Diffraction by a Dirichlet right angle on a discrete planar lattice
DDiffraction by a Dirichlet right angle on a discrete planarlattice
A. V. Shanin, A. I. KorolkovSeptember 9, 2020
Abstract
A problem of scattering by a Dirichlet right angle on a discrete square lattice is studied.The waves are governed by a discrete Helmholtz equation. The solution is looked for inthe form of the Sommerfeld integral. The Sommerfeld transformant of the field is builtas an algebraic function. The paper is a continuation of [1]. a r X i v : . [ m a t h - ph ] S e p OTATIONS C , C complex plane and Riemann sphere K wavenumber parameter of equation (1) u ( m, n ) wave field on the lattice φ in , φ angle of propagation of the incident wave, angle of scattering m , n indexes of nodes in the discrete physical plane x , y wavenumber parameters S branched discrete plane, introduced in [1] w m,n ( x, y ) plane wave (3) x in , y in wavenumber parameters of the incident waveˆ D ( x, y ) dispersion function (5)Ξ( x ) root of dispersion equation (8) defined by (10)Υ( x ) irrationality of Ξ( x ), (14) R Riemann surface of Ξ( x ) or Υ( x ) R a 3-sheet covering of R P , P projections between R , R , and C D dispersion surface D a 3-sheet covering of D ˜ x , ˆ x , x notations for points on R , R , and C linked by natural projections η , , η , , η , , η , branch points of R , defined by (12), (13)Λ, Π, Π (cid:48) symmetries of R , see (21), (23) A (˜ x ) Sommerfeld transformant of the field (see representation (17)) A , A , A the components of A having different properties with respect to Λ˜ x , ˜ x , ˜ x , ˜ x prescribed poles of A on R Y , . . . , Y residues of Aw m,n ( x, y ) discrete plane wave, (7)Γ , Γ contours for the Sommerfeld integral J , J , J (cid:48) , J (cid:48) , J (cid:48) contours encircling zero / infinity points on R K , K , K fields of functions meromorphic on C , on R , and on R , respectivelyΩ j : l basis of extension K j over K l F ( x ), F ( x ) nontrivial elements of the basis Ω (cid:36) exp { πi/ } , cubic root of 1ˆ b , b an important point on R used for building F and its affix χ (ˆ x ), T α , T β Abelian integral of the first kind on R (see (46)) and its periods α , β natural coordinates on the torus R ψ mapping χ → ˆ x Introduction
This paper continues the research presented in [1]. A 2D discrete square lattice is underconsideration. The lattice bears a discrete Helmholtz equation with a 5-point stencil. Thefirst quadrant of the lattice is blocked by setting the field equal to zero there. The problem ofdiffraction of an incident plane wave by the blocked angle is studied. The motivation and theliterature review for such a problem can be found in [1].A new formalism has been developed for this problem. Similarly to continuous problemsof diffraction in angular domains, a branching surface S is introduced in the physical discreteplane, and the diffraction problem is reformulated as a propagation problem on this surface byusing the reflection principle. An analog of the Sommerfeld integral for field representation isintroduced. This integral is a contour integral on a complex manifold D that is the dispersiondiagram for waves on the discrete plane. Topologically, this dispersion diagram is a torus. Theintegrand is a differential form that is multivalued on the dispersion diagram and possessesprescribed poles corresponding to the incident wave and reflected waves. The contour of inte-gration depends on the position of the observation point, and “slides” along the surface as theobservation point moves.The integrand form contains an unknown function referred to as the Sommerfeld trans-formant of the field. It obeys a certain functional problem. In [1] the authors found thistransformant in terms of elliptic functions. However, such a representation is not convenient.Moreover, it can be proven that such a transformant should be an algebraic function, thus,a representation through the elliptic functions is somewhat unnecessarily complicated. Theaim of the current paper is to build the Sommerfeld transformant of the field as an algebraicfunction, and then to study the properties of the field.The paper is organized as follows. In Section 2 the initial diffraction problem is formulated.Then, the main result of [1] in application to the angle diffraction problem is written down.Namely, a functional problem for the Sommerfeld transformant is set. In Section 3 the processof finding the transformant is outlined. Namely, it is proposed first to find the functionalfield K to which the transformant belongs, and then to specify the transformant in this field.The field is presented by its basis, and the particular element in it is constructed as a linearcombination of the basis elements. In Section 4 the basis of the field K is constructed. Thisis the most tricky part of the paper. In Section 5 the coefficients of representation of theSommerfeld transformant through the basis of the field are found. Thus, the representationof the wave field in the form of the Sommerfeld integral becomes obtained. In Section 6 theconstructed field representation is carefully checked. We demonstrate that the result obeysall condition imposed by the initial diffraction problem. This section can be considered as adouble check of the argument of [1] used for developing of the Sommerfeld integral formalism.In Section 7 some numerical checks are performed. We demonstrate the process of building thebasis function and, finally, compute the solution of the diffraction problem.3 Problem formulation
Consider a planar square lattice whose nodes have integer indices ( m, n ). Let the homogeneousdiscrete Helmholtz equation u ( m, n −
1) + u ( m, n + 1) + u ( m − , n ) + u ( m + 1 , n ) + ( K − u ( m, n ) = 0 (1)be valid in the domain m < n < K has a positive real part and a small positiveimaginary part corresponding to an energy absorption.The set of nodes with ( m = 0 and n ≥
0) or ( n = 0 and m ≥ u ( m, n ) = 0 (2)on it.Fig. 1: Geometry of the problem of diffraction by an angle. Black circles show the position ofthe Dirichlet boundary (blocked nodes)The total wave is a sum of the incident wave and the scattered wave: u ( m, n ) = u in ( m, n ) + u sc ( m, n ) , where u in ( m, n ) = x m in y n in , (3)4here x in and y in are wavenumber parameters. These parameters obey the dispersion equation:ˆ D ( x in , y in ) = 0 , (4)ˆ D ( x, y ) ≡ x + x − + y + y − + K − . (5)We assume that the wave travels into the direction of positive m and n . This means that | x in | < , | y in | < . Besides, we can introduce the angle of incidence by the relation φ in ≡ arctan (cid:18) y − y − x − x − (cid:19) . (6)We assume that angle φ in is real, and 0 < φ in < π/ . The scattered wave u sc should obey the radiation condition, i. e. it should decay at infinity.The aim is to find u sc . In [1] the authors developed the Sommerfeld integral technique for the lattice diffraction prob-lem formulated above. The unknown function describing the field is the Sommerfeld transfor-mant. The problem for finding the Sommerfeld transformant is formulated as a problem offinding a meromorphic function of a certain Riemann surface. The transformant should haveprescribed poles and residues.Plane waves on the lattice have form w m,n = w m,n ( x, y ) = x m y n , (7)provided that the pair of wavenumber parameters ( x, y ) obey the dispersion equationˆ D ( x, y ) = 0 . (8)One can see that (8) guarantees fulfillment of the homogeneous Helmholtz equation (1) by w .The set of all complex pairs ( x, y ) ∈ C obeying (8) is referred to as the dispersion diagramof the lattice. Denote this set by D . This is a complex manifold embedded in C . The simplestway to represent D is to express y as a function of x using (8): y ( x ) = Ξ( x ) or y ( x ) = Ξ − ( x ) , (9) A complex manifold [2] is a union of possibly intersecting neighborhoods in each of which a local com-plex variable can be introduced, describing the neighborhood in a trivial way. The transition between thelocal variables in intersecting neighborhoods is holomorphic. We assume also that the coordinates x and y areholomorphic functions of local variables. For D , one can take x as a local variable everywhere except the neigh-borhoods of the points η j,l (see (12), (13)) and except the neighborhoods of the infinities. In the neighborhoodsof the points η j,l one can choose y as a local variable. At the infinities one can use x − . x ) = − K − x + x − i (cid:112) − ( K − x + x − ) , (10)Ξ − ( x ) = − K − x + x − − i (cid:112) − ( K − x + x − ) , (11)and study the Riemann surface R of Ξ( x ). This Riemann surface is a projection of the manifold D onto the coordinate x . Topologically, D and R are similar.The complex structure of D is pulled to R by the projection, thus R is also a complexmanifold.One can see that R has two sheets over C . The branch points of the surface are η , , η , , η , , η , : η , = − d − i √ − d , η , = − d i √ − d , d = K − , (12) η , = − d √ d − , η , = − d − √ d − , d = K − . (13)All branch points are of second order. One can see that y = ± R is shown in Fig. 2. The branch points are connected by cuts (shown bybold lines). The sides of the cuts marked by the same Roman numbers should be attached toeach other. For definiteness, the cuts on R are conducted in such a way that | y ( x ) | = 1 onthem. The Riemann surface R is compactified, i. e. two infinite points are added to the sheets.These infinite points are not branch points. We select the physical sheet (or sheet 1) of R asthe sheet on which | Ξ( x ) | < | x | = 1. The unit circle on the physical sheetis shown by a dashed line in the figure.Introduce the functionΥ( x ) = x (Ξ( x ) − Ξ − ( x )) = (cid:113) ( x − η , )( x − η , )( x − η , )( x − η , ) = (14) x (cid:112) ( K − x + x − ) − , Ξ( x ) = − K − x + x − x )2 x (15)The first and the second representation in (14) are equivalent (one can check this), but the firstone links the the branch of the square root of Υ to that of Ξ. Indeed, Υ( x ) is the irrationalityof Ξ( x ), thus the Riemann surface of Υ( x ) is the same as of Ξ( x ), i. e. this surface is R .Topologically, R is a torus, and thus D is a torus as well. This fact has been heavilycommented and exploited in [1].One of the main ideas of [1] is as follows. Application of the reflection principle to thephysical configuration shown in Fig. 1 leads to a branched discrete physical plane having threesheets (it is referred to as S in [1]). To construct a Sommerfeld integral for the field, one needsa three-sheet covering of the dispersion diagram D . There exist several three-sheet coverings6 ,1 h h h h Re[ ] x Im[ ] x I I II II III
III I V I V Sheet 1 Sheet 2
Fig. 2: Riemann surface R of D . One should select the particular covering that triples the “real wave” contour and leavesunchanged the contour of integration for the Green’s function. These two selected contoursare homotopic to the loops σ α and σ β , respectively, introduced below. Corresponding coveringdenoted by D is built in [1] (see Fig. 13 there).Here we specify the covering D over D by its projection onto the variable x . The result isthe covering R over R . R is a Riemann surface whose scheme is shown in Fig. 3. Note thatthere is no function having such a Riemann surface is known a priori .The complex structure of D remains valid on D , thus D is a complex manifold immersed in C . Indeed, topologically D (or R ) is also a torus. The Riemann surface R is a 3-sheetcovering of R without branching, therefore any function meromorphic on R is also meromorphicon R . Sheets 1, 3, and 5 of R shown in Fig. 3 correspond to the physical sheet of R .Introduce notations for the points of the compactified complex plane ¯ C of x , and for theRiemann surfaces R , R . The points of R will be indicated by the ˜ · decoration, the points of R will be indicated by the ˆ · decoration, and the points of ¯ C will exist without decorations. Forexample, ˜ x ∈ R , ˆ x ∈ R , x ∈ ¯ C . There exist natural projections that constitute the definitions of coverings:˜ x P −→ ˆ x P −→ x. The projections ˜ x → x and ˆ x → x are taking an affix of a point of a Riemann surface.It is important to notice that we keep the following convention in the whole paper: every-where ˆ x is the projection of ˜ x , and x is a projection of ˆ x and ˜ x . Indeed, this is valid for anyletter instead of x (it may be, say, a or b ). This can be written asˆ · ≡ P (˜ · ) , · ≡ P (ˆ · ) ≡ P ( P (˜ · )) , (16)where · stays for any letter, possibly with indexes, but without a decoration.If function f (ˆ x ) is single valued on R , there is no difficulty to define it on R as a single-valued function f (˜ x ) = f ( P (˜ x )) = P (ˆ x ) . ,1 h h h h Re[ ] x Im[ ] x I II III I V Sheet 1Sheet 3Sheet 5 Sheet 2Sheet 4Sheet 6 I II III VV I V V I I V V II V III V II V III I X X I XX X I X II X I X II JJ J'J'J'
Fig. 3: Riemann surface R and integration contours on itConversely, if f (˜ x ) is single-valued on R , the function f (ˆ x ) is generally three-valued on R ,and f ( x ) is six-valued on ¯ C . Functions Υ(ˆ x ) and Ξ(ˆ x ) are single-valued; functions Υ( x ) andΞ( x ) are two-valued.Let ˜ x , . . . ˜ x be the points of R which are specified by the affixes x j , the values y = y (˜ x )and the sheets of R to which they belong:˜ x : ( x in , y in ) , sheet 3 , ˜ x : ( x in , y − ) , sheet 4 , ˜ x : ( x − , y − ) , sheet 6 , ˜ x : ( x − , y in ) , sheet 1 . As it is shown in [1], the point ˜ x corresponds to the incident plane wave, the points ˜ x and ˜ x correspond to the plane waves reflected by the horizontal and the vertical parts of the boundary,and the point ˜ x corresponds to the mirror image of the reflected wave.8et be Y j = Υ(ˆ x j ) , j = 1 , . . . , . One can see that Y = − Y , Y = − Y .Following [1], let us formulate the functional problem for the Sommerfeld transfor-mant A (˜ x ) of the wave u ( m, n ):1. A (˜ x ) is meromorphic on R . As it follows from this condition, the branch points of A ( x )can be located only at η , , η , , η , , η , .2. Function A (˜ x ) are regular as | x | → ∞ on each of the six sheets of R .3. Function A (˜ x ) has four poles on R . The position of the poles and the residues at themare listed in the following table:˜ x x y ( x ) Sheet Residue˜ x x in y in − (2 πi ) − Y ˜ x x in y − − (2 πi ) − Y ˜ x x − y − − (2 πi ) − Y ˜ x x − y in − (2 πi ) − Y Indeed, once function A (˜ x ) is found, one can say that R is the Riemann surface of A ( x ).Besides, being defined on R , the function A can be pulled back to D . So one can considerfunction A as being a 3-valued function defined on the dispersion diagram of the discrete plane.Function A , formally, has been found in [1] (see (70) there), however it has been expressedin elliptic functions, and this solution is hardly practical. However, the formulation of theproblem is of algebraic nature, so one can expect a purely algebraic solution.Once A (˜ x ) is found, the wave field can be built using the Sommerfeld integral introducedin [1] and having the form u ( m, n ) = (cid:90) Γ j w m,n (ˆ x, Ξ(ˆ x )) A (˜ x ) dx Υ(ˆ x ) . (17)The contours of integration Γ j are drawn on R . The indexing of the contours is kept similarto that of [1]. For m ≤ :Γ = J + J + J (cid:48) + J (cid:48) . (18)For n ≤ :Γ = J + J (cid:48) + J (cid:48) + J (cid:48) . (19)The contours J , J , J (cid:48) , J (cid:48) , J (cid:48) are shown in Fig. 3. Contours J and J (cid:48) encircle the point x = 0on corresponding sheets. Contours J , J (cid:48) , J (cid:48) encircle the inifinities on corresponding sheets(i. e. they encircle all finite singularities). 9he domain with m ≤ n ≤ or contourΓ . It is demonstrated in [1] that these representations are equivalent.In Section 6, to make a trustful presentation, we prove that the field (17) obeys all conditionimposed on the solution of the diffraction problem.The aim of the current paper is to construct the function A (˜ x ) obeying the properties 1–3listed above. A ( x ) Let ˜ x ( j ) , j ∈ { , . . . , } denote a point of R having affix x and lying on the sheet number j (the numbering of the sheets is kept as in Fig. 3).Consider the cyclical change of sheets1 → → → , → → → . (20)This change generates a symmetry (referred to as Λ) of R , Namely,Λ(˜ x ( j ) ) = ˜ x ( j (cid:48) ) , (21)where the point ˜ x ( j (cid:48) ) has the same affix as ˜ x ( j ) , and the sheet j (cid:48) is obtained from j by applyingthe function (20).One can see that if a function f (˜ x ) is meromorphic on R then the same is valid for f (Λ(˜ x )).Moreover, Λ does not change the value of Υ(˜ x ):Υ(Λ(˜ x )) = Υ(˜ x ) . (22)Another symmetry (referred to as Π) is as follows:Π(˜ x ( j ) ) = ˜ x (7 − j ) (23)(again, the affix remains the same). One can see that if a function f (˜ x ) is meromorphic on R then the same is valid for f (Π( x )). A direct check shows thatΥ(Π(˜ x )) = − Υ(˜ x ) . (24)As it is shown in Section 6, the symmetry Π of the Riemann surface R corresponds to thegeometrical symmetry n → − n of the initial physical system. The symmetry Π can be elevatedto D . It has form x → x, y → y − . x )) = ˜ x, Λ(Λ(Λ(˜ x ))) = ˜ x. The symmetries Π and Λ can be used to simplify the formulation of the functional problemfor A (˜ x ). One can see that A (˜ x ) can be chosen symmetrical: A (Π(˜ x )) = A (˜ x ) . (25)since the combination ( A (˜ x ) + A (Π(˜ x )) / A . Besides, one canapply a symmetrization with respect to Λ (i. e. a discrete Fourier transform of dimension 3 foreach point ˆ x ), i. e. represent A as the sum of three components A (˜ x ) = A (˜ x ) + A (˜ x ) + A (˜ x ) , (26)such that A (Λ(˜ x )) = A (˜ x ) , A (Λ(˜ x )) = (cid:36)A (˜ x ) , A (Λ(˜ x )) = (cid:36) − A (˜ x ) , (27)where (cid:36) ≡ e πi/ . (28)The components of A can be found by the transform A (˜ x ) = 13 ( A (˜ x ) + A (Λ(˜ x )) + A (Λ(Λ(˜ x )))) , (29) A (˜ x ) = 13 (cid:0) A (˜ x ) + (cid:36) A (Λ(˜ x )) + (cid:36)A (Λ(Λ(˜ x ))) (cid:1) , (30) A (˜ x ) = 13 (cid:0) A (˜ x ) + (cid:36)A (Λ(˜ x )) + (cid:36) A (Λ(Λ(˜ x ))) (cid:1) . (31)Each of the functions A , A , A is meromorphic on R , is regular at the infinite points, andpossesses 12 poles at the points with affixes x in and x − . The residues of the poles for each ofthe functions is given by the following table (common for all three functions):˜ x x Sheet Residue˜ x x in − (6 πi ) − Y ˜ x x in − (6 πi ) − Y ˜ x x − − (6 πi ) − Y ˜ x x − − (6 πi ) − Y A , A , A rather than that of for A . Remark
There is another symmetry Π (cid:48) of the Riemann surface R . This symmetry will be used oncein Section 6. The choice of x as the independent variable is not unique. In a very similar way,one could use y as an independent variable. In this case, x = Ξ( y ) is a double-valued functionhaving a Riemann surface R (cid:48) . Indeed, R (cid:48) is similar to R . Moreover, R (cid:48) is a projection of thesame manifold D onto another variable.One can define the Riemann surface R (cid:48) as a projection of D onto y . R (cid:48) has the samestructure as R , but the independent variable is y .There exists a symmetry of R (cid:48) constructed the same way as Π. Denote it by Π (cid:48) . Thissymmetry has form x → x − , y → y and corresponds to the symmetry m → − m of the physical plane. One can demonstrate that A (Π (cid:48) (˜ x )) = A (˜ x ) . The set of functions f ( x ) meromorphic on ¯ C is a field ( [3], chapter 11). (Here “meromorphic”means also that the function has a finite number of poles/zeros.) This field consists of allrational functions of x . Denote this field by K .Consider the field of functions meromorphic on R , i. e. on the Riemann surface of thefunction Ξ( x ) or of Υ( x ). Denote this field by K . Note that we specify the field by indicatinga Riemann surface on which its elements are meromorphic. One can prove a nontrivial theoremthat all elements of K are rational functions of x and Υ( x ), so K is an extension of K bya single element Υ( x ). K is an algebraic extension of K since the function Υ( x ) obeys analgebraic equation with coefficients belonging to K :Υ ( x ) − ( x − η , )( x − η , )( x − η , )( x − η , ) = 0 . (32)The fields of functions meromorphic on certain Riemann surfaces are studied in more details,for example, in the monograph [4].One can prove (see [3], chapter 23) that such an algebraic extension has a basis , i. e. a setΩ = [ ω , . . . , ω j ] of elements of K such that any element z of K can be uniquely representedas z (ˆ x ) = q ( x ) ω (ˆ x ) + · · · + q j ( x ) ω j (ˆ x ) , (33)where q j ∈ K , i. e. they are rational functions of x . A basis of the extension K over K canbe easily found: Ω = [1 , Υ(ˆ x )] . (34)12n other words, the following statement is valid: any function meromorphic on R can berepresented uniquely in the form q (ˆ x ) = z ( x ) + z ( x )Υ(ˆ x ) , z , z ∈ K . (35)The number of elements of the basis is referred to as the degree of extension of K over K .This degree is equal to 2.According to the functional problem for the Sommerfed’s transformant A (˜ x ), this transfor-mant should belong to the set of functions meromorphic on R . This set is also a field. Denoteit by K . As it can be proven (see the next subsection), K is an algebraic extension of K ,and the degree of the extension is equal to 3. Since K ⊂ K , one of the elements of the basiscan be taken equal to 1. Thus, we are looking for three-valued functions F (ˆ x ), F (ˆ x ) (or, thesame, six-valued functions F ( x ) and F ( x )) such thatΩ = [1 , F , F ] (36)is the basis of K over K . The Sommerfeld transformant of the field A (˜ x ) should be uniquelyrepresented as A (˜ x ) = q (ˆ x ) + q (ˆ x ) F (˜ x ) + q (ˆ x ) F (˜ x ) , (37)where q j (ˆ x ) belong to K being rational functions of x and Υ(ˆ x ).Finding the functions F (˜ x ) and F (˜ x ) is an unusual problem since no function whoseRiemann surface is R is given a priori . Finding the coefficients of q j ( x ) is, conversely, analmost trivial task when the basis (36) is built. They are constructed by using the knowledgeof poles and residues of the Sommerfeld transformant, i. e. by using the condition 3 imposedon A .Note that the concept of a pole / zero at some point on the complex manifolds R or on R differs from that on C if the affix of the point is equal to any of the branch points η j,l .Consider a function f (ˆ x ) ∈ K such that f ( η j,l ) = 0 or f ( η j,l ) = ∞ . To find the multiplicity ofthe zero / pole at η j,l one should introduce the local variable τ in the neighborhood of η j,l insuch a way that the neighborhood of η j,l on R is described by a small circle in the τ -plane ina topologically trivial way, i. e. such that ˆ x ( τ ) is a bijection. One can use for example τ = (cid:112) x − η j,l as such a variable (another choice is to take τ = Υ(ˆ x )). Then, one should express locally f (ˆ x )in terms of τ , i. e. build a function θ ( τ ) = f (ˆ x ( τ )) . The multiplicity of zero / pole of f is, by definition, the multiplicity of corresponding zero /pole of θ ( τ ).For example, the function Υ(ˆ x ) has simple zeros on R at each of the points η j,l , although itmay seem surprising. Moreover, the function1Υ (ˆ x ) = 1( x − η , )( x − η , )( x − η , )( x − η , )13as double poles at each η j,l on R , while the same function, but considered on C , has simplepoles at those points.The poles / zeros on R are defined in the same way as for R .For all three surfaces C , R , and R , the poles / zeros at the infinity can be studied byintroducing the local variable τ = 1 /x .The plan for building the function A (˜ x ) is is as follows. First, we construct the functions F (˜ x ), F (˜ x ) and form the basis Ω (see (36)). Second, we find the coefficients q (ˆ x ), q (ˆ x ), q (ˆ x ) of (37). Remark
In [1] we studied functions meromorphic on a two-sheets covering of R , namely on R . Thefunctions meromorpic on this surface form a field K . This field can be considered as as anextension over K of degree 2. The basis of this extension isΩ = [1 , Q ( x )] , Q ( x ) = (cid:113) ( x − η , )( x − η , ) . (38)The field K can be also considered as an extension over K . In this case, the basis consist offour elements listed in [1], (50), (51):Ω = [1 , Υ( x ) , Q ( x ) , Q ( x )Υ( x )] . (39)In this case, the coefficients should belong to K . over K According to the standard argument of Galois theory [4], the symmetries of R are linked withthe structure of the field K . Let us sketch out the proof that the degree of the the extension K over K is equal to 3 and that this extension is algebraic.Let a function g (˜ x ) ∈ K have the property g (Λ(˜ x )) = g (˜ x ). Then this function is single-valued on R , and, thus, g ∈ K .Consider any function f (ˆ x ) meromorphic on R but not meromorphic on R . This meansthat f ∈ K is not single-valued on R . Consider the combinations similar to (29), (30), (31): f (˜ x ) = 13 ( f (ˆ x ) + f (Λ(ˆ x )) + f (Λ(Λ(ˆ x )))) , (40) f (˜ x ) = 13 (cid:0) f (ˆ x ) + (cid:36) − f (Λ(ˆ x )) + (cid:36)f (Λ(Λ(ˆ x ))) (cid:1) , (41) f (˜ x ) = 13 (cid:0) f (ˆ x ) + (cid:36)f (Λ(ˆ x )) + (cid:36) − f (Λ(Λ(ˆ x ))) (cid:1) , (42)Obviously, if f , f , f are known, the function f can be reconstructed by f (˜ x ) = f (˜ x ) + f (˜ x ) + f (˜ x ) . (43)14ote that f (Λ(˜ x )) = f (˜ x ) , f (Λ(˜ x )) = (cid:36)f (˜ x ) , f (Λ(˜ x )) = (cid:36) − f (˜ x ) . (44)One can conclude from (44) that the functions f , ( f ) , ( f ) are single-valued on R , and thusbelong to K , i. e. they are represented by (35). Obviously, functions f and f obey cubicequations with coefficients from K .Take the basis Ω = [1 , f , f ] . (45)The function f is represented through this basis by using the coefficients f , 1, 1. Let us showthat any other function f (cid:48) (˜ x ) ∈ K can be represented as (37) using the basis (45). Perform thesame procedure as above. Get the functions f (cid:48) , f (cid:48) , f (cid:48) having the properties (44). Function f (cid:48) belongs to K and can be used as the first coefficient of the expansion. Consider the functions f (cid:48) j /f j ( x ), j = 1 ,
2. They are meromorphic and single-valued on R , thus they belong to K .These functions can be used as corresponding coefficients. F , F As a tool, we use the Abelian integral of the first kind on R . The detailed description of thissubject can be found, e.g., in [3], chapter 12. Since R is a torus, there is one Abelian integralregular everywhere (up to a constant factor and a constant additive term): χ (ˆ x ) = ˆ x (cid:90) η , dx (cid:48) Υ(ˆ x (cid:48) ) . (46)The integral is assumed to be taken along some contour γ drawn on R . The contour starts at η , and ends at some point ˆ x ∈ R . Indeed, the starting point is arbitrary, and the value η , ischosen for convenience. The value of the integral depends not only on the point ˆ x , but also onthe homotopic class of the contour γ .Consider the contours σ α and σ β on R (see Fig. 4). These contours have a common point η , .Being cut along the contours σ α and σ β , the surface R becomes (topologically) a parallelogram.Contours σ α and σ β play an important role in [1] and here. The contour σ α is homotopicto the “real waves” contour on R . This contour is a set of points ˆ x such that an expression x m Ξ n (ˆ x ) can be treated as a usual plane wave on the discrete plane. The real waves can beeasily defined if Im[ K ] = 0: they possess ( x, y = Ξ(ˆ x )) with | x | = | y | = 1 (such waves do notdecay in any direction). Thus, the real waves line is composed of two copies of an arc of theunit circle connecting η , and η , . It is shown in Fig. 5 by a dashed line.If Im[ K ] (cid:54) = 0, the definition of the real wave may be ambiguous since all waves possess decay/ growth in some directions. We fix “real waves” as the connected set of points ( x, y ) ∈ D ,1 h h h h Re[ ] x Im[ ] x Sheet 1 Sheet 2 s b s a s a Fig. 4: Contours σ α and σ β on R Re[ ] x Im[ ] x f =0 f p = f p/2 = Re[ ] x Im[ ] x f p =2 =0 f p = f p/2 =3sheet 1 sheet 2 h h Fig. 5: “Real waves” line on R for real K (dashed line)obeying the property Im (cid:20) y − y − x − x − (cid:21) = 0 (47)and passing through the branch points η , and η , . This set tends to the line shown in Fig. 5as Im[ K ] →
0. One can show, e. g. numerically or asymptotically, that the set of all such “realwaves” is a closed contour on R homotopical to σ α .The contour σ β is homotopic to the unit circle on the physical sheet of R . Being equippedwith an orientation, it becomes the integral path for the Green’s function of a discrete plane(see [1]).Comparing Fig. 4 with Fig. 3 one can conclude that the covering R of R is such that thepreimage of σ β is a set of three copies of σ β , while the preimage of σ α is a three-sheet coveringof σ α . Note that a bypass along the contour σ α , being elevated onto R , changes the sheets of R cyclically according to the transformation Λ introduced above.The integrals of the form (46) taken along the closed contours σ α and σ β on R are the periods of χ (ˆ x ) referred to as T α and T β : T α = (cid:90) σ α dx Υ(ˆ x ) , T β = (cid:90) σ β dx Υ(ˆ x ) . (48)16he mapping ˆ x → χ maps the surface R cut along the contours σ α and σ β onto an elemen-tary parallelogram in the complex χ -plane. This parallelogram is shown in Fig. 6, left. Re[ ] c Im[ ] c T a T b ab p p a bc ( ) P - Fig. 6: Elementary parallelogram in the χ -plane and coordinates ( α, β ) on a torusConsider an inverse mapping ψ : χ → ˆ x . As it is known, this mapping has the followingproperties: ψ ( χ + T α ) = ψ ( χ + T β ) = ψ ( χ ) , (49)i. e. it is bi-periodic. This property can be used for introduction of coordinates α and β on R ,revealing the structure of R as the structure of a torus. Namely, the coordinates α and β canbe introduced as linear combinations α = c , Re[ χ ] + c , Im[ χ ] , β = c , Re[ χ ] + c , Im[ χ ] , (50)with the coefficients c j,k found from the following equations c , Re[ T α ] + c , Im[ T α ] = 2 π, c , Re[ T β ] + c , Im[ T β ] = 0 , (51) c , Re[ T α ] + c , Im[ T α ] = 0 , c , Re[ T β ] + c , Im[ T β ] = 2 π. (52)The coordinates ( α, β ) on R are shown in Fig. 6, right. The surface R is displayed schemat-ically as a torus, i. e. R is deformed in an appropriate way. The resulting surface is compact,thus, the infinities are represented as two points on it. The coordinate lines of α and β on theinitial representation of R are close to those shown in Fig. 4 of [1].According to the schemes in Fig. 2 and Fig. 3, the torus R corresponds to the parallelogram R : 0 ≤ α < π, ≤ β < π, while the torus R corresponds to the parallelogram R : 0 ≤ α < π, ≤ β < π. Coordinates α and β are close to the coordinates α and β defined in [1], but not exactly the same. Notethat the requirement that β = π on the “real waves” line is not fulfilled in the new formulation. α, β ) ∈ R is the image of three points ( α, β ) , ( α + 2 π, β ) , ( α + 4 π, β ) of the cover-ing R .The symmetries Λ and Π have the following representations in the coordinates ( α, β ):Λ : α → α + 2 π, β → β, Π : α → π − α, β → − β. We are starting with an auxiliary problem. For any 4 points ˆ a , ˆ a , ˆ b , ˆ b on R find out whetherthere exists a function M (ˆ x ) ∈ K having simple poles ˆ b , ˆ b only, and simple zeros ˆ a , ˆ a only,and, indeed, build such a function if it exists.A criterion of existence of such a function M is known. This criterion is the Abel’s theorem(see [5], chapter 10) : On R , there should exist a contour γ going from ˆ a to ˆ b and a contour γ going from ˆ a to ˆ b such that (cid:90) γ dx Υ(ˆ x ) + (cid:90) γ dx Υ(ˆ x ) = 0 . (53)The criterion has a transcendent (non-algebraic) character. In this subsection we are going toderive an algebraic version of it. A general case
Assume that a , a , b , b are all distinct values not equal to infinity or to η j,l . This is thegeneral case. Some important particular cases will be considered below.Let us try to construct function M (ˆ x ) explicitly. First, construct function M (ˆ x ) (possiblydepending on parameters) having poles at ˆ b and ˆ b . An obvious Ansatz (up to a commonconstant factor) is as follows: M (ˆ x ) = Υ(ˆ x )( x − b )( x − b ) + g x − b + g x − b + c. (54)for some complex values g , g , c . As above, x, a , a , b , b are the affixes of the pointsˆ x, ˆ a , ˆ a , ˆ b , ˆ b , respectively.For arbitrary g , g , this function has poles at four points of R : at P − ( b ) and P − ( b ).Choose the values of g , g such that they suppress the poles that have affixes b , b , but thatare not ˆ b and ˆ b . One can see that the appropriate function is as follows: M (ˆ x ) = Υ(ˆ x )( x − b )( x − b ) + Υ(ˆ b )( x − b )( b − b ) + Υ(ˆ b )( x − b )( b − b ) + c. (55)Now let us fix the zeros. Choose parameter c in such a way that M (ˆ a ) = 0: c = − (cid:34) Υ(ˆ a )( a − b )( a − b ) + Υ(ˆ b )( a − b )( b − b ) + Υ(ˆ b )( a − b )( b − b ) (cid:35) . (56)18inally, the condition guaranteeing that ˆ a is also a zero is the equationΥ(ˆ a )( a − b )( a − b ) + Υ(ˆ b )( a − b )( b − b ) + Υ(ˆ b )( a − b )( b − b ) =Υ(ˆ a )( a − b )( a − b ) + Υ(ˆ b )( a − b )( b − b ) + Υ(ˆ b )( a − b )( b − b ) . (57)This is an equation linking ˆ a , ˆ a , ˆ b , ˆ b that guarantees existence of a function M (ˆ x ) meromor-phic on R , having simple zeros at ˆ a , ˆ a and simple poles at ˆ b , ˆ b . Thus, (57) is an algebraicanalog of the analytic equation (53). The function M itself is given by (55), (56).Another (more symmetrical) form of (57) is( b − b )[( a − b )( a − b ) Υ(ˆ a ) − ( a − b )( a − b ) Υ(ˆ a )] = (58)( a − a )[( a − b )( a − b ) Υ(ˆ b ) − ( a − b )( a − b ) Υ(ˆ b )] . A special case: poles ˆ b and ˆ b coincide The coincidence of ˆ b and ˆ b means that ˆ b = ˆ b = ˆ b is a pole of order 2. The Ansatz for M that should replace (55) is as follows: M (ˆ x ) = Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + c, (59)where ˙Υ(ˆ x ) ≡ d Υ(ˆ x ) dx . (60)The constant c is chosen in such a way that M (ˆ a ) = 0: c = − (cid:34) Υ(ˆ a )( a − b ) + Υ(ˆ b )( a − b ) + ˙Υ(ˆ b ) a − b (cid:35) (61)Finally, this function is zero at ˆ a ifΥ(ˆ a )( a − b ) + Υ(ˆ b )( a − b ) + ˙Υ(ˆ b ) a − b = Υ(ˆ a )( a − b ) + Υ(ˆ b )( a − b ) + ˙Υ(ˆ b ) a − b . (62)Thus, (62) is the condition of existence of a meromorphic function on R having zeros at ˆ a andˆ a , and a double pole at ˆ b . The same condition should be valid for existence of a function witha double zero at ˆ b and simple poles at ˆ a and ˆ a . A special case: poles ˆ b and ˆ b coincide with a branch point η j,l If ˆ a and ˆ a are different points of R having the same (arbitrary) affix a = a = a , then nocondition is needed. One can draw the contours γ and γ on different sheets of R such that19 ( γ ) = P ( γ ), and (53) will be valid automatically. The function M (ˆ x ) is then as follows: M (ˆ x ) = x − ax − η j,l . (63)Note that η j,l is a double pole of M on R according to a comment at the end of Section 3.A detailed study based on the bijection between the elementary parallelogram in the χ -planeand R shows that if ˆ a and ˆ a have different affixes, a corresponding function M cannot exist. A special case: a double pole at ˆ b , a simple zero at ˆ b , another simple zero at η j,l ,affixes of ˆ b and ˆ b coincide: b = b = b A function with a double pole at ˆ b and regular at ˆ b is as follows: M (ˆ x ) = Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + c. (64)Function M (ˆ x ) has a zero at ˆ b if c = ¨Υ(ˆ b )2 , (65)where ¨Υ(ˆ x ) ≡ d Υ(ˆ x ) dx . Since Υ( η j,l ) = 0, the condition M ( η j,l ) = 0 reads asΥ(ˆ b )( η j,l − b ) + ˙Υ(ˆ b ) η j,l − b + ¨Υ(ˆ b )2 = 0 . (66)Indeed, equations (62) and (66) are also algebraic versions of (53) in the correspondingspecial cases. F and F of the basis Ω In this subsection we describe the main result of the paper, namely, we build functions F and F .The difficulty of building the basis Ω is as follows. The reasoning made above shows thatconstructing of, say, F (˜ x ) should include taking a cubic radical of a function belonging to K .Let this function be G (ˆ x ), i. e. let be F (˜ x ) = G / (ˆ x ) . (67)Since function F is not allowed to have branch points on R , all poles and zeros of G (ˆ x ) on R should have order 3 ν , ν ∈ Z . At the same time, if function G (ˆ x ) is a cube of another functionfrom K , say G (ˆ x ) = g (ˆ x ) , G / (ˆ x ) are just g (ˆ x ), (cid:36)g (ˆ x ), (cid:36) − g (ˆ x ). All of these functions belong to K , and thus, cannot contribute to a basis Ω . Therefore, it is necessary to find a function G (ˆ x ) ∈ K , having poles and zeros of order 3 ν , but that is not a cube of a function from K . Remark.
This can be illustrated by the example solved in [1], where a basis Ω has been constructed.A non-trivial element of this basis is Q ( x ) defined by (38). One can see that Q ( x ) = G / ( x ) , where G ( x ) = ( x − η , )( x − η , ) , The function G has double zeros at x = η , and x = η , (they are double in the sense of localvariable τ on the complex manifold R , sew above), and two double poles at the infinities of thetwo sheets of R . However, G ( x ) is not a square of any function meromorphic on R .To get F (˜ x ) having the Riemann surface R , one should impose two additional restrictionson G (ˆ x ). Namely, the variation of the argument of G along σ β should be equal to 6 πν , ν ∈ Z ,while its variation along σ α should be 2 π (3 ν + 1) or 2 π (3 ν + 2).Let us describe the procedure of constructing the function G (ˆ x ). First, we describe it interms of the Abel’s criterion. Consider the period T β defined by (48). Take an arbitrary pointˆ a ∈ R . Find the points ˆ b, ˆ c on R such that (cid:90) γ dx Υ( x ) = (cid:90) γ dx Υ( x ) = (cid:90) γ dx Υ( x ) = T β . (68)The contours γ , , cyclically connecting the points ˆ a , ˆ b , ˆ c are shown in Fig. 7, right. TheRiemann surface R is shown schematically as a torus in this figure, i. e. it is deformed homo-topically to emphasize its topology. The contours σ α and σ β introduced by Fig. 4 are shown inFig. 7, left. We take the contours γ , γ , γ such that their concatenation is homotopic to σ β .To find such points, ˆ b , ˆ c , one can build explicitly the coordinate line α = const startingfrom ˆ a , i. e. the line along which the integral ˆ x (cid:90) ˆ a dx (cid:48) Υ(ˆ x (cid:48) )changes linearly from 0 to T β . Then one should split this segment into 3 equal parts in the χ -plane. Contours γ , γ , γ can be taken as corresponding straight segments in the χ -planesubject to the mapping ψ .Construct two functions M (ˆ x ) and M (ˆ x ) such that: • M has a double pole at ˆ a and simple zeros at ˆ b and ˆ c ; • M has a double pole at ˆ b and simple zeros at ˆ a and ˆ c .21 g g s s a b s ' b a ˆ b ˆ c ˆ Fig. 7: Contours σ α and σ β on R (left). Points ˆ a , ˆ b , ˆ c on the torus R (right)Both functions do exist according to the Abel’s criterion.Function G is constructed as G (ˆ x ) = M (ˆ x ) M (ˆ x ) . (69)By construction, G (ˆ x ) has a triple pole at ˆ a and a triple zero at ˆ b . There are no other poles orzeros of G .Function F defined by (67) is three-valued on R and has no branch points. Moreover, the F cannot be a meromorphic function on R , since there is no function meromorphic on R and having a single simple pole / zero on it.Let us find the change of the argument of the function G (ˆ x ) along the contour σ β . Computethe phase change on the contour σ (cid:48) β shown in Fig. 7, right, not passing through the points ˆ a ,ˆ b , ˆ c , and taken parallel to the β -axis in the ( α, β )-plane. What is important, this contour goesnot cross the contours γ , γ , γ since they are taken also parallel to the β -axis by construction.Consider a family of functions M (ˆ z ; ˆ x ) (ˆ x is considered as an argument and ˆ z is a param-eter, both belong to D ). The functions are as follows. Construct points z and z in such away that ˆ z (cid:90) ˆ z dx Υ( x ) = ˆ z (cid:90) ˆ z dx Υ( x ) = ˆ z (cid:90) ˆ z dx Υ( x ) = T β . (70)and demand that M (ˆ z ; ˆ x ) has a double pole at ˆ z and simple zeros in ˆ z and ˆ z . Normalize M in any invariant way (say, by the residue of M − at ˆ z ). By construction, the family M (ˆ z ; ˆ x )is continuous with respect to ˆ z . Moreover, M (ˆ x ) = M (ˆ a ; ˆ x ) , M (ˆ x ) = M (ˆ b ; ˆ x ) . Carry ˆ z from ˆ a to ˆ b continuously along γ . The argument variation of M (ˆ z ; ˆ x ) is continuousduring this change (since the points ˆ z , ˆ z , ˆ z do not hit the contour σ (cid:48) β ), thus this argumentvariation remains constant. Therefore the phase change of G on σ (cid:48) β is zero.Homotopic deformation of σ (cid:48) β into some σ (cid:48)(cid:48) β may lead to the crossing of a triple zero ora triple pole of G , thus the argument variation of G along σ (cid:48)(cid:48) β is 3 πν for some ν ∈ Z . Theparameter ν depends on the particular choice of σ β .22 similar but slightly more complicated consideration can be applied to computation of theargument variation of G along σ α . Consider the crossing of the contours σ α and σ β . The pairof the coordinate vectors ( e α , e β ), being considered on the local complex coordinate plane τ is oriented as the pair (Im[ τ ] , Re[ τ ]), and this fact is a topological invariant, i. e. it remainsunchanged with any homotopic deformation of σ α and σ β leaving their intersection transversal.One can show that in this case the argument variation of G along σ α is 2 π (3 ν + 1), ν ∈ Z . Theparameter ν depends on the particular choice of σ α .Fix the point ˆ a as η , for simplicity. Then the points ˆ b and ˆ c should have the same affix,still unknown (see the second special case in the previous subsection). Denote correspondingpoint ˆ b by ˆ b and its affix by b . This point plays an important role in what follows. Denote thepoint ˆ c having the same affix b , but located on another sheet of R , by ˆ c .Function M is given by the formula (63): M (ˆ x ) = x − b x − η , . (71)Function M (ˆ x ) is constructed in (64), (65): M (ˆ x ) = Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2 (72)provided that equation (66) is valid for ˆ b = ˆ b and η j,l = η , . We discuss finding of ˆ b in detailsin the next subsection.Thus, when the point ˆ b is found, the function F (˜ x ) can be written as: F (˜ x ) = (cid:32) Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2 (cid:33) − / (cid:18) x − b x − η , (cid:19) / . (73)Note that the function F is defined by (73) ambiguously. This ambiguity follows from thatof the cubic radical, i. e. the result can be multiplied by (cid:36) or (cid:36) − . This ambiguity is partlyaddressed below.Let us list the properties of the function F (˜ x ): • It has Riemann surface R . • It has simple poles at three points of R having affixes η , . There are no other poles(including infinities). • It has simple zeros at three points of R that are P − (ˆ b ). There are no other zeros(including infinities). • As it follows from the argument change of G along σ α , F (Λ(˜ x )) = (cid:36)F (˜ x ) . (74)23et us build the function F . Note thatΥ(ˆ c ) = − Υ(ˆ b ) , ˙Υ(ˆ c ) = − ˙Υ(ˆ b ) , ¨Υ(ˆ c ) = − ¨Υ(ˆ b ) . Construct a function M (ˆ x ) having a double pole at ˆ c and simple zeros at η , and at ˆ b . Thentake F (˜ x ) = ( M (ˆ x ) /M (ˆ x )) / . (75)Similarly to (73), F (˜ x ) = (cid:32) − Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2 (cid:33) − / (cid:18) x − b x − η , (cid:19) / . (76)The properties of F (˜ x ) are as follows: • It has Riemann surface R . • It has simple poles at three points of R having affixes η , . There are no other poles(including the infinities). • It has simple zeros at three points of R that are P − (ˆ c ). The affixes of these points areequal to b . There are no other zeros (including the infinities). • Similarly to F , F (Λ(˜ x )) = (cid:36) − F (˜ x ) . (77)This property of F and the similar property of F guarantee that the elements of thebasis (36) are linearly independent.Besides, there are some properties linking F and F .The symmetry Π converts F into F : F (Π(˜ x )) = δF (˜ x ) , δ ∈ { , (cid:36), (cid:36) − } . (78)To prove this, use (24) and note thatΥ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2 Π −→ − Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2To remove some of the ambiguity of determining F and F , fix the value δ = 1, thus fixing F (Π(˜ x )) = F (˜ x ) . (79)The product of functions F and F is rational. Namely, it belongs to K since F (Λ(˜ x )) F (Λ(˜ x )) = F (˜ x ) F (˜ x ) , K since F (Π(˜ x )) F (Π(˜ x )) = F (˜ x ) F (˜ x ) . One can easily see that F (˜ x ) F (˜ x ) should have a simple pole at x = η , and a simple zero at x = b . Studying the function at infinity, one can find that F (˜ x ) F (˜ x ) = 1(( ¨Υ(ˆ b )) / − / x − b x − η , . (80)Indeed, some ambiguity is still left in the choice of the branch of the cubic root. This ambiguity,however, does not affect the final formulae. Here our aim is to replace the transcendent equation (68) by an algebraic equation. As before,we consider a special case when ˆ a = η , .Let us use the condition (65) of existence of a function M having a double pole at ˆ b andsimple zeros η , and ˆ c , provided that ˆ b and ˆ c have the same affix b :Υ(ˆ b )( η , − b ) + ˙Υ(ˆ b ) η , − b + ¨Υ(ˆ b )2 = 0 . (81)Note that ratios ˙Υ(ˆ b ) / Υ(ˆ b ) and ¨ Y (ˆ b ) /Y (ˆ b ) are rational functions of b . Thus, (81) is analgebraic equation for b . After some algebra, equation (81) becomes reduced to the fourthorder equation: h + h b + h b + h b + h b = 0 , (82)where h = η , + 3 η , − η , η , + η , η , ,h = − η , η , + η , ) ,h = 6( η , + η , + η , η , + η , η , ) ,h = − η , ( η , + 2 η , + η , η , ) ,h = η , − η , + 3 η , η , + η , η , . The algebraic condition (82) is slightly weaker than the transcendent condition (70). Namely,equation (82) is fulfilled if there exist any contours γ , γ , γ cyclically connecting the points η , and two points on R having affix b , such that (cid:90) γ dx Υ(ˆ x ) = (cid:90) γ dx Υ(ˆ x ) = (cid:90) γ dx Υ(ˆ x ) . (83)The concatenation of the contours γ + γ + γ is not necessarily homotopic to σ β , thus, eachof the integrals is not necessarily equal to T β / b , b , b , b , such that b (cid:90) η , dx Υ(ˆ x ) = ± T β µT β + νT α , (84) b (cid:90) η , dx Υ(ˆ x ) = ± T α µT β + νT α , (85) b (cid:90) η , dx Υ(ˆ x ) = ± T α + T β µT β + νT α , (86) b (cid:90) η , dx Υ(ˆ x ) = ± T α − T β µT β + νT α . (87)The integrals are defined up to the sign and up to the integers µ , ν , which depend on theparticular choice of the integration contour. One can see that only b fits the condition (70),i. e. b = b .For practical computations, we propose two following algorithms for computing the value b with a high accuracy. Algorithm 1:
1. Compute T β by numerical integration.2. Find b approximately from the condition b (cid:90) η , dx Υ( x ) = T β . (88)For this, solve numerically the ordinary differential equation dxdχ = Υ( x ) (89)on the segment χ ∈ [0 , T β / x (0) = η , . The value x ( T β /
3) is the approximationfor b . Denote it by b (cid:48) .3. Using b (cid:48) as a starting approximation, solve (82) by Newton’s method. As a result, afterseveral iterations, get a refined value of b .26ince the first two steps are necessary only to obtain the starting approximation for Newton’smethod used on the third step, very coarse meshes can be used for numerical integration andfor solving the ordinary differential equation. The Newton’s method is very cheap, and, severaliterations provide the value of b having the machine accuracy.Another algorithm can be developed, taking the algebraic equation (82) as the startingpoint. The algorithm is as follows. Algorithm 2:
1. Solve equation (82) and find four values: b , b , b , b .2. For each value b j construct the function G (ˆ x ) by (69) and check the variation of Arg[ G ]along the contour σ β . There should exist only one value of b (among the four valuesfound on step 1), for which the variation of Arg[ G ] is equal to zero. This value of b iswhat we are looking for.Indeed, Algorithm 1 and Algorithm 2 should yield the same result. Remark
To solve the ordinary differential equation (89) near the branch point η , one can use thelocal variable on R , namely, τ = τ ( x ) = (cid:112) x − η , . One can rewrite (89) as an ODE for τ ( χ ): dτdχ = 12 (cid:113) ( τ + η , − η , )( τ + η , − η , )( τ + η , − η , ) (cid:18) = Υ2 τ (cid:19) . (90)Thus, one can solve (90) in some small neighborhood of η , , and then solve (89). A ( x ) Here we assume that x in and x − are not equal to b or to η j,l .Let us build the Sommerfeld transformant A (˜ x ) obeying the functional problem formulatedin Subsection 2.1. For this, we use the representation (37) expressing A through the basis Ω .The functions F and F are built above.The coefficients q j (ˆ x ) of the representation (37) belong to K . Thus, q j (ˆ x ) = q (cid:48) j ( x ) + q (cid:48)(cid:48) j ( x )Υ(ˆ x ) , j = 0 , , , (91)where q (cid:48) j ( x ) and q (cid:48)(cid:48) j ( x ) belong to K , i. e. they are rational functions. The aim of this sectionis to find the functions q (cid:48) j ( x ) and q (cid:48)(cid:48) j ( x ).According to the representation (26) with properties (27) and (74), (77), A (˜ x ) = q (cid:48) ( x ) + q (cid:48)(cid:48) ( x )Υ(ˆ x ) , (92)27 (˜ x ) = ( q (cid:48) ( x ) + q (cid:48)(cid:48) ( x )Υ(ˆ x )) F (˜ x ) , (93) A (˜ x ) = ( q (cid:48) ( x ) + q (cid:48)(cid:48) ( x )Υ(ˆ x )) F (˜ x ) . (94)To build the functions q (cid:48) j ( x ) and q (cid:48)(cid:48) j ( x ), one should study poles and zeros of these functions.Let us formulate a series of statements. Let x be not equal to ∞ , η j,l , b , x in , or x − . Then all functions q (cid:48) j ( x ), q (cid:48)(cid:48) j ( x ), j = 0 , , x .The proof is as follows. Consider some particular j . Let ν be the highest pole order of thefunctions q (cid:48) j ( x ), q (cid:48)(cid:48) j ( x ) at x . Note that F , ( x ) (cid:54) = 0 and Υ( x ) (cid:54) = 0. Thus, the pole of order ν will appear on the sheet 1 or 2 (corresponding residues cannot be compensated both). Thiscontradicts to the functional problem for A j The statements 2, 3 and 4 are similar to statement 1, so we omit their proofs. The functions q (cid:48) k ( x ), q (cid:48)(cid:48) k ( x ), k = 0 , , η , , η , , η , . The functions q (cid:48) k ( x ), x q (cid:48)(cid:48) k ( x ), k = 0 , , The functions q (cid:48) k ( x ), q (cid:48)(cid:48) k ( x ) have simple poles at x in and x − .Slightly more subtle consideration is needed for the values x equal to η , and b , sincefunctions F and F have poles and zeros at these affixes. The following statements can beproven: The functions q (cid:48) ( x ), q (cid:48)(cid:48) ( x ), ( x − η , ) − q (cid:48) ( x ), q (cid:48)(cid:48) ( x ), ( x − η , ) − q (cid:48) ( x ), q (cid:48)(cid:48) ( x ) are regular at x = η , . Functions q (cid:48) and q (cid:48)(cid:48) are regular at b . q (cid:48) j ( x ), q (cid:48)(cid:48) j ( x ), j = 1 , b . Thefollowing identities should be valid:lim x → b [ q (cid:48) ( x ) − Υ(ˆ b ) q (cid:48)(cid:48) ( x )] = 0 , lim x → b [ q (cid:48) ( x ) + Υ(ˆ b ) q (cid:48)(cid:48) ( x )] = 0 , Using these statements, one can find the most general form for the functions q (cid:48) j ( x ), q (cid:48)(cid:48) j ( x )obeying statements 1–6: q (cid:48) ( x ) = s x − x in + s x − x − + s , (95) q (cid:48)(cid:48) ( x ) = s x − x in − s x − x − , (96) q (cid:48) ( x ) = ( x − η , ) (cid:32) s x − x in + s x − x − − ( s + s )Υ(ˆ b ) x − b (cid:33) , (97) q (cid:48)(cid:48) ( x ) = ( b − η , ) (cid:18) s x − x in + s x − x − − s + s x − b (cid:19) , (98)28 (cid:48) ( x ) = ( x − η , ) (cid:32) s x − x in + s x − x − + ( s + s )Υ(ˆ b ) x − b (cid:33) , (99) q (cid:48)(cid:48) ( x ) = ( b − η , ) (cid:18) s x − x in + s x − x − − s + s x − b (cid:19) . (100)One can see that these functions contain 12 scalar parameters s , . . . , s . The parameter s can be chosen arbitrarily (we assume further that s = 0). The rest 11 parameters can be foundfrom the known residues of A j at ˜ x , ˜ x , ˜ x , ˜ x .Using these residues given by the formulation of the functional problems for A j (˜ x ), build asystem of equations for s , . . . , s : s = iY / (6 π ) , s = iY / (6 π ) , s = 0 , (101)( x in − η , ) s F (˜ x ) + ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (102)( x in − η , ) s F (˜ x ) − ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (103)( x in − η , ) s F (˜ x ) + ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (104)( x in − η , ) s F (˜ x ) − ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (105)( x − − η , ) s F (˜ x ) + ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (106)( x − − η , ) s F (˜ x ) − ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (107)( x − − η , ) s F (˜ x ) + ( b − η , ) s Y F (˜ x ) = iY / (6 π ) , (108)( x − − η , ) s F (˜ x ) − ( b − η , ) s Y F (˜ x ) = iY / (6 π ) . (109)The equations (102)–(109) can be easily solved. The result can be written using (80) asfollows: s = Z Υ(ˆ x ) 1 x in − b ( F (˜ x ) + F (˜ x )) , (110) s = Z (cid:18) b − η , + 1 x in − b (cid:19) ( F (˜ x ) − F (˜ x )) , (111) s = Z Υ(ˆ x ) 1 x in − b ( F (˜ x ) + F (˜ x )) , (112)29 = Z (cid:18) b − η , + 1 x in − b (cid:19) ( F (˜ x ) − F (˜ x )) , (113) s = Z Υ(ˆ x ) 1 x − − b ( F (˜ x ) + F (˜ x )) , (114) s = Z (cid:18) b − η , + 1 x − − b (cid:19) ( F (˜ x ) − F (˜ x )) , (115) s = Z Υ(ˆ x ) 1 x − − b ( F (˜ x ) + F (˜ x )) , (116) s = Z (cid:18) b − η , + 1 x − − b (cid:19) ( F (˜ x ) − F (˜ x )) , (117)where Z = i (( ¨Υ(ˆ b )) / − / π . (118)Finally, the Sommerfeld transformant is found. The formulae that should be used forcomputations are (26), (92), (93), (94), (101), (110)–(117). u ( m, n ) obeys the diffraction problem? Here we check directly that the Sommerfeld integral (17) defines the wave u ( m, n ) obeying allconditions imposed on it. Consistency of the Sommerfeld integral
One can see that u ( m, n ) is defined in a different way for the domains m ≤ n ≤ and Γ are different (see (18), (19)). Let us show thatthese representations yield the same result in the intersection of the domains, namely, in thequadrant m ≤ n ≤
0. For this, let us show that (cid:90) J w m,n ( x, Ξ(ˆ x )) A (˜ x ) dx Υ(ˆ x ) = (cid:90) J (cid:48) w m,n ( x, Ξ(ˆ x )) A (˜ x ) dx Υ(ˆ x ) = 0 for m ≤ n ≤ . (119)The proof is straightforward. One can see that at the infinities of the sheets 2 and 4 of thescheme shown in Fig. 3 both x and Ξ(ˆ x ) tend to ∞ . Thus, the function w m,n ( x, Ξ(ˆ x )) does notgrow as | x | → ∞ . According to the conditions imposed on A , the integrals are equal to zero. Validity of the discrete Helmholtz equation
Substitute the representation (17) with (18) for m <
0, or with (19) for n <
0. Substitutethis representation into the equation (1). Note that the the discrete Laplace operator acts onlyon w . A direct check shows that (1) is valid. 30 adiation condition Let us demonstrate that u ( m, n ) obeys the radiation condition formulated in the form ofthe limiting absorption principle. For this, deform the contours Γ and Γ homotopically asfollows: Γ = λ + λ + λ , Γ = λ + λ + λ , (120)where contours λ , . . . , λ are shown in Fig. 8. Sheets 5 and 6 are not shown. Contours λ and λ are drawn around corresponding cuts (we remind that the cuts are conducted along the setsof x for which | y (ˆ x ) | = 1). Contours λ and λ are unit circles. Contour λ encircles x in . Notethat | x in | <
1. Fig. 8: Contours of integration λ , . . . , λ One can see that (cid:90) λ w m,n (ˆ x, y (ˆ x )) A (˜ x ) dx Υ(ˆ x ) = u in . (121)As the result, the following representations of the field are obtained: u ( m, n ) = u in ( m, n ) + (cid:90) λ + λ w m,n (ˆ x, y (ˆ x )) A (˜ x ) dx Υ(ˆ x ) for m ≤ , (122) u ( m, n ) = u in ( m, n ) + (cid:90) λ + λ w m,n (ˆ x, y (ˆ x )) A (˜ x ) dx Υ(ˆ x ) for n ≤ . (123)31onsider the exponential factor w m,n = x m y n of the representation (122). For each pointof the representation contours, | y | = 1 and | x | >
1. Since m ≤
0, the result should decay forlarge negative m . Besides, the field should decay for constant negative m and growing positive n due to the oscillatory nature of factor w on the contours λ and λ .Similarly, for the representation (123), | x | = 1 and | y | > n .Thus, we obtain that the total field is a sum of the incident field and a decaying field. Boundary conditions
Let us check the boundary condition u = 0 on the side m ≥ n = 0. For this, use therepresentation (17), (19).On the boundary m ≥ n = 0, the contour of the Sommerfeld integral can be deformedinto two unit circles drawn in sheet 3 and 4 (see Fig. 9). Namely, u ( m,
0) = (cid:90) λ + λ x m A (˜ x ) dx Υ(ˆ x ) . (124)Due to the symmetry (25), this integral is zero. Re[ ] x Im[ ] x I V Sheet 1Sheet 3Sheet 5 Sheet 4
III VV I V V I V II V III Sheet 2Sheet 6 l x in 4 l Fig. 9: Contours of integration λ and λ The situation is slightly more subtle with the boundary m = 0, n ≥
0. As it follows from theconsideration performed in [1], the Sommerfeld integral is introduced invariantly with respectto the choice of the independent variable. For example, one can choose y as an independentvariable, and repeat the whole consideration based on this variable. This would lead to anintegral representation u ( m, n ) = − (cid:90) Γ j w m,n (Ξ(ˆ y ) , y ) A (cid:48) (˜ y ) dy Υ(ˆ y ) . (125)The new Sommerfeld transformant A (cid:48) is linked with the old transformant by the relation A (cid:48) (˜ y ) = A (˜ x (˜ y )) . (126)32ince the whole consideration can be repeated in the variable y , the symmetry argument de-scribed here can be also reproduced. The symmetry Π (cid:48) mentioned above should be used. Theaxes m and n become swapped, so now this argument works for the boundary m = 0, n ≥ In this section we are demonstrating the ideas of the paper using some numerical examples. Inall cases we take real values of K , having in mind the limit Im[ K ] → +0. ˆ b Take K = 0 . b . Use Algorithm 1 for this. First, find the period T β (see(48)). For the numerical integration, use contour σ β shown in Fig. 10, left. The positions ofthe branch point η j,l are shown by stars. −1 0 1 2 3 4 5 6−1−0.8−0.6−0.4−0.200.20.40.60.81 Re[x] I m [ x ] s b I m [ x ] h h h h Fig. 10: Contour for finding T β (left), solution of equation (89), rightFind the correct values of Υ(ˆ x ) on this contour. The contour passes through the point x = 1 on sheet 1, thus, one can fix Υ(1), and then utilize the continuity. One can see thatΥ(1) = ± . i , and one should choose the correct sign. Take K close to 0.5, but havinga small positive imaginary part, say K = 0 . . i . The values of Υ(1) for this K are ± ( − . . i ), and they correspond to the values y = 0 . . y =0 . − . | y | < | y | >
1. Thus, one should chooseΥ(1) = − . . i for K = 0 . . i . By continuity, Υ(1) = 0 . i for K = 0 .
5. Thisreasoning yields that Im[Υ(1)] > < K < T β can be easily computed for K = 0 . T β = − . . i. χ ∈ [0 , T β ], taking x (0) = η , . The result is the trajectory going from η , to η , alongone of the sheets of R and returning back along another sheet. The trajectory ends almostexactly at η , . The projection of this trajectory onto the x -plane is shown in Fig. 10, right.According to Algorithm 1, solve equation (89) for χ ∈ [0 , T β / x (0) = η , . As theresult, get the position of the point ˆ b , i. e. the affix b and the value Υ(ˆ b ). We use this valueas a starting approximation for b and refer to it as b (cid:48) . If the ODE is solved by the simplestEuler’s scheme on a mesh of 100 nodes, b (cid:48) = 0 . . i, Υ(ˆ b (cid:48) ) = − . . i. The position of b (cid:48) is shown in Fig. 10, right, by a circle. The value of Υ(ˆ b (cid:48) ) is needed toconclude that ˆ b (cid:48) belongs to sheet 2 of R .The value of b (cid:48) obtained so far can be considered as a rough approximation for this param-eter. According to Algorithm 1, one can solve (82) by the Newton’s method to refine the value.The process stabilizes after 4 steps, and the result is: b = 0 . . i. One can see that the starting approximation b (cid:48) happens to be quite close to the exact root of(82). This is a clear demonstration of consistency of our approach.Taking K belonging to a dense grid covering the segment [10 − ,
1] and repeating the pro-cedure described above, one can obtain the values ˆ b ( K ). They are presented graphically inFig. 11. The affix b is shown by its real and imaginary part. The value Υ(ˆ b ) is necessary onlyto select a correct sheet of R , so we displayed the imaginary part of it.An important conclusion that can be made from Fig. 11 (and of course one can prove thisanalytically) is that b → K → , and ˆ b is located on sheet 2. G = ( F ) Fix K = 0 . b found in the previous subsection. Let us construct thefunction G (ˆ x ) by the formulae (69), (71), (72), i. e. G (ˆ x ) = (cid:32) Υ(ˆ x )( x − b ) + Υ(ˆ b )( x − b ) + ˙Υ(ˆ b ) x − b + ¨Υ(ˆ b )2 (cid:33) − x − b x − η , . Let us check numerically the validity of the non-trivial condition imposed on G , i. e. that thearguments variation of G on σ β is zero, while the argument variation of G on σ α is equal to 2 π .To check this, we build hodographs of G on σ α and on σ β , i. e. we plot the values of G (ˆ x )for ˆ x running along the contours σ α and σ β . As the result, we get oriented contours in thecomplex plane of G . 34 ¡ ] Fig. 11: The values of Re[ b ], Im[ b ], and Im[Υ(ˆ b )] as functions of K The contour homotopic to σ β has been already built (see Fig. 10, left). The contour homo-topic to σ α and convenient for numerical computations is shown in Fig. 12. The contour passesthe value x = 1 on sheet 1 on the way down.The hodographs of G (ˆ x ) on σ α and σ β are shown in Fig. 13, left and right, respectively. Theorigin is marked by letter O in both graphs. One can see that the hodograph for σ α encirclesthe origin for a single time in the positive direction, and the hodograph for σ β does not encirclethe origin at all. Thus, the conditions for G are valid.Indeed, a similar check can be performed for ( F ) . u ( m, n ) Take the value of K equal to 0 . φ in = π/
4. By symmetry, x in = y in and we arelooking for the solution of the equation ˆ D ( x in , x in ) = 0 corresponding to the wave traveling in35 I m [ x ] h h h h s a Fig. 12: Contour homotopic to σ α −1.5 −1 −0.5 0 0.5 1 1.5 2−1.5−1−0.500.51 Re[G] I m [ G ] O −2 −1.5 −1 −0.5 0 0.5−2−1.5−1−0.500.51 Re[G] I m [ G ] O Fig. 13: Hodographs of G on σ α (left) and on σ β (right)the positive direction with respect to m and n : x in = y in = 4 − K + iK √ − K . . i. We compute the total field by the formulae (122), (123) with the transformant A representedby (26).The components A j , j ∈ { , , } are computed by (92), (93), (94). We took 50000nodes on each contour for integration.The real part of the total field is shown in Fig. 14.The field pattern corresponds to what can be expected. The field is zero at the boundary,and there are visible zones of the reflected waves.In Fig. 15 we plot the scattered field u sc only. The real part is in the left, while the imaginarypart is in the right. One can see cylindrical wave scattered by the angle vertex.36 m −2−1.5−1−0.500.511.52 Fig. 14: The real part of u ( m, n ) −40−30−20−10010203040−40 −30 −20 −10 0 10 20 30 40 n m −1−0.8−0.6−0.4−0.200.20.40.60.81 −40−30−20−10010203040−40 −30 −20 −10 0 10 20 30 40 n m −1−0.8−0.6−0.4−0.200.20.40.60.81 Fig. 15: The scattered wave u sc ( m, n ). The real part (left); the imaginary part (right)37 Conclusion
Let us summarize the process of solving the problem of diffraction by a Dirichlet right angle ona discrete plane. Note that the problem is characterized by two parameters: by the wavenumberparameter K of the Helmholtz equation (1) and by the incident angle φ in defined by (6). Theprocedure is as follows:1. The consideration is based on the structure of the Riemann surface R . This surface isdescribed by four branch points η , , η , , η , , η , . These branch points depend only on K , and they are found from (12), (13).2. One should find the period T β . This can be done by computing the corresponding integralin (48). The contour is shown in Fig. 4, but it is practical to perform the integration alongthe unit circle in the negative direction. Function Υ( x ) is given by the last expression of(14). The branch of Υ( x ) is chosen in such a way that Ξ( x ) defined by (15) has property | Ξ( x ) | < b should be found. This is the point on R , thus it is characterized by theaffix b and the branch of Υ(ˆ b ).Algorithm 1 described in Subsection 4.4 can be used for this. According to this algorithm,first, the differential equation (89) is solved numerically on the segment χ ∈ [0 , T β / b (cid:48) of the parameter b becomes obtained. Besides, the sheet of R onwhich ˆ b is located becomes determined. Second, an algebraic equation (82) is solvediteratively using ˆ b (cid:48) as the starting approximation.4. The functions F (˜ x ) and F (˜ x ) are constructed by (73) and (76). Note that these functionsdepend on K as on a parameter.5. The Sommerfeld transformant of the field A (˜ x ) is built using (26), (92), (93), (94), (101),(110)–(117).6. The function u ( m, n ) is built using the Sommerfeld integral (17).The whole consideration is held in the framework of the Sommerfeld integral. The structureof the integral may seem slightly unusual, however, as we demonstrate in [1], it is a naturalgeneralization of the Sommerfeld integral for angular domains for the discrete case.We first build the functional field K to which the Sommerfeld transfomant of the fieldbelongs. Note that this field is common for all incident angles. The functional field K isrepresented as the basis Ω composed of three functions or by the basis Ω composed of sixfunctions. The construction of the basis is a non-trivial procedure. Second, for a particularangle of incidence φ in we find the Sommerfeld transformant A (˜ x ). This task is tedious, but quitesimple. The coefficients are rational functions, and one should find these functions obeying someknown restrictions and having some known poles. This structure of solution seems to be deeplylinked with the embedding procedure [6, 7]. The elements of the basis are either solutions on38he branched physical surface, having some fixed configurations of the incident fields, or theyare oversingular solutions. This, possibly, gives a new view on the embedding procedure. Authors are grateful to Anastasia Kisil for valuable discussions. The work is supported by theRFBR grant 19-29-06048.
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