]> 6.4.2 Example 2

6.4.2 Example 2

The Mach number control system

ż1(t) = az1(t) + kaz2(t r), t 0 ż2(t) = z3(t), t 0 ż3(t) = ω2z 2(t) 2ζωz3(t) + ω2u(t), t 0 z1(0) = z10 z2(0) = z20 z3(0) = z30 z2(θ) = φ2(θ), r θ 0,φ2 L2(r, 0) (6.50)

where a1 = 1.964, ω = 0.6, ζ = 0.8, r = 0.33, k = 0.0117, has been investigated in several papers – see [89, Section 4, pp. 17-20 and Section 6, p. 7] and the references therein. The problem is to find an optimal stabilizing lq controller minimizing the performance index

J =0[1002z 12(t) + u2(t)]dt (6.51)

Introduction of the new variables

ξ1(t) = z1(t + r),ξ2(t) = z2(t),ξ3(t) = z3(t) (6.52)

reduces our lq problem to a finite–dimensional one

ξ̇1(t) = aξ1(t) + kaξ2(t) ξ̇2(t) =ξ3(t) ξ̇3(t) = ω2ξ 2(t) 2ζωξ3(t) + ω2u(t) ξ1(0) =ξ10 = z 1(r) ξ2(0) =ξ20 ξ3(0) =ξ30 (6.53)
J =0r1002z 12(t)dt +0[1002ξ 12(t) + u2(t)]dt (6.54)

which can be solved both by classical methods and the spectral factorization. Here the restriction of z1 to an interval [0,r] is the solution of the nonhomogeneous differential equation

ż1(t) = az1(t) + kaφ2(t r),0 t r z1(0) =z10 .

In a Hilbert space H = 3 L2(1, 0; 3), (6.50) can be written as an abstract equation (6.1), 𝒜 defined by (6.42) and with

A = a 0 0 0 0 1 0 ω2 2ζω ,B = 0ka0 0 0 0 0 0 0 ,n = 3 ,

x = z1 z2 z3 ψ1 ψ2 ψ3 ,b = 0 0 ω2 0 0 0 ,c = 100 0 0 0 0 0 .

Now, det[sI A esrB] = (s2 + 2ζωs + ω2)(s + a) and the semigroup generated by 𝒜 is EXS. The pair (𝒜,b) is stabilizable, (𝒜,c) is detectable and the lq problem has a unique solution.

The rank condition (6.44) is not satisfied and thus (𝒜,b) is not approximately controllable. As we have mentioned at the end of Section 6.3, the full form of a controller can be found with the aid of a Lyapunov operator equation, however our intent is to find the controller without invoking the use of such an equation. It turns out that despite the lack of approximate controllability, this is still possible. To do this, we evaluate

(sI𝒜)1b = ω2 s2 + 2ζωs + ω2 ka s + aesr 1 s ka s + aes(θr) esθ sesθ (6.55)
c(sI 𝒜)1b = 100ω2kaesr (s2 + 2ζωs + ω2)(s + a) (6.56)

Taking f = 0 H, we find via (6.24), (6.25)

φ(s) = s3 + αs2 + βs + γ (s2 + 2ζωs + ω2)(s + a) (6.57)

where α,β,γ are positive solutions, αβ > γ satisfying the equations

a2 2ω2 + 4ζ2ω2 =α2 2β 2ω2a2 ω4 4a2ζ2ω2 =2αγ β2 1002a2k2ω4 + a2ω4 =γ2 (6.58)

Thus the values found in [89, Section 4, pp. 17 - 20] are confirmed to be: α = 1.5939, β = 1.0398, γ = 0.2821. By Riesz representation theorem the controller is assumed to be a vector

g = α1 α2 α3 β1() β2() β3() H .

Multiplying both sides of the identity (6.23) on the characteristic polynomial, we convert (6.23) into an identity for entire functions

kaω2esrα 1 + ω2(s + a)α 2 + ω2s(s + a)α 3 +r0kaω2es(θr)β 1(θ)dθ+ +r0ω2(s + a)esθβ 2(θ)dθ +r0ω2s(s + a)esθβ 3(θ)dθ = = s2(α a 2ζω) + s(β ω2 2aζω) + (γ aω2) (6.59)

Since the system is not approximately controllable one can try to apply Theorem 6.3.1 to find the optimal feedback. This requires verification whether (6.30) is satisfied. By (6.21) and the Paley–Wiener theorem (support arguments) we get

R(𝒜,c) = NO(c,𝒜) = x = z ψ 𝕄2 : kaesr(s + 2ζω)z 2 + kaesrz 3+ + s2 + 2ζωs + ω2 z 1 + kar0es(θ+r)ψ 2(θ)dθ = 0s = = x 𝕄2 : x = 0 0 0 ψ1 0 ψ3 = x 𝕄2 : z = z1 z2 z3 0 ψ2 0 (6.60)

Similarly, by (6.22)

NO(b,𝒜) = x = z ψ 𝕄2 : kaesrz 1 + (s + a)z2 + s(s + a)z3+

+r0kaes(θr)ψ 1(θ)dθ +r0(s + a)esθψ 2(θ)dθ+

+r0s(s + a)esθψ 3(θ)dθ = 0s .

Integrating–by–parts the second integral and applying the Paley–Wiener theorem one obtains the nontrivial closed subspace

NO(b,𝒜) = x = 0 z2 0 0 ψ2 ψ3 𝕄2 : ψ 3(θ) =rθψ 2(τ)dτ,z2 = r0ψ 2(θ)dθ (6.61)

and thus its orthogonal complement

R(𝒜,b) = x 𝕄2 : x = z1 z2 z3 ψ1 ψ2 ψ3 ,ψ2 = z2 +0θψ 3(τ)dτ

is not equal to 𝕄2. It follows from (6.60) and (6.61) that (6.30) holds contrary to (6.29) which is not satisfied.

As we already know, the left–hand side of (6.59) vanishes for g NO(b,𝒜). The system reduced to R(𝒜,c) arises by zeroing the first and third function components and hence β1 = 0, β3 = 0 in (6.59). Now, to eliminate integrals and exponential terms from the left–hand side of (6.59) and in order to get polynomial identity, the following equations should hold

β2(θ) = aβ 2(θ),β2(r) = akα1 , ω2 0 1 0 1 a0 a0 1 α2 α3 β2(0) = α a2ζω β ω2 2ζωa γ aω2 (6.62)

(6.62) has a unique solution, determining the optimal lq controller

α1 = ear 1 kaβ2(0), β2(0) =αa2 a3 aβ + γ ω2 α2 = aα + a2 + β ω2 ω2 , α3 =α a 2ζω ω2 β1 = β3 = 0, β2(θ) =eaθβ 2(0) (6.63)

(6.63) confirms the results obtained by Kappel and Salamon – see [89, Section 4, pp. 17 - 20] with the aid of the Riccati operator equation.

Observe also, that eigenvectors of 𝒜 corresponding to eigenvalues λ¯1, λ¯2, a, where λ1, λ2 are the roots of trinomial λ2 + 2ζωλ + ω2 are, respectively

fi = 0 ω2 λ¯i 0 0 0 (i = 1, 2),f3 = 1 ak(2ζω a) a2 2ζωa + ω2ear ak a2 2ζωa + ω2ear 0 akea(θ+r) 0

while eigenvectors of 𝒜 corresponding to λ1, λ2, a are, respectively

ei = ak λi + aeλir 1 λi ak λi + aeλi(θr) eλiθ λieλiθ (i = 1, 2),e3 = 1 0 0 eaθ 0 0 .

The next result shows that the optimal controller can be expressed as a linear combination of eigenvectors of 𝒜.

Lemma 6.4.1. We have:

span{f1,f2,f3}R(𝒜,c) , span{e1,e2,e3} R(𝒜,c) = {0} , span{e1,e2,e3} R(𝒜,b) ,

and there exist constants c1, c2, and c3 such that:

g = i=13c ifi (6.64)

Proof. All above properties of eigensystems of 𝒜 and 𝒜 are a consequence of the fact that their 3 components are linearly independent, i.e. the original system as well as its dual system are 3complete. Thus, only a proof of (6.64) will be given. We have to prove that there exist constants c1, c2, c3 such that

c3 ω2(c 1 + c2) + c3 ak(2ζω a) a2 2ζωa + ω2ear λ¯1c1 λ¯2c2 + c3 ak a2 2ζωa + ω2ear 0 c3akea(θ+r) 0 = g = α1 α2 α3 0 β2(0)eaθ 0 .

We take c3 = α1 = 1 akβ2(0)ear. Then c3akea(θ+r) = β 2(0)eaθ and the remaining constants c1, c2 can be uniquely (as λ1λ2 or equivalently 3 components of f1, f2, f3 are linearly independent) determined from the linear system of equations

ω2 ω2 λ¯1 λ¯2 c1 c2 = α2 2ζω a a2 2ζωa + ω2β2(0) α3 1 a2 2ζωa + ω2β2(0) .

Remark 6.4.1. Korytowski ([55]) discovered by generalization of formulae (6.33) and (6.52), a class of lq problems for degenerated time–delay systems which admit a reduction to a finite–dimensional lq problem. Both our examples are representatives of that class of retarded systems.