Existence and Iterative Algorithms of Positive Solutions for a Higher Order Nonlinear Neutral Delay Differential Equation

Lemma 1. Let τ > 0,   c ≥ 0, F ∈ C([c, +∞) ³, ℝ⁺) and G ∈ C([c, +∞) ², ℝ⁺). Then

• (a)

  $$;

• (b)

  $$;

• (c)

  if $$, then

  $$ ()

• (d)

  if $$, then

  $$ ()

Proof. Let [t] denote the largest integral number not exceeding t ∈ ℝ. Note that

Clearly (12) means that Thus (a) follows from (11) and (13).

Assume that $$. As in the proof of (a), we infer that

that is, (c) holds.

Similar to the proofs of (a) and (c), we conclude that (b) and (d) hold. This completes the proof.

Theorem 2. Assume that there exist three constants M, N, and b₀ and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying

Then

(a) for any L ∈ (b₀M + N, (1 − b₀)M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the following error estimate: where $$ is an arbitrary sequence in [0,1] such that

(b) Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Firstly, we prove that (a) holds. Set L ∈ (b₀M + N, (1 − b₀)M). From (15) and (18), we know that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by It is obvious that S_Lx is continuous for each x ∈ Ω₁(N, M). By means of (16), (22), (23), and (25), we deduce that for any x, y ∈ Ω₁(N, M) and t ≥ T which yields that On the basis of (17), (22), (24), and (25), we acquire that for any x ∈ Ω₁(N, M) and t ≥ T which guarantee that S_L(Ω₁(N, M))⊆Ω₁(N, M). Consequently, (27) gives that S_L is a contraction mapping in Ω₁(N, M) and it has a unique fixed point x ∈ Ω₁(N, M). It is easy to see that x ∈ Ω₁(N, M) is a positive solution of (5).

It follows from (19), (25), and (27) that

which yields that That is, (20) holds. Thus (20) and (21) ensure that lim _k→∞x_k = x.

Secondly, we show that (b) holds. Let L₁, L₂ ∈ (b₀M + N, (1 − b₀)M) with L₁ ≠ L₂. In light of (15) and (18), we know that for each p ∈ {1,2}, there exist θ_p ∈ (0,1), T_p and T^* with T_p > 1+|t₀ | + τ+|γ| and T^* > max  {T₁, T₂} satisfying (22)–(24) and

where θ and T are replaced by θ_p and T_p, respectively. Let the mapping $$ be defined by (25) with L and T replaced by L_p and T_p, respectively. As in the proof of (a), we deduce easily that the mapping $$ possesses a unique fixed point z_p ∈ Ω₁(N, M), that is, z_p is a positive solution of (5) in Ω₁(N, M). In order to prove (b), we need only to show that z₁ ≠ z₂. In fact, (25) means that for each t ≥ T^* and p ∈ {1,2} It follows from (16), (22), (31), and (32) that for each t ≥ T^* which implies that that is, z₁ ≠ z₂. This completes the proof.

Theorem 3. Assume that there exist three constants M, N, and b₀ and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16)–(18) and

Then

• (a)

  for any L ∈ (b₀M + N, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by (19) converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] satisfying (21);

• (b)

  Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Let L ∈ (b₀M + N, M). Equations (18) and (36) ensure that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying (23),

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by (25). Obviously, S_Lx is continuous for every x ∈ Ω₁(N, M). Using (16), (23), (25), and (36), we conclude that for any x, y ∈ Ω₁(N, M) and t ≥ T which implies that (27) holds. In light of (17), (25), (36), and (37), we know that for any x ∈ Ω₁(N, M) and t ≥ T which mean that S_L(Ω₁(N, M))⊆Ω₁(N, M). Equation (27) guarantees that S_L is a contraction mapping in Ω₁(N, M) and it possesses a unique fixed point x ∈ Ω₁(N, M). As in the proof of Theorem 2, we infer that x ∈ Ω₁(N, M) is a positive solution of (5). The rest of the proof is similar to that of Theorem 2 and is omitted. This completes the proof.

Theorem 4. Assume that there exist three constants M, N, and b₀ and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16)–(18) and

Then

• (a)

  for any L ∈ (N, (1 − b₀)M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by (19) converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] satisfying (21);

• (b)

  Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Set L ∈ (N, (1 − b₀)M). It follows from (18) and (40) that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying (23),

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by (25). Distinctly, S_Lx is continuous for each x ∈ Ω₁(N, M). In terms of (16), (23), (25), and (41), we reason that for any x, y ∈ Ω₁(N, M) and t ≥ T which means that (27) holds. Owing to (17), (25), (41), and (42), we earn that for any x ∈ Ω₁(N, M) and t ≥ T which yield that S_L(Ω₁(N, M))⊆Ω₁(N, M). Thus (27) ensures that S_L is a contraction mapping in Ω₁(N, M) and it owns a unique fixed point x ∈ Ω₁(N, M). As in the proof of Theorem 2, we infer that x ∈ Ω₁(N, M) is a positive solution of (5). The rest of the proof is parallel to that of Theorem 2, and hence is elided. This completes the proof.

Theorem 5. Assume that there exist three constants M, N, and b₀ and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (18) and

Then

(a) for any L ∈ (N + M/b₀, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₂(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₂(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] with (21);

(b) Equation (5) has uncountably many positive solutions in Ω₂(N, M).

Proof. First of all, we show that (a) holds. Set L ∈ (N + M/b₀, M). It follows from (18) and (45) that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that

Define a mapping S_L : Ω₂(N, M) → CB([γ, +∞), ℝ) by In light of (46), (49), (50), and (52), we conclude that for x, y ∈ Ω₂(N, M) and t ≥ T which yields that In view of (47), (49), (51), and (52), we obtain that for any x ∈ Ω₂(N, M) and t ≥ T which imply that S_L(Ω₂(N, M))⊆Ω₂(N, M). It follows from (50) and (54) that S_L is a contraction mapping in Ω₂(N, M) and it has a unique fixed point x ∈ Ω₂(N, M). It is clear that x ∈ Ω₂(N, M) is a positive solution of (5).

Note that (48), (52), and (54) undertake that

which indicates that (20) holds. Thus (20) and (21) assure that lim _k→∞x_k = x.

Next we prove that (b) holds. Let L₁, L₂ ∈ (N + M/b₀, M) with L₁ ≠ L₂. As in the proof of (a) we infer that for each p ∈ {1,2} there exist θ_p ∈ (0,1), T_p > 1+|t₀ | + τ+|γ| and $$ satisfying (49)–(52), where L, θ, T, and S_L are replaced by L_p, θ_p, T_p, and $$, respectively, and $$ has a unique fixed point z_p ∈ Ω₂(N, M), which is a positive solution of (5) in Ω₂(N, M). It follows that for each t ≥ T_p and p ∈ {1,2}

On behalf of proving (b), we need only to show that z₁ ≠ z₂. Notice that (18) guarantees that there exits T₃ > max  {T₁, T₂} satisfying Due to (46), (51), (57), and (58), we conclude that for each t ≥ T₃ which yields that z₁ ≠ z₂. This completes the proof.

Theorem 6. Assume that there exist three constants M, N, and b₀ and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (18), (46), (47), and

Then

(a) for any L ∈ (N, (1 − 1/b₀)M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₃(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₃(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] satisfying (21);

(b) Equation (5) has uncountably many positive solutions in Ω₃(N, M).

Proof. Put L ∈ (N, (1 − 1/b₀)M). It follows from (18) and (60) that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying (50) and

Define a mapping S_L : Ω₃(N, M) → CB([γ, +∞), ℝ) by By virtue of (47), (62), and (63), we know that for any x ∈ Ω₃(N, M) and t ≥ T which imply that S_L(Ω₃(N, M))⊆Ω₃(N, M). The rest of the proof is identical with the proof of Theorem 5 and hence is omitted. This completes the proof.

Theorem 7. Let m ≥ 2. Assume that there exist two constants M, N with M > N > 0 and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16)–(18) and

Then

(a) for any L ∈ (N, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] with (21);

(b) Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Let L ∈ (N, M). It follows from (18) and (65) that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by With a view to (16), (68), and (70), we derive that for any x, y ∈ Ω₁(N, M) and t ≥ T which gives (27). By virtue of (17), (69), and (70), we deduce that for any x ∈ Ω₁(N, M) and t ≥ T which mean that S_L(Ω₁(N, M))⊆Ω₁(N, M). Coupled with (27) and (68), we get that S_L is a contraction mapping in Ω₁(N, M) and it possesses a unique fixed point x ∈ Ω₁(N, M). Clearly, x ∈ Ω₁(N, M) is a positive solution of (5).

From (27), (66), and (70), we gain that

which yields (20). It follows from (20) and (21) that lim _k→∞x_k = x.

Now we prove that (b) holds. Let L₁, L₂ ∈ (N, M) and L₁ ≠ L₂. As in the proof of (a), we conclude that for each p ∈ {1,2}, there exist θ_p ∈ (0,1), T_p > 1+|t₀ | + τ+|γ| and $$ satisfying (69)–(77), where L, θ, T, and S_L are replaced by L_p, θ_p, T_p, and $$, respectively, and $$ has a unique fixed point z_p ∈ Ω₁(N, M), which is a positive solution of (5) in Ω₁(N, M), that is,

For purpose of proving (b), we just need to show that z₁ ≠ z₂. It follows from (16), (27), (68), and (74) that which yields that that is, z₁ ≠ z₂. This completes the proof.

Theorem 8. Let m = 1. Assume that there exist two constants M, N with M > N > 0 and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16), (17), (65), and

Then

(a) for any L ∈ (N, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] with (21);

(b) Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Let L ∈ (N, M). It follows from (65) and (77) that there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying (67),

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by By virtue of (16), (79), and (81), we derive that for any x, y ∈ Ω₁(N, M) and t ≥ T which gives (27). It follows from (17), (80), and (81) that for any x ∈ Ω₁(N, M) and t ≥ T which mean that S_L(Ω₁(N, M))⊆Ω₁(N, M). Combined with (27) and (79), we know that S_L is a contraction mapping in Ω₁(N, M) and it possesses a unique fixed point x ∈ Ω₁(N, M). Obviously, x ∈ Ω₁(N, M) is a positive solution of (5).

In light of (27), (78), and (81), we gain that

which yields (20). It follows from (20) and (21) that lim _k→∞x_k = x.

Now we prove that (b) holds. Let L₁, L₂ ∈ (N, M) and L₁ ≠ L₂. As in the proof of (a), we conclude that for each p ∈ {1,2}, there exist θ_p ∈ (0,1), T_p > 1+|t₀ | + τ+|γ| and $$ satisfying (67) and (79)–(81), where L, θ, T, and S_L are replaced by L_p, θ_p, T_p, and $$, respectively, and $$ has a unique fixed point z_p ∈ Ω₁(N, M), which is a positive solution of (5) in Ω₁(N, M), that is,

In order to prove (b), we just need to show that z₁ ≠ z₂. In view of (16), (27), (79), and (85), we get that which implies that that is, z₁ ≠ z₂. This completes the proof.

Theorem 9. Let m ≥ 2. Assume that there exist two constants M, N with M > N > 0 and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16), (17),

Then

(a) for any L ∈ (N, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] with (21);

(b) Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Set L ∈ (N, M). In view of (88) and (89), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by By virtue of (16), (92), (94), and Lemma 1, we acquire that for any x, y ∈ Ω₁(N, M) and t ≥ T which yields that (27) holds. From (17), (94), (98), and Lemma 1, we obtain that for any x ∈ Ω₁(N, M) and t ≥ T which means that S_L(Ω₁(N, M))⊆Ω₁(N, M). It follows from (27) and (92) that S_L is a contraction mapping and it has a unique fixed point x ∈ Ω₁(N, M). It is clear that x ∈ Ω₁(N, M) is a positive solution of (5).

On the basis of (27), (90), and (94), we deduce that

which signifies that (20) holds. It follows from (20) and (21) and that lim _k→∞x_k = x.

Now we show that (b) holds. Let L₁, L₂ ∈ (N, M) and L₁ ≠ L₂. As in the proof of (a), we conclude that for each p ∈ {1,2}, there exist θ_p ∈ (0,1), T_p > 1+|t₀ | + τ+|γ| and $$ satisfying (91)–(94), where L, θ, T, and S_L are replaced by L_p, θ_p, T_p, and $$, respectively, and $$ has a unique fixed point z_p ∈ Ω₁(N, M), which is a positive solution of (5) in Ω₁(N, M), that is,

In order to prove (b), it is sufficient to show that z₁ ≠ z₂. Note that (16), (92), (98), and Lemma 1 lead to which means that that is, z₁ ≠ z₂. This completes the proof.

Theorem 10. Let m = 1. Assume that there exist two constants M, N with M > N > 0 and four functions P, Q, R, W ∈ C([t₀, +∞), ℝ⁺) satisfying (16), (17), (89), and

Then

(a) for any L ∈ (N, M), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that for each x₀ ∈ Ω₁(N, M), the Mann iterative sequence $$ generated by the following scheme

converges to a positive solution x ∈ Ω₁(N, M) of (5) and has the error estimate (20), where $$ is an arbitrary sequence in [0,1] with (21);

(b) Equation (5) has uncountably many positive solutions in Ω₁(N, M).

Proof. Set L ∈ (N, M). Due to (101), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| satisfying (91),

Define a mapping S_L : Ω₁(N, M) → CB([γ, +∞), ℝ) by In view of (16), (103), (105), and Lemma 1, we achieve that for any x, y ∈ Ω₁(N, M) and t ≥ T which means that (27) holds. It follows from (17), (104), (105), and Lemma 1 that for any x ∈ Ω₁(N, M) and t ≥ T which means that S_L(Ω₁(N, M))⊆Ω₁(N, M). Coupled with (27), we know that S_L is a contraction mapping and it has a unique fixed point x ∈ Ω₁(N, M). It follows that x ∈ Ω₁(N, M) is a positive solution of (5).

In view of (27), (102), and (105), we deduce that

which signifies that (20) holds. It follows from (20) and (21) that lim _k→∞x_k = x.

Now we show that (b) holds. Let L₁, L₂ ∈ (N, M) and L₁ ≠ L₂. As in the proof of (a), we conclude that for each p ∈ {1,2}, there exist θ_p ∈ (0,1), T_p > 1+|t₀ | + τ+|γ| and $$ satisfying (91) and (103)–(105), where L, θ, T, and S_L are replaced by L_p, θ_p, T_p and $$, respectively, and $$ has a unique fixed point z_p ∈ Ω₁(N, M), which is a positive solution of (5) in Ω₁(N, M). It follows that for any t ≥ T_p and p ∈ {1,2}

In order to prove (b), we just need to show that z₁ ≠ z₂. Notice that (16), (103), (109), and Lemma 1 ensure that which yields that that is, z₁ ≠ z₂. This completes the proof.

Remark 11. Theorems 2–10 extend, improve, and unifies Theorems 1–4 in [1], the theorem in [2], Theorems 2.1–2.4 in [4], Theorems 2.1–2.5 in [5, 8], Theorems 1–3 in [9], and Theorems 1–4 in [11], respectively. The examples below prove that Theorems 2–10 extend substantially the corresponding results in [1, 2, 4, 5, 8, 9, 11]. Note that none of the known results can be applied to these examples.

Example 12. Consider the higher order nonlinear neutral delay differential equation

where τ > 0 and i ≤ n − m − 1. Let l = 2, t₀ = 2, γ = min  {2 − τ, −2}, M = 10, N = 1, b₀ = 2/5 and It is easy to verify that the conditions of Theorem 2 are satisfied. Thus Theorem 2 ensures that (112) has uncountably many positive solutions in Ω₁(1,10), and for any L ∈ (5,6), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (19) and (21) converges to a positive solution x ∈ Ω₁(1,10) of (112) and has the error estimate (20).

Example 13. Consider the higher order nonlinear neutral delay differential equation

where τ > 0 and i ≤ n − m − 1. Let l = 2, t₀ = 1, γ = min  {1 − τ, 0}, M = 6, N = 1, b₀ = 3/4 and It is easy to check that the conditions of Theorem 3 are satisfied. Therefore (114) has uncountably many positive solutions in Ω₁(1,6), and for any L ∈ (11/2, 6), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (19) and (21) converges to a positive solution x ∈ Ω₁(1,6) of (114) and has the error estimate (20).

Example 14. Consider the higher order nonlinear neutral delay differential equation

where τ > 0 and i ≤ n − m − 1. Let l = 2, t₀ = 0, γ = min  {−τ, −1}, M = 8, N = 1/2, b₀ = 7/8 and It is easy to prove that the conditions of Theorem 4 are satisfied. Hence (116) has uncountably many positive solutions in Ω₁(1/2, 8), and for any L ∈ (1/2, 1), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (19) and (21) converges to a positive solution x ∈ Ω₁(1/2, 8) of (116) and has the error estimate (20).

Example 15. Consider the higher order nonlinear neutral delay differential equation

where τ > 0 and i ≤ n − m − 1. Let l = 3, t₀ = 0, γ = min  {3 − τ, −9}, M = 12, N = 5, b₀ = 2 and It is easy to verify that the conditions of Theorem 5 are satisfied. Hence Theorem 5 ensures that (118) has uncountably many positive solutions in Ω₂(5,12), and, for any L ∈ (11,12), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (48) and (21) converges to a positive solution x ∈ Ω₂(5,12) of (118) and has the error estimate (20).

Example 16. Consider the higher order nonlinear neutral delay differential equation

where τ > 0, and i ≤ n − m − 1. Let l = 2, t₀ = 1, γ = min  {1 − τ, −3}, M = 6, N = 2, b₀ = 3 and It is easy to check that the conditions of Theorem 6 are satisfied. Thus Theorem 6 ensures that (120) has uncountably many positive solutions in Ω₃(2,6), and, for any L ∈ (2,4), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (61) and (21) converges to a positive solution x ∈ Ω₃(2,6) of (120) and has the error estimate (20).

Example 17. Consider the higher order nonlinear neutral delay differential equation

where τ > 0, m ≥ 2 and i ≤ n − m − 1. Let l = 3, t₀ = 4, γ = min  {4 − τ, 0}, M = 100, N = 1 and It is easy to check that the conditions of Theorem 7 are satisfied. Thus (122) has uncountably many positive solutions in Ω₁(1,100), and, for any L ∈ (1,100), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (66) and (21) converges to a positive solution x ∈ Ω₁(1,100) of (122) and has the error estimate (20).

Example 18. Consider the higher order nonlinear neutral delay differential equation

where τ > 0, m = 1 and i ≤ n − 2. Let l = 2, t₀ = 3, γ = min  {3 − τ, 0}, M = 10, N = 9 and It is easy to check that the conditions of Theorem 8 are satisfied. Thus Theorem 8 ensures that (124) has uncountably many positive solutions in Ω₁(9,10), and, for any L ∈ (9,10), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (78) and (21) converges to a positive solution x ∈ Ω₁(9,10) of (124) and has the error estimate (20).

Example 19. Consider the higher order nonlinear neutral delay differential equation

where τ > 0, m ≥ 2 and i ≤ n − m − 1. Let l = 1, t₀ = 4, γ = min  {4 − τ, 1}, M = 7, N = 5 and It is easy to check that the conditions of Theorem 9 are satisfied. Thus Theorem 9 ensures that (126) has uncountably many positive solutions in Ω₁(5,7), and, for any L ∈ (5,7), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (90) and (21) converges to a positive solution x ∈ Ω₁(5,7) of (126) and has the error estimate (20).

Example 20. Consider the higher order nonlinear neutral delay differential equation

where τ > 0, m = 1 and i ≤ n − 2. Let l = 2, t₀ = 5, γ = min  {5 − τ, 1}, M = 4, N = 2 and It is easy to check that the conditions of Theorem 10 are satisfied. Thus Theorem 10 ensures that (128) has uncountably many positive solutions in Ω₁(2,4), and, for any L ∈ (2,4), there exist θ ∈ (0,1) and T > 1+|t₀ | + τ+|γ| such that the Mann iterative sequence $$ generated by (102) and (21) converges to a positive solution x ∈ Ω₁(2,4) of (128) and has the error estimate (20).