Existence of 2m − 1 Positive Solutions for Sturm-Liouville Boundary Value Problems with Linear Functional Boundary Conditions on the Half-Line

Lemma 2.1. Under the condition $$ and ρ > 0, the boundary value problem

has a unique solution for any y ∈ L[0, +∞). Moreover, this unique solution can be expressed in the form where G(t, s),   A(y), and B(y) are defined by

Proof. a(t) and b(t) in (1.3) are two linear independent solutions of the equation (p(t)u′(t))′ = 0, so the general solutions for the equation (p(t)u′(t))′ + y(t) = 0 can be expressed in the form

where C, D are undetermined constants. Through verifying directly, when C and D satisfy (a) and (b) separately, u(t) in (2.4) is a solution of BVP(2.1).

Now we need to prove that when u(t) in (2.4) is a solution of BVP(2.1), C and D satisfy (a) and (b) separately.

Let $$ be a solution of BVP(2.1), then

That is, (p(t)u′(t))′ + y(t) = 0.

By (2.4), we have

then From (2.7), we obtain that C and D satisfy (a) and (b) separately. The proof is completed.

Remark 2.2. Assume that (H₂) holds. Then 0 ≤ A(y)<+∞,   0 ≤ B(y)<+∞ for any y ≥ 0 and any solution u(t) of BVP(2.1) is nonnegative.

Lemma 2.3. From (1.3) and (2.3), it is easy to get the following properties.

• (1)

  G(t, s)/ρ⁻¹[1 + a(t)b(t)] ≤ 1,   a(t)/1 + a(t)b(t) < 1/b(t) ≤ 1/β₂,   b(t)/1 + a(t)b(t) < 1/a(t) ≤ 1/β₁.

• (2)

  $$.

• (3)

  G(t, s) ≤ G(s, s) ≤ a(s)b(s)/ρ < +∞.

Lemma 2.4. For any constant 0 < a^* < b^* < ∞, there exists 0 < c^* < 1, such that, for τ, s ∈ [0, ∞),   G(t, s)/ρ⁻¹[1 + a(t)b(t)] ≥ c^*G(τ, s)/ρ⁻¹[1 + a(τ)b(τ)],   a(t)/ρ⁻¹[1 + a(t)b(t)] ≥ c^*a(τ)/ρ⁻¹[1 + a(τ)b(τ)],   b(t)/ρ⁻¹[1 + a(t)b(t)] ≥ c^*b(τ)/ρ⁻¹[1 + a(τ)b(τ)],   t ∈ [a^*, b^*].

Proof. By (1.3), it is obvious that a(t) is increasing, and b(t) is decreasing on t ∈ [0, +∞); therefore, by (2.3), we have

We take c^* = min {a(a^*)β₂/(1 + a(b^*)b(a^*)), b(b^*)β₁/(1 + a(b^*)b(a^*))}, then 0 < c^* < 1; this is because that By Lemma 2.3(1), we have G(τ, s)/ρ⁻¹[1 + a(τ)b(τ)] ≤ 1, then Similarly, we can obtain that b(t)/(1 + a(t)b(t)) ≥ c^*(b(τ)/(1 + a(τ)b(τ))). The proof is completed.

Lemma 2.5 (see [10].)Let M⊆C_l(R⁺, R) = {x ∈ C(R⁺, R)∣lim _t→+∞x(t) exists}, then M is precompact if the following conditions hold:

• (a)

  M is bounded in C_l;

• (b)

  the functions belonging to M are locally equicontinuous on any interval of R⁺;

• (c)

  the functions from M are equiconvergent; that is, given ɛ > 0, there corresponds T(ɛ) > 0 such that |x(t) − x(∞)| < ɛ for any t ≥ T(ɛ) and x ∈ M.

Theorem 2.6 (see [11].)Let $$ be completely continuous and α a nonnegative continuous concave functional on P with α(x)≤∥x∥ for any $$. Suppose that there exist 0 < a < b < d ≤ c such that

•  

  c₁ {x ∈ P(α, b, d)∣ α(x) > b} ≠ ϕ, and α(Tx) > b, for x ∈ P(α, b, d);

•  

  c₂ ∥Tx∥<a, for $$;

•  

  c₃ α(Tx) > b for x ∈ P(α, b, c) with ∥Tx∥>d.

Then T has at least three fixed points x₁, x₂, x₃, with

Theorem 3.1. Suppose that (H₁), (H₂) hold, and assume there exist 0 < r₁ < b₁ < l₁ < r₂ with l₁ = max {b₁/c^*, sup _t∈[0,+∞)(b₁/c^*p(t))}, such that

•  

  (H₃) $$, $$, 0 ≤ y₁ ≤ r₂, |y₂| ≤ r₂,

•  

  (H₄) Ψ(t, y₁, y₂) > b₁/δ,   t ∈ [a^*, b^*],   b₁ ≤ y₁ ≤ r₂,    | y₂ | ≤ r₂,

•  

  (H₅) $$, $$, 0 ≤ y₁ ≤ r₁, |y₂ | ≤ r₁.

Then BVP(1.1) has at least three positive solutions u₁, u₂, and u₃ with

Proof. Firstly we prove that T : P → P is continuous.

We will show that T : P → P is well defined and T(P) ⊂ P. For all u(t) ∈ P, by (H₂), Φ(t) and f are nonnegative functions, and we have Tu(t) ≥ 0. From (H₁), (H₂), we obtain

In the same way, we have By Lemma 2.3(1), (A), (B), and (H₁), for all u(t) ∈ P, we have Hence, T : P → P is well defined. By (3.1), (H₁), the Lebesgue dominated convergence theorem and the continuity of p(t), for any u ∈ P,   t₁, t₂ ∈ R⁺, we have That is, $$; therefore, (Tu)(t) ∈ E.

By Lemma 2.4, we have

therefore T : P → P.

We show that T : P → P is continuous. In fact suppose {u_m}⊆P, u₀ ∈ P and u_m → u₀(m → +∞), then there exists M > 0, such that ∥u_m∥≤M. By (H₁), we have

Therefore, by Lemma 2.3(1), the continuity of f and the Lebesgue dominated convergence theorem imply that Thus, ∥Tu_m − Tu₀∥→0(m → +∞). Therefore T : P → P is continuous.

Secondly we show that T : P → P is compact operator.

For any bounded set B ⊂ P, there exists a constant L > 0 such that ∥u∥≤L, for all u ∈ B. By Lemma 2.3(1), (A), (B), and (H₁), we have

Therefore, (Tu)(t)⊆C_l(R⁺, R).

By (3.4) and (3.5), we have

so TB is bounded.

Given T > 0,   t₁,   t₂ ∈ [0, T], by (H₁) and Lemma 2.3(1), we have

Therefore for any u ∈ B, by (3.1), the Lebesgue dominated convergence theorem and the continuity of G(t, s), a(t), and b(t), we have By a similar proof as (3.6), we obtain |(Tu)′(t₁)−(Tu)′(t₂)| → 0, as t₁ → t₂. Thus, TB is equicontinuous on [0, T]. Since T > 0 is arbitrary, TB is locally equicontinuous on [0, +∞).

By Lemma 2.3(2), (H₂) and the Lebesgue dominated convergence theorem, we obtain

By (3.5), we know that lim _t→+∞ | (Tu)′(t)|<+∞, then Therefore, TB is equiconvergent at ∞. By Lemma 2.5, TB is completely continuous.

Finally we will show that all conditions of Theorem 2.6 hold.

From the definition of α, we can get α(u)≤∥u∥ for all u ∈ P. For all $$, we have ∥u∥≤r₂; therefore 0 ≤ y₁ ≤ r₂, |y₂ | ≤ r₂. By (3.4), (3.5), and (H₃), we have

that is, ∥Tu∥≤r₂ for $$. Thus $$.

Similarly for any $$, we have ∥Tu∥<r₁, which means that condition (c₂) of Theorem 2.6 holds.

In order to apply condition (c₁) of Theorem 2.6, we choose $$, then ∥u∥≤l₁; this is because

and $$, which means that {u ∈ P(α, b₁, l₁) | α(u) > b₁} ≠ ϕ. For all u ∈ P(α, b₁, l₁), we have α(u) ≥ b₁ and ∥u∥≤l₁, thus b₁ ≤ u(t)/ρ⁻¹[1 + a(t)b(t)] ≤ l₁, |u′(t)| ≤ l₁, that is, b₁ ≤ y₁ ≤ l₁, |y₂ | ≤ l₁. By (H₄), we can get Consequently condition (c₁) of Theorem 2.6 holds.

We will prove that condition (c₃) of Theorem 2.6 holds. If u ∈ P(α, b₁, r₂), and ∥Tu(t)∥>l₁, by (H₄), we have

Therefore, condition (c₃) of Theorem 2.6 is satisfied. Then we can complete the proof of this theorem by Leggett-Williams fixed point theorem.

Theorem 3.2. Suppose that (H₁), (H₂) hold, and assume there exist 0 < r₁ < b₁ < l₁ < r₂ < b₂ < l₂ < r₃ < ···<r_m with l_i = max {b_i/c^*, sup _t∈[0,+∞)b_i/c^*p(t)}, such that

•  

  (H₆) $$, $$, 0 ≤ y₁ ≤ r_i, |y₂| ≤ r_i, 1 ≤ i ≤ m,

•  

  (H₇) Ψ(t, y₁, y₂) > b_i/δ, t ∈ [a^*, b^*], b_i ≤ y₁ ≤ r_i+1, |y₂| ≤ r_i+1, 1 ≤ i ≤ m − 1.

Then BVP(1.1) has at least 2m − 1 positive solutions.

Proof. When m = 1, it follows from (H₆) that T has at least one positive solution by the Schauder fixed point theorem. When m = 2, it is clear that Theorem 3.1 holds. Then we can obtain three positive solutions. In this way, we can finish the proof by the method of induction.