1. Introduction
During these last decades, see [6] [7], functional analysis and the theory of monotone operators, defined on Banach spaces have played, an important role in the study and analysis of systems. Reference [4], introduced the feedback systems described by monotone operators, defined on appropriate spaces. He has established, a series of existence and uniqueness results, of the solutions of this system on a Hilbert space H. These types of systems find their uses in several fields such as: control theory, network theory, solving the Hammerstein equation... etc. [8]. The techniques used are based, on the surjectivity theorem, of the monotonic and coercive maximal operators on a reflective Banach space. References [4] [5] introduced, the notion of extended Hilbert space He, and obtained, among others, a normality and linearization results for a feedback system, on this space.
One of our fundamental results is that, the behavior of the FS
is completely determined, by the inverse of some application
(see (2)). Note that, in the case where the operators A and B are not linear, and if
, then
. If one of the two operators is linear, the writing of the solution
, can take forms that do not necessarily depend, on the inverse of the operator
, see (4). This approximation allows us, to obtain suitable estimates of the solutions, in the sense of Section 3, these estimates have a significant effect, on the study of robust stability and sensitivity [1]. For more details, on the study of the inverse of such an operator, which is non linear, one consult [9] [10] [11] [12].
The subject of our work, is to proceed to the approximation method. Therefore, to find an approximate solution of
, supposed nonlinear by one linearizes, in the neighborhood of zero. We then consider a linear
, and prove that, if
;
, with
,
(
) and
,
the respective solutions of
and
corresponding to the given
in
. There exists,
,
,
,
, positive real constants such that
and
.
The paper is organized as follows. In Section 2, we recall some definitions, and we demonstrate results of normality of the FS
, according to whether the two operators are nonlinear or one of the operators is linear. Section 3, is reserved for our results of normality, Lipschitz continuity and approximate solution of
, supposed nonlinear, by one linearizes, in the neighborhood of zero. An example is presented, at the end of this section.
2. Definitions and Preliminary Results
Let E be a real vector space,
the set of all parts of E,
the algebraic dual of E. Let
and
, the pair
is said feedback system and it is noted FS
or
.
, the domain of A. We say that, A is an operator, if
and
is a singleton. A is said simple if, for every
,
we have
. Note that, if A is a simple operator it is injective.
The meaning of the following definition, can be understood, by looking at Figure 1.
Figure 1. 1 − (u, v) input; 2 − (y, g) output; 3 − (e, f) solution, 4 − [A, B] a feedback system.
Definition 2.1. We say that, an element
of
is a solution of the FS
, corresponding to the given
(input) in
and we write
, if there exists
(output) in
such that:
(1)
Definition 2.2. We say that the FS
is:
1) Resoluble, if for all
, there exists
, checking (1).
2) Unambiguous, if each solution is unique.
3) Normal, if it is resoluble and unambiguous.
The existence and uniqueness results of the solutions of
, are based on the mapping
defined for all
, by
(2)
Proposition 2.1. The FS
is resoluble, iff
,
.
Proof. Since
is resoluble, for
, there are
and
verifying:
and
. So
therefore
. Reciprocally, let
, since
, it exists
such that
, satisfying
, where
, then
, with
. Therefore, there are
verifying
.
Proposition 2.2.
1) If
is unambiguous, then for every
,
is simple.
2) If A is an operator and for every
,
is simple. Then
is unambiguous.
Proof.
1) Suppose that, there are
,
such that
. Let
, for
, it exists
checking
, since
where
, with
, then
. Likewise, when
, there are
, as
then
so, for every
,
is simple.
2) Assume that, for
there exists two solutions
,
of
, related to
. Then, there are
,
such that:
, and
. So,
, also
hence
. As, for every
,
is simple, we have
and
.
Corollary 2.1. Let A and B be two operators, the FS
is:
1) Resoluble, iff
is surjective.
2) Unambiguous, iff
is injective.
3) Normal, iff
is invertible. In this case, if
then:
(3)
Proof.
1) Let
, so
,
, (see, proposition 2.1), then for
, it exists
such that
,
. Reciprocally, if
,
as
, there is
, such that
. Since
, it follows that
,
the reverse inclusion is obvious, hence
is resoluble.
2) If,
is unambiguous, from the proposition 2.2 (1),
is simple, hence it is injective. Conversely, since
is injective, then for
in E,
we have
, then
, so
is simple, from proposition 2.2 (2)
is unambiguous.
3) Direct consequence of 1) and 2).
Let’s demonstrate the Formula (3). If
it exists
,
satisfying
, because B is an operator, and
. Then,
, from where
, this implies that
.
Proposition 2.3. Let A and B be two operators,
the identity. If B is linear, then
is bijective iff
,
is bijective.
Proof.
Proof that,
is surjective
,
is surjective. Let
, since
, B is linear and
,
is surjective, it exists
such that
,
, then
and
is surjective. Reciprocally, since
, and
is surjective, it exists
such that
,
, this implies
,
therefore
,
is surjective.
Proof that,
is injective
,
is injective. Let
in E, with
, then
,
, as
,
is injective, then
. Conversely, Let
such that
,
that is to say
,
, so
, since
is injective then
Corollary 2.2. Let A and B be two operators, and
the identity. If B is linear, the FS
is:
1) Resoluble iff
is surjective.
2) Unambiguous iff
is injective.
3) Normal iff
is invertible. In this case, if
then:
(4)
Proof. Direct consequence of proposition 2.3 and corollary 2.1. To demonstrate (4), let
, there are
and
satisfying
and
, from where
, so
, hence
and
.
3. Linearization Results
Let
be, the dual and the bidual of a real normed space E. Since the canonical application
, defined by: for every
;
is linear and isometric, then it's continuous and injective. If the range
, we say that E is reflexive, then E is topologically identical at
and
can be considered as a linear form on
, it is natural to write for any
;
. Since
is a Banach space, then E is also Banach. Note that, if E is the real Hilbert space then
. In the sequel, we assume that E is reflexive, and we denote indifferently by
the scalar product in the duality between these spaces, and
their norms.
Definition 3.1. A is said:
1) Monotone if, for every
,
;
or
if A is an operator. It’s strictly monotone if
implies
or
, whenever
.
2) Maximal monotone, if A is monotone and the following property holds: (
;
, S monotone) then
, where
the graph of A.
Definition 3.2. B is said:
1) Monotone if, for every
,
;
or
if A is an operator. It’s strictly monotone if
implies
, or
, whenever
.
2) Maximal monotone, if B is monotone and the following property holds: (
;
, T monotone) then
, where
the graph of B.
Corollary 3.1. If A is strictly monotone, or B is an operator strictly monotone. Then
is simple.
Proof. Let
in E, and
, there are
and
such that
, hence it exists
verifying
, which implies
and
. As
and
because A and B are monotone, then
and
. Therefore, if A is strictly monotone or B is strictly monotone, we have
which implies,
, replacing in y we get
.
Corollary 3.2. If A or B is an operator strictly monotone,
is unambiguous.
Proof. According to proposition 2.2, and corollary 3.1, this amounts to demonstrating that, for every
,
(
) is simple. It suffices to note that
is monotone (respectively strictly monotone) iff A is strictly monotone (respectively strictly monotone).
Definition 3.3. An operator
is said:
1) Coersive, if
.
2) Hemicontinuous, if for any
,
we have
weakly converges to
.
Note that hemicontinuity is operational, continuity implies hemicontinuity, and see [13] [14] [15] [16].
If A is monotone, bounded and hemicontinuous, then A is maximal monotone.
If A is maximal monotone and coercive, then
.
If A is monotone, hemicontinuous and coercive, then
.
If A is hemicontinuous, and there exists
such that
, for all
. Then, A is invertible and the inverse
is monotone continuous.
To simplify the statements of linearization theorems, we note by:
and
The following assertions (which are also valid for
and
) are true:
Proposition 3.1.
1)
.
2)
,
,
,
,
and
.
3)
, N is monotone (respectively strictly monotone) iff
(respectively
).
4)
,
and
iff N is constant.
5)
,
,
,
,
,
.
6) If
, and N is linear, then
iff N is linear. In this case
.
7) If
and
, then
and
.
The numbers
and
can be interpreted crudely as a “gain” and “minimal slope” of the operator N, respectively.
Lemma 3.1. Let
with
; if N is hemicontinuous, then N is invertible,
,
and
. If in addition
, then
.
Proof. Since,
,
(*), as N is hemicontinuous then N is invertible. From (*) and because
,
, we have
, then
,
,
,
,
witch implies
and
. If
,
, returning to (*) we obtain
. It follows that,
,
, which leads to
.
Lemma 3.2. Let
which is hemicontinuous and let
, with
. If,
, then
and
are invertible,
,
(5)
Proof. Since
, then
,
, so A is hemicontinuous, with
. By lemma 3.1
exists,
,
and
. Thus
is hemicontinuous with
, using again lemma 3.1,
exists,
and
. As
. then
is invertible,
and
.
Lemma 3.3. Let
be linear, with
and let
be hemicontinuous, with
. If,
, then
and
are invertible,
,
(6)
Proof. Since
is linear,
, then B is bounded
,
, where
is the conjugate of B. Hence B is hemicontinuous, by lemma 3.1
exists,
, and
. The open mapping theorem ensures the continuity of
, and hence its hemicontinuity. To continue, let
, since for any
,
we have
weakly converges to
, taking into account the continuity of B, we have
weakly converges to
, which gives the hemicontinuity of D. Moreover, for
. As,
;
and
, then
, hence
, so
. Lemma 3.1 confirms that, D is invertible,
and
. As
is invertible,
; then
is invertible,
, so
, which give (6).
Lemma 3.4. Let
be linear, with
and let
be hemicontinuous, with
. If,
, then
and
are invertible,
,
(7)
Proof. Since
is linear,
, then A is bounded
,
, where
is the conjugate of A. Hence A is hemicontinuous, by lemma 3.1
exists,
, and
. The open mapping theorem ensures the continuity of
, and hence its hemicontinuity. To continue, let
, since for any
,
we have
weakly converges to
, taking into account the continuity of A, we have
weakly converges to
, which gives the hemicontinuity of C. Moreover, for
. Since, for any
;
then,
;
and
, then
, hence
;
, so
. Lemma 3.1 confirms that, C is invertible,
and
. As
is invertible,
; then
is invertible,
, so
, which give (7).
Lemma 3.5. Let A be linear, and let A,
be invertible, then
, the operator
is invertible and
(8)
Proof. Indeed,
, we have
and,
Definition 3.4. We say that, a normal FS
:
1) is Lipschitz continuous for the first inputs, if there are positive numbers
and
such that
and
where
,
.
2) is Lipschitz continuous for both inputs, if there are positive numbers
,
,
and
such that:
and
where
,
.
New, let a non linear FS
be. The main idea in this section is to linearize
in the neighborhood of the zero. We then consider a linear FS
and prove that, if
and
where
with
(
) and
,
the respective solutions of
and
, corresponding to the given
. There exists
and
positive real constants such that
and
.
The inequalities above are given by theorem 3.1. To have suitable estimates, in the sense that the solutions of the two systems become sufficiently close. It is assumed that, one of the two operators of
is linear, this is the subject of theorems 3.2 and 3.3.
Before establishing the first linearization result of this paper, let us denote by
the closed ball of E, and
the closed ball of
, which are centered in zero and of radius
.
Theorem 3.1. Assume that:
1)
, such that there exist a linear
,
verifying
and
(9)
2)
, such that there exist a linear
,
verifying
(10)
where
.
3)
.
Then
a)
and
are normal and Lipschitz continuous in the first input.
b) if
, and
for
;
for
, we have
(11)
and
(12)
where
;
;
;
with
.
Proof. Beginning by demonstrating (a). The linearity of
and (9) implies that
, and
,
, hence
is bounded and
. As
,
;
,
, then
,
. Therefore
, returning to the lemma 3.1,
is invertible. By the same arguments and since for any
;
, we have,
is bounded,
and
. Now, let’s pose for
,
;
and
;
. It is clear that
,
,
,
,
and
, these with (3) then give,
. By lemma
3.2,
is invertible
and
. Then corollary 2.1, implies that the FS
is normal. Since see (3), for
,
we have
;
, then
and
where,
and
in definition 3.1, (a); i.e.
is Lipschitz continuous for the first inputs. Using the same for
, we obtain
. By lemma 3.2,
is invertible
and
.
Then corollary 2.1 with (3) imply that, the linear FS
is normal, and for
,
, we have
;
, then
and
where,
and
in definition 3.1, (a); i.e.
is Lipschitz continuous for the first inputs.
To demonstrate (b), let
, since
is linear with
,
is linear and
. By lemma 3.2,
exists,
,
therefore
. Let now,
,
, it is obvious that
, witch implies that
. By lemma 3.5 and (8), we have
, then
As,
, then
hence
(13)
New, if
, and
for
for
, by (3) we have,
and
, to get (11) just replace x by u and z by v in (13). Finally, to have (12) and complete the demonstration of (b), it suffices to notice that
The estimates (11) and (12) in theorem 3.1, can be improved if one of the operators of FS
is linear. Starting with
Theorem 3.2. Assume that:
1)
with
be linear and
with
, such that there exist a linear
,
verifying
and
(14)
where
.
2)
.
Then
a)
and
are normal and Lipschitz continuous in both inputs.
b) if
, and
for
;
for
, we have
(15)
and
(16)
where
.
Proof. Let
,
, an in the proof of theorem 3.1 (a) we have
, therefore
. Also,
is bounded,
,
, then
, since by (2)
we have see lemma 3.3, N is invertible,
, and
. Since B is linear, then by the corollary 2.2,
is normal. By using (4), and because
,
and
, then
is Lipschitz continuous in both inputs. On the other hand, by (2),
, so lemma 3.3 implies that
is invertible,
, and
. Always by the corollary 2.2 ,
is normal and Lipschitz continuous in both inputs. To demonstrate (b), let
then
, hence
, by using (14) we have
(17)
Now, if
, and
for
;
for
, we have
, then
so by (4), in corollary 2.2 and (17) we get
and
The last linearization result in this work is to assume that the operator A in the FS
is linear.
Theorem 3.3. Assume that
1) Let
with
be linear and
with
, such that there exist a linear
,
verifying
and
(18)
where
.
2)
.
Then
a)
and
are normal and Lipschitz continuous in both inputs.
b) if
, and
for
;
for
, we have
(19)
and
(20)
where
.
Proof. By (18), we have
,
furthermore B is bounded, and
by lemma 3.1, A is bounded, then it is invertible,
,
and
. The operators
,
, they are such that:
(see (2)) then, lemma 3.4 with (7) imply that N is invertible,
and
. Likewise,
(see (2)) then, lemma 3.4 with (7) imply
that
is invertible,
and
. Now, let's for
,
and
. Since
exist, A is linear and it is invertible, lemma 3.5 implies that,
and
exist. Moreover by (8), we have
and
. It’s easy to see that,
;
and
. Returning to corollary 2.1,
and
are normal. Now, assume that for
,
and
it follows, by (3) that
and
So,
is Lipschitz continuous in both inputs. We prove in the same way that
is Lipschitz continuous in both inputs, the proof of a) is then complete.
Now, if
, and
for
;
for
. Let
then
and
then
. Using, (3) and (18) we have
therefore (19) is checked. Finally,
hence (20) is established and the proof is finished.
Example Reference [4]. Let
,
be, where
is the Lebesgue space, equipped with the natural inner product, then E is the Hilbert space. Let D be a real
matrix, denote by
and
, where
is the transposed of
. Let
be
matrix, with
and let
be the Fourier transform of
, (defined as 0 if
). Denote
and
, where
denotes the square root of the largest eigenvalue of the matrix
, where
is the complex conjugate martix of M (note that
and
). Furthermore, let
be defined by: it exists
such that
, for every
;
(*).
And
. Now define operators A and B as follows, for any
,
;
, and
. Moreover, let
, where F is the
constant
matrix, suppose that, it exists
such that
,
(**) and define
by:
when
. Clearly, A is linear and bounded, using Parseval’s equality and the number k, we have
,
. Also, it is known [12] that
. On the other hand (*) shows that B is continuous, it is also easy to see that,
,
. Thus, if
, we have,
. In the other hand,
,
by virtue of (**), then by
theorem 3.3,
and
are normal and Lipschitz continuous in both inputs, with
and
whenever
, and
for
;
for
, with
.
4. Conclusion
The aim of this work is to extend the results obtained in [4] [5], concerning the normality, Lipschitz continuity, of a non linear feedback system described by the monotone maximal operators, defined on real reflexive Banach spaces. In addition, the results of approximation of the solutions of the feedback system assumed to be nonlinear, by solutions of another linear are established. These types of systems find their uses in several fields such as: control theory, network theory, solving the Hammerstein equation... etc. The techniques used are based, on the surjectivity theorem, of the monotone maximal operator and hemicontinuous, defined on real reflexive Banach spaces [14].