# Qualitative Analysis for a Predator Prey System with Holling Type III Functional Response and Prey Refuge

Authors: Xia Liu; Yepeng Xing
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

A predator prey system with Holling III functional response and constant prey refuge is considered. By using the Dulac criterion, we discuss the global stability of the positive equilibrium of the system. By transforming the system to a Liénard system, the conditions for the existence of exactly one limit cycle for the system are given. Some numerical simulations are presented.


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## Body

## 1. Introduction

Recently, the qualitative analysis of predator prey systems with Holling II or III types functional response and prey refuge has been done by several papers, see [1–5]. Their main objective is to discuss under what conditions the positive equilibrium of the corresponding system is stable or unstable and the existence of exactly one limit cycles. In general, the prey refuge has two types, one is the so-called constant proportion prey refuge: (1-m)x, where m∈(0,1), the other type is called constant prey refuge: (x-m).In [2], the authors considered the following system with a constant proportion prey refuge:
(1.1)dxdt=ax-bx2-α(1-m)2x2yβ2+(1-m)2x2,dydt=-cy+kα(1-m)2x2yβ2+(1-m)2x2,
where x and y denote the prey and predator density, respectively, at time t, the parameters a,b,α,β,c,k are positive constants, and their biological meanings can be seen in [2]. The main result is that when 0<m<(1+2bcβ/(a(kα-2c)))c/(kα-c) system (1.1) admits only one limit cycle which is globally asymptotically stable.In paper [4], the authors only gave the local stability analysis to the following system with a constant prey refuge:
(1.2)dxdt=ax-bx2-α(x-m)2yβ2+(x-m)2,dydt=-cy+kα(x-m)2yβ2+(x-m)2.
In this paper, we will research under what conditions that the positive equilibrium is globally asymptotically stable and the existence of exactly one stable limit cycle of system (1.2). For ecological reason, we only consider system (1.2) in Ω0={(x,y)∣x>m,y>0} or Ω¯0.It easy to obtain the following lemma.Lemma 1.1.
Any solution(x(t),y(t)) of system (1.2) with initial condition x(0)>m,y(0)>0 is positive and bounded for all t≥0.


## 2. Basic Results

Letx-=x-m,y-=αy,dt=(β2+x-2)dt-, then system (1.2) changes (still denote x-, y-,t- as x,y,t)
(2.1)dxdt=(x+m)(a-b(x+m))(β2+x2)-x2y,dydt=-cβ2y+(kα-c)x2y.
Then Ω0 transforms to Ω={(x,y)∣x>0,y>0} and system (2.1) is bounded.Clearly, if(H1)  0<m<a/b holds, system (2.1) has positive boundary equilibrium E0((a/b)-m,0); if (H2)   kα>c,0<m<(a-bx*)/b, system (2.1) has a positive equilibrium E*(x*,y*), where
(2.2)x*=βckα-c,y*=kαc(x*+m)(a-b(x*+m)).It is easy to obtain the following lemma.Lemma 2.1.
Let(H1) hold. Further assume that (H3)  kα≤c and (H4)kα>c, m>max{0,(a-bx*)/b}. Then E0 is locally asymptotically stable, if any of (H3) and (H4) holds. When kα>c,0<m<(a-bx*)/b,E0 is unstable, furthermore, E0 is a saddle point.About the properties of the positive equilibrium, we have the following theorem.Theorem 2.2.
Assumekα>c. Then(I)
E* is locally asymptotically stable for 0<m<(a-bx*)/b if a(2c-kα)≤2bcx* holds.(II)
E* is locally asymptotically stable for m1<m<(a-bx*)/b and E* is locally unstable for 0<m<m1 if a(2c-kα)>2bcx* holds, where
(2.3)m1=bx*3+β2(a-bx*)-Δ/42bβ2,(III)
system (2.1) undergoes Hopf bifurcation at m=m1 if a(2c-kα)>2bcx* holds.Proof.
The Jacobian matrix of system (2.1) at E* is
(2.4)J(E*)=(-Px*-x*22(kα-c)x*y*0),
where P=2bx*4+(2bm-a)x*3+β2(a-2bm)x*+2mβ2(a-bm). Then tr(J(E*))=-P/x*=R(m)/x*, where R(m)=2bβ2m2+2(bβ2x*-bx*3-aβ2)m-aβ2x*+ax*3-2bx*4, the discriminant of R(m)=0 is Δ=4(b2x*6+2b2x*4β2+b2β4x*2+4a2β4)>0. Hence, the equation R(m)=0 has two roots m1 and m2, where m1=(bx*3+β2(a-bx*)-Δ/4)/2bβ2,m2=(bx*3+β2(a-bx*)+Δ/4)/2bβ2.
Note that(2.5)(bx*3+β2(a-bx*))2-Δ4=2bx*β2(ax*2-aβ2-2bx*3)=2bx*β2(aβ22c-kαkα-c-2bx*3)=2bx*β4a(2c-kα)-2bcx*kα-c,
and a(2c-kα)>(≤)2bcx* implies m1>(≤)0. Consider
(2.6)m2>a-bx*2b+β2a2+b2x*2-2abx*+2abx*2bβ2>a-bx*b.
Then (I)
Ifa(2c-kα)≤2bcx* holds, then m1≤0,R(m)<0 holds for m1<m<m2. Considering (H2) and m2>(a-bx*)/b, for 0<m<(a-bx*)/b, tr(J(E*))<0, which implies E* is locally asymptotically stable.(II)
Ifa(2c-kα)>2bcx* holds, then m1>0, for m1<m<m2,R(m)<0, since m1-(a-bx*)/b=(bx*3-β2(a-bx*)-Δ/4)/2bβ2, by (bx*3-β2(a-bx*))2-(Δ/4)=-2abβ2x*(x*2+β2)<0, we obtain m1<(a-bx*)/b. Together with (H2), for m1<m<(a-bx*)/b,tr(J(E*))<0, which means E* is locally asymptotically stable. On the other hand, for 0<m<m1,tr(J(E*))>0,E* is locally unstable.(III)
We have(2.7)det(J(E*))>0,tr(J(E*)|m1)=R(m1)x*=0,∂tr(J(E*))∂m|m1≠0,
these satisfy Liu’s Hopf bifurcation criterion (see [6], page 255); hence, the Hopf bifurcation occurs at m=m1. This ends the proof.


## 3. Global Stability of the Positive Equilibrium

Denotem3:=(9a-23βb-81a2+12b2β2)/18b<0,m4:=(9a-23βb+81a2+12b2β2)/18b>0.Theorem 3.1.
IfE*(x*,y*) is locally stable. Further assume that max{0,(a-4bβ)/2b}<m<m4, then the positive equilibrium E*(x*,y*) of system (2.1) is globally asymptotically stable.Proof.
Take the Dulac functionB(x,y)=x-2y-1, for system (2.1) we have
(3.1)T=∂(BP)∂x+∂(BQ)∂y=-ϕ(x)x3y,
where
(3.2)ϕ(x)=2bx4+(2bm-a)x3+β2(a-2bm)x+2mβ2(a-bm).
Ifa=2bm,ϕ(x)=2b(x4+m2β2)>0 for x>0.
On the other hand, there exist(3.3)ϕ′(x)=8bx3+3(2bm-a)x2+β2(a-2bm),ϕ′′(x)=24bx2+6(2bm-a)x.
The equation ϕ′′(x)=0 has two roots x1=0,x2=(a-2bm)/4b.
Case  1. If a-2bm>0, then for 0<x<x2, ϕ′′(x)<0; for x>x2, ϕ′′(x)>0. Hence, x=x2 is the least value of the function ϕ′(x). If β>x2,ϕ′(x2)=(a-2bm)(β-x2)(β+x2)>0, it has ϕ′(x)>0 for all x>0, then ϕ(x) is increasing for x>0, notice that ϕ(0)>0. Therefore, ϕ(x)>0 for x>0. Since, T<0 for x>0, system (2.1) does not exist limit cycle.
Case  2. If bm<a<2bm, then x2<0, for x>0,ϕ′′(x)>0, hence, for x>0,ϕ′(x) is increasing. Evidently, ϕ′(0)<0,ϕ′(β/3)>0, then there exists 0<x0<β/3 such that ϕ′(x0)=0, where ϕ′(x0)=8bx03+3(2bm-a)x02+β2(a-2bm), hence, when 0<x<x0,ϕ′(x)<0, when x>x0,ϕ′(x)>0. We know that ϕ(x) takes the least value at x=x0, that is, ϕ(x)>ϕ(x0). According to ϕ′(x0)=0, for x>0 we obtain ϕ(x)>ϕ(x0)=(2bm-a)(x03-3β2x0+(8mβ2(a-bm))/(2bm-a)), where 0<x0<β/3.
To proveϕ(x)>0 for x>0, it suffices to prove ϕ~(x)=x3-3β2x+(8mβ2(a-bm))/(2bm-a)>0 for 0<x<β/3. Clearly, ϕ~(x) takes the least value at x=β, and ϕ~(x) is strictly decreasing at the interval (0,β). Hence, for 0<x<β/3,ϕ~(x)>ϕ~(β/3) holds. Since ϕ~β/3>0⇔(m(a-bm))/(2bm-a)>β/33⇔-33bm2+(33a-2βb)m+βa>0⇔m3<m<m4. Therefore, for 0<x0<β/3,ϕ~(x)>0 holds if m3<m<m4 holds, then for x>0,ϕ(x)>0 holds.
In sum, if one of the following three conditions holds (1)m=a/2b; (2) 0<m<a/2b,x2<β⇒max{0,(a-4bβ)/2b}<m<a/2b; (3) a/2b<m<a/b, m3<m<m4⇒a/2b<m<m4, the function T does not change the sign for x>0, then system (2.1) does not exist limit cycle. It is easy to see that the conditions (1),(2), and (3) are equal to max{0,(a-4bβ)/2b}<m<m4. The proof is completed.


## 4. Existence and Uniqueness of Limit Cycle

Theorem 4.1.
Ifa(2c-kα)>2bcx* holds, for 0<m<m1 system (2.1) admits at least one limit cycle in Ω.Proof.
We construct a Bendixson loopOABCD^ which includes E* of system (2.1). Let OA¯ be a length of the line L1:y=0,AB¯ be a length of line L2:b(x+m)-a=0. Define
(4.1)x˙=x2(a0-y),y˙=(-cβ2+(kα-c)x2)y,
where a0=maxx*≤x≤(a/b)-m{((x+m)(a-b(x+m))(β2+x2))/x2}. The orbit of system (4.1) with initial value ((a/b)-m,a0) intersects with the line x=x* and the intersection point C(x*,y1), we obtain the orbit arc BC^. Let CD¯ be a length of line L3:y=y1,DO¯ be a length of line L4:x=0. Because OA¯ is a length of orbit line of system (2.1) and (dL2/dt)|(2.1)=-b((a/b)-m)2y<0(y>0), (dL3/dt)|(2.1)=y1(-cβ2+(kα-c)x2)<0(0<x<x*),(dL4/dt)|(2.1)=mβ2(a-bm)>0, the orbits of system (2.1) tend to the interior of the Bendixson loop from the outer of AB¯,CD¯, and BC^, by comparing system (2.1) to system (4.1): dx/dt|(2.1)<dx/dt|(4.1)<0 and dy/dt|(2.1)=dy/dt|(4.1)>0. Then the orbits of system (2.1) tend to the interior of the Bendixson loop from the outer of BC^. On the other hand, under the condition of Theorem 4.1, E*(x*,y*) is unstable, by Poincaré-Bendixson Theorem, system (2.1) admits at least one limit cycle in the region OABCD^∈Ω. This ends the proof.Lemma 4.2 (see [7]).
Letf(x), g(x) be continuously differentiable functions on the open interval (r1,r2), and φ(y) be continuously differentiable functions on R in
(4.2)dxdt=φ(y)-∫x0x‍f(u)du,dydt=-g(x),
such that(1)
dφ(y)/dy>0,(2)
having a uniquex0∈(r1,r2), such that (x-x0)g(x-x0)>0 for x≠x0 and g(x0)=0,(3)
f(x0)d/dx(f(x)/g(x))<0 for x≠x0,then system (4.1) has at most one limit cycle.Theorem 4.3.
Ifa(2c-kα)>2bcx* holds, for 0<m<min{m1,a/2b-(83x*3)/(9(x*2-β2))} system (2.1) exists exactly one limit cycle which is globally asymptotically stable in Ω.Proof.
Letu=x,v=lny,τ=-x2t,  still denote u,v,τ,  as x,y,t, then system (2.1) becomes
(4.3)dxdt=ey-(x+m)(a-b(x+m))(β2+x2)x2,dydt=-(kα-c)x2-cβ2x2,
the positive equilibrium E*(x*,y*) changes E~*(x*,lny*).
Letx-=x-x*,y-=y-lny*, then E~* transform to the origin O(0,0), still denote x-,y-, as x,y yield
(4.4)dxdt=y*ey-y*-(x+x*+m)(a-b(x+x*+m))(β2+(x+x*)2)(x+x*)2+y*:=φ(y)-F(x),(x>-x*),dydt=-(kα-c)(x+x*)2-cβ2(x+x*)2:=-g(x),
where F(x)=((x+x*+m)(a-b(x+x*+m))(β2+(x+x*)2))/(x+x*)2-y*.
Clearly,F(0)=0. It is easy to see that the conditions (1) and (2) of Lemma 4.2 for x0=0 are satisfied. Consider
(4.5)f(x)=F′(x)=2b(x+x*)4+(2bm-a)(x+x*)3+β2(a-2bm)(x+x*)+2mβ2(a-bm)(x+x*)3.
Note that by the assumption of Theorem 4.3, E* is unstable equilibrium and
(4.6)tr(J(E*))=-1x*[2bx*4+(2bm-a)x*3+β2(a-2bm)x*+2mβ2(a-bm)]>0,
then f(0)=-x*-2tr(J(E*))<0. Consider
(4.7)ddx(f(x)g(x))=2ψ(x)(x+x*)2((kα-c)(x+x*)2-cβ2)2,
where
(4.8)ψ(x)=b(kα-c)(x+x*)6-3bcβ2(x+x*)4+β2(2c-kα)(a-2bm)(x+x*)3-3mβ2(kα-c)(a-bm)(x+x*)2+mcβ4(a-bm)=(kα-c)ψ~(x),
where
(4.9)ψ~(x)=b(x+x*)6-3bx*2(x+x*)4+β2(2c-kα)(a-2bm)kα-c(x+x*)3-3mβ2(a-bm)(x+x*)2+mβ2x*2(a-bm).
Then, we have(4.10)ψ~′(x)=6b(x+x*)5-12bx*2(x+x*)3+3β2(2c-kα)(a-2bm)kα-c(x+x*)2-6mβ2(a-bm)(x+x*)=(x+x*)ϕ~(x),
where
(4.11)ϕ~(x)=6b(x+x*)4-12bx*2(x+x*)2+3β2(2c-kα)(a-2bm)kα-c(x+x*)-6mβ2(a-bm).
By a simple computation, we obtain(4.12)ϕ~′(x)=24b(x+x*)3-24bx*2(x+x*)+3β2(2c-kα)(a-2bm)kα-c,ϕ~′′(x)=24b(3(x+x*)2-x*2).
It is easy to verify that ϕ~′′(-x*)<0 and ϕ~′′(x)=0 has two roots x-1 and x-2 defined by, respectively,
(4.13)x-1=(-1-33)x*,x-2=(-1+33)x*.
Obviously, x-1<-x*<x-2. Therefore, ϕ~′′(x)<0 for -x*⩽x<x-2 and ϕ~′′(x)>0 for x>x-2 which indicates that x-2 is the minimum point of the function ϕ~'(x) when x⩾-x*. Substituting x-2 into ϕ~'(x), we obtain
(4.14)minx⩾-x*ϕ~′(x)=ϕ~′(x-2)=-163bx*33+3β2(2c-kα)(a-2bm)kα-c=-163bx*33+3(x*2-β2)(a-2bm).
It is easy to see that if 0<m<(a/2b)-(83x*3)/(9(x*2-β2)), then ϕ~'(x-2)>0, which implies ϕ~'(x)>0 for all x⩾-x*. That is, the function ϕ~(x) is a strictly increasing function for x⩾-x*.
Note thatϕ~(-x*)=-6mβ2(a-bm)<0 for 0<m<a/b and limx→+∞ϕ~(x)=+∞. It follows from (4.6) that
(4.15)ϕ~(0)=-3(2bx*4+(2bm-a)x*3+(a-2bm)β2x*+2mβ2(a-bm))>0.
Hence, there exists a point-x*<x^<0, such that ϕ~(x^)=0, that is,
(4.16)b(x^+x*)4-2bx*2(x^+x*)2+β2(2c-kα)(a-2bm)2(kα-c)(x^+x*)-mβ2(a-bm)=0.
This, together with the monotonicity of ϕ~(x) when x⩾-x*, we may conclude that ψ~'(x)=(x+x*)ϕ~(x)<0 for x∈(-x*,x^) and ψ~'(x)>0 for x∈(x^,∞). Therefore, x^ is the minimum point of the function ψ~(x) for -x*<x<∞.
Together with (4.16), we obtain
(4.17)minx⩾-x*ψ~(x)=ψ~(x^)=-bx*2(x^+x*)4+β2(2c-kα)(a-2bm)2(kα-c)(x^+x*)3-2mβ2(a-2bm)(x^+x*)2+mβ2(a-bm)x*2=-bx*2(x^+x*)4+12(x*2-β2)(a-2bm)(x^+x*)3-2mβ2(a-2bm)(x^+x*)2+mβ2(a-bm)x*2=-2bx*4(x^+x*)2+12(x*2-β2)(a-2bm)(x^+x*)(2x*2+2x^x*+x^2)-2mβ2(a-2bm)(x^+x*)2=-2bx*4(x^+x*)2+(x*2-β2)(a-2bm)x*(x^+x*)2-2mβ2(a-2bm)(x^+x*)2+12(x*2-β2)(a-2bm)(x^+x*)x^2=-(x^+x*)2(2bx*4+(2bm-a)x*3+β2(a-2bm)x*+2mβ2(a-2bm))+12(x*2-β2)(a-2bm)(x^+x*)x^2.
It follows from (4.6), we have minx⩾-x*ψ~(x)>0. This indicates ψ~(x)>0 for all x>-x*.
Then all the conditions of Lemma4.2 are satisfied, considering Theorem 4.1, we obtain the conclusion of this theorem. The proof is completed.


## 5. Numerical Simulations

Takeα=0.5, k=0.2, β=0.5, a=1, b=0.1, and c=0.09. Then a(2c-kα)-2bcx*=0.053, and m1≈1.986121812. One can see a Hopf bifurcation occurring at m=1.955 and the bifurcated periodic solution is stable in Figure 1.Figure 1
The bifurcated periodic solution is stable.When takingm=4.5, then x*=1.5, y*≈2.666666667, a(2c-kα)-2bcx*=0.053,m1≈1.986121812,(a-bx*)/b=8.5,(a-4bβ)/2b=4,a/2b=5. Theorem 3.1 is satisfied; the equilibrium E* of system (2.1) is globally asymptotically stable. See Figure 2.Figure 2
The positive equilibriumE* of system (2.1) is globally asymptotically stable.Takem=1, we obtain E*(1.5,2.083333333), a(2c-kα)-2bcx*=0.053,m1≈1.986121812,(a/2b)-(83x*3/9(x*2-β2))≈2.401923788. The conditions in Theorem 4.1 are satisfied; hence, system (2.1) exists exactly one limit cycle which is globally asymptotically stable. One can see Figure 3.The dynamical behaviors of system (2.1) when α=0.5, k=0.2, β=0.5, a=1, b=0.1, c=0.09, m=1. (a) The existence of unique limit cycle. (b) The global stability of the limit cycle.
(a)
(b)



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