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The stability of local strong solutions for a shallow water equation

Abstract

We establish the L 1 stability of local strong solutions for a shallow water equation which includes the Degasperis-Procesi equation provided that its initial value lies in the Sobolev space H s (R) with s> 3 2 . The key element in our analysis is that the L norm of the solutions keeps finite for all finite time t.

MSC:35G25, 35L05.

1 Introduction

From the propagation of shallow water waves over a flat bed, Constantin and Lannes [1] derived the equation

g t + g x + 3 2 ρg g x +μ(α g x x x +β g t x x )=ρμ(γ g x g x x +δg g x x x ),
(1)

where the constants α, β, γ, δ, ρ and μ satisfy certain restrictions. As illustrated in [1], using suitable mathematical transformations turns Eq. (1) into the form

g t g t x x +k g x +mg g x =a g x g x x +bg g x x x ,
(2)

where a, b, k and m are constants. We know that the Camassa-Holm and Degasperis-Procesi models are special cases of Eq. (2). Lai and Wu [2] established the well-posedness of local strong solutions and obtained the existence of local weak solutions for Eq. (2).

The aim of this paper is to investigate a special case of Eq. (2). Namely, we study the shallow water equation

g t g t x x +k g x +mg g x =3 g x g x x +g g x x x ,
(3)

where k0 and m>0 are constants. Letting y=g x x 2 g, v= ( m x x 2 ) 1 g and using Eq. (3), we derive the conservation law

R yvdx= R 1 + ξ 2 m + ξ 2 | g ˆ ( t , ξ ) | 2 dξ= R 1 + ξ 2 m + ξ 2 | g ˆ 0 ( ξ ) | 2 dξ g 0 L 2 ( R ) ,
(4)

where g 0 =g(0,x) and g ˆ (t,ξ) is the Fourier transform of g(t,x) with respect to variable x. In fact, the conservation law (4) plays an important role in our further investigations of Eq. (3).

For m=4, k=0, Eq. (3) reduces to the Degasperis-Procesi equation [3]

g t g t x x +4g g x =3 g x g x x +g g x x x .
(5)

Various dynamic properties for Eq. (5) have been acquired by many scholars. Escher et al. [4] and Yin [5] studied the global weak solutions and blow-up structures for Eq. (5), while the blow-up structure for a generalized periodic Degasperis-Procesi equation was obtained in [6]. Lin and Liu [7] established the stability of peakons for Eq. (5) under certain assumptions on the initial value. For other dynamic properties of the Degasperis-Procesi (5) and other shallow water models, the reader is referred to [819] and the references therein.

The objective of this work is to establish the L 1 (R) stability of local strong solutions for the generalized Degasperis-Procesi equation (3) under the condition that we let the initial value g 0 belong to the space H s (R) with s> 3 2 . Here we address that the L 1 stability of local strong solutions for Eq. (3) has never been established in the literature. Our main approaches come from those presented in [20].

This paper is organized as follows. Section 2 gives several lemmas. The main result and its proof are presented in Section 3.

2 Several lemmas

The Cauchy problem of Eq. (3) is written in the form

{ g t g t x x + k g x + m g g x = 3 g x g x x + g g x x x , g ( 0 , x ) = g 0 ( x ) ,
(6)

which is equivalent to

{ g t + g g x + k Λ 2 g x + m 1 2 Λ 2 ( g 2 ) x = 0 , g ( 0 , x ) = g 0 ( x ) ,
(7)

where Λ 2 f= 1 2 R e | x y | fdy for any f L 2 (R) or L (R).

Let Q g (t,x)= m 1 2 Λ 2 ( g 2 )+k Λ 2 g and J g = x ( m 1 2 Λ 2 ( g 2 )+k Λ 2 g), we have

g t + 1 2 ( g 2 ) x + J g =0.
(8)

Lemma 2.1 For problem (6) with m>0, it holds that

R yvdx= R 1 + ξ 2 m + ξ 2 | g ˆ ( t , ξ ) | 2 dξ= R 1 + ξ 2 m + ξ 2 | g ˆ 0 ( ξ ) | 2 dξ g 0 L 2 ( R ) .
(9)

In addition, there exist two positive constants c 1 and c 2 depending only on m such that

c 1 g 0 L 2 ( R ) c 1 g L 2 ( R ) c 2 g 0 L 2 ( R ) .

Proof Letting y=g x x 2 g and v= ( m x x 2 ) 1 g and using Eq. (3), we have g=mv x x 2 v and

d d t R y v d x = R y t v d x + R y v t d x = 2 R v y t d x = 2 R [ ( m 2 g 2 ) x + k g x + 1 2 x x x 3 g 2 ] v d x = 2 R [ ( m 2 g 2 ) x v + k g x v + 1 2 x g 2 x x 2 v ] d x = R [ ( m g 2 ) x v + k g x v + ( g 2 ) x ( m v g ) ] d x = R g ( g 2 ) x d x + k R ( m v x v x x x ) v d x = k R v x x v x d x = 0 ,

from which we complete the proof. □

Lemma 2.2 ([2])

If g 0 H s (R) with s> 3 2 , there exist maximal T=T( u 0 )>0 and a unique local strong solution g(t,x) to problem (6) such that

g(t,x)C ( [ 0 , T ) ; H s ( R ) ) C 1 ( [ 0 , T ) ; H s 1 ( R ) ) .

Firstly, we study the differential equation

{ p t = g ( t , p ) , t [ 0 , T ) , p ( 0 , x ) = x .
(10)

Lemma 2.3 Let g 0 H s (R), s>3 and let T>0 be the maximal existence time of the solution to problem (10). Then problem (10) has a unique solution p C 1 ([0,T)×R,R). Moreover, the map p(t,) is an increasing diffeomorphism of R with p x (t,x)>0 for (t,x)[0,T)×R.

Proof From Lemma 2.2, we have g C 1 ([0,T); H s 1 (R)) and H s 1 (R) C 1 (R). Thus we conclude that both functions g(t,x) and g x (t,x) are bounded, Lipschitz in space and C 1 in time. Using the existence and uniqueness theorem of ordinary differential equations derives that problem (10) has a unique solution p C 1 ([0,T)×R,R).

Differentiating (10) with respect to x yields

{ d d t p x = g x ( t , p ) p x , t [ 0 , T ) , p x ( 0 , x ) = 1 ,
(11)

which leads to

p x (t,x)=exp ( 0 t g x ( τ , p ( τ , x ) ) d τ ) .
(12)

For every T <T, using the Sobolev embedding theorem yields

sup ( τ , x ) [ 0 , T ) × R | g x ( τ , x ) | <.

It is inferred that there exists a constant K 0 >0 such that p x (t,x) e K 0 t for (t,x)[0,T)×R. This completes the proof. □

Lemma 2.4 Assume g 0 H s (R) with s> 3 2 . Let T be the maximal existence time of the solution g to Eq. (3). Then we have

g ( t , x ) L c 0 g 0 L 2 2 t+ g 0 L t[0,T],
(13)

where constant c 0 depends on m, k.

Proof Let Z(x)= 1 2 e | x | , we have ( 1 x 2 ) 1 f=Zf for all f L 2 (R) and g=Zy(t,x). Using a simple density argument presented in [6], it suffices to consider s=3 to prove this lemma. If T is the maximal existence time of the solution g to Eq. (3) with the initial value g 0 H 3 (R) such that gC([0,T), H 3 (R)) C 1 ([0,T), H 2 (R)). From (7), we obtain

g t +g g x =(m1)Z(g g x )kZ g x .
(14)

Since

Z ( g g x ) = 1 2 e | x y | g g y d y = 1 2 x e x + y g g y d y 1 2 x + e x y g g y d y = 1 4 x e | x y | g 2 d y 1 4 x e | x y | g 2 d y
(15)

and

d g ( t , p ( t , x ) ) d t = g t ( t , p ( t , x ) ) + g x ( t , p ( t , x ) ) d p ( t , x ) d t = ( g t + g g x ) ( t , p ( t , x ) ) ,
(16)

from (16), we have

d g d t = m 1 4 p ( t , x ) e | p ( t , x ) y | g 2 dy m 1 4 p ( t , x ) e | p ( t , x ) y | g 2 dykZ g x ,
(17)

from which we get

| d g ( t , p ( t , x ) ) d t | | m 1 | 4 e | p ( t , x ) y | g 2 d y + | k Z g x | | m 1 | 4 g 2 d y + k | 1 2 e | x y | g y d y | | m 1 | 4 g L 2 2 + k g L 2 c g L 2 ( R ) c g 0 L 2 ( R ) ,
(18)

where c is a positive constant independent of t. Using (18) results in

ct g 0 L 2 ( R ) + g 0 g ( t , p ( t , x ) ) ct g 0 L 2 ( R ) + g 0 .
(19)

Therefore,

| g ( t , p ( t , x ) ) | g ( t , p ( t , x ) ) L ct g 0 L 2 ( R ) + g 0 L .
(20)

Using the Sobolev embedding theorem to ensure the uniform boundedness of g x (s,η) for (s,η)[0,t]×R with t[0, T ), from Lemma 2.3, for every t[0, T ), we get a constant C(t) such that

e C ( t ) p x (t,x) e C ( t ) ,xR.

We deduce from the above equation that the function p(t,) is strictly increasing on R with lim x ± p(t,x)=± as long as t[0, T ). It follows from (20) that

g ( t , x ) L = g ( t , p ( t , x ) ) L ct g 0 L 2 ( R ) + g 0 L .
(21)

 □

Lemma 2.5 Assume g 0 L 2 (R). Then

Q g L ( R + × R ) , J g L ( R + × R ) c 0 g 0 L 2 2 ,
(22)

where c 0 is a constant independent of t.

Proof Using (7), we get

Q g (t,x)= m 1 4 R e | x y | g 2 (t,y)dy+ k 2 R e | x y | gdy,
(23)
J g ( t , x ) = m 1 4 R e | x y | sign ( y x ) g 2 ( t , y ) d y + k 2 R e | x y | sign ( y x ) g ( t , y ) d y .
(24)

It follows from (23)-(24) and Lemma 2.1 that (22) holds. □

Lemma 2.6 Assume that g 1 (t,x) and g 2 (t,x) are two local strong solutions of equation (3) with initial data g 10 , g 20 H s (R), s> 3 2 , respectively. Then, for any f(t,x) C 0 ([0,)×R), it holds that

| J g 1 ( t , x ) J g 2 ( t , x ) | | f ( t , x ) | dx c 0 (1+t) | g 1 g 2 |dx,
(25)

where c 0 >0 depends on t, f, g 10 L 2 ( R ) , g 20 L 2 ( R ) , g 10 L ( R ) and g 20 L ( R ) .

Proof We have

| J g 1 ( t , x ) J g 2 ( t , x ) | | f ( t , x ) | d x | m 1 | 2 | x Λ 2 ( g 1 2 g 2 2 ) | | f ( t , x ) | d x + k 2 e | x y | | sign ( x y ) | | g 1 g 2 | | f ( t , x ) | d y d x = | m 1 | 4 | e | x y | | sign ( x y ) | | g 1 2 g 2 2 | d y | f ( t , x ) | d x | + c 0 | g 1 g 2 | d y e | x y | | f ( t , x ) | d x | m 1 | 4 | ( g 1 g 2 ) ( g 1 + g 2 ) | d y | e | x y | | f ( t , x ) | d x | + c 0 | g 1 g 2 | d y c 0 ( 1 + t ) | g 1 g 2 | d y ,

in which we have used the Tonelli theorem and Lemma 2.4. The proof is completed. □

We define δ(σ) to be a function which is infinitely differentiable on (,+) such that δ(σ)0, δ(σ)=0 for |σ|1 and δ(σ)dσ=1. For any number h>0, we let δ h (σ)= δ ( h 1 σ ) h . Then we know that δ h (σ) is a function in C (,) and

{ δ h ( σ ) 0 , δ h ( σ ) = 0 if  | σ | h , | δ h ( σ ) | c h , δ h ( σ ) = 1 .
(26)

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

u h (x)= 1 h δ ( x y h ) u(y)dy,h>0.
(27)

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

lim h 0 1 h | x x 0 | h | u ( x ) u ( x 0 ) | dx=0.

At any Lebesgue points x 0 of the function u(x), we have lim h 0 u h ( x 0 )=u( x 0 ). Since the set of points which are not Lebesgue points of u(x) has measure zero, we get u h (x)u(x) as h0 almost everywhere.

We introduce notation connected with the concept of a characteristic cone. For any R 0 >0, we define N> max t [ 0 , T ] g L <. Let designate the cone {(t,x):|x|< R 0 Nt,0t T 0 =min(T, R 0 N 1 )}. We let S τ designate the cross section of the cone by the plane t=τ, τ[0, T 0 ].

Let K r + 2 ρ ={x:|x|r+2ρ}, where r>0, ρ>0 and π T =[0,T]×R for an arbitrary T>0. The space of all infinitely differentiable functions f(t,x) with compact support in [0,T]×R is denoted by C 0 ( π T ).

Lemma 2.7 ([20])

Let the function u(t,x) be bounded and measurable in cylinder Ω T =[0,T]× K r . If for ρ(0,min[r,T]) and any number h(0,ρ), then the function

V h = 1 h 2 | t τ 2 | h , ρ t + τ 2 T ρ , | x y 2 | h , | x + y 2 | r ρ | u ( t , x ) u ( τ , y ) | dxdtdydτ

satisfies lim h 0 V h =0.

Lemma 2.8 ([20])

Let | G ( u ) u | be bounded. Then the function

H(u,v)=sign(uv) ( G ( u ) G ( v ) )

satisfies the Lipschitz condition in u and v, respectively.

Lemma 2.9 Let g be the strong solution of problem (7), f(t,x) C 0 ( π T ) and f(0,x)=0. Then

π T { | g k | f t + sign ( g k ) 1 2 [ g 2 k 2 ] f x sign ( g k ) J g ( t , x ) f } dxdt=0,
(28)

where k is an arbitrary constant.

Proof Let Φ(g) be an arbitrary twice smooth function on the line <g<. We multiply the first equation of problem (7) by the function Φ (g)f(t,x), where f(t,x) C 0 ( π T ). Integrating over π T and transferring the derivatives with respect to t and x to the test function f, for any constant k, we obtain

π T { Φ ( g ) f t + [ k g Φ ( z ) z d z ] f x Φ ( g ) J g ( t , x ) f } dxdt=0,
(29)

in which we have used [ k g Φ (z)zdz] f x dx= [f Φ (g)g g x ]dx.

Integration by parts yields

[ k g Φ ( z ) z d z ] f x d x = [ 1 2 ( g 2 k 2 ) Φ ( g ) 1 2 k g ( z 2 k 2 ) Φ ( z ) d z ] f x d x .
(30)

Let Φ h (g) be an approximation of the function |gk| and set Φ(g)= Φ h (g). Using the properties of sign(gk), (29), (30) and sending h0, we have

π T { | g k | f t + sign ( g k ) 1 2 [ g 2 k 2 ] f x sign ( g k ) J g ( t , x ) f } dxdt=0,
(31)

which completes the proof. □

In fact, the proof of (28) can also be found in [20].

For g 10 H s (R) and g 20 H s (R) with s> 3 2 , using Lemma 2.2, we know that there exists T>0 such that two local strong solutions g 1 (t,x) and g 2 (t,x) of Eq. (3) satisfy

g 1 (t,x), g 2 (t,x)C ( [ 0 , T ) ; H s ( R ) ) C 1 ( [ 0 , T ) ; H s 1 ( R ) ) ,t[0,T).
(32)

3 Main result

Now, we give the main result of this work.

Theorem 3.1 Assume that g 1 and g 2 are two local strong solutions of Eq. (3) with initial data g 10 , g 20 L 1 (R) H s (R), s> 3 2 . For T>0 in (32), it holds that

g 1 ( t , ) g 2 ( t , ) L 1 ( R ) c e c t | g 10 ( x ) g 20 ( x ) | dx,t[0,T],
(33)

where c depends on g 10 L ( R ) , g 20 L ( R ) , g 10 L 2 ( R ) , g 20 L 2 ( R ) and T.

Proof For arbitrary T>0 and f(t,x) C 0 ( π T ), we assume that f(t,x)=0 outside the cylinder

= { ( t , x ) } =[ρ,T2ρ]× K r 2 ρ ,0<2ρmin(T,r).
(34)

We set

η=f ( t + τ 2 , x + y 2 ) δ h ( t τ 2 ) δ h ( x y 2 ) =f() λ h (),
(35)

where ()=( t + τ 2 , x + y 2 ) and ()=( t τ 2 , x y 2 ). The function δ h (σ) is defined in (26). Note that

η t + η τ = f t () λ h (), η x + η y = f x () λ h ().
(36)

Using the Kruzkov device of doubling the variables [20] and Lemma 2.9, we have

π T × π T { | g 1 ( t , x ) g 2 ( τ , y ) | η t + sign ( g 1 ( t , x ) g 2 ( τ , y ) ) ( g 1 2 ( t , x ) 2 g 2 2 ( τ , y ) 2 ) η x sign ( g 1 ( t , x ) g 2 ( τ , y ) ) J g 1 ( t , x ) η } d x d t d y d τ = 0 .
(37)

Similarly, we have

π T × π T { | g 2 ( τ , y ) g 1 ( t , x ) | η τ + sign ( g 2 ( τ , y ) g 1 ( t , x ) ) ( g 2 2 ( τ , y ) 2 g 1 2 ( t , x ) 2 ) η y sign ( g 2 ( τ , y ) g 1 ( t , x ) ) J g 2 ( τ , y ) η } d x d t d y d τ = 0 ,
(38)

from which we obtain

0 π T × π T { | g 1 ( t , x ) g 2 ( τ , y ) | ( η t + η τ ) + sign ( g 1 ( t , x ) g 2 ( τ , y ) ) ( g 1 2 ( t , x ) 2 g 2 2 ( τ , y ) 2 ) ( η x + η y ) } d x d t d y d τ + | π T × π T sign ( g 1 ( t , x ) g 2 ( t , x ) ) ( J g 1 ( t , x ) J g 2 ( τ , y ) ) η d x d t d y d τ | = I 1 + I 2 + | π T × π T I 3 d x d t d y d τ | .
(39)

We will show that

0 π T { | g 1 ( t , x ) g 2 ( t , x ) | f t + sign ( g 1 ( t , x ) g 2 ( t , x ) ) ( g 1 2 ( t , x ) 2 g 2 2 ( t , x ) 2 ) f x } d x d t + | π T sign ( g 1 ( t , x ) g 2 ( t , x ) ) [ J g 1 ( t , x ) J g 2 ( t , x ) ] f d x d t | .
(40)

In fact, the first two terms in the integrand of (39) can be represented in the form

A h =A ( t , x , τ , y , g 1 ( t , x ) , g 2 ( τ , y ) ) λ h ().

From Lemma 2.4 and the assumptions on solutions g 1 , g 2 , we have g 1 L < C T and g 2 L < C T . From Lemma 2.8, we know that A h satisfies the Lipschitz condition in g 1 and g 2 , respectively. By the choice of η, we have A h =0 outside the region

{ ( t , x ; τ , y ) } = { ρ t + τ 2 T 2 ρ , | t τ | 2 h , | x + y | 2 r 2 ρ , | x y | 2 h }
(41)

and

π T × π T A h d x d t d y d τ = π T × π T [ A ( t , x , τ , y , g 1 ( t , x ) , g 2 ( τ , y ) ) A ( t , x , t , x , g 1 ( t , x ) , g 2 ( t , x ) ) ] λ h ( ) d x d t d y d τ + π T × π T A ( t , x , t , x , g 1 ( t , x ) , g 2 ( t , x ) ) λ h ( ) d x d t d y d τ = K 11 ( h ) + K 12 .
(42)

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

| K 11 ( h ) | c [ h + 1 h 2 × | t τ 2 | h , ρ t + τ 2 T ρ , | x y 2 | h , | x + y 2 | r ρ | g 2 ( t , x ) g 2 ( τ , y ) | d x d t d y d τ ] ,
(43)

where the constant c does not depend on h. Using Lemma 2.7, we obtain K 11 (h)0 as h0. The integral K 12 does not depend on h. In fact, substituting t=α, t τ 2 =β, x=ζ, x y 2 =ξ and noting that

h h λ h (β,ξ)dξdβ=1,
(44)

we have

K 12 = 2 2 π T A h ( α , ζ , α , ζ , g 1 ( α , ζ ) , g 2 ( α , ζ ) ) { h h λ h ( β , ξ ) d ξ d β } d ζ d α = 4 π T A ( t , x , t , x , g 1 ( t , x ) , g 2 ( t , x ) ) d x d t .
(45)

Hence

lim h 0 π T × π T A h dxdtdydτ=4 π T A ( t , x , t , x , g 1 ( t , x ) , g 2 ( t , x ) ) dxdt.
(46)

Since

I 3 =sign ( g 1 ( t , x ) g 2 ( τ , y ) ) ( J g 1 ( t , x ) J g 2 ( τ , y ) ) f λ h ()= I ¯ 3 (t,x,τ,y) λ h ()
(47)

and

π T × π T I 3 d x d t d y d τ = π T × π T [ I ¯ 3 ( t , x , τ , y ) I ¯ 3 ( t , x , t , x ) ] λ h ( ) d x d t d y d τ + π T × π T I ¯ 3 ( t , x , t , x ) λ h ( ) d x d t d y d τ = K 21 ( h ) + K 22 ,
(48)

we obtain

| K 21 ( h ) | c ( h + 1 h 2 × | t τ 2 | h , ρ t + τ 2 T ρ , | x y 2 | h , | x + y 2 | r ρ | J g 2 ( t , x ) J g 2 ( τ , y ) | d x d t d y d τ ) .
(49)

Using Lemma 2.7, we have K 21 (h)0 as h0. Using (44), we have

K 22 = 2 2 π T I ¯ 3 ( α , ζ , α , ζ , g 1 ( α , ζ ) , g 2 ( α , ζ ) ) { h h λ h ( β , ξ ) d ξ d β } d ζ d α = 4 π T I ¯ 3 ( t , x , t , x , g 1 ( t , x ) , g 2 ( t , x ) ) d x d t = 4 π T sign ( g 1 ( t , x ) g 2 ( t , x ) ) ( J g 1 ( t , x ) J g 2 ( t , x ) ) f ( t , x ) d x d t .
(50)

From (42), (46), (48), (49) and (50), we prove that inequality (40) holds.

Let

μ(t)= | g 1 ( t , x ) g 2 ( t , x ) | dx.
(51)

We define

θ h = σ δ h (σ)dσ ( θ h ( σ ) = δ h ( σ ) 0 )
(52)

and choose two numbers ρ and τ(0, T 0 ), ρ<τ. In (40), we choose

f= [ θ h ( t ρ ) θ h ( t τ ) ] χ(t,x),h<min(ρ, T 0 τ),
(53)

where

χ(t,x)= χ ε (t,x)=1 θ ε ( | x | + N t R + ε ) ,ε>0.
(54)

We note that the function χ(t,x)=0 outside the cone and f(t,x)=0 outside the set . For (t,x), we have the relations

0= χ t +N| χ x | χ t +N χ x .
(55)

Applying (53)-(55) and (40), we have the inequality

0 π T 0 { [ δ h ( t ρ ) δ h ( t τ ) ] χ ε | g 1 ( t , x ) g 2 ( t , x ) | } d x d t + 0 T 0 [ θ h ( t ρ ) θ h ( t τ ) ] | [ J g 1 ( t , x ) J g 2 ( t , x ) ] χ ( t , x ) | d x d t .
(56)

Using Lemma 2.6 and letting ε0 and R 0 , we obtain

0 0 T 0 { [ δ h ( t ρ ) δ h ( t τ ) ] | g 1 ( t , x ) g 2 ( t , x ) | d x } d t + c 0 ( 1 + T 0 ) 0 T 0 [ θ h ( t ρ ) θ h ( t τ ) ] | g 1 ( t , x ) g 2 ( t , x ) | d x d t .
(57)

By the properties of the function δ h (σ) for hmin(ρ, T 0 ρ), we have

| 0 T 0 δ h ( t ρ ) μ ( t ) d t μ ( ρ ) | = | 0 T 0 δ h ( t ρ ) | μ ( t ) μ ( ρ ) | d t | c 1 h ρ h ρ + h | μ ( t ) μ ( ρ ) | d t 0 as  h 0 ,
(58)

where c is independent of h. Letting

L(ρ)= 0 T 0 θ h (tρ)μ(t)dt= 0 T 0 t ρ δ h (σ)dσμ(t)dt,
(59)

we get

L (ρ)= 0 T 0 δ h (tρ)μ(t)dtμ(ρ),as h0,
(60)

from which we obtain

L(ρ)L(0) 0 ρ μ(σ)dσas h0.
(61)

Similarly, we have

L(τ)L(0) 0 τ μ(σ)dσas h0.
(62)

It follows from (61) and (62) that

L(ρ)L(τ) ρ τ μ(σ)dσas h0.
(63)

Send ρ0, τt, and note that

| g 1 ( ρ , x ) g 2 ( ρ , x ) | | g 1 ( ρ , x ) g 10 ( x ) | + | g 2 ( ρ , x ) g 20 ( x ) | + | g 10 ( x ) g 20 ( x ) | .
(64)

Thus, from (57), (58), (63)-(64), we have

| g 1 ( t , x ) g 2 ( t , x ) | d x | g 10 g 20 | d x + c 0 ( 1 + T 0 ) 0 t | g 1 ( t , x ) g 2 ( t , x ) | d x d t ,
(65)

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

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Acknowledgements

This work is supported by the National Natural Science Foundation of China (11471263).

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Correspondence to Shaoyong Lai.

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Lai, S., Yan, H., Chen, H. et al. The stability of local strong solutions for a shallow water equation. J Inequal Appl 2014, 410 (2014). https://doi.org/10.1186/1029-242X-2014-410

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Keywords

  • L 1 stability
  • strong solutions
  • shallow water equation