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

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 nth-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.

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   1   /  1 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 [  ,  +1 ].Since the aforementioned inequalities have to be satisfied for all  ∈ [, ], in these cases we need to introduce a security margin  > 0 in the comparison of the discrete values   1   (  )/  1 and   1   (  )/  1 for all  (i.e., to enforce that either 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,  <  1 ; compare with the original  ≤  1 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 , V ∈ , then  + V ∈  for any real numbers ,  ≥ 0.
We will denote the interior of the cone  by  0 , and we will say that the cone  is reproducing if any  ∈  can be expressed as  =  − V with , V ∈ .The existence of a cone in a Banach space  allows defining a partial ordering relationship in that Banach space by setting  ≤ V if and only if V −  ∈ .
Accordingly, we will say that the operator  is  0 -positive if there exists a  0 ∈  such that for any V ∈  \ {0} one can find positive constants  1 and  2 such that  1  0 ≤ V ≤  2  0 (note that the constants  1 and  2 do not need to be the same for all V).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  0 -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.
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.

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 problem with  defined as in (7).Let us define the Banach space  as if  = 0, and as if  > 0; in both cases the associated norm is Let us also define the cone  by From the definition of  it is clear that (−1) −  () () ≥ 0,  ∈ [, ], 0 ≤  ≤ .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  0 -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 notation for  ∈ [, ], it is clear that If  = 0 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  0 -positivity of , let us consider the same auxiliar Banach space B defined in [2, Theorem 8]; namely, and the auxiliar cone P defined by whose interior, if we denote by  the lowest integer that satisfies  >  1 and  ̸ =  2 , . . .,  − , 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 \{0} into P 0 , that is, V ∈ P 0 for any V ∈  \ {0}.Since P ⊂  (this follows from the facts that B ⊂  and  ≤  1 ), one has that () ⊂ , and also that for any  0 ∈ P 0 there must be an  1 > 0 such that International Journal of Differential Equations which implies Likewise, there must be an  2 > 0 such that which implies Combining ( 26) and (28) one gets As  0 ∈ P 0 ⊂ P ⊂ , (29) proves that  is  0 -positive in .
This completes the proof.
The next theorem will allow us to exploit Theorem 2 as explained in Section 1.
To apply statement (1) of Theorem 3 we can consider, for instance, the function which satisfies And in order to apply statement (2) of Theorem 3 we can pick any  ∈], [ and define, for example, the function which satisfies  () () = 0 for 0 ≤  ≤  − 1, and Remark 5.As pointed out in [2], (1)-( 6) is just a way of representing a set of problems of the type in a way that allows the existence and calculation of the Green function of the problem  = 0 with boundary conditions (42)-( 43), while guaranteeing the right disfocality of  = 0 on [, ] and at the same time yielding functions   (), 0 ≤  ≤ , which satisfy the conditions for the application of the method.But there still exists some freedom in the choice of   () and   () as long as   () =   () −   (), 0 ≤  ≤ , 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   (), 0 ≤  ≤  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 P 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 by as well as the cone Since 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).
and there exists an integer  > 0 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 (1) of Theorem 7 is, for instance, International Journal of Differential Equations which verifies And a candidate function which fulfils the conditions of the statement (2) of Theorem 7 is for any  ∈], [, which satisfies  () () = 0 for 0 ≤  ≤  − 1, and

Calculation of the Security Margin for the General Case of [2, Theorem 8]
As mentioned in Section 1, the application of [2, to the problem ( 1)-( 6) requires the determination of the sign of the function   1   ()/  1 −   1   ()/  1 for  >  ≥ 0 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 (−1) − , 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 (−1) − , 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   1   ()/  1 −   1   ()/  1 at a partition {  } of [, ] for  >  ≥ 0, in order to guarantee that the sign of that function holds for all  ∈ [, ].It is, therefore, a problem of assessing the value of   1   ()/  1 −   1   ()/  1 in each subinterval [  ,  +1 ] based on the knowledge that we have of   1   (  )/  1 −   1   (  )/  1 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 [  ,  +1 ] and takes the value (  ) at each of the points {  }.Numerically that can be expressed as Applying (57) to   1   ()/  1 −   1   ()/  1 one gets the spline The spline () of (58) interpolates the function   1   ()/   1 −   1   ()/  1 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 [  ,  +1 ] as where  is a value in [  ,  +1 ] 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.
International Journal of Differential Equations 7 then the sign of (−1) − (  1   ()/  1 −  1   ()/  1 ) is the same as the sign of (−1) − ( In order to apply Theorem 8 we need to calculate a bound for 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 |  1 +2   ()/  1 +2 −  1 +2   ()/  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   .
given that  1 ≤  − 1.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  >  ≥ 0 (the method of Section 2 assumes that  >  = 0), just replacing  1 th derivatives by th derivatives.This implies that we can make use of the same functions  ∈  to apply both methods and compare results.

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 International Journal of Differential Equations progressively better upper and lower bounds of the extremes that make (1)-( 6) has a nontrivial solution, for the cases  = 3 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 [  ,  +1 ] of the mesh.This also includes the calculation of the derivatives (  V) () (), 0 ≤  ≤ , 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.
The application of Theorem 3 to the functions () defined in (37) and (39) (considering two different cases  = /2 and  = /3 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  = /2 instead of  = /3 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  6  −  5  belongs to the cone  of (18), gives a value of  between 2.166 and 2.167, much is more precise than the bounds shown in the Table 1 for  = 5, 6.And this occurs regardless of () being that of (37), that of (39) with  = /2, or that of (39) with  = /3.This implies that the use of the mentioned security margin to evaluate whether   − −1  belong to the cone yields better bounds at the expense of complicating the calculations.

Remark 9 .
Both in the first subinterval [,  1 ] and the last subinterval [  , ] of [, ] the application of Theorem 8 is not possible since