The exact traveling wave solutions of a class of generalized Black-Scholes equation

Weiping Gao; Yanxia Hu; Weiping Gao; Yanxia Hu

doi:10.3934/Math.2017.3.385

AIMS Mathematics

2017, Volume 2, Issue 3: 385-399. doi: 10.3934/Math.2017.3.385

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Research article

The exact traveling wave solutions of a class of generalized Black-Scholes equation

Weiping Gao ,
Yanxia Hu ^,

School of Mathematics and Physics, North China Electric Power University, Beijing, 102206, China

Received: 19 May 2017 Accepted: 13 June 2017 Published: 27 June 2017

In this paper, the traveling wave solutions of a class of generalized Black-Scholes equation are considered. By using the first integral method and the G'/G-expansion method, several exact traveling wave solutions of the equation are obtained.

Keywords:

Citation: Weiping Gao, Yanxia Hu. The exact traveling wave solutions of a class of generalized Black-Scholes equation[J]. AIMS Mathematics, 2017, 2(3): 385-399. doi: 10.3934/Math.2017.3.385

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Abstract

1. Introduction

In financial mathematics and financial engineering, the classical Black-Scholes equation is a practical partial differential equation. In 1973, Black and Scholes derived the famous Black-Scholes Option Pricing Model ^[1]. In ^[2], Sunday O. Edeki etc successfully calculated the European option valuation using the Projected Differential Transformation Method. The results obtained converge faster to their associated exact solutions. In ^[3], the author studied the Black-Scholes equation in stochastic volatility model. In ^[4], the author considered to deal with the Black-Scholes equation in financial mathematics by the volatility of a variable and the abstract boundary conditions.

In this paper, we consider to obtain the traveling wave solutions of a class of generalized Black-Scholes equation by using the first integral method and the $G'/G$ -expansion method. In 2002, Feng first proposed the first integral method ^[5]. This method has been widely used to solve the exact solutions of some partial differential equations. In ^[6], authors applied first integral method and functional variable method to obtain optical solitons from the governing nonlinear Schr¨ odinger equation with spatio-temporal dispersion. In ^[7], the first integral method is applied for solving the system of nonlinear partial differential equations which are (2 + 1)-dimensional Broer-Kaup-Kupershmidt system and (3 + 1)-dimensional Burgers equations exactly. In ^[8], authors applied first integral method to construct travelling wave solutions of modified Zakharov-Kuznetsov equation and ZK-MEW equation. This method can also be applied to other systems of nonlinear differential equations ^[9,10,11,12]. The advantage of the first integral method is that the calculation is concise. A more accurate traveling wave solution can be obtained by the first integral method. In 2008, Mingliang Wang et al introduced the $G'/G$ -expansion method in detail ^[13]. In ^[14], the $G'/G$ -expansion method is applied to address the resonant nonlinear Schr $\ddot{o}$ dinger equation with dual-power law nonlinearity. In ^[15], the author constructed the traveling wave solutions involving parameters for some nonlinear evolution equations in mathematical physics via the (2 + 1)-dimensional Painlev $\acute{e}$ integrable Burgers equations, the (2 + 1)-dimensional Nizhnik-Novikov-Vesselov equations, the (2 + 1)-dimensional Boiti-Leon-Pempinelli equations and the (2 + 1)-dimensional dispersive long wave equations by using the $G'/G$ -expansion method. In ^[16], a generalized $G'/G$ -expansion method is proposed to seek exact solutions of the mKdV equation with variable coefficients. The $G'/G$ -expansion method has been proposed to construct more explicit travelling wave solutions to many nonlinear evolution equations ^[17,18,19]. The performance of this method is effective, simple, convenient and gives many new solutions.

The classical celebrated Black-Scholes option pricing model is as follows:

$\frac{\partial f}{\partial \tau}+\frac{1}{2}S^2\delta^2\frac{\partial^2f}{\partial S^2}+rS\frac{\partial f}{\partial S}-rf=0.$

We consider the class of generalized Black-Scholes equation is as follows:

$v_t+\frac{1}{2}A^2x^2v_{xx}+Bxv_x-Cv+Dv^3=0,$

(1)

where $v=v(t, x), A, B, C, D\neq0$ are arbitrary constants, when $D=0,$ (1) changes to the classical Black-Scholes equation.

Using the wave transformation $v(t, x)=v(\xi),$ $\xi=\ln x-at,$ and $a$ is wave velocity, we have the following ordinary differential equation,

$-\frac{1}{2}A^2v''+(\frac{1}{2}A^2-B+a)v'+Cv-Dv^3=0.$

(2)

Letting $w=v',$ (2) is equivalent to the autonomous system,

$\left\{\begin{array}{lll} {\displaystyle v'=w, }\\ \\{\displaystyle w'=\frac{A^2-2B+2a}{A^2}w+\frac{2C}{A^2}v-\frac{2D}{A^2}v^3.}\\ \end{array}\right.$

(3)

2. Traveling wave solutions of (1) by using the first integral method

In this section, we apply the first integral method to obtain the traveling wave solution to (1). We assume that $p(v, w)=\sum_{i=0}^N\alpha_i(v)w^i$ is an irreducible polynomial in $C[v, w]$ , and $p(v, w)=\sum_{i=0}^N\alpha_i(v)w^i=0$ is a first integral of (3), such that,

$\frac{dP}{d\xi}\bigg\vert_{(3)}=0.$

Owing to Division Theorem, there exists a polynomial $g(v)+h(v)w$ in the complex domain $C(v, w)$ , such that,

$\frac{dP}{d\xi}=\frac{\partial P}{\partial v}\frac{dv}{d\xi}+\frac{\partial P}{\partial w}\frac{dw}{d\xi}=[g(v)+h(v)w]\sum\limits_{i=0}^N\alpha_i(v)w^i=0.$

(4)

Here, we mainly consider (4) in two cases: $N=1$ and $N=2.$

2.1. $N=1$

At present,

$P(v, w)=\alpha_0(v)+\alpha_1(v)w=0.$

From (4), we have

$\frac{dP}{d\xi}=\frac{d\alpha_0}{dv}w+\frac{d\alpha_1}{dv}w^2+\alpha_1(v)(\frac{A^2-2B+2a}{A^2}w+\frac{2C}{A^2}v-\frac{2D}{A^2}v^3) \\ \quad=g(v)\alpha_0(v)+[g(v)\alpha_1(v)+h(v)\alpha_0(v)]w+h(v)\alpha_1(v)w^2.$

(5)

By observing the coefficients of $w^i (i = 0, 1)$ of the two sides of (5), obviously, we have

$\left\{\begin{array}{lll} {\displaystyle \frac{d\alpha_{1}}{dv}=h(v)\alpha_{1}(v), }\\ \\{\displaystyle \frac{d\alpha_{0}}{dv}=g(v)\alpha_{1}(v)+h(v)\alpha_{0}(v)-\frac{A^2-2B+2a}{A^2}\alpha_1(v)}, \\ \\{\displaystyle \alpha_{0}(v)g(v)=\alpha_{1}(v)(\frac{2C}{A^2}v-\frac{2D}{A^2}v^{3})}. \end{array}\right.$

(6)

Since $\alpha_i(v)(i=0, 1)$ are polynomials, then from the first equation of (6), we deduce that $\alpha_1(v)$ is constant and $h(v)=0$ . For simplification, taking $\alpha_1(v)=1$ . In order to keep balancing the degree of $g(v),$ $\alpha_1(v)$ and $\alpha_0(v)$ in the second and the third equations of (6), one can conclude that deg $g(v)=1$ . Suppose that $g(v)=g_0v+d_0,$ where $g_0$ and $d_0$ are arbitrary constants. By solving the above equations, one can obtain

$\left\{\begin{array}{ll} {\displaystyle \alpha_1(v)=1, }\\ \\{\displaystyle \alpha_0(v)=\frac{1}{2}g_0v^2+(d_0-\frac{A^2-2B+2a}{A^2})v+d_1, } \end{array}\right.$

where $d_1$ is an integration constant. Substituting $g(v),$ $\alpha_1(v)$ and $\alpha_0(v)$ into the third equation of (6) and setting all the coefficients of $v^i(i=0, 1, 2, 3)$ to be zeros, then one can get

$\left\{\begin{array}{llll} {\displaystyle \frac{1}{2}g_0^2=-\frac{2D}{A^2}, }\\ \\{\displaystyle \frac{3}{2}g_0d_0-g_0\frac{A^2-2B+2a}{A^2}=0}, \\ \\{\displaystyle g_0d_1+d_0^2-d_0\frac{A^2-2B+2a}{A^2}=\frac{2C}{A^2}, }\\ \\{\displaystyle d_0d_1=0. } \end{array}\right.$

(7)

From the last equation of (7), we consider to solve (7) in two cases $d_0=0$ or $d_0\neq0.$

Case 1: $d_0=0$

By solving (7), we have

$\left\{\begin{array}{lll} {\displaystyle g_0=\pm2\sqrt{\frac{-D}{A^2}}, }\\ \\{\displaystyle d_0=0}, \\ \\{\displaystyle d_1=\pm\frac{C}{A^2}\sqrt{\frac{A^2}{-D}}, } \end{array}\right.$

and $D < 0, A^2-2B+2a=0,$ such that,

$\left\{\begin{array}{ll} {\displaystyle \alpha_0(v)=\pm\sqrt{\frac{-D}{A^2}}v^2\pm\frac{C}{A^2}\sqrt{\frac{A^2}{-D}} }, \\\\ {\displaystyle \alpha_1(v)=1. } \end{array}\right.$

So,

$P(v, w)=\pm\sqrt{\frac{-D}{A^2}}v^2\pm\frac{C}{A^2}\sqrt{\frac{A^2}{-D}}+w=0.$

One can obtain the following one order ordinary differential equation,

$w=\frac{dv}{d\xi}=\mp\sqrt{\frac{-D}{A^2}}v^2\mp\frac{C}{A^2}\sqrt{\frac{A^2}{-D}}.$

(8)

Integrating (8) once with respect to $\xi$ , we can get the following results.

Ⅰ: $\frac{C}{D}>0,$ we have

$v(\xi)=\frac{\sqrt{\frac{C}{D}}(1+\xi_0e^{\mp2\sqrt{\frac{-D}{A^2}}\xi})}{1-\xi_0e^{\mp2\sqrt{\frac{-D}{A^2}}\xi}},$

where $\xi_0$ is an integration constant.

The traveling wave solution to (1) can be got as follows:

$v(t, x)=\frac{\sqrt{\frac{C}{D}}(1+\xi_0e^{\mp2\sqrt{\frac{-D}{A^2}}(\ln x-at)})}{1-\xi_0e^{\mp2\sqrt{\frac{-D}{A^2}}(\ln x-at)}}.$

Ⅱ: $\frac{C}{D} < 0,$ we have

$v(\xi)=\sqrt{\frac{C}{-D}}\tan[\mp\sqrt{\frac{C}{A^2}}\xi+\xi_1],$

where $\xi_1$ is an arbitrary integration constant.

The traveling wave solution to (1) can be got as follows:

$v(t, x)=\sqrt{\frac{C}{-D}}\tan[\mp\sqrt{\frac{C}{A^2}}(\ln x-at)+\xi_1].$

Case 2: $d_0 \neq 0$

From (7), we have

$\left\{\begin{array}{lll} {\displaystyle g_0=\pm2\sqrt{\frac{-D}{A^2}}, }\\\\ {\displaystyle d_0=\frac{2(A^2-2B+2a)}{3A^2}}, \\\\ {\displaystyle d_1=0, } \end{array}\right.$

and $D < 0,$ $6C+(A^2-2B+2a)=0,$ such that,

$\left\{\begin{array}{ll} {\displaystyle \alpha_0(v)=\pm\sqrt{\frac{-D}{A^2}}v^2-\frac{A^2-2B+2a}{3A^2}v }, \\\\ {\displaystyle \alpha_1(v)=1. } \end{array}\right.$

So,

$P(v, w)=\pm\sqrt{\frac{-D}{A^2}}v^2-\frac{A^2-2B+2a}{3A^2}v+w=0.$

One can obtain the following one order ordinary differential equation,

$w=\frac{dv}{d\xi}=\mp\sqrt{\frac{-D}{A^2}}v^2+\frac{A^2-2B+2a}{3A^2}v.$

(9)

Integrating equation (9) once with respect to $\xi$ , we can get the following results,

$v(\xi)=\frac{\frac{A^2-2B+2a}{3A^2}\xi_2e^{\frac{A^2-2B+2a}{3A^2}\xi}}{1\pm\sqrt{\frac{-D}{A^2}}\xi_2e^{\frac{A^2-2B+2a}{3A^2}\xi}},$

where $\xi_2$ is an integration constant.

The exact traveling wave solution to (1) can be got as follows:

$v(t, x)=\frac{\frac{A^2-2B+2a}{3A^2}\xi_2e^{\frac{A^2-2B+2a}{3A^2}(\ln x-at)}}{1\pm\sqrt{\frac{-D}{A^2}}\xi_2e^{\frac{A^2-2B+2a}{3A^2}(\ln x-at)}}.$

2.2. $N=2$

At present,

$P(v, w)=\alpha_0(v)+\alpha_1(v)w+\alpha_2(v)w^2=0.$

(10)

From (4), we have

$\frac{dP}{d\xi}=\frac{d\alpha_0}{dv}w+\frac{d\alpha_1}{dv}w^2+\alpha_1(v)(\frac{A^2-2B+2a}{A^2}w+\frac{2C}{A^2}v-\frac{2D}{A^2}v^3)\\ \quad\quad+\frac{d\alpha_2}{dv}w^3+2\alpha_2(v)w(\frac{A^2-2B+2a}{A^2}w+\frac{2C}{A^2}v-\frac{2D}{A^2}v^3)\\\\ \quad=g(v)\alpha_0(v)+[g(v)\alpha_1(v)+h(v)\alpha_0(v)]w+[g(v)\alpha_2(v)+h(v)\alpha_1(v)]w^2+h(v)\alpha_2(v)w^3.$

(11)

By observing the coefficients of $w^i(i = 0, 1, 2, 3)$ of the two sides of (11), obviously, we have

$\left\{\begin{array}{lll} {\displaystyle \frac{d\alpha_{2}}{dv}=h(v)\alpha_{2}(v), }\\ {\displaystyle \frac{d\alpha_{1}}{dv}=g(v)\alpha_{2}(v)+h(v)\alpha_{1}(v)-2\frac{A^2-2B+2a}{A^2}\alpha_2(v)}, \\ {\displaystyle \frac{d\alpha_{0}}{dv}=g(v)\alpha_{1}(v)+h(v)\alpha_{0}(v)-\frac{A^2-2B+2a}{A^2}\alpha_1(v)-2\alpha_2(v)(\frac{2C}{A^2}v-\frac{2D}{A^2}v^3), } \\{\displaystyle \alpha_{0}(v)g(v)=\alpha_{1}(v)(\frac{2C}{A^2}v-\frac{2D}{A^2}v^{3})}. \end{array}\right.$

(12)

Similarly, from the first equation of (12), we deduce that $\alpha_2(v)$ is constant and $h(v)=0.$ From the second and the third equations of (12), we assume that deg $g(v)=k$ , then deg $\alpha_1(v)=k+1$ and deg $\alpha_0(v)=2k+2.$ By the last equation of (12), we have the degree of the polynomial $\alpha_1(v)(\frac{2C}{A^2}v-\frac{2D}{A^2}v^3)$ is $k+4$ , and the degree of the polynomial $\alpha_0(v)g(v)$ is $3k + 2$ . By balancing the degree of the last equation of (12), we have $k+4=3k+2,$ obviously, $k=1$ . Specially, we conclude that deg $g(v)=0$ , then deg $\alpha_1(v) = 1$ . The degree on both sides of the last equation of (12) is still true.

Case 1: deg $g(v)=0$

Assume $g(v)=g_1.$ From (12), we have

$\left\{\begin{array}{llll} {\displaystyle \alpha_0(v)=\frac{D}{A^2}v^4+(\frac{1}{2}g_1^2-\frac{3}{2}g_1\frac{A^2-2B+2a}{A^2}+\frac{(A^2-2B+2a)^2}{A^4}-\frac{2C}{A^2})v^2 }\\\\ {\displaystyle +d_2(g_1-\frac{A^2-2B+2a}{A^2})v+d_3, }\\\\ {\displaystyle \alpha_1(v)=(g_1-2\frac{A^2-2B+2a}{A^2})v+d_2 }, \\\\ {\displaystyle \alpha_2(v)=1, } \end{array}\right.$

(13)

where $d_2, d_3$ are integration constants. From the last one of equation (12), we have

$\left\{\begin{array}{llll} {\displaystyle g_1\frac{D}{A^2}=-\frac{2D}{A^2}(g_1-2\frac{A^2-2B+2a}{A^2}), }\\ {\displaystyle -d_2\frac{2D}{A^2}=0, }\\ {\displaystyle g_1(\frac{1}{2}g_1^2-\frac{3}{2}g_1\frac{A^2-2B+2a}{A^2}+\frac{(A^2-2B+2a)^2}{A^4}-\frac{2C}{A^2})=\frac{2C}{A^2}(g_1-2\frac{A^2-2B+2a}{A^2}) }, \\ {\displaystyle g_1d_2(g_1-\frac{A^2-2B+2a}{A^2})=d_2\frac{2C}{A^2}, }\\ {\displaystyle g_1d_3=0.} \end{array}\right.$

(14)

Solving (14), we have

$\left\{\begin{array}{llll} {\displaystyle g_1=\frac{4(A^2-2B+2a)}{3A^2}, }\\ {\displaystyle 9CA^2+(A^2-2B+2a)^2=0, }\\ {\displaystyle d_2=0 }, \\ {\displaystyle d_3=0}. \end{array}\right.$

(15)

Then from (15), (13) and (10), one can obtain the following equation,

$P(v, w)=\frac{D}{A^2}v^4+\frac{(A^2-2B+2a)^2}{9A^4}v^2-\frac{2(A^2-2B+2a)}{3A^2}vw+w^2=0.$

Solving the above algebraic equation with respect to the variable $w$ , we have

$w=\frac{dv}{d\xi}=\pm\sqrt{\frac{-D}{A^2}}v^2+\frac{A^2-2B+2a}{3A^2}v.$

Integrating (16) once with respect to $\xi$ , then we have

$v(\xi ) = \frac{{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;{\xi _3}\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^2}}}\;\;\;\;\xi }}}}{{1 \mp \sqrt {\frac{{ - D}}{{{A^2}}}\;\;} {\xi _3}\;{e^{\frac{{{A^2} - 2B + 2a}}{{3{A^2}}}\;\;\;\xi }}}}\;\;\;\;,$

where $\xi_3$ is an integration constant.

The exact traveling wave solution to (1) can be got as follows:

$v(t,x) = \frac{{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\;{\xi _3}\;\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\;(\ln x - at)}}}}{{1 \mp \sqrt {\frac{{ - D}}{{{A^2}}}} \;\;{\xi _3}\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\;(\ln x - at)}}}}\;\;\;\;.$

Case 2: deg $g(v)=1$

Assume $g(v)=g_2v+d_4.$ From (12), we have

$\left\{\begin{array}{llll} {\displaystyle \alpha_0(v)=\frac{1}{4}(\frac{1}{2}g_2^2+\frac{4D}{A^2})v^4+\frac{1}{3}(\frac{3}{2}g_2d_4-\frac{5}{2}g_2\frac{A^2-2B+2a}{A^2})v^3+(d_4d_5-d_5\frac{A^2-2B+2a}{A^2})v }\\ {\displaystyle +\frac{1}{2}(g_2d_5+d_4^2-3d_4\frac{A^2-2B+2a}{A^2}+2\frac{(A^2-2B+2a)^2}{A^4}-\frac{4C}{A^2})v^2+d_6, }\\ {\displaystyle \alpha_1(v)=\frac{1}{2}g_2v^2+(d_4-2\frac{A^2-2B+2a}{A^2})v+d_5 }, \\ {\displaystyle \alpha_2(v)=1, } \end{array}\right.$

where $d_5, d_6$ are integration constants.

From the last equation of (12), we have

$\left\{\begin{array}{llllll} {\displaystyle -g_2\frac{D}{A^2}=\frac{1}{8}g_2^3+g_2\frac{D}{A^2}, }\\\\ {\displaystyle -\frac{2D}{A^2}(d_4-2\frac{A^2-2B+2a}{A^2})=\frac{5}{8}g_2^2d_4-\frac{5}{6}g_2^2\frac{A^2-2B+2a}{A^2}+d_4\frac{D}{A^2}, }\\\\ {\displaystyle g_2\frac{C}{A^2}-d_5\frac{2D}{A^2}=g_2d_4^2-\frac{7}{3}g_2d_4\frac{A^2-2B+2a}{A^2}+\frac{1}{2}g_2^2d_5+g_2\frac{(A^2-2B+2a)^2}{A^4}-g_2\frac{2C}{A^2}, }\\\\ {\displaystyle \frac{-4C(A^2-2B+2a)}{3A^4}=\frac{3}{2}g_2d_4d_5+\frac{1}{2}d_4^3-\frac{3(A^2-2B+2a)}{2A^2}d_4^2}\\\\ {\displaystyle +d_4(\frac{(A^2-2B+2a)^2}{A^4}-\frac{2C}{A^2})-g_2d_5\frac{A^2-2B+2a}{A^2}, }\\ \\{\displaystyle d_5\frac{2C}{A^2}=g_2d_6+d_4^2d_5-d_4d_5\frac{A^2-2B+2a}{A^2}, }\\\\ {\displaystyle d_4d_6=0.} \end{array}\right.$

(17)

From the last equation of (17), we need to discuss (17) in two cases $d_4=0$ or $d_4 \neq 0.$

Case Ⅰ: $d_4=0$

Solving (17), we have

$\left\{\begin{array}{lll} {\displaystyle g_2=\pm4\sqrt{\frac{-D}{A^2}}, }\\ {\displaystyle d_4=0, }\\ {\displaystyle d_5=\mp2\sqrt{\frac{-D}{A^2}}\frac{C}{D}, }\\ {\displaystyle d_6=-\frac{C^2}{A^2D}, } \end{array}\right.$

and $D < 0, A^2-2B+2a=0,$ such that,

$\left\{\begin{array}{lll} {\displaystyle \alpha_0(v)=\frac{-D}{A^2}v^4+\frac{2C}{A^2}v^2-\frac{C^2}{A^2D}, }\\\\ {\displaystyle \alpha_1(v)=\pm2\sqrt{\frac{-D}{A^2}}v^2\mp2\sqrt{\frac{-D}{A^2}}\frac{C}{D} }, \\\\ {\displaystyle \alpha_2(v)=1.} \end{array}\right.$

So,

$P(v, w)=\frac{-D}{A^2}v^4+\frac{2C}{A^2}v^2-\frac{C^2}{A^2D}+(\pm2\sqrt{\frac{-D}{A^2}}v^2\mp2\sqrt{\frac{-D}{A^2}}\frac{C}{D})w+w^2=0.$

Solving the above algebraic equation with respect to the variable $w$ , we have

$w=\frac{dv}{d\xi}=\mp\sqrt{\frac{-D}{A^2}}v^2\pm\sqrt{\frac{-D}{A^2}}\frac{C}{D}.$

Integrating (18) once with respect to $\xi$ , we can get the following results.

Ⅰ: $\frac{C}{D}>0.$

$v(\xi)=\frac{\sqrt{\frac{C}{D}}(1+\xi_4e^{\mp2\sqrt{\frac{-C}{A^2}}\xi})}{1-\xi_4e^{\mp2\sqrt{\frac{-C}{A^2}}\xi}},$

where $\xi_4$ is an integration constant.

The traveling wave solution to (1) can be got as follows:

$v(t, x)=\frac{\sqrt{\frac{C}{D}}(1+\xi_4e^{\mp2\sqrt{\frac{-C}{A^2}}(\ln x-at)})}{1-\xi_4e^{\mp2\sqrt{\frac{-C}{A^2}}(\ln x-at)}}.$

Ⅱ: $\frac{C}{D} < 0.$

$v(\xi)=\sqrt{\frac{C}{-D}}\tan[\mp\sqrt{\frac{C}{A^2}}\xi+\xi_5],$

where $\xi_5$ is an integration constant.

The traveling wave solution to (1) can be got as follows:

$v(t, x)=\sqrt{\frac{C}{-D}}\tan[\mp\sqrt{\frac{C}{A^2}}(\ln x-at)+\xi_5].$

Case Ⅱ: $d_4 \neq0$

By solving (17), we have

$\left\{\begin{array}{lll} {\displaystyle g_2=\pm4\sqrt{\frac{-D}{A^2}}, }\\\\ {\displaystyle d_4=\frac{4(A^2-2B+2a)}{3A^2} }, \\\\ {\displaystyle C=-\frac{(A^2-2B+2a)^2}{9A^2}, }\\ {\displaystyle d_5=0, }\\ {\displaystyle d_6=0.} \end{array}\right.$

(19)

Substituting (19) into (13), we have

$\left\{\begin{array}{lll} {\displaystyle \alpha_0(v)=-\frac{D}{A^2}v^4\mp\frac{2}{3}\sqrt{\frac{-D}{A^2}}\frac{A^2-2B+2a}{A^2}v^3+\frac{(A^2-2B+2a)^2}{9A^4}v^2, }\\\\ {\displaystyle \alpha_1(v)=\pm2\sqrt{\frac{-D}{A^2}}v^2-\frac{2(A^2-2B+2a)}{3A^2}v }, \\\\ {\displaystyle \alpha_2(v)=1.} \end{array}\right.$

So,

$P(v, w)=-\frac{D}{A^2}v^4\mp\frac{2}{3}\sqrt{\frac{-D}{A^2}}\frac{A^2-2B+2a}{A^2}v^3+\frac{(A^2-2B+2a)^2}{9A^4}v^2\\ \quad\quad+(\pm2\sqrt{\frac{-D}{A^2}}v^2-\frac{2(A^2-2B+2a)}{3A^2}v)w+w^2=0.$

Solving the above algebraic equation with respect to the variable $w$ , we have

$w=\frac{dv}{d\xi}=\mp\sqrt{\frac{-D}{A^2}}v^2+\frac{A^2-2B+2a}{3A^2}v.$

(20)

Integrating (20) once with respect to $\xi$ , then we have

$v(\xi ) = \frac{{\frac{{{A^2} - 2B + 2a}}{{3{A^2}}}\;\;\;\;{\xi _6}\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\xi }}}}{{1 \pm \sqrt {\frac{{ - D}}{{{A^{\;2}}}}} \;\;{\xi _6}\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\xi }}}}\;\;\;,$

where $\xi_6$ is an integration constant.

The traveling wave solution to (1) can be got as follows:

$v(t,x) = \frac{{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\;\;{\xi _6}\;{e^{\frac{{{A^2} - 2B + 2a}}{{3{A^2}}}\;\;\;\;(\ln x - at)}}}}{{1 \pm \sqrt {\frac{{ - D}}{{{A^{\;2}}}}\;\;} {\xi _6}\;{e^{\frac{{{A^{\;2}} - 2B + 2a}}{{3{A^{\;2}}}}\;\;\;\;(\ln x - at)}}}}\;\;\;\;.$

3. Traveling wave solutions of (1) by the $G'/G$ -expansion method

In this section, we get the traveling wave solutions to (1) using the $G'/G$ -expansion method. Introducing the solution $v(\xi)$ of equation (2) in a form of finite series:

$v(\xi)=\sum\limits_{l=0}^Na_l\bigg(\frac{G'(\xi)}{G(\xi)}\bigg)^l,$

(21)

where $a_l$ are real constants with $a_N\neq0$ and $N$ is a positive integer that needs to be determined. The function $G(\xi)$ is the solution of the auxiliary linear ODE

$G''(\xi)+\lambda G'(\xi)+\mu G(\xi)=0,$

(22)

where $\lambda$ and $\mu$ are real constants to be determined. By balancing $v''$ with $v^3$ in the equation (2), we have $N+2=3N$ , obviously, $N=1.$ Therefore, (21) is written as

$v(\xi)=a_0+a_1(\frac{G'(\xi)}{G(\xi)}).$

(23)

Correspondingly,

$v'(\xi)=-a_1\lambda\frac{G'}{G}-a_1(\frac{G'}{G})^2-a_1\mu,$

(24)

$v''(\xi)=(a_1\lambda^2+2a_1\mu)\frac{G'}{G}+3a_1\lambda(\frac{G'}{G})^2+2a_1(\frac{G'}{G})^3+a_\lambda\mu.$

(25)

Substituting (23), (24), (25) and $v^3$ into (2), we have

$[-\frac{1}{2}A^2(a_1\lambda^2+2a_1\mu)-(\frac{1}{2}A^2-B+a)a_1\lambda+Ca_1-3Da_0^2a_1]\frac{G'}{G}\\ \quad\quad-[\frac{3}{2}A^2a_1\lambda+(\frac{1}{2}A^2-B+a)a_1+3Da_0a_1^2](\frac{G'}{G})^2\\ \quad\quad-(A^2a_1+Da_1^3)(\frac{G'}{G})^3-[\frac{1}{2}A^2a_1\lambda\mu+(\frac{1}{2}A^2-B+a)a_1\mu-Ca_0+Da_0^3]=0.$

Setting all the coefficients of $(\frac{G'}{G})^i (i=0, 1, 2, 3)$ to be zeros, we yield a set of algebraic equations for $a_0,$ $a_1,$ $\lambda$ and $\mu$ as follows:

$\left\{\begin{array}{llll} {\displaystyle -\frac{1}{2}A^2(a_1\lambda^2+2a_1\mu)-(\frac{1}{2}A^2-B+a)a_1\lambda+Ca_1-3Da_0^2a_1=0, }\\\\ {\displaystyle \frac{3}{2}A^2a_1\lambda+(\frac{1}{2}A^2-B+a)a_1+3Da_0a_1^2=0 }, \\\\ {\displaystyle A^2a_1+Da_1^3=0, }\\\\ {\displaystyle \frac{1}{2}A^2a_1\lambda\mu+(\frac{1}{2}A^2-B+a)a_1\mu-Ca_0+Da_0^3.} \end{array}\right.$

Solving the above algebraic equations, we have

$\left\{\begin{array}{llll} {\displaystyle a_0=\pm(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C}), }\\\\ {\displaystyle a_1=\pm\sqrt{\frac{A^2}{-D}} }, \\\\ {\displaystyle \lambda=-(\frac{16(\frac{1}{2}A^2-B+a)^3}{27A^4C}+\frac{A^2-2B+2a}{A^2}), }\\\\ {\displaystyle \mu=\frac{(A^2-2B+2a)^6}{729A^8C^2}+\frac{(A^2-2B+2a)^4}{27A^6C}+\frac{5(A^2-2B+2a)^2}{12A^4}+\frac{C}{A^2} .} \end{array}\right.$

(26)

From equations (23) and (26), the solution of (2) can be written as the following equation,

$V(\xi)=\mp(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C})\pm\sqrt{\frac{A^2}{-D}}(\frac{G'}{G}).$

(27)

By solving the second order linear ODE (22), then (27) gives three types traveling wave solutions.

Case 1: $\Delta=\lambda^2-4\mu=-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}>0,$ we obtain

$\frac{G'}{G}=-\frac{\lambda}{2}+\frac{\sqrt{\lambda^2-4u}}{2}\frac{C_1e^{\frac{\sqrt{\lambda^2-4u}}{2}\xi}-C_2e^{-\frac{\sqrt{\lambda^2-4u}}{2}\xi}}{C_1e^{\frac{\sqrt{\lambda^2-4u}}{2}\xi}+C_2e^{-\frac{\sqrt{\lambda^2-4u}}{2}\xi}},$

where $C_1, C_2$ are integration constants. The traveling wave solution to (1) can be got as follows:

$v(t, x)=\mp(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C})\pm\sqrt{\frac{A^2}{-D}} \{\frac{(A^2-2B+2a)^3}{27A^4C}+\frac{A^2-2B+2a}{2A^2}+\\\\ \quad\quad\frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}} [\frac{C_1e^{H_1(\ln x-at)}-C_2e^{-H_1(\ln x-at)}}{C_1e^{H_1(\ln x-at)}+C_2e^{-H_1(\ln x-at)}}]\}.$

where $H_1=\frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}}.$ Specially, when $C_1=C_2,$ we obtain traveling wave solution of (1) as follows:

$v(t, x)=\mp(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C})\pm\sqrt{\frac{A^2}{-D}} \{\frac{(A^2-2B+2a)^3}{27A^4C}+\frac{A^2-2B+2a}{2A^2}+\\\\ \quad\quad\frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}}\tanh \{\frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}}(\ln x-at)\}.$

Case 2: $\Delta=\lambda^2-4\mu=-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2} < 0,$ we obtain

$\frac{G'}{G}=-\frac{\lambda}{2}+\frac{\sqrt{\lambda^2-4u}}{2}\frac{C_2\cos \frac{\sqrt{\lambda^2-4u}}{2}\xi-C_1\sin \frac{\sqrt{\lambda^2-4u}}{2}\xi }{C_1\cos \frac{\sqrt{\lambda^2-4u}}{2}\xi+C_2\sin \frac{\sqrt{\lambda^2-4u}}{2}\xi},$

where $C_1, C_2$ are integration constants. The traveling wave solution to (1) can be got as follows:

$v(t, x)=\mp(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C})\pm\sqrt{\frac{A^2}{-D}} \{\frac{(A^2-2B+2a)^3}{27A^4C}\\\\ \quad+\frac{A^2-2B+2a}{2A^2}+ \frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}} [\frac{C_2\cos{H_2(\ln x-at)}-C_1\sin{H_2(\ln x-at)}}{C_1\cos{H_2(\ln x-at)}+C_2\sin{H_2(\ln x-at)}}]\}.$

where $H_2=\frac{1}{2}\sqrt{-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}}.$

Case 3: $\Delta=\lambda^2-4\mu=-\frac{2(A^2-2B+2a)^2}{3A^4}-\frac{4C}{A^2}=0,$ we obtain

$\frac{G'}{G}=-\frac{\lambda}{2}+\frac{C_2}{C_1+C_2\xi},$

where $C_1, C_2$ are integration constants. The traveling wave solution to (1) can be got as follows:

$v(t, x)=\mp(\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)}{3A^2}+\frac{\sqrt{\frac{A^2}{-D}}(A^2-2B+2a)^3}{27A^4C})\\ \quad\quad \pm\sqrt{\frac{A^2}{-D}}\{\frac{(A^2-2B+2a)^3}{27A^4C}+\frac{A^2-2B+2a}{2A^2}+\frac{C_2}{C_1+C_2(\ln x-at)}\}.$

4. Conclusion

We have obtained several traveling wave solutions of a class of generalized Black-Scholes equation under certain parametric conditions by using the first integral method and the $G'/G$ -expansion method. From the above investigation, we see that the two methods are effective for obtaining the exact traveling wave solutions of the class of generalized Black-Scholes equation. The first integral method is based on the ring theory of commutative algebra, and supposes that (2) has rational first integrals of the variables $v$ and $v'.$ The $G'/G$ -expansion method is to obtain a special class of solutions of (2) based on the balance principle. When $D=0,$ the class of generalized Black-Scholes equation is changed to the classical Black-Scholes equation, that is, the classical Black-Scholes equation is a special case of (2). The results obtained in this paper can provide some useful reference and help for the relevant financial research.

Conflict of Interest

All authors declare no conflicts of interest in this paper.

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AIMS Mathematics

The exact traveling wave solutions of a class of generalized Black-Scholes equation

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Abstract

1. Introduction

2. Traveling wave solutions of (1) by using the first integral method

2.1. $N=1$

2.2. $N=2$

3. Traveling wave solutions of (1) by the $G'/G$ -expansion method

4. Conclusion

Conflict of Interest

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AIMS Mathematics

The exact traveling wave solutions of a class of generalized Black-Scholes equation

Related Papers:

Abstract

1. Introduction

2. Traveling wave solutions of (1) by using the first integral method

2.1. N=1N=1

2.2. N=2N=2

3. Traveling wave solutions of (1) by the G′/GG'/G-expansion method

4. Conclusion

Conflict of Interest

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2.1. $N=1$

2.2. $N=2$

3. Traveling wave solutions of (1) by the $G'/G$ -expansion method