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

Abstract

In this paper, we consider the existence of positive solutions for the fully fourth-order boundary value problem , where 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 of positive solutions of such BVP.

1. Introduction

In this paper, we study the existence of positive solutions for the fully fourth-order boundary value problem:where is continuous. This 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 has its physical meaning: is the slope, is the bending moment stiffness, is the shear force stiffness, and is the load density stiffness. The nonlinear fourth-order 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 [1–11].

As the nonlinear term does not contain the derivative term of the unknown function, equation (1) becomes

If is superlinear or sublinear growth on , 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 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 . 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).

2. Preliminaries

In this section, we give some assumptions that are important to our main results:(i): is continuous(ii): there exist nonnegative constants and such that , .(iii): there exist nonnegative constants and such that

Let denote the Banach space of continuous functions from into with norm . Let . Then, is a positive cone on .

The functions on cone have the following properties:

Lemma 1 (see [12]). Every function on the cone is differentiable almost everywhere on and satisfies

Let , then the differential equation BVP (1) can be transformed into the following two second-order differential equations:

Let , which is the corresponding Green’s function of BVP (7). Thus, BVP (7) can be transformed into an equivalent integral equation:

By using (9), BVP (8) can be reduced towhere operator is defined as . Thus, BVP (10) can be reduced to the equivalent integral equation:

From the expression of Green’s function , we know that is continuous on , and we have

From the standard proof, we can easily obtain the following statement.

Lemma 2. is a completely continuous operator.

For , we define an operator . It follows from Lemma 1 and condition that

It shows that is an operator which is defined on to . It is easy to conclude that is the continuous bounded operator mapping from to . Therefore, is a completely continuous operator. So, BVP (1) is equivalent to the operator equation , where the operator is given by

Let with . We define an operator on cone :

It is easy to see that is a linear operator.

Let , by (15), we obtain

Then, if scalars , not all equal to zero, there exists a constant satisfies . Thus, from Krein–Rutman theorem, we know , and there is such that , where is the first eigenvalue of the operator .

Now, we estimate the range of . For any , from (15), we getthereby

By the expression of Green’s function and (15), we infer that

According to the above two inequalities, we have

Recalled that . By (20), we have

Note the positivity of operator , we have . Hence, for any , using the recursion method to the above inequality, we get . Therefore, we have

Thus,

From this inequality and Gelfand formula on spectral radius, we obtain

From (24), we get

By , we can get

Now, we calculate and as follows. We have known that

We first calculate . From (15), we get

Through integration, we have

Next, we calculate . Because

Through integration, we have

Combining this with (29), we can obtain

Now, we consider the special case with . We make a definition that

Because of , it is equivalent to and satisfying the differential equation:

We notice that the function is nonnegative. After solving the above differential equations, we may take . Hence, we conclude the value of is .

To prove our main results, we also need the two following lemmas.

Lemma 3 (see [13, 14]). Let is a Banach space, is a cone in , and is a bounded open set in . Assume that is a completely continuous operator. If there exists such thatThen, .

Lemma 4 (see [13, 14]). Let is a Banach space, is a cone in , is a bounded open set in , and . Assume that is a completely continuous operator. Ifthen .

3. Main Results

Theorem 1. Suppose that conditions , and are satisfied, and

and are the first eigenvalues of operator and operator , respectively. Then, BVP (1) has at least one positive solution.

Proof. It follows from and that, for some , the function satisfies

It is known that is the positive eigenvector of corresponding to , and we have . According to (14), (38), and (39), for any , we have

Thus, we get

Let us suppose that has no fixed point on ; otherwise, Theorem 1 is proved.

Now, we provewhere . If otherwise, there exists and such that . Then, we have and . Let , we can know and . By , we get . Based on (41), we have

Then, by the definition of , we conclude that . By (15), we notice that and . So, the above inequality contradicts the definition of . Hence, (42) holds. By Lemma 3, we have

From , we have

Note that . Take

According to Gelfand formula, , and there is a natural number , so that, for , .

For any , we definewhere is an identity operator. It is not difficult to verify that is a norm in , and we havewhere is a constant. Take

Because of , we can take so that when , we have .

Next, we prove

If there exists and , such that , by Lemma 1 and (45), we can obtain

Hence, we have

Furthermore, , and we haveand we also get

Therefore, from (47), we have

When , we have . Hence, from (48) and (55), it follows that

By and (56), we can get . But it is contradictory to the definition of . Hence, (50) holds. According to Lemma 4, we obtain that

Now, from (44) and (57), it follows that

Hence, has at least one fixed point in . That is to say, BVP (1) has at least one positive solution. The proof of Theorem 1 is completed.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The research was supported by the National Natural Science Foundation of China (11571207), Shandong Natural Science Foundation (ZR2018MA011), and the Tai’shan Scholar Engineering Construction Fund of Shandong Province of China.