B(x,y)=3√r1(x)+r2(x)Υ(x), (71)

where , some rational functions. Thus, using (71) we can build Sommerfeld integral for three-sheeted surface with two unknown rational functions.

One can construct (71) by studying polynomial

 P[z,w]=z6+a5z5+a4z4+a3z3+a2z2+a1z+w. (72)

Namely, suppose that the roots of this polynomial define function , and this function has Riemann surface . Thus, there are exactly four points in which has exactly three roots of order two, i.e. it can be represented as:

 P[z,ηj]=((z−b1j)(z−b2j)(z−b3j))2, (73)

where are unknown parameters. These parameters should be determined from the following system of equations:

 ((z−b1j)(z−b2j)(z−b3j))2=z6+a5z5+a4z4+a3z3+a2z2+a1z+ηj. (74)

Equating coefficients at the same powers of we will obtain system of equation for unknown parameters , . Solving this system one can obtain exact expression for function (71).

## 5 Conclusion

In this paper we applied Sommerfeld integral approach to several diffraction problem for discrete Helmholtz equation (1). We showed that the field is represented as integral on a manifold. This manifold is torus, and corresponding integrals are Abelian integrals. For point source problem we proposed recursive procedure of field calculations which reduces integral computation to integral computation. For half-plane problem we constructed solution using Sommerfeld integral and showed it is equivalent to the Wiener—Hopf Solution. For the problem of diffraction by a right-angled wedge we showed that the problem can be reduced to the solution of nonlinear equation.

## Appendix A. Abelian integrals

Indeed, are Abelian differentials on . The form is an Abelian differential of the first kind, while all other are Abelian differentials of the third kind.

Moreover, since is a torus, the Abelian integrals in this case are elliptic functions.

The classical framework of study of the Abelian integrals is as follows. The surface is cut by several cuts (by two cuts in our case) such that the surface becomes mapped onto a polygon with an edge. These cuts are and (the latter is shown in Fig. 13).

The integrals

 u(m,n)=∫σψm,n,andu′(m,n)≡∫σ′ψm,n, (75)

are cyclic periods of an Abelian differential .

Note that is a solution of the problem based on equation (1) but with another radiation condition (one should take with a negative imaginary part).

All theorems related to Abelian and elliptic integrals can be applied to and . This properties can be found in [Bateman1955].

## Appendix B. Sommerfeld integral as an integral on the dispersion manifold

Let us build an analogy between Sommerefeld integral for continuous problem and Somerfeld integral for discrete problem.

let the Helmholtz equation

 Δusc(x,y)+k2usc(x,y)=0. (76)

be satisfied everywhere except half-plane

 y=0,x>0

where Dirichlet boundary condition is satisfied:

 usc(x,0)=−uin(x,0). (77)

Here is an incident wave:

 uin=exp{ikinxx+ikinyy}. (78)

Plane wave should satisfy the following dispersion equation:

 (kinx)2+(kiny)2=k2. (79)

Also, radiation and Meixner conditions should be satisfied. Let us introduce a plane wave decomposition. Following the idea of field representation by integration on a manifold one should first study (79). This manifold is a Riemann sphere with two punctured points. In this points field has exponential growth. It is more natural to study this manifold as a tube (see figure 14).

Also, equation (79) can be written in a parametrical form:

 kinx=kcosθin,kiny=ksinθin. (80)

Thus, this tube can be mapped to the strip:

 −∞

Thus, the polar coordinates

 x=rcosθ,y=rsinθ (82)

should be introduced and the plane wave decomposition should be some integral on . It was shown by Sommerfeld that it has the following form:

 uz(x,y,θin)=∫Γ1+Γ2A(r,θ)exp{ikrcos(θ−θin)}dθ. (83)

Contours of integration are showed in the figure 15.

On -branch surface function should be periodical with respect to with period , and should have a pole with unity residue corresponding to the incident wave. For a plane wave on a plane we have:

 A=12πexp{iθ}exp{iθ}−exp{iθin}. (84)

For a plane wave on 2-sheeted surface we have:

 A=12πexp{iθ/2}exp{iθ/2}−exp{iθin/2}. (85)

Integral (83) with (85) is called Sommerfeld integral. The solution of original problem is obtained using reflection principle:

 u(x,y)≡usc+uin=uz(x,y,θin)−uz(x,y,−θin). (86)

So, the following analogies between discrete and continuous solutions can be seen:

1. The field is represented as an integral on some manifold defined by dispersion equation. In the discrete case this manifold is torus, and in contentious case it is tube.

2. There are two contours of integration in Sommerefeld integral. Both contours do not cross the line of propagating waves (the line of real wavenumbers).

3. To obtain the solution for half-plane problem one need to construct a function that covers the manifold twice.

## Appenidx C. Sliding plane wave decomposition

For the discrete problem we can mimic this sliding. Namely, consider torus . There are 4 contours along which the integration can be held, and 4 possible “infinites” at which there can be singularities. They are shown in Fig. 4. The contours are , , , and .

A simpler scheme of the same surface with contours and infinities is shown in Fig. 6. A torus is shown as a torus in a usual sense. The relative position of the contours and the infinity points is drawn.

Consider an integral

 u(m,n)=∫σxmymΦ (87)

Let be some 1-form on having poles only at the four infinities. Let the order of the poles there is for some integer . According to the consideration made above, the form is regular at the infinites under the following conditions:

Take the observation point such that . Move this point about the origin in the direction of the red arrow in the figure. One can see that one can slide the contour sequentially in the order

 σ1→σ2→σ3→σ4→σ1…

(In fact, here we have in mind that, for example, in order to slide contour to we should ensure that the form is analytic at the point Inf 1.)

Corresponding contours provide a necessary decay of the solution in corresponding sectors of the plane .