A New Solution to the Fractional Black–Scholes Equation Using the Daftardar-Gejji Method

Abstract

The main objective of this study is to determine the existence and uniqueness of solutions to the fractional Black–Scholes equation. The solution to the fractional Black–Scholes equation is expressed as an infinite series of converging Mittag-Leffler functions. The method used to discover the new solution to the fractional Black–Scholes equation was the Daftardar-Geiji method. Additionally, the Picard–Lindelöf theorem was utilized for the existence and uniqueness of its solution. The fractional derivative employed was the Caputo operator. The search for a solution to the fractional Black–Scholes equation was essential due to the Black–Scholes equation’s assumptions, which imposed relatively tight constraints. These included assumptions of a perfect market, a constant value of the risk-free interest rate and volatility, the absence of dividends, and a normal log distribution of stock price dynamics. However, these assumptions did not accurately reflect market realities. Therefore, it was necessary to formulate a model, particularly regarding the fractional Black–Scholes equation, which represented more market realities. The results obtained in this paper guaranteed the existence and uniqueness of solutions to the fractional Black–Scholes equation, approximate solutions to the fractional Black–Scholes equation, and very small solution errors when compared to the Black–Scholes equation. The novelty of this article is the use of the Daftardar-Geiji method to solve the fractional Black–Scholes equation, guaranteeing the existence and uniqueness of the solution to the fractional Black–Scholes equation, which has not been discussed by other researchers. So, based on this novelty, the Daftardar-Geiji method is a simple and effective method for solving the fractional Black–Scholes equation. This article presents some examples to demonstrate the application of the Daftardar-Gejji method in solving specific problems.

1. Introduction

The Black–Scholes model, which Fisher Black and Myron Scholes created in 1973, calculates the pricing options. The model changed to address mathematical financial issues, particularly when calculating insurance premium values. The Black–Scholes model is commonly represented by a partial differential equation, presenting a challenging mathematical problem.

The Black–Scholes model evolved into the fractional Black–Scholes model and became a special case. The fractional Black–Scholes model is a special form of the diffusion equation. The difference is that the Black–Scholes equation describes stock market performance. The solution to the fractional Black–Scholes equation is the value of the option, which is an important concept in financial mathematics.

This article will discuss the existence, solutions, and simulation of option values using simulation. Several methods for solving the equation include Sumudu transformation, homotopy perturbation, Adomian decomposition, Laplace Adomian decomposition, and differential transformation. The study results of [1] found that the solution can be determined by combining the homotopy perturbation method, Sumudu transformation, and He’s polynomials. The column–row decomposition method in [2] can also be used to solve the equation. At the same time, [3] found that the solution to the modified model can be obtained by combining Sumudu transformation and fractional calculus. In 2018, the Adomian decomposition method was employed to solve the fractional Black–Scholes equation [4].

In 2019, Ref. [5] determined the solution to the Black–Scholes equation using the homotopy perturbation combination method and the Laplace transform. The results are the solution to the fractional Black–Scholes equation, along with numerical simulations for various fractional orders. Then, Ref. [6] determined the solution to the fractional Black–Scholes equation using the homotopy perturbation combination method and the generalized Laplace transform. The fractional derivative used is the Caputo–Kartugampola type. The result is an analytical solution to the fractional Black–Scholes equation with two assets. Another study by Kaya and Yilmas [7] solved the partial differential equation using Sumudu transformation properties in 2019. Meanwhile, Uddin [8] used a combination of Laplace transformation and radial basis function kernel methods to determine the solution. Based on these studies, the analytical solution to the fractional Black–Scholes equation is an infinite series or a Mittag-Leffler function. In [9,10,11], the homotopy analysis method (HAM) was applied to determine the European call option (ECO) with the time-fractional Black–Scholes equation (TFBSE). Stock prices are assumed to follow a geometric Brownian motion and do not pay dividends. Consequently, the HAM produced a series of solutions for the TFBSE. On the other hand, Ref. [12] determined the solution to the Black–Scholes equation using the homotopy perturbation combination method and the Laplace transform. Then, Ref. [13] determined the solution to the fractional Black–Scholes equation using the homotopy perturbation combination method and the generalized Laplace transform. The fractional derivative used is the Caputo–Kartugampola type. The results are an analytical solution to the Fractional Black–Scholes equation with two assets. In 2020, Refs. [14,15,16] successively solved the fractional Black–Scholes equation using the homotopy analysis method as well as the Crank–Nicholson method, generalized homotopy analysis (a combination of homotopy analysis methods and homotopy perturbation), and the Elzaki transform homotopy perturbation algorithm (this method is a combination of the homotopy perturbation method and Elzaki transform). The results are an analytical solution to the fractional Black–Scholes equation with two assets.

In 2022, a Batiha study [17] was conducted to determine the solution to the Sine-Gordon equation using the Daftardar-Gejji method and to prove the level of accuracy. It compared the Daftardar-Gejji method and the variational iteration method. The results concluded that the Daftardar-Gejji method is more promising for solving non-linear differential equations than the variational iteration method. The Daftardar-Gejji method is relatively new and simpler compared to previous ones. Therefore, this current study examined the solution to the fractional Black–Scholes equation using both the Daftardar-Gejji and Jafari methods.

The general method to find the solution to the fractional Black–Scholes equation combines Sumudu transforms and homotopy perturbation. Finding a solution to the fractional Black–Scholes equation is necessary to form He’s polynomial, looking for the coefficients of He’s polynomial and forming an infinite series. The solution obtained is an infinite series of the Mittag-Lefler function. He’s polynomials use the inverse property of the Sumudu transformation to obtain the polynomial coefficient. It makes it difficult to compare it with the Daftardar-Gejji method, which simply uses the properties of fractional integrals. Therefore, finding a fractional differential equation solution is easier than using the combination method of Sumudu transformation and homotopy perturbation. This paper aimed to determine the solution to the fractional Black–Scholes equation using the Daftardar-Gejji method and look for its estimated solutions and errors. The initial procedure for finding a solution to the fractional Black–Scholes equation is to determine the existence and uniqueness of the solution using the Picard– Lindelöf theorem, i.e., finding a solution using the fractional integral property. Research on solutions to the Black–Scholes equation is important because the Black–Scholes model does not correspond to reality. This is because there are weaknesses in the Black–Scholes model. The novelty of this article is the discussion of the existence and uniqueness of solutions to the fractional Black–Scholes equation as well as the Daftardar-Geiji method, which is simple and effective for finding solutions to the fractional Black–Scholes equation. Simulations are also used to clarify the methods and results obtained. In addition, this study discussed the error when a fractional solution to the Black–Scholes equation was used to estimate the classical Black–Scholes equation. Graph simulation is given to strengthen the results obtained. The contribution of this paper is that a fractional Black–Scholes equation solution can determine European call and put options. The values of the call and put options are applied to determine the insurance premium value, which is very useful in financial mathematics.

It is crucial to study the fractional Black–Scholes equation as it can provide solutions and formulas for call and put options. Despite numerous studies, most are limited to finding solutions, leaving the search for formulas to call and put options from the fractional Black–Scholes equation largely unaddressed. Obtaining these formulas can enable the analysis of the optimal time to buy, hold, or sell assets for profit-making. It implies that the formula of call and put options from the fractional Black–Scholes equation can be leveraged by investors in making informed decisions.

Future studies need to focus on creating a modified form of the fractional Black–Scholes equation and its applications to option value. The procedures begin with the telegraph diffusion equation and its standard form based on Fick’s first law. To solve this telegraph diffusion equation, Fourier and Laplace transformation was used. However, the equation was altered due to the limitations of Fick’s first law. One of the modifications to the fractional Black–Scholes equation is achieved by combining the model with telegraph diffusion equations. This leads to a more comprehensive form of the equation.

2. Fractional Black–Scholes Equation and Contractive Mapping

The fractional Black–Scholes equation is expressed as follows:

$$\frac{{\partial }^{\mathrm{q}}\mathsf{Ѱ}}{\partial {\sigma }^{\mathrm{q}}}+\frac{\sigma x^2}{2}\frac{{\partial }^2\mathsf{Ѱ}}{{\partial \mathsf{ҳ}}^2}r\left(\tau \right)x\frac{\partial \mathsf{Ѱ}}{\partial \mathsf{ҳ}}-r\left(\sigma \right)\psi =0;(x\tau ),\in {\mathbb{R}}^{+}\times \left(0,T\right),0<q\le 1$$

where $\mathsf{Ѱ}(\mathsf{ҳ},\tau )$ is the value of the European option at the time, the asset price $x$ and time to $\tau ,T$ represents maturity, $r\left(\tau \right)$ denotes the risk-free interest rate, and $\sigma (x,r)$ indicates the volatility function of the underlying asset [13]. The payoff function is expressed as ${\mathsf{Ѱ}}_c(\mathsf{ҳ},\tau$) =$\mathrm{m}\mathrm{a}\mathrm{x}(-E,0)$ and ${\mathsf{Ѱ}}_p(\mathsf{ҳ},$ $\tau$) = $\mathrm{m}\mathrm{a}\mathrm{x}(E-\mathrm{x},0)$, where ${\psi }_c$and ${\psi }_p$ represent the European price for the call and put options, while$E$ is the exercise price for the option.

Definition 1

([18,19,20,21,22]). Caputo’s fractional derivative$D^{\mathrm{q}}$ of a function is defined for real numbers  $q$ with  $m-1<\mathrm{q}\le m.,wherem\in N$ is a positive integer and  $x>0and\psi \in C_{-1}^m$, thus fulfilling the following conditions:

$$D^q\psi \left(x\right)=\left\{\begin{array}{c}\frac{1}{\Gamma (m-q)}{\int }_0^{\mathsf{ҳ}}(\mathsf{ҳ}-\tau {)}^{m-q-1}{\psi }^{\left(m\right)}\left(\tau \right)d\tau ,\\ \frac{{\partial }^m\psi \left(x\right)}{\partial x^m}q=m.\end{array}\right.$$

Definition 2.

The right Riemann–Liouville fractional integral with the order is defined as follows:

$${}_{x}D_b^{-\alpha }f=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_x^b(t-x{)}^{\alpha -1}f\left(t\right)dt$$

For $\mathfrak{R}\left(\alpha \right)>0$.

Definition 3.

The basic Mittag-Leffler function is denoted and defined as  $E_{\alpha }\left(z\right)$:

$$E_{\alpha }\left(z\right)={\sum }_{k=0}^{\infty }\frac{z^k}{\mathsf{\Gamma }(1+\alpha k)},\mathfrak{R}\left(\alpha \right)>0$$

Specifically, to obtain $\alpha =1,$

$${\sum }_{k=0}^{\infty }\frac{z^k}{\mathsf{\Gamma }(1+\alpha k)}={\sum }_{k=0}^{\infty }\frac{z^k}{k!}=e^z,$$

So, it is a generalization of the exponential function $E_{\alpha }\left(z\right)$.

Definition 4

([23]). $Tx=x$ is the fixed point of mapping; the set  $TX\to X$ into itself is for every  $x$ that is mapped into itself [23].

Definitions 5

([24]). Given the metric space $X=(X,d)$, mapping is called a contraction in  $X$ if there is a real number  $0<\alpha <1$ such that each  $x,y\in X$ implies the following:

$$d(Tx,Ty)\le \alpha d(x,y)$$

Theorem 1

([25]). $X=\left(X,d\right)$ metric space. Suppose that  $X$ is complete and  $T:X\to X$ contraction in  $X$; hence,  $T$ has one fixed point.

Theorem 2

([26]). $f:\left[t_0-\alpha ,t_0+\alpha \right]\times \mathcal{B}(\alpha ,\beta )\to {\mathbb{R}}^n$ is continuous and bounded by  $M$. Suppose that  $f(t,·)$ Lipschitz continuous with the Lipschitz constant is  $L$ for each  $I\in [t_0-\alpha ,t_0+\alpha ]$ and then  $\dot{x}=f(t,x)$ with initial conditions  $x\left(t_0\right)$ has a unique solution defined in  $[t_0-b,t_0+b]$, with  $b=min\left[\alpha ,\frac{\beta }{M}\right]$.

3. Daftardar-Gejji Method

The Daftardar-Gejji method was developed from the Adomian decomposition and is more efficient because mathematicians are not required to formulate Adomian polynomials beforehand. According to Ref. [27], the Daftardar-Gejji method was applied to solve the multispecies Lotka–Volterra Equation.

When $u$ is written as a series of $u=\sum _{n=0}^{\infty }{u}_n$ and has a linear property, then $L(\sum _{n=0}^{\infty }{u}_n)=\sum _{n=0}^{\infty }L\left(u_n\right).$ The non-linear decomposition is defined below:

$$N\left({\sum }_{n=0}^{\infty }{u}_n\right)=N\left(u_0\right)+{\sum }_{i=1}^{\infty }\left\{N({\sum }_{j=0}^i{u}_j)-N({\sum }_{j=0}^{i-1}{u}_j)\right\}={\sum }_{i=0}^{\infty }{G}_i$$

where $G_0=N\left(u_0\right)$; the equation becomes:

$$G_i=\left\{N\left({\sum }_{j=0}^i{u}_j\right)-N\left({\sum }_{j=0}^{i-1}{u}_j\right)\right\},i\ge 1$$

Since $u$ can be defined as an addition from the non-linear and linear operator, then:

$${\sum }_{i=0}^{\infty }{u}_i=f+{\sum }_{i=0}^{\infty }L\left(u_i\right)+{\sum }_{i=0}^{\infty }{G}_i$$

Based on the procedure above, the following recursive formula can be attained:

$$u_0=f,u_1=L\left(u_0\right)+G_{0;}{u}_{m+1}=L\left(u_m\right)+G_m,m=\mathrm{1,2},\dots$$

The following is an example of how the Daftardar-Gejji method is used to effectively solve a fractional differential equation.

Example: solve the following fractional diffusion equation:

$$\frac{{\partial }^{\alpha }u(x,t)}{\partial t^{\alpha }}=\frac{{\partial }^2}{\partial x^2}u\left(x,t\right),t>0,x\in \mathbb{R},0<\alpha \le 1;$$

(1)

with initial conditions $u\left(x,0\right)=sinx$

Answer.

By integrating the Riemann–Liouvile integral, both fractional sides of Equation (1) are obtained

$$u\left(x,t\right)=sinx+{}_0{}^t{D}_x^{-\alpha }\frac{{\partial }^2}{{\partial x}^2}u(x,t)$$

Using the Daftardar-Gejji method obtained, $u_0=sinx{;u}_1=\frac{{\partial }^2}{{\partial x}^2}sinx{\int }_0^t\frac{{(t-y)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy$

So that

$$u_1=\left(-sinx\right)t^{\alpha }\frac{{\int }_0^1(1-z{)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}=-sinx\frac{t^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}$$

$$u_2=\frac{{\partial }^2}{\partial x^2}\left(-sinx\right)\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_0^t(t-y{)}^{\alpha -1}\frac{y^{\alpha }}{\mathsf{\Gamma }\left(1+\alpha \right)}dy=sinx\frac{t^{2\alpha }}{\mathsf{\Gamma }(1+2\alpha )}$$

In the same way, it is obtained that

$$u_n=(-1{)}^{n}sinx\frac{t^{n\alpha }}{\mathsf{\Gamma }(1+n\alpha )}$$

So, the following solution is obtained:

$$u\left(x,t\right)={\sum }_{k=0}^{\infty }(-1{)}^k\sin x\frac{t^{k\alpha }}{\mathsf{\Gamma }(1+k\alpha )}=sinx{E}_{\alpha }(-t^{\alpha })$$

(2)

Before attempting to solve the fractional Black–Scholes equation, it is important to first explore the existence and uniqueness of its solution. This is explained in Theorem 2.

Theorem 3.

(solution existence). The fractional Black–Scholes equation $\frac{{\partial }^{q}v}{\partial {\tau }^q}=\frac{{\partial }^{2}v}{\partial x^2}+$ ($k-1)\frac{\partial v}{\partial x}-kv$ imposed on the Caputo operator has a unique solution.

Proof. 

The idea of the proof concerns the method of the Picard–Lindelöf theorem. The existence of a solution to the fractional Black–Scholes equation must prove that $f\left(\tau ,V\left(x,\tau \right)\right)$ must satisfy the Lipschitz condition of continuous functioning. Next, the $\mu$ operator given as

$$\mu \left(u\left(\tau \right)\right)=u_0+\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }f(\tau ,u\left(\tau ,s\right)(\tau -s{)}^{\alpha -1}ds$$

is bounded. Meanwhile, the uniqueness of the solution must be shown $\mu$ that it is a contractive mapping. The following is a step to prove the existence and uniqueness of the solution to the fractional Black–Scholes equation. □

Given the fractional Black–Scholes equation $\frac{{\partial }^{q}v}{\partial {\tau }^q}=\frac{{\partial }^{2}v}{\partial x^2}+$($k-1)\frac{\partial v}{\partial x}-kv$, according to the Caputo operator, the equation is defined as follows:

$$D^{\alpha }\left(x,\tau \right)=\frac{1}{\mathsf{\Gamma }(1-\alpha )}{\int }_0^t(t-\tau {)}^{\alpha }\frac{\partial V(x,\tau )}{\partial \tau }dt$$

The equation further becomes

$$f\left(\tau ,V\left(x,\tau \right)\right)=\frac{{\partial }^{2}V}{{\partial x}^2}+\left(k-1\right)\frac{\partial V}{\partial x}-kV$$

Based on the definition above and using Riesz–Minkowski inequality [18], the following was formulated:

$$‖f({\tau }_1,V_1\left(x,{\tau }_1\right)-f({\tau }_0,V_0(x,{\tau }_0)‖=‖\frac{{\partial }^2{V}_1}{\partial x^2}-\frac{{\partial }^2{V}_0}{\partial x^2}+k\left(V_0-V_1\right)+(k-1)\frac{\partial }{\partial x}(V_1-V_0)‖$$

$$\le ‖\frac{{\partial }^2}{\partial x^2}\left(V_1-V_0\right)‖+k‖V_0-V_1‖+(k-1)‖\frac{\partial }{\partial x}(V_1-V_0)‖$$

$$\le k‖V_1-V_0‖+k_1{k}_2‖V_1-V_0‖+k_3(k-1)‖V_1-V_0‖$$

$$=(k+k_1{k}_2+k_{3}k-k_3)‖V_1-V_0‖$$

with $k_1,k_2,k_3\in \mathbb{R}$.

where $f(\tau ,V\left(x,\tau \right))$ is a Lipschitz continuous function.

The equation was further defined as follows:

$$j_a\left({\tau }_0\right)=[{\tau }_0-a,{\tau }_0+a]$$

$$t_b\left(u\right)=[u_0-b,u_0+b]$$

$$C_{a,b}=j_a\left({\tau }_0\right)\times t_b\left(u\right)$$

Let $M=\underset{x\in C_{a,b}}{\sup }‖f\left(x\right)‖$ and $‖f‖=\int \left|f\left(x\right)\right|dx$

When there is $\mu :u(j_a\left({\tau }_0\right),t_b\left(u\right))\to u(j_a\left({\tau }_0\right),t_b\left(u\right))$ with definition:

$$\mu \left(u\left(\tau \right)\right)=u_0+\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }f(\tau ,u\left(\tau ,s\right)(\tau -s{)}^{\alpha -1}ds$$

The value of $\mu$ bounded needs to be proven first:

$$‖\mu \left(u\left(\tau \right)\right)-u_0‖=‖\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }f(\tau ,u\left(\tau ,s\right)(\tau -s{)}^{\alpha -1}ds‖$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}‖{\int }_{{\tau }_0}^{\tau }f(\tau ,u\left(\tau ,s\right)(\tau -s{)}^{\alpha -1}ds‖$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }\left|f(\tau ,u\left(\tau ,s\right)(\tau -s{)}^{\alpha -1}\right|ds$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }\left|f(\tau ,u\left(\tau ,s\right)\right|{\left|\tau -s\right|}^{\alpha -1}ds$$

$$\le \frac{M}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }{\left|\tau -s\right|}^{\alpha -1}ds\le \frac{M}{\mathsf{\Gamma }\left(\alpha \right)}\frac{{\left|\tau -{\tau }_0\right|}^{\alpha -1}}{\alpha }\le \frac{M}{\mathsf{\alpha }\mathsf{\Gamma }\left(\alpha \right)}{a}^{\alpha -1}$$

$$=\frac{M}{\mathsf{\Gamma }(\alpha +1)}{a}^{\alpha -1}$$

Take $u_1,u_2\in u(j_a\left({\tau }_0\right),t_b\left(u\right))$

$$‖(\mu u_1-\mu u_2)\left(\tau \right)‖=‖\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }(\tau -s{)}^{\alpha -1}(\left(f\right(\tau ,u_1\left(\tau ,s\right)-f\left(\tau ,u_2\left(\tau ,s\right)\right)ds‖$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }\left|(\tau -s{)}^{\alpha -1}(\left(f\right(\tau ,u_1\left(\tau ,s\right)-f\left(\tau ,u_2\left(\tau ,s\right)\right)\right|ds$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }{‖(\tau -s)‖}^{\alpha -1}((f(\tau ,u_1\left(\tau ,s\right)-f\left(\tau ,u_2\left(\tau ,s\right)\right)d$$

$$=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }{\left|\tau -s\right|}^{\alpha -1}‖f(\tau ,u_1\left(\tau ,s\right)-f\left(\tau ,u_2\left(\tau ,s\right)\right)‖ds$$

$$\le \frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }{\left|\tau -s\right|}^{\alpha -1}{k}_1‖u_1-u_2‖ds$$

$$=\frac{k_1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_{{\tau }_0}^{\tau }{\left|\tau -s\right|}^{\alpha -1}‖u_1-u_2‖ds$$

$$=\frac{k_1}{\mathsf{\Gamma }\left(\alpha \right)}‖u_1-u_2‖\frac{(\tau -{\tau }_0{)}^{\alpha }}{a}$$

$\mu$ contractive mapping when:

$$\frac{k_1}{a\mathsf{\Gamma }\left(\alpha \right)}(\tau -{\tau }_0{)}^{\alpha }<1$$

$$\frac{k_1}{\mathsf{\Gamma }\left(\alpha \right)}{a}^{\alpha -1}<1$$

$$a<\sqrt[\alpha -1]{\frac{\mathsf{\Gamma }\left(\alpha \right)}{k_1}}$$

The following is an example of searching for the fractional Black–Scholes equation solution using the Daftardar-Gejji method.

The Black–Scholes equation with initial conditions for a valued European call option $C(S,\tau )$ is:

$$\frac{\partial C}{\partial \tau }+\frac{1}{2}{\sigma }^2{S}^2\frac{{\partial }^{2}C}{\partial S^2}+rS\frac{\partial C}{\partial S}-rC=0$$

where $C\left(0,\tau \right)=0,C\left(S,\tau \right)=S$ for $S\to \infty$ and $C\left(S,T\right)=max\left\{S-E,0\right\},$ where $\sigma$ is the volatility of the underlying asset $E$, $T$ is the exercise price, $r$ is the expiry time and is the risk-free interest rate. By using an analogy and the chain rule $S=E{e}^x$, $\tau =T-\frac{1}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\sigma }^2},C=Ev\left(x,t\right),k=\frac{2r}{{\sigma }^2}$, another form of the Black–Scholes equation is obtained, namely:

$$\frac{\partial v}{\partial t}=\frac{{\partial }^{2}v}{{\partial x}^2}+\left(k-1\right)\frac{\partial v}{\partial x}-kv$$

Example 1.

Given the fractional Black–Scholes equation:

$$\frac{{\partial }^{\alpha }v}{\partial t^{\alpha }}=\frac{{\partial }^{2}v}{\partial x^2}+\left(k-1\right)\frac{\partial v}{\partial x}-kv$$

where the following initial conditions are:

$$v\left(x,0\right)=\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}$$

Based on the Daftardar-Gejji method, the following equation was obtained:

$$v_0=\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}$$

For

$$v_1\left(x,t\right)=\frac{{\partial }^2}{\partial x^2}\max \left\{e^x-1,0\right\}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy+\left(k-1\right)\frac{\partial }{\partial x}\max \left\{e^x-1,0\right\}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy$$

$k\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy$.

Before solving $v_1(x,t)$, the following needs to determined$:$ $v_1(x,t){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy$

$${\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy=-{\int }_t^0\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}d(t-y)$$

$$=-{\left[\frac{1}{\mathsf{\Gamma }\left(\mathsf{\alpha }\right)}\frac{1}{\alpha }{\left(t-y\right)}^{\alpha }\right]}_t^0=\frac{t^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}$$

Based on the Daftardar-Gejji method obtained:

$$v_1\left(x,t\right)=\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}\frac{t^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}+\left(k-1\right)\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}\frac{t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}-k\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}\frac{t^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}$$

$$=\frac{k{t}^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}\left(\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}\right).$$

$$\begin{array}{c}{v}_2(x,t)=\frac{{\partial }^2}{\partial x^2}\left(\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\})\right){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{k{y}^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}dy+\\ \left(k-1\right)\frac{\partial }{\partial x}\left(\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}\right){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{k{y}^{\alpha }}{\mathsf{\Gamma }(1+\alpha )}dy-\\ k\left(\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}\right){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{k{y}^{\alpha }}{\mathsf{\Gamma }(\alpha +1)}dy\end{array}$$

$$=-k^2(\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\})\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}\left(\frac{t^{2\alpha }}{\mathsf{\Gamma }(1+\alpha )}\frac{\mathsf{\Gamma }\left(\alpha \right)\mathsf{\Gamma }(1+\alpha )}{\mathsf{\Gamma }(1+2\alpha )}\right)$$

$$=\frac{t^{2\alpha }}{\mathsf{\Gamma }(1+2\alpha )}\left(k^2\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}-k^2\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}\right).$$

$$v_n\left(x,t\right)=\frac{t^{n\alpha }}{\mathsf{\Gamma }(1+n\alpha )}\left[{(-k)}^n\max \left\{e^x-1,0\right\}-{\left(-k\right)}^n\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}\right].$$

Subsequently, the fractional Black–Scholes equations were solved as follows:

$$v\left(x,t\right)={\sum }_{n=0}^{\infty }{v}_n\left(x,t\right)=\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x-1,0\right\}{E}_{\alpha }\left(-k{t}^{\alpha }\right)+\mathrm{m}\mathrm{a}\mathrm{x}\left\{e^x,0\right\}(1-E_{\alpha }\left(-k{t}^{\alpha }\right))$$

(3)

By taking the exact solution obtained from the Black–Scholes equation, namely$\alpha =1$

$$v\left(x,t\right)=max\left[e^x-\mathrm{1,0}\right]e^{-kt}+max\left\{e^x,0\right\}\left(1-e^{-kt}\right)$$

(4)

Using Python 3.7 software, the simulation graph of Equation (3) is obtained for the values given in Figure 1, where $\alpha =0.25$.

Figure 1 shows the graph obtained in the form of a cylinder. If the value is fixed at [0.5] and the value runs at [0.5], the monotonic graph decreases sharply. The maximum value in Figure 1 is $v\left(x,t\right)=$298, obtained when $x=5$ and $t=0$; the minimum value $v\left(x,t\right)=0$, obtained when $x=0$ and $t\in \left[\mathrm{0,5}\right]$.

Figure 2 shows the graph for $\alpha =0.5$, which is similar to graphics $v(x,t)$ for $\alpha =0.25$. However, the graph for $\alpha =0.5$ tends to be more concave than the graph $\left(x,t\right)$ with $\alpha =0.25$.

Figure 2 shows that the monotonic graph decreases sharply to 0, less than or equal to t less, than or equal to 5, and the value is fixed at [0.5]. The maximum value $x$ fixed at [0.5] is a monotone graph that drops sharply. A maximum value of $v\left(x,t\right)=$298 was obtained when $=5andt=0$, while the minimum value $v\left(x,t\right)=0$ when $x=0$ and $t\in \left[\mathrm{0,5}\right]$.

From Figure 1 and Figure 2, it can be concluded that in the initial conditions, $t=0$, and the value $x$ is large enough; the value $v(x,t)$ also had a significant value. And vice versa, for the initial value, $x=0$, and as time goes by, the value $v\left(x,t\right)=0$.

Based on Figure 1 and Figure 2, different values of α will determine different solutions and graphs. Then, based on the α values, a graph drawn in three dimensions and cut for certain $x$ and $t$ values obtained graphs in two dimensions. So, the graphs shown in Figure 3, Figure 4, Figure 5, Figure 6, Figure 7 and Figure 8 could be obtained. For example, in Figure 3, α = 0.0, 0.25, 0.5, 0.75, and 1.0, $k$= 0.1, x = 5.0, and 0 ≤$t$≤ 1.0.

Figure 3 shows that the maximum value at $v\left(x,t\right)=295$. What is quite interesting here is that for $\alpha =0.25$ and $\alpha =0.5$, the graph $v(x,t)$ for $\alpha =0.5$ shows that the interval [0.0.85] is above the graph $v(x,t)$ for $\alpha =0.25$. When expressed in economic language, the value of the option for $\alpha =0.5$ is greater than that $=0.25$ of the interval [0.0.85]. What happens for $t>0.85$ is the opposite. The value $v(x,t)$for $\alpha =0.25$ is actually greater than the value $v(x,t)$ for $\alpha =0.5$, so the value of the option $\alpha =0.25$ is greater compared with $\alpha =0.5$ on $t>0.85$.

The same is the case for $\alpha =0.25$and $\alpha =0.75$ values on the interval [0, 0.975] and $t>0.975.$ On the other hand, the value $v(x,t)$ on the interval [0, 0.95] for $\alpha =1.0$ is greater than the value $v(x,t)$ for the values $\alpha$ = 0.0, 0.25, 0.5, and 0.75, while the $t>0.95$ value $v(x,t)$ is lower than the value $v\left(x,t\right)$ with $\alpha$ = 0.0, 0.25, 0.5, and 0.75.

Figure 4 shows that the t-axis is expanded to $0\le \mathrm{t}\le 10.$ Based on Figure 3, the behavior of the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ becomes more interesting. In Figure 3, at intervals $0\le t\le 1$, graph $v\left(x,t\right)for\alpha =1.0$, $0\le t\le 1$ graph $v\left(x,t\right)for\alpha =1.0$ is located at the top. Then, at interval $t\ge 1.5$, the graph $v\left(x,t\right)$ is at the bottom. Meanwhile, graphics $pv\left(x,t\right)for\alpha =0.75$ at the interval $0\le t\le 1$ are located above the graph $v\left(x,t\right)on\alpha =0.5\mathrm{a}\mathrm{n}\mathrm{d}\alpha =0.25.$Meanwhile, at interval $t\ge 1.75$, the graph is located at the bottom. Likewise, for graph $v\left(x,t\right)for\alpha =0.5,$ at interval [0, 0.85], it is located above graph $v\left(x,t\right)for\alpha =0.25.$ Meanwhile, at interval $0.85\le t\le 10.0$, it is quite the opposite. Graphics $v\left(x,t\right)for\alpha =0.25$ are precisely above the graph $v\left(x,t\right)for\alpha =0.5$.

It is interesting to show the behavior $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for the value of the environment $\mathrm{x}=1$. Figure 5 and Figure 6 demonstrate the behavior for values of $\mathrm{x}$ close to 1. Figure 5 is taken for values of $\mathrm{x}=0.9$, and Figure 6 is taken for the value $\mathrm{x}=0.8$, with values $\mathsf{\alpha }$ = 0.0, 0.25, 0.5, 0.75, and 1.0.

Figure 5 shows that, at interval 0 $\le t\le 1.0$, it is clear that the maximum value $v\left(x,t\right)=3.0$at $x=0$, in which $v\left(x,t\right)$, is interesting enough. For $\alpha =0.25$ and $\alpha =0.5$, the graph $v(x,t)$ for $\alpha =0.5$ at the interval [0, 0.85] is above the graph for $v(x,t)$ for $\alpha =0.25$. When expressed in economic language, the option value for $\alpha =0.5$ is greater than $\alpha =0.25$ at the interval [0, 0.85]. At the interval [0.85, 1], the opposite happens, i.e., the value of $v(x,t)$ for $\alpha =0.25$ is greater than the value $v(x,t)$ for $\alpha =0.5.$ Therefore, the option value is greater for $\alpha =0.25$ compared to $\alpha =0.5$ at the interval [0.85, 1].

For $\alpha =0.25$and $\alpha =0.75$ at the interval $[0,0.975]$, the value of $v(x,t)$ for $\alpha =0.75$ is greater than the value of $v(x,t)$ for $\alpha =0.25$. However, for the value of $t$ at the interval [0.975, 1], the value of $v\left(x,t\right)$ for $\alpha =0.25$ is much larger than the value of $v(x,t)$ for $\alpha =0.75.$ The same method is used for $\alpha =0.5$and $\alpha =0.75$ values for the interval [0, 1.0] and $t\ge 1.$On the other hand, the value of $v(x,t)$ at the interval [0, 0.95] for $\alpha =1.0$ is larger than the value of $v(x,t)$ for the values $\alpha$= 0.0, 0.25, 0.5, and 0.75; while at the interval (0.95,1.0), the value of $v(x,t)$ is lower than the value of $v\left(x,t\right)with\alpha$= 0.0, 0.25, 0.5, and 0.75. For the value of $\alpha =0.0$, the value of $v\left(x,t\right)=2.65$.

To more clearly see the behavior of the graph $v\left(x,t\right),$ the value of t is expanded to$\left[0,100.0\right]$. It is quite interesting to note that based on Figure 6 obtained for $0\le t\le 100$, the graph $v(x,t)$, which was originally at the interval [0, 0.975] with $\alpha =1.0$, is at the top. However, at the interval [0.975, 100], the graph is the lowest and $v(x,t)$ is the lowest. Meanwhile, the graph $v(x,t)$ for $\alpha =0.75$at the interval [0, 0.985] was originally at the top compared to the graph for $v(x,t)$ for $\alpha =$ 0.25 and $\alpha =0.5$. At the interval [0.985, 100], the graph is below the graph of $v\left(x,t\right)$ for $\alpha =0.0,\alpha =0.25$ and $\alpha =0.5$. Figure 6 shows clearer behavior of the graph $v(x,t)$.

Figure 7 shows the maximum value $v\left(x,t\right)=2.48$ at $x=0$. The behavior of $v\left(x,t\right)$is interesting enough for $\alpha =0.25$ and $\alpha =0.5$; the graph $v(x,t)$ for $\alpha =0.5$ at the interval [0, 0.87] is above the graph for $v(x,t)$ for $\alpha =0.25$. This means that the option value for $\alpha =0.5$ is greater than $\alpha =0.25$ at the interval [0, 0.87]. At the interval [0.87, 2], the opposite occurs, namely the value of $v(x,t)$ for $\alpha =0.25$ is actually greater than the value $v(x,t)$ for $\alpha =0.5.$ Thus, the option value is greater for $\alpha =0.25$ compared to $\alpha =0.5$ at the interval [0.87, 2].

The same method was used for values $\alpha =0.25$and $\alpha =0.75$ at the intervals [0,1] and [1,2] as for values of $\alpha =0.5$ and $\alpha =0.75$ at the intervals [0, 1.06] and [1.06, 2]. On the other hand, the value of $v(x,t)$ at the interval [0, 1.06] for $\alpha =1.0$ is larger than the value of $v(x,t)$ for the values $\alpha$= 0.25, 0.5, and 0.75; at the interval [1.06, 2], the value of $v(x,t)$ is lower than the value of $v\left(x,t\right)\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\alpha$ = 0.0, 0.25, 0.5, and 0.75.

To further clarify the behavior of graph $v\left(x,t\right)$ on the axis $t$, the interval is expanded to $0\le t\le 100.$Based on Figure 8, it is clear that the maximum value $v\left(x,t\right)=2.48$ at the moment $x=0$. The graph shows that the highest $v(x,t)$ is for the value of $\alpha =0.0$. As for the graph $v(x,t)$ with the value of $\alpha =1.0,$ when $t\ge 0.95$, a very sharp decline was experienced. So, the graph $v(x,t)$ for the value $\alpha =1.0$ in the interval $\left[0.95,100\right]$ is at the bottom. The values of $v\left(x,t\right)=0$. For the graph $v(x,t)$ with value $\alpha =0.25$, after $1.05\le t\le 100$, it monotonically decreases towards the value of $v\left(x,t\right)=1.8$.

Here is the graph for $v(x,t)$ by taking $0\le x\le 10$ with the value of $t=25.$ Somewhat different from the graph $v\left(x,t\right)$ when x is fixed, graph $v\left(x,t\right)$ for fixed t values $t=25$ tends to coincide at the interval [0,6]; then, at the interval $6\le x\le 10$, graph $v\left(x,t\right)$ tends to spread and rise monotonously. Graph $v(x,t)$ for the value of $\alpha =0.0$ occupies the lowest position compared to the others. The maximum value for $v(x,t)$ with value $\alpha =0.0$ on $0\le x\le 10$ is 40,000. Following the value $v(x,t)$ for the value of $\alpha =0.25$, the position of the graph is slightly lower than the graph of the $v\left(x,t\right)\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\alpha =0.0.$ The maximum value $v\left(x,t\right)=3500$. Furthermore, the graph $v\left(x,t\right)$ for the value of $\alpha =0.5$. The height of this graph occupies the third position below the graph $v\left(x,t\right)$ for $\alpha =0.0$ and $\alpha =0.25.$ The behavior of these graphs is relatively the same, i.e., they overlap first and then begin to spread out $6\le x\le 10.$ The difference is that this graph is more convex compared to the graph $v(x,t)$ above it with the maximum value $v\left(x,t\right)=2750.$ It is followed by the graph $v(x,t)$ for the value of $\alpha =0.75$ with the maximum value of $v\left(x,t\right)$ = 15,000. The last is graph $v(x,t)$ for the value of $\alpha =1.0$, which is the lowest compared to the graph $v(x,t)$ with the maximum value of $v\left(x,t\right)=4000.$ Figure 9 shows the same pattern of behavior for graphs $v(x,t)$, i.e., the graph is rising and cupped.

The Daftardar-Gejji method was used to solve several problems involving the fractional Black–Scholes equation. The following examples illustrate the procedure for using the method to determine the solution.

Example 2.

The Daftardar-Gejji method was used to solve the fractional Black–Scholes equation of $\frac{{\partial }^{\alpha }v}{\partial t^{\alpha }}+0.08(2+sinx{)}^2{x}^2\frac{{\partial }^{2}v}{\partial x^2}+0.06x\frac{\partial v}{\partial x}-0.06v=0$,$0<\alpha \le 1$with the following initial condition of $v\left(x,0\right)=max\left\{x-25e^{-0.06},0\right\}$Solution:

$$\frac{{\partial }^{\alpha }v}{\partial t^{\alpha }}=-0.08{(2+\sin x)}^2{x}^2\frac{{\partial }^{2}v}{\partial x^2}-0.06x\frac{\partial v}{\partial x}+0.06v$$

$$\begin{array}{ll}{v}_1\left(x,t\right)=-0.08& (2+\sin x{)}^2{x}^2\frac{{\partial }^{2}max\left\{x-25e^{-0.06},0\right\}}{\partial x^2}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy\\ & -0.06x\frac{\partial max\left\{x-25e^{-0.06},0\right\}}{\partial x}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy\\ & +0.06max\left\{x-25e^{-0.06},0\right\}{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}dy\end{array}$$

$$=-0.06x\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}+0.06max\left\{x-25e^{-0.06},0\right\}\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}$$

$$=\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}\left[-0.06x+0.06max\left\{x-25e^{-0.06},0\right\}\right].$$

$$\begin{array}{ll}{v}_2\left(x,t\right)=& -0.08(2+\sin x{)}^2{x}^2\frac{{\partial }^2}{{\partial x}^2}\left(-0.06x-0.06max\left\{x-25e^{-0.06},0\right\}\right){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}dy\\ & -0.06x\frac{\partial }{\partial x}\left[-0.06x-0.06max\left\{x-25e^{-0.06},0\right\}\right]{\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}dy\\ & +0.06(-0.06x+0.06max\left\{x-25e^{-0.06},0\right\}){\int }_0^t\frac{{\left(t-y\right)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}\frac{t^{\alpha }}{\mathsf{\Gamma }\left(\alpha +1\right)}dy\end{array}$$

$$=[-(0.06{)}^{2}x+(0.06{)}^{2}max\left\{x-25e^{-0.06},0\right\}]\frac{t^{2\alpha }}{\mathsf{\Gamma }(1+2\alpha )}$$

$$v_n\left(x,t\right)=\frac{t^{n\alpha }}{\mathsf{\Gamma }(1+n\alpha )}[-(0.06{)}^{n}x+\left(0.06{)}^{n}max\left\{x-25e^{-0.06},0\right\}\right].$$

Furthermore, the solution to the fractional Black–Scholes equation is:

$$v\left(x,t\right)=\sum _{n=0}^{\infty }{v}_n\left(x,t\right)=max\left\{x-25e^{-0.06},0\right\}{E}_{\alpha }\left(0.06t^{\alpha }\right)+x(1-E_{\alpha }(0.06t^{\alpha }))$$

(5)

Using Python 3.7 software, the graphical representation of the solution in Example 2 is expressed as follows:

Figure 10 shows that for $0\le t\le 5$, with the value of $x$ fixed at [0,5], the graph is obtained to be steadily increasing. The maximum value of $v\left(x,t\right)=$4.9 is obtained when $x=5andt=5$ and the minimum value $v\left(x,t\right)=0$ when $x=0$and $t=0.$ Based on Figure 10, different values of α will determine different solutions and graphs. Then, based on the α values, a graph drawn in three dimensions and cut for certain x and t values obtained graphs in two dimensions. Thus, the graphs in Figure 11, Figure 12, Figure 13, Figure 14, Figure 15 and Figure 16 can be obtained. For example, in Figure 11, α = 0.0, 0.25, 0.5, 0.75, and 1.0, k = 0.5, x = 5.0, and 0 ≤ t ≤ 5.0.

Figure 11 shows that the maximum value $v\left(x,t\right)=4.8$at $x=5$. Figure 11 shows the graph of $v\left(x,t\right)$for $\alpha =1.0$ and $\alpha =0.75$; of note, at the interval $[0,1]$, the graph $v(x,t)$ for $\alpha =1.0$is above the graph $v(x,t)$ for $\alpha =0.75$. The graph $v\left(x,t\right)$for $\alpha =1.0$is marked with a light blue color, whereas the graph $v\left(x,t\right)$ for $\alpha =0.75$is marked with red color. This means that the option value for $\alpha =1.0$is greater than $\alpha =0.75$ at the interval [0, 1]. Meanwhile, at the interval [1,5], the opposite is true, namely the value of $v(x,t)$ for $\alpha =1.0$is actually greater than the value $v(x,t)$ for $\alpha =0.75.$ This is indicated by the blue graph, which is higher than the red graph. Therefore, the option value is greater for $\alpha =1.0$compared to $\alpha =0.75$ at the interval [1,5].

For $\alpha =0.25$and $\alpha =0.5$ at the interval [0, 0.7], the value of $v(x,t)$ for $\alpha =0.25$ is greater than the value $v(x,t)$ for $\alpha =0.5$. However, for the value $t$ at the interval [0.7, 5], the value of $v\left(x,t\right)$ for $\alpha =0.5$ is much larger than the value of $v(x,t)$ for $\alpha =0.25.$ This is indicated by the green graph, which is higher than the yellow graph. As for the values of $\alpha =0.5$ and $\alpha =0.75$ at the interval [0, 1], the value of $v\left(x,t\right)$ for $\alpha =0.5$ is greater than $v\left(x,t\right)$ for $\alpha =0.75.$However, at the interval [1,5], the opposite happens, namely the value of $v\left(x,t\right)$ for $\alpha =0.75$ is greater than the value of $v(x,t)$ for $\alpha =0.5.$On the other hand, the value of $v(x,t)$ at the interval [0, 0.8] for $\alpha =0.0$ is larger than the value of $v(x,t)$ for the values $\alpha$= 0.25, 0.5, 0.75, and 1.0. Meanwhile, at the interval [0.8,5], the value of $v(x,t)$ is lower than the value of $v\left(x,t\right)with\alpha$= 0.25, 0.5, 0, and 1.0, as indicated by the blue graph, which is the lowest compared to other graphs.

Figure 12 shows that the graph is made by expanding the interval $\le t\le 25$. It aims to see the behavior of the graph $v\left(x,t\right)$at interval t, which is longer. It shows that the maximum value $v\left(x,t\right)=5$at $x=25forvalue\alpha =1.0$, which is marked with a light blue graph, and the location of the graph is the highest. Moreover, for $\alpha =0.5$ and $\alpha =0.75$, the graph $v(x,t)$ for $\alpha =0.5$ at the interval [0, 1] is above the graph for $v(x,t)$ for $\alpha =0.75$. Thus, it can be interpreted that the option value for $\alpha =0.5$ is greater than $\alpha =0.75$ at the interval [0, 1]. However, at the interval [1,25], the opposite occurs, namely the value of $v(x,t)$ for $\alpha =0.75$ is actually greater than the value $v(x,t)$ for $\alpha =0.5.$ This is indicated by the red graph, which is higher than the green graph at the interval [1,25]. Therefore, the value of the option is greater for $\alpha =0.75$ compared to $\alpha =0.5$ in the interval [1,25].

By using the same method for values $\alpha =0.25$and $\alpha =0.75$ values on the interval [0, 0.7] and [0.7, 25]$.$This also applies to the values $\alpha =0.5$ and $\alpha =0.75$ at the intervals [0, 1] and [1,25]. On the other hand, the value of $v(x,t)$ at the interval [0, 0.8] for $\alpha =0.0$ is larger than the value of $v(x,t)$ for the values $\alpha$= 0.0, 0.25, 0.5, and 0.75; at the interval [0.8,25], the value of $v(x,t)$ is lower than the value of $v\left(x,t\right)with\alpha$= 0.0, 0.25, 0.5, and 0.75.

The conclusion drawn from Figure 11 and Figure 12 is that for a long period of time and if the value of $x$ is sufficiently large enough, the value of $v(x,t)$ is also large.

Then, $v\left(x,t\right)fork=1.0wasgraphed.$Figure 13 shows that the maximum value of $v\left(x,t\right)=5$ at $0\le t\le 5$. It is quite interesting that for $\alpha =0.25$ and $\alpha =0.5$, the graph $v(x,t)$ for $\alpha =0.5$ at the interval [0, 0.8] is below the graph of $v(x,t)$ for $\alpha =0.25$. This means that the option value for $\alpha =0.25$ is greater than $\alpha =0.5$ at the interval [0, 0.8]. Meanwhile, at the interval [0.8, 5], the opposite occurs, namely the value of $v(x,t)$ for $\alpha =0.25$ is actually smaller than the value $v(x,t)$ for $\alpha =0.5.$ Therefore, the option value is greater for $\alpha =0.5$ compared to $\alpha =0.25$ in the interval [0.8, 5].

For $\alpha =0.25$and $\alpha =0.75$ at the interval [0, 0.7], the value of $v(x,t)$ for $\alpha =0.25$ is greater than the value $v(x,t)$ for $\alpha =0.75$. However, for the value of $t$ at the interval [0.7, 5], the value of $v\left(x,t\right)$ for $\alpha =0.75$ is larger than the value of $v(x,t)$ for $\alpha =0.25.$ As for the values of $\alpha =0.5$ and $\alpha =0.75$ at the interval [0, 1], the value of $v\left(x,t\right)$ for $\alpha =0.5$ is greater than $v\left(x,t\right)$ for $\alpha =0.75.$However, at the interval [1,5], the opposite happens, namely the value of $v\left(x,t\right)$ for $\alpha =0.75$ is greater than the value of $v(x,t)$ for $\alpha =0.5.$On the other hand, the value of $v(x,t)$ at the interval [0, 1] for $\alpha =0.0$ is greater than the values of other $v(x,t)$, namely for the values $\alpha$= 0.25, 0.5, 0.75, and 1.0.

Figure 14 shows that the axis $t$was expanded to $0\le t\le 10.$Furthermore, the graph behavior $v\left(x,t\right)$ becomes more interesting. If at interval $0\le t\le 5$ the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.0$ lies topmost, then at interval $\mathrm{t}\ge 10$, the precise graphic $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.25$ will exceed the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ with value $\mathsf{\alpha }=0.0.$ Meanwhile, the graphic $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.75$ at the interval $0\le \mathrm{t}\le 1$ is located above the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ on $\mathsf{\alpha }=0.5$ at interval $1.75\le \mathrm{t}\le 10;$ instead, the graph is located below. Likewise for the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.5,$ at interval $\left[\mathrm{0,0.8}\right]$, it is located above the graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $0.8\le \mathrm{t}\le 10.0,$ quite the opposite. Graph $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.75$ is precisely above graphic $\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)$ for $\mathsf{\alpha }=0.5.$

Figure 15 shows a graphic image $v(x,t)$ with $k=0.5$, $0\le x\le 20$ with the value of $t$ being fixed, namely $t=10.$The graph $v(x,t)$ with $\alpha =1.0$ is a straight line. The graph $v(x,t)$ is the highest compared to the graph $v\left(x,t\right)$ with other values. The maximum value of $v\left(x,t\right)$= 20, with a value of $\alpha =1.0$ achieved when $x=20.0.$ Then, the graph $v(x,t)$ for $\alpha =0.75$ is a straight line. The graph is located below the graph $v(x,t)$ for $\alpha =1.0$. The maximum value of $v\left(x,t\right)=17.5$ is reached when $x=20.0.$Furthermore, the graph $v(x,t)$ for the value of $\alpha =0.5$is a straight line. The location of the graph is, respectively, located below the graph $v(x,t)$ for values $\alpha =1.0$ and $\alpha =0.25$. For the graph $v(x,t)$ for $\alpha =0.5,$ the shape of the graph is a straight line. The maximum value of $v\left(x,t\right)=13.6$ is reached when $x=20.$ For the graph $v(x,t)$ for $\alpha =0.25,$the graph is a straight line with the maximum value $v\left(x,t\right)=10$ achieved when $x=20.0$. Then, the last graph has the lowest height compared to the height of the other graphs, namely the graph $v(x,t)$ with value $\alpha =0.0$, which has a maximum value $v\left(x,t\right)=6.75$ achieved when $x=20.0.$

Figure 16 shows a graphic image $v(x,t)$ with $k=0.5$, $0\le x\le 75$ in which the value of $t$ is made fixed, namely $t=25.$The graph $v(x,t)$ with $\alpha =1.0$ is a straight line. The location of the graph $v(x,t)$ is the highest compared to the graph $v\left(x,t\right)$ with other values. The maximum value of $v\left(x,t\right)$= 78, with a value of $\alpha =1.0$, is achieved when $x=70.0$. The graph $v(x,t)$ for $\alpha =0.75$ is a straight line. The graph is located below the graph $v(x,t)$ for $\alpha =1.0$. The maximum value of $v\left(x,t\right)=75.0$ is reached when $x=70.0.$The graph $v(x,t)$ for the value of $\alpha =0.5$is a straight line. The location of the graph is located below the graph $v(x,t)$ for values $\alpha =1.0$ and $\alpha =0.25$. The maximum value of $v\left(x,t\right)=65$ is reached when the value of $x=75.$ The graph $v(x,t)$ for the value of $\alpha =0.25$, unlike the other graphs $v\left(x,t\right),$ is a straight line. The graph $v(x,t)$ with the value of $\alpha =0.25$ at the interval $0\le x\le 40$ is a straight line. At $x=40,$ the graph turns slightly up, although the shape of the graph remains a straight line. The maximum value of $v\left(x,t\right)=55$ is reached when $x=75$. The last graph $v(x,t)$ for $\alpha =0.0$ has the lowest height compared to the other graphs $v(x,t)$. This graph has similarities with the graph $v\left(x,t\right)$ for the value of $\alpha =0.25$. The behavior of the graph $v(x,t)$ for the value of $\alpha =0.0$ at the interval $0\le x\le 40$ is a straight line. When $x=40$, the graph turns up in the form of a straight line. The maximum value $v\left(x,t\right)=$ 45 is achieved when $x=75.$

From the fractional Black–Scholes equation, if taken for $\alpha =1$, the solution to the Black–Scholes equation approach is obtained. The solution is an iterative solution, in the form of an infinite series of the Mittag-Lefler function. The following shows a graph of the solution to the fractional Black–Scholes equation approximation by taking $\alpha =1$ and a graph of the exact solution to the Black–Scholes equation.

It can be clearly seen from Figure 17 and Figure 18 that the obtained solutions almost coincide. Therefore, the resulting error is very small, at approximately 0 percent. This means that the exact solution to the Black–Scholes equation can be approached as a solution to the fractional Black–Scholes equation, namely in the form of an infinite series of the Mittag-Leffler function by taking $\alpha =1$.

Table 1. Approximate solutions, exact solutions, and error values for $\alpha =1$.

$\mathit{x}$	$\mathit{t}$	$\mathit{v}$	$\overline{\mathit{v}}$	$\mathit{e}$ (%)
0.1	0.1	0.2003335000396882	0.20033350003968808	$5.541874047052588\times {10}^{-14}$
0.2	0.1	0.31656534012421034	0.3165653401242102	3.507089639660298 × ${10}^{-14}$
0.1	0.2	0.2864401649976659	0.2864401649976658	$3.875933476836963\times {10}^{-14}$
0.2	0.2	0.40267200508218803	0.4026720050821879	$2.7571398324513585\times {10}^{-14}$
1.5	1.5	4.258558910189635	4.258558910189635	0.0
50	1.5	5.184705528587072 × ${10}^{21}$	5.184705528587072 × ${10}^{21}$	0.0
100	1.5	2.6881171418161356 × ${10}^{43}$	2.6881171418161356 × ${10}^{43}$	0.0
100	2.5	2.6881171418161356 × ${10}^{43}$	2.6881171418161356 × ${10}^{43}$	1.8420923999599334 × ${10}^{-14}$
150	2.5	1.3937095806663795 × ${10}^{65}$	1.3937095806663797 × ${10}^{65}$	1.677826322045785 × ${10}^{-14}$

shows $v(x,t)$ values for the exact solution to the Black–Scholes equation, approximate solutions, and error values. The error formula used is $e=\left|\frac{v-\overline{v}}{v}\right|\times 100$%.

Table 1 shows that the value of $x$ is made to change, namely $x=0.1$ and $x=0.2,$ while for the value of $t=0.1,$the error values obtained are $5.541874047052588\times {10}^{-14}$ and 3.507089639660298 × ${10}^{-14}$.

The value $t$ fixed is the value of $t=0.2$, while the values of $x$ changed are $x=0.1$and $x=0.2$. The error values $t=0.2$ obtained are $3.875933476836963\times {10}^{-14}$ and $2.7571398324513585\times {10}^{-14}$. Meanwhile, for the value of $x=1.5,x=50,\mathrm{a}\mathrm{n}\mathrm{d}x=100$ for the value of $t=1.5$, the same error value is 0.0%. For the value of $x=100$and $x=150$ for the value of $t=2.5$, the error values obtained are 1.8420923999599334 × ${10}^{-14}$ and 1.677826322045785 × ${10}^{-14}$. Therefore, it can be concluded that the solution to the fractional Black–Scholes equation using the Daftardar-Geiji method is very good since the error value obtained is very small, almost 0.0%. It is interesting if the value of $\alpha$ is taken close to 1.

The following table shows approximate solutions, exact solutions, and error values by taking the value of $\alpha$, which is close enough to 1: $\alpha =0.97$.

From

Table 2. Approximate solutions, exact solutions, and error values for $\alpha =0.97$.

$\mathit{x}$	$\mathit{t}$	$\mathit{v}$	$\overline{\mathit{v}}$	$\mathit{e}$(%)
0.1	0.1	0.2003335000396882	0.2078185428941125	3.7362911609598197
0.2	0.1	0.31656534012421034	0.32405038297863464	2.364454318178803
0.3	0.1	0.44502138954004367	0.452506432394468	1.6819512568060038
0.1	0.2	0.2864401649976659	0.2960476582031938	3.35
0.1	0.3	0.36435269739392984	0.37416666992705205	2.69
1.5	1.5	4.258558910189635	4.248745537260461	0.23043882064642066
50	1.4	5.184705528587072  $\times {10}^{21}$	5.184705528587072 × ${10}^{21}$	0.0
50	1.5	5.184705528587072 × ${10}^{21}$	5.184705528587072 × ${10}^{21}$	0.0
100	1.5	2.6881171418161356 × ${10}^{43}$	2.6881171418161356 × ${10}^{43}$	0.0
100	2.0	2.6881171418161356 × ${10}^{43}$	2.6881171418161356 × ${10}^{43}$	0.0
100	2.5	2.688117141816135 × ${10}^{43}$	2.6881171418161356 × ${10}^{43}$	1.8420923999599334 × ${10}^{-14}$
100	50	2.6881171418161356 × ${10}^{43}$	2.6895918281430976 × ${10}^{43}$	0.05485945177097765

, with $x$ values that change, namely $x=0.1,0.2,\mathrm{a}\mathrm{n}\mathrm{d}0.3$, and $t$ values kept constant, namely $t=0.1$, it is found that error values tend to decrease, namely three in a row: 74%, 2.36%, and 1.68%. Likewise, if the value of $x$ is fixed, namely $x=0.1,$and the value of $t$ is varied, namely $t=0.1,0.2,\mathrm{a}\mathrm{n}\mathrm{d}0.3,$the error values obtained are 3.74% 3.35%, and 2.69%, respectively. The error obtained is still very good since it is less than 5%.

From

Table 3. Approximate solutions, exact solutions, and error values for $\alpha =0.95$.

$\mathit{x}$	$\mathit{t}$	$\mathit{v}$	$\overline{\mathit{v}}$	$\mathit{e}$ (%)
0.1	0.1	0.2003335000396882	0.21306186281832695	6.35
0.2	0.1	0.31656534012421034	0.3292937029028491	4.02
0.3	0.1	0.44502138954004367	0.4577497523186824	2.86
0.1	0.2	0.2864401649976659	0.3025940917164728	5.64
0.1	0.3	0.36435269739392984	0.3807229933900663	4.49

, with $x$ values that change, namely $x=0.1,0.2,$ and $0.3,$ and with the $t$ value being fixed, namely $t=0.1$, it is found that the error value tends to decrease, namely 6.35%, 4.02%, and 2.86%. Likewise, if the value of $x$ is fixed, namely $x=0.1$, and the value of $t$ is varied, namely $t=0.1,0.2,\mathrm{a}\mathrm{n}\mathrm{d}0.3,$error values of 6.35%, 5.64%, and 4.49%, respectively, are obtained. The error values obtained tend to decrease, but it can be said that this is still good since it is less than 10%.

From

Table 4. Approximate solutions, exact solutions and error values for $\mathit{\alpha }=\mathbf{0.9}$.

$\mathit{x}$	$\mathit{t}$	$\mathit{v}$	$\overline{\mathit{v}}$	$\mathit{e}$(%)
0.1	0.1	0.2003335000396882	0.33729694259672105	13.35
0.2	0.1	0.31656534012421034	0.3433066351345849	8.45
0.3	0.1	0.44502138954004367	0.47176268455041825	6.01
0.1	0.2	0.2864401649976659	0.31940849935827975	11.51
0.1	0.3	0.36435269739392984	0.3970963571315984	8.99
0.1	0.4	0.4348508720400084	0.464071891044021	6.72

, for the $x$ values that change, namely $x=0.1,0.2,$ and 0.3, and with the $t$ values made constant, namely $t=0.1,$it is found that the error values tend to decrease, namely 13.35%, 8.45%, and 6.01%, respectively. Even though the error tends to decrease, an error value of 13.35% is not good for estimating the value of $v(x,t)$ because it is more than 10%. Likewise, if the value of $x$ is fixed, namely $x=0.1,$ and the value of $t$ is varied, namely $t=0.1,0.2,0.3,and0.4$, the error values of 13.35%, 11.51%, 8.99%, and 6.72%, respectively, are obtained. For an error value of 13.35%, 11.51% is not good enough to estimate the value of $v(x,t)$ since it is above 10%.

From Table 1, Table 2, Table 3 and Table 4, it can be concluded that the closer the value of $\alpha$ is to 1, the closer the graph obtained is to the graph of the solution to the Black–Scholes equation. This is reinforced by the graph below:

Figure 19 shows that the graph $v(x,t)$ for the value of $\alpha$, which is closer to 1, is becoming closer to the exact solution. This can be shown through the red graph that is close to the Black–Scholes equation solution graph. The red graph is the graph $v(x,t)$ for the value of $\alpha =0.975$. Meanwhile, the graph $v\left(x,t\right)=1$ is shown in purple. It is clear that the graph with the value $\alpha =0.9$ shown in blue is still far from approaching the exact solution, namely the solution to the Black–Scholes equation. Then, the graph with the value of $\alpha =0.925$, which is yellow, is located below the blue graph. Meanwhile, the graph with the value $\alpha =0.95$, which is green, moves closer to the graph of the exact solution to the fractional Black–Scholes equation. Finally, the red graph has the value $\alpha =0.975.$ The error comparison produced using the analysis Shehu transform method [28] is as follows:

According to

Table 5. Error comparison of the Daftardar-Geijji method and homotopy Shehu transform.

	$\mathit{x}$	$\mathit{t}$	${\mathit{v}}_{\mathit{e}\mathit{x}\mathit{c}}$	${\mathit{v}}_{\mathit{D}\mathit{a}\mathit{f}\mathit{t}}$	${\mathit{v}}_{\mathit{H}\mathit{A}\mathit{S}\mathit{T}\mathit{M}}$	${\mathit{e}}_{\mathit{D}\mathit{a}\mathit{f}\mathit{t}}$ (%)	${\mathit{e}}_{\mathit{H}\mathit{A}\mathit{S}\mathit{T}\mathit{M}}$ (%)
0	0.000000	0.0	0.000000	0.000000	0.000000
1	0.000000	0.5	0.393469	0.398521	0.398521	1.283859	1.283859
2	0.000000	1.0	0.632121	0.630747	0.630747	0.217324	0.217324
3	0.333333	0.0	0.395612	0.395612	0.395612	0.000000	0.000000
4	0.333333	0.5	0.789082	0.794133	0.794133	0.640186	0.640186
5	0.333333	1.0	1.027733	1.026359	1.026359	0.133668	0.133668
6	0.666667	0.0	0.947734	0.947734	0.947734	0.000000	0.000000
7	0.666667	0.5	1.341203	1.346255	1.346255	0.376646	0.376646
8	0.666667	1.0	1.579855	1.578481	1.578481	0.086954	0.086954
9	1.000000	0.0	1.718282	1.718282	1.718282	0.000000	0.000000
10	1.000000	0.5	2.111751	2.116803	2.116803	0.239213	0.239213
11	1.000000	1.0	2.350402	2.349029	2.349029	0.058447	0.058447

, both the Daftardar-Geijji method approach and the homotopy analysis Shehu transform method produce the same error, which is relatively minimal. The two strategies were shown to be successful in solving the fractional Black–Scholes equation.

The value of European call options can be expressed as $E_C=Kv\left(x,t\right)$with $K$ being the exercise price. As a result, the produced graph is shown in Figure 20 and Figure 21.

The following is a theorem about the convergence of fractional Black–Scholes equation solutions and the upper limit for the absolute maximum error.

Theorem 4.

Let $\mathcal{H}$ Banach spaces and ${\mathrm{v}}_{\mathrm{p}}\left(\mathsf{\zeta },\mathcal{l}\right)$ , ${\mathrm{v}}_{\mathrm{n}}\left(\mathsf{\zeta },\mathcal{l}\right)\in \mathcal{H}$. The approximation solution in the form of a series $\sum _{p=0}^{\infty }{v}_p\left(\zeta ,\mathcal{l}\right)$ convergences to the solution to the Black–Scholes equation $v(\zeta ,\mathcal{l})$ if $v_p\left(\zeta ,\mathcal{l}\right)\le \lambda v_{p-1}\left(\zeta ,\mathcal{l}\right)$ for $\lambda \in \left(\mathrm{0,1}\right)$, i.e., for every $ϵ>0,$there is $E\in \mathbb{N}$ such that $‖v_{p+n}\left(\zeta ,\mathcal{l}\right)‖<ϵ$ for every $p,n>E$.

Proof. 

Suppose that the partial sum $\mathrm{m}$ terms ${\mathrm{u}}_{\mathrm{m}}\left(\mathsf{\zeta },\mathcal{l}\right)=\sum _{\mathrm{i}=1}^{\mathrm{m}}{\mathrm{v}}_{\mathrm{i}}\left(\mathsf{\zeta },\mathcal{l}\right)$.

$$‖u_{m+1}-u_m‖=‖v_{m+1}\left(\zeta ,\mathcal{l}\right)‖\le \lambda ‖v_m\left(\zeta ,\mathcal{l}\right)‖$$

$$\le {\lambda }^2‖v_{m-1}\left(\zeta ,\mathcal{l}\right)‖\le {\lambda }^3‖v_{m-2}\left(\zeta ,\mathcal{l}\right)‖\dots \le {\lambda }^{m+1}‖v_0\left(\zeta ,\mathcal{l}\right)‖$$

$$‖u_m\left(\zeta ,\mathcal{l}\right)-u_n\left(\zeta ,\mathcal{l}\right)‖=‖v_{m+n}‖$$

$$=‖\left(u_p-u_{p-1}\right)+\left(u_{p-1}-u_{p-2}\right)+\dots (u_{m+1}-u_m)‖$$

$$\le {\lambda }^m‖v_0‖+{\lambda }^{m-1}‖v_0‖+{\lambda }^{m-2}‖v_0‖\dots +{\lambda }^{m+1}‖v_0‖$$

$$=‖v_0‖\left({\lambda }^m+{\lambda }^{m-1}+{\lambda }^{m-2}+\dots {\lambda }^{m+1}\right)$$

$$=‖v_0‖\left(\frac{1-{\lambda }^{m-n}}{1-\lambda }\right){\lambda }^{n+1}<M\left(\frac{1-{\lambda }^{m-n}}{1-\lambda }\right){\lambda }^{n+1}$$

Because $0<\lambda <1$ and $M=\underset{(\zeta ,\mathcal{l})}{\sup }‖v_0\left(\zeta ,\mathcal{l}\right)‖$, take $\epsilon =M\left(\frac{1-{\lambda }^{m-n}}{1-\lambda }\right){\lambda }^{n+1}$. It is proved that $‖v_{m+n}‖\le \epsilon$. This implies that $\underset{i\to \mathrm{\infty }}{\lim }{v}_i\left(\zeta ,\mathcal{l}\right)=v\left(\zeta ,\mathcal{l}\right)$. □

Theorem 5.

Suppose that $\sum _{\mathrm{i}=0}^{\mathrm{m}}{\mathrm{v}}_{\mathrm{i}}\left(\mathsf{\zeta },\mathcal{l}\right)$ is finite and the exact solution is $\mathrm{v}(\mathsf{\zeta },\mathcal{l})$. If $‖v_{m+1}\left(\zeta ,\mathcal{l}\right)‖\le \lambda ‖v_0\left(\zeta ,\mathcal{l}\right)‖$ for $0<\lambda <1$, then the maximum absolute error is $‖v\left(\zeta ,\mathcal{l}\right)-\sum _{i=0}^m{v}_i\left(\zeta ,\mathcal{l}\right)‖\le \frac{{\lambda }^{m+1}}{1-\lambda }‖v_0\left(\zeta ,\mathcal{l}\right)‖$.

Proof .

Based on the exact and the fractional Black–Scholes approach solutions, the difference norm is taken to find the upper bound of the absolute error value.

$$‖v\left(\zeta ,\mathcal{l}\right)-{\sum }_{i=0}^m{v}_i\left(\zeta ,\mathcal{l}\right)‖=‖{\sum }_{i=0}^{\infty }{v}_i\left(\zeta ,\mathcal{l}\right)‖$$

$$\le {\sum }_{i=m+1}^{\infty }‖v_i\left(\zeta ,\mathcal{l}\right)‖$$

$$\le {\sum }_{i=m+1}^{\infty }{\lambda }^i‖v_0\left(\zeta ,\mathcal{l}\right)‖$$

$$={\lambda }^{m+1}(1+\lambda +{\lambda }^2+\dots )‖v_0\left(\zeta ,\mathcal{l}\right)‖$$

$$\le \frac{{\lambda }^{m+1}}{1-\lambda }‖v_0\left(\zeta ,\mathcal{l}\right)‖.$$

Based on Theorems 4 and 5, it can be concluded that the approximate solution to the fractional Black–Scholes equation converges to the exact solution and has an upper bound for the absolute error. □

4. Conclusions

The Daftardar-Gejji method is simple and produces small errors for solving the fractional Black–Scholes equation. It involves fractional integrals to find the solution. The result obtained in this paper are the guaranteed existence and uniqueness of the solution to the fractional Black–Scholes equation. The solution form of the fractional Black–Scholes equation is an infinite series of Mittag-Lefler functions. If the fractional order of the solution to the fractional Black–Scholes equation is closer to one, the graph obtained is closer to the graph of the Black–Scholes equation. In financial mathematics, the solution to the Black–Scholes equation is the value of an option. This research is very important and useful for investors to determine whether to sell, hold or buy shares within a certain time period to gain profits.

5. Further Research

One way to create a modified Black–Scholes equation is by combining fractional telegraph diffusion and fractional Black–Scholes equations, which leads to a more general form of the model. Further research can focus on exploring the existence and uniqueness of solutions, as well as their application to option values through analytical and numerical solutions. Additionally, selecting a suitable fractional derivative operator and stochastic process with consideration for the underlying asset can aid in constructing a modified fractional Black–Scholes equation formula.