The general formula for the sixth-order ODE’s initial value problem can be expressed as follows ⁷ : Ψ ( ζ , u ( ζ ) , u ′ ( ζ ) , u ′ ′ ( ζ ) , u ′ ′ ′ ( ζ ) , u ( 4 ) ( ζ ) , u ( 5 ) ( ζ ) , u ( 6 ) ( ζ ) ) = 0 , a ≤ ζ ≤ b (1) with the initial-conditions (ICs), u ( j ) ( a ) = ζ j , j = 0 , 1 , 2 , … , 5 . (2)

while the initial-values problem of the special quasi-linear ODEs of sixth-order can be given as follows: u ( 6 ) ( ζ ) = ϕ ( ζ , u ( ζ ) , u ′ ( ζ ) , u ′ ′ ( ζ ) , u ′ ′ ′ ( ζ ) , u ( 4 ) ( ζ ) , u ( 5 ) ( ζ ) ) , a ≤ ζ ≤ b (3) with the ICs in Equation (2). Definition 2.1:

General Class of Sixth-Order FDEs

Below is the definition of the general class of sixth-order FDEs. ϕ ( ζ , u ( ζ ) , u ( j ) ( ζ ) , u ( jα ) ( ζ ) ) = 0 , j = 1 , 2 , … , 6 ; 0 < α < 1 , a ≤ ζ ≤ b , (4) with the ICs, u ( j α ) ( a ) = ζ j , j = 0 , 1 , 2 , … , 5 . (5)

In this subsection, we introduced the α-fractional derivative of a function Definition 2.3:

α-Fractional-derivative ¹⁵

Mechee et al. ³² introduced a novel α-fractional-derivative definition for the real function ϕ ( ζ ) : [ a , ∞ ) → R by following: T α ( ϕ ( ζ ) ) = ϕ ( α ) ( ζ ) = lim ϵ → 0 ϕ ( ζ + ϵ ζ 1 − α ) − ϕ ( ζ − ϵ ζ 1 − α ) 2 ϵ , (7) for α ∈ ( 0 , 1 ) .

From the α-fractional derivative in Definition 2.3, we get the following property of α-fractional derivative of the function ϕ ( ζ ) .

The form of a Hermit polynomial is defined as follows: P 2 ( ξ ) = ∑ i = 0 m ∑ k = 0 m i − 1 ϕ i ( k ) L ik ( ξ ) .

Consider h = b − a m and ξ i = a + ih and f i = f ( ξ i , y i , y i ′ , y i ′ ′ ) , for i = 0 , 1 , … , m where m = number of partitions in domain interval which is equal to the number of Lagrange’s polynomials and m i − 1 is the degree of Lagrange polynomial. In this case, L ik ( ξ ) is the generalized Lagrange polynomial for m = 0,1,2, …, m i , with a step-size equal to 2 h, and y m + 1 and y m + 2 are the approximation generates the approximated solutions at two-points ξ m + 1 and ξ m + 2 resp. In the block interval [ ξ m , ξ m + 2 ] , we have, ξ m is the starting-point while ξ m + 2 is the end-point. However, the initial value for all next iterations should be the approximated solution by y m + 2 at the end-point ξ m + 2 P 2 ( ξ ) = f 0 L 00 ( ξ ) + f 1 L 10 ( ξ ) + f 2 L 20 ( ξ ) + g 0 L 01 ( ξ ) + g 1 L 11 ( ξ ) + g 2 L 21 ( ξ ) , (9)

3.1.1 Lagrange polynomials

Lagrange polynomials are defined as follows in this article: L 00 ( λ ) = ( λ − λ n ) ( 1 − χ 1 ) ( ( λ n + 2 − λ ) ( λ n + 1 − λ ) ( λ n − λ n + 2 ) ( λ n − λ n + 1 ) ) 2 , (10) L 10 ( λ ) = ( λ − λ n + 1 ) ( 1 − χ 2 ) ( ( λ − λ n ) ( λ n + 2 − λ ) ( λ n + 1 − λ n ) ( λ n + 1 − λ n + 2 ) ) 2 , (11) L 20 ( λ ) = ( λ − λ n + 1 ) ( 1 − χ 3 ) ( ( λ − λ n ) ( λ n + 1 − λ ) ( λ n + 2 − λ n ) ( λ n + 2 − λ n + 1 ) ) 2 , (12) L 01 ( λ ) = ( λ − λ n ) ( λ n + 1 − λ λ n − λ n + 1 ) 2 ( λ n + 2 − λ λ n − λ n + 2 ) 2 , (13) L 11 ( λ ) = ( λ − λ n + 1 ) ( ( λ − λ n ) ( λ n + 2 − λ ) ( λ n + 1 − λ n ) ( λ n + 1 − λ n + 2 ) ) 2 , (14) L 21 ( λ ) = ( λ − λ n + 2 ) ( ( λ n − λ ) ( λ n + 1 − λ ) ( λ n + 2 − λ n ) ( λ n + 2 − λ n + 1 ) ) 2 , (15) where, χ 1 = 2 λ n − λ n + 2 + 2 λ n − λ n + 1 , χ 2 = 2 λ n + 1 − λ n + 2 + 2 λ n + 1 − λ n , χ 3 = 2 λ n + 2 − λ n + 1 + 2 λ n + 2 − λ n + 1 ,

The expressions for Lagrange polynomials in the independent variable s are as follows, assuming that s = λ − λ n + 2 h and, L 00 ( λ ) = 1 4 ( β 1 β 2 ) 2 + 3 β 3 , (16) L 10 ( λ ) = ( λ β 3 ) 2 , (17) L 20 ( λ ) = 1 4 β 3 ( β 2 ) 2 β 4 (18) L 01 ( λ ) = h 4 β 3 ( β 1 β 2 ) ) 2 , (19) L 11 ( λ ) = h ( λ β 3 ) 2 β 2 , (20) and L 21 ( λ ) = h 4 ( β 3 β 2 ) 2 . (21)

where β 1 = λ , β 2 = λ + 1 , β 3 = λ + 2 , β 4 = 1 − 3 λ .

3.1.2 Construction of the numerical GIBM2P approach

The approximated solution u n ( ζ ) of Equation (3) at the initial point ζ n can be obtained by multiple-integrating Equation (3) up to five times with respect to the independent variable ξ along the range [ ζ n ; ζ n + 1 ] , then, compute these integrations to obtain the formulas in Equation (22). u n + 1 ( 6 − j ) = ∑ k = 6 − j 5 h 6 − j − k ( 6 − j − k ) ! v n ( k ) + ∫ t n t n + 1 ψ ( λ , u ( λ ) d λ , j = 1 , 2 , … , 6 (22)

By assuming that t n + 2 = t − hs , the coordinate t is changed to s.

The following six formulas were created by integrating the Equation (22) u n + 1 ( j ) = ∆ 6 − j + ∑ k = 1 3 a jk f n + k + a jk + 3 g n + k , for j = 5,4,3,2,1,0 where, A = ( a ij ) = ( 101 240 8 15 11 240 13 42 1 6 1 42 817 6720 4 105 47 6720 13 h 240 − h 6 − h 80 59 1680 8 105 11 1680 83 6720 1 48 13 6720 67 2016 1 144 1 672 1690 241920 1 945 61 241920 577 475200 1 7200 17 475200 37 12069 4 945 5 12096 143 241920 1 1440 17 241920 179 1900800 1 10395 19008 ) (23) and, ∆ j = ∑ k = 6 − j 5 h 6 − j − k ( 6 − j − k ) ! v n ( k ) , for j = 1 , 2 , … , 6 .

In this subsection, the analysis of the proposed method for solving FDEs of sixth-order is introduced and studied. In Equation (7), Mechee et al. presented a new definition of the α-fractional-derivative for the real function f ( τ ) : [ a , ∞ ) → R . From the α-fractional- derivative in Definition 2.3, we get the property of α-fractional-derivative of the function ϕ ( ζ ) in Equation (8). Consequently, by deriving the two sides of Equations (8) five times to get the sixth α-fractional-derivative of the function f ( τ ) . Thus, the α-fractional-derivative of sixth-order for the real function f ( τ ) in the domain I α , denoted by T 6 α , has the following definition: T 6 α ( f ( ζ ) ) = ζ 1 − 6 α ( ζ 5 f ( 6 ) ( ζ ) + 15 ( 1 − α ) ζ 4 f ( 5 ) ( ζ ) + 5 ( 1 − α ) ( 13 − 7 α ) ζ 3 f ( 4 ) ( ζ ) + ( 1 − α ) ( 15 ( 5 − 7 α ) + 2 ( 1 − 2 α ) ( 7 − 11 α ) ) ζ 2 f ( 3 ) ( ζ ) + 2 ( 1 − α ) ( 1 − 2 α ) ( 7 − 11 α ) ( 2 − 5 α ) ζ f ′ ′ ( ζ ) + ( 1 − α ) ( 1 − 2 α ) ( 1 − 3 α ) ( 1 − 4 α ) ( 1 − 5 α ) f ′ ( ζ ) ) , which is simplified to the following T 6 α ( f ( ζ ) ) = ζ 1 − 6 α ( ζ 5 f ( 6 ) ( ζ ) + 15 ( 1 − α ) ζ 4 f ( 5 ) ( ζ ) + ( 1 − α ) ( 5 ( 13 − 7 α ) ζ 3 f ( 4 ) ( ζ ) + ( 15 ( 5 − 7 α ) + 2 ( 1 − 2 α ) ( 7 − 11 α ) ) ζ 2 f ( 3 ) ( ζ ) + 2 ( 1 − 2 α ) ( 7 − 11 α ) ( 2 − 5 α ) ζ f ′ ′ ( ζ ) + ( 1 − 2 α ) ( 1 − 3 α ) ( 1 − 4 α ) ( 1 − 5 α ) f ′ ( ζ ) ) ) . (24)

To prove Equation (24), by using α -fractional-derivative five times for two sides of Equation (8) to obtain the formula in Equation (24)

In this subsection, using the α-fractional derivative property from Equation (24), the FDE in Equation (6) with ICs in Equation (5) has been transformed into an ODE in Equation (3) with ICs in Equation (2) and then, using GIBM2P approach to obtain the numerical solution of the IVP.

4.2.1 The proposed method’s algorithm

The solution of FDE in Equation (6) with the ICs in Equation (5) has been introduced the proposed method using the following algorithm:

4.2.1.1 The Proposed Algorithm for Solving Sixth-Order FDEs

Step I: Convert the FDE in Equation (6) with ICs in Equation (5) into an ODE in Equation (3) with ICs in Equation (2).

Step II: Utilizing an appropriate analytical technique, or the proposed GIBM2P approach, solve the ODE in Equation (3) with the ICs in Equation (2).

Step III: The analytical or numerical solution for the IVP in step II is consequently exactly the same solution of given FDE problem in Equation (6) with the ICs in Equation (5).