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

Authors: Hanze Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The variable-coefficients partial differential equations (vc-PDEs) in finance are investigated by Lie symmetry analysis and the 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.


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## Body

## 1. Introduction

Gazizov and Ibragimov [1] studied the Black-Scholes equation of option pricing by Lie equivalence transformations. By the optimal system method, some invariant solutions to heat and Black-Scholes equations are obtained [2]. In [3–5], the fundamental solutions to the bond pricing equations are considered by Lie symmetry analysis and the integral transform method. In [6], the invariance properties of the bond pricing equation are studied by the group classification method. In [7], the finite element method was adopted to solve the bond pricing type of PDE system, and the numerical implementation was provided, such as system that models the TF convertible bonds with credit risk in bond pricing theory. However, the similarity reductions and exact solutions to such variable-coefficient equations are not considered generally in the aforementioned papers. Recently, we studied some nonlinear PDEs by Lie symmetry analysis and the dynamical system method [8–13]; for example, in [8], we considered Lie group classifications and exact solutions to the space-dependent coefficients hanging chain equation and the simplified bond pricing equation. In [9], we investigated the integrable condition and exact solutions to the time-dependent coefficient Gardner equations by the Painlevé test and Lie group analysis method. In [10–13], we developed the generalized power series method for dealing with exact solutions to some nonlinear PDEs based on the symmetry analysis method.It is known that the Lie symmetry analysis is a systematic and powerful method for dealing with symmetries and exact solutions to partial differential equations (see, e.g., [1–6, 8–18] and the references therein). Furthermore, we find that the combination of Lie symmetry analysis and the power series method is a feasible approach to investigating exact solutions to nonlinear PDEs [8–13]. On the other hand, under the perspective of mathematical physics and Lie symmetry analysis, the space-time dependent coefficients system differs greatly from its time-dependent counterpart, and it is more complicated than the latter. However, most of the studies are related to the time-dependent coefficient systems. Moreover, the determination of exact solutions to the variable-coefficients PDEs is a complicated problem that challenges researchers greatly. In the present paper, we consider the symmetry reductions and exact solutions to the general space-dependent coefficients PDEs in finance as follows:
(1)ut+αx2uxx+βxux+γxνu=0,x>0,
where u=u(x,t) denotes the unknown function of the space variable x and time t and the parameters α,β,γ,ν∈ℝ are arbitrary constants, ν≥0 and α≠0.We first note that (1) is the general form of the bond pricing types of equations [1–7]. In particular, if ν=0, then this equation becomes the following Black-Scholes equation of option pricing:
(2)ut+αx2uxx+βxux+γu=0,x>0.Ifν=1, then (1) is the general bond pricing equation given by
(3)ut+αx2uxx+βxux+γxu=0,x>0.Such equations are called bond pricing types of equations, which are of great importance in financial mathematics and bond pricing theory [3–7]. For dealing with exact solutions to the variable-coefficients PDEs, we will introduce the generalized power series method [10–13] in the present paper. By a generalized power series solution, we mean a generalized power series is of the form
(4)f(ξ)=A(ξ)+∑n=0∞‍cnξn,
which is a solution to a system with respect to the variable ξ, where cn (n=0,1,2,…) are constant coefficients to be determined and A(ξ) is the undetermined function with respect to the variable ξ. In particular, if A(ξ)≡0, then (4) is the regular power series solution. So, the generalized power series solution is the generalization of the regular power series solution and it naturally includes the latter as its special case. If we obtained a generalized power series solution (4) to a system and the convergence of this power series is shown, then the exact generalized power series solution is obtained. This solution sometimes is called the exact analytic solution [10–13, 19].The main purpose of this paper is to develop the combination of Lie symmetry analysis and the generalized power series method for dealing with symmetries and exact solutions to the variable-coefficients PDEs in finance. The remainder of this paper is organized as follows. In Section2, we perform Lie symmetry analysis on the bond pricing types of (2) and (3) and give all of the geometric vector fields of the equations in terms of the arbitrary parameters. In Section 3, we consider the symmetry reductions of the equations and provide the exponentiated solutions and the similarity solutions to the equations. In Section 4, we investigate the exact analytic solutions to the variable-coefficient equations by the generalized power series method. In Section 5, we deal with the vector fields and exact solutions to the bond pricing type of (1) for the general case ν≠0,1. Finally, the conclusions and some remarks are given in Section 6.


## 2. Lie Symmetry Analysis for (2) and (3)

In this section, we will present a complete list of all possible Lie symmetry algebras for the bond pricing types of equations of the forms (2) and (3).Recall that the geometric vector fields of such equations are as follows:(5)V=ξ(x,t,u)∂x+τ(x,t,u)∂t+ϕ(x,t,u)∂u,
where ξ(x,t,u), τ(x,t,u), and ϕ(x,t,u) are coefficient functions of the vector field to be determined. The symmetry groups of (2) and (3) will be generated by the vector field of the form (5), respectively. Applying the second prolongation pr(2)V of V to (2) and (3), we find that the coefficient functions ξ, τ, and ϕ must satisfy the following Lie symmetry condition:
(6)pr(2)V(Δ)|Δ=0=0,
where Δ=ut+αx2uxx+βxux+γu for (2) and Δ=ut+αx2uxx+βxux+γxu for (3), respectively. Then, the Lie symmetry group calculation method leads to the following conditions on the coefficient functions ξ, τ, and ϕ:
(7)ξ=12xτtlogx+xρ,ϕ=r(x,t)u+s(x,t),r=18ατttlog2x+α-β4ατtlogx+12αρtlogx+σ,
for some functions τ, ρ, and σ. Now the functions τ, ρ, and σ depend only on t. Moreover, for (2), we have
(8)18ατtttlog2x+12αρttlogx+14τtt-(α-β)2-4αγ4ατt-α-β2αρt+σt=0;
for (3), we have
(9)18ατtttlog2x+12γτtxlogx+12αρttlogx+γ(ρ+τt)x+14τtt-(α-β)24ατt-α-β2αρt+σt=0.
These equations fix the functions ξ, τ, ρ, σ, and ϕ. Solving the equations, we obtain the vector field of (2) as follows:
(10)V1=∂t,V2=x∂x,V3=u∂u,V4=2αxt∂x+[logx+(α-β)t]u∂u,V5=2αxt(logx)∂x+4αt∂t+[(α-β)logx+((α-β)2-4αγ)t]u∂u,V6=4αxt(logx)∂x+4αt2∂t+[2(α-β)tlogx+log2x-2αt2222222+((α-β)2-4αγ)t2]u∂u,Vs=s∂u,
where the parameters α≠0, β, γ are arbitrary constants and the function s=s(x,t) satisfies (2).For (3), we have the vector field as follows:
(11)V1=∂t,V2=u∂u,Vs=s∂u,
where the function s=s(x,t) satisfies (3).Clearly, for (2), a basis of the Lie algebra is {V1,…,V6,Vs}. For (3), a basis for the Lie algebra is {V1,V2,Vs}. Thus, the new symmetries cannot be derived from the Lie brackets for the two equations.Moreover, we can obtain the one-parameter groups generated byVi, respectively. In fact, for (2), the one-parameter groups Gi generated by Vi (i=1,…,6,s) are given in the following:
(12)G1:(x,t,u)⟶(x,t+ϵ,u),G2:(x,t,u)⟶(eϵx,t,u),G3:(x,t,u)⟶(x,t,eϵu),G4:(x,t,u)⟶(xe2αϵt,t,xϵuexp[(α-β)ϵt+αϵ2t]),G5:(x,t,u)⟶((α-β)2-4αγ4α(δ2-1)txδ,δ2t,uexp[(α-β)2-4αγ4α(δ2-1)tα-β2α(δ-1)logx00000000000000000000000000000+(α-β)2-4αγ4α00000000000000000000000000000×(δ2-1)t(α-β)2-4αγ4α(δ2-1)tα-β2αα-β2α]α-β2α),G6:(x,t,u)⟶((α-β)2-4αγ4α4αϵt1-4αϵt(α-β)2-4αγ4α}),x1/(1-4αϵt),t1-4αϵt,000000000u1-4αϵtexp{[(α-β)2-4αγ4αtα-β2αlogx00000000000000000000000000+14αtlog2x00000000000000000000000000+(α-β)2-4αγ4αt]00000000000000000000000000×4αϵt1-4αϵt(α-β)2-4αγ4α}),Gs:(x,t,u)⟶(x,t,u+ϵs),
where δ=e2αϵ, ϵ≪1, and the function s=s(x,t) is an arbitrary solution to (2). For (3), the one-parameter groups are Gi (i=1,3,s) as above, while s=s(x,t) is an arbitrary solution to (3).From the above, we observe thatG1 is a time translation and G2 and G3 are trivial scaling transformations, while Gi (i=4,5,6) are nontrivial local groups of transformations. Their appearances are far from obvious from basic physical principles, but they are important for us to investigate the exact solutions to PDEs (see, e.g., [3–5, 10]).


## 3. Symmetry Reductions and Exact Solutions to the Bond Pricing Types of Equations

In the preceding section, we obtained the symmetries and symmetry groups of (2) and (3). Now, we deal with the symmetry reductions and exact solutions to the equations.


### 3.1. The Exponentiated Solutions

Since eachGi (i=1,…,6,s) is a symmetry group, it implies that if u=f(x,t) is a solution to (2), then u(i)(i=1,…,6,s) are all solutions to the following equation as well: (13a)u(1)=f(x,t-ϵ),(13b)u(2)=f(e-ϵx,t),(13c)u(3)=eϵf(x,t),(13d)u(4)=xϵexp[(α-β)ϵt-αϵ2t]f(e-2αϵtx,t),(13e)u(5)=exp[(α-β)2-4αγ4αα-β2α(1-δ-1)logx0000000000+(α-β)2-4αγ4α(1-δ-2)t]00010×f(x1/δ,δ-2t),(13f)u(6)=11+4αϵtexp{[(α-β)2-4αγ4αα-β2αlogx+14αtlog2x000000000000000000000+(α-β)2-4αγ4αt]4αϵt1+4αϵt}00010×f(x1/(1+4αϵt),t1+4αϵt),(13g)u(s)=f(x,t)+ϵs,whereδ=e2αϵ, ϵ is an arbitrary real number, and the function s=s(x,t) satisfies (2).For (3), the exponentiated solutions are u(i) (i=1,3,s) as above while s=s(x,t) satisfies (3).Such exponentiated solutions are one of group-invariant types of solutions to the PDEs, which are generated from the one-parameter groups and are of importance for studying the exact solutions and investigating the properties of solutions (see Remark2).Next, we investigate the symmetry reductions and exact explicit solutions to the two bond pricing equations. Firstly, we consider (2).


### 3.2. Similarity Solution forV1

For the generatorV1, we have the following reduced ordinary differential equation (ODE):
(14)αξ2f′′+βξf′+γf=0,
where f′=df/dξ. This is an Euler equation; the corresponding characteristic equation is αK2-(α-β)K+γ=0. Solving this equation, we have K=((α-β)±Δ)/2α, where Δ=(α-β)2-4αγ.WhenΔ>0, (14) has the general solution f=c1ξK1+c2ξK2. Thus, we obtain the exact solution to (2) as follows:
(15)u(x,t)=c1xK1+c2xK2,
where c1 and c2 are arbitrary constants and K1,2=((α-β)±Δ)/2α are two real roots to the characteristic equation, respectively.WhenΔ=0, (14) has the general solution f=ξK(c1+c2logξ). Thus, we obtain the exact solution to (2) as follows:
(16)u(x,t)=xK(c1+c2logx),
where c1 and c2 are arbitrary constants, K=(α-β)/2α are the real root to the characteristic equation.WhenΔ<0, (14) has the general solution f=ξK(c1cos(-Δ/2α)logξ+c2sin(-Δ/2α)logξ). Thus, we obtain the exact solution to (2) as follows:
(17)u(x,t)=xK(c1cos-Δ2αlogx+c2sin-Δ2αlogx),
where c1 and c2 are arbitrary constants, K=(α-β)/2α.


### 3.3. Similarity Solution forV2

For the generatorV2, we have the following reduced ODE:
(18)f′+γf=0,
where f′=df/dξ. Solving this equation, we have f=ce-γξ. Thus, we obtain the exact solution to (2) as follows:
(19)u(x,t)=ce-γt,
where c is an arbitrary constant.


### 3.4. Similarity Solution forV4

For the generatorV4, we have the following similarity transformation:
(20)ξ=t,ω=logu-14αt(logx+at)2,
and the similarity solution is ω=f(ξ); that is,
(21)u=exp[f(t)+14αt(logx+at)2].
Substituting (21) into (2), we reduce the bond pricing equation to the following ODE:
(22)2ξf′+2γξ+1=0,
where f′=df/dξ. It implies that if ω=f(ξ) is a solution to (22), then (21) is a solution to (2). Solving (22), we get f(ξ)=-(1/2)logξ-γξ+c1. Thus, we obtain the solution to (2) as follows:
(23)u(x,t)=cexp[14αt(logx+at)2-12logt-γt],
where c is an arbitrary constant.


### 3.5. Similarity Solution forV5

For the generatorV5, we have the following similarity transformation:
(24)ξ=t-1/2logx,ω=logu-α-β2αlogx-(α-β)2-4αγ4αt,
and the similarity solution is ω=f(ξ); that is,
(25)u=exp[f(t-1/2logx)+α-β2αlogx+(α-β)2-4αγ4αt].
Substituting (25) into (2), we reduce the bond pricing equation to the following ODE:
(26)αf′′+αf′2-12ξf′=0,
where f′=df/dξ.Lettingf′=y, we get the Bernoulli equation
(27)dydξ=12αξy-αy2.
Clearly, y=0; that is, f=c is a solution to (26). Thus, we get a solution to (2) as follows:
(28)u(x,t)=exp[α-β2αlogx+(α-β)2-4αγ4αt+c],
for an arbitrary constant number c.Wheny≠0, solving the Bernoulli equation, we get y=e(1/4α)ξ2/(∫‍e(1/4α)ξ2dξ+c1). Thus, we obtain the solution to (26) as follows:
(29)f(ξ)=∫‍e(1/4α)ξ2∫‍e(1/4α)ξ2dξ+c1dξ+c2,
where c1 and c2 are constants of integration. Substituting (29) into (25), we obtain the exact solution to (2) immediately.


### 3.6. Similarity Solution forV6

For the generatorV6, we have the following similarity transformation:
(30)ξ=t-1logx,ω=logu+12logt-α-β2αlogx-(α-β)2-4αγ4αt-14αt-1log2x,
and the similarity solution is ω=f(ξ); that is,
(31)u=exp[(α-β)2-4αγ4αf(t-1logx)-12logt+α-β2αlogx2222222+(α-β)2-4αγ4αt+14αt-1log2x].
Substituting (31) into (2), we reduce the bond pricing equation to the following ODE:
(32)f′′+f′2=0,
where f′=df/dξ.Solving (32), we get f(ξ)=log|ξ+c1|+c3. Thus, we obtain the solution to (2) as follows:
(33)u(x,t)=c2(1tlogx+c1)2222222×exp[α-β2αlogx+(α-β)2-4αγ4αt3233333333333+14αt-1log2x-12logt(α-β)2-4αγ4α],
where c1, c2 are arbitrary constants.


### 3.7. Similarity Solution forV1+vV2

For the linear combinationV=V1+vV2 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(34)ξ=logx-vt,ω=u,
and the similarity solution is ω=f(ξ); that is,
(35)u=f(logx-vt).
Substituting (35) into (2), we reduce the bond pricing equation to the following ODE:
(36)αf′′-(v+α-β)f′+γf=0,
where f′=df/dξ.This is a second-order linear ODE; the corresponding characteristic equation isαλ2-(v+α-β)λ+γ=0. Solving the algebraic equation, we have λ1=(v+α-β+Δ)/2α, λ2=(v+α-β-Δ)/2α, where Δ=(v+α-β)2-4αγ.WhenΔ>0, (36) has the solution f(ξ)=c1eλ1ξ+c2eλ2ξ. Thus, we obtain the solution to (2) as follows:
(37)u(x,t)=c1xλ1e-λ1vt+c2xλ2e-λ2vt,
where c1, c2 are arbitrary constants.WhenΔ=0, (36) has the solution f(ξ)=(c1+c2ξ)eλξ, where λ=(v+α-β)/2α. Thus, we obtain the solution to (2) as follows:
(38)u(x,t)=[c1+c2(logx-vt)]xλe-λvt,
where c1, c2 are arbitrary constants.WhenΔ<0, (36) has the solution f(ξ)=(c1cos(-Δ/2α)ξ+c2sin(-Δ/2α)ξ)e((v+α-β)/2α)ξ. Thus, we obtain the solution to (2) as follows:
(39)u(x,t)=x(v+α-β)/2αe-(v(v+α-β)/2α)t2222222×[c1cos-Δ2α(logx-vt)11111111111+c2sin-Δ2α(logx-vt)],
where c1, c2 are arbitrary constants.


### 3.8. Similarity Reduction forV1+vV3

For the linear combinationV=V1+vV3 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(40)ξ=x,ω=logu-vt,
and the similarity solution is ω=f(ξ); that is,
(41)u=exp{f(x)+vt}.
Substituting (41) into (2), we reduce the bond pricing equation to the following ODE:
(42)αξ2f′′+αξ2f′2+βξf′+v+γ=0,
where f′=df/dξ. This is a nonlinear second-order ODE. In the next section, we will deal with such an equation by the special transformation technique.


### 3.9. Similarity Solution forV2+vV3

For the linear combinationV=V2+vV3 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(43)ξ=t,ω=x-vu,
and the similarity solution is ω=f(ξ); that is,
(44)u=xvf(t).
Substituting (44) into (2), we reduce the bond pricing equation to the following ODE:
(45)f′+αv(v-1)f+βvf+γf=0,
where f′=df/dξ.Solving (45), we get f=cexp{-[αv2-(α-β)v+γ]ξ}. Thus, we obtain the solution to (2) as follows
(46)u(x,t)=cxvexp{-[αv2-(α-β)v+γ]t},
where c is an arbitrary constant.


### 3.10. Similarity Reduction forV1+vV4

For the linear combinationV=V1+vV4 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(47)ξ=logx-vαt2,ω=logu-vtlogx+23αv2t3-12(α-β)vt2,
and the similarity solution is ω=f(ξ); that is,
(48)u=exp[f(logx-vαt2)+vtlogx-23αv2t322222222+12(α-β)vt2].
Substituting (48) into (2), we reduce the bond pricing equation to the following ODE:
(49)αf′′+αf′2-(α-β)f′+vξ+γ=0,
where f′=df/dξ. This is a nonlinear second-order ODE also. In the next section, we will deal with the exact solutions to such equations.Secondly, we consider (3). In fact, for this equation, we have the nontrivial cases as follows only.


### 3.11. Similarity Reduction forV1 of (3)

For the generatorV1, we have the following reduced ordinary differential equation (ODE):
(50)αξf′′+βf′+γf=0,
where f′=df/dξ. This is a nonlinear second-order ODE as well; there is no general method for tackling it yet. In Section 4, we will deal with such equations by the power series method.


### 3.12. Similarity Reduction forV1+vV2 of (3)

For the linear combinationV=V1+vV2 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(51)ξ=x,ω=logu-vt,
and the similarity solution is ω=f(ξ); that is,
(52)u=exp[f(x)+vt].
Substituting (52) into (3), we reduce the second bond pricing equation to the following ODE:
(53)αξ2f′′+αξ2f′2+βξf′+γξ+v=0,wheref′=df/dξ. This is a nonlinear second-order ODE also. Similar to the above equations, we will deal with such equations by the generalized power series method in the next section.


## 3.1. The Exponentiated Solutions

Since eachGi (i=1,…,6,s) is a symmetry group, it implies that if u=f(x,t) is a solution to (2), then u(i)(i=1,…,6,s) are all solutions to the following equation as well: (13a)u(1)=f(x,t-ϵ),(13b)u(2)=f(e-ϵx,t),(13c)u(3)=eϵf(x,t),(13d)u(4)=xϵexp[(α-β)ϵt-αϵ2t]f(e-2αϵtx,t),(13e)u(5)=exp[(α-β)2-4αγ4αα-β2α(1-δ-1)logx0000000000+(α-β)2-4αγ4α(1-δ-2)t]00010×f(x1/δ,δ-2t),(13f)u(6)=11+4αϵtexp{[(α-β)2-4αγ4αα-β2αlogx+14αtlog2x000000000000000000000+(α-β)2-4αγ4αt]4αϵt1+4αϵt}00010×f(x1/(1+4αϵt),t1+4αϵt),(13g)u(s)=f(x,t)+ϵs,whereδ=e2αϵ, ϵ is an arbitrary real number, and the function s=s(x,t) satisfies (2).For (3), the exponentiated solutions are u(i) (i=1,3,s) as above while s=s(x,t) satisfies (3).Such exponentiated solutions are one of group-invariant types of solutions to the PDEs, which are generated from the one-parameter groups and are of importance for studying the exact solutions and investigating the properties of solutions (see Remark2).Next, we investigate the symmetry reductions and exact explicit solutions to the two bond pricing equations. Firstly, we consider (2).


## 3.2. Similarity Solution forV1

For the generatorV1, we have the following reduced ordinary differential equation (ODE):
(14)αξ2f′′+βξf′+γf=0,
where f′=df/dξ. This is an Euler equation; the corresponding characteristic equation is αK2-(α-β)K+γ=0. Solving this equation, we have K=((α-β)±Δ)/2α, where Δ=(α-β)2-4αγ.WhenΔ>0, (14) has the general solution f=c1ξK1+c2ξK2. Thus, we obtain the exact solution to (2) as follows:
(15)u(x,t)=c1xK1+c2xK2,
where c1 and c2 are arbitrary constants and K1,2=((α-β)±Δ)/2α are two real roots to the characteristic equation, respectively.WhenΔ=0, (14) has the general solution f=ξK(c1+c2logξ). Thus, we obtain the exact solution to (2) as follows:
(16)u(x,t)=xK(c1+c2logx),
where c1 and c2 are arbitrary constants, K=(α-β)/2α are the real root to the characteristic equation.WhenΔ<0, (14) has the general solution f=ξK(c1cos(-Δ/2α)logξ+c2sin(-Δ/2α)logξ). Thus, we obtain the exact solution to (2) as follows:
(17)u(x,t)=xK(c1cos-Δ2αlogx+c2sin-Δ2αlogx),
where c1 and c2 are arbitrary constants, K=(α-β)/2α.


## 3.3. Similarity Solution forV2

For the generatorV2, we have the following reduced ODE:
(18)f′+γf=0,
where f′=df/dξ. Solving this equation, we have f=ce-γξ. Thus, we obtain the exact solution to (2) as follows:
(19)u(x,t)=ce-γt,
where c is an arbitrary constant.


## 3.4. Similarity Solution forV4

For the generatorV4, we have the following similarity transformation:
(20)ξ=t,ω=logu-14αt(logx+at)2,
and the similarity solution is ω=f(ξ); that is,
(21)u=exp[f(t)+14αt(logx+at)2].
Substituting (21) into (2), we reduce the bond pricing equation to the following ODE:
(22)2ξf′+2γξ+1=0,
where f′=df/dξ. It implies that if ω=f(ξ) is a solution to (22), then (21) is a solution to (2). Solving (22), we get f(ξ)=-(1/2)logξ-γξ+c1. Thus, we obtain the solution to (2) as follows:
(23)u(x,t)=cexp[14αt(logx+at)2-12logt-γt],
where c is an arbitrary constant.


## 3.5. Similarity Solution forV5

For the generatorV5, we have the following similarity transformation:
(24)ξ=t-1/2logx,ω=logu-α-β2αlogx-(α-β)2-4αγ4αt,
and the similarity solution is ω=f(ξ); that is,
(25)u=exp[f(t-1/2logx)+α-β2αlogx+(α-β)2-4αγ4αt].
Substituting (25) into (2), we reduce the bond pricing equation to the following ODE:
(26)αf′′+αf′2-12ξf′=0,
where f′=df/dξ.Lettingf′=y, we get the Bernoulli equation
(27)dydξ=12αξy-αy2.
Clearly, y=0; that is, f=c is a solution to (26). Thus, we get a solution to (2) as follows:
(28)u(x,t)=exp[α-β2αlogx+(α-β)2-4αγ4αt+c],
for an arbitrary constant number c.Wheny≠0, solving the Bernoulli equation, we get y=e(1/4α)ξ2/(∫‍e(1/4α)ξ2dξ+c1). Thus, we obtain the solution to (26) as follows:
(29)f(ξ)=∫‍e(1/4α)ξ2∫‍e(1/4α)ξ2dξ+c1dξ+c2,
where c1 and c2 are constants of integration. Substituting (29) into (25), we obtain the exact solution to (2) immediately.


## 3.6. Similarity Solution forV6

For the generatorV6, we have the following similarity transformation:
(30)ξ=t-1logx,ω=logu+12logt-α-β2αlogx-(α-β)2-4αγ4αt-14αt-1log2x,
and the similarity solution is ω=f(ξ); that is,
(31)u=exp[(α-β)2-4αγ4αf(t-1logx)-12logt+α-β2αlogx2222222+(α-β)2-4αγ4αt+14αt-1log2x].
Substituting (31) into (2), we reduce the bond pricing equation to the following ODE:
(32)f′′+f′2=0,
where f′=df/dξ.Solving (32), we get f(ξ)=log|ξ+c1|+c3. Thus, we obtain the solution to (2) as follows:
(33)u(x,t)=c2(1tlogx+c1)2222222×exp[α-β2αlogx+(α-β)2-4αγ4αt3233333333333+14αt-1log2x-12logt(α-β)2-4αγ4α],
where c1, c2 are arbitrary constants.


## 3.7. Similarity Solution forV1+vV2

For the linear combinationV=V1+vV2 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(34)ξ=logx-vt,ω=u,
and the similarity solution is ω=f(ξ); that is,
(35)u=f(logx-vt).
Substituting (35) into (2), we reduce the bond pricing equation to the following ODE:
(36)αf′′-(v+α-β)f′+γf=0,
where f′=df/dξ.This is a second-order linear ODE; the corresponding characteristic equation isαλ2-(v+α-β)λ+γ=0. Solving the algebraic equation, we have λ1=(v+α-β+Δ)/2α, λ2=(v+α-β-Δ)/2α, where Δ=(v+α-β)2-4αγ.WhenΔ>0, (36) has the solution f(ξ)=c1eλ1ξ+c2eλ2ξ. Thus, we obtain the solution to (2) as follows:
(37)u(x,t)=c1xλ1e-λ1vt+c2xλ2e-λ2vt,
where c1, c2 are arbitrary constants.WhenΔ=0, (36) has the solution f(ξ)=(c1+c2ξ)eλξ, where λ=(v+α-β)/2α. Thus, we obtain the solution to (2) as follows:
(38)u(x,t)=[c1+c2(logx-vt)]xλe-λvt,
where c1, c2 are arbitrary constants.WhenΔ<0, (36) has the solution f(ξ)=(c1cos(-Δ/2α)ξ+c2sin(-Δ/2α)ξ)e((v+α-β)/2α)ξ. Thus, we obtain the solution to (2) as follows:
(39)u(x,t)=x(v+α-β)/2αe-(v(v+α-β)/2α)t2222222×[c1cos-Δ2α(logx-vt)11111111111+c2sin-Δ2α(logx-vt)],
where c1, c2 are arbitrary constants.


## 3.8. Similarity Reduction forV1+vV3

For the linear combinationV=V1+vV3 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(40)ξ=x,ω=logu-vt,
and the similarity solution is ω=f(ξ); that is,
(41)u=exp{f(x)+vt}.
Substituting (41) into (2), we reduce the bond pricing equation to the following ODE:
(42)αξ2f′′+αξ2f′2+βξf′+v+γ=0,
where f′=df/dξ. This is a nonlinear second-order ODE. In the next section, we will deal with such an equation by the special transformation technique.


## 3.9. Similarity Solution forV2+vV3

For the linear combinationV=V2+vV3 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(43)ξ=t,ω=x-vu,
and the similarity solution is ω=f(ξ); that is,
(44)u=xvf(t).
Substituting (44) into (2), we reduce the bond pricing equation to the following ODE:
(45)f′+αv(v-1)f+βvf+γf=0,
where f′=df/dξ.Solving (45), we get f=cexp{-[αv2-(α-β)v+γ]ξ}. Thus, we obtain the solution to (2) as follows
(46)u(x,t)=cxvexp{-[αv2-(α-β)v+γ]t},
where c is an arbitrary constant.


## 3.10. Similarity Reduction forV1+vV4

For the linear combinationV=V1+vV4 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(47)ξ=logx-vαt2,ω=logu-vtlogx+23αv2t3-12(α-β)vt2,
and the similarity solution is ω=f(ξ); that is,
(48)u=exp[f(logx-vαt2)+vtlogx-23αv2t322222222+12(α-β)vt2].
Substituting (48) into (2), we reduce the bond pricing equation to the following ODE:
(49)αf′′+αf′2-(α-β)f′+vξ+γ=0,
where f′=df/dξ. This is a nonlinear second-order ODE also. In the next section, we will deal with the exact solutions to such equations.Secondly, we consider (3). In fact, for this equation, we have the nontrivial cases as follows only.


## 3.11. Similarity Reduction forV1 of (3)

For the generatorV1, we have the following reduced ordinary differential equation (ODE):
(50)αξf′′+βf′+γf=0,
where f′=df/dξ. This is a nonlinear second-order ODE as well; there is no general method for tackling it yet. In Section 4, we will deal with such equations by the power series method.


## 3.12. Similarity Reduction forV1+vV2 of (3)

For the linear combinationV=V1+vV2 (v≠0 is an arbitrary constant), we have the following similarity transformation:
(51)ξ=x,ω=logu-vt,
and the similarity solution is ω=f(ξ); that is,
(52)u=exp[f(x)+vt].
Substituting (52) into (3), we reduce the second bond pricing equation to the following ODE:
(53)αξ2f′′+αξ2f′2+βξf′+γξ+v=0,wheref′=df/dξ. This is a nonlinear second-order ODE also. Similar to the above equations, we will deal with such equations by the generalized power series method in the next section.


## 4. Exact Analytic Solutions in terms of the Generalized Power Series Method

In Section3, we considered the symmetry reductions and exact solutions to the bond pricing types of (2) and (3). In this section, we will deal with the nonlinear ODEs (42), (49), (50), and (53) by the special transformation technique and generalized power series method. Thus, the exact analytic solutions to (2) and (3) are obtained.


### 4.1. Exact Solution to (2)

Firstly, we consider the ODE (42). Letting f′=y, we get the Riccati equation
(54)dydξ=-y2-βαξy-v+γαξ2.
Now, we solve the equation by the transformation technique directly. Suppose that (54) has the solution of the form
(55)y=pξ-1,
where p is a constant to be determined. Substituting (55) into (54), we have αp2-(α-β)p+v+γ=0. Solving the algebraic equation, we get
(56)p=(α-β)±Δ2α,
where Δ=(α-β)2-4α(v+γ).Settingy=z+pξ-1 and plugging it into (54), we get
(57)dzdξ=-z2-qzξ,q=2p+βα.
This is a Bernoulli equation. Solving the equation, we have the following results.Whenq=1, we get f(ξ)=plogξ+log(logξ+c1)+c2. Thus, the exact solution to (2) is
(58)u(x,t)=c2xp(logx+c1)evt,
where c1 and c2 are arbitrary constants; p and q are given by (56) and (57).Whenq≠1, we get f(ξ)=plogξ+(1-q)∫‍dξ/(ξ+c1ξq)+c2. Thus, the exact solution to (2) is
(59)u(x,t)=c2xpexp[(1-q)∫‍dxx+c1xq+vt],
where c1 and c2 are arbitrary constants and p and q are given by (56) and (57).


### 4.2. Exact Analytic Solution to (2)

Through the transformation technique, we solve the Riccati equation (54), so the exact solutions to (2) are obtained. But for the other equations such as (49), (50), and (53), we cannot get the exact solutions by such special transformation technique. However, we know that the power series can be used to solve nonlinear ODEs, including many complicated differential equations with nonconstant coefficients [10–13, 19, 20]. Now, we consider the power series solution to the reduced equation (49). Letting f′=y, we get the following Riccati equation:
(60)αy′+αy2-(α-β)y+vξ+γ=0.We will seek a solution of (60) in a power series of the form
(61)y=∑n=0∞‍cnξn=p+∑n=1∞‍cnξn,p=c0,
where the coefficients cn (n=0,1,2,…) are constants to be determined.Substituting (61) into (60) and comparing coefficients, we obtain
(62)c1=-p2+α-βαp-γα,c2=-pc1+α-β2αc1-v2α.
Generally, for n≥2, we have
(63)cn+1=1(n+1)α[(α-β)cn-α∑k=0n‍ckcn-k].Thus, for arbitrarily choosing the parameterc0, from (62), we can get c1 and c2. Furthermore, in view of (63), we have
(64)c3=α-β3αc2-13(2pc2+c12),c4=α-β4αc3-12(pc3+c1c2),
and so on.Therefore, the other terms of the sequence{cn}n=0∞ can be determined successively from (63) in a unique manner. This implies that for (60) there exists a power series solution (61) with the coefficients given by (62) and (63). Furthermore, we can show the convergence of the power series solution (61) with the coefficients given by (62) and (63) (see, e.g., [10, 12, 13, 19]); the details are omitted here. So, this solution (61) to (60) is an exact analytic solution.Hence, the exact power series solution to (49) can be written as follows:
(65)f(ξ)=c+pξ+12c1ξ2+13c2ξ3+∑n=2∞‍1n+2cn+1ξn+2.Substituting (65) into (39), we obtain the exact analytic solution to (2) as follows:
(66)u(x,t)=qexp[23p(logx-αvt2)22222222+12c1(logx-αvt2)222222222+13c2(logx-αvt2)322222222+∑n=2∞‍1n+2cn+1(logx-αvt2)n+200000000+12(α-β)vt2-23αv2t3+vtlogx],
where p=c0 and q are arbitrary constants and the other coefficients cn (n=1,2,…) are given by (62) and (63) successively.Similarly, we can give the exact power series solution to (50) in the power series form (61). So, the exact analytic solution to (3) is obtained. The details are omitted here.


### 4.3. Exact Analytic Solution to (3)

In Section4.2, we construct the exact analytic solution to (49) by the power series method and obtain the exact analytic solution to (2). Now, we consider (53). Firstly, let f′=y; then we get the following Riccati type of equation:
(67)αξ2y′+αξ2y2+βξy+γξ+v=0.We will seek a solution of (67) in a generalized power series of the form
(68)y=Aξ-1+∑n=0∞cnξn,
where the parameters A and cn (n=0,1,2,…) are constants to be determined.Substituting (68) into (67) and comparing coefficients, we obtain
(69)A=(α-β)±Δ2α,
where Δ=(α-β)2-4αv, and
(70)c0=-γ2αA+β,2αA+β≠0.Generally, forn≥0, we have
(71)cn+1=-α(n+1)α+2αA+β∑k=0n‍ckcn-k,n=0,1,2,….Thus, from (69) and (70), we can get A and c0. Furthermore, in view of (71), we have
(72)c1=-αc02α+2αA+β,c2=-2αc0c12α+2αA+β,c3=-α(2c0c2+c12)3α+2αA+β,
and so on (see Remark 3).Therefore, the other terms of the sequence{cn}n=0∞ can be determined successively from (71) in a unique manner. This implies that for (67) there exists a generalized power series solution (68) with the coefficients given by (69)–(71). The convergence of the generalized power series solution (68) to (67) is similar to that in Section 4.2; we omit it in this paper. Thus, the power series solution (68) to (67) is also an exact analytic solution.Hence, the power series solution of (53) can be written as follows:
(73)f(ξ)=c-+Alog|ξ|+c0ξ+12c1ξ2+∑n=1∞‍1n+2cn+1ξn+2.Substituting (73) into (52), we get the exact analytic solution to (3) as follows:
(74)u(x,t)=cxAexp[∑n=1∞‍c0x+12c1x21111111111111111+∑n=1∞‍1n+2cn+1xn+2+vt],
where c is an arbitrary constant and A and cn (n=0,1,2,…) are given by (69)–(71) successively.Remark 1.
We note that the generalized power series solution (68) differs from the regular form (61) since A≠0 in (68). In other words, there is no exact power series solution of the form (61) for (67). In particular, the determination of parameter A depends on the equation greatly (cf. [10, 11] for details).


## 4.1. Exact Solution to (2)

Firstly, we consider the ODE (42). Letting f′=y, we get the Riccati equation
(54)dydξ=-y2-βαξy-v+γαξ2.
Now, we solve the equation by the transformation technique directly. Suppose that (54) has the solution of the form
(55)y=pξ-1,
where p is a constant to be determined. Substituting (55) into (54), we have αp2-(α-β)p+v+γ=0. Solving the algebraic equation, we get
(56)p=(α-β)±Δ2α,
where Δ=(α-β)2-4α(v+γ).Settingy=z+pξ-1 and plugging it into (54), we get
(57)dzdξ=-z2-qzξ,q=2p+βα.
This is a Bernoulli equation. Solving the equation, we have the following results.Whenq=1, we get f(ξ)=plogξ+log(logξ+c1)+c2. Thus, the exact solution to (2) is
(58)u(x,t)=c2xp(logx+c1)evt,
where c1 and c2 are arbitrary constants; p and q are given by (56) and (57).Whenq≠1, we get f(ξ)=plogξ+(1-q)∫‍dξ/(ξ+c1ξq)+c2. Thus, the exact solution to (2) is
(59)u(x,t)=c2xpexp[(1-q)∫‍dxx+c1xq+vt],
where c1 and c2 are arbitrary constants and p and q are given by (56) and (57).


## 4.2. Exact Analytic Solution to (2)

Through the transformation technique, we solve the Riccati equation (54), so the exact solutions to (2) are obtained. But for the other equations such as (49), (50), and (53), we cannot get the exact solutions by such special transformation technique. However, we know that the power series can be used to solve nonlinear ODEs, including many complicated differential equations with nonconstant coefficients [10–13, 19, 20]. Now, we consider the power series solution to the reduced equation (49). Letting f′=y, we get the following Riccati equation:
(60)αy′+αy2-(α-β)y+vξ+γ=0.We will seek a solution of (60) in a power series of the form
(61)y=∑n=0∞‍cnξn=p+∑n=1∞‍cnξn,p=c0,
where the coefficients cn (n=0,1,2,…) are constants to be determined.Substituting (61) into (60) and comparing coefficients, we obtain
(62)c1=-p2+α-βαp-γα,c2=-pc1+α-β2αc1-v2α.
Generally, for n≥2, we have
(63)cn+1=1(n+1)α[(α-β)cn-α∑k=0n‍ckcn-k].Thus, for arbitrarily choosing the parameterc0, from (62), we can get c1 and c2. Furthermore, in view of (63), we have
(64)c3=α-β3αc2-13(2pc2+c12),c4=α-β4αc3-12(pc3+c1c2),
and so on.Therefore, the other terms of the sequence{cn}n=0∞ can be determined successively from (63) in a unique manner. This implies that for (60) there exists a power series solution (61) with the coefficients given by (62) and (63). Furthermore, we can show the convergence of the power series solution (61) with the coefficients given by (62) and (63) (see, e.g., [10, 12, 13, 19]); the details are omitted here. So, this solution (61) to (60) is an exact analytic solution.Hence, the exact power series solution to (49) can be written as follows:
(65)f(ξ)=c+pξ+12c1ξ2+13c2ξ3+∑n=2∞‍1n+2cn+1ξn+2.Substituting (65) into (39), we obtain the exact analytic solution to (2) as follows:
(66)u(x,t)=qexp[23p(logx-αvt2)22222222+12c1(logx-αvt2)222222222+13c2(logx-αvt2)322222222+∑n=2∞‍1n+2cn+1(logx-αvt2)n+200000000+12(α-β)vt2-23αv2t3+vtlogx],
where p=c0 and q are arbitrary constants and the other coefficients cn (n=1,2,…) are given by (62) and (63) successively.Similarly, we can give the exact power series solution to (50) in the power series form (61). So, the exact analytic solution to (3) is obtained. The details are omitted here.


## 4.3. Exact Analytic Solution to (3)

In Section4.2, we construct the exact analytic solution to (49) by the power series method and obtain the exact analytic solution to (2). Now, we consider (53). Firstly, let f′=y; then we get the following Riccati type of equation:
(67)αξ2y′+αξ2y2+βξy+γξ+v=0.We will seek a solution of (67) in a generalized power series of the form
(68)y=Aξ-1+∑n=0∞cnξn,
where the parameters A and cn (n=0,1,2,…) are constants to be determined.Substituting (68) into (67) and comparing coefficients, we obtain
(69)A=(α-β)±Δ2α,
where Δ=(α-β)2-4αv, and
(70)c0=-γ2αA+β,2αA+β≠0.Generally, forn≥0, we have
(71)cn+1=-α(n+1)α+2αA+β∑k=0n‍ckcn-k,n=0,1,2,….Thus, from (69) and (70), we can get A and c0. Furthermore, in view of (71), we have
(72)c1=-αc02α+2αA+β,c2=-2αc0c12α+2αA+β,c3=-α(2c0c2+c12)3α+2αA+β,
and so on (see Remark 3).Therefore, the other terms of the sequence{cn}n=0∞ can be determined successively from (71) in a unique manner. This implies that for (67) there exists a generalized power series solution (68) with the coefficients given by (69)–(71). The convergence of the generalized power series solution (68) to (67) is similar to that in Section 4.2; we omit it in this paper. Thus, the power series solution (68) to (67) is also an exact analytic solution.Hence, the power series solution of (53) can be written as follows:
(73)f(ξ)=c-+Alog|ξ|+c0ξ+12c1ξ2+∑n=1∞‍1n+2cn+1ξn+2.Substituting (73) into (52), we get the exact analytic solution to (3) as follows:
(74)u(x,t)=cxAexp[∑n=1∞‍c0x+12c1x21111111111111111+∑n=1∞‍1n+2cn+1xn+2+vt],
where c is an arbitrary constant and A and cn (n=0,1,2,…) are given by (69)–(71) successively.Remark 1.
We note that the generalized power series solution (68) differs from the regular form (61) since A≠0 in (68). In other words, there is no exact power series solution of the form (61) for (67). In particular, the determination of parameter A depends on the equation greatly (cf. [10, 11] for details).


## 5. Further Discussion about the General Bond Pricing Type of (1)

In the above sections, we considered the symmetries, symmetry reductions, and exact solutions to the general bond pricing type of equation for the casesν=0 and ν=1, which are the common forms in many practical applications, such as in financial mathematics. In this section, we discuss the generalized bond pricing type of equation of the form
(75)ut+αx2uxx+βxux+γxνu=0,
where ν≠0,1 is an arbitrary positive number. Firstly, by the group classification method, we get the geometric vector field of (75) as follows:
(76)V1=∂t,V2=u∂u,Vs=s∂u,
where the function s=s(x,t) satisfies (75).Moreover, through the similarity transformation (42), we can reduce this equation to the following equation (ODE):
(77)αξ2f′′+αξ2f′2+βξf′+γξν+v=0,
where f′=df/dξ. Similarly, we can consider the symmetry reductions and exact solutions to the equation. Now, as an example, we study the special case ν=2. In this case, we have
(78)ut+αx2uxx+βxux+γx2u=0.
Referring to (77) and setting f′=y, then we get the following reduced ODE of (78):
(79)αξ2y′+αξ2y2+βξy+γξ2+v=0.Suppose that (79) has the power series solution of the generalized form (68). Then, substituting (68) into (79) and comparing coefficients, we obtain
(80)A=(α-β)±Δ2α,
where Δ=(α-β)2-4αv,
(81)(2αA+β)c0=0,(82)c1=-αc02-γα+2αA+β.Generally, forn≥1, we have
(83)cn+1=-α(n+1)α+2αA+β∑k=0n‍ckcn-k,n=1,2,….In view of (81), we have two special cases as follows.When2αA+β≠0, from (81), we have c0=0. Furthermore, from (82) and (83), we have
(84)c1=-γα+2αA+β,c2=0,c3=-αc123α+2αA+β,c4=0,
and so on. In this case, by induction method, we have
(85)c2n=0,n=0,1,2,….When2αA+β=0, from (81), we get that c0 is an arbitrary constant. Furthermore, from (82) and (83), we have
(86)c1=-αc02-γα+2αA+β,c2=-2αc0c12α+2αA+β,c3=-α(2c0c2+c12)3α+2αA+β,c4=-2α(c0c3+c1c2)4α+2αA+β,
and so on (see Remark 3).Thus, the exact power series solutions to (79) are obtained. In view of (42), the exact analytic solutions to (78) are provided in power series form, respectively. More generally, for ν is an arbitrary positive integer, the exact power series solutions to (75) can be considered similarly by the generalized power series method; the details are omitted here.


## 6. Conclusion and Remarks

In this paper, we investigate the symmetry classifications and exact solutions to the bond pricing types of equations by the combination of Lie symmetry analysis and the generalized power series method; all of the exponentiated solutions and similarity solutions are obtained explicitly for the first time in the literature. Furthermore, for the generalized bond pricing type of equation, the vector field and exact solutions are provided simultaneously. These similarity solutions possess significant features in both financial problems and physical applications. On the other hand, it is known that tackling exact solutions to the vc-PDEs is a difficult problem; from the above discussion, we can see that the combination of Lie symmetry analysis and generalized power series method is a feasible approach and is worthy of further study.Remark 2.
Since there is no space translation(x,t,u)→(x+ϵ,t,u), the bond pricing equations have no traveling wave solutions. However, based on the exponentiated solutions, we can consider the other types of solutions, such as the fundamental solutions and sometimes iterative solutions [3–5, 10].Remark 3.
In general, we cannot get the exact explicit solutions to the nonlinear equations such as (49), (53), and (79) by the classical analysis method. To tackle these equations, the generalized power series method and special techniques are necessary sometimes. For getting the exact analytic solutions in Sections 4.3 and 5, the condition (n+1)α+2αA+β≠0 is necessary for n=0,1,2,….



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