Benjamin-Ono Equation on a Half-Line

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.


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: 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∂ 2 x ∂ x −∂ 2 x 1/2 since the operator H can be written as follows: H 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 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 ∂ β x u. Thus, for example, in the case of K p p 3/2 we get the following definition of the fractional derivative ∂ 3/2 x : Here and below p β is the main branch of the complex analytic function in the complex halfplane Re p ≥ 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 Re p > 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| σ u.
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 halfline 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 . International Journal of Mathematics and Mathematical Sciences 3 To state precisely the results of the present paper we give some notations. We denote t Sobolev space is We define a linear functional f: Now we state the main results.
Theorem 1.1. Suppose that the initial data u 0 ∈ Z ≡ H 1 R ∩ L 1,a 1 R with a ∈ 0, 1 are such that the norm is sufficiently small. Then there exists a unique global solution to the initial-boundary value problem 1.1 . Moreover the following asymptotic is valid in L ∞ R :

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: Then Cauchy type integral 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 Re p 0, on approaching the contour from the left and from the right, respectively. These limiting values are expressed by Sokhotzki-Plemelj formula:

2.3
International Journal of Mathematics and Mathematical Sciences 5 Subtracting and adding the formula 2.3 we obtain the following two equivalent formulas: which will be frequently employed hereafter.
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 Theorem 2.3. An arbitrary function ϕ p given on the contour Re p 0, satisfying the Hölder condition, and having zero index, 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 We consider the following linear initial-boundary value problem on half-line

2.9
Setting K q − q q, where the function G x, y, t is given by formula for ε > 0, x > 0, y > 0, t > 0. Here and below Γ p, ξ and Γ − p, ξ are a left and right limiting values of sectionally analytic function Γ z, ξ given by 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 Re z 0. Also by Sokhotzki-Plemelj formula we have for Re p 0

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 Since L{u} is analytic for all Re q > 0, we have u q, t L{u} P u p, t .

2.21
Therefore applying the Laplace transform with respect to x to problem 2.9 we obtain for t > 0 We rewrite 2.22 in the form Applying the Laplace transformation with respect to time variable to problem 2.24 we find for Re p > 0 Here the functions u p, ξ , Φ p, ξ , u 0, ξ , and u x 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 u in the right-half complex planes Re p > 0 and Re ξ > 0. We have for Re p 0 In view of Sokhotzki-Plemelj formula via 2.27 the condition 2.28 can be written as where the sectionally analytic functions Θ z, ξ and Λ z, ξ are given by Cauchy type integrals: To perform the condition 2.29 in the form of nonhomogeneous Riemann problem we introduce the sectionally analytic function: International Journal of Mathematics and Mathematical Sciences 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 Also observe that from 2.30 and 2.32 by formula 2.4 Substituting 2.29 and 2.34 into this equation we obtain nonhomogeneous Riemann problem It is required to find two functions for some fixed point ξ, Re ξ > 0: Ω z, ξ , analytic in Re z < 0 and Ω − z, ξ , analytic in Re z > 0, which satisfy on the contour Re p 0 the relation 2.36 . Here, for some fixed point ξ, Re ξ > 0, the functions 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 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 Re p 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 International Journal of Mathematics and Mathematical Sciences we introduce the function where for some fixed point k ξ Re k ξ > 0 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 , K 1 z are analytic for Re z > 0 and the function w z is analytic for Re z < 0. We observe that the function W p, ξ given on the contour Re p 0 satisfies the Hölder condition and under the assumption Re K 1 p > 0 does not vanish for any Re ξ > 0. Also we have Therefore in accordance with Theorem 2.3 the function W p, ξ can be represented in the form of the ratio Now we return to the nonhomogeneous Riemann problem 2.36 . Multiplying and dividing the expression K p ξ /ξ by 1/ K 1 p ξ w − p /w p and making use of the formula 2.43 we get 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 Now we perform the function Λ z, ξ given by formula 2.31 as Firstly we calculate the left limiting value Λ 1 p, ξ . Since there exists only one root k ξ of equation K 1 z −ξ such that Re k ξ > 0 for all Re ξ > 0, therefore, taking limit z → p from the left-hand side of complex plane, by Cauchy theorem we get The last relation implies that K 1 p ξ Λ 1 p, ξ can be expressed by the function Λ 3 z, ξ which is analytic in Re z > 0 : 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 Replacing in 2.47 −K p Λ p, ξ by 2.56 , we reduce the nonhomogeneous Riemann problem 2.47 in the form where

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
also satisfy this condition.
Since g 1 p, ξ satisfies on Re p 0 the Hölder condition, on basis of this Lemma the function 1/Y p, ξ g 1 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 International Journal of Mathematics and Mathematical Sciences 13 Therefore the problem 2.58 takes the form The last relation indicates that the function Ω 1 /Y U , analytic in Re z < 0, and the function Ω − 1 /Y − U − , analytic in Re z > 0, constitute the analytic continuation of each other through the contour Re z 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

2.63
Since there exists only one root k ξ of equation K 1 z −ξ such that Re k ξ > 0 for all Re ξ > 0, therefore, in the expression for the function Ω − z, ξ the factor ξ/ K 1 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 We reduce 2.64 to the form 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: 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 , Now we prove that the coefficient of u x 0, ξ does not vanish for all Re ξ > 0. We represent the function I k ξ , ξ in the form where

2.70
Since, for Re z / 0, making use of analytic properties of the function 1/Y q, ξ − 1 by Cauchy Theorem we have where ξ is some fixed point, Re ξ > 0. To calculate the function I 2 we will use the identity 2.43 . Observe that the function 1/Y − q, ξ is analytic for all Re q > 0. Therefore, setting the relation 2.43 into definition of I 2 and making use of Cauchy Theorem we find Thus, from 2.72 and 2.73 we obtain the following relation for the function I : Substituting this formula into 2.67 we get

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

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 Re z > 0 P{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 u :

2.80
where the function g 1 p, ξ is given by formula 2.55 : Now we prove that, in accordance with last relation, the function u p, ξ constitutes the limiting value of an analytic function in Re z > 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 Λ 2 z, ξ , and making use of Sokhotzki-Plemelj formula 2.3 we have

2.82
Substituting these relations into 2.80 we express the function u in the following form: If it is taken into account that Λ z, ξ Λ 1 z, ξ Λ 2 z, ξ by virtue of the relation 2.45 , the last expression agrees with formula Expressing the function U in the last equation in terms of U − we arrive at the following relation: where by virtue of 2.49 and 2.66 , Thus the function u is the limiting value of an analytic function in Re z > 0. Note the fundamental importance of the proven fact that the solution u constitutes an analytic function in Re z > 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 where u x 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 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 C 1 p ∈ ∞e −i π/2 ε , 0 0, ∞e i π/2 ε , 2.90 where ε > 0 can be chosen such that all functions under integration are analytic and Re k ξ > 0 for ξ ∈ C 1 .

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

2.97
Here

2.99
Firstly we consider the sectionally analytic function Υ 1 z, ξ, y given by Cauchy type integral 2.98 .

2.101
Making use of analytic properties of the functions 1/Y q, ξ − 1 , for Re q < 0, and e −qy , for Re q > 0, by Cauchy theorem we have

2.102
where ξ is some fixed point, Re ξ > 0. To calculate the function I 2 p, ξ, y we will use the following identity: Observe that the function 1/Y − q, ξ is analytic for all Re q > 0. Therefore, setting the relation 2.103 into definition of I 2 p, ξ, y and making use of Cauchy theorem we find In the same way see also proof of relation 2.74 we can prove that 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 J 3 x, y, t in the form

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 J 4 x, y, t in the form

2.112
International Journal of Mathematics and Mathematical Sciences 21 Also, note that since

2.113
we obtain

2.116
Using relation we rewrite last formula in the following form: i∞ −i∞ e px−K p t e −py dp K p ξ I p, ξ, y dp.

2.120
Now we consider for Re ξ > 0

2.127
Thus using relation 2.121 we get relation 2.93 . Lemma is proved.

2.144
Here γ ∈ 0, 1 , K q −q|q| −q 2 exp −iθ , θ arg q and After this observation in accordance with the integral representation 2.133 by the Hölder inequality we have arrived at the following estimate for n 0, 1 where Here we used that since w q, ξ e Γ q,ξ is analytic for p ∈ C 2 , ξ ∈ C 1 , and Re k ξ > 0 by Cauchy theorem,

2.149
Substituting 2.147 into definition of the function J 2 x, y, t see 2.133 we get e px Y p, ξ p − k ξ K p ξ I 1 p, ξ, y dp e px Y p, ξ K p ξ dp.

2.150
Since by Cauchy theorem e px Y p, ξ p − k ξ K p ξ I 1 p, ξ, y dp, 2.155 F j x, y, t φ y dy.

2.157
Now we prove estimate We have e |p|p e pxt −1/2 e −pyt −1/2 − 1 dp Here we used the following estimations: Thus after the change of variables ξt q 1 , p z √ t, q q 2 √ t we get

Proof of Theorem 1.1
By Proposition 2.4 we rewrite the initial-boundary value problem 1.1 as the following integral equation: where G is the Green operator of the linear problem 2.9 . We choose the space with 0 < a < 1 being small and the space X ∂ n x φ ∈ C 0, ∞ ; L 2 R 0, ∞ ; L 2,n a/2 R : φ X < ∞ , n 0, 1, 3.3 where now the norm μ ∈ 0, a φ X sup t≥0 1 n 0 t 1−μ /4 t 1/2 ∂ n x φ t L 2,n μ/2 t 1/2 n 3/2 ∂ n x φ t L 2 3.4 for all t > 1. Also in view of the definition of the norm X we have f N u τ ≤ N u τ L 1,1 ≤ C{τ} −1/2 τ −3/2 u 2 X .

3.23
By a direct calculation we have for some small γ 1 > 0, γ > 0 for all t > 1. By virtue of the integral equation 3.1 we get

3.27
The all summands in the right-hand side of 3.27 are estimated by C u 0 Z C u 2 X via estimates 3.24 -3.26 . Thus by 3.27 the asymptotic 3.19 is valid. Theorem is proved.