# Some New Oscillation Results for Fourth-Order Neutral Differential Equations with a Canonical Operator

Authors: Omar Bazighifan
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

By this work, our aim is to study oscillatory behaviour of solutions to 4th-order differential equation of neutral typeLy′+∑j=1kqjyzβgjy=0 where Ly=ξyw‴yα,wy:=zy+ryzg˜y. By using the comparison method with first-order differential inequality, we find new oscillation conditions for this equation.


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## Body

## 1. Introduction

Neutral differential equations are getting an increasing importance among the practitioners in the recent past, since they are used in numerous applications in technology and natural science [1]. During the last one and half decade, a multidimensional application of differential equation with delay has been observed in the field of statistics, science, and engineering [2–6].In this paper, we are concerned with the oscillation of solutions of the even-order nonlinear differential equation:(1)Ly′+∑j=1kqjyzβgjy=0,where Ly=ξyw′′′yα,wy:=zy+ryzg˜y and αandβ are quotient of odd positive integers and(2)α:=1,ifβ≤1,2β−1,ifβ>1.The operatorLy is said to be in canonical form if ∫y0∞1/ξ1/αsds=∞; otherwise, it is called noncanonical. The main results are obtained under the following conditions:F1.ξ∈Cy0,∞, ξy>0, ξ′y≥0,r∈Cy0,∞,0≤ry<r0<∞,F2.g˜,gj,qj∈Cy0,∞,qjy>0, g˜y≤y, limy⟶∞g˜y=limy⟶∞gjy=∞, j=1,2,…,k,F3. For someμ∈0,1the differential inequality(3)η′y+1αμy36ξ1/αyβg0g˜0g˜0+r0ββ/α∑j=1kq^jyηβ/αg˜−1gjy≤0is oscillatory, where∑j=1kq^jy:=min∑j=1kqjgj−1y,∑j=1kqjgj−1g˜y.F4. For someμ∈0,1the differential inequality(4)ϑ′y+1αμy36ξ1/αyβg0g˜0g˜0+r0β∑j=1kq^jyϑβ/αgjy≤0is oscillatory.F5. For someμ∈0,1the equation(5)ψ′y+1−r0βμgj3y6ξ1/αgjyβ∑j=1kqjyψβ/αgjy=0is oscillatory.By a solution of (1), we mean a function z∈C3y,∞,y≥y0which has the property ξyw‴yα∈C1y0,∞ and satisfies (1) on y0,∞.Over the years, several studies have been carried out on oscillation of neutral differential equations [7–13].Li et al. [14] have studied oscillation fourth-order differential equations with neutral type.In [15], Bazighifan has considered differential equation with positive and negative coefficients and studied oscillatory behaviour of equation:(6)ξyw‴yα′+qyfwgy=0,under the condition(7)∫y0∞1ξ1/αsds=∞.In [16], Liu et al. have considered an even-order differential equation of the form(8)ξyFwn−1y′+pyFwn−1y+qyFwgy=0,where F=sp−2s and obtained some new oscillation results for this equation.Zafer [17], Zhang and Yan [18] proved that the equation(9)wjy+qywgy=0is oscillatory if(10)liminft⟶∞∫σyyBsds>2j−1j−2j−1e,(11)liminft⟶∞∫σttBsds>j−1!e,where Bt:=gn−1y1−rgyqy and j is even.Xing et al. [19] and Moaaz et al. [20] proved that the equation(12)ξywn−1yα′+qyzβgy=0is oscillatory if(13)liminfy⟶∞∫g˜−1gyyq^sξssn−1αds>1g0+p0αg0g˜0n−1!αe,(14)liminfy⟶∞∫g˜−1ηyyg˜−1ηsn−1ξ1/αg˜−1ηsαqsPnαgsds>n−1!αe,where η∈C1y0,∞,R,j is an even and q^y:=minqg−1y,qg−1g˜y.Now, we consider the equation(15)zy+78zye4+q0y−4zye2=0,y≥1.By applying conditions (10), (11), (13), and (14) to equation (15), we obtain Table 1.Table 1
Comparison of results.
The condition(10)(11)(13)(14)The criterionq0>113981.3q0>3561.9q0>3008.5q0>587.93From above, we get that [20] enriched the results in [17–19].Our aim of this paper is complement results in [17–20].By the comparison method with first-order differential inequality, we obtain new conditions of oscillation for this equation. Some examples are provided to show effectiveness of the obtained results.


## 2. Main Results

Lemma 1.
Reference [21] If w satisfies wiy>0, i=0,1,…,j, and wj+1y<0 eventually. Then, for every ε1∈0,1,wy/w′y≥ε1y/j eventually.Lemma 2 (see [22], Lemma 2.2.3).
Letw∈Cjy0,∞,0,∞ and wj−1ywjy≤0. If limy⟶∞wy≠0, then(16)wy≥μj−1!yj−1wj−1yfory≥yμ,for every μ∈0,1.Lemma 3 (see [23], Lemma 1 and 2).
Letm1,m2≥0, then(17)m1+m2β≤2β−1m1β+m2β,forβ≥1,m1β+m2β,forβ≤1,where β is a positive real number.Theorem 1.
Assume that(18)gj−1y′≥gj0>0andg˜′y≥g˜0>0.IfF3 holds, then (1) is oscillatory. Proof. Assume to the contrary that (1) has a nonoscillatory solution in y0,∞. Without loss of generality, we let y be an eventually positive solution of (1). Then, there exists a y1≥y0 such that zy>0, yg˜y>0 and ygjy>0 for y≥y1. Since ξ′y>0, we have(19)wy>0,w′y>0,w‴y>0,w4y<0andξyw‴yα′≤0,for y≥y1. From (1), we get(20)1gj−1y′ξgj−1yw‴gj−1yα′+∑j=1kqjgj−1yyβy=0.It follows from definition ofw and Lemma 3 that(21)wβy=zy+ryzg˜yβ≤αzβy+r0βzβg˜y.From (20) and (21), we obtain(22)0=1gj−1y′ξgj−1yw‴gj−1yα′+∑j=1kqjgj−1yzβy+r0β1gj−1g˜y′ξgj−1g˜yw‴gj−1g˜yα′+∑j=1kqjgj−1g˜yzβg˜y=ξgj−1yw‴gj−1yα′gj−1y′+r0βξgj−1g˜yw‴gj−1g˜yα′gj−1g˜y′+∑j=1kqjgj−1yzβy+r0β∑j=1kqjgj−1g˜yzβg˜y≥ξgj−1yw‴gj−1yα′gj−1y′+r0βξgj−1g˜yw‴gj−1g˜yα′gj−1g˜y′+1α∑j=1kq^jywβy,which with (18) gives(23)1g0ξgj−1yw‴gj−1yα′+r0βg0g˜0ξgj−1g˜yw‴gj−1g˜yα′+1α∑j=1kq^jywβy≤0.Sincew′y>0, we get that limy⟶∞wy≠0. Thus, from Lemma 2, we get(24)wy≥μ6y3w‴y.Combining (23) and (24), we see that(25)1g0ξgj−1yw‴gj−1yα′+r0βg0g˜0ξgj−1g˜yw‴gj−1g˜yα′+1α∑j=1kq^jyμ6y3βw‴yβ≤0.If we set(26)ηy:=1g0ξgj−1yw‴gj−1yα+r0βg0g˜0ξgj−1g˜yw‴gj−1g˜yα,then it is easy to see that(27)ηg˜−1gjy≤g˜0+r0βg0g˜0ξyw‴yα.Thus, from (25), we get that η is a positive solution of(28)η′y+1αμy36ξ1/αyβg0g˜0g˜0+r0ββ/α∑j=1kq^jyηβ/αg˜−1gjy≤0.which is a contradiction. The proof is complete.Theorem 2.
Assume that (18) holds. If F4 holds, then (1) is oscillatory.Proof. Proceeding as in the proof of Theorem 1, we get (25). If we set ϑy:=ξgj−1yw‴gj−1yα, then ϑ is a positive solution of (4), which is a contradiction. The proof is complete.Corollary 1.
Letα=β and (18) holds. If ξy≤y and(29)liminfy⟶∞∫ξyys3αξs∑j=1kq^jsds>g˜0+r0αg0g˜0α6αe,where ξy=g˜−1gjy or gjy, then (1) is oscillatory.Proof. It is well-known (see, e.g., [24], Theorem 2.1.1]) that condition (29) implies the oscillation of (3) and (5).Theorem 3.
Assume thatr0<1 and gjy≤y. If F5 holds, then (1) is oscillatory.Proof. Proceeding as in the proof of Theorem 1, we get (19). From definition of w, we get(30)zy≥wy−r0zg˜y≥wy−r0wg˜y≥1−r0wy,which with (1) gives(31)ξyw‴yα′+∑j=1kqjy1−r0βwβgjy≤0.From Lemma2, we obtain(32)wy≥μ6y3w‴y.Combining (31) and (32), we get(33)ξyw‴yα′+∑j=1kqjy1−r0βμ6gj3yβw‴gjyβ≤0.Hence, if we setψ:=ξw‴α, then we get that ψ is a positive solution of the inequality(34)ψ′y+1−r0βμgj3y6ξ1/αgjyβ∑j=1kqjyψβ/αgjy≤0.In view of ([12], Corollary 1), the associated delay differential equation (5) also has a positive solution, which is a contradiction. The proof is complete.Corollary 2.
Letα=β, r0<1 and gjy≤y . If(35)liminfy⟶∞∫gjyygj3αsξgjs∑j=1kqjsds>6α1−r0αe,then (1) is oscillatory.Proof. It is well-known (see, e.h., [24], Theorem 2.1.1) that condition (35) implies the oscillation of (5).Theorem 4.
Assume thatr0<1 and gjy≤y. If there exists a positive functions ϖ1,ϖ∈C1y0,∞ such that(36)∫y0∞Ψs−2αα+1α+1ξsϖ1′sα+1μ1αs2αϖ1αsds=∞,(37)∫y0∞Ψ1s−ϖ′s24ϖsds=∞,for some μ1,μ2∈0,1 and every M1,M2>0, where(38)Ψy:=M1β−αϖ1y∑j=1kqjy1−r0βgjyy3β,Ψ1y:=1−r0β/αϖyM2β−α/α∫y∞1ξt∫t∞∑j=1kqjsgjβssβds1/αdt,then (1) is oscillatory.Proof. Proceeding as in the proof of Theorem 3, we find (19) and (31). From (19), we have w″ is of one sign.In the case wherew″y>0, we define a generalized Riccati substitution by(39)ϕ1y:=ϖ1yξyw‴yαwαy>0.By differentiating and using (31), we obtain(40)ϕ1′y≤ϖ1′yϖ1yϕ1y−ϖ1y∑j=1kqjy1−r0βwβgjywαy−αϖ1yξyw‴yαwα+1yw′y.From Lemma1, we have that wy≥y/3w′y, and hence,(41)wgjywy≥gj3yy3.Using Lemma2, we get(42)w′y≥μ12y2w‴y,ϕ for all μ1∈0,1. Thus, by (40)–(42), we obtain(43)ϕ1′y≤ϖ1′yϖ1yϕ1y−ϖ1y∑j=1kqjy1−r0βwβ−αygjyy3β−αμ1y22ξ1/αyϖ11/αyϕ1α+1/αy.Sincew′y>0, there exist y2≥y1 and a constant M>0 such that(44)wy>M,for all y≥y2. Using the inequality(45)Ex−Fxα+1/α≤ααα+1α+1Eα+1F−α,F>0,with E=ϖ1′y/ϖ1y,F=αμy2/2ξ1/αyϖ111/αy and x=ϕ1, we find(46)ϕ1′y≤−Ψy+2αα+1α+1ξyϖ1′yα+1μ1αy2αϖ1αy.This implies that(47)∫y1yΨs−2αα+1α+1ξsϖ1′sα+1μ1αs2αϖ1αsds≤ϕ1y1,which contradicts (36).Forw″y<0, integrating (31) from y to t, we obtain(48)ξtw‴tα−ξyw‴yα≤−∫yt∑j=1kqjs1−r0βwβgjsds.From Lemma1, we see that wy≥yw′y, and hence,(49)wgjy≥gjyywy.For (48), letting t⟶∞and using (49), we get(50)ξyw‴yα≥1−r0βwβy∫y∞∑j=1kqjsgjβssβds.Integrating (50) from y to ∞, we get(51)w″y≤−1−r0β/αwβ/αy∫y∞1ξt∫t∞∑j=1kqjsgjβssβds1/αdt,for all μ2∈0,1. Now, we define(52)ϕ2y=ϖyw′ywy.Thenϑy>0 for y≥y1. By using (44) and (51), we obtain(53)ϕ2′y=ϖ′yϖyϕ2y+ϖyw′′ywy−ϖyw′ywy2≤ϖ′yϖyϕ2y−1ϖyϕ22y−1−r0β/αϖywβ/α−1y∫y∞1ξt∫t∞∑j=1kqjsgjβssβds1/αdt.Thus, we find(54)ϕ2′y≤−Ψ1y+ϖ′yϖyϑy−1ϖyϕ22y,and so(55)ϕ2′y≤−Ψ1y+ϖ′y24ϖy.Then, we obtain(56)∫y1yΨ1s−ϖ′y24ϖyds≤ϕ2y1,which contradicts (37). This completes the proof.Example 1.
Consider the differential equation(57)z+12z13y′′′′+q0y4z12y=0,y≥1,where q0>0. We note that α=β=1, ξy=1, ry=1/2, g˜y=1/3y, gy=1/2y, and qy=q0/y4. Hence, it is easy to see that(58)q^y=q016y4.
Using Corollary1, the equation (57) is oscillatory if(59)q0>0.995.From Corollary 2, the equation (57) is oscillatory if(60)q0>50.951.Example 2.
Consider the differential equation(61)z+r0zϖy′′′α′+q0y3α+1zλy=0,y≥1,where ϖ,λ∈0,1 and r0,q0>0. We note that α=β, ξy=1, ry=r0, g˜y=ϖy, gy=λy and qy=q0/y3α+1. Hence, it is easy to see that(62)q^y=q0λ3α+11y3α+1.
Using Corollary1, the equation (61) is oscillatory if(63)q0ln1λ>αϖ+r0αϖ6αλ3αe.
From Corollary2, if(64)q0ln1λ>11−r0α6αλ3αe,then (61) is ocillatory.
Finally, if we setϖ1s:=y3α and ϖy:=y2, then we have(65)Ψy=q01−r0αλ3α1s,Ψ1y:=12q03α1/α1−r0λ,
Thus, from Theorem4, equation (61) is oscillatory if(66)q01−r0αλ3α>2α3α+1αα+1α+1,q0>21−r0λα3α.



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*Source: 8825413-2020-10-17.xml*