Qualitative Analysis of a Diffusive Ratio-Dependent Holling-Tanner Predator-Prey Model with Smith Growth

Theorem 1. All the solutions of model (6) are nonnegative and defined for all t > 0. Furthermore, the nonnegative solution (N, P) of model (6) satisfies

Proof. The nonnegativity of the solution of model (6) is clear since the initial value is nonnegative. We only consider the latter of the theorem.

Note that N satisfies

Let z(t) be a solution of the ordinary differential equation:

As a result, for any ε > 0, there exists t₀ > 0, such that N(x, t) ≤ K + ε for all $$ and t ≥ t₀. Hence, P(x, t) is a lower solution

Let P(t) be the unique positive solution of problem

Definition 2 (see [24].)The spatial model (6) is said to have the persistence property if for any nonnegative initial data (N₀(x), P₀(x)), there exists a positive constant ε = ε(N₀, P₀), such that the corresponding solution (N, P) of model (6) satisfies

Theorem 3. If m(1 + c) < ar, then model (6) has the persistence property.

Proof. Let N(x, t) be an upper solution of the following problem:

Let N(t) be the unique positive solution to the following problem:

Due to m(1 + c) < ar, we have that lim _t→∞w(t) = K(ar − m(1 + c))/ar. By comparison, it follows that

Similarly, by the second equation of model (6), we have that P(x, t) is an upper solution of problem

Let P(t) be the unique positive solution to the following problem:

Theorem 4. Assume that

Proof. Define $$ by

For each i = 0,1, 2, …, X_i is invariant under the operator ℒ, and λ is an eigenvalue of this operator on X_i if and only if it is an eigenvalue of the following matrix:

In view of (27) and (28), we have det (A_i) > 0 > tr (A_i) for any i ≥ 0. Therefore, the eigenvalues of the matrix A_i have negative real parts.

In the following, we prove that there exists δ > 0 such that

Let λ = μ_iξ, then

By the Routh-Hurwitz criterion, it follows that the two roots ξ₁, ξ₂ of $$ all have negative real parts. Thus, let $$, we have that $$. By continuity, we see that there exists i₀ such that the two roots ξ_i1, ξ_i2 of $$ satisfy $$,  $$, for all i ≥ i₀. In turn, $$, for all i ≥ i₀.

Let $$, then $$ and (33) hold for $$. Consequently, the spectrum of ℒ which consists of eigenvalues, lies in {Reλ ≤ −δ}. In the sense of [25], we obtain that the positive constant solution E^* = (N^*, P^*) of model (6) is uniformly asymptotically stable. This ends the proof.

Theorem 5. Assume that the following hold:

• (A1)

  m(1 + c) < ar;

• (A2)

  h(K + cN^*)(N^* + 2P^*)+(K + cN^*)(N^* + aP^*)(a + h) ≤ 2ηr(N^* + aP^*)(a + h);

• (A3)

  hKN^* ≤ η(2h − 1)(N^* + aP^*)(a + h),

Proof. In order to give the proof, we need to construct a Lyapunov function. Define

Set φ = N − N^*, ϕ = P − P^*. We have

By virtue of Theorems 1 and 3 and under the assumption of Theorem, we have

As a result, we have I(t) ≤ 0. Thus dE(t)/dt ≤ 0, which implies the desired assertion. The proof is completed.

Lemma 6 (Harnack’s inequality [26]). Assume that $$ and let $$ be a positive solution to

Lemma 7 (maximum principle [27]). Let Ω be a bounded Lipschitz domain in ℝ² and $$.

• (a)

  Assume that $$ and satisfies

  $$ (43)

•   

  If $$, then g(x₀, w(x₀)) ≥ 0.

• (b)

  Assume that $$ and satisfies

  $$ (44)

•  

  If $$, then g(x₀, w(x₀)) ≤ 0.

Theorem 8. For any positive solution (N, P) of model (9),

Proof. Assume that (N, P) is a positive solution of model (9). Set

Since 0 < P(x)≤(1/h)∥N(x)∥_∞, we have P(x) ≤ K/h in $$.

Theorem 9. Let d be a fix positive constant. Then there exists positive constant $$ such that if d₁, d₂ > d, any positive solution (N, P) of model (6) satisfies

Proof. Let

By Lemma 7, it is clear that

Since m(K + cN(x₀))/(N(x₀) + aP(x₀)) ≤ C with C > 0, then, by virtue of (52), we derive

Theorem 10. Let D > s/μ₁ be a fixed positive constant. Then there exists a positive constant d^* = d^*(Λ, D) such that model (9) has no positive nonconstant solution provided that d₁ > d^* and d₂ > D.

Proof. Let (N, P) be any positive solution of model (9) and denote $$. Then

By the ε-Young inequality and the Poincaré inequality, we obtain that

In view of d₂ > D > s/μ₁, we can find a sufficiently small ϵ₀ > 0 such that d₂μ₁ ≥ s + ϵ₀M. Let d^* = (1/μ₁)(r + M/ϵ), then

Lemma 11. Assume that, for all i ≥ 0, the matrix μ_iI − 𝒜 is nonsingular, then

Theorem 12. Assume that m(ar + cm + 2hr − acr) > r²(a + h) ² and $$. If μ⁻ ∈ (μ_i, μ_i+1) and μ⁺ ∈ (μ_j, μ_j+1) for some 0 ≤ i < j, and $$ is odd, then model (9) has at least one nonconstant solution.

Proof. By Theorem 10, we can fix $$ and $$ such that model (9) with diffusion coefficients $$ and $$ has no nonconstant solutions.

By virtue of Theorems 8 and 9, there exists a positive constant $$,  $$ such that $$,  $$.

Set

Since Ψ(w, t) is compact, the Leray-Schauder topological degree deg (I − Ψ(w, t), ℳ, 0) is well defined. From the invariance of Leray-Schauder degree at the homotopy, we deduce

Clearly, I − Ψ(w, 1) = ℱ. Thus, if model (9) has no other solutions except the constant one w₀, then Lemma 11 shows that

On the contrary, by the choice of $$ and $$, we have that w₀ is the only solution of Ψ(w, 0) = 0. Furthermore, we have

From (79)–(81), we get a contradiction, and the proof is completed.