“…Moreover, we have α (x) + q(x) f (x, α(x), α (x)) = 2(1 + x) e x arctan(x + 1) > 0, x ∈ (0, +∞), ) + q(x) f (x, β(x), β (x)) = 2(1 + x) e x arctan(−x − 1) < 0, x ∈ (0, +∞),Thus α, β are lower and upper solutions of (30), respectively. Furthermore, α, β ∈ X, α(x) ≤ β(x), x ∈ [0, +∞).Clearly, f is continuous, moreover, f satisfies the Naguma's condition with respect to α(x) = −x − 1 and β(x) = x + 1; that is, when 0 ≤x < +∞, −x − 1 ≤ u ≤ x + 1 and v ∈ R, it holds | f (x, u, v)| ≤ φ(x)h(|v|),whereφ(x) = (1 + x) and h(v) + x) γ q(x)φ(x) = sup x∈[0,+∞) 65 < ∞,that is,(12) and(13) are satisfied. Therefore, from Theorem 3.1, the boundary problem (30) has at least one solution u such thatα(x) = −x − 1 ≤ u(x) ≤ x + 1 = β(x), |u (x)| < N for all x ∈ [0, +∞),where N > e 108π e 2 (r 2 + 1) − 1, r ≥ 1 with γ = 2.…”