By Prössdorf, Silbermann
Publication by way of Prössdorf, Silbermann
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Extra info for Numerical analysis for integral and related operator equations
Example text
We will consider two cases, namely γ 0 and −2 < γ < 0. Case (i). γ 0. 20 with g(y) = Ay −θ q(t) = 1, and h(y) = μ2 . 99) hold. Let n0 = 1, ρn = Notice for n ∈ {1, 2, . }, A 2(n+1)γ (μ2 + 1) 1 2n+1 t 1/θ 1 and 0 < y and k0 = 1. 96) is satisfied. Finally notice for c > 0 that c 1 {1 + h(c) g(c) } 0 1 cθ+1 du = , g(u) θ + 1 A + μ2 cθ so lim c→∞ c 1 {1 + h(c) g(c) } 0 du = ∞. 105) holding. 20. Case (ii). −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.
96) holds. 74) holding. 23 to obtain our next result. 24. Let n0 ∈ {1, 2, . 112) hold. 62) has a solution y ∈ C[0, 1] ∩ C 2 (0, 1) with y(t) > 0 for t ∈ (0, 1). 96) we replace holds. 25. Let n0 ∈ {1, 2, . 96) (with 2n+1 t 1 replaced by 2n+1 t 1 − 2n+1 ) hold. 62) has a solution 2 y ∈ C[0, 1] ∩ C (0, 1) with y(t) > 0 for t ∈ (0, 1). 4. 121) with A > 0, κ > −1, θ > 0, 0 γ < 1, 0 a0 < π 2 , b0 0 and μ ∈ R. 121) has a solution y ∈ C[0, 1] ∩ C 2 (0, 1) with y(t) > 0 for t ∈ (0, 1). P. Agarwal and D.
36) hold. 25) has two solutions y1 , y2 ∈ C[0, 1] ∩ C 2 (0, 1) with y1 > 0, y2 > 0 on (0, 1) and |y1 |0 < r < |y2 |0 R. P ROOF. 6. 1. 61) has two solutions y1 , y2 ∈ C[0, 1] ∩ C 2 (0, 1) with y1 > 0, y2 > 0 on (0, 1) and |y1 |0 < 1 < |y2 |0 . P. Agarwal and D. 33) hold. Also note 2 b0 = max α+1 1 2 t (1 − t) dt, 0 2 α+1 1 α+1 , 1 1 2 g(u) = u−α and h(u) = uβ + 1. t (1 − t) dt = 1 . 34) holds (with r = 1) since r 1 {1 + h(r) g(r) } 0 1 r α+1 du = g(u) (1 + r α+β + r α ) α + 1 = 1 1 > b0 = . 36) holding.