Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift
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288 D. Guidetti
(I) there exists C>0, depending only on T and α, such that
K
k
(t)g
α,p,Ω
≤ Ct
−
α−β
2m
(g
2m+β−μ
k
,p,Ω
+ t
−
2m+β−μ
k
−θ
2m
g
θ,p,Ω
);
(II) K
k
∈ C
1
(R
+
; L(N
2m+β−μ
k
p
(Ω); N
α
p
(Ω)) and
K
k
(t)=(2πi)
−1
γ
e
λt
λN
k
(λ)gdλ; (3.8)
(III) K
k
(t)g
α,p,Ω
≤ Ct
−1−
α−β
2m
(g
2m+β−μ
k
,p,Ω
+ t
−
2m+β−μ
k
−θ
2m
g
θ,p,Ω
);
(IV) ∀t ∈ R
+
[A(., ∂
x
)+P ]K
k
(t)g = K
k
(t)g;
(V) ∀t ∈ R
+
, ∀i ∈{1,...,m}, B
i
(., ∂
x
)K
k
(t)g
|∂Ω
=0;
(VI) if ξ is positive and sufficiently large, ξ ∈ ρ(A) and
(ξ − A)
−1
K
k
(t)g =(2πi)
−1
γ
e
λt
(ξ − λ)
−1
N
k
(λ)gdλ.
Proof. Modifying γ, if necessary, we may assume, owing to Cauchy’s theorem, that
|λ|≥Tr
0
∀λ ∈ γ. So, by Remark 2.9,
K
k
(t)g
α,p,Ω
= (2πi)
−1
t
−1
γ
e
λt
N
i
(λ)gdλ
α,p,Ω
= (2πit)
−1
γ
e
λ
N
i
(t
−1
λ)gdλ
α,p,Ω
≤ Ct
−1
γ
e
Re(λ)
|t
−1
λ|
α−β
2m
−1
(g
2m+β−μ
k
,p,Ω
+ |t
−1
λ|
2m+β−μ
k
−θ
2m
g
θ,p,Ω
)|dλ|,
which implies the conclusion.
(II) is obvious.
(III) can be obtained from (II), with the same method of (I).
(IV) follows from
K
k
(t)g =(2πi)
−1
γ
e
λt
[A(., ∂
x
)+P ]N
k
(λ)gdλ
=(2πi)
−1
γ
e
λt
λN
k
(λ)gdλ.
(V) follows from
B
i
(., ∂
x
)K
k
(t)g
|∂Ω
=(2πi)
−1
γ
e
λt
[B
i
(., ∂
x
)N
k
(λ)g]
|∂Ω
dλ
=(2πi)
−1
γ
e
λt
δ
ik
g
|∂Ω
dλ =0.
(VI) The first statement is a consequence of the fact that A is a sectorial
operator. Concerning the formula, observe first that, if ξ = λ,
(ξ − A)
−1
N
k
(λ)g =(ξ − λ)
−1
[N
k
(λ) −N
k
(ξ)]g.
On Linear Elliptic and Parabolic etc. 289
It follows
(ξ − A)
−1
K
k
(t)g =(2πi)
−1
γ
e
λt
(ξ − λ)
−1
N
k
(λ)gdλ − (2πi)
−1
×
γ
e
λt
(ξ − λ)
−1
N
k
(ξ)gdλ,
and the second integral is 0.
We deduce from Lemma 3.1, by choosing θ>β− ν, that, at least as an
element of L(N
2m+β−μ
k
p
(Ω),N
β
p
(Ω)), we can define the following operator
K
(−1)
k
(t):=
t
0
K
k
(s)ds =(2πi)
−1
γ
e
λt
λ
N
k
(λ)dλ. (3.9)
We have:
Lemma 3.2. Let T ∈ R
+
, k ∈{1,...,m}, g ∈ N
2m+β−μ
k
p
(Ω), β
≤ α ≤ 2m + β,
θ ∈ (0,p
−1
),withν + θ ≥ β.Then:
(I) there exists C>0, depending only on T and α, such that
K
(−1)
k
(t)g
α,p,Ω
≤ Ct
1−
α−β
2m
(g
2m+β−μ
k
,p,Ω
+ t
−
2m+β−μ
k
−θ
2m
g
θ,p,Ω
);
(II) ∀t ∈ R
+
[A(., ∂
x
)+P ]K
(−1)
k
(t)g = K
k
(t)g;
(III) ∀t ∈ R
+
, ∀i ∈{1,...,m}, B
i
(., ∂
x
)K
(−1)
k
(t)g
|∂Ω
= δ
ik
g
|∂Ω
.
Proof. (I) can be obtained with the same arguments in the proof of Lemma 3.1.
(II) follows from
K
(−1)
k
(t)g =(2πi)
−1
γ
e
λt
λ
[A(., ∂
x
)+P ]N
k
(λ)gdλ
=(2πi)
−1
γ
e
λt
N
k
(λ)gdλ.
Concerning (III),
B
i
(., ∂
x
)K
(−1)
k
(t)g
|∂Ω
=(2πi)
−1
γ
e
λt
λ
[B
i
(., ∂
x
)N
k
(λ)g]
|∂Ω
dλ
=(2πi)
−1
γ
e
λt
λ
δ
ik
g
|∂Ω
dλ = δ
ik
g
|∂Ω
.
After these preliminaries, we pass to consider system (3.1). We begin with
the case u
0
=0andg
k
≡ 0foreachk =1,...,m. Here we have ([7], Theorem 2.7)
that
N
β
p
(Ω) = (N
β
p
(Ω),D(A
β
))
β−β
2m
. (3.10)
290 D. Guidetti
As A
β
is a sectorial operator, the following characterization of N
β
p
(Ω) holds (see
[9], Proposition 2.2.2): is [ξ
0
, ∞) ⊆ ρ(A
β
),
N
β
p
(Ω) =
f ∈ N
β
p
(Ω) : sup
ξ≥ξ
0
ξ
β−β
2m
A
β
(ξ − A
β
)
−1
f
β
,p,Ω
< ∞
. (3.11)
Moreover, an equivalent norm in N
β
p
(Ω) is
f →f
β
,p,Ω
+sup
ξ≥ξ
0
ξ
β−β
2m
A
β
(ξ − A
β
)
−1
f
β
,p,Ω
.
Employing Theorem 4.3.8 in [9], we deduce the following
Lemma 3.3. Assume that the assumptions (I1)–(I4) are satisfied. Consider the
system (3.1) in case f ∈ C([0,T]; N
β
p
(Ω)) ∩ B([0,T]; N
β
p
(Ω)), g
k
≡ 0 for each
k =1,...,m,andu
0
=0. Then there is a unique solution u belonging to C
1
([0,T];
N
β
p
(Ω))∩B([0,T]; N
2m+β
p
(Ω)),withD
t
u ∈ B([0,T]; N
β
p
(Ω)). u can be represented
by the variation of parameter formula
u(t)=
t
0
T (t − s)f(s)ds, t ∈ [0,T]. (3.12)
Next, we consider the case f ≡ 0, u
0
=0,g
i
≡ 0ifi = k (k ∈{1,...,m}).
We shall often use the following
Lemma 3.4. Let T ∈ R
+
,letβ
0
, β
1
be real numbers, such that β
0
<β
1
, ρ ∈ R
+
,
p ∈ [1, +∞],andletg ∈ B([0,T]; N
β
1
p
(Ω))∩C
ρ
([0,T]; N
β
0
p
(Ω)).Then,ifξ ∈ (0, 1)
and (1 − ξ)ρ ∈ Z, g ∈ C
(1−ξ)ρ
([0,T]; N
(1−ξ)β
0
+ξβ
1
p
(Ω)). Moreover, there exists
C>0, independent of g, such that
g
C
(1−ξ)ρ
([0,T ];N
(1−ξ)β
0
+ξβ
1
p
(Ω))
≤ C(g
B([0,T ];N
β
1
p
(Ω))
+ g
C
ρ
([0,T ];N
β
0
p
(Ω))
).
Proof. Employing the characterization of C
(1−ξ)ρ
([0,T]; N
(1−ξ)β
0
+ξβ
1
p
(Ω)) by hi-
gher-order differences (see [14], 3.4.2), we have, taking l ∈ N, l ≥ β
1
, h ∈ (0,T/l]
and t ∈ [0,T − lh],
h
−(1−ξ)ρ
#
#
#
#
l
!
j=0
l
j
(−1)
l−j
g(t + jh)
#
#
#
#
(1−ξ)β
0
+ξβ
1
,p,Ω
≤ C
0
h
−ρ
#
#
#
#
l
!
j=0
l
j
(−1)
l−j
g(t + jh)
#
#
#
#
β
0
,p,Ω
1−ξ
#
#
#
#
l
!
j=0
l
j
(−1)
l−j
g(t + jh)
#
#
#
#
ξ
β
1
,p,Ω
≤ C
1
g
1−ξ
C
ρ
([0,T ];N
β
0
p
(Ω))
g
ξ
B([0,T ];N
β
1
p
(Ω))
.
Now we consider the first case with nonhomogeneous boundary conditions.
Lemma 3.5. Assume that the assumptions (I1)–(I4) are satisfied. Consider the
system (3.1) in case f ≡ 0, u
0
=0, g
i
≡ 0 for each i =1,...,m and i = k,
On Linear Elliptic and Parabolic etc. 291
g
k
= g. We assume that g ∈ B([0,T]; N
2m+β−μ
k
p
(Ω))∩C
2m+β−μ
k
−θ
2m
([0,T]; N
θ
p
(Ω)))
for some θ ∈ (0,p
−1
),andg(0) = 0. Consider the function
u(t):=
t
0
K
k
(t − s)g(s)ds. (3.13)
Then:
(I) u is well defined in [0,T];
(II) u belongs to
C
1
([0,T]; N
β
p
(Ω)) ∩ B([0,T]; N
2m+β
p
(Ω))),D
t
u ∈ B([0,T]; N
β
p
(Ω)));
(III) u satisfies (3.1).
Proof. By Lemma 3.4, it is not restrictive to assume that
θ>β−ν ≥ β − μ
k
, (3.14)
in such a way that
2m+β−μ
k
−θ
2m
< 1. Now we observe that
u(t)=K
(−1)
k
(t)g(t)+
t
0
K
k
(t −s)[g(s) − g(t)]ds := v
0
(t)+v
1
(t). (3.15)
From Lemma 3.2 and the assumptions on g, we deduce that v
0
∈ B([0,T]; N
2m+β
p
(Ω)) and
[A(., ∂
x
)+P ]v
0
(t)=K
k
(t)g(t), (3.16)
∀i ∈{1,...,m},
B
i
(., ∂
x
)v
0
(t)
|∂Ω
= δ
ik
g(t)
|∂Ω
. (3.17)
Now we consider v
1
. By Lemma 3.1, we have that, at least, v
1
∈ B([0,T]; N
2m+β
p
(Ω)). Moreover,
[A(., ∂
x
)+P ]v
1
(t)=
t
0
K
k
(t −s)[g(s) − g(t)]ds, (3.18)
and, for each i ∈{1,...,m},
B
i
(., ∂
x
)v
1
(t)
|∂Ω
=0. (3.19)
In order to prove that v
1
is bounded with values in B([0,T]; N
2m+β
p
(Ω)), we
recall again (3.11). Owing to (3.19), recalling that the domain of A is a sub-
space of N
2m+β
p
(Ω), we can try to show that A
β
v
1
is bounded with values in
(N
β
p
(Ω),D(A
β
))
β−β
2m
. This is equivalent to prove that, for some ξ
0
> 0,
ξ
β−β
2m
A
β
(ξ − A
β
)
−1
A
β
v
1
(t)
β
,p,Ω
≤ C, (3.20)
292 D. Guidetti
with C independent of ξ ≥ ξ
0
and t ∈ [0,T]. We have from Lemma 3.1
A
β
(ξ − A
β
)
−1
A
β
v
1
(t)=ξ
2
(ξ − A
β
)
−1
v
1
(t) − A
β
v
1
(t) − ξv
1
(t)
=(2πi)
−1
t
0
γ
e
λ(t−s)
λ
2
ξ − λ
N
k
(λ)[g(s) − g(t)]dλ
ds
=
1
!
j=0
(2πi)
−1
t
0
γ
j
e
λ(t−s)
λ
2
ξ − λ
N
k
(λ)[g(s) − g(t)]dλ
ds :=
1
!
j=0
I
j
(ξ,t).
with γ
0
describing {λ ∈ C \{0} : |Arg(λ)| = θ
1
, |λ|≥r
0
}, γ
1
describing {λ ∈
C \{0} : |Arg(λ)|≤θ
1
, |λ| = r
0
}.
If λ ∈ γ, we have from Remark 2.9, recalling our assumptions on g,
N
k
(λ)[g(s) − g(t)]
β
,p,Ω
≤ C|λ|
β
−β
2m
−1
g
B([0,T ];N
2m+β−μ
k
p
(Ω))
+ |λ|
2m+β−μ
k
−θ
2m
(t −s)
2m+β−μ
k
−θ
2m
g
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
.
So we can easily deduce that
I
1
(ξ, t)
β
,p,Ω
≤ Cξ
−1
g
B([0,T ];N
2m+β−μ
k
p
(Ω))
,
if ξ ≥ ξ
0
,withξ
0
suitably large. Moreover,
I
0
(ξ,t)
β
,p,Ω
≤ C
0
t
0
R
+
e
r(t−s)cos(θ
1
)
r
1+
β
−β
2m
r + ξ
(g
B([0,T ];N
2m+β−μ
k
p
(Ω))
+ r
2m+β−μ
k
−θ
2m
(t − s)
2m+β−μ
k
−θ
2m
g
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
)dr
ds
≤ C
1
R
+
r
β
−β
2m
r + ξ
dr
g
B([0,T ];N
2m+β−μ
k
p
(Ω))
+ g
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
≤ C
2
ξ
β
−β
2m
g
B([0,T ];N
2m+β−μ
k
p
(Ω))
+ g
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
.
Next, we show that u ∈ C
1
([0,T]; N
β
p
(Ω)) and D
t
u =[A(., ∂
x
)+P]u.To
this aim, we follow a standard argument: let ∈ (0,T] and define, for t ∈ [, T ],
u
(t):=
t−
0
K
k
(t −s)g(s)ds. (3.21)
Then we have
D
t
u
(t)=K
k
()g(t − )+
t−
0
K
k
(t − s)g(s)ds (3.22)
= K
k
(t)g(t)+
t−
0
K
k
(t − s)[g(s) − g(t)]ds + K
k
()[g(t − ) − g(t)],
On Linear Elliptic and Parabolic etc. 293
which, by Lemma 3.1, converges in B([δ, T ]; N
β
p
), for every δ ∈ (0,T], to
K
k
(t)g(t)+
t
0
K
k
(t −s)[g(s) − g(t)]ds =[A(., ∂
x
)+P ]u(t), (3.23)
by (3.16) and (3.18). This can be extended to t = 0, observing that the final
expression in (3.23) converges to 0 in N
β
p
(Ω) as t → 0.
Inconclusion,wehaveprovedthatu ∈ C
1
([0,T]; N
β
p
(Ω)) ∩B([0,T]; N
2m+β
p
(Ω)). By interpolation, u ∈ C([0,T]; N
ρ
p
(Ω)) for every ρ<2m + β. Finally, from
D
t
u =[A(., ∂
x
)+P ]u we deduce that D
t
u ∈ B([0,T]; N
β
p
(Ω)).
The proof is complete.
Now we are able to prove the main result of existence and uniqueness of a
solution in the autonomous case.
Theorem 3.6. Assume that the assumptions (I1)–(I4) hold. We consider system
(2.1) and assume that:
(I) for some β
<β, f ∈ B([0,T]; N
β
p
(Ω)) ∩ C([0,T]; N
β
p
(Ω));
(II) for each k ∈{1,...,m},andforsomeθ ∈ (0,p
−1
), g
k
∈ B([0,T]; N
2m+β−μ
k
p
(Ω)) ∩ C
2m+β−μ
k
−θ
2m
([0,T]; N
θ
p
(Ω));
(III) u
0
∈ N
2m+β
p
(Ω);
(IV) for each k ∈{1,...,m}, (B
k
(., ∂
x
)u
0
)
|∂Ω
= g
k
(0)
|∂Ω
.
Then (3.1) has a unique solution u belonging to C
1
([0,T]; N
β
p
(Ω))∩B([0,T];
N
2m+β
p
(Ω))),withD
t
u ∈ B([0,T]; N
β
p
(Ω))). u can be represented by the variation
of parameter formula
u(t)=T (t)u
0
+
t
0
T (t − s)f(s)ds +
m
!
k=1
t
0
K
k
(t − s)g
k
(s)ds. (3.24)
Proof. The uniqueness follows from well-known properties of sectorial operators.
Concerning the existence, we already know, from Lemmas 3.3 and 3.5, that it holds
if u
0
=0andg
k
(0) = 0 for each k =1,...,m. Now we consider the general case.
Subtracting to the unknown u the initial value u
0
, and setting v(t):=u(t) − u
0
,
we are reduced to the system
D
t
v(t, x)=A(x, ∂
x
)v(t, x)+Pv(t, x)+f(t, x)+A(x, ∂
x
)u
0
(x)+Pu
0
(x),
(t, x) ∈ [0,T] × Ω,
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
B
k
(x
,∂
x
)v(t, x
) − (g
k
(t, x
) − B
k
(x
,∂
x
)u
0
(x
)) = 0,k=1,...,m,
(t, x
) ∈ [0,T] ×∂Ω
v(0,x)=0,x∈ Ω. (3.25)
From assumption (IV), we have that, if we set ˜g
k
:= g
k
−B
k
(., ∂
x
)u
0
,˜g
k
(0) vanishes
in ∂Ω. So, if we replace ˜g
k
with ˜g
k
− ˜g
k
(0), (3.25) does not change and we are
294 D. Guidetti
reduced to a case already treated. We deduce that (3.1) has a unique solution
with the desired properties.
It remains to show that the representation (3.24) holds. From (3.25) observing
that K
k
(t)g =0ifg
|∂Ω
≡ 0, we have
u(t)=u
0
+
t
0
T (t −s)f(s)ds +
m
!
k=1
t
0
K
k
(t − s)g
k
(s)ds (3.26)
+ T
(−1)
(t)[A(., ∂
x
)u
0
+ Pu
0
] −
m
!
k=1
K
(−1)
k
(t)B
k
(., ∂
x
)u
0
.
We observe that for every λ ∈ ρ(A)
u
0
=(λ − A)
−1
(λ −A(., ∂
x
) − P )u
0
+
m
!
k=1
N
k
(λ)B
k
(., ∂
x
)u
0
. (3.27)
So,
u
0
+ T
(−1)
(t)[A(., ∂
x
)u
0
+ Pu
0
] −
m
!
k=1
K
(−1)
k
(t)B
k
(., ∂
x
)u
0
= u
0
+(2πi)
−1
γ
e
λt
λ
{(λ − A)
−1
[A(., ∂
x
)u
0
+ Pu
0
] −
m
!
k=1
N
k
(λ)B
k
(., ∂
x
)u
0
}dλ
= u
0
+ T (t)u
0
− (2πi)
−1
γ
e
λt
λ
u
0
dλ = T (t)u
0
. (3.28)
So the proof is complete.
We precise Theorem 3.6:
Proposition 3.7. Consider the system (3.1), with the assumptions (I1)–(I4).Let
T
0
∈ R
+
.Then:
(I) if 0 <T≤ T
0
and θ>β−ν, the solution u satisfies the estimate
u
B([0,T ];N
2m+β
p
(Ω)
+ D
t
u
B([0,T ];N
β
p
(Ω)
≤ C(T
0
)
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
+ u
0
2m+β,p,Ω
.
(3.29)
(II) Assume that u
0
=0, θ>β−ν, β ≤ α ≤ 2m + β, 0 ≤ γ ≤ 1 −
α−β
2m
.Then,
u
B([0,T ];N
α
p
(Ω))
≤ C(α, T
0
)T
1−
α−β
2m
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
,
(3.30)
On Linear Elliptic and Parabolic etc. 295
u
C
γ
([0,T ];N
α
p
(Ω))
≤ C(α, T
0
)T
1−
α−β
2m
−γ
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
.
(3.31)
Proof. We extend f and g
k
to [0,T
0
], setting
˜
f(t)=
f(t)if 0≤ t ≤ T,
f(T )ifT ≤ t ≤ T
0
,
˜g
k
(t)=
g
k
(t)if 0≤ t ≤ T,
g
k
(T )ifT ≤ t ≤ T
0
.
As θ>β−ν,
2m+β−μ
k
−θ
2m
< 1foreachk =1,...,m,sothat
˜g
k
C
2m+β−μ
k
−θ
2m
([0,T
0
];N
θ
p
(Ω))
= g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
.
We call ˜u the corresponding solution (again with initial datum u
0
)in[0,T
0
]. Of
course, u is the restriction of ˜u to [0,T]. We deduce that
u
B([0,T ];N
2m+β
p
(Ω))
+ D
t
u
B([0,T ];N
β
p
(Ω))
≤˜u
B([0,T ];N
2m+β
p
(Ω))
+ D
t
˜u
B([0,T ];N
β
p
(Ω))
≤ C(T
0
)
˜
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
˜g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
˜g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
+ u
0
2m+β,p,Ω
= C(T
0
)
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
+ u
0
2m+β,p,Ω
.
(3.32)
Concerning (II), by (I) we have
u
B([0,T ];N
β
p
(Ω))
≤ T D
t
u
B([0,T ];N
β
p
(Ω))
≤ C(T
0
)T
f
B([0,T ];N
β
p
(Ω))
+
m
!
k=1
g
k
B([0,T ];N
2m−μ
k
+β
p
(Ω))
+
m
!
k=1
g
k
C
2m+β−μ
k
−θ
2m
([0,T ];N
θ
p
(Ω))
.
(3.33)
The intermediate cases follow from the interpolation inequalities.
296 D. Guidetti
We conclude extending Theorem 3.6 to the nonautonomous case. We shall
use the following
Theorem 3.8. Let Ω be a bounded open subset of R
n
, with smooth boundary. Then:
(I) if β,ρ ∈ R and ρ>|β|, C
ρ
(Ω) is a space of pointwise multipliers for N
β
p
(Ω).
(II) In case β>0, C
β
(Ω) is a space of pointwise multipliers for N
β
p
(Ω); moreover,
there exists C ∈ R
+
such that, ∀a ∈C
β
(Ω), ∀f ∈ N
β
p
(Ω), ∀α ∈ (0,β),
af
β,p,Ω
≤ C
0
(a
L
∞
(Ω)
f
β,p,Ω
+ a
β,∞,Ω
f
L
p
(Ω)
)
≤ C(α)(a
L
∞
(Ω)
f
β,p,Ω
+ a
β,∞,Ω
f
α,p,Ω
).
Proof. Concerning (I), see [14], 3.3.2.
(II) is a particular case of Lemma 1(II) in [10, Chapter 5.3.7]. The constant
σ
p
in the reference is defined in Chapter 2.1.3.
We pass to consider the following system in the unknown u:
D
t
u(t, x)=A(t, x, ∂
x
)u(t, x)+P (t)u(t, .)(x)+f(t, x), (t, x) ∈ [0,T] × Ω,
⎧
⎪
⎨
⎪
⎩
B
k
(t, x
,∂
x
)u(t, x
) − g
k
(t, x
)=0,k=1,...,m, (t, x
) ∈ [0,T] × ∂Ω,
u(0,x)=u
0
(x),x∈ Ω, (3.34)
with the following assumptions:
(J1) Ω is an open bounded subset of R
n
with smooth boundary ∂Ω, for every t ∈
[0,T] the operators A(t, x, ∂
x
) and B
k
(t, x
,∂
x
)(1≤ k ≤ m) satisfy the
conditions (H1)–(H3), with m, μ
1
,...,μ
m
independent of t;
(J2) p ∈ [1, +∞), β ∈ (μ + p
−1
− 2m, ν + p
−1
),withν := min
1≤k≤m
μ
k
, μ :=
max
1≤k≤m
μ
k
;
(J3) the coefficients a
s
(t, x) of A(t, x, ∂
x
)(|s|≤2m) belong to C([0,T]; C
|β|+
(Ω))
for some ∈ R
+
if β ≤ 0,toB([0,T]; C
β
(Ω)) ∩ C([0,T]; C(Ω)) if β>0;
(J4) for each k =1,...,m, the coefficients b
ks
of B
k
(t, x, ∂
x
)(|s|≤2m − μ
k
)
belong to B([0,T]; C
2m+β−μ
k
(Ω)) ∩ C
2m+β−μ
k
−θ
2m
([0,T]; C
θ
(Ω)) for some θ ∈
(0,p
−1
);
(J5) f ∈ B([0,T]; N
β
p
(Ω)) ∩ C([0,T]; N
β
p
(Ω)),forsomeβ
<β;
(J6) for each k ∈{1,...,m}, g
k
∈ B([0,T]; N
2m+β−μ
k
p
(Ω)) ∩ C
2m+β−μ
k
−θ
2m
([0,T];
N
θ
p
(Ω));
(J7) u
0
∈ N
2m+β
p
(Ω) and, for each k =1,...,m,
(B
k
(0,.,∂
x
)u
0
)
|∂Ω
= g
k
(0)
|∂Ω
;
(J8) For some β
<β,
P ∈ B([0,T]; L(N
2m+β
p
(Ω),N
β
p
(Ω))) ∩ C([0,T]; L(N
2m+β
p
(Ω),N
β
p
(Ω))).
On Linear Elliptic and Parabolic etc. 297
Remark 3.9. Owing to the interpolation inequalities, it is not restrictive to assume
that
θ>β− ν, β
∨ (μ + p
−1
− 2m) <β
. (3.35)
We want to prove the following
Theorem 3.10. Assume that the assumptions (J1)–(J8) are satisfied. Then the
system (3.34) has a unique solution u belonging to C
1
([0,T]; N
β
p
(Ω)) ∩ B([0,T];
N
2m+β
p
(Ω)),withD
t
u ∈ B([0,T]; N
β
p
(Ω)).
Proof. It is convenient to define A(t, x, ∂
x
):=A(T,x,∂
x
), P (t):=P (T ), B
k
(t, x,
∂
x
):=B
k
(T,x,∂
x
)ift>T.
As [0,T] is compact, the conclusion will be proved if we show the following:
(P) let s
0
∈ [0,T] and consider the problem
D
t
v(t, x)=A(s + t, x, ∂
x
)v(t, x)+P (s + t)v(t, .)(x)+r(t, x),
(t, x) ∈ [0,δ] × Ω,
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
B
k
(s + t, x
,∂
x
)v(t, x
) − r
k
(t, x
)=0,k=1,...,m, (t, x
) ∈ [0,δ] × ∂Ω,
v(0,x)=v
0
(x),x∈ Ω, (3.36)
with δ ∈ R
+
, s ∈ (s
0
−δ, s
0
+δ)∩[0, +∞), r ∈ B([0,δ]; N
β
p
(Ω))∩C([0,δ]; N
β
p
((Ω))
for some β
<β,foreachk ∈{1,...,m}, and for a fixed θ ∈ (0 ∨ (β − ν),p
−1
),
r
k
∈ B([0,δ]; N
2m+β−μ
k
p
(Ω)) ∩ C
2m+β−μ
k
−θ
2m
([0,δ]; N
θ
p
(Ω)), v
0
∈ N
2m+β
p
(Ω) and,
for each k ∈{1,...,m}, (B
k
(s, ., ∂
x
)v
0
)
|∂Ω
= r
k
(0)
|∂Ω
. Then there exists δ>
0, depending only on s
0
, such that (3.36) has a unique solution v in C
1
([0,δ];
N
β
p
((Ω)) ∩ B([0,δ]; N
2m+β
p
(Ω)),withD
t
v ∈ B([0,δ]; N
β
p
(Ω)).
By subtracting the constant function v
0
, we may limit ourselves to study (3.36) in
the case v
0
= 0. We consider the family of problems
D
t
v(t, x)=A(s
0
,x,∂
x
)v(t, x)+τ{[A(s + t, x, ∂
x
) −A(s
0
,x,∂
x
)]v(t, x)
+ P (s + t)v(t, .)(x)} + r(t, x), (t, x) ∈ [0,δ] ×Ω,
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
B
k
(s
0
,x
,∂
x
)v(t, x
) − τ{B
k
(s
0
,x
,∂
x
)v(t, x
) − B
k
(s + t, x
,∂
x
)v(t, x
)}
− r
k
(t, x
)=0,k=1,...,m, (t, x
) ∈ [0,δ] × ∂Ω,
v(0,x)=0,x∈ Ω, (3.37)
depending on the parameter τ ∈ [0, 1]. As the problem is uniquely solvable in case
τ = 0, by the continuation method, it suffices to obtain an a priori estimate of a
solution, which is independent of τ We are going to show that this is possible if δ
is sufficiently small.
We recall that we are assuming θ>β−ν and β
∨ (μ + p
−1
− 2m) <β
.