2 Formulation of the problem
We are interested in finding the problem of the minimum of the functional
J
α
(
v
)
=
∥
ψ
(
⋅
,
L
)
−
y
∥
L
2
(
0
,
l
)
2
+
α
∥
v
−
ω
∥
H
2
in the set
V
≡
{
v
=
(
v
0
,
v
1
,
φ
0
,
φ
1
)
,
v
m
∈
L
2
(
0
,
l
)
,
v
1
(
z
)
≥
0
,
∀
0
z
∈
(
0
,
L
)
,
∥
v
m
∥
L
2
(
0
,
l
)
≤
b
m
,
φ
m
∈
L
2
(
0
,
l
)
,
∥
φ
m
∥
L
2
(
0
,
l
)
≤
d
m
,
m
=
0
,
1
}
under the condition
![]()
![]()
![]()
where i
=
−
1
,
a
0
>
0
, l
>
0
, L
>
0
, α
≥
0
,
b
0
≥
0
,
b
1
>
0
,
d
0
>
0
,
d
1
>
0
are numbers, x
∈
[
0
,
l
]
, z
∈
[
0
,
L
]
,
Ω
z
=
(
0
,
l
)
×
(
0
,
z
)
, Ω
=
Ω
L
, y
(
x
)
, f
(
x
,
z
)
are complex valued measurable functions and satisfy the conditions
![]()
![]()
respectively, ω
=
(
ω
0
,
ω
1
,
ϖ
0
,
ϖ
1
)
and H
=
(
L
2
(
0
,
l
)
)
2
×
(
L
2
(
0
,
L
)
)
2
.
L
2
(
0
,
l
)
is a Hilbert space that consists of all functions in (
0
,
l
)
, which are measurable and square-integrable.
L
2
(
Ω
)
is the well-known Lebesgue space consisting of all functions in Ω, which are measurable and square-integrable.
The problem of finding a function ψ
=
ψ
(
x
,
z
)
≡
ψ
(
x
,
z
;
v
)
under the condition (2)-(4) for each ∀
v
∈
V
, which is a boundary value problem, is a function for Eq. (2).
Generalized solution of this problem is a function ψ=ψ(x,z)≡ψ(x,z;v) belonging to the
C
0
(
[
0
,
L
]
,
L
2
(
0
,
l
)
)
, and it satisfies the integral identity
![]()
for ∀
η
∈
C
0
(
[
0
,
L
]
,
L
2
(
0
,
l
)
)
.
3 Existence and uniqueness of a solution of the optimal control problem
In this section, we prove the optimal control problem using the Galerkin method and the existence and uniqueness of a solution of the problem (1)-(4).
Theorem 1 Suppose that a function f satisfies the condition (5). So, for each ∀v∈V, the problem (2)-(4) has a unique solution, and for this solution, the estimate
∥
ψ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
0
(
∥
φ
∥
L
2
(
0
,
l
)
2
+
∥
f
∥
L
2
(
Ω
)
2
)
is valid for ∀
z
∈
[
0
,
L
]
. Here, the number
c
0
>
0
is independent of z.
Proof Proof can be done by processes similar to those given in 7. □
Theorem 2 Let us accept that the conditions of Theorem 1 hold and y
∈
L
2
(
0
,
l
)
is a given function. Then there is such a set G dense in H
≡
[
L
2
(
0
,
L
)
]
2
×
[
L
2
(
0
,
l
)
]
2
that the optimal control problem (1)-(4) has a unique solution ∀
ω
∈
G
and α
>
0
.
Proof Firstly, let us show that
J
0
(
v
)
=
∥
ψ
(
⋅
,
L
)
−
y
∥
L
2
(
0
,
l
)
2
is continuous on the set V. Let us take an arbitrary ∈V, and let v
+
Δ
v
be an increment of the v for the Δ
v
∈
H
. Then the solution ψ
(
x
,
z
;
v
)
of the problem (1)-(4) will have an increment Δ
ψ
=
Δ
ψ
(
x
,
z
)
=
ψ
(
x
,
z
;
v
+
Δ
v
)
−
ψ
(
x
,
z
;
v
)
. Here, the function
ψ
Δ
(
x
,
z
)
=
ψ
(
x
,
z
;
v
+
Δ
v
)
is the solution of (2)-(4). On the basis of the assumptions and conditions (2)-(4), it can be shown that the function Δ
ψ
(
x
,
z
)
is a solution of the following boundary value problem:
![]()
![]()
![]()
Because the problem (10)-(12) and the problem (2)-(4) are the same type problems, we can write the following estimate the same as (8):
∥
Δ
ψ
(
⋅
,
z
)
∥
2
≤
c
4
∥
Δ
ψ
∥
L
2
(
0
,
l
)
2
+
∥
Δ
v
0
ψ
+
i
Δ
v
1
ψ
∥
L
2
(
Ω
)
2
,
∀
z
∈
[
0
,
L
]
.
If we use estimate (13) then we can write the following estimate:
∥
Δ
ψ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
5
∥
Δ
v
∥
H
2
,
∀
z
∈
[
0
,
L
]
.
c
5
>
0
is constant that does not depend on Δv.
Now, let us evaluate the increment of the functional
J
0
(
v
)
on v
∈
V
. Using formula (9) we can write the equality as
Δ
J
0
(
v
)
=
J
0
(
v
+
Δ
v
)
−
J
0
(
v
)
=
2
∫
0
l
Re
(
ψ
(
x
,
L
)
−
y
(
x
)
)
Δ
ψ
¯
(
x
,
L
)
d
x
+
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
2
.
Using the Cauchy-Bunyakowski inequality and estimates (8) and (14), we write the inequality as
|
Δ
J
0
(
v
)
|
≤
c
6
∥
Δ
v
∥
H
2
,
∀
v
∈
V
,
where
c
6
>
0
is a constant that does not depend on Δv. This inequality shows that the functional J0(v) is continuous on the set V. On the other hand,
J
0
(
z
)
≥
0
for ∀
z
∈
V
; therefore, J0(v) is bounded on V. The set V is closed, bounded on a Hilbert space H. According to Theorem (Goebel) in 8, there is such a set G dense in H that optimalcontrol problem (1)-(4) has a unique solution for α>0 and ∀ω∈G. Theorem 2 is proven. □
3.1 Fréchet diffrentiability of the functional
In this section, we prove the Fréchet differentiability of a given functional. For this purpose, we consider the following adjoint boundary value problem:
![]()
![]()
![]()
Here, the function ψ
=
ψ
(
x
,
z
)
≡
ψ
(
x
,
z
;
v
)
is a solution of (2)-(4) for v∈V. The solution of the boundary value problem (17)-(19) corresponding to v∈V is a function φ
=
φ
(
x
,
z
)
that belongs to the space
C
0
(
[
0
,
L
]
,
L
2
(
0
,
L
)
)
and satisfies the integral identity
![]()
As seen, the problem (17)-(19) is an initial boundary value problem. This can easily be obtained by a transform θ
=
L
−
z
. Actually, if we do a variable transform θ=L−z, we obtain the boundary problem as
![]()
![]()
![]()
where
![]()
If we write the complex conjugate of this boundary value problem, we obtain the following boundary value problem:
![]()
![]()
![]()
where
F
(
x
,
θ
)
=
φ
˜
(
x
,
θ
)
¯
,
h
(
x
)
=
−
2
i
(
ψ
¯
(
x
,
L
)
−
y
¯
(
x
)
)
.
This problem is a type of (2)-(4) boundary value problem. As the right-hand side is equal to zero, and initial function h
(
x
)
belongs to the space L2(0,l) for ψ
∈
C
0
(
[
0
,
L
]
,
L
2
(
0
,
l
)
)
, y∈L2(0,l). By using Theorem 2, it follows that the solution of the bounded value problem (24)-(26) existing in the space C0([0,L],L2(0,l)) is unique, and the following estimate is obtained:
∥
F
(
⋅
,
θ
)
∥
L
2
(
0
,
l
)
2
≤
c
7
∥
h
∥
L
2
(
0
,
l
)
2
,
∀
θ
∈
[
0
,
L
]
.
If we use the problem (24)-(26) as a type of the conjugate problem (17)-(19), we obtain the initial bounded value problem (17)-(19) has a unique solution belonging to the space C0([0,L],L2(0,l)), and the following estimate is obtained:
∥
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
8
∥
ψ
(
⋅
,
L
)
−
y
∥
L
2
(
0
,
l
)
2
,
∀
z
∈
[
0
,
L
]
.
Here, the number
c
8
>
0
is independent of ψ and y. Now, using estimate (8) in this inequality, we easily write the following estimate:
∥
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
9
(
∥
φ
∥
L
2
(
0
,
l
)
2
+
∥
y
∥
L
2
(
0
,
l
)
2
+
∥
φ
∥
L
2
(
Ω
)
2
)
,
∀
z
∈
[
0
,
L
]
.
Here, the number
c
9
>
0
is constant.
Theorem 3 Let us accept that the conditions of Theorem 2 hold and ω
∈
H
is given. Then the functional
J
α
(
v
)
can be Frechet differentiable in the set V and the formula below for a gradient of the functional is valid:
![]()
Proof Let us evaluate the increment of the functional Jα(v) for the element ∀v∈V. We can write the following equation for the increment of the functional:
Δ
J
α
(
v
)
=
J
α
(
v
+
Δ
v
)
−
J
α
(
v
)
=
2
∫
0
l
Re
[
(
ψ
(
x
,
L
)
−
y
(
x
)
)
Δ
ψ
¯
(
x
,
L
)
]
d
x
+
2
α
∫
0
l
(
φ
0
(
x
)
−
ω
˜
0
(
x
)
)
Δ
φ
0
(
x
)
d
x
+
2
α
∫
0
l
(
φ
1
(
x
)
−
ω
˜
1
(
x
)
)
Δ
φ
1
(
x
)
d
x
+
2
α
∫
0
T
(
v
0
(
z
)
−
ω
0
(
z
)
)
Δ
v
0
(
z
)
d
z
+
2
α
∫
0
T
(
v
1
(
z
)
−
ω
1
(
z
)
)
Δ
v
1
(
z
)
d
z
+
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
2
+
α
∥
Δ
v
∥
H
2
.
The last formula can be written as follows:
Δ
J
α
(
v
)
=
J
α
(
v
+
Δ
v
)
−
J
α
(
v
)
=
∫
0
L
(
∫
0
l
Re
(
ψ
φ
¯
)
d
x
+
2
α
(
v
0
(
z
)
−
ω
0
(
z
)
)
Δ
v
0
(
z
)
d
z
−
∫
0
l
Im
(
ψ
,
φ
¯
)
d
x
Δ
v
1
(
z
)
)
+
∫
0
L
(
−
∫
0
l
Im
(
ψ
φ
¯
)
d
x
+
2
α
(
v
1
(
z
)
−
ω
1
(
z
)
)
)
Δ
v
1
(
z
)
d
z
+
∫
0
l
[
Im
(
φ
¯
(
x
,
0
)
)
+
2
α
(
φ
0
(
x
)
−
ω
˜
0
(
x
)
)
]
Δ
φ
0
(
x
)
d
x
+
∫
0
l
[
Re
(
φ
¯
(
x
,
0
)
)
+
2
α
(
φ
1
(
x
)
−
ω
˜
1
(
x
)
)
]
Δ
φ
1
(
x
)
d
x
+
R
(
Δ
v
)
,
where R
(
Δ
v
)
is defined as the formula
R
(
Δ
v
)
=
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
2
+
α
∥
Δ
v
∥
H
2
+
∫
Ω
Re
(
Δ
ψ
φ
¯
)
Δ
v
0
(
z
)
d
x
d
z
−
∫
Ω
Im
(
Δ
ψ
φ
¯
)
Δ
v
1
(
z
)
d
x
d
z
.
Applying the Cauchy-Bunyakowski inequality, we obtain:
|
R
(
Δ
v
)
|
≤
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
2
+
α
∥
Δ
v
∥
H
2
+
(
∥
Δ
v
1
∥
L
2
(
0
,
T
)
+
∥
Δ
v
0
∥
L
2
(
0
,
T
)
)
max
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
∥
φ
∥
L
2
(
0
,
L
)
.
If we use estimates (13) and (28) in this inequality, we obtain
|
R
(
Δ
v
)
|
≤
c
10
∥
Δ
v
∥
H
2
.
Here,
c
10
>
0
is a constant that does not depend on Δv. Hence, we write
R
(
Δ
v
)
=
o
(
∥
Δ
v
∥
H
)
.
By using equality (33), the increment of the functional can be written as
![]()
Considering this equality (34), and by using the definition of Fréchet differentiable, we can easily obtain the validity of the rule. Theorem 3 is proved. □
3.2 A necessary condition for an optimal solution
In this section, we prove the continuity of a gradient and state a necessary condition to an optimal solution in the variational inequality form using the gradient.
Theorem 4 Accept that the conditions of Theorem 3 hold and
v
∗
∈
V
is an optimal solution of the problem (1)-(4). Then the following inequality is valid for ∀v∈V:
![]()
Here, the functions
ψ
∗
(
x
,
z
)
≡
ψ
(
x
,
z
;
v
∗
)
,
φ
∗
(
x
,
z
)
≡
φ
(
x
,
z
;
v
∗
)
are solutions of the problems (2)-(4) and a conjugate problem corresponding to v∗∈V, respectively.
Proof Now, we prove that the gradient
J
α
′
(
v
)
is continuous at V. For this we show
![]()
![]()
![]()
![]()
for
∥
Δ
v
∥
H
→
0
.
In order to show (36), using the formula
J
α
v
0
′
(
v
)
=
∫
0
l
Re
(
ψ
φ
¯
)
d
x
+
2
α
(
v
0
(
z
)
−
ω
0
(
z
)
)
in (29), we can write the following equation:
![]()
Here, Δ
ψ
=
Δ
ψ
(
x
,
z
)
is the solution of the problem (9)-(11) and Δ
φ
=
Δ
φ
(
x
,
z
)
is the solution of the following problem:
![]()
![]()
![]()
This bounded value problem is a type of a conjugate problem. For this solution, the following estimate is valid:
∥
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
11
(
∥
Δ
v
0
φ
+
i
Δ
v
0
φ
∥
L
2
(
Ω
)
2
+
∥
Δ
ψ
(
⋅
,
L
)
∥
L
2
(
0
,
l
)
2
)
,
∀
z
∈
(
0
,
L
)
.
Here, the number
c
11
is constant.
Using (13) and (28), we write
∥
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
12
(
∥
Δ
v
∥
H
2
)
,
∀
z
∈
(
0
,
L
)
.
Here, the number
c
12
is constant. Using (13) and (45) and applying the Cauchy-Bunyakovski inequality, we obtain
|
J
α
v
0
′
(
v
+
Δ
v
)
−
J
α
v
0
′
(
v
)
|
≤
∥
ψ
Δ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
∥
Δ
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
+
∥
Δ
ψ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
∥
φ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
+
2
α
|
Δ
v
0
(
z
)
|
,
∀
z
∈
(
0
,
L
)
,
and then
![]()
If we use estimate (8), we can write the following inequality:
∥
ψ
Δ
(
⋅
,
z
)
∥
L
2
(
0
,
l
)
2
≤
c
13
,
∀
z
∈
[
0
,
L
]
.
Using this inequality and estimates (13), (28), and (45), we obtain
∥
J
α
v
0
′
(
v
+
Δ
v
)
−
J
α
v
0
′
(
v
)
∥
L
2
(
0
,
L
)
≤
c
14
∥
Δ
v
∥
H
.
Here, the number of
c
14
is constant. Similarly, we can prove the following inequality:
∥
J
α
v
1
′
(
v
+
Δ
v
)
−
J
α
v
1
′
(
v
)
∥
L
2
(
0
,
L
)
≤
c
15
∥
Δ
v
∥
H
.
If we use inequalities (48) and (49), we see that the correlations limit (36) and (37) is valid.
Now, we prove (38). To prove this using the formula
J
α
φ
1
′
(
v
)
=
Re
(
φ
¯
(
x
,
0
)
)
+
2
α
(
φ
1
(
x
)
−
ω
˜
1
(
x
)
)
in (29), we can write the following inequality:
J
α
v
0
′
(
v
+
Δ
v
)
−
J
α
v
0
′
(
v
)
=
I
m
(
Δ
φ
¯
(
x
,
0
)
)
+
2
α
Δ
φ
0
.
Here, Δ
φ
(
x
,
z
)
is a solution of the problem (41). Estimate (45) is valid for ∀z∈[0,L]. Therefore, the following estimate can be written at z
=
0
:
∥
φ
(
⋅
,
0
)
∥
L
2
(
0
,
l
)
2
≤
c
12
(
∥
Δ
v
∥
H
2
)
.
If this inequality is used in (49), we easily can write
∥
J
α
φ
0
′
(
v
+
Δ
v
)
−
J
α
φ
0
′
(
v
)
∥
L
2
(
0
,
l
)
≤
c
16
(
∥
Δ
v
∥
H
)
.
Similarly, if we use (39), we obtain
∥
J
α
φ
1
′
(
v
+
Δ
v
)
−
J
α
φ
1
′
(
v
)
∥
L
2
(
0
,
l
)
≤
c
17
(
∥
Δ
v
∥
H
)
.
We can see that (38) and (39) are valid by inequalities (51) and (52). That is,
J
α
∈
C
1
(
V
)
. On the other hand, V is a convex set according to the definition. So, the functional Jα(v) holds by the condition of Theorem (Goebel) in 8 at V. Therefore, considering Theorem 3, we obtain
〈
J
α
′
(
v
∗
)
,
v
−
v
∗
)
〉
H
≥
0
for ∀z∈V. Here, using (29), it is seen that the statement of Theorem 4 is valid. □