NEW EXISTING RESULTS FOR A SYSTEM OF NONLINEAR THIRD-ORDER DIFFERENTIAL EQUATION VIA FIXED POINT INDEX∗

By using fixed point index theory, we investigate a system of nonlinear third-order differential equation. We give some sufficient conditions for the existence of at least one or two positive solutions to the system of nonlinear third-order differential equation. As applications, we also present two examples to demonstrate the main results.


Introduction
There are some areas of applied mathematics and physics involving third-order differential equations with different boundary conditions, such as the deflection of a curved beam having a constant or varying cross section, electromagnetic waves, three-layer beams, gravity driven flows see [9]. So third-order differential equations with different boundary conditions have been paid much attention during the past several decades. Especially, the existence of positive solutions for third-order boundary value problems has been studied widely, see [1-8, 10, 12-28] and the references therein. Recently, In [15] the authors established an existence result for positive solutions to the following system of third-order differential equation −v (t) = b(t)g(t, u), t ∈ (0, 1), The method is Krasnoselskii's fixed point theorem. However, we found that the authors replaced u(t), v(t) by some integral expressions in the proof of the main result. Evidently, this is false. So we need continue to study this kind of system for nonlinear third-order differential equation. In addition, there are very few works on a system of nonlinear third-order differential equation in literature. In this paper, we will use fixed point index theory to study the following system: Our purpose here is to give the existence of single and multiple positive solutions to the problem (1.1). By a positive solution of (1.1) we understand a function (u, v) which is positive on 0 < t < 1 and satisfies the differential equation and boundary conditions in (1.1). Assuming that (H 1 ) a(t), b(t) are continuous and do not vanish identically on any subinterval of [0, 1]; (

Preliminary results
In this section we summarize some lemmas which will be used throughout this paper.
Lemma 2.1 (see [11]). Let X be a Banach space and K be a cone in X. For r > 0, define K r = {x ∈ K : x ≤ r}. Assume that T : K r → K is a compact map such that T x = x for x ∈ ∂K r .
(i) If x ≤ T x for x ∈ ∂K r , then i(T, K r , K) = 0; (ii) If x ≥ T x for x ∈ ∂K r , then i(T, K r , K) = 1, where i denotes the fixed point index.

Main results
For convenience, we set The main results of this paper are as follows: Then the problem (1.1) has at least one positive solution.
Theorem 3.2. Assume that (H 1 ), (H 2 ) hold and the following conditions are satisfied: Then the problem (1.1) has at least two positive solutions ( Theorem 3.3. Assume that (H 1 ), (H 2 ) hold and the following conditions are satisfies (H 6 ) There exists a constant ρ 1 > 0 such that where k is given as in Lemma 2.3. Then the problem (1.1) has at least two positive solutions.
To prove the above theorems, we need seek some fixed points of the relative nonlinear operators. So we define .
where k is given as in Lemma 2.3.
Proof. For u, v ∈ K, by Lemma 2.2, and in consequence, for 0 u, v r 1 . In addition, we know that τ 1 satisfies τ 1 l > 1 2 . Hence, By using the same way, It is also easily to prove that Note that r 1 < r 2 , and from the additivity of the fixed point index and (3.1), (3.2), we have Therefore, T has a fixed point (u, v) in K r2 \K r1 . Evidently, (u, v) is a positive solution for problem (1.1) with r 1 < (u, v) < r 2 .

Proof of Theorem 3.2. Let
Hence, Similarly, we can prove that Hence, By the similar way, we can prove Hence, Using the same proof, we get Because ρ 1 < ρ 2 < ρ 3 , from (3.3)-(3.5) and the additivity of the fixed point index, we have Therefore, we can conclude that T has two points (u 1 , v 1 ) and ( That is, these are the positive solutions for the problem (1.1) which satisfy 0 < (u 1 , v 1 ) < ρ 2 < (u 2 , v 2 ) .

Proof of Theorem 3.3.
Firstly, since f 0 = g 0 = 0, there exists r 1 ∈ (0, ρ) such that Then for any (u, v) ∈ ∂K r1 , by using the same calculation in the proof of Theorem 3.2, we have and thus Similarly, we can get Secondly, in view of f ∞ = g ∞ = 0, there exists R > ρ 1 , such that for (u + v) R, where τ 2 > 0 with τ 2 h < 1 4 . We divide the proof into two cases: f is bounded and f is unbounded. Case (i), suppose f is bounded, which implies that there exists M 1 > 0 such that Hence, in either case, we may always set K r2 = {(u, v) ∈ K : (u, v) < r 2 }, such that Similarly, we can prove Therefore, by Lemma 2.1 implies that (u(t) + v(t)) k( u + v ) = kρ 1 . Hence, for any (u, v) ∈ ∂Kρ 1 , by
This shows that T has two fixed points, and consequently, the problem (1.1) has two positive solutions. This completes the proof.