Qualitative analysis of a producer–scrounger model

J. Math. Anal. Appl. 440 (2016) 33–47

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Qualitative analysis of a producer–scrounger model ✩ Jun Zhou School of Mathematics and Statistics, Southwest University, Chongqing, 400715, PR China

a r t i c l e

i n f o

Article history: Received 5 September 2015 Available online 15 March 2016 Submitted by Y. Yamada Keywords: Producer–scrounge model Positive steady state solutions Uniqueness Stability

a b s t r a c t In this paper, we consider a producer–scrounger model proposed in [6]. We prove that when the trivial steady state solution or the semi-trivial steady state solution to the system is stable, then the system admits no positive steady state solution, while if both the trial steady state solution and the semi-trivial steady state solution are unstable, then the system admits at least one positive steady state solution. Furthermore, in the one dimension case, we prove that the positive steady state solution, if exists, is unique. © 2016 Elsevier Inc. All rights reserved.

1. Introduction Let the habitat Ω be a bounded domain in RN (N = 1, 2, 3, · · · ) with smooth boundary ∂Ω. Let P(x, t)  t) denote the producer and scrounger density, respectively, at location x ∈ Ω and time t ≥ 0. We and S(x, suppose that producers and scroungers move randomly in their environment according to the equations (see [6]) ⎧    ∂P ⎪  −d− ⎪ − d1 ΔP = P φ(S)m aP , x ∈ Ω, t > 0, ⎨ ∂t (1.1)   ⎪ ⎪ ⎩ ∂ S − d ΔS = S θψ(S)m  P − e − bS , x ∈ Ω, t > 0. 2 ∂t N ∂ 2 Here, Δ denotes the Laplace operator (Δ = ∇2 = i=1 ∂x 2 ). The diffusion (or dispersal) rates d1 and d2 i are positive constants. The function m(x) represents the producer per capita rate of resource discovery at x ∈ Ω, and the producer and scrounger share functions  = φ(S)

 c  = 1 , and ψ(S) S +  c S +  c

(1.2)

✩ This work is partially supported by the Fundamental Research Funds for the Central Universities grant XDJK2015A16, NSFC grant 11201380, Project funded by China Postdoctoral Science Foundation grants 2014M550453, 2015T80948 and the Second Foundation for Young Teachers in Universities of Chongqing. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2016.03.030 0022-247X/© 2016 Elsevier Inc. All rights reserved.

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J. Zhou / J. Math. Anal. Appl. 440 (2016) 33–47

which denote the proportion of this resource that is ultimately acquired by each producer and scrounger. The monopolization factor  c is a positive constant representing a producer’s ability to defend resources that it uncovers from theft by scroungers. The units of the resource m(x) are scaled to match that of the producer birth rate, and the corresponding resource-to-scrounger unit conversion factor θ is a positive constant. The parameters  a, b, d and e are positive constant density-independent and density-dependent death rates. Problem (1.1)–(1.2) with homogeneous Neumann boundary was studied by Cosner and Nevai in [6], and they found that (i) both species can persist when the habitat has high productivity, (ii) neither species can persist when the habitat has low productivity, and (iii) slower dispersal of both the producer and scrounger is favored when the habitat has intermediate productivity. In this work we shall assume that the region outside the habitat is immediately lethal and, consequently, we shall subsequently impose that  t) = 0, x ∈ ∂Ω, t > 0. P(x, t) = S(x,

(1.3)

In other words, we are assuming that the habitat is entirely surrounded by an absorbing boundary. Terrestrial-aquatic edges are absorbing boundaries for seeds of plant species incapable of surviving in both habitats [10], as is the legislative Boundary of Yellowstone National Park for bison dispersing into Montana, where they are shot to control the spread of brucellosis [9]. We will consider (1.1)–(1.3) in a simple case, i.e., d1 = d2 = 1 and m is a positive constant. Let P =  aP,      a = b S = bS, cm, b = mθb/ c, c = b c, then we get

 ⎧ a ∂P ⎪ ⎪ − ΔP = P − d − P , x ∈ Ω, t > 0, ⎪ ∂t ⎪ S+c ⎪ ⎪ ⎪

 ⎪ ⎪ ⎨ ∂S bP − ΔS = S −e−S , x ∈ Ω, t > 0, ∂t S+c ⎪ ⎪ ⎪ ⎪ ⎪ P = S = 0, x ∈ ∂Ω, t > 0, ⎪ ⎪ ⎪ ⎪ ⎩ P (x, 0) = P0 (x), S(x, 0) = S0 (x), x ∈ Ω,

(1.4)

where P0 (x) and S0 (x) are nonnegative, nontrivial continuous functions. The steady state problem associated with (1.4) is 

⎧ a ⎪ ⎪ − d − P , x ∈ Ω, −ΔP = P ⎪ ⎪ S+c ⎪ ⎪ ⎨ 

bP − e − S , x ∈ Ω, −ΔS = S ⎪ ⎪ S+c ⎪ ⎪ ⎪ ⎪ ⎩ P = S = 0, x ∈ ∂Ω.

(1.5)

It is obvious that model (1.5) has, in addition to the trivial solution (0, 0), a semi-trivial solution (θr , 0) if r > λ1 , where r :=

a − d, c

(1.6)

where λ1 is the principal eigenvalue of −Δ with zero Dirichlet boundary condition, which is given in (2.2), θr (x) > 0 for x ∈ Ω, θr (x) = 0 for x ∈ ∂Ω. The definition and the properties of θr are shown in Theorem A.3 in the Appendix A. The main task of this paper is to study the stability of the trivial solution (0, 0), and the semi-trivial solution (θr , 0); the nonexistence, existence and uniqueness of positive solutions to (1.5). We say (P, S) is

J. Zhou / J. Math. Anal. Appl. 440 (2016) 33–47

35

a positive solution if P (x), S(x) > 0 for x ∈ Ω and satisfies (1.5). Let Di , i = 1, 2, 3, 4, be the four sets in (r, e)-plane, defined as in (2.12) (see Fig. 1). We found that 1. (0, 0) is globally asymptotic stable if (r, e) ∈ D1 , while it is unstable if (r, e) ∈ / D1 (see Theorem 2.1). 2. (θr , 0) is globally asymptotically stable if (r, e) ∈ D2 ; (θr , 0) is locally asymptotic stable if (r, e) ∈ D2 ∪ D3 ; (θr , 0) is unstable if (r, e) ∈ D4 (see Theorem 2.1). 3. Problem (1.5) admits no positive solution if (r, e) ∈ D1 ∪ D2 ∪ D3 (see Theorem 3.1). 4. Problem (1.5) admits at least one positive solution if (r, e) ∈ D4 (see Theorem 3.4). 5. Problem (1.5) with N = 1 admits exactly one positive solution if (r, e) ∈ D4 (see Theorem 3.8). Remark 1.1. From the above results we get (i) if (0, 0) or (θr , 0) is stable, then (1.5) admits no positive solution; (ii) if (0, 0) and (θr , 0) are both unstable, then (1.5) admits at least one positive solution. Furthermore, the positive solution is unique if N = 1. The organization of the remaining part of the paper is as follows. The stability of the trivial and semitrivial solutions to (1.5) is studied in Section 2. In Section 3, we consider the nonexistence, existence and unique of positive solutions to (1.5). Finally, in Appendix A, we list some basic results which are used in this paper. 2. Analysis of the trivial and semi-trivial steady state solutions In this section, we mainly study the stability/instability of the trivial solution and semi-trivial solution to problem (1.5) with respect to (1.4). For each q ∈ C(Ω), let λ1 (q) be the principle eigenvalue of

−ΔU + q(x)U = λU, x ∈ Ω, x ∈ ∂Ω.

U = 0,

(2.1)

As is well known, the principal eigenvalue λ1 (q) is given by the following variational characterization:

λ1 (q) =

(|∇φ|2 + q(x)φ2 )dx.

inf 1

φ∈H0 , φ2 =1

(2.2)

Ω

We denote λ1 (0) by λ1 and let φ1 (x) be the positive eigenfunction corresponding to λ1 with φ1 2 = 1. The main result is the following theorem. Theorem 2.1. Let r be the constant defined as (1.6). Assume (P (x, t), S(x, t)) is the solution to model (1.4). (i) If r ≤ λ1 , then (P (x, t), S(x, t)) converges to the trivial steady state solution (0, 0) uniformly on Ω as t → +∞, while (0, 0) is unstable if r > λ1 .   (ii) Assume r > λ1 , then the semi-trivial solution (θr , 0) is locally asymptotic stable if e + λ1 − cb θr > 0,   while it is unstable if e + λ1 − cb θr < 0. Moreover if e ≥ br c − λ1 , (θr , 0) is globally asymptotic stable, i.e., (P (x, t), S(x, t)) converges to (θr , 0) uniformly on Ω as t → +∞. Proof. (i) By the first equation of (1.4), we get Pt − ΔP ≤ P (r − P ). Then 0 ≤ P (x, t) ≤ U (r; P0 (x)) in Ω × [0, +∞) by comparison principle for parabolic equation, where U (r; P0 (x)) is the function given in (A.4). By Theorem A.4, it follows from r ≤ λ1 that U (r; P0 (x)) → 0 uniformly on Ω as t → +∞, which

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combines with 0 ≤ P (x, t) ≤ U (r; P0 (x)) shows that P (x, t) → 0 uniformly on Ω as t → +∞. Then there exists a constant t0 1 such that P (x, t) ≤ δ :=

c(2e + λ1 ) , x ∈ Ω, t ≥ t0 . 2b

(2.3)

It follows from the second equation of (1.4) and (2.3) that St − ΔS ≤ S (bδ/c − e − S) for x ∈ Ω and t ≥ t0 . Then 0 ≤ S(x, t) ≤ U (bδ/c − e; S(x, t0 )) in Ω × [t0 , +∞). Since bδ/c − e ≤ λ1 /2 < λ1 , by Theorem A.4, we get U (bδ/c − e; S(x, t0 )) → 0 uniformly on Ω as t → +∞. Then S(x, t) → 0 uniformly on Ω as t → +∞. To study the instability of (0, 0), we need to consider the following eigenvalue problem, which is got by linearizing the problem (1.5) at (0, 0) ⎧ Δφ + rφ = λφ, x ∈ Ω, ⎪ ⎪ ⎨ Δψ − eψ = λψ, x ∈ Ω, ⎪ ⎪ ⎩ P = S = 0, x ∈ ∂Ω.

(2.4)

Then r − λ1 is an eigenvalue, which is positive if r > λ1 . So, (0, 0) is unstable if r > λ1 . (ii) We first consider the global asymptotic stable of (θr , 0) under the condition e ≥ from (1.6) and r > λ1 that 0 :=

br c

− λ1 . It follows

c a − > 0. 2(d + λ1 ) 2

As in the proof of (i), we get 0 ≤ P (x, t) ≤ U (r; P0 (x)) in Ω × [0, +∞). Since r > λ1 , we get from Theorem A.4 that lim sup P (x, t) ≤ lim U (r; P0 (x)) = θr < r, x ∈ Ω. t→+∞

t→+∞

(2.5)

Then there exists t1 1 such that P (x, t) ≤ r in Ω × [t0 , +∞). In view of e ≥ br c − λ1 , we can get S(x, t) → 0 uniformly on Ω as t → +∞ by analysis similar to the proof of (i). Then for any  ∈ (0, 0 ), there exists t2  1 such that 0 < S(x, t) ≤  in Ω × [t2 , +∞). By the first equation of (1.4) again, we obtain  a a Pt − ΔP ≥ P +c − d − P in Ω × [t2 , +∞), which implies P (x, t) ≥ U +c − d; P (x, t1 ) . By  < 0 , we a get +c − d > λ1 , then it follows from Theorem A.4 that

lim inf P (x, t) ≥ lim U t→+∞

t→+∞



a − d; P (x, t1 ) +c

a = θ +c −d , x ∈ Ω.

(2.6)

a Since θ +c −d → θr uniformly on Ω as  → 0 (see Theorem A.3), letting  → 0 in (2.6), we get

lim inf P (x, t) ≥ θr . t→+∞

(2.7)

In view of (2.5) and (2.7), we get (P (x, t), S(x, t)) converges to (θr , 0) uniformly on Ω as t → +∞. Next we consider the local stability and instability of (θr , 0) by studying the following eigenvalue problem, which is got by linearizing the problem (1.5) at (θr , 0) ⎧ a ⎪ ⎪ Δφ + (r − 2θr ) φ − 2 θr ψ = λφ, x ∈ Ω, ⎪ c ⎪ ⎪

 ⎨ b Δψ + θ − e ψ = λψ, x ∈ Ω, r ⎪ c ⎪ ⎪ ⎪ ⎪ ⎩ φ = ψ = 0, x ∈ ∂Ω.

(2.8)

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Let λ be an eigenvalue of (2.8) and let (φ, ψ) be its corresponding eigenfunction. 1. Assume ψ = 0, then λ must be an eigenvalue of Δ + (r − 2θr ) with zero Dirichlet boundary condition. Then it follows from Theorem A.3 that λ ≤ −λ1 (2θr − r) < −λ1 (θr − r) = 0. 2. If ψ = 0, then λ is an eigenvalue of the second equation of (2.8). The largest eigenvalue λ∗ among such eigenvalues is given by



 b b λ∗ = −λ1 e − θr = −e − λ1 − θr . c c   Combining the above results one can prove that, if e + λ1 − cb θr > 0, then all eigenvalues of (2.8) are  b  negative, which implies (θr , 0) is stable; while if e + λ1 − c θr < 0, then (2.8) has at least one positive eigenvalue, which means (θr , 0) is unstable. 2 In order to understand the meaning of Theorem 2.1, we define a function:

 b − θr , r ≥ λ1 . c

e = f (r) = −λ1

(2.9)

Then as in the proof of [16, Lemmas 1.2–1.4], we get the following results. Lemma 2.2. The function f (r) defined as (2.9) is a strictly increasing function of class C 1 and satisfies (i) f (λ1 ) = −λ1 , lim f (r) = +∞. r→+∞

b (ii) f (λ1 ) = . c 

Since θr < r, we get from Theorem A.1 that

e = f (r) < −λ1

br − c

 =

br − λ1 . c

(2.10)

From Lemma 2.2 and (2.10), it is obvious that there exists a unique positive constant r∗ such that f (r∗ ) = 0, where r∗ = f −1 (0) >

c b λ1 ,

if b < c; λ1 , if b ≥ c.

Based on above analysis, we define some sets in (r, e)-plane as follows (see Fig. 1): D1 := {(r, e) : −∞ < r ≤ λ1 , 0 < e < +∞}, ⎧  cλ ⎪ ⎨ (r, e): λ1 < r ≤ b 1 , 0 < e < +∞  D2 := < +∞ , if b < c; ∪ (r, e) : cλb 1 < r < +∞, br c − λ1 ≤ e   ⎪ ⎩ (r, e) : λ < r < +∞, br − λ ≤ e < +∞ , if b ≥ c, 1 1 c

(2.11)

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38

Fig. 1. Illusions of the sets in Corollary 2.3. Left is the case b < c; right is the case b ≥ c.

⎧  ⎪ (r, e) : cλb 1 < r ≤ r∗ , 0 < e < br ⎪ c − λ1  ⎪ ⎨ ∪ (r, e) : r∗ < r < +∞, f (r) < e <  D3 :=  ⎪ − λ1 (r, e) : λ1 < r ≤ r∗ , 0 < e < br ⎪ c  ⎪ ⎩ ∪ (r, e) : r∗ < r < +∞, f (r) < e <

br c

 − λ1 , if b < c;

br c

 − λ1 , if b ≥ c,

D4 := {(r, e) : r∗ < r < +∞, 0 < e < f (r)}.

(2.12)

Then the following results follows from Theorem 2.1. Corollary 2.3. Let Di , i = 1, 2, 3, 4, be the sets defied as in (2.12). Then the following results hold. (i) (0, 0) is globally asymptotic stable if (r, e) ∈ D1 , while it is unstable if (r, e) ∈ / D1 . (ii) (θr , 0) is globally asymptotically stable if (r, e) ∈ D2 ; (θr , 0) is locally asymptotic stable if (r, e) ∈ D2 ∪ D3 ; (θr , 0) is unstable if (r, e) ∈ D4 . 3. Positive solutions In this section, we will consider the nonexistence, existence, and uniqueness of positive solutions to the problem (1.5). 3.1. Nonexistence of positive solutions In this subsection, we mainly consider the nonexistence of positive solutions to problem (1.5), and the main result is the following theorem. Theorem 3.1. Let Di , i = 1, 2, 3, 4, be the sets defied as in (2.12). Then problem (1.5) has no positive solution if (r, e) ∈ D1 ∪ D2 ∪ D3 . Proof. Let (P, S) be a positive solution to (1.5). Taking L2 -inner product to the first equation of (1.5), it follows from (2.2) that



λ1 P 22 ≤ ∇P 22 =

P2

a −d−P S+c

 dx < rP 22 .

Ω

Hence a necessary condition for (1.5) admits a positive solution is r > λ1 , and so (1.5) admits no positive solution if r ≤ λ1 , i.e., (r, e) ∈ D1 .

J. Zhou / J. Math. Anal. Appl. 440 (2016) 33–47

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Next we assume (r, e) ∈ D2 ∪ D3 , i.e., r > λ1 , e > f (r) and (1.5) admits a positive solution (P, S). It is easy to see P ≤ θr . Then we get from the second equation of (1.5) that 

bθr S ≤ −S 2 ≤ 0. −ΔS + e − c   Since e > f (r), i.e., λ1 e − bθcr > 0, then it follows from comparison principle that S(x) ≤ 0 for x ∈ Ω, which contradicts to the assumption that S(x) > 0 for x ∈ Ω. So (1.5) admits no positive solution if (r, e) ∈ D2 ∪ D3 . 2 3.2. Existence of positive solutions In this subsection, we consider the existence of positive solutions to problem (1.5) by degree theory prescribed in Appendix A. In view of Theorem 3.1, we only need to consider the case (r, e) ∈ D4 , where D4 is defined as in (2.12). Then it is easy to see any positive solution (P, S) of (1.5) satisfies P (x) < r :=

br a − d, S(x) < δ := − e, x ∈ Ω. c c

(3.1)

Throughout this subsection, we will use the notations in Theorem A.5 and E := C0 (Ω) × C0 (Ω), where C0 (Ω) = {P (x) ∈ C(Ω) : P (x) = 0, x ∈ ∂Ω}. It is obvious that E is a Banach space with the norm (P, S)E = maxx∈Ω |P (x)| + maxx∈Ω |S(x)|. W = K × K, where K := {P ∈ C0 (Ω) : P (x) ≥ 0 for x ∈ Ω}. D := {(P, S) ∈ W : P (x) ≤ r, S(x) ≤ δ for x ∈ Ω}. ˚ In order to apply the degree theory, By (3.1), any positive solutions to (1.5) lie in the interior of D (=D). we choose a sufficient large number α such that α+

bP a − d − p > 0 and α + − e − S > 0 for (P, S) ∈ D. S+c S+c

Define a mapping A : E → E by  P α+ A(P, S) = (−Δ + αI)−1 ⎝  S α+ ⎛

a S+c bP S+c

⎞   −1 −d−P (αP + F (P, S)) (−Δ + αI) ⎠= , (−Δ + αI)−1 (αS + G(P, S)) −e−S

(3.2)

where

F (P, S) := P

a −d−P S+c



, G(P, S) := S

 bP −e−S . S+c

(3.3)

By (3.2) and the maximum principle for elliptic equations, A maps D into W, and the strong maximum principle implies that A maps D/(0, 0) into the demi-interior of W (see [7,8]). Moreover, the regularity theory for elliptic equations tells us that A is completely continuous in E. Therefore, one can define the ˚ for A with respect to W because A has no fixed point on the boundary of D with degree, degW (I − A, D), respect to W by (3.1). ˚ = 1. Lemma 3.2. degW (I − A, D)

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Proof. For t ∈ [0, 1], we define a completely continuous mapping on E by ⎛

 P α+ At (P, S) = (−Δ + αI)−1 ⎝  S α+

at S+c

−d−P

btP S+c

−e−S

⎞  ⎠.

As in the studies of A, one can show that At maps D into W and At has no fixed point on the boundary ˚ is independent of D with respect to W. Hence it follows from the homotopy invariance that degW (I − At , D) of t ∈ [0, 1] (see [1, Theorem 11.1]); so it follows from (0, 0) is the unique fixed point of A0 in D that ˚ = degW (I − A1 , D) ˚ = degW (I − A0 , D) ˚ = indexW (A0 , (0, 0)). degW (I − A, D)

(3.4)

Let A0 (0, 0) be the Fréchet derivative of A0 , which is given by A0 (0, 0)(P, S) = (−Δ + αI)−1 ((α − d)P, (α − e)S)T . Observe that r(A0 (0, 0))

 = max

α−d α−e , λ1 + α λ1 + α

 < 1.

Then indexW (A0 , (0, 0)) = 1 is obtained by Theorem A.5. Thus the conclusion follows from (3.4).

2

As said at the beginning of this subsection, we assume (r, e) ∈ D4 to study the existence of positive solutions. Then (1.5) admits a trivial solution (0, 0) and a semi-trivial solution (θr , 0) in this case. Next we will compute the indexes of (0, 0) and (θr , 0). Lemma 3.3. indexW (A, (0, 0)) = indexW (A, (θr , 0)) = 0 if (r, e) ∈ D4 , where D4 is the set defined as in (2.12). Proof. A direct calculation shows that A (0, 0)(P, S) = (−Δ + αI)−1 ((α + r)P, (α − e)S) . Clearly,  (0, 0) is identical with A (0, 0). Observe that W(0,0) = W and S(0,0) = {(0, 0)}; so that A T

r(A (0, 0)) = max



α−e α+r , λ1 + α λ1 + α

 =

α+r . λ1 + α

Since (r, e) ∈ D4 , we get r > λ1 . Then it is easy to see that A (0, 0)y = y on W/{(0, 0)} and that  (0, 0)) > 1. Hence Theorem A.5 yields index (A, (0, 0)) = 0. r(A (0, 0)) = r(A W Next we consider the index of (θr , 0). A direct calculation shows that  

−1

A (θr , 0)(P, S) = (−Δ + αI)

(α + r − 2θr )P − ca2 θr S   α − e + cb θr S



 (θ , 0) with T := (−Δ + Since W(θr ,0) = C0 (Ω) × K and S(θr ,0) = C0 (Ω) × {0}, we can identify A r   b −1 α − e + c θr . αI) Next we show A (θr , 0) admits no fixed point on W(θr ,0) /{(0, 0)}. We assume there exists (ξ, η) ∈ W(θr ,0) such that A (θr , 0)(ξ, η) = (ξ, η). Then (ξ, η) satisfies

⎧ a ⎪ ⎪ −Δξ = (r − 2θr )ξ − 2 θr η, x ∈ Ω, ⎪ c ⎪ ⎪ 

⎨ b η, x ∈ Ω, −Δη = −e + θ r ⎪ c ⎪ ⎪ ⎪ ⎪ ⎩ ξ = η = 0, x ∈ ∂Ω.

(3.5)

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  Since (e, r) ∈ D4 , we get e < f (r), i.e., λ1 e − cb θr < 0, this combines with the second equation of (3.5) and η ∈ K implies η = 0. Since λ1 (2θr − r) > λ1 (θr − r) > 0, we get from the first equation of (3.5) and η = 0 that ξ = 0. So A (θr , 0) admits no fixed point on W(θr ,0) /{(0, 0)}.    (θ , 0)) > 1. Then it follows from Since λ1 e − cb θr < 0, we get from Theorem A.2 that r(T ) = r(A r Theorem A.5 that indexW (A, (θr , 0)) = 0. 2 The main result of this subsection is the following theorem. We should note that the existence of positive solutions and the persistence of both populations provided the semi-trivial solution (θr , 0) exists and is unstable; it follows from results of [4]. Theorem 3.4. Problem (1.5) admits at least one positive solution if (r, e) ∈ D4 , where D4 is given as in (2.12). Proof. We will prove this theorem by the degree theory. Assume on the contrary that (1.5) admits no positive solution. It follows from the definition of degree that ˚ = indexW (A, (0, 0)) + indexW (A, (θr , 0)). degW (I − A, D)

(3.6)

By Lemma 3.2, the left-hand side of (3.6) is equal to 1. On the other hand, Lemma 3.3 implies the right-hand side of (3.6) is equal to zero. This contradicts to (3.6); so that (1.5) must possess at least one positive solution. 2 3.3. Uniqueness of positive solutions In this subsection, we consider the uniqueness of positive solutions to (1.5) for N = 1, i.e., 

⎧ a ⎪  ⎪ −d−P , 0 < x < , −P = P ⎪ ⎪ S+c ⎪ ⎪ ⎨ 

bP  − e − S , 0 < x < , −S = S ⎪ ⎪ S+c ⎪ ⎪ ⎪ ⎪ ⎩ P (0) = P ( ) = S(0) = E( ) = 0,

(3.7)

where is a positive constant. By Theorems 3.1 and 3.4, we need to assume (r, e) ∈ D4 to ensure (3.7) admits at least one positive solution, where D4 is defined as in (2.12). To study the uniqueness result, we first consider the following problem: ⎧ 0 < x < , −P  = P (r − S − P ) , ⎪ ⎪ ⎪ ⎪ 

⎨ b P −e−S , 0 < x < , −S  = S ⎪ c ⎪ ⎪ ⎪ ⎩ P (0) = P ( ) = S(0) = E( ) = 0,

(3.8)

where r = a/c − d and  is a positive constant.   Lemma 3.5. Assume (r, e) ∈ D4 with D4 given in (2.12), i.e., r > λ1 and 0 < e < f (r) = −λ1 − cb θr . Then for  > 0 small enough, problem (3.8) admits a unique positive solution (P∗ , S∗ ). Proof. It is easy to see any positive solution (P, S) to (3.8) satisfies

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P (x) < r and S(x) < δ :=

br − e > 0, x ∈ Ω. c

  Since r > λ1 and e < f (r) = −λ1 − cb θr , by Theorem A.3, we can choose  > 0 small enough such that

 b γ := r − δ > λ1 and e < −λ1 − θγ . c

(3.9)

For S ∈ C 1 (Ω) satisfying 0 ≤ S(x) ≤ δ, x ∈ Ω, we consider the following problem −P  + SP = P (r − P ), 0 < x < , P (0) = P ( ) = 0.

(3.10)

By (3.9) and Theorem A.1, we know that λ1 (S) ≤ λ1 (δ) = λ1 + δ < r. Then it follows from Theorem A.3 that problem (3.10) admits a unique positive solution, which we denote by P (S). By [2] or [13, page 7, Lemma 2.2], the following conclusions hold. (i) The mapping S → P (S) considered as a function from C 1 (Ω) to C 1 (Ω) is continuous; (ii) if S1 ≥ S2 in Ω, then P (S1 ) ≤ P (S2 ) in Ω. Since 0 ≤ S ≤ δ, we get from (3.9) and (ii) of above conclusions that θγ (x) = P (δ)(x) ≤ P (S)(x) ≤ P (0)(x) = θr (x) < r, x ∈ Ω.

(3.11)

Next we consider the following problem 



−S = S

 b P (S) − e − S , 0 < x < , S(0) = S( ) = 0, c

(3.12)

which is equivalent to 

b −S  + 2e − P (S) S = S(e − S), 0 < x < , S(0) = S( ) = 0. c

(3.13)

By (3.9), (3.11) and Theorem A.1, we know that

λ1



b 2e − P (S) c

≤ 2e + λ1

b − θγ c

 < e.

Then it follows from Theorem A.3 that (3.13) or (3.12) has a unique positive solution S∗ . It is obvious that 0 ≤ S∗ ≤ δ. Let P∗ = P (S∗ ), then P∗ is the unique positive solution to (3.10) with S = S∗ . Finally, we get that problem (3.8) admits a unique positive solution (P∗ , S∗ ). 2 Next we consider a perturbation problem of (3.7): 

⎧ a ⎪  ⎪ − (1 − τ )S − d − P , 0 < x < , −P = P ⎪ ⎪ τS + c ⎪ ⎪ ⎨ 

bP  −S − e − S , 0 < x < , = S ⎪ ⎪ τS + c ⎪ ⎪ ⎪ ⎪ ⎩ P (0) = P ( ) = S(0) = E( ) = 0,

(Proτ )

where  > 0 is a positive constant and τ ∈ [0, 1] is a parameter. It is easy to see (Pro0 ) is the problem (3.8) and (Pro1 ) is the problem (3.7).

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Lemma 3.6. Let (P0 , S0 ) be arbitrary positive solution to (Proτ ), then the linearized system of (Proτ ) at (P0 , S0 ) has only the trivial solution (0, 0). Hence, any positive solution to (Proτ ) is not degenerate. Proof. The linearized system of (Proτ ) at (P0 , S0 ) is ⎧ 0 < x < , −φ + L1 φ = Aψ, ⎪ ⎪ ⎨ −ψ  + L2 ψ = Bφ, 0 < x < , ⎪ ⎪ ⎩ φ(0) = φ( ) = ψ(0) = ψ( ) = 0, where L1 = d + 2P0 + (1 − τ )S0 − A=−

a bcP0 , L2 = e + 2S0 − , τ S0 + c (τ S0 + c)2

aτ P0 bS0 > 0. − (1 − τ )P0 < 0, B = (τ S0 + c)2 τ S0 + c

Since (P0 , S0 ) is a positive solution to (Proτ ), it follows from the well-known Krein–Rutman Theorem that

λ1

a d + P0 + (1 − τ )S0 − τ S0 + c



 = λ1

bP0 e + S0 − τ S0 + c

 = 0.

(3.14)

Since a a > d + P0 + (1 − τ )S0 − , τ S0 + c τ S0 + c bP0 c bP0 L2 = e + 2S0 − > e + S0 − , τ S0 + c τ S0 + c τ S0 + c

L1 = d + 2P0 + (1 − τ )S0 −

then it follows from (3.14) and Theorem A.1 that λ1 (Li ) > 0, i = 1, 2. Let     X := W 2,p (Ω) ∩ W01,p (Ω) × W 2,p (Ω) ∩ W01,p (Ω) , Y := Lp (Ω) × Lp (Ω),   ∂w (x) < 0 for x ∈ ∂Ω , P = w ∈ C01 (Ω), w(x) > 0 for x ∈ Ω and ∂ν where ν is the unit outward normal on ∂Ω. We define a linear operator  = (1 , 2 ) : X → Y, where i = − + Li , i = 1, 2. Let −1 be the inverse operator of i . It is obvious that −1 is compact and i i ˚ In this settings, we can strictly order-preserving operator with respect to P. Moreover −1 (P/{0}) ⊂ P. i show that φ = ψ = 0 using a similar proof as in [5,15], which completes the proof. 2 A perturbation argument can be used to show that if (Proτ ) has exactly one positive solution, which is assumed to be non-degenerate, then (Proτ +σ ) also has exactly one positive solution provided |σ| is small enough. For that purpose, we state the following lemma. Since the proof is basically same as [5, Lemma 5.4], we omit its proof. Lemma 3.7. Assume (Proτ ) has exactly one positive solution (P0 , S0 ), which is not degenerate. Then there exists σ0 = σ0 (τ ) > 0 such that for every σ ∈ (−σ0 , σ0 ), problem (Proτ +σ ) has exactly one positive solution (P (σ), S(σ)). Moreover (P (0), S(0)) = (P0 , S0 ) and the mapping σ → (u(σ), v(σ)), from (−σ0 , σ0 ) to P × P, is of class C 1 .

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Assume (r, e) ∈ D4 , and choose  > 0 small enough such that (3.9) is satisfied. Then it follows from Lemma 3.5, problem (Pro0 ), i.e., problem (3.8) admits exactly one positive solution (P∗ , S∗ ). Since (Pro1 ) is the problem (3.7), by Lemmas 3.6 and 3.7, we can state the following uniqueness result. Again the proof is similar to [5, Theorem 5.1], thus we omit its proof. Theorem 3.8. Assume (r, e) ∈ D4 with D4 given in (2.12). Then problem (3.7) admits a unique positive solution. Appendix A In this section we list the well-known facts which are used in this paper. The principal eigenvalue λ1 (q) defined in (2.2) has some useful properties as follows (see [12, Proposition A.1] or [16, Proposition 1.1]). Theorem A.1. The following conclusions hold. (i) If qi ∈ C(Ω) (i = 1, 2) satisfy q1 ≥ q2 and q1 ≡ q2 , then λ1 (q1 ) > λ1 (q2 ). (ii) For qn ∈ C(Ω) and q ∈ C(Ω), let φn ∈ H01 (Ω) and φ ∈ H01 (Ω) be the corresponding eigenfunctions of (2.1) satisfying φn 2 = φ2 = 1, where n ∈ N . If limn→∞ qn − q∞ = 0, then limn→∞ λ1 (qn ) = λ1 (q) and limn→∞ φn = φ strongly in H01 (Ω). (iii) Let (k, ) be an open interval and assume that a mapping α → qα is continuously differentiable from (k, ) to C(Ω) with respect to supremum norm. If φα ∈ H01 (Ω) with φα 2 = 1 is the unique positive eigenfunction corresponding to λ1 (qα ), then α → λ1 (qα ) is continuously differentiable from (k, ) to R and d λ1 (qα ) = dα

∂qα 2 φ dx. ∂α α

Ω

Let α be a sufficiently large constant such that α − q(x) > 0 for any x ∈ Ω. Define a bounded linear operator T : C0 (Ω) → C0 (Ω) by U = T V = (−Δ + αI)−1 (α − q(x))V , where U ∈ C0 (Ω) := {U ∈ C(Ω) : U |∂Ω = 0} is the unique solution to the following problem

−ΔU + αU = (α − q(x))V, x ∈ Ω, U = 0,

x ∈ ∂Ω.

(A.1)

Let r(T ) be the spectral radius of T . Then the relationship between λ1 (q) and r(T ) can be given as follows (see [8, Proposition 1] or [14, Lemmas 2.1 and 2.3]). Theorem A.2. Let q ∈ C(Ω) and let α be a sufficiently large number such that α > q(x) for any x ∈ Ω. Then we have (i) λ1 (q) > 0 if and only if r((−Δ + αI)−1 (α − q(x))) < 1; (ii) λ1 (q) < 0 if and only if r((−Δ + αI)−1 (α − q(x))) > 1; (iii) λ1 (q) = 0 if and only if r((−Δ + αI)−1 (α − q(x))) = 1. We consider the following steady-state problem for logistic equation with linear diffusion

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45

⎧ −ΔU = U (ρ − u), x ∈ Ω, ⎪ ⎪ ⎨ U ≥ 0, x ∈ Ω, ⎪ ⎪ ⎩ U = 0, x ∈ ∂Ω,

(A.2)

where ρ is a positive constant and Ω ⊂ RN is a bounded open set with smooth boundary ∂Ω. Then the following results are well known (see [8, Lemma 1] and [11, Propositions 6.1–6.4]). Theorem A.3. Let λ1 := λ1 (0) be the constant defined in (2.2). Then the following conclusions hold. (i) If ρ ≤ λ1 , then (A.2) has no nontrivial solutions. θ (ii) If ρ > λ1 , then there exists a unique positive solution θρ of (A.2) such that θρ (x) and ρρ (x) are strictly increasing with respect to ρ and 0 < θρ (x) < ρ for every x ∈ Ω. Furthermore, limρ→+∞ θρ (x) = +∞ θ and limρ→+∞ ρρ (x) = 1 uniformly in K, where K is any compact subset of Ω. (iii) limρ→λ+ θρ = 0 uniformly in Ω. More precisely, 1 ⎛

θρ = ⎝

⎞−1 φ31 dx⎠

(ρ − λ1 )φ1 + o(ρ − λ1 )

as ρ → λ+ 1.

Ω

(iv) ρ → θρ is a C 1 -mapping from (λ1 , ∞) to C0 (Ω) := {U ∈ C(Ω) : U |∂Ω = 0} and precisely,

∂θρ ∂ρ

> 0 in Ω. More

∂θρ = (−Δ + (2θρ − ρ)I)−1 θρ , ∂ρ where (−Δ +(2θρ −ρ)I)−1 denotes the inverse operator of −Δ +(2θρ −ρ)I with zero Dirichlet boundary condition. We consider the following semi-linear parabolic equation ⎧ x ∈ Ω, t > t0 ≥ 0 Ut − ΔU = U (ρ − U ), ⎪ ⎪ ⎨ U = 0, x ∈ ∂Ω, t > t0 ≥ 0, ⎪ ⎪ ⎩ U (x, t0 ) = Ut0 (x) ≥ 0, ≡ 0, x ∈ ∂Ω,

(A.3)

where Ω ⊂ RN (N = 1, 2, · · · ) is a bounded domain with smooth boundary ∂Ω, ρ is a positive constant. It is well known (A.3) has a unique solution U (x, t), which is denoted by U (ρ; Ut0 (x)) (x, t).

(A.4)

Furthermore, we have (see [3]) Theorem A.4. Let U be the unique solution to (A.3) given in (A.4), then U > 0 for x ∈ Ω and t > t0 . Moreover, (1) (2)

lim U (x, t) = 0 uniformly on Ω if ρ ≤ λ1 ;

t→+∞

lim U (x, t) = θρ (x) uniformly on Ω if ρ > λ1 .

t→+∞

46

J. Zhou / J. Math. Anal. Appl. 440 (2016) 33–47

We summarize the index theory on a positive cone, which has been developed by Amann [1] and Dancer [7,8] to study positive solutions for nonlinear elliptic equations. Let E be a real Banach space and let W be a closed convex set in E. We use the notation following the paper of Dancer [7]. Let y be any element of W and define Wy by Wy := {x ∈ E : y + γx ∈ W for some γ > 0}, which is also a convex set. Define Sy = Wy ∩ (−Wy ) for y ∈ W, which is also a closed subspace of E. Assume that T : E → E is a compact linear operator such that T (Wy ) ⊂ Wy . It is easy to say Sy is invariant under T . This fact implies that T induces a compact linear mapping T y an image of Wy under the quotient mapping E → E/Sy . Since from E/Sy into itself. We denote by W    T (Wy ) ⊂ Wy , it follows that T Wy ⊂ Wy . Let A : W → W be a compact and Fréchet differentiable mapping. Denote by A(x) the Fréchet derivative of A at x ∈ W. Let y ∈ W be any fixed point of A, and assume that A (y) is compact. By [7, §2, lemma 1], A (y) maps Wy into itself. Then one can define Leray–Schauder degree degW (I − A, U, 0) for any open subset U in W if A has no fixed points on ∂U . For each isolated fixed y ∈ W, indexW (A, y) means degW (I − A, N (y), 0), where N (y) is a suitable neighborhood of y in W. Moreover, it is known that, if degW (I − A, 0) = 0, then A has at least one fixed point in U . We know give an index formula which is essentially due to Dancer [7]. Theorem A.5. Let y ∈ W be a fixed point of A. If (I − A (y))x = 0 for every x ∈ Wy /{0}, then  (y)) > 1;  (i) indexW (A, y) = 0 if r(A (ii) indexW (A, y) = 1 if r(A (y)) < 1.

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[13] A.W. Leung, Nonlinear Systems of Partial Differential Equations: Applications to Life and Physical Sciences, World Scientific Pub Co Inc, 2009. [14] L. Li, Coexistence theorems of steady states for predator–prey interacting systems, Trans. Amer. Math. Soc. 305 (1) (1988) 143–166. [15] J. López-Gómez, R. Pardo, Existence and uniqueness of coexistence states for the predator–prey model with diffusion: the scalar case, Differential Integral Equations 6 (5) (1993) 1025–1031. [16] Y. Yamada, Positive solutions for Lotka–Volterra systems with cross-diffusion, in: Handbook of Differential Equations: Stationary Partial Differential Equations, vol. 6, 2008, pp. 411–501.