# Oscillation Criteria of Fourth-Order Differential Equations with Delay Terms

Authors: Shoura Ahmed Balatta; Ishak Hashim; Ahmad Sami Bataineh; Eddie Shahril Ismail
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The aim of this paper is to derive oscillation criteria of the following fourth-order differential equation with delay termrxz′′′xγ′+∑i=1nqixfzηix=0, under the assumption ∫x0∞r−1/γsds=∞. The results are based on comparison with the oscillatory behaviour of second-order delay equations and the generalised Riccati transformation. Not only do the provided theorems provide an entirely new technique but also they vastly improve on a number of previously published conclusions. We give three examples to illustrate our findings.


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## Body

## 1. Introduction

Higher-order neutral differential equations have recently been recognized as being sufficient to describe a variety of real applications [1–4]. As a result, many researchers have studied the qualitative behaviour of solutions of these equations (see [5–8]). The research of oscillation and oscillatory behaviour of these equations, which has been investigated using multiple approaches and techniques, has received special attention (see [9–11]). The attempt to improve the work and obtain a generalised platform that covers all special cases inspires the investigation of fourth- and higher-order equations.In this work, we are concerned with oscillation of fourth-order delay differential equations of the form(1)rxz′′′xγ′+∑i=1nqixfzηix=0,wherex≥x0. Throughout this work, we suppose the following:
(i)
r∈C1x0,∞,R and γ is a quotient of odd positive integers(ii)
The following condition holds:(2)∫x0∞1r1/γsds=∞,forrx>0, r′x>0, and
(iii)
qi,ηi∈Cx0,∞,R, qix≥0, ηix≤x, limx⟶∞ηix=∞ (i=1,2,⋯), and f∈CR,R such that(3)fxxγ≥ℓ>0,forx≠0.By a solution of (1), we mean a function z∈C3xz,∞, xz≥x0, that has the property rxz′′′xγ∈C1xz,∞ and fulfills (1) on xz,∞. If a solution of (1) has arbitrarily large zeros on xz,∞, then it is considered oscillatory; otherwise, it is called nonoscillatory. Equation (1) is said to be oscillatory if all its solutions are oscillatory.Next, we give some previous findings in the literature that are relevant to the present work. Grace [12] has studied the equation
(4)rxzn−vxv+qxfzgx=0,in addition to Agarwal et al. [13] and Xu and Xia [14] who have studied the equation
(5)zn−1xzn−1xγ−1′+fx,zηx=0,subject to condition (2). Zhang et al. [15] obtained oscillatory criteria of the equation
(6)rxz′′′xγ′+qxzγηx=0,with the condition(7)∫x0∞1r1/γudu<∞.Baculikova et al. [16] used the comparison theory to prove that if
(8)y′x+qxfδηn−1xn−1!r1/γηxfy1/γηx=0is oscillatory, then(9)rxzn−1xγ′+qxfzηx=0is oscillatory for evenn. Grace et al. [7] presented oscillation criteria for fourth-order delay differential equations of the form
(10)r3r2r1z′′′′x+qxzηx=0,under the assumption(11)∫x0∞dtrix<∞,i=1,2,3.Using the Riccati transformation, an oscillation criterion for fourth-order neutral delay differential equation of the form(12)rxzx+pxzηx′′′α′+∫abqx,ξfzgx,ξdξ=0was obtained by Chatzarakis et al. [17]. By using the technique of the Riccati transformation and the theory of comparison with first-order delay equations, Bazighifan and Abdeljawad [18] established some new oscillation criteria for fourth-order advanced differential equations with p-Laplacian-like operator of the form
(13)bxz′′′xp−2z′′′x′+∑i=1jqixgzηix=0.Very recently, Bazighifan et al. [19] established new criteria for the oscillatory behaviour of the following fourth-order differential equations with middle term
(14)rxz′′′xp1−2z′′′x′+σxz′′′xp1−2z′′′x+qxzτxp2−2zτx=0,by the comparison technique and employing the Riccati transformation under the condition(15)∫x0∞1rsexp−∫x0sσηrηdη1/p1−1ds=∞.For convenience, in the present work, we denote(16)δx=∫x∞1r1/γsds,ψx=πx∑i=1nℓqixηi3xx3γ+μx2−2γ2r1/γxδγ+1x,φx=π′xπx+γ+1μx22r1/γxδx,φ∗x=τ′xτx+2δx,ψ∗=τx∫x∞ℓrx∫x∞∑i=1nqisηiγssγds1/γdx+1−r−1/γxδ2x,where π,τ∈C1x0,∞,0,∞. The generalised Riccati transformation is defined as
(17)ωx≔πxrxz′′′γxzγx+1δγx,(18)ϑx≔τxz′xzx+1δx.We remark that in the study of the asymptotic behaviour of the positive solutions of (1), there are only two cases:Case 1.
zjx>0forj=1,2,3,Case 2.
zjx>0forj=1,3andz′′x<0.In this work, using the Riccati approach and a comparison with a second-order equation, we shall obtain oscillation criteria for (1).


## 2. Some Significant Auxiliary Lemmas

The following lemmas serve as a basis for our findings.Lemma 1 (see [20]).
Letα be a ratio of two odd numbers; H>0 and K are constants. Then,
(19)Pα+1/α−P−Qα+1/α≤PαQ1/α+Q1/αP−1αQ1+α/α,PQ≥0,α≥1,ααα+1α+1Kα+1Hα≥Km−Hmα+1/α,H>0.Lemma 2 (see [17]).
Letfj>0 and fn+1<0 for all j=0,1,⋯,n. Then,
(20)n!xnfx≥n−1!xn−1ddxfx.Lemma 3 (see [21]).
The equation(21)axm′xγ′+qxmγx=0,where a∈Cx0,∞, ax>0, and qx>0, is nonoscillatory if and only if there exist x≥x0 and σ∈C1x,∞ such that
(22)σ′x+γr1/γxσ1+1/γx+qx≤0,
forx≥x0.Lemma 4 (see [22]).
Suppose thath∈Cnx0,∞,0,∞; then,
(23)hn−1xhnx≤0,
for everyλ∈0,1 and x≥xλ.


## 3. Oscillation Criteria

In this section, we shall obtain some oscillation criteria for equation (1).Lemma 5.
Suppose thatz is a solution of (1) such that z>0 and zj>0 for all j=1,2,3. If we have the function ω∈C1x,∞ defined in (17), where π∈C1x0,∞,0,∞, then
(24)ω′x≤−ψx+φxωx−γμx22rxπx1/γωγ+1/γx,
for allx>x1, where x1 is large enough.Proof.
Letz be a solution of (1) where z>0 and zjx>0 for all j=1,2,3. Thus, from Lemma 4, we get
(25)z′x≥μ2x2z′′′x,
for allμ∈0,1 and for every large x. From (17), we have that ωx>0 for x≥x1, and
(26)ω′x=π′xrxz′′′xγxzγx+1δγx+πxrxz′′′xγ′zγx−γπxzγ−1xz′xrxz′′′xγz2γx+γπxr1/γxδγ+1x.
Using (25) and (17), we acquire
(27)ω′x≤π′xπxωx+πxrxz′′′xγ′zγx−γπxμ2x2rxz′′′xγ+1zγ+1x+γπxr1/γxδγ+1x≤π′xπxωx+πxrxz′′′xγ′zγx−γπxμ2x2rxωxπxrx−1rxδγxγ+1/γ+γπxr1/γxδγ+1x.
LettingP=ωx/πxrx, Q=1/rxδγx, and α=γ and by using Lemma 1, we get
(28)ωxπxrx−1rxδγxγ+1/γ≥ωxπxrxγ+1/γ−1γr1/γxδxγ+1ωxπxrx−1rxδγx.
From Lemma2, we obtain zx≥x/3z′x, and hence,
(29)zηixzx≥ηi3xx3.
From (1), (27), and (28), we obtain
(30)ω′x≤π′xπxωx−ℓπx∑i=1nqixηi3xx3γ−γπxμ2x2rxωxπxrxγ+1/γ−γπxμ2x2rx−1γr1/γxδxγ+1ωxπxrx−1rxδγx+γπxr1/γxδγ+1x.
This implies that(31)ω′x≤π′xπx+γ+1μx22r1/γxδxωx−γμx22r1/γxπ1/γxωγ+1/γx−πx∑i=1nℓqixηi3xx3γ+μx2−2γ2r1/γxδγ+1x.
Thus,(32)ω′x≤−ψx+φxωx−γμx22rxπx1/γωγ+1/γx.
The proof is completed.Lemma 6.
Letz be a solution of (1) such that z>0 and zjx>0 for j=1,3 and z′′x<0. If the function ϑ∈C1x,∞ is defined in (18) such that τ∈C1x0,∞,0,∞, then
(33)ϑ′x≤φ∗xϑx−ψ∗x−1τxϑ2x,
for allx>x1, where x1 is large enough.Proof.
Letz be a solution of (1) where z>0 and zjx>0 for j=1,3 and z′′x<0. From Lemma 2, we have that zx≥xz′x. Integrating this inequality from ηx to x, we obtain
(34)zηix≥ηixxzx.
Hence, from (3) we have
(35)fzηix≥ℓηiγxxγzγx.
By integrating (1) from x to u and since z′x>0, we get
(36)ruz′′′uγ−rxz′′′xγ=−∫xu∑i=1nqisfzηisds≤−ℓzγx∫xu∑i=1nqisηiγssγds.
Now lettingu⟶∞ yields
(37)rxz′′′xγ≥ℓzγx∫x∞∑i=1nqisηiγssγds,and so
(38)z′′′x≥zxℓrx∫x∞∑i=1nqisηiγssγds1/γ.
Integrating this fromx to ∞ gives
(39)z′′x≤−zx∫x∞ℓrx∫x∞∑i=1nqisηiγssγds1/γdx.
From (18), we have that ϑx>0 for x≥x1 and by differentiating, we get
(40)ϑ′x=τ′xτxϑx+τxz′′xzx−τxϑxτx−1δx2+τxr1/rxδ2x.
Now, using Lemma1 with P=ϑx/τx, Q=1/δx, and α=1 yields
(41)θxτx−1δx2≥θxτx2−1δx2ϑxτx−1δx.
From (1), (40), and (41), we have the following:
(42)ϑ′x≤τ′xτxϑx−τx∫x∞ℓrx∫x∞∑i=1nqisηiγssγds1/γdx−τxϑxτx2−1δx2ϑxτx−1δx+τxr1/γxδ2x.
This implies that(43)ϑ′x≤τ′xτx+2δxϑx−1τxϑ2x−τx∫x∞ℓrx∫x∞∑i=1nqisηiγssγds1/γdx+1−r−1/γxδ2x.
Thus,(44)ϑ′x≤φ∗xϑx−ψ∗x−1τxϑ2x.
The proof is completed.Lemma 7.
Letz be a solution of (1) with z>0. If π∈Cx0,∞ such that
(45)∫x0∞ψs−2μs2γrsπsφsγ+1γ+1γ+1ds=∞,
for someμ∈0,1, then z does not fulfill Case 1.Proof.
Letz be a solution of (1) such that z>0. From Lemma 5, we obtain that (24) holds. Using Lemma 1 with
(46)K=φx,H=γμx22rxπx1/γ,
andm=ω, we get
(47)ω′x≤−ψx+2μx2γrxπxφxγ+1γ+1γ+1.
Now, integrating fromx1 to x yields
(48)∫x1∞ψs−2μs2γrsπsφsγ+1γ+1γ+1ds≤ωx1,
which contradicts (45). So, the proof is complete.Lemma 8.
Letz be a solution of (1) with z>0 and zjx>0 for j=1,3 and z′′x<0. If τ∈Cx0,∞ such that
(49)∫x0∞ψ∗s−14τsφ∗s2ds=∞,
thenz does not fulfill Case 2.Proof.
Letz be a solution of (1) such that z>0. From Lemma 6, we get that (33) holds. Using Lemma 1 with
(50)H=φ∗x,K=1τx,γ=1,m=ϑ,we obtain
(51)ω′x≤−ψ∗x+14τxφ∗x2.
Integrating fromx1 to x gives
(52)∫x1∞ψ∗s−14τsφ∗s2ds≤ωx1,
which contradicts (49). This completes the proof.Theorem 9.
Letπ,τ∈Cx0,∞ such that (45) and (49) hold for some μ∈0,1. Then, equation (1) is oscillatory.Proof.
The proof is very similar to the proofs of Lemmas7 and 8.
Now, by using the comparison method, we develop additional oscillation results for (1) in the following theorem:Theorem 10.
Let (2) hold and assume that
(53)rxx2γz′xγ′+ψxzγx=0,(54)z′′x+zx∫x∞ℓrx∫x∞∑i=1nqisηiγssγds1/γdx=0,
are both oscillatory; then, (1) is oscillatory.Proof.
Assume the contrary that (1) has a positive solution z, and by virtue of Lemma 3 and if we set πx=1 in (24), then we get
(55)ω′x+γμx22r1/γxωγ+1/γ+ψx≤0.
Hence, we have that (53) is nonoscillatory, which is a contradiction. If we set τx=1 in (33), then we obtain
(56)ϑ′x+ψ∗x+ϑ2x≤0.
Thus, equation (54) is nonoscillatory, which is a contradiction. The proof is now complete.It is well known (see Řehák [23])) that if
(57)∫x0∞1rxdx=∞,limx⟶∞inf∫x0x1rsds∫x∞qsds>14,then equation (21) with γ=1 is oscillatory.Theorem 11.
Let (2) hold. Assume that
(58)∫x0∞x2rxdx=∞,
and(59)limx⟶∞inf∫x0xs2rsds∫x∞ψsds>12λ1,
for some constantλ1∈0,1 and
(60)limx⟶∞inf∫x∞∫x∞ℓrx∫x∞∑i=1nqisηiγssγdsdxds>14,
then every solution of (1) is oscillatory.
The proof is obvious.


## 4. Examples

In this section, we provide some examples to prove that the results of Section3 are valid.Example 1.
Consider(61)z4x+q0−x2x4+1x2zx=0,x≥1,where q0>0.
Letγ=1, rx=1, qx=q0−x2/x4+1/x2=q0/x, and ηx=x.
Hence, we have(62)δx0=∞,ψx=q0x,φx=3x,φ∗x=1x,ψ∗x=q06x.
If we setπx=x3, τx=x, and ℓ=1, then condition (45) becomes
(63)∫x0∞ψs−2μs2γrsπsφsγ+1γ+1γ+1ds=∫x0∞q0s−92μsds=q0−92μ∫x0∞1sds=∞,ifq0>92μ.
Therefore, from Lemma7, if q0>9/2μ, then (61) has no positive solution z satisfying z′′x>0. Also, condition (49) becomes
(64)∫x0∞ψ∗s−14τsφ∗s2ds=∫x0∞q06s−14sds=∞,ifq0>32.
From Lemma8, if q0>3/2, then (61) has no positive solution z satisfying z′′x<0. Thus, from Theorem 9 every solution of (61) is oscillatory if q0>max9/2μ,3/2.Example 2.
Consider the following differential equation representing equation (1),
(65)x3z′′′x3′+c−x4x7+1x3z3εx=0,x≥1,
wherec>0 and 0<ε<1 are constants.
Here,γ=3, rx=x3, qx=c−x4/x7+1/x3=c/x7, and ηx=εx. Hence,
(66)δx=∞,ψx=cε9x,φx=6x,φ∗x=1x,ψ∗x=cε3481/31x.
If we setπx=x6, τx=x, and ℓ=1, then condition (45) yields
(67)∫x0∞ψs−2μs2γrsπsφsγ+1γ+1γ+1ds=∫x0∞cε9s−812μ3sds=cε9−812μ3∫x0∞1sds.
Therefore, from Lemma7, if c>3/4ε3, then (65) has a solution z>0 satisfying z′′x>0. Also, from condition (49) we have
(68)∫x0∞ψ∗s−14τsφ∗s2ds=∫x0∞cε3481/31s−14sds=cε3481/3−14∫x0∞1sds.
Thus, from Theorem9, every solution of (65) is oscillatory if
(69)c>max34ε3,812ε9μ3.Example 3.
Consider(70)z4x+q0x−x2x5+1x3zx2=0,x≥1,
whereq0>0.
Letγ=1, rx=1, qx=q0x−x2/x5+1/x3=q0/x4, and ηx=x/2. When ℓ=1 is used, condition (59) becomes
(71)limx⟶∞inf∫x0xs2rsds∫x∞q0s4ds=limx⟶∞infx33∫x∞q0s4ds=q09>14,and condition (60) gives
(72)limx⟶∞inf∫x∞∫x∞ℓrx∫x∞qsηγssγdsdxds=limx⟶∞inf∫x∞∫x∞∫x∞q02s4dsdxds=q06>14.
Therefore, from Theorem11, all solutions of (70) are oscillatory if q0>2.25.


## 5. Conclusion

In this paper, we have established some new sufficient criteria which ensure that every solution of the fourth-order differential equations (1) is oscillatory. The approach we used was based on comparisons with the oscillatory behaviour of second-order delay equations and the Riccati transformation. Several illustrative examples have also been presented.



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