On the Solvability of a Resonant Third-Order Integral m-Point Boundary Value Problem on the Half-Line

Abstract

In this work, the existence of at least one solution for the following third-order integral and -point boundary value problem on the half-line at resonance will be investigated. The Mawhin’s coincidence degree theory will be used to obtain existence results while an example will be used to validate the result obatined.

1. Introduction

This work studies the existence of solution for a resonant third-order boundary value problem with integral and -point boundary conditions on the half-line where is an -Carathéodory function, , , , , for , and the resonance condition is .

Boundary value problems on the half-line arise in the modeling of various physical processes like the flow of fluid over semi-infinite porous media and wet surfaces [1].

Boundary value problem (1)-(2) is said to be at resonance since the corresponding homogeneous problem has a nontrivial solution , where is a constant. Boundary value problems at resonance can be expressed in the abstract form as , where is a linear differential operator that is not invertible. Mawhin's coincidence degree theory [2] is an excellent tool for studying resonant problems of this type.

The problem of existence of solutions for resonant boundary value problems has received the attention of many authors recently, both in the bounded domain and on the half-line. For instance, the authors in [3] studied the third-order problem under the resonance conditions: and . They applied the coincidence degree arguments to obtain existence results in a bounded domain. For other works on a bounded domain, see [4–7].

In [8], the authors considered the multipoint boundary value problem on the half-line where and . They applied a perturbation technique in obtaining existence results under the resonant condition .

Iyase [9] used coincidence degree arguments to study existence of solutions for the multipoint boundary value problem at resonance on the half-line under the resonant condition which is different from ours. For other literature on resonant multipoint boundary value problems on the half-line, see [10–14]. Motivated by the results mentioned above, we study the existence of solutions for the resonant third-order boundary value problem with integral and -point boundary conditions on the half-line. In Section 2 of this work, necessary lemmas, theorems, and definitions will be given; Section 3 will be dedicated to stating and proving the condition for existence of solutions. An example will be given in Section 4 to corroborate the result obtained.

2. Preliminaries

In this section, we will give definitions, theorems, and lemmas that are required for this work.

Definition 1 (see [6]). Let and be Banach spaces. A linear operator is called a Fredholm mapping of index zero if and are finite dimensional.

Take , to be normed spaces, a Fredholm mapping of zero index and projectors that are continuous such that , , and , , then is invertible. The inverse of the mapping will be denoted by while the generalized inverse, is defined as .

Definition 2 (see [13]). A map is -Carathéodory, if the following conditions are satisfied: (i)For each , the mapping is Lebesgue measurable(ii)For a.e. , the mapping is continuous on (iii)For each , there exists , with such that, for a.e. and every , we have

Definition 3. Let be a Fredholm mapping, a metric space, and a nonlinear mapping. is said to be -compact on if and are compact on . Also, is -completely continuous if it is -compact on every bounded .

Theorem 4 (see [9]). Let be the space of all bounded continuous vector-valued functions on and . Then, is relatively compact on if the following conditions hold: (i) is a bounded subset of (ii)The functions in are equicontinuous on any compact interval of (iii)The functions from are equiconvergent; that is, if given , there exists such that , for all and

Theorem 5 (see [2]). Let be a Fredholm map of index zero and let be -compact on . Assume that the following conditions are satisfied: (i) for every (ii) for every (iii), where is a projection such that

Then, the abstract equation has at least one solution in .

Let where is the set of absolutely continuous functions. The norm defined on is where .

Let and define the norm where , and is the supremum norm on . The linear operator will be defined by where

Also, the nonlinear operator will be defined by , ; thus, problem (1)-(2) may be written in the form .

Lemma 6. The following conditions hold: (i)(ii) is a Fredholm operator of index zero, while the continuous linear projector may be defined as where (iii)The generalized inverse of may be written as Furthermore, where

Proof. Let then ☐

From (2), we have ; then, (17) becomes

Therefore, ; hence, (i) holds. Next, we show that (ii), (iii), and (iv) also hold.

For , consider the problem which has a solution defined as

Applying the boundary conditions and using , one obtains where is an arbitrary constant and is a solution of (19) satisfying (2). Therefore, condition (ii) holds.

For any , let the projector

Let , then , where is the identity operator. Since then and . From , we have . Therefore, . Thus, is a Fredholm operator of index zero and (iii) holds.

Given defined as the generalized inverse of can then be written as

In fact, for any , and for , it follows that

Since , then thus, . In addition,

Hence, (29), (30), and (31) give

Lemma 7. The nonlinear operator is -compact if is an -Carathéodory function.

Proof. To prove this lemma, we have to show that Theorem 4 satisfied the operator . This we will do in three steps. Given that is bounded and , consider . Since satisfies the -Carathéodory conditions with respect to , there exists a Lebesgue integrable function , with such that ☐

Hence, for all ,

Hence, is bounded. Next, we prove that is compact. For , we have

Hence, is uniformly bounded in ; thus, (i) of Also, we will prove that (ii) of Theorem 4 holds. Also, we will prove that (ii) of Theorem 4 holds, that is, is equicontinuous on . Let , where . Then,

Thus, is equicontinuous every compact subset of . Finally, we show that (iii) of Theorem 4 holds; that is, is equiconvergent at :

Hence, is equiconvergent at . Thus, from Definition 3 and Theorem 4, we see that the nonlinear operator is -compact.

3. Existence Result

Theorem 8. Let be an -Carathéodory function. If the following hold:
()There exists constant and functions , satisfying , such that for all and , ()There exists a constant , such that for , if , for all , then ()There exists a constant , such that for any , if , then Then, problem (1) subject to (2) has at least one solution if where .

In order to prove Theorem 8, the following lemmas are required. These lemmas will help us prove that condition (iii) of Theorem 5 holds.

Lemma 9. The set is bounded.

Proof. Assume that and , then , and by condition () of Theorem 8, there exists , such that . By the absolute continuity of , it follows from that From , one gets and from , we obtain ence, from (43) and (45) Also, for , , then and ; then, from Lemma 6, one gets Hence, from (46) and (47), Setting gives From (38), we see that From (49), we obtain Hence, is bounded.☐

Lemma 10. Given that . Suppose that (40) holds, for all where is the linear isomorphism defined by , then is bounded.

Proof. For all , one obtains When , then . However, when , then in view of (40), one gets contradicting . Then, , implying that . Hence, is bounded.☐

Proof of Theorem 8. We will now show that all the conditions of Theorem 5 hold. Take as an open bounded subset of . Lemma 6 shows that is a Fredholm operator of index zero, and from Lemma 7, is -compact on . Thus, conditions (i) and (ii) of Theorem 5 are satisfied based on the definition of . Finally, we show that condition (iii) of Theorem 5 holds. Set . Since , then for every , . Therefore, the homotopy property of the Brouwer degree gives ☐

Hence, condition (iii) of Theorem 5 holds, and problem (1)-(2) has at least one solution in .

4. Example

Consider the following problem where

Given that , then and hence, condition () is satisfied. For any such that , where ,

Now, . Taking , , , , , then satisfy condition (). Finally,

All the conditions are satisfied, and we conclude from Theorem 8 that there exists at least one solution for problem (55) subject to (56).

Data Availability

N/A.

Conflicts of Interest

The authors declare they have no competing interest.

Acknowledgments

The authors acknowledge Covenant University for the support received from them.