On Six Solutions for m-Point Differential Equations System with Two Coupled Parallel Sub-Super Solutions

Theorem 1.1. Suppose that $$ holds, u₁ and u₂ are two strict lower solutions of (1.2), v₁ and v₂ are two strict upper solutions of (1.2), and u₁ < v₁,   u₂ < v₂,   u₂≰v₁,   u₁≰v₂. Moreover, assume

for some ξ₀ > 0. Then, the three-point boundary value problem (1.2) has at least six solutions.

Definition 2.1 (see [29].)Let B be a positive linear operator. The operator B is said to satisfy condition H, if there exist $$, and δ > 0 such that (2.1) holds, and B maps P into P(g^*, δ).

Lemma 2.2 (see [10].)Suppose that $$. Then, the BVP

has Green′s function where

Theorem 3.1. Assume (H₀), (H₁)–(H₄) hold, then BVP (1.1) has at least six distinct continuous solutions.

Proof. It is easy to check that BVP (1.1) is equivalent to the following integral equation systems:

where G(t, s) is defined as in Lemma 2.2. By (H₀), we know that G(t, s) ≥ 0,   for  all  t, s ∈ [0,1].

Let E = C[0,1] × C[0,1], define the norm in E as ∥(φ, ψ)∥ = ∥φ∥+∥ψ∥. Then, E is a Banach space with this norm. Let P = {(φ, ψ) ∈ E∣φ(t) ≥ 0, ψ(t) ≥ 0, for  all  t ∈ [0,1]}, Q = {φ ∈ C[0,1]∣φ(t) ≥ 0,    for  all    t ∈ [0,1]}. Then, P = Q × Q is a normal and solid cone. Set T : E → E, such that

it is clear that the solutions of (1.1) are equivalent to the fixed points of T.

Set 0 < ξ₁ ≤ λ₁, let

where H₁ = (λ₁ + ξ₁)H, then T = Kf. It is easy to see that H is a strongly positive completely continuous operator, and it follows from G(t, s) ≥ 0 and the continuity of G(t, s) that K is a strongly positive completely continuous operator. Since f_i,   g_i : R¹ → R¹(i = 1,2) are strictly increasing continuous functions, we know that f is a strictly increasing continuous bounded operator. By T = Kf, we can prove that T is completely continuous. We infer from the increasing properties of K and f that T is increasing.

Let φ₁ ≡ c₁,   ψ₁ ≡ −k,     φ₂ ≡ k,     ψ₂ ≡ c₂, φ₃ ≡ −k,     ψ₃ ≡ c₃,     φ₄ ≡ c₄,     ψ₄ ≡ k, then (φ_i, ψ_i)(i = 1,2, 3,4) satisfy

By (H₂)(iii) and (H₃)(i), we have It follows from (H₂)(ii),(iv), and the increasing property of g₂ that Equations (3.2), (3.5), and (3.6) imply that Similarly, by (H₁)–(H₃), we obtain

By [20, Lemma 3], we get that H₁ satisfies condition H. Therefore, there exist $$, such that

By the definition of spectral radius of completely continuous operator, we have r(K) = r(H₁), and combining (3.9), we infer that Let $$, then j^* ∈ P^*∖{θ}. According to the proof in [18], we can get that K satisfies condition H.

By condition (H₄), we obtain that there exists C > 0, such that

Equations (3.12)–(3.15) imply Since f₁(φ), g₁(ψ) are continuous in [−(c₁ + k), C], so they are bounded, then there exists h > 0 such that By virtue of (3.18) and the increasing properties of f₂ and g₂, one shows

In addition, if φ, ψ satisfy φ ≥ C,    − ∥(φ₁, ψ₁)∥≤ψ ≤ C, then it follows from (3.12), (3.18), and the increasing property of g₂ that

where Similarly, if φ,   ψ satisfy ψ ≥ C,    − ∥(φ₁, ψ₁)∥≤φ ≤ C, then combining the increasing property of f₂ with (3.14) and (3.18), we know that where d₂ = (λ₁ + ξ₁)C + h+|f₂(−∥(φ₁, ψ₁)∥)|. Let d = max {d₁, d₂}. By (3.21) and (3.22), we get that if φ ≥ C,    − ∥(φ₁, ψ₁)∥≤ψ ≤ C or ψ ≥ C,    − ∥(φ₁, ψ₁)∥≤φ ≤ C, it is obvious that It follows from (3.16), (3.19), and (3.23) that In a similar way, from (3.12) and (3.14), we can show that there exists e > 0 such that Let a = max {d, e}. It follows from (3.24) and (3.25) that if φ, ψ ≥ −∥(φ₁, ψ₁)∥ or φ, ψ ≥ −∥(φ₃, ψ₃)∥, then In a similar way, from (3.13) and (3.15), we can prove that there exists constant $$ such that if φ, ψ ≤ ∥(φ₂, ψ₂)∥ or φ, ψ ≤ ∥(φ₄, ψ₄)∥, then Let P((φ₁, ψ₁)) = {(u, v) ∈ E∣(u, v)≥(φ₁, ψ₁)}, P((φ₃, ψ₃)) = {(u, v) ∈ E∣(u, v)≥(φ₃, ψ₃)}, P((φ₂, ψ₂)) = {(u, v) ∈ E∣(u, v)≤(φ₂, ψ₂)}, P((φ₄, ψ₄)) = {(u, v) ∈ E∣(u, v)≤(φ₄, ψ₄)}, for  all  (φ, ψ) ∈ P((φ₁, ψ₁)) ∪ P((φ₃, ψ₃)), then φ(t), ψ(t)≥−∥(φ₁, ψ₁)∥ or φ(t), ψ(t)≥−∥(φ₃, ψ₃)∥, for  all  t ∈ [0,1]; therefore, in virtue of expression of B, f, and (3.26), we have where $$. Since 0 < ξ₁ ≤ λ₁, one can show

This implies that there exist (u₁, v₁) ∈ E and 0 < ξ₂ ≤ r⁻¹(K) such that

Similarly, we get by (3.27) that there exist (u₂, v₂) ∈ E and 0 < ξ₃ ≤ r⁻¹(K) such that We get by (3.7) that Let E₁ = {(u, v) ∈ E∣(φ₁, ψ₁)≤(u, v)≤(φ₂, ψ₂)}. Since P is normal, then E₁ is bounded (see [28]). Choose M₁ > 0 such that Let Ω₁ = {(u, v) ∈ P((φ₁, ψ₁))∣∥(u, v)−(φ₁, ψ₁)∥<M₁, (u, v)≱(φ₃, ψ₃)}, then E₁ ⊂ Ω₁ and Ω₁ is a bounded open set. By the proof of Theorem 2.1 in [18], we can show that where $$ satisfies $$.

Equation (3.34) implies that T has no fixed point on ∂Ω₁. It is easy to prove that P((φ₁, ψ₁)) is a retract of E, which together with (3.32) implies that the fixed point index i(T, Ω₁, P((φ₁, ψ₁))) over Ω₁ with respect to P((φ₁, ψ₁)) is well defined, and a standard proof yields

Set $$, then for any $$, we have $$. It follows from (3.34) that H(t, (u, v)) ≠ (u, v),   for  all  (t, (u, v))∈[0,1] × ∂Ω₁, and by (3.35) and the homotopy invariance of the fixed point index, we get Let W₁ = {(u, v) ∈ P((φ₁, ψ₁))∣(u, v)≪(φ₂, ψ₂)}. By means of usual method (see [30]), we get that It is evident that A has no fixed point on ∂W₁, by (3.36), (3.37), and the additivity of the fixed point index, we have Set E₂ = {(u, v) ∈ E∣(φ₃, ψ₃)≤(u, v)≤(φ₄, ψ₄)}, and choose M₂ > 0 such that Let Similarly to the proof of (3.37) and (3.38), we get that Choose M₃, M₄ > 0 such that Set By virtue of (3.31) and the same method as that for (3.34), we have By (3.44), similarly to the proof of (3.38), we can prove that Equations (3.37)–(3.41), (3.45) imply that T has at least six distinct fixed points, that is, the system of differential equations (1.1) has at least six solution in C[0,1] × C[0,1].