Directory UMM :Journals:Journal_of_mathematics:EJQTDE:

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 Spa
e
R. BOUKHAMLA(1) & S. MAZOUZI(2)
(1) Centre Universitaire de Souk-Ahras, Souk-Ahras 41000, Algeria.
(2) Département de Mathématiques, Université d'Annaba, BP 12, ANNABA 23000, Algeria.

Abstra
t

We shall establish a ne
essary and su
ient
ondition under whi
h
we have the null
ontrollability of some rst order impulsive evolution
equation in a Hilbert spa

e.
MSC(2000) : 34A37, 93B05, 93C15.
Keywords: Null-
ontrollability, impulsive

onditions, mild solutions,

evolution equation.

1

Introdu
tion

The problem of exa
t
ontrollability of linear systems represented by innite
onservative systems has been extensively studied by several authors A.
Haraux [8℄, R.Triggiani [16℄, Z.H. Guan, T.H. Qian, and X.Yu [7℄, see also
the referen

es [1, 2, 6, 10,15℄. In the sequel, we shall be
on
erned with the
problem of null
ontrollability of some rst order evolution equation subje
t to impulsive
onditions and so we shall derive a ne
essary and su
ient
ondition under whi
h null
ontrollability o

urs. A
tually, we shall establish an equivalen
e between the null-
ontrollability and some observability
inequality in somehow more general framework than that proposed by A Haraux [8℄. Regarding the literature on the impulsive dierential equations we
refer the reader to the works of D.D. Bainov and P.S. Simeonov [3, 4℄ and

EJQTDE, 2007 No. 19, p. 1

the referen
es [5, 9,11, 12, 13℄. We are going to study the following problem
′

y (t) + Ay (t) = Bu (t) ,

t ∈ (0, T ) \ {tk }k∈σm ,
1

0

y (0) = y ,
∆y (tk ) = Ik y (tk ) + Dk vk , k ∈ σ1m ,

(1)
(1k )

where the nal time T is a positive number, y 0 is an initial

ondition in a
Hilbert spa
e H endowed with an inner produ
t h., .iH , y (t) : [0, T ] → H is a
ve
tor fun
tion, σ1m is a subset of N given by σ1m = {1, 2, ..., m}, and nally,
{tk }k∈σm is an in
reasing sequen
e of numbers in the open interval (0, T ) ,
1
and ∆y (tk ) denotes the jump of y (t) at t = tk , i.e.,







−
∆y (tk ) = y t+
k − y tk

where y t+
and y t−
represent the right and left limits of y (t) at t = tk
k
k
respe
tively. On the other hand, the operators A, B, Ik , Dk : H → H are
given linear bounded operators. Moreover, we set the following assumptions:
(H1) A∗ = −A,
(H2) Ik∗ = −Ik , for every k ∈ σ1m , and for ea
h k ∈ σ1m , the operator
Ik = Ik + I is invertible,
(H3) B ∗ = B ≥ 0 and there is d0 > 0 su
h that
(Bu, u)H ≤ d0 kuk2H , for all u ∈ H,
(H4) Dk∗ = Dk ≥ 0,

dk > 0 su

h that

for every k ∈ σ1m , and for ea
h k ∈ σ1m there is

(Dk u, u)H ≤ dk kuk2H , for all u ∈ H.

In the sequel we shall designate by

h

the fun
tion



h (t) = u (t) , {vk }k∈σm ,
1





where u (t) ∈ L2 (0, T ) \ {tk }k∈σ1m ; H and
{vk }k∈σm ∈ l
1

2

(σ1m ; H)

n
o
+ {vk }k∈σm , vk ∈ H .
1

EJQTDE, 2007 No. 19, p. 2





We point out that the spa
e Km = L2 (0, T ) \ {tk }k∈σ1m ; H × l2 (σ1m ; H) is
a Hilbert spa
e with respe
t to the inner produ
t



h, e
h

Km

=

Z

T
0

(u (t) , u
e (t))H dt +

m
X
k=1

(vk , e
vk )H ,

e
dened for all h = (u (t) , {vk }m
u (t) , {e
vk }m
k=1 ) and h = (e
k=1 ) ∈ Km .
We shall denote by B the

ontrol operator given by





 
B = B, {Dk }k∈σm ∈ L L2 (0, T ) \ {tk }k∈σm ; H × l2 (σ1m ; H) ,
1

1

so that



Bh (t) = Bu (t) , {Dk vk }k∈σm .
1

We have for every h =

(u(t), {vk }m
k=1 )

(Bh, h)Km =
=

Z

∈ Km

T

(Bu (t) , u (t))H dt +

0

Z

m
X

(Dk vk , vk )H

k=1

T

(u (t) , Bu (t))H dt +

0

m
X

(vk , Dk vk )H

k=1

= (h, Bh)Km ,

whi
h shows that B∗ = B, that is, B is self-adjoint. On the other hand, we
have
(Bh, h)Km =

Z

T
0

≤ d0
≤ δ

(Bu (t) , u (t))H dt +
Z

m
X

(Dk vk , vk )H

k=1

T
0

ku (t)k2H dt +

khk2Km

,

m
X
k=1

dk kvk k2H

where δ = max {d0 , d1 , ..., dm } . Thus, the operator is B bounded in Km .
Next, we
onsider the homogeneous system asso
iated with (1) :
′

ϕ (t) + Aϕ (t) = 0, t ∈ (0, T ) \ {tk }k∈σm ,
1

ϕ (0) = ϕ0 ,
∆ϕ (tk ) = Ik ϕ (tk ) , k ∈ σ1m .

(2)
(2k )

EJQTDE, 2007 No. 19, p. 3

We point out that on ea
h interval [tk , tk+1 ), for k = 0, ..., m, the solution ϕ
is left
ontinuous at ea
h time tk .
Consider the
orresponding homogeneous ba
kward problem :
(3)

′

−ϕ˜ (t) + Aϕ(t)
˜
= 0, , t ∈ (0, T ) \ {tk }k∈σm ,
1

0

ϕ(T
˜ ) = ϕ ,
m
∆ϕ(t
˜ m−(k−1) ) = −I˜m−(k−1) ϕ(t
˜ +
m−(k−1) ), k ∈ σ1 ,

(3k )

where
∗
A = A∗ = −A, I˜m−(k−1) = Im−(k−1)
= −Im−(k−1) , k ∈ σ1m .

We observe that the problem (3) on the interval [tm , T ] is equivalent to the
lassi
al ba
kward problem
′

−ϕ˜ (t) + Aϕ(t)
˜ = 0, t ∈ [tm , T ] ,
0
ϕ(T
˜ )=ϕ .

We introdu
e the following spa
e : PC ([0, T ] ; H) = {y, y : [0, T ] → H
su
h that y(t) is
ontinuous at t 6= tk , and has dis
ontinuities of rst kind at
t = tk , for every k ∈ σ1m }.
Evidently, PC ([0, T ] ; H) is a Bana
h spa
e with respe
t to the norm
kykPC = sup ky(t)k .
t∈(0,T )

On the other hand, we dene the subspa
es PLC , (respe
tively, PRC )=
{y, y ∈ PC su
h that y(t) is left ( respe
tively, right)
ontinuous at t = tk ,
for every k ∈ σ1m }.

Remark 1 1) The spa
e PLC , (respe
tively, PRC )
an be identied to a
subspa
e of Km . That is, to ea
h y ∈ PLC , (respe
tively, y˜ ∈ PRC ) is
assigned the fun
tion h (respe
tively, eh) dened by




h (t) = y (t) , {y (tk )}k∈σ1m ,

and



e
h (t) = y˜ (t) , {˜
y (tk )}k∈σm .
1

The mapping y 7→ h (t) (respe
tively, y˜ 7→ he) is a linear inje
tion.

EJQTDE, 2007 No. 19, p. 4

2) Let ye ∈ PRC , the fun
tion y
an be written as :

 ye[0] (t)
ye[k] (t)
ye(t) =

ye[m] (t)

if t ∈ [t0 , t1 )
if t ∈ [tk , tk+1 )
if t ∈ [tm , T ] .

Next, let τk = tk − tk−1 , we dene the operator T : D(T ) = PRC ⊂ Km →

Km by





τ1
+ t0 )
ye[0] ((T − t) τm+1
if
t ∈ [tm , T ] ,



τk+1
ye[k]((tm−(k−1) − t) τm−(k−1) + tk ) if t ∈ tm−k , tm−(k−1) , k ∈ σ1m−1 ,
(T ye)(t) =


+ tm )
if
t ∈ (0, t1 ] .
ye[m] ((t1 − t) τm+1
τ1

(4)
We note that the range of T is exa
tly PLC . The fun
tion (T ye)(t)
an be
written as follows:

 y[0] (t)
y[k](t)
(T ye)(t) =

y[m] (t)

if t ∈ [t0 , t1 ] ,
if t ∈ (tk , tk+1] , k ∈ σ1m−1 ,
if t ∈ (tm , T ] .

Let X (t) be the resolvent solution of the operator system
′

X (t) + AX(t) = 0, 0 = t0 < t < tm+1 = T, t 6= tk , k = 1, 2, ..., m,
X(0) = I,
X(tk + 0) − X(tk − 0) = Ik X(tk ), k = 1, 2, ..., m,

where I : H → H is the identity operator. We shall suppose that the operator
Ik = Ik + I has a bounded inverse.

Denition 1 A fun
tion y ∈ PC ([0, T ] ; H)is a mild solution to the impul-

sive problem (1) if the impulsive
onditions are satised and
y(t) =

Rt
+ 0
G(t,
0
)y
+
G(t, s)Bu (s) ds
0
P
+ 0 0.

Claim 1 Assume that there is α > 0 su
h that
(Z

T
0

kBu(t)k2H dt +

k=m
X
k=1

kDk vk k2H

)

≥α

(Z

T
0

ku(t)k2H dt +

k=m
X
k=1

kvk k2H

)

then, for every y 0 ∈ H there is ϕ0 ∈ H su
h that the mild solution of (1)
with
e
ϕ(t
e 1 ), .., ϕ(t
e k ).., ϕ(t
e m )) ∈ Km and y(0) = y 0
h(t) = (ϕ(t),
satises y(T ) = 0.

To prove this Claim, we
onsider for every z ∈ H the solution ϕ of (2)
satisfying ϕ(T ) = z and the unique mild solution y to the problem
/
′
y (t) + Ay(t) = B ϕ(t),
e
t ∈ (0, T ) {tk }k∈σm ,
1

∆y(tk ) = Ik y(tk ) + Dk ϕ(t
e k ),
y(T ) = 0.

Next, we introdu
e a bounded linear operator Λ : H → H dened by
Λz = −y(0).

A

ording to formula (11) and the Corollary 1 we have
Z

k=m
 T

X


|hΛz, zi| = |−hy(0), ϕ(0)i|
e
=
hB ϕ(t),
e
ϕ(t)idt
e
+
hDk ϕ(t
e k ), ϕ(t
e k )i
 0

k=1
Z

k=m
 T

X


= 
hDk ϕ(tm−(k−1) ), ϕ(tm−(k−1) )i
hBϕ(t), ϕ(t)idt +
 0

k=1
)
(Z
k=m
T
X
kϕ(tk )k2 ,
≤ ς
kϕ(t)k2 dt +
0

k=1

EJQTDE, 2007 No. 19, p. 11

where
ς = sup {dk } < ∞.
k∈σ0m

We have
Z

T

0

2

kϕ(t)k dt =

Z

t1

2

kϕ(t)k dt +

0

Z

t2

t1

2

kϕ(t)k dt + ... +

Z

T

tm

kϕ(t)k2 dt.

Sin
e there is no impulse in the interval [tk , tk+1 ) we have


, for every t ∈ [tk , tk+1 ) , k ∈ σ m ,
kϕ(t)k =
ϕ(t+
)
0
k
−

+

ϕ(t )
=
ϕ(t )
, k ∈ σ m .
0
k−1
k+1

(13)

Therefore, there are τk+1 = tk+1 − tk > 0, k ∈ σ0m su
h that
R tk+1
tk


2


= τk+1
Ik ϕ(t− ) + ϕ(t− )
2 , k ∈ σ m . (14)
kϕ(t)k2 dt ≤ ρk
ϕ(t+
)
1
k
k
k

On the other hand, the
ontinuity of Ik implies that

+
2




ϕ(t )
=
(Ik + I)ϕ(t− )
2 ≤ (1 + L(Ik ))2
ϕ(t− )
2 , k ∈ σ m .
1
k
k
k

(15)

It follows from (14) and (15) that
R tk+1
tk


2
, k ∈ σ1m .
kϕ(t)k2 dt ≤ τk+1 (1 + L(Ik ))2
ϕ(t−
k)

(16)

Sin
e m is nite, and due to (13),(16), then there is a
onstant 0 < µ < ∞
su
h that hΛz, zi ≤ µ kzk2 , and thus, Λ is bounded.
Now, as B is nonnegative in Km , we have
kBξ (t)k ≥ α {(ξ (t) , ξ (t))Km }1/2

for all ξ ∈ Km ; thus, by virtue of Lemma 2, we have
(Z

T

(Bu(t), u(t))H dt +
0

≥ α kBk

(Z

T
0

k=m
X

ku(t)k2H dt +

(Dk vk , vk )H

k=1
k=m
X
k=1

kvk k2H

)

)

(17)

.

EJQTDE, 2007 No. 19, p. 12

It follows from (11), (17) and Corollary 1 that
hΛz, zi = −hy(0), ϕ(0)i
e
Z T
k=m
X
=
hBϕ(t), ϕ(t)idt +
hDk ϕ(tm−(k−1) ), ϕ(tm−(k−1) )i
0

≥ α kBk
≥ α kBk

(Z
Z

0

0

t1

k=1

T

2

kϕ(t)k dt +

k=m
X
k=1

kϕ(tk )k

2

)

kϕ(t)k2 dt = kBk αt1 kzk2 = θ kzk2 ,

be
ause there is no impulse before time t1 . Therefore, Λ is
oer
ive on H .
To show that there is a bije
tion from H onto H, it su
es to prove that
Λ + I is a bije
tion from H onto H. Clearly, Λ + I is inje
tive sin
e
hΛz + z, zi = hΛz, zi + hz, zi ≥ (θ + 1) kzk2 .

On the other hand, let y 0 ∈ H , as the form a(f, g) + hf, gi = hΛf, gi + hf, gi
is symmetri
and
oer
ive, then, by virtue of Lax-Milgram Theorem, there
is an element f ∈ H su
h that
a(f, g) + hf, gi = hy 0, gi, for all g ∈ H.

This implies that Λ(H) = H . Thus, for every y 0 ∈ H , there is a unique
z ∈ H su
h that Λ(z) = −y 0 , whi
h
ompletes the proof of Claim 1.
Step 3: Assume that B, Dk ≥ 0, then B ≥ 0,
We dene for ε > 0,

e 2 = B, D
e 2 = Dk .
B
k
e 2 + εI,
βε + B
e 2 + εI,
δε + D
k

and

k

ε
e 2 + εI; D
e 2 + εI, .., D
e 2 + εI).
Bε + (β ε ; δ1ε , .., δm
) = (B
1
m

A

ording to Claim 1, there is ϕ˜0,ε ∈ H su
h that the mild solution yε of (1)
with yε (0) = y 0 satises yε (T ) = 0; where B(h) has been repla
ed by
Bε (ϕ(t),
e
ϕ(t
e 1 ), .., ϕ(t
e k ).., ϕ(t
e m )) ∈ Km .

EJQTDE, 2007 No. 19, p. 13

We obtain from (11) and Corollary 1

−hy(0), ϕ
eε (0)i =

Z

T

hβεε ϕ(t),
e
ϕ
eε (t)idt +

0

and (7) gives

−hy(0), ϕ
eε(0)i ≤ C

(Z

T
0

When
e,

−hy(0), ϕ
eε(0)i ≤ C

(Z

e 2 ϕε (t), ϕε (t)idt +
hB

T
ε

0

hβ ϕε (t), ϕε (t)idt +

k=m
X

hδkε ϕ
eε (tk ), ϕ
eε (tk )i,

k=1

k=m
X
k=1

e 2 ϕε (tm−(k−1) ), ϕε (tm−(k−1) )i
hD
k

hδkε ϕε (tm−(k−1) ), ϕε (tm−(k−1) )i
(20)

It follows at on
e from (18), (19) and (20) that

ε


RT

=

0

RT
0

2

kϕε (t)k dt +
+

RT
0

k=m
P
k=1

kϕε (tk )k



k=m
e ε (t), Bϕ
e ε (t)idt + P hD
e k ϕε (tm−(k−1) ), D
e k ϕε (tm−(k−1) )i
hBϕ

(β ε ϕε (t), ϕε (t))dt +

k=1

k=m
P
k=1

Step 4:

2

hδkε ϕε (tm−(k−1) ), ϕε (tm−(k−1) )) ≤ C 2 .

(21)

A

ording to the estimate (20) the family

bε = Bε (ϕ
eε (t); ϕ
eε (t1 )..., ϕ
eε (tm )
2
2
e ϕ(t);
e ϕ
emϕ
= (B
e
D
eε (t1 )..., D
eε (tm )) + ε(ϕ
eε (t); ϕ
eε (t1 )..., ϕ
eε (tm ))
ε

1

is
ontained in a bounded subset
Thus, both of the families

√

ε(ϕ
eε (t); ϕ
eε (t1 )..., ϕ
eε (tm ))

are bounded in

Km .

Km .

and

(B ϕ
eε (t); D1 ϕ
eε (t1 )..., Dm ϕ
eε (tm ))

Therefore, we may extra
t a subfamily, say

(B ϕ
eε (t); D1 ϕ
eε (t1 )..., Dm ϕ
eε (tm )) ⇀ h,

)1/2

(19)

k=m
X
k=1

(18)

weakly in

Km .

EJQTDE, 2007 No. 19, p. 14

)1/2

.

.

Then
learly
e2ϕ
e 2ϕ
e 2 eε (tm ))+ε(ϕ
(B
eε (t); D
eε (t); ϕ
eε (t1 )..., ϕ
eε (tm )) ⇀ Bh, weakly in Km .
1 eε (t1 )..., Dm ϕ

Taking the limit as ε → 0, we see that the solution y of (1) with
initial
ondition y(0) = y 0 , h being as in step 4 satises y(T ) = 0. This
ompletes the proof of Theorem 1.

As an immediate appli
ation of the foregoing Theorem we give the following example.
Example. One dimensional impulsive S
hrödinger equation :
We
onsider the problem
Step 5:

∂2y
∂y (t, x)
+ i 2 (t, x)
∂t
∂x
y(t, 0)
y (0, x)
∆y (tk , x)

= χω0 u (t, x) , t ∈ (0, T ) \ {tk }k∈σm , x ∈ Ω = (0, 2π),
1

(22)

= y(t, 2π) = 0,
= y0,
= iαk y (tk , x) + χωk vk (x), k ∈ σ1m ,

where
tk+1 − tk > 2π, ωk = (ak1 , ak2 ) ⊂ Ω, k ∈ σ0m , {αk }k∈σm ⊂ R+ .
1

Let

n
o
2
2
H = L2 (Ω, C), Aw(x) = i ∂∂xw2 (x), D(A) = w ∈ H, ∂∂xw2 ∈ H, w(0) = w(π) = 0 ,

and Ik w(x) = iαk w(x) and the
ontrol operator is given by B = χω0 , Dk =
χωk , then the system (22) be
omes an abstra
t formulation of (1). As a
onsequen
e of Theorem 1, the initial state y 0 ∈ L2 (Ω, C) = H of the
solution of (22) is null-
ontrollable at t = T, if and only if, there is C > 0
su
h that

Z


0

 y 0(x)ϕ
e
(x)dx


Ω
(Z Z
T

≤ C

0

2

ω0

|ϕ| (t, x)dxdt +

(23)
m Z
X
k=1

2

ωk

|ϕ| (tm−(k−1) , x)

) 21

, ∀ϕ
e0 ∈ L2 (Ω, C),

EJQTDE, 2007 No. 19, p. 15

where

ϕ
e0 (x) = ϕ(T, x)

and

ϕ

∂ϕ(t, x)
∂ 2 ϕ(t, x)
+i
∂t
∂x2
ϕ(t, 0)
ϕ (0, x)
∆ϕ (tk , x)

Here

ϕ

where

is the mild solution of

= 0,

t ∈ (0, T ) \ {tk }k∈σm , x ∈ Ω,
1

= ϕ(t, 2π) = 0,
= ϕ0 (x), x ∈ Ω,
= iαk ϕ (tk , x) , x ∈ Ω, k ∈ σ1m .

is given by

ϕ[k](t)


 ϕ[0] (t) ,
ϕ[k] (t) ,
ϕ(t) =

ϕ[m] (t) ,

if
if
if

t ∈ [t0 , t1 )

t ∈ tk , tk+1)
t ∈ [tm , T ] ,

is a solution of the
lassi
al S
hrödinger equation

∂ 2 ϕ[k]
∂ϕ[k] (t, x)
+i
(t, x) = χω0 u (t, x) , t ∈ (t0 , t1 ) , x ∈ Ω = (0, 2π),
∂t
∂x2
ϕ[k] (t, 0) = ϕ[k] (t, 2π) = 0,
ϕ[0] (t0 , x) = ϕ0 (x), x ∈ Ω,
and

∂ 2 ϕ[k]
∂ϕ[k] (t, x)
+i
(t, x) = χω0 u (t, x) , t ∈ (tk , tk+1 ) , x ∈ Ω = (0, 2π),
∂t
∂x2
ϕ[k] (t, 0) = ϕ[k] (t, 2π) = 0,
ϕ[k] (tk , x) = (1 + iαk )ϕ[k−1] (tk , x) , x ∈ Ω, k ∈ σ1m .
Then a standard appli
ation of a variant of Ingham's Inequality [8℄ shows
that

Z

tk+1
tk

Z

w0

 
ϕ[k]  (t, x)dtdx ≥ c(τk , w0 )

for some positive
onstants

m Z
X
k=0

tk+1

tk

Z

w0

c(τk , w0 ) > 0.

 
ϕ[k] (t, x)dtdx =

Z

Ω

  +
ϕ[k]  (t , x)dx,
k

Summing up we get

Z

0

T

Z

ω0

|ϕ|2 (t, x)dxdt

m Z
X
  +
ϕ[k]  (t , x)dx,
≥ c1
k
k=1

Ω

EJQTDE, 2007 No. 19, p. 16

where

c1 = minm c(τk , w0 ) > 0.
k∈σ0

On the other hand, there is a positive
onstant

m Z
X

ωk

k=1

It follows that

Z

0

T

Z

ω0

c2 > 0

su
h that

m Z
X
 2 +
ϕ[k]  (t , x)dx.
|ϕ| (tm−(k−1) , x) ≥ c2
k
2

k=1

|ϕ|2 (t, x)dxdt

m Z
X

+

k=1

ωk

Ω

|ϕ|2 (tm−(k−1) , x)

m Z
X
 2 +
ϕ[k]  (t , x)dx
≥ (c1 + c2 )
k
Ω

Zk=1

2
≥ (c1 + c2 ) ϕ[m]  (t+
m , x)dx
Ω
Z
= (c1 + c2 ) |ϕ|2 (T, x)dx.
Ω

ϕ
e (x) = ϕ(0,
e x) = ϕ(T, x), then,
Z
m Z
X
 0 2
2
2
|ϕ| (tm−(k−1) , x) ≥ m(c1 + c2 ) ϕ
e  (x)dx,
|ϕ| (t, x)dxdt +

Now, sin
e

Z

T

0

Z

ω0

0

k=1

ωk

Ω

from whi
h we get

Z

Ω

 0 2
ϕ
e  (x)dx ≤

1
m(c1 + c2 )

Z

T

0

Z

ω0

2

|ϕ| (t, x)dxdt +

m Z
X
k=1

ωk

2

|ϕ| (tm−(k−1) , x) .

We
on
lude by Cau
hy-S
hwarz inequality that


Z
1/2
Z
Z
 0 2
 0 2


0





 y 0(x)ϕ
y (x)dx
ϕ
e (x)dx
e (x)dx ≤

Ω
Ω
Ω
(R
)1/2 Z Z
0 2
T
|y
|
(x)dx
Ω
≤
|ϕ|2 (t, x)dxdt
m(c1 + c2 )
0
ω0
!1/2
m Z
X
|ϕ|2 (tm−(k−1) , x)dx
,
+
k=1

!

ωk

EJQTDE, 2007 No. 19, p. 17

whi
h establishes the ne
essary and su
ient
ondition of null
ontrollability
stated in Theorem 1.
We
on
lude our paper by a spe
ial
ase when our initial state is an
eigensolution of the following linear operator Γ : H → H dened by
Γ(ψ) =

Z

T

X

−1

2

(s)B X(s)ψds +

0

k=m
X

X −1 (tk )Dk2 X(tk )ψ.

k=1

We have the following result of null-
ontrollability.

Proposition 1 Let λ > 0 be an eigenvalue of Γ with eigenve
tor ψ ∈ H.
Then, the solution y to the problem

/
′

t ∈ (0, T ) {tk }k∈σm ,
 y (t) + Ay(t) = − λ1 B 2 (X(t)ψ),
1
1 2
m
D
(X(t
)ψ),
k
∈
σ
∆y(t
)
=
I
y(t
)
−
k
k
k
k
1
λ k


y(0) = ψ,

(24)

satises

y(T ) = 0.

Proof.
Write system (24) into the form

/
′
1 2

t ∈ (0, T ) {tk }k∈σm ,
 y (t) + Ay(t) = − λ B (X(t)ψ),
1
+
1 2
m
D
(X(t
)ψ),
k
∈
σ
y(t
)
=
I
y(t
)
−
k
k
k
1
k
λ k


y(0) = ψ.

Therefore, this impulsive problem has a solution whi
h
an be represented
expli
itly as follows
y(t) = X(t)ψ+

Z

0

t





X
1 2
1 2
G(t, tk ) − Dk X(tk )ψ ,
G(t, s) − B (X(s)ψ ds+
λ
λ
0