Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations
The local existence and the uniqueness of some multidimensional inverse problems for the second-order hyperbolic integro-di erential equations in the class of functions having certain smoothness on time variable and analyticity on a part of spatial variables are proven.
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2007
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| Цитувати: | Some multidimensional inverse problems of memory determination in hyperbolic equations / D.K. Durdiev // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 411-423. — Бібліогр.: 5 назв. — англ. |
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irk-123456789-76162010-04-07T12:01:05Z Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations Durdiev, D.K. The local existence and the uniqueness of some multidimensional inverse problems for the second-order hyperbolic integro-di erential equations in the class of functions having certain smoothness on time variable and analyticity on a part of spatial variables are proven. Доведено теореми локального існування і єдності у цілому деяких багатовимірних обернених задач відтворення пам'яті для інтегро-диференціальних рівнянь гіперболічного типу другого порядку у класі функцій, що мають скінченну гладкість відносно часової змінної та аналітичних за частиною просторових змінних. 2007 Article Some multidimensional inverse problems of memory determination in hyperbolic equations / D.K. Durdiev // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 411-423. — Бібліогр.: 5 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/7616 en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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The local existence and the uniqueness of some multidimensional inverse problems for the second-order hyperbolic integro-di erential equations in the class of functions having certain smoothness on time variable and analyticity on a part of spatial variables are proven. |
| format |
Article |
| author |
Durdiev, D.K. |
| spellingShingle |
Durdiev, D.K. Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| author_facet |
Durdiev, D.K. |
| author_sort |
Durdiev, D.K. |
| title |
Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| title_short |
Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| title_full |
Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| title_fullStr |
Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| title_full_unstemmed |
Some Multidimensional Inverse Problems of Memory Determination in Hyperbolic Equations |
| title_sort |
some multidimensional inverse problems of memory determination in hyperbolic equations |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2007 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/7616 |
| citation_txt |
Some multidimensional inverse problems of memory determination in hyperbolic equations / D.K. Durdiev // Журн. мат. физики, анализа, геометрии. — 2007. — Т. 3, № 4. — С. 411-423. — Бібліогр.: 5 назв. — англ. |
| work_keys_str_mv |
AT durdievdk somemultidimensionalinverseproblemsofmemorydeterminationinhyperbolicequations |
| first_indexed |
2025-07-02T10:25:50Z |
| last_indexed |
2025-07-02T10:25:50Z |
| _version_ |
1836530480505159680 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2007, vol. 3, No. 4, pp. 411�423
Some Multidimensional Inverse Problems of Memory
Determination in Hyperbolic Equations
D.K. Durdiev
Bukhara State University
11 M. Ikbol Str., Bukhara, 705018, Uzbekistan
E-mail:durdiev65@mail.ru
Received March 21, 2007
The local existence and the uniqueness of some multidimensional inverse
problems for the second-order hyperbolic integro-di�erential equations in the
class of functions having certain smoothness on time variable and analyticity
on a part of spatial variables are proven.
Key words: inverse problem, integro-di�erential equation, hyperbolic
equation, agreement condition, uniqueness.
Mathematics Subject Classi�cation 2000: 35L10.
In this paper the local existence and the uniqueness of some multidimensional
inverse problems for the second-order hyperbolic integro-di�erential equations in
the class of functions having certain smoothness on time variable and analyticity
on a part of spatial variables are proven. Unlike in paper [1], where the memory
is multiplied by the solution, we consider the equations in which the memory
is multiplied by the second derivative of time solution or the �rst-order di�er-
ential operator with analytical coe�cients. Problems of this type often arise in
applications. A distinctive feature of problems of memory determination is the
dependence of unknown function both on time and on spatial variables (a multi-
dimensional problem). Among the problems of �nding memory in hyperbolic
integro-di�erential equations there should be mentioned papers [2, 3], where the
problem with the sources distributed over the whole region is considered. In [4]
the questions of uniqueness of memory determination in the wave equation on the
measurement of di�used wave at the location of the point source are studied.
1. We consider an initial-boundary problem for the wave equation with
memory
utt � uzz �4u =
tZ
0
k(x; �)utt(x; z; t� �)d�; (x; t) 2 Rn+1
; z 2 R+; (1:1)
c
D.K. Durdiev, 2007
D.K. Durdiev
ujt<0 � 0; uz jz=0 = �Æ
0(t) + g(x; t)�(t); (x; t) 2 Rn+1
: (1:2)
Here 4 is the Laplace operator in variables (x1; : : : ; xn) := x; Æ0(t) is a deriva-
tive of the Dirac delta-function; �(t) is the Heavyside function, R+ := fz 2 Rjz >
0g; g is a given smooth function. For the given function k(x; t) �nding of the func-
tion u(x; t) satisfying equations (1,1),(1,2) is a well-posed problem in the space
of generalized functions. We formulate the inverse problem: to determine k(x; t)
via the trace of solution of problem (1,1), (1,2) on the hyperplane z = 0 for all
x 2 R
n, t < T , T > 0, i.e.
ujz=0 = F (x; t); x 2 Rn
; t < T: (1:3)
In equation (1.1) by integrating by parts we "relocate" time derivative and
introducing a new function v according to the formula
u = �(x; t)v(x; z; t); �(x; t) := exp[k0(x)t=2]; k0(x) := k(x; 0)
from equalities (1.1), (1.2), we have
vtt � vzz = 4v + trk0rv +H(x; t)v +
tZ
0
h(x; t� �)v(x; z; �)d�;
(x; t) 2 Rn+1
; z 2 R+; (1:4)
vjt<0 � 0; vz jz=0 = �Æ
0(t)� [k0(x)=2]Æ(t) +G(x; t)�(t); (x; t) 2 Rn+1
; (1:5)
where the equation exp[�k0(x)t=2]Æ
0(t) = Æ
0(t) + [k0(x)=2]Æ(t) is used as well as
the notations are introduced
H(x; t) := k0t(x) + k
2
0(x)=4 + t4k0(x)=2 + (t2=2)
nX
i=1
k
2
0xi
(x);
h(x; t) := exp[�k0(x)t=2]ktt(x; t); (1:6)
G(x; t) = exp[�k0(x)t=2]g(x; t):
From the theory of hyperbolic equations we conclude that v � 0, t < z, x 2 Rn,
z 2 R+. We represent the solution of (1.4), (1.5) as
v(x; z; t) = Æ(t� z) + ~v(x; z; t)�(t� z):
Using the method of separation of variables, it is not di�cult to �nd
v(x; z; z + 0) = (1=2)
�
k0(x) +
zZ
0
H(x; �)d�
�
=: �(x; z):
412 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
That is why the function F (x; t) in (1.3) should be represented as
F (x; t) = Æ(t) + f(x; t)�(t); (x; t) 2 Rn+1
: (1:7)
It is clear that ~v = v when t > z: For the regular part of function v(x; z; t) in the
region t > z; x 2 R
n, the inverse problem (1.4), (1.5), (1.3) is equivalent to the
problem
vtt � vzz = 4v + trk0rv +H(x; t)v +
tZ
0
h(x; t� �)v(x; z; �)d�; (1:8)
vjz=0 = f(x; t); vzjz=0 = G(x; t); (1:9)
vjt=z+0 = �(x; z): (1:10)
In equations (1.8)�(1.10) we replace the variables z; t by z1; t1 by the formulas
z1 = t+ z; t1 = t� z:
Then v(x; z; t) = v(x; (z1 � t1)=2; (z1 + t1)=2) := v1(x; z1; t1): The problem (1.8)�
(1.10) in new variables is rewritten as the problem of �nding the functions v; k
from the equations
@
2
v1(x; z1; t1)
@t1@z1
= �
1
4
�
4v1 + (t1 + z1)rk0rv1=2 +H(x; (t1 + z1)=2)v1 + h(x; t1)
+
t1Z
0
h(x; �)v1(x; z1 � �; t1 � �)d�
�
;
(x; z1; t1) 2 f(x; z1; t1) j x 2 R
n
; 0 � t1 � z1g := D; (1:11)
v1 jt1=z1= f(x; z1);
@
@z1
v1 jt1=z1=
1
2
ft(x; t) jt=z1 +
1
2
G(x; z1); z1 2 R+; x 2 R
n
;
(1:12)
v1 jt1=0= �[x; (1=2)z1 ]; z1 2 R+; x 2 R
n
: (1:13)
We recall that h is related to k according to the second formula of (1.6).
We introduce the function
w(x; z1; t1) =
@
@z1
v1(x; z1; t1); t1 < z1: (1:14)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 413
D.K. Durdiev
Demanding the continuity of functions v(x; z; t); w(x; z; t); when z1 = t1 = 0; x 2
R
n
; from (1.12), (1.13) it is not di�cult to express k0(x); k0t(x) by the known
functions:
k0(x) = 2f(x; 0); k0t(x) = 2ft(x; t)jt=0 � f
2(x; 0) + 2g(x; 0):
Further we will assume that in the equalities for H(x; z); G(x; t); �(x; z) instead
of functions k0(x); k0t(x) their expressions by means of the latter equations are
used. For simplicity we will omit index 1 in z1; t1; v1:
Following [5, p. 92], we introduce the Banach space As(s > 0; r > 0) of
analytical functions �(x); x 2 Rn
; with the norm
k�ks := sup
jxj�r
1X
j�j=0
s
j�j
�!
�����
@
j�j
�(x)
@x1
�1 : : : @xn
�n
�����; � := (�1; : : : ; �n);
j�j := �1 + : : :+ �n; �! := (�1)! : : : (�)!:
The following properties are obvious: if �(x) 2 As, then �(x) 2 As0 for all s0 2
(0; s), therefore, As � As0 if s0 < s. Besides, if �(x) 2 As, then
@j�j�(x)=@x1�1 : : :
@xn
�n
ks0 � C�jj�(x)jjs=(s� s
0)j�j, where the constant C� depends only on �. We
denote by Ci
z (As; G) a class of functions with values in As which are continuously
di�erentiable i-times in z and continuous in t in the region G. For �xed (z; t) the
norm of the function !(x; z; t) in As0 we denote by k!ks0(z; t). The norm of the
function ! in C (As; G) is de�ned by the equality
k!kC(As;G) = sup
(z;t)2G
k!ks0(z; t):
Theorem 1. Let f(x; 0), ft(x; t)jt=0, fxi(x; 0), i = 1; : : : ; n, 4f(x; 0), g(x; 0)
belong to As0, so > 0, and f(x; t), ft(x; t), ftt(x; t), g(x; t), gt(x; t) belong to
C(As0 ; [0; T ]), and max[kf(x; t)ks0(t); kw0(x; z)ks0(z); kh0(x; z)ks0 (z); kk0(x)ks0 ]
= R, for (z; t) 2 GT := f(z; t)j0 � t � z � Tg: Then for any � > 0 we can �nd
the number a = a(s0; T;R; n), as0 < T so that for any s 2 (0; s0) there exists
a unique solution of the problem (1.11)�(1.13) v(x; z; t) 2 C1
z (As0 ;Ds) ; k(x; t) 2
C
2
t (As0 ; [0; a(s0 � s)]) ; where Ds is the region on the plane z; t : Ds := f(z; t)j0 �
t � z < a(s0 � s)g; and solution satis�es the following inequalities:
kv � v0ks(z; t) � �; kk � k
0
ks(z) �
2�
(s0 � s)
; (1:15)
where
h0(x; z) := 2f(x; 0) exp[�f(x; 0)z]g(x; z) � 2 exp[�f(x; 0)z]gt(x; z)jt=z
414 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
+(1=2)4f(x; 0) �H(x; z)f(x; z) �4f(x; z)� 2zrf(x; 0)rf(x; z)
�2ftt(x; t)jt=z + (z=2)
nX
i=1
f
2
xi
(x; 0);
v0 := f(x; z); k0 := k0(x) + zk0t; (z; t) 2 Ds:
P r o o f. In the beginning the problem (1.11)�(1.13) is reduced to the closed
system of the Volterra type integro-di�erential equations in the area (x; z; t) 2 D.
The equation for v is rewritten with the help of equation (1.14) while the equation
for k is rewritten with the help of the second equation of (1.6):
v(x; z; t) = v0(x; z) +
zZ
t
w(x; �; t)d�; (1:16)
k(x; z) = k
0(x) +
zZ
0
(z � �) exp [f(x; 0)�] h(x; �)d�: (1:17)
For �xed x 2 R
n by integrating equality (1.11) on the plane (�; �) along the
line � = z from the point (t; z) to the point (z; z) and using the second condition
in (1.12), we can get the equation for w(x; z; t):
w(x; z; t) = w0(x; z) +
1
4
zZ
t
[4v(x; z; �) + (z + �)rf(x; 0)rv(x; z; �)
+H[x; (z + �)=2]v(x; z; �) + h(x; �) +
�Z
0
h(x; �)v(x; z � �; � � �)d�]d�; (1:18)
where w0(x; z) = (1=2)ft(x; t)jt=z + (1=2)G(x; z):
Using equation (1.13) di�erentiated by z, when t = 0, from (1.18), we �nd
zZ
0
[4v(x; z; �) + (z + �)rf(x; 0)rv(x; z; �)
+H[x; (z + �)=2]v(x; z; �) + h(x; �) +
�Z
0
h(x; �)v(x; z � �; � � �)d�]d� = 2ft(x; 0)
+2g(x; 0)+(z=2)4f(x; 0)+(z2=4)
nX
i=1
f
2
xi
(x; 0)�2ft(x; z)�2 exp[�f(x; 0)z]g(x; z):
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 415
D.K. Durdiev
The equation for h(x; t) can be easily found by di�erentiation of the latter equation
on z and by using the �rst condition from (1.12):
h(x; t) = h0(x; t)�
zZ
0
[4w(x; z; �) + (z + �)rf(x; 0)rw(x; z; �)
+�rf(x; 0)rv(x; z; �) +Hz[x; (z + �)=2]v(x; z; �) +H[x; (z + �)=2]w(x; z; �)
+h(x; �)f(x; z � �) +
�Z
0
h(x; �)w(x; z � �; � � �)d�]d�; (1:19)
where the function h0(x; t) is determined in Th. 1.
The equations (1.16)�(1.19) represent a closed system of integro-di�erential
equations for the functions v, k, w, h in the area DT = GT �R
n.
For convenience we introduce the function vector
�(x; z; t) = (�1; �2; �3; �4) := (v; k; w; h):
We rewrite the system (1.16)�(1.19) as an operator equation
� =M�; (1:20)
where M = (M1; : : : ;M4) is de�ned by the right sides of equations (1.16)�(1.19),
and
�
0 =M(0); �0 = (�01; �
0
2; �
0
3�
0
4) := (v0; k
0
; w0; h0):
We de�ne the iterations for equation (1.20):
�
i+1 =M�
i
; i = 0; 1; 2; : : : ; �i = (�i1; : : : ; �
i
4) (1:21)
and
�
i+1
� �
i :=
i
; i = 0; 1; 2; : : : ; i = ( i
1; : : : ;
i
4):
Let the sequence of numbers a0; a1; : : : ; ai; : : : be determined by the expressions
ai+1 = ai=(1 + (i + 1)�2), i = 0; 1; 2; : : : . Here a0 is a �xed positive number.
The number a0 < T=s0 will be chosen later. With the numerical sequence a�
we link the sequence of enclosed �elds Fi = f(z; t; s)j0 < s < s0; 0 � t � z <
ai(s0 � s)g:
The following lemma is valid.
Lemma. If the conditions of Th. 1 for any �xed � > 0 and any i = 0; 1; 2; : : :
are ful�lled, then there exist a0 2 (0; T=s0), a0 = a0(R; s0; �; n) and
�i = �i(R; s0; �; n) > 0; such that for each s 2 (0; s0)
�
i
1;
i
3
�
2 C(As;Dsi),
416 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
�
i
2;
i
4
�
2 C (As; [0; ai(s0 � s)]), Dsi := f(z; t)j0 � t � z < ai(s0 � s)g, and the
following inequalities are valid:
k
i
1ks(z; t) �
�iz
ai(s0 � s)� z
; k
�
j ks(z; t) �
�iaiz
[ai(s0 � s)� z]2
; j = 2; 3;
k
i
4ks(z) �
�ia
2
i z
[ai(s0 � s)� z]3
; (z; t; s) 2 Fi; (1:22)
k�
i+1
1 � �
0
1ks(z; t) � �; k�
i+1
j � �
0
jks(z; t) �
2�
s0 � s
j = 2; 3;
k�
i+1
4 � �
0
4ks(z) �
4�
(s0 � s)2
; (z; t; s) 2 Fi+1: (1:23)
We use the Nirenberg method and its modi�cation, developed in [5] to prove
the lemma. Using estimate [5, p. 92]
@
j�j
�(x)
@x1
�1 :::@xn
�n
s0
� C�
k�(x)ks
(s� s0)j�j
; (1:24)
it is not di�cult to check that inequalities (1.22), (1.23) are satis�ed when i = 0,
besides �0 is proportional to a0. We determine the validity of these inequalities
for any i using the induction method. We assume that if the statement of the
lemma is valid for i � � and prove that, then it is valid for i = � + 1, as well.
Using the inductive assumption, we �nd that �+1
2 C(As;Ds(�+1)). Besides,
from (1.20) we �nd
k
�+1
1 ks(z; t) �
zZ
t
k
�
3 ks(�; t)d�
�
zZ
t
��a��d�
[a�(s0 � s)� �]2
�
��a�z
[a�(s0 � s)� z]
� a0
��z
a�+1(s0 � s)� z
;
k
�+1
2 ks(z) �
zZ
0
k
�
4 ks(�)d� � a0
��a�+1z
[a�+1(s0 � s)� z]2
;
k
�+1
3 ks(z; t) �
1
4
zZ
t
fk4
�
1 ks(z; �) + 2TR
nX
j=1
k
@
@xj
�
1 ks(z; �)
+Rk �
1 ks(z; �) + k
�
4 ks(�) +
�Z
0
[k �
4 ks(�)k�
�+1
1 js(z � �; � � �)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 417
D.K. Durdiev
+k��4ks(�)k
�
1 ks(z � �; � � �)]d�gd�:
Using inequality (1.24) to estimate k4 �
1 ks; k
@
@xj
�
1 ks; j = 1; 2; : : : ; n; and
assuming that s0 = s
0(�) = (s+ s0 � �=an)=2; we get
k
�+1
3 ks(z; t) �
1
4
zZ
t
f
16na2�
[a�(s0 � s)� �]2
�
2��z
a�(s0 � s)� z
+
8TRna�
a�(s0 � s)� �
�
2��z
a�(s0 � s)� z
+
2R��z
a�(s0 � s)� z
+
��a
2
��
[a�(s0 � s)� �]3
+
�Z
0
[
��a
2
��(�+R)
(a�(s0 � s)� �)3
+
4�+Rs
2
0
(s0 � s)2
�
2��z
a�(s0 � s)� z
]d�gd�
� c1a0
a�+1��z
[a�+1(s0 � s)� z]2
;
where
c1 = c1(s0; n; a0; �;R; T ):
In the latter inequality it is used that a2� � a�+1a0 are valid for any � � 1.
Moreover, a3� � a
2
�+1a0 for � � 1. These inequalities can be easily checked by the
method of mathematical induction. The latter inequalities are used to estimate
the function �+1
4 . Similarly, for �+1
4 we obtain
k
�+1
4 ks(z) � c2a0
a
2
�+1��z
[a�+1(s0 � s)� z]3
;
where
c2 = c2(s0; n; a0; �;R; T ):
It is obvious that ci; i = 1; 2, in the latter estimates are monotonically nonde-
creasing functions of the s0; a0 parameters. From the estimations made above it
follows that (1.22) is valid when i = � + 1; if we assume that
��+1 = a0c��;
where
c = max(1; s0; c1; c2): (1:25)
Now we show that inequalities (1.23) are also satis�ed when i = � + 1, if the
number a0 is chosen properly. For (x; z; t) 2 F�+2
k�
�+2
1 � �
0
1ks(z; t) �
�+1X
j=0
k
j
1ks(z; t) �
�+1X
j=0
�jz
aj(s0 � s)� z
418 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
� �0
�+1X
j=0
(a0c)
j
�
aj
aj+1
� 1
�
� �0
�+1X
j=0
(a0c)
j(j + 1)2;
k�
�+2
� �
0
ks(z; t) �
�+1X
j=0
k
j
ks(z; t) �
2�0
s0 � s
�+1X
j=0
(a0c)
j(j + 1)4;
= 2; 3;
k�
�+2
4 � �
0
4ks(z) �
�+1X
j=0
k
j
4ks(z) �
4�0
(s0 � s)2
�+1X
j=0
(a0c)
j(j + 1)6:
That is why inequalities (1.23) are satis�ed when i = � + 1; if the number a0 is
chosen so that
a0c < 1; �0
1X
j=0
(a0c)
j(j + 1)6 � �: (1:26)
It is clear that a0 can be always chosen to be so small that inequalities (1.26) are
satis�ed. Thus the validity of the lemma is proven. Further we assume that the
number a0 is chosen from the conditions (1.26).
In order to complete the proof of Th. 1, we should note that under the
chosen value of a0 the �i sequence uniformly converges in the norm of space
C(As;Ds); a = limai; the limiting function belongs to C(As;Ds) and provides the
solution of the operator equation (1.20). The limit transition in the �rst two in-
equalities (1.23), when (x; z; t) 2 F = f(z; t; s)j0 < s < s0; 0 � t � z < a(s0 � s)g
leads to the values coinciding with (1.15). The uniqueness of the constructed so-
lution is established by using the described above technics based on the standard
method [5, p. 103].
We indicate a few settings of inverse problems of memory determination which
could be studied by the above methods.
2. Find the functions u(x; z; t); k(x; t) that satisfy the following equalities:
utt � uzz �4u =
tZ
0
k(x; �)ut(x; z; t � �)d�; (x; t) 2 Rn+1
; z 2 R+; (2:1)
ujt<0 � 0; uz jz=0 = �g(x)Æ0(t); (x; t) 2 Rn+1
; (2:2)
ujz=0 = g(x)Æ(t) + f(x; t)�(t); (x; t) 2 Rn+1
; (2:3)
where g(x); f(x; t) are given smooth functions.
The solution of problem (2.1),(2.2) is represented as
u(x; z; t) = g(x)Æ(t � z) + ~u(x; z; t)�(t� z): (2:4)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 419
D.K. Durdiev
We introduce the function ~u(x; (z � t)=2; (z + t)=2) := v(x; z; t): As in Part 1,
the problem (2.1)�(2.3) can be replaced by the equivalent problem of �nding the
functions v(x; z; t); k(x; t) in the region z > t > 0; x 2 Rn from the equations
@
2
v(x; z; t)
@t@z
= �
1
4
�
4v + k0(x)v � g(x)h(x; t) �
tZ
0
h(x; �)v(x; z � �; t� �)d�
�
;
(x; z; t) 2 f(x; z; t) j x 2 Rn
; 0 � t � zg := D; (2:5)
v jt=z= f(x; z);
@
@z
v jt=z=
1
2
ft(x; t)jt=z ; z 2 R+; x 2 R
n
; (2:6)
v jt=0=
1
2
k0(x)z; z 2 R+; x 2 R
n
; (2:7)
in which k0 := k(x; 0) = 1=g(x)[ft(x; t)jt=0 +4g(x)]; h(x; t) := kt(x; t); t > 0:
With respect to problem (2.5)�(2.7) the analogue of Th. 1 is valid.
Theorem 2. Let
�
g(x); 1=g(x); (@=@z)f(x; z)jz=0 ; (@
3
=@
2
xi@z)f(x; z)jz=0
2 As0, s0 > 0, i = 0; 1; : : : ; n;
�
f(x; z); (@=@z)f(x; z); (@2=@2z)f(x; z)
2
C (As0 ; [0; T ]), when some �xed s0 > 0, T > 0, whereas
max [kg(x)ks0 ; kf(x; z)ks0 ; kfz(x; z)ks0 ; 1=kg(x)ks0 ; k4g(x)ks0 ; kfzz(x; z)ks0 ; ] = R;
for t 2 [0; T ] and the consistency condition is satis�ed:
fz(x; z)jz=0 +4g(x) = g(x)fz(x; z)jz=0:
Then for any � > 0 the number a 2 (0; T=s0) can be found such that for
any s 2 (0; s0) there exists the unique solution of problem (2.5)�(2.7) v(x; z; t) 2
C
1
z (As0 ; Ds) ; k(x; t) 2 C
1
t (As0 ; [0; a(s0 � s)]), where the Ds �eld is determined in
Th. 1, and for the solution inequalities (1.15) are valid with the functions
v0 = f(x; z); k0(x) =
1
g(x)
[ft(x; t)jt=0 +4g(x)]:
3. Find the functions u(x; z; t); k(x; t) that satisfy the following equalities:
utt � uzz � Lu =
tZ
0
k(x; �)L0u(x; z; t� �)d�; (x; t) 2 Rn+1
; z 2 R+; (3:1)
ujt<0 � 0; uzjz=0 = �g(x)Æ0(t); (x; t) 2 Rn+1
; (3:2)
420 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
ujz=0 = g(x)Æ(t) + f(x; t)�(t); (x; t) 2 Rn+1
: (3:3)
Here
L =
nX
i=1;j=1
aij(x)
@
2
@xi@xj
+ L0; L0 =
nX
i=1
bi(x)
@
@xi
+ c(x);
aij , bi, c, g, f (1 � i; j � n) are given smooth functions.
Problem (3.1)�(3.3), when aij = Æij , bi = 0, where Æij is the Kronecker sym-
bol, was studied in paper [1]. In the case of inequality (1.24) let us estimate
the di�erential expressions Lu;L0u with analytical coe�cients. We assume that
functions aij, bi, c(x), 1 � i, j � n, are the elements of As0 , s0 > 0, and
d := max
1�i;j�n
kaijks0 ; b0 := max
1�i;j�n
kbiks0 ; c0 := kcks0:
For the operators L0; L from (1.24) it follows:
kL0�ks0 �
c1
s� s0
k�ks; kL�ks0 �
c2
(s� s0)2
k�ks; s
0
2 (0; s);
c1 := ne
�1
b0 + s0c0; c2 := 4n2e�2d+ c1s0:
The solution of problem (3.1), (3.2) is represented as
u(x; z; t) = �(x; z)Æ(t � z) + ~u(x; z; t)�(t� z); (3:4)
and we make use of the method of separation of variables [5, p. 29]. Denoting
~u(x; z; z + 0) =: �(x; z); we put (3.4) in (3.1), (3.2) and �nd
�(x; z) � �(x) = g(x); �(x; z) = (z=2)Lg(x); x 2 Rn
; z 2 R+: (3:5)
It is not di�cult, using equalities (3.4), (3.5), to replace the inverse problem
(3.1)�(3.3) by initial-characteristic problem with Cauchy data on z = 0. It should
be noted that when t > z the equality u = ~u takes place, then the functions
u(x; z; t); k(x; z) satisfy the equations
utt � uzz � Lu = L0g(x)k(x; t � z)
+
t�zZ
0
k(x; �)L0u(x; z; t��)d�; (x; z; t) 2 G := f(x; z; t)jx 2 Rn
; t > z > 0g; (3:6)
ujz=0 = f(x; t); uzjz=0 = 0; x 2 Rn
; t 2 R+; (3:7)
ujz=0 =
1
2
zLg(x); x 2 Rn
; t 2 R+: (3:8)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 421
D.K. Durdiev
Further, the problem (3.6)�(3.8) is reduced to the following system of integro-
di�erential equations for u; ut; k:
u(x; z; t) = u0(x; z; t) +
1
2
Z Z
4(z;t)
[Lu(x; �; �) � L0g(x)k(x; � � �)
+
���Z
0
k(x;
)L0u(x; �; � �
)d
]d�d�; (3:9)
ut(x; z; t) = u0t(x; z; t) +
1
2
Z z
�z
[Lu(x; z � j�j; t+ �)� L0g(x)k(x; t � z + j�j+ �)
+
t�z+j�j+�Z
0
k(x;
)L0u(x; z � j�j; t+ � �
)d
]sgn�d�; (3:10)
k(x; z) = k0(x; z) +
2
L0g(x)
z=2Z
0
[Lut(x; �; z � �) +
1
2
�L0g(x)k(x; z � 2�)
+
z�2�Z
0
k(x;
)L0ut(x; �; z � � �
)d
]d�; (3:11)
where
u0(x; z; t) =
1
2
[f(x; t+ z) + f(x; t� z)];
k0(x; z) =
1
L0g(x)
[(1=4)zL2
g(x) � 2ftt(x; t)jt=z ];
4(z; t) = f(�; �)j0 � � � z; t� z + � � � � t+ z � �; t > zg:
For the system (3.9)�(3.11) the following theorem is valid.
Theorem 3. Assume that the consistency conditions of f(x; 0) = 0,
ft(x; t)jt=0 = (1=2)Lg(x), are satis�ed, besides
(aij ; bi; c; g)(x) 2 As0 ; 1 � i; j � n;
[f(x; t); ft(x; t); ftt(x; t)] 2 C(As0 ; [0; T ]);
max[kfks0(t); kftks0(t); kfttks0(t); kL0gks0 ; 1=kL0gks0 ; kL
2
gks0 ] = R;
422 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4
Some Multidimensional Inverse Problems of Memory Determination
for t 2 [0; T ], R > 0. Then a 2 (0; T=2); as0 < T=2 can be found such that for
any s 2 (0; s0) in the region GT \f(x; z; t)jx 2 R
n
; 0 � z � a(s0� s)g there exists
the unique solution of the system (3.9)�(3.11), for which
(u(x; z; t); ut(x; z; t)) 2 C(As0 ; PsT )
k(x; z) 2 C(As0 ; [0; a(s0 � s)]); PsT := GT \ f(z; t)j0 � z � a(s0 � s)g;
moreover,
ku� u0ks(z; t) � R0; kk � k0ks(t) �
R0
(s0 � s)2
; kut � u0tks(z; t) �
R0
(s0 � s)
;
(z; t) 2 PsT ; R0 = R0(R;T; s0; n) is a constant.
Theorems 2 and 3 are proved similarly to Th. 1.
R e m a r k 1. The solution of the inverse problem suggests the unique con-
tinuation on the variable t from the interval [0; a(s0 � s)] onto the interval [0; T ]
for any T (see [1]).
R e m a r k 2. Under appropriate conditions similar results hold when the
operator 4 from Parts 1,2 is replaced by the operator L, de�ned in Part 3.
References
[1] D.K. Durdiev, A Multi-Dimensional Inverse Problem for Equation with Memory. �
Sib. Mat. J. 35 (1994), No. 3, 574�582. (Russian)
[2] A. Lorensi, An Identi�cation Problem Related to a Nonlinear Hyperbolic Integro-
Di�erential Equation. � Nonlinear Analysis: Theory, Meth., Appl. 22 (1994), 297�
321.
[3] J. Janno and L. Von Welfersdorf, Inverse Problems for Identi�cation of Memory
Kernels in Viscoelasticity. � Math. Meth. Appl. Sci. 20 (1997), No. 4, 291�314.
[4] A.L. Bukhgeim and G.V. Dyatlov, A Uniqueness in one Inverse Problem of Memory
Determination. � Sib. Mat. J. 37 (1996), No. 3, 52�53. (Russian)
[5] V.G. Romanov, Stability in Inverse Problems. Scienti�c World, Moscow, 2005.
(Russian)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 4 423
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