Existence and stability of traveling curved fronts for nonlocal dispersal equations with bistable nonlinearity

Abstract

This paper is concerned with existence and stability of V-shaped traveling fronts for a class of nonlocal dispersal equations with unbalanced bistable nonlinearity. The main tool is sub- and supersolution technique combined with a comparison principle.

1 Introduction

We consider the following equation

$$\begin{aligned} u_{t}=J*u-u+f(u),\quad (\varvec{x},t)\in \mathbb {R}^{2}\times (0,+\infty ) \end{aligned}$$

(1.1)

with the nonlocal dispersal operator $(J*u-u)(\varvec{x},t)=\int \limits _{\mathbb {R}^{2}}J(\varvec{x}-\varvec{y})\left( u(\varvec{y},t)-u(\varvec{x},t)\right) \mathrm{d}\varvec{y}.$ The kernel $J\in C^{1}(\mathbb {R}^{2})$ satisfies

$$\begin{aligned}&(\mathbf{J1} )\ J\ge 0 \text { is radial symmetric and has unit integral};&\\&(\mathbf{J2} )\int \limits _{0}^{\infty }J(r)e^{\lambda r}\mathrm{d}r<\infty ~~ \mathrm{for\ some}\ \lambda >0.&\end{aligned}$$

The nonlinearity $f\in C^{2}(\mathbb {R})$ has only three zeros 0, a and 1, and satisfies

$$\begin{aligned}&(\mathbf{F1} )\ f'(0)<0, f'(a)>0, f'(1)<0 {~\mathrm and~}\int \limits _{0}^{1}f(u)\mathrm{d}u\ne 0;&\\&(\mathbf{F2} )\ \sup _{s\in [0,1]}f'(s)<1. \end{aligned}$$

Obviously, if $J(\varvec{x}) = \frac{1}{4\pi \lambda } e^{-\frac{|\varvec{x}|^{2}}{4\lambda }}$ for any given $\lambda > 0$ or $J(\varvec{x})$ is compactly supported with symmetric property, then it satisfies (J1)–(J2). Condition (F2) guarantees that the solution of the corresponding Cauchy problem of (1.1) has the same regularity with its initial function [13].

Traveling waves of (1.1) are widely used to model nonlocal diffusion phenomena in fields such as physics, chemistry, ecology and epidemiology. In one-dimensional space, traveling wave solutions have the form $u(x,t)=U(\xi )$, $\xi =x+ct$, where U is the wave profile and c is the wave speed. It is referred to [1, 6, 8, 9, 19, 29] for the mathematical study on traveling waves of (1.1).

Let

$$\begin{aligned} J_{1}(x)=\int \limits _{\mathbb {R}}J(x,x_{2})\mathrm{d}x_{2}. \end{aligned}$$

(1.2)

Then under the condition (J1), $J_{1}$ is nonnegative and even, with unit integral on $x\in \mathbb {R}$. It is also not difficult to prove that $J'_{1}(x)\in L^{1}(\mathbb {R})$ with the aid of condition (J2). Then under the condition (F1), the following equation

$$\begin{aligned} -(J_{1}*U-U)(\xi )+cU'(\xi )-f(U(\xi ))=0,\ U'(\xi )>0, \ \xi \in \mathbb {R} \end{aligned}$$

(1.3)

admits a unique (up to translation) solution U connecting 0 and 1. Moreover, U is of class $C^{k+1}$ if f is of class $C^{k}$ for some $k\ge 1$, and its speed c is given by

$$\begin{aligned} c=\frac{\int \limits _{0}^{1}f(u)\mathrm{d}u}{\int \limits _{-\infty }^{\infty }\left( U'(\xi )\right) ^{2}\mathrm{d}\xi }, \end{aligned}$$

which can be positive or negative [1]. We assume that $c>0$ in the present paper, and the case $c<0$ can be dealt with by a same way. Furthermore, the wave profile U and its derivative $U'$ have exponential behaviors near $\pm \infty$:

$$\begin{aligned} \begin{aligned} B_{1}e^{-\delta _{1}\xi }\le 1-U(\xi )\le A_{1}e^{-\lambda _{1}\xi },\ \ \&\mathrm{when}\ \xi \rightarrow +\infty ,\\ B_{2}e^{\delta _{2}\xi }\le U(\xi )\le A_{2}e^{\lambda _{2}\xi },\ \ \ \ \&\mathrm{when}\ \xi \rightarrow -\infty ,\\ U'(\xi )\le A_{3}e^{-\lambda _{3}|\xi |}, \ \ \ \ \ \ \ \&\mathrm{when}\ |\xi |\rightarrow +\infty ,\\ \end{aligned} \end{aligned}$$

(1.4)

where $A_{i},B_{i},\lambda _{i},\delta _{i}(i\in \{1,2,3\})$ are positive constants, see [9, 10]. Following the technique as in [9, Section 1.5], we can also get that $U''$ has exponential behavior near $\pm \infty$.

Studies on the existence and stability of nonplanar traveling waves for the classical reaction diffusion equations or systems are already quite a lot, see [2,3,4,5, 11, 12, 16, 17, 20,21,22,23, 26] and references therein for scalar equations, and see [15, 18, 24, 25, 27, 28] and references therein for reaction–diffusion systems. While there are still few studies on nonplanar traveling waves of nonlocal dispersal equations. Chan and Wei proved the existence of pyramidal traveling wave solutions for the fractional bistable equation [7], and Li et al. proved the existence of pyramidal traveling wave solutions for the bistable nonlocal equation [14]. However, to the best of our knowledge, there is still no result about the stability of nonplanar traveling wave solutions for the nonlocal dispersal equations. In the current paper, we aim to prove the existence and stability of V-shaped traveling fronts for (1.1).

Since the curvature accelerates propagation of waves, it is naturally to assume that the speed s of nonplanar traveling waves satisfies $s>c$. Without loss of generality, we also assume that the solutions travel towards the $x_{2}-$direction; thus, they have the form $u(\varvec{x},t)=\widehat{u}(x_{1},x_{2}+st,t)$, and $\widehat{u}$ satisfies

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \widehat{u}_{t}-(J*\widehat{u}-\widehat{u})+s\widehat{u}_{x_{2}}-f(\widehat{u})=0,\quad (\varvec{x},t)\in \mathbb {R}^{2}\times (0,+\infty ),&{}\\ \widehat{u}(\varvec{x},0)=u_{0}(\varvec{x}), ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \varvec{x}\in \mathbb {R}^{2}.&{} \end{array}\right. } \end{aligned} \end{aligned}$$

(1.5)

Throughout this paper, we denote the solution of (1.5) by $\widehat{u}(\varvec{x},t;u_{0})$. In this paper, we first find a nontrivial steady-state function $V(\varvec{x})$ of (1.5), i.e., $V(\varvec{x})$ satisfies

$$\begin{aligned} \mathscr {L}[V]:=-(J*V-V)+sV_{x_{2}}-f(V)=0 \ \ \mathrm{in}\ \ \mathbb {R}^{2}, \end{aligned}$$

(1.6)

and then prove its stability.

Let $m_{*}=\sqrt{s^2-c^2}/c$ and

$$\begin{aligned} v^{-}(\varvec{x})=U\left( \frac{c}{s}(x_{2}+m_{*}|x_{1}|)\right) =\max \limits _{1\le j\le n}\left\{ U\left( \frac{c}{s}(x_{2}+m_{*}x_{1})\right) ,U\left( \frac{c}{s}(x_{2}-m_{*}x_{1})\right) \right\} . \end{aligned}$$

(1.7)

Then, $v^{-}(\varvec{x})$ is a subsolution to (1.6). Actually, denote $v_{j}^{1}(\varvec{x}):= U(A_{j}\cdot \varvec{x})$ with $A_{j}=\frac{c}{s}(m_{*}, 1)$. If we let ${\varvec{\xi }}=\varvec{A}\varvec{y}$ with $\varvec{A}$ a $2\times 2$ orthonormal matrix whose first row is $A_{j}$, then we have

$$\begin{aligned} \int \limits _{\mathbb {R}^{2}}J(\varvec{x}-\varvec{y})U(A_{j}\cdot \varvec{y})\mathrm{d}\varvec{y}= & {} \int \limits _{\mathbb {R}^{2}}J(\varvec{y})U(A_{j}\cdot (\varvec{x}-\varvec{y}))\mathrm{d}\varvec{y}=\int \limits _{\mathbb {R}^{2}}J(\varvec{A}^{-1}{\varvec{\xi }})U(A_{j}\cdot \varvec{x}-\xi _{1})\mathrm{d}{{\varvec{\xi }}}\\= & {} \int \limits _{\mathbb {R}^{2}}J({\varvec{\xi }})U(A_{j}\cdot \varvec{x}-\xi _{1})d{{\varvec{\xi }}}=\int \limits _{\mathbb {R}}J_{1}(\xi _{1})U(A_{j}\cdot \varvec{x}-\xi _{1})\mathrm{d}\xi _{1}. \end{aligned}$$

See (1.2) for the definition of $J_{1}$. It follows that

$$\begin{aligned} \mathscr {L}[v_{j}^{1}(\varvec{x})]= & {} -\left[ (J*U(A_{j}\cdot ))(\varvec{x})-U(A_{j}\cdot \varvec{x})\right] +sU_{x_{2}}(A_{j}\cdot \varvec{x})-f(U(A_{j}\cdot \varvec{x}))\\= & {} -(J_{1}*U-U)(A_{j}\cdot \varvec{x})-cU'(A_{j}\cdot \varvec{x})-f(U(A_{j}\cdot \varvec{x}))=0, \end{aligned}$$

which means that $v^{1}_{j}(\varvec{x})$ is a planar wave to (1.6). Similarly, denote $v_{j}^{2}(\varvec{x}):= U(A_{j}\cdot \varvec{x})$ with $A_{j}=\frac{c}{s}(-m_{*}, 1)$, and then, we can get $\mathscr {L}[v_{j}^{2}(\varvec{x})]=0$. Thus, $v^{-}(\varvec{x})$ is a subsolution.

Now we give the main results.

Theorem 1.1

(Existence) Assume (J1)–(J2) and (F1)–(F2) hold. For each $s>c$, (1.6) admits a solution $V_{*}(\varvec{x})$ with $\partial _{x_{2}}V_{*}(\varvec{x})>0$ in $\mathbb {R}^{2}$ and

$$\begin{aligned}&\lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}\left| V_{*}(\varvec{x})-v^{-}(\varvec{x})\right| =0, \end{aligned}$$

(1.8)

$$\begin{aligned}&v^{-}(\varvec{x})< V_{*}(\varvec{x})<1, \ \ \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

(1.9)

Then $u(\varvec{x},t)=V_{*}(x_{1},x_{2}+st)$ is a traveling front of (1.1), whose global average speed tends to c along with time, i.e.,

$$\begin{aligned} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{\mathrm{dist}(L_{t_{1}},L_{t_{2}})}{|t_{1}-t_{2}|}=c, \end{aligned}$$

(1.10)

where $L_{t}$ represents the level set of $u(\varvec{x},t)$ at time t.

Theorem 1.2

(Stability) Let $v_{0}\in C(\mathbb {R}^{2},[0,1])$ satisfy $v_{0}-v^{-}\in L^{1}(\mathbb {R}^{2})$ and

$$\begin{aligned}&\lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}|v_{0}(\varvec{x})-v^{-}(\varvec{x})|=0,\\&v^{-}(\varvec{x})\le v_{0}(\varvec{x}),~~~~\forall \ \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

Then, we have

$$\begin{aligned} \lim _{t\rightarrow +\infty }||v(\cdot ,t;v_{0})-V_{*}(\cdot )||_{L^{\infty }(\mathbb {R}^{2})}=0. \end{aligned}$$

Remark 1.3

(1.8) implies that $V_{*}(\varvec{x})$ has V-shaped level sets and behaves like planar traveling waves far away in the space. The speed of $V_{*}(\varvec{x})$ is a semi-continuum, that is, $s\in (c,+\infty )$, which is quite different to classical bistable case, while (1.10) tells that the average speed of $V_{*}(\varvec{x})$ is unique and always equals the planar wave’s speed c.

In the next section, we establish the existence result. The proof looks simpler than that of [14] but is a little different. In Sect. 3, we obtain the stability result.

2 Existence of V-shaped traveling fronts

For any $s>c$, the equation

$$\begin{aligned} s=\frac{\varphi _{xx}}{1+\varphi _{x}^{2}}+c\sqrt{1+\varphi _{x}^{2}} \end{aligned}$$

admits a unique solution $\varphi (x)$ whose asymptotic line is $y=m_{*}|x|$ satisfying $m_{*}|x|\le \varphi (x)$, see [16]. Furthermore, there exist positive constants $K_{0},K_{1},K_{2},K_{3}$ such that for all $x\in \mathbb {R}$,

$$\begin{aligned}&\max |\varphi '(x)|\ \le \ K_{0}, \end{aligned}$$

(2.1)

$$\begin{aligned}&\max \{|\varphi ''(x)|, |\varphi '''(x)|\}\ \le \ K_{1}\mathrm{sech}(\gamma x), \end{aligned}$$

(2.2)

$$\begin{aligned}&K_{2}\mathrm{sech}(\gamma x)\ \le \frac{s}{\sqrt{1+\varphi '(x)^{2}}}-c \le K_{3}\mathrm{sech}(\gamma x), \end{aligned}$$

(2.3)

$$\begin{aligned}&\mu _{-}\ \le \mu (x)=\frac{ s(\varphi (x)-m_{*}|x|)}{s-c\sqrt{1+\varphi '(x)^{2}}}\le \mu _{+}, \end{aligned}$$

(2.4)

where $\mu _{\pm }>0$ are constants and $\gamma =sm_{*}=\frac{s\sqrt{s^{2}-c^{2}}}{c}>0.$

2.1 Construction of a supersolution

By the assumption (F1), there exists a positive constant $\delta _{0}\in (0,\frac{1}{4})$ such that

$$\begin{aligned} -f'(u)\ge \kappa _{1} \ \mathrm{if}\ |u|\le 2\delta _{0} \ \mathrm{or}\ |1-u|\le 2\delta _{0}, \end{aligned}$$

(2.5)

where $\kappa _{1}:=\frac{1}{2}\min \{-f'(0),-f'(1)\}>0.$

Lemma 2.1

There exist a positive constant $\varepsilon _{0}^{+}$ and a positive function $\alpha _{0}^{+}(\varepsilon )$ such that for any $\varepsilon \in (0, \varepsilon _{0}^{+})$ and $\alpha \in (0, \alpha _{0}^{+}(\varepsilon ))$, the function

$$\begin{aligned} v^{+}(\varvec{x};\varepsilon ,\alpha )=U\left( \frac{x_{2}+\frac{1}{\alpha }\varphi (\alpha x_{1})}{\sqrt{1+(\varphi '(\alpha x_{1}))^{2}}}\right) +\varepsilon \mathrm{sech}(\gamma \alpha x_{1}), \end{aligned}$$

(2.6)

is a supersolution to (1.6) and

$$\begin{aligned}&\lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R }|v^{+}(\varvec{x};\varepsilon ,\alpha )-v^{-}(\varvec{x})|\le 2\varepsilon , \end{aligned}$$

(2.7)

$$\begin{aligned}&v^{-}(\varvec{x})<v^{+}(\varvec{x};\varepsilon ,\alpha ) \ \ \mathrm{for}\ \ \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

(2.8)

Proof

Assume $\alpha \in (0,1)$ and write $v^{+}(\varvec{x})$ instead of $v^{+}(\varvec{x};\varepsilon ,\alpha )$ throughout the proof. Denote

$$\begin{aligned} \zeta (\varvec{x})=\frac{x_{2}+\frac{1}{\alpha }\varphi (\alpha x_{1})}{\sqrt{1+(\varphi '(\alpha x_{1}))^{2}}}~~~\mathrm{and}~~~\sigma (x_{1})=\mathrm{sech}(\gamma \alpha x_{1}), \end{aligned}$$

then $v^{+}(\varvec{x})$ can be rewritten as $v^{+}(\varvec{x})=U(\zeta (\varvec{x}))+\varepsilon \sigma (x_{1}).$ By the equation (1.3) and the definition of $\mathscr {L}$, we have

$$\begin{aligned} \mathscr {L}[v^{+}(\varvec{x})]= & {} -(J*v^{+}-v^{+})(\varvec{x})+s\partial _{x_{2}}v^{+}(\varvec{x})-f(v^{+}(\varvec{x}))\\= & {} -(J*U(\zeta (\cdot )))(\varvec{x})-\varepsilon (J*\sigma )(\varvec{x})+(J_{1}*U)(\zeta (\varvec{x}))+\varepsilon \sigma (x_{1})\\&+\frac{ s}{\sqrt{1+(\varphi '(\alpha x_{1}))^{2}}}U'(\zeta (\varvec{x}))-cU'(\zeta (\varvec{x}))+f(U(\zeta (\varvec{x})))-f(v^{+}(\varvec{x})). \end{aligned}$$

Denote

$$\begin{aligned}&I :=-(J*U(\zeta (\cdot )))(\varvec{x})+(J_{1}*U)(\zeta (\varvec{x})),~~II:=-\varepsilon (J*\sigma )(\varvec{x})+\varepsilon \sigma (x_{1}),\\&III :=\left( \frac{ s}{\sqrt{1+(\varphi '(\alpha x_{1}))^{2}}}-c\right) U'(\zeta (\varvec{x})),~~IV:=f(U(\zeta (\varvec{x})))-f(v^{+}(\varvec{x})). \end{aligned}$$

Now we estimate the four terms.

(1) Estimate of term I. Since J is radially symmetric, it is easy to see that

$$\begin{aligned} I =-\int \limits _{\mathbb {R}^{2}}J(\varvec{y})U(\zeta (\varvec{x}-\varvec{y}))\mathrm{d}\varvec{y} +\int \limits _{\mathbb {R}}J_{1}(\mu )U(\zeta (\varvec{x})-\mu )\mathrm{d}\mu . \end{aligned}$$

Let

$$\begin{aligned} A= \left( \begin{array}{ccc} \frac{\varphi '(\alpha x_{1})}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}} &{} \frac{1}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}}\\ \frac{1}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}} &{} -\frac{\varphi '(\alpha x_{1})}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}} \end{array} \right) , \end{aligned}$$

and ${{\varvec{\xi }}}=A\varvec{y}$, where ${{\varvec{\xi }}}=(\mu ,\nu )^{T}$. Such an orthogonal transformation gives that $\mu =\frac{\varphi '(\alpha x_{1})}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}}y_{1}+\frac{1}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}}y_{2}$. Then

$$\begin{aligned} \int \limits _{\mathbb {R}}J_{1}(\mu )U(\zeta (\varvec{x})-\mu )\mathrm{d}\mu= & {} \int \limits _{\mathbb {R}^{2}}J({{\varvec{\xi }}})U(\zeta (\varvec{x})-\mu )\mathrm{d}\mu \mathrm{d}\nu \\= & {} \int \limits _{\mathbb {R}^{2}}J(A\varvec{y}^{T})U\left( \zeta (\varvec{x})-\frac{\varphi '(\alpha x_{1})y_{1}+y_{2}}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}}\right) \mathrm{d}y_{1}\mathrm{d}y_{2}\\= & {} \int \limits _{\mathbb {R}^{2}}J(\varvec{y})U\left( \frac{x_{2}-y_{2}+\frac{1}{\alpha }\varphi (\alpha x_{1})-\varphi '(\alpha x_{1})y_{1}}{\sqrt{1+\left( \varphi '(\alpha x_{1})\right) ^{2}}}\right) \mathrm{d}y_{1}\mathrm{d}y_{2}. \end{aligned}$$

Let

$$\begin{aligned} \mu ^{*}(t)=\frac{x_{2}-y_{2}+(1-t)\left( \frac{1}{\alpha }\varphi (\alpha x_{1})-\varphi '(\alpha x_{1})y_{1}\right) +t\cdot \frac{1}{\alpha }\varphi (\alpha (x_{1}-y_{1}))}{\sqrt{1+\left( \varphi '(\alpha (x_{1}-ty_{1}))\right) ^{2}}}, \end{aligned}$$

and define $F(t)=-U(\mu ^{*}(t)), t\in (0,1)$, and then, (2.9) can be written as

$$\begin{aligned} I=\int \limits _{\mathbb {R}^{2}}J(\varvec{y})\left( -U(\mu ^{*}(1))+U(\mu ^{*}(0))\right) \mathrm{d}\varvec{y} =\int \limits _{\mathbb {R}^{2}}J(\varvec{y})\left( F(1)-F(0)\right) \mathrm{d}\varvec{y}. \end{aligned}$$

(2.9)

Denote $y_{1}(t)=\alpha (x_{1}-ty_{1})$. Then, $\mu ^{*}_{t}(t)=A(t)+B(t)\mu ^{*}(t)$, where $B(t)=\frac{\alpha \varphi '(y_{1}(t))\varphi ''(y_{1}(t))y_{1}}{1+\left( \varphi '(y_{1}(t))\right) ^{2}}$ and

$$\begin{aligned} A(t)=\frac{-\left( \frac{1}{\alpha }\varphi (\alpha x_{1})-\varphi '(\alpha x_{1})y_{1}\right) +\frac{1}{\alpha }\varphi (\alpha (x_{1}-y_{1}))}{\sqrt{1+\left( \varphi '(y_{1}(t))\right) ^{2}}}=\frac{\frac{\alpha }{2}\varphi ''(y_{1}(\tau ))y_{1}^{2}}{\sqrt{1+\left( \varphi '(y_{1}(t))\right) ^{2}}},~\tau \in (0,1). \end{aligned}$$

Furthermore, $A_{t}(t)=A(t)B(t)$, and thus, $\mu ^{*}_{tt}(t)=2A(t)B(t)+(B_{t}(t)+B^{2}(t))\mu ^{*}(t)$, where

$$\begin{aligned} B_{t}(t)=\frac{-\alpha ^{2} y_{1}^{2}\left[ \left( \varphi ''(y_{1}(t))\right) ^{2}\left( 1-\left( \varphi '(y_{1}(t))\right) ^{2}\right) +\varphi '(y_{1}(t))\varphi '''(y_{1}(t))\left( 1+\left( \varphi '(y_{1}(t))\right) ^{2}\right) \right] }{\left[ 1+\left( \varphi '(y_{1}(t))\right) ^{2}\right] ^{2}}. \end{aligned}$$

Following from (2.1)–(2.2), we have

$$\begin{aligned}&|A(t)|\le \alpha K_{1}y_{1}^{2}\mathrm{sech}(\gamma \alpha (x_{1}-\tau y_{1})),\\&|B(t)|\le \alpha K_{0}K_{1} |y_{1}|\mathrm{sech}(\gamma \alpha (x_{1}-t y_{1})),\\&|B_{t}(t)|\le \alpha ^{2}K_{1}(1+K_{0})(1+K_{0}^2) y_{1}^{2}\mathrm{sech}(\gamma \alpha (x_{1}-t y_{1})). \end{aligned}$$

Since $F(1)-F(0)=F'(0)+\int \limits _{0}^{1}(1-t)F''(t)\mathrm{d}t$ and

$$\begin{aligned} F'(t)=-U'(\mu ^{*}(t))\mu ^{*}_{t}(t),~~ F''(t)=-U''(\mu ^{*}(t))(\mu ^{*}_{t}(t))^{2}-U'(\mu ^{*}(t))\mu ^{*}_{tt}(t), \end{aligned}$$

we have

$$\begin{aligned}&\left| -U(\mu ^{*}(1))+U(\mu ^{*}(0))\right| \nonumber \\&\le \left| -U'(\mu ^{*}(0))\left( A(0)+B(0)\mu ^{*}(0)\right) \right| \nonumber \\&\quad +\bigg |\int \limits _{0}^{1}(1-t)\Big [-U''(\mu ^{*}(t))\left( A^{2}(t)+2A(t)B(t) \mu ^{*}(t)+B^{2}(t)(\mu ^{*}(t))^{2}\right) \nonumber \\&~~~~~~~~~~~~~~~~~~~~-U'(\mu ^{*}(t))\left( 2A(t)B(t)+(B_{t}(t) +B^{2}(t))\mu ^{*}(t)\right) \Big ]\mathrm{d}t\bigg | \nonumber \\&\le C_{U}\sigma (x_{1})\left( \alpha K_{1}y_{1}^{2}e^{\gamma |y_{1}|} +\alpha K_{0}K_{1}|y_{1}|\right) +C_{U}\sigma (x_{1})\int \limits _{0}^{1}(1-t)\Big (\alpha ^{2} K_{1}^{2}y_{1}^{4} \nonumber \\&\quad +4\alpha ^{2} K_{0}K_{1}^{2}|y_{1}|^{3}+2\alpha ^{2} K_{0}^{2}K_{1}^{2}y_{1}^{2} +\alpha ^{2}K_{1}(1+K_{0})(1+K_{0}^{2}) \Big )e^{\gamma |y_{1}|}\mathrm{d}t \nonumber \\&\le C_{U}K^{*}\alpha \sigma (x_{1})\left( |y_{1}|+4y_{1}^{2}+4|y_{1}|^{3} +y_{1}^{4}\right) e^{\gamma |y_{1}|}, \end{aligned}$$

(2.10)

where $K^{*}=\max \{K_{1},K_{1}^{2},K_{0}K_{1},K_{0}K_{1}^{2},K_{0}^{2}K_{1}^{2},K_{1}(1+K_{0})(1+K_{0}^{2})\}$ and

$$\begin{aligned} C_{U}=\max \{|| U'(\mu )||_{\infty },|| U''(\mu )||_{\infty },|| U'(\mu )\mu ||_{\infty },|| U''(\mu )\mu ||_{\infty },|| U''(\mu )\mu ^{2}||_{\infty }\}. \end{aligned}$$

Here, $||\cdot ||_{\infty }$ is the supremum norm about $\mu \in \mathbb {R}$. And in the above estimate we use the inequality $\mathrm{sech}(x_{1}+y_{1})\le \mathrm{sech}x_{1}\cdot e^{|y_{1}|}$. Combining (2.9)–(2.10) and the above estimates, we have

$$\begin{aligned} I|\le C_{U}K^{*}\alpha \sigma (x_{1})\int \limits _{\mathbb {R}^{2}}J(\varvec{y})\left( |y_{1}|+4y_{1}^{2}+4|y_{1}|^{3}+y_{1}^{4}\right) e^{\gamma |y_{1}|}\mathrm{d}\varvec{y}. \end{aligned}$$

Under the condition (J2), the integral $\int \limits _{\mathbb {R}^{2}}J(\varvec{y})\left( |y_{1}|+4y_{1}^{2}+4|y_{1}|^{3}+y_{1}^{4}\right) e^{\gamma |y_{1}|}\mathrm{d}\varvec{y}$ is bounded for some $\lambda >0$, and thus, there exists a constant $C_{1}>0$ such that

$$\begin{aligned} |I|\le C_{1}\alpha \sigma (x_{1}). \end{aligned}$$

(2) Estimate of term II.

$$\begin{aligned} II= & {} -\varepsilon \int \limits _{\mathbb {R}^{2}}J(\varvec{y})\left( \sigma (x_{1}-y_{1})-\sigma (x_{1})\right) \mathrm{d}\varvec{y}\\= & {} \varepsilon \gamma \alpha \int \limits _{\mathbb {R}^{2}}J(\varvec{y})\sigma '(x_{1}-\theta y_{1})y_{1}\mathrm{d}\varvec{y}, \end{aligned}$$

where $\theta \in (0,1)$. Since $\mathrm{sech}'x=\mathrm{sech} x\cdot \frac{e^{-x}-e^{x}}{e^{x}+e^{-x}}$, we have

$$\begin{aligned} \left| II\right|\le & {} \varepsilon \gamma \alpha \sigma (x_{1})\int \limits _{\mathbb {R}^{2}}J(\varvec{y})|y_{1}|e^{\gamma |y_{1}|}\mathrm{d}\varvec{y}. \end{aligned}$$

Again the assumption (J2) guarantees that $\int \limits _{\mathbb {R}^{2}}J(\varvec{y})|y_{1}|e^{\gamma |y_{1}|}\mathrm{d}\varvec{y}$ is bounded for some $\lambda >0$, and thus, there exists a constant $C_{2}>0$ such that

$$\begin{aligned} \left| II\right| \le C_{2}\varepsilon \gamma \alpha \sigma (x_{1}). \end{aligned}$$

(3) Estimate of terms III and IV. By (2.3), we have

$$\begin{aligned} 0<K_{2}\sigma (x_{1})U'(\zeta (\varvec{x}))\le \left( \frac{s}{\sqrt{1+(\varphi '(\alpha x_{1}))^{2}}}-c\right) U'(\zeta (\varvec{x})). \end{aligned}$$

About the fourth term, we have

$$\begin{aligned} IV=-f'\left( U(\zeta (\varvec{x}))+\tilde{\theta }\varepsilon \sigma (x_{1})\right) \cdot \varepsilon \sigma (x_{1}), \end{aligned}$$

where $\tilde{\theta }\in (0,1).$

In order to prove $\mathscr {L}[v^{+}(\varvec{x})]\ge 0$, we consider two cases.

Case 1. $U(\zeta (\varvec{x}))\ge 1-\delta _{0}$ or $U(\zeta (\varvec{x}))\le \delta _{0}$.

Then, $-f'\left( U(\zeta (\varvec{x}))+\tilde{\theta }\varepsilon \sigma (x_{1})\right) \ge \kappa _{1}$ by (2.5), provided that $0<\varepsilon <\delta _{0}$. And thus

$$\begin{aligned} \mathscr {L}[v^{+}(\varvec{x})]\ge -\left( C_{1}+C_{2}\gamma \varepsilon \right) \alpha \sigma (x_{1})+\kappa _{1}\varepsilon \sigma (x_{1})\ge 0 \end{aligned}$$

for any $\alpha$ satisfying $\alpha \le \frac{\kappa _{1}\varepsilon }{C_{1}+C_{2}\gamma }$.

Case 2. $\delta _{0}\le U(\zeta (\varvec{x}))\le 1-\delta _{0}$.

Denote $U_{*}=\min _{U(x)\in \left[ \delta _{0},1-\delta _{0}\right] }U'(x)$ and $f^{*}=\max _{x\in [-1,2]}|f'(x)|$. We have

$$\begin{aligned} \mathscr {L}[v^{+}(\varvec{x})]\ge & {} -\left( C_{1}+C_{2}\gamma \varepsilon \right) \alpha \sigma (x_{1})+K_{2}\sigma (x_{1})U_{*}-\varepsilon f^{*}\sigma (x_{1})\ge 0 \end{aligned}$$

provided that $\left( C_{1}+C_{2}\gamma \right) \alpha +\varepsilon f^{*}\le K_{2}U_{*}.$ Let

$$\begin{aligned} \varepsilon _{0}^{+}:=\min \left\{ 1,\delta _{0},\frac{K_{2}U_{*}}{2f^{*}}\right\} ,\quad \alpha _{0}^{+}(\varepsilon ):=\min \left\{ 1,\frac{\kappa _{1}\varepsilon }{C_{1}+C_{2}\gamma },\frac{K_{2}U_{*}}{2(C_{1}+C_{2}\gamma )}\right\} , \end{aligned}$$

then $v^{+}(\varvec{x})$ is a supersolution if $0<\varepsilon <\varepsilon _{0}^{+}$ and $0<\alpha <\alpha _{0}^{+}(\varepsilon )$. A similar argument to that of Taniguchi [20, Lemma 7] yields (2.7)-(2.8). The proof is complete. $\square$

2.2 Proof of the existence result

First, we establish the comparison principle. Define

$$\begin{aligned} BUC(\mathbb {R}^{2}):=\{u:\mathbb {R}^{2}\rightarrow \mathbb {R}, u\mathrm{~is~bounded~and~uniformly~ continuous~in}~\mathbb {R}^{2} \}. \end{aligned}$$

Theorem 2.2

Assume that (J1) and (F1)–(F2) hold. Let $u_{0}(\varvec{x})$ and $\partial _{x_{2}}u_{0}(\varvec{x})$ belong to $BUC(\mathbb {R}^{2})$, and then, the following Cauchy problem

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \ \widehat{u}_{t}=J*\widehat{u}-\widehat{u}+b\widehat{u}_{x_{2}}+f(\widehat{u}),\quad (\varvec{x},t)\in \mathbb {R}^{2}\times [0,\infty ),&{}\\ \ \widehat{u}(\varvec{x},0)=u_{0}(\varvec{x}),\qquad \qquad \qquad \quad \varvec{x}\in \mathbb {R}^{2},&{}\\ \end{array}\right. } \end{aligned} \end{aligned}$$

has a unique solution $\widehat{u}(\varvec{x},t;u_{0})\in C(\mathbb {R}^{2}\times [0,\infty ),[0,2])$, which is also differentiable with respect to $x_{2}$. Here, $b\in \mathbb {R}$ is a nonzero constant. Moreover, if $u_{0}(\varvec{x})$ is globally Lipschitz continuous, then $\widehat{u}(\varvec{x},t;u_{0})$ is also a globally Lipschitz solution which is uniform in time.

Lemma 2.3

(Maximum principle) Assume that (J1) hold and that $\widehat{u}\in C(\mathbb {R}^{2}\times [0,\infty ))$ is bounded and differentiable with respect to $x_{2}$. If $\widehat{u}$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \ \widehat{u}_{t}\ge J*\widehat{u}-\widehat{u}+b\widehat{u}_{x_{2}} +K(\varvec{x},t)\widehat{u},\quad (\varvec{x},t)\in \mathbb {R}^{2}\times [0,\infty ),&{}\\ \ \widehat{u}(\varvec{x},0)\ge 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \varvec{x}\in \mathbb {R}^{2},&{}\\ \end{array}\right. } \end{aligned}$$

where $K(\varvec{x},t):\mathbb {R}^{2}\times [0,\infty )\rightarrow \mathbb {R}$ is continuous and uniformly bounded, and b is a nonzero constant, then $\widehat{u}(\varvec{x},t)\ge 0$ for $(\varvec{x},t)\in \mathbb {R}^{2}\times [0,\infty )$. Furthermore, if $\widehat{u}(\varvec{x},0)\not \equiv 0$ for $\varvec{x}\in \mathbb {R}^{2}$, then $\widehat{u}(\varvec{x},t)> 0$ for $(\varvec{x},t)\in \mathbb {R}^{2}\times (0,\infty )$.

Lemma 2.4

(Comparison principle) Assume that (J1) holds and $u_{1},u_{2}\in C(\mathbb {R}^{2}\times [0,\infty ))$ are both bounded and differentiable with respect to $x_{2}$. Denote $\mathscr {L}_{t}[u]=u_{t}-(J*u-u)+b u_{x_{2}} -f(u)$, where f is continuously differentiable with respect to u, $f'(u)$ is uniformly bounded and b is a nonzero constant. If $u_{1},u_{2}$ satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} \ \mathscr {L}_{t}[u_{1}]\ge \mathscr {L}_{t}[u_{2}],\quad \quad (\varvec{x},t)\in \mathbb {R}^{2}\times [0,\infty ),&{}\\ \ u_{1}(\varvec{x},0)\ge u_{2}(\varvec{x},0),\quad \varvec{x}\in \mathbb {R}^{2},&{}\\ \end{array}\right. } \end{aligned}$$

then $u_{1}\ge u_{2}$ on $(\varvec{x},t)\in \mathbb {R}^{2}\times [0,\infty )$. Furthermore, if $u_{1}(\varvec{x},0)\not \equiv u_{2}(\varvec{x},0)$ for $\varvec{x}\in \mathbb {R}^{2}$, then $u_{1}> u_{2}$ for $(\varvec{x},t)\in \mathbb {R}^{2}\times (0,\infty )$.

The proof of the above three results can be referred to [14]. Now we prove Theorem 1.1.

Proof of Theorem 1.1

After making a slight modification of the proof of [14, Theorem 1.1], we can prove the existence of $V_{*}$ and (1.8)–(1.9). Now we focus on the proof of (1.10).

By the comparison principle, we have

$$\begin{aligned} v^{-}(x_{1},x_{2}+st)< u(\varvec{x},t):=V_{*}(x_{1},x_{2}+st)< v^{+}(x_{1},x_{2}+st),~~~\forall \varvec{x}\in \mathbb {R}^{2}, t\in \mathbb {R}, \end{aligned}$$

(2.11)

where $v^{-}$ and $v^{+}$ are defined by (1.7) and (2.6), respectively. Fix a constant $\delta \in (0,1)$ and denote the level sets of $v^{\pm }(x_{1},x_{2}+st)=\delta$ at time t by $L^{\pm }_{t}$. Due to (2.11), the level sets $L^{\pm }_{t}$ and $L_{t}$ do not intersect each other at any time t. We know

$$\begin{aligned} L^{-}_{t}=\left\{ (x_{1},g^{-}(x_{1},t))\in \mathbb {R}^{2}: g^{-}(x_{1},t)=\frac{s}{c}U^{-1}(\delta )-m_{*}|x_{1}|-st\right\} \end{aligned}$$

and

$$\begin{aligned} L^{+}_{t}=\left\{ (x_{1},g^{+}(x_{1},t))\!\in \!\mathbb {R}^{2}: g^{+}(x_{1},t)\!=\!\sqrt{1\!+\!\varphi '(\alpha x_{1})^{2}}U^{-1}(\delta \!-\!\varepsilon \sigma (x_{1}))-\frac{\varphi (\alpha x_{1})}{\alpha }-st \right\} . \end{aligned}$$

For convenience, we also denote

$$\begin{aligned} L_{t}=\left\{ (x_{1},g(x_{1},t))\in \mathbb {R}^{2}: V_{*}(x_{1},g(x_{1},t)+st)=\delta \right\} . \end{aligned}$$

Since

$$\begin{aligned} \delta =v^{-}(x_{1},g^{-}(x_{1},t)+st)=v^{+}(x_{1},g^{+}(x_{1},t)+st)>v^{-}(x_{1},g^{+}(x_{1},t)+st), \end{aligned}$$

we know $g^{-}(x_{1},t)>g^{+}(x_{1},t)$ for all $x_{1}\in \mathbb {R}$ and $t\in \mathbb {R}$ by the monotonicity of $v^{-}(\varvec{x})$ on $x_{2}$. Similarly, there hold $g^{-}(x_{1},t)>g(x_{1},t)$ and $g(x_{1},t)>g^{+}(x_{1},t)$ for all $x_{1}\in \mathbb {R}$ and $t\in \mathbb {R}$. In summary, there is

$$\begin{aligned} g^{-}(x_{1},t)>g(x_{1},t)>g^{+}(x_{1},t)~~~\mathrm{for~} x_{1}\in \mathbb {R}, t\in \mathbb {R}. \end{aligned}$$

(2.12)

Moreover, we know that

$$\begin{aligned} \begin{aligned}&\mathrm{dist}(L^{-}_{t},L^{+}_{t})=\inf _{\varvec{x}\in L^{-}_{t},\varvec{y}\in L^{+}_{t}}|\varvec{x}-\varvec{y}|\le \inf _{x_{1}\in \mathbb {R}}\left| g^{+}(x_{1},t)-g^{-}(x_{1},t)\right| \\&\le \inf _{x_{1}\in \mathbb {R}}\left( \left| \frac{s}{c}U^{-1}(\delta )-\sqrt{1+|\varphi '(\alpha x_{1})|^{2}}U^{-1}(\delta -\varepsilon \sigma (x_{1}))\right| + \frac{\varphi (\alpha x_{1})}{\alpha } -m_{*}|x_{1}|\right) =0 \end{aligned} \end{aligned}$$

holds for each fixed $\alpha >0$. Consequently, $\mathrm{dist}(L^{\pm }_{t},L_{t})=0$. Define

$$\begin{aligned} M^{*}:=\sup _{x_{1}\in \mathbb {R},t\in \mathbb {R}}\left| g^{+}(x_{1},t)-g^{-}(x_{1},t)\right| >0. \end{aligned}$$

Obviously, $M^{*}<+\infty$ for each fixed $\alpha$. Now we take two moments $t_{1}$, $t_{2}\in \mathbb {R}$ and assume $s(t_{2}-t_{1})>M_{*}$ without loss of generality. Under the assumption $s(t_{2}-t_{1})>M_{*}$, there holds $g^{-}(x_{1},t_{2})<g^{+}(x_{1},t_{1})$, and thus, $L^{+}_{t_{1}}$ does not intersect $L^{-}_{t_{2}}$.

The inequality (2.12) means that the level set $L_{t}$ of $u(\varvec{x},t)$ is between those of $L^{\pm }_{t}$ for all $x_{1}\in \mathbb {R}$ at any time $t\in \mathbb {R}$. Thus by the definition of $M^{*}$ and the choice of $t_{i}~(i=1,2)$, it is straightforward that

$$\begin{aligned} \mathrm{dist}(L_{t_{1}},L_{t_{2}})\le \mathrm{dist}(L^{-}_{t_{1}},L^{+}_{t_{2}}). \end{aligned}$$

(2.13)

To obtain $\mathrm{dist}(L^{-}_{t_{1}},L^{+}_{t_{2}})$, it is sufficient to consider all the perpendicular segments from $L^{+}_{t_{2}}$ to $L^{-}_{t_{1}}$ except those intersecting $L^{+}_{t_{2}}$ once more. Now take an arbitrary point $\varvec{x}^{+}_{2}\in L^{+}_{t_{2}}$ and find the corresponding point $\varvec{x}^{-}_{1}\in L^{-}_{t_{1}}$ such that the segment generated by $\varvec{x}^{+}_{2}$ and $\varvec{x}^{-}_{1}$ is perpendicular to $L^{-}_{t_{1}}$. We denote this perpendicular segment by $\textit{l}_{\varvec{x}^{-}_{1}\varvec{x}^{+}_{2}}$. Obviously, $\textit{l}_{\varvec{x}^{-}_{1}\varvec{x}^{+}_{2}}$ is also perpendicular to $L^{-}_{t_{2}}$, and we denote their intersection point by $\varvec{x}^{-}_{2}$. It follows immediately that $\left| \varvec{x}^{-}_{2}-\varvec{x}^{-}_{1}\right| =c|t_{2}-t_{1}|$. Then for any $\varvec{x}^{+}_{2}\in L^{+}_{t_{2}}$ and $\varvec{x}^{-}_{i}~(i=1,2)$ chosen by this way, we have

$$\begin{aligned} \mathrm{dist}(L^{-}_{t_{1}},L^{+}_{t_{2}})= & {} \inf _{\varvec{x}^{+}_{2}\in L^{+}_{t_{2}}}\left| \varvec{x}^{+}_{2}-\varvec{x}^{-}_{1}\right| \\= & {} \inf _{\varvec{x}^{+}_{2}\in L^{+}_{t_{2}}}\left( \left| \varvec{x}^{+}_{2}-\varvec{x}^{-}_{2}\right| +\left| \varvec{x}^{-}_{2}-\varvec{x}^{-}_{1}\right| \right) \\= & {} \inf _{\varvec{x}^{+}_{2}\in L^{+}_{t_{2}}}\left( \left| \varvec{x}^{+}_{2}-\varvec{x}^{-}_{2}\right| +c|t_{2}-t_{1}|\right) \\= & {} \mathrm{dist}(L^{-}_{t_{2}},L^{+}_{t_{2}})+c|t_{2}-t_{1}|=c|t_{2}-t_{1}|. \end{aligned}$$

Here, $|\cdot |$ denotes the Euclidean norm in $\mathbb {R}^{N}~(N\ge 1)$. This fact and (2.13) yield that

$$\begin{aligned} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{\mathrm{dist}(L_{t_{1}},L_{t_{2}})}{|t_{1}-t_{2}|}\le c. \end{aligned}$$

Now we prove that the average speed is not less than c. For any given point $\varvec{x}_{i}\in L_{t_{i}}~(i=1,2)$, draw a line passing through $\varvec{x}_{i}$ and parallel to the $x_{2}$ axis. Necessarily, this line intersects $L_{t_{i}}^{-}$ at a point, which is still denoted by $\varvec{x}^{-}_{i}$. Since $\varvec{x}_{i}\in L_{t_{i}}$ is arbitrary, $\varvec{x}^{-}_{i}$ is also arbitrary, and vice versa. Obviously,

$$\begin{aligned} \begin{aligned} |\varvec{x}_{2}-\varvec{x}_{1}|&\ge |\varvec{x}_{2}-\varvec{x}^{-}_{1}|-|\varvec{x}^{-}_{1}-\varvec{x}_{1}|\\&\ge |\varvec{x}^{-}_{1}-\varvec{x}^{-}_{2}|-|\varvec{x}^{-}_{2}-\varvec{x}_{2}|-|\varvec{x}^{-}_{1}-\varvec{x}_{1}|\\&\ge |\varvec{x}^{-}_{1}-\varvec{x}^{-}_{2}|-2M_{*}\\&\ge c|t_{2}-t_{1}|-2M_{*}. \end{aligned} \end{aligned}$$

(2.14)

It follows from (2.14) that

$$\begin{aligned} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{\mathrm{dist}(L_{t_{1}},L_{t_{2}})}{|t_{1}-t_{2}|}= & {} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{\inf _{\varvec{x}_{1}\in L_{t_{1}},\varvec{x}_{2}\in L_{t_{2}}}\left| \varvec{x}_{2}-\varvec{x}_{1}\right| }{|t_{1}-t_{2}|}\\\ge & {} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{\inf _{\varvec{x}^{-}_{1}\in L^{-}_{t_{2}},\varvec{x}^{-}_{2}\in L^{-}_{t_{2}}}\left| \varvec{x}^{-}_{2}-\varvec{x}^{-}_{1}\right| -2M^{*}}{|t_{1}-t_{2}|}\\\ge & {} \lim _{|t_{1}-t_{2}|\rightarrow \infty }\frac{c|t_{2}-t_{1}|-2M^{*}}{|t_{1}-t_{2}|}=c. \end{aligned}$$

This implies the average speed is larger than or equal to c. This completes the proof. $\square$

3 Stability of V-shaped traveling fronts in $\mathbb {R}^{2}$

This section establishes the stability result.

Lemma 3.1

For any $M>0$, there exists a constant $C>0$ such that

$$\begin{aligned} \partial _{x_{2}}V_{*}(\varvec{x})\ge C~~\mathrm{and}~~\partial _{x_{2}}v^{+}(\varvec{x})\ge C~~~~\mathrm{if}~~~\left| x_{2}+m_{*}|x_{1}|\right| \le M, \end{aligned}$$

where $V_{*}$ and $v^{+}$ are given by Theorem 1.1 and (2.6), respectively. Moreover, we have

$$\begin{aligned} \lim _{R\rightarrow \infty }\sup _{\left| x_{2}+m_{*}|x_{1}|\right| \ge R}\partial _{x_{2}}v^{+}(\varvec{x})=\lim _{R\rightarrow \infty }\sup _{\left| x_{2}+m_{*}|x_{1}|\right| \ge R}\partial _{x_{2}}V_{*}(\varvec{x})=0. \end{aligned}$$

Proof

The assertions about $v^{+}$ are very straightforward. Now we prove the assertions about $V_{*}$. Since $\partial _{x_{2}}V_{*}>0$ in $\mathbb {R}^{2}$, it suffices to prove that for any sequence $\{(x_{n},z_{n})\}_{n\ge 1}\subseteq \mathbb {R}^2$ satisfying $|(x_{n},z_{n})|\rightarrow \infty$ and $\left| z_{n}+m_{*}|x_{n}|\right| \le M$, there is

$$\begin{aligned} \lim _{n\rightarrow \infty } \partial _{x_{2}}V_{*}(x_{n},z_{n})>0. \end{aligned}$$

Now we prove this result by contradiction. Assume $\lim _{n\rightarrow \infty }\partial _{x_{2}}V_{*}(x_{n}^{*},z_{n}^{*})=0$ for a certain sequence $\{(x_{n}^{*},z_{n}^{*})\}_{n\ge 1}$. Since $v^-<V<v^+$ in $\mathbb {R}^{2}$ and notice that $v^{-}$ is a subsolution, we have

$$\begin{aligned} \begin{aligned} s \partial _{x_{2}}V_{*}|_{(x_{n}^{*},z_{n}^{*})}&=J*V_{*}-V_{*}+f(V_{*})|_{(x_{n}^{*},z_{n}^{*})}\\&> J*v^{-}-v^{+}+f(v^-)+f(V_{*})-f(v^-)|_{(x_{n}^{*},z_{n}^{*})}\\&=J*v^{-}-v^{-}+f(v^-)+f'(V_{\tau })(V_{*}-v^-)+v^{-}-v^{+}|_{(x_{n}^{*},z_{n}^{*})}\\&\ge s\partial _{x_{2}}v^{-}+f'(V_{\tau })(V_{*}-v^-)+v^{-}-v^{+}|_{(x_{n}^{*},z_{n}^{*})}, \end{aligned} \end{aligned}$$

(3.1)

where $V_{\tau }$ is between $V_{*}$ and $v^{-}$. Under the condition $\left| z_{n}^{*}+m_{*}|x_{n}^{*}|\right| \le M$ and $|(x_{n}^{*},z_{n}^{*})|\rightarrow \infty$, there must be $|x_{n}^{*}|\rightarrow \infty$, and thus,

$$\begin{aligned} \lim _{n\rightarrow \infty }\mathrm{sech}(\gamma \alpha x_{n}^{*})=0~~~\mathrm{and}~~~\lim _{n\rightarrow \infty }\left[ \frac{z_{n}^{*}+\varphi (\alpha x_{n}^{*})/\alpha }{\sqrt{1+\left( \varphi '( \alpha x_{n}^{*})\right) ^{2}}}-\frac{c}{s}(z_{n}^{*}+m_{*}|x_{n}^{*}|)\right] =0. \end{aligned}$$

Then, it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }(v^{-}(x_{n}^{*},z_{n}^{*})-v^{+}(x_{n}^{*},z_{n}^{*}))=0, \end{aligned}$$

which further implies that

$$\begin{aligned} \lim _{n\rightarrow \infty }(V_{*}(x_{n}^{*},z_{n}^{*})-v^{-}(x_{n}^{*},z_{n}^{*}))=0. \end{aligned}$$

Let $n\rightarrow \infty$ in (3.1), and then, we have

$$\begin{aligned} 0\ge \liminf _{n\rightarrow \infty }c U'\left( \frac{c}{s}(z_{n}^{*}+m_{*}|x_{n}^{*}|\right) >0, \end{aligned}$$

which is a contradiction. Similarly,

$$\begin{aligned} \begin{aligned} s\partial _{x_{2}}V_{*}&=J*V_{*}-V_{*}+f(V_{*})\\&< J*v^{+}-v^{-}+f(v^+)+f(V_{*})-f(v^+)\\&=J*v^{+}-v^{+}+f(v^+)+f'(V_{\tau })(V_{*}-v^+)+v^{+}-v^{-}\\&\le s\partial _{x_{2}}v^{+}+f'(V_{\tau })(V_{*}-v^+)+v^{-}-v^{+}, \end{aligned} \end{aligned}$$

and thus $\partial _{x_{2}}V_{*}\rightarrow 0$ as $\left| \frac{c}{s}(x_{2}+m_{*}|x_{1}|)\right| \rightarrow \infty$. This completes the proof. $\square$

Lemma 3.2

There exist positive constants $\rho>0, \kappa >0$ and $\delta \in \left( 0,\delta _{0}\right)$ such that

$$\begin{aligned} u^{+}(\varvec{x},t)=V_{*}(x_{1},x_{2}+\xi +\rho \delta (1-e^{-\kappa t}))+\delta e^{-\kappa t} \end{aligned}$$

is a supersolution, where $\xi \in \mathbb {R}$ is a constant and $\delta _{0}$ is defined in (2.5).

Proof

Let $\mathscr {\tilde{L}}[u]=u_{t}-(J*u-u)+s\partial _{x_{2}}u-f(u)$. Then, we have

$$\begin{aligned} \begin{aligned} \mathscr {\tilde{L}}[u^{+}]&=\rho \delta \kappa e^{-\kappa t}\partial _{x_{2}}V_{*}-\delta \kappa e^{-\kappa t}-(J*V_{*}-V_{*})+s\partial _{x_{2}}V_{*}-f(u^{+})\\&=\rho \delta \kappa e^{-\kappa t}\partial _{x_{2}}V_{*}-\delta \kappa e^{-\kappa t}+f(V_{*})-f(u^{+})\\&=\delta e^{-\kappa t}\left( \rho \kappa \partial _{x_{2}}V_{*}-\kappa -f'(V_{*}+\tau \delta e^{-\kappa t}) \right) , \end{aligned} \end{aligned}$$

where $\tau \in (0,1)$. To prove the lemma, we argue as follows.

Case 1. $|x_{2}+m_{*}|x_{1}||> R_{0}$ for some $R_{0}>0$ large enough.

Without loss of generality, assume that $R_{0}>0$ is large enough such that $V_{*}< \delta _{0}$ or $V_{*}>1-\delta _{0}$. Then, we have

$$\begin{aligned} 0<V_{*}+\tau \delta e^{-\kappa t}<2\delta _{0}~~~\mathrm{or}~~~1-\delta _{0}<V_{*}+\tau \delta e^{-\kappa t}<1+\delta _{0}. \end{aligned}$$

It then follows by (2.5) that

$$\begin{aligned} -f'(V_{*}+\tau \delta e^{-\kappa t})>\kappa _{1}. \end{aligned}$$

Thus,

$$\begin{aligned} \mathscr {\tilde{L}}[u^{+}]>\delta e^{-\kappa t}\left( -\kappa -f'(V_{*}+\tau \delta e^{-\kappa t}) \right)>\delta e^{-\kappa t}\left( -\kappa +\kappa _{1} \right) >0 \end{aligned}$$

provided that $\kappa <\kappa _{1}$.

Case 2. $|x_{2}+m_{*}|x_{1}||\le R_{0}$.

In this case, it follows from Lemma 3.1 that $\partial _{x_{2}}V_{*}\ge C:=C(R_{0})$. Thus,

$$\begin{aligned} \begin{aligned} \mathscr {\tilde{L}}[u^{+}]&>\left( \rho C-\kappa -\max _{u\in [0,1+\delta _{0}]}|f'(u)|\right) \\&>\delta \kappa e^{-\kappa t}\left( \rho C-1-\frac{\max _{u\in [0,1+\delta _{0}]}|f'(u)|}{\kappa }\right) \\&>0 \end{aligned} \end{aligned}$$

provided that $\rho >1+\frac{\max _{u\in [0,1+\delta _{0}]}|f'(u)|}{\kappa }$. Taking $0<\kappa <\kappa _{1}$, $\rho >1+\frac{\max _{u\in [0,1+\delta _{0}]}|f'(u)|}{\kappa }$ and combining the above two cases, we know $\mathscr {\tilde{L}}[u^{+}]>0$ in $\mathbb {R}^{2}$. This completes the proof. $\square$

The proof of the following lemma is very similar to that of Lemma 3.2, and we omit it here.

Lemma 3.3

There exist positive constants $\rho>0, \kappa >0$ and $\delta \in \left( 0,\delta _{0}\right)$ such that

$$\begin{aligned} w^{+}(\varvec{x},t)=v^{+}(x_{1},x_{2}+\xi +\rho \delta (1-e^{-\kappa t}))+\delta e^{-\kappa t} \end{aligned}$$

is a supersolution, where $\xi \in \mathbb {R}$ is a constant $\delta _{0}$ is defined in (2.5).

Lemma 3.4

If $u(\varvec{x},t;u_{0})$ is the solution of the Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned}&u_{t}=J*u-u-su_{x_{2}}+f(u),~~~\varvec{x}\in \mathbb {R}^{2},~t>0\\&u(\varvec{x},0)=u_{0}(\varvec{x}),~~~~~~~~~~~~~\varvec{x}\in \mathbb {R}^{2}, \end{aligned} \right. \end{aligned}$$

where the initial function $u_{0}\in C(\mathbb {R}^{2})$ satisfies $u_{0}-v^{-}\in L^{1}(\mathbb {R}^{2})$ and

$$\begin{aligned} \lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}|u_{0}(\varvec{x})-v^{-}(\varvec{x})|=0, \end{aligned}$$

then for any fixed $T>0$, we have

$$\begin{aligned} \lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}|u(\varvec{x},T;u_{0})-v^{-}(\varvec{x})|=0. \end{aligned}$$

Proof

Let $w(\varvec{x})=U\left( \frac{c}{s}(x_{2}+\varphi (x_{1}))\right)$. First we show that $w-v^{-}\in L^{1}(\mathbb {R}^{2})$. In fact,

$$\begin{aligned} \int \limits _{\mathbb {R}^{2}}|w(\varvec{x})-v^{-}(\varvec{x})|\mathrm{d}\varvec{x}= & {} \int \limits _{\mathbb {R}^{2}}\left| U\left( \frac{c}{s}(x_{2}+\varphi (x_{1}))\right) -U\left( \frac{c}{s}(x_{2}+m_{*}|x_{1}|)\right) \right| \mathrm{d}\varvec{x}\\= & {} \int \limits _{\mathbb {R}^{2}}\left| U'(\tau \varsigma (\varvec{x})+(1-\tau )\eta (\varvec{x}))\right| (\varsigma (\varvec{x})-\eta (\varvec{x}))\mathrm{d}\varvec{x}\\\le & {} \int \limits _{\mathbb {R}^{2}}A_{3}e^{-\lambda _{3}|\tau \varsigma (\varvec{x})+(1-\tau )\eta (\varvec{x})|}(\varsigma (\varvec{x})-\eta (\varvec{x}))\mathrm{d}\varvec{x}\\\le & {} \frac{c}{s}\left( 1-\frac{c}{s}\right) \mu _{+}A_{3}\int \limits _{\mathbb {R}^{2}}e^{-\lambda _{3}|\tau \varsigma (\varvec{x})+(1-\tau )\eta (\varvec{x})|}d\varvec{x}<+\infty , \end{aligned}$$

where $\tau \in (0,1)$, $\varsigma (\varvec{x})=\frac{c}{s}(x_{2}+\varphi (x_{1}))$ and $\eta (\varvec{x})=\frac{c}{s}(x_{2}+m_{*}|x_{1}|).$ See (1.4) for $A_{3}, \lambda _{3}$ and see (2.4) for $\mu _{+}$. It follows that $w-u_{0}\in L^{1}(\mathbb {R}^{2})$. It is also not difficult to prove that

$$\begin{aligned} \lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}|w(\varvec{x})-v^{-}(\varvec{x})|=0. \end{aligned}$$

Thus, it suffices to prove that

$$\begin{aligned} \lim _{R\rightarrow +\infty }\sup _{|\varvec{x}|\ge R}|u(\varvec{x},T;u_{0})-w(\varvec{x})|=0 \end{aligned}$$

for any given $T>0$. Let $\Phi (\varvec{x},t):=u(\varvec{x},t;u_{0})-w(\varvec{x})$. Then, it satisfies

$$\begin{aligned} \left\{ \begin{aligned}&\Phi _{t}=J*\Phi -\Phi -s\partial _{x_{2}}\Phi +f'(\Phi _{\tau })\Phi ,~~~\varvec{x}\in \mathbb {R}^{2},~t>0,\\&\Phi (\varvec{x},0)=u_{0}(\varvec{x})-w(\varvec{x}),~~~~~\varvec{x}\in \mathbb {R}^{2}, \end{aligned} \right. \end{aligned}$$

where $\Phi _{\tau }=\tau u+(1-\tau )w$ with $\tau \in (0,1)$. Let $\hat{\Phi }$ be the solution of the following Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned}&\hat{\Phi }_{t}=J*\hat{\Phi }-\hat{\Phi }-s\partial _{x_{2}}\hat{\Phi }+M\hat{\Phi },~~~\varvec{x}\in \mathbb {R}^{2},~t>0,\\&\hat{\Phi }(\varvec{x},0)=|u_{0}(\varvec{x})-w(\varvec{x})|,~~\varvec{x}\in \mathbb {R}^{2}, \end{aligned} \right. \end{aligned}$$

where $M:=\max _{u\in [0,1]}|f'(u)|$. By the maximum principle, $\hat{\Phi }\ge 0$ in $\mathbb {R}^{2}$. By the comparison principle, it is easy to verify that

$$\begin{aligned} |\Phi (\varvec{x})|\le \hat{\Phi }(\varvec{x}), ~~~~\forall \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

(3.2)

In the following, we estimate $\hat{\Phi }(\varvec{x},t)$. Let $\Psi (\varvec{x},t)=\hat{\Phi }(x_{1},x_{2}+st,t)$, then $\Psi$ satisfies

$$\begin{aligned} \left\{ \begin{aligned}&\Psi _{t}=J*\Psi -\Psi +M\Psi ,~~~~~\varvec{x}\in \mathbb {R}^{2},~t>0,\\&\Psi (\varvec{x},0)=|u_{0}(\varvec{x})-w(\varvec{x})|,~~\varvec{x}\in \mathbb {R}^{2}. \end{aligned} \right. \end{aligned}$$

(3.3)

The solution of (3.3) can be expressed as

$$\begin{aligned} \Psi (\varvec{x},t)= & {} e^{Mt}\int \limits _{\mathbb {R}^{2}}S(x_{1}-x,x_{2}-z,t)\Psi (x,z,0)\mathrm{d}x\mathrm{d}z\\= & {} e^{Mt}\int \limits _{\mathbb {R}^{2}}S(x,z,t)\Psi (x_{1}-x,x_{2}-z,0)\mathrm{d}x\mathrm{d}z, \end{aligned}$$

where $S(\varvec{x},t)=e^{-t}\delta _{0}(\varvec{x})+K_{t}(\varvec{x})$ is the fundamental solution of (3.3) with initial data $\delta _{0}$, the Dirac measure at zero and $K_{t}(\varvec{x})=\int \limits _{\mathbb {R}^{2}}e^{-t}(e^{\hat{J}(\varvec{y})t}-1)e^{i(\varvec{x})\cdot \varvec{y}}\mathrm{d}\varvec{y}$ with $\hat{J}$ the Fourier transform of J. It is not difficult to verify that $||S(\varvec{x},t)||_{L^{1}(\mathbb {R}^{2})}\le 3$. Then for any given $T>0$,

$$\begin{aligned} \Psi (\varvec{x},T)\le e^{MT}\left( \int \limits _{|(x,z)|\le R}+\int \limits _{|(x,z)|> R}\right) |S(x,z,T)|\Psi (x_{1}-x,x_{2}-z,0)\mathrm{d}x\mathrm{d}z. \end{aligned}$$

For any $\epsilon >0$ small enough, there exist $R_{1}>0$ and $R_{2}>0$ big enough such that

$$\begin{aligned}&\int \limits _{|(x,z)|>R_{1}}|S(x,z,T)|\Psi (x_{1}-x,x_{2}-z,0)\mathrm{d}x\mathrm{d}z\\&<\sup _{\varvec{x}\in \mathbb {R}^{2}}\Psi (\varvec{x},0)\int \limits _{|(x,z)|> R_{1}}|S(x,z,T)|\mathrm{d}x\mathrm{d}z<\frac{\epsilon }{2e^{MT}} \end{aligned}$$

and

$$\begin{aligned} \int \limits _{|(x,z)|\le R_{1}}|S(x,z,T)|\Psi (x_{1}-x,x_{2}-z,0)\mathrm{d}x\mathrm{d}z<\frac{\epsilon }{2e^{MT}},~~~\forall x_{1}^{2}+x_{2}^{2}\ge R_{2}^{2}. \end{aligned}$$

This implies that $\Psi (\varvec{x},T)<\epsilon$ for $x_{1}^{2}+x_{2}^{2}\ge R_{2}^{2}$ and thus $\lim _{R\rightarrow +\infty }\sup _{x_{1}^{2}+x_{2}^{2}\ge R^{2}}\Psi (\varvec{x},T)=0$. Recall $\hat{\Phi }(\varvec{x},t)=\Psi (x_{1},x_{2}-st,t)$ and we have $\lim _{R\rightarrow +\infty }\sup _{x_{1}^{2}+x_{2}^{2}\ge R^{2}}\hat{\Phi }(\varvec{x},T)=0$. Then, the proof completes following (3.2). $\square$

Lemma 3.5

The solution $u(\varvec{x},t;u_{0})$ of the Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned}&u_{t}=J*u-u-s\partial _{x_{2}}u+f(u),~~~\varvec{x}\in \mathbb {R}^{2},~t>0,\\&u(\varvec{x},0)=u_{0}(\varvec{x}),~~~~~~~~~\varvec{x}\in \mathbb {R}^{2}, \end{aligned} \right. \end{aligned}$$

(3.4)

depends continuously on the initial function $u_{0}(\varvec{x})$. That is, if $u_{1}(\varvec{x},t;u_{0,1})$ and $u_{2}(\varvec{x},t;u_{0,2})$ are two solutions of (3.4) with initial values $u_{0,1}$ and $u_{0,2}$, respectively, then we have

$$\begin{aligned} \sup _{\varvec{x}\in \mathbb {R}^{2}}|u_{1}(\varvec{x},t;u_{0,1})-u_{1}(\varvec{x},t;u_{0,2})|\le A(t)\sup _{\varvec{x}\in \mathbb {R}^{2}}|u_{0,1}(\varvec{x})-u_{0,2}(\varvec{x})| \end{aligned}$$

for some A(t) depending only on t.

Proof

Let $v(\varvec{x},t)=u(x_{1},x_{2}+st)$. Then $v(\varvec{x},t)$ satisfies

$$\begin{aligned} \left\{ \begin{aligned}&v_{t}(\varvec{x},t)=J*v(\varvec{x},t)-v(\varvec{x},t)+f(v(\varvec{x},t)),~~~\varvec{x}\in \mathbb {R}^{2},~t>0,\\&v(\varvec{x},0)=u_{0}(\varvec{x}),~~~~~~~~~~\varvec{x}\in \mathbb {R}^{2}. \end{aligned} \right. \end{aligned}$$

The above problem is equivalent to the following integral equation

$$\begin{aligned} v(\varvec{x},t)=e^{-\mu t}u_{0}(\varvec{x})+\int \limits _{0}^{t}e^{-\mu (t-s)}(J*v-v)(\varvec{x},s)+\mu v(\varvec{x},s) +f(v(\varvec{x},s))\mathrm{d}s, \end{aligned}$$

where $\mu >0$ is a constant. Let $w(\varvec{x},t):=v_{2}(\varvec{x},t)-v_{1}(\varvec{x},t)$, then it satisfies

$$\begin{aligned} \begin{aligned} w(\varvec{x},t)&=e^{-\mu t}\left( u_{0,1}(\varvec{x})-u_{0,2}(\varvec{x})\right) \\&\quad +\int \limits _{0}^{t}e^{-\mu (t-s)}(J*w-w)(\varvec{x},s)+\mu w(\varvec{x},s)+f'(w_{\tau })w(\varvec{x},s)\mathrm{d}s\\&\le \sup _{\varvec{x}\in \mathbb {R}^{2}}|u_{0,1}(\varvec{x})-u_{0,2}(\varvec{x})|\\&\quad +\int \limits _{0}^{t}e^{-\mu (t-s)}\left( \mu +||f'||_{L^{\infty }(\mathbb {R}^{2})}\right) ||w(\cdot ,s)||_{L^{\infty }(\mathbb {R}^{2})}\mathrm{d}s \end{aligned} \end{aligned}$$

It then follows that

$$\begin{aligned} \begin{aligned} ||w(\cdot ,t)||_{L^{\infty }(\mathbb {R}^{2})}\le ||u_{0,1}\!-\!u_{0,2}||_{L^{\infty }(\mathbb {R}^{2})}\!+\!\int \limits _{0}^{t}e^{-\mu (t-s)}\left( \mu \!+\!||f'||_{L^{\infty }(\mathbb {R}^{2})}\right) ||w(\cdot ,s)||_{L^{\infty }(\mathbb {R}^{2})}\mathrm{d}s \end{aligned} \end{aligned}$$

By the Gronwall’s inequality, we have

$$\begin{aligned} ||w(\cdot ,t)||_{L^{\infty }(\mathbb {R}^{2})}\le ||u_{0,1}-u_{0,2}||_{L^{\infty }(\mathbb {R}^{2})}\left( 1+C_{1}te^{C_{1}t}\right) , \end{aligned}$$

where $C_{1}=\mu +||f'||_{L^{\infty }(\mathbb {R}^{2})}$. Let $A(t)=1+C_{1}te^{C_{1}t}$. Then, the proof is complete. $\square$

Now, define

$$\begin{aligned} V^{*}(\varvec{x})=\lim _{t\rightarrow \infty }v(\varvec{x},t;v^{+}). \end{aligned}$$

Then, $V^{*}$ satisfies

$$\begin{aligned} -(J*V^{*}-V^{*})+s\partial _{x_{2}}V^{*}-f(V^{*})=0~~~~\mathrm{in}~~\mathbb {R}^2. \end{aligned}$$

And by the comparison principle, there holds

$$\begin{aligned} v^{-}(\varvec{x})<V_{*}(\varvec{x})\le V^{*}(\varvec{x})<\min \{1,v^{+}(\varvec{x})\}. \end{aligned}$$

Lemma 3.6

$V_{*}(\varvec{x})\equiv V^{*}(\varvec{x})$ in $\mathbb {R}^{2}$.

Proof

Assume on the contrary that $V_{*}(\varvec{x})\not \equiv V^{*}(\varvec{x})$ in $\mathbb {R}^{2}$. Then, they must be $V_{*}(\varvec{x})<V^{*}(\varvec{x})$. By the aid of (2.7), we can find a $\delta >0$ small enough and a proper $\xi >0$ such that

$$\begin{aligned} V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\xi )+\delta . \end{aligned}$$

Then, the comparison principle yields

$$\begin{aligned} V^{*}(\varvec{x})\le u^{+}(\varvec{x},t), ~~~~\forall \varvec{x}\in \mathbb {R}^2,~t>0. \end{aligned}$$

Letting $t\rightarrow \infty$ in the above inequality, we have

$$\begin{aligned} V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\xi +\rho \delta ). \end{aligned}$$

Define

$$\begin{aligned} \Lambda :=\inf \{\lambda >0: V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\lambda ), \forall \varvec{x}\in \mathbb {R}^{2}\}. \end{aligned}$$

Obviously $\Lambda \ge 0$ and $V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\Lambda )$. If $\Lambda =0$, then the proof is done. Thus, we assume $\Lambda >0$ to derive a contradiction. It follows from $v^{-}(\varvec{x})<V_{*}(\varvec{x})<V^{*}(\varvec{x})<v^{+}(\varvec{x})$ that

$$\begin{aligned} \lim _{x_{1}\rightarrow \infty }V^{*}(x_{1},-m_{*}x_{1})=U(0)~~\mathrm{and}~~\lim _{x_{1}\rightarrow \infty }V_{*}(x_{1},-m_{*}x_{1}+\Lambda )=U\left( \frac{c}{s}\Lambda \right) >U(0), \end{aligned}$$

which implies that there must be

$$\begin{aligned} V^{*}(\varvec{x})<V_{*}(x_{1},x_{2}+\Lambda ),~~~\forall \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

By Lemma 3.1, there exists a constant $R^{*}>0$ large enough such that

$$\begin{aligned} \sup _{|x_{2}+m_{*}|x_{1}||\ge R^{*}-\Lambda }\partial _{x_{2}}V_{*}(\varvec{x})<\frac{1}{4\rho }. \end{aligned}$$

(3.5)

Define

$$\begin{aligned} \Omega :=\{\varvec{x}\in \mathbb {R}^{2}||x_{2}+m_{*}|x_{1}||\le R^{*}\}. \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned}&\lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R,\varvec{x}\in \Omega }\left( V_*(x_{1},x_{2}+\Lambda )-V^{*}(\varvec{x})\right) \\&\ge \lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R,\varvec{x}\in \Omega }\left( v^{-}(x_{1},x_{2}+\Lambda )-v^{+}(\varvec{x})\right) \\&=\lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R,\varvec{x}\in \Omega }\left( v^{-}(x_{1},x_{2}+\Lambda )-v^{-}(\varvec{x})+v^{-}(\varvec{x})-v^{+}(\varvec{x})\right) \\&=\lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R,\varvec{x}\in \Omega }\left( v^{-}(x_{1},x_{2}+\Lambda )-v^{-}(\varvec{x})\right) \\&=\lim _{R\rightarrow \infty }\sup _{|\varvec{x}|\ge R,\varvec{x}\in \Omega }U'\left( \frac{c}{s}(x_{2}+m_{*}|x_{1}|+\tau \Lambda ) \right) \frac{c}{s}\Lambda >0, \end{aligned} \end{aligned}$$

there exists a $\sigma \in (0,\delta _{0})$ small enough such that

$$\begin{aligned} V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\Lambda -2\rho \sigma ),~~~\forall \varvec{x}\in \Omega . \end{aligned}$$

For $\varvec{x}\in \mathbb {R}\backslash \Omega$, by (3.5) we have

$$\begin{aligned} \begin{aligned} V_{*}(x_{1},x_{2}+\Lambda -2\rho \sigma )-V_{*}(x_{1},x_{2}+\Lambda )&=\partial _{x_{2}}V_{*}(x_{1},x_{2}+\Lambda -2\tau \rho \sigma )\cdot (-2\rho \sigma )\\&\ge \frac{1}{4\rho }\cdot (-2\rho \sigma )=-\frac{1}{2}\sigma . \end{aligned} \end{aligned}$$

To sum up, there is

$$\begin{aligned} V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\Lambda -2\rho \sigma )+\frac{1}{2}\sigma ,~~~\forall \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

Then by the comparison principle, we have

$$\begin{aligned} V^{*}(\varvec{x})\le u^{+}(\varvec{x},t) \end{aligned}$$

with $\xi =\Lambda -2\rho \sigma$. Let $t\rightarrow \infty$ in the above inequality, then we have

$$\begin{aligned} V^{*}(\varvec{x})\le V_{*}(x_{1},x_{2}+\Lambda -\rho \sigma ), \end{aligned}$$

which contradicts to the definition of $\Lambda$. Thus, $\Lambda =0$ and the proof is complete. $\square$

Now we show that the curved fronts $V_{*}$ are asymptotically stable under the condition that the initial perturbation is positive.

Proof of Theorem 1.2

Denote $v(\varvec{x},t;v_{0})$ by $v(\varvec{x},t)$ for simplicity. On the one hand, since $v^{-}(\varvec{x})\le v_{0}(\varvec{x})$, by the comparison principle, we have

$$\begin{aligned} v^{-}(\varvec{x})\le v(\varvec{x},t;v^{-})\le v(\varvec{x},t)<1,~~~~\forall \varvec{x}\in \mathbb {R}^{2},t>0. \end{aligned}$$

(3.6)

On the other hand, a similar argument as [13, proposition 2.5] can deduce that

$$\begin{aligned} |v(\varvec{x},t;v^{-})-v(\varvec{y},t;v^{-})|\le L|\varvec{x}-\varvec{y}|, ~~~~|v_{t}(\varvec{x},t;v^{-})-v_{t}(\varvec{y},t;v^{-})|\le L|\varvec{x}-\varvec{y}| \end{aligned}$$

for any $\varvec{x}, \varvec{y}\in \mathbb {R}^{2}$ and $|v_{t}|\le C$, where L and C are constants independent of $\varvec{x},\varvec{y}$ and t. Thus, we have

$$\begin{aligned} \lim _{t\rightarrow +\infty }||v(\cdot ,t;v^{-})-V_{*}(\cdot )||_{L^{\infty }(\mathbb {R}^{2})}=0. \end{aligned}$$

(3.7)

Similarly, we have

$$\begin{aligned} \lim _{t\rightarrow +\infty }||v(\cdot ,t;v^{+})-V^{*}(\cdot )||_{L^{\infty }(\mathbb {R}^{2})}=0. \end{aligned}$$

(3.8)

(3.6) and (3.7) imply that it suffices to prove that for any $\epsilon >0$, there exists $T^{*}>0$ such that

$$\begin{aligned} v(\varvec{x},t)\le V_{*}(\varvec{x})+\epsilon \quad \mathrm{for }~~t\ge T^{*}. \end{aligned}$$

Step 1. Let $A^{*}=\sup _{\varvec{x}\in \mathbb {R}^{2}}\partial _{x_{2}}V_{*}(\varvec{x})$, and take $\rho>0, \kappa >0$ as in Lemma 3.2. Then,

$$\begin{aligned} V_*(x_{1},x_{2}+\rho \delta )-V_{*}(\varvec{x})=\rho \delta \int \limits _{0}^{1}\partial _{x_{2}}V_{*}(x_{1},x_{2}+\rho \delta \tau )d\tau<\rho \delta A^{*}<\frac{\epsilon }{3} \end{aligned}$$

provided that $\delta <\frac{\epsilon }{3\rho A^{*}}$. In other words,

$$\begin{aligned} V(x_{1},x_{2}+\rho \delta )<V_{*}(\varvec{x})+\frac{\epsilon }{3}, ~~~~\forall \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

(3.9)

Step 2. Fix $\delta >0$ in step 1. By (3.6), for any $T_{\delta }>0$, we have

$$\begin{aligned} v(\varvec{x},t;v^{-})\le v(\varvec{x},t)<1,~~~~\forall \varvec{x}\in \mathbb {R}^{2},t\ge T_{\delta }>0. \end{aligned}$$

Following from Lemma 3.4, there exists a $R_{\delta }>0$ such that

$$\begin{aligned} v(\varvec{x},T_{\delta })<v^{-}(\varvec{x})+\frac{\delta }{2} ~~~~\mathrm{for~}|\varvec{x}|\ge R_{\delta }. \end{aligned}$$

(3.10)

Let $\alpha >0$ be small enough such that

$$\begin{aligned} U\left( \frac{x_{2}+\frac{1}{\alpha }\varphi (\alpha x_{1})}{\sqrt{1+\varphi '(\alpha x_{1})^{2}}}\right) >U\left( -R_{\delta }+\frac{c\varphi (0)}{s\alpha }\right) \ge 1-\frac{\delta }{2}~~~~\mathrm{for~}|\varvec{x}|\le R_{\delta }. \end{aligned}$$

In other words, if $\alpha$ is chosen to satisfy

$$\begin{aligned} 0<\alpha <\min \left\{ \alpha ^{+}(\beta ,\delta ), \frac{c\varphi (0)}{s\left[ U^{-1}\left( 1-\frac{\delta }{2}\right) +R_{\delta }\right] } \right\} , \end{aligned}$$

then

$$\begin{aligned} v^{+}(\varvec{x})\ge 1-\frac{\delta }{2}~~~~\mathrm{for~}|\varvec{x}|\le R_{\delta }. \end{aligned}$$

(3.11)

Combining the inequalities (3.10) and (3.11), we have

$$\begin{aligned} v(\varvec{x},T_{\delta })<v^{+}(\varvec{x})+\delta ,~~~~\forall \varvec{x}\in \mathbb {R}^{2}. \end{aligned}$$

Then, Lemma 3.3 and the comparison principle yield that

$$\begin{aligned} v(\varvec{x},t+T_{\delta };v^{-})\le v(\varvec{x},t+T_{\delta })<w^{+}(\varvec{x},t) \end{aligned}$$

for $t\ge 0$. Denote $w^{t}_{+}:=w^{+}(\varvec{x},t)$ and applying the comparison principle again, we have

$$\begin{aligned} v(\varvec{x},t'+t+T_{\delta };v^{-})\le v(\varvec{x},t'+t+T_{\delta })<v(\varvec{x},t';w^{t}_{+}). \end{aligned}$$

(3.12)

Since $v(\varvec{x},t;v^{+})$ converges monotonically to $V^{*}(\varvec{x})$ as $t\rightarrow +\infty$, it follows from (3.8) that there exists a $t_{1}>0$ such that

$$\begin{aligned} \sup _{\varvec{x}\in \mathbb {R}^{2}}|v(\varvec{x},t_{1};v^{+,\delta })-V^{*}(x_{1},x_{2}+\rho \delta )|\le \frac{\epsilon }{3}, \end{aligned}$$

where $v^{+,\delta }=v^{+}(x_{1},x_{2}+\rho \delta )$. On the other hand, Lemma 3.5 yields that

$$\begin{aligned} \sup _{\varvec{x}\in \mathbb {R}^{2}}|v(\varvec{x},t_{1};u_{0})-v(\varvec{x},t_{1};v^{+,\delta })|\le A(t_{1})\sup _{\varvec{x}\in \mathbb {R}^{2}}|u_{0}(\varvec{x})-v^{+,\delta }|. \end{aligned}$$

From the definition of $w^{+}$, we know that there exists a $T_{1}>0$ such that

$$\begin{aligned} A(t_{1})\sup _{\varvec{x}\in \mathbb {R}^{2}}|w^{t}_{+}-v^{+,\delta }|\le \frac{\epsilon }{3}~~~~\mathrm{for}~t\ge T_{1}. \end{aligned}$$

Combining the above facts, we obtain that

$$\begin{aligned}&|v(\varvec{x},t_{1};w^{t}_{+})-V^{*}(x,x_{2}+\rho \delta )|\\&\le |v(\varvec{x},t_{1};w^{t}_{+})-v(\varvec{x},t_{1};v^{+,\delta })|+|v(\varvec{x},t_{1};v^{+,\delta })-V^{*}(x_{1},x_{2}+\rho \delta )|\\&\le \frac{\epsilon }{3}+\frac{\epsilon }{3}=\frac{2\epsilon }{3} \end{aligned}$$

for $t\ge T_{1}$ and $\varvec{x}\in \mathbb {R}^{2}$. It follows from (3.12) that

$$\begin{aligned} v(\varvec{x},t_{1}+t+T_{\delta })<v(\varvec{x},t_{1};w^{t}_{+})\le V^{*}(x_{1},x_{2}+\rho \delta )+\frac{2\epsilon }{3} \end{aligned}$$

for $t\ge T_{1}$ and $\varvec{x}\in \mathbb {R}^{2}$. Take $T^{*}=t_{1}+T_{1}+T_{\delta }$. By Lemma 3.6, we have

$$\begin{aligned} v(\varvec{x},t)\le V^{*}(x_{1},x_{2}+\rho \delta )+\frac{2\epsilon }{3}=V_{*}(x_{1},x_{2}+\rho \delta )+\frac{2\epsilon }{3} \end{aligned}$$

for $t\ge T^{*}$ and $\varvec{x}\in \mathbb {R}^{2}$. Combining the above inequality and (3.9), we obtain

$$\begin{aligned} v(\varvec{x},t)\le V_{*}(\varvec{x})+\epsilon \quad \mathrm{for~}\varvec{x}\in \mathbb {R}^{2},\ t\ge T^{*}. \end{aligned}$$

This completes the proof. $\square$