Existence of positive solutions for a class of higher-order m-point boundary value problems, Electron

EXISTENCE OF POSITIVE SOLUTIONS FOR A CLASS OFHIGHER-ORDER m-POINT BOUNDARY VALUEPROBLEMSRodica LucaDepartment of Mathematics, Gh. Asachi Technical University11 Blvd. Carol I, Iasi 700506, RomaniaE-mail: rluca@math.tuiasi.roAbstract. We investigate the existence of positive solutions with respect to a conefor a higher-order nonlinear diﬀerential system, subject to some boundary conditionswhich involve m points.Keywords: Higher-order diﬀerential system, boundary conditions, positive solu-tions, ﬁxed point theorem.AMS Subject Classiﬁcation: 34B10, 34B18.

In Section 2, we shall present several auxiliary results which investigate a boundary value problem for a n-th order equation (the problem (1), (2) below), some of them from the paper [15].In Section 3, we shall give sufficient conditions on λ and µ such that positive solutions with respect to a cone for our problem (S), (BC) exist.In Section 4, we shall present an example that illustrates the obtained results.Our main results (Theorem 2 and Theorem 3) are based on the Guo-Krasnoselskii fixed point theorem, presented below.
Theorem 1.Let X be a Banach space and let C ⊂ X be a cone in X. Assume Ω 1 and Then A has a fixed point in C ∩ (Ω 2 \ Ω 1 ).
EJQTDE, 2010 No. 74, p. 2 In this section, we shall study the n-th order differential equation with the boundary conditions We denote by , then the solution of (1), ( 2) is given by Proof.By (1) and the first relations from (2) we deduce From the above relation and the condition u(T and so Therefore from (4) and the above expression for C we obtain the relation (3).
Therefore, we obtain then the unique solution u of problem (1), ( 2) Proof.We first show that u(T ) ≥ 0. Indeed we have Using a result from [6] (see also Theorem 1.1 from [15]), we deduce that u(t) ≥ 0 for Remark 1.Under the assumptions of Lemma 3, by using Lemma 4 and the expres- EJQTDE, 2010 No. 74, p. 5 Proof.By (3) we have 2 From the proof of Lemma 2.2 in [15] we obtain the following result.Lemma 6.We assume that 0 Then the solution of problem (1), ( 2) Remark 2. From the above expression for γ, we see that γ < 1.

The existence of positive solutions
In this section we shall give sufficient conditions on λ and µ such that positive solutions with respect to a cone for problem (S), (BC) exist.
We present the assumptions that we shall use in the sequel.
x exist and are positive numbers.
Using the Green's function given in Lemma We consider the Banach space X = C([0, T ]) with supremum norm • and define the cone C ⊂ X by where γ is defined in Lemma 6.
Proof.Let λ, µ ∈ (L 1 , L 2 ) and we choose a positive number We now define the operator A : C → X, by By Lemma 6, we have A(C) ⊂ C. By using the Arzela-Ascoli theorem we deduce that the operator A is completely continuous (compact and continuous).By definitions of f 0 and g 0 there exists K 1 > 0 such that EJQTDE, 2010 No. 74, p. 7 Using (H3) we have f (0) = g(0) = 0 and the above inequalities are also valid for x = 0.
satisfies the problem (1), (2) with y(t) = µc(t)g(u(t)), t ∈ [0, T ], then by (6) and the above property of g, we deduce for t ∈ [0, T ] By using once again Lemma 5 (relations ( 6)) and the properties of the function f , we have Next, by the definitions of f ∞ and g ∞ , there exists K2 > 0 such that We consider now K 2 = max 2K 1 , K2 /γ .For u ∈ C with u = K 2 , we obtain by using Lemma 6, that Then, by using (6), Lemma 6, and the above relations, we obtain for t ≥ ξ m−2 and EJQTDE, 2010 No. 74, p. 8 We denote by We now apply Theorem 1 i) and we deduce that A has a fixed point u ∈ C ∩ ( Ω2 \ Ω 1 ).This element together with positive solution of (S), (BC) with respect to cone C, for the given λ and µ.
Proof.Let λ and µ with λ, µ ∈ (L 3 , L 4 ).We select a positive number ε such that EJQTDE, 2010 No. 74, p. 9 We also consider the operator A defined in the proof of Theorem 2. From the definitions of f 0 and g 0 , we deduce that there exists K3 > 0 such that Using the properties of f and g the above inequalities are also valid for x = 0.
In addition, because g is a continuous function with g 0 > 0, then g(0) = 0 and there exists K 3 ∈ (0, K3 ) such that For u ∈ C with u = K 3 , by (6) and the above inequality, we deduce that for all By using (6), Lemma 6 and the properties of f, g we then obtain EJQTDE, 2010 No. 74, p. 10 Hence, A(u) ≥ A(u)(ξ m−2 ) ≥ u , for u ∈ C with u = K 3 .We denote by Ω 3 = {u ∈ C, u < K 3 }, and then we have A(u) ≥ u for all u ∈ C ∩ ∂Ω 3 .
We now consider the functions f * , g . By (H2) we obtain for f * and g * the relations lim We also have f (x) ≤ f * (x), g(x) ≤ g * (x), for all x ≥ 0. Then there exists K4 > 0 such that By Theorem 1 ii) we deduce that A has a fixed point u ∈ C ∩( Ω4 \Ω 3 ), which together with v(t) = µ T n−1 .
We also have Therefore if the above condition is verified, then by Theorem 2 we deduce that for all numbers λ, µ ∈ (L 1 , L 2 ) the problem (S 0 ), (BC 0 ) has positive solutions.