Dynamic behavior analysis of a prey-predator model with ratio-dependent Monod-Haldane functional response

Lemma 2.1

Any coexistence state (u, v) of(2)has a priori boundary, i.e.,

Lemma 2.2

([29, 30]). Suppose thatI − Lis invertible onW_y. Then the following conclusions hold.

1. IfLhas propertyαonW_y, then index_W(F, y) = 0.

2. IfLdoes not have propertyαonW_y, then index_W(F, y) = (−1)^σ, whereσis the sum of algebra multiplicities of the eigenvalue ofLwhich are greater than 1.

Lemma 2.3

Assumea > λ₁.

1. deg_W(I − F, D′) = 1, here deg_W(I − F, D′) is the degree ofT − F inD′ onW.

2. Ifc ≠ λ₁, then index_W(F, (0, 0)) = 0.

3. Ifc>λ1(−dθa1+mθa+θa2),then index_W(F, (θ_a, 0)) = 0.

4. Ifc<λ1(−dθa1+mθa+θa2),then index_W(F, (θ_a, 0)) = 1.

Lemma 2.4

Assumec > λ₁.

1. Ifa>λ1(bθc1+kθc),then index_W(F, (0, θ_c)) = 0.

2. Ifa<λ1(bθc1+kθc),then index_W(F, (0, θ_c)) = 1.

Theorem 2.5

1. Ifa ≤ λ₁, then(2)does’t have any coexistence state; ifa ≤ λ₁andc ≤ λ₁, then(2)does’t exist non-negative non-zero solution.

2. Ifc ≤ λ₁and for(2)there exists a coexistence state, thena>λ1,c+da1+ma>λ1.

3. Ifc > λ₁and for(2)there exists a coexistence state, thena>λ1(bθc1+mθa+θa2+kθc).

Proof

1. Suppose (u, v) is a coexistence state to (2), then (u, v) satisfies

   −Δu=u(a−u−bv1+mu+u2+kv)x∈Ω,u=0x∈∂Ω,

   and a=λ1(u+bv1+mu+u2+kv). With the aid of the comparison principle, we get a > λ₁, a contradiction. We suppose that (u, v) is a non-negative non-zero solution of (2). Obviously, u ≢ 0 and v ≡ 0, then a > λ₁ by the above proof. Similarly, we can deduce c > λ₁ as u ≡ 0 and v ≢ 0, which derives a contradiction again.

2. Let (u, v) be a coexistence state of (2). By (i), we know that a > λ₁, and (2) has the positive semi-trivial solution θ_a. Due to

   −Δu=u(a−u−bv1+mu+u2+kv)≤u(a−u)x∈Ω,u=0x∈∂Ω,

   u is a lower solution to (2). Owing to the uniqueness of θ_a, u ≤ θ_a. It follows that v meets

   −Δv=v(c−v+du1+mu+u2+kv)x∈Ω,v=0x∈∂Ω,

   then 0=λ1(−c+v−du1+mu+u2+kv)>λ1(−c−da1+ma), which gives the result.

3. Suppose (u, v) is a coexistence state of (2), then (2) has the unique positive solution θ_a with u ≤ θ_a. Similarly, c > λ₁ implies the existence of positive solution θ_c of (2) with θ_c ≤ v. According to the same method of (i), we directly obtain

   a=λ1(u+bv1+mu+u2+kv)>λ1(bθc(1+mθa+θa2+kθc)).

   Since the function bv1+mu+u2+kv has a minimum at u = θ_a and v = θ_c (for u ≤ θ_a and v ≥ θ_c). So the result holds.

Theorem 2.6

1. Ifc > λ₁anda>λ1(bθc1+kθc).Then(2)has at least a coexistence state.

2. Supposec < λ₁. Then(2)has a coexistence state if and only ifa > λ₁andc>λ1(−dθa1+mθa+θa2).

Proof

1. By Lemma 2.3-2.4 and the fixed point index theory, we have

   degW(I−F,D)−indexW(f,(0,0))−indexW(f,(θa,0))−indexW(f,(0,θc))=1.

   Then (2) has at least a coexistence state.

2. Firstly, we demonstrate the sufficiency. Due to c < λ₁, (2) does’t have the solution of the form (0, v) with v > 0. If a > λ₁ and c>λ1(−dθa1+mθa+θa2), since c < λ₁, by Lemma 2.4 and the fixed point index theory, we get

   degW(I−F,D)−indexW(f,(0,0))−indexW(f,(θa,0))=1.

   Hence (2) has at least a coexistence state.

   Conversely, we suppose that (u, v) is a coexistence state of (2). Then a > λ₁, and u < θ_a. Thanks to (u, v) satisfies

   −Δv=v(c−v+du1+mu+u2+kv)x∈Ω,v=0x∈∂Ω.

   Hence, 0=λ1(−c+v−du1+mu+u2+kv)>λ1(−c−da1+ma).

Theorem 2.7

If one of the following two conditions holds, then(2)does’t have any non-negative non-zero solution:

1. b ≥ 1 + ma + a² + kBanda ≤ c;

2. b < 1 + ma + a² + kBandc−a≥(1−b(1+ma+a2+kB))B.

Proof

1. Assume that (u, v) is a coexistence state of (2) as b ≥ 1 + ma + a² + kB and a ≤ c. Note that u(x) ≤ a by Lemma 2.1 and 1 + mu(x) + u²(x) + kv(x) > 0 in Ω̄. we get

   0=λ1(−a+u+bv1+mu+u2+kv)≥λ1(−c+v−du1+mu+u2+kv−v+bv1+mu+u2+kv+u+du1+mu+u2+kv)>λ1(−c+v−du1+mu+u2+kv−(1−b1+mu+u2+kv)v)>λ1(−c+v−du1+mu+u2+kv−(1−b1+ma+a2+kB)v)≥λ1(−c+v−du1+mu+u2+kv).

   Which deduces a contradiction.

2. Based on the proof of (i), thanks to the fact b < 1 + ma + a² + kB and c−a≥(1−b(1+ma+a2+kB))B, we can obtain the following inequality:

   0=λ1(−a+u+bv1+mu+u2+kv)>λ1(−c+v−du1+mu+u2+kv−a+c−(1−b1+ma+a2+kB)B)≥λ1(−c+v−du1+mu+u2+kv),

   which is a contradiction. The proof is completed.

Theorem 3.1

Supposec > λ₁. Then there exists a coexistence state of(4) (or(2)) bifurcate from the point (ã; 0, θ_c), anda~=λ1(bθc1+kθc).

Remark 3.2

The proof of Theorem 3.1 refers to Theorem 2 in [19], more precisely, the bifurcation branch neara~=λ1(bθc1+kθc).is determined byC¹continuous curve (a(s); ϕ(s), φ(s)) : (− δ, δ) → R × Z, for someδ > 0 such that

meetsG(a(s); ω(s), χ(s)) = 0, hereX = Z ⊕ span{(Φ, Ψ)}. Thus (a(s); U(s), V(s)) (|s| < δ) is a bifurcation solution of(4)(or(2)), hereU(s) = s(Φ + ϕ(s)), V(s) = θ_c + s(Ψ + φ(s)), Ψ=(−Δ−c+2θc)−1(dθc1+kθcΦ).

Theorem 3.3

Ifc > λ₁, then the local bifurcation solution {(a(s); U(s), V(s)) : 0 < s < δ} can be extended to the global solution which is denoted byCand unbounded by going to infinity in P.

Proof

Define

Let μ ≥ 1 be an eigenvalue of T′(a). Then

Obviously, ω ≢ 0, otherwise, ω ≡ 0, due to the fact that all eigenvalues of (−μΔ − c + 2θ_c) are greater than 0, so χ ≡ 0, a contradiction. Thus, a = a_i(μ) is the eigenvalue of the following problem

So a_i(μ) is increasing along with μ on [1,+∞) and can be ordered by

Conversely, if μ ≥ 1, we demonstrate that all eigenvalues of (−μΔ − c + 2θ_c) are greater than 0, then χ = (−μΔ − c + 2θ_c)⁻¹(dθc1+kθc). Hence, μ ≥ 1 is an eigenvalue of T′(a) if and only if a = a_i(μ) for some i = 1, 2, ….

If a < ã, then a < a₁(1) ≤ a_i(μ) for any μ ≥ 1, i ≥ 1. It follows that T′(a) has no eigenvalue greater than 1, meanwhile, index(T(a; ⋅), 0) = 1 for a < ã.

If ã < a < a₂(1), then a < a_i(μ). Since a₁(1) = ã for any μ ≥ 1, i ≥ 2, limμ→∞a₁(μ) = +∞, and a₁(μ) is increasing along with μ. Hence, there exists a unique μ₁ > 1 such that a = a₁(μ₁). Thus N(μ₁I − T′(a)) = span{(ω, χ)}, dimN(μ₁I − T′(a)) = 1, here ω > 0 is the principal eigenvalue of the following equation

here χ = (−μ₁Δ − c + 2θ_c)⁻¹(dθc1+kθcω¯).

Next, we shall demonstrate that R(μ₁I − T′(a)) ∩ N(μ₁I − T′(a)) = 0. As a matter of fact, suppose the assertion is false, we may assume that (ω, χ) ∈ R(μ₁I − T′(a)). Then there exists (ω, χ) ∈ X such that (μ₁I − T′(a))(ω, χ) = (ω, χ), i.e.,

Multiplying the above equation by ω, and integrating over Ω, by Green’s formula, we have

Thus ∫Ω1μ1(a−bθc1+kθc)ω¯2=0, a contradiction, which has proved the claim. Then the multiplicity of μ₁ is one and index(T(a; ⋅), 0) = −1 for ã < a < a₂(1). With the aid of global bifurcation theory [23], we deduce that there exists a continuum C₀ of zero points of G(a; ω, χ) = 0 in R⁺ × X bifurcating from (ã; 0, 0) and all zero points of G(a; ω, χ) nearby (ã; 0, 0) lie on the curve for which the existence was established by Theorem 3.1.

Define the maximal continuum C₁ as C₁ = C₀ − {(a(s); s(Φ + ϕ(s)), s(Ψ + φ(s))) : − δ < s < 0}, then C₁ consists of the curve {(a(s); s(Φ + ϕ(s)), s(Ψ + φ(s))):− δ < s < 0} in the neighborhood of (ã; 0, 0). Suppose C̃ = {(a; u, v) : U = ω, V = θ_c + χ, (ω, χ) ∈ C₁}. It follows that C̃ is the solution branch of (2) which bifurcates from (ã; 0, θ_c) and keeps the positive near (ã; 0, θ_c) and C̃ ⊂ P. Thus, the continuum C̃ − {(ã; 0, θ_c)} must satisfy one of the three alternatives:

1. joining up with a bifurcation point of the form (ā; 0, θ_c), which I − T′(ā) is not invertible, ã ≠ ā.

2. joining up from (ã; 0, θ_c) to ∞ in R × X.

3. containing points of the form (a; u, θ_c + v) and (a; −u, θ_c − v), here (u, v) ≠ (0, 0).

Now we claim that C̃−{(ã; 0, θ_c)} ⊂ P. Suppose C̃−{(ã; 0, θ_c)} ⊈ P, it follows that there exists (â; û, v̂) ∈ (C̃−{(ã; 0, θ_c)}) ∩ ∂P and the sequence {(a_n; u_n, v_n)} ⊂ C̃ ∩ P, u_n > 0, v_n > 0 such that (a_n; u_n, v_n) → (â; û, v̂) as n → ∞. We can obtain that û ∈ ∂P₁ or v̂ ∈ ∂P₁. If û ∈ ∂P₁, then û ≥ 0, x ∈ Ω. Thus, we find either x₀ ∈ Ω such that û(x₀) = 0 or x₀ ∈ ∂Ω such that ∂u^∂n|x0=0. Since û satisfies

By maximum principle, we can get û ≡ 0. Similarly, we can deduce that v̂ ≡ 0 for v̂ ∈ ∂P₁.

Thus, we will investigate the following three cases:

1. If (û, v̂) ≡ (θ_â, 0), then (a_n; u_n, v_n) → (â; θ_â, 0) as n → ∞. Set Vn=vn||vn||∞, then V_n meets

   −ΔVn=(c−vn+dun(1+mun+un2+kvn))Vn,Vn|∂Ω=0.(7)

   With the aid of L^P estimates and Sobolev embedding theorem, we deduce that there exists a convergent sub-sequence of V_n, which be still denoted by V_n, it follows that V_n → V in C01(Ω) as n → ∞, and V ≥ 0, ≢ 0, x ∈ Ω. Taking the limit in (7) as n → ∞, we get

   −ΔV=(c+dθa^1+mθa^+θa^2)V,V|∂Ω=0.

   By the maximum principle, we get V > 0, x ∈ Ω, which implies c=λ1(−dθa^1+mθa^+θa^2). A contradiction to c > λ₁.

2. If (û, v̂) ≡ (0, θ_c). Then (a_n; u_n, v_n) → (â; 0, θ_c) as n → ∞. Set Un=un||un||∞, hence, U_n meets

   −ΔUn=(an−un−bvn(1+mun+un2+kvn))Un,Un|∂Ω=0.(8)

   Similarly, using L^P estimates and Sobolev embedding theorem, we can get a convergent sub-sequence of U_n, which be still denoted by U_n, then U_n → U in C01(Ω) as n → ∞, and U ≥ 0, ≢ 0, x ∈ Ω. Taking limit in (8) by n → ∞, we get

   −ΔU=(a^−bθc1+kθc)U,U|∂Ω=0.

   Due to maximum principle, we get U > 0, x ∈ Ω. Thus a^=λ1(bθc1+kθc). A contradiction to â ≠ ã.

3. If (û, v̂) ≡ (0, 0), then similarly as in the method from the case above, we also arrive at a contradiction. Hence, C̃−{(ã; 0, θ_c)} ⊂ P. It follows from Lemma 2.1 that 0 ≤ U ≤ a, θ_c ≤ V ≤ c+da1+αa. With the aid of L^P estimates and Sobolev embedding theorem, we prove that there exists a constant M > 0 such that ||U||_C¹,||V||_C¹ ≤ M. Then, the global bifurcation solution branch C̃ of positive solutions of (2) bifurcating at (ã; 0, θ_c) contains points with a arbitrarily large in P.

Lemma 4.1

0 is the eigenvalue ofL₁with the largest real part, and all the other eigenvalues ofL₁lie in the left half complex plane.

Proof

The proof of Lemma 4.1 can be found in [9, 18, 24, 26], here we omit it. □

Lemma 4.2

There existsC¹−function: a → (M(a), γ(a)) ands → (N(s), π(s)), which are defined by the mapping from the neighborhood ofãand 0 intoX₁ × R, respectively, satisfying the formsγ(ã) = π(0) = 0, M(ã) = N(0) = (Φ, Ψ) and

where M(a) = (ϕ₁(a), ϕ₂(a)), N(s) = (φ₁(a)φ₂(a)). Meanwhile, γ′(ã) ≠ 0, furthermore, for |s| ≪ 1, π(s) ≠ 0, π(s) and −sa′(s)γ′(ã) have the same sign, where γ′(ã) is the derivative of γ(a) at a = ã, and a′(s) is the derivative of a(s) on the point s.

Lemma 4.3

γ′(ã) > 0.

Proof

By calculating L(a; 0, θ_c)M(a) = γ(a)M(a), for |a−ã| ≪ 1, we have

Note that |a−ã| ≪ 1, then | γ(a)| ≪ 1. Obviously, ϕ₁ ≢ 0, otherwise, ϕ₁ ≡ 0, then ϕ₂ ≡ 0, a contradiction. Thus γ(a) is an eigenvalue of (Δ+(a−bθc1+kθc)I). So Φ > 0, implies ϕ₁ = ϕ₁(a) > 0 as |a−ã| ≪ 1. It follows that γ(a) is the principal eigenvalue of (Δ+(a−bθc1+kθc)I), and γ(a) is increasing along with a for |a−ã| ≪ 1. Meanwhile, γ′(ã) ≠ 0. Thus γ′(ã) > 0. □

Lemma 4.4

a′(0) satisfies

Proof

By substituting (u(s), v(s), a(s)) into (2), differentiating on s, and let s = 0, we get

where ϕ′(0) is the derivative of ϕ on the point s = 0.

Multiplying the above equation by Φ, and applying Green’s formula and the definition of Φ, we obtain

 □

Theorem 4.5

Letσ=∫Ω(1−bmθc(1+kθc)2)Φ3dx+∫ΩbΨ(1+kθc)2Φ2dx.Ifσ > 0, then local bifurcation solution (u(s), v(s)) is stable; ifσ < 0, then local bifurcation solution (u(s), v(s)) is unstable.

Remark 4.6

In section 2, by the sufficient condition and necessary condition on coexistence states of(2)established in Theorem 2.5-2.6, we find that there exists a gap betweena>λ1(bθc1+kθc)anda>λ1(bθc(1+mθa+θa2+kθc))asc > λ₁.

Theorem 4.7

Suppose thatc>λ1and∫Ω(1−bmθc(1+kθc)2)Φ3dx<0.If there exists a sufficiently smallε > 0 and d ≪1, then coexistence state (u(s), v(s)) is non-degenerate and unstable fora ∈ (ã − ε, ã), moreover, (2)has at least two coexistence states.

Proof

Firstly, we will demonstrate the coexistence state (u(s), v(s)) is non-degenerate and unstable. To this end, we only need to prove that there exists a sufficiently small ε > 0 such that any coexistence state (u(s), v(s)) of (2) is non-degenerate for a ∈ (ã − ε, ã), and the linearized eigenvalue problem

has a unique eigenvalue μ_* and Re(μ_*) < 0 with algebra multiplicity one.

Let {ε_i > 0} and {d_i > 0} be sequences which converge to 0 by i → ∞. Owing to a = ã + a′(0)s + O(s²), we can get the sequence {ε_i > 0} and {a_i} yield a_i ∈ (ã − ε_i, ã) and s_i → 0 as i → ∞. It is easy to see that (u_i, v_i) is one solution of (2). Hence, the corresponding linearized problem (9) can be written in the following form:

where (ξ_i, η_i) ≢ (0, 0) and

Observe that, as i → ∞,Li(ξiηi) converges to

It follows that 0 is a simple eigenvalue of L₀ with the corresponding eigenfunction (ξ, η)^T = (Φ, 0)^T. Meanwhile, all the other eigenvalues of L₀ are positive and keep away from 0. Furthermore, using perturbation theory [31, 33], for large i, the operator L_i has a unique eigenvalue μ_i which is close to zero. Moreover, all other eigenvalues of L_i have positive real parts and keep away from 0. Because μ_i is simple real eigenvalue which tends to zero, we denote the corresponding eigenfunction by (ξ_i, η_i)^T which satisfies (ξ_i, η_i) → (Φ, 0) as i → ∞.

Now we claim that Reμ_i < 0 for large i. Multiplying Φ on the first equation of L_i(ξ_i,η_i)^T = μ_i(ξ_i,η_i)^T and integrating over Ω, we obtain

Multiplying two sides of the first equation of (2) with (a, u, v) = (a_i, u_i, v_i) by ξ_i and integrating, we obtain

Taking u_i = s_iΦ + O(si2) into the above equation we have