Communications In Mathematical Physics - Volume 282 by M. Aizenman (Chief Editor)

By M. Aizenman (Chief Editor)

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N + 1. Thus, z N +1 z N z 1 ... 24) h( p) + E 1 ( p)β0 ( p) + · · · + E N +1 ( p)β0 ( p) . . β N ( p) We do not expand β terms in p as a Taylor series. However, we can Taylor expand E i terms in p. In this way, we obtain h( p) = C1 p 2 + 1 M d+κ + O( p 4 ), h( p) + E 1 ( p)β0 ( p) + · · · + E N +1 ( p)β0 ( p) . . 26) with the real valued function h 1 ( p) satisfying h 1 ( p) > C > 0 for all p. 24) becomes 1 e2 p=0 log(C 1 p 1 = e2 2+ 1 M d+κ p=0 log(C 1 p +O( p4 ))−log(h 1 ( p) p2 + 2 +O( p 4 ))−log(h 1 ( p) p 1 M d+κ +O( p4 )) 2 +O( p 4 ))+log(1+ 1 1 )−log(1+ d+κ ) M d+κ (C1 p 2 +O( p 4 )) M (h 1 ( p) p 2 +O( p 4 )) Towards a Nonperturbative Renormalization Group Analysis =e e 1 2 49 C1 O( p4 ) O( p4 ) p=0 log( h 1 ( p) )+log(1+ C p2 )−log(1+ h ( p) p2 ) 1 1 1 2 1 1 p=0 log(1+ M d+κ (C p 2 +O( p4 )) )−log(1+ M d+κ (h ( p) p2 +O( p4 )) ) 1 1 .

If we take now σ = σn ≡ − log(T (xn ) − τn ), then we see from (39) that s = σn ≡ − log(−τn ) and from (34) and (30) that as n → ∞, eσn (d(0)xn − T (xn )) → − λ(δ1 − δ0 ) , δ1 (λ + d(0) + δ0 ) δ1 − δ0 δ1 (δ0 + d(0) + λ) eσn xn → and 1 + eσn T (xn ) → >0 δ0 (δ1 + d(0) + λ) δ0 (δ1 + d(0) + λ) (41) because of (31). ) 0 − 1 H 1 ×L 2 (Jn ) , (42) by (33). 2 and (42). 2.

1. Our goal is to prove that T (x) is differentiable when x = 0 and that T (0) = d(0). We proceed by contradiction. From the fact that T (x) is 1-Lipschitz, we assume that there is a sequence xn such that xn → 0 and T (xn ) → d(0) + λ with λ = 0 as n → ∞. xn (30) Up to extracting a subsequence and to considering u(−x, t) (also solution to (1)), we can assume that xn > 0. Since 0 is non-characteristic, we see from (20) and (21) that λ + d(0) ≥ −δ0 > −δ0 . 4 Let δ1 = wxn (σn ) ∂s wxn (σn ) 63 1+δ0 2 .

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