Whitham equation with Landau damping on a half-line

J. Math. Anal. Appl. 387 (2012) 359–373

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Whitham equation with Landau damping on a half-line

Felipe Benitez a, Elena I. Kaikina b,∗a Instituto Tecnológico de Morelia, CP 58120, Morelia, Michoacán, Mexicob Instituto de Matemáticas, UNAM Campus Morelia, AP 61-3 (Xangari), Morelia CP 58089, Michoacán, Mexico

a r t i c l e i n f o a b s t r a c t

Article history:Received 15 June 2011Available online 12 September 2011Submitted by K. Nishihara

Keywords:Initial-boundary value problemGreen functionFractional derivative

We consider the initial-boundary value problem on a half-line for the evolution equation

(∂t + R

12 ∂2

x + K)u(x, t) = −uxu,

where

Rαφ = 1

2Γ (α) sin( π2 α)

+∞∫0

φ(y)

|x − y|1−αdy

is the modified Riesz potential and

Ku = 1

2π iθ(x)

i∞∫−i∞

epx p

√tanh |p|

|p|(

u(p, t) − u(0, t)

p

)dp,

where θ(x) is the Heaviside step function. We study traditionally important problems ofthe theory of nonlinear partial differential equations, such as global in time existence ofsolutions to the initial-boundary value problem and the asymptotic behavior of solutionsfor large time.

© 2011 Elsevier Inc. All rights reserved.

1. Introduction

Our aim in the present paper is to study the global existence and large time asymptotic behavior of solutions to theinitial-boundary value problem for the nonlinear Whitham equation with Landau damping on a half-line⎧⎨

⎩(∂t + R

12 ∂2

x + K)u(x, t) = −uxu, t > 0, x > 0,

u(x,0) = u0(x), x > 0,

u(0, t) = 0, t > 0,

(1.1)

where

Rαφ = 1

2Γ (α) sin(π2 α)

+∞∫0

φ(y)

|x − y|1−αdy

* Corresponding author.E-mail addresses: [email protected] (F. Benitez), [email protected] (E.I. Kaikina).

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360 F. Benitez, E.I. Kaikina / J. Math. Anal. Appl. 387 (2012) 359–373

is the modified Riesz potential (see [14, p. 214]) and

Ku = 1

2π iθ(x)

i∞∫−i∞

epx p

√tanh |p|

|p|(

u(p, t) − u(0, t)

p

)dp.

Here and below the Heaviside step function θ(x) = 0 for x < 0 and θ(x) = 1 for x � 0.Due to the intensive development of the theory of nonlinear nonlocal equations itself as well as its applications, the

initial-boundary value problem (1.1) plays an important role in the modern science. Apart from diverse areas of mathematics,evolution nonlocal partial differential equations arise in modern mathematical physics and many other branches of science,such as, for example, chemical physics and electrical networks (see for details, for example, [1–3,9,10,13,15,16]). G. Whithamwas the first (in 1967) to draw general attention to mathematical aspects of nonlinear nonlocal equations. He proposed(see [17]) the following nonlinear nonlocal equation

ut + uux + Ku = 0, (1.2)

where operator K is given by

Ku = 1

2π

∞∫−∞

eipx K (p)u(p, t)dp,

here u(p, t) is the Fourier transform of u(x, t) and the symbol K (p) = p√

tanh |p||p| . This equation is suitable for describing

such typical water wave phenomena as sharp crests and breaking, which the famous Korteweg–de Vries (KdV) equationut + uux + αuxxx = 0 fail to do [8]. Whitham equation is of great interest for physical application, it is applied, for instance,in the theory of capillary waves. In the book [9] it was shown, that Whitham equation (1.2) with various symbols describesall the characteristic properties of waves in a finite time interval, existence of the wave solutions global in time, eventualsmoothing of discontinue initial perturbations, etc.

On the other hand there have appeared lots of works, in which evolution partial differential equations with fractionalderivative are used for a better description of considered material properties. For example, Ott, Sudan and Ostrovskiy [11,12]proposed the following generalizations of the KdV equation

ut + uux + u + αuxxx + R12 ∂2

x u = 0, (1.3)

describing the ion-acoustic waves in plasma with Landau damping. This equation combines the shallow water equationwith the integral term that produces a wave damping. Note that the term with the third derivative ∂3

x appears in the

shallow water theory as a first approximation of the complete dispersion relation p√

tanh |p||p| . So the next natural step is

to generalize (1.3) replacing the third derivative operator ∂3x by Whitham integral operator Ku, which better describes the

complete dispersion of the linear theory of water waves. Therefore instead of Eq. (1.3) one should consider the followingequation

ut + uux + Ku + R12 ∂2

x u = 0.

With this approach, the resulting equation would describe typical water wave phenomenas, such as the peaking, breaking,damping, and also in other aspects. The Cauchy problem for Eq. (1.1) can be considered by the methods of book [4]. Theapplied problem naturally leads to the necessity of the formulation of the initial-boundary value problem to this equation.It should be noted that most papers and books on fractional calculus are devoted to solvability of linear initial fractionalordinary differential equations (ODE). However, few works have considered the initial-boundary value problems for evolutionpartial differential equations with a fractional derivative.

For the general theory of nonlinear equations on a half-line we refer to the book [5]. This book is the first attempt todevelop systematically a general theory of the initial-boundary value problems for evolution equations with pseudodiffer-ential operators on a half-line. The pseudodifferential operator K on a half-line was introduced by virtue of the inverseLaplace transformation of the product of the symbol K (p) = O (pα) which is analytic in the right complex half-plane, andthe Laplace transform of the derivative ∂

[α]x u. Thus, for example, in the case of K (p) = pα we get the following definition of

the fractional derivative ∂αx ,

∂αx = L−1

{pα

(L −

[α]∑j=1

limx→0+ ∂j−1

x

p j

)}. (1.4)

Here and below pα is the main branch of the complex analytic function in the complex half-plane Re p � 0, so that 1α = 1(we make a cut along the negative real axis (−∞,0)). Note that due to the analyticity of pα for all Re p > 0 the inverse

F. Benitez, E.I. Kaikina / J. Math. Anal. Appl. 387 (2012) 359–373 361

Laplace transform gives us the function which is equal to 0 for all x < 0. To obtain an explicit form of the Green functionit was used an approach based on the Laplace transformation with respect to the spatial variable contrary to the standardapplication of the Laplace transformation with respect to the time variable. It was proved that the amount of boundary datawhich we need to put in the problem for its well-posedness is equal to the integer part of [α

2 ], where α is the order ofthe operator K, which is not equal to an odd integer (in the case of odd integer order of operator K the amount of theboundary data depends also on the sign of the highest derivative). Methods of this book can be applied directly to studythe initial-boundary value problem for differential equations with fractional Riemann–Liouville derivative

∂αx = 1

Γ (α − [α])x∫

0

∂[α]+1y

(x − y)α−[α] dy.

In spite of the importance and actuality there are few results about the initial-boundary value problem for pseudodifferentialequations with nonanalytic symbols. For example, in paper [6] it was considered the case of rational symbol K (p) whichhas some poles in the right complex half-plane. It was proposed a new method for constructing the Green operator basedon the introduction of some necessary condition at the singularity points of the symbol K (p). In the paper [7] it was

considered the initial-boundary value problem for a pseudodifferential equation with symbol K (p) = |p| 12 . As far as we

know the case of general nonanalytic symbols K (p) was not studied previously. In the present paper we fill this gap,

considering as example of evolution partial differential equation (1.1) with a nonanalytic symbol K (p) = p√

tanh |p||p| + |p| 3

2 .

There are many natural open questions which should be studied. First we need to answer the following question: how manyboundary data should be posed in problem (1.1) for its correct solvability? Also we study traditionally important problemsof a theory of partial differential equations, such as existence and uniqueness of solution. The main difficulty for Eq. (1.1)on a half-line is that the symbol K (p) is nonanalytic in the complex plane. Therefore we cannot apply the Laplace theorydirectly. To construct Green operator we propose a new method based on the integral representation for the sectionallyanalytic function and theory of singular integro-differential equations with Hilbert kernel and the discontinuous coefficients.We prove that the amount of the boundary data which we need to put in the problem for its well-posedness is equal toone.

We believe that the results of this paper could be applicable to study a wide class of dissipative nonlinear nonlocalequations on half-line by the use of techniques of nonlinear analysis (estimations of function Green, fixed point theorems,etc., see [4]).

To state precisely the results of the present paper we give some notations. We denote 〈t〉 = 1 + t , {t} = t〈t〉 . Direct Laplace

transformation Lx→ξ is

u(ξ) ≡ Lx→ξ u =+∞∫0

e−ξxu(x)dx

and the inverse Laplace transformation L−1ξ→x is defined by

u(x) ≡ L−1ξ→xu = (2π i)−1

i∞∫−i∞

eξxu(ξ)dξ.

Weighted Lebesgue space Lq,a(R+) = {ϕ ∈ S ′; ‖ϕ‖Lq,a < ∞}, where

‖ϕ‖Lq,a =( +∞∫

0

(1 + x)aq∣∣ϕ(x)

∣∣qdx

) 1q

for a > 0, 1 � q < ∞ and

‖ϕ‖L∞ = ess.supx∈R+

∣∣ϕ(x)∣∣.

We introduce Λ(s) ∈ L∞(R+) by formula

Λ(s) = (−1)34

2π i

i∞∫eps−|p| 3

2 p− 34 dp. (1.5)

−i∞


We define the linear operator f :

f (φ) =+∞∫0

φ(y)dy. (1.6)

Now we state the main results.

Theorem 1. Suppose that for small a > 0 the initial data u0 ∈ Z = L1(R+) ∩ L1,a(R+) ∩ L∞(R+) are such that the norm

‖u0‖Z � ε

is sufficiently small. Then there exists a unique global solution

u ∈ C([0,∞);L1(R+) ∩ L1,a(R+)) ∩ C

((0,∞);L∞(

R+))to the initial-boundary value problem (1.1). Moreover the following asymptotic is valid

u = AΛ((x − t)t− 2

3)t− 2

3 + O(t− 2

3 −γ), (1.7)

for t → ∞ in L∞ , where γ ∈ (0,min(1,a)) and

A = f (u0).

2. Preliminaries

In subsequent consideration we shall have frequently to use certain theorems of the theory of functions of complexvariable, the statements of which we now quote. The proofs may be found in all text-book of the theory.

Theorem 2. Let φ(q) be a complex function in Re q = 0, which obeys the Hölder condition for all finite q and tends to a definite limitas q → ±i∞. Then Cauchy type integral

F (z) = 1

2π i

i∞∫−i∞

φ(q)

q − zdq

constitutes a function analytic in the left and right semi-planes. Here and below these functions will be denoted by 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 thecontour from the left and from the right, respectively. These limiting values are expressed by Sokhotzki–Plemelj formula

F ±(p) = limz→p

∓Re z<0

1

2π i

i∞∫−i∞

φ(q)

q − z

= 1

2π i

i∞−∫

−i∞

φ(q)

q − pdq ± 1

2φ(p). (2.1)

Subtracting and adding the formula (2.1) we obtain the following two equivalent formulas

F +(p) − F −(p) = φ(p),

F +(p) + F −(p) = 1

π i

i∞−∫

−i∞

φ(q)

q − pdq, (2.2)

which will be frequently employed hereafter. All the integrals are understood in the sense of the principal values.

Lemma 1. An arbitrary function φ(p) given on the contour Re p = 0, satisfying the Hölder condition, can be uniquely represented inthe form

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 aredetermined by formula

U (z) = 1

2π i

i∞∫−i∞

1

q − zφ(q)dq.

Lemma 2. An arbitrary function ϕ(p) given on the contour Re p = 0, satisfying the Hölder condition, and having zero index,

indϕ(t) := 1

2π i

i∞∫−i∞

d lnϕ(p) = 0,

is uniquely representable as the ratio of the functions X+(p) and X−(p), constituting the boundary values of functions, X+(z) andX−(z), analytic in the left and right complex semi-plane and having in these domains no zero. These functions are determined to withinan arbitrary constant factor and given by formula

X±(z) = eΓ ±(z), Γ (z) = 1

2π i

i∞∫−i∞

1

q − zlnϕ(q)dq.

Lemma 3. If φ(q) is a function that satisfies the Hölder condition on Re q = 0, then the limiting values of the Cauchy type integral

Φ(z) = 1

2π i

i∞∫−i∞

1

q − zφ(q)dq

also satisfy this condition.

3. Method of solution. Uniqueness theorem

In this section we get closed form of the solution of the linear initial-boundary value problem⎧⎨⎩

(∂t + R

12 ∂2

x + K)u(x, t) = 0, t > 0, x > 0,

u(x,0) = u0(x), x > 0,

u(0, t) = 0, t > 0.

(3.1)

Let

K (p) = p

√tanh |p|

|p| + |p| 32 , K1(p) = −p

32 + p.

Denote

Gφ =+∞∫0

G(x, y, t)φ(y)dy,

where

G(x, y, t) = − 1

2π i

1

2π i

i∞∫−i∞

eξt dξ

i∞∫−i∞

epx eΓ +(p,ξ)

(K (p) + ξ)w+(p, ξ)

×(

I−(p, ξ, y) − I−(k(ξ), ξ, y

) Y −(p, ξ)

Y −(k(ξ), ξ)

)dp, (3.2)

where

I(z, ξ, y) =i∞∫

−i∞

1

q − ze−Γ +(q,ξ)w+(q, ξ)e−qy dq, (3.3)

Y(z, ξ) = 1∫

1e−Γ +(q,ξ)w+(q, ξ)

|q| 32

2dq, (3.4)

2π i q − z q


the sectionally analytic function Γ (z, ξ) given by formula

Γ (z, ξ) = 1

2π i

i∞∫−i∞

1

q − zln

(K (q) + ξ

K1(q) + ξ

)w+

w− dq (3.5)

and

w−(z) = z34

(1

z + k(ξ)

) 34

, w+ = z34

(1

z − k(ξ)

) 34

,

k(ξ) is root of equation K1(p) + ξ = 0, such that Rek(ξ) > 0 for Re ξ > 0.Here and below by Φ± we denote the left and right limiting values of sectionally analytic function Φ given by integral

of Cauchy type

Φ(z) = 1

2π i

i∞∫−i∞

φ(q)

q − zdq.

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

Theorem 3. Let

u0 ∈ L1(R+).

There exists the unique solution u(x, t) of the problem (3.1)

u(x, t) ∈ C0([0, T ],L1 ∩ C) ∩ C0((0, T ],L1 ∩ C1).

Moreover

u(x, t) = Gu0.

Proof. To derive an integral representation for the solutions of the problem (1.1) firstly we suppose that there exists asolution u(x, t) ∈ C0([0, T ],L1 ∩ C) ∩ C0((0, T ],L1 ∩ C1) of problem (3.1), which is continued by zero outside of x > 0:

u(x, t) = 0 for all x < 0.

Let φ(p) be a function of the complex variable Re p = 0, which obeys the Hölder condition for all finite p and tends to0 as p → ±i∞. We define the operator

Pφ(z) = − 1

2π i

i∞∫−i∞

1

p − zφ(p)dp.

We have for the Laplace transform

L{Ku} = P

{p

√tanh |p|

|p|(

u − u(0, t)

p

)},

L{

R12 ∂2

x u} = P

{|p| 3

2

(u − u(0, t)

p− ux(0, t)

p2

)}.

Since L{u} = u(p, t) is analytic for all Re q > 0 we have

u(p, t) = P{

u(p, t)}. (3.6)

Therefore applying the Laplace transform to problem (1.1) with respect to x and to ξ variables via u(0, t) = 0 we obtainfor Re p > 0, Re ξ > 0

P

{ξ û − u0(p) + K (p) û(p, ξ) − |p| 3

2ux(0, t)

p2

}= 0. (3.7)

We rewrite (3.7) in the form

û(p, ξ) = 1(

u0(p) + Φ(p, ξ) + |p| 32

ux(0, ξ)

2

), (3.8)

K (p) + ξ p


with some function Φ(p, ξ) such that for all Re p > 0 and any fixed point ξ

PΦ = 0 (3.9)

and for |p| > 1

∣∣Φ(p, ξ)∣∣ � C

1

|p|γ .

We will find the function Φ(p, ξ) using the analytic properties of function û in the right-half complex planes Re p > 0 andRe ξ > 0. We have for Re p = 0

û(p, ξ) = − 1

π i

i∞−∫

−i∞

1

q − pû(q, ξ)dq. (3.10)

In view of Sokhotzki–Plemelj formula via (3.8) the condition (3.10) can be written as nonhomogeneous Riemann problem

Ω+(p, ξ) = K (p) + ξ

ξΩ−(p, ξ) − K (p)Λ+(p, ξ), (3.11)

where the sectionally analytic functions Ω(z, ξ) and Λ(z, ξ) are given by Cauchy type integrals

Ω(z, ξ) = 1

2π i

i∞∫−i∞

1

q − z

K (q)

K (q) + ξΦ(p, ξ)dq, (3.12)

and

Λ(z, ξ) = 1

2π i

i∞∫−i∞

1

q − z

1

K (q) + ξ

(u0(q) + |p| 3

2ux(0, ξ)

p2

)dq. (3.13)

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), analyticin Re z > 0, which satisfy on the contour Re p = 0 the relation (3.11).

Note that bearing in mind formula (3.12) we can find unknown function Φ(p, ξ) which involved in the formula (3.8) bythe relation

Φ(p, ξ) = K (p) + ξ

K (p)

(Ω+(p, ξ) − Ω−(p, ξ)

). (3.14)

The method for solving the Riemann problem (3.11) is based on Lemmas 1 and 2.In the formulations of Lemmas 1 and 2 the coefficient K (p)+ξ

ξof the Riemann problem is 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 functionsdo not have limiting value as p → ±i∞. The principal task now is to get an expression equivalent to the boundary valueproblem (3.11), such that the conditions of lemmas are satisfied.

First, let us introduce some notation and let us establish certain auxiliary relationships. Denote

K1(z) = −z32 + z,

such that Re K1(z) > 0 for Re z = 0. The equation K1(p) = −ξ has one root k(ξ), which is analytic function for Re ξ > 0 andtransforms the half-complex plane Re ξ > 0 to domain, where

Re k(ξ) > 0.

We introduce the function

W (p, ξ) =(

K (p) + ξ

K1(p) + ξ

)w+

w− = 0,

where

w−(z) = zμ

(1

z + k(ξ)

)μ

, w+ = zμ

(1

z − k(ξ)

)μ

,

μ = 3.

4


Here w−(z) is analytic for Re z > 0 and w+(z) is analytic for Re z < 0. We observe that the function W (p, ξ) given on thecontour Re p = 0, satisfies the Hölder condition and under the assumption Re K1ε(p) = 0 does not vanish for any Re ξ > 0.Also we have

Ind.W (p, ξ) = 1

2π i

i∞∫−i∞

d ln W (p, ξ) = 0.

Therefore in accordance with Lemma 2 the function W (p, ξ) can be represented in the form of the ratio

W (p, ξ) = K (p) + ξ

K1(p) + ξ= Y +(p, ξ)

Y −(p, ξ), (3.15)

where

Y ±(p, ξ) = 1

w± eΓ ±(p,ξ), Γ (z, ξ) = 1

2π i

i∞∫−i∞

1

q − zln W (q, ξ)dq.

Now we return to the nonhomogeneous Riemann problem (3.11).By (3.15) we reduce the nonhomogeneous Riemann problem (3.11) to the form

1

Y +(Ω+(p, ξ) − ξΛ+(p, ξ)

) =(

K1(p) + ξ

ξ

)Ω−(p, ξ) − ξΛ−(p, ξ)

Y −(p, ξ)

− 1

Y +(p, ξ)

(u0(p) + |p| 3

2

p2ux(0, ξ)

). (3.16)

Since u0(p)+ |p| 32

p u(0, ξ) satisfies on Re p = 0 the Hölder condition, on the basis of Lemma 3 the function 1Y +(p,ξ)

(u0(p)+|p| 3

2

p u(0, ξ)) also satisfies this condition. Therefore in accordance with Lemma 1 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.

Therefore the problem (3.16) takes the form

Ω+(p, ξ) − ξΛ+(p, ξ)

Y +(p, ξ)+ U+(p, ξ) =

(K1(p) + ξ

ξ

)Ω−

1 (p, ξ) − ξΛ−(p, ξ)

Y −(p, ξ)+ U−(p, ξ),

where

U (z, ξ) = 1

2π i

i∞∫−i∞

1

q − z

1

Y +(q, ξ)

(u0(q) + |q| 3

2

q2ux(0, ξ)

)dq. (3.17)

The last relation indicates that the function Ω+(p,ξ)−ξΛ+(p,ξ)

Y +(p,ξ)+ U+(p, ξ), analytic in Re z < 0, and the function

(K1(p)+ξ

ξ)

Ω−1 (p,ξ)−ξΛ−(p,ξ)

Y −(p,ξ)+ U−(p, ξ), analytic in Re z > 0, except to 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, whichhas zero in p = ∞. According to generalized Liouville theorem this function is zero. Thus, we get

Ω+(p, ξ) = −Y +U+ + ξΛ+,

Ω−(p, ξ) = − ξ

K1(p) + ξY −U− + ξΛ−. (3.18)

Since Ω−(z, ξ) is analytic in Re z > 0 we need to put necessary condition

Y −(k(ξ), ξ

)U−(

k(ξ), ξ) = 0, (3.19)

where k(ξ) is root of the equation K1(p) + ξ = 0, such that Rek(ξ) > 0 for Re ξ > 0. From this condition we find unknownboundary function ux(0, ξ):

ux(0, ξ) = − I−(k(ξ), ξ)

Y −(k(ξ), ξ), (3.20)

where


I(z, ξ) = 1

2π i

∫1

q − z

1

Y + u0(q)dq,

Y(z, ξ) = 1

2π i

∫1

q − z

1

Y +|q| 3

2

q2dq.

From (3.18) with the help of the integral representations (3.17) and (3.13), for sectionally analytic functions U (z, ξ) andΛ(z, ξ), making use of Sokhotzki–Plemelj formula (2.1) and relation (3.15) we can express the difference limiting values ofthe function Ω(z, ξ) in the form

Ω+(p, ξ) − Ω−(p, ξ) = −Y +(

U+ − ξ

K (p) + ξU−

)+ ξ

(Λ+ − Λ−)

= −Y + K (p)

K (p) + ξU+.

We now proceed to find the unknown function Φ(p, ξ) involved in the formula (3.8) for the solution û(p, ξ) of theproblem (1.1). Replacing the difference Ω+(p, ξ) − Ω−(p, ξ) in the relation (3.14) we get

Φ(p, ξ) = K (p) + ξ

K (p)

(Ω+(p, ξ) − Ω−(p, ξ)

)= −Y +U+.

It is easy to observe that Φ(p, ξ) is boundary value of the function analytic in the left complex semi-plane and thereforesatisfies our basic assumption for all Re z > 0

P{Φ} = 0.

Having determined the function Φ(p, ξ) bearing in mind formula (3.8) and (2.1) determine required function û,

û(p, ξ) = 1

K (p) + ξ

(u0(p) + |p| 3

2ux(0, ξ)

p2− Y +(p, ξ)U+

)

= − 1

K1(p) + ξY −(p, ξ)U−(p, ξ). (3.21)

By condition (3.19) the function û is the limiting value of an analytic function in Re z > 0. Note, the fundamental impor-tance of the proven fact, that the solution û constitutes an analytic function in Re z > 0 and, as a consequence, its inverseLaplace transform vanish for all x. Taking inverse Laplace transform of (3.21) with respect to ξ and p variables via (3.20)we get closed form for solution

u(x, t) = Gu0,

where operator G was defined by (3.2). We now prove the uniqueness of the solution. On the contrary we consider twodifferent solutions u1 and u2. Then the difference u1 −u2 satisfies linear problem (1.1) with homogeneous data u0 = 0. Thenby (3.21) we get û1 − û2 = 0; hence u1 = u2. From Lemma 4 we can prove that obtained solution u(x, t) ∈ C0([0, T ],L1 ∩C) ∩ C0((0, T ],L1 ∩ C1). Theorem is proved. �4. Preliminaries

Now we collect some preliminary estimates of the Green operator G(t).

Lemma 4. The following estimates are true, provided that the right-hand sides are finite∥∥∂nx G(t)φ

∥∥Ls,μ � Ct− 2

3 ( 1r − 1+μ

s −n)∥∥φ(·)∥∥Lr + t− 2

3 ( 1r − 1

s −n)∥∥φ(·)∥∥Lr,μ , (4.1)∥∥Gφ − t− 2

3 Λ(xt− 2

3)

f (φ)∥∥

L∞ � Ct− 1+2μ3

∥∥〈x〉μφ∥∥

L1 , (4.2)

where small μ > 0, 1 � r � s � ∞, n = 0,1, Λ(s) and f (φ) are given by (1.5) and (1.6).

Proof. On the basis of definitions (3.3), (3.4) and in accordance with the Sokhotzki–Plemelj formula (2.1) we have

I−(p, ξ, y) = I+(p, ξ, y) − e−py

Y +(p, ξ),

Y −(p, ξ) = Y +(p, ξ) − 1+

|p| 32

2, (4.3)

Y (p, ξ) p


where

Y +(p, ξ) = eΓ +(p,ξ)

w+(p, ξ).

Introducing these expressions into definition (3.2) we find

G(x, y, t) =4∑

j=1

J j(x, y, t), (4.4)

where

J1(x, y, t) = 1

2π i

i∞∫−i∞

ep(x−y)−K (p)t dp, (4.5)

J2(x, y, t) = − 1

2π i

1

2π i

i∞∫−i∞

dξ eξt

i∞∫−i∞

epx Y +(p, ξ)

K (p) + ξI+(p, ξ, y)dp, (4.6)

J3(x, y, t) = − 1

2π i

1

2π i

i∞∫−i∞

eξt dξ

i∞∫−i∞

epx 1

K (p) + ξ

|p| 32

p2

I−(k(ξ), ξ, y)

Y −(k(ξ), ξ)dp, (4.7)

J4(x, y, t) = − 1

2π i

1

2π i

i∞∫−i∞

eξt dξ

i∞∫−i∞

epx Y +

K (p) + ξI−

(k(ξ), ξ, y

) Y +(p, ξ)

Y −(k(ξ), ξ)dp. (4.8)

Denote

J j(t)φ =+∞∫0

J j(x, y, t)φ(y)dy, x > 0.

A similar consideration to that in the book [4] proves that for n = 0,1,∥∥∂nx J1(t)φ

∥∥Ls,μ � Ct− 2

3 ( 1r − 1+μ

s )−n∥∥φ(·)∥∥Lr + t− 2

3 ( 1r − 1

s )−n∥∥φ(·)∥∥Lr,μ . (4.9)

Now we estimate J2(t)φ. Let the contours Ci be defined as

C1 = {p ∈ (∞e−i( π

2 +ε1),0) ∪ (

0,∞ei( π2 +ε1)

)}, (4.10)

C2 = {q ∈ (∞e−i( π

2 −ε2),0) ∪ (

0,∞ei( π2 −ε2)

)}, (4.11)

C3 = {ξ ∈ (∞e−i( π

2 +ε3),0) ∪ (

0,∞ei( π2 +ε3)

)}, (4.12)

where ε j > 0 are small enough, can be chosen such that all functions under integration are analytic and Re k(ξ) > 0 forξ ∈ C3. In particular, for example, K (p) + ξ = 0 outside the origin for all p ∈ C1 and ξ ∈ C3.

From the integral representation (3.3), making use of (4.3) and estimate∣∣Y ±∣∣ � C

we have for y > 0, ξ ∈ C3 and p ∈ C1

I+(z, ξ, y) = 1

2π i

∫C2

1

q − pe−qy(e−Γ +(q,ξ) − e−Γ +(∞,ξ)

) q34

(q − k(ξ))34

dq

+ 1

2π ie−Γ +(∞,ξ)

∫C2

1

q − pe−qy

(q

34

(q − k(ξ))34

− q34

(q + k(ξ))34

)dq.

Using for γ ∈ [0, 23 )

∣∣e−Γ +(q,ξ) − e−Γ +(∞,ξ)∣∣ � C |q|− 3γ

2 |ξ |γ


and for δ ∈ [0,1)

∣∣∣∣ q34

(q − k(ξ))34

− q34

(q + k(ξ))34

∣∣∣∣ � C

∣∣∣∣k(ξ)

〈q〉∣∣∣∣δ

{q} 34

we get

∣∣I+(z, ξ, y)∣∣ � C

∫C2

1

|q − p|e−C |q|y|q|− 3γ2 |ξ |γ {q} 3

4 dq + C

∫C2

1

|q − p|e−C |q|y

∣∣∣∣k(ξ)

〈q〉∣∣∣∣δ

{q} 34 dq.

Also we have k(ξ) = O (〈ξ〉 23 ). After this observation in accordance with the integral representation (3.4) by Hölder inequality

we have arrived at the following estimate for r > 1, s > 1, l−1 = 1 − r−1, small μ � 0, n = 0,1

∥∥∥∥∥∂nx

+∞∫0

J2(·, y, t)φ(y)dy

∥∥∥∥∥Ls,μ

� C

∫C3

e−C |ξ |t∫

C1

dp|p|n− 1+μ

s

|K (p) + ξ | {p}− 34

×∫

C2

dq

|q − p| {q} 34

(|q|− 3γ

2 |ξ |γ +∣∣∣∣k(ξ)

〈q〉∣∣∣∣δ)∥∥e−C |q|·∥∥

Ll ‖φ‖Lr

� C‖φ‖Lr

∫C3

e−C |ξ |t∫

C1

dp|p|n− 1+μ

s −1+ 1r

||p| 32 + ξ |

(|p|− 3γ

2 |ξ |γ +∣∣∣∣k(ξ)

〈p〉∣∣∣∣δ)

×∫

C2

dq1

|q − ei arg p| |q|−1+ 1r {q} 3

4

(|q|− 3γ

2 |ξ |γ +∣∣∣∣ 1

〈q〉∣∣∣∣δ)

� Ct− 23 ( 1

r − 1+μs +n)‖φ‖Lr . (4.13)

Using k(ξ) = O (〈ξ〉 23 ) can be obtained in the same manner for j = 3,4, r = 1,∞

∥∥∥∥∥∂nx

+∞∫0

J j(·, y, t)φ(y)dy

∥∥∥∥∥Ls,μ

� Ct− 23 ( 1

r − 1+μs +n)

∥∥φ(·)∥∥Lr . (4.14)

Therefore according (4.13) and (4.14) we obtain the estimate (4.1) of lemma.Now we prove that uniformly with respect to x > 0 for small μ > 0

G(x, y, t) = t− 23 Λ

((x − t)t− 2

3) + yμ O

(t− 2+μ

3), (4.15)

where the function Λ(s) was defined by (1.5).To obtain asymptotics for G(x, y, t) we use method of stationary phase. For large t , integrand eξt oscillate rapidly and

cancel themselves over most of the range. Cancellation does not occur, however, in the neighborhood of zero, because eξt

changes relatively slowly near stationary point. Therefore since k(ξ) = 1 + O (ξ) for small |ξ | < 1 we rewrite the Greenfunction as

G(x, y, t) = M(x, y, t) + R(x, y, t),

where

M(x, y, t) = − 1

2π i

1

2π i

i∞∫−i∞

eξt dξ

i∞∫−i∞

epx Y +(p,0)

K (p) + ξ

×(

I−(p,0, y) − I−(1,0, y)Y −(p,0)

Y −(1,0)

)dp (4.16)

and

R(x, y, t) = G(x, y, t) − M(x, y, t). (4.17)


Firstly we estimate the function M(x, y, t). By Cauchy theorem taking residue in the point ξ = −K (p) we get

M(x, y, t) = − 1

2π i

i∞∫−i∞

epx−K (p)t Y +(p,0)

×(

I−(p,0, y) − I−(1,0, y)Y −(p,0)

Y −(1,0)

)dp.

Since we have the following representation K (p) = p + p3

6 + |p| 32 + O (p3+γ ) in the neighborhood of p = 0. Thus near the

point p = 0 integrand has the following representation

t− 23 ep(x−t)t−

23 −|p| 3

2

(p − 1

p

) 34

e X+(p,0)

(I−1

(p,0, yt− 2

3) − I−1

(t

23 ,0, yt− 2

3) Y −

1 (p,0)

Y −1 (t

23 ,0)

),

where

X+(z, ξ) = eΓ +1 (z,ξ),

Γ1(z, ξ) = 1

2π i

i∞∫−i∞

1

q − zln

|q| 32

q32

[(q + 1)

(q − 1)

] 34

dq,

I1(z,0, y) = 1

2π i

i∞∫−i∞

e−qy

q − z

(q

q − 1

) 34 1

X+(q,0)dq,

Y1(z,0) = 1

2π i

i∞∫−i∞

1

q − z

(q

q − 1

) 34 1

X+(q,0)

1√|q| dq.

On integrating this function the neighborhood of p = 0 by extending the limits to −i∞ and +i∞ we obtain the contribu-tions to M(x, y, t) from the neighborhood of p = 0

M(x, y, t) = − 1

2π it− 2

3

i∞∫−i∞

ep(x−t)t−23 −|p| 3

2

(p − 1

p

) 34

X+(p,0)

×(

I−1(

p,0, yt− 23) − I−1

(t

23 ,0, yt− 2

3) Y −

1 (p,0)

Y −1 (t

23 ,0)

)dp. (4.18)

We have by direct calculation

Γ +1 (z, ξ) = 1

2π i

i∞∫−i∞

1

q − zln

|q| 32

q32

[(q + 1)

(q − 1)

] 34

dq

= 3

4ln(1 − p)

and therefore

I−1(

p,0, yt− 23) = 1

2π ilim

z→p,Re z<0

i∞∫−i∞

1

q − z

(q

q − 1

) 34 1

(1 − q)34

dq + O(

yμt− 2μ3

),

I−1(t

23 ,0, yt− 2

3) = 1

2π i

i∞∫−i∞

e−qyt−23

q − t23

(q

q − 1

) 34 1

(1 − q)34

dq

= 1

2π i

i∞∫e−qyt−

23

q − t23

(q

q + 1

) 34 1

(1 − q)34

dq

−i∞


+ 1

2π i

i∞∫−i∞

e−qyt−23

q − t23

1

(1 − q)34

((q

q − 1

) 34

−(

q

q + 1

) 34)

dq

= −e−yt− 12 + O

(t− 2

3).

Also by definition (3.4)

Y −1

(t

23 ,0

) = 1

2π i

i∞∫−i∞

1

q

(q

q − 1

) 34 1

(1 − q)34

1√|q| dq

+ 1

2π i

i∞∫−i∞

t23

(q − t23 )q

(q

q − 1

) 34 1

(1 − q)34

1√|q| dq

= O (1).

Substituting the obtained relations into definition of M(x, y, t) we obtain

M(x, y, t) = (−1)34

2π it− 2

3

i∞∫−i∞

ep(x−t)t−23 −|p| 3

2 p− 34 dp + yμ

(t− 1+2μ

3). (4.19)

To estimate the contribution R1 to M(x, y, t) over the rest of the range of integration we can use Watson’s lemma. Due tothe smoothening properties of the integrand function by integrating by parts and using that in the domain of integrationx − K ′(p)t = 0 we get that for μ ∈ (0,1)

R1(x, y, t) = yμ O(t− 1+2μ

3). (4.20)

Having elucidated the principles by way of the proof of the estimates (4.13) and (4.14) similar results can be found for

R(x, y, t) = G(x, y, t) − M(x, y, t)

= yμ O(t− 1+2μ

3). (4.21)

Asymptotics of G(x, y, t) is obtained by expression (4.19) over a stationary point in the range of integration and adding(4.20) and (4.21). Substituting this asymptotics of G(x, y, t) into definition of operator Gφ we prove the estimate (4.2).Lemma is proved. �5. Proof of theorem

By Theorem 3 we rewrite the initial-boundary value problem (1.1) as the following integral equation

u(t) = G(t)u0 −t∫

0

G(t − τ )N(u(τ )

)dτ ,

N(u(τ )

) = uxu, (5.1)

where G is the Green operator of the linear problem (3.1). We choose the space

Z = {φ ∈ L1,a(R+)} ∩ L∞(

R+)with a > 0 small and the space

X = {φ ∈ C

([0,∞);Z) ∩ C

((0,∞);H1∞

(R+))

: ‖φ‖X < ∞},

where now the norm

‖φ‖X = supt�0

({t}− a

3 〈t〉− 23 a

∥∥φ(t)∥∥

L1,a + ∥∥φ(t)∥∥

L1 +1∑

n=0

t23 (n+1)

∥∥∂nx φ(t)

∥∥L∞

)

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

ρ = 1 ‖u0‖Z > 0.
2C


For u ∈ Xρ we define the mapping M(u) by formula

M(u) = G(t)u0 −t∫

0

G(t − τ )N(u(τ )

)dτ . (5.2)

By Lemma 4 following the method of the proof of Theorem 1.19 from book [4] we prove that the mapping M(u) is acontraction mapping and therefore there exists a unique solution v ∈ C([0,∞);L1(R+) ∩ L1,a(R+)) ∩ C((0,∞);H1∞) to theinitial-boundary value problem (1.1).

Now we can prove asymptotic formula

u(x, t) = A1Λ(xt−2)t− 2

3 + O(t− 2

3 −γ), (5.3)

where

A1 = f (u0).

Denote

G0(t) = t− 23 Λ

((x − t)t− 2

3).

From Lemma 4 we have

t23∥∥G(t)φ − G0(t) f (φ)

∥∥L∞ � C‖φ‖Z (5.4)

for all t > 1. Also in view of the definition of N (u(τ )) we have

f(

N(u(τ )

)) = 0.

By a direct calculation we have

∥∥∥∥∥t2∫

0

(G(t − τ )N

(u(τ )

) − G0(t − τ ) f(

N(u(τ )

)))dτ

∥∥∥∥∥L∞

+∥∥∥∥∥

t∫t2

G(t − τ )N(u(τ )

)dτ

∥∥∥∥∥L∞

� C

t2∫

0

(t − τ )−2+μ

3∥∥N

(u(τ )

)∥∥L1,μ dτ + C

t∫t2

∥∥N(u(τ )

)∥∥L1 dτ

� Ct− 23 −γ ‖u‖2

X (5.5)

for all t > 1. By virtue of the integral equation (5.1) we get

〈t〉γ + 23∥∥(

u(t) − AG0(t))∥∥

X �∥∥(

G(t)u0 − G0(t))

f (u0)∥∥

L∞

+ 〈t〉γ + 23

∥∥∥∥∥t2∫

0

(G(t − τ )N

(u(τ )

) − G0(t − τ ) f(

N(u(τ )

)))dτ

∥∥∥∥∥L∞

+ 〈t〉γ + 23

∥∥∥∥∥t∫

t2

G(t − τ )N(u(τ )

)dτ

∥∥∥∥∥L∞

. (5.6)

The all summands in the right-hand side of (5.6) are estimated by C‖u0‖Z + C‖v‖2X via estimates (5.4)–(5.5). Thus by (5.6)

the asymptotic (5.3) is valid. Theorem is proved.

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Whitham equation with Landau damping on a half-line