The Langevin equation, in its fractional issue, is a mathematical dynamic model fundamental in Brownian motion applications to characterize the emergence of physical episodes in oscillating mediums. It is a popularization of the conventional model that utilizes a fractional Gaussian procedure formalized by two vertices, which is much more adaptable to the parameterization of the fractal evolution processes [1–4]. The applications of the FLM can be seen in the stock market, motor control system modeling, photoelectron counting, fluid suspensions, deuteron-cluster dynamics, protein dynamics, evacuation process modeling, financial markets, single-file diffusion, and anomalous transport [5–8]. Commonly, FDMs have wide applications in the formation of many engineering and fluid physics phenomena. In the second half of the last century, the search for more serious numerical algorithms began to control many of the nonlinear problems that appeared with the emergence of fractional derivatives and their emerging applications. However, many studies came into existence, as in the following literature [9–13]. To date, effective analytical and numerical schemes have been developed and successfully applied to deal with different classes of FDMs [14–20].

Although finding accurate solutions to FDMs of different orders in Brownian particle motion placements is an important problem for understanding the dynamic attitudes of oscillatory environments in trifling media, in this article, we contemplate creating an accurate numerical solution to FLMs utilizing the CFD with appropriate boundary data using a new renewal in the RKHA. Here, we will generate effective and straightforward numerical solutions without imposing any restrictions on the nature of the proposed FLM and obtain sufficient convergence while reducing the computation time by exposure to the following model [1–8]: ∂ β ∂ x ∂ α ∂ x + μ E x = H x , E x , (1)

equipped with the posterior boundary condition: E 0 − E 0 = 0 , ∂ α ∂ x E 1 − E 1 α = 0 . (2)

Typically, the FDM (Eqs 1, 2) consists of the posterior parameters, functions, and variable effects as attached:

1) x ∈ Ξ : 0,1 stands for the time-coordinate independent domain.

Ordinarily, no conventional schemes produce an accurate prototype solution for FLMs. Thus, there is a great need for the RKHA, which satisfies the purpose, as usual, and has reached the desired and satisfactory numerical results with fully stochastic properties. Here, the relevant theories and facts have been confirmed by numeric emulations, graphical representation, and scale tables for three types of FLMs. However, the tenor of the paper is arranged as follows: Section 1: Presentation: FDMs and CFDs. Section 2: RKHA: Preliminaries and definitions. Section 3: RKHA: Construction and properties. Section 4: RKHA: Solutions and convergence. Section 5: Error: Estimation and bound. Section 6: Justifications: Algorithms, applications, and analyses. Section 7: Outline: Conclusion and outlook.

The approach of the reproducing kernel is a novel solver built to find solutions to FDMs emerging in physics, waves, statistics, and engineering [21–23]. This technique is based on the Gram–Schmidt process and Fourier expansion approach for an arbitrary order and is used for optimizing an orthogonal basis to detect unknown compounds. The RKHA has many motivational aspects and a great ability to handle complicated problems without imposing any restrictions on the style of the models. Therefore, it has been gaining a lot of solicitude and examination lately [24–32].

First, a reproducing kernel induced from a given Hilbert space is called the RKHA. Here, C Ξ is a set of maps that are continuous absolutely on Ξ . At the outset, some requirements that are necessary to go further in our RKHA scheme will be sought.

Remark 1

[24] The frameworks of Σ 0 Ξ are as follows: Σ 0 Ξ = E : E ∈ C Ξ ∧ E ∈ L 2 Ξ , E 1 x , E 2 x Σ 0 = E 1 0 E 2 0 + ∫ Ξ E 1 ′ x E 2 ′ x d x , E Σ 0 2 = E x , E x Σ 0 , Π x 1 s = 1 + s , s ≤ x , x , s > x . (4)

Remark 2

[24] The frameworks of Σ 1 Ξ are as follows: Σ 1 Ξ = E : E o ∈ C Ξ ∧ E o ∈ L 2 Ξ ∧ E 0 = 0 , E 1 x , E 2 x Σ 1 = ∑ o = 0 1 E 1 o 0 E 2 o 0 + E 1 1 E 2 1 + ∫ Ξ E 1 ‴ x E 2 ‴ x d x , E Σ 1 2 = E x , E x Σ 1 , Π x 1 s = 1 120 Π s , x , s ≤ x , Π x , s , s > x , (5) with Π s , x = s − 120 − 1 + x x − x 120 − 246 x + 10 x 2 − 5 x 3 + x 4 s + 5 − 1 + x x s 3 − − 1 + x 2 s 4 .

Definition 1 The framework of Σ 2 Ξ is as follows: Σ 2 Ξ = E : E ∈ Σ 1 Ξ ∧ ∂ α ∂ x E 1 = 0 . (6)

Here, one can find that Σ 2 Ξ is a subset in Σ 1 Ξ and is closed. Next, to generate the kernel function of Σ 2 Ξ , we put Λ E x = / E x x = 1 .

Theorem 1

If Λ x Λ s Π x 1 s ≠ 0 , then the framework function of Σ 2 Ξ is as follows: Π x 2 s = Π x 1 s − Λ x Π x 1 s Λ s Π x 1 s Λ x Λ s Π x 1 s . (7)

Proof: Because ∂ α / ∂ s Π 1 2 x = 0 , Π x 2 s ∈ Σ 2 Ξ . However, for each E s in Σ 2 Ξ , one can obtain E s , Π x 2 s = E s , Π x 1 s = E x , (8) or Π x 2 s is the reproducing kernel of Σ 2 Ξ .

Theorem 2

If E ∈ Σ 2 Ξ , then E x ≤ 3.5 E Σ 2 , E ′ x ≤ 3 E Σ 2 , and E ″ x ≤ 2 E Σ 2 .

Proof: Since E , E ′ , E ″ , E ‴ ∈ C Ξ , applying successive integration from 0 to x for E ‴ , E ″ , and E ′ , one can obtain E ″ x − E ″ 0 = ∫ 0 x E ‴ p d p , (9) E ′ x − E ′ 0 − E ″ 0 x = ∫ 0 x ∫ 0 x 1 E ‴ p d p d x 1 , (10) E x − E 0 − E ′ 0 x − 0.5 E ″ 0 x 2 = ∫ 0 x ∫ 0 x 2 ∫ 0 x 1 E ‴ p d p d x 1 d x 2 . (11) Taking ∙ values and using the fact 0 ≤ x , x 2 ≤ 1 , we can obtain E x ≤ E 0 + E ′ 0 + 0.5 E ″ 0 + ∫ Ξ E ‴ p d p . (12)

To complete, one can get E 0 = E 2 0 ≤ E Σ 2 , E ′ 0 = E ′ 0 2 ≤ E Σ 2 , E ″ 0 = E ″ 0 2 ≤ E Σ 2 , and ∫ Ξ ‍ E ‴ x d x ≤ ∫ Ξ ‍ E ‴ x 2 d x ≤ E Σ 2 .

Herein, the boundaries in Eq. 2 will first be homogenized to zero to obtain easy-to-access modeling in the proposed Σ 2 Ξ space. The form of the operator that formulates the required solution will also be determined in addition to some of what is needed for the scheme.

To achieve this, one must first carry out the following: ∂ β ∂ x ∂ α ∂ x + μ E x = H ¯ x , E x , (13)

equipped with the posterior boundary condition: E 0 = 0 , ∂ α ∂ x E 1 = 0 . (14) For the briefing, the normalizing modified version in Eqs 13, 14 was obtained from the posterior underlying conversion, taking into account that all the extra terms transformed into H ¯ x , E x as E x : → E x − 0.5 Γ 3 − α E 1 α x 2 + E 0 , (15) H ¯ x , E x ≔ Γ 3 − α Γ 3 − α − β + μ Γ 3 − α Γ 3 − β x α E 1 α x 2 − α − β + H x , E x − 0.5 Γ 3 − α E 1 α x 2 − E 0 . (16)

The conversions in Eqs 15, 16 are needful to insert the equipped boundaries in Eq. 2 inside Σ 2   Ξ . Indeed, we will denote E to new and old solutions always.

1) Define the map Ƌ such that

Theorem 3

Ƌ : Σ 2 Ξ → Σ 0 Ξ is a bounded linear operator.

Proof: From Remark 1, one can obtain Ƌ E x Σ 0 2 = Ƌ E x , Ƌ E x Σ 0 = Ƌ E 0 2 + ∫ Ξ Ƌ E ′ x 2 d x . (20)

Using the reproducing property of Π x 2 s , one can obtain E x = E s , Π x 2 s Σ 2 , Ƌ E o x = E s , Ƌ Π x 2 o s Σ 2 , o = 0,1 . (21) With the use of the Schwarz inequality, one can obtain Ƌ E o x = E x , Ƌ Π x 2 o x Σ 2 ≤ Ƌ Π x 2 o x Σ 0 E Σ 2 ≤ C o E Σ 2 ,   o = 0,1 . (22) So Ƌ E Σ 0 2 ≤ C 0 2 + ∫ Ξ C 1 2 d x E Σ 2 2 or Ƌ E Σ 0 ≤ C 0 2 + C 1 2 E Σ 2 . By picking out a countable dense subset x o o = 1 ∞ in Ξ , defining ω o x = Π x o 1 x , and setting Θ o x = Ƌ * ω o x , one can fit the orthogonal function system of Σ 2 Ξ . Furthermore, by the Gram–Schmidt process, one can fit the orthonormal function systems Θ ¯ o x o = 1 ∞ on Σ 2 Ξ as Θ ¯ o x = ∑ k = 1 o a o k Θ k x . (23)

Theorem 4

Θ o x o = 1 ∞ is the complete function system of Σ 2 Ξ with Θ o x = Ƌ s Π x 2 s s = x o . (24)

Proof: First, Ƌ s indicates that Ƌ applies to a function of s . Certainly, Θ o x = Ƌ * ω o x = Ƌ * ω o s , Π x 2 s Σ 2 = ω o s , Ƌ s Π x 2 s Σ 0 = Ƌ s Π x 2 s s = x o . (25) So Θ o x can be written as Ƌ s Π x 2 s s = x o in Σ 2 Ξ .

To illustrate more effective properties in Σ 2 Ξ , the set Π x o 2 x o = 1 ∞ is linearly independent as c o o = 1 η agrees well with ∑ o = 1 η c o Π x o 2 x = 0 , and for l = 1,2 , ⋯ , η , taking E k x in Σ 2 Ξ as E k x l = δ l , k , then 0 = E k x , ∑ o = 1 η c o Π x o 2 x Σ 2 = ∑ o = 1 η b o E k x , Π x o 2 x Σ 2 = ∑ o = 1 η c o E k x o = c o , k = 1,2 , ⋯ , η . (26) So Π x o 2 x o = 1 η is linearly independent for each η ≥ 1 .

This section aims to construct exact and RKHA numeric solutions of Eq. 19, together with some convergence theories, to ensure this analysis is more efficient. Here, C Ξ , R and C Ξ × R , R indicate a set of continuous maps on the group inside parentheses.

If E ∈ C Ξ , R and Θ ¯ o x o = 1 ∞ are orthonormal, then E x , Θ ¯ o x Σ 2 , o = 1,2 , ⋯ are the Fourier maps of E for Θ ¯ o x o = 1 ∞ and its Fourier expansion E x = ∑ o = 1 ∞ E x , Θ ¯ o x Σ 2 Θ ¯ o x , wherein x o o = 1 ∞ is dense throughout Ξ .

Theorem 5

Suppose that a o k o , k = 1,1 ∞ , o are orthogonalization coefficients of Θ ¯ o x o = 1 ∞ and a unique solution of Eq. 19 exists, then the posteriors are accomplished:

1) As ϰ → ∞ , the exact solution, E x , of Eqs 1, 2 is as follows:

Proof: For the first side: Through Theorem 4, Θ ¯ o x o = 1 ∞ is an orthonormal basis in Σ 2 Ξ , which is complete. Using ∑ o = 1 ∞ E x , Θ ¯ o x Σ 2 Θ ¯ o x as the Fourier expansion concerning Θ ¯ o x o = 1 ∞ , one can obtain ∑ o = 1 ∞ E x , Θ ¯ o x Σ 2 Θ ¯ o x < ∞ in ⋅ Σ 2 . So E x = ∑ o = 1 ∞ E x , Θ ¯ o x Σ 2 Θ ¯ o x = ∑ o = 1 ∞ E x , ∑ k = 1 o a o k Θ k x Σ 2 Θ ¯ o x = ∑ o = 1 ∞ ∑ k = 1 o a o k E x , Ƌ * E k x Σ 2 Θ ¯ o x = ∑ o = 1 ∞ ∑ k = 1 o a o k Ƌ E x , ω j x Σ 0 Θ ¯ o x = ∑ o = 1 ∞ ∑ k = 1 o a o k H ¯ x , E x , ω j x Σ 0 Θ ¯ o x = ∑ o = 1 ∞ ∑ k = 1 o a o k H ¯ x k , E x k Θ ¯ o x . (29) For the second side: Because ∑ o = 1 ∞ E x , Θ ¯ o x Σ 2 Θ ¯ o x < ∞ and Σ 2 Ξ are the Hilbert space, one can truncate Eq. 27 using the ϰ -idiom RKHA numeric solution of E x to generate Eq. 28.

Remark 3

Assume that E ϰ − 1 Σ 2 < ∞ and x o o = 1 ∞ are dense throughout Ξ . Then, according to Eqs 27–28, the portrayal effect can be determined as follows: E x = ∑ o = 1 ∞ V o Θ ¯ o x , V o = ∑ k = 1 o a o k H ¯ x k , E x k . (30) E ϰ x = ∑ o = 1 ϰ V o Θ ¯ o x , V o = ∑ k = 1 o a o k H ¯ x k , E k − 1 x k . (31)

Theorem 6

Assume that E ϰ − 1 − E Σ 2 → 0 , x ϰ → s as ϰ → ∞ , E ϰ − 1 Σ 2 < ∞ , and H ¯ x , E x ∈ C Ξ × R , R . Then, H ¯ x ϰ , E ϰ − 1 x ϰ → H ¯ s , E s as ϰ → ∞ .

Proof: It all started where E ϰ − 1 x ϰ → E s as ϰ → ∞ . Evidently, one can obtain E ϰ − 1 x ϰ − E s = E ϰ − 1 x ϰ − E ϰ − 1 s + E ϰ − 1 s − E s ≤ E ϰ − 1 x ϰ − E ϰ − 1 s + E ϰ − 1 s − E s ≤ E ϰ − 1 ′ σ x ϰ − s + E ϰ − 1 s − E s , (32) with σ in between x ϰ and s . Applying Theorem 2, one can obtain E ϰ − 1 s − E s ≤ 3.5 E ϰ − 1 − E Σ 2 or equivalently E ϰ − 1 s − y s → 0 as ϰ → ∞ and E ϰ − 1 ′ σ ≤ 3 E ϰ − 1 Σ 2 . Using E ϰ − 1 Σ 2 < ∞ and x ϰ → s , one can obtain equivalently E ϰ − 1 x ϰ − E ϰ − 1 s → 0 as ϰ → ∞ . By H ¯ x , E x ∈ C Ξ × R , R , it gives a glimpse that H ¯ x ϰ , E ϰ − 1 x ϰ → H ¯ s , u s as ϰ → ∞ .

Theorem 7

One gains E ϰ x → E x as ϰ → ∞ .

Proof: Because E ϰ + 1 x = E ϰ x + V ϰ + 1 Θ ¯ ϰ + 1 x and the orthogonality of Θ ¯ o x o = 1 ∞ , one can obtain E ϰ + 1 Σ 2 2 = E ϰ Σ 2 2 + V ϰ + 1 2 = E ϰ − 1 Σ 2 2 + V ϰ 2 + V ϰ + 1 2 = ⋮ = E 0 Σ 2 2 + ∑ o = 1 ϰ + 1 V o 2 . (33) So E ϰ + 1 Σ 2 ≥ E ϰ Σ 2 and ∃ κ ∈ R such that ∑ o = 1 ∞ V o 2 = κ or V o 2 o = 1 ∞ ∈ l 2 . To get E ℷ x − E ℷ − 1 x ⊥ E ℷ − 1 x − E ℷ − 2 x ⊥ ⋯ ⊥ E ϰ + 1 x − E ϰ x , (34) it is adequate to have for ℷ > ϰ that E ℷ − E ϰ Σ 2 2 = E ℷ − E ℷ − 1 + E ℷ − 1 − ⋯ + E ϰ + 1 − E ϰ Σ 2 2 = E ℷ − E ℷ − 1 Σ 2 2 + E ℷ − 1 − E ℷ − 2 Σ 2 2 + ⋯ + E ϰ + 1 − E ϰ Σ 2 2 , (35) with E ℷ − E ℷ − 1 Σ 2 2 = V ℷ 2 . However, on the other line, E ℷ − E ϰ Σ 2 2 = ∑ l = ϰ + 1 ℷ V o 2 → 0 as ϰ , ℷ → ∞ . Using completeness, ∃ E ϰ x ∈ Σ 2 Ξ such that E ϰ x → E x as ϰ → ∞ .

Theorem 8

Assume that E ϰ − 1 Σ 2 < ∞ and x o o = 1 ∞ are dense throughout Ξ . Then, E x = ∑ o = 1 ∞ V o Θ ¯ o x as ϰ → ∞ .

Proof: Applying lim ϰ → ∞ on Eq. 31, one gets E x = ∑ o = 1 ∞ V o Θ ¯ o x . So Ƌ E x k = ∑ o = 1 ∞ V o Ƌ Θ ¯ o x , ω k x Σ 0 = ∑ o = 1 ∞ V o Θ ¯ o x , Ƌ * S k x Σ 2 = ∑ o = 1 ∞ V o Θ ¯ o x , Θ k x Σ 2 . , (36) ∑ k = 1 l a l k Ƌ Θ x k = ∑ o = 1 ∞ V o Θ ¯ o x , ∑ k = 1 l a l k Θ k x Σ 2 = ∑ o = 1 ∞ V o Θ ¯ o x , Θ ¯ l x Σ 2 = V l . (37) Sequentially, if l = 1 , then Ƌ E x 1 = H ¯ x 1 , E 0 x 1 , and if l = 2 , then Ƌ E x 2 = H ¯ x 2 , E 1 x 2 . As a rule, one can obtain Ƌ E x ϰ = H ¯ x ϰ , E ϰ − 1 x ϰ . By the condition of density, ∀ x ∈ Ξ ; ∃ x ϰ q q = 1 ∞ such that x ϰ q → x as q → ∞ or equivalently Ƌ E x ϰ q = H ¯ x ϰ q , E ϰ q − 1 x ϰ q . Letting q → ∞ , one gets Ƌ E x = H ¯ E , E x . Because Θ ¯ o x ∈ Σ 2 Ξ , E x fulfills (Eq. 19).

As a matter of fact, the exact solution of FDMs (Eqs 1, 2) restricted with the CFD depends on α and β . In most cases, this real problem is complicated to be solved traditionally because it lacks an exact solution. This induces us to develop numerical schemes like RKHA that generate approximated realizations of the exact solution depending on the values of α and β .

Here, we will fix T = x o o = 1 ϰ ⊂ Ξ − 0,1 with x 1 ≤ x 2 ≤ ⋯ ≤ x ϰ ⊆ Ξ , ℏ = max 0 ≤ o ≤ ϰ x o + 1 − x o , ℘ ∞ = max x ∈ x o , x o + 1 ℘ x , and Ƌ − 1 = sup 0 ≠ E ∈ Σ 2 E Σ 0 − 1 Ƌ − 1 Σ 2 . Furthermore, we will fix R E ϰ x = Ƌ E ϰ x − H ¯ x , E x to the residual truncated error at x ∈ Ξ .

Lemma 1

One gains Ƌ E ϰ x j = Ƌ E x j , x j ∈ T .

Proof: Set Π ϰ : Σ 2 Ξ → ∑ j = 1 ϰ b j E j x , b j ∈ R . So Ƌ E ϰ x j = E ϰ x , Ƌ x k Π x k 2 x Σ 2 = E ϰ x , E k x Σ 2 = Π ϰ E x , E j x Σ 2 = E x , Π ϰ E j x Σ 2 = E x , E j x Σ 2 = E x , Ƌ x j Π x j 2 x Σ 2 = Ƌ x j E x , Π x j 2 x Σ 2 = Ƌ x j E x j = Ƌ E x j . (38)

In other words, Ƌ E ϰ x j = Ƌ E x j .

Lemma 2

Assume that ℘ ∈ C ℷ Ξ , R , ℘ ℷ + 1 ∈ L 2 Ξ for a fixed ℷ ≥ 1 , and ℘ → 0 at T with ϰ ≥ ℷ + 1 . So ℘ ∈ Σ 0 Ξ and a parameter A exist with ℘ Σ 0 ≤ A ℏ ℷ max x ∈ Ξ ℘ ℷ + 1 x . (39)

Proof: Clearly, ℘ ∈ Σ 0 Ξ and ∀ x ∈ x o , x o + 1 , o = 1,2 , ⋯ ϰ , one can obtain ℘ x = ℘ x − ℘ τ o = ∫ τ o x ℘ ′ τ d τ ≤ x − τ o max x ∈ x o , x o + 1 ℘ ′ x ,   ≤ ℏ ℘ ′ ∞ . (40) On x o , x o + 1 , applying Roll’s theorem on ℘ yields ℘ ′ τ o = 0 with τ o ∈ x o , x o + 1 , o = 1,2 , ⋯ , ϰ − 1 . So for fixed x , ∃ τ o such that x − τ o < 2 ℏ . Similarly, one can write ℘ ′ x = ℘ ′ x − ℘ ′ τ o = ∫ τ o x ℘ ″ τ d τ ≤ x − τ o max x ∈ x o , x o + 1 ℘ ″ x ,   ≤ 2 ℏ ℘ ″ ∞ . (41) So ℘ x ≤ 2 ℏ 2 ℘ ″ ∞ . Sequentially, a parameter C 1 exists with ℘ x ≤ C 1 ℏ ℷ + 1 ℘ ℷ + 1 ∞ and ℘ ′ x ≤ C 1 ℏ ℷ ℘ ℷ + 1 ∞ . By compiling the previous results, one can obtain ℘ Σ 0 = ℘ 0 2 + ∫ Ξ ℘ ′ τ 2 d τ 1 2 ≤ A ℏ ℷ max x ∈ Ξ ℘ ℷ + 1 x , (42)

with A = C 1 ℏ + 1 .

Theorem 9

A parameter B exists with E o − E ϰ o ∞ ≤ B ℏ ℷ max x ∈ Ξ R E ϰ ℷ + 1 x , o = 0,1,2 . (43)

Proof: Utilizing Lemma 3, one can obtain R E ϰ W 2 1 ≤ A ℏ ℷ max x ∈ Ξ R E ϰ ℷ + 1 x . (44) However, since R E ϰ x = Ƌ E ϰ x − H ¯ E , E x = Ƌ E ϰ x − E x , then E − E ϰ = Ƌ − 1 R E ϰ and a parameter C 2 exists with E − E ϰ Σ 2 = Ƌ − 1 R E ϰ Σ 2 ≤ Ƌ − 1 R E ϰ Σ 0 ≤ A C 2 ℏ ℷ max x ∈ Ξ R E ϰ ℷ + 1 x . (45)