PSEUDO Differential Operations and NEUMANN
Problems
S. J. Monaquel
Mathematics Department, Faculty of Science, King Abdul Aziz University, Jeddah, Saudi Arabia
Abstract-- Our main purpose of this paper is to find the corresponding set of inequalities defining an optimal control of a system governed by Neumann problem for a class of pseudo differential operators with symbols defined in terms of conditionally exponential convex functions also we formulate the boundary control problem for a system governed by Neumann problem.
Index Term--- Pseudodifferential operators, conditionally exponential convex, optimal controls.
I. INTRODUCTION
Consider a class of pseudo differential operators
,
,
(1.1) nx R
L x D u x
e L x u dwhere L: Rn Rn R is a real valued continuous symbol such that Lx, .:Rn R is coditionally exponential convex functions.
Definition 1.1
A real valued function L :Rn R is said to be coditionally exponential convex function if for any x1,x2, . . .xn Rn and C1,C2, . . .Cn R, we have
, 1
0
(1.2) n
j k j k j k
j k
L x L x L x x C C
see 2, 4.
Under suitable conditions Lx,D extends from C0
Rn to
a generator of a symmetric Dirichlet form B,DB with domain DBL2Rn and
2 0
, , , ,for , .
(1.3)
n
n L R
B u v L x D u v u v C R
In this paper we are interested in boundary value problems for
Lx,D on some open set Rn.
We find the corresponding set of inequalities defining an optimal control
of a system governed by Neumann problem for Lx,D on
,
in(1.4) on
n
L x D u f R
u h
II. A CLASS OF PSEUDO DIFFERENTIAL
OPERATORS
Let us Recall some results from 1, 2, 10, see also
4, 5. Let a2 :Rn R
be a real valued continuous conditionally exponential convex function, see 2, 11, that is a2 is a continuous function such that a200 for all t 0 , the function eta2 is exponentially convex. We define the norm
2
2 2
2 2
, 1 , for 0.
(2.1) n
s a s
R
u
a u d s a
nd the sobolev spaces
2
2 ,
2 : , . (2.2)
a s n n
a s
H R uL R u
The space Ha2,sRn is a real Hilbert space with the scalar product
2
2 2 ,
, 1
(2.3)
n
s a s
R
u
a u da
nd C0
Rn
is a dense subspace of Ha2,sRn. For a2||2, the space Ha2,sRn
coincides with the usual sobolev space H2sRn.
It is known that a real valued continuous conditionally exponential convex function a2 satisfies the estimate
2
20a C 1 , for someC 0 (2.4)
Moreover, by 8, 9, we can construct a chain
2, 2,
2 (2.5)
a s n n a s n
H R L R H R
In the following we will always suppose that
L :Rn Rn R
is a real continuous symbol such that for any fixed x Rn, the function
Lx, .:Rn R
is coditionally exponential convex and Lx, has the decomposition
,
1
2
,
(2.6)L x L L x
where for a suitable m N, we have
1) L1
C
1a2
, for some C 0 and Rn;
2) L2. , CmRn and for all N0 n
,
2
2
, , 1
(2.7)
x
m L x x a
h
old for all Rn with some L1Rn;
3) L120a2 for some 0 0 and all
Rn
, || R 0;
4)
1
L m
is small with respect to 0, see 3.Then the operator Lx,D as defined in 1. 1 maps
C0Rn
into the space CRn and the bilinear form associated with Lx,D
,
,
.
(2.8) nR
B u
L x D u x x dxis defined for u, C0
Rn.
In the following we suppose that the operator
Lx,D is symmetric on C0
Rn,
then Lx,D has a selfadjoint extension on L2Rn with domain
Ha2,1
Rn
. The bilinear form B extends to a continuous symmetric Dirichlet form with domain
Ha2,12Rn, see 4, 5 for the general theory of
Dirichlet forms and their properties. In particular,
the form B is positive definite on Ha
2,1
2Rn, i.e. Bu,u0 , for all u Ha2,12Rn.
Moreover, the form B satisfies Gãrding inequality, see 6
2 2 1
2 2
2
0 , 0
, a L Rn (2.9)
B u u u u
III. FORMULATION OF THE PROBLEM
Let Rn be an open set with smooth boundary
. By 2. 9 , the bilinear form B is a continuous and coercive bilinear form on Ha2,1RnL2Rn. Thus, by the Lax Milgram Theorem, see 9, for each f L2Rn we find a weak solution
y Ha2,1
Rn
satisfying Neumann problem relative to the operator L, defined by 1. 1, and enables us to obtain the state of our system.
Theorem 3.1
If 2. 9 is satisfied then there exists a unique element y H1,a2Rn satisfying Neumann problem
,
,(3.1)
on , L x D y f in
u h
where
1
cos , n
k k k
u u
n x
v x
on ,cosn,xkkth direction cosine of n, n being
the normal at exterior to Rn.
Proof.
Let us choose L to be of the form
1 2
2, 2
, (3.2)
where , .
n
R
a n
L f dx h d
f L R h H
We note that 3. 2 defines a continuous linear form on H1,a2Rn, (see 5, 6, 7), from the coerciveness condition 2. 9there exists a unique element
y H1,a2Rn
such that
,
(3.3)B y L
This equation is equivalent to
,
, in n (3.4)L x D y f R
Multiply both sides by and apply Green's formula, we get
, (3.5)
n n
R R
L x D y dx f dx
,
(3.6)n
R
y
B y d f dx
,
, (3.7) nR
B y f dx h d
then,
0 (3.8)
y
h d
on , (3.9)
y
h
n
ow the space L2Rn, being the space of controls,
is given.
For a control u the state of the system yu is given by the solution of
, , in
(3.10)
, on
n
L x D y xu f u R
y
u h
An observation equation Zuyu is also given, and N LL2Rn,L2Rn, where N is
Hermitian positive definite, satisfying,
2 2
2
, L Rn L Rn . (3.11)
Nu u u
The cost function Ju is the same, and given by
2 2
2 2
2
,
(3.12)
, n n
n n
d L R L R
d L R
R
J u y u Z Nu u
y u Z dx Nu u
where Zd is a given element in L2Rn.
The problem is to find inf, Uad, where Uad
(the set of admissible controls) is a closed convex susbset of L2Rn. Under this consideration , we
have the following theorem.
Theorem 3.2
Assume that 2. 9 holds, the cost function being given by 3. 12, a necessary and sufficient condition for u L2Rn to be an optimal control
is that the following equations and inequalities be satisfied
in , on ,
(3.13)
in , 0 on
n
n d
u
Ly u f u R h
p
Lp u y u Z R u
and
0 (3.14)n
R
p u Nu u dx
for all u, Uad, where pu is the adjoint state
of yu.
Proof
The control v Uad is optimal if and only if
0, for all ad (3.15)J u v u U
That is
2
2
,
, 0. (3.16)
n
n
d L R
L R
y u Z y y u
Nu u
If we set
2
2
2
, 0 , 0
, (3.17)
0 , 0
n
n
n
L R L R
d L R
u y u y y y
Nu u
L Z y y y
The form u, is a continuous bilinear form and
L is a continuous linear form on L2Rn, then if
we set
2 2
, 2 d 0 L Rn (3.18)
J L Z y
since
2 2
2
, y y 0 L Rn N , L Rn .(3.19)
Then from 3. 11, we have
2 2
2
, n , for every (3.20)
n
L R L R
As in 8, 9, there exists a unique element u in Uad
such that
inf
(3.21)ad
U
J u J
and this element is characterized by
0, for all ad. (3.22)J u u U
Since L is a canonical isomorphism from H1,a2Rn into H1,a2Rn, we may write
1
(3.23)
y u L f u
2
2
2 , 0
,
(3.24) n
n
d L R
L R
J u u y u Z y u y
Nu u
But
0
, (3.25)y u y y y u
then
2
2
2 ,
,
(3.26) n
n
d L R
L R
J u u y u Z y y u
Nu u
Therefore, after dividing by 2, 3. 15 is equivalent to
2
2
,
, 0 (3.27)
n
n
d L R
L R
y u Z y y u
Nu u
for the control u L2Rn the adjoint state
1,a2
np u H R is defined by
, in
(3.28)
0, in
n d
Lp u y u Z R
p u
Now, multiplying the first equation in 3. 28 by
yyu and applying Green's formula, we
obtain
2 2
, , 0
(3.29)
n
n L R
L R
Lp u y y u Nu u
and
2
2
2 ,
,
, 0 (3.30)
n
n
L R
L L R
p u L y y u
p u y y u
Nu u
from 3. 1 , we obtain
2 2
2
2 2
, ,
,
, , 0
(3.31)
n
n
n n
L R L
L R L R L R
p u f f u p u h h
Nu u
p u u Nu u
that is u Uad,
p u Nu
u dx
0, for all Uad (3.32)
which completes the proof.
IV. BOUNDARY CONTROL FOR A SYSTEM GOVERNED
BY NEUMANN PROBLEM
Consider the space H
1 2,a
2
U (the space of
controls), for every control uH12,a2
, the state of the system yu is given by the solution of
in
(4.1)
on
n
Ly u f R
u
u h u
a
nd the observation is given by Zuyu. Finally the cost function is given by
2 2
, (4.2)
n
d L R U
J y Z N
where Zd is a given element in L2Rn and,
N LU,U, N is Hermitian, positive definite,
2, , 0. (4.3)
U U
N C C
We wish to find infJ, Uad, where Uad (the
set of admissible controls) is a closed convex subset of U.
Under the given considerations, we have the following theorem.
Theorem 4.1
in
on
(4.4) in
0 on n
n d
Ly u f R
y u h u
Lp u y u Z R
p u
and
2
, , 0 (4.5)
U L
p u u Nu u
f
or all u, Uad, where pu is the adjoint of the
state yu.
Outline of proof.
Using 8, 9, the control u Uad is optimal if and
only if
.
0, for all ad (4.6)J u u U t
hat is
2
, , 0
(4.7)
n
d L R U
y u Z y y u Nu u
t
he adjoint state is given by the solution of the adjoint Neumann problem
in
(4.8)
0 on
n d
Lp u y u Z R
p u
f
rom 4. 7 and 4. 8, we have
2
, , 0
(4.9)
n U
L R
Lp u y y u Nu u
b
y applying Green's formula, we obtain
2
2
2 ,
,
,
, 0 (4.10)
n
L R
L
L U
p u L y y u
p u y y u
p u y y u
Nu u
from 4. 1 and 4. 8, we obtain
2
2
2
,
,
0,
, 0 (4.11)
n
L R L L U
p u Ly Ly u
p u h h u
y y u
Nu u
It follows
2 2
, ,
, 0 (4.12)
n
L R L
U
p u f f p u u
Nu u
which is equivalent to
2
, , U 0 (4.13)
L
p u u Nu u
which completes the proof.
ACKNOWLEDGEMENTS
The author is grateful to prof. Hoda A. Ali for substantial assistance through the paper.
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