Solvability of Third-Order Three-Point Boundary Value Problems

and Applied Analysis 3 where G(t, s) is defined by

Despite the success in the study of ( 1)-( 2) and ( 3)-( 4), it has been recognized that for the resonant cases, that is, (1)-( 2) with  = 1, no much work has been known except recent work [12,17,18] and the references therein where the so-called coincidence-degree method was employed.
More recently, Ouyang and Li [19] have discussed a class of fractional order differential equations of the following three-point boundary value problem with resonance: Using a new method, they obtained some sufficient conditions for the existence of solutions for the fractional order boundary value problem (8).
The purpose of this paper is to study the problem (1)-(2) for the nonresonant case and the resonant case.In the nonresonant case, the Krasnosel'skii fixed point theorem is used to prove the existence of positive solutions for the problem (1)-( 2).In the resonant case, a completely new method is incorporated; we transform the problem into an integral equation with an undetermined parameter.The Intermediate Value Theorem is applied to determine the particular value of the parameter so that true solutions exist.Not only the existence conditions of the solutions, but also the prove of the main results are more simple than [12].
We introduce two lemmas as follows.
Lemma 1 (the Krasnosel'skii fixed point theorem [20]).Let  be a cone in a Banach space .The paper is arranged as follows.In Section 2, we discuss the existence of the positive solutions of the problem (1)-( 2) in nonresonant case.Section 3 is devoted to the existence of the solutions of the problem (1)-( 2) in resonant case.Finally, we give some examples to illustrate our results.

The Nonresonant Case (𝛼 < 1)
Throughout this paper, we suppose that  [0, 1] is a space of continuous functions in [0, 1], equipped with the norm In this section, we consider the nonresonant case, that is, (1) with the boundary value problem (2) with 0 ≤  < 1.We have the following theorem.Theorem 3. Assume that ( 1 ) holds and 0 ≤  < 1.Then problem (1)-( 2) has at least one positive solution.

The Resonant Case (𝛼 = 1)
In this case, the boundary value condition ( 2) can be rewritten as We have the following main theorem.
Proof of Theorem 5. Using a similar method as in the proof of Theorem 3, the problem (1)-( 34) is equivalent to the following integral equation: where (, ) is defined by (36).It is obvious that Green's function (, ) is decreasing with respect to  when  ≥ , and it is independent on the parameter  when  < ; thus, the function (, ) is not increasing with respect to  ∈ [0, 1] and 0 ≤ (, ) ≤ (, ). Let Given any value (1), () satisfies the Hammerstein integral equation by (37): To obtain the solvability of (39), we replace (1) by a real constant ; that is, for any real number .From the condition (H 2 ), it is easy to know that here, we have made use of for any 0 ≤  ≤ 1.By (35), that is, sup we then see that the operator  maps the ball   onto itself, where It is easy to show that the operator  is a compact operator.From Lemma 2 and using a similar method of Theorem 3.6 in [22], we obtain that the operator  has a fixed point   () for any real number .Let   be the fixed point of  with a given parameter ; that is,   =   .For the solvability of (40), we need to find a  0 so that   0 (1) = 0; that is, It is obvious that () is continuously dependent on the parameter ; this would help us to show the existence of  0 such that ( 0 ) = 0. Now, we show that () → ∞ as  → ∞.