Positive Solutions for Two-Point Boundary Value Problems for Fourth-Order Differential Equations with Fully Nonlinear Terms

In this paper, consider the existence of positive solutions for the fully fourth-order boundary value problem ∞ , + ∞ ) ⟶ [ 0 , + ∞ ] is continuous. This equation can simulate the deformation of an elastic beam simply supported at both ends in a balanced state. By using the ﬁxed-point index theory and the cone theory, we discuss the existence of positive solutions of the fully fourth-order boundary value problem. We transform the fourth-order diﬀerential equation into a second-order diﬀerential equation by order reduction method. And then, we examine the spectral radius of linear operators and the equivalent norm on continuous space. After that, we obtain the existence


Introduction
In this paper, we study the existence of positive solutions for the fully fourth-order boundary value problem: is boundary value problem can simulate the deformation of an elastic beam, whose one end is fixed and the other end is free in a balanced state. In mechanics, BVP (1) is called a cantilever beam equation. In this equation, each derivative of u(t) has its physical meaning: u ′ (t) is the slope, u ″ (t) is the bending moment stiffness, u ‴ (t) is the shear force stiffness, and u (4) (t) is the load density stiffness. e nonlinear fourthorder differential equation boundary value problem can simulate the deformation of an elastic beam under external force, and different boundary value conditions can show its force under different conditions. Because of its importance in mechanics, many scholars have done a lot of research on the existence of solutions for fourth-order ordinary differential equations using various nonlinear methods [1][2][3][4][5][6][7][8][9][10][11].
As the nonlinear term does not contain the derivative term of the unknown function, equation (1) becomes u (4) (t) � f(t, u(t)), 0 ≤ t ≤ 1, If f(t, u) is superlinear or sublinear growth on u, the authors in [1] used the fixed-point theorem on the cone to obtain the existence of the positive solution of equation (2). In [3], the author used the fixed-point theorem and topological degree theory to study the existence of one or two positive solutions for the fourth-order differential equation boundary value problem: Under the fixed-point index method on the cone, the authors of [11] discussed the existence of positive solutions for fourth-order boundary value problems with two parameters. Among them, the assumption condition of the nonlinear term f is related to the first eigenvalue of the corresponding linear operator. It is noteworthy that the nonlinear term in the abovementioned boundary value problems does not include the higher-order derivative u ‴ . When the nonlinear term contains the higher-order derivative of the unknown function, the authors of [8,9] used the upper and lower solution method to study the existence of solutions for fully fourth-order nonlinear boundary value problems with nonlinear boundary conditions. In [10], the author discussed a fourth-order boundary value problem with fully form: When the nonlinear term satisfies superlinear growth and sublinear growth, the author used the fixed-point index method, combined with the positivity of linear operators and spectral radius, to get the positive solutions for the boundary value problem. But the linear operator in [10] does not involve the first and second derivatives of unknown functions.
In this paper, by using cone theory and the fixed-point index, combined with the spectral radius of linear integral operators, and the application of equivalent norms, we discuss the existence of positive solutions for boundary value problems (1).

Preliminaries
In this section, we give some assumptions that are important to our main results: r], and x 4 ∈ (−∞, +∞).

(6)
Let v(t) � −u ″ (t), then the differential equation BVP (1) can be transformed into the following two second-order differential equations: , which is the corresponding Green's function of BVP (7). us, BVP (7) can be transformed into an equivalent integral equation: By using (9), BVP (8) can be reduced to where operator A is defined as (Av)(t) � 1 0 G(t, s)v(s)ds. us, BVP (10) can be reduced to the equivalent integral equation: From the standard proof, we can easily obtain the following statement.
. It follows from Lemma 1 and condition (H 3 ) that 2 Mathematical Problems in Engineering It shows that F is an operator which is defined on P to L(0, 1). It is easy to conclude that F is the continuous bounded operator mapping from P to L(0, 1). erefore, T � A°F: P ⟶ P is a completely continuous operator. So, BVP (1) is equivalent to the operator equation u � Tu, where the operator T � P ⟶ P is given by Let a � (a, b, c) with a, b, c ≥ 0. We define an operator T a on cone P: It is easy to see that T a : P ⟶ P is a linear operator.
en, if scalars a, b, c ≥ 0, not all equal to zero, there exists a constant α > 0 satisfies T a u 0 ≥ αu 0 .
us, from Krein-Rutman theorem, we know r(T a ) > 0, and there is v a ∈ P/ θ { } such that λ a T a v a � v a , where λ a � (r(T a )) − 1 is the first eigenvalue of the operator T a . Now, we estimate the range of λ a � (r(T a )) − 1 . For any u ∈ P, from (15), we get s)u(s)ds, a, b, c ≥ 0, a � (a, b, c), By the expression of Green's function G(t, s) and (15), we infer that �(a, b,c).

(19)
According to the above two inequalities, we have Recalled that u 0 (t) � t(1 − t). By (20), we have Note the positivity of operator T a , we have T 2 a u 0 ≥ ‖T a u 0 ‖ 2 u 0 . Hence, for any n ∈ N, using the recursion method to the above inequality, we get T n a u 0 ≥ ‖T a u 0 ‖ n u 0 . erefore, we have us,

Mathematical Problems in Engineering
From this inequality and Gelfand formula on spectral radius, we obtain From (24), we get By λ a � (r(T a )) − 1 , we can get Now, we calculate ‖T a ‖ and ‖T a u 0 ‖ as follows. We have known that We first calculate ‖T a ‖. From (15), we get rough integration, we have (29) Next, we calculate ‖T a u 0 ‖. Because rough integration, we have Combining this with (29), we can obtain 48 a + 4b + 6c ≤ λ a ≤ 960 4a + 16b + 25c  , a, b, c > 0, a � (a, b, c).

(32)
Now, we consider the special case with b � (a 1 , 0, c 1 ). We make a definition that We notice that the function v b (t) is nonnegative. After solving the above differential equations, we may take v b (t) � sin πt. Hence, we conclude the value of λ b is λ b � π 4 /a 1 + π 2 c 1 .
To prove our main results, we also need the two following lemmas.
Lemma 3 (see [13,14]). Let E is a Banach space, P is a cone in E, and Ω(P) is a bounded open set in P. Assume that A: Ω(P) ⟶ P is a completely continuous operator. If there Then, i(A, Ω(P), P) � 0.

Main Results
λ a and λ b are the first eigenvalues of operator T a and operator T b , respectively. en, BVP (1) has at least one positive solution.
Proof. It follows from (H 1 ) and (H 2 ) that, for some r > 0, the function f satisfies According to (14), (38), and (39), for any v ∈zB r ∩ P, we have (40) us, we get Let us suppose that T has no fixed point on zB r ∩ P; otherwise, eorem 1 is proved.