When α ′ > 0, the flow equation ( k = − 1), has two solutions for b ′ $^{2}$( λ ′ ) > 1 and b ′ $^{2}$( λ ′ ) < 1. These are given as: It can be seen that b$_{0}$ = 1 corresponds to a soliton. If we consider α ′ < 0, which essentially produces an immortal solution, we can see that it expands along the flow. The conclusions mentioned in the above proposition are demionstrated in Fig.4 Proposition III.5 Inclusion of the 3 rd order terms in the flow on hyperbolic space, with any choice of initial radius ( b ′ 0 ≥ 1 ) and any choice of α ′ gives rise to an immortal solution. 10 5. 3rd order flow on H 2 FIG. 4: b $^{$^{′}$2}$( λ ′ ) vs λ ′ for 2 nd order flow GLYPH<Minus> 50 0 50 100 150 200 0 100 200 300 400 Λ ' b ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (a) b $^{′}$( λ $^{′}$) 2 > 1 , b ′ 0 = 8, α ′ > 0 GLYPH<Minus> 2.0 GLYPH<Minus> 1.5 GLYPH<Minus> 1.0 GLYPH<Minus> 0.5 0.0 0.2 0.4 0.6 0.8 1.0 Λ ' b ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (b) b ( λ ) 2 < 1 , b ′ 0 = 0 . 5, α ′ > 0 0 2 4 6 8 10 0.0 0.5 1.0 1.5 Λ ' b ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (c) b ′ 0 = 1, α ′ > 0,soliton 0.0 0.5 1.0 1.5 2.0 0 2 4 6 8 10 Λ ' b ' GLYPH<Space> 2 GLYPH<LParen1> Λ ' GLYPH<RParen1> (d) b ′ 0 = 0 . 4, α ′ < 0 b ′ ( λ ′ ) 2 + ln( b ′ ( λ ′ ) 2 − 1) − b ′ 2 0 − ln( b ′ 2 0 − 1) = 2 λ ′ , for b ′ $^{2}$( λ ′ ) > 1 (18) b ′ ( λ ′ ) 2 + ln(1 − b ′ ( λ ′ ) $^{2}$) − b ′ 2 0 − ln(1 − b ′ 2 0 ) = 2 λ ′ for b ′ $^{2}$( λ ′ ) < 1 (19)