Ann.
Acad.
Rom.
Sci.
Ser.
Math.
Appl.
ISSN
2066-6594
Vol.
18,
No.
3/2026
FEEDBACK
NULL
CONTROLLABILITY
IN
MINIMUM
TIME
FOR
NONLINEAR
SYSTEMS
∗
Ovidiu
Cˆ
arj˘
a
†
Alina
I.
Lazu
‡
Dedicated
to
the
memory
of
Professor
Mihail
Megan
DOI
10.56082/annalsarscimath.2026.3.37
Abstract
For
a
nonlinear
control
system,
by
considering
appropriate
feed-
back
laws,
we
prove
results
on
null
controllability
and
regularity
of
the
minimum
time
function.
Keywords:
feedback
control,
null
controllability,
minimum
time
func-
tion,
nonlinear
control
system.
MSC:
93B52,
93B05.
1
Introduction
Let
X
be
a
Banach
space,
A
:
D
(
A
)
⊆
X
⇒
X
be
a
(possible)
multivalued
m
-dissipative
operator
and
f
:
D
(
A
)
→
X
a
given
function.
Let
U
be
a
Hilbert
space,
B
:
U
→
X
a
linear
bounded
operator
and
consider
the
control
system
y
(
t
)
∈
Ay
(
t
)
+
f
(
y
(
t
))
+
Bu
(
t
)
y
(0)
=
x,
(1)
where
x
∈
D
(
A
)
and
u
(
·
)
is
the
control.
∗
Accepted
for
publication
on
February
16,
2026
†
ocarja@uaic.ro
,
Octav
Mayer
Mathematics
Institute,
Romanian
Academy,
Ia¸
si
700505,
Romania
‡
vieru
alina@yahoo.com
,
Department
of
Mathematics,
Gheorghe
Asachi
Technical
University,
Ia¸
si
700506,
Romania
37
Feedback
null
controllability
38
Let
r
>
0
and
denote
by
U
the
set
of
all
admissible
controls,
i.e.
mea-
surable
functions
u
(
·
)
satisfying
u
(
t
)
≤
r
a.e.
Denote
by
C
(
t
)
the
null
controllable
set
at
time
t
>
0
,
i.e.
the
set
of
points
x
∈
X
for
which
there
exist
u
∈
U
and
a
mild
solution
y
(
·
)
to
(
1
)
which
transfers
x
to
0
at
time
t.
Set
C
=
t>
0
C
(
t
)
called
the
null
controllable
set,
and
define
the
minimum
time
function
T
:
X
→
[0
,
+
∞
]
by
T
(
x
)
=
inf
{
t
≥
0;
x
∈
C
(
t
)
}
,
if
x
∈
C
+
∞
,
otherwise.
The
aim
of
this
paper
is
to
provide
a
feedback
control
which
leads
to
null
controllability
and
to
get
estimates
for
the
minimum
time
function
around
the
target.
When
B
is
the
identity
operator,
this
topic
was
studied
in
[
5
],
in
Hilbert
spaces
setting,
in
[
6
],
in
Banach
spaces,
in
the
case
when
A
is
the
generator
of
a
C
0
-semigroup,
using
feedback
controls
of
type
−
ry
(
t
)
/
y
(
t
)
.
See
also
[
1
],
[
2
].
For
the
semilinear
case
with
B
a
surjective
linear
bounded
operator,
we
recall
[
7
],
[
8
],
where
null
controllability
and
estimates
for
the
associated
minimum
time
function
were
obtained,
considering
appropriate
feedback
laws.
For
the
case
when
A
is
m
-dissipative
and
B
is
the
identity
operator,
we
recall
[
9
],
where
it
was
used
the
same
feedback
as
in
[
5
].
In
this
paper,
we
extend
some
results
of
[
9
]
and
[
8
]
to
the
control
system
(
1
)
,
where
A
is
m
-dissipative
and
B
is
a
surjective
linear
operator,
by
means
of
Lyapunov
pairs
characterizations.
2
Preliminaries
First,
we
present
some
concepts
and
results
concerning
Lyapunov
pairs,
which
will
be
used
to
get
the
main
results
of
the
paper.
Let
X
be
a
Banach
space,
A
:
D
(
A
)
⊆
X
⇒
X
a
m
-dissipative
operator,
let
M
be
a
nonempty
subset
of
D
(
A
),
F
:
M
→
X
a
function
and
consider
the
Cauchy
problem
y
(
t
)
∈
Ay
(
t
)
+
F
(
y
(
t
))
y
(0)
=
ξ,
(2)
where
ξ
∈
M.
By
a
mild
solution
to
(
2
)
on
[0
,
T
]
,
with
T
>
0
,
we
mean
a
function
y
:
[0
,
T
]
→
M
which
renders
t
→
ϕ
(
t
)
=
F
(
y
(
t
))
integrable,
O.
Cˆ
arj˘
a,
A.I.
Lazu
39
satisfies
y
(0)
=
ξ
and
is
an
integral
(or
mild)
solution
on
[0
,
T
]
to
the
equation
y
(
t
)
∈
Ay
(
t
)
+
ϕ
(
t
)
.
(3)
We
recall
that
y
:
[0
,
T
]
→
X
is
an
integral
solution
to
(
3
)
on
[0
,
T
]
if
y
is
continuous
on
[0
,
T
],
y
(
t
)
∈
D
(
A
)
for
each
t
∈
[0
,
T
]
,
and
y
(
t
)
−
x
≤
y
(
s
)
−
x
+
t
s
[
y
(
τ
)
−
x,
ϕ
(
τ
)
+
y
]
+
dτ,
for
each
x
∈
D
(
A
)
,
y
∈
Ax
and
0
≤
s
≤
t
≤
T.
We
denoted
by
[
x,
y
]
+
the
right
directional
derivative
of
the
norm
calculated
at
x
in
the
direction
y,
i.e.
[
x,
y
]
+
=
lim
h
↓
0
x
+
hy
−
x
h
.
We
recall
that
[
x,
y
+
ax
]
+
=
[
x,
y
]
+
+
a
x
for
any
x,
y
∈
X
and
a
∈
R
.
For
more
properties
of
[
.,
.
]
+
see,
e.g.,
[
11
].
It
is
known
that,
for
each
ξ
∈
D
(
A
)
and
ϕ
∈
L
1
(0
,
T
;
X
)
,
there
exists
a
unique
integral
solution
y
:
[0
,
T
]
→
D
(
A
)
to
(
3
)
which
satisfies
y
(0)
=
ξ
(see
[
11
,
Theorem
3.6.1]).
Also,
we
recall
the
following
result
of
Benilan
(see,
e.g.,
[
12
]).
Theorem
1.
Let
A
:
D
(
A
)
⊆
X
⇒
X
be
a
m
-dissipative
operator,
let
ϕ,
ψ
∈
L
1
(0
,
T
;
X
)
and
let
y
(
·
)
,
z
(
·
)
be
two
integral
solutions
to
(
3
)
corresponding
to
ϕ
and
ψ
,
respectively,
with
y
(0)
=
ξ
and
z
(0)
=
η,
ξ,
η
∈
D
(
A
)
.
Then
y
(
t
)
−
z
(
t
)
≤
ξ
−
η
+
t
0
[
y
(
s
)
−
z
(
s
)
,
ϕ
(
s
)
−
ψ
(
s
)]
+
ds,
for
any
t
∈
[0
,
T
]
.
By
a
mild
solution
to
(
2
)
on
[0
,
T
)
we
mean
a
function
y
:
[0
,
T
)
→
M
which
is
a
mild
solution
to
(
2
)
on
[0
,
T
]
,
for
each
T
∈
(0
,
T
)
.
A
solution
y
:
[0
,
T
)
→
M
to
(
2
)
is
called
noncontinuable
if
there
is
no
other
solution
y
:
[0
,
T
)
→
M
with
T
<
T
and
y
(
t
)
=
y
(
t
)
on
[0
,
T
)
.
Following
[
10
]
and
[
9
],
we
give
the
following
definition.
Definition
1.
The
functions
V,
g
:
D
(
A
)
→
(
−∞
,
+
∞
]
form
a
Lyapunov
pair
for
the
problem
(
2
)
if
dom
(
V
)
⊆
M
and
for
every
ξ
∈
dom
(
V
)
there
exist
T
>
0
and
a
solution
y
:
[0
,
T
]
→
M
to
(
2
)
such
that
t
→
g
(
y
(
t
))
is
integrable
on
[0
,
T
]
and
we
have
V
(
y
(
t
))
+
t
0
g
(
y
(
s
))
ds
≤
V
(
ξ
)
,
(4)
for
all
t
∈
[0
,
T
]
.
Feedback
null
controllability
40
We
denoted
by
dom
(
V
)
the
domain
of
V,
i.e.
dom
(
V
)
=
x
∈
D
(
A
);
V
(
x
)
<
+
∞
.
Definition
2.
The
A
-contingent
derivative
D
A
V
(
x
)
(
v
)
associated
with
the
m
-dissipative
operator
A
on
a
Banach
space
X
of
the
function
V
:
D
(
A
)
→
(
−∞
,
+
∞
]
at
x
∈
dom
(
V
)
in
the
direction
v
∈
X
is
defined
by
D
A
V
(
x
)
(
v
)
=
lim
inf
t
↓
0
,
w
→
0
1
t
[
V
(
S
A
+
v
(
t
)
x
+
tw
)
−
V
(
x
)]
,
where
S
A
+
v
(
t
)
:
D
(
A
)
→
D
(
A
)
,
t
≥
0
,
stands
for
the
semigroup
of
nonex-
pansive
mappings
generated
by
A
+
v
.
In
[
9
]
it
was
proved
the
following
result.
Theorem
2.
Let
X
be
a
Banach
space,
let
A
:
D
(
A
)
⊆
X
⇒
X
be
a
m
-dissipative
operator
and
let
F
:
M
→
X
be
a
locally
Lipschitz
function,
where
M
is
a
nonempty
subset
of
D
(
A
)
.
Let
g
:
D
(
A
)
→
R
be
a
locally
Lipschitz
function
and
let
V
:
D
(
A
)
→
(
−∞
,
+
∞
]
be
a
proper
function
with
dom
(
V
)
⊆
M
and
having
a
locally
closed
epigraph.
Then
V
and
g
form
a
Lyapunov
pair
for
the
problem
(
2
)
if
and
only
if
D
A
V
(
x
)
F
(
x
)
+
g
(
x
)
≤
0
,
for
all
x
∈
dom
(
V
)
.
Recall
that
the
epigraph
of
the
function
V
is
defined
by
epi
(
V
)
=
(
x,
µ
)
∈
D
(
A
)
×
R
;
V
(
x
)
≤
µ
.
3
Main
results
First,
we
present
the
hypotheses
we
shall
refer
to
in
what
follows.
(H1)
X
is
a
Banach
space
and
U
is
a
Hilbert
space.
(H2)
The
operator
A
:
D
(
A
)
⊆
X
⇒
X
is
m
-dissipative
and
0
∈
A
0
.
(H3)
The
function
f
:
D
(
A
)
→
X
is
locally
Lipschitz
and
satisfies
[
x,
f
(
x
)]
+
≤
λ
x
,
(5)
for
any
x
∈
D
(
A
).
O.
Cˆ
arj˘
a,
A.I.
Lazu
41
(H4)
B
:
U
→
X
a
linear
bounded
surjective
operator.
Remark
1.
Since
U
is
a
Hilbert
space,
the
operator
B
admits
a
right
inverse,
i.e.
there
exists
Λ
:
X
→
U
a
linear
bounded
operator
such
that
B
◦
Λ
=
I
X
(see,
e.g.,
[
4
,
Theorem
2.12]).
Moreover,
since
Λ
is
linear
and
continuous,
there
exists
γ
>
0
such
that
Λ
x
≤
γ
x
(6)
for
any
x
∈
X.
We
provide
an
admissible
control
that
will
be
used
to
reach
the
origin
starting
from
the
initial
point
x
∈
D
(
A
)
in
some
time
T
,
by
mild
solutions
to
(
1
)
.
More
exactly,
we
use
a
feedback
control
of
the
form
u
(
t
)
=
−
r
Λ
y
(
t
)
Λ
y
(
t
)
.
First,
we
prove
the
following
result.
Theorem
3.
Assume
(H1)
-
(H4)
.
Then,
for
any
x
∈
D
(
A
)
,
x
=
0
,
there
exist
T
>
0
and
y
:
[0
,
T
]
→
D
(
A
)
solution
to
y
(
t
)
∈
Ay
(
t
)
+
f
(
y
(
t
))
−
ry
(
t
)
Λ
y
(
t
)
y
(0)
=
x
(7)
which
satisfies
y
(
t
)
≤
x
−
r
γ
t
+
λ
t
0
y
(
s
)
ds
(8)
and
y
(
t
)
=
0
for
every
t
∈
[0
,
T
]
.
Proof.
Define
the
functions
V
:
D
(
A
)
→
(
−∞
,
+
∞
]
by
V
(
x
)
=
x
for
x
∈
D
(
A
)
\
{
0
}
and
V
(
x
)
=
+
∞
otherwise
and
g
:
D
(
A
)
→
R
,
g
(
x
)
=
r
γ
−
λ
x
for
x
∈
D
(
A
)
.
Clearly,
epi
(
V
)
is
locally
closed
and
g
is
Lipschitz
on
D
(
A
)
.
We
shall
apply
Theorem
2
to
prove
that
V
and
g
form
a
Lyapunov
pair
for
the
problem
(
2
)
with
F
:
D
(
A
)
\
{
0
}
→
X,
defined
by
F
(
x
)
=
f
(
x
)
−
rx
Λ
x
,
for
x
∈
D
(
A
)
\
{
0
}
.
Let
us
remark
that
the
function
x
−
→
rx
Λ
x
is
locally
Lipschitz
continuous
on
D
(
A
)
\
{
0
}
.
Further,
we
have
to
check
that
D
A
V
(
x
)
f
(
x
)
−
rx
Λ
x
+
g
(
x
)
≤
0
Feedback
null
controllability
42
for
any
x
∈
dom
(
V
)
=
D
(
A
)
\
{
0
}
.
Let
x
∈
dom
(
V
)
and
define
z
(
t
)
=
S
A
+
f
(
x
)
−
rx
Λ
x
(
t
)
x,
t
≥
0
.
By
Theorem
1
and
assumption
0
∈
A
0
,
we
have
z
(
t
)
≤
x
+
t
0
z
(
s
)
,
f
(
x
)
−
rx
Λ
x
+
ds
(9)
for
all
t
≥
0
.
Definition
2
,
together
with
(
9
)
,
(
5
)
,
(
6
)
and
the
properties
of
[
·
,
·
]
+
implies
D
A
V
(
x
)
f
(
x
)
−
rx
Λ
x
≤
lim
inf
t
↓
0
1
t
[
z
(
t
)
−
x
]
≤
lim
sup
t
↓
0
1
t
t
0
z
(
s
)
,
f
(
x
)
−
rx
Λ
x
+
ds
≤
x,
f
(
x
)
−
rx
Λ
x
+
=
[
x,
f
(
x
)]
+
−
r
Λ
x
x
≤
λ
x
−
r
γ
.
Therefore,
by
Theorem
2
,
V
and
g
form
a
Lyapunov
pair
for
the
problem
(
7
)
.
This
proves
the
existence
of
a
solution
y
(
·
)
to
(
7
)
on
some
interval
[0
,
T
]
satisfying
(
8
)
for
any
t
∈
[0
,
T
]
.
Moreover,
y
(
t
)
∈
D
(
A
)
\
{
0
}
for
any
t
∈
[0
,
T
]
.
By
a
continuation
argument
we
can
prove
the
following
result.
Corollary
1.
Assume
(H1)
-
(H4)
.
Then,
for
any
x
∈
D
(
A
)
,
x
=
0
,
there
exists
y
:
[0
,
σ
)
→
D
(
A
)
\
{
0
}
a
noncontinuable
solution
to
(
7
)
which
satisfies
(
8
)
for
every
t
∈
[0
,
σ
)
.
The
solution
is
unique.
Now
we
can
state
the
main
result
of
the
paper,
which
leads
to
null
controllability
of
the
system
(
1
).
Theorem
4.
Assume
(H1)
-
(H4)
.
Moreover,
assume
that
f
is
bounded
on
bounded
sets.
The
following
properties
hold.
(i)
In
case
λ
≤
0
,
for
any
x
∈
D
(
A
)
,
x
=
0
,
there
exist
a
control
u
(
·
)
and
a
mild
solution
y
(
·
)
to
(
1
)
that
reaches
the
origin
of
X
in
some
time
T
≤
γ
r
x
and
satisfies
y
(
t
)
≤
x
−
r
γ
t
(10)
O.
Cˆ
arj˘
a,
A.I.
Lazu
43
for
any
t
∈
[0
,
T
]
.
Therefore,
T
(
x
)
≤
γ
r
x
for
any
x
∈
D
(
A
)
.
(ii)
In
case
λ
>
0
,
for
every
x
∈
D
(
A
)
satisfying
0
<
x
<
r
γλ
,
there
exist
a
control
u
(
·
)
and
a
mild
solution
y
(
·
)
to
(
1
)
that
reaches
the
origin
of
X
in
some
time
T
≤
1
λ
log
r
r
−
γλ
x
and
satisfies
y
(
t
)
≤
e
λt
x
−
r
γλ
+
r
γλ
(11)
for
any
t
∈
[0
,
T
]
.
Considering
0
<
ρ
<
r
γλ
and
defining
µ
=
γ
r
−
γλρ
we
have
T
(
x
)
≤
µ
x
(12)
for
any
x
∈
D
(
A
)
with
0
<
x
≤
ρ.
Proof.
Since
the
first
situation
is
simpler,
we
present
the
proof
for
the
second
case.
Let
x
∈
D
(
A
)
with
0
<
x
<
r
γλ
.
From
Corollary
1
,
we
get
a
noncontinuable
solution
y
(
·
)
to
(
1
)
on
[0
,
σ
)
,
with
the
feedback
con-
trol
u
(
t
)
=
−
r
Λ
y
(
t
))
/
Λ
y
(
t
))
,
which
satisfies
(
8
)
and
y
(
t
)
=
0
for
any
t
∈
[0
,
σ
)
.
Let
us
remark
that
σ
<
∞
.
Indeed,
from
(
8
)
,
using
the
Gron-
wall
inequality,
we
obtain
that
(
11
)
holds
for
any
t
∈
[0
,
σ
)
,
which
leads
to
σ
<
1
λ
log
r
r
−
γλ
x
.
Moreover,
it
follows
that
y
(
·
)
is
bounded
on
[0
,
σ
)
.
Let
F
(
x
)
=
f
(
x
)
−
rx
Λ
x
,
for
x
∈
D
(
A
)
\
{
0
}
.
Since
f
is
bounded
on
bounded
sets
and
rx
Λ
x
≤
r
B
for
any
x
∈
D
(
A
)
\
{
0
}
(we
have
used
that
B
◦
Λ
=
I
X
)
,
we
obtain
that
F
maps
bounded
subsets
in
D
(
A
)
\
{
0
}
into
bounded
subsets
in
X,
hence
F
(
y
(
·
))
is
bounded
on
[0
,
σ
).
So,
there
exists
lim
t
↑
σ
y
(
t
)
which
belongs
to
D
(
A
)
.
If
lim
t
↑
σ
y
(
t
)
=
0
,
then,
by
Theorem
3
,
we
obtain
that
y
(
·
)
can
be
continued
to
the
right
of
σ,
which
is
false.
Therefore,
lim
t
↑
σ
y
(
t
)
=
0
.
To
get
(
12
)
we
use
the
elementary
inequality
log
(1
+
z
)
<
z
for
any
z
>
0
.
Proposition
1.
Assume
(H1)
-
(H4)
.
Moreover,
assume
that
f
is
globally
Lipschitz
of
constant
L
>
0
.
(i)
If
λ
≤
0
,
then,
for
any
x,
z
∈
D
(
A
)
we
have
T
(
z
)
≤
T
(
x
)
+
γ
r
e
L
T
(
x
)
x
−
z
.
Feedback
null
controllability
44
Moreover,
T
(
·
)
is
locally
Lipschitz
continuous
on
C
.
(ii)
Suppose
λ
>
0
,
let
0
<
ρ
<
r
γλ
and
define
µ
=
γ
r
−
γλρ
.
If
x
∈
C
and
z
∈
D
(
A
)
is
such
that
z
−
x
≤
ρe
−
L
T
(
x
)
,
then
z
∈
C
and
T
(
z
)
≤
T
(
x
)
+
µe
L
T
(
x
)
x
−
z
.
Moreover,
C
is
open
and
T
(
·
)
is
locally
Lipschitz
continuous
on
C
.
Proof.
We
present
the
proof
for
(ii).
We
use
the
same
idea
as
in
[
5
,
Propo-
sition
3.3]
(see
also
[
7
,
Proposition
2.1]).
Let
x
∈
C
and
z
∈
D
(
A
)
such
that
z
−
x
≤
ρe
−
L
T
(
x
)
.
To
simplify
the
exposition
we
assume
the
existence
of
an
optimal
control
u
for
x,
i.e.
y
(
T
(
x
))
=
0
,
where
y
(
·
)
is
the
solution
to
(
1
)
starting
from
x.
Denote
by
y
(
·
)
the
solution
to
(
1
)
with
the
initial
data
y
(0)
=
z.
By
Theorem
1
,
we
have
that
y
(
t
)
−
y
(
t
)
≤
e
Lt
x
−
z
,
for
any
t
≥
0
.
Hence,
y
(
T
(
x
))
≤
e
L
T
(
x
)
x
−
z
≤
ρ.
Using
now
Theorem
4
,
we
get
that
y
(
T
(
x
))
∈
C
and
T
(
y
(
T
))
≤
µ
y
(
T
(
x
))
≤
µe
L
T
(
x
)
x
−
z
.
From
the
dynamic
programming
principle
(see
[
3
]),
we
have
that,
for
any
t
∈
[0
,
T
(
z
)]
,
T
(
z
)
≤
t
+
T
(
y
(
t
))
.
It
follows
that
T
(
z
)
≤
T
(
x
)
+
T
(
y
(
T
(
x
)))
≤
T
(
x
)
+
µe
L
T
(
x
)
x
−
z
.
Let
x
0
∈
C
and
take
δ
=
ρ
2
e
−
L
(
T
(
x
0
)+
µρ
)
.
By
the
first
part
of
the
proof,
we
obtain
T
(
x
1
)
−
T
(
x
2
)
≤
µe
L
(
T
(
x
0
)+
µρ
)
x
1
−
x
2
,
for
any
x
1
,
x
2
∈
B
X
(
x
0
,
δ
)
.
Remark
2.
If
X
is
a
Hilbert
space
too,
then
we
can
also
take
the
feedback
law
−
B
∗
y
(
t
)
/
B
∗
y
(
t
)
,
where
B
∗
is
the
adjoint
operator
of
B,
with
a
similar
proof.
O.
Cˆ
arj˘
a,
A.I.
Lazu
45
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Optimal
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