By Hans Sagan

ISBN-10: 0395170907

ISBN-13: 9780395170908

**Read Online or Download Advanced calculus: Of real-valued functions of a real variable and vector-valued functions of a vector variable PDF**

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**Additional info for Advanced calculus: Of real-valued functions of a real variable and vector-valued functions of a vector variable**

**Example text**

62) has a solution y ∈ C[0, 1] ∩ C 2 (0, 1) with y(t) α(t) for t ∈ [0, 1]. P ROOF. In this case let en = 1 2n+1 ,1 − 1 2n+1 and θn (t) = max 1 , min t, 1 − 2n+1 1 2n+1 . 12. Let n0 ∈ {1, 2, . 93) and the following hold: ⎧ there exists a function β ∈ C[0, 1] ∩ C 2 (0, 1) ⎪ ⎪ ⎪ ⎪ with β(t) ρn0 for t ∈ [0, 1] and with ⎪ ⎪ ⎨ q(t)f t, β(t) + β (t) 0 for t ∈ (0, 1) and ⎪ 1 1 ⎪ ⎪ q(t)f 2n0 +1 , β(t) + β (t) 0 for t ∈ 0, 2n0 +1 and ⎪ ⎪ ⎪ ⎩ q(t)f 1 − 1 , β(t) + β (t) 0 for t ∈ 1 − 1 , 1 . 95) 2 α(t) for t ∈ [0, 1].

2n0 +1 2 Rn0 = sup q(x)f (x, y) : x ∈ β(t) for t ∈ [0, 1]. 88) Let 1 2n0 +1 ,1 − 1 2n0 +1 and α(x) y yn0 (x) . A survey of recent results for initial and boundary value problems The mean value theorem implies that there exists τ ∈ ( n01+1 , 1 − 2 2 sup[0,1] yn0 (t). 89) . The Arzela–Ascoli theorem guarantees the existence of a subsequence Nn0 of integers and a function zn0 ∈ C[ n01+1 , 1 − n01+1 ] with yn converging uniformly to zn0 on [ n01+1 , 1 − 1 2n0 +1 2 2 2 ] as n → ∞ through Nn0 . Similarly {yn }∞ n=n0 +1 is a bounded, equicontinuous family on 1 , 1 − n01+2 2n0 +2 2 , so there is a subsequence Nn0 +1 of Nn0 and a function zn0 +1 ∈ C 1 2n0 +2 ,1 − 1 2n0 +2 with yn converging uniformly to zn0 +1 on [ Note zn0 +1 = zn0 on [ n0 +1 , 1 − 2 subsequences of integers 1 1 2n0 +1 1 ,1 2n0 +2 − 1 ] 2n0 +2 as n → ∞ through Nn0 +1 .

2 < γ < 0. 20 with q(t) = t γ , g(y) = Ay −θ and h(y) = μ2 . P. Agarwal and D. 99) hold. 105) holding. Let n0 = 1, A 2 n(μ + 1) ρn = Notice for n ∈ {1, 2, . }, 1 2n+1 t and k0 = 1. 1 and 0 < y At γ − μ2 ρnθ q(t)f (t, y) 1/θ ρn that we have since γ < 0, A − μ2 = n μ2 + 1 − μ2 ρnθ μ2 + 1 − μ2 = 1. 96) is satisﬁed. 20. 3. 108) y(0) = y(1) = 0, where μ2 1. 108) has a solution y ∈ C[0, 1] ∩ C 2 (0, 1) with y(t) > 0 for t ∈ (0, 1). 14 with q ≡ 1, ρn = 1 n+1 2 (μ2 + a) 1 2 and k0 = a; here a > 0 is chosen so that a 18 .