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International Journal of Mathematics and Mathematical Sciences
Volume 2010 (2010), Article ID 714534, 38 pages
Research Article

Benjamin-Ono Equation on a Half-Line

1Department of Mathematics, Graduate School of Science, Osaka University, Osaka, Toyonaka 560-0043, Japan
2Instituto de Matemáticas, UNAM, Campus Morelia, AP 61-3 (Xangari), Morelia, CP 58089 Michoacán, Mexico

Received 18 September 2009; Accepted 28 January 2010

Academic Editor: Michael Tom

Copyright © 2010 Nakao Hayashi and Elena I. Kaikina. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We consider the initial-boundary value problem for Benjamin-Ono equation on a half-line. We study traditionally important problems of the theory of nonlinear partial differential equations, such as global in time existence of solutions to the initial-boundary value problem and the asymptotic behavior of solutions for large time.

1. Introduction

In this paper we study the large time asymptotic behavior of solutions to the initial-boundary value problem for the Benjamin-Ono equation on a half-line:

𝑢𝑡+𝑢𝑢𝑥+𝑢𝑥𝑥𝑢=0,𝑥>0,𝑡>0,(𝑥,0)=𝑢0(𝑥),𝑥>0,𝑢(0,𝑡)=0,𝑡>0,(1.1) where 𝑢=PV0+𝑢((𝑦,𝑡)/(𝑦𝑥))𝑑𝑦 is the Hilbert transformation, and PV means the principal value of the singular integral. We note that in the case of the whole line we have the relations 𝜕2𝑥=𝜕𝑥(𝜕2𝑥)1/2 since the operator can be written as follows: =1(𝑖𝜉/|𝜉|)=(𝜕2𝑥)1/2𝜕𝑥, where (𝜑)(𝜉)=(1/2𝜋)𝜑(𝑥)e𝑖𝑥𝜉𝑑𝑥 is the usual Fourier transform, and 1 denotes the inverse Fourier transform. This equation is of great interest in many areas of Physics (see [1, 2]). The Cauchy problem (1.1) was studied by many authors. The existence of solutions in the usual Sobolev spaces 𝐇𝑠,0 was proved in [39] and the smoothing properties of solutions were studied in [1014]. In paper [15] it was proved that for small initial data in 𝐇2,0𝐇1,1 solutions decay as 𝑡 in 𝐋 norm at the same rate 1/𝑡 as for the case of the linear Benjamin-Ono equation, where


The initial-boundary value problem (1.1) plays an important role in the contemporary mathematical physics. For the general theory of nonlinear equations on a half-line we refer to the book [16], where it was developed systematically a general theory of the initial-boundary value problems for nonlinear evolution equations with pseudodifferential operators on a half-line, where pseudodifferential operator 𝕂 on a half-line was introduced by virtue of the inverse Laplace transformation of the product of the symbol 𝐾(𝑝)=𝑂(𝑝𝛽) which is analytic in the right complex half-plane, and the Laplace transform of the derivative 𝜕𝑥[𝛽]𝑢. Thus, for example, in the case of 𝐾(𝑝)=𝑝3/2 we get the following definition of the fractional derivative 𝜕𝑥3/2:

𝜕𝑥3/2𝜙=1𝑝3/2𝜙𝜙(0)𝑝.(1.3) Here and below 𝑝𝛽 is the main branch of the complex analytic function in the complex half-plane Re𝑝0, so that 1𝛽=1 (we make a cut along the negative real axis (,0)). Note that due to the analyticity of 𝑝𝛽 for all Re𝑝>0 the inverse Laplace transform gives us the function which is equal to 0 for all 𝑥<0. In spite of the importance and actuality there are few results about the initial-boundary value problem for pseudodifferential equations with nonanalytic symbols. For example, in paper [17] there was considered the case of rational symbol 𝐾(𝑝) which have some poles in the right complex half-plane. There was proposed a new method for constructing the Green operator based on the introduction of some necessary condition at the singularity points of the symbol 𝐾(𝑝). In the paper [18] one of the authors considered the initial-boundary value problem for a pseudodifferential equation with symbol 𝐾(𝑝)=|𝑝|1/2 and nonlinearity |𝑢|𝜎𝑢.

As far as we know the case of nonanalytic conservative symbols 𝐾(𝑝) was not studied previously. In the present paper we fill this gap, considering as example the Benjamin-Ono equation (1.1) with a symbol 𝐾(𝑝)=𝑝|𝑝|. There are many natural open questions which we need to study. First we consider the following question: how many boundary data we should pose on problem (1.1) for its correct solvability? Also we study traditionally important problems of a theory of nonlinear partial differential equations, such as global in time existence of solutions to the initial-boundary value problem and the asymptotic behavior of solutions for large time. We adopt here the approach of book [16] based on the estimates of the Green function. The main difficulty for nonlocal equation (1.1) on a half-line is that the symbol 𝐾(𝑝)=𝑝|𝑝| is non analytic in the complex plane. Therefore we cannot apply the Laplace theory directly. To construct Green operator we proposed a new method based on the integral representation for sectionally analytic function and theory of singular integrodifferential equations with Hilbert kernel and the discontinues coefficients (see [18, 19]).

To state precisely the results of the present paper we give some notations. We denote 𝑡=1+𝑡2,{𝑡}=𝑡/𝑡. Direct Laplace transformation 𝑥𝜉 is

̂𝑢(𝜉)𝑥𝜉𝑢=0+𝑒𝜉𝑥𝑢(𝑥)𝑑𝑥,(1.4) and the inverse Laplace transformation 1𝜉𝑥 is defined by

𝑢(𝑥)1𝜉𝑥̂𝑢=(2𝜋𝑖)1𝑖𝑖𝑒𝜉𝑥̂𝑢(𝜉)𝑑𝜉.(1.5) Weighted Lebesgue space is 𝐋𝑞,𝑎(𝐑+)={𝜑𝒮;𝜑𝐋𝑞,𝑎<}, where

𝜑𝐋𝑞,𝑎=0+𝑥𝑎𝑞||||𝜑(𝑥)𝑞𝑑𝑥1/𝑞(1.6) for 𝑎>0, 1𝑞< and

𝜑𝐋=esssup𝑥𝐑+||||.𝜑(𝑥)(1.7) Sobolev space is

𝐇1𝐑+=𝜑𝒮;𝜕𝑥𝜑𝐋2.<(1.8) We define a linear functional 𝑓:

𝑓(𝜙)=0+𝑦𝜙(𝑦)𝑑𝑦.(1.9) Now we state the main results.

Theorem 1.1. Suppose that the initial data 𝑢0𝐙𝐇1(𝐑+)𝐋1,𝑎+1(𝐑+) with 𝑎(0,1) are such that the norm 𝑢0𝐙𝜀(1.10) is sufficiently small. Then there exists a unique global solution [𝑢𝐂0,);𝐇1𝐑+(1.11) to the initial-boundary value problem (1.1). Moreover the following asymptotic is valid in 𝐋(𝐑+)1𝑢=𝑡𝐴Λ𝑥𝑡1/2𝑥+min1,𝑡𝑂𝑡1(𝑎/2)(1.12) for 𝑡, where Λ(𝑥𝑡1/2)𝐋(𝐑+),Λ(0)=0 is defined below by the formula (2.191), and the constant 𝑢𝐴=𝑓00+𝑓(𝒩(𝑢))𝑑𝜏,𝒩(𝑢)=𝑢𝑥𝑢.(1.13)

Remark 1.2. Note that the time decay rate of the solution is faster comparing with the case of the corresponding Cauchy problem. So the nonlinearity 𝑢𝑢𝑥 in (1.1) is not the super critical case for our problem.

Remark 1.3. In the case of the negative half line 𝑥<0 we expect that the solutions have an oscillation character, and the time decay rate of the solution is the same as the case of the corresponding Cauchy problem. so the nonlinearity 𝑢𝑢𝑥 in (1.1) will be the super critical case.

2. Preliminaries

In subsequent consideration we will have frequently to use certain theorems of the theory of functions of complex variable, the statements of which we now quote. The proofs can be found in [19].

Theorem 2.1. Let 𝜙(𝑞) be a complex function, which obeys the Hölder condition for all finite 𝑞 and tends to a definite limit 𝜙 as |𝑞|, such that for large 𝑞 the following inequality holds: ||𝜙(𝑞)𝜙||||𝑞||𝐶𝜇,𝜇>0.(2.1) Then Cauchy type integral 1𝐹(𝑧)=2𝜋𝑖𝑖𝑖𝜙(𝑞)𝑞𝑧𝑑𝑞(2.2) constitutes a function analytic in the left and right semiplanes. Here and below these functions will be denoted 𝐹+(𝑧) and 𝐹(𝑧), respectively. These functions have the limiting values 𝐹+(𝑝) and 𝐹(𝑝) at all points of imaginary axis Re𝑝=0, on approaching the contour from the left and from the right, respectively. These limiting values are expressed by Sokhotzki-Plemelj formula: 𝐹+(𝑝)=lim𝑧𝑝,Re𝑧<012𝜋𝑖𝑖𝑖𝜙(𝑞)1𝑞𝑧𝑑𝑞=2𝜋𝑖𝑃𝑉𝑖𝑖𝜙(𝑞)1𝑞𝑝𝑑𝑞+2𝐹𝜙(𝑝),(𝑝)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖𝜙(𝑞)1𝑞𝑧𝑑𝑞=2𝜋𝑖𝑃𝑉𝑖𝑖𝜙(𝑞)1𝑞𝑝𝑑𝑞2𝜙(𝑝).(2.3)

Subtracting and adding the formula (2.3) we obtain the following two equivalent formulas:

𝐹+(𝑝)𝐹𝐹(𝑝)=𝜙(𝑝),+(𝑝)+𝐹1(𝑝)=𝜋𝑖𝑃𝑉𝑖𝑖𝜙(𝑞)𝑞𝑝𝑑𝑞,(2.4) which will be frequently employed hereafter.

Theorem 2.2. An arbitrary function 𝜙(𝑝) given on the contour Re𝑝=0, satisfying the Hölder condition, can be uniquely represented in the form 𝜙(𝑝)=𝑈+(𝑝)𝑈(𝑝),(2.5) where 𝑈±(𝑝) are the boundary values of the analytic functions 𝑈±(𝑧) and the condition 𝑈±=0 holds. These functions are determined by formula 1𝑈(𝑧)=2𝜋𝑖𝑖𝑖𝜙(𝑞)𝑞𝑧𝑑𝑞.(2.6)

Theorem 2.3. An arbitrary function 𝜑(𝑝) given on the contour Re𝑝=0, satisfying the Hölder condition, and having zero index, 1ind𝜑(𝑡)=2𝜋𝑖𝑖𝑖𝑑ln𝜑(𝑝)=0,(2.7) is uniquely representable as the ratio of the functions 𝑋+(𝑝) and 𝑋(𝑝), constituting the boundary values of functions, 𝑋+(𝑧) and 𝑋(𝑧), analytic in the left and right complex semiplane and having in these domains no zero. These functions are determined to within an arbitrary constant factor and given by formula 𝑋±(𝑧)=𝑒Γ±(𝑧)1,Γ(𝑧)=2𝜋𝑖𝑖𝑖1𝑞𝑧ln𝜑(𝑞)𝑑𝑞.(2.8)

We consider the following linear initial-boundary value problem on half-line 𝑢𝑡𝑃𝑉0+𝑢𝑦𝑦(𝑦,𝑡)𝑥𝑦𝑑𝑦=0,𝑡>0,𝑥>0,𝑢(𝑥,0)=𝑢0(𝑥),𝑥>0,𝑢(0,𝑡)=0,𝑡>0.(2.9)



where Re𝑘(𝜉)>0 for Re𝜉>0, we define


where the function 𝐺(𝑥,𝑦,𝑡) is given by formula

1𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑑𝑝𝑒𝑝𝑥1𝐾1(𝑝)+𝜉(𝑒𝑝𝑦1+Ψ(𝜉,𝑦))12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑑𝑝𝑒𝑝𝑥1𝐾1𝑌(𝑝)+𝜉(𝑝,𝜉)×lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+(𝐾𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉(𝑒𝑞𝑦+Ψ(𝜉,𝑦))𝑑𝑞(2.12) for 𝜀>0,𝑥>0,𝑦>0,𝑡>0. Here and below

𝑌±=𝑒Γ±𝑤±.(2.13)Γ+(𝑝,𝜉) and Γ(𝑝,𝜉) are a left and right limiting values of sectionally analytic function Γ(𝑧,𝜉) given by

1Γ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝑞𝑧ln𝐾(𝑞)+𝜉𝐾1𝑤(𝑞)+𝜉(𝑞)𝑤+(𝑞)𝑑𝑞,(2.14) where


All the integrals are understood in the sense of the principal values.

Proposition 2.4. Let the initial data be 𝑢0𝐋1(𝐑+). Then there exists a unique solution 𝑢(𝑥,𝑡) of the initial-boundary value problem (2.9), which has integral representation 𝑢(𝑥,𝑡)=𝒢(𝑡)𝑢0.(2.16)

Proof. To derive an integral representation for the solutions of the problem (2.9) we suppose that there exists a solution 𝑢(𝑥,𝑡) of problem (2.9), which is continued by zero outside of 𝑥>0:
Let 𝜙(𝑝) be a function of the complex variable 𝑝, which obeys the Hölder condition for all finite 𝑝 and tends to 0 as 𝑝±𝑖. We define the operator 1𝜙(𝑧)=2𝜋𝑖𝑖𝑖1𝑞𝑧𝜙(𝑞)𝑑𝑞.(2.18)
Since the operator is defined by a Cauchy type integral, it is readily observed that 𝜙(𝑧) constitutes a function analytic in the entire complex plane, except for points of the contour of integration Re𝑧=0. Also by Sokhotzki-Plemelj formula we have for Re𝑝=0+1𝜙=2𝜋𝑖𝑃𝑉𝑖𝑖11𝑞𝑝𝜙(𝑞)𝑑𝑞+2𝜙(𝑝),1𝜙=2𝜋𝑖𝑃𝑉𝑖𝑖11𝑞𝑝𝜙(𝑞)𝑑𝑞2𝜙(𝑝).(2.19) Here +𝜙 and 𝜙 are limits of 𝜙 as 𝑧 tends to 𝑝 from the left and right semi-plane, respectively.
We have for the Laplace transform 𝑢𝑥𝑥||𝑝||𝑝={𝑢}𝑢(0,𝑡)𝑝𝑢𝑥(0,𝑡)𝑝2.(2.20) Since {𝑢} is analytic for all Re𝑞>0, we have ̂𝑢(𝑞,𝑡)={𝑢}=̂𝑢(𝑝,𝑡).(2.21) Therefore applying the Laplace transform with respect to 𝑥 to problem (2.9) we obtain for 𝑡>0̂𝑢𝑡+𝐾(𝑝)̂𝑢(𝑝,𝑡)𝐾(𝑝)𝑝𝑢(0,𝑡)𝐾(𝑝)𝑝2𝑢𝑥(0,𝑡)=0,̂𝑢(𝑝,0)=̂𝑢0(𝑝),(2.22) where ||𝑝||𝐾(𝑝)=𝑝.(2.23) We rewrite (2.22) in the form ̂𝑢𝑡+𝐾(𝑝)̂𝑢(𝑝,𝑡)𝐾(𝑝)𝑝𝑢(0,𝑡)𝐾(𝑝)𝑝2𝑢𝑥(0,𝑡)=Φ(𝑝,𝑡),̂𝑢(𝑝,0)=̂𝑢0(𝑝),(2.24) with some function Φ(𝑝,𝑡) such that for all Re𝑝>0{Φ(𝑝,𝑡)}=0(2.25) and for |𝑝|>1||||1Φ(𝑝,𝑡)𝐶||𝑝||.(2.26) Applying the Laplace transformation with respect to time variable to problem (2.24) we find for Re𝑝>0̂1̂𝑢(𝑝,𝜉)=𝐾(𝑝)+𝜉̂𝑢0(𝑝)+𝐾(𝑝)𝑝̂𝑢(0,𝜉)+𝐾(𝑝)𝑝2̂𝑢𝑥(0,𝜉)+Φ(𝑝,𝜉).(2.27) Here the functions ̂̂𝑢(𝑝,𝜉),Φ(𝑝,𝜉),̂𝑢(0,𝜉), and ̂𝑢𝑥(0,𝜉) are the Laplace transforms for ̂𝑢(𝑝,𝑡),Φ(𝑝,𝑡),𝑢(0,𝑡), and 𝑢𝑥(0,𝑡) with respect to time, respectively. We will find the function Φ(𝑝,𝜉) using the analytic properties of function ̂̂𝑢 in the right-half complex planes Re𝑝>0 and Re𝜉>0. We have for Re𝑝=0̂1̂𝑢(𝑝,𝜉)=𝜋𝑖𝑃𝑉𝑖𝑖1̂𝑞𝑝̂𝑢(𝑞,𝜉)𝑑𝑞.(2.28) In view of Sokhotzki-Plemelj formula via (2.27) the condition (2.28) can be written as Θ+(𝑝,𝜉)=Λ+(𝑝,𝜉),(2.29) where the sectionally analytic functions Θ(𝑧,𝜉) and Λ(𝑧,𝜉) are given by Cauchy type integrals: 1Θ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑧1𝐾(𝑞)+𝜉Φ(𝑞,𝜉)𝑑𝑞,(2.30)Λ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝐾(𝑞)+𝜉̂𝑢0(𝑞)+𝐾(𝑞)𝑞̂𝑢(0,𝜉)+𝐾(𝑞)𝑞2̂𝑢𝑥(0,𝜉)𝑑𝑞.(2.31) To perform the condition (2.29) in the form of nonhomogeneous Riemann problem we introduce the sectionally analytic function: 1Ω(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝑞𝑧Ψ(𝑞,𝜉)𝑑𝑞,(2.32) where Ψ(𝑝,𝜉)=𝐾(𝑝)𝐾(𝑝)+𝜉Φ(𝑝,𝜉).(2.33) Taking into account the assumed condition (2.25) and making use of Sokhotzki-Plemelj formula (2.3) we get for limiting values of the functions Ω(𝑧,𝜉) and Θ(𝑧,𝜉)Ω(𝑝,𝜉)=𝜉Θ(𝑝,𝜉).(2.34) Also observe that from (2.30) and (2.32) by formula (2.4) 𝐾Θ(𝑝)+(𝑝,𝜉)Θ(𝑝,𝜉)=Ψ(𝑝,𝜉)=Ω+(𝑝,𝜉)Ω(𝑝,𝜉).(2.35) Substituting (2.29) and (2.34) into this equation we obtain nonhomogeneous Riemann problem Ω+(𝑝,𝜉)=𝐾(𝑝)+𝜉𝜉Ω(𝑝,𝜉)𝐾(𝑝)Λ+(𝑝,𝜉).(2.36)
It is required to find two functions for some fixed point 𝜉, Re𝜉>0: Ω+(𝑧,𝜉), analytic in Re𝑧<0 and Ω(𝑧,𝜉), analytic in Re𝑧>0, which satisfy on the contour Re𝑝=0 the relation (2.36). Here, for some fixed point 𝜉, Re𝜉>0, the functions 𝑊(𝑝,𝜉)=𝐾(𝑝)+𝜉𝜉,𝑔(𝑝,𝜉)=𝐾(𝑝)Λ+(𝑝,𝜉)(2.37) are called the coefficient and the free term of the Riemann problem, respectively.
Note that bearing in mind formula (2.33) we can find unknown function Φ(𝑝,𝜉) which involved in the formula (2.27) by the relation Φ(𝑝,𝜉)=𝐾(𝑝)+𝜉Ω𝐾(𝑝)+(𝑝,𝜉)Ω(.𝑝,𝜉)(2.38) The method for solving the Riemann problem 𝐴+(𝑝)=𝜑(𝑝)𝐴(𝑝)+𝜙(𝑝) is based on the Theorems 2.2 and 2.3.
In the formulations of Theorems 2.2 and 2.3 the coefficient 𝜑(𝑝) and the free term 𝜙(𝑝) of the Riemann problem are required to satisfy the Hölder condition on the contour Re𝑝=0. This restriction is essential. On the other hand, it is easy to observe that both functions 𝑊(𝑝,𝜉) and 𝑔(𝑝,𝜉) do not have limiting value as 𝑝±𝑖. The principal task now is to get an expression equivalent to the boundary value problem (2.36), such that the conditions of theorems are satisfied. First, let us introduce some notation and let us establish certain auxiliary relationships. Setting 𝐾1(𝑝)=𝑝2,(2.39) we introduce the function 𝑊(𝑝,𝜉)=𝐾(𝑝)+𝜉𝐾1𝑤(𝑝)+𝜉(𝑝)𝑤+(𝑝),(2.40) where for some fixed point 𝑘(𝜉) (Re𝑘(𝜉)>0) 𝑤𝑧(𝑧)=𝑧+𝑘(𝜉)1/2,𝑤+𝑧(𝑧)=𝑧𝑘(𝜉)1/2.(2.41) We make a cut in the plane 𝑧 from point 𝑘(𝜉) to point through 0. Owing to the manner of performing the cut the functions 𝑤(𝑧), 𝐾1(𝑧) are analytic for Re𝑧>0 and the function 𝑤+(𝑧) is analytic for Re𝑧<0.
We observe that the function 𝑊(𝑝,𝜉) given on the contour Re𝑝=0 satisfies the Hölder condition and under the assumption Re𝐾1(𝑝)>0 does not vanish for any Re𝜉>0. Also we have 1Ind.𝑊(𝑝,𝜉)=2𝜋𝑖𝑖𝑖𝑑ln𝑊(𝑝,𝜉)=0.(2.42) Therefore in accordance with Theorem 2.3 the function 𝑊(𝑝,𝜉) can be represented in the form of the ratio 𝑋𝑊(𝑝,𝜉)=+(𝑝,𝜉)𝑋,(𝑝,𝜉)(2.43) where 𝑋±(𝑝,𝜉)=𝑒Γ±(𝑝,𝜉)1,Γ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝑞𝑧ln𝑊(𝑞,𝜉)𝑑𝑞.(2.44) Now we return to the nonhomogeneous Riemann problem (2.36). Multiplying and dividing the expression (𝐾(𝑝)+𝜉)/𝜉 by (1/(𝐾1(𝑝)+𝜉))(𝑤(𝑝)/𝑤+(𝑝)) and making use of the formula (2.43) we get 𝑊(𝑝,𝜉)=𝐾(𝑝)+𝜉𝜉=𝑌+(𝑝,𝜉)𝑌𝐾(𝑝,𝜉)1(𝑝)+𝜉𝜉,(2.45) where 𝑌±(𝑝,𝜉)=𝑋±(𝑝,𝜉)𝑤±(𝑝).(2.46) Replacing in (2.36) the coefficient of the Riemann problem 𝑊(𝑝,𝜉) by (2.45) we reduce the nonhomogeneous Riemann problem (2.36) to the form Ω+(𝑝,𝜉)𝑌+=𝐾(𝑝,𝜉)1(𝑝)+𝜉𝜉Ω(𝑝,𝜉)𝑌1(𝑝,𝜉)𝑌+(𝑝,𝜉)𝐾(𝑝)Λ+(𝑝,𝜉).(2.47) Now we perform the function Λ(𝑧,𝜉) given by formula (2.31) as Λ(𝑧,𝜉)=Λ1(𝑧,𝜉)+Λ2(𝑧,𝜉),(2.48) where Λ11(𝑧,𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝐾1(𝑞)+𝜉̂𝑢0𝐾(𝑞)+1(𝑞)𝑞𝐾̂𝑢(0,𝜉)+1(𝑞)𝑞2̂𝑢𝑥Λ(0,𝜉)𝑑𝑞,(2.49)21(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝐾𝑞𝑧1(𝑞)𝐾(𝑞)𝐾(𝐾(𝑞)+𝜉)1(𝑞)+𝜉̂𝑢0𝜉(𝑞)𝑞̂𝑢(0,𝜉)̂𝑢𝑥(0,𝜉)𝑑𝑞.(2.50) Firstly we calculate the left limiting value Λ+1(𝑝,𝜉). Since there exists only one root 𝑘(𝜉) of equation 𝐾1(𝑧)=𝜉 such that Re𝑘(𝜉)>0 for all Re𝜉>0, therefore, taking limit 𝑧𝑝 from the left-hand side of complex plane, by Cauchy theorem we get Λ+1𝑘(𝑝,𝜉)=(𝜉)𝑝𝑘(𝜉)̂𝑢0𝜉(𝑘(𝜉))𝑝̂𝑢(0,𝜉)̂𝑢𝑥(0,𝜉).(2.51) The last relation implies that (𝐾1(𝑝)+𝜉)Λ+1(𝑝,𝜉) can be expressed by the function Λ3(𝑧,𝜉) which is analytic in Re𝑧>0𝐾1Λ(𝑝)+𝜉+1(𝑝,𝜉)=Λ3(𝑝,𝜉),(2.52) where Λ3(𝑧,𝜉)=𝑘𝐾(𝜉)1(𝑧)+𝜉𝑧𝑘(𝜉)̂𝑢0𝜉(𝑘(𝜉))𝑝̂𝑢(0,𝜉)̂𝑢𝑥(0,𝜉).(2.53) By Sokhotzki-Plemelj formula (2.4) we express the left limiting value Λ+2(𝑝,𝜉) in the term of the right limiting value Λ2(𝑝,𝜉) as Λ+2(𝑝,𝜉)=Λ2(𝑝,𝜉)+̃𝑔1(𝑝,𝜉),(2.54) where ̃𝑔1𝐾(𝑝,𝜉)=1(𝑝)𝐾(𝑝)𝐾(𝐾(𝑝)+𝜉)1(𝑝)+𝜉̂𝑢0𝜉(𝑝)𝑝̂𝑢(0,𝜉)̂𝑢𝑥.(0,𝜉)(2.55) Bearing in mind the representation (2.48) and making use of (2.52), (2.55), and (2.45) after simple transformations we get 𝐾(𝑝)Λ+𝑌=+𝑌Λ3+𝐾1(Λ𝑝)+𝜉2+𝜉Λ+𝑔1(𝑝,𝜉),(2.56) where 𝑔1(𝑝,𝜉)=(𝐾(𝑝)+𝜉)̃𝑔1𝐾(𝑝,𝜉)=1(𝑝)𝐾(𝑝)𝐾1(𝑝)+𝜉̂𝑢0𝜉(𝑝)𝑝̂𝑢(0,𝜉)̂𝑢𝑥(0,𝜉).(2.57) Replacing in (2.47) 𝐾(𝑝)Λ+(𝑝,𝜉) by (2.56), we reduce the nonhomogeneous Riemann problem (2.47) in the form Ω+1(𝑝,𝜉)𝑌+=Ω(𝑝,𝜉)1(𝑝,𝜉)𝑌1(𝑝,𝜉)𝑌+𝑔(𝑝,𝜉)1(𝑝,𝜉),(2.58) where Ω+1(𝑝,𝜉)=Ω+(𝑝,𝜉)𝜉Λ+Ω(𝑝,𝜉),1𝐾(𝑝,𝜉)=1𝜉(𝑝)+𝜉1Ω(𝑝,𝜉)Λ2(𝑝,𝜉)Λ3(𝑝,𝜉).(2.59) In subsequent consideration we will have to use the following property of the limiting values of a Cauchy type integral, the statement of which we now quote. The proofs may be found in [19].Lemma 2.5. If 𝐿 is a smooth closed contour and 𝜙(𝑞) a function that satisfies the Hölder condition on 𝐿, then the limiting values of the Cauchy type integral 1Φ(𝑧)=2𝜋𝑖𝐿1𝑞𝑧𝜙(𝑞)𝑑𝑞(2.60) also satisfy this condition.
Since 𝑔1(𝑝,𝜉) satisfies on Re𝑝=0 the Hölder condition, on basis of this Lemma the function (1/𝑌+(𝑝,𝜉))𝑔1(𝑝,𝜉) also satisfies this condition. Therefore in accordance with Theorem 2.2 it can be uniquely represented in the form of the difference of the functions 𝑈+(𝑝,𝜉) and 𝑈(𝑝,𝜉), constituting the boundary values of the analytic function 𝑈(𝑧,𝜉), given by formula 1𝑈(𝑧,𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+𝑔(𝑞,𝜉)1(𝑞,𝜉)𝑑𝑞.(2.61) Therefore the problem (2.58) takes the form Ω+1(𝑝,𝜉)𝑌+(𝑝,𝜉)+𝑈+Ω(𝑝,𝜉)=1(𝑝,𝜉)𝑌(𝑝,𝜉)+𝑈(𝑝,𝜉).(2.62) The last relation indicates that the function (Ω+1/𝑌+)+𝑈+, analytic in Re𝑧<0, and the function (Ω1/𝑌)+𝑈, analytic in Re𝑧>0, constitute the analytic continuation of each other through the contour Re𝑧=0. Consequently, they are branches of unique analytic function in the entire plane. According to generalize Liouville theorem this function is some arbitrary constant 𝐴. Thus, bearing in mind the representations (2.59) and (2.52) we get Ω+(𝑝,𝜉)=𝑌+𝐴𝑈++𝜉Λ+,Ω𝜉(𝑝,𝜉)=𝐾1𝑌(𝑝)+𝜉(𝐴𝑈Λ)+𝜉+1+Λ2.(2.63) Since there exists only one root 𝑘(𝜉) of equation 𝐾1(𝑧)=𝜉 such that Re𝑘(𝜉)>0 for all Re𝜉>0, therefore, in the expression for the function Ω(𝑧,𝜉) the factor 𝜉/(𝐾1(𝑧)+𝜉) has a pole in the point 𝑧=𝑘(𝜉). Also the function 𝜉Λ+1 has a pole in the point 𝑧=𝑘(𝜉). Thus in general case the problem (2.36) is insolvable. It is soluble only when the functions 𝑈(𝑧,𝜉) and 𝜉Λ+1 satisfy additional conditions. For analyticity of Ω(𝑧,𝜉) in points 𝑧=𝑘(𝜉) it is necessary that Res𝑝=𝑘(𝜉)1𝐾1𝑌(𝑝)+𝜉(𝐴𝑈)+Λ+1=0.(2.64) We reduce (2.64) to the form 𝐴𝑌(𝑘(𝜉),𝜉)𝑌(𝑘(𝜉),𝜉)𝑈𝜉(𝑘(𝜉),𝜉)+𝑘(𝜉)̂𝑢(0,𝜉)+̂𝑢0(𝑘(𝜉))̂𝑢𝑥(0,𝜉)=0.(2.65) Multiplying the last relation by 1/𝑌(𝑘(𝜉),𝜉) and taking limit 𝜉 we get that 𝐴=0. This implies that for solubility of the nonhomogeneous problem (2.36) it is necessary and sufficient that the following condition is satisfied: 𝑌(𝑘(𝜉),𝜉)𝑈𝜉(𝑘(𝜉),𝜉)+𝑘(𝜉)̂𝑢(0,𝜉)̂𝑢0(𝑘(𝜉))+̂𝑢𝑥(0,𝜉)=0.(2.66)
Therefore, we need to put in the problem (2.9) one boundary data and the rest of boundary data can be found from (2.66). Thus, for example, if we put 𝑢(0,𝑡)=0 from (2.66) we obtain for the Laplace transform of 𝑢𝑥(0,𝑡),[𝑌](𝑘(𝜉),𝜉)𝐼(𝑘(𝜉),𝜉)1̂𝑢𝑥(0,𝜉)=̂𝑢0(𝑘(𝜉))𝑌1(𝑘(𝜉),𝜉)2𝜋𝑖𝑖𝑖1𝑞𝑘(𝜉)̂𝑢0(𝑞)𝑌+𝐾(𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉𝑑𝑞,(2.67) where 1𝐼(𝑘(𝜉),𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑌+𝐾(𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉𝑑𝑞.(2.68) Now we prove that the coefficient of ̂𝑢𝑥(0,𝜉) does not vanish for all Re𝜉>0. We represent the function 𝐼(𝑘(𝜉),𝜉) in the form 𝐼=𝐼1+𝐼2,(2.69) where 𝐼1=12𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑌+𝐼(𝑞,𝜉)𝑑𝑞,21=2𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑌+(𝑞,𝜉)𝐾(𝑞)+𝜉𝐾1(𝑞)+𝜉𝑑𝑞.(2.70) Since, for Re𝑧0,𝑖𝑖1𝑞𝑧𝑑𝑞=𝜋𝑖sgn(Re𝑧),(2.71) making use of analytic properties of the function ((1/𝑌+(𝑞,𝜉))1) by Cauchy Theorem we have 𝐼1=12𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑑𝑞+2𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑌+1(𝑞,𝜉)1𝑑𝑞=2,(2.72) where 𝜉 is some fixed point, Re𝜉>0. To calculate the function 𝐼2 we will use the identity (2.43). Observe that the function 1/𝑌(𝑞,𝜉) is analytic for all Re𝑞>0. Therefore, setting the relation (2.43) into definition of 𝐼2 and making use of Cauchy Theorem we find 𝐼21=2𝜋𝑖𝑖𝑖11𝑞𝑘(𝜉)𝑌1(𝑞,𝜉)𝑑𝑞=𝑌+1(𝑘(𝜉),𝜉)12.(2.73) Thus, from (2.72) and (2.73) we obtain the following relation for the function 𝐼1𝐼(𝑘(𝜉),𝜉)=𝑌(𝑘(𝜉),𝜉)1.(2.74) Substituting this formula into (2.67) we get ̂𝑢𝑥(0,𝜉)=̂𝑢0(𝑘(𝜉))𝑌1(𝑘(𝜉),𝜉)2𝜋𝑖𝑖𝑖1𝑞𝑘(𝜉)̂𝑢0(𝑞)𝑌+𝐾(𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉𝑑𝑞.(2.75) Now we return to problem (2.36). From (2.63) under the conditions 𝑢(0,𝑡)=0 and (2.75) the limiting values of solution of (2.36) are given by Ω+(𝑝,𝜉)=𝑌+𝑈++𝜉Λ+2,Ω𝜉(𝑝,𝜉)=𝐾1(𝑌𝑝)+𝜉𝑈+𝜉Λ2.(2.76) From (2.76) with the help of the integral representations (2.61) and (2.50), for sectionally analytic functions 𝑈(𝑧,𝜉) and Λ2(𝑧,𝜉), making use of Sokhotzki-Plemelj formula (2.3) and relation (2.45) we can express the difference limiting values of the function Ω(𝑧,𝜉) in the form Ω+(𝑝,𝜉)Ω(𝑝,𝜉)=𝑌+𝑈++𝜉𝐾1(𝑌𝑝)+𝜉𝑈Λ+𝜉+2Λ2=𝑌+𝑈+𝜉𝑈𝐾(𝑝)+𝜉Λ+𝜉+2Λ21=2𝐾(𝑝)𝑔𝐾(𝑝)+𝜉1(𝑝,𝜉)𝐾(𝑝)𝑌𝐾(𝑝)+𝜉+1(𝑝,𝜉)2𝜋𝑖𝑃𝑉𝑖𝑖11𝑞𝑝𝑌+𝑔(𝑞,𝜉)1(𝑞,𝜉)𝑑𝑞.(2.77) We now proceed to find the unknown function Φ(𝑝,𝜉) involved in the formula (2.27) for the solution ̂̂𝑢(𝑝,𝜉) of the problem (2.9). Replacing the difference Ω+(𝑝,𝜉)Ω(𝑝,𝜉) in the relation (2.38) by formula (2.77) we get Φ(𝑝,𝜉)=𝐾(𝑝)+𝜉ΩK(𝑝)+(𝑝,𝜉)Ω(1𝑝,𝜉)=2𝑔1(𝑝,𝜉)𝑌+1(𝑝,𝜉)2𝜋𝑖𝑃𝑉𝑖𝑖11𝑞𝑝𝑌+𝑔(𝑞,𝜉)1(𝑞,𝜉)𝑑𝑞.(2.78) It is easy to observe that Φ(𝑝,𝜉) is boundary value of the function analytic in the left complex semi-plane and therefore satisfies our basic assumption for all Re𝑧>0{Φ}=0.(2.79) Having determined the function Φ(𝑝,𝜉) bearing in mind formula (2.27) and conditions 𝑢(0,𝑡)=0 we determine required function ̂̂𝑢̂1̂𝑢(𝑝,𝜉)=𝐾(𝑝)+𝜉̂𝑢0(𝑝)̂𝑢𝑥1(0,𝜉)21𝐾𝑔(𝑝)+𝜉11(𝑝,𝜉)𝑌𝐾(𝑝)+𝜉+(1𝑝,𝜉)2𝜋𝑖𝑃𝑉𝑖𝑖11𝑞𝑝𝑌+𝑔(𝑞,𝜉)1(𝑞,𝜉)𝑑𝑞,(2.80) where the function 𝑔1(𝑝,𝜉) is given by formula (2.55): 𝑔1𝐾(𝑝,𝜉)=1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉̂𝑢0(𝑞)̂𝑢𝑥.(0,𝜉)(2.81) Now we prove that, in accordance with last relation, the function ̂̂𝑢(𝑝,𝜉) constitutes the limiting value of an analytic function in Re𝑧>0.
With the help of the integral representations (2.61), (2.31), and (2.50) for sectionally analytic functions 𝑈(𝑧,𝜉),Λ(𝑧,𝜉), and Λ2(𝑧,𝜉), and making use of Sokhotzki-Plemelj formula (2.3) we have 1𝐾(𝑝)+𝜉̂𝑢0(𝑝)̂𝑢𝑥(0,𝜉)=Λ+Λ,1𝐾𝑔(𝑝)+𝜉1(𝑝,𝜉)=Λ+2Λ2,12𝜋𝑖PV𝑖𝑖11𝑞𝑝𝑌+𝑔(𝑞,𝜉)1(1𝑞,𝜉)𝑑𝑞=2𝑈++𝑈.(2.82) Substituting these relations into (2.80) we express the function ̂̂𝑢 in the following form: ̂Λ̂𝑢=+Λ12Λ+2Λ2121𝐾𝑌(𝑝)+𝜉+𝑈++𝑈.(2.83) If it is taken into account that Λ(𝑧,𝜉)=Λ1(𝑧,𝜉)+Λ2(𝑧,𝜉) by virtue of the relation (2.45), the last expression agrees with formula ̂̂𝑢=Λ+1Λ1+12Λ+2Λ2121𝐾𝑌(𝑝)+𝜉+𝑈+121𝐾1𝑌(𝑝)+𝜉𝑈.(2.84) Expressing the function 𝑈+ in the last equation in terms of 𝑈𝑈+=𝑈+1𝑌+Λ(𝐾(𝑝)+𝜉)+2Λ2,(2.85) we arrive at the following relation: ̂̂𝑢=Λ+1Λ11𝐾1𝑌(𝑝)+𝜉𝑈,(2.86) where by virtue of (2.49) and (2.66), Λ+1Λ1=1𝐾1(𝑝)+𝜉̂𝑢0(𝑝)̂𝑢𝑥.(0,𝜉)(2.87) Thus the function ̂̂𝑢 is the limiting value of an analytic function in Re𝑧>0. Note the fundamental importance of the proven fact that the solution ̂̂𝑢 constitutes an analytic function in Re𝑧>0 and, as a consequence, its inverse Laplace transform vanishes for all 𝑥<0. We now return to solution 𝑢(𝑥,𝑡) of the problem (2.9).
Under assumption 𝑢(0,𝑡)=0 the integral representation (2.61) takes form 1𝑈(𝑧,𝜉)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+𝐾(𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1(𝑞)+𝜉̂𝑢0(𝑞)̂𝑢𝑥(0,𝜉)𝑑𝑞,(2.88) where ̂𝑢𝑥(0,𝜉) is defined by (2.75). Substituting this relation into (2.86) and taking inverse Laplace transform with respect to time and inverse Fourier transform with respect to space variables we obtain 𝑢(𝑥,𝑡)=𝒢(𝑡)𝑢0=0𝐺(𝑥,𝑦,𝑡)𝑢0(𝑦)𝑑𝑦,(2.89) where the function 𝐺(𝑥,𝑦,𝑡) was defined by formula (2.12). Proposition 2.4 is proved.

Now we collect some preliminary estimates of the Green operator 𝒢(𝑡). Let the contours 𝒞𝑖 be defined as

𝒞1=𝑝𝑒𝑖(𝜋/2+𝜀),00,𝑒𝑖(𝜋/2+𝜀)𝒞,(2.90)2=𝑞𝑒𝑖((𝜋/2)+2𝜀),00,𝑒𝑖((𝜋/2)+2𝜀)𝒞,(2.91)3=𝑞𝑒𝑖((𝜋+𝜀)/2),00,𝑒𝑖((𝜋+𝜀)/2),(2.92) where 𝜀>0 can be chosen such that all functions under integration are analytic and Re𝑘(𝜉)>0 for 𝜉𝒞1.

Lemma 2.6. The function 𝐺(𝑥,𝑦,𝑡) given by formula (2.12) has the following representation: 1𝐺(𝑥,𝑦,𝑡)=2𝜋𝑖𝑖𝑖𝑒𝑝𝑥𝐾(𝑝)𝑡𝑒𝑝𝑦1𝑑𝑝12𝜋𝑖2𝜋𝑖𝒞1𝑑𝜉𝑒𝜉𝑡𝒞2𝑒𝑝𝑥𝑒Γ(𝑝,𝜉)𝑤+(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾(𝑝)+𝜉𝐼(𝑝,𝜉,𝑦)𝑑𝑝,(2.93) where 1𝐼(𝑝,𝜉,𝑦)=2𝜋𝑖𝑖𝑖11(𝑞𝑝)(𝑞𝑘(𝜉))𝑒Γ+(𝑞,𝜉)𝑤+𝑒(𝑞,𝜉)𝑞𝑦1𝑑𝑞,Γ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝑞𝑧ln𝐾(𝑞)+𝜉𝐾1𝑤(𝑞)+𝜉(𝑞)𝑤+(𝑞)𝑑𝑞.(2.94) The functions 𝑤±(𝑞,𝜉),𝑘(𝜉) were defined in formulas (2.13) and (2.10).

Proof. We rewrite formula (2.12) in the form 𝐺(𝑥,𝑦,𝑡)=𝐽1(𝑥𝑦,𝑡)+𝐽2(𝑥,𝑦,𝑡)+𝐽3(𝑥,𝑦,𝑡)+𝐽4(𝑥,𝑦,𝑡),(2.95) where 𝐽11(𝑥𝑦,𝑡)=2𝜋𝑖𝑖𝑖𝑒𝑝(𝑥𝑦)𝐾1(𝑝)𝑡𝐽𝑑𝑝,21(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖+𝜀𝑖+𝜀𝑑𝜉𝑒𝜉𝑡Ψ(𝜉,𝑦)𝑖𝑖𝑒𝑝𝑥1𝐾1(𝐽𝑝)+𝜉𝑑𝑝,31(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖+𝜀𝑖+𝜀𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾1Υ(𝑝)+𝜉1𝐽(𝑝,𝜉,𝑦)𝑑𝑝,(2.96)41(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖+𝜀𝑖+𝜀𝑑𝜉𝑒𝜉𝑡Ψ(𝜉,𝑦)𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾1Υ(𝑝)+𝜉1(𝑝,𝜉,0)𝑑𝑝.(2.97) Here Υ11(𝑧,𝜉,𝑦)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+𝐾(𝑞,𝜉)1(𝑞)𝐾(𝑞)𝐾1𝑒(𝑞)+𝜉𝑞𝑦𝑒𝑑𝑞,Ψ(𝜉,𝑦)=𝑘(𝜉)𝑦𝑌(𝑘(𝜉),𝜉)+Υ1𝑌(𝑘(𝜉),𝜉,𝑦),±=𝑒Γ±𝑤±,1(2.98)Γ(𝑧,𝜉)=2𝜋𝑖𝑖𝑖1𝐾𝑞𝑧ln(𝑞)+𝜉𝐾1(𝑤𝑞)+𝜉(𝑞)𝑤+(𝑞)𝑑𝑞.(2.99) Firstly we consider the sectionally analytic function Υ1(𝑧,𝜉,𝑦) given by Cauchy type integral (2.98).
On basis of the definition (2.98) its limiting value can be represent in the form Υ1(𝑝,𝜉,𝑦)=𝐼1(𝑝,𝜉,𝑦)+𝐼2(𝑝,𝜉,𝑦),(2.100) where 𝐼1(𝑝,𝜉,𝑦)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+𝑒(𝑞,𝜉)𝑞𝑦𝐼𝑑𝑞,2(𝑝,𝜉,𝑦)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+(𝑞,𝜉)𝐾(𝑞)+𝜉𝐾1𝑒(𝑞)+𝜉𝑞𝑦𝑑𝑞.(2.101) Making use of analytic properties of the functions (1/𝑌+(𝑞,𝜉)1), for Re𝑞<0, and 𝑒𝑞𝑦, for Re𝑞>0, by Cauchy theorem we have 𝐼1(𝑝,𝜉,𝑦)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖1𝑒𝑞𝑧𝑞𝑦𝑑𝑞+lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+(𝑞,𝜉)1(𝑒𝑞𝑦1)𝑑𝑞+lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+(𝑞,𝜉)1𝑑𝑞=𝑒𝑝𝑦+lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+(𝑞,𝜉)1(𝑒𝑞𝑦1)𝑑𝑞,(2.102) where 𝜉 is some fixed point, Re𝜉>0.
To calculate the function 𝐼2(𝑝,𝜉,𝑦) we will use the following identity: 1𝑌+(𝑞,𝜉)𝐾(q)+𝜉𝐾1=1(𝑞)+𝜉𝑌.(𝑞,𝜉)(2.103) Observe that the function 1/𝑌(𝑞,𝜉) is analytic for all Re𝑞>0. Therefore, setting the relation (2.103) into definition of 𝐼2(𝑝,𝜉,𝑦) and making use of Cauchy theorem we find 𝐼2(𝑝,𝜉,𝑦)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧𝑌𝑒(𝑞,𝜉)𝑞𝑦1𝑑𝑞=𝑌𝑒(𝑝,𝜉)𝑝𝑦.(2.104) Thus from (2.102) and (2.104) we obtain the following relation: Υ1(𝑝,𝜉,𝑦)=𝑒𝑝𝑦+Υ1(𝑝,𝜉,𝑦)+𝑌𝑒(𝑝,𝜉)𝑝𝑦,(2.105) where 1Υ(𝑧,𝜉,𝑦)=2𝜋𝑖𝑖𝑖11𝑞𝑧𝑌+𝑒(𝑞,𝜉)1𝑞𝑦𝑑𝑞.(2.106) In the same way (see also proof of relation (2.74)) we can prove that Υ11(𝑝,𝜉,0)=1+𝑌.(𝑝,𝜉)(2.107) Also we observe that Ψ(𝜉,𝑦)=𝑒𝑘(𝜉)𝑦+Υ(𝑘(𝜉),𝜉,𝑦).(2.108) Inserting into definition (2.96) the expression (2.105) for Υ1(𝑝,𝜉,𝑦) we obtain the function 𝐽3(𝑥,𝑦,𝑡) in the form 𝐽31(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾1(𝑝)+𝜉(𝑒𝑝𝑦+Υ(𝑝,𝜉,𝑦))𝑑𝑝𝐽1(𝑥𝑦,𝑡).(2.109) Replacing in formula (2.97) the functions Ψ(𝜉,𝑦) and Υ1(𝑝,𝜉,0) by (2.108) and (2.107), respectively, we reduce the function 𝐽4(𝑥,𝑦,𝑡) in the form 𝐽41(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖𝑖𝑑𝜉𝑒𝜉𝑡Ψ(𝜉,𝑦)𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾11(𝑝)+𝜉1+𝑌1(𝑝,𝜉)𝑑𝑝=12𝜋𝑖2𝜋𝑖𝑖𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾1(𝑒𝑝)+𝜉𝑘(𝜉)𝑦Υ(𝑘(𝜉),𝜉,𝑦)𝑑𝑝𝐽2(𝑥,𝑦).(2.110) Therefore inserting into definition (2.95) expressions (2.109) and (2.110), for 𝐽3(𝑥,𝑦,𝑡) and 𝐽4(𝑥,𝑦,𝑡), respectively, we obtain the function 𝐺(𝑥,𝑦,𝑡) in the form 1𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝑖𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)𝐾1Ξ(𝑝)+𝜉1(𝑝,𝜉,𝑦)𝑑𝑝,(2.111) where Ξ1𝑒(𝑝,𝜉,𝑦)=𝑝𝑦𝑒𝑘(𝜉)𝑦(Υ(𝑝,𝜉,𝑦)Υ(𝑘(𝜉),𝜉,𝑦)).(2.112) Also, note that since Υ(𝑝,𝜉,𝑦)Υ(𝑘(𝜉),𝜉,𝑦)=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖11𝑞𝑧1𝑞𝑘(𝜉)𝑌+𝑒(𝑞,𝜉)1𝑞𝑦𝑑𝑞=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖𝑧𝑘(𝜉)(1𝑞𝑧)(𝑞𝑘(𝜉))𝑌+(𝑒𝑞,𝜉)1𝑞𝑦𝑑𝑞,(2.113) we obtain 𝑒𝑝𝑦𝑒𝑘(𝜉)𝑦(Υ(𝑝,𝜉,𝑦)Υ(𝑘(𝜉),𝜉,𝑦))=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖1+1𝑞𝑧𝑒𝑞𝑘(𝜉)𝑞𝑦+1𝑌+𝑒(𝑞,𝜉)1𝑞𝑦𝑑𝑞=lim𝑧𝑝,Re𝑧>012𝜋𝑖𝑖𝑖𝑘(𝜉)𝑧1(𝑞𝑧)(𝑞𝑘(𝜉))𝑌+𝑒(𝑞,𝜉)𝑞𝑦𝑑𝑞.(2.114) So, 1𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾1𝐼(𝑝)+𝜉(𝑝,𝜉,𝑦)𝑑𝑝,(2.115) where 1𝐼(𝑧,𝜉,𝑦)=2𝜋𝑖𝑖𝑖11(𝑞𝑧)(𝑞𝑘(𝜉))𝑌+𝑒(𝑞,𝜉)𝑞𝑦𝑑𝑞.(2.116) Using relation 𝐾(𝑝)+𝜉𝐾1=𝑌(𝑝)+𝜉+𝑌,(2.117) we rewrite last formula in the following form: 1𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌+(𝑝,𝜉)(𝑝𝑘(𝜉))𝐼𝐾(𝑝)+𝜉(𝑝,𝜉,𝑦)𝑑𝑝.(2.118) On the basis of definitions (2.116) and in accordance with the Sohkotzki-Plemelj formula (2.3) we have 𝐼(𝑝,𝜉,𝑦)=𝐼+1(𝑝,𝜉,𝑦)1(𝑝𝑘(𝜉))𝑌+𝑒(𝑝,𝜉)𝑝𝑦.(2.119) So we get 1𝐺(𝑥,𝑦,𝑡)=2𝜋𝑖𝑖𝑖𝑒𝑝𝑥𝐾(𝑝)𝑡𝑒𝑝𝑦1𝑑𝑝12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑌+(𝑝,𝜉)(𝑝𝑘(𝜉))𝐼𝐾(𝑝)+𝜉+(𝑝,𝜉,𝑦)𝑑𝑝.(2.120) Now we consider for Re𝜉>0Γ+1(𝑝,𝜉)=2𝜋𝑖lim𝑧𝑝Re𝑧<0𝑖𝑖1𝑞𝑧ln𝐾(𝑞)+𝜉𝐾1𝑤(𝑞)+𝜉(𝑞)𝑤+=1(𝑞)𝑑𝑞2𝜋𝑖𝑒𝑖((𝜋/2)+3𝜀)01𝑞𝑝ln𝑖𝑞2+𝜉𝑞2𝑤+𝜉(𝑞)𝑤++1(𝑞)𝑑𝑞2𝜋𝑖0𝑒𝑖((𝜋/2)3𝜀)1𝑞𝑝ln𝑖𝑞2+𝜉𝑞2𝑤+𝜉(𝑞)𝑤+(𝑞)𝑑𝑞+ln𝐾(𝑝)+𝜉𝐾1𝑤(𝑝)+𝜉(𝑝)𝑤+(𝑝)=Γ1(𝑝,𝜉)+ln𝐾(𝑝)+𝜉𝐾1𝑤(𝑝)+𝜉(𝑝)𝑤+.(𝑝)(2.121) Note that Γ1(𝑝,𝜉) is analytic in domain 0arg𝜉<(𝜋/2)+3𝜀, (𝜋/2)3𝜀<arg𝜉0, and 𝜋/2arg𝑝<(𝜋/2)+3𝜀 and (𝜋/2)3𝜀<arg𝑝𝜋/2. So we get 1𝐺(𝑥,𝑦,𝑡)=2𝜋𝑖𝑖𝑖𝑒𝑝𝑥𝐾(𝑝)𝑡𝑒𝑝𝑦1𝑑𝑝12𝜋𝑖2𝜋𝑖𝜀+𝑖𝜀𝑖𝑑𝜉𝑒𝜉𝑡𝑖𝑖𝑒𝑝𝑥𝑒Γ1(𝑝,𝜉)𝑤(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾1(𝐼𝑝)+𝜉+(𝑝,𝜉,𝑦)𝑑𝑝,(2.122) where 𝐼+1(𝑝,𝜉,𝑦)=2𝜋𝑖lim𝑧𝑝Re𝑧<0𝑖𝑖11(𝑞𝑧)(𝑞𝑘(𝜉))𝑤𝑒Γ1(𝑞,𝜉)𝐾1(𝑞)+𝜉𝑒𝐾(𝑞)+𝜉𝑞𝑦𝑑𝑞.(2.123) Changing the contour of integration with respect to 𝜉 by Cauchy theorem we get 1𝐺(𝑥,𝑦,𝑡)=2𝜋𝑖𝑖𝑖𝑒𝑝𝑥𝐾(𝑝)𝑡𝑒𝑝𝑦1𝑑𝑝12𝜋𝑖2𝜋𝑖𝒞1d𝜉𝑒𝜉𝑡𝒞2𝑒𝑝𝑥𝑒Γ1(𝑝,𝜉)𝑤(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾1(𝐼𝑝)+𝜉+(𝑝,𝜉,𝑦)𝑑𝑝res𝜉=𝐾(𝑞)𝒞2𝑒𝑝𝑥𝑒Γ1(𝑝,𝜉)𝑤(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾1𝐼(𝑝)+𝜉+,(𝑝,𝜉,𝑦)𝑑𝑝(2.124) where res𝜉=𝐾(𝑞)𝒞2𝑒𝑝𝑥𝑒Γ1(𝑝,𝜉)𝑤(𝑝,𝜉)(𝑝𝑘(𝜉))𝐾1𝐼(𝑝)+𝜉+=1(𝑝,𝜉,𝑦)𝑑𝑝2𝜋𝑖𝒞2𝑒𝑝𝑥𝑑𝑝𝒞3𝑒𝑑𝑞𝑞𝑦𝐾(𝑞)𝑡𝑒𝑞𝑝𝜙(𝑝,𝑞),𝜙(𝑝,𝑞)=Γ1(𝑝,𝐾(𝑞))𝑤(𝑝)(𝑝𝑘(𝐾(𝑞)))𝐾1𝐾(𝑝)𝐾(𝑞)1(𝑞)𝐾(𝑞)