Smallest Eigenvalues for a Right Focal Boundary Value Problem

1 Introduction

We consider the eigenvalue problems

satisfying the boundary conditions

where 1 < α ≤ 2 is a real number, D0+α is the standard Riemann-Liouville derivative, and p(t) and q(t) are continuous nonnegative functions on [0,1] where neither p(t) nor q(t) vanishes identically on any nondegenerate compact subinterval of [0,1]. In this paper, we modify an approach developed by the authors in [5] to show the existence of smallest eigenvalues (1.1), (1.3)and (1.2),(1.3). We will then compare these smallest eigenvalues under the assumption that p(t) ≤ q(t).

Using Krein-Rutman theory [14] to show the existence of and compare smallest eigenvalues for boundary value problems has been a well-studied area (see [1, 2, 3, 4, 7, 8, 10, 15, 16] for some examples). However, just recently, the existence and comparison of smallest eigenvalues and applications of these results have been studied for fractional boundary value problems in [5, 6, 9].In [5] the authors studied the second order linear fractional eigenvalue problems with conjugate boundary conditions. The standard approach used in the papers same above was modified to account for the unbounded slope of the Green’s function for −D0+α=0,u(0)=u(1)=0at 0. The Green’s function for −D0+α=0,u(0)=u′(1)=0 also has an unbounded slope at 0. Therefore, in this paper, we use the approach established in [5] to show the existence of and compare smallest eigenvalues for the right focal fractional boundary value problems. The difference inthis analysis to the work done on the conjugate problem is that the Banach space for the right focal problem does not involve C⁽¹⁾ functions since showing the interior of the cone used in this paper does not involve the mean value theorem or the sign of the derivative at 1.

2 Preliminary Definitions and Theorems

Cones give rise to a natural partial ordering on a Banach space.

The following two results are fundamental to our comparison results and are attributed to Krasnosel’skii [13]. The proof of Theorem 2.1 can be found in [13], and the proof of Theorem 2.2 is provided by Keener and Travis [12] as an extension of Krasnosel’skii’s results.

3 Comparison of Smallest Eigenvalues

In [11], Kaufmann and Mboumi showed that the Green’s function for −D0α+u(t)=0 satisfying (1.3) is given by

Therefore, if u(t)=λ1∫01G(t, s)p(s)u(s)ds,u(t) solves (1.1), (1.3). Similarly, if u(t)=λ2∫01G(t, s)q(s)u(s)ds,u(t) solves (1.2),(1.3). Notice that G(t, s) ≥ 0 on [0, 1] × [0, 1) and G(t, s)>0 on (0,1]×(0,1).

Define the Banach space

with the norm

where |v|0 = supt∈[0,1]|v(t)|denotes the usual supremum norm. Notice that for u ∈ 𝓑

implying

Define the linear operators

and

Now,

Notice that since α > 1,

Therefore, the first term inside the parentheses is well-defined.

Set

In the proof of Theorem 3.1 in [5], it was shown g ∈ C[0, 1]. Therefore, M : 𝓑 → 𝓑. An application of the Arzelà Ascoli theorem shows M is compact. A similar argument can be made for N. Thus, we have the following result.

Next, we define the cone

Proof. Define

We will show Ω ⊂ 𝓟⁰. Let u ∈ Ω. Since, v(0) > 0, there exists an ϵ₁ > 0 such that v(0)– ϵ₁ > 0. Since v ∈ C[0,1], there exists an a ∈(0,1) such that v(t) > ϵ₁ for all t ∈ (0,a). So u(t) = t^α−1v(t) > ϵ₁t^α−1 for all t ∈ (0,a). Also, since u(t)>0 on [a 1], there exists an ϵ₂ > 0 with u(t) – ϵ₂ > 0 for all t ∈ [a, 1].

Let ϵ=min{ϵ12, ϵ22} . Define Bϵ(u)={u^ϵB:‖u−u^‖<ϵ}.Let u^ϵBε(u) . So u^=tα−1v^, where v^∈C[0,1]. Now |u^(t)−u(t)|≤tα−1‖u^−u‖<εtα−1. So for t ∊ (0, a) u^(t)>u(t)−tα−1ϵ>tα−1ϵ1−tα−1ϵ1/2=tα−1ϵ1/2. So u^(t)>0 for t ∊ (0, a). Also, |u^(t)−u(t)|≤‖u^−u‖<ε . So for t ∊ [a, 1] u^(t)>u(t)−ε>ε2−ε2/2>0. So u^∈P for all t ∊ [a, 1] . So u^(t)>0 and thus Ω⊂P∘ . So Bϵ(u)⊂P∘.

Proof. First, we show M: 𝓟\{0}→Ω⊂𝓟0. Let u ∊ 𝓟. So u(t) ≥ 0. Then since G(t, s) ≥ 0 on [0, 1] × [0, 1) and p(t) ≥ 0 on [0, 1], Mu(t)=∫01G(t, s)p(s)u(s)ds≥0, for 0 ≤ t ≤ 1 So M : P → P

Now let u ∊ 𝓟\{0}. So there exists a compact interval [α, β] ⊂ [0,1] such that u(t) > 0 and p(t) > 0 for all t ϵ[α, β].Then, since G(t, s) > 0 on (0, 1] × (0, 1),

for 0 < t ≤ 1. Now,

Let

So,

, thus M:𝓟\{0}→Ω⊂𝓟o.

Now choose u⁰ ϵ𝓟\{0}, and let uϵ𝓟\{0} . So MuϵΩ⊂𝓟o . Choose k₁>0sufficiently small and k₂sufficiently large so that Mu−k1u0ϵ𝓟o and u0−1k2Muϵ𝓟o . So k1u0≤Mu with respect to 𝓟 and Mu≤k2u0 with respect to 𝓟 . Thus k₁u₀ ≤Mu≤k₂u₀ with respect to 𝓟 and M is u₀-positive with respect to 𝓟. A similar argument shows N is u₀-positive.

Proof. Let Λ be an eigenvalue of M with corresponding eigenvector u(t). Notice that

if and only if

if and only if

With

So

is an eigenvalue of (1.1),(1.3). A similar argument can be made for eigenvalues of N.

Proof. Since M is a compact linear operator that is u₀-positive with respect to 𝓟, by Theorem 2.1, M has an essentially unique eigenvector, say, u ϵ 𝓟 and eigenvalue Λ with the above properties. Since u ≠ 0 MuϵΩ⊂𝓟oand u=M(1Λu)ϵ𝓟o.

Proof. Let p(t) ≤ q(t) on [0, 1]. So for any u ∊ 𝓟 and t ϵ [0,1],

So Nu−Mu ∊ 𝓟 for all u ∊ 𝓟, or M ≤ N with respect to 𝓟 . Then by Theorem 2.2 Λ₁ ≤ Λ₂

If p(t) = q(t), then Λ₁ = Λ₂. Now suppose p(t) ≠ q(t). So p(t) < q(t) on some subinterval [α,β] ⊂ [0,1] .Then, using an argument similar to the proof that N was u₀-positive, (N−M)u₁ ∊ Ω ⊂P⁰ and so there exists ε > 0 such that (N−M)u₁ − εu₁ϵP. So Λ₁u₁ + εu₁ = Mu₁ + εu₁≤Nu₁, implying Nu₁≥(Λ₁+ε)u₁. Since N≤N and Nu₂ = Λ₂u₂ by Theorem 2.2 Λ₁ + ε ≤ Λ₁, or Λ₁ < Λ₂.

Since the eigenvalues of (1.1),(1.3) are reciprocals of eigenvalues of Mand conversely, and the eigenvalues (1.2),(1.3) of are reciprocals of eigenvalues of N and conversely, the following theorem is an immediate consequence of Theorems 3.2 and 3.3