A REMARK ON THE CAUCHY PROBLEM FOR THE ... - Project Euclid

Differential and Integral Equations

Volume 21, Numbers 9-10 (2008), 859–890

A REMARK ON THE CAUCHY PROBLEM FOR THE GENERALIZED BENNEY-LUKE EQUATION ´ Rau ´ l Quintero Jose Departamento de Matemáticas, Universidad del Valle A. A. 25360. Cali, Colombia (Submitted by: Gustavo Ponce) In memory of my professor and colleague Jurgen K. A. Tischer (1943–2007) Abstract. In this article, we address the well posedness of the Cauchy problem associated with the generalized Benney–Luke equation in R1+2 : “ ˆ ˆ` ´p ˜ ˆ` ´p ˜˜ Φtt − ∆Φ + a∆2 Φ − b∆Φtt + θ Φt ∂x ∂x Φ + ∂y ∂y Φ ˆ` ´p ` ´p ˜” ` ´ + 2 ∂x Φ Φxt + ∂y Φ Φyt + β∇ · |∇Φ|m ∇Φ = 0, under a reasonable “physical” initial condition, which is imposed from the formal derivation of the Benney-Luke water wave model.

1. Introduction In this article, we study the well posedness of the Cauchy Problem in R1+2 associated with the nonlinear generalized Benney-Luke equation, p p Φtt − ∆Φ + a∆2 Φ − b∆Φtt + θ Φt ∂x ∂x Φ + ∂y ∂y Φ p p + 2 ∂x Φ Φxt + ∂y Φ Φyt + β∇ · |∇Φ|m ∇Φ = 0, (1.1) where a, b > 0, p = p1 /p2 with (p1 , p2 ) = 1 (p2 odd), m ∈ R+ , and θ, β, ∈ R. We begin the discussion noting that, in the case β = 0 and p = 1, equation (1.1) is the re-scaled version of the Benney-Luke equation Φtt − ∆Φ + µ a∆2 Φ − b∆Φtt + (Φt ∆Φ + 2∇Φt · ∇Φ) = 0. (1.2) Regarding this water wave model, J. Quintero and R. Pego in ([5]) (also see [6], [7]) showed that the evolution of three-dimensional water waves with surface tension can be reduced to studying the solution Φ(x, y, t) of the isotropic equation (1.2), where represents the amplitude parameter, µ represents the long-wave or dispersion parameter and a − b = σ − 1/3 with Accepted for publication: April 2008. AMS Subject Classifications: 35L30, 35Q35, 76B15. 859

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´ Rau ´ l Quintero Jose

− a, b > 0 (σ is called the Bond number). For instance, if → u is the velocity of a particle in an irrotational, three-dimensional flow of an inviscid, incompressible fluid which at rest occupies the region −∞ < x < ∞, −∞ < y < ∞, − 0 < z < h0 , then, for some distribution φ, the velocity potential → u takes → − the form u = (∇φ, ∂z φ), where ∇ = (∂x , ∂y ). Moreover, the study of the water waves with surface tension reduces to finding solutions of the linear equation ∆φ + φzz = 0

for 0 < z < h0 + η,

(∆ = ∂x2 + ∂y2 ),

(1.3)

with the boundary and interface nonlinear conditions φz = 0

at z = 0,

ηt + ∇η · ∇φ − φz = 0 at z = h0 + η, 1 2T H 1 = 0 at z = h0 + η, φt + |∇φ|2 + φ2z + gη − 2 2 ρ

(1.4) (1.5) (1.6)

where z = 0 represents the solid boundary, z = h0 + η is the disturbed free surface, T is the coefficient of surface tension, ρ is density (assumed constant), g is the gravitational acceleration, and the mean curvature of the free surface z = η(x, y, t) is given by 1 ∇η . H = ∇· p 2 1 + |∇η|2 In order to study long water waves with small amplitude, we introduce the amplitude parameter and the long wave parameter µ ˆ = (h0 /L)2 , where L stands for the horizontal length of motion. The long-wave regime corresponds to µ ˆ 1. The system for long water waves with small amplitude is introduced through the following rescaling of the variables x, y, and z: x = Lˆ x,

y = Lˆ y,

z = h0 zˆ,

1

t = L(gh0 )− 2 tˆ,

1 h0 ˆ and η = h0 ηˆ. and the definition of the functions φˆ and ηˆ as φ = √ (gh0 ) 2 φ, µ ˆ Note that a simple computation shows that 1 ˆ ∂y φ = (gh0 ) 12 ∂yˆφ, ˆ ∂x φ = (gh0 ) 2 ∂xˆ φ,

implying that ∂x φ and ∂y φ are of order O(), as long as ∂xˆ φˆ and ∂yˆφˆ are of order O(1) with respect to . Taking T = h20 ρgσ, and after dropping hats, we obtain that the couple (φ, η) satisfies the non-linear system µ∆φ + φzz = 0 φz = 0

for 0 < z < 1 + η,

at z = 0,

(1.7) (1.8)

Cauchy problem for the generalized Benney-Luke equation

1 φz = 0 at z = 1 + η, µ 2 ∇η φz + η − µσ∇ · p =0 φt + |∇φ|2 + 2 2µ 1 + 2 µ|∇η|2

ηt + ∇η · ∇φ −

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(1.9) at z = 1 + η, (1.10)

where ∆ and ∇ are the Laplacian and the gradient with respect to the variables x, y, respectively. Now, to derive the family of Benney-Luke equations, which describes long water waves with small amplitude, we have to assume that ∇φ, η, and its derivatives with respect to the variables x and y are O(1) with respect to . Defining Φ(x, y, t) = φ(x, y, z = 0, t), and taking the Taylor expansion of the velocity potential at the bottom z = 0, one can see that µz 2 µ2 z 4 2 ∆Φ + ∆ Φ + O(µ3 ). (1.11) 2 4! Plugging this in the previous equations, we have that Φt + η = O(, µ) (for details see [5]) and Φtt −∆Φ+µ 16 ∆2 Φ+ σ − 21 ∆Φtt +(Φt ∆Φ+(∇Φ)2t ) = O(µ2 , 2 ). (1.12) φ=Φ−

In particular, note that Φtt = ∆Φ + O(µ, ). Therefore, if we choose a, b such that (1.13) a = b + σ − 31 , we find that Φtt − ∆Φ + µ(a∆2 Φ − b∆Φtt ) + (Φt ∆Φ + 2∇Φ · ∇Φt ) = O(µ2 , 2 ). Neglecting terms of order O(µ2 , 2 ), we obtain a family of Benney-Luke equations. We want to point out that the relevant information from the physical viewpoint in the formal derivation of this water wave model is not in the distribution Φ(x, y), but in the velocity potential at the bottom ∇Φ(x, y, t) = ∇φ(x, y, z = 0, t). It is also important to mention that C. Christov in [2] derived some dispersive shallow-water equations (named dispersive shallow-water system) which are asymptotically correct to the first order of all small dispersion parameters, showing that it is Galilean invariant and possesses Hamiltonian structure. It can be seen directly that equation (1.1) is obtained from the system (2.12 - 2.13) in [2], when p = 1 and m = 2.

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As is well known, one of the main features of water wave models is that they come equipped with a Hamiltonian structure. Moreover, as a principle, results of existence and uniqueness for the associated Cauchy problems follow from the existence of some conserved quantities and the use of energy estimates. It is also a fact that the natural space in which to consider the well posedness of such Cauchy problems is dictated by the definition of either the Hamiltonian or the energy. For our particular problem, the Hamiltonian and the energy are well defined when ∇Φ and Φt are in H 1 (R2 ), being in a perfect concordance with the formal derivation of these models from the full water wave problem after some approximations, since the physical assumptions are not in the distribution Φ, but on the velocity potential ∇Φ evaluated at some level, as we noted above. We are aware of the existence of some recent works related to the Cauchy problem considered in this article. For β = 0, A. González in two spatial dimensions, with θ = 1 and p ≥ 1, established global well posedness in the energy space, but in three spatial dimensions and p ≥ 2 proved the result in a smoother space than the energy space. In both cases, the author used Strichartz inequalities and properties of the commutators of Kato-Ponce type (see [3]). L. Paumond in two spatial dimensions, with θ = 1 and p = 1, obtained global well posedness (see [8]) in a space with high regularity, which is smaller than the energy space. For β ∈ R, S. Wang, G. Xu, G. Chen in spatial dimensions 2, 3, 4, θ ∈ R and p = 1 obtained global well posedness and a non-existence result (see [4]). The authors used a Sobolev multiplicative law. In works by L. Paumond in ([8]) and S. Wang, G. Xu, G. Chen in ([4]), the space for the well posedness of the Cauchy problem controls Φ in L2 , which is mathematically correct, but seems to be unnatural from the physical viewpoint. Our goal is to present some results of local/global existence, and nonexistence of solutions for the Cauchy problem associated with the generalized Benney-Luke equation (1.1), under the reasonable physical assumption on the initial condition (Φ0 , r0 ): ∇Φ0 ∈ (H 1 (R2 ))2 , r0 ∈ H 1 (R2 ), which is associated with the formal derivation of the Benney–Luke model. The article is organized as follows. In section 2, we establish the semigroup estimates, and the non-linear estimates to prove a local existence and uniqueness result for the Cauchy problem associated with the Benney-Luke equation, via a standard fixed point argument. Instead of using Strichartz inequalities, linear and non-linear estimates will be done by using essentially


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a Sobolev multiplicative law and some simple properties of the derivatives of the function g(Φ) = Φp , where Φ is a distribution with ∇Φ ∈ H 1 (R2 ). In section 3, we prove that the Hamiltonian energy is preserved in time in classical solutions of the generalized Benney-Luke equation. In section 4, we establish a global existence and uniqueness result for the Cauchy problem associated with equation (1.1) for β ≤ 0 and θ ∈ R, and for β > 0 and θ ∈ R, imposing some restrictions on the initial conditions. We also prove a non-existence result of solutions of equation (1.1) for β > 0 and θ = 0. In case β > 0, results of existence/non-existence are generalizations for p ≥ 1 of the work by S. Wang, G. Xu, G. Chen in ([4]). 2. Local existence We start by noting that the generalized Benney-Luke equation (1.1) can be written formally as Φ + B −1 F (Φx , Φy , Φt ) = 0, where the linear operator and the function F are defined as Φ = ∂tt − ∆B −1 A, with A = I − a∆ and B = I − b∆ defined on H 2 (R2 ) via the Fourier transform, and h F (φ, ψ, r) = − r [(∂x (φ)p ) + (∂y (ψ)p )] + 2[φp rx + ψ p ry ] i m φ + β∇ · k(φ, ψ)k ψ = − [G1 (φ, r) + G2 (ψ, r) + H1 (φ, r) + H2 (ψ, r) + βK(φ, ψ)] , where Gi (φ, r) = ∂i (rφp ), Hi (φ, r) = φp ∂i r, and m φ K(φ, ψ) = ∇ · k(φ, ψ)k . ψ In order to establish the existence and uniqueness of mild solutions for the Cauchy problem associated with equation (1.1), we consider the following space. Definition 1. Let k ∈ R+ . Vk denotes the closure of C0∞ (R2 ) with respect to the norm given by kψk2Vk := kψx k2H k−1 + kψy k2H k−1 .

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Note that (Vk , k.k) is a Hilbert space with inner product (u, v)Vk = (∂x u, ∂x v)H k−1 (R2 ) + (∂y u, ∂y v)H k−1 (R2 ) . The first observation is that the generalized Benney–Luke equation can be written as a first-order system in the variables Φ and Φt = r : Φ Φ Φ =M +G , (2.1) r t r r where M : Vk+1 × H k → Vk × H k−1 is a linear operator, and G is a function defined on Vk+1 × H k given by 0 I Φ 0 M= and G = . r F (Φx , Φy , r) ∆B −1 A 0 It is not hard to verify that associated with the linear operator M there exists a semigroup T defined in Vk × H k−1 := X k as   sin(|ζ|Λ(ζ)t) −1 cos(|ζ|Λ(ζ)t)  F 0 F 0 |ζ|Λ(ζ)    T(t) =  0 F , 0 F−1  −|ζ|Λ(ζ) sin(|ζ|Λ(ζ)t) cos(|ζ|Λ(ζ)t) q 2 where F stands for the Fourier transform on R2 and Λ(ζ) = 1+a|ζ| . 1+b|ζ|2 Hereafter, we say that a couple (Φ, r) ∈ C 0 (Rt ; X k ) is a mild solution of the generalized Benney-Luke equation (1.1) with initial data (U 0 )t = (Φ0 , r0 ), if (Φ, r) satisfies the integral equation Z t Φ Φ0 Φ Φ (t) = T(t) + T(t − y)G (y) dy := S (t). (2.2) r r0 r r 0 The first result regarding this problem is to establish that T is a bounded semigroup. Hereafter, without any other indication, k ∈ R+ . Lemma 2.1. T(t) ∈ Lb (X k ) for t ∈ R. Moreover, T is a bounded semigroup. Proof. Let U t = (Φ, r) ∈ X k . Then   sin(|ζ|Λ(ζ)t) −1 ˆ rˆ F cos(|ζ|Λ(ζ)t)Φ +   |ζ|Λ(ζ) Φ   T(t) = . r   ˆ + cos(|ζ|Λ(ζ)t)ˆ F−1 −|ζ|Λ(ζ) sin(|ζ|Λ(ζ)t)Φ r


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Since | sin(|ζ|Λ(ζ)t)| ≤ 1, | cos(|ζ|Λ(ζ)t)| ≤ 1, and for some positive constant C(a, b) > 1, we have that C −1 ≤ Λ(ζ) ≤ C, we conclude that

2 Φ 2

ˆ + sin(|ζ|Λ(ζ)t) rˆ

T(t)

2

k = (1 + |ζ|2 )k/2 |ζ|) cos(|ζ|Λ(ζ)t)Φ r |ζ|Λ(ζ) L X

2

ˆ + cos(|ζ|Λ(ζ)t)ˆ + (1 + |ζ|2 )k/2 −|ζ|Λ(ζ) sin(|ζ|Λ(ζ)t)Φ r 2 L

2 Φ

≤ C 2 (a, b)

k. r X Proposition 2.1. For any T > 0, the operator S maps C 0 ([0, T ]; X k ) into itself. We will use the following Sobolev multiplication law (SML) to establish this result. Lemma 2.2. ([10]) Let s, s1 , s2 be real numbers such that (1) s < s1 , s2 , s1 + s2 > 0, s ≤ s1 + s2 − 1, (2) s ≤ s1 , s2 , s1 + s2 ≥ 0, s < s1 + s2 − 1. Then kuvkH s (R2 ) ≤ CkukH s1 (R2 ) kvkH s2 (R2 ) .

(2.3)

Before we proceed to estimate the nonlinear terms, we present some basic results. Lemma 2.3. Suppose that Φ ∈ Vk . (1) If k ≥ 2 and p = 1, then ∂i Φ ∈ H k−1 and k∂i ΦkH k−1 ≤ kΦkVk . (2) If k > 2 and p ≥ 1, (∂i Φ)p ∈ H 1 and k(∂i Φ)p kH 1 ≤ CkΦkpV2 . (3) If p ≥ l ≥ 2 and k ≥ l + 1, then (∂i Φ)p ∈ H l and for some constant C = C(l, p) k(∂i Φ)p kH l ≤ CkΦkpVl+1 . (2.4) Proof. We first observe that, if Φ ∈ Vk , then 2 ∂i Φ ∈ H k−1 (R2 ) ,→ H 1 (R2 ) ,→ Lq (R2 ) (q ≥ 2), ∂ij Φ ∈ H k−2 (R2 ).

In particular, for k ≥ 2 and p = 1 we conclude that k∂i ΦkH k−1 ≤ kΦkVk . Moreover, (∂i Φ)p ∈ Lq (R2 ) for q ≥ 2 and p ≥ 1. Now assume that k > 2. First note that (∂i Φ)p ∈ H 1 (R2 ). In fact, (∂i Φ)p ∈ L2 (R2 ). On the other hand, 2 ∂s (∂i Φ)p = p(∂i Φ)p−1 ∂is Φ.

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In this case, ∂i Φ ∈ H k−1 (R2 ) ,→ L∞ (R2 ) and so it follows directly that 2 Φ ∈ L2 (R2 ), since ∂ 2 Φ ∈ L2 (R2 ). It is easy to see that (∂i Φ)p−1 ∂is is k(∂i Φ)p kH 1 ≤ C k∂i ΦkpH 1 ≤ CkΦkpV2 . Now take l ≥ 2 and suppose that k ≥ l + 1. Note that  p−2 2 2 p−1 3 2 2  p(p − 1)(∂i Φ) ∂js Φ∂is Φ + p(∂i Φ) (∂isj Φ) ∈ L (R ), 2 ∂js ((∂i Φ)p ) = p≥2   3 ∂irs Φ ∈ L2 (R2 ), p = 1, since ∂isj Φ ∈ H k−3 (R2 ) ,→ L2 (R2 ), and ∂is Φ ∈ H k−2 (R2 ) ,→ L∞ (R2 ); thus it is easy to see that (∂i Φ)p ∈ H 2 (R2 ) for either p = 1 or p ≥ 2. In general, for p ≥ l we have that ∂jl1 j2 ···jl ((∂i Φ)p ) has terms of the form (∂i Φ)p−t Dρ1 ΦDρ2 Φ · ·Dρt Φ, 1 ≤ t ≤ l, where 2 ≤ |ρn | ≤ l + 2 − t and |ρ1 | + |ρ2 | + · · +|ρt | = t + l. In particular, for t = 1 we have the main term (∂i Φ)p−1 ∂jl+1 Φ. In order to guarantee 1 j2 ···jl i l+1 p l 2 that (∂i Φ) ∈ H , we require that ∂j1 j2 ···jl i Φ ∈ L , since ∂i Φ ∈ H k−1 (R2 ) ,→ H 2 (R2 ) ,→ L∞ (R2 ). But this is assured because we are taking k ≥ l + 1, and ∂jl+1 Φ ∈ H k−(l+1) (R2 ), for Φ ∈ Vk . On the other hand, Dρ Φ ∈ 1 j2 ···jl i H 1 (R2 ) ,→ Lq (R2 ) (q ≥ 2), with 1 ≤ |ρ| ≤ l. Moreover, using the Hölder inequality appropriately, we have that kΦktVl+1 , k(∂i Φ)p−t Dρ1 ΦDρ2 Φ · ·Dρt Φk2 ≤ k∂i Φkp−t H1 which implies that for some constant C(l, p) k(∂i Φ)p kH l ≤ CkΦkpVl+1 . Corollary 2.1. Suppose that Φ, Ψ ∈ Vk . (1) If k ≥ 2 and p = 1, then k∂i Φ − ∂i ΨkH k−1 ≤ CkΦ − ΨkVk . (2) If k > 2 and p ≥ 2, then p−1 p−1 k(∂i Φ)p − (∂i Ψ)p kH 1 ≤ C1 (p)kΦ − ΨkV2 kΦkV + kΨk . 2 V2 (3) If p ≥ l ≥ 2 and k ≥ l + 1, then for some constant C1 = C1 (l, p) p−1 k(∂i Φ)p − (∂i Ψ)p kH l−1 ≤ C1 (l, p)kΦ − ΨkVl+1 kΦkp−1 + kΨk . l+1 l+1 V V In particular, for p ≥ k − 1 ≥ 2 we have that p−1 k(∂i Φ)p − (∂i Ψ)p kH k−2 ≤ C1 (k, p)kΦ − ΨkVk kΦkp−1 + kΨk . k k V V


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Proof. For k ≥ 2 and p = 1 the conclusion follows directly from the previous lemma. Now assume that k > 2 and p ≥ 2. Then |∂j [(∂i Φ)p − (∂i Ψ)p ] | i h 2 2 ≤ |∂i Φ|p−1 |∂ij Φ − ∂ij Ψ| + |∂ij Ψ| (∂i Φ)p−1 − (∂i Ψ)p−1 h i 2 2 ≤ C(p) |∂i Φ|p−1 |∂ij Φ − ∂ij Ψ| + |∂ij Ψ| |∂i Φ − ∂i Ψ| (|∂i Φ| + |∂i Ψ|)p−2 . Then we conclude that h p−1 k∂j [(∂i Φ)p − (∂i Ψ)p ] kL2 ≤ C(p) k∂i ΦkL ∞ kΦ − ΨkV2 + kΨkV2 k∂i Φ − ∂i ΨkL∞ (k∂i ΦkL∞ + k∂i ΨkL∞ )p−2 h i p−1 p−1 ≤ C(p)kΦ − ΨkV2 kΦkV , 2 + kΨkV2

i

where we have used the fact that H s (R2 ) ,→ L∞ (R2 ), for s > 1. Moreover, we also have in a similar way that i h p−1 . k(∂i Φ)p − (∂i Ψ)p kL2 ≤ C(p)kΦ − ΨkV2 kΦkp−1 + kΨk 2 2 V V Putting together the last two inequalities, we obtain for some constant C = C(p) that p−1 p−1 k∂j [(∂i Φ)p − (∂i Ψ)p ]kH 1 ≤ C(p)kΦ − ΨkV2 kΦkV + kΨk . V2 2 Now, take l ≥ 2 and suppose that k ≥ l + 1. From the previous result, we have for 1 ≤ q ≤ l − 1 that ∂jq1 j2 ···jq [(∂i Φ)p − (∂i Ψ)p ] has for 1 ≤ s ≤ q terms of the form (∂i Φ)p−s Dρ1 ΦDρ2 Φ · ·Dρs Φ − (∂i Ψ)p−s Dρ1 ΨDρ2 Ψ · ·Dρs Ψ, where 2 ≤ |ρn | ≤ q + 2 − s and |ρ1 | + |ρ2 | + · · +|ρs | = s + q. For s = 1, we have that k(∂i Φ)p−1 ∂jq+1 Φ − (∂i Ψ)p−1 ∂jq+1 Ψk2 1 j2 ···jq i 1 j2 ···jq i

q+1 p−1 p−1 ≤ k∂i Φkp−1 kΦ − Ψk + k∂ Ψk (∂ Φ) − (∂Ψ) q+1 2

i V L ∞ j1 j2 ···jq i ∞ h i p−1 p−1 ≤ C(p)kΦ − ΨkVq+1 kΦkVq+1 + kΦkVq+1 . Note for 1 < s ≤ q that ∂i Φ, ∂i Ψ, Dρn Φ, and Dρn Ψ belong to H k−|ρn | (R2 ) ,→ L∞ (R2 ), since k − |ρn | ≥ k − l − 1 + s > 1. Moreover, k∂i Φk∞ ≤ k∂i ΦkH 1 , kDρn Φk∞ ≤ kΦkVq+2 , k∂i Ψk∞ ≤ k∂i ΨkH 1 , kDρn Ψk∞ ≤ kΨkVq+2 .

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Then using the mean value inequality applied to the function g(x, y1 , ··, ys ) = xp−s y1 · ·ys , we conclude that k(∂i Φ)p−s Dρ1 ΦDρ2 Φ · ·Dρs Φ − (∂i Ψ)p−s Dρ1 ΨDρ2 Ψ · ·Dρs Ψk2 s X ≤ C(p, s) (kΦkVq+2 + kΨkVq+2 )p−1 k∂i (Φ − Ψ)kL2 + kDρj (Φ − Ψ)kL2 j=1 p−1

≤ C1 (p, s)kΦ − ΨkVq+2 (kΦkVq+2 + kΨkVq+2 )

,

and so, k∂jq1 j2 ···jq [(∂i Φ)p − (∂i Ψ)p ] k2 ≤ C(p, q)kΦ−ΨkVq+2 (kΦkVq+2 + kΨkVq+2 )p−1 . In particular, we have shown for p ≥ l and k ≥ l + 1 that k(∂i Φ)p −(∂i Ψ)p kH l−1 ≤ C(p, l)kΦ−ΨkVl+1 (kΦkVl+1 + kΨkVl+1 )p−1 .

Lemma 2.4. Suppose that k and p are such that (1) k ≥ 2 and p = 1, or (2) k = 2 and p > 1, or (3) 2 < k ≤ 4 and p ≥ 2, or (4) p ≥ k − 1 > 2. If r ∈ H k−1 and Φ, Ψ ∈ Vk , then for some constant C = C(p) p−1 p−1 kr ((∂i Φ)p − (∂i Ψ)p )kH k−2 ≤ CkrkH k−1 kΦ − ΨkVk kΦkV . k + kΨkVk (2.5) In particular, we have for r1 , r2 ∈ H k−1 (R2 ) and Φ ∈ Vk that k (∂i Φ)p (r1 − r2 ) kH k−2 ≤ CkΦkpVk kr1 − r2 kH k−1 . Proof. Suppose p = 1 and k ≥ 2. Then we apply the Sobolev multiplication law with s = k − 2, s1 = k − 1 and s2 = k − 1. In this case, kr (∂i Φ − ∂i Ψ)kH k−2 ≤ CkrkH k−1 kΦ − ΨkVk . Suppose now that p > 1 and k = 2. For r ∈ H 1 , and Φ, Ψ ∈ V2 , we choose ρ > 1 with ρ(p − 1) > 1 and γ > 2 such that 2/γ + 1/ρ = 1. Then we have that kr ((∂i Φ)p − (∂i Ψ)p )kL2 ≤ C1 (p)k|r| |∂i Φ − ∂i Ψ| [|∂i Φ| + |∂i Ψ|]p−1 kL2 p−1 ≤ C2 (p)krkL2γ k∂i Φ − ∂i ΨkL2γ k∂i Φkp−1 + k∂ Ψk i 2ρ(p−1) 2ρ(p−1) L L p−1 p−1 ≤ C3 (p)krkH 1 kΦ − ΨkV2 kΦkV2 + kΨkV2 .


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Now for k > 2 we know that (∂i Φ)p ∈ H 2 (R2 ) for p ≥ 2. Thus, using the Sobolev multiplication law with s = k − 2, s1 = k − 1 and s2 = 2, for k ≤ 4, we have s ≤ si and k − 2 = s < s1 + s2 − 1 = k. As a consequence of this, kr ((∂i Φ)p − (∂i Ψ)p )kH k−2 ≤ CkrkH k−1 k (∂i Φ)p − (∂i Ψ)p kH 2 .

(2.6)

But from Corollary 2.1 we have that p−1 k (∂i Φ)p − (∂i Ψ)p kH 1 ≤ C1 (p)kΦ − ΨkVk kΦkp−1 + kΨk , Vk Vk and from (2.6) that h i p−1 p−1 kr ((∂i Φ)p − (∂i Ψ)p )kH k−2 ≤ CkrkH k−1 k∂i Φ − ∂i ΨkVk kΦkV . k + kΨkVk Now suppose that p ≥ k − 1 > 2. In this case, (∂i Φ)p ∈ H k−1 . Then applying the Sobolev multiplication law with s = k − 2, s1 = k − 1 and s2 = k − 2, we have that kr ((∂i Φ)p − (∂i Ψ)p )kH k−2 ≤ CkrkH k−1 k (∂i Φ)p − (∂i Ψ)p kH k−2 . But we obtained in a previous result that p−1 p−1 , + kΨk k (∂i Φ)p − (∂i Ψ)p kH k−2 ≤ C2 (p)kΦ − ΨkVk kΦkV k Vk and so, p

p

kr ((∂i Φ) − (∂i Ψ) )kH k−2 ≤ CkrkH k−1 kΦ − ΨkVk

h

kΦkp−1 Vk

+

p−1 kΨkV k

i

.

Using the same type of estimates and similar arguments, we are able to establish the following: Lemma 2.5. Suppose that k and p are as in Lemma 2.4. If r ∈ H k−1 and Φ, Ψ ∈ Vk , then for some constant C = C(p), p−1 k∂i r [(∂i Φ)p − (∂i Ψ)p ] kH k−3 ≤ CkrkH k−1 kΦ − ΨkVk kΦkp−1 + kΨk . Vk Vk (2.7) In particular, we have for r1 , r2 ∈ H k−1 (R2 ) and Φ ∈ Vk that k (∂i Φ)p (∂i r1 − ∂i r2 ) kH k−3 ≤ C(p)kΦkpVk kr1 − r2 kH k−1 . Lemma 2.6. Suppose that k and m satisfy (1) k = 2 and m > 0, or (2) 2 < k ≤ 4 and either m = 1 or m ≥ 2, or (3) m ≥ k − 1 > 2.

870


If Φ, Ψ ∈ Vk , there exists C = C(m) such that m k|∇Φ|m ∇Φ − |∇Ψ|m ∇ΨkH k−2 ≤ CkΦ − ΨkVk kΦkm Vk + kΨkVk ,

(2.8)

and so, for m and k as in (1), (2), and (3), k|∇Φ|m ∇ΦkH k−2 ≤ CkΦkm+1 . Vk

(2.9)

Proof. Let k = 2. First note that |∇Φ|m ∇Φ − |∇Ψ|m ∇Ψ| ≤ C(m)|∇Φ − ∇Ψ| (|∇Φ|m + |∇Ψ|m ) . Since we know that ∇Φ, ∇Ψ ∈ H 1 (R2 ) ,→ Lq (R2 ), for q ≥ 2, then after applying the H¨ older inequality with appropriate exponents, we obtain that m m k|∇Φ| ∇Φ − |∇Ψ|m ∇ΨkL2 ≤ C(m)kΦ − ΨkV2 kΦkm V2 + kΨkV2 . Now assume that k > 2. In this case we have that ∇Φ ∈ (H k−1 (R2 ))2 ,→ (L∞ (R2 ))2 , and that |∇Φ|m ∈ H 1 (R2 ) for m ≥ 1. Moreover, for 2 < k ≤ 3, we can use the Sobolev multiplication law with s = k − 2, s1 = k − 1 and s2 = 1. Then, s ≤ si and k − 2 = s < s1 + s2 − 1 = k − 1. Thus, k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 ≤ Ck∇ΦkH k−1 k |∇Φ|m − |∇Ψ|m kH 1 m

m

≤ CkΦkVk k |∇Φ| − |∇Ψ| kH 1 .

(2.10) (2.11)

If m = 1, then we conclude directly that k∇Φ (|∇Φ| − |∇Ψ|)kH k−2 ≤ kΦkVk kΦ − ΨkVk . Now suppose that m ≥ 2. As in Lemma 2.4, it is easy to check that . (2.12) + kΨkm−1 k |∇Φ|m − |∇Ψ|m kH 1 ≤ C1 (m)kΦ − ΨkVk kΦkm−1 Vk Vk Then from (2.10) we have that i h m−1 . k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 ≤ CkΦkH k−1 kΦ − ΨkVk kΦkm−1 + kΨk k k V V Now observe that k|∇Φ|m ∇Φ − |∇Ψ|m ∇ΨkH k−2 ≤ k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 + k|∇Ψ|m (∇Φ − ∇Ψ)kH k−2 . Then, combining previous inequalities, we conclude, as desired, that k|∇Φ|m ∇Φ − |∇Ψ|m ∇ΨkH k−2 ≤ CkΦkVk kΦ − ΨkVk kΦkm−1 + kΨkm−1 Vk Vk m m ≤ CkΦ − ΨkVk kΦkVk + kΨkVk .

Again as in Lemma 2.4, it follows directly for 3 < k ≤ 4 and either m = 1 or m ≥ 2 that, |∇Φ|m ∈ H 2 (R2 ) and k |∇Φ|m − |∇Ψ|m kH 2 ≤ C2 (p)kΨ − ΨkVk kΦkm−1 + kΨkm−1 . Vk Vk


871

Moreover, using again the Sobolev multiplication law with s = k − 2, s1 = k − 1 and s2 = 2, we have that k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 ≤ Ck∇ΦkH k−1 k |∇Φ|m − |∇Ψ|m kH 2 m−1 ≤ CkΦkVk kΦ − ΨkVk kΦkm−1 + kΨk k k V V m . ≤ CkΦ − ΨkVk kΦkm + kΨk k k V V As above we obtain the desired inequality. Finally, for m ≥ k − 1 > 2, we know that |∇Φ|m , |∇Ψ|m ∈ H k−1 (see Corollary 2.1). Using the fact that H k−2 is an algebra, we conclude that k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 ≤ Ck∇ΦkH k−2 k |∇Φ|m − |∇Ψ|m kH k−2 ≤ CkΦkVk kΦ − ΨkVk kΦkm−1 + kΨkm−1 Vk Vk m m ≤ CkΦ − ΨkVk kΦkVk + kΨkVk .

Then we have that k|∇Φ|m ∇Φ − |∇Ψ|m ∇ΨkH k−2 ≤ k∇Φ (|∇Φ|m − |∇Ψ|m )kH k−2 + k|∇Ψ|m (∇Φ − ∇Ψ)kH k−2 m ≤ CkΦ − ΨkVk (kΦkm Vk + kΨkVk ) .

Now we are able to handle the non-linear terms. Theorem 2.1. Let k, p, m be as in Lemmas 2.4, 2.5, and 2.6. If Φ ∈ C([0, T ], Vk ) and r ∈ C([0, T ], H k−1 (R2 )), we have for q, s ∈ [0, T ] the following estimates kB −1 [Gi (∂i Φ(q), r(q)) − Gi (∂i Φ(s), r(s))] kH k−1 ≤ Λ1 k (Φ(q) − Φ(s), r(q) − r(s)) kX k , kB −1 [Hi (∂i Φ(q), r(q)) − Hi (∂i Φ(s), r(s))] kH k−1 ≤ Λ2 k (Φ(q) − Φ(s), r(q) − r(s)) kX k , kB −1 ∇ · [(|∇Φ|m ∇Φ) (q) − (|∇Φ|m ∇Φ) (s)] kH k−1 ≤ Λ3 kΦ(q) − Φ(s)kVk , where Λi = Λi (kΦkL∞ ([0,T ],Vk ) , krkL∞ ([0,T ],H 1 ) , b, p) (i = 1, 2), Λ3 = Λ3 (kΦkL∞ ([0,T ],Vk ) , b, m). Proof. First, we observe that Gi (∂i Φ(q), r(q)) − Gi (∂i Φ(s), r(s)) = ∂i [(∂i Φ(q))p (r(q) − r(s)) + r(s) ((∂i Φ(q))p − (∂i Φ(s))p )] . (2.13)

872


Then we have for a generic constant C depending only on p that kGi (∂i Φ(q), r(q)) − Gi (∂i Φ(s), r(s))kH k−3 ≤ k (∂i Φ(q))p (r(q) − r(s)) + r(s) ((∂i Φ(q))p − (∂i Φ(s))p ) kH k−2 ≤ k (∂i Φ(q))p (r(q) − r(s))kH k−2 + kr(s) ((∂i Φ(q))p − (∂i Φ(s))p ) kH k−2 h ≤ C kΦ(s)kpVk kr(q) − r(s)kH k−1 + kr(s)kH k−1 kΦ(q) h ii p−1 − Φ(s)kVk k∂i Φ(q)kp−1 + kΦ(s)k k k V V h p−1 ≤ CkΦkL∞ ([0,T ],Vk ) kΦkL∞ ([0,T ],Vk ) kr(q) − r(s)kH k−1 i + krkL∞ ([0,T ],H k−1 ) kΦ(q) − Φ)(s)kVk . Since B −1 is an operator of order -2, we may conclude from the previous inequality that

−1

B [Gi (∂i Φ(q), r(q)) − Gi (∂i Φ(s), r(s))] k−1 H

≤ Λ1 k (Φ(q) − Φ(s), r(q) − r(s)) kX k , where Λ1 = Λ1 (kΦkL∞ ([0,T ],Vk ) , krkL∞ ([0,T ],Vk ) , b, p). On the other hand, in a similar fashion we are able to obtain that kB −1 [Hi (∂i Φ(q), r(q)) − Hi (∂i Φ(s), r(s))] kH k−1 ≤ Ck (∂i r (∂i Φ)p ) (q) − (∂i r (∂i Φ)p ) (s)kH k−3 h ≤ CkΦkp−1 pkrkL∞ ([0,T ],H k−1 ) kΦ(q) − Φ)(s)kVk L∞ ([0,T ],Vk ) i + kΦkL∞ ([0,T ],Vk ) kr(q) − r(s)kH k−1 ≤ Λ2 k (Φ(q) − Φ(s), r(q) − r(s)) kX k , where Λ2 = Λ2 (kΦkL∞ ([0,T ],Vk ) , krkL∞ ([0,T ],Vk ) , b, p). Finally, we also have that ∇ · (|∇Φ|m ∇Φ) (q) − ∇ · (|∇Φ|m ∇Φ) (s) = ∇ · {|∇Φ|m (q) [∇Φ(q) − ∇Φ(s)]) + ∇Φ(s) (|∇Φ(q)|m − |∇Φ(s)|m } . Since ∇ has order one, we also have for a generic constant C = C(b, m) independent of T that kB −1 (∇ · (|∇Φ|m ∇Φ) (q) − ∇ · (|∇Φ|m ∇Φ) (s)) kH k−1 ≤ Ck (|∇Φ|m ∇Φ) (q) − (|∇Φ|m ∇Φ) (s)kH k−2 ≤ Ck (|∇Φ|m (q) [∇Φ(q) − ∇Φ(s)]) kH k−2


873

+ k∇Φ(s) (|∇Φ(q)|m − |∇Φ(s)|m ) kH k−2 m ≤ C kΦ(q)km Vk + kΦ(s)kVk kΦ(q) − Φ(s)kVk ≤ Λ3 (kΦkL∞ ([0,T ],Vk ) , b, m)kΦ(q) − Φ(s)kVk .

Proof of Proposition 2.1. From Lemma 2.1, we only need to study the continuity of the operator Z t K(t) = T(t − y)G(U )(y) dy. 0

Let t0 be fixed and t ∈ R near t0 . To prove the continuity of S at t0 , we need to estimate K(t) − K(t0 ) in X k . Note that, after a change of variables in both integrals, Z t Z t0 K(t) − K(t0 ) = T(y)G(U )(t − y) dy − T(y)G(U )(t0 − y) dy 0 0 Z t0 Z t = T(y) (G(U )(t − y) − G0 (U )(t0 − y)) dy + T(y)G(U )(t − y) dy. 0

t0

Thus, we obtain that kK(t) − K(t0 )kX k ≤

Z

t0

kT(y) (G(U )(t − y) − G(U )(t0 − y)) kX k dy

0 Z t

kT(y)G(U )(t − y)kX k dy.

+

(2.14)

t0

As we showed above, kT(y) [G(U )(t − y) − G(U )(t0 − y)] kX k ≤ C(a, b)kG(U )(t − y) − G(U )(t0 − y)kX k ≤ C(a, b)kF (Φx , Φy , r)(t − y) − F (Φx , Φy , r)(t0 − y)kH k−3 . Using this fact and estimates in Theorem 2.1, we have that kT(y) [G(U )(t − y) − G(U )(t0 − y)] kX k ≤ (Λ1 + Λ2 + Λ3 )k(Φ(t − y) − Φ(t0 − y), r(t − y) − r(t0 − y))kX k . Moreover, we also have that kT(y) (G(U )) (t − y)kX k ≤ (Λ1 + Λ2 + Λ3 )k(Φ(t − y), r(t − y))kX k . Recall that we are assuming that the following functions are continuous: t → kΦ(t − y) − Φ(t0 − y)kVk ,

and

t → k(r(t − y) − r(t0 − y))kH k−1 .

874


Then the dominated convergence theorem implies that Z t0 kΦ(t − y) − Φ(t0 − y)kV dy = 0, lim t→t0

0

and

t0

Z

k(r(t − y) − r(t0 − y))kH 1 = 0.

lim

t→t0

0

Moreover, Z

t

kT(y) (G(U )) (t − y)kX k ≤ (t − t0 )(Λ1 + Λ2 + Λ3 ). t0

Using previous estimates in (2.14), we conclude that lim kK(t)−K(t0 )kX k = 0.

t→t0

Now we are in position to establish the local existence and uniqueness result for the Cauchy problem associated with the generalized Benney-Luke equation. Theorem 2.2. Let k, p, m be as in Lemmas 2.4, 2.5, and 2.6. If Φ0 ∈ Vk and r0 ∈ H k−1 (R2 ), then there exists T = T (Φ0 , r0 ) > 0 such that the integral equation (2.2) has a unique solution (Φ, r) such that Φ ∈ C 0 ([0, T ], Vk ),

r ∈ C 0 ([0, T ], H k−1 (R2 )) ∩ C 1 ([0, T ], H k−2 (R2 )).

Moreover, equation (1.1) has a unique classical solution Φ ∈ C 0 ([0, T ], Vk ), with Φt ∈ C 0 ([0, T ], H k−1 (R2 )) ∩ C 1 ([0, T ], H k−2 (R2 )), that satisfies the initial conditions ∇Φ(0, ·) = ∇Φ0 ,

Φt (0, ·) = r0 .

Proof. The strategy of the proof will be to show that for some R > 0, S is a contraction on BR ⊂ C 0 [0, T ], X k . From the estimate in Lemma 2.1, if (U 0 )t = (Φ0 , r0 )) ∈ Vk × H k−1 (R2 ), then

T(t)U 0 k ≤ C(a, b) U 0 k . X

X

Moreover, if U t = (Φ1 , r1 ) and V t = (Φ2 , r2 ), then following the same computations as in the proof of Proposition 2.1 we have for some constant C = C(p, m, β, b) (independent of T ) that kS(U ) − S(V )kX k ≤ T CkU − V kL∞ ([0,T ],X k ) h p × kU kL∞ ([0,T ],X k ) + kV kL∞ ([0,T ],X k )


875

m i , + kU kL∞ ([0,T ],X k ) + kV kL∞ ([0,T ],X k ) and h i m+1 kS(U )kX k ≤ C(a, b)kU 0 kL∞ ([0,T ],X k ) +T C kU kp+1 +kU k . L∞ ([0,T ],X k ) L∞ ([0,T ],X k ) Let R = 2C(a, b)kU 0 kL∞ ([0,T ],X k ) and choose T > 0 satisfying 2(Rp + Rm )T C6 < 1. Under those conditions we have that S maps B into B. Then, the contraction mapping theorem guarantees the existence of a fixed point. In other words, there exists a local mild solution for the integral equation (2.2). In order to establish that a mild solution is already a classical solution, we have to use the regularizing effect due to the good behavior of the nonlinear part, since G already maps X k into X k . 3. Energy estimates As it is well known, a principle for establishing global existence in time for problems with a Hamiltonian structure is that this follows by local existence and the use of energy estimates. For this particular problem, we will see that a solution Φ of (1.1) conserves the Hamiltonian energy, which has the form Z β 1 2 2 2 2 m+2 Φt + b|∇Φt | + |∇Φ| + a|∆Φ| − |∇Φ| dx dy E(Φ, Φt ) = 2 R2 m+2 Z 1 β = H(Φ, Φt ) − |∇Φ|m+2 dx dy. (3.1) 2 m + 2 R2 Theorem 3.1. Let k, p, m be as in Lemmas 2.4, 2.5, and 2.6. Then if Φ ∈ C 0 ([0, T0 ), Vk ) is a maximal solution of equation (1.1) with Φt ∈ C 0 ([0, T0 ), H k−1 (R2 )) ∩ C 1 ([0, T0 ), H k−2 (R2 )), with ∇Φ(0, ·) = ∇Φ0 , and Φt (0, ·) = r0 , it follows that E(Φ, Φt ) = E(Φ0 , r0 ),

for all t ∈ [0, T0 ).

Before we proceed, we note that H 1 (R2 ) ,→ Ll (R2 ), for l ≥ 2 and ,→ H −1 (R2 ) for 1 < q ≤ 2. Moreover, for w ∈ L2 (R2 ) and 1 r ∈ H (R2 ) we have, via a duality argument, that wr ∈ Lq (R2 ) for 1 < q < 2, and that Lq (R2 )

kwrkH −1 ≤ kwrkLq ≤ kwkL2 krkL2q/(2−q) ≤ kwkL2 krkH 1 . In particular, for p = 1, kr∂ii2 ΦkH −1 ≤ kr∂ii2 ΦkLq ≤ k∂ii2 ΦkL2 krkL2q/(2−q) ≤ kΦkVk krkH 1 .

(3.2)

876


On the other hand, for p > 1 we have the following result. Lemma 3.1. Let p > 1 and 1 < α < 2 such that p > 2/α. If r ∈ H k−1 , Φ ∈ Vk and ψ ∈ L2 (R2 ), then we have that (∂i Φ)p−1 ψ ∈ Lα (R2 ) with k (∂i Φ)p−1 ψkLα ≤ k∂i Φkp−1 kψkL2 ≤ kΦkp−1 kψkL2 , Vk L2α/(2−α) and that r (∂i Φ)p−1 ψ ∈ Lα (R2 ) with p−1 kr (∂i Φ)p−1 ψkH −1 ≤ krkH k−1 kΦkV k kψkL2 .

Proof. Let γ = α/(α − 1) and take w ∈ Lγ (R2 ). Then, since p > 2/α if and only if 2α(p−1) older inequality that 2−α > 2, we have from the H¨ Z (∂i Φ)p−1 ψw dx 2 R 2−α Z 1/2 Z 1/γ Z 2α(p−1) 2α 2 2−α dx |ψ| dx |w|γ dx ≤ |∂i Φ| R2

R2

≤ k∂i Φk

p−1 L

2α(p−1) 2−α

kψkL2 kwkLγ ≤

R2 p−1 kΦkVk kψkL2 kwkLγ .

To prove the second inequality, we proceed as before. Let 1 < q < 2 be such that q < α. From the H¨ older inequality, kr (∂i Φ)p−1 ψkH −1 ≤ Ckr (∂i Φ)p−1 ψkLq Z α−q qα αq ≤C |r| α−q dx k (∂i Φ)p−1 ψkLα ≤ CkrkH k−1 kΦkp−1 kψkL2 , Vk R2

since qα/(α−q) > 2 if and only if q > 2α/(2+α), and q > 1 > 2α/(2+α). As a consequence of the previous result, we have the following. Corollary 3.1. Let either p = 1 and α = 2 or 1 < α < 2 and p > 2/α. If r ∈ H k−1 and Φ ∈ Vk , we then have that (1) (∂i Φ)p−1 ∂ii Φ ∈ Lα (R2 ), and k (∂i Φ)p−1 ∂ii ΦkLα ≤ k∂i Φkp−1 2α(p−1) k∂ii ΦkL2 L

(only for p > 1)

2−α

≤ kΦkpVk . (2) r (∂i Φ)p−1 ∂ii Φ ∈ Lα (R2 ), and kr (∂i Φ)p−1 ∂ii ΦkH −1 ≤ Ckr (∂i Φ)p−1 ∂ii ΦkLα ≤ krkH k−1 k∂i Φkp−1 2α(p−1) k∂ii ΦkL2 L

≤ krkH k−1 kΦkpVk ,

2−α

(only for p > 1)


877

(3) r (∂i Φ)p ∈ Lα (R2 ) and kr (∂i Φ)p kH −1 ≤ kr (∂i Φ)p kLα ≤ krkH k−1 kΦkpVk . On the other hand, we have the following. Lemma 3.2. Let either l > 1 and γ < 2, or l = 1 and γ = 2 be such that lγ ≥ 2. Then the function L : Vk → Lγ defined as L(Φ) = (∂i Φ)l is continuous. Moreover, kL(Φ) − L(Ψ)kγ (3.3) (l−1) (l−1) ≤ C(l) k∂i ΦkL2γ(l−1)/(2−γ) + k∂i ΨkL2γ(l−1)/(2−γ) k∂i Φ − ∂i ΨkL2 l−1 l−1 kΦ − ΨkVk . ≤ C(l, γ) kΦkV k + kΨkVk Proof. The proof for the case l = 1 and γ = 2 follows directly since H 1 (R2 ) ,→ L2 (R2 ). Now assume that l > 1 and γ < 2. Then, from the mean value inequality, |(∂i Φ)l − (∂i Ψ)l | ≤ C(l)(|∂i Φ|l−1 + |∂i Ψ|l−1 )|∂i Φ − ∂i Ψ|. Then, by applying the H¨ older inequality and using the fact that 2γ(l−1)/(2− γ) ≥ 2 if and only if lγ ≥ 2, we get the desired result. Proof of Theorem 3.1. Hereafter, for l ∈ R, we set h·, ·i−l,l as the dual pairing between H l (R2 ) and its dual space H −l . As we established in previous results, we have for Φ ∈ Vk that ∂i Φt (∂i Φ)p = ∂i [Φt (∂i Φ)p ] − Φt ∂i (∂i Φ)p ∈ H −1 , and also K(∂x Φ, ∂y Φ) = ∇ · (|∇Φ|m ∇Φ) ∈ H −1 (see Lemma 2.6). Moreover, for either p = 1 and α = 2 or 1 < α < 2 and p > 2/α, F (Φx , Φy , Φt ) ∈ C 0 ([0, T0 ), Lα (R2 )) ⊂ C 0 ([0, T0 ), H −1 (R2 )).

(3.4)

So, we have that hΦt [∂x [(∂x Φ)p ] + ∂y [(∂y Φ)p ]] + 2 [(∂x Φ)p Φxt + (∂y Φ)p Φyt ] , Φt i−1,1 Z (Φx )p 2 =− ∇ Φt · dx dy = 0, (Φy )p R2 and 1 d m + 2 dt In other words, we have already shown that h∇ · (|∇Φ|m ∇Φ) , Φt i−1,1 = −

hF (∂x Φ, ∂y Φ, ∂t Φ), ∂t Φi−1,1

Z

1 d =− m + 2 dt

|∇Φ|m+2 dx dy.

R2

Z R2

|∇Φ|m+2 dx dy.

878


Since Φtt − ∆Φ ∈ H k−2 (R2 ) ,→ L2 (R2 ), we also have that Z 2 1d Φt + |∇Φ|2 dx dy. hΦtt − ∆Φ, Φt i−1,1 = 2 dt R2 Now, we observe that ∆2 Φ and ∆Φtt are elements of H −2 . Then by the Banach theorem there exist v, w ∈ H −1 such that v|H 2 = ∆2 Φ and w|H 2 = ∆Φtt . Moreover, if g ∈ H 1 (R2 ) and (gn ) ⊂ H 2 (R2 ) is such that gn → g in H 1 (R2 ), then

v(g) = lim ∆2 Φ, gn −2,2 , w(g) = limn→∞ h∆Φtt , gn i−2,2 , n→∞

and so for g ∈ H 1 (R2 ) and (gn ) ⊂ H 2 (R2 ) such that gn → g in H 1 (R2 ),

(av − bw)(g) = limn→∞ a∆2 Φ − b∆Φt , gn −2,2 . Finally, to evaluate av − bw in Φt ∈ H 1 (R2 ), we will use a regularization R procedure. Take a positive test function φ ∈ C0∞ (R2 ) such that R2 φ = 1, and define the approximation of unity φ(j) (x, y) = j 2 φ(jx, jy). The first observation is that, for l ≥ 1, we have φ(j) ∗∂iΦ ∈ C 1 ([0, T0 ), H l (R2 )), and φ(j) ∗ ∂t Φ ∈ C 0 ([0, T0 ), H l (R2 )) , where ∗ denotes the convolution in space. As a consequence of this, for Ψ(j) = φ(j) ∗ Φ, E D E D (j) (j) + lim ∆2 Ψ(j) , Ψt v(Φt ) = lim ∆2 (Φ − Ψ(j) ), Ψt j→∞ j→∞ −2,2 −2,2 Z i E hD d 1 (j) + ∆2 (Φ − Ψ(j) ), Ψt = lim |∆Ψ(j) |2 dx dy j→∞ 2 dt R2 −2,2 Z 1d = |∆Φ|2 dx dy, 2 dt R2 since we have that D E (j) lim ∆2 (Φ − Ψ(j) ), Ψt j→∞

−2,2

(j) ≤ lim k∆2 (Φ − Ψ(j) )kH −2 kΨt kH 2 j→∞

(j)

≤ lim k∇Φ − ∇Ψ(j) kH 1 kΨt kH 2 = 0. j→∞

In the same way, w(Φt ) =

1d 2 dt

Z

|∇Φt |2 dx dy.

R2

In other words, Φtt − ∆Φ + a∆2 Φ − b∆Φtt + F (Φx , Φy , Φt ) (Φt )


d1 = dt 2

Z R2

h Φ2t + b|∇Φt |2 + |∇Φ|2 + a|∆Φ|2 −

879

i β |∇Φ|m+2 dx dy = 0. m+2

So, we have shown that E(Φ, Φt ) is constant, and so, E(Φ, Φt ) = E(Φ0 , r0 ), for 0 ≤ t < T0 .

4. Global existence Now, we are ready to establish the existence of a global solution. For β ≤ 0, we use the standard energy method. For β > 0 and p ≥ 1, we extend the results obtained by S. Wang, G. Xu, G. Chen in ([4]) when p = 1. Case 1: β ≤ 0. Theorem 4.1. Let k be an integer, p, m real numbers such that (1) k = 2, p ≥ 1 and m > 0, or (2) k = 3, 4 and m, p ≥ 2 (3) m, p ≥ k − 1 ≥ 3. Let T0 > 0, Φ0 ∈ Vk , r0 ∈ H k−1 (R2 ) and β ≤ 0. Then equation (1.1) has a unique global solution Φ ∈ C 0 ([0, ∞), Vk ) such that Φt ∈ C 0 ([0, ∞), H k−1 (R2 )) ∩ C 1 ([0, ∞), H k−2 (R2 )), with ∇Φ(x, y, 0) = ∇Φ0 (x, y), and Φt (x, y, 0) = r0 (x, y). Moreover, H(Φ, Φt ) is bounded independent of T0 . Proof. The first observation is that there exists a positive constant M = M (a, b) > 1 such that

2

2

φ −1 φ M (4.1)

≤ H(φ, ψ) ≤ M

. ψ X2 ψ X2 p In other words, H(·, ·) is an equivalent norm to the norm in k · kX 2 . Let us assume that k = 2 and suppose that there exist 0 < Tmax < ∞ such that the initial-value problem associated with equation (1.1) has a unique local classical solution Φ on the interval [0, Tmax ). In this case, we have that

Φ lim (4.2)

2 = ∞, Φt t↑Tmax X But from the previous lemma, we have that E(·, ·) is conserved in time along classical solutions, and H(q0 , r0 ) −

2β 2β k∇Φ0 km+2 k∇Φkm+2 m+2 = H (q(t, ·), r(t, ·)) − m+2 , m+2 m+2

880


for all t ∈ [0, Tmax ). Since β ≤ 0, then we have that 2|β| k∇Φ0 |m+2 m+2 , for all t ∈ [0, Tmax ). m+2 As a consequence of this and inequality (4.1)

2

Φ 2|β| m+2

≤ M H(q0 , r0 ) + k∇Φ0 km+2

Φt m+2 V×H 1

2

m

Φ0

Φ0

≤ C(m, β) 1+ , r0 V×H 1 r0 V×H 1 H (q(t, ·), r(t, ·)) ≤ H(q0 , r0 ) +

which contradicts the limit (4.2). In other words, Φ is a global classical solution of the initial-value problem associated with equation (1.1). Now, assume that k ≥ 3. From the local result Proposition 2.2, there exists a maximal solution Φ ∈ C 0 ([0, T1 ), Vk ) such that Φt ∈ C 0 ([0, T1 ), H k−1 (R2 )) ∩ C 1 ([0, T1 ), H k−2 (R2 )), with ∇Φ(x, y, 0) = ∇Φ0 (x, y), and Φt (x, y, 0) = r0 (x, y). Suppose that T1 < T0 , and that k∇ΦkH k−2 and kΦt kH k−2 are bounded on [0, T1 ). Let α ∈ N2 be any index with |α| = k − 2. Then, by assumption, we have that Dα Φt ∈ H 1 (R2 ) and that Dα (Φtt ), Dα (∆Φ) ∈ L2 . Moreover, we also have that Z hDα (Φtt − ∆Φ), Dα Φt i−1,1 = Dα (Φtt − ∆Φ)Dα Φt dxdy 2 R Z α 2 1d |D Φt | + |Dα ∇Φ|2 dx dy. = 2 dt R2 On the other hand, Dα ∆2 Φ and Dα ∆Φtt are elements of H −2 . Then by the Hahn-Banach theorem there exist vα , wα ∈ H −1 such that vα |H 2 = ∆2 Φ and wα |H 2 = ∆Φtt . Moreover, if g ∈ H 1 (R2 ) and (gn ) ⊂ H 2 (R2 ) are such that gn → g in H 1 (R2 ),

(avα − bwα )(g) = limn→∞ Dα (a∆2 Φ − b∆Φt ), gn −2,2 . Now to evaluate avα −bwα in Dα Φt ∈ H 1 (R2 ), we proceed as in Theorem 2.2. If we use the same notation Ψ(j) = Φ(j) ∗ Φ and the same type of arguments, we have that Z 1 d (j) 2 α α (j) 2 α (avα − bwα )(D Φt ) = lim a|D ∆Ψ | + b|D ∇Ψt | dx dy 2 j→∞ dt R2 Z 1d = |aDα ∆Φ|2 + b|Dα ∇Φt |2 dx dy. 2 dt R2


881

In other words, α D Φtt − ∆Φ + a∆2 Φ − b∆Φtt (Dα Φt ) Z α 2 1d = |D Φt | + b|Dα ∇Φt |2 + |Dα ∇Φ|2 + a|Dα ∆Φ|2 dx dy. 2 dt R2 We also have that Dα F (Φx , Φy , Φt ) ∈ H −(k−1) . Thus we assure the existence of Fα ∈ H −1 such that Fα |H k−1 = Dα F (Φx , Φy , Φt ). Moreover, D E α (j) (j) α (j) D F (Φx , Φy , Φt ) − Dα F (Ψ(j) x , Ψy , Ψt ), D Ψt −(k−1),k−1

≤ kF (Φx , Φy , Φt ) −

(j) (j) α (j) F (Ψ(j) x , Ψy , Ψt )kH −1 kD Ψt kH k−1 .

If we assume that p = 1, then using the fact that ∂i Φ, ∂s Φ ∈ H k−1 (R2 ) ,→ H 2 (R2 ) ,→ L∞ (R2 ), we conclude directly that k∂ii2 ΦΦs − ∂ii2 Ψ(j) Ψ(j) s kL2 ≤ k∂ii2 ΦkL2 kΦs − Ψs kH 1 + kΨs kH 1 k∂ii2 Φs − ∂ii2 Ψs kL2 , k∂i Φ∂i Φs − ∂i Ψ(j) ∂i Ψ(j) s kL2 ≤ k∂i ΦkH 1 k∂i Φs − ∂i Ψs kL2 + k∂i Ψs kH 1 k∂i Φs − ∂i Ψs kL2 . Now, for p > 1, we choose 1 < α < 2 such that p > 2/α. Then since Φs , Ψs ∈ H k−1 (R2 ) ,→ H 2 (R2 ) ,→ L∞ (R2 ), we have that k(∂i Φ)p−1 ∂ii2 ΦΦs − (∂i Ψ(j) )p−1 (∂ii2 Ψ(j) )(Ψ(j) s )kLα ≤ k(∂i Φ)p−1 ∂ii2 Φ(Φs − Ψ(j) s )kLα i h + kΨ(j) (∂i Φ)p−1 ∂ii2 Φ − (∂i Ψ(j) )p−1 ∂ii2 Ψ(j) kLα s ≤ k(∂i Φ)p−1 ∂ii2 ΦkLα kΦs − Ψ(j) s kH 1 p−1 2 + kΨ(j) ∂ii Φ − (∂i Ψ(j) )p−1 ∂ii2 Ψ(j) kLα . s kH 1 k(∂i Φ)

Since we have that p > 2/α is equivalent to estimates:

2(p−1)α 2−α

≥ 2, we have the following

k(∂i Φ)p−1 ∂ii2 Φ − (∂i Ψ(j) )p−1 ∂ii2 Ψ(j) kLα ≤ k(∂i Φ)p−1 (∂ii2 Φ − ∂ii2 Ψ(j) )kLα + k∂ii2 Ψ(j) ((∂i Φ)p−1 − (∂i Ψ(j) )p−1 )kLα ≤ k∂i Φk ≤

2α(p−1)

L 2−α p−1 kΦkVk k∂ii2 Φ

k∂ii2 Φ−∂ii2 Ψ(j) kL2 +k∂ii2 Ψ(j) kL2 k(∂i Φ)p−1 −(∂i Ψ(j) )p−1 k − ∂ii2 Ψ(j) kL2 + k∂ii2 Ψ(j) kL2 k(∂i Φ)p−1 − (∂i Ψ(j) )p−1 k

2α

L 2−α

2α

L 2−α

.

882

´ Rau ´ l Quintero Jose qα

(j)

(j)

But we have that Ψs → Φs in L q−α , and ∂ii2 Ψs inequality (3.3) in Lemma 3.2 implies that

→ ∂ii2 Φs in L2 . Then

k(∂i Φ)p−1 ∂ii2 ΦΦs − (∂i Ψ(j) )p−1 (∂ii2 Ψ(j) )(Ψ(j) s )kLα → 0, as j → ∞. On the other hand, for p > 1 and p > 2/α, k(∂i Φ)p ∂i Φs − (∂i Ψ(j) )p (∂i Ψ(j) s )kLα h i (j) p (j) p α + k∂i Ψ ≤ k(∂i Φ)p (∂i Φs − ∂i Ψ(j) )k (∂ Φ) − (∂ Ψ ) kLα i i L s s ≤ k∂i Φkp

2pα L 2−α

(j) p (j) p k∂i Φs − ∂i Ψ(j) s kL2 + k∂i Ψs kL2 k(∂i Φ) − (∂i Ψ ) k

2pα

L 2−α

(j) p (j) p ≤ kΦkpVk k∂i Φs − ∂i Ψ(j) s kL2 + k∂i Ψs kL2 k(∂i Φ) − (∂i Ψ ) k

2pα

L 2−α

.

(j)

Again, using that ∂i Ψs → ∂i Φs in L2 , inequality (3.3) in Lemma 3.2 implies that k(∂i Φ)p ∂i Φs − (∂i Ψ(j) )p (∂i Ψ(j) s )kLα → 0, as j → ∞. In other words, for either p = 1 and α = 2 or p > 1 and p > 2/α, kGi (∂i Φ, Φs ) − Gi (φ(j) ∗ ∂i Φ, φ(j) ∗ Φs )kLα → 0, as j → ∞. In a similar fashion, it can be established that kHi (∂i Φ, ∂i Φs ) − Hi (φ(j) ∗ ∂i Φ, φ(j) ∗ ∂i Φs )kLα → 0, as j → ∞. On the other hand, in Lemma 2.6 we obtain for m > 0 and k = 2 that (j) m + kΨ k kΦ − Ψ(j) kV2 . k∇Φ|m ∆Φ − |∇Ψ(j) |m ∆Ψ(j) kL2 ≤ kΦkm 2 2 V V In other words, we have established that kK(∂x Φ, ∂y Φ) − K(φ(j) ∗ ∂x Φ, φ(j) ∗ ∂y Φ)kL2 → 0, as Since then

Lα (R2 )

,→

H −1

j → ∞.

for either p = 1 and α = 2 or p > 1 and 1 < α < 2,

kGi (∂i Φ, ∂s Φ) − Gi (∂i Ψ(j) , ∂s Ψ(j) )kH −1 → 0, as (j)

kHi (∂i Φ, ∂i ∂s Φ) − Hi (∂x Ψ

(j)

j → ∞.

)kH −1 → 0, as

j → ∞,

kK(∂x Φ, ∂y Φ) − K(∂x Ψ(j) , ∂y Ψ(j) )kH −1 → 0, as

j → ∞.

, ∂i ∂s Ψ

As a direct consequence of the previous estimates, we are able to conclude that D E (j) Fα (Dα Φt ) = lim Dα F (Φx , Φy , Φt ), Dα Ψt j→∞ −(k−1),k−1 D E (j) (j) α (j) = lim Dα F (Ψ(j) , Ψ , Ψ ), D Ψ x y t t j→∞

−(k−1),k−1


Z

(j)

883

(j)

(j) α Dα F (Ψ(j) x , Ψy , Ψt )D Ψt dx dy.

= lim

j→∞ R2

Thus, we have that 1 dHα (Φ, Φt ) = − lim j→∞ 2 dt where

Z

(j)

R2

(j)

(j) α Dα F (Ψ(j) x , Ψy , Ψt )D Ψt dx dy,

(4.3)

Hα (Φ, Φt ) = kDα Φt k2L2 + bkDα ∇Φt k2L2 + kDα ∇Φk2L2 + akDα ∆Φk2L2 . Now assume that p ≥ k − 1 ≥ 3. In this case, we have that H |α| (R2 ) is an algebra, since |α| ≥ 2. Recall that we are assuming that kΦt k|α| and k∇Φk|α| are bounded in [0, T1 ). First, we observe that (j)

(j) (j) m (j) kF (Ψ(j) x , Ψy , Ψt )kH |α| ≤ k|∇Ψ | ∇Ψ kH |α| + 2 X

(j) (j) pkΨt kH |α| k(∂i Ψ(j) )p−1 ∂ii2 Ψ(j) kH |α| + 2k∂i Ψt kH |α| k(∂i Ψ(j) )p kH |α| .

i=1

Using the fact that H |α| (R2 ) is an algebra we conclude that k(∂i Ψ(j) )p−1 ∂ii2 Ψ(j) kH |α| ≤ k(∂i Ψ(j) )p−1 kH |α| k∆Ψ(j) kH |α| , k(∂i Ψ(j) )p kH |α| ≤ k(∂i Ψ(j) )p−1 kH |α| k∇Ψ(j) kH |α| , k|∇Ψ(j) |m ∇Ψ(j) kH |α| ≤ k|∇Ψ(j) |m kH |α| k∇Ψ(j) kH |α| . But for some constant independent of k we have that k(∂i Ψ(j) )m kH |α| , k(∂i Ψ(j) )p−1 kH |α| ≤ C, since, for |ρ| ≤ |α|, we have that Dρ (∂i Ψ(j) )p−1 and Dρ |∇Ψ(j) |m are controlled by k∇Ψ(j) kH |α| . These facts show that, for some constant C = C(p), (j)

(j) kF (Ψ(j) x , Ψy , Ψt )kH |α| (j) (j) ≤ C kΨt kH |α| k∆Ψ(j) kH |α| + k∇Ψt kH |α| + C k∇Ψ(j) kH |α| .

Then we have for t ∈ [0, T1 ) that Z α (j) (j) α (j) , Ψ , Ψ )D Ψ D F (Ψ(j) x y t t dx dy R2

(j)

(j)

(j) ≤ kF (Ψ(j) x , Ψy , Ψt )kH |α| kΨt kH |α| (j) (j) ≤ C1 1 + kDα Ψt k2L2 + bkDα ∇Ψt k2L2 + kDα ∇Ψ(j) k2L2 + akDα ∆Ψ(j) k2L2 .

884


So, for t ∈ [0, T1 ), and after taking the limit as j → ∞ in inequality (4.3), dHα (Φ, Φt ) ≤ C2 (1 + Hα (Φ, Φt )), dt and so integrating from 0 to t ∈ (0, T1 ), Z t Hα (Φ, Φs ) ds. Hα (Φ, Φt ) ≤ (Hα (Φ0 , r0 ) + C2 t) + C2 0

From Gronwall’s inequality we conclude that Hα (Φ, Φt ) is bounded in [0, T1 ) by a constant depending only on Hα (Φ0 , r0 ) and T0 . The case k = 3 follows by a regularization process as in the discussion in Proposition 3.1. Case 2: β > 0. In this section, we will adapt to the general Benney-Luke model the results obtained in [4], which correspond to p = 1. For the sake of completeness we will include the proofs of some results. Before we go further, we adapt our notation to that in [4]: For u ∈ V2 , we define Z 2β 1 J(u) = a|∆u|2 + |∇u|2 − |u|m+2 dx dy, 2 R2 m+2 Z I(u) = a|∆u|2 + |∇u|2 − β|u|m+2 dx dy. R2

Note in particular that J(u) =

1 m ak∆uk22 + k∇uk22 + I(u). 2(m + 2) m+2

(4.4)

We also observe for m > 0 that the Sobolev embedding from H 1 (R2 ) to Lm+2 (R2 ) implies, for any u ∈ V2 , that, k∇ukLm+2 (R2 ) ≤ Ck∇ukH 1 (R2 ) ≤ C(a) ak∆ukL2 (R2 ) + k∇ukL2 (R2 ) . The right side of the previous inequality is obtained easily by recalling that the H s -norm is defined through the Fourier transform. Moreover, C(a) is characterized in terms of k∇ukm+2 C(a, m) = sup . 1/2 u∈V2 ,∇u6=0 (ak∆uk2 + k∇uk) In other words, we have, for u ∈ V2 , k∇ukLm+2 (R2 ) ≤ C(a, m) ak∆ukL2 (R2 ) + k∇ukL2 (R2 ) .

(4.5)

As shown in [4], we have the following variational characterizations of C(a, m): Let d be defined as d = inf{sup J(λu) : u ∈ V2 , ∇u 6= 0}, λ≥0


885

then we have that d = inf{J(u) : u ∈ V2 , ∇u 6= 0, I(u) = 0} m = β −2/m C(a, m)−2(m+2)/m > 0. 2(m + 2)

(4.6)

Using this characterization of C(a, m) and the Sobolev embedding from H 1 (R2 ) to Lm+2 (R2 ), we obtain some properties of the stable and unstable set given, respectively, by W = u ∈ V2 : J(u) < d, I(u) > 0 ∪ u ∈ V2 : ∇u = 0 , V = u ∈ V2 : J(u) < d, I(u) < 0 . Lemma 4.1. Let β > 0, Φ(x, t) be a local solution of (1.1) with initial condition Φ0 ∈ V2 and r0 ∈ H 1 (R2 ) on [0, T0 ), with E(Φ0 , r0 ) < d. If I(Φ(·, 0)) > 0 or ∇Φ0 (·, 0) = 0, then Φ(·, t) ∈ W and E(Φ(·, t), r(·, t)) < d for all t ∈ [0, T0 ). Moreover, for t ∈ [0, T0 ), E(Φ , r ) m 2 0 0 m+2 βk∇Φ(·, t)kLm+2 (R2 ) ≤ ak∆Φ(·, t)kL2 (R2 ) + k∇Φ(·, t)kL2 (R2 ) . d (4.7) Proof. Note that, for t ∈ [0, T0 ), J(Φ(·, t)) ≤ E(Φ(·, t), r(·, t)) = E(Φ0 , r0 ) < d. Now, suppose that there exists t0 ∈ [0, T0 ) such that, for some t1 ∈ (t0 , T0 ), we have that I(Φ(·, t1 )) = 0 and ∇Φ(·, t1 ) 6= 0. In other words, Φ(·, t) 6∈ W, for t ∈ [t0 , T0 ). As a consequence of this and from (4.4), 2(m + 2) 2 J(Φ(·, t1 )) − I(Φ(·, t1 )) m m 2(m + 2) 2(m + 2) E(Φ(·, t1 )) < d. (4.8) ≤ m m

0 < ak∆Φ(·, t1 )k22 + k∇Φ(·, t1 )k22 =

But, this is a contradiction because this inequality implies that I(Φ(·, t1 )) > 0. In fact, from (4.5), and (4.6), m+2 βk∇Φ(·, t1 ))km+2 (ak∆Φ(·, t1 )k22 + k∇Φ(·, t1 )k22 )m+2/2 m+2 ≤ β(C(a, m)) m/2 m+2 2(m + 2)d < β(C(a, m)) ak∆Φ(·, t1 )k22 + k∇Φ(·, t1 )k22 m

< ak∆Φ(·, t1 )k22 + k∇Φ(·, t1 )k22 .

(4.9)

886


Thus, we have shown that I(Φ(·, t)) > 0 for t ∈ [t0 , T0 ), and so, I(Φ(·, t)) > 0 for t ∈ [0, T0 ), since I(Φ(·, 0)) > 0. Note that the condition I(Φ(·, 0)) > 0 implies from (4.8) that 2(m + 2) d. 2 We now assume that ∇Φ(·, 0) = 0. If ∇Φ(·, t) = 0 for t ∈ [0, T0 ), then we already have Φ(·, t) ∈ W, for t ∈ [0, T0 ). By continuity, we have the existence of 0 < t1 < T0 such that for t ∈ [0, t1 ), we have that ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 6≡ 0, and that 0 < ak∆Φ(·, 0)k22 + k∇Φ(·, 0)k22
0 and J(Φ(·, t)) < d. Then we have that 0 ≤ ak∆Φ(·, t)k22 + k∇Φ(·, t)k22
0, and let Φ0 ∈ V2 and r0 ∈ H 1 be such that E(Φ0 , r0 ) < d and I(Φ0 ) > 0 or ∇Φ0 = 0. Then, there exists a unique weak global solution Φ ∈ C([0, ∞), V2 ) of equation (1.1) such that Φt ∈ C([0, ∞), H 1 (R2 )) ∩ C 1 ([0, ∞), L2 (R2 )), satisfying the initial conditions ∇Φ(·, 0) = ∇Φ0 , Φt (·, 0) = r0 .


887

Proof. Note that kΦt k22 + bk∇Φt k22 m ak∆Φk22 + k∇Φk22 I(Φ) E(Φ0 , r0 ) = E(Φ, Φt ) = + + . 2 2(m + 2) m+2 From the previous result, we have that I(Φ(·, t)) ≥ 0. Then we have that m kΦt k22 + bk∇Φt k22 + ak∆Φk22 + k∇Φk22 ≤ 2E(Φ0 , r0 ). (m + 2) But for some positive constant C = C(a, b, m) we have that

h i m

Φ ak∆Φk22 + k∇Φk22

≤ C kΦt k22 + bk∇Φt k22 + Φt (m + 2) X ≤ 2CE(Φ0 , r0 ). This fact implies that, for any finite time T0 < ∞,

Φ(·, t)

< ∞. lim t→T0 Φt (·, t) X 2 So, any local solution can be extended in time. A non-existence result on global solutions. As in the proof of the existence of global solutions, the proof of non-existence is based on the invariance of a set under the flow for the generalized Benney-Luke equation: the unstable set V. Lemma 4.2. Let β > 0, Φ(x, t) be a local solution of (1.1) with initial condition Φ0 ∈ V2 and r0 ∈ H 1 (R2 ) on [0, T0 ). If E(Φ0 , r0 ) < d and I(Φ0 ) < 0, then Φ(·, t) ∈ V and E(Φ(·, t), r(·, t)) < d for all t ∈ [0, T0 ). Moreover, for t ∈ [0, T0 ), 2(m + 2) d. (4.10) ak∆Φ(·, 0)k22 + k∇Φ(·, 0)k22 > m Proof. For t ∈ [0, T0 ), J(Φ(·, t)) ≤ E(Φ(·, t), r(·, t)) = E(Φ0 , r0 ) < d. By hypothesis, I(Φ(·, 0)) < 0. Then if we assume that Φ(·, t) 6∈ V, using the continuity of I(Φ(·, t)), there exists t0 > 0 such that, for t ∈ [0, t0 ), I(Φ(·, t)) < 0, and I(Φ(·, t0 )) = 0. Then we have that ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 < βk∇Φ(·, t))km+2 m+2 ≤ β(C(a, m))m+2 (ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 )m+2/2 .

888


In other words, for t ∈ [0, t0 ), ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 > β −2/m (C(a, m))−2(m+2)/m =

2(m + 2) d. m

In particular, J(Φ(·, t0 )) =

m ak∆Φ(·, t0 )k22 + k∇Φ(·, t0 )k22 ≥ d. 2(m + 2)

This contradiction shows that I(Φ(·, t)) < 0, for t ∈ [0, T0 ). Using this result, we are able to show that any local solution cannot be a global solution. Corollary 4.2. Let p ≥ 1, β > 0, and θ = 0. Let Φ0 ∈ V2 and r0 ∈ H 1 be such that E(Φ0 , r0 ) < d and I(Φ0 ) < 0. If Φ ∈ C([0, T0 ), V2 ) is a solution of equation (1.1) with Φt ∈ C([0, T0 ), H 1 (R2 )) ∩ C 1 ([0, T0 ), L2 (R2 )) satisfying the initial conditions ∇Φ(·, 0) = ∇Φ0 and Φt (·, 0) = r0 , then T0 < ∞. The proof of this result requires establishing the following concavity result: Lemma 4.3. ([9]) Let P ∈ C 2 ([0, T0 )) be a non-negative function such that, 0 for some 0 < a < T0 , P is positive in [a, T0 ) and P (a) > 0. If for some α > 0 we have that 00

0

P P − (α + 1)(P ) > 0 in (a, T0 ), then T0 < ∞. Proof. Assume that T0 = ∞. We will see that for some a < t0 < T0 , lim P (t) = ∞.

t→t− 0

Define the function g = P −α . Then a simple computation implies that, in [a, ∞), h i 00 00 0 g = −αP −(α+2) P P − (α + 1)(P ) ≤ 0. In other words, g is a concave function in [a, ∞). Thus, for t ∈ (a, ∞), 0

g(t) ≤ g(a) + g (a)t, which is equivalent to h i 0 0 0 < P −α (t) ≤ P −α (a) − αP −(α+1) P (a)t = P −(α+1) (a) P (a) − αP (a)t .


889

0

Thus, we have for t ∈ (a, ∞) that P (a) − αP (a)t > 0 and h i−1 0 . P α (t) ≥ P (α+1) (a) P (a) − αP (a)t 0

0

Since P (a) > 0 we have for t0 = P (a)/αP (a) that lim P (t) = ∞,

t→t− 0

as claimed above. As a consequence of this, we conclude that T0 < ∞. Proof Corollary 4.2. Suppose that T0 = ∞. Let P be the function defined in [0, ∞) as P (t) = kΦ(·, t)k2L2 + bk∇Φ(·, t)k2L2 . It is not hard to see that we have the identity d2 P (t) = kΦt (·, t)k22 + bk∇Φt (·, t)k22 + hΦtt (·, t) − a∆Φtt (·, t), Φ(·, t)i−2,2 . dt2 On the other hand, multiplying equation (1.1) by Φ(·, t) we obtain that hΦtt (·, t) − a∆Φtt (·, t), Φ(·, t)i−2,2 = βk∇Φ(·, t)km+2 − ak∆Φ(·, t)k22 − k∇Φ(·, t)k22 . Lm+2 But, from E(Φ(·, t), Φt (·, t)) = E(Φ0 , r0 ), we conclude that (see Theorem 3.1) 2β k∇Φ(·, t)km+2 − ak∆Φ(·, t)k22 − k∇Φ(·, t)k22 + 2E(Φ0 , r0 ) Lm+2 m+2 = kΦt (·, t)k22 + bk∇Φt (·, t)k22 , or equivalently, m+2 kΦt (·, t)k22 + bk∇Φt (·, t)k22 2 m + ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 + I(Φ(·, t)) 2 = (m + 2)E(Φ0 , r0 ). Using those previous facts, d2 P (t) = 2kΦt (·, t)k22 + 2bk∇Φt (·, t)k22 − 2I(Φ(·, t)) dt2 m+2 = kΦt (·, t)k22 + bk∇Φt (·, t)k22 2 + m ak∆Φ(·, t)k22 + k∇Φ(·, t)k22 − 2(m + 2)E(Φ0 , r0 ) ≥ 2(m + 2)(d − E(Φ0 , r0 )) > 0.

(4.11)

890


Integrating from 0 to t, we obtain dP (t) dP (t) ≥ |t=0 + 2(m + 2)(d − E(Φ0 , r0 ))t dt dt ≥ 2 hΦ0 , r0 i + 2 h∇Φ0 , ∇r0 i + 2(m + 2)(d − E(Φ0 , r0 ))t. This fact implies that, for some a > 0, Note from Young’s inequality that

dP (t) dt

> 0, and P (t) > 0 in [a, ∞).

0

|P (t)|2 ≤ 4 (kΦ(·, t)kkΦt (·, t) + bk∇Φ(·, t)kk∇Φt (·, t))2 ≤ 4P (t) kΦt (·, t)k2 + bk∇Φt (·, t)k2 . Using this and inequality (4.11), we are able to show for t > a that m+4 0 00 |P (t)|2 ≥ 2(m + 2)(d − E(Φ0 , r0 ))P (t) > 0. P (t)P (t) − 4 From Lemma 4.3 with α = m+4 4 , we conclude that T0 must be finite. In other words, any local solution can not be extended in time. Acknowledgments. This work was supported by the Universidad del Valle, Cali-Colombia. References [1] J.R. Quintero, Existence and analyticity of lumps for a generalized Benney-Luke equation, Revista Colombiana de Matematicas, 36 (2002), 71–95. [2] C.I. Christov, An energy-consistent dispersive shallow-water-wave model, Wave motions, 734 (2002), 161–174. [3] A. Gonz´ alez, The Cauchy problem for Benney-Luke equation and generalized BenneyLuke equations, Differential and Integral Equations, 20 (2007), 1341–1362. [4] S. Wang, G. Xu, and G. Chen, The Cauchy Problem for the Generalized Benney-Luke Equation, J. Math. Phys., to appear. [5] R.L. Pego and J.R. Quintero, Two–dimensional solitary Waves for a Benney-Luke equation, Physica D, 132 (1999), 476–496. [6] D.J. Benney and J.C. Luke, Interactions of permanent waves of finite amplitude, J. Math. Phys., 43 (1964), 309–313. [7] P.A. Milewski and J. B. Keller, Three dimensional water waves, Studies Appl. Math., 37 (1996), 149–166. [8] L. Paumond, A rigorous link between Kp and a Benney-Luke equation, Differential and Integral Equations, 16 (2003), 1039–1064. [9] H. Levine, Instability and nonexistence of global solutions to nonlinear wave equations of the form Putt = −Au + F(u), Trans. Amer. Math. Soc., 192 (1974), 1–21. [10] C. Tao, Multilinear weighted convolution of L2 functions and applications to nonlinear dispersive equations, Amer J. Math., 122 (2001), 839–908. [11] A. Pazy, “Semigroups of Linear Operator and Applications to Partial Differential Equations,” Springer Verlag, New York, 1983.