On the monotonic properties and oscillatory behavior of solutions of neutral differential equations

Definition 1.1

[2] A nontrivial solution x of the differential equation is said to be oscillatory if x has arbitrarily large zeros, that is, there exists an infinite sequence { u n } n = 0 ∞ such that x ( u n ) = 0 and lim n → ∞ u n = ∞ . Otherwise, it is said to be nonoscillatory. A differential equation is said to be oscillatory if all of its solutions are oscillatory.

Notation 1.1

A solution of (1) is said to be oscillatory if it has arbitrarily large zeros on [ U x , ∞ ) . Otherwise, it is said to be nonoscillatory. Equation (1) is said to be oscillatory if all of its solutions are oscillatory.

Theorem 1.1

([13], Theorem 4) Let (3) hold. Assume that, for some δ ∈ ( 0 , 1 ) and every u 1 ≥ u 0 , both

and

are oscillatory. Assume further that there exists ζ ( u ) ∈ C ( [ u 0 , ∞ ) ) such that

is oscillatory for every u 1 ≥ u 0 and

where

and ζ ( u ) ∈ C ( [ u 0 , ∞ ) ) . Then, equation (2) is oscillatory.

Theorem 1.2

([14], Theorem 2.1) Let (3) hold. Assume that the differential equation

is oscillatory for some constant λ 0 ∈ ( 0 , 1 ) . If

and

hold for some constant λ 1 ∈ ( 0 , 1 ) and for every constant M > 0 , where δ ( u ) = ∫ u ∞ 1 ℓ 1 ∕ r ( b ) d b and

then, (4) is oscillatory.

Theorem 1.3

([15], Theorem 2.1) Let (3) hold. Assume that there exists a positive function ρ ∈ C 1 [ u 0 , ∞ ) such that

for some constant k 1 ∈ ( 0 , 1 ) . Assume further that there exists a positive function θ ∈ C 1 [ u 0 , ∞ ) such that

If

and

for some constant k 2 ∈ ( 0 , 1 ) , ℓ ( u ) ≔ ∫ l ∞ 1 ℓ 1 ∕ γ ( b ) d b , θ + ′ ( l ) ≔ max { 0 , θ ′ ( l ) } , and ρ + ′ ( l ) ≔ max { 0 , ρ ′ ( l ) } , then every solution of (5) is oscillatory.

Theorem 1.4

([21], Theorem 8) Let 0 < r = β ≤ 1 and (3) hold, and there exist three functions η 1 , η 2 , η 3 ∈ C ( [ u 0 , ∞ ) , R ) such that

Suppose also that conditions

and

hold. If

then (6) is oscillatory.

Note that

Lemma 2.1

([22], Lemma 2.2.3) Suppose that f ∈ C m ( [ u 0 , ∞ ) , ℓ + ) , f ( m ) ( u ) is of fixed sign and not identically zero on [ u 0 , ∞ ) and that there exists u 1 ≥ u 0 such that f ( m − 1 ) ( u ) f ( m ) ( u ) ≤ 0 for all u 1 ≥ u 0 . If lim u → ∞ f ( u ) ≠ 0 , then, for every δ ∈ ( 0 , 1 ) , there exists u δ ∈ [ u 1 , ∞ ) such that

for u ∈ [ u δ , ∞ ) .

Lemma 2.2

[23] Let A and B be real numbers and A > 0 . Then,

Lemma 2.3

Assume that x is a positive solution to equation (1). Then, ℓ ( u ) ( w ( n − 1 ) ( u ) ) r is nonincreasing and w satisfies one of the following cases:

eventually.

Lemma 2.4

Assuming x belongs to Ω , we obtain the following, eventually

1. x ( u ) ≥ ( 1 − p ( u ) ) w ( u ) ;

2. w ( u ) ≥ μ 0 ( n − 2 ) ! u n − 2 w ( n − 2 ) ( u ) , for all μ 0 ∈ ( 0 , 1 ) ;

3. ( ℓ ( u ) ( w ( n − 1 ) ( u ) ) r ) ′ ≤ − Q ( u ) w r ( ϱ ( u ) ) ;

4. w ( n − 2 ) ( u ) ≥ − L 0 ( u ) ℓ 1 ∕ r ( u ) w ( n − 1 ) ( u ) ;

5. w ( n − 2 ) ( u ) ∕ L 0 ( u ) is increasing.

Proof

( B 1 , 1 ) : As a result of the facts x ∈ Ω and τ ( u ) ≤ u , we find w ′ ( u ) > 0 and x ( τ ( u ) ) ≤ w ( τ ( u ) ) ≤ w ( u ) . Consequently, we obtain

( B 1 , 2 ) : Using Lemma 2.1 with m = n − 1 and g = w , we have

for all μ 0 ∈ ( 0 , 1 ) .

( B 1 , 3 ) : Equation (1) with ( B 1 , 1 ) becomes

( B 1 , 4 ) : Since ℓ ( u ) ( w ( n − 1 ) ( u ) ) r is decreasing, we obtain

( B 1 , 5 ) : From ( B 1 , 4 ) , we obtain

The proof of this lemma is now complete.□

Lemma 2.5

Assuming x belongs to Ω and there are δ > 0 and u 1 ≥ u 0 such that

we obtain, for u ≥ u 1 ,

1. lim u → ∞ w ( n − 2 ) ( u ) = 0 ;

2. w ( n − 2 ) ( u ) ∕ L 0 β 0 ( u ) is decreasing;

3. lim u → ∞ w ( n − 2 ) ( u ) ∕ L 0 β 0 ( u ) = 0 ;

4. w ( n − 2 ) ( u ) ∕ L 0 1 − β 0 ( u ) is increasing,

Proof

( B 2 , 1 ) : Since x ∈ Ω , we can conclude that ( B 1 , 1 ) − ( B 1 , 5 ) in Lemma 2.4 are satisfied for all u ≥ u 1 , u 1 large enough. Now, since w ( n − 2 ) ( u ) is a positive decreasing function, we conclude that lim u → ∞ w ( n − 2 ) ( u ) = c 1 ≥ 0 . We claim that c 1 = 0 . If not, then w ( n − 2 ) ( u ) ≥ c 1 > 0 eventually, which with ( B 1 , 2 ) gives

for all μ 0 ∈ ( 0 , 1 ) . Thus, from ( B 1 , 3 ) , we obtain

which with (9) gives

Integrating the previous inequality from u 2 to u , we have

Since L 0 − r ( u ) → ∞ as u → ∞ , there is a u 3 ≥ u 2 such that L 0 − r ( u ) − L 0 − r ( u 2 ) ≥ ε L 0 − r ( u ) for all ε ∈ ( 0 , 1 ) . Hence, (10) becomes

for all u ≥ u 3 . Integrating the last inequality from u 3 to u , we find

which is a contradiction. Then, c 1 = 0 .

( B 2 , 2 ) From (9), ( B 1 , 2 ) and ( B 1 , 3 ) , we obtain

By integrating the last inequality from u 1 to u and using the fact w ( n − 1 ) ( u ) < 0 , we obtain