We consider multitarget linear-quadratic control problem on semi-infinite interval. We show that the problem can be reduced to a simple convex optimization problem on the simplex.

1. Introduction

Let be a Hilbert space, be its closed vector subspace, , and be vectors in . Consider the following optimization problem:

Here is the norm in induced by the scalar product . In [1], we analyzed (1.1) using duality theory for infinite-dimensional second-order cone programming. We obtained a reduction of this problem to a finite-dimensional second-order cone programming and applied this result to a multitarget linear-quadratic control problem on a finite time interval. In this paper, we consider a reduction (1.1) to even simpler optimization problem of minimization of convex quadratic function on the dimensional simplex. We then apply this result to the analysis of a multitarget linear-quadratic control problem on semi-infinite time interval. We show that the coefficients of the quadratic function admit a simple expressions in term of the original data.

2. Reduction to a Simple Quadratic Programming Problem

Let . It is obvious that (1.1) is equivalent to the following optimization problem:

Consider the Lagrange function

Notice that despite the fact that our original problem is infinite dimensional, the usual KKT theorem holds true (see e.g., [2], page 72). It is also clear that Slater conditions are satisfied. Hence, optimality condition for (2.1) takes the form where is the orthogonal complement of in . Conditions (2.3) lead to

Here is the orthogonal projection. Let us form the Lagrange dual of (2.1). Consider

Using (2.4), we obtain that where

Notice that for any . Here is the orthogonal projection of onto orthogonal complement of . To further simplify (2.6), introduce the notation


Hence, according to (2.6), we have the following:

We, hence, arrive at the following expression of :

We can simplify (2.11) somewhat. Notice that

Consequently, Here, Hence, the Lagrange dual to (2.1) takes the following form:

If is an optimal solution to (2.15), we can recover the optimal solution of the original problem using the relation (2.7), and gives the optimal value for the original problem (1.1).

3. Linear-Quadratic Case

Denoted by , the vector space of square integrable functions . Let , and Here (respectively ) is an by (respectively by ) matrix. Observe that

In this setting, the problem (1.1) admits a natural interpretation as a linear-quadratic multitarget control problem. An interesting solution for this problem for is described in [3]. In our approach, we need an explicit computation of the coefficients of the objective function (2.13) which in turn requires an explicit description of orthogonal projection . Such a description has been found in [4]. We briefly describe it here.

Theorem 3.1. Let be an antistable by matrix (i.e., real parts of all eigenvalues of are positive). Consider the following system of linear differential equations: where . Then there exists a unique solution of (3.3) belonging to . Moreover, the map is linear and bounded. Explicitly,

For the proof, see [4].

Consider the algebraic Riccati equation We assume that (3.5) has a real symmetric solution such that the matrix is stable (i.e., real parts of all eigenvalues of are negative). Notice that such a solution exists if and only if the pair is stabilizable. See, for example, [5].

Theorem 3.2. We have the following:
Given that , we have where is the solution of the differential equation and is a unique solution to the differential equation belonging to .
In particular, , and consequently is a closed subspace in with

Remark 3.3. The required solution exists and unique by Theorem 3.1, since the matrix is antistable.

Sketch of the Proof
Let be absolutely continuous and such that . Suppose that . Then But , as (see e.g., [4] for details) and . Hence,
Let us now show that the decomposition (3.5) and (3.9) takes place for an arbitrary . Indeed, using (3.12),
Hence by (3.10) and (3.13), Combining all terms with and all terms with in two separate groups, we obtain that Using now the fact that satisfies (3.5), we obtain that which is (3.8). Using (3.11) and (3.12), we obtain that which is (3.9). Finally, it is clear that for and defined by (3.11) and (3.12), we have and consequently . This completes the proof of Theorem 3.2.
Looking at (2.13), we see that the evaluation of coefficients of the quadratic function requires the knowledge of expressions of the type , where .

Theorem 3.4. Let , and is the function entering the decomposition (3.8) and (3.9) and described in (3.13). Then

Proof. Let . Let, further, Here for simplicity of notations, we suppressed the dependence on . Then where Since , we have Hence, Notice that . Hence, Using the fact that is a solution to (3.5) and (3.13), we obtain that Integrating (3.31) from 0 to and using the fact that as , we obtain that Notice that and provided . See (3.11). Consequently, (3.32) implies that Hence, This completes the proof of Theorem 3.4.

We can now easily compute the coefficients of the objective function (2.11). Assuming that and noticing that by Theorem 3.4 where is the solution of the differential equation belonging to and Consequently, where and are solutions of differential equations respectively.

Hence, where which allows us to easily express the objective function (2.13) in terms of integrals of and .

4. Concluding Remarks

In this paper, we have shown that multitarget linear-quadratic control problem on semi-infinite interval can be reduced to solving a simple convex optimization on the simplex. The reduction involves solving one standard algebraic Riccati equation and linear differential equations, where is the number of targets. Notice that our results can be easily extended to discrete-time systems.


The research of L. Faybusovich was partially supported by NSF Grant DMS07-12809. The research of T. Mouktonglang was partially supported by the Commission on Higher Education and Thailand Research Fund under Grant MRG5080192.