1.1. The method of characteristics 7 3. If A1 and A2 satisfy Hypothesis 1.1.2. then so does A1 0 0 A2 with M := M1 0 0 M2 . In this way one constructs examples with variable multiplicity. Exercise 1.1.2. Prove assertions 1 and 2. Since A = MDM −1 with D = diag (c1,...,cN), L = ∂t + MDM −1 ∂x + B . Define v by u = Mv so M −1 Lu = M −1 ∂t + MDM −1 ∂x + B M v . When the derivatives on the right fall on v, the product M −1 M = I simpli- fies. This shows that M −1 Lu = ∂t + D ∂x + B u := L v, where B := M −1 B M + M −1 Mt + M −1 A Mx . This change of variable converts the equation Lu = f to Lv = M −1 f where L has the same form as L but with leading part that is a set of directional derivatives. Theorem 1.1.1. Suppose that Hypothesis 1.1.2 is satisfied, k ≥ 1, f ∈ Ck([0,T] × R), g ∈ Ck(R), and for all α, β with |α| ≤ k and |β| ≤ k, ∂t.xf α ∈ L∞([0,T] × R) and ∂xg β ∈ L∞([0,T] × R) . Then there is a unique solution u ∈ Ck([0,T]×R) to the initial value problem Lu = f, u|t=0 = g so that all partial derivatives of u of order ≤ k are in L∞([0,T] × R). The crux is the following estimate called Haar’s inequality. For a vector valued function w(x) = (w1(x),...,wN(x)) on R, the L∞ norm is taken to be w L∞(R) := max 1≤j≤N wj(x) L∞(R) . Haar’s Lemma 1.1.2. Suppose that Hypothesis 1.1.2 is satisfied. i. There is a constant C = C(T, L) so that if u and Lu are bounded continuous functions on [0,T] × R, then for t ∈ [0,T] u(t) L∞(R) ≤ C u(0) L∞(R) + t 0 Lu(σ) L∞(R) dσ .

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