69
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 , and has bounded partial derivatives such
that and i , , for
To equation (1) we add the initial conditions
(2) the boundary conditions
for , , (3)
for , , (4)
and integral condition
(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
, (6)
where with domain of definition consisting of functions , such that, and, satisfies conditions in problems (3)-(5);
the operator is considered from to , where is the Banach space consisting of all functions 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:
! (9) and we show that the operator has the closure .
A solution of the operator equation is called a strong solution of Definition 1:
the problem in equations (1)-(5).
Inequality (9) can be extended to , i.e.,
! " .
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 .
(2)
Lemma 1: For any function satisfying the condition , for $ % &
exp $
& exp $ (10)
Proof: Starting from exp $ and integrating by parts, we obtain equation (10).
We note the above lemma holds for weaker conditions on . Remark 1:
Lemma 2: For ,
(11) Proof: Starting from and using elementary inequalities yields (11).
(2)
Lemma 3: For satisfying the condition in equation ,
exp $ (x,
exp $ (12)
Proof: Integrating the expression exp $ by parts for using elementary inequalities and Lemma 2, we obtain expression (12).
3. An Energy Inequality and Its Consequences
Theorem 1: For any function we have the inequality
! , (13)
where constant ! exp$ maxmin(' ( )& , with the constant satisfying$
*'
$ % & and $ % * * (14)
Denote Proof:
+ , 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 $ (16)
By substituting the expression of + in formula (15), using elementary inequalities and the inequality
, ' where ,
yields:
Re exp $ + ( ' exp $
exp $ . (17)
From equation (1) we have:
. (18)
Using elementary inequalities, we obtain
' exp $
exp $
* 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 $
$ ** exp $ (20)
Using elementary inequalities and condition (14), we obtain:
( ' )&
% exp $ &
.
(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.
Suppose that is a sequence such that
Proof: 0
01 in (22)
and
0 1 in (23)
We must show that The fact that 2 results directly from
the continuity of the trace operators Introducing the operator
3 3
3
defined on the domain of functions 3 verifying
3 3 3 3
we note that is dense in the Hilbert space obtained from the completion of with respect to the norm
3 3
Additionally, since
0 0
3 0 1 4lim 3 0 1 4lim 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
(1)-(5)
Corollary 1: A strong solution of the problem in equations is unique if it exists,
and depends continuously on .
The range of the operator is closed in , and
Corollary 2: # # #
4 Solvability of the Problem
To prove the solvability of the problem in equations (1)-(5), it is sufficient to show that # is 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 , 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] , 8 and , 9, then these operators provide the solutions of the problems
: : :
g (26)
and
: :
:9 9
(27) :9
and also have the following properties: for any : , the functions : , : and :9 , 9: are in such that : and :9 . Moreover, , commutes
with , so : : 1 and :9 : 1 for 1 .
Now, for given , we introduce the function
3
Integrating by parts, we obtain
3 , 3 and 3 (28)
Then from equality (25), we have
; 3 < 3 (29)
where ; 3 3 , 3, and < Replacing in
equation (29) by the smoothed function , , and using the relation
< , , < , < , we get
3 9 < 9
; 9 < 3 3 (30)
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
< : (32)
Hence the function can be represented in the form , < < Then,
< < < , where
< : = : =
?
?
@
?
? ? @
? @ ? @
@
!
? A
? A
? >
@ A
A A @ ?
A @ A @
@ (33)
Consequently, equation (30) becomes
3 9 9 9
; 9 < 3 < 3 (34)
in which <9 is the adjoint of the operator < .
The left-hand side of equation 34) is a continuous linear functional of Hence the
function B 39 < 39 9has the derivatives B B
and the following conditions are satisfied
B B (35)
The operators <9 are bounded in For sufficiently small, we have <9 8 <
, hence the operator, 9 has a bounded inverse in In addition, <9 , are bounded operators in . So from the equality,
B 9 3 ? < 3
?
?
9 ? 9 ? 9
8 < ? (36)
We conclude that the function has the derivatives 39 39 , . Taking into account equations (36) and (35), we have
8 <9 3 ? < 3
?
9 ? 9 ? 9
? ? (37)
and
. (38)
8 <9 3 ? < 3
?
9 ? 9 ? 9
? ?
Similarly, for sufficiently small, in each fixed point . /, the operators <9
are bounded in , and the operator 8 <9 is continuous inversible in . And so from equations (37) and (38), satisfies the conditions39
39
39 . (39)
So, for sufficiently small, the function 39has 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
9 3
exp $ 3 ; 3 < exp $ < 9 (40)
Integrating each term in the left-hand side of equation (40) by parts and taking the real parts, we have
Re < exp $ % $ exp $
exp $ (41)
Re < 39 % exp $ (42)
Now, using inequalities (41) and (42) in equation (40) with the choice of indicated above, we$ have 2Re exp $ 3 ; 39 ; 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 and , implies that , and .
Putting in relation (43), we obtain Taking
, and using Lemma 5, we obtain , then . Consequently, " we have
(44)
The range of the trace operator is everywhere dense in Hilbert space with the norm
hence .
References
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[5] Cannon, J.R., Perez Esteva, S., and Van Der Hoek, J., A Galerkin procedure for the diffusion equation subject to the specification of mass, SIAM. J. Numer. Anal. (1987),24 499-515.
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[9] Kamynin, N.I., A boundary value problem in the theory of the heat condition with nonclassical boundary condition, USSR Comput. Math. and Math. Phys. (1964), 33-59.4 [10] Kartynnik, A.V., Three-point boundary value problem with an integral space-variable
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