0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.0 0.2 0.4 0.6 0.8 1.0 Λ ' a ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (a) a ′ 0 = 1 , λ ′ s = 0 . 153, α ′ > 0 0 2 4 6 8 10 0.0 0.5 1.0 1.5 2.0 Λ ' a ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (b) a ′ 0 = 1, α ′ < 0, soliton 0 5 10 15 20 0 5 10 15 20 25 30 35 Λ ' a ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (c) a $^{′}$( λ $^{′}$) 2 > 1 , a ′ 0 = 6, α ′ < 0, 0 5 10 15 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Λ ' a ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (d) a $^{′}$( λ $^{′}$) 2 < 1 , a ′ 0 = 0 . 5, α ′ < 0 FIG. 1: a ′ $^{2}$( λ ′ ) vs λ ′ for 2 nd order flow 2. 3rd order flow on S 2 Proposition III.2 Inclusion of the 3 rd order terms for the flow on the canonical two- sphere, with any choice of initial radius ( a ′ 0 ≥ 1 ) and α ′ always gives rise to an ancient solution. The solution of the flow equation, taken up to 3 rd order, will be: where C = a ′ 2 0 ∓ 1 2 ln(4 a ′ 4 0 ± 4 a ′ 2 0 +5) ∓ 3 4 tan − 1 (2 a ′ 2 0 ± 1) 2 .The ± signs refer to α ′ > 0 and α ′ < 0 respectively. From eqn.14 and the figures the proposition can be proved without much difficulty. Fig.2 illustrates the conclusions in the proposition. Figures on the left and right are for α ′ > 0 and α ′ < 0 respectively. 7 a ′ ( λ ′ ) 2 ∓ 1 2 ln(4 a ′ ( λ ′ ) 4 ± 4 a ′ ( λ ′ ) 2 + 5) ∓ 3 4 tan − 1 (2 a ′ ( λ ′ ) 2 ± 1) 2 = − 2 λ ′ + C (14)