# Several Dynamic Properties of Solutions to a Generalized Camassa-Holm Equation

Authors: Zheng Yin
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

For a nonlinear generalization of the Camassa-Holm equation,
we investigate the dynamic properties of solutions for the equation under
the assumption that the initial valueu0(x) lies in the space H1(R). A one-sided upper bound estimate on the first-order spatial derivative, Lp bound estimate, and a space-time higher-norm estimate for the solutions
are obtained.


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## Body

## 1. Introduction

Hakkaev and Kirchev [1] investigated the following generalized Camassa-Holm equation:
(1)ut-utxx+(2ku+n+22un+1)x=(nun-1ux22+unuxx)x,
where n is an integer. When n=1, (1) becomes the Camassa-Holm model (see [2]). The local well-posedness in the Sobolev space Hs with s>3/2 is established, and sufficient conditions for the stability and instability of the solitary wave solutions are given in [1]. However, the Lp estimate of strong solutions and one-sided upper bound estimate on the first-order spatial derivative for the solutions are not discussed in [1]. This constitutes the objective of this work.Like the Camassa-Holm equation (see [1, 2]), (1) has the conservation law
(2)∫R‍(u2(t,x)+ux2(t,x))dx=∫R‍(u2(0,x)+ux2(0,x))dx,
which plays an important role in our further investigations.In fact, many scholars have paid their attentions to the study of the Camassa-Holm equation. The existence of global weak solutions is established in Constantin and Escher [3], Constantin and Molinet [4], Xin and Zhang [5], and Coclite et al. [6]. It was shown in Constantin and Escher [7] that the blowup occurs in the form of breaking waves. Namely, the solution remains bounded, but its slope becomes unbounded in finite time. After wave breaking, the solution is continued uniquely either as a global conservative weak solution [8, 9] or as a global dissipative solution [10, 11]. Exact traveling wave solutions for the Camassa-Holm equation are presented in [12]. For other methods to investigate the problems involving various dynamic properties of the Camassa-Holm equation, the reader is referred to [13–16] and the references therein.In this paper, we investigate several dynamic properties of strong solutions for the generalized Camassa-Holm equation (1) in the case where n is an odd natural number and the assumption u0(x)∈H1(R). The results obtained in this work include a one-sided upper bound estimate on the first-order derivatives of the solution, a space-time higher-norm estimate, and the Lp(2≤p<∞) bound estimate.The rest of this paper is organized as follows. Section2 states the main result. Several lemmas are given in Section 3 where the proof of main result is completed.


## 2. Main Result

Letm be a nonnegative integer and n=2m+1. In this case, the Cauchy problem for (1) is written in the form
(3)ut-utxx+∂x(2ku+2m+32u2m+2)=12m+2u2m+2∂x3u2m+2-2m+12∂x(u2mux2),u(0,x)=u0(x),
which is equivalent to
(4)ut+u2m+1ux+∂H∂x=0,∂H∂x=Λ-2∂x[2ku+2m2+5m+22(m+1)u2m+2+2m+12u2mux2],u(0,x)=u0(x),
where the operator Λ2=1-∂2/∂x2 and
(5)Λ-2f(x)=12∫R‍e-|x-y|f(y)dyforf(x)∈L2(R).We introduce a result presented in [1] for problem (3).Lemma 1 (see [1]).
Suppose thatu0∈Hs with constant s>3/2. Then there is a real number T>0 such that the problem (3) has a unique solution u(t,x) satisfying
(6)u(t,x)∈C([0,T);Hs(R))⋂‍C1([0,T);Hs-1(R)).Now we state the main result of this paper.Theorem 2.
Letu0∈Hs(R) with s>3. Then the solution of problem (3) has the following properties.(a)
There exists a positive constantc0 depending on m and ∥u0∥H1(R) such that the one-sided L∞ norm estimate on the first-order spatial derivative holds
(7)∂u(t,x)∂x≤c0(1+t),for(t,x)∈[0,∞)×R.(b)
Let0<γ<1, T>0, and a,b∈R,a<b. Then there exists a positive constant c1 depending only on ∥u0∥H1(R),γ,T,a,b, and m such that the following estimate holds:
(8)∫0t‍∫ab‍u2m|∂u(t,x)∂x|2+γdx≤c1.(c)
There exists a constantc2 depending only on ∥u0∥H1(R),andm,p such that
(9)∥u∥Lp(R)≤c2(1+t),t∈[0,∞),2≤p<∞.


## 3. Proof of Main Result

From the conservation law (2), we have
(10)∥u∥L∞(R)≤∥u∥H1(R)=∥u0∥H1(R).
Differentiating the first equation of problem (4) with respect to x and writing ∂u/∂x=q, we obtain
(11)∂q∂t+u2m+1∂q∂x+2m+12u2mq2=2ku+2m2+5m+22(m+1)u2m+2-Λ-2[2ku+2m2+5m+22(m+1)u2m+2+2m+12u2mq2]=Q(t,x).Lemma 3.
Let0<γ<1,T>0, and a,b∈R,a<b. Then there exists a positive constant c1 depending only on ∥u0∥H1(R),γ,T,a,b, and m, such that the space higher integrability estimate holds
(12)∫0T‍∫ab‍u2m|∂u(t,x)∂x|2+γdx≤c1,
where u=u(t,x) is the unique solution of problem (3).Proof.
The proof is a variant of the proof presented in Xin and Zhang [5] (or see Coclite et al. [6]). Let ϕ∈C∞(R) be a cut-off function such that 0<ϕ<1 and
(13)ϕ(x)={1,ifx∈[a,b],0,ifx∈(-∞,a-1]⋃‍[b+1,∞).
Letting θ(ξ)=ξ(1+|ξ|)γ,ξ∈R, and 0<γ<1 yields
(14)θ′(ξ)=(1+(1+γ)|ξ|)(1+|ξ|)γ-1,θ′′(ξ)=γsign(ξ)(1+|ξ|)γ-2(2+(1+γ)|ξ|)=γ(1+γ)sign(ξ)(1+|ξ|)γ-1+(1-γ)γsign(ξ)(1+|ξ|)γ-2,
from which we get
(15)|θ(ξ)|≤|ξ|+|ξ|1+γ,|θ′(ξ)|≤1+(1+γ)|ξ|,|θ′′(ξ)|≤2γ,(16)ξθ(ξ)-12ξ2θ′(ξ)=1-γ2ξ2(1+|ξ|)γ+γ2ξ2(1+|ξ|)γ-1≥1-γ2ξ2(1+|ξ|)γ.
Multiplying (11) by ϕθ′(q), using the chain rule, and integrating over πT:=[0,T]×R, we have
(17)(2m+1)∫πT‍ϕ(x)u2mqθ(q)dtdx-2m+12∫πT‍ϕ(x)u2mq2θ′(q)dtdx=∫R‍ϕ(x)(θ(q(t,x))-θ(q(0,x)))dx-∫πT‍u2m+1ϕ′(x)θ(q)dtdx-∫πT‍Q(t,x)ϕ(x)θ′(q)dtdx.
From (16), we get
(18)(2m+1)∫πT‍ϕ(x)u2mqθ(q)dtdx-2m+12∫πT‍ϕ(x)u2mq2θ′(q)dtdx=∫πT‍(2m+1)ϕ(x)u2m(qθ(q)-12q2θ′(q))dtdx≥(1-γ)2∫πT‍ϕ(x)u2mq2(1+|q|)γdtdx.
Using the Hölder inequality, (2), and (15) yields
(19)|∫R‍ϕ(x)θ(q)dx|≤∫R‍ϕ(x)(|q|1+γ+|q|)dx≤∥ϕ∥L2/(1-γ)(R)∥q∥L2(R)1+γ+∥ϕ∥L2(R)∥q∥L2(R)≤(b-a+2)(1-γ)/2∥u0∥H1(R)1+γ+(b-a+2)1/2∥u0∥H1(R),|∫πT‍u2m+1ϕ′(x)θ(q)dtdx|≤∫πT‍|u|2m+1|ϕ′(x)|(|q|1+γ+|q|)dtdx≤∥u0∥H1(R)2m+1∫0T‍(∥ϕ′∥L2/(1-γ)(R)∥q∥L2(R)1+γ∥ϕ′∥L2/(1-γ)(R)∥q∥L2(R)1+γ≤∥u0∥H1(R)2m+1∫0T‍+∥ϕ′∥L2(R)∥q∥L2(R)∥ϕ′∥L2/(1-γ)(R)∥q∥L2(R)1+γ)dt≤cT(∥ϕ′∥L2/(1-γ)(R)∥u0∥H1(R)1+γ+∥ϕ′∥L2(R)∥u0∥H1(R)).
Using (10), we obtain
(20)∥2ku+2m2+5m+22(m+1)u2m+2∥L∞≤c,∥Λ-2[2ku+2m2+5m+22(m+1)u2m+2]∥L∞≤c.
Applying (10), the Hölder inequality, and ∫R‍e-|x-y|dy=2, we have
(21)|Λ-2[u2mq2]|=|∫R‍12e-|x-y|u2muy2dy|≤c∥u∥H1(R)≤c,
where c is a constant depending on ∥u0∥H1(R) and m.
From (20) and (21), we deduce that there exists a positive constant c depending on ∥u0∥H1(R) and m, such that
(22)∥Q(t,x)∥L∞(R)≤c,
from which we get
(23)|∫πT‍Q(t,x)ϕ(x)θ′(q)dtdx|≤c∫πT‍|ϕ(x)|((1+γ)|q|+1)dtdx≤cT(+∫R‍|ϕ(x)|dx(1+γ)∥ϕ(x)∥L2(R)∥u0∥H1(R)≤cT+∫R‍|ϕ(x)|dx).
From inequalities (18)–(19) and (23), we obtain (12).Lemma 4.
There exists a positive constantc depending only on ∥u0∥H1(R) and m such that
(24)∥Q(t,·)∥L∞(R)≤c,∥∂H(t,·)∂x∥L∞(R)≤c.
If k=0, it holds that
(25)∥Q(t,·)∥L1(R)≤c,∥∂H(t,·)∂x∥L1(R)≤c,(26)∥Q(t,·)∥L2(R)≤c,∥∂H(t,·)∂x∥L2(R)≤c.Proof.
We have(27)Q(t,x)=2ku+2m2+5m+22(m+1)u2m+2-Λ-2[2ku+2m2+5m+22(m+1)u2m+2+2m+12u2mq2],∂H∂x=Λ-2∂x[2ku+2m2+5m+22(m+1)u2m+2+2m+12u2mux2].
The first inequality of (24) is proved in Lemma 3 (see (22)). For the second inequality in (24), we have
(28)|∂H(t,·)∂x|=|[2ku+2m2+5m+22(m+1)u2m+2+2m+12u2muy2]y12∫Re-|x-y|=12∫R×[2ku+2m2+5m+22(m+1)u2m+2=12∫R×+2m+12u2muy22m2+5m+22(m+1)]ydy12∫R|=|12e-x∫-∞x‍ey[2ku+2m2+5m+22(m+1)u2m+2=12e-x∫-∞x‍ey+2m+12u2muy22m2+5m+22(m+1)]ydy+12ex∫x∞‍e-y[2ku+2m2+5m+22(m+1)u2m+2+12ex∫x∞‍e-y+2m+12u2muy22m2+5m+22(m+1)]ydy|=|-12e-x∫-∞x‍ey[2ku+2m2+5m+22(m+1)u2m+2=-12e-x∫-∞x‍ey+2m+12u2muy22m2+5m+22(m+1)]dy=+12ex∫x∞‍e-y[2ku+2m2+5m+22(m+1)u2m+2=+12ex∫x∞‍e-y+2m+12u2muy22m2+5m+22(m+1)]dy|≤∫R‍e-|x-y||2ku+2m2+5m+22(m+1)u2m+2≤∫R‍e-|x-y|+2m+12u2muy22m2+5m+22(m+1)|dy≤c∥u∥H1(R)≤c∥u0∥H1(R),
which together with (22) results in (24).
In fact, we have(29)∥u2m+2∥L1(R)≤c∥u∥L2(R)2≤c∥u∥H1(R)2≤c∥u0∥H1(R)2,∥Λ-2[2m2+5m+22(m+1)u2m+2+2m+12u2mq2]∥L1(R)≤c∥u∥H1(R)2≤c∥u0∥H1(R)2,∥Λ-2[∂xu2m+2]∥L1(R)≤∬Re-|x-y|u2m+2dxdy≤c∥u∥H1(R)2≤c∥u0∥H1(R)2,∥Λ-2∂x[u2mux2]∥L1(R)≤∬Re-|x-y|u2muy2dxdy≤c∥u∥H1(R)2≤c∥u0∥H1(R)2.
From (29)–(31), we obtain (25).
Using(30)∥Q(t,·)∥L2(R)2≤∥Q(t,·)∥L∞∥Q(t,·)∥L1(R)≤c,∥∂H(t,·)∂x∥L2(R)2≤∥∂H(t,·)∂x∥L∞∥∂H(t,·)∂x∥L1(R)≤c
completes the proof of (26).From Lemma1, we know that for any u0∈Hs(R) with s>3/2, there exist a maximal T=T(u0)>0 and a unique strong solution u to problem (3) such that
(31)u∈C([0,T);Hs(R))C1([0,T);Hs-1(R)).Consider the differential equation(32)ξt=u2m+1(t,ξ),t∈[0,T),ξ(0,x)=x.Lemma 5.
Assumeu0∈Hs,s>3, and let T>0 be the maximal existence time of the solution to problem (3). Then there exists a unique solution ξ∈C1([0,T)×R,R) to problem (32). In addition, the map p(t,·) is an increasing diffeomorphism of R with ξx(t,x)>0 for (t,x)∈[0,T)×R.Proof.
Using Lemma1, we obtain u∈C1([0,T);Hs-1(R)) and Hs-1∈C1(R). Therefore, we know that functions u(t,x) and ux(t,x) are bounded, Lipschitz in space, and C1 in time. The existence and uniqueness theorem for differential equations guarantees that problem (32) has a unique solution ξ∈C1([0,T)×R,R).
From (32), we get
(33)ddtξx=(2m+1)u2muξ(t,ξ)ξx,t∈[0,T),px(0,x)=1.
Furthermore,
(34)ξx(t,x)=exp(∫0t‍(2m+1)u2muξ(τ,ξ(τ,x))dτ).
For every T′<T, the Sobolev imbedding theorem gives rise to
(35)sup(τ,x)∈[0,T′)×R|ux(τ,x)|<∞.
Therefore, there exists a constant K0>0 such that ξx(t,x)≥e-K0t for (t,x)∈[0,T)×R. The proof is completed.Using (11) and (33), we get
(36)duξ(t,ξ(t,x))dt=duξ(t,ξ(t,x))dt+duξ(t,ξ(t,x))dξξt=duξ(t,ξ(t,x))dt+duξ(t,ξ(t,x))dξu2m+1=Q(t,ξ(t,x))-2m+12u2mq2≤∥Q(t,x)∥L∞≤c.
It follows from (36) that
(37)uξ(t,ξ(t,x))≤c+ct.
Using Lemma 5 and (37), we have
(38)ux(t,x)≤c+ct,t∈[0,∞).Using the first equation of problem (4) and Lemma 4, for an arbitrary integer K≥2, we have
(39)ddt∫R‍u2Kdx=2K∫R‍u2K-1dudtdx=2K∫R‍u2K-1[-u2m+1ux-∂H∂x]dx=-2K∫R‍u2K+2muxdx-2K∫R‍u2K-1∂H∂xdx=-2K∫R‍u2K-1∂H∂xdx≤2K∥u∥L∞(R)2K-3∥∂H∂x∥L∞(R)∥u∥L2(R)2≤c∥u0∥H1(R)≤c,
from which we get
(40)∫R‍u2Kdx≤ct+∫R‍u02Kdx≤ct+c∥u0∥L∞(R)∥u0∥H1(R)2.By theLp interpolation theorem, for all 2≤p<∞, we obtain
(41)∥u∥Lp(R)≤c(1+t),t∈[0,∞),
where c depends on ∥u0∥H1(R) and p.From Lemma3 and (38) and (41), we complete the proof of Theorem 2.



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