1. THE LOCAL THEORY OF THE CAUCHY PROBLEM 3 unique solution in R3 × I, with (u, ∂tu) ∈ C I ˙ 1 × L2 , so that u S(I) 2δ, D 1 2 u W (I) + D− 1 2 ∂tu W (I) ∞, sup t∈I (u(t),∂tu(t)) ˙ 1 ×L2 ≤ C(A), and in addition u L5L10 I x ∞. Also, if (u0,k,u1,k) k→∞ → (u0,u1) in ˙ 1 × L2, then (uk,∂tuk) → (u, ∂tu) in C I ˙ 1 × L2 . For the proof, we use the Strichartz estimates and the consequence to the Chain Rule, combined with the fact that |F (u)| |u|4 , |F (u)| |u|3, to obtain the Theorem using the contraction mapping principle, in the space where v S(I) ∞ and D 1 2 v W (I) ∞. To obtain the fact that u L5L10 I x ∞, we use the Strichartz estimate (S1). Remark 1.5. Note that if u(1),u(2) are solutions of (NLW) on I, (u(1)(0),∂tu(1)(0)) = u(2)(0),∂tu(2)(0) , then u(1) ≡ u(2) on R3 × I. Indeed, let supt∈I ( u(i)(t),∂tu(i)(t) ) ˙ 1 ×L2 ≤ A. Partition now I into intervals Ij so that u(i) S(Ij) ≤ a, D 1 2 u(i) W (Ij) ≤ b, where a = a(A),b = b(A) are so small that the fixed point argument applies. Then if 0 ∈ Ij 0 , using the uniqueness of the fixed point in Ij 0 , we have u(1) = u(2) on Ij 0 . Moving in j our conclusion follows. Thus, there exists a maximal interval I = I(u0,u1) = (T−(u0,u1),T1(u0,u1)) with u ∈ C I ˙ 1 × L2 ∩ D 1 2 u ∈ W (I ) ∩ {u ∈ S(I )} for each I ⊂⊂ I, where the solution is defined. I is called the maximal interval of existence. Moreover, it is easy to see, using the Strichartz estimates that we also have ∂tD− 1 2 u ∈ W (I ),I ⊂⊂ I, since D 1 2 F (u) ∈ L 4 3 I Lx 4 3 . Remark 1.6. There exists δ such that if (u0,u1) ˙ 1 ×L2 δ, the assumption in Theorem 1.4 holds with I = R. This is a consequence of the Strichartz estimates. Also, given (u0,u1) ∈ ˙ 1 × L2, there exists I(0 ∈ I) such that the hypothesis of Theorem 1.4 holds on I. Again this is from Strichartz. Remark 1.7 (Higher regularity). If (u0,u1) ∈ ˙ 1 × L2 ∩ ˙ 1+μ × ˙ μ , 0 ≤ μ, then (u, ∂tu) ∈ C I ˙ 1 ∩ ˙ 1+μ × L2 ∩ ˙ μ and D 1 2 +μ u W (I ) + ∂tD− 1 2 +μ u W (I ) ∞. Remark 1.8 (Standard finite blow-up criterion). If T+(u0,u1) +∞, we must have u S[0,T+(u0,u1)) = +∞. If not, M = u S[0,T+(u0,u1)) . For ε 0 to be chosen, partition [0,T+(u0,u1) = ∪γ(ε,M)Ij, j=1 so that u S(Ij) ≤ ε. Let Ij = [tj,tj+1) and use the integral equation, the Strichartz estimates and the chain rule to obtain, supt∈I j (u(t),∂tu(t)) ˙ 1 ×L2 + D 1 2 u W (Ij) + u S(Ij) ≤ C (u(tj),∂tu(tj)) ˙ 1 ×L2 + C u 4 S(Ij) D 1 2 u W (Ij) . If Cε4 ≤ 1 2 , we show induc- tively that sup t∈[0,T+(u0,u1)) (u(t),∂tu(t))) ˙ 1 ×L2 + D 1 2 u W ([0,T+(u0,u1)) + u S([0,T+(u0,u1)) ≤ C(M).

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2015 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.