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THE INVERSE PROBLEM FOR EULER’S
EQUATION ON LIE GROUPS
Wayne Lawton
Department of Mathematics
National University of Singapore
2 Science Drive 2
Singapore 117543
Email [email protected]
Tel (65) 874-2749
RIGID BODIES
Euler’s equation

A  (A)  
for their inertial motion
g : R  SO(3)
angular velocity in the body
  v  g g v
inertia operator (from mass distribution)
1
A
Theoria et ad motus corporum solidorum seu rigodorum
ex primiis nostrae cognitionis principiis stbilita onmes
motus qui inhuiusmodi corpora cadere possunt
accommodata, Memoirs de l'Acad'emie des Sciences
Berlin, 1765.
IDEAL FLUIDS
u n  0
u  u u  grad p,   u  0
Euler’s equation
g : R  SDiff (D)
1
u  g g pressure p
n of domain D
for their inertial motion
velocity in space
outward normal
Commentationes mechanicae ad theoriam corporum
fluidorum pertinentes, M'emoirs de l'Acad'emie des
Sciences Berlin, 1765.
GEODESICS
Moreau observed that these classical equations
describe geodesics, on the Lie groups that
parameterize their configurations, with respect to
the left, right invariant Riemannian metric
determined by the inertia operator (determined from
kinetic energy) on the associated Lie algebra
Une method de cinematique fonctionnelle en
hydrodynamique, C. R. Acad. Sci. Paris 249(1959),
2156-2158
EULER’S EQUATION ON LIE GROUPS
Arnold derived Euler’s equation
*
Au   adu Au
that describe geodesics on Lie groups with respect to
left, right
  1,  1 invariant Riemannian metrics
Mathematical Methods of Classical Mechanics,
Springer, New York, 1978
LAGRANGIAN FORMULATION
A trajectory
g:R G
is a geodesic
for a left, right invariant Riemannian metric iff
the associated angular velocity/momentum
1
c  g g , s  g g
Mc  Ac , Ms 
satisfies
1
*
Ad g1 Mc

Mc  Ad g Ms (0)
The momentum lies within a coadjoint orbit which
has a sympletic structure and thus even dimension
GENERAL ASSUMPTIONS
FOR THE INVERSE PROBLEM
G is a connected Lie group with Lie algebra G
*
A : G G is an inertia operator
(self-adjoint and positive definite)
c : R  G satisfies Euler’s equation wrt A
Problem Compute A ,up to multiplication by a
constant, from the values of c over an interval
A GENERAL SOLUTION
1
1
c  g g and s  g g

are nondegnerate, then ,  : G  G
Theorem If

b
a
*
cc ,  
b
a
*
ss
are invertible and A is determined, up to multiplication
by a constant, from the following two equations
*
*
*
b
b
Ms   2E a s , A  a ( Adg Ms )c
SOLUTION FOR RIGID BODIES
Theorem (Lyle-Noakes, JMP, 2001)
For G=SO(3), A can be determined iff c is
nondegenerate (not contained in a proper subspace)
*
is degenerate then there exists v  G
c
such that  v, c   0
Proof If
then
c
satisfies
~
*
Euler’s equation for the inertia operator A  A  vv
To complete the proof it suffices to show that if
s is degenerate then c is degenerate. Consider
 c  Ag g 
s
 M s , s   2 E , c  Ag g s , 
THREE DIMENSIONAL PROBLEM
Define the scalar product ( x , y)   Ax , y 
and choose an orientation on a three dimensional G
Let
 denote the corresponding vector cross product
(u, v  w )  (u  v, w ) Choose a basis
{e1,e 2,e 3 } so ei  ei 1  ei 2
Construct a linear operator Lei  2  ade ei 1
i
Let y, [L] denote c , L wrt this basis
Theorem Euler’s equation for y is y
   y  [ L]y
Then
THREE DIMENSIONAL PROBLEM
G
1
Construct the operator C  A B : G G
Let  denote the corresponding vector cross product
Assume that
B
is also an inertia operator on
T
T
T
Lemma [C]  [C], y y  y [C]y  2E
[C]  I or [C] - I is nonsingular
   y  [L][C]y
Lemma [C]y
Lemma [C]  I   Hom. Pol. R 0 , R 2 , R 4
2
on R , R 2  0,
R i ( y 2 , y3 )  0
and either

THREE DIMENSIONAL PROBLEM
Proof
P(x)  x x  2E 
T
2
2
2
x1  x 2  x 3
Q( x)  x [C]x  2E 
i, j1
P(x) 
 2E
T
Q(x)
Clearly
2
p0 x1  p1x1  p2
3
2
 q0 x1  q1x1  q 2
cij x i x j
 2E
P( y1, y 2 , y3 )  Q( y1, y 2 , y3 )  0
p 0
0
R  det 
q 0

0
p1
p0
q1
q0
p 2  2E
0 

p1
p 2  2E

q 2  2E
0 

q1
q 2  2E 
2
2
R  ( q 2  c11p 2  ( c11  1) 2E )  ( p 2  2E ) q1
2 2
R 0  ( c11  1) 4 E
2
R 2  ( c11  1) 4 E ( q 2  c11p 2 )  2 Eq1
2
2
R 4  (q 2  c11p 2 )  p 2 q1
UNIMODULAR GROUPS
Theorem (Milnor) G is unimodular iff [ L]T  [ L]
Then an orientation and basis can be chosen so that
L = is diagonal and the signs determine G as below

SO (3)
compact , simple
   SL ( 2, R )
simple
 0
E ( 2)
solvable
 0
 00
000
E (1,1)
H (1)
3
R
solvable
nilpotent
commutativ e
NONUNIMODULAR GROUPS
Theorem (Milnor) If G is unimodular for some basis
0 0 0 


[L]  0    ,     0
0    
  
D

  
 y 2 
 y 2 (0)
 y   exp( f ( t )D)  y (0) 
 3
 3 