The global solution and blow-up phenomena to a modified Novikov equation

Abstract

A modified Novikov equation with symmetric coefficients is investigated. Provided that the initial value $u_0\in H^s(R)$ ($s>\frac{3}{2}$), $(1-{\partial }_x^2)u_0$ does not change sign and the solution u itself belongs to $L^1(R)$, the existence and uniqueness of the global strong solutions to the equation are established in the space $C([0,\mathrm{\infty });H^s(R))\cap C^1([0,\mathrm{\infty });H^{s-1}(R))$. A blow-up result to the development of singularities in finite time for the equation is acquired.

MSC:35G25, 35L05.

1 Introduction

Many scholars have paid attention to the integrable equation

$$u_t-u_{txx}+4u^2{u}_x=3u{u}_x{u}_{xx}+u^2{u}_{xxx},$$

(1)

which was derived by Novikov [1]. Well-posedness of the Novikov equation in the Sobolev spaces on the torus was first done by Tiglay in [2], and was completed on both the line and the circle by Himonas and Holliman in [3]. Its Hölder continuity properties were studied in Himonas and Holmes [4]. The periodic and the non-periodic Cauchy problem for Eq. (1) and continuity results for the data-to-solution map in the Sobolev spaces are discussed in Grayshan [5]. A matrix Lax pair for Eq. (1) is acquired in [6] and is shown to be related to a negative flow in the Sawada-Kotera hierarchy. The scattering theory is applied to find non-smooth explicit soliton solutions with multiple peaks for Eq. (1) in [7]. Sufficient conditions on the initial data to guarantee the formation of singularities in finite time for Eq. (1) are given in Jiang and Ni [8]. This multiple peak property is common with the Camassa-Holm and Degasperis-Procesi equations [9–11]. Mi and Mu [12] established many dynamic results for a modified Novikov equation with peak solution. It is shown in Ni and Zhou [13] that the Novikov equation associated with initial value has locally well-posedness in a Sobolev space $H^s$ with $s>\frac{3}{2}$ by using the abstract Kato theorem. Two results about the persistence properties of the strong solution for Eq. (1) are established in [13]. Using the Littlewood-Paley decomposition and nonhomogeneous Besov spaces, Yan et al. [14] proved the global existence and blow-up phenomena for the weakly dissipative Novikov equation. For other methods to handle the Novikov equation and the related partial differential equations, the reader is referred to [15–22] and the references therein.

Observing the coefficients of the Novikov equation (1), we see that the coefficient of $u^2{u}_x$ is equal to the coefficient of $u{u}_x{u}_{xx}$ plus the coefficient of $u^2{u}_{xxx}$. That is,

$$4=3+1.$$

Indeed, this relationship among the coefficients plays important roles in the study of the essential dynamical properties of the Novikov model [1, 2, 11–13]. This motivates us to study the following equation:

$$u_t-u_{txx}+(a+b)u^2{u}_x=au{u}_x{u}_{xx}+b{u}^2{u}_{xxx},$$

(2)

where $a>0$ and $b>0$ are arbitrary constants. Clearly, letting $a=3$ and $b=1$, Eq. (2) becomes the Novikov equation (1). The essential difference between Eq. (2) and the Novikov equation (1) is that Eq. (2) does not conform with the following conservation law:

$${\int }_R(u^2+u_x^2)dx={\int }_R(u_0^2+u_{0x}^2)dx,$$

which results in the bounds of ${\parallel u(t,\cdot )\parallel }_{L^{\mathrm{\infty }}(R)}$ for Eq. (1).

Making use of $u_0\in H^s(R)$, $s>\frac{3}{2}$, the assumption that $(1-{\partial }_x^2)u_0$ does not change sign, and the assumption that the solution of Eq. (2) satisfies $u\in L^1(R)$, we prove the global existence theorem of Eq. (2) in the Sobolev space,

$$u(t,x)\in C([0,\mathrm{\infty });H^s(R))\cap C^1([0,\mathrm{\infty });H^{s-1}(R)).$$

The objective of this work is to investigate Eq. (2). Since $a>0$ and $b>0$ are arbitrary constants, we cannot obtain the boundedness of the solution u for Eq. (2) although the initial data satisfy the sign condition. To overcome this, assuming that the solution itself satisfies $u\in L^1(R)$ and the initial data satisfy the sign condition, we adopt the methods used in Rodriguez-Blanco [16] to derive that ${\parallel \frac{\partial u(t,x)}{\partial x}\parallel }_{L^{\mathrm{\infty }}(R)}$ possesses bounds for any time $t>0$. This leads us to establish the well-posedness of the global strong solutions to Eq. (2). Parts of the main results in [17, 18] are extended. In addition, we acquire a blow-up result to the development of singularities in finite time, which includes the blow-up result in [12].

The rest of this paper is organized as follows. Section 2 states the main results of this work. Section 3 proves the global existence result. The proof of a blow-up result is given in Section 4.

2 Main results

We let $L^p=L^p(R)$ ($1\le p<+\mathrm{\infty }$) be the space of all measurable functions h such that ${\parallel h\parallel }_{L^p}^p={\int }_R{|h(t,x)|}^{p}dx<\mathrm{\infty }$. We define $L^{\mathrm{\infty }}=L^{\mathrm{\infty }}(R)$ with the standard norm ${\parallel h\parallel }_{L^{\mathrm{\infty }}}={inf}_{m(e)=0}{sup}_{x\in R\mathrm{\setminus }e}|h(t,x)|$. For any real number s, we let $H^s=H^s(R)$ denote the Sobolev space with the norm defined by

$${\parallel h\parallel }_{H^s}={\left({\int }_R{(1+{|\xi |}^2)}^s{|\hat{h}(t,\xi )|}^{2}d\xi \right)}^{\frac{1}{2}}<\mathrm{\infty },$$

where $\hat{h}(t,\xi )={\int }_R{e}^{-ix\xi }h(t,x)dx$. Here we note that the norms ${\parallel \cdot \parallel }_{L^p}^p$, ${\parallel \cdot \parallel }_{L^{\mathrm{\infty }}}$ and ${\parallel \cdot \parallel }_{H^s}$ depend on variable t.

For $T>0$ and nonnegative number s, $C([0,T);H^s(R))$ denotes the Frechet space of all continuous $H^s$-valued functions on $[0,T)$. We set $\mathrm{\Lambda }={(1-{\partial }_x^2)}^{\frac{1}{2}}$.

In order to study the existence of solutions for Eq. (2), we consider its Cauchy problem in the form

$$\{\begin{array}{l}{u}_t-u_{txx}+(a+b)u^2{u}_x=au{u}_x{u}_{xx}+b{u}^2{u}_{xxx},\\ u(0,x)=u_0(x),\end{array}$$

(3)

which is equivalent to

$$\{\begin{array}{l}{u}_t+b{u}^2{u}_x={\mathrm{\Lambda }}^{-2}[(-a{u}^2{u}_x+\frac{a-6b}{2}{(u{u}_x^2)}_x+\frac{2b-a}{2}{u}_x^3],\\ u(0,x)=u_0(x),\end{array}$$

(4)

where $a>0$ and $b>0$ are arbitrary constants. Now we give the main results for problem (3).

Theorem 1 Assume that the solution of problem (3) satisfies $u(t,x)\in L^1(R)$ and let $u_0(x)\in H^s$, $s>\frac{3}{2}$ and $(1-{\partial }_x^2)u_0\ge 0$ for all $x\in R$ (or equivalently $(1-{\partial }_x^2)u_0\le 0$ for all $x\in R$). Then problem (3) has a unique solution satisfying

$$u(t,x)\in C([0,\mathrm{\infty });H^s(R))\cap C^1([0,\mathrm{\infty });H^{s-1}(R)).$$

Theorem 2 Assume that $u_0(x)\in H^s(R)$ with $s>\frac{3}{2}$. If $a=b$, then every solution of problem (3) exists globally in time. If $a>b$, then the solution blows up in finite time if and only if $u{u}_x$ becomes unbounded from below in finite time. If $a<b$, then the solution blows up in finite time if and only if $u{u}_x$ becomes unbounded from above in finite time.

3 Global strong solutions

For proving the global existence for problem (3), we cite the local well-posedness result presented in [18].

Lemma 3.1 ([18])

Let $u_0(x)\in H^s(R)$ with $s>\frac{3}{2}$. Then the Cauchy problem (3) has a unique solution $u(t,x)\in C([0,T);H^s(R))\cap C^1([0,T);H^{s-1}(R))$ where $T>0$ depends on ${\parallel u_0\parallel }_{H^s(R)}$.

Assume $u_0\in H^s(R)$ with $s>\frac{3}{2}$. Then there exists a unique solution $u(t,x)$ to problem (3) and

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

with the maximal existence time $T>0$. First, we study the differential equation

$$\{\begin{array}{l}{p}_t=b{u}^2(t,p),t\in [0,T),\\ p(0,x)=x.\end{array}$$

(5)

Lemma 3.2 Let $u_0\in H^s$, $s>3$ and let $T>0$ be the maximal existence time of the solution to problem (3). Then problem (5) has a unique solution $p\in C^1([0,T)\times R,R)$. Moreover, the map $p(t,\cdot )$ is an increasing diffeomorphism of R with $p_x(t,x)>0$ for $(t,x)\in [0,T)\times R$.

Proof From Lemma 3.1, we have $u\in C^1([0,T);H^{s-1}(R))$ and $H^{s-1}\in C^1(R)$. Thus we conclude that both functions $u(t,x)$ and $u_x(t,x)$ are bounded, Lipschitz in space and $C^1$ in time. Using the existence and uniqueness theorem of ordinary differential equations derives that problem (5) has a unique solution $p\in C^1([0,T)\times R,R)$.

Differentiating Eq. (5) with respect to x yields

$$\{\begin{array}{l}\frac{d}{dt}{p}_x=2bu{u}_x(t,p)p_x,t\in [0,T),b\ne 0,\\ p_x(0,x)=1,\end{array}$$

(6)

which leads to

$$p_x(t,x)=exp\left({\int }_0^{t}2bu{u}_x(\tau ,p(\tau ,x))d\tau \right).$$

(7)

For every $T^{\prime }<T$, using the Sobolev imbedding theorem yields

$$\underset{(\tau ,x)\in [0,T^{\prime })\times R}{sup}|u{u}_x(\tau ,x)|<\mathrm{\infty }.$$

It is inferred that there exists a constant $K_0>0$ such that $p_x(t,x)\ge e^{-K_{0}t}$ for $(t,x)\in [0,T)\times R$. It completes the proof. □

Lemma 3.3 Let $u_0\in H^s$ with $s>3$, and let $T>0$ be the maximal existence time of the problem (3). We have

$$y(t,p(t,x))p_x^2(t,x)=y_0(x)e^{-(a-4b){\int }_0^{t}u{u}_{x}d\tau },$$

(8)

where $(t,x)\in [0,T)\times R$ and $y:=u-u_{xx}$.

Proof Using Eqs. (2) and (6)-(8), we have

$$\begin{array}{r}\frac{d}{dt}[y(t,p(t,x))p_x^2(t,x)]\\ =y_t{p}_x^2+2y{p}_x{p}_{xt}+y_x{p}_t{p}_x^2\\ =y_t{p}_x^2+4byu{u}_x{p}_x^2+b{u}^2{y}_x{p}_x^2\\ =(u_t-u_{txx}+au{u}_x(u-u_{xx})+b{u}^2(u_x-u_{xxx}))p_x^2-au{u}_{x}y{p}_x^2+4bu{u}_{x}y{p}_x^2\\ =(u_t-u_{txx}+(a+b)u^2{u}_x-au{u}_x{u}_{xx}-b{u}^2{u}_{xxx})p_x^2-(a-4b)u{u}_{x}y{p}_x^2\\ =-(a-4b)u{u}_{x}y{p}_x^2.\end{array}$$

(9)

Using $p_x(0,x)=1$ and solving the above equation, we complete the proof of the lemma. □

Remark 1 From Lemma 3.3, we conclude that, if $u_0-u_{0xx}=(1-{\partial }_x^2)u_0\ge 0$, then $(1-{\partial }_x^2)u(t,x)\ge 0$. Since the operator ${(1-{\partial }_x^2)}^{-1}$ preserves positivity, we get $u\ge 0$. Similarly, if $(1-{\partial }_x^2)u_0\le 0$, we have $(1-{\partial }_x^2)u\le 0$ and $u\le 0$.

Lemma 3.4 If $u_0\in H^s$, $s>\frac{3}{2}$, such that $(1-{\partial }_x^2)u_0\ge 0$ (or $(1-{\partial }_x^2)u_0\le 0$) and ${\int }_R|u|dx<\mathrm{\infty }$, then there exists a constant $K>0$ such that the solution of problem (3) satisfies ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}\le K$.

Proof We will prove this lemma to assume $u_0\in H^{\mathrm{\infty }}$ which results in $u\in H^{\mathrm{\infty }}$ from Lemma 3.1. For $(1-{\partial }_x^2)u_0\ge 0$, from Lemma 3.3, we have $(1-{\partial }_x^2)u\ge 0$. Then $u\ge 0$ does not change sign. From the assumption ${\int }_R|u|dx<\mathrm{\infty }$ one derives

$$-u_x+{\int }_{-\mathrm{\infty }}^{x}udx={\int }_{-\mathrm{\infty }}^x(u-u_{xx})dx\le {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(u-u_{xx})dx=c,$$

(10)

where c is a positive constant. Then

$$-u_x\le c-{\int }_{-\mathrm{\infty }}^{x}udx\le c+{\int }_{-\mathrm{\infty }}^{x}udx\le 2c.$$

(11)

On the other hand, we have

$$u_x+{\int }_x^{\mathrm{\infty }}udx={\int }_x^{\mathrm{\infty }}(u-u_{xx})dx\le {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(u-u_{xx})dx=c,$$

(12)

which results in

$$u_x\le c-{\int }_x^{\mathrm{\infty }}udx\le c+{\int }_x^{\mathrm{\infty }}udx\le 2c.$$

(13)

We conclude from Eqs. (11) and (13) that ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}\le K$. To complete the proof, we use a simple density argument [16]. Setting $u_0^{\epsilon }=e^{\epsilon {\partial }_x^2}{u}_0$, we have $u_0^{\epsilon }\in H^{\mathrm{\infty }}$ and ${\parallel u_x^{\epsilon }\parallel }_{L^{\mathrm{\infty }}}\le 2{\int }_R|u|dx<K$. Applying ${\parallel u_x^{\epsilon }-u_x\parallel }_{L^{\mathrm{\infty }}}\le {sup}_{[0,T]}{\parallel u_x^{\epsilon }-u_x\parallel }_{H^s}\to 0$ when $\epsilon \to 0$, we have ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}\le K$. □

Using the first equation of system (3) one derives

$$\frac{d}{dt}{\int }_R(u^2+u_x^2)dx+2(a-3b){\int }_{R}u{u}_x^{3}dx=0,$$

from which we have the conservation law

$${\int }_R(u^2+u_x^2)dx+2(a-3b){\int }_0^t{\int }_{R}u{u}_x^{3}dx={\int }_R(u_0^2+u_{0x}^2)dx.$$

(14)

Lemma 3.5 (Kato and Ponce [23])

If $r\ge 0$, then $H^r\cap L^{\mathrm{\infty }}$ is an algebra. Moreover

$${\parallel uv\parallel }_r\le c({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel v\parallel }_r+{\parallel u\parallel }_r{\parallel v\parallel }_{L^{\mathrm{\infty }}}),$$

where c is a constant depending only on r.

Lemma 3.6 (Kato and Ponce [23])

Let $r>0$. If $u\in H^r\cap W^{1,\mathrm{\infty }}$ and $v\in H^{r-1}\cap L^{\mathrm{\infty }}$, then

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

Lemma 3.7 Let $s>\frac{3}{2}$ and the function $u(t,x)$ is a solution of problem (3) and the initial data $u_0(x)\in H^s(R)$. Then the following results hold:

$${\parallel u\parallel }_{L^{\mathrm{\infty }}}\le {\parallel u\parallel }_{H^1}\le {\parallel u_0\parallel }_{H^1(R)}{e}^{\frac{|a-3b|}{2}{\int }_0^t{\parallel u_x\parallel }_{L^{\mathrm{\infty }}(R)}^{2}d\tau }.$$

(15)

For $q\in (0,s-1]$, there is a constant c only depending on a and b such that

$$\begin{array}{rl}{\int }_R{\left({\mathrm{\Lambda }}^{q+1}u\right)}^{2}dx\le & {\int }_R\left[{\left({\mathrm{\Lambda }}^{q+1}{u}_0\right)}^2\right]dx\\ +c{\int }_0^t{\parallel u\parallel }_{H^{q+1}}^2({\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2)d\tau .\end{array}$$

(16)

Proof Using $|2u{u}_x|\le (u^2+u_x^2)$, the Gronwall inequality and Eq. (14), one derives Eq. (15).

Using ${\partial }_x^2=-{\mathrm{\Lambda }}^2+1$ and the Parseval equality gives rise to

$${\int }_R{\mathrm{\Lambda }}^{q}u{\mathrm{\Lambda }}^q{\partial }_x^3\left(u^3\right)dx=-3{\int }_R\left({\mathrm{\Lambda }}^{q+1}u\right){\mathrm{\Lambda }}^{q+1}\left(u^2{u}_x\right)dx+3{\int }_R\left({\mathrm{\Lambda }}^{q}u\right){\mathrm{\Lambda }}^q\left(u^2{u}_x\right)dx.$$

For $q\in (0,s-1]$, applying $({\mathrm{\Lambda }}^{q}u){\mathrm{\Lambda }}^q$ to both sides of the first equation of system (3) and integrating with respect to x by parts, we have the identity

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}{\int }_R[{\left({\mathrm{\Lambda }}^{q}u\right)}^2+{\left({\mathrm{\Lambda }}^q{u}_x\right)}^2]dx=& -a{\int }_R\left({\mathrm{\Lambda }}^{q}u\right){\mathrm{\Lambda }}^q\left(u^2{u}_x\right)dx\\ -b{\int }_R\left({\mathrm{\Lambda }}^{q+1}u\right){\mathrm{\Lambda }}^{q+1}\left(u^2{u}_x\right)dx-2b{\int }_R{\mathrm{\Lambda }}^{q}u{\mathrm{\Lambda }}^q{u}_x^{3}dx\\ +(a-6b){\int }_R{\mathrm{\Lambda }}^{q}u{\mathrm{\Lambda }}^q(u{u}_x{u}_{xx})dx.\end{array}$$

(17)

We will estimate the terms on the right-hand side of Eq. (17) separately. For the first term, by using the Cauchy-Schwartz inequality and Lemmas 3.5 and 3.6, we have

$$\begin{array}{rcl}\left|{\int }_R\left({\mathrm{\Lambda }}^{q}u\right){\mathrm{\Lambda }}^q\left(u^2{u}_x\right)dx\right|& =& |{\int }_R\left({\mathrm{\Lambda }}^{q}u\right)[{\mathrm{\Lambda }}^q\left(u^2{u}_x\right)-u^2{\mathrm{\Lambda }}^q{u}_x]dx+{\int }_R\left({\mathrm{\Lambda }}^{q}u\right)u^2{\mathrm{\Lambda }}^q{u}_{x}dx|\\ \le & c{\parallel u\parallel }_{H^q}(2{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{H^q}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{H^q})\\ +{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{q}u\parallel }_{L^2}^2\\ \le & c{\parallel u\parallel }_{H^q}^2{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}.\end{array}$$

(18)

Using the above estimate to the second term yields

$$\left|{\int }_R\left({\mathrm{\Lambda }}^{q+1}u\right){\mathrm{\Lambda }}^{q+1}\left(u^2{u}_x\right)dx\right|\le c{\parallel u\parallel }_{H^{q+1}}^2{\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}.$$

(19)

Using the Cauchy-Schwartz inequality and Lemma 3.5, we obtain

$$\begin{array}{rl}\left|{\int }_R\left({\mathrm{\Lambda }}^q{u}_x\right){\mathrm{\Lambda }}^q\left(u{u}_x^2\right)dx\right|& \le {\parallel {\mathrm{\Lambda }}^q{u}_x\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^q\left(u{u}_x^2\right)\parallel }_{L^2}\\ \le c{\parallel u\parallel }_{H^{q+1}}({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x^2\parallel }_{H^q}+{\parallel u\parallel }_{H^q}{\parallel u_x^2\parallel }_{L^{\mathrm{\infty }}})\\ \le c{\parallel u\parallel }_{H^{q+1}}^2({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2).\end{array}$$

(20)

For the last term in Eq. (17), using $u{(u_x^2)}_x={(u{u}_x^2)}_x-u_x{u}_x^2$ results in

$$\begin{array}{r}|{\int }_R\left({\mathrm{\Lambda }}^{q}u\right){\mathrm{\Lambda }}^q(u{u}_x{u}_{xx})dx|\\ \le \frac{1}{2}\left|{\int }_R{\mathrm{\Lambda }}^q{u}_x{\mathrm{\Lambda }}^q\left(u{u}_x^2\right)dx\right|+\frac{1}{2}\left|{\int }_R{\mathrm{\Lambda }}^{q}u{\mathrm{\Lambda }}^q\left[u_x{u}_x^2\right]dx\right|\\ =K_1+K_2.\end{array}$$

(21)

For $K_1$, it follows from Eq. (20) that

$$K_1\le c{\parallel u\parallel }_{H^{q+1}}^2({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2).$$

(22)

For $K_2$, applying Lemma 3.5 derives

$$\begin{array}{rl}{K}_2& \le c{\parallel u\parallel }_{H^q}{\parallel u_x{u}_x^2\parallel }_{H^q}\\ \le c{\parallel u\parallel }_{H^q}({\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x^2\parallel }_{H^q}+{\parallel u_x\parallel }_{H^q}{\parallel u_x^2\parallel }_{L^{\mathrm{\infty }}})\\ \le c{\parallel u\parallel }_{H^{q+1}}^2{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2.\end{array}$$

(23)

It follows from Eqs. (18)-(23) that there exists a constant c such that

$$\frac{1}{2}\frac{d}{dt}{\int }_R[{\left({\mathrm{\Lambda }}^{q}u\right)}^2+{\left({\mathrm{\Lambda }}^q{u}_x\right)}^2]dx\le c{\parallel u\parallel }_{H^{q+1}}^2({\parallel u_x\parallel }_{L^{\mathrm{\infty }}}{\parallel u\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2).$$

(24)

Integrating both sides of the above inequality with respect to t results in inequality (16). □

Proof of Theorem 1 Using Eq. (16) with $q=s-1$, we obtain

$${\parallel u\parallel }_{H^s}^2\le {\parallel u_0\parallel }_{H^s}^2+c{\int }_0^t{\parallel u\parallel }_{H^s}^2({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2)d\tau .$$

(25)

Applying the Gronwall inequality, we get

$${\parallel u\parallel }_{H^s}^2\le {\parallel u_0\parallel }_{H^s}^2{e}^{2c{\int }_0^t({\parallel u\parallel }_{L^{\mathrm{\infty }}}{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}+{\parallel u_x\parallel }_{L^{\mathrm{\infty }}}^2)d\tau }.$$

(26)

Using Eq. (15) and Lemma 3.4, we complete the proof of Theorem 1. □

Remark 2 In fact, using ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}\le {\parallel u\parallel }_{H^s}$ with $s>\frac{3}{2}$, Eqs. (15) and (26), we derive that the solution of Eq. (2) in space $H^s(R)$ blows up in finite time if and only if ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}=+\mathrm{\infty }$.

4 Proof of Theorem 2

Multiplying Eq. (2) by $y=u-u_{xx}$ and integrating by parts, we get

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}{\int }_R{y}^{2}dx& ={\int }_{R}y{y}_{t}dx\\ ={\int }_{R}y(au{u}_x{u}_{xx}+b{u}^2{u}_{xxx}-(a+b)u^2{u}_x)dx\\ ={\int }_{R}y(au{u}_x(u-y)+b{u}^2(u_x-y_x)-(a+b)u^2{u}_x)dx\\ ={\int }_{R}y(-au{u}_{x}y-b{u}^2{y}_x)dx\\ =(b-a){\int }_{R}u{u}_x{y}^{2}dx.\end{array}$$

(27)

When $a=b$, from Eq. (27), we derive ${\parallel u_x\parallel }_{L^{\mathrm{\infty }}}$ is bounded. From Lemma 3.7 and Remark 2, we see that problem (3) has a global solution in the space

$$C([0,\mathrm{\infty });H^s(R))\cap C^1([0,\mathrm{\infty });H^{s-1}(R)).$$

Assume that the solution $u=u(\cdot ,u_0)$ of problem (3) blows up in finite time in the space $H^s(R)$ with $s>\frac{3}{2}$. If $b-a<0$, we assume that $u{u}_x$ is bounded from below on $[0,T)\times R$, i.e., there exists a constant $M>1$ such that

$$(b-a)u{u}_x(t,x)\le M\text{on }[0,T)\times R.$$

From Eq. (27), we get

$${\parallel u\parallel }_{H^2}\le c{\parallel u_0\parallel }_{H^2}{e}^{Mt},$$

(28)

from which we derive that the $H^2$ norm of the solution to problem (3) does not blow up in finite time. From Remark 2, we know that this is impossible. Therefore, we have ${lim}_{t\to T}inf\{{inf}_{x\in R}u{u}_x(t,x)\}=-\mathrm{\infty }$.

Similar to the above, we know that if $b-a>0$, the solution of problem (3) blows up if and only if ${lim}_{t\to T}inf\{{inf}_{x\in R}u{u}_x(t,x)\}=\mathrm{\infty }$.