# 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  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 , 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  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 $k≥0$ 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 

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

Various dynamic properties for Eq. (5) have been acquired by many scholars. Escher et al.  and Yin  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 . Lin and Liu  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  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 .

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 ()

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)

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

For every $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 $( 1 − ∂ x 2 ) − 1 f=Z⋆f$ for all $f∈ L 2 (R)$ and $g=Z⋆y(t,x)$. Using a simple density argument presented in , 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 $g∈C([0,T), H 3 (R))∩ C 1 ([0,T), H 2 (R))$. From (7), we obtain

$g t +g g x =−(m−1)Z⋆(g g x )−kZ⋆ g x .$
(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

$d g d t = m − 1 4 ∫ − ∞ p ( t , x ) e − | p ( t , x ) − y | g 2 dy− m − 1 4 ∫ p ( t , x ) ∞ e − | p ( t , x ) − y | g 2 dy−kZ⋆ g x ,$
(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

$e − C ( t ) ≤ p x (t,x)≤ e C ( t ) ,x∈R.$

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

(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 $h→0$ 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 ${(t,x):|x|< R 0 −Nt,0≤t≤ T 0 =min(T, R 0 N − 1 )}$. 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 ()

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 ()

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

$H(u,v)=sign(u−v) ( G ( u ) − G ( v ) )$

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 $−∞. 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 $|g−k|$ and set $Φ(g)= Φ h (g)$. Using the properties of $sign(g−k)$, (29), (30) and sending $h→0$, 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 .

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

$⊎= { ( t , x ) } =[ρ,T−2ρ]× K r − 2 ρ ,0<2ρ≤min(T,r).$
(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  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 $h→0$. 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 $h→0$. 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
(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 $ε→0$ 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 $h≤min(ρ, T 0 −ρ)$, we have

(58)

where c is independent of h. Letting

$L(ρ)= ∫ 0 T 0 θ h (t−ρ)μ(t)dt= ∫ 0 T 0 ∫ − ∞ t − ρ δ h (σ)dσμ(t)dt,$
(59)

we get

(60)

from which we obtain

(61)

Similarly, we have

(62)

It follows from (61) and (62) that

(63)

Send $ρ→0$, $τ→t$, 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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The article is a joint work of four authors who contributed equally to the final version of the paper. All authors read and approved the final manuscript.

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