Two Positive Periodic Solutions for a Neutral Delay Model of Single-Species Population Growth with Harvesting

Lemma 2.1 (see[8], Proposition XI.2.). Let L be a closed Fredholm mapping of index zero, and let $$ be k′-set contractive with

Then $$ is a L-k-set contraction with constant k = k′/l(L) < 1.

Lemma 2.2. Let L be a Fredholm mapping of index zero, and let $$ be L-k-set contractive on $$. Suppose that

• (i)

  Lx ≠ λNx for every x ∈ dom L∩∂Ω and every λ ∈ (0,1);

• (ii)

  QNx ≠ 0 for every x ∈ ∂Ω∩Ker L;

• (iii)

  Brouwer degree deg _B(JQN, Ω∩Ker L, 0) ≠ 0.

Then Lx = Nx has at least one solution in $$.

Theorem 3.1. In addition to (H₁), (H₂),   (H₃), assume further that the following condition holds:

•  

  (H₄) $$.

Then (1.2) has at least two positive T-periodic solutions.

Lemma 3.2 (see [11].)L is a Fredholm map of index 0 and satisfies

Lemma 3.3. Under the assumptions of Theorem 3.1, let

where and 0 < δ < l⁻ such that Then $$ is a k₀-set-contractive map.

Proof. The proof is similar to that of lemma  3.3 in [9], but for the sake of completeness we give the proof here. Let $$ be a bounded subset and let η = Γ_X(A). Then for any ε > 0, there is a finite family of subsets {A_i} with A = ⋃_i A_i and diam₁(A_i) ≤ η + ε.

Set the following:

Now it follows from the fact that g(t, x, x₁, x₂) is uniformly continuous on any compact subset of R × R³, and from the fact A and A_i are precompact in $$ with norm |·|₀, that there is a finite family of subsets {A_ij} of A_i such that A_i = ⋃_j A_ij with for any x, u ∈ A_ij. Therefore, for x, u ∈ A_ij we have That is Γ_Y(N(A)) ≤ k₀Γ_X(A). The proof is complete.

Lemma 3.4. If the assumptions of Theorem 3.1 hold, then every solution x ∈ X of the problem

satisfies

Proof. Let Lx = λNx for x ∈ X, that is,

Therefore, we have where c₀(t) = c(t)/(1 − τ^′(t)).

By (3.19), we have

Integrating this identity leads to From (3.20),(3.21), we have By (3.21), we have It follows that for some ξ ∈ [0, T].

Therefore, we have

From (3.22) and (3.25), we see that Hence, we have x(t) < R₁.

Let s = t − τ(t). It follows from (3.19) that

Then from (3.27) and the inequality x(t) < R₁, we obtain that So that Since $$, we have Choose t_M,   t_m ∈ [0, T], such that Then, it is clear that From this and (3.27), we obtain that It follows from (3.33) that By (H₃), we have It also follows from (3.33) that By (H₃), we have Similarly, it follows from (3.34) that By (H₃), we have Hence, it follows from (3.36), (3.38), and (3.40) that From the above inequality and (3.18), we obtain that So that Since $$, we have The proof is complete.

The Proof of Theorem 3.1 Clearly, l^±, u^± are independent of λ. Now, let us consider QN(x) with x ∈ R. Note that

It is easy to see that QN(x) = 0 has two distinct solutions: By (3.8), one can take v⁻, v⁺ > 0 such that Let Then Ω₁  and  Ω₂ are bounded open subsets of X. Clearly, Ω_i ⊂ Ω    (i = 1,2). It follows from Lemma 3.3 that $$ is a k₀-set-contractive map (i = 1,2). Therefore, it follows from Lemmas 2.1 and 3.2 that $$ is L-k-set contractive on $$ with k = k₀/l(L) ≤ k₀ < 1.

It follows from (3.8) and (3.46) that $$. From (3.8), (3.47) and Lemma 3.4, it is easy to see that $$ and Ω_i satisfies (i) in Lemma 2.2 for i = 1,2. Moreover, QN(x) ≠ 0 for x ∈ ∂Ω_i∩Ker L    (i = 1,2).

A direct computation gives the following:

Here, J is taken as the identity mapping since ImQ = Ker L. So far we have proved that Ω_i satisfies all the assumptions in Lemma 2.2  (i = 1,2). Hence, (3.2) has at least two T-periodic solutions: $$ and $$. Obviously, $$ are different. Let $$. Then $$ are two different positive T-periodic solutions of (1.2). The proof is complete.

Example 3.5. Take the following:

where the constant ϵ > 0.

Clearly, we have

Therefore, we have $$. Moreover, it is easy to see that R₁, M₀,   and  l⁺ are bounded with respect to ϵ. Hence, for some sufficiently small ϵ > 0, we have In this case, all necessary conditions of Theorem 3.1 hold. By Theorem 3.1, (1.2) has at least two positive 2π-periodic solutions.