On the Linearization of the Ricatti s equation via the Sundman Transformation

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1 On the Linearization of the Ricatti s equation via the Sundman Transformation Abbas Mohamed Sherif 1 and Fawaz Shittu Jimoh 2 1 Stellenbosch University 2 New Mexico State University January 25, 2015 Abstract A Ricatti equation is any equation of the form x = a 2 x 2 +a 1 x+a 0 where a i for i = 0, 1, 2 are functions of the independent variable. In this paper, we show that in general the Ricatti s equation is not linearizable using the Sundman transformation. We use a substitution to convert the equation into a linearizable second order non-linear form and then we determine completely the class of the Ricatti s equation that are linearizable using this transformation under the particular substitution). 1 Introduction Equations of the form x = a 2 x 2 + a 1 x + a 0 where a i for i = 0, 1, 2 are functions of t, are known as the Ricattis equations are first-order non-linear ordinary differential equations that normally require a prior solution in order to obtain a second solution. If a prior solution is not given, there is no known method of direct application to obtain any solution for such equations. In this paper the method of linearization of second order non-linear equations via the Sundman Transformation is used. We go through a step by step process and see that the conditions for linearization inherently constrains the functions a i. 2 The Sundman Transformation For a second order non-linear ODE of the form 1

2 x + λ 2 x ) 2 + λ1 x + λ 0 = 0 1) to be linearized and solved using the Sundman Transformation the following conditions are sufficient to be satisfied. For λ 3 = 0, if λ 6 = 0 then we require λ 2tt = λ 2t λ 1 If λ 6 0 then we require λ 6t = 3λ 0tλ 6 2λ 0 λ 6x = λ 0xλ 6 + 2λ 2 0 λ 2t λ 0 λ 4x = 2λ 1 λ 4 2) where λ 2tt = λ 2t λ 1 λ 6t = 3λ 0tλ 6 2λ 0 λ 6x = λ 0xλ 6 + 2λ 2 0 λ 2t λ 0 λ 4t = λ 4λ λ2 3 λ2 2t 4λ 0λ 1 λ 4 λ 6 2λ 0 λ 6 3) λ 3 = λ 1x 2λ 2t λ 6 = λ 0t + 2λ 0 λ 1 4) λ 4 = 2λ 0xx 2λ 1tx + 2λ 0 λ 2x λ 1 λ 1x + 2λ 0x λ 2 + 2λ 2tt For λ 3 0, if λ 5 = 0 then are sufficient. If λ 5 0 then we require λ 0t = 2λ 0 λ 1 λ 2t = λ 3 5) λ 2ttt = λ 1 3 λ 5 λ 5 + λ 3t ) 2λ 1 λ 5 + λ 2 ) 1 λ 1t λ3 + λ 2t ) λ 3tt λ 0t = 2λ 0 λ 5 λ 1 λ 3 ) λ 3 d λ 2ttx = 2λ 2t λ 3 + λ 2t ) λ 3tx λ 1 dx λ 3 + λ 2t ) + λ 1 3 λ 3xλ 5 6) λ 5 λ 3 6λ ox λ 2t + 2λ 2tx λ 0 + 4λ 2t λ 0 λ 2 + 2λ 3x λ 0 + 4λ 2 λ 2 λ 3 + λ 1 λ 5 ) 6λ 2 3 λ 2 2t λ 0 λ 0x λ 5 + λ 0 λ 3 λ 1x ) λ 2 5 λ 4 2λ 5 ) = 0 2

3 where λ 5 = λ 2tt + λ 1 λ 3 + λ 2t ) + λ 3t 7) 3 Conditions for Linearizing the Ricatti s equation The Ricatti s equation is any first order non-linear ODE of the form x = a 2 x 2 + a 1 x + a 0 8) where a i for i = 0, 1, 2 are functions of t only). We make the substitution so that x = p x = p and x 2 = p ) 2 Then equation 8 is transformed into p a 2 p ) 2 a1 p a 0 = 0 9) which is the form of the second order non-linear ODE where We now investigate the conditions. We have λ 2 = a 2 λ 1 = a 1 10) λ 0 = a 0 λ 3 = λ 1p 2λ 2t = a 1p + 2a 2t = 2a 2t 0 11) We now check λ 5. λ 5 = λ 2tt + λ 1 λ 3 + λ 2t ) + λ 3t = a 2tt a 1 2a 2t a 2t ) + 2a 2tt = λ 2tt a 1 a 2t 0 12) Since λ 5 0 we require the conditions in equation 6 to be satisfied. We start by checking the third equation of equation 6 and see that 3

4 2a 2 2t = 0 = a 2t = 0 = a 2 = b = constant 13) Checking the rest of the equations in equation 6 shows that they are satisfied. But the condition that a 2t = 0 puts a further constraint on λ 3 since this would imply Then we check λ 6 and see that λ 3 = 0 λ 6 = 2a 0 a 1 a 0t 14) If λ 6 = 2a 0 a 1 a 0t 0 then we check the conditions in equation 3. The first, third and fourth equations in equation 3 are clearly satisfied. The second equation of equation 3 gives the requirement 2a 0 a 1 a 0t + a 0tt 3a 2 ot 4a 2 0a 1t = 0 15) which is sufficient. If λ 6 = 2a 0 a 1 a 0t = 0 this would imply 2a 0 a 1 = a 0t. Substituting this into equation 15 gives the requirement a 0 = 0. All other conditions are satisfied. In summary, The classes of Ricatti s equations solvable via the Sundman transformation are provided equation 15 holds. Also all equations of the form x = bx 2 + a 1 x + a 0 16) x = bx 2 + a 1 x 17) for any time dependent function a 1 are linearizable, where b is a real constant 0 since b = 0 reduces the equation to a linear one anyway). We recognize the class in 17 as a Bernoulli s equation with n = 2. We will later see how we use our method to solve such equations. Now for the first class of equations x = bx 2 + a 1 x + a 0 the set of partial differential equations to be solved [2] in order for such class to be linearized are given by F pp = G pf p + F p Gλ 2 G G p = 0 18) G t = G λ 0t 2λ 0 = G a 0t 2a 0 4

5 The second equation of equation 18 can be solved to give For the class in equation 17 where λ 6 = 0, a 0t = 2a 0 a 1. reduces the third equation of equation 18 to Solving equation 19 gives This G t = Ga 1 19) G = e a1 dt Since G x = 0 the first equation of equation 18 reduces to which is the other equation to be solved. Solving equation 21 gives 20) F pp = bf p 21) F = A b e bp + c 22) for arbitrary constants A and c. The third equation of equation 18 can be solved without further ristriction) to give G = Ba 0 23) for some arbitrary constant B. In summary, for the class in 16, F and G are given by equations 22 and 23 respectively. For the class in 17, F and G are given by equations 22 and 20 respectively. 4 The Solution Once our functions F and G are known we set out to find our parameters α, β and γ given by For the class in 16, α = a 1G + G t G 2 β = λ 0λ 2 G + λ 0x G λ 0 G x G 3 γ = βf G2 λ 0 F x G 2 α = a 0t a 0 a 1 β = Ba 2 0 b B 2 a 0 γ = 0 24) 25) 5

6 Our linearized equation then becomes ) ) u a0t a 0 a 1 b + u + B 2 u = 0 26) a 0 Ba 2 0 for arbitrary constant B where B 0 and a 0 0. If it so happens that a 0 and a 1 are such that the coefficients in equation 26 are constants, immediately solutions can be found using normal methods for solving linear ODE with constant coefficients. If the coefficients are non constant, then there are known methods to solve such equations which sometimes for which a solution will be required and then we could use reduction of order to solve. However, for a particular case where 26 is exact. Then a first integral can be completely obtained. Suppose we have an equation of the form p 0 u + p 1 u + p 2 u = 0 27) where p i for i = 0, 1, 2 are functions of the independent variable T only. We can rewrite equation 27 as p0 u p 0u ) + p1 u) p 0 p 1 + p 2 ) u = 0 28) Then equation 27 is called exact if p 0 p 1 + p 2 = 0. This reduces equation 28 to the first order equation p 0 u p 1 p 0) u = c 29) for some constant c whose solution is given by u = e p 1 p 0 p1 p dt 0 dt p 0 e p 0 dt 30) For the case ) under consideration equation 26), p 0 = 1 and p 1 = a0t a 0 a 1 where we have written the a Ba 2 is in terms of T using relation T = B a 0 dt from the tramsformation equation dt = Gdt and 0 equation 23. Then the condition of exactness in equation reduces to p 2 p 1 = 0 = p 1 = p 2 31) Once equation 31 is satisfied an exact solution to 26 is given by u = e p 1 dt p1 e dt dt 32) Using the transformation equations u = F = u = Ae p and dt = Gdt [?] where G is given by equation 23 and writing p 1 in terms of t gives the solution of equation 16 as p = x = 1 b d dt ln bae B p 1 a 0 dt 6 e B )) p 1 a 0 dt dt 33)

7 for constants A, B. Otherwise other methods requiring a solution may be used to solve 26. For the class in 17 α = 0 β = 0 γ = 0 34) Our linearized equation becomes 4.1 Examples 1. Consider the equation u = 0 35) x = x ) Clearly this falls into the first class 16 as b = 1 and a 0 0. Here a 2 = 1, a 1 = 0, a 0 = 1. This reduces 26 to the form the general solution which is given by u 1 B 2 u = 0 37) u = c 1 e 1 B )T + c 2 e 1 B )T 38) Now from the transformation equations u = F = u = Ae p and dt = Gdt = dt = Bdt = T = Bt where we keep ignoring the constant in F since it will vanish for all differentiations). Then p = ln u ) = ln c 1e t + c 2 e t ) A A p = x = c 1e t c 2 e t 39) c 1 e t + c 2 e t for A and B 0. It can be easily verified that equation 39 is a solution to the differential equation Consider another differential equation x = x t x 40) 7

8 We see that this is an equation of the class in 17. We have now shown that for all equations in the class in 17 the linearized form is the form in equation 35 whose solution is given by u = ct 41) for some constant c. Again from the transformation equations u = F = u = Ae p and dt = Gdt = dt = e 1 t dt dt = dt = Btdt = T = B 2 t2 where we again ignore the constant in F ). Then p = ln u ) = ln cb ) A 2A t2 = p = x = 2 t 42) Again the above equation can easily be verified to be a solution to the differential equation 40. For the class in 17 we can write explicitly the solution as x = 1 b d ln ba dt )) a1 e dt dt 43 x = 1 b d ln dt e a1 dt dt or )) 43) for some arbitrary constant A. We neglect the constants since they will cancel out upon differentiation. 5 Conclusion We conclude that in general the Ricatti s equation is not solvable using the Sundman transformation. Only particular classes of such equations can be linearized 16 and 17) for which exact solutions can be directly extracted. 8

9 References [1] Meleshko S.V., Methods for constructing exact solutions of partial differential equations, Mathematical and Analytical Techniques with Applications to Engineering, Springer, New York, [2] Nakpim, W. and Meleshko, S.V., Linearization of Second-Order Ordinary Differential Equations By Generalized Sundman Transformations. Symmetry, Integrability and Geometry: Methods and Applications SIGMA).6051). 2010) [3] Berkovich L.M., The integration of ordinary differential equations: factorization and transformations, Math. Comput. Simulation ),

Linearization of Second-Order Ordinary Dif ferential Equations by Generalized Sundman Transformations

Symmetry, Integrability and Geometry: Methods and Applications SIGMA 6 (2010), 051, 11 pages Linearization of Second-Order Ordinary Dif ferential Equations by Generalized Sundman Transformations Warisa