Electronic Journal of Qualitative Theory of Differential Equations 2005, No. 15, 1-9, http://www.math.u-szeged.hu/ejqtde/

COMPARISON OF EIGENVALUES FOR A FOURTH-ORDER FOUR-POINT BOUNDARY VALUE PROBLEM

BASANT K. KARNA^{†}, ERIC R. KAUFMANN^{‡} AND JASON NOBLES^{‡}

Abstract. We establish the existence of a smallest eigenvalue for the fourthorder four-point boundary value problem (ϕ_{p}(u′′(t)))′′ = λh(t)u(t), u′(0) = 0, β_{0}u(η_{0}) = u(1), ϕ′_{p}(u′′(0)) = 0, β_{1}ϕ_{p}(u′′(η_{1})) = ϕ_{p}(u′′(1)), p > 2, 0 < η_{1}, η_{0} < 1, 0 < β_{1}, β_{0} < 1, using the theory of u_{0}-positive operators with respect to a cone in a Banach space. We then obtain a comparison theorem for the smallest positive eigenvalues, λ_{1} and λ_{2}, for the differential equations (ϕ_{p}(u′′(t)))′′ = λ_{1}f(t)u(t) and (ϕ_{p}(u′′(t)))′′ = λ_{2}g(t)u(t) where 0 ≤ f(t) ≤ g(t), t ∈ [0, 1].

1. Introduction

In this paper, we will compare the smallest eigenvalues for the eigenvalue problems,

(ϕ_{p}(u′′(t)))′′ = λ_{1}f(t)u(t),  (1.1)
(ϕ_{p}(u′′(t)))′′ = λ_{2}g(t)u(t),  (1.2)

t ∈ [0, 1], with eigenvectors satisfying the nonlocal boundary conditions,

u′(0) = 0, β_{0}u(η_{0}) = u(1),  (1.3)
ϕ′_{p}(u′′(0)) = 0, β_{1}ϕ_{p}(u′′(η_{1})) = ϕ_{p}(u′′(1)).  (1.4)

Throughout this paper we assume that 0 < η_{0}, η_{1} < 1, 0 < β_{0}, β_{1} < 1, p > 2, ϕ_{p}(z) = z|z|^{p−2}, and f, g : [0, 1] → [0, +∞) are continuous and do not vanish on any nontrivial compact subsets of [0, 1].

We use sign properties of Green’s functions and the theory of u_{0}-positive operators with respect to a cone in a Banach space to establish our results. The theory of u_{0}-positive operators is developed in the books by Krasnosel’skiĭ [9] and Deimling [2] as well as in the manuscript by Keener and Travis [8]. Many authors have used cone theoretic techniques to compare smallest eigenvalues for a pair of differential equations; see, for example [1, 3, 4, 5, 6, 7, 8, 10] and references therein. In particular, Eloe and Henderson [3] compared smallest eigenvalues for a class of multi-point boundary value problems while Karna [5, 6] considered the comparison of smallest eigenvalues for nonlocal three-point and m-point boundary value problems. Finally, we mention the paper by Lui and Ge [12] who considered the p-Laplacian differential equation

(ϕ_{p}(u′′(t)))′′ = a(t)f(u(t)), t ∈ (0, 1),

EJQTDE, 2005, No. 15, p. 1 with solutions satisfying one of the following two sets of boundary conditions

2000 Mathematics Subject Classification. 3410.

Key words and phrases. Eigenvalue problem, cone u_{0}-positive operator.

u(0) − λu′(η) = u′(1) = 0, u′′′(0) = α_{1}u′′′(ξ), u′′(1) = β_{1}u′′(ξ),

or

u(1) + λu′(η) = u′(0) = 0, u′′′(0) = α_{1}u′′′(ξ), u′′(1) = β_{1}u′′(ξ).

In section 2, we present preliminary definitions and fundamental results from the theory of u_{0}-positive operators with respect to a cone in a Banach space. In section 3, we apply the theorems in section 2 to obtain a comparison theorem for the smallest eigenvalues, λ_{1} andλ_{2} of (1.1), (1.3), (1.4) and (1.2), (1.3), (1.4), when 0 ≤ f(t) ≤ g(t). In section 4, we compare eigenvalues for the 2m+2 order problem.

2. Banach Spaces, Cones and Preliminary Results

In this section, we state some definitions and theorems from the theory of u_{0}−positive operators that we will apply in the next sections to obtain our comparison theorems. Most of the discussion of this section, involving the theory of cones in a Banach space, can be found in [9].

Let B be a Banach space over the reals. A closed, nonempty set P ⊂ B is said to be a cone provided, (i) αu + βv ∈ P, for all u, v ∈ P and all α, β ≥ 0, and, (ii) u, −u ∈ P implies u ≡ 0. A cone, P, is said to be reproducing, if, for each w ∈ B, there exists u, v ∈ P such that w = u − v. A cone, P, is said to be solid, if P^{∘} ≠ ∅, where P^{∘} is the interior of P.

Remark: Krasnosel’skiĭ [9] proved that every solid cone is reproducing.

A Banach space B is called a partially ordered Banach space, if there exists a partial ordering, ⪯, on B such that, (i) u ⪯ v, for all u, v ∈ B, implies tu ⪯ tv, for all t ≥ 0, and tv ⪯ tu, for all t < 0, where tv ≺ tu means tv ⪯ tu and, tv ≠ tu, and (ii) u_{1} ⪯ v_{1} and u_{2} ⪯ v_{2}, for all u_{1}, u_{2}, v_{1}, v_{2} ∈ B, imply that u_{1} + u_{2} ⪯ v_{1} + v_{2}.

Let P ⊂ B be a cone and define u ⪯ v if, and only if, v − u ∈ P. Then ⪯ is a partial ordering on B, and we say that ⪯ is the partial ordering induced by P. Moreover, B is a partially ordered Banach space with respect to ⪯.

Let M, N : B → B be bounded, linear operators. We say that M ⪯ N with respect to P, if Mu ⪯ Nu for all u ∈ P. A bounded, linear operator M : B → B, is said to be u_{0}−positive with respect to P, if there exists a u_{0} ∈ P, u_{0} ≠ 0, such that for every nonzero u ∈ P, there exist positive constants, k_{1}(u), k_{2}(u) ∈ ℝ, such that k_{1}u_{0} ⪯ Mu ⪯ k_{2}u_{0}.

Of the next two results, the first can be found in Krasnosel’skiĭ [9] and the second was proved by Keener and Travis [8] as an extension of results from [9].

Theorem 2.1. Let B be a Banach space over the reals and let P ⊂ B be a reproducing cone. Let M : B → B be a compact, linear operator which is u_{0}−positive with respect to P. Then M has an essentially unique eigenvector in P, and the corresponding eigenvalue is simple, positive, and larger than the absolute value of any other eigenvalue.

Theorem 2.2. Let B be a Banach space over the reals and let P ⊂ B be a cone. Let M, N : B → B be bounded, linear operators, and assume that at least one of the operators is u_{0}−positive with respect to P. If M ⪯ N with respect to P, and if there exists nonzero u_{1}, u_{2} ∈ P and positive real numbers λ_{1} and λ_{2}, such that

EJQTDE, 2005, No. 15, p. 2 λ_{1}u_{1} ⪯ Mu_{1} and Nu_{2} ⪯ λ_{2}u_{2}, then λ_{1} ≤ λ_{2}. Moreover, if λ_{1} = λ_{2}, then u_{1} is a scalar multiple of u_{2}.

Remark: It is well known that the function ϕ_{p} is invertible and that its inverse is ϕ_{q} where p and q satisfy \frac{1}{p} + \frac{1}{q} = 1. Furthermore both ϕ_{p} and ϕ_{q} are increasing function.

3. Comparison of Eigenvalues

In this section, we apply the results in section 2 to compare the smallest positive eigenvalues of (1.1), (1.3), (1.4) and (1.2), (1.3), (1.4). We do so by defining Hammerstien integral operators, M and N, associated with (1.1), (1.3), (1.4) and (1.2), (1.3), (1.4). Let α ≠ 1 and consider the second order linear eigenvalue problem,

−y′′ = λh(t),  (3.1)
y′(0) = 0, αy(ξ) = y(1).  (3.2)

It is well known that y is a solution of (3.1), (3.2) if, and only if, y is a solution of

y(t) = λ ∫_{0}^{1}G(t, s; α, ξ)h(s) ds,  (3.3)

where G(t, s; α, ξ) is the Green’s function for −y′′ = 0, (3.2) and is given by

G(t, s; α, ξ) = \frac{1 − s}{1 − α} − \{ \begin{matrix}\frac{α(ξ−s)}{1−α}, & s ≤ ξ \\ 0, & s > ξ \end{matrix} − \{ \begin{matrix}t − s, & s ≤ t \\ 0, & s > t. \end{matrix}  (3.4)

Note that if 0 ≤ α < 1 then

G(t, s; α, ξ) > 0

for all (t, s) ∈ (0, 1) × (0, 1). As noted in Karna [5],

\frac{∂}{∂t}G(t, s; α, ξ) = −1 < 0 for s < t, and

\frac{∂}{∂t}G(t, s; α, ξ) = 0 for s > t.

Let y = −ϕ_{p}(u′′(t)) in (3.3) and (3.2), and set α = β_{1}, ξ = η_{1} to obtain,

−ϕ_{p}(u′′(t)) = λ ∫_{0}^{1}G(t, s; β_{1}, η_{1})h(s) ds,
ϕ′_{p}(u′′(0)) = 0, β_{1}ϕ_{p}(u′′(η_{1})) = ϕ_{p}(u′′(1)).

Rewrite the differential equation as

−u′′(t) = ϕ_{q} (λ ∫_{0}^{1}G(t, s; β_{1}, η_{1})h(s) ds) .  (3.5)

We now consider the second order linear boundary value problem (3.5), (1.3). Again, we see that u is a solution of (3.5), (1.3) if, and only if, u satisfies

u(t) = ϕ_{q}(λ) ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})h(s) dτ) ds.

EJQTDE, 2005, No. 15, p. 3 We define, for our Banach space,

B = {u ∈ C^{3}[0, 1] : u satisfies the boundary conditions (1.3), (1.4) }

with norm

∥u∥ = max_{0≤i≤3} { sup_{t∈[0,1]} |u^{(i)}(t)|} .

Define P ⊂ B by

P = {u ∈ B : u(t) ≥ 0 and u′(t) ≤ 0 for t ∈ [0, 1]}.

Then P is a cone in B. To prove that P is solid we employ an auxiliary set, Θ, defined as follows,

Θ = {u ∈ B : u(t) > 0 for t ∈ [0, 1] and u′(t) < 0 for t ∈ (0, 1]}.

Lemma 3.1. The cone P is solid and hence reproducing.

Proof. We show that Θ ⊂ P^{∘} from which we have P^{∘} ≠ ∅.

Clearly, Θ ⊂ P. Let u ∈ Θ. Then u(t) > 0 on [0, 1] and u′(t) < 0 on (0, 1]. Consider the open ball B_{ε} = {v ∈ B : ∥v − u∥ < ε}. Let v ∈ B_{ε}. Since ∥v − u∥ < ε then |v′(t) − u′(t)| < ε for all t ∈ (0, 1]. Hence u′(t) − ε < v′(t) < u′(t) + ε < 0 for ε sufficiently small. Likewise ∥v − u∥ < ε implies u(t) + ε > v(t) > u(t) − ε > 0 for ε sufficiently small. Consequently, for ε sufficiently small B_{ε} ⊂ Θ. Since u ∈ Θ was arbitrary, Θ is open in P. Hence P^{∘} ≠ ∅ and the proof is complete. □

Define the integral operators M, N : B → B as follows,

Mu(t) = ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds, t ∈ [0, 1],

Nu(t) = ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})g(τ)u(τ) dτ)ds, t ∈ [0, 1].

Standard arguments are used to show that M and N are completely continuous. Our first theorem states that the operators M and N are u_{0}-positive with respect to P.

Theorem 3.2. The operators M and N are u_{0}−positive with respect to the cone P.

Proof. We will prove the theorem for the operator M. The proof for the operator N is similar. We first show that M : P → P. Next we show that M : P \ {0} → Θ. Finally, given a u ∈ P \{0}, we determine constants k_{1}, k_{2} such that the appropriate inequalities hold.

Let u ∈ P. Then u(t) ≥ 0 and u′(t) ≤ 0 for all t ∈ [0, 1]. Since f(t) ≥ 0 and since G(t, s; α, ξ) ≥ 0 for all (t, s) ∈ [0, 1] × [0, 1], then Mu(t) ≥ 0 for all t ∈ [0, 1]. Also, since \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) = −1 for s < t, then

∫_{0}^{t} \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ)dτ)ds

= − ∫_{0}^{t}ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds ≤ 0.

Since \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) = 0 for s > t, then

∫_{t}^{1} \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds = 0.

Consequently, (Mu)′(t) ≤ 0 for all t ∈ [0, 1] and so, M(P) ⊆ P.

Now consider u ∈ P \{0}. Since u′(t) ≤ 0 for all t ∈ [0, 1] then u is non-increasing over [0, 1]. Suppose that u(0) = 0 then either u ≡ 0 or u(t) ≤ 0 for all t ∈ [0, 1]. In either case, u ∉ P \ {0}. Hence u(0) > 0. By continuity, there exists c ∈ (0, 1] such that u(t) > 0 for all t ∈ [0, c). Since f does not vanish on any nontrivial compact subsets of [0, 1], there exists [α, β] ⊂ [0, c) such that f(t) > 0 for all t ∈ [α, β]. So, for t ∈ [0, 1]

Mu(t) = ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds

≥ ∫_{α}^{β}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{α}^{β}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds
> 0.

Also, if t ∈ (0, 1] then

(Mu)′(t) = ∫_{0}^{1} \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds

≤ − ∫_{0}^{t}ϕ_{q} (∫_{α}^{β} G(s, τ; β_{1}, η_{1})f(τ)u(τ) dτ)ds
< 0.

Consequently, if u ∈ P \ {0}, then Mu ∈ Θ ⊂ P^{∘}. That is, M : P \ {0} → P^{∘}.

To complete the proof, fix u_{0} ∈ P \ {0} and let u ∈ P \ {0}. From the above we know that Mu ∈ Θ ⊂ P^{∘} and so, there exists k_{1} sufficiently small so that Mu − k_{1}u_{0} ∈ P. Similarly, there exists k_{2} sufficiently large so that u_{0} − \frac{1}{k_{2}} Mu ∈ P. Thus,

Mu − k_{1}u_{0} ∈ P ⇒ k_{1}u_{0} ⪯ Mu,

u_{0} − \frac{1}{k_{2}} Mu ∈ P ⇒ Mu ⪯ k_{2}u_{0}.

That is, given u_{0} ∈ P \ {0}, for each u ∈ P \ {0}, there exists k_{1}, k_{2} such that

k_{1}u_{0} ⪯ Mu ⪯ k_{2}u_{0}.

The operator M is u_{0}-positive with respect to the cone P and the proof is complete. □

Now we apply Theorems 2.1 and 2.2 to obtain results concerning the eigenvectors and eigenvalues of M and N.

Theorem 3.3. The operator M (N) has an essentially unique eigenvector, u ∈ P^{∘}, and the corresponding eigenvalue, Λ_{1}, (Λ_{2}), is simple, positive, and larger than the absolute value of any other eigenvalue.

EJQTDE, 2005, No. 15, p. 5 Proof. From Theorem 3.2, we have that the compact, linear operator M is u_{0}- positive with respect to P. By Theorem 2.1, M has an essentially unique eigenvector, u_{1} ∈ P, and the corresponding eigenvalue, Λ_{1} is simple, positive, and larger than the absolute value of any other eigenvalue. Since u_{1} ≠ 0 then, Mu_{1} ∈ Θ ⊂ P^{∘}. Now Mu_{1} = Λ_{1}u_{1} and so u_{1} = \frac{1}{Λ_{1}} Mu_{1} ∈ P^{∘} and the proof is complete. □

Theorem 3.4. Assume that 0 ≤ f(t) ≤ g(t) for all t ∈ [0, 1]. Let Λ_{1} and Λ_{2} be the largest positive eigenvalues of M and N respectively with corresponding essentially unique eigenvectors u_{1} and u_{2}. Then Λ_{1} ≤ Λ_{2}. Furthermore, Λ_{1} = Λ_{2}, if, and only if, f(t) = g(t) for all t ∈ [0, 1].

Proof. Since 0 ≤ f(t) ≤ g(t) for all t ∈ [0, 1] and since G(t, s; α, ξ) ≥ 0 for (t, s) ∈ [0, 1] × [0, 1], then

(Nu − Mu)(t) = ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})(g(τ) − f(τ))u(τ) dτ)ds
≥ 0.

Since \frac{∂}{∂t}G(t, s; α, ξ) = −1 if s < t and \frac{∂}{∂t}G(t, s; α, ξ) = 0 if s > t, then

(Nu − Mu)′(t)

= ∫_{0}^{1} \frac{∂}{∂t}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})(g(τ) − f(τ))u(τ) dτ)ds

= − ∫_{0}^{t} ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})(g(τ) − f(τ))u(τ) dτ)ds

≤ 0.

Thus, Nu − Mu ∈ P and so, M ⪯ N with respect to P. From Theorem 2.2, if u_{1} and u_{2} are eigenvectors of M and N respectively, with corresponding eigenvalues Λ_{1} and Λ_{2} then Λ_{1} ≤ Λ_{2}.

Note if f(t) = g(t) then by Theorem 2.2 Λ_{1} ≤ Λ_{2} and Λ_{1} ≥ Λ_{2}. In this case Λ_{1} = Λ_{2}.

To finish the proof, we need to show that Λ_{1} = Λ_{2} implies f(t) = g(t) for all t ∈ [0, 1]. Suppose that f(t) < g(t) on some interval [α, β] ⊂ [0, 1]. As in Theorem 3.2, (N − M)u_{1} ∈ Θ ⊆ P^{∘}. Since u_{1} ∈ P^{∘}, there exists an ε > 0 sufficiently small so that εu_{1} ⪯ (N − M)u_{1} = Nu_{1} − Mu_{1} = Nu_{1} − Λu_{1}. Thus (Λ_{1} + ε)u_{1} ⪯ Nu_{1}. Since N ⪯ N, (Λ_{1} + ε)u_{1} ⪯ Nu_{1}, and Nu_{2} ⪯ Λ_{2}u_{2}, then by Theorem 2.2 we have Λ_{1} + ε ≤ Λ_{2}. Hence Λ_{1} < Λ_{2}. By the contrapositive, Λ_{1} = Λ_{2} implies f(t) = g(t) for all t ∈ [0, 1] and the proof is complete. □

Let Λ_{1} be an eigenvector of M with corresponding eigenvector u_{1} and let λ_{1} = ϕ_{p} (\frac{1}{Λ_{1}} ). Then for all t ∈ [0, 1],

Λ_{1}u_{1}(t) = Mu_{1}(t) = ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u_{1}(τ) dτ)ds.

EJQTDE, 2005, No. 15, p. 6 Since \frac{1}{Λ_{1}} = ϕ_{q}(λ), then

u_{1}(t) = \frac{1}{Λ_{1}} ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u_{1}(τ) dτ)ds
= ϕ_{q}(λ_{1}) ∫_{0}^{1}G(t, s; β_{0}, η_{0}) ϕ_{q} (∫_{0}^{1}G(s, τ; β_{1}, η_{1})f(τ)u_{1}(τ) dτ)ds.

That is, λ_{1} is an eigenvalue corresponding to (1.1), (1.3), (1.4). The converse also holds. Thus, Λ_{1}(Λ_{2}) is an eigenvalue of M(N) if, and only if λ_{1} = ϕ (\frac{1}{Λ_{1}} ) (λ_{2} =

ϕ (\frac{1}{Λ_{2}} ) ) is an eigenvalue of (1.1), (1.3), (1.4), ( (1.2), (1.3), (1.4) ). Furthermore,

since ϕ_{p} is an increasing one-to-one function, then \frac{1}{ϕ(λ_{1})} = Λ_{1} ≤ Λ_{2} = \frac{1}{ϕ(λ_{2})} implies that λ_{2} ≤ λ_{1} and Λ_{1} = Λ_{2} if, and only if λ_{1} = λ_{2}.

Theorem 3.5. Let 0 ≤ f(t) ≤ g(t) for all t ∈ [0, 1]. Then there exist smallest positive eigenvalues of (1.1), (1.3), (1.4) and (1.2), (1.3), (1.4), respectively, each of which is simple, positive, and less than the absolute value of any other eigenvalue of the corresponding problems. Also, the corresponding essentially unique eigenvectors may be chosen to belong to P^{∘}. Finally, λ_{2} ≤ λ_{1}, and λ_{1} = λ_{2} if and only if f(t) = g(t), 0 ≤ t ≤ 1.

4. The 2m + 2 Order Problem

Let m > 1 be a fixed integer. Define L_{0}u(t) ≡ u′′(t) and for k = 1, 2, …, m

L_{k}u(t) ≡ ϕ_{p_{k}}((L_{k−1}u)′′(t)).

In this section we compare eigenvalues for the 2m + 2 order problems,

(−1)^{m+1}(L_{m}u)′′(t) = λ_{1}f(t)u(t),  (4.1)
(−1)^{m+1}(L_{m}u)′′(t) = λ_{2}g(t)u(t),  (4.2)

t ∈ [0, 1], with eigenvectors satisfying the nonlocal boundary conditions,

u′(0) = 0, (L_{k}u)′(0) = 0,  (4.3)
β_{0}u(η_{0}) = u(1), β_{k}L_{k}u(η_{k}) = L_{k}u(1),  (4.4)

k = 1, …, m, where 0 < η_{i} < 1, 0 < β_{i} < 1, and p_{i} > 2 for 1 ≤ i ≤ m.

For each k = 1, 2, …, m define the operator 𝔾_{k} by

𝔾_{k}h(s_{k+1}) ≡ ϕ_{q_{m+1−k}} (∫_{0}^{1}G(s_{k+1}, s_{k}; β_{m+1−k}, η_{m+1−k})h(s_{k}) ds_{k}) ,

where G(t, s; α, ξ) is defined in (3.4). Using the technique outlined in the beginning of section 3, we see that u(t) is a solution of (−1)^{m+1} (L_{m}u)′′ (t) = λh(t), (4.3), (4.4), if, and only if,

\begin{matrix}u(t) = & ϕ_{q_{1}}(ϕ_{q_{2}}( ⋯ (ϕ_{q_{m}}(λ)) ⋯))× \\ ∫_{0}^{1} G(t, s_{m+1}; β_{0}, η_{0}) [(𝔾_{m} ∘ 𝔾_{m−1} ∘ ⋯ ∘ 𝔾_{1})(h)] (s_{m+1}) ds_{m+1}. \end{matrix} (4.5)

Our Banach space is

B = {u ∈ C^{2m+1}[0, 1] : u satisfies the boundary conditions (4.3), (4.4) }

with norm

∥u∥ = max {|u(t)|_{0}, |u′(t)|_{0}, |L_{k}u(t)|_{0}, |(L_{k}u)′(t)|_{0}, k = 1, 2, …, m} ,

where |z|_{0} = sup_{t∈[0,1]} |z(t)|. Define the cone P_{2} ⊂ B by

P_{2} = {u ∈ B : u(t) ≥ 0, (−1)u′(t) ≥ 0, (−1)^{k}L_{k}u(t) ≥ 0, k = 1, 2, …, m,
and (−1)^{k+1}(L_{k}u)′(t) ≥ 0, k = 1, 2, …, m, for t ∈ [0, 1]}

and the auxiliary set, Θ_{2}, as follows,

Θ_{2} = {u ∈ B : u(t) > 0, (−1)^{k}L_{k}u(t) > 0, k = 1, 2, …, m, for t ∈ [0, 1]

and − u′(t) > 0, (−1)^{k+1}(L_{k}u)′(t) > 0, k = 1, 2, …, m, for t ∈ (0, 1]}.

A modification of the proof of Lemma 3.1 yields that Θ_{2} ⊂ P_{2}^{∘}. Hence the cone P_{2} is solid and reproducing. We define operators M, N : B → B as follows,

Mu(t) = ∫_{0}^{1} G(t, s_{m+1}; β_{0}, η_{0}) [(𝔾_{m} ∘ 𝔾_{m−1} ∘ ⋯ ∘ 𝔾_{1})(f)] (s_{m+1}) ds_{m+1},
N u(t) = ∫_{0}^{1} G(t, s_{m+1}; β_{0}, η_{0}) [(𝔾_{m} ∘ 𝔾_{m−1} ∘ ⋯ ∘ 𝔾_{1})(g)] (s_{m+1}) ds_{m+1},

for all t ∈ [0, 1]. The proofs of the following theorems are similar to those of their counterparts in section 3 and are omitted.

Theorem 4.1. The operators M, N : P_{2} → P_{2} are completely continuous and u_{0}−positive with respect to the cone P_{2}.

Theorem 4.2. The operator M (N) has an essentially unique eigenvector, u ∈ P_{2}^{∘}, and the corresponding eigenvalue, Λ_{1}, (Λ_{2}), is simple, positive, and larger than the absolute value of any other eigenvalue.

Theorem 4.3. Assume that 0 ≤ f(t) ≤ g(t) for all t ∈ [0, 1]. Let Λ_{1} and Λ_{2} be the largest positive eigenvalues of M and N respectively with corresponding essentially unique eigenvectors u_{1} and u_{2}. Then Λ_{1} ≤ Λ_{2}. Furthermore, Λ_{1} = Λ_{2}, if, and only if, f(t) = g(t) for all t ∈ [0, 1].

Let Λ_{1} be an eigenvector of M with corresponding eigenvector u_{1} and let λ_{1} = ϕ_{p_{m}} (⋯ ϕ_{p_{1}}(\frac{1}{Λ_{1}} ) ⋯). Then λ_{1} is an eigenvalue corresponding to (4.1), (4.3), (4.4). The converse also holds. Thus, Λ_{1}(Λ_{2}) is an eigenvalue of M (N) if, and only if λ_{1} = ϕ_{p_{m}} (⋯ ϕ_{p_{1}}(\frac{1}{Λ_{1}} ) ⋯) (ϕ_{p_{m}} (⋯ ϕ_{p_{1}}(\frac{1}{Λ_{2}} ) ⋯) ) is an eigenvalue of

(4.1), (4.3), (4.4), ( (4.2), (4.3), (4.4) ). Furthermore, since each ϕ_{p_{k}}, k = 1, …, m, is an increasing one-to-one function, and since Λ_{1} ≤ Λ_{2}, then λ_{2} ≤ λ_{1} and Λ_{1} = Λ_{2} if, and only if, λ_{1} = λ_{2}.

EJQTDE, 2005, No. 15, p. 8 Theorem 4.4. Let 0 ≤ f(t) ≤ g(t) for all t ∈ [0, 1]. Then there exist smallest positive eigenvalues of (4.1), (4.3), (4.4) and (4.2), (4.3), (4.4), respectively, each of which is simple, positive, and less than the absolute value of any other eigenvalue of the corresponding problems. Also, the corresponding essentially unique eigenvectors may be chosen to belong to P_{2}^{∘}. Finally, λ_{2} ≤ λ_{1}, and λ_{1} = λ_{2} if and only if f(t) = g(t), 0 ≤ t ≤ 1.

Remark: We would like to thank the reviewers for their suggestions that improved this paper.

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(Received October 12, 2004)

†Marshall University, Department of Mathematics, WV 25755-2560, USA

E-mail address: karna@marshall.edu

‡Department of Mathematics & Statistics, University of Arkansas at Little Rock, Little Rock, AR 72204, USA

E-mail address: erkaufmann@mail.ualr.edu, jwnobles@ualr.edu