Examensarbete. Separation of variables for ordinary differential equations. LiTH - MAT - EX / SE

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1 Examensarbete Separation of variables for ordinary differential equations Anna Måhl LiTH - MAT - EX / SE

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3 Separation of variables for ordinary differential equations Department of Applied Mathematics, Linköpings Universitet Anna Måhl LiTH - MAT - EX / SE Examensarbete: 0 p Level: D Supervisor: Stefan Rauch, Department of Applied Mathematics, Linköpings Universitet Examiner: Stefan Rauch, Department of Applied Mathematics, Linköpings Universitet Linköping: February 006

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5 Avdelning, Institution Division, Department Matematiska Institutionen LINKÖPING SWEDEN Datum Date February 006 Språk Language x Svenska/Swedish Engelska/English Rapporttyp Report category x Licentiatavhandling Examensarbete C-uppsats D-uppsats Övrig rapport ISBN ISRN LiTH - MAT - EX / SE Serietitel och serienummer Title of series, numbering ISSN URL för elektronisk version Titel Title Separation of variables for ordinary differential equations Författare Author Anna Måhl Sammanfattning Abstract In case of the PDE s the concept of solving by separation of variables has a well defined meaning. One seeks a solution in a form of a product or sum and tries to build the general solution out of these particular solutions. There are also known systems of second order ODE s describing potential motions and certain rigid bodies that are considered to be separable. However, in those cases, the concept of separation of variables is more elusive; no general definition is given. In this thesis we study how these systems of equations separate and find that their separation usually can be reduced to sequential separation of single first order ODE s. However, it appears that other mechanisms of separability are possible. Nyckelord Keyword ODE, Separation of variables, Potential motion, Heavy symmetric top, Cofactor systems, Direct separability.

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7 Abstract In case of the PDE s the concept of solving by separation of variables has a well defined meaning. One seeks a solution in a form of a product or sum and tries to build the general solution out of these particular solutions. There are also known systems of second order ODE s describing potential motions and certain rigid bodies that are considered to be separable. However, in those cases, the concept of separation of variables is more elusive; no general definition is given. In this thesis we study how these systems of equations separate and find that their separation usually can be reduced to sequential separation of single first order ODE s. However, it appears that other mechanisms of separability are possible. Keywords: ODE, Separation of variables, Potential motion, Heavy symmetric top, Cofactor systems, Direct separability. Måhl, 006. vii

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9 Acknowledgements I would like to thank my supervisor and examiner Stefan Rauch. You have been a dedicated, and skillful, educationalist. By dividing problems into pieces big enough for me to calculate and comprehend, you have taught me not to be afraid of approaching non familiar problems and you have encouraged me to try new ways. These things I will treasure and bring with me in the future. I would like to thank my dad Per Måhl. All my life you have read and given critique on my writing, but this must be the first time that you have helped me by knowing less! It is because you have not understood majority of the mathematical preliminaries that you have been an invaluable partner for discussion. I would like to thank my opponent Peter Brommesson, and also Mattias Hansson, for reading this report and giving critique. I would also like to thank Zebastian Zaar for creating my pictures of HST and rolling disk. Moreover, I would like to thank Daniel Petersson, in the office next to mine, for helping me to understand some mechanics. Finally, I would like to thank the rest of my family and my friends. You all mean a lot to me. Måhl, 006. ix

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11 Contents 1 Introduction 1 Preliminaries 3.1 Separation of variables for one dimensional potential motion Potential Newton Equations Two dimensional potential Newton equations Separation of variables for two dimensional potential motion 9 3. Three dimensional potential motion Separation of variables for three dimensional potential motion Vector equations of a rigid body Motion of a heavy symmetric top Separation of variables for the HST equations The equations of motion of the rolling disk Separation of variables for the RD equations Triangular Newton equations Two dimensional triangular cofactor system Separation of variables for two dimensional triangular Newton equations The Hénon-Heiles Hamiltonian with cubic potential The Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian with cubic potential Separation of variables for the Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian Direct separability An example of second order Separation of variables for this example Conclusion and discussion 39 Måhl, 006. xi

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13 Chapter 1 Introduction An ordinary differential equation (ODE) is an algebraic equation f(x,y, dy dx ) = 0 involving derivatives of some unknown function with respect to one independent variable. A separable ordinary differential equation is an ODE which can be written in such a way that the dependent variable and its differential appear on one side of the equals sign and the independent variable and its differential appear on the other side. The method of separation of variables is best known for partial differential equations of mathematical physics, for which it is the main method of solving these equations. The basic idea behind the method of separating variables in the theory of partial differential equations (PDE) is to consider an ansatz for a solution (additive or multiplicative) that allows reducing the problem to a system of uncoupled ordinary differential equations for functions of one variable. Example 1. The heat equation. Consider u t = u xx (1.1) for an initial boundary value problem on the interval 0 < x < L and with the boundary conditions u t = u xx,0 < t,0 < x < L u(0,t) = u(l,t) = 0,0 < t u(x,0) = f(x),0 < x < L One usually makes an ansatz assuming the solution to be a product of two functions. Substitution of u(x, t) = X(x)T(t) into (1.1) reduces this partial differential equation to two uncoupled ordinary differential equations { t T = k T xx X = k X These two ordinary differential equations have solutions { T k (t) = Ae k t X k (x) = B cos kx + C sin kx Måhl,

14 Chapter 1. Introduction and due to linearity of (1.1) any formal linear combination u(x,t) = k c kx k (x)t t (t) is also a solution. This solution can be further specified according to the boundary conditions u(0,t) = u(l,t) = 0. If we put k n = nπ L, where n = 1,,..., the complete solution reads u(x,t) = c n e k n t sink n x (1.) n=1 where the coefficients c n are given by the integrals c n = L ([3],[6]). L 0 f(x)sin k nxdx Example. The Hamilton-Jacobi equation of natural Hamiltonian Another example of an equation known to be solvable by separating variables is the Hamilton-Jacobi equation for the natural Hamiltonian ([4],[6]). One method of solving the canonical Hamilton equations ẋ = H(x,y) y, ẏ = H(x,y), x = (x 1,...,x n ), y = (y 1,...,y n ) (1.3) x where H(x,y) is a Hamiltonian, is to find a function W(x,α), given by the canonical transformation y = W(x,α) x, β = W(x,α) α, for which (1.3) is transformed into a set of simple linear equations for new variables β, α. This function W(x,α) satisfies the first order nonlinear PDE ( H x, W(x,α) ) = α 1 (1.4) x which is known as the Hamilton-Jacobi equation. In order to find it, one makes an additive ansatz W(x,α) = n k=1 W k(x k,α), where the n functions W k (x k,α) each depend on a single variable x k. This ansatz works well for the Hamiltonians of Stäckel class with H = G+V = 1 i gii (x)yi. It simplifies the problem; instead of integrating the nontrivial Hamilton-Jacobi equation the problem reduces to integrating n uncoupled first order ODE s for functions W k (x k,α). The purpose of this thesis is to investigate how variables can be separated in case of ODE s. In order to do so we put together known examples of Newton equations solved in mechanics and describe the features of their separability. In chapter we present basic concepts needed for this work; we give a definition of separability in general and we explain why we consider the Newton equation of one dimensional potential motion to be separable. In the following chapters we study a variety of mechanical problems, mainly of Newtonian form, to see how these equations are solved by separating variables. In chapter 3 we describe the problem of potential motion. In chapter 4 we look at the equations of motion of rigid bodies and we show the problems of the heavy symmetric top and of the rolling disk. In chapter 5 we describe how separation of variables takes place when solving triangular systems of Newton equations and in chapter 6 we consider the Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian with cubic potential. Finally, in chapter 7, we give an example of a directly separable second order system and in chapter 8 we formulate conclusions of our work.

15 Chapter Preliminaries In this chapter we state the definition of separability ([1]). Moreover, we explain how the Newton equation of one dimensional potential motion is consistent with this definition and therefore is separable ([4]). Definition 1 An ordinary first order differential equation y = f(x,y) is said to be separable if it can be written as g(y)y = h(x) (.1) where g and h are known continuous functions depending on one real variable. A separable equation can be solved by integrating both sides w. r. t. x ( G(y) = g(y) dy ) dx = (h(x)) dx = H(x) + C dx where g has the primitive function G and h has the primitive function H. The solution has the implicit form G(y) = H(x) + C Example 3. The logistic equation. Consider dx (1 dt = µx x ) k where µ and k are positive constants. It is separable since it can be factorised into 1 µx ( 1 x k ) dx dt = 1 (.) for x k and x 0. By integrating both sides, w. r. t. time, we obtain an implicit solution ( ) 1 1 t + C = µ x + 1 k ( ) 1 x dx = 1 (ln x ln 1 x ) µ k k Måhl,

16 4 Chapter. Preliminaries and an explicit solution x(t) reads x = keµt C + e µt In the separation procedure we had to assume that x k and x 0. A direct check confirms that constant solutions x = k and x = 0 also satisfy the equation (.). Example 4. The homogeneous equation. Take dy dx = f(y x ) (.3) This equation can not be written directly in the separable form (.1), but by defining a new variable z = y dy x we get dx = x dz dx + z and f( y x ) = f(z). By substituting these expressions into the equation (.3) we get an equation x dz dx + z = f(z) which is separable since it can be written as 1 dz f(z) z dx = 1 x.1 Separation of variables for one dimensional potential motion Let us consider the second order ODE of the form dv (x) mẍ = dx (.4) where each dot over x denotes a time derivative. It is not separable directly, but we observe that it can be integrated once by multiplying with ẋ. Then ( ) d mẋ dv (x) mẋ = and + V (x) = E = constant dt dt We can solve this first order ODE for x(t) by separating variables since the equation of energy mẋ + V (x) = E can be written which gives us t = m m 1 (E V (x)) dx dt = 1 dx (E V (x)) =: G(x) + C Notice that in solving (.4) by separation of variables we have used the energy integral of motion. Our solution is given in an implicit form since we need to invert the equation t = G(x) + C to find the solution x(t,c).

17 .1. Separation of variables for one dimensional potential motion 5 Definition An integral of motion for a second order differential equation ẍ = M(x) is a function of position and velocity K(x,ẋ) such that it takes constant value on solutions x(t). That is As a consequence we have K(x(t), ẋ(t)) = constant K(x,ẋ) = dk dt = K x ẋ + K ẋ ẍ = K x ẋ + K ẋ M(x) = 0 for all (x,ẋ). So, since the second order Newton equation of one dimensional potential motion (.4) admits an integral of motion (the energy integral) that constitutes a first order separable equation, (the equation which can be factorised like (.1)) we can consider it to be separable.

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19 Chapter 3 Potential Newton Equations In this chapter we study how potential Newton equations in three dimensions are solved by separating variables ([4]). A general potential Newton equation has the form m r = grad V(r), r R 3 (3.1) where the function V (r) is called a potential. By multiplying both sides with ṙ we see that (3.1) always admits the energy integral of motion E = 1 mṙ + V (r) (3.) For a central potential V (r) = V (r) with r = r, the force F = grad V(r) = dv (r) r dr r is parallel to r. Then the equation (3.1) admits an additional vector integral of motion, namely the angular momentum L = r mṙ. By differentiating we find ( ) L dv (r) r = ṙ mṙ + r m r = r m r = r F = dr r r = 0 By using the angular momentum and the energy integrals of motion we can solve potential Newton equations with central potentials V (r) = V (r) through separation of variables. 3.1 Two dimensional potential Newton equations Let us first consider the two dimensional potential Newton equation with the central potential V (r) m r = grad V(r) (3.3) where r = (x 1,x ) T, that takes the coordinate form V (r) mẍ 1 = = V (r) x 1 x 1 r V (r) mẍ = = V (r) x x r (3.4) (3.5) Måhl,

20 8 Chapter 3. Potential Newton Equations This equation, (3.3), admits the energy integral E = mṙ + V (r). Moreover, by multiplying the first coordinate equation (3.4) by x, and the second one (3.5) by x 1 and subtracting we find d dt (mẍ 1x mẍ x 1 ) = 0 which we recognize as the derivative of the third component of the angular momentum L 3 = m(x 1 ẋ x ẋ 1 ). The energy and the angular momentum integrals of motion reduce the second order system (ẍ 1,ẍ ) into a first order system L 3 = m(x 1 ẋ x ẋ 1 ) (3.6) E = mṙ + V (r) (3.7) where L 3 and E denote constant values of these integrals of motion. The system ( ) is not separable yet, but can be solved by separation of variables in the polar coordinates r and θ. By substituting x 1 = r cos θ, ẋ 1 = ṙ cos θ r θ sinθ, x = r sin θ and ẋ = ṙ sinθ + r θ cos θ into ( ) we get L 3 = mr θ (3.8) E = m ( ) ṙ + r θ + V (r) (3.9) When we resolve θ = L3 mr from (3.8) and substitute in (3.9) we get the separable equation of one dimensional motion w. r. t. variable r (.4) ( E = m ( ) ) ṙ + r L3 mr + V (r) (3.10) We factorise it 1 m (E V (r)) L 3 m r dr dt = 1 (3.11) and a solution is obtained by integrating both sides w. r. t. time t dr t = (3.1) m (E V (r)) L 3 m r Finally to get a complete solution in terms of r and θ, we have to invert (3.1) to get r(t) explicitly. Then by substituting r(t) into (3.8) we get a solution for θ(t) like θ = L 3 dt (3.13) mr(t)

21 3.1. Two dimensional potential Newton equations Separation of variables for two dimensional potential motion Naively one would expect that the aim of separation of variables always is, as we described in our examples of PDE, to transform the original system into a new system where the equations, for functions of one variable, are uncoupled. However, in general this may not be possible. For example, when we have solved the problem of two dimensional potential motion, we have transformed our original system into polar coordinates. We have obtained a new system which we have solved by separating variables, even though the equations have remained coupled. When we solve a coupled system by separation of variables we use a procedure we have chosen to call stepwise. The stepwise separation of variables for two dimensional potential motion can be symbolically represented by a diagram L 3 θ(r) E L 3 r(t) θ(t) Figure 3.1: Stepwise separation of two dimensional potential motion Start reading the diagram (Figure 3.1) in the upper left corner. Let the horisontal arrows indicate that the quantity is resolved from the expression, while the vertical arrows indicate that the quantity is substituted into the expression. From L 3 resolve θ, follow the arrows and substitute the expression of θ into E. From E we get the solution r(t). To obtain a complete solution keep following the arrows upwards until reaching the final solution θ(t) in the upper right corner. Let us describe where and how the separation of variables have been taking place and lead to a solution for our original second order system of equations (3.3). Due to the energy and angular momentum integrals of motion, we have reduced the system of two second order equations ( ) into a new system of the two first order integral equations ( ). Subsequently, we have transformed this first order system into polar coordinates and obtained the new coupled system of equations ( ) which we have solved by successively eliminating and stepwise separating variables. Let us symbolically denote this system ( ) as { L 3 = f 1 (r, θ) E = f (r,ṙ, θ) We have, as described in the diagram (Figure 3.1), resolved θ = g 1 (r) from L 3 (3.8) and substituted it into the expression of E (3.9). When we have eliminated θ variable from the energy E we have got an equation (3.10) on the form E = f (r,ṙ,g 1 (r)) = f 3 (r,ṙ) This equation depends on r only. It is separable and, in (3.11), we have written it like (.1). By integrating this equation (3.11) we have got an implicit solution

22 10 Chapter 3. Potential Newton Equations for r (3.1). Then we have calculated an explicit solution r(t). This solution we have substituted into (3.8). That has given us one more equation (3.13) to integrate, which is also consistent with (.1), and finally we have calculated a solution for θ(t). Summing up, in two dimensional potential Newton equations we have two integrals of motion that constitutes a system of two first order equations. When we transform this system we obtain a new system of equations, which we can solve by stepwise separation of variables since we successively can, after eliminating certain variables, factorise each equation like (.1). In short, we have obtained a complete solution by substituting in the following manner: E r(t) L3 θ(t) 3. Three dimensional potential motion Let us solve the three dimensional problem of motion in the central potential V (r) by separation of variables. As in the two dimensional case, we start with the Newton equation m r = grad V(r), r = (x 1,x,x 3 ) T R 3 (3.14) and we know that E and L are integrals of motion. Then, we conclude that the third component L 3 and L are also integrals of motion L 3 = (Lê 3 ) = Lê 3 + L ê 3 = Lê 3 = 0 L = (L L) = L L + L L = L L = 0 As a result, for three dimensional potential motion, the system m r = grad V(r), has three integrals of motion: E, L 3 and L. In Cartesian coordinates, these integrals of motion read L 3 = m(x 1 ẋ x ẋ 1 ) (3.15) L = m(x ẋ 3 x 3 ẋ ) + m(x 3 ẋ 1 x 1 ẋ 3 ) + m(x 1 ẋ x ẋ 1 ) (3.16) E = mṙ + V (r) (3.17) We can solve this system of equations ( ) by separation of variables if we introduce the spherical coordinates r, ϕ and θ. We substitute x 1 = r sinθ cos ϕ, x = r sinθ sin ϕ, x 3 = r cos θ, and ẋ 1 = ṙ sinθ cos ϕ + r( θ cos θ cos ϕ ϕsin θ sin ϕ), ẋ = ṙ sin θ sinϕ + r( θ cos θ sin ϕ + ϕ sin θ cos ϕ), ẋ 3 = ṙ cos θ r θ sinθ which gives us a new system We resolve ϕ from (3.18) L 3 = mr ϕsin θ (3.18) L = m r 4 ( θ + ϕ sin θ) (3.19) E = m ( ) ṙ + r θ + r ϕ sin θ + V (r) (3.0) ϕ = L 3 mr sin θ (3.1)

23 3.. Three dimensional potential motion 11 and substitute ϕ into (3.19) to get θ = L m r 4 L 3 m r 4 sin θ (3.) By substituting (3.1) and (3.) into the energy equation (3.0) we get ( E = m ( L ṙ + r m r 4 L ) ( ) 3 m r 4 sin + r L 3 θ mr sin sin θ) + V (r) θ which miraculously simplifies to a one dimensional energy integral (.4) since the dependence on θ cancels out and leaves us with the expression E = m ) (ṙ + L m r + V (r) (3.3) This equation is separable and we factorise it with respect to r and t m 1 (E V (r)) L m r A solution is then obtained by integrating both sides dr t = (E V (r)) L m dr dt = 1 (3.4) (3.5) m r To get an explicit solution r(t), ϕ(t) and θ(t) we invert (3.5) to get r(t). Then we substitute r(t) into (3.) to find a solution for θ(t) dθ dt = L L 3 mr(t) (3.6) sin θ Finally we substitute r(t) and θ(t) into (3.1) and get ϕ(t) L 3 dt dϕ = mr(t) sin θ(t) (3.7)

24 1 Chapter 3. Potential Newton Equations 3..1 Separation of variables for three dimensional potential motion When we have solved this three dimensional problem of potential motion, we have separated variables stepwise. Let us draw a diagram similar to Figure 3.1: L 3 ϕ(r,θ) L θ (r,θ) E L r(t) L 3 θ(t) ϕ(t) Figure 3.: Stepwise separation of three dimensional potential motion Start reading the diagram (Figure 3.) in the upper left corner. Let the horisontal arrows represent resolving the expression w. r. t. the variable and the vertical arrows signify substituting the variable into the expression. Resolve ϕ from L 3 and substitute in L and E. Keep following the arrows, resolving, substituting and obtaining solutions until reaching the upper right corner. Now let us learn more about where and how the separation of variables has been taking place by looking at the features that have enabled us to obtain a solution for (3.14). Since the energy and the angular momentum are integrals of motion we have reduced the original second order system, the three coordinate equations (3.14), into a system of first order ( ). Then having transformed this first order system into the spherical coordinates, we have obtained the system ( ) that we have solved by stepwise separation of variables. Let us, as in the case of two dimensional potential motion, denote this system ( ) as L 3 = f 1 (r, ϕ,θ) L = f (r, ϕ,θ, θ) E = f 3 (r,ṙ, ϕ,θ, θ) As can be read in the diagram (Figure 3.), we have started by resolving ϕ in terms of r and θ from L 3 (3.18) like ϕ = g 1 (r,θ) Then, by resolving θ from L (3.19) we have got θ = g (r,θ) By substituting the expressions of ϕ and θ into E (3.0) we have obtained an equation (3.3) in the form E = f 3 (r,ṙ,g 1 (r,θ),θ,g (r,θ)) = f 4 (r,ṙ) This equation is separable. In (3.4), we have factorised it as (.1) and we have integrated it to get an implicit solution for r (3.5). We have calculated the

25 3.. Three dimensional potential motion 13 explicit solution r(t). Then we have substituted this solution into the expression θ = g (r(t),θ) (3.) and we have obtained an equation which is separable with respect to t and θ (3.6). We have integrated this equation and we have inverted the solution to get θ(t). At last, we have substituted our solutions, r(t) and θ(t), into our expression for ϕ = g 1 (r(t),θ(t)) (3.1) and we have obtained a third separable equation (3.7) thus the solution ϕ(t). In short, let us illustrate this stepwise solving procedure as follows: E r(t) L θ(t) L3 ϕ(t) Remark. Notice that it is due to the fact that our variable θ miraculously disappears in (3.3) that separation of r and t becomes possible.

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27 Chapter 4 Vector equations of a rigid body In this chapter we study how the heavy symmetric top equations ([5]) and the equations of the rolling disk ([10]) are solved by separating variables. The vector equations of motion of a rigid body are m s = F (4.1) L = K (4.) where F is the force, K denotes the torque, L is the angular momentum and s denotes the position of center of mass w. r. t. an inertial reference frame ([4]). 4.1 Motion of a heavy symmetric top A heavy symmetric top, a HST, is a symmetric rigid body with one point on its symmetry axis fixed in space. The center of mass s of the HST is located on Figure 4.1: The heavy symmetric top Måhl,

28 16 Chapter 4. Vector equations of a rigid body the symmetry axis at the distance l from the origin. When describing motion of the HST we choose an inertial coordinate system K 0 = [ˆx,ŷ,ẑ] and a body-fixed coordinate system K = [ˆ1,ˆ,ˆ3] such that both have their origins at a point fixed in space (see Figure 4.1). The equations of motion for the HST is then derived by choosing the symmetry axis of the top as the ˆ3-axis of the body-fixed coordinate system. The total force F acting on the HST is the sum of the gravitational force acting on the center-of-mass and of the reaction force F r acting on the support point. The torque K and the angular momentum L are both vectors defined with respect to the common origin of both coordinate systems. The equations of motion, (4.1-4.), in case of the HST read d(ml ˆ3) dt = mgẑ + F r (4.3) dl dt = lˆ3 ( mgẑ) (4.4) These equations are complemented by the kinematic equation describing the motion of ˆ3 w. r. t. the inertial coordinate system K 0 ˆ3(t) = ω ˆ3 where ω is the instantaneous angular velocity of the HST. By using the relations L = (I x Ω x,i y Ω y,i z Ω z ) T where I x,i y,i z are moments of inertia w. r. t. the fixed point, and that ω ˆ3 = L I x ˆ3 for an axially symmetric rigid body, we get the vector equations of motion for the HST ˆ3 = L I x ˆ3 (4.5) dl dt = mglˆ3 ẑ (4.6) This system is a closed system of six ODE s, for six scalar unknowns; i.e. the components of ˆ3 and L. The vector equations of the HST ( ) admit three integrals of motion: the total energy E, the projection of the angular momentum L z onto the vertical axis and the projection of angular momentum L 3 onto the symmetry axis. We can show that L z and L 3 are integrals of motion by differentiating them d dt (L ẑ) = ( mglˆ3 ẑ) ẑ = 0 (4.7) d dt (L ˆ3) = ( mglˆ3 ẑ) ˆ3 + L ( L ˆ3) = 0 (4.8) I x Moreover, if we use the relation ω L = ω L (valid for any axial symmetric rigid body), we see that the total energy of the system is also an integral of motion d dt (E) = d dt (1 ω L + mglˆ3 ẑ) = 1 ω L + 1 ω L + mgl ˆ3 ẑ = ω L + mgl(ω ˆ3) ẑ = ω ( mglˆ3 ẑ) + mgl(ω ˆ3) ẑ = 0

29 4.1. Motion of a heavy symmetric top 17 As a result, the problem of motion of a HST is reduced to solving three equations given by these three integrals of motion L z = Lẑ (4.9) L 3 = Lˆ3 (4.10) E = 1 ω L + mglˆ3 ẑ (4.11) To obtain a system which we can solve by separating variables we express ( ) in the coordinates θ, φ and ψ, known as Euler angles Figure 4.: The Euler Angles Figure 4. gives us the Euler coordinates ˆ1 = (cos ψ cos φ cos θ sin φsin ψ)ˆx + (cos ψ sinφ + cos θ cos φsin ψ)ŷ + (sin θ sin ψ)ẑ ˆ = (sin ψ cos φ + cos θ sin φcos ψ)ˆx (sin ψ sinφ cos θ cos φcos ψ)ŷ + (sin θ cos ψ)ẑ ˆ3 = (sin θ sin φ)ˆx (sin θ cos φ)ŷ + (cos θ)ẑ We express the angular velocity, ω = ω xˆx + ω y ŷ + ω z ẑ, by ω x = ψ sinθ sin φ + θ cos φ ω y = ψ sin θ cos φ + θ sin φ ω z = θ + ψ cos θ and angular momentum, L = I ω, by L x = I x ( θ cos φ φ sinθ cos θ sin φ) + I z sin θ sinφ( ψ + φcos θ) L y = I x ( θ sin φ + φ sinθ cos θ cos φ) I z sin θ cos φ( ψ + φcos θ) L z = I x φsin θ + I z cos θ( ψ + φcos θ)

30 18 Chapter 4. Vector equations of a rigid body We get our system ( ) expressed in new coordinates L z = I x φsin θ + I z cos θ( ψ + φcos θ) (4.1) ( L 3 = I z ψ + φcos ) θ (4.13) E = I ( x θ + φ sin θ) + I ( z ψ + φcos ) θ + mgl cos θ (4.14) In order to solve this system by separating variables, we start with equations (4.1) and (4.13) and express φ and ψ in terms of θ ψ = L ( 3 Lz cos θ L 3 cos θ ) I z I x sin (4.15) θ ( Lz cos θ L 3 cos φ θ ) = I x sin (4.16) θ Substituting (4.15) and (4.16) into (4.14) we get a one dimensional energy integral equation (.4) E = I x θ + (L z L 3 cos θ) I x sin θ This equation can be factorised, and separated, as where + L 3 + mgl cos θ = I θ x + U(θ) (4.17) I z 1 (E U(θ)) I x U(θ) = (L z L 3 cos θ) I x sin θ dθ dt = 1 (4.18) + L 3 I z + mgl cos θ When we integrate both sides we get t = 1 I x (E U(θ)) dθ (4.19) If we invert (4.19) we get the explicit solution θ(t). Further, if we substitute θ(t) into, and integrate right hand side of, (4.15) and (4.16), we get φ(t) and ψ(t).

31 4.1. Motion of a heavy symmetric top Separation of variables for the HST equations We have solved the problem of the HST by separating variables stepwise. Let us draw a diagram: L z L 3 φ(θ), ψ(θ) φ(θ) ψ(θ) φ(t),ψ(t) E θ(t) Figure 4.3: Stepwise separation of variables for motion of a heavy symmetric top Start reading the diagram (Figure 4.3) in the upper left corner and read that we have used both the equation for L z and the equation for L 3 to resolve φ and ψ simultaneously and that we secondly have substituted them into the expression for E to get a solution for θ(t). To solve the equations of the HST we have used three integrals of motion, derived from the energy and the angular momentum integrals, and we have reduced the original system of second order ODE s to three first order ODE s. By transforming these first order ODE s into Euler angle variables we have got a system that we have separated stepwise. Let us denote it ( ) as L z = f 1 ( φ, ψ,θ) L 3 = f ( φ, ψ,θ) E = f 3 ( φ, ψ,θ, θ) First we have, as described in Figure 4.3, resolved φ and ψ from L z (4.1) and L 3 (4.13) and we have got φ = g 1 (θ) ψ = g (θ) Then we have eliminated φ and ψ from (4.14) and got an equation in the form E = f 3 (θ, θ) This equation we have solved by separating variables since we have factorised it like (.1) in (4.18). Then we have substituted the solution for θ(t) into the expressions of φ (4.15) and ψ (4.16) and they have become directly integrable equations. Let us illustrate the stepwise procedure leading to a complete solution as E θ(t) L z L 3 φ(t),ψ(t)

32 0 Chapter 4. Vector equations of a rigid body 4. The equations of motion of the rolling disk Another problem in classical mechanics of rigid bodies, that can be considered separable, is the problem of motion of a uniform disk with radius a rolling, without slipping, on a horisontal plane ([10]). Figure 4.4: The rolling disk When describing motion of the rolling disk one uses two coordinate systems; a fixed system K 0 = [ˆx,ŷ,ẑ] and a moving with the body K = [ˆ1,ˆ,ˆ3] (see Figure 4.4). The fixed coordinate system ˆx,ŷ,ẑ is such that the ˆx,ŷ axes lie in the horisontal plane and the ẑ axis points vertically upwards. The moving system has its origin at the center of the disk. Here ˆ3 is the symmetry axis, ˆ the horisontal axis defined by ˆ = ˆ3 ˆ1 and ˆ1 is the line from the center to the point of contact with the plane. The vector from the center to the point of contact is then a = aˆ1. The angle between the axes ˆ3 and ẑ is denoted θ and varies between 0 to π. Moreover, the angular velocity of the disk about the ˆ3 axis is called ω, and the angular velocity of K about the ẑ axis is called Ω. The vector equations of motion for the rigid body, (4.1-4.), for this disk read m d r cm dt = F mgẑ (4.0) dl cm dt = a F (4.1) where g is acceleration of gravity. Here the unknown force F, acting at the point of contact A, can be eliminated between (4.0) and (4.1). We get 1 dl cm + d r cm ma dt dt 1 = g(ˆ1 ẑ) (4.) In the case of pure rolling motion the velocity of the contact point A is zero: 0 = v A = drcm dt + ω a, and we can eliminate drcm dt = a ω from the equation (4.). Using known kinematic relations one can calculate expressions of angular velocity about the triad ˆ1,ˆ,ˆ3 and obtain the time rates of the moving axes as well

33 4.. The equations of motion of the rolling disk 1 as that of the radius vector r, in terms of the components ˆ1,ˆ,ˆ3. Furthermore, one can compute the velocity of the center of the disk and the angular momentum itself. When substituting all expressions into (4.) we get the coordinate equations for ˆ1,ˆ,ˆ3 (k + 1) ω + θωsin θ = 0 (4.3) kω sin θ cos θ + (k + 1)ωΩsin θ (k + 1) θ = g cos θ a (4.4) Ω sin θ + θωcos θ + ω = 0 (4.5) where k is defined by the inertia tensor. This system, ( ), consists of the variables θ, θ, ω and Ω and has the total order four. It admits one integral of motion, namely the energy E = 1 mv cm + 1 ω I ω + mgẑ (4.6) which expressed in the variables ω, Ω, θ and θ has the form E = [(k ma + 1)ω + (k + 1) θ + kω sin θ + ga ] sinθ (4.7) Thus we have a problem of three component equations ( ) that admits one integral of motion (4.7). However, we can simplify equations (4.3), (4.5) and reduce them to the Legendre equation (w. r. t. the new variable z = cos θ) which is solvable. Take ω = dω dt = dω dθ dθ dt = dω dθ θ Ω = dω dt = dω dθ dθ dt = dω dθ θ and substitute into (4.3) and (4.5) to get (k + 1) dω + Ωsin θ dθ = 0 (4.8) dω sin θ + Ωcos θ + ω dθ = 0 (4.9) Then we simplify (4.8) and (4.9). We define z = cosθ and substitute into (4.8) and (4.9) and obtain dω dz = 1 k + 1 Ω (4.30) d dz [(1 z )Ω] = ω (4.31) Now we have a problem we can solve. We resolve Ω = (k+1) dω dz from (4.30) and substitute it in (4.31). By doing this we get the Legendre equation d dz [(1 z ) dω dz ] k + 1 ω = 0 (4.3)

34 Chapter 4. Vector equations of a rigid body The Legrendre equation is not elementary, but it is well studied and a solution for ω is given as a convergent power series in z. Furthermore, since (4.30) expresses Ω as a derivative of ω, we obtain Ω as a function of z = cos θ as well. We proceed by eliminating Ω(cos θ) and ω(cos θ) (that now are known functions of θ) from our the energy integral (4.7) and we get (k + 1) θ + k(ω(cos θ)) sin θ + (k + 1) (ω(cos θ)) + g sinθ = h (4.33) a This equation is time dependent and we factorise it with respect to θ and t k+1 1 [ ] h k(ω(cos θ) ) sin θ (k+1) (ω(cos θ)) g a sinθ and we integrate it w. r. t. time t t = k+1 dθ dt = 1 dθ [ ] (4.34) h kω sin θ (k+1) ω g a sinθ We get a complete solution by first resolving θ(t) explicitly. Then, by substituting θ(t) into the expressions for ω(cos θ) and Ω(cos θ) we can determine ω(t) and Ω(t) Separation of variables for the RD equations Let us draw how we have solved the problem of the rolling disk by stepwise separating variables in a diagram: f 1 f Ω(z),ω(z) Ω(cos(θ(t))) ω(cos(θ(t))) E θ(t) Figure 4.5: Stepwise separation of variables for motion of a rolling disk Start reading the diagram (Figure 4.5) in the upper left corner. It says that we have used the first and the third component equation to resolve both ω and Ω in terms of z = cosθ and that we have substituted these expressions into the energy integral E in order to get the solution θ(t). Here we have started out with two component equations and one integral of motion and we have rewritten the two component equations, (4.3) and (4.5), to get the system we have separated stepwise. Let us denote the system consisting of (4.30), (4.31) and (4.7) as 0 = f 1 (ω,ω) 0 = f (ω,ω,ω ) E = f 3 (ω,ω,θ, θ)

35 4.. The equations of motion of the rolling disk 3 where the symbol denotes differentiation with respect to z = cos θ. Initially we have resolved Ω from (4.30) and substituted it into (4.31). By doing this we have obtained an equation (4.3) like g (ω,ω ) = 0 which is known as the Legendre equation. We have solved this second order non autonomous equation and we have got a solution for ω(cos θ) and moreover for Ω(cos θ). By substituting these solutions into E (4.7) we have got E in the form E = f 3 (θ, θ) and we have solved it by separating variables, and we have obtained a solution for θ(t). Then we have substituted the solution θ(t) into the expressions ω(cos θ) and Ω(cos θ) and we have got the solutions ω(t) and Ω(t). In short, we can draw the stepwise solving procedure like E θ(t) Ω(cos θ) ω(cos θ) Ω(t) ω(t) Clearly, the problem of RD, with two component equations (4.3), (4.5), and one integral of motion, energy E (4.7), has similar structure to the three equations of the HST and when we have stepwise separated variables to solve RD it is similar to when we have solved HST. Both in RD and HST we have obtained Ω, ω and φ, ψ as functions of θ. Then, for both RD and HST, we have obtained a separable equation (.1) by substituting our expressions in terms of θ into the energy integral, (4.14) and (4.7). However, in RD it was more complicated to obtain our energy integral in a separable form, because we solved the non elementary second order Legendre equation (4.3) to obtain Ω and ω.

36 4 Chapter 4. Vector equations of a rigid body

37 Chapter 5 Triangular Newton equations In this chapter we will see how triangular systems of cofactor Newton equations can be solved by separating variables ([9]). A system of Newton equations is triangular if i M j = 0 for all pairs i > j, i,j = 1,...,n. It has the form q 1 = M 1 (q 1 ) q = M (q 1,q ). q n = M n (q 1,q,...,q n ) A Newton equation is called quasi-potential if it admits an energy type integral of motion E(q, q) = 1 qt A(q) q + k(q) depending quadratically on velocities and with non degenerate n n matrix deta(q) 0. Besides, if this matrix is of the form A = cof(g), where cof(g) =, with the entries of G(q) given by G det(g) G ij = αq i q j + q i β j + q j β i + γ ij, α,β i,β j,γij = γji R, i,j = 1,...,n. then we say that the triangular Newton equation is cofactor. 5.1 Two dimensional triangular cofactor system Let us see how we can separate variables for two dimensional triangular systems. A two dimensional triangular system has the form q 1 = M 1 (q 1 ) q = M (q 1,q ) (5.1) Generally, for such systems we know that the first equation q 1 = M 1 (q 1 ) always admits the energy integral E = 1 q 1 M 1 (q 1 )dq 1 (5.) Måhl,

38 6 Chapter 5. Triangular Newton equations and a solution for q 1 (t) can be obtained by quadratures. If we can find a second integral of motion depending quadratically on velocities we will be able to solve the entire system (5.1) by separation of variables. Cofactor triangular systems admit such a second integral of motion. Example 5. First let us consider q 1 = 4q 1 4 q = 6q1 + 1q 1 4q + 10 This system is cofactor since it can be written like d ( ) ( ) q1 4q dt = 1 4 q 6q1 = M(q) = (cof(g)) 1 k () (q) + 1q 1 4q + 10 where and ( g = 1 1 q 1 1 q 1 6 q k () (q) = 34q 1 14q + 13q 1 3 q4 1 1q 1 q q 3 1 q 1q + q ) (5.3) Hence it has two integrals of motion E (1) = 1 q 1 M 1 (q 1 )dq 1 = 1 q 1 + 4q 1 + q1 (5.4) and E () = 1 qt (cofg) q + k () = (3 q ) q 1 + (q 1 1) q 1 q + 1 q + k () (q) (5.5) Define separation variables as u 1 = q 1 and the variable u through the equation ( ) u (q) = ρ(q)(a 1,A ) T q1 1 = 1 1 where A = cof(g), and ρ(q) = 1 is an integrating factor. Then and u (q) = q 1 + q 1 + q q = u + u 1 u 1, q = u + u 1 u 1 u 1 When substituting u 1,u into E 1 (5.4) and E (5.5) we get and E (1) = 1 u 1 + 4u 1 + u 1 (5.6) E () = 5 u 1 u 1u + 1 u + 0u 1 14u + 10u 1 8u 1 u 4u 1u + u = 1 u + u 14u u ( 1 u 1 + 4u 1 + u 1) + 5( 1 u 1 + 4u 1 + u 1) = 1 u + u 14u u E 1 + 5E 1 (5.7)

39 5.1. Two dimensional triangular cofactor system 7 Since E 1 is an integral of motion we get a solution u 1 (t) (and thus for q 1 (t)) by quadratures dq 1 t = (E1 4q 1 q1 ) (5.8) Analogously, we get a solution for u (t) (and thus q (t)) du t = (E u + 14u + u E 1 5E 1 ) (5.9) Example 6. Now let us consider q 1 = 3q 1 q = (3q5 1 +5q3 1 +)q (q 1 +1) (5.10) This system is cofactor since it can be written like d ( ) ( ) q1 3q1 dt = = M(q) = (cof(g)) 1 k () (q) q (3q5 1 +5q3 1 +)q (q1 +1) where g is the elliptic matrix and g = ( (q 1 + 1) q 1 q q 1 q (q + 1) k () (q) = 1 q 1 + 1(q3 1detg + q ) = q q q 1 + 1q Thus it has two integrals of motion and ) E (1) = 1 q 1 + q 3 1 (5.11) E () = 1 q 1q + 1 q q 1 q + 1 q q 1 q q 1 q + q q q 1 + 1q (5.1) Now, in order to write it in a separable form, we define q 1 = u 1 and u = which satisfies q q 1 +1 with the integrating factor From this we get ( u (q) = ρ(q)(a 1,A ) T q1 q = ρ(q) q1 + 1 ρ(q) = (q 1 + 1) 3 q 1 = u 1, q 1 = u 1 q = u u u1u u , q = + u u u )

40 8 Chapter 5. Triangular Newton equations Our integrals of motion ( ) expressed through u 1,u take the form and E (1) = 1 u 1 + u 3 1 (5.13) E () = 1 u 1u (u 1 + 1) + 1 u 1 u 1 u 1 (u u u1 u1u 1 + 1)( + u u u )+ + u u u ) u u 1 + 1) 1 u1 u1u ( u1 u1u ( + u u u ) + u 3 1+ u 3 1 u +1(u 1 = u ( 1 u 1 + u ) + 1 u 1 + u u (u 1 + 1) = u (E 1 + 1) + E 1 + u (u 1 + 1) (5.14) In order to solve this system we can obtain a solution for u 1 (t) (thus q 1 ) since (5.13) separates to t = du 1 (E1 u 3 1 ) (5.15) By substituting u 1 (t) into (5.14) we get E in a separable form. We then get a solution for u (t) by quadratures dt (u 1 (t) + 1) = du (E E 1 (E 1 + 1)u ) (5.16) Once we get u (t) we can get a solution for q (t) Separation of variables for two dimensional triangular Newton equations We have solved the two dimensional triangular system in the previous example (example 6) by stepwise separation of variables. Let us draw that in a diagram E 1 E u1 (t) u (t) Figure 5.1: Stepwise separation of variables for two dimensional triangular equations Read, from the diagram, that we have obtained a solution u 1 (t) from the equation E 1 and by substituting u 1 (t) into the expression for E we have obtained a solution u (t). The two dimensional triangular cofactor system (5.10) admit two integrals of motion and therefore we can reduce the order of each equation by one. When we have transformed our reduced equations ( ) using the variable u 1, u we have obtained a stepwise separable system ( ) which we can denote { E 1 = f 1 (u 1, u 1 ) E = f (u 1, u 1,u, u )

41 5.1. Two dimensional triangular cofactor system 9 When we have solved this system we have started by solving the equation for E 1 (5.13), and we have got a solution q 1 (t) = u 1 (t) in (5.15). Then we have substituted u 1 (t) into E (5.14) and we have obtained E in the form E = f (u 1 (t),u, u ) This equation is separable (.1) and quadratures have given us u (t) implicitly in (5.16). Our first example (example 5), however, is different. Here one can see that, when we have transformed our reduced system ( ), we have got a completely separated system of equations for u 1 and u ( ). That means we have got a decoupled system where all equations depend on one variable each. We can symbolically represent this complete separability in a diagram such as E 1 E u1 (t) u (t) Figure 5.: Complete separation of example 5 Figure 5. illustrates that we, from E 1, have got u 1 (t) and, from E, u (t) independently.

42 30 Chapter 5. Triangular Newton equations

43 Chapter 6 The Hénon-Heiles Hamiltonian with cubic potential In this chapter we will study how the Kaup-Kuperschmidt integrable case of the Hénon-Heiles Hamiltonian is solved by separating variables ([7], [8]). The Hénon-Heiles Hamiltonian H : R 3 R with cubic potential is defined by H(x,y,p x,p y ) = 1 (p x + p y) + 1 Ax + 1 By + αx y 1 3 βy3 (6.1) where A, B, α, β are constant parameters. The corresponding Hamiltonian system is ẋ = H p x, ẏ = H p y, ṗ x = H x, ṗ y = H y If we calculate the derivatives ẋ = H p x = p x ẏ = H p y = p y and eliminate p x and p y we get ẍ = Ax αxy ÿ = By αx + βy i.e. these equations have the same Newtonian form as the equations we have studied previously in this thesis. Måhl,

44 3 Chapter 6. The Hénon-Heiles Hamiltonian with cubic potential 6.1 The Kaup-Kuperschmidt case of the Hénon- Heiles Hamiltonian with cubic potential The Hénon-Heiles system is well studied and there are only three cases where it is known to be integrable. By integrable we mean that, for certain values of the parameters A, B, α and β, the system admits a second integral of motion besides the Hamiltonian. One of these integrable cases is known as the Kaup- Kuperschmidt (K-K) case. Then the parameters are B = 16A β = 16α and the second integral of motion takes the form I = x 4 ( (r + A + αy) 4αr(p y ry) 16αy(A + αy) α x ) (6.) where r = px x. When we want to solve the K-K case of the H-H Hamiltonian with cubic potential we start out with the two integral equations (6.1-6.) and we define the separating variables u = P 0 (x,y,p x,p y ) Q 0 (x,y,p x,p y ), u = P 1 (x,y,p x,p y ) Q 1 (x,y,p x,p y ) v = P 0 (x,y,p x,p y ) + Q 0 (x,y,p x,p y ), v = P 1 (x,y,p x,p y ) + Q 1 (x,y,p x,p y ) where Q 0 (x,y,p x,p y ) = x I, P 0 (x,y,p x,p y ) = r + A + αy, Q 1 (x,y,p x,p y ) = rx I P 1 (x,y,p x,p y ) = αp y r(r + A + 6αy) We express x, y, p x and p y in new variables u, v, u and v as x = I u v p x u v = x (u v) [ ] u + v αy = + p x x + A u + v αp y = + r [ r + A + 6αy ] and when we rewrite H (6.1) and I (6.) in new coordinates we get two equations H = I = 1 [ v 16α + u + (v 3 + u 3 ) + 8A(v + u ) ] (6.3) [ 1 [ v 8α u + (v 3 u 3 ) + 8A(v u ) ] ] (6.4) If we rearrange these equations, ( ), they decouple u = u (u + 4A) + 4α (H I) (6.5) v = v (v + 4A) + 4α (H + I) (6.6)

45 6.1. The Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian with cubic potential 33 Therefore the system ( ) is completely separable. We get a complete solution straight away since both equations can be factorised like (.1) directly and integrated w. r. t. time simultaneously. The implicit solutions are du t = u (u + 4A) + 4α (H (6.7) I) dv t = v (v + 4A) + 4α (H + (6.8) I) and by inverting these expressions; (6.7) and (6.8), we get u(t) and v(t) explicitly Separation of variables for the Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian We have solved the K-K case of the H-H Hamiltonian by complete separation of variables. We can illustrate this by a diagram f 3 f 4 u(t) v(t) Figure 6.1: Complete separation of the K-K case of the H-H Hamiltonian with cubic potential The K-K case of the H-H Hamiltonian with cubic potential admits two integrals of motion which constitute a set of two first order equations for x(t) and y(t). When we have transformed this first order system into a new set of coordinates we have obtained (6.1-6.) in the form { H = f 1 (u, u,v, v) I = f (u, u,v, v) We have calculated H I and H + I and obtained the uncoupled equations { H I = f 3 (u, u) H + I = f 4 (v, v) that separate directly. The separation of the K-K case of the H-H Hamiltonian has essentially new features. This transformation is more complicated than the transformations we have used earlier in this thesis since the separation variables involve both the positions x and y and velocities p x = ẋ, p y = ẏ.

46 34 Chapter 6. The Hénon-Heiles Hamiltonian with cubic potential

47 Chapter 7 Direct separability Separation of variables, in almost all examples considered in this thesis, has been based on the use of integrals of motion that reduced our system of second order Newton equations to a (coupled) system of first order ODE s. Then a suitable change of variables have decoupled the equations to such an extent that we have been able to integrate them in stepwise manner. In this chapter we give an example of two second order equations that can be separated as second order equations without using integrals of motion. 7.1 An example of second order Example 8. Consider ẍ = 1 (10ẋ 4ey ẏ 14x + 10e y ) (7.1) ÿ = 1 e y ( 4ẋ + 10ey ẏ + 10x 14e y e y ẏ ) (7.) This is a coupled system of two second order ODE. Define new variables u = x + e y v = x e y and transform (7.1) and (7.) by substituting x, y, ẋ, ẏ, ẍ and ÿ as x = u + v ( ) u v, y = ln ẋ = u + v u v, ẏ = u v Then ẍ = ü + v, ÿ = ü v u v ( ) u v u v ü + v 3 u 7 v + u + 1v = 0 (7.3) ü v 3 u + 7 v + u 1v = 0 (7.4) Måhl,

48 36 Chapter 7. Direct separability Now we rewrite these equations ( ) and find that they decouple ü 3 u + u = 0 (7.5) v 7 v + 1v = 0 (7.6) We get a system of two linear second order ODE s with constant coefficients which has the solutions u(t) = C 1 e t + C e t v(t) = C 3 e 3t + C 4 e 4t so that x(t) = C 1e t + C e t + C 3 e 3t + C 4 e 4t ( C1 e t + C e t C 3 e 3t C 4 e 4t ) y(t) = ln Separation of variables for this example We have solved a second order system (7.1-7.) by direct separation of variables. Let us draw a diagram f 3 f 4 u(t) v(t) Figure 7.1: Direct separation Figure 7.1 illustrates that each equation has been solved independently of the other. Then, when we have solved the second order system (example 8) we have transformed two second order equations (7.1-7.) directly into new coordinates for which we have got a transformed second order system ( ) in the form { f 1 (u, u,ü,v, v, v) = 0 f (u, u,ü,v, v, v) = 0 We have calculated the functions f 3 = 1 (f 1 + f ) and f 4 = 1 (f 1 f ) and we have obtained a decoupled system ( ) like { f 3 (u, u,ü) = 0 f 4 (v, v, v) = 0 where each equation have been solved independently. Remark. This directly separable second order system also possesses two time independent integrals of motion depending on u, u, v and v. In order to find them we resolve the linear system u(t) = C 1 e t + C e t u(t) = C 1 e t + C e t

49 7.1. An example of second order 37 with respect to C 1 e t and C e t : and we eliminate time. Thus C 1 e t = u u C e t = u u u u (u u) = C C 1 = A Analogously, v(t) = C 3 e 3t + C 4 e 4t and v(t) = 3C 3 e 3t + 4C 4 e 4t gives ( v 3v) 3 (4v v) 4 = C3 4 C 4 3 = B where A and B are new constants. These integrals of motion, how ever, are complicated and it is more difficult to calculate solutions using them than solving our directly separated system of linear equations ( ).

50 38 Chapter 7. Direct separability

51 Chapter 8 Conclusion and discussion The concept of separation of variables for a simple ODE of first order is well defined (.1). Certain systems of second order ODE s, describing mechanical systems, are also recognised as separable. Our aim has been to study what this kind of separability means and how it refers to the definition of separability for a simple first order ODE. We have studied several systems known as separable; two and three dimensional potential motion, motion of the heavy symmetric top, motion of the rolling disk, triangular Newton equations and the Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian, in order to describe the procedure of separation and to see common features and differences. Our general conclusion is that the process of separation usually uses integrals of motion and a subsequent transformation into new variables. The integrals of motion reduce the order of the system so that all equations effectively become of first order. The change of variables is needed for decoupling the equations to such extent that the resulting equations can be integrated by quadratures using the elementary method of separation of the equation (.1). This decoupling is usually not complete. Instead, solving is performed sequentially, that is certain equations have to be solved first and then the remaining equations can be integrated in a specific order. This is the case of the two and three dimensional potential motion, motion of the HST and of triangular systems of Newton equations in general. On the other hand, there are certain systems of equations that admit to be completely separated (decoupled) as in the case of the triangular system (example 5) and the K-K case of the H-H Hamiltonian system. These may be considered as special cases of stepwise separation. The transformation to separation variables has usually been pointwise (not involving velocities) so that the new position variables depend only on the old position variables. An interesting exception has been the K-K case of the H- H Hamiltonian system where the transformation involved velocities. The H-H system also differed from the previous ones since it had an integral depending quartically on velocities while all the other systems; potential motion, HST, RD and triangular Newton equations, have had integrals depending quadratically on velocities at most and have been separated through pointwise separation only. It probably reflects a general feature that pointwise separable systems usually have integrals of motion which depend quadratically on velocities. However, Måhl,