# Optimal control problem for stationary quasi-optic equations

- Yusuf Koçak
^{1}Email author and - Ercan Çelik
^{2}

**2012**:151

https://doi.org/10.1186/1687-2770-2012-151

© Koçak and Çelik; licensee Springer 2012

**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

## 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 [1–5]. Optimal control problem for nonstationary Schrödinger equation of quasi optics was investigated for the first time in [6].

## 2 Formulation of the problem

respectively, $\omega =({\omega}_{0},{\omega}_{1},{\varpi}_{0},{\varpi}_{1})$ and $H={({L}_{2}(0,l))}^{2}\times {({L}_{2}(0,L))}^{2}$. ${L}_{2}(0,l)$ is a Hilbert space that consists of all functions in $(0,l)$, which are measurable and square-integrable. ${L}_{2}(\mathrm{\Omega})$ is the well-known Lebesgue space consisting of all functions in Ω, which are measurable and square-integrable.

The problem of finding a function $\psi =\psi (x,\mathrm{z})\equiv \psi (x,z;v)$ under the condition (2)-(4) for each $\mathrm{\forall}v\in V$, which is a boundary value problem, is a function for Eq. (2).

for $\mathrm{\forall}\eta \in {C}^{0}([0,L],{L}_{2}(0,l))$.

## 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*$\mathrm{\forall}v\in V$,

*the problem*(2)-(4)

*has a unique solution*,

*and for this solution*,

*the estimate*

*is valid for* $\mathrm{\forall}z\in [0,L]$. *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* $y\in {L}_{2}(0,l)$ *is a given function*. *Then there is such a set* *G* *dense in* $H\equiv {[{L}_{2}(0,L)]}^{2}\times {[{L}_{2}(0,l)]}^{2}$ *that the optimal control problem* (1)-(4) *has a unique solution* $\mathrm{\forall}\omega \in G$ *and* $\alpha >0$.

*Proof*Firstly, let us show that

*V*. Let us take an arbitrary ∈

*V*, and let $v+\mathrm{\Delta}v$ be an increment of the

*v*for the $\mathrm{\Delta}v\in H$. Then the solution $\psi (x,z;v)$ of the problem (1)-(4) will have an increment $\mathrm{\Delta}\psi =\mathrm{\Delta}\psi (x,z)=\psi (x,z;v+\mathrm{\Delta}v)-\psi (x,z;v)$. Here, the function ${\psi}_{\mathrm{\Delta}}(x,z)=\psi (x,z;v+\mathrm{\Delta}v)$ is the solution of (2)-(4). On the basis of the assumptions and conditions (2)-(4), it can be shown that the function $\mathrm{\Delta}\psi (x,z)$ is a solution of the following boundary value problem:

${c}_{5}>0$ is constant that does not depend on Δ*v*.

where ${c}_{6}>0$ is a constant that does not depend on Δ*v*. This inequality shows that the functional ${J}_{0}(v)$ is continuous on the set *V*. On the other hand, ${J}_{0}(z)\ge 0$ for $\mathrm{\forall}z\in V$; therefore, ${J}_{0}(v)$ 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 $\alpha >0$ and $\mathrm{\forall}\omega \in G$. Theorem 2 is proven. □

### 3.1 Fréchet diffrentiability of the functional

*ψ*and

*y*. Now, using estimate (8) in this inequality, we easily write the following estimate:

Here, the number ${c}_{9}>0$ is constant.

**Theorem 3**

*Let us accept that the conditions of Theorem*2

*hold and*$\omega \in H$

*is given*.

*Then the functional*${J}_{\alpha}(v)$

*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 ${J}_{\alpha}(v)$ for the element $\mathrm{\forall}v\in V$. We can write the following equation for the increment of the functional:

*v*. Hence, we write

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*${v}^{\ast}\in V$

*is an optimal solution of the problem*(1)-(4).

*Then the following inequality is valid for*$\mathrm{\forall}v\in V$:

*Here*, *the functions* ${\psi}^{\ast}(x,z)\equiv \psi (x,z;{v}^{\ast})$, ${\phi}^{\ast}(x,z)\equiv \phi (x,z;{v}^{\ast})$ *are solutions of the problems* (2)-(4) *and a conjugate problem corresponding to* ${v}^{\ast}\in V$, *respectively*.

*Proof*Now, we prove that the gradient ${J}_{\alpha}^{\prime}(v)$ is continuous at

*V*. For this we show

for ${\parallel \mathrm{\Delta}v\parallel}_{H}\to 0$.

Here, the number ${c}_{11}$ is constant.

If we use inequalities (48) and (49), we see that the correlations limit (36) and (37) is valid.

*V*is a convex set according to the definition. So, the functional ${J}_{\alpha}(v)$ holds by the condition of Theorem (Goebel) in [8] at

*V*. Therefore, considering Theorem 3, we obtain

for $\mathrm{\forall}z\in V$. Here, using (29), it is seen that the statement of Theorem 4 is valid. □

## Declarations

## Authors’ Affiliations

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## Copyright

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