Open Access

Optimal control problem for stationary quasi-optic equations

Boundary Value Problems20122012:151

DOI: 10.1186/1687-2770-2012-151

Received: 1 October 2012

Accepted: 30 November 2012

Published: 28 December 2012

Abstract

In this paper, an optimal control problem was taken up for a stationary equation of quasi optic. For the stationary equation of quasi optic, at first judgment relating to the existence and uniqueness of a boundary value problem was given. By using this judgment, the existence and uniqueness of the optimal control problem solutions were proved. Then we state a necessary condition to an optimal solution. We proved differentiability of a functional and obtained a formula for its gradient. By using this formula, the necessary condition for solvability of the problem is stated as the variational principle.

Keywords

stationary equation of quasi optic boundary value problem optimal control problem variational problem

1 Introduction

Optimal control theory for the quantum mechanic systems described with the Schrödinger equation is one of the important areas of modern optimal control theory. Actually, a stationary quasi-optics equation is a form of the Schrödinger equation with complex potential. Such problems were investigated in [15]. Optimal control problem for nonstationary Schrödinger equation of quasi optics was investigated for the first time in [6].

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ1_HTML.gif
(1)
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 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equa_HTML.gif
under the condition
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ2_HTML.gif
(2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ3_HTML.gif
(3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ4_HTML.gif
(4)
where i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq1_HTML.gif, a 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq2_HTML.gif, l > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq3_HTML.gif, L > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq4_HTML.gif, α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq5_HTML.gif, b 0 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq6_HTML.gif, b 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq7_HTML.gif, d 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq8_HTML.gif, d 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq9_HTML.gif are numbers, x [ 0 , l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq10_HTML.gif, z [ 0 , L ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq11_HTML.gif, Ω z = ( 0 , l ) × ( 0 , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq12_HTML.gif, Ω = Ω L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq13_HTML.gif, y ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq14_HTML.gif, f ( x , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq15_HTML.gif are complex valued measurable functions and satisfy the conditions
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ5_HTML.gif
(5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ6_HTML.gif
(6)

respectively, ω = ( ω 0 , ω 1 , ϖ 0 , ϖ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq16_HTML.gif and H = ( L 2 ( 0 , l ) ) 2 × ( L 2 ( 0 , L ) ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq17_HTML.gif. L 2 ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq18_HTML.gif is a Hilbert space that consists of all functions in ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq19_HTML.gif, which are measurable and square-integrable. L 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq20_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq21_HTML.gif under the condition (2)-(4) for each v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq22_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq21_HTML.gif belonging to the C 0 ( [ 0 , L ] , L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq23_HTML.gif, and it satisfies the integral identity
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ7_HTML.gif
(7)

for η C 0 ( [ 0 , L ] , L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq24_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq22_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ8_HTML.gif
(8)

is valid for z [ 0 , L ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq25_HTML.gif. Here, the number c 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq26_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq27_HTML.gif is a given function. Then there is such a set G dense in H [ L 2 ( 0 , L ) ] 2 × [ L 2 ( 0 , l ) ] 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq28_HTML.gif that the optimal control problem (1)-(4) has a unique solution ω G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq29_HTML.gif and α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq30_HTML.gif.

Proof Firstly, let us show that
J 0 ( v ) = ψ ( , L ) y L 2 ( 0 , l ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ9_HTML.gif
(9)
is continuous on the set V. Let us take an arbitrary V, and let v + Δ v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq31_HTML.gif be an increment of the v for the Δ v H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq32_HTML.gif. Then the solution ψ ( x , z ; v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq33_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq34_HTML.gif. Here, the function ψ Δ ( x , z ) = ψ ( x , z ; v + Δ v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq35_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq36_HTML.gif is a solution of the following boundary value problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ10_HTML.gif
(10)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ11_HTML.gif
(11)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ12_HTML.gif
(12)
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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ13_HTML.gif
(13)
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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ14_HTML.gif
(14)

c 5 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq37_HTML.gif is constant that does not depend on Δv.

Now, let us evaluate the increment of the functional J 0 ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq38_HTML.gif on v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq39_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ15_HTML.gif
(15)
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ16_HTML.gif
(16)

where c 6 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq40_HTML.gif is a constant that does not depend on Δv. This inequality shows that the functional J 0 ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq38_HTML.gif is continuous on the set V. On the other hand, J 0 ( z ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq41_HTML.gif for z V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq42_HTML.gif; therefore, J 0 ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq38_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq30_HTML.gif and ω G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq29_HTML.gif. 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:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ17_HTML.gif
(17)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ18_HTML.gif
(18)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ19_HTML.gif
(19)
Here, the function ψ = ψ ( x , z ) ψ ( x , z ; v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq43_HTML.gif is a solution of (2)-(4) for v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq39_HTML.gif. The solution of the boundary value problem (17)-(19) corresponding to v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq39_HTML.gif is a function φ = φ ( x , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq44_HTML.gif that belongs to the space C 0 ( [ 0 , L ] , L 2 ( 0 , L ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq45_HTML.gif and satisfies the integral identity
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ20_HTML.gif
(20)
As seen, the problem (17)-(19) is an initial boundary value problem. This can easily be obtained by a transform θ = L z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq46_HTML.gif. Actually, if we do a variable transform θ = L z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq46_HTML.gif, we obtain the boundary problem as
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ21_HTML.gif
(21)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ22_HTML.gif
(22)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ23_HTML.gif
(23)
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equb_HTML.gif
If we write the complex conjugate of this boundary value problem, we obtain the following boundary value problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ24_HTML.gif
(24)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ25_HTML.gif
(25)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ26_HTML.gif
(26)
where
F ( x , θ ) = φ ˜ ( x , θ ) ¯ , h ( x ) = 2 i ( ψ ¯ ( x , L ) y ¯ ( x ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equc_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq47_HTML.gif belongs to the space L 2 ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq18_HTML.gif for ψ C 0 ( [ 0 , L ] , L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq48_HTML.gif, y L 2 ( 0 , l ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq27_HTML.gif. By using Theorem 2, it follows that the solution of the bounded value problem (24)-(26) existing in the space C 0 ( [ 0 , L ] , L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq23_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ27_HTML.gif
(27)
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 C 0 ( [ 0 , L ] , L 2 ( 0 , l ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq23_HTML.gif, 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equd_HTML.gif
Here, the number c 8 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq49_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ28_HTML.gif
(28)

Here, the number c 9 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq50_HTML.gif is constant.

Theorem 3 Let us accept that the conditions of Theorem 2 hold and ω H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq51_HTML.gif is given. Then the functional J α ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq52_HTML.gif can be Frechet differentiable in the set V and the formula below for a gradient of the functional is valid:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ29_HTML.gif
(29)
Proof Let us evaluate the increment of the functional J α ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq52_HTML.gif for the element v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq22_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ30_HTML.gif
(30)
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Eque_HTML.gif
where R ( Δ v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq53_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ31_HTML.gif
(31)
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equf_HTML.gif
If we use estimates (13) and (28) in this inequality, we obtain
| R ( Δ v ) | c 10 Δ v H 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ32_HTML.gif
(32)
Here, c 10 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq54_HTML.gif is a constant that does not depend on Δv. Hence, we write
R ( Δ v ) = o ( Δ v H ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ33_HTML.gif
(33)
By using equality (33), the increment of the functional can be written as
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ34_HTML.gif
(34)

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq55_HTML.gif is an optimal solution of the problem (1)-(4). Then the following inequality is valid for v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq22_HTML.gif:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ35_HTML.gif
(35)

Here, the functions ψ ( x , z ) ψ ( x , z ; v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq56_HTML.gif, φ ( x , z ) φ ( x , z ; v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq57_HTML.gif are solutions of the problems (2)-(4) and a conjugate problem corresponding to v V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq55_HTML.gif, respectively.

Proof Now, we prove that the gradient J α ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq58_HTML.gif is continuous at V. For this we show
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ36_HTML.gif
(36)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ37_HTML.gif
(37)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ38_HTML.gif
(38)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ39_HTML.gif
(39)

for Δ v H 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq59_HTML.gif.

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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq60_HTML.gif in (29), we can write the following equation:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ40_HTML.gif
(40)
Here, Δ ψ = Δ ψ ( x , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq61_HTML.gif is the solution of the problem (9)-(11) and Δ φ = Δ φ ( x , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq62_HTML.gif is the solution of the following problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ41_HTML.gif
(41)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ42_HTML.gif
(42)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ43_HTML.gif
(43)
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ44_HTML.gif
(44)

Here, the number c 11 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq63_HTML.gif is constant.

Using (13) and (28), we write
φ ( , z ) L 2 ( 0 , l ) 2 c 12 ( Δ v H 2 ) , z ( 0 , L ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ45_HTML.gif
(45)
Here, the number c 12 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq64_HTML.gif 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equg_HTML.gif
and then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ46_HTML.gif
(46)
If we use estimate (8), we can write the following inequality:
ψ Δ ( , z ) L 2 ( 0 , l ) 2 c 13 , z [ 0 , L ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ47_HTML.gif
(47)
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ48_HTML.gif
(48)
Here, the number of c 14 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq65_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ49_HTML.gif
(49)

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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq66_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ50_HTML.gif
(50)
Here, Δ φ ( x , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq67_HTML.gif is a solution of the problem (41). Estimate (45) is valid for z [ 0 , L ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq25_HTML.gif. Therefore, the following estimate can be written at z = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq68_HTML.gif:
φ ( , 0 ) L 2 ( 0 , l ) 2 c 12 ( Δ v H 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equh_HTML.gif
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ51_HTML.gif
(51)
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equ52_HTML.gif
(52)
We can see that (38) and (39) are valid by inequalities (51) and (52). That is, J α C 1 ( V ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq69_HTML.gif. On the other hand, V is a convex set according to the definition. So, the functional J α ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq52_HTML.gif holds by the condition of Theorem (Goebel) in [8] at V. Therefore, considering Theorem 3, we obtain
J α ( v ) , v v ) H 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_Equi_HTML.gif

for z V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-151/MediaObjects/13661_2012_Article_261_IEq42_HTML.gif. Here, using (29), it is seen that the statement of Theorem 4 is valid. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Ağrı İbrahim Çeçen University Faculty of Science and Art
(2)
Department of Mathematics, Atatürk University Faculty of Science

References

  1. İskenderov AD, Yagubov GY: Optimal control of non-linear quantum-mechanical systems. Autom. Remote Control 1989, 50: 1631-1641.
  2. İskenderov AD, Yagubov GY: A variational method for solving the inverse problem of determining the quantumnmechanical potential. Sov. Math. Doklady (Engl. Trans.) Am. Math. Soc. 1989, 38: 637-641.
  3. Yagubov GY, Musayeva MA: About the problem of identification for nonlinear Schrödinger equation. J. Differ. Equ. 1997, 33(12):1691-1698.
  4. Yagubov, GY: Optimal control by coefficient of quasilinear Schrödinger equation. Abstract of these doctors sciences, Kiev, p. 25 (1994)
  5. Yetişkin H, Subaşı M: On the optimal control problem for Schrödinger equation with complex potential. Appl. Math. Comput. 2010, 216: 1896-1902. 10.1016/j.amc.2010.03.039MathSciNetView Article
  6. Yıldız B, Yagubov G: On the optimal control problem. J. Comput. Appl. Math. 1997, 88: 275-287. 10.1016/S0096-3003(96)00335-9View Article
  7. Ladyzhenskaya OA, Solonnikov VA, Uralsteva NN Translation of Mathematical Monograps. In Linear and Quasi-Linear Equations of Parabolic Type. Am. Math. Soc., Providence; 1968.
  8. Goebel M: On existence of optimal control. Math. Nachr. 1979, 93: 67-73. 10.1002/mana.19790930106MathSciNetView Article

Copyright

© Koçak and Çelik; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.