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

−1Lu

= M

−1

∂t + MDM

−1∂x

+ B M v .

When the derivatives on the right fall on v, the product M

−1M

= I simpli-

fies. This shows that

M

−1Lu

= ∂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

−1f

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 ∂x

βg

∈

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 .