Key words and phrases : The target control problem, the Markov moment problem, the Hausdorff moment problem

https://doi.org/10.14317/jami.2020.239

MOMENT APPROACH TO THE TARGET CONTROL PROBLEM FOR LINEAR SYSTEM

CHUNJI LI^∗AND HAN YAO

Abstract. In this paper, we consider the target control problem for the linear systems by using the solution of the Hausdorff moment problem.

AMS Mathematics Subject Classification : 92B05, 34D20.

Key words and phrases : The target control problem, the Markov moment problem, the Hausdorff moment problem.

1. From beginning to obtaining the data Consider the following linear continuous system







˙

x (t) = Ax (t) + Bu (t) , y (t) = C (t) x (t) ,

s.t. x (0) = x0, and r (θ) = y (θ) ,

(1) where x (t) ∈ Rⁿ is the state vector, u (t) ∈ R^mis the input vector, y (t) ∈ R^pis the output vector, r (t) ∈ R^p is the target signal, A ∈ R^n×n, B ∈ R^n×m, C (t) ∈ R^p×n. Given an initial condition x0∈ Rⁿ, and a time θ, find one of the controls

|u (t)| ≤ 1 such that the trajectory from x0of the system (1) arrives to the target signal at time θ. This problem is called the target control (TC) problem.

Moment problem is related to operator theory and has many applications (see [1], [2], [4], and [5], etc). In this paper, we consider the target control problem for the linear systems by using the solution of the Hausdorff moment problem.

Let C0,Lbe the set of all measurable functions on [0, θ] such that 0 ≤ f (τ ) ≤ L for all τ ∈ [0, θ] . Then the L-Markov moment problem (MMP) for the interval [0, θ] is stated as follows: Given a finite sequence of real numbers c₀, c₁, . . . , c_k, find the set of functions f ∈ C_0,L such that

c_j = Z θ

0

τ^jf (τ ) dτ, j = 0, 1, . . . , k.

Received February 8, 2020. Revised February 26, 2020. Accepted March 10, 2020.

∗Corresponding author.

2020 KSCAMc .

239

Let M [0, θ] be the set of all nonnegative measures on [0, θ] . Then the Haus- dorff moment problem (HMP) for an interval [0, θ] is stated as follows: Given a finite sequence of real numbers s0, s1, . . . , sk, find the set of measures σ ∈ M [0, θ]

such that

s_j = Z θ

0

τ^jdσ (τ ) , j = 0, 1, . . . , k.

Recall from [5] that there is a bijection between the set C0,L and the set of measures σ ∈ M [0, θ] satisfyingRθ

0dσ (τ ) = 1. This bijection is given by Z θ

0

dσ (τ ) τ − z = −1

zexp 1 L

Z θ 0

f (τ ) dτ z − τ

! ,

which determines the relation between (cj)^k−1_j=0 and (sj)^k_j=0: s0= 1, s1= c1, and

sk= 1 k!

c1 −1 · · · 0

2c₂ c₁ . .. 0

... . .. . .. ... (k − 1) c_k−1 (k − 2) c_k−2 . .. − (k − 1)

kc_k (k − 1) c_k−1 · · · c₁             

, k ≥ 2. (2)

By [5, Theorem 2.1], we know that the MMP is solvable with moments (cj)ⁿ⁻¹_j=0 if and only if the HMP with (s_j)ⁿ_j=0 is solvable. As usual, we let δ_ij be the Kronecker symbol.

Theorem 1. Let A = (δi,j+1)ⁿ_i,j=1, B = 1 0 0 · · · 0
T

∈ Rⁿ, and C (θ) be invertible. And let

Φ (θ) := e^−AθC⁻¹(θ) r (θ) = Φ1(θ) , Φ2(θ) , · · · , Φn(θ)
^T . Then the TC problem (1) is solvable if and only if the Markov moment problem with L = 1 is solvable with entries c_i(i = 1, 2, . . . n) as the following

ci=(Φ_i(θ) − x_0i) i! + (−1)ⁱ⁻¹θⁱ

2 (−1)ⁱ⁻¹i . (3)

Proof. Since C (θ) is invertible, we have C⁻¹(θ) r (θ) = x (θ) = e^Aθ x0+

Z θ 0

e^−AτBu (τ ) dτ

! . That is,

e^−AθC⁻¹(θ) r (θ) − x₀= Z θ

0

e^−AτBu (τ ) dτ.

Since u (τ ) = 2f (τ ) − 1, Z θ

0

e^−AτBu (τ ) dτ = 2 Z θ

0

e^−AτBf (τ ) dτ − Z θ

0

e^−AτBdτ, and

e^−AτB =

















1 0 0 · · · 0 0

−τ 1 0 · · · 0 0

τ²

2! −τ 1 · · · 0 0

... ... . .. . .. 1 0

(−1)ⁿ⁻¹τⁿ⁻¹

(n−1)! · · · −τ 1

































 1 0 0 ... 0 0



















=















 1

−τ

τ² 2!...

(−1)ⁿ⁻¹τⁿ⁻¹ (n−1)!















 ,

we obtain

(Φ_i(θ) − x_0i) i! + (−1)ⁱ⁻¹θⁱ 2 (−1)ⁱ⁻¹i =

Z θ 0

τⁱ⁻¹f (τ ) dτ, i = 1, 2, . . . n.

Thus, if the Markov moment problem with L = 1 is solvable with entries ci

as in (3), then the TC problem (1) is solvable, i.e., the trajectory x(t) satisfies x (0) = x0, and r (θ) = y (θ). The inverse implication can be reserved step by

step. 

In [10], the authors introduced the method for obtaining the admissible control of the system (1). We summerize that as following algorithm.

I. Calculate all data ci of (3).

II. Calculate s_i by (2).

III. Calculate Hi, ui, vi, T and RT(z) by the following relations

if n = 2k + 1 if n = 2k

H1









s1 · · · sk+1

..

. . .. ... sk+1 · · · s2k+1

















s0 · · · sk

..

. . .. ... sk · · · s2k









H2









θs0− s1 · · · θsk− s_k+1 ..

. . .. ... θsk− sk+1 · · · θs2k− s2k+1

















θs1− s2 · · · θsk− s_k+1 ..

. . .. ... θsk− sk+1 · · · θs2k−1− s2k







 u1 (−s0, −s1, . . . , −sk)^T (0, −s0, . . . , −sk−1)^T

T (δi,j+1)^k_i,j=0 (δi,j+1)^k_i,j=0

u2 (θT − 1) u1 (s1− θs0, s2− θs1, . . . , sk− θs_k−1)^T v1 (1, 0, . . . , 0)^T∈ R^k+1 (1, 0, . . . , 0)^T∈ R^k+1

v2 (1, 0, . . . , 0)^T∈ R^k+1 (1, 0, . . . , 0)^T∈ R^k T1 (δi,j+1)^k_i,j=0 (δi,j+1)^k_i,j=0 T2 (δi,j+1)^k_i,j=0 (δi,j+1)^k_i,j=0

RT(z) (I − zT )⁻¹ (I − zT )⁻¹

IV. Calculate U11, U12, U21, and U22 by the following relations

if n is odd if n is even

U11(z) 1 − zu^∗₂R_T^∗

1 (z) H₂⁻¹v1 1 − zu^∗₁R_T^∗

1(z) H₁⁻¹v1

U12(z) u^∗₁R_T^∗

1 (z) H₁⁻¹u1 M − zu^∗₁R_T^∗

1(z) H₁⁻¹v1M + zu^∗₁R_T^∗

1(z) H₁⁻¹u1

U21(z) − (θ − z) zv^∗₁RT₁^∗(z) H₂⁻¹v1 −zv^∗₁RT₁^∗(z) H₁⁻¹v1

U22(z) 1 + zv^∗₁R_T^∗

1(z) H₁⁻¹u1 1 − zv^∗₁R_T^∗

1(z) H₁⁻¹v1M + zv₁^∗R_T^∗

1 (z) H⁻¹₁ u1

M

 1 + θ



u^∗₁H₁⁻¹v1− u^∗₂H₂⁻¹v2

 

θv^∗₁H₁⁻¹v1

−1

V. Let z = t + i, and calculate −zs (z) by the following relations

−zs (z) = U11(θ − (t + i)) (F + iG + iπ) + U12

U21(θ − (t + i)) (F + iG + iπ) + U22

, where

F = 1

2ln(θ − t)²+ ²

t²+ ² , G = arctan θ

t²− T t + ².

VI. Let  = 0, and calculate the real part X and the imaginary part Y of

−zs (z) .

VII. Calculate X and Y, and obtain u (t) = −2

πargY X − 1.

Theorem 2. The TC problem (if n = 1)

˙

x = u (t) , |u| ≤ 1, x (0) = x₀, s.t. x (θ) = x_f, (4) is admissible if and only if |x0− xf| ≤ θ. In this case, u (t) = −²_πarg^Y_X − 1, where

X = 1

8

nt (t − θ)³(θ − x₀+ x_f)²(2t − θ − x₀+ x_f) F² +1

2(t − θ)

θ²− (x₀− x_f)² 

θ²− (x₀− x_f)²+ 8t (t − θ) F +π²t (t − θ)³(θ − x0+ xf)²(2t − θ − x0+ xf)

+ (θ + x0− xf)²(2t − θ + x0− xf)o

, with F = lnθ − t t ,

Y = 1

16π (t − θ) (θ − x0+ xf)²(θ + x0− xf)²≤ 0.

Proof. By the algorithm. 

Corollary 3. The TC problem (if n = 1)

˙

x = ax + u (t) , a < 0, |u| ≤ 1, x (0) = x0, s.t. x (θ) = xf, (5) is admissible if and only if 

x0− xfe^−aθ

≤ θ. In this case, u (t) = −²_πarg_X^Y − 1
e^at, where X and Y are as in Theorem 2.

Proof. Let z = xe^−at, then by Theorem 2, we know that

˙

z = u (t) e^−at:= ˜u (t) , |˜u| ≤ 1, z (0) = z0, s.t. z (θ) = zf,

is admissible if and only if |z0− zf| ≤ θ, that is,

x0− xfe^−aθ

≤ θ. In this case,

˜

u (t) = −_π²arg^Y_X − 1, that is, u (t) = −_π²arg_X^Y − 1
e^at.  2. Examples

In this section, we give some interesting examples, for n = 2 and n = 3.

Example 1. We consider











˙ x (t) =

 0 0 1 0

 x (t) +

 1 0

 u (t) , y (t) =

 1 sin^πt₃

0 1

 x (t) ,

(6)

with initial state vector point x (0) = (x01, x02)^T and output terminal point y (3) = (2, 3)^T, i.e., θ = 3. In this case, we can find the admissible region

Rad=



(x01, x02)^T   1 4x²₀₁−5

2x01−5

4 ≤ x02≤ −1 4x²₀₁−1

2x01+5 4

 .

Fig. 1. The admissible region R_adof system (6).

Green line is x₀₂=¹₄x²₀₁−⁵₂x₀₁−⁵₄, red line is x₀₂= −¹₄x²₀₁−¹₂x₀₁+⁵₄. In particular, we take x₀₁= 0, x₀₂= 1. Then (0, 1)^Tis in the above admissible region and according to the algorithm, we finally obtain

X = 1

81t (t − 3) (t (t − 3) (20t − 59) (20t − 9))

 ln3 − t

t

2

+1

81t (t − 3) 1776t²− 480t³− 1062t + 81

 ln3 − t

t



+1

81t (t − 3) π²t (t − 3) (20t − 59) (20t − 9) + 9 (4t − 1) (4t − 3)
, Y = πt (t − 3) .

The numerical roots of X in [0, 3] are t₁ ≈ 0.0064797, t2 ≈ 0.459614. Hence the control is given by the following

u (t) =







−_π² arctan _X^Y
− π
− 1, if 0 ≤ t ≤ 0.0064797,

−_π² arctan _X^Y

− 1, if 0.0064797 ≤ t ≤ 0.459614,

−_π² arctan _X^Y
− π
− 1. if 0.459614 ≤ t ≤ 3,

Fig. 2. The plot of u (t) for system (6)

The plots of state vector and output curve of system (6) are as following.

Fig. 3. The state vector (pink) and output curve (green) of system (6) Example 2. We consider



















˙ x (t) =





0 0 0 1 0 0 0 1 0



x (t) +



 1 0 0



u (t) ,

y (t) =





1 sin^πt₃ 0

0 1 0

0 0 1



x (t) ,

(7)

with initial state vector point x (0) = 0,¹₂, 0
^T

and output terminal point y (3) = (1, 2, 3)^T, i.e., θ = 3. In this case, we can obtain

X = 1

4t²(3t − 5) −8t + 3t²+ 2
(t − 3)³

 ln3 − t

t

²

+1

5t (t − 3) −60t + 101t²− 54t³+ 9t⁴+ 5

 ln3 − t

t



+ 1

4π²t²(3t − 5) −8t + 3t²+ 2
(t − 3)³+ 1

25t (3t − 4) −10t + 3t²+ 5

 , Y = πt (t − 3) .

The numerical roots of X in [0, 3] are t1 ≈ 0.005692, t2 ≈ 0.289924, t3 ≈ 1.66483, t4≈ 2.34597, and t5≈ 2.9001. Hence the control is given by the follow- ing

u (t) =



















−²_π arctan ^Y_X
− π
− 1, if 0 ≤ t ≤ 0.005692,

−²_π arctan ^Y_X

− 1, if 0.005692 ≤ t ≤ 0.289924,

−²_π arctan ^Y_X
− π
− 1, if 0.289924 ≤ t ≤ 1.66483,

−²_π arctan ^Y_X

− 1, if 1.66483 ≤ t ≤ 2.34597,

−²_π arctan ^Y_X
− π
− 1, if 2.34597 ≤ t ≤ 2.9001,

−²_π arctan ^Y_X

− 1. if 2.9001 ≤ t ≤ 3,

Fig. 4. The plot of u (t)

The plots of state vector and output vector of system (7) are as following.

Fig. 5. The plots of state vector and output vector of system (7)

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Chunji Li received Ph.D. degree from Kyungpook National University, Korea. His research interests focus on the control theory, moment method, and unilateral weighted shifts.

Department of Mathematics, Northeastern University, Shenyang 110004, R. R. China.

e-mail: lichunji@mail.neu.edu.cn

Han Yao is a Master course student of Northeastern University. Her research interests focus on the control theory.

Department of Mathematics, Northeastern University, Shenyang 110004, R. R. China.

e-mail: Alice9312@sina.com