SOLVABILITY OF A NONLINEAR SECOND ORDER CONJUGATE EIGENVALUE PROBLEM ON A TIME SCALE

In this paper, we are concerned with determining intervals of eigenvalues for boundary value problems for certain second order nonlinear differential equations on a time scale (also referred to, in the literature, as a measure chain). Much recent attention has been given to differential equations on time scales, and we refer the reader to [2, 4, 8] for some historical works as well as to the more recent papers [1, 5, 6] and the book [10] for excellent references on these types of equations. Before introducing the problems of interest for this paper, we present some definitions and notation which are common to the recent literature. Our sources for this background material are the two papers by Erbe and Peterson [5, 6].


Introduction
In this paper, we are concerned with determining intervals of eigenvalues for boundary value problems for certain second order nonlinear differential equations on a time scale (also referred to, in the literature, as a measure chain).Much recent attention has been given to differential equations on time scales, and we refer the reader to [2,4,8] for some historical works as well as to the more recent papers [1,5,6] and the book [10] for excellent references on these types of equations.Before introducing the problems of interest for this paper, we present some definitions and notation which are common to the recent literature.Our sources for this background material are the two papers by Erbe and Peterson [5,6].Definition 1.1.Let T be a closed subset of R, and let T have the subspace topology inherited from the Euclidean topology on R. The set T is referred to as a time scale or a measure chain.For t < sup T and r > inf T, define the forward jump operator, σ, and the backward jump operator, ρ, respectively, by If σ (t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be left scattered.If σ (t) = t, t is said to be right dense, and if ρ(r) = r, r is said to be left dense.
We refer to (1) as the continuous case and (2) as the discrete case for obvious reasons.However, we are not limited to these "extremes."Examples (3) and ( 4) are viable time scales as well.Note that (1) is both left dense and right dense for each t ∈ T while ( 2) is left scattered and right scattered for each t ∈ T. On the other hand, (3) is left dense except at the left endpoints of each interval and right dense except at the right endpoints of each interval.Finally, ( 4) is left scattered for each t ∈ (2, 3] and right scattered for each t ∈ [2, 3).Definition 1.3.For x : T → R and t ∈ T (if t = sup T, assume t is not left scattered), define the delta derivative of x(t), denoted by x (t), to be the number (when it exists), with the property that, for any > 0, there is a neighborhood, U, of t such that (1. 2) The second delta derivative of x(t) is defined by Remark 1.4.If T = R, then for t ∈ R, and any differentiable f : R → R, we have Hence, differential equations on this time scale are ordinary differential equations.
On the other hand, if T = Z, then for t ∈ Z and any sequence f : Z → R, we have Here f (t) = f (t + 1) − f (t) is the usual forward difference operator, and hence differential equations on this time scale are finite difference equations.We want to consider the nonlinear conjugate eigenvalue problem ) We make the following assumptions throughout: (A1) a(t) is a nonnegative, continuous function defined on (0, σ ( 1)) such that where We remark that by a solution of (1.8) and (1.9), we mean a function u : [0, σ 2 (1)] → R, where u satisfies (1.8) on [0, 1] and the boundary conditions (1.9).We further remark that, if u is a nonnegative solution of (1.8) and (1.9), then u (t) ≤ 0 on [0, 1], and we will say u is concave on [0, σ 2 (1)].
We note that (A1) allows for a(t) ≡ 0 on some subinterval(s) of [0, σ ( 1)] and allows a(t) to have a singularity at t = 0 and/or t = 1.Jiang and Liu [9] provided (1.12) as an example.Assumption (A1) is important because it admits a larger class of functions than those allowed in [3].This paper constitutes an extension of the recent work by Erbe and Peterson [6] and Chyan and Henderson [3] in which they obtained positive solutions of (1.8) and (1.9) for all 0 < λ < ∞ assuming that f is either superlinear or sublinear.The solutions obtained in [3,6] were found to belong to the intersection of a cone with an annular type region.Now we state a Green's function inequality which is fundamental in the proof of our main result.satisfying the boundary conditions (1.9).Define Proof.Erbe and Peterson [6] have found the explicit form of the Green's function to be (1.16) From the above inequality, we can see for t ∈ [0, s], 1) . (1.17) where g(t) = min{t, σ (1) − t} and h(s) = min{σ (s),σ (1) − σ (s)}.(We remark that in the case when s is right dense, h(s) = g(s).) Theorem 1.9 (Guo and Lakshmikantham [7, page 94]).Let E be a Banach space, K ⊆ E be a cone, and suppose that 1 , 2 are open subsets of E with 0 ∈ 1 and ¯ 1 ⊂ 2 .Suppose further that A :
Take our Banach space to be (2.1) Define the cone ᏼ ⊂ Ꮾ by g(t). (2.4) We are now ready to state our first theorem which establishes an open interval of values of λ for which (1.8) and (1.9) has a positive solution.
(2.6) We note that y(t) is a solution of (1.8) and (1.9) if and only if Motivated by this, we define the operator T : ᏼ → Ꮾ by We seek a fixed point of T in ᏼ.We prove this by showing that the conditions in Theorem 1.9 hold.First, if y ∈ ᏼ then (2.9) and so Next, if y ∈ ᏼ, then by (1.15) and inequality (2.10) we see that σ ( 1) T y . (2.11) Hence T : ᏼ → ᏼ.Standard arguments show that T is completely continuous.We begin with f 0 .By the definition of f 0 , there exists an Using (1.15) and the assumed right density of σ (1) (2.12) The last inequality follows from the right side of (2.6).Therefore, T y ≤ y and in particular Now we turn our attention to f ∞ .By the definition of f ∞ , there exists an (2.15) then we have shown that (2.17) An application of Theorem 1.9 yields the conclusion of our theorem.

.18)
there exists at least one solution of (1.8) and (1.9) in ᏼ.
Proof.Let λ be given as in (2.18) and let η be given such that

.19)
Let T be the cone preserving, completely continuous operator defined in (2.8).
Beginning with f 0 , there exists an (2.20) Using inequality (2.20), and the left side of (2.19), we obtain (2.21) Thus, T y ≥ y .So, if we define then we have shown that

.23)
Now we turn our attention to f ∞ .By the definition of f ∞ , there exists H2 > H 1 such that f (x) ≤ (f ∞ +η)x for x ≥ H2 .We have two cases: when f is bounded and when f is not bounded.First, suppose f is bounded, then there exists N > 0 such that f (x) ≤ N for all 0 < x < ∞.Let  (2.28) An application of Theorem 1.9 yields the conclusion of our theorem.

then define the integral byRemark 1 . 6 .Definition 1 . 7 .
If T = R, then the integral is the usual Riemann integral.If T = Z, then the integral on this time scale is the usual summation operator.Define the closed interval, [a, b] ⊂ T by [a, b] := t ∈ T | a ≤ t ≤ b .(1.7) Other closed, open, half-open, and half-closed intervals in T are similarly defined.John M. Davis et al. 93

Theorem 1 . 8 .
Let G(t, s) denote the Green's function for the homogeneous problem