Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.2. pp. 85-91, 2010 Applied Mathematics
Lie Symmetries of the Black-Scholes Equation Refet Polat
Yasar University Department of Mathematics, Faculty of Science and Letters Bornova, ˙Izmir, Türkiye
e-mail: refet.p olat@ yasar.edu.tr
Received Date: January 15, 2010 Accepted Date: February 3, 2010
Abstract. In this paper symmetry expansions for Black-Scholes equation are studied. Differently then the other studies in the literature for Black-Scholes equation, we expanded the equation into a parametric form by adding a coeffi-cient a. By using this expansion the dimension of the solution spaces is increased by one and symmetry reduction will be done with the deterministic equations in the new increased solution spaces.
Key words: Black-Scholes, Lie Symmetries, Nonlinear. 2000 Mathematics Subject Classification. 35Q91. 1. Introduction
In 1893 Sophus Lie developed the theory of Lie Symmetry groups of differential equations. Symmetry analysis is a unique theory for solving differential equa-tions in an algorithmic approach. Since lie theory can be applied to all types of differential equations it can be used in many research areas. In 1998 Ibragimov and Gazizov computed the Lie (point) symmetries of Black-Sholes equation [3], later, Silberberg changed the generators to simplify the computations and gave the expanded commutator table in 2005 [5]. Also some other researchers devel-oped distinct commutator tables for Black- Scholes equation [6,7]. In this work Mathlie package programme which has been developed for Mathematica is used for new computations [1].
2. The Symmetry Expansions of Black-Scholes Equations The Black-Scholes equation in general form can be given as
(1) +
1 2
22
Symmetry expansions of Boussineq equations applied by Burde [2] and Kiraz [4]. In this work the application of symmetry expansions will be applied to (1). Add a coefficient a into the equation (1) , hence
(2) +
1 2
22
+ − = 0
In the (x,t,u,a) expanded space
(3) = 1( ) + 2( ) + 3() + ( ) Write the Lie group with one parameter defined by the generator (3) and the generator must be in the following prolongation form (4)
(4) (2) Pr = 1 + 2 + 1 + 3 + 10 + 01 + 20
The prolongation applied to equation (2) to get the symmetry condition then we get (5) 1£2+ ¤ +2[0] +1[−] +3[] +10[] +01[] +20 £ 22¤= 0 Then we find (6) h −1− (1)− (1)+ (1)−2122(1)+ 22(1) i + £ −22( 2) ¤ + h −2− (2)− (2)+ (1)−1222(2)+ 3 i +£−22(2)¤+ h −(1)− (2)− 22(2) i + £ 2( 1) + 22(1)+1222(1) ¤ + 2 £ −1 2 22( 2) ¤ + 2 £ −(1)−1222(1)− 22(1)+1222(1) ¤ + 2 h −(2) i + (1)+ (1)+1222(1)+ £ −3 2 22( 1) ¤ +£−1222(2)¤= 0
The terms underlined by shows the different terms from prolongation in which obtained by using Mathlie Package program [1]. For getting deterministic equations. Let equal the coefficients of all the derivations in (6) to zero. Since
(1)= 0 (2)= 0 (2)= 0 (2)= 0
2= 2() with respect to t. From −(1)− 1 2 22( 1)− 22(1)+ 1 2 22( 1)= 0
We get (1)= 0 and we get 1function as
1= ( ) + ( ) And then we get
−1− (1)− (1)+ (1)−1222(1)+ 22(1)= 0 −2− (2)− (2)+ (1)−1222(2)+ 3= 0 (1)+ (1)+2122(1)= 0 2 1+ 22(1)+1222(1)= 0 −2− (2)+ (1)+ 3= 0
Take (1)= 0 then we have
(7) 1= [1()] + 1() 1= [1() + 1()] and (8) (1)+ (1)+ 1 2 22( 1)= 0
The equation (8) becomes
(9) (1)+ (1)= 0
Substitute (7) in (9)
[1() + 1()] + [1() + 1()] = 0
1+ 1+ 1+ 1= 0
(11) 1= −1
We obtain two ordinary differential equations (10) and (11). After solving (10) and (11) we find 1= 1 − and 1= 1 − Substitute (7) , we have (12) 1= h 1 − + 2− i 1( ) = − (1 + 2)
Hence we have equations (13).
(13) −1−(1)−(1)+(1)− 1 2 22( 1)+ 22( 1)= 0 −2− (2)− (2)+ (1)+ 3= 0 2 1+ 22(1)+1222(1)= 0
Substitute (12) in equations (13) and we get
−1− (1)− (1)+ 12 − −1 2 22( 1)+ 122 − = 0 −2− (2)+ 1 − + 3= 0 2 1+ 22(1)+12213 − = 0
From the equation (14)
(14) 21+ 22(1)+ 1 2 2 13 − = 0 we get (15) 1+ (1)+ 1 21 2− = 0
Arranging the equation (15) we get
(16) (1)+ 1 1= − 1 21 −
Linear ordinary differential equation
(17) 1( ) = −1
61
− 2+1
Hence we have equations (18) (18) −1−(1)−(1)+1 2− −1 222(1)+122 − = 0 −2− (2)+ 1 − + 3= 0
Substitute (17) in (18) then we have
(19) −h−161− 2+1 3 i −h161 − 2 i +h131− + 1 23 i +12 − −1 2 22h −131− + 2 33 i + 122 − = 0 12 − ∙ 1 6 + 7 6 + 13 6 2 ¸ +1 3 £ − + − 2¤= 0 Under the condition = 3
4 equation (19) will be obtained. For the rest
equation (20) (20) −2− (2)+ 1 − + 3= 0 when (2)= 0 ( w.r.t t ), taking (21) 3= −1 − Then (22) −2− (2)= 0
From the ordinary differential equation (22)
(23) 2() = 4−
can be found. As a result, infinitesimal generators of the Black-Scholes equations will be found as follows
1( ) = −1612 − +1 3 2() = 4 − 3( ) = −1 − 1( ) = − (1 + 2)
In Table I the infinitesimal generators found in this study compared with the infinitesimal generators gained from MathLie package program.
Table 1. Infinitesimal Generators Hence symmetry generator of the equations is
=³−1612 − +1 3 ´ + ³ 4 − ´ + ³ −1 − ´ +³− (1 + 2) ´
For this generator 1= 1; 1= µ −162− ¶ + ³ −− ´ + − 2= 1; 2= − 3= 1; 3= 1 4= 1; 4= ³ − ´
Table 2. Commutator Table References
1. Baumann, G., MathLie a program of doing Symmetry Analysis, Math. Comput. Simulation 48. no.2,205-223(2002).
2. Burde, G.I., Expand Lie Group Transformations and Similarity Reductions of Differential Equations, Proc.Ins.Math.NAS of Ukraine,Vol.43,PartI,93-101(2002). 3. Ibragimov, StateN.H., GazizovR.K. Lie Symmetry Analysis of Differential Equa-tions in Finance., Nonlinear Dynamics 17:387-407,(1998).
4. Kiraz, F.A., Kısmi Türevli Diferansiyel Denklemlerin Lie Simetrileri Üzerine, Dok-tora Tezi, Ege Üniversitesi, 108 p, (2007) (unpublished).
5. Silberberg G., Discrete Symmetries of the Black Scholes Proceedings of 10th Inter-national Conference in Modern Analysis
6. Singh J.P., Parabakaran S., Group Properties of the Black-Scholes Equation & its Solution Electronic Journal of Theoritical Physics EJTP 5 No.18 (2008) 51-60. 7. Yan Xuan Liu, Zhang Shun-Li, QU Chang Zheng. Symmetry Breaking for Black-Scholes Equations. Commun.Theor.Phys. 47 (2007) pp. 995-1000
