# Linearizability of Nonlinear Third-Order Ordinary Differential Equations by Using a Generalized Linearizing Transformation

Authors: E. Thailert; S. Suksern
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We discuss the linearization problem of third-order ordinary differential equation under the generalized linearizing transformation. We identify the form of the linearizable equations and the conditions which allow the third-order ordinary differential equation to be transformed into the simplest linear equation. We also illustrate how to construct the generalized linearizing transformation. Some examples of linearizable equation are provided to demonstrate our procedure.


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## Body

## 1. Introduction

There has been major interest in the nonlinear problems, since most equations are inherently nonlinear in nature. In general, the nonlinear problems are very difficult to solve explicitly. It is of interest to provide general criteria for the linearizability of nonlinear ordinary differential equations, as they can then be reduced to easily solvable equations. Therefore, the approach of investigating nonlinear ordinary differential equations via transforming to simpler ordinary differential equations becomes important and has been quite plentiful in analysis of physical problems. This includes the classical linearization problem of finding transformations that linearize a given ordinary differential equation. The linearization problem has been studied in many aspects. A short review can be found in [1, 2]. The tools commonly used for solving the linearization problem are the transformations such as point transformation, contact transformation, reduction of order, differential substitution, and generalized Sundman transformation. For this paper, we employ the extension of the generalized Sundman transformations.The linearization problem for a second-order ordinary differential equation was investigated with respect to a generalized Sundman transformation(1)X=F(t,x),dT=G(t,x)dt
by Duarte et al. [3] earlier. They obtained the form of the linearizable equations and the conditions which allow the second-order ordinary differential equation to be transformed into the free particle equation. A characterization of these equations that can be linearized by means of generalized Sundman transformations in terms of first integral and procedure for construction of linearizing transformations has been given by Muriel and Romero [4]. In [5], Mustafa et al. gave a new characterization of linearizable equations in terms of the coefficients of ordinary differential equations and one auxiliary function. In [6], Nakpim and Meleshko pointed out that the solution of the linearization problem for a second-order ordinary differential equation via the generalized Sundman transformation considered earlier by Duarte et al. [3] using the Laguerre form is not complete.The linearization problem for a third-order ordinary differential equation was also investigated with respect to a generalized Sundman transformation [7, 8]. Criteria for a third-order ordinary differential equation to be equivalent to the linear equation X′′′(T)=0 with respect to a Sundman transformation were presented in [8]. The generalized Sundman transformation was also applied for obtaining necessary and sufficient conditions for a third-order ordinary differential equation to be equivalent to a linear equation in the Laguerre form [6]. Some applications of the generalized Sundman transformation to ordinary differential equations were considered in [9] and earlier papers, summarized in the book [10].The linearization problem of a fourth-order ordinary differential equation with respect to generalized Sundman transformations was studied in [11]. They found the necessary and sufficient conditions which allow the fourth-order ordinary differential equation to be transformed into the simplest linear equation.In this work, we expose a more general transformation, that is, the extension of the generalized Sundman transformation(2)X=F(t,x),dT=G(t,x,x′)dt.
This transformation was studied in [12–14] where they designated the transformation as thegeneralized linearizing transformation. They showed that this transformation can be utilized to linearize a wider class of nonlinear ordinary differential equations and, in particular, certain equations which cannot be linearized by the nonpoint and invertible point transformations. If the function G in (2) is independent of the variable x′, then it becomes a nonpoint transformation (vide (1)). On the other hand, if G is a differentiable function, then it becomes an invertible point transformation. So (2) is a unified transformation as it includes nonpoint and invertible point transformations as special cases. An example of an equation which can be linearized by a transformation of the form (2) is given in [13]. It is worth noting that any second-order equation x′′=f(t,x,x′) can be transformed by a transformation (2) into the free particle equation and that this is not so for third-order ordinary differential equations. Hence, the linearization problem using generalized linearizing transformations becomes interesting for ordinary differential equations of order greater than 2. In [12], the authors applied a particular class of transformations (2), where the function G(t,x,x′) is linear with respect to x′.We are now paying attention to the case whereG is a polynomial function in x′ and in particular where it is linear in x′ with coefficients which are arbitrary functions of t and x. To be specific, we focus here on the case
(3)X=F(t,x),dT=(G1(t,x)x′+G2(t,x))dt.
Notice that for the case G1=0, the generalized linearizing transformation becomes a generalized Sundman transformation, so that we assume G1≠0.The paper is organized as follows. In Section2, the necessary conditions of linearization of a third-order ordinary differential equation are presented. In Section 3, we get the theorems that yield criteria for a third-order ordinary differential equation to be linearizable via generalized linearizing transformations. Examples which illustrate the procedure of using the linearization theorems are presented in Section 4.


## 2. Necessary Conditions of Linearization

Here we consider a nonlinear third-order ordinary differential equation(4)x′′′=f(t,x,x′,x′′).
Our goal in this section is to describe all equations (4) which are equivalent with respect to generalized linearizing transformations
(5)X=F(t,x),dT=(G1(t,x)x′+G2(t,x))dt
to a linear equation
(6)X′′′(T)=0.
We begin with investigating the necessary conditions for linearization, that is, the general form of third-order equation (4) that can be obtained from a linear equation (6) by any generalized linearizing transformation (5).Applying a generalized linearizing transformation (5), one obtains the following transformation of the third-order derivatives:
(7)X′(T)=DtFG1x′+G2=Ft+x′FxG1x′+G2=P(t,x,x′),X′′(T)=DtPG1x′+G2=Pt+x′Px+x′′Px′G1x′+G2=-((G2xx′+G1x′′+G2t+G1xx′2+G1tx′)Ft-(Ftt+Fxxx′2+2Ftxx′)(G1x′+G2)+(G2xx′2-G2x′′+G2tx′+G1xx′3+G1tx′2)Fx)×(G1x′+G2)-3=Q(t,x,x′,x′′),X′′′(T)=((G1x′+G2)-5(3G2x2x′3+G22x′′′+3G2t2x′+3G1x2x′5+3G1t2x′3+(x′x′′′-3x′′2)G1G2+2(G1x′-2G2)G2xx′x′′-(G2tt+G2xxx′2+2G2txx′+G1xxx′3+G1ttx′+2G1txx′2)(G1x′+G2)x′+3((G1x′-G2)x′′+2G2xx′2)G2t+6(G2xx′2-G2x′′+G2tx′)G1xx′2+((G1x′-5G2)x′′+6G2xx′2+6G2tx′+6G1xx′3)G1tx′)Fx-(3(G2t+G2xx′+G1xx′2+G1tx′+G1x′′)Ftt-((3Fttx+Fxxxx′2)x′+Fttt+3Ftxxx′2)×(G1x′+G2)+3((G2t+G2xx′+G1xx′2+G1tx′)x′-G2x′′)×Fxxx′+3(2(G2t+G2xx′+G1xx′2+G1tx′)x′+(G1x′-G2)x′′)Ftx)(G1x′+G2)-((G1x′+G2)-5(G2tt+G2xxx′2+2G2txx′+G1xxx′3+G1ttx′+2G1txx′2)×(G1x′+G2)-(3G2x2x′2-G1G2x′′′+3G2t2+3G1x2x′4+3G1t2x′2-(x′x′′′-3x′′2)G12)-(5G1x′-G2)G2xx′′-6(G2xx′+G1x′′)G2t-3((G1x′-G2)x′′+2G2xx′2+2G2tx′)×G1xx′-2((G1x′+G2)-5(2G1x′-G2)x′′+3G2xx′2+3G2tx′+3G1xx′3(G1x′+G2)-5)G1t(G1x′+G2)-5)Ft(G1x′+G2)-5)×(G1x′+G2)-5=R(t,x,x′,x′′,x′′′),
where Dt=∂/∂t+x′(∂/∂x)+x′′(∂/∂x′)+x′′′(∂/∂x′′)+⋯ is a total of derivatives. Substituting the resulting expression in linear equation (6) and setting r=G2/G1,K=Ft-Fxr, we arrive at the following equation:
(8)x′′′+1x′+r×[-3x′′2+(A2x′2+A1x′+A0)x′′+B5x′5+B4x′4+B3x′3+B2x′2+B1x′+B0]=0,
where Ai(i=0,1,2) and Bj(j=0,1,…,5) are functions of t and x determined as follows:
(9)A2=(3((Ftx-Fxxr)G1-FtG1x)+(2(2G1xr-rxG1)-G1t)Fx)/(KG1),A1=-((KG12)-1(2G1xr+5rxG1+4G1t)Ft-3(Ftt-Fxxr2)G1+((KG12)-1(3rt-4rxr)G1-4G1xr2-2G1tr(KG12)-1)Fx(KG12)-1)/(KG1),(10)A0=-((KG12)-13FtxG1r2-3FttG1r+4FtG1tr-FtG1xr2+6FtrtG1-FtrxG1r-3FxG1tr2-3FxrtG1r(KG12)-1)/(KG1),(11)B5=-((FxxxG1-3FxxG1x)G1-(G1xxG1-3G1x2)Fx)(KG12),(12)B4=(3(G1t+2G1xr+rxG1)FxxG1+(G1xxG1-3G1x2)Ft+((KG12)-12G1txG1-6G1tG1x+2G1xxG1r-6G1x2r-4G1xrxG1+rxxG12(KG12)-1)Fx+(2(3FtxG1x-FxxxG1r)-3FtxxG1)G1)×(KG12)-1,(13)B3=(((KG12)-13FttG1x-FxxxG1r2-3FttxG1-6FtxxG1r+6(G1t+2G1xr+rxG1)Ftx(KG12)-1)G1+((KG12)-12G1txG1-6G1tG1x+2G1xxG1r-6G1x2r-4G1xrxG1+rxxG12(KG12)-1)Ft+3((G1xr+rxG1+G1t)r+G1tr+rtG1)FxxG1+((KG12)-1(2G1txr+G1tt)G1-3(G1xr+rxG1)2-6(G1tr+rtG1)G1x-3(2(G1xr+rxG1)+G1t)G1t+(2G1xrx+rxxG1+G1xxr)G1r+2(G1xrt+rtxG1+G1trx+G1txr)G1(KG12)-1)Fx(KG12)-1)×(KG12)-1,(14)B2=(((KG12)-1(2G1txr+G1tt)G1-3(G1xr+rxG1)2-6(G1tr+rtG1)G1x-3(2(G1xr+rxG1)+G1t)G1t+(2G1xrx+rxxG1+G1xxr)G1r+2(G1xrt+rtxG1+G1trx+G1txr)G1(KG12)-1)Ft-(((KG12)-16(G1xr+rxG1+G1t)×(G1tr+rtG1)-G1ttG1r-(2G1trt+rttG1+G1ttr)G1-2(G1xrt+rtxG1+G1trx+G1txr)G1r(KG12)-1)Fx+((KG12)-1(3Ftxxr2+Fttt)G1-3(G1t+2G1xr+rxG1)Ftt-3((G1tr+rtG1)Fxx-2FttxG1)r-6((G1xr+rxG1+G1t)r+G1tr+rtG1)×Ftx(KG12)-1)G1(KG12)-1))×(KG12)-1,(15)B1=-((KG12)-1(6(G1xr+rxG1+G1t)(G1tr+rtG1)-G1ttG1r-(2G1trt+rttG1+G1ttr)G1-2(G1xrt+rtxG1+G1trx+G1txr)G1r)Ft-((KG12)-1(3((G1xr+rxG1+G1t)r+G1tr+rtG1)Ftt-((2Fttt+3Fttxr)G1-6(G1tr+rtG1)Ftx)r)G1+((KG12)-1(2G1trt+rttG1+G1ttr)G1r-3(G1tr+rtG1)2(KG12)-1)Fx(KG12)-1))×(KG12)-1,B0=((KG12)-1((KG12)-1(2G1trt+rttG1+G1ttr)G1r-3(G1tr+rtG1)2(KG12)-1)Ft+(3(G1tr+rtG1)Ftt-FtttG1r)G1r(KG12)-1)/(KG12).
Thus, we proved the theorem.Theorem 1.
Any third-order ordinary differential equation (4) obtained from a linear equation (6) by a generalized linearizing transformation (5) has to be in the form (8).


## 3. Formulation of the Linearization Theorem

We have shown in the previous section that every linearizable third-order ordinary differential equation belongs to the class of equations (8). In this section, we formulate the main theorems containing necessary and sufficient conditions for linearization as well as the methods for constructing the linearizing transformations.For obtaining sufficient conditions, one has to solve the compatibility problem. Consider the representations of the coefficientsAi and Bi through the unknown functions F and G1. According to our notation K=Ft-Fxr, we define the derivative Ft as
(16)Ft=Fxr+K.
From (9), one can find the derivatives
(17)Kx=(FxG1t-FxG1xr-FxrxG1+3G1xK+A2G1K)(3G1),Kt=((3FxxG1xG1+FxG1xxG1-3FxG1x2-B5G12K)G12FxG1tr-FxG1xr2-FxrxG1r+4G1tK-G1xKr+G1K(5rx+A1-A2r)(3FxxG1xG1+FxG1xxG1-3FxG1x2-B5G12K)G12)/(3G1).
From (10), one obtains the condition
(18)rt=(6rxr-A0+A1r-A2r2)6.
Equation (11) defines the derivative
(19)Fxxx=(3FxxG1xG1+FxG1xxG1-3FxG1x2-B5G12K)G12.
So that equation (12) becomes
(20)6FxxG1tG1-6FxxG1xG1r-6FxxrxG12+3FxG1txG1-12FxG1tG1x-FxG1tA2G1-3FxG1xxG1r+12FxG1x2r+FxG1xG1(6rx+A2r)+FxG12(-3rxx+rxA2)-6G1xxG1K+9G1x2K+G12K(-3A2x-A22-3B4+15B5r)=0.
The compatibility analysis depends on the value of Fx. A complete study of all cases is given here.


### 3.1. CaseFx=0

In this case, the forms of derivativesFt,Kx, and Kt become
(21)Ft=K,Kx=(3G1x+A2G1)K(3G1),Kt=(4G1t-G1xr+G1(5rx+A1-A2r))K(3G1).
Substituting Fx into Fxxx, one arrives at the condition
(22)B5=0.
Comparing the mixed derivatives (Fx)t=(Ft)x, one gets the derivative
(23)G1x=-(A2G1)3.
In this case, (Fxxx)t=(Ft)xxx is satisfied. Equations (12) and (13) give the conditions
(24)A2x=(-2A22-9B4)3,rxx=(-9A1x+6A2t+3rxA2-3A1A2-2A22r-9B3)36.
Comparing the mixed derivative (Kt)x=(Kx)t, one obtains the condition
(25)A1x=(-6A2t-3rxA2-5A1A2+2A22r-15B3+24B4r)3.
Equation (14) provides the derivative
(26)G1tt=(2250G1t2+150G1tG1h1+G12h2)(1350G1),
where
(27)h1=15rx+3A1-2A2r,h2=-225A0x-1350A1t-1350A2tr-1050A0A2-477A12+516A1A2r+33A1h1-432A22r2-57A2h1r-4050B2+4275B3r-4275B4r2-8h12.
The relation (rx)x=rxx gives the condition
(28)h1x=4A2t.
Comparing the mixed derivative (G1tt)x=(G1x)tt, one arrives at the condition
(29)A2tt=(50A2th1-h2x)450.
Solving (15), one finds the conditions
(30)A0t=(15930A1tr+15930A2tr2-1260h1xr-1575A0A1+11970A0A2r+5517A12r-5697A1A2r2-558A1h1r+4986A22r3+504A2h1r2-8100B1+48600B2r-48600B3r2+48600B4r3+148h12r+8h2r)/1350,(31)B0=(-3240A1tr2-3240A2tr3+180h1xr2-135A02+270A0A1r-2430A0A2r2-1107A12r2+1134A1A2r3+108A1h1r2-999A22r4-108A2h1r3+1620B1r-9720B2r2+9720B3r3-9720B4r4-28h12r2-2h2r2)/1620.


### 3.2. CaseFx≠0

From (20) and (13), one obtains the derivatives
(32)G1tx=((3A2x+A22+3B4-15B5r)-6FxxG1tG1+6FxxG1xG1r+6FxxrxG12+12FxG1tG1x+FxG1tA2G1+3FxG1xxG1r-12FxG1x2r+FxG1xG1(-6rx-A2r)+FxG12(3rxx-rxA2)+6G1xxG1K-9G1x2K+G12K(3A2x+A22+3B4-15B5r))×(3FxG1)-1,(33)G1tt=(G1x2K2+4G12K2(3A2x+A22+3B4-15B5r)-24FxxFxG1tG1r+24FxxFxG1xG1r2+24FxxFxrxG12r-24FxxG1tG1K+24FxxG1xG1Kr+24FxxrxG12K+14Fx2G1t2+20Fx2G1tG1xr+2Fx2G1tG1(rx+A1)+6Fx2G1xxG1r2-34Fx2G1x2r2+Fx2G1xG1(-26rxr-A0-A1r-A2r2)+Fx2G12(-A0x+A1xr-A2xr2+12rxxr-4rx2-rxA1-2rxA2r)+24FxG1tG1xK+24FxG1xxG1Kr-60FxG1x2Kr-24FxG1xrxG1K+2FxG12K(3A1x+18rxx-3rxA2+A1A2+3B3-6B4r)+24G1xxG1K2-36G1x2K2+4G12K2(3A2x+A22+3B4-15B5r))×(6Fx2G1)-1.
Comparing the mixed derivative (Kt)x=(Kx)t, one obtains
(34)G1xx=(G12K2(-3A2x-A22-3B4+15B5r)6FxxG1tG1K-6FxxG1xG1Kr-6FxxrxG12K-Fx2G1t2+2Fx2G1tG1xr+2Fx2G1trxG1-Fx2G1x2r2-2Fx2G1xrxG1r-Fx2rx2G12-6FxG1tG1xK+6FxG1x2Kr+6FxG1xrxG1K+FxG12K((6G1K2)-1-3A2t+3A2xr-A1A2+2A22r-3B3+12B4r-30B5r2(6G1K2)-1)+9G1x2K2+G12K2(-3A2x-A22-3B4+15B5r))×(6G1K2)-1.
Equation (14) becomes
(35)Fxs1+2Ks2=0,
where
(36)s1=-6A1t+6A1xr+12A2tr-12A2xr2-5A0A2-2A12+13A1A2r-13A22r2-18B2+54B3r-108B4r2+180B5r3,s2=-3A1x+6A2t-18rxx+3rxA2+A1A2-2A22r+3B3-12B4r+30B5r2.
Further analysis of the compatibility depends on value of s1 in (35): it is separated into two cases; that is, s1=0 and s1≠0.


#### 3.2.1. Cases1≠0

From (35), one finds
(37)Fx=-(2Ks2)s1.
Since this case Fx≠0, then s2≠0 too. Comparing the mixed derivatives (Fx)t=(Ft)x, one gets the derivative
(38)G1t=(3G1xs1(2rs2-s1)+G1s3)(6s1s2),
where
(39)s3=-6rxs1s2+6s1ts2-6s1xrs2-6s2ts1+6s2xrs1-2A1s1s2+4A2rs1s2-A2s12.
Substituting Fx into Fxxx, G1t into G1tx and G1tt, one arrives at the conditions
(40)s2xx=(-12A2ts12s22+12A2xrs12s22-6A2xs13s2+36rxs1xs1s22-36rxs2xs12s2-12rxA2s12s22+18s1xxs12s2-36s1x2s1s2+36s1xs2xs12+12s1xA2s12s2-6s1xs2s3-12s2xA2s13+6s2xs1s3-4A1A2s12s22+8A22rs12s22-2A22s13s2+2A2s1s2s3-12B3s12s22+48B4rs12s22-120B5r2s12s22+9B5s14)×(18s13)-1,s3x=(-6A1xs13s2-6A2ts13s2+18A2xrs13s2-9A2xs14+36rx2s12s22-36rxs1xs12s2+36rxs2xs13+6rxA2s13s2-12rxs1s2s3+12s1xs1s3-4A1A2s13s2+8A22rs13s2-3A22s14-12B3s13s2+48B4rs13s2-9B4s14-120B5r2s13s2+45B5rs14-2s13s22+s32)/(6s12),s3t=(-6A0xs13s2-3A1xs14-6A2trs13s2-12A2ts14+12A2xr2s13s2+9A2xrs14+36rx2rs12s22-36rx2s13s2-36rxs1xrs12s2+36rxs2xrs13-6rxA1s13s2+18rxA2rs13s2-3rxA2s14-12rxrs1s2s3+12s1ts1s3-4A1A2rs13s2-5A1A2s14+8A22r2s13s2+7A22rs14-12B3rs13s2-15B3s14+48B4r2s13s2+51B4rs14-120B5r3s13s2-105B5r2s14-2rs13s22+rs32+5s14s2)/(6s12).
Equation (15) provides the conditions
(41)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3-rs1)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
Comparing the mixed derivatives (G1tt)x=(G1tx)t, (G1xx)t=(G1tx)x, and (Fxxx)t=(Ft)xxx, one gets the conditions
(42)A1xx=(-33A1xA2s12-18A2txs12-108A2trxs1s2+24A2tA2s12+18A2ts3+54A2xxrs12+108A2xrxrs1s2+18A2xrxs12-30A2xA1s12+102A2xA2rs12-18A2xrs3-90B3xs12+54B4ts12+306B4xrs12-270B5trs12-630B5xr2s12-36rxA1A2s1s2+72rxA22rs1s2+27rxA22s12-108rxB3s1s2+432rxB4rs1s2+252rxB4s12-1080rxB5r2s1s2-1260rxB5rs12+36rxs1s22-18s1xs1s2+30s2xs12+45A0B5s12-5A1A22s12+6A1A2s3-45A1B5rs12+10A23rs12-12A22rs3-15A2B3s12+60A2B4rs12-105A2B5r2s12+5A2s12s2+18B3s3-72B4rs3+180B5r2s3-6s2s333A1xA2s12)/(18s12),A2tt=(36A2txrs1+72A2trxs1-18A2xxr2s1-72A2xrxrs1-3A2xA0s1+3A2xA1rs1-3A2xA2r2s1-18B3ts1+18B3xrs1+72B4trs1-72B4xr2s1-180B5tr2s1+180B5xr3s1+24rxA1A2s1-48rxA22rs1+72rxB3s1-288rxB4rs1+720rxB5r2s1-6rxs1s2-3s1xs1+3A0A22s1-12A0B4s1+60A0B5rs1+4A12A2s1-19A1A22rs1+6A1B3s1-12A1B4rs1+19A23r2s1+18A2B2s1-66A2B3rs1+144A2B4r2s1-240A2B5r3s1+2A2s12+s3)/(18s1),A2tx=(-6A1xA2s12s2-72A2trxs1s22+90A2ts1xs1s2-90A2ts2xs12-24A2tA2s12s2+12A2ts2s3+18A2xxrs12s2-9A2xxs13+72A2xrxrs1s22+18A2xrxs12s2-90A2xs1xrs1s2+90A2xs2xrs12-6A2xA1s12s2+48A2xA2rs12s2-18A2xA2s13-12A2xrs2s3-18B3xs12s2+72B4xrs12s2-27B4xs13+54B5ts13-180B5xr2s12s2+81B5xrs13-24rxA1A2s1s22+48rxA22rs1s22+12rxA22s12s2-72rxB3s1s22+288rxB4rs1s22+72rxB4s12s2-720rxB5r2s1s22-360rxB5rs12s2+135rxB5s13+30s1xA1A2s1s2-60s1xA22rs1s2+90s1xB3s1s2-360s1xB4rs1s2+900s1xB5r2s1s2-30s2xA1A2s12+60s2xA22rs12-90s2xB3s12+360s2xB4rs12-900s2xB5r2s12-8A1A22s12s2+4A1A2s2s3+18A1B5s13+16A23rs12s2-4A23s13-8A22rs2s3-24A2B3s12s2+96A2B4rs12s2-18A2B4s13-240A2B5r2s12s2+54A2B5rs13+12B3s2s3-48B4rs2s3+120B5r2s2s3)/(18s12s2).


#### 3.2.2. Cases1=0

From (35), one finds the condition
(43)s2=0.
Equation (15) gives the conditions
(44)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
From the mixed derivative (G1xx)t=(G1tx)x, one finds the condition
(45)A2tt=((12rxs4+3s4xr+A1s4-2A2rs4)336A2txr+72A2trx-18A2xxr2-72A2xrxr-3A2xA0+3A2xA1r-3A2xA2r2-18B3t+18B3xr+72B4tr-72B4xr2-180B5tr2+180B5xr3+24rxA1A2-48rxA22r+72rxB3-288rxB4r+720rxB5r2+3A0A22-12A0B4+60A0B5r+4A12A2-19A1A22r+6A1B3-12A1B4r+19A23r2+18A2B2-66A2B3r+144A2B4r2-240A2B5r3(12rxs4+3s4xr+A1s4-2A2rs4)3)/18.
The relation (G1tt)x=(G1tx)t becomes
(46)18(FxG1t-FxG1xr-FxrxG1-G1xK)s4+G1Ks5=0,
where
(47)s4=3A2t-3A2xr+A1A2-2A22r+3B3-12B4r+30B5r2,s5=18A1xx+27A1xA2-36A2xxr+24A2xA1-102A2xA2r+72B3x-54B4t-234B4xr+270B5tr+450B5xr2-15rxA22-180rxB4+900rxB5r+6s4x-45A0B5+13A1A22+45A1B5r-26A23r+39A2B3-156A2B4r+345A2B5r2-8A2s4.
The relation (A2t)t-A2tt=0 provides the condition
(48)s4t=(12rxs4+3s4xr+A1s4-2A2rs4)3.
Further study depends on s4.(i) Cases4≠0From (46), one gets the derivative
(49)g1t=(18(FxG1xr+FxrxG1+G1xK)s4-G1Ks5)(18Fxs4).
Differentiating g1t with respect to x, one obtains the derivative
(50)Fx=(Ks6)(108s43),
where
(51)s6=324A2xs42-36s4xs5+36s5xs4+108A22s42+324B4s42-1620B5rs42-s52.
The relations (Fx)t=(Ft)x,(G1t)t=G1tt,(Fxxx)t=(Ft)xxx,and(Fx)xx=Fxxx provide the conditions
(52)s6t=(30rxs6+3s6xr+2A1s6-4A2rs6+108A2s43+18s42s5)/3,s5t=(-108A1xs42-108A2xrs42+108rxA2s42+180rxs4s5+36s4xrs5-36A1A2s42+12A1s4s5-36A22rs42-24A2rs4s5-108B3s42+108B4rs42+540B5r2s42+rs52+rs6-144s43)/(36s4),A2xx=(-5832A2xA2s43-8748B4xs43+17496B5ts43+26244B5xrs43+43740rxB5s43-126s4xs6+45s6xs4+5832A1B5s43-1296A23s43-5832A2B4s43+17496A2B5rs43+12A2s4s6-s5s6)/(2916s43),s6xx=(-324A2xs42s6+2916s4xxs4s6-11664s4x2s6+5832s4xs6xs4+1944s4xA2s4s6-162s4xs5s6-648s6xA2s42+54s6xs4s5-108A22s42s6+18A2s4s5s6-104976B5s45+s62)/(972s42).(ii)
Cases4=0From (46), one gets the condition
(53)s5=0.
Comparing the mixed derivative (Fxxx)t=(Ft)xxx, one arrives at the condition
(54)A2xx=(-18A2xA2-27B4x+54B5t+81B5xr+135rxB5+18A1B5-4A23-18A2B4+54A2B5r)/9.
All obtained results can be summarized in the following theorems.Theorem 2.
Sufficient conditions for (8) to be linearizable via the generalized linearizing transformation (5) with Fx=0 are equations (18), (22), (24), (25), (28), (29), (30), and (31).Corollary 3.
Provided that the sufficient conditions in Theorem2 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the compatible system of equations (21), (23), and (26) for the functions F(t),G1(t,x), and G2(t,x).Theorem 4.
Sufficient conditions for equation (8) to be linearizable via the generalized linearizing transformation (5) with Fx≠0 are as follows. (a)
Ifs1≠0, then the conditions are (18), (40), (41), and (42).(b)
Ifs1=0,s4≠0, then the conditions are (18), (43), (44), (48), and (52).(c)
Ifs1=0,s4=0, then the conditions are (18), (43), (44), (53), and (54).Corollary 5.
Provided that the sufficient conditions in Theorem4 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the following compatible system of equations for the functions F(t,x),G1(t,x), and G2(t,x): (a)
(16), (17), (34), (37), and (38);(b)
(16), (17), (34), (49), and (50);(c)
(16), (17), (19), (32), (33), and (34).


## 3.1. CaseFx=0

In this case, the forms of derivativesFt,Kx, and Kt become
(21)Ft=K,Kx=(3G1x+A2G1)K(3G1),Kt=(4G1t-G1xr+G1(5rx+A1-A2r))K(3G1).
Substituting Fx into Fxxx, one arrives at the condition
(22)B5=0.
Comparing the mixed derivatives (Fx)t=(Ft)x, one gets the derivative
(23)G1x=-(A2G1)3.
In this case, (Fxxx)t=(Ft)xxx is satisfied. Equations (12) and (13) give the conditions
(24)A2x=(-2A22-9B4)3,rxx=(-9A1x+6A2t+3rxA2-3A1A2-2A22r-9B3)36.
Comparing the mixed derivative (Kt)x=(Kx)t, one obtains the condition
(25)A1x=(-6A2t-3rxA2-5A1A2+2A22r-15B3+24B4r)3.
Equation (14) provides the derivative
(26)G1tt=(2250G1t2+150G1tG1h1+G12h2)(1350G1),
where
(27)h1=15rx+3A1-2A2r,h2=-225A0x-1350A1t-1350A2tr-1050A0A2-477A12+516A1A2r+33A1h1-432A22r2-57A2h1r-4050B2+4275B3r-4275B4r2-8h12.
The relation (rx)x=rxx gives the condition
(28)h1x=4A2t.
Comparing the mixed derivative (G1tt)x=(G1x)tt, one arrives at the condition
(29)A2tt=(50A2th1-h2x)450.
Solving (15), one finds the conditions
(30)A0t=(15930A1tr+15930A2tr2-1260h1xr-1575A0A1+11970A0A2r+5517A12r-5697A1A2r2-558A1h1r+4986A22r3+504A2h1r2-8100B1+48600B2r-48600B3r2+48600B4r3+148h12r+8h2r)/1350,(31)B0=(-3240A1tr2-3240A2tr3+180h1xr2-135A02+270A0A1r-2430A0A2r2-1107A12r2+1134A1A2r3+108A1h1r2-999A22r4-108A2h1r3+1620B1r-9720B2r2+9720B3r3-9720B4r4-28h12r2-2h2r2)/1620.


## 3.2. CaseFx≠0

From (20) and (13), one obtains the derivatives
(32)G1tx=((3A2x+A22+3B4-15B5r)-6FxxG1tG1+6FxxG1xG1r+6FxxrxG12+12FxG1tG1x+FxG1tA2G1+3FxG1xxG1r-12FxG1x2r+FxG1xG1(-6rx-A2r)+FxG12(3rxx-rxA2)+6G1xxG1K-9G1x2K+G12K(3A2x+A22+3B4-15B5r))×(3FxG1)-1,(33)G1tt=(G1x2K2+4G12K2(3A2x+A22+3B4-15B5r)-24FxxFxG1tG1r+24FxxFxG1xG1r2+24FxxFxrxG12r-24FxxG1tG1K+24FxxG1xG1Kr+24FxxrxG12K+14Fx2G1t2+20Fx2G1tG1xr+2Fx2G1tG1(rx+A1)+6Fx2G1xxG1r2-34Fx2G1x2r2+Fx2G1xG1(-26rxr-A0-A1r-A2r2)+Fx2G12(-A0x+A1xr-A2xr2+12rxxr-4rx2-rxA1-2rxA2r)+24FxG1tG1xK+24FxG1xxG1Kr-60FxG1x2Kr-24FxG1xrxG1K+2FxG12K(3A1x+18rxx-3rxA2+A1A2+3B3-6B4r)+24G1xxG1K2-36G1x2K2+4G12K2(3A2x+A22+3B4-15B5r))×(6Fx2G1)-1.
Comparing the mixed derivative (Kt)x=(Kx)t, one obtains
(34)G1xx=(G12K2(-3A2x-A22-3B4+15B5r)6FxxG1tG1K-6FxxG1xG1Kr-6FxxrxG12K-Fx2G1t2+2Fx2G1tG1xr+2Fx2G1trxG1-Fx2G1x2r2-2Fx2G1xrxG1r-Fx2rx2G12-6FxG1tG1xK+6FxG1x2Kr+6FxG1xrxG1K+FxG12K((6G1K2)-1-3A2t+3A2xr-A1A2+2A22r-3B3+12B4r-30B5r2(6G1K2)-1)+9G1x2K2+G12K2(-3A2x-A22-3B4+15B5r))×(6G1K2)-1.
Equation (14) becomes
(35)Fxs1+2Ks2=0,
where
(36)s1=-6A1t+6A1xr+12A2tr-12A2xr2-5A0A2-2A12+13A1A2r-13A22r2-18B2+54B3r-108B4r2+180B5r3,s2=-3A1x+6A2t-18rxx+3rxA2+A1A2-2A22r+3B3-12B4r+30B5r2.
Further analysis of the compatibility depends on value of s1 in (35): it is separated into two cases; that is, s1=0 and s1≠0.


### 3.2.1. Cases1≠0

From (35), one finds
(37)Fx=-(2Ks2)s1.
Since this case Fx≠0, then s2≠0 too. Comparing the mixed derivatives (Fx)t=(Ft)x, one gets the derivative
(38)G1t=(3G1xs1(2rs2-s1)+G1s3)(6s1s2),
where
(39)s3=-6rxs1s2+6s1ts2-6s1xrs2-6s2ts1+6s2xrs1-2A1s1s2+4A2rs1s2-A2s12.
Substituting Fx into Fxxx, G1t into G1tx and G1tt, one arrives at the conditions
(40)s2xx=(-12A2ts12s22+12A2xrs12s22-6A2xs13s2+36rxs1xs1s22-36rxs2xs12s2-12rxA2s12s22+18s1xxs12s2-36s1x2s1s2+36s1xs2xs12+12s1xA2s12s2-6s1xs2s3-12s2xA2s13+6s2xs1s3-4A1A2s12s22+8A22rs12s22-2A22s13s2+2A2s1s2s3-12B3s12s22+48B4rs12s22-120B5r2s12s22+9B5s14)×(18s13)-1,s3x=(-6A1xs13s2-6A2ts13s2+18A2xrs13s2-9A2xs14+36rx2s12s22-36rxs1xs12s2+36rxs2xs13+6rxA2s13s2-12rxs1s2s3+12s1xs1s3-4A1A2s13s2+8A22rs13s2-3A22s14-12B3s13s2+48B4rs13s2-9B4s14-120B5r2s13s2+45B5rs14-2s13s22+s32)/(6s12),s3t=(-6A0xs13s2-3A1xs14-6A2trs13s2-12A2ts14+12A2xr2s13s2+9A2xrs14+36rx2rs12s22-36rx2s13s2-36rxs1xrs12s2+36rxs2xrs13-6rxA1s13s2+18rxA2rs13s2-3rxA2s14-12rxrs1s2s3+12s1ts1s3-4A1A2rs13s2-5A1A2s14+8A22r2s13s2+7A22rs14-12B3rs13s2-15B3s14+48B4r2s13s2+51B4rs14-120B5r3s13s2-105B5r2s14-2rs13s22+rs32+5s14s2)/(6s12).
Equation (15) provides the conditions
(41)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3-rs1)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
Comparing the mixed derivatives (G1tt)x=(G1tx)t, (G1xx)t=(G1tx)x, and (Fxxx)t=(Ft)xxx, one gets the conditions
(42)A1xx=(-33A1xA2s12-18A2txs12-108A2trxs1s2+24A2tA2s12+18A2ts3+54A2xxrs12+108A2xrxrs1s2+18A2xrxs12-30A2xA1s12+102A2xA2rs12-18A2xrs3-90B3xs12+54B4ts12+306B4xrs12-270B5trs12-630B5xr2s12-36rxA1A2s1s2+72rxA22rs1s2+27rxA22s12-108rxB3s1s2+432rxB4rs1s2+252rxB4s12-1080rxB5r2s1s2-1260rxB5rs12+36rxs1s22-18s1xs1s2+30s2xs12+45A0B5s12-5A1A22s12+6A1A2s3-45A1B5rs12+10A23rs12-12A22rs3-15A2B3s12+60A2B4rs12-105A2B5r2s12+5A2s12s2+18B3s3-72B4rs3+180B5r2s3-6s2s333A1xA2s12)/(18s12),A2tt=(36A2txrs1+72A2trxs1-18A2xxr2s1-72A2xrxrs1-3A2xA0s1+3A2xA1rs1-3A2xA2r2s1-18B3ts1+18B3xrs1+72B4trs1-72B4xr2s1-180B5tr2s1+180B5xr3s1+24rxA1A2s1-48rxA22rs1+72rxB3s1-288rxB4rs1+720rxB5r2s1-6rxs1s2-3s1xs1+3A0A22s1-12A0B4s1+60A0B5rs1+4A12A2s1-19A1A22rs1+6A1B3s1-12A1B4rs1+19A23r2s1+18A2B2s1-66A2B3rs1+144A2B4r2s1-240A2B5r3s1+2A2s12+s3)/(18s1),A2tx=(-6A1xA2s12s2-72A2trxs1s22+90A2ts1xs1s2-90A2ts2xs12-24A2tA2s12s2+12A2ts2s3+18A2xxrs12s2-9A2xxs13+72A2xrxrs1s22+18A2xrxs12s2-90A2xs1xrs1s2+90A2xs2xrs12-6A2xA1s12s2+48A2xA2rs12s2-18A2xA2s13-12A2xrs2s3-18B3xs12s2+72B4xrs12s2-27B4xs13+54B5ts13-180B5xr2s12s2+81B5xrs13-24rxA1A2s1s22+48rxA22rs1s22+12rxA22s12s2-72rxB3s1s22+288rxB4rs1s22+72rxB4s12s2-720rxB5r2s1s22-360rxB5rs12s2+135rxB5s13+30s1xA1A2s1s2-60s1xA22rs1s2+90s1xB3s1s2-360s1xB4rs1s2+900s1xB5r2s1s2-30s2xA1A2s12+60s2xA22rs12-90s2xB3s12+360s2xB4rs12-900s2xB5r2s12-8A1A22s12s2+4A1A2s2s3+18A1B5s13+16A23rs12s2-4A23s13-8A22rs2s3-24A2B3s12s2+96A2B4rs12s2-18A2B4s13-240A2B5r2s12s2+54A2B5rs13+12B3s2s3-48B4rs2s3+120B5r2s2s3)/(18s12s2).


### 3.2.2. Cases1=0

From (35), one finds the condition
(43)s2=0.
Equation (15) gives the conditions
(44)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
From the mixed derivative (G1xx)t=(G1tx)x, one finds the condition
(45)A2tt=((12rxs4+3s4xr+A1s4-2A2rs4)336A2txr+72A2trx-18A2xxr2-72A2xrxr-3A2xA0+3A2xA1r-3A2xA2r2-18B3t+18B3xr+72B4tr-72B4xr2-180B5tr2+180B5xr3+24rxA1A2-48rxA22r+72rxB3-288rxB4r+720rxB5r2+3A0A22-12A0B4+60A0B5r+4A12A2-19A1A22r+6A1B3-12A1B4r+19A23r2+18A2B2-66A2B3r+144A2B4r2-240A2B5r3(12rxs4+3s4xr+A1s4-2A2rs4)3)/18.
The relation (G1tt)x=(G1tx)t becomes
(46)18(FxG1t-FxG1xr-FxrxG1-G1xK)s4+G1Ks5=0,
where
(47)s4=3A2t-3A2xr+A1A2-2A22r+3B3-12B4r+30B5r2,s5=18A1xx+27A1xA2-36A2xxr+24A2xA1-102A2xA2r+72B3x-54B4t-234B4xr+270B5tr+450B5xr2-15rxA22-180rxB4+900rxB5r+6s4x-45A0B5+13A1A22+45A1B5r-26A23r+39A2B3-156A2B4r+345A2B5r2-8A2s4.
The relation (A2t)t-A2tt=0 provides the condition
(48)s4t=(12rxs4+3s4xr+A1s4-2A2rs4)3.
Further study depends on s4.(i) Cases4≠0From (46), one gets the derivative
(49)g1t=(18(FxG1xr+FxrxG1+G1xK)s4-G1Ks5)(18Fxs4).
Differentiating g1t with respect to x, one obtains the derivative
(50)Fx=(Ks6)(108s43),
where
(51)s6=324A2xs42-36s4xs5+36s5xs4+108A22s42+324B4s42-1620B5rs42-s52.
The relations (Fx)t=(Ft)x,(G1t)t=G1tt,(Fxxx)t=(Ft)xxx,and(Fx)xx=Fxxx provide the conditions
(52)s6t=(30rxs6+3s6xr+2A1s6-4A2rs6+108A2s43+18s42s5)/3,s5t=(-108A1xs42-108A2xrs42+108rxA2s42+180rxs4s5+36s4xrs5-36A1A2s42+12A1s4s5-36A22rs42-24A2rs4s5-108B3s42+108B4rs42+540B5r2s42+rs52+rs6-144s43)/(36s4),A2xx=(-5832A2xA2s43-8748B4xs43+17496B5ts43+26244B5xrs43+43740rxB5s43-126s4xs6+45s6xs4+5832A1B5s43-1296A23s43-5832A2B4s43+17496A2B5rs43+12A2s4s6-s5s6)/(2916s43),s6xx=(-324A2xs42s6+2916s4xxs4s6-11664s4x2s6+5832s4xs6xs4+1944s4xA2s4s6-162s4xs5s6-648s6xA2s42+54s6xs4s5-108A22s42s6+18A2s4s5s6-104976B5s45+s62)/(972s42).(ii)
Cases4=0From (46), one gets the condition
(53)s5=0.
Comparing the mixed derivative (Fxxx)t=(Ft)xxx, one arrives at the condition
(54)A2xx=(-18A2xA2-27B4x+54B5t+81B5xr+135rxB5+18A1B5-4A23-18A2B4+54A2B5r)/9.
All obtained results can be summarized in the following theorems.Theorem 2.
Sufficient conditions for (8) to be linearizable via the generalized linearizing transformation (5) with Fx=0 are equations (18), (22), (24), (25), (28), (29), (30), and (31).Corollary 3.
Provided that the sufficient conditions in Theorem2 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the compatible system of equations (21), (23), and (26) for the functions F(t),G1(t,x), and G2(t,x).Theorem 4.
Sufficient conditions for equation (8) to be linearizable via the generalized linearizing transformation (5) with Fx≠0 are as follows. (a)
Ifs1≠0, then the conditions are (18), (40), (41), and (42).(b)
Ifs1=0,s4≠0, then the conditions are (18), (43), (44), (48), and (52).(c)
Ifs1=0,s4=0, then the conditions are (18), (43), (44), (53), and (54).Corollary 5.
Provided that the sufficient conditions in Theorem4 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the following compatible system of equations for the functions F(t,x),G1(t,x), and G2(t,x): (a)
(16), (17), (34), (37), and (38);(b)
(16), (17), (34), (49), and (50);(c)
(16), (17), (19), (32), (33), and (34).


## 3.2.1. Cases1≠0

From (35), one finds
(37)Fx=-(2Ks2)s1.
Since this case Fx≠0, then s2≠0 too. Comparing the mixed derivatives (Fx)t=(Ft)x, one gets the derivative
(38)G1t=(3G1xs1(2rs2-s1)+G1s3)(6s1s2),
where
(39)s3=-6rxs1s2+6s1ts2-6s1xrs2-6s2ts1+6s2xrs1-2A1s1s2+4A2rs1s2-A2s12.
Substituting Fx into Fxxx, G1t into G1tx and G1tt, one arrives at the conditions
(40)s2xx=(-12A2ts12s22+12A2xrs12s22-6A2xs13s2+36rxs1xs1s22-36rxs2xs12s2-12rxA2s12s22+18s1xxs12s2-36s1x2s1s2+36s1xs2xs12+12s1xA2s12s2-6s1xs2s3-12s2xA2s13+6s2xs1s3-4A1A2s12s22+8A22rs12s22-2A22s13s2+2A2s1s2s3-12B3s12s22+48B4rs12s22-120B5r2s12s22+9B5s14)×(18s13)-1,s3x=(-6A1xs13s2-6A2ts13s2+18A2xrs13s2-9A2xs14+36rx2s12s22-36rxs1xs12s2+36rxs2xs13+6rxA2s13s2-12rxs1s2s3+12s1xs1s3-4A1A2s13s2+8A22rs13s2-3A22s14-12B3s13s2+48B4rs13s2-9B4s14-120B5r2s13s2+45B5rs14-2s13s22+s32)/(6s12),s3t=(-6A0xs13s2-3A1xs14-6A2trs13s2-12A2ts14+12A2xr2s13s2+9A2xrs14+36rx2rs12s22-36rx2s13s2-36rxs1xrs12s2+36rxs2xrs13-6rxA1s13s2+18rxA2rs13s2-3rxA2s14-12rxrs1s2s3+12s1ts1s3-4A1A2rs13s2-5A1A2s14+8A22r2s13s2+7A22rs14-12B3rs13s2-15B3s14+48B4r2s13s2+51B4rs14-120B5r3s13s2-105B5r2s14-2rs13s22+rs32+5s14s2)/(6s12).
Equation (15) provides the conditions
(41)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3-rs1)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
Comparing the mixed derivatives (G1tt)x=(G1tx)t, (G1xx)t=(G1tx)x, and (Fxxx)t=(Ft)xxx, one gets the conditions
(42)A1xx=(-33A1xA2s12-18A2txs12-108A2trxs1s2+24A2tA2s12+18A2ts3+54A2xxrs12+108A2xrxrs1s2+18A2xrxs12-30A2xA1s12+102A2xA2rs12-18A2xrs3-90B3xs12+54B4ts12+306B4xrs12-270B5trs12-630B5xr2s12-36rxA1A2s1s2+72rxA22rs1s2+27rxA22s12-108rxB3s1s2+432rxB4rs1s2+252rxB4s12-1080rxB5r2s1s2-1260rxB5rs12+36rxs1s22-18s1xs1s2+30s2xs12+45A0B5s12-5A1A22s12+6A1A2s3-45A1B5rs12+10A23rs12-12A22rs3-15A2B3s12+60A2B4rs12-105A2B5r2s12+5A2s12s2+18B3s3-72B4rs3+180B5r2s3-6s2s333A1xA2s12)/(18s12),A2tt=(36A2txrs1+72A2trxs1-18A2xxr2s1-72A2xrxrs1-3A2xA0s1+3A2xA1rs1-3A2xA2r2s1-18B3ts1+18B3xrs1+72B4trs1-72B4xr2s1-180B5tr2s1+180B5xr3s1+24rxA1A2s1-48rxA22rs1+72rxB3s1-288rxB4rs1+720rxB5r2s1-6rxs1s2-3s1xs1+3A0A22s1-12A0B4s1+60A0B5rs1+4A12A2s1-19A1A22rs1+6A1B3s1-12A1B4rs1+19A23r2s1+18A2B2s1-66A2B3rs1+144A2B4r2s1-240A2B5r3s1+2A2s12+s3)/(18s1),A2tx=(-6A1xA2s12s2-72A2trxs1s22+90A2ts1xs1s2-90A2ts2xs12-24A2tA2s12s2+12A2ts2s3+18A2xxrs12s2-9A2xxs13+72A2xrxrs1s22+18A2xrxs12s2-90A2xs1xrs1s2+90A2xs2xrs12-6A2xA1s12s2+48A2xA2rs12s2-18A2xA2s13-12A2xrs2s3-18B3xs12s2+72B4xrs12s2-27B4xs13+54B5ts13-180B5xr2s12s2+81B5xrs13-24rxA1A2s1s22+48rxA22rs1s22+12rxA22s12s2-72rxB3s1s22+288rxB4rs1s22+72rxB4s12s2-720rxB5r2s1s22-360rxB5rs12s2+135rxB5s13+30s1xA1A2s1s2-60s1xA22rs1s2+90s1xB3s1s2-360s1xB4rs1s2+900s1xB5r2s1s2-30s2xA1A2s12+60s2xA22rs12-90s2xB3s12+360s2xB4rs12-900s2xB5r2s12-8A1A22s12s2+4A1A2s2s3+18A1B5s13+16A23rs12s2-4A23s13-8A22rs2s3-24A2B3s12s2+96A2B4rs12s2-18A2B4s13-240A2B5r2s12s2+54A2B5rs13+12B3s2s3-48B4rs2s3+120B5r2s2s3)/(18s12s2).


## 3.2.2. Cases1=0

From (35), one finds the condition
(43)s2=0.
Equation (15) gives the conditions
(44)A0t=(6A0xr+6A2tr2-6A2xr3-7A0A1+9A0A2r+5A12r-8A1A2r2+A22r3-36B1+54B2r-54B3r2+36B4r3)/6,B0=(-A02+2A0A1r-2A0A2r2-A12r2+2A1A2r3-A22r4+12B1r-12B2r2+12B3r3-12B4r4+12B5r5)/12.
From the mixed derivative (G1xx)t=(G1tx)x, one finds the condition
(45)A2tt=((12rxs4+3s4xr+A1s4-2A2rs4)336A2txr+72A2trx-18A2xxr2-72A2xrxr-3A2xA0+3A2xA1r-3A2xA2r2-18B3t+18B3xr+72B4tr-72B4xr2-180B5tr2+180B5xr3+24rxA1A2-48rxA22r+72rxB3-288rxB4r+720rxB5r2+3A0A22-12A0B4+60A0B5r+4A12A2-19A1A22r+6A1B3-12A1B4r+19A23r2+18A2B2-66A2B3r+144A2B4r2-240A2B5r3(12rxs4+3s4xr+A1s4-2A2rs4)3)/18.
The relation (G1tt)x=(G1tx)t becomes
(46)18(FxG1t-FxG1xr-FxrxG1-G1xK)s4+G1Ks5=0,
where
(47)s4=3A2t-3A2xr+A1A2-2A22r+3B3-12B4r+30B5r2,s5=18A1xx+27A1xA2-36A2xxr+24A2xA1-102A2xA2r+72B3x-54B4t-234B4xr+270B5tr+450B5xr2-15rxA22-180rxB4+900rxB5r+6s4x-45A0B5+13A1A22+45A1B5r-26A23r+39A2B3-156A2B4r+345A2B5r2-8A2s4.
The relation (A2t)t-A2tt=0 provides the condition
(48)s4t=(12rxs4+3s4xr+A1s4-2A2rs4)3.
Further study depends on s4.(i) Cases4≠0From (46), one gets the derivative
(49)g1t=(18(FxG1xr+FxrxG1+G1xK)s4-G1Ks5)(18Fxs4).
Differentiating g1t with respect to x, one obtains the derivative
(50)Fx=(Ks6)(108s43),
where
(51)s6=324A2xs42-36s4xs5+36s5xs4+108A22s42+324B4s42-1620B5rs42-s52.
The relations (Fx)t=(Ft)x,(G1t)t=G1tt,(Fxxx)t=(Ft)xxx,and(Fx)xx=Fxxx provide the conditions
(52)s6t=(30rxs6+3s6xr+2A1s6-4A2rs6+108A2s43+18s42s5)/3,s5t=(-108A1xs42-108A2xrs42+108rxA2s42+180rxs4s5+36s4xrs5-36A1A2s42+12A1s4s5-36A22rs42-24A2rs4s5-108B3s42+108B4rs42+540B5r2s42+rs52+rs6-144s43)/(36s4),A2xx=(-5832A2xA2s43-8748B4xs43+17496B5ts43+26244B5xrs43+43740rxB5s43-126s4xs6+45s6xs4+5832A1B5s43-1296A23s43-5832A2B4s43+17496A2B5rs43+12A2s4s6-s5s6)/(2916s43),s6xx=(-324A2xs42s6+2916s4xxs4s6-11664s4x2s6+5832s4xs6xs4+1944s4xA2s4s6-162s4xs5s6-648s6xA2s42+54s6xs4s5-108A22s42s6+18A2s4s5s6-104976B5s45+s62)/(972s42).(ii)
Cases4=0From (46), one gets the condition
(53)s5=0.
Comparing the mixed derivative (Fxxx)t=(Ft)xxx, one arrives at the condition
(54)A2xx=(-18A2xA2-27B4x+54B5t+81B5xr+135rxB5+18A1B5-4A23-18A2B4+54A2B5r)/9.
All obtained results can be summarized in the following theorems.Theorem 2.
Sufficient conditions for (8) to be linearizable via the generalized linearizing transformation (5) with Fx=0 are equations (18), (22), (24), (25), (28), (29), (30), and (31).Corollary 3.
Provided that the sufficient conditions in Theorem2 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the compatible system of equations (21), (23), and (26) for the functions F(t),G1(t,x), and G2(t,x).Theorem 4.
Sufficient conditions for equation (8) to be linearizable via the generalized linearizing transformation (5) with Fx≠0 are as follows. (a)
Ifs1≠0, then the conditions are (18), (40), (41), and (42).(b)
Ifs1=0,s4≠0, then the conditions are (18), (43), (44), (48), and (52).(c)
Ifs1=0,s4=0, then the conditions are (18), (43), (44), (53), and (54).Corollary 5.
Provided that the sufficient conditions in Theorem4 are satisfied, the transformation (5) mapping equation (8) to a linear equation (6) is obtained by solving the following compatible system of equations for the functions F(t,x),G1(t,x), and G2(t,x): (a)
(16), (17), (34), (37), and (38);(b)
(16), (17), (34), (49), and (50);(c)
(16), (17), (19), (32), (33), and (34).


## 4. Examples

For understanding the procedure of using the linearization theorems, we consider the following examples.Example 1.
Consider the nonlinear third-order ordinary differential equation(55)3x′4t2+2x′3t(3t+2x)+3x′2x′′t2x+x′2(3t2+8tx+3x2)+2x′x′′tx(t+2x)-x′x′′′t2x2+2x′x(2t+3x)+3x′′2t2x2+x′′tx(-t+4x)-x′′′t2x2+3x2=0.
It is an equation of the form (8) in Theorem 1 with the coefficients
(56)A2=-3x,A1=-2(t+2x)tx,A0=t-4xtx,B5=0,B4=-3x2,B3=-2(3t+2x)tx2,B2=-3t2+8tx+3x2t2x2,B1=-2(2t+3x)t2x,B0=-3t2,r=1,h1=-12t,h2=-450t2.
One can check that these coefficients obey the conditions in Theorem 2. Thus, (55) is linearizable via a generalized linearizing transformation. For finding the functions F,G1, and G2, we have to solve equations in Corollary 3, which become
(57)Fx=0,Ft=K,(58)G1x=G1x,G1tt=(5G1t2t2-4G1tG1t-G12)(3G1t2),(59)Kx=0,Kt=(4KG1tt-G1)(3G1t).
From the first equation of system (58), we get G1=xf(t), and choosing f(t)=t, we have
(60)G1=xt
and this solution satisfies the second equation. Since r=1, then we obtain
(61)G2=xt.
System (59) becomes
(62)Kx=0,Kt=0,
and one can take the simplest solution
(63)K=1.
System (90) becomes
(64)Fx=0,Ft=1,
so that we get the particular solution
(65)F=t.
Thus, one obtains the linearizing transformation
(66)X=t,dT=tx(x′+1)dt.
Hence, (55) is mapped by the transformation of (66) into the linear equation (6).Example 2.
Consider the nonlinear third-order ordinary differential equation(67)3x′5t2+x′4t(3t+4x)+x′3x(4t+3x)+x′2x′′tx(3t+x)+3x′2x2+4x′x′′tx2-x′x′′′t2x2+3x′′2t2x2=0.
It is an equation of the form (8) in Theorem 1 with the coefficients
(68)A2=-(3t+x)tx,A1=-4t,A0=0,B5=-3x2,B4=-(3t+4x)tx2,B3=-(4t+3x)t2x,B2=-3t2,B1=0,B0=0,r=0,s1=-2t2,s2=1t2,s3=12(t-x)t5x.
One can check that these coefficients obey the conditions in Theorem 4(a). Thus, (67) is linearizable via a generalized linearizing transformation. For finding the functions F,G1, and G2, we have to solve equations in Corollary 5(a), which become
(69)Fx=K,Ft=K,(70)G1t=(G1xtx-G1t+G1x)(tx),G1xx=(G1x2tx2+4G1xG1tx-4G1xG1x2-5G12t+4G12x)/(3G1tx2),(71)Kx=(4K(G1xx-G1))(3G1x),Kt=(4K(G1xx-G1))(3G1x).
From the first equation of system (70), one can take the particular solution
(72)G1=tx
and this solution satisfies the second equation. Since r=0, then we obtain
(73)G2=0.
System (71) becomes
(74)Kx=0,Kt=0,
and one can take the simplest solution
(75)K=1.
System (69) becomes
(76)Fx=1,Ft=1,
so that we get the particular solution
(77)F=t+x.
Thus, one obtains the linearizing transformation
(78)X=t+x,dT=txx′dt.
Hence, (67) is mapped by the transformation of (78) into the linear equation (6).Example 3.
Consider the nonlinear third-order ordinary differential equation(79)3x′′2x2-3x′4-3x′2x′′x-x′x′′′x2=0.
Note that this equation can be reduced to an autonomous equation by the substitution
(80)x=tv(s),s=ln⁡(t),
and then to the second-order ordinary differential equation
(81)y′′z2y2(z+y)=y′2z2y(-2z+y)-3y′zy(z2+y2)-3z4-14z3y-20z2y2-15zy3-3y4,
where y=y(z). However, the latter equation is not linearizable by point transformations.Equation (79) is an equation of the form (8) in Theorem 1 with the coefficients
(82)A2=3x,A1=0,A0=0,B5=0,B4=3x2,B3=0,B2=0,B1=0,B0=0,r=0,s1=0,s2=0,s4=0,s5=0.
One can check that these coefficients obey the conditions in Theorem 4(c). Thus, (79) is linearizable via a generalized linearizing transformation. For finding the functions F,G1, and G2, we have to solve equations in Corollary 5(c), which become
(83)Ft=K,Fxxx=(6FxxFxG1tG1Kx2+18FxxG1xG1K2x2-Fx3G1t2x2-6Fx2G1tG1xKx2-9FxG1x2K2x2-9FxG12K2)/(6G12K2x2),(84)G1tt=(5G1t2)(3G1),G1tx=(G1t(-FxG1tx+6G1xKx+3G1K))(3G1Kx),G1xx=(6FxxG1tG1Kx2-Fx2G1t2x2-6FxG1tG1xKx2+9G1x2K2x2-9G12K2)/(6G1K2x2),(85)Kx=(FxG1tx+3G1xKx+3G1K)(3G1x),Kt=(4G1tK)(3G1).
From the first equation of system (84), one can take the particular solution
(86)G1=x
and this solution satisfies the second and third equations. Since r=0, then we obtain
(87)G2=0.
System (85) becomes
(88)Kx=(2K)x,Kt=0,
and one can take the particular solution
(89)K=x2.
System (83) becomes
(90)Fx=x2,Fxxx=(3(Fxxx-Fx))x2,
so that one obtains the particular solution of the first equation as
(91)F=tx2
and this solution satisfies the second equation. Then we get the linearizing transformation
(92)X=tx2,dT=xx′dt.
Hence, equation (79) is mapped by the transformation of (92) into the linear equation (6).


## 5. Conclusion

This paper is devoted to find the conditions which allow the third-order ordinary differential equation to be transformed into the simplest linear equation. Necessary conditions which guarantee that the third-order ordinary differential equation can be linearized are found in Theorem1. Theorems 2 and 4 are sufficient conditions for the linearization problem. The linearizing transformation can be found by solving the compatible system in Corollaries 3 and 5. Finally, some examples are provided to demonstrate our procedure.



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*Source: 486717-2014-08-14.xml*