On controllability problems for the wave equation on a half-plane
Necessary and sufficient conditions for null-controllability and approximate null-controllability are obtained for the wave equation on a half-plane. Controls solving these problems are found explicitly. Moreover bang-bang controls solving the approximate null-controllability problem are constructed...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2005
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irk-123456789-1065672016-10-01T03:01:43Z On controllability problems for the wave equation on a half-plane Fardigola, L.V. Necessary and sufficient conditions for null-controllability and approximate null-controllability are obtained for the wave equation on a half-plane. Controls solving these problems are found explicitly. Moreover bang-bang controls solving the approximate null-controllability problem are constructed with the aid of the Markov power moment problem. 2005 Article On controllability problems for the wave equation on a half-plane / L.V. Fardigola // Журнал математической физики, анализа, геометрии. — 2005. — Т. 1, № 1. — С. 93-115. — Бібліогр.: 12 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106567 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Necessary and sufficient conditions for null-controllability and approximate null-controllability are obtained for the wave equation on a half-plane. Controls solving these problems are found explicitly. Moreover bang-bang controls solving the approximate null-controllability problem are constructed with the aid of the Markov power moment problem. |
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Article |
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Fardigola, L.V. |
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Fardigola, L.V. On controllability problems for the wave equation on a half-plane Журнал математической физики, анализа, геометрии |
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Fardigola, L.V. |
| author_sort |
Fardigola, L.V. |
| title |
On controllability problems for the wave equation on a half-plane |
| title_short |
On controllability problems for the wave equation on a half-plane |
| title_full |
On controllability problems for the wave equation on a half-plane |
| title_fullStr |
On controllability problems for the wave equation on a half-plane |
| title_full_unstemmed |
On controllability problems for the wave equation on a half-plane |
| title_sort |
on controllability problems for the wave equation on a half-plane |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2005 |
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http://dspace.nbuv.gov.ua/handle/123456789/106567 |
| citation_txt |
On controllability problems for the wave equation on a half-plane / L.V. Fardigola // Журнал математической физики, анализа, геометрии. — 2005. — Т. 1, № 1. — С. 93-115. — Бібліогр.: 12 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT fardigolalv oncontrollabilityproblemsforthewaveequationonahalfplane |
| first_indexed |
2025-07-07T18:39:03Z |
| last_indexed |
2025-07-07T18:39:03Z |
| _version_ |
1837014496693977088 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2005, v. 1, No. 1, p. 93�115
On controllability problems for the wave equation
on a half-plane
L.V. Fardigola
Department of Mechanics and Mathematics, V.N. Karazin Kharkov National University
4 Svobody Sq., Kharkov, 61077, Ukraine
E-mail:fardigola@univer.kharkov.ua
Mathematical Division, B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkov, 61103, Ukraine
E-mail:fardigola@ilt.kharkov.ua
Received September 20, 2004
Necessary and su�cient conditions for null-controllability and approxi-
mate null-controllability are obtained for the wave equation on a half-plane.
Controls solving these problems are found explicitly. Moreover bang-bang
controls solving the approximate null-controllability problem are constructed
with the aid of the Markov power moment problem.
0. Introduction
Controllability problems for hyperbolic partial di�erential equation were in-
vestigated in a number of papers (see, e.g., the references in [1]).
One of the most generally accepted ways to study control systems with dis-
tributed parameters is their interpretation in the form
dw
dt
= Aw+Bu; t 2 (0; T ); (0.1)
where T > 0, w : (0; T ) �! H is an unknown function, u : (0; T ) �! H is
a control, H, H are Banach spaces, A is an in�nitesimal operator in H, B :
H �! H is a linear bounded operator. An important advantage of this approach
is a possibility to employ ideas and technique of the semigroup operator theory.
At the same time it should be noticed that the most substantial and important for
Mathematics Subject Classi�cation 2000: 93B05, 35B37, 35L05.
Key words: wave equation, control problem.
c
L.V. Fardigola, 2005
L.V. Fardigola
applications results on operator semigroups deal with the case when the semigroup
generator A has a discrete spectrum or a compact resolvent and therefore the
semigroup may be treated by means of eigenelements of A. These assumptions
correspond to di�erential equations in bounded domains only.
In this paper we consider the wave equation on a half-plane. We should note
that most of papers studied controllability problems for the wave equation dealt
with this equation on bounded domains and controllability problems considered in
context of L2-controllability or, more generally, Lp-controllability (2 � p < +1)
[2�6]. But only L1-controls can be realized practically. Moreover, such controls
should be bounded by a hard constant (like in restriction (0.4)) for practical
purposes. Furthermore classical control theory started precisely from this point
view as switching controls are the ones realized in a concrete system. That is why
we build also bang-bang controls solving approximate null-controllability problem
in this paper.
Controllability problems for the wave equation on a half-axis in context of
bounded of a hard constant controls were investigated in [9, 10].
Consider the wave equation on a half-plane
@2w
@t2
= �w; x1 2 R; x2 > 0; t 2 (0; T ); (0.2)
controlled by the boundary condition
w(x1; 0; t) = Æ(x1)u(t); x1 2 R; t 2 (0; T ); (0.3)
where T > 0. We also assume that the control u satis�es the restriction
u 2 B(0; T ) =
�
v 2 L2
(0; T ) j jv(t)j � 1 almost everywhere on (0; T )
: (0.4)
All functions appearing in the equation (0.2) are de�ned for x1 2 R, x2 � 0.
Further, we assume everywhere that they are de�ned for x 2 R
2 and vanish for
x2 < 0.
Let us give de�nitions of the spaces used in our work. Let S be the Schwartz
space [7]
S =
n
' 2 C1
(R
n
) j 8m 2 N
8l 2 N sup
n���D�'(x)
��� �1 + jxj2
�l j x 2 R
n ^ j�j � m
o
< +1
o
;
S+ = f' 2 S j supp' 2 R � (0;+1)g
and let S0, S0+ be the dual spaces, here D = (�i@=@x1; : : : ;�i@=@xn), � =
(�1; : : : �n) is multi-index, j�j = �1 + � � �+ �n, j � j is the Euclidean norm.
94 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
Denote by Hs
l
the following Sobolev spaces:
Hs
l =
n
' 2 S0 j
�
1 + jxj2
�l=2 �
1 + jDj2
�s=2
' 2 L2
(R
n
)
o
;
k'ksl =
0@Z
Rn
����1 + jxj2
�l=2 �
1 + jDj2
�s=2
'(x)
���2 dx
1A1=2
:
Let F : S0 �! S0 be the Fourier transform operator. For ' 2 S we have
(F') (�) = (2�)�n=2
Z
Rn
e�ihx;�i'(x) dx;
where h�; �i is the scalar product in R
n corresponding to the Euclidean norm. It
is well known [8, Ch. 1] that FHs
0 = H0
s and k'ks0 = kF'k0
s
, if ' 2 Hs
0 .
A distribution f 2 S0 is said to be odd if (f; '(�)) = �(f; '(��)), ' 2 S.
Further, we assume throughout the paper that s � 0 and use the spaces
H
s
=
�
' 2 Hs
0 �Hs�1
0 j ' 2 S0+ ^ 9'(+0) 2 R
;eHs
=
�
' 2 Hs
0 �Hs�1
0 j ' is odd with resp. to x2
with the norm jjj'jjjs =
��
k'0ks0
�2
+
�
k'1ks�1
0
�2�1=2
and also the space
bHs =
�
' 2 H0
s �H0
s�1 j ' is odd with resp. to �2
with the norm [[j'j]]
s
=
��
k'0k0s
�2
+
�
k'1k0s�1
�2�1=2
.
Denote by A the following operator
A =
�
0 1
� 0
�
; A : eHs�2 �! eHs�2; D(A) = eHs (0.5)
and by B the operator
B =
�
0
�2Æ(x1)Æ0(x2)
�
; B : R �! eHs�2; D(B) = R; (0.6)
where Æ is the Dirac function. Then the system (0.2), (0.3) is reduced to the form
(0.1) with these operators A and B.
In Section 1 we obtain necessary and su�cient conditions for null-controllabi-
lity and approximate null-controllability for the system (0.2), (0.3) with restric-
tions (0.4) on the control. Controls solving the problems of null-controllability
and approximate null-controllability are found explicitly. But these controls may
have a rather complicated form.
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 95
L.V. Fardigola
The main goal of the Section 2 is to build bang-bang controls solving the
approximate null-controllability problem. We show that this problem can be
reduced to a system of Markov power moment problems. They may be solved
by the method given in [9]. Further, we prove that solutions of the Markov
power moment problems give us solutions of the approximate null-controllability
problem (Theorems 2.3, 2.4).
In Sections 3 and 4 some auxiliary statements are proved.
1. Null-controllability problems
Consider the control system (0.2), (0.3) with the initial conditions�
w(x; 0) = w0
0(x)
@w(x; 0)=@t = w0
1(x)
; x1 2 R; x2 > 0; (1.1)
and the steering conditions�
w(x; T ) = wT
0 (x)
@w(x; T )=@t = wT
1 (x)
; x1 2 R; x2 > 0; (1.2)
where w0
=
�
w0
0
w0
1
�
2 Hs, wT
=
�
wT
0
wT
1
�
2 Hs. We consider solutions of the
problem (0.2), (0.3) in the space Hs.
Let T > 0, w0 2 Hs. Denote by RT (w
0
) the set of states wT 2 Hs for which
there exists a control u 2 B(0; T ) such that the problem (0.2), (0.3), (1.1), (1.2)
has a unique solution.
De�nition 1.1. A state w0 2 Hs
is called null-controllable at a given time
T > 0 if 0 belongs to RT (w
0
) and approximately null-controllable at a given time
T > 0 if 0 belongs to the closure of RT (w
0
) in Hs
.
Let w
0
=
2w
0, w
T
=
2w
T , w(�; t) =
2
�
w(�; t)
@w(�; t)=@t
�
, where
2 is
the odd-extension operator with respect to x2. Evidently, w0 2 eHs, wT 2 eHs,
w(�; t) 2 eHs
(t 2 (0; T )). It is easy to see that control problem (0.2), (0.3), (1.1),
(1.2) is equivalent to the following problem for system (0.1):
w(x; 0) = w
0; (1.3)
w(x; T ) = w
T : (1.4)
Let us investigate this new problem. First we analyze the following auxiliary
Cauchy problem: system (0.1) with an arbitrary parameter u 2 B(0; T ) under
initial condition (1.3).
96 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
Applying the Fourier transform with respect to x to problem (0.1), (1.3), we
obtain the following Cauchy problem in bHs:
dv
dt
=
�
0 1
�j�j2 0
�
v � i�2
�
�
0
1
�
u; t 2 (0; T ); (1.5)
v(�; 0) = v
0; (1.6)
where v(�; t) = Fw(�; t), t 2 [0; T ], v0 = Fw0. Then the function
v(�; t) = �(j�j; t)
0@v
0
(�)� i�2
�
tZ
0
�(j�j;��)
�
0
1
�
u(�) d�
1A ; t 2 [0; T ]; (1.7)
where
�(�; t) �
0@ cos(�t)
sin(�t)
�
�� sin(�t) cos(�t)
1A �
�
@=@t 1
(@=@t)2 @=@t
�
sin(�t)
�
is a unique solution of (1.5), (1.6) in bHs.
Put E(jxj; t) = F�1
� �(j�j; t)=(2�). It is well known that
F�1
�
sin(j�jt)
j�j
�
(x) =
sign t H (jtj � jxj)p
t2 � jxj2
; (1.8)
where H is the Heaviside function: H(�) = 1 if � � 0 and H(�) = 0 otherwise.
Then we have
E(r; t) =
1
2�
�
@=@t 1
(@=@t)2 @=@t
�
sign t H (jtj � jxj)p
t2 � jxj2
:
It follows from (1.7) that
w(x; T ) = E(jxj; T )�
24w0
(x)� 1
�
@
@x2
F
�1
0@ TZ
0
0@ �sin(j�jt)
j�j
cos(j�jt)
1Au(t) dt
1A35 : (1.9)
Here and further � is the convolution with respect to x. With regard to Lemma
4.1 we get
w(x; T ) = E(jxj; T ) �
�
w
0
(x)� 1p
2�
x2
jxj�
�
U
U 0
�
(jxj)
�
; (1.10)
where U(t) = u(t) (H(t)�H(t� T )), t 2 R.
Denote for w0 2 eHs
RT (w
0
) =
�
E(jxj; T ) �
�
w
0
(x)� 1p
2�
x2
jxj�
�
U
U 0
�
(jxj)
�
j u 2 B(0; T )
�
:
Then De�nition 1.1 is equivalent to
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 97
L.V. Fardigola
De�nition 1.2. A state w
0 2 eHs
is called null-controllable at a given time
T > 0 if 0 belongs to RT (w
0
) and approximately null-controllable at a given time
T > 0 if 0 belongs to the closure of RT (w
0
) in eHs
.
Obviously, the following two statements are true.
Statement 1.1. A state w0 2 Hs
is null-controllable at a given time T > 0
i� the state w
0
=
2w
0
is null-controllable at this time.
Statement 1.2. A state w0 2Hs
is approximately null-controllable at a given
time T > 0 i� the state w
0
=
2w
0
is approximately null-controllable at this time.
Further we consider the (approximate) null-controllability problem for the
system (0.1) where w
0 is an odd function with respect to x2.
The following theorem give us su�cient conditions for (approximate) null-
controllability.
Theorem 1.1. For a state w
0 2 eHs
assume that there exists w
0 2 S0 such
that following conditions hold:
w
0
=
x2
jxjw
0
(jxj) in Hs
0 �Hs�1
0 ; (1.11)
suppw
0 � [0; T ]; (1.12)��w0
0(r)
�� � T
�r
p
T 2 � r2
a.e. on (0; T ); (1.13)
w
0
1(r) =
d
dr
24w0
0(r) +
1Z
�1
w
0
0(�)k(�; r) d�
35 ; (1.14)
where k(�; r) =
2
�
H (�(� � r))
�=2Z
0
sin
2 � d�p
�2 sin2 �+ r2 cos2 �
. Then the state w
0
is
null-controllable at the time T . Moreover, the solution of the null-controllability
problem (the control u) is unique and
u(t) = 2t
TZ
t
w
0
0(r) drp
r2 � t2
a.e. on (0; T ):
98 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
P r o o f. Put
U =
1p
2�
�w
0
0: (1.15)
It follows from (1.12) and Lemma 3.2 that suppU � (0; T ) and
U(t) = 2t
TZ
t
w
0
0(r) drp
r2 � t2
a.e. on (0; T ):
Denote u(t) = U(t), t 2 (0; T ). Due to (1.13) we obtain ju(t)j � 1 a.e. on (0; T ).
Applying Lemma 4.2 and (1.15), we have
w
0
1 =
d
dr
24w0
0 +
1Z
�1
w
0
0(�)k(�; �) d�
35 = �
d
dt
�
�1
w
0
0 =
1p
2�
�U 0:
Finally, taking into account (1.10), (1.11), (1.15), we get that w(x; T ) = 0 for
the found control u where w is a solution of the Cauchy problem (0.1), (1.3).
Invertibility of the operator � (see Sect. 4) implies uniqueness of the control u
solving the null-controllability problem.
Thus the state w0 is null-controllable at the time T that was to be proved.
The following theorem asserts that conditions (1.11)�(1.14) are not only suf-
�cient but also necessary for (approximate) null-controllability.
Theorem 1.2. If a state w
0 2 eHs
is approximately null controllable at a given
time T > 0 then there exists w
0 2 S0 such that conditions (1.11)�(1.14) hold.
P r o o f. For each n 2 N there exists a state w
n 2 RT (w
0
) such that
jjjwnjjjs < 1=n. With regard to (1.10) for some un 2 B(0; T ) we have
w
n
(x) = E(jxj; T ) �
�
w
0
(x)� 1p
2�
x2
jxj�
�
Un
U 0
n
�
(jxj)
�
; t 2 R;
where Un(t) = un(t) (H(t)�H(t� T )). Using Lemma 4.4, we obtain
1p
2�
x2
jxj�
�
Un
U 0
n
�
(jxj) �! w
0 as n �!1 in eHs: (1.16)
Therefore w
0
=
x2
jxjw
0
(jxj). According to the Lemma 3.2 suppw
0
0 � [0; T ]. Thus
(1.11), (1.12) are true. Denote �Un = hn0 , �U 0
n = hn1 . Taking into account Lemma
4.3, we obtain
jhn0 j �
T
�r
p
T 2 � r2
; r 2 (0; T ): (1.17)
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 99
L.V. Fardigola
Let an arbitrary " > 0 be �xed, V (") =
�
x 2 R
2 j jxj < "
. It follows from (1.16)
that
hn(jxj) �! w
0
(jxj) as n �!1 in S0: (1.18)
Since hn0 (jxj) 2 L2
�
R
2nV (")
�
and S is dense in L2
�
R
2
�
we obtain
hn0 (jxj) �! w
0
0(jxj) as n �!1 in
�
L2
�
R
2nV (")
��0
:
By the Riesz theorem we conclude that w
0
0(jxj) 2 L2
�
R
2nV (")
�
and
w
0
0 2 L2
(";+1). Taking into account arbitrariness of " > 0 and (1.17), we get
(1.13). We have hn1 = �
d
dt
�
�1hn0 . Due to Lemmas 3.1, 4.2 and (1.17) we get
(1.14). The theorem is proved.
2. Bang-bang controls and the Markov power moment problem
The solution of the null-controllability problem (i.e., the control) found in
Sect. 1 may be too complicated for the practical purposes. In this section we
�nd bang-bang controls solving the approximate null-controllability problem.We
consider a system of Markov power moment problems and show that their bang-
bang solutions are solutions of the approximate null-controllability problem.
Consider control system (0.1), (1.3) and assume that for T > 0 and w
0 2 eHs
conditions (1.11)�(1.14) hold. According to Theorem 1.1 there exists eu 2 B(0; T )
such that
w
0
=
1p
2�
�
�eU� (r); (2.1)
where eU(t) = eu(t) [H(t)�H(t� T )]. With regard to Lemma 4.1 and (1.11) we
get
v
0
(�) =
1
�
i�2
TZ
0
0@ �sin(j�jt)
j�j
cos(j�jt)
1Aeu(t) dt;
where v
0
= F
2w
0. Put
h(�; u) =
1
�
TZ
0
0@ �sin(�t)
�
cos(�t)
1A (eu(t)� u(t)) dt: (2.2)
Then for system (1.5), (1.6) we get
v(�; T ) = �(j�j; T )i�2h(j�j; u):
With regard to (1.7) and Lemma 4.4 we conclude that
[[jv(�; T )j]]
s
�
p
4T 2 + 6 [[ji�2h(j�j; u)j]]s :
100 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
We have
�
ki�2hj(j�j; u)k0s�j
�2
= �
1Z
0
�
1 + �2
�s�j jhj(j�j; u)j2 �3 d�; j = 0; 1:
Hence
[[jv(�; T )j]]
s
�
p
4T 2 + 6 �
0@ 1X
j=0
1Z
0
�
1 + �2
�s�j jhj(j�j; u)j2 �3 d�
1A1=2
: (2.3)
Thus we have proved
Theorem 2.1. Assume that T > 0 and for a state w
0 2 eHs
conditions (1.11)�
(1.14) are ful�lled. Then the following two assertions hold:
i. w
0
is null-controllable at the time T i� there exists u 2 B(0; T ) such that
h(�; u) � 0 on R;
ii. w
0
is approximately null-controllable at the time T i� for each " > 0 there
exists u" 2 B(0; T ) such that
1Z
0
�
1 + �2
�s�j jhj(j�j; u")j2 �3 d� < "2; j = 0; 1: (2.4)
Moreover, if estimate (2.4) is true then
jjjw(�; T )jjjs = [[jv(�; T )j]]
s
� �"
p
4T 2 + 6; (2.5)
where w and v are solutions of (0.1), (1.3) and (1.5), (1.6), respectively.
Due to the Wiener�Paley theorem we conclude that h(�; u) is an entire func-
tion with respect to �. Let us expand it in the Taylor series. To do this we
calculate h(m)
(0; u) (we consider the derivatives with respect to �). Put
ev0(�) = 1
�
TZ
0
0@ �sin(�t)
�
cos(�t)
1Aeu(t) dt; ew0
(jxj) = F
�1ev0(j�j): (2.6)
Evidently ev0 is also entire. With regard to (1.11) and Lemma 4.1 we get
w
0
= ew00: (2.7)
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 101
L.V. Fardigola
According to (1.11), (1.12) and (1.14), we conclude that
ew0
0(r) = (H(r)�H(r � T ))
TZ
r
eu(t) dtp
t2 � r2
; (2.8)
ew0
1(r) = w
0
0(r) +
1Z
�1
w
0
0(�)k(�; r) d�: (2.9)
Obviously, supp ew0
0 � [0; T ]. It follows from (1.12) that supp ew0
1 � [0; T ]. Taking
into account
2T
�
TZ
r
1
�
p
T 2 � �2
�=2Z
0
sin
2 � d�p
�2 sin2 �+ r2 cos2 �
d�
� T
r
TZ
r
1
�
p
T 2 � �2
=
1
2r
ln
�����T +
p
T 2 � r2
T �
p
T 2 � r2
����� ; r 2 (0; T ); (2.10)
and (1.13), (1.14), we get
��ew0
0(r)
�� �
TZ
r
dtp
t2 � r2
= � ln
0@T
r
�
s�
T
r
�2
� 1
1A ; r 2 (0; T ); (2.11)
��ew0
1(r)
�� � T
�r
p
T 2 � r2
+
1
�
TZ
r
T
��
p
T 2 � �2
�=2Z
0
d�p
�2 sin2 �+ r2 cos2 �
d�
=
T
�r
p
T 2 � r2
+
1
2r
ln
�����T +
p
T 2 � r2
T �
p
T 2 � r2
����� ; r 2 (0; T ): (2.12)
Taking into account (2.11), (2.12), (2.6), we obtain ev0(2m+1)
(0) = 0:
ev0(2m)
(0) =
1
�
d2m
d�2m
1Z
0
0@ �Z
0
e�ir� cos' d'
1A ew0
(r) dr
������
�=0
=
(�1)m
�
1Z
0
0@ �Z
0
cos
2m 'd'
1A r2m+1ew0
(r) dr
=
(�1)m
�
B
�
m+
1
2
;
1
2
� 1Z
0
r2m+1ew0
(r) dr; (2.13)
102 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
where B(�; �) is the Euler beta-function. Therefore
ev00(2m)
(0) =
(�1)m
�(2m+ 2)
B
�
m+
1
2
;
1
2
� 1Z
0
r2m+2
w
0
0(r) dr:
With regard to (2.9) we have
ev01(2m)
(0) =
(�1)m
�
B
�
m+
1
2
;
1
2
�24 1Z
0
r2m+1
w
0
0(r) dr
+
1Z
0
r2m+1
1Z
r
w
0
0(�)k(�; r) d� dr
35 :
Since
�=2Z
0
q
�2 sin2 �+ r2 cos2 � d� =
�Z
r
t2 dtp
�2 � t2
p
t2 � r2
then
�Z
0
r2m+1k(�; r) dr = ��2m+1
+
2
�
1
�
d
d�
�Z
0
r2m+1
�Z
r
t2 dtp
�2 � t2
p
t2 � r2
dr
= ��2m+1
+
2
�
1
�
d
d�
�=2Z
0
�2m+3
sin
2m+3 d
�=2Z
0
sin
2m+1 'd'
= ��2m+1
+
2m+ 3
2�
B
�
m+ 1;
1
2
�
B
�
m+ 2;
1
2
�
:
Therefore
ev01(2m)
(0) =
(�1)m(2m+ 3)
2�2
B
�
m+
1
2
;
1
2
�
B
�
m+ 1;
1
2
�
B
�
m+ 2;
1
2
�
�
1Z
0
r2m+1
w
0
0(r) dr:
Put
!n =
1Z
0
rn+1
w
0
0(r) dr: (2.14)
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 103
L.V. Fardigola
Hence
ev00(2m)
(0) = (�1)m+1 (2m� 1)!!
(2m+ 2)!!
; (2.15)
ev01(2m)
(0) =
(�1)m
�
(2m+ 2)!!
(2m+ 1)!!
: (2.16)
We have
h(2m+1)
(0; u) = 0; (2.17)
and h(2m)
(0; u) = 0 i�
0 = ev00(2m)
(0) +
(�1)m
2m+ 1
TZ
0
t2m+1u(t) dt; (2.18)
0 = ev01(2m)
(0)� (�1)m
TZ
0
t2mu(t) dt: (2.19)
Thus
h(n)(0; u) = 0; (2.20)
i�
TZ
0
tnu(t) dt = !n; n = 0;1; (2.21)
where
!2m =
(2m+ 2)!!
�(2m+ 1)!!
!2m ; (2.22)
!2m+1 =
(2m+ 1)!!
(2m+ 2)!!
!2m+1 : (2.23)
According to Theorem 2.1, we obtain that the state w0 is null-controllable at
the time T i� (2.21) is valid.
The problem of determination of a function u 2 B(0; T ) satisfying condition
(2.21) for a given f!ng1n=0 and T > 0 is called a Markov power moment problem
on (0; T ) for the in�nite sequence f!ng1n=0.
Uniqueness of the solution of the null-controllability problem yields uniqueness
of the solution of the Markov moment problem (2.21) (see Theorem 1.1). Hence
u = u is the unique solution of this Markov moment problem.
Thus we have proved
104 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
Theorem 2.2. Assume that T > 0 and for a state w
0 2 eHs
conditions (1.11)�
(1.14). Assume also that f!ng1n=0 is de�ned by (2.14), (2.22), (2.23). Then
Markov power moment problem (2.21) on (0; T ) for f!ng1n=0 has a unique solu-
tion. Moreover, this solution is a solution of the null-controllability problem for
w
0
at the time T .
Consider (2.21) for a �nite set of n:
TZ
0
tnu(t) dt = !n; n = 0; N: (2.24)
The problem of determination of a function u 2 B(0; T ) satisfying condition
(2.24) for a given f!ngNn=0 and T > 0 is called a Markov power moment problem
on (0; T ) for the �nite sequence f!ngNn=0.
Obviously, u = u is a solution of this problem, but it is not unique.
Let us show that solutions of moment problem (2.24) for various N give us
controls solving the approximate null-controllability problem.
Theorem 2.3. Let T > 0, w
0 2 eHs
, s < �1. Let also conditions (1.11)�
(1.14) be ful�lled and f!ng1n=0 be de�ned by (2.14), (2.22), (2.23). Then 8" > 0
there exists N > 0 such that for each solution uN 2 B(0; T ) of moment problem
(2.24) the corresponding solution w of control system (0.1), (1.3) satis�es the
condition jjjw(�; T )jjjs < ".
P r o o f. Let N = 2K + 1, uN 2 B(0; T ) be a solution of problem (2.24).
With regard to (2.20) and (2.21) for the function h(�; u) de�ned by (2.2) we get
h(n)(0; uN ) = 0; n = 0; 2K + 1:
By the Taylor formula for j�j < a we obtain
���1�jhj(�; uN )�� � a2K+2
(2K + 2)!
sup
j�j�a
�����1�jhj�(2K+2)
(�; uN )
��� ; j = 0; 1:
Taking into account (2.2), we conclude that�����1�jhj(�; uN )�(2K+2)
��� � T 2K+3
�(2K + 3)
; j = 0; 1:
Hence ���1�jhj(�; uN )�� � T
�
(Ta)2K+2
(2K + 3)!
; j = 0; 1; j�j � a:
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 105
L.V. Fardigola
Then
aZ
0
�
1 + �2
�s�j jhj(�; uN )j2 �3 d� � a
�
(Ta)2K+3
(2K + 3)!
; j = 0; 1: (2.25)
With regard to (2.2) we get���1�jhj(�; uN )�� � T
�
; j = 0; 1; � > 0:
Therefore
1Z
a
�
1 + �2
�s�j jhj(�; uN )j2 �3 d�
� T
�
1Z
a
�
1 + �2
�s
� d� � � Ta2(s+1)
2�(s+ 1)
; j = 0; 1:
Taking into account (2.25), we obtain
�
1Z
0
�
1 + �2
�s�j jhj(�; uN )j2 �3 d� � a(Ta)2K+3
(2K + 3)!
� Ta2(s+1)
2(1 + s)
; j = 0; 1:
Due to Theorem 2.1 and (2.3) we conclude that
jjjw(�; T )jjjs �
p
2T 2 + 3
"
a(Ta)2K+3
(2K + 3)!
� Ta2(s+1)
2(1 + s)
#
: (2.26)
Applying the Stirling formula, we have
(Ta)2K+3
(2K + 3)!
�
�
Tae
2K + 3
�2K+3
1p
2�(2K + 3)
:
Setting a = (2K + 3)=(2Te), we obtain from (2.26) that
jjjw(�; T )jjjs �
p
2T 2 + 3
"p
2K + 3
Te4K+2
� T
2s+ 2
�
2K + 3
2Te
�2s+2
#
! 0 as K !1:
(2.27)
The theorem is proved.
Denote
B
N
(0; T ) = fu 2 B(0; T ) j 9T� 2 (0; T )(ju(t)j = 1 a.e. on (0; T�))
^ (u(t) = 0 a.e. on (T�; T ))
^ (u has no more than N discontinuity points on (0; T�))g:
106 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
It is well known [11, 12] that if Markov power moment problem (2.24) is solv-
able then there exists its solution u 2 BN
(0; T ). Taking into account Theorem 2.3,
we conclude that under the conditions ot this theorem we can �nd a solution
uK 2 B2K+1
(0; T ) of Markov power moment problem (2.24) for N = 2K + 1
and such solutions fuKg1K=1 give us bang-bang controls solving the approximate
null-controllability problem (see also (2.27)).
Thus the following theorem is true.
Theorem 2.4. Let T > 0, w
0 2 eHs
, s < �1. Let also conditions (1.11)�(1.14)
be ful�lled and f!ng1n=0 be de�ned by (2.14), (2.22), (2.23). Then 8K 2 N there
exists a solution uK 2 B2K+1
(0; T ) of moment problem (2.24) with N = 2K + 1.
Moreover, for this uK the corresponding solution w of control system (0.1), (1.3)
satis�es the estimate
jjjw(�; T )jjjs �
p
2T 2 + 3
"p
2K + 3
Te4K+2
� T
2s+ 2
�
2K + 3
2Te
�2s+2
#
: (2.28)
Let us show that the condition s < �1 of Theorems 2.3, 2.3 is essential.
Precisely if �1=2 � s � 0 then 9w0 2 eHs 8T > 0 8u 2 [N2NB
N
(0; T ) 9"0 > 0
such that for a solution w of (0.1), (1.3), corresponding to the control u we have
jjjw(�; T )jjjs � "0. Thus the state w
0 is not approximate null-controllable at the
time T by bang-bang controls in space eHs, if �1=2 � s � 0.
E x a m p l e 2.1. Let �1=2 � s � 0, T > 0,
w
0
0(x) =
x2T
2�jxj2
p
T 2 � jxj2
[H(jxj)�H(jxj � T )] ;
w
0
1(x) =
x2
2�
q
(T 2 � jxj2)3
[H(jxj)�H(jxj � T )] :
Obviously, w
0
(x) =
1p
2�
�
eUeU 0
!
(jxj), where eU(t) =
1
2
[H(t)�H(t� T )].
Therefore w
0 2 eHs satis�es (1.11)�(1.14). Let u 2 BN
(0; T ), n 2 N. Hence
u(t) = �
NX
k=0
(�1)k [H(t� tk)�H(t� tk+1)] ;
where � = �1, 0 = t0 < t1 < t2 � � � < tN+1 = T� � T , U(t) = [H(t)�H(t� T )].
Let w be a solution of (0.1), (1.3) corresponding to the control u. According to
(1.10), we have
p
2�E(jxj;�T ) � w(x; T ) = x2
jxj�
eU � UeU 0 � U 0
!
(jxj):
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 107
L.V. Fardigola
Put a = �=(12T ). With regard to Lemma 4.4 we get
jjjw(x; T )jjjs � 1p
2�
p
4T 2 + 6
�����
�����
����� x2jxj�
eU � UeU 0 � U 0
!
(jxj)
�����
�����
�����
s
� 1
p
�
p
4T 2 + 6
0B@ 1Z
0
�
1 + �2
��1=2
������
TZ
0
sin(�t)
� eU(t)� U(t)
�
dt
������
2
� d�
1CA
1=2
�
p
a
p
�
4
p
1 + a2
p
4T 2 + 6
0B@ 1Z
a
������
TZ
0
sin(�t)
� eU(t)� U(t)
�
dt
������
2
d�
1CA
1=2
�
p
ap
2
4
p
1 + a2
p
4T 2 + 6
264
0@ 1Z
�1
���F
�eU � U
�
(�)
���2 d�
1A1=2
�
0@ aZ
�a
���F
�eU � U
�
(�)
���2 d�
1A1=2
375 : (2.29)
We have
1Z
�1
���F
�eU � U
�
(�)
���2 d� � 1
4
1Z
�1
j(H(t+ T )�H(t� T ))j2 dt = T
2
: (2.30)
On the other hand
aZ
�a
���F
�eU � U
�
(�)
���2 d� � 3
�
aZ
�a
1
�
NX
k=0
jcos(tk�)� cos(tk+1�)j
!2
d�
=
6
�
aZ
0
2
�
NX
k=0
����sin��tk+1 � tk
2
�
sin
�
�
tk+1 + tk
2
�����
!2
d�
� 6
�
aZ
0
NX
k=0
t2
k+1 � t2
k
2
!2
d� � 3T 2a
2�
=
T
8
: (2.31)
Comparing (2.29), (2.31), we obtain
jjjw(�; T )jjjs �
p
ap
2
4
p
1 + a2
p
4T 2 + 6
"r
T
2
�
r
T
8
#
� T
4(4T 2 + 6)3=4
= "0: (2.32)
That was to be proved.
108 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
3. Operators � and ��
In this section we introduce and study operators � and �
�.
Let the operator ��
: S �! S be de�ned by the rule
(�
�') (t) = �
r
2
�
1Z
�1
H(r(t� r))p
t2 � r2
'0(r) dr; ' 2 S: (3.1)
Obviously, (��') (t) = �
r
2
�
Z
�=2
0
'0(t sin�) d�, ' 2 S. Hence ��' 2 S, if ' 2 S.
It is easy to see that ���1
: S �! S can be de�ned by the rule
�
�
��1
�
(t) =
r
2
�
1Z
�1
H(t(r � t))p
r2 � t2
t (t) dt; 2 S: (3.2)
It is clear that
�
�
��1
�
(t) =
r
2
�
t
Z
�=2
0
(t sin�) sin� d�, ' 2 S, and �
��1 2
S, if 2 S. Thus
�
�
(S) = S = �
��1
(S):
Let the operator � : S0 �! S0 be de�ned by the rule
(�f; ') = (f;��') ; ' 2 S; f 2 S0:
Obviously, ��1 is de�ned by�
�
�1f; '
�
=
�
f;���1'
�
; ' 2 S; f 2 S0:
Thus
�(S
0
) = S
0
= �
�1
(S
0
):
One can easily show that the following three lemmas are true.
Lemma 3.1. If fn �! f as n �! 1 in S0 then �fn �! �f and �
�1fn �!
�
�1f as n �!1 in S0.
Lemma 3.2. Let 0 < A � +1, f 2 S0, supp f � [0; A] and 8a 2 (0; A)
f 2 L1
(a;A). Then supp�f � [0; A] and
(�f) (r) = �
r
2
�
d
dr
AZ
r
f(t) dtp
t2 � r2
; r 2 (0; A):
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 109
L.V. Fardigola
Lemma 3.3. Let 0 < A � +1, g 2 S0, supp g � [0; A] and 8a 2 (0; A)
g 2 L1
(a;A). Then supp�
�1g � [0; A] and
�
�
�1g
�
(t) =
r
2
�
t
AZ
t
g(r) drp
r2 � t2
; t 2 (0; A):
4. Auxiliary statements
In this section we denote by Sn the space of functions ' 2 S de�ned on Rn , if
we want to indicate the dimension. For each functional f 2 S01, supp f � [0;+1),
we can de�ne f(jxj) 2 S02 by the rule
(f(jxj); (x)) = (f(r); rSr[ ]) ; (4.1)
where Sr[ ] =
R 2�
0
(r cos�; r sin�) d�, r 2 R. Obviously, if 2 S2 then Sr[ ] 2
S1.
To prove conditions for (approximate) null-controllability we need the follow-
ing four lemmas.
Lemma 4.1. Let T > 0, u 2 B(0; T ), U(t) = u(t) [H(t)�H(t� T )], � 2 R
2
,
x 2 R
2
. Then
F
�1
24i�2 TZ
0
0@ �sin(j�jt)
j�j
cos(j�jt)
1Au(t) dt
35 =
r
�
2
x2
jxj�
�
U
U 0
�
(jxj): (4.2)
P r o o f. Denote h(�; t) =
0@ �sin(�t)
�
cos(�t)
1AH(�). We have
F
�1
24i�2 TZ
0
h(j�j; t)u(t) dt
35 =
@
@x2
F
�1
24 TZ
0
h(j�j; t)u(t) dt
35 : (4.3)
For each ' 2 S2 we get0@F�1
24 TZ
0
h(j�j; t)u(t) dt
35 ; '
1A =
0@ TZ
0
h(�; t)u(t) dt; �S� [F']
1A
=
1Z
0
0@ TZ
0
h(�; t)u(t) dt
1A �S� [F']d�;
110 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
where z means the complex conjugation of z. Since �S� [F'] 2 S1 we obtain0@F�1
24 TZ
0
h(j�j; t)u(t) dt
35 ; '
1A =
1Z
0
U(t)
1Z
0
h(�; t)u(t)�S� [F']d� dt
=
0@U(t); 1Z
0
h(�; t)u(t)�S� [F'] d�
1A
= �
0@� U(t)
U 0
(t)
�
;
1Z
0
sin(�t)S� [F'] d�
1A
= �
0@� U(t)
U 0
(t)
�
;
Z
R2
sin(j�jt)
j�j (F') (�) d�
1A
= �
0@� U(t)
U 0
(t)
�
;
Z
R2
F
�1
�
sin(j�jt)
j�j
�
(x)'(x) dx
1A : (4.4)
With regard to (4.4) and (1.8) that gives0@F�1
24 TZ
0
h(j�j; t)u(t) dt
35 ; '
1A = �
0@� U(t)
U 0
(t)
�
;
1Z
�1
H (t(t� r))p
t2 � r2
rSr['] dr
1A :
(4.5)
Consider the operator �
: S �! S such that
(
��) =
1Z
�1
H (t(t� r))p
t2 � r2
�(r) dr =
�=2Z
0
�(t sin�) d�; � 2 S:
It is clear that if � 2 S then
�� 2 S. Denote by the operator : S0 �! S0
such that
( f; �) = (f; ��) ; � 2 S; f 2 S0:
Evidently, if suppf � [0;+1) (f 2 S0) then supp f � [0;+1). One can see
that � = �
r
�
2
d
dr
. All this implies that
F
�1
24 TZ
0
h(j�j; t)u(t) dt
35 = � @
@x2
�
U
U 0
�
(jxj) =
r
�
2
x2
jxj�
�
U
U 0
�
(jxj):
That was to be proved.
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 111
L.V. Fardigola
Lemma 4.2. Let f 2 S0, supp f � [0;+1) and 8a > 0 f 2 L1
(a;+1). Then�
�
d
dt
�
�1f
�
(r) =
d
dr
24f(r) + 1Z
�1
f(�)k(�; r) d�
35 ; (4.6)
where k(�; r) =
2
�
H (�(� � r))
Z
�=2
0
sin
2 �d�p
�2 sin2 �+ r2 cos2 �
.
P r o o f. For each ' 2 S we have�
�
d
dt
�
�1f; '
�
= �
�
f;���1 d
dt
�
�'
�
: (4.7)
With regard to (3.1), (3.2) for � > 0 we get
�
�
��1 d
dt
�
�'
�
(�) =
2
�
�Z
0
1p
�2 � t2
d
dt
24t tZ
0
'0(r) drp
t2 � r2
35 dt
=
2
�
1
�
d
d�
�Z
0
p
�2 � t2
d
dt
24t tZ
0
'0(r) drp
t2 � r2
35 dt
=
2
�
1
�
d
d�
Z
�
0
'0(r)
�Z
r
t2 dtp
�2 � t2
p
t2 � r2
dr
=
2
�
1
�
d
d�
�Z
0
'0(r)
�=2Z
0
q
�2 sin2 �+ r2 cos2 � d� dr
= '0(�) +
2
�
Z
�
0
'0(r)
�=2Z
0
sin
2 �d�p
�2 sin2 �+ r2 cos2 �
d�dr:
Taking into account (4.7), we obtain�
�
d
dt
�
�1f; '
�
= �
0@f; '0(�) + 1Z
�1
'0(r)k(�; r) dr
1A
=
0@ d
dr
24f(r) + 1Z
�1
f(�)k(�; r) d�
35 ; '
1A : (4.8)
Hence (4.6) holds, and the lemma is proved.
112 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1
On controllability problems for the wave equation on a half-plane
Lemma 4.3. Let u 2 B(0; T ), U(t) = u(t) [H(t)�H(t� T )]. Then
supp�U � [0; T ] and
j(�U) (r)j �
p
2T
p
�r
p
T 2 � r2
; r 2 (0; T ): (4.9)
P r o o f. According to the Lemma 3.2, we obtain supp�U � [0; T ]. We also
have that
(�U) (r) =
r
2
�
d
dr
TZ
r
u(t) dtp
t2 � r2
; r 2 (0; T ):
Denote fn(r) =
Z
T
r
u(t) dtp
t2 � (r � 1=n)2
; f(r) =
Z
T
r
u(t) dtp
t2 � r2
(r 2 (0; T ]).
One can see that
fn(r)! f(r) as n!1; r 2 (0; T ]: (4.10)
First let us prove that 8r0 2 (0; T ) 8" 2 (0; T � r0) we have
f 0n(r)� f 0(r) as n �!1; on [r0; T � "]: (4.11)
Let 8r0 2 (0; T ) 8" 2 (0; T � r0) be �xed. We have
f 0n(r) = � u(r)p
r2 � (r � 1=n)2
+ (r � 1=n)
TZ
r
u(t) dt
(t2 � (r � 1=n)2)3=2
: (4.12)
Let n > m > 0 be large enough. Denote
gr(�) = � u(r)p
r2 � (r � 1=n)2
+ (r � 1=n)
TZ
r
u(t) dt
(t2 � (r � 1=n)2)3=2
; � 2 [r � 1=n; r � 1=m]:
Applying the mean value theorem to gr(�) (with respect to �), we get
jfn(r)� fm(r)j = jgr(r � 1=n)� gr(r � 1=m)j
� sup
�2[r� 1
n
;r�
1
m
]
24 2�
(r2 � �2)3=2
+
TZ
r
t2 + 2�2
(r2 � �2)5=2
35� 1
m
� 1
n
�
� sup
�2[r� 1
n
;r�
1
m
]
"
2�
�
r2 � �2
�
+ (T � r)
�
T 2
+ 2�2
�
(r2 � �2)5=2
#
2
m
� 14T 3
m7=2r0
! 0 as m!1; r 2 [r0; T � "]:
Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 113
L.V. Fardigola
With regard to (4.10) we conclude that the consequence ff 0ng1n=1 uniformly
converges on [r0; T � "] and (4.11) is true.
Finally let us prove (4.9). Due to (4.12) we have 8r 2 (0; T )��f 0n(r)�� � � 1
p
n(r � 1=n)
p
2r � 1=n
+
T
(r � 1=n)
p
T 2 � (r � 1=n)2
! T
r
p
T 2 � r2
as n!1:
Taking into account (4.11), we conclude that (4.9) holds that was to be proved.
Lemma 4.4. If f 2 Hs
0 �Hs�1
0 and g = Ff then
jjjE(jxj; t) � f jjjs = [[j�(j�j; t)gj]]
s
�
p
4t2 + 6 [[jgj]]
s
=
p
4t2 + 6 jjjf jjjs ; t 2 R:
(4.13)
P r o o f. For all t 2 R we have
jjjE(jxj; t) � f jjjs = [[j�(j�j; t)gj]]
s
�
������� cos(j�jt)
�j�j sin(j�jt)
�
g0
������
s
+
2424������
0@ sin(j�jt)
j�j
cos(j�jt)
1A g1
������
3535
s
�
p
2 kg0k0s +
0@
sin(j�jt)j�j g1
0
s
!2
+
�
kg1k0s�1
�21A1=2
:
Since
�
1 + j�j2
� ����sin(j�jt)j�j
����2 � 2
�
t2 + 1
�
we obtain (4.13). The lemma is proved.
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