Abstract

This paper presents a method to obtain lower and upper bounds for the minimum distance between points and of the solution of the second order linear differential equation satisfying general separated boundary conditions of the type and . 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 and as the recursivity index grows. The method covers conjugacy and disfocality as particular cases.

1. Introduction

In a recent paper of the authors (see [1]) it was shown that the recursive application of the operator defined by where is continuous on and strictly positive almost everywhere on that same interval, provided a method to determine if the second order linear differential equation was either left disfocal or left nondisfocal in the interval , the concept of left disfocal alluding to the nonexistence of a nontrivial solution of (2) with zeroes in such that (a similar definition exists for right disfocal). The method was based on two features of the operator , namely, (i)the fact that is positive and monotonic, so that for on one has on ;(ii)the fact that if (2) is left nondisfocal in an interval interior to and for any value of as long as (2) is left disfocal in .

The purpose of this paper is to extend the results of [1] to (2) with the more general boundary conditions: which contain conjugacy ( ) and nondisfocality as particular cases, by means of the recursive application of the operator defined by where (or , as we will name it in some sections) is the Green function of the problem In consequence this paper will yield criteria (in fact infinitely many) to obtain lower and upper bounds for the minimum distance between points and satisfying (3) when 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 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 of (2) such that satisfied with which implied that is, Beesack [3], Hinton [4], Levin [5, 6], and Reid [79] 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 and to obtain more complex versions of (9), the fact is that the iterative application of has rarely been proposed in any papers, with the exception of Harris [11], who suggested its application for the disfocal case— or —without getting to prove that it guaranteed any improvement. We will show in this paper that under certain conditions (3) the function 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 gets to be negative it is still sometimes possible to obtain upper and lower bounds for the distance between and , which are as close to the real distance between and 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 ( ) and by the authors in [1, 13, 14] for the nondisfocal case. References [12, 1520] are notable exceptions to this trend.

As indicated in the first paragraph, throughout the paper we will assume that is continuous on an interval such that and that is strictly positive almost everywhere on . This allows defining the internal product for , continuous on (it is easy to prove that (10) satisfies all the conditions required by an internal product) and the associated norm defined by

Likewise, we will use the notation to name the operator defined in (4), or to name the function with domain resulting from the application of to , and or to name the value of the function at the point and when other extremes of integration , potentially different from , are used in (4). In that latter case, any norm appearing in the same formula will be assumed to be calculated with and as integration extremes in (11).

We will say that the points , are interior to an interval if and .

The organization of the paper is as follows. Section 2 will present the main properties of the operator . Sections 3 and 4 will apply these properties in different ways to find upper and lower bounds for the minimum distance between points and 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 Operator

The purpose of this section will be to present the main properties of the operator defined in (4) for 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 operator is linear. In addition, if is positive almost everywhere for , then is positive and monotonic.

Proof. The linearity of is quite evident. If is positive almost everywhere for , the positiveness (and in consequence monotonicity) of is a consequence of being positive almost everywhere on by hypothesis.

Lemma 2. The operator is compact.

Proof. From [21, Theorem 7.2.6] and the fact that is continuous on it is straightforward to show that is compact with the norm and therefore with the norm defined in (11).

Lemma 3. If any of the following conditions are met, then the operator is self-adjoint.

Proof. Let us define as the solution of and as the solution of A straightforward calculation shows that if any of the conditions (12) are met, then the Wronskian of and , namely, is not zero, which implies that and are not linearly dependent and therefore no nontrivial solutions of (2)-(3) exist. In consequence from [22, Theorem IX.3] one has Now, in order to prove self-adjointness, we need to prove that, given , . Thus, from (10) we have Combining (17) and (16) one gets to And integrating by parts the right-hand side of (18) one finally has

Lemma 4. The operator is bounded with the norm and verifies where the 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 has a countably infinite number of eigenvalues and associated orthonormal eigenfunctions , which allow expressing , , as Moreover, (i)if no nontrivial solution of (2) satisfies (3) either at , or at any , interior to , then (ii)if there is a nontrivial solution of (2) that satisfies (3) at some , interior to , then In addition, if and is such that , then the sign corresponding to that of .

Proof. Let us consider the eigenvalue problem 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 which form an increasing sequence with , each of which has its corresponding orthonormal (with the norm (11)) eigenfunction , and that the set of eigenfunctions forms an orthonormal basis of . Applying the operator to these eigenfunctions and integrating by parts it is easy to show that which implies that are also the eigenfunctions of the operator with corresponding eigenvalues . Since from Lemmas 1, 2 and 3 is linear, compact, and self-adjoint, we can apply [21, Theorem 7.5.2] and represent in the canonical form Applying again to (30) it yields given that and for . Applying recursively to (31) one gets to which is in fact (22). And the application of Parseval’s identity (see [21, Lemmas 1.5.14]) to (32) leads to Since form an increasing sequence, from (33) it is clear that Now, let us note that if (2) does not have any nontrivial solution that satisfies (3) either at , or at any , interior to , the first eigenvalue (and therefore all the others) must be strictly greater than 1. In that case, from (34) one has (23) and From (35) and Lemma 4 one gets (24).
Likewise, if there is a nontrivial solution of (2) that satisfies (3) at some , interior to , then must be such that . From this and (33) one gets (25). If, in addition, , from (33) we get to which is (26). On the other hand, we can write We can divide both sides of (37) by to yield Applying Parseval’s identity to (38) one gets Since , from (39) one yields which implies that there exists an index such that From Lemma 4 and (41) one has for ; that is, for . Since by hypothesis and , (43) leads to (27).

Remark 6. Theorem 5 provides two types of methods to obtain upper and lower bounds for the minimum distance between points satisfying (3), one based on comparing the norm of with some other constants and another based on comparing the value of at concrete points of for different functions . These will be addressed separately in the next two sections.

3. Bounds for the Distance between and Based on

As stated before, Theorem 5 provides methods to get upper and lower bounds for the minimum distance between points , satisfying (3), which are based on the comparison of with some thresholds for different values of the extremes of integration and of (4) and (11).

Thus, on the one hand, if , are such that there is a nontrivial solution of (2) which satisfies (3) at some , interior to , from (26) (and as long as , i.e., is “close” to ) it is clear that will grow with the index regardless of the choice of and accordingly there will be an index 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 , and calculating the minimum values of the extremes which give for different values of , as the following theorem shows.

Theorem 7. Assume that there exists a nontrivial solution of (2) satisfying (3) at , . Let the sequence be defined by (44), where is a continuous function on such that and , fulfill any of conditions (12) for each . Then and tends to as .

Proof. The fact that is obvious from (23). Now, let us pick a . From (26) , which means that there exists an index such that Given that for , from (23) and the continuity of on (continuity guaranteed by the hypothesis), there must exist a value such that for each . This proves the second assertion of the theorem.

The application of the method based on Theorem 7 is quite straightforward given that the right-hand side of (44), that is, , is easy to calculate once fixed . However, it is worth remarking that, from (34), the closer the selected function is to , the smaller the terms for will be and the faster the sequence will converge to . Therefore, although the method can work with any continuous function such that , it is desirable to select one that may be as close as possible to the expected associated to the problem (28).

On the other hand, if , are such that no nontrivial solution of (2) satisfies (3) either at , or at some , interior to , from (24) it is clear that will shrink with the index regardless of the choice of and accordingly there will be an index 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 , and calculating the values of the extremes which give where is any lower bound of the right-hand side of (25), which has (in turn) a positive lower bound that does not depend on . This is shown in the following theorem.

Theorem 8. Assume that there exists a nontrivial solution of (2) satisfying (3) at , . Let the sequence be defined by (46), where is a continuous function on and , fulfill any of conditions (12) for each . Then and tends to as .

Proof. Again, the fact that is obvious from (25) and (46). Now, let us pick a . From (24), . This and the fact that is bounded below by a positive amount which does not vary with imply that there exists an index such that Given that for , from (25) and the continuity of on , there must exist a value such that for each . This proves the second assertion of the theorem.

Unlike what happens with the method based on Theorem 7, the method based on Theorem 8 presents some difficulties in its application due to the fact that neither nor the eigenvalues 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 given that the discarded terms are positive (in fact may be a possible positive lower bound for some types of terms used in (46), which does not depend on ). The resulting method to obtain lower bounds for the distance between and will work in the same way as the one described in the previous paragraphs, at the expense of requiring greater values of the index to violate (48). But even with such a simplification there is still a need to obtain a lower bound for which is not evident at all.

The following lemmas aim at finding cases (fundamentally choices of and boundary conditions (3)) where the values of , , and 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. Let and be defined as in Theorem 5. Assume that , satisfy any of conditions (12) of Lemma 3. Fixed and choosing and one has

Proof. From (10) one has And since satisfies (28) one yields As is the Green function of the problem (5), is the value at of the function satisfying (3) whose second derivative is , that is, . This proves (49). As for (50), note simply that

Lemma 10. Assume that either or and that either or . Let and be defined as in Theorem 5. Let one suppose that on and that it can be decomposed as with and being nondecreasing functions. Depending on the values of , , , and , one has (i)if , then (ii)if , then (iii)if , then (iv)if , then

Proof. First of all, let us note that can always be decomposed in the mentioned manner, given that and therefore exists in and can be expressed as the difference between two increasing functions and ; that is, which, taking exponentials, becomes with and nondecreasing. Now let us consider the functional defined by Using integration by parts and the fact that is orthonormal with respect to the norm (11) it is easy to prove that From (3), (61), and the hypothesis one gets And given that and are increasing on and that verifies (28) for , one has that is, is increasing on , and that is, is decreasing on . The application of (60), (63), and (64) to (62) leads to From (3), (60), and (65) it is straightforward to obtain (54)–(57).

Lemma 11. Let and be defined as in Theorem 5. Then one has with

Proof. From [23, Theorem 8], if we define the function angle by where is a real constant, then satisfies the equation being, therefore, a decreasing function. Let us fix as the range of the function. In that case, (i)if we will set . Else, we will set ;(ii)if we will define . Else,(a)if we will define ;(b)we will define . With this in mind, we will also define is a constant which depends only on and the boundary conditions (3) and is related to the angle distance between and due to (3). Since is a function satisfying (3) with one more zero in than , integrating (69) and taking (70) into account we obtain Taking in (71) one has The problem for (72) to be used to get upper bounds for is the dependence of on . We can eliminate it by obtaining an upper bound for . Accordingly we will define From (70), (72), and (73) one yields Taking squares in (74) one finally gets to (66).

Finally, by (25) and Lemmas 911 we are in a position to prove the next theorem, which allows a straightforward application of the method based on Theorem 8.

Theorem 12. Let and be defined as in Theorem 5. Let and be defined as in Lemma 10. Let and be defined as in Lemma 11. Assume that there is a nontrivial solution of (2) that satisfies (3) at some , interior to . Assume also that , satisfy any of conditions (12) of Lemma 3, that either or , and that either or . Then the following inequalities hold. (i)If then Otherwise, if is the first integer greater than 1 such that , then (ii)If then Otherwise, if is the first integer greater than 1 such that , then

Proof. Let us focus first on (75). From (25), the fact that (from the hypothesis) and Lemmas 9-10 it is straightforward to show that On the other hand, from (66) one has But given that from (79)–(81) one gets (75).
As for (76), again from (25) and Lemmas 9-10 it is straightforward to show that On the other hand, from (66) one has From the definition of , it is clear that And given that from (82)–(85) and the fact that 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 Theorem 12 by using better lower bounds for , , , and than the ones displayed in Lemma 10 and better upper bounds for 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 between and Based on Concrete Values of

This section will elaborate on the results of Section 2 in order to obtain upper and lower bounds for the distance between the extremes , in a similar way as it was done in [1]. A critical condition for that is the positivity of the Green function , which is ensured under certain boundary conditions by the next lemma.

Lemma 14. Let , be any real numbers such that and . If the boundary conditions (3) verify any of the following hypotheses for , , then is positive almost everywhere for except if or in case (86), in case (87), and in case (88).

Proof. As was mentioned in the proof of Lemma 3, if we define and as the solutions of respectively, then we can represent as follows: We can divide the analysis of (91) in four cases. (i) and . In this case one has which is clearly positive almost everywhere for as long as and . This yields condition (86).(ii) and . In this case, one has From (93) it is clear that will be positive almost everywhere for as long as and which gives condition (87).(iii) and . In this case, one has From (95) it is clear that will be positive almost everywhere for as long as and which gives condition (88).(iv) and . In this case, one has From (97) it is clear that will be positive almost everywhere for if , , and . Condition (89) is a consequence of this and the fact that and imply and that and imply (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 , , they will also be met by any other extremes , such that . This is straightforward in the case of conditions and . In the case of the condition simply note that is a linear function positive at the extremes 0 and by the hypothesis, and therefore it must be positive for any .

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 are enclosed intervals such that and that any of conditions (86)–(89) are met in . Assume also that is positive almost everywhere on and that is such that is not identically zero. Then the operator satisfies

Proof. Since and are positive almost everywhere on by hypothesis, and and are also positive almost everywhere on and , respectively, by the hypothesis and Lemma 14, it suffices to prove that for . We will do it only for the case (the proof of the rest of cases is very similar), for which the Green function is given by (97). If we denote by the expression , a routine calculation yields From (89), which is the hypothesis applicable if , one has if and if . In other words, fixing , , and , either increases ( ) or stays constant ( ) as decreases.
In a similar manner we can show that ; that is, fixing , , and , either increases ( ) or stays constant ( ) as increases. This proves (100) and the lemma.

Lemma 17. Assume that any of conditions (86)–(89) holds in . If (i)either ,(ii)or and ,(iii)or and ,(iv)or , then one has

Proof. We will only prove the last case (the proof of the other cases is very similar). Thus, from (97) it follows that, fixing , is a function increasing with for and decreasing with for . Therefore it will have a maximum at , which is exactly (102).

Theorem 18. Let one suppose that there exists a nontrivial solution of (2) that satisfies (3) at , . Let be defined by . Assume that any of conditions (86)–(89) holds in extremes , such that . Assume also that almost everywhere on and almost everywhere on . Then Moreover, (i)if and either or , then (ii)if and either or , then (iii)if any of the conditions of Lemma 17 are met, then

Proof. From the hypothesis and Lemma 14, is either positive almost everywhere or identically zero on , depending on the value of . In consequence we can apply Lemma 1 recursively to yield And given that almost everywhere on , from Lemma 16 and (107) one gets (103). Equations (104) and (105) are consequences of and having maxima at and , in the first case, and and , 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 solution of (2) that satisfies (3) at , . Let be defined by . Assume that any of conditions (86)–(89) holds at , and that the extremes , are such that . Assume also that almost everywhere on . Then Moreover, (i)if and either or , then (ii)if and either or , then (iii)otherwise

Proof. Applying Lemmas 1 and 16 recursively one has This gives (108) and also (111). As before, (109) and (110) are consequences of and having maxima at and , in the first case, and and , in the second case, respectively.

Corollary 20. Let one suppose that there exists a nontrivial solution of (2) that satisfies (3) at , . Assume that any of conditions (86)–(89) holds at the extremes , . (i)If and either or , then (ii)If and either or , then (iii)If any of the conditions of Lemma 17 are met and is any real number in , then where is defined by

Proof. Equations (113), (115), and (117) can be easily obtained by setting in (104)–(106). In turn, (114) can be obtained by setting (with ) in (108) and taking into account that is concave (its second derivative is negative almost everywhere on since and are positive almost everywhere on 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).
Theorems 18 and 19 and Corollary 20 provide conditions to assess if there exists a solution satisfying (3) at extremes inner or outer to . 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 , . That will be the purpose of the next theorems.

Theorem 22. Let one suppose that there exists a nontrivial solution of (2) that satisfies (3) at , . Let be defined by . Assume that any of conditions (86)–(89) holds at , . Let be such that almost everywhere on and let be a sequence of real values whose definition depends on the boundary conditions (3) as follows: (i)if and either or , then (ii)if and either or , then (iii)if any of the conditions of Lemma 17 are met, then with almost everywhere on . Then for and tends to as .

Proof. From the fact that almost everywhere on and almost everywhere on and from (104)–(106) and (120)–(122), it is clear that (i.e., ) for . Now, let us assume that does not have a limit in . In that case there exist a and a subsequence of such that . But then, from Theorem 5 and Lemma 16 one has In both cases the convergence is uniform for . Therefore, for every there will exist infinitely many such that which contradicts (120) and (121) and, taking into account that also contradicts (122).

Theorem 23. Let one suppose that there exists a nontrivial solution of (2) that satisfies (3) at , . Let be defined by . Let be a sequence of real values whose definition depends on the boundary conditions (3) as follows: (i)if and either or , then (ii)if and either or , then (iii)if any of the conditions of Lemma 17 are met, let one pick any and set where is a family of continuous functions on such that almost everywhere in . Let one suppose that there exists a continuous function such that almost everywhere in . Assume that any of conditions (86)–(89) holds at , . Then for and tends to as .

Proof. From the fact that almost everywhere on , (109)–(111), and (126)–(128), it is clear that (i.e., ) for . Now, let us assume that does not have a limit in . In that case there exist a and a subsequence of such that . But then, from Theorem 5 and Lemmas 1 and 16, for those such that , one has the sign depending on that of . Let us suppose that such a sign is positive (the negative case can be treated in the same manner). In that case for every there will exist infinitely many such that Since the values at the left-hand side of (126)-(127) are upper bounds of , (130) leads to a contradiction and proves the assertion.

Remark 24. Theorems 22 and 23 guarantee that, given , one can determine if there exists a nontrivial solution of (2) satisfying (3) at , by calculating the sequences defined by (120)–(122) and (126)–(128) and checking if they converge to .

5. The Calculation of

A common key point of Sections 3 and 4 is the importance of the calculation of for different functions . 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 .

Theorem 25. The operator defined in (4) verifies

Proof. It is straightforward from the definition of given in (4) and the definition of the internal product (10).

Theorem 26. If , satisfy any of conditions (12), then the operator defined in (4) verifies And in particular

Proof. From (131) one has From the hypothesis and Lemma 3, is self-adjoint. We can apply that property recursively to (134) to obtain Equation (133) is a particular case of (132) for .

Remark 27. The advantage of (133) is that, given (2) and fixing the value we want to apply to , it allows testing easily different functions , (as many as we want) in Theorems 7 and 1823 in a simple way, just leaving the complication of the method to the calculation of .

6. Some Examples

Throughout this section we will introduce examples where Theorems 7 and 12 and Corollary 20 will be used to determine upper and lower bounds for the distance between extremes and of a solution of (2) for different functions and boundary conditions (3). For the sake of simplicity, the analysis will fix the value of the starting point (in all cases zero) and will search for upper and lower bounds of the adjacent right extreme . 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 of 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 for different values of the constant .
The application of Theorem 7 to the function , together with Theorem 12 and Corollary 20 (already applied to this problem in [1] with and equal to the function of [1, Equation ]) gives Table 1.

As can be seen from Table 1, Theorem 7 (i.e., ) provides better upper bounds of than Corollary 20 (that is, ) with a lesser number of iterations for all values of the constant . On the other hand, Theorem 12 yields lower bounds of 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: for different values of the constant .
The application of Theorem 7 to the function , 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.

Example 3. Let us consider the following boundary value problem: for different values of the constant .
The application of Theorem 7 to the function , together with Theorem 12 and Corollary 20 (already applied to this problem in [1]), gives Table 3.

In this third example one can observe that Theorem 12 finally gets to yield greater (i.e., better) lower bounds of than those provided by Corollary 20 and Brown and Hinton’s method [24] with a lesser number of iterations .

Example 4. Let us consider the following boundary value problem: for different values of the constant .
The application of Theorem 7 and Corollary 20—(118)—to the function defined by together with Theorem 12 and Corollary 20—(117)—gives Table 4.

In this example the methods based on comparison of norms yield better results than those based on comparison of values of . 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 to be used in the comparison (note that in the three previous examples the maximum was always placed at ), causing the selection of as comparison value in the case of Corollary 20—(118)—and the use of 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 for different values of the constant .
The application of Theorem 7 and Corollary 20—(118)—to the function 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.

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 , inner, equal, or outer to other fixed extremes , . Both rely on an iterative application of the operator defined in (4) and allow obtaining sequences of extremes , that converge to the exact values , for which 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 Theorem 7 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 . In general it seems that boundary conditions aiming for a maximum in different from or 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 at either or .

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 function and can be low in some cases, many iterations being necessary to approximate the values of the extremes and that satisfy (3). On the other hand, as they have been presented, the methods cannot deal with functions 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 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.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This work has been supported by the Spanish Ministry of Science and Innovation Project DPI2010-C02-01.