This section discusses singular third-order boundary value problem (1.1).

Let
. Obviously,
is a normal solid cone of Banach space
; by [16, Lemma 2.1.2], we have that
is a generating cone in
.

Theorem 3.1.

Suppose that

, and there exist two positive linear bounded operators

and

with

such that for any

,

,

,

, we have

and there exists

, such that

Then (1.1) has a unique solution

in

. And moreover, for any

, the iterative sequence

converges to
.

Remark 3.2.

Recently, in the study of BVP (1.1), almost all the papers have supposed that the Green's function
is nonnegative. However, the scope of
is not limited to
in Theorem 3.1, so, we do not need to suppose that
is nonnegative.

Remark 3.3.

The function
in Theorem 3.1 is not monotone or convex; the conclusions and the proof used in this paper are different from the known papers in essence.

Proof.

It is easy to see that, for any

,

can be divided into finite partitioned monotone and bounded function on

, and then by (3.2), we have

For any

, let

,

, then

, by (3.1), we have

Following the former inequality, we can easily have

Similarly, by

and

being converged, we have that

Define the operator

by

Then

is the solution of BVP (1.1) if and only if

. Let

By (3.1) and (3.10), for any

,

,

, we have

so we can choose an

, which satisfies

, and so there exists a positive integer

such that

Since

is a generating cone in

, from Lemma 2.1, there exists

such that every element

can be represented in

By (3.16), we know that

is well defined for any

. It is easy to verify that

is a norm in

. By (3.15)–(3.17), we get

On the other hand, for any

which satisfies

, we have

. Thus

, where

denotes the normal constant of

. Since

is arbitrary, we have

It follows from (3.18) and (3.19) that the norms
and
are equivalent.

Now, for any

and

which satisfies

, let

then
,
,
,
, and
.

It follows from (3.12) that

Subtracting (3.22) from (3.21) + (3.23), we obtain

Let

; then we have

As

and

are both positive linear bounded operators, so,

is a positive linear bounded operator, and therefore

. Hence, by mathematical induction, it is easy to know that for natural number

in (3.14), we have

Since

, we see that

which implies by virtue of the arbitrariness of

that

By
, we have
. Thus the Banach contraction mapping principle implies that
has a unique fixed point
in
, and so
has a unique fixed point
in
; by the definition of
has a unique fixed point
in
, that is,
is the unique solution of (1.1). And, for any
, let
; we have
. By the equivalence of
and
again, we get
. This completes the proof.

Example 3.4.

In this paper, the results apply to a very wide range of functions, we are following only one example to illustrate.

Consider the following singular third-order boundary value problem:

where

and there exists

, such that for any

,

,

, we have

Applying Theorem 3.1, we can find that (3.29) has a unique solution

provided

. And moreover, for any

, the iterative sequence

converges to
.

Then (3.1) is satisfied for any
,
,
, and
.

In fact, if

, then

If

, then

Next, for any

, by (3.30) and (3.32), we get

Then, from (3.32) and (3.36), we have

so it is easy to know by induction, for any

, we get

Let

; then

Thus all conditions in Theorem 3.1 are satisfied.