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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.