Research Article | Open Access

# Distributions of Zeros of Solutions for Third-Order Differential Equations with Variable Coefficients

**Academic Editor:**Zhen-Lai Han

#### Abstract

For the third-order linear differential equations of the form , we will establish lower bounds for the distance between zeros of a solution and/or its derivatives. The main results will be proved by making use of Hardy’s inequality and some generalizations of Opial and Wirtinger type inequalities.

#### 1. Introduction

The mathematical description of some physical systems demands that we often solve linear differential equations subject to some boundary conditions. Some of these problems are the mathematical models of the deflection of beams. These beams, which appear in many structures, deflect under their own weight or under the influence of some external forces. For example, if a load is applied to the beam in a vertical plane containing the axis of symmetry, the beam undergoes a distortion, and the curve connecting the centroids of all cross sections is called the deflection curve or elastic curve. In elasticity, it is shown that the deflection of the curve, say measured from the -axis, approximates the shape of the beam and satisfies a linear fourth-order differential equation on an interval, say , with some boundary conditions. The distribution of boundary conditions which is the distribution of zeros of solutions of differential equations has been started by Picard [1, 2], who derived some uniqueness results for solutions of the second-order nonlinear differential equation with two-point boundary conditions when the nonlinear function satisfies the Lipschitz condition. The distribution of zeros of th-order differential equations with more than two points has been considered by Niccoletti [3]. Motivated by the work of Picard and Niccoletti, de la Vallée Poussin [4] considered the general th-order linear differential equation with real coefficients that are locally integrable inside and studied the disconjugacy of solutions.

Equation (1) is said to be disconjugate on an interval if every nontrivial solution has less than zeros on , with multiple zeros being counted according to their multiplicity. More precisely, disconjugate on a connected set means that the number of zeros of a nontrivial solution cannot equal the order of the equation. Equation (1) is said to be -disconjugate on an interval if no nontrivial solution has a zero of order followed by a zero of order . This means that, for every pair of points , , , there does not exist a nontrivial solution of (1) which satisfies The last value of , such that there exists a nontrivial solution which satisfies (2), is called the -conjugate point of . The first work that has been published by de la Vallée Poussin in 1929 was on the evaluation of the length of the interval in which the boundary problem , (), for the linear differential equation (1) only admits the null solution.

Following the way indicated by de la Vallée Poussin, one has tried to evaluate the length in a function of the upper bounds of the coefficients. The precise evaluation of the maximal length of the considered interval has only been obtained for second-order differential equations [5, 6] with the best possible constant . Coming then to (1), de la Vallée Poussin proved that this equation is disconjugate in if the coefficients and the length of the interval satisfy the inequality This is the part of his paper which has motivated the many refinements and generalizations described in the analysis. The importance of Poussin’s work has been emphasized and testified by many authors in the literature; we refer the reader to the papers [7–16] and the references cited therein. For completeness, we recall some of the related results which motivate the contents of this paper. Lasota [8] considered the third-order differential equation and proved that if then (4) is disconjugate, where and . Mathsen [9] proved that (4) is disconjugate if on and where , on , and . Casadei [7] also proved that if then (4) is disconjugate. Agarwal and Krishnamurthy [17] also proved that if then (4) is disconjugate.

Motivated by these papers, we will study the distribution of zeros of the solutions of the third-order differential equation where is an interval of reals and , , and are real-valued functions defined on such that .

By a solution of (9) on the interval , we mean a nontrivial real-valued function , which has the property that and satisfies (9) on . The nontrivial solution of (9) is said to oscillate or to be oscillatory if it has arbitrarily large zeros. Equation (9) is oscillatory if one of its nontrivial solutions is oscillatory. An equation of the form (9) is said to be disconjugate on an interval if no nontrivial solution has more than two zeros on counting multiplicities. We say that (9) is right disfocal (left disfocal) on the interval , () if the solutions of (9) such that , (, and ) do not have two zeros counting multiplicities in ().

The paper is organized as follows. In Section 2, we present Hardy’s inequality, the extensions of Opial’s inequality, and Wirtinger’s inequality that will be used to prove our main results. In Section 3, we are concerned with the lower bounds of the distance between zeros of a nontrivial solution and/or its derivatives for the third-order differential equation (9) subject to two sets of boundary conditions. In particular, we will prove the following.(i)Obtain lower bounds for the spacing , where is a nontrivial solution of (9) which satisfies (ii)Obtain lower bounds for the spacing , where is a solution of (9) which satisfies

#### 2. Hardy, Opial, and Wirtinger Inequalities

In this section, we present Hardy’s inequality and some generalizations of Opial and Wirtinger type inequalities that will be needed in the proof of the main results. The generalizations of Opial and Wirtinger type inequalities are adapted from Agarwal and Pang [18, 19] and the papers due to Beesack and Das [20], Clark and Hinton [21], and Fink [22]. The Hardy inequality is adapted from the book due to Kufner and Persson [23]. We begin with Hardy’s inequality which states the following. If is absolutely continuous on , such that or (), then where the weighted functions , are positive functions defined on and , are real parameters that satisfy and . The constant is given by where Note that inequality (12) has an immediate application to the case when . In this case, inequality (12) is satisfied if and only if exists and is finite. The Opial type inequality due to Beesack which is a generalization of Opial’s inequality states that if is absolutely continuous on with , then the following inequality holds: where , are real numbers such that , , , and are nonnegative, measurable functions defined on such that , and If instead , then (16) holds, where is replaced by The Opial type inequality due to Agarwal and Pang states that if and satisfies , () and absolutely continuous on , then where and are nonnegative and measurable function defined on , , are real numbers such that , and If instead , (), then (19) holds, where is replaced by We also need the following inequality, which is the special case of an inequality proved by Agarwal and Pang [18] with two functions: where and satisfies , , (), and is absolutely continuous on , , and , being nonnegative measurable functions defined on , and If , , then (22) holds, where is replaced: The Wirtinger type inequality due to Agarwal and Pang states that if , then The Opial type inequality due to Clark and Hinton inequality states that if , where , then The Opial type inequality due to Fink inequality states that if , , , , then where , (), and

#### 3. Main Results

In this section, we state and prove the main results. For simplicity, we introduce the following notations: where and , , Now, we are ready to state and prove the main results.

Theorem 1. *Assume that is a nontrivial solution of (9). If , then
**
If , then
*

*Proof. *We prove (35). Multiplying (9) by and integrating the new equation from to , we obtain
Integrating by parts the left-hand side, we get that
Using the assumptions that and , we have
Integrating the term by parts and using the assumption that , we obtain
Substituting (40) into (39), we get
where . Applying inequality (19) on the integral
with , , , , , , and , we get
where is defined as in (29). Applying inequality (12) on the integral with and , we have that
where is defined as in (31). Substituting (44) and (43) into (41), we get
Cancelling the term , we get
which is the desired inequality (35). The proof of (36) is similar to the proof of (35) by replacing by and by . The proof is complete.

In the following, we apply the Clark and Hinton inequality (26) to get a new result.

Theorem 2. *Assume that is a nonincreasing function and suppose that is a solution of (9). If , then
**
where is defined as in (31) for and is defined as in (29).*

*Proof. *Multiply (9) by and proceed as in the proof of Theorem 1 to get
where . Applying the Schwarz inequality
on the integral , we have that
Applying inequality (26) and using the assumption , we get that
Substituting (51) into (50) and using the assumption that is a nonincreasing function, we have
Applying Hardy’s inequality (12) on the integral , we obtain
where . Substituting (52) and (53) into (48), we obtain
The desired inequality (47) followed by cancelling the term . The proof is complete.

The proof of the following theorem is similar to the proof of Theorem 2.

Theorem 3. *Assume that is a nonincreasing function and suppose that is a solution of (9). If , then
**
where is defined as in (31) for and is defined as in (30).*

One can use Theorems 2 and 3 to obtain some different special cases. For example, as a special case of Theorem 2, when and , we have the following result.

Corollary 4. *If is a nontrivial solution of
**
which satisfies , then
**
where is defined as in (29).*

Now, we will prove a new result when .

Theorem 5. *Suppose that and assume that is a solution of (9). If , or , then
**
where and .*

*Proof. *Multiplying (9) by and integrating by parts the left-hand side, we have
Using the assumptions that , we obtain
By using the maximum values of and , we can write (60) as
For the first term on the right-hand side , we apply inequality (25) with (note that ) and , to obtain
Applying Fink inequality (27) on , with , , , and , we have
Substituting (62) and (63) into (61), we get, after cancelling the term , that
which is the desired inequality (58). The proof is complete.

As a special case of Theorem 5, when , we have the following result.

Corollary 6. *Let be a nontrivial solution of (56). If , then
*

In the following, we prove some results related to the boundary conditions presented in .

Theorem 7. *Assume that is a nontrivial solution of (9). If , then
**
If , then
*

*Proof. *Multiplying (9) by and integrating by parts the left-hand side from to , we get
Using the assumption that , we have
Applying inequality (12) on the integral with , we get
where is defined as in (32). Again applying inequality (22) on the integral , we get
where is defined as in (32). Substituting (70) and (71) into (69), we obtain
By cancelling the term , we get that
which is the desired result (66). The proof (67) is similar to the proof of (66) by using and instead of and and hence is omitted. The proof is complete.

In the following, we apply an inequality due to Boyd [24] to obtain new results. The Boyd inequality states that if , with (or ), then where , , , and Note that an inequality of type (34) also holds when . Choose and apply (34) to and and by addition we obtain where is defined as in (75). An inequality of type (74) holds when and (or ). In this case, (74) becomes where and is the gamma function.

Theorem 8. *Assume that is nonincreasing function and is a nontrivial solution of (9). If , then
**
If , then
*

*Proof. *Multiplying (9) by and integrating by parts the left-hand side, we have that
Using the assumption , we get that
Applying inequality (12) on the integral with , we get
where is defined as in (34). Applying Schwarz’s inequality on the term , we get that
Applying again inequality (78) on the integral with (note that ) we obtain
where is a nonincreasing function. Substituting (86) into (85), we get
Applying inequality (12) on the integral with and , we get
Substituting (88) into (87), we have
Substituting (84) and (89) into (83), we obtain
Cancelling the term , we obtain
which is the desired inequality (80). The proof of (81) is similar to the proof of (80) by using instead of . The proof is complete.

#### Conflict of Interests

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

#### References

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