Direct Energy, 2018a

14 LIE ANALYSIS 327
equation must also be zero for an innitesimal generator U that describes
a continuous symmetry.
pr (n) U (ÿ) =0. (14.75)
Using Eqs. 14.42, 14.43, and 14.44, we can write this symmetry condition
in terms of ξ and η.
η tt =0 (14.76)
η tt =0 =∂ tt η +2ẏ∂ yt η +ÿ∂ y η +ẏ 2 ∂ yy η − 2ÿ∂ t ξ
−ẏ∂ tt ξ − 2ẏ 2 ∂ yt ξ − ẏ 3 ∂ yy ξ − 3ẏÿ∂ y ξ
Use ÿ =0, and regroup the terms.
(14.77)
(∂ tt η)+ẏ (2∂ yt η − ∂ tt ξ)+ẏ 2 (∂ yy η − 2∂ yt ξ) − ẏ 3 (∂ yy ξ)=0 (14.78)
The above equation is true for all y only if all of the quantities in
parentheses are zero.
∂ tt η =0 (14.79)
2∂ yt η − ∂ tt ξ =0 (14.80)
∂ yy η − 2∂ yt ξ =0 (14.81)
∂ yy ξ =0 (14.82)
The next step is to solve the above set of equations for all possible solutions
of ξ and η which will determine the innitesimal generators of all possible
continuous symmetry transformations.
We will consider three cases: case 1 with η =0, case 2 with ξ =0, and
case 3 with both ξ and η nonzero.
Case 1 with η =0:Assume η =0. What solutions can be found for ξ?
Equation 14.79 to Eq. 14.82 can be reduced.
∂ tt ξ =0 (14.83)
∂ yy ξ =0 (14.84)
∂ yt ξ =0 (14.85)