first integrals integrating factors and invariant solutions of the path equation based on noether and symmetries

Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 284653, 15 pages http://dx.doi.org/10.1155/2013/284653 Research Article First Integrals, Integrating Factors, and Invariant Solutions of the Path Equation Based on Noether and 𝜆-Symmetries Gülden Gün1 and Teoman Özer2 Department of Mathematics, Faculty of Science and Letters, Istanbul Technical University, Maslak, 34469 Istanbul, Turkey Division of Mechanics, Faculty of Civil Engineering, Istanbul Technical University, Maslak, 34469 Istanbul, Turkey ă Correspondence should be addressed to Teoman Ozer; tozer@itu.edu.tr Received 13 March 2013; Accepted 28 April 2013 Academic Editor: Nail Migranov ¨ Copyright © 2013 G G¨un and T Ozer 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 We analyze Noether and 𝜆-symmetries of the path equation describing the minimum drag work First, the partial Lagrangian for the governing equation is constructed, and then the determining equations are obtained based on the partial Lagrangian approach For specific altitude functions, Noether symmetry classification is carried out and the first integrals, conservation laws and group invariant solutions are obtained and classified Then, secondly, by using the mathematical relationship with Lie point symmetries we investigate 𝜆-symmetry properties and the corresponding reduction forms, integrating factors, and first integrals for specific altitude functions of the governing equation Furthermore, we apply the Jacobi last multiplier method as a different approach to determine the new forms of 𝜆-symmetries Finally, we compare the results obtained from different classifications Introduction In a fluid medium, drag forces are the major sources of energy loss for moving objects Fuel consumption may have reduced to minimize the drag work This can be achieved by the selection of optimum path The drag force depends on the density of fluid, the drag coefficient, the cross-sectional area, and the velocity These parameters are the combination of the altitude-dependent parameters which can be expressed as a single arbitrary function If all parameters are assumed to be constants, then the minimum drag work path would be a linear path But these parameters change during the motion And all parameters can be defined as the function of altitude [1, 2] The main purpose of the work is to study Noether and 𝜆-symmetry classifications of the path equation for the different forms of arbitrary function of the governing equation [3–7] Based on Noether’s theorem, if Noether symmetries of an ordinary differential equation are known, then the conservation laws of this equation can be obtained directly by using Euler-Lagrange equations [8] However, in order to apply this theorem, a differential equation should have standard Lagrangian Thus, an important problem in such studies is to determine the standard Lagrangian of the differential equation In fact, for many problems in the literature, it may not be possible to determine the Lagrangian function of the equation To overcome this problem, partial Lagrangian method can be used alternatively and the Noether symmetries and first integrals can be obtained in spite of the fact that the differential equation does not have a standard Lagrangian [9] Here, we examine the partial Lagrangian of path equation and classify the Noether symmetries and first integrals corresponding to special forms of arbitrary function in the governing equation The second type of classification that is called 𝜆symmetries is carried out by using the relation with Lie point symmetries as a direct method For second-order ordinary differential equation, the method of finding 𝜆-symmetries has been investigated extensively by Muriel and Romero [10, 11] They have demonstrated that integrating factors and the integrals from 𝜆-symmetries for a second-order ordinary differential equation can be determined algorithmically [12] In their studies, for the sake of simplicity, the 𝜆-symmetry is assumed to be a linear form as 𝜆(𝑥, 𝑦) = 𝜆1 (𝑥, 𝑦)𝑦󸀠 +𝜆2 (𝑥, 𝑦) However, it is possible to show that the 𝜆-symmetry cannot be chosen generally in this linear form Therefore, we propose in this study to use the relation between Lie point symmetries and 𝜆-symmetries for the classification 2 Abstract and Applied Analysis The other classification that we discuss in our study is how to obtain 𝜆-symmetries with the Jacobi last multiplier approach Recently, Nucci and Levi [13] have shown that 𝜆symmetries and corresponding invariant solutions can be algorithmically obtained by using the Jacobi last multiplier This new approach includes the new determining equation including 𝜆-function that can be obtained from the divergence of the ordinary differential equation In the 𝜆symmetries approach based on a new form of the prolongation formula, the determining equations are difficult to solve since they include three unknown variables to determine and then the determining equation cannot be reduced to a simpler form However, by considering the Jacobi last multiplier approach, first we determine the 𝜆-function, which reduces to two the number of unknown functions, and then the other functions called infinitesimals functions can be calculated easily Taking into account these ideas we analyze 𝜆-symmetries of the path equation for different cases of the altitude function The outline of this work is as follows In the next section, we present the necessary preliminaries In Section 3, Noether symmetries, first integrals, and some invariant solutions of path equation are obtained In Section 4, firstly we introduce some fundamental information about 𝜆-symmetries, integration factors, and first integrals, and then 𝜆-symmetries corresponding to different choice of the arbitrary function are investigated Also for some cases the reduced forms of path equation are found and the new solutions of path equation are established Section is devoted to introduce another approach that is called Jacobi last multiplier to investigate the 𝜆-symmetries The conclusions and results are discussed in Section where 𝛼 , 𝜁𝛼𝑠 = 𝐷𝑥𝑠 (𝑊𝛼 ) + 𝜉𝑦𝑠+1 𝑠 ≥ 2, 𝛼 = 1, 2, , 𝑚, (5) 𝛼 in which 𝑊 is the Lie characteristic function 𝑊𝛼 = 𝜂𝛼 − 𝜉𝑦𝑥𝛼 , 𝛼 = 1, 2, , 𝑚 (6) For convenience the generalized operator (4) can be rewritten by using characteristic function such as 𝑋 = 𝜉𝐷𝑥 + 𝑊𝛼 𝜕 𝜕 + ∑𝐷𝑠 (𝑊𝛼 ) 𝛼 , 𝜕𝑦𝛼 𝑠≥1 𝑥 𝜕𝑦𝑠 (7) and the Noether operator associated with a generalized operator 𝑋 can be defined 𝑁 = 𝜉 + 𝑊𝛼 𝜕 𝜕 + ∑𝐷𝑠 (𝑊𝛼 ) 𝛼 ⋅ 𝜕𝑦𝛼 𝑠≥1 𝑥 𝜕𝑦𝑠 (8) Definition Let us consider an 𝑛th-order ordinary differential equation system 𝐹𝛼 (𝑥, 𝑦, 𝑦(1) , 𝑦(2) , , 𝑦(𝑛) ) = 0, 𝛼 = 1, 2, , 𝑚, (9) then the first integral of this system is a differential function (9) 𝐼 ∈ A, the universal space and the vector space of all differential functions of all finite orders, which is given by the following formula: 𝐷𝑥 𝐼 = 0, (10) and this equality is valid for every solution of (9) The first integral is also referred to as the local conservation law Definition Let (9) be in the following form Preliminaries 𝐹𝛼 ≡ 𝐹𝛼0 + 𝐹𝛼1 = 0, Let us assume that 𝑥 be the independent variable and 𝑦 = (𝑦1 , , 𝑦𝑚 ) be the dependent variable with functions 𝑦𝛼 The derivatives of 𝑦𝛼 with respect to 𝑥 are given by 𝑦𝑥𝛼 = 𝑦1𝛼 = 𝐷𝑥 (𝑦𝛼 ) , 𝑦𝑠𝛼 = 𝐷𝑥𝑠 (𝑦𝛼 ) , 𝑠 ≥ 2, 𝛼 = 1, 2, , 𝑚, (1) where 𝐷𝑥 is the total derivative operator [14–18] with respect to 𝑥, which can be defined as 𝐷𝑥 = 𝜕 𝜕 𝛼 𝜕 + 𝑦𝑥𝛼 𝛼 + 𝑦𝑥𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦𝑥𝛼 (2) Definition For each 𝛼 we can define the operator 𝜕 𝛿 𝑠 𝜕 = + ∑(−𝐷𝑥 ) , 𝛿𝑦𝛼 𝜕𝑦𝛼 𝑠≥1 𝜕𝑦𝑠𝛼 𝛼 = 1, 2, , 𝑚, (3) which is called the Euler-Lagrange operator 𝜕 𝜕 𝜕 + 𝜂𝛼 𝛼 + ∑𝜁𝛼𝑠 𝛼 , 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝑠 𝑠≥1 (4) (11) and 𝐿 = 𝐿(𝑥, 𝑢, 𝑢(1) , 𝑢(2) , , 𝑢(𝛼) ) ∈ A, 𝛼 ≤ 𝑘 and then nonzero functions 𝑓𝛼𝛽 ∈ A satisfy the relations 𝛿𝐿/𝛿𝑢𝛼 = 𝑓𝛼𝛽 𝐹𝛽1 , 𝐹𝛽1 ≠ 0, in which 𝐿 is called partial Lagrangian of (11) Otherwise, 𝐿 is a standard Lagrangian On the other hand the Euler-Lagrange equations can be defined as following form 𝛿𝐿 = 0, 𝛿𝑢𝛼 𝛼 = 1, 2, , 𝑚, (12) and similarly the form of partial Euler-Lagrange equations is 𝛿𝐿 = 𝑓𝛼𝛽 𝐹𝛽1 𝛿𝑢𝛼 (13) Definition Let 𝐵 ∈ A be a vector that satisfies 𝐵 ≠ 𝑁𝐿 + 𝐶, where 𝐶 is a constant Then 𝑋(𝛼) represents 𝛼th prolongation of the generalized operator (7), and partial Noether operator corresponding to a partial Lagrangian is formulated as 𝑋(𝛼) 𝐿 + 𝐿𝐷𝑥 (𝜉) = 𝑊𝛼 Definition Generalized operator can be formulated as 𝑋=𝜉 𝛼 = 1, 2, , 𝑚 𝛿𝐿 + 𝐷𝑥 (𝐵) , 𝛿𝑦𝛼 (14) in which 𝑊 = (𝑊1 , , 𝑊𝑚 ), 𝑊𝛼 ∈ A, is the characteristic of 𝑋 Also 𝐵(𝑥, 𝑦) is called the gauge function Abstract and Applied Analysis Definition If 𝑋 is a partial Noether operator corresponding to partial Lagrangian 𝐿, then the gauge function 𝐵(𝑥, 𝑦) exists Hence, the first integral is given by 󸀠 𝐼 = 𝜉𝐿 + (𝜂 − 𝑦 𝜉) 𝐿 𝑦󸀠 − 𝐵 (15) Noether Symmetries of Path Equation The differential equation describing the path of the minimum drag work is given in the form 𝑦󸀠󸀠 − 𝑓󸀠 (𝑦) 𝑓󸀠 (𝑦) − 𝑦󸀠2 = 0, 𝑓 (𝑦) 𝑓 (𝑦) (16) where 𝑦 = 𝑦(𝑥) is the altitude function In this section we use partial Lagrangian approach to analyze Noether symmetries Firstly, we can determine the Euler-Lagrange operator (3) for the path equation (16) such as 𝜕 𝜕 𝜕 𝛿 = 𝛼 − 𝐷𝑥 + 𝐷𝑥2 , 𝛼 𝛼𝑦 𝜕𝑦 𝜕𝑦𝑥 𝜕𝑦𝑥𝑥 (18) 𝑓󸀠 (𝑦) = 0, 𝜉𝑦 + 𝜉 𝑓 (𝑦) (19) 𝑓󸀠 (𝑦) = 0, 𝜂 𝑦 − 𝜉𝑥 + 𝜂 𝑓 (𝑦) (20) 𝜂𝑥 + 𝜉𝑦 ln 𝑓 (𝑦) − 𝐵𝑦 = 0, (21) 𝑓󸀠 (𝑦) = 𝑓 (𝑦) If the infinitesimals 𝜉 (23) and 𝜂 (24) are inserted into (26) then one can find the following classification relationship in terms of 𝑓(𝑦): 𝑏 (𝑥) (4𝑓(𝑦) − 2𝑓 (𝑦) 𝑓󸀠󸀠 (𝑦)) + 𝑎󸀠 (𝑥) (−3𝑓󸀠 (𝑦) + ∫ 𝑓 (𝑦) 𝑑𝑦 −𝑓 (𝑦) ∫ 𝜉= 𝑓(𝑦) , 𝑑𝑦 𝑎󸀠 (𝑥) + 𝑏 (𝑥)) ( ∫ 𝑓 (𝑦) 𝑓 (𝑦) 3.1 𝑓(𝑦) = 𝑘 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 For this case the solution of (27) gives to the following infinitesimals: (28) 𝑐2 𝑦 + 2𝑥𝑦𝑐3 + 2𝑘𝑐4 + 2𝑘𝑥𝑐5 , 𝜂= 2𝑘2 where 𝑐𝑖 are constants 𝑖 = 1, , Integrating (21) with respect to 𝑦 gives 𝑦2 𝑐3 + 2𝑦𝑐5 + 2𝑥𝑐2 ln 𝑘 + 2𝑥2 𝑐3 ln 𝑘 (29) 2𝑘2 The associated infinitesimal generators turn out to be 𝐵 (𝑥, 𝑦) = 𝑋1 = 𝑋2 = 𝐵𝑦𝑥 = 𝜂𝑥𝑥 + 𝜉𝑦𝑥 ln 𝑓 (𝑦) , 𝑓󸀠 (𝑦) 𝑓󸀠 (𝑦) 𝑓󸀠 (𝑦) = 𝜉𝑥𝑦 ln 𝑓 (𝑦) + 𝜉𝑥 + 𝜂𝑦 + 𝜂( ) 𝑓 (𝑦) 𝑓 (𝑦) 𝑓 (𝑦) (25) 𝜕 , 𝑘2 𝜕𝑥 𝑦 𝜕 𝑥 𝜕 + , 𝑘2 𝜕𝑥 2𝑘2 𝜕𝑦 𝑥𝑦 𝜕 𝑥2 𝜕 𝑋3 = + , 𝑘 𝜕𝑥 𝑘 𝜕𝑦 (23) (24) 𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥2 , 𝑘2 𝜉= (22) Differentiating (21)-(22) with respect to 𝑥 and 𝑦, respectively, gives 𝐵𝑥𝑦 𝑑𝑦 ) = 𝑓 (𝑦) (27) and then substituting (23) into (20) and solving for 𝜂 yield 𝜂= 𝑑𝑦 󸀠󸀠 𝑓 (𝑦)) 𝑓 (𝑦) + 𝑓(𝑦) (2𝑏󸀠󸀠 (𝑥) + 𝑎󸀠󸀠󸀠 (𝑥) ∫ To find the infinitesimals 𝜉 and 𝜂, (19)–(22) should be solved together First, (19) is integrated as 𝑎 (𝑥) (26) Here several cases should be examined separately for different forms of 𝑓(𝑦) Then the application of (18) to (14) and separation with respect to powers of 𝑦󸀠 and arranging yield the set of determining equations, the over-system of partial differential equations 𝜉𝑥 ln 𝑓 (𝑦) − 𝐵𝑥 + 𝜂 𝑓󸀠 (𝑦) 𝑓󸀠 (𝑦) ) + (𝜉𝑥 − 𝜂𝑦 ) − 𝜂𝑥𝑥 = 𝜂( 𝑓 (𝑦) 𝑓 (𝑦) (17) and the partial Lagrangian 𝐿 for the path equation (16) is 𝐿 = 𝑦󸀠2 + ln 𝑓 (𝑦) Using (25) and eliminating 𝐵, we find that 𝑋4 = 𝜕 , 𝑘 𝜕𝑦 𝑋5 = (30) 𝑥 𝜕 𝑘 𝜕𝑦 Thus, the first integrals by Definition are given as follows: 𝐼1 = 𝑐1 (2 ln 𝑘 − 𝑦󸀠 ) 2𝑘2 𝐼3 = 𝐼4 = 𝑦󸀠 , 𝑘 , 𝐼2 = 𝑦𝑦󸀠 − 𝑥𝑦󸀠2 , 2𝑘2 −𝑦2 + 2𝑥𝑦𝑦󸀠 − 𝑥2 𝑦󸀠2 , 2𝑘2 𝐼5 = −2𝑘𝑦 + 2𝑘𝑥𝑦󸀠 2𝑘2 (31) Abstract and Applied Analysis 3.2 𝑓(𝑦) = 𝑦 For the linear case of 𝑓(𝑦), we obtain 𝜉= 𝑐1 , 𝑦2 𝜂 = 0, 𝐵 (𝑥, 𝑦) = 𝑐1 ( ln 𝑦 + ), 2𝑦2 𝑦 (32) 𝐼3 = − 𝑒−2𝑦𝛼 (1 + 𝑦󸀠 ) , 2𝑘 𝛼 𝐼5 = 𝐼4 = 𝑒−𝑦𝛼 󸀠 (𝑦 cos 𝑥𝛼 − sin 𝑥𝛼) , 𝑘 𝑒−𝑦𝛼 (cos 𝛼𝑥 + 𝑦󸀠 sin 𝑥𝛼) 𝑘 (38) where 𝑐1 is a constant The partial Noether operator is 𝑋1 = 𝜕 , 𝑦2 𝜕𝑥 (33) 3.4 𝑓(𝑦) = 1/(𝑚𝑦 + 𝑛) For this case, the infinitesimal functions read and the first integral is 𝜉 = (𝑚𝑦 + 𝑛) (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥2 ) , 𝐼1 = − (1 + 𝑦󸀠 ) 2𝑦2 (34) 𝜂 = (𝑚𝑦 + 𝑛) (− 𝑚𝑥2 𝑐2 − 𝑚𝑥3 𝑐3 + 𝑦 (2𝑛 + 𝑚𝑦) (𝑐2 + 2𝑥𝑐3 ) + 𝑐4 + 𝑥𝑐5 ) , (39) 3.3 𝑓(𝑦) = 𝑘𝑒𝛼 𝑦 The solution of determining equations for the form of 𝑓(𝑦) = 𝑘𝑒𝛼 𝑦 gives the following infinitesimals 𝜉= 𝜂= 𝑒−2𝑦𝛼 (𝑐 sin 2𝑥𝛼 − 𝑐2 cos 2𝑥𝛼 + 2𝑐3 𝛼) , 2𝑘2 𝛼 where 𝑐𝑖 are constants 𝑖 = 1, , 5, and the gauge function is 𝑒−2𝑦𝛼 (2𝑐4 𝑒𝛼𝑦 𝑘𝛼 cos 𝑥𝛼 2𝑘2 𝛼 𝐵 (𝑥, 𝑦) = −𝑐1 cos 2𝑥𝛼 + 2𝑐5 𝑒𝛼𝑦 sin 𝑥𝛼 − 𝑐2 sin 2𝑥𝛼) , (35) where 𝑐𝑖 are constants 𝑖 = 1, , 5, and the gauge function is 𝐵 (𝑥, 𝑦) = 𝑒−2𝑦𝛼 (2𝑐3 𝛼 + 4𝑦𝑐3 𝛼2 − 4𝑒𝑦𝛼 𝑐5 cos 𝑥𝛼 + 4𝑐3 𝛼 ln 𝑘 4𝑘2 𝛼 1 ( (2𝑛 + 𝑚𝑦) × (2𝑛𝑦𝑐3 + 𝑚 (4𝑐1 −2𝑥𝑐2 −2𝑥2 𝑐3+𝑦2 𝑐3 ) + 4𝑐5 ) + 2𝑚𝑦 (2𝑛 + 𝑚𝑦) (𝑐1 + 𝑥 (𝑐2 + 𝑐3 )) ln −2𝑛2 (𝑐1 + 𝑥 (𝑐2 + 𝑐3 )) ln (𝑚𝑦 + 𝑛) ) + 𝑥 (𝑚 (2𝑚𝑥2 𝑐2 + 𝑚𝑥3 𝑐3 − 8𝑐4 − 4𝑥𝑐5 ) − 𝑐2 cos 2𝑥𝛼 (2 ln 𝑘 + 2𝑦𝛼 − 1) ×4𝑐4 𝑒𝑦𝛼 𝑘𝛼 sin 𝑥𝛼 − 𝑐1 sin 2𝑥𝛼 + 8𝑛2 (𝑐2 + 𝑥𝑐3 ) ln +2𝑐1 𝑦𝛼 sin 2𝑥𝛼 + 2𝑐1 ln 𝑘 sin 2𝑥𝛼) (36) 𝑚𝑦 + 𝑛 +8𝑛2 (𝑐2 + 𝑥𝑐3 ) ln (𝑚𝑦 + 𝑛)) (40) The associated five-parameter symmetry generators take the form 𝑋1 = 𝑒−2𝑦𝛼 𝜕 𝑋3 = , 𝑘 𝛼 𝜕𝑥 𝑋5 = 𝑒−𝑦𝛼 cos 𝑥𝛼 𝜕 𝑋4 = , 𝑘 𝜕𝑦 𝑒−𝑦𝛼 sin 𝑥𝛼 𝜕 , 𝑘 𝜕𝑦 and the corresponding first integrals are 𝑒−2𝑦𝛼 𝐼1 = (−2 cos 2𝑥𝛼𝑦󸀠 − sin 2𝛼𝑥 (𝑦󸀠 − 1)) , 4𝑘 𝛼 𝐼2 = The corresponding Noether symmetry generators are 𝑒−2𝑦𝛼 sin 2𝑥𝛼 𝜕 𝑒−2𝑦𝛼 cos 2𝑥𝛼 𝜕 − , 2𝑘 𝛼 𝜕𝑥 2𝑘2 𝛼 𝜕𝑥 𝑒−2𝑦𝛼 cos 2𝑥𝛼 𝜕 𝑒−2𝑦𝛼 sin 2𝑥𝛼 𝜕 − , 𝑋2 = − 2𝑘2 𝛼 𝜕𝑥 2𝑘2 𝛼 𝜕𝑥 𝑒−2𝑦𝛼 (−2 sin 2𝑥𝛼𝑦󸀠 + cos 2𝛼𝑥 (𝑦󸀠 − 1)) , 4𝑘 𝛼 𝑚𝑦 + 𝑛 𝑋1 = (𝑚𝑦 + 𝑛) (37) 𝜕 , 𝜕𝑥 𝜕 + (𝑚𝑦 + 𝑛) 𝜕𝑥 𝑋2 = 𝑥(𝑚𝑦 + 𝑛) × (− 𝑚𝑦2 3𝑚𝑥2 𝜕 + (𝑛𝑦 + )) , 2 𝜕𝑦 𝑋3 = 𝑥2 (𝑚𝑦 + 𝑛) × (− (41) 𝜕 + (𝑚𝑦 + 𝑛) 𝜕𝑥 𝑚𝑦2 𝑚𝑥3 𝜕 + 𝑥 (𝑛𝑦 + )) , 𝜕𝑦 𝑋4 = (𝑚𝑦 + 𝑛) 𝜕 , 𝜕𝑦 𝑋5 = 𝑥 (𝑚𝑦 + 𝑛) 𝜕 𝜕𝑦 Abstract and Applied Analysis (b) For the same 𝑓(𝑦) function, the conservation law is And the conservation laws are 𝐼1 = 1 (−8𝑚𝑛𝑦 − 4𝑚2 𝑦2 + 8𝑛2 ln 𝑚𝑦 + 𝑛 𝐼= + 8𝑛2 ln (𝑚𝑦 + 𝑛) − (4𝑚𝑦 + 𝑛) 𝑦󸀠 ) , 𝐼2 = + 2𝑚𝑦2 (𝑚𝑦 + 𝑛) 𝑦󸀠 − 4𝑥(𝑚𝑦 + 𝑛) 𝑦󸀠 ) , 1 𝑦 (𝑥) = − ln (𝑐𝑘𝛼3 cos 𝑥𝛼 (−𝑐1 − tan 𝑥𝛼)) , 𝛼 𝛼 Case Let us consider 𝑓(𝑦) = 1/(𝑚𝑦 + 𝑛), then the first integral yields 𝐼 = − (−𝑚2 𝑥4 + 4𝑚𝑛𝑥2 𝑦 − 4𝑛2 𝑦2 + 2𝑚2 𝑥2 𝑦2 − 4𝑚𝑛𝑦3 − 𝑚2 𝑦4 − 4𝑚𝑥3 (𝑚𝑦 + 𝑛) 𝑦󸀠 + 8𝑛𝑥𝑦 (𝑚𝑦 + 𝑛) 𝑦󸀠 + 4𝑚𝑥𝑦2 (𝑚𝑦 + 𝑛) 𝑦󸀠 − 4𝑚𝑛𝑦3 − 𝑚2 𝑦4 − 4𝑚𝑥3 (𝑚𝑦 + 𝑛) 𝑦󸀠 󸀠2 − 4𝑥2 (𝑚𝑦 + 𝑛) 𝑦 ) , and the solution of this equation gives (42) 𝑦𝑛 For this choice of 𝑓(𝑦), we find the (43) where 𝑐1 is constant, and we have the first integral 𝐼 = − 𝑦−2𝑛 (1 + 𝑦󸀠 ) (44) For convenience all Noether symmetries and first integrals are presented in Table 3.6 Invariant Solutions Invariant solutions that satisfy the original path equation can be obtained by first integrals according to the relation 𝐷𝑥 𝐼 = We here determine some special cases and investigate the corresponding invariant solutions Case (a) For the case of 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 , the conservation law is 𝑒−2𝑦𝛼 𝐼 = − (1 + 𝑦󸀠 ) ; 2𝑘 𝛼 (45) by using the relation 𝐷𝑥 𝐼 = 0, then the invariant solution of path equation (16) is ln (− 𝛼 √−1 − tan (𝛼 (𝑥 + 𝑐1 ))2 where 𝑐1 , 𝑐 are constants 𝑘√2𝑐 𝑦 (𝑥) = −𝑛 − √−2𝑚√−2𝑐 + 𝑛2 − 𝑚2 𝑥2 − 2𝑚2 𝑥𝑐1 𝑚 , (50) where 𝑐1 , 𝑐 are constants, in which it is obvious that the invariant solution (50) satisfies the original path equation 𝜂 = 0, 𝐵 (𝑥, 𝑦) = 𝑐1 𝑦−2𝑛 (1 + ln (𝑦𝑛 )) , 𝑦 (𝑥) = + 4𝑚𝑥𝑦2 (𝑚𝑦 + 𝑛) 𝑦󸀠 − 4𝑥2 (𝑚𝑦 + 𝑛) 𝑦󸀠 ) , (4𝑚𝑥2 − 8𝑛𝑦 − 4𝑚𝑦2 + 8𝑥 (𝑚𝑦 + 𝑛) 𝑦󸀠 ) 𝜉 = 𝑐1 𝑦−2𝑛 , (49) + 8𝑛𝑥𝑦 (𝑚𝑦 + 𝑛) 𝑦󸀠 𝐼4 = 𝑚𝑥 + (𝑚𝑦 + 𝑛) 𝑦󸀠 , 3.5 𝑓(𝑦) = infinitesimals (48) where 𝑐1 , 𝑐 are constants 𝐼3 = − ( − 𝑚2 𝑥4 + 4𝑚𝑛𝑥2 𝑦 − 4𝑛2 𝑦2 + 2𝑚2 𝑥2 y2 𝐼5 = (47) and the invariant solution similar to previous one is (−2𝑚2 𝑥3 + 4𝑚𝑛𝑥𝑦 + 2𝑚2 𝑥𝑦2 − 6𝑚𝑥2 (𝑚𝑦 + 𝑛) 𝑦󸀠 𝑒−𝑦𝛼 (cos 𝑥𝛼𝑦󸀠 − sin 𝑥𝛼) , 𝑘 ), (46) 𝜆-Symmetries of Path Equation The relationship between 𝜆-symmetries, integration factors and first integrals of second-order ordinary differential equation is very important from the mathematical point of view [10–12] Let us consider first the second-order differential equation of the form 𝑦󸀠󸀠 = 𝜙 (𝑥, 𝑦, 𝑦󸀠 ) , (51) and let vector field of (51) be in the form of 𝐴 = 𝜕𝑥 + 𝑦󸀠 𝜕𝑦 + 𝜙 (𝑥, 𝑦, 𝑦󸀠 ) 𝜕𝑦󸀠 (52) In terms of 𝐴, a first integral of (51) is any function in the form of 𝐼(𝑥, 𝑦, 𝑦󸀠 ) providing equality of 𝐴(𝐼) = An integrating factor of (51) is any function satisfying the following equation: 𝜇 [𝑦󸀠󸀠 − 𝜙 (𝑥, 𝑦, 𝑦󸀠 )] = 𝐷𝑥 𝐼, (53) where 𝐷𝑥 is total derivative operator in the form of 𝐷𝑥 = 𝜕𝑥 + 𝑦󸀠 𝜕𝑦 + 𝑦󸀠󸀠 𝜕𝑦󸀠 + ⋅ ⋅ ⋅ (54) Thus 𝜆-symmetries of second-order differential equation (51) can be obtained directly by using Lie symmetries of this same equation Secondly, let 𝜐 = 𝜉 (𝑥, 𝑦) 𝜕 𝜕 + 𝜂 (𝑥, 𝑦) 𝜕𝑥 𝜕𝑦 (55) Abstract and Applied Analysis Table 1: Noether symmetry classification table of path equation Function Infinitesimals and first integrals 𝑐 𝑦 + 2𝑥𝑦𝑐3 + 2𝑘𝑐4 + 2𝑘𝑥𝑐5 𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥2 ,𝜂= 𝜉= 𝑘󸀠22 2𝑘2 𝑐1 (2 ln 𝑘 − 𝑦 ) 𝑦𝑦󸀠 − 𝑥𝑦󸀠2 −𝑦 + 2𝑥𝑦𝑦󸀠 − 𝑥2 𝑦󸀠2 𝐼1 = , 𝐼2 = , 𝐼3 = 2 2𝑘 2𝑘 2𝑘2 𝑦󸀠 −2𝑘𝑦 + 2𝑘𝑥𝑦󸀠 𝐼4 = , 𝐼5 = 𝑘 2𝑘2 (1 + 𝑦󸀠2 ) 𝑐 𝜉 = 12 , 𝜂 = 0, 𝐼1 = − 𝑦 2𝑦2 −2𝑦𝛼 𝑒 𝜉 = (𝑐1 sin 2𝑥𝛼 − 𝑐2 cos 2𝑥𝛼 + 2𝑐3 ) 2𝑘 𝛼 𝑒−2𝑦𝛼 𝑦𝛼 𝜂 = (2𝑐4 𝑘𝛼𝑒 cos 𝑥𝛼 − 𝑐1 cos 2𝑥𝛼 + 2𝑐5 𝑘𝛼𝑒𝑦𝛼 sin 𝑥𝛼 − 𝑐2 sin 2𝑥𝛼) 2𝑘 𝛼 𝑒−2𝑦𝛼 𝐼1 = (−2 cos 2𝑥𝛼𝑦󸀠 − sin 2𝛼𝑥(𝑦󸀠2 − 1)) 4𝑘 𝛼 𝑒−2𝑦𝛼 𝐼2 = (−2 sin 2𝑥𝛼𝑦󸀠 + cos 2𝛼𝑥(𝑦󸀠2 − 1)) 4𝑘 𝛼−𝑦𝛼 −2𝑦𝛼 𝑒 𝑒 𝑒−𝑦𝛼 󸀠2 𝐼3 = − (1 + 𝑦 ), 𝐼4 = (cos 𝑥𝛼𝑦󸀠 − sin 𝑥𝛼), 𝐼5 = (cos 𝑥𝛼 + sin 𝑥𝛼𝑦󸀠 ) 2𝑘 𝛼 𝑘 𝑘 𝑓(𝑦) = 𝑘 𝑓(𝑦) = 𝑦 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 𝑓(𝑦) = 𝜉 = (𝑚𝑦 + 𝑛)2 (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥2 ) 1 𝜂 = (𝑚𝑦 + 𝑛)(− 𝑚𝑥 𝑐2 − 𝑚𝑥3 𝑐3 + 𝑦(2𝑛 + 𝑚𝑦)(𝑐2 + 2𝑥𝑐3 ) + 𝑐4 + 𝑥𝑐5 ) 4 1 𝐼1 = (−8𝑚𝑛𝑦 − 4𝑚2 𝑦2 + 8𝑛2 ln + 8𝑛2 ln (𝑚𝑦 + 𝑛) − 4(𝑚𝑦 + 𝑛)2 𝑦󸀠2 ) 𝑚𝑦 + 𝑛 𝐼2 = (−2𝑚2 𝑥3 + 4𝑚𝑛𝑥𝑦 + 2𝑚2 𝑥𝑦2 − 6𝑚𝑥2 (𝑚𝑦 + 𝑛) 𝑦󸀠 +2𝑚𝑦2 (𝑚𝑦 + 𝑛) 𝑦󸀠 − 4𝑥(𝑚𝑦 + 𝑛)2 𝑦󸀠2 ) 𝐼3 = − (−𝑚 𝑥 + 4𝑚𝑛𝑥2 𝑦 − 4𝑛2 𝑦2 + 2𝑚2 𝑥2 𝑦2 − 4𝑚𝑛𝑦3 − 𝑚2 𝑦4 − 4𝑚𝑥3 (𝑚𝑦 + 𝑛) 𝑦󸀠 +8𝑛𝑥𝑦 (𝑚𝑦 + 𝑛) 𝑦󸀠 + 4𝑚𝑥𝑦2 (𝑚𝑦 + 𝑛) 𝑦󸀠 − 4𝑥2 (𝑚𝑦 + 𝑛)2 𝑦󸀠2 ) 𝐼4 = 𝑚𝑥 + (𝑚𝑦 + 𝑛)𝑦, 𝐼5 = (4𝑚𝑥2 − 8𝑛𝑦 − 4𝑚𝑦2 + 8𝑥 (𝑚𝑦 + 𝑛) 𝑦󸀠 ) −2𝑛 𝜉 = 𝑐1 𝑦 , 𝜂 = 0, 𝐼 = − 𝑦−2𝑛 (1 + 𝑦󸀠2 ) 𝑚𝑦 + 𝑛 𝑓(𝑦) = 𝑦𝑛 be a Lie point symmetry of (51), and then the characteristic of 𝜐 is O ¸ = 𝜂 − 𝜉𝑦󸀠 , (56) and for the path equation (16) the total derivative operator can be written as 𝐴= 󸀠 𝑓 (𝑦) 𝜕 𝜕 𝜕 ; + 𝑦󸀠 + (1 + 𝑦󸀠 ) 𝜕𝑥 𝜕𝑦 𝑓 (𝑦) 𝜕𝑦󸀠 (57) thus the vector field 𝜕𝑦 is called 𝜆-symmetry of (16) if the following equality is satisfied where 𝜐[𝜆,(1)] is the first-order 𝜆-prolongation of the vector field 𝜐 (2) The solution of (59) will be in terms of first order derivative of 𝑦 To write equation of (51) in terms of the reduced equation of 𝑤, we can obtain the firstorder derivative the solution of (59) and we can write (51) equation in terms of 𝑤 (3) Let 𝐺 be an arbitrary constant of the solution of the reduced equation written in terms of 𝑤 Therefore, 𝜇 = 𝐺𝑤 𝑤𝑦󸀠 (60) (58) is an integrating factor of (51) (4) The solution of 𝑤(𝑥, 𝑦, 𝑦󸀠 ) is the first integral of 𝜐[𝜆,(1)] The following four steps can be defined for finding 𝜆symmetries and first integrals 4.1 𝜆-Symmetries Using Lie Symmetries of Path Equation Let us consider an 𝑛th-order ODE as follows: 𝜆= 𝐴 (O ¸) O ¸ (1) Find a first integral 𝑤(𝑥, 𝑦, 𝑦󸀠 ) of 𝜐[𝜆,(1)] , that is, a particular solution of the equation 𝑤𝑦 + 𝜆𝑤𝑦󸀠 = 0, (59) 𝑦(𝑛) = 𝑓 (𝑥, 𝑦, 𝑦󸀠 , 𝑦󸀠󸀠 , , 𝑦(𝑛−1) ) (61) Thus the invariance criterion of (61) is 󵄨 pr𝑋 (𝑦(𝑛) − 𝑓 (𝑥, 𝑦, 𝑦󸀠 , 𝑦󸀠󸀠 , , 𝑦(𝑛−1) ) 󵄨󵄨󵄨󵄨 𝑦(𝑛) =𝑓 = (62) Abstract and Applied Analysis Taking derivative of (70) with respect to 𝑥 gives The expansion of relation (62) gives the determining equation related to path equation, which is the system of partial differential equations In this system there are three unknowns, namely, 𝜆, 𝜉, and 𝜂, which are difficult to solve because they are highly nonlinear In the literature [10–12], for the convenience the 𝜆 function are chosen generally in the form 𝜆 (𝑥, 𝑦, 𝑦󸀠 ) = 𝜆1 (𝑥, 𝑦) 𝑦󸀠 + 𝜆2 (𝑥, 𝑦) (63) In addition, for solving the remaining determining equations, the infinitesimal functions 𝜉 and 𝜂 are chosen specifically as 𝜉 = and 𝜂 = [10–12] Therefore, the number of unknowns in the equation is reduced to find 𝜆1 (𝑥, 𝑦) and 𝜆2 (𝑥, 𝑦) functions, and finally, 𝜆-symmetries can be determined explicitly However, for the path equation (16), it is possible to check that 𝜆-symmetries of this equation cannot be determined by taking the form of 𝜆 in (63) Thus, we study 𝜆-symmetries of path equation by using the relation with the Lie point symmetries of the same equation [2, 19] Here Lie point symmetries of path equation are examined by considering four different cases of function 𝑓(𝑦) 𝑦󸀠󸀠 = 𝑒2𝑤(𝑥) 𝑓 (𝑦 (𝑥)) × (𝑓󸀠 (𝑦 (𝑥)) + 𝜂 = 0, (64) 𝑤󸀠 (𝑥) = (65) Applying this generator (56), we obtain the characteristic O ¸ = −𝑎𝑦󸀠 󸀠 󸀠 𝐴 (O ¸ ) (1 + 𝑦 ) 𝑓 (𝑦) 𝜆= = O ¸ 𝑓 (𝑦) 𝑦󸀠 (67) 𝑤 (𝑥) = 𝐺, 𝑤𝑦 + 𝐺 ∈ R (73) According to (60),we find the integration factor 𝜇 to be of the form 𝜇= 𝑦󸀠 + 𝑦󸀠2 (74) Then the conserved form satisfies the following equality: + 𝑦󸀠 (𝑥)2 𝐷𝑥 ( ln ( )) = 0, 2 𝑓(𝑦 (𝑥)) (75) which gives the original path equation Thus the reduced equation is + 𝑦󸀠 (𝑥)2 ) − 𝑘 = 0, ln ( 2 𝑓(𝑦 (𝑥)) (76) where 𝑘 is a constant, and the solution of (76) is determined for two different cases of arbitrary 𝑓(𝑦) function (i) For 𝑓(𝑦) = 𝑦, 𝑘 𝑘 𝑘 𝑘 𝑦 (𝑥) = 𝑒−𝑘−𝑒 𝑥−𝑒 𝑐1 (4𝑒2𝑒 + 𝑒2𝑒 𝑐1 ) , (77) where 𝑐1 is a constant, is the solution of original path equation (16) (ii) For 𝑓(𝑦) = 𝑒𝑦 , If we substitute 𝜆-symmetry (67) in (59), then we have 󸀠2 (72) It is easy to see that the general solution of this equation (66) Using (58), the 𝜆-symmetry is obtained in the following form: (71) is and the generator is 𝜕 𝑋=𝑎 𝜕𝑥 √−1 + 𝑒2𝑤(𝑥) 𝑓(𝑦 (𝑥))2 ), and by using 𝑦󸀠 and 𝑦󸀠󸀠 , (16) becomes 4.1.1 Arbitrary 𝑓(𝑦) For arbitrary 𝑓(𝑦) the one-parameter Lie group of transformations is 𝜉 = 𝑎, 𝑓 (𝑦 (𝑥)) 𝑤󸀠 (𝑥) 𝑦 (𝑥) = −𝑘 + ln (−cot (𝑥 − 𝑐1 )) √1 + tan (𝑥 − 𝑐1 ) (78) is the other solution of the same equation 󸀠 (1 + 𝑦 ) 𝑓 (𝑦) 𝑓 (𝑦) 𝑦󸀠 𝑤𝑦󸀠 = (68) It is clear that a solution of (68) is 4.1.2 𝑓(𝑦) = 𝑘 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 For another case 𝑓(𝑦) = 𝑘, the infinitesimal generators are 𝑋1 = 𝑥𝑦 𝜕 𝜕 + 𝑦2 , 𝜕𝑥 𝜕𝑦 𝑋3 = 𝑥2 𝜕 𝜕 + 𝑥𝑦 , 𝜕𝑥 𝜕𝑦 󸀠2 (1 + 𝑦 ) 𝑤 (𝑥, 𝑦, 𝑦 ) = ln ( ) 2 𝑓(𝑦) 󸀠 (69) 󸀠 To write (16) in terms of {𝑥, 𝑤, 𝑤 }, we can express the following equality using (69): 𝑦󸀠 = √−1 + 𝑒2𝑤(𝑥) 𝑓(𝑦 (𝑥)) (70) 𝜕 𝑋5 = 𝑥 , 𝜕𝑥 𝑋7 = 𝜕 , 𝜕𝑦 𝑋2 = 𝑦 𝑋4 = 𝑥 𝜕 𝑋6 = 𝑦 , 𝜕𝑦 𝑋8 = 𝜕 𝜕𝑥 𝜕 , 𝜕𝑥 𝜕 , 𝜕𝑥 𝜕 𝑋6 = 𝑥 , 𝜕𝑦 (79) Abstract and Applied Analysis Thus, we can calculate 𝜆-symmetry of path equation using, for example, 𝑋1 Lie symmetry generator For this generator 𝑋1 the infinitesimals are 𝜂 = 𝑦2 𝜉 = 𝑥𝑦, (80) 4.1.3 𝑓(𝑦) = 1/(𝑚𝑦 + 𝑛) For this case the eight-parameter symmetry generators are obtained as follows: 𝑋1 = ( Therefore, the characteristic is written as 󸀠 O ¸ = 𝑦 − 𝑥𝑦𝑦 𝑚𝑦2 𝑚𝑥3 𝜕 + 𝑥 (𝑛𝑦 + )) 2 𝜕𝑥 +( (81) 𝑚2 𝑥4 (𝑚𝑦 + 𝑛) By using (58) we obtain the 𝜆-symmetry + 𝑦󸀠 𝜆= 𝑦 𝑋2 = (𝑛𝑦 + 𝑦󸀠 , 𝑦 𝑦󸀠󸀠 = 𝑤2 𝑦 + 𝑦𝑤󸀠 𝑤󸀠 + 𝑤2 = 0, (85) in which the general solution is , 𝑥−𝐺 𝑤 (𝑥) = 𝐺 ∈ R (86) To find the integration factor one can write above equation in terms of 𝐺 as 𝑤𝑥 − , 𝐺= 𝑤 𝑋3 = (84) By using these equalities (84) we find the following equation: (87) 𝑤2 𝑦 If we substitute 𝑤 = 𝑦󸀠 /𝑦 in (87), then the reduced equation in terms of 𝑦󸀠 is 𝑦 (𝑥 − 󸀠 ) − 𝑐 = 0, 𝑦 (89) 𝑋6 = 𝜕 𝑥 , 𝑚𝑦 + 𝑛 𝜕𝑦 𝑋7 = 𝜕 , 𝑚𝑦 + 𝑛 𝜕𝑦 𝑋8 = 𝑚𝑦2 𝑚𝑥3 + 𝑥 (𝑛𝑦 + ), 2 𝑛2 𝑦2 + 𝑚𝑛𝑦3 + (𝑚2 𝑦4 /4) 𝑚2 𝑥4 + 𝜂=− ⋅ 𝑚𝑦 + 𝑛 (𝑚𝑦 + 𝑛) (93) Using these infinitesimals we find the characteristic O ¸ = ((2𝑛𝑦 + 𝑚 (𝑥2 + 𝑦2 )) (90) (94) −1 × (4 (𝑚𝑦 + 𝑛)) , and the 𝜆-symmetry is (91) which is the first integral of equation that provides the path equation (16) 𝜕 ⋅ 𝜕𝑥 (92) Now let us consider 𝑋1 operator, and then the corresponding infinitesimals 𝜉 and 𝜂 are × (2𝑛 (𝑦 − 𝑥𝑦󸀠 ) + 𝑚 (−𝑥2 + 𝑦2 − 2𝑥𝑦𝑦󸀠 ))) where 𝑐 and 𝑐3 are constants It is clear that this solution satisfies the original path equation (16) Also, one can write 𝑦 (𝑥) 𝐷𝑥 (𝑥 − 󸀠 ) = 0, 𝑦 (𝑥) 𝑦 (2𝑛 + 𝑚𝑦) 𝜕 𝜕 +( ) , 𝜕𝑥 𝑚𝑦 + 𝑛 𝜕𝑦 2𝑛𝑦 + 𝑚 (𝑥2 + 𝑦2 ) 𝜕 𝑋5 = , 𝜕𝑦 (𝑚𝑦 + 𝑛) and the solution of (89) is 𝑦 (𝑥) = (𝑥 − 𝑐) 𝑐3 , 𝑚𝑦2 𝜕 ) 𝜕𝑥 𝜉= (88) 𝜕 , 𝜕𝑦 𝑥 (𝑛𝑦 + (𝑚𝑦2 /2)) 𝑥2 𝜕 𝜕 𝑚𝑥3 + ) + (− , 𝜕𝑥 𝜕𝑦 (𝑚𝑦 + 𝑛) (𝑚𝑦 + 𝑛) 𝑋4 = 𝑥 and then the integration factor becomes 𝜇= ) 3𝑚𝑥 (𝑛𝑦 + (𝑚𝑦2 /2)) 𝜕 𝑚2 𝑥3 − ) , + (− 𝜕𝑦 (𝑚𝑦+𝑛) (𝑚𝑦 + 𝑛) (83) and we can write 𝑤 = 𝑦󸀠 /𝑦, then to obtain path equation in terms of {𝑥, 𝑤, 𝑤󸀠 } one can have 𝑦󸀠 = 𝑤𝑦, 𝑚𝑦 + 𝑛 (82) A solution of (59) for this case is 𝑤 (𝑥, 𝑦, 𝑦󸀠 ) = 𝑛2 𝑦2 + 𝑚𝑛𝑦3 + (𝑚2 𝑦4 /4) 𝜆= 2𝑛2 𝑦󸀠 + 2𝑚𝑛 (𝑥 + 𝑦𝑦󸀠 ) + 𝑚2 (2𝑥𝑦 − 𝑥2 𝑦󸀠 + 𝑦2 𝑦󸀠 ) (𝑚𝑦 + 𝑛) (2𝑛𝑦 + 𝑚 (𝑥2 + 𝑦2 )) ⋅ (95) Abstract and Applied Analysis where 𝑐 is a constant By the solution of (104), we obtain the solution that satisfies the original path equation (16) as By using (95) the equation (59) becomes 𝑤𝑦 + 2𝑛2 𝑦󸀠 + 2𝑚𝑛 (𝑥 + 𝑦𝑦󸀠 ) + 𝑚2 (2𝑥𝑦 − 𝑥2 𝑦󸀠 + 𝑦2 𝑦󸀠 ) (𝑚𝑦 + 𝑛) (2𝑛𝑦 + 𝑚 (𝑥2 + 𝑦2 )) 𝑦 (𝑥) = (−2𝑛 + 𝑚√− × 𝑤𝑦󸀠 = 4𝑛2 ×√2𝑐2 − +4𝑚𝑥 (𝑥 − 2𝑐3 )+4𝑐𝑚 (𝑐3 − 𝑥)) 𝑚 (96) A solution of (96) is 𝑚𝑥 + 𝑛𝑦󸀠 + 𝑚𝑦𝑦󸀠 𝑤 (𝑥, 𝑦, 𝑦 ) = 𝑚𝑥2 + 2𝑛𝑦 + 𝑚𝑦2 󸀠 × (2𝑚)−1 , −𝑚𝑥 + 𝑚𝑤𝑥2 + 2𝑛𝑤𝑦 + 𝑚𝑤𝑦2 ⋅ 𝑚𝑦 + 𝑛 (105) (97) where 𝑐3 is a constant, and the corresponding conservation law is This equation can be written as 𝑦󸀠 = (𝑐 − 2𝑥) 𝑚(𝑐 − 2𝑥)2 (98) 𝐷𝑥 ( (−𝑚𝑦(𝑥)2 +𝑥 (𝑚𝑥 + 2𝑛𝑦󸀠 (𝑥) − 2𝑦 (𝑥) (𝑛 − 𝑚𝑥𝑦󸀠 (𝑥)))) By differentiation of (98) we have −𝑚(𝑚𝑥 − 𝑤 (𝑚𝑥2 + 2𝑛𝑦 + 𝑚𝑦2 )) = (𝑚𝑦 + 𝑛) + −𝑚+2𝑤2 (𝑚𝑥2 +2𝑛𝑦+𝑚𝑦2 )+(𝑚𝑥2 +2𝑛𝑦+𝑚𝑦2 ) 𝑤󸀠 𝑚𝑦+𝑛 , (99) and if we substitute (98) and (99) into the path equation, we obtain 𝑤󸀠 + 2𝑤2 = 0, , 2𝑥 − 𝐺 𝐺 ∈ R (101) 𝑤2 2𝑤𝑥 − 𝑤 (102) 𝑚𝑦 + 𝑛 ⋅ + 2𝑛𝑦 + 𝑚𝑦2 ) (𝑚𝑥2 (103) If we rewrite (102) in terms of 𝑦󸀠 and then we substitute this expression into integration factor, the reduced equation of path equation becomes 𝜕 𝜕 + 𝑒−𝛼𝑦 sin 𝛼𝑥 , 𝜕𝑥 𝜕𝑦 𝑋2 = 𝑒−𝛼𝑦 sin 𝛼𝑥 𝜕 𝜕 − 𝑒−𝛼𝑦 cos 𝛼𝑥 , 𝜕𝑥 𝜕𝑦 󸀠 ( − 𝑚𝑦(𝑥) + 𝑥 (𝑚𝑥 + 2𝑛𝑦 (𝑥) −2𝑦 (𝑥) (𝑛 − 𝑚𝑥𝑦󸀠 (𝑥)))) × (𝑚𝑥 + (𝑚𝑦 (𝑥) + 𝑛) 𝑦󸀠 (𝑥)) −1 − 𝑐 = 0, (104) 𝜕 𝜕 + sin 2𝛼𝑥 , 𝜕𝑥 𝜕𝑦 (107) 𝜕 𝜕 𝑋4 = sin 2𝛼𝑥 − cos 2𝛼𝑥 , 𝜕𝑥 𝜕𝑦 𝑋5 = Therefore, by using the relation (60) we find the integration factor 𝜇= 𝑋1 = 𝑒−𝛼𝑦 cos 𝛼𝑥 𝑋3 = cos 2𝛼𝑥 To define 𝐺, one can write 𝐺= 4.1.4 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 For this case the infinitesimal generators of path equation are (100) and the solution of (100) is 𝑤 (𝑥) = (106) × (𝑚𝑥 + (𝑚𝑦 (𝑥) + 𝑛) 𝑦 (𝑥)) ) 𝑦󸀠󸀠 = −1 󸀠 𝜕 , 𝜕𝑥 𝑋6 = 𝑋7 = 𝑒𝛼𝑦 cos 𝛼𝑥 𝜕 , 𝜕𝑦 𝜕 , 𝜕𝑥 𝑋8 = 𝑒𝛼𝑦 sin 𝛼𝑥 𝜕 ⋅ 𝜕𝑥 If we consider, for example, 𝑋1 symmetry generator and then 𝜉 and 𝜂 are 𝜉 = 𝑒−𝛼𝑦 cos 𝛼𝑥, 𝜂 = 𝑒𝛼𝑦 sin 𝛼𝑥, (108) then the characteristic by (56) is O ¸ = −𝑒−𝑦𝛼 𝑦󸀠 cos 𝑥𝛼 − 𝑒−𝑦𝛼 sin 𝑥𝛼 (109) If we apply the operator 𝐴 (52) to this characteristic (109), we obtain 𝐴(O ¸ ) = 0, and the 𝜆-symmetry is equal to zero For 𝑋2 symmetry generator we find also 𝜆 = similar to previous one Hence, we can use another symmetry generator, for example, 𝑋7 to obtain 𝜆-symmetry For this case, 𝜉 = 0, 𝜂 = 𝑒𝑦𝛼 cos 𝑥𝛼, (110) 10 Abstract and Applied Analysis are infinitesimals, and the corresponding characteristic is O ¸ = 𝑒𝑦𝛼 cos 𝑥𝛼 4.1.5 𝑓(𝑦) = 𝑦𝑛 If 𝑓(𝑦) is assumed in the polynomial form and then Lie symmetry generators are (111) We find the 𝜆-symmetry from (58) as in the following form: 𝜆 = 𝛼 (𝑦󸀠 − tan 𝑥𝛼) (112) By applying (112) to (59) we obtain the solution 𝑋1 = 𝑥 𝜕 𝜕 +𝑦 , 𝜕𝑥 𝜕𝑦 𝑋2 = 𝜕 ⋅ 𝜕𝑥 (123) 𝑋2 , for example, can be used to obtain 𝜆-symmetry, and for this generator the infinitesimals are 𝜉 = 1, 𝜂 = (124) By using 𝜉, 𝜂 the characteristic function is written as 𝑤 (𝑥, 𝑦, 𝑦󸀠 ) = 𝑒𝑦𝛼 (𝑦󸀠 − tan 𝑥𝛼) O ¸ = −𝑦󸀠 (113) And we write this expression (113) in terms of {𝑥, 𝑤, 𝑤󸀠 } as By considering (58), the 𝜆-symmetry becomes (114) 𝜆= By differentiating 𝑦󸀠 (114) with respect to 𝑦󸀠 one can write The solution of (59) is 𝑦󸀠 = 𝑒𝑦𝛼 𝑤 + tan 𝑥𝛼 𝑦󸀠󸀠 = 𝛼sec(𝑥𝛼)2 + 𝑒𝑦𝛼 𝛼𝑤 (tan 𝑥𝛼 + 𝑒𝑦𝛼 𝑤) + 𝑒𝑦𝛼 𝑤󸀠 , 𝑛 (1 + 𝑦󸀠 ) 𝑦𝑦󸀠 (126) ⋅ (115) and by substituting 𝑦󸀠 and 𝑦󸀠󸀠 to the original path equation we obtain 𝑤󸀠 − 𝛼 tan (𝑥𝛼) 𝑤 = 0, (125) (116) where the solution of (116) is (1 + 𝑦󸀠 ) 𝑤 (𝑥, 𝑦, 𝑦 ) = ln ( )⋅ 𝑦2𝑛 󸀠 (127) To write (16) in terms of {𝑥, 𝑤, 𝑤󸀠 }, we can express the following equality: 𝑦󸀠 = −√−1 + 𝑒2𝑤(𝑥) 𝑦(𝑥)2𝑛 (128) By taking derivative (128) with respect to 𝑥, then we have 𝑤 (𝑥) = sec (𝑥𝛼) (117) To define this equality in terms of variable 𝑤 then 𝐺 is defined as follows: 𝐺 = 𝑤 cos (𝑥𝛼) , (118) 𝑦󸀠󸀠 = 𝑒2𝑤(𝑥) 𝑦(𝑥)(2𝑛−1) (𝑛 − 𝑦 (𝑥) 𝑤󸀠 (𝑥) √−1 + 𝑒2𝑤(𝑥) 𝑦(𝑥)2𝑛 If we substitute 𝑦󸀠 and 𝑦󸀠󸀠 into the path equation, then one can find 𝑤󸀠 (𝑥) = 0, so we obtain the integration factor using (60) 𝜇 = 𝑒−𝛼𝑦 cos 𝑥𝛼 (119) Finally one can write the conservation law ) ⋅ (129) (130) and a solution of this equation (130) is 𝑤 (𝑥) = 𝐺, 𝐺 ∈ R (131) By using (60) we find the integration factor 𝜇 of the form 𝐷𝑥 (𝑒−𝛼𝑦(𝑥) cos 𝑥𝛼 (𝑦󸀠 (𝑥) − tan 𝑥𝛼)) = 0, (120) 𝜇= 𝑦󸀠 ⋅ + 𝑦󸀠2 (132) which gives the original path equation And thus we can express the first integral, which is reduced form of the path equation It is easy to see that the conserved form satisfies the following equality: 𝑒−𝛼𝑦(𝑥) cos 𝑥𝛼 (𝑦󸀠 (𝑥) − tan 𝑥𝛼) − 𝑐 = 0, + 𝑦󸀠 (𝑥)2 𝐷𝑥 ( ln ( )) = 0, 𝑦(𝑥)2𝑛 (121) where 𝑐 is a constant Integrating (121) we obtain the solution that satisfies the original equation 𝑦 (𝑥) = − ln (𝑐𝛼3 cos (𝑥𝛼) (−𝑐1 − (tan 𝑥𝛼/𝛼3 ))) where 𝑐1 is a constant 𝛼 , (122) (133) and this equality gives the original path equation Thus the reduced form of path equation is + 𝑦󸀠 (𝑥)2 ) − 𝑘 = 0, ln ( 𝑦(𝑥)2𝑛 (134) where 𝑘 is a constant And all results are summarized in Table Abstract and Applied Analysis 11 Table 2: Table of 𝜆-symmetry classification with Lie symmetry of path equation Function Lie symmetries Arbitrary 𝑓(𝑦) 𝜉 = 𝑎, 𝜂 = 𝑓(𝑦) = 𝑘 𝜉 = 𝑥𝑦, 𝜂 = 𝑦2 𝑓(𝑦) = 𝑦𝑛 𝜉 = 1, 𝜂 = 𝜉= 𝑓(𝑦) = 𝑚𝑦 + 𝑛 𝑓(𝑦) = 𝑘Exp(𝛼𝑦) 𝑚𝑦2 𝑚𝑥3 + 𝑥(𝑛𝑦 + ) 2 𝑚𝑥 𝜂=− 4(𝑚𝑦 + 𝑛) 𝑛2 𝑦2 + 𝑚𝑛𝑦3 + (𝑚2 𝑦4 /4) + 𝑚𝑦 + 𝑛 𝜉 = 0, 𝜂 = 𝑒𝑦𝛼 cos 𝑥𝛼 + 𝜂(1) (𝑥, 𝑦, 𝑦󸀠 , 𝑦󸀠󸀠 , , 𝑦(𝑛−1) ) 𝜕𝑦󸀠 (135) + 𝜂(2) (𝑥, 𝑦, 𝑦󸀠 , 𝑦󸀠󸀠 , , 𝑦(𝑛−1) ) 𝜕𝑦󸀠󸀠 , with 𝜂(𝑛+1) = [(𝐷𝑥 + 𝜆) 𝜂(𝑛) − 𝑦󸀠 (𝐷𝑥 + 𝜆) 𝜉] , (136) where 𝐷𝑥 is total derivative operator with respect to 𝑥 such that 𝐷𝑥 = 𝜕𝑥 + ∑ 𝑦 𝑘=0 𝜕𝑦(𝑘) , 𝑦 (0) + 𝑚2 (2𝑥𝑦 − 𝑥2 𝑦1 + 𝑦2 𝑦󸀠 ) (𝑚𝑦 + 𝑛)(2𝑛𝑦 + 𝑚(𝑥2 + 𝑦2 )) 𝜆 = 𝛼(𝑦󸀠 − tan 𝑥𝛼) 𝑛 (1) pr𝑋 = 𝜉 (𝑥, 𝑦) 𝜕𝑥 + 𝜂 (𝑥, 𝑦) 𝜕𝑦 (𝑘+1) 𝑚𝑦 + 𝑛 𝑤2 (𝑚𝑥2 + 2𝑛𝑦 + 𝑚𝑦2 ) where, namely, 𝑀 is Definition of 𝜆 ∈ 𝐶 (𝑀 )-Symmetry Let V be a vector field on 𝑀 which is open subset, and has the property of 𝑀 ⊂ 𝑋 × 𝑌 For 𝑘 ∈ N, 𝑀(𝑘) ⊂ 𝑋 × 𝑌(𝑘) denotes the corresponding 𝑘jet space, and their elements are (𝑥, 𝑦(𝑘) ) = (𝑥, 𝑦, 𝑦1 , , 𝑦𝑘 ), where, for 𝑖 = 1, , 𝑘, 𝑦𝑖 denotes the derivative of order 𝑖 of 𝑦 with respect to 𝑥 In addition let 𝑋 = 𝜉(𝑥, 𝑦)𝜕𝑥 + 𝜂(𝑥, 𝑦)𝜕𝑦 be a vector field defined on 𝑀, and let 𝜆 ∈ 𝐶∞ (𝑀(1) ) be an arbitrary function Then the 𝜆-prolongation of 𝑋 is 𝑛 𝜇= 2𝑛2 𝑦󸀠 + 2𝑚𝑛(𝑥 + 𝑦𝑦󸀠 ) (𝑚𝑦 + 𝑛)(2𝑛𝑦 + 𝑚(𝑥2 + 𝑦2 )) 𝜆= 𝜇 = 𝑒−𝛼𝑦 cos 𝑥𝛼 𝜆-Symmetries and Jacobi Last Multiplier Approach ∞ 𝜆-Symmetries (1 + 𝑦󸀠2 ) 𝑓󸀠 (𝑦) 𝜆= 𝑓 (𝑦) 𝑦󸀠 𝑦󸀠 𝜆= 𝑦 𝑛(1 + 𝑦󸀠2 ) 𝜆= 𝑦𝑦󸀠 Integration factor 𝑦󸀠 𝜇= + 𝑦󸀠2 𝜇= 𝑤𝑦 𝑦󸀠 𝜇= + 𝑦󸀠2 ≡ 𝑦, (0) 𝜂 ≡ 𝜂 (137) In this section we analyze 𝜆-symmetries of path equation by using Jacobi last multiplier as another approach First (61) can be written by using system of first-order equations, which is equivalent to the expression 𝑤𝑖󸀠 = 𝑊𝑖 (𝑡, 𝑤1 , , 𝑤𝑛 ) , (138) 𝜕𝑊𝑖 𝑑𝑡) 𝑖=1 𝜕𝑤𝑖 𝑀 = exp (− ∫ ∑ The nonlocal approach [13, 20] to 𝜆-symmetries is analyzed to seek 𝜆-symmetries such that 𝑛 𝜕𝑊𝑖 ⋅ 𝑖=1 𝜕𝑤𝑖 𝑤󸀠 = 𝜆 = ∑ (141) With this idea 𝑤 always can be considered to be of the form such as 𝑤 = log (1/𝑀) But this relation cannot be considered if the divergence of (138) Div ≡ ∑𝑛𝑖=1 (𝜕𝑊𝑖 /𝜕𝑤𝑖 ) is equal to zero So 𝑤 is chosen like this form because any Jacobi last multiplier is a first integral of (138) In this section we again consider different choices of 𝑓(𝑦) for 𝜆-symmetry classification 5.1 𝑓(𝑦) = 𝑘 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 For this case the divergence of the path equation yields 𝜆𝑗 = (142) Substituting 𝜆𝑗 into (135) then from the solution of the determining equations (62) we obtain eight-parameter 𝜆infinitesimals 𝜉(𝜆) = 𝑦 (𝑐2 + 𝑐1 𝑥) + 𝑐3 𝑥2 + 𝑐8 𝑥 + 𝑐7 , 𝜂(𝜆) = 𝑦 (𝑐4 + c3 𝑥) + 𝑐5 𝑥 + 𝑐1 𝑦2 + 𝑐6 , and the generators are 𝜕 𝜕 + 𝑦2 , 𝜕𝑥 𝜕𝑦 𝑋2(𝜆) = 𝑦 𝜕 , 𝜕𝑥 𝜕 𝜕 + 𝑦𝑥 , 𝜕𝑥 𝜕𝑥 𝑋4(𝜆) = 𝑦 𝜕 , 𝜕𝑦 and by solving the following differential equation, the Jacobi last multiplier of (138) 𝑀 is found: 𝑋1(𝜆) = 𝑦𝑥 𝑑 log (𝑀) 𝑛 𝜕𝑊𝑖 = 0, +∑ 𝑑𝑡 𝑖=1 𝜕𝑤𝑖 𝑋3(𝜆) = 𝑥2 (139) (140) (143) 12 Abstract and Applied Analysis 𝑋5(𝜆) = 𝑥 𝑋7(𝜆) = 𝜕 , 𝜕𝑦 𝜕 , 𝜕𝑥 sin 2𝑥𝛼 𝜕 cos 2𝑥𝛼 𝜕 − 𝑒−2𝑦𝛼 , 2𝛼 𝜕𝑥 2𝛼 𝜕𝑦 𝑋6(𝜆) = 𝜕 , 𝜕𝑦 𝑋6(𝜆) = 𝑒−2𝑦𝛼 𝑋8(𝜆) = 𝑥 𝜕 , 𝜕𝑥 𝑋7(𝜆) = 𝑒−𝑦𝛼 cos 𝑥𝛼 𝜕 , 𝜕𝑦 𝑋8(𝜆) = 𝑒−𝑦𝛼 sin 𝑥𝛼 𝜕 𝜕𝑦 (144) which corresponds to the classical Lie point symmetries since 𝜆𝑗 is equal to zero 5.2 𝑓(𝑦) = 𝑦 Another special form we consider here is 𝑓(𝑦) = 𝑦 For this case we obtain the divergence of (16) in the form 𝜆𝑗 = 2𝑦󸀠 , 𝑦 (145) and by substituting 𝜆𝑗 into the prolongation formula, the 𝜆infinitesimals can be found as follows: 𝑐 𝜉(𝜆) = 12 , 𝜂(𝜆) = 0, (146) 𝑦 5.4 𝑓(𝑦) = 1/(𝑚𝑦 + 𝑛) For this case we find that 𝜆𝑗 = − 2𝑚𝑦󸀠 , 𝑚𝑦 + 𝑛 and it is clear that 𝜆-infinitesimal functions are 𝜉(𝜆) = (𝑚𝑦 + 𝑛) (𝑐1 𝑚2 𝑥3 + 3𝑐2 𝑚2 𝑥2 + 2𝑐3 𝑥2 + 𝑐6 𝑥 + 𝑐5 ) +(𝑚𝑦 + 𝑛) (𝑐1 𝑥 + 𝑐2 ) , 𝜂(𝜆) = (𝑚𝑦 + 𝑛) × (2𝑐7 + 𝜕 ⋅ 𝑦2 𝜕𝑥 𝑐1 (𝑚𝑦 + 𝑛) (𝑚2 𝑦2 + 2𝑚𝑛𝑦 − 𝑛2 ) 𝑚 (147) + 2(𝑚𝑦 + 𝑛) (𝑎4 + 𝑎3 𝑥) which is a new 𝜆-symmetry − 𝑥 (−2𝑐8 + 𝑚𝑥 (2𝑐6 − 2𝑚𝑐4 5.3 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 For this case of 𝑓(𝑦) the divergence of (16) gives 𝜆𝑗 = 2𝛼𝑦󸀠 , + 2𝑎1 𝑛2 + 2𝑐3 𝑚𝑥 (148) +4𝑐2 𝑚2 𝑥 + 𝑐1 𝑚2 𝑥2 )) ) and the corresponding 𝜆-infinitesimals are (152) 𝜉(𝜆) = 𝑒−3𝑦𝛼 (𝑐1 cos 𝑥𝛼 + 𝑐2 sin 𝑥𝛼 𝑒𝑦𝛼 + (2𝑐4 𝛼 − 𝑐5 cos 2𝑥𝛼 + 𝑐6 sin 2𝑥𝛼)) , 𝛼 𝜂(𝜆) = 𝑒−3𝑦𝛼 (𝑐1 sin 𝑥𝛼 − 𝑐2 cos 𝑥𝛼 while the corresponding new 𝜆-symmetries are found to be as follows 𝜕 𝜕 𝑋1(𝜆) = 𝑒−3𝑦𝛼 cos 𝑥𝛼 + 𝑒−3𝑦𝛼 sin 𝑥𝛼 , 𝜕𝑥 𝜕𝑦 𝑋3(𝜆) = 𝑒−2𝑦𝛼 𝑋5(𝜆) = −𝑒−2𝑦𝛼 𝜕 , 𝜕𝑦 𝜕 𝜕 − 𝑒−3𝑦𝛼 cos 𝑥𝛼 , 𝜕𝑥 𝜕𝑦 𝑋4(𝜆) = 𝑒−2𝑦𝛼 𝑋1(𝜆) = (𝑚𝑦 + 𝑛) 𝑥 + 𝑚2 (𝑚𝑦 + 𝑛) 𝑥3 𝜕 + (𝑚𝑦 + 𝑛) 𝜕𝑥 2 𝑐 cos 2𝑥𝛼 + 𝑐5 sin 2𝑥𝛼)) , 2𝛼 (149) 𝑋2(𝜆) = 𝑒−3𝑦𝛼 sin 𝑥𝛼 Therefore, we find new 𝜆-symmetries as follows: ×( + 𝑒2𝑦𝛼 (𝑐7 cos 𝑥𝛼 + 𝑐8 sin 𝑥𝛼) +𝑒𝑦𝛼 (𝑐3 − (151) and the corresponding generator is 𝑋1(𝜆) = (150) 𝜕 , 𝜕𝑥 cos 2𝑥𝛼 𝜕 sin 2𝑥𝛼 𝜕 − 𝑒−2𝑦𝛼 , 2𝛼 𝜕𝑥 2𝛼 𝜕𝑦 (𝑚𝑦 + 𝑛) (−𝑛2 + 2𝑚𝑛𝑦 + 𝑚2 𝑦2 ) 𝑚 −𝑚𝑥2 (2𝑛2 + 𝑚2 𝑥2 ) ) 𝜕 , 𝜕𝑦 𝑋2(𝜆) = ((𝑚𝑦 + 𝑛) + 3𝑚2 (𝑚𝑦 + 𝑛) 𝑥2 ) − 2𝑚3 (𝑚𝑦 + 𝑛) 𝑥3 𝑋3(𝜆) = 2𝑚(𝑚𝑦 + 𝑛) 𝑥2 𝜕 𝜕𝑥 𝜕 , 𝜕𝑦 𝜕 𝜕𝑥 + (𝑚𝑦 + 𝑛) ((𝑚𝑦 + 𝑛) 𝑥 − 𝑚2 𝑥3 ) 𝑋4(𝜆) = (𝑚𝑦 + 𝑛) ((𝑚𝑦 + 𝑛) + 𝑚2 𝑥2 ) 𝜕 , 𝜕𝑦 𝜕 , 𝜕𝑦 Abstract and Applied Analysis 𝑋5(𝜆) = (𝑚𝑦 + 𝑛) 13 𝜕 , 𝜕𝑥 5.6 Invariant Solutions In this section we present some invariant solutions based on Jacobi multiplier approach 𝑋6(𝜆) = (𝑚𝑦 + 𝑛) 𝜕 𝜕 − 𝑚 (𝑚𝑦 + 𝑛) 𝑥2 , 𝜕𝑥 𝜕𝑦 𝑋7(𝜆) = (𝑚𝑦 + 𝑛) 𝜕 , 𝜕𝑥 𝑋8(𝜆) = (𝑚𝑦 + 𝑛) Case For the case 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 we can investigate 𝑋1𝜆 to find the invariant solution of path equation The first prolongation of 𝑋1𝜆 is 𝜕 𝜕𝑥 (153) 5.5 𝑓(𝑦) = 𝑦𝑛 , 𝑛 ≠ 1/3, 1/2, The divergence of the path equation yields 2𝑛𝑦󸀠 𝜆𝑗 = 𝑦 (154) Pr𝑋1𝜆 = 𝑒−𝑦𝛼 cos 𝑥𝛼𝜕𝑦 + 𝑒−𝑦𝛼 (𝛼𝑦󸀠 cos 𝑥𝛼 − 𝛼 sin 𝑥𝛼) 𝜕𝑦󸀠 , (163) and the Lagrange equations are 𝑑𝑦 𝑑𝑦󸀠 𝑑𝑥 , = −𝑦𝛼 = −𝑦𝛼 𝑒 cos 𝑥𝛼 𝑒 (𝛼𝑦󸀠 cos 𝑥𝛼 − 𝛼 sin 𝑥𝛼) (164) If we substitute (154) 𝜆𝑗 into (135), we obtain 𝜆infinitesimals √(𝑛−1)2 (3𝑛−1)/√3𝑛−1)+(1/2)(1−5𝑛−√(𝑛−1)2 (3𝑛−1)/√3𝑛−1)) 𝜉(𝜆) = 𝑐1 𝑦(( , 𝜂(𝜆) = (155) gives the first order invariants 𝑥̃ = 𝑥, 𝑋 =𝑦 𝜕 𝜕𝑥 (156) 𝑑𝑦̃ ̃ tan 𝑥𝛼, = 𝑦𝛼 𝑑𝑥̃ ((√(𝑛−1)2 (3𝑛−1)/√3𝑛−1)+(1/2)(1−5𝑛−√(𝑛−1)2 (3𝑛−1)/√3𝑛−1)) It is clear that we should analyze two specific values for 𝑛 Case (𝑛 = 1/3) The divergence of path equation for this value of 𝑛 is 2𝑦󸀠 𝜆𝑗 = , 3𝑦 the 𝜆-infinitesimals can be written as 3𝑐1 𝜉(𝜆) = − 2/3 , 𝜂(𝜆) = 0, 2𝑦 (157) (158) and the 𝜆-generator is 𝑦󸀠 − tan 𝑥𝛼 , 𝑒𝑦𝛼 (165) that replaced into path equation generate the first-order equation and the corresponding one-parameter 𝜆-generator (𝜆) 𝑦̃ = (166) the solution of this equation yields ̃ = 𝑐1 , 𝑦̃ cos 𝑥𝛼 (167) and the first integral is 𝐷𝑥 (( 𝑦󸀠 − tan 𝑥𝛼 ) cos 𝑥𝛼) = 0; 𝑒𝑦𝛼 (168) this equality gives the original path equation (16) The reduced form of path equation is ( 𝑦󸀠 − tan 𝑥𝛼 ) cos 𝑥𝛼 − 𝑐 = 0, 𝑒𝑦𝛼 (169) in which the solution of (169) is 𝑋(𝜆) = − 𝜕 2𝑦2/3 𝜕𝑥 (159) Case (𝑛 = 1/2) For another specific value of 𝑛 the divergence is 𝜆𝑗 = 𝑦󸀠 , 𝑦 the 𝜆-infinitesimals are found as follows: 𝑐 𝜉(𝜆) = , 𝜂(𝜆) = 𝑦 (160) tan 𝑥𝛼 𝑦 (𝑥) = − ln (𝑐𝛼3 cos 𝑥𝛼 (−𝑐1 − )) , 𝛼 𝛼3 where 𝑐1 and 𝑐 are constants It is clear that (170) is similar to the solutions (48) and (122) If we apply similar process for the 𝑋2 symmetry generator, we obtain first-order invariants for this case as 𝑥̃ = 𝑥, (161) and the 𝜆-generator is 𝜕 , 𝑦 𝜕𝑥 (162) In summary all new 𝜆-symmetries are presented in Table 𝑦̃ = 𝑦󸀠 + cot 𝑥𝛼 , 𝑒𝑦𝛼 (171) and the first integral is 𝐷𝑥 (( 𝑋(𝜆) = (170) 𝑦󸀠 + cot 𝑥𝛼 ) sin 𝑥𝛼) = 0; 𝑒𝑦𝛼 (172) another reduced form of path equation (16) is ( 𝑦󸀠 + cot 𝑥𝛼 ) sin 𝑥𝛼 − 𝑐 = 𝑒𝑦𝛼 (173) 14 Abstract and Applied Analysis Table 3: 𝜆-Symmetry classification table of path equation with Jacobi last multiplier 𝜆-Symmetries with Jacobi last multiplier Function (𝜆) 𝜉 𝑓(𝑦) = 𝑘 𝜂(𝜆) = 𝑦(𝑐4 + 𝑐3 𝑥) + 𝑐5 𝑥 + 𝑐1 𝑦2 + 𝑐6 𝑐1 (𝜆) ,𝜂 =0 𝑦2 𝑒𝑦𝛼 = 𝑒−3𝑦𝛼 (𝑐1 cos 𝑥𝛼 + 𝑐2 sin 𝑥𝛼 + (2𝑐4 𝛼 − 𝑐5 cos 2𝑥𝛼 + 𝑐6 sin 2𝑥𝛼)) 𝛼 𝜉 𝑓(𝑦) = 𝑦 𝜉 (𝜆) 𝑓(𝑦) = 𝑘𝑒𝛼𝑦 𝜂(𝜆) = (𝑚𝑦 + 𝑛)2 (𝑐1 𝑚2 𝑥3 + 3𝑐2 𝑚2 𝑥2 + 2𝑐3 𝑥2 + 𝑐6 𝑥 + 𝑐5 ) + (𝑚𝑦 + 𝑛)4 (𝑐1 𝑥 + 𝑐2 ) 𝑐 (𝑚𝑦 + 𝑛)2 (𝑚2 𝑦2 + 2𝑚𝑛𝑦 − 𝑛2 ) = (𝑚𝑦 + 𝑛)(2𝑐7 + + 2(𝑚𝑦 + 𝑛)2 (𝑎4 + 𝑎3 𝑥)) 𝑚 (𝜆) 𝜉 = 𝑐1 𝑦( √(𝑛−1)2 (3𝑛−1)/√3𝑛−1)+(1/2)(1−5𝑛−(√(𝑛−1)2 (3𝑛−1)/√3𝑛−1)) (𝜆) 𝜉 𝑓(𝑦) = 𝑦𝑛 , 𝑛 = 1/3 (𝜆) The solution of (173) is given by 3𝑐1 , 𝜂(𝜆) = 2𝑦2/3 𝑐 = , 𝜂(𝜆) = 𝑦 The solution of (178) is cot 𝑥𝛼 )) 𝑦 (𝑥) = − ln (𝑐𝛼3 sin 𝑥𝛼 (−𝑐1 − 𝛼 𝛼3 2𝑚𝑥̃𝑦̃ − = 𝑐1 , 𝑦̃ (174) Case As another case 𝑓(𝑦) = 1/(𝑚𝑦 + 𝑛), we can analyze 𝑋4𝜆 generator to find the invariant solution of path equation The first prolongation of 𝑋4𝜆 is written as 2 𝐷𝑥 ( 𝑚2 𝑥2 − 𝑛2 − 𝑚2 𝑦2 + 2𝑚𝑛𝑥𝑦󸀠 + 2𝑚𝑦 (𝑚𝑥𝑦󸀠 − 𝑛) 𝑦󸀠 (𝑚𝑦 + 𝑛) + 𝑚𝑥 (175) 2 󸀠 + (𝑛 + 2𝑚𝑛𝑦 + 𝑚 (𝑦 − 𝑥 )) 𝑦 ) 𝜕𝑦󸀠 which is equal to the original path equation (16) The new reduced form is 𝑚2 𝑥2 − 𝑛2 − 𝑚2 𝑦2 + 2𝑚𝑛𝑥𝑦󸀠 + 2𝑚𝑦 (𝑚𝑥𝑦󸀠 − 𝑛) and the Lagrange equation are 𝑦󸀠 (𝑚𝑦 + 𝑛) + 𝑚𝑥 𝑑𝑦 𝑑𝑥 = (𝑚𝑦 + 𝑛) ((𝑚𝑦 + 𝑛) + 𝑚2 𝑥2 ) and the solution of (181) is 󸀠 and the corresponding first-order invariants become 𝑦̃ = 𝑦󸀠 (𝑚𝑦 + 𝑛) + 𝑚𝑥 , 𝑛2 + 𝑚2 𝑥2 + 2𝑚𝑛𝑦 + 𝑚2 𝑦2 𝑦 (𝑥) = 2𝑚√1/𝑚(𝑐 − 2𝑚𝑥)2 (2𝑚𝑥 − 𝑐) × (2𝑐𝑛√ 1 − 4𝑚𝑛𝑥√ 𝑚(𝑐 − 2𝑚𝑥) 𝑚(𝑐 − 2𝑚𝑥)2 (177) ±√ − that replaced into path equation generate the first order equation 𝑑𝑦̃ + 2𝑚𝑦̃2 = 𝑑𝑥̃ − 𝑐 = 0, (181) 𝑑𝑦 , 𝑚 (2𝑚𝑥 (𝑚𝑦 + 𝑛) + (𝑛2 + 2𝑚𝑛𝑦 + 𝑚2 (𝑦2 − 𝑥2 )) 𝑦󸀠 ) (176) 𝑥̃ = 𝑥, ) = 0, (180) + 𝑚 (2𝑚𝑥 (𝑚𝑦 + 𝑛) (179) and the first integral is = (𝑚𝑦 + 𝑛) ((𝑚𝑦 + 𝑛) + 𝑚 𝑥 ) 𝜕𝑦 , 𝜂(𝜆) = =− 𝜉 𝑓(𝑦) = 𝑦𝑛 , 𝑛 = 1/2 = = (𝜆) 𝜉 𝑓(𝑦) = 𝑦𝑛 , 𝑛 > Pr𝑋4𝜆 (𝜆) 𝜂(𝜆) = 𝑒−3𝑦𝛼 (𝑐1 sin 𝑥𝛼 − 𝑐2 cos 𝑥𝛼 + 𝑒2𝑦𝛼 (𝑐7 cos 𝑥𝛼 + 𝑐8 sin 𝑥𝛼) +𝑒𝑦𝛼 (𝑐3 − 𝑐 cos 2𝑥𝛼 + 𝑐5 sin 2𝑥𝛼)) 2𝛼 𝑚𝑦 + 𝑛 𝑓(𝑦) = = 𝑦(𝑐2 + 𝑐1 𝑥) + 𝑐3 𝑥2 + 𝑐8 𝑥 + 𝑐7 (178) 2𝑐2 + 4𝑐𝑥 − 4𝑚𝑥2 + 4𝑚𝑐1 − 8𝑚2 𝑥𝑐1 ) , 𝑚 (182) where 𝑐1 and c are constants, and it is clear that (182) is similar to the solution (105) Abstract and Applied Analysis Conclusion The aim of this study is to classify Noether and 𝜆-symmetries of path equation describing the minimum drag work The symmetry classification of the equation is analyzed with respect to different choices of altitude-dependent arbitrary function 𝑓(𝑦) of the governing equation, which represents a combination of the density, the drag coefficient, the cross sectional area, and the velocity It is a fact that an ordinary differential equation should have a Lagrangian function to obtain Noether symmetries In this study we consider the partial Lagrangian approach for obtaining Noether symmetries and constructing a classification in the problem Thus, new first integrals (conserved forms) are obtained directly by using each Noether symmetry given by symmetry of the equation With this point of view we find and classify the new forms of first integrals, and then the invariant solutions of path equation are constructed for specific forms of 𝑓(𝑦) In the literature, as a different and a new concept, 𝜆symmetries of the second order ordinary differential equations are analyzed by assuming 𝜆-function in the linear form However, in our study, we prove that it is not possible to obtain 𝜆-symmetries of the drag equation by selecting 𝜆function in a linear form So we study another approach to obtain 𝜆-symmetries based on using Lie point symmetries of the path equation Thus, we have derived 𝜆-symmetries, integrating factors, first integrals, and the reduced form of the original path equation Based on using these new 𝜆symmetries, we present some new different invariant solutions by calculating new reduced forms and first integrals In our study, additionally, the Jacobi last multiplier concept is presented as a new and an alternative approach to construct 𝜆-symmetries of the path equation algorithmically In this method, first, 𝜆-function is determined by taking divergence of the governing equation and then the infinitesimals functions 𝜉 and 𝜂 are determined from the determining equations, then we calculate new 𝜆-symmetries In this study we generate first-order equations by using these new symmetries, which provide invariant solutions of path equation 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