ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER. 1. Introduction

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1 t m Mathematical Publications DOI: /tmmp Tatra Mt. Math. Publ , ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER Vladimir B. Vasilyev ABSTRACT. One considers two boundary value problems for the Laplacian and biharmonic operator in a plane sector with boundary conditions of the Dirichlet and the Neumann type on the angle sides. The system of difference equations of first order to which this problem is reduced, is explicitly written out. Certain cases of solvability for such difference equations of first order are described; it will be useful for studying similar equations. 1. Introduction In 90th the author has suggested the wave factorization method for studying solvability of pseudo differential equations in non-smooth domains. It has permitted to obtain a solvability picture for model pseudo differential equations in canonical non-smooth domains of cone type. Besides, for roughly speaking, positive index of wave factorization starting from form of general solution one can describe certain correct statements of boundary value problems for pseudo differential equations and to obtain for them the analogue of algebraic Shapiro- -Lopatinskii condition [1], [4]. But using the reduction to the boundary for a lot of cases we obtain the system of linear difference equations with variable coefficients of arbitrary order n instead of the system of linear algebraic equations [7]. In some simple cases there were the difference equations of first order, which can be solved by the special methods [5], [6]. We will give here some calculations based on the wave factorization related to boundary value problems for the Laplacian, and give the obtained difference equations. In conclusion we give some results devoted to solvability of simplest difference equations of first order. Here we use also the factorization idea. c 013 Mathematical Institute, Slovak Academy of Sciences. 010 M a t h e m a t i c s Subject Classification: 35J05, 39A06. K e y w o r d s: boundary value problem, wave factorization, difference equation. 165

2 VLADIMIR B. VASILYEV The difference equations given in Sections, 3 look very hard, but the author hopes, it is possible their further simplification and studying solvability. Some calculations were done by my postgraduate student M. I. K h o d o t o v a.. Difference equations in the oblique derivative problem for the Laplacian in a plane sector We consider the following problem finding the function u +, which is defined in C a + = { x R : x >a x 1 } for simplicity we take a = 1 and satisfy the Laplace equation Δu + x =0, x C a +, 1 and boundary Neumann condition on one angle side C+ a u n 0 x =x 1,x 1 >0= and boundary condition a u + b u + cu x x 1 g, 3 x = x 1,x 1 <0= on the other one, where n is the normal vector to the straight line x = x 1. Using wave factorization method [4] and changing variables in the problem 1 3 x 1 = x 1+x, x = x 1 x, going to Fourier images, we have the following general form for the solution of 1: Ũ + t 1,t =a 1 t 1,t c t 1 + d t. 4 The formula 4 is a special case of the formula for general solution of elliptic pseudodifferential equation with symbol Aξ, which admits the homogeneous wave factorization in the form see [4] for details Aξ =A ξa = ξ. We use the following notations: Ũ+ t 1,t =ũ + t + t 1 /, t t 1 /, a t 1,t =A t + t 1 /, t t 1 /, c t 1 = c 0 t 1, d t = d 0 t, where the unknown functions c 0, d 0 satisfy the conditions: 166 t 1 Ũ + t 1,t dt 1 =0,

3 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER a t 1 Ũ + t 1,t dt + b t Ũ + t 1,t dt + c Ũ + t 1,t dt = g t 1. 5 Substituting 4 into 5 we obtain the following system of linear integral equations with respect to c, d: t 1 a 1 t 1,t c t 1 dt 1 + d t c t 1 + at 1 + bt + c a 1 t 1,t dt t 1 a 1 t 1,t dt 1 =0, 6 at 1 + bt + c a 1 t 1,t d t dt = g t 1. To simplify this system we need to calculate the following integrals: b t 1 def = a 1 t 1,t dt, b 1 t 1 def = t a 1 t 1,t dt, We remind [4] b t def = t 1 a 1 t 1,t dt 1. a t 1,t = { t t 1 / + t 1 t sgn t t 1, t 1 t < 0, t 1 t, t t 1 / +i t 1 t, t 1 t > 0. One can see the function b t 1 is homogeneous of order zero, and hence it is sufficient to calculate b ±1. Let us find the value b 1. Taking into account the definition for the function we will represent a 1,t=t 1 / t for all t Rtakingsuch branch z of the square root for which its image is in lower half-plane. We have e ixt dt a 1, t = t + t

4 VLADIMIR B. VASILYEV Let us expand the fraction t + t +1 1 into the simplest ones and obtain e ixt dt a 1,t = i e ixt dt. t +1 i / t +1+i / The first integral is vanishing, because the function under integral sign is holomorphic in upper and lower half-planes. The second integral for x<0isvanish- ing in view of residue theorem, but for x>0 i t +1+i / = i πi Res z=i [ e ixz z +1+i / ] = 4πe iπ/4 x, from which b 1 = 4πe iπ/4. The same calculations give b 1 = 4πe iπ/4. The second integral b 1 t 1 is a function homogeneous of order 1, and for its calculation we do the following: first we calculate b 1 ±1, and then b 1 t 1 =b 1 sgnt 1 t 1 sgnt 1. At the end we obtain that b 1 1 = 4πie iπ/4 =4πe iπ/4, b 1 1 = 4πie iπ/4 =4πe iπ/4. The function represented by the integral b t is homogeneous of order 1 also, and for its calculation it is sufficient to find b ±1, and therefore, b t =b sgnt t sgnt. We have b 1 = 4πie iπ/4 =4πe iπ/4, b 1 = 4πie iπ/4 =4πe iπ/4. The system 6 can be rewritten as K t 1,t c t 1 dt 1 + d t =0, 7 a c t 1 + bb 1 t 1 b 1 t 1 +c t 1 1 c t 1 + a + ct 1 1 L t 1,t d t dt + M t 1,t d t dt = g 1 t 1, where we use the following notations: K t 1,t =a 1 t 1,t b 1 t t 1, L t 1,t =a 1 t 1,t b 1 t 1, M t 1,t =a 1 t 1,t b 1 t 1 t t 1 1, g 1 t 1 = g t 1 b 1 t 1 t

5 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER For all t 1 > 0, t > 0wedenote K 11 t 1,t =Kt 1,t, K 1 t 1,t =K t 1,t, K 1 t 1,t =K t 1, t, K t 1,t =Kt 1, t. Analogously, we define the kernels M ij t 1,t, L ij t 1,t, i, j =1,. Let us note all kernels are the functions homogeneous of order 1. Denote c 0 t 1 the restriction c t 1 on0;, c 1 t 1 the restriction c t 1 on 0; ; the same we define the functions d 0 t, d 1 t. Finally, we denote g 10 t 1 the restriction g 1 t 1 on0;, g 11 t 1 the restriction g 1 t 1 on 0;. As a result instead of the system of two linear integral equation 7 with respect to two unknowns c t 1, d t we obtain the system of four linear integral equations with respect to four unknown functions c 0 t 1, c 1 t 1, d 0 t, d 1 t on the positive half-axis, for which their kernels are homogeneous of the order 1. It permits to apply the Mellin transform [3] for studying solvability of this system. We remind the Mellin transform is defined by the formula ˆfs = 0 fxx s 1 dx, s = σ + iτ, at least for functions fx C0 R +. The integral converges for all complex s and it is an entire analytic function. If we change variable x = e t, then the Mellin transform passes into the Fourier transform of function fe t ˆfs = e ts fe t dt, s = σ + iτ. Thus, all properties of the Mellin transform can be obtained from corresponding properties of the Fourier transform. Particularly, the inversion formula of the Mellin transform for fx C0 R has the following form fx = 1 π ˆfst s dτ, s = σ + iτ. The Parseval equality for the Mellin transform t σ 1 ft dt = 1 π ˆfs dτ, s = σ + iτ, 0 169

6 particularly, for σ =1/ wehave or, in other words, 0 VLADIMIR B. VASILYEV ft dt = 1 π ˆfs dτ, s =1/+iτ, 0 ft dt = 1 πi meaning the right integral as lim 1/+i 1/ i 1/+iy y 1/ iy ˆfs ds, ˆfs ds. The last transform reduces the 4 4-system mentioned to a system of linear difference equations with respect to unknowns ĉ 0 λ, ĉ 1 λ, d 0 λ, d 1 λ: K 11 λ ĉ 0 λ+ K 1 λ ĉ 1 λ+ d 0 λ =0, K 1 λ ĉ 0 λ+ K λ ĉ 1 λ+ d 1 λ =0, aĉ 0 λ+ bb 1 1 b c ĉ 0 λ 1 + a L 11 λ+b M 11 λ d0 λ + a L 1 λ+b M 1 λ d1 λ+c L 11 λ d 0 λ 1 + c L 1 λ d 1 λ 1 = ĝ 10 λ, aĉ 1 λ bb 1 1 b c ĉ 1 λ 1 + a L 1 λ+b M 1 λ d0 λ + a L λ+b M λ d1 λ c L 1 λ d 0 λ 1 c L λ d 1 λ 1 = ĝ 0 λ, 8 where denotes the Mellin transform for corresponding functions, K ij λ, M ij λ, L ij λ are the Mellin transforms for the functions K ij 1,t, M ij t, 1, L ij t, 1, i, j = 1,, respectively. So, it is necessary to calculate the Mellin transforms for the functions K ij 1,t, M ij t, 1, L ij t, 1: K 11 1,t=a 1 K 1 1,t= a ,t b 1 1 t =4π 1 e iπ/4 t 1 t 1 / +i 1, t 1,t b 1 1 t = 4π 1 e iπ/4 t 1 t +1/ + 1, t

7 K 1 1,t= a 1 K 1,t= a 1 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER 1, t b 1 1 t = 4π 1 e iπ/4 t 1 1 t / +i 1, t 1, t b 1 1 t = 4π 1 e iπ/4 t 1 t +1/ + t. By analogous calculations we find M ij t, 1, L ij t, 1. Further we have: L 11 t, 1 = 4π 1 e iπ/4 1 t / +i t 1= itk1 1,t, L 1 t, 1 = 4π 1 e iπ/4 t +1/ + t 1 = tk 1,t, L 1 t, 1 = 4π 1 e iπ/4 t 1 / +i t 1 = itk11 1,t, L 1,t=4π 1 e iπ/4 t +1/ + t 1= tk1 1, t; M 11 t, 1 = t 1 L 11 t, 1, M 1 t, 1 = t 1 L 1 t, 1, M 1 t, 1 = t 1 L 1 t, 1, M t, 1 = t 1 L t, 1. From computational point of view it is more convenient to find the Mellin transform of the functions L ij t, 1, i, j =1,, and then, taking into account property of the Mellin transform, t 1 f t = f λ 1, and then to obtain the Mellin transform of the functions M ij t, 1, K ij 1,t. We have: t λ 1 t 1 / +i t = y λ 1 y + iy Let us note, the following representation is valid y + 1 i +1 iy 1 = Hence, we obtain 0 t λ 1 dt a 1,t = i i 1 y 1 i 1 1+ i 1 π λ cos ec πλ / i +1 π λ cos ec πλ / i 1 1 i +1 y. = π cos ec πλ e iπ/4λ 1 e i3π/4λ 1, 171

8 which implies VLADIMIR B. VASILYEV L 1 λ = 1/ e iπ/4 cos ec πλ e iπ/4λ 1 e i3π/4λ 1, and then K 11 = i L 1 λ 1, K 11 λ =i/ e iπ/4 cos ec π λ 1 e iπ/4λ 3 e i3π/4λ 3 =1/ e iπ/4 cos ec πλ e iπ/4λ 1 e i3π/4λ 1. M 1 t, 1 = L 1 λ 1, hence, M 1 λ = 1/ e 3iπ/4 cos ec πλ e iπ/4λ 1 e i3π/4λ 1. Analogously, L λ = e iπ/4 cos ec πλsin π/4 λ 1, K 1 λ = M λ = L λ 1 = e iπ/4 cos ec πλcos π/4 λ 1, L 1 λ = e iπ/4 cos ec πλsin π/4 λ 1, K λ = M 1 λ = L 1 λ 1 = e iπ/4 cos ec πλcos π/4 λ 1, L 11 λ =1/ e iπ/4 cos ec πλ e i3π/4λ 1 e iπ/4λ 1, K 1 λ =i L 11 λ 1 =1/ e iπ/4 cos ec πλ e i3π/4λ 1 + e iπ/4λ 1, M 1 λ = L 11 λ 1 = 1/ e iπ/4 cos ec πλ e i3π/4λ 1 + e iπ/4λ 1. It is not hard work to reduce the system 8 by elementary transformations to the system of two difference equations with two unknowns ĉ 0 λ, ĉ 1 λ. So, the solving the problem 1 with boundary conditions, 3 reduces to solving the following system: { Â λ ĉ0 λ + B λ ĉ 1 λ + P λ ĉ 0 λ 1 + R λ ĉ 1 λ 1 = ĝ 10 λ, Â 1 λ ĉ 0 λ+ B 1 λ ĉ 1 λ+ P 1 λ ĉ 0 λ 1 + R 1 λ ĉ 1 λ 1 = ĝ 0 λ, 17

9 where ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER Â λ =a 1 K 11 λ L 11 λ K 1 λ L 1 λ b K11 λ M 11 λ+ K 1 λ M 1 λ, B λ = a K1 λ L 11 λ+ K λ L 1 λ b K1 λ M 11 λ+ K λ M 1 λ, P λ =bb 1 1 b c 1 K 11 λ L 11 λ K 1 λ L 1 λ, R λ = c K1 λ L 11 λ+ K λ L 1 λ, Â 1 λ =a K1 λ L λ K 11 λ L 1 λ + b K1 λ M λ K 11 λ M 1 λ, B 1 λ =a 1 K 1 λ L 1 λ K λ L λ b K1 λ M 1 λ+ K λ M λ, P 1 λ =c K11 λ L 1 λ+ K 1 λ L λ, R 1 λ =c K λ L λ+ K 1 λ L 1 λ 1 bb 1 1 b Difference equations in the oblique derivative problem for the biharmonic equation in a plane sector Here we consider the following problem: finding the function u +,whichis defined in C 1 + = { x R : x > x 1 } and satisfy the biharmonic equation Δ u + x =0, x C+, 1 9 and boundary conditions on angle sides u + n = g 1, x =x 1,x 1 >0 10 u + x =x 1,x 1 >0 = g on one angle side, and u + n = g 3, x = x 1,x 1 <0 11 u + x = x 1,x 1 <0 = g 4, 173

10 VLADIMIR B. VASILYEV on the other one, where n is unit normal vector for the straight line x = x 1. Using the wave factorization we have the formula for general solution of the equation 9 and changing variables x 1 = x 1 + x, x = x 1 x in the problem 9 11 and going to Fourier images, we have the equation Ũ + t 1,t =a t 1,t c 0 t 1 + c 1 t 1 t + d 0 t + d 1 t t 1 1 and four conditions for determining the unknown functions c 0, d 0, c 1, d 1 : t 1 Ũ + t 1,t dt 1 = g 1 t, Ũ + t 1,t dt 1 = g t, t Ũ + t 1,t dt = g 3 t 1, Ũ + t 1,t dt = g 4 t 1, 13 for which we use the following notations: Ũ + t 1,t =ũ + t + t 1 /, t t 1 /, a t 1,t =A t + t 1 /, t t 1 /, c t 1 = c 0 t 1, d t = d 0 t. Substituting 1 into 13 we obtain the following system of linear integral equations with respect to c, d : 174 t 1 a t 1,t c 0 t 1 dt 1 + t + d 0 t + d 1 t a t 1 a t 1,t dt 1 t 1a t 1,t dt 1 = g 1 t, t 1,t c 0 t 1 dt 1 + t + d 0 t a t 1,t dt 1 + d 1 t t 1 a t 1,t dt 1 = g t, t 1 a t 1,t c 1 t 1 dt 1 a t 1,t c 1 t 1 dt 1

11 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER c 0 t 1 + +t 1 c 0 t 1 t a t 1,t dt + c 1 t 1 t a t 1,t d 0 t dt a + +t 1 t a t 1,t d 1 t dt = g 3 t 1, t 1,t dt + c 1 t 1 a t 1,t d 0 t dt a t 1,t d 1 t dt = g 4 t 1. As in the previous case we write the expression t a t 1,t dt t a t 1,t dt { t t a t 1,t = 1 / + t 1 t sgn t t 1, t1 t < 0, t 1 t, t t 1 / +i t 1 t, t 1 t > 0. For simplifying the system above and taking into account the concrete form of the symbol a t 1,t, we will calculate the following integrals: b t 1 def = a t 1,t dt, b 1 t 1 def = b t 1 def = t a t 1,t dt, b 3 t def = b 4 t def = t 1 a t 1,t dt 1, b 5 t def = t a t 1,t dt, a t 1,t dt 1, t 1a t 1,t dt 1. These integrals can be calculated by the same way as above for the Laplacian. Let us calculate bt 1 = a t 1,t dt. It is sufficient to calculate b ±1. Let us find the value b 1. We have e ixt dt a 1,t = t + t

12 VLADIMIR B. VASILYEV If we expand the fraction t + t +1 into the simplest ones, we will obtain: e ixt dt a 1,t = t +1 i/ + i i from which it implies that i t +1 i/ t +1+i/ t +1+i / = 4 πe iπ 4 x, t +1+i / =πi Res z=i b 1 = πe iπ/4 By the similar way we find that 176 b 1 = πe iπ/4 Let us calculate the second integral b 1 t 1 = [ t +1+i/, ] e ixz z +1+i / πie iπ 4 x = e iπ 4 + +, e iπ b 1 1 = πe iπ/4 i + b 1 1 = πe iπ/4 i + b t 1 = i. e iπ/4 +1+e iπ/ i. e iπ/4 +1+e iπ/ t a t 1,t dt : 1, e iπ/4 +1+e iπ/ 1. e iπ/4 +1+e iπ/ t a t 1,t dt :

13 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER b 1 = πe iπ/4 + b 1 = πe iπ/4 i, e iπ/4 +1+e iπ/ i. e iπ/4 +1+e iπ/ b 3 t = a t 1,t dt 1 : b 3 1 = πe i iπ/4 1 +, e iπ/4 +1 e iπ/ b 3 1 = πe i iπ/ e iπ/4 +1 e iπ/ b 4 t = t 1 a t 1,t dt 1 : b 4 1 = πe + iπ/4 i, e iπ/4 +1 e iπ/ b 4 1 = πe + iπ/4 i. e iπ/4 +1 e iπ/ b 5 t def = t 1 a t 1,t dt 1 : b 5 1 = πe i iπ/4 1, e iπ/4 +1 e iπ/ b 5 1 = πe i iπ/4 1. e iπ/4 +1 e iπ/ The previous system takes the form: t 1 a t 1,t c 0 t 1 dt 1 + t t 1 a t 1,t c 1 t 1 dt 1 + d 0 t b 4 t + d 1 t b 1 5 t = g 1 t, b 1 3 t a t 1,t c 0 t 1 dt 1 + t + d 0 t + d 1 t b 1 3 t b 4 t = g t, b 1 3 t a t 1,t c 1 t 1 dt 1 177

14 VLADIMIR B. VASILYEV c 0 t 1 b 1 t 1 + c 1 t 1 b t t 1 t a t 1,t d 1 t dt = g 3 t 1, c 0 t 1 + c 1 t 1 b 1 t 1 b t t 1 t a t 1,t d 0 t dt b 1 t 1 a t 1,t d 0 t dt b 1 t 1 a t 1,t d 1 t dt = g 4 t 1, and can be reduced as above to a system of 8 8-system of linear integral equations, and further by Mellin transform to the 8 8-system of linear difference equations. 4. One solvability case for a system of linear difference equations of first order Let us consider a system of two linear difference equations of first order with two unknowns c 1 λ, c λ: a 11 λ c 1 λ+a 1 λ c λ b 11 λ c 1 λ 1 b 1 λ 1 c λ =d 1 λ, 14 a 1 λ c 1 λ+a λ c λ b 1 λ c 1 λ 1 b λ c λ 1 = d λ, where d 1 λ, d λ are given functions. Our main goal is to find the solution of the system 14. For this purpose we rewrite the system in a matrix form: A λ C λ B λ C λ 1 = D λ, 15 where a11 λ a 1 λ A λ = a 1 λ a λ are matrices of order, and c1 λ C λ = c λ b11 λ b 1 λ, Bλ = b 1 λ b λ d1 λ, Dλ = d λ are vectors. Multiply the quantity 15 by A 1 λ from left. We obtain A 1 λ A λ C λ A 1 λ B λ C λ 1 = A 1 λ D λ 178

15 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER or C λ G λ C λ 1 = R λ, where G λ =A 1 λ B λ is matrix of order, R λ =A 1 λ D λ is vector. If we suppose that G λ admits factorization G λ =P1 1 λ P 1 λ 1, 16 then substituting 16 into last matrix equation we have: C λ P1 1 λ P 1 λ 1 C λ 1 = R λ or P 1 λ C λ P 1 λ 1 C λ 1 = P 1 λ R λ. It is easily seen the last expression is a system of linear difference equation of first order, and its solution will be in the following form see [] for details and notations P 1 λ C λ = S λ P 1 z R zδz + ϖ, c where ϖ is periodic function with period 1. Hence, C λ =P 1 1 λ λ S c P 1 z R zδz + ϖ = P 1 1 λ λ S c P 1 z A 1 z D zδz + ϖ. So, the problem is reduced to finding P 1 λ. Let us consider the equation 16, which we rewrite in the form: p 11 λ g 11 λ+p 1 λ g 1 λ =p 11 λ 1, p 11 λ g 1 λ+p 1 λ g λ =p 1 λ 1, p 1 λ g 11 λ+p λ g 1 λ =p 1 λ 1, p 1 λ g 1 λ+p λ g λ =p λ 1, where p 11 λ, p 1 λ, p 1 λ, p λ are the unknown functions. Such system decomposes into the two systems of linear difference equations of first order with variable coefficients g 11 λ, g 1 λ, g 1 λ, g λ: { p11 λ g 11 λ+p 1 λ g 1 λ =p 11 λ 1, 17 p 11 λ g 1 λ+p 1 λ g λ =p 1 λ 1, { p1 λ g 11 λ+p λ g 1 λ =p 1 λ 1, p 1 λ g 1 λ+p λ g λ =p λ We will seek the solution of the system 17 in the form p 11 λ =α 1 e rλ, p 1 λ =α e rλ. 179

16 VLADIMIR B. VASILYEV Then our system can be rewritten in the form: { g11 λ α 1 e rλ + g 1 λ α e rλ = α 1 e rλ 1, g 1 λ α 1 e rλ + g λ α e rλ = α e rλ 1, or, in the other terms, { g11 λ e r α 1 + g 1 λ α =0, g 1 λ α 1 + g λ e r α =0. The last system has a non-trivial solution in such case only, if its determinant is equal to zero. Let us compose the determinant and find the non-trivial solutions of the system. g 11 λ e r g 1 λ g 1 λ g λ e r =detg e r tr G + e r =0, 19 tr G is trace of the matrix G. Changing variables δ = e r, a = tr G, b =detg, we obtain quadratic equation δ aδ + b = 0, which has always two solutions δ 1 and δ, and, hence, we have two values r, i.e., r 1 and r. Take α 1 = 1. Then the solutions of the systems 17 and 18 can be p 11 λ =e r1λ, p 1 λ =ae r1λ, p 1 λ =e rλ, p λ =be rλ, where r 1 and r are solutions of our quadratic equation. Theorem 1. If the elements of the matrix G such that under certain r the determinant 19 is equal zero, then G λ admits the factorization. The method for solving -system of linear difference equations of first order mentioned above is valid for a system of three linear difference equations of first order with three unknowns. Theorem. If the elements of the matrix G, whereg is corresponding matrix of order 3 3 such that under certain r the following equation det G e r tr G e 3r e r g 11 λg 33 λ+g λg 33 λ+g 11 λg λ g 31 λg 13 λ g 3 λg 3 λ g 1 λg 1 λ = 0 0 is valid, then G λ admits the factorization. Concerning to n n-system of linear difference equations of first order, reasoning by the same way one can obtain an algebraic equation of order n of type 0 and formulate the following result. 180

17 ON SOME DIFFERENCE EQUATIONS OF FIRST ORDER Theorem 3. If the factorization for the matrix G exists, then it is unique up to periodic function. The proof of Theorem 3 is easily obtained by taking logarithm of both sides of the equation 16 and reducing to the problem of type Rλ Rλ 1 = Qλ with given right-hand side Qλ. Such problem is solved by the formula [3] Rλ = λ S c QzΔz + ϖ, where ϖ is periodic function with the period 1. REFERENCES [1] VASILYEV, V. B.: Elliptic equations and boundary value problems in non-smooth domains, in: Oper. Theory Adv. Appl., Vol. 13, Birkhäuser, Basel, 011, pp [] MILNE-THOMSON, L. M.: The Calculus of Finite Differences. Chelsea Publ. Comp., New York, [3] TITCHMARSH, E.: Introduction to the Theory of Fourier Integrals. Chelsea Publ. Comp., New York, [4] VASIL EV, V. B.: Wave Factorization of Elliptic Symbols: Theory and Applications. Introduction to the theory of boundary value problems in non-smooth domains. Kluwer Acad. Publ., Dordrecht, 000. [5] SOLONNIKOV, V. A. FROLOVA, E. V.: On the third value problem for the Laplace equation in a plane sector and application to parabolic problems, AlgebraAnal. 1990, [6] BAZALIY, B. V. VASYLYEVA, N.: On the solvability of a transmission problem for the Laplace operator with a dynamic boundary condition on a nonregular interface, J. Math. Anal. Appl , [7] VASILYEV, V. B.: Discrete convolutions and difference equations, in: Proc. of the 5th Internat. Conf., Morehouse College, Atlanta, GA, USA, 007 G. S. Ladde et al., eds., Dynam. Systems Appl., Vol. 5, Dynamic Publishers, Atlanta, USA, 008, pp Received October 8, 01 Chair of Pure Mathematics Lipetsk State Technical University Moskovskaya 30 Lipetsk RUSSIA vbv57@inbox.ru 181