Comparison of eigenvalues for a fourth-order four-point boundary value problem, Electron

We establish the existence of a smallest eigenvalue for the fourth- order four-point boundary value problem ( p(u 00 (t))) 00 = h (t)u(t); u 0 (0) = 0; 0u( 0) = u(1); 0(u 00 (0)) = 0; 1 p(u 00 ( 1)) = p(u 00 (1)), p > 2, 0 < 1; 0 < 1; 0 < 1; 0 < 1, using the theory of u0-positive operators with respect to a cone in a Banach space. We then obtain a comparison theorem for the smallest positive eigenvalues, 1 and 2, for the dieren tial equations ( p(u 00 (t))) 00 = 1f(t)u(t) and ( p(u 00 (t))) 00 = 2g(t)u(t) where 0 f(t) g(t); t 2 (0; 1).


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.
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) 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 M u N u 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, Of the next two results, the first can be found in Krasnosel'skiȋ [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 Remark: It is well known that the function φ p is invertible and that its inverse is φ q where p and q satisfy 1 p + 1 q = 1.Furthermore both φ p and φ q are increasing function.
Let y = −φ p (u (t)) in (3.3) and (3.2), and set α = β 1 , ξ = η 1 to obtain, Rewrite the differential equation as 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 Define P ⊂ B by 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 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.