Deterministic Kalman Filtering on Semi-Infinite Interval

Abstract

We relate a deterministic Kalman filter on semi-infinite interval to linear-quadratic tracking control model with unfixed initial condition.

1. Introduction

In [1], Sontag considered the deterministic analogue of Kalman filtering problem on finite interval. The deterministic model allows a natural extension to semi-infinite interval. It is of a special interest because for the standard linear-quadratic stochastic control problem extension to semi-infinite interval leads to complications with the standard quadratic objective function (see, e.g., [2]). According to [1], the model which we are going to consider has the following form: 𝐽𝑥,𝑢,𝑥₀=₀^+∞𝑢^𝑇𝑅𝑢+𝐶𝑥−𝑦^𝑇𝑄𝐶𝑥−𝑦𝑥(0)𝑑𝑡,𝑥̇=𝐴𝑥+𝐵𝑢,=𝑥₀. (1.1)(1.2)(1.3) Here we assume that the pair (𝑥,𝑢)∈𝑎(𝑥₀)+𝑍, where 𝑍 is a vector subspace of the Hilbert space 𝐿^𝑛₂[0,+∞)×𝐿^𝑚₂[0,+∞) (with 𝐿^𝑛₂[0,+∞) a Hilbert space of 𝑅^𝑛-value square integrable functions) is defined as follows: 𝑍=(𝑥,𝑢)∈𝐿^𝑛₂[0,+∞)×𝐿^𝑚₂[0,+∞)∶𝑥isabsolutelycontinuous,𝑥̇∈𝐿ⁿ₂[0,+∞),𝑥̇=𝐴𝑥+𝐵𝑢,𝑥(0)=0. (1.4) Here 𝐴 is an 𝑛 by 𝑛 matrix; 𝐵 is an 𝑛 by 𝑚 matrix; 𝑅=𝑅^𝑇 is an 𝑛 by 𝑛 and positive definite; 𝑄=𝑄^𝑇 is an 𝑟 by 𝑟 and positive definite; 𝐶 is an 𝑟 by 𝑛 matrix; 𝑦∈𝐿^𝑟₂[0,+∞). Notice that in (1.1)–(1.3) 𝑥₀ is not fixed and we minimize over all triple (𝑥,𝑢,𝑥₀)∈𝐿^𝑛₂[0,+∞)×𝐿^𝑚₂[0,+∞)×𝑅^𝑛 satisfying our assumption.

Notice also that we interpret (1.1)–(1.3) as an estimation problem of the form 𝑥̇=𝐴𝑥+𝐵𝑢,𝑦=𝐶𝑥+𝑣, (1.5) where we try to estimate 𝑥 with the help of observation 𝑦 by minimizing perturbations 𝑢, 𝑣 and choosing an appropriate initial condition 𝑥₀.

2. Solution of the Deterministic Problem

Consider the algebraic Riccati equation 𝐾𝐴+𝐴^𝑇𝐾+𝐾𝐿𝐾−𝐶^𝑇𝑄𝐶=0, (2.1) where 𝐿=𝐵𝑅⁻¹𝐵^𝑇. Assuming that the pair (𝐴,𝐵) is stabilizable and the pair (𝐶,𝐴) is detectable, there exists a negative definite symmetric solution 𝐾_𝑠𝑡 to (2.1) such that the matrix 𝐴+𝐿𝐾_𝑠𝑡 is stable (see, e.g., Theorem  12.3 in [3]). According to [4], we have described a complete solution of the linear-quadratic control problem on a semi-infinite interval with the linear term in the objective function. The major motivation for this extension comes from [5] where we consider applications of primal-dual interior-point algorithms to the computational analysis of multicriteria linear-quadratic control problems in mini-max form. To compute a primal-dual direction it is required to solve linear-quadratic control problems with the same quadratic and different linear parts on each iteration. Using the results in [5], we can describe the optimal solution to (1.1)–(1.3) with fixed 𝑥₀ as follows.

There exists a unique solution 𝜌₀∈𝐿^𝑛₂[0,∞) satisfying the differential equation 𝜌̇=−𝐴+𝐿𝐾_𝑠𝑡^𝑇𝜌−𝐶^𝑇𝑄𝑦. (2.2) Moreover, 𝜌₀ can be explicitly described as follows: 𝜌₀(𝑡)=₀^+∞exp𝐴+𝐿𝐾_𝑠𝑡^𝑇𝜏𝐶^𝑇𝑄𝑦(𝑡+𝜏)𝑑𝜏. (2.3) The optimal solution (𝑥,𝑢) to (1.1)–(1.3) has the form 𝑥̇=𝐴+𝐿𝐾_𝑠𝑡𝑥+𝐿𝜌₀,𝑥(0)=𝑥₀,𝑢=𝑅⁻¹𝐵^𝑇𝐾_𝑠𝑡𝑥+𝜌₀. (2.4)(2.5) For details see [5].

Notice that 𝜌₀ does not depend on 𝑥₀. To solve the original problem (1.1)–(1.3) we need to express the minimal value of the functional (1.1) in term of 𝑥₀.

Theorem 2.1. Let (𝑥,𝑢) be an optimal solution of (1.1)–(1.3) with fixed 𝑥₀ given by (2.2)–(2.5). Then 𝐽𝑥,𝑢,𝑥₀=&−𝑥^𝑇₀𝐾_𝑠𝑡𝑥₀−2𝜌₀(0)^𝑇𝑥₀+₀^+∞𝑦^𝑇𝑄𝑦−𝜌^𝑇₀𝐿𝜌₀𝑑𝑡. (2.6)

Remark 2.2. Notice that 𝐽(𝑥,𝑢,𝑥₀) is a strictly convex function of 𝑥₀ and hence minimum of 𝐽 as a function of 𝑥₀ is attained at 𝑥^opt₀=−𝐾⁻¹_𝑠𝑡𝜌₀(0). (2.7) Hence (2.2)–(2.5) gives a complete solution of the original problem (1.1)–(1.3).

Proof. Let (𝑦,𝑤)∈𝑎(𝑥₀)+𝑍 be feasible solution to (1.1)–(1.3), where 𝑥₀ is fixed. Consider Δ𝑦,𝑤=𝑤−𝑅⁻¹𝐵^𝑇𝐾_𝑠𝑡𝑦+𝜌₀^𝑇⋅𝑅⋅𝑤−𝑅⁻¹𝐵^𝑇𝐾_𝑠𝑡𝑦+𝜌₀, (2.8) where we suppressed an explicit dependence on time. Notice that by (2.5) Δ()Δ𝑥,𝑢≡0,𝑦,𝑤≡0, (2.9) for any feasible solution (𝑦,𝑤) implies that (𝑦,𝑤)=(𝑥,𝑢). Furthermore, let Δ(𝑦,𝑤)=Δ₁+Δ₂+Δ₃, where Δ₁=𝑤^𝑇Δ𝑅𝑤,₂=−2𝐾_𝑠𝑡𝑦+𝜌₀^𝑇Δ𝐵𝑤,₃=𝐾_𝑠𝑡𝑦+𝜌₀^𝑇𝐿𝐾_𝑠𝑡𝑦+𝜌₀. (2.10) Now 𝐵𝑤=𝑦̇−𝐴𝑦, and consequently Δ₂=−2𝑦^𝑇𝐾_𝑠𝑡+𝜌^𝑇₀𝑦̇−𝐴𝑦=𝑦^𝑇𝐾_𝑠𝑡𝐴+𝐴^𝑇𝐾_𝑠𝑡𝑦−2𝑦^𝑇𝐾_𝑠𝑡𝑦̇−2𝜌^𝑇₀𝑦̇+2𝜌^𝑇Δ𝐴𝑦,₃=𝑦^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝑦+𝜌^𝑇₀𝐿𝜌₀+2𝜌^𝑇₀𝐿𝐾_𝑠𝑡𝑦. (2.11) Consequently, Δ𝑦,𝑤=𝑤^𝑇𝑅𝑤+𝜌^𝑇₀𝐿𝜌₀+𝑦^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡+𝐾_𝑠𝑡𝐴+𝐴^𝑇𝐾_𝑠𝑡𝑑𝑦−𝑦𝑑𝑡^𝑇𝐾_𝑠𝑡𝑦𝑑−2𝜌𝑑𝑡^𝑇0𝑦+2𝜌^𝑇₀̇𝑦+2𝜌^𝑇₀𝐿𝐾_𝑠𝑡𝑦+2𝜌^𝑇₀𝐴𝑦. (2.12) Using (2.1) and (2.2), we obtain Δ𝑦,𝑤=𝑤^𝑇𝑅𝑤+𝜌^𝑇₀𝐿𝜌₀+𝑦^𝑇𝐶^𝑇𝑑𝑄𝐶𝑦−2𝜌𝑑𝑡^𝑇0𝑦−𝑑𝑦𝑑𝑡^𝑇𝐾_𝑠𝑡𝑦−2𝐶^𝑇𝑄𝑦^𝑇𝑦=𝑤^𝑇𝑅𝑤+𝜌^𝑇₀𝐿𝜌₀𝑑−2𝜌𝑑𝑡^𝑇0𝑦−𝑑𝑦𝑑𝑡^𝑇𝐾_𝑠𝑡𝑦+𝑦−𝐶𝑦^𝑇𝑄𝑦−𝐶𝑦−𝑦^𝑇𝑄𝑦. (2.13) Hence, taking into account that 𝜌₀(𝑡)→0,𝑦(𝑡)→0,𝑡→+∞ (see, for details [5]), we obtain₀^+∞Δ𝑦,𝑤𝑑𝑡=₀^+∞𝑤^𝑇𝑅𝑤+𝑦−𝐶𝑦^𝑇𝑄+𝑦−𝐶𝑦𝑑𝑡_0+∞𝜌^𝑇₀𝐿𝜌₀−𝑦^𝑇𝑄𝑦𝑑𝑡+2𝜌₀(0)^𝑇𝑥₀+𝑥₀𝐾_𝑠𝑡𝑥₀=𝐽𝑦,𝑤,𝑥₀+2𝜌₀(0)^𝑇𝑥₀+𝑥₀𝐾_𝑠𝑡𝑥₀+𝑐, (2.14) where 𝑐=∫₀^+∞[𝜌^𝑇₀𝐿𝜌₀−𝑦^𝑇𝑄𝑦]𝑑𝑡.
Notice, that Δ(𝑦,𝑤)≥0 and Δ(𝑥,𝑢)≡0. This shows that, indeed, (𝑥,𝑢) is an optimal solution to (1.1)–(1.3) (with fixed 𝑥₀) and proves (2.6).

Remark 2.3. By (2.14) and Δ(𝑥,𝑢)≡0, we have 𝐽(𝑦,𝑤,𝑥₀)≥𝐽(𝑥,𝑢,𝑥₀) and the equality occurs if and only if (𝑦,𝑤)≡(𝑥,𝑢) (see also (2.9)). Hence (𝑥,𝑢) is a unique solution to the problem (1.1)–(1.3). Similary reasoning works in discrete time case.

3. Steady-State Deterministic Kalman Filtering

In light of (2.7), it is natural to consider the process 𝑧(𝑡)=−𝐾⁻¹_𝑠𝑡𝜌₀(𝑡),𝑡∈[0,+∞) (3.1) as a natural estimate for the optimal solution to problem (1.1)–(1.3). Let us find the differential equation for 𝑧.

Proposition 3.1. One has 𝑧̇=𝐴𝑧+𝐾⁻¹_𝑠𝑡𝐶^𝑇Q𝑦−𝐶𝑧. (3.2)

Remark 3.2. Notice that 𝐾⁻¹_𝑠𝑡 is a solution to the algebraic equation 𝐿−𝑃𝐶^𝑇𝑄𝐶𝑃+𝐴𝑃+𝑃𝐴^𝑇=0. (3.3) In other words, the differential equation (3.2) is a precise deterministic analogue for the stochastic differential equation describing the optimal (steady-state) estimation in Kalman filtering problem. See, for example, [2].

Proof. Using (2.2) and (3.1), we obtain 𝑧̇=𝐾⁻¹_𝑠𝑡𝐴+𝐿𝐾_𝑠𝑡^𝑇𝜌₀+𝐾⁻¹_𝑠𝑡𝐶^𝑇𝑄𝑦=−𝐾⁻¹_𝑠𝑡𝐴^𝑇+𝐿𝐾_𝑠𝑡z+𝐾⁻¹_𝑠𝑡𝐶^𝑇𝑄𝑦. (3.4) Since 𝐾_𝑠𝑡 is a solution to (2.1), we have −𝐾⁻¹_𝑠𝑡𝐴^𝑇𝐾_𝑠𝑡−𝐿𝐾_𝑠𝑡=𝐴−𝐾⁻¹_𝑠𝑡𝐶^𝑇𝑄𝐶. (3.5) Hence, 𝑧̇=𝐴𝑧−𝐾⁻¹_𝑠𝑡𝐶^𝑇𝑄𝐶𝑧+𝐾⁻¹_𝑠𝑡𝐶^𝑇𝑄𝑦. (3.6) Hence, we obtain (3.2).

Remark 3.3. Notice that due to (3.1) Δ(𝑧,0)≡0 and consequently (𝑧,0) would be an optimal solution to (1.1)–(1.3) if it were feasible for this problem.

4. The Solution of the Discrete Deterministic Problem

It is natural to consider the discrete version for the problem (1.1)–(1.3). In this case, the problem can be reformulated as follows: 𝐽𝑥,𝑢,𝑥₀=12^∞_𝑘=1𝑢^𝑇_𝑘𝑅𝑢_𝑘+𝐶𝑥_𝑘−𝑦_𝑘^𝑇𝑄𝐶𝑥_𝑘−𝑦_𝑘𝑥⟶min,_𝑘+1=𝐴𝑥_𝑘+𝐵𝑢_𝑘,𝑥₀=𝑥_𝑜. (4.1)(4.2)(4.3)

Here we let 𝑥 denote a sequence {𝑥_𝑘}⊂ℝ^𝑛 for 𝑘=0,…,∞. We say that 𝑥∈𝑙^𝑛₂(ℕ) if ∑^∞_𝑖=1‖𝑥_𝑖‖²<∞, where ‖⋅‖ is a norm induced by an inner product ⟨,⟩ in ℝ^𝑛. Let (𝑥,𝑢)∈𝑙^𝑛₂(ℕ)×𝑙^𝑚₂(ℕ).

Like in the continuous case, we assume that the pair (𝑥,𝑢)∈𝑎(𝑥₀)+𝑍, where 𝑍 is a vector subspace of the Hilbert space 𝑙^𝑛₂(ℕ)×𝑙^𝑚₂(ℕ).

Observe now the inner product in 𝐻 has the following form: 𝑥,𝑦,(𝑢,𝑣)_𝐻=^∞_𝑘=0𝑥_𝑘,𝑢_𝑘+𝑦_𝑘,𝑣_𝑘. (4.4)

The vector subspace 𝑍 now takes the following form: 𝑍=(𝑥,𝑢)∈𝐻∶𝑥_𝑘+1=𝐴𝑥_𝑘+𝐵𝑢_𝑘,𝑘=0,1,…,𝑥₀=0. (4.5) Here 𝐴 is an 𝑛 by 𝑛 matrix. 𝐵 is an 𝑛 by 𝑚 matrix. 𝑅=𝑅^𝑇 is an 𝑛 by 𝑛 and positive definite. 𝑄=𝑄^𝑇 is an 𝑟 by 𝑟 and positive definite. 𝐶 is an 𝑟 by 𝑛 matrix and 𝑦∈𝑙^𝑟₂(ℕ).

As in the continuous case, we interpret (4.1)–(4.3) as an estimation problem of the form 𝑥_𝑘+1=𝐴𝑥_𝑘+𝐵𝑢_𝑘,𝑦_𝑘=𝐶𝑥_𝑘+𝑣_𝑘, (4.6) where we try to estimate 𝑥 with the help of observation 𝑦 by minimizing perturbations 𝑢, 𝑣 and choosing an appropriate initial condition 𝑥₀.

According to [4], a general cost function for a discrete linear-quadratic control problem with linear term on the cost function has the following form: 𝐽𝑥,𝑢,𝑥₀=^∞_𝑘=112𝑥^𝑇_𝑘𝑄𝑥_𝑘+𝑢^𝑇_𝑘𝑅𝑢_𝑘+𝑥^𝑇_𝑘𝜓_𝑘+𝑢^𝑇_𝑘𝜙_𝑘⟶min, (4.7) where 𝜓∈𝑙^𝑛₂(ℕ) and 𝜙∈𝑙^𝑚₂(ℕ). The solution to the particular class of problems can be completely described by solving several system of recurrence relations and the following discrete algebraic Riccati equation (DARE): 𝐾=𝐴^𝑇𝐾𝐴−𝐴^𝑇𝐾𝐵𝑅+𝐵^𝑇𝐾𝐵⁻¹𝐴^𝑇𝐾𝐵^𝑇+𝑄. (4.8) We assume that this equation has a positive definite stabilizing solution 𝐾_𝑠𝑡. For sufficient conditions, see [6].

In our situation, we have 𝐽𝑥,𝑢,𝑥₀=12^∞_𝑘=1𝑢^𝑇_𝑘𝑅𝑢_𝑘+𝐶𝑥_𝑘−𝑦_𝑘^𝑇𝑄𝐶𝑥_𝑘−𝑦_𝑘⟶min. (4.9) It is easy to see that 𝜓_𝑘=−𝐶^𝑇𝑄𝑦_𝑘 and 𝜙_𝑘=0,𝑘=0,1,…. By [4], there is a unique solution 𝜌={𝜌_𝑘}∈𝑙^𝑛₂(ℕ) of the following recurrence relations 𝜌_𝑘=𝐴^𝑇−A^𝑇𝐾_𝑠𝑡𝐵𝑅+𝐵^𝑇𝐾_𝑠𝑡𝐵⁻¹𝐵^𝑇𝜌_𝑘+1+𝐶^𝑇𝑄𝑦_𝑘. (4.10) For details on an explicit solution of the above recurrence relation, see [4]. For simplicity, we let 𝑅=𝑅+𝐵^𝑇𝐾_𝑠𝑡𝐵, (4.11) and we also let 𝐿=𝐵𝑅⁻¹𝐵^𝑇. (4.12) So our recurrence relation for 𝜌 now takes the form 𝜌_𝑘=𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿𝜌_𝑘+1+𝐶^𝑇𝑄𝑦_𝑘 (4.13) with the corresponding DARE 𝐾=𝐴^𝑇𝐾𝐴−𝐴^𝑇𝐾𝐵𝑅⁻¹𝐴^𝑇𝐾𝐵^𝑇+𝐶^𝑇𝑄𝐶,𝐾=𝐴^𝑇𝐾𝐴−𝐴^𝑇𝐾𝐿𝐾𝐴+𝐶^𝑇𝑄𝐶. (4.14) The optimal solution to (4.1)–(4.3) has the following form: 𝑥_𝑘+1=𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿^𝑇𝑥_𝑘+𝐿𝜌_𝑘+1,𝑢_𝑘=−𝑅⁻¹𝐵^𝑇𝐾_𝑠𝑡𝐴𝑥_𝑘+𝑅⁻¹𝐵^𝑇𝜌_𝑘+1. (4.15)(4.16) For details, see [4]. To solve the original problem (4.1)–(4.3) we need to express the minimal value of the functional (4.1) in terms of 𝑥₀.

Theorem 4.1. Let (𝑥,𝑢) be an optimal solution of (4.1)–(4.3) with fixed 𝑥₀ given by (4.15)-(4.16). Then 𝐽𝑥,𝑢,𝑥₀=12𝑥^𝑇₀𝐾_𝑠𝑡𝑥₀−𝜌^𝑇₀𝑥₀+12^∞_𝑘=02𝑦^𝑇_𝑘𝑄𝑦_𝑘−𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1. (4.17)

Proof. For simplicity of notation, we use 𝐾 for 𝐾_𝑠𝑡. Let Δ𝑦_𝑘,𝑤_𝑘=𝑤_𝑘+𝑅⁻¹𝐵^𝑇𝐾𝐴𝑦_𝑘−𝜌_𝑘+1^𝑇⋅𝑅⋅𝑤_𝑘+𝑅⁻¹𝐵^𝑇𝐾𝐴𝑦_𝑘−𝜌_𝑘+1=Δ₁+Δ₂+Δ₃, (4.18) where Δ₁=𝑤^𝑇_𝑘𝑅𝑤_𝑘,Δ₂=2𝐾𝐴𝑦_𝑘−𝜌_𝑘+1^𝑇𝐵𝑤_𝑘=2𝐾𝐴𝑦_𝑘−𝜌_𝑘+1^𝑇𝑦_𝑘+1−𝐴𝑦_𝑘=2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝜌^𝑇_𝑘+1𝐴𝑦_𝑘,Δ₃=𝐾𝐴𝑦_𝑘−𝜌_𝑘+1^𝑇𝐿𝐾𝐴𝑦_𝑘−𝜌_𝑘+1=𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝜌_𝑘+1. (4.19) We assume that (𝑦,𝑤)∈𝑎(𝑥₀)+𝑍. Since 𝐴^𝑇𝐾𝐴−𝐴^𝑇𝐾𝐿𝐾𝐴=𝐾−𝐶^𝑇𝑄𝐶 and [𝐴^𝑇−𝐴^𝑇𝐾𝐿]𝜌_𝑘+1=𝜌_𝑘−𝐶^𝑇𝑄𝑦_𝑘, we have Δ𝑦_𝑘,𝑤_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐴−𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝜌^𝑇_𝑘+1𝐴𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝜌_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1=𝑤^𝑇_𝑘𝑅𝑤_𝑘−2𝑦^𝑇_𝑘𝐾−𝐶^𝑇𝑄𝐶𝑦_𝑘−𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇𝜌_𝑘+1+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝜌_𝑘+1+2𝑦^𝑇_𝑘A^𝑇𝐾𝑦_𝑘+1=𝑤^𝑇_𝑘𝑅𝑤_𝑘−2𝑦^𝑇_𝑘𝐾−𝐶^𝑇𝑄𝐶𝑦_𝑘−𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇−𝐴^𝑇𝐾𝐿𝜌_𝑘+1+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1=𝑤^𝑇_𝑘𝑅𝑤_𝑘−2𝑦^𝑇_𝑘𝐾−𝐶^𝑇𝑄𝐶𝑦_𝑘−𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝑦^T_𝑘𝜌_𝑘−𝐶^𝑇𝑄𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1=𝑤^𝑇_𝑘𝑅𝑤_𝑘−2𝑦^𝑇_𝑘𝐾−𝐶^𝑇𝑄𝐶𝑦_𝑘−𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐿𝐾𝐴𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝑦^𝑇_𝑘𝜌_𝑘−2𝑦^𝑇_𝑘𝐶^𝑇𝑄𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1. (4.20) By recalling now the definition of 𝑅, we have 𝑤^𝑇_𝑘𝑅𝑤_𝑘=𝑤^𝑇_𝑘𝑅+𝐵^𝑇𝐾𝐵𝑤_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘+𝑤^𝑇_𝑘𝐵^𝑇𝐾𝐵𝑤_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘+𝑦_𝑘+1−𝐴𝑦_𝑘^𝑇𝐾𝑦_𝑘+1−𝐴𝑦_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘+𝑦^𝑇_𝑘+1𝐾𝑦_𝑘+1+𝑦^𝑇_𝑘𝐴^𝑇𝐾𝐴𝑦_𝑘−2𝑦^𝑇_𝑘𝐴^𝑇𝐾𝑦_𝑘+1. (4.21) Therefore, Δ𝑦_𝑘,𝑤_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘−𝑦^𝑇_𝑘𝐾𝑦_𝑘−𝑦^𝑇_𝑘𝐶^𝑇𝑄𝐶𝑦_𝑘−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝑦^𝑇_𝑘𝜌_𝑘−2𝑦^𝑇_𝑘𝐶^𝑇𝑄𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+𝑦^𝑇_𝑘+1𝐾𝑦_𝑘+1. (4.22) We then rearrange the terms and complete the square to obtain a useful expression for Δ: Δ𝑦_𝑘,w_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘−𝑦^𝑇_𝑘𝐾𝑦_𝑘+𝑦^𝑇_𝑘+1𝐾𝑦_𝑘+1−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝜌^𝑇_𝑘𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+𝑦^𝑇_𝑘𝐶^𝑇𝑄𝐶𝑦_𝑘−2𝑦^𝑇_𝑘𝐶^𝑇𝑄𝑦_𝑘=𝑤^𝑇_𝑘𝑅𝑤_𝑘−𝑦^𝑇_𝑘𝐾𝑦_𝑘+𝑦^𝑇_𝑘+1𝐾𝑦_𝑘+1−2𝜌^𝑇_𝑘+1𝑦_𝑘+1+2𝜌^𝑇_𝑘𝑦_𝑘+𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1+𝐶𝑦_𝑘−𝑦_𝑘^𝑇𝑄𝐶𝑦_𝑘−𝑦_𝑘−2𝑦^𝑇_𝑘𝑄𝑦_𝑘. (4.23) Notice, since we fixed 𝑥₀, we let 𝑦₀=𝑥₀ and take summation of both sides: ^∞_𝑘=0Δ𝑦_𝑘,𝑤_𝑘=−𝑥^𝑇₀𝐾𝑥₀+2𝜌^𝑇₀𝑥₀+^∞_𝑘=0𝑤^𝑇_𝑘𝑅𝑤_𝑘+𝐶𝑦_𝑘−𝑦_𝑘^𝑇𝑄𝐶𝑦_𝑘−𝑦_𝑘+^∞_𝑘=0𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1−2𝑦^𝑇_𝑘𝑄𝑦_𝑘. (4.24) By the definition of Δ(𝑦_𝑘,𝑤_𝑘), Δ(𝑥_𝑘,𝑢_𝑘)=0. Therefore, 0=−𝑥^𝑇₀𝐾𝑥₀+2𝜌^𝑇₀𝑥₀+2𝐽𝑥,𝑢,𝑥₀+^∞_𝑘=0𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1−2𝑦^𝑇_𝑘𝑄𝑦_𝑘. (4.25) As a result, 𝐽𝑥,𝑢,𝑥₀=12𝑥^𝑇₀𝐾𝑥₀−𝜌^𝑇₀𝑥₀+12^∞_𝑘=02𝑦^𝑇_𝑘𝑄𝑦_𝑘−𝜌^𝑇_𝑘+1𝐿𝜌_𝑘+1. (4.26) Then the proof is completed.

As in continuous case, for the discrete case, 𝐽(𝑥,𝑢,𝑥₀) is a strictly convex function of 𝑥₀ and hence minimum of 𝐽 as a function of 𝑥₀ is attained at 𝑥^opt₀=𝐾⁻¹_𝑠𝑡𝜌₀, (4.27) where 𝜌 is the unique 𝑙₂ solution to (4.13).

Since we have (4.27), it is natural to consider the process 𝑧_𝑘=𝐾⁻¹_𝑠𝑡𝜌_𝑘, (4.28) as an estimate for the optimal solution to problem (4.1)−(4.3). Let us find the recurrence relation for 𝑧_𝑘.

Proposition 4.2. Assuming that the closed loop matrix 𝐴-𝐿𝐾𝐴 is invertible, one has 𝑧_𝑘+1=𝐴𝑧_𝑘−𝐾⁻¹_𝑠𝑡𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿⁻¹𝑦_𝑘−𝐶𝑧_𝑘. (4.29)

Proof. We can rewrite (4.13) in the form 𝐾_𝑠𝑡𝑧_𝑘=𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝑧_𝑘+1+𝐶^𝑇𝑄𝑦_𝑘. (4.30)
Using the algebraic Riccati equation 𝐾_𝑠𝑡=𝐴^𝑇𝐾_𝑠𝑡𝐴−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝐴+𝐶^𝑇𝑄𝐶, (4.31) we can rewrite (4.30) in the form 𝐶^𝑇𝑄𝐶𝑧_𝑘+𝐴^𝑇𝐾_𝑠𝑡−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝐴𝑧_𝑘=𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝑧_𝑘+1+𝐶^𝑇𝑄𝑦_𝑘, (4.32) which is equivalent to 𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝑧_𝑘+1=𝐴^𝑇−𝐴^𝑇𝐾_𝑠𝑡𝐿𝐾_𝑠𝑡𝐴𝑧_𝑘−𝐶^𝑇𝑄𝑦_𝑘−𝐶𝑧_𝑘. (4.33) The result follows.

Remark 4.3. Notice that (4.29) is the analogue of the “limiting” discrete Kalman filter [6, Page 384, (17.6.1)].

5. Concluding Remarks

In this paper, we relate a deterministic Kalman filter on semi-infinite interval to linear-quadratic tracking control model with unfixed initial condition. Solutions of the deterministic problems both continuous and discrete cases are described. This extends the result of Sontag to semi-infinite interval.

Acknowledgments

The research of L. Faybusovich was partially supported by National science foundation. Grant DMS07-12809. The research of T. Mouktonglang was partially supported by the Centre of Excellence in Mathematics and the Commission for Higher Education (CHE), Sri Ayutthaya Road, Bangkok, Thailand.