The Distribution of Zeroes and Critical Points of Solutions of a Second Order Half-Linear Differential Equation

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Introduction
In a paper dated in 1981 (see [1]), Kwong suggested an idea to calculate Lyapunov inequalities for solutions of the second order linear differential equation as follows: for the case () > 0, which was based in the concave nature of the solution ().Years later, in [2], among many other interesting results, he applied such an idea to the determination of a sufficient condition for a solution of (1) to be oscillatory.Although the idea can also be employed to obtain upper bounds for the distance between zeroes and adjacent critical points of these solutions, seemingly Kwong did not get to elaborate it in that way (although very little effort had been needed to do it from the proof of Theorem 7 of [2] indeed), as a detailed analysis of his works shows (see [3] for a list of such works and links to the actual papers).Likewise, no other authors seem to have rediscovered it and used in the mentioned problem, to the knowledge of the authors.
Bearing that in mind, the purpose of this paper is to apply such an idea to the determination of upper bounds for the distance between zeroes and adjacent critical points of the solutions of the more general half-linear differential equation ( () Φ (  )) +  () Φ () = 0,  >  0 , where (), () > 0, Φ() = || −2  and  is a real number such that  > 1, of which ( 1) is a very particular case.The analysis of the maximum distance between zeroes and critical points is a problem which has received little or no attention in the last decades, apart from the paper [4], where the authors used Prüfer transformation techniques to elaborate two possible methods to tackle that problem.The reason for such little interest is unclear, anyway, given the success that its sibling problem (the determination of lower bounds or Lyapunov inequalities) has enjoyed in recent years (see [5][6][7][8][9][10][11]).As this paper will show, the method explained here improves those described in [4] in a significant amount of cases.
The organization of the paper is as follows.Section 2 will prove the main results and Section 3 will apply them in several examples so as to compare them with other methods from the authors.Finally Section 4 will state several conclusions.

Main Results
We will start our analysis on the equation where () > 0, Φ() = || −2  and  is a real number such that  > 1, and we will extend it later to the more general equation (2).Thus, let us prove the following lemma first.
Proof.From the definition of Φ(), that is, it is straightforward to show that Combining ( 3), (4), and (5) one has which implies that   has an opposite sign to that of  and completes the proof.
With the aid of Lemma 1 it is possible to prove the next theorem.
Theorem 2. Let () be a solution of (3) with () piecewise continuous and positive on an interval  ⊂ R. Let , ,   , and   be real numbers such that [, ] ⊂  and  ≤   ≤   ≤ .
Then, if () =   () = 0 one has Proof.Let us focus first on proving (7).For the sake of simplicity let us assume that () > 0 on ], ] (the case () < 0 can be treated in the same manner).
Let us focus now on (8).Again, for the sake of simplicity we will only consider the case () ≥ 0 on [, ].
Remark 3. From Theorem 2 it becomes clear that, fixed   = , any value of the extreme   that violates either (7) or (8), depending on the case, will be an upper bound for the zero or critical point , given that all values of   between  and  must satisfy such formulae.Likewise, if we fix   = , any value of the extreme   that violates either (7) or (8), depending on the case, will be a lower bound for the zero or critical point   .However, it is important to remark that there is no guarantee that an extreme that violates either (7) or (8), depending on the case, can be found in all cases, as one of the examples of Section 3 will show.
For the particular linear case  = 2 it is possible to obtain the following corollary.Then, if () =   () = 0 one has Proof.Let us focus first on proving (21).Setting  = 2 (the linear case) in (7) one gets Integrating by parts (23) one gets (21).The proof of ( 22) is similar and will not be repeated.Corollary 4 allows to tackle the problem of getting upper bounds for the distance between consecutive zeroes of the solution of (1), as the next corollary shows.Corollary 6.Let () be a solution of (1) with () piecewise continuous and positive on an interval  ⊂ R. Let ,  be adjacent zeroes of () such that [, ] ⊂  and let   ,   be such that  ≤   ≤   ≤ .
If we define () as the function then one has  ( If we minimize the left hand side of (26) on , it is easy to prove that the minimum occurs at  = (  +  )/2.Setting such a value of  on (26) and taking into account the definition of (), one gets (25).
Remark 7.For the same reasons as before, fixed   = , any value of the extreme   that violates (26) will be an upper bound for the value of the zero , and fixed   = , any value of the extreme   that violates (26) will be a lower bound for the value of the zero .In general, for any values   ,   that violate (26) there must be at least one zero  of the solution of (1) such that  ∈]  ,   [.Now we can turn our eyes into the more general equation ( 2), which we will do in the next theorem.Theorem 8. Let () be a solution of (2) with (), () piecewise continuous and positive on an interval  ⊂ R. Let , ,   , and   be real numbers such that [, ] ⊂  and  ≤   ≤   ≤ .
As happened before, for the particular linear case  = 2 it is possible to obtain the following corollaries.Corollary 9. Let () be a solution of (2) with  = 2 and (), () piecewise continuous and positive on an interval  ⊂ R. Let , ,   , and   be real numbers such that [, ] ⊂  and  ≤   ≤   ≤ .
Then, if () =   () = 0 one has and () as the function then one has Proof.The proof follows the same steps of that of Corollary 6 and will be omitted.
Remark 11.As happened before, for any values   ,   that violate (38) there must be at least one zero  of the solution of (2) with  = 2 such that  ∈]  ,   [.

Some Examples
Throughout this section we will introduce examples where upper bounds for the distance between a zero and an adjacent critical point of a solution of (2) will be provided by means The application of Theorem 2 and [4, Corollary 2.6 and Theorem 3.1] to this example yields Table 1.

Corollary 4 .
Let () be a solution of(1) with () piecewise continuous and positive on an interval  ⊂ R. Let , ,   , and   be real numbers such that [, ] ⊂  and  ≤   ≤   ≤ .

Remark 5 .
Equation (21), in fact, was obtained by Kwong as an intermediate result in the proof of [2, Theorem 7], focused on the determination of the oscillatory nature of a solution of (1).

Table 1 :
Comparison of upper bounds for  in Example 12.

Table 2 :
Comparison of upper bounds for  in Example 13.

Table 3 :
Comparison of upper bounds for  in Example 14.

Table 4 :
Comparison of upper bounds for  in Example 15.