Research on artificial intelligence of accounting information processing based on image processing

1.   Introduction

The relationship among different biological populations is complex and very important and is an essential part of the research on the development of ecology. Due to the prevalence and importance of predation in nature, studying the dynamic relationship between predator and prey has always been one of the dominant topics.

In the 1920s and 1930s, as the pioneers of mathematical ecology, Lotka ^[1] and Volterra ^[2] proposed the famous Lotka-Volterra model independently, which is used to discribe the interaction between two groups composed of predators and preys:

Here, we assume an ecosystem that includes two groups of predators and preys. The predator survives on prey, and the system has no population exchange relationship with the outside world. In order to establish a mathematical model describing the system, the prey and predator population are regarded as the basic variables, which are represented by $u\left(t\right)$ and $v\left(t\right)$, respectively. The natural increment of the prey population is proportional to the number of itself, and if the proportional constant is $a_{1} > 0$, and if the mortality rate of predator population is proportional to its own number, the proportional constant is $a_{2} > 0$, and $b_{1} > 0$ and $b_{2} > 0$ are positive constant. Lotka-Volterra model is a basic model to describe the predator-prey relationship between predator and prey.

In 1975, Landahl and Hanson ^[3] and Tognetti ^[4] proposed a stage structure model and used different equations to describe individual behavior at different stages. In the last two decades, Zhang et al. ^[5] proposed and discussed a delayed predator-prey model with stage structure and nonlocal diffusion, and they studied the existence and exact asymptotic behavior of traveling wave solutions. Zhang and Xu ^[6,7] considered the predator-prey model with nonlocal delay and stage structure, and further studied the global stability. One can also see ^{[8,9,10,11,12,13,14]}.

Recently, Hong and Weng ^[15] studied the delayed predator-prey model with local diffusion and nonlocal spatial effects, and they investigated the stability of the equilibria and the existence of traveling wave solutions connecting the zero equilibrium point and the unique positive equilibrium point.

where $G\left(x, y, \tau\right) = \frac{1}{\sqrt{4D_{1}\pi\tau}}e^{-\frac{\left(x-y\right) ^{2}}{4D_{1}\tau}}$. The model considered the Holling Ⅱ functional response function. Although Holling type functional response functions are widely used, they do not consider the effect of predator density on predation rate. For this reason, some scholars have proposed a ratio dependent functional response function, and the results are also supported by many experimental facts. For results about stage structure, we refer to ^[16,17].

In 2001, Skalaski and Gilliam ^[18] compared the statistical data in some predator-prey systems, and found that the predator-dependent functional response function model has a high degree of fit with the data. The Beddington-DeAngelis functional response function is more practical in reality. This function maintains all the characteristics of the proportional dependent functional response function and avoids the singular behavior caused by the low density state, so it can better reflect the predator-prey effect (we refer to ^[19,20,21] for details).

In 2017, Khajanchi and Banerjee ^[22] introduced a persistent prey refuge in a stage structured predator-prey model with a ratio dependent functional response and obtained sufficient conditions for permanence and global asymptotic stability by constructing a suitable Lyapunov function.

where $\alpha$ represents the growth rate of juvenile prey. The conversion coefficient from juvenile prey to adult prey is proportional to the existing juvenile prey, and the proportional constant is $\beta$. $\gamma$ represents the intraspecific competition rate of adult prey. $\delta_{1}$, $\delta_{2}$, and $\delta_{3}$ represent the natural mortality of juvenile prey, adult prey, and predator, respectively. We introduced an adult prey shelter $\theta x_{m}$, $\theta\in\left(0, 1\right)$, which measures the strength of the prey shelter. For related work, Cheng and Yuan ^[23] considered the existence and stability of traveling wave solutions of Holling-Tanner predator-prey model with nonlocal diffusion and Holling type I functional response.

The local Laplacian operator to represent the spatial diffusion phenomenon cannot accurately describe the spatial and temporal behavior of species. In fact, spatial nonlocal effects are ubiquitous in nature. As for a biological population, it will move in a large spatial range than be limited to a small range, which leads to the occurrence of spatial nonlocal effects. Accordingly, many researchers have introduced convolution operators into the research models to describe the movement of individuals in the whole space and used convolution operators to describe the spatial diffusion process (see ^[24,25,26]).

In this paper, motivated by the results in ^[15], we consider the influence of Beddington-DeAngelis functional response function on the existence of traveling wave solutions of the model and consider the stage structure of the prey population and divide the prey population into two categories: Juvenile and adult. For many mammals, the juvenile prey is hidden in the cave and fed by their parents, so they do not have to go out to find food; thus, we have reason to think that the juvenile prey is not at risk of being attacked by predators. Our model is as follows:

where $G\left(x, y, \tau\right) = \frac{1}{\sqrt{4\pi D_{1}\tau}}e^{-\frac{\left(x-y\right) ^{2}}{4D_{1}\tau}}$, $\frac{\beta_{1}}{\beta}$ is the rate at which nutrients are converted to predators for reproduction. $u_{1}\left(x, t\right)$, $u_{2}\left(x, t\right)$ and $v\left(x, t\right)$ are the population density of juvenile prey population, adult prey population and predator population at position $x$ and moment $t$, respectively. $\hat{a}e^{-d_{1}\tau}\int_{-\infty}^{+\infty}G\left(x, y, \tau\right) u_{2}\left(y, t-\tau\right) dy$ represents the number of prey species converted from juvenile to adult at position $x$ and moment $t$. Here, the application of nonlocal Fourier transform and convolution shows that the function value at position $x$ is not only related to this point, but also affected by the surrounding area. $\tau > 0$ is a time delay, indicating that the change rate of the unit population at moment $t$ depends on the number of populations at moment $t-\tau$. $D_{1} > 0$, $D_{2} > 0$ and $D_{5} > 0$ are the diffusion coefficients. $a_{1} > 0$ and $a_{2} > 0$ are the birth rates of juvenile prey and predator populations respectively. $d_{1} > 0$, $d_{2} > 0$ and $b_{2} > 0$ are the mortality of juvenile prey population, adult prey population and predator population, respectively. $a_{12} > 0$ and $a_{55} > 0$ are the overcrowding rates of adult prey population and predator population respectively. $q_{2}e_{2}u_{2}\left(x, t\right) > 0$ and $q_{5}e_{5}v\left(x, t\right) > 0$ represent capture items of adult prey population and predator population, respectively, and $m$ and $w$ are positive constant.

We take the intial condition

Based on the above discussion, we first study the stability of equilibrium points of the delayed predator-prey model with stage structure and Beddington-DeAngelis functional response function using the linear stability method. Then, we establish the existence of traveling wave solutions of (1.1) by constructing a new pair of upper and lower solutions, combined with the Schauder fixed point theorem.

2.   Stability of equilibrium points

Note that the second and third equations of system (1.1) are independent of $u_{1}\left(x, t\right)$, and only related to themselves and each other. Thus, it is sufficient to consider the last two equations on their own. For simplicity of notation, we denote $u_{2}\left(x, t\right)$ by $u\left(x, t\right)$. Then, we consider the following system:

In order to facilitate the discussion of subsequent issues, we write here

Obviously, the system (2.1) has three equilibrium points, which are expressed as

For any constant equilibrium point $\left(u^{*}, v^{*}\right)$, we linearize the system (2.1) in $\left(u^{*}, v^{*}\right)$, and obtain

The system (2.4) has a non-trivial solution in the form of $\left(c_{1}, c_{2} \right)^{T} e^{\lambda t+i \sigma x}$ (see ^[27]) if and only if the corresponding determinant of the system (2.4) coefficient matrix is 0, where $\lambda$ is a complex number and $\sigma$ is a real number.

equal to

where

2.1.   The zero equilibrium point

Theorem 2.1. Suppose that $\vartheta_{1}\leq0$ and $\vartheta_{2}\leq0$, then the zero equilibrium point $E_{0}\left(0, 0\right)$ is stable; conversely, assume that either $\vartheta_{1} > 0$ or $\vartheta_{2} > 0$, then the zero equilibrium point $E_{0}\left(0, 0\right)$ is unstable.

Proof. Substituting $E_{0}\left(0, 0\right)$ into (2.5), where $u^{*} = v^{*} = 0$, we get

equal to

then

If $\lambda+D_{2}\sigma^{2}-\hat{a}e^{-d_{1}\tau-\lambda\tau}e^{-D_{2}\sigma^{2}\tau}+d_{2}+q_{2}e_{2} = 0$, according to the Eq (2.2), we can get

then when $\vartheta_{1}\leq0$, $\lambda_{1} < 0$.

If $\lambda+D_{5}\sigma^{2}-a_{2}+b_{2}+q_{5}e_{5} = 0$, according to the Eq (2.3), we can get

then when $\vartheta_{2}\leq0$, $\lambda_{2} < 0$.

Accordingly, if $\vartheta_{1}\leq0$ and $\vartheta_{2}\leq0$, the zero equilibrium point $E_{0}\left(0, 0\right)$ is stable; if either $\vartheta_{1} > 0$ or $\vartheta_{2} > 0$, we see that there exists at least a $\left(\lambda_{0}, \sigma_{0}\right)$ satisfying (2.9) such that $\lambda_{0} > 0$. Therefore, the zero equilibrium point $E_{0}\left(0, 0\right)$ are unstable. □

2.2.   The boundary equilibrium point

Theorem 2.2. Suppose that $\vartheta_{1}\geq0$ and $\vartheta_{2}+\frac{\beta_{1}\frac{\vartheta_{1}}{a_{12}}}{ 1+m\frac{\vartheta_{1}}{a_{12}}}\leq0$, then the boundary equilibrium point $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$ is stable; conversely, assume that either $\vartheta_{1} < 0$ or $\vartheta_{2}+\frac{\beta_{1}\frac{\vartheta_{1}}{a_{12}}}{ 1+m\frac{\vartheta_{1}}{a_{12}}} > 0$, then the boundary equilibrium point $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$ is unstable.

Proof. Substituting $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$ into (2.5), where $u^{*} = \frac{\vartheta_{1}}{a_{12}} = \overline{u}$, $v^{*} = 0$, we get

equal to

then

If $\lambda+D_{2}\sigma^{2}-\hat{a}e^{-d_{1}\tau-\lambda\tau}e^{-D_{2}\sigma^{2}\tau}+d_{2}+q_{2}e_{2}+2a_{12}\overline{u} = 0$, according to the Eq (2.2), we can get

then when $\vartheta_{1}\geq0$, $\lambda_{1} < 0$.

If $\lambda+D_{5}\sigma^{2}-a_{2}+b_{2}+q_{5}e_{5}-\frac{\beta_{1}\overline{u}}{ 1+m\overline{u}} = 0$, according to the Eq (2.3), we can get

then when $\vartheta_{2}+\frac{\beta_{1}\frac{\vartheta_{1}}{a_{12}}}{ 1+m\frac{\vartheta_{1}}{a_{12}}}\leq0$, $\lambda_{2} < 0$.

Accordingly, if $\vartheta_{1}\geq0$ and $\vartheta_{2}+\frac{\beta_{1}\frac{\vartheta_{1}}{a_{12}}}{ 1+m\frac{\vartheta_{1}}{a_{12}}}\leq0$, the boundary equilibrium $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$ is stable; if either $\vartheta_{1} < 0$ or $\vartheta_{2}+\frac{\beta_{1}\frac{\vartheta_{1}}{a_{12}}}{ 1+m\frac{\vartheta_{1}}{a_{12}}} > 0$, we see that there exists at least a $\left(\lambda_{0}, \sigma_{0}\right)$ satisfying (2.10) such that $\lambda_{0} > 0$. Therefore, the boundary equilibrium $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$ is unstable. □

Theorem 2.3. Suppose that $\vartheta_{1}-\frac{\beta\frac{\vartheta_{2}}{a_{55}}}{1+w\frac{\vartheta_{2}}{a_{55}}}\leq0$ and $\vartheta_{2}\geq0$, then the boundary equilibrium point $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ is stable; conversely, assume that either $\vartheta_{1}-\frac{\beta\frac{\vartheta_{2}}{a_{55}}}{1+w\frac{\vartheta_{2}}{a_{55}}} > 0$ or $\vartheta_{2} < 0$, then the boundary equilibrium point $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ is unstable.

Proof. Substituting $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ into (2.5), where $u^{*} = 0$, $v^{*} = \frac{\vartheta_{2}}{a_{55}} = \overline{v}$, we get

equal to

then

If $\lambda+D_{2}\sigma^{2}-\hat{a}e^{-d_{1}\tau-\lambda\tau}e^{-D_{2}\sigma^{2}\tau}+d_{2}+q_{2}e_{2}+\frac{\beta\overline{v}}{1+w\overline{v}} = 0$, according to the Eq (2.2), we can get

then when $\vartheta_{1}-\frac{\beta\frac{\vartheta_{2}}{a_{55}}}{1+w\frac{\vartheta_{2}}{a_{55}}}\leq0$, $\lambda_{1} < 0$.

If $\lambda+D_{5}\sigma^{2}-a_{2}+b_{2}+q_{5}e_{5}+2a_{55}\overline{v} = 0$, according to the Eq (2.3), we can get

then when $\vartheta_{2}\geq0$, $\lambda_{2} < 0$.

Accordingly, if $\vartheta_{1}-\frac{\beta\frac{\vartheta_{2}}{a_{55}}}{1+w\frac{\vartheta_{2}}{a_{55}}}\leq0$ and $\vartheta_{2}\geq0$, the boundary equilibrium $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ is stable; if either $\vartheta_{1}-\frac{\beta\frac{\vartheta_{2}}{a_{55}}}{1+w\frac{\vartheta_{2}}{a_{55}}} > 0$ or $\vartheta_{2} < 0$, we see that there exists at least a $\left(\lambda_{0}, \sigma_{0}\right)$ satisfying (2.11) such that $\lambda_{0} > 0$. Therefore, the boundary equilibrium $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ is unstable. □

2.3.   The unique positive equilibrium point

Considering the actual background of our model, we will assume that $\vartheta_{1} > 0$ and $\vartheta_{2} > 0$ in the following discussion, so the above three equilibrium points $E_{0}\left(0, 0\right)$, $E_{1}\left(\frac{\vartheta_{1}}{a_{12}}, 0\right)$, $E_{2}\left(0, \frac{\vartheta_{2}}{a_{55}}\right)$ are non-negative equilibrium points. Now we shall discuss the possibility of the positive equilibrium point.

The positive equilibrium point $E_{3}\left(u^{+}, v^{+}\right)$ of system (2.1) satisfies the system

From the Eq (2.12), we have form

Substituting the above equation into the Eq (2.13), we can get

expand and simplify to get the function

where

Next we shall analyze the existence of positive roots of the function $f\left(u\right)$, and assume that $f\left(u\right) = A_{0}u^{3}+A_{1}u^{2} +A_{2}u+A_{3}$ has a unique positive root $u^{+}$. Obviously, the main part of the function $f\left(u\right)$ is $A_{0} = m^{2}a_{55}a_{12}\beta+w^{2}a_{12}^{2}\beta_{1} > 0$, so we assume $A_{3} = \beta^{2}\vartheta_{2}-a_{55}\beta\vartheta_{1}-w\beta\vartheta_{1}\vartheta_{2}\leq0$. Therefore,

The discriminant of the derivative $f^{\prime}\left(u\right)$ is $\Delta = \left(2A_{1}\right) ^{2}-4\times3A_{0}A_{2} = 4A_{1}^{2}-12A_{0}A_{2}$, let $\Delta_{0} = A_{1}^{2}-3A_{0}A_{2}$, thus $\Delta = 4\Delta_{0}$. The system (2.1) has the unique positive equilibrium $E_{3}\left(u^{+}, v^{+}\right)$ if and only if the function $f\left(u\right)$ has a unique positive root $u^{+}$.

1) If $\Delta_{0} > 0$, then the function $f\left(u\right)$ has two zero roots $u_{1}$ and $u_{2}$, which are equivalent to

(a). If $A_{1} > 0$ and $A_{2}\geq0$, then $u_{1} < u_{2}\leq0$, $f\left(u\right)$ is increasing in $\left[ 0, +\infty\right)$. If $f\left(0\right) = A_{3} < 0$, then $f\left(u\right) = 0$ has a unique positive root; if $f\left(0\right) = A_{3} = 0$, then $f\left(u\right) = 0$ has no positive root.

(b). If $A_{1} > 0$ and $A_{2} < 0$, then $u_{1} < 0$, $u_{2} > 0$, $f\left(u\right)$ is decreasing in $\left[ 0, u_{2}\right)$, and is increasing in $\left[ u_{2}, +\infty\right)$. Since $f\left(0\right) = A_{3}\leq0$, $f\left(u\right) = 0$ has a unique positive root.

(c). If $A_{1} < 0$ and $A_{2}\leq0$, then $u_{1}\leq0$, $u_{2} > 0$, $f\left(u\right)$ is decreasing in $\left[ 0, u_{2}\right)$, and is increasing in $\left[ u_{2}, +\infty\right)$. Since $f\left(0\right) = A_{3}\leq0$, $f\left(u\right) = 0$ has a unique positive root.

(d). If $A_{1} < 0$ and $A_{2} > 0$, then $u_{1} > u_{2} > 0$, $f\left(u\right)$ is increasing in $\left[ 0, u_{1}\right)$ and $\left[ u_{2}, +\infty\right)$, and is decreasing in $\left[ u_{1}, u_{2}\right)$. If $f\left(0\right) = A_{3} < 0$, $f\left(u_{1}\right) < 0$, $f\left(u_{2}\right) < 0$, then $f\left(u\right) = 0$ has a unique positive root; if $f\left(0\right) = A_{3} = 0$, $f\left(u_{1}\right) > 0$, $f\left(u_{2}\right) = 0$, then $f\left(u\right) = 0$ has a unique positive root; otherwise, $f\left(u\right) = 0$ has two positive roots or no positive roots.

(e). If $A_{1} = 0$ and $A_{2} < 0$, then $u_{1} < 0$, $u_{2} > 0$, $f\left(u\right)$ is decreasing in $\left[ 0, u_{2}\right)$, and is increasing in $\left[ u_{2}, +\infty\right)$. Since $f\left(0\right) = A_{3}\leq0$, $f\left(u\right) = 0$ has a unique positive root.

2) If $\Delta_{0} < 0$, $f\left(0\right) = A_{3} < 0$, then the function $f\left(u\right)$ is monotonically increasing in $\left[ 0, +\infty\right)$, thus $f\left(u\right) = 0$ has a unique positive root.

3) If $\Delta_{0} = 0$, $f\left(0\right) = A_{3} < 0$, then the function $f\left(u\right)$ is monotonically increasing in $\left[ 0, +\infty\right)$, thus $f\left(u\right) = 0$ has a unique positive root.

Summarizing the above discussion, we get the following conclusions.

Lemma 2.1. Suppose that $A_{0} > 0$, $A_{3}\leq 0$, equation $f\left(u\right) = 0$ has a unique positive root $u^{+}$ if and only if one of the following six conditions holds:

B1). $\Delta_{0} > 0$, $A_{1} > 0$, $A_{2}\geq0$, $A_{3} < 0$;

B2). $\Delta_{0} > 0$, $A_{1} > 0$, $A_{2} < 0$, $A_{3}\leq0$;

B3). $\Delta_{0} > 0$, $A_{1} < 0$, $A_{2}\leq0$, $A_{3}\leq0$;

B4). $\Delta_{0} > 0$, $A_{1} < 0$, $A_{2} > 0$, $A_{3}\leq0$;

B5). $\Delta_{0} > 0$, $A_{1} = 0$, $A_{2} < 0$, $A_{3}\leq0$;

B6). $\Delta_{0}\leq0$, $A_{3}\leq0$.

$f\left(u\right)$ has a unique positive root $u^{+}$, through the Eq (2.13) we can get

As $\frac{\beta}{\vartheta_{1}-a_{12}u^{+}} > w$, that is, $u^{+} > \frac{\vartheta_{1}-w\beta}{a_{12}}$, there exists a unique corresponding $v^{+}$. Thus, the system (2.1) has a unique positive equilibrium point $E_{3}\left(u^{+}, v^{+}\right)$.

Theorem 2.4 (The existence condition of $E_{3}$). Suppose $\vartheta_{1} > 0$, $\vartheta_{2} > 0$ and $a_{12}u^{+} < \vartheta_{1} < \frac{\beta}{w}+a_{12}u^{+}$, then the system (2.1) has a unique positive equilibrium point $E_{3}\left(u^{+}, v^{+}\right)$, where $u^{+} > 0$ is the only positive root of $f\left(u\right) = 0$.

Theorem 2.5 (The stability of $E_{3}$). Assume that the unique positive equilibrium point $E_{3}\left(u^{+}, v^{+}\right)$ exists, if $a_{12}\geq\frac{m\beta v^{+}}{\left(1+mu^{+}+wv^{+}\right) ^{2}}$, then $E_{3}\left(u^{+}, v^{+}\right)$ is stable.

Proof. Substituting $E_{3}\left(u^{+}, v^{+}\right)$ into (2.5), where$u^{*} = u^{+}$, $v^{*} = v^{+}$, we get

Here we introduce some representations

such that

where $\gamma_{1}, \; \gamma_{2}, \; u^{+}, \; v^{+} > 0$.

Due to $\lambda+D_{2}\sigma^{2}-\hat{a}e^{-d_{1}\tau-\lambda\tau}e^{-D_{2}\sigma^{2}\tau}+d_{2}+q_{2}e_{2}+2a_{12}u^{+} +\gamma_{2}v^{+}\neq0$, with the help of Eqs (2.3) and (2.13), the Eq (2.14) can be transformed into

$\lambda = \mu+i\omega$ is a complex number, that is, $Re\lambda = \mu$, $Im\lambda = \omega$, $\left(\lambda, \sigma\right) = \left(\mu+i\omega, \sigma \right)$. The real part of $\lambda$ is $Re\lambda = \mu < 0$. Otherwise, $Re\lambda = \mu\geq0$ is not established, and the counter-proof method proves as follows.

Suppose that there exists $\left(\lambda_{1}, \sigma_{1}\right) = \left(\mu_{1}+i\omega_{1}, \sigma_{1} \right)$, $\mu_{1}\geq0$. Using Euler formula to split the real and imaginary parts of $\lambda$. Let

then

as $a_{12}\geq\frac{m\beta v^{+}}{\left(1+mu^{+}+wv^{+}\right) ^{2}}$. This is a contradiction.

Consequently, assume that the unique positive equilibrium point $E_{3}\left(u^{+}, v^{+}\right)$ exists, if $a_{12}\geq\frac{m\beta v^{+}}{\left(1+mu^{+}+wv^{+}\right) ^{2}}$, then $E_{3}\left(u^{+}, v^{+}\right)$ is stable. □

3.   Existence of traveling wave solutions

In this section, by using the Schauder fixed point theorem and the method of constructing upper and lower solutions by cross iteration, the existence of traveling wave solutions of the connecting equilibrium points $E_{0}$ and $E_{3}$ of the system (2.1) is obtained. The traveling wave solution of the system (2.1) is a special translation invariant solution in the form of $\left(u\left(x, t\right), v\left(x, t\right) \right) = \left(\phi\left(x+ct\right), \psi\left(x+ct\right) \right)$, where the wave velocity $c > 0$, $\phi$ and $\psi$ are wave profile functions, and the wave profile propagates in one-dimensional space domain at a constant speed $c > 0$. Substituting $\left(u\left(x, t\right), v\left(x, t\right) \right) = \left(\phi\left(x+ct\right), \psi\left(x+ct\right) \right)$ into system (2.1) and replacing $x+ct$ with $t$, we get

satisfy the following asymptotic boundary value conditions

where

3.1.   The construction of upper and lower solutions of the system

In this section, we discuss the existence of upper and lower solutions. Firstly, we give the definition of the upper and lower solutions of the system (3.1).

Definition 3.1. Let $\overline{\rho}\left(t\right) = \left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$, $\underline{\rho}\left(t\right) = \left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$, $t\in\mathbb{R}$ be two continuous functions, then $\overline{\rho}\left(t\right) = \left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\underline{\rho}\left(t\right) = \left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$, $t\in\mathbb{R}$ are the upper and lower solutions of the system (3.1), respectively. If there exists a finite set of points $S = \left\lbrace s_{i}\in\mathbb{R}, i = 1, 2, \cdots, n \right\rbrace$, where $s_{1} < s_{2} < \cdots < s_{n}$, such that $\overline{\rho}\left(t\right)$ and $\underline{\rho}\left(t\right)$ are twice continuously differentiable on $\mathbb{R}\backslash S$, and for any $t\in\mathbb{R}\backslash S$, satisfy

and

Now linearizing the system (3.1) at $\left(0, 0\right)$, we obtain

Substituting $\phi\left(t\right) = e^{\lambda t}$ and $\psi\left(t\right) = e^{\lambda t}$ into the system (3.2), due to $\int_{-\infty}^{+\infty}\frac{1}{\sqrt{4\pi D_{1}\tau}}e^{-\frac{y^{2}}{4D_{1}\tau}}e^{-\lambda\left(y+c\tau\right)} dy = e^{\left(D_{1}\lambda^{2}-c\lambda\right) \tau}$, we get

where

Lemma 3.1. If $\hat{a}e^{-d_{1}\tau}e^{\left(D_{1}\lambda^{2}-c\lambda\right) \tau}- d_{2}-q_{2}e_{2} > 0$, let $c_{1}^{*} = \sqrt{4D_{2}\left(\hat{a}e^{-d_{1}\tau}e^{\left(D_{1}\lambda^{2}-c\lambda\right) \tau}- d_{2}-q_{2}e_{2}\right)}$, then the following conclusions hold.

1). If $c > c_{1}^{*}$, then $\Delta_{1}\left(\lambda, c \right) = 0$ has two different positive roots $\lambda_{1}\left(c\right)$ and $\lambda_{2}\left(c\right)$, we may set $0 < \lambda_{1}\left(c\right) < \lambda_{2}\left(c\right)$;

2). If $0 < c < c_{1}^{*}$, then $\Delta_{1}\left(\lambda, c \right) = 0$ has no real root.

Lemma 3.2. If $a_{2}-b_{2}-q_{5}e_{5} > 0$, write $c_{2}^{*} = \sqrt{4D_{5}\left(a_{2}-b_{2}-q_{5}e_{5}\right) }$, then the following conclusions hold.

1). If $c > c_{2}^{*}$, then $\Delta_{2}\left(\lambda, c \right) = 0$ has two different positive roots $\lambda_{3}\left(c\right)$ and $\lambda_{4}\left(c\right)$, we may set $0 < \lambda_{3}\left(c\right) < \lambda_{4}\left(c\right)$;

2). If $0 < c < c_{2}^{*}$, then $\Delta_{2}\left(\lambda, c \right) = 0$ has no real root.

Proof. We regard $\Delta_{1}\left(\lambda, c \right) = 0$ and $\Delta_{2}\left(\lambda, c \right) = 0$ as a quadratic equation with one variable $\lambda$, and consider the existence of the solution of the equation according to the size of the respective discriminant $\Delta$ and $0$. □

Lemma 3.3. Assume that $a_{12}u^{+}\geq\frac{\left(3+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$ and $a_{55}v^{+}\geq\frac{2\sqrt{2} \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$ hold, there exist $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$ and $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$ such that

where $\varepsilon_{0} > 0$ is a constant.

Proof. Let

$g_{1}\left(\varepsilon_{1}\right)$ is a quadratic function with respect to $\varepsilon_{1}$. The image opens down through the origin, and the symmetry axis is $x = \left(\sqrt{2}-1\right)u^{+} > 0$, so that $g_{1}\left(\varepsilon_{1}\right)$ is increasing in $\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$. Thus, $g_{1}\left(0\right) = 0$ and the maximum value is $max\left\lbrace g_{1}\left(\varepsilon_{1}\right)\right\rbrace = g_{1}\left(\left(\sqrt{2}-1\right)u^{+}\right) = \left(3-2\sqrt{2} \right)a_{12}\left(u^{+}\right) ^{2}$; $g_{2}\left(\varepsilon_{1}\right)$ is decreasing with respect to $\varepsilon_{1}$ and the maximum value is $max\left\lbrace g_{2}\left(\varepsilon_{1}\right)\right\rbrace < g_{2}\left(0\right) = \frac{\beta u^{+}v^{+}}{1+mu^{+}+2wv^{+}}$. According to the assumption of $a_{12}u^{+}\geq\frac{\left(3+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$, then $\left(3-2\sqrt{2} \right)a_{12}\left(u^{+}\right) ^{2}\geq\frac{\beta u^{+}v^{+}}{1+mu^{+}+2wv^{+}}$, there exists $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$, so that $g_{1}\left(\varepsilon_{1}\right) > g_{2}\left(\varepsilon_{1}\right)$. The first inequality is proved.

$g_{3}\left(\varepsilon_{2}\right)$ is a quadratic function with respect to $\varepsilon_{2}$. The image opening goes down through the origin, and the symmetry axis is $x = \frac{v^{+}}{2} > 0$, so that $g_{3}\left(\varepsilon_{2}\right)$ is increasing in $\left(0, \frac{v^{+}}{2} \right)$. Thus, $g_{3}\left(0\right) = 0$ and the maximum value is $max\left\lbrace g_{3}\left(\varepsilon_{2}\right)\right\rbrace = g_{3}\left(\frac{v^{+}}{2}\right) = \frac{1}{4}a_{55}\left(v^{+}\right) ^{2}$; $g_{4}\left(\varepsilon_{2}\right)$ is increasing in $\left(0, \frac{v^{+}}{2} \right)$ with respect to $\varepsilon_{2}$, then the maximum value is $max\left\lbrace g_{4}\left(\varepsilon_{2}\right)\right\rbrace < g_{4}\left(\frac{v^{+}}{2} \right) = \frac{\beta_{1} u^{+}v^{+}}{1+mu^{+}+wv^{+}}-\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}\right) \frac{v^{+}}{2} }{1+m\left(u^{+}-\varepsilon_{1}\right)+w\frac{v^{+}}{2}}$, here $-\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}\right) \frac{v^{+}}{2} }{1+m\left(u^{+}-\varepsilon_{1}\right)+w\frac{v^{+}}{2}}$ is increasing for $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$ with respect to $\varepsilon_{1}$, such that

According to the assumption of $a_{55}v^{+}\geq\frac{2\sqrt{2} \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$, then $\frac{1}{4}a_{55}\left(v^{+}\right) ^{2}\geq\frac{\frac{\sqrt{2}}{2}\beta_{1} u^{+}v^{+}}{1+mu^{+}+wv^{+}}$, there exists $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$, so that $g_{3}\left(\varepsilon_{2}\right) > g_{4}\left(\varepsilon_{2}\right)$. The second inequality is proved. □

For the unique positive equilibrium $\left(u^{+}, v^{+} \right)$, we know that $\vartheta_{1}-a_{12}u^{+}-\frac{\beta v^{+}}{1+mu^{+}+wv^{+}} = 0$ and $\vartheta_{2}-a_{55}v^{+}+\frac{\beta_{1}u^{+}}{1+mu^{+}+wv^{+}} = 0$, thereby

1). If $\vartheta_{1} > \frac{\left(4+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$ holds, then $a_{12}u^{+} = \vartheta_{1}-\frac{\beta v^{+}}{1+mu^{+}+wv^{+}} > \frac{\left(3+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$;

2). If $\vartheta_{2} > \frac{\left(2\sqrt{2}-1\right) \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$ holds, then $a_{55}v^{+} = \vartheta_{2}+\frac{\beta_{1}u^{+}}{1+mu^{+}+wv^{+}} > \frac{2\sqrt{2} \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$.

Remark 3.1. Suppose $\vartheta_{1} > \frac{\left(4+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$ and $\vartheta_{2} > \frac{\left(2\sqrt{2}-1\right) \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$ hold, Lemma 3.3 holds.

In addition, from $a_{12}u^{+}\geq\frac{\left(3+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$ we can deduce

As $a_{12}\geq\frac{m\beta v^{+} }{\left(1+mu^{+}+wv^{+}\right)^{2}}$, the unique positive equilibrium $\left(u^{+}, v^{+} \right)$ is stable.

Let $c^{*} = max\left\lbrace c_{1}^{*}, c_{2}^{*}\right\rbrace$. For fixed $c > c^{*}$, take constant $\eta\in\left(1, min\left\lbrace 2, \frac{\lambda_{2}}{\lambda_{1}}, \frac{\lambda_{4}}{\lambda_{3}}, \frac{\lambda_{1}+\lambda_{3}}{\lambda_{1}}, \frac{\lambda_{1}+\lambda_{3}}{\lambda_{3}}\right\rbrace \right)$, then there are $\Delta_{1}\left(\eta\lambda_{1}, c \right) < 0$ and $\Delta_{2}\left(\eta\lambda_{3}, c \right) < 0$.

Let $\eta > 1$, $q > 1$ be large enough and $\lambda > 0$ be small enough. Here $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$ and $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$. We write $\lambda_{i} = \lambda_{i}\left(c\right) > 0\left(i = 1, 2, 3, 4\right)$. The continuous functions $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$ are defined as follows:

It is easy to know that $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$ have the following properties:

1) There are two constants $N_{1} > 0$ and $N_{2} > 0$ such that $\left(0, 0\right) \leq\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)\leq\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)\leq\left(N_{1}, N_{2}\right)$;

2) $\lim\limits_{t\to-\infty}\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right) = \left(0, 0\right), \lim\limits_{t\to+\infty}\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right) = \lim\limits_{t\to+\infty}\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right) = \left(u^{+}, v^{+}\right)$;

3) For all $t\in\mathbb{R}$, $\overline{\phi}^{\prime}\left(t+\right)\leq\overline{\phi}^{\prime}\left(t-\right)$, $\underline{\phi^{\prime}}\left(t+\right)\geq\underline{\phi^{\prime}}\left(t-\right)$.

Remark 3.2. According to the definition of $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$, we know

Remark 3.3. If $q > 1$ is large enough, then it is clear that $t_{1}\geq max\left\lbrace t_{2}, t_{3}, t_{4}\right\rbrace$.

Lemma 3.4. Assume that $\vartheta_{1} > \frac{\left(4+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$ and $\vartheta_{2} > \frac{\left(2\sqrt{2}-1\right) \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$ hold, and $q > 1$ is large enough, then $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$ are a pair of upper and lower solutions of system (3.1).

Proof. We first consider $\overline{\phi}\left(t\right)$. According to the function definition, we have

According to the definition of upper and lower solutions Definition 3.1, we want to prove that $D_{2}\overline{\phi}^{\prime\prime}\left(t \right)-c\overline{\phi}^{\prime}\left(t \right)+f_{2}\left(\overline{\phi}, \underline{\psi} \right) \left(t \right)\leq0$, where $\underline{\psi}\left(t\right)\geq max\left\lbrace e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t}, v^{+}-\varepsilon_{2}e^{-\lambda t}\right\rbrace$.

If $t\leq t_{1}$, then $\overline{\phi}\left(t\right) = e^{\lambda_{1}t}$, $\overline{\phi}^{\prime}\left(t\right) = \lambda_{1}e^{\lambda_{1}t}$ and $\overline{\phi}^{\prime\prime}\left(t\right) = \lambda_{1}^{2}e^{\lambda_{1}t}$. In addition, we note that if $t-y-c\tau\leq t_{1}$, then $\overline{\phi}\left(t-y-c\tau\right) = e^{\lambda_{1}\left(t-y-c\tau\right) }$; if $t-y-c\tau\geq t_{1}$, then $\overline{\phi}\left(t-y-c\tau\right)\leq e^{\lambda_{1}\left(t-y-c\tau\right) }$. Thus,

If $t\geq t_{1}$, then $\overline{\phi}\left(t\right) = u^{+}+u^{+}e^{-\lambda t}$, here $\underline{\psi}\left(t\right)\geq v^{+}-\varepsilon_{2}e^{-\lambda t}$, $\overline{\phi}^{\prime}\left(t\right) = -\lambda u^{+}e^{-\lambda t}$ and $\overline{\phi}^{\prime\prime}\left(t\right) = \lambda^{2}u^{+}e^{-\lambda t}$. In addition, we note that if $t-y-c\tau\geq t_{1}$, then $\overline{\phi}\left(t-y-c\tau\right) = u^{+}+u^{+}e^{-\lambda \left(t-y-c\tau\right) }$; if $t-y-c\tau\leq t_{1}$, then $\overline{\phi}\left(t-y-c\tau\right)\leq u^{+}+u^{+}e^{-\lambda\left(t-y-c\tau\right)}$. Thus,

According to the premise assumption, $\Delta_{1}\left(0, c \right)-2a_{12}u^{+} = \vartheta_{1}-2a_{12}u^{+} = \frac{\beta u^{+}v^{+}}{1+mu^{+}+wv^{+}}-a_{12}u^{+} < 0$ can be obtained, so there is a constant $\widetilde{\lambda}_{1} > 0$, which makes $\Delta_{1}\left(-\lambda, c \right)-2a_{12}u^{+} < 0$ for $\forall\lambda\in\left(0, \widetilde{\lambda}_{1}\right)$.

Let $I_{1}\left(\lambda, t \right): = a_{12}u^{+} e^{-2\lambda t}+\frac{\beta \left(1+e^{-\lambda t}\right)\underline{\psi}\left(t\right)}{1+m\left(u^{+}+u^{+}e^{-\lambda t}\right)+w\underline{\psi}\left(t\right)}-\frac{\beta v^{+}}{1+mu^{+}+wv^{+}}$, where $t\geq t_{1}$ and $\underline{\psi}\left(t\right)\geq v^{+}-\varepsilon_{2}e^{-\lambda t} \geq0$. $\frac{\beta \left(1+e^{-\lambda t}\right)\underline{\psi}\left(t\right)}{1+m\left(u^{+}+u^{+}e^{-\lambda t}\right)+w\underline{\psi}\left(t\right)}$ is increasing with respect to $\underline{\psi}\left(t\right)$, thus

Here $t\geq t_{1}$, $\left(t_{1} < 0 \right)$. Therefore, $t$ is divided into $t\in\left[ t_{1}, 0\right]$ and $t > 0$ for discussion.

If $t\in\left[ t_{1}, 0\right]$, from the hypothesis we know that

If $t > 0$, here $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$, we have that

Since $a_{12}u^{+} e^{-2\lambda t}$ and $\frac{\beta \left(1+e^{-\lambda t}\right) \frac{v^{+}}{2}}{1+mu^{+}\left(1+e^{-\lambda t}\right)+w\frac{v^{+}}{2}}$ are decreasing about the variable $t$ on $t > 0$, furthermore, $I_{1}\left(\lambda, 0 \right) = a_{12}u^{+}+\frac{2\beta \left(v^{+}-\varepsilon_{2}\right) }{1+2mu^{+}+w\left(v^{+}-\varepsilon_{2}\right)}-\frac{\beta v^{+}}{1+mu^{+}+wv^{+}} > \frac{\left(2+2\sqrt{2} \right)\beta v^{+}}{1+mu^{+}+wv^{+}}+\frac{2\beta \left(v^{+}-\varepsilon_{2}\right) }{1+2mu^{+}+w\left(v^{+}-\varepsilon_{2}\right)} > 0$ and $I_{1}\left(\lambda, +\infty \right) = 0$, then $I_{1}\left(\lambda, t \right) > 0$ for $\forall t\geq t_{1}$.

In consequence, $\overline{\phi}$ satisfies the upper solution definition, that is, $D_{2}\overline{\phi}^{\prime\prime}\left(t \right)-c\overline{\phi}^{\prime}\left(t \right)+f_{2}\left(\overline{\phi}, \underline{\psi} \right) \left(t \right)\leq0$.

Next we consider $\overline{\psi}\left(t\right)$. According to the function definition, we have

According to the definition of upper and lower solutions Definition 3.1, we want to prove that $D_{5}\overline{\psi}^{\prime\prime}\left(t \right)-c\overline{\psi}^{\prime}\left(t \right)+f_{5}\left(\overline{\phi}, \overline{\psi} \right) \left(t \right)\leq0$, where $\overline{\phi}\left(t\right)\leq min\left\lbrace e^{\lambda_{1}t}, u^{+}+u^{+}e^{-\lambda t}\right\rbrace$.

If $t\leq t_{2}$, then $\overline{\psi}\left(t\right) = e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t}$, here $\overline{\phi}\left(t\right)\leq e^{\lambda_{1}t}$, $\overline{\psi}^{\prime}\left(t\right) = \lambda_{3}e^{\lambda_{3}t}+q\eta\lambda_{3}e^{\eta\lambda_{3}t}$ and $\overline{\psi}^{\prime\prime}\left(t\right) = \lambda_{3}^{2}e^{\lambda_{3}t}+q\eta^{2}\lambda_{3}^{2}e^{\eta\lambda_{3}t}$. Thus,

Here $q > 1$ is large enough, then $-t_{2} > 0$ is also large enough. $q\left[ \Delta_{2}\left(\eta\lambda_{3}, c \right)+\beta_{1}e^{\lambda_{1}t}\right] +\beta_{1}$ is increasing about the variable $t$ on $t\leq t_{2}$ for $\forall t\leq t_{2}$, so there exists $q\left[ \Delta_{2}\left(\eta\lambda_{3}, c \right)+\beta_{1}e^{\lambda_{1}t}\right] +\beta_{1} < 0$, thus $e^{\eta\lambda_{3}t}\left\lbrace q\left[ \Delta_{2}\left(\eta\lambda_{3}, c \right)+\beta_{1}e^{\lambda_{1}t}\right] +\beta_{1} \right\rbrace < 0$ for $\forall t\leq t_{2}$.

If $t\geq t_{2}$, then $\overline{\psi}\left(t\right) = v^{+}+v^{+}e^{-\lambda t}$, here $\overline{\phi}\left(t\right)\leq u^{+}+u^{+}e^{-\lambda t}$, $\overline{\psi}^{\prime}\left(t\right) = -\lambda v^{+}e^{-\lambda t}$ and $\overline{\psi}^{\prime\prime}\left(t\right) = \lambda^{2}v^{+}e^{-\lambda t}$. Thus,

Because of $\Delta_{2}\left(0, c \right)-a_{55}v^{+} = \vartheta_{2}-a_{55}v^{+} = -\frac{\beta_{1} u^{+}}{1+mu^{+}+wv^{+}} < 0$, there is a constant $\widetilde{\lambda}_{2} > 0$, which makes $\Delta_{2}\left(-\lambda, c \right)-a_{55}v^{+} < 0$ for $\forall\lambda\in\left(0, \widetilde{\lambda}_{2}\right)$.

Let $I_{2}\left(\lambda, t \right): = a_{55}v^{+}e^{-\lambda t}+a_{55} v^{+}e^{-2\lambda t}-\frac{\beta_{1}u^{+}\left(1+e^{-\lambda t}\right)^{2}}{1+mu^{+}\left(1+e^{-\lambda t}\right) +wv^{+}\left(1+e^{-\lambda t}\right) }+\frac{\beta_{1}u^{+}}{1+mu^{+}+wv^{+}}$, we have that

According to the premise hypothesis, $a_{55}v^{+} > \frac{2\sqrt{2}\beta_{1}u^{+}}{1+mu^{+} +wv^{+}}$ can be obtained, thus $I_{2}\left(\lambda, t \right) > 0$ for $\forall t\geq t_{2}$.

In consequence, $\overline{\psi}$ satisfies the upper solution definition, that is, $D_{5}\overline{\psi}^{\prime\prime}\left(t \right)-c\overline{\psi}^{\prime}\left(t \right)+f_{5}\left(\overline{\phi}, \overline{\psi} \right) \left(t \right)\leq0$.

We next consider $\underline{\phi}\left(t\right)$. According to the function definition, we have

According to the definition of upper and lower solutions Definition 3.1, we want to prove that $D_{2}\underline{\phi}^{\prime\prime}\left(t \right)-c\underline{\phi}^{\prime}\left(t \right)+f_{2}\left(\underline{\phi}, \overline{\psi} \right) \left(t \right)\geq0$, where $\overline{\psi}\left(t\right)\leq min\left\lbrace e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t}, v^{+}+v^{+}e^{-\lambda t}\right\rbrace$.

If $t\leq t_{3}$, then $\underline{\phi}\left(t\right) = e^{\lambda_{1}t}-qe^{\eta\lambda_{1}t}$, here $\overline{\psi}\left(t\right)\leq e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t}$, $\underline{\phi}^{\prime}\left(t\right) = \lambda_{1}e^{\lambda_{1}t}-q\eta\lambda_{1}e^{\eta\lambda_{1}t}$ and $\underline{\phi}^{\prime\prime}\left(t\right) = \lambda_{1}^{2}e^{\lambda_{1}t}-q\eta^{2}\lambda_{1}^{2}e^{\eta\lambda_{1}t}$. In addition, we note that if $t-y-c\tau\geq t_{3}$, then $\overline{\phi}\left(t-y-c\tau\right) = e^{\lambda_{1}\left(t-y-c\tau\right)}-qe^{\eta\lambda_{1}\left(t-y-c\tau\right)}$; if $t-y-c\tau\leq t_{3}$, then $\overline{\phi}\left(t-y-c\tau\right)\geq e^{\lambda_{1}\left(t-y-c\tau\right)}-qe^{\eta\lambda_{1}\left(t-y-c\tau\right)}$. Thus,

Here $q > 1$ is large enough, then $-t_{3} > 0$ is also large enough. $q\left[ \Delta_{1}\left(\eta\lambda_{1}, c \right)+\beta e^{\left(\lambda_{1}+\eta\lambda_{3}-\eta\lambda_{1}\right)t}\right]+\left(a_{12}+\beta\right)$ is increasing about the variable $t$ on $t\leq t_{3}$ for $\forall t\leq t_{3}$, so there exists $q\left[ \Delta_{1}\left(\eta\lambda_{1}, c \right)+\beta e^{\left(\lambda_{1}+\eta\lambda_{3}-\eta\lambda_{1}\right)t}\right]+\left(a_{12}+\beta\right) < 0$, thus $-e^{\eta\lambda_{1}t}\left\lbrace q\left[ \Delta_{1}\left(\eta\lambda_{1}, c \right)+\beta e^{\left(\lambda_{1}+\eta\lambda_{3}-\eta\lambda_{1}\right)t}\right]+\left(a_{12}+\beta\right) \right\rbrace > 0$ for $\forall t\leq t_{3}$.

If $t > t_{3}$, then $\underline{\phi}\left(t\right) = u^{+}-\varepsilon_{1}e^{-\lambda t}$, here $\overline{\psi}\left(t\right)\leq v^{+}+v^{+}e^{-\lambda t}$, $\underline{\phi}^{\prime}\left(t\right) = \lambda\varepsilon_{1}e^{-\lambda t}$ and $\underline{\phi}^{\prime\prime}\left(t\right) = -\lambda^{2}\varepsilon_{1}e^{-\lambda t}$. In addition, we note that if $t-y-c\tau\geq t_{3}$, then $\underline{\phi}\left(t-y-c\tau\right) = u^{+}-\varepsilon_{1}e^{-\lambda \left(t-y-c\tau\right)}$; if $t-y-c\tau\leq t_{3}$, then $\underline{\phi}\left(t-y-c\tau\right)\geq u^{+}-\varepsilon_{1}e^{-\lambda \left(t-y-c\tau\right)}$. Thus,

According to the premise assumption, we can obtain that $-\Delta_{1}\left(-\lambda, c \right)+\left(4-2\sqrt{2}\right) a_{12}u^{+} = -\vartheta_{1}+\left(4-2\sqrt{2}\right) a_{12}u^{+} = \left(3-2\sqrt{2}\right) a_{12}u^{+}-\frac{\beta v^{+}}{1+mu^{+}+wv^{+}} > 0$, so there is a constant $\widetilde{\lambda}_{3} > 0$, which makes $-\Delta_{1}\left(-\lambda, c \right)+\left(4-2\sqrt{2}\right) a_{12}u^{+} > 0$ for $\forall\lambda\in\left(0, \widetilde{\lambda}_{3}\right)$.

Let $I_{3}\left(\lambda, t \right) = \left(2\sqrt{2}-2\right)a_{12}u^{+}\varepsilon_{1}e^{-\lambda t}-a_{12}\varepsilon_{1}^{2}e^{-2\lambda t}+\frac{\beta u^{+}v^{+}}{1+mu^{+}+wv^{+}}-\frac{\beta \left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)\overline{\psi}\left(t\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)+w\overline{\psi}\left(t\right)}$, we have that

where $t\geq t_{3}$. Therefore, $I_{3}\left(\lambda, 0 \right) = -a_{12}\varepsilon_{1}^{2}+\left(2\sqrt{2}-2\right) a_{12}u^{+}\varepsilon_{1}+\frac{\beta u^{+}v^{+}}{1+mu^{+}+wv^{+}}-\frac{2\beta\left(u^{+}-\varepsilon_{1}\right) v^{+}}{1+m\left(u^{+}-\varepsilon_{1}\right)+2wv^{+}}$.

From Remark 3.1, we can see that $\vartheta_{1} > \frac{\left(4+2\sqrt{2} \right)\beta v^{+} }{1+mu^{+}+wv^{+}}$is established, there is $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$, making $-a_{12}\varepsilon_{1}^{2}+\left(2\sqrt{2}-2\right) a_{12}u^{+}\varepsilon_{1}+\frac{\beta u^{+}v^{+}}{1+mu^{+}+wv^{+}}-\frac{2\beta\left(u^{+}-\varepsilon_{1}\right) v^{+}}{1+m\left(u^{+}-\varepsilon_{1}\right)+2wv^{+}} > \varepsilon_{0} > 0$. We have $I_{3}\left(\lambda, 0 \right) > 0$, here $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$. We can choose a small enough $\delta_{1} > 0$, such that $\delta^{*}: = \varepsilon_{1}+\delta_{1}$ for $\forall\delta\in\left[\varepsilon_{1}, \delta^{*} \right]$ satisfying

We want to prove that $I_{3}\left(\lambda, t \right) > 0$ for $\forall t > t_{3}$. Here $t > t_{3}\left(t_{3} < 0 \right)$, therefore, $t$ is divided into $t\in\left(t_{3}, 0\right]$ and $t > 0$ two parts to discuss.

If $t\in\left(t_{3}, 0\right]$, let $\nu\left(t \right): = \varepsilon_{1}e^{-\lambda t}$, $\mu\left(t \right): = v^{+}+v^{+}e^{-\lambda t}$. Select $\widetilde{\lambda}_{3} > 0$ small enough such that for any given $\lambda\in\left(0, \widetilde{\lambda}_{3}\right)$, we have

which leads to $\varepsilon_{1}\leq\nu\left(t\right)\leq\delta^{*}$ and $\varepsilon_{1}\leq\mu\left(t\right)\leq\delta^{*}$. So we get $I_{3}\left(\lambda, t \right) > 0$ for $\forall t\in\left(t_{3}, 0\right]$.

If $t\geq0$, here $-\frac{\beta \left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)\overline{\psi}\left(t\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)+w\overline{\psi}\left(t\right)} < 0$, based on the assumption that $\lambda > 0$ is small enough,

where $\overline{\psi}\left(t\right) > 0$ and $\overline{\psi}^{\prime}\left(t\right) < 0$. Thus we have the function $-\frac{\beta \left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)\overline{\psi}\left(t\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right)+w\overline{\psi}\left(t\right)}$is increasing about the variable $t$ on $t\geq0$; the function $\left(2\sqrt{2}-2\right)a_{12}u^{+}\varepsilon_{1}e^{-\lambda t}-a_{12}\varepsilon_{1}^{2}e^{-2\lambda t} > 0$ is decreasing about the variable $t$ on $t\geq0$. We can get that

So we get $I_{3}\left(\lambda, t \right) > 0$ for $\forall t\geq0$. Thus, $I_{3}\left(\lambda, t \right) > 0$ for $\forall t > t_{3}$. In consequence, $\underline{\phi}$ satisfies lower solution definition, that is, $D_{2}\underline{\phi}^{\prime\prime}\left(t \right)-c\underline{\phi}^{\prime}\left(t \right)+f_{2}\left(\underline{\phi}, \overline{\psi} \right) \left(t \right)\geq0$.

We finally consider $\underline{\psi}\left(t\right)$. According to the function definition, we have

According to the definition of upper and lower solutions Definition 3.1, we want to prove that $D_{5}\underline{\psi}^{\prime\prime}\left(t \right)-c\underline{\psi}^{\prime}\left(t \right)+f_{5}\left(\underline{\phi}, \underline{\psi} \right) \left(t \right)\geq0$, where $\underline{\phi}\left(t\right)\geq max\left\lbrace e^{\lambda_{1}t}-qe^{\eta\lambda_{1}t}, u^{+}-\varepsilon_{1}e^{-\lambda t}\right\rbrace$.

If $t\leq t_{4}$, then $\underline{\psi}\left(t\right) = e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t}$, $\underline{\psi}^{\prime}\left(t\right) = \lambda_{3}e^{\lambda_{3}t}-q\eta\lambda_{3}e^{\eta\lambda_{3}t}$ and $\underline{\psi}^{\prime\prime}\left(t\right) = \lambda_{3}^{2}e^{\lambda_{3}t}-q\eta^{2}\lambda_{3}^{2}e^{\eta\lambda_{3}t}$. Thus,

here $\eta\in\left(1, min\left\lbrace 2, \frac{\lambda_{2}}{\lambda_{1}}, \frac{\lambda_{4}}{\lambda_{3}}, \frac{\lambda_{1}+\lambda_{3}}{\lambda_{1}}, \frac{\lambda_{1}+\lambda_{3}}{\lambda_{3}}\right\rbrace \right)$ and $q > 1$is large enough.

If $t\geq t_{4}$, then $\underline{\psi}\left(t\right) = v^{+}-\varepsilon_{2}e^{-\lambda t}$, here $\underline{\phi}\left(t\right)\geq u^{+}-\varepsilon_{1}e^{-\lambda t}$, $\underline{\psi}^{\prime}\left(t\right) = \lambda \varepsilon_{2}v^{+}e^{-\lambda t}$ and $\underline{\psi}^{\prime\prime}\left(t\right) = -\lambda^{2}\varepsilon_{2}v^{+}e^{-\lambda t}$. Thus,

Because of $-\Delta_{2}\left(0, c \right)+a_{55}v^{+} = -\vartheta_{2}+a_{55}v^{+} = \frac{\beta_{1} u^{+}}{1+mu^{+}+wv^{+}} > 0$, there is a constant $\widetilde{\lambda}_{4} > 0$, which makes $-\Delta_{2}\left(-\lambda, c \right)+a_{55}v^{+} > 0$ for $\forall\lambda\in\left(0, \widetilde{\lambda}_{4}\right)$.

Let $I_{4}\left(\lambda, t \right): = a_{55}\varepsilon_{2}\left(v^{+}\right)^{2}e^{-\lambda t}-a_{55}\left(\varepsilon_{2}\right)^{2}e^{-2\lambda t}-\frac{\beta_{1}u^{+}v^{+}}{1+mu^{+}+wv^{+} }+\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) \left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) +w\left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)}$, where $t\geq t_{4}$. Therefore, $I_{4}\left(\lambda, 0 \right) = -a_{55}\varepsilon_{2}^{2}+a_{55}v^{+}\varepsilon_{2}-\frac{\beta_{1} u^{+}v^{+}}{1+mu^{+}+wv^{+}}+\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}\right) \left(v^{+}-\varepsilon_{2}\right) }{1+m\left(u^{+}-\varepsilon_{1}\right)+w\left(v^{+}-\varepsilon_{2}\right)} > 0$ and $I_{4}\left(\lambda, +\infty \right) = 0$.

From Remark 3.1, we can see that $\vartheta_{2} > \frac{\left(2\sqrt{2}-1\right) \beta_{1} u^{+} }{1+mu^{+}+wv^{+}}$, there are $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$ and $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$, making $-a_{55}\varepsilon_{2}^{2}+a_{55}v^{+}\varepsilon_{2}-\frac{\beta_{1} u^{+}v^{+}}{1+mu^{+}+wv^{+}}+\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}\right) \left(v^{+}-\varepsilon_{2}\right) }{1+m\left(u^{+}-\varepsilon_{1}\right)+w\left(v^{+}-\varepsilon_{2}\right)} > \varepsilon_{0} > 0$. We have that $I_{4}\left(\lambda, 0 \right) > 0$, here $\varepsilon_{1}\in\left(0, \left(\sqrt{2}-1\right)u^{+} \right)$ and $\varepsilon_{2}\in\left(0, \frac{v^{+}}{2}\right)$. We can choose a small enough $\delta_{2} > 0$ such that $\delta^{**}: = \varepsilon_{2}+\delta_{2}$ for $\forall\delta\in\left[\varepsilon_{2}, \delta^{**} \right]$ satisfying

We want to prove that $I_{4}\left(\lambda, t \right) > 0$ for $\forall t\geq t_{4}$. Here $t\geq t_{4}$, $\left(t_{4} < 0 \right)$, therefore, $t$ is divided into $t\in\left[ t_{4}, 0\right]$ and $t > 0$ to discuss.

If $t\in\left(t_{4}, 0\right]$, let $\nu\left(t \right): = \varepsilon_{2}e^{-\lambda t}$. Select $\widetilde{\lambda}_{4} > 0$ small enough such that for any given $\lambda\in\left(0, \widetilde{\lambda}_{4}\right)$, we have $\nu\left(t_{4} \right) = \varepsilon_{2}e^{-\lambda t_{4}} = \delta^{**}$, which leads to $\varepsilon_{2}\leq\nu\left(t\right)\leq\delta^{**}$. So we get $I_{4}\left(\lambda, t \right) > 0$ for $\forall t\in\left(t_{4}, 0\right]$.

If $t\geq0$, here $\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) \left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) +w\left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)} > 0$, according to the assumption that $\lambda > 0$ is small enough, we have the function $\frac{\beta_{1}\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) \left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)}{1+m\left(u^{+}-\varepsilon_{1}e^{-\lambda t}\right) +w\left(v^{+}-\varepsilon_{2}e^{-\lambda t}\right)}$ is increasing about the variable $t$ on $t\geq0$; the function $a_{55}\varepsilon_{2}\left(v^{+}\right)^{2}e^{-\lambda t}-a_{55}\left(\varepsilon_{2}\right)^{2}e^{-2\lambda t} > 0$ is decreasing about the variable $t$ on $t\geq0$. We can get

So we get $I_{4}\left(\lambda, t \right) > 0$ for $\forall t\geq0$. Thus, $I_{4}\left(\lambda, t \right) > 0$ for $\forall t > t_{4}$. In consequence, $\underline{\psi}$ satisfies the following solution definition, that is, $D_{5}\underline{\psi}^{\prime\prime}\left(t \right)-c\underline{\psi}^{\prime}\left(t \right)+f_{5}\left(\underline{\phi}, \underline{\psi} \right) \left(t \right)\geq0$.

Let $\widetilde{\lambda} = min\left\lbrace \widetilde{\lambda}_{1}, \widetilde{\lambda}_{2}, \widetilde{\lambda}_{3}, \widetilde{\lambda}_{4}\right\rbrace$, select $\lambda\in\left(0, \widetilde{\lambda}\right)$, so that $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$ satisfy the upper and lower solution definition Definition 3.1, which proves the existence of the upper and lower solutions of the system (3.1). □

3.2.   Existence of traveling wave solutions

For $\mu > 0$, define

and

it is easy to know that $\left(B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right), \left|\ \cdot\ \right|_{\mu} \right)$ is a Banach space.

Define $\Omega = \left\lbrace \left(\phi, \psi\right)\in C\left(\mathbb{R}, \mathbb{R}^{2} \right): 0\leq\phi\left(t\right)\leq N_{1}, 0\leq\psi\left(t\right)\leq N_{2}, t\in \mathbb{R} \right\rbrace$, let $\eta_{1}$ and $\eta_{2}$ be two constants, satisfying

Define the operator $\mathit{\boldsymbol{H}} = \left(H_{1}, H_{2}\right):\Omega\longrightarrow C\left(\mathbb{R}, \mathbb{R}^{2} \right)$ as

the system (3.1) becomes the following form

Let

In this paper, we take the above $\mu$ to satisfy $0 < \mu < \min\left\lbrace -\lambda_{11}, \lambda_{12}, -\lambda_{21}, \lambda_{22}\right\rbrace$.

Define the operator $\mathit{\boldsymbol{F}} = \left(F_{1}, F_{2}\right):\Omega\longrightarrow C\left(\mathbb{R}, \mathbb{R}^{2} \right)$ as

Define the set $\Gamma = \left\lbrace \left(\phi, \psi\right)\in\Omega:\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)\leq\left(\phi\left(t\right), \psi\left(t\right) \right)\leq \left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)\right\rbrace$. Obviously, $\Gamma$ is not empty and is a bounded closed convex set. The operator $\mathit{\boldsymbol{F}} = \left(F_{1}, F_{2}\right)$ for $\forall\left(\phi, \psi\right)\in\Gamma$ satisfying

The fixed point of $\mathit{\boldsymbol{F}}$ is the solution of system (3.5), which is the solution of system (3.1).

Previously, we find a pair of upper and lower solutions $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$ of the system (3.1) satisfying properties $P1$), $P2$) and $P3$). We will find the traveling wave solutions of the system (3.1) in the profile set $\Gamma$.

Lemma 3.5. For sufficiently large $\eta_{1}$ and $\eta_{2}$ satisfying (3.4), we have

for $t\in \mathbb{R}$ with $0\leq\phi_{2}\left(t\right)\leq\phi_{1}\left(t\right)\leq N_{1}, 0\leq\psi_{2}\left(t\right)\leq\psi_{1}\left(t\right)\leq N_{2}$.

Proof. According to the definition of operator $\mathit{\boldsymbol{H}} = \left(H_{1}, H_{2}\right)$, we know that

The derivative of $H_{2}\left(\phi, \psi\right)\left(t\right)$ with respect to variable $\phi$ is obtained

the derivative $\frac{\partial H_{2}\left(\phi, \psi\right)}{\partial \phi} > 0$, so that $H_{2}$ is increasing with respect to the variable $\phi$.

The derivative of $H_{2}\left(\phi, \psi\right)\left(t\right)$ with respect to variable $\psi$ is obtained

Since (3.4) knows $\eta_{2}\geq 2a_{55}N_{2}+b_{2}+q_{5}e_{5}-a_{2} > 2a_{55}\psi\left(t\right)+b_{2}+q_{5}e_{5}-a_{2}-\frac{\beta_{1}\phi\left(1+m\phi \right) }{\left[ 1+m\phi+w\psi\right]^{2}}$, the derivative $\frac{\partial H_{2}\left(\phi, \psi\right)}{\partial \psi} > 0$, thus $H_{2}$ is increasing with respect to the variable $\psi$.

The derivative of $H_{1}\left(\phi, \psi\right)\left(t\right)$ with respect to variable $\psi$ is obtained

the derivative $\frac{\partial H_{1}\left(\phi, \psi\right)}{\partial \psi} < 0$, so that $H_{1}$is decreasing with respect to the variable $\psi$.

The derivative of $H_{1}\left(\phi, \psi\right)\left(t\right)$ with respect to variable $\phi$ is obtained

where $\left[\hat{a}e^{-d_{1}\tau}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{4\pi D_{1}\tau}}e^{-\frac{y^{2}}{4D_{1}\tau}}\phi\left(t-y-c\tau\right) dy \right]_{\phi}^{\prime} > 0$, $0\leq\phi\left(t\right)\leq N_{1}$, $0\leq\psi\left(t\right)\leq N_{2}$ and $\frac{\beta\psi\left(1+\omega\psi \right) }{\left[ 1+m\phi+w\psi\right]^{2}}\leq\beta\psi\left(1+\omega\psi \right) \leq \beta N_{2}\left(1+\omega N_{2}\right)$. Since (3.4) knows $\eta_{1}\geq d_{2}+q_{2}e_{2}+2a_{12}N_{1}+\beta N_{2}\left(1+\omega N_{2}\right) > d_{2}+q_{2}e_{2}+2a_{12}\phi\left(t\right)+\frac{\beta\psi\left(1+\omega\psi \right) }{\left[ 1+m\phi+w\psi\right]^{2}}-\left[\hat{a}e^{-d_{1}\tau}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{4\pi D_{1}\tau}}e^{-\frac{y^{2}}{4D_{1}\tau}}\phi\left(t-y-c\tau\right) dy \right]_{\phi}$, the derivative $\frac{\partial H_{1}\left(\phi, \psi\right)}{\partial \phi} > 0$, thus $H_{1}$ is increasing with respect to the variable $\phi$.

In consequence, $H_{1}$ is increasing with respect to the variable $\phi$ and is decreasing with respect to the variable $\psi$; $H_{2}$ is increasing with respect to the variable $\phi$ and is increasing with respect to the variable $\psi$. The above lemma holds. □

Lemma 3.6. For sufficiently large $\eta_{1}$ and $\eta_{2}$ satisfying (3.4), we have

for $t\in \mathbb{R}$ with $0\leq\phi_{2}\left(t\right)\leq\phi_{1}\left(t\right)\leq N_{1}, 0\leq\psi_{2}\left(t\right)\leq\psi_{1}\left(t\right)\leq N_{2}$.

Proof. According to the definition of the operator $\mathit{\boldsymbol{F}} = \left(F_{1}, F_{2}\right)$ and the lemma 3.5, we know that

thus $F_{1}\left(\phi_{1}, \psi_{1} \right) \left(t \right)\geq F_{1}\left(\phi_{2}, \psi_{1} \right) \left(t \right)$. Similarly, other conclusions can be proved. The above lemma is established. □

Lemma 3.7. $\mathit{\boldsymbol{F}}:\Gamma\longrightarrow\Gamma$ is completely continuous.

Proof. The proof process is divided into three parts.

Step1. $\mathit{\boldsymbol{F}} = \left(F_{1}, F_{2}\right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$ on $B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$. We first prove the continuity of $\mathit{\boldsymbol{H}}$.

We can notice that

If $\Phi = \left(\phi_{1}, \psi_{1} \right)$, $\Psi = \left(\phi_{2}, \psi_{2} \right)\in B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$. We get

where $\kappa_{1} = \hat{a}e^{-d_{1}\tau}e^{\left(D_{1}\mu^{2}+c\mu\right)\tau }+d_{2}+q_{2}e_{2}+2a_{12}N_{1}+\beta N_{1}+m\beta N_{1}^{2}+\beta N_{2}+m\beta N_{2}^{2}+\eta_{1}$. Thus, $H_{1}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$.

Similarly, we can prove that $H_{2}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$.

where $\kappa_{2} = a_{2}+b_{2}+q_{5}e_{5}+2a_{55}N_{2}+\beta_{1} N_{1}+m\beta_{1} N_{1}^{2}+\beta_{1} N_{2}+m\beta_{1} N_{2}^{2}+\eta_{2}$. Thus $H_{2}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$.

Next we prove that $F_{1}$ and $F_{2}$ are continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$ on $B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$. Because $t\in \mathbb{R}$, we divide it into $t > 0$ and $t\leq0$ for discussion.

For $t > 0$, due to $\left|H_{1}\left(\phi_{1}, \psi_{1} \right) \left(t \right)- H_{1}\left(\phi_{2}, \psi_{2} \right) \left(t \right) \right| e^{-\mu\left|t\right|}\leq\kappa_{1}\left|\Phi-\Psi\right|_{\mu}$, thus

For $t\leq0$, due to $\left|H_{1}\left(\phi_{1}, \psi_{1} \right) \left(t \right)- H_{1}\left(\phi_{2}, \psi_{2} \right) \left(t \right) \right| e^{-\mu\left|t\right|}\leq\kappa_{1}\left|\Phi-\Psi\right|_{\mu}$, thus

Therefore, $F_{1}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$ on $B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$.

Similarly, $F_{2}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$ on $B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$.

Step2. $\mathit{\boldsymbol{F}}\left(\Gamma\right)\subset\Gamma$, that is, for any $\left(\phi, \psi\right)\in\Gamma$, we have $\mathit{\boldsymbol{F}}\left(\phi, \psi\right)\in\Gamma$.

Since $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)\leq\left(\phi\left(t\right), \psi\left(t\right) \right)\leq \left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$, according to the Lemma 3.6 can get

Next we prove that $F_{1}\left(\overline{\phi}, \underline{\psi}\right)\leq\overline{\phi}$.

Without losing generality, we assume the finite point set $S = \left\lbrace s_{i}\in\mathbb{R}, i = 1, 2, \ldots, n \right\rbrace$, where $s_{1} < s_{2} < \cdots < s_{n}$, and define $s_{0} = 0$, $s_{n+1} = +\infty$.

According to the definition of Definition 3.1, we have $H_{1}\left(\overline{\phi}, \underline{\psi} \right) \left(t \right)\leq-D_{2}\overline{\phi}^{\prime\prime}\left(t \right)+c\overline{\phi}^{\prime}\left(t \right)+\eta_{1}\overline{\phi}\left(t\right)$ for $t\in\mathbb{R}\backslash S$.

Due to the properties $P3)$ of $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$, that is, $\overline{\phi^{\prime}}\left(t+\right)\leq\overline{\phi^{\prime}}\left(t-\right)$ and $\underline{\phi^{\prime}}\left(t+\right)\geq\underline{\phi^{\prime}}\left(t-\right)$ for $\forall t\in\mathbb{R}$. Then,

In fact, through the continuity of $F_{1}\left(\overline{\phi}, \underline{\psi}\right)\left(t \right)$ and $\overline{\phi}\left(t\right)$, the above inequality holds for $t\in\mathbb{R}$.

Similarly, we can get $F_{1}\left(\underline{\phi}, \overline{\psi}\right)\geq\underline{\phi}$, $F_{2}\left(\underline{\phi}, \underline{\psi}\right)\geq\underline{\psi}$, $F_{2}\left(\overline{\phi}, \overline{\psi}\right)\leq\overline{\psi}$, then $\underline{\phi}\leq F_{1}\left(\phi, \psi\right)\leq\overline{\phi}$, $\underline{\psi}\leq F_{2}\left(\phi, \psi\right)\leq\overline{\psi}$. Therefore, $\mathit{\boldsymbol{F}}\left(\phi, \psi\right)\in\Gamma$ for $\forall\left(\phi, \psi\right)\in\Gamma$.

Step3. $\mathit{\boldsymbol{F}}:\Gamma\longrightarrow\Gamma$ is compact.

For any $\left(\phi, \psi\right)\in\Gamma$,

Therefore, we have

If $t > 0$, then

If $t < 0$, then

Therefore, $F_{2}:B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)\longrightarrow B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ is continuous with respect to the norm $\left|\ \cdot\ \right|_{\mu}$, and the set $\Gamma$ is uniformly bounded.

Thus there exists the constant $M_{2}$ such that $\left| F_{2}^{\prime}\left(\phi, \psi\right)\left(t\right)\right|_{\mu} \leq M_{2}$; Similarly, there exists constant $M_{1}$ such that $\left| F_{1}^{\prime}\left(\phi, \psi\right)\left(t\right)\right|_{\mu}\leq M_{1}$. Therefore, $\mathit{\boldsymbol{F}}$ is equicontinuous on $\Gamma$ and $\mathit{\boldsymbol{F}}\left(\Gamma\right)$ is uniformly bounded.

Next we prove that $\mathit{\boldsymbol{F}}:\Gamma\longrightarrow\Gamma$ is compact. Define $\mathit{\boldsymbol{F}}^{n}\left(\phi, \psi\right)$ as follows

For any $n\geq1$, $\mathit{\boldsymbol{F}}^{n}\left(\Gamma\right)$ is uniformly bounded equicontinuous.

Now, in the interval $\left[ -n, n\right]$, it follows from Ascoli-Arzela Theorem that $\mathit{\boldsymbol{F}}^{n}$ is compact.

In addition, in $B_{\mu}\left(\mathbb{R}, \mathbb{R}^{2} \right)$ we have $\mathit{\boldsymbol{F}}^{n}\longrightarrow\mathit{\boldsymbol{F}}$, as $n\longrightarrow +\infty$. For any $\left(\phi, \psi\right)\in\Gamma$,

In consequence, $\mathit{\boldsymbol{F}}$ is compact. □

Theorem 3.1. Assume that the unique positive equilibrium $E_{3}\left(u^{+}, v^{+}\right)$ exists and satisfies

then for every $c > c^{*}$, There is always a traveling wave solution $\left(\phi^{*}\left(t\right), \psi^{*}\left(t\right) \right)$ connecting the equilibrium points $\left(0, 0\right)$ and $\left(u^{+}, v^{+}\right)$ with the wave velocity $c$ in the system (2.1). Moreover

Proof. According to the Lemma 3.5, Lemma 3.6 and Schauder fixed point theorem, it can be concluded that the operator $\mathit{\boldsymbol{F}}$ has a fixed point $\left(\phi^{*}\left(t\right), \psi^{*}\left(t\right) \right)$ in $\Gamma$, so $\left(\phi^{*}\left(t\right), \psi^{*}\left(t\right) \right)$ is the solution of the system (3.1).

In order to prove that the solution is a traveling wave solution, only the asymptotic conditions need to be verified. According to the property $P2$) of $\left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$ and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)$, and $\left(\underline{\phi}\left(t\right), \underline{\psi}\left(t\right) \right)\leq\left(\phi^{*}\left(t\right), \psi^{*}\left(t\right) \right)\leq \left(\overline{\phi}\left(t\right), \overline{\psi}\left(t\right) \right)$, we have

Because of $\underline{\phi}\leq\phi^{*}\leq\overline{\phi}$ and $\underline{\psi}\leq\psi^{*}\leq\overline{\psi}$, then

Consequently,

The above conclusion is proved. □

Use of AI tools declaration

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The authors declare that there is no conflicts of interest.