Second-Order Differential Equation with Multiple Delays: Oscillation Theorems and Applications

Diﬀerential equations of second order appear in physical applications such as ﬂuid dynamics, electromagnetism, acoustic vi-brations, and quantum mechanics. In this paper, necessary and suﬃcient conditions are established of the solutions to second-order half-linear delay diﬀerential equations of the form ( ς ( y )( u ′ ( y )) a ) ′ + 􏽐 mj � 1 p j ( y ) u c j ( ϑ j ( y )) � 0for y ≥ y 0 , under the assumption 􏽒 ∞ ( ς ( η )) − 1/ a d η � ∞ . We consider two cases when a < c j and a > c j , where a and c j are the quotient of two positive odd integers. Two examples are given to show eﬀectiveness


Introduction
e differential equation of second order appears in models concerning biological, physical, and chemical phenomena, optimization, the mathematics of networks, and dynamical systems, see [1].

Remark 1.
When the domain is not specified explicitly, all functional inequalities considered in this paper are assumed to hold eventually, i.e., they are satisfied for all y large enough.  (1). en, there exist y 1 ≥ y 0 and d > 0 such that

Necessary and Sufficient Conditions
for y ≥ y 1 .
Proof. Let u be eventually positive solution of (1). en, by (A1), there exists a y * such that u(y) > 0 and u(ϑ j (y)) > 0 for all y ≥ y * and j � 1, 2, . . . , m. From (1), it follows that erefore, ς(y)(u ′ (y)) a is nonincreasing for y ≥ y * . Next, we show the ς(y)(u ′ (y)) a is positive. By contradiction, assume that ς(y)(u ′ (y)) a ≤ 0 at a certain time y ≥ y * . Using that p j is not identically zero on any interval [b, ∞) and by (6), there exist y 1 ≥ y * such that Recall that a is the quotient of two positive odd integers. en, Integrating from y 1 to y, we have By (A3), the right-hand side approaches −∞; then, lim t⟶∞ u(y) � −∞. is is a contradiction to the fact that u(y) > 0. erefore, ς(y)(u ′ (y)) a > 0 for all y ≥ y * . From ς(y)(u ′ (y)) a being nonincreasing, we have integrating this inequality from y 1 to y, and using that u is continuous, Since lim y⟶∞ Y(y) � ∞, there exists a positive constant d such that (4) holds.
Letting limit as b ⟶ ∞, we obtain en, Since u(y 1 ) > 0, integrating the above inequality yields Since the integrand is positive, we can increase the lower limit of integration from η to y and then use the definition of Υ(y) to obtain which yields (5).

Complexity
Proof. On the contrary, we assume that a solution u is eventually positive. So, Lemma 1 holds, and then there exists where Computing the derivative of w, we have us, w is nonnegative and nonincreasing. Since u > 0, by (A2), it follows that m j�1 p j (y)u c j (ϑ j (y)) cannot be identically zero in any interval [b, ∞); thus, w ′ cannot be identically zero, and w cannot be constant on any interval [b, ∞).
erefore, w(y) > 0 for y ≥ y 1 . Computing the derivative, Integrating (21) from y 2 to y and using that w > 0, we have Next, we find a lower bound for the right-hand side of (22), independent of the solution u. By (4) and (19), we have Since w is nonincreasing, b 1 /a > 0 and ϑ j (η) < η, and it follows that Going back to (22), we have Since (17), the right-hand side approaches +∞ as y ⟶ ∞.
is contradicts (25) and completes the proof of sufficiency for eventually positive solutions.
e eventually negative solution can be dealt with similarly by introducing the variables v � −u.
Next, we show the necessity part by a contrapositive argument. If (17) does not hold, then, for each κ > 0, there exists y 1 ≥ y 0 such that for all η ≥ y 1 . We define the set of continuous functions: We define an operator Ω on S by Note that when u is continuous, Ωu is also continuous on [0, ∞). If u is a fixed point of Ω, i.e., Ωu � u, then u is a solution of (1).
First, we estimate (Ωu)(y) from below. By (A3), we have Now, we estimate (Ωu)(y) from above. For u in S, we have u c j (ϑ j (ζ)) ≤ (κ 1/a Y(ϑ j (ζ))) c j . en, by (26), erefore, Ω maps S to S. Next, we find a fixed point for Ω in S. Let us define a sequence of functions in S by the recurrence relation: v n+1 (y) � Ωv n (y), for n ≥ 1, y ≥ y 1 .

(31)
Note that, for each fixed y, we have v 1 (y) ≥ v 0 (y). Using mathematical induction, we can show that v n+1 (y) ≥ v n (y). erefore, the sequence v n converges pointwise to a function v. Using the Lebesgue Dominated Convergence eorem, we can show that v is a fixed point of Ω in S. is shows under assumption (26) that there is a nonoscillatory solution that does not converge to zero. is completes the proof. □ Theorem 2. Assume that there exists a constant b 2 and the quotient of two positive odd integers such that 0 < a < b 2 < c j .

If (A1)-(A4) hold and ς(y) is nondecreasing, then each solution of (1) is oscillatory if and only if
Proof. On the contrary, we assume that u is an eventually positive solution that does not converge to zero. Using the same argument as in Lemma 1, there exists y 1 ≥ y 0 such that u(ϑ j (y)) > 0 and ς(y)(u ′ (y)) a is positive and nonincreasing. Since ς(y) > 0, so u(y) is increasing for y ≥ y 1 . Using u(y) ≥ u(y 1 ), we have and hence Using (34) and ϑ j (y) ≥ ϑ 0 (y), from (13), we have for y ≥ y 2 . From ς(y)(u ′ (y)) a being nonincreasing and ϑ 0 (y) ≤ y, we have We use this in the left-hand side of (35). en, dividing by ς(ϑ 0 (y))u b 2 (ϑ 0 (y)) > 0, raising both sides to the 1/a power, we have u ′ ϑ 0 ((y)) for y ≥ y 2 . Multiplying the left-hand side by ϑ 0 ′ (y)/ϑ 0 ≥ 1 and integrating from y 2 to y, On the left-hand side, since a < b 2 , integrating, we have On the right-hand side of (38), we use that ς(ϑ 0 (η)) ≤ ς(η), to conclude that (32) implies the right-hand side approaching +∞, as y ⟶ ∞, which is a contradiction. Hence, the solution u cannot be eventually positive.
For eventually negative solutions, we use the same change of variables as in eorem 1 and proceed as above.
To prove the necessity part, we assume that (32) does not hold and obtain an eventually positive solution that does not converge to zero. If (32) does not hold, then for each κ > 0 there exists y 1 ≥ y 0 such that We define the set of continuous function en, we define the operator: Note that if u is continuous, Ωu is also continuous at y � y 1 . Also note that if Ωu � u, then u is solution of (1).

(44)
Note that, for each fixed y, we have v 1 (y) ≥ v 0 (y). Using mathematical induction, we can prove that v n+1 (y) ≥ v n (y). erefore, v n converges pointwise to a function v in S. en, v is a fixed point of Ω and a positive solution of (1). e proof is completed. Here, So, every condition of eorem 1 holds true. erefore, all solution of (45) is oscillatory.

Conclusion
is work aims to study the oscillatory behavior of secondorder neutral nonlinear differential equation. e obtained oscillation theorems complement the well-known oscillation results present in the literature.

Data Availability
e data used to support the findings of the study are available from the corresponding author upon request.

Conflicts of Interest
e authors declare that they have no conflicts of interest.