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Existence and monotone iteration of concave positive symmetric solutions for a three-point second-order boundary value problems with integral boundary conditions.

1. INTRODUCTION

The multi-point boundary value problems for ordinary differential equations arise in variety of different areas of applied mathematics and physics. The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated Il'in and Moiseev [5]. Since then, nonlinear multi-point boundary value problems have been studied by many authors. We refer the reader to [2- 4,12] and their references.

At the same time, boundary value problems with integral boundary conditions for ordinary differential equations represent a very interesting and important class of problems. For an overview of the literature on integral boundary value problems, see [1, 6, 11, 14].

In [14], J. Tariboon and T. Sitthiwirattham considered the second-order three-point differential equation

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

They showed the existence of at least one positive solutions if f is either superlinear or sublinear by applying the fixed point theorem in cones.

In [11], H. Pang and Y. Tong considered second-order boundary value problem

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

They investigated the existence of concave symmetric positive solutions and established corresponding iterative schemes for a second-order boundary value problem with integral boundary conditions.

Motivated by the results above, in this paper, we are interested in the existence of the concave symmetric positive solutions for the following second-order three-point boundary value problems with integral boundary conditions

(1.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [eta] [member of] (0, 1), 0 < [alpha] < 1/[eta], and f [member of] C((0,1) x [0, +[infinity]) x R, [0, +[infinity])).

The organization of the paper is as follows. In Section 2, we present definitions and some necessary lemmas that will be used to prove our main result. In Section 3, we apply the monotone iterative technique to obtain the existence of concave symmetric positive solutions for BVP (1.1). Monotone iterative technique has been successfully used to prove to existence of a positive solutions of boundary value problems, see [7-11, 13, 15, 16]. In Section 4, we give example to illustrate our result.

2. PRELIMINARIES

Definition 2.1. Let E be a real Banach Space. A nonempty closed convex set P [subset] E is called a cone if it provides the following two conditions:

(i) u [member of] P, [lambda] [greater than or equal to] 0 implies [lambda]u [member of] P;

(ii) u [member of] P, -u [member of] P implies u = 0.

Definition 2.2. Let E be a real Banach Space. A function u [member of] E is said to be symmetric on [0, 1] if

u(x) = u(1 - x), x [member of] [0, 1].

Definition 2.3. Let (E, [less than or equal to]) be an ordered real Banach Space. An operator a : E [right arrow] E is said to be nondecreasing provided that [alpha](u)[less than or equal to] [alpha](v) for all u, v [member of] E with u [less than or equal to] v. If the inequality is strict, then a is said to be strictly nondecreasing.

Definition 2.4. Let E be a real Banach Space, u [member of] E is said to be concave on [0,1] if

u([lambda][x.sub.i] + (1 - [lambda])[x.sub.2]) [greater than or equal to] [lambda]u([x.sub.i]) + (1 - [lambda])u([x.sub.2])

for any [x.sub.i], [x.sub.2] [member of] [0, 1] and [lambda] [member of] [0, 1].

We consider the Banach space E = [C.sup.2][0, 1] equipped with norm [parallel]u[parallel] = max{[[parallel]u[parallel].sub.infinity], [[parallel]u'[parallel].sub.infinity]]} where [[parallel]u'[parallel].sub.infinity]]} Throughout this paper, we always assume that the following assumptions are satisfied:

(H1) f [member of] C((0, 1) x [0, +[infinity]) x R, [0, +[infinity])), f(x, u, v) = f(l-x, u, -v) for x [member of] (0, 1/2], and f (x, u, v) [greater than or equal to] 0 for all (x, u, v) [member of] (0, 1) x [0, +infinity]) x R.

(H2) f(x, ., v) is nondecreasing for each (x, v) [member of] (0, 1/2], x R, f(x, u,.) is nondecreasing for (x, u) [member of] (0, 1/2] x [0, +[infinity]).

Define the cone P [subset] E by

P = {u [member of] E : u(x) [greater than or equal to] 0 is concave and u(x) = u(1 - x),x e [0, 1]}.

Lemma 2.5. For any u [member of] [C.sup.2][0, 1], suppose that u is the solution of the following BVP

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then we can easily get the solution

(2.1) u(x)= [[integral].sup.1.sub.0] (H(s) + G(x,s))f(s,u(s),u'(s))ds,

where

(2.2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(2.3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Proof. Suppose that u [member of] [C.sup.2] [0,1] is a solution of problem (1.1). Then we have

u"(x) = -f (x, u(x), u'(x)).

For x [member] [0,1], by integration from 0 to 1, we have

u'(x)= u'(0) - [[integral].sup.x.sub.0] f(s, u(s), u'(s))ds.

For x [member] [0,1], by integration again from 0 to 1, we have

u(x) = u'(0)x - [[integral].sup.x.sub.0]([[integral].sup.[tau].sub.0] f(s, u(s), u'(s))ds)d[tau].

That is,

(2.4) u(x) = u(0) + u'(0)x - [[integral].sup.x.sub.0] (x - s)f(s,u(s),u'(s))ds,

therefore,

u(1) = u(0)+ u'(0) - [[integral].sup.1.sub.0] (1 - s)f(s,u(s),u'(s))ds.

From condition (1.1), we have

u'(0) = [[integral].sup.1.sub.0](1 - s)f(s,u(s),u'(s))ds.

By integrating (2.4) from 0 to [eta], where [eta] [member of] (0,1), we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and from u(0) = [alpha] [[integral].sup.[eta].sub.0] u(s)ds, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Therefore, (1.1) has a unique solution

From (2.2) and (2.3), we obtain

u(x) = [[integral].sup.1.sub.0](H(s) + G(x, s))f(s,u(s),u'(s))ds.

The proof is complete.

The functions H and G have the following properties.

Lemma 2.6. If [eta] [member of] (0,1) and 0 < [alpha] < 1/[eta], then we have H(s) [greater than or equal to] 0; for s [member of] [0,1].

Proof. From the definition of H(s), s [member of] (0,1), [eta] [member of] (0,1), and 0 < [alpha], we have H(s) [greater than or equal to] 0.

Lemma 2.7. G(1 - x, 1 - s) = G(x, s), 0 [less than or equal to] G(x, s) [less than or equal to] G(s,s) for x, s [member of] [0,1].

Proof. From the definition of G(x,s), we get G(1 - x, 1 - s) = G(x, s) and 0 [less than or equal to] G(x,s) [less than or equal to] G(s,s) for x,s [member of] [0,1],

Lemma 2.8. Let [eta] [member of] (0,1) and 0 < [alpha] < 1/[eta]. If f(x, u{x), u'{x)) [member of] C((0,1) x [0, +[infinity]) x R, [0, +[infinity])); then the unique solution u of BVP (1.1) satisfies u(x) [greater than or equal to] 0 for x [member of] [0,1].

Proof. From the definition of u(x), f (x, u(x), u'(x)) [member of] C((0,1) x [0, +[infinity])x R, [0, +[infinity])), Lemma 2.6, and Lemma 2.7, we have u(x) [greater than or equal to] 0.

Lemma 2.9. Let [alpha][lambda] > 1. If f (x,u(x),u'(x)) [member of]C((0,1) x [0, +[infinity]) x R, [0, +[infinity])) then BVP (1.1) has no positive solution.

Proof. Suppose that problem (1.1) has a positive solutions u satisfying u(x) > 0, x [member of] (0, 1). If u(0) = u(1) > 0, by the concavity of u

(2.5) u(s) [greater than or equal to] u(1) for s [member of] (0,1),

by integrating (2.5) from 0 to [eta], where [eta] [member of] (0,1), we have

[[integral].sup.[eta].sub.0] u(s)ds [greater than or equal to] [eta]u(1),

and from u(1) = [alpha][[integral].sup.[eta].sub.0]u(s)ds, we have

u(1)(1 - [alpha][eta]) [greater than or equal to] 0,

which is a contradiction to the u(1) > 0 and (1 - [alpha][eta]) < 0. So, no positive solutions exist.

For any u [member of] [C.sup.2][0,1], T : P [right arrow] E is defined

(2.6) (Tu)(x)= [[integral].sup.1.sub.0](H(s) + G(x,s))f(s,u(s),u'(s))ds, for x [member of] [0,1].

Clearly, u is the solution of BVP (1.1) if and only if u is fixed point of T.

Lemma 2.10. Assume that (H1) and (H2) are satisfied, and let [eta] [member of] (0,1), 0 < [alpha] < 1/[eta]. Then the operator T is completely continuous in [C.sup.2][0,1] and T is nondecreasing.

Proof. For any u [member of] P, from the expression of Tu, we know

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Clearly, Tu is concave. From the definition of Tu, Lemma 2.6, and Lemma 2.7 we see that Tu is nonnegative on [0,1]. We now show that Tu is

symmetric about 1/2. From Lemma 2.7 and (H1), for x [member of] [0,1], we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Therefore, TP [subset] P.

The continuity of T with respect to u(x) [member of] [C.sup.2][0,1] is clear. We now show that T is compact. Suppose that D [subset] P is a bounded set. Then there exists r such that

D = {u [member of] P || [parallel]u[parallel] [less than or equal to] r}.

For any u [member of] D, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So, we have from (2.6)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

These equations imply that the operator T is uniformly bounded. Now we show that Tu is equi-continuous. We separate these three conditions:

Case (i). 0 [less than or equal to] [s.sub.2] [less than or equal to] 1/2;

Case (ii). 1/2 [less than or equal to][x.sub.1] [less than or equal to] [x.sub.2] < 1;

Case (iii). 0 [less than or equal to] [x.sub.1] [less than or equal to] 1/2 [less than or equal to] [sub.2] [less than or equal to] 1.

We solely need to think Case (i) since the proofs of the other two are like. For 0 [less than or equal to] [x.sub.1] [less than or equal to] [x.sub.2] [less than or equal to] 1/2, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

In addition

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

By applying the Arzela-Ascoli theorem, we can guarantee that T(D) is relatively compact, which means T is compact. Then we have T is completely continuous.

Finally, we show T is noncecreasing with respect to u(x) [member of] [C.sup.2][0,1].

Let [u.sub.i](x) [member of] P (i = 1, 2) and [u.sub.1](x) [less than or equal to] [u.sub.2] (x) then, we have [u.sub.2](x) - [u.sub.i](x) [member of] P and [u.sub.2](x) - [u.sub.1](x) [greater than or equal to] 0 is concave, symmetric about 1/2. Therefore

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So, for x [member of] [0,1], by applying (H1), (H2), and the definition of Tu, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus T is nondecreasing. These complete the proof.

3. EXISTENCE OF TWO CONCAVE SYMMETRIC POSITIVE SOLUTIONS FOR BVP (1.1)

Now we find the existence of two concave symmetric positive solutions and corresponding iterative scheme for BVP (1.1).

Theorem 3.1. Suppose that (H1) and (H2) are provided, and let [eta] [member of] (0,1), 0 < [alpha] < 1/[eta]. If there exist two positive number [b.sub.1] < b such that

(3.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where b and [b.sub.1] satisfy,

(3-2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

then BVP (1.1) has a concave symmetric positive solutions [w.sup.*], [v.sup.*] [member of] P with

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Proof. We show [P.sub.b] = {w G[member of] P: [parallel]w[parallel] [less than or equal to] b}. In what follows, we now show T[P.sub.b] [subset]

[P.sub.b]. Let w [member of] [P.sub.b], then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. On the other hand, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By using (3.1) and ([H.sub.2]), for x [member of] [0,1/2], we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Let x [member of] [1/2, 1], then (1 - x) [member of] [0, 1/2], by using ([H.sub.1]), ([H.sub.2]), and (3.1), we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Then

(3.3) f(x, w(x), w'(x)) [less than or equal to] [b.sub.1], for x [member of] [0,1].

For any w(x) [member of] [P.sub.b], from Lemma 2.10, we obtain T(w) [member of] and, thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

So, [parallel]Tw[parallel] [less than or equal to] b. Then, we obtain T[P.sub.b] [subset] [P.sub.b]. Let [w.sub.0](x) = bx(l - x) + b/4 for x [member of] [0,1], then [parallel][w.sub.0][parallel] = b and [w.sub.0](x) [member of] [P.sub.b]. Let [w.sub.1] = T[w.sub.0], then [w.sub.1] [member of] [P.sub.b]. We denote

(3.4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Because T[P.sub.b] [subset] [P.sub.b], we have [w.sub.n] [member of] [P.sub.b] (n = 0,1, 2,...). According to Lemma 2.10, T is compact, we claim that [{[w.sub.n]}.sup.[infinity].sub.n=1] has a convergent subsequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and there exist [w.sup.*] [member of] [P.sub.b] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. From the definition of T, (3.2), and (3.3), we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus, [w.sub.0](x) - [w.sub.1](x) [member of] [P.sub.b]. By using Lemma 2.10, we obtain T[w.sub.1] [less than or equal to] T[w.sub.0], which means [w.sub.2] [less than or equal to] [w.sub.1], x [member of] [0,1]. By induction, [w.sub.n+1] [less than or equal to] [w.sub.n], x [member of] [0,1], (n = 0,1,2,...).

Now we show that [absolute value of [w'.sub.n+1](x)] [less than or equal to] [absolute value of [w'.sub.n](x)], x [member of] [0,1]. We separate these two conditions:

Case (i). Let x [member of] [0, 1/2], then [w'.sub.n](x) [greater than or equal to] 0.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Then, [absolute value of [w'.sub.1](x)] [less than or equal to] [absolute value of [w'.sub.0](x)], by using Lemma 2.10, we obtain [absolute value of [T[w'.sub.1](x)] [less than or equal to] [absolute value of T[w'.sub.0](x)], which means [absolute value of [w'.sub.2](x)] [less than or equal to] [absolute value of [w'.sub.1](x)], x [member of] [0, 1/2] By the induction [absolute value of [w'.sub.n+1](x)] [less than or equal to] [absolute value of [w'.sub.n](x)], x [member of] [0, 1/2].

Case (ii). Let x [member of] [1/2, 1], then [w'.sub.n](x) [less than or equal to] 0.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then, [absolute value of [w'.sub.1](x)] [less than or equal to] [absolute value of [w'.sub.0](x)], by using Lemma 2.10, we obtain [absolute value of [T[w'.sub.1](x)] [less than or equal to] [absolute value of T[w'.sub.0](x)] which means [absolute value of [w'.sub.2](x)] [less than or equal to] [absolute value of [w'.sub.1](x)], x [member of] [1/2, 1]. By the induction [absolute value of [w'.sub.n+1](x)] [less than or equal to] [absolute value of [w'.sub.n](x)], x [member of] [1/2, 1]. Consequently, [absolute value of [w'.sub.n+1](x)] [less than or equal to] [absolute value of [w'.sub.n](x)], x [member of] [0, 1].

So, we claim that [w.sub.n] [right arrow] [w.sup.*] in norm [parallel]x[parallel]. Let n [right arrow] [infinity] in (3.4) to get T[w.sup.*] = [w.sup.*] because T is continuous. It is clear that the fixed point of the operator T is the solution of BVP (1.1). Hence, [w.sup.*] is concave symmetric positive solution (1.1). And since [w.sup.*] [member of] [P.sub.b], we have [parallel][w.sup.*][parallel] [less than or equal to] b.

Let [v.sub.0](x) = 0, 0 [less than or equal to] x [less than or equal to] 1, then [v.sub.0] [member of] [P.sub.b]. Let [v.sub.1] = T[v.sub.0], then [v.sub.1] [member of] [P.sub.b], we denote

(3.5) [v.sub.n+1] = T[v.sub.n] = [T.sup.n+1][v.sub.0] (n = 0,1, 2,...).

Likely to [{v.sub.n]}.sup.[infinity].sub.n=1], we claim that [{v.sub.n]}.sup.[infinity].sub.n=1] has a convergent subsequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and there exist [v.sup.*] [member of] [P.sub.b] such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Because [v.sub.1] [greater than or equal to] [v.sub.0], by using Lemma 2.10, we obtain T[v.sub.1] [greater than or equal to] [v.sub.0], which means [v.sub.2] [greater than or equal to] [v.sub.1], x [member of] [0,1]. By induction, [v.sub.n+1] [greater than or equal to] [v.sub.n], x [member of] [0,1] (n = 0,1, 2,...). And [absolute value of [v'.sub.2](x)] [greater than

or equal to] [absolute value of [v'.sub.0](x)], by using Lemma 2.10, we obtain

[absolute value of T[v'.sub.1](x)] [greater than or equal to] [absolute value of T [v'.sub.0](x)], which means [absolute value of [v.sub.2'](x)] [greater than or equal to] [absolute value of [v'.sub.1](x)], x [member of] [0,1]. By the induction, [absolute value of [v'.sub.n+1]] [greater than or equal to] [absolute value of [v'.sub.n]], x [member of] [0,1] (n = 0,1,2,...). So we claim that [v.sub.n] [right arrow] [v.sup.*] in norm [parallel]x[parallel] and then T[v.sup.*] = [v.sup.*] and [v.sup.*] [greater than or equal to] 0, 0 [less than or equal to] x [less than or equal to] 1. Hence, [v.sup.*] is concave symmetric positive solution of BVP (1.1). And since [v.sup.*] [member of] [P.sub.b], we have [parallel][v.sup.*][parallel] [less than or equal to] b. Therefore, our proof is complete.

4. EXAMPLE

Example 4.1. We consider the following three-point second-order boundary value problem with integral boundary conditions:

(4.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

It is not difficult to check that the assumptions (H1) and (H2) hold. Let b = 33 and b = 32, then conditions (3.1) and (3.2) are confirmed. Then applying Theorem 3.1, BVP (4.1) has a concave symmetric positive solutions [w.sup.*], [v.sup.*] [member of] P with

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Received April 2, 2015

UMMAHAN AKCAN AND NUKET AYKUT HAMAL

Department of Mathematics, Faculty of science, Anadolu University 26470 Eskisehir, Turkey

Department of Mathematics, Ege University, 35100 Bornova, Izmir, Turkey

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Date:Sep 1, 2015
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