Blow-up phenomena and global existence for the weakly dissipative generalized periodic Degasperis-Procesi equation

Abstract

In this paper, we investigate the Cauchy problem of a weakly dissipative generalized periodic Degasperis-Procesi equation. The precise blow-up scenarios of strong solutions to the equation are derived by a direct method. Several new criteria guaranteeing the blow-up of strong solutions are presented. The exact blow-up rates of strong solutions are also determined. Finally, we give a new global existence results to the equation.

MSC:35G25, 35Q35, 58D05.

1 Introduction

Recently, the following generalized periodic Degasperis-Procesi equation (μ DP) was introduced and studied in [1–3]

$$\mu {(u)}_t-u_{txx}+3\mu (u)u_x=3u_x{u}_{xx}+u{u}_{xxx},$$

where $u(t,x)$ is a time-dependent function on the unite circle $\mathbb{S}=\mathbb{R}/\mathbb{Z}$ and $\mu (u)={\int }_{\mathbb{S}}u(t,x)dx$ denotes its mean. The μ DP equation can be formally described as an evolution equation on the space of tensor densities over the Lie algebra of smooth vector fields on the circle $\mathbb{S}$. In [2], the authors verified that the periodic μ DP equation describes the geodesic flows of a right-invariant affine connection on the Fréchet Lie group ${Diff}^{\mathrm{\infty }}(\mathbb{S})$ of all smooth and orientation-preserving diffeomorphisms of the circle $\mathbb{S}$.

Analogous to the generalized periodic Camassa-Holm (μ CH) equation [4–6], μ DP equation possesses bi-Hamiltonian form and infinitely many conservation laws. Here we list some of the simplest conserved quantities:

$$H_0=-\frac{9}{2}{\int }_{\mathbb{S}}ydx,H_1=\frac{1}{2}{\int }_{\mathbb{S}}{u}^{2}dx,H_2={\int }_{\mathbb{S}}(\frac{3}{2}\mu (u){\left(A^{-1}{\partial }_{x}u\right)}^2+\frac{1}{6}{u}^3)dx,$$

where $y=\mu (u)-u_{xx}$, $A=\mu -{\partial }_x^2$ is an isomorphism between $H^S$ and $H^{s-1}$. Moreover, it is easy to see that ${\int }_{\mathbb{S}}u(t,x)dx$ is also a conserved quantity for the μ DP equation.

Obviously, under the constraint of $\mu \equiv 0$, the μ DP equation is reduced to the μ Burgers equation [7].

It is clear that the closest relatives of the μ DP equation are the DP equation [8–11]

$$u_t-u_{txx}+4u{u}_x=3u_x{u}_{xx}+u{u}_{xxx},$$

which was derived by Degasperis and Procesi in [8] as a model for the motion of shallow water waves, and its asymptotic accuracy is the same as for the Camassa-Holm equation.

Generally speaking, energy dissipation is a very common phenomenon in the real world. It is interesting for us to study this kind of equation. Recently, Wu and Yin [12] considered the weakly dissipative Degasperis-Procesi equation. For related studies, we refer to [13] and [14]. Liu and Yin [15] discussed the blow-up, global existence for the weakly dissipative μ-Hunter-Saxton equation.

In this paper, we investigate the Cauchy problem of the following weakly dissipative periodic Degasperis-Procesi equation [16]:

$$\{\begin{array}{ll}\mu {(u)}_t-u_{txx}+3\mu (u)u_x=3u_x{u}_{xx}+u{u}_{xxx}-\lambda (\mu (u)-u_{xx}),& t>0,x\in \mathbb{R},\\ u(0,x)=u_0(x),& x\in \mathbb{R},\\ u(t,x+1)=u(t,x),& t\ge 0,x\in \mathbb{R},\end{array}$$

(1.1)

the constant λ is a nonnegative dissipative parameter and the term $\lambda y=\lambda (\mu (u)-u_{xx})$ models energy dissipation. Obviously, if $\lambda =0$ then the equation reduces to the μ DP equation. we can rewrite the system (1.1) as follows:

$$\{\begin{array}{ll}{y}_t+u{y}_x+3u_{x}y+\lambda y=0,& t>0,x\in \mathbb{R},\\ y=\mu (u)-u_{xx},& t>0,x\in \mathbb{R},\\ u(0,x)=u_0(x),& x\in \mathbb{R},\\ u(t,x+1)=u(t,x),& t\ge 0,x\in \mathbb{R}.\end{array}$$

(1.2)

Let $G(x):=\frac{1}{2}{x}^2-\frac{1}{2}|x|+\frac{13}{12}$, $x\in \mathbb{R}$ be the associated Green’s function of the operator $A^{-1}$, then the operator can be expressed by its associated Green’s function,

$$A^{-1}f(x)=(G\ast f)(x),f\in L^2,$$

where ∗ denotes the spatial convolution. Then equation (1.1) takes the equivalent form of a quasi-linear evolution equation of hyperbolic type:

$$\{\begin{array}{ll}{u}_t+u{u}_x+3\mu (u)A^{-1}{\partial }_{x}u+\lambda u=0,& t>0,x\in \mathbb{R},\\ u(0,x)=u_0(x),& x\in \mathbb{R},\\ u(t,x+1)=u(t,x),& t\ge 0,x\in \mathbb{R}.\end{array}$$

(1.3)

It is easy to check that the operator $A=\mu -{\partial }_x^2$ has the inverse

$$\begin{array}{rl}\left(A^{-1}f\right)(x)=& (\frac{1}{2}{x}^2-\frac{1}{2}x+\frac{13}{12})\mu (f)+(x-\frac{1}{2}){\int }_0^1{\int }_0^{y}f(s)dsdy\\ -{\int }_0^x{\int }_0^{y}f(s)dsdy+{\int }_0^1{\int }_0^y{\int }_0^{s}f(r)drdsdy.\end{array}$$

(1.4)

Since $A^{-1}$ and ${\partial }_x$ commute, the following identities hold:

$$\left(A^{-1}{\partial }_{x}f\right)(x)=(x-\frac{1}{2}){\int }_0^{1}f(x)dx-{\int }_0^{x}f(y)dy+{\int }_0^1{\int }_0^{x}f(y)dydx$$

(1.5)

and

$$\left(A^{-1}{\partial }_x^{2}f\right)(x)=-f(x)+{\int }_0^{1}f(x)dx.$$

(1.6)

The paper is organized as follows. In Section 2, we briefly give some needed results, including the local well-posedness of equation (1.1), and some useful lemmas and results which will be used in subsequent sections. In Section 3, we establish the precise blow-up scenarios and blow-up criteria of strong solutions. In Section 4, we give the blow-up rate of strong solutions. In Section 5, we give two global existence results of strong solutions.

Remark 1.1 Although blow-up criteria and global existence results of strong solutions to equation (1.1) are presented in [16], our blow-up results improve considerably earlier results.

2 Preliminaries

In this section we recall some elementary results which we want to use in this paper. We list them and skip their proofs for conciseness. Local well-posedness for equation (1.1) can be obtained by Kato’s theory [17], in [16] the authors gave a detailed description on well-posedness theorem.

Theorem 2.1 [16]

Let $s>3/2$ and $u_0\in H^s(\mathbb{S})$; then there is a maximal time T and a unique solution

$$u\in C([0,T);H^s(\mathbb{S}))\cap C^1([0,T);H^{s-1}(\mathbb{S}))$$

of the Cauchy problems (1.1) which depends continuously on the initial data, i.e. the mapping

$$H^s(\mathbb{S})\to C([0,T);H^s(\mathbb{S}))\cap C^1([0,T);H^{s-1}(\mathbb{S})),u_0\mapsto u(\cdot ,u_0),$$

is continuous.

Remark 2.1 The maximal time of existence $T>0$ in Theorem 2.1 is independent of the Sobolev index $s>3/2$.

Next we present the Sobolev-type inequalities, which play a key role to obtain blow-up results for the Cauchy problem (1.1) in the sequel.

Lemma 2.2 [18]

If $f\in H^1(\mathbb{S})$ is such that ${\int }_{\mathbb{S}}f(x)dx=0$, then we have

$$\underset{x\in \mathbb{S}}{max}{f}^2(x)\le \frac{1}{12}{\int }_{\mathbb{S}}{f}_x^2(x)dx.$$

Lemma 2.3 [19]

If $r>0$, let $\mathrm{\Lambda }={(1-{\partial }_x^2)}^{1/2}$, then

$${\parallel [{\mathrm{\Lambda }}^r,f]g\parallel }_{L^2}\le c({\parallel {\partial }_{x}f\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{r-1}g\parallel }_{L^2}+{\parallel {\mathrm{\Lambda }}^{r}f\parallel }_{L^2}{\parallel g\parallel }_{L^{\mathrm{\infty }}}),$$

where c is a constant depending only on r.

Lemma 2.4 [20]

Let $t_0>0$ and $v\in C^1([0,t_0);H^2(\mathbb{R}))$, then for every $t\in [0,t_0)$ there exists at least one point $\xi (t)\in \mathbb{R}$ with

$$m(t):=\underset{x\in \mathbb{R}}{inf}{v}_x(t,x)=v_x(t,\xi (t)),$$

and the function m is almost everywhere differentiable on $(0,t_0)$ with

$$\frac{d}{dt}m(t)=v_{tx}(t,\xi (t))\mathit{\text{a.e. on }}(0,t_0).$$

We also need to introduce the classical particle trajectory method which is motivated by McKean’s deep observation for the Camassa-Holm equation in [21]. Suppose $u(x,t)$ is the solution of the Camassa-Holm equation and $q(x,t)$ satisfies the following equation:

$$\{\begin{array}{ll}{q}_t=u(q,t),& 0<t<T,x\in \mathbb{R},\\ q(x,0)=x,& x\in \mathbb{R},\\ q(x+1,t)=x,& 0<t<T,x\in \mathbb{R},\end{array}$$

(2.1)

where T is the maximal existence time of solution, then $q(t,\cdot )$ is a diffeomorphism of the line. Taking the derivative with respect to x, we have

$$\frac{d{q}_x}{dt}=q_{tx}=u_x(q,t)q_x,t\in (0,T).$$

Hence

$$q_x(x,t)=exp({\int }_0^t{u}_x(q,s)ds)>0,q_x(x,0)=1,$$

(2.2)

which is always positive before the blow-up time.

In addition, integrating both sides of the first equation in equation (1.1) with respect to x on $\mathbb{S}$, we obtain

$$\frac{d}{dt}\mu (u)=-\lambda \mu (u),$$

it follows that

$$\mu (u)=\mu (u_0)e^{-\lambda t}:={\mu }_0{e}^{-\lambda t},$$

(2.3)

where

$${\mu }_0:=\mu (u_0)={\int }_{\mathbb{S}}{u}_0(x)dx.$$

(2.4)

3 Blow-up solutions

In this section, we are able to derive an import estimate for the $L^{\mathrm{\infty }}$-norm of strong solutions. This enables us to establish precise blow-up scenario and several blow-up results for equation (1.1).

Lemma 3.1 Let $u_0\in H^s$, $s>3/2$ be given and assume the T is the maximal existence time of the corresponding solution u to equation (1.1) with the initial data $u_0$. Then we have

$${\parallel u(t,x)\parallel }_{L^{\mathrm{\infty }}}\le e^{-\lambda t}(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}),\mathrm{\forall }t\in [0,T).$$

(3.1)

Proof The first equation of the Cauchy problem (1.1) is

$$u_t+u{u}_x+3\mu (u)A^{-1}{\partial }_{x}u+\lambda u=0.$$

In view of equation (1.5), we have

$$\left|A^{-1}{\partial }_{x}u\right|\le \frac{1}{2}|{\mu }_0|e^{-\lambda t}+2{\left({\int }_{\mathbb{S}}{u}^{2}dx\right)}^{\frac{1}{2}}.$$

A direct computation implies that

$$\begin{array}{rl}\frac{d}{dt}{\int }_{\mathbb{S}}{u}^{2}dx& =2{\int }_{\mathbb{S}}2u{u}_{t}dx\\ =-2{\int }_{\mathbb{S}}2u(u{u}_x+3\mu (u)A^{-1}{\partial }_{x}u+\lambda u)dx\\ =-2\lambda {\int }_{\mathbb{S}}{u}^{2}dx.\end{array}$$

It follows that

$${\int }_{\mathbb{S}}{u}^{2}dx={\int }_{\mathbb{S}}{u}_0^{2}dx\cdot e^{-2\lambda t}:={\mu }_2^2{e}^{-2\lambda t}.$$

(3.2)

So we have

$$|A^{-1}{\partial }_x(u)|\le (\frac{1}{2}|{\mu }_0|+2{\mu }_2)e^{-\lambda t}.$$

In view of equation (2.1) we have

$$\frac{du(t,q(t,x))}{dt}=u_t(t,q(t,x))+u_x(t,q(t,x))\frac{dq(t,x)}{dt}=(u_t+u{u}_x)(t,q(t,x)).$$

Combing the above relations, we arrive at

$$|\frac{du(t,q(t,x))}{dt}+\lambda u(t,q(t,x))|\le 3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)e^{-2\lambda t}.$$

Integrating the above inequality with respect to $t<T$ on $[0,t]$ yields

$$|e^{\lambda t}u(t,q(t,x))-u_0(x)|\le \frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }.$$

Thus

$$\left|u(t,q(t,x))\right|\le {\parallel u(t,q(t,x))\parallel }_{L^{\mathrm{\infty }}}\le e^{-\lambda t}(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}).$$

In view of the diffeomorphism property of $q(t,\cdot )$, we can obtain

$$|u(t,x)|\le {\parallel u(t,x)\parallel }_{L^{\mathrm{\infty }}}\le e^{-\lambda t}(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}).$$

This completes the proof of Lemma 3.1. □

Theorem 3.2 Let $u_0\in H^s$, $s>3/2$ be given and assume that T is the maximal existence time of the corresponding solution $u(t,x)$ to the Cauchy problem (1.1) with the initial data $u_0$. If there exists $M>0$ such that

$${\parallel u_x(t,\cdot )\parallel }_{L^{\mathrm{\infty }}}\le M,t\in [0,T),$$

then the $H^s$-norm of $u(t,\cdot )$ does not blow up on $[0,T)$.

Proof We assume that c is a generic positive constant depending only on s. Let $\mathrm{\Lambda }={(1-{\partial }_x^2)}^{1/2}$. Applying the operator ${\mathrm{\Lambda }}^s$ to the first one in equation (1.3), multiplying by ${\mathrm{\Lambda }}^{s}u$, and integrating over $\mathbb{S}$, we obtain

$$\frac{d}{dt}{\parallel u\parallel }_{H^s}^2=-2{(u{u}_x,u)}_{H^s}-6{(u,A^{-1}{\partial }_x(\mu (u)u))}_{H^s}-2\lambda {(u,u)}_{H^s}.$$

(3.3)

Let us estimate the first term of the above equation,

$$\begin{array}{rl}\left|{(u{u}_x,u)}_{H^s}\right|& =\left|{({\mathrm{\Lambda }}^s(u{u}_x),{\mathrm{\Lambda }}^{s}u)}_{L^2}\right|=|{([{\mathrm{\Lambda }}^s,u]u_x,{\mathrm{\Lambda }}^{s}u)}_{L^2}+{(u{\mathrm{\Lambda }}^s{u}_x,{\mathrm{\Lambda }}^{s}u)}_{L^2}|\\ \le {\parallel [{\mathrm{\Lambda }}^s,u]u_x\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{s}u\parallel }_{L^2}+\frac{1}{2}\left|{(u_x{\mathrm{\Lambda }}^{s}u,{\mathrm{\Lambda }}^{s}u)}_{L^2}\right|\\ \le 2{\parallel (u,v)\parallel }_{H^1\times H^1}^2\left(2{\parallel (u,v)\parallel }_{H^1\times H^1}^2\right)\\ \le c{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{H^s}^2,\end{array}$$

(3.4)

where we used Lemma 2.3 with $r=s$. Furthermore, we estimate the second term of the right hand side of equation (3.3) in the following way:

$$\begin{array}{rl}\left|{(u,A^{-1}{\partial }_x(\mu (u)u))}_{H^s}\right|& =\left|{(u,A^{-1}{\partial }_x\left(e^{-\lambda t}{\mu }_{0}u\right))}_{H^s}\right|\\ \le e^{-\lambda t}|{\mu }_0|{\parallel u\parallel }_{H^s}{\parallel A^{-1}{\partial }_{x}u\parallel }_{H^s}\\ \le c|{\mu }_0|{\parallel u\parallel }_{H^s}^2.\end{array}$$

(3.5)

Combing equations (3.4) and (3.5) with equation (3.3) we arrive at

$$\frac{d}{dt}{\parallel u\parallel }_{H^s}^2\le c(|{\mu }_0|+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}+2\lambda )|{\parallel u\parallel }_{H^s}^2.$$

An application of Gronwall’s inequality and the assumption of the theorem yield

$${\parallel u\parallel }_{H^s}^2\le e^{c(|{\mu }_0|+M+2\lambda )t}{\parallel u_0\parallel }_{H^s}^2.$$

This completes the proof of the theorem. □

The following result describes the precise blow-up scenario. Although the result which is proved in [16], our method is new, concise, and direct.

Theorem 3.3 Let $u_0\in H^s$, $s>3/2$ be given and assume that T is the maximal existence time of the corresponding solution $u(t,x)$ to the Cauchy problem (1.1) with the initial data $u_0$. Then the corresponding solution blows up in finite time if and only if

$$\underset{t\to T}{lim\hspace{0.17em}inf}\{\underset{x\in \mathbb{S}}{inf}{u}_x(t,x)\}=-\mathrm{\infty }.$$

Proof Since the maximal existence time T is independent of the choice of s by Theorem 2.1, applying a simple density argument, we only need to consider the case $s=3$. Multiplying the first one in equation (1.2) by y and integrating over $\mathbb{S}$ with respect to x yield

$$\begin{array}{rl}\frac{d}{dt}{\int }_{\mathbb{S}}{y}^{2}dx& =2{\int }_{\mathbb{S}}y{y}_{t}dx=-2{\int }_{\mathbb{S}}y(u{y}_x+3u_{x}y+\lambda y)dx\\ =-2{\int }_{\mathbb{S}}uy{y}_{x}dx-6{\int }_{\mathbb{S}}{u}_x{y}^{2}dx-2\lambda {\int }_{\mathbb{S}}{y}^{2}dx\\ =-5{\int }_{\mathbb{S}}{u}_x{y}^{2}dx-2\lambda {\int }_{\mathbb{S}}{y}^{2}dx.\end{array}$$

If $u_x$ is bounded from below on $[0,T)\times \mathbb{S}$, then there exists $N>\lambda >0$ such that

$$u_x(t,x)\ge -N,\mathrm{\forall }(t,x)\in [0,T)\times \mathbb{S},$$

then

$$\frac{d}{dt}{\int }_{\mathbb{S}}{y}^{2}dx\le (5N-2\lambda ){\int }_{\mathbb{S}}{y}^{2}dx.$$

Applying Gronwall’s inequality then yields for $t\in [0,T)$

$${\int }_{\mathbb{S}}{y}^{2}dx\le e^{(5N-2\lambda )t}{\int }_{\mathbb{S}}{y}^2(0,x)dx.$$

Note that

$${\int }_{\mathbb{S}}{y}^{2}dx={\mu }^2(u)+{\int }_{\mathbb{S}}{u}_{xx}^{2}dx\ge {\parallel u_{xx}\parallel }_{L^2}^2.$$

Since $u_x\in H^2\subset H^1$ and ${\int }_{\mathbb{S}}{u}_x=0$, Lemma 2.2 implies that

$${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}\le \frac{1}{2\sqrt{3}}{\parallel u_{xx}\parallel }_{L^2}\le e^{\frac{(5N-2\lambda )t}{2}}{\parallel y(0,x)\parallel }_{L^2}.$$

Theorem 3.1 ensures that the solution u does not blow up in finite time. On the other hand, by the Sobolev embedding theorem it is clear that if

$$\underset{t\to T}{lim\hspace{0.17em}inf}\{\underset{x\in \mathbb{S}}{inf}{u}_x(t,x)\}=-\mathrm{\infty },$$

then $T<\mathrm{\infty }$. This completes the proof of the theorem. □

We now give first sufficient conditions to guarantee wave breaking.

Theorem 3.4 Let $u_0\in H^s$, $s>3/2$ and T be the maximal time of the solution $u(t,x)$ to equation (1.1) with the initial data $u_0$. If

$$\underset{x\in \mathbb{S}}{inf}{u}_0^{\prime }(x)<-\frac{1}{2}\lambda -\frac{1}{2}\sqrt{{\lambda }^2+4\alpha },$$

then the corresponding solution to equation (1.1) blow up in finite time in the following sense: there exists $T_0$ satisfying

$$0<T_0\le \frac{1}{\sqrt{{\lambda }^2+4\alpha }}ln\left(\frac{2{inf}_{x\in \mathbb{S}}{u}_0^{\prime }(x)+\lambda -\sqrt{{\lambda }^2+4\alpha }}{2{inf}_{x\in \mathbb{S}}{u}_0^{\prime }(x)+\lambda +\sqrt{{\lambda }^2+4\alpha }}\right),$$

where $\alpha =3|{\mu }_0|(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}})$, such that

$$\underset{t\to T_0}{lim\hspace{0.17em}inf}\{\underset{x\in \mathbb{S}}{inf}{u}_x(t,x)\}=-\mathrm{\infty }.$$

Proof As mentioned early, we only need to consider the case $s=3$. Let

$$m(t):=\underset{x\in \mathbb{S}}{inf}[u_x(t,x)],t\in [0,T)$$

and let $\xi (t)\in \mathbb{S}$ be a point where this minimum is attained by using Lemma 2.4. It follows that

$$m(t)=u_x(t,\xi (t)).$$

Differentiating the first one in equation (1.3) with respect to x, we have

$$u_{tx}+u_x^2+u{u}_{xx}+3\mu (u)A^{-1}{\partial }_x^{2}u+\lambda u_x=0.$$

From equation (1.6) we deduce that

$$u_{tx}=-u_x^2-u{u}_{xx}+3\mu (u)(u-{\mu }_0)-\lambda u_x.$$

(3.6)

Obviously $u_{xx}(t,\xi (t))=0$ and $u(t,\cdot )\in H^3(\mathbb{S})\subset C^2(\mathbb{S})$. Substituting $(t,\xi (t))$ into equation (3.6), we get

$$\begin{array}{rl}\frac{dm(t)}{dt}& =-m^2(t)-\lambda m(t)+3\mu (u)u(t,\xi (t))-3{\mu }^2(u)\\ =-m^2(t)-\lambda m(t)+3{\mu }_0{e}^{-\lambda t}u(t,\xi (t))-3{\mu }_0^2{e}^{-2\lambda t}\\ \le -m^2(t)-\lambda m(t)+3|{\mu }_0|(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}).\end{array}$$

Set

$$\alpha =3|{\mu }_0|(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}).$$

Then we obtain

$$\begin{array}{rl}\frac{dm(t)}{dt}& \le -m^2(t)-\lambda m(t)+\alpha \\ \le -\frac{1}{4}(2m(t)+\lambda +\sqrt{{\lambda }^2+4\alpha })(2m(t)+\lambda -\sqrt{{\lambda }^2+4\alpha }).\end{array}$$

Note that if $m(0)<-\frac{1}{2}\lambda -\frac{1}{2}\sqrt{{\lambda }^2+4\alpha }$, then $m(t)<-\frac{1}{2}\lambda -\frac{1}{2}\sqrt{{\lambda }^2+4\alpha }$ for all $t\in [0,T)$. From the above inequality we obtain

$$\frac{2m(0)+\lambda +\sqrt{{\lambda }^2+4\alpha }}{2m(0)+\lambda -\sqrt{{\lambda }^2+4\alpha }}{e}^{\sqrt{{\lambda }^2+4\alpha }t}-1\le \frac{2\sqrt{{\lambda }^2+4\alpha }}{2m(t)+\lambda -\sqrt{{\lambda }^2+4\alpha }}\le 0.$$

Since

$$0<\frac{2m(0)+\lambda +\sqrt{{\lambda }^2+4\alpha }}{2m(0)+\lambda -\sqrt{{\lambda }^2+4\alpha }}<1,$$

then there exists $T_0$,

$$0<T_0\le \frac{1}{\sqrt{{\lambda }^2+4\alpha }}ln\left(\frac{2m(0)+\lambda -\sqrt{{\lambda }^2+4\alpha }}{2m(0)+\lambda +\sqrt{{\lambda }^2+4\alpha }}\right)$$

such that ${lim}_{t\to T_0}m(t)=-\mathrm{\infty }$. Theorem 3.3 implies that the solution u blows up in finite time. □

We give another blow-up result for the solutions of equation (1.1).

Theorem 3.5 Let $u_0\in H^s$, $s>3/2$ and T be the maximal time of the solution $u(t,x)$ to equation (1.1) with the initial data $u_0$. If $u_0$ is odd satisfies $u_0^{\prime }<-\lambda$, then the corresponding solution to equation (1.1) blows up in finite time.

Proof By $\mu (u(t,-x))={\mu }_0(t,-x)e^{-\lambda t}=-{\mu }_0(t,x)e^{-\lambda t}=-\mu (u(t,x))$, we can check the function

$$v(t,x):=-u(t,-x),t\in [0,T),x\in \mathbb{R},$$

is also a solution of equation (1.1), therefore $u(x,t)$ is odd for any $t\in [0,T)$. By continuity with respect to x of u and $u_{xx}$, we get

$$u(t,0)=u_{xx}(t,0)=0,\mathrm{\forall }t\in [0,T).$$

Define $h(t):=u_x(t,0)$ for $t\in [0,T)$. From equation (3.6), we obtain

$$\begin{array}{rl}\frac{dh(t)}{dt}& =-h^2(t)-\lambda h(t)-3{\mu }^2(u)\\ \le -h^2(t)-\lambda h(t)\\ =-h(t)(h(t)+\lambda ).\end{array}$$

Note that if $h(0)<-\lambda$, then $h(t)<-\lambda$ for all $t\in [0,T)$. From the above inequality we obtain

$$(1+\frac{\lambda }{h(0)})e^{\lambda t}-1\le \frac{\lambda }{h(t)}\le 0.$$

Since

$$0<\frac{h(0)+\lambda }{h(0)}<1,$$

there exists $T_0$,

$$0<T_0\le \frac{1}{\lambda }ln\frac{h(0)}{h(0)+\lambda }$$

such that ${lim}_{t\to T_0}m(t)=-\mathrm{\infty }$. Theorem 3.3 implies that the solution u blows up in finite time. □

4 Blow-up rate

In this section, we consider the blow-up profile; the blow-up rate of equation (1.1) with respect to time can be shown as follows.

Theorem 4.1 Let $u_0\in H^s$, $s>3/2$ and T be the maximal time of the solution $u(t,x)$ to equation (1.1) with the initial data $u_0$. If T is finite, then

$$\underset{t\to T}{lim}\{(T-t)\underset{x\in \mathbb{S}}{min}{u}_x(x,t)\}=-1.$$

Proof It is inferred from Lemma 2.4 that the function

$$m(t):=\underset{x\in \mathbb{S}}{min}{u}_x(x,t)=u_x(t,\xi (t))$$

is locally Lipschitz with $m(t)<0$, $t\in [0,T)$. Note that $u_{xx}=0$, a.e. $t\in [0,T)$. Then we deduce that

$$\begin{array}{rl}|m^{\prime }(t)+m^2(t)+\lambda m(t)|& =|3\mu (u)u(t,\xi (t))-3{\mu }^2(u)|\\ =|3{\mu }_0{e}^{-\lambda t}u(t,\xi (t))-3{\mu }_0^2{e}^{-2\lambda t}|\\ \le 3|{\mu }_0|(\frac{3|{\mu }_0|(\frac{1}{2}|{\mu }_0|+2{\mu }_2)}{\lambda }+{\parallel u_0\parallel }_{L^{\mathrm{\infty }}}+|{\mu }_0|):=K.\end{array}$$

It follows that

$$-K\le m^{\prime }(t)+m^2(t)+\lambda m(t)\le K\text{a.e. on }(0,T).$$

(4.1)

Thus,

$$-K-\frac{1}{4}{\lambda }^2\le m^{\prime }(t)+{(m(t)+\frac{1}{2}\lambda )}^2\le K+\frac{1}{4}{\lambda }^2\text{a.e. on }(0,T).$$

Now fix any $\epsilon \in (0,1)$. In view of Theorem 3.1, there exists $t_0\in (0,T)$ such that $m(t_0)<-\sqrt{(K+\frac{1}{4}{\lambda }^2)(1+\frac{1}{\epsilon })}-\frac{1}{2}\lambda$. Being locally Lipschitz, the function $m(t)$ is absolutely continuous on $[0,T)$. It then follows from the above inequality that $m(t)$ is decreasing on $[t_0,T)$ and satisfies

$$m(t)<-\sqrt{(K+\frac{1}{4}{\lambda }^2)(1+\frac{1}{\epsilon })}-\frac{1}{2}\lambda ,t\in [t_0,T).$$

Since $m(t)$ is decreasing on $[t_0,T)$, it follows that

$$\underset{t\to T}{lim}m(t)=-\mathrm{\infty }.$$

It is found from equation (4.1) that

$$1-\epsilon \le \frac{d}{dt}{(m(t)+\frac{1}{2}\lambda )}^{-1}=-\frac{m^{\prime }(t)}{{(m(t)+\frac{1}{2}\lambda )}^2}\le 1+\epsilon .$$

(4.2)

Integrating both sides of equation (4.2) on $(t,T)$, we obtain

$$(1-\epsilon )(T-t)\le -\frac{1}{(m(t)+\frac{1}{2}\lambda )}\le (1+\epsilon )(T-t),t\in [t_0,T),$$

(4.3)

that is,

$$\frac{1}{(1+\epsilon )}-\le (m(t)+\frac{1}{2}\lambda )(T-t)\le \frac{1}{(1-\epsilon )},t\in [t_0,T).$$

(4.4)

By the arbitrariness of $\epsilon \in (0,\frac{1}{2})$, we have

$$\underset{t\to T}{lim}(T-t)(m(t)+\lambda )=-1.$$

(4.5)

This completes the proof of the theorem. □

5 Global existence

In this section, we will present some global existence results. Let us now prove the following lemma.

Lemma 5.1 Let $u_0\in H^s$, $s>3/2$ be given and assume that $T>0$ is the maximal existence time of the corresponding solution $u(t,x)$ to the Cauchy problem (1.1). Let $q\in C^1([0,T)\times \mathbb{R};\mathbb{R})$ be the unique solution of equation (2.1). Then we have

$$y(t,q(t,x))q_x^3=y_0(x)e^{-\lambda t},$$

where $y=\mu (u)-u_{xx}$.

Proof By the first one in equation (1.2) and equation (2.1) we have

$$\begin{array}{rl}\frac{d}{dt}y(t,q(t,x))q_x^3& =(y_t+y_x{q}_t)q_x^3+3y{q}_x{q}_{xt}\\ =(y_t+y_{x}u)q_x^3+3y{q}_x{q}_{xt}\\ =(y_t+u{y}_x+3y{u}_x{y}_{x}u)q_x^3\\ =-\lambda y{q}_x^3.\end{array}$$

Therefore

$$y(t,q(t,x))q_x^3=y_0(x)e^{-\lambda t}.$$

 □

Lemma 5.1 and equation (2.2) imply that y and $y_0$ have the same sign.

Theorem 5.2 Let $u_0\in H^s$, $s>3/2$. If $y_0={\mu }_0-u_{0,xx}\in H^1$ does not change sign, then the corresponding solution $u(t,x)$ to equation (1.1) with the initial data $u_0$ exists globally in time.

Proof By equation (2.1), we know that $q(t,\cdot )$ is diffeomorphism of the line and the periodicity of u with respect to spatial variable x, given $t\in [0,T)$, there exists a $\xi (t)\in \mathbb{S}$ such that $u_x(t,\xi (t))=0$.

We first consider the case that $y_0\ge 0$ on $\mathbb{S}$, in which case Lemma 5.1 ensures that $y\ge 0$. For $x\in [\xi (t),\xi (t)+1]$, we have

$$\begin{array}{rl}-u_x(t,x)& =-{\int }_{\xi (t)}^x{u}_{xx}(t,x)dx={\int }_{\xi (t)}^x(y-\mu (u))dx\\ ={\int }_{\xi (t)}^{x}ydx-\mu (u)(x-\xi (t))\le {\int }_{\mathbb{S}}ydx-\mu (u)(x-\xi (t))\\ =\mu (u)(1-x+\xi (t))\le |{\mu }_0|.\end{array}$$

It follows that $u_x(t,x)\ge -|{\mu }_0|$.

On the other hand, if $y_0\le 0$ on $\mathbb{S}$, then Lemma 5.1 ensures that $y\le 0$. Therefore, for $x\in [\xi (t),\xi (t)+1]$, we have

$$\begin{array}{rl}-u_x(t,x)& =-{\int }_{\xi (t)}^x{u}_{xx}(t,x)dx={\int }_{\xi (t)}^x(y-\mu (u))dx\\ ={\int }_{\xi (t)}^{x}ydx-\mu (u)(x-\xi (t))\\ \le -\mu (u)(x-\xi (t))\le |{\mu }_0|.\end{array}$$

It follows that $u_x(t,x)\ge -|{\mu }_0|$. By using Theorem 3.2, we immediately conclude that the solution is global. This completes the proof of the theorem. □

Corollary 5.3 If the initial value $u_0\in H^3$ such that

$${\parallel {\partial }_x^3{u}_0\parallel }_{L^2}\le 2\sqrt{3}|{\mu }_0|,$$

then the corresponding solution u of the initial value $u_0$ exists globally in time.

Proof Since ${\int }_{\mathbb{S}}{\partial }_x^2{u}_{0}dx=0$, by Lemma 2.2, we obtain

$${\parallel {\partial }_x^2{u}_0\parallel }_{L^{\mathrm{\infty }}}\le \frac{1}{2\sqrt{3}}{\parallel {\partial }_x^3{u}_0\parallel }_{L^2}.$$

If ${\mu }_0\ge 0$, we have

$$y_0={\mu }_0-{\partial }_x^2{u}_0\ge {\mu }_0-\frac{1}{2\sqrt{3}}{\parallel {\partial }_x^3{u}_0\parallel }_{L^2}\ge {\mu }_0-|{\mu }_0|=0.$$

If ${\mu }_0<0$, we have

$$y_0={\mu }_0-{\partial }_x^2{u}_0\le {\mu }_0+{\parallel {\partial }_x^2{u}_0\parallel }_{L^{\mathrm{\infty }}}\le {\mu }_0+\frac{1}{2\sqrt{3}}{\parallel {\partial }_x^3{u}_0\parallel }_{L^2}\le {\mu }_0+|{\mu }_0|=0.$$

 □

Thus the theorem is proved by using Theorem 5.2.