On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions
We prove an additional regularity of time derivative of the trace of solution to the wave equation on the 3D half space with the homogeneous Neumann boundary conditions. Доведено додаткову регулярність похідної за часом від сліду розв'язку хвильового рівняння у тривимірному півпросторі з однорі...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2007
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| Cite this: | On trace regularity of solutions to a wave equation with homogeneous Neumann boundary conditions / I.A. Ryzhkova // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 468-489. — Бібліогр.: 13 назв. — англ. |
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| citation_txt | On trace regularity of solutions to a wave equation with homogeneous Neumann boundary conditions / I.A. Ryzhkova // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 468-489. — Бібліогр.: 13 назв. — англ. |
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| description | We prove an additional regularity of time derivative of the trace of solution to the wave equation on the 3D half space with the homogeneous Neumann boundary conditions.
Доведено додаткову регулярність похідної за часом від сліду розв'язку хвильового рівняння у тривимірному півпросторі з однорідними крайовими умовами Неймана.
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Journal of Mathematical Physics, Analysis, Geometry
2007, vol. 3, No. 4, pp. 468�489
On Trace Regularity of Solutions to a Wave Equation
with Homogeneous Neumann Boundary Conditions
I.A. Ryzhkova
Department of Mechanics and Mathematics, V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:Iryna.A.Ryzhkova@univer.kharkov.ua
Received January 31, 2007
We prove an additional regularity of time derivative of the trace
of solution to the wave equation on the 3D half space with the homogeneous
Neumann boundary conditions.
Key words: wave equation, boundary trace regularity.
Mathematics Subject Classi�cation 2000: 35L05, 35B65.
1. Introduction
In this paper we consider the following equation on R
3
+ :
(@t + U@x1)
2� = ��; x = (x1; x2; x3) 2 R
3
+ = fx : x3 > 0g; (1)
@�
@x3
����
x3=0
= 0; (2)
�(0) = �0; �t(0) = �1: (3)
The equations of this type arise, for example, in aerodynamics of potential gas
�ows. We consider the following motivating model. Gas occupies the half-space
R
3
+ and moves along x1-axis with the velocity U � 0; U 6= 1. Then the potential
of the velocity of a perturbed gas �ow � satis�es (1)�(3).
The problem of the trace regularity of solutions to hyperbolic equations fre-
quently arises in hybrid systems theory. In particular, for the purposes of [1, 2]
we need to study the regularity of the function (@t + U@x1)�(x1; x2; 0), where �
is a solution to (1)�(3). In the present paper we will prove several new results on
the smoothness concerned with the equation (1)�(3).
c
I.A. Ryzhkova, 2007
On Trace Regularity of Solutions to a Wave Equation...
The regularity of solutions to general hyperbolic equations and of their traces
on the boundary were studied by I. Lasiecka and R. Triggiani (see [3, 4] and
references therein). Their results [3, 4] give the following trace regularity for the
problem (1)�(3).
Theorem 1. Let �(t) be a solution to (1)�(3) with the initial conditions
(�0; �1), �T = R
2� [0; T ] and
[�] be the Sobolev trace of a function de�ned on R
3
+
onto the plane fx : x3 = 0g. Then the mapping (�0; �1) 7�!
[�] is a continuous
operator from H1(R3
+) � L2(R3
+) to H3=4(�T ) and from L2(R3
+ ) � (H1(R3
+))0 to
H�1=4(�T ) for every 0 < T < +1.
Set U = 0 and denote ~� = �t. Formally di�erentiating (1), we obtain that ~�
satis�es (1)�(2) with the initial conditions ~�(0) = �1; ~�t(0) = ��0. If (�0; �1) 2
H1(R3
+)� L2(R3
+), then (�1;��0) 2 L2(R3
+)� (H1(R3
+))0. Thus, Th. 1 can give
us only @t
[�] 2 H�1=4(�).
Our main result improves Th. 1 in two directions. First, we prove that @t
[�] 2
L2(0; T ;H�1=4(R2)) provided initial conditions (�0; �1) 2 H1(R3
+)�L2(R3
+). Sec-
ond, we can in some sense improve this result, �nding an appropriate decompo-
sition of @t
[�] as a sum @t
[�] = f1 + f2. This new idea allows us to prove
that f1 2 L1(0; T ;H�1=4��(R2 )) and f2 2 L2(0; T ;L2(R2)), provided initial con-
ditions (�0; �1) 2 H1(R3
+) � L2(R3
+ ). We also study, how the trace regularity
can be improved when more smooth initial conditions are considered, and what
happens if �0 lies in a homogeneous Sobolev space.
In the proof of Th. 3 we rely on some ideas borrowed from [3], and �rst study
a trace regularity of nonomogeneous problem with zero initial conditions. These
results are collected in Th. 4, which, we believe, is an interest on itself.
The structure of the paper is as follows. In Section 2 we introduce the de�-
nitions and notations we need and state our main results. In Section 3 we study
the properties of solutions to wave equation (1)�(3) with smooth initial condi-
tions and prove some results on the interpolation of functional spaces we use. In
Section 4 we prove Th. 4. In Section 5 we use Th. 4 to prove Th. 3. In Section 6
we prove a "local" version of Th. 3.
2. Notations and Main Results
To describe the behaviour of a solution � to (1)�(3) we use a homogeneous
Sobolev space H1(R3
+). We de�ne the space H1(R3) (see, e.g., [5]) as the closure
of C1
0 (R3 ) with respect to the norm jjujjH1(R3) = jjrujjR3 . For H1(R3
+) de�ned as
the space of restrictions of functions from H1(R3) onto R
3
+ we use the equivalent
norm jjr�jjR3+ .
If we consider the system (1)�(3) as a model for a perturbed gas �ow, the
norm for � introduced above is natural. Indeed, in this case �(t) represents the
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 469
I.A. Ryzhkova
potential of velocity of a perturbed gas �ow, and has no physical meaning itself,
while r�(t) is the �eld of velocity of the perturbed �ow and gives the complete
information about the �ow. For (�0; �1) 2 H1(R3
+ ) � L2(R3
+) we de�ne a local
energy by
ER(�0; �1) =
Z
B
+
R
�
jr�0(x)j2 + j�1(x)j2
�
dx;
where B+
R = fx = (x1; x2; x3) : jxj < R; x3 > 0g.
Now we �nd the appropriate spaces for initial data which are "more smooth"
than H1(R3
+) � L2(R3
+) and introduce a suitable notion of local energy for these
spaces. First, we consider the wave equation in the whole space R3 with smooth
initial conditions
@2t � = ��; x 2 R
3 ; (4)
�(0) = �0; �t(0) = �1: (5)
Formally di�erentiating (4), we see that ��(t) = �t(t) satis�es (4) with the initial
conditions ��(0) = �1; ��t(0) = ��0. Thus energy conservation law for ��(t) gives
us the following energy relation for �:
jjr�t(t)jj20;R3 + jj��(t)jj20;R3 = jjr�1jj20;R3 + jj��0jj20;R3 :
The classical conservation law for (4)�(5), together with the previous relation
give us
jjr�(t)jj20;R3+jj��(t)jj20;R3+jj�t(t)jj21;R3 = jjr�0jj20;R3+jj��0jj20;R3+jj�1jj21;R3 : (6)
Hence, if we de�ne W1;1(R3) as the closure of C1
0 (R3) with respect to the norm
jj � jj2
W1;1(R3)
= jj� � jj2
0;R3
+ jjr � jj2
0;R3
, we can easily verify that for initial data
(�0; �1) 2 W1;1(R3 )�H1(R3 ) the problem (4)�(5) possesses precisely one solution
(�(t); �t(t)) 2 C(0; T ;W1;1(R3) � H1(R3)) for any T > 0 for which the energy
relation (6) holds.
We consider initial data from the spaces that are "intermediate" between
H1(R3 ) and W1;1(R3). These spaces can be de�ned via Fourier transform (see
Prop. 2). The space W1;s(R3 ); s � 0 consists of all distributions f 2 S0(R3) such
that its Fourier transform ~f is a regular distribution and the integral
jjf jj2
W1;s(R3) =
Z
R3
d�(1 + j�j)2sj�j2j ~f(�)j2
is �nite. It is easy to see that there is another description of W1;s(R3):
W1;s(R3 ) = ff 2 H1(R3 ) : rf 2 (Hs(R3 ))3g (7)
with the equivalent norm jjf jjW1;s(R3) = jjrf jjs;R3 .
470 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Now we return to the wave equation in the half-space. De�ne the space
W1;s(R3
+), s 2 [0; 1], as the space of restrictions of functions from W1;s(R3 )
onto R
3
+ . Due to description (7) we see that jjrf jjs;R3+ is an equivalent norm in
W1;s(R3
+). Similarly as in the case of H1(R3
+) � L2(R3
+), we can de�ne a local
energy in the space W1;s(R3
+ )�Hs(R3
+):
EsR(�0; �1) = jjr�0jj2s;B+
R
+ jj�1jj2s;B+
R
; s 2 [0; 1]:
In order to obtain smooth solutions to (1)�(3) we need not only the smooth ini-
tial data but also consistency conditions imposed on these initial data. Therefore
we need the spaces
W1;s
(R3
+) =
(
f 2 W1;s(R3
+) :
@f
@x3
����
x3=0
= 0
)
; s > 1=2;
with the same norm as in W1;s(R3
+).
The following interpolation lemma is valid.
Lemma 1. [W1;1
(R3
+);H1(R3
+)][1��] = G�, where
G� =W1;�(R3
+); 0 � � < 1=2;
G1=2 = ff 2 H1(R3
+) : rf 2 (L2(R+ ;H1=2(R2)))2 �H
1=2
00 (R+ ;L2(R2))g; (8)
G� =W1;�
(R3
+); 1=2 < � � 1;
where H
1=2
00 (R+ ;L2(R2 )) = ff 2 H1=2(R+ ;L2(R2)) : x
�1=2
3 f 2 L2(R3
+)g with the
norm jjf jj2
H
1=2
00 (R+;L2(R2))
= jjf jj2
H1=2(R+;L2(R2))
+ jjx�1=2
3 f jj2
L2(R3+)
.
Now we can state the following existence theorem for the wave equation in
the half-space.
Theorem 2. Assume that initial data (�0; �1) 2 G� �H�(R3
+). Then:
(i) For every T > 0 there exists precisely one solution to (1)-(3) (�; �t)(t) 2
C(0; T ;G� �H�(R3
+)).
(ii) The norm of the solution does not increase. For � 2 [0; 1]; � 6= 1=2 the
following inequality is valid:
jjr�(t)jj2
�;R3+
+ jj�t(t)jj2�;R3+ � jjr�0jj
2
�;R3+
+ jj�1jj2�;R3+ : (9)
(iii) For every R > 0 and � 2 [0; 1] the local energy E�R(�(t); �t(t))! 0, when
t! +1.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 471
I.A. Ryzhkova
We will use weight spaces of X-valued functions. The space L2(
;X; d�)
consists of all functions mapping
into X such that
jjf jj2L2(
;X;d�) =
Z
jjf(t)jj2Xd� <1
Here X is a normed space and d� is a measure on
. In the case X = R we omit
X in the notation.
Further we state our main results.
Theorem 3. Assume that �(t) is a solution to (1)�(3) with the initial condi-
tions �0 2 G�(R3
+); �1 2 H�(R3
+), � 2 [0; 1]; � 6= 1=2. Then:
(i) (@t + U@x1)
[�] 2 L2(0; T ;H
�1=4+�
loc
(R2)). The following estimate takes
place
jj(@t + U@x1)
[�]jjL2(0;T ;H�1=4+�(B)) � C(T;U;B)
�
jjr�0jj�;R3+ + jj�1jj�;R3+
�
:
for any bounded set B � R
2 .
(ii) (@t + U@x1)
[�] = f1 + f2, where f1 2 L1(0; T ;H
�1=4+���
loc
(R2)) for any
� > 0, f2 2 L2(0; T ;H�
loc(R
2)), and
jjf1jjL1(0;T ;H�1=4+���(B)) � C(T;U; �;B)
�
jjr�0jj�;R3+ + jj�1jj�;R3+
�
;
jjf2jjL2(0;T ;H�(B)) � C(T;U;B)
�
jjr�0jj�;R3+ + jj�1jj�;R3+
�
for any bounded set B � R
2 .
Remark 1. Since the inclusion G� � L2(R3
+) does not take place, estimates
for (@t + U@x1)
[�] have only local character. However, if �0 2 G�
T
L2(R3
+) �
H1+�(R3
+ ) we have the estimate
jj(@t + U@x1)
[�]jjL2(0;T ;H�1=4+�(R2)) � C(T;U)
�
jj�0jj1+�;R3+ + jj�1jj�;R3+
�
in point (i) of Th. 3 and the estimates
jjf1jjL1(0;T ;H�1=4+���(R2)) � C(T;U; �)
�
jj�0jj1+�;R3+ + jj�1jj�;R3+
�
;
jjf2jjL2(0;T ;H�(R2)) � C(T;U)
�
jj�0jj1+�;R3+ + jj�1jj�;R3+
�
in point (ii) of Th. 3.
The theorem implies the following local energy estimate.
472 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Corollary 1. Assume that �(t) is a solution to (1)�(3) with the initial condi-
tions �0 2 G�(R3
+); �1 2 H�(R3
+), � 2 [0; 1]; � 6= 1=2. Let
be a bounded smooth
domain in R
2 , and r
be the operator of restriction of a function de�ned on R
2
to
. Then:
(i) r
(@t + U@x1)
[�] 2 L2(0; T ;H�1=4+�(
)). The following estimate takes
place
jjr
(@t + U@x1)
[�]jj2L2(0;T ;H�1=4+�(
))
� CE�R(�0; �1):
The constants C;R depend only on T ,
, and U .
(ii) r
(@t + U@x1)
[�] = f1 + f2, where f1 2 L1(0; T ;H�1=4+���(
)) for any
� > 0, f2 2 L2(0; T ;H�(
)), and
jjf1jj2L1(0;T ;H�1=4+���(
))
� CE�R(�0; �1);
jjf2jj2L2(0;T ;H�(
)) � CE�R(�0; �1):
The constants C;R depend on T ,
, and U , the constant C in the �rst inequality
depends also on �.
Remark 2. Theorem 3 and Corollary 1 allow us to justify the stabilization
results of [1, 2] for the initial data (�(0); �t(0)) 2 G� � H�(R3
+), � 2 (0; 1],
� 6= 1=2.
The theorem below deals with the trace regularity of the nonhomogeneous
wave equation. Following [3], we choose a certain function f in (10) and obtain
Th. 3 as a consequence of the following theorem.
Theorem 4. Consider the problem
(@t + U@x1)
2� = ��+ f(t; x); x 2 R
3
+ ; (10)
@�
@x3
����
x3=0
= 0; �(0) = �t(0) = 0: (11)
(i) Let f 2 L2(0; T ;H�(R3
+)). Then (@t + U@x1)
[�] 2 L2(0; T ;H�1=4+�(R2))
and the following estimate takes place
jj(@t + U@x1)
[�]jjL2(0;T ;H�1=4+�(R2)) � C(T;U)jjf jjL2(0;T ;H�(R3+)):
(ii) Let f 2 L1(R+ ;H�(R3
+)). Then (@t + U@x1)
[�] = f1 + f2, where
f1 2 L1(0; T ;H�1=4+���(R2 )), � > 0, f2 2 L2(0; T ;H�(R2 )), and the follow-
ing estimates are valid
jjf1jjL1(0;T ;H�1=4+���(R2)) � C(T;U; �)jjf jjL1(R+;H�(R3+));
jjf2jjL2(0;T ;H�(R2)) � C(T;U)jjf jjL1(R+;H�(R3+)):
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 473
I.A. Ryzhkova
3. Smooth Solutions to the Wave Equation
and Interpolation Spaces
For the proof of Th. 3 we need to generalize spaces W1;s. De�ne space
W�;�(Rn) as the space of distributions f 2 S0(Rn) such that their Fourier trans-
forms ~f are regular distributions and
jjf jj2
W�;�(Rn) =
Z
Rn
j�j2�(1 + j�j)2� j ~f(�)j2d� <1:
The Sobolev trace theorem can be easily generalized for these spaces.
Lemma 2. The Sobolev trace operator is continuous from W�;�(Rn) to
W��1=2;�(Rn�1 ), if � > 1=2.
The following embedding takes place.
Lemma 3. If � < n=2 then W�;0(Rn) is continuously embedded in Lp(Rn),
p = 2n=(n� 2�).
P r o o f. f 2 W�;0(Rn) if and only if g = F�1j�j�Ff 2 L2(Rn). Equivalently,
f = F�1j�j��Fg 2 Lp(Rn) if and only if F�1j�j��F is a continuous operator from
L2(Rn) to Lp(Rn). Thus, the assertion of the lemma follows from Th. 1.11 [6].
This means that for � < n=2 functions formW�;0(Rn) are locally L2-integrable.
First, we prove interpolation Lem. 1 which will be used in the proofs of Th. 2
on smooth solutions to the wave equation and in the main Th. 3.
P r o o f o f L e m m a 1. Let the following propositions be proved.
Proposition 1.
W1;�(R3
+) =
n
f 2 H1(R3
+) : rf 2 (L2(R+ ;H�(R2 )))2 �H�(R+ ;L2(R2))
o
;
if 0 � � � 1, and
W1;�
(R3
+) =
n
f 2 H1(R3
+) : rf 2 (L2(R+ ;H�(R2 )))2 �H�
0 (R+ ;L2(R2))
o
;
if 1=2 < � � 1.
Proposition 2.
[W1;1(R3);H1(R3)][1��] =W1;�(R3); � 2 [0; 1]:
474 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Proposition 3. Let f be a continuous function on R
3
+ . Then we can de�ne
an operator of even extension as
~f(x1; x2; x3) =
(
f(x1; x2; x3); x3 � 0;
f(x1; x2;�x3); x3 < 0:
The operator of even extension ~� can be extended to a continuous operator
~� : G� !W1;�(R3 ); � 2 [0; 1];
where G� is de�ned by (8).
Proposition 4. There exists an operator R such that
R 2 L(H1(R3);H1(R3
+)); R 2 L(W1;1(R3 );W1;1
(R3
+));
R~u = u; 8u 2 H1(R3
+); 8u 2 W1;1
(R3
+);
where ~� is the operator of even extension de�ned in Prop. 3.
In this proof we assume that G� is de�ned by the equalities (8).
First we prove that [W1;1
(R3
+);H1(R3
+)][1��] � G�. Due to Prop. 1 we have
that
r : H1(R3
+)! (L2(R+ ;L2(R2)))2 � L2(R+ ;L2(R2));
r : W1;1
(R3
+)! (L2(R+ ;H1(R2)))2 �H1
0 (R+ ;L2(R2))
is a continuous operator. Thus Ths. 11.6, 11.7 [7, Ch. 1] imply that
r : [W1;1
(R3
+);H1(R3
+)][1��] ! (L2(R+ ;H�(R2 )))2 �H�(R+ ;L2(R2 ));
0 � � < 1=2;
r : [W1;1
(R3
+ );H1(R3
+)][1=2] ! (L2(R+ ;H1=2(R2)))2 �H
1=2
00 (R+ ;L2(R2 ));
� = 1=2;
r : [W1;1
(R3
+);H1(R3
+)][1��] ! (L2(R+ ;H�(R2 )))2 �H�
0 (R+ ;L2(R2 ));
1=2 < � � 1;
is also a continuous operator. Since u 2 [W1;1
(R3
+);H1(R3
+)][1��] implies
u 2 H1(R3
+), using Prop. 1 we obtain the desired embedding.
Now we prove the embedding G� � [W1;1
(R3
+);H1(R3
+)][1��]. Consider u 2 G�.
Using Props. 2 and 3, we have ~u 2 W1;�(R3) = [W1;1(R3 );H1(R3)][1��]. Using
Prop. 4 and interpolation, we obtain R~u = u 2 [W1;1
(R3
+);H1(R3
+)][1��]. The
proof of the lemma will be complete when we prove Props. 1�4.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 475
I.A. Ryzhkova
P r o o f o f P r o p o s i t i o n 1. Using the de�nition of W1;�(R3) via
Fourier transform, we can easily verify that
W1;�(R3) = L2(R;W1;� (R2 ))
\
W1;�(R;L2(R2)):
Thus, we obtain the �rst representation
W1;�(R3
+) = L2(R+ ;W1;�(R2))
\
W1;�(R+ ;L2(R2)); 0 � � � 1;
W1;�
(R3
+) = L2(R+ ;W1;�(R2 ))
\
W1;�
(R+ ;L2(R2)); 1=2 < � � 1:
Using representation (7), we �nish the proof.
P r o o f o f P r o p o s i t i o n 2. Following [7], in the space H1(R3) we
construct the operator A = F�1(1+j�j2)1=2F with D(A) =W1;1(R3). It is easy to
verify that we can de�ne the operator A� = F�1(1+j�j2)�=2F , where F is a Fourier
transform, with D(A�) = W1;�(R3). Obviously, A� is really A to power �. Since
A is a closed maximal accretive operator, D(A�) = [W1;1(R3 );H1(R3)][1��] [8].
P r o o f o f P r o p o s i t i o n 3. Since C1
0 (R
3
+) is dense in H1(R3
+) and
C0(R
3)
T
C1(R3
+)
T
C1(R3
�) is dense in H1(R3), it is easy to prove that the even
extension is a continuous operator from H1(R3
+) in H1(R3 ). Prop. 1 implies that
the even extension of a function f is continuous if and only if the odd extension of
@x3f is continuous. Evidently, the odd extension of @x3f is a continuous operator:
H�(R+ ;L2(R2))! H�(R;L2(R2)); 0 � � < 1=2;
H�
00(R+ ;L2(R2 ))! H�(R;L2(R2 )); � = 1=2;
H�
0 (R+ ;L2(R2))! H�(R;L2(R2)); 1=2 < � � 1;
since the extension by zero is a continuous operator between the above mentioned
spaces (see [7, Ch. 1], Th. 11.4). The proof is complete.
P r o o f o f P r o p o s i t i o n 4. Set (Ru)(x) = 1=2(f(x) + f(�x)). It is
easy to verify that this operator satis�es all assertions of the proposition.
Now we can prove Th. 2.
P r o o f o f T h e o r e m 2. The existence and uniqueness of solution to (1)�
(3) with the initial data (�0; �1) 2 H1(R3
+)�L2(R3
+) is a well-known fact, as well
as the energy conservation law and the decay of local energy E0R(�(t); �t(t)) for
this case. Using formal di�erentiation with respect to t, we can easily prove points
(i), (ii), and (iii) of the theorem for the initial data (�0; �1) 2 W1;1
(R3
+)�H1(R3
+).
Using interpolation Lem. 1, we complete the proof of points (i) and (ii). To prove
(iii) for all � 2 [0; 1] we need the following criterium of the pointwise convergence
of an operator sequence.
476 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Proposition 5. Let H1 be a Banach space and H2 be a pseudonormed space
with a pseudonorm p(�). Assume that fAng1n=1 is a sequence of the linear operators
An : H1 ! H2 such that p(Anh) � CjjhjjH1
8h 2 H1, and there exists a dense
set D � H1 such that 8h 2 D p(Anh) ! 0, when n ! +1. Then 8g 2 H1
p(Ang)! 0, when n! +1.
When H1 and H2 are both Banach spaces, the result is well known. The proof
of Prop. 5 is similar to the proof of this result.
Now we �x R and the set H1 = G� � H�(R3
+), H2 = H1+�(B+
R ) � H�(B+
R )
with the pseudonorm p('0; '1) = (jjr'0jj2�;B+
R
+ jj'1jj2�;B+
R
)1=2, An(�0; �1) =
r
B
+
R
(�(tn); �t(tn)), where (�(tn); �t(tn)) is the solution to (1)�(3) with the ini-
tial conditions (�0; �1) at the moment tn such that tn ! +1 when n ! +1.
The operator of restriction on B+
R r
B
+
R
is continuous from G� to H1+�(B+
R ) (see
[5] and interpolation Lem. 1). We chose D =W1;1
(R3
+)�H1(R3
+). Using the in-
terpolation inequality jjujj[X;Y ][�]
� jjujj1��X jjujj�Y (see, e.g., [7, Ch. 1]), we obtain
that for (�0; �1) 2 D
p(An(�0; �1)) = E�R(�(tn); �t(tn)) = jjr�(tn)jj2�;B+
R
+ jj�t(tn)jj2�;B+
R
� jjr�(tn)jj2�1;B+
R
jjr�(tn)jj2�2�
0;B+
R
+ jj�t(tn)jj2�1;B+
R
jj�t(tn)jj2�2�
0;B+
R
! 0;
when n ! +1. It is easy to see that all assumptions of Prop. 5 are satis�ed.
Thus we prove (iii) for all � 2 [0; 1]. The proof of Th. 2 is complete.
4. Proof of Theorem 4
In this section we need some facts on the Laplace and Fourier transforms.
Denote by D0(R; X) the space of distributions on R with the values in a Hilbert
space X and by S0(R;X) the space of X-valued temperate distributions. We also
set D0
+(X) = ff 2 D0(R; X) : suppf � fx : x � 0gg and D0
+(a;X) = ff 2
D0
+(X) : e�atf(t) 2 S0(R;X)g. For the functions from S(R;X) (test functions
for S0(R; X)) we de�ne the Fourier transform as
F [f(t)](�) =
Z
R
dte�it�f(t) (12)
and for functions from S0(R;X) by (F [f ]; �) = (f; F [�]), respectively. Further we
de�ne the Laplace transform for functions from D0
+(a;X) by
F(s) = L[f(t)](s) = F [e��tf(t)](�); s = �+ i�; � > a: (13)
The Laplace transform of a function from D0
+(a;X) is an analytic in the complex
half-plane C a = fs 2 C : Res > ag X-valued function.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 477
I.A. Ryzhkova
Next we de�ne a space Ha(X). This space consists of the analytic
in C a X-valued functions F(s) that satisfy the following growth condition: for
every � > 0 and �0 > a there exist constants C�(�0) � 0; m = m(�0) � 0 such
that
jjF(s)jjX � C�(�0)e
��Res(1 + jsjm); Res > �0: (14)
The following fundamental theorem takes place (see, e.g., [9]).
Theorem 5. A function f(t) 2 D0
+(a;X) if and only if its Laplace transform
F(s) 2 Ha(X).
Due to de�nition (13) L[f(t)](�+i�) 2 S0(R; X) as a function of the variable �,
provided � to be �xed, so the next representation takes place
L�1[F(� + i�)](t) = e�tF�1
� [F(�+ i�)](t): (15)
Lemma 4. Let B be a linear mapping that maps a function from Ha(X)
to a function from Ha(Y ), where X and Y are Hilbert spaces, and let there exist
a constant
> a such that the operator B
: F(
+i�) 7! (BF)(
+i�) is a linear
bounded operator from L2(R;X) to L2(R; Y ). Then the operator A = L�1 ÆB ÆL
is a linear bounded operator from L2(R+ ;X; e�2
tdt) to L2(R+ ;Y; e�2
tdt).
P r o o f. Since BF 2 Ha(Y ), we can use representation (15) for the inverse
Laplace transform, thus
[Af ](t) = e
tF�1
� [B
Ft[e
�
tf(t)](�)](t):
Using Plansherel's theorem (the Fourier transform is an isometry on L2(R;X)),
one can easily verify the assertion of the lemma.
Corollary 2. If A is a linear bounded operator from L2(R+ ;X; e�2
tdt) to
L2(R+ ;Y; e�2
tdt) then r(0;T )A is a linear bounded operator from L2(R+ ;X) to
L2(0; T ;Y ) and form L1(R+ ;X) to L2(0; T ;Y ), where r(0;T ) is the operator of
restriction of functions from L2(R+ ;Y; e�2
tdt) to (0; T ). The operator norm
jjr(0;T )Ajj � C(T;
).
P r o o f o f T h e o r e m 4.
Remark 3. We can apply the following change of variables to (1)�(3) (and
also to (10)�(11)):
s = t; x1 = y1 + Ut; x2 = y2; x3 = y3:
Then equation (1) changes to @2s� = �y�, the operator (@t + U@x1)
[�] changes
to @s
[�], and initial conditions (3) remain unchanged. Thus, without loss of
generality, we give the proofs of all the theorems above only for the case U = 0.
478 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
To prove part (i) of the theorem we consider the nonhomogeneous problem
(10)�(11) with f 2 L2(R+ ;H�(R3
+)) and U = 0. Similarly as in [3], we use the
Fourier transform in x1; x2 and the Laplace transform in t. We denote^= LtFx1x2 .
Applying the Fourier�Laplace transform to (10)�(11), we obtain
LtFx1x2(@t
[�]) = s�̂(s; �; x3 = 0) = � sp
s2 + j�j2
�
1Z
0
e�y
p
s2+j�j2 f̂(s; �; y)dy:
(16)
Thus
js�̂(s; �; x3 = 0)j
� jsj
j
p
s2 + j�j2j
� (1 + j�j2)��=2q
2Re
p
s2 + j�j2
�
0
@(1 + j�j2)�
1Z
0
dyjf̂(s; �; y)j2
1
A
1=2
:
Now our aim is to �nd an appropriate Sobolev spaceH�(R2 ) such that @t
[�] 2
L2(R+ ;H�(R2); e�2
tdt), provided f 2 L2(R+ ;H�(R3
+); e�2
tdt). That is, we
should �nd � and such that
K(s; �) =
jsj2(1 + j�j2)���
js2 + j�j2jRe
p
s2 + j�j2
� C� (17)
for some �xed � > 0. Here we denote s = �+ i�.
Similarly as in [3], we divide the �rst quarter of the half-plane (�; j�j) into four
domains:
R0 = f�2 + j�j2 � 1g;
R1 = f�=2 � j�j � 2�; �2 + j�j2 � 1g;
R2 = fj�j � 2�; �2 + j�j2 � 1g;
R3 = fj�j � �=2; �2 + j�j2 � 1g:
and estimate K(s; �) in each domain separately. We use the following inequalities
for complex numbers:
p
2jzj1=2 �
p
2Re
p
z =
p
Rez + jzj � jzj1=2; Rez � 0; (18)
p
2jzj1=2 �
p
2Re
p
z =
p
jzj � jRezj = 2��p
jzj �Rez
; Rez < 0: (19)
Further we denote z = s2 + j�j2 = �2 � �2 + j�j2 + 2i�� and D = jzjRepz.
Domain R0. It is easy to see that for every � > 0 (17) takes place without
any restriction on �.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 479
I.A. Ryzhkova
Domain R2. Here Rez � �2 + 3�2 � 0, therefore due to (18)
Re
p
z � Cjzj1=2 = C
�
(�2 � �2 + j�j2)2 + 4�2�2
�1=4 � C�j�j: (20)
Thus
K(s; �) � C
jsj2(1 + j�j2)���
jzj3=2 � C
(�2 + 1=2j�j2)(1 + j�j2)���
j�j3 � C�
for � � 1=2 + �.
Domain R1. In this domain we have jzj � j�j. If Rez � 0, then
Re
p
z � C�jzj1=2 � C�j�j1=2 (21)
and D � C�jzj3=2 � C�j�j3=2. If Rez < 0, then (19) implies
Re
p
z �
p
2��p
jzj �Rez
� C�
j�jp
jzj
� C�j�j1=2 (22)
and D � C�j�j3=2. Thus, in R1
K(s; �) � C�
(�2 + �2)(1 + j�j2=4)���
j�j3=2 � C�
for � � �1=4 + �.
Domain R3. In this domain Rez = �2 � �2 + j�j2 � �2 � 3=4�2. If Rez � 0,
then
Re
p
z � Cjzj1=2 � C�j�j (23)
and D � Cjzj3=2 � Cj�j3. If Rez < 0, (19) implies
2jRe
p
zj2 � 4�2�2
�Rez +
p
(Rez)2 + 4�2�2
� C�: (24)
Since in R3 D � C�j�j2 we have K(s; �) � C�.
Combining the inequalities for R0�R4, we get K(s; �) � C� for � = �1=4+�.
The proof of part (i) of Th. 4 is complete.
To prove (ii) we use a rather di�erent technique. Here and further in this
section we assume f 2 L1(R+ ;H�(R3
+)). Denote Fx1x2 =~. Formally applying
the inverse Laplace transform to (16), we get
@t ~�(t; �; x3 = 0)
= �L�1
"
sp
s2 + j�j2
#
(t; �)
t� L�1
2
4 1Z
0
e�y
p
s2+j�j2 f̂(s; �; y)dy
3
5 (t; �): (25)
480 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
De�ne
[Bf̂ ](s; �) =
1Z
0
e�y
p
s2+j�j2 f̂(s; �; y)dy: (26)
If [Bf̂ ](s; �) 2 Ha(Y ), as soon as f̂ 2 Ha(X) for certain spaces X; Y , the equality
(25) is meaningful due to Th. 5. In fact, we can prove even a stronger assertion
on operator B.
Lemma 5. The operator B has the following properties:
(i) The function [Bf̂ ](s; �) 2 H0(L
2(R2 ; (1+j�j)2�d�)) provided f̂ 2 H0(L
2(R3
+ ;
(1 + j�j)2�d�dy)).
(ii) Denote [B
f̂ ](�; �) = [Bf̂ ](
+ i�; �). There exists
> 0 such that B
:
L2(R;L2(R3
+ ; (1 + j�j)2�d�dy))! L2(R;L2(R2 ; (1 + j�j)2�d�)) is a bounded linear
operator.
P r o o f. First we prove that [Bf̂ ](s; �) is a holomorphic function of the
variable s in the complex half-plane fs : Res > 0g. Due to the Dunford theorem
it is enough to prove that [Bf̂ ](s; �) is weakly holomorphic. Let g(�) 2 L2(R2 ;
(1 + j�j)�2�d�). Then
(Bf̂; g) =
Z
R2
d�g(�)
1Z
0
dy � e�y
p
s2+j�j2 f̂(s; �; y):
Consider the following expression
d
ds
Z
j�j<Q
d�g(�)
RZ
0
dy � e�y
p
s2+j�j2 f̂(s; �; y)
= �
Z
j�j<Q
d�g(�)
RZ
0
dy � ye�y
p
s2+j�j2 sp
s2 + j�j2
f̂(s; �; y)
+
Z
j�j<Q
d�g(�)
RZ
0
dy � e�y
p
s2+j�j2 d
ds
f̂(s; �; y) = A1(s;Q;R) +A2(s;Q;R): (27)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 481
I.A. Ryzhkova
Estimate the �rst term. Using the Schwartz inequality, we get
jA1(s;Q;R)j
� C
Z
j�j<Q
d�g(�)
����� sp
s2 + j�j2
����� (1 + j�j)��
(Re
p
s2 + j�j2)3=2
0
@(1 + j�j)2�
RZ
0
dyjf̂(s; �; y)j2
1
A
1=2
� C
0
B@ Z
j�j<Q
d�(1 + j�j)2�
RZ
0
dyjf̂(s; �; y)j2
1
CA
1=2
�
0
B@ Z
j�j<Q
d�jg(�)j2 � jsj2(1 + j�j)�2�
js2 + j�j2j(Re
p
s2 + j�j2)3
1
CA
1=2
:
If we prove that
K1(s; �) =
jsj2
js2 + j�j2j(Re
p
s2 + j�j2)3
� C(s); Res > a; (28)
this will imply
jA1(s;Q;R)j � C(s)jjgjjL2(R2;(1+j�j)�2�d�)jjf(s)jjL2(R3+;(1+j�j)
2�d�dy); Res > a:
(29)
We use the method of the proof of part (i) of the theorem: denote s = � + i�,
z = s2 + j�j2, D = js2 + j�j2j(Re
p
s2 + j�j2)3, divide the �rst quarter of the
half-plane (�; j�j) into four domains, and use inequalities (18), (19) .
For the domain R0 (28) is evident with a = 0. In the domain R2 D �
C(�)jzj5=2 � C(�)j�j5, therefore K1(s; �) � C(�), � > 0. In the domain R3 (23),
(24) imply D � C(�)jzj2 � C(�)j�j4, hence, K1(s; �) � C(�), � > 0. In the
domain R1 D � C(�)jzj5=2 � C(�)j�j5=2 if Rez � 0 and, provided Rez < 0, (22)
yields D � C(�)j�j5=2, too. Thus, K1(s; �) � C(�), � > 0 for all s; j�j : Res =
� > 0 and (28) together with (29) are proved.
Now we estimate the second term in (27). Using the Schwartz inequality, we
obtain
jA2(s;Q;R)j � C
0
B@ Z
j�j<Q
d�(1 + j�j)2�
RZ
0
dy
���� ddsf̂(s; �; y)
����
2
1
CA
1=2
�
0
B@ Z
j�j<Q
d�jg(�)j2 � (1 + j�j)�2�
Re
p
s2 + j�j2
1
CA
1=2
:
482 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Estimates (20)�(22) imply
K2(s; �) =
1
Re
p
s2 + j�j2
� C(�); � = Res > 0; (30)
and, consequently,
jA2(s;Q;R)j � C(s)jjgjjL2(R2;(1+j�j)�2�d�)
ddsf(s)
L2(R3+;(1+j�j)
2�d�dy)
: (31)
Now, letting Q; R to tend to +1 in (27) and using (29), (31), we prove week
(and, consequently, strong) analyticity of [Bf̂ ](s; �).
It is easy to see that the constants C(s) in (29), (31) grow as jsjm, jsj ! +1
for some m, and therefore the growth condition (14) is ful�lled. Thus, part (i) of
the lemma is proved.
The estimate (30) guarantees that part (ii) of the lemma is also true.
Corollary 3. The operator A = L�1 ÆB Æ L is bounded from L1(R+ ;L2(R3
+ ;
(1 + j�j)2�d�dy)) to L2(0; T ;L2(R2 ; (1 + j�j)2�d�)).
The assertion of the lemma follows directly from Lems. 4, 5 and Cor. 2.
Returning to (25), we note (see, e.g., [10]) that
L�1
� sp
s2 + j�j2
!
= �(t)j�jJ1(j�jt)� Æ(t)J0(0):
Thus @t ~�(t; �; 0) can be represented as a sum [A1
~f ](t; �) + [A2
~f ](t; �), where
[A1
~f ](t; �) = (�(t)j�jJ1(j�jt))
t� [L�1 ÆB Æ L ~f ](t; �)
[A2
~f ](t; �) = (�Æ(t)J0(0))
t� [L�1 Æ B Æ L ~f ](t; �) = �J0(0)[L�1 Æ B Æ L ~f ](t; �):
It is easy to verify that
jjF�1
x1x2
Æ A2 Æ Fx1x2f jjL2(0;T ;H�(R2)) � Cjjf jjL1(R+;H�(R3+)); � 2 [0; 1]: (32)
As for [A1
~f ](t; �), denoting g(t; �) = [L�1 Æ B Æ L ~f ](t; �) and using the Schwartz
inequality one can get
jj[A1
~f ](t; �)jj2L2(R2;(1+j�j)�2�d�)
=
Z
R2
d�
1
(1 + j�j)2�
������
tZ
0
j�jJ1(j�j(t � �))g(�; �)
������
2
�
Z
R2
d�
j�j1��
(1 + j�j)2(�+�)
�
tZ
0
d�
1
(t� �)2�
tZ
0
jj�j�(t� �)�J1(j�j(t� �))j2(1 + j�j)2�jg(�; �)j2;
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 483
I.A. Ryzhkova
for all t < T . The asymptotical properties of the Bessel functions imply that
x�J1(x) is bounded on R+ provided 0 � � � 1=2. The integral
R t
0
(t � �)�2�d�
converges for � < 1=2, and for ��� = (1��)=2 the function j�j1��=(1+ j�j)2(���)
is bounded for all � 2 R
2 . Thus, using Cor. 2, we obtain
jjA1
~f jjL1(0;T ;L2(R2;(1+j�j)�1+�+2�d�)) � C(T; �)jjgjjL2(0;T ;L2(R2;(1+j�j)2�d�))
� C(T; �)jj ~f jjL1(R+;L2(R3+;(1+j�j)
2�d�dy)); � < 1=2:
Hence, we have
jjF�1
x1x2
ÆA1 Æ Fx1x2f jjL1(0;T ;H�1=4+���(R2)) � C(T; �)jjf jjL1(R+;H�(R3+)): (33)
Combining (32) and (33), we obtain part (ii) of the theorem.
5. Proof of Theorem 3
Let us consider the problem (1)�(3). Due to Remark 3 we can prove this
theorem only for the case U = 0, i. e., we consider the problem
@tt� = ��; x 2 R
3
+ ; (34)
@�
@x3
����
x3=0
= 0; �(0) = �0; �t(0) = �1: (35)
We introduce the following operators concerning this problem.
An operator A on L2(R3
+) is de�ned as
A = ��; D(A) =
�
f 2 H2(R3
+) :
@f
@x3
����
x3=0
= 0
�
: (36)
The operator �A generates a s.c. cosine operator C(t) on L2(R3
+) with the
sine operator S(t) =
tR
0
d�C(�) [11]. For the problem (34)�(35) we can write them
down explicitly:
C(t)�0(x) =
1
(2�)3=2
Z
R3
ei�x cos (j�jt)[Fx�̂0](�)d�; (37)
S(t)�0(x) =
1
(2�)3=2
Z
R3
ei�x
sin (j�jt)
j�j [Fx�̂0](�)d�; (38)
where �̂0 is an even extension of �0 from R
3
+ onto R
3 .
484 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
The solution to (34)�(35) can be represented as �(t) = C(t)�0 + S(t)�1. It is
well known that
dC(t)x
dt
= �AS(t)x; x 2 D(A1=2): (39)
The following lemma is also true.
Lemma 6. Denote G�(R3
+) = [W1;1
(R3
+);H1(R3
+)][1��]. Then
C(t) : H�(R3
+)! C(0; T ;H�(R3
+)); (40)
AS(t) : G�(R3
+)! C(0; T ;H�(R3
+)) (41)
are continuous operators.
P r o o f. The continuity of cosine operator is evident. For the sine operator
and �0 smooth enough we have the following:
AS(t)�0 = S(t)A�0 = F�1
� (sin (j�jt)j�j[Fx�̂0](�));
where �̂0 is an even extension of �0 from R
3
+ onto R3 . Assume that �0 2 H1(R3
+).
Proposition 3 implies jj�̂0jjH1(R3) � Cjj�0jjH1(R3+) and therefore jjAS(t)�0jjL2(R3+)
� Cjj�0jjH1(R3+) for all t > 0. If �0 2 W1;1
(R3
+), then in the same way we obtain
that jjAS(t)�0jjH1(R3+) � Cjj�0jj
W
1;1
(R3+)
for all t > 0. Continuity with respect to
t can be proved in the standard way. Finally, we use the interpolation Lem. 1 and
prove the second assertion of the present lemma.
We will need the Neumann map N connected with the operator A. Since
Ker(A) = f0g (due to (36) it can be easily veri�ed), N is de�ned as follows:
u = Nh ,
�
�Au = 0 in R
3
+ ;
@u
@x3
����
x3=0
= h
�
:
The following fact also holds true [12]: N�A�y =
[y], where
[y] denotes the
Sobolev trace of function y from R
3
+ onto R
2 and A� is the L2(R3
+ )-adjoint of A.
Now we are ready to prove Th. 3. The idea of the proof was also borrowed
from [3]. It was already mentioned that a solution to (34)�(35) can be represented
as �(t) = C(t)�0 + S(t)�1. Then due to (39)
@t
[�](t) = N�A�AS(t)�0 +N�A�C(t)�1: (42)
To analyze each term in (42), we consider the problem
@tt� = ��+ f(x; t); x 2 R
3
+ ; (43)
@�
@x3
����
x3=0
= 0; �(T ) = �t(T ) = 0 (44)
with a certain function f .
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 485
I.A. Ryzhkova
Step 1. Term N�A�C(t)�1. Let in (43)�(44) f(t; x) = C(t)�1. Due to
Lemma 6 f 2 L1(R+ ;H�(R3
+)) provided �1 2 H�(R3
+). Then for this problem
@t
[�](t) = N�A�
TZ
t
C(� � t)C(�)�1:
Using the formula C(t1) � C(t2) = 1=2 (C (t1 + t2) + C (t1 � t2)), we obtain that
@t
[�](t) =
1
2
N�A�C(t)�1(T � t) +
1
4
N�A�(S(2T � t)� S(t))�1: (45)
Theorem 4 (i) and (40) imply
jj@t
[�](t)jjL2(0;T ;H�1=4+�(R2)) � CT jjC(t)�1jjL1(R+;H�(R3+)) � CT jj�1jjH�(R3+):
(46)
Lemma 2 yields to the following estimate of the second term in (45):
jjN�A�(S(2T � t)� S(t))�1jjC(0;T ;W1=2;�(R2)) � Cjj�1jjH�(R3+); (47)
where C does not depend on T . Estimates (46) and (47) together with Lem. 3
give us that N�A�C(t)�1 2 L2(0; T � 1;H
�1=4��
loc (R2)) and
jjN�A�C(t)�1jjL2(0;T�1;H�1=4��(B)) � C(T;B)jj�1jjH�(R3+) (48)
for any bounded set B � R
2 . Form the other hand, using Th. 4 (ii) and (47), we
get
N�A�C(t)�1 = A1 +A2; (49)
where A1 2 L1(0; T ;H
�1=4+���
loc (R2 )), A2 2 L2(0; T ;H�
loc(R
2 )), and
jjA1jjL1(0;T ;H�1=4+���(B)) � C(T;B; �)jj�1jjH�(R3+); (50)
jjA2jjL2(0;T ;H�(B)) � C(T;B)jj�1jjH�(R3+) (51)
for any bounded set B � R
2 .
Step 2. Term N�A�AS(t)�0. Let in (43)�(44) f(t; x) = AS(t)�0. Due to
Lemma 6 f 2 L1(R+ ;H�(R3
+)) provided �0 2 G�(R3
+). Then for this problem
@t
[�](t) = N�A�
TZ
t
C(� � t)AS(�)�0:
Using the formula S(t1) � C(t2) = 1=2 (S (t1 + t2)� S (t1 � t2)), we obtain
@t
[�](t) =
1
2
N�A�AS(t)�0(T � t) +
1
4
N�A�(C(2T � t)� C(t))�0:
486 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
On Trace Regularity of Solutions to a Wave Equation...
Theorem 4 yields
jj@t
[�](t)jjL2(0;T ;H�1=4+�(R2)) � CT jjAS(t)�0jjL1(R+;H�(R3+)) � CT jj�0jjG�(R3+):
(52)
Similarly as in Step 1,
jjN�A�(C(2T � t)� C(t))�0jjC(0;T ;W1=2;�(R2)) � Cjj�0jjG�(R3+); (53)
where C does not depend on T , and, �nally, (52), (53), and Lem. 3 imply
N�A�AS(t)�0 2 L2(0; T � 1;H
�1=4+�
loc (R2 )) and
jjN�A�AS(t)�0jjL2(0;T�1;H�1=4+�(B)) � C(T;B)jj�0jjG�(R3+) (54)
for any bounded set B � R
2 . From the other hand, Th. 4 and (53) give us
N�A�AS(t)�0 = B1 +B2; (55)
where B1 2 L1(0; T ;H
�1=4+���
loc (R2)), B2 2 L2(0; T ;H�
loc(R
2 )), and
jjB1jjL1(0;T ;H�1=4+���(B)) � C(T;B; �)jj�0jjG�(R3+); (56)
jjB2jjL2(0;T ;H�(B)) � C(T;B)jj�0jjG�(R3+) (57)
for any bounded set B � R
2 .
Since T was chosen arbitrary, (48) and (54) give us assertion (i) of the theorem,
and (49)�(51), (55)�(57) give us assertion (ii) of the same theorem.
Remark 4. The norm in G1=2 is not equivalent to jjr � jj1=2;R3+ (see [7, Ch.
1], Th. 11.7), therefore the case � = 1=2 is excluded from Th. 3 and Cor. 1.
6. Proof of Corollary 1
The representation of a solution to (34)�(35) given by Th. 3.3 [13] implies,
that the solution (�; �t)(t0) in a half-ball of radius R depends only on the initial
data values in the ball of radius R1(R; t0). Moreover, r�(t0) depends only on
r�0 in this ball. Chose new initial data. For �0 the Poincar�e inequality holds
jj�0jj21;B+
R1
� C
0
BB@jjr�0jj20;B+
R1
+
��������
Z
B
+
R1
�0(x)dx
��������
21
CCA :
We set ~�0 = �0 � C1, where C1 = 1=mes(B+
R1
)
R
B
+
R1
�0(x)dx. Thus, for ~�0 we
have jj~�0jj21;B+
R1
� Cjjr~�0jj20;B+
R1
and, consequently, jj~�0jj21+�;B+
R1
� Cjjr~�0jj2�;B+
R1
,
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 487
I.A. Ryzhkova
� 2 [0; 1]. There exists an operator of �nitary extension L� from H�(B+
R1
) to
H�(R3
+), � 2 [0; 2] such that suppLu � B+
2R1
. Since for � > 3=2 we need boundary
conditions (2) to be satis�ed by the initial datum �0, we apply the operator R
de�ned in Prop. 4, to L� ~�0. We keep the same notation for the obtained operator.
Hence, we have
jjrL~�0jj�;B+
2R1
� CjjL~�0jj�+1;R3+
� Cjj~�0jj�+1;B+
R1
� Cjjr~�0jj�;B+
R1
= Cjjr�0jj�;B+
R1
;
for � 2 [0; 1]. We de�ne new initial data as follows:
�(0) = L1+�
~�0; �t(0) = L��1; � 2 [0; 1]; � 6= 1=2:
Denote by (��; ��t)(t) the solution to (34)�(35) with this new initial data. Obvi-
ously, (��; ��t)(t0) = (�; �t)(t0) in B+
R . Thus, Th. 3 implies Cor. 1.
References
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Flow. � Z. Angew. Math. Phys. 58 (2007), 246�261.
[3] I. Lasiecka and R. Triggiani, Trace Regularity of the Solution of the Wave Equation
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[6] L. H�ormander, Estimates for translation invariant operators in Lp spaces. � Acta
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[7] J.-L. Lions and E. Magenes, Probl�emes aux Limites Non Homog�enes et Applica-
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488 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
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[10] Tables of integral trasforms. (A. Erdelyi, Ed.). McGraw-Hill Book Company, Inc.,
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|
| id | nasplib_isofts_kiev_ua-123456789-7619 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1812-9471 |
| language | English |
| last_indexed | 2025-12-07T17:55:35Z |
| publishDate | 2007 |
| publisher | Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| record_format | dspace |
| spelling | Ryzhkova, I.A. 2010-04-06T09:39:26Z 2010-04-06T09:39:26Z 2007 On trace regularity of solutions to a wave equation with homogeneous Neumann boundary conditions / I.A. Ryzhkova // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 468-489. — Бібліогр.: 13 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/7619 We prove an additional regularity of time derivative of the trace of solution to the wave equation on the 3D half space with the homogeneous Neumann boundary conditions. Доведено додаткову регулярність похідної за часом від сліду розв'язку хвильового рівняння у тривимірному півпросторі з однорідними крайовими умовами Неймана. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions Про крайову регулярність розв'язків хвильового рівняння з однорідними крайовими умовами Неймана Article published earlier |
| spellingShingle | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions Ryzhkova, I.A. |
| title | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions |
| title_alt | Про крайову регулярність розв'язків хвильового рівняння з однорідними крайовими умовами Неймана |
| title_full | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions |
| title_fullStr | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions |
| title_full_unstemmed | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions |
| title_short | On Trace Regularity of Solutions to a Wave Equation with Homogeneous Neumann Boundary Conditions |
| title_sort | on trace regularity of solutions to a wave equation with homogeneous neumann boundary conditions |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/7619 |
| work_keys_str_mv | AT ryzhkovaia ontraceregularityofsolutionstoawaveequationwithhomogeneousneumannboundaryconditions AT ryzhkovaia prokraiovuregulârnístʹrozvâzkívhvilʹovogorívnânnâzodnorídnimikraiovimiumovamineimana |