# Benjamin-Ono Equation on a Half-Line

Authors: Nakao Hayashi; Elena I. Kaikina
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## 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.


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## Body

## 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:(1.1)ut+uux+ℋuxx=0,x>0,t>0,u(x,0)=u0(x),x>0,u(0,t)=0,t>0,
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∂x2=∂x(-∂x2)1/2 since the operator ℋ can be written as follows: ℋ=-ℱ-1(iξ/|ξ|)ℱ=(-∂x2)-1/2∂x, where (ℱφ)(ξ)=(1/2π)∫φ(x)e-ixξdx 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 Hs,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 H2,0∩H1,1 solutions decay as t→∞ in L∞ norm at the same rate 1/t as for the case of the linear Benjamin-Ono equation, where(1.2)Hm,s={ϕ∈L2:∥ϕ∥m,s=∥(1+x2)s/2(1-∂x2)m/2ϕ∥L2<∞}.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:(1.3)∂x3/2ϕ=ℒ-1{p3/2(ℒϕ-ϕ(0)p)}.
Here and below pβ is the main branch of the complex analytic function in the complex half-plane Rep≥0, 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 symbolsK(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 ℒx→ξ is(1.4)û(ξ)≡ℒx→ξu=∫0+∞e-ξxu(x)dx,
and the inverse Laplace transformation ℒξ→x-1 is defined by(1.5)u(x)≡ℒξ→x-1û=(2πi)-1∫-i∞i∞eξxû(ξ)dξ.
Weighted Lebesgue space is Lq,a(R+)={φ∈𝒮′;∥φ∥Lq,a<∞}, where(1.6)∥φ∥Lq,a=(∫0+∞xaq|φ(x)|qdx)1/q
for a>0, 1≤q<∞ and(1.7)∥φ∥L∞=esssupx∈R+|φ(x)|.
Sobolev space is(1.8)H1(R+)={φ∈𝒮′;∥〈∂x〉φ∥L2<∞}.
We define a linear functional f:(1.9)f(ϕ)=∫0+∞yϕ(y)dy.
Now we state the main results.Theorem 1.1.
Suppose that the initial datau0∈Z≡H1(R+)∩L1,a+1(R+) with a∈(0,1) are such that the norm
(1.10)∥u0∥Z≤ε
is sufficiently small. Then there exists a unique global solution
(1.11)u∈C([0,∞);H1(R+))
to the initial-boundary value problem (1.1). Moreover the following asymptotic is valid in L∞(R+):(1.12)u=1tAΛ(xt-1/2)+min(1,xt)O(t-1-(a/2))
for t→∞, where Λ(xt-1/2)∈L∞(R+),Λ(0)=0 is defined below by the formula (2.191), and the constant
(1.13)A=f(u0)-∫0+∞f(𝒩(u))dτ,𝒩(u)=uxu.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 nonlinearityuux in (1.1) is not the super critical case for our problem.Remark 1.3.
In the case of the negative half linex<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.
Letϕ(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:
(2.1)|ϕ(q)-ϕ∞|≤C|q|-μ,μ>0.
Then Cauchy type integral
(2.2)F(z)=12πi∫-i∞i∞ϕ(q)q-zdq
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:
(2.3)F+(p)=limz→p,Rez<012πi∫-i∞i∞ϕ(q)q-zdq=12πiPV∫-i∞i∞ϕ(q)q-pdq+12ϕ(p),F-(p)=limz→p,Rez>012πi∫-i∞i∞ϕ(q)q-zdq=12πiPV∫-i∞i∞ϕ(q)q-pdq-12ϕ(p).Subtracting and adding the formula (2.3) we obtain the following two equivalent formulas:(2.4)F+(p)-F-(p)=ϕ(p),F+(p)+F-(p)=1πiPV∫-i∞i∞ϕ(q)q-pdq,
which will be frequently employed hereafter.Theorem 2.2.
An arbitrary functionϕ(p) given on the contour Rep=0, satisfying the Hölder condition, can be uniquely represented in the form
(2.5)ϕ(p)=U+(p)-U-(p),
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
(2.6)U(z)=12πi∫-i∞i∞ϕ(q)q-zdq.Theorem 2.3.
An arbitrary functionφ(p) given on the contour Rep=0, satisfying the Hölder condition, and having zero index,
(2.7)indφ(t):=12πi∫-i∞i∞dlnφ(p)=0,
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
(2.8)X±(z)=eΓ±(z),Γ(z)=12πi∫-i∞i∞1q-zlnφ(q)dq.We consider the following linear initial-boundary value problem on half-line(2.9)ut-PV∫0+∞uyy(y,t)x-ydy=0,t>0,x>0,u(x,0)=u0(x),x>0,u(0,t)=0,t>0.Setting(2.10)K(q)=-|q|q,K1(q)=-q2,k(ξ)=|ξ|1/2e(1/2)iargξ,whereRek(ξ)>0 for Reξ>0, we define(2.11)𝒢(t)ϕ=∫0+∞G(x,y,t)ϕ(y)dy,where the functionG(x,y,t) is given by formula(2.12)G(x,y,t)=12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞dpepx1K1(p)+ξ(e-py+Ψ(ξ,y))-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞dpepx1K1(p)+ξY-(p,ξ)×limz→p,Rez>012πi∫-i∞i∞1q-z1Y+(q,ξ)K1(q)-K(q)K1(q)+ξ(e-qy+Ψ(ξ,y))dq
for ε>0,x>0,y>0,t>0. Here and below(2.13)Y±=eΓ±w±.Γ+(p,ξ) and Γ-(p,ξ) are a left and right limiting values of sectionally analytic function Γ(z,ξ)  given by(2.14)Γ(z,ξ)=12πi∫-i∞i∞1q-zln{(K(q)+ξK1(q)+ξ)w-(q)w+(q)}dq,
where(2.15)w-(z)=(zz+k(ξ))1/2,w+(z)=(zz-k(ξ))1/2,Ψ(ξ,y)=-e-k(ξ)yY-(k(ξ),ξ)+12πi∫-i∞i∞1q-k(ξ)1Y+(q,ξ)K1(q)-K(q)K1(q)+ξe-qydq.All the integrals are understood in the sense of the principal values.Proposition 2.4.
Let the initial data beu0∈L1(R+). Then there exists a unique solution u(x,t) of the initial-boundary value problem (2.9), which has integral representation
(2.16)u(x,t)=𝒢(t)u0.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:
(2.17)u(x,t)=0,∀x<0.
Letϕ(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
(2.18)ℙϕ(z)=-12πi∫-i∞i∞1q-zϕ(q)dq.
Since the operatorℙ 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=0(2.19)ℙ+ϕ=12πiPV∫-i∞i∞1q-pϕ(q)dq+12ϕ(p),ℙ-ϕ=12πiPV∫-i∞i∞1q-pϕ(q)dq-12ϕ(p).
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(2.20)ℒ{ℋuxx}=ℙ{-|p|p(ℒ{u}-u(0,t)p-ux(0,t)p2)}.
Since ℒ{u} is analytic for all Req>0, we have
(2.21)û(q,t)=ℒ{u}=ℙû(p,t).
Therefore applying the Laplace transform with respect to x to problem (2.9) we obtain for t>0(2.22)ℙ{ût+K(p)û(p,t)-K(p)pu(0,t)-K(p)p2ux(0,t)}=0,û(p,0)=û0(p),
where
(2.23)K(p)=-|p|p.
We rewrite (2.22) in the form
(2.24)ût+K(p)û(p,t)-K(p)pu(0,t)-K(p)p2ux(0,t)=Φ(p,t),û(p,0)=û0(p),
with some function Φ(p,t) such that for all Rep>0(2.25)ℙ{Φ(p,t)}=0
and for |p|>1(2.26)|Φ(p,t)|≤C1|p|.
Applying the Laplace transformation with respect to time variable to problem (2.24) we find for Rep>0(2.27)û̂(p,ξ)=1K(p)+ξ(û0(p)+K(p)pû(0,ξ)+K(p)p2ûx(0,ξ)+Φ̂(p,ξ)).
Here the functions û̂(p,ξ),Φ̂(p,ξ),û(0,ξ), and ûx(0,ξ) are the Laplace transforms for û(p,t),Φ(p,t),u(0,t), and ux(0,t) with respect to time, respectively. We will find the function Φ̂(p,ξ) using the analytic properties of function û̂ in the right-half complex planes Rep>0 and Reξ>0. We have for Rep=0(2.28)û̂(p,ξ)=1πiPV∫-i∞i∞1q-pû̂(q,ξ)dq.
In view of Sokhotzki-Plemelj formula via (2.27) the condition (2.28) can be written as
(2.29)Θ+(p,ξ)=-Λ+(p,ξ),
where the sectionally analytic functions Θ(z,ξ) and Λ(z,ξ) are given by Cauchy type integrals:
(2.30)Θ(z,ξ)=12πi∫-i∞i∞1q-z1K(q)+ξΦ̂(q,ξ)dq,(2.31)Λ(z,ξ)=12πi∫-i∞i∞1q-z1K(q)+ξ(û0(q)+K(q)qû(0,ξ)+K(q)q2ûx(0,ξ))dq.
To perform the condition (2.29) in the form of nonhomogeneous Riemann problem we introduce the sectionally analytic function:
(2.32)Ω(z,ξ)=12πi∫-i∞i∞1q-zΨ(q,ξ)dq,
where
(2.33)Ψ(p,ξ)=K(p)K(p)+ξΦ̂(p,ξ).
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)(2.34)Ω-(p,ξ)=-ξΘ-(p,ξ).
Also observe that from (2.30) and (2.32) by formula (2.4)
(2.35)K(p)(Θ+(p,ξ)-Θ-(p,ξ))=Ψ(p,ξ)=Ω+(p,ξ)-Ω-(p,ξ).
Substituting (2.29) and (2.34) into this equation we obtain nonhomogeneous Riemann problem
(2.36)Ω+(p,ξ)=K(p)+ξξΩ-(p,ξ)-K(p)Λ+(p,ξ).
It is required to find two functions for some fixed pointξ, 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
(2.37)W(p,ξ)=K(p)+ξξ,g(p,ξ)=-K(p)Λ+(p,ξ)
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 Φ̂(p,ξ) which involved in the formula (2.27) by the relation
(2.38)Φ̂(p,ξ)=K(p)+ξK(p)(Ω+(p,ξ)-Ω-(p,ξ)).
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 Theorems2.2 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
(2.39)K1(p)=-p2,
we introduce the function
(2.40)W̃(p,ξ)=(K(p)+ξK1(p)+ξ)w-(p)w+(p),
where for some fixed point k(ξ) (Rek(ξ)>0)
(2.41)w-(z)=(zz+k(ξ))1/2,w+(z)=(zz-k(ξ))1/2.
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 functionW̃(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
(2.42)Ind.W̃(p,ξ)=12πi∫-i∞i∞dlnW̃(p,ξ)=0.
Therefore in accordance with Theorem 2.3  the function W̃(p,ξ) can be represented in the form of the ratio
(2.43)W̃(p,ξ)=X+(p,ξ)X-(p,ξ),
where
(2.44)X±(p,ξ)=eΓ±(p,ξ),Γ(z,ξ)=12πi∫-i∞i∞1q-zlnW̃(q,ξ)dq.
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
(2.45)W(p,ξ)=K(p)+ξξ=Y+(p,ξ)Y-(p,ξ)(K1(p)+ξξ),
where
(2.46)Y±(p,ξ)=X±(p,ξ)w±(p).
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
(2.47)Ω+(p,ξ)Y+(p,ξ)=(K1(p)+ξξ)Ω-(p,ξ)Y-(p,ξ)-1Y+(p,ξ)K(p)Λ+(p,ξ).
Now we perform the function Λ(z,ξ) given by formula (2.31) as
(2.48)Λ(z,ξ)=Λ1(z,ξ)+Λ2(z,ξ),
where
(2.49)Λ1(z,ξ)=12πi∫-i∞i∞1q-z1K1(q)+ξ(û0(q)+K1(q)qû(0,ξ)+K1(q)q2ûx(0,ξ))dq,(2.50)Λ2(z,ξ)=12πi∫-i∞i∞1q-zK1(q)-K(q)(K(q)+ξ)(K1(q)+ξ)(û0(q)-ξqû(0,ξ)-ûx(0,ξ))dq.
Firstly we calculate the left limiting value Λ1+(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
(2.51)Λ1+(p,ξ)=-k′(ξ)p-k(ξ)(û0(k(ξ))-ξpû(0,ξ)-ûx(0,ξ)).
The last relation implies that (K1(p)+ξ)Λ1+(p,ξ) can be expressed by the function Λ3(z,ξ) which is analytic in Rez>0:(2.52)(K1(p)+ξ)Λ1+(p,ξ)=Λ3-(p,ξ),
where
(2.53)Λ3(z,ξ)=-k′(ξ)(K1(z)+ξz-k(ξ))(û0(k(ξ))-ξpû(0,ξ)-ûx(0,ξ)).
By Sokhotzki-Plemelj formula (2.4) we express the left limiting value Λ2+(p,ξ) in the term of the right limiting value Λ2-(p,ξ) as
(2.54)Λ2+(p,ξ)=Λ2-(p,ξ)+g̃1(p,ξ),
where
(2.55)g̃1(p,ξ)=K1(p)-K(p)(K(p)+ξ)(K1(p)+ξ)(û0(p)-ξpû(0,ξ)-ûx(0,ξ)).
Bearing in mind the representation (2.48) and making use of (2.52), (2.55), and (2.45) after simple transformations we get
(2.56)-K(p)Λ+=-Y+Y-[Λ3-+(K1(p)+ξ)Λ2-]+ξΛ+-g1(p,ξ),
where
(2.57)g1(p,ξ)=(K(p)+ξ)g̃1(p,ξ)=K1(p)-K(p)K1(p)+ξ(û0(p)-ξpû(0,ξ)-ûx(0,ξ)).
Replacing in (2.47) -K(p)Λ+(p,ξ) by (2.56), we reduce the nonhomogeneous Riemann problem (2.47) in the form
(2.58)Ω1+(p,ξ)Y+(p,ξ)=Ω1-(p,ξ)Y-(p,ξ)-1Y+(p,ξ)g1(p,ξ),
where
(2.59)Ω1+(p,ξ)=Ω+(p,ξ)-ξΛ+(p,ξ),Ω1-(p,ξ)=(K1(p)+ξ)(ξ-1Ω-(p,ξ)-Λ2-(p,ξ))-Λ3-(p,ξ).
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.
IfL 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
(2.60)Φ(z)=12πi∫L1q-zϕ(q)dq
also satisfy this condition.
Sinceg1(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
(2.61)U(z,ξ)=12πi∫-i∞i∞1q-z1Y+(q,ξ)g1(q,ξ)dq.
Therefore the problem (2.58) takes the form
(2.62)Ω1+(p,ξ)Y+(p,ξ)+U+(p,ξ)=Ω1-(p,ξ)Y-(p,ξ)+U-(p,ξ).
The last relation indicates that the function (Ω1+/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
(2.63)Ω+(p,ξ)=Y+(A-U+)+ξΛ+,Ω-(p,ξ)=ξK1(p)+ξY-(A-U-)+ξ(Λ1++Λ2-).
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 ξΛ1+ 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 ξΛ1+ satisfy additional conditions. For analyticity of Ω-(z,ξ) in points z=k(ξ) it is necessary that
(2.64)Resp=k(ξ){1K1(p)+ξY-(A-U-)+Λ1+}=0.
We reduce (2.64) to the form
(2.65)AY-(k(ξ),ξ)-Y-(k(ξ),ξ)U-(k(ξ),ξ)+(-ξk(ξ)û(0,ξ)+û0(k(ξ))-ûx(0,ξ))=0.
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:
(2.66)Y-(k(ξ),ξ)U-(k(ξ),ξ)+ξk(ξ)û(0,ξ)-û0(k(ξ))+ûx(0,ξ)=0.
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),(2.67)-[Y-(k(ξ),ξ)I(k(ξ),ξ)-1]ûx(0,ξ)=û0(k(ξ))-Y-(k(ξ),ξ)12πi∫-i∞i∞1q-k(ξ)û0(q)Y+(q,ξ)K1(q)-K(q)K1(q)+ξdq,
where
(2.68)I(k(ξ),ξ)=12πi∫-i∞i∞1q-k(ξ)1Y+(q,ξ)K1(q)-K(q)K1(q)+ξdq.
Now we prove that the coefficient of ûx(0,ξ) does not vanish for all Reξ>0. We represent the function I(k(ξ),ξ) in the form
(2.69)I=I1+I2,
where
(2.70)I1=12πi∫-i∞i∞1q-k(ξ)1Y+(q,ξ)dq,I2=-12πi∫-i∞i∞1q-k(ξ)1Y+(q,ξ)K(q)+ξK1(q)+ξdq.
Since, for Rez≠0,(2.71)∫-i∞i∞1q-zdq=-πisgn(Rez),
making use of analytic properties of the function ((1/Y+(q,ξ))-1) by Cauchy Theorem we have
(2.72)I1=12πi∫-i∞i∞1q-k(ξ)dq+12πi∫-i∞i∞1q-k(ξ)(1Y+(q,ξ)-1)dq=-12,
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
(2.73)I2=-12πi∫-i∞i∞1q-k(ξ)1Y-(q,ξ)dq=(1Y-(k(ξ),ξ)-1)+12.
Thus, from (2.72) and (2.73) we obtain the following relation for the function I:(2.74)I(k(ξ),ξ)=1Y-(k(ξ),ξ)-1.
Substituting this formula into (2.67) we get
(2.75)ûx(0,ξ)=û0(k(ξ))Y-(k(ξ),ξ)-12πi∫-i∞i∞1q-k(ξ)û0(q)Y+(q,ξ)K1(q)-K(q)K1(q)+ξdq.
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
(2.76)Ω+(p,ξ)=-Y+U++ξΛ2+,Ω-(p,ξ)=-ξK1(p)+ξY-U-+ξΛ2-.
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
(2.77)Ω+(p,ξ)-Ω-(p,ξ)=-Y+U++ξK1(p)+ξY-U-+ξ(Λ2+-Λ2-)=-Y+(U+-ξK(p)+ξU-)+ξ(Λ2+-Λ2-)=-12K(p)K(p)+ξg1(p,ξ)-K(p)K(p)+ξY+(p,ξ)12πiPV∫-i∞i∞1q-p1Y+(q,ξ)g1(q,ξ)dq.
We now proceed to find the unknown function Φ̂(p,ξ) involved in the formula (2.27) for the solution û̂(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)Φ̂(p,ξ)=K(p)+ξK(p)(Ω+(p,ξ)-Ω-(p,ξ))=-12g1(p,ξ)-Y+(p,ξ)12πiPV∫-i∞i∞1q-p1Y+(q,ξ)g1(q,ξ)dq.
It is easy to observe that Φ̂(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(2.79)ℙ{Φ}=0.
Having determined the function Φ̂(p,ξ) bearing in mind formula (2.27) and conditions u(0,t)=0 we determine required function û̂:(2.80)û̂(p,ξ)=1K(p)+ξ(û0(p)-ûx(0,ξ))-121K(p)+ξg1(p,ξ)-1K(p)+ξY+(p,ξ)12πiPV∫-i∞i∞1q-p1Y+(q,ξ)g1(q,ξ)dq,
where the function g1(p,ξ) is given by formula (2.55):
(2.81)g1(p,ξ)=K1(q)-K(q)K1(q)+ξ(û0(q)-ûx(0,ξ)).
Now we prove that, in accordance with last relation, the function û̂(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
(2.82)1K(p)+ξ(û0(p)-ûx(0,ξ))=Λ+-Λ-,1K(p)+ξg1(p,ξ)=Λ2+-Λ2-,12πiPV∫-i∞i∞1q-p1Y+(q,ξ)g1(q,ξ)dq=12(U++U-).
Substituting these relations into (2.80) we express the function û̂ in the following form:
(2.83)û̂=(Λ+-Λ-)-12(Λ2+-Λ2-)-121K(p)+ξY+(U++U-).
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
(2.84)û̂=Λ1+-Λ1-+12(Λ2+-Λ2-)-121K(p)+ξY+U+-121K1(p)+ξY-U-.
Expressing the function U+ in the last equation in terms of U-(2.85)U+=U-+1Y+(K(p)+ξ)(Λ2+-Λ2-),
we arrive at the following relation:
(2.86)û̂=Λ1+-Λ1--1K1(p)+ξY-U-,
where by virtue of (2.49) and (2.66),
(2.87)Λ1+-Λ1-=1K1(p)+ξ(û0(p)-ûx(0,ξ)).
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 assumptionu(0,t)=0 the integral representation (2.61) takes form
(2.88)U(z,ξ)=12πi∫-i∞i∞1q-z1Y+(q,ξ)K1(q)-K(q)K1(q)+ξ(û0(q)-ûx(0,ξ))dq,
where ûx(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
(2.89)u(x,t)=𝒢(t)u0=∫0∞G(x,y,t)u0(y)dy,
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 operator𝒢(t). Let the contours 𝒞i be defined as(2.90)𝒞1={p∈(∞e-i(π/2+ɛ),0)⋃(0,∞ei(π/2+ɛ))},(2.91)𝒞2={q∈(∞e-i((π/2)+2ɛ),0)⋃(0,∞ei((π/2)+2ɛ))},(2.92)𝒞3={q∈(∞e-i((π+ɛ)/2),0)⋃(0,∞ei((π+ɛ)/2))},
where ɛ>0 can be chosen such that all functions under integration are analytic and Rek(ξ)>0 for ξ∈𝒞1.Lemma 2.6.
The functionG(x,y,t) given by formula (2.12) has the following representation:
(2.93)G(x,y,t)=12πi∫-i∞i∞epx-K(p)te-pydp-12πi12πi∫𝒞1dξeξt∫𝒞2epxeΓ(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξI(p,ξ,y)dp,
where
(2.94)I(p,ξ,y)=12πi∫-i∞i∞1(q-p)(q-k(ξ))1eΓ+(q,ξ)w+(q,ξ)e-qydq,Γ(z,ξ)=12πi∫-i∞i∞1q-zln{(K(q)+ξK1(q)+ξ)w-(q)w+(q)}dq.
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
(2.95)G(x,y,t)=J1(x-y,t)+J2(x,y,t)+J3(x,y,t)+J4(x,y,t),
where
(2.96)J1(x-y,t)=12πi∫-i∞i∞ep(x-y)-K1(p)tdp,J2(x,y,t)=12πi12πi∫-i∞+ɛi∞+ɛdξeξtΨ(ξ,y)∫-i∞i∞epx1K1(p)+ξdp,J3(x,y,t)=-12πi12πi∫-i∞+ɛi∞+ɛdξeξt∫-i∞i∞epxY-(p,ξ)K1(p)+ξΥ1-(p,ξ,y)dp,(2.97)J4(x,y,t)=-12πi12πi∫-i∞+ɛi∞+ɛdξeξtΨ(ξ,y)∫-i∞i∞epxY-(p,ξ)K1(p)+ξΥ1-(p,ξ,0)dp.
Here
(2.98)Υ1(z,ξ,y)=12πi∫-i∞i∞1q-z1Y+(q,ξ)K1(q)-K(q)K1(q)+ξe-qydq,Ψ(ξ,y)=-e-k(ξ)yY-(k(ξ),ξ)+Υ1(k(ξ),ξ,y),Y±=eΓ±w±,(2.99)Γ(z,ξ)=12πi∫-i∞i∞1q-zln{(K(q)+ξK1(q)+ξ)w-(q)w+(q)}dq.
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
(2.100)Υ1-(p,ξ,y)=I1(p,ξ,y)+I2(p,ξ,y),
where
(2.101)I1(p,ξ,y)=limz→p,Rez>012πi∫-i∞i∞1q-z1Y+(q,ξ)e-qydq,I2(p,ξ,y)=-limz→p,Rez>012πi∫-i∞i∞1q-z1Y+(q,ξ)K(q)+ξK1(q)+ξe-qydq.
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
(2.102)I1(p,ξ,y)=limz→p,Rez>012πi∫-i∞i∞1q-ze-qydq+limz→p,Rez>012πi∫-i∞i∞1q-z(1Y+(q,ξ)-1)(e-qy-1)dq+limz→p,Rez>012πi∫-i∞i∞1q-z(1Y+(q,ξ)-1)dq=-e-py+limz→p,Rez>012πi∫-i∞i∞1q-z(1Y+(q,ξ)-1)(e-qy-1)dq,
where ξ is some fixed point, Reξ>0.
To calculate the functionI2(p,ξ,y) we will use the following identity:
(2.103)1Y+(q,ξ)K(q)+ξK1(q)+ξ=1Y-(q,ξ).
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
(2.104)I2(p,ξ,y)=-limz→p,Rez>012πi∫-i∞i∞1q-z1Y-(q,ξ)e-qydq=1Y-(p,ξ)e-py.
Thus from (2.102) and (2.104) we obtain the following relation:
(2.105)Υ1-(p,ξ,y)=-e-py+Υ-(p,ξ,y)+1Y-(p,ξ)e-py,
where
(2.106)Υ(z,ξ,y)=12πi∫-i∞i∞1q-z(1Y+(q,ξ)-1)e-qydq.
In the same way (see also proof of relation (2.74)) we can prove that
(2.107)Υ1-(p,ξ,0)=-1+1Y-(p,ξ).
Also we observe that
(2.108)Ψ(ξ,y)=-e-k(ξ)y+Υ(k(ξ),ξ,y).
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
(2.109)J3(x,y,t)=-12πi12πi∫-i∞i∞dξeξt∫-i∞i∞epxY-(p,ξ)K1(p)+ξ(-e-py+Υ-(p,ξ,y))dp-J1(x-y,t).
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
(2.110)J4(x,y,t)=-12πi12πi∫-i∞i∞dξeξtΨ(ξ,y)∫-i∞i∞epxY-(p,ξ)K1(p)+ξ(-1+1Y-(p,ξ))dp=-12πi12πi∫-i∞i∞dξeξt∫-i∞i∞epxY-(p,ξ)K1(p)+ξ(e-k(ξ)y-Υ(k(ξ),ξ,y))dp-J2(x,y).
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
(2.111)G(x,y,t)=12πi12πi∫-i∞i∞dξeξt∫-i∞i∞epxY-(p,ξ)K1(p)+ξΞ1(p,ξ,y)dp,
where
(2.112)Ξ1(p,ξ,y)=(e-py-e-k(ξ)y)-(Υ-(p,ξ,y)-Υ(k(ξ),ξ,y)).
Also, note that since
(2.113)Υ-(p,ξ,y)-Υ(k(ξ),ξ,y)=limz→p,Rez>012πi∫-i∞i∞(1q-z-1q-k(ξ))(1Y+(q,ξ)-1)e-qydq=limz→p,Rez>012πi∫-i∞i∞(z-k(ξ)(q-z)(q-k(ξ)))(1Y+(q,ξ)-1)e-qydq,
we obtain
(2.114)(e-py-e-k(ξ)y)-(Υ-(p,ξ,y)-Υ(k(ξ),ξ,y))=limz→p,Rez>012πi∫-i∞i∞(-1q-z+1q-k(ξ))[e-qy+(1Y+(q,ξ)-1)e-qy]dq=limz→p,Rez>012πi∫-i∞i∞(k(ξ)-z(q-z)(q-k(ξ)))1Y+(q,ξ)e-qydq.
So,
(2.115)G(x,y,t)=-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞epxY-(p,ξ)(p-k(ξ))K1(p)+ξI-(p,ξ,y)dp,
where
(2.116)I(z,ξ,y)=12πi∫-i∞i∞1(q-z)(q-k(ξ))1Y+(q,ξ)e-qydq.
Using relation
(2.117)K(p)+ξK1(p)+ξ=Y+Y-,
we rewrite last formula in the following form:
(2.118)G(x,y,t)=-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞epxY+(p,ξ)(p-k(ξ))K(p)+ξI-(p,ξ,y)dp.
On the basis of definitions (2.116) and in accordance with the Sohkotzki-Plemelj formula (2.3) we have
(2.119)I-(p,ξ,y)=I+(p,ξ,y)-1(p-k(ξ))1Y+(p,ξ)e-py.
So we get
(2.120)G(x,y,t)=12πi∫-i∞i∞epx-K(p)te-pydp-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞epxY+(p,ξ)(p-k(ξ))K(p)+ξI+(p,ξ,y)dp.
Now we consider for Reξ>0(2.121)Γ+(p,ξ)=12πilimz→pRez<0∫-i∞i∞1q-zln{(K(q)+ξK1(q)+ξ)w-(q)w+(q)}dq=12πi∫0∞ei((π/2)+3ɛ)1q-pln{(iq2+ξ-q2+ξ)w-(q)w+(q)}dq+12πi∫∞ei((-π/2)-3ɛ)01q-pln{(-iq2+ξ-q2+ξ)w-(q)w+(q)}dq+ln{(K(p)+ξK1(p)+ξ)w-(p)w+(p)}=Γ1(p,ξ)+ln{(K(p)+ξK1(p)+ξ)w-(p)w+(p)}.
Note that Γ1(p,ξ) is analytic in domain 0≤argξ<(π/2)+3ɛ, (-π/2)-3ɛ<argξ≤0, and π/2≤argp<(π/2)+3ɛ and (-π/2)-3ɛ<argp≤-π/2. So we get
(2.122)G(x,y,t)=12πi∫-i∞i∞epx-K(p)te-pydp-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞epxeΓ1(p,ξ)w-(p,ξ)(p-k(ξ))K1(p)+ξI+(p,ξ,y)dp,
where
(2.123)I+(p,ξ,y)=12πilimz→pRez<0∫-i∞i∞1(q-z)(q-k(ξ))1w-eΓ1(q,ξ)K1(q)+ξK(q)+ξe-qydq.
Changing the contour of integration with respect to ξ by Cauchy theorem we get
(2.124)G(x,y,t)=12πi∫-i∞i∞epx-K(p)te-pydp-12πi12πi∫𝒞1dξeξt∫𝒞2epxeΓ1(p,ξ)w-(p,ξ)(p-k(ξ))K1(p)+ξI+(p,ξ,y)dp-resξ=-K(q){∫𝒞2epxeΓ1(p,ξ)w-(p,ξ)(p-k(ξ))K1(p)+ξI+(p,ξ,y)dp},
where
(2.125)resξ=-K(q){∫𝒞2epxeΓ1(p,ξ)w-(p,ξ)(p-k(ξ))K1(p)+ξI+(p,ξ,y)dp}=12πi∫𝒞2epxdp∫𝒞3dqe-qy-K(q)tq-pϕ(p,q),ϕ(p,q)=eΓ1(p,-K(q))w-(p)(p-k(-K(q)))K1(p)-K(q)(K1(q)-K(q)eΓ1(q,-K(q))w-(q)(q-k(-K(q)))).
Since for q∈𝒞3,Re(-K(q))<0, from (2.121)
(2.126)e-Γ1(q,-K(q))=e-Γ+(q,-K(q))(K(q)-K(q)K1(q)-K(q))w-(q)w+(q)=0,
and therefore
(2.127)ϕ(p,q)=0.
Thus using relation (2.121) we get relation (2.93). Lemma is proved.Lemma 2.7.
The estimates are true, provided that the right-hand sides are finite:(2.128)∥𝒢(t)ϕ∥H1≤∥ϕ∥H1,(2.129)∥∂xn𝒢ϕ∥Ls,μ≤Ct(-1/2)(n+1+δ-(1/s)-μ)∥ϕ∥L1,δ,(2.130)𝒢(t)ϕ-t-1Λ(xt-1/2)f(ϕ)=min(xt-1/2,1)t-1-a∥ϕ∥L1,1+a,a>0,
where μ>0, (1/s)+μ<n+1,n=0,1,δ∈[0,1], and f(ϕ) is given by (1.9). The function Λ(xt-1/2)∈L∞(R+),Λ(0)=0 is defined below (see (2.191)).Proof.
From Lemma2.6 we have
(2.131)G(x,y,t)=J1(x,y,t)+J2(x,y,t),
where
(2.132)J1(x,y,t)=12πi∫-i∞i∞epx-K(p)te-pydp,(2.133)J2(x,y,t)=-12πi12πi∫𝒞1dξeξt∫𝒞2epxw+(p,ξ)eΓ(p,ξ)(p-k(ξ))K(p)+ξI(p,ξ,y)dp,(2.134)I(p,ξ,y)=12πi∫-i∞i∞1(q-p)(q-k(ξ))1w+eΓ+e-qydq,(2.135)Γ(z,ξ)=12πi∫-i∞i∞1q-zln{(K(q)+ξK1(q)+ξ)w-(q)w+(q)}dq.
The functions w±(q,ξ),k(ξ) were defined in formulas (2.13) and (2.10). The contours 𝒞1,𝒞2 was defined by (2.90), and (2.91).
For subsequent considerations it is required to investigate the behavior of the functionΓ(z,ξ). Set
(2.136)ϕ(p,ξ)=ln{(K(p)+ξK1(p)+ξ)w-(p)w+(p)}≠0,Rep=0,Reξ<0.
Observe that the function ϕ(p,ξ) obeys the Hölder condition for all finite p and tends to a definite limit ϕ∞(ξ) as p→±i∞:(2.137)ϕ(∞,ξ)=limp→±i∞ln{(K(p)+ξK1(p)+ξ)w-(p)w+(p)}=limp→±i∞ln|(K(p)+ξK1(p)+ξ)w-(p)w+(p)|+ilimp→±i∞arg{(K(p)+ξK1(p)+ξ)w-(p,ξ)w+(p,ξ)}=-iπ2.
Also there can be easily obtained that for large p and some fixed ξ the following inequality holds:
(2.138)|ϕ(p,ξ)-ϕ∞(ξ)|≤C(|ξ|μ|p|2μ),μ>0.
Therefore
(2.139)|eΓ±(z,ξ)|≤C
for all ξ∈𝒞1 and
(2.140)eΓ+(p,ξ)-e1/2=O(|ξ|μ|p|2μ),μ∈[0,1].
Denote
(2.141)𝒥j(t)ϕ=θ(x)∫0+∞Jj(x,y,t)ϕ(y)dy,
By Plansherel Theorem it is easily to see that
(2.142)∥𝒥1(t)ϕ∥H1≤C∥ϕ∥H1.
Now we estimate𝒥2(t)ϕ.
From the integral representation (2.134), making use the relation
(2.143)eΓ+(p,ξ)-Γ-(p,ξ)w+w-=K(p)+ξK1(p)+ξ
and the estimate (2.138) we have for y>0,ξ∈𝒞1, and p∈𝒞2(2.144)I(p,ξ,y)=12πi∫-i∞i∞1(q-p)1(q-k(ξ))1w+eΓ+e-qydq=limz→p,Rez<012πi∫-i∞i∞1(q-z)1(q-k(ξ))1w-eΓ-(K1(q)+ξ)(K(q)+ξ)e-qydq=O(y-1+γ∫𝒞3|q|-1+γ|q-p|1|q-k(ξ)|dq).
Here γ∈[0,1),K(q)=-q|q|=-q2exp(-iθ),θ=argq and
(2.145)𝒞3={q∈(∞e-i((π-ε)/2),0)⋃(0,∞ei((π-ε)/2))},ɛ>0.
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(2.146)∥∂xn∫0+∞J2(·,y,t)ϕ(y)dy∥L2≤C∥ϕ∥H1∫𝒞1dξe-C|ξ|t∫𝒞2dp|p|n|p-k(ξ)||K(p)+ξ|∫𝒞3|q|-1+γ|q-p|1|q-k(ξ)|dq≤C∥ϕ∥H1.
Therefore according to (2.142) and (2.146) we obtain the estimate (2.129) of the lemma. Now we prove the second estimate of the lemma.
We rewrite the functionI(p,ξ,y) (see (2.134)) for p∈𝒞2,ξ∈𝒞1 in the form
(2.147)I(p,ξ,y)=I1(p,ξ,y)+1(p-k(ξ))1w+(p,ξ)eΓ(p,ξ)+e-1/21(p-k(ξ))(e-k(ξ)y-1),
where
(2.148)I1(p,ξ,y)=12πi∫-i∞i∞1(q-p)(q-k(ξ))(1w+(q,ξ)eΓ+(q,ξ)-e-1/2)(e-qy-1)dq.
Here we used that since w+(q,ξ)eΓ+(q,ξ) is analytic for p∈𝒞2,ξ∈𝒞1, and Rek(ξ)>0 by Cauchy theorem,
(2.149)12πi∫-i∞i∞1(q-p)(q-k(ξ))1w+(q,ξ)eΓ+(q,ξ)dq=1(p-k(ξ))1w+(p,ξ)eΓ+(p,ξ)e-1/22πi∫-i∞i∞1(q-p)(q-k(ξ))(e-qy-1)dq=e-1/21(p-k(ξ))(e-k(ξ)y-1).
Substituting (2.147) into definition of the function J2(x,y,t) (see (2.133)) we get
(2.150)J2(x,y,t)=-12πi12πi∫𝒞1dξeξt∫𝒞2epxY+(p,ξ)(p-k(ξ))K(p)+ξI1(p,ξ,y)dp-12πi12πi∫𝒞1dξeξt∫𝒞2epx1K(p)+ξdp-e-1/22πi∫𝒞1dξeξt(e-k(ξ)y-1)∫𝒞2epxY+(p,ξ)K(p)+ξdp.
Since by Cauchy theorem
(2.151)-12πi12πi∫𝒞1dξeξt∫𝒞2epx1K(p)+ξdp=-12πi∫-i∞i∞epx-K(p)tdp,
we obtain the following form for Green function G(x,y,t) (see (2.131)):
(2.152)G(x,y,t)=∑j=13Fj(x,y,t),
where
(2.153)F1(x,y,t)=12πi∫-i∞i∞epx-K(p)t(e-py-1)dp,(2.154)F2(x,y,t)=12πie-1/22πi∫𝒞1dξeξt(e-k(ξ)y-1)∫𝒞2epxY+(p,ξ)K(p)+ξdp,(2.155)F3(x,y,t)=-12πi12πi∫𝒞1dξeξt∫𝒞2epxY+(p,ξ)(p-k(ξ))K(p)+ξI1(p,ξ,y)dp,(2.156)I1(z,ξ,y)=12πi∫-i∞i∞1(q-z)(q-k(ξ))(1Y+(q,ξ)-e-1/2)(e-qy-1)dq.
Denote
(2.157)ℱjϕ=∫0+∞Fj(x,y,t)ϕ(y)dy.
Now we prove estimate
(2.158)∥∂xnℱ1ϕ∥Ls,μ≤Ct(-1/2)(n+1+δ-(1/s)-μ)∥ϕ∥L1,δ.
We have(2.159)F1(x,y,t)=12πit∫-i∞i∞e|p|pepxt-1/2(e-pyt-1/2-1)dp=1πtRe∫0∞eip2-ipyt-1/2eipxt-1/2(1-eipyt-1/2)dp=1πtRe∫0y/teip2eipxt-1/2(e-ipyt-1/2-1)dp+1πtRe∫y/t∞eiϕeip2-ipyt-1/2eipxt-1/2(1-eipyt-1/2)dp=I11+I12,
where ϕ>0 is such that Reip2>0.
By Bonnet Theorem we get(2.160)|∂xnI11|≤Ct(-1/2)(1+n)|1-eiy2t-1||∫ky/teip2eipxt-1/2pndp|≤Ct(-1/2)(1+n)|1-eiy2t-1||∫k∞eip2eipxt-1/2pndp-∫y/t∞eip2eipxt-1/2pndp|≤Cyδt(-1/2)(1+δ+n)∫0∞eiϕe-C|p|2e-C|p|xt-1/2|p|n|dp|,
where k∈[0,y/t] is intermediate point. Therefore
(2.161)∥∂xnI11∥Ls,μ≤Cyδt(-1/2)(1+δ+n-(1/s)-μ)∫0∞eiϕe-C|p|2|p|n-(1/s)-μdp,
where
(2.162)1s+μ<n+1.
Now we get estimate for I12 from (2.159).
Using forp=|p|eiϕ,|p|>yt-1/2,C1=sinϕ<1,(2.163)|eip2-ipyt-1/2|≤C1||p|2-C1|p|yt-1/2|μ1,μ1≥0,
we get
(2.164)|∂xnI12(x,y,t)|≤Ct(-1/2)(1+n)∫yt-1/2∞1|p2-C1pyt-1/2|μ1e-C|p|xt-1/2|p|n|e-C|p|yt-1/2-1|dp.
Therefore from (2.170) we obtain
(2.165)∥∂xnI12(·,y,t)∥Ls,μ≤Ct(-1/2)(n+1-(1/s)-μ+δ)yδ∫yt-1/2∞pn-(1/s)-μ+δ|p2-Cpyt-1/2|μ1dp≤Ct(-1/2)(n+2-(1/s)-μ)yδ(∫01pn+δ-(1/s)-μdp+∫1∞pn+δ-(1/s)-μp2μ1dp)≤Ct(-1/2)(n+1-(1/s)-μ+δ)yδ,
where
(2.166)1s+μ<n+1+δ.
Therefore
(2.167)∥∂xnℐ12ϕ∥Ls,μ≤Ct(-1/2)(n+1+δ-(1/s)-μ)∥ϕ∥L1,δ.
From (2.161) and (2.167) it follows estimate (2.158).
Now we estimateℱ2ϕ.
Making the change of variablesξ=qt and p=tz in (2.154) we get
(2.168)|∂xnF2(x,y,t)|≤Ct(-1/2)(1+δ+n)∫𝒞1dqe-C|q|O((k(q)y)δ)∫𝒞2e-C|z|xt-1/2|z|n{|z|}1/2|K(z)+q||dz|.
Here we used the following estimations:
(2.169)e-k(ξ)y-1=O((k(ξ)y)δ),δ∈[0,1],|w+eΓ(p,ξ)|≤C{|p|}1/2.
Since
(2.170)∥e-C|z|xt-1/2∥Ls,μ≤C(∫0+∞xμse-Cst-1/2|z|dx)1/s≤C(|z|t-1/2)(-1/s)-μ(∫0+∞x1μse-Csx1dx1)1/s,
we get
(2.171)∥F2∥Ls,μ≤Ct(-1/2)(1+δ+n-(1/s)-μ)yδ∫𝒞1e-C|q|kδ(|q|)dq∫𝒞2|z|n-(1/s)-μ{|z|}1/2|K(z)+q||dz|≤Ct(-1/2)(1+δ+n-(1/s)-μ)yδ,
where
(2.172)1s+μ<n+32,(δ,μ)∈[0,1].
From the estimate (2.171) we have under condition (2.172)
(2.173)∥∂xnℱ2ϕ∥Ls,μ≤Ct(-1/2)(1+δ+n-(1/s)-μ)∥ϕ∥L1,δ,
Now we estimate ℱ3ϕ. From estimate (2.138) for γ∈(0,1) we have
(2.174)1eΓ+w+-e(-1/2)=O(|ξ|γ|p|2γ),
and therefore for p∈𝒞2(2.175)|I1+(p,ξ,y)|≤Cyδ∫-i∞i∞|q|δ|q-p||q-k(ξ)||ξ|γ|q|2γdq.
Thus after the change of variables ξt=q1,p=zt,q=q2t we get
(2.176)∂xnF3(x,y,t)≤Cyδt(-1/2)(1+δ+n)∫𝒞1dq1e-C|q1||q1|γ×∫𝒞2dze-C|z|xt-1/2|z-k(q1)||z|n{|z|}1/2|K(z)+q1|×∫-i∞i∞|q2|δ-2γ|q2-z||q2-k(q1)|dq2.
From (2.170) we obtain for γ∈[0,(1+δ)/2),δ∈[0,1](2.177)∥∂xnℱ3ϕ∥Ls,μ≤Ct(-1/2)(1+δ+n-(1/s)-μ)∥yδϕ∥L1×∫𝒞1e-C|q1||q1|γdq1∫𝒞2dz|z-k(q1)||K(z)+q1||z|n-(1/s)-μ{|z|}1/2×∫-i∞i∞|q2|δ-2γ|q2-z||q2-k(q1)|dq2≤Ct(-1/2)(1+δ+n-(1/s)-μ)∥ϕ∥L1,δ,
where
(2.178)1s+μ<n+32,(δ,μ)∈[0,1].
Thus from (2.171), (2.177), and (2.158) it follows the second estimate (2.129) of the lemma.
Now we prove the asymptotic formula (2.130).
We will use the formula (2.115) from Lemma 2.6:
(2.179)G(x,y,t)=-12πi12πi∫ɛ-i∞ɛ+i∞dξeξt∫-i∞i∞epxY-(p,ξ)(p-k(ξ))K1(p)+ξI-(p,ξ,y)dp,
where
(2.180)I(z,ξ,y)=12πi∫-i∞i∞1(q-z)(q-k(ξ))1Y+(q,ξ)(e-qy-1)dq.
Since for Reξ>0(2.181)∫-i∞i∞Y-(p,ξ)(p-k(ξ))K1(p)+ξI-(p,ξ,y)dp=0,
we have G(0,y,t)=0 and therefore
(2.182)G(x,y,t)=∑j=13F̃j(x,y,t),
where
(2.183)F̃1(x,y,t)=12πi∫-i∞i∞e-K(p)t(epx-1)(e-py-1)dp,F̃2(x,y,t)=12πie-1/22πi∫𝒞1dξeξt(e-k(ξ)y-1)∫𝒞2(epx-1)eΓ+(p,ξ)w+(p,ξ)K(p)+ξdp,F̃3(x,y,t)=-12πi12πi∫𝒞1dξeξt∫𝒞2(epx-1)eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξI1(p,ξ,y)dp.
Here the function I1(p,ξ,y) was defined by (2.148). Since
(2.184)I1(p,ξ,y)=Ĩ1+(p,ξ,y)-yp(p-k(ξ))(1eΓ+(p,ξ)w+(p,ξ)-e-1/2),
where
(2.185)Ĩ1(z,ξ,y)=12πi∫-i∞i∞1(q-z)(q-k(ξ))(1eΓ+(q,ξ)w+(q,ξ)-e-1/2)×(e-qy-1+qy)dq,
we have
(2.186)F̃3(x,y,t)=-12πi12πi∫𝒞1dξeξt∫𝒞2(epx-1)eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξĨ1+(p,ξ,y)dp+12πiy2πi∫𝒞1dξeξt∫𝒞2(epx-1)Y+(p,ξ)pK(p)+ξ(1eΓ+(p,ξ)w+(p,ξ)-e-1/2)dp.
So
(2.187)G(x,y,t)=ytΛ(xt-1/2)+∑k=13Rk(x,y,t),
where
(2.188)Λ(xt-1/2)=e-1/22π2i∫𝒞1dξeξ∫𝒞2(epxt-1/2-1)eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξdp,R1(x,y,t)=12πi∫-i∞i∞e-K(p)t(epx-1)(e-py-1+py)dp,R2(x,y,t)=12πie-1/22πi∫𝒞1dξeξt(e-k(ξ)y-1+k(ξ)y)×∫𝒞2(epx-1)eΓ+(p,ξ)w+(p,ξ)K(p)+ξdp,R3(x,y,t)=-12πi12πi∫𝒞1dξeξt×∫𝒞2(epx-1)eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξĨ1+(p,ξ,y)dp.
Using e-py-1+py=O(p1+μy1+μ),e-k(ξ)y-1+k(ξ)y=O(k(ξ)1+μy1+μ),μ∈(0,1) by the same method as in the proof of estimate (2.129) we obtain that
(2.189)∑j=13|∫0∞dyRj(x,y,t)ϕ(y)|≤Ct-1-(μ/2)min(xt-1/2,1)∥ϕ∥L1,1+μ.
Since
(2.190)eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξ-e1/2p2+1=O(ξp2),∫𝒞1dξeξ∫𝒞2(epxt-1/2-1)1p2+1dp=0,
we rewrite
(2.191)Λ(xt-1/2)=e-1/22π2i∫𝒞1dξeξ∫𝒞2(epxt-1/2-1)(eΓ+(p,ξ)w+(p,ξ)(p-k(ξ))K(p)+ξ-e1/2p2+1)dp.
So Λ(xt-1/2)∈L∞(R+) and Λ(0)=0. From (2.189),(2.187) it follows (2.130) and consequently,
(2.192)∥𝒢ϕ-t-1Λ(xt-1/2)f(ϕ)∥L∞≤Ct-1-(μ/2)∥x1+μϕ∥L1.
The lemma is proved.


## 3. Proof of Theorem1.1

By Proposition2.4 we rewrite the initial-boundary value problem (1.1) as the following integral equation:(3.1)u(t)=𝒢(t)u0-∫0t𝒢(t-τ)𝒩(u(τ))dτ,𝒩(u)=uxu,
where 𝒢 is the Green operator of the linear problem (2.9). We choose the space(3.2)Z={ϕ∈H1(R+)∩L1(R+)∩L1,1+a(R+)}
with 0<a<1 being small and the space(3.3)X={∂xnϕ∈C([0,∞);L2(R+)⋂(0,∞);L2,n+(a/2)(R+)):∥ϕ∥X<∞},n=0,1,
where now the norm μ∈[0,a](3.4)∥ϕ∥X=supt≥0(∑n=01(t(1-μ)/4〈t〉1/2∥∂xnϕ(t)∥L2,n+(μ/2)+〈t〉(1/2)(n+(3/2))∥∂xnϕ(t)∥L2))
reflects the optimal time decay properties of the solution. We apply the contraction mapping principle in a ball Xρ={ϕ∈X:∥ϕ∥X≤ρ}in the space X of a radius(3.5)ρ=12C∥u0∥Z>0.
For v∈Xρ we define the mapping ℳ(u) by formula(3.6)ℳ(u)=𝒢(t)u0-∫0t𝒢(t-τ)𝒩(u(τ))dτ.
We first prove that(3.7)∥ℳ(u)∥X≤ρ,
where ρ>0 is sufficiently small. By virtue of Lemma 2.7 we have(3.8)∥𝒢(t)ϕ∥H1≤∥ϕ∥H1,∥∂xn𝒢ϕ∥L2,μ≤Ct(-1/2)(n+δ+(1/2)-μ)∥ϕ∥L1,δ
for all t≥0,μ<n+(1/2). Therefore for t<1(3.9)∥𝒢u0∥H1≤C∥u0∥Z,
(3.10)∥∂xn𝒢u0∥L2,n+(a/2)≤Ct(-1/2)(n+(1/2)-n-(a/2))∥u0∥L1≤Ct-(1-a)/4∥u0∥Z.
For t>1(3.11)∥∂xn𝒢u0∥L2≤Ct(-1/2)(n+(3/2))∥u0∥L1,1≤Ct(-1/2)(n+(3/2))∥u0∥Z,∥∂xn𝒢u0∥L2,n+(a/2)≤Ct(-1/2)(n+(3/2)-n-(a/2))∥ϕ∥L1,1≤Ct-(3-a)/4∥u0∥Z.
Thus(3.12)∥𝒢u0∥X≤C∥u0∥Z.
Also since v∈Xρ we get for all τ>0(3.13)∥𝒩(u(τ))∥L1,1≤C∥u∥L2∥ux∥L2,1≤C{τ}-1/4〈τ〉-3/2∥u∥X2,∥𝒩(u(τ))∥L1≤C∥u∥L2∥ux∥L2≤C〈τ〉-2∥u∥X2.
Therefore by Lemma 2.7 we get(3.14)∥∂xn∫0t𝒢(t-τ)𝒩(u(τ))dτ∥L2≤∫0t/2〈t-τ〉(-1/2)(n+(3/2))∥𝒩(u(τ))∥L1,1dτ+∫t/2t(t-τ)(-1/2)(n+1-(1/2))∥𝒩(u(τ))∥L1dτ≤C∥u∥X2[∫0t/2〈t-τ〉(-1/2)(n+2-(1/2)){τ}-1/2〈τ〉-(2-(1/2))dτ+∫t/2t(t-τ)(-1/2)(n+1-(1/2))〈τ〉-2dτ]≤C∥u∥X2[〈t〉(-1/2)(n+2-(1/2))+〈t〉-2t(-1/2)(n+1-(1/2))+1]≤C∥u∥X2〈t〉(-1/2)(n+(3/2))
for all t≥0. In the same manner by virtue of Lemma 2.7 we have(3.15)∥∂xn∫0t𝒢(t-τ)𝒩(u(τ))dτ∥L2,n+(a/2)≤∫0t/2(t-τ)-(3-a)/4∥𝒩(u(τ))∥L1,1dτ+∫t/2t(t-τ)-1/4∥𝒩(u(τ))∥L1dτ≤C∥u∥X2[∫0t/2(t-τ)-(3-a)/4{τ}-1/4〈τ〉-3/2dτ+∫t/2t(t-τ)-1/4〈τ〉-2dτ]≤C∥u∥X2[〈t〉(-1/2)(n+2-(1/2)-((1+a)/2))+〈t〉-2t(-1/2)(n+1-(1/2))+1]≤C∥u∥X2〈t〉(-(3-a))/4{t}-1/4
for all t≥0. Thus we get(3.16)∥∫0t𝒢(t-τ)𝒩(u(τ))dτ∥X≤C∥u∥X2,
and hence in view of (3.6) and (3.9)(3.17)∥ℳ(u)∥X≤∥𝒢u0∥X+∥∫0t𝒢(t-τ)𝒩(u(τ))dτ∥X≤C∥u0∥Z+C∥u∥X2≤ρ2+Cρ2<ρ
since ρ>0 is sufficiently small. Hence the mapping ℳ transforms a ball Xρ into itself. In the same manner we estimate the difference(3.18)∥ℳ(w)-ℳ(v)∥X≤12∥w-v∥X,
which shows that ℳ is a contraction mapping. Therefore we see that there exists a unique solution u∈C([0,∞);L1(R+)∩L1,a(R+))∩C((0,∞);H∞1) to the initial-boundary value problem (1.1). Now we can prove asymptotic formula(3.19)u(x,t)=A1Λ(xt-1/2)t-1+min(xt-1/2,1)O(t-1-γ),
where
(3.20)A1=f(u0)-∫0+∞dτ∫0+∞yuyudy.
Denote(3.21)G0(t)=t-1Λ(xt-1/2).
From Lemma 2.7 we have(3.22)t∥𝒢(t)ϕ-G0(t)f(ϕ)∥L∞≤C∥ϕ∥Z
for all t>1. Also in view of the definition of the norm X we have(3.23)|f(𝒩(u(τ)))|≤∥𝒩(u(τ))∥L1,1≤C{τ}-1/2〈τ〉-3/2∥u∥X2.
By a direct calculation we have for some small γ1>0,γ>0(3.24)∥∫0t/2e-τ|G0(t-τ)-G0(t)|f(𝒩(u(τ)))dτ∥L∞≤〈t〉-1C∥u∥X2∫0t/2∥(G0(t-τ)+G0(t))∥L∞{τ}-γ1〈τ〉-γ2dτ≤C〈t〉-2∫0t/2{τ}-γ1〈τ〉-γ2dτ≤C〈t〉-γ-1
and in the same way(3.25)∥〈t〉γG0(t)∫t/2∞f(𝒩(u(τ)))dτ∥L∞≤C∥u∥X2.
Also we have(3.26)∥∫0t/2(𝒢(t-τ)𝒩(u(τ))-G0(t-τ)f(𝒩(u(τ))))dτ∥L∞+∥∫t/2t𝒢(t-τ)𝒩(u(τ))dτ∥L∞≤C∫0t/2(t-τ)-1∥𝒩(u(τ))∥L1dτ+C∫t/2t∥𝒩(u(τ))∥L1dτ≤Ct-1-γ∥u∥X2
for all t>1. By virtue of the integral equation (3.1) we get(3.27)〈t〉γ+1∥(u(t)-AG0(t))∥X≤∥(𝒢(t)u0-G0(t)f(u0)∥L∞+〈t〉γ+1∥∫0t/2(𝒢(t-τ)𝒩(a(τ))-G0(t-τ)f(𝒩(u(τ))))dτ∥L∞+〈t〉γ+1∥∫t/2t𝒢(t-τ)𝒩(a(τ))dτ∥L∞+〈t〉γ+1∥G0(t)∫t/2∞f(𝒩(u(τ)))dτ∥L∞+〈t〉γ+1∥∫0t/2(G0(t-τ)-G0(t))f(𝒩(u(τ)))dτ∥L∞.
The all summands in the right-hand side of (3.27) are estimated by C∥u0∥Z+C∥u∥X2 via estimates (3.24)–(3.26). Thus by (3.27) the asymptotic (3.19) is valid. Theorem is proved.



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*Source: 714534-2010-03-11.xml*