JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 735910 10.1155/2014/735910 735910 Research Article On the Fiber Preserving Transformations for the Fifth-Order Ordinary Differential Equations Suksern S. Pinyo W. Jovanoski Zlatko Department of Mathematics Faculty of Science Naresuan University Phitsanulok 65000 Thailand nu.ac.th 2014 342014 2014 15 01 2014 11 03 2014 3 4 2014 2014 Copyright © 2014 S. Suksern and W. Pinyo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper is devoted to the study of the linearization problem of fifth-order ordinary differential equations by means of fiber preserving transformations. The necessary and sufficient conditions for linearization are obtained. The procedure for obtaining the linearizing transformations is provided in explicit form. Examples demonstrating the procedure of using the linearization theorems are presented.

1. Introduction 1.1. The Research Problem and Its Significance

In mathematics, a nonlinear equation is an equation which is not linear; that is, an equation which does not satisfy the superposition principle, or whose output is not directly proportional to its input. Less technically, a nonlinear equation is any problem where the variables to be solved for cannot be written as a linear combination of independent components.

Nonlinear problems are of interest to engineers, physicists, and mathematicians because most physical systems are inherently nonlinear in nature. Nonlinear equations are difficult to solve and give rise to interesting phenomena. While solving problems related to nonlinear ordinary differential equations, it is often expedient to simplify equations by a suitable change of variables. One of the fundamental methods to solve this relies upon the transformation of a given equation to another equation of standard form. The transformation may be to an equation of equal order or of greater or lesser order. In particular, the possibility that a given equation could be linearized, that is, transformed to a linear equation, was a most attractive proposition due to the special properties of linear differential equations. The reduction of an ordinary differential equation to a linear ordinary differential equation besides simplification allows us to construct an exact solution of the original equation.

One type of the classification problem is the equivalence problem. Two equations of differential equations are said to be equivalent if there exists an invertible transformation which transforms any solution of one equation to a solution of the other equation and vice versa. The linearization problem is a particular case of the equivalence problem, where one of the equations is a linear equation. It is one of the essential parts in the study of nonlinear equations.

The main difficulty in solving the linearization problem comes from the large number of complicated calculations. Because of this difficulty, no one attempts to solve this problem for nonlinear equations are higher than fourth. However if we can solve the linearization problem of fifth-order ordinary differential equations, then we should set a new process to solve the problems in Physics or Engineering.

1.2. Historical Review

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 Lie and Cartan. The first linearization problem for ordinary differential equations was solved by Lie [1, 2]. He found the general form of all ordinary differential equations of second-order that can be reduced to a linear equation by changing the independent and dependent variables. He showed that any linearizable second-order equation should be at most cubic in the first-order derivative and provided a linearization test in terms of its coefficients. The linearization criterion is written through relative invariants of the equivalence group. Liouville  and Tresse  treated 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 . The idea of his approach was to associate with every differential equation a uniquely defined geometric structure of a certain form.

In 1993, Bocharov et al.  considered the linearization problem of third-order with respect to point transformations. Grebot  studied the linearization of third-order ordinary differential equations by means of a restricted class of point transformations, namely, t=φ(x) and u=ψ(x,y). However, the problem was not completely solved. Complete criteria for linearization by means of point transformations were obtained by Ibragimov and Meleshko .

In 2008, Ibragimov et al.  solved the linearization problem for fourth-order ordinary differential equations by using point transformation.

Nowadays, the linearization problem of fifth-order ordinary differential equations via point transformations still is an unsolved one.

1.3. The Mapping of a Function by a Point Transformation Definition 1.

A transformation, (1)t=φ(x,y),u=ψ(x,y), where φ and ψ are sufficiently smooth functions, is called a point transformation. If φy=0, a transformation (1) is called a fiber preserving transformation.

Let us explain how a point transformation maps one function into another.

Assume that y0(x) is a given function, the transformed function u0(t) is defined by the following two steps. On the first step one has to solve with respect to x the equation (2)t=φ(x,y0(x)). Using the Inverse Function Theorem we find that x=α(t) is a solution of this equation. The transformed function is determined by the formula (3)u0(t)=ψ(α(t),y0(α(t))). Conversely, if one has the function u0(t), then for finding the function y0(x) one has to solve the ordinary differential equation (4)u0(φ(x,y0(x)))=ψ(x,y0(x)).

2. Necessary Conditions of Linearization

We begin with investigating the necessary conditions for linearization. Recall that according to the Laguerre theorem a linear fifth-order ordinary differential equation has the form (5)u(5)+α(t)u′′+β(t)u+γ(t)u=0. Here we consider the fifth-order ordinary differential equations (6)y(5)=f(x,y,y′′,y′′′,y(4)), which can be transformed to the linear equation (5) with α=β=γ=0 under the fiber preserving transformation (7)t=φ(x),u=ψ(x,y). So we arrive at the following theorem.

Theorem 2.

Any fifth-order ordinary differential equation (6) obtained from a linear equation (5) with α=β=γ=0 by a fiber preserving transformation (7) has to be the form (8)y(5)+(A1y+A0)y(4)+(B3y+B2y2+B1y+B0)y+(C1y+C0)y(2)+(D3y3+D2y2+D1y+D0)y+E5y5+E4y4+E3y3+E2y2+E1y+E0=0, where (9)A1=5ψyyψy,(10)A0=-5(2φxxψy-φxψxy)(φxψy),(11)B3=10ψyyψy,(12)B2=10ψyyyψy,(13)B1=-20(2φxxψyy-φxψxyy)(φxψy),(14)B0=-5(2φxxxφxψy-9φxx2ψy+8φxxφxψxy-2φx2ψxxy)(φx2ψy)-1,(15)C1=15ψyyyψy,(16)C0=-15(2φxxψyy-φxψxyy)(φxψy),(17)D3=10ψyyyyψy,(18)D2=-30(2φxxψyyy-φxψxyyy)(φxψy),(19)D1=(-2φxxxφxψyy)15(9φxx2ψyy-8φxxφxψxyy+2φx2ψxxyy-2φxxxφxψyy))(φx2ψy)-1,(20)D0=-5(φxxxxφx2ψy-12φxxxφxxφxψy+6φxxxφx2ψxy+21φxx3ψy-27φxx2φxψxy+12φxxφx2ψxxy-2φx3ψxxxy)×(φx3ψy)-1,(21)E5=ψyyyyyψy,(22)E4=-5(2φxxψyyyy-φxψxyyyy)(φxψy),(23)E3=(-5((2φxxxφx-9φxx2)ψyyy+2(4φxxψxyyy-φxψxxyyy)φx)×(φx2ψy)-1,(24)E2=5((6(2φxxψyy-φxψxyy)φxxx-φxxxxφxψyy)φx-(21φxx3ψyy-27φxx2φxψxyy+12φxxφx2ψxxyy-2φx3ψxxxyy))×(φx3ψy)-1,(25)E1=(5(21φxx4ψy-42φxx3φxψxy+27φxx2φx2ψxxy-8φxxφx3ψxxxy+φx4ψxxxxy+2φxxx2φx2ψy)-ψxxxxxφx3ψy+5(3φxxψy-2φxψxy)φxxxxφx2-15(7φxx2ψy-8φxxφxψxy+2φx2ψxxy)×φxxxφx21φxx4ψy-42φxx3φxψxy+27φxx2φx2ψxxy)(φx4ψy)-1,(26)E0=-(ψxxxxxφx3ψx-15φxxxxφxxφx2ψx+5φxxxxφx3ψxx-10φxxx2φx2ψx+105φxxxφxx2φxψx-60φxxxφxxφx2ψxx+10φxxxφx3ψxxx-105φxx4ψx+105φxx3φxψxx-45φxx2φx2ψxxx+10φxxφx3ψxxxx-φx4ψxxxxx)×(φx4ψy)-1.

Proof.

Applying a fiber preserving transformation (7), one obtains the following transformation of derivatives: (27)u(t)  =DxψDxφ=ψx+yψyφx=P(x,y,y),(28)u′′(t)=DxPDxφ=Px+yPy+y′′Pyφx=1φx3[(φxψy)y′′+(φxψyy)y2+(-φxxψy+2φxψxy)y-φxxψx+φxψxx]=Q(x,y,y,y′′),(29)u′′′(t)=DxQDxφ=Qx+yQy+y′′Qy+y′′′Qy′′φx=1φx5[(φx2ψy)y′′′+(3φx2ψyy)yy′′+3φx(-φxxψy+φxψxy)y′′+]=R(x,y,y,y′′,y′′′),(30)u(4)(t)=DxRDxφ=Rx+yRy+y′′Ry+y′′′Ry′′+y(4)Ry′′′φx=1φx7[(φx3ψy)y(4)+(4φx3ψyy)yy′′′+2φx2(-3φxxψy+2φxψxy)y′′′+]=S(x,y,y,y′′,y′′′,y(4)),(31)u(5)(t)=DxSDxφ=Sx+ySy+y′′Sy+y′′′Sy′′+y(4)Sy′′′+y(5)Sy(4)φx=1φx9[(φx4ψy)y(5)+(5φx4ψyy)yy(4)+5φx3(-2φxxψy+φxψxy)y(4)+], where Dx=(/x)+y(/y)+y′′(/y)+y′′′(/y′′)+y(4)(/y′′′)+y(5)(/y(4))+ is a total derivative. Substituting u(5)(t) into the linear equation (5), we have (32)y(5)+((5ψyyψy)y-5(2φxxψy-φxψxy))y(4)+((10ψyyψy)y′′+(10ψyyyψy)y2-(20(2φxxψyy-φxψxyy)(φxψy))y+)y′′′+((15ψyyyψy)y+)y′′2+=0. Denoting Ai, Bi, Ci, Di, and Ei as (10)–(21), we obtain the necessary form (8). These prove the theorem.

3. Formulation of the Linearization Theorem

We have shown in the previous section that every linearizable fifth-order ordinary differential equation belongs to the class of (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.

Theorem 3.

Sufficient conditions for (8) to be linearizable via a fiber preserving transformation are as follows: (33)A0y=A1x,(34)B3=2A1,(35)A1y=-(2A12-5B2)10,(36)A1x=-(4A0A1-5B1)20,(37)B2=2C13,(38)B1=4C03,(39)C1y=-(2A1C1-15D3)10,(40)C0y=-(2A1C0-5D2)10,(41)D3y=-(A1D3-50E5)5,(42)D2y=-(A1D2-30E4)5,(43)B0y=-2(15A0xA1-25C0x+3A02A1-5A0C0)75,(44)C0x=(30A0xA1+6A02A1-10A0C0-15A1B0+25D1)50,(45)A0xx=-(60A0xA0-75B0x+8A03-30A0B0+50D0)50,(46)D1y=-(A1D1-15E3)5,(47)D0y=(60A0xA0A1-100A0xC0-75B0xA1+125D1x+12A03A1-20A02C0-45A0A1B0+25A0D1-75A1D0+50B0C0+375E2)(750)-1,(48)B0xx=(-65A02B0+100A0D0+100B02-625E1))2(175A0x2+70A0xA02-325A0xB0-75B0xA0+500D0x+7A04-65A02B0+100A0D0+100B02-625E1))×(375)-1.

Proof.

For obtaining sufficient conditions, one has to solve the compatibility problem. Considering the representations of the coefficients Ai, Bi, Ci, Di, and Ei through the unknown functions φ and ψ. We first rewrite the expressions (9) and (10) for A1 and A0 in the following form: (49)ψyy=ψyA15,(50)φxx=(5ψxy-ψyA0)φx(10ψy). Differentiating (50) with respect to y, one obtains the condition (33). Substituting the expressions of ψyy and φxx into (11), (12), (13), (15), (16), (17), (18), (21), and (22) one gets conditions (34)–(42), respectively. From (14) we have (51)ψxxy=-(20A0xψy2-125ψxy2+10ψxyψyA0+7ψy2A02-20ψy2B0)(100ψy)-1. Comparing the mixed derivative (ψxxy)y=(ψyy)xx one obtains the condition (43). Equations (19), (20), (23), (24), and (25) provide the conditions (44)–(48), respectively.

Consider the form of ψyy:ψyy=(ψyA1/5) one can solve that (52)ψy=ω1(x,y)ψ1(x), where ω1(x,y)=e(A1/5)dy and ψ1(x)=eK1(x). Since ψy0 then ψ1 and ω1 cannot be zero. From ω1=e(A1/5)dy, we found the relation (53)A1=5ω1yω1. Relations (A1)x=A1x and (A1)y=A1y provide the conditions (54)ω1xy=(15ω1xω1y-3A0ω1yω1+C0ω12)(15ω1),ω1yy=C1ω115, respectively, and the relation (52) satisfied the condition (ψy)y=ψyy. cComposing the relation (ψy)xx=ψxxy, one has the equation (55)ψ1xx=(-20A0xω12ψ12-100ω1xxω1ψ12+125ω1x2ψ12+50ω1xψ1xω1ψ1-10ω1xA0ω1ψ12+125ψ1x2ω12-10ψ1xA0ω12ψ1-7A02ω12ψ12+20B0ω12ψ12)(100ω12ψ1)-1. Consider ψy=ω1(x,y)ψ1(x); one can solve that (56)ψ=ψ1(x)ω2(x,y)+ψ2(x), where ω2(x,y)=ω1(x,y)dy. Because of ω1(x,y)dy=ω2(x,y) then (57)ω1(x,y)=ω2y. Since ω10 then ω2y0. Substituting ω1 into ω1xy and ω1yy we obtain the additional conditions (58)ω2xyy=((-3ω2yyA0+ω2yC0)ω2y+15ω2xyω2yy)(15ω2y),ω2yyy=ω2yC115, respectively, and these satisfied the relations (ψ)y=ψy, (ψ)yy=ψyy, and (ψ)xxy=ψxxy. From (26), setting μ1(x,y), μ2(x,y), and μ3(x,y) as (A.1), (see Appendix) then we obtain (59)ψ2xxxxx=(-140625ψ1x4ψ2xω2y5+1125000ψ1x3ψ2xxω2y5ψ1+112500ψ1x3ψ2xω2y4ψ1(-5ω2xy+ω2yA0)-2250000ψ1x2ψ2xxxω2y5ψ12+675000ψ1x2ψ2xxω2y4ψ12(5ω2xy-ω2yA0)+11250ψ1x2ψ2xω2y3ψ12×(-40A0xω2y2-75ω2xy2+30ω2xyω2yA0-11ω2y2A02+20ω2y2B0)+1500000ψ1xψ2xxxxω2y5ψ13+900000ψ1xψ2xxxω2y4ψ13(-5ω2xy+ω2yA0)+75000ψ1xψ2xxω2y3ψ13×(16A0xω2y2+45ω2xy2-18ω2xyω2yA0+5ω2y2A02-8ω2y2B0)+1500ψ1xψ2xω2y2ψ13×(200A0xω2xyω2y2-40A0xω2y3A0+375ω2xy3-225ω2xy2ω2yA0+85ω2xyω2y2A02-100ω2xyω2y2B0-11ω2y3A03+20ω2y3A0B0-2μ3)+300000ψ2xxxxω2y4ψ14×(5ω2xy-ω2yA0)+30000ψ2xxxω2y3ψ14×(-20A0xω2y2-75ω2xy2+30ω2xyω2yA0-7ω2y2A02+10ω2y2B0)+3000ψ2xxω2y2μ3ψ14+25ψ2xω2yμ2ψ14+16μ1ψ15)×(300000ω2y5ψ14)-1. By the proof of Theorem 3, we arrive at the following Corollary.

Corollary 4.

Provided that the sufficient conditions in Theorem 3 are satisfied, the transformation (7) of mapping equation (8) to a linear equation u(5)(t) is obtained by solving the compatible system of equations (49), (50), (56), (55), and (59) for the functions φ(x) and ψ(x,y).

4. Examples Example 1.

Consider the nonlinear ordinary differential equation (60)x2yy(5)+5(x2y+2xy)y(4)+10(x2y+4xy+2y)y+30xy2+60yy=0. It is an equation of the form (8) with the coefficients (61)A1=5y,A0=10x,B3=10y,B2=0,B1=40xy,B0=20x2,C1=0,  C0=30xy,  D3=0,D2=0,D1=60x2y,  D0=0,E5=0,  E4=0,  E3=0,E2=0,E1=0,E0=0,ω1=y,ω2=y22,μ1=0,μ2=-90000y4x4,μ3=-3000y3x3. Applying Theorem 3 for checking the linearity, the coefficients in (61) obey all the conditions in Theorem 3, so that one concludes that equation (60) is linearizable. Applying Corollary 4, the linearizing transformation is found by solving the following equations: (62)φxx=(φxψ1xx-2ψ1)(2ψ1x),(63)ψ1xx=(5ψ1x2x2-4ψ1xψ1x-4ψ12)(4ψ1x2),(64)ψ2xxxxx=(5(-3ψ1x4ψ2xx4+24ψ1x3ψ2xxψ1x4+24ψ1x3ψ2xψ1x3-48ψ1x2ψ2xxxψ12x4-144ψ1x2ψ2xxψ12x3-72ψ1x2ψ2xψ12x2+32ψ1xψ2xxxxψ13x4+192ψ1xψ2xxψ13x3+288ψ1xψ2xxψ13x2+96ψ1xψ2xψ13x-64ψ2xxxxψ14x3-192ψ2xxxψ14x2-192ψ2xxψ14x-48ψ2xψ14))×(32ψ14x4)-1(65)ψ=(ψ1y2+2ψ2)2. Since φy=0, then one can take the simplest solution (66)φ=x. Thus, (62) becomes ψ1xx-2ψ1=0; the solution for this equation is (67)ψ1=Cx2. Choosing C=2, we have (68)ψ1=2x2. This solution satisfied (63). Equation (64) becomes (69)ψ2xxxxx=0. Choosing the particular solution (70)ψ2=0. Hence (65) is in the form (71)ψ=x2y2. So one obtains the linearizing transformation (72)t=x,u=x2y2. Thus, the nonlinear equation (60) can be mapped by transformation of (72) into the linear equation u(5)(t)=0. Next, we will find the solution of (60). Since (73)u(5)(t)=0, then we get the general solution (74)u(t)=C0t424+C1t36+C2t22+C3t+C4, where C0, C1, C2, C3, and C4 are arbitrary constants. Substituting (72) into (74) we get (75)x2y2=C0x424+C1x36+C2x22+C3x+C4.

Example 2.

Consider the nonlinear ordinary differential equation (76)16y4y(5)-40(y3y+4y4)y(4)-40(2y3y′′-3y2y2-8y3y-14y4)y′′′+60(3yy2+4y3)y′′2-20(15yy3+36y2y2+42y3y+40y4)y′′×105y5+300yy4+420y2y3+400y3y2+384y4y=0. It is an equation of the form equation (8) with the coefficients (77)A1=-52y,  A0=-10,  B3=-5y,B2=152y2,B1=20y,B0=35,C1=454y2,C0=15y,  D3=-754y3,D2=-45y2,D1=-1052y,D0=-50,E5=10516y4,  E4=754y3,  E3=1054y2,E2=25y,E1=24,  E0=0,ω1=1y,  ω2=2y,μ1=0,  μ2=-288000y2,μ3=5000yy. Applying Theorem 3 for checking the linearity, the coefficients in (77) obey all conditions in Theorem 3, so that one concludes that (76) is linearizable. Applying Corollary 4, the linearizing transformation is found by solving the following equations: (78)φxx=φx(ψ1x+2ψ1)(2ψ1),(79)ψ1xx=ψ1x(5ψ1x+4ψ1),(80)ψ2xxxxx=(-15ψ1x4ψ2x+120ψ1x3ψ2xxψ1-120ψ1x3ψ2xψ1-240ψ1x2ψ2xxxψ12+720ψ1x2ψ2xxψ12-480ψ1x2ψ2xψ12+160ψ1xψ2xxxxψ13-960ψ1xψ2xxxψ13+1760ψ1xψ2xxψ13-960ψ1xψ2xψ13+320ψ2xxxxψ14-1120ψ2xxxψ14+1600ψ2xxψ14-768ψ2xψ14)×(32ψ14)-1,(81)ψ=(ψ1y2+2ψ2)2. From (78), (82)φxxφx=ψ1x2ψ1+1. Taking the particular solution φ=ex, then the solution of (82) is (83)ψ1=C. Choosing C=(1/2), we have (84)ψ1=12. This solution satisfied (79). Equation (80) becomes (85)ψ2xxxxx=10ψ2xxxx-35ψ2xxx+50ψ2xx-24ψ2x. Choosing the particular solution (86)ψ2=0. Hence (81) is in the form (87)ψ=y. So one obtains the linearizing transformation (88)t=ex,u=y. Thus, the nonlinear equation (76) can be mapped by transformation of (88) into the linear equation u(5)(t)=0. Next, we will find the solution of (76). Since (89)u(5)(t)=0, then we get the general solution (90)u(t)=C0t424+C1t36+C2t22+C3t+C4, where C0, C1, C2, C3, and C4 are arbitrary constants. Substituting (88) into (90) we get (91)y=C0e4x24+C1e3x6+C2e2x2+C3ex+C4.

Appendix Equations in Section <xref ref-type="sec" rid="sec3">3</xref>

Consider(A.1)μ1=700A0x2ω2y5A0ω2-1750A0xB0xω2y5ω2+280A0xω2y5A03ω2-1050A0xω2y5A0B0ω2+2750A0xω2y5D0ω2-350B0xω2y5A02ω2+875B0xω2y5B0ω2+1250D0xxω2y5ω2+500D0xω2y5A0ω2-6250E1xω2y5ω2+2250000ω2xy5ω2-2250000ω2xy4ω2xω2y-450000ω2xy4ω2yA0ω2-4500000ω2xy3ω2xxyω2yω2+1125000ω2xy3ω2xxω2y2+450000ω2xy3ω2xω2y2A0+112500ω2xy3ω2y2B0ω2+1125000ω2xy2ω2xxxyω2y2ω2-375000ω2xy2ω2xxxω2y3+3375000ω2xy2ω2xxyω2xω2y2+675000ω2xy2ω2xxyω2y2A0ω2-225000ω2xy2ω2xxω2y3A0-112500ω2xy2ω2xω2y3B0-37500ω2xy2ω2y3D0ω2-187500ω2xyω2xxxxyω2y3ω2+93750ω2xyω2xxxxω2y4-750000ω2xyω2xxxyω2xω2y3-150000ω2xyω2xxxyω2y3A0ω2+75000ω2xyω2xxxω2y4A0+1687500ω2xyω2xxy2ω2y2ω2-1125000ω2xyω2xxyω2xxω2y3-450000ω2xyω2xxyω2xω2y3A0-112500ω2xyω2xxyω2y3B0ω2+56250ω2xyω2xxω2y4B0+37500ω2xyω2xω2y4D0+18750ω2xyω2y4E1ω2+18750ω2xxxxxyω2y4ω2-18750ω2xxxxxω2y5+93750ω2xxxxyω2xω2y4+18750ω2xxxxyω2y4A0ω2-18750ω2xxxxω2y5A0-375000ω2xxxyω2xxyω2y3ω2+187500ω2xxxyω2xxω2y4+75000ω2xxxyω2xω2y4A0+18750ω2xxxyω2y4B0ω2+187500ω2xxxω2xxyω2y4-18750ω2xxxω2y5B0-562500ω2xxy2ω2xω2y3-112500ω2xxy2ω2y3A0ω2+112500ω2xxyω2xxω2y4A0+56250ω2xxyω2xω2y4B0+18750ω2xxyω2y4D0ω2-18750ω2xxω2y5D0-18750ω2xω2y5E1+18750ω2y6E0+28ω2y5A05ω2-210ω2y5A03B0ω2+550ω2y5A02D0ω2+350ω2y5A0B02ω2-1250ω2y5A0E1ω2-1250ω2y5B0D0ω2,μ2=-3200A0x2ω2y4-18000A0xω2xy2ω2y2-2400A0xω2xyω2y3A0-2000A0xω2y4A02+5600A0xω2y4B0+12000B0xω2xyω2y3-4000D0xω2y4-5625ω2xy4+4500ω2xy3ω2yA0-4950ω2xy2ω2y2A02+9000ω2xy2ω2y2B0-300ω2xyω2y3A03+3600ω2xyω2y3A0B0-12000ω2xyω2y3D0-281ω2y4A04+1480ω2y4A02B0-800ω2y4A0D0-2000ω2y4B02+8000ω2y4E1,μ3=400A0xω2xyω2y2-100B0xω2y3+375ω2xy3-225ω2xy2ω2yA0+125ω2xyω2y2A02-200ω2xyω2y2B0-3ω2y3A03-20ω2y3A0B0+100ω2y3D0.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was financially supported by Naresuan University, Thailand.

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