The Existence of Positive Solutions for a Nonlinear Sixth-Order Boundary Value Problem

Definition 2.1. We say μ is a generalized eigenvalue of the linear problem

if (2.1) has nontrivial solutions.

Lemma 2.2. For u ∈ X, then ∥u∥_∞ ≤ ∥u^′∥_∞ ≤ ∥u^′′∥_∞ ≤ ∥u^′′′∥_∞ ≤ ∥u⁽⁴⁾∥_∞ ≤ ∥u∥_X.

Proof. (1) By u(0) = u(1), there is a ξ ∈ (0,1) such that u^′(ξ) = 0, and so $$. Hence $$. By u^′′(0) = u^′′(1), there is a η ∈ (0,1), which makes u^′′′(η) = 0, thereby we come to $$. And we have $$.

(2) Because of u(0) = 0, we come to $$. Correspondingly, ∥u∥_∞ ≤ ∥u^′∥_∞ and we can obtain ∥u^′′∥_∞ ≤ ∥u^′′′∥_∞ for the same reason.

(3) By the definition of X, we know |u⁽⁴⁾(t)| ≤ ɛe(t) ≤ ɛ, that is, |u⁽⁴⁾|_∞ ≤ ɛ. Correspondingly, we come to |u⁽⁴⁾|_∞≤|u|_X.

Based on the combination of (1), (2) and (3), the conclusion can be reached and the lemma is thus proved.

Theorem 2.3. Assume that (A₄) holds, let r(t) be the spectral radius of T, then Problem (2.1) has an algebraically simple eigenvalue, μ₁(A, B, C) = (r(t)) ⁻¹, with a positive eigenfunction $$. Moreover, there is no other eigenvalue with a positive eigenfunction.

Proof. It is easy to check that Problem (2.1) is equivalent to the integral equation u(t) = μTu(t).

We define T : X → X.

In fact, for u ∈ X, we have

Combining this with the fact [A(s)u(s) − B(s)u^′′(s) + C(s)u⁽⁴⁾(s)] ∈ Y, it can be concluded that where Thus, there exist corresponding constants ρ₁, ρ₂, which make Consequently, we obtain Tu ∈ X, thus T(X)⊆X. The assertion is proved.

If u ∈ P, then [A(s)u(s) − B(s)u^′′(s) + C(s)u⁽⁴⁾(s)] ≥ 0,  s ∈ [0,1], and correspondingly,

So, Tu ∈ P, and correspondingly T(P)⊆P.

Because T(X) ⊂ C⁶[0,1]∩X, and C⁶[0,1]∩X is compactly embedded in X, thus we obtain that T : X → X is completely continuous.

Next, we will prove that T : X → X is strongly positive.

(1) For any u ∈ P∖{0}, if A(t) > 0 on [0,1], then there exists a constant k > 0 such that

It is easy to verify that there exists r₁ > 0, such that T₁u(t) ≥ r₁e(t) on [0,1]. Thus Tu(t) ≥ r₁e(t).

For any u ∈ P∖{0}, if B(t) > 0 on [0,1], then there exists a constant k₁ > 0 such that

Then there exists r₂ > 0, which makes T₂u(t) ≥ r₂e(t) on [0,1]. Thus, Tu(t) ≥ r₂e(t).

For any u ∈ P∖{0}, if C(t) > 0 on [0,1], similarly, we can verify that there exists r₃ > 0, which makes Tu(t) ≥ r₃e(t) on [0,1].

Hence we obtain Tu(t) ≥ re(t), for all t ∈ [0,1], in which r = min {r₁, r₂, r₃}.

(2) Thanks to the definition of T, we come to

For any u ∈ P∖{0}, if A(t) > 0 on [0,1], there exist k > 0 and r₄ > 0 such that For any u ∈ P∖{0}, if B(t) > 0 on [0,1], there exist k₁ > 0 and r₅ > 0 such that To u ∈ P∖{0}, if C(t) > 0 on [0,1], there exist k₂ > 0 and r₆ > 0 such that Hence we obtain (Tu) ⁽⁴⁾(t) ≥ r^′e(t), for all t ∈ [0,1], where r^′ = min {r₄, r₅, r₆}.

(3) It is easy to come to

for all u ∈ P∖{0}, if A(t) > 0 on [0,1], there exist constants k > 0 and r₇ > 0 such that for all u ∈ P∖{0}, if B(t) > 0 on [0,1], there exist constants k₁ > 0 and r₈ > 0 such that for all u ∈ P∖{0}, if C(t) > 0 on [0,1], there exist constants k₂ > 0 and r₉ > 0 such that Hence we obtain −(Tu) ^′′(t) ≥ r^′′e(t), for all t ∈ [0,1], where r^′′ = min {r₇, r₈, r₉}.

By (1), (2), and (3), we have $$.

According to Krein-Rutman theorem, we know that T has a single algebraic eigenvalue r(T) > 0 which corresponds to the eigenvector $$. Furthermore, there is no other eigenvalues with corresponding positive eigenfunctions. Correspondingly, μ₁(A, B, C) = (r(t)) ⁻¹ is an algebraic single eigenvalue of Problem (2.1) with a corresponding positive eigenvector $$, and there is no any other eigenvalues which have corresponding positive feature vector. The theorem is thus proved.

Theorem 3.1. Let (A₁), (A₂), and (A₃) hold. Assume that either

Then (1.4) has at least one positive solution.

Proof. Define that L : D(L) → Y by L(u): = −u⁽⁶⁾, u ∈ D(L), where

It is easy to verify that L⁻¹ : Y → X is compact.

Let ζ, ξ ∈ C([0,1]×[0, ∞)×(−∞, 0]×[0, ∞)), and satisfy

Obviously, by the condition (A₁), we have Let It is easy to see the fact that $$ is monotone, not decreasing and

Let us consider

as a bifurcation problem from the trivial solution u ≡ 0.

It is easy to verify that (3.7) is equivalent to equation

From the proof of Theorem 2.3, we know that $$ is strong positive and compact: Define F : [0, ∞) × X → X by then by (3.4) and Lemma 2.2, we know that when ∥u∥_X → 0, Based on Theorem  2 in literature [11], we come to the following conclusion.

There exists an unbounded connected subset Γ for the following set:

such that (μ₁(a, b, c), 0) ∈ Γ.

Next, we will verify the result of this theorem.

Obviously, any solution of (3.7), such as (1, u), is the solution of problem (1.4). If we want to verify Γ passing through hyperplane {1} × X, we only need to verify that Γ connects (μ₁(a, b, c), 0) and (μ₁(d, m, n), ∞).

Let (η_n, y_n) ∈ Γ and satisfy

Since (0,0) is the only solution of (3.7) when λ = 0 and Γ∩({0} × X) = ⌀, we have η_n > 0 for all n ∈ N.

By (3.13), when n → ∞, ∥y_n∥_X → ∞, and we divide the two sides of equation

with ∥y_n∥_X at the same time. Let $$, then $$ is bounded in X, and we have already known that L⁻¹ is compact and the bounded set is mapped as bicompact set, so there exists convergent subsequence in $$ we might as well still mark it as $$; and satisfy Furthermore, because according to (3.6) and Lemma 2.2, we obtain Hence there is where $$. Hence $$. Combined with Theorem 2.3, it is easy to obtain To sum up, Γ connects (μ₁(a, b, c), 0) with (μ₁(d, m, n), ∞).

Thanks to the Lemma  2.1 in [8], we only need to verify that nonlinear operator R(λ, u) has linear function V, and there exists (η, y)∈(0, ∞) × P such that ∥y∥_X = 1 and ηVy ≥ y. It follows from (A₃) that there exist a₀, b₀, c₀ ∈ [0, ∞) such that $$, and

To u ∈ X, let Then V is the linear function of R(λ, u).

Again, as

that is, by the Lemma  2.1 in [8], we obtain The conclusion is thus proved.