1.1. The method of characteristics 9 Gronwall’s lemma implies ii. Proof of Theorem 1.1.1. The change of variable shows that it suffices to consider the case of A = diag(c1,...,cN). The solution u is constructed as a limit of approximate solutions un. The solution u0 is defined as the solution of the initial value problem ∂tu0 + A ∂xu0 = f, u0|t=0 = g . The solution is explicit by the method of characteristics. It is Ck with bounded derivatives on [0,T] × R, so (1.1.10) ∃C1, ∀t [0,T], mk(u0,t) C1 . For n 0 the solution un is again explicit by the method of character- istics in terms of un−1, (1.1.11) ∂tun + A ∂xun + B un−1 = f , un−1| t=0 = g . Using (1.1.10) and Haar’s inequality yields, (1.1.12) ∃C2, ∀t [0,T], mk(u1,t) C2 . For n 2 estimate un un−1 by applying Haar’s inequality to ˜ (un un−1) + B (un−1 un−2) = 0, (un un−1) t=0 = 0 , to find (1.1.13) mk(un un−1,t) C t 0 mk(un−1 un−2,σ) . For n = 2, this together with (1.1.10) and (1.1.12) yields mk(u2 u1,t) (C1 + C2)Ct . Injecting this in (1.1.13) yields mk(u3 u2,t) (C1 + C2)C2t2/2 . Continuing yields (1.1.14) mk(un un−1,t) (C1 + C2)Cn−1tn−1/(n 1)! . The summability of the right-hand side implies (Weierstrass’s M-test) that un and all of its partials of order k converge uniformly on [0,T] × R. The limit u is Ck with bounded partials. Passing to the limit in (1.1.11) shows that u solves the initial value problem. To prove uniqueness, suppose that u and v are solutions. Haar’s inequal- ity applied to u v implies that u v = 0.
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