Journal of Applied Mathematics and Stochastic Analysis, 16:1 (2003), 69-79.
Printed in the U.S.A. ©2003 by North Atlantic Science Publishing Company
MIXED PROBLEM WITH INTEGRAL BOUNDARY
CONDITION FOR A HIGH ORDER MIXED TYPE
PARTIAL DIFFERENTIAL EQUATION
M. DENCHE and A.L. MARHOUNE
Université Mentouri Constantine
Institut de Mathematiques
25000 Constantine, Algeria
(Received April, 2000; Revised March, 2002)
In this paper, we study a mixed problem with integral boundary conditions for a high order partial
differential equation of mixed type. We prove the existence and uniqueness of the solution. The
proof is based on energy inequality, and on the density of the range of the operator generated by the
considered problem.
Key words: Integral Boundary Condition , Energy Inequalities, Equation of Mixed Type,
Sobolev Spaces.
AMS subject classifications: 35B45, 35G10, 35M10.
1 Introduction
In the rectangle
, we consider the equation
(1)
where
is bounded for
that
and
, and has bounded partial derivatives such
i
,
, for
To equation (1) we add the initial conditions
(2)
the boundary conditions
for
,
,
(3)
for
,
,
(4)
and integral condition
69
70
M. DENCHE and A.L. MARHOUNE
(5)
where
and
are known functions which satisfy the compatibility conditions given in
equations (3)-(5).
Integral boundary conditions for evolution problems have various applications in
chemical engineering, thermoelasticity, underground water flow and population dynamics; see
for example Choi and Chan [6], Ewing and Lin [7], Shi [12], and Shi and Shillor [13].
Boundary value problems for parabolic equations with integral boundary conditions are
investigated by Batten [1], Bouziani and Benouar [2], Cannon [3, 4], Cannon, Perez Esteva and
Van der Hoeck [5], Ionkin [8], Kamynin [9], Kartynnik [10], Shi [12], Yurchuk [14] and many
references therein. A method was developed in [2], [10], and [14] based on energy
inequalities. In this paper, our objective is to extend this method to high order partial
differential equations of mixed type.
2 Preliminaries
In this paper, we prove existence and uniqueness of a strong solution of the problem stated in
equations (1)-(5). For this, we consider the problem in equations (1)-(5) as a solution of the
operator equation
,
where
such that,
the operator
functions
(6)
with domain of definition
consisting of functions
,
and, satisfies conditions in problems (3)-(5);
is considered from
to , where
is the Banach space consisting of all
satisfying equations (3)-(5), with the finite norm
sup
(7)
and is the Hilbert space of vector-valued functions
obtained by completion of
the space
with respect to the norm
(8)
where is an arbitrary number such that
We then establish an energy inequality:
.
Mixed Problem with Integral Boundary Condition
71
!
(9)
and we show that the operator has the closure .
Definition 1: A solution of the operator equation
the problem in equations (1)-(5).
Inequality (9) can be extended to
, i.e.,
!
is called a strong solution of
.
"
From this inequality, we obtain the uniqueness of a strong solution if it exists, and the equality
of sets #
and #
Thus, to prove the existence of a strong solution of the problem in
equations (1)-(5) for any
, it remains to prove that the set #
is dense in .
Lemma 1: For any function
satisfying the condition (2), for $ % &
exp
$
&
exp
(10)
$
Proof: Starting from
exp
$
and integrating by parts, we
obtain equation (10).
Remark 1: We note the above lemma holds for weaker conditions on .
Lemma 2: For
,
(11)
Proof: Starting from
(11).
Lemma 3: For
and using elementary inequalities yields
satisfying the condition in equation (2),
exp
$
(x,
exp
(12)
$
Proof: Integrating the expression exp
$
by parts for
elementary inequalities and Lemma 2, we obtain expression (12).
using
3. An Energy Inequality and Its Consequences
Theorem 1: For any function
we have the inequality
!
,
(13)
72
M. DENCHE and A.L. MARHOUNE
exp $
where constant !
)
&
max ' (
min( *'
, with the constant $ satisfying
$ % & and $
%
(14)
*
*
Proof: Denote
+
where ,
,
We consider the quadratic formula
Re
exp
$
(15)
+
with the constant $ satisfying condition (14); obtained by multiplying equation (1) by
exp
$ + ; integrating over
, where
, with
; and by
taking the real part. Integrating by parts times in formula (15) with the use of boundary
conditions in equations (3), (4), and (5), we obtain
Re
exp
$
exp
+
Re
exp
By substituting the expression of +
inequality
,
$
(16)
$
in formula (15), using elementary inequalities and the
where
'
,
yields:
Re
exp
$
(
+
exp
exp
'
$
.
$
(17)
From equation (1) we have:
.
Using elementary inequalities, we obtain
'
exp
$
(18)
Mixed Problem with Integral Boundary Condition
exp
73
$
exp
*
,
$
(19)
where By integrating the last term on the right-hand side of expression (16) and combining the
obtained results with Lemmas 1 and 3, and the inequalities in equations (17), (18), and (19), we
get:
(
'
exp
)
&
$
exp
%
+&
exp
exp
+1
$
*'
$
+
exp
exp
$
*
*
exp
Using elementary inequalities and condition (14), we obtain:
(
'
)
&
% exp
$
$
$
exp
$
$
&
$
(20)
74
M. DENCHE and A.L. MARHOUNE
.
(21)
As the left-hand side of equation (21) is independent of , by replacing the right-hand side by
its upper bound with respect to in the interval . /, we obtain the desired inequality.
Lemma 4: The operator from to admits a closure.
Proof: Suppose that 0
is a sequence such that
0
1
0
1
in
(22)
and
We must show that
the continuity of the trace operators
Introducing the operator
The fact that
results directly from
2
3
of functions 3
3
3
(23)
3
3
defined on the domain
in
verifying
3
3
we note that
is dense in the Hilbert space obtained from the completion of
respect to the norm
3
with
3
Additionally, since
3
lim
01 4
lim
3
0
0
01 4
3
this holds for every function 3
, and yields
.
Theorem 1 is valid for strong solutions, i.e., we have the inequality
!
, "
,
hence we obtain
Corollary 1: A strong solution of the problem in equations (1)-(5) is unique if it exists,
and depends continuously on
.
Corollary 2: The range #
of the operator is closed in , and #
#
4 Solvability of the Problem
Mixed Problem with Integral Boundary Condition
75
To prove the solvability of the problem in equations (1)-(5), it is sufficient to show that #
dense in . The proof is based on the following lemma.
Lemma 5: Suppose that
is also bounded. Let
5
6
7 If, for
and some
is
, we have
(24)
where is an arbitrary number such that
, then
Proof: The equality (24) can be written as follows:
.
(25)
If we introduce the smoothing operators with respect to [14] ,
then these operators provide the solutions of the problems
:
:
and ,
8
9
,
:
(26)
g
and
:9
:9
:
(27)
:9
and also have the following properties: for any :
, the functions :
:9
, 9 : are in
such that :
and :9
. Moreover, ,
with , so
and
for 1 .
:
:
1
:9 :
1
Now, for given
, we introduce the function
, : and
commutes
3
Integrating by parts, we obtain
3
and
,3
(28)
3
Then from equality (25), we have
; 3
where ; 3
3
, 3, and <
equation (29) by the smoothed function ,
<
(29)
3
Replacing
, and using the relation
in
76
M. DENCHE and A.L. MARHOUNE
<
,
,
<
,
<
,
we get
39
;
<
39
<
(30)
39
By passing to the limit, we satisfy the equality (30) for all functions satisfying the conditions in
equations (2)-(5), such that
for
.
The operator <
has, on
, a continuous inverse defined by the relation
<
:
=
=
!
:
=
(31)
=
>
where
<
Hence the function
<
<
<
(32)
:
can be represented in the form
, where
,
<
<
Then,
?
<
?
:
@
?
?
? @
@
=
? @
:
=
@
?
A
A
!?
?
A
@
A
A
A @
@
@
A @
?
?
(33)
>
Consequently, equation (30) becomes
;
39
39
<
in which < 9 is the adjoint of the operator <
.
The left-hand side of equation 34) is a continuous linear functional of
B
function B
39
< 9 39 has the derivatives
and the following conditions are satisfied
B
B
(34)
< 9 39
Hence the
B
(35)
Mixed Problem with Integral Boundary Condition
The operators < 9
are bounded in
, hence the operator, 8
<9
, are bounded operators in
B
For sufficiently small, we have
has a bounded inverse in
. So from the equality,
39
<9
8
77
?
?
<9
<9
In addition,
? 9
3
?
<9
(36)
?
?
39
We conclude that the function 39 has the derivatives
account equations (36) and (35), we have
39
<9
8
?
?< 9
,
. Taking into
? 39
?
(37)
?
?
and
39
<9
8
?
?
<9
? 9
3
?
.
?
(38)
?
Similarly, for sufficiently small, in each fixed point
. /, the operators
are bounded in
, and the operator 8
< 9 is continuous inversible in
from equations (37) and (38), 39 satisfies the conditions
<9
. And so
39
39
.
(39)
So, for sufficiently small, the function 39 has the same properties as B
In addition, 39
satisfies the integral condition in equation (28).
Putting
exp $ 39
in equation (29), where the constant $ satisfies
$
%
and using equation (27), we obtain
exp $ 39 ; 3
exp
<
$
<
39
(40)
Integrating each term in the left-hand side of equation (40) by parts and taking the real parts,
we have
Re
<
exp
$
%
exp
Re
<
39
$
exp
$
(41)
$
%
exp
$
(42)
78
M. DENCHE and A.L. MARHOUNE
Now, using inequalities (41) and (42) in equation (40) with the choice of $ indicated above, we
have 2Re exp $ 39 ; 3
; then for 1 , we obtain 2Re exp $ 3; 3
,
i.e., 2Re exp $
C3C
2Re exp $ 3, 3
.
Since Re exp $ 3, 3
, we conclude that 3
; hence
, which ends the
proof of this lemma.
Theorem 2: The range #
of coincides with .
Proof: Since is a Hilbert space, we have #
if and only if the relation
(43)
for arbitrary
Putting
and
, implies that
in relation (43), we obtain
, and using Lemma 5, we obtain
Consequently, "
we have
,
and
.
Taking
, then
.
(44)
The range of the trace operator
hence
is everywhere dense in Hilbert space with the norm
.
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Mixed Problem with Integral Boundary Condition
79
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