Direct Energy, 2018a

14 LIE ANALYSIS 333
dG
[ ]
dt = ξ L−ẏ∂ ˙ y L−ÿ∂ẏL + η∂ y L + [ d
(η∂ dt ẏL) − η d (∂ dt ẏL) ]
− [ d
(ξẏ∂ dt ẏL) − ξÿ∂ẏL−ξẏ d ∂ dt ẏL ] + ˙ξL
Two terms cancel.
dG
dt = ξ L−ξẏ∂ ˙ y L + η∂ y L + d (η∂ dt ẏL) − η d (∂ dt ẏL)
− d (ξẏ∂ dt ẏL)+ξẏ d ∂ dt ẏL + ˙ξL
(14.135)
(14.136)
Regroup terms.
dG
dt = ( ∂ y L− d ∂ dt ẏL ) [(
(η − ẏξ)
+ ξL ˙ + L ˙ξ
)
]
+ d (η∂ dt ẏL) − d (ξẏ∂ dt ẏL)
(14.137)
The rst term in parentheses is zero because the Lagrangian L satises the
Euler-Lagrange equation.
dG
dt = d dt (ξL +(η∂ ẏL) − (ξẏ∂ẏL)) (14.138)
d
dt [ξL +(η∂ ẏL) − (ξẏ∂ẏL) − G] =0 (14.139)
Therefore, if we can nd G, then the quantity in brackets Υ must be invariant.
Υ=ξL +(η∂ẏL) − (ξẏ∂ẏL) − G = invariant (14.140)
14.5.4 Line Equation Invariants Example
Let us apply Noether's theorem to some examples. First, consider the line
equation ÿ =0which results from application of calculus of variations with
Lagrangian
L = 1 2ẏ2 . (14.141)
A continuous symmetry of this equation is described by the innitesimal
generator U = ∂ y with ξ =0and η =1. The prolongation of the generator
acting on the Lagrangian is zero.
pr (n) UL = η t ẏ =ẏ
Using Eq.14.117, we see that G =0.
dG
dt
( ) dη
=0 (14.142)
dt
=0+L·0=0 (14.143)