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Theory and Modern Applications

Four periodic solutions for a food-limited two-species Gilpin-Ayala type predator-prey system with harvesting terms on time scales

Abstract

By using Mawhin’s coincidence degree theory, this paper establishes a new criterion on the existence of four periodic solutions for a food-limited two-species Gilpin-Ayala type predator-prey system with harvesting terms on time scales. An example is given to illustrate the effectiveness of the result.

1 Introduction

The theory of calculus on time scales was initiated by Hilger [1] in order to unify continuous and discrete analysis, and it has become an effective approach to the study of mathematical models involving the hybrid discrete-continuous processes. Since the population dynamics in the real world usually involves the hybrid discrete-continuous processes, it may be more realistic to consider population models on time scales [2].

In recent years, some researchers studied the existence of periodic solutions for some population models on time scales under the assumption of periodicity of the parameters by using Mawhin’s coincidence degree theory (see [37]). To our best knowledge, few papers deal with the existence of multiple periodic solutions for population models with harvesting terms on time scales. The main difficulty is that the techniques used in continuous population models with harvesting terms are generally not available to population models with harvesting terms on time scales. Indeed, almost all papers involving continuous population models with harvesting terms used Fermat’s theorem on local extrema of differentiable functions in real analysis; for example, see [811]. However, Fermat’s theorem is not true in time scales calculus.

In this paper, we consider a food-limited two-species Gilpin-Ayala type predator-prey system with harvesting terms on time scales:

{ u 1 Δ ( t ) = 1 k 1 ( t ) + c 1 ( t ) e u 1 ( t ) [ a 1 ( t ) a 11 ( t ) e θ 1 u 1 ( t ) a 12 ( t ) e θ 2 u 2 ( t ) ] H 1 ( t ) e u 1 ( t ) , u 2 Δ ( t ) = 1 k 2 ( t ) + c 2 ( t ) e u 2 ( t ) [ a 2 ( t ) + a 21 ( t ) e θ 1 u 1 ( t ) a 22 ( t ) e θ 2 u 2 ( t ) ] H 2 ( t ) e u 2 ( t ) .
(1.1)

In system (1.1), let x i (t)=exp{ u i (t)}, i=1,2. If the time scale T=R (the set of all real numbers), then system (1.1) reduces to

{ x 1 ( t ) = x 1 ( t ) k 1 ( t ) + c 1 ( t ) x 1 ( t ) [ a 1 ( t ) a 11 ( t ) x 1 θ 1 ( t ) a 12 ( t ) x 2 θ 2 ( t ) ] H 1 ( t ) , x 2 ( t ) = x 2 ( t ) k 2 ( t ) + c 2 ( t ) x 2 ( t ) [ a 2 ( t ) + a 21 ( t ) x 1 θ 1 ( t ) a 22 ( t ) x 2 θ 2 ( t ) ] H 2 ( t ) ,
(1.2)

where x 1 and x 2 denote the prey and the predator, respectively; a i (t) (i=1,2), a i i (t) (i=1,2), H i (t) (i=1,2) are all positive continuous functions denoting the intrinsic growth rate, the intra-specific competition rates and the harvesting rates, respectively; a 12 (t) is the predation rate of the predator and a 21 (t) represents the conversion rate; k i (t) (i=1,2) are the population numbers of two species at saturation, respectively. θ i (i=1,2) represent a nonlinear measure of interspecific interference. When c i (t)0 (i=1,2), a i ( t ) k i ( t ) c i ( t ) (i=1,2) are the rate of replacement of mass in the population at saturation (including the replacement of metabolic loss and of dead organisms). In this case, system (1.2) is a food-limited population model. For other food-limited population models, we refer to [1217] and the references cited therein. When θ i 1 (i=1,2), system (1.2) is a Gilpin-Ayala type population model. Gilpin-Ayala type population models were firstly proposed by Gilpin and Ayala in [18]. For some recent work, we refer to [17, 1922]. When c i (t)0, θ i =1 (i=1,2), system (1.2) was consider by Zhao and Ye [9].

Motivated by the work of Bohner et al. [3] and Chen [8], we study the existence of multiple periodic solutions of (1.1) by using Mawhin’s coincidence degree.

2 Preliminaries from calculus on time scales

In this section, we briefly present some foundational definitions and results from the calculus on time scales so that the paper is self-contained. For more details, one can see [1, 23, 24].

Definition 2.1 [23]

A time scale is an arbitrary nonempty closed subset of the real numbers .

Let ω>0. Throughout this paper, the time scale is assumed to be ω-periodic, i.e., tT implies t+ωT. In particular, the time scale under consideration is unbounded above and below.

Definition 2.2 [23]

We define the forward jump operator σ:TT, the backward jump operator ρ:TT, and the graininess μ:T R + =[0,+) by

σ(t):=inf{sT:s>t},ρ(t):=sup{sT:s<t},μ(t)=σ(t)tfor tT,

respectively. If σ(t)=t, then t is called right-dense (otherwise, right-scattered), and if ρ(t)=t, then t is called left-dense (otherwise, left-scattered).

Definition 2.3 [23]

Assume that f:TR is a function, and let tT. Then we define f Δ (t) to be the number (provided it exists) with the property that given any ε>0, there is a neighborhood U of t (i.e., U=(tδ,t+δ)T for some δ>0) such that

| [ f ( σ ( t ) ) f ( s ) ] f Δ ( t ) [ σ ( t ) s ] | ε | σ ( t ) s | for all sU.

In this case, f Δ (t) is called the delta (or Hilger) derivative of f at t. Moreover, f is said to be delta or Hilger differentiable on if f Δ (t) exists for all tT.

Definition 2.4 [23]

A function F:TR is called an antiderivative of f:TR provided F Δ (t)=f(t) for all tT. Then we define

r s f(t)Δt=F(s)f(r)for r,sT.

Definition 2.5 [23]

A function f:TR is said to be rd-continuous if it is continuous at right-dense points in and its left-sided limits exist (finite) at left-dense points in . The set of rd-continuous functions f:TR will be denoted by C rd (T).

The following notation will be used throughout this paper.

Let

T κ = { T ( ρ ( sup T ) , sup T ] , if  sup T < + , T , if  sup T = + , κ = min { [ 0 , + ) T } , I ω = [ κ , κ + ω ] T , g u = sup t I ω g ( t ) , g l = inf t I ω g ( t ) , g ¯ = 1 ω I ω g ( s ) Δ s = 1 ω κ κ + ω g ( s ) Δ s ,

where g C rd (T) is a nonnegative ω-periodic real function, i.e., g(t+ω)=g(t) for all tT.

Lemma 2.1 [23]

Every rd-continuous function has an antiderivative.

Lemma 2.2 [23]

Assume that f:TR is a function, and let t T κ . Then we have:

  1. (i)

    If f is continuous at t and t is right-scattered, then f is differential at t with

    f Δ (t)= f ( σ ( t ) ) f ( t ) μ ( t ) .
  2. (ii)

    If f is right-dense, then f is differential at t iff the limit

    lim s t f ( t ) f ( s ) t s

exists as a finite number. In this case,

f Δ (t)= lim s t f ( t ) f ( s ) t s .

Lemma 2.3 [3]

Let t 1 , t 2 I ω and tT. If g:TR is ω-periodic, then

g(t)g( t 1 )+ κ κ + ω | g ( s ) | Δsandg(t)g( t 2 ) κ κ + ω | g ( s ) | Δs.

Lemma 2.4 [6]

Assume that { f n } n N is a function sequence on I ω such that

  1. (i)

    { f n } n N is uniformly bounded on I ω ;

  2. (ii)

    { f n Δ } n N is uniformly bounded on I ω .

Then there is a subsequence of { f n } n N converging uniformly on I ω .

3 Existence of multiple periodic solutions

We first briefly state Mawhin’s coincidence degree theory (see [25]).

Let X, Z be normed vector spaces, L:domLXZ be a linear mapping, N:XZ be a continuous mapping. The mapping L will be called a Fredholm mapping of index zero if dimKerL=codimImL<+ and ImL is closed in Z. If L is a Fredholm mapping of index zero, then there exist continuous projectors (i.e., linear and idempotent linear operators) P:XX and Q:ZZ such that ImP=KerL, ImL=KerQ=Im(IQ). If we define L P :domLKerPImL as the restriction L | dom L Ker P of L to domLKerP, then L P is invertible. We denote the inverse of that map by K P . If Ω is an open bounded subset of X, the mapping N will be called L-compact on Ω ¯ if QN( Ω ¯ ) is bounded and K P (IQ)N: Ω ¯ X is compact, i.e., continuous and such that K P (IQ)N( Ω ¯ ) is relatively compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQKerL.

For convenience, we introduce Mawhin’s continuation theorem [25] as follows.

Lemma 3.1 Let L be a Fredholm mapping of index zero, and let N: Ω ¯ ×[0,1]Z be L-compact on Ω ¯ ×[0,1]. Suppose

  1. (a)

    LuλN(u,λ) for every udomLΩ and every λ(0,1);

  2. (b)

    QN(u,0)0 for every uΩKerL;

  3. (c)

    Brouwer degree deg B (JQN(,0) | Ker L ,ΩKerL,0)0.

Then Lu=N(u,1) has at least one solution in domL Ω ¯ .

Set

h(x)=ba x α c x ,x(0,+).

Lemma 3.2 [17]

Assume that a, b, c, α are positive constants and

b>(1+α) a 1 1 + α ( c α ) α 1 + α .

Then there exist 0< x < x + such that

h ( x ) = h ( x + ) = 0 , h ( x ) > 0 for  x ( x , x + ) , h ( x ) < 0 for  x ( 0 , x ) ( x + , + )

and

h ( x ) >0, h ( x + ) <0.

Set

N 1 = ln ( a 1 a 11 ) u θ 1 + 2 ω ( a 1 k 1 ) u , N 2 = ln [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] θ 2 + 2 ω ( a 2 k 2 ) u + 2 ω ( a 21 k 2 ) u e θ 1 N 1 .

From now on, we always assume that:

(H1) c i (t), k i (t), a i (t), H i (t) (i=1,2), a i j (t) (i,j=1,2) are positive continuous ω-periodic functions, θ i (i=1,2) are positive constants.

(H2) 1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ]>(1+ θ 1 ) [ ( a 11 k 1 ) u ] 1 1 + θ 1 [ H 1 u θ 1 ] θ 1 1 + θ 1 .

(H3) ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 >(1+ θ 2 ) [ ( a 22 k 2 ) u ] 1 1 + θ 2 [ H 2 u θ 2 ] θ 2 1 + θ 2 .

Set

M 1 ( x ) = 1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] ( a 11 k 1 ) u x θ 1 H 1 u x , M 2 ( x ) = ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 ( a 22 k 2 ) u x θ 2 H 2 u x , p i ( x ) = ( a i k i ) ¯ ( a i i k i ) ¯ x θ i H ¯ i x ( i = 1 , 2 ) , m 1 ( x ) = ( a 1 k 1 ) u ( a 11 k 1 ) l 1 + ( c 1 k 1 ) u [ ( a 1 a 11 ) u ] 1 θ 1 x θ 1 H 1 l x , m 2 ( x ) = ( a 2 k 2 ) u + ( a 21 k 2 ) u e θ 1 N 1 ( a 22 k 2 ) l 1 + ( c 2 k 2 ) u [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] 1 θ 2 x θ 2 H 2 l x .

Lemma 3.3 Assume that (H1)-(H3) hold. Then the following assertions hold:

  1. (1)

    There exist 0< u i < u i + such that

    M i ( u i ) = M i ( u i + ) =0, M i (x)>0for x ( u i , u i + )

and

M i (x)<0for x ( 0 , u i ) ( u i + , + ) ,i=1,2.
  1. (2)

    There exist 0< x i < x i + such that

    p i ( x i ) = p i ( x i + ) = 0 , p i ( x ) > 0 for  x ( x i , x i + ) , p i ( x ) < 0 for  x ( 0 , x i ) ( x i + , + )

and

p i ( x i ) >0, p i ( x i + ) <0,i=1,2.
  1. (3)

    There exist 0< l i < l i + such that

    m i ( l i ) = m i ( l i + ) =0, m i (x)>0for x ( l i , l i + )

and

m i (x)<0for x ( 0 , l i ) ( l i + , + ) ,i=1,2.
  1. (4)
    l i < x i < u i < u i + < x i + < l i + ,i=1,2.
  2. (5)
    ln u i + < N i ,i=1,2.

Proof It follows from (H1)-(H3) and Lemma 3.2 that assertions (1)-(3) hold. Noticing that

1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] < ( a 1 k 1 ) ¯ ( a 1 k 1 ) u , ( a 11 k 1 ) l 1 + ( c 1 k 1 ) u [ ( a 1 a 11 ) u ] 1 θ 1 < ( a 11 k 1 ) ¯ ( a 11 k 1 ) u , ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 < ( a 2 k 2 ) ¯ < ( a 2 k 2 ) u + ( a 21 k 2 ) u e θ 1 N 1 , ( a 22 k 2 ) l 1 + ( c 2 k 2 ) u [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] 1 θ 2 < ( a 22 k 2 ) ¯ ( a 22 k 2 ) u , H i l H ¯ i H i u ( i = 1 , 2 ) ,

we have

M i (x)< p i (x)< m i (x),i=1,2.

By assertions (1)-(3), assertion (4) holds.

It follows from M i ( u i + )=0 (i=1,2) that

( a 11 k 1 ) u [ u 1 + ] θ 1 < 1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] < ( a 1 k 1 ) l , ( a 22 k 2 ) u [ u 2 + ] θ 2 < ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 < ( a 2 k 2 ) l .

Therefore, we have

u 1 + < [ ( a 1 k 1 ) l ( a 11 k 1 ) u ] 1 θ 1 [ ( a 1 a 11 ) u ] 1 θ 1 < e N 1 , u 2 + < [ ( a 2 k 2 ) l ( a 22 k 2 ) u ] 1 θ 2 [ ( a 2 a 22 ) u ] 1 θ 2 < e N 2 ,

which implies that assertion (5) also holds. □

Now, we are ready to state the main result of this paper.

Theorem 3.1 Assume that (H1)-(H3) hold. Then system (1.1) has at least four ω-periodic solutions.

Proof Take

X=Z= { u = ( u 1 , u 2 ) T : u i C rd ( T , R 2 )  and  u i ( t + ω ) = u i ( t )  for all  t T , i = 1 , 2 } ,

and define

u= [ i = 1 2 ( max t I ω | u i ( t ) | ) 2 ] 1 2 ,uX(or Z).

Equipped with the above norm , it is easy to verify that X and Z are both Banach spaces.

Set

Δ 1 ( u , t , λ ) = [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] Δ 1 ( u , t , λ ) = × [ a 1 ( t ) k 1 ( t ) a 11 ( t ) e θ 1 u 1 ( t ) k 1 ( t ) λ a 12 ( t ) e θ 2 u 2 ( t ) k 1 ( t ) ] H 1 ( t ) e u 1 ( t ) , Δ 2 ( u , t , λ ) = [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] Δ 2 ( u , t , λ ) = × [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) a 22 ( t ) e θ 2 u 2 ( t ) k 2 ( t ) ] H 2 ( t ) e u 2 ( t ) .

Define the mappings L:XZ, N:X×[0,1]Z, P:XX and Q:ZZ as follows:

N ( u , λ ) = ( Δ 1 ( u , t , λ ) Δ 2 ( u , t , λ ) ) , L u = ( u 1 Δ u 2 Δ ) , P u = 1 ω κ κ + ω u ( t ) Δ t = Q u , u X ( or  Z ) .

We first show that L is a Fredholm mapping of index zero and N is L-compact on Ω ¯ ×[0,1] for any open bounded set ΩX. The argument is standard, one can see [35]. But for the sake of completeness, we give the details here.

It is easy to see that KerL={uX: ( u 1 ( t ) , u 2 ( t ) ) T = ( h 1 , h 2 ) T R 2  for tT}, ImL={uX: κ κ + ω u(t)Δt=0} is closed in Z, and dimKerL=codimImL=2. Therefore, L is a Fredholm mapping of index zero. Clearly, P and Q are continuous projectors such that

ImP=KerL,KerQ=ImL=Im(IQ).

On the other hand, K p :ImLdomLKerP, the inverse to L, exists and is given by

K p (u)= κ t u(s)Δs 1 ω κ κ + ω κ η u(s)ΔsΔη.

Obviously, QN and K p (IQ)N are continuous. By Lemma 2.4, it is not difficult to show that K p ( I Q ) N ( Ω ¯ × [ 0 , 1 ] ) ¯ is compact for any open bounded set ΩX. Moreover, QN( Ω ¯ ×[0,1]) is bounded. Hence, N is L-compact on Ω ¯ ×[0,1] for any open bounded set ΩX.

In order to apply Lemma 3.1, we need to find at least four appropriate open, bounded subsets Ω 1 , Ω 2 , Ω 3 , Ω 4 in X.

Corresponding to the operator equation Lu=λN(u,λ), λ(0,1), we have

u 1 Δ ( t ) = λ [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] u 1 Δ ( t ) = × [ a 1 ( t ) k 1 ( t ) a 11 ( t ) e θ 1 u 1 ( t ) k 1 ( t ) λ a 12 ( t ) e θ 2 u 2 ( t ) k 1 ( t ) ] λ H 1 ( t ) e u 1 ( t ) ,
(3.1)
u 2 Δ ( t ) = λ [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] u 2 Δ ( t ) = × [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) a 22 ( t ) e θ 2 u 2 ( t ) k 2 ( t ) ] λ H 2 ( t ) e u 2 ( t ) .
(3.2)

Suppose that ( u 1 ( t ) , u 2 ( t ) ) T is an ω-periodic solution of (3.1), (3.2) for some λ(0,1). Since u(t)=( u 1 (t), u 2 (t))X, there exist t i M , t i m [κ,κ+ω]T, i=1,2, such that

u i ( t i M ) = max t [ κ , κ + ω ] T u i ( t ) , u i ( t i m ) = min t [ κ , κ + ω ] T u i ( t ) , i = 1 , 2 .

By Lemma 2.2, it is easy to see that

u i Δ ( t i M ) 0, u i Δ ( t i m ) 0,i=1,2.

From this and (3.1), (3.2), we obtain that

0 [ k 1 ( t 1 M ) + ( 1 λ ) c 1 ( t 1 M ) e u 1 ( t 1 M ) k 1 ( t 1 M ) + c 1 ( t 1 M ) e u 1 ( t 1 M ) ] 0 × [ a 1 ( t 1 M ) k 1 ( t 1 M ) a 11 ( t 1 M ) e θ 1 u 1 ( t 1 M ) k 1 ( t 1 M ) λ a 12 ( t 1 M ) e θ 2 u 2 ( t 1 M ) k 1 ( t 1 M ) ] H 1 ( t 1 M ) e u 1 ( t 1 M ) ,
(3.3)
0 [ k 2 ( t 2 M ) + ( 1 λ ) c 2 ( t 2 M ) e u 2 ( t 2 M ) k 2 ( t 2 M ) + c 2 ( t 2 M ) e u 2 ( t 2 M ) ] 0 × [ a 2 ( t 2 M ) k 2 ( t 2 M ) + λ a 21 ( t 2 M ) e θ 1 u 1 ( t 2 M ) k 2 ( t 2 M ) a 22 ( t 2 M ) e θ 2 u 2 ( t 2 M ) k 2 ( t 2 M ) ] H 2 ( t 2 M ) e u 2 ( t 2 M )
(3.4)

and

0 [ k 1 ( t 1 m ) + ( 1 λ ) c 1 ( t 1 m ) e u 1 ( t 1 m ) k 1 ( t 1 m ) + c 1 ( t 1 m ) e u 1 ( t 1 m ) ] 0 × [ a 1 ( t 1 m ) k 1 ( t 1 m ) a 11 ( t 1 m ) e θ 1 u 1 ( t 1 m ) k 1 ( t 1 m ) λ a 12 ( t 1 m ) e θ 2 u 2 ( t 1 m ) k ( t 1 m ) ] H 1 ( t 1 m ) e u 1 ( t 1 m ) ,
(3.5)
0 [ k 2 ( t 2 m ) + ( 1 λ ) c 2 ( t 2 m ) e u 2 ( t 2 m ) k 2 ( t 2 m ) + c 2 ( t 2 m ) e u 2 ( t 2 m ) ] 0 × [ a 2 ( t 2 m ) k 2 ( t 2 m ) + λ a 21 ( t 2 m ) e θ 1 u 1 ( t 2 m ) k 2 ( t 2 m ) a 22 ( t 2 m ) e θ 2 u 2 ( t 2 m ) k 2 ( t 2 m ) ] H 2 ( t 2 m ) e u 2 ( t 2 m ) .
(3.6)

Claim A

u 1 ( t 1 M ) < N 1 : = ln ( a 1 a 11 ) u θ 1 + 2 ω ( a 1 k 1 ) u , u 2 ( t 2 M ) < N 2 : = ln [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] θ 2 + 2 ω ( a 2 k 2 ) u + 2 ω ( a 21 k 2 ) u e θ 1 N 1 .

From (3.1), we obtain that

κ κ + ω [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] a 1 ( t ) k 1 ( t ) Δ t = κ κ + ω [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] [ a 11 ( t ) e θ 1 u 1 ( t ) k 1 ( t ) + λ a 12 ( t ) e θ 2 u 2 ( t ) k 1 ( t ) ] Δ t + κ κ + ω H 1 ( t ) e u 1 ( t ) Δ t .

Therefore, we have

κ κ + ω | u 1 Δ ( t ) | Δ t < κ κ + ω [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] [ a 1 ( t ) k 1 ( t ) + a 11 ( t ) e θ 1 u 1 ( t ) k 1 ( t ) + λ a 12 ( t ) e θ 2 u 2 ( t ) k 1 ( t ) ] Δ t + κ κ + ω H 1 ( t ) e u 1 ( t ) Δ t = 2 κ κ + ω [ k 1 ( t ) + ( 1 λ ) c 1 ( t ) e u 1 ( t ) k 1 ( t ) + c 1 ( t ) e u 1 ( t ) ] a 1 ( t ) k 1 ( t ) Δ t < 2 κ κ + ω a 1 ( t ) k 1 ( t ) Δ t 2 ω ( a 1 k 1 ) u .
(3.7)

By (3.5), we have

a 11 ( t 1 m ) e θ 1 u 1 ( t 1 m ) k 1 ( t 1 m ) < a 1 ( t 1 m ) k 1 ( t 1 m ) ,

which implies

e θ 1 u 1 ( t 1 m ) < a 1 ( t 1 m ) a 11 ( t 1 m ) ( a 1 a 11 ) u .

Therefore, we have

u 1 ( t 1 m ) < ln ( a 1 a 11 ) u θ 1 .
(3.8)

From (3.7), (3.8) and Lemma 2.3, we have

u 1 ( t 1 M ) u 1 ( t 1 m ) + κ κ + ω | u 1 Δ ( t ) | Δt< ln ( a 1 a 11 ) u θ 1 +2ω ( a 1 k 1 ) u := N 1 .
(3.9)

From (3.2), we obtain that

κ κ + ω [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) ] Δ t = κ κ + ω [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] a 22 ( t ) e θ 2 u 2 ( t ) k 2 ( t ) Δ t + κ κ + ω H 2 ( t ) e u 2 ( t ) Δ t .

Therefore, we have

κ κ + ω | u 2 Δ ( t ) | Δ t < κ κ + ω [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) + a 22 ( t ) e θ 2 u 2 ( t ) k 2 ( t ) ] Δ t + κ κ + ω H 2 ( t ) e u 2 ( t ) Δ t = 2 κ κ + ω [ k 2 ( t ) + ( 1 λ ) c 2 ( t ) e u 2 ( t ) k 2 ( t ) + c 2 ( t ) e u 2 ( t ) ] [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) ] Δ t < 2 κ κ + ω [ a 2 ( t ) k 2 ( t ) + λ a 21 ( t ) e θ 1 u 1 ( t ) k 2 ( t ) ] Δ t .

From this and (3.9), we have

κ κ + ω | u 2 Δ ( t ) | Δt<2ω ( a 2 k 2 ) u +2ω ( a 21 k 2 ) u e θ 1 N 1 .
(3.10)

By (3.6), we have

a 22 ( t 2 m ) e θ 2 u 2 ( t 2 m ) k 2 ( t 2 m ) < a 2 ( t 2 m ) k 2 ( t 2 m ) + λ a 21 ( t 2 m ) e θ 1 u 1 ( t 2 m ) k 2 ( t 2 m ) ,

which implies

e θ 2 u 2 ( t 2 m ) < a 2 ( t 2 m ) a 22 ( t 2 m ) + a 21 ( t 2 m ) a 22 ( t 2 m ) e θ 1 N 1 ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 .

Therefore, we have

u 2 ( t 2 m ) < ln [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] θ 2 .
(3.11)

From (3.10), (3.11) and Lemma 2.3, we have

u 2 ( t 2 M ) u 2 ( t 2 m ) + κ κ + ω | u 2 Δ ( t ) | Δ t < ln [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] θ 2 + 2 ω ( a 2 k 2 ) u + 2 ω ( a 21 k 2 ) u e θ 1 N 1 : = N 2 .
(3.12)

Claim B

ln l i < u i ( t i m ) <ln l i + ,i=1,2.

From (3.5) and noticing that

k 1 ( t 1 m ) k 1 ( t 1 m ) + c 1 ( t 1 m ) e u 1 ( t 1 m ) < k 1 ( t 1 m ) + ( 1 λ ) c 1 ( t 1 m ) e u 1 ( t 1 m ) k 1 ( t 1 m ) + c 1 ( t 1 m ) e u 1 ( t 1 m ) <1,

we have

0< ( a 1 k 1 ) u a 11 ( t 1 m ) k 1 ( t 1 m ) + c 1 ( t 1 m ) e u 1 ( t 1 m ) e θ 1 u 1 ( t 1 m ) H 1 l e u 1 ( t 1 m ) .

Therefore, by (3.8), we have

0< ( a 1 k 1 ) u ( a 11 k 1 ) l 1 + ( c 1 k 1 ) u [ ( a 1 a 11 ) u ] 1 θ 1 e θ 1 u 1 ( t 1 m ) H 1 l e u 1 ( t 1 m ) .

From assertion (3) of Lemma 3.3 and the above inequality, we have

ln l 1 < u 1 ( t 1 m ) <ln l 1 + .
(3.13)

Similarly, from (3.6) and noticing that

k 2 ( t 2 m ) k 2 ( t 2 m ) + c 2 ( t 2 m ) e u 2 ( t 2 m ) < k 2 ( t 2 m ) + ( 1 λ ) c 2 ( t 2 m ) e u 2 ( t 2 m ) k 2 ( t 2 m ) + c 2 ( t 2 m ) e u 2 ( t 2 m ) <1,

we have

0< ( a 2 k 2 ) u + ( a 21 k 2 ) u e θ 1 N 1 a 22 ( t 2 m ) k 2 ( t 2 m ) + c 2 ( t 2 m ) e u 2 ( t 2 m ) e θ 2 u 2 ( t 2 m ) H 2 l e u 2 ( t 2 m ) .

Therefore, by (3.11), we have

0< ( a 2 k 2 ) u + ( a 21 k 2 ) u e θ 1 N 1 ( a 22 k 2 ) l 1 + ( c 2 k 2 ) u [ ( a 2 a 22 ) u + ( a 21 a 22 ) u e θ 1 N 1 ] 1 θ 2 e θ 2 u 2 ( t 2 m ) H 2 l e u 2 ( t 2 m ) .

From assertion (3) of Lemma 3.3 and the above inequality, we have

ln l 2 < u 2 ( t 2 m ) <ln l 2 + .
(3.14)

Claim C

u i ( t i M ) >ln u i + or u i ( t i M ) <ln u i ,i=1,2.

From (3.3), (H2) and noticing that

k 1 ( t 1 M ) k 1 ( t 1 M ) + c 1 ( t 1 M ) e u 1 ( t 1 M ) < k 1 ( t 1 M ) + ( 1 λ ) c 1 ( t 1 M ) e u 1 ( t 1 M ) k 1 ( t 1 M ) + c 1 ( t 1 M ) e u 1 ( t 1 M ) <1,

we have

0> k 1 ( t 1 M ) k 1 ( t 1 M ) + c 1 ( t 1 M ) e u 1 ( t 1 M ) [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] ( a 11 k 1 ) u e θ 1 u 1 ( t 1 M ) H 1 u e u 1 ( t 1 M ) .

Therefore, we have

0> 1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] ( a 11 k 1 ) u e θ 1 u 1 ( t 1 M ) H 1 u e u 1 ( t 1 M ) .
(3.15)

From assertion (1) of Lemma 3.3 and (3.15), we have

u 1 ( t 1 M ) >ln u 1 + or u 1 ( t 1 M ) <ln u 1 .
(3.16)

By a similar argument, it follows from (3.4) that

0> ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 ( a 22 k 2 ) u e θ 2 u 2 ( t 2 M ) H 2 u e u 2 ( t 2 M ) .
(3.17)

From assertion (1) of Lemma 3.3 and (3.17), we have

u 2 ( t 2 M ) >ln u 2 + or u 2 ( t 2 M ) <ln u 2 .
(3.18)

It follows from (3.9), (3.13), (3.16) and assertions (4)-(5) of Lemma 3.3 that

u 1 ( t 1 M ) ( ln l 1 , ln u 1 ) ( ln u 1 + , N 1 ) ,
(3.19)
u 1 ( t 1 m ) ( ln l 1 , ln l 1 + ) .
(3.20)

It follows from (3.12), (3.14), (3.18) and assertions (4)-(5) of Lemma 3.3 that

u 2 ( t 2 M ) ( ln l 2 , ln u 2 ) ( ln u 2 + , N 2 ) ,
(3.21)
u 2 ( t 2 m ) ( ln l 2 , ln l 2 + ) .
(3.22)

Clearly, l i ± , u i ± (i=1,2) are independent of λ. Now, let us consider QN(u,0) with u= ( u 1 , u 2 ) T R 2 . Note that

QN(u,0)= ( ( a 1 k 1 ) ¯ ( a 11 k 1 ) ¯ e θ 1 u 1 H ¯ 1 e u 1 ( a 2 k 2 ) ¯ ( a 22 k 2 ) ¯ e θ 2 u 2 H ¯ 2 e u 2 ) .

Therefore, it follows from assertion (2) of Lemma 3.3 that QN(u,0)=0 has four distinct solutions:

u ˜ 1 = ( ln x 1 + , ln x 2 + ) T , u ˜ 2 = ( ln x 1 + , ln x 2 ) T ,
(3.23)
u ˜ 3 = ( ln x 1 , ln x 2 + ) T , u ˜ 4 = ( ln x 1 , ln x 2 ) T .
(3.24)

Let

Ω 1 = { u = ( u 1 , u 2 ) T X | max t I ω u 1 ( t ) ( ln u 1 + , max { ln l 1 + , N 1 } ) , min t I ω u 1 ( t ) ( ln l 1 , ln l 1 + ) , max t I ω u 2 ( t ) ( ln u 2 + , max { ln l 2 + , N 2 } ) , min t I ω u 2 ( t ) ( ln l 2 , ln l 2 + ) } , Ω 2 = { u = ( u 1 , u 2 ) T X | max t I ω u 1 ( t ) ( ln u 1 + , max { ln l 1 + , N 1 } ) , min t I ω u 1 ( t ) ( ln l 1 , ln l 1 + ) , max t I ω u 2 ( t ) ( ln l 2 , ln u 2 ) , min t I ω u 2 ( t ) ( ln l 2 , ln l 2 + ) } , Ω 3 = { u = ( u 1 , u 2 ) T X | max t I ω u 1 ( t ) ( ln l 1 , ln u 1 ) , min t I ω u 1 ( t ) ( ln l 1 , ln l 1 + ) , max t I ω u 2 ( t ) ( ln u 2 + , max { ln l 2 + , N 2 } ) , min t I ω u 2 ( t ) ( ln l 2 , ln l 2 + ) } , Ω 4 = { u = ( u 1 , u 2 ) T X | max t I ω u 1 ( t ) ( ln l 1 , ln u 1 ) , min t I ω u 1 ( t ) ( ln l 1 , ln l 1 + ) , max t I ω u 2 ( t ) ( ln l 2 , ln u 2 ) , min t I ω u 2 ( t ) ( ln l 2 , ln l 2 + ) } .

Then Ω 1 , Ω 2 , Ω 3 , Ω 4 are bounded open subsets of X. It follows from assertion (4) of Lemma 3.3, (3.23) and (3.24) that u ˜ i Ω i (i=1,2,3,4). From assertion (4) of Lemma 3.3, (3.19)-(3.22), it is easy to see that Ω ¯ i Ω ¯ j = (i,j=1,2,3,4, ij) and Ω i satisfies (a) in Lemma 3.1 for i=1,2,3,4. Moreover, QN(u,0)0 for u Ω i KerL. By assertion (2) of Lemma 3.3, a direct computation gives

deg { J Q N ( , 0 ) , Ω 1 Ker L , 0 } = 1 , deg { J Q N ( , 0 ) , Ω 2 Ker L , 0 } = 1 , deg { J Q N ( , 0 ) , Ω 3 Ker L , 0 } = 1 , deg { J Q N ( , 0 ) , Ω 4 Ker L , 0 } = 1 .

Here, J is taken as the identity mapping since ImQ=KerL. So far we have proved that Ω i satisfies all the assumptions in Lemma 3.1. Hence, (1.1) has at least four ω-periodic solutions ( u 1 i ( t ) , u 2 i ( t ) ) T (i=1,2,3,4) with ( u 1 i , u 2 i ) T domL Ω ¯ i . Obviously, ( u 1 i ( t ) , u 2 i ( t ) ) T (i=1,2,3,4) are different.  □

Example 3.1 In system (1.1), take

ω=2, T 0 ={0,0.5}[1,2],T={ t 0 +2k| t 0 T 0 ,kZ},

where is the integer set. Clearly, the time scale is ω-periodic, i.e., tT implies t+ωT. In this case, we have

κ=min { [ 0 , + ) T } =0, I ω =[κ,κ+ω]T= T 0 ={0,0.5}[1,2].

Take

θ 1 = θ 2 = 0.5 , H 1 ( t ) = 2 + sin ( π t ) 54 , 000 , H 2 ( t ) = 2 + sin ( π t ) 3 , 000 , k 1 ( t ) = 100 ( 2 + sin ( π t ) ) , k 2 ( t ) = 100 ( 3 + sin ( π t ) ) , a 1 ( t ) = [ 2 + sin ( π t ) ] 2 , a 11 ( t ) = [ 2 + sin ( π t ) ] 2 10 , a 12 ( t ) = [ 2 + sin ( π t ) ] 2 9 ( 20 + 0.5 e 0.06 ) e ( 0.08 + 0.002 e 0.06 ) , a 2 ( t ) = [ 3 + sin ( π t ) ] 2 , a 22 ( t ) = [ 3 + sin ( π t ) ] 2 20 , a 21 ( t ) = [ 3 + sin ( π t ) ] 2 400 , c 1 ( t ) = 2 + sin ( π t ) e 0.12 , c 2 ( t ) = 3 + sin ( π t ) [ 2 + 0.05 e 0.06 ] 2 e ( 0.16 + 0.004 e 0.06 ) .

Then

H 1 u = 1 18 , 000 , H 2 u = 1 1 , 000 , ( a 1 k 1 ) l = 1 100 , ( a 1 k 1 ) u = 3 100 , ( a 11 k 1 ) l = 1 1 , 000 , ( a 11 k 1 ) u = 3 1 , 000 , ( a 12 k 1 ) u = 1 300 ( 20 + 0.5 e 0.06 ) e ( 0.08 + 0.002 e 0.06 ) , ( a 2 k 2 ) l = 1 50 , ( a 2 k 2 ) u = 1 25 , ( a 22 k 2 ) l = 1 1 , 000 , ( a 22 k 2 ) u = 1 500 , ( a 21 k 2 ) u = 1 10 , 000 , ( a 1 a 11 ) u = 10 , ( a 2 a 22 ) u = 20 , ( a 21 a 22 ) u = 1 20 , ( c 1 k 1 ) u = 1 100 e 0.12 , ( c 2 k 2 ) u = 1 [ 20 + 0.5 e 0.06 ] 2 e ( 0.16 + 0.004 e 0.06 ) , N 1 = 2 ln 10 + 0.12 , N 2 = 2 ln [ 20 + 0.5 e 0.06 ] + 0.16 + 0.004 e 0.06 .

Therefore, we have

1 1 + ( c 1 k 1 ) u e N 1 [ ( a 1 k 1 ) l ( a 12 k 1 ) u e θ 2 N 2 ] = 1 300 , ( a 2 k 2 ) l 1 + ( c 2 k 2 ) u e N 2 = 1 100 , ( 1 + θ 1 ) [ ( a 11 k 1 ) u ] 1 1 + θ 1 [ H 1 u θ 1 ] θ 1 1 + θ 1 = 1.5 × ( 3 1 , 000 ) 2 3 × ( 1 9 , 000 ) 1 3 = 1.5 1 , 000 , ( 1 + θ 2 ) [ ( a 22 k 2 ) u ] 1 1 + θ 2 [ H 2 u θ 2 ] θ 2 1 + θ 2 = 1.5 × ( 1 500 ) 2 3 × ( 1 500 ) 1 3 = 3 1 , 000 .

Hence, all the conditions in Theorem 3.1 are satisfied. By Theorem 3.1, system (1.1) has at least four 2-periodic solutions.

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Acknowledgements

This research is supported by the National Natural Science Foundation of China (Grant No. 10971085).

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Fang, H., Wang, Y. Four periodic solutions for a food-limited two-species Gilpin-Ayala type predator-prey system with harvesting terms on time scales. Adv Differ Equ 2013, 278 (2013). https://doi.org/10.1186/1687-1847-2013-278

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