1. THE LOCAL THEORY OF THE CAUCHY PROBLEM 13 so that v x lt √1 1 l2 , x , t lx1 1 l2 = e i(x 1 √−lt)ξ1 1−l 2 e i(t−lx 1 ) ξ2+|ξ 1 |2 1−l 2 · eix ξ f(ξ)dξ1dξ = e ix 1 ( √ξ1−l|ξ|) 1−l 2 · e √iltξ1 1−l 2 · e it|ξ| 1−l 2 eix ξ f(ξ)dξdξ = e ix 1 1 −l|ξ|) 1−l 2 eix ·ξ gt(ξ)dξ1dξ , where gt(ξ) = e √iltξ1 1−l2 e it|ξ| 1−l2 f(ξ). Define now η1 = ξ1−l|ξ| 1−l2 , η = ξ , and compute = det 1− 1 |ξ| 1−l2 lξ2 |ξ| 1−l2 lξ3 |ξ| 1−l2 0 1 0 0 0 1 = 1 lξ1 |ξ| 1 l2 1, since |l| 1. The result now follows from the Plancherel Theorem. Remark 1.24. A density argument shows that, in fact, t w x1−lt 1−l2 , x , t−lx1 1−l2 C R ˙ 1 and similarly for ∂tw. We now let l (−1, 1), define for (y, s) R3 × R, (x, t) = Φl(y, s) = y + ls √1 1 l2 , y , s + ly1 1 l2 , where x = (x2,x3) , y = (y2,y3). Thus, (y, s) = Φ−1(x, l t) = x1−lt 1−l2 , x , t−lx1 1−l2 . We then define, for a globally defined solution u of (NLW) (i.e., I(u) = (−∞, +∞)), its Lorentz transform ul by ul(x, t) = u x lt √1 1 l2 , x , t lx1 1 l2 = u ( Φ−1(x, l t) ) . Since u is global in time, ul is well defined, say as an element of L8 loc (R4). (It was this fact that led us in [62] to the notion of solution based on L8). We will now prove that ul is indeed a global solution of (NLW) Lemma 1.25. Let u be a global, finite-energy solution of (NLW). Then, ul is also a global, finite-energy solution of (NLW). We start out with a preliminary Lemma: Lemma 1.26. Let u be a global solution of (NLW), which scatters in both time directions. Let ul be defined as above. Then, ul is a global solution of (NLW), scattering in both time directions. Proof. Since u scatters in both time directions, then by a previous remark, u L5L10 t x ∞. By the integral equation and Lemma 1.23 and Remark 1.24, (ul,∂tul) C R ˙ 1 × L2 . Furthermore, since u L8 ( R3 × R ) , a change of variables gives ul L8 ( R3 × R ) . If (u0,u1) C0 ( R3 ) × C0 (R3), it is easy to see
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