Traveling wave solutions to some reaction diffusion equations with fractional Laplacians

Abstract

We show the nonexistence of traveling wave solutions in the combustion model with fractional Laplacian $\displaystyle (-\Delta )^s$ when $\displaystyle s\in (0,1/2]$. Our method can be used to give a direct and simple proof of the nonexistence of traveling fronts for the usual Fisher-KPP nonlinearity. Also we prove the existence and nonexistence of traveling wave solutions for different ranges of the fractional power $s$ for the generalized Fisher–KPP type model.

1 Introduction

The paper is concerned with the traveling fronts of the reaction diffusion equation:

$$\begin{aligned} u_t+(-\Delta )^su=f(u),\ \quad \mathrm{in }\,\mathbb {R}\times \mathbb {R}, \end{aligned}$$

for $f\in C^1(\mathbb {R})$, namely the solution to the following equation:

$$\begin{aligned} \left\{ \begin{array}{ll} (-\Delta )^su(x)+\mu u'(x)=f(u(x)),\quad \forall x\in \mathbb {R}\\ \displaystyle \lim _{x\rightarrow -\infty }\ u(x)=0,\quad \lim _{x\rightarrow \infty }\ u(x)=1 \end{array} \right. \end{aligned}$$

(1.1)

where $\mu$ is the speed of propagation of the front and the operator $(-\Delta )^s$ denotes the fractional power of the Laplacian in one dimension with $0<s<1$. Recall the fractional Laplacian is defined as follows:

$$\begin{aligned} (-\Delta )^su(x)=C_{1,s}{(\mathrm P.V.)}\int _{\mathbb {R}}\frac{u(x)-u(y)}{|x-y|^{1+2s}}dy, \end{aligned}$$

where (P.V.) stands for Cauchy principal value and $\displaystyle C_{1,s}=\frac{2^{2s}s\Gamma ((1+2s)/2)}{\pi ^{1/2}\Gamma (1-s)}$, see for example [8]. The original models with the standard Laplacian $\displaystyle (-\Delta )$ arise in applied sciences such as population genetics, combustion, and nerve pulse propagation, etc. The detailed formulations of the models were discussed by Fisher in [5], Kolmogorov et al. in [7] and Aronson and Weinberger in [1], etc. The classical results of the existence and nonexistence of traveling fronts for the models can be found therein.

By a compactness argument, we know that if (1.1) has a solution $u(x)$ then

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\ u'(x)=0\quad \mathrm{and}\quad f(0)=f(1)=0 \end{aligned}$$

(1.2)

Multiplying $u'(x)$ on both sides in (1.1) and integrating over $\mathbb {R}$, we can get the Hamiltonian identity as in [6]:

$$\begin{aligned} \mu \int _\mathbb {R} |u'(x)|^2\ dx=\int _0^1 f(u)\ du \end{aligned}$$

(1.3)

Roquejoffre et al. [9] studied the combustion model, i.e, there exists some $\theta \in (0,1)$ such that $f\in C^1(\mathbb {R})$ satisfies

$$\begin{aligned} f(u)=f(1)=0,\ \ \forall u\in [0,\theta ],\qquad f(u)>0,\ \ \forall u\in (\theta ,1),\qquad \mathrm{and}\qquad f'(1)<0.\nonumber \\ \end{aligned}$$

(1.4)

They have shown that when $s\in (1/2, 1)$ and $f$ satisfies (1.4), there exists an unique $(\mu , u)$ with $\mu >0$ to (1.1).

In this paper, we will show that when $s\in (0,1/2]$ and $f$ satisfies (1.4), there is no traveling wave solution for the combustion model, i.e., (1.1) has no solution. In fact, we shall show the following:

Theorem 1.1

Suppose that there exists some $\theta \in (0,1)$ such that $f\in C^1(\mathbb {R})$ satisfies

$$\begin{aligned} \int _0^1 f(u)\ du>0,\qquad \mathrm{and}\qquad f'(u)\ge 0,\ \ \forall u\in (0,\theta ]. \end{aligned}$$

(1.5)

Then there is no solution to (1.1) if $0<s\le \frac{1}{2}$.

Obviously this theorem applies to the combustion model. For the Fisher-KPP model, i.e, $f\in C^1(\mathbb {R})$ satisfies

$$\begin{aligned} f(u)>0=f(0)=f(1),\ \ \forall u\in (0,1),\qquad f'(0)>0,\qquad \mathrm{and}\qquad f'(1)<0, \end{aligned}$$

(1.6)

Theorem 1.1 implies that if $0<s\le 1/2$, (1.1) has no solution.

We shall also study the generalized Fisher-KPP model and prove nonexistence and existence of solutions to (1.1) for different ranges of $s\in (0,1)$. We shall point out that there is a delicate balance between the diffusion factor $s$ and the reaction power $p$ in order to have a traveling wave solution. In fact, we shall prove the following theorem.

Theorem 1.2

Assume there exist some $\theta \in (0,1)$, $0<p<\infty$, $A_1>0$ and $A_2>0$ such that

$$\begin{aligned} \left\{ \begin{array}{l} f(u)>0=f(0)=f(1),\quad \forall u\in (0,1), \quad f'(1)<0, \\ A_1 u^p\le f(u)\le A_2u^p,\quad \forall u\in [0, \theta ],\\ f'(u)\ge A_1 u^{p-1},\quad \forall u\in (0,\theta ). \end{array} \right. \end{aligned}$$

(1.7)

Then (1.1) has a solution if and only if $p>2$ and $s\ge \frac{p}{2(p-1)}$.

We note that the condition $A_1 u^p\le f(u)\le A_2u^p,\quad \forall u\in [0, \theta ]$ in the above theorem is not needed for the nonexistence result. We include it in (1.7) for the simplicity of the statement.

We also obtain the asymptotics of solutions as $x\rightarrow \pm \infty$ when they exist. Indeed we show the following asymptotic behaviors.

Theorem 1.3

Assume that $f$ satisfies (1.7), let $(\mu ,u)$ be a solution to (1.1) with $\mu >0$. Then there exists some constant $C>0$ such that

$$\begin{aligned} \frac{1}{C|x|^{1+2s}}\le u'(x)\le \frac{C}{|x|^{1+2s}}, \quad \mathrm{and }\quad \frac{1}{C|x|^{2s}}\le 1-u(x)\le \frac{C}{|x|^{2s}},\qquad \forall x\ge 1. \end{aligned}$$

$$\begin{aligned} \frac{1}{C|x|^{2s}}\le u'(x),\quad \mathrm{and}\quad \frac{1}{C|x|^{2s-1}}\le u(x)\le \frac{C}{|x|^{2s-1}},\qquad \forall x\le -1. \end{aligned}$$

Note in Theroem 1.3, $s$ is always bigger than $1/2$ by Theorem 1.2.

We would also like to point out that Cabré and Roquejoffre in [4] already proved that when $0<s<1$, there is no traveling wave solution for the Fisher-KPP model by studying the exponential speed of the front propagation. Theorem 1.3, in particular, shows directly that (1.1) has no solution for the Fisher-KPP model.

The plan of the paper is the following. In Sect. 2, we prove Theorem 1.1 by considering the $s$-harmonic extension of the fractional Laplacian given in [3]. The key ingredient is to show certain asymptotic rates of solutions at $-\infty$. Section 3 is devoted to prove Theorem 1.2. The proof of nonexistence follows the same idea as in Sect. 2 by using $s$-harmonic extension of the fractional Laplacian. We use an iterative argument to obtain an accurate asymptotic behavior of possible solutions. The proof of existence relies on the truncation of domain, asymptotic behavior of solutions and a sliding argument as in [9]. In Sect. 4, detailed asymptotical behavior of solutions will be given.

2 Nonexistence in the combustion and Fisher–KPP models when $0<s\le 1/2$

In this section, we assume $0<s\le 1/2$ and $f\in C^1(\mathbb {R})$ satisfies condition (1.5). We prove Theorem 1.1 by contradiction. Assume that $(\mu , u)$ is a solution to (1.1). By (1.3) and (1.5), we have $\mu >0$. Let $\overline{u}$ be the $s$-harmonic extension of $u$ on $\mathbb {R}^2_+$, that is,

$$\begin{aligned} \overline{u}(x,y)=P_y*u(x),\quad \forall (x,y)\in \mathbb {R}^2_+ \end{aligned}$$

(2.1)

with

$$\begin{aligned} P_y(x)=\frac{a_sy^{2s}}{[y^2+x^2]^{\frac{1+2s}{2}}},\ \ \forall (x,y)\in \mathbb {R}^2_+\quad \mathrm{and}\quad a_s=\frac{\Gamma \left( \frac{1+2s}{2}\right) }{\pi ^{\frac{1}{2}}\Gamma (s)}. \end{aligned}$$

(2.2)

Let $v(x,y)=\overline{u}_x(x,y)=P_y*u'(x)$ for all $(x,y)\in \mathbb {R}^2_+$, Caffarelli and Silvestre [3] proved that $v$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} \mathrm{div}[y^{1-2s}\nabla v(x,y)]=0, \quad \forall (x,y)\in \mathbb {R}^2_+,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}v_y(x,y)=(-\Delta )^su'(x), \quad \forall x\in \mathbb {R},\\ v(x,0)=u'(x), \quad \forall x\in \mathbb {R}. \end{array} \right. \end{aligned}$$

where $\displaystyle d_s=\frac{2^{1-2s}\Gamma (1-s)}{\Gamma (s)}$.

By the standard maximal principle arguments, it is easy to see that $u'(x)>0$ for all $x\in \mathbb {R}$ and $\displaystyle \lim _{|x|\rightarrow \infty }\ u'(x)=0$ (see, e.g., [6, 9]). Then we know that

$$\begin{aligned} v(x,y)>0,\ \ \forall (x,y)\in \overline{\mathbb {R}^2_+},\qquad \mathrm{and}\qquad \lim _{|(x,y)|\rightarrow \infty }\ v(x,y)=0. \end{aligned}$$

By (1.1), without loss of generality, we can assume $u(-1)=\theta$. Since $u'(x)>0$ for all $x\in \mathbb {R}$, we have $f'(u(x))\ge 0$ for all $x\le -1$. In summary, we know that $v$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} \mathrm{div}[y^{1-2s}\nabla v(x,y)]=0, \quad \forall (x,y)\in \mathbb {R}^2_+,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}v_y(x,y)+\mu v_x(x,0)=f'(u(x))u'(x)\ge 0, \quad \forall x\le -1,\\ \displaystyle v(x,y)>0, \ \ \forall (x,y)\in \overline{\mathbb {R}^2_+}\qquad \mathrm{and}\qquad \lim _{|(x,y)|\rightarrow \infty }\ v(x,y)=0. \end{array} \right. \end{aligned}$$

(2.3)

Define the auxiliary function

$$\begin{aligned} \varphi (x,y)=\frac{y^{2s}}{[x^2+y^2]^{\frac{1+2s}{2}}}+\frac{s d_s}{\mu }\cdot \frac{1}{[x^2+y^2]^{\frac{1}{2}}}\quad \forall x\le -1,\ y\ge 0. \end{aligned}$$

Direct computations tell us that for all $x\le -1$ and all $y\ge 0$, we have

$$\begin{aligned} \frac{s d_s}{\mu }\cdot \frac{1}{|(x,y)|}\le \varphi (x,y)&\le \left( 1+\frac{sd_s}{\mu }\right) \cdot \frac{1}{|(x,y)|},\\ \mathrm{div }[y^{1-2s}\nabla \varphi (x,y)]&= \frac{2s^2 d_s}{\mu }\cdot \frac{y^{1-2s}}{[x^2+y^2]^{\frac{3}{2}}}\ge 0,\\ \lim _{y\searrow 0}\ -d_s y^{1-2s}\varphi _y(x,y)&= d_s\lim _{y\searrow 0} \ \left[ \frac{y^2-2s x^2}{[x^2+y^2]^{\frac{3}{2}+s}}+\frac{sd_s}{\mu }\cdot \frac{y^{2-2s}}{[x^2+y^2]^{\frac{3}{2}}}\right] =-\frac{2sd_s}{|x|^{1+2s}},\\ \varphi _x(x,0)&= \frac{sd_s}{\mu }\cdot \frac{1}{|x|^2}. \end{aligned}$$

Since $0<s\le \frac{1}{2}$, we have $\frac{1}{|x|^2}\le \frac{1}{|x|^{1+2s}}$ for all $x\le -1$. Hence for all $x\le -1$, we have

$$\begin{aligned}&\lim _{y\searrow 0}-d_sy^{1-2s}\varphi _y(x,y)+\mu \varphi _x(x,0)=-\frac{2sd_s}{|x|^{1+2s}}+\frac{sd_s}{|x|^2}\\&\quad \le -\frac{2sd_s}{|x|^{1+2s}}+\frac{sd_s}{|x|^{1+2s}}=-\frac{sd_s}{|x|^{1+2s}}<0. \end{aligned}$$

For any $\delta >0$, let $w_\delta (x,y)=v(x,y)-\delta \varphi (x,y)$ for all $x\le -1$ and all $y\ge 0$, then $w_\delta$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} \mathrm{div}[y^{1-2s}\nabla w_\delta (x,y)]\le 0, \quad \forall x\le -1, \ y>0,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}D_yw_\delta (x,y)+\mu D_xw_\delta (x,0)\ge 0, \quad \forall x\le -1,\\ \displaystyle \lim _{|(x,y)|\rightarrow \infty }\ w_\delta (x,y)=0. \end{array} \right. \end{aligned}$$

(2.4)

Lemma 2.1

There exists some $\delta _0>0$ such that $w_{\delta _0}(-1,y)\ge 0$ for all $y\ge 0$.

Proof

First we see that

$$\begin{aligned} \lim _{y\rightarrow \infty }\ \frac{\varphi (-1,y)}{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}}=\lim _{y\rightarrow \infty }\ \frac{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}+\frac{s d_s}{\mu }\cdot \frac{1}{[1+y^2]^{\frac{1}{2}}}}{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}}=1+\frac{sd_s}{\mu }<\infty . \end{aligned}$$

Since $u'(x)>0$ for all $x\in \mathbb {R}$, then $u(0)>u(-1)$, which implies that there exists some constant $B_1>0$ such that

$$\begin{aligned} a_s[u(0)-u(-1)]\cdot \frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}\ge B_1\varphi (-1,y), \quad \forall y\ge 1. \end{aligned}$$

(2.5)

Since $v(x,y)=P_y*u'(x)$ for all $(x,y)\in \mathbb {R}^2_+$, by (2.2), for all $y\ge 1$ we have

$$\begin{aligned} v(-1,y)&= \int _{\mathbb {R}} \frac{a_sy^{2s}}{[(-1-x)^2+y^2]^{\frac{1+2s}{2}}}\cdot u'(x)\ dx\\&\ge \frac{a_sy^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}\int _{-1}^0u'(x)\ dx\\&= a_s[u(0)-u(-1)]\cdot \frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}\\&\ge B_1\varphi (-1,y). \end{aligned}$$

On the other hand, since $v(x,y)>0$ for all $(x,y)\in \overline{\mathbb {R}^2_+}$, there exists some $B_2>0$ such that

$$\begin{aligned} \inf _{0\le y\le 1}\ v(-1,y)\ge B_2\cdot \sup _{0\le y\le 1}\ \varphi (-1,y). \end{aligned}$$

Let $\delta _0=\min \{B_1,B_2\}>0$, we know that

$$\begin{aligned} w_{\delta _0}(-1,y)\ge 0,\quad \forall y\ge 0. \end{aligned}$$

$\square$

Lemma 2.2

For the above $\delta _0$ in Lemma 2.1, there holds

$$\begin{aligned} w_{\delta _0}(x,y)\ge 0, \quad \forall x\le -1,\ y\ge 0. \end{aligned}$$

Proof

Assume $w_{\delta _0}(x_0,y_0)<0$ for some $x_0\le -1$ and some $y_0\ge 0$. Since $w_{\delta _0}(x,y)\rightarrow 0$, as $|(x,y)|\rightarrow \infty$, by Lemma 2.1, we know that there exists some $x_1<-1$ and some $y_1\ge 0$ such that

$$\begin{aligned} w_{\delta _0}(x_1,y_1)=\inf _{x\le -1,\ y\ge 0}\ w_{\delta _0}(x,y)<0. \end{aligned}$$

By the strong maximum principle for uniformly elliptic equations, we know that $y_1=0$. Applying Hopf lemma as in [2], we have

$$\begin{aligned} \lim _{y\searrow 0}\ -d_sy^{1-2s}D_yw_{\delta _0}(x_1,y)<0. \end{aligned}$$

Since $x_1$ is an interior minimum of $w_{\delta _0}(x,0)$ in $x<-1$, then we have $D_xw_{\delta _0}(x_1,0)=0$. By (2.4), we get

$$\begin{aligned} \lim _{y\searrow 0}-d_sy^{1-2s}D_yw_{\delta _0}(x_1,y)=\lim _{y\searrow 0}-d_sy^{1-2s}D_yw_{\delta _0}(x_1,y)+\mu D_xw_{\delta _0}(x_1,0)\ge 0. \end{aligned}$$

We get a contradiction. Therefore

$$\begin{aligned} w_{\delta _0}(x,y)\ge 0,\quad \forall x\le -1,\ y\ge 0. \end{aligned}$$

$\square$

Proof of Theorem 1.1

Assume $(\mu ,u)$ is a solution to (1.1). By Lemma 2.2, we know that

$$\begin{aligned} w_{\delta _0}(x,0)\ge 0,\quad \forall x\le -1. \end{aligned}$$

Since $\displaystyle \varphi (x,0)\ge \frac{sd_s}{\mu }\cdot \frac{1}{|x|}$ for all $x\le -1$, we know that

$$\begin{aligned} u'(x)\ge \frac{\delta _0sd_s}{\mu }\cdot \frac{1}{|x|},\quad \forall x\le -1. \end{aligned}$$

On the other hand, we know that $\displaystyle \int _\mathbb {R}u'(x)\ dx=1$. This is a contradiction which implies that there is no solution to (1.1). $\square$

3 Generalized Fisher-KPP model when $1/2<s<1$

In this section, we assume that $\frac{1}{2}<s<1$ and $f\in C^1(\mathbb {R})$ satisfies condition (1.7). One example for (1.7) is the following:

$$\begin{aligned} f(u)=u^p(1-u),\quad \forall u\in \mathbb {R}, \end{aligned}$$

where $p>0$ is the reaction power.

Our goal is to find the critical exponent $s=s(p)$ such that a solution of (1.1) exists if and only if $1> s\ge s(p)$. In this section, we provide the proof of nonexistence of solutions for (1.1) when $s<s(p)$ by studying the asymptotics of solutions related to (1.1). By Theorem 1.1, it is readily seen that the solution to (1.1) does not exist when $\displaystyle 0< s\le 1/2$. Later, we shall discuss the existence of solutions to (1.1) by a similar argument as in [9].

3.1 Nonexistence results

The following lemma is important in the proof of nonexistence, and has already been proven in [9]. For completeness, we list the proof here.

Lemma 3.1

Let $\frac{1}{2}<s<1$ and $u\in C^2(\mathbb {R})$ such that $\displaystyle \lim _{|x|\rightarrow \infty }u'(x)=0$ and $\displaystyle \lim _{x\rightarrow \pm \infty }u(x)=L^{\pm }$ for some $L^-,L^+\in \mathbb {R}$, then we have

$$\begin{aligned} \displaystyle \lim _{R\rightarrow \infty }\int _{-R}^{R}(-\Delta )^su(y)\ dy=0. \end{aligned}$$

Proof

For any $R>0$, we have

$$\begin{aligned} \int _{-R}^R(-\Delta )^su(y)\ dy&= C_{1,s}\left[ \,\int _{-R}^R \int _{|w|\ge 1}\ \frac{u(y)-u(y+w)}{|w|^{1+2s}}\ dwdy\right. \\&\qquad \quad \left. +\int _{-R}^R\mathrm{(P.V.)}\int _{|w|<1}\frac{u(y)-u(y+w)}{|w|^{1+2s}}\ dwdy\right] \\&= C_{1,s}\left[ \,\int _{-R}^R\int _{|w|\ge 1}\frac{u(y)-u(y+w)}{|w|^{1+2s}}\ dwdy\right. \\&\left. \quad \qquad -\int _{-R}^R\int _{|w|<1}\frac{u(y+w)-u(y)-u'(y)w}{|w|^{1+2s}}\ dwdy\right] . \end{aligned}$$

For $\displaystyle \int _{-R}^R\int _{|w|\ge 1}\frac{u(y)-u(y+w)}{|w|^{1+2s}}\ dwdy$, since $\frac{1}{2}<s$, by Fubini–Tonelli’s theorem and the dominated convergence theorem, we know that

$$\begin{aligned}&\int _{-R}^R\int _{|w|\ge 1}\frac{u(y)-u(y+w)}{|w|^{1+2s}}\ dwdy\\&\qquad = -\int _{-R}^R\int _{|w|\ge 1} \int _0^1 \frac{u'(y+tw)\cdot w}{|w|^{1+2s}}\ dtdwdy\\&\qquad =-\int _{|w|\ge 1}\frac{w}{|w|^{1+2s}}\int _0^1 \int _{-R}^Ru'(y+tw)\ dydtdw\\&\qquad =-\int _{|w|\ge 1}\frac{w}{|w|^{1+2s}}\int _0^1[u(R+tw)-u(-R+tw)]\ dtdw\\&\qquad \rightarrow \int _{|w|\ge 1}\frac{w}{|w|^{1+2s}}\cdot (L^--L^+)\ dtdw=0,\quad \text {as } R\rightarrow \infty \end{aligned}$$

For $\displaystyle \int _{-R}^R\int _{|w|<1}\frac{u(y+w)-u(y)-u'(y)w}{|w|^{1+2s}}\ dwdy$, since $s<1$, by Fubini–Tonelli’s theorem and the dominated convergence theorem, we know that

$$\begin{aligned}&\int _{-R}^R\int _{|w|<1}\frac{u(y+w)-u(y)-u'(y)w}{|w|^{1+2s}}\ dwdy\\&\quad =\int _{-R}^R\int _{|w|<1}\int _0^1\int _0^1\frac{1}{|w|^{2s-1}}\cdot u''(y+rtw)\ drdtdwdy\\&\quad =\int _{|w|\le 1} \frac{1}{|w|^{2s-1}}\int _0^1\int _0^1\int _{-R}^Ru''(y+rtw)\ dydrdtdw\\&\quad =\int _{|w|\le 1} \frac{1}{|w|^{2s-1}}\int _0^1\int _0^1[ u'(R+rtw)-u'(-R+rtw)]\ drdtdw\\&\quad \rightarrow 0,\quad \text {as } R\rightarrow \infty . \end{aligned}$$

Therefore, we can conclude that $\displaystyle \int _{-R}^R(-\Delta )^su(y)\ dy\rightarrow 0$, as $R\rightarrow \infty$. $\square$

Remark 3.1

If $(\mu ,u)$ is a solution to (1.1), since $u'\in L^1(\mathbb {R})$, by Lemma 3.1 and $f(u)\ge 0$ for all $u\in [0,1]$ we know that $f(u)\in L^1(\mathbb {R})$. In particular, if we know that there exists some constants $C>0$ and $r>0$ such that

$$\begin{aligned} u'(x)\ge \frac{C}{|x|^r},\quad \forall x\le -1, \end{aligned}$$

then we have $r>1$ by the integrability of $u'$. On the other hand, by (1.7), we know that $f(u(x))\ge A\left( \frac{C}{r-1}\cdot \frac{1}{|x|^{r-1}}\right) ^p$ for all $x\le -1$. Hence it necessarily holds that $(r-1)p>1$, i.e., $r>\frac{p+1}{p}$.

In the following, we assume that $(\mu ,u)$ is a solution to (1.1) with $\mu >0$ and $u(-1)=\theta$. Let $\overline{u}$ be the $s$-harmonic extension of $u$ on $\mathbb {R}^2_+$ and $v(x,y)=\overline{u}_x(x,y)=P_y*u'(x)$ for all $(x,y)\in \mathbb {R}^2_+$, by the same discussion as in Sect. 2, we know that $v$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \mathrm{div}[y^{1-2s}\nabla v(x,y)]=0, \quad \forall (x,y)\in \mathbb {R}^2_+,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}v_y(x,y)+\mu v_x(x,0)=f'(u(x))u'(x),\quad \forall x\in \mathbb {R},\\ \displaystyle v(x,y)>0,\ \ \forall (x,y)\in \overline{\mathbb {R}^2_+},\qquad \mathrm{and}\qquad \lim _{|(x,y)|\rightarrow \infty }\ v(x,y)=0. \end{array} \right. \end{aligned}$$

(3.1)

For any $\alpha \in [1,2s]$ and $\beta >0$, we consider the auxiliary functions

$$\begin{aligned} \varphi _{\alpha ,\beta }(x,y)=\frac{y^{2s}}{[x^2+y^2]^{\frac{1+2s}{2}}}+\frac{2 \beta sd_s}{\alpha \mu }\cdot \frac{1}{[x^2+y^2]^{\frac{\alpha }{2}}},\qquad \forall x\le -1,\ y\ge 0. \end{aligned}$$

By direct computations, for all $x\le -1$ and all $y\ge 0$ we know that

$$\begin{aligned} \frac{2\beta sd_s}{2s\mu }\cdot \frac{1}{[x^2+y^2]^{\frac{\alpha }{2}}}&\le \varphi _{\alpha ,\beta }(x,y)\le \left( 1+\frac{2\beta sd_s}{\alpha \mu }\right) \cdot \frac{1}{|(x,y)|},\\ \mathrm{div }[y^{1-2s}\nabla \varphi _{\alpha ,\beta }(x,y)]&= \frac{2\beta sd_s}{\mu }\cdot \frac{(2s-1+\alpha )y^{1-2s}}{[x^2+y^2]^{\frac{\alpha +2}{2}}}\ge 0,\\ \lim _{y\searrow 0}\ -d_sy^{1-2s}D_y\varphi _{\alpha ,\beta }(x,y)&= d_s\lim _{y\searrow 0}\ \left[ \frac{y^2-2sx^2}{[x^2+y^2]^{\frac{3}{2}+s}}+\frac{2\beta sd_s}{\mu }\cdot \frac{ y^{2-2s}}{[x^2+y^2]^{\frac{\alpha +2}{2}}}\right] \\&= -\frac{2sd_s}{|x|^{1+2s}}, \text {and} \\ D_x\varphi _{\alpha ,\beta }(x,0)&= \frac{2\beta sd_s}{\mu }\cdot \frac{1}{|x|^{1+\alpha }}. \end{aligned}$$

Hence for all $x\le -1$, we have

$$\begin{aligned} \lim _{y\searrow 0}\ -d_sy^{1-2s}D_y\varphi _{\alpha ,\beta }(x,y)+\mu D_x\varphi _{\alpha ,\beta }(x,0)&= -\frac{2sd_s}{|x|^{1+2s}}+\frac{2 \beta sd_s}{|x|^{1+\alpha }}. \end{aligned}$$

For any $\delta \in (0,1)$, let

$$\begin{aligned} w_{\delta ,\alpha ,\beta }(x,y)=v(x,y)-\delta \varphi _{\alpha ,\beta }(x,y),\quad \forall x\le -1,\ y\ge 0. \end{aligned}$$

Then $w_{\delta ,\alpha ,\beta }$ satisfies

$$\begin{aligned} \quad \left\{ \begin{array}{ll} \mathrm{div}[y^{1-2s}\nabla w_{\delta ,\alpha ,\beta }(x,y)]\le 0, \quad \forall x\le -1, \ y>0,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}D_yw_{\delta ,\alpha ,\beta }(x,y)+\mu D_xw_{\delta ,\alpha ,\beta }(x,0)\\ \qquad =f'(u(x))u'(x)-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}+\frac{2\delta sd_s}{|x|^{1+2s}},\quad \forall x\le -1,\\ \displaystyle \lim _{|(x,y)|\rightarrow \infty }\ w_{\delta ,\alpha ,\beta }(x,y)=0. \end{array} \right. \end{aligned}$$

(3.2)

Lemma 3.2

For any fixed $\alpha \in [1,2s]$ and $\beta >0$, for all $\delta \in (0,1]$, if we have

$$\begin{aligned} f'(u(x))u'(x)-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}+\frac{2\delta sd_s}{|x|^{1+2s}}\ge 0,\quad \forall x\le -1, \end{aligned}$$

then there exists some constant $C>0$ such that

$$\begin{aligned} u'(x)\ge \frac{C}{|x|^\alpha },\quad \mathrm{and}\quad u(x)\ge \frac{C}{|x|^{\alpha -1}},\qquad \forall x\le -1. \end{aligned}$$

Proof

Since $\alpha \ge 1$, we know that $\displaystyle \frac{1}{[1+y^2]^{\frac{\alpha }{2}}}\le \frac{1}{[1+y^2]^{\frac{1}{2}}}$ for all $y\ge 0$. By taking the limit of the ratio, one can get

$$\begin{aligned} \lim _{y\rightarrow \infty }\ \frac{\varphi _{\alpha ,\beta }(-1,y)}{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}}&\le \lim _{y\rightarrow \infty }\ \frac{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}+\frac{2\alpha \beta s d_s}{\mu }\cdot \frac{1}{[1+y^2]^{\frac{1}{2}}}}{\frac{y^{2s}}{[1+y^2]^{\frac{1+2s}{2}}}}=1+\frac{2\alpha \beta sd_s}{\mu }>0. \end{aligned}$$

By the same arguments as in Lemma 2.1 and Lemma 2.2, we know that there exists some small $\delta _0>0$ such that

$$\begin{aligned} w_{\delta _0,\alpha ,\beta }(x,y)\ge 0,\quad \forall x\le -1,\ y\ge 0. \end{aligned}$$

Since $\displaystyle \varphi _{\alpha ,\beta }(x,0)\ge \frac{2\beta sd_s}{\mu }\cdot \frac{1}{|x|^\alpha }$ for all $x\le -1$, we have

$$\begin{aligned} u'(x)=v(x,0)\ge \frac{2\delta _0\beta sd_s}{\mu }\cdot \frac{1}{|x|^\alpha },\quad \forall x\le -1. \end{aligned}$$

$\square$

Lemma 3.3

(Initial Asymptotic Rate) There exists some constant $C_0>0$ such that

$$\begin{aligned} u'(x)\ge \frac{C_0}{|x|^{2s}},\quad \mathrm{and}\quad u(x)\ge \frac{C_0}{|x|^{2s-1}},\qquad \forall x\le -1. \end{aligned}$$

Proof

Let $\alpha =2s$ and $\beta =1$ in Lemma 3.2. Observe that

$$\begin{aligned} f'(u(x))u'(x)-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}+\frac{2\delta sd_s}{|x|^{1+2s}}&= f'(u(x))u'(x)-\frac{2\delta sd_s}{|x|^{1+2s}}+\frac{2\delta sd_s}{|x|^{1+2s}}\\&= f'(u(x))u'(x)\ge 0,\quad \forall x\le -1. \end{aligned}$$

Then Lemma 3.2 leads to the conclusion. $\square$

Remark 3.2

Lemma 3.3 provides an alternative proof of Proposition 4.2 in [9].

As an immediate consequence of Lemma 3.3 and Remark 3.1, we have the following

Theorem 3.1

Let $\displaystyle \frac{1}{2}<s\le \frac{p+1}{2p}$, then there is no solution to (1.1). In particular, for all $0<p\le 1$ and $\displaystyle \frac{1}{2}<s<1$, there is no solution to (1.1).

Lemma 3.4

(Asymptotic Rate Lifting) Let $\frac{p+1}{2p}<s<1$ and $r\in (\frac{p+1}{p},2s]$, we assume there exists some constant $B_0>0$ such that

$$\begin{aligned} u'(x)\ge \frac{B_0}{|x|^r},\quad \mathrm{and}\quad u(x)\ge \frac{B_0}{|x|^{r-1}},\qquad \forall x\le -1. \end{aligned}$$

Let $\alpha \in [1,2s]$ be such that $\alpha \ge p(r-1)$, then there exists some constant $C>0$ such that

$$\begin{aligned} u'(x)\ge \frac{C}{|x|^\alpha },\quad \mathrm{and}\quad u(x)\ge \frac{C}{|x|^{\alpha -1}},\qquad \forall x\le -1. \end{aligned}$$

Proof

By the assumption and (1.7), for all $\beta >0$, all $\delta \in (0,1]$ and all $x\le -1$, we know that

$$\begin{aligned} f'(u(x))u'(x)-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}+\frac{2\delta sd_s}{|x|^{1+2s}}&\ge A_1 |u(x)|^{p-1} u'(x)-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}\\&\ge A_1\left( \frac{B_0}{|x|^{r-1}}\right) ^{p-1}\cdot \frac{B_0}{|x|^r}-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}\\&= \frac{A_1 B_0^p}{|x|^{r+(p-1)(r-1)}}-\frac{2\delta \beta sd_s}{|x|^{1+\alpha }}\\&\ge \frac{A_1 B_0^p-2\delta \beta sd_s}{|x|^{1+\alpha }}. \end{aligned}$$

Let $\displaystyle \beta =\frac{A_1B_0^p}{2\delta sd_s}>0$, by Lemma 3.2, we have completed the proof. $\square$

Remark 3.3

If $\frac{p+1}{p}<r<\frac{p}{p-1}$, by letting $\rho (r):=p(r-1)$, we know that

$$\begin{aligned} 1<\rho (r)<\frac{r}{r-1}(r-1)=r. \end{aligned}$$

We shall show the following theorem.

Theorem 3.1

Let $p>1$ and $\frac{1}{2}<s<\min \left\{ 1,\frac{p}{2(p-1)}\right\}$, then (1.1) has no solution.

Proof

By Lemma 3.4, we have the following

Claim : if

$$\begin{aligned} u'(x)\ge \frac{B_0}{|x|^{r}},\quad \mathrm{and}\quad u(x)\ge \frac{B_0}{|x|^{r}},\qquad \forall x\le -1. \end{aligned}$$

for some $r \in \left( \frac{p+1}{p}, 2s\right] ,$ then

$$\begin{aligned} u'(x)\ge \frac{C}{|x|^{\alpha }},\quad \mathrm{and}\quad u(x)\ge \frac{C}{|x|^{\alpha }},\qquad \forall x\le -1 \end{aligned}$$

with $\alpha \in \left( \frac{p+1}{p}, p(r-1)\right)$. This is a consequence of the fact that the function $\rho (r)=p(r-1)$ has a unique fixed point at $r = \frac{p}{p-1}$ and $\rho (r) <r$ for $r <\frac{p}{p-1}$, which implies that, for $\alpha$ and $r$ as above, there holds $r<2s<\frac{p}{p-1}$ and then $\alpha < r \le 2s$. The claim then follows from Lemma 3.4. Now, one can apply recursively the claim, starting with $r = 2s < \frac{p}{p-1}$ and after a finite number of steps, get $\alpha =\frac{p+1}{p}$, because $\rho ^{(n)}(r):=\rho \circ \rho \cdots \rho (r) = \frac{ p^n[p(r-1)-r]+p}{p-1} \rightarrow -\infty$ as $n \rightarrow \infty$. This is a contradiction to Remark 3.1. $\square$

Note that $\frac{p}{2(p-1)}\ge 1$ if $1<p\le 2$, and $\frac{p}{2(p-1)}<1$ if $2<p$. Therefore, there is no solution to (1.1) for all $s \in (0, 1)$ if $p \le 2$.

3.2 Existence results

In this subsection, we assume that $f$ satisfies (1.7), $p>2$ and $\frac{p}{2(p-1)}\le s<1$, we will show that a solution to (1.1) exists. Mellet et al. [9] have shown the existence of traveling fronts for the non local combustion model when $\frac{1}{2}<s<1$. The proof for the generalized Fisher-KPP model follows a similar argument to that in [9]. For any $\mu \in \mathbb {R}$ and $b>0$, we first consider the following truncated problem:

$$\begin{aligned} \left\{ \begin{array}{l} (-\Delta )^s u(x)+\mu u'(x)=f(u(x)),\quad \forall x\in (-b,b),\\ u(x)=0,\quad \forall x\le -b,\\ u(x)=1,\quad \forall x\ge b. \end{array} \right. \end{aligned}$$

(3.3)

Proposition 3.1

Assume $\displaystyle s\ge \frac{p}{2(p-1)}$ and $f$ satisfies (1.7). Then there exists a constant $M$ such that if $b>M$ the truncated problem 3.3 has a solution $(u_b, \mu _b)$. Furthermore, the following properties hold:

• (1) There exists $K$ independent of $b$ such that $-K\le \mu _b\le K$;

• (2) $u_b$ is non-decreasing with respect to $x$ and satisfies $0<u_b(x)<1$ for all $x\in (-b, b)$.

To prove this Proposition, we need the construction of sub- and super-solutions. The construction is based on the following lemmas, same as in [9]. We would like to present the proof of the following second lemma, and especially elaborate on the sliding method mentioned in [9].

Lemma 3.5

For any $\mu \in \mathbb {R}$ and $b>0$, (3.3) has a solution $u_{\mu ,b}$ such that $0\le u_{\mu ,b}(x)\le 1$ in $\mathbb {R}$, $u_{\mu ,b}$ is non-decreasing in $\mathbb {R}$ and $\mu \rightarrow u_{\mu ,b}$ is continuous.

Proof

The proof is the same as the proof of Lemma 2.4 in [9]. $\square$

Lemma 3.6

There exists some constants $M,K>0$ such that for all $b>M$, we have

• a. If $\mu >K$, then $u_{\mu ,b}(0)<\theta$;

• b. If $\mu <-K$, then $u_{\mu ,b}(0)>\theta$.

Together with Lemma 3.5, Lemma 3.6 implies that there exists $\mu _b\in [-K, K]$ such that $u_{\mu ,b}(0)=\theta$.

Proof

Consider the function

$$\begin{aligned} \varphi (x)=\left\{ \begin{array}{l} \frac{1}{|x|^{2s-1}},\quad \forall x\le -1,\\ 1,\quad \forall x>-1. \end{array} \right. \end{aligned}$$

Since $2s>1$, by Lemma 2.2 in [9], we have

$$\begin{aligned} (-\Delta )^s\varphi (x)+\mu \varphi '(x)=-\frac{C_{1,s}}{2s|x|^{2s}}+\frac{\mu (2s-1)}{|x|^{2s}}+O\left( \frac{1}{|x|^{4s-1}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

Moreover, by (1.7), we get

$$\begin{aligned} f(\varphi (x))\le A_2|\varphi (x)|^p\le \frac{A_2}{|x|^{(2s-1)p}},\quad \forall x\le -1. \end{aligned}$$

Since $\frac{p}{p-1}\le 2s$, we have $(2s-1)p\ge 2s$, which implies that for all $\displaystyle \mu \ge \frac{C_{1,s}}{2s(2s-1)}+\frac{A_2+1}{2s-1}$,

$$\begin{aligned} (-\Delta )^s\varphi (x)+\mu \varphi '(x)-f(\varphi (x))\ge \frac{1}{|x|^{2s}}+O\left( \frac{1}{|x|^{4s-1}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

Since $4s-1>2s$, we know that there exists some large $A>0$ , which is independent of $\mu$, such that for all $\displaystyle \mu \ge \frac{C_{1,s}}{2s(2s-1)}+\frac{A_2+1}{2s-1}$, we have

$$\begin{aligned} (-\Delta )^s\varphi (x)+\mu \varphi '(x)\ge f(\varphi (x)),\quad \quad x\le -A. \end{aligned}$$

For $-A<x<-1$, we know that $(-\Delta )^s\varphi (x)$ is bounded, but $\displaystyle \varphi '(x)=\frac{2s-1}{|x|^{2s}}\ge \frac{2s-1}{A^{2s}}$. So there exists some $K>0$ such that for all $\mu \ge K$,

$$\begin{aligned} (-\Delta )^s\varphi (x)+\mu \varphi '(x)\ge \sup _{x\in [-A,-1]}\ f(\varphi (x)). \end{aligned}$$

Hence for all $\mu \ge K$, we have

$$\begin{aligned} (-\Delta )^s\varphi (x)+\mu \varphi '(x)\ge f(\varphi (x)),\quad \forall x\le -1. \end{aligned}$$

On the other hand, by the definition of $\varphi (x)$ and (1.2), we know that for all $x\ge -1$, $(-\Delta )^s\varphi (x)>0$, $\varphi '(x)=0$ and $f(\varphi (x))=0$. In summary, for all $\mu \ge K$, we have $\varphi (x)$ is a super-solution for (3.3). Now fix some large $M>0$ such that $\varphi (-M)=\frac{1}{M^{2s-1}}<\theta$.

Claim: For all $\mu \ge K$ and all $b\ge M$, we have $u_{\mu ,b}(x)\le \varphi (x-M)$ for all $x\in \mathbb {R}$, in particular, $u_{\mu ,b}(0)<\theta$.

Let $\phi (x)=\varphi (x-M)$ and define

$$\begin{aligned} \Psi _t(x)=\phi (x+t)-u_{\mu }(x),\quad \quad x\in \mathbb {R}. \end{aligned}$$

Let

$$\begin{aligned} \mathcal {O}=\{t\ge 0: \Psi _t(x)=\phi (x+t)-u_{\mu }(x)\ge 0, \quad \quad x\in \mathbb {R}\}, \end{aligned}$$

then $\mathcal {O}$ is nonempty since $\{t\ge 2b\}\subset \mathcal {O}$. $\mathcal {O}$ is clearly closed. Take a convergent sequence $\{t_n\}\subset \mathcal {O}$, $t_n\rightarrow t$ as $n\rightarrow \infty$, then

$$\begin{aligned} \lim _{n\rightarrow \infty }\Psi _{t_n}(x)=\lim _{n\rightarrow \infty }\phi (x+t_n)-u_{\mu }(x)\ge 0, \quad \quad x\in \mathbb {R}. \end{aligned}$$

Therefore $t\in \mathcal {O}$.

Next we show that for any $t\in \mathcal {O}$,

$$\begin{aligned} \Psi _t(x)=\phi (x+t)-u_{\mu }(x)> 0\, \mathrm{{for\, all} }\, x\in (-b, b). \end{aligned}$$

In fact, if there exists $x_0\in (-b, b)$ such that $\Psi _t(x_0)=\phi (x_0+t)-u_{\mu }(x_0)= 0$, then

$$\begin{aligned} 0>(-\Delta )^s\Psi _t(x_0)+\mu \Psi _t'(x_0)\ge f(\phi (x_0+t))-f(u_{\mu }(x_0))=0. \end{aligned}$$

This is a contradiction.

It follows that $\mathcal {O}$ is open. Together with the fact that $\mathcal {O}$ is closed, we get $\mathcal {O}=[0, \infty )$. By the above sliding argument we know

$$\begin{aligned} u_{\mu }(0)\le \varphi (-M)<\theta . \end{aligned}$$

Similarly, for a lower bound we define $\varphi _1(x)=1-\varphi (-x)$. Then if $\mu \le -K$, $x>1$,

$$\begin{aligned} (-\Delta )^s\varphi _1(x)+\mu \varphi '_1(x)=-[(-\Delta )^s\varphi (-x)-\mu \varphi '(x)]\le 0\le f(\varphi _1). \end{aligned}$$

Moreover $\varphi _1(x)=0$ for $x\le 1$. Take $M$ so that $\varphi _1(-M)=1-t_0$, then $\varphi _1(x)>\theta$ for $x\ge M$. Define $\varphi _{1,M}(x)=\varphi _1(x+M)$, then $\varphi _{1,M}$ is a sub-solution to 3.3. Therefore by the same argument as above $u_{\mu }(0)\ge \varphi _{1,M}(0)>\theta$ for $\mu <-K$. $\square$

Theorem 3.2

Under the conditions of Proposition 3.1, there exists a subsequence $b_n\rightarrow \infty$ such that $\displaystyle u_{b_n}\rightarrow u_0$ and $\mu _{b_n}\rightarrow \mu _0$. Furthermore, $\mu _0\in (0,K]$ and $u_0$ is a monotone increasing solution of (1.1).

Proof

By Lemma 3.6, $\mu _b\in [-K, K]$ we have the elliptic estimate for $u_b$:

$$\begin{aligned} \Vert u_b\Vert _{C^{2, \alpha }}\le C \end{aligned}$$

for some $\alpha \in (0,1)$. Thus there exists a subsequence $b_n\rightarrow \infty$ such that

$$\begin{aligned} \mu _n&:= \mu _{b_n}\rightarrow \mu _0\in [-K,K]\\ u_n&:= u_{b_n}\rightarrow u_0,\quad \text {as } n\rightarrow \infty . \end{aligned}$$

Thus $u_0$ satisfies $(-\Delta )^s u_0+\mu _0u'_0=f(u_0)$. Also we know $u_0$ is monotone increasing, $u_0(0)=\theta$ and $u_0$ is bounded. By a compactness argument, there exist $\gamma _0$, $\gamma _1$ such that $\displaystyle \lim _{x\rightarrow -\infty }u_0(x)=\gamma _0$ and $\displaystyle \lim _{x\rightarrow \infty }u_0(x)=\gamma _1$ with

$$\begin{aligned} 0\le \gamma _0\le \theta \le \gamma _1\le 1. \end{aligned}$$

We know both $\gamma _0$ and $\gamma _1$ satisfy $f(\gamma _0)=0$ and $f(\gamma _1)=0$ which implies $\gamma _0=0,\ \gamma _1=1$. Moreover, by integrating $(-\Delta )^{s}u_0+\mu _0 u'_0=f(u_0)$ over $\mathbb {R}$, together with Lemma 3.1, we know

$$\begin{aligned} \mu _0=\int _{\mathbb {R}}f(u_0(x))dx>0. \end{aligned}$$

$\square$

4 Asymptotic rate at $\pm \infty$

In this section, we will study asymptotic behaviors of solutions to (1.1) when $x\rightarrow \pm \infty$. Let $f\in C^1(\mathbb {R})$ satisfy (1.7) and $(\mu ,u)$ be a solution to (1.1). First we investigate the asymptotic behavior of $u$ when $x\rightarrow \infty$. Let $M=\Vert f\Vert _{C^1([0,1])}>0$, by (3.1), we know that

$$\begin{aligned} \left\{ \begin{array}{lll} \mathrm{div}[y^{1-2s}\nabla v(x,y)]=0, \quad \forall (x,y)\in \mathbb {R}^2_+,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}v_y(x,y)+\mu v_x(x,0)+Mv(x,0)\\ \qquad =[M+f'(u(x))]u'(x)\ge 0,\quad \forall x\in \mathbb {R},\\ \displaystyle v(x,y)>0,\ \ \forall (x,y)\in \overline{\mathbb {R}^2_+},\qquad \mathrm{and}\qquad \lim _{|(x,y)|\rightarrow \infty }\ v(x,y)=0. \end{array} \right. \end{aligned}$$

(4.1)

We consider the auxiliary function

$$\begin{aligned} \varphi (x,y)=\frac{y^{2s}}{[x^2+y^2]^{\frac{1+2s}{2}}}+\frac{2 sd_s}{M}\cdot \frac{1}{[x^2+y^2]^{\frac{1+2s}{2}}},\qquad \forall x\ge 1,\ y\ge 0. \end{aligned}$$

By direct computations, for all $x\ge 1$ and all $y\ge 0$, we know that

$$\begin{aligned} \frac{2sd_s}{M}\cdot \frac{1}{[x^2+y^2]^{\frac{1+2s}{2}}}&\le \varphi (x,y)\le \left( 1+\frac{2 sd_s}{M}\right) \cdot \frac{1}{|(x,y)|},\\ \mathrm{div }[y^{1-2s}\nabla \varphi (x,y)]&= \frac{2 sd_s}{M}\cdot \frac{(4s)(1+2s)y^{1-2s}}{[x^2+y^2]^{\frac{2s+3}{2}}}\ge 0,\\ \lim _{y\searrow 0}\ -d_sy^{1-2s}\varphi _y(x,y)&= d_s\lim _{y\searrow 0}\ \left[ \frac{y^2-2sx^2}{[x^2+y^2]^{\frac{3}{2}+s}}+\frac{2 sd_s}{M}\cdot \frac{ y^{2-2s}}{[x^2+y^2]^{\frac{\alpha +2}{2}}}\right] \\&= -\frac{2sd_s}{|x|^{1+2s}},\quad \text {and} \\ D_x\varphi (x,0)&= -\frac{2 sd_s}{M}\cdot \frac{2s}{|x|^{2+2s}}. \end{aligned}$$

Hence for all $x\ge 1$, we have

$$\begin{aligned}&\lim _{y\searrow 0}\ -d_sy^{1-2s}\varphi _y(x,y)+\mu \varphi _x(x,0)+M\varphi (x,0)\\&\quad =-\frac{2sd_s}{|x|^{1+2s}}-\frac{2\mu sd_s}{M}\cdot \frac{2s}{|x|^{2+2s}}+M\cdot \frac{2 sd_s}{M}\cdot \frac{1}{|x|^{1+2s}},\\&\quad =-\frac{2\mu sd_s}{M}\cdot \frac{2s}{|x|^{2+2s}}\le 0. \end{aligned}$$

For any $\delta >0$, let

$$\begin{aligned} w_{\delta }(x,y)=v(x,y)-\delta \varphi (x,y),\quad \forall x\ge 1,\ y\ge 0. \end{aligned}$$

Then $w_{\delta }$ satisfies

$$\begin{aligned} \quad \left\{ \begin{array}{ll} \mathrm{div}[y^{1-2s}\nabla w_{\delta }(x,y)]\le 0, \quad \forall x\ge 1, \ y>0,\\ \displaystyle \lim _{y\searrow 0}-d_sy^{1-2s}D_yw_{\delta }(x,y)+\mu D_xw_{\delta }(x,0)+Mw_{\delta }(x,0)\ge 0,\quad \forall x\ge 1,\\ \displaystyle \lim _{|(x,y)|\rightarrow \infty }\ w_{\delta }(x,y)=0. \end{array} \right. \end{aligned}$$

(4.2)

We have the following

Proposition 4.1

There exists some constant $C>0$ such that

$$\begin{aligned} u'(x)\ge \frac{C}{|x|^{1+2s}},\quad \forall x\ge 1. \end{aligned}$$

Proof

By the same argument as in Lemma 2.2, we know that there is a positive constant $\delta _0$ such that

$$\begin{aligned} v(x,y)\ge \delta _0\varphi (x,y),\quad \forall x\ge 1,\ y\ge 0. \end{aligned}$$

In particular, we know that

$$\begin{aligned} u'(x)=v(x,0)\ge \delta _0\varphi (x,0)=\frac{2\delta _0 sd_s}{|x|^{1+2s}},\quad \forall x\ge 0. \end{aligned}$$

$\square$

Lemma 4.1

Let $\beta >0$, we consider the function

$$\begin{aligned} \psi _\beta (x)=\left\{ \begin{array}{l} \displaystyle \frac{1}{|x|^{\beta }},\quad \forall x<-1,\\ 0,\quad \forall x\ge -1. \end{array} \right. \end{aligned}$$

Then

• a. If $0<\beta <1$, we have

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}\cdot B(2s+\beta ,1-\beta )}{x^{2s+\beta }}+o\left( \frac{1}{x^{2s+\beta }}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

• b. If $\beta >1$, we have

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}}{\beta -1}\cdot \frac{1}{x^{1+2s}}+o\left( \frac{1}{x^{1+2s}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

• c. If $\beta =1$, we have

  $$\begin{aligned} (-\Delta )^s\psi _1(x)=-\frac{C_{1,s}\ln x}{x^{2s+1}}+o\left( \frac{\ln x}{x^{2s+1}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

Proof

In fact, for all $x\ge 2$, by changing of variables, we know that

$$\begin{aligned} (-\Delta )^s\psi _\beta (x)&= C_{1,s}\left[ \int _{-\infty }^{-x-1}\frac{\psi _\beta (x)-\psi _\beta (x+y)}{|y|^{1+2s}}\ dy\right. \\&\left. +\mathrm{(P.V.)}\int _{-x-1}^\infty \frac{\psi _\beta (x)-\psi _\beta (x+y)}{|y|^{1+2s}}\ dy\right] \\&= C_{1,s}\int _{-\infty }^{-x-1}\frac{-1}{|x+y|^{\beta }|y|^{1+2s}}\ dy=-\frac{C_{1,s}}{x^{2s+\beta }}\int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz. \end{aligned}$$

• a. When $0<\beta <1$, we have

  $$\begin{aligned} \displaystyle \int _{-2}^{-1}\frac{1}{|z+1|^\beta }\ dz<\infty . \end{aligned}$$

  By the dominated convergence theorem, we know that

  $$\begin{aligned} \displaystyle \int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz\rightarrow \int _{-\infty }^{-1}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz, \quad \text {as } x\rightarrow \infty . \end{aligned}$$

  On the other hand, we know that

  $$\begin{aligned} \int _{-\infty }^{-1}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz&= \int _0^1 \frac{y^{1+2s}}{\left| -\frac{1}{y}+1\right| ^\beta }\cdot \frac{1}{y^2}\ dy \quad \left( \hbox {by letting } z=-\frac{1}{y}\right) \\&= \int _0^1 y^{2s+\beta -1}(1-y)^{-\beta }\ dy=B(2s+\beta ,1-\beta )>0. \end{aligned}$$

  So we know that

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}\cdot B(2s+\beta ,1-\beta )}{x^{2s+\beta }}+o\left( \frac{1}{x^{2s+\beta }}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

• b. When $\beta >1$, we know that $\displaystyle \int _{-2}^{-1}\frac{1}{|z+1|^{\beta }}\ dz=\infty$, which implies that

  $$\begin{aligned} \displaystyle \int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz\rightarrow \infty , \quad \text { as } x\rightarrow \infty . \end{aligned}$$

  By L’Hospital rule, we have

  $$\begin{aligned} \lim _{x\rightarrow \infty }\ \frac{\int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1|^{\beta }|z|^{1+2s}}\ dz}{x^{\beta -1}}&= \lim _{x\rightarrow \infty }\frac{x^{\beta }\cdot \left| 1+\frac{1}{x}\right| ^{-1-2s}\cdot \frac{1}{x^2}}{(\beta -1) x^{\beta -2}}=\frac{1}{\beta -1}. \end{aligned}$$

  So we derive

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}}{\beta -1}\cdot \frac{1}{x^{1+2s}}+o\left( \frac{1}{x^{1+2s}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

• c. When $\beta =1$, we know that $\displaystyle \int _{-2}^{-1}\frac{1}{|z+1|}\ dz=\infty$, which implies that

  $$\begin{aligned} \displaystyle \int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1||z|^{1+2s}}\ dz\rightarrow \infty , \quad \text {as } x\rightarrow \infty . \end{aligned}$$

  By L’Hospital rule, we know that

  $$\begin{aligned} \lim _{x\rightarrow \infty }\ \frac{\int _{-\infty }^{-1-\frac{1}{x}}\frac{1}{|z+1||z|^{1+2s}}\ dz}{\ln x}&= \lim _{x\rightarrow \infty }\ \frac{|x|\cdot \left| 1+\frac{1}{x}\right| ^{-1-2s}\cdot \frac{1}{x^2}}{\frac{1}{x}}=1. \end{aligned}$$

  Therefore we have

  $$\begin{aligned} (-\Delta )^s\psi _1(x)=-\frac{C_{1,s}\ln x}{x^{2s+1}}+o\left( \frac{\ln x}{x^{2s+1}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

$\square$

Lemma 4.2

Let $\beta >0$, $\psi _\beta (x)$ be defined as in Lemma 4.1, then we have the following estimates:

• a. If $0<\beta <1$, there holds that

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}\cdot A(s,\beta )}{|x|^{2s+\beta }}+o\left( \frac{1}{|x|^{2s+\beta }}\right) ,\quad \text {as } x\rightarrow -\infty ; \end{aligned}$$

  where

  $$\begin{aligned} A(s,\beta )=\int _1^\infty \frac{1}{|z|^{1+2s}|z+1|^\beta }\ dz-\frac{1}{s}+\int _{0}^{1} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz; \end{aligned}$$

• b. If $\beta >1$, we have

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}}{\beta -1}\cdot \frac{1}{|x|^{1+2s}}+o\left( \frac{1}{|x|^{2s+1}}\right) ,\quad \text {as } x\rightarrow -\infty ; \end{aligned}$$

• c. If $\beta =1$, we have

  $$\begin{aligned} (-\Delta )^s \psi _1(x)=-\frac{C_{1,s}\ln |x|}{|x|^{2s+1}}+o\left( \frac{\ln |x|}{|x|^{2s+1}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

Proof

For all $x<-2$, we know that $x+1<-x-1$ and

$$\begin{aligned} (-\Delta )^s\psi _\beta (x)&= -\frac{C_{1,s}}{2}\int _{\mathbb {R}}\frac{\psi _\beta (x+y)+\psi _\beta (x-y)-2\psi _\beta (x)}{|y|^{1+2s}}\ dy\\&= -\frac{C_{1,s}}{2}\left[ \int _{-\infty }^{x+1} \frac{\frac{1}{|x+y|^\beta }-\frac{2}{|x|^\beta }}{|y|^{1+2s}}\ dy+\int _{x+1}^{-x-1} \frac{\frac{1}{|x+y|^{\beta }}+\frac{1}{|x-y|^\beta }-\frac{2}{|x|^\beta }}{|y|^{1+2s}}\ dy\right. \\&\qquad \qquad \left. +\int _{-x-1}^\infty \frac{\frac{1}{|x-y|^\beta }-\frac{2}{|x|^\beta }}{|y|^{1+2s}}\ dy \right] \\&= -\frac{C_{1,s}}{2|x|^{2s+\beta }}\left[ \int _{-\infty }^{-1-\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta } -2}{|z|^{1+2s}}\ dz +\int _{-1-\frac{1}{x}}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz\right. \\&\qquad \qquad \left. +\int _{1+\frac{1}{x}}^\infty \frac{\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz \right] \quad \mathrm{Let} y=-xz\\&= -\frac{C_{1,s}}{|x|^{2s+\beta }} \left[ \int _{1+\frac{1}{x}}^\infty \frac{\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz+\int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz\right] \end{aligned}$$

For the first term inside the bracket, we know that

$$\begin{aligned} \lim _{x\rightarrow -\infty }\ \int _{1+\frac{1}{x}}^\infty \frac{\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz&= \int _1^\infty \frac{1}{|z|^{1+2s}|z+1|^\beta }\ dz-\frac{1}{s}. \end{aligned}$$

• a. Since $\beta \in (0,1)$, we know that $\displaystyle \int _0^1 \frac{1}{|z-1|^\beta }\ dz<\infty$, which implies that

  $$\begin{aligned} \lim _{x\rightarrow -\infty }\ \int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz=\int _{0}^{1} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz. \end{aligned}$$

  Let

  $$\begin{aligned} \displaystyle A(s,\beta )=\int _1^\infty \frac{1}{|z|^{1+2s}|z+1|^\beta }\ dz-\frac{1}{s}+\int _{0}^{1} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz, \end{aligned}$$

  then we have

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}\cdot A(s,\beta )}{|x|^{2s+\beta }}+o\left( \frac{1}{|x|^{2s+\beta }}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

• b. Since $\beta >1$, we know that $\displaystyle \int _0^1\frac{1}{|z-1|^\beta }\ dz=\infty$, which implies that

  $$\begin{aligned} \displaystyle \int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz\rightarrow \infty , \quad \text {as } x\rightarrow -\infty . \end{aligned}$$

  By L’Hospital rule, we know that

  $$\begin{aligned} \lim _{x\rightarrow -\infty }\ \frac{\int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz}{(-x)^{\beta -1}}&= \lim _{x\rightarrow -\infty }\frac{\left[ |x|^\beta +\frac{1}{2^\beta }-2\right] \cdot \left( -\frac{1}{x^2}\right) }{-(\beta -1)(-x)^{\beta -2}}=\frac{1}{\beta -1}. \end{aligned}$$

  Hence we have

  $$\begin{aligned} (-\Delta )^s\psi _\beta (x)=-\frac{C_{1,s}}{\beta -1}\cdot \frac{1}{|x|^{2s+1}}+o\left( \frac{1}{|x|^{1+2s}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

• c. Since $\beta =1$, we know that $\displaystyle \int _0^1\frac{1}{|z-1|}\ dz=\infty$, which implies that

  $$\begin{aligned} \displaystyle \int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|^\beta }+\frac{1}{|z+1|^\beta }-2}{|z|^{1+2s}}\ dz\rightarrow \infty , \quad \text { as } x\rightarrow -\infty . \end{aligned}$$

  By L’Hospital rule, we know that

  $$\begin{aligned} \lim _{x\rightarrow -\infty }\ \frac{\int _{0}^{1+\frac{1}{x}} \frac{\frac{1}{|z-1|}+\frac{1}{|z+1|}-2}{|z|^{1+2s}}\ dz}{\ln (-x)}&= \lim _{x\rightarrow -\infty }\frac{\left[ |x|+\frac{1}{2}-2\right] \cdot \left( -\frac{1}{x^2}\right) }{\frac{1}{x}}=1. \end{aligned}$$

  Hence we have

  $$\begin{aligned} (-\Delta )^s \psi _1(x)=-\frac{C_{1,s}\ln |x|}{|x|^{2s+1}}+o\left( \frac{\ln |x|}{|x|^{2s+1}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

$\square$

Lemma 4.3

Consider the function

$$\begin{aligned} \phi (x)=\left\{ \begin{array}{l} 1, \quad \forall x\le -1, \\ 0,\quad \forall x>-1. \end{array} \right. \end{aligned}$$

Then

$$\begin{aligned} (-\Delta )^s\phi (x)=-\frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow \infty , \quad \text {and} \end{aligned}$$

$$\begin{aligned} (-\Delta )^s\phi (x)=\frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

Proof

• a. In fact, for all $x\ge 2$, we have

  $$\begin{aligned} (-\Delta )^s\phi (x)&= C_{1,s}\left[ \,\int _{-\infty }^{-x-1}\frac{\phi (x)-\phi (x\!+\!y)}{|y|^{1+2s}}\ dy\!+\!\mathrm{(P.V.)}\!\int _{-x-1}^\infty \frac{\phi (x)-\phi (x\!+\!y)}{|y|^{1+2s}}\ dy\!\!\right] \\&= -C_{1,s}\int ^{-x-1}_{-\infty } \frac{1}{|y|^{1+2s}}\ dy\\&= -\frac{C_{1,s}}{2s}\cdot \frac{1}{|x+1|^{2s}}\\&= -\frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

• b. If $x\le -2$, we have

  $$\begin{aligned} (-\Delta )^s\phi (x)&= C_{1,s}\left[ \!\mathrm{(P.V.)}\!\int _{-\infty }^{-x-1}\frac{\phi (x)-\phi (x+y)}{|y|^{1+2s}}\ dy\!+\!\!\int _{-x-1}^\infty \frac{\phi (x)-\phi (x+y)}{|y|^{1+2s}}\ dy\!\!\right] \\&= C_{1,s}\int _{-x-1}^\infty \frac{1}{|y|^{1+2s}}\ dy\\&= \frac{C_{1,s}}{2s}\cdot \frac{1}{|x+1|^{2s}}\\&= \frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

$\square$

Below we show a form of the maximal principle which is a slight variation of those in [2, 6].

Lemma 4.4

(The Maximum Principle) Let $H$ be a nonempty open subset of $\mathbb {R}$, assume $d(x)\ge 0$ for all $x\in H$ and $w\in C^1(\overline{H})$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} (-\Delta )^s w(x)+\mu w'(x)+d(x)w(x)\ge 0,\quad \forall x\in H,\\ \displaystyle \lim _{|x|\rightarrow \infty }\ w(x)=0,\\ w(x)\ge 0,\quad \forall x\notin H. \end{array} \right. \end{aligned}$$

Then $w(x)\ge 0$ for all $x$ in $\mathbb {R}$.

Proof

Assume $w(x_0)<0$ for some $x_0\in \mathbb {R}$, since $w(x)\ge 0$ for all $x\notin H$, $\displaystyle \lim _{|x|\rightarrow \infty }\ w(x)=0$, and $w\in C^1(\overline{H})$, then there exists some $x_1\in H$ such that

$$\begin{aligned} w(x_1)=\inf _{x\in \mathbb {R}}\ w(x)<0. \end{aligned}$$

Since $x_1$ is a global minimum of $w$ in $\mathbb {R}$, $x_1\in H$ and $w\in C^1(H)$, then

$$\begin{aligned} (-\Delta )^sw(x_1)<0,\quad \mathrm{and}\quad w'(x_1)=0. \end{aligned}$$

Since $d(x)\ge 0$ for all $x\in H$, and $x_1\in H$, so we have

$$\begin{aligned} (-\Delta )^s w(x_1)+\mu w'(x_1)+d(x_1)w(x_1)<0, \end{aligned}$$

which contradicts with the assumption. $\square$

The following two propositions give suitable lower and upper bounds of the asymptotic decay rates of $u'$ and $1-u$ at $\infty$, which are expected to be a power of $1+2s$ and $2s$, respectively.

Proposition 4.2

Let $\displaystyle \frac{1}{2}<s<1$ and $(\mu ,u)$ be a solution to (1.1) with $\mu >0$. Assume that $f'(1)<0$, then there exists some constant $C>0$ such that

$$\begin{aligned} u'(x)\le \frac{C}{|x|^{2s}},\quad \mathrm{and}\quad 1-u(x)\le \frac{C}{|x|^{2s}},\qquad x\ge 1. \end{aligned}$$

Proof

Since $f'(1)<0$, there exists some $m>0$ and $\theta _0\in (0,1)$ such that $f'(u)\le -m$ for all $u\in [\theta _0,1]$. Let $\epsilon >0$ be such that $\displaystyle -\frac{C_{1,s}}{2s}+m\epsilon ^{-2s}=\frac{m}{2}\epsilon ^{-2s}$, that is, $\displaystyle \epsilon =\left( \frac{sm}{C_{1,s}}\right) ^{\frac{1}{2s}}$. Consider

$$\begin{aligned} \displaystyle \Psi (x)=\phi \left( x-\epsilon ^{-1}-1\right) +\psi _{2s}(-\epsilon x) \quad \forall x\in \mathbb {R}. \end{aligned}$$

we know that

$$\begin{aligned} \Psi (x)=\epsilon ^{-2s}\cdot \frac{1}{|x|^{2s}},\quad \mathrm{and}\quad \Psi '(x)=-2s\epsilon ^{-2s}\cdot \frac{1}{|x|^{1+2s}},\qquad \forall x>\frac{1}{\epsilon }. \end{aligned}$$

By Lemma 4.3 and Lemma 4.2, we know that

$$\begin{aligned} (-\Delta )^s\Psi (x)=-\frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

Hence we have

$$\begin{aligned} (-\Delta )^s\Psi (x)+\mu \Psi '(x)+m\Psi (x)&= \left[ -\frac{C_{1,s}}{2s}+m\epsilon ^{-2s}\right] \cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) \\&= \frac{m}{2}\epsilon ^{-2s}\cdot \frac{1}{|x|^{2s}}+o\left( \frac{1}{|x|^{2s}}\right) ,\quad \text {as } x\rightarrow \infty . \end{aligned}$$

So there exists some large $R>0$ such that

$$\begin{aligned} (-\Delta )^s\Psi (x)+\mu \Psi '(x)+m\Psi (x)\ge 0,\quad \forall x\ge R. \end{aligned}$$

Up to a translation, without loss of generality, we assume $u(0)=\theta _0$. Notice that $v(x)=u'(x)>0$ in $\mathbb {R}$ satisfies

$$\begin{aligned} (-\Delta )^s v(x)+\mu v'(x)+mv(x)=[m+f'(u(x))]v(x)\le 0,\quad \forall x\ge R. \end{aligned}$$

Since $\Psi (x)>0$ for all $x\in \mathbb {R}$, there exists some $C>0$ such that

$$\begin{aligned} C>\Vert v\Vert _{C(\mathbb {R})}\quad \mathrm{and}\quad C\inf _{x\in [\epsilon ^{-1},R]}\ \Psi (x)\ge \Vert v\Vert _{C(\mathbb {R})}. \end{aligned}$$

Since $\displaystyle \Psi (x)=\phi \left( x-\epsilon ^{-1}-1\right) =1$ for all $\displaystyle x\le \epsilon ^{-1}$, we get $\displaystyle C\Psi (x)=C\ge \Vert v\Vert _{C(\mathbb {R})}$ for all $x\le \epsilon ^{-1}$. In summary, we know that

$$\begin{aligned} C\Psi (x)\ge v(x),\quad \forall x\le R. \end{aligned}$$

Let $w(x)=C\Psi (x)-v(x)$ for all $x\in \mathbb {R}$, we have

$$\begin{aligned} \left\{ \begin{array}{l} (-\Delta )^s w(x)+\mu w'(x)+mw(x)\ge 0,\quad \forall x\ge R,\\ \displaystyle \lim _{x\rightarrow \infty }\ w(x)=0,\\ w(x)\ge 0,\quad \forall x\le R. \end{array} \right. \end{aligned}$$

By Lemma 4.4, we have $w(x)\ge 0$ in $\mathbb {R}$, which implies that

$$\begin{aligned} \frac{C}{|x|^{2s}}\ge v(x)=u'(x),\quad \forall x\ge 1. \end{aligned}$$

$\square$

Proposition 4.3

Let $\displaystyle \frac{1}{2}<s<1$, assume that $f'(1)<0$, let $(\mu ,u)$ be a solution to (1.1) with $\mu >0$. Then there exists some constant $C>0$ such that

$$\begin{aligned} u'(x)\le \frac{C}{|x|^{1+2s}},\quad \mathrm{and}\quad 1-u(x)\le \frac{C}{|x|^{2s}},\qquad x\ge 1. \end{aligned}$$

Proof

Since $f'(1)<0$, there exists some $m>0$ and $\theta _0\in (0,1)$ such that $f'(u)\le -m$ for all $u\in [\theta _0,1]$. Let $\epsilon >0$ be such that

$$\begin{aligned} -\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s-1}-\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s}+m\epsilon ^{-1-2s} =\frac{m}{2}\cdot \epsilon ^{-1-2s}. \end{aligned}$$

That is, we have

$$\begin{aligned} \frac{\epsilon ^{2s}}{2s}+\frac{\epsilon ^{2s}}{2s-1}=\frac{m}{2C_{1,s}}. \end{aligned}$$

By considering the function $\Psi (x)=\psi _{2s}(\epsilon x-2)+\psi _{1+2s}(-\epsilon x)$ for all $x\in \mathbb {R}$, we know that

$$\begin{aligned} \Psi (x)=\epsilon ^{-1-2s}\cdot \frac{1}{|x|^{1+2s}},\quad \mathrm{and}\quad \Psi '(x)=-\epsilon ^{-1-2s}\cdot \frac{1+2s}{|x|^{2+2s}},\quad \forall x>\epsilon ^{-1}. \end{aligned}$$

By Lemma 4.1 and Lemma 4.2, we know that

$$\begin{aligned} (-\Delta )^s\Psi (x)\!=\!-\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s-\!1}\cdot \frac{1}{|x|^{1+2s}}-\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s}\cdot \frac{1}{|x|^{1+2s}}\!+\!o\left( \frac{1}{|x|^{1+2s}}\right) ,\quad \text {as } x\!\rightarrow \!\infty . \end{aligned}$$

So we get

$$\begin{aligned} (-\Delta )^s\Psi (x)+\mu \Psi '(x)+m\Psi (x)&= \left[ -\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s-1}-\epsilon ^{-1}\cdot \frac{C_{1,s}}{2s}+m\epsilon ^{-1-2s} \right] \cdot \frac{1}{|x|^{1+2s}}\\&+o\left( \frac{1}{|x|^{1+2s}}\right) \\&= \frac{m}{2}\cdot \epsilon ^{-1-2s}\cdot \frac{1}{|x|^{1+2s}}+o\left( \frac{1}{|x|^{1+2s}}\right) , \quad \text {as } x\rightarrow \infty . \end{aligned}$$

Hence there exists some large $R>0$ such that

$$\begin{aligned} (-\Delta )^s\Psi (x)+\mu \Psi '(x)+m\Psi (x)\ge 0,\quad \forall x\ge R. \end{aligned}$$

Without loss of generality, we assume $\displaystyle u\left( \epsilon ^{-1}\right) =\theta _0$, we know that $v=u'$ satisfies

$$\begin{aligned} (-\Delta )^s v(x)+\mu v'(x)+mv(x)=[m+f'(u(x))]v(x)\le 0,\quad \forall x\ge \epsilon ^{-1}. \end{aligned}$$

For all $x\le \epsilon ^{-1}$, we have $\epsilon x-2\le -1$ and $-\epsilon x\ge -1$, which implies that

$$\begin{aligned} \Psi (x)=\psi _{2s}(\epsilon x-2)=\frac{1}{|\epsilon x-2|^{2s}}. \end{aligned}$$

By Proposition (4.2), we know that there exists some constant $C_1>0$ such that

$$\begin{aligned} u'(x)=v(x)\le C_1\Psi (x),\quad \forall x\le \epsilon ^{-1}. \end{aligned}$$

Notice that for all $\displaystyle x\ge \epsilon ^{-1}$, $\displaystyle \Psi (x)\ge \psi _{1+2s}(-\epsilon x)>0$, which implies that there exists some $C_2>0$ such that

$$\begin{aligned} C_2\inf _{x\in [\frac{1}{\epsilon }, R]}\ \Psi (x)\ge \sup _{x\in [\epsilon ^{-1}, R]}\ v(x). \end{aligned}$$

Let $C=\max \{C_1,C_2\}>0$ and $w(x)=C\Psi (x)-v(x)$ for all $x\in \mathbb {R}$, then

$$\begin{aligned} \left\{ \begin{array}{l} (-\Delta )^s w(x)+\mu w'(x)\ge 0,\quad \forall x\ge R,\\ \displaystyle \lim _{x\rightarrow \infty }\ w(x)=0,\\ w(x)\ge 0,\quad \forall x\le R. \end{array} \right. \end{aligned}$$

By Lemma 4.4, we know that $w(x)\ge 0$ in $\mathbb {R}$, which implies

$$\begin{aligned} \frac{C}{|x|^{1+2s}}\ge v(x)=u'(x),\quad \forall x\ge 1. \end{aligned}$$

The inequality for $1-u(x)$ follows immediately. $\square$

Proposition 4.4

Let $\displaystyle \frac{1}{2}<s<1$,let $(\mu ,u)$ be a solution to (1.1) with $\mu >0$ in Theorem 3.2. Then there exists some constant $C>0$ such that

$$\begin{aligned} \frac{1}{C|x|^{2s}}\le u'(x),\quad \mathrm{and}\quad \frac{1}{C|x|^{2s-1}}\le u(x)\le \frac{C}{|x|^{2s-1}},\qquad \forall x\le -1. \end{aligned}$$

Proof

We have shown in the proof of Theorem 3.2 that there exists some constant $C>0$ such that

$$\begin{aligned} u(x)\le \frac{C}{|x|^{2s-1}},\qquad \forall x\le -1. \end{aligned}$$

Now it suffices to show that there exists some constant $C>0$ such that

$$\begin{aligned} \frac{1}{C|x|^{2s}}\le u'(x),\qquad \forall x\le -1. \end{aligned}$$

Let $\epsilon >0$ be such that $\displaystyle -\frac{C_{1,s}}{2s-1}+2s\mu \epsilon ^{1-2s}=-\frac{C_{1,s}}{2(2s-1)}$, that is,

$$\begin{aligned} \epsilon ^{1-2s}=\frac{C_{1,s}}{4s\mu (2s-1)}. \end{aligned}$$

Let $\Phi (x)=\psi _{2s}(\epsilon x)$ in $\mathbb {R}$, then

$$\begin{aligned} \Phi (x)=\epsilon ^{-2s}\cdot \frac{1}{|x|^{2s}},\quad \mathrm{and}\quad \Phi '(x)=\epsilon ^{-2s}\cdot \frac{2s}{|x|^{1+2s}},\qquad \forall x\le -\epsilon ^{-1}. \end{aligned}$$

By Lemma 4.2, we have

$$\begin{aligned} (-\Delta )^s\Phi (x)=-\frac{C_{1,s}}{2s-1}\cdot \frac{\epsilon ^{-1}}{|x|^{1+2s}}+o\left( \frac{1}{|x|^{1+2s}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

So we get

$$\begin{aligned} (-\Delta )^s\Phi (x)+\mu \Phi '(x)&= -\frac{C_{1,s}}{2s-1}\cdot \frac{\epsilon ^{-1}}{|x|^{1+2s}}+\mu \epsilon ^{-2s}\cdot \frac{2s}{|x|^{1+2s}}+o\left( \frac{1}{|x|^{1+2s}}\right) \\&= \left[ -\frac{C_{1,s}}{2s-1}+2s\mu \epsilon ^{1-2s}\right] \frac{\epsilon ^{-1}}{|x|^{2s}} +o\left( \frac{1}{|x|^{1+2s}}\right) \\&= -\frac{C_{1,s}}{2(2s-1)}\cdot \frac{\epsilon ^{-1}}{|x|^{1+2s}}+o\left( \frac{1}{|x|^{1+2s}}\right) ,\quad \text {as } x\rightarrow -\infty . \end{aligned}$$

Therefore there exists some large $R>0$ such that

$$\begin{aligned} (-\Delta )^s\Phi (x)+\mu \Phi '(x)\le 0,\quad \forall x\le -R. \end{aligned}$$

Since $f'(t)\ge 0$ for all $t\in [0,\theta _0]$, without loss of generality, we may assume $u(-\epsilon ^{-1})=\theta _0$. Notice that $v(x)=u'(x)>0$ in $\mathbb {R}$ satisfies

$$\begin{aligned} (-\Delta )^s v(x)+\mu v'(x)=f'(u(x))(x)\ge 0,\quad \forall x\le -\epsilon ^{-1}. \end{aligned}$$

Since $\Phi (x)=0$ for all $x\ge -\epsilon ^{-1}$, we get $\Phi (x)\le v(x)$ for all $x\ge -\epsilon ^{-1}$. Since $v(x)>0$ in $\mathbb {R}$, there exists some $C>1$ such that

$$\begin{aligned} C\inf _{x\in [-R,-\epsilon ^{-1}]}\ v(x)\ge \sup _{x\in [-R,-\epsilon ^{-1}}\ \Phi (x). \end{aligned}$$

Let $w(x)=Cv(x)-\Phi (x)$ for all $x\in \mathbb {R}$, we have

$$\begin{aligned} \left\{ \begin{array}{ll} (-\Delta )^s w(x)+\mu w'(x)\ge 0,\quad \forall x\le -R,\\ \displaystyle \lim _{x\rightarrow -\infty }\ w(x)=0,\\ w(x)\ge 0,\quad \forall x\ge -R. \end{array} \right. \end{aligned}$$

By Lemma 4.4, we have $w(x)\ge 0$ in $\mathbb {R}$, which implies that

$$\begin{aligned} \frac{C}{|x|^{2s}}\le v(x)=u'(x),\quad \forall x\le -1. \end{aligned}$$

$\square$