1 Introduction

This paper is concerned with the following bistable nonlocal dispersal equation

$$\begin{aligned} u_t(x,t)=\int _{\Omega }J(x-y)[u(y,t)-u(x,t)]dy+f(u(x,t)),~ x\in \Omega , \end{aligned}$$
(1.1)

where \(\Omega ={\mathbb {R}}^N\backslash K\) and K is a compact subset of \({\mathbb {R}}^N\) and the dispersal kernel function J is nonnegative. Throughout the paper, we make the following assumptions.

  1. (J)

    The kernel function \(J\in C^1({\mathbb {R}}^N)\) is radially symmetric and compactly supported such that

    $$\begin{aligned}J(x)\ge 0~\text {for}~x\in {\mathbb {R}}^N,~J(0)>0~\text {and}~\int _{{\mathbb {R}}^N}J(y)dy=1.\end{aligned}$$
  2. (F)

    \(f\in C^{1,1}([0,1])\) and there exists \(\theta \in (0,1)\) such that

    $$\begin{aligned}&f(0)=f(1)=f(\theta )=0,\,\,f(s)<0 \text { in }(0,\theta ),\,\,\,f(s)>0 \text { in }(\theta ,1), \\&\int _0^1f(s)ds>0,\,\,f'(0)<0,\,\,f'(1)<0,\,\,f'(\theta )>0, \end{aligned}$$

    and

    $$\begin{aligned} f'(s)<\inf \limits _{x\in \Omega }\int _{\Omega }J(x-y)dy<1 \end{aligned}$$
    (1.2)

    for \(s\in [0,1]\).

It follows from the assumption (F) that there exists \(L_f>0\) such that

$$\begin{aligned} |f(u+v)-f(u)-f(v)|\le L_fuv~\text {for}~0\le u,v\le 1. \end{aligned}$$

On the other hand, we know that \(\inf \limits _{x\in \Omega }\int _{\Omega }J(x-y)dy\ge 1/2\) when K is convex. In this case, we can see that (1.2) is automatically satisfied if \(f'(s)\le 1/2\) for all \(s\in [0,1]\). Moreover, under the assumption (1.2), some nonlocal Liouville type results were established by Brasseur et al. [7]. Besides, note that K is compact, without loss of generality, we may assume that

$$\begin{aligned} K\subset \{x\in {\mathbb {R}}^N:x_1\le 0\} \text { or } K\subset \{{\mathbb {R}}^N\setminus \text {supp}(J)\}\cap \{x\in {\mathbb {R}}^N: x_1\le 0\}. \end{aligned}$$

It is well-known that the classical diffusion problem in exterior domain is established by the seminal works of Berestycki et al. [4] and Bouhours [6]. In order to study how a planar wave front propagates around an obstacle, they considered the following semi-linear parabolic problem

$$\begin{aligned} \left\{ \begin{aligned}&u_t=\Delta u+f(u),~ x\in \Omega ,\\&\nu \cdot \nabla u=0,~x\in \partial \Omega , \end{aligned}\right. \end{aligned}$$
(1.3)

where \(\nu \) denotes the outward unit normal to the smooth exterior domain \(\Omega \). More precisely, they proved how a planar traveling front can eventually recover its profile after disturbed by an obstacle K, leaving the obstacle behind. Bouhours [6] further obtained the robustness for the Liouville type results in [4]. Later on, Guo et al. [21] showed that the global mean speed of the entire solution constructed in [4] is the speed of traveling waves in homogeneous environment. More recently, Guo and Monobe [22] extended the results in [4] to V-shaped front. Hoffman et al. [27] considered a similar problem for two dimensional lattice differential equations with directionally convex obstacles.

The nonlocal dispersal equation has got numerous scholars interested, in view of its extensive use to describe the long range effects of spatial structure in biology, physics and chemistry [1, 2, 11, 16,17,18,19]. Moreover, the problems of nonlocal dispersal equations in exterior domains have attracted much attention recently. In particular, Cortázar et al. [12,13,14] considered the asymptotic behaviors of the solutions to linear equations. Brasseur et al. [7, 8] have established some Liouville results for such nonlocal obstacle problems and found that the stationary solutions of (1.1) converging to 1 as \(|x|\rightarrow +\infty \) is indeed 1 for compact convex obstacle K.

When the obstacle K is empty, there have been many works devoted to the traveling wave solutions and entire solutions for (1.1) and its local dispersal counterpart in recent decades. In particular, the authors of [2, 17, 34, 40] have obtained the monotone traveling wave and its asymptotic behaviors for (1.1) with bistable type nonlinearities f, and Chen [17] showed the uniqueness and stability of the traveling wave. For other types of nonlinear terms in (1.1), one can refer to [11, 18, 19, 29, 36, 37, 39] and references therein. For the results on local dispersal equations, readers can consult [15, 20, 23, 24, 28, 35, 38, 42, 43] and references therein.

If \(\Omega ={\mathbb {R}}^N\) in (1.1), let \(u(x,t)=\phi (x_1+ct)\) and \(z=x_1\), we have

$$\begin{aligned} \left\{ \begin{aligned}&\int _{{\mathbb {R}}}J_1(z-y)[\phi (y)-\phi (z)]dy-c\phi '(z)+f(\phi (z))=0,~z\in {\mathbb {R}},\\&\phi (-\infty )=0,~\phi (+\infty )=1,\\&0<\phi (z)<1,~z\in {\mathbb {R}}, \end{aligned}\right. \end{aligned}$$
(1.4)

where \(c>0\) and \(J_1(x_1)=\int _{{\mathbb {R}}^{N-1}}J(x_1,y_2,y_3,\cdots ,y_{N})dy\). Then it follows from [2, 34] that there exists a unique real number \(c>0\) such that (1.4) admits a solution \(\phi \). In fact, the solution \(\phi \) is the unique monotone planar traveling wave solution of (1.1). Besides, \(\phi (z)\) satisfies

$$\begin{aligned} \left\{ \begin{aligned}&\alpha _0e^{\lambda z}\le \phi (z)\le \beta _0e^{\lambda z},~z\le 0,\\&\alpha _1e^{-\mu z}\le 1-\phi (z)\le \beta _1e^{-\mu z},~z>0, \end{aligned} \right. \end{aligned}$$
(1.5)

where \(\alpha _0,~\alpha _1,~\beta _0\) and \(\beta _1\) are some positive constants, \(\lambda \) and \(\mu \) are the positive roots of

$$\begin{aligned} c\lambda =\int _{{\mathbb {R}}}J_1(y)e^{-\lambda y}dy-1+f'(0),~c\mu =\int _{{\mathbb {R}}}J_1(y)e^{-\mu y}dy-1+f'(1), \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned}&\gamma _0e^{\lambda z}\le \phi '(z)\le \delta _0e^{\lambda z},~z\le 0, \\&\gamma _1e^{-\mu z}\le \phi '(z)\le \delta _1e^{-\mu z},~z>0 \end{aligned} \right. \end{aligned}$$
(1.6)

for some constants \(\gamma _0,~\gamma _1,~\delta _0\) and \(\delta _1>0\). However, if the domain is not the whole space (such as (1.1), (1.3)), there is no classical traveling wave front. Therefore, it is naturally to consider the generalization of traveling fronts. In fact, such extensions have been introduced in [3, 5, 30]. In particular, the transition wave front, as a fully general notion of traveling front, has been widely established in many works [10, 25, 26, 31,32,33, 44, 45]. It is interesting to point out that the entire solution constructed in [4, 27] is indeed a generalized transition front.

In the present paper, we are interested to consider the nonlocal dispersal problem (1.1) in exterior domains. The main ingredient of this paper is to obtain a unique entire solution of (1.1) which behaves as planar wave fronts for large negative and positive time. More precisely, we first prove the existence and uniqueness of the entire solution like a planar wave front for large negative time by sub- and super-solutions method. Moreover, we find that the entire solution also approaches planar wave fronts as x is far away from K. Finally, we shall investigate the procedure how the front goes through K and eventually recovers its shape. Due to the lack of compactness of nonlocal operators, which is necessary to establish the uniqueness and asymptotic behaviors as \(|x|\rightarrow +\infty \) of the entire solution, the methods and techniques adopted here are different from that in [4, 27], and additional difficulties appear when the entire solution is constructed. So motivated by the recent work of Li et al. [29], we establish the Lipschitz continuity in space variable x of entire solutions to the nonlocal problem (1.1) in exterior domains. Then we can discuss the uniqueness and asymptotic behaviors of entire solutions of (1.1). In addition, the appearance of convolution term and interior domain leads to the fact that planar wave fronts are not the solutions of (1.1), which causes much trouble in verifying the sub- and super-solutions when we construct the entire solution. Particularly, nonlocal dispersal equations admit no explicit fundamental solutions as Laplacian dispersal equations. Therefore, the sub- and super-solutions in [4, 27] are not suitable here to study the asymptotic behaviors of such an entire solution for large positive time. Consequently, we have to construct new sub- and super-solutions inspired by [4, 27] to investigate the asymptotic behaviors of the entire solution to (1.1). At last, our results show that the geometric shape of the interior domain affects the propagation of planar wave fronts.

Now we are ready to state the main result of this paper.

Theorem 1.1

Assume that (F) and (J) hold. Let \((\phi ,c)\) be the unique solution of (1.4), and K be a compact subset of \({\mathbb {R}}^N\). Then there exists an entire solution u(x, t) to (1.1) with \(0<u(x,t)<1\) and \(u_t(x,t)>0\) for all \((x,t)\in {{\overline{\Omega }}}\times {\mathbb {R}}\), and satisfying that

$$\begin{aligned} u(x,t)-\phi (x_1+ct)\rightarrow 0~\text {as}~t\rightarrow -\infty ~\text {uniformly in}~x\in {{\overline{\Omega }}}, \end{aligned}$$
(1.7)

and as \(|x|\rightarrow +\infty \) uniformly in \(t\in {\mathbb {R}}\). Moreover, if K is convex, then we have

$$\begin{aligned} u(x,t)-\phi (x_1+ct)\rightarrow 0~\text {as}~t\rightarrow +\infty ~\text {uniformly in}~x\in {\overline{\Omega }}. \end{aligned}$$

In particular, the condition (1.7) determines a unique entire solution of (1.1).

Remark 1.2

It follows from Brasseur and Coville [9, Thorem 10] that the entire solution constructed in Theorem 1.1 is a generalized transition almost-planar front with global mean speed c.

Remark 1.3

The techniques and ideas developed in this paper can be modified to treat a much more general case for deformations of K (see [7, Definition 1.2]). Let \(K\subset {\mathbb {R}}^N\) be a compact convex set with non-empty interior and let \(\{K_\epsilon \}_{0<\epsilon \le 1}\) be a family of \(C^{0,\alpha }\) (\(\alpha \in (0,1]\)) deformations of K. Assume that

$$\begin{aligned} \max \limits _{s\in [0,1]}f'(s)<\inf \limits _{0<\epsilon \le 1}\inf \limits _{x\in {\mathbb {R}}^N\setminus K_\epsilon } \Vert J(x-\cdot )\Vert _{L^1({\mathbb {R}}^N\setminus K_\epsilon )}. \end{aligned}$$

Then there exists \(\epsilon _0\in (0,1]\) such that the conclusions in Theorem 1.1 also hold true with K replaced by \(K_\epsilon \) for \(\epsilon \in (0,\epsilon _0]\).

In this paper, under assumptions that K is a compact convex set and the nonlocal dispersal kernel is compactly supported and radially symmetric, we establish the existence of an entire solution for the nonlocal dispersal equation (1.1) in the exterior domain \(\Omega \), which behaves like a planar traveling front for large negative and positive time. It is naturally to ask if the kernel function is not compactly supported, whether the entire solution we construct in this paper exists. In addition, we conjecture that when the obstacle K just being a compact subset of \({\mathbb {R}}^N\) is not convex, the entire solution of (1.1) can not recover its shape uniformly in space, but it converges to the nonconstant stationary solution in any bounded subset of \({\mathbb {R}}^N\) containing K. We shall study these in a future work.

This paper is organized as follows. In Sect. 2, we consider the Cauchy problem and establish the comparison principle for (1.1). Then the entire solution is constructed in Sect. 3. In Sect. 4, we study the behaviors of the entire solution far away from K in space for finite time. Section 5 is devoted to discussing the asymptotic behavior of the entire solution as time goes to positive infinity.

2 Preliminaries

2.1 The Nonlocal Cauchy Problem

We first consider the nonlocal Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned}&u_t(x,t)=\int _{\Omega }J(x-y)[u(y,t)-u(x,t)]dy+f(u(x,t)),~x\in \Omega ,~t\ge 0, \\&u(x,0)=u_0(x),~x\in \Omega , \end{aligned}\right. \end{aligned}$$
(2.1)

where \(\Omega \) is a subset of \({\mathbb {R}}^N\). We call u(x, t) a solution of (2.1), if it satisfies

$$\begin{aligned} u(x,t)=u_0(x)+\int _0^t\int _{\Omega }J(x-y)[u(y,s)-u(x,s)]dyds+\int _0^tf(u(x,s))ds \end{aligned}$$
(2.2)

for \(x\in \Omega \) and \(t\ge 0\). Then we have the following theorem.

Theorem 2.1

Suppose that (J) holds and \(f\in C^{1,1}({\mathbb {R}})\). Then, for any \(u_0\in L^1(\Omega )\), there exists a unique solution \(u\in C([0,t_0],L^1(\Omega ))\) to (2.2) for some \(t_0>0\).

Proof

For every \(\omega \in C([0,t_0],L^1(\Omega ))\), we define the norm

$$\begin{aligned} |||\omega |||=\max _{0\le t\le t_0}\Vert \omega (\cdot ,t)\Vert _{L^1(\Omega )}, \end{aligned}$$

and the operator

$$\begin{aligned} {\mathcal {T}}w(x,t)=u_0(x)+\int _0^t\int _{\Omega }J(x-y)[w(y,s)-w(x,s)]dyds+\int _0^tf(w(x,s))ds. \end{aligned}$$

It is easily seen that

$$\begin{aligned} |||{\mathcal {T}}w|||\le \Vert u_0\Vert _{L^1(\Omega )}+[2+L]t_0|||w|||, \end{aligned}$$

here

$$\begin{aligned} L=\sup \limits _{\tau \in [-|||\omega |||,|||\omega |||]} |f'(\tau )|, \end{aligned}$$

which means \({\mathcal {T}}\) maps \(C([0,t_0],L^1(\Omega ))\) into \(C([0,t_0],L^1(\Omega ))\). On the other hand, note that

$$\begin{aligned} {\mathcal {T}}u(x,t)-{\mathcal {T}}v(x,t) =&\int _0^t\int _{\Omega }J(x-y)[u(y,s)-v(y,s)+v(x,s)-u(x,s)]dyds\\&+\int _0^t[f(u(x,s))-f(v(x,s))]ds. \end{aligned}$$

It follows that

$$\begin{aligned} |||{\mathcal {T}}u-{\mathcal {T}}v||| \le 2t_0|||u-v||| +t_0L|||u-v||| \le (2+L)t_0|||u-v|||. \end{aligned}$$

In fact, let \(t_0\) be sufficiently small such that \((2+L)t_0<1\). Then one can obtain that \({\mathcal {T}}\) is a strict contraction mapping in \(C([0,t_0],L^1(\Omega ))\).

To extend the solution to \([0,+\infty )\), we can take \(u(x,t_0)\in L^1(\Omega )\) as the initial datum and further obtain a solution in \([t_0,2t_0]\). Then by iterating this procedure, we get a solution in \([0,+\infty )\). \(\square \)

2.2 Comparison Principle

Theorem 2.2

Suppose that the assumptions of Theorem 1.1 hold and \(u(x,0),v(x,0),u_0(x)\in L^{\infty }(\Omega )\). Furthermore, if u(x, t), \(v(x,t)\in C^1([0,+\infty ),L^\infty (\Omega ))\) are uniformly bounded and satisfy

$$\begin{aligned}&\left\{ \begin{aligned}&\frac{\partial u(x,t)}{\partial t}-\left( \int _{\Omega }J(x-y)[u(y,t)-u(x,t)]dy\!\right) +f(u(x,t))\ge 0,~(x,t)\in \Omega \times (0,+\infty ),\\&u(x,0)\ge u_0(x),~x\in \Omega , \end{aligned}\right. \\&\left\{ \begin{aligned}&\frac{\partial v(x,t)}{\partial t}-\left( \int _{\Omega }J(x-y)[v(y,t)-v(x,t)]dy\!\right) +f(v(x,t))\le 0,~(x,t)\in \Omega \times (0,+\infty ),\\&v(x,0)\le u_0(x),~x\in \Omega , \end{aligned}\right. \end{aligned}$$

respectively, then

$$\begin{aligned} u(x,t)\ge v(x,t)~\text {in}~ \Omega \times [0,+\infty ). \end{aligned}$$

Proof

Define \(W(x,t)=u(x,t)-v(x,t)\), it follows that \(W(x,0)\ge 0\) and

$$\begin{aligned} \begin{aligned} W_t(x,t)&\ge \int _{\Omega }J(x-y)[W(y,t)-W(x,t)]dy+f(u(x,t))-f(v(x,t))\\&=\int _{\Omega }J(x-y)[W(y,t)-W(x,t)]dy+F(x,t)W(x,t), \end{aligned} \end{aligned}$$
(2.3)

where

$$\begin{aligned} F(x,t)=\int _0^1f'(v(x,t)+\theta W(x,t))d\theta . \end{aligned}$$

Suppose that there exist \(t_*>0\) and \(x_*\in \Omega \) such that \(W(x_*,t_*)<0\). Denote \(\theta _*=-W(x_*,t_*)\), we can take \(\epsilon >0\) and \(K'>0\) such that \(\theta _*=\epsilon e^{2K't_*}\). Let

$$\begin{aligned} T_*:=\sup \left\{ \tau \ge 0\mid W(x,t)>-\epsilon e^{2K't}~\text {for all}~x\in \Omega ,~0\le t\le \tau \right\} , \end{aligned}$$

then we have \(0<T_*\le t_*\) since the facts \(W(x,\cdot )\in C^1(0,\infty )\) and \(W(x,0)\ge 0\). Moreover, it follows that

$$\begin{aligned} \inf \limits _{\Omega }W(x,T_*)=-\epsilon e^{2K'T_*}. \end{aligned}$$

Without loss of generality, we may assume that \(0\in \Omega \) and \(W(0,T_*)<-\frac{7}{8}\epsilon e^{2K'T_*}\).

Consider now the function

$$\begin{aligned} W^-(x,t,\beta )=-\epsilon \left( \frac{3}{4}+\beta Z(x)\right) e^{2K't}, \end{aligned}$$

in which \(\beta >0\) is a parameter and \(Z\in L^{\infty }\left( {\mathbb {R}}^N\right) \) with \(Z(0)=1\), \(\lim \limits _{|x|\rightarrow +\infty }Z(x)=3,~1\le Z(x)\le 3\). Take \(\beta _*\in \left( \frac{1}{8},\frac{1}{4}\right] \) as the minimal value of \(\beta \) for which \(W(x,t)\ge W^-(x,t)\) holds for all \((x,t)\in \Omega \times [0,T_*]\). Since

$$\begin{aligned} \lim \limits _{|x|\rightarrow +\infty }W^-(x,t,\beta _*)=-\epsilon \left( \frac{3}{4}+3\beta _*\right) e^{2K't}<-\frac{9}{8}\epsilon e^{2K't}, \end{aligned}$$

there exist \(x^*\in \Omega \) and \(0<t_0<T_*\) such that \(W(x^*,t_0)=W^-(x^*,t_0,\beta _*)\). The definition of \(\beta _*\) now implies that

$$\begin{aligned} W_t(x^*,t_0)\le W_t^-(x^*,t_0,\beta _*). \end{aligned}$$

In addition, by the fact that \(W(x,t)\ge W^-(x,t,\beta _*)\) for all \((x,t)\in \Omega \times [0,T_*]\), we have

$$\begin{aligned} \int _\Omega J(x^*-y)[W(y,t_0)-W(x^*,t_0)]dy\ge \int _\Omega J(x^*-y)[W^-(y,t_0,\beta _*)-W^-(x^*,t_0,\beta _*)]dy. \end{aligned}$$

It follows from (2.3) that

$$\begin{aligned} \begin{aligned} -\frac{7}{4}\epsilon K'e^{2K't_0}\ge&W^-_t(x^*,t_0,\beta _*)\ge W_t(x^*,t_0)\\ \ge&\int _\Omega J(x^*-y)[W^-(y,t_0,\beta _*)-W^-(x^*,t_0,\beta _*)]dy\\&+F(x^*,t_0)W^-(x^*,t_0,\beta _*). \end{aligned} \end{aligned}$$

In particular, it follows from the assumptions of Theorem 2.2 that u(x, t) and v(x, t) are uniformly bounded and \(f'(v(x,t)+\theta W(x,t))\) is also bounded, which means there is some \({\mathfrak {M}}>0\) such that \(|f'(v(x,t)+\theta W(x,t))|<{\mathfrak {M}}\) and \(|F(x,t)|<{\mathfrak {M}}\) for \(x\in \Omega \) and \(t>0\). Then we obtain

$$\begin{aligned} \begin{aligned} -\frac{7}{4}\epsilon K'e^{2K't_0}&\ge \beta _*\epsilon \int _\Omega J(x^*-y)[Z(x^*)-Z(y)]dye^{2K't_0}+F(x^*,t_0)W^-(x^*,t_0,\beta _*)\\&\ge -\epsilon \left[ 2\beta _*+\left( 3\beta _*+\frac{3}{4}\right) {\mathfrak {M}}\right] e^{2K't_0}\\&\ge -\left( \frac{1}{2}+\frac{3{\mathfrak {M}}}{2}\right) \epsilon e^{2K't_0}. \end{aligned} \end{aligned}$$

This leads to a contradiction upon choosing \(K'\) to be sufficiently large. \(\square \)

Corollary 2.3

Under the assumptions of Theorem 2.2, let u(x, t) and v(x, t) be solutions of (1.1) with initial values u(x, 0) and v(x, 0), respectively. If \(u(x,0)\ge v(x,0)\) and \(u(x,0)\not \equiv v(x,0)\), then \(u(x,t)>v(x,t)\) for all \(x\in \Omega \) and \(t>0\).

Proof

It is sufficient to show that \(u(x,t)>v(x,t)\) for all \(x\in \Omega \) and \(t\in (0,t_0]\) for some \(t_0>0\). In fact, if \(u(x,t)>v(x,t)\) for all \(x\in \Omega \) and \(t\in (0,t_0]\), we can similarly have \(u(x,t)>v(x,t)\) for all \(x\in \Omega \) and \(t\in [t_0,2t_0]\). Then by repeating this process, we can obtain the results of this lemma. Now, denote

$$\begin{aligned} w(x,t)=u(x,t)-v(x,t),\quad {\tilde{w}}(x,t)=e^{pt}w(x,t)-\epsilon t, \end{aligned}$$

where \(\epsilon ,~p>0\) are real numbers and p is sufficiently large such that

$$\begin{aligned} p+F(x,t)-2\ge 0 ~\text {for all}~ (x,t)\in \Omega \times [0,+\infty ), \end{aligned}$$

with F(x, t) being defined as that in the proof of Theorem 2.2. Suppose that, by contradiction, there is \((x_*,t_*)\in \Omega \times (0,t_0]\) such that \(w(x_*,t_*)=0\). It then follows that \(\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)<0\). Furthermore, one can find a sequence \((x_n,t_n)\) such that \(t_n\rightarrow {\tilde{t}}_*\) and

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }{\tilde{w}}(x_n,t_n)=\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)<0. \end{aligned}$$

Observe that

$$\begin{aligned} {\tilde{w}}_t(x,t)=&pw(x,t)e^{pt}+e^{pt}w_t(x,t)-\epsilon \\ \ge&pw(x,t)e^{pt}+e^{pt}\left( \int _{\Omega }J(x-y)[w(y,t)-w(x,t)]dy+F(x,t)w(x,t)\right) -\epsilon \\ =&\int _{\Omega }J(x-y)[{\tilde{w}}(y,t)-{\tilde{w}}(x,t)]dy+(p+F(x,t))[{\tilde{w}}(x,t)-\epsilon t]-\epsilon . \end{aligned}$$

Then we have that

$$\begin{aligned}&{\tilde{w}}(x_n,t_n)-{\tilde{w}}(x_n,0)\\&\quad \ge \int _0^{t_n}\left[ \int _{\Omega }J(x_n-y){\tilde{w}}(y,s)dy-{\tilde{w}}(x_n,s)+(p+F(x,s))[{\tilde{w}}(x_n,s) -\epsilon s]\right] ds-\epsilon t_n\\&\quad \ge \int _0^{t_n}\int _{\Omega }J(x_n-y){\tilde{w}}(y,s)dyds+t_0\Bigg [p-1+\sup _{x\in \Omega ,t\in (0,t_0]}F(x,s)\\&\qquad -\frac{\epsilon \left[ \frac{t_0}{2}\left( \sup _{x\in \Omega ,t\in (0,t_0]}F(x,s)+p\right) +1\right] }{\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)}\Bigg ] \inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t). \end{aligned}$$

Letting \(n\rightarrow \infty \), it follows that

$$\begin{aligned}&\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)\\&\quad \ge t_0\left[ p+\sup \limits _{x\in \Omega ,t\in (0,t_0]}F(x,s) -\frac{\epsilon \left[ \frac{t_0}{2}\left( \sup _{x\in \Omega ,t\in (0,t_0]}F(x,s)+p\right) +1\right] }{\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)}\right] \\&\qquad \inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t). \end{aligned}$$

Choose \(t_0>0\) being sufficiently small such that

$$\begin{aligned} t_0\left[ p+\sup _{x\in \Omega ,t\in (0,t_0]}F(x,s)-\frac{\epsilon \left[ \frac{t_0}{2}\left( \sup _{\Omega \times (0,t_0]}F(x,s)+p\right) +1\right] }{\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)}\right] <1, \end{aligned}$$

which implies that \(\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)>\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}{\tilde{w}}(x,t)\), since \(\inf _{x\in {\overline{\Omega }},t\in (0,t_0]}<0\). Thus we have finished the proof. \(\square \)

3 Existence and Uniqueness of the Entire Solution

This section is devoted to establishing the existence and uniqueness of an entire solution to (1.1) which behaves as a planar traveling front until it approaches the interior domain K. Since the profile \(\phi \) in (1.4) is monotone increasing and unique in the translation sense, without loss of generality, we further assume that \(\phi (0)\le \theta \) and \(\phi ''(\xi )\ge 0\) for \(\xi \le 0\). The main result of this section is stated as follows.

Theorem 3.1

Assume that (F) and (J) hold and let \((\phi ,c)\) be the unique solution of (1.4). If \(K\subset \{x\in {\mathbb {R}}^N:~x_1\le 0\}\cap {\mathbb {R}}^N\setminus \)supp(J), then there exists an entire solution U(x, t) of (1.1) satisfying

$$\begin{aligned} 0<U(x,t)<1,\quad U_t(x,t)>0~\text {for all}~(x,t)\in {\overline{\Omega }}\times {\mathbb {R}} \end{aligned}$$

and

$$\begin{aligned} U(x,t)\rightarrow \phi (x_1+ct) ~\text {as}~t\rightarrow -\infty ~\text {uniformly in}~x\in {\overline{\Omega }}. \end{aligned}$$
(3.1)

Moreover, condition (3.1) determines a unique entire solution of (1.1).

In this section, the radial symmetry of \(J(\cdot )\) can be released to \(J(x)=J(-x)\). Moreover, the convexity and compactness of the obstacle are not required, while the boundedness of K is necessary. We prove Theorem 3.1 by constructing sub- and super-solutions.

3.1 Construction of the Entire Solution

To establish the entire solution, we shall construct some suitable sub- and super-solutions. Inspired by [4], we define the sub-solution

$$\begin{aligned} W^-(x,t)=\left\{ \begin{aligned}&\phi (x_1+ct-\xi (t))-\phi (-x_1+ct-\xi (t)),~x_1\ge 0,\\&0,~x_1<0, \end{aligned}\right. \end{aligned}$$

and the super-solution

$$\begin{aligned} W^+(x,t)=\left\{ \begin{aligned}&\phi (x_1+ct+\xi (t))+\phi (-x_1+ct+\xi (t)),~x_1\ge 0,\\&2\phi (ct+\xi (t)),~x_1<0, \end{aligned}\right. \end{aligned}$$

here \(\xi (t)\) is the solution of the following equation

$$\begin{aligned} {\dot{\xi }}(t)=Me^{\lambda _0(ct+\xi )}, ~t<-T,~\xi (-\infty )=0, \end{aligned}$$

where \(M,~\lambda _0\) and T are positive constants to be specified later. A direct calculation yields that

$$\begin{aligned} \xi (t)=\frac{1}{\lambda _0}\ln \frac{1}{1-c^{-1}Me^{\lambda _0 ct}}. \end{aligned}$$

For the function \(\xi (t)\) to be defined, one must have \(1-c^{-1}Me^{\lambda _0 ct}>0\). In addition, we suppose that

$$\begin{aligned} ct+\xi (t)\le 0 ~\text {for}~-\infty <t\le T. \end{aligned}$$

Thus set \(T:=\frac{1}{\lambda _0 c}\ln \frac{c}{c+M}\). Moreover, it follows from (1.5) that there exist two positive numbers \(K_\phi \) and \(k_\phi \) such that

$$\begin{aligned} \left| \phi (x_1)-C_{\phi }e^{\lambda x_1}\right| \le K_{\phi }e^{(k_{\phi }+\lambda )x_1} \text { for all}~ x_1\le 0. \end{aligned}$$

Then the following proposition holds.

Proposition 3.2

Assume that \(\lambda _0<\min \{\lambda ,k_\phi \}\) and \(K\subset \{x\in {\mathbb {R}}^N:~x_1\le 0\}\cap {\mathbb {R}}^N\setminus \)supp(J). Then there exists a sufficiently large number \(M>0\) such that \(W^-(x,t)\) and \(W^+(x,t)\) are sub- and super-solutions of (1.1) in the time range \(-\infty <t\le T_1\) for some \(T_1\in (-\infty , T]\).

The proof of this lemma will be given in Appendix for the coherence of this paper.

Now we are in a position to construct the entire solution. Let \(u_n(x,t)\) be the unique solution of (1.1) for \(t\ge -n\) with initial data

$$\begin{aligned} u_n(x,-n)=W^-(x,-n). \end{aligned}$$

Since \(W^-(x,t)\) is a sub-solution, it is not difficult to show that the sequence \(\{u_n(x,t)\}_{n=1}^\infty \) is nondecreasing in n. Choose some constant \(T^*>0\) such that \(c>{\dot{\xi }}(t)\) for \(t\le -\max \{T^*,T_1\}\). In the following discussing, without loss of generality, we assume that \(n^*\ge T^*\). Then, we have

$$\begin{aligned} \frac{\partial u_n(x,t)}{\partial t}=\int _\Omega J(x-y)[u_n(y,t)-u_n(x,t)]dy+f(u_n(x,t)) \end{aligned}$$
(3.2)

for \(n\ge n^*\), \(t\ge -n\) and \(x\in \Omega \). Since \(\frac{\partial W^-(x,t)}{\partial t}=0\) for \(x_1\le 0\) and

$$\begin{aligned} W^-_t(x,t)=(c-{\dot{\xi }}(t))(\phi '(x_1+ct-\xi (t))-\phi '(-x_1+ct-\xi (t)))\ge 0 \end{aligned}$$

for \(0<x_1\le |ct-\xi (t)|\) and \(t<-T^*\), it follows that

$$\begin{aligned} \begin{aligned} \frac{\partial u_n(x,-n)}{\partial t}&=\int _\Omega J(x-y)[u_n(y,-n)-u_n(x,-n)]dy+f(u_n(x,-n)) \\&\ge \frac{\partial W^-(x,-n)}{\partial t}\\&\ge 0 \end{aligned} \end{aligned}$$

for all \(x_1\le |cn+\xi (-n)|\). Furthermore, by Corollary 2.3, \(u_n(x,t)\) satisfies

$$\begin{aligned} \frac{\partial u_n(x,t)}{\partial t}>0~\text {for all}~x_1\le |cn+\xi (-n)|,~0<u_n(x,t)<1~\text {for all}~t>-n,~x\in \Omega , \end{aligned}$$

and

$$\begin{aligned} W^-(x,t)<u_n(x,t)<W^+(x,t)~\text {for all}~-n< t\le T^*~\text {and}~x\in \Omega . \end{aligned}$$

In particular, since the sequence \(\{u_n(x,t)\}_{n=n^*}^{\infty }\) being uniformly bounded in n satisfies (3.2) with \(f\in C^{1,1}([0,1])\), we have that the sequence \(\left\{ \frac{\partial u_n(x,t)}{\partial t}\right\} _{n=n^*}^{\infty }\) is uniformly bounded. Then it follows that \(\{u_n(x,t)\}_{n=n^*}^{\infty }\) is well-defined for each n and equicontinuous in t. Similarly, \(\left\{ \frac{\partial u_n(x,t)}{\partial t}\right\} _{n=n^*}^{\infty }\) is equicontinuous in t. Therefore, by Arzela-Ascolit theorem, for each fixed \(x\in \Omega \), there exists a subsequence, still denoted by \(\left\{ u_n(x,t),\frac{\partial u_n(x,t)}{\partial t}\right\} _{n=n^*}^{\infty }\) such that

$$\begin{aligned} \left( u_n(x,t),\frac{\partial u_n(x,t)}{\partial t}\right) \rightarrow (u(x,t),u_t(x,t))~\text {as}~n\rightarrow +\infty , \end{aligned}$$
(3.3)

where the convergence is locally uniform in \(t\in {\mathbb {R}}\). Moreover, via diagonalization, take a subsequence of \(\left\{ u_n(x,t),\frac{\partial u_n(x,t)}{\partial t}\right\} _{n=n^*}^{\infty }\), which converges to some function U(x, t). Since that \(u_n(x,t)\) is the solution to (3.3) with initial value \(u_n(x,-n)=W^-(x,-n)\) and that U(x, t) is the limit of \(u_n(x,t)\) as \(n\rightarrow \infty \), we have that U(x, t) is well defined for \(t\in {\mathbb {R}}\). Then it follows from Lebesgue’s dominated convergence theorem that

$$\begin{aligned} U_t(x,t)=\int _\Omega J(x-y)(U(y,t)-U(x,t))dy+f(U(x,t)), \end{aligned}$$

and

$$\begin{aligned} U_t(x,t)\ge 0,~0\le U(x,t)\le 1. \end{aligned}$$

Besides, it follows from the definition of \(W^-(x,t)\) and \(W^+(x,t)\) that

$$\begin{aligned} \sup \limits _{x\in \Omega }|U(x,t)-\phi (x_1+ct)|\rightarrow 0~\text {as}~t\rightarrow -\infty . \end{aligned}$$

Note that U(x, t) is not a constant, by applying Corollary 2.3 to \(U_t(x,t)\) and U(x, t), one have that

$$\begin{aligned} U_t(x,t)>0~\text {and}~0<U(x,t)<1. \end{aligned}$$

In addition, inspired by [29], we can show that U(x, t) satisfies the following proposition.

Proposition 3.3

Let U(x, t) be the entire solution in Theorem 3.1. Then U(x, t) satisfies

$$\begin{aligned} |U(x+\eta ,t)-U(x,t)|\le M'\eta ~\text {for all}~(x,t)\in {\bar{\Omega }}\times {\mathbb {R}}, \end{aligned}$$
(3.4)

and

$$\begin{aligned} \left| \frac{\partial U(x+\eta ,t)}{\partial t}-\frac{\partial U(x,t)}{\partial t}\right| \le M''\eta ~\text {for all}~(x,t)\in {\bar{\Omega }}\times {\mathbb {R}} \end{aligned}$$
(3.5)

with \(M'>0\) and \(M''>0\) being two real numbers.

Proof

Since \(\int _{{\mathbb {R}}^N}J(x)dx=1,~J(x)\ge 0\) and J(x) is compactly supported, we have \(J'\in L^1({\mathbb {R}}^N)\). Furthermore, we get

$$\begin{aligned} \int _{\Omega }|J(x+\eta )-J(x)|dx=\int _{\Omega }\int _0^1|\nabla J(x+\theta \eta )\eta |d\theta dx \le L_1|\eta |~\text {for some constant}~L_1>0. \end{aligned}$$

Let

$$\begin{aligned} m=\inf \limits _{u\in [0,1]}\left( \inf \limits _{x\in \Omega }\int _\Omega J(x-y)dy-f'(u)\right) >0 \end{aligned}$$

and v(t) be a solution of the equation

$$\begin{aligned}\left\{ \begin{aligned}&v'(t)=L_1|\eta |-mv(t)~\text {for any}~t>-n,\\&v(-n)=M|\eta | \end{aligned} \right. \end{aligned}$$

for some \(M\ge 2\sup \limits _{\xi \in {\mathbb {R}}}|\phi '(\xi )|\). In addition, denote \(V(x,t)=u_n(x+\eta ,t)-u_n(x,t)\), where \(u_n(x,t)\) is the solution of (1.1) with initial value \(u_n(x,-n)=W^-(x,-n)\). Then

$$\begin{aligned} V_t(x,t)\le&\int _\Omega [J(x+\eta -y)-J(x-y)][u_n(y,t)-u_n(x+\eta ,t)]dy\\&-\inf \limits _{x\in \Omega }\int _\Omega J(x-y)dyV(x,t) +f'({\overline{V}})V(x,t), \end{aligned}$$

where \({\overline{V}}\) is between \(u_n(x,t)\) and \(u_n(x+\eta ,t)\). Consequently, V(x, t) satisfies

$$\begin{aligned} V_t(x,t)\le L_1|\eta |-mV(t) ~\text {for}~t>-n ~\text {and}~V(x,-n)\le M|\eta |. \end{aligned}$$

Moreover, \(|V(x,t)|\le v(t)\le M^*|\eta |\) for any \(x\in \Omega ,~t\ge -n\) and \(M^*=M+\frac{L_1}{m}\). Indeed,

$$\begin{aligned} 0<v(t)=e^{-m(t+n)}M|\eta |+\frac{L_1\eta }{m}\left( 1-e^{-m(t+n)}\right)<\left( M+\frac{L_1}{m}\right) |\eta |<M^*|\eta | \end{aligned}$$

for any \(x\in {\mathbb {R}}^N\) and \(t\ge -n\). In particular, in view of \(f'(s)<1\) for \(s\in [0,1]\), there holds

$$\begin{aligned} \begin{aligned}&\left| \frac{\partial u_n(x+\eta ,t)}{\partial t}-\frac{\partial u_n(x,t)}{\partial t}\right| \\&\quad \le \bigg |\int _{\Omega }[J(x+\eta -y)-J(x-y)]u_n(y,t)dy\bigg | +\big |[u_n(x+\eta ,t)-u_n(x,t)]\big |\\&\qquad +\big |f'({\overline{V}})[u_n(x+\eta ,t)-u_n(x,t)]\big |\\&\quad \le \int _\Omega \big |J(x+\eta -y)-J(x-y)\big |dy+(1+\max \limits _{s\in [0,1]}f'(s))\big |u_n(x+\eta ,t)-u_n(x,t)\big |\\&\quad \le [L_1+2M^*]|\eta |.\\ \end{aligned} \end{aligned}$$

At last, since \(u_n(x,t)\rightarrow U(x,t)\) locally uniformly in \(t\in {\mathbb {R}}\) as \(n\rightarrow +\infty \), we have

$$\begin{aligned} |U(x+\eta ,t)-U(x,t)|&\le |U(x+\eta )-u_n(x+\eta ,t)|+|u_n(x+\eta )-u_n(x,t)|\\&\quad +|u_n(x,t)-U(x,t)|\\&\le (M^*+2)|\eta |. \end{aligned}$$

Now take \(M'=M^*+2\), we can show that (3.4) and (3.5) hold by taking \(M''=L_1+2+2M^*\). \(\square \)

3.2 Uniqueness of the Entire Solution

Now, we show the uniqueness of the entire solution constructed in Theorem 3.1. First, the following lemma is valid.

Lemma 3.4

Assume that the settings of Theorem 1.1 hold. Then for any \(\varphi \in (0,\frac{1}{2}]\), there exist constants \(T_\varphi =T_\varphi (\varphi )>1\) and \(K_\varphi =K_\varphi (\varphi )>0\) such that

$$\begin{aligned} U_t(x,t)\ge K_\varphi ~\text {for any}~t\le -T_\varphi ~\text {and}~x\in \Omega _{\varphi }(t), \end{aligned}$$

where

$$\begin{aligned} \Omega _{\varphi }(t)=\left\{ x\in \Omega :~\varphi \le U(x,t)\le 1-\varphi \right\} . \end{aligned}$$

Proof

It is easy to choose \(T_\varphi \) and \(M_\varphi \) such that \(\Omega _{\varphi }(t)\subset \left\{ x\in \Omega :~|x_1+ct|\le M_\varphi \right\} \subset \{x\in {\mathbb {R}}^N:~x_1\ge 1\}\). Now suppose there exist sequences \(t_k\in (-\infty ,-T_\varphi ]\) and \(x^k:=(x^k_1,x^k_2,...x^k_N)\in \Omega _{\varphi }(t)\) such that

$$\begin{aligned} U_t(t_k,x^k)\rightarrow 0~\text {as}~k\rightarrow +\infty . \end{aligned}$$

There are only two cases that can happen: \(t_k\rightarrow -\infty \) or \(t_k\rightarrow t_*\) for some \(t_*\in (-\infty ,-T_\varphi ]\) as \(k\rightarrow +\infty \). For the former case, denote

$$\begin{aligned} U_k(x,t)=U(x+x^k,t+t_k). \end{aligned}$$

By Lemma 3.3, \(\{U_k(x,t)\}_{k=1}^\infty \) is equicontinuous in \(x\in \Omega \) and \(t\in {\mathbb {R}}\). It follows that there exists a subsequence still denoted by \(\{U_k(x,t)\}_{k=1}^\infty \) such that

$$\begin{aligned} U_k\rightarrow U_*~\text {as}~k\rightarrow +\infty ~\text {locally uniformly in}~(x,t)\in \Omega \times {\mathbb {R}} \end{aligned}$$

by Arzela-Ascolit theorem. Furthermore, \(U_*\) satisfies \(\frac{\partial U_*(0,0)}{\partial t}=0\). Applying strong maximum principle theorem to \(\frac{\partial U_*(x,t)}{\partial t}\), We further have

$$\begin{aligned} \frac{\partial U_*(x,t)}{\partial t}\equiv 0~\text {for all}~t\le 0~\text {and}~x\in \Omega . \end{aligned}$$

However, this is impossible because

$$\begin{aligned} U_*(x,t)=\phi (x_1+ct+a)~\text {for some}~a\in [-M_\eta ,M_\eta ]. \end{aligned}$$

For the second case, \(x^k_1\) remains bounded by the definition of \(\Omega _{\varphi }(t)\). Therefore, we assume that \(x^k_1\rightarrow x^*_1\) as \(k\rightarrow +\infty \) and let

$$\begin{aligned} U_k(x,t):=U(x+x^k,t). \end{aligned}$$

Then, each \(U_k(x,t)\) is defined for all \((x,t)\in (-\infty ,-T_\varphi ]\times \{x\in {\mathbb {R}}^N\mid x_1\ge -1\}\) by the definition of \(\Omega _\varphi (t)\). Similarly, there exists a subsequence, again denoted by \(\{U_k\}^{\infty }_{k=1}\), such that

$$\begin{aligned} U_k\rightarrow U^*~\text {as}~k\rightarrow +\infty ~\text {locally uniformly in}~(x,t)\in \Omega \times {\mathbb {R}}~\text {with}~x_1\ge -1, \end{aligned}$$

for some function \(U^*\) satisfying (1.1) on \(\{x\in {\mathbb {R}}^N\mid x_1\ge -1\}\times (-\infty ,-T_\varphi ]\). Note that \(U_t(x,t)>0\), we have

$$\begin{aligned} \frac{\partial U^*}{\partial t}(0,t_*)=0,~\frac{\partial U^*}{\partial t}(x,t)\ge 0 ~\text {for}~(x,t)\in \{x\in {\mathbb {R}}^N\mid x_1\ge -1\}\times (-\infty ,-T_\varphi ]. \end{aligned}$$

Then we obtain \(\frac{\partial U^*}{\partial t}(x,t)\equiv 0\) for \(t\le t_*\) by strong maximum principle, but this is impossible since

$$\begin{aligned} U^*(x,t)-\phi (x_1+x^*_1+ct)\rightarrow 0~\text {as}~t\rightarrow -\infty ~\text {uniformly in} ~{\{x\in {\mathbb {R}}^N\mid x_1\ge -1\}}. \end{aligned}$$

This ends the proof. \(\square \)

Now we are ready to show the uniqueness of the entire solution. Suppose that there exists another entire solution V(x, t) of (1.1) satisfying (3.1). Extend the function f as

$$\begin{aligned} f(s)=f'(0)s~\text {for}~s\le 0,\quad f(s)=f'(1)(s-1)~\text {for}~s\ge 1 \end{aligned}$$

and choose \(\eta >0\) being sufficiently small such that

$$\begin{aligned} f'(s)\le -\omega ~\text {for}~s\in [-2\eta ,2\eta ]\cup [1-2\eta ,1+2\eta ]~\text {and}~\omega >0. \end{aligned}$$

Then for any \(\epsilon \in (0,\eta )\), we can find \(t_0\in {\mathbb {R}}\) such that

$$\begin{aligned} \Vert V(\cdot ,t)-U(\cdot ,t)\Vert _{L^{\infty }(\Omega )}<\epsilon ~\text {for}~-\infty <t\le t_0. \end{aligned}$$
(3.6)

For each \(t_0\in (-\infty ,T_\varphi -\sigma \epsilon ]\), define

$$\begin{aligned} {\tilde{U}}^+(x,t):=U\left( x,t_0+t+\sigma \epsilon \left( 1-e^{-\omega t}\right) \right) +\epsilon e^{-\omega t} \end{aligned}$$

and

$$\begin{aligned} {\tilde{U}}^-(x,t):=U\left( x,t_0+t-\sigma \epsilon \left( 1-e^{-\omega t}\right) \right) -\epsilon e^{-\omega t}, \end{aligned}$$

where the constant \(\sigma >0\) is specified later. Then by (3.6),

$$\begin{aligned} {\tilde{U}}^-(x,0)\le V(x,t_0)\le {\tilde{U}}^+(x,0)~\text { for all}~ x\in \Omega . \end{aligned}$$
(3.7)

Next we show that \({\tilde{U}}^-(x,t)\) and \({\tilde{U}}^+(x,t)\) are sub- and super-solutions in \(t\in [0,T_\varphi -t_0-\sigma \epsilon ]\), respectively. Let

$$\begin{aligned} {\mathcal {L}}w(x,t):=w_t(x,t)-\int _{\Omega }J(x-y)[w(y,t)-w(x,t)]dy+f(w(x,t)) ~\text {for all}~(x,t)\in {\bar{\Omega }}\times {\mathbb {R}}. \end{aligned}$$

A straightforward calculation implies that

$$\begin{aligned} \begin{aligned} {\mathcal {L}}{\tilde{U}}^+(x,t)=&\sigma \epsilon \omega e^{-\omega t}U_t-\epsilon \omega e^{-\omega t}+f(U)-f\left( U+\epsilon e^{-\omega t}\right) \\ =&\epsilon e^{-\omega t}\left( \sigma \omega U_t-\omega -f'\left( U+\theta \epsilon e^{-\omega t}\right) \right) , \end{aligned} \end{aligned}$$

where \(0<\theta <1\). For any \(x\not \in \Omega _\eta \left( x,t_0+t+\sigma \epsilon \left( 1-e^{-\omega t}\right) \right) \), we can see

$$\begin{aligned} U+\theta \epsilon e^{-\omega t}\in [0,2\eta ]\cup [1-\eta ,1+\eta ]. \end{aligned}$$

Consequently, \(f'\left( U+\theta \epsilon e^{-\omega t}\right) \le -\omega \), which implies that

$$\begin{aligned} {\mathcal {L}}{\tilde{U}}^+(x,t)\ge \epsilon e^{-\omega t}(-\omega +\omega )=0. \end{aligned}$$

For \(x\in \Omega _\eta \left( x,t_0+t+\sigma \epsilon \left( 1-e^{-\omega t}\right) \right) \), by Lemma 3.4, there holds

$$\begin{aligned} {\mathcal {L}}{\tilde{U}}^+(x,t)\ge \epsilon e^{-\omega t}\left( \sigma \omega K_\eta -\omega -\max \limits _{0\le s\le 1}f'(s)\right) . \end{aligned}$$

As a consequence, \({\mathcal {L}}{\tilde{U}}^+(x,t)\ge 0\) provided that \(\sigma \) is a sufficiently large number.

Similarly, we can show \({\mathcal {L}}{\tilde{U}}^-(x,t)\le 0\) in \(\Omega \times [0,T_\varphi -t_0-\sigma \epsilon ]\). In view of this and (3.7), we see that

$$\begin{aligned} {\tilde{U}}^-(x,t)\le V(x,t+t_0)\le {\tilde{U}}^+(x,t)~\text {for all}~(x,t)\in \Omega \times [0,T_\varphi -t_0 -\sigma \epsilon ]. \end{aligned}$$

Letting \(t+t_0\) be replaced by t, the inequality above can be rewritten as

$$\begin{aligned}&U\left( x,t-\sigma \epsilon \left( 1-e^{-\omega (t-t_0)}\right) \right) -\epsilon e^{-\omega (t-t_0)}\\&\quad \le V(x,t)\le U\left( x,t+\sigma \epsilon \left( 1-e^{-\omega (t-t_0)}\right) \right) +\epsilon e^{-\omega (t-t_0)} \end{aligned}$$

for all \((x,t)\in \Omega \times [t_0,T_\varphi -\sigma \epsilon ]\) and \(t_0\in (-\infty ,T_\varphi -\sigma \epsilon ]\). As \(t_0\rightarrow -\infty \), we obtain that

$$\begin{aligned} U(x,t-\sigma \epsilon )\le V(x,t)\le U(x,t+\sigma \epsilon ) ~\text {for all}~(x,t)\in \Omega \times (-\infty ,T_\varphi -\sigma \epsilon ]. \end{aligned}$$

By the comparison principle, the inequality holds for \(t\in {\mathbb {R}}\) and \(x\in \Omega \). Letting \(\epsilon \rightarrow 0\), since \(\sigma \) is independent of the choice of \(\epsilon \), we have \(V(x,t)\equiv U(x,t)\).

4 Behaviors Far Away from the Interior Domain

In this section, we are going to figure out what the entire solution, constructed in Theorem 3.1, is like far away from the interior domain.

Theorem 4.1

Assume that (F) and (J) hold. Let \((\phi ,c)\) be the unique solution of (1.4) and u(x, t) be a solution of

$$\begin{aligned} \left\{ \begin{aligned}&u_t(x,t)=\int _\Omega J(x-y)[u(y,t)-u(x,t)]dy+f(u(x,t)),\quad (x,t)\in {\overline{\Omega }}\times {\mathbb {R}},\\&0\le u(x,t)\le 1,\quad (x,t)\in {\overline{\Omega }}\times {\mathbb {R}} \end{aligned}\right. \end{aligned}$$
(4.1)

such that

$$\begin{aligned} \sup \limits _{x\in {\bar{\Omega }}}|u(x,t)-\phi (x_1+ct)| \rightarrow 0~\text {as}~t\rightarrow -\infty . \end{aligned}$$

Then, for any sequence \((x'_n)_{n\in {\mathbb {N}}}\in {\mathbb {R}}^{N-1}\) such that \(|x'_n|\rightarrow +\infty \) as \(n\rightarrow +\infty \), there holds

$$\begin{aligned} u(x_1,x'+x'_n,t)\rightarrow \phi (x_1+ct)~\text {as}~n\rightarrow +\infty , \end{aligned}$$

locally uniformly with respect to \((x,t)=(x_1,x',t)\in {\mathbb {R}}^N\times {\mathbb {R}}\).

Proof

In order to prove this theorem we need to extend the function f as that in Sect. 3, and let \(A>0\) be sufficiently large such that

$$\begin{aligned} \phi (\xi )\le \frac{\eta }{2}~\text {for}~\xi \le -A,~~ \phi (\xi )\ge 1-\frac{\eta }{2}~\text {for}~\xi \ge A. \end{aligned}$$

Denote \(\delta =\min \limits _{[-A,A]}\phi '(\xi )>0\) and take \(T_\phi >0\) such that for any \(\epsilon \in (0,\frac{\eta }{2})\), we have that

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\le \epsilon ~\text {for all}~t\le -T_\phi ~\text {and}~x\in \Omega . \end{aligned}$$

Now we are in a position to show this theorem. Under the assumptions of Theorem 4.1, let \((x'_n)_{n\in {\mathbb {N}}}\in {\mathbb {R}}^{N-1}\) be a sequence such that \(|x'_n|\rightarrow +\infty \) as \(n\rightarrow +\infty \). And denote \(u_n(x,t)=u(x_1,x'+x'_n,t)\) for each \(t\in {\mathbb {R}}\) and \(x=(x_1,x')\in \Omega -(0,x'_n)\). Since \(0\le u\le 1\), K is compact and \(\left\{ u_n(x,t),\frac{\partial u_n(x,t)}{\partial t}\right\} _{n=1}^{\infty }\) is equicontinuous in \(x\in \Omega \) and \(t\in {\mathbb {R}}\) by Proposition 3.3, then there exists a subsequence, still denoted by \(\left\{ u_n(x,t),\frac{\partial u_n(x,t)}{\partial t}\right\} _{n=1}^\infty \), such that

$$\begin{aligned} u_n(x,t)\rightarrow {\mathfrak {u}}(x,t),~\frac{\partial u_n(x,t)}{\partial t}\rightarrow {\mathfrak {u}}_t(x,t)~\text {as}~n\rightarrow +\infty , \end{aligned}$$

locally uniformly in \((x,t)\in {\mathbb {R}}^N\times {\mathbb {R}}\). In addition, for all \((x,t)\in {\mathbb {R}}^N\times {\mathbb {R}}\), we further have that \(0\le {\mathfrak {u}}(x,t)\le 1\) and

$$\begin{aligned} {\mathfrak {u}}_t(x,t)=\int _{{\mathbb {R}}^N}J(x-y)[{\mathfrak {u}}(y,t)-{\mathfrak {u}}(x,t)]dy +f({\mathfrak {u}}(x,t)), \end{aligned}$$
(4.2)

since J is compactly supported, K is compact and \(|x'_n|\rightarrow +\infty \) as \(n\rightarrow \infty \). In addition, recall that

$$\begin{aligned} u_n(x,t)-\phi (x_1+ct)\rightarrow 0~\text {as}~t\rightarrow -\infty , \end{aligned}$$

uniformly in \(x\in {\overline{\Omega }}\), the function \({\mathfrak {u}}(x,t)\) satisfies

$$\begin{aligned} {\mathfrak {u}}(x,t)-\phi (x_1+ct)\rightarrow 0~\text {as}~t\rightarrow -\infty ~\text {locally uniformly in}~{\mathbb {R}}^N. \end{aligned}$$

Now define two functions \({\underline{u}}(x,t)\) and \({\overline{u}}(x,t)\) as follows

$$\begin{aligned} {\underline{u}}(x,t)=\phi (\xi _-(x,t))-\epsilon e^{-\omega (t-t_0)},~{\overline{u}}(x,t)=\phi (\xi _+(x,t))+\epsilon e^{-\omega (t-t_0)},~t\ge t_0,~x\in {\mathbb {R}}^N, \end{aligned}$$

where

$$\begin{aligned} t_0\le -T,~\xi _\pm (x,t)=x_1+ct\pm 2\epsilon \Vert f'\Vert \delta ^{-1}\omega ^{-1} \left[ 1-e^{-\omega (t-t_0)}\right] . \end{aligned}$$

Then the following lemma holds, whose proof is left to Appendix as a regular argument. \(\square \)

Lemma 4.2

The functions \({\underline{u}}(x,t)\) and \({\overline{u}}(x,t)\) are sub- and super-solutions to (4.2) for \(t\ge t_0\), respectively.

By the comparison theorem and let \(\epsilon \rightarrow 0\), we have \({\mathfrak {u}}(x,t)\equiv \phi (x_1+ct)\). Since the limit is uniquely determined, the sequence \(\{u_n(x,t)\}_{n=1}^{\infty }\) converges to \(\phi (x_1+ct)\) locally uniformly in \((x,t)\in {\mathbb {R}}^N\times {\mathbb {R}}\) as \(n\rightarrow +\infty \). Then the proof of Theorem 4.1 is completed. \(\square \)

Theorem 4.3

Suppose that all the assumptions in Theorem 4.1 hold. Then the solution u(x, t) of (4.1) in Theorem 4.1 satisfies

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\rightarrow 0~\text {as}~|x|\rightarrow +\infty ~\text {locally uniformly in}~ t\in {\mathbb {R}}. \end{aligned}$$

Proof

Extend f as that in Sect. 3. Then define \(f_\delta (u)=f(u-\delta )\), \((c_\delta ,\phi _\delta )\) satisfies

$$\begin{aligned} c_\delta (\phi _\delta )'(x)=\int _{{\mathbb {R}}^N}J(x-y) [\phi _\delta (y)-\phi _\delta (x)]dy+f_\delta (\phi _\delta (x)), \end{aligned}$$

and

$$\begin{aligned} \phi _\delta (-\infty )=\delta ,~\phi _\delta (+\infty )=1+\delta , \end{aligned}$$

where \(c_\delta >0\) and \(\delta >0\) is sufficiently small. Now we are going to show the theorem in three steps.

Step 1. For \(x_1\gg 1\), since

$$\begin{aligned} \sup \limits _{x\in {\bar{\Omega }}}|u(x,t)-\phi (x_1+ct)| \rightarrow 0~\text {as}~t\rightarrow -\infty , \end{aligned}$$

there exists a sufficiently large number \(T^*_1\ge 0\) such that

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\le \frac{\epsilon }{2}~\text {for all}~x\in \Omega ~\text {and}~t\le -T^*_1. \end{aligned}$$

In particular, for any \(x\in \Omega \), let \(N_1\gg 1\) such that

$$\begin{aligned} \phi (x_1-cT^*_1)\ge 1-\frac{\epsilon }{2},~u(x,-T^*_1)\ge 1-\epsilon ~\text {for all}~x_1\ge N_1. \end{aligned}$$

Since \(\phi '>0,~u_t(x,t)>0\), we have

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|<\epsilon ~\text {for all}~x_1\ge N_1~\text {and}~t\ge -T^*_1. \end{aligned}$$

Step 2. For \(x_1\ll -1\), let \(\delta =\frac{1}{2}\epsilon \). Similarly as the first step, choose \(T_2>0\) and \(N_2\gg 1\) such that

$$\begin{aligned} u(x,t)\le \delta ~\text {for all}~x_1\le -N_2~\text {and}~t\le -T_2,~\text {particularly},~u(x,-T_2)\le \delta . \end{aligned}$$

Obviously, there exists \(x_0\in {\mathbb {R}}\) such that \(\phi _\delta (x_0)=1\). Then

$$\begin{aligned} u(x,-T_2)\le \phi _\delta (x_1+x_0+N_2)~\text {for all}~x\in \Omega . \end{aligned}$$

Since that \(\phi _\delta (x,t)\) is increasing, we have

$$\begin{aligned} \phi _\delta (x_1+c_\delta (t+T_2)+x_0+N_2) \ge \phi _\delta (-N_2+x_0+N_2)=1~\text {for all}~x_1\ge -N_2~\text {and}~t\ge -T_2. \end{aligned}$$

Then, by applying the comparison principle (see [41]) for \(t\ge -T_2\) and \(x_1\le -N_2\), one have that

$$\begin{aligned} u(x,t)\le \phi _\delta (x_1+x_0+N_2+c_\delta (t+T_2))~\text {for all}~x_1\le -N_2 ~\text {and}~t\ge -T_2. \end{aligned}$$

In particular, since \(\phi _\delta (-\infty )=\delta =\frac{1}{2}\epsilon \), for any \(\tau \ge 0\), there exists a \(N_3\gg 1\) such that

$$\begin{aligned} 0\le u(x,t)\le 2\delta<\epsilon ,~0<\phi (x_1+ct)\le \epsilon ~\text {for all}~x_1\le -N_3~\text {and}~\tau \ge t\ge -T_2, \end{aligned}$$

which again shows

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\le \epsilon ~\text {for all}~x_1\le -N_3~\text {and}~t\le \tau . \end{aligned}$$

Step 3. For \(|x'|\gg 1\), it follows from the Theorem 4.1 that one can choose \(N_4>0\) being sufficiently large such that

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\le \epsilon ~\text {holds true for }t\text { in any bounded interval}, \end{aligned}$$

whenever \(|x'|>N_4\) and \(x_1\in [-N_3,N_1]\). This finishes the proof. \(\square \)

5 The Behavior for the Large Time

In this section we intend to investigate the behavior of the solution constructed in Theorem 3.1 for large positive time. For this goal, we establish the following result.

Theorem 5.1

Suppose that (F) and (J) hold. Let \((\phi ,c)\) be the unique solution of (1.4), \(t_0\in {\mathbb {R}}\) and u(x, t) be a solution of

$$\begin{aligned} \left\{ \begin{aligned} u_t(x,t)=\int _\Omega J(x-y)[u(y,t)-u(x,t)]dy+f(u(x,t)),~&(x,t)\in {\overline{\Omega }}\times [t_0,+\infty ),\\ 0\le u(x,t)\le 1,~\qquad \qquad \qquad \qquad \qquad \qquad \qquad&(x,t)\in {\overline{\Omega }}\times [t_0,+\infty ). \end{aligned} \right. \end{aligned}$$
(5.1)

And assume that, for any \(\epsilon >0\), there is a number \(t_\epsilon \ge t_0\) and a compact set \(K_\epsilon \subset {\overline{\Omega }}\) such that

$$\begin{aligned} |u(x,t_\epsilon )-\phi (x_1+ct_\epsilon )|\le \epsilon ~\text {for all}~x\in \overline{\Omega \backslash K_\epsilon }, \end{aligned}$$

and

$$\begin{aligned} u(x,t)\ge 1-\epsilon ~\text {for all}~t\ge t_\epsilon ~\text {and}~x\in \partial \Omega =\partial K_\epsilon . \end{aligned}$$

Then

$$\begin{aligned} \sup \limits _{x\in {\overline{\Omega }}}|u(x,t)-\phi (x_1+ct)|\rightarrow 0~\text {as}~t\rightarrow +\infty . \end{aligned}$$

The most important ingredient of the proof is to construct suitable sub- and super-solutions. This process is such cumbersome that will be divided several parts. Enlightened by Hoffman [27], we first construct a function z(t) in the following lemma, which plays an important role in the construction of sub- and super-solutions.

Lemma 5.2

For any \(0<\eta _z<\ln 2\), there are two constants \({\mathcal {I}}={\mathcal {I}}(\eta _z)>0\) and \(K_0=K_0(\eta _z)>0\) such that for any \(t_1\ge 0\), there exists a \(C^1\)-smooth function \({\tilde{z}}(t):[0,+\infty )\rightarrow {\mathbb {R}}\) that satisfies the following properties.

  1. (i)

    For all \(t\ge 0\), the inequalities \({\tilde{z}}'(t)\ge -\eta _z{\tilde{z}}(t)\) and \(0<{\tilde{z}}(t)\le {\tilde{z}}(0)=1\) hold.

  2. (ii)

    In addition, \({\tilde{z}}(t)\ge K_0(1+t-t_1)^{-\frac{3}{2}}\) for all \(t\ge t_1\) and \(\int _0^{+\infty }{\tilde{z}}(t)dt<{\mathcal {I}}\).

Proof

First we define

$$\begin{aligned} P_-(x)=-\frac{1}{3}\eta _z^2\left( x+\eta _z^{-1}\right) ^2+1,~P_+(x) =\frac{\nu \eta _z}{3}\left( x-\eta _z^{-1}\right) ^2+\frac{2}{3}-\frac{\nu }{3\eta _z},~0<\nu <\eta _z. \end{aligned}$$

Let \(l_P(\eta _z)=1/\eta _z\). Then, by a direct calculation, it is easy to show that for any fixed \(0<\eta _z<\ln 2\) and \(0<\nu \le \eta _z\), \(P_-(x)\) satisfies that

$$\begin{aligned} P_-(-l_P)=1,\quad P_-'(-l_P)=0,\quad P_-(0)=\frac{2}{3},\quad P_-'(0)=-\frac{2}{3}\eta _z, \end{aligned}$$

with

$$\begin{aligned} -\eta _zP_-(x)\le P'_-(x)\le 0,~-l_P\le x\le 0, \end{aligned}$$

and \(P_+(x)\) satisfies that

$$\begin{aligned} P_+(0)=\frac{2}{3},\quad P'_+(0)=-\frac{2}{3}\nu ,\quad P'_+(l_P)=0, \end{aligned}$$

\(P_+(l_P)\ge \frac{1}{3}\) with

$$\begin{aligned} -\eta _zP_+(x)\le P'_+(x)\le 0,~0\le x\le l_P. \end{aligned}$$

In addition, denote

$$\begin{aligned} z_1(t)=\left\{ \begin{aligned} e^{-\eta _zt},\qquad \qquad \qquad \qquad&0\le t\le \frac{3}{2}\eta _z^{-1}-1, \\ \eta _z^{-\frac{3}{2}}\left( \frac{3}{2}\right) ^{\frac{3}{2}}e^{\eta _z-\frac{3}{2}}(1+t)^{-\frac{3}{2}}, \quad&t\ge \frac{3}{2}\eta _z^{-1}-1. \end{aligned} \right. \end{aligned}$$

Now if \(0<t_1<3\eta _z^{-1}\), then let \({\tilde{z}}(t)=z_1(t)\), otherwise we define the function \({\tilde{z}}(t)\) on five different intervals. In particular, define \(\nu =\frac{-z_1'(t_1-3\eta _z^{-1})}{z_1(t_1-3\eta _z^{-1})}\), which implies \(0<\nu \le \eta _z\). Then, let

$$\begin{aligned} {\tilde{z}}(t)= {\left\{ \begin{array}{ll} z_1(t), &{}\quad 0\le t\le t_1-3\eta _z^{-1},\\ z_1\left( t_1-3\eta _z^{-1}\right) P_+\left( t-\left( t_1-3\eta _z^{-1}\right) \right) , &{}\quad ~t_1-3\eta _z^{-1}\le t\le t_1-2\eta _z^{-1},\\ P_-(t-t_1), &{}\quad ~t_1-\eta _z^{-1}\le t\le t_1,\\ \frac{2}{3}z_1(t-t_1), &{}\quad t\ge t_1. \end{array}\right. } \end{aligned}$$

It remains to specify \({\tilde{z}}(t)\) in \(\left[ t_1-2\eta _z^{-1}, t_1-\eta _z^{-1}\right] \). This can be done by choosing an arbitrary \(C^1\)-smooth function, under the constraints

$$\begin{aligned} {\tilde{z}}\left( t_1-2\eta _z^{-1}\right)= & {} z_1\left( t_1-3\eta _z^{-1}\right) P_+\left( \eta _z^{-1}\right) , ~z\left( t_1-\eta _z^{-1}\right) =1,~{\tilde{z}}'\left( t_1-2\eta _z^{-1}\right) \\= & {} z'\left( t_1-\eta _z^{-1}\right) =0 \end{aligned}$$

and

$$\begin{aligned} {\tilde{z}}'(t)\ge 0,\quad t_1-2\eta _z^{-1}\le t\le t_1-\eta _z^{-1}. \end{aligned}$$

Then we finish the proof since the properties (i) and (ii) are valid by a direct calculation. \(\square \)

Remark 5.3

It is not difficult to see that \(z(1)\ge \frac{1}{2}\) from \(0<\eta _z<\ln 2\) and the first statement in Lemma 5.2.

Now, we are in the position to construct the sub- and super-solutions.

5.1 Sub-solution

In this part, we construct a sub-solution to (5.1). Now define

$$\begin{aligned} {\tilde{u}}(x,t)=u(x,t-1+t_\epsilon ), ~u^-(x,t)=\phi (x_1+c(t-1+t_\epsilon )-\theta (x',t)-Z(t))-z(t), \end{aligned}$$

where \(\theta (x',t)=\beta t^{-\alpha }e^{\frac{-|x'|^2}{\gamma t}}\) with \(\alpha ,~\gamma >1\) being two real numbers, \(z(t)=\epsilon _1{\tilde{z}}(t)\) with \(\epsilon _1=2\epsilon \) and \(Z(t)=K_z\int _0^tz(\tau )d\tau \) with \(K_z>0\) being a large number. z(t) is defined in Lemma 5.2. It follows from the definition of \(u^-(x,t)\) that

$$\begin{aligned} u^-(x,1)\le \phi \left( x_1+ct_\epsilon -\beta e^{\frac{-|x'|^2}{\gamma }} -Z(1)\right) \le {\tilde{u}}(x,1)=u(x,t_\epsilon )~\text {for all}~x\in K_\epsilon . \end{aligned}$$

Thanks to \(\phi (-\infty )=0,~\phi '>0\), \(0<u(x,t)<1\) and \(\min \limits _{x\in K_\epsilon }u(x,t)>0\), the last inequality holds provided that \(\beta \) is sufficiently large. If \(x\in {\mathbb {R}}^N\backslash K_\epsilon \), then by \(z(1)\ge \frac{1}{2}\epsilon _1=\epsilon \) (from Remark 5.3), we have that

$$\begin{aligned} u^-(x,1)\le \phi (x_1+ct_\epsilon )-z(1)\le \phi (x_1+ct_\epsilon )-\epsilon \le u(x,t_\epsilon )={\tilde{u}}(x,1). \end{aligned}$$

As a consequence, there holds \(u^-(x,1)\le {\tilde{u}}(x,1)\) for any \(x\in {\overline{\Omega }}\).

Lemma 5.4

The inequality \({\mathcal {L}}u^-(x,t)\le 0\) holds for all \(x\in \Omega \) and \(t\ge 1\), where

$$\begin{aligned} {\mathcal {L}}u^-(x,t)=u^-_t(x,t)-\int _\Omega J(x-y)[u^-(y,t)-u^-(x,t)]dy-f(u^-(x,t)). \end{aligned}$$

Proof

Since \(u^-(x,t)=\phi (\xi (x,t))-z(t)\), where \(\xi (x,t)=x_1+c(t-1+t_\epsilon ) -\beta t^{-\alpha }e^{\frac{-|x'|^2}{t\gamma }}-Z(t)\), we have

$$\begin{aligned} u^-_t(x,t)=\phi '(\xi (x,t))(c-\theta _t(x',t)-Z')-z'(t), \end{aligned}$$

and

$$\begin{aligned} \int _\Omega J(x-y)[u^-(y,t)-u^-(x,t)]dy=\int _\Omega J(x-y)[\phi (\xi (y,t))-\phi (\xi (x,t))]dy. \end{aligned}$$

Denote \({\mathcal {D}}\phi =\int _{{\mathbb {R}}^N}J(x-y)[\phi (\xi (y,t))-\phi (\xi (x,t))]dy\). Then, applying mean value theorem, we get that

$$\begin{aligned} {\mathcal {D}}\phi =&\int _{{\mathbb {R}}^N}J(y)[\phi (\xi (x,t)-y_1)-\phi (\xi (x,t))]dy+\int _{{\mathbb {R}}^N}J(y) \bigg [\phi \bigg (x_1-y_1+c(t-1+t_\epsilon )\\&-\beta t^{-\alpha }e^{\frac{-|x'-y'|^2}{t\gamma }}-Z(t)\bigg )-\phi (\xi (x,t)-y_1)\bigg ]dy\\ \ge&c\phi '(\xi (x,t))-f(\phi (\xi (x,t))) -C^0\phi '(\xi (x,t))\beta t^{-\alpha } \\&\int _{{\mathbb {R}}^N}J(y)|y'|\frac{2|x'-{\tilde{\theta }}y'|}{t\gamma } e^{-\frac{|x'-{\tilde{\theta }}y'|^2}{t\gamma }}dy, \end{aligned}$$

where \(0<{\tilde{\theta }}<1\). In particular,

$$\begin{aligned} {\mathcal {D}}\phi \ge c\phi '(\xi (x,t))-f(\phi (\xi (x,t)))-C't^{-\alpha -\frac{1}{2}}\phi '(\xi (x,t)). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned} {\mathcal {L}}u^-(x,t)=&u^-_t(x,t)-\int _\Omega J(x-y)[u^-(y,t)-u^-(x,t)]dy-f(u^-(x,t))\\ \le&f(\phi (\xi (x,t)))-f(\phi (\xi (x,t))-z(t))+\left( C't^{-\alpha -\frac{1}{2}}-Z'(t)-\theta _t(x',t)\right) \phi '(\xi (x,t))\\&+\int _KJ(x-y)[\phi (\xi (y,t))-\phi (\xi (x,t))]dy-z'(t). \end{aligned} \end{aligned}$$

Similarly as previous, one can get

$$\begin{aligned} \int _KJ(x-y)[\phi (\xi (y,t)-\phi (\xi (x,t))]\ge -C^Kt^{-\alpha -\frac{1}{2}}\phi '(\xi (x,t))~\text {for some}~C^K>0. \end{aligned}$$

It follows that

$$\begin{aligned} {\mathcal {L}}u^-(x,t)\le&f(\phi (\xi (x,t)))-f(\phi (\xi (x,t))-z(t))\\&+\left[ (C'+C^K)t^{-\alpha -\frac{1}{2}}-Z'(t)-\theta _t(x',t) \right] \phi '(\xi (x,t))-z'(t). \end{aligned}$$

Now we go further to show \({\mathcal {L}}u^-(x,t)\le 0\) in two cases.

Case 1. We assume that \(|\xi (x,t)|\gg 1\) such that \(\phi (\xi (x,t))\in [0,\eta ]\cup [1-\eta ,1]\), where \(\eta \) is sufficiently small to ensure that \(f'(s)\le -\sigma <0\) for any \(s\in [0,\eta ]\cup [1-\eta ,1]\) and \(\sigma >2\eta _z\) with \(\eta _z\) defined as that in Lemma 5.2. Then, since the function z(t) constructed in Lemma 5.2 satisfies

$$\begin{aligned} z'(t)\ge -\eta _zz(t),\quad z(t)\ge K_0(1+t-t_1)^{-\frac{3}{2}}~\text {for}~t_1\ge 0, \end{aligned}$$

there holds

$$\begin{aligned} {\mathcal {L}}u^-(x,t)\le&(-\sigma +\eta _z)z(t)-\phi '(\xi (x,t))\\&\left[ K_zz(t)+\left( \frac{|x'|^2}{\gamma t}-\alpha \right) t^{-1}\theta (x',t)-(C'+C^{K})t^{-\alpha -\frac{1}{2}}\right] \\ \le&-\eta _zK_0t^{-\frac{3}{2}}+\left( \alpha \beta +C^K+C'\right) t^{-\alpha -\frac{1}{2}}\phi '(\xi (x,t))\\ \le&-\frac{1}{2}\eta _zK_0t^{-\frac{3}{2}}. \end{aligned}$$

Indeed, since \(\alpha >1\) and \(|\xi (x,t)|\) is sufficiently large such that \((\alpha \beta +C^K+C')\phi '(\xi )\le \frac{1}{2}\eta _zK_0\), the last inequality above holds obviously.

Case 2. Let \(\phi (\xi (x,t))\in [\eta ,1-\eta ]\) in this case. Since \(\phi '(\xi )>0\) for all \(\xi \in {\mathbb {R}}\), we may choose \(\tau _0>0\) being sufficiently small such that \(\phi '(\xi )\ge \tau _0>0\). Denote \(\max \limits _{[\eta ,1-\eta ]}f'(s)=\delta ^0>0\). Then

$$\begin{aligned} {\mathcal {L}}u^-(x,t)\le&\delta ^0z(t)+\eta _zz(t)-\left[ K_zz(t)+\left( \frac{|x'|^2}{t\gamma } -\alpha \right) t^{-1}\theta (x',t)-\left( C^K+C'\right) t^{-\alpha -\frac{1}{2}}\right] \\&\phi '(\xi (x,t))\\ \le&\left( -K_z\tau _0+\delta ^0+\eta _z\right) z(t)+\left( \alpha \beta +C^K+C'\right) t^{-\alpha -\frac{1}{2}}\phi '(\xi (x,t))\\ \le&\left[ -K_z\tau _0+\delta ^0+\eta _z+\left( \alpha \beta +C^K+C'\right) \frac{1}{K_0}\Vert \phi '\Vert _\infty \right] z(t). \end{aligned}$$

Let \(K_z\) be sufficiently large such that \(-K_z\tau _0+\delta ^0+\eta _z+\left( \alpha \beta +C^K+C'\right) \frac{1}{K_0}\Vert \phi '\Vert _\infty \le -\frac{1}{2}\eta _z\). One hence have that \({\mathcal {L}}u^-(x,t)\le -\frac{1}{2}\eta _zz(t)\). The proof is finished. \(\square \)

5.2 Super-solution

This part is devoted to verifying the super-solution defined as

$$\begin{aligned} u^+(x,t)=\phi (\psi (x,t))+z(t), \end{aligned}$$

where

$$\begin{aligned} \psi (x,t)=x_1+c(t-1+t_\epsilon )+\theta ^1(x',t)+Z(t),~\theta ^1(x',t)=\beta ^+t^{-\alpha ^+} e^{-\frac{|x'|^2}{t\gamma }} \end{aligned}$$

with \(\alpha ^+>1\) and \(\beta ^+>0\) being a large number. It follows from the definition of \(u^+(x,t)\) that

$$\begin{aligned} u^+(x,1)\ge \phi \left( x_1+ct_\epsilon +\beta ^+ e^{\frac{-|x'|^2}{\gamma }} +Z(1)\right) \ge {\tilde{u}}(x,1)=u(x,t_\epsilon ) \end{aligned}$$

for \(x\in K_\epsilon .\) Since \(\phi (-\infty )=0,~\phi '>0\), \(0<u(x,t)<1\) and \(\max \limits _{x\in K_\epsilon }u(x,t)<1\), the last inequality holds provided that \(\beta ^+\) is sufficiently large. If \(x\in {\mathbb {R}}^N\backslash K_\epsilon \), then by \(z(1)\ge \frac{1}{2}\epsilon _1=\epsilon \) (from Remark 5.3), we have

$$\begin{aligned} u^+(x,1)\ge \phi (x_1+ct_\epsilon )+z(1)\ge \phi (x_1+ct_\epsilon )+\epsilon \ge u(x,t_\epsilon )={\tilde{u}}(x,1). \end{aligned}$$

As a consequence, there holds \(u^+(x,1)\ge {\tilde{u}}(x,1)\) for any \(x\in {\overline{\Omega }}\).

Lemma 5.5

The inequality \({\mathcal {L}}u^+(x,t)\ge 0\) holds for all \(x\in \Omega \) and \(t\ge 1\), where

$$\begin{aligned} {\mathcal {L}}u^+(x,t)=u^+_t(x,t)-\int _\Omega J(x-y)[u^+(y,t)-u^+(x,t)]dy-f(u^+(x,t)). \end{aligned}$$

Proof

It follows from a direct calculation that

$$\begin{aligned} u^+_t(x,t)=\left( c+\theta ^1_t+Z'(t)\right) \phi '(\psi )+z'(t), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _\Omega J(x-y)[u^+(y,t)-u^+(x,t)]dy \\&\quad =\int _{{\mathbb {R}}^N}J(x-y)[\phi (\psi (y,t))-\phi (\psi (x,t))]dy\\&\quad -\int _KJ(x-y)[\phi (\psi (y,t))-\phi (\psi (x,t))]dy. \end{aligned} \end{aligned}$$

Note that

$$\begin{aligned} c\phi '(\psi (x,t))=\int _{{\mathbb {R}}^N}J(y)[\phi (\psi (x,t)-y_1)-\phi (\psi (x,t))]dy+f(\phi (\psi (x,t))), \end{aligned}$$

we have

$$\begin{aligned} u^+_t(x,t)= & {} (\theta '_t+Z'(t))\phi '(\psi (x,t))\\&+\int _{{\mathbb {R}}^N}J(y)[\phi (\psi (x,t)-y_1) -\phi (\psi (x,t))]dy+f(\phi (\psi (x,t)))+z'(t). \end{aligned}$$

Then it follows that

$$\begin{aligned} {\mathcal {L}}u^+(x,t) =&(\theta '_t+Z'(t))\phi '(\psi (x,t))+\int _{{\mathbb {R}}^N}J(y)[\phi (\psi (x,t)-y_1) -\phi (\psi (x,t))]dy\\&+f(\phi (\psi (x,t)))+z'(t)-\int _{{\mathbb {R}}^N}J(x-y)[\phi (\psi (y,t))-\phi (\psi (x,t))]dy\\&+\int _KJ(x-y)[\phi (\psi (y,t))-\phi (\psi (x,t))]dy-f(u^+(x,t)), \end{aligned}$$

and we obtain

$$\begin{aligned} {\mathcal {L}}u^+(x,t) =&(\theta '_t+Z'(t))\phi '(\psi (x,t))+\int _{{\mathbb {R}}^N}J(y)[\phi (\psi (x,t)-y_1) -\phi (\psi (x-y,t))]dy\\&+\int _KJ(x-y)[\phi (\psi (y,t))-\phi (\psi (x,t))]dy\\&+f(\phi (\psi (x,t)))+z'(t)-f(u^+(x,t)). \end{aligned}$$

Now we focus on all the integral items above denoted by

$$\begin{aligned} I&:=\int _{{\mathbb {R}}^N}J(y)[\phi (\psi (x,t)-y_1)-\phi (\psi (x-y,t))]dy\\&\quad +\int _KJ(x-y) [\phi (\psi (y,t))-\phi (\psi (x,t))]dy. \end{aligned}$$

By the same progress as the calculation of \(u^-(x,t)\), we have

$$\begin{aligned} I\ge -M't^{-\alpha ^+-\frac{1}{2}}\phi '(\psi (x,t)). \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathcal {L}}u^+(x,t)&\ge \left( \theta ^1_t+Z'(t)-M't^{-\alpha ^+-\frac{1}{2}}\right) \phi '(\psi (x,t))+z'(t)\\&\quad +f(\phi (\psi (x,t)))-f(u^+(x,t)). \end{aligned}$$

Next we are going to show \({\mathcal {L}}u^+(x,t)\ge 0\) in two cases.

Case 1. Let \(|\psi (x,t)|\gg 1\) such that \(\phi (\psi (x,t))\in [0,\eta ]\cup [1-\eta ,1]\), where \(\eta >0\) is sufficiently small to ensure that \(f'(s)\le -\sigma <0\) for any \(s\in [0,\eta ]\cup [1-\eta ,1]\). Since

$$\begin{aligned} \left( K_z+\frac{\beta ^+}{\gamma }\right) z(t)\ge \theta ^1_t+Z'(t)-M't^{-\alpha ^+-\frac{1}{2}}\ge K_zz(t)-(\alpha ^+\beta ^++M')t^{-\alpha ^+-\frac{1}{2}} \end{aligned}$$

and \(|\psi (x,t)|\gg 1\), we obtain that

$$\begin{aligned} \left| \left( \theta ^1_t+Z'(t)-M't^{-\alpha ^+-\frac{1}{2}}\right) \phi '(\psi (x,t))\right| \le \frac{1}{2}\eta _zz(t), \end{aligned}$$

which implies

$$\begin{aligned} {\mathcal {L}}u^+(x,t)\ge -\frac{1}{2}\eta _zz(t)-\eta _zz(t)+\sigma z(t)\ge 0, \end{aligned}$$

since \(\eta _z<\frac{1}{2}\sigma \) and \(\alpha ^+>1\).

Case 2. Since \(\phi (\psi (x,t))\in [\eta ,1-\eta ]\) we have \(\phi '(\psi (x,t))\ge \tau _0>0\). Denote \(\min \limits _{s\in [0,1]}f'(s)=-\delta '<0.\) Then, we obtain

$$\begin{aligned} \begin{aligned} {\mathcal {L}}u^+(x,t)\ge&\left[ K_zz(t)-(\alpha ^+\beta ^++M')t^{-\alpha ^+-\frac{1}{2}}\right] \tau _0-\delta 'z(t) -\eta _zz(t)\\ \ge&\left( K_z-\frac{\alpha ^+\beta ^++M'}{K_0}\right) \tau _0z(t)-\delta 'z(t)-\eta _zz(t). \end{aligned} \end{aligned}$$

Recall the fact \(\alpha ^+>1\), it follows that \(K_0t^{-\alpha ^+-\frac{1}{2}}<z(t)\). Moreover, one can take \(K_z>0\) being sufficiently large such that \(\left( K_z-\frac{\alpha ^+\beta ^++M'}{K_0}\right) \tau _0-\delta '-\eta _z\ge 0\). Then we have that \({\mathcal {L}}u^+(x,t)\ge 0\). The proof is finished. \(\square \)

The proof of Theorem 5.1

From the Lemmas 5.4 and 5.5, for \(t\ge t_{\epsilon }\), we have that

$$\begin{aligned}&\inf \limits _{x\in {\bar{\Omega }}}[u(x,t)-\phi (x_1+ct)]\\&\quad =\inf \limits _{x\in {\bar{\Omega }}}[{\tilde{u}}(x,t+1-t_{\epsilon }) -\phi (x_1+ct)]\\&\quad \ge \inf \limits _{x\in {\bar{\Omega }}}[\phi (x_1+ct-\theta (x',t+1-t_{\epsilon }) -Z(t+1-t_{\epsilon }))-z(t+1-t_{\epsilon }) -\phi (x_1+ct)]\\&\quad \ge -[\beta (t+1-t_{\epsilon })^{-\alpha }+Z(t+1-t_{\epsilon })]\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}-z(t+1-t_{\epsilon })\\&\quad \ge -\beta (t+1-t_{\epsilon })^{-\alpha }\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}-z(t+1-t_{\epsilon }) -\epsilon _1{\mathcal {I}}\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}, \end{aligned}$$

and

$$\begin{aligned}&\sup \limits _{x\in {\bar{\Omega }}}[u(x,t)-\phi (x_1+ct)]\\&\quad =\sup \limits _{x\in {\bar{\Omega }}}[{\tilde{u}}(x,t+1-t_{\epsilon }) -\phi (x_1+ct)]\\&\quad \le \sup \limits _{x\in {\bar{\Omega }}}[\phi (x_1+ct+\theta ^1(x',t+1-t_{\epsilon }) +Z(t+1-t_{\epsilon }))+z(t+1-t_{\epsilon }) -\phi (x_1+ct)]\\&\quad \le [\beta ^+(t+1-t_{\epsilon })^{-\alpha }+Z(t+1-t_{\epsilon })]\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}+z(t+1-t_{\epsilon })\\&\quad \le \beta ^+(t+1-t_{\epsilon })^{-\alpha }\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}+z(t+1-t_{\epsilon }) +\epsilon _1{\mathcal {I}}\Vert \phi '\Vert _{L^{\infty }({\mathbb {R}})}. \end{aligned}$$

For any sufficiently small \(\eta _{s}>0\), take \(0<\epsilon _1<\frac{\eta _s}{{\mathcal {I}}\Vert \phi '\Vert _{L^{\infty }}}\). Then by the construction of z(t), we see that

$$\begin{aligned} \liminf \limits _{t\rightarrow \infty }\inf \limits _{x\in {\bar{\Omega }}}[u(x,t)-\phi (x_1+ct)]\ge -\eta _s, \end{aligned}$$

and

$$\begin{aligned} \limsup \limits _{t\rightarrow \infty }\sup \limits _{x\in {\bar{\Omega }}}[u(x,t)-\phi (x_1+ct)]\le \eta _s. \end{aligned}$$

Since \(\eta _s\) is arbitrary, we have \(\lim \limits _{t\rightarrow \infty }\sup \limits _{x\in {\bar{\Omega }}}|u(x,t)-\phi (x_1+ct)|=0\). Thus, we finish the proof of Theorem 5.1. \(\square \)

5.3 Proofs of Theorem 1.1

It follows from [7, Theorems 2.4 and 2.6] that under the conditions (F) and (J), the unique solution of the stationary problem of (1.1)

$$\begin{aligned} \left\{ \begin{aligned}&\int _{{\mathbb {R}}^N\setminus K}J(x-y)[v(y)-v(x)]dy+f(v(x))=0,~x\in {\mathbb {R}}^N\setminus K(~\text {or}~K_{\epsilon }),\\&0\le v(x)\le 1,~~x\in {\mathbb {R}}^N\setminus K(~\text {or}~K_{\epsilon }),\\&\sup \limits _{{\mathbb {R}}^N\setminus K}v(x)=1 \end{aligned}\right. \end{aligned}$$
(5.2)

is \(v(x)\equiv 1 ~\text {in}~\overline{{\mathbb {R}}^N \backslash K}(~\text {or}~\overline{{\mathbb {R}}^N \backslash K_{\epsilon }}).\)

Now we are in position to show the main theorem. It is obvious that if the solution U(x, t) of (1.1) constructed in Theorem 3.1 satisfies the conditions in Theorem 5.1 together with Theorem 4.3, then the conclusions in Theorem 1.1 hold. Indeed, since \(U_t(x,t)>0\) and U(x, t) is Lipschitz continuous in \(x\in {\mathbb {R}}^N\backslash K\), we know that as time tends to positive infinity, U(x, t) locally uniformly converges to some continuous function V(x), without loss of generality, which can be deemed as a uniformly continuous function by [7, Lemma 3.2] because of the assumption (F). Furthermore, we claim that V(x) satisfies (5.2). Therefore, we have \(V(x)\equiv 1\) for all \(x\in {\mathbb {R}}^N\backslash K\). In fact, it is sufficient to show \(\sup _{{\mathbb {R}}^N\setminus K}V(x)=1\). In view of \(u(x,t)-\phi (x_1+ct)\rightarrow 0\) as \(t\rightarrow -\infty \) in Theorem 3.1, for any small \(\epsilon '>0\), there exist \(t_{\epsilon '}<0\) and \(X_1>0\) being sufficiently large such that

$$\begin{aligned} \phi (x+ct_{\epsilon '})\ge 1-\frac{\epsilon '}{2}~\text {for all}~x_1\ge X_1, \end{aligned}$$

and

$$\begin{aligned} |u(x,t)-\phi (x_1+ct)|\le \frac{\epsilon '}{2}~\text {for all}~x\in {\mathbb {R}}^N\backslash K~\text {and}~t\le t_{\epsilon '}. \end{aligned}$$

Thus

$$\begin{aligned} u(x,t)\ge 1-\epsilon '~\text {for all}~x_1\ge X_1~\text {and}~t\ge t_\epsilon ', \end{aligned}$$

which implies that \(V(x)\ge 1-\epsilon '\) due to \(u_t(x,t)>0\). Since that \(\epsilon '\) is actually arbitrary, one has that

$$\begin{aligned} \sup \limits _{{\mathbb {R}}^N\setminus K}V(x)=1. \end{aligned}$$

Therefore, \(V(x)\equiv 1\) for all \(x\in {\mathbb {R}}^N\backslash K\). Hence, \(u(x,t)\rightarrow 1\) as \(t\rightarrow +\infty \) for all \(x\in {\mathbb {R}}^N\backslash K\). It follows that for any \(\epsilon >0\), there are some \(t_\epsilon >0\) being sufficiently large and \(K_\epsilon \subset {\overline{\Omega }}\) with \(K\subset K_\epsilon \) such that

$$\begin{aligned} u(x,t)\ge 1-\epsilon ,~\text {for all}~t\ge t_\epsilon ,~x\in \partial K_\epsilon . \end{aligned}$$

In addition, it follows from Theorem 4.3 that

$$\begin{aligned} |u(x,t_\epsilon )-\phi (x_1+ct_\epsilon )|<\epsilon ,~\text {for all}~x\in \overline{{\mathbb {R}}^N\setminus K_\epsilon }. \end{aligned}$$

Thus, from Theorem 5.1, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\sup _{x\in {\overline{\Omega }}}|u(x,t)-\phi (x_1+ct)|=0. \end{aligned}$$

Moreover, since \(u(x,t)-\phi (x_1+ct)\rightarrow 0\) as \(t\rightarrow \pm \infty \) uniformly in \(x\in \Omega \) and Theorem 4.3, we have that \(u(x,t)-\phi (x_1+ct)\rightarrow 0\) as \(|x|\rightarrow +\infty \) uniformly in \(t\in {\mathbb {R}}.\) The proof of Theorem 1.1 is finished. Similarly, we know the results of Remark 1.2 hold.