Examensarbete Separation of Variables for Ordinary Differential

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Separation of variables for ordinary diﬀerential equations Anna M˚ahl

LiTH - MAT - EX - - 06 / 01 - - SE

Separation of variables for ordinary diﬀerential equations

Department of Applied Mathematics, Link¨opings Universitet

Anna M˚ahl

LiTH - MAT - EX - - 06 / 01 - - SE

Examensarbete: 20 p

Level: D

Supervisor: Stefan Rauch, Department of Applied Mathematics, Link¨opings Universitet

Examiner: Stefan Rauch, Department of Applied Mathematics, Link¨opings Universitet

Link¨oping: February 2006

Datum Avdelning, Institution Date Division, Department

Matematiska Institutionen February 2006 581 83 LINKOPING¨ SWEDEN

Spr˚ak Rapporttyp ISBN Language Report category ISRN Svenska/Swedish Licentiatavhandling LiTH - MAT - EX - - 06 / 01 - - SE x Engelska/English x Examensarbete C-uppsats Serietitel och serienummer ISSN D-uppsats Title of series, numbering 0348-2960 Ovrig¨ rapport

URL f¨or elektronisk version http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva- 5620

Titel Separation of variables for ordinary diﬀerential equations Title

F¨orfattare Anna M˚ahl Author

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. vi 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.

M˚ahl, 2006. vii viii 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 M˚ahl. 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.

M˚ahl, 2006. ix x Contents

1 Introduction 1

2 Preliminaries 3 2.1 Separation of variables for one dimensional potential motion . . . 4

3 Potential Newton Equations 7 3.1 Two dimensional potential Newton equations ...... 7 3.1.1 Separation of variables for two dimensional potential motion 9 3.2 Three dimensional potential motion ...... 10 3.2.1 Separation of variables for three dimensional potential motion...... 12

4 Vector equations of a rigid body 15 4.1 Motion of a heavy symmetric top ...... 15 4.1.1 Separation of variables for the HST equations ...... 19 4.2 The equations of motion of the rolling disk ...... 20 4.2.1 Separation of variables for the RD equations ...... 22

5 Triangular Newton equations 25 5.1 Two dimensional triangular cofactor system ...... 25 5.1.1 Separation of variables for two dimensional triangular New- tonequations...... 28

6 The Hénon-Heiles Hamiltonian with cubic potential 31 6.1 The Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian withcubicpotential ...... 32 6.1.1 Separation of variables for the Kaup-Kuperschmidt case of the Hénon-Heiles Hamiltonian ...... 33

7 Direct separability 35 7.1 An example of second order ...... 35 7.1.1 Separation of variables for this example ...... 36

8 Conclusion and discussion 39

M˚ahl, 2006. xi xii Contents Chapter 1

Introduction

dy An ordinary differential equation (ODE) is an algebraic equation f(x,y, 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 = u (1.1) ∂t xx for an initial boundary value problem on the interval 0

∂u ∂t = uxx, 0 < t, 0

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

2 ∂tT = k T − 2 ∂xxX = k X ( − These two ordinary diﬀerential equations have solutions

−k2t Tk(t) = Ae (Xk(x) = B cos kx + C sin kx

M˚ahl, 2006. 1 2 Chapter 1. Introduction

and due to linearity of (1.1) any formal linear combination u(x,t) = k ckXk(x)Tt(t) is also a solution. This solution can be further speciﬁed according to the bound- nπ P ary conditions u(0,t) = u(L, t) = 0. If we put kn = L , where n = 1, 2,..., the complete solution reads

∞ 2 −knt u(x,t) = cne sin knx (1.2) n=1 X 2 L where the coeﬃcients cn are given by the integrals cn = L 0 f(x) sin knxdx ([3],[6]). R Example 2. 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

∂H(x,y) ∂H(x,y) x˙ = , y˙ = , x = (x1,...,xn), y = (y1,...,yn) (1.3) ∂y − ∂x where H(x,y) is a Hamiltonian, is to find a function W (x, α), given by the ∂W (x,α) ∂W (x,α) canonical transformation y = ∂x , β = ∂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

∂W (x, α) H x, = α (1.4) ∂x 1 which is known as the Hamilton-Jacobi equation. In order to find it, one makes n k k an additive ansatz W (x, α) = k=1 Wk(x , α), where the n functions Wk(x , α) each depend on a single variable xk. This ansatz works well for the Hamiltoni- P 1 ii 2 ans of Stäckel class with H = G + V = 2 i g (x)yi . It simplifies the problem; instead of integrating the nontrivial Hamilton-Jacobi equation the problem re- P k duces to integrating n uncoupled first order ODE’s for functions Wk(x , 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 2 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. Chapter 2

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) (2.1) 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 dy G(y) = g(y) dx = (h(x)) dx = H(x) + C dx Z Z 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 x = µx 1 (2.2) dt − k where µ and k are positive constants. It is separable since it can be factorised into 1 dx = 1 µx 1 x dt − k for x = k and x = 0. By6 integrating6 both sides, w. r. t. time, we obtain an implicit solution

1 1 1 1 x t + C = + x dx = ln x ln 1 µ x k 1 k ! µ | | − | − k | Z − M˚ahl, 2006. 3 4 Chapter 2. Preliminaries and an explicit solution x(t) reads keµt x = C + eµt

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 = 06 also satisfy6 the equation (2.2).

Example 4. The homogeneous equation. Take dy y = f( ) (2.3) dx x This equation can not be written directly in the separable form (2.1), but by y dy dz y deﬁning a new variable z = x we get dx = x dx + z and f( x ) = f(z). By substituting these expressions into the equation (2.3) we get an equation dz x + z = f(z) dx which is separable since it can be written as 1 dz 1 = f(z) z dx x − 2.1 Separation of variables for one dimensional potential motion

Let us consider the second order ODE of the form dV (x) mx¨ = (2.4) − dx 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 withx ˙. Then d mx˙ 2 dV (x) mx˙ 2 = and + V (x) = E = constant dt 2 − dt 2 We can solve this ﬁrst order ODE for x(t) by separating variables since the equation of energy mx˙ 2 + V (x) = E 2 can be written 1 dx = 1 2 (E V (x)) dt m − which gives us q m dx t = =: G(x) + C 2 (E V (x)) r Z − Notice that in solving (2.4)p 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). 2.1. Separation of variables for one dimensional potential motion 5

Deﬁnition 2 An integral of motion for a second order diﬀerential equation

x¨ = M(x) is a function of position and velocity K(x, x˙) such that it takes constant value on solutions x(t). That is

K(x(t), x˙(t)) = constant

As a consequence we have dK ∂K ∂K ∂K ∂K K˙ (x, x˙) = = x˙ + x¨ = x˙ + M(x) = 0 dt ∂x ∂x˙ ∂x ∂x˙ for all (x, x˙). So, since the second order Newton equation of one dimensional potential motion (2.4) admits an integral of motion (the energy integral) that constitutes a ﬁrst order separable equation, (the equation which can be factorised like (2.1)) we can consider it to be separable. 6 Chapter 2. Preliminaries 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 R3 (3.1) − ∈ where the function V (r) is called a potential. By multiplying both sides with r˙ we see that (3.1) always admits the energy integral of motion

1 2 E = mr˙ + V (r) (3.2) 2

For a central potential V (r) = V (r) with r = r , the force F = grad V(r) = | | − dV (r) r is parallel to r. Then the equation (3.1) admits an additional vector in- − dr r tegral of motion, namely the angular momentum L = r mr˙. By diﬀerentiating we ﬁnd × dV (r) r L˙ = r˙ mr˙ + r m¨r = r m¨r = r F = r = 0 × × × × dr r × 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) − T where r = (x1,x2) , that takes the coordinate form

∂V (r) ′ x1 mx¨1 = = V (r) (3.4) − ∂x1 − r ∂V (r) ′ x2 mx¨2 = = V (r) (3.5) − ∂x2 − r

M˚ahl, 2006. 7 8 Chapter 3. Potential Newton Equations

2 mr˙ This equation, (3.3), admits the energy integral E = 2 + V (r). Moreover, by multiplying the first coordinate equation (3.4) by x2, and the second one (3.5) by x1 and subtracting we find d (mx¨ x mx¨ x ) = 0 dt 1 2 − 2 1 which we recognize as the derivative of the third component of the angular momentum L = m(x x˙ x x˙ ). The energy and the angular momentum 3 1 2 − 2 1 integrals of motion reduce the second order system (¨x1, x¨2) into a first order system

L3 = m(x1x˙ 2 x2x˙ 1) (3.6) 2 − mr˙ E = + V (r) (3.7) 2 where L3 and E denote constant values of these integrals of motion. The system (3.6 - 3.7) is not separable yet, but can be solved by separation of variables in the polar coordinates r and θ. By substituting x1 = r cos θ, x˙ 1 =r ˙ cos θ rθ˙ sin θ, x2 = r sin θ andx ˙ 2 =r ˙ sin θ + rθ˙ cos θ into (3.6 - 3.7) we get −

2 L3 = mr θ˙ (3.8) m E = r˙2 + r2θ˙2 + V (r) (3.9) 2 ˙ L3 When we resolve θ = mr2 from (3.8) and substitute in (3.9) we get the separable equation of one dimensional motion w. r. t. variable r (2.4)

m L 2 E = r˙2 + r2 3 + V (r) (3.10) 2 mr2 ! We factorise it 1 dr 2 = 1 (3.11) 2 L3 dt (E V (r)) 2 2 m − − m r q and a solution is obtained by integrating both sides w. r. t. time t dr t = 2 (3.12) 2 L3 Z (E V (r)) 2 2 m − − m r q Finally to get a complete solution in terms of r and θ, we have to invert (3.12) to get r(t) explicitly. Then by substituting r(t) into (3.8) we get a solution for θ(t) like

L θ = 3 dt (3.13) mr(t)2 Z 3.1. Two dimensional potential Newton equations 9

3.1.1 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

- - L3 θ˙(r) L3 θ(t) ? 6 E - r(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 L3 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 (3.4 - 3.5) into a new system of the two first order integral equations (3.6 - 3.7). Subsequently, we have transformed this first order system into polar coordinates and obtained the new coupled system of equations (3.8 - 3.9) which we have solved by successively eliminating and stepwise separating variables. Let us symbolically denote this system (3.8 - 3.9) as

L3 = f1(r, θ˙) ˙ (E = f2(r, r,˙ θ)

We have, as described in the diagram (Figure 3.1), resolved θ˙ = g1(r) from L3 (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 = f2(r, r,˙ g1(r)) = f3(r, r˙) This equation depends on r only. It is separable and, in (3.11), we have written it like (2.1). By integrating this equation (3.11) we have got an implicit solution 10 Chapter 3. Potential Newton Equations for r (3.12). 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 (2.1), 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 (2.1). In short, we have obtained a complete solution by substituting in the following manner: - - - E r(t) L3 θ(t)

3.2 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 ,x ,x )T R3 (3.14) − 1 2 3 ∈ and we know that E and L are integrals of motion. Then, we conclude that the 2 third component L3 and L are also integrals of motion ˙ ˙ L˙ 3 = (Leˆ3)˙= Leˆ3 + Leˆ˙3 = Leˆ3 = 0 2 L˙ = (L L)˙= L L˙ + L˙ L = 2L L˙ = 0 · · · · As a result, for three dimensional potential motion, the system m¨r = grad V(r), 2 − has three integrals of motion: E, L3 and L . In Cartesian coordinates, these integrals of motion read L = m(x x˙ x x˙ ) (3.15) 3 1 2 − 2 1 2 2 2 2 L = m(x2x˙ 3 x3x˙ 2) + m(x3x˙ 1 x1x˙ 3) + m(x1x˙ 2 x2x˙ 1) (3.16) 2 − − − mr˙ E = + V (r) (3.17) 2 We can solve this system of equations (3.15 - 3.17) by separation of variables if we introduce the spherical coordinates r, ϕ and θ. We substitute x1 = r sin θ cos ϕ, x2 = r sin θ sin ϕ, x3 = r cos θ, andx ˙ 1 =r ˙ sin θ cos ϕ + r(θ˙ cos θ cos ϕ ϕ˙ sin θ sin ϕ),x ˙ 2 =r ˙ sin θ sin ϕ + r(θ˙ cos θ sin ϕ +ϕ ˙ sin θ cos ϕ), x˙ =r ˙ cos θ −rθ˙ sin θ which gives us a new system 3 −

2 2 L3 = mr ϕ˙ sin θ (3.18) L2 = m2r4(θ˙2 +ϕ ˙ 2 sin2 θ) (3.19) m E = r˙2 + r2θ˙2 + r2ϕ˙ 2 sin2 θ + V (r) (3.20) 2 We resolveϕ ˙ from (3.18) L ϕ˙ = 3 (3.21) mr2 sin2 θ 3.2. Three dimensional potential motion 11 and substituteϕ ˙ into (3.19) to get

L2 L2 θ˙2 = 3 (3.22) m2r4 − m2r4 sin2 θ By substituting (3.21) and (3.22) into the energy equation (3.20) we get

m L2 L2 L 2 E = r˙2 + r2 3 + r2 3 sin2 θ + V (r) 2 m2r4 − m2r4 sin2 θ mr2 sin2 θ ! which “miraculously” simpliﬁes to a one dimensional energy integral (2.4) since the dependence on θ cancels out and leaves us with the expression

m L2 E = r˙2 + + V (r) (3.23) 2 m2r2 This equation is separable and we factorise it with respect to r and t 1 dr = 1 (3.24) 2 L2 dt (E V (r)) 2 2 m − − m r q A solution is then obtained by integrating both sides dr t = (3.25) 2 L2 Z (E V (r)) 2 2 m − − m r q To get an explicit solution r(t), ϕ(t) and θ(t) we invert (3.25) to get r(t). Then we substitute r(t) into (3.22) to ﬁnd a solution for θ(t)

dθ dt = (3.26) 2 2 2 L3 mr(t) Z L 2 Z − sin θ q Finally we substitute r(t) and θ(t) into (3.21) and get ϕ(t)

L dt dϕ = 3 (3.27) mr(t)2 sin2 θ(t) Z Z 12 Chapter 3. Potential Newton Equations

3.2.1 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:

- - L3 ϕ˙(r, θ) L3 ϕ(t) ? 6 L2 - θ˙2(r, θ) L2 - θ(t) ?? 6 E - r(t)

Figure 3.2: Stepwise separation of three dimensional potential motion

Start reading the diagram (Figure 3.2) 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 2 ϕ˙ from L3 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 (3.15 - 3.17). Then having transformed this ﬁrst order system into the spherical coordinates, we have obtained the system (3.18 - 3.20) that we have solved by stepwise separation of variables. Let us, as in the case of two dimensional potential motion, denote this system (3.18 - 3.20) as

L3 = f1(r, ϕ,θ˙ ) 2 ˙ L = f2(r, ϕ,θ,˙ θ) E = f3(r, r,˙ ϕ,θ,˙ θ˙)

As can be read in the diagram (Figure 3.2), we have started by resolvingϕ ˙ in terms of r and θ from L3 (3.18) like

ϕ˙ = g1(r, θ)

Then, by resolving θ˙2 from L2 (3.19) we have got

2 θ˙ = g2(r, θ)

By substituting the expressions ofϕ ˙ and θ˙2 into E (3.20) we have obtained an equation (3.23) in the form

E = f3(r, r,˙ g1(r, θ), θ, g2(r, θ)) = f4(r, r˙)

This equation is separable. In (3.24), we have factorised it as (2.1) and we have integrated it to get an implicit solution for r (3.25). We have calculated the 3.2. Three dimensional potential motion 13 explicit solution r(t). Then we have substituted this solution into the expression 2 θ˙ = g2(r(t),θ) (3.22) and we have obtained an equation which is separable with respect to t and θ (3.26). 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 forϕ ˙ = g1(r(t),θ(t)) (3.21) and we have obtained a third separable equation (3.27) thus the solution ϕ(t). In short, let us illustrate this stepwise solving procedure as follows:

- - 2 - - - 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 (3.23) that separation of r and t becomes possible. 14 Chapter 3. Potential Newton Equations 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 (4.2) 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

M˚ahl, 2006. 15 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 K0 = [ˆx, y,ˆ zˆ] and a body-fixed coordinate system K = [1ˆ, 2ˆ, 3]ˆ 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 3-axisˆ 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 - 4.2), in case of the HST read

d(ml3)ˆ˙ = mgzˆ + F r (4.3) dt − dL = l3ˆ ( mgzˆ) (4.4) dt × − These equations are complemented by the kinematic equation describing the ˙ motion of 3ˆ w. r. t. the inertial coordinate system K0

3(ˆ˙ t) = ω 3ˆ × where ω is the instantaneous angular velocity of the HST. By using the relations T L = (IxΩx,IyΩy,IzΩz) where Ix,Iy,Iz are moments of inertia w. r. t. the ω ˆ L ˆ ﬁxed point, and that 3 = Ix 3 for an axially symmetric rigid body, we get the vector equations of× motion for× the HST L 3ˆ˙ = 3ˆ (4.5) Ix × dL = mgl3ˆ zˆ (4.6) dt − × This system is a closed system of six ODE’s, for six scalar unknowns; i.e. the components of 3ˆ and L. The vector equations of the HST (4.5 - 4.6) admit three integrals of motion: the total energy E, the projection of the angular momentum Lz onto the vertical axis and the projection of angular momentum L3 onto the symmetry axis. We can show that Lz and L3 are integrals of motion by diﬀerentiating them d (L zˆ) = ( mgl3ˆ zˆ) zˆ = 0 (4.7) dt · − × · d L (L 3)=(ˆ mgl3ˆ zˆ) 3ˆ + L ( 3)ˆ = 0 (4.8) dt · − × · · Ix × 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 d 1 1 1 (E) = ( ω L + mgl3ˆ zˆ) = ω˙ L + ω L˙ + mgl3ˆ˙ zˆ dt dt 2 · · 2 · 2 · · = ω L˙ + mgl(ω 3)ˆ zˆ = ω ( mgl3ˆ zˆ) + mgl(ω 3)ˆ zˆ = 0 · × · · − × × · 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

Lz = Lzˆ (4.9)

L3 = L3ˆ (4.10) 1 E = ω L + mgl3ˆ zˆ (4.11) 2 · ·

To obtain a system which we can solve by separating variables we express (4.9 - 4.11) in the coordinates θ, φ and ψ, known as Euler angles

Figure 4.2: The Euler Angles

Figure 4.2 gives us the Euler coordinates

1ˆ = (cos ψ cos φ cos θ sin φ sin ψ)ˆx + (cos ψ sin φ + cos θ cos φ sin ψ)ˆy + (sin θ sin ψ)ˆz − 2ˆ = (sin ψ cos φ + cos θ sin φ cos ψ)ˆx (sin ψ sin φ cos θ cos φ cos ψ)ˆy + (sin θ cos ψ)ˆz − − − 3ˆ = (sin θ sin φ)ˆx (sin θ cos φ)ˆy + (cos θ)ˆz −

We express the angular velocity, ω = ωxxˆ + ωyyˆ + ωzzˆ, by

ωx = ψ˙ sin θ sin φ + θ˙ cos φ

ωy = ψ˙ sin θ cos φ + θ˙ sin φ − ωz = θ˙ + ψ˙ cos θ and angular momentum, L = I ω, by ·

Lx = Ix(θ˙ cos φ φ˙ sin θ cos θ sin φ) + Iz sin θ sin φ(ψ˙ + φ˙ cos θ) − Ly = Ix(θ˙ sin φ + φ˙ sin θ cos θ cos φ) Iz sin θ cos φ(ψ˙ + φ˙ cos θ) − 2 Lz = Ixφ˙ sin θ + Iz cos θ(ψ˙ + φ˙ cos θ) 18 Chapter 4. Vector equations of a rigid body

We get our system (4.9 - 4.11) expressed in new coordinates

2 Lz = Ixφ˙ sin θ + Iz cos θ(ψ˙ + φ˙ cos θ) (4.12)

L3 = Iz ψ˙ + φ˙ cos θ (4.13) I I 2 E = x θ˙2 + φ˙2 sin2 θ + z ψ˙ + φ˙ cos θ + mgl cos θ (4.14) 2 2 In order to solve this system by separating variables, we start with equations (4.12) and (4.13) and express φ˙ and ψ˙ in terms of θ

2 L Lz cos θ L3 cos θ ψ˙ = 3 − (4.15) I 2 z − Ix sin θ L θ L 2 θ ˙ z cos 3 cos φ = − 2 (4.16) Ix sin θ Substituting (4.15) and (4.16) into (4.14) we get a one dimensional energy integral equation (2.4)

˙2 2 2 ˙2 Ixθ (Lz L3 cos θ) L3 Ixθ E = + − 2 + + mgl cos θ = + U(θ) (4.17) 2 2Ix sin θ 2Iz 2 This equation can be factorised, and separated, as 1 dθ = 1 (4.18) 2(E−U(θ)) dt Ix q where 2 2 (Lz L3 cos θ) L3 U(θ) = − 2 + + mgl cos θ 2Ix sin θ 2Iz When we integrate both sides we get 1 t = dθ (4.19) 2(E−U(θ)) Z Ix q 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). 4.1. Motion of a heavy symmetric top 19

4.1.1 Separation of variables for the HST equations We have solved the problem of the HST by separating variables stepwise. Let us draw a diagram:

Lz φ˙(θ) - φ˙(θ), ψ˙(θ) - φ(t), ψ(t) L3 ψ˙(θ) 6 ? 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 Lz and the equation for L3 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 (4.12 - 4.14) as

Lz = f1(φ,˙ ψ,θ˙ ) ˙ ˙ L3 = f2(φ, ψ,θ) E = f3(φ,˙ ψ,θ,˙ θ˙)

 First we have, as described in Figure 4.3, resolved φ˙ and ψ˙ from Lz (4.12) and L3 (4.13) and we have got

φ˙ = g1(θ)

ψ˙ = g2(θ) Then we have eliminated φ˙ and ψ˙ from (4.14) and got an equation in the form

E = f3(θ, θ˙)

This equation we have solved by separating variables since we have factorised it like (2.1) 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 - Lz E - θ t - φ t , ψ t ( ) - ( ) ( ) L3 20 Chapter 4. Vector equations of a rigid body

4.2 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 K0 = [ˆx, y,ˆ zˆ] and a moving with the body K = [1ˆ, 2ˆ, 3]ˆ (see Figure 4.4). The fixed coordinate systemx, ˆ y,ˆ zˆ is such that thex, ˆ yˆ axes lie in the horisontal plane and thez ˆ axis points vertically upwards. The moving system has its origin at the center of the disk. Here 3ˆ is the symmetry axis, 2ˆ the horisontal axis defined by 2ˆ = 3ˆ 1ˆ and 1ˆ 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 = a1.ˆ The angle between the axes 3ˆ andz ˆ is denoted θ and varies between 0 to π. Moreover, the angular velocity of the disk about the 3ˆ axis is called ω, and the angular velocity of K about thez ˆ axis is called Ω. The vector equations of motion for the rigid body, (4.1 - 4.2), for this disk read d2r m cm = F mgzˆ (4.20) dt2 − dL cm = a F (4.21) dt × where g is acceleration of gravity. Here the unknown force F , acting at the point of contact A, can be eliminated between (4.20) and (4.21). We get

1 dL d2r cm + cm 1 = g(1ˆ zˆ) (4.22) ma dt dt2 × × In the case of pure rolling motion the velocity of the contact point A is zero: drcm drcm 0 = vA = dt + ω a, and we can eliminate dt = a ω from the equation (4.22). × × Using known kinematic relations one can calculate expressions of angular velocity about the triad 1ˆ, 2ˆ, 3ˆ and obtain the time rates of the moving axes as well 4.2. The equations of motion of the rolling disk 21 as that of the radius vector r, in terms of the components 1ˆ, 2ˆ, 3.ˆ Furthermore, one can compute the velocity of the center of the disk and the angular momentum itself. When substituting all expressions into (4.22) we get the coordinate equations for 1ˆ, 2ˆ, 3ˆ

(2k + 1)ω ˙ + θ˙Ω sin θ = 0 (4.23) g kΩ2 sin θ cos θ + (2k + 1)ωΩ sin θ (k + 1)θ¨ = cos θ (4.24) − a Ω˙ sin θ + 2θ˙Ω cos θ + 2ω = 0 (4.25) where k is deﬁned by the inertia tensor. This system, (4.23 - 4.25), consists of the variables θ, θ˙, ω and Ω and has the total order four. It admits one integral of motion, namely the energy 1 1 E = mv2 + ω I ω + mgzˆ (4.26) 2 cm 2 · · which expressed in the variables ω, Ω, θ and θ˙ has the form

ma2 2g E = (2k + 1)ω + (k + 1)θ˙2 + kΩ2 sin2 θ + sin θ (4.27) 2 a Thus we have a problem of three component equations (4.23 - 4.25) that admits one integral of motion (4.27). However, we can simplify equations (4.23), (4.25) and reduce them to the Legendre equation (w. r. t. the new variable z = cos θ) which is solvable. Take dω dω dθ dω ω˙ = = = θ˙ dt dθ dt dθ dΩ dΩ dθ dΩ Ω˙ = = = θ˙ dt dθ dt dθ and substitute into (4.23) and (4.25) to get dω (2k + 1) + Ω sin θ = 0 (4.28) dθ dΩ sin θ + 2Ω cos θ + 2ω = 0 (4.29) dθ Then we simplify (4.28) and (4.29). We deﬁne z = cosθ and substitute into (4.28) and (4.29) and obtain dω 1 = Ω (4.30) dz 2k + 1 d [(1 z2)Ω] = 2ω (4.31) dz −

dω Now we have a problem we can solve. We resolve Ω = (2k+1) dz from (4.30) and substitute it in (4.31). By doing this we get the Legendre equation d dω 2 [(1 z2) ] ω = 0 (4.32) dz − dz − 2k + 1 22 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 (4.27) and we get

(k + 1) k(Ω(cos θ))2 (2k + 1) g θ˙2 + sin2 θ + (ω(cos θ))2 + sin θ = h (4.33) 2 2 2 a This equation is time dependent and we factorise it with respect to θ and t 1 dθ = 1 2 dt 2 h k(Ω(cos θ) ) sin2 θ (2k+1) (ω(cos θ))2 g sin θ k+1 − 2 − 2 − a r h i and we integrate it w. r. t. time t dθ t = (4.34) 2 Z 2 h kΩ sin2 θ (2k+1) ω2 g sin θ k+1 − 2 − 2 − a r h i 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).

4.2.1 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:

- f1 Ω(cos(θ(t))) - Ω(z), ω(z) - f2 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, (4.23) and (4.25), to get the system we have separated stepwise. Let us denote the system con- sisting of (4.30), (4.31) and (4.27) as

′ 0 = f1(ω , Ω) ′ 0 = f2(ω, Ω, Ω ) E = f3(ω, Ω, θ, θ˙)

 4.2. The equations of motion of the rolling disk 23 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 (4.32) like

′′ g2(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 (4.27) we have got E in the form E = f3(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

- Ω(cos θ) - Ω(t) E - θ(t) - ω(cos θ) - ω(t)

Clearly, the problem of RD, with two component equations (4.23), (4.25), and one integral of motion, energy E (4.27), 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 (2.1) by substituting our expressions in terms of θ into the energy integral, (4.14) and (4.27). 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 (4.32) to obtain Ω and ω. 24 Chapter 4. Vector equations of a rigid body 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 ∂iMj = 0 for all pairs i>j, i, j = 1,...,n. It has the form

q¨1 = M1(q1)

q¨2 = M2(q1, q2) . .

q¨n = Mn(q1, q2,...,qn) A Newton equation is called quasi-potential if it admits an energy type integral of motion 1 E(q, q˙) = q˙T A(q)q ˙ + k(q) 2 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) = G 6 det(G) , with the entries of G(q) given by

Gij = αqiqj + qiβj + qjβi + γij , α,βi, βj ,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¨ = M (q ) 1 1 1 (5.1) q¨2 = M2(q1, q2)

Generally, for such systems we know that the ﬁrst equationq ¨1 = M1(q1) always admits the energy integral 1 E = q˙2 M (q )dq (5.2) 2 1 − 1 1 1 Z M˚ahl, 2006. 25 26 Chapter 5. Triangular Newton equations

and a solution for q1(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¨ = 4q 4 1 1 (5.3) q¨ =− 6q2 +− 12q 4q + 10 2 1 1 − 2 This system is cofactor since it can be written like 2 d q1 4q1 4 −1 (2) = 2 − − = M(q) = (cof(g)) k (q) dt2 q2 6q + 12q1 4q2 + 10 − ∇ 1 − where 1 1 q g = 1 1 q 6 − q − 1 − 2 and 3 k(2)(q) = 34q 14q + 13q2 q4 12q q 2q3 2q2q + 2q2 1 − 2 1 − 2 1 − 1 2 − 1 − 1 2 2 Hence it has two integrals of motion 1 1 E = q˙2 M (q )dq = q˙2 + 4q + 2q2 (5.4) (1) 2 1 − 1 1 1 2 1 1 1 Z and 1 1 E = q˙T (cofg)q ˙ + k(2) = (3 q )q ˙2 + (q 1)q ˙ q˙ + q˙2 + k(2)(q) (5.5) (2) 2 − 2 1 1 − 1 2 2 2

Deﬁne separation variables as u1 = q1 and the variable u2 through the equation T q1 1 u2(q) = ρ(q)(A12, A22) = 1 − ∇ · 1 where A = cof(g), and ρ(q) = 1 is an integrating factor. Then q2 u (q) = q + 1 + q 2 − 1 2 2 and u2 q = u + u 1 , q˙ =u ˙ +u ˙ u u˙ 2 2 1 − 2 2 2 1 − 1 1 When substituting u1,u2 into E1 (5.4) and E2 (5.5) we get 1 E = u˙ 2 + 4u + 2u2 (5.6) (1) 2 1 1 1 and 5 1 E = u˙ 2 u˙ 2u + u˙ 2 + 20u 14u + 10u2 8u u 4u2u + 2u2 (2) 2 1 − 1 2 2 2 1 − 2 1 − 1 2 − 1 2 2 1 1 1 = u˙ 2 + 2u2 14u 2u ( u˙ 2 + 4u + 2u2) + 5( u˙ 2 + 4u + 2u2) 2 2 2 − 2 − 2 2 1 1 1 2 1 1 1 1 = u˙ 2 + 2u2 14u 2u E + 5E (5.7) 2 2 2 − 2 − 2 1 1 5.1. Two dimensional triangular cofactor system 27

Since E1 is an integral of motion we get a solution u1(t) (and thus for q1(t)) by quadratures dq t = 1 (5.8) 2(E 4q 2q2) Z 1 − 1 − 1 Analogously, we get a solution forpu2(t) (and thus q2(t)) du t = 2 (5.9) 2(E 2u2 + 14u + 2u E 5E ) Z 2 − 2 2 2 1 − 1 Example 6. Now let us consider

2 q¨1 = 3q1 5 3 − (3q1 +5q1 +2)q2 (5.10) q¨2 = 2 2 − (q1 +1) This system is cofactor since it can be written like

2 2 d q 3q1 − 1 5− 3 M q g 1 k(2) q 2 = (3q1 +5q1 +2)q2 = ( ) = (cof( )) ( ) dt q2 2 2 − ∇ − (q1 +1) ! where g is the elliptic matrix

(q2 + 1) q q g = 1 1 2 q q (q2 + 1) 1 2 2 and 3 (2) 1 3 2 3 q1 + 1 2 k (q) = 2 (q1detg + q2) = q1 + 2 q2 q1 + 1 q1 + 1 Thus it has two integrals of motion 1 E = q˙2 + q3 (5.11) (1) 2 1 1 and

3 1 2 2 1 2 1 2 2 1 2 3 q1 + 1 2 E(2) = q˙1q2 + q˙1 + q1q˙2 + q˙2 q1q2q˙1q˙2 + q1 + 2 q2 (5.12) 2 2 2 2 − q1 + 1

Now, in order to write it in a separable form, we deﬁne q1 = u1 and u2 = q2 2 which satisﬁes √q1 +1

T q1q2 u2(q) = ρ(q)(A12, A22) = ρ(q) − ∇ q2 + 1 1 with the integrating factor

3 2 − 2 ρ(q) = (q1 + 1) From this we get

q1 = u1, q˙1 =u ˙ 1 2 u1u2u˙ 1 2 q2 = u2 u1 + 1, q˙2 = 2 +u ˙ 2 u1 + 1 √u1+1 p p 28 Chapter 5. Triangular Newton equations

Our integrals of motion (5.11 - 5.12) expressed through u1,u2 take the form

1 2 3 E(1) = 2 u˙ 1 + u1 (5.13) and

1 2 2 2 1 2 2 u1u˙ 1u2 2 E(2) = 2 u˙ 1u2(u1 + 1) + 2 u˙ 1 u1u˙ 1(u2 u1 + 1)( 2 +u ˙ 2 u1 + 1)+ − √u1+1 1 u1u˙ 1u2 2 2 2 1 u1u˙ 1u2 2 2 3 2 ( 2 +u ˙ 2 u1 + 1) u1 + 2 (p 2 +u ˙ 2 u1 + 1)p+ u1+ √u1+1 √u1+1 3 u1+1 2 2 2 (u2 u +p 1) p u1+1 1 2 2 1 2 3 1 2 3 u˙ 2 2 2 = u2( u˙ 1 +pu1 + 1) + u˙ 1 + u1 + (u1 + 1) 2 22 2 2 u˙ 2 2 2 = u2(E1 + 1) + E1 + 2 (u1 + 1) (5.14) In order to solve this system we can obtain a solution for u1(t) (thus q1) since (5.13) separates to du t = 1 (5.15) 2(E u3) Z 1 − 1 By substituting u1(t) into (5.14) wep get E2 in a separable form. We then get a solution for u2(t) by quadratures dt du = 2 (5.16) (u (t)2 + 1) 2(E E (E + 1)u2) Z 1 Z 2 − 1 − 1 2 Once we get u2(t) we can get a solution for q2(t).

5.1.1 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

- E2 u2(t) 6 - E1 u1(t)

Figure 5.1: Stepwise separation of variables for two dimensional triangular equations

Read, from the diagram, that we have obtained a solution u1(t) from the equation E1 and by substituting u1(t) into the expression for E2 we have obtained a solution u2(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 (5.11 - 5.12) using the variable u1, u2 we have obtained a stepwise separable system (5.13 - 5.14) which we can denote

E1 = f1(u1, u˙ 1) (E2 = f2(u1, u˙ 1,u2, u˙ 2) 5.1. Two dimensional triangular cofactor system 29

When we have solved this system we have started by solving the equation for E1 (5.13), and we have got a solution q1(t) = u1(t) in (5.15). Then we have substituted u1(t) into E2 (5.14) and we have obtained E2 in the form

E2 = f2(u1(t),u2, u˙ 2)

This equation is separable (2.1) and quadratures have given us u2(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 (5.4 - 5.5), we have got a completely separated system of equations for u1 and u2 (5.6 - 5.7). 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

- E1 u1(t) - E2 u2(t)

Figure 5.2: Complete separation of example 5

Figure 5.2 illustrates that we, from E1, have got u1(t) and, from E2, u2(t) independently. 30 Chapter 5. Triangular Newton equations Chapter 6

The H´enon-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 : R3 R with cubic potential is defined by →

1 2 2 1 2 1 2 2 1 3 H(x,y,px,py) = (p + p ) + Ax + By + αx y βy (6.1) 2 x y 2 2 − 3 where A, B, α, β are constant parameters. The corresponding Hamiltonian system is ∂H ∂H ∂H ∂H x˙ = , y˙ = , p˙x = , p˙y = ∂px ∂py − ∂x − ∂y

If we calculate the derivatives ∂H x˙ = = px ∂px ∂H y˙ = = py ∂py and eliminate px and py we get

x¨ = Ax 2αxy − − y¨ = By αx2 + βy2 − − i.e. these equations have the same Newtonian form as the equations we have studied previously in this thesis.

M˚ahl, 2006. 31 32 Chapter 6. The H´enon-Heiles Hamiltonian with cubic potential

6.1 The Kaup-Kuperschmidt case of the H´enon- Heiles Hamiltonian with cubic potential

The H´enon-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

4 2 2 2 2 I = x (r + 2A + 2αy) 4αr(py 2ry) 16αy(A + αy) 2α x (6.2) − − − − px where r = 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 - 6.2) and we deﬁne the separating variables u = P (x,y,px,py) Q (x,y,px,py), u˙ = P (x,y,px,py) Q (x,y,px,py) − 0 − 0 − 1 − 1 v = P (x,y,px,py) + Q (x,y,px,py), v˙ = P (x,y,px,py) + Q (x,y,px,py) − 0 0 − 1 1 where

−2 −2 Q (x,y,px,py) = x √I, Q (x,y,px,py) = 2rx √I 0 − 1 2 2 P (x,y,px,py) = r + 2A + 2αy, P (x,y,px,py) = 2αpy 2r(r + 2A + 6αy) 0 1 −

We express x, y, px and py in new variables u, v,u ˙ andv ˙ as

2√I x2 = u v − px u˙ v˙ = − x −2(u v) − u + v p2 2αy = + x + 2A − 2 x2 u˙ +v ˙ 2 2αpy = + 2r r + 2A + 6αy − 2 and when we rewrite H (6.1) and I (6.2) in new coordinates we get two equations 1 H = v˙ 2 +u ˙ 2 + 2(v3 + u3) + 8A(v2 + u2) (6.3) 16α2 1 2 I = v˙ 2 u˙ 2 + 2(v3 u3) + 8A(v2 u2) (6.4) 8α2 − − − If we rearrange these equations, (6.3 - 6.4), they decouple

u˙ 2 = 2u2(u + 4A) + 4α2(2H √I) (6.5) − − v˙ 2 = 2v2(v + 4A) + 4α2(2H + √I) (6.6) − 6.1. The Kaup-Kuperschmidt case of the H´enon-Heiles Hamiltonian with cubic potential 33

Therefore the system (6.5 - 6.6) is completely separable. We get a complete solution straight away since both equations can be factorised like (2.1) directly and integrated w. r. t. time simultaneously. The implicit solutions are du t = (6.7) Z 2u2(u + 4A) + 4α2(2H √I) − q dv t = (6.8) Z 2v2(v + 4A) + 4α2(2H + √I) q and by inverting these expressions; (6.7) and (6.8), we get u(t) and v(t) explicitly.

6.1.1 Separation of variables for the Kaup-Kuperschmidt case of the H´enon-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

- f3 u(t) - f4 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 - 6.2) in the form

H = f1(u, u,v,˙ v˙) (I = f2(u, u,v,˙ v˙)

We have calculated 2H √I and 2H + √I and obtained the uncoupled equations −

2H √I = f (u, u˙) − 3 (2H + √I = f4(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 px =x ˙, py =y ˙. 34 Chapter 6. The H´enon-Heiles Hamiltonian with cubic potential 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 1 x¨ = (10x ˙ 4eyy˙ 14x + 10ey) (7.1) 2 − − 1 y¨ = ( 4x ˙ + 10eyy˙ + 10x 14ey 2eyy˙2) (7.2) 2ey − − − This is a coupled system of two second order ODE. Deﬁne new variables u = x + ey v = x ey − and transform (7.1) and (7.2) by substituting x, y,x ˙,y ˙,x ¨ andy ¨ as u + v u v x = , y = ln − 2 2 u˙ +v ˙ u˙ v˙ x˙ = , y˙ = − 2 u v − u¨ +v ¨ u¨ v¨ u˙ v˙ 2 x¨ = , y¨ = − − 2 u v − u v − − Then u¨ +v ¨ 3u ˙ 7v ˙ + 2u + 12v = 0 (7.3) − − u¨ v¨ 3u ˙ + 7v ˙ + 2u 12v = 0 (7.4) − − − M˚ahl, 2006. 35 36 Chapter 7. Direct separability

Now we rewrite these equations (7.3 - 7.4) and ﬁnd that they decouple

u¨ 3u ˙ + 2u = 0 (7.5) − v¨ 7v ˙ + 12v = 0 (7.6) − We get a system of two linear second order ODE’s with constant coeﬃcients which has the solutions

t 2t u(t) = C1e + C2e 3t 4t v(t) = C3e + C4e so that C et + C e2t + C e3t + C e4t x(t) = 1 2 3 4 2 C et + C e2t C e3t C e4t y(t) = ln 1 2 − 3 − 4 2 7.1.1 Separation of variables for this example We have solved a second order system (7.1 - 7.2) by direct separation of variables. Let us draw a diagram

- f3 u(t) - f4 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 - 7.2) directly into new coordinates for which we have got a transformed second order system (7.3 - 7.4) in the form

f1(u, u,˙ u,v,¨ v,˙ v¨) = 0 (f2(u, u,˙ u,v,¨ v,˙ v¨) = 0

1 1 We have calculated the functions f3 = 2 (f1 + f2) and f4 = 2 (f1 f2) and we have obtained a decoupled system (7.5 - 7.6) like −

f3(u, u,˙ u¨) = 0 (f4(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 andv ˙. In order to ﬁnd them we resolve the linear system

t 2t u(t) = C1e + C2e t 2t u˙(t) = C1e + 2C2e 7.1. An example of second order 37

t 2t with respect to C1e and C2e :

t C1e = 2u u˙ C e2t =u ˙ −u 2 − and we eliminate time. Thus u˙ u C − = 2 = A (2u u˙)2 C2 − 1 3t 4t 3t 4t Analogously, v(t) = C3e + C4e andv ˙(t) = 3C3e + 4C4e gives

(v ˙ 3v)3 C3 − = 4 = B (4v v˙)4 C4 − 3 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 (7.5 - 7.6). 38 Chapter 7. Direct separability Chapter 8

Conclusion and discussion

The concept of separation of variables for a simple ODE of first order is well defined (2.1). 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 (2.1). This decoupling is usually not complete. Instead, solving is performed se- quentially, that is certain equations have to be solved first and then the re- maining 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,

M˚ahl, 2006. 39 40 Chapter 8. Conclusion and discussion such a general statement may be diﬃcult to prove. In order to show that this sequential separability, obtained through a set of integrals of motion, is not the only mechanism available we have presented an example (example 8) of direct separation of two second order ODE’s. In this example (example 8) we have, by a pointwise change of variables, reduced a system of two second order equations into another second order system of two linear equations with constant coeﬃcients. Moreover, we have shown that this second order system also possesses two integrals of motion. However, they are not quadratic with respect to velocities and therefore less useful for sequential separation than in the case of the other examples in this thesis. Example 8 shows that there are other possible mechanisms of separation than the ones studied here. For instance, one can perform direct separation with the use of transformations into new variables which involve position variables and velocities as well. Summarizing, our study has shown that there is a typical mechanism of pointwise, and stepwise, separation which works for several classical mechanical examples. However, there are other mechanisms of separation for systems of second order ODE’s possible. These mechanisms would be an interesting subject of further studies. Bibliography

[1] Andersson Karl Gustav, Böiers Lars-Christer, (1992), Ordinära differen- tialekvationer, Studentlitteratur, Lund.

[2] Boyce William E, DiPrima Richard C, (2001), Elementary diﬀerential equations and boundary value problems. Seventh edition., John Wiley & Sons, Inc., USA.

[3] Strauss, Walter A, (1992), Partial diﬀerential equations. An introduction., John Wiley & Sons, Inc., USA.

[4] Landau, Lifschitz, (1960), Course of theoretical physics vol 1. Mechanics, Pergamon Press Ldt., London.

[5] Sköldstam Markus, (2004), Analysis of the phase space, asymptotic behavior and stability for heavy symmetric top and tippe top, Linköpings Universitet, Linköping.

[6] Rauch-Wojciechowski Stefan, Marciniak Krzysztof, (2006), “Separation of variables for diﬀerential equations”, (forthcoming in) Encyclopedia of Math- ematical Physics, Elsevier.

[7] Romeiras Filipe J., (1995), “Separability and lax pairs for the two dimensional Hamiltonian system with a quartic potential”, J. Math. Phys., 36(7).

[8] Maciej, Blaszak, Rauch-Wojciechowski Stefan, (1994), “A generalized Henon-Heiles system and related integrable Newton equations”, J. Math. Phys., 35(4).

[9] Marciniak Krzysztof, Rauch-Wojciechowski Stefan, (2005), Separable sys- 2 d qi tems of coordinates for triangular Newton equations dt2 = Mi(q1,...,qi), Department of Science and Technology (ITN), ISSN 1650-2612, Linköping University, Norrköping. Linköping.

[10] Stanislavsky A. A., Weron K., (2001), “Nonlinear oscillations in the rolling motion of Euler disk”, Physica D, 156, pp. 247-259.

M˚ahl, 2006. 41 42 Bibliography LINKÖPING UNIVERSITY ELECTRONIC PRESS

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M˚ahl, 2006. 43