Coexistence for a Resource-Based Growth Model with Two Resources ^{1}

Yijie Meng Yifu Wang Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, P.R. China

Abstract

We investigate the coexistence of positive steady-state solutions to a parabolic system, which models a single species on two growth-limiting, non-reproducing resources in an un-stirred chemostat with diffusion. We establish the existence of a positive steady-state solution for a range of the parameter (m, n), the bifurcation solutions and the stability of bifurcation solutions. The proof depends on the maximum principle, bifurcation theorem and perturbation theorem.

Keywords: chemostat; coexistence; local bifurcation; maximum principle.

1 Introduction

Consider the following parabolic system

\begin{matrix}S_{t} = d_{1}S_{xx} − muf(S, R), & 0 < x < 1, t > 0, \\ R_{t} = d_{2}R_{xx} − nug(S, R), \\ u_{t} = d_{3}u_{xx} + u(mf(S, R) + cng(S, R)), \end{matrix}  (1.1)

with the boundary conditions

\begin{matrix}S_{x}(0, t) = −1, R_{x}(0, t) = −1, u_{x}(0, t) = 0, \\ S_{x}(1, t) + γS(1, t) = 0, R_{x}(1, t) + γR(1, t) = 0, u_{x}(1, t) + γu(1, t) = 0, \end{matrix}  (1.2)

and initial conditions

S(x, 0) = S_{0}(x) ≥ 0, R(x, 0) = R_{0}(x) ≥ 0, u(x, 0) = u_{0}(x) ≥ 0, ≢ 0,  (1.3)

where f(S, R) = S/(1 + aS + bR), g(S, R) = R/(1 + aS + bR), m > 0 is the maximal growth rate of species u on resource S in the absence of resource R, the constant n is defined similarly, constant c denotes the ratio of the growth yield constant of S and R. The constant a > 0 and b > 0 are the Michaelis-Menten constants, γ > 0.

Since we are only concerned with the nonnegative solutions (S, R, u) of (1.1), we can redefine the response functions f, g for S ≤ 0, R ≤ 0 without affecting our results. The un-stirred chemostat with one resource has been considered by many authors in the past decade(see [1][2][3]). Just as pointed out in [4], the un-stirred chemostat with two resources is more realistic and thus of interest, and the system (1.1) with equal diffusion rates is investigated in paper [5]. Without the assumption of equal diffusion rates, we obtain some estimates on the size of the coexistence region near a bifurcation point in the parameter space. The existence of positive steady-state solution of the system (1.1) is established by the maximum principle and the theorem of bifurcation, which appears in [6] to study the local solutions. The stability of bifurcation solutions is also studied via the perturbation theorem.

1Research supported by the NSF of China(Grant # 19971004, 10226013)

2 Extinction

In this section we use the maximum principle to establish conditions under which the species become extinct.

Lemma 2.1. The region (S ≥ 0, R ≥ 0, u ≥ 0) is invariant.

Proof. Consider the nutrient equation

\begin{matrix}S_{t} = d_{1}S_{xx} − muf(S, R), & 0 < x < 1, t > 0, \\ S_{x}(0, t) = −1, S_{x}(1, t) + γS(1, t) = 0, S(x, 0) = S_{0}(x) ≥ 0. \end{matrix}

For the fixed u, S̄(x, t) ≡ 0 is a solution of the differential equation above, S(x, 0) ≥ S̄(x, 0), and −S_{x}(0, t) = 1 ≥ 0 = S̄_{x}(0). By the comparison theorem for the parabolic equation (for example see [7]), we can show that S(x, t) ≥ 0 for all (x, t). Moreover, the boundary condition S ≢ 0 implies that S(x, t) > 0 for t > 0. Similarly, we can prove that R(x, t) > 0 and u(x, t) > 0 for all t > 0, and thus the proof is completed.

Let λ_{0}^{(i)} > 0(i = 1, 2, 3) be the principle eigenvalue of the following problem

\begin{matrix}d_{i}ϕ_{xx} + λϕ = 0, & 0 < x < 1, \\ ϕ_{x}(0) = 0, ϕ_{x}(1) + γϕ(1) = 0. \end{matrix}

with the eigenfunction ϕ_{0}^{(i)} > 0(i = 1, 2, 3) on [0, 1].

Let S̄(x) be the unique positive solution of the following problem

\begin{matrix}S_{xx} = 0, & 0 < x < 1, \\ S_{x}(0) = −1, S_{x}(1) + γS(1) = 0. \end{matrix}

The existence and uniqueness of S̄(x) is standard, and by the maximum principle it is easy to show that S̄ > 0 on [0, 1].

Lemma 2.2. There are positive constants α_{i} and K_{i}(i = 1, 2) such that S(x, t) ≤ S̄(x) + K_{1}e^{−α_{1}t}, R(x, t) ≤ S̄(x) + K_{2}e^{−α_{2}t}, for all x ∈ [0, 1], t > 0.

Proof. Let ω(x, t) = S(x, t) − S̄(x), then ω satisfies

\begin{matrix}ω_{t} ≤ d_{1}ω_{xx}, & 0 < x < 1, t > 0, \\ ω_{x}(0, t) = 0, ω_{x}(1, t) + γω(1, t) = 0, & t > 0. \end{matrix}

Then, by the comparison theorem, we have ω(x, t) ≤ W (x, t), where W (x, t) is the unique solution of the linear problem

W_{t} = d_{1}W_{xx},  0 < x < 1, t > 0,
W_{x}(0, t) = 0, W_{x}(1, t) + γW (1, t) = 0,
W (x, 0) = S(x, 0) − S̄(x).

In order to estimate W, let 0 < α_{1} < λ_{0}^{(1)}, W (x, t) = ϕ_{0}^{(1)}(x)h(x, t)e^{−α_{1}t}. Then we have

h_{t} = d_{1}h_{xx} + \frac{2d_{1}}{ϕ_{0}^{(1)}} ϕ_{0x}^{(1)}h_{x} + (α_{1} − λ_{0}^{(1)})h, 0 < x < 1, t > 0,

h_{x}(0, t) = 0, h_{x}(1, t) = 0,

h(x, 0) = \frac{S(x, 0) − S̄(x)}{ϕ_{0}^{(1)}(x)}.

The maximum principle ([7]) implies that

|h(x, t)| ≤ max_{[0,1]} \frac{|S(x, 0) − S̄(x)|}{ϕ_{0}^{(1)}(x)}

and this leads to

S(x, t) ≤ S̄(x) + K_{1}e^{−α_{1}t}, x ∈ [0, 1], t > 0,

for some constants K_{1} > 0. Similarly result holds for R.

Lemma 2.3. Let (S, R, u) be a solution of system (1.1)-(1.3), and suppose that \frac{m + cn}{min(a, b)} < λ_{0}^{(3)}. Then there are positive constants K, α such that u(x, t) ≤ Ke^{−αt}.

Proof. Straightforward computation leads to

u_{t} = d_{3}u_{xx} + u(mf(S, R) + cng(S, R)) ≤ d_{3}u_{xx} + \frac{m + cn}{min(a, b)}u, 0 < x < 1, t > 0.

Let V (x, t) be the unique solution of the following problem

V_{t} = d_{3}V_{xx} + \frac{m + cn}{min(a, b)}V,  0 < x < 1, t > 0,
V_{x}(0, t) = 0, V_{x}(1, t) + γV (1, t) = 0,
V (x, 0) = u(x, 0).

By the comparison principle, we have u(x, t) ≤ V (x, t). Let V (x, t) = ϕ_{0}^{(3)}(x)h(x, t)e^{−αt}, where α > 0 is small enough so that α + \frac{m + cn}{min(a, b)} − λ_{0}^{(3)} < 0. then

h_{t} = d_{3}h_{xx} + \frac{2d_{3}}{ϕ_{0}^{(3)}} ϕ_{0x}^{(3)}h_{x} + (α + \frac{m + cn}{min(a, b)} − λ_{0}^{(3)})h, 0 < x < 1, t > 0,

h_{x}(0, t) = 0, h_{x}(1, t) = 0,

h(x, 0) = \frac{u(x, 0)}{ϕ_{0}^{(3)}(x)}.

As in the previous lemma, it follows that |h(x, t)| ≤ max_{[0,1]} \frac{|u(x, 0)|}{ϕ_{0}^{(3)}(x)} and the lemma follows.

3 Coexistence.

In this section we consider the coexistence of the positive steady-state solutions of the system (1.1). So we consider the elliptic system

\begin{matrix}d_{1}S_{xx} − muf(S, R) = 0, & 0 < x < 1, \\ d_{2}R_{xx} − nug(S, R) = 0, \\ d_{3}u_{xx} + u(mf(S, R) + cng(S, R)) = 0, \end{matrix}  (3.1)

with the boundary conditions

\begin{matrix}S_{x}(0) = −1, R_{x}(0) = −1, u_{x}(0) = 0, \\ S_{x}(1) + γS(1) = 0, R_{x}(1) + γR(1) = 0, u_{x}(1) + γu(1) = 0. \end{matrix}  (3.2)

Let z = (d_{1}S + cd_{2}R + d_{3}u)/(d_{1} + cd_{2}), then z satisfies

\begin{matrix}z_{xx} = 0, & 0 < x < 1, \\ z_{x}(0) = −1, z_{x}(1) + γz(1) = 0, \end{matrix}  (3.3)

and we have z = (1 + γ)/γ − x.

First we give some estimates about the nonnegative solution of (3.1)-(3.2). The similar proof can be found in [4,8]. We omit the detail here.

Lemma 3.1. Suppose that (S, R, u) is a nonnegative solution of (3.1)-(3.2), then S > 0, R > 0, and either 0 < S < z, 0 < R < z or S = R = z. Furthermore, d_{1}S + cd_{2}R + d_{3}u = (d_{1} + cd_{2})z.

Let s = z − S, r = z − R, then by lemma 3.1, either 0 < s, r < z or s = r = 0, and

\begin{matrix}d_{1}d_{3}s_{xx} + m(d_{1}s + cd_{2}r)f(z − s, z − r) = 0, & 0 < x < 1, \\ d_{2}d_{3}r_{xx} + n(d_{1}s + cd_{2}r)g(z − s, z − r) = 0, & 0 < x < 1, \end{matrix}  (3.4)

with the boundary conditions

\begin{matrix}s_{x}(0) = 0, r_{x}(0) = 0, \\ s_{x}(1) + γs(1) = 0, r_{x}(1) + γr(1) = 0. \end{matrix}  (3.5)

3.1. The special case of d_{2}m = d_{1}n

In this subsection, we consider the case of d_{2}m = d_{1}n and discuss the existence of a positive solution of (3.4).

Let ω = s − r, then ω satisfies

ω_{xx} − C(x)ω = 0, 0 < x < 1, ω_{x}(0) = 0, ω_{x}(1) + γω(1) = 0,

where C(x) = m(d_{1}s + cd_{2}r)/(d_{1}d_{3}(1 + a(z − s) + b(z − r))). It follows from the maximum principle that ω = 0, which leads to s = r on [0, 1]. Substituting s = r into (3.4), we have

\begin{matrix}d_{1}d_{3}s_{xx} + m(d_{1} + cd_{2})sf(z − s, z − s) = 0, & 0 < x < 1, \\ s_{x}(0) = 0, s_{x}(1) + γs(1) = 0. \end{matrix}  (3.6)

Let λ_{1} > 0 and ϕ_{1} > 0 be the principle eigenvalue and eigenfunction of the following problem, with ϕ normalized so that ∫_{0}^{1} f(z, z)ϕ^{2}dx = 1

ϕ_{xx} + λ_{1}f(z, z)ϕ = 0, ϕ_{x}(0) = 0, ϕ_{x}(1) + γϕ(1) = 0.  (3.7)

By the result in [8,9], we have

Theorem 3.1. There exists a unique positive solution s̄ of (3.6), if and only if m > d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}), moreover 0 < s̄ < z, s̄ is continuous with respect to m ∈

[d_{1}d_{3}λ_{1}/(d_{1} +cd_{2}), ∞], and lim_{m→( \frac{d_{1}d_{3}λ_{1}}{d_{1}+cd_{2}} )^{+}} s̄ = 0 uniformly in (0, 1), lim_{m→∞} s̄ = z a.e. x ∈ (0, 1).

Clearly, if m > d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}), then (S̄, R̄, ū) = (z − s̄, z − s̄, (d_{1} + cd_{2})s̄/d_{3}) is the unique positive steady-state solution of (3.1)-(3.2) in the case d_{2}m = d_{1}n.

3.2. The case of d_{2}m ≠ d_{1}n

In this subsection, we discuss the existence and nonexistence of a positive solution of (3.4)(3.5). First we give a basic estimate for (s, r).

Lemma 3.2. If d_{2}m ≥ d_{1}n, then the solution (s, r) of (3.4)(3.5) satisfies r ≤ s ≤

\frac{d_{2}m}{d_{1}n}r.

Proof. Let ω = s − r, then

ω_{xx} − C(x)ω ≤ 0, 0 < x < 1, ω_{x}(0) = 0, ω_{x}(1) + γω(1) = 0,

where C(x) = n(d_{1}s + cd_{2}r)/(d_{2}d_{3}(1 + a(z − s) + b(z − r))) ≥ 0. It follows from the maximum principle that ω ≥ 0, and thus r ≤ s.

Again, let ω = d_{1}ns − d_{2}mr, then

ω_{xx} = \frac{mn(d_{1}s + cd_{2}r)}{d_{3}}(g(z − s, z − r) − f(z − s, z − r))

= \frac{mn(d_{1}s + cd_{2}r)}{d_{3}(1 + a(z − s) + b(z − r))}(s − r)

≥ 0

ω_{x}(0) = 0,  ω_{x}(1) + γω(1) = 0,

it follows that ω ≤ 0, i.e. s ≤ \frac{d_{2}m}{d_{1}n}r. Similarly, if d_{2}m ≤ d_{1}n, then we have s ≤ r ≤ \frac{d_{1}n}{d_{2}m}s.

The following theorem shows that a positive solution of (3.4)(3.5) cannot exist if both m and n are too small.

Theorem 3.2. Suppose m ≤ d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}) and n ≤ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}), then (s, r) = (0, 0) is the unique nonnegative solution of (3.4)(3.5).

Proof. If m ≤ d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}) and n ≤ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}), and (s, r) is a nontrivial nonnegative solution of (3.4)(3.5). Then it follows from the maximum principle that s > 0, r > 0. If \frac{d_{1}n}{d_{2}} ≤ m ≤ \frac{d_{1}d_{3}λ_{1}}{d_{1} + cd_{2}} , multiplying the first equation in (3.4) by s,

integrating over (0, 1) and using Green formula, we find

\begin{aligned}d_{1}d_{3}(∫_{0}^{1} s_{x}^{2}dx + γs^{2}(1)) & = m ∫_{0}^{1}(d_{1}s + cd_{2}r)sf(z − s, z − r)dx \\ & ≤ m(d_{1} + cd_{2}) ∫_{0}^{1} s^{2}f(z, z)dx. \end{aligned}

By the variational property of the principle eigenvalue, we have

∫_{0}^{1} s_{x}^{2}dx + γs^{2}(1) ≥ λ_{1} ∫_{0}^{1} s^{2}f(z, z)dx.

Hence (d_{1}d_{3}λ_{1} − m(d_{1} + cd_{2})) ∫_{0}^{1} s^{2}f(z, z)dx ≤ 0, which leads to s = 0, a contradiction. A similar result holds if m ≤ d_{1}n/d_{2}. This completes the proof.

Thus if m ≤ d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}) and n ≤ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}), then the washout solution (z, z, 0) is the unique nontrivial nonnegative solution of (3.1)(3.2).

Theorem 3.3 Suppose m ≥ d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}) and n ≥ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}). Then there exists a positive solution of (3.4)(3.5).

Proof. It is easy to check that (3.4)(3.5) is a quasi-monotone increasing system. Let (s̄, r̄) = (z, z) and (\underline{s}, \underline{r}) = (δϕ, δϕ), where ϕ is the principle eigenfunction defined by (3.7) and δ > 0 is small enough. Obviously (s̄, r̄) = (z, z) is the upper solution of (3.4)(3.5). Again

d_{1}\underline{s}_{xx} + m(\frac{d_{1}\underline{s} + cd_{2}\underline{r}}{d_{3}})f(z − \underline{s}, z − \underline{r})

= δϕ((\frac{m(d_{1} + cd_{2})}{d_{3}} − d_{1}λ_{1})f(z, z) − \frac{m(d_{1} + cd_{2})}{d_{3}}(f(z, z) − f(z − δϕ, z − δϕ)))
≥ δϕ((\frac{m(d_{1} + cd_{2})}{d_{3}} − d_{1}λ_{1})\frac{1}{γ + a + b} − \frac{m(d_{1} + cd_{2})δϕ}{d_{3}(1 + (a + b)(z − θδϕ))^{2}} ) (0 < θ < 1)

as long as δ is sufficiently small, we have

d_{1}\underline{s}_{xx} + m(\frac{d_{1}\underline{s} + cd_{2}\underline{r}}{d_{3}})f(z − \underline{s}, z − \underline{r}) > 0.

Similarly we have

d_{2}\underline{r}_{xx} + n(\frac{d_{1}\underline{s} + cd_{2}\underline{r}}{d_{3}})g(z − \underline{s}, z − \underline{r}) > 0.

Thus, for sufficiently small δ > 0, the pair (s̄, r̄) and (\underline{s}, \underline{r}) are the ordered upper and lower solutions of (3.4)(3.5). From [7], there exists a solution (s, r) satisfies (δϕ, δϕ) ≤ (s, r) ≤ (z, z).

Theorem 3.4. Suppose that either m > d_{3}λ_{1}, n ≤ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}) or m ≤ d_{1}d_{3}λ_{1}/(d_{1} + cd_{2}), n > d_{3}λ_{1}/c. Then there exists a positive solution of (3.4)(3.5).

Proof. We consider the former case, the other case can be done similarly. Let (s̄, r̄) = (z, z) and (\underline{s}, \underline{r}) = (δϕ, 0), then

\begin{matrix}d_{1}\underline{s}_{xx} + m(\frac{d_{1}\underline{s} + cd_{2}\underline{r}}{d_{3}})f(z − \underline{s}, z − \underline{r}) \\ = δϕ((\frac{md_{1}}{d_{3}} − d_{1}λ_{1})f(z, z) − \frac{md_{1}}{d_{3}} (f(z, z) − f(z − δϕ, z))). \end{matrix}

For sufficiently small δ > 0, we note that (s̄, r̄) and (\underline{s}, \underline{r}) are the ordered upper and lower solution of (3.4)(3.5). Hence there exists solution (s, r) of (3.4)(3.5) such that (δϕ, 0) ≤ (s, r) ≤ (z, z). So s > 0. It follows that r > 0 from lemma 3.2. This completes the proof.

3.3. Bifurcation Theorem

Now, for fixed n ≤ d_{2}d_{3}λ_{1}/(d_{1} + cd_{2}), we treat m as a bifurcation parameter to obtain the local bifurcation which corresponds to the positive solution of (3.4)(3.5).

At first, we rewrite (3.4)(3.5) as

\begin{matrix}s_{xx} + m(\frac{d_{1}s + cd_{2}r}{d_{1}d_{3}})f(z, z) + F_{1}(s, r) = 0, & 0 < x < 1, \\ r_{xx} + n(\frac{d_{1}s + cd_{2}r}{d_{2}d_{3}})g(z, z) + F_{2}(s, r) = 0, & 0 < x < 1, \end{matrix}  (3.8)

with the same boundary conditions, where

\begin{matrix}F_{1}(s, r) = m(\frac{d_{1}s + cd_{2}r}{d_{1}d_{3}})(f(z − s, z − r) − f(z, z)), \\ F_{2}(s, r) = n(\frac{d_{1}s + cd_{2}r}{d_{2}d_{3}})(g(z − s, z − r) − g(z, z)). \end{matrix}

Let K be the inverse operator of −\frac{d^{2}}{dx^{2}} , then

\begin{matrix}s − mK((\frac{d_{1}s + cd_{2}r}{d_{1}d_{3}})f(z, z)) − KF_{1}(s, r) = 0, & 0 < x < 1, \\ r − nK((\frac{d_{1}s + cd_{2}r}{d_{2}d_{3}})g(z, z)) − KF_{2}(s, r) = 0, & 0 < x < 1, \end{matrix}

Let T (m, s, r) = (mK((\frac{d_{1}s + cd_{2}r}{d_{1}d_{3}})f(z, z))+KF_{1}(s, r), nK((\frac{d_{1}s + cd_{2}r}{d_{2}d_{3}})g(z, z))+KF_{2}(s, r)),

and G(m, s, r) = (s, r) − T (m, s, r). Then the zeros of G(m, s, r) are the solutions of (3.4)(3.5).

Let C_{B}^{1}[0, 1] = {u ∈ C^{1}[0, 1] : u_{x}(0) = 0, u_{x}(1) + γu(1) = 0}, endowed with the usual norm ∥ · ∥, and X = C_{B}^{1}[0, 1] × C_{B}^{1}[0, 1]. Then we have the following theorem

Theorem 3.5. Suppose n ≤ \frac{d_{2}d_{3}λ_{1}}{d_{1} + cd_{2}} . Then (m_{0}, 0, 0) is a bifurcation point of

G(m, s, r) = 0, and in the neighborhood of (m_{0}, 0, 0), part of the bifurcation branch corresponds to the positive solution of (3.4)(3.5), where m_{0} = d_{3}λ_{1} − cn.

Proof. Let L(m, 0, 0) = DG_{(s,r)}(m, 0, 0) is the Frechet derivative of G(m, s, r) with respect to (s, r) at (0,0). Straightforward computation gives

L(m_{0}, 0, 0)(ω, χ) = (ω − m_{0}K((\frac{d_{1}ω + cd_{2}χ}{d_{1}d_{3}})f(z, z)), χ − nK((\frac{d_{1}ω + cd_{2}χ}{d_{2}d_{3}})g(z, z))).

Then L(m_{0}, 0, 0)(ω, χ) = 0 leads to

\begin{matrix}ω_{xx} + m_{0}(\frac{d_{1}ω + cd_{2}χ}{d_{1}d_{3}})f(z, z) = 0, & 0 < x < 1, \\ χ_{xx} + n(\frac{d_{1}ω + cd_{2}χ}{d_{2}d_{3}})g(z, z) = 0, & 0 < x < 1, \\ ω_{x}(0) = 0, χ_{x}(0) = 0, \\ ω_{x}(1) + γω(1) = 0, χ_{x}(1) + γχ(1) = 0. \end{matrix}  (3.9)

Noting that m_{0} = d_{3}λ_{1} − cn and f(z, z) = g(z, z), we have

(d_{1}ω + cd_{2}χ)_{xx} + λ_{1}(d_{1}ω + cd_{2}χ)f(z, z) = 0,

so d_{1}ω + cd_{2}χ = ϕ, and putting this into (3.9), we find

ω_{xx} + \frac{m_{0}}{d_{1}d_{3}} f(z, z)ϕ = 0, χ_{xx} + \frac{n}{d_{2}d_{3}} g(z, z)ϕ = 0.

It is easy to show that there exists a unique positive solution (ω_{1}, χ_{1}) of the above problem. Moreover ω_{1} ≥ χ_{1} and d_{1}ω_{1} + cd_{2}χ_{1} = ϕ. Hence the null space of L(m_{0}, 0, 0), N(L(m_{0}, 0, 0)) = spans{(ω_{1}, χ_{1})}. This, dimN(L(m_{0}, 0, 0)) = 1. Let R(L(m_{0}, 0, 0)) be the range of the operator L(m_{0}, 0, 0). If (h_{1}, h_{2}) ∈ R(L(m_{0}, 0, 0)), then there exists (Φ, Ψ) ∈ X satisfies

Φ_{xx} + m_{0}(\frac{d_{1}Φ + cd_{2}Ψ}{d_{1}d_{3}})f(z, z) = h_{1xx},  0 < x < 1,
Ψ_{xx} + n(\frac{d_{1}Φ + cd_{2}Ψ}{d_{2}d_{3}})g(z, z) = h_{2xx},  0 < x < 1,
Φ_{x}(0) = 0, Ψ_{x}(0) = 0,
Φ_{x}(1) + γΦ(1) = 0, Ψ_{x}(1) + γΨ(1) = 0.

Thus, we find

(d_{1}Φ + cd_{2}Ψ)_{xx} + λ_{1}(d_{1}Φ + cd_{2}Ψ)f(z, z) = (d_{1}h_{1} + cd_{2}h_{2})_{xx},

multiplying the above equation by ϕ, and integrating over (0, 1), shows

− ∫_{0}^{1} λ_{1}ϕ(d_{1}h_{1} + cd_{2}h_{2})dx = 0,

which implies R(L(m_{0}, 0, 0)) = {(h_{1}, h_{2}) ∈ X : ∫_{0}^{1} ϕ(d_{1}h_{1} + cd_{2}h_{2})dx = 0} and codimR(L (m_{0}, 0, 0)) = 1.

Now Let L_{1}(m_{0}, 0, 0) = D^{2}G_{m(s,r)}(m_{0}, 0, 0), then

L_{1}(m_{0}, 0, 0)(m_{0}, 0, 0)(ω_{1}, χ_{1}) = (−K(\frac{d_{1}ω_{1} + cd_{2}χ_{1}}{d_{1}d_{3}} )f(z, z), 0).

It is easy to see that L_{1}(m_{0}, 0, 0)(m_{0}, 0, 0)(ω_{1}, χ_{1}) ∉ R(L(m_{0}, 0, 0)). According to Theorem1.7 in [10], there exists a δ > 0 and a C^{1} function (m(τ), ω(τ), χ(τ)) : (−τ, τ) → R×X, such that m(0) = m_{0}, ω(0) = 0, χ(0) = 0 and (m(τ), s(τ), r(τ)) = (m(τ), τ(ω_{1} + ω(τ)), τ(χ_{1} + χ(τ)))(|τ| < δ) satisfies G(m(τ), s(τ), r(τ)) = 0. Point on the curve {(m(τ), z − τ(ω_{1} + ω(τ)), z − τ(χ_{1} + χ(τ))) : |τ| < δ} with τ > 0 corresponds to the positive solutions of (3.1)(3.2).

3.4. Stability of the Bifurcation Solution

In this section we shall determine the stability of the bifurcation solutions.

Lemma 3.3. 0 is a i − simple eigenvalue of L(m_{0}, 0, 0).

Proof. Suppose L(m_{0}, 0, 0) = 0. From the proof of Theorem 3.5, we have N(L(m_{0}, 0, 0))= spans{(ω_{1}, χ_{1})}, codimR(L(m_{0}, 0, 0)) = dimN(L(m_{0}, 0, 0)) = 1. We say i(ω_{1}, χ_{1}) ∉ R(L(m_{0}, 0, 0)), otherwise

∫_{0}^{1} f(z, z)ϕ(d_{1}ω_{1} + cd_{2}χ_{1})dx = 0,

which is impossible. Thus we complete the proof of the lemma.

Let L(m(τ), s(τ), r(τ)) be the linearized operator of (5.1) at (m(τ), s(τ), r(τ)). Then the corollary 1.13 and Theorem 1.16 in[11] can be applied and we have the following lemma.

Lemma 3.4. There exist C^{1} function m → (ξ(m), U(m)), τ → (η(τ), V (τ)) defined on the neighborhoods of m_{0} and 0, respectively, into R × X, such that (ξ(m_{0}), U(m_{0})) = (0, (ω_{1}, χ_{1})) = (η(0), V (0)) and on these neighborhoods

\begin{matrix}L(m, 0, 0)U(m) = ξ(m)U(m), & |m − m_{0}| << 1, \\ L(m(τ), s(τ), r(τ))V (τ) = η(τ)v(τ), & |τ| << 1 \end{matrix}  (3.10)

where U(m) = (u_{1}(m), u_{2}(m)), V (τ) = (v_{1}(τ), v_{2}(τ)), and ξ′(m_{0}) ≠ 0, η(τ) and −τm′(τ)ξ′(m_{0}) have the same sign if η(τ) ≠ 0.

Theorem 3.6 The differential ξ′(m_{0}) > 0.

Proof. By (3.10) we have

\begin{matrix}u_{1xx} + m(\frac{d_{1}u_{1} + cd_{2}u_{2}}{d_{1}d_{3}} )f(z, z) = ξ(m)u_{1}, & 0 < x < 1, \\ u_{2xx} + n(\frac{d_{1}u_{1} + cd_{2}u_{2}}{d_{2}d_{3}} )g(z, z) = ξ(m)u_{2}, \\ u_{1x}(0) = 0, u_{2x}(0) = 0, \\ u_{1x}(1) + γu_{1}(1) = 0, u_{2x}(1) + γu_{2}(1) = 0. \end{matrix}  (3.11)

Clearly,u_{1} ≢ 0, u_{2} ≢ 0. Since U(m) is continuous and U(m_{0}) = (ω_{1}, χ_{1}), u_{1}(m) > 0, u_{2}(m) > 0 for |m − m_{0}| << 1. By (3.11) we have

(d_{1}u_{1} + cd_{2}u_{2})_{xx} + \frac{m + cn}{d_{3}}(d_{1}u_{1} + cd_{2}u_{2})f(z, z) = ξ(m)(d_{1}u_{1} + cd_{2}u_{2}).

Since d_{1}u_{1} + cd_{2}u_{2} > 0, it follows that ξ(m) is the principle eigenvalue of L_{2} = \frac{d^{2}}{dx^{2}} +

\frac{m + cn}{d_{3}}f(z, z), and increases in m for |m−m_{0}| << 1. Again ξ′(m_{0}) ≠ 0, so we must have

ξ′(m_{0}) > 0.

Theorem 3.7. The differential of m(τ) at τ = 0 satisfies

m′(0) ∫_{0}^{1} ϕ^{2}f(z, z)dx = ∫_{0}^{1} ϕ^{2} \frac{m_{0}ω_{1} + cnχ_{1} + (bm_{0} − acn)(ω_{1} − χ)z}{(1 + (a + b)z)^{2}} dx.

Proof. Substitute (m(τ), s(τ), r(τ)) into the equation of (3.4), divide by τ, differential with respect to τ, and set τ = 0, this gives

d_{1}ω′(0)_{xx} + m′(0)(\frac{d_{1}ω_{1} + cd_{2}χ_{1}}{d_{3}} )f(z, z) + m_{0}(\frac{d_{1}ω′(0) + cd_{2}χ′(0)}{d_{3}})f(z, z)

+m_{0}(\frac{d_{1}ω_{1} + cd_{2}χ_{1}}{d_{3}} )\frac{−ω_{1} + b(χ_{1} − ω_{1})z}{(1 + (a + b)z)^{2}} = 0

d_{2}χ′(0)_{xx} + n(\frac{d_{1}ω′(0) + cd_{2}χ′(0)}{d_{3}})f(z, z) + n(\frac{d_{1}ω_{1} + cd_{2}χ_{1}}{d_{3}} )\frac{−χ_{1} + a(ω_{1} − χ_{1})z}{(1 + (a + b)z)^{2}} = 0.

Now, multiplying the first equation by ϕ, adding to the second equation which is multiplied by cϕ, integrating over (0, 1) to get

\begin{aligned}& m′(0) ∫_{0}^{1} \frac{d_{1}ω_{1} + cd_{2}χ_{1}}{d_{3}} ϕf(z, z)dx \\ = & ∫_{0}^{1} \frac{d_{1}ω_{1} + cnχ_{1}}{d_{3}} ϕ\frac{m_{0}ω_{1} + cnχ_{1} + (bm_{0} − acn)(ω_{1} − χ)z}{(1 + (a + b)z)^{2}} dx \end{aligned}

i.e.

m′(0) ∫_{0}^{1} ϕ^{2}f(z, z)dx = ∫_{0}^{1} ϕ^{2} \frac{m_{0}ω_{1} + cnχ_{1} + (bm_{0} − acn)(ω_{1} − χ)z}{(1 + (a + b)z)^{2}} dx.

Now, we have

Theorem 3.8. Suppose n ≤ \frac{d_{2}d_{3}λ_{1}}{d_{1} + cd_{2}} and bm_{0} ≥ acn. Then the bifurcation solutions

defined by Theorem 3.5 are stable for τ > 0.

Proof. From Theorem 3.5 and 3.7, m′(0) > 0, m′(τ) > 0 for |τ| << 1. By Lemma 3.4 and Theorem 3.6, we have η(τ) < 0 for τ > 0, which completes the proof of Theorem.

Acknowledgement: The research is supported by the National Natural Science Foundation of China (19971004, 10226013)

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