Full-Newton step polynomial-time methods for LO based on locally ...

Proof of Theorem 4 (3)Now assuming z ≠ 0, we can eliminate λ by substituting 2 ξ(v) vy = λ ψb ′ (v) into (12),which givesψ b ′ [ (v) ψ′′′b (v) vy2 − ξ(v) z 2] = λ ψ b ′ (v) ψ′′ b (v) y = 2 ξ(v)ψ′′ b (v) vy2 .Rearranging the terms, and using (10) we obtain−ψ ′ b (v)ξ(v) z2 = [ 2 ξ(v)ψ ′′b (v) − ψ′ b (v)ψ′′′ b (v) ]vy 2 = (2−ρ(v)) ξ(v)ψ ′′b (v) vy2 ,yielding−ψ ′ b (v) z2 = (2 − ρ(v)) ψ ′′b (v) vy2 , (14)Since −ψb ′ (v) > 0 and ψ′′b(v) > 0, this equation has no nonzero solution <strong>for</strong> y if ρ(v) > 2,and hence κ is then given by (13).If ρ(v) ≤ 2, substitution of (14) into the constraint ψ ′′b (v) v2 y 2 − ψ ′ b (v) vz2 = 1 yieldsor, equivalently,Henceψ ′′b (v) v2 y 2 + (2 − ρ(v)) ψ ′′b (v) v2 y 2 = 1,[3 − ρ(v)] ψ ′′b (v) v2 y 2 = 1. (15)vy =±1√[3 − ρ(v)] ψb ′′(16)(v).31