Direct Energy, 2018a

324 14.4 Derivation of the Innitesimal Generators
We would like to identify continuous symmetries of Eq. 14.46. These
symmetries will be specied by innitesimal generators of the form
U = ξ∂ t + η∂ y (14.47)
where ξ and η have the form ξ(t, y) and η(t, y). Solutions of the equation
satisfy
(ÿ
− y 3/2 t −1/2) =0. (14.48)
For innitesimal generators that describe symmetries of this equation, the
prolongation is also zero.
pr (n) U ( ÿ − y 3/2 t −1/2) =0. (14.49)
Eq. 14.49 can be solved for all generators U corresponding to continuous
symmetries of the Thomas Fermi equation. Eqs. 14.42 and 14.49 can be
combined.
η tt + 1 2 ξy3/2 t −3/2 − 3 2 ηy1/2 t −1/2 =0 (14.50)
Next, Eq. 14.44 is used.
∂ 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 + 1 2 ξy3/2 t −3/2 − 3 2 ηy1/2 t −1/2 =0
(14.51)
Substitute the original equation for ÿ.
∂ tt η +2ẏ∂ yt η + y 3/2 t −1/2 ∂ y η +ẏ 2 ∂ yy η − 2y 3/2 t −1/2 ∂ t ξ − ẏ∂ tt ξ − 2ẏ 2 ∂ yt ξ
−ẏ 3 ∂ yy ξ − 3ẏy 3/2 t −1/2 ∂ y ξ + 1 2 ξy3/2 t −3/2 − 3 2 ηy1/2 t −1/2 =0
(14.52)
Regroup terms.
(
∂tt η + y 3/2 t −1/2 ∂ y η − 2y 3/2 t −1/2 ∂ t ξ + 1 2 ξy3/2 t −3/2 − 3 2 ηy1/2 t −1/2)
+ẏ ( 2∂ yt η − ∂ tt ξ − 3y 3/2 t −1/2 ∂ y ξ ) +ẏ 2 (∂ yy η − 2∂ yt ξ) − ẏ 3 (∂ yy ξ)=0
(14.53)
Each of the terms in parentheses in Eq. 14.53 must be zero.
∂ tt η + y 3/2 t −1/2 ∂ y η − 2y 3/2 t −1/2 ∂ t ξ + 1 2 ξy3/2 t −3/2 − 3 2 ηy1/2 t −1/2 =0 (14.54)
2∂ yt η − ∂ tt ξ − 3y 3/2 t −1/2 ∂ y ξ =0 (14.55)
∂ yy η − 2∂ yt ξ =0 (14.56)
∂ yy ξ =0 (14.57)