]> 8.7 Formal derivation of the optimal cost

8.7 Formal derivation of the optimal cost

It is known, see (6.11), (6.14), (6.15) that the Lyapunov operator equations for the closed–loop and open–loop systems are respectively
𝒜cx,x + x,𝒜cx = c#x2 g#x2x D(𝒜 c) (8.27)
𝒜x,x + x,𝒜x = c#x2 + g#x2x D(𝒜) (8.28)

In (8.27) the closed–loop operator 𝒜c is given by (8.20), while in (8.28) the open–loop operator 𝒜 is defined as in (8.8). In both formulae (8.27) and (8.28) the functional g# is determined by (8.18). Solutions to (8.27) and (8.28) will be sought in the form

(x)(θ) = Hx(θ),H = H = h1h2 h2h3 L(2) ,

i.e. is the bounded linear operator of multiplication by a constant matrix H. Introducing all data into (8.27) and taking into account that x D(𝒜c), we obtain the linear system of equations characterizing h1, h2, h3:

h2 =0 h1 h3 = a2 q2 p2(p2 1)2 =h 3q2 p2 h1 (8.29)

Hence the solution of (8.27) takes the form

(x)(θ) = Hx(θ),H = q2 p2 q2 a2 + (p2 1)2 0 0 p2 q2a2 + (p2 1)2 (8.30)

Clearly, is a coercive self–adjoint bounded operator since qp < 1 is the necessary and sufficient condition for EXS of the closed–loop system. By EXS, is also a unique solution of (8.27).

Similar considerations for (8.28) with x D(𝒜c) replaced by x D(𝒜) lead to the linear system

h2 =0 h1 h3 = a2 q2(p2 1)2 =h 3b2 h 1 (8.31)

instead of (8.29). Hence, a unique solution to (8.28) turns out to be

(x)(θ) = Hx(θ),H = q2 1 b2 p2a2 (p2 1)2 0 0 a2 q2 (p2 1)2 (8.32)

Using the identities

pq = b,a2 = p2 + q2 1 p2q2 ,

which follow from (8.12), one can reduce both (8.30) and (8.32) to the form

(x)(θ) = Hx(θ),H = (p21) q20 0 1 .

Observe that the solution of (8.27) and (8.28) is not a HS operator, contrary to the standard lq problem for a SISO system – see Section 2.4.

Another interesting feature is that, again contrary to the standard lq problem for a SISO system, the inclusion D(𝒜) D(𝒜) does not hold for our special system. Indeed, by (8.8),

𝒜v = v,D(𝒜) = {v W1,2(r, 0) W1,2(r, 0) : v(r) = AT v(0)} .

Moreover,

D(𝒜) = {v W1,2(r, 0) W1,2(r, 0) : HA1H1v(0) = v(r)} .

If the inclusion D(𝒜) D(𝒜) were true, then AT HA = H would hold. But, since λ(A) < 1, this is impossible in view of the discrete Lyapunov matrix equation theory. Thus

D(𝒜) D(𝒜) (8.33)