# Multitarget Linear-Quadratic Control Problem: Semi-Infinite Interval

Authors: L. Faybusovich; T. Mouktonglang
Journal: Mathematical Problems in Engineering
 (2012)

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## 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.


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## Body

## 1. Introduction

Let(H,〈,〉) be a Hilbert space, Z be its closed vector subspace, h1,…,hm, and c be vectors in H. Consider the following optimization problem:(1.1)max⁡1≤i≤m‖h-hi‖⟶min⁡,h∈c+Z.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

Letfi(h)=∥h-hi∥2,i=1,2,…,m. It is obvious that (1.1) is equivalent to the following optimization problem:(2.1)z⟶min⁡,fi(h)≤z,i=1,2,…,m,h∈c+Z.Consider the Lagrange function(2.2)L(λ1,…,λm,h,z)=z+∑i=1mλi(fi(h)-z)=z(1-∑i=1mλi)+∑i=1mλifi(h).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(2.3)λi≥0,λi(fi(h)-z)=0,i=0,1,2,…,m,∂L∂z=0,∑i=1mλi∇fi(h)∈Z⊥,
where ∇fi(h)=2(h-hi),i=1,2,…,m,Z⊥ is the orthogonal complement of Z in H. Conditions (2.3) lead to(2.4)∑i=0mλi=1,λi≥0,i=1,2,…,m,πZ(h)=∑i=1mλi(πZhi).HereπZ:H→Z is the orthogonal projection. Let us form the Lagrange dual of (2.1). Consider(2.5)φ(λ1,λ2,…,λm)=min⁡{L(λ1,…,λm,h,z):h∈c+Z,z∈Z}.Using (2.4), we obtain that(2.6)φ(λ1,λ2,…,λm)=∑i=1mλifi(h(λ1,…,λm)),
where(2.7)h(λ1,…,λm)=πZ⊥(c)+∑i=1mλiπZ(hi).Notice that for anyh∈c+Z,πZ⊥(h)=πZ⊥(c). Here πZ⊥:H→Z⊥ is the orthogonal projection of H onto orthogonal complement Z⊥ of Z. To further simplify (2.6), introduce the notation(2.8)h(λ)=∑i=1mλihi.Then(2.9)fj(h(λ1,…,λm))=‖πZ(h(λ)-hj)+πZ⊥(c-hj)‖2=‖πZ(h(λ)-πZ(hj))‖2+‖πZ⊥(c-hj)‖2=‖πZ(h(λ))‖2+‖πZ(hj)‖2-2〈πZ(h(λ)),πZ(hj)〉+‖πZ⊥(c-hj)‖2.Hence, according to (2.6), we have the following:(2.10)φ(λ1,…,λm)=‖πZ(h(λ))‖2+∑j=1mλj‖πZ(hj)‖2-2〈πZ(h(λ)),πZ(h(λ))〉+∑j=1mλj‖πZ⊥(c-hj)‖2.We, hence, arrive at the following expression ofφ:(2.11)φ(λ1,…,λm)=-‖πZ(∑i=1mλihi)‖2+∑j=1mλj(‖πZ(hj)‖2+πZ⊥(c-hj)‖2).We can simplify (2.11) somewhat. Notice that(2.12)‖πZ⊥(c-hj)‖2=‖πZ⊥(c)‖2+‖πZ⊥(hj)‖2-2〈πZ⊥(c),πZ⊥(hj)〉.Consequently,(2.13)φ(λ1,…,λm)=-‖πZ(h(λ))‖2+∑j=1mλj‖hj‖2-2〈πZ⊥(c),πZ⊥(h(λ))〉+‖πZ⊥(c)‖2=-‖h(λ)‖2+‖πZ⊥(h(λ)-c)‖2+∑j=1mλj‖hj‖2.
Here,(2.14)h(λ)=∑i=1mλihi.
Hence, the Lagrange dual to (2.1) takes the following form:(2.15)φ(λ1,…,λm)⟶max⁡,∑i=1mλi=1,λi≥0,i=1,2,…,m.If(λ1*,…,λm*) is an optimal solution to (2.15), we can recover the optimal solution of the original problem using the relation (2.7), and φ(λ1*,…,λm*) gives the optimal value for the original problem (1.1).


## 3. Linear-Quadratic Case

Denoted byL2n[0,∞), the vector space of square integrable functions f:[0,∞)→Rn. Let H=L2n[0,∞)×L2m[0,∞), and(3.1)Z={(α,β)∈H:αisabsolutelycontinuouson⁡[0,∞),α̇=Aα+Bβ,α(0)=0}.
Here A (respectively B) is an n by n (respectively n by m) matrix. Observe that(3.2)〈(α1,β1),(α2,β2)〉=∫0∞[α1(t)Tα2(t)+β1(t)Tβ2(t)]dt,(αi,βi)∈H,i=1,2.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 πZ. Such a description has been found in [4]. We briefly describe it here.Theorem 3.1.
LetC 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:
(3.3)ẋ=Cx+f,
where f∈L2n[0,∞). Then there exists a unique solution L(f) of (3.3) belonging to L2n[0,∞). Moreover, the map L:L2n[0,∞)→L2n[0,∞) is linear and bounded. Explicitly,
(3.4)L(f)(t)=-∫0∞e-Cτf(t+τ)dτ.For the proof, see [4].Consider the algebraic Riccati equation(3.5)KBBTK+ATK+KA-I=0.
We assume that (3.5) has a real symmetric solution Kst such that the matrix(3.6)F=A+BBTKst
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:(3.7)Z⊥={(ṗ+ATp,BTp);p∈L2n[0,∞),pisabsolutelycontinuous,ṗ∈L2n[0,∞)}.
Given that(ψ,φ)∈H, we have(3.8)ψ=x-(ṗ+ATp),(3.9)φ=u-BTp,
where x is the solution of the differential equation
(3.10)ẋ=(A+BBTKst)x+BBTρ+Bφ,x(0)=0,(3.11)u=BTKstx+BTρ+φ,(3.12)p=Kstx+ρ,
and ρ is a unique solution to the differential equation
(3.13)ρ̇=-(A+BBTKst)Tρ-KstBφ-ψ
belonging to L2n[0,∞).
In particular,(x,u)∈Z,-(ṗ+ATp,BTp)∈Z⊥, and consequently Z is a closed subspace in H with
(3.14)πZ(ψ,φ)=(x,u),πZ⊥(ψ,φ)=-(ṗ+ATp,BTp).Remark 3.3.
The required solutionρ exists and unique by Theorem 3.1, since the matrix -(A+BBTKst) is antistable.Sketch of the Proof
Letp∈L2n[0,∞) be absolutely continuous and such that ṗ∈L2n[0,∞). Suppose that (x,u)∈Z. Then
(3.15)〈(x,u),(ṗ+ATp,BTp)〉=∫0∞(xTṗ+xTATp+uBTp)dt=∫0∞[xTṗ+(Ax+Bu)Tp]dt=∫0∞(xTṗ+ẋTp)dt=∫0∞ddt(xTp)dt=lim⁡τ→∞xT(τ)p(τ)-x(0)Tp(0).
But x(τ),p(τ)→0, as τ→∞ (see e.g., [4] for details) and x(0)=0. Hence,
(3.16)〈(x,u),(ṗ+ATp,BTp)〉=0.
Let us now show that the decomposition (3.5) and (3.9) takes place for an arbitrary (ψ,φ)∈H. Indeed, using (3.12),
(3.17)ṗ=Kstẋ+ρ̇.
Hence by (3.10) and (3.13),
(3.18)ṗ=Kst(A+BBTKst)x+KstBBTρ+KstBφ-(A+BBTKst)Tρ-KstBφ-ψ.
Combining all terms with x and all terms with ρ in two separate groups, we obtain that
(3.19)ṗ+ATp=ṗ+ATKstx+ATρ=(KstA+KstBBTKst+ATKst)x+(KstBBT-AT-KstBBT+AT)ρ-ψ.
Using now the fact that Kst satisfies (3.5), we obtain that
(3.20)ṗ+ATp=x-ψ
which is (3.8). Using (3.11) and (3.12), we obtain that
(3.21)u-BTp=BTKstx+BTρ+φ-BTKstx-BTρ=φ,
which is (3.9). Finally, it is clear that for x and u defined by (3.11) and (3.12), we have
(3.22)ẋ=Ax+Bu
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 ∥πZ⊥(h)∥2, where h∈H.Theorem 3.4.
Leth=(ψ,φ)∈H, and ρ∈L2n[0,∞) is the function entering the decomposition (3.8) and (3.9) and described in (3.13). Then
(3.23)‖πZ(h)‖2=‖BTρ+φ‖2,(3.24)‖πZ⊥(h)‖2=‖h‖2-‖BTρ+φ‖2.Proof.
Let(y,ν)∈Z. Let, further,
(3.25)Δ(y,ν)=(ν-BTKsty-BTρ-φ)T(ν-BTKsty-BTρ-φ).
Here for simplicity of notations, we suppressed the dependence on t. Then
(3.26)Δ(y,ν)=Δ1+Δ2+Δ3,
where
(3.27)Δ1=(ν-φ)T(ν-φ),Δ2=(Ksty+ρ)TBBT(Ksty+ρ),andΔ3=-2(ν-φ)T(BTKsty+ρ).
Since (y,ν)∈Z, we have
(3.28)ẏ=Ay+Bν,y(0)=0.
Hence,
(3.29)Δ2=yT(KstBBTKst)y+ρTBBTρ+2ρTBBTKsty,Δ3=-2(Bν-Bφ)T(Ksty+ρ)=-2(ẏ-Ay-Bφ)T(Ksty+ρ)=-2ẏKsty+yT(ATKst+KstA)y+2(Bφ)TKsty-2ẏTρ+2(Ay)Tρ+2(Bφ)Tρ.
Notice that ẏTρ+yTρ̇=(d/dt)(yTρ). Hence,
(3.30)Δ(y,ν)=(ν-φ)T(ν-φ)+yT(KstBBTKst+ATKst+KstA)y+2yT(ρ̇+KstBφ+KstBBTρ+ATρ)+(BTρ)T(BTρ)+2φT(BTρ)-ddt(yTρ)-ddt(yTKsty).
Using the fact that Kst is a solution to (3.5) and (3.13), we obtain that
(3.31)Δ(y,ν)=(ν-φ)T(ν-φ)+yTy-2yTψ+(BTρ+φ)T(BTρ+φ)-φTφ-ddt(yTρ)-ddt(yTKsty)=(ν-φ)T(ν-φ)+(y-ψ)T(y-ψ)+(BTρ+φ)T(BTρ+φ)-φTφ-ψTψ-ddt(yTρ)-ddt(yTKsty).
Integrating (3.31) from 0 to +∞ and using the fact that y(0)=0,y(t),ρ(t)→0 as t→∞, we obtain that
(3.32)∫0∞Δ(y,ν)dt=‖(y-ψ,ν-φ)‖2-‖(ψ,φ)‖2+‖BTρ+φ‖2.
Notice that Δ(y,ν)≥0 and Δ(y,ν)=0 provided (y,ν)=πZ(ψ,φ). See (3.11). Consequently, (3.32) implies that
(3.33)‖(ψ,φ)‖2=‖BTρ+φ‖2+∥πZ⊥(ψ,φ)∥2.
Hence,
(3.34)‖πZ(ψ,φ)‖2=‖BTρ+φ‖2.
This completes the proof of Theorem 3.4.We can now easily compute the coefficients of the objective function (2.11). Assuming that hi=(ψi,φi)∈L2n[0,∞)×L2m[0,∞),i=1,2,…,m,c=(α,β)∈L2n[0,∞)×L2m[0,∞) and noticing that by Theorem 3.4(3.35)‖πZ(h(λ)-c)‖2=∫0∞[BTρ(λ)+φ(λ)]T[BTρ(λ)+φ(λ)]dt,
where ρ(λ) is the solution of the differential equation(3.36)ddtρ(λ)=-(A+BBTKst)Tρ(λ)-KstB(φ(λ)-ψ(λ)),
belonging to L2n[0,∞) and(3.37)φ(λ)=∑i=1mλi(φi-β),ψ(λ)=∑i=1mλi(ψi-α).
Consequently,(3.38)ρ(λ)=∑i=1mλi(ρi-ρc),
where ρi and ρc are L2n[0,∞) solutions of differential equations(3.39)ρ̇i=-(A+BBTKst)ρi-KstBφi-ψi,i=1,2,…,m,ρ̇c=-(A+BBTKst)ρc-KstBβ-α,
respectively.Hence,(3.40)‖πZ(h(λ)-c)‖2=∫0∞Γ(λ)TΓ(λ)dt,
where(3.41)Γ(λ)=∑i=1mλi[BT(ρi-ρc)+(φi-β)],
which allows us to easily express the objective function (2.13) in terms of integrals of ρi and ρc.


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



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