8 1. Simple Examples of Propagation ii. More generally, there is a constant C(k, T, L) so that if for all |α| k, ∂α t,x u and ∂α t,x Lu are bounded continuous functions on [0,T] × R, then mk(u, t) := |α|≤k ∂α t,x u(t) L∞(R) satisfies for t [0,T], mk(u, t) C mk(u, 0) + t 0 mk(Lu, σ) . Proof of Lemma 1.1.2. The change of variable shows that it suffices to consider the case of a real diagonal matrix A = diag (c1(t, x),...,cN(t, x)). i. For t [0,T] and 0 choose j and x so that u(t) L∞(R) uj(t,x) L∞(R) + . Choose (t, x(t)) the integral curve of x = cj(t, x) passing through t,x. Then uj(t,x) = uj(0,x(0)) + t 0 (∂t + cj(t, x)∂x)uj(σ, x(σ)) . Therefore u(t) L∞(R) u(0) L∞(R) + t 0 (Lu B u)(σ) L∞(R) + . Since this is true for all , one has u(t) L∞(R) u(0) L∞(R) + t 0 (Lu)(σ) L∞(R) + C u(σ) L∞(R) , and i follows using Gronwall’s Lemma 2.1.3. ii. Apply the inequality of i to ∂t,xu α with |α| k. ∂αu(t) L∞(R) C ∂αu(0) L∞(R) + t 0 L∂αu(σ) L∞(R) dσ. Compute L ∂αu = ∂αLu + [L, ∂α] u . The commutator is a differential operator of order k with bounded coeffi- cients, so [L, ∂α] u(σ) L∞(R) C mk(u, σ) . Therefore, ∂αu(t) L∞(R) C ∂αu(0) L∞(R) + t 0 ∂αLu(σ) L∞(R) + mk(u, σ) . Sum over |α| k to find mk(u, t) C mk(u, 0) + t 0 mk(u, σ) + mk(Lu, σ) .
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