Symmetry Analysis and Exact Solutions to the Space-Dependent Coefficient Pdes in Finance

Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 156965, 10 pages http://dx.doi.org/10.1155/2013/156965

Research Article Symmetry Analysis and Exact Solutions to the Space-Dependent Coefficient PDEs in Finance

Hanze Liu1,2 1 School of Mathematical Sciences, Liaocheng University, Liaocheng, Shandong 252059, China 2 Department of Mathematics, Binzhou University, Binzhou, Shandong 256603, China

Correspondence should be addressed to Hanze Liu; [email protected]

Received 20 June 2013; Accepted 25 August 2013

Academic Editor: Mariano Torrisi

Copyright © 2013 Hanze Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The variable-coefficients partial differential equations (vc-PDEs) in finance are investigated by Lie symmetry analysis andthe generalized power series method. All of the geometric vector fields of the equations are obtained; the symmetry reductions and exact solutions to the equations are presented, including the exponentiated solutions and the similarity solutions. Furthermore, the exact analytic solutions are provided by the transformation technique and generalized power series method, which has shown that the combination of Lie symmetry analysis and the generalized power series method is a feasible approach to dealing with exact solutions to the variable-coefficients PDEs.

1. Introduction the generalized power series method for dealing with exact solutions to some nonlinear PDEs based on the symmetry Gazizov and Ibragimov [1] studied the Black-Scholes equa- analysis method. tion of option pricing by Lie equivalence transformations. It is known that the Lie symmetry analysis is a systematic By the optimal system method, some invariant solutions to and powerful method for dealing with symmetries and exact heat and Black-Scholes equations are obtained [2]. In [3– solutions to partial differential equations (see, e.g.,1 [ –6, 8– 5], the fundamental solutions to the bond pricing equations 18] and the references therein). Furthermore, we find that areconsideredbyLiesymmetryanalysisandtheintegral the combination of Lie symmetry analysis and the power transform method. In [6], the invariance properties of the series method is a feasible approach to investigating exact bond pricing equation are studied by the group classification solutions to nonlinear PDEs [8–13]. On the other hand, under method. In [7], the finite element method was adopted the perspective of mathematical physics and Lie symmetry to solve the bond pricing type of PDE system, and the analysis, the space-time dependent coefficients system differs numerical implementation was provided, such as system greatly from its time-dependent counterpart, and it is more that models the TF convertible bonds with credit risk in complicatedthanthelatter.However,mostofthestudies bond pricing theory. However, the similarity reductions and are related to the time-dependent coefficient systems. More- exact solutions to such variable-coefficient equations are not over, the determination of exact solutions to the variable- considered generally in the aforementioned papers. Recently, coefficients PDEs is a complicated problem that challenges we studied some nonlinear PDEs by Lie symmetry analysis researchers greatly. In the present paper, we consider the and the dynamical system method [8–13]; for example, in [8], symmetry reductions and exact solutions to the general we considered Lie group classifications and exact solutions to space-dependent coefficients PDEs in finance as follows: the space-dependent coefficients hanging chain equation and 𝑢 +𝛼𝑥2𝑢 +𝛽𝑥𝑢 +𝛾𝑥]𝑢=0, 𝑥>0, the simplified bond pricing equation. In9 [ ], we investigated 𝑡 𝑥𝑥 𝑥 (1) the integrable condition and exact solutions to the time- where 𝑢=𝑢(𝑥,𝑡)denotes the unknown function of the space dependent coefficient Gardner equations by the Painlevetest´ variable 𝑥 and time 𝑡 and the parameters 𝛼, 𝛽, 𝛾, ] ∈ R are and Lie group analysis method. In [10–13], we developed arbitrary constants, ] ≥0and 𝛼 =0̸ . 2 Abstract and Applied Analysis

We first note that (1) is the general form of the bond where 𝜉(𝑥, 𝑡,, 𝑢) 𝜏(𝑥, 𝑡,,and 𝑢) 𝜙(𝑥, 𝑡, 𝑢) are coefficient func- pricing types of equations [1–7]. In particular, if ] =0,then tions of the vector field to be determined. The symmetry this equation becomes the following Black-Scholes equation groups of (2)and(3) will be generated by the vector field of of option pricing: the form (5), respectively. Applying the second prolongation (2)𝑉 𝑉 𝑢 +𝛼𝑥2𝑢 +𝛽𝑥𝑢 + 𝛾𝑢 = 0, 𝑥 >0. pr of to (2)and(3), we find that the coefficient functions 𝑡 𝑥𝑥 𝑥 (2) 𝜉, 𝜏,and𝜙 must satisfy the following Lie symmetry condition: If ] =1,then(1) is the general bond pricing equation 󵄨 (2)𝑉 (Δ)󵄨 =0, given by pr 󵄨Δ=0 (6) 2 𝑢𝑡 +𝛼𝑥 𝑢𝑥𝑥 +𝛽𝑥𝑢𝑥 +𝛾𝑥𝑢=0, 𝑥>0. (3) 2 where Δ=𝑢𝑡 +𝛼𝑥𝑢𝑥𝑥 +𝛽𝑥𝑢𝑥 +𝛾𝑢for (2)andΔ= 2 Such equations are called bond pricing types of equations, 𝑢𝑡 +𝛼𝑥 𝑢𝑥𝑥 +𝛽𝑥𝑢𝑥 +𝛾𝑥𝑢for (3), respectively. Then, the Lie which are of great importance in financial mathematics and symmetry group calculation method leads to the following bond pricing theory [3–7]. For dealing with exact solutions conditions on the coefficient functions 𝜉, 𝜏,and𝜙: to the variable-coefficients PDEs, we will introduce the 1 generalized power series method [10–13] in the present paper. 𝜉= 𝑥𝜏𝑡 log 𝑥+𝑥𝜌,( 𝜙=𝑟 𝑥,) 𝑡 𝑢+𝑠(𝑥,) 𝑡 , By a generalized power series solution, we mean a generalized 2 (7) powerseriesisoftheform 1 𝛼−𝛽 1 𝑟= 𝜏 2𝑥+ 𝜏 𝑥+ 𝜌 𝑥+𝜎, ∞ 𝑡𝑡log 𝑡 log 𝑡 log 𝑛 8𝛼 4𝛼 2𝛼 𝑓 (𝜉) =𝐴(𝜉) + ∑𝑐𝑛𝜉 , (4) 𝑛=0 for some functions 𝜏, 𝜌,and𝜎.Nowthefunctions𝜏, 𝜌,and𝜎 𝑡 which is a solution to a system with respect to the variable depend only on .Moreover,for(2), we have 𝜉,where𝑐𝑛 (𝑛 = 0, 1, 2, .) . are constant coefficients to 1 2 1 1 be determined and 𝐴(𝜉) is the undetermined function with 𝜏 𝑥+ 𝜌 𝑥+ 𝜏 8𝛼 𝑡𝑡𝑡log 2𝛼 𝑡𝑡 log 4 𝑡𝑡 respect to the variable 𝜉.Inparticular,if𝐴(𝜉),then( ≡0 4)is (8) the regular power series solution. So, the generalized power (𝛼 − 𝛽)2 −4𝛼𝛾 𝛼−𝛽 − 𝜏 − 𝜌 +𝜎 =0; series solution is the generalization of the regular power series 4𝛼 𝑡 2𝛼 𝑡 𝑡 solutionanditnaturallyincludesthelatterasitsspecialcase. If we obtained a generalized power series solution (4)toa for (3), we have system and the convergence of this power series is shown, 1 1 then the exact generalized power series solution is obtained. 𝜏 2𝑥+ 𝛾𝜏 𝑥 𝑥 This solution sometimes is called the exact analytic solution 8𝛼 𝑡𝑡𝑡log 2 𝑡 log [10–13, 19]. 1 1 The main purpose of this paper is to develop the + 𝜌 𝑥+𝛾(𝜌+𝜏)𝑥+ 𝜏 2𝛼 𝑡𝑡 log 𝑡 4 𝑡𝑡 (9) combination of Lie symmetry analysis and the generalized power series method for dealing with symmetries and exact (𝛼 − 𝛽)2 𝛼−𝛽 solutions to the variable-coefficients PDEs in finance. The − 𝜏 − 𝜌 +𝜎 =0. 4𝛼 𝑡 2𝛼 𝑡 𝑡 remainder of this paper is organized as follows. In Section 2, we perform Lie symmetry analysis on the bond pricing types These equations fix the functions 𝜉, 𝜏, 𝜌, 𝜎,and𝜙.Solvingthe of (2)and(3) and give all of the geometric vector fields of the equations, we obtain the vector field of2 ( ) as follows: equations in terms of the arbitrary parameters. In Section 3, we consider the symmetry reductions of the equations and 𝑉1 =𝜕𝑡,𝑉2 =𝑥𝜕𝑥,𝑉3 =𝑢𝜕𝑢, provide the exponentiated solutions and the similarity solutions to the equations. In Section 4, we investigate the exact 𝑉4 =2𝛼𝑥𝑡𝜕𝑥 +[log 𝑥+(𝛼−𝛽)𝑡]𝑢𝜕𝑢, analytic solutions to the variable-coefficient equations by the generalized power series method. In Section 5,wedealwith 𝑉5 =2𝛼𝑥𝑡(log 𝑥)𝑥 𝜕 +4𝛼𝑡𝜕𝑡 the vector fields and exact solutions to the bond pricing type 2 of (1) for the general case ] =0,1̸ . Finally, the conclusions +[(𝛼−𝛽)log 𝑥+((𝛼−𝛽) −4𝛼𝛾)𝑡]𝑢𝜕𝑢, and some remarks are given in Section 6. 2 (10) 𝑉6 =4𝛼𝑥𝑡(log 𝑥)𝑥 𝜕 +4𝛼𝑡 𝜕𝑡 2. Lie Symmetry Analysis for and 2 (2) (3) +[2(𝛼−𝛽)𝑡log 𝑥+log 𝑥−2𝛼𝑡 In this section, we will present a complete list of all possible +((𝛼−𝛽)2 −4𝛼𝛾)𝑡2]𝑢𝜕 , Liesymmetryalgebrasforthebondpricingtypesofequations 𝑢 of the forms (2)and(3). 𝑉 =𝑠𝜕, Recall that the geometric vector fields of such equations 𝑠 𝑢 are as follows: where the parameters 𝛼 =0̸ , 𝛽, 𝛾 are arbitrary constants and 𝑉=𝜉(𝑥, 𝑡, 𝑢) 𝜕𝑥 +𝜏(𝑥, 𝑡, 𝑢) 𝜕𝑡 +𝜙(𝑥, 𝑡, 𝑢) 𝜕𝑢, (5) the function 𝑠=𝑠(𝑥,𝑡)satisfies (2). Abstract and Applied Analysis 3

For (3), we have the vector field as follows: principles, but they are important for us to investigate the exact solutions to PDEs (see, e.g., [3–5, 10]). 𝑉1 =𝜕𝑡,𝑉2 =𝑢𝜕𝑢,𝑉𝑠 =𝑠𝜕𝑢, (11) where the function 𝑠=𝑠(𝑥,𝑡)satisfies (3). 3. Symmetry Reductions and Exact Solutions {𝑉 ,...,𝑉,𝑉} Clearly, for (2), a basis of the Lie algebra is 1 6 𝑠 . to the Bond Pricing Types of Equations For (3), a basis for the Lie algebra is {𝑉1,𝑉2,𝑉𝑠}.Thus,thenew symmetries cannot be derived from the Lie brackets for the In the preceding section, we obtained the symmetries and two equations. symmetry groups of (2)and(3). Now, we deal with the Moreover, we can obtain the one-parameter groups gen- symmetry reductions and exact solutions to the equations. erated by 𝑉𝑖, respectively. In fact, for (2), the one-parameter groups 𝐺𝑖 generated by 𝑉𝑖 (𝑖 = 1,...,6,𝑠)aregiveninthe 3.1. The Exponentiated Solutions. Since each 𝐺𝑖 (𝑖= following: 1,...,6,𝑠) is a symmetry group, it implies that if 𝑢=𝑓(𝑥,𝑡) 𝑢(𝑖) (𝑖= 1,...,6,𝑠) 𝐺1 : (𝑥, 𝑡, 𝑢) 󳨀→ (𝑥, 𝑡 + 𝜖,𝑢) , is a solution to (2), then are all solutions 𝜖 to the following equation as well: 𝐺2 : (𝑥, 𝑡, 𝑢) 󳨀→ ( 𝑒 𝑥, 𝑡, 𝑢), 𝜖 𝐺3 : (𝑥, 𝑡, 𝑢) 󳨀→(𝑥,𝑡,𝑒 𝑢) , (1) 𝑢 =𝑓(𝑥, 𝑡) −𝜖 , (13a) 𝐺 : (𝑥, 𝑡, 𝑢) 󳨀→ ( 𝑥 𝑒 2𝛼𝜖𝑡,𝑡,𝑥𝜖𝑢 [(𝛼 − 𝛽) 𝜖𝑡2 +𝛼𝜖 𝑡]) , 4 exp (2) −𝜖 𝑢 =𝑓(𝑒 𝑥, 𝑡) , (13b)

𝛿 2 𝛼−𝛽 (3) 𝜖 𝐺 : (𝑥, 𝑡, 𝑢) 󳨀→ ( 𝑥 ,𝛿 𝑡, 𝑢 [ (𝛿−1) 𝑥 𝑢 =𝑒𝑓 (𝑥,) 𝑡 , (13c) 5 exp 2𝛼 log (4) 𝜖 2 −2𝛼𝜖𝑡 𝑢 =𝑥 exp [(𝛼−𝛽) 𝜖𝑡 − 𝛼𝜖 𝑡] 𝑓 (𝑒 𝑥,) 𝑡 , (13d) (𝛼 − 𝛽)2 −4𝛼𝛾 + 4𝛼 𝛼−𝛽 𝑢(5) = [ (1 − 𝛿−1) 𝑥 exp 2𝛼 log ×(𝛿2 −1)𝑡]), 2 (𝛼 − 𝛽) −4𝛼𝛾 (13e) + (1 − 𝛿−2)𝑡] 4𝛼 𝐺6 : (𝑥, 𝑡, 𝑢) 1/𝛿 −2 𝑡 ×𝑓(𝑥 ,𝛿 𝑡) , 󳨀→ ( 𝑥 1/(1−4𝛼𝜖𝑡), , 1 − 4𝛼𝜖𝑡 (6) 1 𝛼−𝛽 1 2 𝑢 = exp {[ log 𝑥+ log 𝑥 𝛼−𝛽 √1 + 4𝛼𝜖𝑡 2𝛼 4𝛼𝑡 𝑢√1 − 4𝛼𝜖𝑡 {[ 𝑥 exp 2𝛼 log (𝛼 − 𝛽)2 −4𝛼𝛾 4𝛼𝜖𝑡 1 + 𝑡] } + 2𝑥 4𝛼 1 + 4𝛼𝜖𝑡 4𝛼𝑡log 1/(1+4𝛼𝜖𝑡) 𝑡 (𝛼 − 𝛽)2 −4𝛼𝛾 ×𝑓(𝑥 , ), + 𝑡] 1 + 4𝛼𝜖𝑡 4𝛼 (13f) (𝑠) (13 ) 4𝛼𝜖𝑡 𝑢 =𝑓(𝑥,) 𝑡 +𝜖𝑠, g × }) , 1 − 4𝛼𝜖𝑡 2𝛼𝜖 where 𝛿=𝑒 , 𝜖 is an arbitrary real number, and the function 𝐺𝑠 : (𝑥, 𝑡, 𝑢) 󳨀→ (𝑥, 𝑡, 𝑢 +𝜖𝑠) , 𝑠=𝑠(𝑥,𝑡)satisfies (2). (𝑖) (12) For (3), the exponentiated solutions are 𝑢 (𝑖 = 1, 3, 𝑠)as above while 𝑠=𝑠(𝑥,𝑡)satisfies (3). 2𝛼𝜖 where 𝛿=𝑒 , 𝜖≪1,andthefunction𝑠=𝑠(𝑥,𝑡)is an Such exponentiated solutions are one of group-invariant arbitrary solution to (2). For (3), the one-parameter groups types of solutions to the PDEs, which are generated from the are 𝐺𝑖 (𝑖 = 1, 3, 𝑠) as above, while 𝑠=𝑠(𝑥,𝑡)is an arbitrary one-parameter groups and are of importance for studying the solution to (3). exact solutions and investigating the properties of solutions Fromtheabove,weobservethat𝐺1 is a time translation (see Remark 2). and 𝐺2 and 𝐺3 are trivial scaling transformations, while 𝐺𝑖 Next, we investigate the symmetry reductions and exact (𝑖 = 4, 5,)arenontriviallocalgroupsoftransformations. 6 explicit solutions to the two bond pricing equations. Firstly, Their appearances are far from obvious from basic physical we consider (2). 4 Abstract and Applied Analysis

󸀠 3.2. Similarity Solution for 𝑉1. For the generator 𝑉1,wehave where 𝑓 =𝑑𝑓/𝑑𝜉.Itimpliesthatif𝜔=𝑓(𝜉)is a solution to thefollowingreducedordinarydifferentialequation(ODE): (22), then (21)isasolutionto(2). Solving (22), we get 𝑓(𝜉) = −(1/2) 𝜉−𝛾𝜉+𝑐 2 󸀠󸀠 󸀠 log 1.Thus,weobtainthesolutionto(2)as 𝛼𝜉 𝑓 +𝛽𝜉𝑓 +𝛾𝑓=0, (14) follows: 󸀠 1 2 1 where 𝑓 =𝑑𝑓/𝑑𝜉. This is an Euler equation; the corre- 𝑢 (𝑥,) 𝑡 =𝑐exp [ (log 𝑥+𝑎𝑡) − log 𝑡−𝛾𝑡], (23) 2 4𝛼𝑡 2 sponding characteristic equation is 𝛼𝐾 −(𝛼−𝛽)𝐾+𝛾=0. Solving this equation, we have 𝐾 = ((𝛼 − 𝛽)± √Δ)/2𝛼,where where 𝑐 is an arbitrary constant. 2 Δ = (𝛼 − 𝛽) −4𝛼𝛾. 𝐾1 𝑉 𝑉 When Δ>0,(14) has the general solution 𝑓=𝑐1𝜉 + 3.5. Similarity Solution for 5. For the generator 5,wehave 𝐾2 𝑐2𝜉 .Thus,weobtaintheexactsolutionto(2) as follows: the following similarity transformation: −1/2 𝐾1 𝐾2 𝜉=𝑡 𝑥, 𝑢 (𝑥,) 𝑡 =𝑐1𝑥 +𝑐2𝑥 , (15) log (𝛼 − 𝛽)2 −4𝛼𝛾 (24) where 𝑐1 and 𝑐2 are arbitrary constants and 𝐾1,2 = ((𝛼 − 𝛼−𝛽 𝜔=log 𝑢− log 𝑥− 𝑡, 𝛽) ± √Δ)/2𝛼 are two real roots to the characteristic equation, 2𝛼 4𝛼 respectively. 𝜔=𝑓(𝜉) 𝐾 and the similarity solution is ;thatis, When Δ=0,(14) has the general solution 𝑓=𝜉 (𝑐1 + 2 𝑐2 log 𝜉). Thus, we obtain the exact solution to (2) as follows: −1/2 𝛼−𝛽 (𝛼 − 𝛽) −4𝛼𝛾 𝑢=exp [𝑓 (𝑡 log 𝑥) + log 𝑥+ 𝑡] . 𝐾 2𝛼 4𝛼 𝑢 (𝑥,) 𝑡 =𝑥 (𝑐1 +𝑐2 log 𝑥) , (16) (25) where 𝑐1 and 𝑐2 are arbitrary constants, 𝐾 = (𝛼 − 𝛽)/2𝛼 are Substituting (25)into(2), we reduce the bond pricing equa- the real root to the characteristic equation. tion to the following ODE: When Δ<0,(14) has the general solution 𝑓= 𝐾 √ √ 𝜉 (𝑐1 cos( −Δ/2𝛼) log 𝜉+𝑐2 sin( −Δ/2𝛼) log 𝜉).Thus,we 󸀠󸀠 󸀠2 1 󸀠 𝛼𝑓 +𝛼𝑓 − 𝜉𝑓 =0, (26) obtain the exact solution to (2) as follows: 2 󸀠 √−Δ √−Δ where 𝑓 =𝑑𝑓/𝑑𝜉. 𝑢 (𝑥,) 𝑡 =𝑥𝐾 (𝑐 𝑥+𝑐 𝑥) , 󸀠 1 cos 2𝛼 log 2 sin 2𝛼 log Letting 𝑓 =𝑦, we get the Bernoulli equation (17) 𝑑𝑦 1 = 𝜉𝑦 − 𝛼𝑦2. 𝑑𝜉 2𝛼 (27) where 𝑐1 and 𝑐2 are arbitrary constants, 𝐾 = (𝛼 − 𝛽)/2𝛼. Clearly, 𝑦=0;thatis,𝑓=𝑐is a solution to (26). Thus, we get 3.3. Similarity Solution for 𝑉2. For the generator 𝑉2,wehave asolutionto(2) as follows: thefollowingreducedODE: 𝛼−𝛽 (𝛼 − 𝛽)2 −4𝛼𝛾 󸀠 𝑢 (𝑥,) 𝑡 = exp [ log 𝑥+ 𝑡+𝑐], (28) 𝑓 +𝛾𝑓=0, (18) 2𝛼 4𝛼

󸀠 −𝛾𝜉 𝑐 where 𝑓 =𝑑𝑓/𝑑𝜉.Solvingthisequation,wehave𝑓=𝑐𝑒 . for an arbitrary constant number . 𝑦 =0̸ 𝑦= Thus, we obtain the exact solution to2 ( ) as follows: When , solving the Bernoulli equation, we get (1/4𝛼)𝜉2 (1/4𝛼)𝜉2 𝑒 /(∫ 𝑒 𝑑𝜉1 +𝑐 ).Thus,weobtainthesolutionto 𝑢 𝑥, 𝑡 =𝑐𝑒−𝛾𝑡, ( ) (19) (26) as follows:

2 where 𝑐 is an arbitrary constant. 𝑒(1/4𝛼)𝜉 𝑓 (𝜉) = ∫ 𝑑𝜉 +𝑐 , (1/4𝛼)𝜉2 2 (29) ∫𝑒 𝑑𝜉1 +𝑐 3.4. Similarity Solution for 𝑉4. For the generator 𝑉4,wehave the following similarity transformation: where 𝑐1 and 𝑐2 are constants of integration. Substituting (29) 1 into (25), we obtain the exact solution to (2)immediately. 𝜉=𝑡, 𝜔= 𝑢− ( 𝑥+𝑎𝑡)2, log 4𝛼𝑡 log (20) 3.6. Similarity Solution for 𝑉6. For the generator 𝑉6,wehave and the similarity solution is 𝜔=𝑓(𝜉);thatis, the following similarity transformation: 1 𝜉=𝑡−1 𝑥, 𝑢= [𝑓 (𝑡) + ( 𝑥+𝑎𝑡)2]. log exp 4𝛼𝑡 log (21) 1 𝛼−𝛽 𝜔=log 𝑢+ log 𝑡− log 𝑥 Substituting (21)into(2), we reduce the bond pricing equa- 2 2𝛼 (30) tion to the following ODE: (𝛼 − 𝛽)2 −4𝛼𝛾 1 󸀠 − 𝑡− 𝑡−1 2𝑥, 2𝜉𝑓 +2𝛾𝜉+1=0, (22) 4𝛼 4𝛼 log Abstract and Applied Analysis 5

𝜆𝜉 and the similarity solution is 𝜔=𝑓(𝜉);thatis, When Δ=0,(36)hasthesolution𝑓(𝜉) =1 (𝑐 +𝑐2𝜉)𝑒 , where 𝜆=(V +𝛼−𝛽)/2𝛼.Thus,weobtainthesolutionto(2) 1 𝛼−𝛽 as follows: 𝑢= [𝑓 −1(𝑡 𝑥) − 𝑡+ 𝑥 exp log 2 log 2𝛼 log 𝜆 −𝜆V𝑡 𝑢 (𝑥,) 𝑡 =[𝑐1 +𝑐2 (log 𝑥−V𝑡)] 𝑥 𝑒 , (38) (31) 2 𝑐 𝑐 (𝛼 − 𝛽) −4𝛼𝛾 1 −1 2 where 1, 2 are arbitrary constants. + 𝑡+ 𝑡 𝑥] . Δ<0 𝑓(𝜉) = 4𝛼 4𝛼 log When ,(36)hasthesolution √ √ ((V+𝛼−𝛽)/2𝛼)𝜉 (𝑐1 cos( −Δ/2𝛼)𝜉2 +𝑐 sin( −Δ/2𝛼)𝜉)𝑒 .Thus,we obtain the solution to (2) as follows: Substituting (31)into(2), we reduce the bond pricing equation to the following ODE: 𝑢 (𝑥,) 𝑡 =𝑥(V+𝛼−𝛽)/2𝛼𝑒−(V(V+𝛼−𝛽)/2𝛼)𝑡

󸀠󸀠 󸀠2 √ 𝑓 +𝑓 =0, (32) −Δ ×[𝑐1 cos (log 𝑥−V𝑡) 2𝛼 (39) 𝑓󸀠 =𝑑𝑓/𝑑𝜉 where . √−Δ Solving (32), we get 𝑓(𝜉) = log |𝜉+𝑐1|+𝑐3.Thus,weobtain +𝑐 ( 𝑥−V𝑡)] , 2 sin 2𝛼 log the solution to (2) as follows: 𝑐 𝑐 1 where 1, 2 are arbitrary constants. 𝑢 (𝑥,) 𝑡 =𝑐 ( 𝑥+𝑐) 2 𝑡 log 1 3.8. Similarity Reduction for 𝑉1+V𝑉3. For the linear combina- 2 𝛼−𝛽 (𝛼 − 𝛽) −4𝛼𝛾 tion 𝑉=𝑉1 + V𝑉3 (V =0̸ is an arbitrary constant), we have the × [ 𝑥+ 𝑡 exp 2𝛼 log 4𝛼 (33) following similarity transformation: 𝜉=𝑥, 𝜔=log 𝑢−V𝑡, (40) 1 1 + 𝑡−1 2𝑥− 𝑡] , 4𝛼 log 2 log and the similarity solution is 𝜔=𝑓(𝜉);thatis,

𝑢=exp {𝑓 (𝑥) + V𝑡} . (41) where 𝑐1, 𝑐2 are arbitrary constants. Substituting (41)into(2), we reduce the bond pricing equation to the following ODE: 3.7. Similarity Solution for 𝑉1+V𝑉2. For the linear combination 𝑉=1 𝑉 + V𝑉2 (V =0̸ is an arbitrary constant), we have 2 󸀠󸀠 2 󸀠2 󸀠 𝛼𝜉 𝑓 +𝛼𝜉 𝑓 +𝛽𝜉𝑓 + V +𝛾=0, (42) the following similarity transformation: 󸀠 where 𝑓 =𝑑𝑓/𝑑𝜉. This is a nonlinear second-order ODE. In 𝜉= 𝑥−V𝑡, 𝜔 = 𝑢, log (34) the next section, we will deal with such an equation by the special transformation technique. and the similarity solution is 𝜔=𝑓(𝜉);thatis,

3.9. Similarity Solution for 𝑉2+V𝑉3. For the linear combina- 𝑢=𝑓( 𝑥−V𝑡) . log (35) tion 𝑉=𝑉2 + V𝑉3 (V =0̸ is an arbitrary constant), we have the following similarity transformation: Substituting (35)into(2), we reduce the bond pricing equa- 𝜉=𝑡, −𝜔=𝑥 V𝑢, tion to the following ODE: (43) 𝜔=𝑓(𝜉) 󸀠󸀠 󸀠 and the similarity solution is ;thatis, 𝛼𝑓 −(V +𝛼−𝛽)𝑓 +𝛾𝑓=0, (36) V 𝑢=𝑥𝑓 (𝑡) . (44) 𝑓󸀠 =𝑑𝑓/𝑑𝜉 where . Substituting (44)into(2), we reduce the bond pricing This is a second-order linear ODE; the corresponding 2 equation to the following ODE: characteristic equation is 𝛼𝜆 −(V +𝛼−𝛽)𝜆+𝛾=0.Solving √ 󸀠 the algebraic equation, we have 𝜆1 =(V +𝛼−𝛽+ Δ)/2𝛼, 𝑓 +𝛼V (V −1) 𝑓+𝛽V𝑓+𝛾𝑓=0, (45) 𝜆 =(V +𝛼−𝛽−√Δ)/2𝛼 Δ=(V +𝛼−𝛽)2 −4𝛼𝛾 2 ,where . 󸀠 𝜆1𝜉 𝜆2𝜉 where 𝑓 =𝑑𝑓/𝑑𝜉. When Δ>0,(36)hasthesolution𝑓(𝜉)1 =𝑐 𝑒 +𝑐2𝑒 . 𝑓=𝑐 {−[𝛼V2 −(𝛼−𝛽)V + 𝛾]𝜉} Thus, we obtain the solution to2 ( ) as follows: Solving (45), we get exp . Thus, we obtain the solution to2 ( )asfollows 𝜆 −𝜆 V𝑡 𝜆 −𝜆 V𝑡 𝑢 (𝑥,) 𝑡 =𝑐𝑥 1 𝑒 1 +𝑐𝑥 2 𝑒 2 , V 2 1 2 (37) 𝑢 (𝑥,) 𝑡 =𝑐𝑥 exp {− [𝛼V −(𝛼−𝛽)V +𝛾]𝑡}, (46) where 𝑐1, 𝑐2 are arbitrary constants. where 𝑐 is an arbitrary constant. 6 Abstract and Applied Analysis

3.10. Similarity Reduction for 𝑉1+V𝑉4. For the linear combi- 4. Exact Analytic Solutions in terms of the 𝑉=𝑉 + V𝑉 V =0̸ nation 1 4 ( is an arbitrary constant), we have Generalized Power Series Method the following similarity transformation: In Section 3, we considered the symmetry reductions and 2 𝜉=log 𝑥−V𝛼𝑡 , exact solutions to the bond pricing types of (2)and(3). In this section, we will deal with the nonlinear ODEs (42), (49), (47) 2 2 3 1 2 𝜔=log 𝑢−V𝑡 log 𝑥+ 𝛼V 𝑡 − (𝛼 − 𝛽) V𝑡 , (50), and (53) by the special transformation technique and 3 2 generalized power series method. Thus, the exact analytic solutions to (2)and(3)areobtained. and the similarity solution is 𝜔=𝑓(𝜉);thatis, 4.1. Exact Solution to (2). Firstly, we consider the ODE (42). 2 2 2 3 󸀠 𝑢= [𝑓 ( 𝑥−V𝛼𝑡 )+V𝑡 𝑥− 𝛼V 𝑡 Letting 𝑓 =𝑦, we get the Riccati equation exp log log 3 (48) 1 𝑑𝑦 2 𝛽 V +𝛾 + (𝛼 − 𝛽) V𝑡2]. =−𝑦 − 𝑦− . (54) 2 𝑑𝜉 𝛼𝜉 𝛼𝜉2

Substituting (48)into(2),wereducethebondpricing Now, we solve the equation by the transformation technique equation to the following ODE: directly. Suppose that (54)hasthesolutionoftheform −1 2 𝑦=𝑝𝜉 , (55) 𝛼𝑓 󸀠󸀠 +𝛼𝑓󸀠 −(𝛼−𝛽)𝑓󸀠 + V𝜉+𝛾=0, (49) where 𝑝 is a constant to be determined. Substituting (55)into 󸀠 2 where 𝑓 =𝑑𝑓/𝑑𝜉. This is a nonlinear second-order ODE (54), we have 𝛼𝑝 −(𝛼−𝛽)𝑝+V +𝛾=. 0 Solving the algebraic also. In the next section, we will deal with the exact solutions equation, we get to such equations. (𝛼 − 𝛽) ± √Δ Secondly, we consider (3). In fact, for this equation, we 𝑝= , (56) have the nontrivial cases as follows only. 2𝛼

2 𝑉 𝑉 where Δ = (𝛼 − 𝛽) −4𝛼(V +𝛾). 3.11. Similarity Reduction for 1 of (3). For the generator 1, 𝑦=𝑧+𝑝𝜉−1 we have the following reduced ordinary differential equation Setting andpluggingitinto(54), we get (ODE): 𝑑𝑧 2 𝑧 𝛽 =−𝑧 −𝑞 ,𝑞=2𝑝+. (57) 󸀠󸀠 󸀠 𝑑𝜉 𝜉 𝛼 𝛼𝜉𝑓 +𝛽𝑓 +𝛾𝑓=0, (50)

󸀠 ThisisaBernoulliequation.Solvingtheequation,wehavethe where 𝑓 =𝑑𝑓/𝑑𝜉. This is a nonlinear second-order ODE following results. as well; there is no general method for tackling it yet. In When 𝑞=1,weget𝑓(𝜉) =𝑝 log 𝜉+log(log 𝜉+𝑐1)+𝑐2. Section 4,wewilldealwithsuchequationsbythepower Thus, the exact solution to2 ( )is series method. 𝑝 V𝑡 𝑢 (𝑥,) 𝑡 =𝑐2𝑥 (log 𝑥+𝑐1) 𝑒 , (58) 𝑉 +V𝑉 3.12. Similarity Reduction for 1 2 of (3). For the linear 𝑐 𝑐 𝑝 𝑞 𝑉=𝑉+ V𝑉 V =0̸ where 1 and 2 are arbitrary constants; and are given by combination 1 2 ( is an arbitrary constant), (56)and(57). we have the following similarity transformation: When 𝑞 =1̸ ,weget𝑓(𝜉) =𝑝 log 𝜉+(1−𝑞)∫𝑑𝜉/(𝜉+ 𝑞 𝑐1𝜉 )+𝑐2. Thus, the exact solution to (2)is 𝜉=𝑥, 𝜔=log 𝑢−V𝑡, (51) 𝑑𝑥 𝑢 (𝑥,) 𝑡 =𝑐𝑥𝑝 [(1 − 𝑞) ∫ + V𝑡] , and the similarity solution is 𝜔=𝑓(𝜉);thatis, 2 exp 𝑞 (59) 𝑥+𝑐1𝑥 𝑢= [𝑓 (𝑥) + V𝑡] . exp (52) where 𝑐1 and 𝑐2 are arbitrary constants and 𝑝 and 𝑞 are given by (56)and(57). Substituting (52)into(3),wereducethesecondbondpricing equation to the following ODE: 4.2. Exact Analytic Solution to (2). Through the transfor- 2 󸀠󸀠 2 󸀠2 󸀠 mation technique, we solve the Riccati equation (54), so 𝛼𝜉 𝑓 +𝛼𝜉 𝑓 +𝛽𝜉𝑓 +𝛾𝜉+V =0, (53) the exact solutions to (2) are obtained. But for the other equations such as (49), (50), and (53), we cannot get the 󸀠 where 𝑓 =𝑑𝑓/𝑑𝜉. This is a nonlinear second-order ODE exact solutions by such special transformation technique. also. Similar to the above equations, we will deal with such However,weknowthatthepowerseriescanbeusedto equations by the generalized power series method in the next solve nonlinear ODEs, including many complicated differen- section. tial equations with nonconstant coefficients10 [ –13, 19, 20]. Abstract and Applied Analysis 7

Now, we consider the power series solution to the reduced Substituting (65)into(39), we obtain the exact analytic 󸀠 equation (49). Letting 𝑓 =𝑦,wegetthefollowingRiccati solution to (2) as follows: equation: 2 𝑢 (𝑥,) 𝑡 =𝑞exp [𝑝 (log 𝑥−𝛼V𝑡 ) 󸀠 2 𝛼𝑦 +𝛼𝑦 −(𝛼−𝛽)𝑦+V𝜉+𝛾=0. (60) 1 2 + 𝑐 ( 𝑥−𝛼V𝑡2) 2 1 log We will seek a solution of (60)inapowerseriesofthe 1 3 form + 𝑐 ( 𝑥−𝛼V𝑡2) 3 2 log ∞ ∞ ∞ 𝑦=∑𝑐 𝜉𝑛 =𝑝+∑𝑐 𝜉𝑛,𝑝=𝑐, 1 2 𝑛+2 𝑛 𝑛 0 (61) + ∑ 𝑐𝑛+1(log 𝑥−𝛼V𝑡 ) 𝑛=0 𝑛=1 𝑛=2 𝑛+2 1 2 𝑐 𝑛 = 0, 1, 2, . . + (𝛼 − 𝛽) V𝑡2 − 𝛼V2𝑡3 + V𝑡 𝑥] , where the coefficients 𝑛 ( )areconstantstobe 2 3 log determined. (66) Substituting (61)into(60) and comparing coefficients, we obtain where 𝑝=𝑐0 and 𝑞 are arbitrary constants and the other coefficients 𝑐𝑛 (𝑛 = 1, 2, .)aregivenby( . 62)and(63) 𝛼−𝛽 𝛾 𝑐 =−𝑝2 + 𝑝− , successively. 1 𝛼 𝛼 Similarly, we can give the exact power series solution to (62) (50)inthepowerseriesform(61). So, the exact analytic 𝛼−𝛽 V 𝑐 =−𝑝𝑐 + 𝑐 − . solution to (3) is obtained. The details are omitted here. 2 1 2𝛼 1 2𝛼 4.3. Exact Analytic Solution to (3). In Section 4.2,wecon- Generally, for 𝑛≥2,wehave struct the exact analytic solution to (49)bythepowerseries method and obtain the exact analytic solution to (2). Now, 󸀠 we consider (53). Firstly, let 𝑓 =𝑦;thenwegetthefollowing 1 𝑛 𝑐 = [(𝛼 − 𝛽) 𝑐 −𝛼∑𝑐 𝑐 ]. Riccati type of equation: 𝑛+1 (𝑛+1) 𝛼 𝑛 𝑘 𝑛−𝑘 (63) 𝑘=0 2 󸀠 2 2 𝛼𝜉 𝑦 +𝛼𝜉 𝑦 +𝛽𝜉𝑦+𝛾𝜉+V =0. (67)

Thus, for arbitrarily choosing the parameter 𝑐0,from (62), we can get 𝑐1 and 𝑐2.Furthermore,inviewof(63), we We will seek a solution of (67)inageneralizedpower have series of the form

∞ −1 𝑛 𝛼−𝛽 1 2 𝑦=𝐴𝜉 + ∑ 𝑐 𝜉 , 𝑐 = 𝑐 − (2𝑝𝑐 +𝑐 ), 𝑛 (68) 3 3𝛼 2 3 2 1 𝑛=0 (64) 𝛼−𝛽 1 𝐴 𝑐 𝑛 = 0, 1, 2, . . 𝑐 = 𝑐 − (𝑝𝑐 +𝑐𝑐 ), where the parameters and 𝑛 ( )areconstants 4 4𝛼 3 2 3 1 2 to be determined. Substituting (68)into(67) and comparing coefficients, we and so on. obtain {𝑐 }∞ Therefore, the other terms of the sequence 𝑛 𝑛=0 can be (𝛼 − 𝛽) ± √Δ determined successively from (63) in a unique manner. This 𝐴= , (69) implies that for (60) there exists a power series solution (61) 2𝛼 with the coefficients given by62 ( )and(63). Furthermore, we 2 can show the convergence of the power series solution (61) where Δ = (𝛼 − 𝛽) −4𝛼V,and with the coefficients given by62 ( )and(63) (see, e.g., [10, 12, −𝛾 13, 19]); the details are omitted here. So, this solution (61)to 𝑐 = ,2𝛼𝐴+𝛽=0.̸ (60) is an exact analytic solution. 0 2𝛼𝐴 +𝛽 (70) Hence, the exact power series solution to (49)canbe written as follows: Generally, for 𝑛≥0,wehave

1 1 ∞ 1 −𝛼 𝑛 𝑓 (𝜉) =𝑐+𝑝𝜉+ 𝑐 𝜉2 + 𝑐 𝜉3 + ∑ 𝑐 𝜉𝑛+2. 𝑐 = ∑𝑐 𝑐 , 𝑛=0,1,2,.... 1 2 𝑛+1 (65) 𝑛+1 (𝑛+1) 𝛼+2𝛼𝐴+𝛽 𝑘 𝑛−𝑘 (71) 2 3 𝑛=2 𝑛+2 𝑘=0 8 Abstract and Applied Analysis

Thus, from (69)and(70), we can get 𝐴 and 𝑐0.Further- where ] =0,1̸ is an arbitrary positive number. Firstly, by the more, in view of (71), we have group classification method, we get the geometric vector field of (75) as follows: −𝛼𝑐2 −2𝛼𝑐 𝑐 𝑐 = 0 ,𝑐= 0 1 , 𝑉 =𝜕,𝑉=𝑢𝜕,𝑉=𝑠𝜕, 1 𝛼+2𝛼𝐴+𝛽 2 2𝛼 + 2𝛼𝐴 +𝛽 1 𝑡 2 𝑢 𝑠 𝑢 (76) (72) −𝛼 (2𝑐 𝑐 +𝑐2) where the function 𝑠=𝑠(𝑥,𝑡)satisfies (75). 𝑐 = 0 2 1 , 3 3𝛼 + 2𝛼𝐴 +𝛽 Moreover, through the similarity transformation (42), we can reduce this equation to the following equation (ODE): andsoon(seeRemark 3). 𝛼𝜉2𝑓󸀠󸀠 +𝛼𝜉2𝑓󸀠2 +𝛽𝜉𝑓󸀠 +𝛾𝜉] + V =0, ∞ (77) Therefore, the other terms of the sequence {𝑐𝑛}𝑛=0 can be determined successively from (71) in a unique manner. This 󸀠 where 𝑓 =𝑑𝑓/𝑑𝜉. Similarly, we can consider the symmetry implies that for (67) there exists a generalized power series reductions and exact solutions to the equation. Now, as an solution (68) with the coefficients given by (69)–(71). The example, we study the special case ] =2.Inthiscase,wehave convergence of the generalized power series solution (68)to (67) is similar to that in Section 4.2;weomititinthispaper. 2 2 Thus, the power series solution68 ( )to(67) is also an exact 𝑢𝑡 +𝛼𝑥 𝑢𝑥𝑥 +𝛽𝑥𝑢𝑥 +𝛾𝑥 𝑢=0. (78) analytic solution. 󸀠 Hence, the power series solution of (53)canbewrittenas Referring to (77)andsetting𝑓 =𝑦,thenwegetthefollowing follows: reduced ODE of (78):

∞ 2 󸀠 2 2 2 󵄨 󵄨 1 2 1 𝑛+2 𝛼𝜉 𝑦 +𝛼𝜉 𝑦 +𝛽𝜉𝑦+𝛾𝜉 + V =0. (79) 𝑓 (𝜉) = 𝑐+𝐴log 󵄨𝜉󵄨 +𝑐0𝜉+ 𝑐1𝜉 + ∑ 𝑐𝑛+1𝜉 . (73) 2 𝑛=1 𝑛+2 Suppose that (79) has the power series solution of the generalized form (68). Then, substituting68 ( )into(79)and Substituting (73)into(52), we get the exact analytic comparing coefficients, we obtain solution to (3) as follows: (𝛼 − 𝛽) ± √Δ 1 𝐴= , (80) 𝑢 (𝑥,) 𝑡 =𝑐𝑥𝐴 [𝑐 𝑥+ 𝑐 𝑥2 2𝛼 exp 0 2 1 2 (74) where Δ = (𝛼 − 𝛽) −4𝛼V, ∞ 1 +∑ 𝑐 𝑥𝑛+2 + V𝑡] , 𝑛+2 𝑛+1 𝑛=1 (2𝛼𝐴 + 𝛽)0 𝑐 =0, (81) −𝛼𝑐2 −𝛾 𝑐 𝐴 𝑐 𝑛 = 0, 1, 2, . . 0 where is an arbitrary constant and and 𝑛 ( ) 𝑐1 = . (82) are given by (69)–(71) successively. 𝛼+2𝛼𝐴+𝛽

Remark 1. We note that the generalized power series solution Generally, for 𝑛≥1,wehave (68) differs from the regular form (61)since𝐴 =0̸ in (68). In other words, there is no exact power series solution of −𝛼 𝑛 𝑐 = ∑𝑐 𝑐 , 𝑛=1,2,.... the form (61)for(67). In particular, the determination of 𝑛+1 (𝑛+1) 𝛼+2𝛼𝐴+𝛽 𝑘 𝑛−𝑘 (83) parameter 𝐴 depends on the equation greatly (cf. [10, 11]for 𝑘=0 details). In view of (81), we have two special cases as follows. When 2𝛼𝐴 +𝛽 =0̸ ,from(81), we have 𝑐0 =0.Further- 5. Further Discussion about the General Bond more, from (82)and(83), we have Pricing Type of (1) −𝛾 𝑐 = ,𝑐=0, In the above sections, we considered the symmetries, sym- 1 𝛼+2𝛼𝐴+𝛽 2 metry reductions, and exact solutions to the general bond 2 (84) pricing type of equation for the cases ] =0and ] =1,which −𝛼𝑐 𝑐 = 1 ,𝑐=0, are the common forms in many practical applications, such 3 3𝛼+2𝛼𝐴+𝛽 4 as in financial mathematics. In this section, we discuss the generalized bond pricing type of equation of the form and so on. In this case, by induction method, we have

2 ] 𝑢𝑡 +𝛼𝑥 𝑢𝑥𝑥 +𝛽𝑥𝑢𝑥 +𝛾𝑥 𝑢=0, (75) 𝑐2𝑛 =0, 𝑛=0,1,2,.... (85) Abstract and Applied Analysis 9

When 2𝛼𝐴+𝛽 =0,from(81), we get that 𝑐0 is an arbitrary References constant. Furthermore, from (82)and(83), we have [1] R. K. Gazizov and N. H. Ibragimov, “Lie symmetry analysis of differential equations in finance,” Nonlinear Dynamics,vol.17, 2 no. 4, pp. 387–407, 1998. −𝛼𝑐0 −𝛾 −2𝛼𝑐0𝑐1 𝑐 = ,𝑐= , [2]C.A.Pooe,F.M.Mahomed,andC.WafoSoh,“Invariant 1 𝛼+2𝛼𝐴+𝛽 2 2𝛼 + 2𝛼𝐴 +𝛽 solutions of the Black-Scholes equation,” Mathematical & Com- (86) putational Applications,vol.8,no.1–3,pp.63–70,2003. −𝛼 (2𝑐 𝑐 +𝑐2) −2𝛼 (𝑐 𝑐 +𝑐𝑐 ) 0 2 1 0 3 1 2 [3] M. Craddock and E. Platen, “Symmetry group methods for 𝑐3 = ,𝑐4 = , 3𝛼 + 2𝛼𝐴 +𝛽 4𝛼 + 2𝛼𝐴 +𝛽 fundamental solutions,” Journal of Differential Equations,vol. 207, no. 2, pp. 285–302, 2004. [4] M. Craddock and K. A. Lennox, “Lie group symmetries as andsoon(seeRemark 3). integral transforms of fundamental solutions,” Journal of Dif- Thus, the exact power series solutions79 to( )are ferential Equations,vol.232,no.2,pp.652–674,2007. obtained. 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Totackle these equations, the [15]G.W.BlumanandS.C.Anco,Symmetry and Integration Meth- generalized power series method and special techniques are ods for Differential Equations,vol.154ofApplied Mathematical necessary sometimes. For getting the exact analytic solutions Sciences, Springer, New York, NY, USA, 2002. in Sections 4.3 and 5,thecondition(𝑛 + 1)𝛼 + 2𝛼𝐴 +𝛽 =0̸ is [16] E. Pucci and G. Saccomandi, “Potential symmetries and solu- necessary for 𝑛 = 0, 1, 2, . . tions by reduction of partial differential equations,” Journal of Physics,vol.26,no.3,pp.681–690,1993. [17] C. Qu and Q. Huang, “Symmetry reductions and exact solutions Acknowledgments of the affine heat equation,” Journal of Mathematical Analysis and Applications,vol.346,no.2,pp.521–530,2008. This work is supported by the National Natural Science [18] T. 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[19] H. Liu and W. Li, “The exact analytic solutions of a nonlinear differential iterative equation,” Nonlinear Analysis: Theory, Methods & Applications,vol.69,no.8,pp.2466–2478,2008. [20] N. Asmar, Partial Differential Equations with Fourier Series and Boundary Value Problems, China Machine Press, Beijing, China, 2nd edition, 2005. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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