# Eigenvalues of complementary Lidstone boundary value problems

## Abstract

We consider the following complementary Lidstone boundary value problem

$\begin{array}{c}{\left(-1\right)}^{m}{y}^{\left(2m+1\right)}\left(t\right)=\lambda F\left(t,y\left(t\right),{y}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)\\ y\left(0\right)=0,\phantom{\rule{1em}{0ex}}{y}^{\left(2k-1\right)}\left(0\right)={y}^{\left(2k-1\right)}\left(1\right)=0,\phantom{\rule{1em}{0ex}}1\le k\le m\end{array}$

where λ > 0. The values of λ are characterized so that the boundary value problem has a positive solution. Moreover, we derive explicit intervals of λ such that for any λ in the interval, the existence of a positive solution of the boundary value problem is guaranteed. Some examples are also included to illustrate the results obtained. Note that the nonlinear term F depends on y' and this derivative dependence is seldom investigated in the literature.

AMS Subject Classification: 34B15.

## 1 Introduction

In this article, we shall consider the complementary Lidstone boundary value problem

$\begin{array}{c}{\left(-1\right)}^{m}{y}^{\left(2m+1\right)}\left(t\right)=\lambda F\left(t,y\left(t\right),{y}^{\prime }\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)\\ y\left(0\right)=0,\phantom{\rule{1em}{0ex}}{y}^{\left(2k-1\right)}\left(0\right)={y}^{\left(2k-1\right)}\left(1\right)=0,\phantom{\rule{1em}{0ex}}1\le k\le m\end{array}$
(1.1)

where m ≥ 1, λ > 0, and F is continuous at least in the interior of the domain of interest. Note that the nonlinear term F involves a derivative of the dependent variable--this is seldom studied in the literature and most research articles on boundary value problems consider nonlinear terms that involve y only.

We are interested in the existence of a positive solution of (1.1). By a positive solution y of (1.1), we mean a nontrivial y C(2m+1)(0, 1) satisfying (1.1) and y(t) ≥ 0 for t (0, 1). If, for a particular λ the boundary value problem (1.1) has a positive solution y, then λ is called an eigenvalue and y is a corresponding eigenfunction of (1.1). We shall denote the set of eigenvalues of (1.1) by E, i.e.,

$E=\left\{\lambda >0|\left(1.1\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{has}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{a}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{positive}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{solution}}\right\}.$

The focus of this article is eigenvalue problem, as such we shall characterize the values of λ so that the boundary value problem (1.1) has a positive solution. To be specific, we shall establish criteria for E to contain an interval, and for E to be an interval (which may either be bounded or unbounded). In addition explicit subintervals of E are derived.

The complementary Lidstone interpolation and boundary value problems are very recently introduced in , and studied by Agarwal et. al. [2, 3] where they consider an odd order ((2m+ 1)th order) differential equation together with boundary data at the odd order derivatives

$y\left(0\right)={a}_{0},\phantom{\rule{1em}{0ex}}{y}^{\left(2k-1\right)}\left(0\right)={a}_{k},\phantom{\rule{1em}{0ex}}{y}^{\left(2k-1\right)}\left(1\right)={b}_{k},\phantom{\rule{1em}{0ex}}1\le k\le m.$
(1.2)

The boundary conditions (1.2) are known as complementary Lidstone boundary conditions, they naturally complement the Lidstone boundary conditions  which involve even order derivatives. To be precise, the Lidstone boundary value problem comprises an even order (2m th order) differential equation and the Lidstone boundary conditions

${y}^{\left(2k\right)}\left(0\right)={a}_{k},\phantom{\rule{1em}{0ex}}{y}^{\left(2k\right)}\left(1\right)={b}_{k},\phantom{\rule{1em}{0ex}}0\le k\le m-1.$
(1.3)

There is a vast literature on Lidstone interpolation and boundary value problems. The Lidstone interpolation has a long history from 1929 when Lidstone  introduced a generalization of Taylor's series that approximates a given function in the neighborhood of two points instead of one. Further characterization can be found in the study of . More research on Lidstone interpolation as well as Lidstone spline is seen in [1, 1723]. On the other hand, the Lidstone boundary value problems and several of its particular cases have been the subject matter of numerous investigations, see [4, 18, 2437] and the references cited therein. It is noted that in most of these studies the nonlinear terms considered do not involve derivatives of the dependent variable, only a handful of articles [30, 31, 34, 35] tackle nonlinear terms that involve even order derivatives. In the present study, our study of the complementary Lidstone boundary value problem (1.1) where F depends on a derivative certainly extends and complements the rich literature on boundary value problems and in particular on Lidstone boundary value problems.

The plan of the article is as follows. In Section 2, we shall state a fixed point theorem due to Krasnosel'skii , and develop some inequalities for certain Green's function which are needed later. The characterization of the set E is presented in Section 3. Finally, in Section 4, we establish explicit subintervals of E.

## 2 Preliminaries

Theorem 2.1.  Let B be a Banach space, and let C( B) be a cone. Assume Ω1, Ω2 are open subsets of B with $0\in {\Omega }_{1},{\stackrel{̄}{\Omega }}_{1}\subset {\Omega }_{2}$, and let $S:C\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)\to C$ be a completely continuous operator such that, either

1. (a)

Syy, y C Ω1, and Syy, y C Ω2, or

2. (b)

Syy, y C Ω1, and Syy, y C Ω2.

Then, S has a fixed point in $C\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)$.

To tackle the complementary Lidstone boundary value problem (1.1), let us review certain attributes of the Lidstone boundary value problem. Let g m (t, s) be the Green's function of the Lidstone boundary value problem

$\begin{array}{c}{x}^{\left(2m\right)}\left(t\right)=0,\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)\\ {x}^{\left(2k\right)}\left(0\right)={x}^{\left(2k\right)}\left(1\right)=0,\phantom{\rule{1em}{0ex}}0\le k\le m-1.\end{array}$
(2.1)

The Green's function g m (t, s) can be expressed as [4, 5]

${g}_{m}\left(t,s\right)=\underset{0}{\overset{1}{\int }}g\left(t,u\right){g}_{m-1}\left(u,s\right)du$
(2.2)

where

${g}_{1}\left(t,s\right)=g\left(t,s\right)=\left\{\begin{array}{c}\hfill t\left(s-1\right),0\le t\le s\le 1\hfill \\ \hfill s\left(t-1\right),0\le s\le t\le 1.\hfill \end{array}\right\$
(2.3)

Further, it is known that

$\left|{g}_{m}\left(t,s\right)\right|={\left(-1\right)}^{m}{g}_{m}\left(t,s\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{g}_{m}\left(t,s\right)={g}_{m}\left(s,t\right),\phantom{\rule{1em}{0ex}}\left(t,s\right)\in \left(0,1\right)×\left(0,1\right).$
(2.4)

We also have the inequality

$\frac{2}{{\pi }^{2}}\mathsf{\text{sin}}\pi t\le t\left(1-t\right)\le \frac{1}{\pi }\mathsf{\text{sin}}\pi t,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(2.5)

The following two lemmas give the upper and lower bounds of |g m (t, s)|, they play an important role in subsequent development.

Lemma 2.1. For (t, s) [0, 1] × [0, 1], we have

$\left|{g}_{m}\left(t,s\right)\right|\le \frac{1}{{\pi }^{2m-1}}\mathsf{\text{sin}}\pi s.$
(2.6)

Proof. For (t, s) [0, 1] × [0, 1], it is clear from (2.3) that

$\left|g\left(t,s\right)\right|\le s\left(1-s\right).$
(2.7)

Using (2.7), (2.4), and (2.5) in (2.2) yields for (t, s) [0, 1] × [0, 1],

$\begin{array}{ll}\hfill \left|{g}_{m}\left(t,s\right)\right|=\underset{0}{\overset{1}{\int }}\left|g\left(t,u\right)\right|\cdot \left|{g}_{m-1}\left(u,s\right)\right|du& \le \underset{0}{\overset{1}{\int }}\left|{g}_{m-1}\left(u,s\right)\right|u\left(1-u\right)du\phantom{\rule{2em}{0ex}}\\ \le \frac{1}{\pi }\underset{0}{\overset{1}{\int }}\left|{g}_{m-1}\left(s,u\right)\right|\mathsf{\text{sin}}\pi udu.\phantom{\rule{2em}{0ex}}\end{array}$
(2.8)

By induction, we can show that

$\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\mathsf{\text{sin}}\pi sds=\frac{1}{{\pi }^{2m}}\mathsf{\text{sin}}\pi t,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(2.9)

Now (2.6) is immediate by applying (2.9) to (2.8).

Lemma 2.2. Let $\delta \in \left(0,\frac{1}{2}\right)$ be given. For (t, s) [δ, 1-δ] × [0, 1], we have

$\left|{g}_{m}\left(t,s\right)\right|\ge \frac{2\delta }{{\pi }^{2m}}\mathsf{\text{sin}}\pi s.$
(2.10)

Proof. For (t, s) [δ, 1-δ] × [0, 1], from (2.3) we find

$\left|g\left(t,s\right)\right|\ge \left\{\begin{array}{cc}\hfill \delta \left(1-s\right),\hfill & \hfill 1\le s\hfill \\ \hfill \left[1-\left(1-\delta \right)\right]s,\hfill & \hfill s\le t\hfill \end{array}\right\}\ge \delta s\left(1-s\right).$
(2.11)

Then, using (2.11), (2.4), and (2.5) in (2.2), we get for (t, s) [δ, 1 - δ ] × [0, 1],

$\begin{array}{ll}\hfill \left|{g}_{m}\left(t,s\right)\right|=\underset{0}{\overset{1}{\int }}\left|g\left(t,u\right)\right|\cdot \left|{g}_{m-1}\left(u,s\right)\right|du& \ge \delta \underset{0}{\overset{1}{\int }}\left|{g}_{m-1}\left(u,s\right)\right|u\left(1-u\right)du\phantom{\rule{2em}{0ex}}\\ \ge \frac{2\delta }{{\pi }^{2}}\underset{0}{\overset{1}{\int }}\left|{g}_{m-1}\left(s,u\right)\right|\mathsf{\text{sin}}\pi udu,\phantom{\rule{2em}{0ex}}\end{array}$

which, in view of (2.9), gives (2.10) immediately.

Remark 2.1. The bounds in Lemmas 2.1 and 2.2 are sharper than those given in the literature [4, 5, 35, 37].

## 3 Eigenvalues of (1.1)

To tackle (1.1) we first consider the initial value problem

$\begin{array}{c}{y}^{\prime }\left(t\right)=x\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)\\ y\left(0\right)=0\end{array}$
(3.1)

whose solution is simply

$y\left(t\right)=\underset{0}{\overset{t}{\int }}x\left(s\right)ds.$
(3.2)

Taking into account (3.1) and (3.2), the complementary Lidstone boundary value problem (1.1) reduces to the Lidstone boundary value problem

$\begin{array}{c}{\left(-1\right)}^{m}{x}^{\left(2m\right)}\left(t\right)=\lambda F\left(t,\underset{0}{\overset{t}{\int }}x\left(s\right)ds,x\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,1\right)\\ {x}^{\left(2k-2\right)}\left(0\right)={x}^{\left(2k-2\right)}\left(1\right)=0,\phantom{\rule{1em}{0ex}}1\le k\le m.\end{array}$
(3.3)

If (3.3) has a positive solution x*, then by virtue of (3.2), ${y}^{*}\left(t\right)={\int }_{0}^{t}{x}^{*}\left(s\right)ds$ is a positive solution of (1.1). Hence, the existence of a positive solution of the complementary Lidstone boundary value problem (1.1) follows from the existence of a positive solution of the Lidstone boundary value problem (3.3). It is clear that an eigenvalue of (3.3) is also an eigenvalue of (1.1), thus

$E=\left\{\lambda >0|\left(1.1\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{has}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{a}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{positive}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{solution}}\right\}=\left\{\lambda >0|\left(3.3\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\mathsf{\text{has}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{a}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{positive}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{solution}}\right\}.$

With the lemmas developed in Section 2 and a technique to handle the nonlinear term F, we shall study the eigenvalue problem (1.1) via (3.3).

For easy reference, we list below the conditions that are used later. In these conditions, f, α, and β are continuous functions with f : (0, ∞) × (0, ∞) → (0, ∞) and α, β : (0, 1) → [0, ∞).

(A1) f is nondecreasing in each of its arguments, i.e., for u, u1, u2, v, v1, v2 (0, ∞) with u1u2 and v1v2, we have

$f\left({u}_{1},v\right)\le f\left({u}_{2},v\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}f\left(u,{v}_{1}\right)\le f\left(u,{v}_{2}\right);$

(A2) for t (0, 1) and u, v (0, ∞),

$\alpha \left(t\right)f\left(u,v\right)\le F\left(t,u,v\right)\le \beta \left(t\right)f\left(u,v\right);$

(A3) α(t) is not identically zero on any nondegenerate subinterval of (0, 1) and there exists a0 (0, 1] such that α(t) ≥ a0β(t) for all t (0, 1);

(A4) $0<\underset{0}{\overset{1}{\int }}\beta \left(t\right)\mathsf{\text{sin}}\pi tdt<\infty$;

(A5) for t (0, 1) and u, u1, u2, v, v1, v2 (0, ∞) with u1u2 and v1v2, we have

$F\left(t,{u}_{1},v\right)\le F\left(t,{u}_{2},v\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}F\left(t,u,{v}_{1}\right)\le F\left(t,u,{v}_{2}\right).$

We shall consider the Banach space B = C[0, 1] equipped with the norm

$∥x∥=\underset{t\in \left[0,1\right]}{\mathsf{\text{sup}}}\left|x\left(t\right)\right|,\phantom{\rule{1em}{0ex}}x\in B.$

For a given $\delta \in \left(0,\frac{1}{2}\right)$, let the cone C δ be defined by

${C}_{\delta }=\left\{x\in B\left|x\left(t\right)\ge 0,\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right]\right;\underset{t\in \left[\delta ,1-\delta \right]}{\mathsf{\text{min}}}x\left(t\right)\ge \gamma ∥x∥\right\}$

where $\gamma =\frac{2\delta }{\pi }{a}_{0}$ (a0 is defined in (A3)). Further, let

${C}_{\delta }\left(M\right)=\left\{x\in {C}_{\delta }|∥x∥\le M\right\}.$

Let the operator S : C δ B be defined by

$\begin{array}{ll}\hfill Sx\left(t\right)& =\lambda \underset{0}{\overset{1}{\int }}{\left(-1\right)}^{m}{g}_{m}\left(t,s\right)F\left(s,\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ =\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].\phantom{\rule{2em}{0ex}}\end{array}$
(3.4)

To obtain a positive solution of (3.3), we shall seek a fixed point of the operator S in the cone C δ .

Further, we define the operators U, V : C δ B by

$Ux\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds$

and

$Vx\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds.$

If (A2) holds, then

$Ux\left(t\right)\le Sx\left(t\right)\le Vx\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left[0,1\right].$
(3.5)

Lemma 3.1. Let (A1)-(A4) hold. Then, the operator S is compact on the cone C δ .

Proof. Let us consider the case when α(t) is unbounded in a deleted right neighborhood of 0 and also in a deleted left neighborhood of 1. Clearly, β(t) is also unbounded near 0 and 1.

For n {1, 2, 3, ...}, let α n , β n : [0, 1] → [0, ∞) be defined by

${\alpha }_{n}\left(t\right)=\left\{\begin{array}{c}\hfill \alpha \left(\frac{1}{n+1}\right),0\le t\le \frac{1}{n+1}\hfill \\ \hfill \phantom{\rule{1em}{0ex}}\alpha \left(t\right),\phantom{\rule{1em}{0ex}}\frac{1}{n+1}\le t\le \frac{n}{n+1}\hfill \\ \hfill \alpha \left(\frac{n}{n+1}\right),\frac{n}{n+1}\le t\le 1\hfill \end{array}\right\$

and

${\beta }_{n}\left(t\right)=\left\{\begin{array}{c}\hfill \beta \left(\frac{1}{n+1}\right),0\le t\le \frac{1}{n+1}\hfill \\ \hfill \phantom{\rule{1em}{0ex}}\beta \left(t\right),\phantom{\rule{1em}{0ex}}\frac{1}{n+1}\le t\le \frac{n}{n+1}\hfill \\ \hfill \beta \left(\frac{n}{n+1}\right),\frac{n}{n+1}\le t\le 1.\hfill \end{array}\right\$

Also, we define the operators U n , V n : C δ B by

${U}_{n}x\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|{\alpha }_{n}\left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds$

and

${V}_{n}x\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|{\beta }_{n}\left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds.$

It is standard that for each n, both U n and V n are compact operators on C δ . Let M > 0 and x C δ (M). For t [0, 1], we get

$\begin{array}{ll}\hfill \left|{V}_{n}x\left(t\right)-Vx\left(t\right)\right|& \le \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\left|{\beta }_{n}\left(s\right)-\beta \left(s\right)\right|f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\frac{1}{n+1}}{\int }}\left|{g}_{m}\left(t,s\right)\right|\left|{\beta }_{n}\left(s\right)-\beta \left(s\right)\right|f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\underset{\frac{n}{n+1}}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\left|{\beta }_{n}\left(s\right)-\beta \left(s\right)\right|f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds.\phantom{\rule{2em}{0ex}}\end{array}$

By the monotonicity of f (see (A1)), we have

$f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\le f\left(\underset{0}{\overset{1}{\int }}Md\tau ,M\right)=f\left(M,M\right)$
(3.6)

Coupling with Lemma 2.1, it follows that

The integrability of β(t) sin πt (see (A4)) ensures that V n converges uniformly to V on C δ (M). Hence, V is compact on C δ . By a similar argument, we see that U n converges uniformly to U on C δ (M) and therefore U is also compact on C δ . It follows immediately from inequality (3.5) that the operator S is compact on C δ .

Remark 3.1. From the proof of Lemma 3.1, we see that if the functions α and β are continuous on the close interval [0, 1], then the conditions (A1) and (A4) are not needed in Lemma 3.1.

The first result shows that E contains an interval.

Theorem 3.1. Let (A1)-(A4) hold. Then, there exists c > 0 such that the interval (0, c] E.

Proof. Let M > 0 be given. Define

$c=\frac{M}{f\left(M,M\right)}{\left[\underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}.$
(3.7)

Let λ (0, c]. We shall prove that S(C δ (M)) C δ (M). Let x C δ (M). First, we shall show that Sx C δ . It is clear from (3.5) that

$Sx\left(t\right)\ge \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\ge 0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.8)

Further, from (3.5) and Lemma 2.1 we get

$\begin{array}{ll}\hfill Sx\left(t\right)& \le \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{sin}}\pi sds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right]\phantom{\rule{2em}{0ex}}\end{array}$

$∥Sx∥\le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{sin}}\pi sds.$
(3.9)

Now, applying (3.5), Lemma 2.2, (A3) and (3.9) successively, we find for t [δ, 1-δ],

$\begin{array}{ll}\hfill Sx\left(t\right)& \ge \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{0}{\overset{1}{\int }}\frac{2\delta }{{\pi }^{2m}}\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{0}{\overset{1}{\int }}\frac{2\delta }{{\pi }^{2m}}{a}_{0}\beta \left(s\right)f\phantom{\rule{2.77695pt}{0ex}}\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \ge \frac{2\delta }{\pi }{a}_{0}∥Sx∥=\gamma ∥Sx∥.\phantom{\rule{2em}{0ex}}\end{array}$

Therefore,

$\underset{t\in \left[\delta ,1-\delta \right]}{\mathsf{\text{min}}}Sx\left(t\right)\ge \gamma ∥Sx∥.$
(3.10)

Inequalities (3.8) and (3.10) imply that Sx C δ .

Next, we shall verify that SxM. For this, an application of (3.5), Lemma 2.1, (3.6) and (3.7) provides

$\begin{array}{ll}\hfill Sx\left(t\right)& \le c\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ \le cf\left(M,M\right)\underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\mathsf{\text{sin}}\pi sds=M,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right]\phantom{\rule{2em}{0ex}}\end{array}$

or equivalently

$∥Sx∥\le M.$

Hence, S(C δ (M)) C δ (M). Also, the standard arguments yield that S is completely continuous. By Schauder fixed point theorem, S has a fixed point in C δ (M). Clearly, this fixed point is a positive solution of (3.3) and therefore λ is an eigenvalue of (3.3). Since λ (0, c] is arbitrary, it follows immediately that the interval (0, c] E.

Remark 3.2. From the proof of Theorem 3.1, we see that (A2) and (A3) lead to S : C δ C δ .

Theorem 3.2. Let (A1)-(A5) hold. Suppose that λ* E, for any λ (0, λ*), we have λ E, i.e., (0, λ*] E.

Proof. Let x* be the eigenfunction corresponding to the eigenvalue λ*. Thus, we have

${x}^{*}\left(t\right)=S{x}^{*}\left(t\right)={\lambda }^{*}\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}{x}^{*}\left(\tau \right)d\tau ,{x}^{*}\left(s\right)\right)ds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.11)

Define

${K}^{*}=\left\{x\in B\left|0\le x\left(t\right)\le {x}^{*}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right]\right\right\}.$

Let λ (0, λ*) and x K*. Using (A5), we get

$\begin{array}{ll}\hfill 0\le Sx\left(t\right)& =\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \le {\lambda }^{*}\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}{x}^{*}\left(\tau \right)d\tau ,{x}^{*}\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ =S{x}^{*}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right]\phantom{\rule{2em}{0ex}}\end{array}$

where the last equality follows from (3.11). This immediately implies that the operator S maps K* into K*. Moreover, the operator S is continuous and completely continuous. Schauder's fixed point theorem guarantees that S has a fixed point in K*, which is a positive solution of (3.3). Hence, λ is an eigenvalue, i.e., λ E.

The following result shows that E is an interval.

Corollary 3.1. Let (A1)-(A5) hold. If $E\ne \varnothing$, then E is an interval.

Proof. Suppose E is not an interval. Then, there exist ${\lambda }_{0},{\lambda }_{0}^{\prime }\in E\left({\lambda }_{0}<{\lambda }_{0}^{\prime }\right)$ and $\tau \in \left({\lambda }_{0},{\lambda }_{0}^{\prime }\right)$ with τ E. However, this is not possible as Theorem 3.2 guarantees that τ E. Hence, E is an interval.

The following two results give the upper and lower bounds of an eigenvalue in terms of some parameters of the corresponding eigenfunction.

Theorem 3.3. Let (A1) and (A2) hold. Assume that m is odd. Let λ be an eigenvalue of (3.3) and x C δ be a corresponding eigenfunction. If x(i)(0) = b i , i = 1, 3, ..., 2m - 1, where b2m-1> 0, then λ satisfies

${M}_{1}\le \lambda \le {M}_{2}$
(3.12)

where

$\begin{array}{c}{M}_{1}=\underset{0\le k\le m-1}{\mathsf{\text{max}}}\left[\sum _{i=k}^{m-1}\frac{{b}_{2i+1}}{\left(2\left(i-k\right)+1\right)!}\right]{\left[f\left(D,D\right)\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}\beta \left(s\right)ds\right]}^{-1},\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{M}_{2}=\underset{0\le k\le m-1}{\mathsf{\text{min}}}\left[\sum _{i=k}^{m-1}\frac{{b}_{2i+1}}{\left(2\left(i-k\right)+1\right)!}\right]{\left[f\left(0,0\right)\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}\alpha \left(s\right)ds\right]}^{-1}\end{array}$

and

$D=\underset{t\in \left[0,1\right]}{\mathsf{\text{max}}}\sum _{i=0}^{m-1}{b}_{2i+1}\frac{{t}^{2i+1}}{\left(2i+1\right)!}.$

Proof. For n {1, 2, 3, ...}, we define f n = f * ω n , where ω n is a standard mollifier  such that f n is Lipschitz and converges uniformly to f.

For a fixed n, let λ n be an eigenvalue and x n (t), with ${x}_{n}^{\left(i\right)}\left(0\right)={b}_{i},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}i=1,3,\dots ,2m-1$ be a corresponding eigenfunction of the following boundary value problem

${\left(-1\right)}^{m}{x}_{n}^{\left(2m\right)}\left(t\right)={\lambda }_{n}{F}_{n}\left(t,\underset{0}{\overset{t}{\int }}{x}_{n}\left(s\right)ds,{x}_{n}\left(t\right)\right),\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right]$
(3.13)
${x}_{n}^{\left(2i\right)}\left(0\right)={x}_{n}^{\left(2i\right)}\left(1\right)=0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0\le i\le m-1$
(3.14)

where F n converges uniformly to F, and for u, v (0, ∞),

${\alpha }_{n}\left(t\right){f}_{n}\left(u,v\right)\le {F}_{n}\left(t,u,v\right)\le {\beta }_{n}\left(t\right){f}_{n}\left(u,v\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left(0,1\right)$
(3.15)

(see the proof of Lemma 3.1 for the definitions of α n (t) and β n (t)).

It is clear that x n (t) is the unique solution of the initial value problem (3.13),

$\begin{array}{ll}\hfill {x}_{n}^{\left(i\right)}\left(0\right)& =0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}i=0,2,\dots ,2m-2\phantom{\rule{2em}{0ex}}\\ \hfill {x}_{n}^{\left(i\right)}\left(0\right)& ={b}_{i},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}i=1,3,\dots ,2m-1.\phantom{\rule{2em}{0ex}}\end{array}$
(3.16)

First, we shall establish an upper bound for x n . Since

${\left(-1\right)}^{m}{x}_{n}^{\left(2m\right)}\left(t\right)={\lambda }_{n}{F}_{n}\left(t,\underset{0}{\overset{t}{\int }}{x}_{n}\left(s\right)ds,{x}_{n}\left(t\right)\right)\ge {\lambda }_{n}{\alpha }_{n}\left(t\right){f}_{n}\left(\underset{0}{\overset{t}{\int }}{x}_{n}\left(s\right)ds,{x}_{n}\left(t\right)\right)\ge 0,$

we have ${x}_{n}^{\left(2m-1\right)}\left(t\right)$ is nonincreasing and hence

${x}_{n}^{\left(2m-1\right)}\left(t\right)\le {x}_{n}^{\left(2m-1\right)}\left(0\right)={b}_{2m-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.17)

In view of the initial conditions (3.16) and also (3.17), we find

${x}_{n}^{\left(2m-2\right)}\left(t\right)=\underset{0}{\overset{t}{\int }}{x}_{n}^{\left(2m-1\right)}\left(s\right)ds\le {b}_{2m-1}t,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.18)

Next, an application of (3.18) gives

${x}_{n}^{\left(2m-3\right)}\left(t\right)={b}_{2m-3}+\underset{0}{\overset{t}{\int }}{x}_{n}^{\left(2m-2\right)}\left(s\right)ds\le {b}_{2m-3}+{b}_{2m-1}\frac{{t}^{2}}{2!},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$

By repeating the process, we get

${x}_{n}\left(t\right)\le {b}_{1}t+{b}_{3}\frac{{t}^{3}}{3!}+\cdots +{b}_{2m-1}\frac{{t}^{2m-1}}{\left(2m-1\right)!}\le D,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.19)

By the monotonicity of f n , we have

${f}_{n}\left(\underset{0}{\overset{t}{\int }}{x}_{n}\left(s\right)ds,{x}_{n}\left(t\right)\right)\le {f}_{n}\left(\underset{0}{\overset{1}{\int }}Dds,D\right)={f}_{n}\left(D,D\right)$

and

${f}_{n}\left(\underset{0}{\overset{t}{\int }}{x}_{n}\left(s\right)ds,{x}_{n}\left(t\right)\right)\ge {f}_{n}\left(\underset{0}{\overset{t}{\int }}0ds,0\right)={f}_{n}\left(0,0\right).$

Coupling with (3.13) and (3.15), it follows that

${\lambda }_{n}{\alpha }_{n}\left(t\right){f}_{n}\left(0,0\right)\le {\left(-1\right)}^{m}{x}_{n}^{\left(2m\right)}\left(t\right)\le {\lambda }_{n}{\beta }_{n}\left(t\right){f}_{n}\left(D,D\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.20)

Once again, using the initial conditions (3.16), repeated integration of (3.20) from 0 to t provides

${\varphi }_{1,k}\left(t\right)\le {x}_{n}^{\left(2k\right)}\left(t\right)\le {\varphi }_{2,k}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right],\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0\le k\le m-1$
(3.21)

where

${\varphi }_{1,k}\left(t\right)=\sum _{i=k}^{m-1}{b}_{2i+1}\frac{{t}^{2\left(i-k\right)+1}}{\left(2\left(i-k\right)+1\right)!}-{\lambda }_{n}{f}_{n}\left(D,D\right)\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}{\beta }_{n}\left(s\right)ds$

and

${\varphi }_{2,k}\left(t\right)=\sum _{i=k}^{m-1}{b}_{2i+1}\frac{{t}^{2\left(i-k\right)+1}}{\left(2\left(i-k\right)+1\right)!}-{\lambda }_{n}{f}_{n}\left(0,0\right)\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}{\alpha }_{n}\left(s\right)ds.$

In order to satisfy the boundary conditions ${x}_{n}^{\left(2k\right)}\left(1\right)=0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0\le k\le m-1$, from inequality (3.21) it is necessary that

${\varphi }_{1,k}\left(1\right)\le 0\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{\varphi }_{2,k}\left(1\right)\ge 0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0\le k\le m-1.$

${M}_{1,n}\le {\lambda }_{n}\le {M}_{2,n}$
(3.22)

where

${M}_{1,n}=\underset{0\le k\le m-1}{\mathsf{\text{max}}}\left[\sum _{i=k}^{m-1}\frac{{b}_{2i+1}}{\left(2\left(i-k\right)+1\right)!}\right]{\left[{f}_{n}\left(D,D\right)\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}{\beta }_{n}\left(s\right)ds\right]}^{-1}$

and

${M}_{2,n}=\underset{0\le k\le m-1}{\mathsf{\text{min}}}\left[\sum _{i=k}^{m-1}\frac{{b}_{2i+1}}{\left(2\left(i-k\right)+1\right)!}\right]{\left[{f}_{n}\left(0,0\right)\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{2\left(m-k\right)-1}}{\left(2\left(m-k\right)-1\right)!}{\alpha }_{n}\left(s\right)ds\right]}^{-1}.$

From (3.20) it is observed (by using the initial conditions (3.16) and repeated integration) that ${\left\{{x}_{n}^{\left(i\right)}\right\}}_{n=1}^{\infty },\phantom{\rule{2.77695pt}{0ex}}0\le i\le 2m-1$ is a uniformly bounded sequence on [0, 1]. Thus, there exists a subsequence, which can be relabeled as ${\left\{{x}_{n}\right\}}_{n=1}^{\infty }$, that converges uniformly (in fact, in C(2m-1)-norm) to some x on [0, 1]. We note that each x n (t) can be expressed as

${x}_{n}\left(t\right)={\lambda }_{n}\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|{F}_{n}\left(s,\underset{0}{\overset{s}{\int }}{x}_{n}\left(\tau \right)d\tau ,{x}_{n}\left(s\right)\right)ds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.23)

Since ${\left\{{\lambda }_{n}\right\}}_{n=1}^{\infty }$ is a bounded sequence (from (3.22)), there is a subsequence, which can be relabeled as ${\left\{{\lambda }_{n}\right\}}_{n=1}^{\infty }$, that converges to some λ. Then, letting n → ∞ in (3.23) yields

$x\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$

This means that x(t) is an eigenfunction of (3.3) corresponding to the eigenvalue λ. Further, x(i)(0) = b i , i = 1, 3, ..., 2m - 1 and inequality (3.12) follows from (3.22) immediately.

Theorem 3.4. Let (A1)-(A4) hold. Let λ be an eigenvalue of (3.3) and x C δ be a corresponding eigenfunction. Further, let x = p. Then,

$\lambda \ge \frac{p}{f\left(p,p\right)}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}$
(3.24)

and

$\lambda \le \frac{p}{f\left(\gamma p\left(\frac{1}{2}-\delta \right),\gamma p\right)}{\left[\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)ds\right]}^{-1}$
(3.25)

where t1 is any number in (0, 1) such that x(t1) ≠ 0.

Proof. Let t0 [0, 1] be such that

$p=∥x∥=x\left({t}_{0}\right).$

Then, using (3.5), Lemma 2.1 and the monotonicity of f, we find

$\begin{array}{ll}\hfill p=x\left({t}_{0}\right)=Sx\left({t}_{0}\right)& \le \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left({t}_{0},s\right)\right|\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{1}{\int }}pd\tau ,p\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ =\lambda \frac{f\left(p,p\right)}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\end{array}$

Next, we employ (3.5), the monotonicity of f and the fact that mint[δ, 1-δ]x(t) ≥ γp to get

$\begin{array}{ll}\hfill p\ge x\left({t}_{1}\right)& \ge \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{\frac{1}{2}}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)f\left(\underset{\delta }{\overset{\frac{1}{2}}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)f\left(\underset{\delta }{\overset{\frac{1}{2}}{\int }}\gamma p\phantom{\rule{2.77695pt}{0ex}}d\tau ,\gamma p\right)ds\phantom{\rule{2em}{0ex}}\\ =\lambda f\left(\gamma p\left(\frac{1}{2}-\delta \right),\gamma p\right)\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}_{1},s\right)\right|\alpha \left(s\right)ds\phantom{\rule{2em}{0ex}}\end{array}$

from which (3.25) is immediate.

The following result gives the criteria for E to be a bounded/unbounded interval.

Theorem 3.5. Define

$\begin{array}{ll}\hfill {W}_{B}& =\left\{f\left|\frac{u}{f\left(u,u\right)}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{is}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{bounded}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}u\in \left(0,\infty \right)\right\right\},\phantom{\rule{2em}{0ex}}\\ \hfill {W}_{0}& =\left\{f\left|\underset{u\to \infty }{\mathsf{\text{lim}}}\frac{u}{f\left(u,u\right)}=0\right\right\},\phantom{\rule{2em}{0ex}}\\ \hfill {W}_{\infty }& =\left\{f\left|\underset{u\to \infty }{\mathsf{\text{lim}}}\frac{u}{f\left(u,u\right)}=\infty \right\right\}.\phantom{\rule{2em}{0ex}}\end{array}$
1. (a)

Let (A1)-(A5) hold. If f W B , then E = (0, c) or (0, c] for some c (0, ∞).

2. (b)

Let (A1)-(A5) hold. If f W0, then E = (0, c] for some c (0, ∞).

3. (c)

Let (A1)-(A4) hold. If f W, then E = (0, ∞).

Proof. (a) This is immediate from (3.25) and Corollary 3.1.

1. (b)

Since W0 W B , it follows from Case (a) that E = (0, c) or (0, c] for some c (0, ∞).

In particular,

$c=\mathsf{\text{sup}}\phantom{\rule{0.3em}{0ex}}E.$

Let ${\left\{{\lambda }_{n}\right\}}_{n=1}^{\infty }$ be a monotonically increasing sequence in E which converges to c, and let ${\left\{{x}_{n}\right\}}_{n=1}^{\infty }$ be a corresponding sequence of eigenfunctions in the context of (3.3). Further, let p n = x n . Then, (3.25) together with f W0 implies that no subsequence of ${\left\{{p}_{n}\right\}}_{n=1}^{\infty }$ can diverge to infinity. Thus, there exists R > 0 such that p n R for all n. So ${\left\{{x}_{n}\right\}}_{n=1}^{\infty }$ is uniformly bounded. This implies that there is a subsequence of ${\left\{{x}_{n}\right\}}_{n=1}^{\infty }$, relabeled as the original sequence, which converges uniformly to some x, where x(t) ≥ 0 for t [0, 1]. Clearly, we have Sx n = x n , i.e.,

${x}_{n}\left(t\right)={\lambda }_{n}\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}{x}_{n}\left(\tau \right)d\tau ,{x}_{n}\left(s\right)\right)ds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$
(3.26)

Since x n converges to x and λ n converges to c, letting n → ∞ in (3.26) yields

$x\left(t\right)=c\underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|F\left(s,\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left[0,1\right].$

Hence, c is an eigenvalue with corresponding eigenfunction x, i.e., c = sup E E. This completes the proof for Case (b).

1. (c)

Let λ > 0 be fixed. Choose ε > 0 so that

$\frac{\lambda }{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\le \frac{1}{\epsilon }.$
(3.27)

By definition, if f W, then there exists M = M(ε) > 0 such that

$f\left(u,u\right)<\epsilon \phantom{\rule{2.77695pt}{0ex}}u,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}u\ge M.$
(3.28)

We shall prove that S(C δ (M)) C δ (M). Let x C δ (M). As in the proof of Theorem 3.1, we have (3.8) and (3.10) and so Sx C δ . Thus, it remains to show that SxM. Using (3.5), Lemma 2.1, (3.6), (3.28), and (3.27), we find for t [0, 1],

$\begin{array}{ll}\hfill Sx\left(t\right)& \le \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \le \lambda f\left(M,M\right)\underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \epsilon M\underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\le M.\phantom{\rule{2em}{0ex}}\end{array}$

It follows that SxM and hence S(C δ (M)) C δ (M). Also, S is continuous and completely continuous. Schauder's fixed point theorem guarantees that S has a fixed point in C δ (M). Clearly, this fixed point is a positive solution of (3.3) and therefore λ is an eigenvalue of (3.3). Since λ > 0 is arbitrary, we have proved that E = (0, ∞).

Example 3.1. Consider the complementary Lidstone boundary value problem

$\begin{array}{c}{y}^{\left(5\right)}=\lambda {\left(\frac{{t}^{5}}{10}+\frac{{t}^{4}}{4}-{t}^{3}+\frac{{t}^{2}}{4}+\frac{t}{2}+2\right)}^{-q}{\left[\frac{y+{y}^{\prime }}{2}+2\right]}^{q},\phantom{\rule{2.77695pt}{0ex}}t\in \left(0,1\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y\left(0\right)={y}^{\prime }\left(0\right)={y}^{‴}\left(0\right)={y}^{\prime }\left(1\right)={y}^{‴}\left(1\right)=0\end{array}$
(3.29)

where λ > 0 and q ≥ 0.

Here, m = 2 and

$F\left(t,y,{y}^{\prime }\right)={\left(\frac{{t}^{5}}{10}+\frac{{t}^{4}}{4}-{t}^{3}+\frac{{t}^{2}}{4}+\frac{t}{2}+2\right)}^{-q}{\left(\frac{y+{y}^{\prime }}{2}+2\right)}^{q}.$

Clearly, F(t, u, v) is nondecreasing in u and v, thus (A5) is satisfied.

Choose

$\alpha \left(t\right)=\beta \left(t\right)={\left(\frac{{t}^{5}}{10}+\frac{{t}^{4}}{4}-{t}^{3}+\frac{{t}^{2}}{4}+\frac{t}{2}+2\right)}^{-q}$

and

$f\left(u,v\right)={\left(\frac{u+v}{2}+2\right)}^{q}.$

We see that (A1)-(A4) are satisfied.

Case 1. 0 ≤ q < 1. Clearly f W. It follows from Theorem 3.5(c) that the set E = (0, ∞). As an example, when λ = 24, the boundary value problem (3.29) has a positive solution given by $y\left(t\right)=\frac{{t}^{5}}{5}-\frac{{t}^{4}}{2}+\frac{{t}^{2}}{2}$.

Case 2. q = 1. Here f W B . By Theorem 3.5(a) the set E is an open or a half-closed interval. Further, from Case 1 and Theorem 3.2 we note that E contains the interval (0, 24].

Case 3. q > 1. Clearly f W0. By Theorem 3.5(b) the set E is a half-closed interval. Again, as in Case 2 we note that (0, 24] E.

## 4 Eigenvalue intervals

In this section, we shall establish explicit subintervals of E. Here, the functions α and β in (A2)-(A4) are assumed to be continuous on the closed interval [0, 1]. Hence, noting Remark 3.1, we shall not require conditions (A1) and (A4) to show the compactness of the operator S. For the function f in (A2), we define

$\begin{array}{cc}\hfill {\stackrel{̄}{f}}_{0}=\underset{u\to 0,\phantom{\rule{2.77695pt}{0ex}}v\to 0}{\mathsf{\text{lim}}\mathsf{\text{sup}}}\frac{f\left(u,v\right)}{v},\hfill & \hfill \phantom{\rule{1em}{0ex}}{\underset{-}{f}}_{0}=\underset{u\to 0,\phantom{\rule{2.77695pt}{0ex}}v\to 0}{\mathsf{\text{lim}}\mathsf{\text{inf}}}\frac{f\left(u,v\right)}{v},\hfill \\ \hfill {\stackrel{̄}{f}}_{\infty }=\underset{u\to \infty ,\phantom{\rule{2.77695pt}{0ex}}v\to \infty }{\mathsf{\text{lim}}\mathsf{\text{sup}}}\frac{f\left(u,v\right)}{v},\hfill & \hfill \phantom{\rule{1em}{0ex}}{\underset{-}{f}}_{\infty }=\underset{u\to \infty ,\phantom{\rule{2.77695pt}{0ex}}v\to \infty }{\mathsf{\text{lim}}\mathsf{\text{inf}}}\frac{f\left(u,v\right)}{v}.\hfill \end{array}$

Let $\delta \in \left(0,\frac{1}{2}\right)$ be given. Define ${t}^{*},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\stackrel{^}{t}\in \left[0,1\right]$ by

$\begin{array}{c}\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds=\underset{t\in \left[0,1\right]}{\mathsf{\text{sup}}}\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)ds,\\ \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds=\underset{t\in \left[0,1\right]}{\mathsf{\text{sup}}}\underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)ds,\end{array}$
(4.1)

Theorem 4.1. Let (A2)-(A4) hold. Then, λ E if λ satisfies

$\frac{1}{{\underset{-}{f}}_{\infty }}{\left[\gamma \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds\right]}^{-1}<\lambda <\frac{1}{{\stackrel{̄}{f}}_{0}}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}.$
(4.2)

Proof. We shall use Theorem 2.1. Let λ satisfy (4.2) and let ε > 0 be such that

$\frac{1}{{\underset{-}{f}}_{\infty }-\epsilon }{\left[\gamma \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds\right]}^{-1}\le \lambda \le \frac{1}{{\stackrel{̄}{f}}_{0}+\epsilon }{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}.$
(4.3)

First, we pick p > 0 so that

$f\left(u,v\right)\le \left({\stackrel{̄}{f}}_{0}+\epsilon \right)v,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0
(4.4)

Let x C δ be such that x = p. Note that for s [0, 1],

$\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau \le \underset{0}{\overset{1}{\int }}p\phantom{\rule{2.77695pt}{0ex}}d\tau =p.$

Then, using (3.5), Lemma 2.1, (4.4) and (4.3) successively, we find for t [0, 1],

$\begin{array}{ll}\hfill Sx\left(t\right)& \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\left({\stackrel{̄}{f}}_{0}+\epsilon \right)x\left(s\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\left({\stackrel{̄}{f}}_{0}+\epsilon \right)∥x∥\mathsf{\text{sin}}\pi sds\le ∥x∥.\phantom{\rule{2em}{0ex}}\end{array}$

Hence,

$∥Sx∥\le ∥x∥.$
(4.5)

If we set Ω1 = {x B |x < p}, then (4.5) holds for x C δ Ω1.

Next, let q > 0 be such that

$f\left(u,v\right)\ge \left({\underset{-}{f}}_{\infty }-\epsilon \right)v,\phantom{\rule{1em}{0ex}}u\ge q,\phantom{\rule{2.77695pt}{0ex}}v\ge q.$
(4.6)

Let x C δ be such that

$∥x∥=\mathsf{\text{max}}\left\{p+1,\frac{q}{\gamma },\frac{q}{\gamma }{\left(\frac{1}{2}-\delta \right)}^{-1}\right\}=\mathsf{\text{max}}\left\{p+1,\frac{q}{\gamma }{\left(\frac{1}{2}-\delta \right)}^{-1}\right\}\equiv {q}_{0}.$

It is clear that

$\left\{\begin{array}{c}x\left(s\right)\ge \gamma ∥x∥\ge q,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}s\in \left[\frac{1}{2},1-\delta \right]\hfill \\ \underset{\delta }{\overset{\frac{1}{2}}{\int }}x\left(\tau \right)d\tau \ge \underset{\delta }{\overset{\frac{1}{2}}{\int }}\gamma ∥x∥d\tau =\left(\frac{1}{2}-\delta \right)\gamma ∥x∥\ge q.\hfill \end{array}\right\$
(4.7)

Then, an application of (3.5), (4.7), and (4.6) gives for t [0, 1],

$\begin{array}{ll}\hfill Sx\left(t\right)& \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{\delta }{\overset{\frac{1}{2}}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)\left({\underset{-}{f}}_{\infty }-\epsilon \right)x\left(s\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)\left({\underset{-}{f}}_{\infty }-\epsilon \right)\gamma ∥x∥ds.\phantom{\rule{2em}{0ex}}\end{array}$

Taking supremum both sides and using (4.3) then provides (see (4.1) for the definition of t*)

$∥Sx∥\ge \lambda \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds\cdot \left({\underset{-}{f}}_{\infty }-\epsilon \right)\gamma ∥x∥\ge ∥x∥.$

Therefore, if we set Ω2 = {x B| x < q0}, then for x C δ Ω2 we have

$∥Sx∥\ge ∥x∥.$
(4.8)

Now that we have obtained (4.5) and (4.8), it follows from Remark 3.2 and Theorem 2.1 that S has a fixed point $x\in {C}_{\delta }\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)$ such that pxq0. Obviously, this x is a positive solution of (3.3) and hence λ E.   □

Theorem 4.2. Let (A2)-(A4) hold. Then, λ E if λ satisfies

$\frac{1}{{\underset{-}{f}}_{0}}{\left[\gamma \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds\right]}^{-1}<\lambda <\frac{1}{{\stackrel{̄}{f}}_{\infty }}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}.$
(4.9)

Proof. We shall apply Theorem 2.1 again. Let λ satisfy (4.9) and let ε > 0 be such that

$\frac{1}{{\underset{-}{f}}_{0}-\epsilon }{\left[\gamma \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds\right]}^{-1}\le \lambda \le \frac{1}{{\stackrel{̄}{f}}_{\infty }+\epsilon }{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}.$
(4.10)

First, we choose r > 0 so that

$f\left(u,v\right)\ge \left({\underset{-}{f}}_{0}-\epsilon \right)v,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0
(4.11)

Let x C δ be such that x = r. Then, on using (3.5), (4.11), and (4.10) successively, we have for t [0, 1],

$\begin{array}{ll}\hfill Sx\left(t\right)& \ge \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{0}{\overset{1}{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)\left({\underset{-}{f}}_{0}-\epsilon \right)x\left(s\right)ds\phantom{\rule{2em}{0ex}}\\ \ge \lambda \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)\left({\underset{-}{f}}_{0}-\epsilon \right)\gamma ∥x∥ds.\phantom{\rule{2em}{0ex}}\end{array}$

Taking supremum both sides and using (4.10) then yields (see (4.1) for the definition of $\stackrel{^}{t}$)

$∥Sx∥\ge \lambda \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds\cdot \left({\underset{-}{f}}_{0}-\epsilon \right)\gamma ∥x∥\ge ∥x∥.$

Hence, if we set Ω1 = {y B| x < r}, then (4.8) holds for x C δ Ω1.

Next, pick w > 0 such that

$f\left(u,v\right)\le \left({\stackrel{̄}{f}}_{\infty }+\epsilon \right)v,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u\ge w,\phantom{\rule{2.77695pt}{0ex}}v\ge w.$
(4.12)

We shall consider two cases - when f is bounded and when f is unbounded.

Case 1. Suppose that f is bounded. Then, there exists some M > 0 such that

$f\left(u,v\right)\le M,\phantom{\rule{1em}{0ex}}u,v\in \left(0,\infty \right).$
(4.13)

Let x C δ be such that

$∥x∥=\mathsf{\text{max}}\left\{r+1,\frac{\lambda M}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right\}\equiv {w}_{0}.$

From (3.5), Lemma 2.1 and (4.13), it is clear for t [0, 1] that

$\begin{array}{ll}\hfill Sx\left(t\right)& \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)M\mathsf{\text{sin}}\pi sds\le {w}_{0}=∥x∥.\phantom{\rule{2em}{0ex}}\end{array}$

Hence, (4.5) holds.

Case 2. Suppose that f is unbounded. Then, there exists w0 > max {r + 1, w} such that

$f\left(u,v\right)\le f\left({w}_{0},{w}_{0}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0
(4.14)

Let x C δ be such that x = w0. Then, applying (3.5), Lemma 2.1, (4.14), (4.12), and (4.10) successively gives for t [0, 1],

$\begin{array}{ll}\hfill Sx\left(t\right)& \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left(\underset{0}{\overset{s}{\int }}x\left(\tau \right)d\tau ,x\left(s\right)\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)f\left({w}_{0},{w}_{0}\right)\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ \le \lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\left({\stackrel{̄}{f}}_{\infty }+\epsilon \right){w}_{0}\mathsf{\text{sin}}\pi sds\phantom{\rule{2em}{0ex}}\\ =\lambda \underset{0}{\overset{1}{\int }}\frac{1}{{\pi }^{2m-1}}\beta \left(s\right)\left({\stackrel{̄}{f}}_{\infty }+\epsilon \right)∥x∥\mathsf{\text{sin}}\pi sds\le ∥x∥.\phantom{\rule{2em}{0ex}}\end{array}$

Thus, (4.5) follows immediately.

In both Cases 1 and 2, if we set Ω2 = {x B| x < w0}, then (4.5) holds for x C δ Ω2.

Now that we have obtained (4.8) and (4.5), it follows from Remark 3.2 and Theorem 2.1 that S has a fixed point $x\in {C}_{\delta }\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)$ such that rxw0. It is clear that this x is a positive solution of (3.3) and hence λ E.

Remark 4.1. In (4.2) and (4.9), although t* and $\stackrel{^}{t}$ can be computed from (4.1), we can circumvent the computation by giving further bounds. Indeed, applying Lemma 2.2 we find

$\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds\ge \underset{t\in \left[\delta ,1-\delta \right]}{\mathsf{\text{sup}}}\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)ds\ge \frac{2\delta }{{\pi }^{2m}}\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\alpha \left(s\right)\mathsf{\text{sin}}\pi sds$

and

$\underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds\ge \underset{t\in \left[\delta ,1-\delta \right]}{\mathsf{\text{sup}}}\underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(t,s\right)\right|\alpha \left(s\right)ds\ge \frac{2\delta }{{\pi }^{2m}}\underset{\delta }{\overset{1-\delta }{\int }}\alpha \left(s\right)\mathsf{\text{sin}}\pi sds.$

The following corollary is immediate from Theorems 4.1, 4.2 and Remark 4.1.

Corollary 4.1. Let (A2)-(A4) hold. Then,

$\begin{array}{ll}\hfill E& \supseteq \left(\frac{1}{{\underset{-}{f}}_{\infty }}{\left[\gamma \underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\left|{g}_{m}\left({t}^{*},s\right)\right|\alpha \left(s\right)ds\right]}^{-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\stackrel{̄}{f}}_{0}}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}\right)\phantom{\rule{2em}{0ex}}\\ \supseteq \left(\frac{1}{{\underset{-}{f}}_{\infty }}{\left[\frac{2\gamma \delta }{{\pi }^{2m}}\underset{\frac{1}{2}}{\overset{1-\delta }{\int }}\alpha \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\stackrel{̄}{f}}_{0}}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}\right)\phantom{\rule{2em}{0ex}}\end{array}$

and

$\begin{array}{ll}\hfill E& \supseteq \left(\frac{1}{{\underset{-}{f}}_{0}}{\left[\gamma \underset{\delta }{\overset{1-\delta }{\int }}\left|{g}_{m}\left(\stackrel{^}{t},s\right)\right|\alpha \left(s\right)ds\right]}^{-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\stackrel{̄}{f}}_{\infty }}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}\right)\phantom{\rule{2em}{0ex}}\\ \supseteq \left(\frac{1}{{\underset{-}{f}}_{0}}{\left[\frac{2\gamma \delta }{{\pi }^{2m}}\underset{\delta }{\overset{1-\delta }{\int }}\alpha \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\frac{1}{{\stackrel{̄}{f}}_{\infty }}{\left[\frac{1}{{\pi }^{2m-1}}\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Remark 4.2. If f is superlinear (i.e., ${\stackrel{̄}{f}}_{0}=0$ and ${\underset{-}{f}}_{\infty }=\infty$) or sublinear (i.e., ${\underset{-}{f}}_{0}=\infty$ and ${\stackrel{̄}{f}}_{\infty }=0$), then we conclude from Corollary 4.1 that E = (0, ∞), i.e., the boundary value problem (3.3) (or (1.1)) has a positive solution for any λ > 0.

Example 4.1. Consider the complementary Lidstone boundary value problem

$\begin{array}{c}{y}^{\left(5\right)}=\lambda {\left[a\left({t}^{5}-\frac{5{t}^{4}}{2}+\frac{5{t}^{2}}{2}\right)+b\left(5{t}^{4}-10{t}^{3}+5t\right)+c\right]}^{-r}{\left(ay+b{y}^{\prime }+c\right)}^{r},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\in \left(0,1\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}y\left(0\right)={y}^{\prime }\left(0\right)={y}^{‴}\left(0\right)={y}^{\prime }\left(1\right)={y}^{‴}\left(1\right)=0\end{array}$
(4.15)

where λ, a, b, c > 0 and r ≤ 1.

Here, m = 2. It is clear that (A2)-(A4) are satisfied with

$\alpha \left(t\right)=\beta \left(t\right)={\left[a\left({t}^{5}-\frac{5{t}^{4}}{2}+\frac{5{t}^{2}}{2}\right)+b\left(5{t}^{4}-10{t}^{3}+5t\right)+c\right]}^{-r}\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}f\left(u,v\right)={\left(au+bv+c\right)}^{r}.$

Case 1. r < 1. It is clear that f is sublinear. Therefore, by Remark 4.2 the boundary value problem (4.15) has a positive solution for any λ > 0. In fact, we note that when λ = 120, (4.15) has a positive solution given by $x\left(t\right)={t}^{5}-\frac{5{t}^{4}}{2}+\frac{5{t}^{2}}{2}$.

Case 2. r = 1, a = b = 0.5, c = 10. Here, ${\underset{-}{f}}_{0}=\infty$ and ${\stackrel{̄}{f}}_{\infty }=1$. It follows from Corollary 4.1 that

$E\supseteq \left(0,\phantom{\rule{2.77695pt}{0ex}}\frac{{\pi }^{3}}{{\stackrel{̄}{f}}_{\infty }}{\left[\underset{0}{\overset{1}{\int }}\beta \left(s\right)\mathsf{\text{sin}}\pi sds\right]}^{-1}\right)=\left(0,528.99\right).$

Once again we note that when λ = 120 (0,528.99), the corresponding eigenfunction is given by $x\left(t\right)={t}^{5}-\frac{5{t}^{4}}{2}+\frac{5{t}^{2}}{2}$.

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Agarwal, R.P., Wong, P.J. Eigenvalues of complementary Lidstone boundary value problems. Bound Value Probl 2012, 49 (2012). https://doi.org/10.1186/1687-2770-2012-49 