8 1. Simple Examples of Propagation
ii. More generally, there is a constant C(k, T, L) so that if for all |α| k,
∂t,xu
α
and ∂t,xLu
α
are bounded continuous functions on [0,T] × R, then
mk(u, t) :=
|α|≤k
∂t,xu(t)
α
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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