The Lagrangian Method Summary: An interesting numerical method
for dynamics foregoes the differential equations of motion working,
instead, directly with the action. Since the object of a numerical
evolution is to find a sequence of points that approximates the
motion of the system, the question comes down to formulating the
least action principle with repsect to a path defined piecewise by
a given interpolation. Assuming the interpolation is fixed, this
reduces to an optimization problem over the point sequence, itself.
Solving this yields a series of iteration equations which can then
be used to numerically evolve the system. Since the interpolation
is fixed, this produces a suboptimal solution, but one optimal with
respect to the constraint.
The process is illustrated with application to the simple
harmonic oscillator and the Kepler problem. A notable feature in
the latter application is that the iteration involves logarithms,
rather than polynomials!
Mark Hopkins
1. Formulation Assume the system is given by an action of the
form
( )( ) ( )( )1 .2T
a b aab a
TS m q q A t q U t dt
+
=
q q
The summation convention is used here and throughout. Here, the
configuration space is assumed to be M -dimensional with coordinate
( )0 1 1, , , Mq q q =q
This is the most general case of an action whose Lagrangian is
quadratic in the velocities (with constant second derivatives). It
is assumed there are no zero-mass modes, so that the mass matrix (
)abm m= is non-singular.
The time interval [ ],T T + is divided into N intervals with
interpolating times +=
Note, in particular, the symmetry ( ) ( ) ( ) ( ), , , , , .a aA
A U U= =q Q Q q q Q Q q
The extremal points occur where the discrete sum is stationary
with respect to the point-sequence. Taking the gradient at each of
the points lead to iteration equations, plus an additional equation
which may be viewed as a constraint or an error check. The
gradients of ( ), , ,S t Tq Q are computed as follows,
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
, , , ,
, , ( ) , .
b bb b
ab a a b aa
b bb b
ab a a b aa
Q qSm A Q q A T t U
T tQq QS
m A Q q A T t UT tq
=
= +
q Q q Q q Q
q Q Q q Q q
The gradient of the averaged functions are taken with respect to
the second coordinate,
( ) ( ) ( )( )10
, , 1 .a a a
z z zdzQ Q = + q Q q Q Q q
The two gradients are related by the identity,
( ) ( ) ( ), , , .a aa Q =
q Q q Q Q q
Thus,
( ) ( ) ( ) ( ), , .b b ba aa a a aAS S UQ q T tQ q Q Q
+ =
q Q q Q
Differentiating the action, one finds
( ) ( )
( ) ( ) ( )0 0 1 1 1 1
0
1 1 1 1
0 , , , , 0 , , , ,
0 , , , , , , 0 .
N N N Na a a aN
n n n n n n n na a a
n
S S S St t t tQ q Q Q
S S St t t t n NQ q Q
+ +
= = = =
= = + < <
Q Q Q Q
Q Q Q Q
The last equation advances the iteration, given two previous
points. The first equation establishes the first two points from
the first point, alone. Finally, the second equation may be viewed
as a constraint on the overall system which establishes a global
optimization for the entire path.
These constraints also provide an optimal setting for the
initial velocity. In the unconstrained version of the problem, one
conceives of the sequence as part of an infinite sequence extending
indefinitely in both directions. This, then, requires setting the
values of the first two points.
2. Examples 2.1 Simple Harmonic Oscillator Here, well look at
the simple harmonic oscillator and the Kepler problem. In the
former case, we have a 1-dimensional system with
20 200
0, 0, .2m
q x A U x
= = =
Without loss of generality, we may take 100 =m . One then
obtains
( ) ( )( )2 2 2 21 20
, 1 ,2 2 3
X xX xU x X zX z x dz + += + = and
( )2
02
, .
2 3X xU x X + =
Thus,
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