Nonlinear Analysis 67 (2007) 1966–1979
www.elsevier.com/locate/na
A strongly coupled predator–prey system with non-monotonic
functional responseI
Xinfu Chen a , Yuanwei Qi b,∗ , Mingxin Wang c
a Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260, USA
b Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA
c Department of Applied Mathematics, Southeast University, Nanjing 210018, PR China
Received 21 March 2006; accepted 17 August 2006
Abstract
The predator–prey system with non-monotonic functional response is an interesting field of theoretical study. In this paper
we consider a strongly coupled partial differential equation model with a non-monotonic functional response—a Holling typeIV function in a bounded domain with no flux boundary condition. We prove a number of existence and non-existence results
concerning non-constant steady states (patterns) of the underlying system. In particular, we demonstrate that cross-diffusion can
create patterns when the corresponding model without cross-diffusion fails.
c 2006 Elsevier Ltd. All rights reserved.
MSC: 35J55; 92D25
Keywords: Cross-diffusion; Predator–prey model; Non-constant positive steady states
1. Introduction
In population dynamics, a spatially homogeneous predator–prey system can be modeled [9,21,30] as
u
ug(u) − vp(u)
ut = G(u),
u=
,
G(u) =
,
v
−bv + vq(u)
(1.1)
where u and v are the population densities of prey and predator respectively, and b > 0 is the death rate of the predator.
Function p(u), the functional response of predator to prey density, refers to the change in the density of prey attached
per unit time per predator as the prey density changes. Function q(u), the rate of conversion of prey to predator, is
assumed to satisfy the same properties as p(u). In Gauss’s model, q(u) is a constant multiple of p(u). In microbial
dynamics or chemical kinetics, the functional response describes the uptake of substrate by the micro-organisms. In
the classical Lotka–Volterra model, a linear functional response p(u) = mu is assumed, which is unbounded.
I Chen thanks the National Science Foundation Grant DMS–9971043 USA. Qi was partially supported by Hong Kong RGC Grant HKUST
6129/009. Wang thanks PRC grant NSFC 19831060 and Hwa Ying Culture & Education Foundation.
∗ Corresponding author. Tel.: +1 407 823 2810.
E-mail addresses: [email protected] (X. Chen), [email protected] (Y. Qi), [email protected] (M. Wang).
c 2006 Elsevier Ltd. All rights reserved.
0362-546X/$ - see front matter doi:10.1016/j.na.2006.08.022
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
1967
To overcome the unboundedness of such a response function, a response function p(u) = mu/(a+u) was proposed
by Michaelis and Menton in the context of studying enzymatic reactions. In it, m > 0 is the maximal growth rate of
the species and a > 0 is the half-saturation constant. Later on, such a model was also used by Holling as one of the
predator functional responses, now popularly referred to as a Holling type-II function.
Another type, known as the Holling type-III function, takes the form p(u) = mu 2 /(a + u 2 ). Similar types of
response functions can be found in Freedman [8]. The main characteristic of these two type of model (Holling type-II,
Holling type-III) is that the function p(u) is monotone and has a finite limit as u → ∞. Many earlier studies make
the monotonic assumption in many predator–prey interaction models; see Hsu [12], Kooji and Zegeling [14], Kuang
and Freedman [15] and May [20].
Nevertheless, evidence from experiment and observation shows that this is not always the case. For instance,
experiments of the microbial dynamics as conducted in Andrews [1], Bonn and Landelout [2], Edwards [7], Yang and
Humphrey [32] indicate that non-monotonic responses happen when the nutrient concentration reaches a high level.
The high concentration of nutrient surprisingly produces an inhibitory effect on the growth rate of micro-organisms.
Such a phenomenon is often seen when micro-organisms are used for waste decomposition or water purification.
Andrews [1] proposed a function p(u) = mu/(a + bu + u 2 ), which is qualitatively similar to the Holling type-II
function at lower concentration levels but exhibits the inhibitory effect at high concentration levels. It is called the
Monod–Haldane function. Collings [4] also used the same type of functions in the modeling of mite predator–prey
interaction model and termed it a Holling type-IV function. In that particular model the prey species is a spider
mite. It uses the webbing to interfere with predators by decreasing their walking speed and reducing their chance of
contacting the prey. In experiments on the uptake of phenol by pure culture of Pseudomonas putida growing on phenol
in continuous culture, Sokol and Howell [28] used a simplified form of p(u) = mu/(a + u 2 ) and found that it fits
their experimental data significantly better and it enjoys the additional benefit of being simpler, since it involves only
two parameters.
Another type of phenomenon in population dynamics where non-monotonic response occurs involves group
defense. Group defense is a term used to describe the ability of prey to better defend or disguise themselves resulting in
a decrease or even prevention of predation when their numbers are large enough. An outstanding and simple example
of this phenomenon is described by Tener [29]. Lone musk OX can be successfully attacked by wolves. Small herds
of two to six OX are attacked but with rare success. No successful attacks have been observed in larger herds. A
similar pattern is also observed by Holmes and Bethel [11] in their study of insect populations. Large swarms of
insects make individual identification difficult for their predators. A third example is supplied by the study of algae
by Davidowicz, Gliwicz and Gulati [5]. Filamentous algae are often qualified as inedible by herbivorous zooplankton.
But the experiments show (see [5]) that Daphnia can consume them at low concentration, but their filtering apparatus
get jammed by algae when concentration is high.
Freedman and Wolkowicz [9], Mischaikow and Wolkowicz [21], and Wolkowicz [30] proposed (1.1) with nonmonotone p(u) to study predator–prey interaction when the prey exhibits group defense. A prototype is
u
mu
σu
g(u) = r u 1 −
,
p(u) =
,
q(u) =
,
(1.2)
k
a + u2
a + u2
where r > 0 is the growth rate of prey at u = 0, k > 0 is the carry capacity of the prey and m, σ, a > 0 are free
parameters. Freedman and Wolkowicz [9] showed for (1.1) and (1.2) that if the carrying capacity k of the prey is
sufficiently large, the predator is almost driven to extinction. This is related to the phenomenon of paradox enrichment
as explained in Rosenzweig [25]. Wolkowicz [30] showed that if the carrying capacity is made sufficiently large by
enrichment of the environment, the model predicts the eventual extinction of the predator.
Ruan and Xiao [26] gave a more detailed study of (1.1) and (1.2) regarding (i) uniqueness of interior equilibrium,
(ii) various kinds of bifurcation and (iii) the global dynamics of the model depending on all parameters. In particular,
they analyzed the existence and uniqueness of the limit cycle or homoclinic loop for a range of parameters, the global
attraction of an equilibrium in the first quadrant and the appearance of Bogdanov–Takens bifurcation. A remarkable
property of system (1.1) and (1.2), as revealed by the study of Ruan and Xiao [26] is that it has more complicated
dynamics than models with monotone response such as Holling type I–III.
To capture the inhomogeneous distribution of both prey and predator in different locations of space at any given
time, and the dynamics of both prey and predator population under such an environment, Pang and Wang [24]
1968
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
considered the following reaction–diffusion version of (1.1) and (1.2):
ut = diag(d1 , d2 )1u + G(u)
(1.3)
in a bounded domain Ω of R n , where d1 > 0 and d2 > 0 are the diffusion coefficients and 4 is the Laplacian. In their
model, the no-flux boundary conditions are used.
For a partial differential equation (PDE) model such as (1.3), an important issue is to find the non-constant steady
state and study its stability. An important criterion to determine the efficiency and success of a particular type of
model, from the modeling point of view, is their ability to generate rich non-constant steady states, referred to as
patterns.
In [24], Pang and Wang discussed the existence, bifurcation and global existence of non-constant positive steady
states of (1.3). In particular, it is shown that if d1 , d2 are suitably large, (1.3) has no non-constant positive steady
states.
One of the observed features in population dynamics is that different concentration levels of prey at different sites
will give different signals to direct the movements of predator and vice versa. In other words, the movement of a
predator at any particular location is influenced by the gradient of the concentration of prey at that location, and that
of the prey is affected by the gradient of the concentration of predator at the same location.
A model taking this effect into account should capture more interesting phenomena and patterns of underlying
biological systems. One such model is
φ (u)
d u(1 + d3 v)
ut = 1Φ(u) + G(u),
Φ(u) = 1
= 1
,
(1.4)
φ2 (u)
d2 v(1 + d4 /(1 + u))
where d3 and d4 are positive constants, and G(u) is determined by (1.1) and (1.2). The diffusion 1Φ(u) of population
can be understood as follows. Writing 1Φ = div(Φu ∇u), we can regard −Φu ∇u as the flux. φ1, v = d1 d3 u > 0
represents that the prey runs away from the predator, whereas φ2, u = −d2 d4 v(1 + u)−2 < 0 indicates that predator
chases the prey, in agreement with observed general pattern of predator–prey movement; see Okubo [23, Chpt.10] for
a more detailed discussion on relevant biological models. The coefficients d3 and d4 (or, d1 d3 and d2 d4 ), called crossdiffusion pressures [23,27], play a central role in producing various patterns in our study. Here we may regard φ1 (u)
and φ2 (u) as the potentials (living conditions) of the prey and predator respectively, so that the diffusion of each is
toward places of better living conditions (i.e., lower potentials). For a recent survey on the mathematical developments
of cross-diffusion equations, see Ni [22].
There are many works on the existence of positive steady states of ecological models with no flux boundary
conditions, but most of them do not include the cross-diffusion. For ecological competition models, see [17–19] and
the references therein. For predator–prey models, see [3–13,24,31] and the references therein.
The main aim of this paper is to study the effects of the cross-diffusion pressures on the existence of non-constant
positive steady states of (1.4). Here, by a positive solution we mean a smooth solution (u, v) with both u and v
being positive. We shall demonstrate that the cross-diffusion pressures d3 and d4 may help forming more patterns, in
comparison with the results of non-constant positive steady states of [24]. Scaling the parameters we may assume that
m = b = r = 1. Thus, we study the elliptic system, for (u, v) = (u(x), v(x)),

uv
u

−
in Ω ,
−d
1(u
+
d
uv)
=
u
1
−

1
3

k
a
+
u2



σu
d4 v
=v
−1
in Ω ,
−d2 1 v +
(1.5)
1+u
a + u2



∂ n u = ∂ n v = 0
on ∂Ω ,


where Ω is a bounded domain in Rn with smooth boundary ∂Ω , and ∂n is the outward directional derivative normal
to ∂Ω .
In the next section, we establish a priori upper and lower bounds for positive solutions of (1.5). As a preparation
for existence, in Section 3 we use a degree theory to develop a general setting to enable one to conclude the existence
of patterns when the indexes of constant steady states change. In Section 4 we establish the existence of non-constant
positive solutions to (1.5) for a large range of diffusion and cross-diffusion coefficients. Finally, in Section 5, we study
the asymptotic behavior of non-constant positive solutions of (1.5) when one or more diffusion coefficients tend to
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
1969
infinity. As an application we use a singular perturbation method to establish the existence of patterns for (1.5) when
the limiting problem possesses non-trivial solutions.
We end this section with the following observations:
√
σs
(i) When σ ≤ 2 a, there holds sups≥0 { a+s
2 − 1} ≤ 0, so that the only non-negative solutions to (1.5) are
u = (0, 0) and u = (k, 0). Consequently, (1.5) does not have any positive solution.
√
(ii) When σ > 2 a, the possible positive constant solutions to (1.5) are (u 1 , v1 ) and (u 2 , v2 ), where

p
1

u 1 = (σ − σ 2 − 4a),
v1 = σ u 1 (k − u 1 )/k if k > u 1 ,
2
(1.6)
p

u 2 = 1 (σ + σ 2 − 4a),
v2 = σ u 2 (k − u 2 )/k if k > u 2 .
2
2. Upper and lower bounds for positive solutions
The main purpose of this section is to give a priori upper and lower positive bounds for positive solutions of (1.5).
To this end, we first cite two known results. The first is due to Lin, Ni and Takagi [16], and the second to Lou and Ni
[18].
Harnack Inequality [16]. Let c ∈ C(Ω ) and w ∈ C 2 (Ω ) ∩ C 1 (Ω ) be a positive classical solution to 1w(x) +
c(x)w(x) = 0 in Ω subject to the homogeneous Neumann boundary condition. Then there exists a positive constant
C = C(Ω , kck∞ ) such that maxΩ w ≤ C minΩ w.
Maximum Principle [18]. If u ∈ C 2 (Ω ) satisfies ∂n u = 0 on ∂Ω and x0 ∈ Ω is a point where u achieves its maximum,
then −1u(x0 ) ≥ 0.
Theorem 2.1. Let p = (σ, a, k) ∈ (0, ∞)3 be given and d > 0 be a fixed constant. Assume that d1 , d2 ≥ d and
d3 , d4 ≥ 0. There exists a positive constant C = C(p, d, Ω ), which does not depend on d3 and d4 , such that any
positive solution (u, v) of (1.5) satisfies
max u(x) + max v(x) +
Ω
Ω
maxΩ u(x) maxΩ v(x)
+
≤ C min{1 + d3 , 1 + d4 }.
minΩ u(x)
minΩ v(x)
Proof. Let φ1 (u) and φ2 (u) be defined as in (1.4), and denote
v
u
σu
g1 (u) = u 1 − −
,
g
(u)
=
v
−
1
,
2
k
a + u2
a + u2
(2.1)
G(u) = (g1 (u), g2 (u))T .
Then (1.5) becomes
−1Φ(u) = G(u)
in Ω ,
∂n u = 0
on ∂Ω .
Let x0 ∈ Ω be a point where φ1 (u(x0 )) = maxx∈Ω φ1 (u(x)). By the Maximum Principle, g1 (u(x0 )) ≥ 0, i.e.,
1 − u(x0 )/k − v(x0 )/(a + u 2 (x0 )) ≥ 0. This implies that
u(x0 ) < k,
v(x0 ) < a + k 2 ,
1
1
max u(x) ≤
max φ1 = φ(u(x0 )) ≤ k[1 + d3 (a + k 2 )].
d1 Ω
d1
Ω
(2.2)
(2.3)
The function φ2 (u) satisfies
1φ2 (u) + c(x)φ2 (u) = 0
in Ω ,
∂n φ2 (u) = 0
on ∂Ω ,
where c(x) = g2 (u)/φ2 (u) ∈ C(Ω ). It is easy to verify that the norm kck∞ is bounded above by a constant depending
only on p and d. By the Harnack Inequality, we have
max φ2 (u(x)) ≤ C min φ2 (u(x)),
Ω
Ω
(2.4)
1970
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
where C = C(Ω , kck∞ ) = C(p, d, Ω ). It follows that
maxΩ v
maxΩ φ2 (u(x)) maxΩ {1 + d4 (1 + u)−1 }
≤
×
minΩ v
minΩ φ2 (u(x))
minΩ {1 + d4 (1 + u)−1 }
≤ C[1 + min{d4 , max u}] ≤ C min{1 + d3 , 1 + d4 }
Ω
by (2.3). It then follows from (2.2) that
max v ≤ C(a + k 2 ) min{1 + d3 , 1 + d4 }.
Ω̄
Similarly, by considering separately the cases 0 ≤ d3 ≤ 1 and d3 ≥ 1 for the equation of φ1 , (2.4) holds for φ1 (u),
and hence
maxΩ u
maxΩ φ1 (u(x)) maxΩ {1 + d3 v}
≤
×
minΩ u
minΩ φ1 (u(x))
minΩ {1 + d3 v}
maxΩ v
≤ C min{1 + d3 , 1 + d4 }.
≤C
minΩ v
Finally, using (2.2) again we then complete the proof.
Theorem 2.2. Assume that σ k 6= a + k 2 . Let d, d̂ > 0 be fixed constants. There exists a constant C =
C(d, d̂, p, Ω ) > 0, such that any positive solution (u, v) of (1.5) satisfies
min u(x) > C,
Ω
min v(x) > C
Ω
provided that d1 , d2 ≥ d and d3 , d4 ≤ d̂.
R
Proof. Since Ω g2 (u)dx = 0, there exists x1 ∈ Ω such that g2 (u(x1 )) = 0, that is
σ u(x1 ) = a + u 2 (x1 ).
(2.5)
It follows that maxΩ u ≥ a/σ . Consequently, by Theorem 2.1,
min u ≥
Ω
1
a
max u ≥
.
C Ω
σC
Denote M2 = maxΩ v. To prove Theorem 2.2, it suffices to show that M2 cannot be very small. Let x2 and x3
be such that φ1 (x2 ) = maxΩ φ1 (u) and φ1 (x3 ) = minΩ φ1 (u). Then, g1 (u(x2 )) ≥ 0 ≥ g1 (u(x3 )) by the Maximum
Principle. This implies
u(x2 ) ≤ k,
u(x3 ) ≥ k(1 − M2 /a),
maxΩ φ1
maxΩ u
(1 + d3 M2 )
≤ (1 + d3 M2 )
≤ (1 + d3 M2 )
.
minΩ u
minΩ φ1
(1 − M2 /a)
(2.6)
(2.7)
Hence, ku − kk∞ = O(M2 ). From (2.5) and the assumption σ k 6= a + k 2 , we then conclude that M2 cannot be very
small. This completes the proof of the theorem. 3. A degree theory
We shall use a degree theory to establish the existence of non-constant positive solutions to (1.5). In this section
we provide a general setting.
Denote d = (d1 , d2 , d3 , d4 ) and p = (σ, a, k). We shall fix p ∈ (0, ∞)3 and take d ∈ (0, ∞)2 × [0, ∞)2 as
bifurcation parameters. The dependence of p will often be suppressed. Define
X = {u ∈ [C 2 (Ω̄ )]2 | ∂n u = 0 on ∂Ω },
X+ = {(u, v) ∈ X | u > 0, v > 0 on Ω̄ },
B(C) = {(u, v) ∈ X | 1/C < u, v < C on Ω̄ }.
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
1971
Since the determinant det Φu (u) of Φu (u) is positive for all non-negative u, Φu−1 exists. Hence, u is a positive
solution to (1.5) if and only if
n
o
F(d; u) := u − (I − 1)−1 Φu (u)−1 [G(u) + ∇u Φuu (u)∇u] + u = 0 in X+
where (I − 1)−1 is the inverse of I − 1 in X. As F(d; ·) is a compact perturbation of an identity operator, the
Leray–Schauder deg(F(d; ·), 0, B) is well-defined if F(d; u) 6= 0 for all u ∈ ∂ B.
Theorem 3.1. Assume that p = (σ, a, k) ∈ (0, ∞)3 with σ 6= k + a/k. Then for every d ∈ (0, ∞)2 × [0, ∞)2 , there
exists a constant C0 (d, p) such that for every C > C0 (d, p),
0 if σ < k + a/k,
deg (F(d; ·), 0, B(C)) =
1 if σ > k + a/k.
We remark that the change of degree when σ passes the borderline σ = k + a/k is due to the appearance
(disappearance) of a positive constant steady state bifurcating from u ≡ (k, 0).
Proof of Theorem 3.1. We divide the proof into several steps.
Step 1. By Theorems 2.1 and 2.2, there exists a positive constant C0 (d, p) such that if F(d; u) = 0 in X+ , then
u ∈ B(C0 ). Note that C0 (d, p) can be taken as a continuous function for σ 6= k + a/k. By the invariance property of
the Leray–Schauder degree, we then conclude that deg(F(d, ·), 0, B(C)) does not depend on d if C > C0 (d, p) and
it also does not depend on p if p changes continuously in (0, ∞)3 without touching the surface σ = k + a/k.
Observe that (1.5) does not have any positive solution when σ ≤ a 2 /4. We then conclude that
deg(F(d; ·), 0, B(C)) = 0
∀ p ∈ {(σ, a, k) ∈ (0, ∞)3 | σ < k + a/k}, C > C0 (d, p).
Step 2. It remains to consider the case σ > k + a/k. By degree invariance, we need only consider a special d,
say d = (d, d, 0, 0) with large d. For this we can use the following non-existence result. In order to compare the
existence regions of (1.5) with and without cross-diffusion, we provide a non-existence result stronger than what is
needed here.
Theorem 3.2. Let (d2 , p) ∈ (0, ∞)4 be given. There exists a positive constant d̂1 (d2 , p) such that when d =
(d1 , d2 , 0, 0) and d1 ≥ d̂1 (d2 , p), (1.5) does not have any non-constant positive solution.
A slightly weaker version (i.e., for fixed suitably large d2 ) of the Theorem was proven in [24].
Proof. Integrating (1.5) multiplied by (1 − ū/u, 1 − v̄/v) (ū is the average of u over Ω ) we obtain
Z Z d1 ū|∇u|2
d2 v̄|∇v|2
v
σu
u
+
− 1 (v − v̄) dx
dx =
1− −
(u − ū) +
k
u2
v2
a + u2
a + u2
Ω
ZΩ h
i
≤
ε(v − v̄)2 + Cε (u − ū)2 dx
Ω
for every ε ∈ (0, 1). Since the upper bounds of u, v,
large we can conclude that u ≡ ū. max u
min u
and
max v
min v
do not depend on d1 ≥ 1, taking ε small and d1
Step 3. It remains to calculate deg(F, 0, B) when all solutions to F = 0 are positive constant solutions. For applications
in the next section, here we shall derive a general formula. For this purpose, we assume that u∗ is a positive root to
G(u∗ ) = 0. We can calculate
Du F(d; u∗ ) = I − (I − 1)−1 {Φu−1 (u∗ )Gu (u∗ ) + I}
in L(X, X).
If Du F does not have any pure imaginary (or zero) eigenvalue, the index of F at u∗ is defined as index (F(d; ·), u∗ ) =
(−1)m , where m is the number of eigenvalues of Du F with negative real parts. The deg(F(d; ·), 0, B) is then equal to
the summation of the indexes over all solutions to F = 0 in B, provided that F 6= 0 on ∂ B.
∞ be a complete set of eigenpairs for −1 on Ω
To calculate m, we use the eigenspaces of −1. Let {µi , φi }i=1
with zero flux boundary condition, ordered such that 0 = µ1 < µ2 ≤ µ3 ≤ · · ·. Note that φ1 ≡ 1. We decompose
∞ X where X = {cφ | c ∈ R2 }. Then for each integer i ≥ 1, X is invariant under D F(d; u∗ ), and
X = ⊕i=1
i
i
i
i
u
1972
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
λ is an eigenvalue of Du F on Xi if and only if λ is an eigenvalue of the matrix I −
1
1+µi
[Φu (u∗ )−1 Gu (u∗ ) + I] =
[µi I − Φu (u∗ )−1 Gu (u∗ )]. Hence, the number of eigenvalues with negative real parts of Du F(d; u∗ ) on Xi is odd
if and only if H (d, u∗ ; µi ) < 0 where
1
1+µi
H (d, u∗ ; µ) := det[µ I − Φu−1 (u∗ )Gu (u∗ )].
In conclusion, we have the following lemma.
Lemma 3.1. Let u∗ be a positive root to G(·) = 0. Assume that H (d, u∗ ; µ j ) 6= 0 for all j. Then
X
1.
index(F(d; ·), u∗ ) = (−1)m , m =
i≥1, H (d, µi )<0
Step 4. Now we can complete the proof of Theorem 3.1. We need only to calculate the degree for the case σ > k +a/k.
In this case, G(·) = 0 has a unique positive solution, denoted by u1 . When d = (d, d, 0, 0), Φu (u1 ) = d I so that
H (d, u1 ; µ) = d1 det(d µ I − Gu (u1 )). As shall be seen in the next section, det(Gu (u1 )) > 0. This implies that when
d is sufficiently large, H (d, u1 ; µi ) > 0 for all i = 1, 2, . . .. Thus, m = 0 and deg(F(d, ·), 0, B(C)) = 1. This
completes the proof. 4. Existence of positive non-constant solutions
In this section, we establish the existence of positive non-constant solutions for (1.5). In particular, we show that
for certain ranges of parameters where (1.3) does not have any positive non-constant steady state, our model can still
produce patterns. The idea is as follows. First we calculate the index of F(d; ·) at positive constant steady states. If the
sum of all these indexes is not equal to the degree stated in Theorem 3.1, then F(d; ·) = 0 in B(C) for C = C0 (d, p)
must have a non-constant positive solution, which also solves (1.5).
√
In what follows we always assume that σ > 2 a so that σ u = a + u 2 has two positive roots given in (1.6).
When u = (u, v) is a positive solution to G(u) = 0, we can calculate, using a + u 2 = σ u and v = (1 − u/k)σ u,
(2u/σ k)(k − u − σ/2)
−1/σ
g11
−1/σ
Gu (u) =
=:
,
(4.1)
g21
0
(a − u 2 )v/(σ u 2 )
0
d1 (1 + d3 v)
d1 d3 u
φ11
φ12
Φu (u) =
=:
,
(4.2)
φ21
φ22
−d2 d4 v/(1 + u)2 d2 [1 + d4 /(1 + u)]
H (d, u; µ) ≡ det(µI − Φu−1 G(u))
g21 /σ
|φ21 |/σ + φ12 g21 − g11 φ22
+
.
= µ2 + µ
φ11 φ22 + |φ21 φ12 |
φ11 φ22 + |φ12 φ21 |
(4.3)
To calculate the roots of H (d, u; ·) = 0, we shall restrict our attention to large |d|. Note that
lim H (d, u; µ) = µ2 − Λi µ
di →∞
∀ i = 1, 2, 3, 4,
(4.4)
where
d3 (u 2 − a)v(1 + u)
,
σ d2 u{(1 + d3 v)(1 + u + d4 ) + d3 d4 uv/(1 + u)}
2u{k − [u + σ/2] − σ d4 (k − u)/[2(1 + u)(1 + u + d4 )]}
,
Λ2 = (d1 , d3 , d4 ; u) =
d1 σ k{1 + d3 v + d3 d4 uv/[(1 + u)(1 + u + d4 )]}
Λ1 = Λ(d2 , d3 , d4 ; u) =
(u 2 − a)(1 + u)
,
σ d2 u{(1 + u + d4 ) + d4 u/(1 + u)}
2u{k − [u + σ/2] − σ (k − u)/[2(1 + u)]}
.
Λ4 = Λ4 (d1 , d3 ; u) :=
d1 σ k{1 + d3 v + d3 uv/(1 + u)}
Λ3 = Λ(d2 , d4 ; u) =
(4.5)
(4.6)
(4.7)
(4.8)
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X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
The sign of the trace tr(Gu ) = g11 is determined by k − σ/2 − u and the sign of the determinant det(Gu ) by
(a − u 2 ). Hence, we shall discuss separately the following cases:
(i) u 1 < k < u 1 + σ/2,
(ii) u 1 + σ/2 < k < u 2 ,
(iii) k > u 2 .
4.1. Local stability of u1 when u 1 < k < u 1 + σ/2
In this subsection, we assume that
u 1 < k < u 1 + σ/2
where u 1 = (σ −
p
σ 2 − 4a)/2.
(4.9)
Then u1 := (u 1 , v1 ), where v1 = σ u 1 (1 − u 1 /k), is a positive solution to G(·) = 0. We consider local stability of the
constant steady state u ≡ u1 for the evolution dynamics
diag(τ1 , τ2 ) ut = 1Φ(u) + G(u)
in Ω × (0, ∞),
∂n u = 0
on ∂Ω .
(4.10)
Theorem 4.1. Let τ1 and τ2 be positive constants and k satisfy (4.9). Then the positive constant solution u(x, t) ≡
(u 1 , v1 ) is asymptotically stable with respect to the dynamics (4.10). Consequently, in a small neighborhood of u1 ,
(1.5) does not have any non-constant positive solution.
Proof. The linearization of (4.10) at u1 takes the form
vt = diag(1/τ1 , 1/τ2 ) {Φu (u1 )1v + Gu (u1 )v}
in Ω × (0, ∞),
∂n v = 0
on ∂Ω .
ceλt φ
For each eigenpair (µ, φ) of −1 on Ω with no flux boundary condition, v =
is a solution if and only if
(λ, c) is an eigenpair of the matrix diag(τ1−1 , τ2−1 ){−µΦu (u1 ) + Gu (u1 )}, which has a positive determinant and a
negative trace, as can be seen from (4.1)–(4.3), g11 < 0, and g21 > 0 (since k < u 1 + σ/2 and a = u 1 u 2 > u 21 ).
Thus, Re(λ) < 0. It then follows from Theorem 5.1.1 of [10] (p. 98) that the constant steady state u(x, t) ≡ u1 is
asymptotically stable to (4.10). 4.2. The case u 1 + σ/2 < k < u 2
In this case u1 = (u 1 , v1 ) is the only positive solution to G(·) = 0. By fixing the diffusion coefficients d1 and
d3 (for prey) and using the diffusion coefficients d2 and d4 (for predator) as bifurcation parameters, we shall show
that a symmetry-breaking occurs, thereby creating non-constant positive solutions to (1.5). We want to emphasize that
it is caused by the presence of cross-diffusion which has a more complex role than that of the diffusion coefficients
∞ all the eigenvalues (counting multiplicity) of −1 on Ω , ordered so that
d1 and d2 . As before, we denote by {µi }i=1
0 = µ1 < µ2 ≤ µ3 ≤ · · ·.
Theorem 4.2 (Existence With Suitable d2 and d4 ). Assume that σ > k + a/k, i.e., u 1 < k < u 2 . Define
Λ2 (d1 , d3 , d4 , u1 ) and Λ4 (d1 , d3 , u1 ) as in (4.6) and (4.8) with (u, v) = (u 1 , v1 ).
(i) Suppose d1 , d3 , and d4 are given such that Λ2 (d1 , d3 , d4 ; u 1 ) ∈ (µ j , µ j+1 ) for some positive even integer j.
There exists a positive constant d2∗ such that if d2 ≥ d2∗ , then (1.5) has at least one non-constant positive solution.
(ii) Suppose d1 and d3 are given such that Λ4 (d1 , d3 ; u1 ) ∈ (µ j , µ j+1 ) for some positive even integer j. Then for
any given d2 > 0, there is a positive constant d4∗ such that when d4 > d4∗ , (1.5) has at least one non-constant positive
solution.
Remark 4.1. (i) For Λ4 (d1 , d3 ; u1 ) to be positive, it is necessary and sufficient to have
k > u1 +
σ (k − u 1 )
σ
+
.
2
2(1 + u 1 )
(4.11)
When this inequality holds, Λ2 (d1 , d3 , d4 ; u1 ) is also positive provided that d4 is large. We then can adjust d1 to
make the assumptions in (i) or (ii) of the theorem hold.
(ii) When both σ and a/σ 2 are small, u 1 /σ ≈ 0 and u 2 /σ ≈ 1. There are plenty of k ∈ (u 1 , u 2 ) such that (4.11)
holds.
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X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
Proof of Theorem 4.2. Denote by µ± (d, u1 ), with Re(µ− ) ≤ Re(µ+ ), the two roots to H (d, u1 ; µ) = 0. Then
µ+ (d, u1 )µ− (d, u1 ) = [det(Φu (u1 ))]−1 det(Gu (u1 )) > 0 since a − u 21 = u 1 u 2 − u 21 > 0. From (4.4), we see that
lim µ− (d, u1 ) = 0,
d2 →∞
lim µ− (d, u1 ) = 0,
d4 →∞
lim µ+ (d, u1 ) = Λ2 (d1 , d3 , d4 ; u1 ),
d2 →∞
lim µ+ (d, u1 ) = Λ4 (d1 , d3 ; u1 ).
d4 →∞
Suppose Λ2 (d1 , d3 , d4 ; u1 ) ∈ (µ j , µ j+1 ) for some positive even j. Then for all d2 1,
0 = µ1 < µ− (d, u1 ) < µ2 ,
µ+ (d, u1 ) ∈ (µ j , µ j+1 ).
Hence if d2 1, H (d, u1 ; µi ) < 0 is equivalent to i ∈ {2, . . . , j}. Since j is even, Lemma 3.1 gives
index(F(d ; ·), u1 ) = (−1) j−1 = −1.
Consequently F(d; u) = 0 has at least another positive solution that is different from the constant function u ≡ u1 ,
since otherwise the degree of F = 0 in B(C) would be −1 for all large enough C, which would contradict Theorem 3.1.
This proves the first assertion of the theorem. The second assertion is similarly proved. 4.3. The case k > u 2
√
Theorem 4.3 (Existence When d1 or d3 are Large). Assume that k > u 2 := 12 (σ + σ 2 − 4a). Let Λ1 (d2 , d3 , d4 ; u2 )
and Λ3 (d2 , d4 ; u2 ) be as in (4.5) and (4.7) with (u, v) = (u 2 , v2 ).
(i) Suppose d2 , d3 , and d4 are given such that Λ1 (d2 , d3 , d4 ; u2 ) ∈ (µ j , µ j+1 ) for some positive even integer j.
Then there exists a positive constant d1∗ such that if d1 > d1∗ , (1.5) admits at least one positive non-constant solution.
(ii) Suppose d2 and d4 are given such that Λ3 (d2 , d4 ; u2 ) ∈ (µ j , µ j+1 ) for some positive even integer j. Then for
each given d1 , there exists a positive constant d3∗ such that if d3 > d3∗ , (1.5) admits at least one positive non-constant
solution.
Proof. In this case, G(·) = 0 has exactly two positive solutions: u1 = (u 1 , v1 ) and u2 = (u 2 , v2 ). We calculate the
indexes of the two constant solutions. Let µ± (d, u) be the two roots to H (d, u; ·) = 0.
The steady state u1 . Since a − u 21 > 0, µ− (d, u1 )µ+ (d, u1 ) = det(Φu−1 (u1 )) det(Gu (u1 )) > 0 for all d. From (4.4)
and the expression of Λ1 and Λ3 , we see that when d1 + d3 is large, µ− (d, u1 ) < µ+ (d, u1 ) < 0, which implies, by
Lemma 3.1,
index(F(d; ·), u1 ) = (−1)0 = 1.
The steady state u2 . In this case, a − u 22 < 0 so that µ+ (d, u2 )µ− (d, u2 ) < 0. Thus, µ− (d, u2 ) < 0 < µ+ (d, u2 )
for all d ∈ (0, ∞)2 × [0, ∞)2 .
Now suppose that Λ1 (d2 , d3 , d4 ; u 2 ) ∈ (µ j , µ j+1 ) for some positive even j. Then for all large enough d1 , we
have H (d, u2 ; µi ) < 0 if and only i ∈ {1, . . . , j}. As j is even, Lemma 3.1 gives us
index(F(d; ·), u2 ) = (−1) j = 1.
In conclusion, when d1 is large enough,
index (F(d; ·), u1 ) + index (F(d; ·), u2 ) = 1 + 1 = 2.
This implies that (1.5) admits at least one non-constant positive solution, and also the first assertion of the theorem.
The proof of the second assertion is similar and is omitted. Remark 4.2. When k > u 2 + σ/2, we can also fix d1 , d3 and take d2 and/or d4 large to establish the existence of
non-constant positive solutions to (1.5). For simplicity we omit the details.
Remark 4.3. When d2 is small and d1 and/or d3 are large, our system (1.5) with cross-diffusion are qualitatively
different from its counterpart (1.3) which does not have cross-diffusion. To see a comparison, let us consider
the case d4 = 0 and k > u 2 . For an arbitrarily fixed d3 > 0, we can find a small positive d2 such that
Λ1 (d2 , d3 , 0; u2 ) =
(u 22 −a)d3 v2
d2 σ (1+d3 v2 )
∈ (µ j , µ j+1 ) for some even positive integer j. Then Theorem 4.3 shows that (1.5)
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
1975
admits at least one positive non-constant solution for all large enough d1 , whereas Theorem 3.1 states that (1.5) with
d3 = 0 does not have any positive non-constant solution for all sufficiently large d1 ! In this sense, we see that the
patterns are totally attributed to the cross-diffusion.
5. Asymptotic limits and singular perturbation
In this section we study the limiting behavior of non-constant positive solutions of (1.5) as one or more of the
parameters d1 , d2 , d3 and d4 tend to infinity. The purpose is to reveal more qualitative and quantitative properties
of the limiting cases since the model will be dramatically simplified, and more detailed analysis can be performed
(e.g., in one space dimension). As an illustration, we establish an existence result by using the asymptotic limit and a
singular perturbation method.
5.1. Asymptotic limits
Since our a priori estimate (2.1) depends on one of the coefficients d3 , d4 , we need to assume that one of them is
bounded. We first consider the case that d4 is bounded.
∞ be a sequence such that each u is a positive solution to (1.5) with d = d . Assume
Theorem 5.1. Let {di , ui }i=1
i
i
that limi→∞ di = (d1 , d2 , d3 , d4 ) ∈ (0, ∞]2 × [0, ∞] × (0, ∞) and limi→∞ d1,i d3,i = ∞. Then by passing to a
subsequence if necessary, one and only one of the followings must hold:
(ia) limi→∞ d3,i kvi k∞ = 0, σ = k + a/k, and limi→∞ ui = (k, 0);
(ib) limi→∞ d3,i kvi k∞ = 0 and limi→∞ ui = u∗ , where u∗ is a positive solution to G(·) = 0;
(ii) limi→∞ d3,i kvi k∞ = ∞, limi→∞ (u i , vi /kvi k∞ ) = (1/w, w/kwk∞ ) in C 1 (Ω̄ ) for some positive solution w
to

σw
d4 w 2
w


−
1
in Ω ,
∂n w = 0 on ∂Ω ,
−1
w
+
=

1+w
d2 1 + aw 2
(5.1)
Z

1
1


1−
dx ≥ 0,
kw
Ω w
where the equality in the last line holds if and only if limi→∞ kvi k∞ = 0.
(iiia) limi→∞ d3,i kvi k∞ ∈ (0, ∞), limi→∞ d1,i = ∞, and limi→∞ (u i , d3,i vi ) → (α/(1 + w), w) in C 1 (Ω̄ ), for
some positive constant α and positive solution w to the problem

σ α(1 + w)
w
d4 w(1 + w)


=
−1
in Ω ,
∂n w = 0 on ∂Ω ,
−1 w +
2
2
1 + α + w
d2 α + a(1 + w)
Z
(5.2)
1
α


1−
dx ≥ 0,

k(1 + w)
Ω 1+w
where equality in the last line holds if and only if limi→∞ vi = 0.
(iiib) limi→∞ d3,i kvi k∞ ∈ (0, ∞), limi→∞ d1,i = d1 ∈ (0, ∞), limi→∞ (u i , d3,i vi ) → (u, w), in C 1 (Ω̄ ), where
(u, w) is a positive solution to

∂n u = 0 on ∂Ω ,
−d1 1(u
u(1 −u/k)
in Ω ,
+ u w) =
w
σu
d4 w
(5.3)
=
−1
in Ω ,
∂n w = 0 on ∂Ω .
−1 w +
1+u
d2 a + u 2
Moreover, if d2 = ∞, then limi→∞ (u i , vi ) = u∗ , where u∗ is a solution to G(·) = 0, or more precisely,
σ u ∗ = a + u ∗ 2 , 1 = u ∗ /k + v ∗ /(a + u ∗ 2 ).
Remark 5.1. The integral inequalities in (5.1) and (5.2) contain a parameter k, which does not appear in the
differential equations that w satisfies. That is, if w solves the PDE, then we can take a large k such that the integral
inequalities are satisfied.
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X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
Proof of Theorem 5.1. From Theorem 2.1, the sequence {ui } is a bounded family in C 0 (Ω̄ ). Upon a subsequence,
limi→∞ d3,i kvi k∞ and limi→∞ kvi k∞ exist. We divide our proof into three cases:
(i) lim d3,i kvi k∞ = 0,
i→∞
(ii) lim d3,i kvi k∞ = ∞,
i→∞
(iii) lim d3,i kvi k∞ ∈ (0, ∞).
i→∞
Case (i): limi→∞ d3,i kvi k∞ = 0. We consider limi→∞ kvi k∞ = 0 and > 0 separately.
(ia) Suppose limi→∞ kvi k∞ = 0. Then by (2.6) and (2.7), we conclude that limi→∞ u i = k. As σ u i /(a + u i2 ) − 1
vanishes at least once in Ω , we see that σ k/(a+k 2 )−1 = 0. Thus, we must have σ = k+a/k and limi→∞ ui = (k, 0).
(ib) Suppose limi→∞ kvi k∞ > 0. Then we must have limi→∞ di,3 = 0. Hence, limi→∞ di,1 = ∞. The equation
1[u i (1 + di,3 vi )] = O(1/d1,i ) then implies that u i approaches a constant, say u ∗ . Since σ u i /(a + u i2 ) − 1 vanishes
at least once, σ u ∗ = a + u ∗ 2 > 0. Consequently, 1[vi (1 + d4,i )/(1 + u i )] → 0, which implies that vi approaches a
constant, say v ∗ , which has to be positive, by the assumption limi→∞ kvi k∞ > 0. As 1 − u i /k − vi /(a + u i2 ) vanishes
at least once, we then see that G(u∗ ) = 0. Thus assertion (ib) holds.
Case (ii): limi→∞ d3,i kvi k∞ = ∞. Set ṽi = vi /kvi k∞ and ϕi = u i [ṽi + (d3,i kvi k∞ )−1 ]. Then
#
"
u i vi
ui
−
in Ω ,
∂n ϕi = 0 on ∂Ω .
−d1,i d3,i kvi k∞ 1ϕi = u i 1 −
k
a + u i2
Because d3,i kvi k∞ → ∞, d1,i → d1 > 0 and ku i k∞ , kvi k∞ are uniformly bounded, we can prove that, subject to
choosing a subsequence, ϕi converges uniformly to some constant τ , and therefore also u i ṽi → τ uniformly. From
(2.1) and (2.5) we see that τ > 0.
The function ψi = ṽi + d4,i ṽi /(1 + u i ) satisfies
!
σ ui
ψi
−1
in Ω ,
∂n ψi = 0 on ∂Ω .
(5.4)
−1ψi =
d2,i [1 + d4,i /(1 + u i )] a + u i2
Since {ψi } is bounded independent of i, by a standard elliptic regularity and Harnack inequality, ψi → ψ in C 1 (Ω̄ )
for some positive function ψ.
Once we know the convergence of (ϕi , ψi ), we then conclude that (u i , ṽi ) → (1/w, τ w) uniformly, for some
positive function w uniquely determined by ψ = τ w(1 + d4 /(1 + 1/w)). Since kṽi k∞ = 1, we see that τ = 1/kwk∞ .
Also taking the limit in (5.4) we obtain, for w, the PDE equation in (5.1). Finally, from the integral identity
Z
Z
u i ṽi
dx,
(5.5)
u i (1 − u i /k)dx = kvi k∞
2
Ω
Ω a + ui
we obtain the integral inequality in (5.1), and conclude that the inequality is an identity if and only if limi→∞ kvi k∞ =
0.
Case (iii): limi→∞ d3i kvi k = α̂ ∈ (0, ∞).
First consider (iiia): limi→∞ d1,i = ∞. In this case, we derive in the same way as in the previous case that ϕi → τ
and ψi → ψ for some positive constant τ and positive function ψ. Note that ϕi = (d3, i kvi k∞ )−1 u i (1 + d3 vi ). It then
follows that (u i , d3 vi ) → (u, w) which is determined uniquely from τ = (1/α̂)u(1+w), ψ = (w/α̂)[1+d4 /(1+u)].
Setting α = τ α̂, we have u = α/(1 + w) and ψ = (w/α̂)(1 + d4 /(1 + u)). From the equation for ψi , we then see that
w satisfies the differential equation in (5.2). The integral inequality in (5.2) follows from (5.5).
Next we consider (iiib) limi→∞ d1,i = d1 ∈ (0, ∞). Then we must have limi→∞ di,3 = ∞ and limi→∞ kvi k∞ = 0
since limi→∞ d1,i d3,i = ∞. In this case, by compactness, (ϕi , ψi ) approaches, in C 1 (Ω̄ ), a limit (ϕ, ψ). This implies
that (u i , vi /kvi k) → (u, v) in C 1 (Ω̄ ) which is uniquely determined by ϕ = u(1/α̂ + v), ψ = v(1 + d4 /(1 + u)).
Upon noting that d3,i vi = (d3,i kvi k∞ )(vi /kvi k∞ ) → α̂v =: w, we then obtain (5.3) by taking the limit of (1.5).
Thus, assertion (iiib) holds.
Finally, we note that if d2 = ∞, then in the previous derivation we can conclude that limi→∞ ψi = ψ = constant.
One can then directly verify that ui approaches a constant, say u∗ . In addition, since 1 − u i /k − vi /(a + u i2 ) and
σ u i /(a + u i2 ) − 1 vanish at least once, we conclude that σ u ∗ = a + u ∗ 2 , 1 = u ∗ /k + v ∗ /(a + u ∗ 2 ). The proof of the
theorem is now complete. In a similar manner, we can consider the case d4 → ∞, keeping d3 bounded. For simplicity, we only state the
result.
X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
1977
∞ be a sequence such that each u is a positive solution to (1.5) with d = d . Assume
Theorem 5.2. Let {di , ui }i=1
i
i
that limi→∞ (d1,i , d2,i , d3,i ) = (d1 , d2 , d3 ) ∈ (0, ∞]2 × (0, ∞) and limi→∞ d4,i = ∞. Then by passing to a
subsequence if necessary, (u i , vi ) → (u, α(1 + u)) for some constant α ≥ 0 and some positive solution u to the
problem

u
α(1 + u)
u


1− −
in Ω ,
∂n u = 0 on ∂Ω ,
−1[u + αd3 u(1 + u)] =
k
a + u2
d1
Z
(5.6)
σu


−
1
dx
=
0.
 (1 + u)
a + u2
Ω
We remark that σ does not appear in the differential equation of (5.6); namely if we have a positive solution to the
differential equation of (5.6), we can select a unique positive σ such that the integral identity in (5.6) holds.
5.2. Singular perturbation
At last, we use a singular perturbation method and solutions to the limit problem to construct non-constant positive
solutions to (1.5).
Theorem 5.3. Let (a, σ, d2 , d4 ) ∈ (0, ∞)3 × [0, ∞) be given. Assume that (5.1) has a positive solution w and it is
non-degenerate, i.e., the following linear problem, for φ,
−1[F 0 (w)φ] = G 0 (w)φ
in Ω ,
∂n φ = 0
on ∂Ω
(5.7)
has only
zero solution,
where F(w) = w + d4 w 2 /(1 + w), G(w) = (w/d2 )[σ w/(1 + aw 2 ) − 1]. Then, for every
R a−2
R
−1
k > Ω w dx/ Ω w dx and every d1 > 0, there exists an ε0 > 0 such that (1.5) has at least one non-constant
positive solution provided that d3 ≥ 1/ε0 .
Remark 5.2. (i) In the one-dimensional case where Ω = (0, `), one can show, by a first integral and a phase plane
analysis, that (5.1) admits a positive non-degenerate solution for all ` R1.
R
(ii) We did not state a corresponding result for the case when k = Ω w −2 dx/ Ω w −1 dx since the conditions
needed are quite messy and very hard to verify. Nevertheless, from the bifurcation point of view, it is a very important
one.
(iii) We can establish the existence of non-trivial solution to (1.5) from non-degenerate solutions to other limit
problems, say (5.2), (5.3), or (5.6). Here we omit the details.
Proof of Theorem 5.3. For simplicity, we take d1
of non-constant positive solutions to the problem

u
v

−1(εu + uv) = εu 1 − −



a + u2
k
σu
−1v = v
−1



a + u2

∂n u = ∂n v = 0
= d2 = 1, d4 = 0 and d3 = 1/ε. Our goal is to prove the existence
in Ω ,
in Ω ,
on ∂Ω .
We make a transformation: For a suitable positive constant A,
ψ = (v + ε)/A
v = Aψ − ε
⇐⇒
φ = [u(v + ε) − A]/(ε A)
u = (1 + εφ)/ψ.
R
We normalize φ such that Ω φ dx = 0. The resulting equation for φ is solvable only if
i
R 1+εφ h
ε
1 − 1+εφ
Ω ψ
kψ + a+(1+εφ)2 /ψ 2 dx
.
A = A(ψ, εφ, ε) :=
R
1+εφ
Ω a+(1+εφ)2 /ψ 2 dx
(5.8)
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X. Chen et al. / Nonlinear Analysis 67 (2007) 1966–1979
R
Then solving (5.8) is equivalent to finding a (φ, ψ) ∈ X := {(φ, ψ) ∈ C 0 (Ω̄ ) × C 0 (Ω̄ ) | Ω φ dx = 0} such that
−1φ = F1 (ψ, εφ, ε) in Ω ,
∂n φ = 0 on ∂Ω ,
−1ψ = F2 (ψ, εφ, ε) in Ω ,
∂n ψ = 0 on ∂Ω ,
where
1 + εφ
εψ 2
(1 + εφ)ψ 2
1 + εφ
1−
+
,
−
2
2
ψA(ψ, εφ, ε)
kψ
aψ + (1 + εφ)
aψ 2 + (1 + εφ)2
σ (1 + εφ)ψ
ε
F2 (ψ, εφ, ε) = ψ −
−
1
.
A(ψ, εφ, ε)
aψ 2 + (1 + εφ)2
F1 (ψ, εφ, ε) =
When ε = 0, there exists a solution (φ, ψ) = (φ0 , ψ0 ), given by
Z
−1
Z
1
1
w2
1−
dx
ψ0 = w,
A=
,
dx
2
kw
Ω w
Ω 1 + aw
Z
−1φ0 = F1 (ψ0 , 0, 0) in Ω ,
∂n φ0 = 0 on ∂Ω ,
φ0 dx = 0.
Ω
Note
that A > 0 since k >
R
F
(ψ0 , 0, 0) dx = 0. We define
1
Ω
o
n
√
B = (φ, ψ) ∈ X | kφ − φ0 k∞ ≤ ε, kψ − ψ0 k∞ ≤ ε2/3 .
R
R
−1
Ω ψ0 dx/ Ω
ψ0−2 dx. The problem for φ0
is well-posed since the definition of A implies
For each (φ, ψ) ∈ B, we define (φ̃, ψ̃) = T (φ, ψ) as follows:
Z
−1φ̃ = F1 (ψ, εφ, ε) in Ω ,
∂n φ̃ = 0 on ∂Ω ,
φ̃ dx = 0,
Ω

−1(ψ̃ − ψ0 ) − Dψ F2 (ψ0 , 0, 0)(ψ̃ − ψ0 )
= F2 (ψ, εφ, ε) − F2 (ψ0 , 0, 0) − Dψ F2 (ψ0 , 0, 0)(ψ − ψ0 ), in Ω ,

∂n ψ̃ = 0
on ∂Ω .
R
Since Ω F1 dx = 0, φ̃ is well-defined. In addition, since −1φ0 = F1 (ψ0 , 0, 0),
kφ̃ − φ0 kC 1 ≤ C1 kF1 (ψ, εφ, ε) − F1 (ψ0 , 0, 0)k∞
√
≤ C{kψ − ψ0 k∞ + εkφk∞ + ε} ≤ ε.
By assumption, (−1 − Dψ F2 (ψ0 , 0, 0)I)−1 is bounded from C 0 to C 1 . Thus,
kψ̃ − ψ0 kC 1 ≤ C1 kF2 (ψ, εφ, ε) − F2 (ψ0 , 0, 0) − Dψ F2 (ψ0 , 0, 0)(ψ − ψ0 )k∞
≤ C{kψ − ψ0 k2∞ + εkφk∞ + ε} ≤ ε 2/3 .
Clearly, T is compact. Since T maps B into itself, by the Schauder fixed point theorem, T has a fixed point, which
gives a solution to our problem. This completes the proof. References
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