Two-point nonlocal nonlinear fractional boundary value problem with Caputo derivative: Analysis and numerical solution

Leyla AhmadSoltani

doi:10.1515/nleng-2022-0009

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Two-point nonlocal nonlinear fractional boundary value problem with Caputo derivative: Analysis and numerical solution

Leyla AhmadSoltani

Published/Copyright: May 30, 2022

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From the journal Nonlinear Engineering Volume 11 Issue 1

Abstract

This work presents the existential and unique results for the solution to a kind of high-order fractional nonlinear differential equations involving Caputo fractional derivative. The boundary condition is of the integral type, which entangles both starting and ending points of the domain. First, the unique exact solution is extracted in terms of Green’s function for the linear fractional differential equation, and then Banach contraction mapping theorem is applied to prove the main result in the case of the general nonlinear source term. Then, our main result is demonstrated by an illustrative example, which shows its legitimacy and applicability. Furthermore, numerical-based semi-analytical technique has been presented to approximate the unique solution to the desired order of precision.

Keywords: high-order fractional differential equation; integral boundary condition; Caputo derivative; fixed point theorem

MSC 2010: 34B10; 34B15; 34B27; 34B99

1 Introduction and the problem formulation

Fractional calculus while standing for differential and integral equations, due to nonlocal property in behind, have various applications in applied branches such as blood flow problems, anomalous diffusion, disease spread, control processing, and population dynamics, for instance, see [1,2, 3,4,5, 6,7,8, 9,10].

The theory of fixed point to study the solution of the boundary value problems is a consequential tool, and it played important role in both proving the existence of the solution and obtaining it approximately; see refs [11,12, 13,14,15, 16,17,18, 19,20]. More interest is shown in researching on the subject of nonlocal nonlinear fractional order in boundary value problems in recent years. In the research line of the study of the existential and unique results for BVPs based on the fixed point theory, finding Green function in a useful way helps to present the unique solution in the linear case. In ref. [21], Cabada and Hamdi have investigated the fractional nonlocal BVP:

(1) D x δ 0 w + g ( x , w ( x ) ) = 0 , x ∈ ( 0 , 1 ) , w ( 0 ) = w ′ ( 0 ) = 0 , w ( 1 ) = χ ∫ 0 1 w ( s ) d s ,

with 2 < δ ≤ 3 , χ > 0 , χ ≠ δ , and D x δ 0 is Rieman-Liouville fractional derivative. The authors in ref. [22] studied very special nonlinear sequential form of fractional differential equation in the presence of nonlocal form of fractional integral conditions and proved the existence of the unique solution. Also, Cabada and Wang [23] have studied almost the same problem, i.e.,

(2) D x α C w + g ( x , w ( x ) ) = 0 , x ∈ ( 0 , 1 ) , w ( 0 ) = w ″ ( 0 ) = 0 , w ( 1 ) = χ ∫ 0 1 w ( s ) d s ,

where D x δ C is Caputo fractional derivative.

In this work, we go forward to deal with the study of existence and uniqueness of the solution to a class of two-point nonlocal nonlinear fractional BVP given by:

(3) D t α C u + f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = u ″ ( 0 ) = u ‴ ( 0 ) = 0 , λ u ′ ( 0 ) + ( 1 − λ ) u ′ ( 1 ) = ∫ 0 1 u ( s ) d s ,

where 3 < α ≤ 4 , 0 ≤ λ ≤ 1 , D t α C u is the Caputo fractional derivative, and f : ( 0 , 1 ) × R × R × R × R → R is a given continuous function. Furthermore, we present a numerical-based semi-analytical technique to approximate the unique solution to the desired order of precision.

2 Preliminaries

In this section, we introduce some basic definitions of fractional calculus to reach the aforementioned aim, and see refs [24,25,26] for more details and also for proofs.

Definition 2.1

The Caputo type differentiation of fractional order δ for a function u : [ 0 , ∞ ] → R is explained as follows:

D δ C u ( x ) ≔ 1 Γ ( m − δ ) ∫ 0 x ( x − τ ) m − δ − 1 u ( m ) ( τ ) d τ , m = [ δ ] + 1 ,

where δ is a real number and [ δ ] is its integer part.

Lemma 2.2

Let δ > 0 , then the unique solution to

D δ C u ( x ) = 0

is given by the following expression:

u ( x ) = ∑ j = 0 [ δ ] u ( j ) ( 0 ) j ! x j .

Definition 2.3

The Riemann–Liouville type differentiation of fractional order δ for a function u : [ 0 , ∞ ] → R is explained as follows:

I δ u ( x ) ≔ 1 Γ ( δ ) ∫ 0 x ( x − τ ) δ − 1 u ( τ ) d τ .

Lemma 2.4

Let δ > 0 , then it holds

(4) I δ [ D δ C u ( x ) ] = u ( x ) − ∑ j = 0 [ δ ] u ( j ) ( 0 ) j ! x j .

3 Solution to the linear equation

We go back to the aimed problem (3). As usual, the approach here is to seek solutions as fixed points of an operator defined by using the Green’s function corresponding to the linear version of the problem, i.e.,

(5) D t α C u ( t ) + y ( t ) = 0 , t ∈ ( 0 , 1 ) ,

with respect to the restrictions as follows:

(6) u ( 0 ) = u ″ ( 0 ) = u ‴ ( 0 ) = 0 , λ u ′ ( 0 ) + ( 1 − λ ) u ′ ( 1 ) = ∫ 0 1 u ( s ) d s .

Theorem 3.1

Let 3 < α ≤ 4 . Assume y ∈ C [ 0 , 1 ] ; hence problems (5) and (6) stand for the unique solution u ∈ C 3 [ 0 , 1 ] , given by

(7) u ( t ) = ∫ 0 1 G ( t , s ) y ( s ) d s ,

where

(8) G ( t , s ) = − 2 t ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] + α ( t − s ) α − 1 Γ ( α + 1 ) , 0 ≤ s ≤ t ≤ 1 ; − 2 t ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] Γ ( α + 1 ) , 0 ≤ t ≤ s ≤ 1 .

Proof

Recalling Lemma 2.4, we observe that Eq. (5) is equivalent to

(9) u ( t ) = I α ( − y ( t ) ) + ∑ j = 0 [ α ] u ( j ) ( 0 ) j ! t j = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + ∑ j = 0 3 u ( j ) ( 0 ) j ! t j .

So, imposing the boundary conditions at t = 0 gives

(10) u ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + t u ′ ( 0 ) ,

as well as

(11) u ′ ( t ) = 1 − α Γ ( α ) ∫ 0 t ( t − s ) α − 2 y ( s ) d s + u ′ ( 0 ) .

Imposing the nonlocal integral condition implies

(12) λ u ′ ( 0 ) + ( 1 − λ ) 1 − α Γ ( α ) ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s + u ′ ( 0 ) = ∫ 0 1 u ( s ) d s ,

and then, we have

(13) u ′ ( 0 ) = ∫ 0 1 u ( s ) d s − ( 1 − λ ) ( 1 − α ) Γ ( α ) ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s ,

which yields:

(14) u ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + t ∫ 0 1 u ( s ) d s − ( 1 − λ ) ( 1 − α ) Γ ( α ) t ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s .

Let us set A = ∫ 0 1 u ( s ) d s , then by integrating both sides of the last equality on interval [ 0 , 1 ] with respect to t , we obtain the following:

A = − 1 Γ ( α ) ∫ 0 1 ∫ 0 t ( t − s ) α − 1 y ( s ) d s d t + A ∫ 0 1 t d t − ∫ 0 1 ( 1 − λ ) ( 1 − α ) Γ ( α ) t ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s d t ,

A = − 1 α Γ ( α ) ∫ 0 1 ( 1 − s ) α y ( s ) d s + A 2 − ( 1 − λ ) ( 1 − α ) 2 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s ,

and therefore, A is extracted as follows:

(15) A = − ( 1 − λ ) ( 1 − α ) Γ ( α ) ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s − 2 Γ ( α + 1 ) ∫ 0 1 ( 1 − s ) α y ( s ) d s .

Replacing this value in Eq. (14), an explicit formulation of u is extracted as follows:

(16) u ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s − ( 1 − λ ) ( 1 − α ) Γ ( α ) t ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s − ( 1 − λ ) ( 1 − α ) Γ ( α ) t ∫ 0 1 ( 1 − s ) α − 2 y ( s ) d s − 2 t Γ ( α + 1 ) ∫ 0 1 ( 1 − s ) α y ( s ) d s = − ∫ 0 t ( t − s ) α − 1 Γ ( α ) y ( s ) d s − t ∫ 0 1 2 ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] Γ ( α + 1 ) y ( s ) d s = − ∫ 0 t 2 t ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] + α ( t − s ) α − 1 Γ ( α + 1 ) y ( s ) d s + − ∫ t 1 2 t ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] Γ ( α + 1 ) y ( s ) d s = ∫ 0 1 G ( t , s ) y ( s ) d s . □

4 Fixed-point iteration

Consider the functional linear space of C 3 [ 0 , 1 ] equipped to its usual norm ‖ u ‖ = ‖ u ‖ ∞ + ‖ u ′ ‖ ∞ + ‖ u ″ ‖ ∞ + ‖ u ‴ ‖ ∞ , which is the Banach space indeed. Replacing y ( x ) by

f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) )

in Theorem 3.1, an operator T : C 3 [ 0 , 1 ] → C 3 [ 0 , 1 ] associated with problem (3) can be defined as follows:

(17) T u ( t ) = ∫ 0 1 G ( t , s ) f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s : ≡ − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s − μ ( t ) ∫ 0 1 ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s ,

where

(18) μ ( t ) = 2 t Γ ( α + 1 ) .

We verify that, from Theorem 3.1, obviously all the fixed points to the operator T are exactly the solutions of problem (3). So it remains to investigate what fixed points of the operator T are. Our main tool is Banach’s contraction principle [27], which says any contraction operator from a closed subset of a Banach space into itself possesses the unique fixed point.

5 Main results

We now present the main existential and unique result. For computational convenience, we put

R = 4 Γ ( α + 2 ) + 1 Γ ( α + 1 ) + 5 − 4 λ Γ ( α ) + 1 Γ ( α − 1 ) + 1 Γ ( α − 2 ) .

Theorem 5.1

Assume the following contraction condition holds: ( H 1 ) There exists L f > 0 such that

(19) ∀ t ∈ [ 0 , 1 ] , u , v , u ¯ , v ¯ , u ˜ , v ˜ , u ˆ , v ˆ ∈ R : ∣ f ( t , u , u ¯ , u ˜ , u ˆ ) − f ( t , v , v ¯ , v ˜ , v ˆ ) ∣ ≤ L f ( ∣ u − v ∣ + ∣ u ¯ − v ¯ + ∣ u ˜ − v ˜ ∣ + ∣ u ˆ − v ˆ ∣ ) .

Then, if L f R < 1 , the BVP (3) admits the unique solution on the ball B r = { u ∈ C 3 [ 0 , 1 ] : ‖ u ‖ ≤ r } , where r ≥ N f R 1 − L f R with N f = sup t ∈ [ 0 , 1 ] ∣ f ( t , 0 , 0 , 0 , 0 ) ∣ .

Proof

First, we show that T : B r → B r , i.e., T ( B r ) ⊂ B r . Take u ∈ B r , t ∈ [ 0 , 1 ] , this means ‖ u ‖ = ‖ u ‖ ∞ + ‖ u ′ ‖ ∞ + ‖ u ″ ‖ ∞ + ‖ u ‴ ‖ ∞ ≤ r . We need to show that ‖ T u ‖ = ‖ T u ‖ ∞ + ‖ ( T u ) ′ ‖ ∞ + ‖ ( T u ) ″ ‖ ∞ + ‖ ( T u ) ‴ ‖ ∞ ≤ r . It is routine to see that

(20) ∣ T u ( t ) ∣ ≤ 1 Γ ( α ) ∫ 0 t ∣ t − s ∣ α − 1 ∣ f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) ∣ d s + ‖ μ ‖ ∞ ∫ 0 1 ∣ 1 − s ∣ α − 2 ∣ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ∣ ∣ f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) ∣ d s ≤ 1 Γ ( α + 1 ) t α ‖ f ‖ ∞ + α ( 1 − λ ) + 1 α + 1 ‖ μ ‖ ∞ ‖ f ‖ ∞ ≤ ‖ f ‖ ∞ 1 Γ ( α + 1 ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ‖ ∞ .

But

∣ f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) ∣ = ∣ f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) + f ( t , 0 , 0 , 0 , 0 ) − f ( t , 0 , 0 , 0 , 0 ) ∣ ≤ ∣ f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) − f ( t , 0 , 0 , 0 , 0 ) ∣ + ∣ f ( t , 0 , 0 , 0 , 0 ) ∣ ≤ L f ( ∣ u ( t ) ∣ + ∣ u ′ ( t ) ∣ + ∣ u ″ ( t ) ∣ + ∣ u ‴ ( t ) ∣ ) + ∣ f ( t , 0 , 0 , 0 , 0 ) ∣ ≤ L f ‖ u ‖ + N f .

Consider supremum of inequality mentioned earlier:

‖ f ‖ ∞ ≤ L f ‖ u ‖ + N f ≤ L f r + N f .

Therefore, the inequality (20) results in

(21) ‖ T u ‖ ∞ ≤ ( L f r + N f ) 1 Γ ( α + 1 ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ‖ ∞ .

Also,

(22) ( T u ) ′ ( t ) = 1 − α Γ ( α ) ∫ 0 t ( t − s ) α − 2 f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s + μ ′ ( t ) ∫ 0 1 ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s .

So, we have

(23) ∣ ( T u ) ′ ( t ) ∣ ≤ t α − 1 Γ ( α ) ‖ f ‖ ∞ + α ( 1 − λ ) + 1 α + 1 ‖ μ ′ ‖ ∞ ‖ f ‖ ∞ ≤ ‖ f ‖ ∞ 1 Γ ( α ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ′ ‖ ∞ .

This yields:

(24) ‖ ( T u ) ′ ‖ ∞ ≤ ( L f r + N f ) 1 Γ ( α ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ′ ‖ ∞ .

In the same way, for the second derivative of T operator after effecting on u , we have

(25) ( T u ) ″ ( t ) = ( 1 − α ) ( α − 2 ) Γ ( α ) × ∫ 0 t ( t − s ) α − 3 f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) ) d s + μ ″ ( t ) ∫ 0 1 ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) d s .

Recalling that μ ″ ( t ) ≡ 0 , then, obviously we have

(26) ∣ ( T u ) ″ ( t ) ∣ ≤ ( α − 1 ) t α − 2 Γ ( α ) ‖ f ‖ ∞ ≤ ‖ f ‖ ∞ 1 Γ ( α − 1 ) .

Therefore, it is concluded

(27) ‖ ( T u ) ″ ‖ ∞ ≤ ( L f r + N f ) 1 Γ ( α − 1 ) .

The same as before, we obtain

(28) ‖ ( T u ) ‴ ‖ ∞ ≤ ( L f r + N f ) 1 Γ ( α − 2 ) .

Now, combining Eqs. (21), (24), (27), and (28), also taking into account ‖ μ ‖ = ‖ μ ‖ ∞ + ‖ μ ′ ‖ ∞ + ‖ μ ″ ‖ ∞ + ‖ μ ‴ ‖ ∞ = 4 Γ ( α + 1 ) , we obtain

(29) ‖ T u ‖ ≤ ( L f r + N f ) 4 Γ ( α + 2 ) + 1 Γ ( α + 1 ) + 5 − 4 λ Γ ( α ) + 1 Γ ( α − 1 ) + 1 Γ ( α − 2 ) = ( L f r + N f ) R .

It is sufficient to choose r ≥ ( L f r + N f ) R , to have T B r ⊂ B r .

Next, we show that T is a contraction as an operator. Notice that, for arbitrary u , v ∈ C 3 [ 0 , 1 ] , we have

(30) ∣ ( T u ) ( t ) − ( T v ) ( t ) ∣ ≤ 1 Γ ( α ) ∫ 0 t ∣ t − s ∣ α − 1 ∣ f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) − f ( s , v ( s ) , v ′ ( s ) , v ″ ( s ) , v ‴ ( s ) ) ∣ d s + ‖ μ ‖ ∞ × ∫ 0 1 ∣ 1 − s ∣ α − 2 ∣ α ( 1 − λ ) ( 1 − α ) − ( 1 − s ) 2 ∣ × ∣ f ( s , u ( s ) , u ′ ( s ) , u ″ ( s ) , u ‴ ( s ) ) − f ( s , v ( s ) , v ′ ( s ) , v ″ ( s ) , v ‴ ( s ) ) ∣ d s ≤ L f ‖ u − v ‖ 1 Γ ( α + 1 ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ‖ ∞ .

Similarly, we can reach into

(31) ∣ ( T u ) ′ ( t ) − ( T v ) ′ ( t ) ∣ ≤ L f ‖ u − v ‖ 1 Γ ( α ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ′ ‖ ∞ ,

(32) ∣ ( T u ) ″ ( t ) − ( T v ) ″ ( t ) ∣ ≤ L f ‖ u − v ‖ 1 Γ ( α − 1 ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ″ ‖ ∞ ,

and

(33) ∣ ( T u ) ‴ ( t ) − ( T v ) ‴ ( t ) ∣ ≤ L f ‖ u − v ‖ 1 Γ ( α − 2 ) + α ( 1 − λ ) + 1 α + 1 ‖ μ ‴ ‖ ∞ .

Inequalities (30), (31), (32), and (33), all together imply

(34) ‖ T u − T v ‖ ≤ L f ‖ u − v ‖ R < ‖ u − v ‖ ,

by the assumption of the theorem. Now, using the Banach contraction mapping theorem, problem (3) admits the unique solution on interval B r , and the proof is complete.□

6 Illustrative example

Consider the fractional differential equation:

(35) D t 3.5 0 u + β u ( t ) 1 + cosh ( u ( t ) ) + β 1 arccot ( u ′ ( t ) ) + β 2 sin 2 ( u ″ ( t ) ) + β 3 cos 2 ( u ‴ ( t ) ) + cos ( t 4 ) = 0 , t ∈ ( 0 , 1 ) , β > 0 u ( 0 ) = u ″ ( 0 ) = u ‴ ( 0 ) = 0 , u ′ ( 0 ) + u ′ ( 1 ) = 2 ∫ 0 1 u ( s ) d s .

As it is shown, α = 3.5 , λ = 1 2 , and

(36) f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) = β u ( t ) 1 + cosh ( u ( t ) ) + β 1 arccot ( u ′ ( t ) ) + β 2 sin 2 ( u ″ ( t ) ) + β 3 cos 2 ( u ‴ ( t ) ) + cos ( t 4 ) .

It is obviously observed that

(37) ∣ f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) , u ‴ ( t ) ) − f ( t , v ( t ) , v ′ ( t ) , v ″ ( t ) , v ‴ ( t ) ) ∣ = β u ( t ) 1 + cosh ( u ( t ) ) + β 1 arccot ( u ′ ( t ) ) + β 2 sin 2 ( u ″ ( t ) ) + β 3 cos 2 ( u ‴ ( t ) ) + cos ( t 4 ) − β v ( t ) 1 + cosh ( v ( t ) ) − β 1 arccot ( v ′ ( t ) ) − β 2 sin 2 ( v ″ ( t ) ) − β 3 cos 2 ( v ‴ ( t ) ) − cos ( t 4 ) = β u ( t ) 1 + cosh ( u ( t ) ) − v ( t ) 1 + cosh ( v ( t ) ) + β 1 ( arccot ( u ′ ( t ) ) − arccot ( v ′ ( t ) ) ) + β 2 ( sin 2 ( u ″ ( t ) ) − sin 2 ( v ″ ( t ) ) ) + β 3 ( cos 2 ( u ‴ ( t ) ) − cos 2 ( v ‴ ( t ) ) ) ∣ ≤ β u ( t ) 1 + cosh ( u ( t ) ) − v ( t ) 1 + cosh ( v ( t ) ) + β 1 ∣ arccot ( u ′ ( t ) ) − arccot ( v ′ ( t ) ) ∣ + β 2 ∣ sin 2 ( u ″ ( t ) ) − sin 2 ( v ″ ( t ) ) ∣ + β 3 ∣ cos 2 ( u ‴ ( t ) ) − cos 2 ( v ‴ ( t ) ) ∣ ≤ β ∣ u ( t ) − v ( t ) ∣ + β 1 ∣ u ′ ( t ) − v ′ ( t ) ∣ + β 2 ∣ u ″ ( t ) − v ″ ( t ) ∣ + β 3 ∣ u ‴ ( t ) − v ‴ ( t ) ∣ ≤ max { β , β 1 , β 2 , β 3 } ( ∣ u ( t ) − v ( t ) ∣ + ∣ u ′ ( t ) − v ′ ( t ) ∣ + ∣ u ″ ( t ) − v ″ ( t ) ∣ + ∣ u ‴ ( t ) − v ‴ ( t ) ∣ ) .

So, by setting L f = max { β , β 1 , β 2 , β 3 } , the first assumption of Theorem 5.1 holds. On the other hand, routine calculation implies the following results:

(38) ‖ μ ( t ) ‖ = 32 105 π ,

(39) N f = sup t ∈ [ 0 , 1 ] ∣ f ( t , 0 , 0 , 0 ) ∣ = 1 ,

(40) R = 5 − 4 λ Γ ( α ) + 1 Γ ( α − 2 ) + 1 Γ ( α − 1 ) + 1 Γ ( α + 1 ) + 4 Γ ( α + 2 ) = 4934 945 π .

Therefore, if we choose β , β 1 , β 2 , β 3 such that L f < 945 π 4934 , then the second condition of Theorem 5.1 also holds, so we conclude the existence of the unique solution in the ball B r = { u ∈ C 3 [ 0 , 1 ] : ‖ u ‖ ≤ r } , where r ≥ R 1 − L f R . For example, if we set β = β 1 = β 2 = β 3 = 1 3 , then L f = 1 3 , and the unique solution belonging to B r = u ∈ C 3 [ 0 , 1 ] : ‖ u ‖ ≤ 14802 2835 π − 4934 is guaranteed. Figure 1 illustrates this issue accurately.

$Figure 1 The feasible region for r r with respect to L f {L}_{f} in the interval 0 , 945 π 4934 \left(0,\frac{945\sqrt{\pi }}{4934}\right) for the example.$

Figure 1

The feasible region for r with respect to L f in the interval 0 , 945 π 4934 for the example.

7 Numerical procedure

In this section, we focus on the numerical solution of the problem (3), which we have already proved the existence and the uniqueness of the solution by some assumptions. The method is somewhat straightforward by remembering Theorems 3.1 and 5.1 and the recurrence formula

(41) u k + 1 ( t ) = ∫ 0 1 G ( t , s ) f ( s , u k ( s ) , u k ′ ( s ) , u k ″ ( s ) , u k ‴ ( s ) ) d s ,

or equivalently

(42) u k + 1 ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s , u k ( s ) , u k ′ ( s ) , u k ″ ( s ) , u k ‴ ( s ) ) d s − μ ( t ) ∫ 0 1 ( 1 − s ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s ) 2 ] f ( s , u k ( s ) , u k ′ ( s ) , u k ″ ( s ) , u k ‴ ( s ) ) d s ,

which comes from operator (17). This recurrence formula can be easily applied by the initial guess belonging to the B r = { u ∈ C [ 0 , 1 ] : ‖ u ‖ ≤ r } , say, for example, u 0 ( t ) ≡ 0 , and then, it can be terminated by a criteria like, for example, ‖ u k + 1 − u k ‖ ≤ ε . In order to easily implementing the iteration process while avoiding the boring and slowly symbolic computations, we suggest the following resolution:

(43) u k + 1 ( t ) = − 1 Γ ( α ) ∑ j = 1 m w j t ( t − s j t ) α − 1 f ( s j t , u k ( s j t ) , u k ′ ( s j t ) , u k ″ × ( s j t ) , u k ‴ ( s j t ) ) − μ ( t ) × ∑ i = 1 m w i ( 1 − s i ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s i ) 2 ] f ( s i , u k ( s i ) , u k ′ ( s i ) , u k ″ ( s i ) , u k ‴ ( s i ) ) ,

where we have used the composite Simpson’s integration rule in both integrals. Notice that w j t and s j t in the first summation denote that Simpson’s weights and nodes depend on t because of the integration interval of the first integral, i.e., ( 0 , t ) . Now, we apply central difference formula for derivatives, which all are of order O ( h 2 ) , where h is the uniform distance between nodes.

(44) u ′ ( t q ) ≈ u ( t q + 1 ) − u ( t q − 1 ) 2 h , u ″ ( t q ) ≈ u ( t q + 1 ) − 2 ( t q ) + u ( t q − 1 ) h 2 , u ‴ ( t q ) ≈ u ( t q + 2 ) − 2 u ( t q + 1 ) + 2 u ( t q − 1 ) − u ( t q − 2 ) 2 h 3 .

Setting the aforementioned formulas in Eq. (43), we obtain

(45) u k + 1 ( t ) = − 1 Γ ( α ) ∑ j = 1 m w j t ( t − s j t ) α − 1 f s j t , u k ( s j t ) , u k ( s j + 1 t ) − u k ( s j − 1 t ) 2 h t , u k ( s j + 1 t ) − 2 u k ( s j t ) + u k ( s j − 1 t ) h t 2 , u k ( s j + 2 t ) − 2 u k ( s j + 1 t ) + 2 u k ( s j − 1 t ) − u k ( s j − 2 t ) 2 h t 3 − μ ( t ) ∑ i = 1 m w i ( 1 − s i ) α − 2 [ α ( 1 − λ ) ( 1 − α ) + ( 1 − s i ) 2 ] f s j , u k ( s j ) , u k ( s j + 1 ) − u k ( s j − 1 ) 2 h , u k ( s j + 1 ) − 2 u k ( s j ) + u k ( s j − 1 ) h 2 , u k ( s j + 2 ) − 2 u k ( s j + 1 ) + 2 u k ( s j − 1 ) − u k ( s j − 2 ) 2 h 3 .

7.1 Numerical experiment

As an example, to verify the validity of the presented the numerical method, consider that example in Section 6. In this example, consider β 1 = β 2 = β 3 = 0 for the simplicity (they can be nonzero without any problem) but suppose β to be a variety of numbers in the feasible range, namely, β = 0.025 , 0.05, 0.075, 0.1, 0.125, 0.15, 0.175, 0.2, 0.225, 0.25. The explicit exact closed-form solution is unknown indeed, but the criteria used for stopping the iteration process is set ‖ u k + 1 − u k ‖ ∞ < 1 0 − 10 , and then, the numerical solution is obtained. The numerical solutions for different β values are shown graphically in Figure 2.

$Figure 2 The diagram of numerical solution for different β \beta values for Example 6.1.$

Figure 2

The diagram of numerical solution for different β values for Example 6.1.

8 Conclusion

In this article, we have applied Banach contraction mapping theorem to prove the existence and the uniqueness of the solution to a kind of high-order nonlinear fractional differential equations involving Caputo fractional derivative whose boundary condition is of integral type in terms of both starting and ending points of the domain. To this aim, we have extracted the unique exact solution in terms of Green’s function for the linear fractional differential equation. An illustrative example has been given to show the legitimacy of the theory. Moreover, it has been obtained numerical approximate solution by defining an iteration process to reach the fixed point as an exact solution.

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Acknowledgements

The author is very grateful to two anonymous reviewers for carefully reading the paper and for their comments and suggestions that have improved the paper very much.

Funding information: The author states no funding involved.
Author contributions: The author have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The author states no conflict of interest.

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Received: 2021-12-12

Revised: 2022-02-02

Accepted: 2022-02-17

Published Online: 2022-05-30

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/nleng-2022-0009

Keywords for this article

high-order fractional differential equation; integral boundary condition; Caputo derivative; fixed point theorem

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