# Existence of Two Positive Periodic Solutions for a Neutral Multi-Delay Logarithmic Population Model with a Periodic Harvesting Rate

Authors: Hui Fang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By using an abstract existence result based on a coincidence degree theory fork-set contractive mapping, a new result is obtained for the existence of at least two positive periodic solutions for a neutral multidelay logarithmic population model with a periodic harvesting rate. An example is given to illustrate the effectiveness of the result.


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## Body

## 1. Introduction

In recent years, many papers have been published on the existence of positive periodic solutions for neutral delay logarithmic population models by using a topological degree theory fork-set contractive mapping (see, e.g., [1–5]). Recently, Xia [6] obtained some new sufficient conditions for the existence and uniqueness of an almost periodic solution of a multispecies logarithmic population model with feedback controls. However, few papers deal with the existence of multiple positive periodic solutions for neutral multidelay logarithmic population models with harvesting. The main difficulty is hard to obtain a priori bounds on solutions for neutral multi-delay models with harvesting.In this paper, we consider the following neutral multi-delay logarithmic population model of single-species population growth with a periodic harvesting rate(1.1)dy(t)dt=y(t)[a(t)-∑j=1n‍bj(t)lny(t-σj(t))-∑i=1m‍ci(t)ddtlny(t-τi(t))]-h(t),
where a(t),bj(t),σj(t)(j=1,2,…,n),ci(t),τi(t)(i=1,2,…,m),h(t) are nonnegative continuous T-periodic functions, and h(t) denotes the harvesting rate. When h(t)≡0, (1.1) was considered by [2–5]. When n=m=1,h(t)≡0, and σ1(t),τ1(t) are constants, (1.1) was considered by [7].The purpose of this paper is to establish the existence of at least two positive periodic solutions for a neutral multi-delay logarithmic population model (1.1) by using a coincidence degree theory for k-set contractions. Motivated by the work of Chen [8], some novel techniques are employed to find a priori bounds on solutions.


## 2. Preliminaries

We now briefly state the part of the coincidence degree theory fork-set contractive mapping developed by Hetzer [9, 10]. For more details, we refer to [11].LetZ be a Banach space. For a bounded subset A⊂Z, let ΓZ(A) denote the (Kuratowski) measure of noncompactness defined by
(2.1)ΓZ(A)=inf{δ>0:∃afinitenumberofsubsetsAi⊂A,A=⋃i‍Ai,diam(Ai)≤δ}.
Here, diam (Ai) denotes the maximum distance between the points in the set Ai.LetX and Z be Banach spaces with norms ∥·∥X and ∥·∥Z, respectively and Ω a bounded open subset of X. A continuous and bounded mapping N:Ω-→Z is called k-set contractive if for any bounded A⊂Ω-, we have
(2.2)ΓZ(N(A))≤kΓX(A).
Also, for a continuous and bounded map T:X→Y, we define
(2.3)l(T)=sup{r≥0:∀boundedsubsetA⊂X,rΓX(A)≤ΓY(T(A))}.LetL:domL⊂X→Z be a linear mapping and N:X→Z 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, there then exist continuous projectors P:X→X and Q:Z→Z such that ImP=KerL,ImL=KerQ=Im(I-Q). If we define LP:domL∩KerP→ImL as the restriction L|domL∩KerP of L to domL∩KerP, then LP is invertible. We denote the inverse of that map by KP. If Ω is an open bounded subset of X, the mapping N will be called L-k-set contractive on Ω- if QN(Ω-) is bounded and KP(I-Q)N:Ω-→X is k-set contractive. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQ→KerL.Lemma 2.1 ([11, Proposition XI.2.]).
LetL be a closed Fredholm mapping of index zero and let N:Ω-→Z be k′-set contractive with
(2.4)0≤k′<l(L).
Then N:Ω-→Z is a L-k-set contraction with constant k=k'/l(L)<1.The following lemma [[11, page 213] will play a key role in this paper.Lemma 2.2.
LetL be a Fredholm mapping of index zero and let N:Ω-→Z be L-k-set contractive on Ω-,k<1. Suppose(i)
Lx≠λNx for every x∈domL∩∂Ω and every λ∈(0,1);(ii)
QNx≠0 for every x∈∂Ω∩KerL;(iii)
Brouwer degreedegB(JQN,Ω∩KerL,0)≠0. Then Lx=Nx has at least one solution in domL∩Ω-.


## 3. Main Result

LetCT0 denote the linear space of real valued continuous T-periodic functions on R. The linear space CT0 is a Banach space with the usual norm for x∈CT0 given by |x|0=maxt∈R|x(t)|. Let CT1 denote the linear space of T-periodic functions with the first-order continuous derivative. CT1 is a Banach space with norm |x|1=max{|x|0,|x′|0}.LetX=CT1 and Z=CT0 and let L:X→Z be given by Lx=dx/dt. Since |Lx|0=|x′|0≤|x|1, we see that L is a bounded (with bound = 1) linear map.Under the transformation ofy(t)=ex(t), (1.1) can be rewritten as
(3.1)x′(t)=a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)(1-τi′(t))x′(t-τi(t))-h(t)ex(t).
Next define a nonlinear map N:X→Z by
(3.2)N(x)(t)=a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)(1-τi′(t))x′(t-τi(t))-h(t)ex(t).
Now, if Lx=Nx for some x∈X, then the problem (3.1) has a T-periodic solution x(t).In the following, we denote(3.3)g-=1T∫0T‍g(t)dt,gl=mint∈[0,T]g(t),gu=maxt∈[0,T]g(t),
where g(t) is a continuous nonnegative T-periodic solution.From now on, we always assume that(H1)
a(t),bj(t)∈C(R,(0,+∞)),σj(t),ci(t),τi(t)∈C1(R,(0,+∞)),forallj∈{1,2,…,n},  for all  i∈{1,2,…,m}.(H2)
σj′(t)<1,τi′(t)<1,forallt∈R, and
(3.4)Γ(t)=∑j=1n‍bj(μj(t))1-σj′(μj(t))-∑i=1m‍ci′(γi(t))1-τi′(γi(t))>0,where μj(t) is the inverse function of t-σj(t),γi(t) is the inverse function of t-τi(t),forallj∈{1,2,…,n},foralli∈{1,2,…,m}.(H3)  Let
(3.5)1+lnhlΓu<auΓl≤hlΓu<1.(H4)  Let
(3.6)T(Γ-+∑j=1n‍b-j)+∑i=1m‍ciu<1,max{1+lnhu∑j=1nbjl,0}<al-∑i=1mciu(1-τi′)uM0∑j=1nbju<R0,where
(3.7)M0=au+∑j=1nbjuR0+hueR01-∑i=1mciu(1-τi′)u,R0=q+p,q=2|lnhlΓu|,p=2Ta-+(Γ-+∑j=1nb-j)Tq1-T(Γ-+∑j=1nb-j)-∑i=1mciu.
We first give some technological lemmas.Set(3.8)f(x)=d-x-re-x,x∈(-∞,+∞).Lemma 3.1.
Assume thatd,r are positive constants such that
(3.9)1+lnr<d.
Then there exist x*-,x*+ such that
(3.10)f(x*-)=f(x*+)=0,x*-<lnr<x*+<d,f(x)>0forx∈(x*-,x*+);f(x)<0forx∈(-∞,x*-)∪(x*+,+∞).
If one assumes further that
(3.11)d≤r<1,
then the following inequalities also hold:
(3.12)2lnr<x*-<lnr<x*+≤0.Proof.
Clearly,f'(x)=r/ex-1=0 if and only if x=lnr. Therefore, noticing that limx→±∞f(x)=-∞, we have
(3.13)supx∈(-∞,+∞)f(x)=d-lnr-1>0.
Set
(3.14)g(r)=2lnr+1r-r.
Since
(3.15)g′(r)=2r-1r2-1=-(1-r)2r2<0for0<r<1,g(r) is monotonically decreasing on (0,1].
Therefore, we have(3.16)g(r)>g(1)=0,
that is,
(3.17)2lnr+1r>r,
which implies
(3.18)f(2lnr)=d-2lnr-1r<d-r≤0.
Again, noticing that
(3.19)f(0)=d-r≤0,
by the monotonicity of the function f(x) on the interval (-∞,lnr) and (lnr,+∞), it is easy to see that the assertion holds.Set(3.20)F(x)=al-∑i=1mciu(1-τi′)uM0∑j=1nbju-x-hu∑j=1nbjle-x,G(x)=a-∑j=1nb-j-x-h-∑j=1nb-je-x,H(x)=au+∑i=1mciu(1-τi′)uM0∑j=1nbjl-x-hl∑j=1nbjue-x.Lemma 3.2.
Assume that(H1),(H2),  and  (H4) hold. Then the following assertions hold.
(1) There existu-,u+ such that
(3.21)F(u-)=F(u+)=0,u-<lnhu∑j=1nbjl<u+<al-∑i=1mciu(1-τi′)uM0∑j=1nbju,F(x)>0forx∈(u-,u+);F(x)<0forx∈(-∞,u-)∪(u+,+∞).
(2) There existx-,x+ such that
(3.22)G(x-)=G(x+)=0,x-<lnh-∑j=1nb-j<x+<a-∑j=1nb-j,G(x)>0forx∈(x-,x+);G(x)<0forx∈(-∞,x-)∪(x+,+∞).
(3) There existl-,l+ such that
(3.23)H(l-)=H(l+)=0,l-<lnhl∑j=1nbju<l+<au+∑i=1mciu(1-τi′)uM0∑j=1nbjl,H(x)>0forx∈(l-,l+);H(x)<0forx∈(-∞,l-)∪(l+,+∞).
(4)(3.24)l-<x-<u-<u+<x+<l+.Proof.
Noticing that(3.25)al-∑i=1mciu(1-τi′)uM0∑j=1nbju<a-∑j=1nb-j<au+∑i=1mciu(1-τi′)uM0∑j=1nbjl,hl∑j=1nbju≤h-∑j=1nb-j≤hu∑j=1nbjl,
we have
(3.26)F(x)<G(x)<H(x).
It follows from (H1),(H2),and(H4), (3.25) that
(3.27)1+lnhu∑j=1nbjl<al-∑i=1mciu(1-τi′)uM0∑j=1nbju,1+lnh-∑j=1nb-j<a-∑j=1nb-j,1+lnhl∑j=1nbju<au+∑i=1mciu(1-τi′)uM0∑j=1nbjl.
Therefore, by Lemma 3.1, the assertions (1)–(3) hold. Furthermore, by (3.26) and the assertions (1)–(3), the assertion (4) also holds.Lemma 3.3 (see [12]).
L is a Fredholm map of index 0 and satisfies
(3.28)l(L)≥1.Lemma 3.4 (see [2]).
Supposeσ∈CT1 and σ'(t)<1,  for all  t∈[0,T]. Then the function t-σ(t) has an inverse function μ(t) satisfying μ∈C(R,R) with μ(a+T)=μ(a)+T.Lemma 3.5.
Assume that (H1)–(H4) hold. Let k0=∑i=1mciu(1-τi′)u, and
(3.29)Ω={x∈X|maxt∈[0,T]x(t)∈(l--δ,R0+δ),mint∈[0,T]x(t)∈(l--δ,l++δ),maxt∈[0,T]|x'(t)|<M0.},
where 0<δ<l-. Then N:Ω-→Z is a k0-set-contractive map.Proof.
The proof is similar to that of Lemma  3.3 in [12], so we omit it.Theorem 3.6.
Assume that (H1)–(H4) hold. Then (1.1) has at least two positive T-periodic solutions.Proof.
LetLx=λNx for x∈X, that is,
(3.30)x′(t)=λ[a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)(1-τi′(t))x′(t-τi(t))-h(t)ex(t)],λ∈(0,1).
Therefore, we have
(3.31)x′(t)=λ[a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)[x(t-τi(t))]'-h(t)ex(t)],λ∈(0,1).
By (3.31), we have
(3.32)[x(t)+λ∑i=1m‍ci(t)x(t-τi(t))]′=λ[a(t)-∑j=1n‍bj(t)x(t-σj(t))+∑i=1m‍ci′(t)x(t-τi(t))-h(t)ex(t)].
Integrating this identity leads to
(3.33)∫0T‍[∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci′(t)x(t-τi(t))+h(t)ex(t)]dt=∫0T‍a(t)dt.
By Lemma 3.4, we have
(3.34)μj(T+s)=T+μj(s),γi(T+s)=T+γi(s),
where t=μj(s) is the inverse function of s=t-σj(t), and t=γi(s) is the inverse function of s=t-τi(t),forallj∈{1,2,…,n}andforalli∈{1,2,…,m}.
Then(3.35)∫0T‍∑j=1n‍bj(t)x(t-σj(t))dt=∑j=1n‍∫-σj(0)T-σj(T)‍bj(μj(s))x(s)1-σj′(μj(s))ds=∑j=1n‍∫0T‍bj(μj(s))1-σj′(μj(s))x(s)ds,∫0T‍∑i=1m‍ci′(t)x(t-τi(t))dt=∑i=1m‍∫-τi(0)T-τi(T)‍ci′(γi(s))x(s)1-τi′(γi(s))ds=∑i=1m‍∫0T‍ci′(γi(s))1-τi′(γi(s))x(s)ds.
From (3.33)–(3.35), we have
(3.36)∫0T‍[a(s)-(∑j=1n‍bj(μj(s))1-σj′(μj(s))-∑i=1m‍ci′(γi(s))1-τi′(γi(s)))x(s)-h(s)ex(s)]ds=0,
which implies
(3.37)a(η)-Γ(η)x(η)-h(η)ex(η)=0,
for some η∈[0,T].
Therefore, by(H2), we have
(3.38)auΓl-x(η)-hl/Γuex(η)≥0.
By (H3) and Lemma 3.1, we have
(3.39)2lnhlΓu<x(η)≤0.
Set
(3.40)q:=2|lnhlΓu|,
then we have
(3.41)|x(t)|≤|x(η)|+∫0T‍|x′(t)|dt≤q+∫0T‍|x′(t)|dt,
which implies
(3.42)|x|0≤q+∫0T‍|x′(t)|dt.
It follows from (3.30) that
(3.43)∫0T‍|x′(t)|dt=λ∫0T‍|a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)(1-τi′(t))x′(t-τi(t))-h(t)ex(t)|dt≤λ∫0T‍|a(t)-∑j=1n‍bj(t)x(t-σj(t))-∑i=1m‍ci(t)(1-τi′(t))x′(t-τi(t))|dt+∫0T‍h(t)ex(t)dt.
By (3.36) and (H2), we have
(3.44)∫0T‍h(t)ex(t)dt≤∫0T‍a(t)dt+∫0T‍Γ(t)dt|x|0=Ta-+TΓ-|x|0.
By this and (3.43), we obtain
(3.45)∫0T‍|x′(t)|dt≤T[2a-+Γ-|x|0+∑j=1n‍b-j|x|0]+∑i=1m‍∫0T‍|ci(t)(1-τi′(t))x′(t-τi(t))|dt.
Meanwhile,
(3.46)∑i=1m‍∫0T‍|ci(t)(1-τi′(t))x′(t-τi(t))|dt=∑i=1m‍∫0T‍ci(γi(s))|x′(s)|ds≤∑i=1m‍ciu∫0T‍|x′(s)|ds.
Substituting (3.42) and (3.46) into (3.45) gives
(3.47)∫0T‍|x′(t)|dt≤T[2a-+Γ-|x|0+∑j=1n‍b-j|x|0]+∑i=1m‍ciu∫0T‍|x′(s)|ds≤T[2a-+(Γ-+∑j=1n‍b-j)(q+∫0T‍|x′(t)|dt)]+∑i=1m‍ciu∫0T‍|x′(s)|ds≤T[2a-+(Γ-+∑j=1n‍b-j)q]+[T(Γ-+∑j=1n‍b-j)+∑i=1m‍ciu]∫0T‍|x′(t)|dt.
Since
(3.48)T(Γ-+∑j=1n‍b-j)+∑i=1m‍ciu<1,
we have
(3.49)∫0T‍|x′(t)|dt<2Ta-+(Γ-+∑j=1nb-j)Tq1-T(Γ-+∑j=1nb-j)-∑i=1mciu∶=p.
Then,
(3.50)|x|0≤q+∫0T‍|x′(t)|dt≤q+p∶=R0.
Again from (3.30), we get
(3.51)|x′|0<au+∑j=1n‍bju|x|0+∑i=1m‍ciu(1-τi′)u|x′|0+hueR0.
Since ∑i=1mciu(1-τi′)u<1, we have
(3.52)|x′|0<au+∑j=1nbjuR0+hueR01-∑i=1mciu(1-τi′)u:=M0.
Choose tM,tm∈[0,T], such that
(3.53)x(tM)=maxt∈[0,T]x(t),x(tm)=mint∈[0,T]x(t).
Then, it is clear that
(3.54)x′(tM)=0,x′(tm)=0.
From this and (3.30), we obtain that
(3.55)a(tM)=∑j=1n‍bj(tM)x(tM-σj(tM))+∑i=1m‍ci(tM)(1-τi′(tM))x′(tM-τi(tM))+h(tM)ex(tM),(3.56)a(tm)=∑j=1n‍bj(tm)x(tm-σj(tm))+∑i=1m‍ci(tm)(1-τi′(tm))x′(tm-τi(tm))+h(tm)ex(tm).
It follows from (3.55) that
(3.57)a(tM)-∑i=1m‍ci(tM)(1-τi′(tM))M0-∑j=1n‍bj(tM)x(tM)-h(tM)ex(tM)≤0,
which implies
(3.58)al-∑i=1mciu(1-τi′)uM0∑j=1nbju-x(tM)-hu∑j=1nbjle-x(tM)≤0.
By the assertion (1) of Lemma 3.2, we have
(3.59)x(tM)≤u-orx(tM)≥u+.
It follows from (3.56) that
(3.60)a(tm)+∑i=1m‍ci(tm)(1-τi′(tm))M0-∑j=1n‍bj(tm)x(tm)-h(tm)ex(tm)≥0,
which implies
(3.61)au+∑i=1mciu(1-τi′)uM0∑j=1nbjl-x(tm)-hl∑j=1nbjue-x(tm)≥0.
By the assertion (3) of Lemma 3.2, we have
(3.62)l-≤x(tm)≤l+.
Hence, it follows from (3.50), (3.59), and (3.62) that
(3.63)x(tM)∈[l-,u-]∪[u+,R0],x(tm)∈[l-,l+].
Clearly, l±,u± are independent of λ. Now, let us consider QN(x) with x∈R. Note that
(3.64)QN(x)=a--∑j=1n‍b-jx-h-ex.
It follows from the assertion (2) of Lemma 3.2 that QN(x)=0 has two distinct solutions:
(3.65)u~1=x-,u~2=x+.
By the assertion (4) of Lemma 3.2, one can take v-,v+>0 such that
(3.66)u-<v-<v+<u+.
Let
(3.67)Ω1={x∈X|maxt∈[0,T]x(t)∈(l--δ,v-),mint∈[0,T]x(t)∈(l--δ,l++δ),maxt∈[0,T]|x′(t)|<M0.},Ω2={x∈X|maxt∈[0,T]x(t)∈(v+,R0+δ),mint∈[0,T]x(t)∈(l--δ,l++δ),maxt∈[0,T]|x′(t)|<M0.}.
Then Ω1,Ω2 are bounded open subsets of X. Clearly, Ωi⊂Ω(i=1,2), where Ω as defined in Lemma 3.5. It follows from Lemma 3.5 that N:Ω-i→Z is a k0-set-contractive map (i=1,2). Therefore, it follows from Lemmas 2.1 and 3.3 that N:Ω-i→Z is L-k-set contractive on Ω-i(i=1,2) with k=k0/l(L)≤k0<1.
By the assertion (4) of Lemma3.2, (3.52), (3.63), (3.65), and (3.66), it is easy to verify that Ωi satisfies the assumptions (i) and (ii) in Lemma 2.2  (i=1,2). By the assertion (2) of Lemma 3.2, a direct computation gives
(3.68)degB{JQN,Ω1∩KerL,0}=sgn(-∑j=1n‍b-j+h-ex-)=1,degB{JQN,Ω2∩KerL,0}=sgn(-∑j=1n‍b-j+h-ex+)=-1.
Here, J is taken as the identity mapping since ImQ=KerL. So far we have proved that Ωi satisfies all the assumptions in Lemma 2.2  (i=1,2). Hence, (3.1) has at least two T-periodic solutions xi*(t) and xi*∈domL∩Ω-i(i=1,2). Since Ω-1∩Ω-2=∅, xi*(i=1,2) are different. Let yi*(t)=exi*(t)(i=1,2). Then yi*(t)(i=1,2) are two different positive T-periodic solutions of (1.1). The proof is complete.Example 3.7.
Consider the following equation:(3.69)dy(t)dt=y(t)[a(t)-b(t)lny(t-σ(t))-c(t)ddtlny(t-τ(t))]-h(t),
where
(3.70)a(t)=(14.2e+14.2esin2t)ϵ,b(t)=(2.35+0.15sint)ϵ,c(t)=(0.9+0.1cost)ϵ,σ(t)≡2π,τ(t)≡4π,h=3ϵ2e+ϵ2esint,
and the constant ϵ>0. Clearly, (H1) is satisfied.
Letμ(t) be the inverse function of t-σ(t), and γ(t) be the inverse function of t-τ(t). Then we have
(3.71)μ(t)=t+2π,γ(t)=t+4π,Γ(t)=b(μ(t))1-σ′(μ(t))-c′(γ(t))1-τ′(γ(t))=b(t+2π)-c′(t+4π)=(2.35+0.25sint)ϵ.
Hence, (H2) is satisfied.
It is easy to see that(3.72)T=2π,au=ϵ2.1e,al=ϵ4.2e,a-=ϵ2.8e,bu=2.5ϵ,bl=2.2ϵ,b-=2.35ϵ,Γu=2.6ϵ,Γl=2.1ϵ,Γ-=2.35ϵ,cu=ϵ,hu=2ϵe,hl=ϵe,h-=3ϵ2e.
Therefore, we have
(3.73)hlΓu=12.6e,auΓl=14.41e,1+lnhlΓu=-ln2.6.
So,
(3.74)1+lnhlΓu<auΓl<hlΓu<1.
Hence, (H3) is satisfied.
Also, it is easy to see that(3.75)al-cu(1-τ′)uM0bu=ϵ/(4.2e)-ϵM02.5ϵ=110.5e-M02.5<1,q=2|lnhlΓu|=2(ln2.6+1)>1,p=2Ta-+(Γ-+b-)Tq1-T(Γ-+b-)-cu=πϵ/(0.7e)+18.8(ln2.6+1)πϵ1-9.4πϵ-ϵ,R0=q+p,M0=au+buR0+hueR01-cu(1-τ′)u=ϵ/(2.1e)+2.5ϵR0+2ϵeR0-11-ϵ.
Therefore, we obtain that
(3.76)R0>q>al-cu(1-τ′)uM0bu.
Noticing that limϵ→0+R0=q, we have limϵ→0+M0=0. Therefore, for some sufficiently small ϵ>0, the following inequalities hold:
(3.77)max{1+lnhubl,0}=max{-ln1.1,0}=0<al-cu(1-τ′)uM0bu,T(Γ-+b-)+cu=9.4πϵ+ϵ<1.
By (3.76)-(3.77), (H4) is also satisfied. Therefore, all necessary conditions of Theorem 3.6 are satisfied. By Theorem 3.6, (3.69) has at least two positive 2π-periodic solutions.



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*Source: 423416-2012-12-24.xml*