Electronic Journal of Qualitative Theory of Differential Equations
2007, No.
19, 1-21;
http://www.math.u-szeged.hu/ejqtde/
Null Controllability of Some Impulsive
Evolution Equation in a Hilbert Spae
R. BOUKHAMLA
(1)
& S. MAZOUZI
(2)
(1)CentreUniversitairedeSouk-Ahras,Souk-Ahras41000,Algeria.
(2)DépartementdeMathématiques,Universitéd'Annaba,BP12,ANNABA23000,Algeria.
Abstrat
Weshallestablishaneessaryandsuientonditionunderwhih
we have thenullontrollabilityofsome rstorderimpulsiveevolution
equation inaHilbertspae.
MSC(2000) : 34A37, 93B05, 93C15.
Keywords: Null-ontrollability, impulsive onditions, mild solutions,
evolution equation.
1 Introdution
The problem of exat ontrollability of linear systems represented by
in-nite onservative systemshas beenextensively studiedby several authorsA.
Haraux [8℄
,
R.Triggiani [16℄,
Z.H. Guan, T.H. Qian, and X.Yu [7℄, see alsothe referenes [1,2, 6, 10,15℄. In the sequel, we shall be onerned with the
problem of null ontrollability of some rst order evolution equation
sub-jet toimpulsiveonditions andso weshall derivea neessary and suient
ondition under whih null ontrollability ours. Atually, we shall
estab-lish anequivalene between the null-ontrollabilityand some observability
inequalityinsomehowmoregeneralframeworkthan thatproposed byA
Ha-raux [8℄. Regarding the literature onthe impulsive dierential equationswe
y
′
(
t
) +
Ay
(
t
) =
Bu
(
t
)
,
t
∈
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
,
(1)
y
(0) =
y
0
,
∆
y
(
t
k
) =
I
k
y
(
t
k
) +
D
k
v
k
, k
∈
σ
1
m
,
(1
k
)
where the nal time
T
is a positive number,y
0
is an initial ondition in a
Hilbertspae
H
endowed withaninnerproduth
., .
iH
,y
(
t
) : [0
, T
]
→
H
isa vetor funtion,σ
m
1
is a subset ofN
given byσ
m
1
=
{
1
,
2
, ..., m
}
, and nally,{
t
k
}k
∈
σ
m
1
is an inreasing sequene of numbers in the open interval
(0
, T
)
,
and
∆
y
(
t
k
)
denotes the jump ofy
(
t
)
att
=
t
k
, i.e.,∆
y
(
t
k
) =
y t
+
k
−
y t
−
k
where
y t
+
k
and
y t
−
k
represent the right and left limits of
y
(
t
)
att
=
t
k
respetively
.
On the other hand, the operatorsA, B, I
k
, D
k
:
H
→
H
aregivenlinearboundedoperators. Moreover, weset thefollowingassumptions:
(H1)
A
∗
=
−
A
,(H2)
I
∗
k
=
−
I
k
,
for everyk
∈
σ
m
1
,
and for eahk
∈
σ
m
1
, the operatorIk
=
I
k
+
I
is invertible,
(H3)
B
∗
=
B
≥
0
and there is
d
0
>
0
suhthat(
Bu, u
)
H
≤
d
0
k
u
k
2
H
,
for allu
∈
H,
(H4)
D
∗
k
=
D
k
≥
0
,
for everyk
∈
σ
m
1
,
and for eahk
∈
σ
m
1
there isd
k
>
0
suh that(
D
k
u, u
)
H
≤
d
k
k
u
k
2
H
,
for allu
∈
H.
In the sequel we shall designate by h the funtion
h
(
t
) =
u
(
t
)
,
{
v
k
}k
∈
σ
m
1
,
where
u
(
t
)
∈
L
2
(0
, T
)
\ {
t
k}k
∈
σ
m
1
;
H
and
{
v
k
}k
∈
σ
m
1
∈
l
2
(
σ
m
1
;
H
)
+
n
{
v
k
}k
∈
σ
m
1
, v
k
∈
H
We pointout that the spae
K
m
=
L
2
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
;
H
×
l
2
(
σ
1
m
;
H
)
isa Hilbert spae with respet to the inner produt
h
,
e
hK
m
=
Z
T
0
(
u
(
t
)
,
e
u
(
t
))
H
dt
+
m
X
k
=1
(
v
k
,
e
v
k
)
H
,
dened for allh
= (
u
(
t
)
,
{
v
k
}
m
k
=1
)
ande
h= (
u
e
(
t
)
,
{
e
v
k
}
m
k
=1
)
∈ K
m
.
We shall denoteby
B
the ontroloperator given byB
=
B,
{
D
k
}k
∈
σ
m
1
∈ L
L
2
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
;
H
×
l
2
(
σ
1
m
;
H
)
,
so that
B
h(
t
) =
Bu
(
t
)
,
{
D
k
v
k
}k
∈
σ
m
1
.
We have for every h
= (
u
(
t
)
,
{
v
k
}
m
k
=1
)
∈ K
m
(
B
h,
h)
K
m
=
Z
T
0
(
Bu
(
t
)
, u
(
t
))
H
dt
+
m
X
k
=1
(
D
k
v
k
, v
k
)
H
=
Z
T
0
(
u
(
t
)
, Bu
(
t
))
H
dt
+
m
X
k
=1
(
v
k
, D
k
v
k
)
H
= (
h,
B
h)
K
m
,
whih shows that
B
∗
=
B
, that is,
B
is self-adjoint. On the other hand, wehave
(
B
h,
h)
K
m
=
Z
T
0
(
Bu
(
t
)
, u
(
t
))
H
dt
+
m
X
k
=1
(
D
k
v
k
, v
k
)
H
≤
d
0
Z
T
0
k
u
(
t
)
k
2
H
dt
+
m
X
k
=1
d
k
k
v
k
k
2
H
≤
δ
k
h
k
2
K
m
,
where
δ
= max
{
d
0
, d
1
, ..., d
m
}
.
Thus, the operator isB
bounded inK
m
.
Next,weonsider the homogeneoussystem assoiatedwith (1) :ϕ
′
(
t
) +
Aϕ
(
t
) = 0
,
t
∈
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
,
(2)
ϕ
(0) =
ϕ
0
,
We point out that on eah interval
[
t
k
, t
k
+1
)
,fork
= 0
, ..., m
, the solutionϕ
is left ontinuous ateahtimet
k
.Consider the orrespondinghomogeneous bakward problem :
−
ϕ
˜
′
(
t
) +
A
ϕ
˜
(
t
) = 0
, ,
t
∈
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
,
(3)
˜
ϕ
(
T
) =
ϕ
0
,
∆ ˜
ϕ
(
t
m
−(
k
−1)
) =
−
I
˜
m
−(
k
−1)
ϕ
˜
(
t
+
m
−(
k
−1)
)
, k
∈
σ
m
1
,
(3
k
)
where
A
=
A
∗
=
−
A,
I
˜
m
−(
k
−1)
=
I
m
∗
−(
k
−1)
=
−
I
m
−(
k
−1)
, k
∈
σ
1
m
.
We observe that the problem (3) onthe interval
[
t
m
, T
]
is equivalent to thelassial bakward problem
−
ϕ
˜
′
(
t
) +
A
ϕ
˜
(
t
) = 0
, t
∈
[
t
m
, T
]
,
˜
ϕ
(
T
) =
ϕ
0
.
We introdue the following spae :
PC
([0
, T
] ;
H
) =
{
y, y
: [0
, T
]
→
H
suh that
y
(
t
)
isontinuous att
=
6
t
k
,
and has disontinuities of rst kind att
=
t
k
, for everyk
∈
σ
m
1
}
.
Evidently,
PC
([0
, T
] ;
H
)
is a Banahspae with respet to the normk
y
k
PC
= sup
t
∈(0
,T
)
k
y
(
t
)
k
.
On the other hand, we dene the subspaes
PLC
, (respetively,PRC
)=
{
y, y
∈ PC
suh thaty
(
t
)
is left (respetively, right) ontinuous att
=
t
k
,for every
k
∈
σ
m
1
}
.
Remark 1 1) The spae
PLC
, (respetively,PRC
) an be identied to asubspae of
K
m
.
That is, to eahy
∈ PLC
, (respetively,y
˜
∈ PRC
) isassigned the funtion h (respetively,
e
h) dened byh
(
t
) =
y
(
t
)
,
{
y
(
t
k
)
}k
∈
σ
m
1
,
and
e
h
(
t
) =
˜
y
(
t
)
,
{
y
˜
(
t
k
)
}k
∈
σ
m
1
.
2) Let
y
e
∈ PRC
, the funtiony
an be written as :e
y
(
t
) =
e
y
[0]
(
t
)
ift
∈
[
t
0
, t
1
)
e
y
[
k
]
(
t
)
ift
∈
[
t
k
, t
k
+1
)
e
y
[
m
]
(
t
)
ift
∈
[
t
m
, T
]
.
Next, let
τ
k
=
t
k
−
t
k
−1
,
we dene the operatorT
:
D
(
T
) =
PRC ⊂ K
m
→
K
m
by(
T
y
e
)(
t
) =
e
y
[0]
((
T
−
t
)
τ
m
τ1
+1
+
t
0
)
if
t
∈
[
t
m
, T
]
,
e
y
[
k
]
((
t
m
−(
k
−1)
−
t
)
τ
k
+1
τ
m
−
(
k
−
1)
+
t
k
)
if
t
∈
t
m
−
k
, t
m
−(
k
−1)
, k
∈
σ
1
m
−1
,
e
y
[
m
]
((
t
1
−
t
)
τ
m
+1
τ1
+
t
m
)
ift
∈
(0
, t
1
]
.
(4)
We note that the range of
T
is exatlyPLC
. The funtion(
T
e
y
)(
t
)
an bewritten asfollows:
(
T
e
y
)(
t
) =
y
[0]
(
t
)
ift
∈
[
t
0
, t
1
]
,
y
[
k
]
(
t
)
ift
∈
(
t
k
, t
k
+1
]
, k
∈
σ
m
−1
1
,
y
[
m
]
(
t
)
ift
∈
(
t
m
, T
]
.
Let
X
(
t
)
be the resolvent solutionof the operator systemX
′
(
t
) +
AX
(
t
) = 0
,
0 =
t
0
< t < t
m
+1
=
T, t
6
=
t
k
, k
= 1
,
2
, ..., m,
X
(0) =
I,
X
(
t
k
+ 0)
−
X
(
t
k
−
0) =
I
k
X
(
t
k
)
, k
= 1
,
2
, ..., m,
where
I
:
H
→
H
istheidentityoperator. WeshallsupposethattheoperatorI
k
=
I
k
+
I
has abounded inverse.Denition 1 A funtion
y
∈ PC
([0
, T
] ;
H
)
is a mild solution to theimpul-sive problem (1) if the impulsive onditions are satised and
y
(
t
) =
G
(
t,
0
+
)
y
0
+
R
t
0
G
(
t, s
)
Bu
(
s
)
ds
+
P
0
<t
k≤
t
G
(
t, t
k
)(
D
k
v
k
)
,
for everyt
∈
(0
, T
)
,
where the evolution operator
G
(
t, s
)
isgiven byItisnothardtohek thattheoperator
G
(
t, t
k
)
satisestheoperatorsystemG
′
(
t, t
k
) +
AG
(
t, t
k
) = 0
, t
∈
[
t
k
, t
k
+1
)
, k
∈
σ
0
m
,
G
(
t
k
, t
k
) =
I,
G
(
t
+
k
+1
, t
k
)
−
G
(
t
−
k
+1
, t
k
) =
I
k
+1
G
(
t
−
k
+1
, t
k
)
.
It iswell known that (1)has a unique solution
y
suh thaty
∈ PLC
([0
, T
] ;
H
)
∩
C
1
[0
, T
]
\ {
t
k
}k
∈
σ
m
1
;
H
.
Now, we dene the onept of mild solution for the bakward impulsive
system (3) assoiatedwith system (2).
Denition 2 We say that
ϕ
˜
∈ PRC
([0
, T
] ;
H
)
is a mild solution for the bakward impulsive system (3) ifT
ϕ
˜
is a mild solution for the homogeneousimpulsive system (2).
Letus introdue the notion of the null ontrollability of the initialstate
as follows:
Denition 3 Wesaythattheinitialstate
y
0
∈
H
isnullontrollable attime
T
, if there is a ontrol funtion h∈ K
m
for whih the solutiony
of system(1) satises
y
(
T
) = 0
.
2 Main Results
First we begin by the following lemma.
Lemma 1 Assume that
ξ
(
t
)
, ζ
(
t
)
∈
L
1
([0
, T
] ;
H
)
and
{
ξ
k
}
m
k
=1
,
{
ζ
k
}
m
k
=1
∈
l
1
(
σ
m
1
, H
)
. Then, for every vetor funtionsγ
(
t
)
∈ PLC
([0
, T
] ;
H
)
∩
C
1
[0
, T
]
\ {
t
k
}k
∈
σ
m
1
;
H
and
η
(
t
)
∈ PRC
([0
, T
] ;
H
)
∩
C
1
[0
, T
]
\ {
t
k}k
∈
σ
m
1
;
H
satisfying the problem
d
dt
h
γ
(
t
)
, η
(
t
)
i
=
h
ξ
(
t
)
, ζ
(
t
)
i
,
t
6
=
t
k
,
fork
∈
σ
m
1
,
h
γ
(
t
)
, η
(
t
)
i|
T
0
=
h
γ
(
T
)
, η
(
T
)
i − h
γ
(0)
, η
(0)
i
(5)=
Z
T
0
h
ξ
(
t
)
, ζ
(
t
)
i
dt
+
m
X
k
=1
h
ξ
k
, ζ
ki
.
Proof. It is straightforward.
We alsoneed the followingLemmas.
Lemma 2
[14]
IfB ∈ L
(
Km
)
is self-adjoint and nonnegative, thenkB
hk ≤ kBk
1
/
2
(
B
h,
h)
1
/
2
K
m
,
h∈ K
m
.
Lemma 3 If
τ
k
+1
=
τ
m
−(
k
−1)
, k
∈
σ
m
−1
0
, then forthe mildsolutionϕ
e
of (3),the identity holds:
Z
T
0
|
B
ϕ
e
|
2
H
dt
+
m
X
k
=1
D
k
ϕ t
e
+
k
2
H
=
Z
T
0
|
Bϕ
|
2
H
dt
+
m
X
k
=1
D
k
ϕ t
m
−(
k
−1)
2
H
.
(6)
Proof. For eah
k
∈
σ
m
0
,
using the hange of variablet
→
(
t
m
−(
k
−1)
−
t
)
τ
k
+1
τ
m
−
(
k
−
1)
+
t
k
we have
Z
t
m
−
(
k
−
1)
t
m
−
k
(
Bϕ
[
m
−
k
]
(
t
)
, Bϕ
[
m
−
k
]
(
t
))
dt
=
Z
t
m
−
(
k
−
1)
t
m
−
k
(
B
ϕ
e
[
k
]
((
t
m
−(
k
−1)
−
t
)
τ
k
+1
τ
m
−(
k
−1)
+
t
k
)
, B
ϕ
e
[
k
]
((
t
m
−(
k
−1)
−
t
)
τ
k
+1
τ
m
−(
k
−1)
+
t
k
))
dt
=
−
τ
m
−(
k
−1)
τ
k
+1
Z
t
k
t
k
+1
(
B
ϕ
e
[
k
]
(
s
)
, B
ϕ
e
[
k
]
(
s
))
ds
=
Z
t
k
+1
t
k
(
B
ϕ
e
[
k
]
(
s
)
, B
ϕ
e
[
k
]
(
s
))
ds.
Summingup with respet to
k
, wegetm
X
k
=0
Z
t
m
−
(
k
−
1)
t
m
−
k
(
Bϕ
[
m
−
k
]
((
t
))
, Bϕ
[
m
−
k
]
(
t
))
dt
=
m
X
k
=0
Z
t
k
+1
t
k
Thus, we obtain
Z
T
0
|
B
ϕ
e
|
2
H
dt
=
Z
T
0
|
Bϕ
|
2
H
dt.
On the other hand,by virtue of the denition of the funtion
ϕ
e
wegetϕ
(
t
m
−
k
) =
ϕ
e
(
t
k
+1
)
, k
∈
σ
m
0
−1
.
Also, we have
ϕ t
m
−(
k
−1)
=
ϕ
e
(
t
k
)
, k
∈
σ
1
m
,
and
e
ϕ
(
t
m
−
k
) =
ϕ
(
t
k
+1
)
, k
∈
σ
m
0
−1
.
This implies that
m
X
k
=1
|
D
k
ϕ
e
(
t
k
)
|
2
H
=
m
X
−1
k
=0
h
D
m
−
k
ϕ
e
(
t
m
−
k
)
, D
m
−
k
ϕ
e
(
t
m
−
k
)
iH
=
m
X
−1
k
=0
h
D
m
−
k
ϕ
(
t
k
+1
)
, D
m
−
k
ϕ
(
t
k
+1
)
iH
=
m
X
l
=1
D
l
ϕ t
m
−(
l
−1)
, D
l
ϕ t
m
−(
l
−1)
H
=
m
X
k
=1
D
k
ϕ t
m
−(
k
−1)
, D
k
ϕ t
m
−(
k
−1)
H
=
m
X
k
=1
D
k
ϕ t
m
−(
k
−1)
2
H
,
whih gives (6).
Corollary 1 If
τ
k
+1
=
τ
m
−(
k
−1)
,
fork
∈
σ
m
−1
0
, andB, D
k
are nonnegativein
H,
then the followingholds:Z
T
0
h
B
ϕ
e
(
t
)
,
ϕ
e
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
k
ϕ
e
(
t
k
)
,
ϕ
e
(
t
k
)
i
=
Z
T
0
h
Bϕ
(
t
)
, ϕ
(
t
)
i
dt
+
k
X
=
m
k
=1
Proof. This follows immediatelyfromLemma 3 if we substitute
B
byB
1
2
,
and
D
k
byD
1
2
k
.
Now, we state and establish the followingTheorem.
Theorem 1 Let
y
0
∈
H
bea giveninitial stateforthe system(1)
,
theny
0
is
null ontrollable at time
T
if and only if there is a positive onstantC
suhthat
h
y
0
,
ϕ
˜
0
i
H
≤
C
(Z
T
0
|
Bϕ
|
2
H
dt
+
m
X
k
=1
D
k
ϕ t
m
−(
k
−1)
2
H
)
1
/
2
,
∀
ϕ
˜
0
∈
H,
(7)
where
ϕ
∈ PLC
([0
, T
] ;
H
)
istheuniquemild solutionto(2)
withϕ
(
T
) = ˜
ϕ
0
.
Proof. Itsues toprovethisTheoremforthespeialase
τ
k
+1
=
τ
m
−(
k
−1)
,
for
k
∈
σ
m
−1
0
,
beause the normk|
.
|k
+
P
m
k
=0
τ
m
−(
k
−1)
τ
k
+1
R
t
k
+1
t
k
|
.
|
2
H
dt
1
/
2
is
equivalent tothe usual norm of
L
2
([0
, T
] ;
H
)
.
We shall proeed inseveral steps.
Step 1: Let
y
andϕ
e
be strong solutions to(1)
and(3)
, respetively.Then, for
t
6
=
t
k
,k
∈
σ
m
1
, wehaved
dt
h
y
(
t
)
,
ϕ
e
(
t
)
i
=
h
y
(
t
)
,
ϕ
e
′
(
t
)
i
+
h
y
′
(
t
)
,
ϕ
e
(
t
)
i
(8)=
h
y
(
t
)
,
−
A
ϕ
e
(
t
)
i
+
h−
Ay
(
t
) +
Bu
(
t
)
,
ϕ
e
(
t
)
i
=
h
y
(
t
)
,
−
A
ϕ
e
(
t
)
i
+
h−
Ay
(
t
)
,
ϕ
e
(
t
)
i
+
h
Bu
(
t
)
,
ϕ
e
(
t
)
i
=
h
Bu
(
t
)
,
ϕ
e
(
t
)
i
.
Multiplying equation(3
k
) in(3) fromthe leftbyy t
m
−(
k
−1)
the solutionof
(1)
,andmultiplyingequation(1k
)in(1)fromtherightbyϕ
e
(
t
k
)
the solutionof (3), and nallyaddingmemberwise we get
∆
h
y
(
t
)
,
ϕ
e
(
t
)
i
|
t
=
t
k
=
h
y
(
t
k
)
,
∆
ϕ
e
(
t
k
)
i
+
h
∆
y
(
t
k
)
,
ϕ
e
(
t
k
)
i
(9)=
h
y
(
t
k
)
, I
k
ϕ
e
(
t
k
)
i
+
h
I
k
y
(
t
k
) +
D
k
v
k
,
ϕ
e
(
t
k
)
i
=
h
y
(
t
k
)
, I
k
ϕ
e
(
t
k
)
i
+
h
I
k
y
(
t
k
)
,
ϕ
e
(
t
k
)
i
+
h
D
k
v
k
,
ϕ
e
(
t
k
)
i
=
h
D
k
v
k
,
ϕ
e
(
t
k
)
i
.
Setting
γ
(
t
) =
y
(
t
)
,η
(
t
) =
ϕ
e
(
t
)
,ξ
(
t
) =
Bu
(
t
)
,ζ
(
t
) =
ϕ
e
(
t
)
,ξ
k
=
D
k
v
k
,h
y
(
T
)
,
ϕ
e
(
T
)
i − h
y
(0)
,
ϕ
e
(0)
i
=
Z
T
0
h
Bu
(
t
)
,
ϕ
e
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
k
v
k
,
ϕ
e
(
t
k
)
i
.
(10)Sine
B
isbounded,self-adjointandB ≥
0
,then bydensitythelatteridentityisstillvalidformildsolutions
y
of(1). Identity(10)anbewrittenasfollowsh
y
(
T
)
,
ϕ
e
(
T
)
i − h
y
(0)
,
ϕ
e
(0)
i
=
Z
T
0
h
u
(
t
)
, B
ϕ
e
(
t
)
i
dt
+
k
X
=
m
k
=1
h
v
k
, D
k
ϕ
e
(
t
k
)
i
.
(11)Next, if there is a ertainh
(
t
)
∈ Km
suh that the mild solutionof (1)withy
(0) =
y
0
satisesy
(
T
) = 0
,
then−h
y
(0)
,
ϕ
e
(0)
i
=
Z
T
0
h
u
(
t
)
, B
ϕ
e
(
t
)
i
dt
+
k
X
=
m
k
=1
h
v
k
, D
k
ϕ
e
(
t
k
)
i
,
and so by Cauhy-Shwarz Inequality we obtain
|h
y
(0)
,
ϕ
e
(0)
i
H
| ≤
(Z
T
0
k
u
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
v
k
k
2
H
)
1
/
2
(12)
×
(Z
T
0
k
B
ϕ
e
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
D
k
ϕ
e
((
t
k
)
k
2
H
)
1
/
2
.
Using Lemma3, and equation (12) we have
|h
y
(0)
,
ϕ
e
(0)
i
H
| ≤
(Z
T
0
k
u
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
v
k
k
2
H
)
1
/
2
×
(Z
T
0
k
Bϕ
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
D
k
ϕ
(
t
m
−(
k
−1)
)
2
H
)
1
/
2
.
Setting
C
=
k
h(
t
)
k
K
m
=
(Z
T
0
k
u
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
v
k
k
2
H
|
(
h
y
(0)
,
ϕ
e
(0)
i
H
| ≤
C
(Z
T
0
k
Bϕ
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
D
k
ϕ
(
t
m
−(
k
−1)
)
2
H
)
1
/
2
.
This shows the neessary ondition of the Theorem.
Step 2: Toprove the suieny we need the following result when
B ≥
α >
0
.
Claim 1 Assume that there is
α >
0
suh that(Z
T
0
k
Bu
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
D
k
v
k
k
2
H
)
≥
α
(Z
T
0
k
u
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
v
k
k
2
H
)
then, for every
y
0
∈
H
there is
ϕ
0
∈
H
suh that the mild solution of (1)
with
h
(
t
) = (
ϕ
e
(
t
)
,
ϕ
e
(
t
1
)
, ..,
ϕ
e
(
t
k
)
..,
ϕ
e
(
t
m
))
∈ Km
andy
(0) =
y
0
satises
y
(
T
) = 0
.
To prove this Claim, we onsider for every
z
∈
H
the solutionϕ
of (2)satisfying
ϕ
(
T
) =
z
and the unique mildsolutiony
tothe problemy
′
(
t
) +
Ay
(
t
) =
B
ϕ
e
(
t
)
, t
∈
(0
, T
)
/
{
t
k
}k
∈
σ
m
1
,
∆
y
(
t
k
) =
I
k
y
(
t
k
) +
D
k
ϕ
e
(
t
k
)
,
y
(
T
) = 0
.
Next, we introdue a bounded linear operator
Λ :
H
→
H
dened byΛ
z
=
−
y
(0)
.
Aording to formula(11) and the Corollary1 wehave
|h
Λ
z, z
i|
=
|−h
y
(0)
,
ϕ
e
(0)
i|
=
Z
T
0
h
B
ϕ
e
(
t
)
,
ϕ
e
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
k
ϕ
e
(
t
k
)
,
ϕ
e
(
t
k
)
i
=
Z
T
0
h
Bϕ
(
t
)
, ϕ
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
k
ϕ
(
t
m
−(
k
−1)
)
, ϕ
(
t
m
−(
k
−1)
)
i
≤
ς
(Z
T
0
k
ϕ
(
t
)
k
2
dt
+
k
X
=
m
k
=1
k
ϕ
(
t
k
)
k
2
ς
= sup
k
∈
σ
m
0
{
d
k
}
<
∞
.
We have
Z
T
0
k
ϕ
(
t
)
k
2
dt
=
Z
t1
0
k
ϕ
(
t
)
k
2
dt
+
Z
t2
t1
k
ϕ
(
t
)
k
2
dt
+
...
+
Z
T
t
m
k
ϕ
(
t
)
k
2
dt.
Sine there is noimpulsein the interval
[
t
k
, t
k
+1
)
we havek
ϕ
(
t
)
k
=
ϕ
(
t
+
k
)
,
for everyt
∈
[
t
k
, t
k
+1
)
, k
∈
σ
m
0
,
ϕ
(
t
−
k
+1
)
=
ϕ
(
t
+
k
−1
)
, k
∈
σ
0
m
.
(13)Therefore, there are
τ
k
+1
=
t
k
+1
−
t
k
>
0
,k
∈
σ
m
0
suh thatR
t
k
+1
t
k
k
ϕ
(
t
)
k
2
dt
≤
ρ
k
ϕ
(
t
+
k
)
2
=
τ
k
+1
I
k
ϕ
(
t
−
k
) +
ϕ
(
t
−
k
)
2
, k
∈
σ
m
1
.
(14)On the other hand,the ontinuity of
I
k
impliesthatϕ
(
t
+
k
)
2
=
(
I
k
+
I
)
ϕ
(
t
−
k
)
2
≤
(1 +
L
(
I
k
))
2
ϕ
(
t
−
k
)
2
, k
∈
σ
m
1
.
(15)It follows from(14) and (15) that
R
t
k
+1
t
k
k
ϕ
(
t
)
k
2
dt
≤
τ
k
+1
(1 +
L
(
I
k
))
2
ϕ
(
t
−
k
)
2
, k
∈
σ
1
m
.
(16)Sine
m
isnite, and due to (13),(16), then there is a onstant0
< µ <
∞
suh that
h
Λ
z, z
i ≤
µ
k
z
k
2
,
and thus,Λ
isbounded.Now, as
B
is nonnegativeinK
m
, we havekB
ξ
(
t
)
k ≥
α
{
(
ξ
(
t
)
, ξ
(
t
))
Km
}
1
/
2
for all
ξ
∈ K
m
; thus, by virtue of Lemma 2,we have(Z
T
0
(
Bu
(
t
)
, u
(
t
))
H
dt
+
k
X
=
m
k
=1
(
D
k
v
k
, v
k
)
H
)
(17)
≥
α
kBk
(Z
T
0
k
u
(
t
)
k
2
H
dt
+
k
X
=
m
k
=1
k
v
k
k
2
H
h
Λ
z, z
i
=
−h
y
(0)
,
ϕ
e
(0)
i
=
Z
T
0
h
Bϕ
(
t
)
, ϕ
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
k
ϕ
(
t
m
−(
k
−1)
)
, ϕ
(
t
m
−(
k
−1)
)
i
≥
α
kBk
(Z
T
0
k
ϕ
(
t
)
k
2
dt
+
k
X
=
m
k
=1
k
ϕ
(
t
k
)
k
2
)
≥
α
kBk
Z
t1
0
k
ϕ
(
t
)
k
2
dt
=
kBk
αt
1
k
z
k
2
=
θ
k
z
k
2
,
beausethere is noimpulse beforetime
t
1
. Therefore,Λ
isoerive onH
.To show that there is a bijetion from
H
ontoH,
it sues to prove thatΛ +
I
is abijetion fromH
ontoH.
Clearly,Λ +
I
is injetive sineh
Λ
z
+
z, z
i
=
h
Λ
z, z
i
+
h
z, z
i ≥
(
θ
+ 1)
k
z
k
2
.
On the other hand,let
y
0
∈
H
,asthe form
a
(
f, g
) +
h
f, g
i
=
h
Λ
f, g
i
+
h
f, g
i
is symmetri and oerive, then, by virtue of Lax-Milgram Theorem, there
is anelement
f
∈
H
suh thata
(
f, g
) +
h
f, g
i
=
h
y
0
, g
i
,for allg
∈
H.
This implies that
Λ(
H
) =
H
. Thus, for everyy
0
∈
H
, there is a unique
z
∈
H
suhthatΛ(
z
) =
−
y
0
, whihompletes the proof of Claim 1.
Step 3: Assume that
B, D
k
≥
0
, thenB ≥
0
,
e
B
2
=
B,
D
e
k
2
=
D
k
.
We denefor
ε >
0
,
β
ε
+
B
e
2
+
εI,
δ
ε
k
+
D
e
2
k
+
εI,
and
B
ε
+
(
β
ε
;
δ
1
ε
, .., δ
ε
m
) = (
B
e
2
+
εI
;
D
e
2
1
+
εI, ..,
D
e
2
m
+
εI
)
.
Aording toClaim1,there is
ϕ
˜
0
,ε
∈
H
suhthat the mildsolution
y
ε
of (1)with
y
ε
(0) =
y
0
satises
y
ε
(
T
) = 0;
whereB
(
h)
has been replaed by−h
y
(0)
,
ϕ
e
ε
(0)
i
=
Z
T
0
h
β
ε
ε
ϕ
e
(
t
)
,
ϕ
e
ε
(
t
)
i
dt
+
k
X
=
m
k
=1
h
δ
ε
k
ϕ
e
ε
(
t
k
)
,
ϕ
e
ε
(
t
k
)
i
,
(18)and (7) gives
−h
y
(0)
,
ϕ
e
ε
(0)
i ≤
C
(Z
T
0
h
e
B
2
ϕ
ε
(
t
)
, ϕ
ε
(
t
)
i
dt
+
k
X
=
m
k
=1
h
D
e
k
2
ϕ
ε
(
t
m
−(
k
−1)
)
, ϕ
ε
(
t
m
−(
k
−1)
)
i
)
1
/
2
.
(19)
Whene,
−h
y
(0)
,
ϕ
e
ε
(0)
i ≤
C
(Z
T
0
h
β
ε
ϕ
ε
(
t
)
, ϕ
ε
(
t
)
i
dt
+
k
X
=
m
k
=1
h
δ
k
ε
ϕ
ε
(
t
m
−(
k
−1)
)
, ϕ
ε
(
t
m
−(
k
−1)
)
i
)
1
/
2
.
(20)
It follows atone from (18), (19) and (20) that
ε
R
T
0
k
ϕ
ε
(
t
)
k
2
dt
+
k
P
=
m
k
=1
k
ϕ
ε
(
t
k
)
k
2
+
R
0
T
h
Bϕ
e
ε
(
t
)
,
Bϕ
e
ε
(
t
)
i
dt
+
k
P
=
m
k
=1
h
e
D
k
ϕ
ε
(
t
m
−(
k
−1)
)
,
D
e
k
ϕ
ε
(
t
m
−(
k
−1)
)
i
=
R
0
T
(
β
ε
ϕ
ε
(
t
)
, ϕ
ε
(
t
))
dt
+
k
P
=
m
k
=1
h
δ
ε
k
ϕ
ε
(
t
m
−(
k
−1)
)
, ϕ
ε
(
t
m
−(
k
−1)
))
≤
C
2
.
(21)
Step 4: Aording to the estimate(20) the family
b
ε
=
B
ε
(
ϕ
e
ε
(
t
);
ϕ
e
ε
(
t
1
)
...,
ϕ
e
ε
(
t
m
)
= (
B
e
2
ε
ϕ
e
(
t
);
D
e
1
2
ϕ
e
ε
(
t
1
)
...,
D
e
m
ϕ
e
ε
(
t
m
)) +
ε
(
ϕ
e
ε
(
t
);
ϕ
e
ε
(
t
1
)
...,
ϕ
e
ε
(
t
m
))
is ontainedin abounded subset
K
m
.
Thus, both of the families
√
ε
(
ϕ
e
ε
(
t
);
ϕ
e
ε
(
t
1
)
...,
ϕ
e
ε
(
t
m
))
and(
B
ϕ
e
ε
(
t
);
D
1
ϕ
e
ε
(
t
1
)
..., D
m
ϕ
e
ε
(
t
m
))
are bounded in
K
m
. Therefore, we may extrat asubfamily, say(
B
e
2
ϕ
e
ε
(
t
);
D
e
1
2
ϕ
e
ε
(
t
1
)
...,
D
e
2
m
ϕ
e
ε
(
t
m
))+
ε
(
ϕ
e
ε
(
t
);
ϕ
e
ε
(
t
1
)
...,
ϕ
e
ε
(
t
m
))
⇀
B
h,
weakly inK
m
.
Step 5: Takingthe limitas
ε
→
0
,wesee thatthe solutiony
of(1)with initial onditiony
(0) =
y
0
, h being as in step 4 satises
y
(
T
) = 0
. Thisompletes the proof of Theorem 1.
As an immediate appliation of the foregoing Theorem we give the
fol-lowingexample.
Example. One dimensionalimpulsiveShrödinger equation :
We onsider the problem
∂y
(
t, x
)
∂t
+
i
∂
2
y
∂x
2
(
t, x
) =
χ
ω0
u
(
t, x
)
, t
∈
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
, x
∈
Ω = (0
,
2
π
)
,
y
(
t,
0) =
y
(
t,
2
π
) = 0
,
(22)y
(0
, x
) =
y
0
,
∆
y
(
t
k
, x
) =
iα
k
y
(
t
k
, x
) +
χ
ω
k
v
k
(
x
)
, k
∈
σ
m
1
,
where
t
k
+1
−
t
k
>
2
π, ω
k
= (
a
k
1
, a
2
k
)
⊂
Ω
, k
∈
σ
0
m
,
{
α
k
}k
∈
σ
m
1
⊂
R
+
.
Let
H
=
L
2
(Ω
,
C
)
, Aw
(
x
) =
i
∂
2
w
∂x
2
(
x
)
, D
(
A
) =
n
w
∈
H,
∂
∂x
2
w
2
∈
H, w
(0) =
w
(
π
) = 0
o
,
and
I
k
w
(
x
) =
iα
k
w
(
x
)
and the ontrol operator is given byB
=
χ
ω0
, D
k
=
χ
ω
k
, then the system (22) beomes an abstrat formulation of (1). As aonsequene of Theorem 1, the initial state
y
0
∈
L
2
(Ω
,
C
) =
H
of the
solution of (22) is null-ontrollable at
t
=
T,
if and only if, there isC >
0
suh that
Z
Ω
y
0
(
x
)
ϕ
e
0
(
x
)
dx
(23)≤
C
(Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
+
m
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
)
1
2
where
ϕ
e
0
(
x
) =
ϕ
(
T, x
)
and
ϕ
is the mildsolutionof∂ϕ
(
t, x
)
∂t
+
i
∂
2
ϕ
(
t, x
)
∂x
2
= 0
,
t
∈
(0
, T
)
\ {
t
k
}k
∈
σ
m
1
, x
∈
Ω
,
ϕ
(
t,
0) =
ϕ
(
t,
2
π
) = 0
,
ϕ
(0
, x
) =
ϕ
0
(
x
)
, x
∈
Ω
,
∆
ϕ
(
t
k
, x
) =
iα
k
ϕ
(
t
k
, x
)
, x
∈
Ω
, k
∈
σ
1
m
.
Here
ϕ
is given byϕ
(
t
) =
ϕ
[0]
(
t
)
,
ift
∈
[
t
0
, t
1
)
ϕ
[
k
]
(
t
)
,
ift
∈
t
k
, t
k
+1)
ϕ
[
m
]
(
t
)
,
ift
∈
[
t
m
, T
]
,
where
ϕ
[
k
]
(
t
)
is asolution of the lassialShrödinger equation∂ϕ
[
k
]
(
t, x
)
∂t
+
i
∂
2
ϕ
[
k
]
∂x
2
(
t, x
) =
χ
ω0
u
(
t, x
)
,
t
∈
(
t
0
, t
1
)
, x
∈
Ω = (0
,
2
π
)
,
ϕ
[
k
]
(
t,
0) =
ϕ
[
k
]
(
t,
2
π
) = 0
,
ϕ
[0]
(
t
0
, x
) =
ϕ
0
(
x
)
, x
∈
Ω
,
and
∂ϕ
[
k
]
(
t, x
)
∂t
+
i
∂
2
ϕ
[
k
]
∂x
2
(
t, x
) =
χ
ω0
u
(
t, x
)
, t
∈
(
t
k
, t
k
+1
)
,
x∈
Ω = (0
,
2
π
)
,
ϕ
[
k
]
(
t,
0) =
ϕ
[
k
]
(
t,
2
π
) = 0
,
ϕ
[
k
]
(
t
k
, x
) = (1 +
iα
k
)
ϕ
[
k
−1]
(
t
k
, x
)
, x
∈
Ω
, k
∈
σ
m
1
.
Then a standard appliation of a variant of Ingham's Inequality [8℄ shows
that
Z
t
k
+1
t
k
Z
w0
ϕ
[
k
]
(
t, x
)
dtdx
≥
c
(
τ
k
, w
0
)
Z
Ω
ϕ
[
k
]
(
t
+
k
, x
)
dx,
for some positiveonstants
c
(
τ
k
, w
0
)
>
0
. Summingup we getm
X
k
=0
Z
t
k
+1
t
k
Z
w0
ϕ
[
k
]
(
t, x
)
dtdx
=
Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
≥
c
1
m
X
k
=1
Z
Ω
ϕ
[
k
]
where
c
1
= min
k
∈
σ
m
0
c
(
τ
k
, w
0
)
>
0
.
On the other hand, there isa positive onstant
c
2
>
0
suh thatm
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
≥
c
2
m
X
k
=1
Z
Ω
ϕ
[
k
]
2
(
t
+
k
, x
)
dx.
It follows that
Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
+
m
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
≥
(
c
1
+
c
2
)
m
X
k
=1
Z
Ω
ϕ
[
k
]
2
(
t
+
k
, x
)
dx
≥
(
c
1
+
c
2
)
Z
Ω
ϕ
[
m
]
2
(
t
+
m
, x
)
dx
= (
c
1
+
c
2
)
Z
Ω
|
ϕ
|
2
(
T, x
)
dx.
Now, sine
ϕ
e
0
(
x
) =
ϕ
e
(0
, x
) =
ϕ
(
T, x
)
,
then,Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
+
m
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
≥
m
(
c
1
+
c
2
)
Z
Ω
e
ϕ
0
2
(
x
)
dx,
from whih we get
Z
Ω
e
ϕ
0
2
(
x
)
dx
≤
1
m
(
c
1
+
c
2
)
Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
+
m
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
!
.
We onlude by Cauhy-Shwarz inequality that
Z
Ω
y
0
(
x
)
ϕ
e
0
(
x
)
dx
≤
Z
Ω
y
0
2
(
x
)
dx
Z
Ω
e
ϕ
0
2
(
x
)
dx
1
/
2
≤
(R
Ω
|
y
0
|
2
(
x
)
dx
m
(
c
1
+
c
2
)
)
1
/
2
Z
T
0
Z
ω0
|
ϕ
|
2
(
t, x
)
dxdt
+
m
X
k
=1
Z
ω
k
|
ϕ
|
2
(
t
m
−(
k
−1)
, x
)
dx
!
1
/
2
stated in Theorem 1.
We onlude our paper by a speial ase when our initial state is an
eigensolution of the followinglinear operator
Γ :
H
→
H
dened byΓ(
ψ
) =
Z
T
0
X
−1
(
s
)
B
2
X
(
s
)
ψds
+
k
X
=
m
k
=1
X
−1
(
t
k
)
D
2
k
X
(
t
k
)
ψ.
We have the following result ofnull-ontrollability.
Proposition 1 Let
λ >
0
be an eigenvalue ofΓ
with eigenvetorψ
∈
H.
Then, the solution
y
to the problem
y
′
(
t
) +
Ay
(
t
) =
−
λ
1
B
2
(
X
(
t
)
ψ
)
,
t
∈
(0
, T
)
/
{
t
k
}k
∈
σ
m
1
,
∆
y
(
t
k
) =
I
k
y
(
t
k
)
−
1
λ
D
2
k
(
X
(
t
k
)
ψ
)
, k
∈
σ
1
m
y
(0) =
ψ,
(24)
satises
y
(
T
) = 0
.
Proof.
Write system (24) intothe form
y
′
(
t
) +
Ay
(
t
) =
−
λ
1
B
2
(
X
(
t
)
ψ
)
,
t
∈
(0
, T
)
/
{
t
k
}k
∈
σ
m
1
,
y
(
t
+
k
) =
I
k
y
(
t
k
)
−
1
λ
D
2
k
(
X
(
t
k
)
ψ
)
, k
∈
σ
1
m
y
(0) =
ψ.
Therefore, this impulsive problem has a solution whih an be represented
expliitlyas follows
y
(
t
) =
X
(
t
)
ψ
+
Z
t
0
G
(
t, s
)
−
λ
1
B
2
(
X
(
s
)
ψ
ds
+
X
0
<t
k≤
t
G
(
t, t
k
)
−
λ
1
D
2
k
X
(
t
k
)
ψ
,
where the evolutionoperator
G
(
t, s
)
isgiven byy
(
T
) =
X
(
T
)
ψ
+
Z
T
0
G
(
T, s
)
−
λ
1
B
2
(
X
(
s
)
ψ
ds
+
X
0
<t
k≤
T
G
(
T, t
k
)
−
λ
1
D
k
2
X
(
t
k
)
ψ
=
X
(
T
)
ψ
+
Z
T
0
X
−1
(
T
)
G
(
T, s
)
−
λ
1
B
2
(
X
(
s
)
ψ
ds
−
λ
1
X
0
<t
k≤
T
X
−1
(
T
)
G
(
T, t
k
)
D
k
2
X
(
t
k
)
ψ
#
=
X
(
T
)
ψ
+
Z
T
0
X
−1
(
s
)
−
λ
1
B
2
(
X
(
s
)
ψ
ds
−
λ
1
X
0
<t
k≤
T
X
−1
(
t
k
)
D
2
k
X
(
t
k
)
ψ
#
=
X
(
T
))
ψ
−
1
λ
Γ (
ψ
)
= 0
.
This shows thatthe initialstate
ψ
isnull-ontrollableattimeT
withontrolh
(
t
) =
u
(
t
)
,
{
v
k}k
∈
σ
m
1
=
−
1
λ
X
(
t
)
ψ,
−
1
λ
X
(
t
k
)
ψ
k
∈
σ
m
1
!
,
whih ompletes the proof of the Proposition.
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