Citation: Elmetwally M. Elabbasy, Amany Nabih, Taher A. Nofal, Wedad R. Alharbi, Osama Moaaz. Neutral differential equations with noncanonical operator: Oscillation behavior of solutions[J]. AIMS Mathematics, 2021, 6(4): 3272-3287. doi: 10.3934/math.2021196
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In this work, we consider the even-order neutral differential equation with several delays
(r(l)(ν(n−1)(l))α)′+k∑i=1qi(l)f(u(gi(l)))=0, | (1.1) |
where l≥l0,ν(l)=u(l)+p(l)u(τ(l)), n≥4 is an even integer, α∈Q+odd:={a/b: a,b∈Z+ are odd} and the following conditions are fulfilled:
(i)r is a differentiable real-valued function and p,τ,qi are continuous real-valued functions on [l0,∞);
(ii)r′(l)≥0,p(l)∈[0,p0],p0 is a constant, τ(l)≤l, and liml→∞τ(l)=∞;
(iii)gi∈C([l0,∞),R),gi(l)≤l,g′i(l)>0 and liml→∞gi(l)=∞;
(iv)f∈C(R,R), f(u)≥ϱuβ for u≠0, ϱ is a positive constant, β is a ratio of odd positive integers; and
∫∞l0r−1/α(s)ds<∞ . | (1.2) |
The function u∈C([lu,∞)) with lu≥l0, is said to be a solution of (1.1) if u has the property v∈Cn−1[lu,∞),r(ν(n−1))α∈C1[lu,∞), and satisfies (1.1) on [lu,∞). We consider only those solutions u of (1.1) which satisfy sup{|u(l)|:l≥l}>0, for all l≥lu. As usual, a solution of (1.1) is called oscillatory if it has arbitrarily large zeros, otherwise, it is called nonoscillatory.
In numerical models of various physical, organic, and intrinsic phenomena, differential equations (even of the fourth order) are usually experienced. In particular, there are many applications of the delay differential equation, for example, in elasticity problems, structural deformation principles, or soil settlement; see [23,24].
The oscillation and nonoscillation of higher-order functional differential equations have concerned many authors, see [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33]. A broad range of methods have been used to investigate the properties of solutions to various groups of equations. As a matter of fact, equation (1.1) (i.e., half-linear/Emden-Fowler differential equation) arises in a variety of real-world problems such as in the study of p-Laplace equations non-Newtonian fluid theory, the turbulent flow of a polytrophic gas in a porous medium, and so forth; see the following papers for more details [5,6,7].
Agarwal et al. [2] and Zhang et al. [29] investigated the oscillatory behavior of a higher-order differential equation
(r(l)(u(n−1)(l))α)′+q(l)uβ(τ(l))=0, | (1.3) |
and considered the both cases (1.2) and
∫∞l0r−1/α(s)ds=∞. | (1.4) |
In particular, assuming that τ(l)<l, α≥β and (1.2) holds, the results obtained by Zhang et al. [29] ensure that every solution u of (1.3) is either oscillatory or satisfies liml→∞u(l)=0.
Meng and Xu [16] established oscillation criteria for even-order neutral differential equations
(a(l)|w(n−1)(l)|α−1w(l))′+q(l)f(u(σ(l)))=0, | (1.5) |
where w(l)=(l)+p(l)u(l−τ),a′(l)≥0,f(u)/|u|α−1u≥k>0,k is a constant and (1.4) holds.
Baculikova et al. [3] considered the equation
[r(l)(u(n−1)(l))α]′+q(l)f(u(τ(l)))=0 |
and proved this equation is oscillatory if the first-order equation
y′(l)+q(l)f(δτn−1(l)(n−1)!r1α(τ(l)))f(y1α(τ(l)))=0 |
is oscillatory when (1.4) holds.
Moaaz et al. [21] investigated the oscillatory behavior of the equation
(a(l)(ν(n−1)(l))α)′+f(l,u(σ(l)))=0, |
where 0≤p(l)≤p0<∞, |f(l,u)|≥q(l)|u|α and under the condition (1.4).
In this work, based on the Riccati substitution technique and comparison with delay equations of first-order, we obtain new sufficient conditions for oscillation of (1.1). Unlike most of the previous related works, we are interested in studying (1.1) in the noncanonical case (1.2). Examples illustrating our new results are also given.
The following lemmas are needed in the proofs of our results:
Lemma 1.1. [1] Let ψ∈Cn([l0,∞),R+),ψ(n) be of fixed sign and not identically zero on a subray of [l0,∞), and ψ(n−1)ψ(n)≤0 for l≥l1∈[l0,∞). If liml→∞ψ(l)≠0, then
ψ≥λ(n−1)!ln−1|ψ(n−1)|, |
for every λ∈(0,1) and l≥lλ∈[l1,∞).
Lemma 1.2. [20] Assume that s≥0,B≥0 and A>0. Then
Bs−As(α+1)/α≤αα(α+1)α+1Bα+1Aα. |
Lemma 1.3. [11,Lemma 1.1] Assume that f∈Cm([l0,∞),(0,∞)) and f(m) is eventually of one sign for all large l. Then, there exists a nonnegative integer h≤m, with m+h even for f(m)≥0, or m+h odd for f(m)≤0 such that
h>0yieldsf(k)(l)>0for k=0,1,...,h−1, |
and
h≤m−1yields(−1)h+kf(k)(l)>0for k=h,h+1,...,m−1, |
eventually.
Lemma 1.4. [11] If φ∈Cm([l0,∞),(0,∞)),φ(k)(l)>0,k=0,1,...,m and φ(m+1)(l)<0. Then,
φ(l)≥λmlφ′(l), |
for every λ∈(0,1) eventually.
Let us define the following:
η(l):={cβ−α1if α≥β;c2 δβ−α0(l)if α<β, |
μ(l):={cβ−α3if α≥β;(c4(n−3)!∫∞l(ϱ−l)n−3δ0(ϱ)dϱ)β−αif α<β, |
δ0(l):=∫∞l1r1/α(ν)dν, | (2.1) |
g(l):=min{gi(l):i=1,2,...,k} |
and
Q(l):=λ1gn−2(l)(n−2)!, |
where c1,c2,c3 and c4 are any positive constants. It is recognized that the identification of the signs of the solution derivatives is required and, before studying the oscillation of the delay differential equations, causes a significant effect.
Lemma 2.1. Assume that u∈C([l0,∞),(0,∞)) is a solution of (1.1). Then, (r(l)(ν(n−1)(l))α)′≤0, and one of the next cases holds, for l large enough
(A)v(l)>0,v′(l)>0,v(n−1)(l)>0andv(n)(l)<0;(B)v(l)>0,v′(l)>0,v(n−2)(l)>0andv(n−1)(l)<0;(C)v(l)>0,(−1)kv(k)(l)>0fork=1,2,…,n−1. |
Proof. Assume that u is an eventually positive solution of (1.1). Then, there exists l1≥l0 such that u(l),u(τ(l)) and u(g(l)) are positive for all l≥l1. Hence, by the definition of ν, we have that ν(l)>0 for l≥l1. It follows from (1.1) that (r(l)(ν(n−1)(l))α)′≤0. Next, using Lemma 1.3 and considering that n is even, we directly get the cases (A)−(C).
Lemma 2.2. Assume that u∈C([l0,∞),(0,∞)) is a solution of (1.1) whose ν satisfies (B). Then (ν(n−2)(l))β−α≥η(l), eventually.
Proof. The proof for this lemma is analogous to the proof of Lemma 2.1 in [18]. Hence, we omit it here.
Lemma 2.3. Assume that u∈C([l0,∞),(0,∞)) is a solution of (1.1) whose ν satisfies (C). Then νβ−α(l)≥μ(l), eventually.
Proof. The proof for this lemma is similar to the proof of Lemma 2.2 in [18]. Hence, we omit it here.
Lemma 2.4. Assume that u∈C([l0,∞),(0,∞)) is a solution of (1.1) whose ν satisfies (C). If
∫∞l0(∫∞l(ξ−l)n−3(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξ)dl=∞, | (2.2) |
then liml→∞u(l)=0.
Proof. Let u∈C([l0,∞),(0,∞)) be a solution of (1.1) and case (C) holds. Then, liml→∞ν(l)=D. We claim that D=0. Indeed, for the sake of a contradiction, suppose that D>0, there exists a l1≥l0 such that u(g(l))≥D for l≥l1. Integrating (1.1) on [l1,l], we have
r(l)(ν(n−1)(l))α=r(l1)(ν(n−1)(l1))α−∫ll1k∑i=1qi(v)f(u(gi(v)))dv≤−ϱDβ∫ll1k∑i=1qi(v)dv, |
that is,
ν(n−1)(l)≤−ϱ1/αDβ/α(1r(l)∫ll1k∑i=1qi(v)dv)1/α. | (2.3) |
Integrating (2.3) twice on [l,∞), we have
−ν(n−2)(l)≤−ϱ1/αDβ/α∫∞l(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξ |
and
ν(n−3)(l)≤−ϱ1/αDβ/α∫∞l∫∞ζ(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξdζ≤−ϱ1/αDβ/α∫∞l(ξ−l)(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξ. |
Similarly, by integrating the above inequality (n−4) times on [l,∞), we get
ν′(l)≤−ϱ1/αDβ/α∫∞l(ξ−l)n−3(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξ. |
Integrating this inequality on [l1,∞), we find
ν(l1)≥ϱ1/αDβ/α∫∞l1(∫∞l(ξ−l)n−3(1r(ξ)∫ξl1k∑i=1qi(v)dv)1/αdξ)dl, |
which is a contradiction with (2.2). Thus, D=0; moreover the inequality u≤ν implies liml→∞u(l)=0. The proof of this lemma is complete.
Theorem 2.1. Assume that (2.2) holds and p0<1. If the first-order delay differential equation
y′(l)+ϱ(λ0gn−1(l)(n−1)!r1/α(g(l)))β(k∑i=1qi(l)(1−p(gi(l)))β)yβ/α(g(l))=0 | (2.4) |
is oscillatory for some and
limsupl→∞∫ll0(ϱη(v)δα0(v)Qβ(v)k∑i=1qi(v)(1−p(gi(v)))β−αα+1r−1/α(v)(α+1)α+1δ0(v))dv=∞ | (2.5) |
holds for some λ,λ0,λ1∈(0,1), then every solution of (1.1) is either oscillatory or converges to zero as l→∞.
Proof. Assume the contrary that there is a nonoscillatory solution u of (1.1). Then, we can assume u(l),u(τ(l)) and u(g(l)) are positive for l≥l1≥l0. It follows from Lemma 2.1 that the behavior of ν and its derivatives is possible in three cases. First, suppose that case (A) holds. Based on the definition of ν, we see that
u(l)=ν(l)−p(l)u(τ(l))≥(1−p(l))ν(l), | (2.6) |
and so
uβ(gi(l))≥(1−p(gi(l)))βνβ(gi(l)), | (2.7) |
from (ⅳ) and (2.7), we have
f(u(gi(l)))≥ϱ(1−p(gi(l)))βνβ(gi(l)), |
which with (1.1) gives
(r(l)(ν(n−1)(l))α)′≤−ϱk∑i=1qi(l)(1−p(gi(l)))βνβ(gi(l))≤−ϱνβ(g(l))k∑i=1qi(l)(1−p(gi(l)))β. | (2.8) |
From Lemma 1.1, we have
ν(g(l))≥λgn−1(l)(n−1)!ν(n−1)(g(l)), | (2.9) |
for every λ∈(0,1). From (2.9) and (2.8), we obtain
(r(l)(ν(n−1)(l))α)′≤−ϱ(λgn−1(l)(n−1)!)β(ν(n−1)(g(l)))βk∑i=1qi(l)(1−p(gi(l)))β. |
Let y(l)=r(l)(ν(n−1)(l))α. Clearly, y is a positive solution of the first-order delay differential inequality
y′(l)+ϱ(λgn−1(l)(n−1)!r1/α(g(l)))β(k∑i=1qi(l)(1−p(gi(l)))β)yβ/α(g(l))≤0. | (2.10) |
It follows from [22,Theorem 1] that the corresponding differential equation (2.4) also has a positive solution for all λ0∈(0,1), which is a contradiction.
Next, we assume that the case (B) holds. We define the function Φ by
Φ(l)=r(l)(ν(n−1)(l))α(ν(n−2)(l))α. | (2.11) |
Then Φ(l)<0 for l≥l1. Since r(l)(ν(n−1)(l))α is decreasing, we get
r1/α(s)ν(n−1)(s)≤r1/α(l)ν(n−1)(l), s≥l≥l1. | (2.12) |
Multiplying (2.12) by r−1/α(s) and integrating it on [l,∞), we obtain
0≤ν(n−2)(l)+r1/α(l)ν(n−1)(l)δ0(l), |
that is,
−r1/α(l)ν(n−1)(l)δ0(l)ν(n−2)(l)≤1. |
From (2.11), we see that
−Φ(l)δα0(l)≤1. | (2.13) |
Differentiating (2.11), we have
Φ′(l)=(r(l)(ν(n−1)(l))α)′(ν(n−2)(l))α−αr(l)(ν(n−1)(l))α+1(ν(n−2)(l))α+1, |
which, in view of (1.1) and (2.11), becomes
Φ′(l)=−k∑i=1qi(l)f(u(gi(l)))(ν(n−2)(l))α−αΦ(α+1)/α(l)r1/α(l). | (2.14) |
Since ν′(l)>0, we get that (2.8) holds. Hence, (2.14) becomes
Φ′(l)≤−−ϱνβ(g(l))k∑i=1qi(l)(1−p(gi(l)))β(ν(n−2)(l))α−αΦ(α+1)/α(l)r1/α(l). | (2.15) |
From Lemma 1.1, we find
ν(g(l))≥λgn−2(l)(n−2)!ν(n−2)(g(l)), |
for all sufficiently large l and for every λ∈(0,1). Then, (2.15) become
Φ′(l)≤−ϱQβ(l)(k∑i=1qi(l)(1−p(gi(l)))β)(ν(n−2)(g(l)))β−α(ν(n−2)(g(l)))α(ν(n−2)(l))α−αΦ(α+1)/α(l)r1/α(l). |
Since l≥g(l) and ν(n−2)(l) is decreasing, we have
Φ′(l)≤−ϱη(l)Qβ(l)(k∑i=1qi(l)(1−p(gi(l)))β)−αΦ(α+1)/α(l)r1/α(l). | (2.16) |
Multiplying (2.16) by δα0(l) and integrating it on [l1,l], we get
0≥Φ(l)δα0(l)−Φ(l1)δα0(l1)+∫ll1αδα−10(v)r1/α(v)Φ(v)dv+∫ll1αδα0(v)r1/α(v)Φ(α+1)/α(v)dv+∫ll1(ϱη(v)δα0(v)Qβ(v)k∑i=1qi(v)(1−p(gi(v)))β)dv. |
Setting A=δα0(s)/r1/α(s), B=δα−10(s)/r1/α(s) and ϑ=−Φ(s), and using Lemma 1.2, we get
∫ll1(ϱη(v)δα0(v)Qβ(v)k∑i=1qi(v)(1−p(gi(v)))β−αα+1r−1/α(v)(α+1)α+1δ0(v))dv≤Φ(l1)δ−α0(l1)+1, |
due to (2.13), that contradicts (2.5).
Finally, suppose that (C) holds. From Lemma 2.4, one can see that liml→∞u(l)=0, which is a contradiction.
This complete the proof.
Theorem 2.2. Suppose that the first-order delay differential equation (2.4) is oscillatory for some λ0∈(0,1) and (2.5) holds for some λ1∈(0,1). If
τ(gi(l))=gi(τ(l)),τ′(l)≥τ0>0,g(l)≤τ(l), |
and
limsupl→∞ϱδαn−2(l)k∑i=1μ(gi(l))∫ll1Ωi(v)dv>κ(1+pβ0τ0), | (2.17) |
then every solution of (1.1) is oscillatory, where Ωi(l)=min{qi(l),qi(τ(l))},
δk+1(l):=∫∞lδk(ϱ)dϱfork=0,1,...,n−3, |
and κ=1 if β∈(0,1]; otherwise, κ=2β−1.
Proof. Assume that there is a nonoscillatory solution u of (1.1). Then, we can assume u(l),u(τ(l)) and u(g(l)) are positive for l≥l1≥l0. It follows from Lemma 2.1 that the behavior of ν and its derivatives is possible in three cases.
The proof of the case where (A) or (B) holds is the same as that of Theorem 2.1.
Suppose that (C) holds. Since (r(l)(ν(n−1)(l))α)′≤0, we have that
r(ζ)(ν(n−1)(ζ))α−r(l)(ν(n−1)(l))α≤0 for all ζ≥l, |
or
ν(n−1)(ζ)≤r1/α(l)ν(n−1)(l)1r1/α(ζ). |
Integrating this inequality from l to ∞ and making use of the fact that ν(n−2) is a positive decreasing function, we arrive at
−ν(n−2)(l)≤r1/α(l)ν(n−1)(l)∫∞l1r1/α(ϱ)dϱ=r1/α(l)ν(n−1)(l)δ0(l). | (2.18) |
Taking into account the behavior of derivatives of ν and integrating (2.18) (n−2) times from l to ∞, we see that
(−1)k+1ν(k)(l)≤r1/α(l)ν(n−1)(l)δn−k−2(l), | (2.19) |
for k=0,1,...,n−3. From (1.1), we get
(r(l)(ν(n−1)(l))α)′+ϱk∑i=1qi(l)uβ(gi(l))≤0 | (2.20) |
and
1τ′(l)(r(τ(l))(ν(n−1)(τ(l)))α)′+ϱk∑i=1qi(τ(l))uβ(gi(τ(l)))≤0, |
that is,
pβ0τ0(r(τ(l))(ν(n−1)(τ(l)))α)′+ϱk∑i=1qi(τ(l))pβ0uβ(gi(τ(l)))≤0. | (2.21) |
From (2.20) and (2.21), we find
0≥(r(l)(ν(n−1)(l))α)′+pβ0τ0(r(τ(l))(ν(n−1)(τ(l)))α)′+ϱk∑i=1qi(l)uβ(gi(l))+ϱk∑i=1qi(τ(l))pβ0uβ(gi(τ(l)))≥(r(l)(ν(n−1)(l))α)′+pβ0τ0(r(τ(l))(ν(n−1)(τ(l)))α)′+ϱk∑i=1Ωi(l)(uβ(gi(l))+pβ0uβ(gi(τ(l))))≥(r(l)(ν(n−1)(l))α)′+pβ0τ0(r(τ(l))(ν(n−1)(τ(l)))α)′+ϱk∑i=1Ωi(l)1κ(u(gi(l))+p(gi(l))u(τ(gi(l))))β=(r(l)(ν(n−1)(l))α+pβ0τ0r(τ(l))(ν(n−1)(τ(l)))α)′+ϱκk∑i=1Ωi(l)νβ(gi(l)). |
By integrating this inequality from l1 to l, we get
r(l)(ν(n−1)(l))α+pβ0τ0r(τ(l))(ν(n−1)(τ(l)))α≤r(l1)(ν(n−1)(l1))α+pβ0τ0r(τ(l1))(ν(n−1)(τ(l1)))α−ϱκ∫ll1k∑i=1Ωi(v)νβ(gi(v))dv≤−ϱκk∑i=1νβ(gi(l))∫ll1Ωi(v)dv. |
Since (r(l)ν(n−1)(l))′≤0, we arrive at
(1+pβ0τ0)r(l)(ν(n−1)(l))α≤−ϱκk∑i=1να(gi(l))νβ−α(gi(l))∫ll1Ωi(v)dv≤−ϱκνα(l)k∑i=1μ(gi(l))∫ll1Ωi(v)dv, |
which, with Lemma 2.3, gives
(1+pβ0τ0)r(l)(ν(n−1)(l))α≤−ϱκνα(l)k∑i=1μ(gi(l))∫ll1Ωi(v)dv. | (2.22) |
Combining [(2.19), k=0] and (2.22), we have that
(1+pβ0τ0)≥ϱκδαn−2(l)k∑i=1μ(gi(l))∫ll1Ωi(v)dv |
which is a contradicts with (2.17). This completes the proof.
In the following theorem, we set new conditions for testing the oscillation of (1.1) when n=4, which apply in the ordinary case.
Theorem 2.3. Assume that n=4,α=β=1,p0<1 and (2.2) hold. Suppose also that
limsupl→∞∫ll0(ϱλ1g2(s)k∑i=1qi(s)(1−p(gi(s)))2!δ(s)−14r(s)δ(s))ds=∞, | (2.23) |
for some constant λ1∈(0,1). Assume further that there exist two positive functions ρ(l),ϑ(l)∈C1[l0,∞), such that
∫∞l0(ϱρ(s)(g(s)s)3/λk∑i=1qi(s)(1−p(gi(s)))−12(ρ′(s))2ρ(s)r(s)λ2s2)ds=∞ | (2.24) |
and
∫∞l0(ϱϑ(ζ)∫∞ζ(1r(v)∫∞vk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds)dv−(ϑ′(ζ))24ϑ(ζ))dζ=∞ | (2.25) |
where λ2∈(0,1). Then every solution of (1.1) is oscillatory or tends to zero as l→∞.
Proof. Assume that Eq (1.1) has a positive solution u(l). It follows from (1.1) and Lemma 2.1 that there exist four possible cases for the behavior of ν and its derivatives:
(i):ν′(l)>0, ν′′(l)>0, ν′′′(l)>0 and ν(4)(l)≤0;(ii):ν′(l)>0, ν′′(l)<0, ν′′′(l)>0 and ν(4)(l)≤0;(iii):ν′(l)<0, ν′′(l)>0 and ν′′′(l)<0 ;(iv):ν′(l)>0, ν′′(l)>0 and ν′′′(l)<0. |
Let (i) hold. Now, we define
ϕ(l)=ρ(l)r(l)ν′′′(l)ν(l). |
Then clearly ϕ(l) is positive for l≥l1 and satisfies
ϕ′(l)=ρ′(l)ρ(l)ϕ(l)+ρ(l)((r(l)ν′′′(l))′ν(l)−r(l)ν′′′(l)ν′(l)ν2(l)). | (2.26) |
From (1.1) and (2.26), we have
ϕ′(l)=ρ′(l)ρ(l)ϕ(l)−ρ(l)∑ki=1qi(l)f(u(gi(l)))ν(l)−ρ(l)r(l)ν′′′(l)ν′(l)ν2(l), | (2.27) |
by using (2.8) and (2.27), we get
ϕ′(l)≤ρ′(l)ρ(l)ϕ(l)−ρ(l)ϱν(g(l))∑ki=1qi(l)(1−p(gi(l)))ν(l)−ρ(l)r(l)ν′′′(l)ν′(l)ν2(l). | (2.28) |
Now, it follows from Lemmas 1.1 and 1.4 that
ν′(l)≥λ2l22ν′′′(l) | (2.29) |
and
ν(g(l))ν(l)≥(g(l)l)3/λ, | (2.30) |
respectively. Substituting (2.29) and (2.30) into (2.28), we get
ϕ′(l)≤ρ′(l)ρ(l)ϕ(l)−ϱρ(l)(g(l)l)3/λk∑i=1qi(l)(1−p(gi(l)))−λ2l22ρ(l)r(l)(ν′′′(l))2ν2(l), |
from the definition of ϕ(l), we obtain
ϕ′(l)≤ρ′(l)ρ(l)ϕ(l)−ϱρ(l)(g(l)l)3/λk∑i=1qi(l)(1−p(gi(l)))−λ2l22ρ(l)r(l)ϕ2(l). |
Set A=λ2l2/2ρ(l)r(l), B=ρ′(l)/ρ(l) and s=ϕ(s). Using Lemma 1.2, we have
ϕ′(l)≤−ϱρ(l)(g(l)l)3/λk∑i=1qi(l)(1−p(gi(l)))+12(ρ′(l))2ρ(l)r(l)λ2l2, | (2.31) |
integrating (2.31) from l1 to l, we have
∫ll1(ϱρ(s)(g(s)s)3/λk∑i=1qi(s)(1−p(gi(s)))−12(ρ′(s))2ρ(s)r(s)λ2s2)ds≤ϕ(l1), |
which contradicts (2.24).
Assume that case (ii) holds. Define the function φ(l) by
φ(l)=ϑ(l)ν′(l)ν(l). |
Then clearly φ(l) is positive for l≥l1 and satisfies
φ′(l)=ϑ′(l)ϑ(l)φ(l)+ϑ(l)(ν′′(l)ν(l)−(ν′(l))2ν2(l)), |
from the definition of φ(l), we obtain
φ′(l)=ϑ′(l)ϑ(l)φ(l)+ϑ(l)ν′′(l)ν(l)−φ2(l)ϑ(l). | (2.32) |
Now integrating (1.1) from l to ∞, we have
−r(l)ν′′′(l)=−∫∞lk∑i=1qi(s)f(u(gi(s)))ds, | (2.33) |
by using (2.8) and (2.33), we get
−r(l)ν′′′(l)≤−ϱ∫∞lk∑i=1qi(s)(1−p(gi(s)))ν(g(l))ds. | (2.34) |
From Lemma 1.4, we get
ν(l)≥lλν′(l), |
that is,
ν(g(l))ν(l)≥(g(l)l)1/λ. | (2.35) |
Combining (2.35) and (2.34), we get
−r(l)ν′′′(l)≤−ϱν(l)∫∞lk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds, |
that is
−ν′′′(l)≤−ϱν(l)r(l)∫∞lk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds, |
integrating the above inequality from l to ∞, we have
ν′′(l)≤−ϱν(l)∫∞l(1r(v)∫∞vk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds)dv. |
Combining above inequality with (2.32), we obtain
φ′(l)≤−ϱϑ(l)∫∞l(1r(v)∫∞vk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds)dv+ϑ′(l)ϑ(l)φ(l)−φ2(l)ϑ(l). |
Thus, we have
φ′(l)≤−ϱϑ(l)∫∞l(1r(v)∫∞vk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds)dv+(ϑ′(l))24ϑ(l), | (2.36) |
integrating (2.36) from l1 to l, we have
∫ll1(ϱϑ(ζ)∫∞ζ(1r(v)∫∞vk∑i=1qi(s)(1−p(gi(s)))(g(s)s)1/λds)dv−(ϑ′(ζ))24ϑ(ζ))dζ≤φ(l1), |
which contradicts (2.25).
The proof of the case where (iii) or (iv) holds is the same as that of Theorem 2.2 and Theorem 2.1 respectively.
This completes the proof.
Example 2.1. Consider the NDDE
(l4(u(l)+p0u(al))′′′)′+q1u(bl)+q2u(cl)=0, | (2.37) |
where a,b∈(0,1) and g1>g2. Then, we note that
n=4,r(l)=l4,p(l)=p0,τ(l)=al,q(l)=q1+q2 and σ(l)=bl. |
Therefore, it is easy to verify that
δ0(l)=13l3,δ1(l)=16l2 and δ2(l)=16l. |
Next, to apply Theorem 2.1. We first check the condition (2.2), (2.4) and (2.5). By substitution and a simple computation, (2.4) becomes
y′(l)+ϱ(q1+q2)λ0(1−p0)6c1ly(cl)=0. | (2.38) |
By applying a well known criterion [12,Theorem 2.1.1] for first-order delay differential equation (2.38) to be oscillatory, the criterion is immediately obtained.
liminfl→∞∫lclϱ(q1+q2)λ0(1−p0)6c1sds>1e, |
that is,
(q1+q2)ln(1c)>6cϱλ0(1−p0)e. | (2.39) |
Now, we note that (2.5) reduces to
limsupl→∞∫ll0(ϱ(q1+q2)(1−p0)λ1c26−34)1vdv=∞, |
which satisfies if
(q1+q2)>184ϱc2(1−p0), | (2.40) |
thus, if condition (2.39) and (2.40) hold, then every solution of (2.37) is oscillatory or tends to zero.
On the other hand, by Theorem 2.2, we see that (2.17) becomes
limsupl→∞ϱ6l∫ll0(q1+q2)dv>(1+p0a) |
and so
(q1+q2)>6ϱ(1+p0a), | (2.41) |
thus, if (2.39), (2.40) and (2.41) hold, then every solution of (2.37) is oscillatory.
Example 2.2. Consider the NDDE
(e3l((u(l)+(1−1l2)u(l−a))′′′)3)′+q1e3lu3(l−b)+q2e3lu3(l−c)=0, | (2.42) |
where l≥1,0<a<b and b>c. Then, we can clearly note that α=β=3,n=4.
r(l)=e3l,p(l)=1−1/l2,τ(l)=l−a,q(l)=(q1+q2)e3l and σ(l)=l−b. |
Therefore, it is easy to verify that
δi(l)=e−l for i=0,1,2. |
By substitution and a simple computation, (2.4) becomes
y′(l)+ϱ(q1+q2)e3l(λ0(l−b)3!el−b)3y(l−b)=0. | (2.43) |
Applying a well-known criterion [12,Theorem 2.1.1], we see that (2.38) is oscillatory. Moreover, (2.5) reduces to
limsupl→∞∫∞l0ϱ((q1+q2)λ3123−(34)4)ds=∞, |
which satisfies if (q1+q2)>81/32. Thus, every solution of (2.42) is oscillatory or tends to zero if (q1+q2)>81/32.
To apply Theorem 2.2, we see that (2.17) becomes
limsupl→∞ϱe−3l∫ll0(q1+q2)e3(v−a)dv>23. |
that is, (q1+q2)>24e3a. Then, every solution of (2.42) is oscillatory if (q1+q2)>max{24e3a,81/32}.
The authors present their sincere thanks to the two anonymous referees. (Taher A. Nofal) Taif University Researchers Supporting Project number (TURSP-2020/031), Taif University, Taif, Saudi Arabia.
There are no competing interests
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