1 Introduction

A quasilinear shallow-water waves equation can be expressed as

$$\begin{aligned}{} & {} u_t+cu_x-\mu \delta ^2u_{txx}-\mu _0\delta ^2u_{xxx}+3\alpha \varepsilon uu_x+\Lambda _1\varepsilon ^2u^2u_x+\Lambda _2\varepsilon ^3u^3u_x\nonumber \\{} & {} \quad =\theta \mu \alpha \varepsilon \delta ^2(2u_xu_{xx}+uu_{xxx}),\ \ t>0,x\in \mathbb {R}, \end{aligned}$$
(1.1)

where u(t, x) represents the horizontal velocity field at a specific depth \(z_0\). After the re-scaling, it is required that \(0\leqslant z_0\leqslant 1\), \(\theta \ne 3\) and

$$\begin{aligned} z_0^2=\frac{3c^8+21c^6+36c^4+9c^2-\left( c^8+7c^6+13c^4+9c^2+3\right) \theta }{3(3-\theta )\left( c^2+c+1\right) \left( c^2-c+1\right) \left( c^2+1\right) ^2}. \end{aligned}$$

In fact, \(\theta \) is a parameter to balance between nonlinear steepening and amplification in fluid convection due to stretching, the parameter A denotes a linear underlying shear flow, which satisfy the relationship \(c=\frac{1}{2}\left( \sqrt{A^2+4}+A\right) \). Moreover,

$$\begin{aligned} \alpha =\frac{c^4+c^2+1}{3\left( c^2+1\right) },\ \ \Lambda _1=\frac{c^5\left( c^2-1\right) \left( c^2+2\right) }{2\left( c^2+1\right) ^3}, \\ \Lambda _2=\frac{c^6(c-1)^2(c+1)^2\left( c^4+4c^2+6\right) }{6\left( c^2+1\right) ^5}, \\ \mu = \left\{ \begin{array}{ll} \frac{c^4+6c^2+3}{(3-\theta )\left( c^2-c+1\right) \left( c^2+c+1\right) \left( c^2+1\right) ^2},&{}\quad \theta \ne 3,\\ 0,&{}\quad \theta =3, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \mu _0= \left\{ \begin{array}{ll} \frac{12c^4+3c^2-3+\left( c^6+2c^4+2c^2+1\right) \theta }{3c(3-\theta )\left( c^3+c+1\right) \left( c^2-c+1\right) \left( c^2+1\right) ^2},&{}\quad \theta \ne 3, \\ -\frac{1}{3c\left( c^2+1\right) },&{}\quad \theta =3. \end{array}\right. \end{aligned}$$

This equation is a model with the effect of underlying shear flow from the incompressible rotational two-dimensional shallow water in the moderately nonlinear regime, which was proposed in [35] according to the method of double asymptotic expansion by Wang, Kang and Liu. In fact, it is inferred from [35] that the presence of the underlying shear flow cannot be neglected, which gives rise to the higher-power nonlinear terms \(u^2u_x\) and \(u^3u_x\) in (1.1).

As mentioned in [35], the role of underlying shear and Coriolis force in the modeling equation is the same, however, the modes of action of these two physical quantities on depth are completely different. For the shallow water, the shear can become a dominant feature in the waves dynamics. On the other hand, the Coriolis force is key for the larger-scale motion. It should be pointed out here that the Coriolis force only in equatorial regions does this act like a wave guide and favours one-directional flow [16].

In recent years, some quasilinear shallow-water waves equations have been widely studied. For instance, Johnson [29] established the Camassa–Holm type equation for the movement of water waves over a shear flow. With the scaling of \(\mu<<1\), \(\varepsilon =O\left( \root 4 \of {\mu }\right) \), a high-order nonlinear shallow water equation was obtained by Quirchmayr in [34]. Li and Liu [32] derived a highly nonlinear shallow water model by using the methods of double asymptotic expansion.

In fact, in the vanishing limit for \(A\rightarrow 0\) with \(\theta \ne 3\), we have \(c\rightarrow 1\), \(\alpha \rightarrow \frac{1}{2}\), \(\mu \rightarrow \frac{5}{6(3-\theta )}\), \(\mu _0\rightarrow \frac{\theta +2}{6(3-\theta )}\), \(\Lambda _1\rightarrow 0\), \(\Lambda _2\rightarrow 0\) and \(z_0^2\rightarrow \frac{23-11\theta }{12(3-\theta )}\). By using the scaling transformation \(u(t,x)\rightarrow \alpha \varepsilon u\left( \sqrt{\mu \delta ^2}t,\sqrt{\mu \delta ^2}x\right) \) and the Galilean transformation \(u(t,x)\rightarrow u\left( t,x-\frac{3}{4}t\right) +\frac{3}{4}\), Eq. (1.1) can be rewritten as the following equation [20, 21]

$$\begin{aligned} u_t-u_{txx}+3uu_x=\theta (2u_xu_{xx}+uu_{xxx}), \ \ t>0, x\in \mathbb {R}, \end{aligned}$$
(1.2)

where u(t, x) describes the radical stretch with respect to a prestressed state [20], which was formerly obtained in the context of compressible hyperelastic rod. Due to the different choices of \(\theta \), Eq. (1.1) can be simply reduced to some classical model equation.

In particular, for \(\theta =1\), Eq. (1.2) is reduced to the classical Camassa–Holm (CH) equation

$$\begin{aligned} u_t-u_{txx}+3uu_x=2u_xu_{xx}+uu_{xxx},\ \ t>0, x\in \mathbb {R}, \end{aligned}$$
(1.3)

where u(t, x) represents the horizontal fluid velocity in the x direction at height \(z_0=\frac{1}{\sqrt{2}}\). The CH equation was firstly proposed according to a formal deduction without physical motivation by Fuchssteiner and Fokas in [25]. However, in view of a physical perspective, the CH equation was derived as a unidirectional model for describing the height of water’s free surface above the flat bottom for a shallow water waves in [4] by Camassa and Holm, which can be considered as the cornerstone of an active field of investigation of nonlocal integrable equation. The CH equation also has a geometrical interpretation which can be associated with the geodesic flow on the infinite demensional Hilbert manifold [8, 17]. Furthermore, as \(\theta =1\), Eq. (2.24) can be viewed as geodesic flow. It is well-known that the CH equation has a bi-Hamiltonian structure [25] and is also completely integrable [4, 9], its solitary wave are peaked [3, 4, 9, 10, 15], they are orbitally stable and interact like solitons [1, 19]. Moreover, the CH equation also has the solutions of multi-peakons [4, 5] and the presence of wave breaking, which means that the solution u(t, x) remains bounded while its slope becomes unbounded in finite time [8, 13].

Meanwhile, the Cauchy problem of (1.3) was widely studied in the past decades and lots of important qualitative properties have been obtained. In particular, local well-posedness [11,12,13,14, 22, 31, 33] and wave-breaking phenomena [2, 6, 11, 13, 14, 18, 33, 36, 39], etc. For instance, the local well-posedness for initial data \(u_0\in H^s, s>\frac{3}{2}\) was proved in [13, 33]. One has studied the wave-breaking criteria for (1.3) by establishing the Riccati-type differential inequality in [6, 36, 39]. Furthermore, due to the structure of CH equation, Brandolese [2] obtained local-in-space blow-up criterion, which describes that the formation of singularity is not locally disturbed by data in the area around that point.

In general, energy dissipation is inevitable in the real word. Ghidaglia [26] investigated the long time behavior of solutions for the KdV equation with the weakly dissipative effect as the finite-dimensional dynamical system. The global existence and wave-breaking phenomena of the weakly dissipative Camassa–Holm equation were considered by Wu and Yin in [37]. The local well-posedness and wave-breaking phenomena of the weakly dissipative Camassa–Holm equation with quadratic and cubic nonlinearities was studied in [24]. Guo [28] focused on the existence of global solutions, persistence properties and propagation speed to the weakly dissipative Degasperis–Procesi equation.

To our best knowledge, the qualitative properties of (1.1) with or without dissipative effect have not been studied yet. In this paper, we mainly consider the following Cauchy problem of a weakly dissipative quasilinear shallow-water waves equation

$$\begin{aligned} \left\{ \begin{array}{ll} u_t+cu_x-\mu \delta ^2u_{txx}-\mu _0\delta ^2u_{xxx}+3\alpha \varepsilon uu_x+\Lambda _1\varepsilon ^2u^2u_x+\Lambda _2\varepsilon ^3u^3u_x\\ \quad +\frac{1}{\sqrt{\mu \delta ^2}}\lambda \left( u-\mu \delta ^2u_{xx}\right) =\theta \mu \alpha \varepsilon \delta ^2(2u_xu_{xx}+uu_{xxx}), \ \ {} &{}\quad t>0, x\in \mathbb {R},\\ u(0,x)=u_0(x), \ \ {} &{}\quad x\in \mathbb {R}, \end{array}\right. \nonumber \\ \end{aligned}$$
(1.4)

where \(\lambda \) is the nonnegative dissipative parameter. First, the local well-posedness of (1.4) is obtained by using Kato’s semigroup theory, and we then get the precise blow-up criterion to (1.4) by using the transport equation theory and Moser-type estimates. Moreover, we study some sufficient conditions of wave-breaking to (1.4) based on the different real-valued intervals in which the dispersive parameter \(\theta \) being located. Specifically, the first wave-breaking criterion with \(\theta \in \mathbb R\setminus \{0\}\) is studied by utilizing the piecewise convolution estimate in Lemma 3.1 and energy method. We then apply Lemma 2.2 and Gagliardo–Nirenberg inequality to give a class of wave-breaking criteria of (1.4) with \(\theta \in \mathbb R_+=(0,+\infty )\), which is inspired by Theorem 2.1 in [38]. Finally, by tracking the dynamics of two linear combinations of u and \(u_x\) along the characteristics, we obtain the wave-breaking criterion with \(\theta \in [1,3)\). It is worth noting that we need to make more subtle convolution estimates in order to overcome the difficulties caused by the complicated nonlocal nonlinear structure and different dispersive parameter ranges. Indeed, the results we obtained include the corresponding results of Eq. (1.1) with and without dispersive effect.

The rest of this paper is organized as follows. In Sect. 2, the local well-posedness result and the precise blow-up criterion are presented. In Sect. 3, the wave-breaking criteria are established in detail.

We denote by \(C, C_i(i=0,1,2...)\) the estimates that hold up to some universal constant which may change from line to line but whose meaning is clear throughout the context.

2 Local well-posedness and precise blow-up criterion

In this section, we give the local well-posedness and precise blow-up criterion of the solution for (1.4). To this end, we first introduce the scaling transforation \(u(t,x)\rightarrow \alpha \varepsilon u\left( \sqrt{\mu \delta ^2}t,\sqrt{\mu \delta ^2}x\right) \), then the Cauchy problem (1.4) turns into

$$\begin{aligned} \left\{ \begin{array}{ll} u_t-u_{txx}+cu_x+3uu_x-\frac{\mu _0}{\mu }u_{xxx}+\frac{\Lambda _1}{\alpha ^2}u^2u_x+\frac{\Lambda _2}{\alpha ^3}u^3u_x+\lambda (u-u_{xx})\\ \quad =\theta (2u_xu_{xx}+uu_{xxx}), &{}\quad t>0, x\in \mathbb R,\\ u(0,x)=u_0(x),\ \ {} &{}\quad x\in \mathbb R, \end{array}\right. \nonumber \\ \end{aligned}$$
(2.1)

equivalents to the following non-local form

$$\begin{aligned} \left\{ \begin{array}{ll} u_t+\left( \frac{\mu _0}{\mu }+\theta u\right) u_x+\lambda u\\ \quad +\partial _xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) =0, &{}\quad t>0, x\in \mathbb R,\\ u(0,x)=u_0(x),\ \ {} &{}\quad x\in \mathbb R, \end{array}\right. \nonumber \\ \end{aligned}$$
(2.2)

where the Green’s function \(p(x)=\frac{1}{2}e^{-|x|}\) and \(*\) represents the spatial convolution. For any \(f\in L^2(\mathbb R)\), \(\left( 1-\partial _x^2\right) ^{-1}f=p*f\) holds.

Theorem 2.1

Given \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\), then there exists a maximal time of existence \(T>0\) and a unique solution u to the Cauchy problem (2.2) such that

$$\begin{aligned} u=u(.,u_0)\in C([0,T);H^s(\mathbb {R}))\cap C^1([0,T);H^{s-1}(\mathbb {R})). \end{aligned}$$

Moreover, the solution u depends continuously on the initial data.

Proof

We can verify the all assumption of Kato’s semigroups theory [30]. As the proof process is standard, we omit it here. \(\square \)

Lemma 2.2

Assume that u is the solution of (2.2) with the initial data \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\). Then for any \(t\in [0,T)\), we have

$$\begin{aligned} ||u||^2_{H^1}=e^{-2\lambda t}||u_0||^2_{H^1}. \end{aligned}$$

Moreover, define

$$\begin{aligned} E_0=\frac{1}{2}\int _{\mathbb R}\left( u_0^2+u_{0,x}^2\right) dx, \end{aligned}$$

then we have \(||u||^2_{H^1}\leqslant ||u_0||^2_{H^1}=2E_0\).

Proof

Multiplying both sides of the first equation of (2.1) by u, we have

$$\begin{aligned}{} & {} uu_t-uu_{txx}+cuu_x+3u^2u_x-\frac{\mu _0}{\mu }uu_{xxx}+\frac{\Lambda _1}{\alpha ^2}u^3u_x+\frac{\Lambda _2}{\alpha ^3}u^4u_x+\lambda u(u-u_{xx})\nonumber \\{} & {} \quad =\theta \left( 2uu_xu_{xx}+u^2u_{xxx}\right) . \end{aligned}$$
(2.3)

Integrating (2.3) with respect to x over \(\mathbb {R}\), we get

$$\begin{aligned}{} & {} \int _\mathbb {R}uu_tdx+\int _\mathbb {R}cuu_xdx-\int _\mathbb {R}uu_{txx}dx-\int _\mathbb {R}\frac{\mu _0}{\mu }uu_{xxx}dx\nonumber \\{} & {} \quad +\int _\mathbb {R}3u^2u_xdx+\int _\mathbb {R}\frac{\Lambda _1}{\alpha ^2}u^3u_xdx\nonumber \\{} & {} \quad +\int _\mathbb {R}\frac{\Lambda _2}{\alpha ^3}u^4u_xdx+\int _\mathbb {R}\lambda u(u-u_{xx})dx-\int _\mathbb {R}\theta \left( 2uu_xu_{xx}+u^2u_{xxx}\right) dx=0,\nonumber \\ \end{aligned}$$
(2.4)

where we have used the relations

$$\begin{aligned} -\int _\mathbb {R}uu_{txx}dx= & {} \int _\mathbb {R}u_xu_{tx}dx,\ \ \int _\mathbb {R} 2uu_xu_{xx}dx=-\int _\mathbb {R}u^2u_{xxx}dx. \end{aligned}$$
(2.5)
$$\begin{aligned} \int _\mathbb {R}uu_{xxx}dx= & {} \int _\mathbb {R}udu_{xx}=-\int _\mathbb {R}u_xu_{xx}dx=0. \end{aligned}$$
(2.6)

Substituting (2.5) and (2.6) into (2.4), we have

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int _\mathbb {R}\left( u^2+u_x^2\right) dx+\lambda \int _\mathbb {R}\left( u^2+u_x^2\right) dx=0. \end{aligned}$$
(2.7)

Integrating (2.7) with respect to t, we obtain

$$\begin{aligned} \int _\mathbb {R} (u^2+u_x^2)dx=e^{-2\lambda t}\int _\mathbb {R}(u_0^2+u_{0,x}^2)dx, \end{aligned}$$

i.e.,\(||u||^2_{H^1}=e^{-2\lambda t}||u_0||^2_{H^1}\leqslant 2E_0\). This completes the proof. \(\square \)

Proposition 2.3

([27]). Let \(s\in (0,1)\), suppose that \(f_0\in H^s(\mathbb {R})\), \(g\in L^1([0,T];H^s(\mathbb {R}))\) and \(v,v_x\in L^1([0,T];L^\infty )\), and that \(f\in L^\infty ([0,T];H^s(\mathbb {R}))\cap C([0,T];S')\) solves the following one-dimensional linear transport equation

$$\begin{aligned} \left\{ \begin{array}{ll} f_t+vf_x=g,\\ f(0,x)=f_0(x). \end{array}\right. \end{aligned}$$

Then \(f\in C([0,t];H^s(\mathbb {R}))\). More precisely, there exists a constant C depending only on s such that the following estimate holds:

$$\begin{aligned} ||f(t)||_{H^s(\mathbb {R})}\leqslant ||f_0||_{H^s(\mathbb {R})}+C\left( \int _0^t||g(\tau )||_{H^s(\mathbb {R})}d\tau +\int _0^t||f(\tau )||_{H^s(\mathbb {R})}V'(\tau )d\tau \right) .\nonumber \\ \end{aligned}$$
(2.8)

Hence

$$\begin{aligned} ||f(t)||_{H^s(\mathbb {R})}\leqslant e^{C V(t)}\left( ||f_0||_{H^s(\mathbb {R})}+C \int _0^t||g(\tau )||_{H^s(\mathbb {R})}d\tau \right) , \end{aligned}$$
(2.9)

where \(V(t)=\int _0^t(||v(\tau )||_{L^\infty }+||v_x(\tau )||_{L^\infty })d\tau \).

Lemma 2.4

([7]). Let \(\Lambda \) be a operator defined as \(\Lambda =\left( 1-\partial _x^2\right) ^\frac{1}{2}\). If f and g are smooth enough, then

$$\begin{aligned} ||[\Lambda ^s,f]g||_{L^2}\leqslant C(||f||_{H^s(\mathbb {R})}||g||_{L^\infty }+||\partial _x f||_{L^\infty }||g||_{H^{s-1}(\mathbb {R})}), \end{aligned}$$

for all \(s>\frac{3}{2}\) and \(C>0\).

Proposition 2.5

([7]). The following Moser-type estimates hold:

  1. (1)

    For \(s\geqslant 0\),

    $$\begin{aligned} ||fg||_{H^s(\mathbb {R})}\leqslant C(||f||_{H^s(\mathbb {R})}||g||_{L^\infty (\mathbb {R})}+||f||_{L^\infty (\mathbb {R})}||g||_{H^s(\mathbb {R})}). \end{aligned}$$
  2. (2)

    For \(s>0\),

    $$\begin{aligned} ||f\partial _x g||_{H^s(\mathbb {R})}\leqslant C(||f||_{H^{s+1}(\mathbb {R})}||g||_{L^\infty (\mathbb {R})}+||f||_{L^\infty (\mathbb {R})}||\partial _xg||_{H^s(\mathbb {R})}). \end{aligned}$$
  3. (3)

    For \(s_1\leqslant \frac{1}{2}\), \(s_2>\frac{1}{2}\) and \(s_1+s_2>0\),

    $$\begin{aligned} ||fg||_{H^{s_1}(\mathbb {R})}\leqslant C||f||_{H^{s_1}(\mathbb {R})}||g||_{H^{s_2}(\mathbb {R})}, \end{aligned}$$

    where the constant C is independent of f and g.

Theorem 2.6

Let u be the solution of (2.2) with initial data, \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\). If T is the maximal time of existence, then \(T<\infty \) such that

$$\begin{aligned} \int _0^T||u_x(\tau )||_{L^\infty }d\tau =\infty . \end{aligned}$$
(2.10)

Proof

We first apply the transformation \(u(t,x)\rightarrow u\left( t,x-\frac{\mu _0}{\mu }t\right) \) to (2.2), we get

$$\begin{aligned} u_t+\theta uu_x+\lambda u+\partial _x p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) =0.\nonumber \\ \end{aligned}$$
(2.11)

Acting both sides of (2.11) on \(\Lambda ^s u\Lambda ^s\), we have

$$\begin{aligned}{} & {} \Lambda ^su\Lambda ^su_t+\theta \Lambda ^su\Lambda ^suu_x+\lambda \Lambda ^su\Lambda ^su+\Lambda ^su\nonumber \\{} & {} \quad \times \Lambda ^s\partial _xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) =0.\nonumber \\ \end{aligned}$$
(2.12)

Integrating (2.12) with respect to x over \(\mathbb {R}\), we get

$$\begin{aligned}{} & {} \int _\mathbb {R}\Lambda ^su\Lambda ^su_tdx+\int _\mathbb {R}\theta \Lambda ^su\Lambda ^suu_xdx+\int _\mathbb {R}\lambda \Lambda ^su\Lambda ^sudx\nonumber \\{} & {} \quad +\int _\mathbb {R}\Lambda ^su\Lambda ^s\partial _xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+ \frac{\Lambda _2}{4\alpha ^4}u^4\right) dx =0.\nonumber \\ \end{aligned}$$
(2.13)

Therefore, combining (2.13) and a direct calculation, we have

$$\begin{aligned}{} & {} \frac{1}{2}\frac{d}{dt}\int _\mathbb {R}(\Lambda ^su)^2dx\nonumber \\{} & {} \quad =-\theta \int _\mathbb {R} (\Lambda ^su)\Lambda ^s uu_xdx-\lambda \int _\mathbb {R}(\Lambda ^s u)^2dx-\int _\mathbb {R}(\Lambda ^s u)\nonumber \\{} & {} \qquad \times \Lambda ^s\partial _xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) dx.\nonumber \\ \end{aligned}$$
(2.14)

A direct calculation yields

$$\begin{aligned} \int _\mathbb {R}(\Lambda ^su)(\Lambda ^s uu_x)dx=\int _\mathbb {R}(\Lambda ^su)([\Lambda ^s,u]u_x)dx+\int _\mathbb {R}(\Lambda ^su)u\Lambda ^su_xdx. \end{aligned}$$
(2.15)

Substituting (2.15) into (2.14), we get

$$\begin{aligned}{} & {} \frac{1}{2}\frac{d}{dt}\int _\mathbb {R}(\Lambda ^su)^2dx\nonumber \\{} & {} \quad =\frac{\theta }{2}\int _\mathbb {R}u\partial _x(\Lambda ^su)^2dx-\theta \int _\mathbb {R}([\Lambda ^s,u]u_x)(\Lambda ^su)dx-\lambda \int _\mathbb {R}(\Lambda ^s u)^2dx\nonumber \\{} & {} \qquad +\int _\mathbb {R}\Lambda ^s\left( -\partial _xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) \right) \Lambda ^s udx \nonumber \\{} & {} \quad \leqslant \frac{\theta }{2}||u_x||_{L^\infty }||\Lambda ^su||^2_{L^2}+\theta ||[\Lambda ^s,u]u_x||_{L^2}||\Lambda ^su||_{L^2}+\lambda ||\Lambda ^su||^2_{L^2}+||\Lambda ^su||_{L^2}\nonumber \\{} & {} \qquad \times \left| \left| \Lambda ^s \left( -\partial _x p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) \right) \right| \right| _{L^2}.\nonumber \\ \end{aligned}$$
(2.16)

By using Lemma 2.4, for \(s>0\), we get

$$\begin{aligned} ||[\Lambda ^s,u]\partial _xu||_{L^2}\leqslant & {} C(||u||_{H^s}||\partial _xu||_{L^\infty }+||\partial _xu||_{H^{s-1}}||u_x||_{L^\infty })\nonumber \\\leqslant & {} C||u||_{H^s}||\partial _xu||_{L^\infty }. \end{aligned}$$
(2.17)

Then from (2.16) and (2.17), we have

$$\begin{aligned}{} & {} \frac{1}{2}\frac{d}{dt}||u||^2_{H^s}\leqslant C||u||_{H^s}\Bigg (||u||_{H^s}||u_x||_{L^\infty }+\lambda ||u||_{H^s}\nonumber \\{} & {} \quad +\left| \left| p_x*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) \right| \right| _{H^s}\Bigg ).\nonumber \\ \end{aligned}$$
(2.18)

Taking integration of (2.18) from 0 to t gives that

$$\begin{aligned}{} & {} ||u||_{H^s}\leqslant ||u_0||_{H^s}+C\int _0^t||u(\tau )||_{H^s}\left( ||\partial _xu(\tau )||_{L^\infty }+\lambda \right) d\tau +C\nonumber \\{} & {} \quad \times \int _0^t\left| \left| p_x*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) \right| \right| _{H^s}d\tau .\nonumber \\ \end{aligned}$$
(2.19)

Applying Proposition 2.5, a direct computation reveals that

$$\begin{aligned}{} & {} \left| \left| p_x*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right) \right| \right| _{H^s}\nonumber \\{} & {} \quad \leqslant C\left| \left| \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^4}u^4\right| \right| _{H^{s-1}}\nonumber \\{} & {} \quad \leqslant C(||u||_{H^{s-1}}+||u||_{H^{s-1}}||u||_{L^\infty }+||u_x||_{L^\infty }||u_x||_{H^{s-1}} +||u||^2_{L^\infty }||u||_{H^{s-1}}\nonumber \\{} & {} \qquad +||u||_{H^{s-1}}||u||^3_{L^\infty }). \end{aligned}$$
(2.20)

It follows from (2.19)–(2.20) that

$$\begin{aligned} ||u||_{H^s}{} & {} \leqslant ||u_0||_{H^s}+C\int _0^t||u(\tau )||_{H^s}(1+\lambda +||u||_{L^\infty }+||u_x||_{L^\infty }+||u||^2_{L^\infty }\nonumber \\{} & {} \qquad +||u||^3_{L^\infty })d\tau . \end{aligned}$$
(2.21)

Then applying the Gronwall’s inequality, Sobolev embedding \(H^s(\mathbb R)\hookrightarrow L^\infty (\mathbb R), s>\frac{1}{2}\), (2.21) and Lemma 2.2, we have

$$\begin{aligned} ||u||_{H^s}\leqslant & {} ||u_0||_{H^s}\exp \left\{ {C\int _0^t(1+\lambda +||u||_{L^\infty }+||u_x||_{L^\infty }+||u||^2_{L^\infty }+||u||^3_{L^\infty })d\tau }\right\} \nonumber \\\leqslant & {} ||u_0||_{H^s}\exp \left\{ C\int _0^t(||u_x||_{L^\infty }+\lambda )d\tau \right\} . \end{aligned}$$
(2.22)

Suppose now the maximal existence time \(T<\infty \) satisfies \(\int _0^T||u_x||_{L^\infty }d\tau <\infty \), it then follows from (2.22) that

$$\begin{aligned} \lim _{t\rightarrow T}\sup _{x\in \mathbb R}||u(t)||_{H^s}<\infty , \end{aligned}$$
(2.23)

which contradicts the assumption on the maximal existence time \(T<\infty \). \(\square \)

Now we consider the following associated Lagrangian scale form of (2.2)

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{\partial q}{\partial t}=\theta u(t,q), &{}(t,x)\in (0,T)\times \mathbb R,\\ q(0,x)=x, &{}x\in \mathbb {R}, \end{array}\right. \end{aligned}$$
(2.24)

where \(u\in C^1\left( [0,T),H^{s-1}\right) \) is the solution of (2.2) with initial data \(u_0\in H^s\), \(s>\frac{3}{2}\), and \(T>0\) is the maximal time of existence. A direct calculation shows that

$$\begin{aligned} q_{tx}(t,x)=\theta u_x(t,q(t,x))q_x(t,x). \end{aligned}$$

Then, for \((t,x)\in [0,T)\times \mathbb R\), we have

$$\begin{aligned} q_x(t,x)=\exp \left\{ {\int _0^t\theta u_x(\tau ,q(\tau ,x))d\tau }\right\} >0, \end{aligned}$$

which implies that \(q(t,\cdot ): \mathbb R\rightarrow \mathbb R\) is a diffeomorphism of the line for each \(t\in [0,T)\).

Lemma 2.7

([13]). Let \(T>0\) and \(v\in C^1([0,T); H^s(\mathbb R)), s>\frac{3}{2}\). Then for every \(t\in [0,T)\), there exists at least one point \(\xi (t)\in \mathbb R\) with

$$\begin{aligned} m(t)=\inf _{x\in \mathbb R}(v_x(t,x))=v_x(t,\xi (t)), \end{aligned}$$

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

$$\begin{aligned} \frac{dm(t)}{dt}=v_{tx}(t,\xi (t)),\ \ a.e.\ on\ (0,T). \end{aligned}$$

In order to obtain the precise blow-up criterion, we first prove that \(\theta u_x\) is uniformly bounded from above in the following lemma. The method of proof is similar with outlined in [27].

Lemma 2.8

Let \(\theta \in \mathbb R\setminus \{0\}\) and u be the solution of (2.2) with initial data \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\) and T the maximal time of existence

(1) If \(\theta >0\), then

$$\begin{aligned} \sup _{x\in \mathbb {R}}u_x(t,x)\leqslant ||u_{0,x}||_{L^\infty }+\frac{C_1}{\sqrt{\theta }}+\frac{\lambda }{\theta }. \end{aligned}$$
(2.25)

(2) If \(\theta <0\), then

$$\begin{aligned} \inf _{x\in \mathbb {R}}u_x(t,x)\geqslant -||u_{0,x}||_{L^\infty }-\frac{C_2}{\sqrt{-\theta }}+\frac{\lambda }{\theta }, \end{aligned}$$
(2.26)

where

$$\begin{aligned} C^2_0= & {} \frac{2|\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\Lambda _2}{2\alpha ^3}E_0^2,\\ C^2_1= & {} \frac{1}{2}\left| c-\frac{\mu _0}{\mu }\right| +\left( \left| c-\frac{\mu _0}{\mu }\right| +\max \{|3-\theta |,|\theta |\}\right) E_0+2C_0^2,\\ C^2_2= & {} \frac{1}{2}\left| c-\frac{\mu _0}{\mu }\right| +\left( \left| c-\frac{\mu _0}{\mu }\right| +3-\theta \right) E_0+2C_0^2. \end{aligned}$$

Proof

Theorem 2.1 and a density argument imply that it suffices to prove the desired estimates for \(s\geqslant 3\), thus, we take \(s=3\) in the proof. Also we may assume \(u_0\ne 0\).

Differentiating (2.11) with respect to x and using the identity \(-\partial _x^2 p*f=f-p*f\), we obtain

$$\begin{aligned}{} & {} u_{tx}+\frac{\theta }{2}u_x^2+\theta uu_{xx}+\lambda u_x\nonumber \\{} & {} \quad =\frac{3-\theta }{2}u^2+\left( c-\frac{\mu _0}{\mu }\right) u+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\nonumber \\{} & {} \qquad -p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\right) . \end{aligned}$$
(2.27)

(1) For \(\theta >0\), it is a fact that

$$\begin{aligned} \sup _{x\in \mathbb {R}}[v_x(t,x)]=-\inf _{x\in \mathbb {R}}[-v_x(t,x)]. \end{aligned}$$
(2.28)

Consider \(m_1(t)\) and \(\alpha (t)\) as follows:

$$\begin{aligned} m_1(t)=u_x(t,\alpha (t))=\sup _{x\in \mathbb R}(u_x(t,x)),\ \ t\in [0,T). \end{aligned}$$

Hence,

$$\begin{aligned} u_{xx}(t,\alpha (t))=0,\ \ a.e.\ t\in [0,T). \end{aligned}$$

Taking the trajectory q(t, x) defined in (2.24), then we know that \(q(t,x): \mathbb R\rightarrow \mathbb R\) is a diffeomorphism for every \(t\in [0,T)\). Therefore, there exists \(x_1(t)\in \mathbb R\) such that

$$\begin{aligned} q(t,x_1(t))=\alpha (t),\ \ t\in [0,T). \end{aligned}$$

Then, along the trajectory \(q(t, x_1)\), we can deduce from (2.27) that

$$\begin{aligned} m'_1(t)+\lambda m_1(t)=-\frac{\theta }{2}m^2_1(t)+f(t,q(t,x)), \end{aligned}$$
(2.29)

for \(t\in [0,T)\), where f(t, q(t, x)) is given by

$$\begin{aligned} f= & {} \frac{3-\theta }{2}u^2+\left( c-\frac{\mu _0}{\mu }\right) u+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\nonumber \\{} & {} -p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\right) . \end{aligned}$$
(2.30)

In order to get the wave-breaking result later, we first derive the upper and lower bounds for f. Using that \(\partial _x^2p*u=p_x*u_x\), we have

$$\begin{aligned} f{} & {} =\frac{3-\theta }{2}u^2-p*\left( \frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2\right) -\left( c-\frac{\mu _0}{\mu }\right) p_x*u_x\nonumber \\{} & {} \quad +\,\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4-p*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\right) \nonumber \\{} & {} \leqslant \frac{|3-\theta |}{2}u^2+\left| p*\left( \frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2\right) \right| +\left| c-\frac{\mu _0}{\mu }\right| |p_x*u_x|\nonumber \\{} & {} \quad +\left| \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4-p*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda ^2}{4\alpha ^3}u^4\right) \right| . \end{aligned}$$
(2.31)

By the Hölder’s inequality and Lemma 2.2, we have

$$\begin{aligned} \frac{3-\theta }{2}u^2\leqslant & {} \frac{|3-\theta |}{4}\int _\mathbb {R}\left( u^2+u_x^2\right) dx\nonumber \\\leqslant & {} \frac{|3-\theta |}{2}E_0, \end{aligned}$$
(2.32)
$$\begin{aligned} \left| p*\left( \frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2\right) \right|\leqslant & {} \frac{1}{2}\left| \left| \frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2\right| \right| _{L^1}\nonumber \\\leqslant & {} \max \left\{ \frac{|3-\theta |}{2},\frac{|\theta |}{2}\right\} E_0, \end{aligned}$$
(2.33)
$$\begin{aligned} \left| c-\frac{\mu _0}{\mu }\right| |p_x*u_x|\leqslant & {} \frac{1}{2}\left| c-\frac{\mu _0}{\mu }\right| ||u_x||_{L^2}\nonumber \\\leqslant & {} \frac{1}{4}\left| c-\frac{\mu _0}{\mu }\right| +\frac{1}{4}\left| c-\frac{\mu _0}{\mu }\right| ||u_x||^2_{L^2}\nonumber \\\leqslant & {} \frac{1}{4}\left| c-\frac{\mu _0}{\mu }\right| +\frac{1}{2}\left| c-\frac{\mu _0}{\mu }\right| E_0, \end{aligned}$$
(2.34)

and

$$\begin{aligned}{} & {} \left| \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4-p*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) \right| \nonumber \\{} & {} \quad \leqslant \frac{|\Lambda _1|}{3\alpha ^2}||u||^3_{L^\infty }+\frac{\Lambda _2}{4\alpha ^3}||u||^4_{L^\infty }+\frac{1}{2}||u||_{L^\infty }\left( \frac{|\Lambda _1|}{3\alpha ^2}||u||^2_{L^2}\right) +\frac{1}{2}||u||^2_{L^\infty }\left( \frac{\Lambda _2}{4\alpha ^3}||u||^2_{L^2}\right) \nonumber \\{} & {} \quad \leqslant \frac{2|\Lambda _1|}{3\alpha ^2}E_0^{\frac{3}{2}}+\frac{\Lambda _2}{2\alpha ^3}E_0^2. \end{aligned}$$
(2.35)

Substituting (2.32)–(2.35) into (2.31), it is natural to bound f(t, x) by

$$\begin{aligned} |f|\leqslant & {} \frac{1}{4}\left| c-\frac{\mu _0}{\mu }\right| +\frac{1}{2}\left( \left| c-\frac{\mu _0}{\mu }\right| +\max \{|3-\theta |,|\theta |\}\right) E_0+\frac{2|\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{|\Lambda _2|}{2\alpha ^3}E_0^2\nonumber \\:= & {} \frac{C^2_1}{2}. \end{aligned}$$
(2.36)

For \(\theta <0\), we have a finer estimate

$$\begin{aligned} -f\leqslant \frac{1}{4}\left| c-\frac{\mu _0}{\mu }\right| +\frac{1}{2}\left( \left| c-\frac{\mu _0}{\mu }\right| +3-\theta \right) E_0+\frac{2|\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+ \frac{|\Lambda _2|}{2\alpha ^3}E_0^2:=\frac{C_2^2}{2}. \end{aligned}$$

In fact, (2.29) can turn into

$$\begin{aligned} m'_1(t)= & {} -\frac{\theta }{2}m^2_1(t)-\lambda m_1(t)+f(t,q(t,x))\nonumber \\= & {} -\frac{\theta }{2}\left( m_1(t)+\frac{\lambda }{\theta }\right) ^2+\left( f(t,q(t,x))+\frac{\lambda ^2}{2\theta }\right) . \end{aligned}$$
(2.37)

Denote

$$\begin{aligned} M_1(t)=m_1(t)+\frac{\lambda }{\theta }. \end{aligned}$$
(2.38)

According to (2.37) and (2.38), we have

$$\begin{aligned} M'_1(t)=-\frac{\theta }{2}M_1^2(t)+f(t,q(t,x))+\frac{\lambda ^2}{2\theta }. \end{aligned}$$
(2.39)

For any given \(x\in \mathbb {R}\), let us define

$$\begin{aligned} P_1(t)=M_1(t)-||u_{0,x}||_{L^\infty }-\frac{C_1}{\sqrt{\theta }}-\frac{\lambda }{\theta }. \end{aligned}$$

Notice that \(P_1(t)\) is a \(C^{1}\)-differential function in [0, T) and satisfies

$$\begin{aligned} P_1(0)= & {} M_1(0)-||u_{0,x}||_{L^\infty }-\frac{C_1}{\sqrt{\theta }}-\frac{\lambda }{\theta } \nonumber \\= & {} m_1(0)-||u_{0,x}||_{L^\infty }-\frac{C_1}{\sqrt{\theta }}\nonumber \\\leqslant & {} m_1(0)-||u_{0,x}||_{L^\infty }\leqslant 0. \end{aligned}$$
(2.40)

We now show that

$$\begin{aligned} P_1(t)\leqslant 0,\ \ t\in [0,T). \end{aligned}$$
(2.41)

If not, assume that there exists \(t_0\in [0,T)\) such that \(P_1(t_0)>0\), let \(t_1=\max \{t<t_0;P_1(t)=0\}\), then \(P_1(t_1)=0\) and \(P'_1(t_1)\geqslant 0\), or equivalently

$$\begin{aligned} M_1(t_1)=||u_{0,x}||_{L^\infty }+\frac{C_1}{\sqrt{\theta }}+\frac{\lambda }{\theta },\ \ M'_1(t_1)\geqslant 0. \end{aligned}$$
(2.42)

On the other hand, we have

$$\begin{aligned} M'_1(t_1)= & {} -\frac{\theta }{2}M_1^2(t_1)+f(t_1,q(t_1,x))+\frac{\lambda ^2}{2\theta } \\\leqslant & {} -\frac{\theta }{2}\left( ||u_{0,x}||_{L^\infty }+\frac{C_1}{\sqrt{\theta }}+\frac{\lambda }{\theta }\right) ^2+\frac{C_1^2}{2}+\frac{\lambda ^2}{2\theta } \\= & {} -\frac{\theta }{2}||u_{0,x}||_{L^\infty }^2-\frac{C_1^2}{2}-\frac{\lambda ^2}{2\theta }-C_1\sqrt{\theta }||u_{0,x}||_{L^\infty }-\lambda ||u_{0,x}||_{L^\infty }\\{} & {} -\frac{C_1\lambda }{\sqrt{\theta }}+\frac{C_1^2}{2}+\frac{\lambda ^2}{2\theta }\\= & {} -\frac{\theta }{2}||u_{0,x}||_{L^\infty }^2-C_1\sqrt{\theta }||u_{0,x}||_{L^\infty }-\lambda ||u_{0,x}||_{L^\infty }-\frac{C_1\lambda }{\sqrt{\theta }} <0, \end{aligned}$$

which contradicts with (2.42). It verifies (2.41). Therefore, for the arbitrary chosen of x, (2.25) holds.

(2) For \(\theta <0\), we derive the lower bound for \(u_x\). Now we consider the function \(m_2(t)\)

$$\begin{aligned} m_2(t):=u_x(t,\alpha _1(t))=\inf _{x\in \mathbb R}(u_x(t,x)),\ \ t\in [0,T). \end{aligned}$$

Hence,

$$\begin{aligned} u_{xx}(t,\alpha _1(t))=0,\ \ a.e.\ t\in [0,T). \end{aligned}$$

Similar as before, we take q(t, x) defined in (2.24), and choose \(x_2(t)\in \mathbb R\) such that

$$\begin{aligned} q(t,x_2(t))=\alpha _1(t),\ \ t\in [0,T). \end{aligned}$$

Let

$$\begin{aligned} M_2(t)=m_2(t)+\frac{\lambda }{\theta }. \end{aligned}$$

Therefore,

$$\begin{aligned} M'_2(t)=-\frac{\theta }{2}M_2^2(t)+f(t,q(t,x))+\frac{\lambda ^2}{2\theta }. \end{aligned}$$

Now for any given \(x\in \mathbb {R}\), we can define

$$\begin{aligned} P_2(t)=M_2(t)+||u_{0,x}||_{L^\infty }+\frac{C_2}{\sqrt{-\theta }}-\frac{\lambda }{\theta }. \end{aligned}$$

Then \(P_2(t)\) is also a \(C^{1}\)-differentiable on [0, T) and satisfies

$$\begin{aligned} P_2(0)= & {} M_2(0)+||u_{0,x}||_{L^\infty }+\frac{C_2}{\sqrt{-\theta }}-\frac{\lambda }{\theta } \nonumber \\= & {} m_2(0)+\frac{\lambda }{\theta }+||u_{0,x}||_{L^\infty }+\frac{C_2}{\sqrt{-\theta }}-\frac{\lambda }{\theta } \nonumber \\\geqslant & {} m_2(0)+||u_{0,x}||_{L^\infty }\geqslant 0. \end{aligned}$$
(2.43)

We now claim that \(P_2(t)\geqslant 0\), for \(t\in [0,T)\). If not, assume there is a \(\bar{t_0}\in [0,T)\) such that \(P_2(\bar{t_0})<0\). Define \(t_2=\max (t<\bar{t_0};P_2(t)=0)\), then \(P_2(t_2)=0\) and \(P'_2(t_2)\leqslant 0\), or equivalently

$$\begin{aligned} M_2(t_2)=-||u_{0,x}||_{L^\infty }-\frac{C_2}{\sqrt{-\theta }}+\frac{\lambda }{\theta },\ \ M'_2(t_2)\leqslant 0. \end{aligned}$$
(2.44)

On the other hand, we have

$$\begin{aligned} M'_2(t_2)= & {} -\frac{\theta }{2}M^2_2(t_2)+f(t_2,q(t_2,x))+\frac{\lambda ^2}{2\theta } \\\geqslant & {} -\frac{\theta }{2}\left( -||u_{0,x}||_{L^\infty }-\frac{C_2}{\sqrt{-\theta }}+\frac{\lambda }{\theta }\right) ^2-\frac{C_2^2}{2}+\frac{\lambda ^2}{2\theta }\\= & {} -\frac{\theta }{2}||u_{0,x}||_{L^\infty }^2-\frac{\theta C_2}{\sqrt{-\theta }}||u_{0,x}||_{L^\infty }+\lambda ||u_{0,x}||_{L^\infty }+\frac{C_2\lambda }{\sqrt{-\theta }}>0. \end{aligned}$$

This is a contradiction. It verifies \(P_2(t)\geqslant 0\) for \(t\in [0,T)\). Therefore, it deduced that (2.26) holds. \(\square \)

Theorem 2.9

Let \(\theta \in \mathbb R\setminus \{0\}\). Assume u be the solution of (2.2) with initial data \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\) and T is the maximal time of existence, then the corresponding solution u blows up in finite time if and only if

$$\begin{aligned} \lim _{t\rightarrow T}\inf _{x\in \mathbb {R}}(\theta u_x)(t,x)=-\infty . \end{aligned}$$
(2.45)

Proof

Assume that \(T<\infty \) and (2.45) is not valid. Then there is some positive constant \(M>0\) such that \((\theta u_x)(t,x)\geqslant -M\), \((t,x)\in [0,T)\times \mathbb {R}\). Then it follows from Lemma 2.8 that \(|u_x|\leqslant C\). Therefore, Theorem 2.6 implies that the maximal existence time \(T=\infty \), which contradicts the assumption that \(T<\infty \).

Conversely, if (2.45) holds, according to the Sobolev embedding theorem \(H^s(\mathbb R)\hookrightarrow L^\infty (\mathbb R), s>\frac{1}{2}\), the corresponding solution blows up in finite time, which completes the proof. \(\square \)

3 Wave-breaking phenomena

In this section, we study the wave-breaking phenomena of (1.4). To this end, we first present the following convolution estimate.

Lemma 3.1

Let \(\theta \in (0,3]\), then

$$\begin{aligned} p*\left( \frac{\theta (3-\theta )}{2}\left( u-\frac{\frac{\mu _0}{\mu }-c}{3-\theta }\right) ^2+\frac{\theta ^2}{2}u_x^2\right) \geqslant \left\{ \begin{array}{ll} \frac{\theta ^2}{2}\left( u-\frac{\frac{\mu _0}{\mu }-c}{3-\theta }\right) ^2, &{}\theta \in (0,1), \\ \frac{\theta (3-\theta )}{4}\left( u-\frac{\frac{\mu _0}{\mu }-c}{3-\theta }\right) ^2, &{}\theta \in [1,3). \end{array}\right. \end{aligned}$$

Proof

The detailed proof of this lemma is similar to [23], so we omit it here. \(\square \)

Theorem 3.2

Let \(\theta \in \mathbb R\setminus \{0\}\) and \(\lambda \geqslant 0\). Assume that u is the solution of (2.2) with initial data \(u_0\in H^s(\mathbb {R}), s>\frac{3}{2}\). Suppose that \(T>0\) be the maximal existence time, and there exists a point \(x_0\in \mathbb {R}\) such that

$$\begin{aligned} u_{0,x}(x_0)=\inf _{x\in \mathbb {R}}u_{0,x}(x), \end{aligned}$$
(3.1)

and

$$\begin{aligned}{} & {} \theta u_{0,x} (x_0)\nonumber \\{} & {} \quad \leqslant \left\{ \begin{array}{ll} -\lambda -\left( \lambda ^2+\frac{\theta \Lambda _2}{\alpha ^3}E_0^2+\frac{4\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\theta (3-2\theta )E_0 \right. \\ \quad \left. +\frac{2|k|\theta (3-2\theta )}{3-\theta }E_0^\frac{1}{2}+\frac{\theta k^2(3-2\theta )}{(3-\theta )^2}\right) ^{\frac{1}{2}},&{}\quad \theta \in (0,1),\\ -\lambda - \left( \lambda ^2 +\frac{\theta \Lambda _2}{\alpha ^3} E_0^2+ \frac{4\theta |\Lambda _1|}{3 \alpha ^2} E_0^{\frac{3}{2}}+ \frac{\theta (3- \theta )}{2} E_0+\theta k E_0^{\frac{1}{2}}+\frac{\theta k^2}{2(3-\theta )}\right) ^{\frac{1}{2}},&{}\quad \theta \in [1,3),\\ -\lambda -\left( \lambda ^2+\frac{\theta \Lambda _2}{\alpha ^3}E_0^2+\frac{4\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+|\theta (3-\theta )|E_0+2|k|\theta E_0^\frac{1}{2}\right) ^{\frac{1}{2}},&{}\quad \theta \in (3,\infty ),\\ -\lambda -\left( \lambda ^2+\frac{|\theta | \Lambda _2}{\alpha ^3}E_0^2+\frac{4|\theta | |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+|\theta (3-\theta )|E_0 +2|k\theta |E_0^\frac{1}{2}\right) ^{\frac{1}{2}},&{}\quad \theta \in (-\infty ,0). \end{array}\right. \nonumber \\ \end{aligned}$$
(3.2)

Then the corresponding solution u to (2.2) blows up in finite time.

Proof

Differentiating (2.11) with respect to x, we get

$$\begin{aligned} u_{tx}+ & {} \theta u^2_x+\theta uu_{xx}+\lambda u_x\nonumber \\+ & {} \partial _x^2 p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2} u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4 \right) =0.\nonumber \\ \end{aligned}$$
(3.3)

Then, by using the identity \(\partial _x^2 p*f=p*f-f\), we have

$$\begin{aligned}{} & {} u_{tx}+\theta uu_{xx}\nonumber \\{} & {} \quad =\frac{3-\theta }{2}\left( u-\frac{k}{3-\theta }\right) ^2-\frac{\theta }{2}u_x^2-\lambda u_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4-\frac{k^2}{2(3-\theta )}\nonumber \\{} & {} \qquad -p*\left( \frac{3-\theta }{2}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4-\frac{k^2}{2(3-\theta )}\right) ,\nonumber \\ \end{aligned}$$
(3.4)

where \(k=\frac{\mu _0}{\mu }-c\).

Multiplying (3.4) by \(\theta \), we get

$$\begin{aligned}{} & {} \theta u_{tx}+\theta ^2 uu_{xx}\nonumber \\{} & {} \quad =\frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2-\frac{\theta ^2}{2}u_x^2-\lambda \theta u_x+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4-\frac{k^2 \theta }{2(3-\theta )}\nonumber \\{} & {} \qquad -p*\left( \frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta ^2}{2}u_x^2+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4-\frac{\theta k^2}{2(3-\theta )}\right) .\nonumber \\ \end{aligned}$$
(3.5)

Define m(t) and \(\xi (t)\) as

$$\begin{aligned} \begin{aligned} m(t)=\theta u_x(t,\xi (t)) =\inf _{x\in \mathbb {R}}\theta u_x(t,x),~~~~~~~t\in [0,T).\\ \end{aligned} \end{aligned}$$
(3.6)

So that \(u_{xx}(t,\xi (t))=0\), almost everywhere \(t\in [0,T)\), since \(q_1(t,.)\) defined by (2.24) is a diffeomorphism of the line for any \(t\in [0,T)\), we see that there exists an \(x_3(t)\in \mathbb {R}\) such that

$$\begin{aligned} q_1(t,x_3(t))=\xi (t),\ \ t\in [0,T). \end{aligned}$$

Therefore, according to (3.5) and (3.6), we get

$$\begin{aligned} m'(t)= & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4-\frac{\theta k^2}{2(3-\theta )}\nonumber \\{} & {} -p*\left( \frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta ^2}{2}u_x^2+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4-\frac{k^2\theta }{2(3-\theta )}\right) .\nonumber \\ \end{aligned}$$
(3.7)

In the following, we divide into the four cases to prove the theorem.

(1) For \(\theta \in (0,1)\), then by (3.7) and the first inequality of Lemma 3.1, we get

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\left( \frac{\theta (3-\theta )}{2}-\frac{\theta ^2}{2}\right) \left( u-\frac{k}{3-\theta }\right) ^2\nonumber \\{} & {} +\,\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4+p*\left( \frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\right) (t,\xi (t)). \end{aligned}$$
(3.8)

By the Hölder’s inequality, Cauchy–Schwarz inequality and Lemma 2.2, we have

$$\begin{aligned} u^2\leqslant & {} \frac{1}{2}\int _\mathbb {R}\left( u^2+u_x^2\right) dx\leqslant E_0, \end{aligned}$$
(3.9)
$$\begin{aligned} |u^3|\leqslant & {} E_0^\frac{3}{2},\ \ |u^4|\leqslant E_0^2, \end{aligned}$$
(3.10)
$$\begin{aligned} |p*u^3|\leqslant & {} ||u||_{L^\infty }|p*u^2|\leqslant E_0^\frac{3}{2}, \end{aligned}$$
(3.11)
$$\begin{aligned} |p*u^4|\leqslant & {} ||u||^2_{L^\infty }|p*u^2|\leqslant E_0^2. \end{aligned}$$
(3.12)

Then, by (3.9)–(3.12), we have

$$\begin{aligned}{} & {} \frac{\theta (3-2\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2\leqslant \frac{\theta (3-2\theta )}{2}E_0+\frac{|k|\theta (3-2\theta )}{3-\theta }E_0^\frac{1}{2}+\frac{\theta k^2(3-2\theta )}{2(3-\theta )^2}, \end{aligned}$$
(3.13)
$$\begin{aligned}{} & {} \left| \frac{\Lambda _1 \theta }{3\alpha ^2}u^3\right| \leqslant \frac{\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2},\ \ \left| p*\left( \frac{\Lambda _1 \theta }{3\alpha ^2}u^3\right) \right| \leqslant \frac{\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}, \end{aligned}$$
(3.14)
$$\begin{aligned}{} & {} \left| \frac{\Lambda _2 \theta }{4\alpha ^3}u^4\right| \leqslant \frac{\theta \Lambda _2}{4\alpha ^3}E_0^2,\ \ \left| p*\left( \frac{\Lambda _2 \theta }{4\alpha ^3}u^4\right) \right| \leqslant \frac{\theta \Lambda _2}{4\alpha ^3}E_0^2. \end{aligned}$$
(3.15)

Plugging (3.13)–(3.15) into (3.8), we get

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\frac{\theta (3-2\theta )}{2}E_0+\frac{|k| \theta (3-2\theta )}{3-\theta }E_0^\frac{1}{2}\nonumber \\{} & {} +\,\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2+\frac{\theta k^2(3-2\theta )}{2(3-\theta )^2}. \end{aligned}$$
(3.16)

Set

$$\begin{aligned} K{} & {} =\frac{\theta (3-2\theta )}{2}E_0+\frac{|k|\theta (3-2\theta )}{3-\theta }E_0^\frac{1}{2}\\{} & {} \quad +\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2+\frac{\theta k^2(3-2\theta )}{2(3-\theta )^2}. \end{aligned}$$

Then, we have

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+K\nonumber \\= & {} -\frac{1}{2}\left( m+\lambda \right) ^2+\left( K+\frac{\lambda ^2}{2}\right) \nonumber \\= & {} -\frac{1}{2}\left( (m+\lambda )^2-\left( \sqrt{\lambda ^2+2K}\right) ^2\right) \nonumber \\= & {} -\frac{1}{2}\left( m+\lambda +\sqrt{\lambda ^2+2K}\right) \left( m+\lambda -\sqrt{\lambda ^2+2K}\right) . \end{aligned}$$
(3.17)

On account of the assumption (3.2), we get

$$\begin{aligned} m(0)<-\lambda -\sqrt{\lambda ^2+2K}. \end{aligned}$$
(3.18)

From the continuity of m(t) with respect to t, we can obtain that for any \(t\in [0,T), m'(t)<0\). As a result, \(m(t)<-\lambda -\sqrt{\lambda ^2+2K}\).

We can infer from (3.17) that

$$\begin{aligned} \left( \frac{m(0)+\lambda +\sqrt{\lambda ^2+2K}}{m(0)+\lambda -\sqrt{\lambda ^2+2K}}\right) e^{t\sqrt{\lambda ^2+2K}}-1\leqslant \frac{2\sqrt{\lambda ^2+2K}}{m(t)+\lambda -\sqrt{\lambda ^2+2K}}\leqslant 0. \end{aligned}$$

According to (3.18), we have

$$\begin{aligned} 0<\frac{m(0)+\lambda +\sqrt{\lambda ^2+2K}}{m(0)+\lambda -\sqrt{\lambda ^2+2K}}<1. \end{aligned}$$

Then there exists \(T>0\), which satisfies

$$\begin{aligned} T\leqslant \frac{1}{\sqrt{\lambda ^2+2K}}\ln \left( \frac{m(0)+\lambda -\sqrt{\lambda ^2+2K}}{m(0)+\lambda +\sqrt{\lambda ^2+2K}}\right) , \end{aligned}$$

such that

$$\begin{aligned} \displaystyle \lim _{t\rightarrow T}m(t)=-\infty . \end{aligned}$$

Therefore, according to Theorem 2.9, the solution u does not exist globally in time.

(2) For \(\theta \in [1,3)\), by (3.7), (3.14)–(3.15) and the second inequality of Lemma 3.1, we have

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\frac{\theta (3-\theta )}{4}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\nonumber \\{} & {} +p*\left( \frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\right) (t,\xi (t)) \nonumber \\\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2\nonumber \\{} & {} +\frac{\theta (3-\theta )}{4}E_0+\frac{|k|\theta }{2}E_0^\frac{1}{2}+\frac{\theta k^2}{4(3-\theta )}. \end{aligned}$$
(3.19)

Set

$$\begin{aligned} K=\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2+\frac{\theta (3-\theta )}{4}E_0+\frac{|k| \theta }{2}E_0^\frac{1}{2}+\frac{\theta k^2}{4(3-\theta )}. \end{aligned}$$

Therefore,

$$\begin{aligned} m'(t)\leqslant -\frac{1}{2}m^2(t)-\lambda m(t)+K. \end{aligned}$$

Following the similar argument as case (1), we deduce that the solution u does not exist globally.

(3) For \(\theta \in (3,\infty )\), noting that \(\theta (3-\theta )<0\), by (3.7), (3.14)–(3.15), we have

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\Bigg [\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\nonumber \\{} & {} +p*\left( -\frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2+\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\right) \Bigg ](t,\xi (t)) \nonumber \\\leqslant & {} -\frac{1}{2} m(t)^2-\lambda m(t)+\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2\nonumber \\{} & {} +\frac{|\theta (3-\theta )|}{2}E_0+|k|\theta E_0^\frac{1}{2}. \end{aligned}$$
(3.20)

Set

$$\begin{aligned} K=\frac{2\theta |\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{\theta \Lambda _2}{2\alpha ^3}E_0^2+\frac{|\theta (3-\theta )|}{2}E_0+|k|\theta E_0^\frac{1}{2}. \end{aligned}$$

Following the similar proof as case (1), we deduce that the solution u does not exist globally.

(4) For \(\theta \in (-\infty ,0)\),

$$\begin{aligned} m'(t)\leqslant & {} -\frac{1}{2}m^2(t)-\lambda m(t)+\Bigg [\frac{\theta \Lambda _1}{3\alpha ^2}u^3+\frac{\theta \Lambda _2}{4\alpha ^3}u^4\nonumber \\{} & {} \left. +p*\left( -\frac{\theta (3-\theta )}{2}\left( u-\frac{k}{3-\theta }\right) ^2 -\frac{\theta \Lambda _1}{3\alpha ^2}u^3-\frac{\theta \Lambda _2}{4\alpha ^3}u^4\right) \right] (t,\xi (t)) \nonumber \\\leqslant & {} -\frac{1}{2} m(t)^2-\lambda m(t)+\frac{2|\theta \Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{|\theta |\Lambda _2}{2\alpha ^3}E_0^2\nonumber \\{} & {} +\frac{|\theta (3-\theta )|}{2}E_0+k|\theta | E_0^\frac{1}{2}. \end{aligned}$$
(3.21)

Set

$$\begin{aligned} K=\frac{2|\theta \Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{|\theta |\Lambda _2}{2\alpha ^3}E_0^2+\frac{|\theta (3-\theta )|}{2}E_0+|k\theta | E_0^\frac{1}{2}. \end{aligned}$$

Following the similar proof as case (1), we deduce that the solution u does not exist globally. \(\square \)

In the following, we give a class of wave-breaking result, which is motivated by the proof of Theorem 2.1 in [38].

Theorem 3.3

Let \(\theta \in \mathbb R_+=(0,+\infty )\) and \(\lambda =0\). Given \(u_0\in H^s(\mathbb {R})\), \(s>\frac{3}{2}\) and \(n\in \mathbb {R}_+\), if the slope of \(u_0\) satisfies

$$\begin{aligned} \int _\mathbb {R} u^{2n+1}_{0,x}dx < -\left( \frac{K_2}{K_1}\right) ^\frac{2n+1}{2n+2}, \end{aligned}$$

then there exists the lifespan \(T<\infty \) such that the solution u of (2.2) blows up. In particular, the lifespan T satisfies

$$\begin{aligned} T\leqslant \int _{-h(0)}^\infty \frac{1}{K_1y^{\frac{2n+2}{2n+1}}-K_2}dy, \end{aligned}$$

where

$$\begin{aligned} h(0)=\int _\mathbb {R} u^{2n+1}_{0,x}dx, \end{aligned}$$
$$\begin{aligned} K_1{} & {} =\frac{\theta }{2}(2n-1)-(2n+1)\left[ \left| c-\frac{\mu _0}{\mu }\right| \frac{6n-1}{6n+4}\eta ^{\frac{6n+4}{6n-1}}+\frac{n\varepsilon ^{\frac{n+1}{n}}}{n+1}\right. \nonumber \\{} & {} \quad \left. \times \Bigg (2\left| c-\frac{\mu _0}{\mu }\right| +|3-\theta |+(1+\sqrt{2})\frac{|\Lambda _1|}{3\alpha ^2}+\frac{\Lambda _2}{4\alpha ^3}\Bigg )\right] , \end{aligned}$$
(3.22)

and

$$\begin{aligned} K_2{} & {} =\frac{5\cdot 2^{\frac{2n+3}{5}}|c-\frac{\mu _0}{\mu }|(2n+1)}{6n+4}\left( \frac{c_1}{\eta }\right) ^\frac{6n+4}{5}E_0^\frac{2n+3}{5}+\frac{(2n+1)}{(n+1)\varepsilon ^{n+1}}\Bigg [\left| c-\frac{\mu _0}{\mu }\right| 2^{\frac{n+1}{2}}\nonumber \\{} & {} \quad \times \left( c_2^{n+1}+||p||_{L^{\frac{2(n+1)}{n+3}}}^{n+1}\right) E_0^{\frac{n+1}{2}}+2^n\cdot |3-\theta |\left( 2^{-n}+||p||^{n+1}_{L^{n+1}}\right) E_0^{n+1}\nonumber \\{} & {} \quad +\frac{|\Lambda _1|}{3\alpha ^2}2^{\frac{3n+3}{2}}\left( 2^{-\frac{3n+1}{2}}+\frac{\sqrt{2}}{2}||p||_{L^{n+1}}^{n+1}\right) E_0^{\frac{3n+3}{2}}+\frac{2\Lambda _2}{ \alpha ^3}E_0^{2n+2}\Bigg ], \end{aligned}$$
(3.23)

with the constant \(\varepsilon ,\eta \in (0,1)\) such that \(K_1>0\).

Proof

For \(\lambda =0\), differentiating (2.11) with respect to x, using the identity \(\partial _x^2 p*f=p*f-f\), we have

$$\begin{aligned}{} & {} u_{tx}+\frac{\theta }{2}u^2_x+\theta uu_{xx}-\left( c-\frac{\mu _0}{\mu }\right) u-\frac{3-\theta }{2}u^2-\frac{\Lambda _1}{3\alpha ^2}u^3-\frac{\Lambda _2}{4\alpha ^3}u^4\nonumber \\{} & {} \quad +p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u^2_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) =0. \end{aligned}$$
(3.24)

Multiplying both sides of (3.24) by (2n+1)\(u^{2n}_x\), we have

$$\begin{aligned}{} & {} (2n+1)u^{2n}_x u_{tx}+\frac{\theta }{2}(2n+1)u^{2n+2}_x+\theta (2n+1)u^{2n}_x uu_{xx}\nonumber \\{} & {} \quad -(2n+1)\left( c-\frac{\mu _0}{\mu }\right) uu^{2n}_x-(2n+1)\frac{3-\theta }{2}u^2u^{2n}_x\nonumber \\{} & {} \quad -(2n+1)\frac{\Lambda _1}{3\alpha ^2}u^3u^{2n}_x-(2n+1)\frac{\Lambda _2}{4\alpha ^3}u^4u^{2n}_x+(2n+1)u^{2n}_x\nonumber \\{} & {} \quad \times p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u^2_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) =0. \end{aligned}$$
(3.25)

Integrating (3.25) with respect to x over \(\mathbb {R}\), we get

$$\begin{aligned}{} & {} \frac{d}{dt}\int _\mathbb {R} u^{2n+1}_x dx\nonumber \\{} & {} \quad +\frac{\theta }{2}(2n+1)\int _\mathbb {R} u^{2n+2}_x dx+\theta (2n+1)\int _\mathbb {R} u^{2n}_x uu_{xx}dx-(2n+1)\left( c-\frac{\mu _0}{\mu }\right) \nonumber \\{} & {} \quad \times \int _\mathbb {R} uu^{2n}_x dx-(2n+1)\frac{3-\theta }{2}\int _\mathbb {R} u^2u^{2n}_x dx-(2n+1)\frac{\Lambda _1}{3\alpha ^2}\int _\mathbb {R} u^3u^{2n}_x dx\nonumber \\{} & {} \quad -(2n+1)\frac{\Lambda _2}{4\alpha ^3}\int _\mathbb {R} u^4u^{2n}_x dx+(2n+1)\int _\mathbb {R} u^{2n}_x p*\Bigg (\left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2\nonumber \\{} & {} \quad +\frac{\theta }{2}u^2_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4dx\Bigg )=0. \end{aligned}$$
(3.26)

Notice that

$$\begin{aligned} \theta (2n+1)\int _\mathbb {R} uu^{2n}_x u_{xx}dx=-\theta \int _\mathbb {R} u^{2n+2}_x dx. \end{aligned}$$
(3.27)

By (3.26), (3.27) and \(\theta , \alpha , \Lambda _2>0\), we have

$$\begin{aligned}{} & {} \frac{d}{dt}\int _\mathbb {R} u^{2n+1}_x dx+\frac{\theta }{2}(2n-1)\int _\mathbb {R} u^{2n+2}_x dx\nonumber \\{} & {} =(2n+1)\left( c-\frac{\mu _0}{\mu }\right) \int _\mathbb {R} uu^{2n}_x dx+(2n+1)\frac{3-\theta }{2}\int _\mathbb {R}u^2u^{2n}_x dx\nonumber \\{} & {} \quad +\,(2n+1)\frac{\Lambda _1}{3\alpha ^2}\int _\mathbb {R}u^3u^{2n}_x dx+(2n+1)\frac{\Lambda _2}{4\alpha ^3}\int _\mathbb {R} u^4u^{2n}_x dx-(2n+1)\nonumber \\{} & {} \quad \times \int _\mathbb {R}u^{2n}_xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u^2_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) dx\nonumber \\{} & {} \leqslant (2n+1)\left( c-\frac{\mu _0}{\mu }\right) \int _\mathbb {R} uu^{2n}_x dx+(2n+1)\frac{3-\theta }{2}\int _\mathbb {R}u^2u^{2n}_x dx\nonumber \\{} & {} \quad +\,(2n+1)\frac{\Lambda _1}{3\alpha ^2}\int _\mathbb {R}u^3u^{2n}_x dx+(2n+1)\frac{\Lambda _2}{4\alpha ^3}\int _\mathbb {R} u^4u^{2n}_x dx-(2n+1)\nonumber \\{} & {} \quad \times \int _\mathbb {R}u^{2n}_xp*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\Lambda _1}{3\alpha ^2}u^3\right) dx. \end{aligned}$$
(3.28)

By the Hölder’s inequality, Young’s inequality with \(\varepsilon \) and \(\eta \) and Lemma 2.2, we have

$$\begin{aligned} \nonumber \left| \int _\mathbb {R} uu_x^{2n}dx\right|\leqslant & {} ||u||_{L^{n+1}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} c_1||u_x||^{\frac{n-1}{3n+2}+2n}_{L^{2n+2}}||u||_{L^2}^{\frac{2n+3}{3n+2}}+c_2||u||_{L^2}||u_x||^{2n}_{L^{2n+2}}\\ \nonumber\leqslant & {} \frac{5}{6n+4}\left( \frac{c_1}{\eta }\right) ^{\frac{6n+4}{5}}||u||^{\frac{4n+6}{5}}_{L^2}+\frac{6n-1}{6n+4}\eta ^{\frac{6n+4}{6n-1}}||u_x||^{2n+2}_{L^{2n+2}}\\ \nonumber{} & {} +\frac{c_2^{n+1}||u||^{n+1}_{L^2}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^{\frac{n+1}{n}}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}.\\ \nonumber\leqslant & {} \frac{5}{6n+4}\left( \frac{c_1}{\eta }\right) ^{\frac{6n+4}{5}}||u_0||^{\frac{4n+6}{5}}_{H^1}+\frac{6n-1}{6n+4}\eta ^{\frac{6n+4}{6n-1}}||u_x||^{2n+2}_{L^{2n+2}}\\ \nonumber{} & {} +\frac{c_2^{n+1}||u_0||^{n+1}_{H^1}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^{\frac{n+1}{n}}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}\\ \nonumber= & {} \frac{5\cdot 2^{\frac{2n+3}{5}}}{6n+4}\left( \frac{c_1}{\eta }\right) ^{\frac{6n+4}{5}}E_0^{\frac{2n+3}{5}}+\frac{6n-1}{6n+4}\eta ^{\frac{6n+4}{6n-1}}||u_x||^{2n+2}_{L^{2n+2}}\\{} & {} +\frac{c_2^{n+1}2^{\frac{n+1}{2}}E_0^{\frac{n+1}{2}}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^{\frac{n+1}{n}}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}, \end{aligned}$$
(3.29)

where the second inequality follows from the following Gagliardo-Nirenbery inequality

$$\begin{aligned} \left\{ \begin{array}{ll} ||u||_{L^{n+1}}=||u||_{L^2},&{} \quad n=1,\\ ||u||_{L^{n+1}}\leqslant c_1||u_x||^{\frac{n-1}{3n+2}}_{L^{2n+2}}||u||^{\frac{2n+3}{3n+2}}_{L^2}+c_2||u||_{L^2},&{}\quad n\geqslant 2. \end{array}\right. \end{aligned}$$

On the other hand, by using the Hölder’s inequality, Young’s inequality with \(\varepsilon \) and Lemma 2.2, we have

$$\begin{aligned} \nonumber \left| \int _\mathbb {R} u^2u_x^{2n}dx\right|\leqslant & {} \left( \int _\mathbb {R} u^{2n+2}dx\right) ^\frac{1}{n+1}\left( \int _\mathbb {R} u_x^{2n+2}dx\right) ^\frac{n}{n+1} \\ \nonumber\leqslant & {} \frac{\int _\mathbb {R} u^{2n+2}dx}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _\mathbb {R} u_x^{2n+2}dx \\ \nonumber\leqslant & {} \frac{||u_0||^{2n+2}_{H^1}}{2^n \varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx\\= & {} \frac{2E_0^{n+1}}{\varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx, \end{aligned}$$
(3.30)
$$\begin{aligned} \nonumber \left| \int _\mathbb {R} u^3u_x^{2n}dx\right|\leqslant & {} \left( \int _\mathbb {R} u^{3n+3}dx\right) ^\frac{1}{n+1}\left( \int _\mathbb {R} u_x^{2n+2}dx\right) ^\frac{n}{n+1} \\ \nonumber\leqslant & {} \frac{\int _\mathbb {R} u^{3n+3}dx}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _\mathbb {R} u_x^{2n+2}dx \\ \nonumber\leqslant & {} \frac{||u_0||^{3n+3}_{H^1}}{2^{\frac{3n+1}{2}} \varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx\\= & {} \frac{2E_0^{\frac{3n+3}{2}}}{\varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx, \end{aligned}$$
(3.31)
$$\begin{aligned} \nonumber \left| \int _\mathbb {R} u^4u_x^{2n}dx\right|\leqslant & {} \left( \int _\mathbb {R} u^{2n+2}dx\right) ^\frac{1}{n+1}\left( \int _\mathbb {R} u_x^{2n+2}dx\right) ^\frac{n}{n+1} \\ \nonumber\leqslant & {} \frac{\int _\mathbb {R} u^{4n+4}dx}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _\mathbb {R} u_x^{2n+2}dx \\ \nonumber\leqslant & {} \frac{||u_0||^{4n+4}_{H^1}}{2^{2n+1} \varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx\\\leqslant & {} \frac{2E_0^{2n+2}}{\varepsilon ^{n+1}(n+1)}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}\int _ \mathbb {R} u_x^{2n+2}dx, \end{aligned}$$
(3.32)
$$\begin{aligned} \nonumber \left| u_x^{2n}p*u\right|\leqslant & {} \left| \left| p*u\right| \right| _{L^{n+1}}||u_x||_{L^{2n+2}}^{2n} \\ \nonumber\leqslant & {} ||u||_{L^2}||p||_{L^{\frac{2(n+1)}{n+3}}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} ||u||_{H^1}||p||_{L^{\frac{2(n+1)}{n+3}}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} \frac{||p||^{n+1}_{L^{\frac{2(n+1)}{n+3}}}||u_0||^{n+1}_{H^1}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}\\= & {} \frac{2^{\frac{n+1}{2}}||p||^{n+1}_{L^{\frac{2(n+1)}{n+3}}}E_0^{\frac{n+1}{2}}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}, \end{aligned}$$
(3.33)
$$\begin{aligned} \nonumber |u_x^{2n}p*u^2|\leqslant & {} \left| \left| p*u^2\right| \right| _{L^{n+1}}||u_x||_{L^{2n+2}}^{2n} \\ \nonumber\leqslant & {} ||u||^2_{L^2}||p||_{L^{n+1}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} ||u||^2_{H^1}||p||_{L^{n+1}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} \frac{||p||^{n+1}_{L^{n+1}}||u_0||^{2n+2}_{H^1}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}\\= & {} \frac{2^{n+1}||p||^{n+1}_{L^{n+1}}E_0^{n+1}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}, \end{aligned}$$
(3.34)

and

$$\begin{aligned} \nonumber \left| u_x^{2n}p*u^3\right|\leqslant & {} \left| \left| p*u^3\right| \right| _{L^{n+1}}||u_x||_{L^{2n+2}}^{2n} \\ \nonumber\leqslant & {} ||u^3||_{L^1}||p||_{L^{n+1}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} \frac{\sqrt{2}}{2}||u||^3_{H^1}||p||_{L^{n+1}}||u_x||^{2n}_{L^{2n+2}} \\ \nonumber\leqslant & {} \frac{\sqrt{2}}{2}\left( \frac{||p||^{n+1}_{L^{n+1}}||u_0||^{3n+3}_{H^1}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}\right) \\= & {} \frac{\sqrt{2}}{2}\left( \frac{2^{\frac{3n+3}{2}}||p||^{n+1}_{L^{n+1}}E_0^{\frac{3n+3}{2}}}{(n+1)\varepsilon ^{n+1}}+\frac{n\varepsilon ^\frac{n+1}{n}}{n+1}||u_x||^{2n+2}_{L^{2n+2}}\right) . \end{aligned}$$
(3.35)

Substituting (3.29)–(3.35) into (3.28), a direct calculation yields

$$\begin{aligned} \frac{d}{dt}\int _\mathbb {R} u^{2n+1}_x dx+K_1\int _\mathbb {R} u_x^{2n+2}dx\leqslant K_2, \end{aligned}$$
(3.36)

where \(K_1\) and \(K_2\) are defined in (3.22) and (3.23).

Again using the Hölder’s inequality, we have

$$\begin{aligned} \left| \int _\mathbb {R} u_x^{2n+1}dx\right| \leqslant \int _\mathbb {R} |u_x|^{2n+1}dx\leqslant \left( \int _\mathbb {R} u_x^{2n+2}dx\right) ^\frac{2n+1}{2n+2}. \end{aligned}$$
(3.37)

Hence, denote \(h(t)=\int _\mathbb {R} u_x^{2n+1}dx\), (3.36) transforms into

$$\begin{aligned} \frac{d}{dt} h(t)+K_1h(t)^\frac{2n+2}{2n+1}\leqslant K_2. \end{aligned}$$
(3.38)

If \(h(0)<-(\frac{K_2}{K_1})^\frac{2n+1}{2n+2}\), we can immediately deduce that \(h'(t)<0\), and h(t) is a decreasing function. So there exists lifespan \(T<\infty \) such that

$$\begin{aligned} \displaystyle \lim _{t\rightarrow T}h(t)=-\infty . \end{aligned}$$

Moreover, the above bound of the lifespan T is estimated by

$$\begin{aligned} 0<T\leqslant \int _{-h(0)}^\infty \frac{d h(t)}{K_1 h(t)^\frac{2n+2}{2n+1}-K_2}. \end{aligned}$$

Note that

$$\begin{aligned} (\inf _{x\in \mathbb {R}}u_x(t,x))^{2n+1}=\inf _{x\in \mathbb {R}}u_x^{2n+1}(t,x)\leqslant \int _\mathbb {R} u_x^{2n+1}dx. \end{aligned}$$

So that

$$\begin{aligned} \displaystyle \lim _{t\rightarrow T}\inf _{x\in \mathbb {R}}u_x(t,x)=-\infty . \end{aligned}$$

It follows from Theorem 2.9 that the solution of (2.2) blows up in finite time. This completes the proof. \(\square \)

Remark 3.4

In fact, when \(\theta \in [0,3]\), according to Lemma 3.1 in [2], we have the estimate

$$\begin{aligned} p*\left( \frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2\right) \geqslant \frac{\sqrt{\theta }}{4}(\sqrt{12-3\theta }-\sqrt{\theta })u^2>0. \end{aligned}$$

As a result, via adjusting the estimate of formula (3.28) and following the similar proof as Theorem 3.3, we can also obtain the wave-breaking result of (2.2).

Finally, we present the wave-breaking criterion with \(\theta \in [1,3)\). To this end, we first give a convolution estimate.

Lemma 3.5

([7]). Let \(\theta \in (0,3)\cup (3,4)\), \(\gamma \in [0, 1]\) and \(\delta _{\theta }=\frac{\sqrt{\theta }}{4}\left( \sqrt{12-3\theta }-\sqrt{\theta }\right) \). Then

$$\begin{aligned} (p\pm \gamma \partial _x p)*\left( \frac{3-\theta }{2}u^2-mu+\frac{\theta }{2}u_x^2\right) \geqslant \delta _{\theta }\left( u-\frac{m}{3-\theta }\right) ^2-\frac{m^2}{2(3-\theta )}. \end{aligned}$$

Theorem 3.6

Let \(\theta \in [1,3)\) and \(\lambda \geqslant 0\). Assume that u be the solution of (2.2) with initial data \(u_0\in H^s(\mathbb R)\), \(s>\frac{3}{2}\) and \(T>0\) be the maximal time of existence, and there exists some \(x_0\in \mathbb {R}\) such that

$$\begin{aligned} u_{0,x}(x_0)<-\gamma _{\theta }\left| u_0(x_0)-\frac{1}{3-\theta }\left( \frac{\mu _0}{\mu }-c\right) \right| -\frac{2\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C_0. \end{aligned}$$
(3.39)

Then the solution u blows up in finite time T. More precisely,

$$\begin{aligned} T\leqslant \frac{1}{\sqrt{\lambda ^2+2\theta C_0^2}}\ln \frac{\sqrt{u_{0,x}^2(x_0)-\gamma _{\theta }^2\left( u_0(x_0)-\frac{k}{3-\theta }\right) ^2}-\frac{\lambda }{\theta }+\sqrt{\frac{2}{\theta }\left( \frac{\lambda ^2}{2\theta }+C_0^2\right) }}{\sqrt{u_{0,x}^2(x_0)-\gamma _{\theta }^2\left( u_0(x_0)-\frac{k}{3-\theta }\right) ^2}-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }\left( \frac{\lambda ^2}{2\theta }+C_0^2\right) }}, \end{aligned}$$

where

$$\begin{aligned} C_0=\left( \frac{2|\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{|\Lambda _2|}{2\alpha ^3}E_0^2+\frac{\gamma _{\theta }|k|\lambda }{3-\theta }\right) ^{\frac{1}{2}},\ \gamma ^2_\theta =\frac{3}{\theta }-\frac{1}{2}-\frac{\sqrt{12-3\theta }}{2\sqrt{\theta }}. \end{aligned}$$

Remark 3.7

In fact, it is noting that the map \(\theta \mapsto \gamma _{\theta }\) is non-increasing on [0, 3] and non-decreasing on (3, 4]. Consequently, \(\gamma _{\theta }\in (0,1]\) as \(\theta \in [1,3)\).

Proof

Along the characteristic q(t, x) defined in (2.24), (2.11) and (2.27) yield

$$\begin{aligned} \frac{\partial u(t,q)}{\partial t}= & {} -\lambda u(t,q)-p_x*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) (t,q), \end{aligned}$$
(3.40)
$$\begin{aligned} \frac{\partial u_x(t,q)}{\partial t}= & {} \left( -\frac{\theta }{2}u_x^2+\frac{3-\theta }{2} u^2+\left( c-\frac{\mu _0}{\mu }\right) u-\lambda u_x+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) (t,q)\nonumber \\{} & {} -p*\left( \left( c-\frac{\mu _0}{\mu }\right) u+\frac{3-\theta }{2}u^2+\frac{\theta }{2}u_x^2+\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) (t,q). \end{aligned}$$
(3.41)

Let

$$\begin{aligned} M(t)=\gamma _\theta \left( u(t,q)-\frac{k}{3-\theta }\right) -u_x(t,q),\ N(t)=\gamma _\theta \left( u(t,q)-\frac{k}{3-\theta }\right) +u_x(t,q),\nonumber \\ \end{aligned}$$
(3.42)

where \(k=\frac{\mu _0}{\mu }-c\).

Differentiating M(t) and N(t) with respect to t, according to (3.40)–(3.42), Lemma 3.5 and Remark 3.7, we have

$$\begin{aligned} \nonumber \frac{\partial M(t)}{\partial t}= & {} \frac{\theta }{2}u_x^2-\frac{3-\theta }{2}u^2+ku-\lambda (\gamma _\theta u-u_x)+(p-\gamma _\theta p_x)*\left( \frac{3-\theta }{2}u^2-ku+\frac{\theta }{2}u_x^2\right) \\ \nonumber{} & {} -\frac{\Lambda _1}{3\alpha ^2}u^3-\frac{\Lambda _2}{4\alpha ^3}u^4+(p-\gamma _\theta p_x)*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) \\ \nonumber= & {} \frac{\theta }{2}u_x^2-\frac{3-\theta }{2}u^2+ku-\lambda \left( M(t)+\frac{\gamma _\theta k}{3-\theta }\right) +(p-\gamma _\theta p_x)\\ \nonumber{} & {} *\left( \frac{3-\theta }{2}u^2-ku+\frac{\theta }{2}u_x^2\right) -\frac{\Lambda _1}{3\alpha ^2}u^3-\frac{\Lambda _2}{4\alpha ^3}u^4+(p-\gamma _\theta p_x)*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) \\ \nonumber\geqslant & {} \frac{\theta }{2}u_x^2-\frac{3-\theta }{2}\left( u-\frac{k}{3-\theta }\right) ^2+\delta _{\theta }\left( u-\frac{k}{3-\theta }\right) ^2-\lambda M(t)-C_0^2 \\ \nonumber= & {} \frac{\theta }{2}\left( u_x^2-\gamma ^2_\theta \left( u-\frac{k}{3-\theta }\right) ^2\right) -\lambda M(t)-C_0^2 \\ \nonumber= & {} -\frac{\theta }{2}M(t)N(t)-\lambda M(t)-C_0^2 \\= & {} -\frac{\theta }{2}M(t)\left( N(t)+\frac{2}{\theta }\lambda \right) -C_0^2, \end{aligned}$$
(3.43)

and

$$\begin{aligned} \nonumber \frac{\partial N(t)}{\partial t}= & {} -\frac{\theta }{2}u_x^2+\frac{3-\theta }{2}u^2-ku-\lambda \left( N(t)+\frac{\gamma _\theta k}{3-\theta }\right) \\{} & {} -(p+\gamma _\theta p_x)*\left( \frac{3-\theta }{2}u^2-ku+\frac{\theta }{2}u_x^2\right) \nonumber \\ \nonumber{} & {} +\frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4-(p+\gamma _\theta p_x)*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) \\ \nonumber\leqslant & {} -\frac{\theta }{2}u_x^2+\frac{3-\theta }{2}\left( u-\frac{k}{3-\theta }\right) ^2-\delta _{\theta }\left( u-\frac{k}{3-\theta }\right) ^2-\lambda N(t)+C_0^2 \\ \nonumber= & {} -\frac{\theta }{2}\left( u_x^2-\gamma ^2_\theta \left( u-\frac{k}{3-\theta }\right) ^2\right) -\lambda N(t)+C_0^2 \\ \nonumber= & {} \frac{\theta }{2}M(t)N(t)-\lambda N(t)+C_0^2 \\= & {} \frac{\theta }{2}N(t)\left( M(t)-\frac{2}{\theta }\lambda \right) +C_0^2, \end{aligned}$$
(3.44)

where \(C_0\) defined as

$$\begin{aligned} \nonumber{} & {} \left| \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4-(p\pm \gamma _\theta p_x)*\left( \frac{\Lambda _1}{3\alpha ^2}u^3+\frac{\Lambda _2}{4\alpha ^3}u^4\right) +\frac{\gamma _\theta k\lambda }{3-\theta }\right| \\ \nonumber{} & {} \quad \leqslant \frac{|\Lambda _1|}{3\alpha ^2}||u||^3_{L^\infty }+\frac{\Lambda _2}{4\alpha ^3}||u||^4_{L^\infty }+\frac{|\Lambda _1|}{3\alpha ^2}||p\pm \gamma _\theta p_x||_{L^\infty }||u||_{L^\infty }||u||^2_{L^2}\\ \nonumber{} & {} \qquad +\frac{\Lambda _2}{4\alpha ^3}||p\pm \gamma _\theta p_x||_{L^\infty }||u||^2_{L^\infty }||u||^2_{L^2}+\frac{\gamma _\theta |k|\lambda }{3-\theta } \\{} & {} \quad \leqslant \frac{2|\Lambda _1|}{3\alpha ^2}E_0^\frac{3}{2}+\frac{|\Lambda _2|}{2\alpha ^3}E_0^2+\frac{\gamma _\theta |k|\lambda }{3-\theta }:=C_0^2, \end{aligned}$$

with the fact that

$$\begin{aligned} p_+*f(x){} & {} =\frac{e^{-x}}{2}\int _{-\infty }^xe^yf(y)dy,\ \ p_-*f(x)=\frac{e^{x}}{2}\int _x^{\infty }e^{-y}f(y)dy \\{} & {} \qquad p=p_++p_-,\ \ p_x=p_--p_+, \end{aligned}$$

and

$$\begin{aligned} ||p_\pm ||_{L^\infty }=\frac{1}{2},\ \ ||p_\pm ||_{L^2}=\frac{1}{2\sqrt{2}}. \end{aligned}$$

By the assumption (3.39), we have

$$\begin{aligned} M(0){} & {} =\gamma _\theta \left( u_0(x_0)-\frac{k}{3-\theta }\right) -u_{0,x}(x_0)>\frac{2\lambda }{\theta }+\sqrt{\frac{2}{\theta }}C_0>0, \\ N(0){} & {} =\gamma _\theta \left( u_0(x_0)-\frac{k}{3-\theta }\right) +u_{0,x}(x_0)<-\frac{2\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C_0<0. \end{aligned}$$

It follows the standard continuity argument that

$$\begin{aligned} \frac{d M(t)}{dt}>0,\ \ \frac{d N(t)}{dt}<0,\ \ t\in [0,T). \end{aligned}$$

That implies that

$$\begin{aligned} M(t)>M(0)>0,\ \ N(t)<N(0)<0,\ \ t\in [0,T). \end{aligned}$$

Let \(g(t)=\sqrt{-M(t)N(t)}\). It then follows from (3.43) and (3.44) that

$$\begin{aligned} \nonumber \frac{d g(t)}{dt}= & {} \frac{-M'N-MN'}{2g} \\ \nonumber\geqslant & {} \frac{\left[ \frac{\theta }{2}M\left( N+\frac{2}{\theta }\lambda \right) +C_0^2\right] N-M\left[ \frac{\theta }{2}N\left( M-\frac{2}{\theta }\lambda \right) +C_0^2\right] }{2g}\\ \nonumber= & {} \left( -\frac{\theta }{2}MN-C_0^2\right) \frac{M-N}{2\sqrt{MN}}-\lambda \sqrt{-MN}\\ \nonumber\geqslant & {} \frac{\theta }{2}g^2-\lambda g-C_0^2\\ \nonumber= & {} \frac{\theta }{2}\left( g-\frac{\lambda }{\theta }\right) ^2-\frac{\lambda ^2}{2\theta }-C_0^2\\:= & {} \frac{\theta }{2}\left( g-\frac{\lambda }{\theta }\right) ^2-C^2, \end{aligned}$$
(3.45)

where we have used the fact \(\frac{M(t)-N(t)}{2g}\geqslant 1\).

Solving (3.45), we have

$$\begin{aligned} 0<\frac{2\sqrt{\frac{2}{\theta }}C}{g(t)-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C}\leqslant \frac{g(0)-\frac{\lambda }{\theta }+\sqrt{\frac{2}{\theta }}C}{g(0)-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C}e^{-\sqrt{2\theta }Ct}-1. \end{aligned}$$

Then,

$$\begin{aligned} g(t)-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C\rightarrow \infty , \end{aligned}$$

as

$$\begin{aligned} t\rightarrow \frac{1}{\sqrt{2\theta }C}\ln \frac{g(0)-\frac{\lambda }{\theta }+\sqrt{\frac{2}{\theta }}C}{g(0)-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C}. \end{aligned}$$

Consequently, when T is estimated by

$$\begin{aligned} 0<T\leqslant \frac{1}{\sqrt{2\theta }C}\ln \frac{g(0)-\frac{\lambda }{\theta }+\sqrt{\frac{2}{\theta }}C}{g(0)-\frac{\lambda }{\theta }-\sqrt{\frac{2}{\theta }}C}, \end{aligned}$$

we have

$$\begin{aligned} g(t)\leqslant -u_x(t,q(t,x))\rightarrow \infty ,\ \ t\rightarrow T. \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{t\rightarrow T}\inf _{x\in \mathbb {R}}u_x(t,x)=-\infty . \end{aligned}$$

It follows from Theorem 2.9 that the solution of (2.2) blows up in finite time. This completes the desired proof. \(\square \)

Remark 3.8

Since the dispersive parameter \(\theta \) belongs to the positive real-value interval in Theorems 3.3 and 3.6, we only need to consider the \(u_x\) unbounded from below in the proof.