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By J. Baker, C. Cleaver, J. Diestel, G. Bennett, S.Y. Chang, D.E. Marshall, J.A. Cima, W. Davis, W.J. Davis, W.B. Johnson, J.B. Garnett, J. Johnson, J. Wolfe, H.E. Lacey, D.R. Lewis, A.L. Matheson, P. Orno, J.W. Roberts

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Example text

I , gi ∈ (0, ∞) and for positive numbers α , α > 0 We suppose first that f0s , gs0 , f∞ 1 2 ∞ such that α1 + α2 = 1. We define the positive numbers L1 , L2 , L3 , and L4 by L1 = α1 L3 = α2 r(αξm +β) d r(γ ηn +δ) e T ξm T ηn −1 i (T − s)c(s)f∞ ds , L2 = α1 −1 (T − s)d(s)gi∞ ds , L4 = α2 αT +β d γT + δ e T 0 −1 (T − s)c(s)f0s ds T 0 , −1 (T − s)d(s)gs0 ds . 1. Assume that (I1)–(I3) hold, f0s , gs0 , f∞ 1 2 ∞ with α1 +α2 = 1, L1 < L2 , and L3 < L4 . Then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ) there exists a positive solution (u(t), v(t)), t ∈ [0, T] for (S0 )–(BC0 ).

So we obtain f (t, x) ≥ C8 xβ1 , g(t, x) ≥ ε1 xβ2 , ∀ (t, x) ∈ [σ , 1 − σ ] × [0, x5 ]. 54) From the assumption q2 (0) = 0 and the continuity of q2 , we deduce that there exists sufficiently small ε2 ∈ (0, min{x5 , 1}) such that q2 (x) ≤ β0−1 x5 for all x ∈ [0, ε2 ]. Therefore, for any u ∈ ∂Bε2 ∩ P0 and s ∈ [0, 1] we have 1 0 G2 (s, τ )g(τ , u(τ )) dτ ≤ β0−1 x5 1 0 J2 (τ )p2 (τ ) dτ = x5 . 7, for any t ∈ [σ , 1 − σ ] we obtain (Du)(t) ≥ C8 1−σ G1 (t, s) σ 1−σ ≥ C8 ν1 σ β 1−σ σ β1 G2 (s, τ )g(τ , u(τ )) dτ 1−σ J1 (s) (ε1 ν2 )β1 β β ≥ C8 ν1 ν2 1 ε1 1 ν β1 β2 θ1 θ2 1 u σ β1 β2 ds J2 (τ )(u(τ ))β2 dτ β1 ds ≥ u .

M − 2; ci , ηi ∈ R for all i = 1, . . , n − 2; 0 < ξ1 < · · · < ξm−2 < T, 0 < η1 < · · · < ηn−2 < T.

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