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Examensarbete
Separation of variables for ordinary differential equations
Anna Mahl
LiTH - MAT - EX - - 06 / 01 - - SE
Separation of variables for ordinary differential equations
Department of Applied Mathematics, Linkopings Universitet
Anna Mahl
LiTH - MAT - EX - - 06 / 01 - - SE
Examensarbete: 20 p
Level: D
Supervisor: Stefan Rauch,Department of Applied Mathematics, Linkopings Universitet
Examiner: Stefan Rauch,Department of Applied Mathematics, Linkopings Universitet
Linkoping: February 2006
Matematiska Institutionen581 83 LINKOPINGSWEDEN
February 2006
x x
http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-5620
LiTH - MAT - EX - - 06 / 01 - - SE
Separation of variables for ordinary differential equations
Anna Mahl
In case of the PDE’s the concept of solving by separation of variables has a welldefined meaning. One seeks a solution in a form of a product or sum and tries tobuild the general solution out of these particular solutions. There are also knownsystems of second order ODE’s describing potential motions and certain rigid bodiesthat are considered to be separable. However, in those cases, the concept of separationof variables is more elusive; no general definition is given.In this thesis we study how these systems of equations separate and find that theirseparation usually can be reduced to sequential separation of single first order ODE´s.However, it appears that other mechanisms of separability are possible.
ODE, Separation of variables, Potential motion, Heavy symmetric top, Cofactor sys-tems, Direct separability.
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vi
Abstract
In case of the PDE’s the concept of solving by separation of variables has a welldefined meaning. One seeks a solution in a form of a product or sum and triesto build the general solution out of these particular solutions. There are alsoknown systems of second order ODE’s describing potential motions and certainrigid bodies that are considered to be separable. However, in those cases, theconcept of separation of variables is more elusive; no general definition is given.
In this thesis we study how these systems of equations separate and findthat their separation usually can be reduced to sequential separation of singlefirst order ODE´s. However, it appears that other mechanisms of separabilityare possible.
Keywords: ODE, Separation of variables, Potential motion, Heavy symmetrictop, Cofactor systems, Direct separability.
Mahl, 2006. vii
viii
Acknowledgements
I would like to thank my supervisor and examiner Stefan Rauch. You havebeen a dedicated, and skillful, educationalist. By dividing problems into piecesbig enough for me to calculate and comprehend, you have taught me not to beafraid of approaching non familiar problems and you have encouraged me to trynew ways. These things I will treasure and bring with me in the future.
I would like to thank my dad Per Mahl. All my life you have read andgiven critique on my writing, but this must be the first time that you havehelped me by knowing less! It is because you have not understood majority ofthe mathematical preliminaries that you have been an invaluable partner fordiscussion.
I would like to thank my opponent Peter Brommesson, and also MattiasHansson, for reading this report and giving critique. I would also like to thankZebastian Zaar for creating my pictures of HST and rolling disk. Moreover, Iwould like to thank Daniel Petersson, in the office next to mine, for helping meto understand some mechanics.
Finally, I would like to thank the rest of my family and my friends. You allmean a lot to me.
Mahl, 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 . . . . . . . . . . . 73.1.1 Separation of variables for two dimensional potential motion 9
3.2 Three dimensional potential motion . . . . . . . . . . . . . . . . 103.2.1 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 . . . . . . . . . . . . . . . . . 154.1.1 Separation of variables for the HST equations . . . . . . . 19
4.2 The equations of motion of the rolling disk . . . . . . . . . . . . 204.2.1 Separation of variables for the RD equations . . . . . . . 22
5 Triangular Newton equations 25
5.1 Two dimensional triangular cofactor system . . . . . . . . . . . . 255.1.1 Separation of variables for two dimensional triangular New-
ton equations . . . . . . . . . . . . . . . . . . . . . . . . . 28
6 The Henon-Heiles Hamiltonian with cubic potential 31
6.1 The Kaup-Kuperschmidt case of the Henon-Heiles Hamiltonianwith cubic potential . . . . . . . . . . . . . . . . . . . . . . . . . 326.1.1 Separation of variables for the Kaup-Kuperschmidt case
of the Henon-Heiles Hamiltonian . . . . . . . . . . . . . . 33
7 Direct separability 35
7.1 An example of second order . . . . . . . . . . . . . . . . . . . . . 357.1.1 Separation of variables for this example . . . . . . . . . . 36
8 Conclusion and discussion 39
Mahl, 2006. xi
xii Contents
Chapter 1
Introduction
An ordinary differential equation (ODE) is an algebraic equation f(x, y, dydx
) = 0involving derivatives of some unknown function with respect to one independentvariable. A separable ordinary differential equation is an ODE which can bewritten in such a way that the dependent variable and its differential appearon one side of the equals sign and the independent variable and its differentialappear on the other side.
The method of separation of variables is best known for partial differentialequations of mathematical physics, for which it is the main method of solv-ing these equations. The basic idea behind the method of separating variablesin the theory of partial differential equations (PDE) is to consider an ansatzfor a solution (additive or multiplicative) that allows reducing the problem to asystem of uncoupled ordinary differential equations for functions of one variable.
Example 1. The heat equation.
Consider∂u
∂t= uxx (1.1)
for an initial boundary value problem on the interval 0 < x < L and with theboundary conditions
∂u∂t
= uxx, 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 twofunctions. Substitution of u(x, t) = X(x)T (t) into (1.1) reduces this partialdifferential equation to two uncoupled ordinary differential equations
∂tT = −k2T
∂xxX = −k2X
These two ordinary differential equations have solutions
Tk(t) = Ae−k2t
Xk(x) = B cos kx+ C sin kx
Mahl, 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-ary conditions u(0, t) = u(L, t) = 0. If we put kn = nπ
L, where n = 1, 2, . . ., the
complete solution reads
u(x, t) =∞∑
n=1
cne−k2
nt sin knx (1.2)
where the coefficients cn are given by the integrals cn = 2L
∫ L
0f(x) sin knxdx
([3],[6]).
Example 2. The Hamilton-Jacobi equation of natural Hamiltonian
Another example of an equation known to be solvable by separating variablesis the Hamilton-Jacobi equation for the natural Hamiltonian ([4],[6]).
One method of solving the canonical Hamilton equations
x =∂H(x, y)
∂y, y = −∂H(x, y)
∂x, x = (x1, . . . , xn), y = (y1, . . . , yn) (1.3)
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 trans-formed into a set of simple linear equations for new variables β, α. This functionW (x, α) satisfies the first order nonlinear PDE
H
(
x,∂W (x, α)
∂x
)
= α1 (1.4)
which is known as the Hamilton-Jacobi equation. In order to find it, one makesan additive ansatz W (x, α) =
∑nk=1Wk(xk, α), where the n functions Wk(xk, α)
each depend on a single variable xk. This ansatz works well for the Hamiltoni-ans of Stackel class with H = G+V = 1
2
∑
i gii(x)y2
i . It simplifies the problem;instead of integrating the nontrivial Hamilton-Jacobi equation the problem re-duces to integrating n uncoupled first order ODE’s for functions Wk(xk, α).
The purpose of this thesis is to investigate how variables can be separatedin case of ODE’s. In order to do so we put together known examples of Newtonequations solved in mechanics and describe the features of their separability.
In chapter 2 we present basic concepts needed for this work; we give adefinition of separability in general and we explain why we consider the Newtonequation of one dimensional potential motion to be separable. In the followingchapters we study a variety of mechanical problems, mainly of Newtonian form,to see how these equations are solved by separating variables. In chapter 3 wedescribe the problem of potential motion. In chapter 4 we look at the equationsof motion of rigid bodies and we show the problems of the heavy symmetric topand of the rolling disk. In chapter 5 we describe how separation of variablestakes place when solving triangular systems of Newton equations and in chapter6 we consider the Kaup-Kuperschmidt case of the Henon-Heiles Hamiltonianwith cubic potential. Finally, in chapter 7, we give an example of a directlyseparable second order system and in chapter 8 we formulate conclusions of ourwork.
Chapter 2
Preliminaries
In this chapter we state the definition of separability ([1]). Moreover, we explainhow the Newton equation of one dimensional potential motion is consistent withthis 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
G(y) =
∫(
g(y)dy
dx
)
dx =
∫
(h(x)) dx = H(x) + C
where g has the primitive function G and h has the primitive function H. Thesolution has the implicit form
G(y) = H(x) + C
Example 3. The logistic equation.
Considerdx
dt= µx
(
1 − x
k
)
(2.2)
where µ and k are positive constants.It is separable since it can be factorised into
1
µx(
1 − xk
)
dx
dt= 1
for x 6= k and x 6= 0.By integrating both sides, w. r. t. time, we obtain an implicit solution
t+ C =
∫
1
µ
(
1
x+
1
k(
1 − xk
)
)
dx =1
µ
(
ln |x| − ln |1 − x
k|)
Mahl, 2006. 3
4 Chapter 2. Preliminaries
and an explicit solution x(t) reads
x =keµt
C + eµt
In the separation procedure we had to assume that x 6= k and x 6= 0. A directcheck confirms that constant solutions x = k and x = 0 also satisfy the equation(2.2).
Example 4. The homogeneous equation.
Takedy
dx= f(
y
x) (2.3)
This equation can not be written directly in the separable form (2.1), but bydefining a new variable z = y
xwe get dy
dx= x dz
dx+ z and f( y
x) = f(z). By
substituting these expressions into the equation (2.3) we get an equation
xdz
dx+ z = f(z)
which is separable since it can be written as
1
f(z) − z
dz
dx=
1
x
2.1 Separation of variables for one dimensional
potential motion
Let us consider the second order ODE of the form
mx = −dV (x)
dx(2.4)
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 x. Then
d
dt
(
mx2
2
)
= −dV (x)
dtand
mx2
2+ V (x) = E = constant
We can solve this first order ODE for x(t) by separating variables since theequation of energy
mx2
2+ V (x) = E
can be written1
√
2m
(E − V (x))
dx
dt= 1
which gives us
t =
√
m
2
∫
dx√
(E − V (x))=: G(x) + C
Notice that in solving (2.4) by separation of variables we have used theenergy integral of motion. Our solution is given in an implicit form since weneed 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
K(x, x) =dK
dt=∂K
∂xx+
∂K
∂xx =
∂K
∂xx+
∂K
∂xM(x) = 0
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 constitutesa 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 dimensionsare solved by separating variables ([4]).
A general potential Newton equation has the form
mr = −grad V(r), r ∈ R3 (3.1)
where the function V (r) is called a potential. By multiplying both sides with rwe see that (3.1) always admits the energy integral of motion
E =1
2mr
2+ V (r) (3.2)
For a central potential V (r) = V (r) with r = |r|, the force F = −grad V(r) =
−dV (r)dr
rr
is parallel to r. Then the equation (3.1) admits an additional vector in-
tegral of motion, namely the angular momentum L = r×mr. By differentiatingwe find
L = r × mr + r × mr = r × mr = r × F =dV (r)
dr
(
r
r× r
)
= 0
By using the angular momentum and the energy integrals of motion we cansolve potential Newton equations with central potentials V (r) = V (r) throughseparation of variables.
3.1 Two dimensional potential Newton equations
Let us first consider the two dimensional potential Newton equation with thecentral potential V (r)
mr = −grad V(r) (3.3)
where r = (x1, x2)T , that takes the coordinate form
mx1 = −∂V (r)
∂x1= −V ′(r)
x1
r(3.4)
mx2 = −∂V (r)
∂x2= −V ′(r)
x2
r(3.5)
Mahl, 2006. 7
8 Chapter 3. Potential Newton Equations
This equation, (3.3), admits the energy integral E = mr2
2 + V (r). Moreover, bymultiplying the first coordinate equation (3.4) by x2, and the second one (3.5)by x1 and subtracting we find
d
dt(mx1x2 −mx2x1) = 0
which we recognize as the derivative of the third component of the angularmomentum L3 = m(x1x2 − x2x1). The energy and the angular momentumintegrals of motion reduce the second order system (x1, x2) into a first ordersystem
L3 = m(x1x2 − x2x1) (3.6)
E =mr
2
2+ V (r) (3.7)
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,x1 = r cos θ − rθ sin θ, x2 = r sin θ and x2 = r sin θ + rθ cos θ into (3.6 - 3.7) weget
L3 = mr2θ (3.8)
E =m
2
(
r2 + r2θ2)
+ V (r) (3.9)
When we resolve θ = L3
mr2 from (3.8) and substitute in (3.9) we get theseparable equation of one dimensional motion w. r. t. variable r (2.4)
E =m
2
(
r2 + r2(
L3
mr2
)2)
+ V (r) (3.10)
We factorise it
1√
2m
(E − V (r)) − L2
3
m2r2
dr
dt= 1 (3.11)
and a solution is obtained by integrating both sides w. r. t. time t
t =
∫
dr√
2m
(E − V (r)) − L2
3
m2r2
(3.12)
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 forθ(t) like
θ =
∫
L3
mr(t)2dt (3.13)
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, aswe described in our examples of PDE, to transform the original system into anew 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 potentialmotion, we have transformed our original system into polar coordinates. Wehave obtained a new system which we have solved by separating variables, eventhough the equations have remained coupled. When we solve a coupled systemby separation of variables we use a procedure we have chosen to call stepwise.The stepwise separation of variables for two dimensional potential motion canbe symbolically represented by a diagram
L3- θ(r)
?E - r(t)
6
L3- 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 ho-risontal arrows indicate that the quantity is resolved from the expression, whilethe 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 follow-ing the arrows upwards until reaching the final solution θ(t) in the upper rightcorner.
Let us describe where and how the separation of variables have been takingplace 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 havereduced the system of two second order equations (3.4 - 3.5) into a new systemof the two first order integral equations (3.6 - 3.7).
Subsequently, we have transformed this first order system into polar coor-dinates and obtained the new coupled system of equations (3.8 - 3.9) which wehave solved by successively eliminating and stepwise separating variables. Letus 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 eliminatedθ 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 writtenit 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 solutionwe have substituted into (3.8). That has given us one more equation (3.13) tointegrate, which is also consistent with (2.1), and finally we have calculated asolution for θ(t). Summing up, in two dimensional potential Newton equationswe have two integrals of motion that constitutes a system of two first orderequations. 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 potentialV (r) by separation of variables. As in the two dimensional case, we start withthe Newton equation
mr = −grad V(r), r = (x1, x2, x3)T ∈ R3 (3.14)
and we know that E and L are integrals of motion. Then, we conclude that the
third component L3 and L2
are also integrals of motion
L3 = (Le3) = Le3 + L ˙e3 = Le3 = 0
L2
= (L · L) = L · L+ L · L = 2L · L = 0
As a result, for three dimensional potential motion, the systemmr = −grad V(r),
has three integrals of motion: E, L3 and L2. In Cartesian coordinates, these
integrals of motion read
L3 = m(x1x2 − x2x1) (3.15)
L2
= m(x2x3 − x3x2)2 +m(x3x1 − x1x3)
2 +m(x1x2 − x2x1)2 (3.16)
E =mr
2
2+ V (r) (3.17)
We can solve this system of equations (3.15 - 3.17) by separation of vari-ables if we introduce the spherical coordinates r, ϕ and θ. We substitutex1 = r sin θ cosϕ, x2 = r sin θ sinϕ, x3 = r cos θ, and x1 = r sin θ cosϕ +r(θ cos θ cosϕ − ϕ sin θ sinϕ), x2 = r sin θ sinϕ + r(θ cos θ sinϕ + ϕ sin θ cosϕ),x3 = r cos θ − rθ sin θ which gives us a new system
L3 = mr2ϕ sin2 θ (3.18)
L2 = m2r4(θ2 + ϕ2 sin2 θ) (3.19)
E =m
2
(
r2 + r2θ2 + r2ϕ2 sin2 θ)
+ V (r) (3.20)
We resolve ϕ from (3.18)
ϕ =L3
mr2 sin2 θ(3.21)
3.2. Three dimensional potential motion 11
and substitute ϕ into (3.19) to get
θ2 =L2
m2r4− L2
3
m2r4 sin2 θ(3.22)
By substituting (3.21) and (3.22) into the energy equation (3.20) we get
E =m
2
(
r2 + r2(
L2
m2r4− L2
3
m2r4 sin2 θ
)
+ r2(
L3
mr2 sin2 θ
)2
sin2 θ
)
+ V (r)
which “miraculously” simplifies to a one dimensional energy integral (2.4) sincethe dependence on θ cancels out and leaves us with the expression
E =m
2
(
r2 +L2
m2r2
)
+ V (r) (3.23)
This equation is separable and we factorise it with respect to r and t
1√
2m
(E − V (r)) − L2
m2r2
dr
dt= 1 (3.24)
A solution is then obtained by integrating both sides
t =
∫
dr√
2m
(E − V (r)) − L2
m2r2
(3.25)
To get an explicit solution r(t), ϕ(t) and θ(t) we invert (3.25) to get r(t). Thenwe substitute r(t) into (3.22) to find a solution for θ(t)
∫
dθ√
L2 − L2
3
sin2 θ
=
∫
dt
mr(t)2(3.26)
Finally we substitute r(t) and θ(t) into (3.21) and get ϕ(t)
∫
dϕ =
∫
L3dt
mr(t)2 sin2 θ(t)(3.27)
12 Chapter 3. Potential Newton Equations
3.2.1 Separation of variables for three dimensional poten-
tial motion
When we have solved this three dimensional problem of potential motion, wehave separated variables stepwise. Let us draw a diagram similar to Figure 3.1:
L3- ϕ(r, θ)
??
?L2 - θ2(r, θ)
E - r(t)
6
L2 - θ(t)
6
L3- ϕ(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 ho-risontal arrows represent resolving the expression w. r. t. the variable andthe vertical arrows signify substituting the variable into the expression. Resolveϕ from L3 and substitute in L2 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 hasbeen taking place by looking at the features that have enabled us to obtain asolution for (3.14). Since the energy and the angular momentum are integrals ofmotion we have reduced the original second order system, the three coordinateequations (3.14), into a system of first order (3.15 - 3.17).
Then having transformed this first order system into the spherical coordi-nates, we have obtained the system (3.18 - 3.20) that we have solved by stepwiseseparation of variables. Let us, as in the case of two dimensional potential mo-tion, denote this system (3.18 - 3.20) as
L3 = f1(r, ϕ, θ)
L2 = 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 resolving ϕin terms of r and θ from L3 (3.18) like
ϕ = g1(r, θ)
Then, by resolving θ2 from L2 (3.19) we have got
θ2 = g2(r, θ)
By substituting the expressions of ϕ and θ2 into E (3.20) we have obtained anequation (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 haveintegrated 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 expressionθ2 = g2(r(t), θ) (3.22) and we have obtained an equation which is separable withrespect to t and θ (3.26). We have integrated this equation and we have invertedthe solution to get θ(t). At last, we have substituted our solutions, r(t) and θ(t),into our expression for ϕ = g1(r(t), θ(t)) (3.21) and we have obtained a thirdseparable equation (3.27) thus the solution ϕ(t). In short, let us illustrate thisstepwise solving procedure as follows:
E - r(t) - L2 - 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 theequations of the rolling disk ([10]) are solved by separating variables.
The vector equations of motion of a rigid body are
ms = F (4.1)
L = K (4.2)
where F is the force, K denotes the torque, L is the angular momentum and sdenotes 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 itssymmetry axis fixed in space. The center of mass s of the HST is located on
Figure 4.1: The heavy symmetric top
Mahl, 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 = [x, y, z] and a body-fixed coordinate system K = [1, 2, 3] such that bothhave their origins at a point fixed in space (see Figure 4.1). The equations ofmotion for the HST is then derived by choosing the symmetry axis of the topas the 3-axis of the body-fixed coordinate system. The total force F actingon the HST is the sum of the gravitational force acting on the center-of-massand of the reaction force F r acting on the support point. The torque K andthe angular momentum L are both vectors defined with respect to the commonorigin of both coordinate systems.
The equations of motion, (4.1 - 4.2), in case of the HST read
d(ml ˙3)
dt= −mgz + F r (4.3)
dL
dt= l3 × (−mgz) (4.4)
These equations are complemented by the kinematic equation describing the
motion of ˙3 w. r. t. the inertial coordinate system K0
˙3(t) = ω × 3
where ω is the instantaneous angular velocity of the HST. By using the relationsL = (IxΩx, IyΩy, IzΩz)
Twhere Ix, Iy, Iz are moments of inertia w. r. t. the
fixed point, and that ω× 3 = LIx
× 3 for an axially symmetric rigid body, we getthe vector equations of motion for the HST
˙3 =L
Ix× 3 (4.5)
dL
dt= −mgl3 × z (4.6)
This system is a closed system of six ODE’s, for six scalar unknowns; i.e. thecomponents of 3 and L. The vector equations of the HST (4.5 - 4.6) admitthree integrals of motion: the total energy E, the projection of the angularmomentum Lz onto the vertical axis and the projection of angular momentumL3 onto the symmetry axis. We can show that Lz and L3 are integrals of motionby differentiating them
d
dt(L · z) = (−mgl3 × z) · z = 0 (4.7)
d
dt(L · 3) = (−mgl3 × z) · 3 + L · ( L
Ix× 3) = 0 (4.8)
Moreover, if we use the relation ω ·L = ω · L (valid for any axial symmetric rigidbody), we see that the total energy of the system is also an integral of motion
d
dt(E) =
d
dt(1
2ω · L+mgl3 · z) =
1
2ω · L+
1
2ω · L+mgl
˙3 · z
= ω · L+mgl(ω × 3) · z = ω · (−mgl3 × z) +mgl(ω × 3) · z = 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 equationsgiven by these three integrals of motion
Lz = Lz (4.9)
L3 = L3 (4.10)
E =1
2ω · L+mgl3 · z (4.11)
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
1 = (cosψ cosφ− cos θ sinφ sinψ)x+ (cosψ sinφ+ cos θ cosφ sinψ)y + (sin θ sinψ)z
2 = −(sinψ cosφ+ cos θ sinφ cosψ)x− (sinψ sinφ− cos θ cosφ cosψ)y + (sin θ cosψ)z
3 = (sin θ sinφ)x− (sin θ cosφ)y + (cos θ)z
We express the angular velocity, ω = ωxx+ ωy y + ωz z, by
ωx = ψ sin θ sinφ+ θ cosφ
ωy = −ψ sin θ cosφ+ θ sinφ
ωz = θ + ψ cos θ
and angular momentum, L = I · ω, by
Lx = Ix(θ cosφ− φ sin θ cos θ sinφ) + Iz sin θ sinφ(ψ + φ cos θ)
Ly = Ix(θ sinφ+ φ sin θ cos θ cosφ) − Iz sin θ cosφ(ψ + φ cos θ)
Lz = Ixφ sin2 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
Lz = Ixφ sin2 θ + Iz cos θ(ψ + φ cos θ) (4.12)
L3 = Iz
(
ψ + φ cos θ)
(4.13)
E =Ix
2
(
θ2 + φ2 sin2 θ)
+Iz
2
(
ψ + φ cos θ)2
+mgl cos θ (4.14)
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
ψ =L3
Iz−(
Lz cos θ − L3 cos2 θ)
Ix sin2 θ(4.15)
φ =
(
Lz cos θ − L3 cos2 θ)
Ix sin2 θ(4.16)
Substituting (4.15) and (4.16) into (4.14) we get a one dimensional energy in-tegral equation (2.4)
E =Ixθ
2
2+
(Lz − L3 cos θ)2
2Ix sin2 θ+L2
3
2Iz+mgl cos θ =
Ixθ2
2+ U(θ) (4.17)
This equation can be factorised, and separated, as
1√
2(E−U(θ))Ix
dθ
dt= 1 (4.18)
where
U(θ) =(Lz − L3 cos θ)
2
2Ix sin2 θ+L2
3
2Iz+mgl cos θ
When we integrate both sides we get
t =
∫
1√
2(E−U(θ))Ix
dθ (4.19)
If we invert (4.19) we get the explicit solution θ(t). Further, if we substituteθ(t) into, and integrate right hand side of, (4.15) and (4.16), we get φ(t) andψ(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. Letus draw a diagram:
Lz
L3
- φ(θ), ψ(θ)
?E - θ(t)
6
φ(θ)
ψ(θ)- φ(t), ψ(t)
Figure 4.3: Stepwise separation of variables for motion of a heavy symmetrictop
Start reading the diagram (Figure 4.3) in the upper left corner and read that wehave used both the equation for Lz and the equation for L3 to resolve φ and ψsimultaneously and that we secondly have substituted them into the expressionfor 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 havereduced 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 havegot 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 factorisedit like (2.1) in (4.18). Then we have substituted the solution for θ(t) into theexpressions of φ (4.15) and ψ (4.16) and they have become directly integrableequations. Let us illustrate the stepwise procedure leading to a complete solutionas
E - θ(t)-
-Lz
L3
- φ(t), ψ(t)
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 consideredseparable, 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 = [x, y, z] and a moving with the body K = [1, 2, 3] (seeFigure 4.4). The fixed coordinate system x, y, z is such that the x, y axes liein the horisontal plane and the z axis points vertically upwards. The movingsystem has its origin at the center of the disk. Here 3 is the symmetry axis,2 the horisontal axis defined by 2 = 3 × 1 and 1 is the line from the center tothe point of contact with the plane. The vector from the center to the point ofcontact is then a = a1. The angle between the axes 3 and z is denoted θ andvaries between 0 to π. Moreover, the angular velocity of the disk about the 3axis is called ω, and the angular velocity of K about the z axis is called Ω.
The vector equations of motion for the rigid body, (4.1 - 4.2), for this diskread
md2rcm
dt2= F −mgz (4.20)
dLcm
dt= a× F (4.21)
where g is acceleration of gravity. Here the unknown force F , acting at thepoint of contact A, can be eliminated between (4.20) and (4.21). We get
1
ma
dLcm
dt+d2rcm
dt2× 1 = g(1 × z) (4.22)
In the case of pure rolling motion the velocity of the contact point A is zero:0 = vA = drcm
dt+ ω × a, and we can eliminate drcm
dt= a× ω from the equation
(4.22).Using known kinematic relations one can calculate expressions of angular ve-
locity about the triad 1, 2, 3 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 1, 2, 3. Furthermore,one can compute the velocity of the center of the disk and the angular momen-tum itself. When substituting all expressions into (4.22) we get the coordinateequations for 1, 2, 3
(2k + 1)ω + θΩsin θ = 0 (4.23)
kΩ2 sin θ cos θ + (2k + 1)ωΩsin θ − (k + 1)θ =g
acos θ (4.24)
Ω sin θ + 2θΩcos θ + 2ω = 0 (4.25)
where k is defined by the inertia tensor. This system, (4.23 - 4.25), consists ofthe variables θ, θ, ω and Ω and has the total order four. It admits one integralof motion, namely the energy
E =1
2mv2
cm +1
2ω · I · ω +mgz (4.26)
which expressed in the variables ω, Ω, θ and θ has the form
E =ma2
2
[
(2k + 1)ω + (k + 1)θ2 + kΩ2 sin2 θ +2g
asin θ
]
(4.27)
Thus we have a problem of three component equations (4.23 - 4.25) that admitsone 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
ω =dω
dt=dω
dθ
dθ
dt=dω
dθθ
Ω =dΩ
dt=dΩ
dθ
dθ
dt=dΩ
dθθ
and substitute into (4.23) and (4.25) to get
(2k + 1)dω
dθ+ Ωsin θ = 0 (4.28)
dΩ
dθsin θ + 2Ωcos θ + 2ω = 0 (4.29)
Then we simplify (4.28) and (4.29). We define z = cosθ and substitute into(4.28) and (4.29) and obtain
dω
dz=
1
2k + 1Ω (4.30)
d
dz[(1 − z2)Ω] = 2ω (4.31)
Now we have a problem we can solve. We resolve Ω = (2k+1)dωdz
from (4.30)and substitute it in (4.31). By doing this we get the Legendre equation
d
dz[(1 − z2)
dω
dz] − 2
2k + 1ω = 0 (4.32)
22 Chapter 4. Vector equations of a rigid body
The Legrendre equation is not elementary, but it is well studied and a solutionfor 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 functionsof θ) from our the energy integral (4.27) and we get
(k + 1)
2θ2 +
k(Ω(cos θ))2
2sin2 θ +
(2k + 1)
2(ω(cos θ))2 +
g
asin θ = h (4.33)
This equation is time dependent and we factorise it with respect to θ and t
1√
2k+1
[
h− k(Ω(cos θ)2)2 sin2 θ − (2k+1)
2 (ω(cos θ))2 − ga
sin θ]
dθ
dt= 1
and we integrate it w. r. t. time t
t =
∫
dθ√
2k+1
[
h− kΩ2
2 sin2 θ − (2k+1)2 ω2 − g
asin θ
]
(4.34)
We get a complete solution by first resolving θ(t) explicitly. Then, by substi-tuting 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 stepwiseseparating variables in a diagram:
f1
f2
- Ω(z), ω(z)
?E - θ(t)
-
- Ω(cos(θ(t)))
ω(cos(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 wehave used the first and the third component equation to resolve both ω and Ωin terms of z = cosθ and that we have substituted these expressions into theenergy integral E in order to get the solution θ(t).
Here we have started out with two component equations and one integral ofmotion 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 ordernon autonomous equation and we have got a solution for ω(cos θ) and moreoverfor Ω(cos θ). By substituting these solutions into E (4.27) we have got E in theform
E = f3(θ, θ)
and we have solved it by separating variables, and we have obtained a solutionfor 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 drawthe 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 (4.23), (4.25),and one integral of motion, energy E (4.27), has similar structure to the threeequations of the HST and when we have stepwise separated variables to solveRD it is similar to when we have solved HST. Both in RD and HST we haveobtained Ω, ω and φ, ψ as functions of θ. Then, for both RD and HST, we haveobtained a separable equation (2.1) by substituting our expressions in termsof θ into the energy integral, (4.14) and (4.27). However, in RD it was morecomplicated to obtain our energy integral in a separable form, because we solvedthe 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 equationscan 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
q1 = M1(q1)
q2 = M2(q1, q2)
...
qn = Mn(q1, q2, . . . , qn)
A Newton equation is called quasi-potential if it admits an energy type integralof motion
E(q, q) =1
2qTA(q)q + k(q)
depending quadratically on velocities and with non degenerate n × n matrixdetA(q) 6= 0. Besides, if this matrix is of the form A = cof(G), where cof(G) =
Gdet(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
q1 = M1(q1)q2 = M2(q1, q2)
(5.1)
Generally, for such systems we know that the first equation q1 = M1(q1) alwaysadmits the energy integral
E =1
2q21 −
∫
M1(q1)dq1 (5.2)
Mahl, 2006. 25
26 Chapter 5. Triangular Newton equations
and a solution for q1(t) can be obtained by quadratures. If we can find a secondintegral of motion depending quadratically on velocities we will be able to solvethe entire system (5.1) by separation of variables. Cofactor triangular systemsadmit such a second integral of motion.
Example 5.
First let us consider
q1 = −4q1 − 4q2 = 6q21 + 12q1 − 4q2 + 10
(5.3)
This system is cofactor since it can be written like
d2
dt2
(
q1q2
)
=
(
−4q1 − 46q21 + 12q1 − 4q2 + 10
)
= M(q) = −(cof(g))−1∇k(2)(q)
where
g =
(
1 1 − q11 − q1 6 − q2
)
and
k(2)(q) = 34q1 − 14q2 + 13q21 − 3
2q41 − 12q1q2 − 2q31 − 2q21q2 + 2q22
Hence it has two integrals of motion
E(1) =1
2q21 −
∫
M1(q1)dq1 =1
2q21 + 4q1 + 2q21 (5.4)
and
E(2) =1
2qT (cofg)q + k(2) = (3 − q2)q
21 + (q1 − 1)q1q2 +
1
2q22 + k(2)(q) (5.5)
Define separation variables as u1 = q1 and the variable u2 through theequation
∇u2(q) = ρ(q)(A12, A22)T = 1 ·
(
q1 − 11
)
where A = cof(g), and ρ(q) = 1 is an integrating factor. Then
u2(q) = −q1 +q212
+ q2
and
q2 = u2 + u1 −u2
1
2, q2 = u2 + u1 − u1u1
When substituting u1, u2 into E1 (5.4) and E2 (5.5) we get
E(1) =1
2u2
1 + 4u1 + 2u21 (5.6)
and
E(2) =5
2u2
1 − u21u2 +
1
2u2
2 + 20u1 − 14u2 + 10u21 − 8u1u2 − 4u2
1u2 + 2u22
=1
2u2
2 + 2u22 − 14u2 − 2u2(
1
2u2
1 + 4u1 + 2u21) + 5(
1
2u2
1 + 4u1 + 2u21)
=1
2u2
2 + 2u22 − 14u2 − 2u2E1 + 5E1 (5.7)
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
t =
∫
dq1√
2(E1 − 4q1 − 2q21)(5.8)
Analogously, we get a solution for u2(t) (and thus q2(t))
t =
∫
du2
2(E2 − 2u22 + 14u2 + 2u2E1 − 5E1)
(5.9)
Example 6.
Now let us consider
q1 = −3q21q2 = − (3q5
1+5q3
1+2)q2
(q2
1+1)2
(5.10)
This system is cofactor since it can be written like
d2
dt2
(
q1q2
)
=
(
−3q21− (3q5
1+5q3
1+2)q2
(q2
1+1)2
)
= M(q) = −(cof(g))−1∇k(2)(q)
where g is the elliptic matrix
g =
(
(q21 + 1) q1q2q1q2 (q22 + 1)
)
and
k(2)(q) =1
q21 + 1(q31detg + q22) = q31 +
q31 + 1
q21 + 1q22
Thus it has two integrals of motion
E(1) =1
2q21 + q31 (5.11)
and
E(2) =1
2q21q
22 +
1
2q21 +
1
2q21 q
22 +
1
2q22 − q1q2q1q2 + q31 +
q31 + 1
q21 + 1q22 (5.12)
Now, in order to write it in a separable form, we define q1 = u1 and u2 =q2√q2
1+1
which satisfies
∇u2(q) = ρ(q)(A12, A22)T = ρ(q)
(
−q1q2q21 + 1
)
with the integrating factor
ρ(q) = (q21 + 1)−3
2
From this we get
q1 = u1, q1 = u1
q2 = u2
√
u21 + 1, q2 = u1u2u1√
u2
1+1
+ u2
√
u21 + 1
28 Chapter 5. Triangular Newton equations
Our integrals of motion (5.11 - 5.12) expressed through u1, u2 take the form
E(1) = 12 u
21 + u3
1 (5.13)
and
E(2) = 12 u
21u
22(u
21 + 1) + 1
2 u21 − u1u1(u2
√
u21 + 1)( u1u1u2√
u2
1+1
+ u2
√
u21 + 1)+
12 ( u1u1u2√
u2
1+1
+ u2
√
u21 + 1)2u2
1 + 12 ( u1u1u2√
u2
1+1
+ u2
√
u21 + 1)2 + u3
1+
u3
1+1
u2
1+1
(u2
√
u21 + 1)2
= u22(
12 u
21 + u3
1 + 1) + 12 u
21 + u3
1 +u2
2
2 (u21 + 1)2
= u22(E1 + 1) + E1 +
u2
2
2 (u21 + 1)2
(5.14)In order to solve this system we can obtain a solution for u1(t) (thus q1)
since (5.13) separates to
t =
∫
du1√
2(E1 − u31)
(5.15)
By substituting u1(t) into (5.14) we get E2 in a separable form. We then get asolution for u2(t) by quadratures
∫
dt
(u1(t)2 + 1)=
∫
du2
2(E2 − E1 − (E1 + 1)u22)
(5.16)
Once we get u2(t) we can get a solution for q2(t).
5.1.1 Separation of variables for two dimensional triangu-
lar 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
E1- u1(t)
6
E2- u2(t)
Figure 5.1: Stepwise separation of variables for two dimensional triangular equa-tions
Read, from the diagram, that we have obtained a solution u1(t) from the equa-tion E1 and by substituting u1(t) into the expression for E2 we have obtaineda solution u2(t).
The two dimensional triangular cofactor system (5.10) admit two integralsof motion and therefore we can reduce the order of each equation by one. Whenwe 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 candenote
E1 = f1(u1, u1)
E2 = f2(u1, u1, u2, u2)
5.1. Two dimensional triangular cofactor system 29
When we have solved this system we have started by solving the equationfor E1 (5.13), and we have got a solution q1(t) = u1(t) in (5.15). Then we havesubstituted u1(t) into E2 (5.14) and we have obtained E2 in the form
E2 = f2(u1(t), u2, u2)
This equation is separable (2.1) and quadratures have given us u2(t) implicitlyin (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 com-pletely separated system of equations for u1 and u2 (5.6 - 5.7). That means wehave 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 Henon-Heiles
Hamiltonian with cubic
potential
In this chapter we will study how the Kaup-Kuperschmidt integrable case of theHenon-Heiles Hamiltonian is solved by separating variables ([7], [8]).
The Henon-Heiles Hamiltonian H : R3 → R with cubic potential is definedby
H(x, y, px, py) =1
2(p2
x + p2y) +
1
2Ax2 +
1
2By2 + αx2y − 1
3βy3 (6.1)
where A, B, α, β are constant parameters. The corresponding Hamiltoniansystem is
x =∂H
∂px
, y =∂H
∂py
, px = −∂H∂x
, py = −∂H∂y
If we calculate the derivatives
x =∂H
∂px
= px
y =∂H
∂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 havestudied previously in this thesis.
Mahl, 2006. 31
32 Chapter 6. The Henon-Heiles Hamiltonian with cubic potential
6.1 The Kaup-Kuperschmidt case of the Henon-
Heiles Hamiltonian with cubic potential
The Henon-Heiles system is well studied and there are only three cases whereit is known to be integrable. By integrable we mean that, for certain values ofthe parameters A, B, α and β, the system admits a second integral of motionbesides 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 = x4(
(r2 + 2A+ 2αy)2 − 4αr(py − 2ry) − 16αy(A+ αy) − 2α2x2)
(6.2)
where r = px
x.
When we want to solve the K-K case of the H-H Hamiltonian with cubicpotential we start out with the two integral equations (6.1 - 6.2) and we definethe separating variables
u = −P0(x, y, px, py) −Q0(x, y, px, py), u = −P1(x, y, px, py) −Q1(x, y, px, py)
v = −P0(x, y, px, py) +Q0(x, y, px, py), v = −P1(x, y, px, py) +Q1(x, y, px, py)
where
Q0(x, y, px, py) = −x−2√I, Q1(x, y, px, py) = 2rx−2
√I
P0(x, y, px, py) = r2 + 2A+ 2αy, P1(x, y, px, py) = 2αpy − 2r(r2 + 2A+ 6αy)
We express x, y, px and py in new variables u, v, u and v as
x2 =2√I
u− vpx
x= − u− v
2(u− v)
2αy = −[
u+ v
2+p2
x
x2+ 2A
]
2αpy = − u+ v
2+ 2r
[
r2 + 2A+ 6αy]
and when we rewriteH (6.1) and I (6.2) in new coordinates we get two equations
H =1
16α2
[
v2 + u2 + 2(v3 + u3) + 8A(v2 + u2)]
(6.3)
I =
[
1
8α2
[
v2 − u2 + 2(v3 − u3) + 8A(v2 − u2)]
]2
(6.4)
If we rearrange these equations, (6.3 - 6.4), they decouple
u2 = −2u2(u+ 4A) + 4α2(2H −√I) (6.5)
v2 = −2v2(v + 4A) + 4α2(2H +√I) (6.6)
6.1. The Kaup-Kuperschmidt case of the Henon-Heiles Hamiltonian with cubic
potential 33
Therefore the system (6.5 - 6.6) is completely separable. We get a completesolution straight away since both equations can be factorised like (2.1) directlyand integrated w. r. t. time simultaneously. The implicit solutions are
t =
∫
du√
2u2(u+ 4A) + 4α2(2H −√I)
(6.7)
t =
∫
dv√
2v2(v + 4A) + 4α2(2H +√I)
(6.8)
and by inverting these expressions; (6.7) and (6.8), we get u(t) and v(t) explic-itly.
6.1.1 Separation of variables for the Kaup-Kuperschmidt
case of the Henon-Heiles Hamiltonian
We have solved the K-K case of the H-H Hamiltonian by complete separationof 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 withcubic potential
The K-K case of the H-H Hamiltonian with cubic potential admits two in-tegrals 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 ofcoordinates 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 = f3(u, u)
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 wehave used earlier in this thesis since the separation variables involve both thepositions x and y and velocities px = x, py = y.
34 Chapter 6. The Henon-Heiles Hamiltonian with cubic potential
Chapter 7
Direct separability
Separation of variables, in almost all examples considered in this thesis, has beenbased on the use of integrals of motion that reduced our system of second orderNewton equations to a (coupled) system of first order ODE’s. Then a suitablechange of variables have decoupled the equations to such an extent that we havebeen able to integrate them in stepwise manner.
In this chapter we give an example of two second order equations that canbe separated as second order equations without using integrals of motion.
7.1 An example of second order
Example 8.
Consider
x =1
2(10x− 4ey y − 14x+ 10ey) (7.1)
y =1
2ey(−4x+ 10ey y + 10x− 14ey − 2ey y2) (7.2)
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 and y as
x =u+ v
2, y = ln
(
u− v
2
)
x =u+ v
2, y =
u− v
u− v
x =u+ v
2, y =
u− v
u− v−(
u− v
u− v
)2
Then
u+ v − 3u− 7v + 2u+ 12v = 0 (7.3)
u− v − 3u+ 7v + 2u− 12v = 0 (7.4)
Mahl, 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 coefficientswhich has the solutions
u(t) = C1et + C2e
2t
v(t) = C3e3t + C4e
4t
so that
x(t) =C1e
t + C2e2t + C3e
3t + C4e4t
2
y(t) = ln
(
C1et + C2e
2t − C3e3t − C4e
4t
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 theother.
Then, when we have solved the second order system (example 8) we havetransformed two second order equations (7.1 - 7.2) directly into new coordinatesfor 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
We have calculated the functions f3 = 12 (f1 + f2) and f4 = 1
2 (f1 − f2) and wehave 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 and v. In order tofind them we resolve the linear system
u(t) = C1et + C2e
2t
u(t) = C1et + 2C2e
2t
7.1. An example of second order 37
with respect to C1et and C2e
2t:
C1et = 2u− u
C2e2t = u− u
and we eliminate time. Thus
u− u
(2u− u)2=C2
C21
= A
Analogously, v(t) = C3e3t + C4e
4t and v(t) = 3C3e3t + 4C4e
4t gives
(v − 3v)3
(4v − v)4=C3
4
C43
= B
where A and B are new constants. These integrals of motion, how ever, arecomplicated and it is more difficult to calculate solutions using them than solvingour 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 welldefined (2.1). Certain systems of second order ODE’s, describing mechanicalsystems, are also recognised as separable. Our aim has been to study what thiskind of separability means and how it refers to the definition of separability fora simple first order ODE.
We have studied several systems known as separable; two and three dimen-sional potential motion, motion of the heavy symmetric top, motion of therolling disk, triangular Newton equations and the Kaup-Kuperschmidt case ofthe Henon-Heiles Hamiltonian, in order to describe the procedure of separationand to see common features and differences.
Our general conclusion is that the process of separation usually uses integralsof motion and a subsequent transformation into new variables. The integrals ofmotion reduce the order of the system so that all equations effectively becomeof first order. The change of variables is needed for decoupling the equations tosuch extent that the resulting equations can be integrated by quadratures usingthe 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 thetwo and three dimensional potential motion, motion of the HST and of trian-gular systems of Newton equations in general. On the other hand, there arecertain 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 theH-H Hamiltonian system. These may be considered as special cases of stepwiseseparation.
The transformation to separation variables has usually been pointwise (notinvolving velocities) so that the new position variables depend only on the oldposition variables. An interesting exception has been the K-K case of the H-H Hamiltonian system where the transformation involved velocities. The H-Hsystem also differed from the previous ones since it had an integral dependingquartically on velocities while all the other systems; potential motion, HST, RDand triangular Newton equations, have had integrals depending quadratically onvelocities at most and have been separated through pointwise separation only.
It probably reflects a general feature that pointwise separable systems usuallyhave integrals of motion which depend quadratically on velocities. However,
Mahl, 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 anexample (example 8) of direct separation of two second order ODE’s. In thisexample (example 8) we have, by a pointwise change of variables, reduced asystem of two second order equations into another second order system of twolinear equations with constant coefficients. Moreover, we have shown that thissecond order system also possesses two integrals of motion. However, they arenot quadratic with respect to velocities and therefore less useful for sequentialseparation than in the case of the other examples in this thesis. Example 8 showsthat there are other possible mechanisms of separation than the ones studiedhere. For instance, one can perform direct separation with the use of trans-formations into new variables which involve position variables and velocities aswell.
Summarizing, our study has shown that there is a typical mechanism ofpointwise, and stepwise, separation which works for several classical mechanicalexamples. However, there are other mechanisms of separation for systems ofsecond order ODE’s possible. These mechanisms would be an interesting subjectof further studies.
Bibliography
[1] Andersson Karl Gustav, Boiers Lars-Christer, (1992), Ordinara differen-
tialekvationer, Studentlitteratur, Lund.
[2] Boyce William E, DiPrima Richard C, (2001), Elementary differential equa-
tions 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] Skoldstam Markus, (2004), Analysis of the phase space, asymptotic behavior
and stability for heavy symmetric top and tippe top, Linkopings Universitet,Linkoping.
[6] Rauch-Wojciechowski Stefan, Marciniak Krzysztof, (2006), “Separation ofvariables for differential equations”, (forthcoming in) Encyclopedia of Math-
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[7] Romeiras Filipe J., (1995), “Separability and lax pairs for the two dimen-sional Hamiltonian system with a quartic potential”, J. Math. Phys., 36(7).
[8] Maciej, Blaszak, Rauch-Wojciechowski Stefan, (1994), “A generalizedHenon-Heiles system and related integrable Newton equations”, J. Math.
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[10] Stanislavsky A. A., Weron K., (2001), “Nonlinear oscillations in the rollingmotion of Euler disk”, Physica D, 156, pp. 247-259.
Mahl, 2006. 41
42 Bibliography
LINKÖPING UNIVERSITY
ELECTRONIC PRESS
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Mahl, 2006. 43