On the Global Well-Posedness and Orbital Stability of Standing Waves for the Schrödinger Equation with Fractional Dissipation

Abstract

In this paper, we are concerned with the nonlinear fractional Schrödinger equation. We extend the result of Guo and Huo and prove that the Cauchy problem of the nonlinear fractional Schrödinger equation is global well-posed in $H^{\frac{3}{2}-\gamma }\left(\mathbb{R}\right)$ with $\frac{1}{2}\le \gamma <1$. In view of the complexity of the nonlinear fractional Schrödinger equation itself, the local smoothing effect and maximal function estimates are not enough for presenting the global well-posedness for the nonlinear fractional Schrödinger equation. In this paper, we use a suitably iterative scheme and complete the global well-posed result for Equation (R). Moreover, we obtain the orbital stability of standing waves for the above equations via establishing the profile decomposition of bounded sequences in $H^s\left({\mathbb{R}}^N\right)$ ($0<s<1$) with $N\ge 2$.

1. Introduction

The Schrödinger equation with fractional dissipation was first introduced by Laskin in [1,2]. It is well known that equations involving the fractional Laplacian operator ${(\wedge )}^{\gamma }$ attract global attention and appear in several areas such as fluid mechanics, solid state physics, plasma physics, ultra-relativistic limits of quantum, biology, water waves modeling, and so on (see [3,4,5,6,7,8,9]).

In [10], the one-dimensional nonlinear fractional Schrödinger equation was studied by Guo and Huo. In their paper, they showed that the Cauchy problem is global well-posed in $L^2\left(\mathbb{R}\right)$ with $\frac{1}{4}<\gamma <\frac{1}{2}$ for the above equation. Afterwards, they proved the local well-posedness of the fractional nonlinear Schrödinger equation (R) in the subcritical Sobolev space $H^s\left(\mathbb{R}\right)$ for $s>n/2-\gamma /2$ in [11]. Gui and Liu [12] studied the Camassa–Holm equation with fractional dissipation, as follows:

$${\partial }_{t}v+u{\partial }_{x}v+2v{\partial }_{x}u+\nu {(-\Delta )}^{\gamma /2}v=0,v=(1-{\alpha }^2{\partial }_x^2)u,(t,x)\in {\mathbb{R}}^{+}\times \mathbb{R}.$$

(1)

The authors proved that Equation (1) is global well-posed. Moreover, they presented a blow-up of solutions for Equation (1). Later, Wang and Zhang [13] considered the fractional BBM equation:

$$\left\{\begin{array}{c}{u}_t+D^{\alpha }{u}_t+u_x+u{u}_x=0,\hfill \\ {u(t,x)|}_{t=0}=u_0\left(x\right),\hfill \end{array}\right.$$

(2)

and the authors used the contraction principle and presented the local well-posedness of problem (2) in $H^s$ with $s\ge max\{0,\frac{3}{2}-\alpha \}$. Moreover, they extended the solution of problem (2) from the local one to a global one by using the I-method. Zhang and Zhu [14] considered the nonlinear fractional Schrödinger equation arising from pseudorelativistic Boson stars

$$i{u}_t={(-\Delta +k^2)}^{\alpha }u-(\frac{1}{|x{\mid }^{\gamma }}\ast \mid u{\mid }^2)u,t\ge 0,x\in {\mathbb{R}}^N,N\ge 2,$$

(3)

where i is the imaginary unit, $u(x,t)$ is a complex-valued function on ${\mathbb{R}}^N\times \mathbb{R}$, and the convolution ∗ is defined by $\frac{1}{{\left|x\right|}^{\gamma }}\ast {\left|u\right|}^2\left(x\right):=\int \frac{{\left|u\left(y\right)\right|}^2}{{|x-y|}^{\gamma }}dy$. They proved that the standing waves are orbitally stable in the case of $0<\gamma <2\alpha$, the ground state solitary waves are strongly unstable to blow up in the case of $\gamma =2\alpha$. Recently, Gui and Wang [15] established the local well-posedness for the Navier–Stokes equations with the 3D vacuum free boundary via the conormal derivative. In [16], Nguyen and Tuan et al. revealed that the Sobolev-Galpern type equations with fractional dissipation is global well-posed by combining the Cauchy sequence technique and Orlicz space. Li and Zheng et al. proved orbital stability of periodic standing waves for the coupled Klein–Gordon–Zakharov equations in [17]. There are other results of global well-posedness and stability of standing waves, such as the derivative nonlinear Schrödinger equation [18], the Moser–Trudinger type Schrödinger equation with fractional dissipation [19], the cubic fractional nonlinear Schrödinger equation [20], the logarithmic Klein–Gordon Equation [21], and so on. For the recent related studies of the fractional Schrödinger equation, we can refer to the works about dissipative character of asymptotics [22], the large time asymptotic behavior of solutions [23], blow-up criteria and instability of normalized standing waves [24], bound state solutions [25], the classification of single traveling wave solutions [26], and so on.

Motivated by the problems described above, we extend the result of Guo and Huo in [11] and prove that the Cauchy problem (R) is global well-posed in $H^{\frac{3}{2}-\gamma }\left(\mathbb{R}\right)$ with $\frac{1}{2}\le \gamma <1$. In the process of our work, the common methods (the local smoothing effect and maximal function estimates) are not enough for presenting the global well-posedness for equation (R) in the space $H^{\frac{3}{2}-\gamma }\left(\mathbb{R}\right)$. Fortunately, the iterative scheme and the crucial energy estimates can replace the common methods (the local smoothing effect and maximal function estimates). Moreover, by establishing the profile decomposition of bounded sequences in $H^s\left({\mathbb{R}}^N\right)$ ($0<s<1$) with $N\ge 2$, the orbital stability of standing waves for problem (R) can be obtained. In the following, we give the concrete process.

This paper is organized as follows. Section 2 introduces some preliminaries. According to a suitably iterative scheme, we obtain the global well-posedness of problem (R) in Section 3. Section 4 proves the orbital stability of standing waves for equation (R), and Section 5 presents the conclusions of our work.

In the sequel, the notations: $c_i$, $C_i$, C denote different positive constants whose exact values are inessential, $L^p:=L^p\left({\mathbb{R}}^n\right)$, $B_{p,r}^s:=B_{p,r}^s\left({\mathbb{R}}^n\right)$, ${\dot{B}}_{p,r}^s:={\dot{B}}_{p,r}^s\left({\mathbb{R}}^n\right)$, $H^s:=H^s\left({\mathbb{R}}^n\right)$, ${\dot{H}}^s:={\dot{H}}^s\left(\mathbb{R}\right)$, $\int f\left(x\right)dx:={\int }_{{\mathbb{R}}^n}f\left(x\right)dx$, ${\parallel f\parallel }_q:={\parallel f\parallel }_{L^q\left({\mathbb{R}}^n\right)}$, $S^{{}^{\prime }}:=S^{{}^{\prime }}\left({\mathbb{R}}^n\right)$, $S_h^{{}^{\prime }}:=S_h^{{}^{\prime }}\left({\mathbb{R}}^n\right)$.

2. Preliminaries

In this section, we fist introduce the Chemin–Lerner type spaces ${\tilde{L}}_T^{\lambda }\left(B_{p,r}^s\left({\mathbb{R}}^n\right)\right)$, the Besov spaces, and Littlewood–Paley decomposition. In addition, we give some other useful results.

Definition 1

([27] Chemin–Lerner spaces). Set $s\in \mathbb{R},r\ge 1,\lambda \ge 1,p\ge 1,T>0$. Then the norm of ${\tilde{L}}_T^{\lambda }\left(B_{p,r}^s\right)$ is defined by

$${\displaystyle {\parallel f\parallel }_{{\tilde{L}}_T^{\lambda }\left(B_{p,r}^s\right)}=(\sum _{n\in \mathbb{Z}}{2}^{nrs}({\int }_0^T\parallel {\Delta }_n{f\left(t\right)\parallel }_{L^p}^{\lambda }dt{{)}^{\frac{r}{\lambda }})}^{\frac{1}{r}}<\infty .}$$

Remark 1.

For short, we denote the Chemin–Lerner type spaces ${\tilde{L}}_T^{\lambda }\left(B_{p,r}^s\right)$ by ${\tilde{L}}_T^{\lambda }\left(H^s\right)$ with $p=r=2$. Similarly, the space ${\tilde{L}}_T^{\lambda }\left(H^s\right)$ replaces the space $C([0,T],B_{p,r}^s)\cap {\tilde{L}}_T^{\lambda }\left(B_{p,r}^s\right)$.

In the following, we insert Bony’s decomposition [28] frequently into the both homogeneous and inhomogeneous context:

$$\begin{array}{c}fg={\dot{T}}_{f}g+\dot{R}(f,g)={\dot{T}}_{f}g+{\dot{T}}_{g}f+\dot{\Re }(f,g),\\ fg=T_{f}g+R(f,g)=T_{f}g+T_{g}f+\Re (f,g),\end{array}$$

(4)

where

$$\begin{array}{c}{\dot{T}}_{f}g=\sum _{n\in \mathbb{Z}}{\dot{S}}_{n-1}f{\dot{\Delta }}_{n}g,\dot{R}(f,g)=\sum _{n\in \mathbb{Z}}{\dot{\Delta }}_{n}f{\dot{S}}_{n+2}g,\\ \dot{\Re }(f,g)=\sum _{n\in \mathbb{Z}}{\dot{\Delta }}_{n}f\widetilde{\dot{{\Delta }_n}}g,\widetilde{\dot{{\Delta }_n}}g=\sum _{|n^{{}^{\prime }}-n|\le 1}{\dot{\Delta }}_{n^{{}^{\prime }}}g,\end{array}$$

and similarly, the inhomogeneous versions of $T_{f}g,R(f,g)$ and $\Re (f,g)$ can be defined.

Remark 2.

Because of the complexity of the nonlinear fractional Schrödinger equation itself, we introduce Chemin–Lerner type spaces, which is crucial in proving Theorem (1).

Definition 2

([29] Besov spaces). The inhomogenous Besov space $B_{p,r}^s$ is defined by

$$B_{p,r}^s=\{f\in S^{{}^{\prime }}{:\parallel f\parallel }_{B_{p,r}^s}<\infty \}.$$

Similarly, the space ${\dot{B}}_{p,r}^s=\{f\in S_h^{{}^{\prime }}{:\parallel f\parallel }_{{\dot{B}}_{p,r}^s}<\infty \}$ denotes homogeneous Besov spaces. Then, we have the norms of f in inhomogenous and homogeneous Besov spaces $B_{p,r}^s,{\dot{B}}_{p,r}^s$, as follows:

$$\begin{array}{cc}\hfill & {\parallel f\parallel }_{B_{p,r}^s}=\left\{\begin{array}{c}(\sum _{j\in \mathbb{Z}}{2}^{jsr}\parallel {\Delta }_{j}f{{\parallel }_{L^p}^r)}^{\frac{1}{r}},r<\infty ,\\ \underset{j\in \mathbb{Z}}{sup}{2}^{js}{\parallel {\Delta }_{j}f\parallel }_{L^p},r=\infty .\end{array}\right.\hfill \\ \hfill & {\parallel f\parallel }_{{\dot{B}}_{p,r}^s}=\left\{\begin{array}{c}(\sum _{j\in \mathbb{Z}}{2}^{jsr}\parallel {\dot{\Delta }}_{j}f{{\parallel }_{L^p}^r)}^{\frac{1}{r}},r<\infty ,\\ \underset{j\in \mathbb{Z}}{sup}{2}^{js}{\parallel {\dot{\Delta }}_{j}f\parallel }_{L^p},r=\infty ,\end{array}\right.\hfill \\ \hfill & S_h^{{}^{\prime }}=\{f\in S^{{}^{\prime }}|\underset{j\to -\infty }{lim}{\dot{S}}_{j}f=0in{S}^{{}^{\prime }}\}.\hfill \end{array}$$

If $s=\infty$, we have $B_{p,r}^{\infty }={\cap }_{s\in \mathbb{R}}{B}_{p,r}^s$.

Definition 3

([29,30] Littlewood–Paley decomposition). Let $(\mu ,\nu )$ be a smooth valued set in $[0,1]$, where μ is supported in $C_1=\{\theta \in {\mathbb{R}}^n:\left|\theta \right|\le 4/3\}$, and ν is supported in $C_2=\{\theta \in {\mathbb{R}}^n:3/4\le \left|\theta \right|\le 8/3\}$. Then, we have

$$\mu \left(\theta \right)+\sum _{q\in \mathbb{N}}\nu \left(2^{-m}\theta \right)=1,\forall \theta \in {\mathbb{R}}^n,$$

and

$$\mathit{supp}\nu (2^{-m}.)\cap \mathit{supp}\nu (2^{-m^{\prime }}.)=\varnothing ,if|m-m^{\prime }|\ge 2,$$

$$\mathit{supp}\mu (.)\cap \mathit{supp}\mu (.)=\varnothing ,ifm\ge 1.$$

Setting $\varphi =:F^{-1}\nu ,\tilde{\varphi }=:F^{-1}\mu$. For all $j\in \mathbb{Z}$, setting the homogeneous and inhomogeneous dyadic blocks ${\dot{\Delta }}_n$, ${\Delta }_n$ by

$$\begin{array}{cc}\hfill & {\displaystyle {\dot{\Delta }}_{n}f=:\nu \left(2^{-n}D\right)f=2^{nd}\int \varphi \left(2^{n}y\right)f(x-y)dy,}\hfill \\ \hfill & {\Delta }_{n}f=:\nu \left(2^{-n}D\right)f=2^{nd}\int \varphi \left(2^{n}y\right)f(x-y)dy,(\forall n\ge 0),\hfill \\ \hfill & {\Delta }_{-1}f=:\mu \left(D\right)f=\int \tilde{\varphi }\left(y\right)f(x-y)dy,(\forall n\le -2).\hfill \end{array}$$

(5)

Similarly, we can define the homogeneous low-frequency cut-off operators ${\dot{S}}_n$ by

$${\dot{S}}_{n}f=:\sum _{n^{{}^{\prime }}\le n-1}{\Delta }_{n^{{}^{\prime }}}f,$$

(6)

and the inhomogeneous operators ${\dot{S}}_n,S_n$ are defined by

$$S_{n}f=:\sum _{n^{{}^{\prime }}\le n-1}{\Delta }_{n^{{}^{\prime }}}f=\mu \left(2^{-n}D\right)f=2^{nd}\int \tilde{\varphi }\left(2^{n}y\right)f(x-y)dy.$$

(7)

Lemma 1

([31] Berstein’s inequality). Let u be a complex polynomial of order n. If $u^{{}^{\prime }}$ is the derivative of u, then we have $ma{x}_{\left|\theta \right|\le 1}\left(\right|u^{{}^{\prime }}\left(\theta \right)\left|\right)\le n·ma{x}_{\left|\theta \right|\le 1}\left(\right|u\left(\theta \right)\left|\right).$

Lemma 2

([32] Gronwall’s inequality). Let $t\in [0,T]$, the nonnegative function $u\left(t\right)$ is absolutely continuous. For nonnegative integrable functions $s\left(t\right),v\left(t\right)$ on $t\in [0,T]$, if

$$u{\left(t\right)}^{{}^{\prime }}\le s\left(t\right)u\left(t\right)+v\left(t\right),$$

we have

$$u\left(t\right)\le e^{{\int }_0^{t}s\left(\tau \right)d\tau }[u\left(0\right)+{\int }_0^{t}v\left(\tau \right)d\tau ].$$

In particular, when $v\left(t\right)=0$ and $u\left(0\right)=0$, one has $u\left(t\right)\equiv 0,\forall t\in [0,T]$.

Lemma 3.

Let $s\in (\frac{1}{2},\frac{5}{4}),\sigma \in (\frac{1}{2},3-2s),T>0$. If $0\le t\le T$, $g\in {\tilde{L}}_T^1({\dot{H}}^{s-\frac{1}{2}}\cap {\dot{H}}^s)$, we have

$$\parallel g^3{\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-1}\right)}\le {\eta \parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-\frac{1}{2}}\right)}+C{\eta }^{-1}{\int }_0^t\parallel g^{}{\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4·{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-1}\right)}d\tau$$

(8)

where C depends continuously on s, and $\eta >0$ is any constant.

Proof. 

Using Bony’s decomposition, we split $g^3$ into

$$g^3=2{\dot{T}}_{g^2}·g+\dot{\Re }(g^2,g).$$

(9)

Thanks to Lemma 1, we deduce that for $\sigma \in (\frac{1}{2},1)$

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{\dot{T}}_{g^2}{·g\parallel }_{L_t^1\left(L^2\right)}=\parallel {\dot{\Delta }}_n\sum _{m\in \mathbb{Z}}{\dot{s}}_{m-1}{g}^2·{\dot{\Delta }}_m{g\parallel }_{L_t^1\left(L^2\right)}\le {\int }_0^t\sum _{|n-m|\le 4}{\parallel {\dot{S}}_{m-1}{g}^2·{\dot{\Delta }}_{m}g\parallel }_{L^2}d\tau \hfill \\ & & =\sum _{|n-m|\le 4}\parallel \sum _{j\le m-2}{\dot{\Delta }}_j{g}^2·{\dot{\Delta }}_m{g\parallel }_{L^2}\le \sum _{|n-m|\le 4}{\int }_0^t\sum _{j\le m-2}\parallel {\dot{\Delta }}_j{g}^2{\parallel }_{L^{\infty }}·{\parallel {\dot{\Delta }}_{m}g\parallel }_{L^2}d\tau \hfill \\ & & \le \sum _{|n-m|\le 4}{\int }_0^t\sum _{j\le m-2}{2}^{\frac{1}{2}j}\parallel {\dot{\Delta }}_j{g}^2{\parallel }_{L^2}·{\parallel {\dot{\Delta }}_{m}g\parallel }_{L^2}d\tau ,\hfill \end{array}$$

(10)

which implies that

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{\dot{T}}_{g^2}{·g\parallel }_{L_t^1\left(L^2\right)}\le \eta \sum _{|n-m|\le 4}{\parallel {\dot{\Delta }}_j{g}^2\parallel }_{L_t^1\left(L^2\right)}{2}^{(\sigma -\frac{1}{2})m}·\sum _{j\le m-2}{2}^{(1-\sigma )j}\hfill \\ & & +{\eta }^{-1}{\int }_0^t\sum _{|n-m|\le 4}\parallel {\dot{\Delta }}_j{g\parallel }_{L^2}{2}^{(\frac{1}{2}-\sigma )m}\sum _{j\le m-2}{2}^{\sigma j}{\parallel {\dot{\Delta }}_m{g}^2\parallel }_{L^2}d\tau \hfill \\ & & \le {\eta }^{-1}{\int }_0^t{\parallel g\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4\sum _{|n-m|\le 4}{\parallel {\dot{\Delta }}_{j}g\parallel }_{L^2}{2}^{(\frac{1}{2}-\sigma )m}·\sum _{j\le m-2}{2}^{(\sigma -\frac{1}{2})j}d\tau \hfill \\ & & +\eta \sum _{|n-m|\le 4}{\parallel {\dot{\Delta }}_{j}g\parallel }_{L_t^1\left(L^2\right)}{2}^{\frac{1}{2}m}.\hfill \end{array}$$

(11)

Then,

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{\dot{T}}_{g^2}{·g\parallel }_{L_t^1\left(L^2\right)}\le \eta c_n\left(\tau \right)2^{-(s-1)n}{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-\frac{1}{2}}\right)}\hfill \\ & & +{\eta }^{-1}{2}^{-(s-1)n}{\int }_0^t{\parallel g\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{c}_n\left(\tau \right){\parallel g\parallel }_{{\dot{H}}^{s-1}}d\tau .\hfill \end{array}$$

(12)

Therefore,

$$\parallel {\dot{\Delta }}_n{\dot{T}}_{g^2}{·g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-1}\right)}\le {\eta \parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-\frac{1}{2}}\right)}+C{\eta }^{-1}{\int }_0^t{\parallel g\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{\parallel g\parallel }_{{\dot{H}}^{s-1}}d\tau .$$

(13)

On the other hand, by Lemma 1, we yield that for $s>\frac{1}{2}$,

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n\dot{\Re }(g^2,g){\parallel }_{L_t^1\left(L^2\right)}=\parallel {\dot{\Delta }}_n\sum _{m\in Z}{\dot{\Delta }}_m{g}^2{\tilde{\dot{\Delta }}}_n{g\parallel }_{L_t^1\left(L^2\right)}\le {\int }_0^t{2}^{\frac{1}{2}n}\sum _{m\ge n-3}{\parallel {\dot{\Delta }}_m{g}^2{\tilde{\dot{\Delta }}}_{m}g\parallel }_{L^1}d\tau \hfill \\ & & \le {\int }_0^t{2}^{\frac{1}{2}n}\sum _{m\ge n-3}\parallel {\dot{\Delta }}_m{g}^2{\parallel }_{L^2}{\parallel {\tilde{\dot{\Delta }}}_{m}g\parallel }_{L^2}d\tau \hfill \\ & & \le {\int }_0^t\sum _{m\ge n-3}(\eta \parallel {\tilde{\dot{\Delta }}}_m{g\parallel }_{L^2}{2}^{\frac{1}{2}n}{)}^{\frac{1}{2}}({\eta }^{-1}\parallel {\dot{\Delta }}_m{g}^2{\parallel }_{L^2}^2\parallel {\tilde{\dot{\Delta }}}_{m}g{{\parallel }_{L^2}{2}^{\frac{1}{2}n})}^{\frac{1}{2}}d\tau .\hfill \end{array}$$

Therefore,

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n\dot{\Re }(g^2,g){\parallel }_{L_t^1\left(L^2\right)}\le {\int }_0^t\sum _{m\ge n-3}\eta \parallel {\tilde{\dot{\Delta }}}_m{g\parallel }_{L^2}{2}^{\frac{1}{2}n}d\tau +{\int }_0^t{\eta }^{-1}\parallel {\dot{\Delta }}_m{g}^2{\parallel }_{L^2}^2{\parallel {\tilde{\dot{\Delta }}}_{m}g\parallel }_{L^2}{2}^{\frac{1}{2}n}d\tau \hfill \\ & & \le {\eta }^{-1}{2}^{-(s-1)n}{\int }_0^t{\parallel g\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{c}_n{\left(\tau \right)\parallel g\parallel }_{{\dot{H}}^{s-1}}d\tau +\eta c_n\left(\tau \right)2^{-(s-1)n}{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-\frac{1}{2}}\right)}.\hfill \end{array}$$

This implies that

$$\parallel {\dot{\Delta }}_n\dot{\Re }(g^2,g){\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-1}\right)}\le {\eta \parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s-\frac{1}{2}}\right)}+C{\eta }^{-1}{\int }_0^t{\parallel g\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{\parallel g\parallel }_{{\dot{H}}^{s-1}}d\tau .$$

(14)

Therefore, we obtain (8) from (13) and (14) and prove Lemma (3). □

Lemma 4.

Let $f_0\left(x\right)\in {\dot{H}}^s,s\in (-\frac{1}{2},\frac{5}{4})$. If f is the smooth solution for the following equation

$$i{f}_t+v{\partial }_{x}f+{\wedge }^{\frac{1}{2}}f=g,(t,x)\in {\mathbb{R}}^{+}\times \mathbb{R},$$

(15)

then we can deduce the a priori estimates, as follows:

$${\parallel f\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{H}}^s\right)}\le \parallel f_0{\parallel }_{{\dot{H}}^s}+{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^s\right)}+{\int }_0^t\parallel {\partial }_x{v\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^2{\parallel f\parallel }_{{\dot{H}}^s},$$

(16)

$${\parallel f\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s+\frac{1}{2}}\right)}\le \parallel 2^{ns}(1-e^{-c_1{2}^{\frac{n}{2}t}})\parallel {\dot{\Delta }}_n{f}_0{\parallel }_{L^2}{\parallel }_{{\ell }^2\left(\mathbb{Z}\right)}+{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^s\right)}+{\int }_0^t\parallel {\partial }_x{v\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^2{\parallel f\parallel }_{{\dot{H}}^s}.$$

(17)

Proof. 

Applying the operator ${\dot{\Delta }}_n(n\in \mathbb{Z})$ to Equation (15), we have

$$i{\partial }_t{\dot{\Delta }}_{n}f+{\dot{S}}_{n-1}v{\dot{\Delta }}_n{\partial }_{x}f+{\wedge }^{\frac{1}{2}}{\dot{\Delta }}_{n}f={\dot{\Delta }}_{n}g+({\dot{S}}_{n-1}v-v){\dot{\Delta }}_n{\partial }_{x}f+[{\dot{\Delta }}_n,v]{\partial }_{x}f.$$

(18)

Multiplying both sides of (18) by ${\dot{\Delta }}_{n}f$, then integrating Equation (18) over $\mathbb{R}$, we have

$$\begin{array}{ccc}& & \frac{d\parallel {\dot{\Delta }}_n{f\parallel }_{L^2}}{dt}+c_1{2}^{\frac{n}{2}}\parallel {\dot{\Delta }}_n{f\parallel }_{L^2}\le \parallel {\dot{S}}_{n-1}{\partial }_x{v\parallel }_{L^{\infty }}{\parallel {\dot{\Delta }}_{n}f\parallel }_{L^2}\hfill \\ & & +\parallel [{\dot{\Delta }}_n,v]{\partial }_x{f\parallel }_{L^2}+\parallel ({\dot{S}}_{n-1}v-v){\dot{\Delta }}_n{\partial }_x{f\parallel }_{L^2}+{\parallel {\dot{\Delta }}_{n}g\parallel }_{L^2}.\hfill \end{array}$$

(19)

In view of Gronwall’s inequality,

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{f\parallel }_{L^2}\le e^{-c_1{2}^{\frac{n}{2}t}}\parallel {\dot{\Delta }}_n{f}_0{\parallel }_{L^2}+C{\int }_0^t{e}^{-c_1{2}^{\frac{n}{2}(t-\tau )}}[\parallel {\dot{S}}_{n-1}{\partial }_x{v\parallel }_{L^{\infty }}{\parallel {\dot{\Delta }}_{n}f\parallel }_{L^2}d\tau \hfill \\ & & +\parallel [{\dot{\Delta }}_n,v]{\partial }_x{f\parallel }_{L^2}+\parallel ({\dot{S}}_{n-1}v-v){\dot{\Delta }}_n{\partial }_x{f\parallel }_{L^2}+\parallel {\dot{\Delta }}_{n}g{\parallel }_{L^2}]d\tau .\hfill \end{array}$$

(20)

Next, integrating (20) on $\tau \in [0,t]$, we have

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{f\left(t\right)\parallel }_{L_t^1\left(L^2\right)}\le 2^{-\frac{n}{2}}(1-e^{-c_1{2}^{\frac{n}{2}t}})\parallel {\dot{\Delta }}_n{f}_0{{\parallel }_{L^2}+2^{-\frac{n}{2}}[\parallel {\dot{\Delta }}_n,v]{\partial }_{x}f\parallel }_{L_t^1\left(L^2\right)}\hfill \\ & & +{\int }_0^t\parallel {\dot{S}}_{n-1}{\partial }_x{v\parallel }_{L^{\infty }}\parallel {\dot{\Delta }}_n{f\parallel }_{L^2}d\tau +\parallel ({\dot{S}}_{n-1}v-v){\dot{\Delta }}_n{\partial }_x{f\parallel }_{L_t^1\left(L^2\right)}+\parallel {\dot{\Delta }}_{n}g{\parallel }_{L_t^1\left(L^2\right)}].\hfill \end{array}$$

(21)

According to Lemma 1 and Minkowski’s inequality, we infer that

$${\parallel f\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s+\frac{1}{2}}\right)}\le \parallel 2^{ns}(1-e^{-c_1{2}^{\frac{n}{2}t}})\parallel {\dot{\Delta }}_n{f}_0{\parallel }_{L^2}{\parallel }_{{\ell }^2\left(\mathbb{Z}\right)}+{\parallel g\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^s\right)}+{\int }_0^t\parallel {\partial }_x{v\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^2{\parallel f\parallel }_{{\dot{H}}^s}.$$

In view of (20),

$$\begin{array}{ccc}& & \parallel {\dot{\Delta }}_n{f\left(t\right)\parallel }_{L_t^{\infty }\left(L^2\right)}\le \parallel {\dot{\Delta }}_n{f}_0{\parallel }_{L^2}+{\int }_0^t\parallel {\dot{S}}_{n-1}{\partial }_x{v\parallel }_{L^{\infty }}{\parallel {\dot{\Delta }}_{n}f\parallel }_{L^2}d\tau \hfill \\ & & +[\parallel {\dot{\Delta }}_n,v]{\partial }_x{f\parallel }_{L_t^1\left(L^2\right)}+\parallel ({\dot{S}}_{n-1}v-v){\dot{\Delta }}_n{\partial }_x{f\parallel }_{L_t^1\left(L^2\right)}+{\parallel {\dot{\Delta }}_{n}g\parallel }_{L_t^1\left(L^2\right)}.\hfill \end{array}$$

This completes the proof of Lemma 4. □

Lemma 5.

Let $N\ge 2,0<s<1$. If $u_0\in H^s$, there exists a unique solution $u(t,x)$ of problem (R) and $u(t,x)\in C([0,T);H^s)\bigcap C^1([0,T);H^{-s})$. Moreover, for any $t\in [0,T)$, $u(t,x)$ satisfies

(a) Conservation of mass ${\parallel u\left(t\right)\parallel }_2={\parallel u_0\parallel }_2$;

(b) Conservation of energy $E\left(u\left(t\right)\right)=E\left(u_0\right)$.

Proof. 

The proof of Lemma 5 is similar to that of Definition 3 in [33] with just a minor modification. □

Lemma 6.

Let $N\ge 2,0<s<1$. If ${\left\{{\omega }_n\right\}}_{n=1}^{+\infty }\in H^s$ is a bounded sequence, there exists a subsequence still denoted ${\left\{{\omega }_n\right\}}_{n=1}^{+\infty }$, a family ${\left\{{\alpha }_n^j\right\}}_{j=1}^{+\infty }$ of sequences in ${\mathbb{R}}^N$, and a sequence ${\left\{{\omega }^j\right\}}_{n=1}^{+\infty }$ of $H^s$ functions satisfying

(a) For any $k\ne j$,

$$|{\alpha }_n^k-{\alpha }_n^j|\to +\infty asn\to +\infty .$$

(22)

(b) For any $l\ge 1,\alpha \in {\mathbb{R}}^N$,

$${\omega }_n\left(\alpha \right)=\sum _{j=1}^l{\omega }^j(\alpha -{\alpha }_n^j)+{\omega }_n^l\left(\alpha \right)$$

(23)

with

$$\underset{l\to +\infty }{lim}\underset{n\to +\infty }{lim}sup{\parallel {\omega }_n^l\parallel }_p=0,\forall p\in (2,\frac{2N}{{(N-2\alpha )}^{+}}),$$

(24)

where $\frac{2N}{{(N-2\alpha )}^{+}}=\frac{2N}{N-2\alpha }$ when $N\ge 3$; $\frac{2N}{{(N-2\alpha )}^{+}}=+\infty$ when $N=2$. Moreover, when $n\to +\infty$, we have

$$\parallel {\omega }_n{\parallel }_2^2=\sum _{j=1}^l\parallel {\omega }^j{\parallel }_2^2+\parallel {\omega }_n^l{\parallel }_2^2+o\left(1\right),\parallel {\omega }_n{\parallel }_{{\dot{H}}^s}^2=\sum _{j=1}^l\parallel {\omega }^j{\parallel }_{{\dot{H}}^s}^2+{\parallel {\omega }_n^l\parallel }_{{\dot{H}}^s}^2+o\left(1\right),$$

(25)

where ${lim}_{n\to \infty }o\left(1\right)=0$.

Proof. 

The proof of Lemma 6 is similar to that of Proposition 3.1 in [34], and we omit it. □

3. Global Well-Posedness of the Nonlinear Fractional Schrödinger Equation

In this section, it is enough to prove the case $\gamma =\frac{1}{2}$, because when $\gamma >\frac{1}{2}$, the energy of problem (R) is subcritical and the solution of problem (R) is smooth, unique, and local. Therefore, we only prove the following equation is global well-posed, as follows:

$$i{\partial }_{t}u+{\wedge }^{\frac{1}{2}}u+{\left|u\right|}^{2}u=0,(t,x)\in {\mathbb{R}}^{+}\times \mathbb{R},$$

(26)

with ${u(t,x)|}_{t=0}=u_0\left(x\right)\in H^s\left(\mathbb{R}\right)$, and we omit the proof for $\gamma \in (\frac{1}{2},1)$.

Theorem 1.

Let $s\ge 1,u_0\left(x\right)\in H^s$. Then there exists a time constant $T>0$, and for any $t\in [0,T]$, the solution $u(t,x)$ of Equation (26) is unique. Moreover, $u(t,x)$ belongs to the set $\tilde{C}([0,T];H^s)\cap {\tilde{L}}^1([0,T];H^{s+\frac{1}{2}})$.

Proof. 

In the following, we present a detailed proof process of the existence and uniqueness of the solution for Equation (26).

Step 1 Constructing the smooth approximate solutions of Equation (26).

Let $u_0^{\left(n\right)}:=S_{n+1}{u}_0$, where $S_{n+1}$ is defined in Definition 2. Then we have $u_0\in H^{\infty }$. Applying the iterative method to problem (26), we can deduce that there exists $u^{\left(n\right)}$ which satisfies the following equations:

$$\left\{\begin{array}{c}i{\partial }_t{u}^{\left(n\right)}+{\wedge }^{\frac{1}{2}}{u}^{\left(n\right)}+{\left|u^{\left(n\right)}\right|}^2{u}^{\left(n\right)}=0,(t,x)\in {\mathbb{R}}^{+}\times \mathbb{R},\hfill \\ u^{\left(n\right)}{(t,x)|}_{t=0}=u_0^{\left(n\right)}\left(x\right):=S_{n+1}{u}_0\left(x\right)\in H^s\left(\mathbb{R}\right)\hfill \end{array}\right.$$

(27)

for some $T^n>0$ and $s\ge 1$. Moreover, u is a unique local-in-time solution and belongs to the set $\tilde{C}([0,T^n];H^s)\cap {\tilde{L}}_{loc}^1([0,T^n];H^{s+2})$. In addition, u satisfies

$$u^{\left(n\right)}\in C((0,T^n)\times \mathbb{R}),\forall n\in \mathbb{N},$$

and

$$\parallel u^{\left(n\right)}{\left(t\right)\parallel }_{H^1}^2+2{\int }_0^t\parallel {\wedge }^{\frac{1}{4}}{u}^{\left(n\right)}{\left(\tau \right)\parallel }_{H^1}^{2}d\tau +2{\int }_0^t\parallel u^{\left(n\right)}{\left(\tau \right)\parallel }_{H^1}^{4}d\tau ={\parallel u^{\left(n\right)}\left(0\right)\parallel }_{H^1}^2,$$

(28)

for any $t>0$.

Step 2 Uniform estimates of $u^{\left(n\right)}$. In view of Lemma 4,

$$\parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+\parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\le C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}+{\parallel g\parallel }_{{\tilde{L}}_t^1\left(H^1\right)}.$$

By Lemma 3, we deduce that

$$\begin{array}{cc}& \parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\hfill \\ & \le C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}+C{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{\frac{3}{2}}\right)}\hfill \\ & +C{\int }_0^t\parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{H^1}d\tau .\hfill \end{array}$$

By inserting the embedding inequality $\parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}\le C{\parallel {\wedge }^{\frac{1}{4}}{u}^{\left(n\right)}\parallel }_{H^1}$ into the above inequality, we can denote that

$$\begin{array}{cc}\hfill & \parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\hfill \\ \hfill & \le C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}+C{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{\frac{3}{2}}\right)}\hfill \\ \hfill & +C{\int }_0^t\parallel {\wedge }^{\frac{1}{4}}{u}^{\left(n\right)}{\parallel }_{H^1}^4{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{H^1}d\tau .\hfill \end{array}$$

In view of (28) and Gronwall’s inequality,

$$\begin{array}{cc}& \parallel u^{\left(n\right)}{\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u^{\left(n\right)}\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\hfill \\ & \le C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}exp\left\{C{\int }_0^t{\parallel {\wedge }^{\frac{1}{4}}{u}^{\left(n\right)}\parallel }_{H^1}^{2}d\tau \right\}\hfill \\ & \le C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}exp\left\{C\parallel u_0^{\left(n\right)}{\parallel }_{H^1}^4\right\}d\tau .\hfill \end{array}$$

Therefore, we present that the approximate solution $u^{\left(n\right)}$ belongs to the set $\tilde{C}([0,T^n];$$H^1)\cap {\tilde{L}}^1([0,T^n];{\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{\frac{3}{2}})$. Moreover, it is also uniformly bounded in the above set.

Step 3 Convergence. In this step, we shall prove that there exists a time $T_1>0$ that satisfies

$${\left\{u^{\left(n\right)}\right\}}_{n\in \mathbb{N}}\mathrm{is}\mathrm{a}\mathrm{Cauchy}\mathrm{sequence}\mathrm{in}\tilde{C}([0,T_1];H^s)\cap L^2([0,T_1];{\dot{H}}^{\frac{1}{4}}\cap {\dot{H}}^{s+\frac{1}{4}}),\forall s\in (\frac{1}{2},\frac{3}{4}).$$

For arbitrary $n_1,n_2\in \mathbb{N}$, let $\omega :=u^{(n_1+n_2)}-u^{\left(n_2\right)}$. It follows from (27) that

$$i{\partial }_t\omega +{\wedge }^{\frac{1}{2}}\omega +|u^{(n_1+n_2)}{|}^2\omega +\left[\right|u^{(n_1+n_2)}{|}^2-|u^{\left(n_2\right)}{|}^2]u^{\left(n_2\right)}=0,$$

(29)

where ${\omega |}_{t=0}={\sum }_{k=n_2+1}^{n_1+n_2}{\dot{\Delta }}_n{u}_0$.

In view of Lemma 4,

$$\begin{array}{cc}& {\parallel \omega \parallel }_{{\tilde{L}}_t^{\infty }\left(H^s\right)}+{\parallel \omega \parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s+\frac{1}{2}}\right)}\le C(\sum _{k=n_2+1}^{n_1+n_2}{2}^{2ks}\parallel {\dot{\Delta }}_n{u}_0{\parallel }_{L^2}^2)\hfill \\ & {\displaystyle +{\int }_0^t\parallel u^{(n_1+n_2)}{\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^2{\parallel \omega \parallel }_{H^s}d\tau +{\int }_0^t\parallel u^{(n_1+n_2)}+u^{\left(n_2\right)}{\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}\parallel u^{\left(n_2\right)}{\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}{\parallel \omega \parallel }_{H^s}d\tau .}\hfill \end{array}$$

Thanks to Gronwall’s inequality and the above inequality, we have

$${\parallel \omega \parallel }_{{\tilde{L}}_t^{\infty }\left(H^s\right)}+{\parallel \omega \parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{s+\frac{1}{2}}\right)}\le (\sum _{k=n_2+1}^{n_1+n_2}{2}^{2ks}\parallel {\dot{\Delta }}_n{u}_0{\parallel }_{L^2}^2)exp\{C\parallel u_0{\parallel }_{H^1}^2\}.$$

(30)

Since $s\in (\frac{1}{2},\frac{3}{4})$ and $u_0\in H^1$, it follows that if $n\to \infty$, the summation ${\sum }_{k=n_2+1}^{n_1+n_2}{2}^{2ks}\parallel$${\dot{\Delta }}_n{u}_0{\parallel }_{L^2}^2$ uniformly goes to zero. From (30), we obtain that ${\left\{u^{\left(n\right)}\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence in the set $\tilde{C}([0,T_1];H^s)\cap L^2([0,T^n];{\dot{H}}^{\frac{1}{4}}\cap {\dot{H}}^{s+\frac{1}{4}})$.

Step 4 Checking that the limit satisfies Equation (26). According to the limit in (27), we know that $u\in {\tilde{L}}^1([0,T_1]$ satisfies Equation (26). Using Definition 3 ([27]) yields that u is a continuity, i.e., $u\in C([0,T_1];H^s)$. From Step 3, we get that ${\left\{u^n\right\}}_{n\in \mathbb{N}}$ is a Cauchy sequence and belongs to the set $\tilde{C}([0,T_1];H^s)\cap L^2([0,T_1];{\dot{H}}^{\frac{1}{4}}\cap {\dot{H}}^{s+\frac{1}{4}})$. Hence, for some limit, u is strongly convergent in

$$\tilde{C}([0,T_1];H^s)\cap L^2([0,T_1];{\dot{H}}^{\frac{1}{4}}\cap {\dot{H}}^{s+\frac{1}{4}}),\forall s\in (\frac{1}{2},\frac{3}{4}).$$

Since ${\left\{u^n\right\}}_{n\in N}\in \tilde{C}({\mathbb{R}}^{+};H^1)\cap {\tilde{L}}^1({\mathbb{R}}^{+};{\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{\frac{3}{2}})$ is uniformly bounded, from Step 3 and Fatou’s property, we have $u\in {\tilde{L}}^{\infty }([0,T_1];H^1)\cap {\tilde{L}}^1([0,T_1];{\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{\frac{3}{2}})$. From Definition 3 ([27]), we can infer that $u\in \tilde{C}([0,T_1];H^1)\cap {\tilde{L}}^1([0,T_1];{\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{\frac{3}{2}})$. Moreover, one has

$$u\in \tilde{C}((0,T_1)\times \mathbb{R})$$

and

$${\displaystyle {\parallel u\left(t\right)\parallel }_{H^1}^2+2{\int }_0^t\parallel {\wedge }^{\frac{1}{4}}{u\left(\xi \right)\parallel }_{H^1}^{2}d\xi +2{\int }_0^t{\parallel u\left(\xi \right)\parallel }_{H^1}^{4}d\xi ={\parallel u\left(0\right)\parallel }_{H^1}^2,\forall t\in [0,T_1].}$$

(31)

Step 5 Uniqueness. Similar to Step 3, it is not difficult to obtain that the solution of Equation (26) is unique.

Step 6 Regularity. For the initial data, $u_0\in H^s$ and $s\in (\frac{1}{2},\frac{3}{4})$. In view of Lemma 3 and Lemma 4,

$$\begin{array}{cc}\hfill & {\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^s\right)}+{\parallel u\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{s+\frac{1}{2}})}\hfill \\ & {\displaystyle \le C\parallel u_0{\parallel }_{H^s}+C{\int }_0^t{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{\parallel u\left(t\right)\parallel }_{H^s}d\tau .}\hfill \end{array}$$

Then, according to Gronwall’s inequality, (31) and (32), one has

$${\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^s\right)}+{\parallel u\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{1}{2}}\cap {\dot{H}}^{s+\frac{1}{2}})}\le C\parallel u_0{\parallel }_{H^1}exp\{\parallel u_0{\parallel }_{H^s}\}.$$

(32)

This proves Theorem 1. □

Theorem 2.

Let $u_0\in H^1$, assume that u is the solution of problem (26) in Theorem 1. If there exists the maximal time ${\tilde{T}}_{u_0}$ that is finite, we have

$${\displaystyle {\int }_0^{{\tilde{T}}_{u_0}}{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\tau =\infty .}$$

Proof. 

Substituting Lemma 4 into (26), we have

$$\begin{array}{cc}\hfill & {\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\hfill \\ & \le C\parallel u_0{\parallel }_{H^1}+{C\parallel \left|u\right|}^2{u\left(t\right)\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^1\right)}.\hfill \end{array}$$

According to (33) and Lemma 3, one has

$$\begin{array}{cc}& {\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\hfill \\ & \le C\parallel u_0{\parallel }_{H^1}+C{\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^1\left({\dot{H}}^{\frac{3}{2}}\right)}\hfill \\ & {\displaystyle +C{\int }_0^t{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^4{\parallel u\left(t\right)\parallel }_{H^1}d\tau .}\hfill \end{array}$$

In view of Gronwall’s inequality,

$${\displaystyle {\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^{\infty }\left(H^1\right)}+{\parallel u\left(t\right)\parallel }_{{\tilde{L}}_t^1({\dot{H}}^{\frac{3}{2}}\cap {\dot{H}}^{\frac{1}{2}})}\le C_0\parallel u_0{\parallel }_{H^1}exp\{C_0{\int }_0^t{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\tau \},\forall t\in [0,{\tilde{T}}_{u_0}.}$$

Hence, if ${\int }_0^{{\tilde{T}}_{u_0}}{\parallel u\left(t\right)\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\tau$ is finite, there exists $Q\left({\tilde{T}}_{u_0}\right)$ such that

$${\displaystyle {\parallel u\left(t\right)\parallel }_{H^1}\le Q\left({\tilde{T}}_{u_0}\right):=C_0\parallel u_0{\parallel }_{H^1}exp\{C_0{\int }_0^t{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\tau \}<\infty .}$$

Under the initial datum $u({\tilde{T}}_{u_0}-\frac{\tilde{T}}{2})$, substituting $T_1=T_1\left(Q\left({\tilde{T}}_{u_0}\right)\right)$ into Theorem 1 yields the solution $\tilde{u}\left(t\right)$ of problem (26) on $[0,\tilde{T})$ via using the local existence theory. By using uniqueness, $u(t+{\tilde{T}}_{u_0}-\frac{\tilde{T}}{2})$ is on $[0,\frac{\tilde{T}}{2})$ so that $\tilde{u}\left(t\right)$ extends $u\left(t\right)$ beyond ${\tilde{T}}_{u_0}$. This deduces a contradiction and proves Theorem 2. □

Theorem 3.

Let $\alpha =\frac{1}{4}$ and $u_0\in H_1$. Then, the time-solution u is unique, global for equation (R). Moreover, u belongs to $C({\mathbb{R}}^{+};H^1)\cap {\tilde{L}}_{loc}^1({\mathbb{R}}^{+};H^{\frac{3}{2}})$; if $t>0$, we have that u belongs to $C^{\infty }((0,\infty )\times \mathbb{R})$.

Proof. 

By contradiction, assume that ${\tilde{T}}_{u_0}<\infty$ and the solution u of equation (R) blows up in $[0,{\tilde{T}}_{u_0})$. Thanks to Theorem 2, we have

$${\displaystyle {\int }_0^{{\tilde{T}}_{u_0}}{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\xi =\infty .}$$

(33)

From (32) in Step 4, for any $0<t<{\tilde{T}}_{u_0}$, we have

$${\displaystyle {\parallel u\left(t\right)\parallel }_{H^1}^2+2{\int }_0^t\parallel {\wedge }^{\frac{1}{4}}{u\left(\xi \right)\parallel }_{H^1}^{2}d\xi +2{\int }_0^t{\parallel u\left(\xi \right)\parallel }_{H^1}^{4}d\xi ={\parallel u\left(0\right)\parallel }_{H^1}^2.}$$

Then, according to the embedding inequality ${\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}\le {\parallel {\wedge }^{\frac{1}{4}}u\left(\xi \right)\parallel }_{H^1}$, we deduce that

$${\displaystyle {\int }_0^{{\tilde{T}}_{u_0}}{\parallel u\parallel }_{{\dot{B}}_{2,\infty }^{\frac{1}{4}}}^{4}d\xi \le {\int }_0^{{\tilde{T}}_{u_0}}\parallel {\wedge }^{\frac{1}{4}}{u\left(\xi \right)\parallel }_{H^1}^{4}d\xi \le C{\parallel u\left(0\right)\parallel }_{H^1}^2\le \infty ,}$$

which contradicts (33). Therefore, u is global and ${\tilde{T}}_{u_0}=\infty$. This proves Theorem 3. □

4. Orbital Stability of Standing Waves

Let $N\ge 2,M>0$. Define the variational problem as follows:

$$d_M:=\underset{\{v\in H^s|\parallel v{\parallel }_2^2=M\}}{inf}E\left(v\right),$$

(34)

where $E\left(v\right):=\frac{1}{2}{\int \left|\xi \right|}^{2s}|\widehat{v}{|}^{2}d\xi -\frac{1}{4}\int {\left|v\right|}^{4}dx$. Define the set

$$S_M:=\{v\in H^s|vistheminimizerofproblem(34)\}.$$

(35)

According to Euler–Lagrange theorem ([35]), we know that for any $v\in S_M$, there exists $\omega \in \mathbb{R}$ such that

$${(-\Delta )}^{\alpha }v+{\left|v\right|}^{2}v-\omega v=0,v\in H^s.$$

(36)

In addition, if $v\in S_M$, we obtain that v satisfying (36) and $u(t,x)=e^{i\omega t}v\left(x\right)$ is a solution of problem (R). Hence, $u(t,x)=e^{i\omega t}v\left(x\right)$ is the orbit of $v\left(x\right)$. It is not difficult to find that for any $t>0$, if $v\left(x\right)$ is the solution of the variational problem (34), $u(t,x)=e^{i\omega t}v\left(x\right)$ is also the solution of problem (34). Therefore, $u(t,x)=e^{i\omega t}v\left(x\right)\in S_M$.

In the following, we give the main result which is concerned with the orbital stability of standing wave for problem (R).

Theorem 4.

Let $N\ge 2,M>0$, $0<\alpha <1$ and the initial data $u_0\in H^s$. Assume that for any $\epsilon >0$, there exists $\eta >0$ such that

$$\underset{v\in S_M}{inf}{\parallel u_0-v\parallel }_{H^s}<\eta ,$$

(37)

then the corresponding solution u of problem (R) satisfies

$$\underset{v\in S_M}{inf}{\parallel u(t,x)-v\left(x\right)\parallel }_{H^s}<\epsilon$$

(38)

for any $t>0$, where $S_M$ is defined in (35).

Proof of Theorem 4.

Firstly, we define the energy function $E\left(v\right)$ of problem (R) in $H^s$:

$$\begin{array}{cc}\hfill E\left(v\right)& =\frac{1}{2}\int \overline{v}{(-\Delta )}^{\alpha }vdx-\frac{1}{4}\int \overline{v}{\left(\right|v|}^{2}v)dx\hfill \\ & =\frac{1}{2}{\int |(-\Delta )}^{\frac{\alpha }{2}}{v|}^{2}dx-\frac{1}{4}\int {\left|v\right|}^{4}dx\hfill \\ & =\frac{1}{2}{\int \left|\xi \right|}^{2\alpha }|\widehat{v}{|}^2-\frac{1}{4}\int {\left|v\right|}^{4}dx.\hfill \end{array}$$

Secondly, by Höld inequality, if $2_s^{\ast }=\frac{2N}{N-2\alpha },1<p<\frac{2N}{N-2\alpha }$, there exists a constant $\theta \in (0,1)$ such that

$$\begin{array}{cc}\hfill {\parallel v\parallel }_p^p& ={\int \left|v\right|}^{\theta p}{\left|v\right|}^{(1-\theta )p}dx\hfill \\ \hfill & \le {\parallel v\parallel }_2^{\theta p}{\parallel v\parallel }_{2_s^{\ast }}^{(1-\theta )p}\hfill \\ \hfill & \le {\theta \parallel v\parallel }_2^p+(1-\theta ){\parallel v\parallel }_{2_s^{\ast }}^p\hfill \\ \hfill & \le {C\{\parallel v\parallel }_2^2+{[\int |\xi |}^{2s}|\widehat{v}\left(\xi \right)\ast \widehat{v}{\left(\xi \right)|}^{2}d\xi {]}^{\frac{p}{2}}\}\hfill \\ \hfill & ={C\parallel v\parallel }_2^2+{C\int \left|\xi \right|}^{2s}{\left|\widehat{v}\left(\xi \right)\right|}^{2}d\xi .\hfill \end{array}$$

(39)

For any $0<\epsilon <\frac{1}{2}$, thanks to (39), we have

$$E\left(v\right)\ge C_1{\parallel v\parallel }_{{\dot{H}}^s}^2-C_2{\parallel v\parallel }_2^2.$$

(40)

It follows that $E\left(v\right)\ge -C(\epsilon ,s,M)$ as ${\parallel v\parallel }_2^2=M$. Hence, the variational problem (34) is well defined. Next, we take $v_n={\rho }_n^{\frac{N}{2}}R\left({\rho }_{n}x\right)$ (${\rho }_n>0$), where R is a function such that $\parallel u_n{\parallel }_2^2={\parallel R\parallel }_2^2=M$. Then we have

$$E\left(v_n\right)=\frac{{\rho }_n^{2\alpha }}{2}{\int |\xi |}^{2\alpha }|\widehat{R}{|}^{2}d\xi -\frac{1}{4}\int {\left|R\right|}^{4}dx.$$

Now, we choose ${\rho }_n$ sufficiently small such that $E\left(v_n\right)\le -C_0<0$ for some $C_0>0$. It follows that

$$d_M\le -C_0<0.$$

(41)

Next, by using the profile decomposition of bounded sequences in $H^s$, we prove that the infimum of the variational problem (34) can be attained. Hence, one can obtain that as $n\to +\infty$,

$$E\left(v_n\right)\to d_{M}and{\parallel v_n\parallel }_2^2\to M.$$

(42)

In view of (41) and (42), $E\left(v_n\right)<d_M+1$.

In addition, by Lemma 6, if $n\to +\infty ,l\to +\infty$, we have

$$E\left(v_n\right)=\sum _{j=1}^{l}E\left(V^j(x-x_n^j)\right)+E\left(v_n^l\right)+o\left(1\right).$$

(43)

Then, for any $V^j(x-x_n^j)$, let $V_{{\rho }_j}^j={\rho }_j{V}^j(x-x_n^j)$ with ${\rho }_j=\frac{\sqrt{M}}{\parallel V^j(x-x_n^j){\parallel }^2}>1$, we have $\parallel V_{{\rho }_j}^j{\parallel }_2^2=M$ and

$$E\left(V^j(x-x_n^j)\right)=\frac{E\left(V_{{\rho }_j}^j\right)}{{\rho }_j^2}-\frac{1}{4}\int {\left|V^j(x-x_n^j)\right|}^{4}dx.$$

(44)

Similarly, when $n\to +\infty ,l\to +\infty$, we have

$$E\left(v_n^l\right)\ge \frac{\parallel v_n^l{\parallel }_2^2}{M}E(\frac{\sqrt{M}}{\parallel v_n^l{\parallel }^2}{v}_n^l)+o\left(1\right),$$

(45)

Noting that $\parallel V_{{\rho }_j}^j{\parallel }_2^2=M={\parallel \frac{\sqrt{M}}{\parallel v_n^l{\parallel }^2}{v}_n^l\parallel }_2^2$, then for any $n\ge 1$ and $j\ge 1$, we obtain

$$E\left(V_{{\rho }_j}^j\right)\ge d_{M}andE\left(\frac{\sqrt{M}}{\parallel v_n^l{\parallel }^2}{v}_n^l\right)\ge d_M.$$

(46)

Since ${\sum }_{j=1}^{+\infty }{\parallel V^j(x-x_n^j)\parallel }_2^2$ is convergent, it follows that there exists $j_0\ge 1$ such that $\parallel V^{j_0}{\parallel }_2^2=sup\{\parallel V_j^j{\parallel }_2^2|j\ge 1\}$. Then, from (41), (43), (45), and (46), we deduce that when $n\to +\infty ,l\to +\infty$,

$$d_M\ge E\left(v_n\right)\ge d_M+C_0(\frac{M}{\parallel V^{j_0}{\parallel }_2^2}-1)+o\left(1\right).$$

It is clear that $\parallel V^{j_0}{\parallel }_2^2\ge M$. By Lemma 2.5, we have $\parallel V^{j_0}{\parallel }_2^2\le M$, which leads to $\parallel V^{j_0}{\parallel }_2^2=M$. Then we have $E\left(V^{j_0}\right)=d_M$, which implies that the infimum of problem (34) is attained at $V^{j_0}$.

Finally, we assume that there exists ${\epsilon }_0>0$ and a series of initial sequence ${\left\{u_{0,n}\right\}}_{n=1}^{+\infty }$ such that

$$\underset{v\in S_M}{inf}{\parallel u_{0,n}-v\parallel }_{H^s}<\frac{1}{n}.$$

(47)

In addition, the corresponding solution sequence ${\left\{u_n(t_n,x)\right\}}_{n=1}^{+\infty }$ of problem (R) with a family ${\left\{t_n\right\}}_{n=1}^{+\infty }$ satisfies

$$\underset{v\in S_M}{inf}{\parallel u_n(t_n,x)-v\parallel }_{H^s}\ge {\epsilon }_0.$$

(48)

From (47) and Lemma 2.4, if $n\to +\infty$,

$$\parallel u_n(t_n,x){\parallel }_2^2=\parallel u_{0,n}{\parallel }_2^2\to {\parallel v\parallel }_2^2=M$$

and

$$E\left(u_n(t_n,x)\right)=E\left(u_{0,n}\right)\to E\left(v\right)=d_M.$$

Hence ${\left\{u_n(t_n,x)\right\}}_{n=1}^{+\infty }$ is the minimizing sequence of problem (34) and there is a minimizer $\omega \left(x\right)\in S_M$ such that

$$\parallel u_n(t_n,x)-\omega \left(x\right){\parallel }_{H^s}<{\epsilon }_0,$$

(49)

which contradicts (48). This completes the proof of Theorem 4. □

5. Conclusions

Up to now, there are many unsolved problems related to global well-posedness of the Schrödinger equation with nonlinear fractional dissipation. For instance, the question left by Guo and Huo in [11] of whether the Cauchy problem of (R) is globally well-posed in $H^{\frac{3}{2}-2\alpha }\left(\mathbb{R}\right)$ or not, the problem of global well-posedness for problem (R) with $N\ge 2$ in $H^{\frac{3}{2}-2\alpha }\left(\mathbb{R}\right)$, the problem of well-posedness of the higher dimension nonlinear fractional Schrödinger equations in the critical space $H^{\frac{3}{2}-2\alpha }\left(\mathbb{R}\right)$ on a bounded domain, and so on. Motivated by the problems described above, we presented some details of the first result that the Schrödinger equation with fractional dissipation in $H^{\frac{3}{2}-\gamma }\left(\mathbb{R}\right)$ with $\gamma \in [\frac{1}{2},1),N=1$. The other problems of equation (R) seem to be difficult and they will be pursued in the future.