Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point
In the paper, the necessary and su±cient conditions of null-controllability and approximate null-controllability are obtained for the some control system. The problems for the control system are considered in the modified Sobolev spaces. The control that solves these problems is found explicitly. Th...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2011
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| Цитувати: | Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point / K.S. Khalina // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 1. — С. 34-58. — Бібліогр.: 22 назв. — англ. |
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irk-123456789-1066632016-10-02T03:03:03Z Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point Khalina, K.S. In the paper, the necessary and su±cient conditions of null-controllability and approximate null-controllability are obtained for the some control system. The problems for the control system are considered in the modified Sobolev spaces. The control that solves these problems is found explicitly. The bang-bang controls solving the approximate null-controllability problem are constructed as the solutions of the Markov trigonometric moment problem. 2011 Article Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point / K.S. Khalina // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 1. — С. 34-58. — Бібліогр.: 22 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106663 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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In the paper, the necessary and su±cient conditions of null-controllability and approximate null-controllability are obtained for the some control system. The problems for the control system are considered in the modified Sobolev spaces. The control that solves these problems is found explicitly. The bang-bang controls solving the approximate null-controllability problem are constructed as the solutions of the Markov trigonometric moment problem. |
| format |
Article |
| author |
Khalina, K.S. |
| spellingShingle |
Khalina, K.S. Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point Журнал математической физики, анализа, геометрии |
| author_facet |
Khalina, K.S. |
| author_sort |
Khalina, K.S. |
| title |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point |
| title_short |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point |
| title_full |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point |
| title_fullStr |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point |
| title_full_unstemmed |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point |
| title_sort |
controllability problems for the non-homogeneous string that is fixed at the right end point and has the dirichlet boundary control at the left end point |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2011 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106663 |
| citation_txt |
Controllability Problems for the Non-Homogeneous String that is Fixed at the Right End Point and has the Dirichlet Boundary Control at the Left End Point / K.S. Khalina // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 1. — С. 34-58. — Бібліогр.: 22 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT khalinaks controllabilityproblemsforthenonhomogeneousstringthatisfixedattherightendpointandhasthedirichletboundarycontrolattheleftendpoint |
| first_indexed |
2025-07-07T18:49:55Z |
| last_indexed |
2025-07-07T18:49:55Z |
| _version_ |
1837015179773083648 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2011, vol. 7, No. 1, pp. 34–58
Controllability Problems for the Non-Homogeneous
String that is Fixed at the Right End Point and has the
Dirichlet Boundary Control at the Left End Point
K.S. Khalina
Mathematics Division, B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv, 61103, Ukraine
E-mail:khalina@meta.ua
Received May 14, 2010
In the paper, the necessary and sufficient conditions of null-controllability
and approximate null-controllability are obtained for the control system
wtt(x, t) = wxx(x, t) − q(x)w(x, t), w(0, t) = u(t), w(d, t) = 0, x ∈ (0, d),
d > 0, t ∈ (0, T ), 0 < T ≤ d, where q(x) ∈ C1[0, d], q(x) ≥ 0, q′+(0) =
q′−(d) = 0, u is a control, |u(t)| ≤ 1 on (0, T ). The problems for the con-
trol system are considered in the modified Sobolev spaces. The control that
solves these problems is found explicitly. The bang-bang controls solving the
approximate null-controllability problem are constructed as the solutions of
the Markov trigonometric moment problem.
Key words: wave equation, controllability problem, Dirichlet control
bounded by a hard constant, modified Sobolev space, Sturm–Liouville prob-
lem, transformation operator.
Mathematics Subject Classification 2000: 93B05, 35B37, 35L05, 34B24.
1. Introduction
In the paper, the wave equation for a non-homogeneous string on a finite
segment is considered. The string is fixed at the right end point. At the left end
point we consider a control that is bounded by a hard constant. We study the
problems of the null- and approximate null-controllability of this system in the
space Hs
Q, s ≤ 0 (the modified Sobolev space under the operator (1+D2+q(x))s/2
instead of (1 + D2)s/2, D = −id/dx). First we consider the Sturm–Liouville
problem on a given segment, find its eigenfunctions that form an orthonormal
basis and expand the functions describing the control system in terms of this basis.
c© K.S. Khalina, 2011
Controllability Problems for the Non-Homogeneous String
Then we apply the transformation operators for the Sturm–Liouville problem.
This method allows to obtain the necessary and sufficient conditions of the null-
and approximate null-L∞-controllability and an explicit formula for the control.
Note that the controllability problems for a hyperbolic partial differential
equation were studied in a number of papers (see, e.g., the references in [1]).
Gugat and Leugering [2] considered the wave equation for the homogeneous string
1
c2
ytt(x, t)−yxx(x, t) = 0 on the segment (0, L). The string was fixed at the left end
point. At the right end point there was considered a control that had a minimal
norm in L∞(0, T ). The problems of exact and approximate null-controllability
at the time T = 2L
c were analyzed. A bang-bang control solving both problems
for all initial states from L∞(0, L)×W−1,∞(0, L) was constructed.
The boundary Lp-controllability (2 ≤ p ≤ ∞) for a homogeneous string on a
finite segment is well studied in [3]–[8] and other papers. Gugat et al. [4] and
Fardigola, Khalina [6] considered the boundary L∞-controllability for the wave
equation for a homogeneous string on a finite segment. Moreover, controls were
bounded by a hard constant for practical purposes in [6]. We should note that
some results obtained in [4, 6] can be found in the present paper, exactly when the
potential q(x) is equal to zero. But if q(x) 6= 0 on the considered segment, then
the studying of the null-controllability problems for the wave equation becomes
more complicated. That is why we use the transformation operators for the
Sturm–Liouville problem on a segment. These operators describe the connection
among the initial state of the controlled system, the control and the corresponding
steering state. We introduce and study the transformation operators in Hs
Q,
s ≤ 0.
Most of [9] is devoted to the controllability and observability theories for
the linear hyperbolic systems of the form wt = A(x)wx + B(x)w, w ∈ En,
t ≥ 0, x ∈ (0, 1). For these systems, a boundary value control is consid-
ered and the conditions of exact and approximate controllability are obtained
when a control, the initial and steering functions are from L2. The author also
considers a problem of the boundary L2-controllability for the wave equation
ρ(x)∂2w
∂t2
− ∑n
i,j=1
∂
∂xi
(
ai
j(x) ∂w
∂xj
)
= 0 in the bounded open connected set Ω in
Rn. The problem is considered in the Sobolev spaces Hm(Ω), m = 0, 1, 2. There
are also obtained the conditions of approximate controllability for this equa-
tion and the conditions of exact controllability for the equation when ρ(x) and
ai
j(x) are constants, i, j = 1, n. We should note that the wave equation for the
non-homogeneous string, studied in the present paper, cannot be reduced to the
systems above.
Emanuilov [10] investigated the controllability problems for the linear hyper-
bolic partial differential second-order equation in the bounded domain Ω ⊂ Rn.
The Dirichlet control is extended on a part of the boundary. The problems
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 35
K.S. Khalina
are considered in L2(Ω) × H−1(Ω). A sufficient condition of the exact L2-
controllability is obtained. Note that the restrictions imposed on the coefficients
of the wave equation in the present paper are weaker than those in [10].
Il’in and Moiseev [11] studied the problem of the boundary controllability
for a string on [0, l] in the case of q = const > 0. A control from W 1
2 (0, T ) is
considered at the left end of the string, and the right end is fixed. The problem is
considered in the class Ŵ 1
2 (0, l), namely in the class W 1
2 (0, l) with an additional
condition of the functions smoothness on the boundary of the considered domain.
The time T = 2l is proved to be minimal when the system is controllable under
the described conditions. The control solving this problem is found in terms of
the initial and steering functions. It is also pointed out that additional conditions
should be imposed on the initial functions when T < 2l. In the present paper, the
potential q is not a constant generally speaking. A boundary control is bounded
by a hard constant. Therefore it is of the class L∞. The problem is considered
in the space Hs
Q, s ≤ 0. In particular, the solution smoothness is smaller than
that in [11]. In the present paper, the condition q(x) 6= const essentially differs
from that in [11]. That is why we have to apply the transformation operators for
the Sturm–Liouville problem on a segment. We also consider only T ∈ (0, l] in
contrast to [11]. We prove that at the time 0 < T ≤ l (from not an arbitrary initial
state) the system becomes (approximately) null-controllable. The description of
these initial functions is given in the present paper. The necessary and sufficient
conditions of the null- and approximate null-controllability for the considered
system are also obtained. For the control, an explicit formula is found in terms
of the initial state by using the transformation operator. Note that for q = const
the results of the present paper are not contained in [11].
The structure of the paper is the following.
In Section 3 we formulate the obtained results: the formula that connects
the initial state, the control and the steering state; the necessary and sufficient
conditions of the null-controllability and approximate null-controllability for a
given control system. The control solving these problems is found explicitly.
Moreover, it is proved that the boundedness of the initial state is the necessary
and sufficient condition of the boundedness of the corresponding control. In this
section the explicit representations for the kernels of the transformation operators
are also obtained in the case when the potential of the wave equation is constant.
The example for this case is given.
In Section 4 the proofs of the theorems formulated in Section 3 are given.
In Section 5 we construct the bang-bang controls that solve the approximate
null-controllability problem. We reduce this problem to the Markov trigonometric
moment problem which can be solved by the algorithm given in [12]. We show
that the solutions of the Markov trigonometric moment problem are the solutions
of the approximate null-controllability problem for s < −1/2.
36 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
In Appendix some auxiliary statements are proved. We describe in detail the
spaces introduced in Section 2. We also introduce and study the transformation
operators for the Sturm–Liouville problem on a segment in these spaces.
2. Notation
Consider the wave equation on a finite segment
wtt(x, t) = wxx(x, t)− q(x)w(x, t), x ∈ (0, d), t ∈ (0, T ), (2.1)
controlled by the boundary conditions
w(0, t) = u(t), w(d, t) = 0, t ∈ (0, T ), (2.2)
where d > 0, 0 < T ≤ d.
Let us introduce the spaces used in this paper. First we consider the Sturm–
Liouville problem on the segment (0, d)
Gv ≡ −v′′(x) + q(x)v(x) = λ2v(x), v(0) = v(d) = 0, x ∈ (0, d), (2.3)
where q ∈ E(0, d) =
{
r ∈ C1[0, d] : r(x) ≥ 0, r′+(0) = r′−(d) = 0
}
.
It is well known that the operator G ≡ − (
d
dx
)2
+ q(x) has a countable set
of eigenvalues {µn = λ2
n}∞n=1 that are real, nonnegative and simple, and λn 6= 0,
n = 1,∞ (see, e.g., [13]). Let {yn(λn, x)}∞n=1 be a system of corresponding
eigenfunctions. They are real and form the orthonormal basis in L2[0, d]. We
have Gyn(λn, x) = λ2
nyn(λn, x) for n = 1,∞.
Let S be the Schwartz space ([14])
S =
{
ϕ ∈ C∞ (R) : ∀m, l ∈ N ∪ 0∃Cml > 0∀x ∈ R
∣∣∣ϕ(m)(x)
(
1 + |x|2)l
∣∣∣ ≤ Cml
}
and S′ be the dual space.
A distribution f ∈ S′ is said to be odd if (f, ϕ(−x)) = − (f, ϕ(x)), ϕ ∈ S. A
distribution f ∈ S′ is said to be even if (f, ϕ(−x)) = (f, ϕ(x)), ϕ ∈ S.
Let Ω : S′ → S′ be the odd extension operator, Ξ : S′ → S′ be the even
extension operator. Thus (Ωf) (x) = f(x) − f(−x), (Ξf) (x) = f(x) + f(−x)
when f ∈ S′. Let Th be the translation operator: Thϕ(x) = ϕ(x + h), ϕ ∈ S and
(Thf, ϕ) = (f, T−hϕ), f ∈ S′, ϕ ∈ S.
Let us assume that q(x) and yn(λn, x), n = 1,∞, are defined on R and
are equal to 0 on R \ [0, d]. Denote by Q(x) the even 2d-periodic extension of
q(x): Q =
∑
k∈Z T2dkΞq. It is obvious that Q ∈ C1(R). Denote by Yn(λn, x)
the odd 2d-periodic extension of yn(λn, x) with respect to x, n = 1,∞. Thus
Yn(λn, ·) =
∑
k∈Z T2dkΩyn(λn, ·). Introduce the operator D2
Q = Q(x)+D2, where
D = −id/dx. Then D2
QYn(λn, x) = λ2
nYn(λn, x), n = 1,∞.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 37
K.S. Khalina
Denote
Hs
Q =
{
f ∈ S′ : f is odd and 2d–periodic,
(
1 + D2
Q
)s/2
f ∈ L2
loc(R)
}
, s ∈ R
with the norm ‖f‖s
Q =
(∫ d
−d |
(
1 + D2
Q
)s/2
f(x)|2 dx
)1/2
. We use the norm
|||f |||sQ =
((
‖f1‖s
Q
)2
+
(
‖f2‖s−1
Q
)2
)1/2
for f =
(
f1
f2
)
∈ Hs
Q ×Hs−1
Q .
Notice that, in fact, the space Hs
Q coincides with the Sobolev space of odd
periodic functions Hs
0,per. It is shown in the proof of Lemma A.2.
It is easy to see that the system
{
1√
2
Yn(λn, x)
}∞
n=1
forms the orthonormal
basis in H0
Q. It is proved in Lemma A.2 (Appendix A) that
Hs
Q =
{
f ∈ S′ : f(x) =
+∞∑
n=1
fnYn(λn, x) and
{
fn(1 + λ2
n)s/2
}+∞
n=1
∈ l2
}
=
{
f ∈ S′ : f(x) =
+∞∑
n=1
f1
n sin
πnx
d
and
{
f1
n(1 + n2)s/2
}+∞
n=1
∈ l2
}
, s ∈ R
with the equivalent norms
‖f‖s
Q =
( ∞∑
n=1
∣∣∣fn(1 + λ2
n)s/2
∣∣∣
2
)1/2
, []f []sQ =
( ∞∑
n=1
∣∣∣f1
n
(
1 + n2
)s/2
∣∣∣
2
)1/2
.
It follows from the above that Hs
Q ⊂ Hs′
Q when s′ < s. Denote for s ∈ R
Hs
Q(a, b) =
{
f ∈ S′ : supp f ⊂ [a, b] and
(
1 + D2
Q
)s/2
f ∈ L2(R)
}
.
One can see that if f ∈ Hs
Q(0, d), then F =
∑
k∈Z T2dkΩf ∈ Hs
Q for s ∈ R.
Evidently,
‖F‖s
Q =
√
2 ‖f‖s
Q .
We also use the following spaces:
Hs
0(−d, d) =
{
f ∈ S′ : supp f ⊂ [−d, d], f(x) =
∞∑
n=1
fne−iλnx
and
{
fn(1 + λ2
n)s/2
}∞
n=1
∈ l2
}
, s ∈ R
with the norm ‖f‖s
0 =
(∑∞
n=1
∣∣∣
(
1 + λ2
n
)s/2
fn
∣∣∣
2
)1/2
, where λn are the arithmeti-
cal roots of eigenvalues of the Sturm–Liouville problem (2.3). For convenience in
further reasoning we will number {λn}∞n=1 in ascending order.
38 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
Note that the series of exponentials in the definition of Hs
0(−d, d) converges
with respect to the norm of the standard Sobolev space Hs
0 that is proved in
Lemma A.4. Note also that Hs
0(−d, d) ⊂ Hs′
0 (−d, d) when s′ < s. In Lemma A.7
it is proved that if f ∈ Hs
0(−d, d), then f ′ ∈ Hs−1
0 (−d, d).
The properties of the functions from the space Hs
0(−d, d) can be found in
Appendix A.
R e m a r k 2.1. The system
{
eiλnx
}∞
n=1
is the Riesz basis in the space
L2(−d, d) (it is proved in Lemma A.1). Hence H0
Q(−d, d) =
{
f ∈ H0
0(−d, d) :
f is odd}.
Further, throughout the paper we will assume that s ≤ 0.
3. Main Results
Consider control system (2.1), (2.2) with the initial conditions
w(x, 0) = w0
0(x), wt(x, 0) = w0
1(x), x ∈ (0, d), (3.1)
where w0 =
(
w0
0
w0
1
)
∈ Hs
Q(0, d)×Hs−1
Q (0, d). We assume that q ∈ E(0, d) and the
control u satisfies the restriction
u ∈ B(0, T ) = {v ∈ L∞(0, T ) : |v(t)| ≤ 1 a. e. on (0, T )} .
We consider the solutions of system (2.1), (2.2), (3.1) in the space Hs
Q.
Extend w(x, t) and the initial functions from the segment (0, d) on the whole
axis. Consider the odd 2d-periodic extensions (with respect to x)
W (·, t) =
∑
k∈Z
T2dkΩw(·, t), W 0 =
∑
k∈Z
T2dkΩw0, t ∈ (0, T ).
One can see that W 0 ∈ Hs
Q ×Hs−1
Q , W (·, t) ∈ Hs
Q (t ∈ (0, T )).
It is easy to prove that control problem (2.1), (2.2), (3.1) is equivalent to the
following problem:
Wtt(x, t) = Wxx(x, t)−Q(x)W (x, t)− 2u(t)
∑
k∈Z
T2dkδ
′(x), x ∈ R, t ∈ (0, T ),
(3.2)
W (x, 0) = W 0
0 (x), Wt(x, 0) = W 0
1 (x), x ∈ R, (3.3)
where δ is the Dirac distribution. Consider (3.2), (3.3) with the steering condition
W (x, T ) = W T
0 (x), Wt(x, T ) = W T
1 (x), x ∈ R. (3.4)
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 39
K.S. Khalina
The solutions of (3.2)–(3.4) are considered in the space Hs
Q.
For the given T > 0, w0 ∈ Hs
Q(0, d) ×Hs−1
Q (0, d) denote by RT (w0) the set
of the states W T ∈ Hs
Q ×Hs−1
Q for which there exists a control u ∈ B(0, T ) such
that problem (3.2)–(3.4) has a unique solution in Hs
Q.
Definition 3.1. A state w0 ∈ Hs
Q(0, d)×Hs−1
Q (0, d) is called null-controllable
at a given time T > 0 if 0 belongs to RT (w0) and approximately null-controllable
at a given time T > 0 if 0 belongs to the closure of RT (w0) in Hs
Q ×Hs−1
Q .
Definition 3.2. Denote by ST : H
p
Q × H
p−1
Q → H
p
Q × H
p−1
Q , p ∈ R, the
operator
(ST f)(x) =
∞∑
n=1
(
cosλnT sin λnT
λn
−λn sinλnT cosλnT
)(
f1
n
f2
n
)
Yn(λn, x),
where f(x) =
(
f1(x)
f2(x)
)
=
∑∞
n=1
(
f1
n
f2
n
)
Yn(λn, x), D(ST ) = R(ST ) = H
p
Q ×H
p−1
Q .
The properties of the operator ST are studied in Lemma A.3.
To proceed further, we define the transformation operators for the Sturm–
Liouville problem on a segment. It is known [15] that the integral operator
K, given by the formula (Kf)(x) = f(x) +
∫ x
0 K(x, t;∞)f(t) dt, transfers the
solution of the equation y′′ + λ2y = 0 on a segment [−d, d] with the initial
conditions y(0) = 0, y′(0) = 1 to the solution of the equation y′′ − q(x)y +
λ2y = 0 with the same initial conditions at the point x = 0. Due to [15],
the operator K has an inverse which is denoted by L (see Appendix B for de-
tails). We determine these operators for p ∈ R in the following spaces: K :
H−p
0 (−d, d) −→ H
−p
Q (−d, d), L : H
−p
Q (−d, d) −→ H−p
0 (−d, d), where D(K) ={
f ∈ H−p
0 (−d, d) : f is odd
}
, D(L) = H
−p
Q (−d, d), R(K) = D(L), R(L) = D(K)
(see Lemma B.1). In Lemma B.2 it is proved that these operators are lin-
ear and continuous. In Appendix B we also determine the adjoint operators
K∗ : H
p
Q(−d, d) → Hp
0(−d, d), L∗ : Hp
0(−d, d) → H
p
Q(−d, d), where D(K∗) =
H
p
Q(−d, d), D(L∗) = {f ∈ Hp
0(−d, d) : f is odd}, R(K∗) = D(L∗), R(L∗) =
D(K∗) that are linear and continuous. The properties of the operators K, L,
K∗, L∗ are studied in Appendix B.
The main results of the work are the following theorems on the null- and
approximate null-controllability of the initial state of (2.1), (2.2), (3.1).
Theorem 3.1. Let 0 < T ≤ d, w0 ∈ Hs
Q(0, d)×Hs−1
Q (0, d), s ≤ 0. Then
RT (w0) =
{
ST
[
W 0 −
∑
k∈Z
T2kd
(
L∗
(
ΩU
ΩU ′
))]
: u ∈ B(0, T )
}
, (3.5)
40 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
where U(t) = u(t) on [0, T ] and U(t) = 0 on R \ [0, T ].
Theorem 3.2. Let 0 < T ≤ d, w0 ∈ Hs
Q(0, d)×Hs−1
Q (0, d), s ≤ 0. Then the
following statements are equivalent:
(i) the state w0 is null-controllable at the time T ;
(ii) the state w0 is approximately null-controllable at the time T ;
(iii) the following conditions hold:
suppw0
0 ⊂ [0, T ], (3.6)
w0
0 ∈ L∞(0, d) and
∣∣(K∗Ωw0
0
)
(x)
∣∣ ≤ 1 a.e. on [−d, d], (3.7)
Ωw0
1 = L∗ (
sign tK∗Ωw0
0
)′ on [−d, d]. (3.8)
In addition, the solution of the controllability problem (the control u) is unique
and given by the formula
u(t) = w0
0(t) +
T∫
t
K(x, t;∞)w0
0(x) dx, t ∈ [0, T ]. (3.9)
We will prove these theorems in Section 4.
R e m a r k 3.1. From (3.9) we have that there exists U > 0 such that
|u(x)| ≤ U on (0, T ) iff there exists V > 0 such that
∣∣w0
0(x)
∣∣ ≤ V on (0, d).
R e m a r k 3.2. Consider the case when q(x) ≡ q = const > 0 on (0, d).
Obviously, Q(x) ≡ q on (−d, d). Find the kernels L(t, x;∞) and K(t, x;∞) of the
transformation operators on (−d, d) × (−d, d). We have K(x, t;∞) = K(x, t) −
K(x,−t), L(x, t;∞) = L(x, t)−L(x,−t), where K(x, t) and L(x, t) are the solutions
of the following systems (see Appendix B):
Kxx(x, t)− Ktt(x, t) = qK(x, t), Lxx(x, t)− Ltt(x, t) = −qL(x, t),
K(x, x) =
1
2
qx, K(x,−x) = 0, L(x, x) = −1
2
qx, L(x,−x) = 0.
We reduce these systems to the Gurs problems and solve by the convergence
method (see, e.g., [13]). As a result, for the kernels we obtain the following
formulas:
K(x, t;∞) = qt
I1
(√
q(x2 − t2)
)
√
q(x2 − t2)
, |t| < |x|, K(x, t;∞) = 0, |t| ≥ |x|;
L(x, t;∞) = −qt
J1
(√
q(x2 − t2)
)
√
q(x2 − t2)
, |t| < |x|, L(x, t;∞) = 0, |t| ≥ |x|;
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 41
K.S. Khalina
where Jm(z) is the Bessel function, Im(z) = i−mJm(iz) is the modified
Bessel function, m ∈ Z. Thus the explicit formula for a control in the case of
q(x) ≡ q > 0 is u(t) = w0
0(t)+qt
∫ T
t
I1
(√
q(x2−t2)
)
√
q(x2−t2)
w0
0(x) dx, t ∈ (0, T ). The inverse
formula is w0
0(t) = u(t)− qt
∫ T
t
J1
(√
q(x2−t2)
)
√
q(x2−t2)
u(x) dx, t ∈ (0, T ). We remark that
the similar formulas for a control and an initial state were obtained for the semi-
infinite string [16] in the case when q = const and the boundary control was
of the Neumann type. It is easy to prove that if
∣∣w0
0(t)
∣∣ ≤ Cw on (0, d), then
|u(t)| ≤ CwI0(
√
qT ) and if |u(t)| ≤ Cu on (0, T ), then
∣∣w0
0(t)
∣∣ ≤ Cu
(
1 +
√
qT
)
.
E x a m p l e 3.1. Set d = 20, T = 15, q(x) ≡ q > 0, x ∈ (0, 20). Set
w0
0(x) = x3
2·153I0(15
√
q)
on (0, 15), w0
0(x) = 0 on (15, 20), w0
1(x), x ∈ (0, 20) such
that Ωw0
1 =
(
L∗ (
sign tK∗Ωw0
0
)′) on (−20, 20). Then
∣∣w0
0(x)
∣∣ ≤ 153
2·153I0(15
√
q)
=
1
2I0(15
√
q) . Therefore, due to Remark 3.2,
∣∣K∗Ωw0
0(x)
∣∣ ≤ 1
2I0(15
√
q)I0(15
√
q) = 1
2 .
Thus, assertion (iii) of Theorem 3.2 holds. Hence the state w0 is null-
controllable at the time T = 15. The control is determined by the formula
obtained in Remark 3.2. Consequently,
u(t) =
t
[
152qI0(
√
q(152 − t2))− 2
√
q(152 − t2)I1(
√
q(152 − t2))
]
2 · 153qI0(15
√
q)
, t ∈ (0, 15).
4. Proofs of Theorems 3.1, 3.2
4.1. Proof of Theorem 3.1
Due to Lemma A.2, we have
W (x, t) =
∞∑
n=1
wn(t)Yn(λn, x),
∑
k∈Z
T2dkδ
′(x) =
∞∑
n=1
δnYn(λn, x), (4.1)
where wn(t) = 1
2 (W (·, t), Yn(λn, ·)) = (w(·, t), yn(λn, ·)), δn = 1
2 (δ′, Yn(λn, ·)) =
−1
2Y ′
n(λn, 0) = −1
2y′n(λn, 0). Note that y′n(λn, 0) 6= 0, n = 1,∞. Substituting
(4.1) into (3.2), we obtain
∞∑
n=1
w′′n(t)Yn(λn, x) =
∞∑
n=1
wn(t)Y ′′
n (λn, x)−
∞∑
n=1
wn(t)Q(x)Yn(λn, x)
+ u(t)
∞∑
n=1
y′n(λn, 0)Yn(λn, x).
42 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
By applying the operator D2
Q, we get
∞∑
n=1
w′′n(t)Yn(λn, x) = −
∞∑
n=1
wn(t)λ2
nYn(λn, x) + u(t)
∞∑
n=1
y′n(λn, 0)Yn(λn, x).
From here we obtain the equation for the coefficients of Yn(λn, x)
w′′n(t) + wn(t)λ2
n = u(t)y′n(λn, 0), t ∈ [0, T ], n = 1,∞. (4.2)
Find the initial and steering conditions for this equation. We have W γ
0 (x) =∑∞
n=1 wγ
0nYn(λn, x) and W γ
1 (x) =
∑∞
n=1 wγ
1nYn(λn, x), where for γ = 0, T wγ
0n =
(wγ
0 (·), yn(λn, ·)), wγ
1n = (wγ
1 (·), yn(λn, ·)). From (3.3) and (3.4) we have
W (x, γ) =
∞∑
n=1
wn(γ)Yn(λn, x) = W γ
0 (x) =
∞∑
n=1
wγ
0nYn(λn, x),
Wt(x, γ) =
∞∑
n=1
w′n(γ)Yn(λn, x) = W γ
1 (x) =
∞∑
n=1
wγ
1nYn(λn, x), γ = 0, T.
Thus we obtain the initial and steering conditions for equation (4.2)
{
wn(0) = w0
0n
∂wn
∂t (0) = w0
1n
,
{
wn(T ) = wT
0n
∂wn
∂t (T ) = wT
1n
.
We reduce (4.2) to the linear system (n = 1,∞)
v′n(t) = Anvn(t) + bn(t), vn(0) =
(
w0
0n
w0
1n
)
, vn(T ) =
(
wT
0n
wT
1n
)
, t ∈ [0, T ], (4.3)
where vn =
(
wn
w′n
)
, An =
(
0 1
−λ2
n 0
)
, bn(t) =
(
0
y′n(λn, 0) · u(t)
)
. The solution of
this system is (n = 1,∞),
vn(t) =
(
cosλnt sin λnt
λn
−λn sinλnt cosλnt
) (
vn(0) +
(
−y′n(λn, 0)
∫ t
0
sin λnτ
λn
u(τ) dτ
y′n(λn, 0)
∫ t
0 cosλnτ · u(τ) dτ
))
.
Taking into account the steering condition in (4.3), for n = 1,∞ we have
(
cosλnT sin λnT
λn
−λn sinλnT cosλnT
)(
w0
0n − y′n(λn, 0)
∫ T
0
sin λnt
λn
u(t) dt
w0
1n + y′n(λn, 0)
∫ T
0 cosλnt · u(t) dt
)
=
(
wT
0n
wT
1n
)
.
(4.4)
Since u ∈ B(0, T ) ⊂ L2(0, d), then U ∈ L2(−d, d). According to Lemmas A.5,
A.6 and A.7, U ∈ H0
0(−d, d), ΩU ∈ H0
0(−d, d), ΞU ∈ H0
0(−d, d), (ΞU)′ = ΩU ′ ∈
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 43
K.S. Khalina
H−1
0 (−d, d). Now we can apply the transformation operators to U(t). Denote
ỹn(λn, x) = Ωyn(λn,x)
y′n(λn,0) (see Appendix B)
y′n(λn, 0)
∫ T
0
sinλnt
λn
u(t) dt =
y′n(λn, 0)
2
∫ d
−d
sinλnt
λn
ΩU(t) dt
=
y′n(λn, 0)
2
(ΩU,L(ỹn)) =
y′n(λn, 0)
2
(L∗ΩU, ỹn) =
1
2
(L∗ΩU,Ωyn) = (L∗ΩU, yn) ,
y′n(λn, 0)
∫ T
0
cosλntu(t) dt =
y′n(λn, 0)
2
∫ d
−d
cosλntΞU(t) dt
=
y′n(λn, 0)
2
(
ΞU,
(
sinλnt
λn
)′)
= −y′n(λn, 0)
2
(
(ΞU)′ ,L(ỹn)
)
= −y′n(λn, 0)
2
(L∗ΩU ′, ỹn
)
= −1
2
(L∗ΩU ′, Ωyn
)
= − (L∗ΩU ′, yn
)
.
Denote un = (L∗ΩU, yn), u′n = (L∗ΩU ′, yn). According to Lemma A.2, we have
∑
k∈Z
T2kd (L∗ΩU) (x) =
∞∑
n=1
unYn(λn, x),
∑
k∈Z
T2kd
(L∗ΩU ′) (x) =
∞∑
n=1
u′nYn(λn, x).
For n = 1,∞, (4.4) is equivalent to the following equation:
(
cosλnT sin λnT
λn
−λn sinλnT cosλnT
)(
w0
0n − un
w0
1n − u′n
)
=
(
wT
0n
wT
1n
)
. (4.5)
From the coefficient equality (4.5) we get the function equality
ST
[
W 0 −
∑
k∈Z
T2kd
(
L∗
(
ΩU
ΩU ′
))]
= W T . (4.6)
Hence (3.5) is true. The theorem is proved.
4.2. Proof of Theorem 3.2
Let assertion (iii) of Theorem 3.2 holds. Put Ũ(t) =
(K∗Ωw0
0
)
(t). It follows
from (3.7) and Definition B.1 (see Appendix B) that Ũ ∈ L∞(−d, d) and Ũ is
odd. By using (3.6) and Lemma B.5, we have that supp Ũ ⊂ [−T, T ]. Denote by
u(t) the restriction of Ũ(t) on [0, T ]. From (3.7) we get that u ∈ B(0, T ). Put
U(t) = u(t) on [0, T ] and U(t) = 0 on R \ [0, T ]. We have
ΩU = K∗Ωw0
0,
(
ΩU ′) = (ΞU)′ = (sign tΩU)′ =
(
sign tK∗Ωw0
0
)′
, on [−d, d].
Due to (3.8), from (4.6) we may conclude that the state w0 is null-controllable
at the time T . In addition, due to Lemma B.5, (3.9) holds.
44 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
Let the state w0 be approximately null-controllable at the time T . This
implies that there exists (W T )m such that (W T )m ∈ RT (w0) and
∣∣∣∣∣∣(W T
)m∣∣∣∣∣∣s
Q
<
1
m , m ∈ N. Thus there exists um ∈ B(0, T ) such that
ST
[∑
k∈Z
T2kdΩw0 −
∑
k∈Z
T2kd
(
L∗Ω
(
Um
(Um)′
))]
=
(
W T
)m
, m = 1,∞,
where Um(t) = um(t) on [0, T ] and Um(t) = 0 on R \ [0, T ].
Obviously, S−T ST f = ST S−T f = f for f ∈ Hs
Q×Hs−1
Q . Therefore, S−1
T = S−T
is inverse to ST . Hence S−1
T is continuous, and we have L∗ΩUm → Ωw0
0 in
Hs
Q(−d, d) and L∗Ω(Um)′ → Ωw0
1 in Hs−1
Q (−d, d) as m → ∞. According to
Lemma B.4, supp Ωw0
0 ∈ [−T, T ], i.e., (3.6) holds. Since L∗ is continuous, we
have
(
ΩUm
Ω(Um)′
)
→
(K∗Ωw0
0
K∗Ωw0
1
)
as m →∞ in Hs
0(−d, d)×Hs−1
0 (−d, d). (4.7)
Put Ũ(t) =
(K∗Ωw0
0
)
(t). According to Lemma B.5 and Definition B.1, supp Ũ ∈
[−T, T ] and Ũ is odd.
Since H−s
0 (−d, d) is dense in H0
0(−d, d), s ≤ 0 ( Lemma A.8), then ΩUm →
K∗Ωw0
0 as m → ∞ in
(
H0
0(−d, d)
)′. According to the Riesz theorem, Ũ ∈
H0
0(−d, d). Therefore, due to Lemma A.5, Ũ ∈ L2(−d, d). Since um ∈ B(0, T ),
m ∈ N, then we have
∣∣∣Ũ(t)
∣∣∣ ≤ 1 a.e. on [−d, d]. Thus Ũ ∈ L∞(−d, d). Since K∗
is continuous, then (3.7) holds.
Put U(t) = Ũ(t) on [0, d] and U(t) = 0 on R \ [0, d]. Taking into account
(4.7), we have
(
ΩUm
Ω(Um)′
)
→
(
ΩU
ΩU ′
)
=
(K∗Ωw0
0
K∗Ωw0
1
)
as m →∞ in Hs
0(−d, d)×Hs−1
0 (−d, d).
From the above we obtain (3.8). Thereby, if the state w0(t) is approximately
null-controllable at the time T , then assertion (iii) of Theorem 3.2 holds. Denote
by u(t) the restriction of U(t) on [0, T ]. It is easy to see that u(t) ∈ B(0, T ).
According to Lemma B.5, formula (3.9) holds. The theorem is proved.
5. The Moment Problem
The control obtained in Theorem 3.2 for the null-controllability problem may
be too complicated for practical use. In this section we consider the Markov
trigonometric moment problem on (0, T ) and prove that there are bang-bang
functions among its solutions. We show that they are the solutions of the ap-
proximate null-controllability problem for s < −1/2.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 45
K.S. Khalina
Consider control system (2.1), (2.2), (3.1). Set 0 < T ≤ d and w0 ∈ Hs
Q(0, d)×
Hs−1
Q (0, d), s ≤ 0. Assume that assertion (iii) of Theorem 3.2 holds. According to
Theorem 3.2, there exists ũ ∈ B(0, T ) which is a solution of the null-controllability
problem. Let us denote Ũ(t) = ũ(t), t ∈ [0, T ], Ũ(t) = 0, t ∈ R \ [0, T ]. Then
W 0 =
∑
k∈Z T2kdL∗
(
ΩŨ
ΩŨ ′
)
. Consider u ∈ B(0, T ) and denote U(t) = u(t),
t ∈ [0, T ], U(t) = 0, t ∈ R \ [0, T ]. By using (3.5), we get
W T = ST
∑
k∈Z
T2kdL∗Ω
(
Ũ − U
Ũ ′ − U ′
)
,
where W is the solution of (3.2)–(3.4). Applying Lemmas A.3, A.6, A.7 and B.3,
we conclude that
∣∣∣∣∣∣W T
∣∣∣∣∣∣s
Q
≤ 2
√
2CSCL∗
∥∥∥
(
Ũ − U
)∥∥∥
s
0
. (5.1)
Put
ωm =
T∫
0
ei πmx
d
(K∗Ωw0
0
)
(x) dx, m = −∞,∞. (5.2)
The problem of determination of a function u ∈ B(0, T ) such that
T∫
0
ei πmx
d u(x) dx = ωm, m = −∞,∞ (5.3)
for the given {ωm}∞m=−∞ and T > 0 is called the Markov trigonometric moment
problem on (0, T ) for the infinite sequence {ωm}∞m=−∞.
Theorem 5.1. Assume that 0 < T ≤ d, w0 ∈ Hs
Q(0, d) ×Hs−1
Q (0, d), s ≤ 0.
Assume that assertion (iii) of Theorem 3.2 holds. Assume also that {ωm}∞m=−∞
is defined by (5.2). Then the state w0 is null-controllable at the time T iff the
Markov trigonometric moment problem (5.3) has a unique solution on (0, T ).
Moreover, this solution is of the form (3.9).
P r o o f. Let the state w0 be null-controllable at the time T . Then due to
Theorem 3.2, the control u(t) =
(K∗Ωw0
0
)
(t) on (0, T ) is the unique solution of
controllability problem of system (2.1), (2.2), (3.1). Consequently, the Markov
trigonometric moment problem (5.3) has a unique solution on (0, T ) of the form
(3.9). Let the Markov trigonometric moment problem (5.3) has a unique solution
on (0, T ). Hence u(t) =
(K∗Ωw0
0
)
(t) on (0, T ). Since assertion (iii) of Theorem
3.2 holds, then the state w0 is null-controllable at the time T . The theorem is
proved.
46 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
R e m a r k 5.1. Theorem 5.1 is close to Theorem 3.2. The system {e−i πmx
d }∞m=−∞
is an orthonormal basis in L2(−T, T ) when T = d. Hence the moment problem
(5.3) has a unique solution in L2(−d, d). If T < d, then the system {e−i πmx
d }∞m=−∞
is complete and the solution is unique if it exists. Therefore the solution of (5.3)
is in B(0, T ). Moreover, it is unique if it exists and coincides with K∗Ωw0
0.
Consider (5.3) for a finite set of m
T∫
0
ei πmx
d u(x) dx = ωm, m = −M, M, M ∈ N. (5.4)
The problem of determination of a function u ∈ B(0, T ) such that (5.4) holds
for the given {ωm}M
m=−M , and T > 0 is called the Markov trigonometric moment
problem on (0, T ) for the finite sequence {ωm}M
m=−M .
Obviously, u of the form (3.9) is a solution of this problem for {ωm}∞m=−∞
given by (5.2), but it is not unique.
Theorem 5.2. Assume that 0 < T ≤ d, w0 ∈ Hs
Q(0, d) × Hs−1
Q (0, d), s <
−1/2. Assume that assertion (iii) of Theorem 3.2 holds. Assume also that
{ωm}∞m=−∞ is defined by (5.2). If uM ∈ B(0, T ) is a solution of the Markov
trigonometric moment problem (5.4) for some M ∈ N, then the corresponding
solution W of control system (3.2) – (3.4) satisfies the estimate
∣∣∣∣∣∣W T
∣∣∣∣∣∣s
Q
≤ 8πsCSCL∗PTM s+ 1
2
ds+1
√−2s− 1
−→ 0 as M →∞,
where P > 0.
P r o o f. Let ũ ∈ B(0, T ) be a solution of the controllability problem of system
(2.1), (2.2), (3.1) and u ∈ B(0, T ) be a solution of the Markov trigonometric
moment problem (5.4) on (0, T ) for the finite sequence {ωm}M
m=−M . Set Ũ(t) =
ũ(t) on [0, T ], Ũ(t) = 0 on R \ [0, T ], U(t) = u(t) on [0, T ], U(t) = 0 on R \ [0, T ].
Consider the following series expansions: Ũ(x) = 1
d
∑∞
m=−∞ ωme−i πmx
d , U(x) =
1
d
∑∞
m=−∞ νme−i πmx
d on (−d, d), where
ωm =
d∫
−d
ei πmx
d Ũ(x) dx =
T∫
0
ei πmx
d
(K∗Ωw0
0
)
(x) dx,
νm =
d∫
−d
ei πmx
d U(x) dx =
T∫
0
ei πmx
d u(x) dx.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 47
K.S. Khalina
Consequently,
Ũ(x)− U(x) =
1
d
∞∑
m=−∞
(ωm − νm) e−i πmx
d . (5.5)
According to Remark A.1 (see Appendix A), we have
∥∥∥Ũ − U
∥∥∥
s
0
=
∥∥∥
(
1 + D2
)s/2
(
Ũ − U
)∥∥∥
0
0
≤ P
∥∥∥
(
1 + D2
)s/2
(
Ũ − U
)∥∥∥
L2
, (5.6)
where P > 0.
Since νm = ωm, m = −M, M , then from (5.5) and (5.6) we have
∥∥∥Ũ − U
∥∥∥
s
0
≤
√
2
P
d
(
+∞∑
m=M+1
(
1 +
(mπ
d
)2
)s
|ωm − νm|2
)1/2
.
Since ũ ∈ B(0, T ) and u ∈ B(0, T ), then |ωm − νm| ≤ 2T . It is easy to see that(
1 +
(
mπ
d
)2
)s
≤ (
mπ
d
)2s for s ≤ 0. Using these two inequalities and the estimate∑∞
m=M+1 m2s ≤ ∫∞
M x2s dx, for s < −1/2 we get
∥∥∥Ũ − U
∥∥∥
s
0
≤ 2
√
2πsPTM s+ 1
2
ds+1
√−2s− 1
.
By continuing estimate (5.1), we have
∣∣∣∣∣∣W T
∣∣∣∣∣∣s
Q
≤ 8πsCSCL∗PTM s+ 1
2
ds+1
√−2s− 1
.
The theorem is proved.
Denote
BM (0, T ) = {u ∈ B(0, T )|∃T∗ ∈ (0, T ) : (|u(t)| = 1 a.e. on (0, T∗)) ,
(u(t) = 0 a.e. on (T∗, T )) ,
and (u(t) has no more than M discontinuities on (0, T∗))} .
Theorem 5.3. Assume that 0 < T ≤ d, w0 ∈ Hs
Q(0, d) × Hs−1
Q (0, d), s <
−1/2. Assume that assertion (iii) of Theorem 3.2 holds. Assume also that
{ωm}∞m=−∞ is defined by (5.2). Then for all ε > 0 there exists M ∈ N such
that for this M there is a solution uM ∈ BM (0, T ) of the Markov trigonometric
moment problem (5.4). Moreover, for this uM the corresponding solution W of
control system (3.2)–(3.4) satisfies the estimate
∣∣∣∣∣∣W T
∣∣∣∣∣∣s
Q
≤ ε. The number M
is defined from the condition
8πsCSCL∗PTM s+ 1
2
ds+1
√−2s− 1
< ε. (5.7)
48 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
P r o o f. It is well known [12, 17] that if the Markov trigonometric moment
problem (5.4) is solvable, then there exists its bang-bang solution u ∈ BM (0, T ).
By given ε > 0 determine M ∈ N from (5.7). Since the conditions of The-
orems 3.2 and 5.2 hold, then one can find a solution uM ∈ BM (0, T ) of the
Markov trigonometric moment problem (5.4) for the obtained M . For all M ,
these solutions {uM}∞M=1 give us the bang-bang controls solving the approximate
null-controllability problem. The theorem is proved.
A. The Spaces Hs
Q and Hs
0(−d, d), s ∈ R
Lemma A.1. The system
{
eiλnx
}∞
n=1
is the Riesz basis in L2(−d, d).
P r o o f. To prove the lemma we use Levin–Golovin’s theorem ([18]):
Theorem (Levin–Golovin). If the set {λn}∞n=1 such that infn 6=m |λm −
λn| > 0 is the set of simple zeros of the sine-type function of an exponential type
d, then
{
eiλnx
}∞
n=1
is the Riesz basis in L2(−d, d).
As noted in Section 2, λn are real, positive, simple and numbered in ascending
order, n = 1,∞. We also use the asymptotic expression λn = nπ
d + O
(
1
n
)
,
n = 1,∞ obtained in [19, chap. V]. Thus,
inf
n 6=m
|λm − λn| = inf
n
∣∣∣∣(n + 1)
π
d
+ O
(
1
n + 1
)
− n
π
d
−O
(
1
n
)∣∣∣∣ =
π
d
> 0.
To find a sine-type function in question we use the corollary to Theorem 1 in [20].
Corollary to Theorem 1. Let S̃(z) =
∏∞
n=1
(
1− z
µn+ψn
)
, where {µn}∞n=1
is the set of zeros of the sine-type function S(z) of an exponential type σ. If
{ψn}∞n=1 ∈ lp, p > 1, then S̃(z) is the sine-type function of an exponential type σ.
It is well known that S(z) = sin dz is a sine-type function of an exponential
type d and {µn = πn
d }∞n=−∞ is a set of its zeros. Put ψn = −πn
d + λn, n = 1,∞.
Then
|ψn| =
∣∣∣∣−
πn
d
+
πn
d
+ O
(
1
n
)∣∣∣∣ =
∣∣∣∣O
(
1
n
)∣∣∣∣ , n = 1,∞.
Hence, {ψn}∞n=1 ∈ lp, where p > 1. Therefore
S̃(z) =
∞∏
n=1
(
1− z
µn + ψn
)
=
∞∏
n=1
(
1− z
πn
d − πn
d + λn
)
=
∞∏
n=1
(
1− z
λn
)
is the sine-type function of an exponential type d. The lemma is proved.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 49
K.S. Khalina
Lemma A.2. The equivalent definitions of the space Hs
Q, s ∈ R are the
following:
Hs
Q =
{
f ∈ S′ : f(x) =
+∞∑
n=1
fnYn(λn, x) and
{
fn(1 + λ2
n)s/2
}+∞
n=1
∈ l2
}
=
{
f ∈ S′ : f(x) =
+∞∑
n=1
f1
n sin
πnx
d
and
{
f1
n(1 + n2)s/2
}+∞
n=1
∈ l2
}
, s ∈ R.
P r o o f. Since
{
Yn(λn,x)√
2
}∞
n=1
is the orthonormal basis in H0
Q, then for any
f ∈ Hs
Q, s ∈ R we have (1+D2
Q)s/2f(x) =
∑+∞
n=1 f̃nYn(λn, x) < ∞, where f̃n ∈ l2.
Use (1 + D2
Q)s/2Yn(λn, ·) = (1 + λ2
n)s/2Yn(λn, ·), n = 1,∞ for the coefficient
f̃n to get f̃n = 1
‖Yn‖2
(
(1 + D2
Q)s/2f, Yn(λn, ·)
)
= 1
2
(
f, (1 + D2
Q)s/2Yn(λn, ·)
)
=
1
2
(
1 + λ2
n
)s/2 (f, Yn(λn, ·)) =
(
1 + λ2
n
)s/2
fn. Here fn = 1
2 (f, Yn(λn, ·)) on (−d, d).
Thus,
f(x) =
+∞∑
n=1
f̃n(1 + D2
Q)
−s
2 Yn(λn, x) =
+∞∑
n=1
fnYn(λn, x).
Since for f ∈ H0
Q the norm can be calculated as ‖f‖0
Q =
(∑∞
n=1 |fn|2
)1/2
, then
the equivalent norm in the space Hs
Q is ‖f‖s
Q =
(∑∞
n=1
∣∣∣
(
1 + λ2
n
)s/2
fn
∣∣∣
2
)1/2
.
It is well known that a function from the space H0
Q can be represented as the
convergent series f(x) =
∑+∞
n=−∞ f1
n sin πnx
d .
Notice that since Q ∈ C1(R) and it is even and 2d-periodic, then (1 +
D2
Q)s/2f ∈ H0
Q iff (1 + D2)s/2f ∈ H0
Q when f ∈ Hs
Q. In other words, f ∈ Hs
Q
iff f ∈ {ϕ ∈ Hs
0,per : ϕ is odd}, where Hs
0,per is the Sobolev space of periodic
functions. Reasoning in the same way as above, with (1 + D2)s/2f we find that
f(x) =
+∞∑
n=1
f1
n sin
πnx
d
and
{
f1
n(1 + n2)s/2
}+∞
n=1
∈ l2, s ∈ R,
where f1
n = 1
d
(
f, sin πnx
d
)
on (−d, d), and []f []sQ =
(∑∞
n=1
∣∣∣
(
1 + n2
)s/2
f1
n
∣∣∣
2
)1/2
is the equivalent norm in the space Hs
Q. The lemma is proved.
Lemma A.3. The operator ST : Hs
Q × Hs−1
Q → Hs
Q × Hs−1
Q , s ∈ R (see
Definition 3.2) is linear and continuous.
50 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
P r o o f. Set f =
(
f1
f2
)
∈ Hs
Q ×Hs−1
Q . Then f(x) =
∑∞
n=1
(
f1
n
f2
n
)
Yn(λn, x),
{
f1
n(1 + λ2
n)s/2
}+∞
n=1
∈ l2 and
{
f2
n(1 + λ2
n)(s−1)/2
}+∞
n=1
∈ l2. By the definition of
ST , we have
(ST f)(x) =
∞∑
n=1
(
f1
n cosλnT + f2
n
sin λnT
λn
−λnf1
n sinλnT + f2
n cosλnT
)
Yn(λn, x).
Taking into account the estimate
∣∣∣ sin λnT
λn
∣∣∣ ≤
√
2T√
1+λ2
n
and the trivial inequality
(|a|+ |b|)2 ≤ 2(a2 + b2), we obtain
{
(1 + λ2
n)s/2
(
f1
n cosλnT + f2
n
sin λnT
λn
)}+∞
n=1
∈
l2,
{
(1 + λ2
n)(s−1)/2
(−λnf1
n sinλnT + f2
n cosλnT
)}+∞
n=1
∈ l2. Therefore, R(ST ) =
Hs
Q ×Hs−1
Q . We also get
|||ST f |||sQ =
( ∞∑
n=1
(1 + λ2
n)s
∣∣∣∣f1
n cosλnT + f2
n
sinλnT
λn
∣∣∣∣
2
+
∞∑
n=1
(1 + λ2
n)(s−1)
∣∣−λnf1
n sinλnT + f2
n cosλnT
∣∣2
)1/2
≤ CS |||f |||sQ ,
where C2
S = 2 max{2, 2T 2 + 1}. The linearity of the operator ST is obvious. The
lemma is proved.
Lemma A.4. f ∈ Hs
0(−d, d), s ∈ R iff
(
1 + D2
)s/2
f ∈ H0
0(−d, d).
P r o o f. Let s ∈ R and f ∈ Hs
0(−d, d). Hence f(x) =
∑∞
n=1 fne−iλnx and{
(1 + λ2
n)s/2fn
}∞
n=1
∈ l2. Then
(
1 + D2
)s/2
f(x) =
∑∞
n=1 fn(1 + D2)s/2e−iλnx =∑∞
n=1(1 + λ2
n)s/2fne−iλnx =
∑∞
n=1 f̂ne−iλnx, where f̂n = (1 + λ2
n)s/2fn, n = 1,∞.
Thus
{
f̂n
}∞
n=1
∈ l2. Therefore,
(
1 + D2
)s/2
f ∈ H0
0(−d, d).
The converse part of the lemma is proved similarly. The lemma is proved.
Lemma A.5. f ∈ H0
0(−d, d) iff f ∈ L2(−d, d).
P r o o f. Let f ∈ H0
0(−d, d). As it is shown in [19, chap. V], λn =
nπ
d + εn, where εn = O
(
1
n
)
, n = 1,∞. Therefore, f(x) =
∑∞
n=1 fne−iλnx =∑∞
n=1 fne−i nπ
d
xe−iεnx, where {fn}∞n=1 ∈ l2. As it is known
∑∞
n=1 fne−i nπ
d
x =
f̃(x), where f̃ is a certain function from L2(−d, d). Thus f ∈ L2(−d, d) iff(
f − f̃
)
∈ L2(−d, d). We have
f(x)− f̃(x) =
∞∑
n=1
fne−i nπ
d
x
(
e−iεnx − 1
)
=
∞∑
n=1
fne−i nπ
d
x (−iεnx + γn(x)) ,
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 51
K.S. Khalina
where γn(x) = o(εnx). Hence |γn(x)| ≤ C|εnx| as n →∞, where C > 0.
Consider
∥∥∥f − f̃
∥∥∥
2
L2(−d,d)
. We have
∥∥∥f − f̃
∥∥∥
2
L2(−d,d)
=
∫ d
−d
∣∣∣∣∣
∞∑
n=1
fne−i nπ
d
x (−iεnx + γn(x))
∣∣∣∣∣
2
dx
≤
∞∑
n,m=1
|fn| · |fm|
∫ d
−d
| − iεnx + γn(x)| · | − iεmx + γm(x)| dx
≤ (1 + C)2
2d3
3
∞∑
n,m=1
|fn| · |fm| · |εn| · |εm| = (1 + C)2
2d3
3
( ∞∑
n=1
|fn| · |εn|
)2
.
Since {fn}∞n=1 ∈ l2 and εn ∼ 1
n as n → ∞, then the last series converges.
Therefore,
∥∥∥f − f̃
∥∥∥
2
L2(−d,d)
< ∞. The lemma is proved.
R e m a r k A.1. From Lemma A.4 it follows that ‖f‖s
0 =
∥∥(1 + D2)s/2f
∥∥0
0
for f ∈ Hs
0(−d, d), s ∈ R. From Lemma A.5 it follows that there exist P, P1 > 0
such that ‖f‖0
0 ≤ P ‖f‖L2 and ‖f‖L2 ≤ P1 ‖f‖0
0 for f ∈ H0
0(−d, d).
Lemma A.6. Let g ∈ Hs
0(−d, d), s ∈ R. Then we have Ωg ∈ Hs
0(−d, d),
Ξg ∈ Hs
0(−d, d) and ‖Ωg‖s
0 = ‖Ξg‖s
0 = 2 ‖g‖s
0.
P r o o f. Since g ∈ Hs
0(−d, d), then g(x) =
∑∞
n=1 gne−iλnx and{
(1 + λ2
n)s/2gn
}∞
n=1
∈ l2. Therefore,
(Ωg) (x) =
∞∑
n=1
gnΩe−iλnx =
∞∑
n=1
gn
(
e−iλnx − eiλnx
)
=
∞∑
n=1
gsin
n sinλnx,
(Ξg) (x) =
∞∑
n=1
gnΞe−iλnx =
∞∑
n=1
gn
(
e−iλnx + eiλnx
)
=
∞∑
n=1
gcos
n cosλnx,
where gsin
n = −2ign, gcos
n = 2gn. Since
{
(1 + λ2
n)s/2gn
}∞
n=1
∈ l2, then{
(1 + λ2
n)s/2gsin
n
}∞
n=1
∈ l2 and
{
(1 + λ2
n)s/2gcos
n
}∞
n=1
∈ l2. Hence Ωg ∈ Hs
0(−d, d)
and Ξg ∈ Hs
0(−d, d). Finally we obtain
‖Ωg‖s
0 =
( ∞∑
n=1
∣∣∣
(
1 + λ2
n
)s/2
gsin
n
∣∣∣
2
)1/2
= 2
( ∞∑
n=1
∣∣∣
(
1 + λ2
n
)s/2
gn
∣∣∣
2
)1/2
= 2 ‖g‖s
0 ,
‖Ξg‖s
0 =
( ∞∑
n=1
∣∣∣
(
1 + λ2
n
)s/2
gcos
n
∣∣∣
2
)1/2
= 2
( ∞∑
n=1
∣∣∣
(
1 + λ2
n
)s/2
gn
∣∣∣
2
)1/2
= 2 ‖g‖s
0 .
The lemma is proved.
52 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
Lemma A.7. If g ∈ Hs
0(−d, d), then g′ ∈ Hs−1
0 (−d, d) and ‖g′‖s−1
0 < ‖g‖s
0,
s ∈ R.
P r o o f. Let g ∈ Hs
0(−d, d), s ∈ R. Hence g(x) =
∑∞
n=1 gne−iλnx and{
gn(1 + λ2
n)
s
2
}
∈ l2, s ∈ R. Taking into account that |λn| <
√
1 + λ2
n, we get
‖g′‖s−1
0 =
(∑∞
n=1
∣∣(1 + λ2
n)(s−1)/2iλngn
∣∣2
)1/2
< ‖g‖s
0. The lemma is proved.
Lemma A.8. Hs
0(−d, d) is dense in H0
0(−d, d), s ≥ 0.
P r o o f. Let f ∈ H0
0(−d, d). Then f(x) =
∑∞
n=1 fne−iλnx and {fn}∞n=1 ∈ l2.
Consider a sequence of the functions {fm(x)}∞m=1 such that fm(x) =
∑∞
n=1 fm
n e−iλnx,
where fm
n = sin(ns/m)
ns/m fn, n,m = 1,∞. One can see that
{
(1 + λ2
n)s/2fm
n
}∞
n=1
∈ l2,
m = 1,∞. Therefore fm ∈ Hs
0(−d, d), m = 1,∞, s ≥ 0. Since fm
n → fn as
m → ∞, we have ‖f − fm‖0
0 =
(∑∞
n=1 |fn − fm
n |2
)1/2
→ 0 as m → ∞. The
lemma is proved.
B. The Transformation Operators for the Sturm–Liouville
Problem on a Segment
We would like to recall some definitions of transformation operators used in
[15] as well as the statements proved there. Denote ỹn(λn, x) = Ωyn(λn,x)
y′n(λn,0) , n =
1,∞. Obviously, ỹn(λn, x) satisfies the following Cauchy problem for n = 1,∞:
−ỹ′′n(λn, x) + Q(x)ỹn(λn, x) = λ2
nỹn(λn, x), x ∈ (−d, d),
ỹn(λn, 0) = 0, ỹ′n(λn, 0) = 1.
According to [15], we have
ỹn(λn, x) = K
(
sinλnt
λn
)
(x) =
sinλnx
λn
+
x∫
0
K(x, t;∞)
sinλnt
λn
dt, n = 1,∞,
sinλnx
λn
= L (ỹn(λn, t)) (x) = ỹn(λn, x) +
x∫
0
L(x, t;∞)ỹn(λn, t) dt, n = 1,∞,
where K(x, t;∞) = K(x, t) − K(x,−t), L(x, t;∞) = L(x, t) − L(x,−t). Under the
condition Q ∈ C1[−d, d] the continuous functions K(x, t) and L(x, t) are the solu-
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 53
K.S. Khalina
tions of the following systems on [−d, d]× [−d, d]:
Kxx(x, t)− Ktt(x, t) = Q(x)K(x, t), Lxx(x, t)− Ltt(x, t) = −Q(x)L(x, t),
K(x, x) =
1
2
x∫
0
Q(ξ) dξ, L(x, x) = −1
2
x∫
0
Q(ξ) dξ,
K(x,−x) = 0, L(x,−x) = 0.
It is also known [15] that the kernels K(x, t) and L(x, t) are bounded functions
with respect to the both arguments on [−d, d]× [−d, d] and K(x, t) = L(x, t) = 0
when |t| ≥ |x|.
Let us determine the transformation operators in the spaces H−s
0 (−d, d) and
H−s
Q (−d, d) via series.
Lemma B.1. The operators K and L act in the spaces
K : H−s
0 (−d, d) −→ H−s
Q (−d, d), L : H−s
Q (−d, d) −→ H−s
0 (−d, d), s ∈ R,
where D(K) =
{
f ∈ H−s
0 (−d, d) : f is odd
}
, D(L) = H−s
Q (−d, d), R(K) = D(L),
R(L) = D(K).
P r o o f. Set s ∈ R. Consider ψ ∈ H−s
Q (−d, d), ϕ ∈ H−s
0 (−d, d) (ϕ is odd).
Then we have ψ(x) =
∑∞
n=1 ψnΩyn(λn, x),
{(
1 + λ2
n
)−s
2 ψn
}∞
n=1
∈ l2, ϕ(x) =
∑∞
n=1 ϕn sinλnx,
{(
1 + λ2
n
)−s
2 ϕn
}∞
n=1
∈ l2. Let ψ̃n = λnϕn
y′n(λn,0) , ϕ̃n = ψny′n(λn,0)
λn
.
Applying the transformation operators, we get
(Kϕ) (x) =
∞∑
n=1
λnϕnK
(
sinλnt
λn
)
(x) =
∞∑
n=1
λnϕnỹn(λn, x)
=
∞∑
n=1
λnϕn
Ωyn(λn, x)
y′n(λn, 0)
=
∞∑
n=1
ψ̃nΩyn(λn, x) = ψ̃(x),
(Lψ) (x) =
∞∑
n=1
ψnL (Ωyn(λn, t)) (x) =
∞∑
n=1
ψny′n(λn, 0)L (ỹn(λn, t)) (x)
=
∞∑
n=1
ψny′n(λn, 0)
sinλnx
λn
=
∞∑
n=1
ϕ̃n sinλnx = ϕ̃(x).
The following asymptotic expressions for the solutions of (2.3) and for their
derivatives are obtained in [19, chap. V]: yn(t) =
√
2
d sinnπ
d t + O
(
1
n
)
, y′n(t) =
nπ
d
√
2
d cosnπ
d t + O (1), n = 1,∞. We also use the expression λn = nπ
d + O
(
1
n
)
.
54 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
Therefore, y′n(λn, 0) ∼ n and λn ∼ n as n →∞. Thus, ψ̃n ∼ nϕn
n = ϕn and ϕ̃n ∼
nψn
n = ψn. Hense,
{(
1 + λ2
n
)−s/2
ψ̃n
}∞
n=1
∈ l2 and
{(
1 + λ2
n
)−s/2
ϕ̃n
}∞
n=1
∈ l2.
Therefore, ϕ̃ ∈ H−s
0 (−d, d) and ψ̃ ∈ H−s
Q (−d, d). The lemma is proved.
Lemma B.2. The operators K and L are linear and continuous on their
domains.
P r o o f. The linearity of the operators is obvious. Let us prove that the oper-
ators K and L are continuous. Let s ∈ R, ϕ ∈ H−s
0 (−d, d) (ϕ is odd). Then Kϕ ∈
H−s
Q (−d, d). We have ϕ(x) =
∑∞
n=1 ϕn sinλnx, (Kϕ) (x) =
∑∞
n=1 ψnΩyn(λn, x),
(‖ϕ‖−s
0
)2
=
∑∞
n=1
∣∣∣ϕn
(
1 + λ2
n
)−s/2
∣∣∣
2
,
(
‖Kϕ‖−s
Q
)2
=
∑∞
n=1
∣∣∣ψn
(
1 + λ2
n
)−s/2
∣∣∣
2
.
Since ψn = λnϕn
y′n(λn,0) (see Lemma B.1), then
(
‖Kϕ‖−s
Q
)2
=
∞∑
n=1
∣∣∣∣
λnϕn
y′n(λn, 0)
(
1 + λ2
n
)−s
2
∣∣∣∣
2
=
∞∑
n=1
∣∣∣∣
λn
y′n(λn, 0)
∣∣∣∣
2 ∣∣∣∣ϕn
(
1 + λ2
n
)−s
2
∣∣∣∣
2
.
It is evident that there exists C >0 such that the estimate
∣∣∣ λn
y′n(λn,0)
∣∣∣= |n π
d
+O( 1
n)|∣∣∣n π
d
√
2
d
+O(1)
∣∣∣
≤ C is valid. Therefore,
(
‖Kϕ‖−s
Q
)2
≤ C2
∑∞
n=1
∣∣∣ϕn
(
1 + λ2
n
)−s/2
∣∣∣
2
= C2
(‖ϕ‖−s
0
)2
.
Thus the operator K is continuous from H−s
0 (−d, d) to H−s
Q (−d, d). Its inverse
operator L is also continuous from H−s
Q (−d, d) to H−s
0 (−d, d). The lemma is
proved.
Definition B.1. Define by K∗ and L∗ the adjoint operators for K and L:
(K∗f, ϕ) = (f,Kϕ), (L∗g, ψ) = (g,Lψ), where f ∈ D(K∗) = Hs
Q(−d, d), ϕ ∈
D(K), g ∈ D(L∗) = {f ∈ Hs
0(−d, d) : f is odd}, ψ ∈ D(L), s ∈ R.
Thus, K∗ : Hs
Q(−d, d) → Hs
0(−d, d), L∗ : Hs
0(−d, d) → Hs
Q(−d, d) and they
are linear and continuous. Moreover, (K∗f) (x) =
∑∞
n=1 Kn sinλnx, (L∗g) (x) =∑∞
n=1 Lnỹn(λn, x). Evidently, R(K∗) = D(L∗), R(L∗) = D(K∗). It is obvious
that K∗f and L∗g are odd when f and g are odd.
Lemma B.3. Let f ∈ Hs
0(−d, d)×Hs−1
0 (−d, d), f be odd, s ∈ R. Then there
exists CL∗ > 0 such that |||L∗f |||sQ ≤ CL∗ |||f |||s0.
P r o o f. Since L∗ is continuous, then there exist B1 > 0, B2 > 0 such that
|||L∗f |||sQ =
((
‖L∗f1‖s
Q
)2
+
(
‖L∗f2‖s−1
Q
)2
)1/2
≤
((
(B1)
2 ‖f1‖s
0
)2
+
(
(B2)
2 ‖f2‖s−1
0
)2
)1/2
≤ CL∗
(
(‖f1‖s
0)
2 +
(
‖f2‖s−1
0
)2
)1/2
= CL∗ |||f |||s0 ,
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1 55
K.S. Khalina
where CL∗ = max {B1, B2}, f =
(
f1
f2
)
. The lemma is proved.
Lemma B.4. Let f ∈ L2(−d, d) be odd, supp f ⊂ [−T, T ]. Then supp (L∗f) ⊂
[−T, T ] and (L∗f) (t) = f(t) +
∫ T
|t| L(x, t;∞)f(x) dx.
P r o o f. Let ψ ∈ L2(−d, d) be odd. Then
(L∗f, ψ) = (f,Lψ) =
d∫
−d
f(x)
ψ(x) +
x∫
0
L(x, t;∞)ψ(t) dt
dx
=
d∫
−d
ψ(x)
f(t) +
d sign t∫
t
L(x, t;∞)f(x) dx
dt.
Taking into account that ỹn(λn, x) and sin λnx
λn
are odd on x, n = 1,∞, and
L(x, t;∞) is odd on t, it is easy to get L(−x, t;∞) = L(x, t;∞). Therefore, due
to the oddness of f , for t < 0 we have
d sign t∫
t
L(x, t;∞)f(x) dx =
−d∫
t
L(x, t;∞)f(x) dx =
d∫
−t
L(−x, t;∞)f(x) dx
=
d∫
|t|
L(x, t;∞)f(x) dx.
Obviously, the support of the last function is contained in [−T, T ]. Consequently,
(L∗f) (t) = f(t) +
∫ T
|t| L(x, t;∞)f(x) dx and we get the assertion of the lemma.
The lemma is proved.
Lemma B.5. Let f ∈ L2(−d, d) be odd, supp f ⊂ [−T, T ]. Then supp (K∗f) ⊂
[−T, T ] and (K∗f) (t) = f(t) +
∫ T
|t| K(x, t;∞)f(x) dx.
The proof of this lemma is similar to the proof of the previous lemma.
56 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 1
Controllability Problems for the Non-Homogeneous String
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