Coexistence of an unstirred chemostat model with B-D functional response by fixed point index theory

Abstract

This paper deals with an unstirred chemostat model with the Beddington-DeAngelis functional response. First, some prior estimates for positive solutions are proved by the maximum principle and the method of upper and lower solutions. Second, the calculation on the fixed point index of chemostat model is obtained by degree theory and the homotopy invariance theorem. Finally, some sufficient condition on the existence of positive steady-state solutions is established by fixed point index theory and bifurcation theory.

1 Introduction

The chemostat is a laboratory apparatus used for the continuous culture of micro-organisms. Mathematical models of the chemostat are surprisingly amenable to analysis. Early results can be found in the articles of Levin [1] and Hsu [2]. For a general discussion of competition, Smith and Waltman [3] discussed the well-unstirred model in detail. Recently, Wu [4–6] and Nie [7, 8] studied the coexistence and asymptotic behavior of chemostat models from the viewpoint of partial differential equations theories. The response functions are mainly the Michaelis-Menten functional response $f(S) = S/(1 + \kappa S)$, κ is a constant. Until very recently, both ecologists and mathematicians chose to base their studies on the Beddington-DeAngelis (denoted by B-D) functional response introduced by Beddington [9] and DeAngelis [10]. The Beddington-DeAngelis functional response has some of the same qualitative features as the Michaelis-Menten form but has an extra term βu in the denominator which models mutual interference between predators. It has been the source of controversy and can provide a better description of predator feeding over a range of predator-prey abundances, which are strongly supported by numerous field and laboratory experiments and observations.

In this paper, we are interested in the following unstirred chemostat model with Beddington-DeAngelis functional response:

$$ \left \{ \textstyle\begin{array}{@{}l} S_{t} = d\Delta S - auf(S,u) - b\upsilon g(S,\upsilon ), \quad x \in \Omega,t > 0, \\ u_{t} = d\Delta u + (1 - q)auf(S,u), \quad x \in \Omega,t > 0, \\ \upsilon_{t} = d\Delta \upsilon + b\upsilon g(S,\upsilon ) + qauf(S,u), \quad x \in \Omega,t > 0, \\ \frac{\partial S}{\partial v} + \gamma S = - S^{0},\qquad\frac{\partial u}{\partial v} + \gamma u = 0,\qquad\frac{\partial \upsilon}{\partial \nu} + \gamma \upsilon = 0,\quad x \in \partial \Omega,t > 0 \\ S(x,0) = S_{0}(x) \ge 0, \quad x \in \Omega, \\ u(x,0) = u_{0}(x) \ge 0, \ne 0, \quad x \in \Omega, \\ \upsilon (x,0) = \upsilon_{0}(x) \ge 0, \ne 0,\quad x \in \Omega, \end{array}\displaystyle \right . $$

(1.1)

where $a, b, m_{i}, k_{i}$, $i = 1, 2$ and γ are positive constants, a and b are the maximal growth rates of the two competitors, respectively. u stands the density of the plasmid-bearing organism; υ denotes plasmid-free organism; The parameter q, $0 < q < 1$ denotes the fraction of plasmid-bearing organism converting into plasmid-free organism. Ω is a bounded domain in $R^{n}$ with smooth boundary ∂Ω, $f(S,u) = \frac{S}{1 + m_{1}S + k_{1}u}$, $g(S,\upsilon ) = \frac{S}{1 + m_{2}S + k_{2}\upsilon}$ are Beddington-DeAngelis functions; $m_{i}$, $i = 1,2$ are the Michaelis-Menton constants; $k_{i}$ ($i = 1,2$) model mutual interference of between predators.

In the present paper, we shall investigate non-negative steady-state solutions of system (1.1). Hence, we will concentrate on the following elliptic system:

$$ \left \{ \textstyle\begin{array}{@{}l} d\Delta S - auf(S,u) - b\upsilon g ( S,\upsilon ) = 0, \quad x \in \Omega, \\ d\Delta u + ( 1 - q )auf ( S,u ) = 0, \quad x \in \Omega, \\ d\Delta \upsilon + b\upsilon g ( S,\upsilon ) + qauf ( S,u ) = 0, \quad x \in \Omega, \\ \frac{\partial S}{\partial \nu} + \gamma S = - S^{0},\qquad\frac{\partial u}{\partial \nu} + \gamma u = 0,\qquad\frac{\partial \upsilon}{\partial \nu} + \gamma \upsilon = 0, \quad x \in \partial\Omega. \end{array}\displaystyle \right . $$

(1.2)

Let $z = S + u + \upsilon$, then $S = z - u - \upsilon$ (refer to [4–7]) and z satisfies

$$\Delta z = 0,\quad x \in \Omega,\qquad\frac{\partial z}{\partial \nu} + \gamma z = - S^{0},\quad x \in \partial \Omega. $$

Then one can argue in the exactly same way as in [4, 5, 7] to obtain the limiting system of (1.2), which can be written as

$$ \left \{ \textstyle\begin{array}{@{}l} d\Delta u + ( 1 - q )auf ( z - u - \upsilon,u ) = 0, \quad x \in \Omega,\\ d\Delta \upsilon + b\upsilon g ( z - u - \upsilon ) + qauf ( z - u - \upsilon,u ) = 0, \quad x \in \Omega,\\ \frac{\partial u}{\partial \nu} + ru = 0,\qquad\frac{\partial \upsilon}{\partial \nu} + r\upsilon = 0, \quad x \in \partial \Omega, \end{array}\displaystyle \right . $$

(1.3)

where $f ( S,u ) = \frac{S}{1 + m_{1}S + k_{1}u}$, $g ( S,\upsilon ) = \frac{S}{1 + m_{2}S + k_{2}\upsilon}$, and $S = z - u - \upsilon$. We only concern on the case that $S(x),u(x),\upsilon (x)$ are non-negative, so we redefine the response function as follows:

$$\hat{f}(S,u) = \left \{ \textstyle\begin{array}{@{}l@{\quad}l} f(S,u), & S \ge 0,u \ge 0, \\ 0, & \mbox{others}, \end{array}\displaystyle \right . \qquad \hat{g}(S,\upsilon ) = \left \{ \textstyle\begin{array}{@{}l@{\quad}l} g(S,\upsilon), & S \ge 0,\upsilon \ge 0 , \\ 0, & \mbox{others}. \end{array}\displaystyle \right . $$

Then $\hat{f}(S,u),\hat{g}(S,\upsilon ) \in C^{1} ( R )$. We will denote $\hat{f}(S,u), \hat{g}(S,\upsilon )$ by $f(S,u), g(S,\upsilon )$ for the sake of simplicity, respectively. Thus, the solution of equation (1.3) satisfies

$$S(x) + u(x) + \upsilon (x) = z(x),\quad x \in \bar{\Omega}. $$

In the following, we set up the fixed point index theory on this paper. Let E be a Banach Space. $W \subset E$ is called a wedge if W is a closed convex set and $\alpha W \subset W$ for all $\alpha \ge 0$.

For $y \in W$, we define $W_{y} = \{ x \in E:\exists r = r(x) > 0, \mbox{s.t.}, y + rx \in W\}$, $S_{y} = \{ x \in \bar{W}_{y}: - x \in \bar{W}_{y}\}$, we always assume that $E = \overline{W - W}$. Let $T:W_{y} \to W_{y}$ be a compact linear operator on E. We say that T has property α on $\bar{W}_{y}$ if there exists $t \in ( 0,1 )$ and $\omega \in \bar{W}_{y}\backslash S_{y}$, such that $\omega - tT\omega \in S_{y}$.

Suppose that $F:W \to W$ is a compact operator, and $y_{0} \in W$ is an isolated fixed point of F, such that $Fy_{0} = y_{0}$, let $L = F'(y_{0})$ is Fréchet differentiable at $y_{0}$, it follows that $L:\overline{W} \to \overline{W}$.

Proposition 1.1

[11] Dancer index theorem

Assume that $I - L$ is invertible on E, then we have

• (1) L has property α on W̄, then $\operatorname{index}_{W}(F,y_{0}) = 0$;

• (2) L does not have property α on W̄, then $\operatorname{index}_{W}(F,y_{0}) = \operatorname{index}_{E}(L,\theta ) = \pm 1$.

Proposition 1.2

[11]

Assume that $F(\theta ) = \theta$, $A_{0} = F'(\theta )$ is Fréchet differentiable of F at θ in W. If the eigenvalue problem

$$ A_{0}h = \lambda h,\quad h \in W $$

(1.4)

has no eigenvalue equal to 1, then θ is isolate fixed point of F, and

• (1) if (1.4) has no eigenvalue larger than 1, then $\operatorname{index}_{W} ( F,\theta ) = 1$;

• (2) if (1.4) has an eigenvalue $\lambda > 1$, then $\operatorname{index}_{W}F ( F,\theta ) = 0$.

Proposition 1.3

[12]

Assuming that T is a positive compact linear operator in ordered Banach spaces, u is a positive element in Banach space, $r(T)$ is the spectral radius of the operator T, then

• (1) if $Tu > u$, then $r ( T ) > 1$;

• (2) if $Tu < u$, then $r ( T ) < 1$;

• (3) if $Tu = u$, then $r ( T ) = 1$.

Proposition 1.4

[13]

Assume that $q(x) \in C(\Omega )$, $q(x) + p > 0$ on Ω̅, p is a positive real constant, $\lambda_{1}$ is the principal eigenvalue of the following problem:

$$- \Delta \phi - q ( x )\phi = \lambda \phi,\quad x \in \Omega,\qquad \frac{\partial \phi}{\partial \nu} + r\phi = 0,\quad x \in \partial \Omega. $$

If $\lambda_{1} > 0$ (or <0), then all eigenvalues of the following problem:

$$- \Delta \phi - p\phi = t \bigl( q ( x ) + p \bigr)\phi,\quad x \in \Omega, \qquad \frac{\partial \phi}{\partial \nu} + r\phi = 0,\quad x \in \partial \Omega $$

are larger than 1 (or less than 1).

The organization of our paper is as follows. In Section 2, some prior estimates for positive solutions are proved by the maximum principle and the upper and lower solution method. In Section 3, the calculations on the fixed point index of chemostat model by degree theory and the homotopy invariance theorem. In Section 4, some sufficient conditions on the existence of positive steady-state solutions is established by the fixed point index theory in cone and bifurcation theory.

2 Some prior estimates for positive solutions

The main purpose of this section is to give prior upper and lower positive bounds for positive solutions of (1.3) by using the maximum principle and the upper and lower solution method.

Let $\lambda_{1},\mu_{1}$ be, respectively, the principal eigenvalue of the following problem:

$$\begin{aligned}& \Delta \varphi + \lambda \varphi f(z,0) = 0,\quad x \in \Omega,\qquad \frac{\partial \varphi}{\partial \nu} + r\varphi = 0,\quad x \in \partial \Omega, \end{aligned}$$

(2.1)

$$\begin{aligned}& \Delta \psi + \mu \psi g(z,0) = 0,\quad x \in \Omega,\qquad\frac{\partial \psi}{ \partial \nu} + r\psi = 0,\quad x \in \partial \Omega, \end{aligned}$$

(2.2)

the corresponding principal eigenvalue function denoted by $\varphi_{1}(x)$, $\psi_{1}(x)$, and

$$\|\varphi_{1}\| = 1,\qquad \|\psi_{1}\| = 1. $$

First, we consider the single species equation as follows:

$$ d\Delta u + a ( 1 - q )uf_{1} ( z - u,u ) = 0,\quad x \in \Omega, \qquad \frac{\partial u}{\partial \nu} + ru = 0,\quad x \in \partial \Omega. $$

(2.3)

By [9, 14, 15], we can directly get the following conclusions.

Lemma 2.1

If $a \le \frac{\lambda_{1}d}{1 - q}$, then 0 is the unique non-negative solution of (2.3); If $a > \frac{\lambda_{1}d}{1 - q}$, (2.3) has a unique positive solution, denoted by Θ, satisfying the following properties:

• (i) $0 < \Theta < z$;

• (ii) Θ is continuously differentiable for $a \in (\frac{\lambda_{1}d}{1 - q}, + \infty)$ and is point wisely increasing when a is increasing.

• (iii) $\lim_{a \to \frac{\lambda_{1}d}{1 - q}}\Theta = 0$ uniformly for $x \in \bar{\Omega}$, $\lim_{a \to \infty} \Theta = z(x)$ for almost every $x \in \Omega$;

• (iv) Let $L_{ ( a,d )} = d\Delta + a(1 - q) ( f(z - \Theta,\Theta ) - \Theta f'_{1}(z - \Theta,\Theta ) + \Theta f'_{2}(z - \Theta,\Theta ) )$ be the linearized operator of (2.3) at Θ, then $L_{ ( a,d )}$ is differentiable in $C_{B}^{2}(\bar{\Omega} ) = \{ u \in C^{2} ( \bar{\Omega} ):\frac{\partial u}{\partial \nu} + ru = 0\}$, and all eigenvalues of $L_{ ( a,d )}$ are strictly negative.

Remark 2.1

For (2.4), we have the same conclusion as Lemma 2.1. Suppose $b > d\mu_{1}$, we denote the unique positive solution by θ for the following problem:

$$ d\Delta \upsilon + b\upsilon g(z - \upsilon,\upsilon ) = 0,\quad x \in \Omega, \qquad \frac{\partial \upsilon}{\partial \nu} + r\upsilon = 0,\quad x \in \partial \Omega, $$

(2.4)

let $L_{ ( b,d )} = d\Delta + b(g(z - \theta,\theta ) - \theta g_{1}^{\prime} (z - \theta,\theta ) + \theta g'_{2}(z - \theta,\theta ))$ be the linearized operator of (2.4) at θ.

Next, let $\hat{\lambda}_{1}$ be the principal eigenvalue of the following eigenvalue problem:

$$ \Delta \varphi + \hat{\lambda} \varphi f ( z - \theta,0 ) = 0,\quad x \in \Omega,\qquad \frac{\partial \varphi}{\partial \nu} + r\varphi = 0,\quad x \in \partial \Omega, $$

(2.5)

the corresponding eigenfunction denoted by $\hat{\varphi}_{1}(x)$ and uniquely determined by normalization $\|\hat{\varphi}_{1}\| = 1$.

In order to accurately estimate the positive solution of (1.3), we consider the following boundary value problem:

$$ d\Delta \upsilon + b\upsilon g(z - \upsilon,\upsilon ) + aq\Theta f(z - \upsilon,0) = 0,\quad x \in \Omega, \qquad\frac{\partial \upsilon}{\partial \nu} + r\upsilon = 0,\quad x \in \partial \Omega. $$

(2.6)

Lemma 2.2

Suppose $a > \frac{\lambda_{1}d}{1 - q}$, there exists a unique positive solution of (2.6). Then $0 < \upsilon < z$, and $\theta < \upsilon < z$, when $b > d\mu_{1}$.

Proof

Let $\omega = z - \upsilon$, then

$$d\Delta \omega + b\upsilon g(\omega,\upsilon ) + aq\Theta f(\omega,0) = 0,\quad x \in \Omega,\qquad \frac{\partial \omega}{\partial \nu} + r\omega = S^{0},\quad x \in \partial \Omega. $$

Suppose $\inf_{x \in \bar{\Omega}} \omega ( x ) = \omega ( x_{0} ) < 0$, then $x_{0} \notin \Omega$. Otherwise, $\Delta \omega ( x_{0} ) \ge 0$. However,

$$d\Delta \omega (x_{0}) = b\upsilon (x_{0})g \bigl( \omega (x_{0}),\upsilon (x_{0}) \bigr) + aq\Theta f \bigl( \omega (x_{0}),0 \bigr) < 0, $$

a contradiction. If $x_{0} \in \partial \Omega$, it follows from the definition of $r ( x )$ and $S^{0} ( x )$ that

$$\frac{\partial \omega}{\partial \nu} \bigg|_{x_{0}} = S^{0}(x_{0}) - r\omega (x_{0}) > 0. $$

This contradicts $\frac{\partial \omega}{\partial \nu} |_{x_{0}} < 0$, hence, $\upsilon \le z$ on Ω̄, and $\upsilon \le z,\upsilon \ne z$, then $\omega = z - \upsilon \ge 0, \ne 0$. If $\omega ( x_{0} ) = 0$, for some point $x_{0} \in \bar{\Omega}$, then it follows from the maximum principle [16] that $x_{0} \in \partial \Omega$. Furthermore, from the Holp lemma [16], we can get $\frac{\partial \omega}{ \partial \nu} |_{x_{0}} < 0$, this contradicts the boundary condition.

Thus, $\upsilon < z$ on Ω̄, and for

$$d\Delta \upsilon + b\upsilon g(z - \upsilon,\upsilon ) + aq\Theta f(z - \upsilon,0) > d\Delta \upsilon + b\upsilon g(z - \upsilon,\upsilon ). $$

Hence $\upsilon > \theta$ when $b > d\mu_{1}$.

Next, we will prove the existence and uniqueness of solutions. For sufficiently small $\delta > 0$, $\delta \phi_{1},z$ are the upper and lower solutions of (2.6). It follows from the comparison principle [16] that (2.6) exists the minimum solution $\upsilon_{1}$ and maximum solution $\upsilon_{2}$, satisfying $\delta \phi {}_{1} \le \upsilon_{1} \le \upsilon_{2} \le z$.

In the following, we prove the uniqueness. Thanking to $\upsilon_{1},\upsilon_{2}$ are the solution of (2.6), we have

$$\begin{aligned}& d\Delta \upsilon_{1} + b\upsilon_{1}g(z - \upsilon_{1},\upsilon_{1}) + aq\Theta f(z - \upsilon_{1},0) = 0, \quad x \in \Omega, \\& d\Delta \upsilon_{2} + b\upsilon_{2}g(z - \upsilon_{2},\upsilon_{2}) + aq\Theta f(z - \upsilon_{2},0) = 0, \quad x \in \Omega. \end{aligned}$$

Multiplying the second equation and first equation by $\upsilon_{1},\upsilon_{2}$, respectively, and applying the Green’s formula, we obtain $I = \int_{\Omega} d( \Delta \upsilon_{1} \cdot \upsilon_{2} - \Delta \upsilon_{2} \cdot \upsilon_{1} ) = 0$. Then

$$\int_{\Omega} b\upsilon_{1}\upsilon_{2}\bigl( g( z - \upsilon_{1},\upsilon_{1} ) - g( z - \upsilon_{2},\upsilon_{2} ) \bigr)\,dx + aq\int_{\Omega} \Theta \bigl[ \upsilon_{2}f( z - \upsilon_{1},0 ) - \upsilon_{1}f( z - \upsilon_{2},0 ) \bigr]\,dx = 0. $$

According to the monotonicity of $f,g$, and $\upsilon_{1} \le \upsilon_{2}$, we get $\upsilon_{1} \equiv \upsilon_{2}$. □

In conclusion, we can get prior estimates on the system (1.3).

Theorem 2.3

Suppose $(u,\upsilon)$ is non-negative solution of (1.3), and $u \ne 0,\upsilon \ne 0$. Then

• (i) $0 < u < \Theta < z$, $0 < \upsilon \le \hat{\upsilon} < z$, $x \in \bar{\Omega}$;

• (ii) $u + \upsilon < z$, $x \in \bar{\Omega}$;

• (iii) $a > \frac{\lambda_{1}d}{1 - q}$.

Proof

The proof is in [8, 14], we omit it. □

3 Calculations of fixed point index

In this section, we will calculate the fixed point index of (1.3) by using the standard fixed point index theory in cone.

Let $C_{0}(\bar{\Omega} ) = \{ y \in C(\bar{\Omega} )|\frac{\partial y}{\partial n} + ry = 0\}$, $E = [C_{0}(\bar{\Omega} ) \times C_{0}(\bar{\Omega} )]$. For a sufficiently large $P > 0$, and $\tau \in [0,1]$, we consider the following equations:

$$ \left \{ \textstyle\begin{array}{@{}l} ( - d\Delta + \tau P)u = \tau (P + a(1 - q)f(z - u - \upsilon,u))u, \quad x \in \Omega, \\ ( - d\Delta + \tau P)\upsilon = \tau (P\upsilon + bg(z - u - \upsilon,\upsilon )\upsilon + qauf(z - u - \upsilon,u)),\quad x \in \Omega, \\ \frac{\partial u}{\partial \nu} + ru = 0,\qquad\frac{\partial \upsilon}{\partial \nu} + r\upsilon = 0, \quad x \in \partial \Omega. \end{array}\displaystyle \right . $$

(3.1)

For $(u,\upsilon )^{T} \in E$, $\tau \in [0,1]$, $(U,V)^{T} \in [C^{1 + \alpha} (\bar{\Omega} )] \times [C^{1 + \alpha} (\bar{\Omega} )]$ is the unique solution of the following linear problem:

$$\left \{ \textstyle\begin{array}{@{}l} ( - d\Delta + \tau P)U = \tau (P + a(1 - q)f(z - u - \upsilon,u))u, \quad x \in \Omega, \\ ( - d\Delta + \tau P)V = \tau ( P\upsilon + bg(z - u - \upsilon,\upsilon )\upsilon + qauf(z - u - \upsilon,u) ),\quad x \in \Omega, \\ \frac{\partial U}{\partial \nu} + rU = 0,\qquad\frac{\partial V}{\partial \nu} + rV = 0, \quad x \in \partial \Omega, \end{array}\displaystyle \right . $$

define $F_{\tau}:[0,1] \times E \to E$, $F_{\tau} (u,\upsilon )^{T} = (U,V)^{T}$, it follows from [17] that $F_{\tau}$ is compact.

It is clear to see that $(u,\upsilon )^{T} \in E$ is fixed point of $F_{\tau}$ if only and if $(u,\upsilon )^{T} \in E$ is a positive solution of (1.3). Let

$$\begin{aligned}& K_{0} = \bigl\{ u \in C_{0}(\bar{\Omega} )|u(x) \ge 0, x \in \Omega \bigr\} , \\& W = \bigl\{ (u,\upsilon ) \in E|u(x) \ge 0,\upsilon (x) \ge 0,x \in \Omega \bigr\} , \\& D = \Bigl\{ (u,\upsilon ) \in W|0 \le u(x) \le \Theta,0 \le \upsilon (x) \le \max_{\bar{\Omega}} \hat{\upsilon} + 1,x \in \Omega \Bigr\} , \\& D' = (\operatorname{int} D ) \cap W. \end{aligned}$$

Then W is a cone in E, D is bounded set in W, let $F = F_{1}$. It follows that there exists $K > 0$ such that $f(z - u - \upsilon,u) \ge f(z - u,u) - K\upsilon$.

Suppose that P is sufficiently large, for all $(u,\upsilon ) \in D$, we can get

$$P + a(1 - q)f(z - u - \upsilon,u) > 0,\qquad P + bg(z - u - \upsilon,\upsilon ) - aquK > 0. $$

Then $F:D' \to W$ is continuously differentiable, hence, (1.3) had non-negative solution if only and if F has fixed point on D. According to the homotopy invariance of degree, we have

$$\deg_{W} \bigl(I - F_{\tau},D',(0,0) \bigr) = \deg _{W} \bigl(I - F,D',(0,0) \bigr),\quad \tau \in [0,1]. $$

In the following, we calculate the index number of $(0,0)$ and $(0,\theta)$ by using the fixed point theory.

Lemma 3.1

For the index number of the operator F at $(0,0)$ the following results hold:

• (i) if $a \ne \frac{\lambda_{1}d}{1 - q}$, $b > \mu_{1}d$, then $\operatorname{index}_{W}(F,(0,0)) = 0$;

• (ii) Suppose that $b < \mu_{1}d$. If $a > \frac{\lambda_{1}d}{1 - q}$, then $\operatorname{index}_{W}(F,(0,0)) = 0$; If $a < \frac{\lambda_{1}d}{1 - q}$, then $\operatorname{index}_{W}(F,(0,0)) = 1$.

Proof

Define the operator at $(0,0)$ of F as follows:

$$L_{0} ( \varphi,\psi ) = F' ( 0,0 ) \begin{pmatrix} \varphi \\ \psi \end{pmatrix} = ( - d\Delta + P )^{ - 1} \begin{pmatrix} P + a ( 1 - q )f ( z,0 ) & 0 \\ aqf ( z,0 ) & P + bg ( z,0 ) \end{pmatrix} \begin{pmatrix} \varphi \\ \psi \end{pmatrix}. $$

Hence, $F'(0,0)\cdot(\varphi,\psi )^{T} = \lambda (\varphi,\psi )^{T}$ is equivalent to

$$ \left \{ \textstyle\begin{array}{@{}l} - d\Delta \varphi + P\varphi = \frac{1}{\lambda} (P\varphi + a(1 - q)f(z,0)\varphi ), \quad x \in \Omega, \\ - d\Delta \psi + P\psi = \frac{1}{\lambda} (P\psi + b\psi g(z,0) + qa\varphi f(z,0)), \quad x \in \Omega, \\ ( \varphi,\psi ) \in W. \end{array}\displaystyle \right . $$

(3.2)

Take $y_{0} = (0,0)$, then $\bar{W}_{y_{0}} = K_{0} \times K_{0}$, $S_{y_{0}} = \{ (0,0)\}$, $\bar{W}_{y_{0}}\backslash S_{y_{0}} = K_{0} \times K_{0}\backslash \{ (0,0)\}$.

If 1 is an eigenvalue of (3.2), then

$$ \left \{ \textstyle\begin{array}{@{}l} d\Delta \varphi + a(1 - q)f(z,0)\varphi = 0, \quad x \in \Omega, \\ d\Delta \psi + b\psi g(z,0) + aq\varphi f(z,0) = 0, \quad x \in \Omega, \end{array}\displaystyle \right . $$

(3.3)

if $a \ne \frac{\lambda_{1}d}{1 - q}$, $b \ne \mu_{1}d$, form the first equation of (3.3) and definition of $\lambda_{1}$, we have $\varphi \equiv 0$. Similarly, it follows from the second equation that $\psi \equiv 0$. Thus 1 is not an eigenvalue of $L_{0}$, it is to see that $(0,0)$ is an isolated fixed point.

(i) Suppose that $a \ne \frac{\lambda_{1}d}{1 - q},b > \mu_{1}d$. If $\hat{\lambda}_{1}$ is the principal eigenvalue of $- d\Delta \psi - bg ( z,0 )\psi = \hat{\lambda} \psi$, then $\hat{\lambda}_{1} < 0$. From Proposition 1.4, there exists $\frac{1}{\hat{\lambda}} < 1$ is an eigenvalue of $( - d\Delta + P)\psi = \frac{1}{\hat{\lambda}} (P + bg(z,0))\psi, \psi$ is the corresponding eigenfunction, then the $L_{0}$ has no eigenvalue greater than 1, so $(0,\psi)$ is the corresponding eigenfunction. It follows from Proposition 1.2 that we have $\operatorname{index}_{W} ( F, ( 0,0 ) ) = 0$.

(ii) Suppose that $b < \mu_{1}d$. If $a > \frac{\lambda_{1}d}{1 - q}$, 1 is not an eigenvalue of (3.2), and $\lambda_{1}( - d\Delta - ( 1 - q )af( z,0 ) ) < 0$. From Proposition 1.4, then there exists $\frac{1}{\hat{\lambda}_{0}} < 1$ which is an eigenvalue of the first equation of (3.2), the corresponding eigenfunction denoted by $\varphi_{1}$, and putting it into the second equation, we have

$$\biggl( - d\Delta + \biggl(1 - \frac{1}{\hat{\lambda}_{0}} \biggr)P - \frac{1}{\hat{\lambda}_{0}}bg(z,0) \biggr)\psi = \frac{1}{\hat{\lambda}_{0}}aqf(z,0) \varphi_{1}. $$

Thanks to $b < \mu_{1}d$ and $\frac{1}{\hat{\lambda}_{0}} < 1$, all eigenvalues of the operator $( - d\Delta + (1 - \frac{1}{\hat{\lambda}_{0}})P - \frac{1}{\hat{\lambda}_{0}}bg(z,0))I$ are larger than 0, that is the operator is inverse, hence, let

$$\psi_{1} = \biggl( - d\Delta + \biggl(1 - \frac{1}{\hat{\lambda}_{0}} \biggr)P - \frac{1}{\hat{\lambda}_{0}}bg(z,0) \biggr)^{ - 1}\frac{1}{\hat{\lambda}_{0}}aqf ( z,0 ) \varphi_{1}. $$

Then there exists the eigenvalue of $L_{0}$ is larger than 1, and the corresponding eigenfunction denoted by $( \varphi_{1},\psi_{1} )$, then from Proposition 1.1, we can get $\operatorname{index}_{W}(F(0,0)) = 0$.

When $a < \frac{\lambda_{1}d}{1 - q}$, 1 is not the eigenvalue of (3.2). Next, we prove that $L_{0}$ does not have the α property on $\bar{W}_{y_{0}}$. Suppose that $L_{0}$ on $\bar{W}_{y_{0}}$ has α property, then there exist $t_{1} \in ( 0,1 )$ and the function $(\phi,\psi ) \in \bar{W}_{y_{0}}\backslash S_{y_{0}} = K_{0} \times K_{0}\backslash \{ (0,0)\}$ such that $(I - t_{1}L_{0})(\varphi,\psi )^{T} \in S_{y_{0}}$, that is,

$$ \left \{ \textstyle\begin{array}{@{}l} ( - d\Delta + P)\varphi = t_{1}(P + a(1 - q)f(z,0))\varphi, \quad x \in \Omega, \\ ( - d\Delta + P)\psi = t_{1}((P + bg(z,0)) \psi + qaf(z,0)\varphi ), \quad x \in \Omega. \end{array}\displaystyle \right . $$

(3.4)

For (3.4), if $\psi \equiv 0$, then $\varphi \equiv 0$, since $( \varphi,\psi )$ is an eigenfunction, hence $\psi \ne 0$, we discuss the following two cases:

(a) If $\varphi \equiv 0$, then from the second equation of (3.4), we see that $\lambda_{1}( - d\Delta - bg(z,0)) > 0$ when $b < \mu_{1}d$, it follows from the Proposition 1.4 that (3.4) has no eigenvalues which are equal to or less than 1. This is a contradiction hypothesis.

(b) If $\varphi\not\equiv 0$, then we can get the following results from equation (3.4), when

$$a < \frac{\lambda_{1}d}{1 - q},\quad \lambda_{1} \bigl( - d\Delta - a ( 1 - q )f ( z,0 ) \bigr) > 0. $$

It follows from Proposition 1.4 that (3.4) has no eigenvalues which are equal to or less than 1, this is a contradiction with the hypothesis. Hence, $L_{0}$ as no property α on $\bar{W}_{y_{0}}$. By Proposition 1.4, we can get that

$$\operatorname{index}_{W}(F,y_{0}) = \operatorname{index}_{W} \bigl(L_{0},(0,0) \bigr) = \pm 1. $$

It is easy to see that $\operatorname{index}_{W} ( L_{0}, ( 0,0 ) ) = ( - 1 )^{\sigma}$, where σ is the sum of the algebraic multiplicities of the eigenvalues of $L_{0}$ which are greater than 1. From above results and (3.2), we see that $L_{0}$ has no eigenvalues which are greater than 1, then $\sigma = 0$, that is, $\operatorname{index}_{W} ( F, ( 0,0 ) ) = 1$. □

Lemma 3.2

$\operatorname{index}_{W} ( F,D' ) = 1$.

Proof

Taking $\tau \in ( 0,1 )$ sufficiently small, such that $\tau a ( 1 - q ) < \lambda_{1}d, \tau b < \lambda_{1}d$, it follows from Lemma 3.1 that $\operatorname{index}_{W} ( F, ( 0,0 ) ) = 1$, so as $0 < \tau << 1$, $\deg_{W} ( I - F_{\tau},D', ( 0,0 ) ) = 1$, by the homotopy invariance, we get $\deg_{W}(I - F_{\tau},D',(0,0)) = 1$, thus $\operatorname{index}_{W}(F,D') = 1$. □

Lemma 3.3

Suppose $b > \mu_{1}d$, the index on the point of $( 0,\theta )$ satisfies the following results:

• (i) If $a < \frac{\hat{\lambda}_{1}d}{1 - q}$, then $\operatorname{index}_{W}(F,(0,\theta )) = 1$; if $a > \frac{\hat{\lambda}_{1}d}{1 - q}$, then $\operatorname{index}_{W}(F,(0,\theta )) = 0$.

• (ii) If $a = \frac{\hat{\lambda}_{1}d}{1 - q}$, then either (1.3) has a positive solution or $\operatorname{index}_{W} ( F, ( 0,\theta ) ) = 1$.

Proof

Defining the linearized operator F on $(0,\theta)$ as follows:

$$L_{1} = F'(0,\theta ) = ( - d\Delta + P)^{ - 1} \begin{pmatrix} P + a(1 - q)f(z - \theta,0) & 0 \\ aqf(z - \theta,0) - b\theta g'_{1}(z - \theta,\theta ) & P + L_{b} \end{pmatrix}, $$

where $L_{b} = b(g(z - \theta,\theta ) - \theta g'_{1}(z - \theta,\theta ) + \theta g'_{2}(z - \theta,\theta ))$. Taking $y_{1} = ( 0,\theta )$, then $\bar{W}_{y_{1}} = K_{0} \times C_{0} ( \Omega ),S_{y_{1}} = \{ 0 \} \times C_{0} ( \bar{\Omega} )$, $\bar{W}_{y_{1}}\backslash S_{y_{0}} = K_{0}\backslash \{ 0 \} \times C_{0} ( \bar{\Omega} )$.

First, we will prove that $I - L_{1}$ is invertible on $\bar{W}_{y_{1}}$ as $a \ne \frac{\hat{\lambda}_{1}d}{1 - q}$. Assume that there exists $\phi, \psi \in C_{0}(\bar{\Omega} )$ such that $L_{1}(\phi,\psi )^{T} = (\phi,\psi )^{T}$, then

$$\begin{aligned}& d\Delta \phi + a ( 1 - q )f ( z - \theta,0 )\phi = 0, \\& d\Delta \psi + L_{b}\psi + qaf ( z - \theta,0 )\phi - b\theta g' ( z - \theta,\theta )\phi = 0. \end{aligned}$$

From the definition of $a \ne \frac{\hat{\lambda}_{1}d}{1 - q}$ and $\lambda_{1}$, we can get $\phi \equiv 0$, and taking $\phi \equiv 0$ into the second equation, then $d\Delta \psi + L_{b}\psi = 0$.

It follows from Lemma 2.1 that $\psi \equiv 0$, hence $I - L_{1}$ is inverse on $\bar{W}_{y_{1}}$.

(i) When $a < \frac{\hat{\lambda}_{1}d}{1 - q}$, we can prove that $L_{1}$ has no property α on $\bar{W}_{y_{1}}$.

We assume that $L_{1}$ has property α on $\bar{W}_{y_{1}}$, then there exist $t_{1} \in (0,1)$ and the function $(\phi,\psi ) \in \bar{W}_{y_{1}}\backslash S_{y_{1}} = K_{0}\backslash \{ 0\} \times C_{0}\{ \bar{\Omega} \}$, such that $I - t_{1}L_{1}(\phi,\psi )^{T} \in S_{y_{1}} = \{ 0\} \times C_{0}(\bar{\Omega} )$, that is

$$ \left \{ \textstyle\begin{array}{@{}l} ( - d\Delta + P)\phi - t_{1}(P + (1 - q)af(z - \theta,0))\phi = 0, \\ ( - d\Delta + P)\psi - t_{1}(aqf(z - \theta,0)\phi - b\theta g'_{1}(z - \theta,\theta )\phi + P\psi + L_{b}\psi ) \in C_{0}(\bar{\Omega} ). \end{array}\displaystyle \right . $$

(3.5)

It follows from the definition of $\phi,\psi$ which satisfies (3.5), let $T = ( - d\Delta + P)^{ - 1}(P + (1 - q)af(z - \theta,0))I$, for (3.5), $T\phi = \frac{1}{t_{1}}\phi > \phi$, from Proposition 1.3, we can obtain $r(T) > 1$. However, from $a < \frac{\hat{\lambda}_{1}d}{1 - q}$, we know that there exists $\phi > 0$ such that

$$d\Delta \phi + a ( 1 - q )f ( z - \theta,0 )\phi = \lambda_{1} \bigl( d\Delta + a ( 1 - q )f ( z - \theta,0 ) \bigr)\phi < 0. $$

Adding both sides with $P\phi > 0$, we can get $(d\Delta + a(1 - q)f(z - \theta,0) + P)\phi < P\phi$, that is

$$T\phi = ( - d\Delta + P)^{ - 1} \bigl(a(1 - q)f(z - \theta,0) + P \bigr) \phi < \phi. $$

From Proposition 1.3, we can obtain $r(T) < 1$, and then get a contradiction. Hence $L_{1}$ has no property α on $\bar{W}_{y_{0}}$. From Proposition 1.1, we know

$$\operatorname{index}_{W}(F,y_{1}) = \operatorname{index}_{W} \bigl(L_{1}, (0,\theta ) \bigr) = \pm 1. $$

It is easy to see that $\operatorname{index}_{W}(L_{1},(0,\theta )) = ( - 1)^{\sigma}$, where σ is the sum of the algebraic multiplicities of the eigenvalues of $L_{1}$ which are greater than 1. Set λ be the eigenvalue of $L_{1}$, the corresponding eigenfunction named by $( \phi,\psi )$, then $L_{1} ( \phi,\psi )^{T} = \lambda ( \phi,\psi )^{T}$, that is

$$\left \{ \textstyle\begin{array}{@{}l} ( - d\Delta + P)^{ - 1}(P + (1 - q)af(z - \theta,0))\phi = \lambda \phi, \\ ( - d\Delta + P)^{ - 1}(aqf(z - \theta,0)\phi - b\theta g'_{1}(z - \theta,\theta )\phi + P\psi + L_{b}\psi ) = \lambda \psi. \end{array}\displaystyle \right . $$

By the definition of the above operator T, we see that $T\phi = \lambda \phi$ and $r ( T ) < 1$, then all eigenvalues of $L_{1}$ are less than 1, so

$$\operatorname{index}_{W} \bigl(F,(0,\theta ) \bigr) = \operatorname{index}_{E} \bigl(L_{1},(0,0) \bigr) = ( - 1)^{\sigma} = ( - 1)^{0} = 1. $$

Next, we prove that $L_{1}$ has property α on $\bar{W}_{y_{1}}$ when $a > \frac{\hat{\lambda}_{1}d}{1 - q}$. Suppose that η is the principal eigenvalue of $d\Delta \phi + a ( 1 - q )f ( z - \theta,0 )\phi = \eta \phi$, thanks to $a > \frac{\hat{\lambda}_{1}d}{1 - q}$, $\eta > 0$, from Proposition 1.4, we know that there exists $t_{1} \in (0,1)$ is eigenvalue of $( - d\Delta + P)\phi = t_{1}(P + a(1 - q)f(z - \theta,0))\phi$, take $\phi \in K_{0}\backslash \{ 0\}$ as the corresponding eigenfunction, then $(\phi, 0) \in \bar{W}_{y_{1}}\backslash S_{y_{1}}$ such that

$$\begin{aligned}& ( - d\Delta + P)\phi - t_{1} \bigl(P + a(1 - q)f(z - \theta,0) \bigr) \phi = 0, \\& - t_{1}( - d\Delta + P)^{ - 1} \bigl(aqf(z - \theta,0)\phi - b \theta g'_{1}(z - \theta,\theta )\phi \bigr) \in C_{0}(\bar{\Omega} ), \end{aligned}$$

that is, $I - t_{1}L_{1}(\phi,\psi )^{T} \in S_{y_{1}}$, thus, $L_{1}$ has property α on $\bar{W}_{y_{1}}$. By the Proposition 1.1, we can obtain $\operatorname{index}_{W}(F,(0,\theta )) = 0$.

(ii) First, we show that $I - L_{1}$ is not invertible on $\bar{W}_{y_{1}}$. That is, there exists $( \omega,\chi ) \in \bar{W}_{y_{1}}$ such that $L_{1}(\omega,\chi )^{T} = (\omega,\chi )^{T}$, it follows that

$$ \left \{ \textstyle\begin{array}{@{}l} d\Delta \omega + \alpha (1 - q)f(z - \theta,0)\omega = 0, \\ d\Delta \chi + L_{b}\chi + qaf(z - \theta,0)\omega - b\theta g'_{1}(z - \theta,\theta )\omega = 0. \end{array}\displaystyle \right . $$

(3.6)

If $a = \frac{\hat{\lambda}_{1}d}{1 - q}$, then we take $\hat{\phi}_{1} > 0$ into the second equation of (3.6) as follows:

$$L_{(a,b)}(d\Delta + L_{b})\chi = - qaf(z - \theta,0) \omega_{1} + b\theta g'_{1}(z - \theta,\theta ) \hat{\phi}_{1}. $$

From Remark 2.1, we see that all the eigenvalues of $L_{(b,d)}$ less than 0, that is, the operator $L_{(b,d)}$ is invertible, then there exists the unique

$$\chi_{1} = L_{ ( a,b )}^{ - 1} \bigl( - qaf(z - \theta,0) \omega_{1} + b\theta g'_{1}(z - \theta, \theta ) \hat{\phi}_{1} \bigr), $$

hence there exists $(\hat{\phi}_{1},\chi_{1}) \in \bar{W}_{y_{1}} = K_{0} \times C_{0}(\bar{\Omega} )$ such that $L_{1}(\omega,\chi )^{T} = (\omega,\chi )^{T}$. Thus, we have proved that $I - L_{1}$ is not invertible on $\bar{W}_{y_{1}}$. Hence, the results cannot be proved by Proposition 1.1 in this case.

In the following, we prove that (1.3) can bifurcate from $(a,0,\theta)$ by the bifurcation theory [18, 19], then we get the result. Defining the function as $F(a,u,\upsilon ) = (d\Delta u + a(1 - q)uf(z - u - \upsilon,u), d\Delta \upsilon + b\upsilon g(z - \upsilon - u,\upsilon ) + aqf(z - u - \upsilon,u)u)$. Clearly, $F(a,0,\theta ) = 0$. Define the operator as follows:

$$L_{1}(a,0,\theta ) = D_{(u,\upsilon )}F(a,0,\theta ) = \begin{pmatrix} d\Delta + a(1 - q)f(z - \theta,0) & 0 \\ - b\theta g'_{1}(z - \theta,\theta ) + aqf(z - \theta,0) & d\Delta + L_{b} \end{pmatrix}, $$

where $D_{(u,\upsilon )}F(a,0,\theta )$ stands for the Fréchet derivatives of F at the point $(u,\upsilon)$.

Let $L_{1}(a,0,\theta )(\omega,\chi )^{T} = (0,0)^{T}$. From the analysis of (3.6), we see that the nuclear space of $L_{1}(a,0,\theta )$ satisfies $N(L_{1}(a,0,\theta )) = \operatorname{span}\{ (\hat{\phi}_{1},\chi_{1})\}$. Hence $\operatorname{dim} N(L_{1}(a,0,\theta )) = 1$.

The adjoint operator of $L_{1}(a,0,\theta )$ can be written as

$$L_{1}^{*}(a,0,\theta ) = \begin{pmatrix} d\Delta + a(1 - q)f(z - \theta,0) & - b\theta g'_{1}(z - \theta,\theta ) + aqf(z - \theta,0) \\ 0 & d\Delta + L_{b} \end{pmatrix}. $$

Let $L_{1}^{*}(a,0,\theta )(\omega,\chi )^{T} = (0,0)^{T}$, it is easy to get $(\omega,\chi ) = (\hat{\phi}_{1},0)$, so $N(L_{1}^{*}(a,0,\theta )) = \operatorname{span}\{ (\hat{\phi}_{1},0)\}$. According to the Fredholm theorem [18], we can obtain $R(L_{1}(a,0,\theta )) = \{ (\omega,\chi ) \in E|\int_{\Omega} \omega \hat{\phi}_{1}\,dx = 0 \}$, so $\operatorname{codim} R ( L ( a,0,\theta ) ) = 1$.

Define $L_{2} ( a,0,\theta ) = D_{a}D_{ ( u,v )}F ( a,0,\theta )$, then

$$L_{2}( a,0,\theta ) \cdot ( \hat{\phi}_{1}, \chi_{1}) = \begin{pmatrix} ( 1 - q )f ( z - \theta,0 ) & 0 \\ qf ( z - \theta,0 ) & 0 \end{pmatrix} \begin{pmatrix} \hat{\phi}_{1} \\ \chi_{1} \end{pmatrix} = \begin{pmatrix} ( 1 - q )f ( z - \theta,0 )\hat{\phi}_{1} \\ qf ( z - \theta,0 )\hat{\phi}_{1} \end{pmatrix}. $$

Owing to $N(L_{1}^{*})$ and $R(L_{1})$ are orthogonal, however $\int_{\Omega} ( 1 - q )f ( z - \theta,0 )\hat{\phi}_{1}^{2} \ne 0$.

Hence $L_{2}(a,0,\theta )(\hat{\phi}_{1},\chi_{1})^{T} \notin R(L_{1})$. To sum up, when $a = \frac{\hat{\lambda}_{1}d}{1 - q}, b > \mu_{1}d$, it follows from the bifurcation theorem [18, 19] that $F(a,u,\upsilon ) = 0$ ($a \in R$) can produce the bifurcation at the point $(a,0,\theta)$. That is, there exist $\vert s \vert < \delta$ ($\delta > 0$) and the $C^{1}$ function as $(a(s),\phi (s),\varphi (s)):( - \delta,\delta ) \to R \times E$, such that $u(s) = s(\hat{\phi}_{1} + \phi (s)), \upsilon (s) = \theta - s(\chi_{1} + \varphi (s))$,

$$a(0) = a,\qquad \phi (0) = \varphi (0) = 0,\qquad \phi (s), \varphi (s) \in Z,\quad Z \oplus N ( L_{1} ) = E, $$

and satisfying $F(a(s),u(s),\upsilon (s)) = 0$ ($|s | < \delta$).

We shall discuss two possible cases as follows:

Case 1: if $a'(s) \equiv 0$, then $a(s) = a$, for s submitting to $\vert s \vert < \sigma$, and $F(a(s),u(s),\upsilon (s)) = 0$, that is, $(u(s),\upsilon (s))$ is the solution (1.3), since $\hat{\phi}_{1} > 0$, and $\vert s \vert$ is very small, then there exists $\varepsilon > 0$ such that $u(s) > 0,\upsilon (s) > 0$, when $0 < s < \varepsilon \le \delta$. Hence, (1.3) have positive solutions.

Case 2: if $a'(s)\not\equiv 0$, then there exists $\delta ' > 0, a'(s)$ is not equal to 0 when $\vert s \vert < \delta \le \delta '$. According to the uniqueness of the bifurcation solution of $F(a(s),u(s),\upsilon (s)) = 0$, $(0,\theta)$ is isolated fixed point of the operator F in the neighborhood of $(a,0,\theta)$. Owing to 1 is the eigenvalue of $F'(0,\theta )$, we cannot calculate $\operatorname{index}_{W}(F,(0,\theta )) = ( - 1)^{\sigma}$. Hence, we can define the operator as

$$F_{t}(u,\upsilon ) = ( - d\Delta + P)^{ - 1} \begin{pmatrix} Pu + (1- q)af(z - u - \upsilon,u)u - tu \\ M + bg(z - u - \upsilon,\upsilon )\upsilon + qaf(z - u - \upsilon,u)u \end{pmatrix}. $$

Obviously, $(0,\theta )$ is the fixed point of $F_{t}$. Defining the derived operator of $F_{t}$ at $( 0,\theta )$ as follows:

$$L_{t} = F'_{t}(0,\theta ) = ( - d\Delta + P)^{ - 1} \begin{pmatrix} p + (1 - q)af(z - \theta,0) - t & 0 \\ - b\theta g'_{1}(z - \theta,\theta ) + aqf(z - \theta,0) & P + L_{b} \end{pmatrix}. $$

The compact operator $F_{t}$ deduce that $F'_{t}$ is compact operator. Suppose that 1 is the eigenvalue of $L_{t}$ when $t \in ( 0,1]$, then $L_{t}(\phi,\psi )^{T} = (\phi,\psi )^{T}$, that is,

$$\left \{ \textstyle\begin{array}{@{}l} d\Delta \phi + a(1 - q)f(z - \theta,0)\phi - t\phi= 0, \\ d\Delta \psi + L_{b}\psi + aqf(z - \theta,0)\phi - b\theta g'_{1}(z - \theta,\theta )\phi = 0. \end{array}\displaystyle \right . $$

Thanks to $a(1 - q) = \lambda_{1}d$, then $\lambda_{1} ( d\Delta + ( 1 - q )af ( z - \theta,0 ) - t ) < 0$, that is, $d\Delta + ( 1 - q )af( z - \theta,0 ) - t$ is invertible, then $\phi \equiv 0$, and from Lemma 2.1, we know that the operator L is invertible. Then, similarly, we see that $\psi \equiv 0$. Thus, 1 is not the eigenvalue of $L_{t}$.

Due to $( 0,\theta )$ is the isolated fixed point of $F_{t}$, $\operatorname{index}_{W} ( L_{t} ( 0,0 ) ) = ( - 1 )^{\sigma}$. To show that $L_{t}$ has no eigenvalue greater than 1, suppose that $\sigma > 1$ is an eigenvalue of $L_{t}$. Let

$$\begin{aligned} &(\omega,\chi ) \ne (0,0),\quad L_{t}(\omega,\chi )^{T} = \sigma (\omega,\chi )^{T}, \quad\mbox{then } \\ &d\Delta \omega + P\omega = \frac{1}{\sigma} \bigl(P\omega + a(1 - q)f(z - \theta,0)\omega - t\omega \bigr). \end{aligned}$$

(3.7)

Equation (3.7) can be rewritten as follows:

$$d\Delta \omega + \frac{1}{\sigma} (1 - q)af(z - \theta,0)\omega = \biggl(P + \frac{t}{\sigma} - \frac{P}{\sigma} \biggr)\omega. $$

Thanks to $\sigma > 1$, then $\nu = P + \frac{t}{\sigma} - \frac{P}{\sigma} > 0$ is some eigenvalue of $d\Delta + \frac{1}{\sigma} (1 - q)af(z - \theta,0)$. However, from

$$\lambda_{1} \biggl(d\Delta + \frac{1}{\sigma} (1 - q)af(z - \theta,0) \biggr) \le \lambda_{1} \bigl(d\Delta + (1 - q)af(z - \theta,0) \bigr), $$

we obtain $\lambda_{1}(d\Delta + \frac{1}{\sigma} (1 - q)af(z - \theta,0)) \le 0$, this contradicts with $\nu > 0$. Hence, $\operatorname{index}_{W}(F_{t}, (0,\theta )) = 1$ as $t \in [0,1]$. Because $(0,\theta)$ is isolated fixed point of (1.3), we can take some neighborhood U of $(0,\theta)$ in $D'$ such that $F_{t}$ has no fixed point at ∂U. Using the homotopy invariance property, we get $\operatorname{index}{}_{W}(F,(0,\theta )) = \operatorname{index}_{W}(F_{t}(0,\theta )) = 1$. □

4 Coexistence of the chemostat model

In this section, by using the fixed point index calculation method, combined with Lemmas 3.1-3.3, we can show that there exists the sufficient condition of existence of non-negative solutions to equation (1.3).

Theorem 4.1

• (i) If $a < \frac{\lambda_{1}d}{1 - q}, b < \mu_{1}d$, then the unique non-negative solution of (1.3) is zero;

• (ii) if $a > \frac{\lambda_{1}d}{1 - q},b < \mu_{1}d$, then (1.3) have at least one positive solution besides $(0,0)$;

• (iii) if $a > \frac{\lambda_{1}d}{1 - q},b > \mu_{1}d$, then (1.3) have at least one positive solution besides $(0,0), (0,\theta )$.

Proof

(i) From Lemma 2.1, (1.3) can have no semi-trivial solution as $( 0,\theta )$ when $a < \frac{\lambda_{1}d}{1 - q}$, $b < \mu_{1}d$ then there exists only a zero solution as $(0,0)$. It follows from Lemma 3.1 that

$$\operatorname{index}_{W} \bigl(F, (0,0) \bigr) = 1, \quad\mbox{then } \operatorname{index}_{W} \bigl(F,D' \bigr) = \operatorname{index}_{W} \bigl(F,(0,0) \bigr). $$

Hence, we cannot prove the existence of positive solution by the fixed point index theory.

In the following, we show it by the upper and lower solution method.

When $a < \frac{\lambda_{1}d}{1 - q}$, we suppose (1.3) have non-negative solution $(\tilde{u},\tilde{\upsilon})$. It follows that either $\tilde{u} > 0$ or $\tilde{u} \equiv 0$. If $\tilde{u} > 0$, then

$$0 = d\Delta \tilde{u} + a ( 1 - q )f ( z - \tilde{u} - \tilde{\upsilon}, \tilde{u} )\tilde{u} \le d\Delta \tilde{u} + a ( 1 - q )f ( z - \tilde{u}, \tilde{u} )\tilde{u},\quad x \in \Omega. $$

Hence ũ is a positive lower solution of (1.3), and there exists an upper solution as $C_{1} > 0$, $\tilde{u} \le C_{1},x \in \bar{\Omega}$. Therefore, (1.3) has a positive solution $u_{ +}$ such that $\tilde{u} \le u{}_{ +} \le C_{1}$, $x \in \bar{\Omega}$. This contradicts Lemma 2.1. Thus $\tilde{u} \equiv 0$, then when $b < \mu_{1}d$, assume that (1.3) have non-negative solution $(0,\tilde{\upsilon})$, and $\tilde{\upsilon} > 0$, this contradicts with Remark 2.1. Hence, the non-negative solution of (1.3) has the only zero.

(ii) If $a > \frac{\lambda_{1}d}{1 - q}, b < \mu_{1}d$, from Lemma 2.1, we know that (1.3) has a semi-trivial solution $(0,\theta)$, then there exists only $(0,0)$. It follows from Lemma 3.1 that

$$\operatorname{index}_{W} \bigl(F,(0,0) \bigr) = 0, $$

and according to Lemma 3.2, $\operatorname{index}_{W}(F,D') = 1$, then

$$\operatorname{index}_{W} \bigl(F,D' \bigr) \ne \operatorname{index}_{W} \bigl(F,(0,0) \bigr). $$

Hence (1.3) has at least a positive solution on $D'$.

(iii) Similar to the proof of (ii), we can prove that $( 0,0 ), ( 0,\theta )$ are the non-negative solutions, if $a > \frac{\hat{\lambda}_{1}d}{1 - q},b > \mu_{1}d$. Thanks to Lemmas 3.1-3.3,

$$\operatorname{index}_{W} \bigl(F,D' \bigr) \ne \operatorname{index}_{W} \bigl(F,(0,0) \bigr) + \operatorname{index}_{W} \bigl(F,(0,\theta ) \bigr). $$

Hence (1.3) has at least a positive solution on $D'$. □

Remark 4.1

When $a = \frac{\lambda_{1}d}{1 - q},b > \mu_{1}d$, it follows from Lemma 3.3 that we can obtain either $\operatorname{index}_{W} ( F, ( 0,\theta ) ) = 1$, or for (1.3) there exists a positive solution.

When $\operatorname{index}_{W} ( F, ( 0,\theta ) ) = 1$,

$$\operatorname{index}_{W} \bigl( F,D' \bigr) = \operatorname{index}_{W} \bigl( F, ( 0,0 ) \bigr) + \operatorname{index}_{W} \bigl( F, ( 0,\theta ) \bigr), $$

the method of the index calculation cannot judge the existence of positive solutions. If another result of Lemma 3.3 holds, then (1.3) has a positive solution. To sum up, we cannot determine whether we have the existence of the positive solution.

5 Conclusion

The coexistence of an unstirred chemostat model with B-D functional response is studied by fixed point index theory in our paper. First of all, some prior estimates for positive solutions are proved by the maximum principle and the upper and lower solution method. Second, the calculations are performed on the fixed point index of chemostat model by degree theory and the homotopy invariance theorem. Finally, some sufficient condition on the existence of positive steady-state solutions is established by fixed point index theory in cone and bifurcation theory.