1 Introduction and Main Results

In this paper, we are interested in the classical two-species Lotka-Volterra competition system with diffusion and two free boundaries:

$$\begin{aligned} \left\{ \begin{array}{ll} u_{t}-du_{xx}=ku(1-u-bv), &{} t>0,\ g(t)<x<h(t),\\ v_{t}-v_{xx}=v(1-v-au), &{} t>0,\ -\infty< x<\infty ,\\ u(t,x)=0, &{} t>0,\ x\in \mathbb {R}\setminus (g(t),h(t)),\\ h(0)=h_{0},\ h'(t)=-\mu u_{x}(t,h(t)), &{} t>0,\\ g(0)=-h_{0},\ g'(t)=-\mu u_{x}(t,g(t)), &{} t>0,\\ u(0,x)=u_{0}(x), &{} -h_{0}\le x\le h_{0},\\ v(0,x)=v_{0}(x), &{} -\infty< x<\infty , \end{array} \right. \end{aligned}$$
(1.1)

where u(tx) and v(tx) are the population densities of two competing species at time t and position x. All the parameters in (1.1) are assumed to be positive: d is the diffusion rate of u; k is the intrinsic growth rate of u; a and b are the competition coefficients for u and v, respectively. The two functions \(x=h(t)\) and \(x=g(t)\) are the moving boundaries to be determined. The initial functions \(u_0\) and \(v_0\) satisfy

$$\begin{aligned} \left\{ \begin{array}{ll} u_0\in C^2([-h_0,h_0]), u_0(-h_0)=u_0(h_0)=0\text { and }u_0(x)>0\text { in }(-h_0,h_0),\\ v_0\in C^2(\mathbb {R})\cap L^{\infty }(\mathbb {R}),~v_0(x)\ge 0\text { in }\mathbb {R}\text { and }\liminf _{\vert x\vert \rightarrow \infty }v_0(x)>0. \end{array} \right. \end{aligned}$$
(1.2)

System (1.1) was first proposed by Du and Lin in [5] to describe the dynamical process of a new species u invading into the habitat of a native species v. It is assumed that the species u, which exists initially in the range \(-h_{0}\le x\le h_{0}\), invades into the new territory through its invading fronts \(x=h(t)\) and \(x=g(t)\). The native species v undergoes diffusion and growth in the available habitat \(-\infty<x<\infty \). The Stefan type condition \(h'(t)=-\mu u_{x}(t,h(t))\) (or \(g'(t)=- \mu u_{x}(t,g(t))\)) means that the invading speed is proportional to the gradient of the population density of u at the invading front \(x=h(t)\) (or \(x=g(t)\)), where the parameter \(\mu \) is a positive constant measuring the intention of the species u to invade into a new territory (see Bunting et al. [2] for an interpretation based on ecological assumptions). One may refer to [5] for more biological background of (1.1).

In the absence of the invasive species u in the environment, (1.1) is reduced to the scalar Fisher-KPP equation:

$$\begin{aligned} \left\{ \begin{array}{ll} v_{t}-v_{xx}=v(1-v), &{} t>0,\ -\infty< x<\infty ,\\ v(0,x)=v_{0}(x), &{} -\infty< x<\infty . \end{array} \right. \end{aligned}$$
(1.3)

It is well known that the spreading properties of (1.3) are determined by the tail behavior of the initial distribution of the species (see [1, 15, 20, 23]). In particular, if the initial function \(v_0\) satisfies the assumptions in (1.2), then v globally exponentially converges to 1 as time goes to infinity. That is, there exist two positive constants \(K^*\) and \(\kappa \) such that \(\Vert v(t,\cdot )-1\Vert _{L^\infty (\mathbb {R})}\le K^*e^{-\kappa t}\) (see the proof of Theorem 1.4).

Nevertheless, in the absence of the native species v in the environment, (1.1) is reduced to the scalar Fisher-KPP equation with diffusion and two free boundaries:

$$\begin{aligned} \left\{ \begin{array}{ll} u_{t}-du_{xx}=ku(1-u), &{} t>0,\ g(t)<x<h(t),\\ u(t,g(t))=u(t,h(t))=0, &{} t>0,\\ h(0)=h_{0},\ h'(t)=-\mu u_{x}(t,h(t)), &{} t>0,\\ g(0)=-h_{0},\ g'(t)=-\mu u_{x}(t,g(t)), &{} t>0,\\ u(0,x)=u_{0}(x), &{} -h_{0}\le x\le h_{0}.\\ \end{array} \right. \end{aligned}$$
(1.4)

Model (1.4) has been extensively studied in the past decade and the dynamical behavior of (1.4) is characterized by a spreading-vanishing dichotomy (see Theorem 5.4 in [4]). Moreover, when the spreading of u happens, it was shown in [4] that there exists an asymptotic spreading speed with the form of \(c^*=\lim _{t\rightarrow \infty }-g(t)/t=\lim _{t\rightarrow \infty }h(t)/t\) satisfying \(c^*=\mu q_{c^*}'(0)\), where \(q_{c^*}\) is the unique solution for following problem with \(c=c^*\):

$$\begin{aligned} \left\{ \begin{array}{ll} dq''-cq'+kq(1-q)=0,&{}y\in (0,\infty ),\\ q(0)=0,\ q(\infty )=1,\ q'(y)>0,&{} y\in (0,\infty ). \end{array} \right. \end{aligned}$$

The sharper long-time dynamical behavior of (1.4) is determined in [7]. More precisely, there exist two constants \(g_*\) and \(h_*\) such that

$$\begin{aligned} \lim _{t\rightarrow \infty }(g(t)+c^*t-g_*)=\lim _{t\rightarrow \infty }(h(t)-c^*t-h_*)=0 \end{aligned}$$
(1.5)

and

$$\begin{aligned} \lim _{t\rightarrow \infty }\sup _{x\in [g(t),0]}|u(t,x)-q_{c^*}(x-g(t))|=\lim _{t\rightarrow \infty }\sup _{x\in [0,h(t)]}|u(t,x)-q_{c^*}(h(t)-x)|=0. \end{aligned}$$
(1.6)

Coming back to (1.1), it is worth noting that the long-time dynamical behaviors of u and v closely depend on the competition coefficients a and b. The case \(a,b\in (0,1)\) is called the weak competition case, while the case \(a,b\in (1,\infty )\) is known as the strong competition case. Moreover, the cases \(a<1<b\) and \(b<1<a\) are known as the weak-strong and strong-weak competition cases, respectively. In these cases, rather different long-time dynamical behaviors are expected. For instance, the native species v vanishes eventually when u invades successfully (namely \((u,v)\rightarrow (1,0)\) as \(t\rightarrow \infty \)) in the strong-weak competition case (see Lemma 4.5 in [5]). This contrasts sharply with the weak competition case, where the two species converge to the co-existence steady state \((u^*,v^*)=(\frac{1-b}{1-ab},\frac{1-a}{1-ab})\) as time goes to infinity (see Lemma 5.3 in [27]).

The strong-weak competition case and the weak competition case of (1.1) have been thoroughly studied in [5] and [27], respectively. It is shown that the spreading-vanishing dichotomy for u always holds. Moreover, when the spreading of u happens, the precise asymptotic spreading speed is established in [9] and [27], which is uniquely determined by a semi-wave solution associated with this system.

It is then natural to investigate whether (1.1) has a sharper dynamical behavior similar to the Fisher-KPP model (1.4). In other words, is it possible to obtain a sharper estimate of the spreading speed of u similar to (1.5)? Does the solution of the free boundary problem (1.1) evolve into a semi-wave solution like (1.6)? The purpose of this paper is to address these problems.

The precise long-time profiles of single equations with free boundaries have been studied by lots of authors in recent years, for instance, see [3, 7, 10, 13, 17] for one-dimensional problems and [8] for high-dimensional problems. However, to the best of our knowledge, there are few papers devoted to the study of sharp threshold for systems with free boundaries, even for the simplest classical Lotka-Volterra competition system (1.1). One of the mathematical difficulties is that different species may spread at different speeds, which brings a nontrivial challenge when one deals with the convergence of solutions. The approach developed in this paper might provide some insights into the precise dynamical behaviors for some free boundary systems.

Recalling from the work due to [5] that if the invasive species u is a weak competitor \((a<1<b)\) and the resident species v is well established initially (i.e. \(v_0\) satisfies the conditions in (1.2)), the invasion of u always fails in the sense that \(\lim _{t\rightarrow \infty }(u(t,\cdot ),v(t,\cdot ))=(0,1)~\text {in}~ L^{\infty }_{\text {loc}}(\mathbb {R})\); whereas if u is a strong competitor \((a>1>b)\), a spreading-vanishing dichotomy holds, that is, either

  • (Spreading of u) \(\lim _{t \rightarrow \infty }h(t)=-\lim _{t \rightarrow \infty }g(t)=+\infty \) and \(\lim _{t \rightarrow \infty }(u(t,x),v(t,x))=(1,0)\) uniformly for x in any compact subset of \(\mathbb {R}\); or

  • (Vanishing of u) \(\lim _{t \rightarrow \infty }\vert h(t)-g(t)\vert <\infty \) and \(\lim _{t \rightarrow \infty }(u(t,x),v(t,x))=(0,1)\) uniformly for x in any compact subset of \(\mathbb {R}\).

The primary goal of the current paper is to derive a sharper dynamical behavior of the solution of (1.1) when the spreading of u happens. To gain a good understanding, this paper focuses only on the strong-weak competition case. That is, hereafter, we always assume that

$$\begin{aligned} a>1>b>0. \end{aligned}$$

In order to capture the precise asymptotic spreading speed of u, the following so-called semi-wave system plays a crucial role:

$$\begin{aligned} \left\{ \begin{array}{ll} d\psi ''-c\psi '+k\psi (1-\psi -b\phi )=0, &{} \xi>0,\\ \phi ''-c\phi '+\phi (1-\phi -a\psi )=0, &{} \xi \in \mathbb {R},\\ \psi \equiv 0\ (\xi \le 0),\ \psi '>0\ (\xi \ge 0),\ \psi (\infty )=1, &{} \\ \phi (-\infty )=1,\ \phi '<0\ (\xi \in \mathbb {R}),\ \phi (\infty )=0. \end{array} \right. \end{aligned}$$
(1.7)

It is shown in [9] that (1.7) has a unique solution when \(c\in [0,s_0)\) and it has no solution when \(c\ge s_0\), where \(s_0\in [2\sqrt{kd(1-b)},2\sqrt{kd}]\) is the minimal speed of the traveling wave (see [16]). More precisely, the following result holds:

Proposition 1.1

(Theorem 1.3 in [9]) Assume that \(a>1>b>0\). Then there exists a positive constant \(s_0\) such that system (1.7) has a unique solution \((\psi _c,\phi _c)\in [C(\mathbb {R})\cap C^{2}([0,\infty ))]\times C^{2}(\mathbb {R})\) for \(c\in [0,s_{0})\) and it has no solution for \(c\ge s_{0}\). Furthermore, for any \(\mu >0\), there exists a unique \(c=s_{\mu }\in (0,s_{0})\) such that

$$\begin{aligned} s_{\mu }=\mu \psi _{s_{\mu }}'(0). \end{aligned}$$

With the help of Proposition 1.1, the asymptotic spreading speed of u is established as follows:

Proposition 1.2

(Theorem 1.1 in [9]) When the species u spreads successfully for (1.1), the asymptotic spreading speed of u satisfies

$$\begin{aligned} \lim \limits _{t \rightarrow \infty }\frac{h(t)}{t}=-\lim \limits _{t \rightarrow \infty }\frac{g(t)}{t}=s_{\mu }. \end{aligned}$$
(1.8)

Our main results for (1.1) are contained in the following two theorems.

Theorem 1.3

(Sharp threshold for spreading) Assume that \(a>1>b>0\) and (uvgh) is the unique solution to (1.1) for which the spreading of u happens. Then there exist two constants \(g^*\) and \(h^*\) such that

$$\begin{aligned}&\lim \limits _{t\rightarrow \infty }(g(t)+s_{\mu }t-g^*)=0,\ \ \lim \limits _{t\rightarrow \infty }g'(t)=-s_{\mu }, \end{aligned}$$
(1.9)
$$\begin{aligned}&\lim \limits _{t\rightarrow \infty }(h(t)-s_{\mu }t-h^*)=0,\ \ \lim \limits _{t\rightarrow \infty }h'(t)=s_{\mu }. \end{aligned}$$
(1.10)

Moreover,

$$\begin{aligned}&\lim \limits _{t\rightarrow \infty }\Big [\Vert u(t,\cdot )-\psi _{s_{\mu }}(\cdot -g(t))\Vert _{L^{\infty }[g(t),0]}+\Vert v(t,\cdot )-\phi _{s_{\mu }}(\cdot -g(t))\Vert _{L^{\infty }(-\infty ,0]}\Big ]=0, \nonumber \\ \end{aligned}$$
(1.11)
$$\begin{aligned}&\lim \limits _{t\rightarrow \infty }\Big [\Vert u(t,\cdot )-\psi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[0,h(t)]}+\Vert v(t,\cdot )-\phi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[0,\infty )}\Big ]=0, \nonumber \\ \end{aligned}$$
(1.12)

where \((\psi _{s_{\mu }},\phi _{s_{\mu }})\) is the unique solution of (1.7) with \(c=s_{\mu }\).

Theorem 1.4

(Exponential stability for vanishing) Assume that \(a>1>b>0\) and (uvgh) is the unique solution to (1.1) for which the vanishing of u happens. Then there exists a positive constant \(\kappa _0\) such that

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\bigg (\sup \limits _{x\in [g(t),h(t)]}\vert u(t,x)\vert e^{\kappa _0 t}+\sup \limits _{x\in \mathbb {R}}\vert v(t,x)-1\vert e^{\kappa _0 t}\bigg )=0. \end{aligned}$$

Remark 1.5

We would like to mention the following three comments:

  1. (i)

    For the estimate of the asymptotic spreading speed of u, it is obvious that (1.9) and (1.10) in Theorem 1.3 are sharper than (1.8) in Proposition 1.2. This is in sharp contrast to the Cauchy problems of (1.3) and (1.1), where a logarithmic correction may occur at the fronts of the level set, please refer to [1, 15, 20, 23] and [22], respectively.

  2. (ii)

    Note that (1.11) and (1.12) in Theorem 1.3 are related to the stability of semi-wave solutions in \(L^{\infty }\) topology. There are much literature aiming at the stability of traveling wave solutions in systems, for instance, one may refer to the monograph [24] and references therein. However, the research on the stability of semi-wave solutions in systems is slightly behind because of technical difficulties.

  3. (iii)

    The exponential stability for vanishing in Theorem 1.4 depends closely on the initial condition (1.2), particularly on the assumption \(\liminf _{\vert x\vert \rightarrow \infty }v_0(x)>0\). If \(v_0\) has a compact support and the vanishing of u happens, then u globally exponentially converges to zero in the long run (see the proof of Theorem 1.4) while v converges eventually to a shifted traveling front with a Bramson correction (see Theorem 2 in [22]).

The mathematical analysis of this paper is motivated by the method of Du et al. in [7, 8]. Since (1.1) is a system (not a single equation) of differential equations, considerable variations and new techniques are needed. As one may see, it will arise two main difficulties when dealing with (1.1). Firstly, we shall use a semi-wave solution to construct a supersolution and a subsolution of (1.1). Since the semi-wave solution contains a semi-infinite solution and an entire solution, the calculations are more tedious and new techniques are involved (see Lemmas 3.4 and 3.5). Secondly, we need to derive that the solution of free boundary problem (1.1) converges to the solution of semi-wave system (1.7), which is nontrivial for such system of differential equations (see Proposition 4.4).

Before ending this section, we mention some further references that form part of the background of this research. Related free boundary problems of two-species competition models have been investigated in many recent works. Apart from [5, 9, 27] mentioned earlier, one may find various interesting results in [11, 14, 18, 25, 26, 28] for a small sampling of such works, where [14, 28] considers the weak competition case, [11, 18] considers the strong-weak competition case and [25, 26] covers more general situations. However, the sharp dynamical behavior in all these works are worthy to be further studied. Finally, it is remarkable that the original problem in [5] is a spherically symmetric model, and hence the sharp dynamical behavior of (1.1) with spherical symmetry is still open.

The organization of this paper is the following. In Sect. 2, we give some basic conclusions that will be needed later, including a comparison principle for (1.1) and some useful properties for (1.1) and (1.7). Section 3 is devoted to proving some bounded results, where the crucial step is to show that \(\vert g(t)+s_{\mu }t\vert \) and \(\vert h(t)-s_{\mu }t\vert \) are bounded for \(t>0\) (see Proposition 3.1) by constructing suitable supersolutions and subsolutions. In Sect. 4, we first prove Theorem 1.3 by the boundedness result obtained in Sect. 3 and then finish the proof of Theorem 1.4 by virtue of one free boundary problem and two ODE problems.

2 Some Basic Results

The global existence and uniqueness of a bounded solution to (1.1) are obtained in [5]; moreover, (1.1) does not has any unbounded solution, which includes the following result:

Lemma 2.1

(Theorem 2.5 in [5]) Problem (1.1) admits a unique and uniformly bounded solution (uvgh). More precisely, there exists a positive constant K independent of t and x such that

$$\begin{aligned} \begin{array}{ll} 0<u(t,x)\le K\text { for }(t,x)\in (0,\infty )\times (g(t),h(t)),\\ 0<v(t,x)\le K\text { for }(t,x)\in (0,\infty )\times \mathbb {R},\\ 0<h'(t),-g'(t)\le K\text { for }t\in (0,\infty ). \end{array} \end{aligned}$$

For the sake of convenience, let us define the following two differential operators

$$\begin{aligned} \begin{aligned} N_{1}[u,v](t,x)&:= u_{t}-du_{xx}-ku(1-u-bv),\\ N_{2}[u,v](t,x)&:= v_{t}-v_{xx}-v(1-v-au). \end{aligned} \end{aligned}$$

We now state a comparison principle for the free boundary problem (1.1), which is a simple variation of Lemma 2.6 in [5].

Lemma 2.2

(Comparison principle) Let (uvgh) be the unique bounded solution of (1.1). Assume that \(T\in (0,\infty ), \overline{g},\overline{h}\in C^{1}([0,T]), \overline{u}\in L^{\infty }(Q_{T})\cap C(\overline{Q_{T}})\cap C^{1,2}(Q_{T}^*), \underline{v}\in L^{\infty }(Q_{T})\cap C(\overline{Q_{T}})\cap C^{1,2}(Q_{T})\) and

$$\begin{aligned} \left\{ \begin{array}{ll} N_1[\overline{u},\underline{v}]\ge 0,\ &{}0<t<T,\ \overline{g}(t)<x<\overline{h}(t),\\ N_2[\overline{u},\underline{v}]\le 0,\ &{}0<t<T,\ -\infty<x<\infty ,\\ \overline{u}(t,x)=0,\ &{}0<t\le T,\ x\ge \overline{h}(t),\\ \overline{u}(t,x)\ge u(t,x),\ &{}0<t\le T,\,x\le \overline{g}(t),\\ \overline{h}'(t)\ge -\mu \overline{u}_{x}(t,\overline{h}(t)),\ \overline{g}(t)\le g(t), \ &{}0<t\le T,\\ \overline{u}(0,x)\ge u_{0}(x),\ &{}-h_0\le x\le h_{0},\\ \underline{v}(0,x)\le v_{0}(x),\ &{}-\infty<x<\infty , \end{array} \right. \end{aligned}$$
(2.1)

where \(Q_{T}^*=\{(t,x)\in \mathbb {R}^{2}:t\in (0,T],x\in (\overline{g}(t),\overline{h}(t))\}\) and \(Q_{T}=\{(t,x)\in \mathbb {R}^{2}:t\in (0,T],x\in \mathbb {R}\}\). Then for \(t\in (0,T]\), it holds

$$\begin{aligned} \overline{h}(t)\ge h(t),\ \overline{u}(t,x)\ge u(t,x)\text { for } x\in [\overline{g}(t),\overline{h}(t)],\ \underline{v}(t,x)\le v(t,x)\text { for }x\in \mathbb {R}. \end{aligned}$$

Remark 2.3

 

  1. (i)

    The vector function \((\overline{u},\underline{v},\overline{g},\overline{h})\) is called a supersolution of (1.1). Meanwhile, if the reverse inequalities in (2.1) hold then we have a subsolution of (1.1).

  2. (ii)

    If both \((\overline{u}_1,\underline{v},\overline{g},\overline{h})\) and \((\overline{u}_2,\underline{v},\overline{g},\overline{h})\) are supersolutions of (1.1), then \((\min \{\overline{u}_1,\overline{u}_2\},\underline{v},\overline{g},\overline{h})\) is a generalized supersolution of (1.1). If both \((\overline{u},\underline{v}_1,\overline{g},\overline{h})\) and \((\overline{u},\underline{v}_2,\overline{g},\overline{h})\) are supersolutions of (1.1), then \((\overline{u},\max \{\underline{v}_1,\underline{v}_2\},\overline{g},\overline{h})\) is a generalized supersolution of (1.1). The proof is standard, for instance, the details can refer to Lemma 2.6 in [5] and Theorem 2.3 in [12].

To describe the asymptotic behavior of \((\psi ,\phi )\) near \(\pm \infty \), let us introduce the characteristic equations

$$\begin{aligned}&d\lambda ^2-c\lambda -k=0, \end{aligned}$$
(2.2)
$$\begin{aligned}&\lambda ^2-c\lambda +1-a=0, \end{aligned}$$
(2.3)
$$\begin{aligned}&\lambda ^2-c\lambda -1=0. \end{aligned}$$
(2.4)

Denote \(\lambda _1=\frac{c-\sqrt{c^2+4kd}}{2d}\), \(\lambda _2=\frac{c-\sqrt{c^2+4(a-1)}}{2}\) and \(\lambda _3=\frac{c+\sqrt{c^2+4}}{2}\) as the roots of (2.2), (2.3) and (2.4), respectively. We now state the exact exponential decays of semi-wave solution of (1.7) (see Lemma 2.2 and the proof of Lemma 2.5 in [9]) as follows:

Lemma 2.4

Suppose \((c,\psi ,\phi )\) is a solution of (1.7) with \(c\in [0,s_0)\). Then there exist positive constants \(l_{i}(i=1,2,3)\) such that

$$\begin{aligned} \lim \limits _{\xi \rightarrow -\infty }\frac{1-\phi (\xi )}{e^{\lambda _{3}\xi }}=l_{1},\, \lim \limits _{\xi \rightarrow \infty }\frac{\phi (\xi )}{e^{\lambda _{2}\xi }}=l_{2} \text{ and } \lim \limits _{\xi \rightarrow \infty }\frac{1-\psi (\xi )}{\xi ^\gamma e^{\Lambda \xi }}=l_3, \end{aligned}$$

where \(\Lambda =\max \{\lambda _1,\lambda _2\},\gamma =0\) if \(\lambda _1\ne \lambda _2\); \(\Lambda =\lambda _1,\gamma =1\) if \(\lambda _1=\lambda _2\).

3 Sharp Bounds

In this section, we show that both \(\vert g(t)+s_{\mu }t\vert \) and \(\vert h(t)-s_{\mu }t\vert \) are bounded functions for \(t\ge 0\) by constructing suitable supersolutions and subsolutions. More precisely, our aim is to show the following result:

Proposition 3.1

There exists a positive constant C such that

$$\begin{aligned} \vert g(t)+s_{\mu }t\vert \le C,\ \vert h(t)-s_{\mu }t\vert \le C\text { for }t\ge 0. \end{aligned}$$

To prove this proposition, we need several lemmas. Firstly, we establish an exponential decay rate of v.

Lemma 3.2

Suppose (uvgh) is the unique bounded solution of (1.1) for which spreading of u happens. Then for any \(\hat{c}\in (0,s_{\mu })\), there exist positive constants \(M_0\) and \(\beta _{0}\) such that

$$\begin{aligned} \vert v(t,x)\vert \le M_0 e^{-\beta _{0}t}\text { for }t>0,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned}$$
(3.1)

Proof

Remark 3.4 in [9] implies that for any \(\hat{c}\in (0,s_{\mu })\) and \(\epsilon \in (0,1)\) small enough, there exists a large positive constant T such that

$$\begin{aligned} \begin{aligned}&1-\epsilon<u(t,x)<1+\epsilon \text { for }t\ge T,\ x\in [-\hat{c}t,\hat{c}t],\\&0<v(t,x)<\epsilon \text { for }t\ge T,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned} \end{aligned}$$
(3.2)

Thanks to \(a>1\), there exists a constant \(\delta \in (0,a(1-\epsilon )-1)\) such that

$$\begin{aligned} u(t,x)>1-\epsilon >(\delta +1)/a \text { for }t\ge T,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned}$$

Thus, for the solution of v of (1.1) we have

$$\begin{aligned} {\left\{ \begin{array}{ll} v_{t}-v_{xx}\le -\delta v, &{} t>T,\ -\hat{c}T<x<\hat{c}T,\\ v(t,\pm \hat{c}T)\le \epsilon , &{} t\ge T,\\ v(T,x)\le \epsilon , &{} -\hat{c}T\le x\le \hat{c}T. \end{array}\right. } \end{aligned}$$

One can repeat the proof of Lemma 2.6 in [22] to obtain that, there exist positive constants \(M_0,\beta _0\) and \(T_0(\ge T)\) such that

$$\begin{aligned} \vert v(t,x)\vert \le M_0 e^{-\beta _{0}t}\text { for }t\ge T_0,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned}$$

In view of \(\vert v(t,x)\vert \le K\) for \(t\in (0,T_0)\) and \(x\in \mathbb {R}\) by virtue of Lemma 2.1, then (3.1) holds by taking \(M_0\) large enough. This completes the proof. \(\square \)

Next, we give an exponential decay rate of u.

Lemma 3.3

Suppose (uvgh) is the unique bounded solution of (1.1) for which spreading of u happens. Then for any \(\hat{c}\in (0,s_{\mu })\), there exist positive constants \(\tilde{M_{0}}\) and \(\tilde{\alpha }_0\) such that

$$\begin{aligned}&u(t,x)\ge 1-\tilde{M_{0}}e^{-\tilde{\alpha }_0t}\text { for }t>0,\ x\in [-\hat{c}t,\hat{c}t], \end{aligned}$$
(3.3)
$$\begin{aligned}&u(t,x)\le 1+\tilde{M_{0}}e^{-\tilde{\alpha }_0t}\text { for }t>0,\ x\in [g(t),h(t)]. \end{aligned}$$
(3.4)

Proof

According to (3.2), for any \(\epsilon \in (0,1)\) small and \(\rho \in (0,1-\epsilon )\), there exists a positive constant T such that

$$\begin{aligned} u(t,x)\ge 1-\epsilon >\rho \text { for }t\ge T,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned}$$

It follows from Lemma 3.2 and the first equation in (1.1) that

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{t}-du_{xx}\ge k[\rho (1-u)-bM_0e^{-\beta _0t}u], &{} t>T,\ -\hat{c}T<x<\hat{c}T,\\ u(t,\pm \hat{c}T)\ge \rho , &{} t\ge T,\\ u(T,x)\ge \rho , &{} -\hat{c}T\le x\le \hat{c}T. \end{array}\right. } \end{aligned}$$

Repeating the proof of Lemma 2.8 in [22], there exist positive constants \(M_0^*, \alpha _{0}\) and \(\tilde{T}_0\) such that

$$\begin{aligned} u(t,x)\ge 1-M_0^* e^{-\alpha _{0}t}\text { for }t\ge \tilde{T}_0,\ x\in [-\hat{c}t,\hat{c}t]. \end{aligned}$$

Let \(\eta (t)\) be the unique solution of the following ODE problem:

$$\begin{aligned} \left\{ \begin{array}{ll} \eta '(t)=k\eta (1-\eta ),&{}t>0,\\ \eta (0)={\Vert u_{0}\Vert }_{\infty }+1.&{} \end{array}\right. \end{aligned}$$

By some elementary calculations, we have

$$\begin{aligned} \eta (t)=\frac{\Vert u_0\Vert _{\infty }+1}{\Vert u_0\Vert _{\infty }+1-\Vert u_0\Vert _{\infty }e^{-kt}}\le 1+\Vert u_0\Vert _{\infty }e^{-kt}\text { for }t\ge 0. \end{aligned}$$

The comparison principle implies that \(u(t,x)\le \eta (t)\) for \(t\ge 0\) and \(x\in [g(t),h(t)]\).

Thanks to \(u(t,x)\ge 0\) for \(t\ge 0\) and \(x\in \mathbb {R}\), both (3.3) and (3.4) hold by taking \(\tilde{M_0}\ge \max \{M_0^*,\Vert u_0\Vert _{\infty }\}\) and \(\tilde{\alpha }_0\le \min \{\alpha _0,k\}\). This completes the proof. \(\square \)

We now construct a supersolution for (1.1). Let \(\tilde{M_0}\) be given in Lemma 3.3 and denote \((\psi ,\phi )=(\psi _{s_{\mu }},\phi _{s_{\mu }})\) for simplicity, where \((\psi _{s_{\mu }},\phi _{s_{\mu }})\) is the unique solution of (1.7) when \(c=s_{\mu }\). Define

$$\begin{aligned} \begin{aligned}&\overline{u}(t,x)= (1+\tilde{M_0}e^{-\delta t})\psi (\overline{h}(t)-x),\ \underline{v}(t,x)=\max \{0,\phi (\overline{h}(t)-x)-M_{1}e^{-\delta (t-T^*)}\},\\&\overline{g}(t)=g(t),\ \overline{h}(t)= s_{\mu }(t-T^*)+h(T^*)+\sigma (e^{-\delta T^*}-e^{-\delta t})+X_{0}, \end{aligned} \end{aligned}$$
(3.5)

where \(\delta ,M_{1},T^{*},\sigma \) and \(X_0\) are positive constants to be determined later.

Lemma 3.4

For appropriately chosen positive constants \(\delta ,M_{1},T^{*},\sigma \) and \(X_0\), (uvh) satisfies

$$\begin{aligned} \begin{aligned}&h(t)\le \overline{h}(t)\text { for }t\in [T^{*},\infty ),\\&u(t,x)\le \overline{u}(t,x)\text { for } (t,x)\in [T^{*},\infty )\times [\overline{g}(t),\overline{h}(t)],\\&v(t,x)\ge \underline{v}(t,x)\text { for }(t,x)\in [T^{*},\infty )\times \mathbb {R}. \end{aligned} \end{aligned}$$

Proof

We check that \((\overline{u},\underline{v},\overline{g},\overline{h})\) is a supersolution for \(t\ge T^{*}\) by taking appropriate parameters \(\delta ,M_{1},T^{*},\sigma \) and \(X_0\), that is,

$$\begin{aligned}&N_{1}[\overline{u},\underline{v}]=\overline{u}_{t}-d\overline{u}_{xx}-k\overline{u}(1-\overline{u}-b\underline{v})\ge 0,\ \ t>T^{*},\ \overline{g}(t)<x<\overline{h}(t), \end{aligned}$$
(3.6)
$$\begin{aligned}&N_{2}[\overline{u},\underline{v}]=\underline{v}_{t}-\underline{v}_{xx}-\underline{v}(1-\underline{v}-a\overline{u})\le 0,\ \ t>T^{*},\ -\infty<x<\infty , \end{aligned}$$
(3.7)
$$\begin{aligned}&\overline{u}(t,\overline{g}(t))\ge u(t,\overline{g}(t)),\ \overline{u}(t,\overline{h}(t))=0,\ \ t>T^{*}, \end{aligned}$$
(3.8)
$$\begin{aligned}&\overline{h}(T^{*})\ge h(T^{*}),\ \overline{h}'(t)\ge -\mu \overline{u}_{x}(t,\overline{h}(t)),\ \ t>T^{*}, \end{aligned}$$
(3.9)
$$\begin{aligned}&\overline{u}(T^{*},x)\ge u(T^{*},x),\ \ \overline{g}(T^{*})< x <h(T^{*}), \end{aligned}$$
(3.10)
$$\begin{aligned}&\underline{v}(T^{*},x)\le v(T^{*},x),\ \ -\infty<x <\infty . \end{aligned}$$
(3.11)

Firstly, we verify (3.6) and (3.7). For convenience, denote

$$\begin{aligned} z=\overline{h}(t)-x,\ q(t)= 1+\tilde{M_{0}}e^{-\delta t},\ p(t)= M_{1}e^{-\delta (t-T^{*})}. \end{aligned}$$

A direct calculation yields

$$\begin{aligned} \begin{aligned} N_{1}[\overline{u},\underline{v}]&=q'\psi +q(s_{\mu }+\sigma \delta e^{-\delta t})\psi '-dq\psi ''-kq\psi +kq^{2}\psi ^{2}+kbq\psi \underline{v}\\&=q'\psi +\sigma \delta q e^{-\delta t}\psi '+kq^{2}\psi ^{2} -kq\psi ^{2}-kbq\psi \phi +kbq\psi \underline{v}~\text {(used } (1.7)). \end{aligned} \end{aligned}$$

If \(\underline{v}=0\), then \(\phi \le p\) and \(-kbq\psi \phi \ge -kbq\psi p\); if \(\underline{v}=\phi -p>0\), then \(-kbq\psi \phi +kbq\psi \underline{v}=-kbq\psi p\). To sum up, it holds

$$\begin{aligned} \begin{aligned} N_{1}[\overline{u},\underline{v}]&\ge q'\psi +\sigma \delta q e^{-\delta t}\psi '+kq^{2}\psi ^{2}-kq\psi ^{2}-kbq\psi p\\&=e^{-\delta t}\psi \big (k\tilde{M_{0}}\psi +k\tilde{M}_{0}^{2}e^{-\delta t}\psi -\tilde{M_{0}}\delta -kbM_{1}e^{\delta T^{*}}-kb\tilde{M_{0}}M_{1}e^{-\delta t}e^{\delta T^{*}}\big )\\&\quad +\sigma \delta (1+\tilde{M_{0}}e^{-\delta t})e^{-\delta t}\psi '. \end{aligned} \end{aligned}$$

Since \(\psi (z)\rightarrow 1\) as \(z\rightarrow \infty \), for any \(z\ge R_1\) with given \(R_1\ge 1\), we have

$$\begin{aligned} \begin{aligned}&k\tilde{M_{0}}\psi +k\tilde{M}_{0}^{2}e^{-\delta t}\psi -\tilde{M_{0}}\delta -kbM_{1}e^{\delta T^{*}}-kb\tilde{M_{0}}M_{1}e^{-\delta t}e^{\delta T^{*}}\\&\quad \ge (k\psi (R_1)-\delta )\tilde{M_{0}}-kb(e^{\delta T^{*}}+\tilde{M_{0}})M_{1}\ge 0 \end{aligned} \end{aligned}$$

by taking \(\delta \in (0,\min \{1/4,k\psi (R_1)\})\) and \(M_{1}<\min \{1/4,\frac{(k\psi (R_1)-\delta )\tilde{M_{0}}}{kb(e^{\delta T^{*}}+\tilde{M_{0}})}\}\). Then \(N_1[\overline{u},\underline{v}]\ge 0\) for \(t>T^*\) and \(z\ge R_1\). On the other hand, for \(z\in [0,R_1]\) and \(t>T^*\), we have

$$\begin{aligned} \begin{aligned} N_{1}[\overline{u},\underline{v}]&\ge e^{-\delta t}\big (\sigma \delta \psi '-\tilde{M_{0}}\delta \psi -kbM_{1}e^{\delta T^{*}}\psi -kb\tilde{M_{0}}M_{1}e^{-\delta t}e^{\delta T^{*}}\psi \big )\\&\ge e^{-\delta t}\big (\sigma \delta \varrho _0-\tilde{M_{0}}\delta \psi (R_1)-kbe^{\delta T^{*}}\psi (R_1)-kb\tilde{M_{0}}\psi (R_1)\big )\ge 0 \end{aligned} \end{aligned}$$

by taking \(\sigma >0\) sufficiently large, where \(\varrho _0:=\min _{z\in [0,R_1]}\psi '(z)>0\). Thus, \(N_{1}[\overline{u},\underline{v}]\ge 0\) for \(t> T^{*}\) and \(z>0\), which means (3.6) holds.

Obviously, \(N_{2}[\overline{u},\underline{v}]=0\) if \(\underline{v}=0\). Hence, it remains to show that \(N_{2}[\overline{u},\underline{v}]\le 0\) when \(\underline{v}>0\). A direct calculation yields

$$\begin{aligned} \begin{aligned} N_{2}[\overline{u},\underline{v}]&=(s_{\mu }+\sigma \delta e^{-\delta t})\phi '-p'-\phi ''-(\phi -p)+(\phi -p)^2+a(\phi -p)q\psi \\&=\sigma \delta e^{-\delta t}\phi '-p'+p-2p\phi +p^{2}+a(q-1)\psi \phi -aq\psi p~\text {(used }(1.7))\\&=e^{-\delta t}\big (\sigma \delta \phi '+\delta M_{1}e^{\delta T^*}+M_{1}e^{\delta T^*}-2M_{1}e^{\delta T^*}\phi +M_{1}^{2}e^{-\delta t}e^{2\delta T^*}\\&\quad +a\tilde{M_0}\psi \phi -aM_{1}e^{\delta T^*}\psi -a\tilde{M_0}M_{1}e^{-\delta t}e^{\delta T^*}\psi \big ):=e^{-\delta t}Q_{1}. \end{aligned} \end{aligned}$$

Recalling that \(\delta ,M_{1}<1/4\) small, \(\phi (z)\rightarrow 1\) as \(z\rightarrow -\infty , \psi (z)=0\) when \(z\le 0\) and \(\phi '<0\), if we take \(R_{2}\) satisfying \(\phi (-R_2)\ge 3/4\), then for \(t>T^*\) and \(z<-R_{2}\),

$$\begin{aligned} Q_{1}\le & {} \delta M_{1}e^{\delta T^{*}}+M_{1}e^{\delta T^{*}}-2M_{1}e^{\delta T^{*}}\phi +M_{1}^{2}e^{-\delta t}e^{2\delta T^{*}}\\\le & {} M_{1}e^{\delta T^{*}}(\delta +1-2\phi (-R_2)+M_{1})\le 0. \end{aligned}$$

Moreover, recalling \(\psi (z)\rightarrow 1, \phi (z)\rightarrow 0\) as \(z\rightarrow \infty ,\phi '<0\) and \(a>1\), if we further demand \(M_1<a\tilde{M_0}e^{-\delta T^*}/2, \delta <(a-1)/2\), and take \(R_3>0\) satisfying \(\phi (R_3)\le \frac{M_{1}\delta }{a\tilde{M_0}}\) and \({\psi (R_3)>\max \{\frac{1}{2},\frac{1+2\delta }{a}\}}\), then for \(t>T^*\) and \(z>R_{3}\),

$$\begin{aligned} \begin{aligned} Q_{1}\le&\big (\delta M_{1}+M_{1}+a\tilde{M_0}\psi \phi -aM_{1}\psi \big )e^{\delta T^*} +\big (M_{1}e^{\delta T^*}-a\tilde{M_0}\psi \big )M_{1}e^{-\delta t}e^{\delta T^*}\\ \le&\big (\delta M_{1}+M_{1}+a\tilde{M_0}\phi (R_3)-aM_{1}\psi (R_3)\big )e^{\delta T^*} \\&+\big (M_{1}e^{\delta T^*}-a\tilde{M_0}\psi (R_3)\big )M_{1}e^{-\delta t}e^{\delta T^*}\le 0. \end{aligned} \end{aligned}$$

In addition, note that there exists \(\varrho _1>0\) such that \(\phi '(z)\le -\varrho _1\) in \([-R_{2},R_{3}]\). Then for such fixed \(R_{2},R_{3},\delta \) and \(T^{*}\), taking \(\sigma \ge \frac{4a\tilde{M_0}}{\delta \varrho _1}\), we have

$$\begin{aligned} \begin{aligned} Q_{1}\le&\sigma \delta \phi '+\delta M_{1}e^{\delta T^*}+M_{1}e^{\delta T^*}+M_{1}^{2}e^{-\delta t}e^{2\delta T^*}+a\tilde{M_{0}}\psi \phi \le -\sigma \delta \varrho _1+4a\tilde{M_0}\le 0 \end{aligned} \end{aligned}$$

for \(t>T^*\) and \(z\in [-R_{2},R_{3}]\). Hence, \(N_{2}[\overline{u},\underline{v}]\le 0\) for \(t>T^{*}\) and \(z\in \mathbb {R}\), which implies that (3.7) holds.

Clearly, \(u(t,\overline{g}(t))=0\le (1+\tilde{M_{0}}e^{-\delta t})\psi (\overline{h}(t)-g(t))=\overline{u}(t,\overline{g}(t))\) and \(\overline{u}(t,\overline{h}(t))=(1+\tilde{M_{0}}e^{-\delta t})\psi (0)=0\) for \(t>T^*\), which means (3.8) holds.

We now turn to prove (3.9). It is obvious that \(\overline{h}(T^*)=h(T^{*})+X_{0}\ge h(T^*)\). Moreover, direct calculations yield

$$\begin{aligned}&{} \overline{h}'(t)=s_{\mu }-\delta \sigma e^{-\delta t}, -\mu \overline{u}_{x}(t,\overline{h}(t))=-\mu (-1+\tilde{M}_{0}e^{-\delta t})\psi '(0)=s_{\mu }\\&-s_{\mu }\tilde{M}_{0}\text {e}^{-\delta t}. \end{aligned}$$

Hence, (3.9) holds by further taking \(\sigma >0\) large enough such that \(\sigma \delta \ge s_{\mu }\tilde{M_{0}}\).

Next, we show (3.10). According to (3.4) in Lemma 3.3 and the monotonicity of \(\psi \), there exist \(X_0>0\) large enough and \(\delta \in (0,\tilde{\alpha }_0)\) such that, for \(x\in (\overline{g}(T^{*}),h(T^{*}))\),

$$\begin{aligned} \overline{u}(T^{*},x)\ge (1+\tilde{M_{0}}e^{-\delta T^{*}})\psi (X_0)\ge 1+\tilde{M_{0}}e^{-\tilde{\alpha }_0 T^{*}}\ge u(T^{*},x). \end{aligned}$$

Finally, we check (3.11). If \(\underline{v}(T^*,x)=0\), (3.11) clearly holds. Otherwise, it follows from Lemma 3.1 in [9] that for any \(\delta _0\in (0,M_1)\), we can find \(M^*,T^*\) large such that \(v(T^*,x)\ge 1-\delta _0\) for \(x\in [M^*,\infty )\). Then

$$\begin{aligned} \underline{v}(T^*,x)=\phi (h(T^*)+X_0-x)-M_1\le 1-\delta _0\le v(T^*,x)\text { for } x\in [M^*,\infty ). \end{aligned}$$

For above \(M^*>0\), one may choose \(X_0\) sufficiently large such that

$$\begin{aligned} \phi (\overline{h}(T^*)-x)=\phi (h(T^*)+X_0-x)\le \phi (h(T^*)+X_0-M^*)\le M_1\text { for }x\in (-\infty ,M^*], \end{aligned}$$

which leads to \(\underline{v}(T^*,x)=0\le v(T^*,x)\) for \(x\in (-\infty ,M^*]\). Hence, (3.11) holds.

Summarizing as above, one can apply Lemma 2.2 to obtain the desired conclusions. This completes the proof. \(\square \)

Next, we construct a subsolution for (1.1). Let \(\tilde{M_0}\) and \((\psi ,\phi )\) be given in (3.5) and define

$$\begin{aligned} \begin{aligned}&\underline{u}(t,x)=(1-\tilde{M_0}e^{-\delta t})\psi (\underline{h}(t)-x),\ \overline{v}(t,x)= \phi (\underline{h}(t)-x)+M_{2}e^{-\delta (t-T_{*})},\\&\underline{g}(t)= -\hat{c}t,\ \underline{h}(t)= s_{\mu }(t-T_*)-\hat{c}T_{*}-\sigma (e^{-\delta T_*}-e^{-\delta t})+h(T_{*}), \end{aligned} \end{aligned}$$

where \(\hat{c}\in (s_{\mu }/2,s_{\mu }),\delta ,M_{2},T_*\) and \(\sigma \) are positive constants to be determined later.

Lemma 3.5

For some suitable choice of positive constants \(\delta ,M_2,T_*\) and \(\sigma \), (uvh) satisfies

$$\begin{aligned} \begin{aligned}&h(t)\ge \underline{h}(t)\text { for }t\in [T_{*},\infty ),\\&u(t,x)\ge \underline{u}(t,x)\text { for } (t,x)\in [T_{*},\infty )\times [\underline{g}(t),\underline{h}(t)],\\&v(t,x)\le \overline{v}(t,x)\text { for } (t,x)\in [T_{*},\infty )\times \mathbb {R}. \end{aligned} \end{aligned}$$

Proof

We check that \((\underline{u},\overline{v},\underline{g},\underline{h})\) is a subsolution for \(t\ge T_{*}\) by taking appropriate parameters \(\delta ,M_{2},T_*\) and \(\sigma \), that is,

$$\begin{aligned}&N_{1}[\underline{u},\overline{v}]=\underline{u}_{t}-d\underline{u}_{xx}-k\underline{u}(1-\underline{u}-b\overline{v})\le 0,\ \ t>T_{*},\ \underline{g}(t)<x<\underline{h}(t), \end{aligned}$$
(3.12)
$$\begin{aligned}&N_{2}[\underline{u},\overline{v}]=\overline{v}_{t}-\overline{v}_{xx}-\overline{v}(1-\overline{v}-a\underline{u})\ge 0,\ \ t>T_{*},\ -\infty<x<\infty , \end{aligned}$$
(3.13)
$$\begin{aligned}&\underline{u}(t,\underline{g}(t))\le u(t,\underline{g}(t)),\ \underline{u}(t,\underline{h}(t))=0,\ \ t>T_{*}, \end{aligned}$$
(3.14)
$$\begin{aligned}&\underline{h}(T_{*})\le h(T_{*}),\ \underline{h}'(t)\le -\mu \underline{u}_{x}(t,\underline{h}(t)),\ \ t>T_{*}, \end{aligned}$$
(3.15)
$$\begin{aligned}&\underline{u}(T_{*},x)\le u(T_{*},x),\ \ \underline{g}(T_{*})< x <h(T_{*}), \end{aligned}$$
(3.16)
$$\begin{aligned}&\overline{v}(T_{*},x)\ge v(T_{*},x),\ \ -\infty< x <\infty . \end{aligned}$$
(3.17)

Firstly, we verify (3.12) and (3.13). For convenience, denote

$$\begin{aligned} \tilde{z}=\underline{h}(t)-x,\ \tilde{q}(t)= 1-\tilde{M_{0}}e^{-\delta t}, \tilde{p}(t)= M_{2}e^{-\delta (t-T_{*})}. \end{aligned}$$

Choosing \(T^*\ge (\ln {\tilde{M_0}})/\delta +1\), then \(\tilde{q}(t)>0\) for \(t\ge T^*\). A direct calculation yields

$$\begin{aligned} \begin{aligned} N_{1}[\underline{u},\overline{v}]&=\tilde{q}(s_{\mu }-\sigma \delta e^{-\delta t})\psi '+\tilde{q}'\psi -d\tilde{q}\psi ''-k\tilde{q}\psi +k\tilde{q}^{2}\psi ^{2}+kb\tilde{q}\psi (\phi +\tilde{p})\\&=-\tilde{q}(1-\tilde{q})k\psi ^{2}+kb\psi \tilde{q}\tilde{p}-\sigma \delta e^{-\delta t}\tilde{q}\psi '+\tilde{q}'\psi ~\text {(used } (1.7))\\&=e^{-\delta t}\big (-\tilde{M_{0}}k\tilde{q}\psi ^{2}+kb\tilde{q}M_{2}e^{\delta T_{*}}\psi -\sigma \delta \tilde{q}\psi '+\tilde{M_{0}}\delta \psi \big ):=e^{-\delta t}Q_{2}. \end{aligned} \end{aligned}$$

Taken \(M_2=\frac{1+b}{2b}>1\), then \(bM_2<1\). It follows from \(\psi (\tilde{z})\rightarrow 1\) when \(\tilde{z}\rightarrow \infty \) and \(\psi '(\tilde{z})>0\) when \(\tilde{z}\ge 0\) that there exists positive constant \(R_{4}\) satisfying \(\psi (R_{4})>bM_2\). It is worthwhile pointing out that Lemma 3.3 still holds by enlarging \(\tilde{M}_0\) or decreasing \(\tilde{\alpha }_0\). Hence, one can take \(\tilde{M_0}=\frac{bM_2+1}{\psi (R_{4})+1}e^{\delta T_{*}}\) and \(\delta \) small enough such that, for \(t\ge T_{*}\) and \(\tilde{z}\in (R_{4},\underline{h}(t)+\hat{c}t)\),

$$\begin{aligned} \begin{aligned} Q_{2}\le&-\tilde{M_0}k\tilde{q}\psi ^{2}+kb\tilde{q}M_{2}e^{\delta T_{*}}\psi +\tilde{M_0}\delta \psi \\ \le&(-\tilde{M_0}\psi ({R_4})+bM_{2}e^{\delta T_{*}})k\tilde{q}(T_*)\psi ({R_4})+\tilde{M_0}\delta \le 0. \end{aligned} \end{aligned}$$

Note that there exists \(\varrho _2>0\) small such that \(\psi '(\tilde{z})\ge \varrho _2\) in \((0,R_4]\). Then for such fixed \(\tilde{M_0},M_{2},\delta ,T_{*}\) and \(R_{4}\), there exists \(\sigma >0\) sufficiently large such that, for \(t\ge T_{*}\) and \(\tilde{z}\in (0,R_{4}]\),

$$\begin{aligned} Q_{2}=-\tilde{M_{0}}k\tilde{q}\psi ^{2}+kb\tilde{q}M_{2}e^{\delta T_{*}}\psi -\sigma \delta \tilde{q}\psi '+\tilde{M_{0}}\delta \psi \le kbM_{2}e^{\delta T_{*}}-\sigma \delta \tilde{q}\varrho _2+\tilde{M_{0}}\delta \le 0. \end{aligned}$$

Thus, \(N_{1}[\underline{u},\overline{v}]\le 0\) for \(t>T_{*}\) and \(\tilde{z}\in (0,\underline{h}(t)+\hat{c}t)\), which means (3.12) holds.

On the other hand, a direct calculation yields

$$\begin{aligned} \begin{aligned} N_{2}[\underline{u},\overline{v}]&=(s_{\mu }-\sigma \delta e^{-\delta t})\phi '+\tilde{p}'-\phi ''-(\phi +\tilde{p})+(\phi +\tilde{p})^{2}+a(\phi +\tilde{p})\tilde{q}\psi \\&=-\sigma \delta e^{-\delta t}\phi '+\tilde{p}'-\tilde{p}+\tilde{p}^{2}+2\tilde{p}\phi -a(1-\tilde{q})\phi \psi +a\tilde{p}\tilde{q}\psi ~\text {(used } (1.7))\\&=e^{-\delta t}\big (-\sigma \delta \phi '-\delta M_{2}e^{\delta T_{*}}-M_{2}e^{\delta T_{*}}+M_{2}^{2}e^{2\delta T_{*}}e^{-\delta t}+2M_{2}e^{\delta T_{*}}\phi \\&\quad -a\tilde{M_{0}}\phi \psi +aM_{2}e^{\delta T_{*}}\psi -a\tilde{M_{0}}M_{2}e^{-\delta (t-T_{*})}\psi \big ):=e^{-\delta t}Q_{3}. \end{aligned} \end{aligned}$$

Since \(\phi (\tilde{z})\rightarrow 1\) as \(\tilde{z}\rightarrow -\infty ,\psi (\tilde{z})=0\) when \(\tilde{z}<0\) and \(\phi '<0\), there exist positive constants \(R_{5}\) satisfying \(\phi (-R_{5})>1/2\) and \(\delta \) small such that, for \(t>T_*\) and \(\tilde{z}\in (-\infty ,-R_{5})\),

$$\begin{aligned} Q_{3}\ge -\delta M_{2}e^{\delta T_{*}}-M_{2}e^{\delta T_{*}}+2M_{2}e^{\delta T_{*}}\phi (-R_{5})\ge 0. \end{aligned}$$

Moreover, due to \(\phi (\tilde{z})\rightarrow 0, \psi (\tilde{z})\rightarrow 1\) as \(\tilde{z}\rightarrow \infty \), there exist positive constants \(R_{6}\) large and \(\delta \) small such that, for \(t>T_*\) and \(\tilde{z}\in (R_{6},\infty )\),

$$\begin{aligned} \begin{aligned} Q_{3}\ge&-\delta M_{2}e^{\delta T_{*}}-a\tilde{M_{0}}\phi \psi -M_{2}e^{\delta T_{*}}+M_{2}^{2}e^{2\delta T_{*}}e^{-\delta t}+aM_{2}e^{\delta T_{*}}\psi -a\tilde{M_{0}}M_{2}e^{-\delta (t-T_{*})}\psi \\ \ge&-\delta M_{2}e^{\delta T_{*}}-a\tilde{M_{0}}\phi (R_6)+{(M_2e^{-\delta (t-T_{*})}-1)M_2e^{\delta T_{*}}}\\&+a M_{2}(\psi (R_6)-\tilde{M_{0}}e^{-\delta T_{*}})e^{\delta T_{*}}\ge 0. \end{aligned} \end{aligned}$$

In addition, note that there exists \(\varrho _3>0\) small such that \(\phi '(\tilde{z})\le -\varrho _3\) in \([-R_5,R_6]\). Then for such fixed \(R_{5},R_{6},\delta \) and \(T_{*}\), there exists \(\sigma >0\) sufficiently large such that, for \(t>T_*\) and \(\tilde{z}\in [-R_{5},R_{6}]\),

$$\begin{aligned} Q_3\ge -\sigma \delta \phi '-\delta M_{2}e^{\delta T_{*}}-M_{2}e^{\delta T_{*}}-a\tilde{M_{0}}\phi \psi +aM_{2}e^{\delta T_{*}}\psi -a\tilde{M_{0}}M_{2}e^{-\delta (t-T_{*})}\psi \ge 0. \end{aligned}$$

Hence, \(N_{2}[\underline{u},\overline{v}]\ge 0\) for \(t> T_{*}\) and \(\tilde{z}\in \mathbb {R}\), which implies that (3.13) holds.

We now show (3.14) and (3.16). Obviously, \(\underline{u}(t,\underline{h}(t))=(1-\tilde{M_{0}}e^{-\delta t})\psi (0)=0\) for \(t>T_{*}\). Thanks to (3.3) in Lemma 3.3 and by choosing \(\delta \in (0,\tilde{\alpha }_0)\), it holds

$$\begin{aligned}&\underline{u}(t,\underline{g}(t))=(1-\tilde{M_{0}}e^{-\delta t})\psi (\underline{h}(t)+\hat{c}t)\le 1-\tilde{M_{0}}e^{-\tilde{\alpha }_0t}\le u(t,\underline{g}(t)) \text { for }t>T_{*},\\&\underline{u}(T_{*},x)=(1-\tilde{M_{0}}e^{-\delta T_{*}})\psi (\underline{h}(T_{*})-x)\le 1-\tilde{M_{0}}\text {e}^{-\tilde{\alpha }_0T_{*}}\\&\le u(T_{*},x)\text { for }x\in (\underline{g}(T_{*}),h(T_{*})), \end{aligned}$$

which implies that (3.14) and (3.16) hold.

Next, we turn to prove (3.15). It is obvious that \(\underline{h}(T_{*})=h(T_{*})-\hat{c}T_{*}\le h(T_{*})\). Moreover, direct calculations yield

$$\begin{aligned}&\underline{h}'(t)=s_{\mu }-\delta \sigma e^{-\delta t}, -\mu \underline{u}_x(t,\underline{h}(t))=-\mu (-1+\tilde{M_{0}}e^{-\delta t})\psi '(0)=s_{\mu }-s_{\mu }\tilde{M_{0}}e^{-\delta t}. \end{aligned}$$

Hence, (3.15) holds by taking \(\sigma >0\) large enough such that \(\sigma \delta \ge s_{\mu }\tilde{M_{0}}\).

Finally, due to \(M_{2}=\frac{1+b}{2b}>1\), there exists \(T_*>0\) large such that

$$\begin{aligned} \overline{v}(T_{*},x)=\phi (\underline{h}(T_{*})-x)+M_{2}\ge v(T_{*},x)\text { for }x\in \mathbb {R}, \end{aligned}$$

which means (3.17) holds.

Summarizing as above, one can apply Remark 2.3 (i) to obtain the desired conclusions. This completes the proof. \(\square \)

Proof of Proposition 3.1

Noting that Lemmas 3.4 and 3.5 hold simultaneously by enlarging \(\sigma \) or decreasing \(\delta \), for \(t>T=\max \{T^{*},T_{*}\}\), we have

$$\begin{aligned} -s_{\mu }T_{*}-\hat{c}T_{*}-\sigma e^{-\delta T_{*}}+h(T_{*})\le h(t)-s_{\mu }t\le -s_{\mu }T^{*}+h(T^{*})+\sigma e^{-\delta T^{*}}+X_{0}. \end{aligned}$$

Denote

$$\begin{aligned}&C^{*}=\max \bigg \{\vert -s_{\mu }T_{*}-\hat{c}T_{*}-\sigma e^{-\delta T_{*}}+h(T_{*})\vert ,-s_{\mu }T^{*}+h(T^{*})+\sigma e^{-\delta T^{*}}+X_{0},\ \\&\qquad \quad \max \limits _{t\in [0,T]}\vert h(t)-s_{\mu }t\vert \bigg \}. \end{aligned}$$

Then \(\vert h(t)-s_{\mu }t\vert \le C^{*}\text { for }t\ge 0\). Moreover, replacing the initial function of (1.1) as \((u_0(-x),v_0(-x))\) and repeating the same arguments as above, there exists a positive constant \(C_{*}\) such that \(\vert g(t)+s_{\mu }t\vert \le C_{*}\) for \(t\ge 0\). Hence, Proposition 3.1 holds by taking \(C=\max \{C^{*},C_{*}\}\). \(\square \)

4 Convergence

This section is devoted to proving Theorems 1.3 and 1.4. Since the proof of Theorem 1.3 is quite involved, it will be divided into three subsections. We first show that \(h(t+t_n)-s_\mu (t+t_n)+2C\rightarrow L(t)\) and \((u,v)(t+t_n,x+s_\mu (t+t_n)-2C)\rightarrow (\hat{H},\hat{M})(t,x)\) locally uniformly along a time sequence \(\{t_n\}\) with \(\lim _{n\rightarrow \infty }t_n=\infty \) in Sect. 4.1. Then we show \(L(t)=L_0\) and \((\hat{H},\hat{M})(t,x)=(\psi _{s_\mu },\phi _{s_\mu })(L_0-x)\) in Sect. 4.2. Finally, the proof of Theorem 1.3 is completed in Sect. 4.3 by constructing a pair of super and sub solution. In the last Sect. 4.4, we finish the proof of Theorem 1.4.

4.1 Convergence Along a Time Sequence \(\{t_{n}\}\).

The crucial step for the proof of Theorem 1.3 is to prove that \(g(t)+s_{\mu }t\) and \(h(t)-s_{\mu }t\) converge to \(g^*\) and \(h^*\) as \(t\rightarrow \infty \), respectively. To this end, some variable substitutions will be needed. Thanks to Proposition 3.1, there exists a positive constant C such that

$$\begin{aligned} -C\le h(t)-s_{\mu }t\le C\text { for }t\ge 0. \end{aligned}$$

Denote

$$\begin{aligned}&r(t)= s_{\mu }t-2C,\ l(t)= h(t)-r(t),\\&H(t,x)= u(t,r(t)+x),\ M(t,x)= v(t,r(t)+x). \end{aligned}$$

Obviously, \(C\le l(t)\le 3C\) for \(t\ge 0\) and (H(tx), M(tx), l(t)) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} H_{t}=dH_{xx}+s_{\mu }H_{x}+kH(1-H-bM), &{} t>0,\ g(t)-r(t)<x<l(t),\\ M_{t}=M_{xx}+s_{\mu }M_{x}+M(1-M-aH), &{} t>0,\ -\infty<x<\infty ,\\ H(t,x)=0, &{} t>0,\ x\in \mathbb {R}\setminus (g(t)-r(t),l(t)),\\ l'(t)=-\mu H_{x}(t,l(t))-s_{\mu }, &{} t>0. \end{array}\right. } \end{aligned}$$

Let \(\{t_{n}\}\) be an arbitrary sequence satisfying \(t_{n}>0\) and \(\lim _{n\rightarrow \infty }t_n=\infty \). Set

$$\begin{aligned} (r_{n},l_{n},g_{n})(t)=(r,l,g)(t+t_{n}),\ (H_{n},M_{n})(t,x)=(H,M)(t+t_{n},x). \end{aligned}$$

Lemma 4.1

Subject to a subsequence if necessary, for some \(\alpha \in (0,1)\), it holds

$$\begin{aligned} {\Vert (H_{n},M_n,l_n)-(\hat{H},\hat{M},L)\Vert }_{C^{\frac{1+\alpha }{2},1+\alpha }_{\text {loc}}(\Omega _{n})\times {C^{\frac{1+\alpha }{2},1+\alpha }_{\text {loc}}(\widetilde{\Omega }_{n})}\times {C^{1+\frac{\alpha }{2}}_{\text {loc}}(\mathbb {R})}}\rightarrow 0\text { as }n\rightarrow \infty , \end{aligned}$$

where \(\Omega _{n}=\{(t,x)\in \Omega :t>-t_{n},-\infty <x\le l_{n}(t)\}\) with \(\Omega =\{(t,x):t\in \mathbb {R},-\infty <x\le L(t)\}\) and \(\widetilde{\Omega }_{n}=\{(t,x)\in \mathbb {R}^2:t>-t_{n},x\in \mathbb {R}\}\). Moreover, \((\hat{H}(t,x),\hat{M}(t,x),L(t))\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \hat{H}_{t}=d\hat{H}_{xx}+s_{\mu }\hat{H}_{x}+k\hat{H}(1-\hat{H}-b\hat{M}), &{} t\in \mathbb {R},\ x<L(t),\\ \hat{M}_{t}=\hat{M}_{xx}+s_{\mu }\hat{M}_{x}+\hat{M}(1-\hat{M}-a\hat{H}), &{} t\in \mathbb {R},\ x\in \mathbb {R},\\ \hat{H}(t,x)=0, &{} t\in \mathbb {R},\ x\ge L(t),\\ L'(t)=-\mu \hat{H}_{x}(t,L(t))-s_{\mu }, &{} t\in \mathbb {R}. \end{array} \right. \end{aligned}$$
(4.1)

Proof

It follows from Lemma 2.1 that there exists a positive constant K such that \(0<h'(t)\le K\) for all \(t>0\), which leads to

$$\begin{aligned} -s_{\mu }<l_{n}'(t)\le K-s_{\mu }\text { for }t+t_{n}\text { large and every }n\ge 1. \end{aligned}$$

Denoting

$$\begin{aligned} \xi =\frac{x}{l_{n}(t)},~(\tilde{H}_{n},\tilde{M}_{n})(t,\xi )=(H_{n},M_{n})(t,x), ~\Xi =\left( \frac{g_{n}(t)-r_{n}(t)}{l_{n}(t)},1\right) , \end{aligned}$$

then \((\tilde{H}_{n}(t,\xi ),\tilde{M}_{n}(t,\xi ),l_{n}(t))\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} (\tilde{H}_{n})_{t}=\frac{d}{l_{n}^{2}(t)}(\tilde{H}_{n})_{\xi \xi }+\frac{\xi l_{n}'(t)+s_{\mu }}{l_{n}(t)}(\tilde{H}_{n})_{\xi }+k\tilde{H}_{n}(1-\tilde{H}_{n}-b\tilde{M}_{n}), &{} t>-t_{n},\ \xi \in \Xi ,\\ (\tilde{M}_{n})_{t}=\frac{1}{l_{n}^{2}(t)}(\tilde{M}_{n})_{\xi \xi }+\frac{\xi l_{n}'(t)+s_{\mu }}{l_{n}(t)}(\tilde{M}_{n})_{\xi }+\tilde{M}_{n}(1-\tilde{M}_{n}-a\tilde{H}_{n}), &{} t>-t_{n},\ \xi \in \mathbb {R},\\ \tilde{H}_{n}(t,\xi )=0, &{} t>-t_{n},\ \xi \in \mathbb {R}\setminus \Xi ,\\ l_{n}'(t)=-\frac{\mu }{l_{n}(t)}(\tilde{H}_{n})_{\xi }(t,1)-s_{\mu }, &{} t>-t_{n}. \end{array} \right. \end{aligned}$$
(4.2)

Moreover, Lemma 2.1 shows that (uv) is uniformly bounded in \((0,T_{0})\times [g(t),h(t)]\) with any \(T_{0}>1\), which means that \((H_{n},M_{n})\) is uniformly bounded in \(\Omega _{n}\). Therefore, for any given \(\tilde{R}>0\) and \(p>1\), one can use the parabolic \(L^{p}\) estimates [21] to (4.2) over \([T_{0}-2,T_{0}+1]\times [-\tilde{R}-2,\tilde{R}+2]\) to obtain that

$$\begin{aligned} {\Vert (\tilde{H}_{n},\tilde{M}_{n})\Vert }_{W^{1,2}_{p}(I)\times W^{1,2}_{p}(J)}\le M_{\tilde{R}}\text { for all large }n, \end{aligned}$$

where \(I=[T_{0}-1,T_{0}+1]\times [-\tilde{R}-1,1]\) and \(J=[T_{0}-1,T_{0}+1]\times [-\tilde{R}-1,\tilde{R}+1]\), \(M_{\tilde{R}}\) is a positive constant depending on \(\tilde{R}\) and p but independent of n and \(T_{0}\). Furthermore, for any \(\alpha '\in (0,1)\), one can take \(p>1\) large enough and use the Sobolev embedding theorem [19] to obtain

$$\begin{aligned} {\Vert (\tilde{H}_{n},\tilde{M}_{n})\Vert }_{C^{\frac{1+\alpha '}{2},1+\alpha '}(I')\times C^{\frac{1+\alpha '}{2},1+\alpha '}(J')}\le \tilde{M}_{\tilde{R}}\text { for all large }n, \end{aligned}$$
(4.3)

where \(I'=[T_{0},\infty )\times [-\tilde{R},1]\) and \(J'=[T_{0},\infty )\times [-\tilde{R},\tilde{R}]\), \(\tilde{M}_{\tilde{R}}\) is a positive constant depending on \(\tilde{R}\) and \(\alpha '\) but independent of n and \(T_{0}\). Combining (4.2) with (4.3), we further conclude that

$$\begin{aligned} {\Vert l_{n}\Vert }_{C^{1+\frac{\alpha '}{2}}([T_{0},\infty ))}\le \tilde{C}\text { for all large }n, \end{aligned}$$

where \(\tilde{C}\) is a positive constant depending on \(\tilde{R}\) and \(\alpha '\) but independent of n and \(T_{0}\) too. Hence, there exists a subsequence (still denote it by \(\{t_{n}\}\)) such that

$$\begin{aligned} (\tilde{H}_{n},\tilde{M}_{n},l_{n})\rightarrow (\overline{H},\overline{M},L)\text { in } C_{loc}^{\frac{1+\alpha }{2},1+\alpha }\big (\mathbb {R}\times (-\infty ,1]\big )\times C_{loc}^{\frac{1+\alpha }{2},1+\alpha }\big (\mathbb {R}\times \mathbb {R}\big )\times C_{loc}^{1+\frac{\alpha }{2}}(\mathbb {R}) \end{aligned}$$

with \(\alpha \in (0,\alpha ')\). Applying the standard regularity theorem to (4.2), it is not hard to see that \((\overline{H}(t,\xi ),\overline{M}(t,\xi ),L(t))\) satisfies the following equations in the classical sense:

$$\begin{aligned} {\left\{ \begin{array}{ll} \overline{H}_{t}=\frac{d}{L^{2}(t)}\overline{H}_{\xi \xi }+\frac{\xi L'(t)+s_{\mu }}{L(t)}\overline{H}_{\xi }+k\overline{H}(1-\overline{H}-b\overline{M}), &{} t\in \mathbb {R},\ \xi <1,\\ \overline{M}_{t}=\frac{1}{L^{2}(t)}\overline{M}_{\xi \xi }+\frac{\xi L'(t)+s_{\mu }}{L(t)}\overline{M}_{\xi }+\overline{M}(1-\overline{M}-a\overline{H}), &{} t\in \mathbb {R},\ \xi \in \mathbb {R},\\ \overline{H}(t,\xi )=0,&{} t\in \mathbb {R},\ \xi \ge 1,\\ L'(t)=-\frac{\mu }{L(t)}\overline{H}_{\xi }(t,1)-s_{\mu }, &{}t\in \mathbb {R}. \end{array}\right. } \end{aligned}$$

Setting \((\hat{H},\hat{M})(t,x)=(\overline{H},\overline{M})(t,x/L(t))\), then \((\hat{H}(t,x),\hat{M}(t,x),L(t))\) satisfies (4.1) and

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert (H_{n},M_n,l_n)-(\hat{H},\hat{M},L) \Vert _{{C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\Omega _{n})}\times {C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\widetilde{\Omega }_{n})}\times {C^{1+\frac{\alpha }{2}}_{loc}(\mathbb {R})}}=0. \end{aligned}$$

This completes the proof. \(\square \)

4.2 Determine the Limit Pair \((\hat{H},\hat{M},L)\).

In this subsection, we will show that \(L(t)=L_{0}\) and \((\hat{H},\hat{M})(t,x)=(\psi _{s_{\mu }},\phi _{s_{\mu }})(L_{0}-x)\) for all \((t,x)\in \mathbb {R}\times \mathbb {R}\), where \(L_{0}\) is a positive constant.

According to \(C\le l(t)\le 3C\) for \(t\ge 0\), it holds

$$\begin{aligned} C\le L(t)\le 3C\text { for }t\in \mathbb {R}. \end{aligned}$$

On one hand, it follows from Lemma 3.4 that for \(t+t_{n}\) large,

$$\begin{aligned} \begin{aligned}&H_{n}(t,x)\le \big (1+\tilde{M_{0}}e^{-\delta (t+t_{n})}\big )\psi _{s_{\mu }}(\overline{h}(t+t_{n})-r(t+t_{n})-x),\,x\in I_n,\\&M_{n}(t,x)\ge \phi _{s_{\mu }}(\overline{h}(t+t_{n})-r(t+t_{n})-x)-M_{1}e^{-\delta (t+t_{n}-T^{*})},\,x\in \mathbb {R}, \end{aligned} \end{aligned}$$
(4.4)

where \(I_n=[g(t+t_{n})-r(t+t_{n}),\overline{h}(t+t_{n})-r(t+t_{n})]\). Moreover,

$$\begin{aligned} \begin{aligned} \overline{h}(t+t_{n})-r(t+t_{n})&=s_{\mu }(t+t_{n})-s_{\mu }T^{*}+h(T^{*})+\sigma (e^{-\delta T^{*}}-e^{-\delta (t+t_{n})})+X_{0}-r(t+t_{n})\\&\le s_{\mu }(t+t_{n})-r(t+t_{n})-s_{\mu }T^{*}+h(T^{*})+\sigma e^{-\delta T^{*}}+X_{0}\le 3C. \end{aligned} \end{aligned}$$

On the other hand, it follows from Lemma 3.5 that for \(t+t_{n}\) large,

$$\begin{aligned} \begin{aligned}&H_{n}(t,x)\ge \big (1-\tilde{M_{0}}e^{-\delta (t+t_{n})}\big )\psi _{s_{\mu }}(\underline{h}(t+t_{n})-r(t+t_{n})-x),\,x\in J_n\\&M_{n}(t,x)\le \phi _{s_{\mu }}(\underline{h}(t+t_{n})-r(t+t_{n})-x)+M_{2}e^{-\delta (t+t_{n}-T_{*})},\,x\in \mathbb {R}, \end{aligned} \end{aligned}$$
(4.5)

where \(J_n=[\underline{g}(t+t_{n})-r(t+t_{n}),\underline{h}(t+t_{n})-r(t+t_{n})]\). Moreover,

$$\begin{aligned} \begin{aligned} \underline{h}(t+t_{n})-r(t+t_{n})&=s_{\mu }(t+t_{n})-s_{\mu }T_{*}-\hat{c}T_{*}-\sigma (e^{-\delta T_{*}}-e^{-\delta (t+t_{n})})+h(T_{*})-r(t+t_{n})\\&\ge s_{\mu }(t+t_{n})-r(t+t_{n})-s_{\mu }T_{*}-\hat{c}T_{*}-\sigma e^{-\delta T_{*}}+h(T_{*})\ge C. \end{aligned} \end{aligned}$$

Letting \(n\rightarrow \infty \) in (4.4) and (4.5), we have

$$\begin{aligned} \begin{aligned}&\psi _{s_{\mu }}(C-x)\le \hat{H}(t,x)\le \psi _{s_{\mu }}(3C-x)\text { for } t\in \mathbb {R}\text { and }x<L(t),\\&\phi _{s_{\mu }}(C-x)\ge \hat{M}(t,x)\ge \phi _{s_{\mu }}(3C-x)\text { for } t\in \mathbb {R}\text { and } x\in \mathbb {R}. \end{aligned} \end{aligned}$$
(4.6)

Define

$$\begin{aligned} \begin{aligned}&R^*:=\inf \left\{ R:\hat{H}(t,x)\le \psi _{s_{\mu }}(R-x),\ \hat{M}(t,x)\ge \phi _{s_{\mu }}(R-x)\text { for } (t,x)\in \mathbb {R}\times \mathbb {R}\right\} ,\\&R_*:=\sup \left\{ R:\hat{H}(t,x)\ge \psi _{s_{\mu }}(R-x),\ \hat{M}(t,x)\le \phi _{s_{\mu }}(R-x)\text { for } (t,x)\in \mathbb {R}\times \mathbb {R}\right\} . \end{aligned} \end{aligned}$$

Obviously, (4.6) implies that both \(R^{*}\) and \(R_*\) are well defined. Furthermore, for \((t,x)\in \mathbb {R}\times \mathbb {R}\),

$$\begin{aligned} \begin{array}{ll} \psi _{s_{\mu }}(R_{*}-x)\le \hat{H}(t,x)\le \psi _{s_{\mu }}(R^{*}-x), \phi _{s_{\mu }}(R_{*}-x)\ge \hat{M}(t,x)\ge \phi _{s_{\mu }}(R^{*}-x) \end{array} \end{aligned}$$
(4.7)

and

$$\begin{aligned} C\le R_{*}\le \inf \limits _{t\in \mathbb {R}}L(t)\le \sup \limits _{t\in \mathbb {R}}L(t)\le R^*\le 3C. \end{aligned}$$
(4.8)

Lemma 4.2

\(R^*=\sup _{t\in \mathbb {R}}L(t)\) and there exists a sequence \(\{t_{n}\}_{n=1}^{\infty }\subset \mathbb {R}\) such that

$$\begin{aligned} L(t+t_{n})\rightarrow R^{*},\ \hat{H}(t+t_{n},x)\rightarrow \psi _{s_{\mu }}(R^{*}-x) \end{aligned}$$
(4.9)

as \(n\rightarrow \infty \) uniformly for (tx) in any compact subset of \(\mathbb {R}\times (-\infty ,R^{*}]\), and

$$\begin{aligned} \hat{M}(t+t_{n},x)\rightarrow \phi _{s_{\mu }}(R^{*}-x) \end{aligned}$$
(4.10)

as \(n\rightarrow \infty \) uniformly for (tx) in any compact subset of \(\mathbb {R}\times \mathbb {R}.\)

Proof

We first prove \(R^*=\sup _{t\in \mathbb {R}}L(t)\). Thanks to (4.8), it remains to show that \(R^{*}\le \sup _{t\in \mathbb {R}}L(t)\). Otherwise, if \(R^{*}>\sup _{t\in \mathbb {R}}L(t)\), then there exists a constant \(\delta >0\) small such that \(L(t)\le R^*-\delta \) for all \(t\in \mathbb {R}\). For clarity, the following arguments are divided into three steps.

Step 1. We claim that

$$\begin{aligned} \begin{aligned} \hat{H}(t,x)<\psi _{s_{\mu }}(R^{*}-x)\, \text { for }(t,x)\in \mathbb {R}\times (-\infty ,L(t)],\\ \hat{M}(t,x)>\phi _{s_{\mu }}(R^{*}-x)\, \text { for }(t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned} \end{aligned}$$
(4.11)

Otherwise, suppose there exists \((t_{0},x_{0})\in \mathbb {R}\times (-\infty ,L(t)]\) such that \(\hat{H}(t_{0},x_{0})=\psi _{s_{\mu }}(R^{*}-x_{0})\) or there exists \((t_{0}',x_{0}')\in \mathbb {R}\times \mathbb {R}\) such that \(\hat{M}(t_{0}',x_{0}')=\phi _{s_{\mu }}(R^{*}-x_{0}')\). Recall that \((\hat{H}(t,x),\hat{M}(t,x))\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \hat{H}_{t}=d\hat{H}_{xx}+s_{\mu }\hat{H}_{x}+k\hat{H}(1-\hat{H}-b\hat{M}), &{}(t,x)\in \mathbb {R}\times (-\infty ,L(t)],\\ \hat{M}_{t}=\hat{M}_{xx}+s_{\mu }\hat{M}_{x}+\hat{M}(1-\hat{M}-a\hat{H}), &{}(t,x)\in \mathbb {R}\times \mathbb {R}. \end{array} \right. \end{aligned}$$
(4.12)

It is easy to check that \((\psi _{s_{\mu }}(R^*-x),\phi _{s_{\mu }}(R^*-x))\) also satisfies (4.12). If \(\hat{H}(t_{0},x_{0})=\psi _{s_{\mu }}(R^{*}-x_{0})\), let us denote \(\vartheta (t,x)=\psi _{s_{\mu }}(R^*-x)-\hat{H}(t,x)\). Obviously, \(\vartheta (t_0,x_0)=0\) and \(\vartheta (t,x)\ge 0\) in \(\mathbb {R}\times (-\infty ,L(t)]\). There exists a positive constant \(\hat{C}\) with \(\hat{C}\ge k(b+1)\) such that

$$\begin{aligned} \begin{aligned} \vartheta _{t}-d\vartheta _{xx}-s_{\mu }\vartheta _{x}+\hat{C}\vartheta&=(-d\psi _{s_{\mu }}''+s_{\mu }\psi _{s_{\mu }}')+(-\hat{H}_{t}+d\hat{H}_{xx}+s_{\mu }\hat{H}_{x})+\hat{C}\vartheta \\&=k\vartheta -k\big (\psi _{s_{\mu }}-\hat{H}\big )\big (\psi _{s_{\mu }}+\hat{H}\big )-kb\psi _{s_{\mu }}\phi _{s_{\mu }} +kb\hat{H}\hat{M}+\hat{C}\vartheta \\&\ge k\vartheta -k\vartheta \big (\psi _{s_{\mu }}+\hat{H}\big )-kb\vartheta \phi _{s_{\mu }}+\hat{C}\vartheta \ge \big (\hat{C}-k-kb\big )\vartheta \ge 0. \end{aligned} \end{aligned}$$

Then the strong maximum principle implies that \(\vartheta (t,x)\equiv 0\) (i.e. \(\hat{H}(t,x)\equiv \psi _{s_{\mu }}(R^*-x)\)) for \(t\le t_0\) and \(x\le L(t)\). In particular,

$$\begin{aligned} \hat{H}(t,L(t))=\psi _{s_{\mu }}(R^*-L(t))\ge \psi _{s_{\mu }}(\delta )>0\text { for }t\le t_0, \end{aligned}$$

which contradicts with \(\hat{H}(t,L(t))=0\) for all \(t\in \mathbb {R}\). On the other hand, if \(\hat{M}(t_{0}',x_{0}')=\phi _{s_{\mu }}(R^{*}-x_{0}')\), one can repeat the same arguments as above to achieve that \(\hat{M}(t,x)=\phi _{s_{\mu }}(R^*-x)\) for \(t\le t_0'\) and \(x\in \mathbb {R}\). Furthermore, using the equations for \(\hat{M}(t,x)\) and \(\phi _{s_{\mu }}(R^*-x)\), it is easy to see that \(\hat{H}(t,x)=\psi _{s_{\mu }}(R^*-x)\) for \(t\le t_0'\) and \(x\le L(t)\), which leads to a same contradiction as above. Hence, the assertion (4.11) holds.

Step 2. Define

$$\begin{aligned} \omega _{1}(x):=\inf \limits _{t\in \mathbb {R}}\big [\psi _{s_{\mu }}(R^*-x)-\hat{H}(t,x)\big ], \, \omega _{2}(x):=\inf \limits _{t\in \mathbb {R}}\big [\hat{M}(t,x)-\phi _{s_{\mu }}(R^*-x)\big ]. \end{aligned}$$

Obviously, \(\omega _{i}(x)\ge 0\) for \(x\in \mathbb {R}\) and \(i=1,2\). Now we claim

$$\begin{aligned} \omega _{1}(x)> 0\text { for }x\in (-\infty ,R^{*}-\delta ],\, \omega _{2}(x)> 0\text { for }x\in \mathbb {R}. \end{aligned}$$
(4.13)

On the contrary, suppose there exists \(x_{0}\in (-\infty ,R^*-\delta ]\) such that \(\omega _{1}(x_{0})=0\) or there exists \(x_{0}'\in \mathbb {R}\) such that \(\omega _{2}(x_{0}')=0\). As a consequence of step 1, neither \(\omega _{1}(x_{0})\) nor \(\omega _{2}(x_{0}')\) achieves zero at any finite time t. If \(\omega _{1}(x_{0})=0\), there exists a sequence \(\{s_{n}\}^{\infty }_{n=1}\subset \mathbb {R}\) with \(\lim _{n\rightarrow \infty }\vert s_{n}\vert =\infty \) such that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\hat{H}(s_{n},x_{0})=\psi _{s_{\mu }}(R^*-x_0). \end{aligned}$$

Define

$$\begin{aligned} (\hat{H}_{n},\hat{M}_{n})(t,x) = (\hat{H},\hat{M})(t+s_{n},x),\ L_{n}(t)=L(t+s_{n}). \end{aligned}$$

Repeating the same arguments used in the proof of Lemma 4.1 and passing to a subsequence if necessary, it holds

$$\begin{aligned} (\hat{H}_{n},\hat{M}_{n},L_{n})\rightarrow (\hat{H}^*,\hat{M}^*,L^*)\text { in } C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\Omega ^{*})\times C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\mathbb {R}\times \mathbb {R})\times C^{1+\frac{\alpha }{2}}_{loc}(\mathbb {R}), \end{aligned}$$

where \(\alpha \in (0,1)\), \(\Omega ^{*}=\{(t,x): t\in \mathbb {R}, x\le L^{*}(t)\}\) and \((\hat{H}^{*},\hat{M}^{*},L^{*})\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \hat{H}^{*}_{t}=d\hat{H}^{*}_{xx}+s_{\mu }\hat{H}^{*}_{x}+k\hat{H}^{*}(1-\hat{H}^{*}-b\hat{M}^{*}), &{}t\in \mathbb {R},\ x<L^*(t),\\ \hat{M}^{*}_{t}=\hat{M}^{*}_{xx}+s_{\mu }\hat{M}^{*}_{x}+\hat{M}^{*}(1-\hat{M}^{*}-a\hat{H}^{*}), &{} t\in \mathbb {R},\ x\in \mathbb {R},\\ \hat{H}^{*}(t,x)=0, &{} t\in \mathbb {R},\ x\ge L^*(t),\\ (L^{*})'(t)=-\mu \hat{H}^{*}_{x}(t,L^{*}(t))-s_{\mu }, &{} t\in \mathbb {R}. \end{array} \right. \end{aligned}$$
(4.14)

Moreover, \(\hat{H}^{*}(0,x_{0})=\psi _{s_{\mu }}(R^*-x_0),~L^*(t)\le R^*-\delta \) for \(t\in \mathbb {R}\) and

$$\begin{aligned} \begin{aligned}&\hat{H}^{*}(t,x)\le \psi _{s_{\mu }}(R^{*}-x)\text { for }(t,x)\in \Omega ^{*},\\&\phi _{s_{\mu }}(R^{*}-x)\le \hat{M}^{*}(t,x)\text { for }(t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned} \end{aligned}$$

Note that \((\psi _{s_{\mu }}(R^*-x),\phi _{s_{\mu }}(R^*-x))\) also satisfies (4.14) with \(L^*(t)\) replaced by \(R^*\). Similar to the arguments used in step 1, one may apply the strong maximum principle to conclude that

$$\begin{aligned} \hat{H}^*(t,x)\equiv \psi _{s_{\mu }}(R^*-x)\text { for }t\le 0,\ x\le L^*(t). \end{aligned}$$

In particular, \(\hat{H}^*(t,L^*(t))=\psi _{s_{\mu }}(R^*-L^*(t))\ge \psi _{s_{\mu }}(\delta )>0\) for \(t\le 0\), which contradicts with \(\hat{H}^*(t,L^*(t))=0\) for all \(t\in \mathbb {R}\). On the other hand, if \(\omega _{2}(x_{0}')=0\), one can repeat the same arguments as above to achieve that \(\hat{M}^*(t,x)=\phi _{s_{\mu }}(R^*-x)\) for \(t\le 0\ \text {and}\ x\in \mathbb {R}\). Furthermore, using the equations for \(\hat{M}^*(t,x)\) and \(\phi _{s_{\mu }}(R^*-x)\), it is easy to see that \(\hat{H}^*(t,x)=\psi _{s_{\mu }}(R^*-x)\) for \(t\le 0\) and \(x\le L^*(t)\), which leads to a same contradiction as above. Hence, the assertion (4.13) holds.

Step 3. We shall reach a contradiction by considering following auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{ll} P_{t}=dP_{xx}+s_{\mu }P_{x}+kP(1-P-bQ), &{} t>0,\ x<R_{0},\\ Q_{t}=Q_{xx}+s_{\mu }Q_{x}+Q(1-Q-aP), &{} t>0,\ x<R_{0},\\ (P,Q)(t,R_{0})=(\psi _{s_{\mu }},\phi _{s_{\mu }})(R^{*}-R_{0}-\epsilon ), &{} t>0,\\ (P,Q)(0,x)=(1,0), &{} x<R_{0}, \end{array} \right. \end{aligned}$$
(4.15)

where \(\epsilon \) is a positive constant and \(R_{0}\) is a negative constant to be determined later. It is not hard to verify that (1, 0) and \((\psi _{s_{\mu }},\phi _{s_{\mu }})(R^{*}-x-\epsilon )\) are a pair of supersolution and subsolution for (4.15), respectively. It follows from the comparison principle that

$$\begin{aligned} \psi _{s_{\mu }}(R^{*}-x-\epsilon )\le P(t,x)\le 1,\, 0\le Q(t,x)\le \phi _{s_{\mu }}(R^{*}-x-\epsilon )\text { for }t\ge 0,\ x\le R_0. \end{aligned}$$

Since (4.15) is a monotone system and \((P,Q)(0,x)=(1,0)\) is a supersolution, there exist the limit functions

$$\begin{aligned} \lim \limits _{t \rightarrow \infty }(P(t,x),Q(t,x))=(P_{*}(x),Q_{*}(x))\text { locally uniformly for }x\le R_{0}. \end{aligned}$$

Obviously, \((P_{*}(x),Q_{*}(x))\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} dP_{*}''+s_{\mu }P_{*}'+kP_{*}(1-P_{*}-bQ_{*})=0, &{} x<R_{0},\\ Q_{*}''+s_{\mu }Q_{*}'+Q_{*}(1-Q_{*}-aP_{*})=0, &{} x<R_{0},\\ (P_{*},Q_{*})(R_{0})=(\psi _{s_{\mu }},\phi _{s_{\mu }})(R^{*}-R_{0}-\epsilon ),\\ (P_{*},Q_{*})(-\infty )=(1,0). \end{array} \right. \end{aligned}$$
(4.16)

It is worth pointing out that \((\psi _{s_{\mu }},\phi _{s_{\mu }})(R^{*}-x-\epsilon )\) also satisfies (4.16) and

$$\begin{aligned} P_{*}(x)\ge \psi _{s_{\mu }}(R^{*}-x-\epsilon ),\ Q_{*}(x)\le \phi _{s_{\mu }}(R^{*}-x-\epsilon )\text { for }x\le R_{0}. \end{aligned}$$

Next, we are going to prove

$$\begin{aligned} (P_{*},Q_{*})(x)=(\psi _{s_{\mu }},\phi _{s_{\mu }})(R^{*}-x-\epsilon )\text { for }x\le R_{0}. \end{aligned}$$
(4.17)

In view of \(\psi _{s_{\mu }}(\infty )=1,~\phi _{s_{\mu }}(\infty )=0\) and \(a>1>b>0\), for any given \(\epsilon _1>0\) small, there exists \(R_0<0\) sufficiently negative such that

$$\begin{aligned} \begin{aligned}&\psi _{s_{\mu }}(R^{*}-x-\epsilon )>1-\epsilon _{1}/a>1-\epsilon _{1}/b\text { for }x<R_0,\\&Q_{*}(x)\le \phi _{s_{\mu }}(R^{*}-x-\epsilon )<\epsilon _{1}\text { for }x<R_0. \end{aligned} \end{aligned}$$
(4.18)

Denote \(\tau (\epsilon )=(a-\epsilon -1)(b-2\epsilon )-ab^{2}\epsilon \). Clearly, \(\tau (0)=(a-1)b>0\) and \(\tau (\epsilon )\) is continuous with respect to \(\epsilon \). Thus, there exists a constant \(\epsilon _{2}>0\) small such that \(\tau (\epsilon _{2})>0\), which is equivalent to

$$\begin{aligned} (a-\epsilon _{2}-1)(b-2\epsilon _{2})>ab^{2}\epsilon _{2}. \end{aligned}$$
(4.19)

Setting \(\epsilon _{0}=\min \{\epsilon _{1},\epsilon _{2}\}\), then (4.18) and (4.19) hold simultaneously with \(\epsilon _1\) and \(\epsilon _2\) being replaced by \(\epsilon _{0}\). Denote

$$\begin{aligned} \hat{P}(x)=P_{*}(x)-\psi _{s_{\mu }}(R^{*}-x-\epsilon ),\ \hat{Q}(x)=\phi _{s_{\mu }}(R^{*}-x-\epsilon )-Q_{*}(x). \end{aligned}$$

Then \(\hat{P}(x)\ge 0\), \(\hat{Q}(x)\ge 0\) for \(x\le R_{0}\) and

$$\begin{aligned} {\left\{ \begin{array}{ll} d\hat{P}''+s_{\mu }\hat{P}'=k[(bQ_{*}+\psi _{s_{\mu }}+P_{*}-1)\hat{P}-b\psi _{s_{\mu }}\hat{Q}],&{} x<R_{0},\\ \hat{Q}''+s_{\mu }\hat{Q}'=(a\psi _{s_{\mu }}+\phi _{s_{\mu }}+Q_{*}-1)\hat{Q}-aQ_{*}\hat{P},&{} x<R_{0},\\ (\hat{P}(-\infty ),\hat{Q}(-\infty ))=(\hat{P}(R_{0}),\hat{Q}(R_{0}))=(0,0).&{} \end{array}\right. } \end{aligned}$$

Define

$$\begin{aligned} \hat{P}(\eta _{1}):=\max \limits _{x\in (-\infty ,R_{0}]}\hat{P}(x),\, \hat{Q}(\eta _{2}):=\max \limits _{x\in (-\infty ,R_{0}]}\hat{Q}(x). \end{aligned}$$

Then \(\hat{P}(\eta _{1})\ge 0\) and \(\hat{Q}(\eta _{2})\ge 0\). Hence, (4.17) is equivalent to \(\hat{P}(\eta _{1})=\hat{Q}(\eta _{2})=0\).

The strong maximum principle means that \(\hat{P}(\eta _{1})>0\) if and only if \(\hat{Q}(\eta _{2})>0\). Therefore, we only suppose \(\hat{P}(\eta _{1})>0\) by a contradiction and distinguish the following two cases:

(i) \((1-2\epsilon _0/b)\hat{P}(\eta _{1})<b\hat{Q}(\eta _{2})\); (ii) \((1-2\epsilon _0/b)\hat{P}(\eta _{1})\ge b\hat{Q}(\eta _{2})\).

When case (i) happens, it follows from (4.18), (4.19) and the equation of \(\hat{Q}(x)\) that

$$\begin{aligned} \begin{aligned} 0\ge \hat{Q}''(\eta _{2})+s_{\mu }\hat{Q}'(\eta _{2})&=(a\psi _{s_{\mu }}+\phi _{s_{\mu }}+Q_{*}-1)\hat{Q}(\eta _{2})-aQ_{*}\hat{P}(\eta _{2})\\&> [a(1-\epsilon _{0}/a)-1]\hat{Q}(\eta _{2})-a\epsilon _{0}\hat{P}(\eta _{2})\\&> (a-\epsilon _{0}-1)(1- 2\epsilon _{0}/b)\hat{P}(\eta _{1})/b-a\epsilon _{0}\hat{P}(\eta _{1})\\&= \frac{(a-\epsilon _{0}-1)(b-2\epsilon _{0})-ab^2\epsilon _0}{b^{2}}\hat{P}(\eta _{1})\ge 0. \end{aligned} \end{aligned}$$

This contradiction implies that the case (i) can not occur. When case (ii) happens, it follows from (4.18) and the equation of \(\hat{P}(x)\) that

$$\begin{aligned} \begin{aligned} 0\ge d\hat{P}''(\eta _{1})+s_{\mu }\hat{P}'(\eta _{1})&=k[(bQ_{*}+\psi _{s_{\mu }}+P_{*}-1)\hat{P}(\eta _{1})-b\psi _{s_{\mu }}\hat{Q}(\eta _{1})]\\&> k[(2(1-\epsilon _{0}/b )-1)\hat{P}(\eta _{1})-b\hat{Q}(\eta _{1})]\\&\ge k[(1-2\epsilon _{0}/b)\hat{P}(\eta _{1})-b\hat{Q}(\eta _{2})]\ge 0. \end{aligned} \end{aligned}$$

This is also a contradiction. Thus, \(\hat{P}(\eta _{1})=0\) (i.e. \(\hat{P}(x)\equiv 0\)) for \(x\le R_{0}\). Furthermore, using the equations for \(P_{*}(x)\) and \(\psi _{s_{\mu }}(R^{*}-x-\epsilon )\), it is easy to see that \(\hat{Q}(x)\equiv 0\) for \(x\le R_{0}\). Hence, (4.17) holds.

Now we focus on \((\hat{H}(t,x),\hat{M}(t,x))\), which satisfies the first and second equations in (4.15). It follows from (4.11) and the monotonicity of \(\psi _{s_{\mu }}\) and \(\phi _{s_{\mu }}\) that, for any given \(\epsilon >0\) sufficiently small,

$$\begin{aligned} \hat{H}(t,x)\le \psi _{s_{\mu }}(R^{*}-x-\epsilon ),\ \hat{M}(t,x)\ge \phi _{s_{\mu }}(R^{*}-x-\epsilon )\text { for }t\in \mathbb {R},\,x\le R_0. \end{aligned}$$

Note that \(\hat{H}(t,R_{0})\le 1=P(0,x)\) and \(\hat{M}(t,R_{0})\ge 0=Q(0,x)\) for \(t\in \mathbb {R}\) and \(x\le R_{0}\). The comparison principle implies that

$$\begin{aligned} \hat{H}(t+s,x)\le P(t,x),\ \hat{M}(t+s,x)\ge Q(t,x)\text { for }(t,x)\in \mathbb {R}^+\times (-\infty ,R_{0}]\text { and }s<0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \hat{H}(t,x)\le P(t-s,x),\ \hat{M}(t,x)\ge Q(t-s,x)\text { for }(t,x)\in (s,\infty )\times (-\infty ,R_{0}]\text { and }s<0. \end{aligned}$$

Letting \(s\rightarrow -\infty \), it holds

$$\begin{aligned} \begin{aligned}&\hat{H}(t,x)\le P_{*}(x)=\psi _{s_{\mu }}(R^{*}-x-\epsilon )\text { for }(t,x)\in \mathbb {R}\times (-\infty ,R_{0}],\\&\hat{M}(t,x)\ge Q_{*}(x)=\phi _{s_{\mu }}(R^{*}-x-\epsilon )\text { for }(t,x)\in \mathbb {R}\times (-\infty ,R_{0}]. \end{aligned} \end{aligned}$$
(4.20)

According to step 2 and the continuity of \(\omega _i(x)~(i=1,2)\), there exists \(\epsilon >0\) small such that \(\omega _{1}(x)\ge \epsilon \) when \(x\in [R_0,R^*-\delta ]\) and \(\omega _{2}(x)\ge \epsilon \) when \(x\in [R_0,\infty )\). Thanks to the monotonicity of \(\psi _{s_{\mu }}\) and \(\phi _{s_{\mu }}\) again, there exists \(\hat{\epsilon }\in (0,\epsilon )\) such that

$$\begin{aligned} \begin{aligned}&\psi _{s_{\mu }}(R^*-x-\hat{\epsilon })\ge \psi _{s_{\mu }}(R^*-x)-\epsilon \text { for }x\in [R_0,R^*-\delta ],\\&\phi _{s_{\mu }}(R^*-x-\hat{\epsilon })\le \phi _{s_{\mu }}(R^*-x)+\epsilon \text { for }x\in [R_0,\infty ), \end{aligned} \end{aligned}$$

which leads to

$$\begin{aligned} \begin{aligned}&\hat{H}(t,x)-\psi _{s_{\mu }}(R^*-x-\hat{\epsilon })\le \epsilon -\omega _1(x)\le 0\text { for }(t,x)\in \mathbb {R}\times [R_0,R^*-\delta ],\\&\hat{M}(t,x)-\phi _{s_{\mu }}(R^*-x-\hat{\epsilon })\ge \omega _2(x)-\epsilon \ge 0\text { for }(t,x)\in \mathbb {R}\times [R_0,\infty ). \end{aligned} \end{aligned}$$

Combining (4.20) with the fact that \(\hat{H}(t,x)=0\) for \(t\in \mathbb {R}\) and \(x\ge L(t)\), it holds

$$\begin{aligned} \hat{H}(t,x)\le \psi _{s_{\mu }}(R^*-x-\hat{\epsilon }),\, \hat{M}(t,x)\ge \phi _{s_{\mu }}(R^*-x-\hat{\epsilon })\text { for }(t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned}$$

This contradicts with the definition of \(R^*\). Hence, the desired conclusion \(R^*=\sup _{t\in \mathbb {R}}L(t)\) holds.

Finally, we prove (4.9) and (4.10) by discussing the following two cases:

  1. (i)

    \(R^{*}=\sup _{t\in \mathbb {R}}L(t)\) is achieved at some finite time \(t_0\);

  2. (ii)

    \(R^{*}>L(t)\) for all \(t\in \mathbb {R}\) and \(L(t_{n})\rightarrow R^{*}\) along some unbounded sequence \(\{t_{n}\}\).

If case (i) occurs, then \(L'(t_0)=0\). From the discussion in steps 1-3 above, it is easy to see that \(\omega _{1}(x)=\omega _{2}(x)=0\) for all \(x\in \mathbb {R}\), in particular for a certain \(x_0\in \mathbb {R}\). The following arguments are divided into two cases: (a) \(x_0=L(t_0)\); (b) \(x_0\ne L(t_0)\).

(a) Suppose that \(\psi _{s_{\mu }}(R^*-x)\ge \hat{H}(t_0,x)>0\) for \(x<L(t_0)\) and \(\hat{H}(t_0,L(t_0))=\psi _{s_{\mu }}(R^*-L(t_0))=0\). Recalling that \(\hat{H}(t,x)\le \psi _{s_{\mu }}(R^*-x)\) for \((t,x)\in \mathbb {R}\times (-\infty ,L(t)]\), Hopf boundary lemma implies that \(\hat{H}_x(t_0,L(t_0))>-\psi _{s_{\mu }}'(0)\). On the other hand, it follows from \(L'(t_0)=0\), (4.1) and Proposition 1.1 that

$$\begin{aligned} \hat{H}_x(t_0,L(t_0))=-(s_{\mu }+L'(t_0))/\mu =-s_{\mu }/\mu =-\psi _{s_{\mu }}'(0). \end{aligned}$$

Hence, above contradiction implies that there must be \(\hat{H}(t,x)=\psi _{s_{\mu }}(R^*-x)\) for \(t\le t_0\) and \(x\le L(t)\). Using the equations for \(\hat{H}(t,x)\) and \(\psi _{s_{\mu }}(R^*-x)\), it is not hard to obtain that \(\hat{M}(t,x)=\phi _{s_{\mu }}(R^*-x)\) for \(t\le t_0\) and \(x\le L(t)\). By the uniqueness of solutions for (4.1) with a given initial value, we can conclude

$$\begin{aligned} (\hat{H},\hat{M})(t,x)=(\psi _{s_{\mu }},\phi _{s_{\mu }})(R^*-x)\text { for }(t,x)\in \mathbb {R}\times (-\infty ,L(t)]. \end{aligned}$$

Furthermore, applying the strong maximum principle again, it holds \(\hat{M}(t,x)=\phi _{s_{\mu }}(R^*-x)\) for \((t,x)\in \mathbb {R}\times \mathbb {R}\). Hence, the desired conclusion holds by taking \(\{t_n=t_0\}^{\infty }_{n=1}\).

(b) Suppose \(x_0\in (-\infty ,L(t_0))\) such that \(\hat{H}(t_0,x_0)=\psi _{s_{\mu }}(R^*-x_0)\) or \(x_0\in \mathbb {R}\setminus L(t_0)\) such that \(\hat{M}(t_0,x_0)=\phi _{s_{\mu }}(R^*-x_0)\). The strong maximum principle implies that

$$\begin{aligned} \begin{aligned}&\hat{H}(t,x)=\psi _{s_{\mu }}(R^*-x)\text { for }(t,x)\in (-\infty ,t_0]\times (-\infty ,L(t)],\\&\hat{M}(t,x)=\phi _{s_{\mu }}(R^*-x)\text { for }(t,x)\in (-\infty ,t_0]\times \mathbb {R}. \end{aligned} \end{aligned}$$

Hence, the desired conclusion holds by taking \(\{t_n=t_0\}^{\infty }_{n=1}\) again.

We now turn to the case (ii). Since \(R^{*}=\sup _{t \in \mathbb {R}}L(t)\), then there exists a sequence \(\{t_{n}\}\) such that \(\lim _{n\rightarrow \infty }L(t_{n})=R^{*}\). Denote

$$\begin{aligned} (\hat{H}_{n},\hat{M}_{n})(t,x)=(\hat{H},\hat{M})(t+t_{n},x),\ L_{n}(t)=L(t+t_{n}). \end{aligned}$$

Similar to the process in the proof of Lemma 4.1, by passing to a subsequence if necessary, it holds

$$\begin{aligned} (\hat{H}_{n},\hat{M}_{n},L_{n})\rightarrow (\tilde{H},\tilde{M},\tilde{L})\text { in } C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\Omega )\times C^{\frac{1+\alpha }{2},1+\alpha }_{loc}(\mathbb {R}\times \mathbb {R})\times C^{1+\frac{\alpha }{2}}_{loc}(\mathbb {R}), \end{aligned}$$

where \((\tilde{H},\tilde{M},\tilde{L})\) satisfies (4.1) with L being replaced by \(\tilde{L}\). Note that

$$\begin{aligned} \tilde{L}(0)=\lim \limits _{n\rightarrow \infty }L(t_{n})=R^{*},\ \tilde{L}(t)=\lim \limits _{n\rightarrow \infty }L(t+t_{n})\le R^{*}\text { for }t\in \mathbb {R}. \end{aligned}$$

Repeating the arguments as in case (i), one can obtain \(\tilde{L}(t)=R^{*}\) for \(t\in \mathbb {R}\) and

$$\begin{aligned} \begin{aligned}&\tilde{H}(t,x)=\psi _{s_{\mu }}(R^{*}-x)\text { for }(t,x)\in \mathbb {R}\times (-\infty ,\tilde{L}(t)],\\&\tilde{M}(t,x)=\phi _{s_{\mu }}(R^{*}-x)\text { for }(t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned} \end{aligned}$$

This completes the proof. \(\square \)

Lemma 4.3

\(R_{*}=\inf _{t\in \mathbb {R}}L(t)\) and there exists a sequence \(\{\tilde{t}_{n}\}_{n=1}^{\infty }\subset \mathbb {R}\) such that

$$\begin{aligned} L(t+\tilde{t}_{n})\rightarrow R_{*},\ \hat{H}(t+\tilde{t}_{n},x)\rightarrow \psi _{s_{\mu }}(R_{*}-x) \end{aligned}$$

as \(n\rightarrow \infty \) uniformly for (tx) in any compact subset of \(\mathbb {R}\times (-\infty ,R_{*}]\), and

$$\begin{aligned} \hat{M}(t+\tilde{t}_{n},x)\rightarrow \phi _{s_{\mu }}(R_{*}-x) \end{aligned}$$

as \(n\rightarrow \infty \) uniformly for (tx) in any compact subset of \(\mathbb {R}\times \mathbb {R}.\)

The proof of Lemma 4.3 is similar to the arguments used in the proof of Lemma 4.2. Hence, the details are omitted here.

Proposition 4.4

\(R^{*}=R_{*}\). Hence \(L(t)=L_{0}\) and \((\hat{H},\hat{M})(t,x)=(\psi _{s_{\mu }},\phi _{s_{\mu }})(L_{0}-x)\), where \(L_0\) is a positive constant.

Proof

Firstly, Lemmas 4.2 and 4.3 imply that for any \(\epsilon >0\), there exist two sequences \(\{t_{n}\}\) and \(\{\tilde{t}_{n}\}\) such that

$$\begin{aligned} L(t)-R^{*}\ge -\epsilon \ (\forall t\ge t_{n}),\ L(t)-R_{*}\le \epsilon \ (\forall t\ge \tilde{t}_{n}). \end{aligned}$$

With the help of Lemmas 4.2 and 4.3, we are going to finish the proof of Proposition 4.4 by refining a pair of supersolution and subsolution. For clarity, the following arguments will be divided into three steps.

Step 1. Construct a supersolution.

It follows from (4.7) that

$$\begin{aligned} \psi _{s_{\mu }}(R_{*}-x)\le \hat{H}(t,x)\le \psi _{s_{\mu }}(R^{*}-x)\text { for } (t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned}$$

Since \(0\le \psi (\xi )<1\) and \(\psi (\xi )\rightarrow 1\) exponentially as \(\xi \rightarrow \infty \) (see Lemma 2.4), there exist two positive constants \(\widetilde{C}\) and \(\beta \) such that

$$\begin{aligned} \vert \hat{H}(t,x)-1\vert \le \widetilde{C} e^{\beta x}\text { for } (t,x)\in \mathbb {R}\times \mathbb {R}, \end{aligned}$$

which leads to

$$\begin{aligned} \hat{H}(t,x)\ge 1- \widetilde{C} e^{\beta x}\ \text {for}\ (t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned}$$
(4.21)

Due to \(\lim _{x\rightarrow -\infty }\psi _{s_{\mu }}(R_{*}-x)=\lim _{x\rightarrow -\infty }\psi _{s_{\mu }}(R^{*}-x)=1\), then \(\lim _{x\rightarrow -\infty }\hat{H}(t,x)=1\) for all \(t\in \mathbb {R}\). Hence, for any \(\epsilon >0\), there exists a constant \(K_0>0\) such that

$$\begin{aligned} \sup \limits _{x\in (-\infty ,-K_0]}\vert \hat{H}(t_n,x)-\psi _{s_{\mu }}(R^{*}-x)\vert<\epsilon ,\ \sup \limits _{x\in (-\infty ,-K_0]}\vert \hat{H}(\tilde{t}_n,x)-\psi _{s_{\mu }}(R_{*}-x)\vert <\epsilon , \end{aligned}$$
(4.22)

where \(t_n\) and \(\tilde{t}_n\) are given in Lemma 4.2 and Lemma 4.3, respectively. It follows from (4.22) and Lemma 4.3 that there exists \(\tilde{n}_0>0\) such that

$$\begin{aligned} \hat{H}(\tilde{t}_{n},x)\le \psi _{s_{\mu }}(R_{*}-x+\epsilon )+\epsilon \text { for }x\le R_*+\epsilon , n>\tilde{n}_0. \end{aligned}$$
(4.23)

Moreover, there exists a constant \(N(>1)\) independent of \(\epsilon \) such that

$$\begin{aligned} \psi _{s_{\mu }}(R_{*}-x+\epsilon )+\epsilon \le (1+N\epsilon )\psi _{s_{\mu }}(R_{*}-x+N\epsilon )\text { for }x\le R_*+\epsilon . \end{aligned}$$
(4.24)

Denote \(\check{L}(t)=L(t)+s_{\mu }t\) and \((\check{H},\check{M})(t,x)=(\hat{H},\hat{M})(t,x-s_{\mu }t)\). Then \((\check{H},\check{M},\check{L})\) satisfies

$$\begin{aligned}\left\{ \begin{array}{ll} \check{H}_{t}=d\check{H}_{xx}+k\check{H}(1-\check{H}-b\check{M}), &{} t\in \mathbb {R},\ x<\check{L}(t),\\ \check{M}_{t}=\check{M}_{xx}+\check{M}(1-\check{M}-a\check{H}), &{} t\in \mathbb {R},\ x\in \mathbb {R},\\ \check{H}(t,x)=0, &{} t\in \mathbb {R},\ x\ge \check{L}(t),\\ {\check{L}'(t)=-\mu \check{H}_{x}(t,\check{L}(t))}, &{} t\in \mathbb {R}. \end{array} \right. \end{aligned}$$

Let \((\psi ,\phi )=(\psi _{s_{\mu }},\phi _{s_{\mu }})\) be the unique solution of (1.7) with \(c=s_{\mu }\) and \(\tilde{t}_{n}\) be given in Lemma 4.3. Define

$$\begin{aligned} \begin{aligned}&\overline{H}(t,x)=\big (1+N\epsilon e^{-\delta (t-\tilde{t}_{n})}\big )\psi (\overline{L}(t)-x),\\&\underline{M}(t,x)=\max \big \{0,\phi (\overline{L}(t)-x)-N_{1}\epsilon e^{-\delta (t-\tilde{t}_{n})}\big \},\\&\overline{L}(t)=R_{*}+s_{\mu }t+N\epsilon +N\epsilon \sigma \big (1-e^{-\delta (t-\tilde{t}_{n})}\big ), \end{aligned} \end{aligned}$$

where \(N,\epsilon ,\delta ,N_1\) and \(\sigma \) are positive constants to be determined later. Since \(\lim _{x\rightarrow -\infty }\overline{H}(t,x)\,>1\), there exists a smooth function \(\overline{G}(t)\) for \(t\ge \tilde{t}_{n}\) such that \(\lim _{t\rightarrow \infty }\overline{G}(t)=-\infty \) and \(\overline{H}(t,\overline{G}(t))\ge 1\). In what follows, we will check that \((\overline{H},\underline{M},\overline{G},\overline{L})\) satisfies the following inequalities:

$$\begin{aligned}&N_{1}[\overline{H},\underline{M}]=\overline{H}_{t}-d\overline{H}_{xx}-k\overline{H}(1-\overline{H}-b\underline{M})\ge 0,\ \ t>\tilde{t}_{n},\ -\infty<x<\overline{L}(t), \end{aligned}$$
(4.25)
$$\begin{aligned}&N_{2}[\overline{H},\underline{M}]=\underline{M}_{t}-\underline{M}_{xx}-\underline{M}(1-\underline{M}-a\overline{H})\le 0,\ \ t>\tilde{t}_{n},\ -\infty<x<\infty , \end{aligned}$$
(4.26)
$$\begin{aligned}&\overline{H}(t,\overline{G}(t))\ge \check{H}(t,\overline{G}(t)),\ \overline{H}(t,\overline{L}(t))=0,\ \ t>\tilde{t}_{n}, \end{aligned}$$
(4.27)
$$\begin{aligned}&\overline{L}(\tilde{t}_{n})\ge \check{L}(\tilde{t}_{n}),\ \overline{L}'(t)\ge -\mu \overline{H}_{x}(t,\overline{L}(t)),\ \ t>\tilde{t}_{n}, \end{aligned}$$
(4.28)
$$\begin{aligned}&\overline{H}(\tilde{t}_{n},x) \ge \check{H}(\tilde{t}_{n},x),\ \ \overline{G}(\tilde{t}_{n})<x<\check{L}(\tilde{t}_{n}), \end{aligned}$$
(4.29)
$$\begin{aligned}&\underline{M}(\tilde{t}_{n},x) \le \check{M}(\tilde{t}_{n},x),\ \ -\infty<x<\infty . \end{aligned}$$
(4.30)

Firstly, we verify (4.25) and (4.26) by setting

$$\begin{aligned} s=\overline{L}(t)-x,\ q_{0}(t)=1+N\epsilon e^{-\delta (t-\tilde{t}_{n})},\ p_{0}(t)= N_{1}\epsilon e^{-\delta (t-\tilde{t}_{n})}. \end{aligned}$$

A direct calculation yields

$$\begin{aligned} \begin{aligned} N_{1}[\overline{H},\underline{M}]&={q_{0}}'\psi +q_{0}\big (s_{\mu }+N\epsilon \sigma \delta e^{-\delta (t-\tilde{t}_{n})}\big )\psi '-dq_{0}\psi ''-kq_{0}\psi +kq_{0}^{2}\psi ^{2}\\&\quad +kbq_{0}\psi \underline{M}\\&={q_{0}}'\psi +N\epsilon \sigma \delta e^{-\delta (t-\tilde{t}_{n})}q_{0}\psi ' +kq_{0}(q_0-1)\psi ^{2}\\&\qquad -kbq_{0}\psi \phi +kbq_{0}\psi \underline{M}~\text {(used }(1.7)). \end{aligned} \end{aligned}$$

If \(\underline{M}=0\), then \(\phi \le p_0\) and \(-kbq_{0}\psi \phi \ge -kbq_{0}\psi p_0\); if \(\underline{M}=\phi -p_0>0\), then \(-kbq_{0}\psi \phi +kbq_{0}\psi \underline{M}=-kbq_{0}\psi p_0\). To sum up, it holds

$$\begin{aligned} \begin{aligned} N_{1}[\overline{H},\underline{M}]&\ge {q_{0}}'\psi +N\epsilon \sigma \delta q_{0}e^{-\delta (t-\tilde{t}_{n})}\psi '+kq_{0}(q_0-1)\psi ^{2}-kbq_{0}\psi p_{0}\\&=e^{-\delta (t-\tilde{t}_{n})}\psi \big (kN\epsilon \psi +kN^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi -N\epsilon \delta \\&\quad -kbN_{1}\epsilon -kbNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\big )\\&\quad +N\epsilon \sigma \delta \big (1+N\epsilon e^{-\delta (t-\tilde{t}_{n})}\big )e^{-\delta (t-\tilde{t}_{n})}\psi '. \end{aligned} \end{aligned}$$

It follows from \(\psi (s)\rightarrow 1\) as \(s\rightarrow \infty \) that there exist positive constants \(D_1, N\) large and \(N_1,\delta \) small such that, for \(t>\tilde{t}_n\) and \(s\ge D_1\),

$$\begin{aligned} kN\epsilon \psi +kN^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi -N\epsilon \delta -kbN_{1}\epsilon -kbNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\ge 0. \end{aligned}$$

Then \(N_1[\overline{H},\underline{M}]\ge 0\) for \(t>\tilde{t}_n\) and \(s\ge D_1\) since \(\psi '(s)>0\) for \(s\ge D_1\). Moreover,

$$\begin{aligned} \begin{aligned} N_{1}[\overline{H},\underline{M}]&\ge e^{-\delta (t-\tilde{t}_{n})}\psi \big (kN\epsilon \psi +kN^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi -N\epsilon \delta -kbN_{1}\epsilon \\&\quad -kbNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\big )\\&\quad +N\epsilon \sigma \delta (1+N\epsilon e^{-\delta (t-\tilde{t}_{n})})e^{-\delta (t-\tilde{t}_{n})}\psi '\\&\ge e^{-\delta (t-\tilde{t}_{n})}\big (N\epsilon \sigma \delta \psi '-N\epsilon \delta \psi -kbN_{1}\epsilon \psi -kbNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi \big ). \end{aligned} \end{aligned}$$

Due to \(\psi '(s)>0\) in \((0,D_1)\), there exists \(\sigma >0\) sufficiently large such that, for \(t>\tilde{t}_n\) and \(s\in (0,D_1)\),

$$\begin{aligned} N\epsilon \sigma \delta \psi '-kbN_{1}\epsilon \psi -kbNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi -N\epsilon \delta \psi \ge 0. \end{aligned}$$

Thus, \(N_{1}[\overline{H},\underline{M}]\ge 0\) for \(t> \tilde{t}_{n}\) and \(s>0\), which means (4.25) holds.

Obviously, \(N_{2}[\overline{H},\underline{M}]=0\) if \(\underline{M}=0\). If \(\underline{M}=\phi -p_{0}>0\), a direct calculation yields

$$\begin{aligned} \begin{aligned} N_{2}[\overline{H},\underline{M}]&=\big (s_{\mu }+N\epsilon \sigma \delta e^{-\delta (t-\tilde{t}_{n})}\big )\phi '-p_{0}'-\phi ''-(\phi -p_0)\\&\quad +(\phi -p_0)^2+a(\phi -p_0)q_0\psi \\&=N\epsilon \sigma \delta e^{-\delta (t-\tilde{t}_{n_{}})}\phi '-p_{0}'+a(q_{0}-1)\psi \phi +p_{0}\\&\quad -2p_{0}\phi +{p_{0}}^{2}-aq_{0}p_{0}\psi ~\text {(used }(1.7))\\&=e^{-\delta (t-\tilde{t}_{n})}\big (N\epsilon \sigma \delta \phi '+N_{1}\epsilon \delta +aN\epsilon \psi \phi +N_{1}\epsilon -2N_{1}\epsilon \phi +N_{1}^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\\&\quad -aN_{1}\epsilon \psi -aNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi \big ):=e^{-\delta (t-\tilde{t}_{n})}P_{1}. \end{aligned} \end{aligned}$$

Since \(\psi (s)\rightarrow 1,\phi (s)\rightarrow 0\) as \(s\rightarrow \infty ,\phi '<0\) and \(a>1\), there exist positive constants \(D_{2},N\) large and \(N_1,\delta \) small such that, for \(t>\tilde{t}_n\) and \(s>D_{2}\),

$$\begin{aligned} P_{1}\le \big (N_{1}\epsilon \delta +aN\epsilon \psi \phi +N_{1}\epsilon -aN_{1}\epsilon \psi \big )+\big (N_{1}^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}-aNN_{1}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\psi \big )\le 0. \end{aligned}$$

Choose \(\epsilon >0\) small such that \(\delta \in (0,1-N_{1}\epsilon )\). Moreover, due to \(\phi (s)\rightarrow 1(s\rightarrow -\infty ),~\psi (s)=0\) when \(s\le 0\) and \(\phi '<0\), there exists a positive constant \(D_{3}\) large such that, for \(t>\tilde{t}_n\) and \(s<-D_{3}\),

$$\begin{aligned} P_{1}\le N_{1}\epsilon \delta +N_{1}\epsilon -2N_{1}\epsilon \phi +{N}_{1}^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\le N_{1}\epsilon (\delta +1+N_{1}\epsilon -2\phi )\le 0. \end{aligned}$$

In addition, for such fixed \(D_{2}\) and \(D_{3}\), it follows from \(\phi '<0\) that there exists \(\sigma >0\) sufficiently large such that, for \(t>\tilde{t}_n\) and \(s\in [-D_{3},D_{2}]\),

$$\begin{aligned} P_{1}\le N\epsilon \sigma \delta \phi '+N_{1}\epsilon \delta +aN\epsilon \psi \phi +N_{1}\epsilon +N_{1}^{2}\epsilon ^{2}e^{-\delta (t-\tilde{t}_{n})}\le 0. \end{aligned}$$

Thus, \(N_{2}[\overline{H},\underline{M}]\le 0\) for \(t>\tilde{t}_{n}\) and \(s\in \mathbb {R}\), which implies that (4.26) holds.

Clearly, \(\check{H}(t,\overline{G}(t))=\hat{H}(t,\overline{G}(t)-s_{\mu }t)\le \psi (R_*-\overline{G}(t)+s_{\mu }t)<\psi (\overline{L}(t)-\overline{G}(t))\le \overline{H}(t,\overline{G}(t))\) and \(\overline{H}(t,\overline{L}(t))=(1+N\epsilon e^{-\delta (t-\tilde{t}_{n})})\psi (0)=0\), which means (4.27) holds.

Next, we show (4.28). Note that \(\check{L}(\tilde{t}_{n})=L(\tilde{t}_{n})+s_{\mu }\tilde{t}_{n}\le R_{*}+s_{\mu }\tilde{t}_{n}+\epsilon < R_{*}+s_{\mu }\tilde{t}_{n}+N\epsilon =\overline{L}(\tilde{t}_{n})\). Direct calculations yield that

$$\begin{aligned}&\overline{L}'(t)=s_{\mu }+N\epsilon \sigma \delta e^{-\delta (t-\tilde{t}_{n})},\\&-\mu \overline{H}_{x}(t,\overline{L}(t))=\mu \psi '(0)(1+N\epsilon e^{-\delta (t-\tilde{t}_{n})})=s_{\mu }+s_{\mu }N\epsilon e^{-\delta (t-\tilde{t}_{n})}. \end{aligned}$$

Then \(\overline{L}'(t)\ge -\mu \overline{H}_{x}(t,\overline{L}(t))\) by taking \(\sigma >0\) large enough such that \(\sigma \delta \ge s_{\mu }\).

It follows from (4.23) and (4.24) that, for \(\overline{G}(\tilde{t}_{n})<x<\check{L}(\tilde{t}_{n})\) with \(n>\tilde{n}_0\),

$$\begin{aligned} \begin{aligned} \overline{H}(\tilde{t}_{n},x)&=(1+N\epsilon )\psi (R_{*}+s_{\mu }\tilde{t}_{n}+N\epsilon -x)\\&\ge \psi (R_{*}+s_{\mu }\tilde{t}_{n}+\epsilon -x)+\epsilon \ge \hat{H}(\tilde{t}_{n},x-s_{\mu }\tilde{t}_{n})=\check{H}(\tilde{t}_{n},x), \end{aligned} \end{aligned}$$

which means (4.29) holds.

Finally, we turn to prove (4.30). Since

$$\begin{aligned} \phi _{s_{\mu }}(R_{*}+s_\mu \tilde{t}_{n}-x)\ge \check{M}(\tilde{t}_{n},x)=\hat{M}(\tilde{t}_{n},x-s_{\mu }\tilde{t}_{n})\ge \phi (R^*+s_\mu \tilde{t}_{n}-x), \end{aligned}$$

there exists a constant \(\widetilde{L}>0\) large enough such that for \(|x-s_{\mu }\tilde{t}_{n}|>\widetilde{L}\),

$$\begin{aligned} \underline{M}(\tilde{t}_{n},x)=\phi (R_*+s_{\mu }\tilde{t}_{n}+N\epsilon -x)-N_1\epsilon \le \phi (R^*+s_\mu \tilde{t}_n-x)\le \check{M}(\tilde{t}_{n},x). \end{aligned}$$
(4.31)

Moreover, Lemma 4.3 shows that, for any \(\epsilon >0\), there exists \(\tilde{n}_1>0\) such that \(|\check{M}(\tilde{t}_{n},x)-\phi _{s_{\mu }}(R_{*}+s_{\mu }\tilde{t}_n-x)|<\epsilon \) for \(n>\tilde{n}_1\) and \(|x-s_{\mu }\tilde{t}_n|\le \widetilde{L}\), which implies that (4.31) still holds for \(|x-s_{\mu }\tilde{t}_n|\le \widetilde{L}\) when \(N_1>1\) and \(n>\tilde{n}_1\). Hence, (4.30) holds.

Summarizing as above, one can apply the comparison principle to obtain that \(\check{L}(t)\le \overline{L}(t)\) for \(t>\tilde{t}_{n}\) with \(n\ge \max \{\tilde{n}_0,\tilde{n}_1\}\), which implies

$$\begin{aligned} L(t)\le R_{*}+N\epsilon +N\epsilon \sigma \big (1-e^{-\delta (t-\tilde{t}_{n})}\big )\le R_{*}+N\epsilon (\sigma +1)\ \text {for}\ t>\tilde{t}_{n}. \end{aligned}$$
(4.32)

Step 2. Construct a subsolution.

It follows from (4.22) and Lemma 4.2 that

$$\begin{aligned} \hat{H}(t_{n},x)\ge \psi _{s_{\mu }}(R^{*}-x-\epsilon )-\epsilon \text { for }x\le R^*-\epsilon ,\ n>\tilde{n}_0. \end{aligned}$$
(4.33)

Moreover, there exists a constant \(N(>1)\) independent of \(\epsilon \) such that

$$\begin{aligned} \psi _{s_{\mu }}(R^{*}-x-\epsilon )-\epsilon \ge (1-N\epsilon )\psi _{s_{\mu }}(R^{*}-x-N\epsilon )\text { for } x\le R^*-\epsilon . \end{aligned}$$
(4.34)

Let \((\psi ,\phi )=(\psi _{s_{\mu }},\phi _{s_{\mu }})\) be the unique solution of (1.7) with \(c=s_{\mu }\) and \(t_{n}\) be given in Lemma 4.2. Define

$$\begin{aligned} \begin{aligned}&\underline{H}(t,x)=\big (1-N\epsilon e^{-\delta (t-t_{n})}\big )\psi (\underline{L}(t)-x),\\&\overline{M}(t,x)=\phi (\underline{L}(t)-x)+N_{2}\epsilon e^{-\delta (t-t_{n})},\\&\underline{G}(t)=-\hat{c}t,\ \underline{L}(t)=R^{*}+s_{\mu }t-N\epsilon -N\epsilon \sigma \big (1-e^{-\delta (t-t_{n})}\big ), \end{aligned} \end{aligned}$$

where \(\hat{c}\) refers to Lemma 3.1 and \(N,\epsilon ,\delta ,N_2\) and \(\sigma \) are positive constants to be determined later. In what follows, we will check that \((\underline{H},\overline{M},\underline{G},\underline{L})\) satisfies the following inequalities:

$$\begin{aligned}&N_{1}[\underline{H},\overline{M}]=\underline{H}_{t}-d\underline{H}_{xx}-k\underline{H}(1-\underline{H}-b\overline{M})\le 0,\ \ t>t_{n},\ -\infty<x<\underline{L}(t), \end{aligned}$$
(4.35)
$$\begin{aligned}&N_{2}[\underline{H},\overline{M}]=\overline{M}_{t}-\overline{M}_{xx}-\overline{M}(1-\overline{M}-a\underline{H})\ge 0,\ \ t>t_{n},\ -\infty<x<\infty , \end{aligned}$$
(4.36)
$$\begin{aligned}&\underline{H}(t,\underline{G}(t))\le \check{H}(t,\underline{G}(t)),\ \underline{H}(t,\underline{L}(t))=0,\ \ t>t_{n}, \end{aligned}$$
(4.37)
$$\begin{aligned}&\underline{L}(t_{n})\le \check{L}(t_{n}),\ \underline{L}'(t)\le -\mu \underline{H}_{x}(t,\underline{L}(t)),\ \ t>t_{n}, \end{aligned}$$
(4.38)
$$\begin{aligned}&\underline{H}(t_{n},x) \le \check{H}(t_{n},x),\ \ \underline{G}(t_{n})<x<\underline{L}(t_{n}), \end{aligned}$$
(4.39)
$$\begin{aligned}&\overline{M}(t_{n},x) \ge \check{M}(t_{n},x),\ \ -\infty<x<\infty . \end{aligned}$$
(4.40)

Firstly, we verify (4.35) and (4.36) by setting

$$\begin{aligned} z=\underline{L}(t)-x,\ q_{1}(t)=1-N\epsilon e^{-\delta (t-t_{n})},\ p_{1}(t)= N_{2}\epsilon e^{-\delta (t-t_{n})}. \end{aligned}$$

Note that \(q_1(t)>0\) for \(t\ge t_n\) when \(2N\epsilon \le 1\). A direct calculation yields

$$\begin{aligned} \begin{aligned} N_{1}[\underline{H},\overline{M}]&=q_{1}'\psi +q_{1}\big (s_{\mu }-N\epsilon \sigma \delta e^{-\delta (t-t_{n})}\big )\psi '-dq_{1}\psi ''\\&\quad -kq_{1}\psi +kq_{1}^{2}\psi ^{2}+kbq_{1}\psi (\phi +p_{1})\\&=-q_{1}(1-q_{1})k\psi ^{2}+kb\psi q_{1}p_{1}-q_{1}N\epsilon \sigma \delta e^{-\delta (t-t_{n})}\psi '+q_{1}'\psi ~\text {(used }(1.7))\\&=e^{-\delta (t-t_{n})}\big (-N\epsilon kq_{1}\psi ^{2}+kbN_{2}\epsilon q_{1}\psi -N\epsilon \sigma \delta q_{1}\psi '{+}N\epsilon \delta \psi \big ){:=}e^{-\delta (t-t_{n})}P_{2}. \end{aligned} \end{aligned}$$

Since \(\psi (z)\rightarrow 1\) as \(z\rightarrow \infty ,\psi '(z)\ge 0\) and \(b<1\), there exist positive constants \(D_{4},N,N_2(N> bN_2)\) large and \(\delta \) small such that, for \(t>t_{n}\) and \(z>D_{4}\),

$$\begin{aligned} P_{2}\le -N\epsilon kq_{1}\psi ^{2}+kbN_{2}\epsilon q_{1}\psi +N\epsilon \delta \psi \le 0. \end{aligned}$$

Moreover, for such fixed \(N,N_{2}\) and \(D_{4}\), it follows from \(\psi '(z)>0\) in \((0,D_{4}]\) that there exists \(\sigma >0\) sufficiently large such that, for \(t>t_{n}\) and \(z\in (0,D_{4}]\),

$$\begin{aligned} P_{2}=-N\epsilon kq_{1}\psi ^{2}+q_{1}kbN_{2}\epsilon -N\epsilon \sigma \delta q_{1}\psi '+N\epsilon \delta \psi \le 0. \end{aligned}$$

Thus, \(N_{1}[\underline{H},\overline{M}]\le 0\) for \(t>t_{n}\) and \(z>0\), which implies (4.35) holds.

On the other hand, a direct calculation yields

$$\begin{aligned} \begin{aligned} N_{2}[\underline{H},\overline{M}]&=\big (s_{\mu }-N\epsilon \sigma \delta e^{-\delta (t-t_{n})}\big )\phi '+p_{1}'-\phi ''-(\phi +p_{1})\\&\quad +(\phi +p_{1})^{2}+a(\phi +p_{1})q_{1}\psi \\&=-N\epsilon \sigma \delta e^{-\delta (t-t_{n})}\phi '+p_{1}'-p_{1}+{p_{1}}^{2}+2p_{1}\phi -a(1-q_1)\psi \phi \\&\quad +ap_1q_1\psi ~\text {(used }(1.7))\\&=e^{-\delta (t-t_{n})}\big (-N\epsilon \sigma \delta \phi '-N_{2}\epsilon \delta -N_{2}\epsilon +{N}_{2}^{2}\epsilon ^{2}e^{-\delta (t-t_{n})} +2N_{2}\epsilon \phi -aN\epsilon \phi \psi \\&\quad +aN_{2}\epsilon \psi -aNN_{2}\epsilon ^{2}e^{-\delta (t-t_{n})}\psi \big ):=e^{-\delta (t-t_{n})}P_{3}. \end{aligned} \end{aligned}$$

Since \(\phi (z)\rightarrow 1(z\rightarrow -\infty ),\psi (z)=0\) when \(z<0\) and \(\phi '<0\), there exist positive constants \(D_{5}\) large and \(\delta \) small such that, for \(t>t_n\) and \(z\in (-\infty ,-D_{5})\),

$$\begin{aligned} P_{3}\ge -N_{2}\epsilon \delta -N_{2}\epsilon +2N_{2}\epsilon \phi =N_2\epsilon (2\phi -\delta -1)\ge 0. \end{aligned}$$

Moreover, due to \(\phi (z)\rightarrow 0,\psi (z)\rightarrow 1\) as \(z\rightarrow \infty \) and \(a>1\), there exist positive constants \(D_{6}\) large and \(\epsilon ,\delta \) small such that, for \(t>t_n\) and \(z\in (D_{6},\infty )\),

$$\begin{aligned} P_{3}\ge -N_{2}\epsilon \delta -N_{2}\epsilon -aN\epsilon \phi \psi +aN_{2}\epsilon \psi +{N}_{2}^{2}\epsilon ^{2}e^{-\delta (t-t_{n})}-aNN_{2}\epsilon ^{2}e^{-\delta (t-t_{n})}\psi \ge 0. \end{aligned}$$

In addition, for such fixed \(D_{5},D_{6}\) and \(\delta \), it follows from \(\phi '<0\) that there exists \(\sigma >0\) sufficiently large such that, for \(t>t_n\) and \(z\in [-D_{5},D_6]\),

$$\begin{aligned} P_3\ge -N\epsilon \sigma \delta \phi '-N_{2}\epsilon \delta -N_{2}\epsilon -aN\epsilon \phi \psi +aN_{2}\epsilon \psi -aNN_{2}\epsilon ^{2}e^{-\delta (t-t_{n})}\psi \ge 0. \end{aligned}$$

Hence, \(N_{2}[\underline{H},\overline{M}]\ge 0\) for \(t>t_{n}\) and \(z\in \mathbb {R}\), which implies (4.36).

Clearly, \(\underline{H}(t,\underline{L}(t)=(1-N\epsilon e^{-\delta (t-t_{n})})\psi (0)=0\). Thanks to (4.21) and by choosing \(\delta >0\) small such that \(\delta \le \beta (s_{\mu }+\hat{c})\), one has

$$\begin{aligned} \begin{aligned} \check{H}(t,\underline{G}(t))&=\hat{H}\big (t,-(s_{\mu }+\hat{c})t\big )\\&\ge 1-\widetilde{C}e^{-\beta (s_{\mu }+\hat{c})t}\ge 1-N\epsilon e^{-\delta (t-t_{n})}\\&\ge (1-N\epsilon e^{-\delta (t-t_{n})})\psi (\underline{L}(t)+ct)=\underline{H}(t,\underline{G}(t)) \end{aligned} \end{aligned}$$

for \(t>t_n\), which implies (4.37) holds.

We now show (4.38). Note that \(\underline{L}(t_{n})=R^{*}-N\epsilon +s_{\mu }t_{n}\le R^{*}-\epsilon +s_{\mu }t_{n}\le L(t_{n})+s_{\mu }t_{n}=\check{L}(t_{n})\). Direct calculations yield

$$\begin{aligned}&\underline{L}'(t)=s_{\mu }-N\epsilon \sigma \delta e^{-\delta (t-t_{n})},\\&-\mu \underline{H}_{x}(t, \underline{L}(t))=\mu (1-N\epsilon e^{-\delta (t-t_{n})})\psi '(0)=s_{\mu }-N\epsilon s_{\mu } e^{-\delta (t-t_{n})}. \end{aligned}$$

Thus, \(\underline{L}'(t)\le -\mu \underline{H}_{x}(t,\underline{L}(t))\) holds by taking \(\sigma >0\) large enough such that \(\sigma \delta \ge s_{\mu }.\)

It follows from (4.33) and (4.34) that, for \(\underline{G}(t_{n})<x<\underline{L}(t_{n})\) with \(n> \tilde{n}_0\),

$$\begin{aligned} \begin{aligned} \underline{H}(t_{n},x)&=(1-N\epsilon )\psi (R^{*}+s_{\mu }t_{n}-N\epsilon -x)\\&\le \psi (R^{*}+s_{\mu }\tilde{t}_{n}-\epsilon -x)-\epsilon \le \hat{H}(t_{n},x-s_{\mu }t_{n})=\check{H}(t_{n},x), \end{aligned} \end{aligned}$$

which means (4.39) holds.

Finally, we turn to prove (4.40). Since

$$\begin{aligned} \phi _{s_{\mu }}(R^{*}-x+s_{\mu }t_{n})\le \check{M}(t_{n},x)=\hat{M}(t_{n},x-s_{\mu }t_{n})\le \phi _{s_{\mu }}(R_{*}-x+s_{\mu }t_{n}), \end{aligned}$$

there exists a positive constant \(L>0\) large such that for \(|x-s_{\mu }t_n|>L\),

$$\begin{aligned} \overline{M}(t_{n},x)=\phi _{s_{\mu }}(\underline{L}(t_{n})-x)+N_{2}\epsilon =\phi _{s_{\mu }}(R^{*}+s_{\mu }t_n-N\epsilon -x)+N_{2}\epsilon \ge \check{M}(t_{n},x). \end{aligned}$$
(4.41)

Moreover, Lemma 4.2 shows that, for any \(\epsilon >0\), there exists \(n_0>0\) such that \(|\check{M}(t_{n},x)-\phi _{s_{\mu }}(R^{*}+s_{\mu }t_n-x)|<\epsilon \) for \(n>n_0\) and \(|x-s_{\mu }t_n|\le L\), which implies that (4.41) still holds for \(|x-s_{\mu }t_n|\le L\) when \(N_2>1\) and \(n>n_0\). Hence, (4.40) holds.

Summarizing as above, one can apply the comparison principle to obtain that \(\check{L}(t)\ge \underline{L}(t)\) for \(t>t_{n}\) with \(n\ge \max \{\tilde{n}_0,n_0\}\), which implies that

$$\begin{aligned} L(t)\ge R^{*}-N\epsilon -N\epsilon \sigma \big (1-e^{-\delta (t-t_{n})}\big )\ge R^{*}-N\epsilon (1+\sigma ). \end{aligned}$$
(4.42)

Step 3. Complete the proof.

Steps 1 and 2 have shown that (4.32) and (4.42) hold for any \(\epsilon >0\). Setting \(T_{0}=\max \{t_{n},\tilde{t}_{n}\}\) with \(n>\max \{n_0,\tilde{n}_0,\tilde{n}_1\}\), it follows from (4.32) and (4.42) that

$$\begin{aligned} \vert R^{*}-R_{*}\vert \le \vert R^{*}-L(t)\vert +\vert L(t)-R_{*}\vert \le 2N\epsilon (1+\sigma )\text { for }t>T_{0}. \end{aligned}$$

Since N and \(\sigma \) are independent of the arbitrarily positive number \(\epsilon \), there must be \(R^{*}=R_{*}\). This completes the proof. \(\square \)

4.3 Proof of Theorem 1.3.

For clarity, we are going to divide the proof into two claims.

Claim 1: Let \(\omega _{h}(t)=h(t)-s_{\mu }t\). Then \(\vert \omega _{h}(t)\vert <C\) for all \(t>0\) and \(\lim _{t\rightarrow \infty }\omega _{h}'(t)=0\), where C refers to Proposition3.1. Moreover, (1.12) holds.

Proposition 3.1 implies that \(\vert \omega _{h}(t)\vert <C\) for all \(t>0\). It follows from Lemma 4.1 and Proposition 4.4 that, for any sequence \(\{t_n\}\) with \(\lim _{n\rightarrow \infty }t_{n}=\infty \) (by passing to a subsequence if necessary), \(l(t+t_{n})=h(t+t_{n})-r(t+t_{n})\rightarrow L_{0}\) in \(C^{1+\frac{\alpha }{2}}_{loc}(\mathbb {R})\) as \(n\rightarrow \infty \). In view of \(r(t)=s_{\mu }t-2C\), we have \(\lim _{n\rightarrow \infty }h'(t+t_{n})=s_{\mu }\) in \(C^{\frac{\alpha }{2}}_{loc}(\mathbb {R})\). The arbitrariness of \(\{t_{n}\}\) means that \(\lim _{t\rightarrow \infty } h'(t)=s_{\mu }\), which leads to \(\lim _{t\rightarrow \infty }\omega _{h}'(t)=0\).

For \(t\ge 0\) and \(x\ge -h(t)\), denote

$$\begin{aligned} (\widetilde{U},\widetilde{V})(t,x)=(u,v)(t,x+h(t)) \end{aligned}$$

and

$$\begin{aligned} h_{n}(t)=h(t+t_{n}),\ (\widetilde{U}_{n},\widetilde{V}_{n})(t,x)=(\widetilde{U},\widetilde{V})(t+t_{n},x). \end{aligned}$$

Then \(\big (\widetilde{U}_{n},\widetilde{V}_{n},h_{n}\big )\) satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} (\widetilde{U}_{n})_{t}=d(\widetilde{U}_{n})_{xx}+{h_{n}}'(t)(\widetilde{U}_{n})_{x}+k\widetilde{U}_{n}(1-\widetilde{U}_{n}-b\widetilde{V}_{n}), &{} t>-t_{n},\ -h_{n}(t)<x<0,\\ (\widetilde{V}_{n})_{t}=(\widetilde{V}_{n})_{xx}+{h_{n}}'(t)(\widetilde{V}_{n})_{x}+\widetilde{V}_{n}(1-\widetilde{V}_{n}-a\widetilde{U}_{n}), &{} t>-t_{n},\ -h_{n}(t)<x<\infty ,\\ \widetilde{U}_{n}(t,x)=0, &{} t>-t_{n},\ x\ge 0,\\ (\widetilde{U}_{n})_{x}(t,0)=-h_{n}'(t)/\mu , &{} t>-t_{n}. \end{array} \right. \end{aligned}$$

Using the standard parabolic regularity theory and Sobolev embedding theorem, one can repeat the arguments used in Lemma 4.1 (by passing to a subsequence if necessary) to obtain

$$\begin{aligned} (\widetilde{U}_{n},\widetilde{V}_{n})\rightarrow (\widetilde{U}_{\infty },\widetilde{V}_{\infty })\text { in } C^{\frac{1+\alpha }{2},1+\alpha }_{loc}\big (\mathbb {R}\times (-\infty ,0]\big )\times C^{\frac{1+\alpha }{2},1+\alpha }_{loc}\big (\mathbb {R}\times \mathbb {R}\big )\text { as } n\rightarrow \infty , \end{aligned}$$

and \((\widetilde{U}_{\infty },\widetilde{V}_{\infty })\) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} (\widetilde{U}_{\infty })_{t}=d(\widetilde{U}_{\infty })_{xx}+s_{\mu }(\widetilde{U}_{\infty })_{x}+k\widetilde{U}_{\infty }(1-\widetilde{U}_{\infty }-b\widetilde{V}_{\infty }), &{} t\in \mathbb {R},\ x<0,\\ (\widetilde{V}_{\infty })_{t}=(\widetilde{V}_{\infty })_{xx}+s_{\mu }(\widetilde{V}_{\infty })_{x}+\widetilde{V}_{\infty }(1-\widetilde{V}_{\infty }-a\widetilde{U}_{\infty }), &{} t\in \mathbb {R},\ x\in \mathbb {R},\\ \widetilde{U}_{\infty }(t,x)=0,&{} t\in \mathbb {R},\ x\ge 0,\\ (\widetilde{U}_{\infty })_{x}(t,0)=-s_{\mu }/\mu , &{} t\in \mathbb {R}, \end{array}\right. } \end{aligned}$$

which is equivalent to (4.1) with \((\hat{H},\hat{M})=(\widetilde{U}_{\infty },\widetilde{V}_{\infty })\) and \(L=0\). Moreover, Proposition 4.4 implies that

$$\begin{aligned} \begin{aligned}&\widetilde{U}_{\infty }(t,x)=\psi _{s_{\mu }}(-x)\text { for } (t,x)\in \mathbb {R}\times (-\infty ,0],\\&\widetilde{V}_{\infty }(t,x)=\phi _{s_{\mu }}(-x)\text { for } (t,x)\in \mathbb {R}\times \mathbb {R}. \end{aligned} \end{aligned}$$

Based on the discussions above, it holds

$$\begin{aligned} \begin{aligned}&u(t+t_{n},x+h(t+t_{n}))\rightarrow \psi _{s_{\mu }}(-x)\text { in } C^{\frac{1+\alpha }{2},1+\alpha }_{loc}\big (\mathbb {R}\times (-\infty ,0]\big )\text { as } n\rightarrow \infty ,\\&v(t+t_{n},x+h(t+t_{n}))\rightarrow \phi _{s_{\mu }}(-x)\text { in } C^{\frac{1+\alpha }{2},1+\alpha }_{loc}\big (\mathbb {R}\times \mathbb {R}\big )\text { as } n\rightarrow \infty . \end{aligned} \end{aligned}$$

Noticing that \(\{t_{n}\}\) is an arbitrary time sequence with \(\lim _{n\rightarrow \infty }t_{n}=\infty \), one has

$$\begin{aligned} \begin{aligned}&\lim \limits _{t \rightarrow \infty }\big (u(t,x+h(t))-\psi _{s_{\mu }}(-x)\big )=0\text { uniformly for }x\text { in any compact set of }(-\infty ,0],\\&\lim \limits _{t \rightarrow \infty }\big (v(t,x+h(t))-\phi _{s_{\mu }}(-x)\big )=0\text { uniformly for }x\text { in any compact set of }\mathbb {R}. \end{aligned} \end{aligned}$$

Then for any positive constant \(\rho \), it holds

$$\begin{aligned} \begin{aligned}&\lim \limits _{t\rightarrow \infty }\Vert u(t,\cdot )-\psi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[h(t)-\rho ,h(t)]}=0,\\&\lim \limits _{t\rightarrow \infty }\Vert v(t,\cdot )-\phi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[h(t)-\rho ,h(t)+\rho ]}=0. \end{aligned} \end{aligned}$$
(4.43)

Next, we turn to prove (1.12). Lemma 3.4 implies that

$$\begin{aligned} \begin{aligned}&u(t,x)\le \overline{u}(t,x)=(1+\tilde{M_{0}}e^{-\delta t})\psi _{s_{\mu }}(\overline{h}(t)-x)\text { for } (t,x)\in [T^*,\infty )\times [g(t),\overline{h}(t)],\\&v(t,x)\ge \underline{v}(t,x)=\max \big \{0,\phi _{s_{\mu }}(\overline{h}(t)-x)-M_{1}e^{-\delta (t-T^{*})}\big \}\text { for }(t,x)\in [T^*,\infty )\times \mathbb {R},\\&h(t)\le \overline{h}(t)=s_{\mu }(t-T^{*})+h(T^{*})+\sigma (e^{-\delta T^{*}}-e^{-\delta t})+X_{0}\text { for }t\ge T^*. \end{aligned} \end{aligned}$$
(4.44)

Meanwhile, Lemma 3.5 implies that

$$\begin{aligned} \begin{aligned}&u(t,x)\ge \underline{u}(t,x)=(1-\tilde{M_{0}}e^{-\delta t})\psi _{s_{\mu }}(\underline{h}(t)-x)\text { for } (t,x)\in [T_*,\infty )\times [-\hat{c}t,\underline{h}(t)],\\&v(t,x)\le \overline{v}(t,x)=\phi _{s_{\mu }}(\underline{h}(t)-x)+M_{2}e^{-\delta (t-T_{*})}\text { for }(t,x)\in [T_*,\infty )\times \mathbb {R},\\&h(t)\ge \underline{h}(t)=s_{\mu }(t-T_{*})-\hat{c}T_{*}-\sigma (e^{-\delta T_{*}}-e^{-\delta t})+h(T_{*})\text { for }t\ge T_*. \end{aligned} \end{aligned}$$
(4.45)

Thanks to the monotonicity of \(\psi _{s_{\mu }}\) and \(\phi _{s_{\mu }}\), for any \(\lambda >0\) large, \(T\ge \max \{T_*,T^*\}\) and \((t,x)\in (T,\infty )\times [0,\underline{h}(t)-\lambda ]\), it holds

$$\begin{aligned} \begin{aligned} \vert u(t,x)-\psi _{s_{\mu }}(h(t)-x)\vert&\le \max \Big \{ \vert (1+\tilde{M_{0}}e^{-\delta t})\psi _{s_{\mu }}(\overline{h}(t)-x)-\psi _{s_{\mu }}(h(t)-x)\vert ,\\&\quad \vert (1-\tilde{M_{0}}e^{-\delta t})\psi _{s_{\mu }}(\underline{h}(t)-x)-\psi _{s_{\mu }}(h(t)-x)\vert \Big \}\\&\le \tilde{M_{0}}e^{-\delta t}+\psi _{s_{\mu }}(\overline{h}(t)-x)-\psi _{s_{\mu }}(\underline{h}(t)-x)\\&\le \tilde{M_{0}}e^{-\delta t}+\psi _{s_{\mu }}(\overline{h}(t))-\psi _{s_{\mu }}(\lambda ). \end{aligned} \end{aligned}$$

Similarly, there exists \(\hat{M}_0>0\) such that

$$\begin{aligned} \vert v(t,x)-\phi _{s_{\mu }}(h(t)-x)\vert \le \hat{M}_0e^{-\delta t}+\phi _{s_{\mu }}(\lambda )-\phi _{s_{\mu }}(\overline{h}(t)). \end{aligned}$$

Note that \(\psi _{s_{\mu }}(\overline{h}(t))\rightarrow 1,\ \phi _{s_{\mu }}(\overline{h}(t))\rightarrow 0\) as \(t\rightarrow \infty \) and \(\psi _{s_{\mu }}(\lambda )\rightarrow 1,\ \phi _{s_{\mu }}(\lambda )\rightarrow 0\) as \(\lambda \rightarrow \infty \). For any \(\epsilon >0\), there exist positive constants \(\hat{T}_{1}(>T)\) and \(\lambda \) large such that

$$\begin{aligned} \begin{aligned}&\vert \psi _{s_{\mu }}(\lambda )-1\vert<\epsilon /4,\ \vert \psi _{s_{\mu }}(\overline{h}(t))-1\vert<\epsilon /4\text { for }t>\hat{T}_{1},\\&\vert \phi _{s_{\mu }}(\lambda )\vert<\epsilon /4,\ \vert \phi _{s_{\mu }}(\overline{h}(t))\vert <\epsilon /4\text { for }t>\hat{T}_{1}, \end{aligned} \end{aligned}$$

which leads to

$$\begin{aligned} \vert \psi _{s_{\mu }}(\overline{h}(t))-\psi _{s_{\mu }}(\lambda )\vert<\epsilon /2,\ \vert \phi _{s_{\mu }}(\overline{h}(t))-\phi _{s_{\mu }}(\lambda )\vert <\epsilon /2\text { for }t>\hat{T}_{1}. \end{aligned}$$

Moreover, since \(e^{-\delta t}\rightarrow 0\) as \(t\rightarrow \infty \), for above \(\epsilon >0\), there exists \(\hat{T}_{2}(>T)\) large such that \(\tilde{M_{0}}e^{-\delta t}<\epsilon /2,\ \hat{M}_0e^{-\delta t}<\epsilon /2\) for \(t>\hat{T}_{2}\). Then for \((t,x)\in (\hat{T},\infty )\times [0,\underline{h}(t)-\lambda ]\) with \(\hat{T}=\max \{\hat{T}_{1},\hat{T}_{2}\}\), it holds

$$\begin{aligned} \vert u(t,x)-\psi _{s_{\mu }}(h(t)-x)\vert<\epsilon ,\ \vert v(t,x)-\phi _{s_{\mu }}(h(t)-x)\vert <\epsilon , \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned}&\lim \limits _{t\rightarrow \infty }\sup \limits _{x\in [0,\underline{h}(t)-\lambda ]}\vert u(t,x)-\psi _{s_{\mu }}(h(t)-x)\vert =0,\\&\lim \limits _{t\rightarrow \infty }\sup \limits _{x\in [0,\underline{h}(t)-\lambda ]}\vert v(t,x)-\phi _{s_{\mu }}(h(t)-x)\vert =0. \end{aligned} \end{aligned}$$
(4.46)

Since \(\rho >0\) is an arbitrary constant in (4.43), then \(h(t)-\rho <\underline{h}(t)-\lambda \) by taking \(\rho > 2C+\lambda \). Combining (4.43) with (4.46), one has

$$\begin{aligned} \begin{aligned}&\lim \limits _{t\rightarrow \infty }\Vert u(t,\cdot )-\psi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[0,h(t)]}=0,\\&\lim \limits _{t\rightarrow \infty }\Vert v(t,\cdot )-\phi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[0,h(t)+\rho ]}=0. \end{aligned} \end{aligned}$$
(4.47)

Moreover, the monotonicity of \(\phi _{s_{\mu }}\) implies that, for \((t,x)\in (T,\infty )\times [h(t)+\rho ,\infty )\),

$$\begin{aligned} \begin{aligned} \vert v(t,x)-\phi _{s_{\mu }}(h(t)-x)\vert&\le \hat{M}_0e^{-\delta t}+\vert \phi _{s_{\mu }}(\underline{h}(t)-x)-\phi _{s_{\mu }}(\overline{h}(t)-x)\vert \\&\le \hat{M}_0e^{-\delta t}+1-\phi _{s_{\mu }}(\underline{h}(t)-x)+1-\phi _{s_{\mu }}(\overline{h}(t)-x)\\&\le \hat{M}_0e^{-\delta t}+1-\phi _{s_{\mu }}(-\rho )+1-\phi _{s_{\mu }}(\overline{h}(t)-\underline{h}(t)-\rho ).\\ \end{aligned} \end{aligned}$$

Noticing that \(\phi _{s_{\mu }}(\xi )\rightarrow 1(\xi \rightarrow -\infty )\), for above \(\epsilon >0\), there exist \(\hat{T}_3(>T)\) and \(\rho \) large such that

$$\begin{aligned} 1-\phi _{s_{\mu }}(-\rho )<\epsilon /4,\ 1-\phi _{s_{\mu }}(\overline{h}(t)-\underline{h}(t)-\rho )<\epsilon /4\text { for }t>\hat{T}_3. \end{aligned}$$

Then for any given \(\hat{T}_0=\max \{\hat{T}_2,\hat{T}_3\}\) and \((t,x)\in (\hat{T}_0,\infty )\times [h(t)+\rho ,\infty )\), it holds

$$\begin{aligned} \vert v(t,x)-\phi _{s_{\mu }}(h(t)-x)\vert <\epsilon , \end{aligned}$$

which leads to

$$\begin{aligned} \lim _{t\rightarrow \infty }\Vert v(t,\cdot )-\phi _{s_{\mu }}(h(t)-\cdot )\Vert _{L^{\infty }[h(t)+\rho ,\infty )}=0. \end{aligned}$$
(4.48)

Combining (4.47) with (4.48), (1.12) clearly holds.

Claim 2: There exists a constant \(h^{*}\in \mathbb {R}\) such that \(\lim _{t \rightarrow \infty }\big (h(t)-s_{\mu }t\big )=h^*\).

It follows from Propositions 3.1, 4.4 and Lemma 4.1 that for any sequence \(\{t_{n}\}\) with \(\lim _{n\rightarrow \infty }t_{n}=\infty \), there exists a subsequence (still denoted it by \(\{t_{n}\}\)) such that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } (h(t_{n})-s_{\mu }t_{n})=h^{*} \end{aligned}$$
(4.49)

and

$$\begin{aligned} \begin{aligned}&\lim \limits _{n\rightarrow \infty }u(t_{n},x+h(t_{n})-s_{\mu }t_{n})=\psi _{s_{\mu }}(h^{*}-x)\text { in }C^{1+\alpha }_{loc}\big ((-\infty ,h^{*}]\big ),\\&\lim \limits _{n\rightarrow \infty }v(t_{n},x+h(t_{n})-s_{\mu }t_{n})=\phi _{s_{\mu }}(h^{*}-x)\text { in }C^{1+\alpha }_{loc}\big (\mathbb {R}\big ). \end{aligned} \end{aligned}$$
(4.50)

Denote

$$\begin{aligned} (\check{u},\check{v})(t,x)=(u,v)(t,x+s_{\mu }t). \end{aligned}$$

Since \((\psi _{s_{\mu }}(\xi ),\phi _{s_{\mu }}(\xi ),\phi _{s_{\mu }}(-\xi ))\rightarrow (1,0,1)\) exponentially as \(\xi \rightarrow \infty \) (see Lemma 2.4), it follows from (4.44) and (4.45) that there exist positive constants \(B,\beta \) and K such that

$$\begin{aligned} \begin{aligned}&\vert \check{u}(t,x)-1\vert \le B(e^{\beta x}+e^{-\delta t})\text { for } (t,x)\in [0,\infty )\times [-(\hat{c}+s_{\mu })t,h(t)-s_{\mu }t],\\&\vert \check{v}(t,x)-1\vert \le B(e^{-\beta x}+e^{-\delta t})\text { for } (t,x)\in [0,\infty )\times [K+s_{\mu }t,\infty ),\\&\vert \check{v}(t,x)\vert \le B(e^{\beta x}+e^{-\delta t})\text { for } (t,x)\in [0,\infty )\times (-\infty ,-K+s_{\mu }t]. \end{aligned} \end{aligned}$$

In particular, it holds

$$\begin{aligned} \begin{aligned}&\vert \check{u}(t_n,x)-1\vert \le B(e^{\beta x}+e^{-\delta t_n})\text { for } x\in [-(\hat{c}+s_{\mu })t_n,h(t_n)-s_{\mu }t_n],\\&\vert \check{v}(t_n,x)-1\vert \le B(e^{-\beta x}+e^{-\delta t_n})\text { for } (t_n,x)\in [0,\infty )\times [K+s_{\mu }t_n,\infty ),\\&\vert \check{v}(t_n,x)\vert \le B(e^{\beta x}+e^{-\delta t_n})\text { for } (t_n,x)\in [0,\infty )\times (-\infty ,-K+s_{\mu }t_n]. \end{aligned} \end{aligned}$$
(4.51)

In view of (4.49) and (4.51), for any \(\epsilon >0\), there exist constants \(\hat{N}_{1},\hat{K}\) and \(\tilde{K}(\ge K)\) such that for any \(n>\hat{N}_{1}\),

$$\begin{aligned} \vert h(t_{n})-s_{\mu }t_{n}-h^{*}\vert <\epsilon \end{aligned}$$

and

$$\begin{aligned}&\sup \limits _{x\in [-(\hat{c}+s_{\mu })t_{n},\hat{K}]}\vert \check{u}(t_{n},x)-\psi _{s_{\mu }}(h^{*}-x)\vert \\&<\epsilon , \sup \limits _{x\in \mathbb {R}\setminus [-\tilde{K}+s_{\mu }t_{n},\tilde{K}+s_{\mu }t_{n}]}\vert \check{v}(t_{n},x)-\phi _{s_{\mu }}(h^{*}-x) \vert <\epsilon . \end{aligned}$$

Moreover, it follows from (4.50) and (4.51) that there exists a constant \(\hat{N}_{2}>0\) large such that, for any \(n>\hat{N}_{2}\),

$$\begin{aligned} \sup \limits _{x\in [\hat{K},h^*]}\vert \check{u}(t_{n},x)-\psi _{s_{\mu }}(h^{*}-x)\vert<\epsilon ,\ \sup \limits _{x\in [-\tilde{K}+s_{\mu }t_{n},\tilde{K}+s_{\mu }t_{n}]}\vert \check{v}(t_{n},x)-\phi _{s_{\mu }}(h^{*}-x)\vert <\epsilon . \end{aligned}$$

Then for any \(n>\max \{\hat{N}_{1},\hat{N}_{2}\}\), it holds

$$\begin{aligned} \sup \limits _{x\in [-(\hat{c}+s_{\mu })t_{n},h^*]}\vert \check{u}(t_{n},x)-\psi _{s_{\mu }}(h^{*}-x)\vert<\epsilon ,\ \sup \limits _{x\in \mathbb {R}}\vert \check{v}(t_{n},x)-\phi _{s_{\mu }}(h^{*}-x)\vert <\epsilon . \end{aligned}$$

One can use the monotonicity of \(\psi _{s_{\mu }}\) and \(\phi _{s_{\mu }}\) to obtain

$$\begin{aligned} \begin{aligned}&\check{u}(t_{n},x)\le \psi _{s_{\mu }}(h^{*}-x)+\epsilon \le \psi _{s_{\mu }}(h^{*}+\epsilon -x)+\epsilon \text { for } x\in [-(\hat{c}+s_{\mu })t_{n},h^{*}+\epsilon ],\\&\check{u}(t_{n},x)\ge \psi _{s_{\mu }}(h^{*}-x)-\epsilon \ge \psi _{s_{\mu }}(h^{*}-\epsilon -x)-\epsilon \text { for } x\in [-(\hat{c}+s_{\mu })t_{n},h^{*}-\epsilon ] \end{aligned} \end{aligned}$$
(4.52)

and

$$\begin{aligned} \begin{aligned}&\check{v}(t_{n},x)\le \phi _{s_{\mu }}(h^{*}-x)+\epsilon \le \phi _{s_{\mu }}(h^{*}-\epsilon -x)+\epsilon \text { for } x\in \mathbb {R},\\&\check{v}(t_{n},x)\ge \phi _{s_{\mu }}(h^{*}-x)-\epsilon \ge \phi _{s_{\mu }}(h^{*}+\epsilon -x)-\epsilon \text { for } x\in \mathbb {R}. \end{aligned} \end{aligned}$$
(4.53)

Moreover, one can find a constant \(N(>1)\) independent of \(\epsilon \) such that

$$\begin{aligned} \begin{aligned}&(1+N\epsilon )\psi _{s_{\mu }}(h^{*}+N\epsilon -x)\ge \psi _{s_{\mu }}(h^{*}+\epsilon -x)+\epsilon \text { for }x\le h^{*}+\epsilon ,\\&(1-N\epsilon )\psi _{s_{\mu }}(h^{*}-N\epsilon -x)\le \psi _{s_{\mu }}(h^{*}-\epsilon -x)-\epsilon \text { for }x\le h^{*}-\epsilon . \end{aligned} \end{aligned}$$
(4.54)

Since \(\lim _{x\rightarrow \infty }\psi _{s_{\mu }}(x)=1\), there exists \(\hat{N}_{3}>\max \{\hat{N}_{1},\hat{N}_{2}\}\) such that

$$\begin{aligned} (1+N\epsilon )\psi _{s_{\mu }}(h^{*}+N\epsilon +s_{\mu }t_{n}+\hat{c}t_{n})\ge 1+\epsilon \text { for }n>\hat{N}_{3}. \end{aligned}$$
(4.55)

For above \(\epsilon>0, t_{n}>t_{\hat{N}_{3}}, N>1\) and \(N_1>1\), let us construct a supersolution for (1.1) with following form:

$$\begin{aligned} \begin{aligned}&\overline{u}(t,x)=\big (1+N\epsilon e^{-\delta (t-t_{n})}\big )\psi _{s_{\mu }}(\overline{h}(t)-x),\ \underline{v}(t,x)=\max \big \{0,\phi _{s_{\mu }}(\overline{h}(t)-x)-N_{1}\epsilon e^{-\delta (t-t_{n})}\big \},\\&\overline{g}(t)= g(t),\ \overline{h}(t)= h^{*}+s_{\mu }t+N\epsilon +N\epsilon \sigma \big (1-e^{-\delta (t-t_{n})}\big ), \end{aligned} \end{aligned}$$

where \(\delta \) and \(\sigma \) are positive constants to be determined. Next, we verify that \((\overline{u},\underline{v},\overline{g},\overline{h})\) satisfies the following inequalities:

$$\begin{aligned}&N_{1}[\overline{u},\underline{v}]=\overline{u}_{t}-d\overline{u}_{xx}-k\overline{u}(1-\overline{u}-b\underline{v})\ge 0,\ \ t>t_{n},\ \overline{g}(t)<x<\overline{h}(t), \end{aligned}$$
(4.56)
$$\begin{aligned}&N_{2}[\overline{u},\underline{v}]=\underline{v}_{t}-\underline{v}_{xx}-\underline{v}(1-\underline{v}-a\overline{u})\le 0,\ \ t>t_{n},\ -\infty<x<\infty , \end{aligned}$$
(4.57)
$$\begin{aligned}&\overline{u}(t,\overline{g}(t))\ge u(t,\overline{g}(t)),\ \overline{u}(t,\overline{h}(t))=0,\ \ t>t_{n}, \end{aligned}$$
(4.58)
$$\begin{aligned}&\overline{h}(t_{n})\ge h(t_{n}),\ \overline{h}'(t)\ge -\mu \overline{u}_{x}(t,\overline{h}(t)),\ \ t>t_{n}, \end{aligned}$$
(4.59)
$$\begin{aligned}&\overline{u}(t_{n},x) \ge u(t_{n},x),\ \ \overline{g}(t_{n})<x<\overline{h}(t_{n}), \end{aligned}$$
(4.60)
$$\begin{aligned}&\underline{v}(t_{n},x) \le v(t_{n},x),\ \ -\infty<x<\infty . \end{aligned}$$
(4.61)

Similar to the ways used in the proofs of (4.25), (4.26) and (4.28), the inequalities (4.56), (4.57) and (4.59) also hold by taking \(\sigma >0\) large enough and \(\delta >0\) small enough. Hence, the details are omitted here.

Clearly, \(\overline{u}(t,\overline{h}(t))=(1+N\epsilon e^{-\delta (t-t_{n})})\psi _{s_{\mu }}(0)=0\) and \(u(t,\overline{g}(t))=u(t,g(t))=0\le \overline{u}(t,\overline{g}(t))\), which means (4.58) holds.

We now show (4.60). It follows from (4.52) and (4.54) that, for \(x\in [-\hat{c}t_{n},\overline{h}(t_{n})]\),

$$\begin{aligned} \begin{aligned} \overline{u}(t_{n},x)&=(1+N\epsilon )\psi _{s_{\mu }}(h^{*}+s_{\mu }t_{n}+N\epsilon -x)\\&\ge \psi _{s_{\mu }}(h^{*}+\epsilon -x+s_{\mu }t_{n})+\epsilon \ge \check{u}(t_{n},x-s_{\mu }t_{n})=u(t_{n},x). \end{aligned} \end{aligned}$$

Moreover, one can use (4.55) and the monotonicity of \(\psi _{s_{\mu }}\) to obtain that

$$\begin{aligned} \overline{u}(t_n,x)\ge \overline{u}(t_{n},-\hat{c}t_{n})=(1+N\epsilon )\psi _{s_{\mu }}(h^{*}+s_{\mu }t_{n}+N\epsilon +\hat{c}t_{n})\ge 1+\epsilon \ge u(t_{n},x) \end{aligned}$$

for \(x\in [\overline{g}(t),-\hat{c}t_{n}]\). Hence, (4.60) holds.

Finally, we prove (4.61). If \(\underline{v}=0\), (4.61) clearly holds. If \(\underline{v}>0\), it follows from (4.53) and the monotonicity of \(\phi _{s_{\mu }}\) that

$$\begin{aligned} \begin{aligned} \underline{v}(t_{n},x)&=\phi _{s_{\mu }}(h^{*}+N\epsilon -x+s_{\mu }t_{n})-N_{1}\epsilon \\&\le \phi _{s_{\mu }}(h^{*}+\epsilon -x+s_{\mu }t_{n})-\epsilon \le \check{v}(t_{n},x-s_{\mu }t_{n})=v(t_{n},x) \end{aligned} \end{aligned}$$

for \(x\in \mathbb {R}\), which implies (4.61).

For above \(\epsilon>0, t_{n}>t_{\hat{N}_{3}},N>1\) and \(N_2>1\), let us construct a subsolution for (1.1) with following form:

$$\begin{aligned} \begin{aligned}&\underline{u}(t,x)=\big (1-N\epsilon e^{-\delta (t-t_{n})}\big )\psi _{s_{\mu }}(\underline{h}(t)-x),\ \overline{v}(t,x)=\phi _{s_{\mu }}(\underline{h}(t)-x)+N_{2}\epsilon e^{-\delta (t-t_{n})},\\&\underline{g}(t)=-\hat{c}t,\ \underline{h}(t)= h^{*}+s_{\mu }t-N\epsilon -N\epsilon \sigma \big (1-e^{-\delta (t-t_{n})}\big ), \end{aligned} \end{aligned}$$

where \(\delta \) and \(\sigma \) are positive constants to be determined. We now check that \((\underline{u},\overline{v},\underline{g},\underline{h})\) satisfies the following inequalities:

$$\begin{aligned}&N_{1}[\underline{u},\overline{v}]=\underline{u}_{t}-d\underline{u}_{xx}-k\underline{u}(1-\underline{u}-b\overline{v})\le 0,\ \ t>t_{n},\ \underline{g}(t)<x<\underline{h}(t), \end{aligned}$$
(4.62)
$$\begin{aligned}&N_{2}[\underline{u},\overline{v}]=\overline{v}_{t}-\overline{v}_{xx}-\overline{v}(1-\overline{v}-a\underline{u})\ge 0,\ \ t>t_{n},\ -\infty<x<\infty , \end{aligned}$$
(4.63)
$$\begin{aligned}&\underline{u}(t,\underline{g}(t))\le u(t,\underline{g}(t)),\ \underline{u}(t,\underline{h}(t))=0,\ \ t>t_{n}, \end{aligned}$$
(4.64)
$$\begin{aligned}&\underline{h}(t_{n})\le h(t_{n}),\ \underline{h}'(t)\le -\mu \underline{u}_{x}(t,\underline{h}(t)),\ \ t>t_{n}, \end{aligned}$$
(4.65)
$$\begin{aligned}&\underline{u}(t_{n},x) \le u(t_{n},x),\ \ \underline{g}(t_{n})<x<\underline{h}(t_{n}), \end{aligned}$$
(4.66)
$$\begin{aligned}&\overline{v}(t_{n},x) \ge v(t_{n},x),\ \ -\infty<x<\infty . \end{aligned}$$
(4.67)

By taking \(\sigma >0\) large enough and \(\delta >0\) small enough again, the inequalities (4.62), (4.63) and (4.65) can be proved in the same ways that (4.35), (4.36) and (4.38) used, respectively. We omit the details here.

Clearly, \(\underline{u}(t,\underline{h}(t))=\big (1-N\epsilon e^{-\delta (t-t_{n})}\big )\psi _{s_{\mu }}(0)=0\). Note that

$$\begin{aligned} \underline{u}(t,\underline{g}(t))=\big (1-N\epsilon e^{-\delta (t-t_{n})}\big )\psi _{s_{\mu }}(\underline{h}(t)+\hat{c}t)\le 1-N\epsilon e^{-\delta (t-t_{n})}\text { for }t>t_n . \end{aligned}$$

Thanks to Lemma 3.3, there exists a constant \(\delta \in (0,\tilde{\alpha }_{0})\) such that

$$\begin{aligned} u(t,\underline{g}(t))\ge 1-\tilde{M_{0}}e^{-\tilde{\alpha }_{0}t}\ge 1-N\epsilon e^{\delta t_{n}}e^{-\delta t}\text { for }t>t_n . \end{aligned}$$

Hence, \(u(t,\underline{g}(t))\ge \underline{u}(t,\underline{g}(t))\) for \(t>t_{n}\), which means (4.64) holds.

We now show (4.66). It follows from (4.52) and (4.54) that

$$\begin{aligned} \begin{aligned} \underline{u}(t_{n},x)&=(1-N\epsilon )\psi _{s_{\mu }}(h^{*}-N\epsilon -x+s_{\mu }t_{n})\\&\le \psi _{s_{\mu }}(h^{*}-\epsilon -x+s_{\mu }t_{n})-\epsilon \le \check{u}(t_{n},x-s_{\mu }t_{n})=u(t_{n},x) \end{aligned} \end{aligned}$$

for \(x\in [\underline{g}(t_{n}),\underline{h}(t_{n})]\), which implies (4.66).

Finally, we prove (4.67). It follows from (4.53) and the monotonicity of \(\phi _{s_{\mu }}\) that

$$\begin{aligned} \begin{aligned} \overline{v}(t_{n},x)&=\phi _{s_{\mu }}(h^{*}-N\epsilon -x+s_{\mu }t_{n})+N_{2}\epsilon \\&\ge \phi _{s_{\mu }}(h^{*}-\epsilon -x+s_{\mu }t_{n})+\epsilon \ge \check{v}(t_{n},x-s_{\mu }t_{n})=v(t_{n},x) \end{aligned} \end{aligned}$$

for \(x\in \mathbb {R}\). Then (4.67) holds.

Summarizing as above, one can apply the comparison principle to obtain that \(\underline{h}(t)\le h(t)\le \overline{h}(t)\) for \(t>t_{n}\), which leads to

$$\begin{aligned} h^{*}-N\epsilon (1+\sigma )\le h(t)-s_{\mu }t\le h^{*}+N\epsilon (1+\sigma )\text { for } t>t_{n}. \end{aligned}$$

Since \(\epsilon \) is an arbitrarily positive number and both N and \(\sigma \) are independent of \(\epsilon \), we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\big (h(t)-s_{\mu }t-h^{*}\big )=0. \end{aligned}$$

This proves (1.10).

Some similar arguments as above, (1.9) and (1.11) can also be proved by considering (1.1) with initial function \((u_0(-x),v_0(-x))\). This completes the proof of Theorem 1.3.

4.4 Proof of Theorem 1.4.

Firstly, we will show that \(\lim _{t\rightarrow \infty }\sup _{x\in [g(t),h(t)]}\vert u(t,x)\vert e^{\delta _2 t}=0\) for some \(\delta _2>0\). Since the vanishing of u happens, then \(\lim _{t\rightarrow \infty }(g(t),h(t))=(g_\infty ,h_\infty )\) and \(\lim _{t\rightarrow \infty }(u(t,\cdot ),v(t,\cdot ))=(0,1)\) locally uniformly in any compact set of \(\mathbb {R}\), where \(0<-g_\infty ,h_\infty <\infty \). Hence, for any \(\epsilon >0\), there exists \(T>0\) such that

$$\begin{aligned} u(t,x)<\epsilon ,\ |v(t,x)-1|<\epsilon \text{ for } t\ge T,\ g_\infty \le x\le h_\infty . \end{aligned}$$
(4.68)

As a consequence, the first equation of (1.1) satisfies

$$\begin{aligned} u_{t}-du_{xx}=ku(1-u-bv)\le ku(1-b+b\epsilon -u) \text{ for } t>T,\ g(t)<x<h(t). \end{aligned}$$

For such T, let us consider the following auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{ll} w_t-dw_{xx}=kw(1-b+b\epsilon -w),\ &{} t>1,\ \tilde{g}(t)< x <\tilde{h}(t),\\ w(t,\tilde{g}(t))=w(t,\tilde{h}(t))=0,\ &{}t>1,\\ \tilde{g}'(t)=-\mu w_x(t,\tilde{g}(t)),\ &{}t>1,\\ \tilde{h}'(t)=-\mu w_x(t,\tilde{h}(t)),\ &{}t>1,\\ w(1,x)=u(T,x),\ &{}\tilde{g}(1)=g(T)\le x\le h(T)=\tilde{h}(1). \end{array} \right. \end{aligned}$$
(4.69)

Then the comparison principle implies that

$$\begin{aligned} u(t+T-1,x)\le w(t,x),\ g(t+T-1)\ge \tilde{g}(t) \text{ and } \ h(t+T-1)\le \tilde{h}(t) \end{aligned}$$

for \(t\ge 1\) and \(g(t+T-1)\le x\le h(t+T-1)\).

It was shown in [4] that the assumption \(\tilde{h}(1)-\tilde{g}(1)<\pi \sqrt{\frac{d}{k(1-b+b\epsilon )}}\) is necessary for the vanishing of w. In fact, for any given \(\mu >0\), the vanishing must happen provided that \(\tilde{h}(1)-\tilde{g}(1)<\pi \sqrt{\frac{d}{k(1-b+b\epsilon )}}\) and w(1, x) small enough (see Theorem 3.2 in [6]). In what follows, we only focus on the vanishing of w by taking u(Tx) small enough. To this end, we construct a supersolution for (4.69) as the form

$$\begin{aligned} \overline{g}(t)&=-H_0^-(1+\delta -\delta e^{-\delta _1 t})+H_0^+,\ \overline{h}(t)=H_0^-(1+\delta -\delta e^{-\delta _1 t})+H_0^+, \\&\quad \overline{w}(t,x)=\epsilon _1 e^{-\delta _1 t}\cos (\omega (t,x)), \end{aligned}$$

where

$$\begin{aligned} H_0^\pm :=\frac{1}{2}(h(T)\pm g(T)),\,\omega (t,x)=\frac{\pi (x-H_0^+)}{2(\overline{h}(t)-H_0^+)}, \end{aligned}$$

and \(\delta ,\delta _1,\epsilon _1\) are positive constants to be determined later.

Some direct calculations yield

$$\begin{aligned} -\overline{g}'(t)=\overline{h}'(t)=H_0^-\delta _1\delta e^{-\delta _1 t},\ \mu \overline{w}_x(t,\overline{g}(t))=-\mu \overline{w}_x(t,\overline{h}(t))=\frac{\mu \pi \epsilon _1}{2(\overline{h}(t)-H_0^+)}e^{-\delta _1t}. \end{aligned}$$

Then \(2(\overline{h}(t)-H_0^+)H_0\delta _1\delta \ge \mu \pi \epsilon _1\) for \(t>1\) by taking \(\epsilon _1>0\) sufficiently small, which leads to \(\overline{h}'(t)\ge -\mu \overline{w}_x(t,\overline{h}(t))\) and \(\overline{g}'(t)\le -\mu \overline{w}_x(t,\overline{g}(t))\) for \(t>1\).

Clearly, \(\overline{w}(t,\overline{g}(t))=\overline{w}(t,\overline{h}(t))=0\) for \(t>1\). Note that \([g(T),h(T)]\subsetneqq (\overline{g}(1),\overline{h}(1))\) and \(\overline{w}(1,x)>0\) for \(x\in [g(T),h(T)]\). Since \(\epsilon _1\) is independent of \(\epsilon \), it follows from (4.68) that there exists \(T>0\) sufficiently large such that

$$\begin{aligned} \overline{w}(1,x)=\epsilon _1e^{-\delta _1}\cos (\omega (1,x))\ge \epsilon \ge u(T,x)\text { for }\tilde{g}(1)\le x\le \tilde{h}(1). \end{aligned}$$

A direct calculation yields that, for \(t>1\) and \(\overline{g}(t)<x<\overline{h}(t)\),

$$\begin{aligned} \begin{aligned}&\overline{w}_t-d\overline{w}_{xx}-k\overline{w}(1-b+b\epsilon -\overline{w}) \\&\quad \ge \overline{w}_t-d\overline{w}_{xx}-k(1-b+b\epsilon )\overline{w}\\&\quad =\epsilon _1 e^{-\delta _1 t}\bigg [-\delta _1\cos (\omega (t,x))+\frac{\pi (x-H_0^+)\overline{h}'(t)}{2(\overline{h}(t)-H_0^+)^2}\sin (\omega (t,x))\\&\qquad \quad +\frac{d\pi ^2}{4(\overline{h}(t)-H_0^+)^2}\cos (\omega (t,x))-k(1-b+b\epsilon )\cos (\omega (t,x))\bigg ]\\&\quad \ge \epsilon _1 e^{-\delta _1 t}\cos (\omega (t,x))\Big [-\delta _1+\frac{d\pi ^2}{4(\overline{h}(t)-H_0^+)^2}-k(1-b+b\epsilon )\Big ]\\&\quad \ge \epsilon _1 e^{-\delta _1 t}\cos (\omega (t,x))\Big [-\delta _1+\frac{d\pi ^2}{4(H_0^-)^2(1+\delta )^2}-k(1-b+b\epsilon )\Big ]. \end{aligned} \end{aligned}$$

Since \(H_0^-<\frac{\pi }{2}\sqrt{\frac{d}{k(1-b+b\epsilon )}}\), there exist small positive constants \(\delta _1\) and \(\delta \) such that

$$\begin{aligned} \frac{d\pi ^2}{4k(1-b+b\epsilon )(H_0^-)^2(1+\delta )^2}-1\ge \frac{\delta _1}{k(1-b+b\epsilon )}>0. \end{aligned}$$

Then \(\overline{w}_t-d\overline{w}_{xx}-k\overline{w}(1-b+b\epsilon -\overline{w})\ge 0\) for \(t>1\) and \(\overline{g}(t)<x<\overline{h}(t)\). Thus, one may apply the comparison principle to obtain that, for \(t>1\) and \(g(t+T-1)\le x\le h(t+T-1)\),

$$\begin{aligned} u(t+T-1,x)\le \overline{w}(t,x)=\epsilon _1 e^{-\delta _1 t}\cos (\omega (t,x))\le \epsilon _1 e^{-\delta _1 t}. \end{aligned}$$

Hence, \(\lim _{t\rightarrow \infty }\sup _{x\in [g(t),h(t)]}\vert u(t,x)\vert e^{\delta _2 t}=0\) by choosing \(\delta _2\in (0,\delta _1).\)

Finally, under the assumptions of the initial condition (1.2) and the vanishing of u, we will prove that there is \(\delta _3>0\) such that \(\vert v(t,x)-1\vert e^{\delta _3 t}\rightarrow 1\) globally uniformly in \(\mathbb {R}\) as \(t\rightarrow \infty \). Consider the following auxiliary problem:

$$\begin{aligned} \left\{ \begin{array}{ll} \eta _t-\eta _{xx}=\eta (1-K_1-aK_2), &{} t>0,\ x\in \mathbb {R},\\ \eta (0,x)=v_0(x), &{}x\in \mathbb {R}, \end{array} \right. \end{aligned}$$
(4.70)

where \(K_1:=\max \{1,\Vert u_0\Vert _{\infty }\}\) and \(K_2:=\max \{1,\Vert v_0\Vert _{\infty }\}\). Since \(u(t,x)\le K_1\) and \(v(t,x)\le K_2\), the comparison principle implies that \(v(t,x)\ge \eta (t,x)\) for \((t,x)\in (0,\infty )\times \mathbb {R}\). Note that (4.70) admits a unique solution

$$\begin{aligned} \eta (t,x)=\frac{e^{(1-K_1-aK_2)t}}{2\sqrt{\pi t}}\int ^{\infty }_{-\infty }v_0(\xi )e^{-\frac{(x-\xi )^2}{4t}}d\xi . \end{aligned}$$

In view of \(\liminf _{|x|\rightarrow \infty }v_0(x)>0\), there exist positive constants A large and \(\epsilon _0\) small such that \(v_0(x)>\epsilon _0\) when \(\vert x\vert \ge A\). Then we have

$$\begin{aligned} \begin{aligned} \eta (t,x)&\ge \epsilon _0\frac{e^{(1-K_1-aK_2)t}}{2\sqrt{\pi t}}\bigg (\int _{-\infty }^{-A}e^{-\frac{(x-\xi )^2}{4t}}d\xi +\int _{A}^{\infty }e^{-\frac{(x-\xi )^2}{4t}}d\xi \bigg )\\&=\epsilon _0\frac{e^{(1-K_1-aK_2)t}}{2\sqrt{\pi t}}\bigg (\int _{-\infty }^{-A-x}e^{-\frac{\zeta ^2}{4t}}d\zeta +\int _{A-x}^{\infty }e^{-\frac{\zeta ^2}{4t}}d\zeta \bigg )\\&\ge \epsilon _0\frac{e^{(1-K_1-aK_2)t}}{2\sqrt{\pi t}}\int _{A}^{\infty }e^{-\frac{\zeta ^2}{4t}}d\zeta :=K_3(t), \end{aligned} \end{aligned}$$

which leads to \(\inf _{x\in \mathbb {R}}\eta (t,x)\ge K_3(t)>0\) for all \(t>0\). Now, for given constants \(T>0\) as above and \(M=\epsilon _1e^{\delta _1(T-1)}\), we consider the following two ODE problems:

$$\begin{aligned} \left\{ \begin{array}{ll} V'_1=V_1(1-V_1), &{}t>0,\\ V_1(0)=\Vert v_0\Vert _{\infty } \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} V'_2=V_2(1-V_2-aMe^{-\delta _1 t}), &{}t>0,\\ V_2(0)=K_3(T). \end{array}\right. \end{aligned}$$

Clearly, it follows from the comparison principle that

$$\begin{aligned} V_2(t)\le v(t+T,x)\le V_1(t+T)\text { for }t\ge 0,\ x\in \mathbb {R}. \end{aligned}$$

Moreover, some elementary calculations yield that

$$\begin{aligned} V_1(t)=\frac{\Vert v_0\Vert _{\infty }}{\Vert v_0\Vert _{\infty }+(1-\Vert v_0\Vert _{\infty })e^{-t}}\text { for }t\ge 0 \end{aligned}$$

and

$$\begin{aligned} V_2(t)=\frac{K_3(T) e^{t+\frac{aM}{\delta _1}(e^{-\delta _1 t}-1)}}{1+K_3(T)\int _0^t e^{\tau +\frac{aM}{\delta _1}(e^{-\delta _1 \tau }-1)}d\tau }\text { for }t\ge 0. \end{aligned}$$

It is not hard to verify that \(\lim _{t\rightarrow \infty }\vert V_i(t)-1\vert e^{\delta _3 t}=0\) for any \(\delta _3\in (0,1)\) and \(i=1,2\). Hence, \(\lim _{t\rightarrow \infty }\sup _{x\in \mathbb {R}}\vert v(t,x)-1\vert \text {e}^{\delta _3 t}=0\). In conclusion, the proof of Theorem 1.4 is completed by taking \(\kappa _0\in (0,\min \{\delta _2,\delta _3\}]\).