# The Distance between Points of a Solution of  a Second Order Linear Differential Equation Satisfying  General Boundary Conditions

Authors: Pedro Almenar; Lucas Jódar
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper presents a method to obtain lower and upper bounds for the minimum distance between pointsa and b of the solution of the second order linear differential equation y′′+q(x)y=0 satisfying general separated boundary conditions of the type a11y(a)+a12y′(a)=0 and a21y(b)+a22y′(b)=0. The method is based on the recursive application of a linear operator to certain functions, a recursive application that makes these bounds converge to the exact distance between a and b as the recursivity index grows. The method covers conjugacy and disfocality as particular cases.


---

## Body

## 1. Introduction

In a recent paper of the authors (see [1]) it was shown that the recursive application of the operator P:C[a,b]→C2[a,b] defined by
(1)Pf=∫ax‍∫tb‍q(s)f(s)dsdt,
where q(x) is continuous on [a,b] and strictly positive almost everywhere on that same interval, provided a method to determine if the second order linear differential equation
(2)y′′+q(x)y=0
was either left disfocal or left nondisfocal in the interval [a,b], the concept of left disfocal alluding to the nonexistence of a nontrivial solution y(x) of (2) with zeroes in [a,b] such that y′(b)=0 (a similar definition exists for right disfocal). The method was based on two features of the operator P, namely, (i)
the fact thatP is positive and monotonic, so that for f>y>g>0 on ]a,b[ one has Pkf(x)>Pky(x)>Pkg(x) on ]a,b];(ii)
the fact thatlim⁡k→∞Pkf(b)=∞ if (2) is left nondisfocal in an interval interior to [a,b] and lim⁡k→∞Pkf(x)=0 for any value of x∈[a,b] as long as (2) is left disfocal in [a,b].The purpose of this paper is to extend the results of [1] to (2) with the more general boundary conditions:
(3)a11y(a)+a12y′(a)=0,a21y(b)+a22y′(b)=0,
which contain conjugacy (y(a)=y(b)=0) and nondisfocality as particular cases, by means of the recursive application of the operator L:C[a,b]→C2[a,b] defined by
(4)Lf=∫ab‍G(x,t)q(t)f(t)dt,
where G(x,t) (or Gab(x,t), as we will name it in some sections) is the Green function of the problem
(5)y′′=-g(x),a11y(a)+a12y′(a)=0,a21y(b)+a22y′(b)=0.
In consequence this paper will yield criteria (in fact infinitely many) to obtain lower and upper bounds for the minimum distance between points a and b satisfying (3) when y(x) is a solution of (2), bounds of which will be shown to converge to the exact values of such a minimum distance as the index k grows. Besides extending the results of [1] to the problem (2)-(3), other results based on the same strategies will also be introduced and it will be shown that they improve the results of [1] in many cases.Operators of the type (4) have often been used in the determination of Lyapunov inequalities for different types of equations and boundary conditions since Nehari [2] first noted that a solution y of (2) such that y(a)=y(b)=0 satisfied
(6)y(x)=∫ab‍G(x,t)q(t)y(t)dt,
with
(7)G(x,t)=min⁡(x-a,t-a)min⁡(b-x,b-t)b-a,
which implied
(8)|y(x)|≤∫ab‍|G(x,t)|q(t)|y(t)|dt<∥y∥∞∫ab‍|G(t,t)|q(t)dt;
that is,
(9)1<∫ab‍|G(t,t)|q(t)dt.
Beesack [3], Hinton [4], Levin [5, 6], and Reid [7–9] are other remarkable examples of such a use, as the excellent monography on Lyapunov inequalities of [10, Chapter 1] shows. Although formulae like (6) can be applied recursively if q(x)≥0 and G(x,t)≥0 to obtain more complex versions of (9), the fact is that the iterative application of L has rarely been proposed in any papers, with the exception of Harris [11], who suggested its application for the disfocal case—y(a)=y′(b)=0 or y′(a)=y(b)=0—without getting to prove that it guaranteed any improvement. We will show in this paper that under certain conditions (3) the function G(x,t) is positive and the recursive application is possible and provides lower and upper bounds which improve all existing results as of today. We will also show that even in the case that G(x,t) gets to be negative it is still sometimes possible to obtain upper and lower bounds for the distance between a and b, which are as close to the real distance between a and b as desired.It is also worth remarking the significant interest that the calculation of lower bounds (i.e., Lyapunov kind-of inequalities) has enjoyed in comparison with the problem of the determination of upper bounds, regardless of the type of boundary conditions (3). This fact was already noted by Došlý in [12] for the conjugate case (y(a)=y(b)=0) and by the authors in [1, 13, 14] for the nondisfocal case. References [12, 15–20] are notable exceptions to this trend.As indicated in the first paragraph, throughout the paper we will assume thatq(x) is continuous on an interval I⊂R such that [a,b]⊂I and that q(x) is strictly positive almost everywhere on [a,b]. This allows defining the internal product
(10)〈f,g〉=∫ab‍q(x)f(x)g(x)dx,
for f, g continuous on [a,b] (it is easy to prove that (10) satisfies all the conditions required by an internal product) and the associated norm ∥·∥2 defined by
(11)∥f∥2=(∫ab‍q(x)f2(x)dx)1/2.Likewise, we will use the notationL to name the operator defined in (4), Lf or L{f} to name the function with domain [a,b] resulting from the application of L to f(x)∈C[a,b], and Lf(x) or L{f}(x) to name the value of the function Lf at the point x and Lα,β when other extremes of integration α, β potentially different from a, b are used in (4). In that latter case, any ∥·∥2 norm appearing in the same formula will be assumed to be calculated with α and β as integration extremes in (11).We will say that the pointsa′, b′ are interior to an interval [a,b] if a′,b′∈[a,b] and [a′,b′]≠[a,b].The organization of the paper is as follows. Section2 will present the main properties of the operator L. Sections 3 and 4 will apply these properties in different ways to find upper and lower bounds for the minimum distance between points a and b satisfying (3). Section 5 will introduce some formulae which simplify the calculations required in Section 3. Section 6 will apply the method to several examples. Finally Section 7 will draw several conclusions.


## 2. The OperatorLf=∫ab‍G(x,t)q(t)f(t)dt

The purpose of this section will be to present the main properties of the operatorL defined in (4) for q(x) as specified in the Introduction. As was done in [1], for the sake of clarity such properties will be presented in several lemmas which will lead to Theorem 5, which can be regarded as the key result of this section.Lemma 1.
The operatorLf is linear. In addition, if G(x,t) is positive almost everywhere for t∈[a,b], then L is positive and monotonic.Proof.
The linearity ofL is quite evident. If G(x,t) is positive almost everywhere for t∈[a,b], the positiveness (and in consequence monotonicity) of L is a consequence of q(x) being positive almost everywhere on [a,b] by hypothesis.Lemma 2.
The operatorLf is compact.Proof.
From [21, Theorem 7.2.6] and the fact that G(x,t) is continuous on [a,b]×[a,b] it is straightforward to show that L is compact with the ∥·∥∞ norm and therefore with the ∥·∥2 norm defined in (11).Lemma 3.
If any of the following conditions(12)a11=0,a21=0,a11=0,a22a21(b-a)+1≠0,-a12a11(b-a)+1≠0,a21=0,a12a11(a22a21(b-a)+1)-a22a21≠0
are met, then the operator Lf is self-adjoint.Proof.
Let us definey1(x) as the solution of
(13)y′′=0,a11y(a)+a12y′(a)=0,
and y2(x) as the solution of
(14)y′′=0,a21y(b)+a22y′(b)=0.
A straightforward calculation shows that if any of the conditions (12) are met, then the Wronskian of y1 and y2, namely,
(15)W=y1(x)y2′(x)-y1′(x)y2(x),
is not zero, which implies that y1 and y2 are not linearly dependent and therefore no nontrivial solutions of (2)-(3) exist. In consequence from [22, Theorem IX.3] one has
(16)G(x,t)={-y1(x)y2(t)y1(a)y2′(a)-y1′(a)y2(a),x<t,-y1(t)y2(x)y1(a)y2′(a)-y1′(a)y2(a),x≥t.
Now, in order to prove self-adjointness, we need to prove that, given f,g∈C[a,b], 〈Lf,g〉=〈f,Lg〉. Thus, from (10) we have
(17)〈Lf,g〉=∫ab‍q(x)Lf(x)g(x)dx=∫ab‍q(x)g(x)∫ab‍G(x,t)q(t)f(t)dtdx.
Combining (17) and (16) one gets to
(18)〈Lf,g〉=-1y1(a)y2′(a)-y1′(a)y2(a)×∫ab‍q(x)g(x)y2(x)∫ax‍q(t)y1(t)f(t)dtdx+-1y1(a)y2′(a)-y1′(a)y2(a)×∫ab‍q(x)g(x)y1(x)∫xb‍q(t)y2(t)f(t)dtdx.
And integrating by parts the right-hand side of (18) one finally has
(19)〈Lf,g〉=-1y1(a)y2′(a)-y1′(a)y2(a)×∫ab‍∫xb‍q(t)g(t)y2(t)dtq(x)y1(x)f(x)dx+-1y1(a)y2′(a)-y1′(a)y2(a)×∫ab∫ax‍q(t)g(t)y1(t)dtq(x)y2(x)f(x)dx‍=∬abq(t)g(t)G(x,t)dtq(x)f(x)dx=〈f,Lg〉.Lemma 4.
The operatorLf is bounded with the ∥·∥∞ norm and verifies
(20)Lf(x)≤max⁡{G(x,t),t∈[a,b]}∥f∥∞∫ab‍q(t)dt,x∈[a,b],(21)Lf(x)≤max⁡{G(x,t),t∈[a,b]}∥f∥2∫ab‍q(t)dt,x∈[a,b],
where the ∥·∥2 norm is defined as in (11).Proof.
Equation (20) is an obvious consequence of (4), and (21) follows from (4) by application of Cauchy-Schwarz inequality.Theorem 5.
If any of conditions (12) are met, then the operator Lf has a countably infinite number of eigenvalues 1/λn and associated orthonormal eigenfunctions Φn(x), which allow expressing Lkf, k≥1, as
(22)Lkf=∑n=1∞‍〈f,Φn〉λnkΦn.
Moreover, (i)
if no nontrivial solution of (2) satisfies (3) either at a, b or at any a′, b′ interior to [a,b], then
(23)∥Lkf∥22<∥f∥22,k≥1;(24)lim⁡k→∞∥Lkf∥2=0,lim⁡k→∞Lkf(x)=0,∥Lkf∥ll2uniformlyforx∈[a,b];(ii)
if there is a nontrivial solution of (2) that satisfies (3) at some a′, b′ interior to [a,b], then
(25)∥Lkf∥22>〈f,Φ1〉2+∑n=2∞‍〈f,Φn〉2λn2k,k≥1.
In addition, if 〈f,Φ1〉≠0 and c∈[a,b] is such that Φ1(c)≠0, then
(26)lim⁡k→∞∥Lkf∥2=∞,(27)lim⁡k→∞Lkf(c)=±∞,
the sign corresponding to that of 〈f,Φ1〉Φ1(c).Proof.
Let us consider the eigenvalue problem(28)y′′+λq(x)y=0,x∈]a,b[,a11y(a)+a12y′(a)=0,a21y(b)+a22y′(b)=0.
From the theory of ordinary differential equations (see [22, Theorems V.8 and V.9]) it is known that there exist a countably infinite number of eigenvalues λn which form an increasing sequence with lim⁡n→∞λn=∞, each of which has its corresponding orthonormal (with the norm (11)) eigenfunction Φn(x), and that the set of eigenfunctions Φn forms an orthonormal basis of C[a,b]. Applying the operator L to these eigenfunctions Φn(x) and integrating by parts it is easy to show that
(29)LΦn=Φnλn,n≥1,
which implies that Φn are also the eigenfunctions of the operator L with corresponding eigenvalues 1/λn. Since from Lemmas 1, 2 and 3  L is linear, compact, and self-adjoint, we can apply [21, Theorem 7.5.2] and represent L in the canonical form
(30)Lf=∑n=1∞‍〈f,Φn〉λnΦn.
Applying L again to (30) it yields
(31)L2f=∑n=1∞‍〈Lf,Φn〉λnΦn=∑n=1∞‍〈∑j=1∞(〈f,Φj〉/λj)Φj,Φn〉λnΦn=∑n=1∞‍〈f,Φn〉λn2Φn,
given that 〈Φi,Φi〉=1 and 〈Φi,Φj〉=0 for i≠j. Applying L recursively to (31) one gets to
(32)Lkf=∑n=1∞‍〈Lk-1f,Φn〉λnΦn=∑n=1∞‍〈∑j=1∞(〈f,Φj〉/λjk-1)Φj,Φn〉λnΦn=∑n=1∞‍〈f,Φn〉λnkΦn,
which is in fact (22). And the application of Parseval’s identity (see [21, Lemmas 1.5.14]) to (32) leads to
(33)∥Lkf∥2=∑n=1∞‍|〈f,Φn〉|2λn2k.
Since {λn} form an increasing sequence, from (33) it is clear that
(34)∥Lkf∥2≤∑n=1∞‍|〈f,Φn〉|2λ12k=∥f∥2λ1k.
Now, let us note that if (2) does not have any nontrivial solution that satisfies (3) either at a, b or at any a′, b′ interior to [a,b], the first eigenvalue λ1 (and therefore all the others) must be strictly greater than 1. In that case, from (34) one has (23) and
(35)lim⁡k→∞∥Lkf∥2=0.
From (35) and Lemma 4 one gets (24).
Likewise, if there is a nontrivial solution of (2) that satisfies (3) at some a′, b′ interior to [a,b], then λ1 must be such that λ1<1. From this and (33) one gets (25). If, in addition, 〈f,Φ1〉≠0, from (33) we get to
(36)lim⁡k→∞∥Lkf∥2=∞,
which is (26). On the other hand, we can write
(37)Lkf-〈f,Φ1〉λ1kΦ1=∑n=2∞‍〈f,Φn〉λnkΦn.
We can divide both sides of (37) by 〈f,Φ1〉/λ1k to yield
(38)Lkf-(〈f,Φ1〉/λ1k)Φ1〈f,Φ1〉/λ1k=∑n=2∞(〈f,Φn〉/λnk)Φn〈f,Φ1〉/λ1k.
Applying Parseval’s identity to (38) one gets
(39)∥Lkf-(〈f,Φ1〉/λ1k)Φ1〈f,Φ1〉/λ1k∥2=∑n=2∞(|〈f,Φn〉|2/λn2k)|〈f,Φ1〉|2/λ12k≤∥f∥2|〈f,Φ1〉|λ1kλ2k.
Since λ1<λ2, from (39) one yields
(40)lim⁡k→∞∥Lkf-(〈f,Φ1〉/λ1k)Φ1〈f,Φ1〉/λ1k∥2=0,
which implies that there exists an index k0 such that
(41)∥Lkf-(〈f,Φ1〉/λ1k)Φ1〈f,Φ1〉/λ1k∥2<ϵ,k>k0.
From Lemma 4 and (41) one has
(42)|Lkf(c)-(〈f,Φ1〉/λ1k)Φ1(c)(〈f,Φ1〉/λ1k)|<ϵmax⁡{G(c,t),t∈[a,b]}∫ab‍q(x)dx,
for k>k0+1; that is,
(43)|〈f,Φ1〉|λ1k×(∫ab‍q(x)dxsign⁡(〈f,Φ1〉)Φ1(c)-ϵmax⁡{G(c,t),t∈[a,b]}×∫ab‍q(x)dx)<Lkf(c)<|〈f,Φ1〉|λ1k×(∫ab‍q(x)dx)sign⁡(〈f,Φ1〉)Φ1(c)+ϵmax⁡{G(c,t),t∈[a,b]}×∫ab‍q(x)dx),
for k>k0+1. Since Φ1(c)≠0 by hypothesis and λ1<1, (43) leads to (27).Remark 6.
Theorem5 provides two types of methods to obtain upper and lower bounds for the minimum distance between points a,b satisfying (3), one based on comparing the norm of Lkf with some other constants and another based on comparing the value of Lkf at concrete points of [a,b] for different functions f. These will be addressed separately in the next two sections.


## 3. Bounds for the Distance betweena and b Based on ∥Lkf∥2

As stated before, Theorem5 provides methods to get upper and lower bounds for the minimum distance between points a, b satisfying (3), which are based on the comparison of ∥Lkf∥2 with some thresholds for different values of the extremes of integration a and b of (4) and (11).Thus, on the one hand, ifa, b are such that there is a nontrivial solution of (2) which satisfies (3) at some a′, b′ interior to [a,b], from (26) (and as long as 〈f,Φ1〉≠0, i.e., f is “close” to Φ1) it is clear that ∥Lkf∥2 will grow with the index k regardless of the choice of f and accordingly there will be an index k0 such that (23) is violated. This allows us to define an algorithm to find progressively better “outer” bounds of the values satisfying (3) by fixing one of them, say a, and calculating the minimum values of the extremes bk which give
(44)∥La,bkkf∥2=∥f∥2,
for different values of k, as the following theorem shows.Theorem 7.
Assume that there exists a nontrivial solutiony(x) of (2) satisfying (3) at a, b. Let the sequence {bk} be defined by (44), where f is a continuous function on [a,bk] such that 〈f,Φ1〉≠0 and a, b′ fulfill any of conditions (12) for each b′∈[b,bk]. Then bk≥b and {bk} tends to b as k→∞.Proof.
The fact thatbk≥b is obvious from (23). Now, let us pick a δ>0. From (26) lim⁡k→∞∥La,b+δkf∥2=∞, which means that there exists an index k1 such that
(45)∥La,b+δkf∥2-∥f∥2>0,k>k1.
Given that ∥La,b-δkf∥2-∥f∥2<0 for k≥1, from (23) and the continuity of ∥La,xkf∥2-∥f∥2 on x (continuity guaranteed by the hypothesis), there must exist a value bk∈[b,b+δ[ such that ∥La,bkkf∥2-∥f∥2=0 for each k>k1. This proves the second assertion of the theorem.The application of the method based on Theorem7 is quite straightforward given that the right-hand side of (44), that is, ∥f∥22, is easy to calculate once fixed f. However, it is worth remarking that, from (34), the closer the selected function f is to Φ1, the smaller the terms 〈f,Φn〉 for n>1 will be and the faster the sequence bk will converge to b. Therefore, although the method can work with any continuous function f such that 〈f,Φ1〉≠0, it is desirable to select one that may be as close as possible to the expected Φ1 associated to the problem (28).On the other hand, ifa, b are such that no nontrivial solution of (2) satisfies (3) either at a, b or at some a′, b′ interior to [a,b], from (24) it is clear that ∥Lkf∥2 will shrink with the index k regardless of the choice of f and accordingly there will be an index k0 such that (25) is violated. As happened before, this allows us to define an algorithm to find progressively better “inner” bounds of the values satisfying (3) by fixing one of them, say a, and calculating the values of the extremes bk which give
(46)∥La,bkkf∥2=Ba,bk,k≥1,
where Ba,bk is any lower bound of the right-hand side of (25), which has (in turn) a positive lower bound that does not depend on k. This is shown in the following theorem.Theorem 8.
Assume that there exists a nontrivial solutiony(x) of (2) satisfying (3) at a, b. Let the sequence {bk} be defined by (46), where f is a continuous function on [a,bk] and a, b′ fulfill any of conditions (12) for each b′∈[bk,b]. Then bk≤b and {bk} tends to b as k→∞.Proof.
Again, the fact thatbk≤b is obvious from (25) and (46). Now, let us pick a δ>0. From (24), lim⁡k→∞∥La,b-δkf∥2=0. This and the fact that Ba,bk is bounded below by a positive amount which does not vary with k imply that there exists an index k2 such that
(47)∥La,b-δkf∥2-Ba,b-δ<0,k>k2.
Given that ∥La,b+δkf∥2-Ba,b+δ>0 for k≥1, from (25) and the continuity of ∥La,xkf∥2-Ba,x on x, there must exist a value bk∈]b-δ,b] such that ∥La,bkkf∥2-Ba,bk=0 for each k>k2. This proves the second assertion of the theorem.Unlike what happens with the method based on Theorem7, the method based on Theorem 8 presents some difficulties in its application due to the fact that neither 〈f,Φn〉 nor the eigenvalues λn are known. We can overcome them partially by discarding, in the right-hand side of (25), the term of the series of the eigenfunctions beyond the first one, that is, by converting (25) into
(48)∥Lkf∥22≥〈f,Φ1〉2,k≥1,
given that the discarded terms are positive (in fact 〈f,Φ1〉2 may be a possible positive lower bound for some types of terms Ba,bk used in (46), which does not depend on k). The resulting method to obtain lower bounds for the distance between a and b will work in the same way as the one described in the previous paragraphs, at the expense of requiring greater values of the index k to violate (48). But even with such a simplification there is still a need to obtain a lower bound for 〈f,Φ1〉 which is not evident at all.The following lemmas aim at finding cases (fundamentally choices off and boundary conditions (3)) where the values of 〈f,Φ1〉, 〈f,Φn〉, and λn are bounded in a way that allows getting a lower bound for the right-hand side of (25) (or for the right-hand side of (48) if there is no way to do it with (25)), overcoming the problem.Lemma 9.
Letλn and Φn be defined as in Theorem 5. Assume that a, b satisfy any of conditions (12) of Lemma 3. Fixed x∈[a,b] and choosing f(t)=G(x,t) and g(t)=(δG(x,t)/δx) one has
(49)〈f,Φn〉=Φn(x)λn,n≥1,(50)〈g,Φn〉=Φn′(x)λn,n≥1.Proof.
From (10) one has
(51)〈f,Φn〉=∫ab‍q(t)G(x,t)Φn(t)dt,n≥1.
And since Φn satisfies (28) one yields
(52)〈f,Φn〉=-∫ab‍G(x,t)Φn′′(t)λndt,n≥1.
As G(x,t) is the Green function of the problem (5), -∫ab‍G(x,t)(Φn′′(t)/λn)dt is the value at x of the function satisfying (3) whose second derivative is Φn′′/λn, that is, Φn(x)/λn. This proves (49). As for (50), note simply that
(53)〈g,Φn〉=∫ab‍q(t)δG(x,t)δxΦn(t)dt=δ∫ab‍q(t)G(x,t)Φn(t)dtδx=1λnδΦn(x)δx,n≥1.Lemma 10.
Assume that either(a11/a12)≥0 or a12=0 and that either -(a21/a22)≥0 or a22=0. Let λn and Φn be defined as in Theorem 5. Let one suppose that q(x)>0 on [a,b] and that it can be decomposed as q(x)=c(x)/d(x) with c(x) and d(x) being nondecreasing functions. Depending on the values of a11, a12, a21, and a22, one has (i)
ifa12=0, then
(54)Φn′2(a)≥2λnc(a)∫ab‍d(t)q(t)dt,n≥1;(ii)
ifa12≠0, then
(55)Φn2(a)≥2d(a)∫ab‍d(t)q(t)dt(1+(a112/a122q(a)λn)),n≥1;(iii)
ifa22=0, then
(56)Φn′2(b)≥2λnd(b)∫ab‍(q(t)/c(t))dt,n≥1;(iv)
ifa22≠0, then
(57)Φn2(b)≥2c(b)∫ab‍(q(t)/c(t))dt(1+(a212/a222q(b)λn)),n≥1.Proof.
First of all, let us note thatq(x) can always be decomposed in the mentioned manner, given that q(x)>0 and therefore ln⁡q(x) exists in [a,b] and can be expressed as the difference between two increasing functions A(x) and B(x); that is,
(58)ln⁡q(x)=A(x)-B(x).
which, taking exponentials, becomes
(59)q(x)=exp⁡ln⁡q(x)=exp⁡(A(x)-B(x))=exp⁡A(x)exp⁡B(x)=c(x)d(x),
with c(x) and d(x) nondecreasing. Now let us consider the functional F defined by
(60)F(Φn,x)=Φn2(x)+(Φn′(x))2λnq(x),x∈[a,b].
Using integration by parts and the fact that Φn is orthonormal with respect to the norm (11) it is easy to prove that
(61)∫ab‍q(x)F(Φn,x)dx=∫ab‍q(x)Φn2(x)dx+∫ab‍(Φn′(x))2λndx=2∫ab‍q(x)Φn2(x)dx+1λn(Φn′(b)Φn(b)-Φn′(a)Φn(a))=2+1λn(Φn′(b)Φn(b)-Φn′(a)Φn(a)).
From (3), (61), and the hypothesis one gets
(62)∫ab‍q(x)F(Φn,x)dx≥2.
And given that c(x) and d(x) are increasing on [a,b] and that Φn(x) verifies (28) for λ=λn, one has
(63)(c(x)F(Φn,x))′=(c(x)Φn2(x)+d(x)(Φn′(x))2λn)′=c′(x)Φn2(x)+d′(x)(Φn′(x))2λn≥0,x∈]a,b[;
that is, c(x)F(Φn,x) is increasing on ]a,b[, and
(64)(F(Φn,x)d(x))′=(Φn2(x)d(x)+(Φn′(x))2λnc(x))′=-d′(x)Φn2(x)d2(x)-c′(x)(Φn′(x))2λnc2(x)≤0,x∈]a,b[;
that is, F(Φn,x)/d(x) is decreasing on ]a,b[. The application of (60), (63), and (64) to (62) leads to
(65)F(Φn,x){c(x)∫ax‍q(t)c(t)dt+1d(x)∫xb‍d(t)q(t)dt}≥∫ax‍c(t)F(Φn,t)q(t)c(t)dt+∫xb‍F(Φn,t)d(t)d(t)q(t)dt≥∫ab‍q(x)F(Φn,x)dx≥2,n≥1.
From (3), (60), and (65) it is straightforward to obtain (54)–(57).Lemma 11.
Letλn and Φn be defined as in Theorem 5. Then one has
(66)λn≤(n+C)2π2(∫ab‍min⁡(1,q(x))dx)2=M(n+C)2,n≥1,
with
(67)C={0,ifa11a12=a21a22ora22=a12=0;-12,ifa12=0,a22≠0,-a21a22≤0;-12,ifa22=0,a12≠0,-a11a12≥0;-1,else.Proof.
From [23, Theorem 8], if we define the function angle θn(x) by
(68)tanθn(x)=sinθn(x)cos⁡θn(x)=Φn′(x)AΦn(x),
where A>0 is a real constant, then θn(x) satisfies the equation
(69)θn′(x)=-Asin2θ(x)-λnq(x)Acos⁡2θ(x),
being, therefore, a decreasing function. Let us fix ]-(π/2),(π/2)[as the range of the arctan function. In that case, (i)
ifa12=0 we will set θn(a)=π/2. Else, we will set θn(a)=arctan(-a11/a12A);(ii)
ifa22=0 we will define ϑn(b)=-(π/2). Else,
(a)
if-a11/a12≥-a21/a22 we will define ϑn(b)=arctan(-a21/a22A);(b)
we will defineϑn(b)=-π+arctan(-a21/a22A). With this in mind, we will also define
(70)K(A)=θn(a)-ϑn(b)π-1.K(A) is a constant which depends only on A and the boundary conditions (3) and is related to the angle distance between a and b due to (3). Since Φn(x) is a function satisfying (3) with one more zero in ]a,b[ than Φn-1(x), integrating (69) and taking (70) into account we obtain
(71)(n+K(A))π=-∫ab‍θn′(x)dx=A∫ab‍sin2θ(x)dx+∫ab‍λnq(x)Acos⁡2θ(x)dx.
Taking A=λn in (71) one has
(72)(n+K(λn))π=λn(∫ab‍sin2θ(x)dx+∫ab‍q(x)cos⁡2θ(x)dx)≥λn∫ab‍min⁡(1,q(x))dx.
The problem for (72) to be used to get upper bounds for λn is the dependence of K(λn) on λn. We can eliminate it by obtaining an upper bound for K(λn). Accordingly we will define
(73)C={0,ifa11a12=a21a22ora22=a12=0;-12,ifa12=0,a22≠0,-a21a22≤0;-12,ifa22=0,a12≠0,-a11a12≥0;-1,else.
From (70), (72), and (73) one yields
(74)(n+C)π≥λn∫ab‍min⁡(1,q(x))dx.
Taking squares in (74) one finally gets to (66).Finally, by (25) and Lemmas 9–11 we are in a position to prove the next theorem, which allows a straightforward application of the method based on Theorem 8.Theorem 12.
Letλn and Φn be defined as in Theorem 5. Let c(x) and d(x) be defined as in Lemma 10. Let M and C be defined as in Lemma 11. Assume that there is a nontrivial solution y(x) of (2) that satisfies (3) at some a′, b′ interior to [a,b]. Assume also that a, b satisfy any of conditions (12) of Lemma 3, that either (a11/a12)≥0 or a12=0, and that either -(a21/a22)≥0 or a22=0. Then the following inequalities hold. (i)
Ifa12=0 then
(75)∥Lk{δG(a,t)δx}∥22>2c(a)∫ab‍d(t)q(t)dt(1+1(4k+1)M2k+1(2+C)4k+1),k≥1.
Otherwise, if Na is the first integer greater than 1 such that M(Na+C)2≥a112/a122q(a), then
(76)∥Lk{G(a,t)}∥22>2d(a)∫ab‍d(t)q(t)dt×(∑n=2Na-111+(a112/a122q(a))+∑n=2Na-1((M(n+c)2+(a112/a122q(a))))M2k+1(n+C)4k+2‍×(M(n+C)2+(a112/a122q(a))))-1+12M2k+2(4k+3)(Na+C)4k+3∑n=2Na-1),k≥1.(ii)
Ifa22=0 then
(77)∥Lk{δG(b,t)δx}∥22>2d(b)∫ab‍(q(t)/c(t))dt×(1+1(4k+1)M2k+1(2+C)4k+1),k≥1.
Otherwise, if Nb is the first integer greater than 1 such that M(Nb+C)2≥a212/a222q(b), then
(78)∥Lk{G(b,t)}∥22>2c(b)∫ab‍(q(t)/c(t))dt×(∑n=2Nb-1‍11+(a212/a222q(b))+∑n=2Nb-1‍((M(n+C)2+(a212/a222q(b)))M2k+1(n+C)4k+2×(M(n+C)2+(a212/a222q(b))))-1+12M2k+2(4k+3)(Nb+C)4k+3∑n=2Nb-1‍),k≥1.Proof.
Let us focus first on (75). From (25), the fact that λ1<1 (from the hypothesis) and Lemmas 9-10 it is straightforward to show that
(79)∥Lk{δG(a,t)δx}∥22>2c(a)∫ab‍d(t)q(t)dt(1+∑n=2∞‍1λn2k+1),k≥1.
On the other hand, from (66) one has
(80)∑n=2∞‍1λn2k+1≥1M2k+1∑n=2∞‍1(n+C)4k+2,k≥1.
But given that
(81)∑n=2∞‍1(n+C)4k+2≥∫2∞‍dx(x+C)4k+2=1(4k+1)(2+C)4k+1,
from (79)–(81) one gets (75).
As for (76), again from (25) and Lemmas 9-10 it is straightforward to show that
(82)∥Lk{G(a,t)}∥22>2d(a)∫ab‍d(t)q(t)dt×(1λ12+λ1(a112/a122q(a))+∑n=2∞‍1λn2k(λn2+λn(a112/a122q(a)))),k≥1.
On the other hand, from (66) one has
(83)∑n=2∞‍1λn2k(λn2+λn(a112/a122q(a)))≥∑n=2∞‍((a112(a122q(a)))M2k+1(n+C)4k+2×(M(n+C)2+a112a122q(a)))-1,k≥1.
From the definition of Na, it is clear that
(84)1M2k+1(n+C)4k+2(M(n+C)2+(a112/a122q(a)))≥12M2k+2(n+C)4k+4,n≥Na,k≥1.
And given that
(85)∑Na∞‍1(n+C)4k+4≥∫Na∞‍dx(x+C)4k+4=1(4k+3)(Na+C)4k+3,
from (82)–(85) and the fact that λ1<1 by the hypothesis, one obtains (76). The proofs of (77) and (78) are very similar mutatis mutandi.Remark 13.
It is possible to improve the results of Theorem12 by using better lower bounds for Φn2(a), Φn2(b), Φn′2(a), and Φn′2(b) than the ones displayed in Lemma 10 and better upper bounds for λn than the one presented in Lemma 11 or calculating more terms in the sums of (75)–(78) before using the integrals to get lower bounds for the remainders of the series.


## 4. Bounds for the Distance betweena and b Based on Concrete Values of Lkf(x)

This section will elaborate on the results of Section2 in order to obtain upper and lower bounds for the distance between the extremes a, b in a similar way as it was done in [1]. A critical condition for that is the positivity of the Green function G(x,t), which is ensured under certain boundary conditions by the next lemma.Lemma 14.
Leta, b be any real numbers such that a≠b and a,b∈[a′,b′]. If the boundary conditions (3) verify any of the following hypotheses for a′, b′,
(86)a12=a22=0,(87)a12=0,1+a21a22(b′-a′)>0,(88)1-a11a12(b′-a′)>0,a22=0,(89)a21a22>a11a12,a21a22-a11a12-a21a22a11a12(b′-a′)>0,either1+a21a22(b′-a′)>0or1-a11a12(b′-a′)>0,
then Gab(x,t) is positive almost everywhere for t∈[a,b] except if x=a or x=b in case (86), x=a in case (87), and x=b in case (88).Proof.
As was mentioned in the proof of Lemma3, if we define y1(x) and y2(x) as the solutions of
(90)y′′=0,a11y(a′)+a12y′(a′)=0,y′′=0,a21y(b′)+a22y′(b′)=0,
respectively, then we can represent Ga′b′(x,t) as follows:
(91)Ga′b′(x,t)={-y1(x)y2(t)y1(a′)y2′(a′)-y1′(a′)y2(a′),x<t,-y1(t)y2(x)y1(a′)y2′(a′)-y1′(a′)y2(a′),x≥t.
We can divide the analysis of (91) in four cases. (i)
a12=0 and a22=0. In this case one has
(92)Ga′b′(x,t)={(x-a′)(t-b′)a′-b′,x<t,(t-a′)(x-b′)a′-b′,x≥t,
which is clearly positive almost everywhere for t∈[a′,b′] as long as x≠a′ and x≠b′. This yields condition (86).(ii)
a12=0 and a22≠0. In this case, one has
(93)Ga′b′(x,t)={(x-a′)(-(a21/a22)(t-b′)+1)-(a21/a22)(a′-b′)+1,x<t,(t-a′)(-(a21/a22)(x-b′)+1)-(a21/a22)(a′-b′)+1,x≥t.
From (93) it is clear that Ga′b′(x,t) will be positive almost everywhere for t∈[a′,b′] as long as x≠a′ and
(94)1+a21a22(b′-a′)>0,
which gives condition (87).(iii)
a12≠0 and a22=0. In this case, one has
(95)Ga′b′(x,t)={-(t-b′)(-(a11/a12)(x-a′)+1)-(a11/a12)(b′-a′)+1,x<t,-(x-b′)(-(a11/a12)(t-a′)+1)-(a11/a12)(b′-a′)+1,x≥t.
From (95) it is clear that Ga′b′(x,t) will be positive almost everywhere for t∈[a′,b′] as long as x≠b′ and
(96)1-a11a12(b′-a′)>0,
which gives condition (88).(iv)
a12≠0 and a22≠0. In this case, one has
(97)Ga′b′(x,t)={(-(a21/a22)(t-b′)+1)(-(a11/a12)(x-a′)+1)(-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12),x<t,(-(a21/a22)(x-b′)+1)(-(a11/a12)(t-a′)+1)(-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12),x≥t.
From (97) it is clear that Ga′b′(x,t) will be positive almost everywhere for t∈[a′,b′] if 1-(a11/a12)(b′-a′)>0, 1-(a21/a22)(a′-b′)>0, and (-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12)>0. Condition (89) is a consequence of this and the fact that (-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12)>0 and 1-(a11/a12)(b′-a′)>0 imply 1-(a21/a22)(a′-b′)>0 and that (-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12)>0 and 1-(a21/a22)(a′-b′)>0 imply 1-(a11/a12)(b′-a′)>0 (both proofs are straightforward and are left to the reader).To complete the theorem it suffices to prove that if (86)–(89) are met by a′, b′, they will also be met by any other extremes a, b such that [a,b]⊂[a′,b′]. This is straightforward in the case of conditions 1-(a11/a12)(b′-a′)>0 and 1+(a21/a22)(b′-a′)>0. In the case of the condition (a21/a22)-(a11/a12)-(a21/a22)(a11/a12)(b′-a′)>0 simply note that
(98)h(x)=a21a22-a11a12-a21a22a11a12x,x∈[0,b′-a′],
is a linear function positive at the extremes 0 and b′-a′ by the hypothesis, and therefore it must be positive for any b-a∈[0,b′-a′].Remark 15.
Conditions (86)–(89) imply conditions (12) of Lemma 3. In consequence they will replace the latter in the rest of results of this section.Once the conditions for the positivity of the Green function have been determined, we can proceed with the key results of this method.Lemma 16.
Assume that[a′,b′]⊂[a′′,b′′] are enclosed intervals such that [a′,b′]≠[a′′,b′′] and that any of conditions (86)–(89) are met in [a′′,b′′]. Assume also that f(x) is positive almost everywhere on [a′′,b′′] and that x∈[a′,b′] is such that G(x,t) is not identically zero. Then the operator L satisfies
(99)La′,b′f(x)<La′′,b′′f(x)<max⁡{La′′,b′′f(t),t∈[a′′,b′′]}.Proof.
Sinceq(x) and f(x) are positive almost everywhere on [a′′,b′′] by hypothesis, and Ga′b′ and Ga′′b′′ are also positive almost everywhere on [a′,b′] and [a′′,b′′], respectively, by the hypothesis and Lemma 14, it suffices to prove that
(100)Ga′b′(x,t)≤Ga′′b′′(x,t)
for (x,t)∈[a′,b′]×[a′,b′]. We will do it only for the case a12,a22≠0 (the proof of the rest of cases is very similar), for which the Green function is given by (97). If we denote by W the expression (-(a11/a12)(b′-a′)+1)(a21/a22)-(a11/a12), a routine calculation yields
(101)δGa′b′(x,t)δa′={-(a11a12W)2(-a21a22(x-b′)+1)×(-a21a22(t-b′)+1),x≥t,-(a11a12W)2(-a21a22(t-b′)+1)×(-a21a22(x-b′)+1),x<t.
From (89), which is the hypothesis applicable if a12,a22≠0, one has δGa′b′(x,t)/δa′<0 if a11≠0 and δGa′b′(x,t)/δa′≡0 if a11=0. In other words, fixing x, t, and b′, either Ga′b′(x,t) increases (a11≠0) or Ga′b′(x,t) stays constant (a11=0) as a′ decreases.
In a similar manner we can show thatδGa′b′(x,t)/δb′≥0; that is, fixing x, t, and a′, either Ga′b′(x,t) increases (a21≠0) or Ga′b′(x,t) stays constant (a21=0) as b′ increases. This proves (100) and the lemma.Lemma 17.
Assume that any of conditions (86)–(89) holds in [a,b]. If (i)
eithera12=a22=0,(ii)
ora12=0 and (a21/a22)>0,(iii)
ora22=0 and (a11/a12)<0,(iv)
or(a11/a12)<0<(a21/a22), then one has
(102)max⁡{G(x,t),x∈[a,b]}=G(t,t),t∈[a,b].Proof.
We will only prove the last case (the proof of the other cases is very similar). Thus, from (97) it follows that, fixing t, G(x,t) is a function increasing with x for x≤t and decreasing with x for t<x. Therefore it will have a maximum at x=t, which is exactly (102).Theorem 18.
Let one suppose that there exists a nontrivial solutiony(x) of (2) that satisfies (3) at a, b. Let ymax⁡ be defined by ymax⁡=max⁡{y(x),x∈[a,b]}. Assume that any of conditions (86)–(89) holds in extremes a′, b′ such that [a,b]⊂[a′,b′]. Assume also that f>y almost everywhere on [a,b] and f>0 almost everywhere on [a′,b′]. Then
(103)La′,b′kf(x)>y(x),k≥1,x∈[a,b].
Moreover, (i)
if(a11/a12)≥0 and either a22=0 or (a21/a22)>0, then
(104)La′,b′kf(a′)>ymax⁡=y(a),k≥1;(ii)
if(a21/a22)≤0 and either a12=0 or (a11/a12)<0, then
(105)La′,b′kf(b′)>ymax⁡=y(b),k≥1;(iii)
if any of the conditions of Lemma17 are met, then
(106)∫a′b′‍q(t)Ga′b′(t,t)La′b′k-1f(t)dt>ymax⁡,k≥1.Proof.
From the hypothesis and Lemma14, Gab(x,t) is either positive almost everywhere or identically zero on t∈[a,b], depending on the value of x. In consequence we can apply Lemma 1 recursively to yield
(107)La,bkf(x)≥La,bky(x)=y(x),k≥1.
And given that f(x)>0 almost everywhere on [a′,b′], from Lemma 16 and (107) one gets (103). Equations (104) and (105) are consequences of y and La′,b′kf having maxima at a and a′, in the first case, and b and b′, in the second case, respectively. As for (106), it results from (103) and Lemma 17.Theorem 19.
Let one suppose that there exists a nontrivial solutiony(x) of (2) that satisfies (3) at a, b. Let ymax⁡ be defined by ymax⁡=max⁡{y(x),x∈[a,b]}. Assume that any of conditions (86)–(89) holds at a, b and that the extremes a′, b′ are such that [a′,b′]⊂[a,b]. Assume also that 0<f(x)<y(x) almost everywhere on [a′,b′]. Then
(108)La′,b′kf(x)<y(x),k≥1,x∈[a′,b′].
Moreover, (i)
if(a11/a12)≥0 and either a22=0 or (a21/a22)>0, then
(109)La′,b′kf(a′)<ymax⁡=y(a),k≥1;(ii)
if(a21/a22)≤0 and either a12=0 or (a11/a12)<0, then
(110)La′,b′kf(b′)<ymax⁡=y(b),k≥1;(iii)
otherwise(111)max⁡{La′,b′kf(x),x∈[a′,b′]}<ymax⁡,k≥1.Proof.
Applying Lemmas1 and 16 recursively one has
(112)ymax⁡≥y(x)=La,bky(x)>La′,b′ky(x)≥La′,b′kf(x),k≥1,x∈[a′,b′].
This gives (108) and also (111). As before, (109) and (110) are consequences of y and La′,b′kf having maxima at a and a′, in the first case, and b and b′, in the second case, respectively.Corollary 20.
Let one suppose that there exists a nontrivial solutiony(x) of (2) that satisfies (3) at a, b. Assume that any of conditions (86)–(89) holds at the extremes a′, b′. (i)
If(a11/a12)≥0 and either a22=0 or (a21/a22)>0, then
(113)La′,b′k{1}(a′)>1,[a,b]⊂[a′,b′],k≥1,(114)La′,b′k{b′-xb′-a′}(a′)<1,[a′,b′]⊂[a,b],k≥1.(ii)
If(a21/a22)≤0 and either a12=0 or (a11/a12)<0, then
(115)La′,b′k{1}(b′)>1,[a,b]⊂[a′,b′],k≥1,(116)La′,b′k{x-a′b′-a′}(b′)<1,[a′,b′]⊂[a,b],k≥1.(iii)
If any of the conditions of Lemma17 are met and c is any real number in ]a′,b′[, then
(117)∫a′b′‍q(t)Ga′b′(t,t)La′,b′k-1{1}(t)dt>1,[a,b]⊂[a′,b′],∫a′b′‍q(t)Ga′b′(t,t)La′,b′k-1{1}(t)dt>1,[a,b]⊂k≥1,(118)La′,b′kg(c)<1,[a′,b′]⊂[a,b],k≥1,
where g is defined by
(119)g(x)={x-a′c-a′x∈[a′,c],b′-xb′-cx∈[c,b′].Proof.
Equations (113), (115), and (117) can be easily obtained by setting f(x)≡ymax⁡ in (104)–(106). In turn, (114) can be obtained by setting f(x)≡y(a′)((b′-x)/(b′-a′)) (with y(a′)>0) in (108) and taking into account that y is concave (its second derivative is negative almost everywhere on [a,b] since q(x) and y(x) are positive almost everywhere on [a,b] due to (86)–(89)). Equations (116) and (118) can also be obtained in a similar manner.Remark 21.
Corollary 1 of [1] is a particular case (disfocality) of Corollary 20, (115)-(116).
Theorems18 and 19 and Corollary 20 provide conditions to assess if there exists a solution y(x) satisfying (3) at extremes inner or outer to [a,b]. However, at least in the way they have been presented, they do not allow determining if there exists a solution satisfying (3) at exactly the extremes a, b. That will be the purpose of the next theorems.Theorem 22.
Let one suppose that there exists a nontrivial solutiony(x) of (2) that satisfies (3) at a, b. Let ymax⁡ be defined by ymax⁡=max⁡{y(x),x∈[a,b]}. Assume that any of conditions (86)–(89) holds at a, b. Let f(x) be such that f(x)>y(x) almost everywhere on [a,b] and let bk be a sequence of real values whose definition depends on the boundary conditions (3) as follows: (i)
if(a11/a12)≥0 and either a22=0 or (a21/a22)>0, then
(120)La,bkkf(a)=y(a),k≥1;(ii)
if(a21/a22)≤0 and either a12=0 or (a11/a12)<0, then
(121)La,bkkf(bk)=y(b),k≥1;(iii)
if any of the conditions of Lemma17 are met, then
(122)∫abk‍q(t)Gabk(t,t)La,bkk-1f(t)dt=ymax⁡,k≥1;with f>0 almost everywhere on [a,bk]. Then bk<b for k≥1 and {bk} tends to b as k→∞.Proof.
From the fact thatf(x)>y(x) almost everywhere on [a,b]and f(x)>0 almost everywhere on [a,bk] and from (104)–(106) and (120)–(122), it is clear that [a,bk]⊆[a,b] (i.e., a<bk≤b) for k≥1. Now, let us assume that {bk} does not have a limit in b. In that case there exist a δ>0 and a subsequence {bkj} of {bk} such that bkj<b-δ. But then, from Theorem 5 and Lemma 16 one has
(123)lim⁡kj→∞La,bkjkjf(x)≤lim⁡kj→∞La,b-δkjf(x)=0,x∈[a,b-δ],lim⁡kj→∞La,bkjkj-1f(x)≤lim⁡kj→∞La,b-δkj-1f(x)=0,x∈[a,b-δ].
In both cases the convergence is uniform for x∈[a,b-δ]. Therefore, for every ϵ>0 there will exist infinitely many bkj such that
(124)La,bkjkj-1f(x),La,bkjkjf(x)<ϵ,x∈[a,bkj]
which contradicts (120) and (121) and, taking into account that
(125)|∫abkj‍q(t)G(t,t)La,bkjkj-1f(t)dt|<ϵ|∫abkj‍q(t)G(t,t)dt|,
also contradicts (122).Theorem 23.
Let one suppose that there exists a nontrivial solutiony(x) of (2) that satisfies (3) at a, b. Let ymax⁡ be defined by ymax⁡=max⁡{y(x),x∈[a,b]}. Let bk be a sequence of real values whose definition depends on the boundary conditions (3) as follows: (i)
if(a11/a12)≥0 and either a22=0 or (a21/a22)>0, then
(126)La,bkkfa,bk(a)=ymax⁡,k≥1;(ii)
if(a21/a22)≤0 and either a12=0 or (a11/a12)<0, then
(127)La,bkkfa,bk(bk)=ymax⁡,k≥1;(iii)
if any of the conditions of Lemma17 are met, let one pick any c∈[a,bk] and set
(128)La,bkkfa,bk(c)=ymax⁡,k≥1, where fa,bk is a family of continuous functions on [a,bk] such that 0<fa,bk(x)<y(x) almost everywhere in [a,bk]. Let one suppose that there exists a continuous function f such that 0<f(x)<fa,bk(x) almost everywhere in [a,bk]. Assume that any of conditions (86)–(89) holds at a, bk. Then bk>b for k≥1 and {bk} tends to b as k→∞.Proof.
From the fact thaty(x)>fa,bk(x)>0 almost everywhere on [a,bk], (109)–(111), and (126)–(128), it is clear that [a,bk]⊇[a,b] (i.e., a<b≤bk) for k≥1. Now, let us assume that {bk} does not have a limit in b. In that case there exist a δ>0 and a subsequence {bkj} of {bk} such that bkj>b+δ. But then, from Theorem 5 and Lemmas 1 and 16, for those x∈[a,b+δ] such that Φ1(x)≠0, one has
(129)lim⁡kj→∞La,bkjkjfa,bkj(x)≥lim⁡kj→∞La,bkjkjf(x)≥lim⁡kj→∞La,b+δkjf(x)=±∞,
the sign depending on that of 〈f,Φ1〉Φ1(x). Let us suppose that such a sign is positive (the negative case can be treated in the same manner). In that case for every M>ymax⁡>0 there will exist infinitely many bkj such that
(130)La,bkjkjfa,bkj(x)>M.
Since the values at the left-hand side of (126)-(127) are upper bounds of La,bkjkjfa,bkj(x), (130) leads to a contradiction and proves the assertion.Remark 24.
Theorems22 and 23 guarantee that, given [a,b], one can determine if there exists a nontrivial solution y(x) of (2) satisfying (3) at a, b by calculating the sequences defined by (120)–(122) and (126)–(128) and checking if they converge to b.


## 5. The Calculation ofLkf(x)

A common key point of Sections3 and 4 is the importance of the calculation of Lkf(x) for different functions f. Bearing this in mind, our aim for this section is to find manners to facilitate such a calculation. That will be done with the following theorems, in all of which the internal product 〈·,·〉 is assumed to be calculated on the variable t.Theorem 25.
The operatorL defined in (4) verifies
(131)Lf(x)=〈G(x,t),f(t)〉,x∈[a,b].Proof.
It is straightforward from the definition ofL given in (4) and the definition of the internal product (10).Theorem 26.
Ifa, b satisfy any of conditions (12), then the operator L defined in (4) verifies
(132)Ln+m+1f(x)=〈LnG(x,t),Lmf(t)〉,Ln+m+1f(xl)x∈[a,b],n,m≥0.
And in particular
(133)Lk+1f(x)=〈LkG(x,t),f(t)〉,x∈[a,b],k≥0.Proof.
From (131) one has
(134)Ln+m+1f(x)=〈G(x,t),Ln+mf(t)〉,Ln+m+1f(x)x∈[a,b],n,m≥0.
From the hypothesis and Lemma 3, L is self-adjoint. We can apply that property recursively to (134) to obtain
(135)Ln+m+1f(x)=〈LG(x,t),Ln+m-1f(t)〉=〈LnG(x,t),Lmf(t)〉,x∈[a,b],n,m≥0.
Equation (133) is a particular case of (132) for m=0.Remark 27.
The advantage of (133) is that, given (2) and fixing the value k we want to apply to Lk, it allows testing easily different functions f, g (as many as we want) in Theorems 7 and 18–23 in a simple way, just leaving the complication of the method to the calculation of Lk-1G(x,t).


## 6. Some Examples

Throughout this section we will introduce examples where Theorems7 and 12 and Corollary 20 will be used to determine upper and lower bounds for the distance between extremes a and b of a solution of (2) for different functions q(x) and boundary conditions (3). For the sake of simplicity, the analysis will fix the value of the starting point a (in all cases zero) and will search for upper and lower bounds of the adjacent right extreme b. The boundary conditions addressed will be disfocality, in the first three examples, and disconjugacy, in the two last ones, and the examples used to illustrate the disfocal case will be the same as those that were introduced in [1], in order to compare the bounds obtained via Theorems 7 and 12 with those coming from Corollary 20, which was already obtained in [1] for the disfocal case. A comparison between these bounds and the bounds obtained via other existing methods, like Brown and Hinton’s one (see [24]) for lower bounds, will also be done.In all examples the calculation ofLk will be done numerically and the number of iterations will be restricted case by case.Example 1.
Let us consider the following boundary value problem(136)y′′+Aex=0,x≥0,y(0)=0,y′(b)=0,
for different values of the constant A.
The application of Theorem7 to the function f(x)=x, together with Theorem 12 and Corollary 20 (already applied to this problem in [1] with f(x)=(x/b) and f(x) equal to the function Φ(x) of [1, Equation (47)]) gives Table 1.Table 1
Comparison of bounds forb in Example 1.
Value ofA
Used formula
Bound
A
=
1
∑
s
d
s
d
s
s
d
f
s
d
L
10
{
x
b
}
(
b
)
<
1
b
<
1.072
L
15
{
Φ
}
(
b
)
>
1
b
>
1.0635
L
17
{
Φ
}
(
b
)
>
1
b
>
1.0637
Brown and Hinton
b
>
1.0635
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
1.067
Theorem12 with k=25
b
>
1.0627
A
=
3
∑
s
d
s
d
s
s
d
d
s
d
L
6
{
x
b
}
(
b
)
<
1
b
<
0.712
L
9
{
Φ
}
(
b
)
>
1
b
>
0.7020
L
12
{
Φ
}
(
b
)
>
1
b
>
0.7025
Brown and Hinton
b
>
0.7020
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.705
Theorem12 with k=24
b
>
0.7023
A
=
5
∑
w
e
w
e
w
q
e
w
q
e
L
6
{
x
b
}
(
b
)
<
1
b
<
0.580
L
8
{
Φ
}
(
b
)
>
1
b
>
0.5707
L
12
{
Φ
}
(
b
)
>
1
b
>
0.5712
Brown and Hinton
b
>
0.5707
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.573
Theorem12 with k=17
b
>
0.5707As can be seen from Table1, Theorem 7 (i.e., ∥Lk{x}∥22<∥x∥22) provides better upper bounds of b than Corollary 20 (that is, Lk{x/b}(b)<1) with a lesser number of iterations for all values of the constant A. On the other hand, Theorem 12 yields lower bounds of b which are either worse than Corollary 20 and Brown and Hinton’s method [24] or, when equal, require a higher number of iterations than [1]. And as was mentioned in [1], it is worth remarking the good accuracy that Brown and Hinton’s method gives for lower bounds with relatively low calculation effort.Example 2.
Let us consider the following boundary value problem:(137)y′′+Bxy=0,x≥0,y(0)=0,y′(b)=0,
for different values of the constant B.
The application of Theorem7 to the function f(x)=x, together with Theorem 12 and Corollary 20 (already applied to this problem in [1]), gives Table 2, where one can observe results and trends similar to those commented in the previous example.Table 2
Comparison of bounds forb in Example 2.
Value ofB
Used formula
Bound
B
=
1
∑
a
s
a
s
s
a
s
a
s
L
5
{
x
b
}
(
b
)
<
1
b
<
1.530
L
22
{
Φ
}
(
b
)
>
1
b
>
1.513
Brown and Hinton
b
>
1.513
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
1.516
Theorem12 with k=24
b
>
1.5106
B
=
3
∑
s
a
s
a
s
a
s
a
s
L
5
{
x
b
}
(
b
)
<
1
b
<
1.062
L
23
{
Φ
}
(
b
)
>
1
b
>
1.049
Brown and Hinton
b
>
1.049
∥
L
3
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
1.0508
Theorem12 with k=23
b
>
1.0473
B
=
5
∑
a
s
a
s
a
a
s
a
s
L
5
{
x
b
}
(
b
)
<
1
b
<
0.894
L
23
{
Φ
}
(
b
)
>
1
b
>
0.884
Brown and Hinton
b
>
0.884
∥
L
6
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.886
Theorem12 with k=25
b
>
0.8835Example 3.
Let us consider the following boundary value problem:(138)y′′+Ce-xy=0,x≥0,y(0)=0,y′(b)=0,
for different values of the constant C.
The application of Theorem7 to the function f(x)=x, together with Theorem 12 and Corollary 20 (already applied to this problem in [1]), gives Table 3.Table 3
Comparison of bounds forb in Example 3.
Value ofC
Used formula
Bound
C
=
5
∑
s
d
s
d
s
d
a
s
d
L
7
{
x
b
}
(
b
)
<
1
b
<
1.007
L
10
{
Φ
}
(
b
)
>
1
b
>
0.969
L
16
{
Φ
}
(
b
)
>
1
b
>
0.972
Brown and Hinton
b
>
0.969
∥
L
3
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.98
Theorem12 with k=7
b
>
0.970
Theorem12 with k=19
b
>
0.974
C
=
10
∑
f
d
f
d
f
r
d
f
L
5
{
x
b
}
(
b
)
<
1
b
<
0.633
L
8
{
Φ
}
(
b
)
>
1
b
>
0.609
L
15
{
Φ
}
(
b
)
>
1
b
>
0.611
Brown and Hinton
b
>
0.609
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.6161
Theorem12 with k=4
b
>
0.609
Theorem12 with k=24
b
>
0.6119
C
=
20
∑
s
a
s
a
s
L
5
{
x
b
}
(
b
)
<
1
b
<
0.415
L
5
{
Φ
}
(
b
)
>
1
b
>
0.401
L
14
{
Φ
}
(
b
)
>
1
b
>
0.403
Brown and Hinton
b
>
0.401
∥
L
2
{
x
}
∥
2
2
<
∥
x
∥
2
2
b
<
0.405
Theorem12 with k=11
b
>
0.403
Theorem12 with k=21
b
>
0.4033In this third example one can observe that Theorem12 finally gets to yield greater (i.e., better) lower bounds of b than those provided by Corollary 20 and Brown and Hinton’s method [24] with a lesser number of iterations k.Example 4.
Let us consider the following boundary value problem:(139)y′′+Aex=0,x≥0,y(0)=0,y(b)=0,
for different values of the constant A.
The application of Theorem7 and Corollary 20—(118)—to the function f(x) defined by
(140)f(x)=v(x)={2xbx∈[0,b2],2(b-x)bx∈[b2,b];
together with Theorem 12 and Corollary 20—(117)—gives Table 4.Table 4
Comparison of bounds forb in Example 4.
Value ofA
Used formula
Bound
A
=
3
∑
a
s
a
s
a
a
s
s
a
s
∫
a
b
‍
q
(
t
)
G
(
t
,
t
)
L
11
{
1
}
(
t
)
d
t
>
1
b
>
1.266
Theorem12 with k=11
b
>
1.279
∥
L
5
{
v
}
∥
2
2
<
∥
v
∥
2
2
b
<
1.287
L
10
{
v
}
(
a
+
b
2
)
<
1
b
<
1.298
A
=
5
∑
s
a
s
a
s
a
s
a
s
∫
a
b
‍
q
(
t
)
G
(
t
,
t
)
L
11
{
1
}
(
t
)
d
t
>
1
b
>
1.044
Theorem12 with k=11
b
>
1.056
∥
L
5
{
v
}
∥
2
2
<
∥
v
∥
2
2
b
<
1.062
L
10
{
v
}
(
a
+
b
2
)
<
1
b
<
1.071In this example the methods based on comparison of norms yield better results than those based on comparison of values ofLk. The reason for this may be that the boundary conditions of this case do not allow knowing in advance the value of the maximum of Lk to be used in the comparison (note that in the three previous examples the maximum was always placed at b), causing the selection of ((a+b)/2) as comparison value in the case of Corollary 20—(118)—and the use of G(t,t) in the integral in the case of Corollary 20 —(117)—both of which require further iterations to surpass the threshold value 1.Example 5.
Let us consider the following boundary value problem(141)y′′+Bxy=0,x≥0,y(0)=0,y(b)=0,
for different values of the constant B.
The application of Theorem7 and Corollary 20—(118)—to the function v(x) defined by (140) together with Theorem 12 and Corollary 20—(117)—gives Table 5, where one can observe results and trends similar to those commented in the previous example.Table 5
Comparison of bounds forb in Example 5.
Value ofB
Used formula
Bound
B
=
3
∑
w
s
w
s
s
∫
a
b
‍
q
(
t
)
G
(
t
,
t
)
L
11
{
1
}
(
t
)
d
t
>
1
b
>
1.825
Theorem12 with k=12
b
>
1.838
∥
L
7
{
v
}
∥
2
2
<
∥
v
∥
2
2
b
<
1.85
L
10
{
v
}
(
a
+
b
2
)
<
1
b
<
1.866
B
=
5
∑
f
g
f
f
g
∫
a
b
‍
q
(
t
)
G
(
t
,
t
)
L
11
{
1
}
(
t
)
d
t
>
1
b
>
1.539
Theorem12 with k=12
b
>
1.55
∥
L
9
{
v
}
∥
2
2
<
∥
v
∥
2
2
b
<
1.56
L
10
{
v
}
(
a
+
b
2
)
<
1
b
<
1.574


## 7. Conclusions

Throughout this paper two different types of methods have been provided to assess whether there are nontrivial solutions of (2) satisfying the boundary conditions (3) at extremes a′, b′ inner, equal, or outer to other fixed extremes a, b. Both rely on an iterative application of the operator L defined in (4) and allow obtaining sequences of extremes ak, bk that converge to the exact values a, b for which y(x) satisfies (3). The second characteristic is by far the most relevant one when one compares them with other methods present in the literature, since these latter only provide bounds (in fact very often only lower bounds or Lyapunov inequalities since the number of methods dealing with upper bounds is quite low) which cannot be improved following the same approach.In addition it is worth remarking that the set of methods presented over almost all possible boundary conditions (3), despite the fact that some of them (namely, the method based on Theorem 12 and those based on Theorems 18 and 19) only apply under a limited subset of boundary conditions (3). The reason for this is, on the one hand, that the only conditions common for all these methods are those of Lemma 3 (which indeed are relatively easy to satisfy) and, on the other hand, that Theorem 7 does not require any additional conditions at all and Theorem 12 requires additional conditions on (3) which complement those required by Theorems 18 and 19.As for what method is better in what case, it is difficult to decide, but the examples show that in the quest for outer bounds the method based on Theorem7 converges normally much faster than the one based on Theorem 18. In the search of inner bounds, as indicated before, Theorems 12 and 19 cannot be applied simultaneously to most of the boundary conditions (3), but in the cases where that is possible, the speed of convergence depends on the concrete boundary conditions and the concrete function q(x). In general it seems that boundary conditions aiming for a maximum in Lk different from a or b tend to favour the method based on Theorem 12 versus the method based on Theorem 19, the opposite occurring (with exceptions, as Example 3 shows) when the boundary conditions force maxima of Lk at either a or b.As for the main drawbacks, we can comment on two. On the one hand, the fact that the speed of convergence of the method depends vastly on the selected starting functionf and can be low in some cases, many iterations being necessary to approximate the values of the extremes a and b that satisfy (3). On the other hand, as they have been presented, the methods cannot deal with functions q(x) which are zero or negative in a subset of positive measure. However it can be shown, as was done in [1], that these methods can be extended to the case q(x) zero or negative in a way similar to most of the criteria published so far for the same problem.All in all and taking the mentioned constraints into consideration, we the authors believe that the advantages of the methods presented in this paper surpass their drawbacks largely and that they can become a very powerful tool to assess problems of conjugacy/disconjugacy, nondisfocality/disfocality, and so forth, of (2), whether directly or by means of other methods based on them which in turn improve them.



---

*Source: 126713-2014-05-15.xml*