1. Introduction
In recent years, a lot of works have been carried out to investigate the Camassa-Holm equation [1],
(1)ut-utxx+kux+3uux=2uxuxx+uuxxx,
which is a completely integrable equation. In fact, the Camassa-Holm equation arises as a model describing the unidirectional propagation of shallow water waves over a flat bottom [1–3]. The equation was originally derived much earlier as a bi-Hamiltonian generalization of the Korteweg-de Vries equation (see [4]). Johnson [2], Constantin and Lannes [5] derived models which include the Camassa-Holm equation (1). It has been found that (1) conforms with many conservation laws (see [6, 7]) and possesses smooth solitary wave solutions if k>0 [3, 8] or peakons if k=0 [3, 9]. Equation (1) is also regarded as a model of the geodesic flow for the H1 right invariant metric on the Bott-Virasoro group if k>0 and on the diffeomorphism group if k=0 (see [10–14]). The well-posedness of local strong solutions for generalized forms of (1) has been given in [15–17]. The sharpest results for the global existence and blow-up solutions are found in Bressan and Constantin [18, 19].

Recently, Li et al. [20] studied the following generalized Camassa-Holm equation:
(2)ut-utxx+kumux+(m+3)um+1ux =(m+2)umuxuxx+um+1uxxx,
where m≥0 is a natural number. Obviously, (2) reduces to (1) if m=0. The authors applied the pseudoparabolic regularization technique to build the local well-posedness for (2) in Sobolev space Hs(R) with s>3/2 via a limiting procedure. Provided that the initial value u0 satisfies a sign condition and u0∈Hs(R)(s>3/2), it is shown that there exists a unique global strong solution for (2) in space C([0,∞);Hs(R))⋂C1([0,∞);Hs-1(R)). However, the existence and uniqueness of the global weak solution for (2) is not investigated in [20].

The objective of this paper is to establish the well-posedness of global weak solutions for (2). Using the estimates in Hq(R) with 0≤q≤1/2, which are derived from the equation itself, we prove that there exists a unique global weak solution to (2) in space Hs(R) with 1≤s≤3/2 if u0∈Hs(R), and (1-∂x2)u0 satisfies an associated sign condition.

The structure of this paper is as follows. The main result is given in Section 2. Several lemmas are given in Section 3. Section 4 establishes the proof of the main result.

2. Main Results
Firstly, we give some notations.

The space of all infinitely differentiable functions ϕ(t,x) with compact support in [0,+∞)×R is denoted by C0∞. Lp=Lp(R) (1≤p<+∞) is the space of all measurable functions h such that ∥h∥Lpp=∫R|h(t,x)|pdx<∞. We define L∞=L∞(R) with the standard norm ∥h∥L∞=infm(e)=0 supx∈R∖e |h(t,x)|. For any real number s, we let Hs=Hs(R) denote the Sobolev space with the norm defined by
(3)∥h∥Hs=(∫R(1+|ξ|2)s|h^(t,ξ)|2dξ)1/2<∞,
where h^(t,ξ)=∫Re-ixξh(t,x)dx.

For T>0 and nonnegative number s, let C([0,T);Hs(R)) denote the Frechet space of all continuous Hs-valued functions on [0,T). We set Λ=(1-∂x2)1/2.

Defining
(4)ϕ(x)={e1/(x2-1),|x|<1,0,|x|≥1,
and letting ϕε(x)=ε-(1/4)ϕ(ε-(1/4)x) with 0<ε<1/4 and uε0=ϕε⋆u0 (convolution of ϕε and u0), we know that uε0∈C∞ for any u0∈Hs with s>0. Notation (1-∂x2)u+k/2(m+1)∈N+(R) (or equivalently (1-∂x2)u+k/2(m+1)∈N-(R)) means that (1-∂x2)u⋆ϕε+k/2(m+1)≥0 (or equivalently (1-∂x2)u⋆ϕε+k/2(m+1)≤0) for an arbitrary sufficiently small ε>0.

For the equivalent form of (2), we consider its Cauchy problem
(5)ut-utxx=-km+1(um+1)x-m+3m+2(um+2)x+1m+2∂x3(um+2)-(m+1)∂x(umux2)+umuxuxx,u(0,x)=u0(x).

Definition 1.
A function u(t,x)∈L2([0,+∞),Hs(R)) is called a global weak solution to problem (5) if for every T>0, u(t,x)∈Hs(R),ut(t,x)∈Hs-1(R), and all ψ(t,x)∈C0∞, it holds that
(6)∫0T∫R[ut-utxx+kumux+(m+3)um+1ux -(m+2)umuxuxx-um+1uxxx]ψ(t,x)dx dt=0
with u(0,x)=u0(x).

Now, we give the main result of this work.

Theorem 2.
Let u0(x)∈Hs(R), 1≤s≤3/2, (1-∂x2)u0+k/2(m+1)∈N+(R), and k≥0 (or equivalently (1-∂x2)u0+k/2(m+1)∈N-(R), k≤0). Then, problem (5) has a unique global weak solution u(t,x)∈L2([0,+∞),Hs(R)) in the sense of distribution, and ux∈L∞([0,+∞)×R).

3. Several Lemmas
Lemma 3 (see [<xref ref-type="bibr" rid="B20">20</xref>]).
Let u0(x)∈Hs(R) with s>3/2. Then, the Cauchy problem (5) has a unique solution
(7)u(t,x)∈C([0,T);Hs(R))⋂C1([0,T);Hs-1(R)),
where T>0 depends on ∥u0∥Hs(R).

Lemma 4 (see [<xref ref-type="bibr" rid="B20">20</xref>]).
Let u0(x)∈Hs, s>3/2, and k≥0,(1-∂x2)u0+k/2(m+1)≥0 (or equivalently k≤0, (1-∂x2)u0+k/2(m+1)≤0). Then, problem (5) has a unique solution satisfying
(8)u(t,x)∈C([0,∞);Hs(R))⋂C1([0,∞);Hs-1(R)).

Using the first equation of system (5) derives
(9)ddt∫R(u2+ux2)dx=0,
from which one has the conservation law
(10)∫R(u2+ux2)dx=∫R(u02+u0x2)dx.

Lemma 5 (see [<xref ref-type="bibr" rid="B20">20</xref>]).
Let s>3/2, and the function u(t,x) is a solution of problem (5) and the initial data u0(x)∈Hs. Then, the following inequality holds:
(11)∥u∥H12≤∫R(u2+ux2)dx=∫R(u02+u0x2)dx.

For q∈(0,s-1], there is a constant c such that
(12)∫R(Λq+1u)2dx≤∫R(Λq+1u0)2dx+c∫0t∥u∥Hq+12(∥ux∥L∞∥u∥L∞m +∥u∥L∞m-1∥ux∥L∞2)dτ.

For q∈[0,s-1], there is a constant c such that
(13)∥ut∥Hq≤c∥u∥Hq+1(∥u∥L∞m∥u∥H1+∥u∥L∞m∥ux∥L∞ +∥u∥L∞m-1∥ux∥L∞2).

For (2), consider the problem
(14)pt=um+1(t,p), t∈[0,T),p(0,x)=x.

Lemma 6 (see [<xref ref-type="bibr" rid="B20">20</xref>]).
Let u0∈Hs, s≥3, and let T>0 be the maximal existence time of the solution to problem (5). Then, problem (14) has a unique solution p∈C1([0,T)×R). Moreover, the map p(t,·) is an increasing diffeomorphism of R with px(t,x)>0 for (t,x)∈[0,T)×R.

Differentiating (14) with respect to x yields
(15)ddtpx=(m+1)umux(t,p)px, t∈[0,T),px(0,x)=1,
which leads to
(16)px(t,x)=exp(∫0t(m+1)umux(τ,p(τ,x))dτ).

The next lemma is reminiscent of a strong invariance property of the Camassa-Holm equation (the conservation of momentum [21]).

Lemma 7 (see [<xref ref-type="bibr" rid="B20">20</xref>]).
Let u0∈Hs with s≥3, and let T>0 be the maximal existence time of the problem (5). It holds that
(17)y(t,p(t,x))px2(t,x)=y0(x)e∫0tmumuxdτ,
where (t,x)∈[0,T)×R and y:=u-uxx+k/2(m+1).

Lemma 8.
If u0∈Hs, s≥3, such that (1-∂x2)u0+k/2(m+1)≥0, k≥0 (or equivalently, (1-∂x2)u0+k/2(m+1)≤0, k≤0), then the solution of problem (5) satisfies
(18)∥ux∥L∞≤∥u∥L∞+|k|2(m+1)≤c.

Proof.
Using u0-u0xx+k/2(m+1)≥0, it follows from Lemma 7 that u-uxx+k/2(m+1)≥0. Letting Y1=u-uxx, we have
(19)u=12e-x∫-∞xeηY1(t,η)dη+12ex∫x∞e-ηY1(t,η)dη,
from which we obtain
(20)∂xu(t,x) =-12(e-x∫-∞xeηY1(t,η)dη+ex∫x∞e-ηY1(t,η)dη) +ex∫x∞e-ηY1(t,η)dη =-u(t,x)+ex∫x∞e-ηY1(t,η)dη =-u(t,x)+ex∫x∞e-η(Y1(t,η)+k2(m+1))dη -k2(m+1)ex∫x∞e-ηdη =-u(t,x)+ex∫x∞e-η(y(t,η))dη-k2(m+1) ≥-u(t,x)-k2(m+1).
On the other hand, we have
(21)∂xu(t,x) =12(e-x∫-∞xeηY1(t,η)dη+ex∫x∞e-ηY1(t,η)dη) -e-x∫-∞xeηY1(t,η)dη =u(t,x)-e-x∫-∞xeηY1(t,η)dη =u(t,x)-e-x∫-∞xeη(Y1(t,η)+k2(m+1))dη +k2(m+1)e-x∫-∞xeηdη =u(t,x)-e-x∫-∞xeηy(t,η)dη+k2(m+1) ≤u(t,x)+k2(m+1).
The inequalities (19), (20), and (21) derive that inequality (18) is valid. Similarly, if (1-∂x2)u0+k/2(m+1)≤0, k≤0, we still know that (18) is valid.

Lemma 9.
For s>0, u0∈Hs, it holds that
(22)∥uε0x∥L∞≤c∥u0x∥L∞,∥uε0∥Hq≤c, if q≤s,∥uε0∥Hq≤cεs-q/4, if q>s,∥uε0-u0∥Hq≤cεs-q/4, if q≤s,∥uε0-u0∥Hs=o(1),
where c is a constant independent of ε.

The proof of this lemma can be found in Lai and Wu [15].

From Lemma 3, it derives that the Cauchy problem
(23)ut-utxx=-m+3m+2(um+2)x+1m+2∂x3(um+2)-(m+1)∂x(umux2)+umuxuxx,u(0,x)=uε0(x), x∈R,
has a unique solution u depending on the parameter ε. We write uε(t,x) to represent the solution of problem (23). Using Lemma 3 derives that uε(t,x)∈C∞([0,T),H∞(R)) since uε0(x)∈C0∞(R).

Lemma 10.
Provided that u0∈Hs, 1≤s≤3/2, k≥0, and (1-∂x2)u0+k/2(m+1)∈N+(R) (or equivalently (1-∂x2)u0+k/2(m+1)∈N-(R), k≤0), then there exists a constant c0>0 independent of ε such that the solution of problem (23) satisfies
(24)∥uεx∥L∞≤∥ux∥L∞+|k|2(m+1)≤c0.

Proof.
Using identity (10) and Lemma 9, if u0∈Hs(R) with 1≤s≤3/2, we have
(25)∥uε∥L∞≤∥uε∥H1=∥uε0∥H1≤c,
where c is independent of ε.

From Lemma 8, we have
(26)∥uεx∥L∞≤∥uε∥L∞+|k|2(m+1)≤c+|k|2(m+1),
which completes the proof.

Lemma 11.
For any f1∈L∞, f2∈Hz with z≤0, it holds that
(27)∥f1f2∥Hz≤c∥f1∥L∞∥f2∥Hz for any z≤0.

The proof of this lemma can be found in [15].

4. Existence and Uniqueness of Global Weak Solution
Provided that 1≤s≤3/2, for problem (23), applying Lemmas 5, 9, and 10, and the Gronwall’s inequality, we obtain the inequalities
(28)∥uε∥H1≤∥uε0∥H1≤c,∥uε∥Hq≤c∥uε0∥Hqexp[∫0t(∥uεx∥+∥uεx∥L∞2)dτ]≤cect,∥uεt∥Hr≤∥uε∥Hr+1(1+ect)≤c(1+ect),
where q∈(0,s],r∈[0,s-1], and c is a constant independent of ε. It follows from the Aubin’s compactness theorem that there is a subsequence of {uε}, denoted by {uεn}, such that {uεn} and their temporal derivatives {uεnt} are weakly convergent to a function u(t,x) and its derivative ut in L2([0,T],Hs) and L2([0,T],Hs-1), respectively, where T is an arbitrary fixed positive number. Moreover, for any real number R1>0, {uεn} is convergent to the function u strongly in the space L2([0,T],Hq(-R1,R1)) for q∈(0,s] and {uεnt} converges to ut strongly in the space L2([0,T],Hr(-R1,R1)) for r∈[0,s-1].

4.1. The Proof of Existence for Global Weak Solution
For an arbitrary fixed T>0, from Lemma 10, we know that {uεnx}(εn→0) is bounded in the space L∞. Thus, the sequences {uεn}, {uεnx}, {uεnx2}, and {uεnx3} are weakly convergent to u, ux, ux2, and ux3 in L2([0,T],Hr(-R1,R1)) for any r∈[0,s-1), separately. Using um(ux2)x=(umux2)x-(um)xux2, we know that u satisfies the equation
(29)-∫0T∫Ru(gt-gxxt)dx dt =∫0T∫R[(m+3m+2um+2+(m+1)umux2)gx -1m+2um+2gxxx-12umux2gx -m2um-1ux3g]dx dt,
with u(0,x)=u0(x) and g∈C0∞. Since X=L1([0,T]×R) is a separable Banach space and {uεnx} is a bounded sequence in the dual space X*=L∞([0,T]×R) of X, there exists a subsequence of {uεnx}, still denoted by {uεnx}, weakly star convergent to a function v in L∞([0,T]×R). As {uεnx} weakly converges to ux in L2([0,T]×R), it results that ux=v almost everywhere. Thus, we obtain ux∈L∞([0,T]×R). Since T>0 is an arbitrary number, we complete the global existence of weak solutions to problem (5).

Proof of Uniqueness.
Suppose that there exist two global weak solutions u(t,x) and v(t,x) to problem (5) with the same initial value u(0,x)∈Hs(R), 1≤s≤3/2, we consider its associated regularized problem (23). Letting wε=uε(t,x)-vε(t,x), from Lemma 10, we get ∥∂uε(t,x)/∂x∥L∞≤c and ∥∂vε(t,x)/∂x∥L∞≤c which is independent of ε. Still denoting u=uε,v=vε, and w=wε, it holds that
(30)wt=(1-∂x2)-1[-∂x(um+2-vm+2) -∂x[∂x(um+1)∂xw +∂x(um+1-vm+1)∂xv] +[umuxuxx-vmvxvxx]-∂x(um+2-vm+2)] -1m+2∂x(um+2-vm+2),w(0,x)=0.

Multiplying both sides of (30) by w, we get
(31)12ddt∫Rw2dx≤c|∫Rw(um+2-vm+2)xdx| +|∫RwΛ-2(um+2-vm+2)xdx|+|∫RwΛ-2[∂x(um+1)∂xw]xdx|+|∫RwΛ-2[∂x(um+1-vm+1)∂xv]xdx|+|∫RwΛ-2[umuxuxx-vmvxvxx]dx|=I1+I2+I3+I4+I5.
Using ∥u∥L∞≤c, ∥v∥L∞≤c, ∥ux∥L∞≤c, ∥vx∥L∞≤c, we have
(32)I1≤c|∫Rw[w∑j=0m+1ujvm+1-j]xdx|=c|∫Rw[wx∑j=0m+1ujvm+1-j+w∑j=0m+1(ujvm+1-j)x]dx|=c|∫R(12w2)x∑j=0m+1ujvm+1-j+w2∑j=0m+1(ujvm+1-j)xdx|=c|∫R(-12w2)∑j=0m+1(ujvm+1-j)x+w2∑j=0m+1(ujvm+1-j)xdx|=c|∫R(12w2)∑j=0m+1(ujvm+1-j)xdx|≤c∥w∥L22∑j=0m+1∥(ujvm+1-j)x∥L∞≤c∥w∥L22.
Applying Lemma 11 repeatedly, we have
(33)I2≤c∥w∥L2∥Λ-2(um+2-vm+2)x∥L2≤c∥w∥L2∥w∑j=0m+1ujvm+1-j∥L2≤c∥w∥L22∑j=0m+1∥u∥L∞j∥v∥L∞m+1-j≤c∥w∥L22,I3≤c∥w∥L2∥Λ-2[∂x(um+1)∂xw]x∥L2≤c∥w∥L2∥∂x(um+1)∂xw∥H-1≤c∥w∥L2∥∂xw∥H-1∥∂x(um+1)∥L∞≤c∥w∥L22,I4≤c∥w∥L2∥∂x(um+1-vm+1)∂xv∥H-1≤c∥w∥L2∥∂xv∥L∞∥∂x(um+1-vm+1)∥H-1≤c∥w∥L2∥um+1-vm+1∥H0≤c∥w∥L2∥w∑j=0mujvm-j∥L2≤c∥w∥L22∑j=0m∥u∥L∞j∥v∥L∞m-j≤c∥w∥L22.
For I5, using Lemma 11 derives
(34)I5≤c∥w∥L2∥(um-vm)(ux2)x+vm[ux2-vx2]x∥H-2≤c∥w∥L2∥(um-vm)(ux2)x∥H-2+∥vm[ux2-vx2]x∥H-2≤c∥w∥L2(∥[(um-vm)(ux2)]x-(um-vm)xux2∥H-2 +∥v∥L∞m∥(u-v)x(ux+vx)∥H-1∥[(um-vm)(ux2)]x-(um-vm)xux2∥H-2)≤c∥w∥L2(∥(um-vm)ux2∥H-1+∥(um-vm)xux2∥H-2+c∥w∥L2)≤c∥w∥L2(∥ux∥L∞2∥w∥L2∑j=0m-1∥u∥L∞j∥v∥L∞m-1-j+c∥w∥L2)≤c∥w∥L22.
Using (32)–(34), we get
(35)12ddt∫Rw2dx≤c∥w∥L22.
Applying w(0)=0 results in ∥w∥L22=0. Consequently, we know that the global weak solution is unique.