# Brownian motion and martingales in analysis by Richard Durrett By Richard Durrett

This booklet could be of curiosity to scholars of arithmetic.

Best calculus books

Extra resources for Brownian motion and martingales in analysis

Example text

2 Positive Periodic Solutions of the Equation x (t) = a(t)x(t) − τ f (t, x(h(t))) 35 c2 MT , . (1 − λ)(T − λ1 ) λ c4 > max Now for x ∞ K c4 , we have t+T ≥Aτ x≥ = sup τ 0≤t≤T G(t, s) f (s, x(h(s))) ds t t+T ≤ sup τ 0≤t≤T G(t, s)(a(s)x(h(s)) λ1 + M) ds t t+T ≤ sup τ 0≤t≤T ⎡ G(t, s)(a(s)≥x≥ λ1 + M) ds t t+T t+T ≤ τ ⎣ λ1 c4 sup G(t, s) ds ⎦ a(s)G(t, s) ds + sup M 0≤t≤T 0≤t≤T t ⎤ t M T 1−λ ≤ τ λ1 c4 + < c4 . Hence, Aτ : K c4 → K c4 . Next, we define a nonnegative concave continuous functional ψ on K by ψ(x) = min x(t).

A(β, x(β)) dβ − 1 0 In view of the above, we define an operator Aτ by t+T Aτ x = τ G(t, s) f (s, x(s − γ1 (s, x(s))), . . 30) t for every x ∞ X and t ∞ R. Clearly Aτ x(t + T ) = Aτ x(t) and Aτ : X → X . The Green’s kernel G(t, s) satisfies the inequality T c(β) dβ exp 1 ψ= T exp c(β) dβ − 1 0 0 ≤ |G(t, s)| ≤ T exp b(β) dβ − 1 0 for every 0 ≤ t ≤ s ≤ t + T . Let k1 = exp ⎜ T exp 0 c(β) dβ . Then, ψ= =ω ⎜ T 0 b(β) dβ and k2 = k2 k2 (k2 − 1) 1 ω , ω= , k1 ≤ k2 , and λ = = > 1. k2 − 1 k1 − 1 ψ (k1 − 1) For any x ∞ X , we have 48 2 Positive Periodic Solutions of Nonlinear Functional Differential Equations τk2 ≥Aτ x≥ ≤ k1 − 1 T f (s, x(s − γ1 (s, x(s))), .

10 Let τ = 1, f˜0 < 1, and f˜∈ < 1. In addition, assume that there exists ρ > 0 such that f (t, x) > a(t)|x| for μρ < |x| < ρ, where μ = T ex p{− 0 a(s) ds}. 5) has at least two positive T -periodic solutions x1 and x2 such that 0 < ≥x1 ≥ < ρ < ≥x2 ≥. 9. 1 Let f˜0 < T and f˜∈ < T , and assume that there exists a constant c2 > 0 such that either (H1 ) or (H8 ) holds. 5) has at least three positive T -periodic solutions for 1 λ−1 <τ < . 10. 10 it is 1. 1. 11 Let f˜∈ < T . Assume that there are constants 0 < c1 < c2 such that (H1 ) holds and (H9 ) f (t, x) < x for 0 ≤ x ≤ c1 and 0 ≤ t ≤ T .