Smooth solutions 293
S : U → Sym
n
such that S(u) is positive-definite, and the matrices S(u)A
j
(u)
are symmetric for all u ∈ U .
Example The system (10.1.4) admits the diagonal symmetrizer
diag (1/(ρc
2
),ρ,···,ρ,1)
in the domain {ρ>0 ,c
2
> 0}, that is, away from vacuum and where the sound
speed is well-defined (which corresponds to hyperbolicity).
In more generality, systems admitting a strictly convex entropy are known
to be Friedrichs-symmetrizable – see [46, 184] for an introduction to the notion
of mathematical entropy and an extensive analysis of systems endowed with an
entropy. For this reason, since mathematical entropies are often related to phys-
ical entropy (or energy), most ‘physical’ systems are Friedrichs-symmetrizable.
In the case of Euler equations, various symmetrizations are discussed in Section
13.2.3. Also, see [71, 167], concerning the symmetrization of more complicated
systems, or systems endowed with non-convex entropies [21,46, 188].
The main result regarding the Cauchy problem for Friedrichs-symmetrizable
systems is the following. It was proved by several authors independently [67,94].
Theorem 10.1 Let U be an open subset of R
n
containing 0. We assume that
A
j
and b are C
∞
functions of u ∈ U , all vanishing at 0, and that (10.1.3) is
Friedrichs-symmetrizable in U . We consider the Cauchy problem associated with
(10.1.3) and initial data g ∈ H
s
(R
d
) with s>
d
2
+1 taking values in U .There
exists T>0 such that (10.1.3) has a unique classical solution u ∈ C
1
(R
d
× [0,T])
achieving the initial data u(0) = g. Furthermore, u belongs to C ([0,T]; H
s
) ∩
C
1
([0,T]; H
s−1
).
Observe that this result does not apply straight to the Euler equations, which
in general are not symmetrizable up to ρ = 0. But a slightly modified version of
Theorem 10.1, allowing solutions that do not vanish at infinity, does apply: it
suffices to replace 0 by some u
0
,takeg ∈ u
0
+ H
s
and obtain u(t) ∈ u
0
+ H
s
.
See Section 13.3 for more details on the Euler equations.
Proof The proof is based on the iteration scheme
S(u
k
) ∂
t
u
k+1
+
d
j=1
S(u
k
) A
j
(u
k
) ∂
j
u
k+1
= S(u
k
) b(u
k
). (10.1.5)
Since we are looking for smooth solutions of (10.1.3), we are only concerned
with smooth solutions of (10.1.5). By Theorem 2.12 in Chapter 2 we know that
(10.1.5) admits a C
∞
solution u
k+1
provided that u
k
and the initial data are
C
∞
. This is why we are going to smooth back the initial data.
We introduce a mollifier ρ
k
, being defined as usual by
ρ
k
(x)=ε
−d
k
ρ
x
ε
k
,