Direct Energy, 2018a

14 LIE ANALYSIS 325
Eqs. 14.54, 14.55, 14.56, and 14.57 can be solved for ξ and η. From Eq.
14.57, ∂ yy ξ =0,soξ must have form
ξ =(c 1 + c 2 y) b(t). (14.58)
The quantities denoted c n are constants. From Eq. 14.56, η must have the
form
η = ( c 3 + c 4 y + c 5 y 2) g(t). (14.59)
Functions b(t) and g(t) only depend on t, not y. The condition of Eq. 14.55
can be rewritten.
(2c 4 ∂ t g − c 1 ∂ tt b)+y(4c 5 ∂ t g − 2c 2 ∂ tt b) − 3y 3/2 t −1/2 c 2 b =0 (14.60)
To satisfy Eq. 14.60, c 2 must be zero, and either c 5 =0or g(t) =0.
From Eqs. 14.55 and 14.56, ∂ y η and ∂ t ξ must be constant. Therefore, the
form of ξ must be
This form can be substituted into Eq. 14.54.
ξ = c 6 + c 7 t. (14.61)
y 3/2 t −1/2 (c 4 +2c 5 y) − 2y 3/2 t −1/2 c 7 + 1 2 (c 6 + c 7 t)y 3/2 t −3/2
− 3 2 (c 3 + c 4 y + c 5 y 2 )y 1/2 t −1/2 =0
(14.62)
y 3/2 t ( −1/2 c 4 − 2c 7 + 1c 2 7 − 3c 2 4)
+
1
c 2 6y 3/2 t −1/2
− 3c 2 3y 1/2 t −1/2 + y 5/2 t ( −1/2 2c 5 − 3c ) (14.63)
2 5 =0
The coecients c 3 , c 5 , and c 6 must be zero. Also, c 4 = −3c 7 . No other
solutions here are possible. Thus, the symmetry condition of Eq. 14.49 can
be satised by ξ = t and η = −3y .
This procedure nds one regular continuous innitesimal symmetry of
the Thomas Fermi equation, with innitesimal symmetry generator
U = t∂ t − 3y∂ y . (14.64)
No other solutions can satisfy the constraints given by Eq. 14.49. Therefore,
this equation has only one continuous symmetry.
Finite transformations are related to innitesimal transformations by
Eq. 14.14. In this case, the independent variable transforms as
t → ˜t = e ε(t∂ t−3y∂ y ) t. (14.65)
[
t → ˜t = 1+ε (t∂ t − 3y∂ y )+ 1 ]
2! ε2 (t∂ t − 3y∂ y ) 2 + ... t (14.66)