Asymptotic behavior of even-order noncanonical neutral differential equations

Lemma 1.1

[34, Lemma 2.2.3] Assume that ϖ ∈ C m ( [ ℓ 0 , ∞ ) , R + ) , ϖ ( m ) is not identically zero on a subray of [ ℓ 0 , ∞ ) and ϖ ( m ) is of fixed sign. Suppose that ϖ ( m − 1 ) ϖ ( m ) ≤ 0 for ℓ ∈ [ ℓ 1 , ∞ ) , where ℓ 1 ≥ ℓ 0 large enough. If lim ℓ → ∞ ϖ ( ℓ ) ≠ 0 , then there exists a ℓ λ ∈ [ ℓ 1 , ∞ ) such that

for every λ ∈ ( 0 , 1 ) and ℓ ∈ [ ℓ λ , ∞ ) .

Lemma 1.2

[35, Lemma 1] Let f ∈ C m ( [ ℓ 0 , ∞ ) , R ) , f ( r ) > 0 , r = 0 , 1 , … , m and f ( m + 1 ) ≤ 0 eventually. Then, for every η ∈ ( 0 , 1 )

Lemma 1.3

[36, Lemma 1.2] Assume that B ≥ 0 , A > 0 , w ≥ 0 and α > 0 . Then,

Lemma 1.4

[20, Lemma 2.1] Let f ∈ C n ( [ t 0 , ∞ ) , ( 0 , ∞ ) ) . If the derivative f ( n ) ( t ) is eventually of one sign for all large t , then there exist a t x such that t x ≥ t 0 and an integer l , 0 ≤ l ≤ n , with n + l even for f ( n ) ( t ) ≥ 0 , or n + l odd for f ( n ) ( t ) ≤ 0 such that

and

Lemma 2.1

Assume that u ∈ C ( [ ℓ 0 , ∞ ) , ( 0 , ∞ ) ) is a solution of (1), eventually. Then, ν > 0 , ( a ν ( m − 1 ) ) ′ ≤ 0 and ν satisfies one of the following:

1. ν ′ , ν ( m − 1 ) and ( − ν ( m ) ) are positive;

2. ν ′ , ν ( m − 2 ) and ( − ν ( m − 1 ) ) are positive;

3. ( − 1 ) k ν ( k ) are positive, for all k = 1 , 2 , … , m − 1 ,

Proof

Assume that u is an eventually positive solution of (1). It follows from (1) that

Now, from above inequality and Lemma 2 that there exist three possible cases (1)–(3) for ℓ ≥ ℓ 1 large enough.□

Lemma 2.2

Assume that u ∈ C ( [ ℓ 0 , ∞ ) , ( 0 , ∞ ) ) is a solution of (1), where ν satisfies case ( 3 ) . If

then, lim ℓ → ∞ u ( ℓ ) = 0 .

Proof

Assume that u is an eventually positive solution of (1), where ν satisfies case ( 3 ) . Then, lim ℓ → ∞ ν ( ℓ ) = D . We claim that D = 0 . Suppose that D > 0 , and so for all ε > 0 , there exists ℓ 1 ≥ ℓ 0 such that u ( g ( ℓ ) ) ≥ D for ℓ ≥ ℓ 1 . Integrating (1) from ℓ 1 to ℓ , we get

that is,

Integrating (5) twice from ℓ to ∞ , we obtain

and

Similarly, integrating (6) m − 4 times from ℓ to ∞ , we find

Integrating this inequality from ℓ 1 to ∞ , we obtain

which is a contradiction with (4). Thus, D = 0 . This completes the proof.□

Theorem 2.1

Let (4) hold. If there exists a λ 0 ∈ ( 0 , 1 ) such that the first-order delay differential equation

is oscillatory and

holds for some constant λ 1 ∈ ( 0 , 1 ) , then every nonoscillatory solution u of (1) satisfies lim t → ∞ u ( ℓ ) = ∞ .

Proof

Suppose that (1) has a positive solution u which satisfies lim ℓ → ∞ u ( ℓ ) ≠ 0 . It follows from (1) that

From Lemma 2.1, there are three possible cases for the behavior of ν and its derivatives.

Let ( 1 ) hold. From Lemma 1.1, we have

for every λ ∈ ( 0 , 1 ) . It follows from the definition of ν ( ℓ ) that

Combining (9) and (11), we get

From (10), we obtain

Now, we define the function y ( ℓ ) = a ( ℓ ) ν ( m − 1 ) ( ℓ ) . Clearly, y is a positive solution of the first-order delay differential inequality

Thus, using [40, Theorem 1], equation (7) has also a positive solution for all λ 0 ∈ ( 0 , 1 ) , this contradicts the assumption that (7) is oscillatory.

Let ( 2 ) hold. We define ω by

Then, ω ( ℓ ) < 0 for ℓ ≥ ℓ 1 . Noting that ( a ( ℓ ) ν ( m − 1 ) ( ℓ ) ) ′ ≤ 0 , we find

Dividing (15) by a and integrating it from ℓ to ∞ , we obtain

which yields

Thus, by (14), we get

Differentiating (14), we arrive at

which follows from (1) and (14) that

From the definition of ν ( ℓ ) and the fact that ν ′ ( ℓ ) > 0 , we get that (11) holds. Hence, it follows from (17) that

Using Lemma 1.1, we get

for every λ ∈ ( 0 , 1 ) and for all sufficiently large ℓ . Then, (18) becomes

Since ℓ ≥ g ( ℓ ) and ν ( m − 2 ) ( ℓ ) are decreasing, we have

Multiplying (19) by δ ( ℓ ) and integrating it from ℓ 1 to ℓ , we have

Setting A = δ ( s ) / a ( s ) , B = 1 / a ( s ) , and w = − ω ( s ) , and using Lemma 1.3, we have

due to (16), which contradicts (8).

Assume that case ( 3 ) holds. From Lemma 2.2 and (4), we see that lim ℓ → ∞ u ( ℓ ) = 0 , which is a contradiction.

This completes the proof.□

Corollary 2.1

Assume that (4) and (8) hold. If

for some λ 1 ∈ ( 0 , 1 ) , then every nonoscillatory solution u of (1) satisfies lim t → ∞ u ( ℓ ) = ∞ .

Proof

By [38, Theorem 2.1.1], assumption (20) ensures that the differential equation (7) has no positive solutions. Application of Theorem 2.1 yields the result.□

Remark 2.1

Combining Theorem 2.1 and the results reported in [39] for the oscillation of equation (7), one can derive various oscillation criteria for equation (1).

Example 2.1

We consider the NDDE

where h 0 > 0 and θ , ε ∈ ( 0 , 1 ) . Note that, a ( ℓ ) = e ℓ , ρ ( ℓ ) = ρ 0 , τ ( ℓ ) = θ ℓ , h ( ℓ ) = h 0 e ℓ and g ( ℓ ) = ε ℓ . It is easy to see that δ ( ℓ ) = e − ℓ .

Now, from Corollary 2.1, we have

and

Thus, (4), (20), and (8) are satisfied. Therefore, every solution of (21) is oscillatory or tends to zero.

Example 2.2

Consider the equation

here h 0 > 0 and θ , ε ∈ ( 0 , 1 ) . We note that m = 4 , a ( ℓ ) = ℓ 2 , ρ ( ℓ ) = ρ 0 , τ ( ℓ ) = θ ℓ , h ( ℓ ) = h 0 / ℓ 2 , and g ( ℓ ) = ε ℓ . It is easy to see that δ ( ℓ ) = 1 / ℓ and (4) holds. Next, (20) reduces to

Moreover, (8) becomes

which is verified if

Using Corollary 2.1, if

then every solution of (22) is oscillatory or tends to zero, where

and

Theorem 2.2

Assume that m = 4 and (4) hold. If

for some constant λ 1 ∈ ( 0 , 1 ) . Assume further that there exist two positive functions ζ ( ℓ ) , ϑ ( ℓ ) ∈ C 1 [ ℓ 0 , ∞ ) , such that

and

for some constant λ 2 ∈ ( 0 , 1 ) . Then, every nonoscillatory solution u of (1) satisfies lim t → ∞ u ( ℓ ) = ∞ .

Proof

Assume that (1) has a nonoscillatory solution u which is eventually positive and lim ℓ → ∞ u ( ℓ ) ≠ 0 . It follows from (1) and Lemma 2.1 that there exist four possible cases for the behavior of ν and its derivatives:

1. ν ′ ( ℓ ) > 0 , ν ″ ( ℓ ) > 0 , ν ‴ ( ℓ ) > 0 and ν ( 4 ) ( ℓ ) ≤ 0 ;

2. ν ′ ( ℓ ) > 0 , ν ″ ( ℓ ) < 0 , ν ‴ ( ℓ ) > 0 and ν ( 4 ) ( ℓ ) ≤ 0 ;

3. ν ′ ( ℓ ) < 0 , ν ″ ( ℓ ) > 0 and ν ‴ ( ℓ ) < 0 ;

4. ν ′ ( ℓ ) > 0 , ν ″ ( ℓ ) > 0 and ν ‴ ( ℓ ) < 0 .

Then, clearly ϕ ( ℓ ) is positive for ℓ ≥ ℓ 1 and satisfies

From (1) and (28), we have

Using (11) and (29), we get

Now, it follows from Lemmas 1.1 and 1.2 that

and so

respectively. Substituting (31) and (32) into (30), we get