The Iterative Technique for a Fourth-Order Three-Point Nonlinear BVP with Changing Sign Green’s Function

The Iterative Technique for a Fourth-Order Three-Point Nonlinear BVP with Changing Sign Green’s Function

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The Iterative Technique for a Fourth-Order Three-Point Nonlinear BVP with Changing Sign Green’s Function Banner

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Abstract

In this paper, we discuss the existence of a monotonic positive solution for the following fourth-order three points Non-linear BVP: which has the sign -changing Green’s function. where and . The point is that although the corresponding Green is changing the sign, by applying iterative methods, We can still obtain the existence of a monotonic positive solution under certain suitable conditions of .

I. INTRODUCTION

{ u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( η ) = 0

In this article, we aim to study the existence of a monotonic positive solution for fourth-order three-point Nonlinear BVP with changing sign Green's function

( 1.1 ) { u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( η ) = 0

where α ∈ [0,2), f ∈ C(...) where α ∈ [ 0 , 2 ) , f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) and η ∈ [ 1 2 , 1 ) . By using iterative methods, We can still obtain the existence of a monotonic positive solution under certain suitable conditions of f .

In recent decades, The differential equations come from various fields of a mathematical applied and physics, for example, in the deflection of curved beams with constant or varying cross-sections, triple-layer beams, electromagnetic waves or gravity-driven currents, etc. [1].

Recently, when the corresponding Green's function is undergoing sign conversion, there has been some work on the positive solution of the second and third-order BVP. For example, in [8] the existence of at least one positive solution of the following second-order periodic BVP with positive and negative transformation Green's function studied by Zhong and An

{ u ′ ′ ( t ) + ρ 2 u = f ( u ) , 0 < t < T , u ( 0 ) = u ( T ) , u ′ ( 0 ) = u ′ ( T ) ,

where η ∈ ( 17 24 , 1 ) , Palamide and Smirlis [9] discussed the existence of at least one positive solution. Their technique is a combination of GuoKrasnosel'ski fixed point theory and the corresponding vector field characteristics. In 2012, Sun and Zhao [10], [11] obtained single or multiple positive solutions with three-point positive and negative BVP by applying the fixed point theory of Guo-Krasnosel'skii

{ u ′ ′ ′ ( t ) = f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = u ′ ′ ( η ) = 0

Motivated. Through the above work, this article will study BVP (1.1)Through an iterative method. Throughout this article, we always assume α ∈ [ 0 , 2 ) and η ∈ [ 1 2 , 1 ) . Although the corresponding Green function is changing its sign, under certain suitable conditions, we can still obtain the existence of the monotonic positive solution of BVP (1.1) on f . Moreover, our iterative scheme starts with a zero function, This means that iterative scheme is feasible.

( 1.1 ) { u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( η ) = 0

In this article, by applying an iterative approach, we always assume that α ∈ [ 0 , 2 ) and η ∈ [ 1 2 , 1 ) .

We first recall the following fixed point of Krasnoselskii's type.

Theorem 1.1. Let E be a Banach space and K be a cone in E . Assume that Ω 1 and Ω 1 are bounded open subsets of E such that 0 ∈ Ω 1 , Ω ― 1 ⊂ Ω 2 , and let A : K ∩ ( Ω 2 ∖ Ω 1 ) → K be a completely continuous operator such that either

(1) ‖ A u ‖ ≤ ‖ A u ‖ for u ∈ K ∩ ∂ Ω 1 and ‖ A u ‖ ≥ ‖ u ‖ for u ∈ K ∩ ∂ Ω 2 or (2) ‖ A u ‖ ≥ ‖ A u ‖ f o r u ∈ K ∩ ∂ Ω 1 and ‖ A u ‖ ≤ ‖ u ‖ for u ∈ K ∩ ∂ Ω 2

  • A. has a fixed point in K ∩ ( Ω ― 2 ∖ Ω 1 ) .

II. PRELIMINARIES

For the BVP

( 2.1 ) { u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( η ) = 0

we have the following lemma

Lemma 2.1. The BVP (2.1) has only trivial solution.

Proof. It is simple to check. for any y ∈ C [ 0 , 1 ] , we consider the boundary value problems

( 2.2 ) { u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( η ) = 0

After a direct computation, one may obtain the expression of Greens function G ( t , s ) of the BVP (2.2) as following:

Proof. Integrating four times the linear problem gives us that

u ′ ′ ′ ( t ) = u ′ ′ ′ ( 0 ) + ∫ 0 t y ( s ) d s ,
u ′ ′ ( t ) = u ′ ′ ( 0 ) + t u ′ ′ ′ ( 0 ) + ∫ 0 t ( t − s ) y ( s ) d s ,
u ′ ( t ) = u ′ ( 0 ) + t u ′ ′ ( 0 ) + t 2 2 u ′ ′ ′ ( 0 ) + 1 2 ∫ 0 t ( t − s ) 2 y ( s ) d s ,
u ( t ) = u ( 0 ) + t u ′ ( 0 ) + t 2 2 u ′ ′ ( 0 ) + t 3 6 u ′ ′ ′ ( 0 ) + 1 6 ∫ 0 t ( t − s ) 3 y ( s ) d s .

The conditions u ′ ( 0 ) = u ‴ ( 0 ) = 0 implies that

u ( t ) = u ( 0 ) + t 2 2 u ′ ′ ( 0 ) + 1 6 ∫ 0 t ( t − s ) 3 y ( s ) d s ,

and the conditions u ( 1 ) = 0 this means

u ( 1 ) = u ( 0 ) + 1 2 u ′ ′ ( 0 ) + 1 6 ∫ 0 1 ( 1 − s ) 3 y ( s ) d s = 0 ,

Next, α u ( 0 ) + u ″ ( η ) = 0 is rewritten Such as

u ′ ′ ( 0 ) + ∫ 0 η ( η − s ) y ( s ) d s − α 2 u ′ ′ ( 0 ) − α 6 ∫ 0 1 ( 1 − s ) 3 y ( s ) d s = 0 ,

whence

( 2.3 ) u ′ ′ ( 0 ) = α 3 ( 2 − α ) ∫ 0 1 ( 1 − s ) 3 y ( s ) d s − 2 2 − α ∫ 0 η ( η − s ) y ( s ) d s .

form The conditions u ( 1 ) = 0 we have

u ( 0 ) = − 1 2 u ′ ′ ( 0 ) − 1 6 ∫ 0 1 ( 1 − s ) 3 y ( s ) d s

If we substitute(1.3) with the expression from above and simplify, we get that

( 2.4 ) u ( 0 ) = 1 2 − α ∫ 0 η ( η − s ) y ( s ) d s − 1 3 ( 2 − α ) ∫ 0 1 ( 1 − s ) 3 y ( s ) d s .

Finally, we obtain that

u ( t ) = 1 2 − α ∫ 0 η ( η − s ) y ( s ) d s − 1 3 ( 2 − α ) ∫ 0 1 ( 1 − s ) 3 y ( s ) d s + α t 2 6 ( 2 − α ) ∫ 0 1 ( 1 − s ) 3 y ( s ) d s − t 2 2 − α ∫ 0 η ( η − s ) y ( s ) d s + 1 6 ∫ 0 t ( t − s ) 3 y ( s ) d s .

As a result, we have that For s ≥ η

G ( t , s ) = { − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) 0 ≤ t ≤ s , ( t − s ) 3 6 − ( 2 − α t 3 ) ( 1 − s ) 3 6 ( 2 − α ) s ≤ t ≤ 1

and s < η

G ( t , s ) = { 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) 0 ≤ t ≤ s , ( t − s ) 3 6 + 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) s ≤ t ≤ 1

Remark 2.1. It is not difficult to verify that G ( t , s ) has the following characteristics:

G ( t , s ) ≥ 0 for 0 ≤ s ≤ η and G ( t , s ) ≤ 0 for η ≤ s ≤ 1.

Moreover, if s ≥ η , then

max G ( t , s ) : t ∈ [ 0 , 1 ] = G ( 1 , s ) = 0 ,
min G ( t , s ) : t ∈ [ 0 , 1 ] = G ( 0 , s ) = − ( 1 − s ) 3 3 ( 2 − α ) ≥ − ( 1 − η ) 3 3 ( 2 − α )

if s < η , then

max G ( t , s ) : t ∈ [ 0 , 1 ] = G ( 0 , s ) = ( s 3 − 3 s 2 ) + ( 3 η − 1 ) 3 ( 2 − α ) ≤ ( η 3 − 3 η 2 ) + ( 3 η − 1 ) 3 ( 2 − α ) ,
min G ( t , s ) : t ∈ [ 0 , 1 ] = G ( 1 , s ) = 0

therefore, if we let δ = max | G ( t , s ) | : t , s ∈ [ 0 , 1 ] then

δ = max { − ( 1 − η ) 3 3 ( 2 − α ) , ( η 3 − 3 η 2 ) + ( 3 η − 1 ) 3 ( 2 − α ) } < η − s ( 2 − α )

Now, let Banach space E = C [ 0 , 1 ] is equipped with the ‖ u ‖ = max t ∈ [ 0 , 1 ] | u ( t ) | .

K = { y ∈ C [ 0 , 1 ] : y ( t ) } is nonnegative and decreasing on [0, 1]. Then K is a cone in C [ 0 , 1 ] .

Note that this order relationship is inducesan in E by defining u v if and only if u − v ∈ K

In the remainder of this paper, we always assume that f : C [ 0 , 1 ] × [ 0 , + ∞ ) → [ 0 , + ∞ ) is continuous and satisfies the following conditions:

(F1) For each u ∈ [ 0 , + ∞ ) , the mapping t ↦ f ( t , u ) is decreasing;

(F2) For each t ∈ [ 0 , 1 ] , the mapping u ↦ f ( t , u ) is increasing.

( 2.5 ) ( A u ) ( t ) = ∫ 0 1 G ( t , s ) f ( s , u ( s ) ) d s , t ∈ [ 0 , 1 ]

Obviously, if u is a fixed point of A in K , then u is a nonnegative and decreasing solution of the BVP (1.1).

Lemma 2.2. Let A : K → K is completely continuous.

Proof. let u ∈ K . Then, for 0 ≤ t ≤ η , we have

( A u ) ( t ) = ∫ 0 t [ ( t − s ) 3 6 + 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] y ( s ) d s + ∫ t η [ 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] y ( s ) d s + ∫ η 1 − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) y ( s ) d s

which together with ( F 1 ) and ( F 2 ) implies that

( A u ) ′ ( t ) = ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 1 2 t ( η − s ) 6 ( 2 − α ) y ( s ) d s + 1 2 ∫ 0 t ( s 2 − 2 t s ) y ( s ) d s + t 2 2 ∫ t η y ( s ) d s + ∫ η 1 α t ( 1 − s ) 3 3 ( 2 − α ) y ( s ) d s ≤ y ( η ) [ ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 1 2 t ( η − s ) 6 ( 2 − α ) + 1 2 ∫ 0 t ( s 2 − 2 t s ) + t 2 2 ∫ t η + ∫ η 1 α t ( 1 − s ) 3 3 ( 2 − α ) ] d s ≤ t y ( η ) [ 4 η t ( 2 − α ) + ( α − 8 η ) 1 2 ( 2 − α ) − 5 t 2 6 + t η 2 ] ≤ b t y ( η ) [ 4 η t ( 2 − α ) + ( α − 8 η ) 1 2 ( 2 − α ) − η 8 6 ] ≤ 0

At the same time, η > 1 2 shows that

( A u ) ′ ′ ( t ) = ∫ 0 η 6 t ( 2 − α ) + 2 α ( 1 − s ) 3 − 1 2 ( η − s ) 6 ( 2 − α ) y ( s ) d s − ∫ 0 t s y ( s ) d s + t ∫ t η y ( s ) d s + ∫ η 1 α ( 1 − s ) 3 3 ( 2 − α ) y ( s ) d s ≤ y ( η ) [ ∫ 0 η 6 t ( 2 − α ) + 2 α ( 1 − s ) 3 − 1 2 ( η − s ) 6 ( 2 − α ) − ∫ 0 t s d s + t ∫ t η + ∫ η 1 α ( 1 − s ) 3 3 ( 2 − α ) ] d s ≤ y ( η ) [ α ( 3 t − 2 η ) + 6 ( η − t ) 2 ( 2 − α ) − 2 η 2 + α ] ≤ y ( η ) [ η ( α − 2 η ) 2 ( 2 − α ) + α ] ≤ 0 t ∈ ( 0 , η )

For t ∈ [ η , 1 ] , we have

( A u ) ( t ) = ∫ 0 η [ ( t − s ) 3 6 − 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] y ( s ) d s + ∫ η t [ ( t − s ) 3 6 − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] y ( s ) d s + ∫ t 1 [ − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] y ( s ) d s

which together with ( F 1 ) and ( F 2 ) implies that

( A u ) ′ ( t ) = ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 12 t ( η − s ) 6 ( 2 − α ) d s + 1 2 ∫ 0 η ( s 2 − 2 t s ) y ( s ) d s + ∫ η t ( t − s ) 2 2 + ∫ η 1 2 α t ( 1 − s ) 3 6 ( 2 − α ) y ( s ) d s ≤ y ( η ) [ ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 12 t ( η − s ) 6 ( 2 − α ) + 1 2 ∫ 0 η ( s 2 − 2 t s ) + ∫ η t ( t − s ) 2 2 + ∫ η 1 α t ( 1 − s ) 3 6 ( 2 − α ) ] d s = t y ( η ) [ ( α − 12 η ) 12 ( 2 − α ) + t η − η 2 2 + 1 + η 3 6 − η ] ≤ t y ( η ) [ ( α − 12 η ) 12 ( 2 − α ) + η 3 6 + η 2 2 − η + 1 ] ≤ 0 t ∈ ( η , 1 )

So, ( A u ) ( t ) is decreasing on [ 0 , 1 ] . At the same time, since ( A u ) ( 1 ) = 0 , we know that ( A u ) ( t ) is nonnegative on [ 0 , 1 ] . This indicates that ( A u ) ( t ) ∈ K . Furthermore, although G ( t , s ) is not continuous, it follows from known text book results, for example, see [12], that A : K → K is completely continuous

Theorem 2.3. Assume that f ( t , 0 ) ≠ 0 for t ∈ [ 0 , 1 ] and there exist two positive constants a and b such that the following conditions are satisfied:

( H 1 ) f ( 0 , a ) ≤ 6 ( 2 − α ) a , ( H 2 ) b ( u 2 − u 1 ) ≤ f ( t , u 2 ) − f ( t , u 1 ) ≥ 2 b ( u 2 − u 1 ) , t ∈ [ 0 , 1 ] , 0 ≤ u 1 ≤ u 2 ≤ a .

If we construct a iterative sequence v n + 1 = A v n and n = 0 , 1 , 2 , … , where v 0 ( t ) ≡ 0 for t ∈ [ 0 , 1 ] , then v n , n = 1 ∞ converges to v † in E and v † is a decreasing positive solution of the BVP (1.1)

Proof. Let K a = u ∈ K : ‖ u ‖ ≤ a . Then we may assert that A : K a → K a .

In fact, if u ∈ K a , then it follows from (H1) that

0 ≤ ( A u ) ( t ) = ∫ 0 1 G ( t , s ) f ( s , u ( s ) ) d s ≤ ∫ 0 1 | G ( t , s ) | f ( 0 , a ) d s ≤ 6 ( 2 − α ) a δ < a , t ∈ [ 0 , 1 ] ,

which indicate that ‖ A u ‖ ≤ a

s o A : K a → K a .

Now, we prove that v n n = 1 ∞ converges to v † in E and v † is a decreasing positive solution of (1.1). Indeed, in view of v 0 ∈ K a and A : K a → K a , we have v n ∈ K a , n = 1 , 2 , … . Since the set v n n = 0 ∞ is bounded and A is completely continuous, we know that the set v n n = 1 ∞ is relatively compact. In what follows, we prove that v n n = 0 ∞ is monotone by induction. First, it is explicit that v 1 − v 0 = v 1 ∈ K , Which indicates this v 1 v 0 . Subsequently, we suppose that v k − 1 v k . Then v k − v k − 1 is decreasing and 0 ≤ v k − 1 ( t ) ≤ v k ( t ) ≤ a , 0 ≤ t ≤ 1 . So, it follows from (H2) that for 0 ≤ t ≤ η

v k + 1 ′ ( t ) − v k ′ ( t ) = 1 6 ( 2 − α ) ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 12 t ( η − s ) [ f ( s , v k ( s ) ) − f ( s , v k − 1 ( s ) ) ] d s + 1 2 ∫ 0 η ( s 2 − 2 t s ) [ f ( s , v k ( s ) ) − f ( s , v k − 1 ( s ) ) ] d s + 1 2 ∫ η t ( t − s ) 2 [ f ( s , v k ( s ) ) − f ( s , v k − 1 ( s ) ) ] d s + α t 3 ( 2 − α ) ∫ η 1 ( 1 − s ) 3 [ f ( s , v k ( s ) ) − f ( s , v k − 1 ( s ) ) ] d s ≤ b 6 ( 2 − α ) ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 12 t ( η − s ) [ v k ( s ) − v k − 1 ( s ) ] d s + b 2 ∫ 0 η ( s 2 − 2 t s ) [ v k ( s ) − v k − 1 ( s ) ] d s + b 2 ∫ η t ( t − s ) 2 [ v k ( s ) − v k − 1 ( s ) ] d s + α t 6 ( 2 − α ) ∫ η 1 ( 1 − s ) 3 [ v k ( s ) − v k − 1 ( s ) ] d s
≤ b [ v k ( η ) − v k − 1 ( η ) ] × [ 1 6 ( 2 − α ) ∫ 0 η 3 t 2 ( 2 − α ) + 2 α t ( 1 − s ) 3 − 1 2 t ( η − s ) + 1 2 ∫ 0 η ( s 2 − 2 t s ) + 1 2 ∫ η t ( t − s ) 2 + α t 3 ( 2 − α ) ∫ η 1 ( 1 − s ) 3 ] d s = b [ v k ( η ) − v k − 1 ( η ) ] t [ ( α − 1 2 η ) 1 2 ( 2 − α ) + t η − η 2 2 + 1 + η 3 6 − η ] ≤ b [ v k ( η ) − v k − 1 ( η ) ] t [ ( α − 1 2 η ) 1 2 ( 2 − α ) + η 3 6 + η 2 2 − η + 1 ] ≤ 0 t ∈ ( η , 1 )

And therefore,

v k + 1 ′ ( t ) − v k ′ ( t ) ≤ 0 , v k + 1 ′ ′ ( t ) − v k ′ ′ ( t ) ≤ 0 t ∈ [ 0 , 1 ]

This together with

v k + 1 ′ ( t ) − v k ′ ( t ) = ∫ 0 1 G ( 1 , s ) [ f ( s , v k ( s ) ) − f ( s , v k − 1 ( s ) ) ] d s , t ∈ [ 0 , 1 ] .

v k + 1 ( t ) − v k ( t ) ≥ 0 , t ∈ [ 0 , 1 ] Subsequently, given the above (1.7) and (1.8) that v k + 1 − v k ∈ K , Which shows v k + 1 v k ∈ K

Thus, we have shown that v k + 1 v k ∈ K , n = 0 , 1 , 2 … . Since v n n = 1 ∞ Relatively compact and monotonous, there exists a v † ∈ K a such that lim n → ∞ v n = v † , which together with the continuity of A and the fact that v n + 1 = A v n It means that v † = A v † . This indicates that v † is a decreasing non-negative solution of (1.1). Moreover, in view of f ( t , 0 ) ≠ 0 , t ∈ [ 0 , 1 ] , we know that zero function is not a solution of (1.1), which indicates that v † is a positive solution of (1.1).

III. AN EXAMPLE

Consider the boundary value problem:

( 3.1 ) { u ( 4 ) ( t ) = 1 8 u 2 ( t ) + u ( t ) + ( 1 − t ) , t ∈ [ 0 , 1 ] , u ′ ( 0 ) = u ‴ ( 0 ) = u ( 1 ) = 0 , α u ( 0 ) + u ″ ( 1 2 ) = 0

If we let η = 1 2 , α = 1 and f ( t , u ) = 1 8 u 2 ( t ) + u ( t ) + ( 1 − t ) , ( t , u ) ∈ [ 0 , 1 ] × [ 0 , + ∞ ) , Then all the assumptions of Theorem 2.2 a = 2 and b = 1 . It follows from Theorem 2.2 that (3.1) has a decreasing positive solution v † . Moreover, the iterative scheme is v 0 ( t ) ≡ 0 for t ∈ [ 0 , 1 ]

v n + 1 ( t ) = { ∫ 0 t [ ( t − s ) 3 6 + 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s + ∫ t η [ 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s + ∫ η 1 − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s if  t ∈ [ 0 , 1 2 ] n = 0 , 1 , 2 , 3 , 4 , … ∫ 0 η [ ( t − s ) 3 6 − 6 ( 1 − t 2 ) ( η − s ) − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s + ∫ η t [ ( t − s ) 3 6 − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s + ∫ t 1 [ − ( 2 − α t 2 ) ( 1 − s ) 3 6 ( 2 − α ) ] [ 1 8 u n 2 ( s ) + u n ( s ) + ( 1 − t ) ] d s if  t ∈ [ 1 2 , 1 ] n = 0 , 1 , 2 , 3 , …

IV. CONCLUSION

in this paper, when α ∈ [ 0 , 2 ) and η ∈ [ 1 2 , 1 ) , we have successfully constructed an animation sequence whose limit is just the ideal monotonic positive solution of boundary value problem (1.1).

In addition, a zero function started with the iterative scheme, which shows that the iterative scheme is feasible.

ACKNOWLEDGMENT

This paper is supported by the National Natural Science Foundation of China(no.11961060), The Key Project of Natural Sciences Foundation of Gansu Province(no.18JR3RA084).

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How to Cite This Article

Alhussein. M.A. Ahmed, Mutasim Abdalmonim Alsiddig, Tarteel Abdalgader, Khalid Ahmed Abbakar, Haroun M. M. Suliman. 2026. "The Iterative Technique for a Fourth-Order Three-Point Nonlinear BVP with Changing Sign Green’s Function". Global Journal of Science Frontier Research - F: Mathematics & Decision GJSFR-F Volume 25 (GJSFR Volume 25 Issue F1).

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Crossref Journal DOI 10.17406/GJSFR

Print ISSN 0975-5896

e-ISSN 2249-4626

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GJSFR-F Classification MSC 2010: 34B15; 34B18; 34B27; 34C23.
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v1.2

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September 3, 2025

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The Iterative Technique for a Fourth-Order Three-Point Nonlinear BVP with Changing Sign Green’s Function

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Khalid Abbakar
Haroun Suliman
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