The r$_{σ}$ part can be absorbed into a redefinition of the σ coordinate and we essentially have r = r ( λ ). These simplify the equations to give (up to 2 nd order), Since the equations are now in a variable separable form, for consistency at the leading order we must have f ′′ $_{σ}$+ f ′ 2 σ = const$_{1}$ and ( f ′′ σ + f ′ 2 σ )+3 f ′ 2 σ = const$_{2}$ ; i.e. f ′ σ = k = const. and f$_{σ}$ = kσ . Note that this condition comes only from the leading order terms. The equations in λ become r 2 e 2 f λ ˙ ( e 2 $^{f$_{λ}$}$) = ˙ ( r $^{2}$), which gives e 2 f$_{λ}$ = r 2 = Ω( λ ) ≥ 0. Thus we have a solution with the metric Eq.(26) which is conformal to the Anti-de Sitter metric, as stated in Prop.IV.1. The Ricci and Kretschmann scalars, R = g $^{ij}$R$_{ij}$ and K = R$_{ijkl}$R ijkl respectively, for such a metric are given by – Thus the curvature diverges and the manifolds become singular at Ω = 0. Using the separable form for the functions in Eq.(5), we get (at third order), where F ( n ) = d $^{n}$F dσ $_{n}$. But, from the separability at leading order, we already have f ′ = k and thus both the terms reduce to e − 2 $^{f}$r $^{6}$β (3) µµ = r $^{4}$β (3) σσ = − 8 k $^{6}$. Similarly all higher derivative terms vanish at order 4 leaving e − 2 $^{f}$r $^{8}$β (4) µµ = r $^{6}$β (4) σσ = 2 (3 + 5 ζ (3)) k $^{8}$. This leads to the following ODE for Ω – We can readily see that k = 0 is a fixed point of the flow. This corresponds to a 5 17 1 8 k 2 d Ω dλ = 1 − α ′ k 2 Ω + 2 ( α ′ k 2 Ω ) 2 − 3 + 5 ζ (3) 2 ( α ′ k 2 Ω ) 3 (36) β (3) µµ = e 2 f 4 r 6 ( − 32 f ′ 6 σ − 16 f ′ 4 σ f ′′ σ − 12 f ′ 2 σ f ′′ 2 σ − 14 f ′′ 3 σ + 4 f ′ 3 σ f (3) σ + 8 f ′ $_{σ}$f ′′ (3) σ + f (3) σ 2 ) β (3) σσ = 1 r 4 − 8 f ′ 6 σ − 12 f ′ 4 σ f ′′ σ + 3 f ′ 2 σ f ′′ 2 σ − 3 f ′′ 3 σ + 5 f ′ 3 σ f (3) σ + 6 f ′ $_{σ}$f ′′ (3) σ + f (3) σ 2 + f ′ 2 σ f (4) σ + f ′′ $_{σ}$f (4) σ ) (35) K = 8 r 4 [ 2 ( f ′′ + f ′ $^{2}$) 2 + 3 f ′ 4 ] = 40 k 4 Ω 2 (34) R = − 4 r 2 [ 2 f ′′ + 5 f ′ $^{2}$] = − 20 k 2 Ω (33) ˙ f$_{λ}$ = 1 r 2 [( f ′′ σ + f ′ 2 σ ) + 3 f ′ 2 σ ] − α ′ r 4 [ ( f ′′ σ + f ′ 2 σ ) 2 + 3 f ′ 4 σ ] (31) ˙ r = 4 r [ f ′′ σ + f ′ 2 σ ] − 4 α ′ r 3 [ f ′′ σ + f ′ 2 σ ] 2 (32)