1. THE LOCAL THEORY OF THE CAUCHY PROBLEM 15 (NLW). Furthermore, by finite speed of propagation, u(y, s) = u(y, s), i.e., ul(x, t) = ul(x, t) for (y, s) such that |y| ≥ A + |s|. We claim that |x| ≥ ClA + |t| =⇒ |y| ≥ A + |s| . Indeed by (1.28) and (1.29), |y| − |s| = |y|2 − |s|2 |y| + |s| = |x|2 − |t|2 |y| + |s| ≥ |x|2 − |t|2 Cl (|x| + |t|) = |x| − |t| Cl . The conclusion (1.30) with v = ul,B = ClA follows. Note that as a consequence of (1.30), we have (1.31) R |x| |t|+B |ul(x, t)|10 dx 1 2 dt ∞. Step 2: Local estimate. Let (X, T ) ∈ R3×R. We show that there exists ε 0 and a scattering solution v of (NLW) such that (1.32) |x − X| ≤ ε − |t − T | =⇒ ul(x, t) = v(x, t). Indeed, let (Y, S) = Φ−1 l (X, T ). Let Ψ ∈ C∞(R3), 0 Ψ(y) = 1 if |y| ≤ 1, Ψ(y) = 0 for |y| ≥ 2. Let (u0, u1) = Ψ y−Y η (u(y, S),∂tu (y, S)). Choose η 0 so small that (u0, u1) ˙ 1 ×L2 ≤ δ, δ as before. Let u be the solution of (NLW) with data: (u(S),∂tu(S)) = (u0, u1) . Then, u is globaly defined and scatters. By Lemma 1.26, its Lorentz transform ul is a scattering solution of (NLW). Note that by finite speed of propagation, |y − Y | η − |s − S| =⇒ ul(x, t) = ul(x, t). Furthermore, by (1.29), |y − Y | + |s − S| ≤ Cl (|x − X| + |t − T |), and thus |x − X| ≤ η Cl − |t − T | =⇒ |y − Y | ≤ η − |s − S| . Thus, the desired conclusion (1.32) follows. Again (1.32) implies (1.33) T +ε Tε |x−X|ε−|t−T | |ul(x, t)|10 dx 1 2 dt ∞. Step 3: End of proof. Combining Step 1 and Step 2, we get that (ul,∂tul) ∈ C R ˙ 1 × L2 . By (1.31) and (1.33), ul ∈ L5(L10), I x ∀I ⊂⊂ R. Furthermore, by Step 1 and 2 and Remark 1.18, ∂t 2 ul − Δul = u5 l in the distributional sense. Hence, Lemma 1.21 yields the result.
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.