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The L 1 stability of strong solutions to a generalized BBM equation

Abstract

A nonlinear generalized Benjamin-Bona-Mahony equation is investigated. Using the estimates of strong solutions derived from the equation itself, we establish the L 1 (R) stability of the solutions under the assumption that the initial value u 0 (x) lies in the space H 1 (R).

MSC:35G25, 35L05.

1 Introduction

Benjamin, Bona and Mahony [1] established the BBM model

u t +a u x b u x x t +k ( u 2 ) x =0,
(1)

where a, b and k are constants. Equation (1) is often used as an alternative to the KdV equation which describes unidirectional propagation of weakly long dispersive waves [2]. As a model that characterizes long waves in nonlinear dispersive media, the BBM equation, like the KdV equation, was formally derived to describe an approximation for surface water waves in a uniform channel. Equation (1) covers not only the surface waves of long wavelength in liquids, but also hydromagnetic waves in cold plasma, acoustic waves in anharmonic crystals, and acoustic gravity waves in compressible fluids (see [2, 3]). Nonlinear stability of nonlinear periodic solutions of the regularized Benjamin-Ono equation and the Benjamin-Bona-Mahony equation with respect to perturbations of the same wavelength is analytically studied in [4]. Unique continuation property and control for the Benjamin-Bona-Mahony equation on a periodic domain are discussed in [5]. The L q (q2) asymptotic property of solutions for the Benjamin-Bona-Mahony-Burgers equations is studied in [6] under certain assumptions on the initial data. The tanh technique is employed in [7] to get the compact and noncompact solutions for KP-BBM and ZK-BBM equations.

Applying the tanh method and the sine-cosine method, Wazwaz [8] obtained compactons, solitons, solitary patterns and periodic solutions for the following generalized Benjamin-Bona-Mahony equation

u t +a u x b u x x t +k ( u m ) x =0,
(2)

where a0, b>0 and k0 are constants, and m1 is an integer.

The objective of this work is to investigate Eq. (2). Using the methods of the Kruzkov’s device of doubling the variables presented in Kruzkov’s paper [9], we obtain the L 1 stability of strong solutions. Namely, for any solutions u 1 (t,x) and u 2 (t,x) satisfying Eq. (2), we will derive that

u 1 ( t , x ) u 2 ( t , x ) L 1 ( R ) c e c t u 1 ( 0 , x ) u 2 ( 0 , x ) L 1 ( R ) ,t[0,T],
(3)

where T is the maximum existence time of solutions u 1 and u 2 and c depends on u 1 ( 0 , x ) H 1 ( R ) and u 2 ( 0 , x ) H 1 ( R ) . From our knowledge, we state that the L 1 stability of strong solutions for Eq. (2) has never been acquired in the literature.

This paper is organized as follows. Section 2 gives several lemmas and Section 3 establishes the proofs of the main result.

2 Several lemmas

Let η T =[0,T]×R for an arbitrary T>0. We denote the space of all infinitely differentiable functions f(t,x) with compact support in [0,T]×R by C 0 ( η T ). We define γ(σ) to be a function which is infinitely differentiable on (,+) such that γ(σ)0, γ(σ)=0 for |σ|1 and γ(σ)dσ=1. For any number ε>0, we let γ ε (σ)= γ ( ε 1 σ ) ε . Then we have that γ ε (σ) is a function in C (,) and

{ γ ε ( σ ) 0 , γ ε ( σ ) = 0 if  | σ | ε , | γ ε ( σ ) | c ε , γ ε ( σ ) d σ = 1 .
(4)

Assume that the function v(x) is locally integrable in (,). We define the approximation of function v(x) as

v ε (x)= 1 ε γ ( x y ε ) v(y)dy,ε>0.

We call x 0 a Lebesgue point of function v(x) if

lim ε 0 1 ε | x x 0 | ε | v ( x ) v ( x 0 ) | dx=0.

At any Lebesgue point x 0 of the function v(x), we have lim ε 0 v ε ( x 0 )=v( x 0 ). Since the set of points which are not Lebesgue points of v(x) has measure zero, we get v ε (x)v(x) as ε0 almost everywhere.

We introduce notations connected with the concept of a characteristic cone. For any M>0, we define N> sup t [ 0 , ) u L ( R ) <. Let denote the cone {(t,x):|x|MNt,0<t< T 0 =min(T,M N 1 )}. We let S τ represent the cross section of the cone by the plane t=τ, τ[0, T 0 ]. Let H r ={x:|x|r}, where r>0.

Lemma 2.1 ([9])

Let the function v(t,x) be bounded and measurable in cylinder Ω=[0,T]× H r . If for any δ(0,min[r,T]) and any number ε(0,δ), then the function

V ε = 1 ε 2 | t τ 2 | ε , δ t + τ 2 T δ , | x y 2 | ε , | x + y 2 | r δ | v ( t , x ) v ( τ , y ) | dxdtdydτ

satisfies lim ε 0 V ε =0.

In fact, for Eq. (2), we have the conservation law

R ( u 2 + b u x 2 ) dx= R ( u 2 ( 0 , x ) + b u x 2 ( 0 , x ) ) dx,
(5)

from which we have

u L ( R ) c u 0 H 1 ( R ) ,
(6)

where c only depends on b.

We write the equivalent form of Eq. (2) in the form

u t + Λ 2 ( a u + k u m ) x =0,
(7)

where the operator Λ 2 g= 1 2 b e 1 b | x y | g(y)dy for any g L 2 (R).

Lemma 2.2 Let u 0 =u(0,x) H 1 (R), K u (t,x)= Λ 2 (au+k u m ) and P u (t,x)= x K u (t,x). For any t[0,), it holds that

K u ( t , x ) L ( R ) <C, P u ( t , x ) L ( R ) <C,

where the constant C is independent of time t.

Proof We have

K u (t,x)= 1 2 b e 1 b | x y | [ a u ( t , y ) + k u m ( t , y ) ] dy
(8)

and

| P u ( t , x ) | = | x 1 2 b e 1 b | x y | [ a u ( t , y ) + k u m ( t , y ) ] d y | = | 1 2 b e 1 b x x e 1 b y [ a u ( t , y ) + k u m ( t , y ) ] d y + 1 2 b e 1 b x x e 1 b y [ a u ( t , y ) + k u m ( t , y ) ] d y | 1 2 b e 1 b | x y | | a u ( t , y ) + k u m ( t , y ) | d y .
(9)

Using (5)-(6), the integral e 1 b | x y | dy=2 b and (8)-(9), we obtain the proof of Lemma 2.2. □

Lemma 2.3 Let u be the strong solution of Eq. (2), f(t,x) C 0 ( η T ). Then

η T { | u k | f t sign ( u k ) P u ( t , x ) f } dxdt=0,
(10)

where k is an arbitrary constant.

Proof Let Φ(u) be an arbitrary twice smooth function on the line <u<. We multiply Eq. (7) by the function Φ (u)f(t,x), where f(t,x) C 0 ( η T ). Integrating over η T and transferring the derivatives with respect to t and x to the test function f, we obtain

η T { Φ ( u ) f t Φ ( u ) P u ( t , x ) f } dxdt=0.
(11)

Let Φ ε (u) be an approximation of the function |uk| and set Φ(u)= Φ ε (u). Letting ε0, we complete the proof. □

In fact, the proof of (10) can also be found in [9].

Lemma 2.4 Assume that u 1 (t,x) and u 2 (t,x) are two strong solutions of Eq. (2) associated with the initial data u 10 = u 1 (0,x) and u 20 = u 2 (0,x). Then, for any f C 0 ( η T ),

| sign ( u 1 u 2 ) [ P u 1 ( t , x ) P u 2 ( t , x ) ] f d x | c | u 1 u 2 | d x ,
(12)

where c depends on u 10 H 1 ( R ) and u 20 H 1 ( R ) and f.

Proof Using (9), we have

| P u 1 ( t , x ) P u 2 ( t , x ) | = | 1 2 b e 1 b x x e 1 b y [ a u 1 ( t , y ) + k u 1 m ( t , y ) a u 2 ( t , y ) k u 2 m ( t , y ) ] d y + 1 2 b e 1 b x x e 1 b y [ a u 1 ( t , y ) + k u 1 m ( t , y ) a u 2 ( t , y ) k u 2 m ( t , y ) ] d y | c e 1 b | x y | | a u 1 ( t , y ) + k u 1 m ( t , y ) a u 2 ( t , y ) k u 2 m ( t , y ) | d y c e 1 b | x y | | u 1 ( t , y ) u 2 ( t , y ) | d y ,
(13)

in which we have used u 1 L u 10 H 1 ( R ) and u 2 L u 20 H 1 ( R ) . Using the Fubini theorem completes the proof. □

3 Main results

Theorem 3.1 Let u 1 and u 2 be two local or global strong solutions of Eq. (2) with initial data u 1 (0,x)= u 10 H 1 (R) and u 2 (0,x)= u 20 H 1 (R), respectively. Let T 0 be the maximum existence time of solutions u 1 and u 2 . For any t[0, T 0 ), it holds that

u 1 ( t , ) u 2 ( t , ) L 1 ( R ) c e c t u 10 u 20 L 1 ( R ) ,
(14)

where c depends on u 10 H 1 ( R ) and u 20 H 1 ( R ) .

Proof For an arbitrary T>0, set η T =[0,T]×R. Let f(t,x) C 0 ( η T ). We assume that f(t,x)=0 outside the cylinder

= { ( t , x ) } =[δ,T2δ]× H r 2 δ ,0<2δmin(T,r).
(15)

We define

g=f ( t + τ 2 , x + y 2 ) γ ε ( t τ 2 ) γ ε ( x y 2 ) =f() λ ε (),
(16)

where ()=( t + τ 2 , x + y 2 ) and ()=( t τ 2 , x y 2 ). The function γ ε (σ) is defined in (4). Note that

g t + g τ = f t () λ ε (), g x + g y = f x () λ ε ().
(17)

Taking u= u 1 (t,x) and k= u 2 (τ,y) and assuming f(t,x)=0 outside the cylinder , from Lemma 2.3, we have

η T × η T { | u 1 ( t , x ) u 2 ( τ , y ) | g t + sign ( u 1 ( t , x ) u 2 ( τ , y ) ) P u 1 ( t , x ) g } d x d t d y d τ = 0 .
(18)

Similarly, it holds

η T × η T { | u 2 ( τ , y ) u 1 ( t , x ) | g τ + sign ( u 2 ( τ , y ) u 1 ( t , x ) ) P u 2 ( τ , y ) g } d x d t d y d τ = 0 ,
(19)

from which we obtain

0 η T × η T | u 1 ( t , x ) u 2 ( τ , y ) | ( g t + g τ ) + | η T × η T sign ( u 1 ( t , x ) u 2 ( τ , y ) ) ( P u 1 ( t , x ) P u 2 ( τ , y ) ) g d x d t d y d τ | = η T × η T I 1 d x d t d y d τ + | η T × η T I 2 d x d t d y d τ | .
(20)

We will show that

0 η T | u 1 ( t , x ) u 2 ( t , x ) | f t + | η T sign ( u 1 ( t , x ) u 2 ( t , x ) ) [ P u 1 ( t , x ) P u 2 ( t , x ) ] f d x d t | .
(21)

We note that the first term in the integrand of (20) can be represented in the form

Y ε =Y ( t , x , τ , y , u 1 ( t , x ) , u 2 ( τ , y ) ) λ ε ().
(22)

By the choice of g, we have Y ε =0 outside the region

{ ( t , x ; τ , y ) } = { δ t + τ 2 T 2 δ , | t τ | 2 ε , | x + y | 2 r 2 δ , | x y | 2 ε }
(23)

and

η T × η T Y ε d x d t d y d τ = η T × η T [ Y ( t , x , τ , y , u 1 ( t , x ) , u 2 ( τ , y ) ) Y ( t , x , t , x , u 1 ( t , x ) , u 2 ( t , x ) ) ] λ ε ( ) d x d t d y d τ + η T × η T Y ( t , x , t , x , u 1 ( t , x ) , u 2 ( t , x ) ) λ ε ( ) d x d t d y d τ = J 1 ( ε ) + J 2 .
(24)

Considering the estimate |λ()| c ε 2 and the expression of function Y ε , we have

| J 1 ( ε ) | c [ 1 ε 2 | t τ 2 | ε , δ t + τ 2 T δ , | x y 2 | ε , | x + y 2 | r δ | u 2 ( t , x ) u 2 ( τ , y ) | d x d t d y d τ ] ,
(25)

where the constant c does not depend on ε. Using Lemma 2.1, we obtain J 1 (ε)0 as ε0. The integral J 2 does not depend on ε. In fact, substituting t=α, t τ 2 =β, x=η, x y 2 =ξ and noting that

ε ε λ ε (β,ξ)dξdβ=1,
(26)

we have

J 2 = 2 2 η T Y ε ( α , η , α , η , u 1 ( α , η ) , u 2 ( α , η ) ) { ε ε λ ε ( β , ξ ) d ξ d β } d η d α = 4 η T Y ( t , x , t , x , u 1 ( t , x ) , u 2 ( t , x ) ) d x d t .
(27)

Hence

lim ε 0 η T × η T Y ε dxdtdydτ=4 η T Y ( t , x , t , x , u 1 ( t , x ) , u 2 ( t , x ) ) dxdt.
(28)

Since

I 2 =sign ( u 1 ( t , x ) u 2 ( τ , y ) ) ( P u 1 ( t , x ) P u 2 ( τ , y ) ) f λ ε ()
(29)

and

η T × η T I 2 d x d t d y d τ = η T × η T [ I 2 ( t , x , τ , y ) I 2 ( t , x , t , x ) ] d x d t d y d τ + η T × η T I 2 ( t , x , t , x ) d x d t d y d τ = K 1 ( ε ) + K 2 ,
(30)

we obtain

| K 1 ( ε ) | c ( ε + 1 ε 2 | t τ 2 | ε , δ t + τ 2 T δ , | x y 2 | ε , | x + y 2 | r δ | P u 2 ( t , x ) P u 2 ( τ , y ) | d x d t d y d τ ) .
(31)

By Lemmas 2.1 and 2.2, we have K 1 (ε)0 as ε0. Using (26), we have

K 2 = 2 2 η T I 2 ( α , η , α , η , u 1 ( α , η ) , u 2 ( α , η ) ) { ε ε λ ε ( β , ξ ) d ξ d β } d η d α = 4 η T I 2 ( t , x , t , x , u 1 ( t , x ) , u 2 ( t , x ) ) d x d t = 4 η T sign ( u 1 ( t , x ) u 2 ( t , x ) ) ( P u 1 ( t , x ) P u 2 ( t , x ) ) f ( t , x ) d x d t .
(32)

From (28) and (32), we prove that inequality (21) holds.

Let

w(t)= | u 1 ( t , x ) u 2 ( t , x ) | dx.
(33)

We define the following increasing function

θ ε (ρ)= ρ γ ε (σ)dσ ( θ ε ( ρ ) = γ ε ( ρ ) 0 )
(34)

and choose two numbers τ 1 and τ 2 (0, T 0 ), τ 1 < τ 2 . In (21), we choose

f= [ θ ε ( t τ 1 ) θ ε ( t τ 2 ) ] χ(t,x),ε<min( τ 1 , T 0 τ 2 ),
(35)

where

χ(t,x)= χ h (t,x)=1 θ h ( | x | + N t M + h ) ,h>0.
(36)

When h is sufficiently small, we note that function χ(t,x)=0 outside the cone and f(t,x)=0 outside the set . For (t,x), we have the relation

χ t +N| χ x |=0,

which derives

χ t 0.
(37)

Applying (21), (34)-(37) and the increasing properties of θ ε , we have the inequality

0 η T 0 { [ γ ε ( t τ 1 ) γ ε ( t τ 2 ) ] χ h | u 1 ( t , x ) u 2 ( t , x ) | } d x d t + | η T 0 [ θ ε ( t τ 1 ) θ ε ( t τ 2 ) ] [ P u 1 ( t , x ) P u 2 ( t , x ) ] B ( t , x ) χ h ( t , x ) d x d t | ,
(38)

where B(t,x)=sign[ u 1 (t,x) u 2 (t,x)].

From (38), we obtain

0 η T 0 { [ γ ε ( t τ 1 ) γ ε ( t τ 2 ) ] χ h | u 1 ( t , x ) u 2 ( t , x ) | } d x d t + 0 T 0 ( θ ε ( t τ 1 ) θ ε ( t τ 2 ) ) | [ P u 1 ( t , x ) P u 2 ( t , x ) ] B ( t , x ) χ h ( t , x ) d x | d t .
(39)

Using Lemma 2.4, we have

0 η T 0 { [ γ ε ( t τ 1 ) γ ε ( t τ 2 ) ] χ h | u 1 ( t , x ) u 2 ( t , x ) | } d x d t + c 0 T 0 ( θ ε ( t τ 1 ) θ ε ( t τ 2 ) ) | u 1 u 2 | d x d t ,
(40)

where c is defined in Lemma 2.4.

Letting h0 in (40) and letting M, we have

0 0 T 0 { [ γ ε ( t τ 1 ) γ ε ( t τ 2 ) ] | u 1 ( t , x ) u 2 ( t , x ) | d x } d t + c 0 T 0 ( θ ε ( t τ 1 ) θ ε ( t τ 2 ) ) ( | u 1 u 2 | d x ) d t .
(41)

By the properties of the function γ ε (σ) for εmin( τ 1 , T 0 τ 1 ), we have

| 0 T 0 γ ε ( t τ 1 ) w ( t ) d t w ( τ 1 ) | = | 0 T 0 γ ε ( t τ 1 ) [ w ( t ) w ( τ 1 ) ] d t | c 1 ε τ 1 ε τ 1 + ε | w ( t ) w ( τ 1 ) | d t 0 as  ε 0 ,
(42)

where c is independent of ε.

Set

L( τ 1 )= 0 T 0 θ ε (t τ 1 )w(t)dt= 0 T 0 t τ 1 γ ε (σ)dσw(t)dt.
(43)

Using the similar proof of (42), we get

L ( τ 1 )= 0 T 0 γ ε (t τ 1 )w(t)dtw( τ 1 )as ε0,
(44)

from which we obtain

L( τ 1 )L(0) 0 τ 1 w(σ)dσas ε0.
(45)

Similarly, we have

L( τ 2 )L(0) 0 τ 2 w(σ)dσas ε0.
(46)

Then we get

L( τ 1 )L( τ 2 ) τ 1 τ 2 w(σ)dσas ε0.
(47)

Letting ε0, τ 1 0 and τ 2 t, from (41), (42) and (47), for any t[0, T 0 ], we have

| u 1 ( t , x ) u 2 ( t , x ) | d x | u 1 ( 0 , x ) u 2 ( 0 , x ) | d x + c 0 t | u 1 ( t , x ) u 2 ( t , x ) | d x d t ,
(48)

from which we complete the proof of Theorem 3.1 by using the Gronwall inequality. □

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Acknowledgements

The authors are very grateful to the reviewers for their helpful and valuable comments, which have led to a meaningful improvement of the paper. This work is supported by both the Fundamental Research Funds for the Central Universities (JBK120504) and the Applied and Basic Project of Sichuan Province (2012JY0020).

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Lai, S., Wang, A. The L 1 stability of strong solutions to a generalized BBM equation. J Inequal Appl 2014, 3 (2014). https://doi.org/10.1186/1029-242X-2014-3

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Keywords

  • generalized BBM equation
  • strong solutions
  • L 1 stability