Existence of Positive Solutions of Neutral Differential Equations

and Applied Analysis 3 We now define two maps S1 and S2 : Ω → C t0,∞ , R as follows: S1x t ⎧⎨ ⎩ a t x t − τ , t ≥ t1, S1x t1 , t0 ≤ t ≤ t1, S2x t ⎪⎨ ⎪⎩ − 1 n − 1 ! ∫∞ t s − t n−1p s f x s − σ ds, t ≥ t1, S2x t1 v t − v t1 , t0 ≤ t ≤ t1. 2.5 We will show that for any x, y ∈ Ω we have S1x S2y ∈ Ω. For every x, y ∈ Ω and t ≥ t1 we obtain S1x t ( S2y ) t ≤ a t v t − τ − 1 n − 1 ! ∫∞ t s − t n−1p s f u s − σ ds ≤ v t . 2.6 For t ∈ t0, t1 we have S1x t ( S2y ) t S1x t1 ( S2y ) t1 v t − v t1 ≤ v t1 v t − v t1 v t . 2.7 Furthermore for t ≥ t1 we get S1x t ( S2y ) t ≥ a t u t − τ − 1 n − 1 ! ∫∞ t s − t n−1p s f v s − σ ds ≥ u t . 2.8 Finally let t ∈ t0, t1 and with regard to 2.2 we get v t − v t1 u t1 ≥ u t , t0 ≤ t ≤ t1. 2.9 Then for t ∈ t0, t1 and any x, y ∈ Ω we get S1x t ( S2y ) t S1x t1 ( S2y ) t1 v t − v t1 ≥ u t1 v t − v t1 ≥ u t . 2.10 Thus, we have proved that S1x S2y ∈ Ω for any x, y ∈ Ω. We will show that S1 is a contraction mapping on Ω. For x, y ∈ Ω and t ≥ t1 we have ∣∣ S1x t − (S1y) t ∣∣ |a t |∣∣x t − τ − y t − τ ∣∣ ≤ c∥∥x − y∥∥. 2.11 This implies that ∥∥S1x − S1y∥∥ ≤ c∥∥x − y∥∥. 2.12 4 Abstract and Applied Analysis Also for t ∈ t0, t1 the inequality above is valid. We conclude that S1 is a contraction mapping on Ω. We now show that S2 is completely continuous. First we will show that S2 is continuous. Let xk xk t ∈ Ω be such that xk t → x t as k → ∞. Because Ω is closed, x x t ∈ Ω. For t ≥ t1 we have | S2xk t − S2x t | ≤ 1 n − 1 ! ∣∣∣ ∫∞ t s − t n−1p s [f xk s − σ − f x s − σ ]ds ∣∣∣ ≤ 1 n − 1 ! ∫∞ t1 s − t1 n−1p s ∣∣f xk s − σ − f x s − σ ∣∣ds. 2.13 According to 2.8 we get ∫∞ t1 s − t1 n−1p s f v s − σ ds < ∞. 2.14 Since |f xk s − σ − f x s − σ | → 0 as k → ∞, by applying the Lebesgue dominated convergence theorem we obtain that lim k→∞ ‖ S2xk t − S2x t ‖ 0. 2.15 This means that S2 is continuous. We now show that S2Ω is relatively compact. It is sufficient to show by the ArzelaAscoli theorem that the family of functions {S2x : x ∈ Ω} is uniformly bounded and equicontinuous on t0,∞ . The uniform boundedness follows from the definition of Ω. For the equicontinuity we only need to show, according to Levitan result 8 , that for any given ε > 0 the interval t0,∞ can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have change of amplitude less than ε. With regard to the condition 2.14 , for x ∈ Ω and any ε > 0 we take t∗ ≥ t1 large enough so that 1 n − 1 ! ∫∞ t∗ s − t1 n−1p s f x s − σ ds < ε 2 . 2.16 Then for x ∈ Ω, T2 > T1 ≥ t∗ we have | S2x T2 − S2x T1 | ≤ 1 n − 1 ! ∫∞ T2 s − t1 n−1p s f x s − σ ds 1 n − 1 ! ∫∞ T1 s − t1 n−1p s f x s − σ ds