Dynamics for a Ratio-Dependent Prey–Predator Model with Different Free Boundaries

Abstract

In this paper, we study the dynamics of the ratio-dependent type prey–predator model with different free boundaries. The two free boundaries, determined by prey and predator, respectively, implying that they may intersect with each other as time evolves, are used to describe the spreading of prey and predator. Our primary focus lies in analyzing the long-term behaviors of both predator and prey. We establish sufficient conditions for the spreading and vanishing of prey and predator. Furthermore, in cases where spread occurs, we offer estimates for the asymptotic spreading speeds of prey and predator, denoted as u and v, respectively, as well as the asymptotic speeds of the free boundaries, denoted by h and g. Our findings reveal that when the predator’s speed is lower than that of the prey, it leads to a reduction in the prey’s asymptotic speed.

1. Introduction

The spreading and vanishing of species is a fundamental content in ecology. In order to study this problem, numerous studies have been conducted from a mathematical viewpoint. In this paper, we consider a ratio-dependent prey–predator model with different double free boundaries. This is a significant topic because the following phenomenon continuously occurs in the real world.

There exists a certain species (local species, prey u) in a bounded region (initial habitat, for example, $H_0$), and at some time (initial time, $t=0$), another species (invasive species, predator v) enters a part $G_0$ of $H_0$. Generally, both species tend to migrate from the boundary to acquire their respective new habitats. That is, as time t increases, $H_0$ and $G_0$ will evolve into extended regions $H(t)$ and $G(t)$ with expanding frontiers $\partial H(t)$ and $\partial G(t)$, respectively. For simplicity, we restrict the problem to the one-dimensional case, and consider the left boundaries $H(t)$ and $G(t)$ to be fixed and coincident. Thus, we can take the initial regions $H_0=(0,h_0)$, $G_0=(0,g_0)$ with $0<g_0\le h_0$, and the expanded regions $H(t)=(0,h(t))$, $G(t)=(0,g(t))$, where $h(t)$, $g(t)$ are free boundary functions as follows:

$$h^{\prime }(t)=-\mu u_x(t,h(t)),g^{\prime }(t)=-\rho v_x(t,g(t)),$$

where the positive constants $\mu$ and $\rho$ denote the moving parameters. $h(t)$ and $g(t)$ represent the spreading fronts of prey and predator, respectively. The free boundary conditions are in accordance with the one-phase Stefan condition, which can be employed in the modeling various applied problems. For instance, it has been utilized in studies concerning the melting of ice [1] and wound healing [2]. Furthermore, it can be applied to investigate the spreading of populations [3,4].

The initial functions $u_0(x)$ and $v_0(x)$ will evolve into positive functions $u(t,x)$ and $v(t,x)$, and both $u(t,x)$ and $v(t,x)$ vanish at the moving boundaries $\partial H(t)$ and $\partial G(t)$, respectively. Hence, the ratio-dependent prey–predator model we considered transforms the following problem

$$\left\{\begin{array}{cc}{u}_t-u_{xx}=u(\lambda -u-\frac{bv}{u+mv}),\hfill & t>0,0<x<h(t),\hfill \\ v_t-d{v}_{xx}=v(1-v+\frac{cu}{u+mv}),\hfill & t>0,0<x<g(t),\hfill \\ u_x=v_x=0,\hfill & t\ge 0,x=0,\hfill \\ u(x)=0forx\ge h(t),v(x)=0forx\ge g(t),\hfill & t\ge 0,\hfill \\ h^{\prime }(t)=-\mu u_x(t,h(t)),g^{\prime }(t)=-\rho v_x(t,g(t)),\hfill & t\ge 0,\hfill \\ u(0,x)=u_0(x)in[0,h_0],v(0,x)=v_0(x)in[0,g_0],\hfill \\ h(0)=h_0\ge g(0)=g_0>0,\hfill \end{array}\right.$$

(1)

where $\lambda$, b, m, d, c, $h_0$ ($g_0$) are positive constants that stand for prey intrinsic growth, capturing rate, half capturing saturation constant, dispersal rate, conversion rate, and the initial habitat of the prey (predator), respectively. u and v stand for prey and predator density, respectively. Term $\frac{u}{u+mv}$ is the functional response. The initial functions $u_0(x)$ and $v_0(x)$ satisfy

$$u_0\in C^2([0,h_0]),u_0(x)>0,x\in [0,h_0),$$

$$v_0\in C^2([0,g_0]),v_0(x)>0,x\in [0,g_0).$$

In [3], Du and Lou made the initial attempt to incorporate the Stefan condition into the study of spreading populations. They investigated a diffusive logistic model with a free boundary in one space dimension, establishing a spreading-vanishing dichotomy, namely the species either successfully spreads throughout the new environment or fails to establish itself and eventually dies out. They provided sharp criteria for spreading and vanishing. Du and Guo later extended these conclusions to higher dimensions [4]. In addition, numerous other studies have explored free boundary problems involving a single equation to describe the spreading mechanism of an invasive species. For example, refer to [5,6,7].

Subsequently, the same spreading mechanism has been adopted in the study of multi-species systems. Wang et al. explored two-species prey–predator system [8,9,10,11,12,13,14,15] as well as competition system [9,16,17]. For more related papers, refer to [17,18,19,20,21,22,23,24,25]. In [14], Wang and Zhang studied a similar spreading mechanism to problem (1) but for a diffusive Lotka–Volterra type prey–predator model. The response function of the model mentioned above depends only on prey, which illustrates that the predator’s behavior is determined solely by prey. Nevertheless, there is mounting evidence that, in some specific ecological environments, especially when a predator has to actively seek, share, and plunder prey, a more rational prey–predator model should be ratio-dependent, that is, the response function should depend on the ratio of prey to predator. Hence, we aim to study the ratio-dependent prey–predator model with different double free boundaries in this paper.

The main objective of this paper was to delve into the dynamics of problem (1) and provide insights into its biological significance. Model (1) represents a ratio-dependent prey–predator model with double different boundaries $h(t)$ and $g(t)$, which may intersect with each other due to their separate independence. This may be a challenge to understand the whole dynamics of model (1). This study can be regarded as a continuation of our earlier works [26,27]. In [26], the authors investigated a ratio-dependent prey–predator model with a free boundary problem where the free boundary $h^{\prime }(t)=-\mu u_x$ was solely determined by the prey, and both prey and predator live in a region enclosed by free boundaries. In [27], the authors studied a similar prey–predator model with a free boundary where the free boundary $h^{\prime }(t)=-\mu (u_x+\rho v_x),(\rho >0)$ was influenced by both the prey and predator. Comparing these previous works enables us to gain a better understanding of the dynamics with different free boundaries.

Same to [Theorem 2.1, Lemma 2.1, Theorem 2.2] in [14], for the global existence, uniqueness, and regularity of solution $(u,v,h,g)$, we have the following theorem.

Theorem 1.

The problem (1) has a unique global solution. Additionally, for any $\alpha \in (0,1)$,

$$(u,v,h,g)\in C^{\frac{1+\alpha }{2},1+\alpha }({\overline{D}}_{\infty }^{h(t)})\times C^{\frac{1+\alpha }{2},1+\alpha }({\overline{D}}_{\infty }^{g(t)})\times {\left[C^{1+\frac{\alpha }{2}}([0,T])\right]}^2,$$

where $D_{\infty }^s=\{(t,x)\in {\mathbb{R}}^2:t>0,0<x<s(t)\}$. Moreover,

$$0<u(t,x)\le max\{\lambda ,||u_0{||}_{\infty }\}:=M_1,t>0,0\le x\le h(t),$$

$$0<v(t,x)\le max\{1+c,||v_0{||}_{\infty }\}:=M_2,t>0,0\le x\le g(t),$$

$$0<h^{\prime }(t)\le 2\mu max\{M_1\sqrt{\lambda /2},-\underset{[0,h_0]}{min}{u}_0^{\prime }(x)\}:=M_3,t>0,$$

$$0<g^{\prime }(t)\le 2\rho max\{M_2\sqrt{(1+c)/2d},-\underset{[0,g_0]}{min}{v}_0^{\prime }(x)\}:=M_4,t>0.$$

This paper is organized as follows. In Section 2, we give some fundamental results which will be used in the following sections. Section 3 is concerned with the long-term behaviors of u and v under different conditions. In Section 4, we provide the conditions for the spreading and vanishing of prey and predator. In Section 5, some estimates of the asymptotic spreading speeds of u, v, and asymptotic speeds of g, h are provided. According to the results above, Section 6 makes a summary and gives some practical significance.

2. Preliminaries

We first consider a diffusive logistic problem with a free boundary

$$\left\{\begin{array}{cc}{\omega }_t-d{\omega }_{xx}=\omega (\theta -\omega ),\hfill & t>0,0<x<s(t),\hfill \\ {\omega }_x(t,0)=0,\omega (t,s(t))=0,\hfill & t\ge 0,\hfill \\ s^{\prime }(t)=-\beta {\omega }_x(t,s(t)),\hfill & t\ge 0,\hfill \\ \omega (0,x)={\omega }_0(x),s(0)=s_0,\hfill & 0\le x\le s_0,\hfill \end{array}\right.$$

(2)

where d, $\theta$, $\beta$, and $s_0$ are positive constants.

Proposition 1

([3]). The problem (2) has a unique global solution $(\omega ,s)$ and $\underset{t\to \infty }{lim}s(t)=s_{\infty }$ exists. Additionally:

(i) If $s_0\ge \frac{\pi }{2}\sqrt{\frac{d}{\theta }}$, then $s_{\infty }=\infty$ for all $\beta >0$;

(ii) If $s_0<\frac{\pi }{2}\sqrt{\frac{d}{\theta }}$, then there is a positive constant $\beta (d,\theta ,s_0,{\omega }_0)$ such that $s_{\infty }=\infty$ if $\beta >\beta (d,\theta ,s_0,{\omega }_0)$, and $s_{\infty }<\infty$ if $\beta <\beta (d,\theta ,s_0,{\omega }_0)$;

(iii) If $s_{\infty }=\infty$, then $\underset{t\to \infty }{lim}\omega (t,x)=\theta$ uniformly in any compact subset of $[0,\infty )$.

In addition, it has been proven that the expanding front $s(t)$ moves at a constant speed for large time, i.e., $s(t)=(c+o(1))t$ as $t\to \infty$ in [3]. And the spreading speed c is determined by the following auxiliary elliptic problem

$$\left\{\begin{array}{c}d{q}^{″}-c{q}^{\prime }+q(\theta -q)=0,0<y<\infty ,\hfill \\ q(0)=0,q^{\prime }(0)=c/\beta ,q(\infty )=\theta ,\hfill \\ c\in (0,2\sqrt{\theta d});q^{\prime }(y)>0,0<y<\infty ,\hfill \end{array}\right.$$

(3)

where d, c, $\theta$, $\beta$ are given positive constants.

Proposition 2

([7]). The problem (3) has a unique solution $(q(y),c)$ and $c(\beta ,d,\theta )$ is strictly increasing in β and θ, respectively. Moreover,

$$\underset{\frac{\theta \beta }{d}\to \infty }{lim}\frac{c(\beta ,d,\theta )}{\sqrt{\theta d}}=2,\underset{\frac{\theta \beta }{d}\to 0}{lim}\frac{c(\beta ,d,\theta )}{\sqrt{\theta d}}\frac{d}{\theta \beta }=\frac{1}{\sqrt{3}}.$$

(4)

Proposition 3

([16]). Let $(q,c)$ be the unique solution of (3). If $(\omega ,s)$ is a solution of (2) for which spreading occurs, then there exists $H\in \mathbb{R}$ such that

$$\underset{t\to \infty }{lim}(s(t)-ct-H)=0,\underset{t\to \infty }{lim}{s}^{\prime }(t)=c,$$

$$\underset{t\to \infty }{lim}||\omega (t,x)-{q(ct+H-x)||}_{L^{\infty }([0,s(t)])}=0.$$

3. Long-Term Behaviors of $\mathit{u}$ and $\mathit{v}$

Since the two free boundaries $h(t)$ and $g(t)$ are monotonically increasing and may intersect with each other, there exist four cases: (i) $h_{\infty }=g_{\infty }=\infty$; (ii) $h_{\infty }=\infty$ and $g_{\infty }<\infty$; (iii) $h_{\infty }<\infty$ and $g_{\infty }=\infty$; (iv) $h_{\infty }<\infty$ and $g_{\infty }<\infty$. In this section, we will study the limits of $u(t,x)$ and $v(t,x)$ as $t\to \infty$ under the above four cases.

Theorem 2.

Assume that $h_{\infty }<\infty$$(g_{\infty }<\infty )$. Then

$$\underset{t\to \infty }{lim}\underset{0\le x\le h(t)}{max}u(t,x)=0,(\underset{t\to \infty }{lim}\underset{0\le x\le g(t)}{max}v(t,x)=0).$$

Proof. 

According to [Lemma 4.1] in [14], we can obtain that, if $h_{\infty }<\infty$ ($g_{\infty }<\infty$), then there exists positive constant $K_1$ ($K_2$), such that

$${||u(t,·)||}_{C^1[0,h(t)]}\le K_1{,(||v(t,·)||}_{C^1[0,g(t)]}\le K_2),\forall t>1,$$

(5)

and

$$\underset{t\to \infty }{lim}{h}^{\prime }(t)=0(\underset{t\to \infty }{lim}{g}^{\prime }(t)=0).$$

(6)

We only prove that $\underset{t\to \infty }{lim}\underset{0\le x\le h(t)}{max}u(t,x)=0$ holds. On the contrary, we assume that there exist $\sigma >0$ and ${\left\{(t_j,x_j)\right\}}_{j=1}^{\infty }$ with $0\le x_j<h(t_j)$ and $t_j\to \infty$ as $j\to \infty$ such that

$$u(t_j,x_j)\ge 4\sigma ,j=1,2,\cdots .$$

(7)

Since $x_j<h_{\infty }<\infty$, there exists the subsequence $\left\{x_j\right\}$, denoted by itself, and $x_0\in [0,h_{\infty }]$, such that $x_j\to x_0$ as $j\to \infty$. We affirm that $x_0\ne h_{\infty }$. If $x_0=h_{\infty }$, then $x_j-h(t_j)\to 0$ as $j\to \infty$. According to the inequalities (5) and (7), we have

$$\begin{array}{cc}|\frac{4\sigma }{x_j-h(t_j)}|\hfill & \le |\frac{u(t_j,x_j)}{x_j-h(t_j)}|\hfill \\ & = |\frac{u(t_j,x_j)-u(t_j,h(t_j))}{x_j-h(t_j)}|\hfill \\ & = |u_x(t_j,{\overline{x}}_j)|\hfill \\ & \le K_1,\hfill \end{array}$$

where ${\overline{x}}_j\in (x_j,h(t_j))$. It is a contradiction since $x_j-h(t_j)\to 0$.

For the inequalities (5) and (7), there is $\delta >0$, such that $x_0+\delta <h_{\infty }$ and

$$u(t_j,x)\ge 2\sigma ,\forall x\in [x_0,x_0+\delta ]$$

for all large j. Since $h(t_j)\to h_{\infty }$ ($j\to \infty$), we can assume that $h(t_j)>x_0+\delta$ for all large j. Let

$$r_j(t)=x_0+\delta +t-t_j.$$

Due to $h^{\prime }(t)>0$ and $h_{\infty }<\infty$, there is only one unique number ${\tau }_j>t_j$ such that $h({\tau }_j)=r_j({\tau }_j)$ and $x_0+\delta +{\tau }_j-t_j=r_j({\tau }_j)=h({\tau }_j)<h_{\infty }$. Define

$${\Omega }_j=(t_j,{\tau }_j)\times [x_0,r_j(t)],$$

$$u_j(t,x)=\sigma e^{-k(t-t_j)}(cos{y}_j(t,x)+cos\theta ),(t,x)\in {\overline{\Omega }}_j,$$

$$y_j(t,x)=(\pi -\theta )\frac{2(x-x_0)-(\delta +t-t_j)}{\delta +t-t_j},$$

where $\theta (0<\theta <\pi /8)$ and k are positive constants to be determined. It is easy to obtain $|y_j(t,x)| \le \pi -\theta$ for $(t,x)\in {\overline{\Omega }}_j$. So $u_j(t,x)\ge 0$ for $(t,x)\in {\Omega }_j$.

Next, we compare $u(t,x)$ with $u_j(t,x)$ in $\overline{{\Omega }_j}$. Obviously,

$$u(t,x_0)\ge 0=u_j(t,x_0),u(t,r_j(t))\ge 0=u_j(t,r_j(t)),\forall t\in (t_j,{\tau }_j),$$

$$u(t_j,x)\ge 2\sigma >u_j(t_j,x),\forall x\in [x_0,x_0+\delta ].$$

So if there are positive constants $\theta$ and k independent of j such that

$${(u_j)}_t-{(u_j)}_{xx}-u_j(\lambda -u_j-b/m)\le 0,(x,t)\in {\Omega }_j,$$

(8)

then we have $u\ge u_j$ in ${\Omega }_j$ by the comparison principle. For $h({\tau }_j)=r_j({\tau }_j)$ and $u({\tau }_j,h({\tau }_j))=u_j({\tau }_j,r_j({\tau }_j))=0$, we have

$$u_x({\tau }_j,h({\tau }_j))\le {(u_j)}_x({\tau }_j,r_j({\tau }_j)).$$

Owing to $\theta <\pi /8$ and $\delta +{\tau }_j-t_j<h_{\infty }$, it leads to

$$\begin{array}{c}{(u_j)}_x({\tau }_j,r_j({\tau }_j))=-\frac{2\sigma (\pi -\theta )}{\delta +{\tau }_j-t_j}{e}^{-k({\tau }_j-t_j)}sin(\pi -\theta )\le -\frac{7\sigma \pi }{4h_{\infty }}{e}^{-k{h}_{\infty }}sin\theta .\hfill \end{array}$$

So

$$h^{\prime }({\tau }_j)\ge -\mu {(u_j)}_x({\tau }_j,r_j({\tau }_j))\ge \frac{7\mu \sigma \pi }{4h_{\infty }}{e}^{-k{h}_{\infty }}sin\theta ,$$

which contradicts $\underset{t\to \infty }{lim}{h}^{\prime }(t)=0$. So $\underset{t\to \infty }{lim}\underset{0\le x\le h(t)}{max}u(t,x)=0$ holds.

In the following, we demonstrate that (8) holds as long as $\theta$ and k satisfy

$$\theta <\frac{\pi }{8},sin\theta <\frac{{(\pi -\theta )}^2{\delta }^2}{\pi h_{\infty }^3},$$

(9)

$$k>\frac{2\pi h_{\infty }}{{\delta }^2(cos\theta -cos2\theta )}+{(\frac{2\pi }{\delta })}^2+2\sigma +\frac{b}{m}.$$

(10)

Remember that $0\le u_j\le 2\sigma$ and $\delta +{\tau }_j-t_j<h_{\infty }$, so for $(t,x)\in \Omega$,

$$\begin{array}{c}{(u_j)}_t-{(u_j)}_{xx}-u_j(\lambda -u_j-\frac{b}{m})\hfill \\ =-k{u}_j-\sigma e^{-k(t-t_j)}{(y_j)}_{t}sin{y}_j+\sigma e^{-k(t-t_j)}{(y_j)}_x^{2}cos{y}_j-u_j(\lambda -u_j-\frac{b}{m})\hfill \\ \le (-k+{(\frac{2\pi }{\delta })}^2+2\sigma +\frac{b}{m})u_j-\sigma e^{-k(t-t_j)}[{(\frac{2(\pi -\theta )}{h_{\infty }})}^{2}cos\theta -\frac{2\pi (x-x_0)}{{\delta }^2}sin{y}_j]\hfill \\ =\sigma e^{-k(t-t_j)}·I\hfill \\ :=\Pi ,\hfill \end{array}$$

where $I=[(-k+{(\frac{2\pi }{\delta })}^2+2\sigma +\frac{b}{m})(cos{y}_j+cos\theta )-{(\frac{2(\pi -\theta )}{h_{\infty }})}^{2}cos\theta +\frac{2\pi (x-x_0)}{{\delta }^2}sin{y}_j]$. Noticing that $-(\pi -\theta )\le y_j\le \pi -\theta$ in ${\overline{\Omega }}_j$, we can decompose ${\Omega }_j={\Omega }_j^1\bigcup {\Omega }_j^2$, where

$$\begin{array}{c}{\Omega }_j^1=\{(t,x)\in {\Omega }_j:t_j<t<{\tau }_j,\pi -2\theta <|y_j(t,x)|\le \pi -\theta \},\hfill \\ {\Omega }_j^2=\{(t,x)\in {\Omega }_j:t_j<t<{\tau }_j,|y_j(t,x)|\le \pi -2\theta \}.\hfill \end{array}$$

It is easy to obtain $|sin{y}_j|\le sin2\theta$ in ${\Omega }_j^1$ and $cos{y}_j\ge -cos2\theta$ in ${\Omega }_j^2$. Due to $x-x_0\le h_{\infty }$ in ${\Omega }_j$ and (9) and (10), we have when $(t,x)\in {\Omega }_j^1$,

$$\Pi \le \sigma e^{-k(t-t_j)}[-{(\frac{2(\pi -\theta )}{h_{\infty }})}^{2}cos\theta +\frac{2\pi h_{\infty }}{{\delta }^2}sin2\theta ]<0;$$

and when $(t,x)\in {\Omega }_j^2$,

$$\Pi \le \sigma e^{-k(t-t_j)}[(-k+{(\frac{2\pi }{\delta })}^2+2\sigma +\frac{b}{m})(cos\theta -cos2\theta )+\frac{2\pi h_{\infty }}{{\delta }^2}]<0.$$

So, (8) holds. Thus, $\underset{t\to \infty }{lim}\underset{0\le x\le h(t)}{max}u(t,x)=0$ is proved. In the same way, $\underset{t\to \infty }{lim}\underset{0\le x\le g(t)}{max}v(t,x)$ $=0$ can be also obtained. □

The author had researched a similar case when $h_{\infty }=g_{\infty }=\infty$ in [Theorem 3.3] in [26]. So, we directly present the results as follows.

Theorem 3.

Assume that $m\lambda -b>max\{2b-m^2,0\}$ and $h_{\infty }=g_{\infty }=\infty$. Then, $\underset{t\to \infty }{lim}(u(t,x),v(t,x))$ is determined by

$$\left\{\begin{array}{c}\lambda -u-\frac{bv}{u+mv}=0,\hfill \\ 1-v+\frac{cu}{u+mv}=0.\hfill \end{array}\right.$$

(11)

Further calculations give, when $m\lambda -b<b/c$,

$$\underset{t\to \infty }{lim}u(t,x)=u^{*}:=\frac{A+\sqrt{{\Delta }_1}}{2(b+c{m}^2)},\underset{t\to \infty }{lim}v(t,x)=v^{*}:=\frac{u^{*}(\lambda -u^{*})}{b-m(\lambda -u^{*})},$$

where $A=\lambda (2c{m}^2+b)-mb(1+2c)$, ${\Delta }_1=A^2+4(b+c{m}^2)[b(1+c)-mc\lambda ](m\lambda -b)$.

Theorem 4.

(i) If $h_{\infty }=\infty$ and $g_{\infty }<\infty$, then

$$\underset{t\to \infty }{lim}u(t,x)=\lambda uniformlyinanycompactsubsetof[0,\infty ).$$

(ii) If $h_{\infty }<\infty$ and $g_{\infty }=\infty$, then

$$\underset{t\to \infty }{lim}v(t,x)=1uniformlyinanycompactsubsetof[0,\infty ).$$

Proof. 

We only prove case (i) as proof of (ii) can be verified in the same way. According to the conclusions of the logistic equation and comparison principle, we have

$$\lambda -b/m\le \underset{t\to \infty }{lim\; inf}u(t,x),\underset{t\to \infty }{lim\; sup}u(t,x)\le \lambda$$

(12)

uniformly in any compact subset of $[0,\infty )$ when $h_{\infty }=\infty$. Note that $\underset{t\to \infty }{lim}\underset{0\le x\le g(t)}{max}v(t,x)=0$ for $g_{\infty }<\infty$ by Theorem 2 and $v(t,x)=0$ for $x>g(t)$. Owing to the former inequality of (12), for any given $0<\epsilon \ll 1$, there exists $T\gg 1$ such that $v(t,x)<\epsilon$ for all $t\ge T$ and $x\ge 0$. Then, u satisfies

$$\left\{\begin{array}{cc}{u}_t-d{u}_{xx}\ge (\lambda -\frac{\epsilon b}{\lambda -b/m+\epsilon m})u-u^2,\hfill & t\ge T,0<x<h(t),\hfill \\ u_x(t,0)=0,u(t,h(t))=0,\hfill & t\ge T,\hfill \end{array}\right.$$

which implies that $\underset{t\to \infty }{lim\; inf}u(t,x)\ge \lambda -\frac{\epsilon b}{\lambda -b/m+\epsilon m}$ uniformly in any compact subset of $[0,\infty )$. By the arbitrariness of $\epsilon$, we have

$$\underset{t\to \infty }{lim\; inf}u(t,x)\ge \lambda uniformlyinanycompactsubsetof[0,\infty ).$$

Combining with the latter inequality of (12), we can obtain case (i). □

4. Conditions for Spreading and Vanishing

In any case, the following inequalities are always true

$$\left\{\begin{array}{c}max\{0,\lambda -\frac{b}{m}\}u-u^2\le \lambda u-u^2-\frac{buv}{u+mv}\le \lambda u-u^2,\hfill \\ v-v^2\le v-v^2+\frac{cuv}{u+mv}\le (1+c)v-v^2.\hfill \end{array}\right.$$

(13)

Denote

$${\mu }^{*}:=\beta (1,\lambda ,h_0,u_0),{\mu }_{*}:=\beta (1,\lambda -b/m,h_0,u_0),$$

and

$${\rho }^{*}:=\beta (d,1+c,g_0,v_0),{\rho }_{*}:=\beta (d,1,g_0,v_0).$$

We first give necessary conditions for spreading and vanishing.

Theorem 5.

Assume that $m\lambda >b$.

(i) If $h_0<\frac{\pi }{2}\sqrt{\frac{1}{\lambda }}$ and $\mu \le {\mu }^{*}$, then $h_{\infty }<\infty$;

(ii) If $h_0\ge \frac{\pi }{2}\sqrt{\frac{m}{m\lambda -b}}$ or $h_0<\frac{\pi }{2}\sqrt{\frac{m}{m\lambda -b}}$ and $\mu >{\mu }_{*}$, then $h_{\infty }=\infty$;

(iii) If $g_0<\frac{\pi }{2}\sqrt{\frac{d}{1+c}}$ and $\rho \le {\rho }^{*}$, then $g_{\infty }<\infty$;

(iv) If $g_0\ge \frac{\pi }{2}\sqrt{d}$ or $g_0<\frac{\pi }{2}\sqrt{d}$ and $\rho >{\rho }_{*}$, then $g_{\infty }=\infty$.

Proof. 

Consider the following auxiliary problem

$$\left\{\begin{array}{cc}{\overline{u}}_t-{\overline{u}}_{xx}=\lambda \overline{u}-{\overline{u}}^2,\hfill & t>0,0<x<\overline{h}(t),\hfill \\ {\overline{u}}_x(t,0)=0,\overline{u}(t,\overline{h}(t))=0,\hfill & t\ge 0,\hfill \\ {\overline{h}}^{\prime }(t)=-\mu {\overline{u}}_x(t,\overline{h}(t)),\hfill & t\ge 0,\hfill \\ \overline{u}(0,x)=u(0,x),\overline{h}(0)=h(0),\hfill & 0\le x\le \overline{h}(t).\hfill \end{array}\right.$$

It follows from the comparison principle that

$$\overline{h}(t)\ge h(t),\overline{u}(t,x)\ge u(t,x)fort>0,h_0<x<\overline{h}(t).$$

Recall that, if $h_0<\frac{\pi }{2}\sqrt{\frac{1}{\lambda }}$ and $\mu <{\mu }^{*}$, then ${\overline{h}}_{\infty }<\infty$ by Proposition 1(ii). Therefore, $h_{\infty }<\infty$. (i) is proved.

On the other hand, we construct another auxiliary problem as follows

$$\left\{\begin{array}{cc}{\underline{u}}_t-{\underline{u}}_{xx}=(\lambda -\frac{b}{m})\underline{u}-{\underline{u}}^2,\hfill & t>0,0<x<\underline{h}(t),\hfill \\ {\underline{u}}_x(t,0)=0,\underline{u}(t,\underline{h}(t))=0,\hfill & t\ge 0,\hfill \\ {\underline{h}}^{\prime }(t)=-\mu {\underline{u}}_x(t,\underline{h}(t)),\hfill & t\ge 0,\hfill \\ \underline{u}(0,x)=u(0,x),\underline{h}(0)=h(0),\hfill & 0\le x\le \underline{h}(t).\hfill \end{array}\right.$$

By the comparison principle, we have

$$\underline{h}(t)\le h(t),\underline{u}(t,x)\le u(t,x)fort>0,h_0<x<\underline{h}(t).$$

By using Proposition 1(i)(ii), it is easy to obtain that, if $h_0\ge \frac{\pi }{2}\sqrt{\frac{m}{m\lambda -b}}$, then ${\underline{h}}_{\infty }=\infty$; And if $h_0<\frac{\pi }{2}\sqrt{\frac{m}{m\lambda -b}}$ and $\mu >{\mu }_{*}$, then ${\underline{h}}_{\infty }=\infty$. Hence, $h_{\infty }=\infty$. (ii) is proved.

Similarly, let

$$\left\{\begin{array}{cc}{\overline{v}}_t-d{\overline{v}}_{xx}=(1+c)\overline{v}-{\overline{v}}^2,\hfill & t>0,0<x<\overline{g}(t),\hfill \\ {\overline{v}}_x(t,0)=0,\overline{v}(t,\overline{g}(t))=0,\hfill & t\ge 0,\hfill \\ {\overline{g}}^{\prime }(t)=-\rho {\overline{v}}_x(t,\overline{g}(t)),\hfill & t\ge 0,\hfill \\ \overline{v}(0,x)=v(0,x),\overline{g}(0)=g(0),\hfill & 0\le x\le \overline{g}(t).\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{cc}{\underline{v}}_t-{\underline{v}}_{xx}=\underline{v}-{\underline{v}}^2,\hfill & t>0,0<x<\underline{g}(t),\hfill \\ {\underline{v}}_x(t,0)=0,\underline{v}(t,\underline{g}(t))=0,\hfill & t\ge 0,\hfill \\ {\underline{g}}^{\prime }(t)=-\rho {\underline{v}}_x(t,\underline{g}(t)),\hfill & t\ge 0,\hfill \\ \underline{v}(0,x)=v(0,x),\underline{g}(0)=g(0),\hfill & 0\le x\le \underline{g}(t).\hfill \end{array}\right.$$

By use of the comparison principle and Proposition 1(i)(ii), Theorem 5(iii)(iv) can be proved. □

Next, we will study the relationship between $h(t)$ and $g(t)$ when $t\to \infty$. In view of Theorem 1, for a constant $s\ge M_4$, we have $0<g(t)\le st+g_0$.

Theorem 6.

Let λ, μ be fixed. Then, there exists $0<\overline{s}<\sqrt{2\lambda }$ such that when

$$0<s<\overline{s},h_0-g_0>\frac{2\pi }{\sqrt{2\lambda -s^2}}:=L_s,$$

we have $h(t)\ge st+g_0+L_s$, which implies $h(t)>g(t)$ for all $t\ge 0$ and $h_{\infty }\to \infty$ as $t\to \infty$.

Proof. 

Inspired by [Theorem 5.3] in [14], for $h(t)>st+g_0$, we define

$$y=x-st-g_0,\phi (t,y)=u(t,x),\eta (t)=h(t)-st-g_0.$$

Then, $\phi (t,y)>0$ for $t\ge 0$ and $0\le y<\eta (t)$. Evidently, $v\ge 0$ implies that $x\ge g(t)$. Note that $v(t,x)=0$ for $x\ge g(t)$, so $\phi$ satisfies

$$\left\{\begin{array}{cc}{\phi }_t-{\phi }_{xx}-s{\phi }_y=\phi (\lambda -\phi ),\hfill & t>0,0<y<\eta (t),\hfill \\ \phi (t,0)=u(t,st+g_0),\phi (t,\eta (t))=0,\hfill & t\ge 0,\hfill \\ \phi (0,y)=u_0(y+g_0),\hfill & 0\le y\le h_0-g_0.\hfill \end{array}\right.$$

Let $\sigma$ be the principle eigenvalue of

$$\left\{\begin{array}{cc}-{\varphi }^{″}-s{\varphi }^{\prime }-\lambda \varphi =\sigma \varphi ,\hfill & 0<x<L,\hfill \\ \varphi (0)=\varphi (L)=0.\hfill \end{array}\right.$$

(14)

Then, the relation between $\sigma$ and L satisfies

$$\frac{\pi }{L}=\frac{\sqrt{4(\lambda +\sigma )-s^2}}{2}.$$

For $0<s<\sqrt{2\lambda }$, choose $\sigma =-\lambda /2$ and define

$$L_s=\frac{2\pi }{\sqrt{2\lambda -s^2}},\varphi (y)=e^{-\frac{s}{2}y}sin\frac{\pi }{L_s}y,$$

and then $(L_s,\varphi )$ satisfies (14) with $\sigma =-\frac{\lambda }{2}$ and $L=L_s$. Assume that $h_0-g_0>L_s$ and define

$${\delta }_s=min\left\{\underset{(0,L_s)}{inf}\frac{\phi (0,y)}{\varphi (y)},\frac{\lambda }{2}\underset{(0,L_s)}{inf}\frac{1}{\varphi (y)}\right\},\psi (y)={\delta }_s\varphi (y).$$

Then, $0<{\delta }_s<\infty$ and $\psi (y)\le \phi (0,y)$ in $[0,L_s]$. Moreover, $\psi (y)$ satisfies

$$\left\{\begin{array}{cc}-{\psi }^{″}-s{\psi }^{\prime }\le \psi (\lambda -\psi ),\hfill & 0<x<L_s,\hfill \\ \psi (0)=\psi (L_s)=0.\hfill \end{array}\right.$$

Take a maximal $\overline{s}\in (0,\sqrt{2\lambda })$ so that

$$s<\mu {\delta }_s\frac{\pi }{L_s}{e}^{-\frac{s}{2}{L}_s},\forall s\in (0,\overline{s}).$$

(15)

For any given $s\in (0,\overline{s})$, we claim that $\eta (t)>L_s$ for $t\ge 0$, which leads to

$$h(t)\ge st+g_0+L_s\to \infty ,t\to \infty .$$

Note that $\eta (0)=h_0-g_0>L_s$, if our claim is invalid, then we can find a $t_0>0$ such that $\eta (t)>L_s$ for all $0<t<t_0$ and $\eta (t_0)=L_s$. So ${\eta }^{\prime }(t_0)\le 0$ and $h^{\prime }(t_0)\le s$. Additionally, by the comparison principle, we have $\phi (t,y)\ge \psi (y)$ in $[0,t_0]\times [0,L_s]$. So, we have $\phi (t_0,y)\ge \psi (y)$ in $[0,L_s]$. Thanks to $\phi (t_0,L_s)=\phi (t_0,\eta (t_0))=0=\psi (L_s)$, we obtain

$${\phi }_y(t_0,L_s)\le {\psi }^{\prime }(L_s)=-{\delta }_s\frac{\pi }{L_s}{e}^{-\frac{s}{2}{L}_s}.$$

A direct calculation deduces that

$$\begin{array}{cc}\hfill s& \ge h^{\prime }(t_0)=-\mu u_x(t_0,h(t_0))=-\mu {\phi }_y(t_0,\eta (t_0))=-\mu {\phi }_y(t_0,L_s)\hfill \\ & \ge \mu {\delta }_s\frac{\pi }{L_s}{e}^{-\frac{s}{2}{L}_s},\hfill \end{array}$$

which contradicts (15). The proof is complete. □

Combining Theorem 5(iv) and Theorem 6, we obtain the following remark immediately.

Remark 1.

Assume that λ, μ, s, $h_0$, $g_0$ satisfy the conditions of Theorem 6. If $g_0\ge \frac{\pi }{2}\sqrt{d}$, then $g_{\infty }=\infty$, which implies that $h_{\infty }=\infty$ for all $\mu >0$.

According to Proposition 2, we have

$$\underset{\mu \to 0}{lim}c(\mu ,1,\lambda )=0,\underset{\rho \to \infty }{lim}c(\rho ,d,1)=2\sqrt{d}.$$

Note that $c(\beta ,d,\theta )$ is strictly increasing in $\beta$, so there exist positive constants $\overline{\mu }$ and $\overline{\rho }$ such that $c(\mu ,1,\lambda )<c(\rho ,d,1)$ for all $0<\mu \le \overline{\mu }$ and $\rho \ge \overline{\rho }$. Define

$$\mathcal{F}:=\{(\mu ,\rho ):\mu ,\rho >0,c(\mu ,1,\lambda )<c(\rho ,d,1)\},$$

then $(0,\overline{\mu }]\times [\overline{\rho },\infty )\subset \mathcal{F}$.

Theorem 7.

Assume that $(\mu ,\rho )\in \mathcal{F}$. If ${\lambda }^2+m\lambda <b$ and $g_{\infty }=\infty$, then $h_{\infty }<\infty$.

Proof. 

As a result of ${\lambda }^2+m\lambda <b$, there exists $0<\epsilon \ll 1$ such that

$${\lambda }^2+(m\lambda -b)(1-\epsilon )\le 0.$$

(16)

Owing to $g_{\infty }=\infty$, there exists $t_1>0$ such that $g(t_1)>\frac{\pi }{2}\sqrt{d}$. Let $(\underline{v},\underline{g})$ be the unique solution of

$$\left\{\begin{array}{cc}{\underline{v}}_t-d{\underline{v}}_{xx}=\underline{v}-{\underline{v}}^2,\hfill & t>t_1,0<x<\underline{g}(t),\hfill \\ {\underline{v}}_x(t,0)=0,\underline{v}(t,\underline{g}(t))=0,\hfill & t>t_1,\hfill \\ {\underline{g}}^{\prime }=-\rho {\underline{v}}_x(t,\underline{g}(t)),\hfill & t\ge t_1,\hfill \\ \underline{v}(t_1,x)=v_0(t_1,x),\underline{g}(t_1)=g(t_1),\hfill & 0\le x\le g(t_1).\hfill \end{array}\right.$$

By the comparison principle, it yields that $v(t,x)\ge \underline{v}(t,x)$, $g(t)\ge \underline{g}(t)$ for $t\ge t_1$, $0\le x\le \underline{g}(t)$. It follows from Proposition 1(i) that $\underline{g}(\infty )=\infty$. In addition, by Proposition 3, there exists ${\mathcal{S}}_1\in \mathbb{R}$ such that

$$\underline{g}(t)-\tilde{c}t\to {\mathcal{S}}_1,||\underline{v}(t,x)-q_1(\tilde{c}t+{\mathcal{S}}_1-x){||}_{L^{\infty }([0,\underline{g}(t)])}\to 0,ast\to \infty ,$$

(17)

where $(q_1(y),\tilde{c})$ is the unique solution of (3) with $(\beta ,d,\theta )=(\rho ,d,1)$, i.e., $\tilde{c}=c(\rho ,d,1)$.

In view of $(\mu ,\rho )\in \mathcal{F}$, we have $\tilde{c}>c(\mu ,1,\lambda )$ which implies that $\underline{g}(t)-h(t)\ge \underline{g}(t)-\overline{h}(t)\to \infty$. So, we obtain that $\underset{0\le x\le h(t)}{min}{q}_1(\tilde{c}t+{\mathcal{S}}_1-x)\to 1$ as $t\to \infty$. Thus, by (17), we have $\underset{t\to \infty }{lim}\underset{0\le x\le h(t)}{min}\underline{v}(t,x)=1$. There exists $t_2>t_1$ such that

$$\underline{v}(t,x)>1-\epsilon ,t\ge t_2,0\le x\le h(t).$$

(18)

Besides, it follows from Theorem 1 that $u(t,x)\le \lambda$ for $t>t_2$, $0\le x\le h(t)$. By the comparison principle and (16), we then have

$$\begin{array}{cc}\hfill \lambda -u-\frac{bv}{u+mv}& <\lambda -\frac{b(1-\epsilon )}{\lambda +m(1-\epsilon )}-u\hfill \\ \hfill & =\frac{{\lambda }^2+(m\lambda -b)(1-\epsilon )}{\lambda +m(1-\epsilon )}-u\hfill \\ \hfill & <0\hfill \end{array}$$

(19)

for all $t\ge t_2$ and $0\le x\le h(t)$. Direct calculation gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{\int }_0^{h(t)}u(t,x)dx& ={\int }_0^{h(t)}{u}_t(t,x)dx+h^{\prime }(t)u(t,h(t))\hfill \\ & ={\int }_0^{h(t)}{u}_{xx}(t,x)dx+{\int }_0^{h(t)}(\lambda u-u^2-\frac{buv}{u+mv})dx\hfill \\ & =-\frac{1}{\mu }{h}^{\prime }(t)+{\int }_0^{h(t)}(\lambda u-u^2-\frac{buv}{u+mv})dx.\hfill \end{array}$$

By firstly integrating from $t_2$ to t and secondly applying (19), we have

$$\begin{array}{cc}\hfill {\int }_0^{h(t)}u(t,x)dx& ={\int }_0^{h(t_2)}u(t_2,x)+\frac{1}{\mu }(h(t_2)-h(t))+{\int }_{t_2}^t{\int }_0^{h(\tau )}(\lambda u-u^2-\frac{buv}{u+mv})dxd\tau \hfill \\ & \le {\int }_0^{h(t_2)}u(t_2,x)+\frac{1}{\mu }(h(t_2)-h(t)),\hfill \end{array}$$

which implies

$$h(t)\le h(t_2)+\mu {\int }_0^{h(t_2)}u(t_2,x)dx.$$

Hence, $h_{\infty }<\infty$. This completes the proof. □

5. Asymptotic Spreading Speeds of $\mathit{u}$, $\mathit{v}$ and Asymptotic Speeds of $\mathit{g}$, $\mathit{h}$

This section is devoted to dealing with the asymptotic spreading speeds of u, v, and asymptotic speeds of g, h when spreading occurs ($h_{\infty }=g_{\infty }=\infty$). In this section, we always assume that $m\lambda >b$.

According to (13), Proposition 2 and the comparison principle, we have

$$\underset{t\to \infty }{lim\; sup}\frac{h(t)}{t}\le c(\mu ,1,\lambda ):={\overline{c}}_{\mu },\underset{t\to \infty }{lim\; inf}\frac{h(t)}{t}\ge c(\mu ,1,\lambda -b/m):={\underline{c}}_{\mu },$$

(20)

$$\underset{t\to \infty }{lim\; sup}\frac{g(t)}{t}\le c(\rho ,d,1+c):={\overline{c}}_{\rho },\underset{t\to \infty }{lim\; inf}\frac{g(t)}{t}\ge c(\rho ,d,1):={\underline{c}}_{\rho }.$$

(21)

Denote

$$c_1=2\sqrt{\lambda -b/m},c_2=2\sqrt{\lambda },c_3=2\sqrt{d},c_4=2\sqrt{d(1+c)}.$$

Theorem 8.

For any given $0<\epsilon \ll 1$, there exist ${\mu }_{\epsilon }$, ${\rho }_{\epsilon }$, $T\gg 1$ such that when $\rho \ge {\rho }_{\epsilon }$ and $\mu \ge {\mu }_{\epsilon }$,

$$u(t,x)=0fort\ge T,x\ge (c_2+\epsilon )t,$$

(22)

$$v(t,x)=0fort\ge T,x\ge (c_4+\epsilon )t,$$

(23)

$$\underset{t\to \infty }{lim\; inf}\underset{0\le x\le (c_1-\epsilon )t}{min}u(t,x)\ge \lambda -b/m,$$

(24)

$$\underset{t\to \infty }{lim\; inf}\underset{0\le x\le (c_3-\epsilon )t}{min}v(t,x)\ge 1.$$

(25)

Proof. 

It follows from Proposition 2 that

$$\underset{\mu \to \infty }{lim}{\underline{c}}_{\mu }=c_1,\underset{\mu \to \infty }{lim}{\overline{c}}_{\mu }=c_2,\underset{\rho \to \infty }{lim}{\underline{c}}_{\rho }=c_3,\underset{\rho \to \infty }{lim}{\overline{c}}_{\rho }=c_4.$$

Take notice of (20) and (21), so for any given $0<\epsilon \ll 1$, there exist ${\mu }_{\epsilon }$, ${\rho }_{\epsilon }\gg 1$ such that

$$c_1-\epsilon /2<{\underline{c}}_{\mu }\le \underset{t\to \infty }{lim\; inf}\frac{h(t)}{t},\underset{t\to \infty }{lim\; sup}\frac{h(t)}{t}\le {\overline{c}}_{\mu }<c_2+\epsilon /4,$$

(26)

$$c_3-\epsilon /2<{\underline{c}}_{\rho }\le \underset{t\to \infty }{lim\; inf}\frac{g(t)}{t},\underset{t\to \infty }{lim\; sup}\frac{g(t)}{t}\le {\overline{c}}_{\rho }<c_4+\epsilon /2.$$

(27)

Thus, there exists a ${\tau }_1\gg 1$ such that, for all $t\ge {\tau }_1$, $\mu \ge {\mu }_{\epsilon }$, and $\rho \ge {\rho }_{\epsilon }$,

$$(c_1-\epsilon )t<h(t)<(c_2+\epsilon /2)t,(c_3-\epsilon )t<g(t)<(c_4+\epsilon )t.$$

(28)

It is obvious that (22) and (23) hold.

The proof of (24). Let $(\underline{u},\underline{h})$ be the unique solution of (2) with $(\rho ,d,a,b)=(\rho ,1,\lambda -b/m,1)$. Then, $h(t)\ge \underline{h}(t)$, $u(t,x)\ge \underline{u}(t,x)$ for $t>0$ and $0<x\le \underline{h}(t)$ by Proposition 1. Make use of [Theorem 3.1] in [16], we obtain

$$\underset{t\to \infty }{lim}(\underline{h}(t)-c^{*}t)=H\in \mathbb{R},\underset{t\to \infty }{lim}||\underline{u}(t,x)-q^{*}(c^{*}t+H-x){||}_{L^{\infty }([0,\underline{h}(t)])}=0,$$

where $(q^{*}(y),c^{*})$ is the unique solution of (3) with $(\beta ,d,\theta )=(\rho ,1,\lambda -b/m)$, i.e., $c^{*}=c(\rho ,1,\lambda -b/m)>c_1-\frac{\epsilon }{2}$. Obviously, ${min}_{[0,(c_1-\frac{\epsilon }{2})t]}(c^{*}t+H-x)\to \infty$ as $t\to \infty$. Owing to $q^{*}(y)\nearrow \lambda -b/m$ as $y\nearrow \infty$, we have ${min}_{[0,(c_1-\frac{\epsilon }{2})t]}{q}^{*}(c^{*}t+H-x)\to \lambda -b/m$ as $t\to \infty$. So ${min}_{[0,(c_1-\frac{\epsilon }{2})t]}\underline{u}(t,x)\to \lambda -b/m$ as $t\to \infty$. Thus, (24) holds due to $u(t,x)\ge \underline{u}(t,x)$ for $t>0$ and $0<x\le \underline{h}(t)$.

Similarity, we can prove that (25) holds. □

In the following, we give a further study on u, h and v, g.

Theorem 9.

If $d(1+c)<\lambda -b/m$, then for any given $\epsilon >0$, there exist ${\rho }_{\epsilon }$, ${\mu }_{\epsilon }$, $T\gg 1$ such that, when $\rho \ge {\rho }_{\epsilon }$ and $\mu \ge {\mu }_{\epsilon }$, we have

$$\underset{\mu \to \infty }{lim}\underset{t\to \infty }{lim}\frac{h(t)}{t}\ge c_5$$

(29)

and

$$\underset{t\to \infty }{lim}\underset{0\le x\le (c_5-\epsilon )t}{max}u(t,x)>0,$$

(30)

where $c_5$ is given by (32).

Proof. 

The inequality $d(1+c)<\lambda -b/m$ implies $c_3<c_4<c_1<c_2$. Choose $\epsilon >0$ such that $c_4+\epsilon <c_1-\epsilon$. So

$$g(t)<c_4+\epsilon <c_1-\epsilon <h(t),\forall \rho \ge {\rho }_{\epsilon },\mu \ge {\mu }_{\epsilon },\tau >{\tau }_1.$$

(31)

Note that $v\le M_2$, where $M_2$ is given by Theorem 1. By (24), there exists ${\tau }_2\gg 1$ such that $v(t,x)\le \frac{2M_2}{\lambda -b/m}u(t,x)$ for all $t\ge {\tau }_2$ and $0\le x\le (c_1-\epsilon )t$. Additionally, by (23) and (31), we have $v(t,x)=0$ for $t\ge 0$ and $x\ge (c_1-\epsilon )t$. Denote $\kappa :=\frac{2M_2}{\lambda -b/m}$ and let $(\underline{u},\underline{h})$ be the unique solution of

$$\left\{\begin{array}{cc}{\underline{u}}_t-{\underline{u}}_{xx}=(\lambda -\frac{b\kappa }{1+m\kappa })\underline{u}-{\underline{u}}^2,\hfill & t>{\tau }_2,0<x<\underline{h}(t),\hfill \\ {\underline{u}}_x(t,0)=0,\underline{u}(t,\underline{h}(t))=0,\hfill & t\ge {\tau }_2,\hfill \\ {\underline{h}}^{\prime }(t)=-\mu {\underline{u}}_x(t,\underline{h}(t)),\hfill & t\ge {\tau }_2,\hfill \\ \underline{u}({\tau }_2,x)=u({\tau }_2,x),\underline{h}({\tau }_2)=h({\tau }_2),\hfill & 0\le x\le \underline{h}(t).\hfill \end{array}\right.$$

By the comparison principle, we have $\underline{u}(t,x)\le u(t,x)$ and $\underline{h}(t)\le h(t)$ for $t\ge {\tau }_2$ and $0\le x\le \underline{h}(t)$. By Proposition 2, we have

$$\underset{\mu \to \infty }{lim}\underset{t\to \infty }{lim}\frac{\underline{h}(t)}{t}=\underset{\mu \to \infty }{lim}c(\mu ,1,\lambda -\frac{b\kappa }{1+m\kappa })=2\sqrt{\lambda -\frac{b\kappa }{1+m\kappa }}:=c_5,$$

(32)

which implies (29) holds. Similarly to the proof of (24), we can show that

$$\underset{t\to \infty }{lim\; inf}\underset{0\le x\le (c_5-\epsilon )t}{min}u(t,x)\ge \frac{b\kappa }{1+m\kappa },$$

which leads to (30). □

Theorem 10.

Suppose $\lambda <d$. Then, the following holds:

(i) For any given $\epsilon >0$, there exists ${\mu }_{\epsilon }$, $T\ge 1$ such that, when $\mu \ge {\mu }_{\epsilon }$,

$$\underset{t\to \infty }{lim}\underset{x\ge (c_3+\epsilon )t}{sup}v(t,x)=0.$$

(33)

Furthermore, if $d\le 2\sqrt{\lambda }+1$, then

$$\underset{t\to \infty }{lim}\underset{(c_2+\epsilon )t\le x\le (c_3-\epsilon )t}{max}|v(t,x)-1|=0.$$

(34)

(ii) There exists ${\rho }_0\gg 1$ such that when $\rho >{\rho }_0$,

$$\underset{\rho \to \infty }{lim}\underset{t\to \infty }{lim}\frac{g(t)}{t}=c_3.$$

(35)

Proof. 

(i) The assumption $\lambda <d$ implies $c_1<c_2<c_3<c_4$. So, there exist ${\tau }_3\gg 1$ and ${\mu }_{*}\gg 1$ such that $h(t)\le c_{3}t$ for all $t\ge {\tau }_3$ and $\mu \ge {\mu }_{*}$, which implies that $u(t,x)=0$ for all $t\ge {\tau }_3$, $x\ge c_{3}t$ and $\mu \ge {\mu }_{*}$. Choose $0<\epsilon \ll 1$ and define

$$r(t)=max\{(c_3+\epsilon )t,g(t)\},fort\ge {\tau }_3.$$

Remember that $v(t,x)=0$ for $x\ge g(t)$ and $v_x(t,g(t))<0$, so it is easy to verify that v satisfies, in the weak sense,

$$\left\{\begin{array}{cc}{v}_t-d{v}_{xx}\le v-v^2,\hfill & t\ge {\tau }_3,c_{3}t\le x\le r(t),\hfill \\ v(t,x)\le M_2,\hfill & t\ge {\tau }_3,c_{3}t\le x\le r(t),\hfill \\ v(t,r(t))=0,\hfill & t\ge {\tau }_3,\hfill \end{array}\right.$$

where $M_2$ is given by Theorem 1. Define

$$\xi (t,x)=M_2{e}^{\frac{r({\tau }_3)-c_3{\tau }_3}{\sqrt{d}}}{e}^{\frac{c_{3}t-x}{\sqrt{d}}},t\ge {\tau }_3,c_{3}t\le x<r(t).$$

Obviously,

$$\underset{x\ge (c_3+\epsilon )t}{sup}\xi (t,x)\le M_2{e}^{\frac{r({\tau }_3)-c_3{\tau }_3}{\sqrt{d}}}{e}^{\frac{-\epsilon t}{\sqrt{d}}}\to 0,t\to \infty ,$$

$$\xi (t,c_{3}t)>M_2,\xi (t,r(t))>0,fort\ge {\tau }_3,$$

$$\xi ({\tau }_3,x)>M_2,for{c}_3{\tau }_3\le x\le r({\tau }_3).$$

Moreover, the direct calculations yield that

$${\xi }_t-d{\xi }_{xx}\ge \xi (1-\xi ),fort\ge {\tau }_3,c_{3}t\le x\le r(t).$$

By the comparison principle, we have $v(t,x)\le \xi (t,x)$ for $t\ge {\tau }_3$ and $c_{3}t\le x\le r(t)$. Thus, (33) is obtained.

The proof of (34). Combined with (25), we just have to prove that

$$\underset{t\to \infty }{lim\; sup}\underset{x\ge c_2+\epsilon }{max}v(t,x)\le 1.$$

(36)

Choose $0<\epsilon \ll 1$ such that $c_2+\epsilon /2\le c_3-\epsilon$. It follows from (22) that v satisfies

$$\left\{\begin{array}{cc}{v}_t-d{v}_{xx}=v-v^2,\hfill & t\ge {\tau }_1,(c_2+\epsilon /2)t\le x<g(t).\hfill \\ v(t,(c_2+\epsilon )t)\le M_2,v(t,g(t))=0,\hfill & t\ge {\tau }_1.\hfill \\ v({\tau }_1,x)\le M_2,\hfill & (c_2+\epsilon /2){\tau }_1\le x<g({\tau }_1).\hfill \end{array}\right.$$

Define

$$\varphi (t,x)=1+M_2{e}^{g({\tau }_1)}{e}^{(c_2+\epsilon /2)t-x},fort\ge {\tau }_1,(c_2+\epsilon /2)t\le x\le g(t),$$

Note that $d\le 2\sqrt{\lambda }+1$ and $\varphi (t,x)\ge 1$ for $t\ge {\tau }_1$, $(c_2+\epsilon /2)t\le x\le g(t)$. Then, $c_2+\frac{\epsilon }{2}-d\ge 2\sqrt{\lambda }-d\ge -1$. Thus, direct calculations yield that

$${\varphi }_t-d{\varphi }_{xx}=(c_2+\frac{\epsilon }{2}-d)(\varphi -1)\ge \varphi (1-\varphi ),t\ge {\tau }_1,(c_2+\epsilon /2)t\le x\le g(t),$$

$$\varphi (t,(c_2+\epsilon /2)t)>1+M_2,\varphi (t,g(t))\ge 1,t\ge {\tau }_1,$$

$$\varphi ({\tau }_1,x)\ge 1+M_2,(c_2+\epsilon /2){\tau }_1\le x\le g({\tau }_1).$$

By the comparison principle, we have $v(t,x)\le \varphi (t,x)$ for $t\ge {\tau }_1$ and $(c_2+\epsilon /2)t\le x\le g(t)$. Thus,

$$\underset{x\ge (c_2+\epsilon )t}{max}v(t,x)=\underset{(c_2+\epsilon )t\le x\le g(t)}{max}v(t,x)\le \underset{(c_2+\epsilon )t\le x\le g(t)}{max}\varphi (t,x)=1+M_2{e}^{g({\tau }_1)}{e}^{-\epsilon t/2},$$

(37)

which implies that (36) holds.

(ii) For any given $0<\sigma \ll 1$ and $\rho \ge {\rho }_{\epsilon }$, where ${\rho }_{\epsilon }$ is given by Theorem 8. Let $(q_{*}(y),c_{*})$ be the unique solution of (3) with $(\beta ,d,\theta )=(\rho ,d,1+\sigma )$. Then $q_{*}^{\prime }(y)>0$, $q_{*}(y)\to 1+\sigma$ as $y\to \infty$ and

$$\underset{\rho \to \infty }{lim}{c}_{*}=2\sqrt{d(1+\sigma )}.$$

(38)

Note that (37), there exist ${\rho }_0>{\rho }_{\epsilon }$, ${\tau }_0>{\tau }_1$, $y_0\gg 1$ such that for all $\rho \ge {\rho }_0$, we have

$$c_{*}>c_2+\epsilon ,g(t)>(c_2+\epsilon )t,\forall t>{\tau }_0,$$

$$v(t,x)<(1+M_2{e}^{g({\tau }_1)}{e}^{-\epsilon t/2})q(y),\forall t\ge {\tau }_0,x\ge (c_2+\epsilon )t,y\ge y_0.$$

Let $K=M_2{e}^{g({\tau }_1)}$ and define

$$\overline{g}(t)=c_{*}t+\vartheta K(e^{-\epsilon {\tau }_0/2}-e^{-\epsilon t/2})+y_0+g({\tau }_0),t\ge {\tau }_0,$$

$$\overline{v}(t,x)=(1+K{e}^{-\epsilon t/2})q_{*}(\overline{g}(t)-x),t\ge {\tau }_0,(c_2+\epsilon )t\le x\le \overline{g}(t),$$

where $\vartheta$ is to be determined. It is obvious that

$$\overline{g}({\tau }_0)>g({\tau }_0),\overline{v}({\tau }_0,x)\ge v({\tau }_0,x),(c_2+\epsilon ){\tau }_0\le x\le g({\tau }_0),$$

$$\overline{v}(t,\overline{g}(t))=0=v(t,g(t)),\overline{v}(t,(c_2+\epsilon )t)>v(t,(c_2+\epsilon )t),t\ge {\tau }_0.$$

Similarly to the arguments of [Lemma 3.5] in [20], we can calculate that, when $\vartheta$ is suitably large,

$${\overline{v}}_t-d{\overline{v}}_{xx}\ge \overline{v}(1-\overline{v}),t\ge {\tau }_0,(c_2+\epsilon )t\le x<\overline{g}(t),$$

$${\overline{g}}^{\prime }(t)\ge -\rho {\overline{v}}_x(t,\overline{v}(t)),t\ge {\tau }_0.$$

Remember that $v_t-d{v}_{xx}=v-v^2$ for $t\ge {\tau }_0$ and $(c_2+\epsilon )t\le x<g(t)$, so we have $v(t,x)\le \overline{v}(t,x)$ and $g(t)\le \overline{g}(t)$ for all $t\ge {\tau }_0$ and $(c_2+\epsilon )t\le x<g(t)$ by the comparison principle. Therefore,

$$\underset{t\to \infty }{lim\; sup}\frac{g(t)}{t}\le c_{*}.$$

By (38) and the arbitrariness of $\sigma$, we have

$$\underset{\rho \to \infty }{lim\; sup}\underset{t\to \infty }{lim\; sup}\frac{g(t)}{t}\le 2\sqrt{d}=c_3.$$

This, combined with the second inequality of (28), leads to (35). □

Theorem 11.

Assume that $0<m\lambda -b<b/c$ and denote $c_0:=min\{c_1,c_3\}$. For any given $0<\epsilon <c_0$, we have

$$\underset{t\to \infty }{lim}\underset{[0,(c_0-\epsilon )t]}{max}|u(t,x)-u^{*}|=0,\underset{t\to \infty }{lim}\underset{[0,(c_0-\epsilon )t]}{max}|v(t,x)-v^{*}|=0,$$

where $u^{*},v^{*}$ is given by Theorem 3.

Proof. 

The proof is inspired by [Lemma 4.6] in [28]. By Theorem 8, it is obvious that

$$\underset{t\to \infty }{lim}\underset{0<x<(c_0-\frac{\epsilon }{2})t}{inf}(u(t,x),v(t,x))>\mathbf{0},\underset{t\to \infty }{lim}\underset{0<x<(c_0-\epsilon )t}{inf}(u(t,x),v(t,x))>\mathbf{0}.$$

Let ${\left\{{\epsilon }_n\right\}}_{n=1}^{\infty }$ be a sequence with

$$\frac{\epsilon }{2}={\epsilon }_0<{\epsilon }_1<{\epsilon }_2<\dots <{\epsilon }_n<\dots ,\underset{n\to \infty }{lim}{\epsilon }_n=\epsilon .$$

Define sequences ${\overline{u}}_n$, ${\overline{v}}_n$, ${\underline{u}}_n$, ${\underline{v}}_n$ as follows,

$${\overline{u}}_n=\underset{t\to \infty }{lim}\underset{0<x<(c_0-{\epsilon }_n)t}{sup}u(t,x),{\overline{v}}_n=\underset{t\to \infty }{lim}\underset{0<x<(c_0-{\epsilon }_n)t}{sup}v(t,x),$$

$${\underline{u}}_n=\underset{t\to \infty }{lim}\underset{0<x<(c_0-{\epsilon }_n)t}{inf}u(t,x),{\underline{v}}_n=\underset{t\to \infty }{lim}\underset{0<x<(c_0-{\epsilon }_n)t}{inf}v(t,x).$$

Then, $\left\{{\overline{u}}_n\right\}$, $\left\{{\overline{v}}_n\right\}$ are monotonically non-increasing and $\left\{{\underline{u}}_n\right\}$, $\left\{{\underline{v}}_n\right\}$ are monotonically non-decreasing. Furthermore, there exist positive constant $\overline{u}$, $\overline{v}$, $\underline{u}$, $\underline{v}$ such that

$$\underset{n\to \infty }{lim}({\overline{u}}_n,{\overline{v}}_n,{\underline{u}}_n,{\underline{v}}_n)=(\overline{u},\overline{v},\underline{u},\underline{v}).$$

By [Theorem 3.2] in [28], there exists a constant $\gamma >0$ such that

$${\overline{u}}_{n+1}\le {\overline{u}}_n+\frac{{\overline{u}}_n}{\gamma }(\lambda -{\overline{u}}_n-\frac{b{\underline{v}}_n}{{\overline{u}}_n+m{\underline{v}}_n}),$$

By letting $\epsilon \to \infty$, we obtain $\lambda -\overline{u}-\frac{b\underline{v}}{\overline{u}+m\underline{v}}\ge 0$. Similarly, we can verify that

$$\lambda -\underline{u}-\frac{b\overline{v}}{\underline{u}+m\overline{v}}\le 0,1-\overline{v}+\frac{c\overline{u}}{\overline{u}+m\overline{v}}\ge 0,1-\underline{v}+\frac{c\underline{u}}{\underline{u}+m\underline{v}}\le 0.$$

Then, $\overline{u}=\underline{u}=u^{*}$, $\overline{v}=\underline{v}=v^{*}$, where $(u^{*},v^{*})$ is the positive equilibrium of (1) determined by (11). The proof is finished. □

6. Discussion

This paper mainly deals with the dynamics of a ratio-dependent type prey–predator model with free boundaries $x=h(t)$ and $x=g(t)$ which describe the spreading fronts of the prey and predator, respectively. Here, the two free boundaries $x=h(t)$ and $x=g(t)$ may intersect with each other as time goes on. So, their dynamics are rather complicated.

In Section 3, we obtain the long-term behaviors of two species as follows:

(i)

If the prey (predator) cannot spread to the whole space, then it will eventually vanish (Theorem 2);

(ii)

If two species can spread successfully, then they will stabilize at a positive equilibrium state (Theorem 3);

(iii)

If the prey (predator) spreads successfully and the predator (prey) cannot spread into $[0,\infty )$, then the former will stabilize at a positive point and the latter will eventually vanish (Theorem 4).

Section 4 gives some conditions for spreading and vanishing about the prey and predator which are summarized as follows.

(i)

If one of the initial habitat and the moving parameter of the prey (predator) is “suitably large”, then it is always able to spread successfully. While, when both the initial habitat and the moving parameter of the prey (predator) are“suitably small”, the prey (predator) will eventually vanish (Theorem 5);

(ii)

If the predator spreads slowly and the initial habitat of prey is much larger than that of predator, then the prey’s territory always covers that of the predator and the prey will spread successfully no matter whether the predator successfully spreads or not (Theorem 6);

(iii)

Assume that the prey spreads slowly and the predator does quickly, if ${\lambda }^2-(b-m\lambda )<0$, and the predator spreads successfully, then the prey will eventually vanish (Theorem 7).

The asymptotic spreading speed and asymptotic speed are studied in Section 5, whose conclusions show the complicated and realistic spreading phenomena of the prey and predator. To better understand these dynamics, we first provide the spreading speed of the prey and predator when the problem (1) is uncoupled ($b=c=0$):

$$\left\{\begin{array}{cc}{u}_t-u_{xx}=\lambda u-u^2,\hfill & t>0,0<x<h(t),\hfill \\ u_x(t,0)=u(t,h(t))=0,\hfill & t\ge 0,\hfill \\ h^{\prime }(t)=-\mu u_x(t,h(t)),\hfill & t\ge 0,\hfill \\ u(0,x)=u_0(x),h(0)=h_0,\hfill & 0\le x\le h_0.\hfill \end{array}\right.$$

(39)

$$\left\{\begin{array}{cc}{v}_t-d{v}_{xx}=v-v^2,\hfill & t>0,0<x<g(t),\hfill \\ v_x(t,0)=v(t,g(t))=0,\hfill & t\ge 0,\hfill \\ g^{\prime }(t)=-\rho v_x(t,g(t)),\hfill & t\ge 0,\hfill \\ v(0,x)=v_0(x),g(0)=g_0,\hfill & 0\le x\le g_0.\hfill \end{array}\right.$$

(40)

By Proposition 2, we have

$$\underset{\mu \to \infty }{lim}\underset{t\to \infty }{lim}\frac{h(t)}{t}=2\sqrt{\lambda },\underset{\mu \to \infty }{lim}\underset{t\to \infty }{lim}\frac{g(t)}{t}=2\sqrt{d},$$

which shows that the asymptotic speed of the prey is $2\sqrt{\lambda }$ and that of the predator is $2\sqrt{d}$ when problem (1) is uncoupled.

(i)

Assume that $d(1+c)<\lambda -b/m$. Then, the asymptotic speed of the prey is between $2\sqrt{\lambda -\frac{b\kappa }{1+m\kappa }}$ and $2\sqrt{\lambda }$ and that of predator is between $2\sqrt{d}$ and $2\sqrt{d(1+c)}$ as $\mu$, $\rho \to \infty$. Those manifest that the predator could decrease the prey’s asymptotic speed; however, conversely, the prey could accelerate that of predator. Additionally, when we are to move to the right at a fixed speed less than $2\sqrt{d}$, we will observe that the two species will stabilize at the unique positive equilibrium; when we do that with a fixed speed between $2\sqrt{d(1+c)}$ and $2\sqrt{\lambda -\frac{b\kappa }{1+m\kappa }}$, we will only see the prey; when we do that with a fixed speed over than $2\sqrt{\lambda }$, we will see neither (Theorems 8, 9 and 11).

(ii)

Assume that $m\lambda >b$ and $\lambda <d$. Then, the asymptotic speed of prey is between $2\sqrt{\lambda -b/m}$ and $2\sqrt{\lambda }$ and that of predator is $2\sqrt{d}$ when $\mu$, $\rho \to \infty$, which illustrates that the predator could decrease the asymptotic speed of the prey, while the prey has no effect on the predator. In addition, if we are to move to the right at a fixed speed less than $2\sqrt{\lambda -b/m}$, then we will see that the two species will stabilize at the unique positive equilibrium; if we do that with a fixed speed between $2\sqrt{\lambda }$ and $2\sqrt{d}$, then we will only observe the predator; if we do that with a fixed speed of more than $2\sqrt{d}$, then we will see neither (Theorems 8, 10 and 11).

By comparing the ratio-dependent type (1) and the logistic type with the same free boundary mechanism in [14], we find some differences. The main difference is that, when the speed of predator is slower than that of the prey (i.e., $d(1+c)<\lambda -\frac{b}{m}$), the predator could decrease the prey’s asymptotic speed in the setting model studied in this paper; while in [14], the former is not harmful to the latter. The phenomenon in this paper seems closer to reality.