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Electronic Journal of Differential Equations, Vol. 2003(2003), No. 21, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp) Minimal and maximal solutions for two-point boundary-value problems ∗ Myron K. Grammatikopoulos & Petio S. Kelevedjiev Abstract In this article we consider a boundary-value problem for the equation f (t, x, x0 , x00 ) = 0 with mixed boundary conditions. Assuming the existence of suitable barrier strips, and using the monotone iterative method, we obtain the minimal and maximal solutions. 1 Introduction We apply the monotone iterative method to obtain minimal and maximal solutions to the nonlinear boundary-value problem (BVP) f (t, x, x0 , x00 ) = 0, 0 ≤ a ≤ t ≤ b, x(a) = A, x0 (b) = B, (1.1) where the scalar function f (t, x, p, q) is continuous and has continuous first derivatives on suitable subsets of [a, b] × R3 . For results, which guarantee the existence of C 2 [a, b]-solutions to BVPs for the equation x00 = f (t, x, x0 , x00 )−y(t) with various linear boundary conditions, see [6, 7, 17, 18, 21, 22, 23]. Concerning the uniqueness results, we refer to [21]. A result, concerning the existence and uniqueness of C 2 [a, b]-solutions to the BVP for the equation x00 = f (t, x, x0 , x00 ), with general linear boundary conditions, can be found in [27]. The results of [19] guarantee the existence of W 2,∞ [a, b]-solutions or of C 2 [a, b]-solutions to the Dirichlet BVP for the equation f (t, x, x0 , x00 ) = 0, where the function f (t, x, p, q)is defined on [a, b] × Rn × Rn × Y , and Y is a non-empty closed connected or locally connected subset of Rn . Finally, the C 2 [a, b]-solvability of BVPs for the equation f (t, x, x0 , x00 ) = 0 with fully nonlinear boundary conditions is studied in [12]. Note that, in the literature, the monotone iterative method is applied on BVPs for equations of the forms x00 = f (t, x, x0 ) and (φ(x0 ))0 = f (t, x, x0 ) with various boundary conditions (see, for example, [2, 3, 4, 5, 9, 10, 11, 13, 15, 20, 26, 28]). The sequences of iterates, considered in [2, 3, 4, 5, 10, 13, 28], ∗ Mathematics Subject Classifications: 34B15. Key words: Boundary-value problems, minimal and maximal solutions, monotone method, barrier strips. c 2003 Southwest Texas State University. Submitted December 12, 2002. Published February 28, 2003. 1 2 Minimal and maximal solutions EJDE–2003/21 converge to the extremal solutions, while the sequences of iterates, considered in[9, 15, 19], converge to the unique solution. The first elements u0 (t) and v0 (t) of such sequences of iterates usually are lower and upper solutions respectively of the problems under consideration (see, for example, [2, 3, 4, 5, 10, 13, 28]. To derive the needed monotone iterates, the authors of [2, 3, 4, 5, 10, 13, 15, 28] use suitable growth conditions. For more applications of the monotone iterative method, see citeb1,l1,l3,s1,y1. In this article, following citek1, we obtain the extremal solutions to (1.1) under assumption of the existence of suitable barrier strips (see Remarks 2.1 and 2.2 below), which immediately imply the first iterates u0 (t) and v0 (t). A version of [12, Theorem 5.1] implies the existence of the next iterates, and a suitable comparison result guarantees the monotone properties for the sequences of iterates. Finally, the Arzela-Askoli’s theorem ensures the existence of the extremal solutions of the problem (1.1) as limits of the sequences of iterates. 2 Basic hypotheses The following four hypotheses will be a tool for obtaining our results. (H1) There are constants K > 0, F, F1 , L, L1 such that F a ≤ A ≤ La, F1 < F ≤ B ≤ L < L1 . For the set T := {(t, x) : a ≤ t ≤ b, F t ≤ x ≤ Lt}, we assume that f (t, x, p, q) + Kq ≥ 0 on (t, x, p, q) : (t, x) ∈ T, p ∈ [L, L1 ], q ∈ (−∞, 0) , and f (t, x, p, q) + Kq ≤ 0 on (t, x, p, q) : (t, x) ∈ T, p ∈ [F1 , F ], q ∈ (0, ∞) . Remark 2.1 Set Φ1 (t, x, p, q) ≡ f (t, x, p, q) + Kq. Then, the strip ∆1 = [a, b] × [L, L1 ], on which Φ1 (t, x, p, q) ≥ 0, and the strip ∆2 = [a, b] × [F1 , F ], on which Φ1 (t, x, p, q) ≤ 0, are such that the graph of the function x0 (t), t ∈ [a, b], does not cross ∆1 and ∆2 , and is located between them. For this reason ∆1 and ∆2 are called barrier strips for x0 (t), t ∈ [a, b]. + − + (H2) There are constants G− i , Gi , Hi , Hi , i = 1, 2, such that + G+ 2 > G1 ≥ 2C, H2+ < H1+ ≤ −2C, − G− 2 > G1 ≥ 2C, H2− < H1− ≤ −2C, where C = max{|L|, |F |}/(b − a), f (t, x, p, q) and fq (t, x, p, q) are continuous and fq (t, x, p, q) < 0 for (t, x, p, q) ∈ [a, b] × [m1 − ε, M1 + ε] × [F − ε, L + ε] × [m2 − ε, M2 + ε], EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 3 where m1 = min{F a, F b} M1 = max{La, Lb}, m2 = min H2+ , H2− , + − M2 = max G2 , G2 , and ε > 0 is fixed and such that H1− > H2− + ε, H1+ > H2+ + ε, + G+ 2 > G1 + ε, − G− 2 > G1 + ε. (2.1) ft (t, x, p, q), fx (t, x, p, q) and fp (t, x, p, q) are continuous for (t, x, p, q) in [a, b] × [m1 , M1 ] × [F, L] × [m2 , M2 ]; ft (t, x, p, q) + fx (t, x, p, q)p + fp (t, x, p, q)q ≥ 0 + for (t, x, p, q) in [a, b] × [m1 , M1 ] × [F, L] × [H2+ , H1+ ] ∪ [G+ 1 , G2 ] , and ft (t, x, p, q) + fx (t, x, p, q)p + fp (t, x, p, q)q ≤ 0 − for (t, x, p, q) in [a, b] × [m1 , M1 ] × [F, L] × [H2− , H1− ] ∪ [G− 1 , G2 ] , where F and L are the constants of H1. Remark 2.2 Set Φ2 (t, x, p, q) ≡ ft (t, x, p, q) + fx (t, x, p, q)p + fp (t, x, p, q)q. + Then, the pair of strips Ω1 = [a, b]×([H2+ , H1+ ]∪[G+ 1 , G2 ]), where Φ2 (t, x, p, q) ≥ − − − 0, and the pair of strips Ω2 = [a, b]×([H2 , H1 ]∪[G1 , G− 2 ]), where Φ2 (t, x, p, q) ≤ 0, are such that the graph of the function x00 (t), t ∈ [a, b], can not cross the outer strips, of the four such ones, defined by Ω1 and Ω2 . For this reason the outer strips of Ω1 and Ω2 are called barrier strips for x00 (t), t ∈ [a, b]. − (H3) For m3 = min{H1+ , H1− } and M3 = max{G+ 1 , G1 } h(λ, t, x, p, m3 − ε)h(λ, t, x, p, M3 + ε) ≤ 0 for (λ, t, x, p) in [0, 1] × [a, b] × [m1 − ε, M1 + ε] × [F − ε, L + ε], where h(λ, t, x, p, q) = (λ − 1)Kq + λf (t, x, p, q), F, L, K are the constants of H1, − and H1+ , H1− , G+ 1 , G1 , C, m1 , M1 , and ε are as in H2. − (H4) For (t, x, p, q) in T ×[F, L]×[min{H1+ , H1− }, max{G+ 1 , G1 }], fx (t, x, p, q) ≥ 0, where the trapezoid T and the constants F and L are as in H1, and − H1+ , H1− , G+ 1 and G1 are the constants in H2, and m3 and M3 are as in H3. 3 Main result For a function y(t) ∈ C[a, b] bounded on [a, b], we define a mapping Ay = x, where x(t) ∈ C 2 [a, b] is a solution to the BVP f (t, y(t), x0 , x00 ) = 0, t ∈ [a, b], x(a) = A, x0 (b) = B. (3.1) 4 Minimal and maximal solutions EJDE–2003/21 We will show that under the hypotheses H1, H2, and H3, the map A is uniquely determined. For this reason, we consider two sequences {un } and {vn }, n = 0, 1, . . . , defined by un+1 = Aun and vn+1 = Avn , where u0 = F t, v0 = Lt, t ∈ [a, b], and F and L are as in H1. Now we formulate our main result. Theorem 3.1 Under hypotheses H1–H4, there are sequences {un } and {vn }, n = 0, 1, . . . , such that for n → +∞: un → um , vn → v M and u0 ≤ u1 ≤ · · · ≤ un ≤ · · · ≤ um ≤ x ≤ v M ≤ · · · ≤ vn ≤ · · · ≤ v1 ≤ v0 , where um (t) and v M (t) are the minimal and maximal solutions of the BVP (1.1) respectively, and x(t) ∈ C 2 [a, b] is a solution of (1.1). The proof of this theorem can be found at the end of this article and is based on the auxiliary results, which we present in the next section. 4 Auxiliary results We begin this section with an existence result, which is a modification of [8, Theorem 6.1, Chapter II]. Namely, we consider the family of BVPs Kx00 = λ Kx00 + f (t, y(t), x0 , x00 ) , t ∈ [a, b], (4.1) x(a) = A, x0 (b) = B, where λ ∈ [0, 1] and K > 0. Lemma 4.1 Assume that there are constants Qi , i = 0, 1, . . . , 5, independent of λ such that (i) For each solution x(t) ∈ C 2 [a, b] of (4.1) it holds Q0 < x(t) < Q1 , Q2 < x0 (t) < Q3 , Q4 < x00 (t) < Q5 , t ∈ [a, b]. Also assume that: (ii) f (t, x, p, q) and fq (t, x, p, q) are continuous, and fq (t, x, p, q) < 0 for all (t, x, p, q) in [a, b] × [Q0 , Q1 ] × [Q2 , Q3 ] × [Q4 , Q5 ] (iii) h(λ, t, x, p, Q4 )h(λ, t, x, p, Q5 ) ≤ 0 for (λ, t, x, p) in Λ := [0, 1] × [a, b] × [Q0 , Q1 ] × [Q2 , Q3 ], where h(λ, t, x, p, q) = (λ − 1)Kq + λf (t, x, p, q). Then the BVP (3.1) has a C 2 [a, b]-solution for each y(t) ∈ C[a, b] such that Q0 < y(t) < Q1 , t ∈ [a, b]. EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 5 Proof In view of (ii) and (iii), we conclude that there is a unique function G(λ, t, x, p) which is continuous on Λ and such that q = G(λ, t, x, p) for (λ, t, x, p) ∈ Λ is equivalent to the equation h(λ, t, x, p, q) = 0 on Λ × [Q4 , Q5 ]. Note that h(0, t, x, p, 0) = 0 yields G(0, t, x, p) = 0 for (t, x, p) ∈ [a, b] × [Q0 , Q1 ] × [Q2 , Q3 ]. (4.2) Thus, the family (4.1) is equivalent to the family of BVPs x00 = G(λ, t, y(t), x0 ), t ∈ [a, b], x(a) = A, x0 (b) = B, (4.3) where λ ∈ [0, 1]. Now, define the set U = x(t) ∈ C 2 [a, b] : x(t) ∈ (Q0 , Q1 ), x0 (t) ∈ (Q2 , Q3 ), x00 (t) ∈ (Q4 , Q5 ) , 2 which is an open subset of the convex set CQ [a, b] of the Banach space C 2 [a, b] 2 and consider the map N : CQ [a, b] → C[a, b], defined by Nx = x00 , 2 [a, b] = {x ∈ C 2 [a, b] : x(a) = A, x0 (b) = B}. It is easy to see that the where CQ 2 [a, b] → C[a, b], defined by map S : CQ 0 Sx = x00 2 with CQ [a, b] = {x ∈ C 2 [a, b] : x(a) = 0, x0 (b) = 0}, is one-to-one and 0 the problem Sx = 0, x(a) = A, x0 (b) = B, has a unique solution l. Then 2 N−1 : C[a, b] → CQ [a, b] exists, is continuous, and moreover N−1 s = S−1 s + l. 2 Let Hλ : U → CQ [a, b] be defined by Hλ x = N−1 Gλ j(x), λ ∈ [0, 1], 2 where j : CQ [a, b] → C 1 [a, b] is defined by jx = x, Gλ : C 1 [a, b] → C[a, b] is defined by (Gλ x) (t) = G λ, t, y(t), x0 (t) , λ ∈ [0, 1] . Clearly, Hλ is a compact homotopy, because j is a completely continuous embeding, and Gλ and N−1 are continuous. Moreover, Hλ x = x implies x = N−1 Gλ j(x). 6 Minimal and maximal solutions EJDE–2003/21 Hence, by the definition of N−1 , we have x = S−1 Gλ j(x) + l. Finally, since Sl = 0, it follows that Sx = Gλ j(x). Thus, the fixed points of Hλ are solutions to (4.3) and obviouslyHλ has no fixed points on ∂U.In view of (4.2), the map H0 , which has the form H0 x = l, is constant. Moreover, l, as the unique solution of (4.1)0 , belongs to the set U . Hence, by [8, Theorem 2.2], the map H0 is essential. The topological transversality theorem of [8] implies that H1 is also essential, i.e. for λ = 1 (4.3) has a solution. Moreover, for λ = 1 (4.3) coincides with (3.1). Therefore, the problem (3.1) has a solution. The proof of the lemma is complete. ♦ To obtain our next auxiliary results, we introduce the following two sets V = {y(t) ∈ C[a, b] : F t ≤ y(t) ≤ Lt, t ∈ [a, b]}, V1 = {y(t) ∈ C 1 [a, b] : F t ≤ y(t) ≤ Lt, F ≤ y 0 (t) ≤ L, t ∈ [a, b]}, where the constants L and F are as in H1. Then we formulate the following results. Lemma 4.2 Let H1 hold and x(t) ∈ C 2 [a, b] be a solution to (4.1) with y(t) ∈ V . Then the following statements hold: (i) If there is an interval T1 ⊆ [a, b] such that L ≤ x0 (t) ≤ L1 for t ∈ T1 , (4.4) then x00 (t) ≥ 0 for t ∈ T1 . (ii) If there is an interval T2 ⊆ [a, b] such that F1 ≤ x0 (t) ≤ F for t ∈ T2 , then x00 (t) ≤ 0 for t ∈ T2 . Proof Since the proofs of (i) and (ii) are similar, it is sufficient to show that (4.4) implies x00 (t) ≥ 0 for t ∈ T1 . Indeed, the assertion is true for λ = 0. Now, let λ ∈ (0, 1] and assume that there is a t0 ∈ T1 such that x00 (t0 ) < 0. Then 0 > Kx00 (t0 ) = λ [Kx00 (t0 ) + f (t0 , x(t0 ), x0 (t0 ), x00 (t0 ))] ≥ 0. This contradiction proves the assertion. ♦ Lemma 4.3 Let H1 hold, and x(t) ∈ C 2 [a, b] be a solution to (4.1) with y(t) ∈ V . Then F t ≤ x(t) ≤ Lt, F ≤ x0 (t) ≤ L for t ∈ [a, b]. EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 7 Proof Consider the sets Y0 = {t ∈ [a, b] : L < x0 (t) ≤ L1 } and Y1 = {t ∈ [a, b] : F1 ≤ x0 (t) < F } and suppose that they are not empty. Then, using the continuity of x0 (t) and the inequality F ≤ x0 (b) ≤ L, we easily conclude that there are closed intervals [t0 , τ0 ] ⊆ Y0 and [t1 , τ1 ] ⊆ Y1 such that x0 (t0 ) > x0 (τ0 ) and x0 (t1 ) < x0 (τ1 ). (4.5) On the other hand, by Lemma 4.2, we have x00 (t) ≥ 0 for t ∈ [t0 , τ0 ] and x00 (t) ≤ 0 for t ∈ [t1 , τ1 ] and therefore, we have x0 (t0 ) ≤ x0 (τ0 ) and x0 (t1 ) ≥ x0 (τ1 ). But this contradicts (4.5). The obtained contradiction shows that Y0 and Y1 are empty, and so we see that F ≤ x0 (t) ≤ L for t ∈ [a, b]. Integrating this expression from a to t and using the fact that F a ≤ A ≤ La, we get F t ≤ x(t) ≤ Lt, t ∈ [a, b] ♦ which concludes the proof. Remark 4.4 Let x(t) ∈ C 2 [a, b] be a solution to (1.1). Then, in view of Lemma 4.3, if F = L, it follows thatx0 (t) = B, t ∈ [a, b]. Now, using F a ≤ A ≤ La, we see that x(t) = Bt, t ∈ [a, b], is the unique C 2 [a, b]-solution to the problem (1.1). Lemma 4.5 Let H1 and H2 hold, and x(t) ∈ C 2 [a, b] be a solution to (4.1) with y(t) ∈ V1 . Then m3 ≤ x00 (t) ≤ M3 , t ∈ [a, b], and there is a constant D independent of λ such that |x000 (t)| ≤ D for t ∈ [a, b]. Proof By the mean value theorem, there is a ξ ∈ (a, b) such that x00 (ξ) = [x0 (b) − x0 (a)]/(b − a). Since Lemma 4.3 implies F ≤ x0 (t) ≤ L for t ∈ [a, b], (4.6) we see that x00 (ξ) ≤ 2C ≤ G+ 1, (4.7) 8 Minimal and maximal solutions EJDE–2003/21 where C = max{|L|, |F |}/(b − a). Now suppose that the set + 00 Y = t ∈ [a, ξ] : G+ 1 < x (t) ≤ G2 is not empty. The continuity of x00 (t) and (4.7) imply that there is a closed interval [t0 , τ0 ] ⊆ Y such that x00 (t0 ) > x00 (τ0 ). (4.8) Since (4.6) holds for t ∈ [t0 , τ0 ] and + 00 G+ 1 < x (t) ≤ G2 for t ∈ [t0 , τ0 ], m1 ≤ F t ≤ y(t) ≤ Lt ≤ M1 for t ∈ [t0 , τ0 ], F ≤ y 0 (t) ≤ L for t ∈ [t0 , τ0 ], (4.9) in view of H2, we have Ψ1 (t) ≡ fq t, y(t), x0 (t), x00 (t) < 0, t ∈ [t0 , τ0 ], and for t ∈ [t0 , τ0 ], Ψ2 (t) ≡ft t, y(t), x0 (t), x00 (t) + fx t, y(t), x0 (t), x00 (t) y 0 (t) + fp t, y(t), x0 (t), x00 (t) x00 (t) ≥ 0. Thus, using the last two inequalities and the continuity of ft , fx , fp and fq on [t0 , τ0 ], we conclude that x000 is continuous on [t0 , τ0 ] and x000 (t) = λΨ2 (t)/[K(1 − λ) − λΨ1 (t)] ≥ 0 for t ∈ [t0 , τ0 ]. (4.10) Consequently, x00 (t0 ) ≤ x00 (τ0 ), which contradicts (4.8). Thus, x00 (t) ≤ G+ 1 for t ∈ [a, ξ]. The inequality H1− ≤ x00 (t), t ∈ [a, ξ] can be obtained in the same manner. Similarly, it is easy to show that H1+ ≤ x00 (t) ≤ G− 1, t ∈ [ξ, b]. Finally, using (4.6), (4.9), the fact that x00 is bounded on [a, b] and the continuity of the partial derivatives of f (t, x, p, q) on the set [a, b] × [m1 , M1 ] × [F, L] × [m3 , M3 ], from (4.10) it follows that there is a constant D independent of λ such that |x000 (t)| ≤ D for t ∈ [a, b]. The proof of the lemma is complete. ♦ Lemma 4.6 Suppose that H1, H2 and H3 hold. Then the BVP (3.1) has a C 2 [a, b]-solution, if y(t) ∈ V1 . EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 9 Proof Let x(t) ∈ C 2 [a, b] be a solution to (4.1)λ . Then, by Lemma 4.3, we have F − ε < x0 (t) < L + ε for t ∈ [a, b] m1 − ε < x(t) < M1 + ε for t ∈ [a, b], while, by Lemma 4.5, m3 − ε < x00 (t) < M3 + ε for t ∈ [a, b], where ε > 0 is as in H2. Thus, the condition (i) of Lemma 4.1 holds for Q0 = m1 − ε, Q1 = M1 + ε, Q2 = F − ε, Q3 = L + ε, Q4 = m3 − ε and Q5 = M3 + ε. Moreover, from (2.1) and H3 it follows that the conditions (ii) and (iii) of Lemma 4.1 are satisfied. Also, m1 − ε < y(t) < M1 + ε for t ∈ [a, b]. So, we can apply Lemma 4.1 to conclude that the problem (3.1) has a solution in C 2 [a, b]. The proof of the lemma is complete. ♦ We need the following two lemmas which are adopted from [24]. Lemma 4.7 ([24, Chapter I, Theorem 1]) Suppose φ(t) satisfies the differential inequality φ00 + g(t)φ0 ≥ 0 for a < t < b, (4.11) with g(t) a bounded function. If φ(t) ≤ M in (a, b) and if the maximum M of φ is attained at an interior point c of (a, b), then φ ≡ M . Lemma 4.8 ([24, Chapter I, Theorem 2]) Suppose φ(t) is a nonconstant function which satisfies the inequality (4.11) and has one-sided derivatives at a and b, and suppose g is bounded on every closed subinterval of (a, b). If the maximum of φ occurs at t = a and g is bounded below at t = a, then φ0 (a) < 0. If the maximum occurs at t = b and g is bounded above at t = b, then φ0 (b) > 0. Lemma 4.9 Suppose that φ ∈ C 2 (a, b) ∩ C 1 [a, b] satisfies the inequality φ00 (t) + g(t)φ0 (t) ≥ 0 for t ∈ (a, b), where g(t) is bounded on (a, b). If φ(a) ≤ 0 and φ0 (b) ≤ 0, (4.12) then φ(t) ≤ 0 for t ∈ [a, b]. (4.13) 10 Minimal and maximal solutions EJDE–2003/21 Proof First, assume that φ(t)achieves its maximum at t0 ∈ (a, b). By Lemma 4.7, for t ∈ [a, b] we obtain φ(t) ≡ φ(t0 ) = φ(a) ≤ 0 and so (4.13) holds. Next, suppose that φ(t) achieves its maximum at the ends of the interval [a, b]. If we assume φ(t) ≤ φ(b), t ∈ [a, b], the application of Lemma 4.8 shows that φ0 (b) > 0, which contradicts (4.12). Thus, by our assumtions, φ(t) ≤ φ(a) ≤ 0, t ∈ [a, b], and so (4.13) follows. The proof is complete. ♦ In the last two lemmas we use the map A defined in the section 3. Lemma 4.10 Under assumptions H1, H2, and H3, for any y ∈ V1 , the image x by the map A exists and it is unique. Proof The existence of the image of x follows from Lemma 4.6. In order to see that x is unique, fix y and assume thatzis an other image of y by A and consider the function φ(t) = x(t) − z(t), t ∈ [a, b]. Then, it is evident that f t, y(t), x0 (t), x00 (t) − f t, y(t), z 0 (t), z 00 (t) = 0, t ∈ [a, b]. Next, we construct the equality f t, y(t), x0 (t), x00 (t) − f t, y(t), z 0 (t), x00 (t) +f t, y(t), z 0 (t), x00 (t) − f t, y(t), z 0 (t), z 00 (t) = 0, which can be rewritten in the form I1 (t)φ0 (t) + I2 (t)φ00 (t) = 0, where Z 1 fp t, y(t), z 0 (t) + θ(x0 (t) − z 0 (t)), x00 (t) dθ, I1 (t) = Z 0 1 I2 (t) = fq t, y(t), z 0 (t), z 00 (t) + θ(x00 (t) − z 00 (t)) dθ. 0 Hence, the function φ(t) is a solution to the BVP I1 (t) 0 φ (t) = 0, t ∈ [a, b], I2 (t) φ(a) = 0, φ0 (b) = 0. φ00 (t) + Moreover, it is easy to conclude that φ(t) ≡ 0, t ∈ [a, b], is the unique solution of the above BVP. Consequently, x(t) ≡ z(t), t ∈ [a, b]. The proof of the lemma is complete. ♦ Lemma 4.11 Under the hypotheses H1–H4, if y1 (t), y2 (t) ∈ V1 are such that y1 (t) ≤ y2 (t) for t ∈ [a, b], then x1 (t) ≤ x2 (t) for where xi = Ayi , i = 1, 2. t ∈ [a, b], EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 11 Proof Observe that, by Lemma 4.3, we have F ≤ x01 (t) ≤ L, t ∈ [a, b], and, by Lemma 4.5, m3 ≤ x1 00 (t) ≤ M3 , t ∈ [a, b]. Moreover, F t ≤ y1 (t) ≤ y2 (t) ≤ Lt, t ∈ [a, b]. Thus, from fx (t, x, p, q) ≥ 0 for (t, x, p, q) in T × [F, L] × [m3 , M3 ] it follows that 0 = f t, y1 (t), x01 (t), x001 (t) ≤ f t, y2 (t), x01 (t), x001 (t) , t ∈ [a, b]. Hence, for t ∈ [a, b] we have f t, y2 (t), x02 (t), x002 (t) − f t, y2 (t), x01 (t), x001 (t) ≤ 0 and then, as in Lemma 4.10, we construct the inequality f t, y2 (t), x01 (t), x001 (t) − f t, y2 (t), x02 (t), x001 (t) +f t, y2 (t), x02 (t), x001 (t) − f t, y2 (t), x02 (t), x002 (t) ≥ 0 from which for φ(t) = x1 (t) − x2 (t), t ∈ [a, b], we find φ00 (t) + J1 (t) 0 φ (t) ≥ 0, J2 (t) t ∈ [a, b], where 1 Z fp t, y2 (t), x02 (t) + θ(x01 (t) − x02 (t)), x001 (t) dθ, J1 (t) = 0 Z J2 (t) = 1 fq t, y2 (t), x02 (t), x002 (t) + θ(x001 (t) − x002 (t)) dθ. 0 Furthermore, φ(a) = 0, φ0 (b) = 0. Finally, applying Lemma 4.9, we see that φ(t) ≤ 0 for t ∈ [a, b], which completes the proof. ♦ 5 Proof of Theorem 3.1 Consider the sequences {un } and {vn }, defined by un+1 = Aun and vn+1 = Avn , n = 0, 1, . . . In view of Lemma 4.6, from Lemma 4.3 it follows that F t = u0 ≤ u1 and v1 ≤ v0 = Lt. Moreover, Lemma 4.11 and induction arguments imply that un−1 ≤ un , vn ≤ vn−1 , n = 1, 2, . . . On the other hand, since u0 ≤ v0 , by Lemma 4.11 and induction arguments, we conclude that un ≤ vn , n = 0, 1, . . . . ¿From this observation it follows that u0 ≤ un ≤ v0 , n = 0, 1, . . . . 12 Minimal and maximal solutions EJDE–2003/21 Therefore, {un } is uniformly bounded. Furthermore, since, by Lemma 4.3, {u0n } is uniformly bounded, we see that {un } is equicontinuous. Finally, since, by Lemma 4.5, {u000 n } is uniformly bounded, it follows that the sequence {u00n } is uniformly bounded and equicontinuous. Thus, we can apply the ArzelaAscoli theorem to conclude that there are a subsequence {uni } and a function u ∈ C 2 [a, b] such that {uni }, {u0ni } and {u00ni } are uniformly convergent on [a, b] to u, u0 and u00 respectively. Now, using the fact that uni = Auni−1 can be rewritten equivalently in the form Z Z 1 t r Ku00ni (s) + f (s, uni−1 (s), u0ni (s), u00ni (s)) ds dr uni (t) = K a b + B(t − a) + A, letting i → +∞, we obtain Z Z 1 t r u(t) = Ku00 (s) + f (s, u(s), u0 (s), u00 (s)) ds dr + B(t − a) + A, K a b from which it follows that u(t) is a solution to the BVP (1.1). Remark that, if x(t) is a solution of (1.1), then, by Lemma 4.3, we have u0 (t) ≤ x(t) for t ∈ [a, b]. Applying Lemma 4.11 (it is possible, because x = Ax), by induction we obtain un (t) ≤ x(t), t ∈ [a, b], n = 0, 1, . . . , and then u(t) ≤ x(t), t ∈ [a, b], which holds for each solution x(t) ∈ C 2 [a, b] of the problem (1.1). Consequently, it follows that u(t) ≡ um (t), t ∈ [a, b]. By similar arguments, we conclude that lim vn = v M (t), t ∈ [a, b]. Thus, the proof is complete. ♦ Acknowledgement P. S. Kelevedjiev would like to thank the Ministry of National Economy of Helenic Republic for providing the NATO Science Fellowship (No. DOO 850/02) and the University of Ioannina for its hospitality. References [1] S. Bernfeld, V. Lakshmikantham, An Introduction to Nonlinear Boundary Value Problems, New York, Academic Press, 1974. 0 [2] A. Cabada, R.L. Pouso, Existence results for the problem (φ(u0 )) = f (t, u, u0 ) with periodic and Neumann boundary conditions, Nonlinear Anal. 30 (1997), 1733. [3] A. Cabada, R.L. Pouso, Extremal solutions of strongly nonlinear discontinuous second-order equations with nonlinear functional boundary conditions, Nonlinear Anal.42 (2000), 1377-1396. EJDE–2003/21 Myron K. Grammatikopoulos & Petio S. Kelevedjiev 13 [4] M. Cherpion, C. De Coster, P Habets, Monotone iterative methods for boundary value problems, Differential Integral Equations 12 (1999), 309338. [5] B.C. Dhage, S.T. Patil, On the existence of extremal solutions of nonlinear discontinuous boundary value problems, Math. Sci. Res. Hot-Line. 2 (1998), 17-29. [6] P.M. Fitzpatrick, Existence results for equations involving noncompact perturbation of Fredholm mappings with applications to differential equations, J. Math. Anal. Appl. 66 (1978), 151-177. [7] P.M. Fitzpatrick, W.V. Petryshyn, Galerkin method in the constructive solvability of nonlinear Hammerstein equations with applications to differential equations, Trans. Amer. Math. Soc. 238 (1978), 321-340. [8] A.Granas, R. B. Guenther, J. W. Lee, Nonlinear boundary value problems for ordinary differential equations, Dissnes Math., Warszawa, 1985. [9] S. Heikkilä, V. Lakshmikantham, Monotone iterative techniques for discontinuous nonlinear differential equations, New York, Dekker, 1994. [10] D. Jiang, L. Kong, A monotone method for constructing extremal solutions to second order periodic boundary value problems, Ann. Polon. Math. 76 (2001), 279-286. [11] P. Kelevedjiev, N. Popivanov, Minimal and maximal solutions of two point boundary value problems for second order differential equations, Nonlinear Oscillations 4 (2001), 216-225. [12] P. Kelevedjiev, N. Popivanov, Existence of solutions of boundary value problems for the equation f (t, x, x0 , x00 ) = 0 with fully nonlinear boundary conditions, Annuaire de l’Universite de Sofia 94, 2000, 65-77. [13] M. Koleva, N. Kutev, Monotone methods and minimal and maximal solutions for nonlinear ordinary differential equations, International series of numerical mathematics 97, Burkhaus Verlag Basel. (1991), 205-209. [14] G. S. Ladde, V. Lakshmikantham, D.S. Vatsala, Monotone Iterative Techniques for Nonlinear Differential Equations, Boston, Pitman, MA, 1985. [15] V. Lakshmikantham, N. Shahzad, Method of quasilinearization for general second order boundary value problems, Nonlinear World. 2 (1995), 133-146. [16] V. Lakshmikantham, N. Shahzad, Nieto, Methods of generalized quasilinearization for periodic boundary value problems, Nonlinear Analysis 27 (1996), 143-151. [17] Y. Mao, J.Lee, Two point boundary value problems for nonlinear differential equations, Rocky Maunt. J. Math. 26 (1996), 1499-1515. 14 Minimal and maximal solutions EJDE–2003/21 [18] Y. Mao, J. Lee, Two point boundary value problems for nonlinear differential equations, Rocky Mount. J. Math. 10 (1980), 35-58. [19] S.A. Marano, On a boundary value problem for the differential equation f (t, x, x0 , x00 ) = 0, J. Math. Anal. Appl. 182 (1994), 309-319. [20] R. N. Mohapatra, K. Vajravelu, Y. Yin, An improved quasilinearization method for second order nonlinear boundary value problems, J. Math. Anal. Appl. 214 (1997), 55-62. [21] W.V. Petryshyn, Solvability of various boundary value problems for the equation x00 = f (t, x, x0 , x00 ) − y, Pacific J. Math. 122 (1986), 169-195. [22] W. V. Petryshyn, Z.S. Yu, Solvability of Neumann BV problems for nonlinear second order ODE’s which need not be solvable for the highest order derivative, J. Math. Anal. Appl. 91 (1983), 244-253. [23] W. V. Petryshyn, Z.S. Yu, Periodic solutions of nonlinear second-order differential equations which are not solvable for the highest-order derivative, J. Math. Anal. Appl. 89 (1982), 462-488. [24] M. N. Proter, H.F. Weinberger, Maximum Principle in Differential Equations, New Jearsy, Prentice-Hall, Inc., 1967. [25] N. Shahzad, S. Malek, Remarks on generalized quasilinearization method for first order periodic boundary problems, Nonlinear World. 2 (1995), 247255. [26] H. B. Thompson, Minimal solutions for two point boundary value problems, Rendiconti del circolo mathematico di Palermo, Serie II, Tomo XXXVII, (1988), 261-281. [27] A. Tineo, Existence of solutions for a class of boundary value problems for the equation x00 = F (t, x, x0 , x00 ), Comment. Math. Univ. Carolin 29 (1988), 285-291. [28] M.X. Wang, A. Cabada, J.J. Nieto, Monotone method for nonlinear second order periodic boundary value problems with Caratheodory functions, Ann. Polon. Math. 58 (1993), 221-235. [29] Y. Yin, Monotone iterative technique and quasilinearization for some antiperiodic problems, Nonlinear World. 3 (1996), 253-266. Myron K. Grammatikopoulos Department of Mathematics, University of Ioannina 451 10 Ioannina, Hellas, Greece e-mail: mgrammat@cc.uoi.gr Petio S. Kelevedjiev Department of Mathematics, Technical University of Sliven 8800 Sliven, Bulgaria e-mail: keleved@mailcity.com