Multiple Positive Solutions for Singular Semipositone Periodic Boundary Value Problems with Derivative Dependence

Lemma 2.1.   G(t, s) has the following properties:

• (G¹)

  G(t, s) is continuous in t and s for all t, s ∈ [0,2π];

• (G²)

  G(t, s) > 0 for all (t, s)∈[0,2π]×[0,2π],  G(0, s) = G(2π, s) and ∂G/∂t| _(0,s) = ∂G/∂t|_(2π,s);

• (G³)

  denote $$ and $$, then l₂ > l₁ > 0;

• (G⁴)

  there exist functions h, H ∈ C²[0,2π] such that

  $$ ()

where α,   β,   c,   d are constants, H,   h are independent solutions of the linear differential equation u^′′ + a(t)u(t) = 0, and H′(t)h(t) − h′(t)H(t) = 1;

• (G⁵)

  $$ is bounded on [0,2π]×[0,2π].

Denote $$, then l₃ > 0.

Remark 2.2. Using paper [5], we can get G(t, s) when a(t) ≡ m² and m ∈ (0, 1/2), obtaining

Let E = {u ∈ C¹[0,2π] : u(0) = u(2π), u^′(0) = u^′(2π)} with norm $$, where $$. Then (E, ∥·∥) is a Banach space. Let $$, from Lemma 2.1, we know that σ, L are both constants and 0 < σ < 1, L > 0.

Define

It is easy to conclude that K is a cone of E and Ω_r is an open set of E.

Lemma 2.3 (see [12].)Let E be a Banach space and P a cone in E. Suppose Ω₁ and Ω₂ are bounded open sets of E such that $$ and suppose that $$ is a completely continuous operator such that

• (1)

  $$ and u ≠ λAu for u ∈ P∩∂Ω₁,   λ ≥ 1; u ≠ λAu for u ∈ P∩∂Ω₂,  0 < λ ≤ 1,   or

• (2)

  $$ and u ≠ λAu for u ∈ P∩∂Ω₂,   λ ≥ 1; u ≠ λAu for u ∈ P∩∂Ω₁,  0 < λ ≤ 1.

Then A has a fixed point in $$.

For convenience, let us list some conditions for later use.

• (H₀)

  a(t) ∈ Λ,   f : (0,2π) × (0, +∞) × R → R is continuous and there exists a constant M > 0 such that

  $$ ()

where g ∈ C((0,2π), R⁺),   h ∈ C((0, +∞) × R, R⁺), and $$;

• (H₁)

  there exist r₁ > σ⁻¹2πMl₂ and a(t) ∈ L[0,2π] with $$ such that

  $$ ()

• (H₂)

  there exists R₁ > r₁ such that

  $$ ()

where $$;

• (H₃)

  there exists [α^*, β^*]⊂(0,2π) such that

  $$ ()

Theorem 3.1. Assume that conditions (H₀)–(H₃) are satisfied, then PBVP (1.1) has at least two positive solutions u₁, u₂ ∈ C¹[0,2π]∩C²(0,2π) such that r₁ < ∥u₁ + Mω∥ < R₁ < ∥u₂ + Mω∥, where $$.

Proof. We consider the following PBVP:

It is easy to see that if u ∈ C¹[0,2π]∩C²(0,2π) and r₁ < ∥u∥ < R₁ is a positive solution of PBVP (3.1) with u(t) > Mω(t) for t ∈ [0,2π], then x(t) = u(t) − Mω(t) is a positive solution of PBVP (1.1) and r₁ < ∥x + Mω∥ < R₁.

As a result, we will only concentrate our study on PBVP (3.1).

Define an operator T : K∖{θ} → E by

where G(t, s) is the Green function to problem (2.3).

•  

  (1) We first show that $$ is completely continuous for any R > r₁.

For any $$, from (H₁), we have u(t) − Mω(t) ≥ σr₁ − 2πMl₂ > 0. So, by Lemma 2.1 and (3.2),

From (3.5), we have $$. Therefore, (Tu)(t) ≥ σ∥Tu∥, $$.

Assume that $$ with ∥u_n − u_*∥ → 0,   n → +∞. Thus, from (H₁), we have

where $$.

Lemma 2.1 and Lebesgue-dominated convergence theorem guarantee that

So, $$ is continuous.

For any bounded $$, From Lemma 2.1 and (H₁), for any u ∈ D, we have

which means the functions belonging to {(TD)(t)} and the functions belonging to {(TD)^′(t)} are uniformly bounded on [0,2π]. Notice that which implies that the functions belonging to {(TD)(t)} are equicontinuous on [0,2π]. From Lemma 2.1, we have where α, β, c, d are constants, h, H ∈ C²[0,2π] are independent solutions of the linear differential equation u^′′ + a(t)u(t) = 0, and H′(t)h(t) − h′(t)H(t) = 1.

It is easy to see that $$ is continuous in t and s for 0 ≤ s ≤ t ≤ 2π and 0 ≤ t < s ≤ 2π. So, for any t₁,   t₂ ∈ [0,2π], t₁ < t₂, we have

Therefore,

Thus, the functions belonging to {TD^′(t)} are equicontinuous on [0,2π]. By Arzela-Ascoli theorem, TD is relatively compact in C¹[0,2π].

Hence, $$ is completely continuous for any R > r₁.

•  

  (2) We now show that

  $$ ()

For any $$, we have

From (H₁) and (3.2),

Suppose that there exist λ₀ ≥ 1 and $$ such that u₀ = λ₀Tu₀, that is, for t ∈ [0,2π],

This is in contradiction with $$ and (3.13) holds.

•  

  (3) Next, we show that

  $$ ()

Suppose this is false, then there exist λ₀ ∈ (0,1] and $$ with u₀ = λ₀Tu₀, that is, for t ∈ [0,2π], we have

From (H₂), we have

Therefore, by (3.18), (3.19), and (H₂), it follows that Thus, ∥u∥ < R₁. This is in contradiction with $$ and (3.17) holds.

•  

  (4) Choose $$. From (H₃), there exists $$ such that

  $$ ()

Now, we show that

where R = (R₂ + 2πMl₂)σ⁻¹.

For any u ∈ K∩∂Ω_R, we have

This and (3.21) together with (3.2) imply

Suppose that there exist λ₀ ≥ 1 and u₀ ∈ K∩∂Ω_R such that u₀ = λ₀Tu₀,then, for t ∈ [α^*, β^*], we have

This is in contradiction with u₀ ∈ K∩∂Ω_R and (3.22) holds.

Now, (3.13), (3.17), (3.22), and Lemma 2.3 guarantee that T has two fixed points $$, $$ with $$. Clear, PBVP (3.1) has at least two positive solutions u₁, u₂ ∈ C¹[0,2π]∩C²(0,2π).

Remark 3.2. From the proof of Theorem 3.1, when f(t, u, v) is nonnegative (i.e., M = 0 in (H₀)), Theorem 3.1 still holds.

Corollary 3.3. Assume that (H₀)–(H₂) hold, then PBVP (1.1) has at least one positive solution u(t) such that r₁ < ∥u + Mω∥ < R₁, where $$.

Corollary 3.4. Assume that (H₀) and (H₃) hold, and

(H₄) there exist R₁ > σ⁻¹2πMl₂ such that

where $$. Then PBVP (1.1) has at least one positive solution u(t) such that ∥u + Mω∥ > R₁, where $$.

Example 3.5. Consider the following second-order singular semipositone PBVP: