Volume 2017, Issue 1 8579065
Research Article
Open Access

On a Singular Second-Order Multipoint Boundary Value Problem at Resonance

S. A. Iyase

S. A. Iyase

Department of Mathematics, Covenant University, PMB 1023, Ota, Ogun State, Nigeria covenantuniversity.edu.ng

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O. F. Imaga

Corresponding Author

O. F. Imaga

Department of Mathematics, Covenant University, PMB 1023, Ota, Ogun State, Nigeria covenantuniversity.edu.ng

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First published: 11 June 2017
Citations: 3
Academic Editor: Qingkai Kong

Abstract

The aim of this paper is to derive existence results for a second-order singular multipoint boundary value problem at resonance using coincidence degree arguments.

1. Introduction

In this paper we derive existence results for the second-order singular multipoint boundary value problem of the form
(1)
where is Caratheodory’s function (i.e., for each the function f(·, x, y) is measurable on [0,1]; for a.e. t ∈ [0,1], the function f(t, ·, ·) is continuous on ). Let ξi ∈ (0,1), i = 1,2, …, m − 2, 0 < ξ1 < ξ2 < ⋯<ξm−2 < 1, ai ∈ (0,1)  i = 1,2, …, m − 2, and , where f and g have singularity at t = 1.

In [1] Gupta et al. studied the above equation when f and g have no singularity and . They obtained existence of a C1[0,1] solution by utilising the Leray-Schauder continuation principle. In [2] Ma and O’Regan derived existence results for the same equation when f and g have a singularity at t = 1 and . They also utilised the Leray-Schauder continuation method. These results correspond to the nonresonance case. The purpose of this article is therefore to derive existence results for (1) when (the resonance case) and when f and g have a singularity at t = 1. We shall employ coincidence degree arguments in obtaining our results. In this case, the methods used in [1, 2] are not valid.

Research on singular differential equations is important because singular differential equations are useful in the modeling of many problems in the physical and engineering sciences; see [3].

In general singular boundary value problems can be difficult to solve because they may blow up near the singularity. The existence and multiplicity of solutions for second-order nonsingular boundary value problems have been extensively studied by many researchers. However to the best of our knowledge the corresponding problem for second-order differential equations at resonance and with a singularity had not received much attention in the literature. For recent results in these directions see [1, 2, 49] and references therein.

The rest of this paper is organised as follows. In Section 2, we present some definitions, lemmas, and theorems necessary for obtaining our main results. In Section 3, we derive some lemmas and the main theorem. In what follows we shall utilise the following assumptions:
  • (A0) For ξi ∈ (0,1), i = 1,2, …, m − 2, 0 < ξ1 < ξ2 < ⋯<ξm−2 < 1 and .

  • (A1) There exist a(t), c(t) ∈ L1[0,1] with (1 − t)a(t), (1 − t)c(t), b(t) ∈ L1[0,1] and |f(t, x, y)| ≤ a(t)|x| + b(t)|y| + c(t), a.e., t ∈ [0,1], .

  • (A2) is such that .

2. Preliminaries

In this section we state some definitions, theorems, and lemmas that will be used in the subsequent section.

Definition 1. Let X and Z be real Banach spaces. One says that the linear operator L : dom⁡LXZ is a Fredholm mapping of index zero if Ker⁡L and Z/Im⁡L are of finite dimension, where Im⁡L denotes the image of L.

As a result of Definition 1, we will require the continuous projections P : XX, Q : ZZ such that Im⁡P = Ker⁡L, Ker⁡Q = Im⁡L, X = Ker⁡L ⊕ Ker⁡P, Z = Im⁡L ⊕ Im⁡Q, and L|dom⁡L∩Ker⁡P : dom⁡L∩Ker⁡P → Im⁡L is an isomorphism.

Definition 2. Let L be a Fredholm mapping of index zero and Ω a bounded open subset of X such that dom⁡LΩϕ. The map N : XZ is called L-compact on , if the map is bounded and Kp(IQ) is compact, where one denotes by Kp : Im⁡L → dom⁡L∩Ker⁡P the generalised inverse of L. In addition N is L-completely continuous if it is L-compact on every bounded ΩX.

Theorem 3 (see [10].)Let L be a Fredholm operator of index zero and let N be L-compact on . Assume that the following conditions are satisfied:

  • (i)

    LxλNx for every (x, λ)∈[(dom⁡L∖Ker⁡L)∩Ω]×(0,1).

  • (ii)

    Nx ∉ Im⁡L, for every x ∈ Ker⁡LΩ.

  • (iii)

    deg⁡(QN|Ker⁡LΩ, Ω∩Ker⁡L, 0) ≠ 0,

with Q : ZZ being a continuous projection such that Ker⁡Q = Im⁡L. Then the equation Lx = Nx has at least one solution in .

In what follows, we shall make use of the following classical spaces, C[0,1], C1[0,1], L1[0,1], and L[0,1]. Let AC[0,1] denote the space of all absolute continuous functions on [0,1], AC1[0,1] = {xC1[0,1] : x(t) ∈ AC[0,1]}, = {x : x|[0,d]L1[0, d]  for  every  compact  interval  [0, d]⊆[0,1)}.

ACloc⁡[0,1) = {x : x|[0,d]AC[0, d]}.

Let Z be the Banach space defined by
(2)
with the norm
(3)
Let X be the Banach space
(4)
with the norm
(5)
We denote the norm in L1[0,1] by ‖·‖1. We define the linear operator L : dom⁡LXZ by
(6)
where and N : XZ is defined by
(7)
Then boundary value problem (1) can be written as
(8)

Lemma 4 (see [2].)Let yZ. Then

  • (i)

    .

  • (ii)

    .

Lemma 5. If then

  • (i)

    ;

  • (ii)

    ;

  • (iii)

    L : dom⁡LXZ is a Fredholm operator of index zero and the continuous operator Q : ZZ can be defined by

    (9)
    where .

  • (iv)

    The linear operator Kp : Im⁡L : →dom⁡L∩Ker⁡P can be defined as

    (10)

  • (v)

    for all yZ.

Proof. (i) It is obvious that

(11)

(ii) We show that

(12)
To do this, we consider the problem
(13)
and we show that (13) has a solution x(t) satisfying x(0) = 0, if and only if
(14)
Suppose (13) has a solution x(t) satisfying x(0) = 0, ; then we obtain from (13) that
(15)
and applying the boundary conditions we get
(16)
since , and using (i) of Lemma 4 we get
(17)
On the other hand if (14) holds, let ; then , where yZ and x(t) ∈ ACloc⁡[0,1). Then from Lemma 4   and = . Hence
(18)

(iii) For yZ, we define the projection Qy as

(19)
where .

We show that Q : ZZ is well defined and bounded.

(20)
In addition it is easily verified that
(21)
We therefore conclude that Q : ZZ is a projection. If y ∈ Im⁡L, then from (14) Qy(t) = 0. Hence Im⁡L⊆Ker⁡Q. Let y1 = yQy; that is, y1 ∈ Ker⁡Q. Then
(22)
Thus, y1 ∈ Im⁡L and therefore Ker⁡Q⊆Im⁡L and hence . It follows that since , then Z = Im⁡L ⊕ Im⁡Q. Therefore
(23)
This implies that L is Fredholm mapping of index zero.

(iv) We define P : XX by

(24)
and clearly P is continuous and linear and P2x = P(Px) = Px(0) = x(0) = Px and Ker⁡P = {xX : x(0) = 0}. We now show that the generalised inverse Kp : Im⁡L → dom⁡L∩Ker⁡P of L is given by
(25)
For y ∈ Im⁡L we have
(26)
and for x ∈ dom⁡L∩Ker⁡P we know that
(27)
since x ∈ dom⁡L∩Ker⁡P, x(0) = 0, and Px = 0.

This shows that Kp = (L|dom⁡L∩Ker⁡P) −1.

(v)

(28)
We conclude that
(29)

Lemma 6. The operator N : XZ defined by

(30)
is L-completely continuous.

Proof. Suppose Ω is an open bounded subset of X. Let R1 = sup⁡{‖xX : xΩ}. From condition (A1) and each xnΩ we have

(31)
We can deduce from (A1) and (A2) that φ(t) ∈ Z:
(32)
This shows that is bounded in Z and QN is continuous by using the Lebesgue Dominated Convergence Theorem. Next we show that is compact.

By using (31) we derive

(33)
This indicates that the sequence is uniformly bounded in C[0,1]. Also for t ∈ [0,1)
(34)
Hence the sequence KP,QNxn(t) is bounded in C[0,1] and . Thus KP,QNxn(t) is bounded in X.

Next we show that the sequence {KP,QNxn(t)} is equicontinuous. Let t1, t2 ∈ [0,1], t1 < t2; then

(35)
for every t1, t2 ∈ [0,1]. By (i) of Lemma 4  . Thus the sequence {KP,QNxn(t)} is equicontinuous on [0,1] and by Arzela-Ascoli Theorem is convergent. Next we prove that the sequence {(1 − t)KP,QNxn) } is also equicontinuous on [0,1]. We have for t ∈ [0,1]
(36)
Using (i) of Lemma 4 and the fact that αr(t) and φ(t) are in Z we conclude that ψ(t) ∈ L1[0,1]. Therefore
(37)
The sequence {(1 − t)(KP,QNxn) (t)} is therefore equicontinuous on [0,1) and therefore converges to some (1 − t)(KP,QNx0) (t) ∈ C[0,1] with , t ∈ [0,1).

We then conclude that KP,Q is relatively compact and since is bounded we conclude from Definition 2 that N is L-compact on every bounded subset Ω of X and hence N is L-completely continuous.

3. Main Result

In this section we will state and prove the main existence results for problem (1).

Theorem 7. Assume that the following conditions are satisfied:

  • (H1) There exists a positive constant B1 such that, for each x ∈ dom⁡L, if |x(t)| > B1 for all t ∈ [0,1] then

    (38)

  •  (H2) There exists a positive constant B2 such that for and |c| > B2 either (i)  QN(c) ≥ 0 or (ii)  QN(c) ≤ 0.

Then (1) has at least one solution in X provided
(39)

To prove Theorem 7, we first establish some lemmas.

Lemma 8. Let Ω1 = {x ∈ dom⁡L∖Ker⁡L : Lx = λNx,   λ ∈ (0,1)} then Ω1 is bounded in X.

Proof. Let xΩ1. We let Lx = λNx, 0 < λ < 1. Since λ ≠ 0 it is clear that Nx ∈ Im⁡L = Ker⁡Q; hence QNx = 0 for all t ∈ [0,1]. Therefore by assumption (H1) there exist t0 ∈ [0,1] such that |x(t0)| < B1. Now

(40)
(41)
We note that (IP)x ∈ dom⁡L∩Ker⁡P:
(42)
From (41) and (42) we get
(43)
From the definition of N we obtain
(44)
From (43) and (44) we get
(45)
Since 1 − 2[‖aZ + ‖b1] > 0 we obtain that
(46)
Therefore Ω1 is bounded in X.

Lemma 9. The set Ω2 = {x ∈ Ker⁡L : Nx ∈ Im⁡L} is a bounded subset of X.

Proof. Let xΩ2 with x(t) = c, . Then QN(c) = 0 implies N(c) ∈ Im⁡L = Ker⁡Q. We therefore derive from (H2) that

(47)

Lemma 10. The sets and are bounded in X provided (H2)(i) and (H2)(ii) are satisfied simultaneously.

Proof. If QN(c) ≥ 0 then, for with x(t) = c, , we have

(48)
If λ = 0, it follows from (48) that N(c) ∈ Ker⁡Q = Im⁡L; that is, N(c) ∈ Ω2, and therefore by Lemma 9. we have ‖xXB2. However if λ ∈ (0,1) and ‖c‖ > B2 then using assumption (H2)(i) we obtain the contradiction
(49)
Thus ‖xX = |c| < B2. Hence is bounded in X. We can use the same argument to prove that is also bounded in X.

Proof of Theorem 7. We show that the conditions of Theorem 3 are satisfied where Ω is an open and bounded set such that . It is easily seen that conditions (i) and (ii) of Theorem 3 are satisfied by using Lemmas 8 and 9. To verify the third condition we set H(x, λ) = ±λx + (1 − λ)QNx. We choose the isomorphism J : Im⁡Q → Ker⁡L defined by J(c) = c, . By Lemma 10, we derive that H(x, λ) ≠ 0 for all (x, λ)∈(Ker⁡LΩ)×[0,1]. Hence

(50)
Therefore problem (1) has at least one solution in X.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work is supported by the Covenant University Centre for Research, Innovation and Discovery (CUCRID).

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