Existence of Nonoscillatory Solutions for a Third-Order Nonlinear Neutral Delay Differential Equation

and Applied Analysis 3 Throughout this paper, we assume that R −∞, ∞ , R 0, ∞ , C t0, ∞ ,R denotes the Banach space of all continuous and bounded functions on t0, ∞ with the norm ‖x‖ supt≥t0 |x t | for each x ∈ C t0, ∞ ,R and A N,M {x ∈ C t0, ∞ ,R : N ≤ x t ≤ M, t ≥ t0} for M > N > 0. 1.8 It is easy to see that A N,M is a bounded closed and convex subset of C t0, ∞ ,R . By a solution of 1.7 , we mean a function x ∈ C t1 − τ, ∞ ,R for some t1 ≥ t0 such that x t p t x t − τ , β t x t p t x t − τ ′ and α t β t x t p t x t − τ ′ ′ are continuously differentiable in t1, ∞ and such that 1.7 is satisfied for t ≥ t1. As is customary, a solution of 1.7 is called oscillatory if it has arbitrarily large zeros, and otherwise it is said to be nonoscillatory. Definition 1.1 see 6 . A family F of functions in C t0, ∞ ,R is equicontinuous on t0, ∞ if for any ε > 0, the interval t0, ∞ can be decomposed into a finite number of subintervals I1, I2, . . . , In such that ∣∣f x − f(y)∣∣ ≤ ε, ∀f ∈ F, x, y ∈ Ii, i ∈ {1, 2, . . . , n}. 1.9 Lemma 1.2 see Krasnoselskii’s fixed point theorem, 4 . Let X be a Banach space. Let Ω be a bounded closed convex subset of X and S1 and S2 mappings from Ω into X such that S1x S2y ∈ Ω for every pair x, y ∈ Ω. If S1 is a contraction and S2 is completely continuous, then the equation S1x S2x x has at least one solution in Ω. Lemma 1.3 see Schauder’s fixed point theorem, 4 . Let Ω be a nonempty closed convex subset of a Banach space X. Let S : Ω → Ω be a continuous mapping such that SΩ is a relatively compact subset of X. Then S has at least one fixed point in Ω. 2. Main Results Now we study those conditions under which 1.7 possesses uncountably many bounded nonoscillatory solutions. Theorem 2.1. Assume that there exist constantsM,N, c1, c2, T0 and a function h ∈ C t0, ∞ ,R satisfying min{c1, c2} ≥ 0, c1 c2 < 1, 0 < N < 1 − c1 − c2 M, 2.1 −c2 ≤ p t ≤ c1, t ≥ T0 > t0; 2.2 ∣∣f t, u1, u2, . . . , un ∣∣ ≤ h t , t ∈ t0, ∞ , ui ∈ N,M , i ∈ {1, . . . , n}; 2.3 ∫ ∞ t0 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < ∞. 2.4 Then 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . 4 Abstract and Applied Analysis Proof. Set k ∈ c1M N, 1 − c2 M . From 2.4 , we pick up T > T0 such that ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < min{ 1 − c2 M − k, k − c1M −N}. 2.5 Define two mappings S1k and S2k : A N,M → C t0, ∞ ,R by S1kx t ⎧ ⎨ ⎩ k − p t x t − τ , t ≥ T, S1kx T , t0 ≤ t < T, 2.6 S2kx t ⎧ ⎪⎨ ⎪⎩ ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v f u, x σ1 u , . . . , x σn u dudsdv, t ≥ T, S2kx T , t0 ≤ t < T 2.7 for x ∈ A N,M . Firstly, we prove that S1k A N,M S2k A N,M ⊆ A N,M . By 2.1 – 2.3 and 2.5 – 2.7 , we get that S1kx t ( S2ky ) t ≤ k c2M ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ k c2M min{ 1 − c2 M − k, k − c1M −N} ≤ M, x, y ∈ A N,M , t ≥ T, S1kx t ( S2ky ) t ≥ k − c1M − ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≥ k − c1M −min{ 1 − c2 M − k, k − c1M −N} ≥ N, x, y ∈ A N,M , t ≥ T, 2.8 which infer that S1k A N,M S2k A N,M ⊆ A N,M for any x, y ∈ A N,M . Secondly, we show that S1k is a contraction mapping. By 2.1 , 2.2 , and 2.6 , we deduce that ∣∣ S1kx t − ( S1ky ) t ∣∣ ≤ ∣∣p t ∣∣∣∣x t − τ − y t − τ ∣∣ ≤ c1 c2 ∥∥x − y∥∥ 2.9 for x, y ∈ A N,M and t ≥ T , which gives that ∥S1kx − S1ky ∥∥ ≤ c1 c2 ∥∥x − y∥∥. 2.10 Abstract and Applied Analysis 5 Thirdly, we show that S2k is completely continuous. Let x ∈ A N,M and {xm}m≥1 ⊆ A N,M with limm→ ∞xm x. By 2.7 , we obtain that | S2kxm t − S2kx t | ≤ ∫ ∞ T ∫ ∞ v ∫ ∞ s 1 α s β v ∣∣f u, xm σ1 u , . . . , xm σn u −f u, x σ1 u , . . . , x σn u ∣∣dudsdv, t ≥ T, m ≥ 1. 2.11and Applied Analysis 5 Thirdly, we show that S2k is completely continuous. Let x ∈ A N,M and {xm}m≥1 ⊆ A N,M with limm→ ∞xm x. By 2.7 , we obtain that | S2kxm t − S2kx t | ≤ ∫ ∞ T ∫ ∞ v ∫ ∞ s 1 α s β v ∣∣f u, xm σ1 u , . . . , xm σn u −f u, x σ1 u , . . . , x σn u ∣∣dudsdv, t ≥ T, m ≥ 1. 2.11 Using 2.3 and 2.4 , we conclude that ∣∣f u, xm σ1 u , . . . , xm σn u − f u, x σ1 u , . . . , x σn u ∣∣ ≤ 2h u , u ∈ T, ∞ , m ≥ 1, ∫ ∞ s ∣∣f u, xm σ1 u , . . . , xm σn u − f u, x σ1 u , . . . , x σn u ∣∣du ≤ 2 ∫ ∞ s h u du, s ∈ T,∞ , m ≥ 1, ∫ ∞ v ∫ ∞ s 1 α s ∣∣f u, xm σ1 u , . . . , xm σn u − f u, x σ1 u , . . . , x σn u ∣∣duds ≤ 2 ∫ ∞ v ∫ ∞ s h u α s duds, v, s ∈ T,∞ , m ≥ 1. 2.12 In light of 2.11 – 2.12 , ∣∣f u, xm σ1 u , . . . , xm σn u − f u, x σ1 u , . . . , x σn u ∣∣ −→ 0 as m −→ ∞, 2.13 and the Lebesgue dominated convergence theorem, we conclude that lim m→ ∞ ‖S2kxm − S2kx‖ 0, 2.14 which means that S2k is continuous in A N,M . Nowwe show that S2k is completely continuous. By virtue of 2.3 , 2.4 , and 2.7 , we get that ‖S2kx‖ ≤ ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv, x ∈ A N,M . 2.15 6 Abstract and Applied Analysis That is, S2k A N,M is uniform bounded. It follows from 2.4 that for each ε > 0, there exists T ∗ > T such that ∫ ∞ T∗ ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < ε 2 . 2.16 In view of 2.3 , 2.7 , and 2.16 , we infer that | S2kx t2 − S2kx t1 | ≤ ∫ ∞ t2 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ∫ ∞ t1 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < ε, x ∈ A N,M , t2 > t1 ≥ T ∗. 2.17 From 2.3 and 2.7 , we get that | S2kx t2 − S2kx t1 | ≤ ∫ t2 t1 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv, x ∈ A N,M , T ≤ t1 ≤ t2 ≤ T ∗, 2.18 which together with 2.4 ensures that there exists δ > 0 satisfying | S2kx t2 − S2kx t1 | < ε, x ∈ A N,M , t1, t2 ∈ T, T ∗ with |t2 − t1| < δ. 2.19 Clearly, | S2kx t2 − S2kx t1 | 0 < ε, x ∈ A N,M , t0 ≤ t1 ≤ t2 ≤ T. 2.20 That is, S2k A N,M is equicontinuous on t0, ∞ . Consequently, S2k is completely continuous. By Lemma 1.2, there is x0 ∈ A N,M such that S1x0 S2x0 x0, which is a bounded nonoscillatory solution of 1.7 . Lastly, we demonstrate that 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . Let k1, k2 ∈ c1M N, 1 − c2 M and k1 / k2. For each j ∈ {1, 2}, we choose Tj > T0 and the mappings Sjk1 and Sjk2 satisfying 2.5 – 2.7 with k and T replaced by kj and Tj , respectively, and some T3 > max{T1, T2} such that ∫ ∞ T3 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < |k1 − k2| 2 . 2.21 Obviously, there are x, y ∈ A N,M such that S1k1x S2k1x x and S1k2y S2k2y y, respectively. That is, x and y are two bounded nonoscillatory solutions of 1.7 in A N,M . Abstract and Applied Analysis 7 In order to prove that 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M , we prove only that x / y. In fact, by 2.6 and 2.7 , we gain that for t ≥ T3 x t k1 − p t x t − τ ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v f u, x σ1 u , . . . , x σn u dudsdv, y t k2 − p t y t − τ ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v f ( u, y σ1 u , . . . , y σn u ) dudsdv, 2.22 which together with 2.1 – 2.3 imply that ∣∣x t − y t ∣∣ ≥ |k1 − k2| − ∣∣p t ∣∣∣∣x t − τ − y t − τ ∣∣ − ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v × ∣∣f u, x σ1 u , . . . , x σn u − f ( u, y σ1 u , . . . , y σn u )∣∣dudsdv ≥ |k1 − k2| − c1 c2 ∥∥x − y∥∥ − 2 ∫ ∞ T3 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv, t ≥ T3, 2.23 which means that ∥∥x − y∥∥ ≥ 1 1 c1 c2 ( |k1 − k2| − 2 ∫ ∞ T3 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ) > 0, 2.24and Applied Analysis 7 In order to prove that 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M , we prove only that x / y. In fact, by 2.6 and 2.7 , we gain that for t ≥ T3 x t k1 − p t x t − τ ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v f u, x σ1 u , . . . , x σn u dudsdv, y t k2 − p t y t − τ ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v f ( u, y σ1 u , . . . , y σn u ) dudsdv, 2.22 which together with 2.1 – 2.3 imply that ∣∣x t − y t ∣∣ ≥ |k1 − k2| − ∣∣p t ∣∣∣∣x t − τ − y t − τ ∣∣ − ∫ ∞ t ∫ ∞ v ∫ ∞ s 1 α s β v × ∣∣f u, x σ1 u , . . . , x σn u − f ( u, y σ1 u , . . . , y σn u )∣∣dudsdv ≥ |k1 − k2| − c1 c2 ∥∥x − y∥∥ − 2 ∫ ∞ T3 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv, t ≥ T3, 2.23 which means that ∥∥x − y∥∥ ≥ 1 1 c1 c2 ( |k1 − k2| − 2 ∫ ∞ T3 ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ) > 0, 2.24 by 2.1 and 2.21 . That is, x / y. This completes the proof. Theorem 2.2. Assume that there exist constantsM,N, c1, c2, T0 and a function h ∈ C t0, ∞ ,R satisfying 2.3 , 2.4 and 0 < 1 − c2 N < 1 − c1 M; 2.25 0 ≤ c2 ≤ p t ≤ c1 < 1, t ≥ T0 > t0. 2.26 Then 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . Proof. Set k ∈ c1M N,M c2N . It follows from 2.4 that there exists T > T0 such that ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < min{M c2N − k, k − c1M −N}. 2.27 8 Abstract and Applied Analysis Define two mappings S1k and S2k : A N,M → C t0, ∞ ,R by 2.6 and 2.7 , respectively. By virtue of 2.3 , 2.6 , 2.7 , 2.26 , and 2.27 , we obtain that S1kx t ( S2ky ) t ≤ k − p t x t − τ ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ k − c2N min{M c2N − k, k − c1M −N} ≤ M, x, y ∈ A N,M , t ≥ T, 2.28 S1kx t ( S2ky ) t ≥ k − p t x t − τ − ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≥ k − c1M −min{M c2N − k, k − c1M −N} ≥ N, x, y ∈ A N,M , t ≥ T, 2.29 which yield that S1k A N,M S2k A N,M ⊆ A N,M . By a similar argument used in the proof of Theorem 2.1, we gain that S1k is a contraction mapping S2k is completely continuous, and 1.7 possesses uncountably many nonoscillatory solutions. This completes the proof. Theorem 2.3. Assume that there exist constantsM,N, c1, c2, T0 and a function h ∈ C t0, ∞ ,R satisfying 2.3 , 2.4 , 2.25 , and −1 < −c1 ≤ p t ≤ −c2 ≤ 0, t ≥ T0 > t0. 2.30 Then 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . Proof. Set k ∈ 1 − c2 N, 1 − c1 M . It follows from 2.4 that there exists T > T0 satisfying ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < min{ 1 − c1 M − k, k − 1 − c2 N}. 2.31 Let S1k and S2k : A N,M → C t0, ∞ ,R be defined by 2.6 and 2.7 , respectively. In view of 2.3 , 2.6 , 2.7 , 2.30 , and 2.31 , we obtain that S1x t ( S2y ) t ≤ k − p t x t − τ ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ k c1M min{ 1 − c1 M − k, k − 1 − c2 N} ≤ M, x, y ∈ A N,M , t ≥ T, S1x t ( S2y ) t ≥ k − p t x t − τ − ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≥ k c2N −min{ 1 − c1 M − k, k − 1 − c2 N} ≥ N, x, y ∈ A N,M , t ≥ T, 2.32 which mean that S1k A N,M S2k A N,M ⊆ A N,M . Abstract and Applied Analysis 9 The rest of the proof is similar to the proof of Theorem 2.1 and is omitted. This completes the proof. Theorem 2.4. Assume that there exist constants M and N with M > N > 0 and a function h ∈ C t0, ∞ ,R satisfying 2.3 and 2.4 . If p t 1 for each t ∈ t0, ∞ , then 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . Proof. Set k ∈ N,M . It follows from 2.4 that there exists T ≥ t0 τ satisfying ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < min{M − k, k −N}. 2.33and Applied Analysis 9 The rest of the proof is similar to the proof of Theorem 2.1 and is omitted. This completes the proof. Theorem 2.4. Assume that there exist constants M and N with M > N > 0 and a function h ∈ C t0, ∞ ,R satisfying 2.3 and 2.4 . If p t 1 for each t ∈ t0, ∞ , then 1.7 possesses uncountably many bounded nonoscillatory solutions in A N,M . Proof. Set k ∈ N,M . It follows from 2.4 that there exists T ≥ t0 τ satisfying ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < min{M − k, k −N}. 2.33 Define a mapping Sk : A N,M → C t0, ∞ ,R by Skx t ⎧ ⎪⎨ ⎪⎩ k ∞ ∑ i 1 ∫ t 2iτ t 2i−1 τ ∫ ∞ v ∫ ∞ s 1 α s β v f u, x σ1 u , . . . , x σn u dudsdv, t ≥ T, Skx T , t0 ≤ t < T 2.34 for x ∈ A N,M . Firstly, we prove that Sk A N,M ⊆ A N,M . By virtue of 2.3 , 2.33 and 2.34 , we obtain that Skx t ≤ k ∞ ∑ i 1 ∫ t 2iτ t 2i−1 τ ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ k ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ M, x ∈ A N,M , t ≥ T, Skx t ≥ k − ∞ ∑ i 1 ∫ t 2iτ t 2i−1 τ ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≥ k − ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≥ N, x ∈ A N,M , t ≥ T, 2.35 which imply that Sk A N,M ⊆ A N,M . 10 Abstract and Applied Analysis Secondly, we show that Sk is continuous inA N,M . Let x ∈ A N,M and {xm}m≥1 ⊆ A N,M with limm→ ∞xm x. By 2.34 , we get that | Skxm t − Skx t | ≤ ∞ ∑ i 1 ∫ t 2iτ t 2i−1 τ ∫ ∞ v ∫ ∞ s 1 α s β v ∣∣f u, xm σ1 u , . . . , xm σn u −f u, x σ1 u , . . . , x σn u ∣∣dudsdv ≤ ∫ ∞ T ∫ ∞ v ∫ ∞ s 1 α s β v ∣∣f u, xm σ1 u , . . . , xm σn u −f u, x σ1 u , . . . , x σn u ∣∣dudsdv, t ≥ T, m ≥ 1. 2.36 In view of 2.12 , 2.13 , 2.36 , and the Lebesgue dominated convergence theorem, we deduce that lim m→ ∞ ‖Skxm − Skx‖ 0, 2.37 which means that Sk is continuous in A N,M . Thirdly, we show that Sk A N,M is relatively compact. From 2.3 , 2.33 , and 2.34 , we gain that ‖Skx‖ ≤ k ∫ ∞ T ∫ ∞ v ∫ ∞ s h u α s β v dudsdv ≤ 2M, x ∈ A N,M , 2.38 which means that Sk A N,M is uniform bounded. Let ε > 0. It follows from 2.4 that there exist T > T ∗ > T such that ∫ ∞ T∗ ∫ ∞ v ∫ ∞ s h u α s β v dudsdv < ε 2 , 2.39