Multitarget Linear-Quadratic Control Problem: Semi-Infinite Interval

Abstract

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 (𝐻,⟨,⟩) be a Hilbert space, 𝑍 be its closed vector subspace, ℎ₁,…,ℎ_𝑚, and 𝑐 be vectors in 𝐻. Consider the following optimization problem:max_{1≤𝑖≤𝑚}ℎ−ℎ_𝑖⟶min,ℎ∈𝑐+𝑍. (1.1)

Here ‖⋅‖ is the norm in 𝐻 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 (𝑚−1) 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 𝑓_𝑖(ℎ)=‖ℎ−ℎ_𝑖‖²,𝑖=1,2,…,𝑚. It is obvious that (1.1) is equivalent to the following optimization problem:𝑓𝑧⟶min,_𝑖(ℎ)≤𝑧,𝑖=1,2,…,𝑚,ℎ∈𝑐+𝑍. (2.1)

Consider the Lagrange functionℒ𝜆₁,…,𝜆_𝑚,ℎ,𝑧=𝑧+^𝑚_𝑖=1𝜆_𝑖𝑓_𝑖(ℎ)−𝑧=𝑧1−^𝑚_𝑖=1𝜆_𝑖+^𝑚_𝑖=1𝜆_𝑖𝑓_𝑖(ℎ). (2.2)

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 form𝜆_𝑖≥0,𝜆_𝑖𝑓_𝑖(ℎ)−𝑧=0,𝑖=0,1,2,…,𝑚,𝜕ℒ𝜕𝑧=0,∑^𝑚_𝑖=1𝜆_𝑖∇𝑓_𝑖(ℎ)∈𝑍^⟂, (2.3) where ∇𝑓_𝑖(ℎ)=2(ℎ−ℎ_𝑖),𝑖=1,2,…,𝑚,𝑍^⟂ is the orthogonal complement of 𝑍 in 𝐻. Conditions (2.3) lead to^𝑚_𝑖=0𝜆_𝑖=1,𝜆_𝑖𝜋≥0,𝑖=1,2,…,𝑚,_𝑍(ℎ)=∑^𝑚_𝑖=1𝜆_𝑖𝜋_𝑍ℎ_𝑖. (2.4)

Here 𝜋_𝑍∶𝐻→𝑍 is the orthogonal projection. Let us form the Lagrange dual of (2.1). Consider𝜑𝜆₁,𝜆₂,…,𝜆_𝑚=minℒ𝜆₁,…,𝜆_𝑚,ℎ,𝑧∶ℎ∈𝑐+𝑍,𝑧∈𝑍. (2.5)

Using (2.4), we obtain that𝜑𝜆₁,𝜆₂,…,𝜆_𝑚=^𝑚_𝑖=1𝜆_𝑖𝑓_𝑖ℎ𝜆₁,…,𝜆_𝑚, (2.6) whereℎ𝜆₁,…,𝜆_𝑚=𝜋_𝑍⟂(𝑐)+^𝑚_𝑖=1𝜆_𝑖𝜋_𝑍ℎ_𝑖. (2.7)

Notice that for any ℎ∈𝑐+𝑍,𝜋_𝑍⟂(ℎ)=𝜋_𝑍⟂(𝑐). Here 𝜋_𝑍⟂∶𝐻→𝑍^⟂ is the orthogonal projection of 𝐻 onto orthogonal complement 𝑍^⟂ of 𝑍. To further simplify (2.6), introduce the notationℎ(𝜆)=^𝑚_𝑖=1𝜆_𝑖ℎ_𝑖. (2.8)

Then𝑓_𝑗ℎ𝜆₁,…,𝜆_𝑚=𝜋_𝑍ℎ(𝜆)−ℎ_𝑗+𝜋_𝑍⟂𝑐−ℎ_𝑗²=𝜋_𝑍ℎ(𝜆)−𝜋_𝑍ℎ_𝑗²+𝜋_𝑍⟂𝑐−ℎ_𝑗²=𝜋_𝑍(ℎ(𝜆))²+𝜋_𝑍ℎ_𝑗²−2𝜋_𝑍(ℎ(𝜆)),𝜋_𝑍ℎ_𝑗+𝜋_𝑍⟂𝑐−ℎ_𝑗². (2.9)

Hence, according to (2.6), we have the following:𝜑𝜆₁,…,𝜆_𝑚=𝜋_𝑍(ℎ(𝜆))²+^𝑚_𝑗=1𝜆_𝑗𝜋_𝑍ℎ_𝑗²−2𝜋_𝑍(ℎ(𝜆)),𝜋_𝑍(ℎ(𝜆))+^𝑚_𝑗=1𝜆_𝑗𝜋_𝑍⟂𝑐−ℎ_𝑗². (2.10)

We, hence, arrive at the following expression of 𝜑:𝜑𝜆₁,…,𝜆_𝑚=−𝜋_𝑍^𝑚_𝑖=1𝜆_𝑖ℎ_𝑖²+^𝑚_𝑗=1𝜆_𝑗𝜋_𝑍ℎ_𝑗²+𝜋_𝑍⟂𝑐−ℎ_𝑗². (2.11)

We can simplify (2.11) somewhat. Notice that𝜋_𝑍⟂𝑐−ℎ_𝑗²=𝜋_𝑍⟂(𝑐)²+𝜋_𝑍⟂ℎ_𝑗²−2𝜋_𝑍⟂(𝑐),𝜋_𝑍⟂ℎ_𝑗. (2.12)

Consequently,𝜑𝜆₁,…,𝜆_𝑚=−𝜋_𝑍(ℎ(𝜆))²+^𝑚_𝑗=1𝜆_𝑗ℎ_𝑗²−2𝜋_𝑍⟂(𝑐),𝜋_𝑍⟂(ℎ(𝜆))+𝜋_𝑍⟂(𝑐)²=−‖ℎ(𝜆)‖²+𝜋_𝑍⟂(ℎ(𝜆)−𝑐)²+^𝑚_𝑗=1𝜆_𝑗ℎ_𝑗². (2.13) Here,ℎ(𝜆)=^𝑚_𝑖=1𝜆_𝑖ℎ_𝑖. (2.14) Hence, the Lagrange dual to (2.1) takes the following form:𝜑𝜆₁,…,𝜆_𝑚⟶max,_𝑚𝑖∑₌₁𝜆_𝑖=1,𝜆_𝑖≥0,𝑖=1,2,…,𝑚. (2.15)

If (𝜆^∗₁,…,𝜆^∗_𝑚) is an optimal solution to (2.15), we can recover the optimal solution of the original problem using the relation (2.7), and 𝜑(𝜆^∗₁,…,𝜆^∗_𝑚) gives the optimal value for the original problem (1.1).

3. Linear-Quadratic Case

Denoted by 𝐿^𝑛₂[0,∞), the vector space of square integrable functions 𝑓∶[0,∞)→𝑅^𝑛. Let 𝐻=𝐿^𝑛₂[0,∞)×𝐿^𝑚₂[0,∞), and𝑍=𝛼,𝛽∈𝐻∶𝛼isabsolutelycontinuouson[0,∞),𝛼̇=𝐴𝛼+𝐵𝛽,𝛼(0)=0. (3.1) Here 𝐴 (respectively 𝐵) is an 𝑛 by 𝑛 (respectively 𝑛 by 𝑚) matrix. Observe that𝛼₁,𝛽₁,𝛼₂,𝛽₂=^∞₀𝛼₁(𝑡)^𝑇𝛼₂(𝑡)+𝛽₁(𝑡)^𝑇𝛽₂(𝑡𝛼)𝑑𝑡,_𝑖,𝛽_𝑖∈𝐻,𝑖=1,2. (3.2)

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 𝑚=2 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 𝜋_𝑍. Such a description has been found in [4]. We briefly describe it here.

Theorem 3.1. Let 𝐶 be an antistable 𝑛 by 𝑛 matrix (i.e., real parts of all eigenvalues of 𝐶 are positive). Consider the following system of linear differential equations: 𝑥̇=𝐶𝑥+𝑓, (3.3) where 𝑓∈𝐿^𝑛₂[0,∞). Then there exists a unique solution 𝐿(𝑓) of (3.3) belonging to 𝐿^𝑛₂[0,∞). Moreover, the map 𝐿∶𝐿^𝑛₂[0,∞)→𝐿^𝑛₂[0,∞) is linear and bounded. Explicitly, 𝐿𝑓(𝑡)=−^∞₀𝑒^−𝐶𝜏𝑓(𝑡+𝜏)𝑑𝜏. (3.4)

For the proof, see [4].

Consider the algebraic Riccati equation𝐾𝐵𝐵^𝑇𝐾+𝐴^𝑇𝐾+𝐾𝐴−𝐼=0. (3.5) We assume that (3.5) has a real symmetric solution 𝐾_st such that the matrix𝐹=𝐴+𝐵𝐵^𝑇𝐾_st (3.6) is stable (i.e., real parts of all eigenvalues of 𝐹 are negative). Notice that such a solution exists if and only if the pair (𝐴,𝐵) is stabilizable. See, for example, [5].

Theorem 3.2. We have the following: 𝑍^⟂=𝑝̇+𝐴^𝑇𝑝,𝐵^𝑇𝑝;𝑝∈𝐿^𝑛₂[0,∞),𝑝isabsolutelycontinuous,𝑝̇∈𝐿^𝑛₂[0,∞). (3.7)
Given that (𝜓,𝜑)∈𝐻, we have𝜓=𝑥−𝑝̇+𝐴^𝑇𝑝,𝜑=𝑢−𝐵^𝑇𝑝, (3.8)(3.9) where 𝑥 is the solution of the differential equation 𝑥̇=𝐴+𝐵𝐵^𝑇𝐾_st𝑥+𝐵𝐵^𝑇𝜌+𝐵𝜑,𝑥(0)=0,𝑢=𝐵^𝑇𝐾_st𝑥+𝐵^𝑇𝜌+𝜑,𝑝=𝐾_st𝑥+𝜌, (3.10)(3.11)(3.12) and 𝜌 is a unique solution to the differential equation 𝜌̇=−𝐴+𝐵𝐵^𝑇𝐾_st^𝑇𝜌−𝐾_st𝐵𝜑−𝜓 (3.13) belonging to 𝐿^𝑛₂[0,∞).
In particular, (𝑥,𝑢)∈𝑍,−(𝑝̇+𝐴^𝑇𝑝,𝐵^𝑇𝑝)∈𝑍^⟂, and consequently 𝑍 is a closed subspace in 𝐻 with 𝜋_𝑍𝜓,𝜑=(𝑥,𝑢),𝜋_𝑍⟂𝜓,𝜑=−𝑝̇+𝐴^𝑇𝑝,𝐵^𝑇𝑝. (3.14)

Remark 3.3. The required solution 𝜌 exists and unique by Theorem 3.1, since the matrix −(𝐴+𝐵𝐵^𝑇𝐾_st) is antistable.

Sketch of the Proof
Let 𝑝∈𝐿^𝑛₂[0,∞) be absolutely continuous and such that 𝑝̇∈𝐿^𝑛₂[0,∞). Suppose that (𝑥,𝑢)∈𝑍. Then (𝑥,𝑢),𝑝̇+𝐴^𝑇𝑝,𝐵^𝑇𝑝=^∞₀𝑥^𝑇𝑝̇+𝑥^𝑇𝐴^𝑇𝑝+𝑢𝐵^𝑇𝑝=𝑑𝑡_∞0𝑥^𝑇𝑝̇+(𝐴𝑥+𝐵𝑢)^𝑇𝑝=𝑑𝑡_∞0𝑥^𝑇𝑝̇+𝑥̇^𝑇𝑝=𝑑𝑡_∞0𝑑𝑥𝑑𝑡^𝑇𝑝𝑑𝑡=lim_𝜏→∞𝑥^𝑇(𝜏)𝑝(𝜏)−𝑥(0)^𝑇𝑝(0). (3.15) But 𝑥(𝜏),𝑝(𝜏)→0, as 𝜏→∞ (see e.g., [4] for details) and 𝑥(0)=0. Hence, (𝑥,𝑢),𝑝̇+𝐴^𝑇𝑝,𝐵^𝑇𝑝=0. (3.16)
Let us now show that the decomposition (3.5) and (3.9) takes place for an arbitrary (𝜓,𝜑)∈𝐻. Indeed, using (3.12), 𝑝̇=𝐾_st𝑥̇+𝜌̇. (3.17)
Hence by (3.10) and (3.13), 𝑝̇=𝐾_st𝐴+𝐵𝐵^𝑇𝐾_st𝑥+𝐾_st𝐵𝐵^𝑇𝜌+𝐾_st𝐵𝜑−𝐴+𝐵𝐵^𝑇𝐾_st^𝑇𝜌−𝐾_st𝐵𝜑−𝜓. (3.18) Combining all terms with 𝑥 and all terms with 𝜌 in two separate groups, we obtain that 𝑝̇+𝐴^𝑇𝑝=𝑝̇+𝐴^𝑇𝐾_st𝑥+𝐴^𝑇𝜌=𝐾_st𝐴+𝐾_st𝐵𝐵^𝑇𝐾_st+𝐴^𝑇𝐾_st𝑥+𝐾_st𝐵𝐵^𝑇−𝐴^𝑇−𝐾_st𝐵𝐵^𝑇+𝐴^𝑇𝜌−𝜓. (3.19) Using now the fact that 𝐾_st satisfies (3.5), we obtain that 𝑝̇+𝐴^𝑇𝑝=𝑥−𝜓 (3.20) which is (3.8). Using (3.11) and (3.12), we obtain that 𝑢−𝐵^𝑇𝑝=𝐵^𝑇𝐾_st𝑥+𝐵^𝑇𝜌+𝜑−𝐵^𝑇𝐾_st𝑥−𝐵^𝑇𝜌=𝜑, (3.21) which is (3.9). Finally, it is clear that for 𝑥 and 𝑢 defined by (3.11) and (3.12), we have 𝑥̇=𝐴𝑥+𝐵𝑢 (3.22) and consequently (𝑥,𝑢)∈𝑍. 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 ‖𝜋_𝑍⟂(ℎ)‖², where ℎ∈𝐻.

Theorem 3.4. Let ℎ=(𝜓,𝜑)∈𝐻, and 𝜌∈𝐿^𝑛₂[0,∞) is the function entering the decomposition (3.8) and (3.9) and described in (3.13). Then 𝜋_𝑍(ℎ)²=𝐵^𝑇𝜌+𝜑²,𝜋_𝑍⟂(ℎ)²=‖ℎ‖²−𝐵^𝑇𝜌+𝜑². (3.23)(3.24)

Proof. Let (𝑦,𝜈)∈𝑍. Let, further, Δ𝑦,𝜈=𝜈−𝐵^𝑇𝐾_st𝑦−𝐵^𝑇𝜌−𝜑^𝑇𝜈−𝐵^𝑇𝐾_st𝑦−𝐵^𝑇𝜌−𝜑. (3.25) Here for simplicity of notations, we suppressed the dependence on 𝑡. Then Δ𝑦,𝜈=Δ₁+Δ₂+Δ₃, (3.26) where Δ₁=𝜈−𝜑^𝑇𝜈−𝜑,Δ₂=𝐾_st𝑦+𝜌^𝑇𝐵𝐵^𝑇𝐾_st𝑦+𝜌,andΔ₃=−2𝜈−𝜑^𝑇𝐵^𝑇𝐾_st𝑦+𝜌. (3.27) Since (𝑦,𝜈)∈𝑍, we have 𝑦̇=𝐴𝑦+𝐵𝜈,𝑦(0)=0. (3.28) Hence, Δ₂=𝑦^𝑇𝐾_st𝐵𝐵^𝑇𝐾_st𝑦+𝜌^𝑇𝐵𝐵^𝑇𝜌+2𝜌^𝑇𝐵𝐵^𝑇𝐾_stΔ𝑦,₃=−2𝐵𝜈−𝐵𝜑^𝑇𝐾_st𝑦+𝜌=−2𝑦̇−𝐴𝑦−𝐵𝜑^𝑇𝐾_st𝑦+𝜌=−2𝑦𝐾̇_st𝑦+𝑦^𝑇𝐴^𝑇𝐾_st+𝐾_st𝐴𝑦+2𝐵𝜑^𝑇𝐾_st𝑦−2𝑦^𝑇̇𝜌+2𝐴𝑦^𝑇𝜌+2𝐵𝜑^𝑇𝜌. (3.29) Notice that 𝑦^𝑇̇𝜌+𝑦^𝑇𝜌̇=(𝑑/𝑑𝑡)(𝑦^𝑇𝜌). Hence, Δ𝑦,𝜈=𝜈−𝜑^𝑇𝜈−𝜑+𝑦^𝑇𝐾_st𝐵𝐵^𝑇𝐾_st+𝐴^𝑇𝐾_st+𝐾_st𝐴𝑦+2𝑦^𝑇𝜌̇+𝐾_st𝐵𝜑+𝐾_st𝐵𝐵^𝑇𝜌+𝐴^𝑇𝜌+𝐵^𝑇𝜌^𝑇𝐵^𝑇𝜌+2𝜑^𝑇𝐵^𝑇𝜌−𝑑𝑦𝑑𝑡^𝑇𝜌−𝑑𝑦𝑑𝑡^𝑇𝐾_st𝑦. (3.30) Using the fact that 𝐾_st is a solution to (3.5) and (3.13), we obtain that Δ𝑦,𝜈=𝜈−𝜑^𝑇𝜈−𝜑+𝑦^𝑇𝑦−2𝑦^𝑇𝜓+𝐵^𝑇𝜌+𝜑^𝑇𝐵^𝑇𝜌+𝜑−𝜑^𝑇𝑑𝜑−𝑦𝑑𝑡^𝑇𝜌−𝑑𝑦𝑑𝑡^𝑇𝐾_st𝑦=𝜈−𝜑^𝑇𝜈−𝜑+𝑦−𝜓^𝑇𝑦−𝜓+𝐵^𝑇𝜌+𝜑^𝑇𝐵^𝑇𝜌+𝜑−𝜑^𝑇𝜑−𝜓^𝑇𝑑𝜓−𝑦𝑑𝑡^𝑇𝜌−𝑑𝑦𝑑𝑡^𝑇𝐾_st𝑦. (3.31) Integrating (3.31) from 0 to +∞ and using the fact that 𝑦(0)=0,𝑦(𝑡),𝜌(𝑡)→0 as 𝑡→∞, we obtain that ^∞₀Δ𝑦,𝜈𝑑𝑡=𝑦−𝜓,𝜈−𝜑²−𝜓,𝜑²+𝐵^𝑇𝜌+𝜑². (3.32) Notice that Δ(𝑦,𝜈)≥0 and Δ(𝑦,𝜈)=0 provided (𝑦,𝜈)=𝜋_𝑍(𝜓,𝜑). See (3.11). Consequently, (3.32) implies that 𝜓,𝜑²=𝐵^𝑇𝜌+𝜑²+‖𝜋_𝑍⟂𝜓,𝜑‖². (3.33) Hence, 𝜋_𝑍𝜓,𝜑²=𝐵^𝑇𝜌+𝜑². (3.34) This completes the proof of Theorem 3.4.

We can now easily compute the coefficients of the objective function (2.11). Assuming that ℎ_𝑖=(𝜓_𝑖,𝜑_𝑖)∈𝐿^𝑛₂[0,∞)×𝐿^𝑚₂[0,∞),𝑖=1,2,…,𝑚,𝑐=(𝛼,𝛽)∈𝐿^𝑛₂[0,∞)×𝐿^𝑚₂[0,∞) and noticing that by Theorem 3.4𝜋_𝑍(ℎ(𝜆)−𝑐)²=^∞₀𝐵^𝑇𝜌(𝜆)+𝜑(𝜆)^𝑇𝐵^𝑇𝜌(𝜆)+𝜑(𝜆)𝑑𝑡, (3.35) where 𝜌(𝜆) is the solution of the differential equation𝑑𝜌(𝜆)𝑑𝑡=−𝐴+𝐵𝐵^𝑇𝐾_st^𝑇𝜌(𝜆)−𝐾_st𝐵𝜑(𝜆)−𝜓(𝜆), (3.36) belonging to 𝐿^𝑛₂[0,∞) and𝜑(𝜆)=^𝑚_𝑖=1𝜆_𝑖𝜑_𝑖−𝛽,𝜓(𝜆)=^𝑚_𝑖=1𝜆_𝑖𝜓_𝑖−𝛼. (3.37) Consequently,𝜌(𝜆)=^𝑚_𝑖=1𝜆_𝑖𝜌_𝑖−𝜌_𝑐, (3.38) where 𝜌_𝑖 and 𝜌_𝑐 are 𝐿^𝑛₂[0,∞) solutions of differential equations𝜌̇_𝑖=−𝐴+𝐵𝐵^𝑇𝐾_st𝜌_𝑖−𝐾_st𝐵𝜑_𝑖−𝜓_𝑖,𝑖=1,2,…,𝑚,𝜌̇_𝑐=−𝐴+𝐵𝐵^𝑇𝐾_st𝜌_𝑐−𝐾_st𝐵𝛽−𝛼, (3.39) respectively.

Hence,𝜋_𝑍(ℎ(𝜆)−𝑐)²=^∞₀Γ(𝜆)^𝑇Γ(𝜆)𝑑𝑡, (3.40) whereΓ(𝜆)=^𝑚_𝑖=1𝜆_𝑖𝐵^𝑇𝜌_𝑖−𝜌_𝑐+𝜑_𝑖−𝛽, (3.41) which allows us to easily express the objective function (2.13) in terms of integrals of 𝜌_𝑖 and 𝜌_𝑐.

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 𝑚+1 linear differential equations, where 𝑚 is the number of targets. Notice that our results can be easily extended to discrete-time systems.

Acknowledgments

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.