Equivalence Group of Lie's Second-Order Equation of Type III Dai Jiayuan [email protected] Thesis for the Degree Master of Science (two years) in Mathematical Modeling and Simulation 30 credit points (30 ECTS credits) February 2012 Blekinge Institute of Technology School of Engineering Deparment of Mathematics and Science Supervisor: Prof. Nail H. Ibragimov Examiner: Dr. Raisa Khamitova Abstract Based on symmetry and invariance principles, Lie group analysis is the only systematic method of analytically solving nonlinear dierential equations. Nonlinear second-order ordinary dierential equations (ODEs) admitting two-dimensional Lie algebras can be transformed into one of the four types of canonical forms via Lie's integration method. In this thesis, Lie's secondorder equation of type III is considered from the point of view of equivalence transformations. The generators of the equivalence group and the principal Lie algebra are calculated. Keywords: equivalence transformation, second-order ordinary dier- ential equations, principal Lie algebra. 2 Acknowledgements First of all, I would like to acknowledge the Master Program in Mathematical Modeling and Simulation at Blekinge Institute of Technology, which enabled us to have such a wonderful experience in Sweden. Secondly, I would like to express my gratitude to my supervisor Prof. Nail H. Ibragimov, who has introduced me into the interesting eld of group analysis in mathematics and helped me with my thesis. His encouragement and inspiration were always present during this research. Furthermore, it is also a pleasure for me to express gratitude to my previous programme manager, Dr. Raisa Khamitova, who has given me valuable advice and shown innite patience. I would also like to thank my current programme manager Mattias Eriksson, whose support has provided me a favorable environment for learning. I would also like to thank all my friends and those who have helped me during my stay and studies in Sweden. Last but not least, I want to thank my parents for all of their understanding and support. Thank you all! 3 Contents 1 2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1 Second-order ODEs . . . . . . . . . . . . . . . . . . . 5 1.2 Lie's integration method using canonical variables . . 5 1.3 Equivalence transformations . . . . . . . . . . . . . . 6 1.4 Lie's second-order equation of type III . . . . . . . . 7 Principal Lie algebra . . . . . . . . . . . . . . . . . . . . . . 7 2.1 Calculation of the principal Lie algebra . . . . . . . . 8 2.2 Checking operators of the equivalence group . . . . . 11 Equivalence transformations . . . . . . . . . . . . . . . . . . 12 3.1 Calculation of equivalence transformations . . . . . . 12 3.2 Checking operators of the equivalence group . . . . . 18 4 Projections of the equivalence Lie algebra . . . . . . . . . . . 19 5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3 4 1 Introduction 1.1 Second-order ODEs Many mathematical models of real world problems involve second-order nonlinear ordinary dierential equations. General method for analytically solving these equations is Lie group analysis, which is particularly simple and ecient for second-order equations. Lie groups provide mathematical tools for revealing and using the symmetry of dierential equations. These methods are based on the theory of continuous groups developed by Sophus Lie. Second-order equations have the following form: u00 = g(t, u, u0 ). (1.1) A decisive step in integrating ODEs involves simplifying the frame by changing the variables. Lie group analysis provides a method of determining a suitable change of variables using symmetries. The process will be further discussed in the subsequent section. 1.2 Lie's integration method using canonical variables Denition 1.1. Let Lr be an r-dimensional linear space spanned by any linearly independent operators of a set of operators. The space Lr r is called a Lie algebra if it is closed under a commutator. Lie's integration method is based on the canonical coordinates for twodimensional Lie algebras L2 . For every L2 , canonical variables provide the simplest form of its basis and therefore reduce a dierential equation admitting L2 to an integrable form. Second-order ordinary dierential equations admitting L2 can be classied into four types introduced by Lie (see, e.g. [1] and [2]). The result is formulated as follows. Theorem 1.1. Any two-dimensional Lie algebra L2 can be transformed into one of the four non-similar standard forms (Table 1.1) through the proper choice of its basis as well as the suitable variables variables. 5 x, y called canonical Table 1.1: Four types of nonlinear second-order equations admitting L2 ∂ ∂ , X2 = X1 = ∂x ∂y ∂ ∂ X1 = , X2 = x ∂y ∂y ∂ ∂ ∂ , X2 = x +y X1 = ∂y ∂x ∂y ∂ ∂ X1 = , X2 = y ∂y ∂y Type Standard form of I II III IV L2 Canonical form of the Equation y 00 = f (y 0 ) y 00 = f (x) y 00 = 1 f (y 0 ) x y 00 = f (x)y 0 Lie proved Theorem 1.1 in order to integrate all second-order equations admitting a two-dimensional Lie algebra. Lie's method consists of classifying these equations into four types according to Table 1.1. By introducing the canonical variables x, y , the admitted Lie algebra L2 is reduced to one of the standard forms given in Table 1.1. Equation (1.1) is then rewritten in the canonical variables, and the resulting equation is as follows: y 00 = f (x, y, y 0 ) (1.2) where Equation (1.2) has one of the four integrable canonical forms given in Table 1.1. The equation is then integrated and rewritten by using the original variables 1.3 t, u to complete the integration procedure. Equivalence transformations Equivalence transformations of dierential equations are useful in investigating various properties of dierential equations. We take one simple example to explain this further. An equivalence transformation of homogeneous linear equations y 00 + a(x)y 0 + b(x)y = 0 (1.3) or L2 [y] = 0 involves a change of variables that preserve the linearity and homogeneity of the equations. The set of all equivalence transformations comprises an arbitrary change 6 of the independent variable, x̄ = φ(x), φ0 6= 0, (1.4) and the linear substitution of the dependent variable, y = σ(x)ȳ, Denition 1.2. σ 6= 0. (1.5) Two equations of a form similar to Equation (1.3) are said to be equivalent if they are connected by a combination of the transformations (1.4) and (1.5). Furthermore, two equations are termed equivalent by function if they can be mapped into each other by using a method of linear substitution (1.5). Theorem 1.2. Any homogeneous linear equation (1.3) is equivalent to the simplest linear equation, ȳ 00 = 0, where (1.6) ȳ 00 = d2 ȳ/dx̄2 . We are going to use this core idea of equivalence transformations to investigate a second-order nonlinear dierential equation. 1.4 Lie's second-order equation of type III In this thesis, the equivalence group of Lie's second-order nonlinear equation of type III is considered. The aim of this thesis is mainly to use the innitesimal method for nding the equivalence Lie algebra. Lie's secondorder equation of type III is written in the form: y 00 = where 2 y 0 = dy/dx, and f 1 f (y 0 ), x (1.7) is a smooth arbitrary function of y0. Principal Lie algebra For a given family of dierential equations, the Lie group classication begins with determining the principal Lie group, the group admitted by any equation of the family in question. The Lie algebra of the principal Lie group is called the principal Lie algebra of the equations and is denoted by 7 Lp (see e.g. [3]). 2.1 Calculation of the principal Lie algebra In order to derive the principal Lie algebra for the Lie's second-order nonlinear equation of type III y 00 = 1 f (y 0 ), x we need to use the invariance condition rstly. The invariance condition for Equation (1.7) can be written as follows: 1 00 0 = 0. X(2) y − f (y ) 1 x y 00 = f (y 0 ) x Here, X(2) is denoted as the second prolongation for the operator has the form X = ξ(x, y) X(2) (2.8) ∂ ∂ + η(x, y) · ∂x ∂y can be considered as the combination of the operator longed terms. Thus, X(2) which (2.9) X and two pro- can be written in the form X(2) = X + ζ1 where the coecients X ∂ ∂ + ζ2 00 , 0 ∂y ∂y (2.10) ζ1 and ζ2 are given by the usual prolongation formulae: ζ1 = Dx (η) − y 0 Dx (ξ), (2.11) 00 ζ2 = Dx (ζ1 ) − y Dx (ξ). The operator Dx in the above equation denotes the operator of total dier- entiation with respect to x: Dx = ∂ ∂ + y0 · ∂x ∂y Thus, the prolongation coecients ζ can be both obtained by formulae (2.11) and (2.12). They have the following form: ζ1 =ηx + y 0 ηy − y 0 (ξx + y 0 ξy ), ζ2 =ηxx + y 0 ηxy + y 00 ηy + y 0 (ηxy + y 0 ηyy ) − y 00 (ξx + y 0 ξy ) − y 0 [ξxx + y 0 ξxy + y 00 ξy + y 0 (ξxy + y 0 ξyy )] − y 00 (ξx + y 0 ξy ), 8 (2.12) or ζ1 =ηx + (ηy − ξx )y 0 − ξy y 02 , (2.13) ζ2 =ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 (2.14) − ξyy y 03 + (ηy − 2ξx − 3ξy y 0 )y 00 . Using (2.10) we derive the following determing equation f0 f ζ2 − ζ1 + 2 ξ 1 = 0. x x y 00 = f (y 0 ) x Substituting ζ1 and ζ2 (2.15) into Equation (2.15) we obtain ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 − ξyy y 03 +(ηy − 2ξx − 3ξy y 0 )y 00 − −ξy y f0 (ηx + (ηy − ξx )y 0 x 02 (2.16) f = 0. + 2ξ 1 x y 00 = f (y 0 ) x Substituting y 00 = 1 f (y 0 ), x we derive the nal form of the determing equation: f0 3f ηxx + 2ηxy − ξxx − ξy − (ηy − ξx ) y 0 x x (2.17) +(ηyy − 2ξxy + The functions ξ and f0 f f0 f ξy )y 02 − ξyy y 03 + (ηy − 2ξx ) − ηx + 2 ξ = 0. x x x x η do not depend on be satised identically for any ξ and η, 9 y0. Since Equation (2.17) should we can split the equation into the following system: y 03 : ξyy = 0, f0 ξy = 0, x f f0 y 0 : 2ηxy − ξxx − 3 ξy − (ηy − ξx ) = 0, x x f ξ f0 : ηxx + (−ηy + 2ξx + ) − ηx = 0. x x x y 02 : ηyy − 2ξxy + Since f is an arbitrary smooth function, the coecient in front of (2.18) f and f0 should be equal to zero. Therefore, the equations can be splitted further. Hence, from (2.18) we obtain the following system: y 03 : ξyy = 0, y 02 : ηyy − 2ξxy = 0, ξy = 0, y 0 : 2ηxy − ξxx = 0, ξy = 0, ξx − ηy = 0, ξ : ηxx = 0, −ηy + 2ξx + = 0, ηx = 0. x From the above equations, ξy = 0 and ηx = 0, (2.19) it follows that ξ = ξ(x), η = η(y). In addition, the functions ξ and η satisfy the equation ηy − ξx = 0. (2.20) It is possible only if ηy = ξx = C1 , Thus the funtions ξ and η C1 = const. have the following form: ξ = C1 x + C2, η = C1 y + C3 . (2.21) The following step is to solve the remaining equation in (2.19), namely, −ηy + 2ξx + ξ = 0. x Using (2.21) we derive C2 = 0, x 10 (2.22) which gives C2 = 0. Thus, ξ = C1 x, η = C1 y + C3 . Hence, Equation (1.7) with an arbitary function dimensional Lie algebra L2 f (y 0 ) admits the two- with the basis ∂ ∂ +y , ∂x ∂y ∂ X2 = · ∂y X1 = x Finally, we can say that L2 (2.23) is the two-dimensional Lie algebra known as the principal Lie algebra for Equation (1.7). 2.2 Checking operators of the equivalence group We have obtained the two-dimensional Lie algebra, we need to check if it is correct. • X1 = x ∂ ∂ +y ∂x ∂y Firstly, we prolong the operator until the second order according to the prolongation formulae (2.11). Here η=y and ξ = x. ζ1 = Dx (y) − y 0 Dx (x) = y 0 − y 0 = 0, ζ2 = −y 00 Dx (x) = −y 00 , therefore X1(2) = x ∂ ∂ ∂ +y − y 00 00 · ∂x ∂y ∂y Then, the invariance condition for Equation (1.7) can be written as 1 00 0 X1(2) y − f (y ) = 0, 1 x y 00 = f (y 0 ) x which yields f −y + = 0. x y00 = 1 f (y0 ) x 00 11 Hence, the operator • X2 = X1 is admitted by Equation (1.7). ∂ ∂y In this case we can not prolong X2 further. Thus, the invariance condition (2.8) is apparently satised. 3 Equivalence transformations Equivalence transformations play a core role in the theory of invariants. The set of all equivalence transformations of a given family of dierential equations forms a group, the so called equivalence group, and is denoted by E. The continuous group of equivalence transformations is a subgroup of the group E and therefore denoted by E c. There are two main methods for calculating the equivalence transformations for a family of equations. The rst one is direct search for equivalence transformations which allows one to calculate the most general equivalence group E. The second method, suggested by L. V. Ovsyannikov, is to de- termine generators of a continuous equivalence group E c. In general, the innitesimal method is simpler for computing than the direct method. 3.1 Calculation of equivalence transformations An equivalence transformation is a non-degenerate change of the variables x, y, f taking any equation of the form (1.7) into an equation of the same form. Thus, the following denition is used: Denition 3.1. An equivalence transformation of the family of Equations (1.7) is a change of variables x̄ = ϕ(x, y), ȳ = ψ(x, y), f¯ = φ(x, y, f ), such that it transfers Equation (1.7) with an arbitrary function (3.24) f (y 0 ) into an equation of the same form y 00 = where the function f, 1¯ 0 f (y ), x̄ (3.25) f¯ may, in general, be dierent from the original function while the equations (1.7) and (3.25) are said to be equivalent. 12 In order to calculate the equivalence transformation (3.24) for Equation (1.7), we write this equation in the following 1 y 00 − f = 0, x Now we look for the generator Y extended form fx = 0, of the group fy = 0. Ec (3.26) of the equivalence trans- formations (3.24). The generator can be written in the space of variables x, y and f as follows: Y = ξ(x, y) ∂ ∂ ∂ + η(x, y) + µ(x, y, f ) · ∂x ∂y ∂f (3.27) The prolongation of the operator (3.27) to all variables involved in Equations (3.26) has the form: ∂ ∂ ∂ ∂ ∂ ∂ ∂ Ye = ξ + ω2 · +η +µ + ζ1 0 + ζ2 00 + ω1 ∂x ∂y ∂f ∂y ∂y ∂fx ∂fy The condition that Y (3.28) is a generator of an equivalence group is equivalent to the statement that Ye satises the innitesimal invariance test for the extended system (3.26). The invariance conditions for the system (3.26) are Ye f 00 y − = 0, x y00 = f x (3.29) and Ye (fx ) = 0, Ye (fy ) = 0. According to the formulae (2.13)-(2.14), calculated in Chapter 2, have the form ζ1 =ηx + (ηy − ξx )y 0 − ξy y 02 , ζ2 =ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 − ξyy y 03 + (ηy − 2ξx − 3ξy y 0 )y 00 . 13 (3.30) ζ1 and ζ2 We use the notation Dx = ∂ ∂ ∂ + y0 + y 00 0 ∂x ∂y ∂y for the usual operator of total dierentiation with respect to x, while the two forms ∂ ∂ + fx ∂x ∂f D̃x = (3.31) and D̃y = ∂ ∂ + fy ∂y ∂f (3.32) denote the new operators of total dierentiations necessary for the extended System (3.26). The other two coecients ω1 and ω2 in (3.28) are obtained by applying the secondary prolongation procedure to the dierential variables 0 the independent variables y , namely, fy 0 with ω1 = D̃x (µ) − fy0 D̃x (ζ1 ), ω2 = D̃y (µ) − fy0 D̃y (ζ1 ). From System (3.26) it follows, in particular, that we can set fx 0 Therefore, f = f (y ) and the operators (3.31), (3.32) as well as = f y = 0. ω1 and ω2 can be simplied: D̃x = ∂ , ∂x ω1 = µx − f 0 ζ1x , (3.33) D̃y = ∂ , ∂y ω2 = µy − f 0 ζ1y . Now we notice that the invariance condition (3.30) can be written as ω1 = 0, ω2 = 0, and by using (3.33) in Equation (3.34), we obtain µx − f 0 ζ1x = 0, µy − f 0 ζ1y = 0. 14 (3.34) These equations are held for an arbitrary function f , which depends only 0 on y . Hence, we can split these equations as follows: This means that µ µx = 0, µy = 0, ζ1x = 0, ζ1y = 0. is a function depending only on f, i.e, µ = µ(f ). (3.35) Recalling the formula (3.28) for the operator Ye , we rewrite the invariance condition (3.29): 1 1 ζ2 − µ + 2 ξf 1 = 0. x x y 00 = f (y 0 ) x (3.36) Thus, we have ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 − ξyy y 03 µ ξf +(ηy − 2ξx − 3ξy y )y − + 2 x x 0 Substituting form y 00 = 1 f x 00 y 00 = 1 0 = 0. f (y ) x (3.37) in (3.37) and simplifying it, we obtain the reduced ηxx + (2ηxy − ξxx )y 0 + (ηyy − 2ξxy )y 02 −ξyy y 03 + (ηy − 2ξx − 3ξy y 0 ) As ζ1 µ ξf f − + 2 = 0. x x x is obtained from the prolongation formula (2.13) ζ1x = ηxx + y 0 ηxy − y 0 (ξxx + y 0 ξxy ), ζ1y = ηxy + y 0 ηyy − y 0 (ξxy + y 0 ξyy ), 15 (3.38) ζ1x = 0 and ζ1y = 0 obtained above, we have the equations yield the follow- ing equations: ηxx + y 0 (ηxy − ξxx ) + y 00 ξxy = 0, (3.39) ηxy + y 0 (ηyy − ξxy ) + y 00 ξyy = 0. (3.40) From Equations (3.39) and (3.40), it is easy to obtain ξxy = 0, ηxy = ξxx , ηxx = 0, (3.41) ξyy = 0, The equations ξxy = 0 and ηyy = ξxy , ηxy = 0 give ξy = C1 , whence, we can determine ξ and η ηxy = 0. ηx = C2 , as ξ = C1 y + A(x), η = C2 x + B(y). Thus, ξxx = ηxy = 0, ηyy = ξxy = 0. (3.42) Using (3.42) we obtain from these equations A00 = 0, B 00 = 0. Based on the above equations, the functions (3.43) A and B can be dened as linear functions: A = C3 x + C5 , B = C4 y + C6 . Hence, we have obtained ξ = C1 y + C3 x + C5 , (3.44) η = C2 x + C4 y + C6 . (3.45) 16 From (3.44) and (3.45) it follows that ξx = C3 , ξxx = 0, ξy = C1 , ξyy = 0, ξxy = 0, (3.46) ηx = C2 , ηxx = 0, ηy = C4 , ηyy = 0, ηxy = 0. According to (3.46), the determing equation (3.38) can be reduced to the following form f µ ξf (ηy − 2ξx − 3ξy y 0 ) − + 2 = 0, x x x or µ f (C1 y + C3 x + C5 ) f (C4 − 2C3 − 3C1 y 0 ) − + = 0. x x x2 (3.47) We rewrite it into a new form µ = (C4 − 2C3 − 3C1 y 0 )f + = C4 − C3 + y x − 3y 0 (C1 y + C3 x + C5 )f x C5 C1 + f. x (3.48) µ = µ(f ), it does not depend on x and y . We can conclude that the constants C1 and C5 in Equation (3.48) should However, from (3.35) we know that be equal to zero. By substituting C1 = 0 and C5 = 0 into System (3.44)-(3.45), we obtain that ξ = C3 x, η = C2 x + C4 y + C6 . Introducing K1 , K2 , K3 and K4 instead of C3 , C2 , C4 (3.49) and C6 , respectively, Equations (3.49) can be rewritten into a new form ξ = K1 x, η = K2 x + K3 y + K4 , (3.50) µ = (K3 − K1 )f (3.51) and where K1 , K2 , K3 and K4 are arbitrary constants. Thus, we have arrived at the following result: 17 Theorem 3.1. The equivalence algebra LE for the equation (1.7) is a four-dimensional Lie algebra spanned by ∂ ∂ −f , ∂x ∂f ∂ = x , ∂y ∂ ∂ = y +f , ∂y ∂f ∂ = · ∂y Y1 = x Y2 Y3 Y4 3.2 Checking operators of the equivalence group We have got four operators for the equivalence Lie algebra, now we check them one by one. • Y1 = x ∂ ∂ −f ∂x ∂f We prolong it to the second order Y1(2) = x ∂ ∂ ∂ ∂ −f − y 0 0 − 2y 00 00 · ∂x ∂f ∂y ∂y According to the invariant condition Y1(2) we have on any solution 1 y − f 1 = 0, x y 00 = f x 00 1 f 1 −2y 00 + f + f x 2 = −2y 00 + 2 = 0 x x x 1 00 of the equation y = f (y 0 ). Hence, Y1 x Equations (3.26). • Y2 = x ∂ ∂y We can prolong it to the rst order only: Y2(1) = x ∂ ∂ + 0· ∂y ∂y 18 is admitted by It satises the invariant condition 1 00 Y2(1) y − f 1 = 0, x y 00 = f x Thus, Y2 is admitted by System (3.26). • Y3 = y Y3 ∂ ∂ +f ∂y ∂f prolonged to the second order has the form Y3(2) = y ∂ ∂ ∂ ∂ +f + y 0 0 + y 00 00 · ∂y ∂f ∂y ∂y The invariant condition is satised, indeed Y3(2) Hence, Y3 • Y4 = f f 00 = y − = 0. y − x y00 = 1 f x y00 = 1 f x x 00 is admitted by Equations (3.26). ∂ ∂y This operator is also admitted by System (3.26). It is the operator X2 obtained in the previous chapter. 4 Projections of the equivalence Lie algebra The Lie algebra of the continuous equivalence group equivalence algebra and is denoted by Ec is the so called LE . Most of the extensions of a principal Lie algebra LF (an algebra admitted by every equation of the family of equations under consideration) are taken from an equivalence algebra LE . These extensions are so called E extensions. Now we use the equivalence Lie algebra to simplify the calculation of symmetry groups without solving the determining equation. If we calculate the equivalence algebra, E extensions for equations of a given family are obtained by solving the simple algebraic equations only. 19 Consider a continuous group of equivalence transformations Y =ξ where the coecients and f. ξ The coordinate ∂ ∂ ∂ ∂ +η +µ + ζ1 0 , ∂x ∂y ∂f ∂y and ζ1 η depend on x and y, while (4.52) µ depends on x, y is given by the prolongation formula: ζ1 = Dx (η) − y 0 Dx (ξ). We denote by Z and X the projections of the equivalence operator (4.52) 0 to the (f, y )-space and to the (x, y)-space, respectively (see [4]). Namely, Z ≡ pr(f,y0 ) (Y ) = µ X ≡ pr(x,y) (Y ) = ξ ∂ ∂ + ζ1 0 , ∂f ∂y ∂ ∂ +η · ∂x ∂y (4.53) The signicance of these projections is determined by the following statement [4], pp. 69-70, Theorem 4.1. An operator X belongs to the principal Lie algebra LF for a system if and only if X = pr(x,y) (Y ) where Y (4.54) is an equivalence generator such that pr(x,y) (Y ) = 0. (4.55) Let us determine the principal Lie algebra by using Theorem 4.1 for the general operator of the equivalence algebra of Equation (1.7), Y =K1 x ∂ ∂ ∂ + (K2 x + K3 y + K4 ) + f (K3 − K1 ) ∂x ∂y ∂f + [K2 + y 0 (K3 − K1 )] Here, Z and X ∂ · ∂y 0 (4.56) are dened according to (4.53). They can be presented as follows: Z = pr(f,y0 ) (Y ) = f (K3 − K1 ) ∂ ∂ + [K2 + y 0 (K3 − K1 )] 0 ∂f ∂y 20 and X = pr(x,y) (Y ) = K1 x Let the operator Z ∂ ∂ + (K2 x + K3 y + K4 ) · ∂x ∂y be equal to zero. Hence, we obtain the relation between the constants: K3 − K1 = 0, whence K1 = K3 . Substituting K1 , K2 K2 = 0, and K3 in Equation (4.56) we obtain the reduced form for the equivalence generator Y = K1 x (4.57) Y: ∂ ∂ + (K1 y + K4 ) · ∂x ∂y (4.58) We rearrange the terms, collecting together the terms with the same constants: Y = K1 (x ∂ ∂ ∂ + y ) + K4 · ∂x ∂y ∂y Thus, we have obtained, that the principal Lie algebra LF is represented by a two-dimensional Lie algebra which is spanned by ∂ ∂ +y , ∂x ∂y ∂ X2 = · ∂y X1 = x This result is exactly the same as the standard form of L2 of type III pre- sented in Table 1.1. 5 Examples The use of the projection about the combination of of Y1 and Z is illustrated here by examples. The rst case is Y1 and Y4 , and the second case is the combination Y2 . Example 1. In the rst case we consider the following equivalence gener- ator: Y = Y1 + Y4 ≡ x The rst prolongation of Y ∂ ∂ ∂ ∂ + −f + ζ1 0 · ∂x ∂y ∂f ∂y can be written as: ∂ ∂ ∂ ∂ Ye = x + −f − y0 0 · ∂x ∂y ∂f ∂y 21 In the rst prolongation equation, the operator tion Y coincides with its projec- pr(y0 ,f ) (Y ) Z1 = pr(y0 ,f ) (Y ) = −f The invariance condition for f ∂ ∂ − y0 0 · ∂f ∂y yields Z1 (f − F (y 0 ))|f =F (y0 ) = 0, which gives the relation between f and F (y 0 ) (5.59) as 0 ∂F −f + y 0 = 0. ∂y f =F (y0 ) Using f = F (y 0 ), we rewrite the above relation into the new form: −F + y 0 ∂F = 0. ∂y 0 Therefore, we obtain the relation between variables y 0 and f in the following form f = Cy 0 , where C (5.60) is an arbitrary constant. Using the additional operator X3 = pr(x,y) (Y ) = x ∂ ∂ + , ∂x ∂y we can easily obtain the form of the relation between x and y dx = dy. x Thus, a non-linear relation between these two variables x and dened as: y = Klnx. Therefore, the rst and second derivatives can be obtained as: y0 = K , x y 00 = − 22 K · x2 y can be By substituting f = xy 00 from Equation (1.7) into Equation (5.60), it is easy to notice that the relationship between x and K is simple: −K = CK, whence C = −1. Thus, we have f = −y 0 . Now we substitute it into Equation (1.7) and obtain y0 y 00 = − · x (5.61) Hence, Equation (5.61) admits along with the operators (2.23) the additional operator X3 = x Example 2. ∂ ∂ + · ∂x ∂y (5.62) In the second case we consider the equivalence generator as Y = Y1 + Y2 ≡ x The rst prolongation of Y ∂ ∂ ∂ ∂ +x −f + ζ1 0 · ∂x ∂y ∂f ∂y can be rewritten as ∂ ∂ ∂ ∂ Ye = x +x −f + (1 − y 0 ) 0 · ∂x ∂y ∂f ∂y The operator as Y coincides with its projection pr(y0 ,f ) (Y ), which we denote it Z2 : Z2 = pr(y0 ,f ) (Y ) = −f The invariance condition for f ∂ ∂ + (1 − y 0 ) 0 · ∂f ∂y has the form: Z2 (f − F (y 0 ))|f =F (y0 ) = 0, which gives the relation between f and −f − (1 − y 0 ) F (y 0 ): ∂F ∂y 0 23 = 0. f =F (y 0 ) (5.63) Based on the condition f = F (y 0 ), the relation we obtained above could be rewritten as F + (1 − y 0 ) ∂F = 0. ∂y 0 y0 Therefore, we derive the relation between and f as f = C(y 0 − 1), where C (5.64) is an arbitrary constant. In order to check this result, let us determine the additional operator X3 = pr(x,y) (Y ) = x The characteristic system for X3 ∂ ∂ +x · ∂x ∂y has the form dx dy = · x x Therefore, we can get a linear relation between y = x + K, The rst and second derivatives of y y 00 = by substituting f and y 00 y are y 00 = 0. f = C(y 0 − 1) second-order equation and K = const. y 0 = 1, Since we have we have x and y 00 = 0, in addition we have Lie's 1 f (y 0 ), x into the equation, then we can obtain 1 0 (y − 1)C = 0. x Finally, we have f = C(y 0 − 1). Substituting this result into Equation (1.7) we have y 00 = C 0 (y − 1)· x 24 (5.65) Hence, the Equation (5.65) admits along with the operators (2.23) the additional operator X3 = x ∂ ∂ +x · ∂x ∂y (5.66) Since both Equations (5.61) and (5.65) are linear, we could make a conclusion that they admit, in fact, an eight-dimensional Lie algebra. 6 Summary The following results have been obtained in this thesis. The innitesimal method was used for nding the equivalence Lie algebra for the second-order equation of type III: y 00 = 1 f (y 0 ). x Its principal Lie algebra is a two-dimensional Lie algebra that spanned by ∂ ∂ +y , ∂x ∂y ∂ X2 = · ∂y X1 = x The equivalence algebra for Lie's second-order equation of type III is a four-dimensional Lie algebra spanned by ∂ ∂ −f , ∂x ∂f ∂ Y2 = x , ∂y ∂ ∂ +f , Y3 = y ∂y ∂f ∂ Y4 = · ∂y Y1 = x Two examples demonstrating the use of the equivalence algebra are presented in Chapter 5. 25 Bibliography A Practical Course in Dierential Equations and Mathematical Modelling: Classical and new methods, nonlinear mathematical models, symmetry and invariance principles, ALGA Publica- [1] Nail H. Ibragimov, tions, Karlskrona, Sweden, 2006 or Higher Education Press, China and World Scientic, Singapore, 2009. [2] Nail H. Ibragimov, Equivalence groups and invariants of linear and non-linear equations, Archives of ALGA, vol. 1, ALGA Publications, Karlskrona, Sweden, pp. 1638, 2004. CRC Handbook of Lie group analysis of differential equations. Vol. 2: Applications in engineering and physical sciences. CRC Press Inc., Boca Raton, 1995. [3] Nail H. Ibragimov, editor. [4] Nail H. Ibragimov, Theorem on projections of equivalence Lie algebras, Selected Works, vol. 2, pp. 6779, ALGA Publications, Karlskrona, Swe- den, 2006. 26
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