Reduction of Sixth-Order Ordinary Differential Equations to Laguerre Form by Fiber Preserving Transformations

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1 Reduction of Sixth-Order Ordinary Differential Equations to Laguerre Form by Fiber Preserving Transformations Supaporn Suksern Abstract This paper is devoted to the study on linearization of sixth-order ordinary differential equations by fiber preserving transformations. The necessary and sufficient conditions for linearization are obtained. The procedure for obtaining the linearizing transformations and the coefficients of linear equation are provided in explicit form. Examples demonstrating the procedure of using the linearization theorems are presented. Keywords: Linearization problem, point transformation, fiber preserving transformation, nonlinear ordinary differential equation 1 Introduction Nonlinear problems are considered to be a core part in many branches of sciences. The exact solutions to these problems are most required. However, it is evidently hard to analyze such problems directily. Precisely, in the nonlinear regime, many of the most basic questions remain unanswered: existence and uniqueness of solutions are not guaranteed; explicit formulae are difficult to come by; linear superposition is no longer available; numerical approximations are not always sufficiently accurate; linear superposition is no longer available; etc. One method to avoid these difficulty is transforming them into the linear differential equations, which is called the linearization. The linearization, that is, mapping a nonlinear differential equation into a linear differential equation, is an important tool in the theory of differential equations. The problem of linearization of ordinary differential equations attracted attention of mathematicians such as S. Lie and E. Cartan. It was admitted that Lie [1] is the first person who solved linearization problem for ordinary differential equations in 188. He found the general form of Manuscript received April, 2017; revised June, This research was financially supported by the National Research Council of Thailand under Grant no. R2560B101. S. Suksern (corresponding author supapornsu@nu.ac.th) is with the Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok 65000, Thailand. S. Suksern is with the Research center for Academic Excellence in Mathematics, Naresuan University, Phitsanulok 65000, Thailand. all ordinary differential equations of second-order that can be reduced to a linear equation by point transformations. He found the conditions for linearization. The linearization criterion is written through relative invariants of the equivalence group. Liouville [2] and Tresse [] attacked the equivalence problem for second-order ordinary differential equations in terms of relative invariants of the equivalence group of point transformations. There are other approaches for solving the linearization problem of a second-order ordinary differential equation. For example, one was developed by Cartan [4], the idea of his approach was to associate with geometric structure. For the third-order ordinary differential equations, Bocharov, Sokolov and Svinolupov [5] considered the linearization problem with respect to point transformations. Grebot [6] studied the linearization by means of a restricted class of point transformations, namely fiber preserving transformation. However, the problem was not completely solved. Complete criteria for linearization by means of point transformations were obtained by Ibragimov and Meleshko [7]. The linearization problem of third-order ordinary differential equations under point transformations were solved by Ibragimov, Meleshko and Suksern [8]. They found the necessary and sufficient conditions for a complete linearization problem. Later, Suksern and Pinyo [9] solved the linearization problem of fifth-order ordinary differential equations by means of fiber preserving transformations. It turns out that not every differential equation of this order can be transformed in that way. Giving a new transformation still be needed and useful. One of main problem in linearization is the complicate calculations. Because of this task, there is no one try to solve the linearization problems for sixth and higher order yet. By the helps of computer algebra system Reduce, we can offer the necessary and sufficient conditions of this type of linearization. Some examples are provided to illustrate the condition. Linearizing transformation and coefficients of linear equation are obtained. A program used

2 for checking the linearity is also produced. 2 Point Transformations Definition 2.1 A transformation t = ϕ(x, y), u = ψ(x, y) (1) where ϕ and ψ are sufficiently smooth functions is called a point transformation. If ϕ y = 0, a transformation (1) is called a fiber preserving transformation. Let us explain how a point transformation maps one function into another. Assume that y 0 (x) is a given function, then the transformed function u 0 (t) is defined by the following two steps. On the first step one has to solve with respect to x the equation t = ϕ(x, y 0 (x)). Using the Inverse Function Theorem we find x = α(t) is a solution of this equation. The transformed function is determined by the formula u 0 (t) =ψ(α(t),y 0 (α(t))). Conversely, if one has the function u 0 (t), then for finding the function y 0 (x) one has to solve the ordinary differential equation u 0 (ϕ(x, y 0 (x)) = ψ(x, y 0 (x)). Necessary Conditions We begin with investigating the necessary conditions for linearization. Recall that according to the Laguerre theorem, a linear sixth-order ordinary differential equation has the form u (6) + α(t)u + β(t)u + γ(t)u + ω(t)u =0. (2) Here we consider the sixth-order ordinary differential equations y (6) = f ( x, y, y,y,y,y (4),y (5)), () which can be transformed to the linear equation (2) by the fiber preserving transformation t = ϕ (x), u = ψ (x, y). So we arrive at the following theorem. (4) Theorem.1 Any sixth-order ordinary differential equations linearizable by a fiber preserving transformation has to be in the form y (6) + (A 1 y + A 0 )y (5) +(B y + B 2 y 2 + B 1 y + B 0 )y (4) + C 0 y 2 +((D 5 y + D 4 )y + D y + D 2 y 2 + D 1 y + D 0 )y + E 0 y + (F 2 y 2 + F 1 y + F 0 )y 2 +(G 4 y 4 + G y + G 2 y 2 + G 1 y + G 0 )y + H 6 y 6 + H 5 y 5 + H 4 y 4 + H y + H 2 y 2 + H 1 y + H 0 =0, (5) where A 1 =(6y )/, (6) A 0 = (5ϕ xx 2ϕ x ψ xy )/(ϕ x ), (7) B =(15y )/, (8) B 2 =(15yy )/, (9) B 1 = 15(5ϕ xx y 2ϕ x ψ xyy )/(ϕ x ), (10) B 0 =5((7ϕ 2 xx 5ϕ xx ϕ x ψ xy + ϕ 2 xψ xxy ) 4ϕ xxx ϕ x )/(ϕ 2 x ), (11) C 0 =(10y )/, (12) D 5 =(60yy )/, (1) D 4 = 0(5ϕ xx y 2ϕ x ψ xyy )/(ϕ x ), (14) D =(20yyy )/, (15) D 2 = 0(5ϕ xx yy 2ϕ x ψ xyyy )/(ϕ x ), (16) D 1 =20((7ϕ 2 xxy 5ϕ xx ϕ x ψ xyy + ϕ 2 xψ xxyy ) 4ϕ xxx ϕ x y )/(ϕ 2 x ), (17) D 0 = (0(14ϕ 2 xx 14ϕ xx ϕ x ψ xy +5ϕ 2 xψ xxy )ϕ xx (ϕ x α +20ψ xxxy )ϕ x 10(21ϕ xx 8ϕ x ψ xy )ϕ xxx ϕ x +15ϕ xxxx ϕ 2 x )/(ϕ x ), (18) E 0 =(15yy )/, (19) F 2 =(45yyy )/, (20) F 1 = 45(5ϕ xx yy 2ϕ x ψ xyyy )/(ϕ x ), (21) F 0 =15((7ϕ 2 xxy 5ϕ xx ϕ x ψ xyy + ϕ 2 xψ xxyy ) 4ϕ xxx ϕ x y )/(ϕ 2 x ), (22) G 4 =(15yyyy )/, (2) G = 0(5ϕ xx yyy 2ϕ x ψ xyyyy )/(ϕ x ), (24) G 2 = 0(4ϕ xxx ϕ x yy 21ϕ 2 xxyy +15ϕ xx ϕ x ψ xyyy ϕ 2 xψ xxyyy )/(ϕ 2 x ), (25) G 1 = (0(14ϕ 2 xxy 14ϕ xx ϕ x ψ xyy +5ϕ 2 xψ xxyy )ϕ xx (ϕ xy α +20ψ xxxyy )ϕ x +15ϕ xxxx ϕ 2 xy 10(21ϕ xx y 8ϕ x ψ xyy )ϕ xxx ϕ x )/(ϕ x ), (26)

3 G 0 = (6ϕ xxxxx ϕ x 105ϕ xxxx ϕ xx ϕ 2 x +45ϕ xxxx ϕ xψ xy 70ϕ 2 xxxϕ 2 x + 840ϕ xxx ϕ 2 xxϕ x 60ϕ xxx ϕ xx ϕ 2 xψ xy + 120ϕ xxx ϕ xψ xxy 945ϕ 4 xx ϕ xxϕ x ψ xy 60ϕ 2 xxϕ 2 xψ xxy +ϕ xx ϕ 6 x α + 150ϕ xx ϕ xψ xxxy ϕ 8 x β ϕ 7 xψ xy α 15ϕ 4 xψ xxxxy )/(ϕ 4 x ), (27) H 6 =(yyyyy )/, (28) H 5 = (5ϕ xx yyyy 2ϕ x ψ xyyyyy )/(ϕ x ), (29) H 4 = 5(4ϕ xxx ϕ x yyy 21ϕ 2 xxyyy +15ϕ xx ϕ x ψ xyyyy ϕ 2 xψ xxyyyy )/(ϕ 2 x ), (0) H = (15ϕ xxxx ϕ 2 xyy 210ϕ xxx ϕ xx ϕ x yy +80ϕ xxx ϕ 2 xψ xyyy + 420ϕ xxyy 420ϕ 2 xxϕ x ψ xyyy + 150ϕ xx ϕ 2 xψ xxyyy ϕ 6 xyy α 20ϕ xψ xxxyyy )/(ϕ x ), (1) H 2 = (((420ϕ xx ϕ 6 xα)ψ xyy 5ϕ xψ xxxxyy ) 70ϕ 2 xxxϕ x y +(ϕ xy α +50ψ xxxyy )ϕ xx ϕ 2 x + 0(28ϕ 2 xxy 21ϕ xx ϕ x ψ xyy +4ϕ 2 xψ xxyy )ϕ xxx )ϕ x ((945ϕ 4 xx + ϕ 8 xβ)y + 60ϕ 2 xxϕ 2 xψ xxyy 6ϕ xxxxx ϕ xy + 15(7ϕ xx y ϕ x ψ xyy )ϕ xxxx ϕ 2 x)/(ϕ 4 x ), (2) H 1 = (15(ϕ 2 xx 6ϕ xx ϕ x ψ xy +4ϕ 2 xψ xxy )ϕ xx (ϕ 5 x γ +2ϕ 4 xψ xy β +ϕ xψ xxy α +6ψ xxxxxy )ϕ 5 x + ϕ xxxxxx ϕ 4 x (ϕ x α + 140ψ xxxy )ϕ 2 xxϕ x +(ϕ 4 x β +6ϕ xψ xy α +75ψ xxxxy )ϕ xx ϕ 4 x + 140(2ϕ xx ϕ x ψ xy )ϕ 2 xxxϕ 2 x (7ϕ xx 4ϕ x ψ xy )ϕ xxxxx ϕ x 5(7ϕ xxx ϕ x 42ϕ 2 xx +42ϕ xx ϕ x ψ xy 9ϕ 2 xψ xxy )ϕ xxxx ϕ 2 x (210(6ϕ 2 xx 8ϕ xx ϕ x ψ xy +ϕ 2 xψ xxy )ϕ xx (ϕ x α +80ψ xxxy )ϕ x)ϕ xxx )ϕ x ))/(ϕ 5 x ), () H 0 = (ϕ xxxxxx ϕ 4 xψ x 21ϕ xxxxx ϕ xx ϕ xψ x +6ϕ xxxxx ϕ 4 xψ xx 5ϕ xxxx ϕ xxx ϕ xψ x + 210ϕ xxxx ϕ 2 xxϕ 2 xψ x 105ϕ xxxx ϕ xx ϕ xψ xx +15ϕ xxxx ϕ 4 xψ xxx + 280ϕ 2 xxxϕ xx ϕ 2 xψ x 70ϕ 2 xxxϕ xψ xx 1260ϕ xxx ϕ xxϕ x ψ x + 840ϕ xxx ϕ 2 xxϕ 2 xψ xx 210ϕ xxx ϕ xx ϕ xψ xxx + ϕ xxx ϕ 7 xψ x α +20ϕ xxx ϕ 4 xψ xxxx + 945ϕ 5 xxψ x 945ϕ 4 xxϕ x ψ xx + 420ϕ xxϕ 2 xψ xxx ϕ 2 xxϕ 6 xψ x α 105ϕ 2 xxϕ xψ xxxx + ϕ xx ϕ 8 xψ x β +ϕ xx ϕ 7 xψ xx α +15ϕ xx ϕ 4 xψ xxxxx ϕ 11 x ωψ ϕ 10 x ψ x γ ϕ 9 xψ xx β ϕ 8 xψ xxx α ϕ 5 x. (4) Proof. Applying a fiber preserving transformation (4), one obtains the following transformation of derivatives u (t) = D xψ D x ϕ = ψ x + y ϕ x = P (x, y, y ), u (t) = D xp D x ϕ = P x + y P y + y P y ϕ x = 1 ϕ [(ϕ x )y +(ϕ x y )y 2 +(2ϕ x ψ xy x ϕ xx )y ϕ xx ψ x + ϕ x ψ xx ] = Q(x, y, y,y ), u (t) = D xq D x ϕ = Q x + y Q y + y Q y + y Q y ϕ x = 1 ϕ 5 [(ϕ 2 x )y +(ϕ 2 xy )y y x +ϕ x (ϕ x ψ xy ϕ xx )y +...] = R(x, y, y,y,y ), u (4) (t) = D xr D x ϕ = R x + y R y + y R y + y R y + y (4) R y ϕ x = 1 ϕ 7 [(ϕ x )y (4) +(4ϕ xy )y y x +2ϕ 2 x(2ϕ x ψ xy ϕ xx )y +...] = S(x, y, y,y,y,y (4) ), u (5) (t) = D xs D x ϕ = (S x + y S y + y S y + y S y S y + y (4) S y +y (5) S y (4))/ϕ x = 1 ϕ 9 [(ϕ 4 x )y (5) +(5ϕ 4 xy )y y (4) x +5ϕ x(ϕ x ψ xy 2ϕ xx )y (4) +...] = V (x, y, y,y,y,y (4),y (5) ), u (6) (t) = D xv D x ϕ = (V x + y V y + y V y + y V y + y (4) V y = +y (5) V y (4) + y (6) V y (5))/ϕ x 1 [(ϕ 5 x )y (6) +(2ϕ 5 xy )y y (5) ϕ 11 x +ϕ 4 x(2ϕ x ψ xy 5ϕ xx )y (5) +...],

4 where y (6) +(( 6ψyy )y ( (5ϕxxψy 2ϕxψxy) (ϕ x) ))y (5) +(( 15ψyy )y +( 15ψyyy +...)y (4) +( 10ψyy )y 2 ( 15(5ϕxxψyy 2ϕxψxyy) (ϕ x) )y )y 2 + ((( (60ψyyy) )y + ( 0(5ϕxxψyy 2ϕxψxyy)) (ϕ x) )y + (20ψyyyy) y +...)y +( 15ψyyy )y +(( 45ψyyyy )y 2 +( 45(5ϕxxψyyy+...) (ϕ x) )y + 15((7ϕ2 xx ψyy 5ϕxxϕxψxyy+ϕ2 x ψxxyy) 4ϕxxxϕxψyy) (ϕ 2 x ψy) )y 2 +(( 15ψyyyyy )y 4 +( 0(5ϕxxψyyyy 2ϕxψxyyyy) (ϕ x) )y +( 0(4ϕxxxϕxψyyy 21ϕ2 xx ψyyy+...) (ϕ 2 x ψy) )y )y +( ψyyyyyy )y 6 +( (5ϕxxψyyyyy 2ϕxψxyyyyy) (ϕ x) )y 5 +( 5(4ϕxxxϕxψyyyy 21ϕ2 xx ψyyyy+...) (ϕ 2 x ψy) )y =0. Denoting A i,b i,c i,d i,e i,f i,g i and H i as equations (6)-(4), so we obtain the necessary form (5). These prove the theorem. 4 Sufficient Conditions and Linearizing Transformations We have shown the previous section that every linearizable sixth-order odinary differential equation belong to the class of equation (5). In this section, we formulate the main theorems containing sufficient conditions for linearization as well as the methods for constructing the linearizing transformations. Theorem 4.1 Sufficient conditions for equation (5) to be linearizable via a fiber preserving tranformation are as follows: A 0y =A 1x, (5) B =(5A 1 )/2, (6) A 1y =( 5A B 2 )/0, (7) A 1x =( 5A 0 A 1 +6B 1 )/0, (8) C 0 =(5A 1 )/, (9) D 5 =4B 2, (40) D 4 =2B 1, (41) B 2y =( 2A 1 B 2 +9D )/12, (42) B 2x =( 2A 0 B 2 +D 2 )/12, (4) B 0y =( 2A 1 B 0 +D 1 )/12, (44) E 0 =B 2, (45) D =(4F 2 )/9, (46) F 1 =(D 2 )/2, (47) F 0 =(D 1 )/4, (48) F 2y =( A 1 F 2 +18G 4 )/6, (49) F 2x =( 2A 0 F 2 +9G )/12, (50) D 1y =( A 1 D 1 +4G 2 )/6, (51) D 1x =( 150A 0x A 0 A A 0x B B 0x A 1 25A 0A 1 +0A 2 0B 1 +90A 0 A 1 B 0 45A 0 D 1 15A 1 D 0 72B 0 B G 1 )/270, (52) G 4y =( A 1 G 4 +90H 6 )/6, (5) G 4x =( A 0 G 4 +15H 5 )/6, (54) G 2y =( A 1 G 2 +6H 4 )/6, (55) G 1y =( A 1 G 1 +18H )/6, (56) G 1x =(150A 0x A 2 0A 1 180A 0x A 0 B 1 180A 0x A 1 B A 0x D 1 180B 0x A 0 A B 0x B D 0x A 1 +25A 4 0A 1 0A 0B 1 120A 2 0A 1 B 0 +45A 2 0D A 0 A 1 D A 0 B 0 B 1 90A 0 G 1 +72A 1 B A 1 G 0 108B 0 D 1 162B 1 D H 2 )/540, (57) λ 1y =λ 2y = λ y = λ 4y = λ 5y = λ 6y,λ 7y = λ 8y =0, (58) where λ 1 = 0A 0x 5A B 0, (59) λ 2 = 72B 0x λ 1x 12A 0 B 0 2A 0 λ 1 +54D 0, (60) λ = 108B 0x 18A 0 B 0 A 0 λ 1 +81D 0 5λ 2, (61) λ 4 =2850D 0x + 210λ 2x + 210λ x A 0 D A 0 λ A 0 λ + 60B 0 λ G 0 +λ 2 1, (62) λ 5 =96900D 0x λ x A 0 D A 0 λ A 0 λ B 0 λ G λ λ 4, (6) λ 6 =11400D 0x A G 0x + λ 4x + λ 5x A 2 0D A 2 0λ A 2 0λ A 0 B 0 λ A 0 G B 0 λ 2 840B 0 λ 780D 0 λ H 1 60λ 1 λ 2 60λ 1 λ, (64) λ 7 =28500D 0x A G 0x 15λ 4x A 2 0D A 2 0λ A 2 0λ + 600A 0 B 0 λ A 0 G B 0 λ B 0 λ 9450D 0 λ H λ 1 λ 2 250λ 1 λ 6λ 6, (65) λ 8 =849000D 0x A D 0x B D 0x λ G 0x A H 0y H 1x + 252λ 7x A 0D A 0λ A 0λ A 2 0B 0 λ A 2 0G A 0 B 0 D 0

5 A 0 B 0 λ A 0 B 0 λ A 0 D 0 λ A 0 H A 0 λ 1 λ A 0 λ 1 λ A 1 H B 2 0λ B 0 G B 0 λ D 0 λ D 0 λ G 0 λ λ 1 147λ 1 λ 4 +λ 1 λ λ λ 2 λ λ 2. (66) Proof. For obtaining sufficient conditions, one has to solve the compatibility problem. Considering the representations of the coefficients A i,b i,c i,d i,e i,f i,g i and H i through the unknown functions ϕ and ψ. We first rewrite the expressions (6) and (7) for A 1 and A 0 in the following forms y =( A 1 )/6, (67) ψ xy =( (15ϕ xx + ϕ x A 0 ))/(6ϕ x ). (68) The mixed derivative (y ) x =(ψ xy ) y provides the condition (5). From equations (8)-(10), one gets the conditions (6)-(8). The relation (A 1x ) y =(A 1y ) x provides the condition B 1y =(12B 2x +2A 0 B 2 A 1 B 1 )/6. From equation (11), we have ϕ xxx =( 0A 0x ϕ 2 x+15ϕ 2 xx 5ϕ 2 xa ϕ 2 xb 0 )/(210ϕ x ). (69) Since ϕ y = 0, then differentiating (69) with respect to y, one arrives at the condition B 1x =(0A 0x A 1 +72B 0y +5A 2 0A 1 6A 0 B 1 )/6. The relation (B 1x ) y =(B 1y ) x provides the condition B 0yy =( B 0y A 1 +B 2xx + B 2x A 0 )/. From equations (12)-(16), one gets the conditions (9)- (4). The relation (B 2x ) y =(B 2y ) x provides the condition D 2y =(18D x +A 0 D A 1 D 2 )/6. From equation (17), one gets the conditions (44). The relation (B 0y ) y = B 0yy provides the condition One can determine α from equation (18). Since ϕ = ϕ(x), then α y = 0, which yields the condition D 0y =(150A 0x A 0 A 1 180A 0x B 1 180B 0x A D 1x +25A 0A 1 0A 2 0B 1 90A 0 A 1 B 0 +45A 0 D 1 +72B 0 B 1 )/810. From equations (19)-(24), one finds the conditions (45)- (50). The relation (F 2x ) y =(F 2y ) x provides the condition G y =(24G 4x +4A 0 G 4 A 1 G )/6. From equation (25), one gets the conditions (51). The relation (D 1x ) y =(D 1y ) x provides the condition G 2x =( 60A 0x A 0 B 2 +90A 0x D 2 +72B 0x B G 1y 10A 0B 2 +15A 2 0D 2 +6A 0 B 0 B 2 0A 0 G 2 +45A 1 G 1 6B 0 D 2 54B 2 D 0 )/180. From equation (26), one obtains the condition (52). One can determine β from equation (27). Since ϕ = ϕ(x), then β y = 0, which yields the conditiion G 0y = (150A 0x A 2 0A 1 180A 0x A 0 B 1 180A 0x A 1 B A 0x D 1 180B 0x A 0 A B 0x B D 0x A 1 540G 1x +25A 4 0A 1 0A 0B 1 120A 2 0A 1 B 0 +45A 2 0D A 0 A 1 D A 0 B 0 B 1 90A 0 G 1 +72A 1 B B 0 D 1 162B 1 D 0 )/2160. Equations (28) and (29) provide the conditions (5) and (54). The relation (G 4x ) y =(G 4y ) x provides the condition H 5y =(6H 6x +6A 0 H 6 A 1 H 5 )/6. From equation (0), one gets the condition (55). relation (G 2y ) x =(G 2x ) y provides the condition The G 1yy =(60A 0x A 0 F 2 270A 0x G 72B 0x F 2 270G 1y A H 4x +10A 0F 2 45A 2 0G 6A 0 B 0 F A 0 H B 0 G 54B 2 G 1 +54D 0 F 2 )/810. From equation (1), one gets the condition (56). relation (G 1y ) x =(G 1x ) y provides the condition The H y =(60A 0x A 0 F 2 270A 0x G 72B 0x F H 4x +10A 0F 2 45A 2 0G 6A 0 B 0 F A 0 H 4 405A 1 H + 108B 0 G +54D 0 F 2 )/240. From equation (2), one finds the condition (57). The relation (G 1x ) y =(G 1y ) x provides the condition H x =(00A 0x A 2 0B 2 450A 0x A 0 D 2 60A 0x B 0 B A 0x G 2 60B 0x A 0 B B 0x D 2 D 2x =(60A 0x B 2 +90D 1y +10A 2 0B 2 15A 0 D 2 +15A 1 D 1 24B 0 B 2 )/90. The relation (D 2x ) y =(D 2y ) x provides the condition D 1yy =( 5D 1y A 1 +45D xx +15D x A 0 +B 0 D B 2 D 1 )/ D 0x B H 2y +50A 4 0B 2 75A 0D 2 240A 2 0B 0 B A 2 0G A 0 B 0 D A 0 B 2 D 0 150A 0 H A 1 H B 2 0B 2 60B 0 G 2 720B 2 G 0 405D 0 D 2 )/8100.

6 The relation (H x ) y =(H y ) x provides the condition H 2yy =(90A 0xx A 0 F 2 405A 0xx G 0A 0x A 2 0F A 0x A 0 G +72A 0x B 0 F 2 240A 0x H 4 108B 0xx F 2 162B 0x G 1620H 2y A H 4xx H 4x A 0 10A 4 0F 2 +45A 0G +48A 2 0B 0 F 2 540A 2 0H 4 162A 0 B 0 G 54A 0 D 0 F 2 6B 2 0F B 0 H 4 24B 2 H D 0 G + 108F 2 G 0 )/4860. One can determine γ from equation (). Since ϕ = ϕ(x), then γ y = 0, which yields the condition H 1y =(750A 0x A 0A 1 900A 0x A 2 0B A 0x A 0 A 1 B A 0x A 0 D A 0x A 1 D A 0x B 0 B A 0x G 1 900B 0x A 2 0A B 0x A 0 B B 0x A 1 B B 0x D D 0x A 0 A D 0x B G 0x A H 2x + 125A 5 0A 1 150A 4 0B 1 750A 0A 1 B A 0D A 2 0A 1 D A 2 0B 0 B 1 450A 2 0G A 0 A 1 B A 0 A 1 G 0 810A 0 B 0 D A 0 B 1 D A 0 H 2 150A 1 B 0 D 0 42B 2 0B B 0 G B 1 G D 0 D 1 )/ One can determine ω from equation (4). Since ϕ = ϕ(x), then ω y = 0, which yields the derivative ψ xxxxxx.forming the mixed derivative (ψ xxxxxx ) y =(ψ xy ) xxxxx, one obtains the condition H 0yy =(1125A 0xx A 0A 1 150A 0xx A 2 0B A 0xx A 0 A 1 B A 0xx A 0 D A 0xx A 1 D A 0xx B 0 B A 0xx G A 0x A 4 0A A 0x A 0B A 0x A 2 0A 1 B A 0x A 2 0D A 0x A 0 A 1 D 0 780A 0x A 0 B 0 B A 0x A 0 G A 0x A 1 B A 0x A 1 G A 0x B 0 D A 0x B 1 D A 0x H 2 150B 0xx A 2 0A B 0xx A 0 B B 0xx A 1 B 0 240B 0xx D B 0x A 0A B 0x A 2 0B B 0x A 0 A 1 B B 0x A 0 D B 0x A 1 D B 0x B 0 B B 0x G D 0xx A 0 A 1 240D 0xx B 1 675D 0x A 2 0A D 0x A 0 B D 0x A 1 B 0 240D 0x D G 0xx A G 0x B H 0y A H 2xx H 2x A 0 250A 6 0A A 5 0B A 4 0A 1 B 0 450A 4 0D A 0A 1 D A 0B 0 B A 0G 1 0A 2 0A 1 B A 2 0A 1 G A 2 0B 0 D A 2 0B 1 D A 2 0H A 0 A 1 B 0 D A 0 A 1 H A 0 B 2 0B 1 240A 0 B 0 G A 0 B 1 G 0 285A 0 D 0 D A 2 1H A 1 B A 1 B 0 G 0 240A 1 D B 2 0D 1 402B 0 B 1 D B 0 H B 1 H B 2 H D 0 G D 1 G 0 )/ One can rewrite the expression of equation (69) as ϕ xxx = (15ϕ 2 xx + ϕ 2 xλ 1 )/(210ϕ x ), (70) where λ 1 is in the form of equation (59). Rewriting the representation of the derivative ψ xxxxxx as ψ xxxxxx =( ϕ 5 xxψ x ϕ 4 xxϕ x ψ xx ϕ xxϕ 2 xψ xxx ϕ xxϕ 2 xψ x λ ϕ xxϕ 2 xλ 2 ψ ϕ 2 xxϕ xψ xxxx ϕ 2 xxϕ xψ xx λ ϕ 2 xxϕ xψ x λ ϕ 2 xxϕ xλ 4 ψ ϕ xx ϕ 4 xψ xxxxx ϕ xx ϕ 4 xψ xxx λ ϕ xx ϕ 4 xψ xx ( 7λ 2 4λ ) ϕ xx ϕ 4 xψ x λ ϕ xx ϕ 4 xλ 6 ψ ϕ 5 xψ xxxx λ ϕ 5 xψ xxx (7λ 2 +λ ) ϕ 5 xψ xx ( 60λ 2 1 7λ 4 2λ 5 ) ϕ 5 xψ x λ ϕ 5 x H 0 + ϕ 5 xλ 8 ψ)/( ϕ 5 x), (71) where λ 2,λ,..., λ 8 are in the form of equations (60)-(66). The relations (A 0x ) y =(A 0y ) x,(λ 1x ) y =(λ 1y ) x,(b 0x ) y =(B 0y ) x,(λ 2x ) y =(λ 2y ) x,(λ x ) y =(λ y ) x,(λ 5x ) y = (λ 5y ) x,(λ 4x ) y =(λ 4y ) x and (λ 7x ) y =(λ 7y ) x provide the conditions (58).

7 Finally, the representations for α, β, γ and ω become α =λ 2 /(54ϕ x), (72) β =( 150ϕ xx λ 2 ϕ x λ 4 )/(7800ϕ 5 x), (7) γ =(94500ϕ 2 xxλ 2 +60ϕ xx ϕ x λ 4 + ϕ 2 xλ 6 )/(567000ϕ 7 x), (74) ω =( ϕ xxλ ϕ 2 xxϕ x λ 4 120ϕ xx ϕ 2 xλ 6 ϕ xλ 8 )/( ϕ 9 x). (75) This completes the proof of Theorem 4.1. Corollary 4.2 Provided that the sufficient conditions in Theorem 4.1 are satisfied, the transformation (4) mapping equation (5) to a linear equation (2) is obtained by solving the compatible system of equations (67), (68), (70) and (71) for the functions ϕ(x) andψ(x, y). Finally, the coefficients α, β, γ and ω of the resulting linear equation (2) are given by equations (72), (7), (74) and (75). 5 Illustration of the Linearization Theorems Example 5.1 Consider the nonlinear sixth-order ordinary differential equation 24y 2 x +12y y x y y + 120y y (4) x +12y y (5) x 2 +24y y + 180y y y x +0y y (4) x 2 +24y xy +20y 2 +4y x 2 y +60y (4) y +24y (5) xy + x 2 y(2y (6) + y) =0. (76) It is an equation of the form (5) in Theorem.1 with the coefficients A 1 = 6 y,a 0 = 12 x,b = 15 y,b 2 =0,B 1 = 60 xy, B 0 = 0 x 2,C 0 = 10 y,d 5 =0,D 4 = 120 xy,d =0, D 2 =0,D 1 = 120 x 2 y,d 0 =0,E 0 =0,F 2 =0, F 1 =0,F 0 = 90 x 2 y,g 4 =0,G =0,G 2 =0, G 1 = 6 y,g 0 = 12 x,h 6 =0,H 5 =0,H 4 =0, H =0,H 2 = 12 xy,h 1 = 12 x 2,H 0 = y 2,λ 1 =0, λ 2 = 180, λ = 78, λ 4 =0,λ 5 =0,λ 6 =0, λ 7 =0,λ 8 = One can check that these coefficients obey the conditions in Theorem 4.1. Hence an equation (76) is linearizable via a fiber preserving transformation. Applying Corollary 4.2, the linearizing transformation is found by solving the following equations y = /y, (77) ψ xy = (5ϕ xx x +4ϕ x )/(2ϕ x x), (78) ϕ xxx =(ϕ 2 xx)/(2ϕ x ), (79) ψ xxxxxx =(45ϕ 5 xxψ x 225ϕ 4 xxϕ x ψ xx + 00ϕ xxϕ 2 xψ xxx +0ϕ xxϕ 2 xψ 150ϕ 2 xxϕ xψ xxxx 60ϕ 2 xxϕ xψ x +0ϕ xx ϕ 4 xψ xxxxx +0ϕ xx ϕ 4 xψ xx 4ϕ 5 xψ xxx + ϕ 5 x y 2ϕ 5 xψ)/(2ϕ 5 x). (80) Consider equation (79), one can choose the particular solution ϕ =x. So that system of equations (77), (78) and (80) are written as y = y, (81) ψ xy = 2 x, (82) ψ xxxxxx = 4ψ xxx + y 2ψ. (8) 2 Consider equations (81) and (82), we have Hence, =Kx 2 y, ψ = Kx2 y 2 2 K = constant. + f(x). Since one can use any particular solution, we set K = 2,f(x) =0andtakeψ = x 2 y 2. One can readily verify that this function solves equation (8) as well. Hence, one obtains the linearizing transformation t = x, u = x 2 y 2. (84) From Corollary 4.2 the coefficients α, β, γ and ω of the resulting linear equation (2) are α =0,β=2,γ=0,ω=1. Hence, the nonlinear equation (76) can be mapped by transformation (84) into the linear equation u (6) +2u + u =0. (85) The solution of equation (85) is u(t) =e t (C 0 + C 1 t)+e t 2 ((C2 + C t) cos( 2 t) +(C 4 + C 5 t)sin( t)), (86) 2

8 where C 0,C 1,C 2,C, C 4 and C 5 are arbitrary constants. Substituting equation (84) into equation (86), we get the solution x 2 y 2 =e x (C 0 + C 1 x)+e x 2 ((C2 + C x) cos( +(C 4 + C 5 x)sin( 2 x)). 2 x) Example 5.2 Consider the nonlinear sixth-order ordinary differential equation y 6 15y 5 +15y 4 y +85y 4 150e y y y +20y y + y (e x 225) + 45y 2 y y 2 y 150y 2 y +15y 2 y (4) + y 2 ( e x + 274) 225y y 2 +60y y y +y y (e x 225) + 40y y 75y y (4) +6y y (5) +2y (e x 60) + 15y + 255y 2 150y y +15y y (4) + y ( e x + 274) + 10y 2 + y (e x 225) + 85y (4) 15y (5) + y (6) =0. (87) It is an equation of the form (5) in Theorem.1 with the coefficients A 1 =6,A 0 = 15, B =15,B 2 =15,B 1 = 75, B 0 =85,C 0 =10,D 5 =60,D 4 = 150, D =20, D 2 = 150, D 1 = 40,D 0 = e x 225, E 0 =15, F 2 =45,F 1 = 225, F 0 = 255, G 4 =15,G = 150, G 2 = 510, G 1 =(e x 225), G 0 = e x + 274, H 6 =1,H 5 = 15, H 4 =85,H = e x 225, H 2 = e x + 274, H 1 =2(e x 60), H 0 =0, λ 1 = 150, λ 2 =54e x,λ = 189e x, λ 4 = e x,λ 5 = (e x +1), λ 6 = e x,λ 7 = e x, λ 8 = e x. One can check that these coefficients obey the conditions in Theorem 4.1. Hence an equation (76) is linearizable via a fiber preserving transformation. Applying Corollary 4.2, the linearizing transformation is found by solving the following equations y =, (88) ψ xy =5 (ϕ xx ϕ x )/(2ϕ x ), (89) ϕ xxx =(ϕ 2 xx ϕ 2 x)/(2ϕ x ), (90) ψ xxxxxx =(45ϕ 5 xxψ x 225ϕ 4 xxϕ x ψ xx + 00ϕ xxϕ 2 xψ xxx + 150ϕ xxϕ 2 xψ x +15e x ϕ xxϕ 2 xψ 150ϕ 2 xxϕ xψ xxxx 00ϕ 2 xxϕ xψ xx 0e x ϕ 2 xxϕ xψ x 45e x ϕ 2 xxϕ xψ +0ϕ xx ϕ 4 xψ xxxxx + 150ϕ xx ϕ 4 xψ xxx +15e x ϕ xx ϕ 4 xψ xx +45e x ϕ xx ϕ 4 xψ x +45ϕ xx ϕ 4 xψ x +45e x ϕ xx ϕ 4 xψ 20ϕ 5 xψ xxxx 2e x ϕ 5 xψ xxx 9e x ϕ 5 xψ xx 2ϕ 5 xψ xx 19e x ϕ 5 xψ x 15e x ϕ 5 xψ)/(2ϕ 5 x). (91) Consider equation (90), one can choose the particular solution ϕ =e x. So that system of equations (88), (89) and (91) as rewritten as y =, (92) ψ xy =0, (9) ψ xxxxxx =15ψ xxxxx 85ψ xxxx e x ψ xxx + 225ψ xxx +e x ψ xx 274ψ xx 2e x ψ x + 120ψ x. (94) Consider equation (92) and (9), we have Hence, =Ke y, K = constant. ψ =Ke y + f(x). Since one can use any particular solution, we set K = 1,f(x) =0andtakeψ = Ke y. One can readily verify that this function solves equation (94) as well. Hence, one obtains the linearizing transformation t = e x, u = e y. (95) From Corollary 4.2 the coefficients α, β, γ and ω of the resulting linear equation (2) are α =1,β= γ = ω =0. Hence, the nonlinear equation (87) can be mapped by transformation (95) into the linear equation The solution of equation (96) is u (6) + u =0. (96) u(t) =(C 0 + C 1 t + C 2 t 2 )+C e t (97) + e t 2 (C4 cos( 2 t)+c 5 sin( 2 t)), where C 0,C 1,C 2,C, C 4 and C 5 are arbitrary constants. Substituting equation (95) into equation (97), we get the solution e y =(C 0 + C 1 e x + C 2 e x2) + C e ex + e ex 2 (C4 cos( 2 ex )+C 5 sin( 2 ex )).

9 References [1] Lie, S. (188). Klassifikation und Integration von gewöhnlichen Differentialgleichungen zwischen x, y, die eine Gruppe von Transformationen gestatten. III. Archiv for Matematik og Naturvidenskab, 8(4), [2] Liouville, R. (1889). Sur les Invariantes de Certaines Equations. J. de lécole Polytechnique, 59, [] Tresse, A. M. (1896). Détermination des Invariants Ponctuels de léquation Différentielle Ordinaire du Second Ordre y = f(x, y, y ). Preisschriften der fürstlichen Jablonowski schen Geselschaft XXXII, Leipzig, S.Herzel. [4] Cartan, E. (1924). Sur les varietes a connexion projective. Bull. Soc. Math. France, 52, [5] Bocharov, A. V., Sokolov, V. V. and Svinolupov, S. I. (199). On some equivalence problems for differential equations. Preprint ESI, 54. [6] Grebot, G. (1997). The characterization of third order ordinary differential equations admitting a transtive fiber-preserving point symmetry group. Journal of Mathematical Analysis and Applications, 206, [7] Ibragimov, N. H. and Meleshko, S. V. (2005). Linearization of third-order ordinary differential equations by point and contact transformations. Journal of Mathematical Analysis and Applications, 08, [8] Ibragimov, N. H., Meleshko, S. V. and Suksern, S. (2008). Linearization of fourth-order ordinary differential equations by point transformations. Journal of Physics A: Mathematical and Theoretical, 41. [9] Suksern, S. and Pinyo, W. (2014). On the fiber preserving transformations for the fifth-orderordinary differential equations. Journal of Applied Mathematics, Volume 2014, Article ID 75910, 8 pages.