Benjamin-Ono Equation on a Half-Line

Abstract

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:

𝑢_𝑡+𝑢𝑢𝑢_𝑥(𝑥+,0ℋ)=𝑢_𝑥𝑥𝑢₀=0,𝑥>0,𝑡>0,𝑢(0,𝑡)=0(𝑥,),𝑡>𝑥0>,0,(1.1) where ℋ𝑢=PV∫^0+∞𝑢((𝑦,𝑡)/(𝑦−𝑥))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 ℋ𝜕^2𝑥=𝜕_𝑥(−𝜕^2𝑥)^1/2 since the operator ℋ can be written as follows: ℋ=−ℱ⁻¹(𝑖𝜉/|𝜉|)ℱ=(−𝜕^2𝑥)^−1/2𝜕_𝑥, where (ℱ𝜑)(𝜉)=(1/√2𝜋)∫𝜑(𝑥)e^{−𝑖𝑥𝜉}𝑑𝑥 is the usual Fourier transform, and ℱ⁻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 𝐇^𝑠,0 was proved in [3–9] and the smoothing properties of solutions were studied in [10–14]. In paper [15] it was proved that for small initial data in 𝐇^2,0∩𝐇^1,1 solutions decay as 𝑡→∞ in 𝐋^∞ norm at the same rate 1/√𝑡 as for the case of the linear Benjamin-Ono equation, where

𝐇^𝑚,𝑠=𝜙∈𝐋²∶‖𝜙‖_𝑚,𝑠=‖‖1+𝑥²^𝑠/21−𝜕^2𝑥^𝑚/2𝜙‖‖_𝐋2<∞.(1.2)

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(𝑝^𝛽) 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)=𝑝^3/2 we get the following definition of the fractional derivative 𝜕^𝑥3/2:

𝜕^𝑥3/2𝜙=ℒ⁻¹𝑝^3/2ℒ𝜙−𝜙𝑝(0).(1.3) Here and below 𝑝^𝛽 is the main branch of the complex analytic function in the complex half-plane Re𝑝≥0, so that 1^𝛽=1 (we make a cut along the negative real axis (−∞,0)). Note that due to the analyticity of 𝑝^𝛽 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 ℒ_𝑥→𝜉 is

𝑢̂(𝜉)≡ℒ_𝑥→𝜉𝑢=^0+∞𝑒^−𝜉𝑥𝑢(𝑥)𝑑𝑥,(1.4) and the inverse Laplace transformation ℒ^{−1𝜉→𝑥} is defined by

𝑢(𝑥)≡ℒ^{−1𝜉→𝑥}𝑢̂=(2𝜋𝑖)⁻¹^{𝑖∞−𝑖∞}𝑒^𝜉𝑥𝑢̂(𝜉)𝑑𝜉.(1.5) Weighted Lebesgue space is 𝐋^𝑞,𝑎(𝐑⁺)={𝜑∈𝒮^;‖𝜑‖_𝐋^𝑞,𝑎<∞}, where

‖𝜑‖_𝐋^𝑞,𝑎=^0+∞𝑥^𝑎𝑞||𝜑(𝑥)||^𝑞𝑑𝑥^1/𝑞(1.6) for 𝑎>0, 1≤𝑞<∞ and

‖𝜑‖_𝐋^∞=esssup_𝑥∈𝐑+||𝜑(𝑥)||.(1.7) Sobolev space is

𝐇¹𝐑⁺=𝜑∈𝒮^;‖‖⟨𝜕_𝑥⟩𝜑‖‖_𝐋2<∞.(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 𝑢₀∈𝐙≡𝐇¹(𝐑⁺)∩𝐋^1,𝑎+1(𝐑⁺) with 𝑎∈(0,1) are such that the norm ‖‖𝑢₀‖‖_𝐙≤𝜀(1.10) is sufficiently small. Then there exists a unique global solution 𝑢∈𝐂[0,∞);𝐇¹𝐑⁺(1.11) to the initial-boundary value problem (1.1). Moreover the following asymptotic is valid in 𝐋^∞(𝐑⁺)∶𝑢=1𝑡𝐴Λ𝑥𝑡^−1/2+min1,√𝑥𝑡𝑂𝑡^{−1−(𝑎/2)}(1.12) for 𝑡→∞, where Λ(𝑥𝑡^−1/2)∈𝐋^∞(𝐑⁺),Λ(0)=0 is defined below by the formula (2.191), and the constant 𝐴=𝑓𝑢₀𝒩−(𝑢)=^0+∞𝑢𝑓_𝑥(𝒩(𝑢))𝑑𝜏,𝑢.(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 𝑢𝑢_𝑥 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 𝑢𝑢_𝑥 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_∞||≤𝐶||𝑞||^−𝜇,𝜇>0.(2.1) Then Cauchy type integral 𝐹(𝑧)=12𝜋𝑖^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑧𝑑𝑞(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)=lim_{𝑧→𝑝,Re𝑧<0}12𝜋𝑖^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑧𝑑𝑞=21𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}𝜙(𝑞)𝐹𝑞−𝑝𝑑𝑞+12𝜙(𝑝),₋(𝑝)=lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑧𝑑𝑞=21𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑝𝑑𝑞−12𝜙(𝑝).(2.3)

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

𝐹𝐹⁺(𝑝)−𝐹⁻(𝑝)=𝜙(𝑝),₊(𝑝)+𝐹⁻(𝑝)=1𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑝𝑑𝑞,(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 𝑈(𝑧)=12𝜋𝑖^{𝑖∞−𝑖∞}𝜙(𝑞)𝑞−𝑧𝑑𝑞.(2.6)

Theorem 2.3. An arbitrary function 𝜑(𝑝) given on the contour Re𝑝=0, satisfying the Hölder condition, and having zero index, ind𝜑(𝑡)∶=12𝜋𝑖^{𝑖∞−𝑖∞}𝑑ln𝜑(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)=𝑒^Γ±(𝑧),Γ(𝑧)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝜑(𝑞)𝑑𝑞.(2.8)

We consider the following linear initial-boundary value problem on half-line 𝑢_𝑡−𝑃𝑉^0+∞𝑢_𝑦𝑦(𝑦,𝑡)𝑢(𝑥,𝑥0)−=𝑦𝑢₀𝑑𝑦=0,𝑡>0,𝑥>0,𝑢(0,𝑡)=0(𝑥,),𝑡>𝑥0>.0,(2.9)

Setting

𝐾(𝑞)=−||𝑞||𝑞,𝐾₁(𝑞)=−𝑞²,𝑘(𝜉)=||𝜉||^1/2𝑒^{(1/2)𝑖arg𝜉},(2.10)

where Re𝑘(𝜉)>0 for Re𝜉>0, we define

𝒢(𝑡)𝜙=^0+∞𝐺(𝑥,𝑦,𝑡)𝜙(𝑦)𝑑𝑦,(2.11)

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

𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑑𝑝𝑒^𝑝𝑥1𝐾₁(𝑝)+𝜉(𝑒^−𝑝𝑦−1+Ψ(𝜉,𝑦))2𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑑𝑝𝑒^𝑝𝑥1𝐾₁(𝑝)+𝜉𝑌⁻×lim_{𝑧→𝑝,Re𝑧>0}(𝑝,𝜉)12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉(𝑒^−𝑞𝑦+Ψ(𝜉,𝑦))𝑑𝑞(2.12) for 𝜀>0,𝑥>0,𝑦>0,𝑡>0. Here and below

𝑌^±=𝑒^Γ±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

Γ(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑤⁻(𝑞)𝑤⁺(𝑞)𝑑𝑞,(2.14) where

𝑤⁻(𝑧)=𝑧𝑧+𝑘(𝜉)^1/2,𝑤⁺(𝑧)=𝑧𝑧−𝑘(𝜉)^1/2Ψ(𝜉,𝑦)=−𝑒_{−𝑘(𝜉)𝑦},𝑌⁻(𝑘(𝜉),𝜉)+21𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑌⁺(1𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑒^−𝑞𝑦𝑑𝑞.(2.15)

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

Proposition 2.4. Let the initial data be 𝑢₀∈𝐋¹(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:
𝑢(𝑥,𝑡)=0,∀𝑥<0.(2.17)
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 ℙ𝜙(𝑧)=−12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝜙(𝑞)𝑑𝑞.(2.18)
Since the operator ℙ is defined by a Cauchy type integral, it is readily observed that ℙ𝜙(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=0ℙ⁺𝜙=12𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1ℙ𝑞−𝑝𝜙(𝑞)𝑑𝑞+12𝜙(𝑝),₋𝜙=12𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1𝑞−𝑝𝜙(𝑞)𝑑𝑞−12𝜙(𝑝).(2.19) Here ℙ⁺𝜙 and ℙ⁻𝜙 are limits of ℙ𝜙 as 𝑧 tends to 𝑝 from the left and right semi-plane, respectively.
We have for the Laplace transform ℒℋ𝑢_𝑥𝑥=ℙ−||𝑝||𝑝ℒ{𝑢}−𝑢(0,𝑡)𝑝−𝑢^𝑥(0,𝑡)𝑝².(2.20) Since ℒ{𝑢} is analytic for all Re𝑞>0, we have 𝑢̂(𝑞,𝑡)=ℒ{𝑢}=ℙ𝑢̂(𝑝,𝑡).(2.21) Therefore applying the Laplace transform with respect to 𝑥 to problem (2.9) we obtain for 𝑡>0ℙ𝑢_𝑡̂+𝐾(𝑝)𝑢̂(𝑝,𝑡)−𝐾(𝑝)𝑝𝑢(0,𝑡)−𝐾𝑝(²𝑝)𝑢̂(𝑝,0)=𝑢₀̂𝑢_𝑥(0,𝑡)=0,(𝑝),(2.22) where 𝐾(𝑝)=−||𝑝||𝑝.(2.23) We rewrite (2.22) in the form 𝑢_𝑡̂+𝐾(𝑝)𝑢̂(𝑝,𝑡)−𝐾(𝑝)𝑝𝑢(0,𝑡)−𝐾𝑝(²𝑝)𝑢̂(𝑝,0)=𝑢₀̂𝑢_𝑥(0,𝑡)=Φ(𝑝,𝑡),(𝑝),(2.24) with some function Φ(𝑝,𝑡) such that for all Re𝑝>0ℙ{Φ(𝑝,𝑡)}=0(2.25) and for |𝑝|>1||Φ(𝑝,𝑡)||≤𝐶1||𝑝||.(2.26) Applying the Laplace transformation with respect to time variable to problem (2.24) we find for Re𝑝>0𝑢̂̂(𝑝,𝜉)=1𝐾(𝑝)+𝜉𝑢⁰̂(𝑝)+𝐾(𝑝)𝑝𝑢̂(0,𝜉)+𝐾𝑝(²𝑝)𝑢_𝑥̂(0,𝜉)+Φ(𝑝,𝜉).(2.27) Here the functions 𝑢̂(̂𝑝,𝜉),Φ(𝑝,𝜉),𝑢(̂0,𝜉), and 𝑢_𝑥̂(0,𝜉) are the Laplace transforms for 𝑢(̂𝑝,𝑡),Φ(𝑝,𝑡),𝑢(0,𝑡), and 𝑢_𝑥(0,𝑡) with respect to time, respectively. We will find the function Φ(𝑝,𝜉) using the analytic properties of function 𝑢̂̂ in the right-half complex planes Re𝑝>0 and Re𝜉>0. We have for Re𝑝=0𝑢̂̂(𝑝,𝜉)=1𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1𝑞−𝑝𝑢̂̂(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: Θ(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1Λ(𝑧,𝜉)=1𝑞−𝑧𝐾(𝑞1)+𝜉Φ(𝑞,𝜉)𝑑𝑞,(2.30)2𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝐾(𝑞1)+𝜉𝑢⁰̂(𝑞)+𝐾(𝑞)𝑞𝑢̂(0,𝜉)+𝐾𝑞(²𝑞)𝑢_𝑥̂(0,𝜉)𝑑𝑞.(2.31) To perform the condition (2.29) in the form of nonhomogeneous Riemann problem we introduce the sectionally analytic function: Ω(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧Ψ(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) 𝐾(𝑝)Θ⁺(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)+𝜉𝐾(𝑝)Ω⁺(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)+𝜉𝐾₁(𝑝)+𝜉𝑤⁻(𝑝)𝑤⁺(𝑝),(2.40) where for some fixed point 𝑘(𝜉) (Re𝑘(𝜉)>0) 𝑤⁻(𝑧)=𝑧𝑧+𝑘(𝜉)^1/2,𝑤⁺(𝑧)=𝑧𝑧−𝑘(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 Ind.𝑊(𝑝,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}𝑑ln𝑊(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 𝑊(𝑝,𝜉)=𝑋⁺(𝑝,𝜉)𝑋⁻(𝑝,𝜉),(2.43) where 𝑋^±(𝑝,𝜉)=𝑒^{Γ±(𝑝,𝜉)},Γ(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝑊(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)𝑌⁻(𝑝,𝜉)𝐾¹(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⁺(𝑝,𝜉)=𝐾¹(𝑝)+𝜉𝜉Ω⁻(𝑝,𝜉)𝑌⁻(𝑝,𝜉)−𝑌⁺(1𝑝,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 Λ₁(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝐾₁(𝑞1)+𝜉𝑢⁰̂(𝑞)+𝐾¹(𝑞)𝑞𝑢̂(0,𝜉)+𝐾¹(𝑞)𝑞²Λ𝑢_𝑥̂(0,𝜉)𝑑𝑞,(2.49)₂(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝐾¹(𝑞)−𝐾(𝑞)(𝐾(𝑞)+𝜉)𝐾₁(𝑞)+𝜉𝑢⁰̂(𝑞)−𝜉𝑞𝑢̂(0,𝜉)−𝑢^𝑥̂(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 𝑧→𝑝 from the left-hand side of complex plane, by Cauchy theorem we get Λ⁺¹(𝑝,𝜉)=−𝑘^(𝜉)𝑝−𝑘(𝜉)𝑢⁰̂(𝑘(𝜉))−𝜉𝑝𝑢̂(0,𝜉)−𝑢^𝑥̂(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>0∶𝐾₁(𝑝)+𝜉Λ⁺¹(𝑝,𝜉)=Λ⁻³(𝑝,𝜉),(2.52) where Λ₃(𝑧,𝜉)=−𝑘^(𝜉)𝐾¹(𝑧)+𝜉𝑧−𝑘(𝜉)𝑢⁰̂(𝑘(𝜉))−𝜉𝑝𝑢̂(0,𝜉)−𝑢^𝑥̂(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 Λ⁻²(𝑝,𝜉) as Λ⁺²(𝑝,𝜉)=Λ⁻²(𝑝,𝜉)+𝑔̃₁(𝑝,𝜉),(2.54) where 𝑔̃₁(𝑝,𝜉)=𝐾¹(𝑝)−𝐾(𝑝)(𝐾(𝑝)+𝜉)𝐾₁(𝑝)+𝜉𝑢⁰̂(𝑝)−𝜉𝑝𝑢̂(0,𝜉)−𝑢^𝑥̂(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⁺𝑌⁻Λ⁻³+𝐾₁(𝑝)+𝜉Λ⁻²+𝜉Λ⁺−𝑔₁(𝑝,𝜉),(2.56) where 𝑔₁(𝑝,𝜉)=(𝐾(𝑝)+𝜉)𝑔̃₁(𝑝,𝜉)=𝐾¹(𝑝)−𝐾(𝑝)𝐾₁(𝑝)+𝜉𝑢⁰̂(𝑝)−𝜉𝑝𝑢̂(0,𝜉)−𝑢^𝑥̂(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⁺(𝑝,𝜉)=Ω⁻¹(𝑝,𝜉)𝑌⁻(𝑝,𝜉)−𝑌⁺(1𝑝,𝜉)𝑔¹(𝑝,𝜉),(2.58) where ΩΩ⁺¹(𝑝,𝜉)=Ω⁺(𝑝,𝜉)−𝜉Λ⁺(𝑝,𝜉),₋₁(𝑝,𝜉)=𝐾₁(𝑝)+𝜉𝜉⁻¹Ω⁻(𝑝,𝜉)−Λ⁻²(𝑝,𝜉)−Λ⁻³(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 Φ(𝑧)=12𝜋𝑖_𝐿1𝑞−𝑧𝜙(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 𝑈(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝑔¹(𝑞,𝜉)𝑑𝑞.(2.61) Therefore the problem (2.58) takes the form Ω⁺¹(𝑝,𝜉)𝑌⁺(𝑝,𝜉)+𝑈⁺(𝑝,𝜉)=Ω⁻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 (Ω⁻¹/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 ΩΩ⁺₋(𝑝,𝜉)=𝑌⁺𝐴−𝑈⁺+𝜉Λ⁺,(𝑝,𝜉)=𝜉𝐾₁(𝑝)+𝜉𝑌⁻(𝐴−𝑈⁻)+𝜉Λ⁺¹+Λ⁻².(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 Res_{𝑝=𝑘(𝜉)}𝐾₁(𝑝1)+𝜉𝑌⁻(𝐴−𝑈⁻)+Λ⁺¹=0.(2.64) We reduce (2.64) to the form 𝐴𝑌⁻(𝑘(𝜉),𝜉)−𝑌⁻(𝑘(𝜉),𝜉)𝑈⁻(𝑘(𝜉),𝜉)+−𝜉𝑘(𝜉)𝑢̂(0,𝜉)+𝑢⁰̂(𝑘(𝜉))−𝑢_𝑥̂(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)𝑈⁻(𝑘(𝜉),𝜉)+𝜉𝑘(𝜉)𝑢̂(0,𝜉)−𝑢⁰̂(𝑘(𝜉))+𝑢_𝑥̂(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 𝑢_𝑥(0,𝑡),−[𝑌=⁻𝑢(𝑘₀̂(𝜉),𝜉)𝐼(𝑘(𝜉),𝜉)−1]𝑢_𝑥̂(0,𝜉)(𝑘(𝜉))−𝑌⁻(𝑘(𝜉),𝜉)12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑢⁰̂(𝑞)𝑌⁺(𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑑𝑞,(2.67) where 𝐼(𝑘(𝜉),𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑌⁺(1𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑑𝑞.(2.68) Now we prove that the coefficient of 𝑢_𝑥̂(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=12𝜋𝑖^{𝑖∞−𝑖∞}1𝐼𝑞−𝑘(𝜉)𝑌⁺(1𝑞,𝜉)𝑑𝑞,₂=−12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑌⁺(1𝑞,𝜉)𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑑𝑞.(2.70) Since, for Re𝑧≠0,^{𝑖∞−𝑖∞}1𝑞−𝑧𝑑𝑞=−𝜋𝑖sgn(Re𝑧),(2.71) making use of analytic properties of the function ((1/𝑌⁺(𝑞,𝜉))−1) by Cauchy Theorem we have 𝐼₁=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑑𝑞+21𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑌⁺(1𝑞,𝜉)−1𝑑𝑞=−21,(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 𝐼₂=−12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑌⁻(1𝑞,𝜉)𝑑𝑞=𝑌⁻(𝑘1(𝜉),𝜉)−1+21.(2.73) Thus, from (2.72) and (2.73) we obtain the following relation for the function 𝐼∶𝐼(𝑘(𝜉),𝜉)=1𝑌⁻(𝑘(𝜉),𝜉)−1.(2.74) Substituting this formula into (2.67) we get 𝑢_𝑥̂(0,𝜉)=𝑢⁰̂(𝑘(𝜉))𝑌⁻(𝑘(𝜉),𝜉)−21𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑘(𝜉)𝑢⁰̂(𝑞)𝑌⁺(𝑞,𝜉)𝐾¹(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₋Ω⁺(𝑝,𝜉)=−𝑌⁺𝑈⁺+𝜉Λ⁺²,(𝑝,𝜉)=−𝜉𝐾₁(𝑝)+𝜉𝑌⁻𝑈⁻+𝜉Λ⁻².(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−)𝑌−₊Ω⁻(𝑝,𝜉)𝑈⁺+𝜉𝐾₁(𝑝)+𝜉𝑌⁻=−𝑌₊𝑈⁻+𝜉Λ⁺²−Λ⁻²𝑈⁺−𝜉𝐾(𝑝)+𝜉𝑈⁻+𝜉Λ⁺²=−1−Λ⁻²2𝐾𝐾(𝑝()𝑝+)𝜉𝑔¹−𝐾(𝑝)(𝑝,𝜉)𝐾(𝑝)+𝜉𝑌⁺(𝑝,𝜉)12𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1𝑞−𝑝𝑌⁺(1𝑞,𝜉)𝑔¹(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 𝑢̂(̂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)+𝜉K(𝑝)Ω⁺=−1(𝑝,𝜉)−Ω⁻(𝑝,𝜉)2𝑔¹(𝑝,𝜉)−𝑌⁺(𝑝,𝜉)12𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1𝑞−𝑝𝑌⁺(1𝑞,𝜉)𝑔¹(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>0ℙ{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 𝑢̂̂∶𝑢̂̂(𝑝,𝜉)=1𝐾(𝑝)+𝜉𝑢⁰̂(𝑝)−𝑢_𝑥̂(0,𝜉)−12𝐾(𝑝1)+𝜉𝑔¹−1(𝑝,𝜉)𝐾(𝑝)+𝜉𝑌⁺(𝑝,𝜉)12𝜋𝑖𝑃𝑉^{𝑖∞−𝑖∞}1𝑞−𝑝𝑌⁺(1𝑞,𝜉)𝑔¹(𝑞,𝜉)𝑑𝑞,(2.80) where the function 𝑔₁(𝑝,𝜉) is given by formula (2.55): 𝑔₁(𝑝,𝜉)=𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑢⁰̂(𝑞)−𝑢_𝑥̂(0,𝜉).(2.81) Now we prove that, in accordance with last relation, the function 𝑢̂(̂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)+𝜉𝑢⁰̂(𝑝)−𝑢_𝑥̂(0,𝜉)=Λ⁺−Λ⁻,1𝐾(𝑝)+𝜉𝑔¹1(𝑝,𝜉)=Λ⁺²−Λ⁻²,2𝜋𝑖PV^{𝑖∞−𝑖∞}1𝑞−𝑝𝑌⁺(1𝑞,𝜉)𝑔¹(𝑞,𝜉)𝑑𝑞=12𝑈⁺+𝑈⁻.(2.82) Substituting these relations into (2.80) we express the function 𝑢̂̂ in the following form: 𝑢̂̂=Λ⁺−Λ⁻−12Λ⁺²−Λ⁻²−12𝐾(𝑝1)+𝜉𝑌⁺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 𝑢̂̂=Λ⁺¹−Λ⁻¹+12Λ⁺²−Λ⁻²−12𝐾(𝑝1)+𝜉𝑌⁺𝑈⁺−12𝐾₁(𝑝1)+𝜉𝑌⁻𝑈⁻.(2.84) Expressing the function 𝑈⁺ in the last equation in terms of 𝑈⁻𝑈⁺=𝑈⁻+1𝑌⁺(𝐾(𝑝)+𝜉)Λ⁺²−Λ⁻²,(2.85) we arrive at the following relation: 𝑢̂̂=Λ⁺¹−Λ⁻¹−1𝐾₁(𝑝)+𝜉𝑌⁻𝑈⁻,(2.86) where by virtue of (2.49) and (2.66), Λ⁺¹−Λ⁻¹=1𝐾₁(𝑝)+𝜉𝑢⁰̂(𝑝)−𝑢_𝑥̂(0,𝜉).(2.87) Thus the function 𝑢̂̂ is the limiting value of an analytic function in Re𝑧>0. Note the fundamental importance of the proven fact that the solution 𝑢̂̂ 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 𝑈(𝑧,𝜉)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑢⁰̂(𝑞)−𝑢_𝑥̂(0,𝜉)𝑑𝑞,(2.88) where 𝑢_𝑥̂(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 𝒞_𝑖 be defined as

𝒞𝒞₁=𝑝∈∞𝑒^{−𝑖(𝜋/2+𝜀)},00,∞𝑒^{𝑖(𝜋/2+𝜀)},(2.90)₂𝒞=𝑞∈∞𝑒^{−𝑖((𝜋/2)+2𝜀)},00,∞𝑒^{𝑖((𝜋/2)+2𝜀)},(2.91)₃=𝑞∈∞𝑒^{−𝑖((𝜋+𝜀)/2)},00,∞𝑒^{𝑖((𝜋+𝜀)/2)},(2.92) where 𝜀>0 can be chosen such that all functions under integration are analytic and Re𝑘(𝜉)>0 for 𝜉∈𝒞₁.

Lemma 2.6. The function 𝐺(𝑥,𝑦,𝑡) given by formula (2.12) has the following representation: 𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖^{𝑖∞−𝑖∞}−1𝑒^{𝑝𝑥−𝐾(𝑝)𝑡}𝑒^−𝑝𝑦𝑑𝑝2𝜋𝑖21𝜋𝑖_𝒞1𝑑𝜉𝑒^𝜉𝑡_𝒞2𝑒^𝑝𝑥𝑒^{Γ(𝑝,𝜉)}𝑤⁺(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾(𝑝)+𝜉𝐼(𝑝,𝜉,𝑦)𝑑𝑝,(2.93) where 𝐼(𝑝,𝜉,𝑦)=12𝜋𝑖^{𝑖∞−𝑖∞}1(𝑞−𝑝)(𝑞−𝑘(𝜉))𝑒^{Γ+(𝑞,𝜉)}𝑤1⁺(𝑞,𝜉)𝑒^−𝑞𝑦Γ(𝑧,𝜉)=1𝑑𝑞,2𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑤⁻(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(𝑥−𝑦,𝑡)+𝐽₂(𝑥,𝑦,𝑡)+𝐽₃(𝑥,𝑦,𝑡)+𝐽₄(𝑥,𝑦,𝑡),(2.95) where 𝐽₁(𝑥−𝑦,𝑡)=12𝜋𝑖^{𝑖∞−𝑖∞}𝐽𝑒^{𝑝(𝑥−𝑦)−𝐾1(𝑝)𝑡}𝑑𝑝,₂(𝑥,𝑦,𝑡)=12𝜋𝑖21𝜋𝑖^{𝑖∞+𝜀−𝑖∞+𝜀}𝑑𝜉𝑒^𝜉𝑡Ψ(𝜉,𝑦)^{𝑖∞−𝑖∞}𝑒^𝑝𝑥1𝐽𝐾₁(𝑝)+𝜉𝑑𝑝,₃(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝑖∞+𝜀−𝑖∞+𝜀}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)𝐾₁(𝑝)+𝜉Υ⁻¹𝐽(𝑝,𝜉,𝑦)𝑑𝑝,(2.96)₄(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝑖∞+𝜀−𝑖∞+𝜀}𝑑𝜉𝑒^𝜉𝑡Ψ(𝜉,𝑦)^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)𝐾₁(𝑝)+𝜉Υ⁻¹(𝑝,𝜉,0)𝑑𝑝.(2.97) Here Υ₁(𝑧,𝜉,𝑦)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝐾¹(𝑞)−𝐾(𝑞)𝐾₁(𝑞)+𝜉𝑒^−𝑞𝑦Ψ(𝜉,𝑦)=−𝑒_{−𝑘(𝜉)𝑦}𝑑𝑞,𝑌⁻(𝑘(𝜉),𝜉)+Υ¹𝑌(𝑘(𝜉),𝜉,𝑦),_±Γ(𝑧,𝜉)=1=𝑒^Γ±𝑤^±,(2.98)2𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑤⁻(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 Υ⁻¹(𝑝,𝜉,𝑦)=𝐼₁(𝑝,𝜉,𝑦)+𝐼₂(𝑝,𝜉,𝑦),(2.100) where 𝐼₁(𝑝,𝜉,𝑦)=lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝑒^−𝑞𝑦𝐼𝑑𝑞,₂(𝑝,𝜉,𝑦)=−lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑒^−𝑞𝑦𝑑𝑞.(2.101) Making use of analytic properties of the functions (1/𝑌⁺(𝑞,𝜉)−1), for Re𝑞<0, and 𝑒^−𝑞𝑦, for Re𝑞>0, by Cauchy theorem we have 𝐼₁(𝑝,𝜉,𝑦)=lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑒^−𝑞𝑦+lim_{𝑧→𝑝,Re𝑧>0}𝑑𝑞12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)−1(𝑒^−𝑞𝑦+lim_{𝑧→𝑝,Re𝑧>0}−1)𝑑𝑞12𝜋𝑖^{𝑖∞−𝑖∞}1=−𝑒_−𝑝𝑦𝑞−𝑧𝑌⁺(1𝑞,𝜉)−1𝑑𝑞+lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)−1(𝑒^−𝑞𝑦−1)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)𝐾𝐾₁((q𝑞))++𝜉𝜉=𝑌⁻(1𝑞,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)=−lim_{𝑧→𝑝,Re𝑧>0}12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁻(1𝑞,𝜉)𝑒^−𝑞𝑦𝑑𝑞=1𝑌⁻(𝑝,𝜉)𝑒^−𝑝𝑦.(2.104) Thus from (2.102) and (2.104) we obtain the following relation: Υ⁻¹(𝑝,𝜉,𝑦)=−𝑒^−𝑝𝑦+Υ⁻(𝑝,𝜉,𝑦)+1𝑌⁻(𝑝,𝜉)𝑒^−𝑝𝑦,(2.105) where Υ(𝑧,𝜉,𝑦)=12𝜋𝑖^{𝑖∞−𝑖∞}1𝑞−𝑧𝑌⁺(1𝑞,𝜉)−1𝑒^−𝑞𝑦𝑑𝑞.(2.106) In the same way (see also proof of relation (2.74)) we can prove that Υ⁻¹(𝑝,𝜉,0)=−1+1𝑌⁻(𝑝,𝜉).(2.107) Also we observe that Ψ(𝜉,𝑦)=−𝑒^{−𝑘(𝜉)𝑦}+Υ(𝑘(𝜉),𝜉,𝑦).(2.108) Inserting into definition (2.96) the expression (2.105) for Υ⁻¹(𝑝,𝜉,𝑦) we obtain the function 𝐽₃(𝑥,𝑦,𝑡) in the form 𝐽₃(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝑖∞−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)𝐾₁(𝑝)+𝜉(−𝑒^−𝑝𝑦+Υ⁻(𝑝,𝜉,𝑦))𝑑𝑝−𝐽₁(𝑥−𝑦,𝑡).(2.109) Replacing in formula (2.97) the functions Ψ(𝜉,𝑦) and Υ⁻¹(𝑝,𝜉,0) by (2.108) and (2.107), respectively, we reduce the function 𝐽₄(𝑥,𝑦,𝑡) in the form 𝐽₄(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝑖∞−𝑖∞}𝑑𝜉𝑒^𝜉𝑡Ψ(𝜉,𝑦)^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)=−1𝐾₁(𝑝)+𝜉−1+𝑌⁻(1𝑝,𝜉)𝑑𝑝2𝜋𝑖21𝜋𝑖^{𝑖∞−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)𝐾₁(𝑝)+𝜉𝑒^{−𝑘(𝜉)𝑦}−Υ(𝑘(𝜉),𝜉,𝑦)𝑑𝑝−𝐽₂(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 𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖21𝜋𝑖^{𝑖∞−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)𝐾₁(𝑝)+𝜉Ξ¹(𝑝,𝜉,𝑦)𝑑𝑝,(2.111) where Ξ₁(𝑝,𝜉,𝑦)=𝑒^−𝑝𝑦−𝑒^{−𝑘(𝜉)𝑦}−(Υ⁻(𝑝,𝜉,𝑦)−Υ(𝑘(𝜉),𝜉,𝑦)).(2.112) Also, note that since Υ⁻(𝑝=,𝜉,𝑦)li−mΥ(_{𝑧→𝑝,Re𝑧>0}𝑘(𝜉),𝜉,𝑦)12𝜋𝑖^{𝑖∞−𝑖∞}𝑞−1𝑧−𝑞−1𝑘(𝜉)𝑌⁺(1𝑞,𝜉)−1𝑒^−𝑞𝑦=lim_{𝑧→𝑝,Re𝑧>0}𝑑𝑞12𝜋𝑖^{𝑖∞−𝑖∞}(𝑞−𝑧𝑧)−(𝑞𝑘−(𝜉𝑘)(𝜉))𝑌⁺(1𝑞,𝜉)−1𝑒^−𝑞𝑦𝑑𝑞,(2.113) we obtain 𝑒=^−𝑝𝑦−𝑒lim^{−𝑘(𝜉)𝑦}_{𝑧→𝑝,Re𝑧>0}−(Υ⁻(𝑝,𝜉,𝑦)−Υ(𝑘(𝜉),𝜉,𝑦))12𝜋𝑖^{𝑖∞−𝑖∞}−𝑞−1𝑧+𝑞−1𝑘(𝜉)𝑒^−𝑞𝑦+1=lim_{𝑧→𝑝,Re𝑧>0}𝑌⁺(𝑞,𝜉)−1𝑒^−𝑞𝑦𝑑𝑞12𝜋𝑖^{𝑖∞−𝑖∞}(𝑞−𝑘𝑧)(𝜉(𝑞)−−𝑧𝑘(𝜉))𝑌⁺(1𝑞,𝜉)𝑒^−𝑞𝑦𝑑𝑞.(2.114) So, 𝐺(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁻(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾₁(𝑝)+𝜉𝐼⁻(𝑝,𝜉,𝑦)𝑑𝑝,(2.115) where 𝐼(𝑧,𝜉,𝑦)=12𝜋𝑖^{𝑖∞−𝑖∞}1(𝑞−𝑧)(𝑞−𝑘(𝜉))𝑌⁺(1𝑞,𝜉)𝑒^−𝑞𝑦𝑑𝑞.(2.116) Using relation 𝐾(𝑝)+𝜉𝐾₁(𝑝)+𝜉=𝑌𝑌⁺⁻,(2.117) we rewrite last formula in the following form: 𝐺(𝑥,𝑦,𝑡)=−12𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁺(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾(𝑝)+𝜉𝐼⁻(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⁺(𝑝,𝜉,𝑦)−1(𝑝−𝑘(𝜉))𝑌⁺(1𝑝,𝜉)𝑒^−𝑝𝑦.(2.119) So we get 𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖^{𝑖∞−𝑖∞}−1𝑒^{𝑝𝑥−𝐾(𝑝)𝑡}𝑒^−𝑝𝑦𝑑𝑝2𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑌⁺(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾(𝑝)+𝜉𝐼⁺(𝑝,𝜉,𝑦)𝑑𝑝.(2.120) Now we consider for Re𝜉>0Γ⁺(𝑝,𝜉)=12𝜋𝑖lim^{𝑧→𝑝Re𝑧<0}^{𝑖∞−𝑖∞}1𝑞−𝑧ln𝐾𝐾₁((𝑞𝑞))++𝜉𝜉𝑤⁻(𝑞)=1𝑤⁺(𝑞)𝑑𝑞2𝜋𝑖^{∞𝑒𝑖((𝜋/2)+3𝜀)}₀1𝑞−𝑝ln𝑖𝑞²+𝜉−𝑞²+𝜉𝑤⁻(𝑞)+1𝑤⁺(𝑞)𝑑𝑞2𝜋𝑖^0∞𝑒_{𝑖((−𝜋/2)−3𝜀)}1𝑞−𝑝ln−𝑖𝑞²+𝜉−𝑞²+𝜉𝑤⁻(𝑞)+ln𝐾(𝑝)+𝜉𝑤⁺(𝑞)𝑑𝑞𝐾₁(𝑝)+𝜉𝑤⁻(𝑝)=Γ₁𝑤⁺(𝑝)(𝑝,𝜉)+ln𝐾(𝑝)+𝜉𝐾₁(𝑝)+𝜉𝑤⁻(𝑝)𝑤⁺(𝑝).(2.121) Note that Γ₁(𝑝,𝜉) is analytic in domain 0≤arg𝜉<(𝜋/2)+3𝜀, (−𝜋/2)−3𝜀<arg𝜉≤0, and 𝜋/2≤arg𝑝<(𝜋/2)+3𝜀 and (−𝜋/2)−3𝜀<arg𝑝≤−𝜋/2. So we get 𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖^{𝑖∞−𝑖∞}−1𝑒^{𝑝𝑥−𝐾(𝑝)𝑡}𝑒^−𝑝𝑦𝑑𝑝2𝜋𝑖21𝜋𝑖^{𝜀+𝑖∞𝜀−𝑖∞}𝑑𝜉𝑒^𝜉𝑡^{𝑖∞−𝑖∞}𝑒^𝑝𝑥𝑒^{Γ1(𝑝,𝜉)}𝑤⁻(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾₁(𝑝)+𝜉𝐼⁺(𝑝,𝜉,𝑦)𝑑𝑝,(2.122) where 𝐼⁺(𝑝,𝜉,𝑦)=12𝜋𝑖lim^{𝑧→𝑝Re𝑧<0}^{𝑖∞−𝑖∞}1(𝑞−𝑧)(𝑞−𝑘(𝜉))𝑤⁻𝑒1^{Γ1(𝑞,𝜉)}𝐾₁(𝑞)+𝜉𝐾(𝑞)+𝜉𝑒^−𝑞𝑦𝑑𝑞.(2.123) Changing the contour of integration with respect to 𝜉 by Cauchy theorem we get 𝐺(𝑥,𝑦,𝑡)=12𝜋𝑖^{𝑖∞−𝑖∞}−1𝑒^{𝑝𝑥−𝐾(𝑝)𝑡}𝑒^−𝑝𝑦𝑑𝑝2𝜋𝑖21𝜋𝑖_𝒞1d𝜉𝑒^𝜉𝑡_𝒞2𝑒^𝑝𝑥𝑒^{Γ1(𝑝,𝜉)}𝑤⁻(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾₁(𝑝)+𝜉𝐼⁺−res_{𝜉=−𝐾(𝑞)}(𝑝,𝜉,𝑦)𝑑𝑝_𝒞2𝑒^𝑝𝑥𝑒^{Γ1(𝑝,𝜉)}𝑤⁻(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾₁(𝑝)+𝜉𝐼⁺(𝑝,𝜉,𝑦)𝑑𝑝,(2.124) where res_{𝜉=−𝐾(𝑞)}_𝒞2𝑒^𝑝𝑥𝑒^{Γ1(𝑝,𝜉)}𝑤⁻(𝑝,𝜉)(𝑝−𝑘(𝜉))𝐾₁(𝑝)+𝜉𝐼⁺(𝑝,𝜉,𝑦)𝑑𝑝=12𝜋𝑖_𝒞2𝑒^𝑝𝑥𝑑𝑝_𝒞3𝑑𝑞𝑒^{−𝑞𝑦−𝐾(𝑞)𝑡}𝜙(𝑝,𝑞)=𝑒_Γ1(𝑝𝑞−𝑝𝜙(𝑝,𝑞),