Research Article | Open Access

# Improving Results on Solvability of a Class of th-Order Linear Boundary Value Problems

**Academic Editor:**Elena Braverman

#### Abstract

This paper presents a modification of a recursive method described in a previous paper of the authors, which yields necessary and sufficient conditions for the existence of solutions of a class of th-order linear boundary value problems, in the form of integral inequalities. Such a modification simplifies the assessment of the conditions on restricting the inequality to be verified to a single point instead of the full interval where the boundary value problem is defined. The paper also provides an error bound that needs to be considered in the integral inequalities of the previous paper when they are calculated numerically.

#### 1. Introduction

Let be a compact interval in , let be such that , and let us consider the th-order boundary value problem:where and are functions piecewise continuous on such thatfor at least one ,and is the right-disfocal differential operator defined bywith , . Examples of these types of problems appear in the study of the deflections of beams, both straight ones with nonhomogeneous cross sections in free vibration, which are subject to the fourth-order linear Euler-Bernoulli equation, and curved ones with different shapes. An account of these and other applications can be found in [1, Chapter IV].

In a recent paper of the authors (see [2, Theorems 8, 9 and 10]) it was shown that the recursive application of the operator is defined bywhere and is the Green function of the problem: to functions belonging to a cone with some properties, providing a procedure to determine the existence (or not) of solutions of the problem (1)–(6) for and interior to , (i.e., for , satisfying or ). The procedure was based on the evaluation of whether either or belonged to the cone for different values of . Such an evaluation implied in practice the comparison of the functions and over the full interval , so that (i)if for all , then there was no solution of (1) satisfying (2)-(3) at extremes interior to , ;(ii) if for all , then there was a solution of (1) satisfying (2)-(3) either at , or at extremes interior to , .

The iterative comparison yielded lower and upper bounds for the extremes and for which (1)–(3) has a nontrivial solution, bounds which converge to the values of these extremes and as the recursivity index grows.

One of the few drawbacks of the method of [2] is that when the calculation of is done numerically using a partition of , very often such a calculation only yields values at the points and not at the interior points of each subinterval . Since the aforementioned inequalities have to be satisfied for all , in these cases we need to introduce a security margin in the comparison of the discrete values and for all (i.e., to enforce that either or ) in a manner that guarantees that the inequalities hold for all .

With this constraint in mind, the purpose of this paper is twofold:(i)To calculate such a security margin that, once taken into account when comparing the mentioned functions over the set , ensures that the same results are obtained on the full interval .(ii) To show that under certain conditions (namely, ; compare with the original of [2]) and upon selection of the proper functions , it is possible to restrict the comparison of derivatives of and to a specific point in instead of the full interval. The argument inspires on an idea from Keener for the focal problem (see [3]).

In terms of nomenclature, we will use the notation to name the operator defined in (7), or to name the function with domain resulting from the application of to , or to name the function with domain resulting from the application of to recursively times, and to name the value of the function at the point . We will use when we want to stress the dependence of with the extremes where it is defined. And we will denote by the set of piecewise continuous functions on .

In order to make this paper self-contained, let us recall that, given a Banach space , a cone is a nonempty closed set defined by the following conditions:(1)If , then for any real numbers .(2)If and , then . We will denote the interior of the cone by , and we will say that the cone is reproducing if any can be expressed as with . The existence of a cone in a Banach space allows defining a partial ordering relationship in that Banach space by setting if and only if . Accordingly, we will say that the operator is -positive if there exists a such that for any one can find positive constants and such that (note that the constants and do not need to be the same for all ). We will denote by the spectral radius of (in other words, is the supremum of the spectrum of ).

The main result of [2] is Theorem 2, which we will state also here for completeness.

Theorem 1 (see [2, Theorem 2]). *Let us suppose that there is a Banach space and a reproducing cone therein for which and is -positive. Then the eigenvalue problem has a solution and its associated eigenvalue is positive, simple, and bigger in absolute value than any other eigenvalue of such a problem.**In addition, if is strictly increasing with the length of the interval (i.e., if is fixed, is increasing with and if is fixed, is decreasing with ) and , one has the following: *(i)*If there is no nontrivial solution of (1)–(6) at extremes , equal or interior to , , then* *for any and for any there exists such that* *and there cannot be any and any such that*(ii) *If there is a nontrivial solution of (1)–(6) at extremes , interior to , , then for any there exists a such that* *and there cannot be any and any such that*

From the historical point of view, let us remark that the use of the theory of cones in boundary value problems dates from the works of Krein and Rutman [4] and Krasnosel’skii [5], which were continued by multiple authors. References [3, 6–20] are a good account of this.

The organization of the paper is as follows. Section 2 will describe a new method to obtain necessary and sufficient conditions for the problem (1)–(6) to have a nontrivial solution, which restricts the comparison of functions to one point. In Section 3 the security margin that needs to be considered in the application of [2, Theorems 8–10] will be calculated using linear splines theory. Section 4 will apply the previous results to several examples. Finally Section 5 will provide some conclusions.

#### 2. A Cone and a Procedure to Reduce the Comparison of Functions to One Point

In this section we will first define a new cone different from that used in [2, Theorem 8] and will prove that it satisfies the properties required by Theorem 1. Then we will show that, for some specific functions belonging to that cone, the application of Theorem 1 implies a comparison of the values of a certain derivative of and at a single point in .

Thus, let us consider the eigenvalue problemwith defined as in (7). Let us define the Banach space asif , and asif ; in both cases the associated norm is Let us also define the cone byFrom the definition of it is clear that , , . With the help of the cone it is possible to prove the following theorem.

Theorem 2. *The conclusions of Theorem 1 are applicable to the problem (1)–(6) and the cone defined in (18).*

*Proof. *We only need to prove that is a reproducing cone, that , and that is -positive in , as this guarantees the existence of an eigenfunction with a positive maximal eigenvalue (e.g., see [3, Theorem 2.1]) and both the monotonicity of this eigenvalue with the extremes and and the compacticity of were already proven in [2, Theorem 8].

Thus, using the notationfor , it is clear that If the reproducing character of is immediate. Otherwise, we can get to the same conclusion by noting that for any , the two terms of the right-hand side of (21) being functions which belong to .

To prove the -positivity of , let us consider the same auxiliar Banach space defined in [2, Theorem 8]; namely, and the auxiliar cone defined bywhose interior, if we denote by the lowest integer that satisfies and , is given by Following exactly the same reasoning used in [2, Theorem 8] and using hypotheses (4) and (15)–(18) it is straightforward to prove that maps into , that is, for any . Since (this follows from the facts that and ), one has that , and also that for any there must be an such thatwhich impliesLikewise, there must be an such thatwhich impliesCombining (26) and (28) one getsAs , (29) proves that is -positive in . This completes the proof.

The next theorem will allow us to exploit Theorem 2 as explained in Section 1.

Theorem 3. *Let us suppose that and that is such that exists and satisfies on those points of where is continuous. One has the following: *(1)*If and there exists an integer such that* *then the problem (1)–(6) cannot have a nontrivial solution at extremes , interior to , .*(2)*If , there exists a single discontinuity point of such that* *and there exists an integer such that* *then the problem (1)–(6) does have a nontrivial solution either at , or at extremes , interior to , .*

*Proof. *If at those points of where is continuous, then from [21, Theorem 2.1], the function defined bysatisfiesat those points of where is continuous.

Thus, from the hypotheses of statement , one has that and , which together with (34) yield that is, with regard to the cone . From Theorems 1 and 2 one gets the first conclusion.

On the other hand, from the hypotheses of statements , (2), (7), and (34), one has that , that is decreasing on except at the point , where it has a discontinuity jump, and that . This implies that which in turn implies that with regard to the cone . Again, from Theorems 1 and 2, one gets the second conclusion.

*Remark 4. *It is possible to extend the results of Theorems 2 and 3 to cones similar to where the condition is replaced by , as long as , since in that case for (this follows from [21, Theorem 2.1], as mentioned in the proof), which implies for those . This latter fact is indeed the key that allows restricting the comparison of functions to one point in Theorem 3.

To apply statement of Theorem 3 we can consider, for instance, the functionwhich satisfies

And in order to apply statement of Theorem 3 we can pick any and define, for example, the functionwhich satisfies for ,and on .

*Remark 5. *As pointed out in [2], (1)–(6) is just a way of representing a set of problems of the typein a way that allows the existence and calculation of the Green function of the problem with boundary conditions (42)-(43), while guaranteeing the right disfocality of on and at the same time yielding functions , , which satisfy the conditions for the application of the method. But there still exists some freedom in the choice of and as long as , , so that it is normal to wonder what choices of such functions give better results than others. [2, Theorem 12] and [2, Remark 13] answered that question for the problem (1)–(6) and a cone different from (18) by proving the faster convergence of the method when the functions , were closer to zero. It is easy to prove that such a behaviour is also applicable to the cone defined in (18) taking into account the fact that maps into the cone of (23).

As was done in [2, Theorem 11 and Remark 15], it is possible to establish theorems similar to Theorems 2 and 3 for the problem symmetric to (1)–(6) and defined bywith , defined as in (6) and left disfocal on ,for at least one such that . To that end one needs to consider the interval and use the Banach spacesif , andif , associated with the norm as well as the coneSince the reasoning to prove them is the same as that used in Theorems 2 and 3, we will state the equivalent theorems without proof.

Theorem 6. *The conclusions of Theorem 1 are applicable to the problem (44)-(45) and the cone defined in (49).*

Theorem 7. *Let us suppose that such that exists and satisfies on those points of where is continuous. One has the following: *(1)*If and there exists an integer such that* *then the problem (44)-(45) cannot have a nontrivial solution at extremes , interior to , .*(2)*If , there exists a single discontinuity point of such that* *and there exists an integer such that* *then the problem (44)-(45) does have a nontrivial solution either at , or at extremes , interior to , .*

A function which satisfies the conditions of the statement of Theorem 7 is, for instance,which verifies

And a candidate function which fulfils the conditions of the statement of Theorem 7 isfor any , which satisfies for ,and on .

#### 3. Calculation of the Security Margin for the General Case of [2, Theorem 8]

As mentioned in Section 1, the application of [2, Theorems 8–10] to the problem (1)–(6) requires the determination of the sign of the function for and belonging to the cone of [2, equation (55)], so that(i)if the sign of such a function over is the opposite to that of , then there is no solution of (1) satisfying (2)-(3) at extremes interior to , ;(ii)if the sign of such a function over is the same as that of , then there is a solution of (1) satisfying (2)-(3) either at , or at extremes interior to , .

With this in mind, this section aims at answering the question of what security margin to consider in the evaluation of at a partition of for , in order to guarantee that the sign of that function holds for all . It is, therefore, a problem of assessing the value of in each subinterval based on the knowledge that we have of for all , a problem of interpolation.

To this end, given the partition of and a function , let us recall the concept of a linear spline (see [22, Sections 6.1 and 6.2]), defined as a function which is a linear polynomial in each of the subintervals and takes the value at each of the points . Numerically that can be expressed as

Applying (57) to one gets the spline The spline of (58) interpolates the function with an interpolation error ; that is,If , from [22, Section 6.5.1], one gets the fact that this interpolation error can be calculated in each subinterval as where is a value in which depends on . Maximizing the right-hand side of (60) one yields

From (59) and (61) one gets the following theorem, which gives the searched security margin.

Theorem 8. *If and have the same sign for some and then the sign of is the same as the sign of for all .*

In order to apply Theorem 8 we need to calculate a bound for on . If , from (7), we know that such a function can be obtained as Otherwise, such a calculation can be done by taking into account, from (6), (7), and (8), that

However, if we are calculating and numerically, in both cases we will probably end up with a problem similar to the one origin of this section: how to calculate/bound the value of from the values of that function on the set . Solutions to this can be as follows:(i)Either note that and, if needed, use the monotonicity properties of all derivatives of and up to ( maps the cone of [2, (55)] into itself and this implies that all the derivatives of up to have the same sign as ; see [2, Proof of Theorem 8]). Such monotonicity properties ensure that the maxima of the absolute value of these derivatives over are precisely at the extremes or .(ii)Or apply the same principles of (57)–(61), decompose as the sum of the corresponding linear spline and another interpolation error, and obtain an upper bound of from the values of that function at and a bound for the mentioned interpolation error, which can be calculated using (61) and (65). This gives for .

As before, the calculation of the maximum of the last term of (66) can be done from (63)-(64) and using the monotonicity properties of the derivatives of and .

*Remark 9. *Both in the first subinterval and the last subinterval of the application of Theorem 8 is not possible since given that . In these subintervals we will need to do the comparison using directly (60).

*Remark 10. *The estimation of the security margin using Theorem 8, (63)-(64) and either (65) or (66) can be a cumbersome exercise. However, as the examples will show, the bounds that they provide for the extremes and for which (1)–(6) has a solution are usually much better than those obtained with the method of Theorem 7 for the same number of iterations.

*Remark 11. *The arguments of this section are also valid when applied to the functions and and the cone of (18) for (the method of Section 2 assumes that ), just replacing th derivatives by th derivatives. This implies that we can make use of the same functions to apply both methods and compare results.

#### 4. Some Examples

In this section we will present a couple of examples where the results of the Sections 2 and 3 will be used to provide progressively better upper and lower bounds of the extremes that make (1)–(6) has a nontrivial solution, for the cases and 4 and different boundary conditions. In all of them the extreme will be fixed to zero (so as to focus the assessment on the extreme ) and the integral calculations will be done numerically using a mesh and applying the trapezoidal rule in each of the subintervals of the mesh. This also includes the calculation of the derivatives , , as these can be written as The maximum number of iterations has been set to 6 in all examples and up to 3 decimal figures have been provided for each bound.

*Example 1. *Let us consider the following boundary value problem:A possible representation of (69) in a manner that yields an equation right disfocal in the interval of interest iswith , , , and . Equation (70) gives the BVP whose Green function can be calculated following Coppel [23] as which coincides with .

The application of Theorem 3 to the functions defined in (37) and (39) (considering two different cases and to test the effect of the choice of in the result of the calculations) gives Table 1.

Table 1 shows that, as expected, the bounds get improved when the number of iterations grows. Additionally the method provides better upper bounds for if we pick instead of in the function of (39), for all values of .

On the other hand, the application of the security margin calculated in Theorem 8, using also (66), to determine if the function belongs to the cone of (18), gives a value of between and , much is more precise than the bounds shown in the Table 1 for . And this occurs regardless of being that of (37), that of (39) with , or that of (39) with . This implies that the use of the mentioned security margin to evaluate whether belong to the cone yields better bounds at the expense of complicating the calculations.

*Example 2. *Let us consider the following boundary value problem: