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Necessary optimality conditions for stationary nonlinear hydrodynamic disrupted problems in a bounded domain.

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UDC 517.917
Necessary Optimality Conditions for Stationary Nonlinear
Hydrodynamic Disrupted Problems in a Bounded Domain
Balé Bailly* , Gozo Yoro† , Richard Assui Kouassi‡
*
†
‡
UFR de Mathématiques et Informatique,
Université de Cocody
22 BP 582, Abidjan 22. Cote d’Ivoire
Laboratoire de Mathématiques et Informatique,
UFR - SFA, Université d’Abobo — Adjamé
02 BP 801, Abidjan 02. Cote d’Ivoire
Département de Mathématiques et Informatique,
Institut National Polytechnique FHB
BP 1083 Yamoussoukro Cote d’Ivoire
In the paper we establish the optimal necessary conditions for guaranteeing uniquely the
resolution of boundary hydrodynamic problems in a bounded domain so that they could
accurately describe the studied hydrodynamic phenomenon.
Key words and phrases: necessary conditions of optimality, Command, optimal command, uniqueness, disruption, linearization, nonlinear.
1.
Introduction
The focus of our research on such problems lies in the fact that for nonlinear
systems of the type of Navier–Stokes in a three-dimensional space, we can not find a
class of spaces where we could uniquely solve the problem at the border. This class
is found by the linearization of systems of Navier–Stokes. However linearized systems
often do not describe accurately the movement of liquid (or fluid). An intermediate
case of investigation was proposed in [1], i.e., to the linearized system, nonlinear terms
are added, which may allow us to more accurately describe the movement of liquid
(or fluid) and at the same time allow the resolution in a unique way of the nonlinear
problem relative to the boundaries, obtained by disrupting our initial system.
Using the Hadamard theorem for infinitely small Lipschitz constant, satisfying the
conditions of these disturbances, the obtained disrupted problem at the borders has a
unique solution  = (, ) where  is the value of the velocity  at the border (with
 = 0 for the studied problem),  is the second member of the perturbed obtained
system, and  satisfies the Lipschitz conditions with respect to  , in the corresponding
functional spaces.
For some smooth conditions on the Nemytsky–Hammerchtéin operator and using
the theorem of Hadamard about strong derivation of inverse functions, the operator
( ) is strongly differentiable in the sense of the corresponding  . This derivation is
weaker than the Fréchet derivation. But it is quite sufficient to establish the necessary
conditions of optimality of problems relative to those equations.
2.
Statement of the Problem of Optimal Control
The physical processes that find their applications in technique, are generally controlled. It means that they can be achieved in many ways at the mercy of man.
Therefore, we must find the best control according to particular criteria, in other
words, the optimal control of the process.
The flow of an incompressible viscous fluid in a not empty and bounded domain
Ω, is characterized by its velocity  = () and pressure  = ().
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Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
Consider the associated system after disruption
 M () +  (, ()) +
∫︁
(, )(, ())d = ∇ () +  (),
(1)
Ω
div () = 0,
(2)
3
 |Ω () = 0,
∈Ω⊂R ,
(3)
and functional with the following form
 ( ) =
∫︁
 (, ()),  ())d,
 = 0, 1, . . . , 1 + 2 ,
(4)
Ω
where  are Caratheodoric functions, that is they are measurable with regard to the
triplet (, ,  ) and continue with regard to the couple (,  ) almost everywhere for
all the elements  of Ω;  is the kinematic coefficient of viscosity (or of tenacity) and
it is considered to be constant. Ω =  is the border of the domain Ω. In addition to
that we have
(︀
)︀
| (, ,  )| 6  () +  ||2 + | |2 ,
(︀
)︀
̂︀ ||2 + | |2
|∇(, )  (, ,  )| 6  () + 
with  ,  = 0, 1, . . . , 1 + 2 , derivable with respect to the pair (,  ),  () ∈ 1 (Ω),
̂︀ are constants. More,  ,  and  verify the Lipschiz
 () ∈ 2 (Ω),  and 
condition from the pair (,  ); 1 and 2 are non negative integers.
According to [1] the following functions:
:
Ω × R3 × R9
(, , )
→ R3
↦
→
 (, , ),
:
Ω × R3 × R9
(, , )
→ R3
↦→ (, , ),
:
Ω×Ω
(, )
→ R9
↦
→
(, ).
are measurable and satisfying the following conditions:
‖ (, , )‖ 6 0 (‖‖ + ‖‖) + 1 ()
|(, , )| 6 1 (‖‖ + ‖‖) + 2 ()
(5)
(6)
where  () ∈ 2 (Ω),  = 1, 2.
Moreover  and  are continuously differentiable with respect to the correspondent
(, ) almost at each fixed point  ∈ Ω , and
|′ | + |′ | 6 2 ,
|′ |
+
|′ |
∀ ∈ R3 , ∀ ∈ R9
3
6 3 ,
∀ ∈ R , ∀ ∈ R
9
(7)
(8)
at almost every  ∈ Ω, where  is a constant for  = 0, 1, 2, 3.
The function  defines a continuous integral operator 2 (Ω) → 2 (Ω), with the
following form:
()() =
∫︁
Ω
(, )().
(9)
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
21
In the same way, the following operators have been defined in [1]:
 : 21 (Ω) → 2 (Ω)
 ↦→ [ ()]() =  (),
(10)
 : 21 (Ω) → 2 (Ω)
 ↦→ [()]() = (),
(11)
[ ()]() =  (, (), ∇())
(12)
[()]() = (, (), ∇())
(13)
and
by the formulae:
and the operators:
 ′ () : 21 (Ω) → 2 (Ω)
ℎ() ↦→ [ ′ ()]() =  ′ ()ℎ,
and
′ : 21 (Ω) → 2 (Ω)
ℎ() ↦→ [′ ()ℎ]() = ′ ()ℎ,
depending on the parameter  ∈ 21 (Ω) by the formulas:
[ ′ ()]ℎ() = ′ (, (), ∇())ℎ() +
3
∑︁
′ 
=1
ℎ()

(14)
and
[′ ()]ℎ() = ′ ℎ() +
3
∑︁
=1
′ 
ℎ()
,

(15)
where ′  , ′ , ′  have for argument (, (), ∇()) (the notations  ′ () and ′ ()
are in [2]).
Let
{︃
}︃
(︂ )︂
=
∈
21 (Ω)
: ∃! ∈
′
∘
21
, ∃! ∈ , 1 () = (, ) ,
‖‖ = ‖ ‖(︀ ∘1 )︀′ + ‖‖ .
2
Assuming that [1 ()]() = △ − ∇ ,
[1 ()]() ≡  (, (), ∇()) +
∫︁
(, )(, (), ∇()).
Ω
It has been prooved in [1] that the operator
1 :
(, ‖ ‖ )

(︀ ∘ )︀′
→ 21 × 
↦
→
1 () = (, )
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Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
∘
is an isomorphism. Where 21 is the Hilbert space of vector functions, obtained by
∙
completing (Ω) according to the standard corresponding scalar product:
(, ) =
∫︁
( +   ),
Ω
∙
(Ω) is the set of infinitely differentiable vector functions and  has been defined by
 ∈  ≡ { ∈ 2 () / ∃  ∈ 21 (Ω),  = 0, | = }
with ‖‖ =  {‖‖21 (Ω) :  ∈ 21 (Ω),  = 0, | = }.
In [1], it was shown that by choosing  = (2 , 3 ) with the operators  and
, satisfying the conditions (5) − (8) and if there is a number 0 > 0 such that for any
, we have 0 <  < 0 then the problem
∫︁
⎧
⎪

()
≡
△()
+

(,
(),
∇())
+
(, )(, ()∇()) = ∇() +  (),
⎪
⎪
⎨
Ω
⎪
div () = 0,
⎪
⎪
⎩
| () = ()
∘
has a unique solution  = (, ) for all  ∈ 21 and  ∈  and more:
∘
∘
1)  : 21 ×  → 21 (Ω) is − continuous and − differentiable on 21 × ;
∘
2) the operator  is strongly differentiable on 21 ×  as a mapping on the space
(21 (Ω), ), where  is a weak topology in 21 (Ω).
We also obtained in [1], that when the solution  = (, ) is − continuous
∘
and − differentiable as a mapping from 21 ×  to  , then  is −continuous and
− differentiable as a mapping from 2 (Ω) ×  in 21 (Ω). This is deduced from the
(︀ ∘ )︀′
(︀
(︀ ∘ )︀′ )︀
continuity of  from  in 21 (Ω) and 2 (Ω) in 21 , this is (Ω) ⊂ 2 (Ω) ⊂ 21 .
To obtain the result above stated, we had to show that the operators  and  are
−continuous and −differentiable on 21 (Ω) and ′ =  * . Similary, it was shown
that, since  is a continuous linear map and that the operators  and  satisfy the
Lipschitz condition, then  ∘  also satisfies the Lipschitz condition.
Therefore, what conditions the command applied to the system (for disruption)
should be submitted to, so that the associated solution to the command coud be
unique?
The problem is to choose a command  from 0 , where 0 is a convex set in 2 (Ω),
such that for the solution () of the system (1)–(3), depending of that command  ,
constraints persist, which are given in the form of inequalities
 ( ) 6 0,
 = 1, 1 ,
(16)
 = 1 + 1, 1 + 2 ,
(17)
given in the form of equalities
 ( ) = 0,
and that in addition to this, the functional 0 ( ) takes the smallest possible value
0 ( ) = inf 0 ( ).
0
(18)
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
23
Such control is called optimal.
Definition 1. The function  = ( ) is called generalized solution of system (1)–
∘
(3) in 21 (Ω) , if it satisfies the integral identity  = 0, ∀ ≡ () ∈ 21 (Ω), where
]︂
∫︁
∫︁ [︂
∫︁
 = −   d
 (, ()) + (, )(, ())d −  () ()d = 0. (19)
Ω
Ω
Ω
Suppose that  is sufficiently smooth. Then
−
)︂
∫︁ ∑︁
∫︁ ∑︁
∫︁
3
3 (︂
3
∑︁
  2 
 

d = 
d
−


 d =

 
 
2
Ω =1
Ω =1
=
∫︁ ∑︁
3
Ω =1
3
∑︁
Ω
 2 
 2 d − 

∫︁
=1
3
∑︁


d = ⟨

Ω
=1
(︂
)︂

 ,  ⟩,
M  −
 
⟨ M  ,  ⟩ = ⟨ M , ⟩ by condition (3).
=1
Thus we obtain
 = ⟨ M  + ( ′ () + ′ ()), ⟩ − ⟨,  ⟩ = 0.
(20)
Theorem 1. Suppose that under the conditions of (4) (see [1])  is a solution of
system (1)–(3), corresponding to the control  () ∈ 0 , where
(︂
)︂

 () =  () +   () −  () , (0 6  6 1),
and   () is a solution relative to the control   () ∈ 0 . Then
⃦
⃦
‖  () − ()‖ 1 (Ω) 6  ⃦  () −  ()⃦2 (Ω) .
2
(21)
Proof. Remark that () =   () − () satisfy the integral identity:
]︂
∫︁
∫︁ [︂
∫︁
∫︁
∘


   −   −  + ( − )d d +  d = 0, ∀ ∈ 21 (Ω). (22)
Ω
Ω
Ω
Ω
So, using condition (16) (see [1]) and the restriction (21) (see [1], for  = 0), we obtain
inequality (21).
3.
Derivation of the Functional
Consider the functional
( ) =
∫︁
 (, ()),  ())d,
Ω
Let’s prove that  is differentiable in 2 (Ω).
(23)
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Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
3.1.
Formula for the Gradient of the Function
Consider the problem (1)–(3) with a disrupted control   ∈ 2 (Ω), which is linked
to the solution   () of the problem and the value of the functional (  ).
Denote the variations by:  =   − ,  =   −  . We have
]︂
∫︁ [︂
M  =M ( ) = (  ) − ( ) =
 (,   ,   ) −  (, ,  ) d,
Ω
 (,   ,   ) −  (, ,  ) =  (,   ,   ) −  (, ,   ) +  (, ,   ) −  (, ,  ) =
=
∫︁1
 (,  + ,   )d +  (, ,   ) −  (, ,  ) =
0

=  (, ,  ) −  (, ,  ) +  (, ,  ) +
∫︁1 [︂
]︂
 (, ,
̂︀  ) −  (, ,  ) d =

0

=  (, ,  ) −  (, ,  ) +  (, ,  ) +
∫︁1 [︂
]︂
 (, ,
̂︀  ) −  (, ,  ) d+


0
[︂
]︂

+  (, ,  ) −  (, ,  ) .
Then
M =
]︂
∫︁
∫︁
 (, ,   ) −  (, ,  ) d +  (, ,  ) + (1 + 2 )d,
∫︁ [︂
Ω
Ω
where
1 =
∫︁1 [︂
]︂
 (, ,
̂︀   ) −  (, ,   ) d,
Ω
̂︀ =  + ,
0
(24)
[︂
]︂

2 =  (, ,  ) −  (, ,  ) .
As the functions  and  satisfy respectively the integral identities (in this case we
use relation (20)), so their difference  =   −  also will satisfy the identity (20).
Taking into account this fact, we have:
M =
∫︁ [︂
]︂
∫︁
 (, ,  ) −  (, ,  ) d +  (, ,  )d+

Ω
Ω
(︂
)︂
∫︁
∫︁


+ ⟨ M  +   −  + ( − )d , ⟩ − ⟨,  ⟩ + (1 + 2 )d. (25)
Ω
Ω
In the last expression, taking into account the conditions on  and , also taking into
account formulas (25) and (26) (see [1]), we rewrite the following:
∫︁
Ω
(︂

)︂
  −  d =
∫︁
Ω
(︂
)︂

  (,  ()) −  (, ()) d =
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
=
∫︁

[︂ ∫︁1
]︂
∫︁
∫︁
 (,  + ) d =  (, ())()d + ()3 d =
0
Ω
25
Ω
=
∫︁
Ω
()[ ′ ()]()()d +
3 =
()3 d, (26)
Ω
Ω
where
∫︁
∫︁1 [︂
]︂
 (, ())
̂︀
−  (, ()) ()d,
̂︀ =  + ,
(27)
0
and
∫︁

Ω
[︂ ∫︁
]︂
[︂ ∫︁
(︂
)︂ ]︂
∫︁
(, ) (,   ()) − (, ()) d d =
(  − )d d = ()
Ω
Ω
=
∫︁
()
Ω
=
∫︁
Ω
(, 
[︂ ∫︁1
]︂ ]︂
 (,  + ) d d =
0
Ω
[︂ ∫︁
]︂
[︂ ∫︁
]︂
∫︁
()
(, ) (, ())()d d + ()
(, )4 d d.
Ω
∫︁
[︂ ∫︁
Ω
Ω
Ω
]︂
]︂
[︂ ∫︁
∫︁ [︂ ∫︁
()(, ) (, ())()d d =
()
(, ) (, ())()d d =
Ω
Ω
Ω
∫︁ [︂ ∫︁
=
Ω
Ω
]︂
]︂
∫︁ [︂ ∫︁
()(, )d  (, ())()d =
()(, ) (, ())d ()d =
Ω
Ω
Ω
∫︁
(′* () *  )()()d,
=
Ω
where ( *  )() =
∫︁
Ω
(, ) ()d. In what follows, as a matter of convenience,
we will simply write  but not  . Thus,
[︂ ∫︁
]︂
]︂
∫︁
∫︁ [︂ ∫︁
∫︁
′*
*

()
(, )4 d ()d.
(, )( − )d d = ( () )()d +
Ω
Ω
Ω
where
4 =
Ω
Ω
∫︁1 [︂
]︂
 (, )
̂︀ −  (, ) d.
(28)
0
 (, ,   ) −  (, ,  ) =  (, ,  ) + 5 ,
where
5 =
∫︁1 [︂
]︂
 (, , ̂︀) −  (, ,  )  d,
with ̂︀ =  + .
0
Taking into account formulas (26), (27), (29) and (25) we obtain
(29)
(30)
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Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
M =
∫︁
 (, ,  ) d +
Ω
∫︁
 (, ,  )d+
Ω
+ ⟨ M  +  ′ (, ) + ′* () * , ⟩ − ⟨,  ⟩+
)︂
∫︁
∫︁
∫︁ (︂ ∫︁
+ (1 + 2 + 5 )d + 3 d +
4  d. (31)
Ω
Ω
Ω
Ω
Remark 1. The transformations in the formula (31) are true only for the functions , sufficiently “smooth” and are issued only by the evidence of obtaining the
conjugate form of the problem. For the following transformations, consider the following conjugate problem:
 M  +  ′ (()) + ′ (()) *  = − .
(32)
From the existence (see Theorem 2 (see [3, p. 54] and Theorem of Hadamard) of
the solution of the conjugate problem (32), we finally have the expression for M 
M =
∫︁
 (, ,  ) d +
Ω
∫︁
 d +
Ω
+
∫︁
3 d +
Ω
where
=
(1 + 2 + 5 )d+
Ω
∫︁ (︂ ∫︁
Ω
∫︁
∫︁
)︂
4 d d =
Ω
Ω
(1 + 2 + 5 )d +
Ω
]︂
 (, ,  ) +   d + ,
∫︁ [︂
∫︁ (︂
3 +
Ω
∫︁
)︂
4 d d.
Ω
(︁
)︁
In assessing the balance of development, one can show that  =  ‖ ‖2 (Ω) .
Due to the fact that  satisfies the Lipschitz condition with respect to the group of
arguments (,  ) and using Theorem 1, we have
⃒
⃒
⃒ ⃒
⃒∫︁
⃒ ⃒∫︁ [︂ ∫︁1 (︂
)︂
]︂ ⃒
⃒
⃒ ⃒
⃒
⃒ 1 d⃒ = ⃒
 (, ,
̂︀   ) −  (, ,   ) d d⃒⃒ 6
⃒
⃒ ⃒
⃒Ω
⃒
⃒ ⃒Ω 0
∫︁ ∫︁1
6
| (, ,
̂︀   ) −  (, ,   )| || dd 6
Ω 0
∫︁
∫︁1
6
=
(︂
)︂


 ‖̂︀
 − ‖ 1 (Ω) + ‖ −  ‖2 (Ω) || dd =
2
Ω 0
∫︁ ∫︁1
1
 ‖‖ 1 (Ω) || d =  ‖‖ 1 (Ω)
2
2
2
Ω 0
1
∫︁
|| d 6
Ω
1
1
1
2
6 (Ω) 2 ‖‖ 1 (Ω) ‖‖ 1 (Ω) = (Ω) 2 ‖‖ 1 (Ω) 6
2
2
2
2
2
(︀
)︀
1
1
2
6 (Ω) 2  ‖‖2 (Ω) =  ‖‖2 (Ω) ,
2
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
27
⃒ ⃒
⃒
⃒
⃒ ⃒∫︁ [︂ ∫︁1 (︂
⃒∫︁
)︂
]︂ ⃒
⃒ ⃒
⃒
⃒
⃒ 5 d⃒ = ⃒
 (, , ̂︀) −  (, ,  )  d d⃒⃒ 6
⃒ ⃒
⃒
⃒ ⃒Ω 0
⃒Ω
⃒
∫︁ ∫︁1 ⃒
⃒
⃒
⃒
⃒ (, , ̂︀) −  (, ,  )⃒ | | dd 6
6
Ω 0
∫︁ ∫︁1 (︂
⃦
⃦
⃦
⃦
 ‖ − ‖ 1 (Ω) + ⃦̂︀ −  ⃦
6
2
2 (Ω)
Ω 0
=
∫︁ ∫︁1
⃦
⃦
⃦
⃦
 ⃦̂︀ −  ⃦
2 (Ω)
Ω 0
1
2
=  ‖ ‖2 (Ω)
∫︁
)︂
| | dd =
∫︁ ∫︁1
| | dd =
 ‖ ‖2 (Ω) dd =
Ω 0
(︀
)︀
1
1
2
| | d 6 (Ω) 2  ‖ ‖2 (Ω) =  ‖ ‖2 (Ω) ,
2
Ω
and
⃒
⃒ ⃒
⃒
⃒∫︁
⃒ ⃒∫︁ (︂
⃒
)︂
⃒
⃒ ⃒
⃒

⃒ 2 d⃒ = ⃒
 (, ,  ) −  (, ,  ) d⃒⃒ 6
⃒
⃒ ⃒
⃒Ω
⃒ ⃒Ω
⃒
∫︁ ⃒
⃒
6 ⃒ (, ,   ) −  (, ,  )⃒ || d 6
Ω
∫︁
6
)︂
(︂
⃦ 
⃦
⃦
⃦
 ‖ − ‖ 1 (Ω) +  −  2 (Ω) || d =
2
Ω
=
∫︁
∫︁
∫︁
⃦ 
⃦
⃦
⃦
  −  2 (Ω) ||  =  ‖ ‖2 (Ω) || d =  ‖ ‖2 (Ω) || d 6
Ω
Ω
Ω
1
1
1
1
2
6 (Ω) 2 ‖ ‖2 (Ω) ‖‖ 1 (Ω) 6 (Ω) 2  ‖ ‖2 (Ω) .
2
2
2
The other members are evaluated in the same way. For the variation of the functional
M , we have finally
]︂
∫︁ [︂
)︀
(︀
M =
 (, ,  ) +   d +  ‖ ‖2 (Ω) .
Ω
Let’s introduce the following function

(, (),  (), ()) = (, , , ) =  (, ,  ) + .
In this case, the formula for the variation will take the following form
M =
∫︁
)︀(︀
)︀
(︀
)︀
 (︀
, ,  ,    −  d +  ‖ ‖2 (Ω) .

Ω
So we’ve just proved the following theorem
Theorem 2. Suppose that all the conditions of paragraph 1 [1] about functions 
and  are satisfied, as well as the requirements of paragraph 1 about  .
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Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
Then the functional ( ) is differentiable with respect to  , and its derivatives at
)︀
 (︀
the point  are expressed by the formulae  ( ) =
, ,  ,  .

4.
Necessary Conditions of Optimality
Let  =  () ∈ 0 , with  () an optimal control. Consider an arbitrary command
 (), with  =  () ∈ 0 .
Let’s find the variation   of the optimal control  in the direction of ( −  ) as
follows:
(︀
)︀
  () =  () +   () −  ()
(33)
(︀
)︀
 =   −  =   −  ,
In the variations,  is always the same and   ∈ 0 . This is satisfied for example,
when  ∈ [0, 1], because 0 is a convex set.
4.1.
First Variation of the Functionals
Consider a family of functions  ,  = 0, 1 + 2 , where
 (, , ,  ) =  (, ,  ) +  .
(34)
The functions  are solutions of the conjugate problems. So
M =
∫︁
)︀(︀
)︀
(︀
)︀
 (︀
, ,  ,    −  d +  ‖ ‖2 (Ω) ,

 = 0, 1 + 2 ,
(35)
Ω
the first variation  of the functional  ( ) at point  is determined as follows:
 =  ( ) = lim
→0
M  ( )
.

(36)
As  −  = ( −  ) and the norm ‖ −  ‖2 (Ω) for any fixed  , are fixed finite
quantities, then
⎧
⎫
⎨∫︁
⎬
)︀
 (︀
1
 = lim
, ,  ,  ( −  )d + () =

→0  ⎩
⎭
Ω
=
∫︁
)︀
 (︀
, ,  ,  ( −  )d

(37)
Ω
where
∘
 =
∫︁

( −  )d,

(38)
Ω
∘
where  is the function  with the arguments related to optimal control  .
4.2.
Establishment of the Necessary Conditions of Optimality
Let  be a set of parameters:
{︀
}︀
 = ( −  ),  > 0,  ∈ 0
(39)
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
29
{︀
}︀
Or simply ,  = ( −  ) , giving the variation of the optimal control
  =  + ( −  ).
)︀
(︀
Then  is the family of variations of the functionals  = 0 , 1 , . . . , 1 +2 ,
where
∘
∫︁
 = 

( −  )d,

 = 0, 1 + 2 .
(40)
Ω

All kinds of , whose form looks like the family of variation of the functional { } =
1 ⊂  1 +2 +1 . Let’s show that 1 is a cone in  1 +2 +1 with its apex at the zero
point.
It is clear that  = 0 ∈  1 +2 +1 corresponds to the family  = {0}, with  = 0.
We have implicitly 0 ∈ 1 .
}︀
{︀
Consider the family (︀ = ( −  ) . For that
)︀ family, there is a vector of variation
of the functionals  = 0 , 1 , . . . , 1 +2 ∈ 1 .
)︀
(︀
Consider  = 0 , 1 , . . . , 1 +2 , where

= 
∫︁
Ω
∘

( −  )d,  > 0.

Consider the family  = {( −  )} too.
Such a family is admissible, like the corresponding vector of variation of the functionals  ∈ 1 . Moreover, it is clear that  =  , as   =  , we conclude
that,  ∈ 1 and 1 is a cone.
We now show that the cone 1 is convex. For this it is sufficient to show that
∀1 ,{︀2 ∈ 1 }︀their sum 1 + 2 ∈ 1 . Consider {︀1 generated
}︀ by the family
1 = 1 ( −  ) , and 1 generated by the family 2 = 2 ( −  ) . Consider the
set
{︀
}︀
1 + 2 = ( −  ) , or  = 1 + 2 ,
 =   1 + (1 −  ) 2 , 0 6  =
1
6 1.
1 + 2
As 0 is convex, the set 1 + 2 is admissible. So is the correspondent vector of
variation of the functionals 1 +2 ∈ 1 .
In expression (40) for 1 +2 ,  = 0, 1 + 2 , consider the expression
∘

∫︁

( −  )d =

(︂
1 + 2
Ω
)︂ ∫︁
∘

(  1 − (1 −  ) 2 −  )d =

Ω
∘
= 1
∫︁
Ω
Thus, 1 +2 = 1 + +2 ,
the cone 1 is convex.
 1
( −  )d + 2

∘
∫︁
 2
( −  )d.

Ω
 = 0, 1 + 2 , then 1 + 2 = 1 +2 ∈ 1 and
Definition 2. The contsraints at the point  , part of restrictions  ( ) 6 0, for
which  ( ) = 0 are called active. Those for which  ( ) < 0 are called inactive at
that point.
30
Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
To begin, suppose that all the restrictions (16) are active.
Consider the set
{︀
}︀
1− =  ∈  1 +2 +1 :  = (0 , 1 , . . . , 1 , 0, . . . , 0),  < 0,  = 0, 1
a negative angle in  1 +2 +1 . It is clear that 1− is a cone in  1 +2 +1 .
Lemma 1. The cone 1 , built for optimal control and the cone 1− are divided
in  1 +2 +1 by the hyperplane Γ, defined by the nontrivial functional * :
1∑︁
+2
(︀
)︀*
* = (0 , 1 , . . . , 1 +2 ) ∈  1 +2 +1 =  1 +2 +1 ,
| | > 0, for  > 0,  = 0, 1 ,
=0
and the rests  ,  = 1 + 1, 1 + 2 may have any sign. The condition of separation
of 1 and 1− takes the following form
∀ ∈ 1 , ∀ ∈ 1− .
⟨* ,  ⟩ 1 +2 +1 > ⟨* , ⟩ 1 +2 +1 ,
(41)
This follows from the known theorem (see [4], p.224 or [5], 3.1).
Theorem 3. Let  be a normed space, 1 a convex set in , * ∈ 1 a local
minimum point in the problem
0 () → inf,
 () 6 0,
 () = 0,
 = 1, 1 ,
 = 1 + 1, 1 + 2 ,
 ∈ 1 ,
where  ,  = 0, 1 + 2 , -differentiable at the point * and  ,  = 1 + 1, 1 + 2
continuous in the neighborhood of the point * .
Then there are numbers 0 , 1 , 2 , . . . , 1 +2 such that
* = (0 , 1 , 2 , . . . , 1 +2 ) ̸= 0,
*
*
*
⟨ℒ( ,  ),  −  ⟩ > 0,
∀ ∈ 1 ,
1 > 0, . . . , 1 > 0,
0 > 0,
*
  ( ) = 0,
 = 1, 1 + 2
here ℒ(* , * ) = 0 0′ (* )+1 1′ (* )+. . .+1 +2 ′ 1 +2 (* ) the gradient of the function
ℒ(, * ) with variable  ∈ 1 at the point  = * .
Then, using inequality (41) in which  → 0, we obtain
⟨* ,  ⟩ 1 +2 +1 > 0, ∀ ∈ 1 ,
(42)
That is, for any family  like in (39).
Inequality (3) is well demonstrated, assuming that all restrictions (16) are actives.
Now consider the general case.
Let  = { : 1 6  6 1 ,  ( ) = 0} be the set of all constraints at the point
 among all the restrictions like (16). The other constraints in the formula (16) are
inactive at the point  , that is  ( ) < 0, 1 6  6 1 , but  ̸∈ . So thanks to
the continuity of the functional  with respect to the control  for small disrupted
controls, non-active constraints are not affected. Therefore, we can not take account
of them. In this case we will examine variations of functionals only for the active
constraints and their vectorial variations
{︀
}︀
 = 0 , { }∈ , 1 +2 , . . . , 1 +2 .
Bailly Balé, Yoro Gozo, Assui Kouassi Richard Necessary Optimality Condi . . .
31
+1 +1
Let’s build the cone 1 = { } ⊂ 
. The corresponding cone is
{︀
}︀
1− =  ∈  +1 +1 :  = (0 , { }∈ , 0, . . . , 0),  < 0 ,
and by taking the above steps till (42), we obtain
0 0 +
∑︁
  +
1∑︁
+2
  > 0, ∀ ∈ 1 .
(43)
=1 +1
∈
Let  = 0 for all  : 1 6  6 1 ,  ̸∈ . Then (43) takes the form of (42).
Thus, for all inactive contraints  ( ) < 0 corresponding to  = 0, and for the
active contraints  ( ) = 0,
  ( ) = 0,  = 1, 1 + 2
(44)
so the condition (42) is verified too. From this we deduce the necessary conditions of
optimality of the control.
∑︁1 +2
  (), where  ()
Let’s introduce the following function: Ψ = Ψ() =
=0
is the solution of the conjugate problem (32). Multiplying (32) by  and making a
summation for all  = 0, 1 + 2 , then the function Ψ() will be a solution of the
problem:
′
′
*
 M Ψ() −  (, ())Ψ() +  (, ()) Ψ = −
1∑︁
+2
  .
(45)
=0
Let’s introduce the functions
ℋ(, , , Ψ) =
1∑︁
+2
  (, (),  (), Ψ()).
(46)
=0
Using the formula (32) for  and taking into account the introduced function (),
we can write formula (44) in expanded form:
ℋ(, , , Ψ) ≡ ℋ(, (),  (), Ψ()) =
1∑︁
+2
  (, (),  ()) + Ψ() ().
(47)
=0
Let’s consider the family  = { −  } for all  ∈ 2 (Ω). To this family we associate
the variation vector of the functional
(︀
)︀
 = 0 , 1 , . . . , 1 +2 ∈ 1 ,
and inequality (43) persists:
1∑︁
+2
  > 0.
=0
Replacing  by their respective expressions from (40) using formulas (47), we
obtain
∫︁
ℋ
(48)
(, (),  (), Ψ())( −  )d > 0, ∀ ∈ 0 .

Ω
So we have just proved the following theorem:
Theorem 4 (The principle of linearized minimum). Suppose that all of the
conditions of theorems 1 and 2 are satisfied. Then, for the optimal control  () ∈ 0
32
Bulletin of PFUR. Series Mathematics. Information Sciences. Physics. No 2, 2012. Pp. 19–32
it is necessary that there exists a nontrivial vector
(︀
)︀
* = 0 , 1 , . . . , 1 +2 ,
1∑︁
+2
| | > 0,
=0
where  > 0 for  = 0, 1 and the conditions (48) as well as the conditions
  ( ) = 0
 = 1, 1 + 2
(conditions (45)),
are satisfied; where function () is the solution of the problem (1)–(3), Ψ() is the
solution of the conjugate problem (46) associated to  () and the function ℋ is defined
in (47).
References
1. Yoro G., Bailly B., Assui K. R. Unicité et dérivabilité de la solution des systumes
hydrodynamiques stationnaires non linéaires et perturbés // Revist. — 2010. —
Vol. 15, B. — Pp. 91–107.
2. Opérateurs intégraux dans les espaces des fonctions additionnables / M. A. Crasnocelski, P. P. Zabrehiko, E. N. Poustilnik, P. E. Sobolevski. — M.: Naouka, 1966. —
496 p.
3. Ladyjenskana O. A. Les questions mathématiques de la dynamique de la ténacité
des liquides non compressibles. — M.: Naouka, 1970.
4. Vassilev F. P. Méthodes numériques pour la résolution des problumes des extrema. — M.: Naouka, 1980. — 520 p.
5. Souhinine M. F. La régle du multiplicateur de Lagrange dans les espaces localement convexes // Revue scientifique de mathématiques de Sibérie. — 1982. —
Vol. 23, No 4. — Pp. 153–165.
УДК 517.917
Необходимые условия оптимальности для стационарной
нелинейной возмущённой задачи гидродинамики в
ограниченной области
*
Бале Байлли , Гозо Йоро† , Ришар Ассюй Куасси‡
*
Факультет математики и информатики
Университет Кокоди Абиджан
Кот Д’Ивуар, 22 BP 582 Абиджан 22
†
Кафедра математики и информатики
Университет Абобо — Аджамэ Абиджан
Кот Д’Ивуар, 02 Вр 801 Абиджан 02 UFR — SFA
‡
Кафедра математики и информатики
Политехнический институт им. Феликса Уфует Буагни Ямуссукро
Кот д’Ивуар, а/я 1083, Ямуссукро
Цель работы состоит в том, чтобы установить оптимальные необходимые условия,
которые могут позволить нам решить задачу относительно границ данной области. В
предлагаемой статье исследуется частный случай, а именно, в линеаризованную систему
добавлены нелинейные члены, позволяющие более точно описать движение жидкости,
и вместе с тем допускающие однозначную разрешимость полученной нелинейной возмущённой краевой задачи.
Ключевые слова: необходимые условия оптимальности, команда, оптимальная команда, уникальность, возмущение, линеаризация, нелинейность.
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