THÈSE Koléhè Abdoulaye COULIBALY-PASQUIER

where G(0) = G(r) = 1. We have:‖∇ t ∂ Ki t (s)‖ 2˜g(t) = (Ġ)2 .∂sBy the index lemma (e.g. [20]), we deduce:andI t (K t i , K t i ) =∫ r0I t (J t i , J t i ) ≤ I t (K t i , K t i ),〈D s K t i , D s K t i 〉˜g(t) − R m,˜g(t) (K t i , ˙γ, ˙γ, K t i )dt,where R m,˜g(t) denote the (4, 0) curvature tensor associated to the metric ˜g(t).Hence:n∑I t (Ki t , Ki t ) =i=2==n∑∫ ri=2 0∫ rn∑i=2∫ r00≤ (n − 1)〈D s K t i , D s K t i 〉˜g(t) − R m,˜g(t) (K t i , ˙γ, ˙γ, K t i )ds‖∇ t ∂ K i (s)‖ 2˜g(t) − R m,˜g(t) (Ki t , ˙γ, ˙γ, Ki t )ds∂s(n − 1)(Ġ)2 − (G) 2 Ric˜g(t) ( ˙γ, ˙γ)ds∫ r0((Ġ)2 − (G) 2 ( 1 − ɛn − 1 ))ds.For performing the computation, we impose to G to satisfy the O.D.E:{G(0) = G(r) = 1..G + ( 1−ɛn−1 )G = 0We notice that:(Ġ)2 − (G) 2 ( 1 − ɛn − 1 )) = (GĠ)′ ,and the solution of this O.D.E is given by the function:√1 − ɛ 1 − cos(G(s) = cos(n − 1 s) + √sin(√1−ɛn−1 r)1−ɛr) n−1√1 − ɛsin(n − 1 s).This function does not explode for r in [0,qπ2 1−ɛn−1], and,(Ġ)(r) − (Ġ)(0) = −2√1 − ɛn − 1 tan( √1−ɛr n−12).8823