The initial-boundary-value problem of a generalized Boussinesq equation on the half line

J. Korean Math. Soc. 45 (2008), No. 1, pp. 79–95

THE INITIAL-BOUNDARY-VALUE PROBLEM OF A GENERALIZED BOUSSINESQ EQUATION

ON THE HALF LINE

Ruying Xue

Reprinted from the

Journal of the Korean Mathematical Society Vol. 45, No. 1, January 2008

c

°2008 The Korean Mathematical Society

THE INITIAL-BOUNDARY-VALUE PROBLEM OF A GENERALIZED BOUSSINESQ EQUATION

ON THE HALF LINE

Ruying Xue

Abstract. The local existence of solutions for the initial-boundary value problem of a generalized Boussinesq equation on the half line is consid- ered. The approach consists of replacing the Fourier transform in the initial value problem by the Laplace transform and making use of mod- ern methods for the study of nonlinear dispersive wave equation

1. Introduction

We consider the initial-boundary value problem for the generalized Boussi- nesq equation on the half line

(1.1)

 



u tt − u xx + u xxxx + (|u| ^κ u) xx = 0, t > 0, x > 0, u(0, t) = h 1 (t), u x (0, t) = h 2 (t),

u(x, 0) = f (x), u t (x, 0) = ∂ x h(x),

where 0 < κ < 4, the velocity is an x-derivative function. Equations of type (1.1) are a class of essential model equations appearing in physics and fluid mechanics. It is derived by Boussinesq to describe two-dimensional irrotational flows of an inviscid liquid in a uniform rectangular channel. And it also arises in a large range of physical phenomena including the propagation of ion-sound waves in a plasma and nonlinear lattice wave.

The study of the initial-value problem for the Boussinesq-type equation has recently attracted considerable attention of many mathematicians and physi- cists ( See [1], [2] and references therein). For instance, Bona and Sachs in [2] proved that the initial-value problem of the Boussinesq equation is lo- cally well posed for smooth data by using Kato’s abstract theory of quasi- linear evolution equation. They proved that for any f (x) ∈ H ^s+2 (R) and h(x) ∈ H ^s+1 (R) with some s > 1/2, there exists a time T > 0 such that the initial-value problem of the Boussinesq equation has a unique solution u ∈ C([0, T ]; H ^s+2 (R)) ∩ C ¹ ([0, T ]; H ^s (R)) ∩ C ² ([0, T ]; H ^s−2 (R)). In the case

Received May 18, 2006; Revised September 26, 2006.

2000 Mathematics Subject Classification. 35Q35, 35G30.

Key words and phrases. Boussinesq equation, existence, initial-boundary-value problem.

Supported by NSFC (10571158) and Zhejiang Provincial NSF of China (Y605076).

c

°2008 The Korean Mathematical Society

79

1 < κ < 4, Linares in [8] established the local and global existence the- ory for the initial-value problem of the Boussinesq equation when (f, ∂ x h) ∈ L ² (R)× ˙ H ⁻¹ (R) or (f, ∂ x h) ∈ H ¹ (R)×L ² (R), by the so-called L ^p −L ^q smooth- ing effect of the Strichartz type. In [12], Xue considered the case κ > 4 and proved some local and global existence results when the initial data belong to some suitable Besov spaces.

The practical, quantitative use of the Boussinesq equation and its rela- tives does not always involve the pure initial-value problem. Instead, initial- boundary value problems either on a finite domain or on the half line often come to the fore. The main difficulty in the study of initial-boundary-value problems is the evaluation of the contribution of the boundary data. In the literature, very few results are available. By using the finite element Galerkin method, Pani and Saranga in [10] got some local existence and uniqueness of the solu- tions of the initial-boundary-value problems of the Boussinesq equation on a finite domain with homogenous boundary conditions. In [11] Varlamov consid- ered the damped Boussinesq equation with the initial-boundary-value problem in a unit disk, and the global-in-time solvability was obtained by using the Fourier-Bessel series. As far as we know the existence of the initial-boundary- value problem of a generalized Boussinesq equation on the half line was not considered previously. In this paper we consider the local existence of the initial-boundary-value problem for the Boussinesq-type equations on the half line. The approach consists of replacing the Fourier transform in the initial- value problem by the Laplace transform and making use of modern methods for the study of nonlinear dispersive wave equation. The main result obtained in this paper is

Theorem 1.1. Assume that f (x)∈ L ² (R ⁺ ), h(x)∈ H ₀ ⁻¹ (R ⁺ ), h 1 (t)∈ H ₀

¹⁴

(R ⁺ ) and h 2 (t)∈ H ⁻ ₀

¹²

^,−

¹⁴

(R ⁺ ). Then there exists a positive constant T > 0, which depends only on

kf k L

²

(R

⁺

) + khk H

⁻¹

(R

⁺

) + kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

^{− 1}₀^{2,− 14}

(R

⁺

) ,

such that the initial-boundary-value problem (1.1) possess a unique local solution u(t, x) satisfying

u ∈ C([0, T ]; L ² _x (R ⁺ )) ∩ L ⁴ _t ([0, T ]; L ^∞ _x (R ⁺ )).

In the sequel, we denote by R ⁺ the open right half-line {x : x > 0} and

denote by C some large constant which may vary from line to line. The notation

A ∼ B means that there exist two harmless positive constants C 1 and C 2 such

that C 1 A ≤ B ≤ C 2 A. For a Banach space X, we denote by k · k X the norm

in X. We also denote by χ(x) the function satisfying χ(x) = 1 for x ∈ R ⁺ and

χ(x) = 0 for x 6∈ R ⁺ .

The rest of this paper is organized as follows. In section 2 we prove some smoothing effects for the linear Boussinesq equation with inhomogeneous initial- boundary data. The existence and the uniqueness of the solution is given in Section 3.

2. Linear estimates

In this section we give some smoothing effects for the linear equation asso- ciated to (1.1). These estimates will be the main ingredient in the proof of local existence of the initial-boundary-value problem (1.1). Consideration is first directed to the linear initial-boundary-value problem

(2.1)

 



u tt − u xx + u xxxx = 0, t > 0, x > 0, u(0, t) = h 1 (t), u x (0, t) = h 2 (t), u(x, 0) = 0, u t (x, 0) = 0.

Denote by

C 0 (R ⁺ ) = {u ∈ C(R) : u = 0 for x ≤ 0}.

Lemma 2.1. Assume that h 1 , h 2 ∈ C 0 (R ⁺ ). Then the solution, W b (t)(h 1 , h 2 ), of the linear problem (2.1) has an explicit formula

u(x, t) = W b (t)(h 1 , h 2 ) = U 1 (x, t) + U 2 (x, t) + U 1 (x, t) + U 2 (x, t), where

U 1 (x, t) = 1 2π

Z _+∞

1

2µ ² − 1 p µ ² − 1(µ + i p

µ ² − 1) e ^iµ

√ µ

²

−1t e ^−µx µZ _+∞

0

³ i p

µ ² − 1h 1 (ξ) − h 2 (ξ)

´ e ^−iµ

√ µ

²

−1ξ dξ

¶ dµ,

U 2 (x, t) = − 1 2π

Z _+∞

1

2µ ² − 1 p µ ² − 1(µ + i p

µ ² − 1) e ^iµ √

µ

²

−1t e ⁱ √

µ

²

−1x

µZ _+∞

0

(µh 1 (ξ) + h 2 (ξ)) e ^−iµ √

µ

²

−1ξ dξ

¶ dµ.

Proof. As a potential global solution of (2.1) is defined on a half-line R ⁺ in each of the two independent variables x and t, it is not unnatural to think of replacing the use of the Fourier transform with the Laplace transform. By taking the Laplace transform with respect to t of both sides of the equation in (2.1), the initial-boundary-value problem is converted to a one-parameter family of four-order, boundary-value problems

(2.2)

½ λ ² u(x, λ) − ˆ ˆ u(x, λ) xx + ˆ u(x, λ) xxxx = 0, Re(λ) ≥ 0, x > 0, ˆ

u(0, λ) = ˆh 1 (λ), ˆ u x (0, λ) = ˆh 2 (λ), ˆ u(+∞, λ) = 0, ˆ u x (+∞, λ) = 0,

where λ is the dual variable, ˆ u(x, λ), ˆh 1 (λ) and ˆh 2 (λ) are the Laplace transform

of u(x, t), h 1 (t) and h 2 (t) with respect to t, respectively. Let γ 1A , γ 2A , γ 3A and

γ 4A be the four roots of the characteristic equation

γ ⁴ − γ ² + λ ² = 0, λ ∈ A = {λ : Re(λ) > 1/2},

ordered so that Re(γ 1A ) < 0, Re(γ 2A ) < 0, Re(γ 3A ) > 0 and Re(γ 4A ) > 0.

It is obvious that γ 1A , γ 2A , γ 3A and γ 4A are analytic for Re(λ) > 1/2 and continuous for Re(λ) ≥ 1/2 except at λ = 1/2. As both ˆ u(x, λ) and ˆ u x (x, λ) tend to zero as x → +∞, it is concluded that for any λ with Re(λ) > 1/2

ˆ

u(x, λ) = 1 γ 2A − γ 1A

h

(γ 2A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1A

^x − (γ 1A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2A

^x i

. Thus, for any fixed p with Re(p) > 1/2, one has the representation for x > 0 and t > 0,

u(x, t) = 1 2πi

Z _p+i∞

p−i∞

e ^λt u(x, λ)dλ ˆ

= 1

2πi Z _p+i∞

p−i∞

e ^λt γ 2A − γ 1A

h

(γ 2A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1A

^x (2.3)

−(γ 1A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2A

^x i dλ.

A little analysis shows that

γ 1A (λ) = γ 2A (¯ λ), γ 3A (λ) = γ 4A (¯ λ) for Re(λ) ≥ 1/2, λ 6= 1/2, and

|γ _1A (λ) − γ _2A (λ)| = O(|λ − 1/2| ^1/2 ) as λ → 1/2, Re(λ) ≥ 1/2.

As this singularity is integrable, we may let p → 1/2 in (2.3), thereby obtaining for x > 0 and t > 0,

u(x, t) = 1 2πi

Z

¹

2

+i∞

12

−i∞

e ^λt γ 2A − γ 1A

h

(γ 2A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1A

^x (2.4)

−(γ 1A ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2A

^x i dλ.

Let γ 1B , γ 2B , γ 3B and γ 4B be the four roots of the characteristic equation γ ⁴ − γ ² + λ ² = 0, λ ∈ B = {λ : 0 < Re(λ) < 1/2},

ordered so that Re(γ 1B ) < 0, Re(γ 2B ) < 0, Re(γ 3B ) > 0 and Re(γ 4B ) > 0. It is obvious that γ 1B , γ 2B , γ 3B and γ 4B are analytic for λ ∈ B and continuous for 0 ≤ Re(λ) ≤ 1/2 except at λ = 1/2 and λ = 0. By the uniqueness and the continuity of the root of the characteristic equation γ ⁴ − γ ² +λ ² = 0 on the half lines Γ + = {λ : Re(λ) = 1/2, Im(λ) > 0} and Γ − = {λ : Re(λ) = 1/2, Im(λ) <

0}, we assume, without loss of generality,

γ 1B = γ 1A , γ 2B = γ 2A , λ ∈ Γ + . Then we have

γ 1B = γ 1A , γ 2B = γ 2A , λ ∈ Γ − ,

or

γ 1B = γ 2A , γ 2B = γ 1A , λ ∈ Γ − . By symmetry we deduce from (2.4) that

u(x, t) = 1 2πi

Z

¹

2

+i∞

1 2

−i∞

e ^λt γ 2B − γ 1B

h

(γ 2B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1B

^x (2.5)

−(γ 1B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2B

^x i

dλ.

A little analysis shows that

γ 1B (λ) = γ 1B (¯ λ), γ 2B (λ) = γ 2B (¯ λ) for 0 < Re(λ) < 1/2

|γ 1B (λ) − γ 2B (λ)| = O(|λ − 1/2| ^1/2 ), as λ → 1/2, 0 < Re(λ) ≤ 1/2,

|γ 1B (λ) − γ 2B (λ)| = O(1), as λ → 0, 0 ≤ Re(λ) < 1/2, and for λ ∈ B, |λ| → +∞,

|γ 1B (λ)| ∼ |λ| ^1/2 , |γ 2B (λ)| ∼ |λ| ^1/2 , Re(γ 1B (λ)) ∼ −|λ| ^1/2 , Re(γ 2B (λ)) ∼ −|λ| ^1/2 .

Hence, we are allowed to change the contour on the basis of Cauchy’s Theorem and thereby determine from (2.5) that

u(x, t) = 1 2πi

Z _0+i∞

0−i∞

e ^λt γ 2B − γ 1B

h

(γ _2B ˆh 1 (λ) − ˆh ₂ (λ))e ^γ

^1B

^x (2.6)

−(γ _1B ˆh 1 (λ) − ˆh ₂ (λ))e ^γ

^2B

^x i dλ.

Denote by

U 1 (x, t) = 1 2πi

Z _0+i∞

0+i0

e ^λt

γ 2B − γ 1B (γ 2B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1B

^x dλ, U 2 (x, t) = − 1

2πi Z _0+i∞

0+i0

e ^λt

γ 2B − γ 1B (γ 1B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2B

^x dλ.

Note that

γ 1B (λ) = γ 1B (¯ λ), γ 2B (λ) = γ 2B (¯ λ) for 0 ≤ Re(λ) ≤ 1/2 with λ 6= 0, λ 6= 1 2 and

ˆh 1 (λ) = ˆh 1 (¯ λ), ˆh 2 (λ) = ˆh 2 (¯ λ) for 0 ≤ Re(λ) ≤ 1/2 with λ 6= 0, λ 6= 1 2 . By direct computation, it follows that for x > 0 and t > 0

1 2πi

Z _0+i0

0−i∞

e ^λt γ 2B − γ 1B

(γ 2B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^1B

^x dλ = U 1 (x, t) and

− 1 2πi

Z _0+i0

0−i∞

e ^λt

γ 2B − γ 1B (γ 1B ˆh 1 (λ) − ˆh 2 (λ))e ^γ

^2B

^x dλ = U 2 (x, t).

Then we get for x > 0 and t > 0

(2.7) u(x, t) = W b (t)(h 1 , h 2 ) = U 1 (x, t) + U 2 (x, t) + U 1 (x, t) + U 2 (x, t).

For λ ∈ {λ : Re(λ) = 0, Im(λ) > 0}, it is convenient to make the change of variables

λ = iµ p

µ ² − 1 for µ ≥ 1

in the characteristic equation γ ⁴ − γ ² + λ ² = 0. In terms of µ, the four roots are

γ 1B = −µ, γ 2B = i p

µ ² − 1, γ 3B = µ and γ 4B = −i p µ ² − 1, and the integral U 1 (x, t) and U 2 (x, t) may be rewritten as

U ₁ (x, t) = 1 2π

Z _+∞

1

2µ ² − 1 p µ ² − 1(µ + i p

µ ² − 1) e ^iµ √

µ

²

−1t e ^−µx µZ _+∞

0

³ i p

µ ² − 1h 1 (ξ) − h 2 (ξ) ´ e ^−iµ √

µ

²

−1ξ dξ

¶ dµ,

U 2 (x, t) = − 1 2π

Z _+∞

1

2µ ² − 1 p µ ² − 1(µ + i p

µ ² − 1) e ^iµ √

µ

²

−1t e ⁱ √

µ

²

−1x

µZ _+∞

0

(µh 1 (ξ) + h 2 (ξ)) e ^−iµ √

µ

²

−1ξ dξ

¶ dµ.

We complete the proof. ¤

For α, β ∈ R we define

H ^α,β (R) = {f : |ξ| ^α (1 + |ξ|) ^β−α f (ξ) ∈ L ˆ ² (R)}

with

kf k _H

α,β

(R) =

° °

°|ξ| ^α (1 + |ξ|) ^β−α f (ξ) ˆ

° °

° L

²

(R) , where ˆ f is the Fourier transform in x of the function f (x). Let

H ^α,β ₀ (R ⁺ ) = {f ∈ H ^α,β (R) : suppf ⊆ [0, ∞)}

with

kf k _H

α,β

0

(R

⁺

) = kf k _H

α,β

(R) , and let

H ^α,β (R ⁺ ) = {f = F | R

⁺

, F ∈ H ^α,β (R)}

with

kf k _H

α,β

(R

⁺

) = inf{kF k _H

α,β

(R) , f = F | _R

⁺

}.

Denote by

H ₀ ^β (R ⁺ ) = H ^0,β ₀ (R ⁺ ), H ^β (R ⁺ ) = H ^0,β (R ⁺ ) and

H ˙ ₀ ^α (R ⁺ ) = H ₀ ^α,0 (R ⁺ ), ˙ H ^α (R ⁺ ) = H ^α,0 (R ⁺ ).

For any α and β with α ≤ 0 and α ≤ β, we have

(2.8) kf k _H

α,β

(R) ∼ kf k H ˙

^α

(R) + kf k _H

β

(R)

because of

|ξ| ^α (1 + |ξ|) ^β−α ∼ |ξ| ^α + (1 + |ξ|) ^β .

Lemma 2.2. Assume α, β and f satisfy one of the following conditions:

• α ∈ (− ¹ ₂ , 0] and β ∈ (− ¹ ₂ , ¹ ₂ ) with α ≤ β, and f ∈ H ^α,β (R);

• α ∈ (− ¹ ₂ , 0] and β ∈ ( ¹ ₂ , ³ ₂ ), f ∈ H ^α,β (R) and f (0) = 0.

Then χ(x)f (x) ∈ H ^α,β ₀ (R ⁺ ), and kχ(x)f (x)k _H

α,β

0

(R

⁺

) ≤ Ckf k _H

α,β

(R) .

Proof. For 0 ≤ β < ¹ ₂ and f 1 (x) ∈ H ^β (R), or for ¹ ₂ < β < ³ ₂ and f 1 (x) ∈ H ^β (R) with f (0) = 0, that χ(x)f 1 (x) ∈ H ₀ ^β (R ⁺ ) and

(2.9) kχ(x)f ₁ (x)k _H

β

0

(R

⁺

) ≤ Ckf ₁ (x)k _H

β

(R)

comes from Lemma 2.3 and Proposition 2.4 in [4].

Note that H ₀ ^β (R ⁺ ) is the dual space of H ^−β (R ⁺ ) for β < 0 ( see Proposition 2.1 in [4]). For any g(x) ∈ H ^−β (R ⁺ ), we can choose F ∈ H ^−β (R) with g = F | R

⁺

and kgk _H

^−β

_(R

⁺

₎ ∼ kF k _H

^−β

_(R) . Thus, for − ¹ ₂ < β < 0 and f 1 (x) ∈ H ^β (R) we have the following chain of inequalities

|hχ(x)f 1 (x), g(x)i| ≤ |hf 1 (x), χ(x)F (x)i|

≤ kf 1 (x)k _H

β

(R) kχ(x)F (x)k _H

−β

0

(R) ≤ Ckf 1 (x)k _H

β

(R) kχ(x)F (x)k _H

−β 0

(R

⁺

)

≤ Ckf 1 (x)k _H

^β

_(R) kF (x)k _H

^−β

_(R) ≤ Ckf 1 (x)k _H

^β

_(R) kg(x)k _H

^−β

_(R

⁺

₎ ,

the last two inequalities hold because of Lemma 2.3 in [4]. Then, χ(x)f 1 (x) ∈ H ₀ ^β (R ⁺ ) and

(2.10) kχ(x)f 1 (x)k _H

^β

0

(R

⁺

) ≤ Ckf 1 (x)k _H

^β

_(R) . For − ¹ ₂ < α ≤ 0 and f 2 ∈ ˙ H ^α (R), we have

kf 2 k _H

^α

_(R) = k(1 + |ξ|) ^α f ˆ 2 (ξ)k _L

²

ξ

(R) ≤ k |ξ| ^α f ˆ 2 (ξ)k _L

²

ξ

(R) = kf 2 k H ˙

^α

(R) , where ˆ f 2 (ξ) is the Fourier transform with respective to x of the function f 2 (x).

The similar argument to that used in (2.10) yields χ(x)f 2 (x) ∈ H ₀ ^α (R ⁺ ) and kχ(x)f 2 (x)k _H

₀^α

_(R

⁺

₎ ≤ Ckf 2 k _H

^α

_(R) ≤ Ckf 2 k H ˙

^α

(R) ,

which, together with Proposition 2.8 in [4], implies χ(x)f 2 (x) ∈ ˙ H ₀ ^α (R ⁺ ) and (2.11) kχ(x)f 2 (x)k H ˙

^α₀

(R

⁺

) ≤ Ckχ(x)f 2 (x)k _H

₀^α

_(R

⁺

₎ ≤ Ckf 2 k H ˙

^α

(R) . Then the lemma follows from a combination of (2.9), (2.10) and (2.11) with

(2.8). The proof is completed. ¤

Lemma 2.3. There exists a positive constant C such that, for h 1 ∈ H ₀

¹⁴

(R ⁺ ) and h 2 ∈ H ⁻ ₀

¹²

^,−

¹⁴

(R ⁺ ),

sup

t≥0 kW b (t)(h 1 , h 2 )k L

²_x

(R

⁺

) ≤ C µ

kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

^{− 1}₀^{2,− 14}

(R

⁺

)

¶ .

Proof. It suffices to prove the lemma for h 1 , h 2 ∈ C ₀ ^∞ (R ⁺ ). It follows from Lemma 3.1 in [3] that

(2.12)

kU 1 (x, t)k _L

²_x

_(R

⁺

₎

≤ C

° °

° °

° µ − i p

µ ² − 1 p µ ² − 1 e ^iµ

√ µ

²

−1t

×

µZ _+∞

0

(i p

µ ² − 1h 1 (ξ) − h 2 (ξ))e ^−iµ √

µ

²

−1ξ dξ

¶° °

° °

L

²_µ

(1,+∞)

≤ C

° °

° °

p 2µ ² − 1 Z _+∞

0

h 1 (ξ)e ^−iµ

√ µ

²

−1ξ dξ

° °

° °

L

²_µ

(1,+∞)

+ C

° °

° °

°

p 2µ ² − 1 p µ ² − 1

Z _+∞

0

h 2 (ξ)e ^−iµ √

µ

²

−1ξ dξ

° °

° °

° L

²_µ

(1,+∞)

≤ C

° °

° °(1 + η)

1 4

Z _+∞

0

h 1 (ξ)e ^−iηξ dξ

° °

° °

L

²_η

(0,+∞)

+ C

° °

° °η ⁻

1

2

(1 + η) ⁻

³⁴

Z _+∞

0

h 2 (ξ)e ^−iηξ dξ

° °

° °

L

²_η

(0,+∞)

≤ C µ

kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

^{− 1}₀^{2,− 14}

(R

⁺

)

¶ .

Note that for U (x) =

Z _∞

1

g(µ)e ⁱ √

µ

²

−1x dµ = Z _∞

0

g(µ(s)) s

µ(s) e ^isx ds where µ(s) ≥ 0 is the solution satisfying s = p

µ ² − 1 for s ≥ 0, we have kU (x)k L

²_x

(0,∞) = kg(µ(s)) s

µ(s) k L

²_s

(0,∞) = kg(µ)(µ ² − 1)

¹⁴

µ ⁻

¹²

k L

²_µ

(1,∞) . Then we have

(2.13)

kU 2 (x, t)k _L

²_x

_(R

⁺

₎

≤ C

° °

° °µ

12

p 2µ ² − 1(µ ² − 1) ⁻

¹⁴

Z _+∞

0

h 1 (ξ)e ^−iµ √

µ

²

−1ξ dξ

° °

° °

L

²_µ

(1,+∞)

+ C

° °

° °µ ⁻

1 2

p 2µ ² − 1(µ ² − 1) ⁻

¹⁴

Z _+∞

0

h 2 (ξ)e ^−iµ √

µ

²

−1ξ dξ

° °

° °

L

²_µ

(1,+∞)

≤ C

° °

° °(1 + η)

1 4

Z _+∞

0

h 1 (ξ)e ^−iηξ dξ

° °

° °

L

²_η

(0,+∞)

+ C

° °

° °(1 + η) ⁻

1 4

Z _+∞

0

h 2 (ξ)e ^−iηξ dξ

° °

° °

L

²_η

(0,+∞)

≤ Ckh 1 k

H

₀¹⁴

(R

⁺

) + Ckh 2 k

H

₀^{− 14}

(R

⁺

) .

The lemma now follows from (2.12) and (2.13) together with Lemma 2.1. We

complete the proof. ¤

Lemma 2.4. There exists a positive constant C such that, for h 1 ∈ H ₀

¹⁴

(R ⁺ ) and h 2 ∈ H ₀ ⁻

¹⁴

(R ⁺ ),

µZ _∞

0

kW b (t)(h 1 , h 2 )k ⁴ _L

∞ x

(R

⁺

) dt

¶

¹

4

≤ C µ

kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

₀^{− 14}

(R

⁺

)

¶ . Proof. For given real functions K ∈ C(R) and h ∈ C 0 (R ⁺ ) satisfying K(−µ) = K(µ), we define

G(K, h) = Z _+∞

1

K(µ)e ^−iµ

√ µ

²

−1t e ^−|µ|x

µZ _+∞

0

h(ξ)e ^iµ

√ µ

²

−1ξ dξ

¶ dµ and

H(K, h) = Z _+∞

1

K(µ)e ^−iµ

√ µ

²

−1t e ^iµ √

1−(1/µ)

²

x

µZ _+∞

0

h(ξ)e ^iµ

√ µ

²

−1ξ dξ

¶ dµ.

Then

G(K, h) + G(K, h)

= Z

|µ|>1

K(µ)e ^−iµ √

µ

²

−1t e ^−|µ|x µZ _+∞

0

h(ξ)e ^iµ √

µ

²

−1ξ dξ

¶ dµ

=

Z _+∞

−∞

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η)

dη e ^iηt e ^−|ϕ

⁻¹

^(η)|x

µZ _+∞

0

h(ξ)e ^−iηξ dξ

¶ dη, and similarly,

H(K, h) + H(K, h)

= Z _+∞

−∞

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η)

dη e ^iηt e ^−iϕ

⁻¹

^(η) √

1−(1/ϕ

⁻¹

(η))

²

x

µZ _+∞

0

h(ξ)e ^−iηξ dξ

¶ dη, where, µ = ϕ ⁻¹ (η) is the inverse function of η = ϕ(µ) = µ p

µ ² − 1. For any g(x, t) ∈ L _t

⁴³

(R, L ¹ _x (R ⁺ )), denoting by

ˆ

g(x, −η) = 1 2π

Z _+∞

−∞

g(x, t)e ^itη dt, ˆh(η) = Z _+∞

0

h(ξ)e ^−iηξ dξ,

we deduce by H¨ older’s inequality and the Sobolev embedding theorem Z _+∞

−∞

Z _+∞

0

[G(K, h) + G(K, h)]g(x, t)dxdt

= 2π Z _+∞

−∞

Z _+∞

−∞

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η)

dη e ^−|ϕ

⁻¹

^(η)|x ˆ g(x, −η)ˆh(η)dηdx

≤ C Z _+∞

−∞

¯ ¯

¯ ¯(1 + |η|)

1

4

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η) dη ˆh(η)

¯ ¯

¯ ¯

° °

° ° ˆ g(x, −η) (1 + |η|)

¹⁴

° °

° °

L

¹_x

(R)

dη

≤ C

° °

° °(1 + |η|)

1

4

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η) dη ˆh(η)

° °

° °

L

²_η

(R)

kg(x, t)k

H

^{− 14}

(R,L

¹_x

(R

⁺

))

≤ C

° °

° °(1 + |η|)

1

4

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η) dη ˆh(η)

° °

° °

L

²_η

(R)

kg(x, t)k

L

_t⁴³

(R,L

¹_x

(R

⁺

)) , which implies that

kG(K, h) + G(K, h)k _L

⁴

t

(R

⁺

,L

^∞_x

(R

⁺

))

≤ kG(K, h) + G(K, h)k _L

⁴

t

(R,L

^∞_x

(R

⁺

))

≤ C

° °

° °(1 + |η|)

14

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η) dη ˆh(η)

° °

° °

L

²_η

(R)

. (2.14)

Similarly,

(2.15)

kH(K, h) + H(K, h)k _L

⁴

t

(R

⁺

,L

^∞_x

(R

⁺

))

≤ C

° °

° °(1 + |η|)

14

K(ϕ ⁻¹ (η)) dϕ ⁻¹ (η) dη ˆh(η)

° °

° °

L

²_η

(R)

.

Rewrite U 1 (x, t) and U 2 (x, t) as U 1 (x, t) = i

2π G(µ, h 1 )+ 1 2π G( p

µ ² − 1, h 1 )− 1

2π G( µ

p µ ² − 1 , h 2 )+ i

2π G(1, h 2 ), and

U 2 (x, t) = − 1

2π H( µ ²

p µ ² − 1 , h 1 )+ i

2π H(µ, h 1 )− 1

2π H( µ

p µ ² − 1 ,h 2 )+ i

2π H(1,h 2 ).

Then we get from (2.14) that kU 1 (x, t) + U 1 (x, t)k _L

⁴

t

(R

⁺

,L

^∞_x

(R

⁺

))

≤ C ³

kη(1 + |η|) ⁻

³⁴

ˆh 1 (η)k _L

²_η

_(R) + kη ² (1 + |η|) ⁻

⁷⁴

ˆh 1 (η)k _L

²_η

_(R) ´ +C ³

k(1 + |η|) ⁻

¹⁴

ˆh 2 (η)k _L

²_η

_(R) + kη(1 + |η|) ⁻

⁵⁴

ˆh 2 (η)k _L

²_η

_(R) ´

≤ C µ

kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

₀^{− 14}

(R

⁺

)

¶

,

(2.16)

and from (2.15),

(2.17) kU 2 (x, t)+U 2 (x, t)k _L

⁴

t

(R

⁺

,L

^∞_x

(R

⁺

)) ≤ C µ

kh 1 k

H

₀¹⁴

(R

⁺

) + kh 2 k

H

₀^{− 14}

(R

⁺

)

¶ . Then the lemma follows from (2.16) and (2.17) together with Lemma 2.1. We

complete the proof. ¤

Let φ(ξ) = ξ(1 + |ξ| ² ) ^1/2 . For f, h ∈ S ⁰ (R) define

(2.18) V (t)(f ) =

Z _+∞

−∞

e ^itφ(ξ) e ^ixξ f (ξ)dξ. ˆ

(2.19) V 1 (t)(f )) = 1 2

Z _+∞

−∞

[e ^{itφ(ξ)+ixξ} + e −itφ(ξ)+ixξ ] ˆ f (ξ)dξ and

(2.20) V 2 (t)(∂ x h) = Z _+∞

−∞

h

A(ξ)e i(−tφ(ξ)+xξ) + B(ξ)e i(tφ(ξ)+xξ) i dξ, where A(ξ) = _2(1+ξ ^{ˆ h(ξ)}

2

)

^1/2

and B(ξ) = − _2(1+ξ ^{h(ξ) ˆ}

2

)

^1/2

, ˆ f and ˆh are the Fourier transform of f and h respectively. Denote by

W R (t)(f, ∂ x h) = V 1 (t)(f ) + V 2 (t)(∂ x h).

Then u(x, t) = W R (t)(f, ∂ x h) is the formal solution to the initial-value problem (2.21)

½ u tt − u xx + u xxxx = 0, t > 0, x ∈ R, u(x, 0) = f (x), u t (x, 0) = ∂ x h.

Lemma 2.5. Let T > 0. There exists a positive constant C such that, for f ∈ L ² (R) and h ∈ H ₀ ⁻¹ (R) with h(x) = −h(−x),

kχ(t)W R (t)(f, ∂ x h)| x=0 k

H

₀¹⁴

(R

⁺

) ≤ C ³

kf k _L

²

_(R) + khk _H

⁻¹

0

(R)

´ , kχ(t)∂ x W R (t)(f, ∂ x h)| x=0 k

H

^{− 1}₀^{2,− 14}

(R

⁺

) ≤ C ³

kf k _L

²

_(R) + khk _H

⁻¹

0

(R)

´

and ÃZ _T

0

kW R (τ )(f, ∂ x h)k ⁴ _L

∞ x

(R

⁺

) dτ

!

¹

4

≤ C(1 + T

¹⁴

) ³

kf k _L

²

_(R) + khk _H

−1 0

(R)

´ . Proof. It suffices to prove the lemma for f, h ∈ C ₀ ^∞ (R) with f (0) = 0 and h(x) = −h(−x). The results in [2] show that u(x, t) = W R (t)(f, ∂ x h) is the classical solution of the initial-value problem

½ ∂ tt u − ∂ xx u + ∂ xxxx u = 0, −∞ < x < +∞, t > 0 u(x, 0) = f (x), ∂ t u(x, 0) = ∂ x h(x), −∞ < x < +∞,

and W R (t)(f, ∂ x h) ∈ C ² (R × R) with W R (t)(f, ∂ x h)| (x,t)=(0,0) = f (0) = 0.

For given α, β ∈ R, sup

x∈R kV (t)(f )k _H

α,β

(R)

= °

°|τ| ^α (1 + |τ |) ^β−α F t→τ (V (t)(f )) (τ ) °

° L

²_τ

(R)

≤ C

° °

°|τ | ^α (1 + |τ |) ^β−α f (ξ) |φ ˆ ⁰ (ξ)| ⁻¹ | _ξ=φ

⁻¹

_{(τ )}

° °

° L

²_τ

(R)

≤ C

° °

°|φ(ξ)| ^α (1 + |φ(ξ)|) ^β−α f (ξ) |φ ˆ ⁰ (ξ)| ⁻

¹²

° °

° L

²_ξ

(R)

≤ C

° °

°|ξ| ^α (1 + |ξ|) ^2β−α−

¹²

f (ξ) ˆ

° °

° L

²_ξ

(R) = C kf k

H

^{α,2β− 1}₀ ²

(R) , which, together with (2.19) and (2.20), implies

sup

x∈R

kW R (t)(f, ∂ x h)k

H

¹⁴

(R) ≤ C ³

kf k _L

2

(R

⁺

) + khk _H

⁻¹

0

(R

⁺

)

´ .

Then, χ(t)W R (t)(f, ∂ x h)| x=0 ∈ H ₀

¹⁴

(R ⁺ ) because of Lemma 2.2, and the first inequality in the lemma holds. To prove the second inequality in the lemma, we define

v(x, t) = Z _t

0

χ(τ )∂ x W R (τ )(f, ∂ x h)dτ, (x, t) ∈ R × R.

Obviously v(0, t) ∈ C 0 (R ⁺ ), ^dv(x,t) _dt = χ(t)∂ x W R (t)(f, ∂ x h), and v(x, t) solves the initial-value problem

½ ∂ tt v − ∂ xx v + ∂ xxxx v = ∂ xx h(x), x ∈ R, t ∈ R ⁺ , v(x, 0) = 0, ∂ t v(x, 0) = ∂ x f (x).

By Duhamel’s principle we can rewrite v(x, t) as

v(x, t) = V 2 (t)(∂ x f (x)) + Z _t

0

V 2 (t − τ )(∂ xx h(x))dτ, t > 0.

Thus, for t ≥ 0, v(0, t)

= V 2 (t)(∂ x f )| x=0 + Z _t

0

V 2 (t − τ )(∂ xx h)| x=0 dτ

= V 2 (t)(∂ x f )| x=0 + Z _t

0

Z _+∞

−∞

iξsgn(ξ)ˆh(ξ) 2(1 + ξ ² )

¹²

×

³

e −itφ(ξ)+iτ φ(ξ) − e itφ(ξ)−iτ φ(ξ) ´

dξdτ

= V 2 (t)(∂ x f )| x=0 + Z _+∞

−∞

ˆh(ξ) 2(1 + ξ ² )

³

2 − e ^itφ(ξ) − e ^−itφ(ξ)

´ dξ

= V 2 (t)(∂ x f )| x=0 + Z _+∞

−∞

ˆh(ξ)

1 + ξ ² dξ − 1 2

Z _+∞

−∞

ˆh(ξ) 1 + ξ ²

³

e ^itφ(ξ) + e ^−itφ(ξ)

´ dξ

= V ₂ (t)(∂ _x f )| _x=0 − 1

2 V (t)(g)| _x=0 − 1

2 V (−t)(g)| _x=0 , with g(x) = R _+∞

−∞

ˆ h(ξ)

1+ξ

²

e ^ixξ dξ, the last inequality holds because of ˆh(−ξ) = ˆh(−ξ). Denote by

J(t) = V 2 (t)(∂ x f )| x=0 − 1

2 V (t)(g)| x=0 − 1

2 V (−t)(g)| x=0 .

It follows from the similar argument to that used in the proof of the first inequality in the lemma that J(t) ∈ H

³⁴

(R) and

kJ(t)k

H

³⁴

(R) ≤ C ³

kf k _L

2

(R) + khk _H

−1 0

(R)

´ .

Note that χ(t)J(t) = v(0, t) and J(0) = v(0, 0) = 0. Lemma 2.2 implies v(0, t) = χ(t)J(t) ∈ H ₀

³⁴

(R ⁺ ) and

(2.22) kv(0, t)k

H

₀³⁴

(R

⁺

) ≤ C ³

kf k _L

²

_(R) + khk _H

−1 0

(R)

´ . Then

kχ(t)∂ x W R (t)(f, ∂ x h)| x=0 k

H

^{− 1}₀^{2,− 14}

(R

⁺

) =

° °

° ° dv(0, t) dt

° °

° °

H

^{− 1}₀^{2,− 14}

(R

⁺

)

≤ kv(0, t)k

H

₀³⁴

(R

⁺

) ≤ C

³

kf k L

²

(R) + khk _H

⁻¹

0

(R)

´ .

The second inequality in the lemma is proved. The last inequality in the lemma comes from the results in [8], the proof is omitted. ¤ For any f ∈ L ² (R ⁺ ) and h ∈ H ₀ ⁻¹ (R ⁺ ) we define ( ˜ f (x), ˜h(x)) = (f (x), h(x)) for x ≥ 0 and ( ˜ f (x), ˜h(x)) = (0, −h(−x)) for x < 0, and define

W C (t)(f, ∂ x h) = W R (t)( ˜ f , ∂ x ˜h) − W b (χ(t)g 1 (t), χ(t)g 2 (t)), where

g 1 (t) = W R (t)( ˜ f , ∂ x ˜h)| x=0 , g 2 (t) = ∂ x W R (t)( ˜ f , ∂ x ˜h)| x=0

are the trace of W _R (t)( ˜ f , ∂ _x ˜h) and ∂ x W _R (t)( ˜ f , ∂ _x ˜h) at x = 0, respectively.

Lemma 2.3 and Lemma 2.5 mean W C (t)(f, ∂ x h) is well-defined and solves the initial-boundary-value problem

(2.23)

 



u tt − u xx + u xxxx = 0, t > 0, x > 0, u(0, t) = 0, u x (0, t) = 0,

u(x, 0) = f (x), u t (x, 0) = ∂ x h(x).

The following lemma comes from a combination of Lemma 2.3 and Lemma 2.4

with Lemma 2.5.

Lemma 2.6. There exists a positive constant C such that, for f ∈ L ² (R ⁺ ) and h ∈ H ₀ ⁻¹ (R ⁺ ),

sup

t≥0 kW C (t)(f, ∂ x h)k _L

2

x

(R

⁺

) ≤ C

³

kf k _L

²

_(R

⁺

₎ + khk _H

⁻¹

0

(R

⁺

)

´ , ÃZ _T

0

kW C (t)(f, ∂ x h)dtk ⁴ _L

∞ x

(R

⁺

)

!

¹

4

≤ C(1 + T

¹⁴

) ³

kf k _L

²

_(R

⁺

₎ + khk _H

⁻¹

0

(R

⁺

)

´ . Lemma 2.7. Let T > 0 and let g(x, t) ∈ L ¹ ([0, T ], L ² _x (R ⁺ )). Then the linear initial-boundary-value problem

(2.24)

 



u tt − u xx + u xxxx = ∂ xx g, t > 0, x > 0, u(0, t) = 0, u x (0, t) = 0,

u(x, 0) = 0, u t (x, 0) = 0,

has a solution W I (t)(g) ∈ C([0, T ], L ² _x (R ⁺ )) ∩ L ⁴ _t ([0, T ], L ^∞ _x (R ⁺ )) such that sup

0≤t≤T

kW I (t)(g)k _L

2

x

(R

⁺

) ≤ Ckgk _L

¹

t

([0,T ],L

²_x

(R

⁺

))

and ÃZ _T

0

kW I (t)(g)k ⁴ _L

∞ x

(R

⁺

) dt

!

¹

4

≤ C(1 + T

¹⁴

)kgk _L

¹_t

_{([0,T ],L}

²_x

_(R

⁺

₎₎ .

Proof. Choose g n (x, t) ∈ C([0, T ], C ₀ ^∞ (R ⁺ )) such that g n → g in L ¹ ([0, T ], L ² _x (R ⁺ )) as n → +∞. It follows from Duhamel’s principle that

u n (x, t) = Z _t

0

W C (t − τ )(0, ∂ xx g n (x, τ ))dτ

is the solution to the initial-boundary-value problem (2.24) replacing g by g n . By Lemma 2.6,

(2.25) sup

0≤t≤T ku n k L

²_x

(R

⁺

) ≤ Z _T

0

kg n (x, τ )k L

²_x

(R

⁺

) dτ, and

(2.26)

ÃZ _T

0

ku n k ⁴ _L

∞ x

(R

⁺

) dt

!

¹

4

≤ C(1 + T

¹⁴

)kgk _L

¹

t

([0,T ],L

²_x

(R

⁺

))

and

(2.27) sup

0≤t≤T

ku n − u m k _L

²_x

_(R

⁺

₎ + ÃZ _T

0

ku n − u m k ⁴ _L

∞ x

(R

⁺

) dt

!

¹

4

≤ C(1 + T

¹⁴

)kg n − g m k _L

¹

t

([0,T ],L

²_x

(R

⁺

)) . Then u n (x, t) is a convergent sequence in

C([0, T ], L ² _x (R ⁺ )) ∩ L ⁴ _t ([0, T ], L ^∞ _x (R ⁺ ))

because of (2.27). Let W I (t)(g) = lim n→+∞ u n (x, t). Obviously W I (t)(g) is a solution of (2.24). Taking n → +∞ in (2.25) and (2.26) we obtain W I (t)(g)

satisfies the estimates in the lemma. ¤

3. The local existence of solutions

In this section we give the proof of Theorem 1.1. To prove the local exis- tence we are going to use a contraction mapping argument. For some positive constant δ > 0 and T > 0 determined below, we define the set Z _T ^δ by

Z _T ^δ =

½ u ∈ C([0, T ]; L ² _x (R ⁺ )) ∩ L ⁴ _t ([0, T ]; L ^∞ _x (R ⁺ )) with sup _{t∈[0,T ]} ku(·, t)k _L

²

_(R

⁺

₎ + kuk _L

⁴

t

([0,T ];L

^∞_x

(R

⁺

)) ≤ δ

¾ .

Let X be the product space

L ² (R ⁺ ) × H ₀ ⁻¹ (R ⁺ ) × H ₀

¹⁴

(R ⁺ ) × H ⁻ ₀

¹²

^,−

¹⁴

(R ⁺ ) with the norm

k(f, h, h 1 , h 2 )k X

= kf (x)k _L

2

(R

⁺

) + kh(x)k _H

−1

0

(R

⁺

) + kh 1 (t)k

H

₀¹⁴

(R

⁺

) + kh 2 (t)k

H

^{− 1}₀^{2,− 14}

(R

⁺

) . For (f, h, h 1 , h 2 ) ∈ X and u ∈ Z _T ^δ we define a mapping Φ(u) by

Φ(u) = W C (t)(f, ∂ x h) + W b (t)(h 1 , h 2 ) − W I (t)((|u| ^κ u(τ )) xx )dτ.

We only need to prove that Φ(u) is a contraction mapping from Z _T ^δ into Z _T ^δ for suitable δ > 0 and T > 0. Denote by δ 0 = k(f, h, h 1 , h 2 )k X and denote by

k|u|k T = sup

0≤t≤T

kuk _L

2 x

(R

⁺

) +

ÃZ _T

0

ku(x, t)k ⁴ _L

∞ x

(R

⁺

) dt

!

¹

4

.

For u, v ∈ Z _T ^δ , it follows from Lemma 2.3, Lemma 2.4, Lemma 2.6 and Lemma 2.7 that

k|Φ(u)|k T

≤ C(1 + T

¹⁴

) Ã

δ 0 + Z _T

0

° ° |u(τ)| ^κ+1 °

° L

²_x

(R

⁺

) dτ

!

≤ C(1 + T

¹⁴

) Ã

δ 0 + T

^4−κ4

sup

t∈[0,T ]

kuk _L

²_x

_(R

⁺

₎ k|u|k ^κ _T

! (3.1)

≤ C(1 + T

¹⁴

)

³

δ 0 + T

^4−κ4

δ ^κ+1

´

,

and (3.2)

k|Φ(u) − Φ(v)|k T

≤ C(1 + T

¹⁴

) Z _T

0

k|u(τ )| ^κ u(τ ) − |v(τ )| ^κ v(τ )k _L

2 x

(R

⁺

) dτ

≤ C(1 + T

¹⁴

) Z _T

0

³

ku(τ )k ^κ _L

∞

x

(R

⁺

) + kv(τ )k ^κ _L

∞ x

(R

⁺

)

´

ku(τ ) − v(τ )k _L

2 x

(R

⁺

) dτ

≤ C(1 + T

¹⁴

)T

^4−κ4

sup

t∈[0,T ]

ku − vk _L

²_x

_(R

⁺

₎ (k|u|k ^κ _T + k|v|k ^κ _T )

≤ Cδ ^κ (1 + T

¹⁴

)T

^4−κ4

sup

t∈[0,T ]

ku − vk _L

²_x

_(R

⁺

₎ .

For given δ 0 > 0 let δ = (2C + 1)δ 0 and choose T > 0 so small that (3.3) C(1 + T

¹⁴

)(δ ₀ + T

^4−κ4

δ ^κ+1 ) ≤ δ, Cδ ^κ (1 + T

¹⁴

)T

^4−κ4

≤ 1

8 .

A combination of (3.1), (3.2) with (3.3) implies that Φ(u) is a contraction map from Z _T ^δ into Z _T ^δ , thus we establish the existence and uniqueness of local solution to the initial-boundary-value problem (1.1) in the set Z _T ^δ . In fact the uniqueness holds in a large class Z = C([0, T ]; L ² (R ⁺ )) ∩ L ⁴ ([0, T ]; L ^∞ _x (R ⁺ )).

Suppose ˜ u ∈ Z satisfying the initial-boundary value data, then it is easy to see that for T ⁰ < T sufficiently small ˜ u ∈ Z _T ^δ

0

. Therefore u = ˜ u in [0, T ⁰ ] × R ⁺ . Reapplying this argument we obtain the desired result.

References

[1] J. L. Bona, M. Chen, and J. C. Saut, Boussinesq equations and other systems for small-amplitude long waves in nonlinear dispersive media II: The nonlinear theory, Nonlinearity 17 (2004), no. 3, 925–952.

[2] J. L. Bona and R. L. Sachs, Global existence of smooth solutions and stability of solitary waves for a generalized Boussinesq equation, Comm. Math. Phys. 118 (1988), no. 1, 15–

29.

[3] J. L. Bona, S. M. Sun, and B. Y. Zheng, A non-homogenous boundary-value problem for the Korteweg-de-Varies equation in a quarter plane, Trans. Amer. Math. Soc. 326 (2001), 427–490.

[4] J. E. Colliander and C. E. Kenig, The generalized Korteweg-de Vries equation on the half line, Comm. Partial Differential Equations 27 (2002), no. 11-12, 2187–2266.

[5] J. Holmer, The initial-boundary value problem for the Korteweg-de Vries equation, Comm. Partial Differential Equations 31 (2006), no. 8, 1151–1190.

[6] C. E. Kenig, G. Ponce, and L. Vega, Oscillatory integrals and regularity of dispersive equations, Indiana Univ. Math. J. 40 (1991), no. 1, 33–69.

[7] C. E. Kenig, G. Ponce, and G. Velo, Well-posedness and scattering results for the gen- eralized Korteweg-de Vries equation via the contraction principle, Comm. Pure Appl.

Math. 46 (1993), no. 4, 527–620.

[8] F. Linares, Global existence of small solutions for a generalized Boussinesq equation, J.

Differential Equations 106 (1993), no. 2, 257–293.

[9] L. Molinet and F. Ribaud, On the Cauchy problem for the generalized Korteweg-de Vries

equation, Comm. Partial Differential Equations 28 (2003), no. 11-12, 2065–2091.

[10] A. K. Pani and H. Saranga, Finite element Galerkin method for the “good” Boussinesq equation, Nonlinear Anal. 29 (1997), no. 8, 937–956.

[11] V. Varlamov, Long-time asymptotics for the damped Boussinesq equation in a disk, Electron. J. Diff. Eqns., Conf. 05 (2000), 285–298.

[12] R. Xue, Local and global existence of solutions for the Cauchy problem of a generalized Boussinesq equation, J. Math. Anal. Appl. 316 (2006), no. 1, 307–327.

Department of Mathematics Zhejiang University

Hangzhou 310027, Zhejiang, P. R. China

E-mail address: [email protected]