Electronic Journal of Qualitative Theory of Differential Equations 212, No. 18, 1-9; http://www.math.u-szeged.hu/ejqtde/ Existence and iteration of monotone positive solutions for third-order nonlocal BVPs involving integral conditions Hai-E Zhang 1, Jian-Ping Sun 2 1. Department of Basic Science, Tangshan College, Tangshan, Hebei 63, People s Republic of China 2. Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 735, People s Republic of China Abstract This paper is concerned with the existence of monotone positive solution for the following third-order nonlocal boundary value problem u (t)+f (t,u(t),u (t)) =, < t < 1; u() =, au () bu () = α[u], cu (1) + du (1) = β[u], where f C([,1] R + R +,R + ), α[u] = u(t)db(t) are linear functionals on C[,1] given by Riemann-Stieltjes integrals. By applying monotone iterative techniques, we not only obtain the existence of monotone positive solution but also establish an iterative scheme for approximating the solution. An example is also included to illustrate the main results. Keywords: Monotone iterative method; Positive solutions; Nonlocal; Integral conditions 2 AMS Subject Classification: 34B1, 34B15 u(t)da(t) and β[u] = 1 Introduction Third-order differential equation arises in a variety of different areas of applied mathematics and physics, e.g., in the deflection of a curved beam having a constant or varying cross section, a three layer beam, electromagnetic waves or gravity driven flows and so on [1]. BVPs with Stieltjes integral boundary condition (BC for short) have been considered recently as both multipoint and integral type BCs are treated in a single framework. For Supported by the National Natural Science Foundation of China (18168). Corresponding author. E-mail: haiezhang@126.com EJQTDE, 212 No. 18, p. 1
more comments on Stieltjes integral BC and its importance, we refer the reader to the papers by Webb and Infante [2, 3, 4] and their other related works. In recent years, thirdorder nonlocal BVPs have received much attention from many authors, see, for example [5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2] and the references therein. For fourthorder or higher-order nonlocal BVPs, one can refer to [3, 4, 21, 22]. In particular, it should be pointed out that Webb and Infante in [2], gave a unified approach for studying the existence of multiple positive solutions of second-order BVPs subject to various nonlocal BCs. In [3], they extended their method to cover equations of order N with any number up to N of nonlocal BCs in a single theory. Recently, iterative methods have been successfully employed to prove the existence of positive solutions of nonlinear BVPs for ordinary differential equations, see [23, 24, 25, 26, 27] and the references therein. It is worth mentioning that, Sun et al. [24] obtained the existence of monotone positive solutions for third-order three-point BVPs, the main tools used were monotone iterative techniques. Inspired by the above mentioned excellent works, the aim of this paper is to investigate the existence and iteration of monotone positive solution for the following BVP u (t) + f (t, u (t),u (t)) =, < t < 1, u () =, au () bu (1.1) () = α[u], cu (1) + du (1) = β[u], where f C([, 1] R + R +, R + ), α[u] = u(t)da(t) and β[u] = u(t)db(t) are linear functionals on C[, 1] given by Riemann-Stieltjes integrals and a, b, c, d are nonnegative constants with ρ := ac + ad + bc >. By a positive solution of BVP (1.1), we understand a solution u(t) which is positive on t (, 1) and satisfies BVP (1.1). By applying monotone iterative techniques, we construct a successive iterative scheme whose starting point is a zero function, which is very useful and feasible for computational purpose. An example is also included to illustrate the main results. 2 Preliminary lemmas In this section, the ideas and the method we will adopt, which have been widely used, are due to Webb and Infante in [2, 3]. In our case, the existence of positive solutions of nonlocal BVP (1.1) with two nonlocal boundary terms α[u], β[u], can be studied, via a perturbed Hammerstein integral equation of the type u(t) = γ(t)α[u] + δ(t)β[u] + Here γ(t), δ(t) are linearly independent and given by G (t, s)f (s, u(s), u (s))ds =: Tu(t). (2.1) γ (t) =, γ () =, aγ () bγ () = 1, cγ (1) + dγ (1) =, δ (t) =, δ () =, aδ () bδ () =, cδ (1) + dδ (1) = 1, EJQTDE, 212 No. 18, p. 2
which imply γ(t) = 2ct+2dt ct2 and δ(t) = at2 +2bt, t [, 1]. A direct calculation shows that 2ρ 2ρ for t [θ, 1], γ(t) c 1 γ and δ(t) c 2 δ ( is the usual supremum norm in C[, 1]), where c 1 = 2cθ+2dθ cθ2 c+2d and c 2 = aθ2 +2bθ a+2b ; G(t, s) is the Green s function for the corresponding problem with local terms when α[u] and β[u] are identically. We first make the following hypotheses on the Green s function: (H1) The kernel G is measurable, non-negative, and for every τ [, 1] satisfies lim G(t, s) G(τ, s) = fors [, 1]. t τ (H2) There exist a subinterval [a, b] [, 1], a measurable function Φ, and a constant c 3 (, 1] such that G(t, s) Φ(s) fort [, 1], s [, 1], G(t, s) c 3 Φ(s) fort [a, b], s [, 1]. (H3) A, B are functions of bounded variation, and K A (s), K B (s) for s [, 1], where K A (s) := G (t, s) da(t) and K B (s) := G (t, s) db(t). In the remainder of this paper, we always assume that α[γ], β[δ] < 1, α[δ], β[γ] and D := (1 α[γ])(1 β[δ]) α[δ]β[γ] >. As shown in Theorem 2.3 in [3], if u is a fixed point of T in (2.1), then u is a fixed point of S, which is now given by Su(t) := γ(t) ( (1 β[δ]) K A (s)f (s, u (s),u (s)) ds + α[δ] D + δ(t) ( β[γ] K A (s)f (s, u (s), u (s)) ds + (1 α[γ]) D + G(t, s)f (s, u (s), u (s))ds =: G S (t, s)f (s, u (s), u (s))ds ) K B (s)f (s, u (s),u (s)) ds ) K B (s)f (s, u (s),u (s)) ds in our case. The kernel G S is the Green s function corresponding to the BVP (1.1). Lemma 2.1 Let ρ := ac + ad + bc >. Then the Green s function G(t, s) satisfies (H1), (H2) with [a, b] = [θ, 1], c 3 = ρ R θ Φ(τ)dτ, < θ < 1. (a+b)(c+d) Proof. A direct calculation shows that, G (t, s) = { (at 2 +2bt)(c(1 s)+d) For any fixed s [, 1], it is easy to see that (t s)2, s t 1, 2ρ 2 (at 2 +2bt)(c(1 s)+d), t s 1. 2ρ G 1 (t, s) := G (t, s) t = 1 ρ { (b + as) (d + c(1 t)), s t 1, (b + at) (d + c(1 s)), t s 1, (2.2) EJQTDE, 212 No. 18, p. 3
which shows that and so, G (t, s) = G 1 (t, s) 1 (b + as) (d + c(1 s)) =: Φ(s), for (t, s) [, 1] [, 1], (2.3) ρ On the other hand, so, Thus, G (t, s) = G 1 (t, s) Φ(s) G 1 (τ, s)dτ = Φ(s)dτ = Φ(s)t Φ(s), for (t, s) [, 1] [, 1]. (2.4) { (b+as)(d+c(1 t)) = (b+at)(d+c(1 t)) (b+as)(d+c(1 s)) (b+at)(d+c(1 s)) (b+at)(d+c(1 s)) (b+as)(d+c(1 s)) = (b+at)(d+c(1 t)) (b+as)(d+c(1 t)) G 1 (t, s) G 1 (τ, s) dτ ρφ(t), s t 1, (a+b)(c+d) ρφ(t), t s 1, (2.5) (a+b)(c+d) ρφ(t) Φ(s), for (t, s) [, 1] [, 1]. (2.6) (a + b)(c + d) ρφ(τ) (a + b)(c + d) Φ(s)dτ ρ θ Φ(τ)dτ Φ(s), for (t, s) [θ, 1] [, 1]. (a + b)(c + d) (2.7) Lemma 2.2 G S (t, s) satisfies (H1), (H2) for a function Φ 1, the same interval [θ, 1], and the constant c = min {c 1, c 2, c 3 }. Proof. Let Φ 1 (s) := γ ((1 β[δ])k D A(s) + α[δ]k B (s))+ δ (β[γ]k D A(s) + (1 α[γ])k B (s))+ Φ(s), s [, 1]. The proof is same to the Theorem 2.4 in [2], so omitted. Moreover, we easily know that G S (t, s) t for a function Φ 2, i.e., for s [, 1], Φ 2 (s) := γ D [(1 β[δ])k A(s) + α[δ]k B (s)] + δ D Φ 2 (s), t, s [, 1] [, 1] (2.8) [β[γ]k A(s) + (1 α[γ])k B (s)] + Φ(s). We will use the classical Banach space E = C 1 [, 1] equipped with the norm u = max { u, u }, where u is the usual supremum norm in C[, 1]. Let P = {u E : u (t) } and let c be same as in Lemma 2.2, then define { } K = u P : min u (t) c u and u (t), t [, 1]. t [θ,1] Then it is to verify that P and K are cones in E. Note that this induces an order relation in E by defining u v if and only if v u K. Similar to the proofs of lemma 2.6, 2.7 and 2.8 in [2], we can get the following lemmas. EJQTDE, 212 No. 18, p. 4
Lemma 2.3 The maps T, S : P E are compact. Lemma 2.4 T : K K and S : P K. Lemma 2.5 T and S have the same fixed points (in K). 3 Main results Now we apply monotone iterative techniques to seek solution of BVP (1.1) as fixed point of the integral operator S. { } Theorem 3.1 Let σ = max max Φ 1(s), max Φ 2(s). Assume that f(t,, ) for t [, 1] s [,1] s [,1] and there exists a constant r > such that f(t, u 1, v 1 ) f(t, u 2, v 2 ) r σ, t 1, u 1 u 2 r, v 1 v 2 r. (3.1) If we construct an iterative sequence v n+1 = Sv n, n =, 1, 2,..., where v (t) = for t [, 1], then {v n } n= converges to v in C 1 [, 1], which is a monotone positive solution of the BVP (1.1) and satisfies < v (t) r for t (, 1], (v ) (t) r for t [, 1]. Proof. Let K r = {u K : u < r}. We assert that S : K r K r. In fact, if u K r, then u(s) u u r, u (s) u u r, for s [, 1], which together with the condition (3.1) and Lemma 2.2 and (2.8) implies that (Su) (t) = (Su) (t) = G S (t, s)f (s, u (s),u (s))ds r, t [, 1], G S (t, s) f (s, u (s), u (s)) ds r, t [, 1]. t Hence, we have shown that S : K r K r. Now, we assert that {v n } n= converges to v in C 1 [, 1], which is a monotone positive solution of the BVP (1.1) and satisfies < v (t) r for t (, 1], (v ) (t) r for t [, 1]. In fact, in view of v K r and S : K r K r, we have that v n K r, n = 1, 2,.... Since the set {v n } n= is bounded and T is completely continuous, we know that {v n} n= is relatively compact. EJQTDE, 212 No. 18, p. 5
In what follows, we prove that {v n } n= is monotone by induction. Firstly, by v = and S : P K, we easily know v 1 v K, which shows that v v 1. Next, we assume that v k 1 v k. Then, in view of Lemma 2.2 and (3.1), we have and v k+1 (t) v k (t) = v k+1 (t) v k (t) = which imply that G S (t, s) [ f (s, v k (s), v k (s)) f ( s, v k 1 (s), v k 1 (s))] ds Φ 1 (s) [ f (s, v k (s), v k(s)) f ( s, v k 1 (s), v k 1(s) )] ds, t [, 1] G S (t, s) [ f (s, v k (s), v k (s)) f ( s, v k 1 (s), v k 1 (s))] ds c Φ 1 (s) [ f (s, v k (s), v k (s)) f ( s, v k 1 (s), v k 1 (s))] ds, t [θ, 1], v k+1 (t) v k (t) c v k+1 v k, t [θ, 1]. (3.2) At the same time, by Lemma 2.2, (2.8) and (3.1), we also have v k+1 (t) v k (t) = G S (t, s) t [ f (s, vk (s), v k (s)) f ( s, v k 1 (s), v k 1 (s))] ds, t [, 1].(3.3) It follows from (3.2) and (3.3) that v k+1 (t) v k (t) K, which shows that v k v k+1. Thus, we have shown that v n v n+1, n =, 1, 2.... Since {v n } n= is relatively compact and monotone, there exists a v K r such that v n v (n ), which together with the continuity of S and the fact that v n+1 = Sv n implies that v = Sv. Moreover, in view of f(t,, ) for t (, 1), we know that the zero function is not a solution of BVP (1.1). Thus, v >. So, it follows from v K r that < v (t) r for t (, 1], (v ) (t) r for t [, 1]. 4 An example Consider the BVP u (t) + 1 2 tu + 1 8 u 2 + 1 =, < t < 1, u () =, u () = α[u], u (1) = β[u], where α[u] = (1 s)u(s)ds and β[u] = su(s)ds are nonlocal BCs of integral type. For this BCs the corresponding γ(t) = 2t t2 and δ(t) = t2. By simple calculation shows that 2 2 α[γ] = 1 8, α[δ] = 1 24, β[γ] = 5 24, β[δ] = 1 19, D = (1 α[γ])(1 β[δ]) α[δ]β[γ] = 8 144, (4.1) EJQTDE, 212 No. 18, p. 6
K A (s) := G (t, s) (1 t)dt = s 8 s2 4 + s3 6 s4 24, K B(s) := G (t, s)tdt = 5s 24 s2 4 + s4 24, Φ 1 (s) = 265s 218 145s2 19 + 13s3 19 s4 218, Φ 2(s) = 156s 19 181s2 19 + 26s3 19 s4 19, { } and σ = max max Φ 1 (s), max Φ 2 (s).333. Then all the hypotheses of Theorem 3.1 are s [,1] s [,1] fulfilled with r = 1. It follows from Theorem 3.1 that the BVP (4.1) has a monotone positive solution v satisfying < v (t) 1 for t (, 1], (v ) (t) 1 for t [, 1]. Moreover, the iterative scheme is where v (t) =, t [, 1], [ ]( 2ts t 2 s s 2 1 v n+1 (t) = + g(t, s) 2 2 sv n(s) + 1 ) 8 (v n(s)) 2 + 1 ds [ ]( t 2 (1 s) 1 + + g(t, s) t 2 2 sv n(s) + 1 ) 8 (v n(s)) 2 + 1 ds, t [, 1], n = 1, 2,..., ( 1 v n+1 (t) = [s(1 t) + g t (t, s)] 2 sv n(s) + 1 ) 8 (v n (s))2 + 1 ds ( 1 + [t(1 s) + g t (t, s)] 2 sv n(s) + 1 ) 8 (v n (s))2 + 1 ds, t [, 1], n = 1, 2,.... g(t, s) = t ( 126t 19 48t2 19 ) ( s 8 s2 4 + s3 6 s4 24 ) ( s 8 s2 4 + s3 6 s4 24 ) ( ) ( 6t + 19 + 6t2 5s 19 24 s2 4 + s4 24 ) ( 6 + 19 + 12t ) ( 5s 19 24 s2 4 + s4 24 ( 126 g t (t, s) = 19 96t 19 for t, s [, 1] [, 1]. The first, second and third terms of the scheme v n and v n are as follows: v (t) =, v 1 (t) = 7 436 t + 142 545 t2 1 6 t3, v 2 (t) = 25546447517 1566467784 t + 5729566951 217564872 t2 56939 341536 t3 497 57288 t4 379849 57288 t5 71 138 t6 + 1 432 t7, v (t) =, v 1(t) = 7 436 + 284 545 t 1 2 t2, v 2(t) = 25546447517 1566467784 + 5729566951 18782436 t 152817 341536 t2 497 142572 t3 379849 114576 t4 71 218 t5 + 1 576 t6. ), ), EJQTDE, 212 No. 18, p. 7
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