Analysis of Stochastic Delay Predator-Prey System with Impulsive Toxicant Input in Polluted Environments

Lemma 1. For system (1), define

• (a)

  If r₁₀ < r₁₁K₁/γ, then lim _t→+∞x_i(t) = 0, i = 1,2.

• (b)

  If r₁₀ > r₁₁K₁/γ and $$, then $$ and x₂(t) goes to extinction.

• (c)

  If $$, then $$,  i = 1,2.

Theorem 2. For system (3), define

• (i)

  If θ₁ < r₁₁K₁/γ, then both x₁ and x₂ go to extinction almost surely (a.s.); that is, lim _t→+∞ x_i(t) = 0  a.s.,   i = 1,2.

• (ii)

  If θ₁ > r₁₁K₁/γ and $$, then x₂(t) goes to extinction and x₁ is stable in time average a.s.; that is,

  $$ ()

• (iii)

  If $$, then both x₁ and x₂ are stable in time average a.s.

  $$ ()

Remark 3. By comparing Lemma 1 with our Theorem 2, we can see that on the one hand, if α₁ = α₂ = 0 and τ₁ = τ₂ = 0, then θ_i = r_i0, $$, i = 1,2, and our stochastic delay system (3) becomes model (1); on the other hand, our results in Theorem 2 improve that in Lemma 1. Lemma 1 shows that the superior limit is positive, while Theorem 2 reveals that the limit exists and gives the explicit form of the limit. The contribution of this paper is therefore clear.

Lemma 4. For any given initial value $$, there is a unique global positive solution x(t) = (x₁(t), x₂(t)) ^T to the first two equations of system (3) a.s.

Proof. The proof is similar to Hung [29] by defining

and hence is omitted.

Lemma 5 (see [13], [15].)System (10) has a unique positive γ-periodic solution $$, and for each solution (C₁₀(t), C₂₀(t), C_e(t)) ^T of (10), $$, $$, and $$ as t → ∞. Moreover, $$ and $$ for all t ≥ 0 if $$ and $$, i = 1,2, where

for t ∈ (nγ, (n + 1)γ] and n ∈ Z⁺. In addition,

Lemma 6 (see [34].)Suppose that x(t) ∈ C[Ω × [0, +∞), R₊].

•  

  (I) If there exist σ and positive constants σ₀,  T such that

  $$ ()

•  

  for t ≥ T, where B_i(t) are independent standard Brownian motions and β_i are constants, 1 ≤ i ≤ n, then one has the following: if σ ≥ 0, then 〈x〉 ^* ≤ σ/σ₀ a.s.; if σ < 0, then lim _t→+∞x(t) = 0  a.s.

•   

  (II) If there exist positive constants σ₀,  T and σ such that

  $$ ()

•  

  for t ≥ T, then 〈x〉 _* ≥ σ/σ₀ a.s.

Lemma 7. If $$, then the solution y(t) of system (15) obeys

Proof. By Lemma 5,

Then, for all  ε > 0, there exists T > 0 such that An application of Itô’s formula to (15) yields That is to say, we have shown that When (18) is used in (20), we can see that for t > T, Let ε be sufficiently small such that θ₁ − r₁₁K₁/γ − r₁₁ε > 0. Making use of (I) and (II) in Lemma 6 to (22) and (23), respectively, we have It then follows from the arbitrariness of ε that Substituting (17) and (25) into (20) and noting that lim _t→+∞ t⁻¹B₁(t) = 0, one can derive that Employing (20) and (21) in the expression a₂₁ln (y₁(t)/y₁(0)) + a₁₁ln (y₂(t)/y₂(0)) yields In view of (25), we get By (17), (26), (27), and (28), for all  ε > 0, there exists T > 0 such that, for t ≥ T, If $$, then we can choose ε sufficiently small such that $$. Then, by (29) and (I) in Lemma 6, we obtain lim _t→+∞ y₂(t) = 0  a.s. If $$, then we can choose ε sufficiently small such that $$. An application of (I) and (II) in Lemma 6 to (29) and (30), respectively, makes one observe that Therefore, using the arbitrariness of ε results in This completes the proof.

Proof of Theorem 2. Applying Itô’s formula to (3) leads to

(i) It follows from (17) and (33) that

for sufficiently large t. Since θ₁ − r₁₁K₁/λ < 0, then we can choose ε sufficiently small such that θ₁ − r₁₁K₁/λ + ε < 0. Then, by (I) in Lemma 6, When (36) is used in (34), one can see that for sufficiently large t, where ε > 0 obeys −θ₂ + ε < 0. In view of Lemma 6 again, lim _t→+∞x₂(t) = 0,  a.s.

(ii) By the stochastic comparison theorem [40], one can observe that

Note that θ₁ > r₁₁K₁/γ and $$; it then follows from Lemma 7 that lim _t→+∞y₂(t) = 0,  a.s. Making use of (38) gives lim _t→+∞x₂(t) = 0,  a.s. Thus, for all  ε > 0, there exists T > 0 such that, for t ≥ T, Substituting the above inequalities into (33) and then using (18), we obtain Let ε be sufficiently small such that θ₁ − r₁₁K₁/γ − ε > 0, and then, applying (I) and (II) in Lemma 6 to (40) and (41), respectively, one can see that An application of the arbitrariness of ε gives

(iii) Clearly, $$ implies θ₁ > r₁₁K₁/γ, and then, by Lemma 7,

Thus, similar to the proof of (28), we get Therefore, by (26), (28), and (38), we can observe that Employing (33) and (34) in the expression a₂₁ln (x₁(t)/x₁(0)) + a₁₁ln (x₂(t)/x₂(0)) yields When (18), (46) and (47), are used in (48), one can obtain for sufficiently large t, where ε > 0 obeys $$. It then follows from (II) in Lemma 6 that By virtue of the arbitrariness of ε, we can see that Consequently, for every $$, there is T > 0 such that Substituting the above inequality into (33) and then using (18) and (47), one can see that for sufficiently large t. Since $$, and then, by Lemma 6 and the arbitrariness of ε, one can observe that When this inequality, (18) and (47), are used in (34), we can see that for sufficiently large t. Then, it follows from Lemma 6 and the arbitrariness of ε that Substituting the above inequality and (18) into (33), we get for sufficiently large t. By (II) in Lemma 6 and the arbitrariness of ε again, we obtain Then, the required assertion follows from (51), (54), (56), and (58).