Blow Up for Klein–Gordon Equation with Nonlocal Operator and Trudinger–Moser Nonlinearity

Abstract

In this paper we prove a blow up result for a class of nonlocal scalar Klein–Gordon equation. We assume that the nonlinearity has critical exponential growth. Additionally, we prove that the ground state solution of the elliptic problem associated to the original problem is unstable. The strategy is to adapt the recent ideas of Carrião et al. (J Dyn Differ Equ 2023. https://doi.org/10.1007/s10884-023-10281-3) to find two regions (stable and unstable regions). Since the one dimensional case combined with the 1/2 Laplacian operator cause lack of the control on the $L^2(\mathbb {R})$ norm of $(-\Delta )^{\frac{1}{4}}u$, new delicate calculations are necessary. We prove also that there exists a subset of $H^{\frac{1}{2}}(\mathbb {R})\times L^2(\mathbb {R})$ such that the solution is global when the initial data is taken into this set.

1 Introduction

In this paper we consider the following Cauchy problem for the nonlinear hyperbolic equation

$$\begin{aligned} \begin{array}{ll} u_{tt}+(-\Delta )^{\frac{1}{2}} u +\lambda _0 u = f(u) &{} \text{ in } \mathbb {R}\times (0,T),\\ u(x,0)=u_0(x),\,\, u_t(x,0)=u_1(x) &{} \text{ in } \mathbb {R}, \end{array} \end{aligned}$$

(1)

where $\lambda _0$ is a real positive constant; $u_0,u_1:\mathbb {R}\rightarrow \mathbb {R}$ are given functions; $f:\mathbb {R}\rightarrow \mathbb {R}$ is a known function with exponential growth, and $(-\Delta )^{\frac{1}{2}}$ is the fractional laplacian operator which is defined as

$$\begin{aligned} (-\Delta )^{\frac{1}{2}} u(x)= -\frac{1}{2\pi }\int _{\mathbb {R}}(u(x+y)+u(x-y)-2u(x))k(y)dy, \end{aligned}$$

for all $x\in \mathbb {R}$, where $k(\xi )=|\xi |_{\mathbb {R}}^{-2}$. Details about the fractional Laplacian can be found in Di Nezza, Palatucci, and Valdinoci [7]. We say that $f:\mathbb {R}\rightarrow \mathbb {R}$ has $\alpha _0$-critical growth at $\pm \infty$ if there exist $\omega \in (0,\pi )$ and $\alpha _0\in (0,\pi )$ such that

$$\begin{aligned} \limsup _{s\rightarrow \pm \infty }\frac{f(s)}{e^{\alpha s^2}-1}=0,\quad \text{ for } \text{ all } \alpha >\alpha _0 \end{aligned}$$

(2)

and

$$\begin{aligned} \limsup _{s\rightarrow \pm \infty }\frac{f(s)}{e^{\alpha s^2}-1}=\pm \infty ,\quad \text{ for } \text{ all } \alpha <\alpha _0. \end{aligned}$$

(3)

We consider the following hypotheses on the function f:

• H1) $f:\mathbb {R}\rightarrow \mathbb {R}$ is an odd and convex function on $\mathbb {R}_+$ and

  $$\begin{aligned} \lim _{s\rightarrow 0}\frac{f(s)}{s}=0. \end{aligned}$$

• H2) the function $s\mapsto \frac{f(s)}{s}$ is increasing on $\mathbb {R_+}$.

• H3) there exists $q>2$ and $C_0>0$ such that

  $$\begin{aligned} F(s)\ge C_0|s|^q, \end{aligned}$$

  for all $s\in \mathbb {R}$, where $F(\xi )=\int _0^{\xi }f(\tau )\,d\tau$.

• H4) there exists $C_A>2$ such that

  $$\begin{aligned} C_AF(s) \le f(s)s, \end{aligned}$$

  for all $s\in \mathbb {R}$.

The assumptions above are well known in the context of elliptic equations. For instance, see do Ó, Miyagaki, and Squassina [18] where it is possible to find an explicit example. See also Iannizzotto and Squassina [12]. The assumptions (H3) and (H4) are used only to prove the existence of the ground state solution. Problems concerning fractional Laplacian in the context of elliptic equations can be found, for instance, in Alves and Torres Ledesma [3], Ferrari and Verbitski [9], Luo and Zhang [15], Bisci, Radulescu, and Servadei [4], and Servadei and Valdinoci [22,23,24] end references therein.

In the classical paper of Shatah [25], the author proved the instability of the ground state solution of the stationary problem associated to the wave equation

$$\begin{aligned} u_{tt}-\Delta u=f(u)\quad \text{ in } \mathbb {R}^N\times (0,T), \end{aligned}$$

(4)

where $\Delta$ is the classical laplacian in the spatial variable, $N\ge 3$, and f is a known function. Shatah combined the use of an appropriate cut-off function and the classical Strauss compactness result. The case $N=2$ was studied by Jeanjean and Le Coz [13] and it is more delicate, because it holds the lose of the control on the $L^2$ norm of $\nabla u$. Concerning the classical wave equation, blow-up results can be found, for instance, in Alves and Cavalcanti [1], Alves et al. [2], Domingos Cavalcanti et al. [8], Gazzola and Squassina [10], Georgiev and Todorova [11], Levine and Todorova [14], Merle and Zaag [16], Messaoudi [17], Ohta and Todorova [19], Todorova [27], and references therein. We would like to cite also the recent work of Yang and Han [28] where the authors studied a damped p-Laplacian type wave equation with logarithmic nonlinearity. They proved a result concerning the finite time blow-up of solutions.

Recently, Carrião, Lehrer, and Vicente [5] introduced a new technique to prove blow-up results for nonlocal hyperbolic operators. Precisely, the authors studied the following problem

$$\begin{aligned} \begin{array}{ll} u_{tt}+(-\Delta )^{s} u +\lambda _0 u = f(u) &{} \text{ in } \mathbb {R}^N\times (0,T),\\ u(x,0)=u_0(x),\,\, u_t(x,0)=u_1(x) &{} \text{ in } \mathbb {R}^N, \end{array} \end{aligned}$$

(5)

where $(-\Delta )^s$ is the fractional laplacian and $N=3$ and $\frac{3}{4}\le s\le 1$ or $N=4$ and $s=1$. This restriction holds only in their existence result. The main result of the paper was proved for $0<s\le 1$ and $N\ge 3$. They introduced a new cut-off function which allows to work with nonlocal operators. This new definition is associated with a function defined with a parameter, $\lambda$, which rediscover some derivative formulas and allows to avoid the use of Green formula. The strategy of the authors is associated to the use of the Pohozaev functional

$$\begin{aligned} J(u)=\frac{N-2s}{2}\int _{\mathbb {R}^N}|(-\Delta u)|^{\frac{s}{2}}\,dx+\frac{N\lambda _0}{2}\int _{\mathbb {R}^N}u^2\,dx-N\int _{\mathbb {R}^N}F(u)\,dx. \end{aligned}$$

This functional allows the authors to define stable and unstable sets for the initial data. They proved also the instability of the ground state solution of the stationary problem associated to the problem (5). Observing the definition of the Pohozaev functional above, we have that the case $N=1$ and $s=\frac{1}{2}$ causes the lose of the control on the $L^2(\mathbb {R})$ norm of $(-\Delta )^{\frac{1}{4}} u$. Thus, in this case the calculations need special attention. Pohozaev identity for the fractional Laplacian can be found in Ros-Oton and Serra [21].

The main objective of this manuscript is extend the result of Carrião, Lehrer, and Vicente [5] to the case $N=1$ and $s=\frac{1}{2}$. Our strategy is combine the ideas of [5] with some ones of Jeanjean and Le Coz [13]. Some additional and delicate estimates are necessary. Summarizing, we prove that there exists a subset of $H^{\frac{1}{2}}(\mathbb {R})\times L^2(\mathbb {R})$ such that the solution blows up when the initial data is taken into this set (in finite or infinite time). Additionally, we prove that the ground state solution of the elliptic problem associated to the problem (1) is unstable. Finally, we prove also that there exists a subset of $H^{\frac{1}{2}}(\mathbb {R})\times L^2(\mathbb {R})$ such that the solution is global when the initial data is taken into this set.

We define the Pohozaev manifold associated to the problem (1) by

$$\begin{aligned} \mathcal {P}= \{u\in H^{\frac{1}{2}}(\mathbb {R})\backslash \{0\};\,\,P(u)=0\}, \end{aligned}$$

where the Pohozaev functional $P: H^{\frac{1}{2}}(\mathbb {R})\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} P(u)=\int _{\mathbb {R}} F(u) dx-\frac{\lambda _0}{2} \int _{\mathbb {R}} u^2 dx. \end{aligned}$$

(6)

We define the functional I by

$$\begin{aligned} I(u) =\frac{1}{2}\int _{\mathbb {R}} |(-\Delta )^{\frac{1}{4}}u|^2 dx + \frac{\lambda _0}{2} \int _{\mathbb {R}} u^2 dx - \int _{\mathbb {R}} F(u)dx \end{aligned}$$

and we consider the elliptic problem

$$\begin{aligned} (-\Delta )^{\frac{1}{2}} u +\lambda _0 u = f(u) \, \text{ in } \mathbb {R}. \end{aligned}$$

(7)

The least energy level of (7) is defined by

$$\begin{aligned} m=\inf \{I(u); u\in H^{\frac{1}{2}}(\mathbb {R})\backslash \{0\} \text{ is } \text{ a } \text{ solution } \text{ of } \text{(7) }\}. \end{aligned}$$

A function $\varphi \in H^{\frac{1}{2}}(\mathbb {R})$ is called ground state or least energy solution of (7) if

$$\begin{aligned} I(\varphi )=m. \end{aligned}$$

Since we will prove an instability result of the stationary problem, we need a result concerning the existence of the ground state solution for (7). This result was proved by do Ó, Miyagaki, and Squassina [18] and it is given by proposition below.

Proposition 1.1

Suppose that f(s) and $f'(s)s$ have $\alpha _0$-critical growth and satisfies (H1)–(H4). Then problem (7) admits a ground state solution $u\in H^{\frac{1}{2}}(\mathbb {R})$ proved $C_0$ in (H3) is large enough.

We introduce the following subset of $H^{\frac{1}{2}}(\mathbb {R})$

$$\begin{aligned} \mathcal {I}=\{(u,v)\in H^{\frac{3}{2}}(\mathbb {R})\backslash \{0\}\times H^1(\mathbb {R});\, E(u,v)<m \text{ and } P(u)>0\}. \end{aligned}$$

where E(u, v) is given by

$$\begin{aligned} E(u,v) = \frac{1}{2}\int _{\mathbb {R}}v^2\,dx +I(u). \end{aligned}$$

Since (H1) and (H2) hold and f has $\alpha _0$-critical growth, for $\alpha >\alpha _0$ and for each $\epsilon >0$, there exists $s_1>0$ such that

$$\begin{aligned} f(s) \le \epsilon s+C(s_1) se^{\alpha s^2}, \end{aligned}$$

(8)

for all $s\in \mathbb {R}$ and

$$\begin{aligned} F(s) \le \frac{\epsilon s^2}{2} +C(s_1) (e^{\alpha s^2}-1), \end{aligned}$$

(9)

for all $s\in \mathbb {R}$. Moreover, since $f'(s)s$ has $\alpha _0$-critical growth it holds

$$\begin{aligned} f'(s) \le \frac{e^{\alpha s^2}-1}{s}+C, \end{aligned}$$

(10)

for all $s\in \mathbb {R}\setminus \{0\}$. Therefore, using (10), for $\alpha >\alpha _0$, we have

$$\begin{aligned} |f(s)-f(\tau )|^2 \le 4 \left( C + \frac{e^{2\alpha s^2}-1}{|s|^2} + \frac{e^{2\alpha \tau ^2}-1}{|\tau |^2} \right) | s-\tau |^2, \end{aligned}$$

(11)

for all $s,\tau \in \mathbb {R}\setminus \{0\}.$

Since $H^{\frac{1}{2}}(\mathbb {R})\hookrightarrow L^q(\mathbb {R})$ continuously for $2\le q<\infty$, we have there exists positive constant $C_q$ such that

$$\begin{aligned} \Vert u\Vert _q\le C_q\Vert u\Vert , \end{aligned}$$

(12)

for all $u\in H^{\frac{1}{2}}(\mathbb {R})$, where $\Vert \cdot \Vert _q$ is the standard norm in $L^q(\mathbb {R})$ and

$$\begin{aligned} \Vert u\Vert =\left( \int _{\mathbb {R}}|(-\Delta )^{\frac{1}{4}}u(x)|^2 \,dx+\lambda _0\int _{\mathbb {R}} |u(x)|^2\,dx\right) ^\frac{1}{2} \end{aligned}$$

is the norm in $H^{\frac{1}{2}}(\mathbb {R})$. Moreover, since $H^{\frac{3}{2}}(\mathbb {R})\hookrightarrow H^{\frac{1}{2}}(\mathbb {R})$ continuously, there exists a positive constant $C_1$ such that

$$\begin{aligned} \Vert u\Vert \le C_1\Vert u\Vert _{H^{\frac{3}{2}}(\mathbb {R})}, \end{aligned}$$

(13)

for all $u\in H^{\frac{3}{2}}(\mathbb {R})$.

We observe that, concerning the nonlinearity f, in present paper unlike [5] (where polinomial growth was considered) the nonlinear term has exponential growth. It is well known that, when exponential growth is in place, it is necessary some kind of Trudinger-Moser inequality. In this manuscript, we use the following result due to Ozawa [20].

Proposition 1.2

There exists $0<\omega \le \pi$ such that, for all $\alpha \in (0,\omega )$, there exists $H_{\alpha }>0$ with

$$\begin{aligned} \int _{\mathbb {R}}(e^{\alpha u^2}-1)\,dx\le H_{\alpha }\Vert u\Vert _2^2, \end{aligned}$$

(14)

for all $u\in H^{\frac{1}{2}}(\mathbb {R})$ with $\Vert (-\Delta )^{\frac{1}{4}}u\Vert _2^2\le 1$.

For Trundiger-Moser inequality on $\mathbb {R}$ see also Takahashi [26]. Now we can enunciate the main results of this paper:

Theorem 1.1

[Blow up Theorem] Suppose that f(s) and $f'(s)s$ have $\alpha _0$-critical growth and satisfies (H1)–(H4). Suposse $\lambda _0>6\alpha C(s_1)$, for some $\alpha \in (\alpha _0,\pi )$. Let $(u_{0},u_{1})\in \mathcal {I}$ be the initial data and let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the unique strong solution of (1) given by Theorem 2.1. Then, either

a) the solution exists locally, i.e. $T<\infty$, and there exists a sequence $(t_k)_{k\in \mathbb {N}}\subset (0,T)$ with $t_k\rightarrow T^-$ such that

$$\begin{aligned} \Vert u(t_k)\Vert _{H^{\frac{3}{2}}(\mathbb {R})} \rightarrow \infty , \,\,\text{ when } t_k\rightarrow T^-; \end{aligned}$$

or

b) the solution exists globally on $[0,\infty )$ and there exists a sequence $(t_k)_{k\in \mathbb {N}}\subset (0,\infty )$ with $t_k\rightarrow \infty$ such that

$$\begin{aligned} \Vert u(t_k)\Vert _{H^{\frac{3}{2}}(\mathbb {R})} \rightarrow \infty , \,\,\text{ when } t_k\rightarrow \infty . \end{aligned}$$

To prove Theorem 1.1, it is necessary a blow up result as Lemma 14 of Jeanjean and Le Coz [13] and Lemma 1.2 of Carrião, Lehrer, and Vicente [5]. This is enunciate in next result. The constant $\delta$ is a known positive real number, which does not depend of t. It is given by Lemma 2.5.

Lemma 1.1

[Blow up Lemma] Suppose that f(s) and $f'(s)s$ have $\alpha _0$-critical growth and satisfies (H1)–(H4). Suposse $\lambda _0>6\alpha C(s_1)$, for some $\alpha \in (\alpha _0,\pi )$. Let $(u_{0},u_{1})\in \mathcal {I}$ be the initial data and let u(t) be the associated strong solution of (1) defined in [0, T). If there exists a constant $K>0$ such that

$$\begin{aligned} \Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {R})} \le K, \end{aligned}$$

(15)

for all $t\in [0,T)$, then for each $T_1\in [0,T)$ it holds

$$\begin{aligned} t \le \frac{1}{2\delta }\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] \left[ \Vert u_t(t)\Vert _2\Vert u_x(t)\Vert _2+\Vert u_1\Vert _2\Vert (u_0)_x\Vert _2\right] +\frac{3}{4}, \end{aligned}$$

for all $t\in [0,T_1]$, where

$$\begin{aligned} C_2=\frac{7\delta }{4\left( K^2C_1^2+2C_3\right) }, \end{aligned}$$

(16)

with

$$\begin{aligned} C_3= & {} 2C_1^2K^2+\frac{1}{2}\left[ 2m+(1+\epsilon \lambda _0^{-1}+2C(s_1)H_{\alpha }\lambda _0^{-1})C_1^2K^2 \right] \nonumber \\{} & {} +2(\epsilon +C(s_1)H_{\alpha }))C_1^2K^2\lambda _0^{-1}. \end{aligned}$$

(17)

We prove the following instability result.

Theorem 1.2

[Instability of the ground state] Suppose that f(s) and $f'(s)s$ have $\alpha _0$-critical growth and satisfies (H1)–(H4). Suposse $\lambda _0>6\alpha C(s_1)$, for some $\alpha \in (\alpha _0,\pi )$. Let $\varphi$ be a ground state solution of (7). Then $\varphi$ viewed as a stationary solution of (1) is strongly unstable. Namelly, for all $\eta >0$ there exist $u_{0,\eta }\in H^{\frac{3}{2}}(\mathbb {R})$, $T_{\eta }\in (0,\infty ]$ and $(t_n)\subset (0,T_{\eta })$ such that $\Vert \varphi -u_{0,\eta }\Vert <\eta$ and $\lim _{t_n\rightarrow T_{\eta }}\Vert u(t_n)\Vert =\infty$, where u is the strong solution of (1) associated to the initial data $(u_{0,\eta },0)$.

Now, we define the stable region by

$$\begin{aligned} \mathcal {S}=\{(u,v)\in H^{\frac{1}{2}}(\mathbb {R})\backslash \{0\}\times L^2(\mathbb {R});\, E(u,v)<m \text{ and } P(u)\le 0\}. \end{aligned}$$

Finally, we prove the following theorem which gives us the existence of global solution when the initial data are taken in $\mathcal {S}$.

Theorem 1.3

[Global existence] Suppose that f(s) and $f'(s)s$ have $\alpha _0$-critical growth and satisfies (H1)–(H4). If $(u_{0},u_{1})$ $\in \mathcal {S}$, then the local generalized solution of (1) given by Theorem 2.1 can be extended for all $t>0$.

Our paper is organized as follows: in Section 2 we give the notations, the preliminaries and we stablish the local existence theorem. In Section 2, we also prove some lemmas. In Section 3 we prove Lemma 1.1. Finally, in Section 4, we prove Theorems 1.1, 1.2 and 1.3.

2 Preliminaries

In this section we establish the notation and give some preliminary results. We denote by C all generic constants and sometimes, to simplify the notation, we write u instead of u(x, t).

We define

$$\begin{aligned} \mathcal {H}= H^{\frac{1}{2}}(\mathbb {R})\times L^2(\mathbb {R}) \end{aligned}$$

with the inner product and norm given by

$$\begin{aligned} ((u,v),(z,w))_{\mathcal {H}}=(u,z)_{H^{\frac{1}{2}}(\mathbb {R})}+(v,w)_{L^2(\mathbb {R})} \end{aligned}$$

and

$$\begin{aligned} \Vert (u,v)\Vert _{\mathcal {H}}^2 = \Vert u\Vert ^2 + \Vert v\Vert _2^2. \end{aligned}$$

We define also the operator $A:D(A)\subset \mathcal {H}\rightarrow \mathcal {H}$ by setting

$$\begin{aligned} A = \left( \begin{array}{cc} 0&{}-I\\ (-\Delta )^{\frac{1}{2}}+\lambda _0 &{} 0 \end{array} \right) \end{aligned}$$

i. e.,

$$\begin{aligned} A \left( \begin{array}{c} u\\ v \end{array} \right) = \left( \begin{array}{cc} -v\\ (-\Delta )^{\frac{1}{2}}u+\lambda _0u \end{array} \right) , \end{aligned}$$

where $D(A)=H^{\frac{3}{2}}(\mathbb {R})\times H^{1}(\mathbb {R})$. We define also $B:\mathcal {H}\rightarrow \mathcal {H}$ by setting

$$\begin{aligned} B \left( \begin{array}{c} u\\ v \end{array} \right) = \left( \begin{array}{cc} 0\\ -f(u) \end{array} \right) . \end{aligned}$$

Therefore, the problem (1) can be write as

$$\begin{aligned}{} & {} \frac{d U}{dt} + AU+BU=0\,\, \text{ in } (0,\infty ), \end{aligned}$$

(18)

$$\begin{aligned}{} & {} U(0)=U_0, \end{aligned}$$

(19)

where $U=(u,u_t)^{\top }$ and $U_0=(u_0,u_1)^{\top }\in D(A)$.

To prove that (18)–(19) has solution, it is suffices to verify that A is a maximal monotone operator and B is a local Lipschitz operator. It is not difficult to verify that

$$\begin{aligned}{} & {} \Big (A\Big (\begin{array}{c} u\\ v \end{array} \Big ) - A\Big (\begin{array}{c} z\\ w \end{array} \Big ),\Big (\begin{array}{c} u\\ v \end{array} \Big ) -\Big (\begin{array}{c} z\\ w \end{array} \Big ) \Big )_{\mathcal {H}} \\{} & {} \quad = -(v-w,u-z)_{H^{\frac{1}{2}}(\mathbb {R})}+((-\Delta )^{\frac{1}{2}}(u-z)+\lambda _0(u-z),v-w)_{L^2(\mathbb {R})} \ge 0, \end{aligned}$$

for all $\Big (\begin{array}{c} u\\ v \end{array} \Big ),\Big (\begin{array}{c} z\\ w \end{array} \Big )\in D(A)$, thus A is monotone. To prove that A is maximal monotone, we show that $I+A$ is into $\mathcal {H}$. Let $h=\Big (\begin{array}{c} h_1\\ h_2\end{array}\Big )$ be in the space $\mathcal {H}$. We are going to prove that there exists $\Big (\begin{array}{c} u\\ v\end{array}\Big )\in D(A)$ such that

$$\begin{aligned} (I+A) \Big (\begin{array}{c} u\\ v\end{array}\Big ) = \Big (\begin{array}{c} h_1\\ h_2\end{array}\Big ). \end{aligned}$$

(20)

We have that (20) can be writen as

$$\begin{aligned}{} & {} u-v=h_1\end{aligned}$$

(21)

$$\begin{aligned}{} & {} v+(-\Delta )^{\frac{1}{2}}u+\lambda _0 u=h_2. \end{aligned}$$

(22)

Combining (21) and (22), we obtain

$$\begin{aligned} (-\Delta )^{\frac{1}{2}}u+(1+\lambda _0) u=h_1+h_2. \end{aligned}$$

(23)

Since $h_1+h_2\in L^2(\mathbb {R})$, from elliptic equation theory, we have that (23) has a weak solution $u\in H^{\frac{1}{2}}(\mathbb {R})$. Moreover, elliptic regularity results give us that the solution is more regular, i.e., $u\in H^{1}(\mathbb {R})$. This regularity and as $h_1\in H^{\frac{1}{2}}(\mathbb {R})$, (21) allows us to conclude that $v\in H^{\frac{1}{2}}(\mathbb {R})$. Using elliptic regularity results one more time, we have that $u\in H^{\frac{3}{2}}(\mathbb {R})$. Thus, (22) gives us that $h_2\in H^{\frac{1}{2}}(\mathbb {R})$. Therefore, from (22) we have that

$$\begin{aligned} (-\Delta )^{\frac{1}{2}}v=\Delta u-\lambda _0 (-\Delta )^{\frac{1}{2}} u+(-\Delta )^{\frac{1}{2}} h_2\in L^2(\mathbb {R}). \end{aligned}$$

Using elliptic regularity, we obtain that $v\in H^1(\mathbb {R})$. Therefore, $\Big (\begin{array}{c} u\\ v\end{array}\Big )\in D(A)$ is solution of (20), consequently A is maximal monotone.

Now we are going to prove that B is a local Lipschitz operator. Indeed, it is enough to prove that $f: H^{\frac{1}{2}}(\mathbb {R})\rightarrow L^2(\mathbb {R})$ is a local Lipschitz function. Using (11) and Hölder inequality, we have

$$\begin{aligned} \Vert f(u)-f(v)\Vert _2^2 \le C\Vert u-v\Vert _2^2 \end{aligned}$$

$$\begin{aligned} +4\left[ \left( \int _{\mathbb {R}}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx\right) ^{\frac{1}{2}} +\left( \int _{\mathbb {R}}\left( \frac{e^{2\alpha v^2}-1}{v^2}\right) ^2\,dx\right) ^{\frac{1}{2}} \right] \Vert u-v\Vert _{L^4(\mathbb {R})}^2. \end{aligned}$$

(24)

Since

$$\begin{aligned} \lim _{s\rightarrow 0}\frac{e^{2\alpha s^2}-1}{s^2}=2\alpha , \end{aligned}$$

there exists $\delta _0>0$ such that

$$\begin{aligned} \alpha ^2\delta _0< \int _{-\delta _0}^{\delta _0} \left( \frac{e^{2\alpha s^2}-1}{s^2}\right) ^2\,ds < 9\alpha ^2\delta _0. \end{aligned}$$

(25)

We define

$$\begin{aligned} D_1^t =\{x\in \mathbb {R};\,|u(x,t)|\le \delta _0\} \quad \text{ and }\quad D_2^t=\mathbb {R}\setminus D_1^t. \end{aligned}$$

We have

$$\begin{aligned} \int _{\mathbb {R}}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx =\int _{D_1^t}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx + \int _{D_2^t}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx. \end{aligned}$$

(26)

From (25), we obtain

$$\begin{aligned} \int _{D_1^t}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx \le C. \end{aligned}$$

(27)

We have that

$$\begin{aligned} \int _{D_2^t}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx \le \frac{1}{\delta _0^4} \int _{\mathbb {R}}(e^{4\alpha u^2}-1)\,dx. \end{aligned}$$

(28)

We suppose that

$$\begin{aligned} \Vert u\Vert \le k, \end{aligned}$$

where k is a positive constant. Thus, using Proposition 2.1 of [18], we infer

$$\begin{aligned} \int _{D_2^t}\left( \frac{e^{2\alpha u^2}-1}{u^2}\right) ^2\,dx \le C. \end{aligned}$$

(29)

Therefore, from (24), (26)–(29), we conclude that

$$\begin{aligned} \int _{\mathbb {R}}\frac{e^{2\alpha u^2}-1}{u^2}\,dx \le C, \end{aligned}$$

(30)

provided $\Vert u\Vert \le k$. Analogously,

$$\begin{aligned} \int _{\mathbb {R}}\frac{e^{2\alpha v^2}-1}{v^2}\,dx \le C, \end{aligned}$$

(31)

provided $\Vert v\Vert \le k$.

Therefore, if $\Vert u\Vert ,\Vert v\Vert \le k$, then (24), (30), and (31) allow us to conclude that

$$\begin{aligned} \Vert f(u)-f(v)\Vert _2 \le C(k)\Vert u-v\Vert , \end{aligned}$$

(32)

i.e., f is local Lipschitz. Thus, using Theorem 7.2 of [6], we can enunciate the following theorem

Theorem 2.1

[Local existence] Assume that (H1)-(H4) hold. If $(u_0,u_1)\in D(A)$, then there exists $T\le \infty$ such that (1) has a unique strong solution $(u,u_t)\in D(A)$ on the interval [0, T). Furthermore, if $(u_0,u_1)\in \mathcal {H}$, then (1) has a unique generalized solution $(u,u_t)\in \mathcal {H}$ on the interval [0, T). In both cases we have that $\lim _{t\rightarrow T^-} \Vert u(t)\Vert =\infty$ provided $T<\infty$.

Let u be the local solution of (1)$_1$ given by Theorem 2.1. We define the energy of (1) by

$$\begin{aligned} E(t)=E(u(t),u_t(t))=\frac{1}{2}\Vert u_t(t)\Vert _2^2+I(u(t)). \end{aligned}$$

Multiplying (1) by $u_t$ and integrating on $\mathbb {R}$, we have the following identity

$$\begin{aligned} E(t)=E(0), \end{aligned}$$

(33)

for all t in the interval of existence of solution u.

Now, we are going to enunciate a lemma which gives us that the ground state holds on the Pohozaev manifold. Its proof is omitted.

Lemma 2.1

Let $\varphi \in H^{\frac{1}{2}}(\mathbb {R})$ be the ground state of (7), then

$$\begin{aligned} I(\varphi )=m=\inf _{u\in \mathcal {P}}I(u)>0. \end{aligned}$$

Next result gives us another characteristic of the ground state level and its proof is an adaptation of the analogous one of Jeanjean and Le Coz [13].

Lemma 2.2

Let $\varphi \in H^{\frac{1}{2}}(\mathbb {R})$ be the ground state of (7), then

$$\begin{aligned} m=\min \{T(v);\,v\in H^{\frac{1}{2}}(\mathbb {R})\backslash \{0\},\,P(v)\ge 0\}, \end{aligned}$$

where $\displaystyle T(v)=\frac{1}{2}\int _{\mathbb {R}}|(-\Delta )^{\frac{1}{4}}v|^2\,dx$.

Proof

Let $v\in H^{\frac{1}{2}}(\mathbb {R})$ be such that $P(v)\ge 0$. If $P(v)=0$, then $v\in \mathcal {P}$. Thus, using Lemma 2.1, we have

$$\begin{aligned} \inf _{P(v)=0}T(v)=\inf _{v\in \mathcal {P}}I(v)=m. \end{aligned}$$

On the other hand, if $P(v)>0$, then for each $\beta >0$, we define

$$\begin{aligned} v_{\beta }(x)=\sqrt{\beta }v(\beta x). \end{aligned}$$

We claim that there exists $\beta _0<1$ such that

$$\begin{aligned} P(v_{\beta _0})=0. \end{aligned}$$

Using (9), we have that

$$\begin{aligned} P(v_{\beta }) \le \frac{\epsilon -\lambda _0}{2} \int _{\mathbb {R}}v_{\beta }^2\,dx +C(s_1)\int _{\mathbb {R}}(e^{\alpha v_{\beta }^2}-1)\,dx. \end{aligned}$$

(34)

Since

$$\begin{aligned} \Vert v_{\beta }\Vert _2=\Vert v\Vert _2, \end{aligned}$$

for all $\beta >0$, then taking $\epsilon =\frac{\lambda _0}{2}$ in (34) we obtain

$$\begin{aligned} P(v_{\beta }) \le -\frac{\lambda _0\Vert v\Vert _2^2}{4} +\frac{C(s_1)}{\beta } \int _{\mathbb {R}}(e^{\alpha \beta v^2}-1)\,dx. \end{aligned}$$

(35)

Observing that

$$\begin{aligned} \frac{1}{\beta }(e^{\alpha \beta v^2}-1) \le e^{\alpha v^2}-1, \end{aligned}$$

(36)

for $0<\beta <1$, and using Proposition 1.2, we have that $e^{\alpha v^2}-1\in L^1(\mathbb {R})$. Moreover, it holds

$$\begin{aligned} \frac{1}{\beta }(e^{\alpha \beta v^2}-1) \rightarrow \alpha v^2, \end{aligned}$$

a.e. in $\mathbb {R}$, as $\beta \rightarrow 0$. Thus, we can use the Lebesgue convergence theorem and to conclude that

$$\begin{aligned} \frac{1}{\beta }\int _{\mathbb {R}}(e^{\alpha \beta v^2}-1)\,dx \rightarrow \int _{\mathbb {R}}\alpha v^2\,dx, \end{aligned}$$

(37)

as $\beta \rightarrow 0$. From (35) and (37), we have

$$\begin{aligned} P(v_{\beta }) \le \left( \frac{3\alpha C(s_1)}{2}-\frac{\lambda _0}{4}\right) \Vert v\Vert _2^2, \end{aligned}$$

(38)

for $\beta$ small enough. Using the assumption $\lambda _0>6\alpha C(s_1)$ in (38), we obtain

$$\begin{aligned} P(v_{\beta })<0, \end{aligned}$$

for $\beta$ small enough. Therefore, there exists $\beta _0$ such that $P(v_{\beta _0})=0$ and the claim is proved. Therefore,

$$\begin{aligned} \inf _{u\in \mathcal {P}}I(u)\le I(v_{\beta _0})=T(v_{\beta _0})\le T(v). \end{aligned}$$

$\square$

Lemma 2.3

Let $(u_{0},u_{1})\in \mathcal {I}$ and let u(t) be the associated solution of (1) defined in [0, T). Then $(u(t),u_t(t))\in \mathcal {I}$ for all $t\in [0,T)$.

Proof

We suppose that the conclusion is not true. Thus, there exists $t_1\in (0,T)$ such that $P(u(t_1))\le 0$. From this and since $P(u_0)>0$, there exists $t_0\in (0,t_1)$ such that

$$\begin{aligned} P(u(t_0))=0. \end{aligned}$$

(39)

On the other hand, Lemma 2.2 gives us that

$$\begin{aligned} T(v) \ge m \end{aligned}$$

on $\{v\in H^{\frac{1}{2}}(\mathbb {R})\backslash \{0\},\, P(v)>0\}$. Therefore, by continuity, we infer

$$\begin{aligned} T(u(t_0)) \ge m>0, \end{aligned}$$

and this implies that $u(t_0)\ne 0$. Thus, (39) gives us that $u(t_0)\in \mathcal {P}$ and since

$$\begin{aligned} I(u(t_0)) \le E(u(t_0),u_t(t_0) )<m, \end{aligned}$$

we have a contradiction with Lemma 2.1. $\square$

Lemma 2.4

Let $(u_{0},u_{1})\in \mathcal {S}$ and let u(t) be the associated solution of (1) defined in [0, T). Then $(u(t),u_t(t))\in \mathcal {S}$ for all $t\in [0,T)$.

Proof

Analogous to the proof of Lemma 2.3. $\square$

Lemma 2.5

Let $(u_{0},u_{1})\in \mathcal {I}$ and let u(t) be the associated solution of (1) defined in [0, T). Then there exists $\delta >0$ such that $P(u(t))>\delta$, for all $t\in [0,T)$.

Proof: See [5] Lemma 2.6. $\square$

Finally, the last lemma of this section, which is used in the proof of the instability result.

Lemma 2.6

Let $\varphi \in H^{\frac{1}{2}}(\mathbb {R})$ be a ground state of (7). For all $\eta >0$ there exists $\varphi _{\eta }\in H^{\frac{1}{2}}(\mathbb {R})$ such that

$$\begin{aligned} \Vert \varphi -\varphi _{\eta }\Vert _{\frac{1}{2}}<\eta ,\quad I(\varphi _{\eta })<I(\varphi ),\quad \text{ and }\quad P(\varphi _{\eta })>0. \end{aligned}$$

Proof: For each $\lambda ,\mu >0$, we define $\displaystyle \varphi _{\lambda ,\mu }(x)=\lambda \varphi \Big (\frac{x}{\mu }\Big )$. Observing the definition of the functional $I(\varphi )$ and that $\varphi \in \mathcal {P}$, we obtain

$$\begin{aligned} \frac{\partial }{\partial \lambda }I(\varphi _{\lambda ,\mu }) =\lambda \int _{\mathbb {R}} |(-\Delta )^\frac{1}{4}\varphi |^2 \,dx +\lambda _0\lambda \mu \int _{\mathbb {R}}\varphi ^2\,dx +\mu \int _{\mathbb {R}}f(\lambda \varphi )\varphi \,dx. \end{aligned}$$

(40)

Multiplying (7) by $\varphi$ and integrating over $\mathbb {R}$, we obtain

$$\begin{aligned} \int _{\mathbb {R}} |(-\Delta )^\frac{1}{4}\varphi |^2 \,dx = \int _{\mathbb {R}} f(\varphi )\varphi \, dx -\lambda _0\int _{\mathbb {R}}\varphi ^2 \,dx. \end{aligned}$$

(41)

Taking $\lambda =1$ in (40) and observing (41), we obtain

$$\begin{aligned} \frac{\partial }{\partial \lambda }I(\varphi _{\lambda ,\mu })_{|_{\lambda =1}} =(1-\mu ) \int _{\mathbb {R}} |(-\Delta )^\frac{1}{4}\varphi |^2 \,dx. \end{aligned}$$

(42)

Thus, for $\mu >1$ there exists $\lambda _{\mu }>0$ such that

$$\begin{aligned} \frac{\partial }{\partial \lambda }I(\varphi _{\lambda ,\mu }) <0, \end{aligned}$$

(43)

for $\lambda \in (1-\lambda _{\mu },1+\lambda _{\mu })$. Therefore, since $P(\varphi )=0$, we have

$$\begin{aligned} I(\varphi _{\lambda ,\mu }) < \frac{1}{2}\int _{\mathbb {R}} |(-\Delta )^\frac{1}{4}\varphi |^2 \,dx =T(\varphi ) =I(\varphi ), \end{aligned}$$

(44)

for $\lambda \in (1,1+\lambda _{\mu })$.

On the other hand, since

$$\begin{aligned} P(\varphi _{\lambda ,\mu }) = \mu \int _{\mathbb {R}}\left( F(\lambda \varphi (x)) -\frac{\lambda _0\lambda ^2}{4}\varphi ^2(x)\right) \,dx \end{aligned}$$

and observing (41), we infer

$$\begin{aligned} \frac{\partial }{\partial \lambda }P(\varphi _{\lambda ,\mu })_{|_{\lambda =1}} =\mu \int _{\mathbb {R}} |(-\Delta )^\frac{1}{4}\varphi |^2 \,dx >0. \end{aligned}$$

(45)

Thus, for all $\mu >0$ there exists $\lambda _{\mu }$ such that

$$\begin{aligned} \frac{\partial }{\partial \lambda }P(\varphi _{\lambda ,\mu }) >0, \end{aligned}$$

(46)

for $\lambda \in (1-\lambda _{\mu },1+\lambda _{\mu })$. Therefore,

$$\begin{aligned} P(\varphi _{\lambda ,\mu }) >0, \end{aligned}$$

(47)

for $\lambda \in (1,1+\lambda _{\mu })$. From (44) and (47) we conclude the proof. $\square$

3 Proof of Lemma 1.1

In this section we prove the main result. The first lemma is analogous to Lemma 3.1 of [5] with one adjust of the level of (49). The strategy to prove Lemma 1.1 is multiply the equation (1)$_1$ by $u_x x\Psi _{\varepsilon ,x}(1)$ and to integrate on $\mathbb {R}$. This gives us the following identity

$$\begin{aligned}{} & {} \int _{\mathbb {R}}u_{tt}(x,t)u_x x\Psi _{\varepsilon ,x}(1)\,dx\nonumber \\{} & {} \quad +\frac{1}{2}\int _{\mathbb {R}^2}(u(x,t)-u(y,t))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdy\nonumber \\{} & {} \quad +\lambda _0\int _{\mathbb {R}}u(x,t)u_x x\Psi _{\varepsilon ,x}(1)\,dx -\int _{\mathbb {R}}f(u)u_x x\Psi _{\varepsilon ,x}(1)\,dx =0. \end{aligned}$$

(48)

After this, it is necessary to estimate each term of the left hand side of (48). The section is organized of the following way: the second integral of (48) is estimate in Lemma 3.6, but some intermediate calculations are made in Lemmas 3.2–3.5. Lemmas 3.2 and 3.3 are not proved here, their proofs are analogous to the ones of [5]. Lemmas 3.4 and 3.5 have some motification in their proofs and we make it here. In Lemma 3.7 we estimate some terms generate by the two last integrals of (48). In Lemma 3.8 we control some terms called here of ‘boundary terms’. They are generated by the use of derivative rules. In Lemma 3.9 we write an additional result (when it is compared with [5]) which is generated by the adaptation of Carrião, Lehrer, and Vicente ideas to the one dimensional case. Finally, the last result of this section is the proof of Lemma 1.1.

For each $\varepsilon >0$, we define $\Phi _{\varepsilon }:\mathbb {R}\rightarrow \mathbb {R}$ by

$$\begin{aligned} \Phi _{\varepsilon }(x)= \left\{ \begin{array}{lll} \displaystyle 1 &{} \text{ if } &{} 0\le |x|\le \exp (\frac{1}{\varepsilon }),\\ \displaystyle 2-\varepsilon ln (|x|)&{} \text{ if } &{} \exp (\frac{1}{\varepsilon })<|x|\le \exp (\frac{2}{\varepsilon }),\\ 0 &{} \text{ if } &{} |x|>\exp (\frac{2}{\varepsilon }). \end{array} \right. \end{aligned}$$

(49)

Now, for each $\varepsilon >0$ and $x\in \mathbb {R}$, we define the function $\varphi _{\varepsilon ,x}:\mathbb {R}^+\rightarrow \mathbb {R}$ which depends on one extra parameter, given by

$$\begin{aligned} \varphi _{\varepsilon ,x}(\lambda )=\Phi _{\varepsilon }(\lambda x). \end{aligned}$$

We define the function $\Psi _{\varepsilon , x}:\mathbb {R}^+\rightarrow \mathbb {R}$ by

$$\begin{aligned} \Psi _{\varepsilon ,x}(\lambda )=\frac{1}{\lambda }\int _0^{\lambda }{\xi }\varphi _{\varepsilon ,x}(\xi )d\xi . \end{aligned}$$

We are going to use the notation $\frac{d}{d\lambda } \Psi _{\varepsilon ,x}(\lambda )=\Psi _{\varepsilon ,x}'(\lambda )$.

Lemma 3.1

For each $\varepsilon >0$, $\lambda >0$ and $x\in \mathbb {R}$, we have

$$\begin{aligned} \Psi _{\varepsilon ,x}'(\lambda )+\frac{1}{\lambda }\Psi _{\varepsilon ,x}(\lambda )=\varphi _{\varepsilon ,x}(\lambda ); \end{aligned}$$

(50)

$$\begin{aligned} \frac{d}{d\lambda }\left( \lambda \Psi _{\varepsilon ,x}(\lambda )\right) =\lambda \varphi _{\varepsilon ,x}(\lambda ); \end{aligned}$$

(51)

$$\begin{aligned} \Vert \Psi _{\varepsilon ,x}'-\frac{1}{\lambda }\Psi _{\varepsilon ,x}\Vert _{L^{\infty }}<\varepsilon ; \end{aligned}$$

(52)

$$\begin{aligned} \left| \int _0^{1}\xi ^2\Phi _{\varepsilon }'(\xi x)x\,d\xi \right| \le \varepsilon . \end{aligned}$$

(53)

Moreover, for each $\varepsilon >0$ and $x\in \mathbb {R}$, we infer

$$\begin{aligned} \Psi _{\varepsilon ,x}'(1)+\Psi _{\varepsilon ,x}(1)=\Phi _{\varepsilon }(x); \end{aligned}$$

(54)

$$\begin{aligned} \Big (\frac{d}{d\lambda }\Psi _{\varepsilon ,\frac{x}{\lambda }}(\lambda )\Big )_{\lambda =1}= \Psi _{\varepsilon ,x}'(1)-\int _0^{1}\xi ^2\Phi _{\varepsilon }' ( \xi x) xd\xi ; \end{aligned}$$

(55)

$$\begin{aligned} |x\Psi _{\varepsilon , x}(1)|\le \frac{1}{4}\left[ 2+\varepsilon \exp \Big (\frac{2}{\varepsilon }\Big )\right] . \end{aligned}$$

(56)

Proof

See [5], Lemma 3.1. $\square$

Let $\lambda$ be a real number. To prove the results below we use an auxiliary function defined by

$$\begin{aligned} u_{\lambda }(x,t)=u(\lambda x,t). \end{aligned}$$

Lemma 3.2

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. It holds

$$\begin{aligned}{} & {} \int _{\mathbb {R}^2}(u(x,t)-u(y,t))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdy\nonumber \\{} & {} \quad =-\frac{1}{2}\int _{\mathbb {R}^2}(u(x,t)-u(y,t))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdy\nonumber \\{} & {} \qquad -\int _{\mathbb {R}^2}(u(x,t)-u(y,t))(\Psi _{\varepsilon ,x}(1) - \Psi _{\varepsilon ,y}(1)) u_x x k(x-y)\,dxdy\nonumber \\{} & {} \qquad +\frac{1}{2}\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )k(z-w)\,dzdw\nonumber \\{} & {} \qquad -\frac{1}{2}\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2\int _0^{1}\xi ^2\varphi _{\varepsilon ,z}' (\xi z) zd\xi \,k(z-w)\,dz\,dw\nonumber \\{} & {} \qquad +\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2\Psi _{\varepsilon ,z}(1) k(z-w)\,dzdw. \end{aligned}$$

(57)

Proof

See [5] Lemma 5.1. $\square$

In next results (Lemma 3.3, 3.4, 3.6, 3.7, and 3.9) $\delta$ is the constant gives by Lemma 2.5.

Lemma 3.3

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. Given $T_1\in [0,T)$ there exists a $\varepsilon _0>0$ such that

$$\begin{aligned} -\frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(x,\tau )-u(y,\tau ))(\Psi _{\varepsilon ,x}(1) - \Psi _{\varepsilon ,y}(1)) u_x x k(x-y)\,dx\,dy\,d\tau \ge -\frac{\delta }{8}t,\nonumber \\ \end{aligned}$$

(58)

for all $t\le T_1$ and for all $\varepsilon <\varepsilon _0$.

Proof

See [5] Lemma 5.2. $\square$

Lemma 3.4

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. Given $T_1\in [0,T)$ there exists a $\varepsilon _1>0$ such that for all $\varepsilon <\varepsilon _1$ it holds

$$\begin{aligned}{} & {} -\frac{1}{4}\int _0^t\int _{\mathbb {R}^2}(u(x,\tau )-u(y,\tau ))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdyd\tau \nonumber \\{} & {} \quad +\frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(z,\tau )-u(w,\tau ))^2\Psi _{\varepsilon ,z}(1) k(z-w)\,dzdwd\tau \ge -\frac{\delta }{8}, \end{aligned}$$

(59)

for all $t\in [0,T_1]$.

Proof

From $\varphi _{\varepsilon ,x}$ definition, we have

$$\begin{aligned} -\frac{1}{4}(u(x,t)-u(y,t))^2\varphi _{\varepsilon ,x}(1)k(x-y) \rightarrow -\frac{1}{4}(u(x,t)-u(y,t))^2k(x-y), \end{aligned}$$

(60)

a.e. in $\mathbb {R}^{2}\times (0,T_1)$, as $\varepsilon \rightarrow 0$. Moreover,

$$\begin{aligned}{} & {} \Big | -\frac{1}{4}(u(x,t)-u(y,t))^2\varphi _{\varepsilon ,x}(1)k(x-y) \Big |\nonumber \\{} & {} \quad \le \Big (\frac{1}{4}(u(x,t)-u(y,t))^2k(x-y)\Big )\in L^1(\mathbb {R}^{2}\times (0,T_1)). \end{aligned}$$

(61)

From (60), (61) and Lebesgue’s dominated convergence theorem, we have

$$\begin{aligned} -\frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdyd\tau \end{aligned}$$

$$\begin{aligned} \rightarrow -\frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)\,dxdyd\tau , \end{aligned}$$

(62)

as $\varepsilon \rightarrow 0$. Therefore, there exists $\tilde{\varepsilon }_1>0$, such that for all $\varepsilon \le \tilde{\varepsilon }_1$, it holds

$$\begin{aligned}{} & {} \frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)[1-\varphi _{\varepsilon ,x}(1)]\,dxdyd\tau \nonumber \\{} & {} \quad =\Big |-\frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdyd\tau \nonumber \\{} & {} \qquad +\frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)\,dxdyd\tau \Big |\nonumber \\{} & {} \quad <\frac{\delta }{16}, \end{aligned}$$

(63)

where $\delta$ is given by Lemma 2.5. We observe that

$$\begin{aligned}{} & {} \Big |-\frac{1}{4}\int _0^{t}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdyd\tau \nonumber \\{} & {} \qquad + \frac{1}{4}\int _0^{t}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)\,dxdyd\tau \Big |\nonumber \\{} & {} \quad \le \frac{1}{4}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)[1-\varphi _{\varepsilon ,x}(1)]\,dxdyd\tau , \end{aligned}$$

(64)

for all $t\in [0,T_1]$. Thus,

$$\begin{aligned}{} & {} -\frac{1}{4}\int _0^{t}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2\varphi _{\varepsilon ,x}(1)k(x-y)\,dxdyd\tau \nonumber \\{} & {} \quad + \frac{1}{4}\int _0^{t}\int _{\mathbb {R}^{2}}(u(x,\tau )-u(y,\tau ))^2k(x-y)\,dxdyd\tau \ge -\frac{\delta }{16}, \end{aligned}$$

(65)

for all $\varepsilon \le \tilde{\varepsilon }_1$.

On the other hand, since $\Psi _{\varepsilon ,z}(1)\rightarrow \frac{1}{2}$, as $\varepsilon \rightarrow 0$, we have

$$\begin{aligned} \frac{1}{2}(u(z,t)-u(w,t))^2k(z-w)\Big |\Psi _{\varepsilon ,z}(1)-\frac{1}{2}\Big | \rightarrow 0, \end{aligned}$$

(66)

a.e. in $\mathbb {R}^{2}\times (0,T_1)$, as $\varepsilon \rightarrow 0$. Moreover,

$$\begin{aligned}{} & {} \Big |\frac{1}{2}(u(z,t)-u(w,t))^2k(z-w)\Big |\Psi _{\varepsilon ,z}(1)-\frac{1}{2}\Big |\Big |\nonumber \\{} & {} \quad \le \frac{1}{2}(u(z,t)-u(w,t))^2k(z-w) \in L^1(\mathbb {R}^{2}\times (0,T_1)). \end{aligned}$$

(67)

From (66), (67) and Lebesgue’s dominated convergence theorem, we have

$$\begin{aligned} \frac{1}{2}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2k (z-w)\Big |\Psi _{\varepsilon ,z}(1)-\frac{1}{2}\Big |\,dzdwd\tau \rightarrow 0, \end{aligned}$$

(68)

as $\varepsilon \rightarrow 0$. Thus, (68) allows us to conclude that there exists $\hat{\varepsilon }_1>0$, such that for all $\varepsilon \le \hat{\varepsilon }_1$, it holds

$$\begin{aligned}{} & {} \left| \frac{1}{2}\int _0^t\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2\Psi _{\varepsilon ,z}(1) k(z-w)\,dzdwd\tau \right. \nonumber \\{} & {} \qquad \left. -\frac{1}{4}\int _0^t\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2k(z-w)\,dzdwd\tau \right| \nonumber \\{} & {} \quad \le \frac{1}{2}\int _0^{T_1}\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2k(z-w)\Big |\Psi _{\varepsilon ,z}(1)-\frac{1}{2}\Big |\,dzdwd\tau \nonumber \\{} & {} \quad <\frac{\delta }{16}, \end{aligned}$$

(69)

for all $t\in [0,T_1]$ and $\varepsilon \le \hat{\varepsilon }_1$, where $\delta$ is given by Lemma 2.5. Thus,

$$\begin{aligned}{} & {} \frac{1}{2}\int _0^t\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2\Psi _{\varepsilon ,z}(1) k(z-w)\,dzdwd\tau \nonumber \\{} & {} \quad -\frac{1}{4}\int _0^t\int _{\mathbb {R}^{2}}(u(z,\tau )-u(w,\tau ))^2k(z-w)\,dzdwd\tau \ge -\frac{\delta }{16}, \end{aligned}$$

(70)

for all $t\in [0,T_1]$. Taking $\varepsilon _1=\min \{\tilde{\varepsilon }_1,\hat{\varepsilon }_1\}$ and combining (65) with (70), we have that (59) is proved. $\square$

Lemma 3.5

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. Given $T_1\in [0,T)$ it holds

$$\begin{aligned}{} & {} \frac{1}{4}\int _0^t\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )k(z-w)\,dzdwdt\nonumber \\{} & {} \qquad -\frac{1}{4}\int _0^t\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2 \int _0^{1}\xi ^2 \varphi _{\varepsilon ,z}' (\xi z) zd\xi \,k(z-w)\,dzdwdt\nonumber \\{} & {} \quad \ge -\frac{1}{2}K^2C_1^2\varepsilon t, \end{aligned}$$

(71)

for all $t\in [0,T_1]$ and for all $\varepsilon >0$.

Proof

The estimate (52) and the assumption (15) allow us to control the first integral of the right hand side of (71), i.e.,

$$\begin{aligned} \Big |\frac{1}{4}\int _0^t\int _{\mathbb {R}^2}(u(z,\tau )-u(w,\tau ))^2 \Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )k(z-w)\,dz\,dw\,d\tau \Big | \le \frac{K^2C_1^2}{4}\varepsilon t.\nonumber \\ \end{aligned}$$

(72)

For the second one, the assumption (15) and (53) allow us to conclude that

$$\begin{aligned} \Big |-\frac{1}{4}\int _0^t\int _{\mathbb {R}^2}(u(z,t)-u(w,t))^2 \int _0^{1}\xi ^2 \varphi _{\varepsilon ,z}' (\xi z) zd\xi \,k(z-w)\,dz\,dw\,dt\Big | \le \frac{K^2C_1^2}{4}\varepsilon t.\nonumber \\ \end{aligned}$$

(73)

$\square$

Lemma 3.6

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. Given $T_1\in [0,T)$ there exists a $\varepsilon _2>0$ such that for all $\varepsilon <\varepsilon _2$ it holds

$$\begin{aligned}{} & {} \frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(x,t)-u(y,t))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdydt\nonumber \\{} & {} \quad \ge -\frac{\delta }{8}t -\frac{\delta }{8}-\frac{1}{2}K^2C_1^2\varepsilon t \end{aligned}$$

(74)

for all $t\in [0,T_1]$.

Proof

Let $T_1\in [0,T)$ be an arbitrarily fixed number. Taking $\varepsilon <\varepsilon _2=\min \{\varepsilon _0,\varepsilon _1\}$ (where $\varepsilon _0$ and $\varepsilon _1$ was given by Lemma 3.3 and 3.4), respectively, we can combine Lemma 3.2, 3.3, and 3.4 to conclude that

$$\begin{aligned}{} & {} \frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(x,\tau )-u(y,\tau ))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdyd\tau \nonumber \\{} & {} \quad \ge \frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(z,\tau )-u(w,\tau ))^2\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )k(z-w)\,dz\,dw\,d\tau \nonumber \\{} & {} \qquad -\frac{1}{2}\int _0^t\int _{\mathbb {R}^2}(u(z,\tau )-u(w,\tau ))^2 \int _0^{1}\xi ^2\varphi _{\varepsilon ,z}' (\xi z) zd\xi \,k(z-w)\,dz\,dw\,d\tau \nonumber \\{} & {} \qquad -\frac{\delta }{8}t-\frac{\delta }{8}-\frac{1}{2}K^2C_1^2\varepsilon t, \end{aligned}$$

(75)

for all $t\in [0,T_1]$ and for all $\varepsilon <\varepsilon _2$. $\square$

Lemma 3.7

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. For each $T_1\in [0,T)$, there exists a $\varepsilon _3>0$ such that for all $\varepsilon <\varepsilon _3$, it holds

$$\begin{aligned} -\frac{\lambda _0}{2} \int _0^t\int _{\mathbb {R}} u^2(x,t) \varphi _{\varepsilon ,x}(1)dx\,d\tau + \int _0^t\int _{\mathbb {R}} F(u) \varphi _{\varepsilon ,x}(1) dx\,d\tau -\int _0^tP(u)\,d\tau \ge -\frac{\delta }{8}, \end{aligned}$$

for all $t\in [0,T_1]$.

Proof

Let $T_1\in [0,T)$ be an arbitrarily fixed number. From $\varphi _{\varepsilon ,x}$ definition, we have

$$\begin{aligned} \Big |-\frac{\lambda _0}{2} u^2(x,t) + F(u)\Big | \big |\varphi _{\varepsilon ,x}(1)-1\big | \rightarrow 0, \end{aligned}$$

(76)

a.e. in $\mathbb {R} \times (0,T_1)$, as $\varepsilon \rightarrow 0$. Moreover,

$$\begin{aligned} \Big |-\frac{\lambda _0}{2} u^2(x,t) + F(u)\Big |\big |\varphi _{\varepsilon ,x}(1)-1 \big | \le 2\Big (\frac{\lambda _0}{2} u^2(x,t) + |F(u)|\Big )\in L^1(\mathbb {R}\times (0,T_1)).\nonumber \\ \end{aligned}$$

(77)

From (76), (77) and Lebesgue’s dominated convergence theorem, we have

$$\begin{aligned} \int _0^{T_1}\int _{\mathbb {R}}\Big |-\frac{\lambda _0}{2} u^2(x,t) + F(u)\Big |\big |\varphi _{\varepsilon ,x}(1)-1\big | \,dx\,d\tau \rightarrow 0, \end{aligned}$$

(78)

as $\varepsilon \rightarrow 0$. Therefore, there exists $\varepsilon _3>0$ such that for all $\varepsilon <\varepsilon _3$, it holds

$$\begin{aligned} \Big |\int _0^t\int _{\mathbb {R}}\Big (-\frac{\lambda _0}{2} u^2(x,\tau ) + F(u)\Big )\varphi _{\varepsilon ,x}(1) \,dx\,d\tau - \int _0^tP(u)\,d\tau \Big | \end{aligned}$$

$$\begin{aligned} \le \int _0^{T_1}\int _{\mathbb {R}}\Big |-\frac{\lambda _0}{2} u^2(x,\tau ) + F(u)\big |\varphi _{\varepsilon ,x}(1)-1\big | \,dx\,d\tau <\frac{\delta }{8}, \end{aligned}$$

(79)

for all $t\in [0,T_1]$, where $\delta$ is given by Lemma 2.5. Thus, the lemma is proved. $\square$

Due the use of product derivative rule, in the proof of Lemma 1.1 arise some terms involving the derivative with respect to the additional variable $\lambda$. These terms are estimate in next lemma.

Lemma 3.8

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. Given $T_1\in [0,T)$, it holds

$$\begin{aligned}{} & {} \frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx\,d\tau -\frac{1}{2}\int _0^t\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}}\,dx\,d\tau \nonumber \\{} & {} \quad -\int _0^t\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx\,d\tau - \int _0^t\int _{\mathbb {R}} F(u) \Psi _{\varepsilon ,z}(1) dz\,d\tau \nonumber \\{} & {} \quad - \frac{1}{2}\int _0^t\int _{\mathbb {R}} u_t^2 \Psi _{\varepsilon ,z}(1) dz\,d\tau +\frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}} u^2 (z,\tau )\Psi _{\varepsilon ,z}(1) dz\,d\tau \ge -C_3\varepsilon t, \end{aligned}$$

(80)

for all $t\in [0,T_1]$ and for all $\varepsilon >0$, where $C_3$ is given by (16).

Proof

We denote the three first integrals of the left hand side of (80) by $I_1,I_2$ and $I_3$, respectively. Now, we are going to estimate these integrals.

Estimate of $I_1$: Making the change of variable $z=\lambda x$, we obtain

$$\begin{aligned} \int _{\mathbb {R}}u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \,dx =\int _{\mathbb {R}}u^2(z,t) \Psi _{\varepsilon ,\frac{z}{\lambda }}(\lambda )\,dz. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\lambda _0}{2}\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx =\frac{\lambda _0}{2}\int _{\mathbb {R}}\Big (u^2(z,t)\frac{d}{d\lambda }\Psi _{\varepsilon ,\frac{z}{\lambda }}(\lambda )\Big )_{|_{\lambda =1}}\,dz. \end{aligned}$$

(81)

Using (55) in (81), we have

$$\begin{aligned}{} & {} \frac{\lambda _0}{2}\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx =\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(z,t)\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )\,dz\nonumber \\{} & {} \quad +\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(z,\tau )\Psi _{\varepsilon ,z}(1)\,dz -\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(z,t)\Big (\int _0^{1}\xi ^2 \Phi _{\varepsilon }' (\xi z) zd\xi \Big )\,dz. \end{aligned}$$

(82)

The estimate (52) and the assumption (15) ($\Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {R})}\le K$, for all $t\in [0,T)$), allow us to control the first integral of the right hand side of (82), i.e.,

$$\begin{aligned} \Big |\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(z,t)\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )\,dz\Big | \le C_1^2K^2\varepsilon . \end{aligned}$$

(83)

For the second one, again the assumption (15) and (53) allow us to conclude that

$$\begin{aligned} \Big |\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(z,t)\Big (\int _0^{1}\xi ^2 \Phi _{\varepsilon }' (\xi z)zd\xi \Big )\,dz\Big | \le C_1^2K^2\varepsilon . \end{aligned}$$

(84)

Combining (82)–(84), we infer

$$\begin{aligned} I_1 \ge -2C_1^2K^2\varepsilon t +\frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}}u^2(z,\tau )\Psi _{\varepsilon ,z}(1)\,dz d\tau . \end{aligned}$$

(85)

Estimate of $I_2$: Making the change of variable $z=\lambda x$, we obtain

$$\begin{aligned} -\frac{1}{2}\int _{\mathbb {R}}u_{\lambda _t}^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \,dx =-\frac{1}{2}\int _{\mathbb {R}}u_t^2(z,t)\Psi _{\varepsilon ,\frac{z}{\lambda }}(\lambda ) \,dz. \end{aligned}$$

Therefore, working as (82), we obtain

$$\begin{aligned}{} & {} -\frac{1}{2}\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}}\,dx =-\frac{1}{2}\int _{\mathbb {R}}u_t^2(z,t)\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big )\,dz \\{} & {} -\frac{1}{2}\int _{\mathbb {R}}u_t^2(z,t)\Psi _{\varepsilon ,z}(1)\,dz +\frac{1}{2}\int _{\mathbb {R}}u_t^2(z,t)\Big (\int _0^{1}\xi ^2 \Phi _{\varepsilon }' (\xi z) zd\xi \Big )\,dz. \end{aligned}$$

Observing the definitions of the energy and I(u), and (9) and (12), we have

$$\begin{aligned} \Vert u_t(t)\Vert _2^2\le & {} 2E(t)+\Vert u(t)\Vert _{H^{\frac{1}{2}}(\mathbb {R})}^2 +2\int _{\mathbb {R}}\Big (\frac{\epsilon }{2}u^2 +C(s_1)(e^{\alpha u^2}-1)\Big )\,dx\nonumber \\\le & {} 2E(t) + [1+(\epsilon +2C(s_1)H_{\alpha })\lambda _0^{-1}]C_1^2\Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {R})}^2 \end{aligned}$$

(86)

a.e. in [0, T). Since $(u(t),u_t(t))\in \mathcal {I}$ for all $t\in [0,T)$, we have that $E(t)<m$. Using this and (15) in (86), we conclude that

$$\begin{aligned} \Vert u_t(t)\Vert _2^2 \le 2m + [1+(\epsilon +2C(s_1)H_{\alpha })\lambda _0^{-1}]C_1^2K^2, \end{aligned}$$

(87)

a. e. in [0, T).

Then, making estimates as (72) and (73), we can conclude that

$$\begin{aligned} I_2\ge & {} -\frac{1}{2}\Big \{2m + [1+(\epsilon +2C(s_1)H_{\alpha })\lambda _0^{-1}]C_1^2K^2 \Big \}\varepsilon t\nonumber \\{} & {} \quad -\frac{1}{2}\int _0^t\int _{\mathbb {R}}u_t^2(z,\tau )\Psi _{\varepsilon ,z}(1)\,dzd\tau . \end{aligned}$$

(88)

Estimate of $I_3$: Making the change of variable $z=\lambda x$, we obtain

$$\begin{aligned} -\int _{\mathbb {R}}F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \,dx =-\int _{\mathbb {R}}F(u(z,t))\Psi _{\varepsilon ,\frac{z}{\lambda }}(\lambda ) \,dz \end{aligned}$$

thus,

$$\begin{aligned}{} & {} -\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}}\,dx =-\int _{\mathbb {R}}F(u(z,t))\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big ) \,dz\nonumber \\{} & {} \quad +\int _{\mathbb {R}}F(u(z,t))\Big (\int _0^{1}\xi ^2 \Phi _{\varepsilon }' (\xi z) zd\xi \Big ) \,dz -\int _{\mathbb {R}}F(u(z,t))\Psi _{\varepsilon ,z}(1) \,dz. \end{aligned}$$

(89)

The assumption (9) and the estimate (52) give us

$$\begin{aligned} \left| \int _{\mathbb {R}}F(u(z,t))\Big (\Psi _{\varepsilon ,z}'(1)-\Psi _{\varepsilon ,z}(1)\Big ) \,dz\right| \le \int _{\mathbb {R}}\frac{\epsilon }{2} |u(z,t)|^2+C(s_1)(e^{\alpha u^2}-1)\,dz \end{aligned}$$

$$\begin{aligned} \le (\epsilon +C(s_1)H_{\alpha })\lambda _0^{-1}C_1^2K^2\varepsilon . \end{aligned}$$

(90)

Analogously, (9) and (53) allow us to infer

$$\begin{aligned} \left| \int _{\mathbb {R}}F(u(z,t))\Big (\int _0^{1}\xi ^2 \Phi _{\varepsilon }' (\xi z)zd\xi \Big ) \,dz \right| \le (\epsilon +C(s_1)H_{\alpha })\lambda _0^{-1}C_1^2K^2\varepsilon . \end{aligned}$$

(91)

Thus, from (89), (90) and (91) we have

$$\begin{aligned} I_3 \ge -2(\epsilon +C(s_1)H_{\alpha })\lambda _0^{-1}C_1^2K^2\varepsilon t -\int _{\mathbb {R}}F(u(z,t))\Psi _{\varepsilon ,z}(1) \,dz. \end{aligned}$$

(92)

$\square$

Lemma 3.9

Let $u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}$ be the solution of (1) given by Theorem 2.1. For each $T_1\in [0,T)$, there exists a $\varepsilon _4>0$ such that for all $\varepsilon <\varepsilon _4$, it holds

$$\begin{aligned} \int _0^t\int _{\mathbb {R}} F(u) \Psi _{\varepsilon ,z}(1) dz\,d\tau -\frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}} u^2 \Psi _{\varepsilon ,z}(1) dz\,d\tau -\frac{1}{2}\int _0^tP(u)d\tau \ge -\frac{\delta }{8}, \end{aligned}$$

for all $t\in [0,T_1]$.

Proof

Analogously to the proof of Lemma 3.7. $\square$

3.1 Proof of Lemma 1.1

Multiplying (1) by $u_x x\Psi _{\varepsilon ,x}(1)$ and integrating on $\mathbb {R}$, we have

$$\begin{aligned} \int _{\mathbb {R}}u_{tt}(x,t)u_x x\Psi _{\varepsilon ,x}(1)\,dx +\lambda _0\int _{\mathbb {R}}u(x,t)u_x x\Psi _{\varepsilon ,x}(1)\,dx -\int _{\mathbb {R}}f(u)u_x x\Psi _{\varepsilon ,x}(1)\,dx \end{aligned}$$

$$\begin{aligned} +\frac{1}{2}\int _{\mathbb {R}^2}(u(x,t)-u(y,t))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdy =0. \end{aligned}$$

(93)

We denote the integrals of (93) by $I_4,I_5,I_6,$ and $I_7$.

Estimate of $I_4$: We observe that

$$\begin{aligned} I_4 = \int _{\mathbb {R}}u_{tt}(x,t)u_x x\Psi _{\varepsilon ,x}(1)\,dx =\frac{\partial }{\partial t}\int _{\mathbb {R}}u_{t}u_x x \Psi _{\varepsilon ,x}(1)\,dx -\int _{\mathbb {R}} u_{t} u_{tx}x \Psi _{\varepsilon ,x}(1)dx.\nonumber \\ \end{aligned}$$

(94)

Since $u_{\lambda }(x,t)=u(\lambda x,t)$, it follows

$$\begin{aligned} -\frac{1}{2}\int _{\mathbb {R}} \Big (\frac{d}{d\lambda }u_{\lambda _t}^2\Big )\Psi _{\varepsilon ,x}(\lambda ) \lambda dx \end{aligned}$$

$$\begin{aligned} = \frac{1}{2}\int _{\mathbb {R}} u_{\lambda _t}^2\frac{d}{d\lambda }\Big (\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )dx -\frac{1}{2}\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )dx. \end{aligned}$$

(95)

We have that

$$\begin{aligned} \Big (\frac{1}{2}\frac{d}{d\lambda }u_{\lambda _t}^2(x,t)\Big )_{|_{\lambda =1}} =u_t(x,t)u_{tx}(x,t)x. \end{aligned}$$

(96)

Taking $\lambda =1$ in (95), using (96) into (95) and combining the resultant equation with (94), we have

$$\begin{aligned} I_4 =\frac{\partial }{\partial t}\int _{\mathbb {R}}u_{t}u_x x \Psi _{\varepsilon ,x}(1)\,dx +\frac{1}{2}\int _{\mathbb {R}} u_t^2 \frac{d}{d\lambda }\Big (\Psi _{\varepsilon ,x}(\lambda )\lambda \Big )_{|_{\lambda =1}} dx \end{aligned}$$

$$\begin{aligned} -\frac{1}{2}\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}} dx. \end{aligned}$$

(97)

From (97) and using (51), we have the following expression

$$\begin{aligned} I_4 =\frac{\partial }{\partial t}\int _{\mathbb {R}}u_{t}u_x x \Psi _{\varepsilon ,x}(1)\,dx +\frac{1}{2}\int _{\mathbb {R}} u_t^2 \varphi _{\varepsilon ,x}(1) dx -\frac{1}{2}\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}} dx.\nonumber \\ \end{aligned}$$

(98)

Estimate of $I_5$: From the product derivative rule, we have

$$\begin{aligned}{} & {} \frac{\lambda _0}{2}\int _{\mathbb {R}}\Big (\frac{d}{d\lambda }u_{\lambda }^2(x,t)\Big )\Psi _{\varepsilon ,x}(\lambda ) \lambda \,dx\\{} & {} =-\frac{\lambda _0}{2}\int _{\mathbb {R}}u_{\lambda }^2(x,t)\frac{d}{d\lambda }\Big (\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )\,dx +\frac{\lambda _0}{2}\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )\,dx. \end{aligned}$$

Taking $\lambda =1$, observing that

$$\begin{aligned} \frac{1}{2}\Big (\frac{d}{d\lambda }u_{\lambda }^2(x,t)\Big )_{|_{\lambda =1}} =u(x,t)u_x(x,t) x \end{aligned}$$

and considering (51), we infer

$$\begin{aligned} I_5 =-\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(x,t)\varphi _{\varepsilon ,x}(1)\,dx +\frac{\lambda _0}{2}\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx. \end{aligned}$$

(99)

Estimate of $I_6$: Observing that $\frac{d}{d\lambda }F(u_{\lambda })=f(u_{\lambda })u_{\lambda x}x$, we have

$$\begin{aligned} \int _{\mathbb {R}}f(u_{\lambda })u_{\lambda x}x \Psi _{\varepsilon ,x}(\lambda ) \lambda \,dx = -\int _{\mathbb {R}}F(u_{\lambda })\frac{d}{d\lambda }\Big (\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )\,dx +\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )\,dx. \end{aligned}$$

Taking $\lambda =1$ and using (51) we obtain the following identity

$$\begin{aligned} I_6 = -\int _{\mathbb {R}}F(u)\varphi _{\varepsilon ,x}(1)\,dx +\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx. \end{aligned}$$

(100)

Replacing (98), (99) and (100) into (93) we infer

$$\begin{aligned}{} & {} \frac{\partial }{\partial t}\int _{\mathbb {R}}u_{t}u_x x \Psi _{\varepsilon ,x}(1)\,dx +\frac{1}{2}\int _{\mathbb {R}} u_t^2 \varphi _{\varepsilon ,x}(1) dx -\frac{1}{2}\int _{\mathbb {R}} \frac{d}{d\lambda }\Big ( u_{\lambda _t}^2\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{|_{\lambda =1}}dx\\{} & {} \quad +\frac{1}{2}\int _{\mathbb {R}^2}(u(x,t)-u(y,t))\Big (u_x x\Psi _{\varepsilon ,x}(1) -u_y y\Psi _{\varepsilon ,y}(1) \Big )k(x-y)\,dxdy\\{} & {} \quad -\frac{\lambda _0}{2}\int _{\mathbb {R}}u^2(x,t)\varphi _{\varepsilon ,x}(1)\,dx +\frac{\lambda _0}{2}\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (u_{\lambda }^2(x,t)\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx \end{aligned}$$

$$\begin{aligned} +\int _{\mathbb {R}}F(u)\varphi _{\varepsilon ,x}(1)\,dx -\int _{\mathbb {R}}\frac{d}{d\lambda }\Big (F(u_{\lambda })\Psi _{\varepsilon ,x}(\lambda ) \lambda \Big )_{\lambda =1}\,dx=0. \end{aligned}$$

(101)

Let $T_1\in [0,T)$ be an arbitrarily fixed number. Integrating from 0 to $t<T_1$ and observing Lemma 3.6 and Lemma 3.8, we infer

$$\begin{aligned}{} & {} -\int _{\mathbb {R}}u_{t}(x,t) u_x(x,t) x \Psi _{\varepsilon ,x}(1)\,dx +\int _{\mathbb {R}}u_{t}(x,0) u_x(x,0) x \Psi _{\varepsilon ,x}(1)\,dx\nonumber \\{} & {} \quad \ge -\frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}}u^2(x,\tau )\varphi _{\varepsilon ,x}(1)\,dx\,d\tau +\int _0^t\int _{\mathbb {R}}F(u)\varphi _{\varepsilon ,x}(1)\,dx\,d\tau \nonumber \\{} & {} \qquad +\int _0^t\int _{\mathbb {R}} F(u) \Psi _{\varepsilon ,z}(1) dz\,d\tau + \frac{1}{2}\int _0^t\int _{\mathbb {R}} u_t^2 \Psi _{\varepsilon ,z}(1) dz\,d\tau -\frac{\lambda _0}{2}\int _0^t\int _{\mathbb {R}} u^2 \Psi _{\varepsilon ,z}(1) dz\,d\tau \nonumber \\{} & {} \qquad +\frac{1}{2}\int _0^t\int _{\mathbb {R}} u_t^2 \varphi _{\varepsilon , x}(1)dx\,d\tau -\frac{\delta }{8}t -\frac{\delta }{8}-\left( \frac{1}{2}C_1^2K^2+C_3\right) \varepsilon t, \end{aligned}$$

(102)

for all $t\in [0,T_1]$ and for all $\varepsilon <\varepsilon _2$. This estimate, Lemma 3.7, and Lemma 3.9 give us

$$\begin{aligned}{} & {} -\int _{\mathbb {R}}u_{t}(x,t)u_x(x,t) x \Psi _{\varepsilon ,x}(1)\,dx +\int _{\mathbb {R}}u_{t}(x,0) u_x(x,0) x \Psi _{\varepsilon ,x}(1)\,dx\nonumber \\{} & {} \quad \ge -\frac{\delta }{8}t-\frac{3\delta }{8} -\left( \frac{1}{2}C_1^2K^2+C_3\right) \varepsilon t +\frac{3}{2}\int _0^tP(u)\,dt, \end{aligned}$$

(103)

for all $t\in [0,T_1]$ and for all $\varepsilon <\min \{\varepsilon _2,\varepsilon _3,\varepsilon _4\}$.

The estimate (103) and Lemma 2.5 allow us to conclude

$$\begin{aligned}{} & {} -\int _{\mathbb {R}}u_{t}(x,t)u_x(x,t) x \Psi _{\varepsilon ,x}(1)\,dx +\int _{\mathbb {R}}u_{t}(x,0)u_x(x,0) x \Psi _{\varepsilon ,x}(1)\,dx\nonumber \\{} & {} \quad \ge -\frac{3\delta }{8}+\left[ \frac{11\delta }{8}-\left( \frac{1}{2}C_1^2K^2+C_3\right) \varepsilon \right] t, \end{aligned}$$

(104)

for all $t\in [0,T_1]$ and for all $\varepsilon <\min \{\varepsilon _2,\varepsilon _3,\varepsilon _4\}$.

Choosing $\varepsilon <\min \left\{ \varepsilon _2,\varepsilon _3,\varepsilon _4,C_2\right\}$, where $C_2$ is the constant defined in (16), we infer

$$\begin{aligned} -\int _{\mathbb {R}}u_{t}(x,t)u_x(x,t) x \Psi _{\varepsilon ,x}(1)\,dx +\int _{\mathbb {R}}u_{t}(x,0) u_x(x,0) x \Psi _{\varepsilon ,x}(1)\,dx \ge -\frac{3\delta }{8}+\frac{\delta }{2}t,\nonumber \\ \end{aligned}$$

(105)

for all $t\in [0,T_1]$. Using the inequality of Cauchy-Schwarz, we obtain

$$\begin{aligned} \frac{\delta t}{2} \le \frac{3\delta }{8}+\Vert u_t(t)\Vert _2\Vert u_x(t) x\Psi _{C_2, x}\Vert _2 +\Vert u_1\Vert _2\Vert (u_0)_x x\Psi _{C_2, x}\Vert _2, \end{aligned}$$

for all $t\in [0,T_1]$. From this and since, from Lemma 3.1,

$$\begin{aligned} |x\Psi _{C_2, x}|_{\mathbb {R}} \le \frac{1}{4}\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] , \end{aligned}$$

we conclude that

$$\begin{aligned} t \le \frac{1}{2\delta }\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] \left[ \Vert u_t(t)\Vert _2\Vert u_x(t)\Vert _2+\Vert u_1\Vert _2\Vert (u_0)_x\Vert _2\right] +\frac{3}{4}, \end{aligned}$$

for all $t\in [0,T_1]$. $\square$

4 Blow Up, Instability and Global Existence

In this section we prove Theorem 1.1, 1.2 and 1.3. They are consequence of Lemma 1.1.

4.1 Proof of Theorem 1.1

The blow up result of the item a) is a consequence of $T<\infty$. Now, we suppose that $T=\infty$. We prove this by contradiction. We suppose that there exists a constant $K>0$ such that

$$\begin{aligned} \Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {R})} \le K, \end{aligned}$$

(106)

for all $t\ge 0$. By the identity of energy (33) we have

$$\begin{aligned} \frac{1}{2}\Vert u_t(t)\Vert _2^2=E(0)-\frac{1}{2}\Vert u(t)\Vert _{H^{\frac{1}{2}}(\mathbb {\mathbb {R}}^n)}^2+\int _{\mathbb {R}}F(u)dx. \end{aligned}$$

(107)

Analogously to (86) we have

$$\begin{aligned} \Vert u_t(t)\Vert _2^2 \le 2E(0) + [1+(\epsilon +2C(s_1)H_{\alpha })\lambda _0^{-1}]C_1^2\Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {R})}^2, \end{aligned}$$

(108)

for all $t\ge 0$. Since $E(0)<m$, from (106) and (108) we obtain

$$\begin{aligned} \Vert u_t(t)\Vert _2^2\le 2m + [1+(\epsilon +2C(s_1)H_{\alpha })\lambda _0^{-1}]C_1^2K^2:=C_5, \end{aligned}$$

(109)

for all $t\ge 0$. Moreover, from the continuous inclusion $H^{\frac{3}{2}}(\mathbb {R})\hookrightarrow H^1(\mathbb {R})$, we have

$$\begin{aligned} \Vert u_x(t)\Vert _2^2\le C_4\Vert u(t)\Vert _{H^{\frac{3}{2}}(\mathbb {\mathbb {R}}^n)}^2 \le C_6K^2, \end{aligned}$$

(110)

for all $t\ge 0$.

On the other hand, using (106) we can apply Lemma 1.1 with

$$\begin{aligned} T_1=2\left\{ \frac{1}{2\delta }\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] \left[ C_5C_6K^2+\Vert u_1\Vert _2\Vert (u_0)_x\Vert _2\right] +\frac{3}{4}\right\} \end{aligned}$$

to infer that

$$\begin{aligned} t \le \frac{1}{2 \delta }\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] \left[ \Vert u_t(t)\Vert _2\Vert u_x(t)\Vert _2+\Vert u_1\Vert _2\Vert (u_0)_x\Vert _2\right] +\frac{3}{4}, \end{aligned}$$

(111)

for all $t\in [0,T_1]$. Combinning (109)–(111) we conclude that

$$\begin{aligned} t \le \frac{1}{2\delta }\left[ 2+C_2\exp \Big (\frac{2}{C_2}\Big )\right] \left[ C_5C_6K^2+\Vert u_1\Vert _2\Vert (u_0)_x\Vert _2\right] +\frac{3}{4}=\frac{T_1}{2}, \end{aligned}$$

(112)

for all $t\in [0,T_1]$, which is a contradiction. $\square$

4.2 Proof of Theorem 1.2

Let $\eta >0$ and $\varphi _{\eta }$ be given by Lemma 2.6. As $H^{\frac{3}{2}}(\mathbb {R})$ is dense in $H^{\frac{1}{2}}(\mathbb {R})$, there exists $u_{0,\eta }\in H^{\frac{3}{2}}(\mathbb {R})$ such that

$$\begin{aligned} \Vert u_{0,\eta }-\varphi _{\eta }\Vert _{H^{\frac{1}{2}}(\mathbb {R})}<\frac{\eta }{2}. \end{aligned}$$

Then

$$\begin{aligned} \Vert u_{0,\eta }-\varphi \Vert _{H^{\frac{1}{2}}(\mathbb {R})}<\eta . \end{aligned}$$

Let u be the solution of (1) associated to the initial data $(u_{0,\eta },0)$. Then the initial energy satisfies

$$\begin{aligned} E(0)<m\quad \text{ and }\quad P(u_{0,\eta })>0. \end{aligned}$$

Therefore, the initial energy is below of the level m and the couple $(u_{0,\eta },0)$ is in $\mathcal {I}$, thus the result follows from Theorem 1.1. $\square$

Finally, we prove our global existence theorem.

4.3 Proof of Theorem 1.3

It is enough to estimate the $H^{\frac{1}{2}}(\mathbb {R})$ norm. We observe that

$$\begin{aligned} \frac{1}{2}\Vert (-\Delta )^{\frac{1}{4}} u(t)\Vert _2^2-P(u)=I(u(t)). \end{aligned}$$

(113)

As $(u_0,u_1)\in \mathcal {S}$, then for all t, $(u(t),u_t(t))\in \mathcal {S}$ and $P(u(t))\le 0$. Therefore, from (113) we have

$$\begin{aligned} \Vert (-\Delta )^{\frac{1}{4}} u(t)\Vert _2^2\le I(u(t)) \le E(t)=E(0)\le m, \end{aligned}$$

(114)

for all $t\ge 0$. Now, we observe that

$$\begin{aligned} \Vert u(t)\Vert _2\le CE(t)=CE(0)\le Cm, \end{aligned}$$

this and (114) give us the result. $\square$