Existence and global attractivity of positive periodic solutions for a Holling II two-prey one-predator system

Abstract

In this paper, a Holling II two-pery one-predator system is investigated. Based on the continuation theorem of coincidence degree theory and by constructing a suitable Lyapunov function, we derive a set of sufficient conditions that guarantee the existence of at least a positive periodic solution and global attractivity of periodic solutions.

Mathematics Subject Classification 2000: 34K20; 34C25

Introduction

In population dynamics, the functional response, which is a key element in all predator-prey interaction, is referred to the number of prey eaten per predator per unit time as a function of prey density. Based on a lot of experiments, Holling [1] suggested the following three different kinds of functional response for different species to model the phenomenon of predation:

$$\left(1\right)p_1\left(x\right)=ax,\left(2\right)p_2\left(x\right)=\frac{ax}{m+x},\left(3\right)p_3\left(x\right)=\frac{a{x}^2}{m+x^2},$$

where x(t) represents the prey density at time t. Functions p₁(x)(i = 1, 2, 3) are referred to the Holling type I, II, and III functional response, respectively. a > 0 denotes the search rate of the predator, m > 0 is the half-saturation constant. Predator-prey systems with Holling type functional response have been investigated extensively, for example, Liu and Chen [2] made a discussion on complex dynamics of Holling type II Lotka-Volterra predator-prey model with impulsive perturbations on the predator. Song and Li [3] studied the linear stability of trivial periodic solution and semi-trivial periodic solutions and the permanence of the periodic predator-prey model with modified Leslie-Gower Holling-type II schemes and impulsive effect. Liu and Xu [4] investigated the existence of periodic solution for a delay one-predator and two-prey system with Holling type-II functional response. Agiza et al. [5] considered the chaotic phenomena of a discrete prey-predator model with Holling type II. Pei et al. [6] analyzed the extinction and permanence for one-prey multi-predators of Holling type II function response system with impulsive biological control. For more knowledge about this theme, one can see [7–18].

In 2007, Song and Li [19] had considered the dynamical behaviors of the following Holling II two-prey one predator system with impulsive effect

$$\left\{\begin{array}{l}\left.\begin{array}{c}{ẋ}_1\left(t\right)=x_1\left(t\right)\left[b_1-x_1\left(t\right)-\alpha x_2\left(t\right)-\frac{\eta z\left(t\right)}{1+{\omega }_1{x}_1\left(t\right),}\right],\hfill \\ {ẋ}_2\left(t\right)=x_2\left(t\right)\left[b_2-\beta x_1\left(t\right)-x_2\left(t\right)-\frac{\eta z\left(t\right)}{1+{\omega }_2{x}_2\left(t\right)}\right],\hfill \\ ż\left(t\right)=z\left(t\right)\left[-b_3+\frac{d\eta x_1\left(t\right)}{1+{\omega }_1{x}_1\left(t\right)}+\frac{d\eta x_2\left(t\right)}{1+{\omega }_2{x}_2\left(t\right)}\right],\hfill \end{array}\right\}t\ne nT,\\ \left.\begin{array}{l}\Delta x_1\left(t\right)=-p_1{x}_1\left(t\right),\\ \Delta x_1\left(t\right)=-p_2{x}_2\left(t\right),\\ \Delta z\left(t\right)=0,\end{array}\right\}t=nT,\end{array}\right.$$

(1)

where x _i (t)(i = 1, 2) is the population size of prey (pest) species and z(t) is the population size of predator (natural enemies) species, b _i > 0(i = 1, 2, 3) are intrinsic rates of increase or decrease, α > 0 and β > 0 are parameters representing competitive effects between two prey, η > 0 and $\mu >0,\frac{\eta x_1\left(t\right)}{1+{\omega }_1{x}_1\left(t\right)}$ and $\frac{\mu x_2\left(t\right)}{1+{\omega }_2{x}_2\left(t\right)}$ are the Holling type II functional responses, d > 0 is the rate of conversing prey into predator. Δx _i (t) = x _i (t⁺)-x _i (t), i = 1, 2, Δz(t) = z(t⁺)-z(t), T is the period of the impulse for predator in order to eradicate both target pests, protect non-target pest (or harmless insect) from extinction and drive target pest to extinction, or control target pests at acceptably low level to prevent an increasing pest populations from causing an economic loss. n ∈ z⁺, z⁺ = {1, 2, ... g, pi > 0(i = 1, 2) is the proportionality constant which represents the rate of mortality due to the applied pesticide. q > 0 is the number of predators released each time. We note that any biological or environmental parameters are naturally subject to fluctuation in time. It is necessary and important to consider models with periodic ecological parameters Thus, the assumption of periodicity of the parameters is a way of incorporating the periodicity of the environment. Furthermore, for simplification, we assume that there is no pulse in system. Based on the point of view, system (1) can be modified as the form:

$$\left\{\begin{array}{c}{ẋ}_1\left(t\right)=x_1\left(t\right)\left[b_1\left(t\right)-x_1\left(t\right)-\alpha \left(t\right)x_2\left(t\right)-\frac{\eta \left(t\right)z\left(t\right)}{1+{\omega }_1\left(t\right)x_1\left(t\right)}\right],\hfill \\ {ẋ}_2\left(t\right)=x_2\left(t\right)\left[b_2\left(t\right)-\beta \left(t\right)x_1\left(t\right)-x_2\left(t\right)-\frac{\eta \left(t\right)z\left(t\right)}{1+{\omega }_2\left(t\right)x_2\left(t\right)}\right],\hfill \\ ż\left(t\right)=z\left(t\right)\left[-b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)x_1\left(t\right)}{1+{\omega }_1\left(t\right)x_1\left(t\right)}+\frac{d\left(t\right)\mu \left(t\right)x_2\left(t\right)}{1+{\omega }_2\left(t\right)x_2\left(t\right)}\right].\hfill \end{array}\right.$$

(2)

Here we give the initial conditions as follows

$$x_i\left(0\right)={\phi }_i\left(0\right)>0\left(i=1,2\right),z\left(0\right)={\phi }_3\left(0\right)>0.$$

(3)

Throughout the paper, we always assume that

(H1) For any t ∈ R, b _i (t)(i = 1, 2, 3), ω _j (t)(j = 1, 2), α(t), β(t), η(t), μ(t), d(t) are all non-negative continuous ω periodic functions, i.e., b _i (t + ω) = b _i (t)(i = 1, 2, 3), ω _j (t + ω) = ω _j (t)(j = 1, 2), α(t + ω) = α(t), β(t + ω) = β(t), η(t + ω) = η (t), μ(t + ω) = μ(t), d(t + ω) = d(t).

The principle object of this article is to find a set of sufficient conditions that guarantee the existence of at least a positive periodic solution and global attractivity of periodic solutions for system (2)-(3). There are some papers which deal with this topic [13, 20–25].

The paper is organized as follows: In Section "Basic lemma", we introduce some basic Lemmas. In Section "Existence of positive periodic solutions", sufficient conditions are established for the existence of positive periodic solutions of system (2)-(3). In Section "Uniqueness and global attractivity", by means of suitable Lyapunov functionals, a set of sufficient conditions are derived for the uniqueness and global attractivity of positive periodic solutions of system (2)-(3).

Basic lemma

In order to explore the existence of positive periodic solutions of (2)-(3) and for the reader's convenience, we shall first summarize below a few concepts and results without proof, borrowing from [11].

Let X, Y be normed vector spaces, L : DomL ∈ X → Y is a linear mapping, N : X → Y is a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL = codimImL < +∞ and ImL is closed in Y . If L is a Fredholm mapping of index zero and there exist continuous projectors P : X → X and Q : Y → Y such that ImP = KerL, ImL = KerQ = Im(I - Q), it follows that L| DomL ∩ KerP : (I - P)X → ImL is invertible. We denote the inverse of that map by K _P . If Ω is an open bounded subset of X, the mapping N will be called L-compact on $\bar{\mathrm{\Omega }}$ if $QN\left(\bar{\mathrm{\Omega }}\right)$ is bounded and $K_P\left(I-Q\right)N:\bar{\mathrm{\Omega }}\to X$ is compact. Since ImQ is isomorphic to KerL, there exist isomorphisms J : ImQ → KerL.

Lemma 1. ([11] Continuation Theorem) Let L be a Fredholm mapping of index zero and let N be L-compact on $\bar{\mathrm{\Omega }}$. Suppose

• (a) for each λ ∈ (0, 1), every solution x of Lx = λNx is such that x ∉ ∂Ω;

• (b) QNx ≠ 0 for each x ∈ KerL∩∂Ω, and deg{JQN, Ω∩KerL, 0} ≠ 0, then the equation Lx = Nx has at least one solution lying in $DomL\cap \bar{\mathrm{\Omega }}$.

Lemma 2. $R_{+}^3=\{\left({\left(x_1\left(t\right),x_2\left(t\right),z\left(t\right)\right)}^T\in R^3|x_1\left(t\right)>0,x_2\left(t\right)>0,z\left(t\right)>0\right\}$ is positive invariant with respect to system (2)-(3).

Proof. In fact,

$$\left\{\begin{array}{c}{ẋ}_1\left(t\right)={\phi }_1\left(0\right)\mathsf{\text{exp}}\left\{{\int }_0^t\left[b_1\left(s\right)-x_1\left(s\right)-\alpha \left(s\right)x_2\left(s\right)-\frac{\eta \left(s\right)z\left(s\right)}{1+{\omega }_1\left(s\right)x_1\left(s\right)}\right]\mathsf{\text{d}}s\right\},\hfill \\ {ẋ}_2\left(t\right)={\phi }_2\left(0\right)\mathsf{\text{exp}}\left\{{\int }_0^t\left[b_2\left(s\right)-\alpha \left(s\right)x_1\left(s\right)-x_2\left(s\right)-\frac{\eta \left(s\right)z\left(s\right)}{1+{\omega }_2\left(s\right)x_2\left(s\right)}\right]\mathsf{\text{d}}s\right\},\hfill \\ ż\left(t\right)={\phi }_2\left(0\right)\mathsf{\text{exp}}\left\{{\int }_0^t\left[-b_3\left(s\right)+\frac{d\left(s\right)\eta \left(s\right)x_1\left(s\right)}{1+{\omega }_1\left(s\right)x_1\left(s\right)}+\frac{d\left(s\right)\mu \left(s\right)x_2\left(s\right)}{1+{\omega }_2\left(s\right)x_2\left(s\right)}\right]\mathsf{\text{d}}s\right\}.\hfill \end{array}\right.$$

Obviously, the conclusion follows.

Existence of positive periodic solutions

For convenience and simplicity in the following discussion, we always use the notations below throughout the paper:

$$ḡ=\frac{1}{\omega }\underset{0}{\overset{\omega }{\int }}g\left(t\right)\mathsf{\text{d}}t,g^L=\underset{t\in \left[0,\omega \right]}{\mathsf{\text{min}}}g\left(t\right),g^M=\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}g\left(t\right),$$

where g(t) is an ω continuous periodic function. In the following, we will ready to state and prove our result.

Theorem 1. Let K₁, K₂, K₄ and K₅ are defined by (19), (23), (32) and (36), respectively. In addition to (H1), if the following conditions (H2) and (H3)

$$\begin{array}{ll}\hfill \left(H2\right)& {\bar{b}}_1>\mathsf{\text{max}}\left\{\mathsf{\text{exp}}\left\{-K_1\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{-K_2\right\},\mathsf{\text{exp}}\left\{K_1\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{K_2\right\},\right.\\ \left.\mathsf{\text{exp}}\left\{-K_5\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\},\mathsf{\text{exp}}\left\{K_5\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{K_4\right\}\right\},\end{array}$$

$$\left(H3\right)b_3^M>\mathsf{\text{max}}\left\{d^M{\eta }^M\mathsf{\text{exp}}\left\{K_1\right\},d^M{\mu }^M\mathsf{\text{exp}}\left\{K_4\right\}\right\}$$

hold, then system (2)-(1.3) has at least one ω periodic solution.

Proof. Since solutions of (2)-(3) remain positive for all t ≥ 0, we let

$$u_1\left(t\right)=\mathsf{\text{ln}}\left[x_1\left(t\right)\right],u_2\left(t\right)=\mathsf{\text{ln}}\left[x_2\left(t\right)\right],u_3\left(t\right)=\mathsf{\text{ln}}\left[z\left(t\right)\right].$$

(4)

Substituting (4) into (2), we obtain

$$\left\{\begin{array}{c}{\dot{u}}_1\left(t\right)=b_1\left(t\right)-\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}},\hfill \\ {\dot{u}}_2\left(t\right)=b_2\left(t\right)-\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}},\hfill \\ {\dot{u}}_3\left(t\right)=-b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}.\hfill \end{array}\right.$$

(5)

It is easy to see that if system (5) has one ω periodic solution ${\left(u_1^{*}\left(t\right),u_2^{*}\left(t\right),u_3^{*}\left(t\right)\right)}^T,$ then ${\left(x_1\left(t\right),x_2\left(t\right),y\left(t\right)\right)}^T={\left(\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\},\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\},\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}\right)}^T$ is a positive solution of system (2). Therefore, to complete the proof, it suffices to show that system (5) has at least one ω periodic solution.

Let X = Z = u(t) = {(u₁(t), u₂(t), u₃(t))^T| u(t) ∈ C(R, R³), u(t + ω) = u(t)}, and define ||u|| = ||(u₁(t); u₂(t), u₃(t))^T || = max_t∈[0,ω]|u₁(t)| + max_t∈[0,ω]|u₂(t)| + max_t∈[0,ω]|u₃(t)|. Then X and Z are Banach spaces when they are endowed with the norm || · ||. Let L : DomL ∈ X → Z and N : X → Z be the following:

$$Nu=\left(\begin{array}{c}\hfill Lu=\dot{u}\left(t\right),\hfill \\ \hfill b_1\left(t\right)-\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\hfill \\ \hfill b_2\left(t\right)-\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\hfill \\ \hfill -b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\hfill \end{array}\right)$$

(6)

Define continuous projective operators P and Q:

$$Pu=\frac{1}{\omega }\underset{0}{\overset{\omega }{\int }}u\left(t\right)\mathsf{\text{d}}t,Qu=\frac{1}{\omega }\underset{0}{\overset{\omega }{\int }}u\left(t\right)\mathsf{\text{d}}t,u\in X,u\in Z.$$

We can see that $\mathsf{\text{Ker}}L=\left\{u\in X|u=h\in R^3\right\},\mathsf{\text{Im}}L=\left\{u\in Z|{\int }_0^{\omega }u\left(t\right)\mathsf{\text{d}}t=0\right\}$ is closed in X and dim(KerL) = 3 = codim(ImL), then it follows that L is a Fredholm mapping of index zero. Moreover, it is easy to check that

$$QNu=\left(\begin{array}{c}\hfill \frac{1}{\omega }{\int }_0^{\omega }\left[b_1\left(t\right)-\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\right]\text{d}t\hfill \\ \hfill \frac{1}{\omega }{\int }_0^{\omega }\left[b_2\left(t\right)-\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\text{d}t\hfill \\ \hfill \frac{1}{\omega }{\int }_0^{\omega }\left[-b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\text{d}t\hfill \end{array}\right)$$

$$\begin{array}{ll}\hfill K_P\left(I-Q\right)Nu& =\left(\begin{array}{c}\hfill {\int }_0^t\left[b_1\left(s\right)-\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\alpha \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{\eta \left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \\ \hfill {\int }_0^t\left[b_2\left(s\right)-\beta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{{\mu }_1\left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \\ \hfill {\int }_0^t\left[-b_3\left(s\right)+\frac{d\left(s\right)\eta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}+\frac{d\left(s\right)\mu \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \end{array}\right)\\ -\left(\begin{array}{c}\hfill \frac{1}{\omega }{\int }_0^{\omega }{\int }_0^t\left[b_1\left(s\right)-\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\alpha \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{\eta \left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}\right]dsdt\hfill \\ \hfill \frac{1}{\omega }{\int }_0^{\omega }{\int }_0^t\left[b_2\left(s\right)-\beta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{{\mu }_1\left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]dsdt\hfill \\ \hfill \frac{1}{\omega }{\int }_0^{\omega }{\int }_0^t\left[-b_3\left(s\right)+\frac{d\left(s\right)\eta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}+\frac{d\left(s\right)\mu \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]dsdt\hfill \end{array}\right)\\ -\left(\begin{array}{c}\hfill \left(\frac{t}{\omega }-\frac{1}{2}\right){\int }_0^{\omega }\left[b_1\left(s\right)-\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\alpha \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{\eta \left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \\ \hfill \left(\frac{t}{\omega }-\frac{1}{2}\right){\int }_0^{\omega }\left[b_2\left(s\right)-\beta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}-\frac{{\mu }_1\left(s\right)\mathsf{\text{exp}}\left\{u_3\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \\ \hfill \left(\frac{t}{\omega }-\frac{1}{2}\right){\int }_0^{\omega }\left[-b_3\left(s\right)+\frac{d\left(s\right)\eta \left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}{1+{\omega }_1\left(s\right)\mathsf{\text{exp}}\left\{u_1\left(s\right)\right\}}+\frac{d\left(s\right)\mu \left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}{1+{\omega }_2\left(s\right)\mathsf{\text{exp}}\left\{u_2\left(s\right)\right\}}\right]\mathsf{\text{d}}s\hfill \end{array}\right).\end{array}$$

(7)

Obviously, QN and K _P (I - Q)N are continuous. Since X is a finite-dimensional Banach space, using the Ascoli-Arzela theorem, it is not difficult to show that $\overline{K_P\left(I-Q\right)N\left(\bar{\mathrm{\Omega }}\right)}$ is compact for any open bounded set Ω ⊂ X. Moreover, $QN\left(\bar{\mathrm{\Omega }}\right)$ is bounded. Thus, N is L-compact on $\bar{\mathrm{\Omega }}$ with any open bounded set Ω ⊂ X.

Now we are at the point to search for an appropriate open, bounded subset Ω for the application of the continuation theorem. Corresponding to the operator equation Lu = λNu, λ ∈ (0, 1), we have

$$\left\{\begin{array}{c}{\dot{u}}_1\left(t\right)=\lambda \left[b_1\left(t\right)-\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\right],\hfill \\ {\dot{u}}_2\left(t\right)=\lambda \left[b_2\left(t\right)-\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right],\hfill \\ {\dot{u}}_3\left(t\right)=\lambda \left[-b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right].\hfill \end{array}\right.$$

(8)

Suppose that u(t) = (u₁(t), u₂(t), u₃(t))^T ∈ X is an arbitrary solution of system (8) for a certain λ ∈ (0, 1), integrating both sides of (8) over the interval [0, ω] with respect to t, we obtain

$$\left\{\begin{array}{c}{\int }_0^{\omega }\left[\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}+\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\right]\mathsf{\text{d}}t={\bar{b}}_1\omega ,\hfill \\ {\int }_0^{\omega }\left[\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}+\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}+\frac{\mu \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\mathsf{\text{d}}t={\bar{b}}_2\omega ,\hfill \\ {\int }_0^{\omega }\left[\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\mathsf{\text{d}}t={\bar{b}}_3\omega .\hfill \end{array}\right.$$

(9)

In view of (8) and (9), we have

$$\begin{array}{ll}\hfill \underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_1\left(t\right)\right|\mathsf{\text{d}}t& =\lambda \underset{0}{\overset{\omega }{\int }}\left|b_1\left(t\right)-\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\right|\mathsf{\text{d}}t\\ \le \underset{0}{\overset{\omega }{\int }}{b}_1\left(t\right)\mathsf{\text{d}}t+\underset{0}{\overset{\omega }{\int }}\left[\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}+\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\right]\mathsf{\text{d}}t\\ =2\underset{0}{\overset{\omega }{\int }}{b}_1\left(t\right)\mathsf{\text{d}}t=2{\bar{b}}_1\omega ,\end{array}$$

(10)

$$\begin{array}{ll}\hfill \underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_2\left(t\right)\right|\mathsf{\text{d}}t& =\lambda \underset{0}{\overset{\omega }{\int }}\left|b_2\left(t\right)-\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}-\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right|\mathsf{\text{d}}t\\ \le \underset{0}{\overset{\omega }{\int }}{b}_1\left(t\right)\mathsf{\text{d}}t+\underset{0}{\overset{\omega }{\int }}\left[\beta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}+\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}+\frac{{\mu }_1\left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\mathsf{\text{d}}t\\ =2\underset{0}{\overset{\omega }{\int }}{b}_2\left(t\right)\mathsf{\text{d}}t=2{\bar{b}}_2\omega ,\end{array}$$

(11)

$$\begin{array}{ll}\hfill \underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_3\left(t\right)\right|\mathsf{\text{d}}t& =\lambda \underset{0}{\overset{\omega }{\int }}\left|-b_3\left(t\right)+\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right|\mathsf{\text{d}}t\\ \le \underset{0}{\overset{\omega }{\int }}{b}_3\left(t\right)\mathsf{\text{d}}t+\underset{0}{\overset{\omega }{\int }}\left[\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}+\frac{d\left(t\right)\mu \left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\right]\mathsf{\text{d}}t\\ =2\underset{0}{\overset{\omega }{\int }}{b}_3\left(t\right)\mathsf{\text{d}}t=2{\bar{b}}_3\omega .\end{array}$$

(12)

Since u = (u₁, u₂, u₃)^T ∈ X, then there exist ξ _i , η _i ∈ [0, ω] such that

$$u_i\left({\xi }_i\right)=\underset{t\in \left[0,\omega \right]}{\mathsf{\text{min}}}{u}_i\left(t\right),u_i\left({\eta }_i\right)=\underset{t\in \left[0,\omega \right]}{\mathsf{\text{min}}}{u}_i\left(t\right),i=1,2.3.$$

It is easy to see that u _i (ξ _i ) = 0, u _i (η _i ) = 0(i = 1, 2, 3). From this and (8), we have

$$\left\{\begin{array}{c}{b}_1\left({\xi }_1\right)-\mathsf{\text{exp}}\left\{u_1\left({\xi }_1\right)\right\}-\alpha \left({\xi }_1\right)\mathsf{\text{exp}}\left\{u_2\left({\xi }_1\right)\right\}-\frac{\eta \left({\xi }_1\right)\mathsf{\text{exp}}\left\{u_3\left({\xi }_1\right)\right\}}{1+{\omega }_1\left({\xi }_1\right)\mathsf{\text{exp}}\left\{u_1\left({\xi }_1\right)\right\}}=0,\hfill \\ b_2\left({\xi }_2\right)-\beta \left({\xi }_2\right)\mathsf{\text{exp}}\left\{u_1\left({\xi }_2\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left({\xi }_2\right)\right\}-\frac{{\mu }_1\left({\xi }_2\right)\mathsf{\text{exp}}\left\{u_3\left({\xi }_2\right)\right\}}{1+{\omega }_2\left({\xi }_2\right)\mathsf{\text{exp}}\left\{u_2\left({\xi }_2\right)\right\}}=0,\hfill \\ -b_3\left({\xi }_3\right)+\frac{d\left({\xi }_3\right)\eta \left({\xi }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\xi }_3\right)\right\}}{1+{\omega }_1\left({\xi }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\xi }_3\right)\right\}}+\frac{d\left({\xi }_3\right)\mu \left({\xi }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\xi }_3\right)\right\}}{1+{\omega }_2\left({\xi }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\xi }_3\right)\right\}}=0\hfill \end{array}\right.$$

(13)

and

$$\left\{\begin{array}{c}{b}_1\left({\eta }_1\right)-\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}-\alpha \left({\eta }_1\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_1\right)\right\}-\frac{\eta \left({\eta }_1\right)\mathsf{\text{exp}}\left\{u_3\left({\eta }_1\right)\right\}}{1+{\omega }_1\left({\eta }_1\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}}=0,\hfill \\ b_2\left({\eta }_2\right)-\beta \left({\eta }_2\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_2\right)\right\}-\mathsf{\text{exp}}\left\{u_2\left({\eta }_2\right)\right\}-\frac{{\mu }_1\left({\eta }_2\right)\mathsf{\text{exp}}\left\{u_3\left({\eta }_2\right)\right\}}{1+{\omega }_2\left({\eta }_2\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_2\right)\right\}}=0,\hfill \\ -b_3\left({\eta }_3\right)+\frac{d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}}{1+{\omega }_1\left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}}+\frac{d\left({\eta }_3\right)\mu \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}}{1+{\omega }_2\left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}}=0.\hfill \end{array}\right.$$

(14)

It follows from the first and the second equation of (13) that

$$\mathsf{\text{exp}}\left\{u_1\left({\xi }_1\right)\right\}<b_1\left({\xi }_1\right)=b_1^L,\mathsf{\text{exp}}\{\left(u_2\left({\xi }_2\right)\right\}<b_2\left({\xi }_2\right)=b_2^L$$

which leads to

$$u_1\left({\xi }_1\right)<\mathsf{\text{ln}}\left[b_1^L\right],u_2\left({\xi }_2\right)<\mathsf{\text{ln}}\left[b_2^L\right].$$

(15)

In the sequel, we consider two cases.

Case 1. If u₁(η₁) ≥ u₂(η₂), then from the third equation of (14), we get

$$\begin{array}{ll}\hfill b_3^M=b_3\left({\eta }_3\right)& <d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}+d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}\\ \le d^M{\eta }^M\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}+d^M{\eta }^M\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}\\ =\left(d^M{\eta }^M+d^M{\mu }^M\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}.\end{array}$$

Then we have

$$u_1\left({\eta }_1\right)>\mathsf{\text{ln}}\left[\frac{b_3^M}{d^M{\eta }^M+d^M{\mu }^M}\right].$$

(16)

By (10), (15) and (16), we can obtain

$$u_1\left(t\right)\le u_1\left({\xi }_1\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_1\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[b_1^L\right]+2{\bar{b}}_1\omega :=B_1,$$

(17)

$$u_1\left(t\right)\ge u_1\left({\eta }_1\right)-\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_1\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{b_3^M}{d^M{\eta }^M+d^M{\mu }^M}\right]-2{\bar{b}}_1\omega :=B_2.$$

(18)

It follows from (17) and (18) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_1\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_1\right|,\left|B_2\right|\right\}:=K_1.$$

(19)

From the third equation of (14), we derive

$$\begin{array}{ll}\hfill b_3^M=b_3\left({\eta }_3\right)& <d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}+d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}\\ \le d^M{\eta }^M\mathsf{\text{exp}}\left\{K_1\right\}+d^M{\eta }^M\mathsf{\text{exp}}\left\{u_2\left({\eta }_2\right)\right\}\end{array}$$

which leads to

$$u_2\left({\eta }_2\right)>\mathsf{\text{ln}}\left[\frac{b_3^M-d^M{\eta }^M\mathsf{\text{exp}}\left\{K_1\right\}}{d^M{\mu }^M}\right].$$

(20)

In view of (11), (15) and (20), we can obtain

$$u_2\left(t\right)\le u_2\left({\xi }_2\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_2\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[b_2^L\right]+2{\bar{b}}_2\omega :=B_3,$$

(21)

$$u_2\left(t\right)\ge u_2\left({\eta }_2\right)-\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_2\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{b_3^M-d^M{\eta }^M\mathsf{\text{exp}}\left\{K_1\right\}}{d^M{\mu }^M}\right]-2{\bar{b}}_2\omega :=B_4.$$

(22)

It follows from (21) and (22) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_2\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_3\right|,\left|B_4\right|\right\}:=K_2.$$

(23)

From the first equation of (9), we get

$$\begin{array}{l}\underset{0}{\overset{\omega }{\int }}\left[\mathsf{\text{exp}}\left\{-K_1\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{-K_2\right\}+\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left({\xi }_3\right)\right\}}{1+{\omega }_1^M\mathsf{\text{exp}}\left\{-K_1\right\}}\right]\mathsf{\text{d}}t<{\bar{b}}_1\omega ,\\ \underset{0}{\overset{\omega }{\int }}\left[\mathsf{\text{exp}}\left\{K_1\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{K_2\right\}+\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left({\eta }_3\right)\right\}\right]\mathsf{\text{d}}t>{\bar{b}}_1\omega ,\end{array}$$

which reduces to

$$\begin{array}{l}\mathsf{\text{exp}}\left\{-K_1\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{-K_2\right\}+\frac{\bar{\eta }\mathsf{\text{exp}}\left\{u_3\left({\xi }_3\right)\right\}}{1+{\omega }_1^M\mathsf{\text{exp}}\left\{-K_1\right\}}<{\bar{b}}_1,\\ \mathsf{\text{exp}}\left\{K_1\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{K_2\right\}+\bar{\eta }\mathsf{\text{exp}}\left\{u_3\left({\eta }_3\right)\right\}>{\bar{b}}_1.\end{array}$$

Therefore, we have

$$u_3\left({\xi }_3\right)<\mathsf{\text{ln}}\left[\frac{\left({\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_1\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_2\right\}\right)\left(1+{\omega }_1^M\mathsf{\text{exp}}\left\{-K_1\right\}\right)}{\bar{\eta }}\right],$$

(24)

$$u_3\left({\eta }_3\right)>\mathsf{\text{ln}}\left[\frac{{\bar{b}}_1-\mathsf{\text{exp}}\left\{K_1\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{K_2\right\}}{\bar{\eta }}\right].$$

(25)

By (3.9), (24) and (25), we can obtain

$$\begin{array}{ll}\hfill u_3\left(t\right)& \le u_3\left({\xi }_3\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_3\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[\frac{\left({\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_1\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_2\right\}\right)\left(1+{\omega }_1^M\mathsf{\text{exp}}\left\{-K_1\right\}\right)}{\bar{\eta }}\right]\\ +2{\bar{b}}_3\omega :=B_5,\end{array}$$

(26)

$$\begin{array}{l}{u}_3\left(t\right)\ge u_3\left({\eta }_3\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_3\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{\left({\bar{b}}_1-\mathsf{\text{exp}}\left\{K_1\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_2\right\}\right)}{\bar{\eta }}\right]-2{\bar{b}}_3\omega :=B_6.\end{array}$$

(27)

It follows from (26) and (27) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_3\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_5\right|,\left|B_6\right|\right\}:=K_3.$$

(28)

Case 2. If u₁(η₁) < u₂(η₂), then from the third equation of (14), we get

$$\begin{array}{ll}\hfill b_3^M=b_3\left({\eta }_3\right)& <d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}+d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}\\ <d^M{\eta }^M\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}+d^M{\eta }^M\mathsf{\text{exp}}\left\{u_2\left({\eta }_2\right)\right\}\\ =\left(d^M{\eta }^M+d^M{\mu }^M\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_2\right)\right\}.\end{array}$$

Then we have

$$u_2\left({\eta }_2\right)>\mathsf{\text{ln}}\left[\frac{b_3^M}{d^M{\eta }^M+d^M{\mu }^M}\right].$$

(29)

By (11), (15) and (29), we can obtain

$$u_2\left(t\right)\le u_2\left({\xi }_2\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_2\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[b_2^L\right]+2{\bar{b}}_2\omega :=B_7,$$

(30)

$$u_2\left(t\right)\ge u_2\left({\eta }_2\right)-\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_2\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{b_3^M}{d^M{\eta }^M+d^M{\mu }^M}\right]-2{\bar{b}}_2\omega :=B_8.$$

(31)

It follows from (30) and (31) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_2\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_7\right|,\left|B_8\right|\right\}:=K_4.$$

(32)

From the third equation of (14), we derive

$$\begin{array}{ll}\hfill b_3^M=b_3\left({\eta }_3\right)& <d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_1\left({\eta }_3\right)\right\}+d\left({\eta }_3\right)\eta \left({\eta }_3\right)\mathsf{\text{exp}}\left\{u_2\left({\eta }_3\right)\right\}\\ \le d^M{\eta }^M\mathsf{\text{exp}}\left\{u_1\left({\eta }_1\right)\right\}+d^M{\eta }^M\mathsf{\text{exp}}\left\{K_4\right\}\end{array}$$

which leads to

$$u_1\left({\eta }_1\right)>\mathsf{\text{ln}}\left[\frac{b_3^M-d^M{\eta }^M\mathsf{\text{exp}}\left\{K_4\right\}}{d^M{\mu }^M}\right].$$

(33)

In view of (10), (15) and (33), we can obtain

$$u_1\left(t\right)\le u_1\left({\xi }_1\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_1\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[b_1^L\right]+2{\bar{b}}_1\omega :=B_9,$$

(34)

$$u_1\left(t\right)\ge u_1\left({\eta }_1\right)-\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_1\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{b_3^M-d^M{\eta }^M\mathsf{\text{exp}}\left\{K_4\right\}}{d^M{\mu }^M}\right]-2{\bar{b}}_1\omega :=B_{10}.$$

(35)

It follows from (34) and (35) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_1\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_9\right|,\left|B_{10}\right|\right\}:=K_5.$$

(36)

From the first equation of (9), we get

$$\begin{array}{l}\underset{0}{\overset{\omega }{\int }}\left[\mathsf{\text{exp}}\left\{-K_5\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{-K_4\right\}+\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left({\xi }_3\right)\right\}}{1+{\omega }_1^M\mathsf{\text{exp}}\left\{K_5\right\}}\right]\mathsf{\text{d}}t<{\bar{b}}_1\omega ,\\ \underset{0}{\overset{\omega }{\int }}[\mathsf{\text{exp}}\left\{K_5\right\}+\alpha \left(t\right)\mathsf{\text{exp}}\left\{K_4\right\}+\eta \left(t\right)\mathsf{\text{exp}}\{u_3({\eta }_3\mathsf{\text{)} ]d}}t>{\bar{b}}_1\omega .\end{array}$$

Then

$$\begin{array}{l}\mathsf{\text{exp}}\left\{-K_5\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\}+\frac{\bar{\alpha }\mathsf{\text{exp}}\left\{u_3\left({\xi }_3\right)\right\}}{1+{\omega }_1^M\mathsf{\text{exp}}\left\{K_5\right\}}<{\bar{b}}_1,\\ \mathsf{\text{exp}}\left\{K_5\right\}+\bar{\alpha }\mathsf{\text{exp}}\left\{K_4\right\}+\bar{\eta }\mathsf{\text{exp}}\left\{u_3\left({\eta }_3\right)\right\}>{\bar{b}}_1.\end{array}$$

Therefore we have

$$u_3\left({\xi }_3\right)<\mathsf{\text{ln}}\left[\frac{\left({\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_5\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\}\right)\left(1+{\omega }_1^M\mathsf{\text{exp}}\left\{K_5\right\}\right)}{\bar{\eta }}\right],$$

(37)

$$u_3\left({\eta }_3\right)>\mathsf{\text{ln}}\left[\frac{{\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_5\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\}}{\bar{\eta }}\right].$$

(38)

By (3.9), (37) and (38), we can obtain

$$\begin{array}{ll}\hfill u_3\left(t\right)& \le u_3\left({\xi }_3\right)+\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_3\left(t\right)\right|\mathsf{\text{d}}t\le \mathsf{\text{ln}}\left[\frac{\left({\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_5\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\}\right)\left(1+{\omega }_1^M\mathsf{\text{exp}}\left\{K_5\right\}\right)}{\bar{\eta }}\right]\\ +2{\bar{b}}_3\omega :=B_{11},\end{array}$$

(39)

$$u_3\left(t\right)\ge u_3\left({\eta }_3\right)-\underset{0}{\overset{\omega }{\int }}\left|{\dot{u}}_3\left(t\right)\right|\mathsf{\text{d}}t\ge \mathsf{\text{ln}}\left[\frac{{\bar{b}}_1-\mathsf{\text{exp}}\left\{-K_5\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{-K_4\right\}}{\bar{\eta }}\right]-2{\bar{b}}_3\omega :=B_{12}.$$

(40)

It follows from (39) and (40) that

$$\underset{t\in \left[0,\omega \right]}{\mathsf{\text{max}}}\left|u_3\left(t\right)\right|\le \mathsf{\text{max}}\left\{\left|B_{11}\right|,\left|B_{12}\right|\right\}:=K_6.$$

(41)

Obviously, B _i (i = 1, 2, 3, ..., 12) are independent of λ ∈ (0, 1). Take M = max{K₁, K₅}+max{K₂, K 4}+ max{K₃, K₆} + K₀, where K₀ is taken sufficiently large such that every solution ${\left({ũ}_1,{ũ}_2,{ũ}_3\right)}^T\in R^3$ of the following algebraic equations

$$\left\{\begin{array}{c}\hfill {\bar{b}}_1-\mathsf{\text{exp}}\left\{u_1\right\}-\bar{\alpha }\mathsf{\text{exp}}\left\{u_2\right\}-\frac{1}{\omega }{\int }_0^{\omega }\frac{\eta \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}\mathsf{\text{d}}t=0,\hfill \\ \hfill {\bar{b}}_2-\bar{\beta }\mathsf{\text{exp}}\left\{u_1\right\}-\mathsf{\text{exp}}\left\{u_2\right\}-\frac{1}{\omega }{\int }_0^{\omega }\frac{\mu \left(t\right)\mathsf{\text{exp}}\left\{u_3\left(t\right)\right\}}{1+{\omega }_2\left(t\right)\mathsf{\text{exp}}\left\{u_2\left(t\right)\right\}}\mathsf{\text{d}}t=0,\hfill \\ \hfill -{\bar{b}}_3+\frac{1}{\omega }{\int }_0^{\omega }\frac{d\left(t\right)\eta \left(t\right)\mathsf{\text{exp}}\left\{u_1\left(t\right)\right\}}{1+{\omega }_1\left(t\right)\mathsf{\text{exp}}\{u_1\left(t\right)}\hfill \end{array}\right.$$