Ngày tải lên :
22/06/2014, 22:20
... q1,τ (η) p0,τ (s) a(s)x(s)ds p0,τ (1) + ≥ q1,τ (η) p0,τ (1) − γ p0,τ (η) (Lτ x)(η) = η p0,τ (η) (2 .18) p0,τ (s)a(s)x(s)ds, η q1,τ (η) p0,τ (s) a(s)x(s)ds p0,τ (1) + p0,τ (η) p0,τ (1) − γ p0,τ (η) ... η p0,τ (η) q1,τ (s) a(s)x(s)ds q1,τ (0) η q1,τ (s)a(s)x(s)ds (2.19) X Xian and D O’Regan By (2 .18) and Lemma 2.6, we have for any t ∈ [0,η], Lτ x (t) = t qη,τ (t) p0,τ (s) a(s)x(s)ds p0,τ (η) ... and (3.17), we have Tλ x (t) ≤ Tλ x0 (t) + τ0 τ LλM z0 (t) ≤ u∗ (t) − LλM z0 (t), 2 t ∈ [0,1], (3 .18) for any x ∈ QλM with x − x0 < δ This implies that x ∈ Ωλ , and so Ωλ is an open set Now we will...