]> 2.7 Solution of the parametric optimization problem

2.7 Solution of the parametric optimization problem

For the evaluation of the performance index (2.6), we put H = L2(0, 1), Y = throughout Section 2.4, and assume K ( cosh π, 1), with K1. From Lemmas 2.6.1 and 2.6.2, and the Remark 2.4.1, it follows that the observation functional 𝒞 is bounded on Hα for α (1 4, 1 2). Then (2.27) is satisfied and, by Theorems 2.4.1, 2.4.2 and 2.4.4, the Lyapunov operator equation
𝒜x1,x2H + x1,𝒜x2H = 𝒞x1𝒞x2¯x1,x2 D(𝒜)

has a unique solution L(H) satisfying = and 0, which is an HS operator. Therefore is the integral operator [86, Theorem 6.11, p. 139] defined by

(x)(θ) =01h(θ,ϑ)x(ϑ)dϑ,x H (2.43)

with a symmetric and positive–definite kernel h L2((0, 1) × (0, 1)) (see [28, Section III, 10.1] for explanation). Moreover, since D(P) = H, (2.23), (2.24) give

J(x0) =0𝒞S(t)x 0 2dt = x 0,x0Hx0 H (2.44)

Theorem 2.4.5 and formulae (2.7) and (2.31) yield an additional characterization of :

x1,x2H = n= Res λ=λnLx1(λ)Lx2(λ) M(λ)M(λ) (2.45)

where

M(λ) = cosh λ K,Lf(λ) = 1 λ01 sinh λθf(θ)dθ,f H ,

with λn (n ) as given in (2.33). M is an entire function and for each fixed f H, the mapping Lf is also entire function with the properties:

M(λ¯)¯ = M(λ¯),Lf(λ¯)¯ = Lf(λ¯) .

We determine the value of J(1) at the constant function 1(θ) = 1 (0 θ 1), since finding the value of J at initial condition (2.5) requires only a slight modification. To do this, we put x1 = x2 = 1 in (2.44), (2.45), which yields

J(1) =1,1H = n= Res λ=λn cosh λ 1 cosh λ 1 λ cosh λ Kλ cosh λ K = =2(1 K) n= 1 λnλn sinh λn 1 1 K cos λn K (2.46)

where λn = μn are given by (2.35). In particular,

J(1) = 2(1 K) 1 K2 n= 1 (φ 2nπ)3 1 1 K cosh(φ 2nπ) cos(φ 2nπ) ,

provided that K < 1 and φ = arccos K. Employing the partial–fraction expansion of the function z(cosh z cos z)1 [56, Problem 5.2.7]:

1 cosh z cos z 1 z2 = k=1 Ak z zk + Ak¯ z zk¯ + Ak z + zk + Ak¯ z + zk¯ = = k=1 2Akzk z2 zk2 + 2Akzk¯ z2 zk¯2 = 4 k=1z2 Re(A kzk) Re(Akzk¯) zk 2 z4 2(Re zk2)z2 + zk 4 = = k=1 4z2(1)kkπ sinh kπ(z4 + 4k4π4) (2.47)

with Ak = 1 sinh zk + sin zk, zk = (1 + j)kπ (k ), we obtain

J(1) =2(1 K) 1 K2 n= 1 (φ 2nπ)3 2(1 K)2 1 K2 n= 1 (φ 2nπ)5 8π(1 K)2 1 K2 k=1(1)kk sinh kπ n= 1 (φ 2nπ)[4k4π4 + (φ 2nπ)4] (2.48)

By the method described in Krzyż [56, Problems 4.5.15, 4.6.4], we have

n= 1 (φ 2nπ)3 = cos φ 2 8 sin 3φ 2 , n= 1 (φ 2nπ)5 = 2 cos φ 2 + cos 3φ 2 96 sin 5φ 2 (2.49)

We now apply the following identity found by the residue method [56, Problem 4.6.1]:

n= 1 (φ 2nπ)[4k4π4 + (φ 2nπ)4] = π m=04 Res z=zm cot πz w(z) = = 1 32k4π4 4 cot φ 2 + 2 Re(cot πz1) + 2 Re(cot πz3) = = 1 16k4π4 2 cot φ 2 m=01 sin(2π Re z2m+1) cosh(2π Im z2m+1) cos(2π Re z2m+1) = = 1 16k4π4 2 cot φ 2 m=01 sin φ (1)mkπ cosh kπ cos φ 2(1)mkπ = = 1 8k4π4 cot φ 2 (1)k sin φ 8k4π4[cosh kπ (1)k cos φ],k (2.50)

where the following facts have been taken into account:

z0 = φ 2π,z1 = φ 2π k 2 jk 2 = z2¯,z3 = φ 2π + k 2 jk 2 = z4¯ , w(z) = (φ 2zπ)5 + 4k4π4(φ 2zπ) , w(z 0) = 8k4π5,w(z m) = 32k4π5form = 1, 2, 3, 4 .

Taking (2.49), (2.50) into account in (2.48), we come to

J(1) = (1 K) cos φ 2 41 K2 sin 3φ 2 (1 K)2(2 cos φ 2 + cos 3φ 2) 481 K2 sin 5φ 2 + + (1 K)2 π31 K2 k=1 (1)k sin φ k3 sinh kπ[(1)k cosh kπ K] k=1(1)k cot φ 2 k3 sinh kπ (2.51)

From the partial–fraction expansion of the function zz2 cosh z + cos z z4 cosh z z4 cos z, which can be easily derived from (2.47), we find

lim z0z2 cosh z + cos z z4 cosh z z4 cos z = 1 360 = 1 π3 k=1 (1)k k3 sinh kπ (2.52)

Eliminating φ from (2.51) and taking (2.52) into account, we obtain the final formula for J(1), valid for all K ( cosh π, 1):

J(1) = K2 17K + 106 360(1 K) + (1 K)2 π3 k=1 1 k3 sinh kπ[cosh kπ (1)kK] .

In comparison with (2.46), the formula (2.53) is more convenient for numerical computation of J(1) as a function of K, since it contains rapidly convergent series.

Now we are in position to solve Problem 2.2.1. To do this, we notice that the value of J at the initial condition x0 given by (2.5) is J(x0) = E2(1 K)2J(1) and therefore

E2J(x 0) = K2 17K + 106 360(1 K)3 + 1 π3 k=1 1 k3 sinh kπ[cosh kπ (1)kK] (2.53)

The plot of E2J(x 0) as a function of K, K ( cosh π, 1) is depicted in Figure 2.4. The minimal value of E2J(x 0) is 0.00193863 and it is achieved at K = 8.2078.


PIC

Figure 2.4: Plot of the performance index for Problem 2.2.1