1.1. The method of characteristics 9
Gronwall’s lemma implies ii.
Proof of Theorem 1.1.1. The change of variable shows that it suffices to
consider the case of A = diag(c1,...,cN ).
The solution u is constructed as a limit of approximate solutions
un.
The solution
u0
is defined as the solution of the initial value problem
∂tu0
+ A
∂xu0
= f,
u0|t=0
= g .
The solution is explicit by the method of characteristics. It is
Ck
with
bounded derivatives on [0,T] × R, so
(1.1.10) ∃C1, ∀t [0,T],
mk(u0,t)
C1 .
For n 0 the solution
un
is again explicit by the method of character-
istics in terms of
un−1,
(1.1.11)
∂tun
+ A
∂xun
+ B
un−1
= f ,
un−1|t=0
= g .
Using (1.1.10) and Haar’s inequality yields,
(1.1.12) ∃C2, ∀t [0,T],
mk(u1,t)
C2 .
For n 2 estimate
un

un−1
by applying Haar’s inequality to
˜
L
(un

un−1)
+ B
(un−1

un−2)
= 0,
(un

un−1)
t=0
= 0 ,
to find
(1.1.13)
mk(un

un−1,t)
C
t
0
mk(un−1

un−2,σ)
.
For n = 2, this together with (1.1.10) and (1.1.12) yields
mk(u2

u1,t)
(C1 + C2)Ct .
Injecting this in (1.1.13) yields
mk(u3

u2,t)
(C1 +
C2)C2t2/2
.
Continuing yields
(1.1.14)
mk(un

un−1,t)
(C1 +
C2)Cn−1tn−1/(n
1)! .
The summability of the right-hand side implies (Weierstrass’s M-test)
that
un
and all of its partials of order k converge uniformly on [0,T] × R.
The limit u is
Ck
with bounded partials. Passing to the limit in (1.1.11)
shows that u solves the initial value problem.
To prove uniqueness, suppose that u and v are solutions. Haar’s inequal-
ity applied to u v implies that u v = 0.
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