66 Linear Cauchy problem with variable coefficients
Regularity If f belongs to C
∞
([0,T]; H
+∞
(R
d
)) and g belongs to H
+∞
(R
d
)
then u belongs to C ([0,T]; H
+∞
(R
d
)) from the above construction applied to
arbitrary large s. The equation ∂
t
u = P (t)u + f then implies that u belongs
to C
k
([0,T]; H
+∞
(R
d
)) for all k ∈ N.Iff and g are less regular, we can use
mollifiers to construct sequences f
k
∈ D (R
d
× [0,T]) and g
k
∈ D (R
d
) such that
f
k
L
2
(0,T ;H
s
)
−−−−−−−→
k→∞
f, g
k
H
s
−−−−→
k→∞
g.
For all k there is a solution u
k
∈ C
∞
([0,T]; H
+∞
(R
d
)) corresponding to the
source term f
k
and the initial data g
k
. Applying the estimate in (2.1.8) to u
k
−
u
m
yields the inequality
u
k
(t) − u
m
(t)
2
H
s
≤ C
g
k
− g
m
2
H
s
+
t
0
f
k
− f
m
(τ)
2
H
s
dτ
for all k, m ∈ N.Thisimpliesthat(u
k
) is a Cauchy sequence in C ([0,T]; H
s
)
and thus converges, say towards u ∈ C ([0,T]; H
s
). In the limit we have Lu = f
and u(0) = g. By uniqueness (in C ([0,T]; H
s−1
)), we have u = u, the solution
constructed by a duality argument. Observe then that u = u satisfies the energy
estimate in (2.1.14) by passing to the limit in (2.1.8) applied to u
k
. This completes
the proof.
Remark 2.8 The end of this proof is sometimes referred to as a weak=strong
argument. Indeed, if we say a strong solution is a solution that is the limit of
infinitely smooth solutions of regularized problems, and a weak solution the one
obtained by duality, it shows that any weak solution is necessarily a strong one.
Remark 2.9 This theorem is also valid backward, that is, prescribing u(T )
instead of u(0). This fact will be used in the application of the Holmgren Principle
below.
Remark 2.10 In the proof above, we have crucially used the infinite smooth-
ness of coefficients. In fact, H
s
-well-posedness is also true for H
s
coefficients
provided that s is large enough. We reproduce from [132] the corresponding
precise result below, which is the continuation of Theorem 2.4 and is useful in
non-linear analysis.
Theorem 2.7 Assume s>
d
2
+1 and take T>0. For a function v belonging
to L
∞
([0,T]; H
s
(R
d
)) ∩ C ([0,T]; H
s
w
(R
d
)) and such that
∂
t
v ∈ L
∞
([0,T]; H
s−1
(R
d
)) ∩ C ([0,T]; H
s−1
w
(R
d
)),
we consider the differential operator
L
v
= ∂
t
+
α
A
α
(v(x, t)) ∂
α
,
where the n × n matrices A
α
are C
∞
functions of their argument v ∈ R
n
.