Lemma 1. Suppose that X(t) is a NOS to the system (1) with λ = 1, and

Proof. Suppose that X(t) is an eventual positive solution to the system (1) (the case X(t) is an eventually negative is similar). Then, from (1), it follows that

That means $$$$ and $$ are nondecreasing; hence, there exists t₁ ≥ t₀ such that $$, and $$ are eventually positive or eventually negative. So, eight cases can be discussed, which are as follows:

Now, we discuss the cases in Table 1 successively:

• (i)

  Since $$ and $$, then $$ is positive nondecreasing, then there exists b_i > 0, t₂ ≥ t₁ such that $$

  $$ (4)

•  

  Integrating (4) from t to δ(t) for some continuous function δ(t) > t, we obtain

  $$ (5)

•  

  We claim that $$ for t ≥ t₃ ≥ t₂, otherwise if $$ for t ≥ t₃ ≥ t₂, then (5) becomes

  $$ (6)

•  

  Integrating (6) from t₃ to t, we get

  $$ (7)

•  

  Letting t⟶∞ the last inequality leads to Lim_t⟶∞y_i(t) = −∞, which is a contradiction. Hence the claim was verified and $$ and $$ this case leads to Lim_t⟶∞y_i(t) = ∞. That is (y₁, y₂, y₃) ∈ K₁.

• (ii)

  Since $$ and $$.

•  

  That is $$, is negative nondecreasing. So there are $$ such that $$ hence $$t ≥ t₂ and so

  $$ (8)

•  

  Integrating (8) from t to δ(t) yields

  $$ (9)

•  

  We have two for $$:

  • (a)

    If $$. For t ≥ t₃ ≥ t₂, Then the last inequality becomes

    $$ (10)

  •  

    Integrating (10) from t₃ to t yields

    $$ (11)

  •  

    As t⟶∞, it follows, either $$ or y_i(t) is bounded away from zero $$

  • (b)

    $$ and $$ leads to Lim_t⟶∞y_i(t) = −∞, which is a contradiction,

  •  

    which means X(t) ∈ K₂.

• (iii)

  Since $$ and $$ that is, $$ are negative nondecreasing, j = 1,2, there exists l_j ≤ 0, such that $$ Then, $$, thus

  $$ (12)

•  

  Integrating (10) from t to δ(t) for some continuous function δ(t) > t, we obtain

  $$ (13)

We claim that $$ for otherwise if $$ and $$ this implies to y_j(t) < 0 and Lim_t⟶∞y_j(t) = −∞ which is a contradiction. Hence, $$ and $$. Now, $$ and $$, so $$ , is positive nondecreasing, then there exists b₃ > 0 and t₂ ≥ t₁ such that

Integrating (14) from t to δ(t), we obtain

We claim that $$ for t ≥ t₃ ≥ t₂, otherwise if $$ for t ≥ t₃ ≥ t₂, then the last inequality becomes

Integrating (16) from t₃ to t

Letting t⟶∞, then inequality (17) leads to Lim_t⟶∞y₃(t) = −∞, which is a contradiction. Hence $$ and $$, this case leads to Lim_t⟶∞y₃(t) = ∞, and so X(t) ∈ K₃. Analogously from the subcases (iv-viii), one can get X(t) ∈ K_n, n = 4,5, …, 8, respectively.

Lemma 2. Assume that X(t) is NOS of (1) with λ = −1 and let (2) hold. Then there are only L₁‒L₈ possible classes.

Proof. Suppose that X(t) be an eventual positive solution of (1), then $$.

This means that $$ is nonincreasing, so from Table 2, eight subcases can be discussed successively.

• (i)

  $$.

•  

  Since $$ is positively nonincreasing, there exists l_i ≥ 0, i = 1,2,3 such that, $$ and then there exists t₂ ≥ t₁ such that $$, therefore,

  $$ (18)

•  

  Integrating (18) from t to δ(t) for some continuous function δ(t) > t, we obtain

  $$ (19)

•  

  We have two cases for $$.

  • (a)

    If $$, for t ≥ t₃ ≥ t₂, in that case, we claim that l_i = 0 otherwise l_i > 0, then (19) becomes

    $$ (20)

  •  

    Integrating (20) from t₃ to t

    $$ (21)

  •  

    As ⟶∞ , it follows that Lim_t⟶∞y_i(t) = −∞, which is a contradiction, hence l_i = 0 .

  • (b)

    If $$ and $$, it follows that Lim_t⟶∞y_i(t) = ∞. Thus, X(t) ∈ L₁.

• (ii)

  $$

•  

  then $$. Since $$ , is negative nonincreasing, then there exists b_i < 0, and t₂ ≥ t₁ such that $$ for t ≥ t₂, therefore,

  $$ (22)

•  

  integrating (22) from t to δ(t), we obtain

  $$ (23)

•  

  We claim that $$ for ≥t₃ ≥ t₂ , otherwise if $$ for t ≥ t₃ ≥ t₂, and $$ implies that Lim_t⟶∞y_i(t) = −∞, which is a contradiction, thus $$ then (23) becomes

  $$ (24)

•  

  Integrating (24) from t₃ to t

  $$ (25)

•  

  As ⟶∞ , it follows that Lim_t⟶∞y_i(t) = ∞. Thus X(t) ∈ L₂.

• (iii)

  $$

•  

  Then, $$ and $$. Since $$ , are negative and nonincreasing, then there exists b_j < 0, and t₂ ≥ t₁ such that $$ for t ≥ t₂, therefore,

  $$ (26)

Integrating (26) from t to δ(t), we obtain

and we claim that $$ for ≥t₃ ≥ t₂ , otherwise if $$ for t ≥ t₃ ≥ t₂, and $$ implies that Lim_t⟶∞y_j(t) = −∞, which is a contradiction, thus $$ then (27) becomes

Integrating the last inequality from t₃ to t yields

As ⟶∞ , it follows that Lim_t⟶∞y_j(t) = ∞, j = 1,2.

Concerning $$ and nonincreasing, so there exists l₃ ≥ 0, such that, $$ then $$ therefore

Integrating the last inequality from t to δ(t) leads to

we have two cases for $$.

• (a)

  If $$ for t ≥ t₃ ≥ t₂, and $$, it follows that Lim_t⟶∞y₃(t) = ∞.

• (b)

  If $$ for t ≥ t₃ ≥ t₂, we claim that l₃ = 0, otherwise l₃ > 0, then (31) reduced to

  $$ (32)

•  

  Integrating (32) from t₃ to t

  $$ (33)

•  

  As ⟶∞, it follows that Lim_t⟶∞y₃(t) = −∞, which is a contradiction. Hence, (y₁, y₂, y₃) ∈ L₃. Analogously from the subcases (iv-viii), one can get X(t) ∈ L_n, n = 4,5, …, 8, respectively.

Theorem 2. Suppose that λ = 1, and (2) holds in addition to

Proof. Suppose that (1) has NOS X(t), so by Lemma 1, from Table 3, there is only the class K₂, can occur for t ≥ t₁ ≥ t₀, that is,

Since y₁(t), y₂(t), y₃(t), are increasing, so there exists c_i > 0, i = 1,2,3 and t₂ ≥ t₁ such that y_i(t) ≥ c_i, t ≥ t₂. Integrating the first equation of system (1) from t to δ(t) for some continuous function δ(t) > t, leads to

Integrating (36) from t to δ(t) yields

Then,

Integrating the above inequality from t₂ to t, we get

As t⟶∞ concerning (34), it follows from (23) that Lim_t⟶∞y₁(t) = ∞, which is a contradiction. Similarly, it can be shown that Lim_t⟶∞y₂(t) = ∞, Lim_t⟶∞y₃(t) = ∞, which is a contradiction.

This leads to the solution X(t) oscillates.

Theorem 3. Suppose λ = −1, (2) and (34) hold. Then every bounded solution of (1) oscillates or tends to zero as t⟶∞.

Proof. Suppose that system (1) has NOS X(t) so by Lemma 1, Table 2, there is only the possible case L₁ to consider for t ≥ t₁ ≥ t₀:

Since y₁(t), y₂(t), y₃(t), are positive and decreasing, so there exists l_i ≥ 0, i = 1,2,3 such that Lim_t⟶∞y_i(t) = l_i, we claim that l_i = 0, otherwise l_i > 0 hence y_i(t) ≥ l_i > 0 for t ≥ t₂ ≥ t₁.

Integrating the first equation of (1) from t to δ(t) yields:

Integrating (42) from t to δ(t), we get

Integrating (43) from t₂ to t, we get

As t⟶∞ we get from (44) Lim_t⟶∞ y₁(t) = −∞, which is a contradiction. Similarly, Lim_t⟶∞ y₂(t) = −∞, Lim_t⟶∞ y₃(t) = −∞. Then Lim_t⟶∞ y_i(t) = 0, i = 1,2,3.

Corollary 4. Suppose that λ = 1, then (2) and (21) hold. Then every solution of system (1) is either oscillatory or Lim_t⟶∞|y_i(t)| = ∞, i = 1,2,3.

Proof. Suppose that system (1) has a nonoscillatory solution X(t), let y_i(t) > 0, t ≥ t₀, i = 1,2,3. So by Lemma 1 and Table 3, there are only the possible classes K₁ − K₈ to consider for t ≥ t₁ ≥ t₀. If X(t) is bounded, then by Theorem 2, it follows that X(t) is oscillatory. Otherwise, X(t) is unbounded.

Case 1. Suppose that X(t) ∈ K₁. By Lemma 1, it follows Lim_t⟶∞y_i(t) = ∞, i = 1,2,3.

Case 2. Suppose that X(t) ∈ K₂. By Theorem 2, Lim_t⟶∞y_i(t) = ∞, i = 1,2,3..

Case 3. Suppose that X(t) ∈ K₃. Since y₁(t), y₂(t), y₃(t), are increasing, so there exists c_i > 0, i = 1,2,3 and t₂ ≥ t₁ such that y_i(t) ≥ c_i, t ≥ t₂.

Integrating the first equation of system (1) from t to δ(t) for some continuous function δ(t) > t, leads to

Integrating the (45) from t to δ(t), yields

Then,

Integrating the last inequality from t₂ to t, we get

As t⟶∞ concerning (34), it follows from (30) that Lim_t⟶∞y₁(t) = ∞.

Now, similarly, it can be shown that Lim_t⟶∞y₂(t) = ∞, by Lemma 2.2, lim_t⟶∞y₃(t) = ∞.

Other cases can be handled in the same way. The proof is complete.

Corollary 5. Suppose that λ = −1, i = 1,2,3, and (2), (40) are held. Then every solution of system (1) is either oscillatory or converges to zero or tends to infinity as t⟶∞.

Proof. Suppose that system (1) has a NOS X(t) so by Lemma 2 Table 2, there are only the possible cases L₁ − L₈ to consider for t ≥ t₁ ≥ t₀. If X(t) is bounded, then by Theorem 2, it follows that X(t) is either oscillatory or X(t)⟶0 as t⟶∞. If X(t) is unbounded, then from Table 2, we conclude that Lim_t⟶∞y_i(t) = ∞, i = 1,2,3.

Example 1. Consider the delay system as follows:

Obviously, to see that

Hence all conditions of Theorem 2 are satisfied, so according to Theorem 2, every bounded solution of system (49) is oscillatory. For instance, (1/2sint/2, 1/4cost/2, 1/4sint/2)^T, has an oscillatory solution, as shown in Figure 1.

Example 2. Consider the delay system as follows:

It is clear that

Hence all conditions of Theorem 3 satisfies, so according to Theorem 3, every bounded solution of (1) oscillates or tends to zero as t⟶∞. The solution $$ has an nonoscillatory solution tends to zero as t⟶∞, as shown in Figure 2.

Example 3. Consider the delay system as follows:

It is clear that

Hence all conditions of Corollary 5 satisfy, so according to Corollary 5, every solution of (1) oscillates or tends to zero or tends to infinity as t⟶∞. The nonoscillatory solution $$ rose to infinity as t⟶∞. as shown in Figure 3.