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)
.
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