This paper has studied the output feedback regulation problem for 1-D anti-stable wave equation with distributed disturbance and a given reference signal generated by a finite-dimensional exosystem. We first design an observer for both exosystem and auxiliary PDE system to recover the state. Then we show the well-posedness of the regulator equations and propose an observer-based feedback control law to regulate the tracking error to zero exponentially and keep all the states bounded.
Output feedback regulation is a classical topic of control theory and engineering practice. A feedback regulator is designed for the controlled system, so that the signal to be regulated can track the target reference signal and the system keeps stable. Many practical problems such as aircraft landing, missile tracking and robot control all depend on output regulation. The wave equation with anti-damping term can simulate many engineering problems like pipeline combustion, acoustic instability or stick slip instability during drilling, which is of great significance to study the output regulation of anti-stable wave model.
In the last few years, a quiet great progress has been made both in the output feedback stabilization [
In this paper, we focus on the output tracking for 1-D anti-stable wave system with in-domain disturbance generated by an exosystem described by
{ w t t ( x , t ) = w x x ( x , t ) + f ( x ) d ( t ) , x ∈ ( 0 , 1 ) , t > 0 , w x ( 0 , t ) = − q w t ( 0 , t ) , t ≥ 0 , w x ( 1 , t ) = U ( t ) , t ≥ 0 , w ( x , 0 ) = w 0 ( x ) , w t ( x , 0 ) = w 1 ( x ) , x ∈ [ 0 , 1 ] , (1.1)
where w ( 0 , t ) is the output to be regulated, the displacement w ( 0 , t ) and its derivative w t ( 0 , t − τ ) , τ ∈ [ 0 , 1 ] with time-delay are the measured output. U ( t ) represents the input (control), q > 0 ( ≠ 1 ) an unknown constant parameter ( q ≠ 1 is to avoid the real part of the plant eigenvalues tending to positive infinity). ( w 0 ( x ) , w 1 ( x ) ) is initial condition, r ( t ) is a given reference signal, f ( x ) ∈ C [ 0 , 1 ] represents the unknown intensity of the distributed external disturbance d ( t ) . d ( t ) , r ( t ) are generated by following exosystem
{ v ˙ ( t ) = S v ( t ) , t > 0 , d ( t ) = P 1 v ( t ) , t ≥ 0 , r ( t ) = P 2 v ( t ) , t ≥ 0 , v ( 0 ) = v 0 ∈ C n , (1.2)
where S is a diagonalizable matrix with all eigenvalues on the imaginary axis. For design purpose, we suppose that the initial value v 0 is unknown and so do the state v ( t ) . Rewrite S as ( S 1 , S 2 ) and d ( t ) , r ( t ) can be written as
{ v ˙ 1 ( t ) = S 1 v 1 ( t ) , t > 0 , v ˙ 2 ( t ) = S 2 v 2 ( t ) , t > 0 , d ( t ) = Q 1 v 1 ( t ) , t ≥ 0 , r ( t ) = Q 2 v 2 ( t ) , t ≥ 0 , v 1 ( 0 ) = v 10 ∈ C n 1 , v 2 ( 0 ) = v 20 ∈ C n 2 . (1.3)
Here r ( t ) is known but d ( t ) is unknown due to the uncertainty of v 10 . We have n 1 + n 2 = n and v = ( v 1 Τ v 2 Τ ) Τ , and assume that the eigenvalues of matrix S 1 are distinct and ∑ ( Q 2 Τ , S 2 ) is observable. The objective of this paper is to design an observer-based output feedback regulator for system (1.1) to regulate the tracking error e ( t ) = w ( 0 , t ) − r ( t ) to zero and simultaneously keep all the states bounded. The advanced nature of our result lies in that the measured output is at the left end and may admit time-delay, which makes the regulation problem of (1.1) challenging.
The rest of the paper is organized as follows. In Section 2, an auxiliary stable system is constructed by introducing a transport equation and a regulator equation, and its observer is derived. We propose the output feedback control law for the auxiliary system and obtain the closed-loop system in Section 3. The main results are presented in Section 4 and Section 5 concludes this paper.
In order to deal with the anti-damping − q w t ( 0 , t ) in system (1.1), we introduce the following transport equation in [
{ δ t ( x , t ) = − δ x ( x , t ) , x ∈ ( 0 , 1 ) , t > 0 , δ ( 0 , t ) = − q + c 0 1 + c 0 w ( 0 , t ) , t ≥ 0 , δ ( x , 0 ) = δ 0 ( x ) , x ∈ [ 0 , 1 ] , (2.1)
where δ 0 is the initial value, c 0 > 0 a tuning parameter.
In the rest of this paper, we omit the obvious domain x ∈ [ 0 , 1 ] and t ∈ [ 0 , + ∞ ) when there is no confusion.
Let
{ v 1 ( t ) = v 1 ( t ) , v 2 ( t ) = v 2 ( t ) , u ( x , t ) = w ( x , t ) + δ ( x , t ) , (2.2)
then u ( x , t ) is governed by
{ v ˙ 1 ( t ) = S 1 v 1 ( t ) , t > 0 , v ˙ 2 ( t ) = S 2 v 2 ( t ) , t > 0 , u t t ( x , t ) = u x x ( x , t ) + f ( x ) Q 1 v 1 ( t ) , u x ( 0 , t ) = c 0 u t ( 0 , t ) , u x ( 1 , t ) = U ( t ) + δ x ( 1 , t ) , u ( x , 0 ) = u 0 ( x ) = w 0 ( x ) + δ 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) = w 1 ( x ) − δ ′ 0 ( x ) . (2.3)
Notice that the parameter − q > 0 in system (1.1) becomes c 0 > 0 in system (2.3). Moreover, we have
u ( 0 , t ) = 1 − q 1 + c 0 w ( 0 , t ) .
To recover the state of (2.3), we now design an observer for (2.3) using the known signal w ( 0 , t ) and w t ( 0 , t − τ ) , τ ∈ [ 0 , 1 ] as
{ v ^ ˙ 1 ( t ) = S 1 v ^ 1 ( t ) + K 1 [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] , t > 0 , v ^ ˙ 2 ( t ) = S 2 v ^ 2 ( t ) + K 2 [ Q 2 v ^ 2 ( t ) − r ( t ) ] , t > 0 , u ^ t t ( x , t ) = u ^ x x ( x , t ) + r 1 ( x ) [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] + f ( x ) Q 1 v ^ 1 ( t ) + r 2 ( x ) [ u ^ t ( 0 , t ) − 1 − q 1 + c 0 w t ( 0 , t ) ] , u ^ x ( 0 , t ) = c 0 u ^ t ( 0 , t ) + ( c 1 + c 0 P K 1 ) [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] , u ^ x ( 1 , t ) = U ( t ) + q + c 0 1 + c 0 w t ( 0 , t − 1 ) , v ^ 1 ( 0 ) = v ^ 10 ∈ C n 1 , v ^ 2 ( 0 ) = v ^ 20 ∈ C n 2 , u ^ ( x , 0 ) = u ^ 0 ( x ) , u ^ t ( x , 0 ) = u ^ 1 ( x ) , (2.4)
where c 1 > 0 is a constant, and r 1 ( x ) , r 2 ( x ) ∈ C [ 0 , 1 ] , K 1 , K 2 are row vectors where K 2 is designed to make the matrix S 2 + K 2 Q 2 Hurwitz and P , K 1 to be determined later.
Let u ˜ = u − u ^ , v ˜ 1 = v 1 − v ^ 1 , v ˜ 2 = v 2 − v ^ 2 , then we have
{ v ˜ ˙ 1 ( t ) = S 1 v ˜ 1 ( t ) + K 1 u ˜ ( 0 , t ) , t > 0 , v ˜ ˙ 2 ( t ) = ( S 2 + K 2 Q 2 ) v ˜ 2 ( t ) , t > 0 , u ˜ t t ( x , t ) = u ˜ x x ( x , t ) + f ( x ) Q 1 v ˜ 1 ( t ) + r 1 ( x ) u ˜ ( 0 , t ) + r 2 ( x ) u ˜ t ( 0 , t ) , u ˜ x ( 0 , t ) = c 0 u ˜ t ( 0 , t ) + ( c 1 + c 0 P K 1 ) u ˜ ( 0 , t ) , u ˜ x ( 1 , t ) = 0 , v ˜ 1 ( 0 ) = v ˜ 10 ∈ C n 1 , v ˜ 2 ( 0 ) = v ˜ 20 ∈ C n 2 , u ˜ ( x , 0 ) = u ˜ 0 ( x ) , u ˜ t ( x , 0 ) = u ˜ 1 ( x ) , (2.5)
Construct the following transformation
{ v ˜ 1 ( t ) = v ˜ 1 ( t ) , v ˜ 2 ( t ) = v ˜ 2 ( t ) , e ( x , t ) = u ˜ ( x , t ) + g ( x ) v ˜ 1 ( t ) ， (2.6)
then the observer error system is found to be
{ v ˜ ˙ 1 ( t ) = ( S 1 − K 1 g ( 0 ) ) v ˜ 1 ( t ) + K 1 e ( 0 , t ) , t > 0 , v ˜ ˙ 2 ( t ) = ( S 2 + K 2 Q 2 ) v ˜ 2 ( t ) , t > 0 , e t t ( x , t ) = e x x ( x , t ) , e x ( 0 , t ) = c 0 e t ( 0 , t ) + c 1 e ( 0 , t ) , e x ( 1 , t ) = 0 , v ˜ 1 ( 0 ) = v ˜ 10 ∈ C n 1 , v ˜ 2 ( 0 ) = v ˜ 20 ∈ C n 2 , e ( x , 0 ) = e 0 ( x ) , e t ( x , 0 ) = e 1 ( x ) , (2.7)
where g ( x ) satisfies the boundary value problem (regulator equation) as
{ d 2 g ( x ) d x 2 = g ( x ) S 1 2 + f ( x ) Q 1 , d g ( x ) d x | x = 0 = c 0 g ( 0 ) S 1 + c 1 g ( 0 ) , d g ( x ) d x | x = 1 = 0. (2.8)
We make P = g ( 0 ) , r 1 ( x ) = − g ( x ) S 1 K 1 , r 2 ( x ) = − g ( x ) K 1 .
Lemma 2.1: Assume that S 1 is a diagonalizable matrix. Then the regulator equation (2.8) admits a unique solution.
Considering our previous assumptions that the matrix S 2 + K 2 Q 2 is Hurwitz and c 1 > 0 , the PDE-part of system (2.7) is exponentially stable. (2.7) will be exponentially stable if we can show that S 1 − K 1 g ( 0 ) is also Hurwitz.
Lemma 2.2: Define a function N 1 Τ ( s ) = − ∫ 0 1 cosh ( s ( 1 − y ) ) f ( y ) d y and ξ i be the eigenvector of matrix S 1 corresponding to the eigenvalue λ i of S 1 . Then ( g ( 0 ) , S 1 ) is observable if and only if N 1 Τ ( λ i ) Q 1 ξ i ≠ 0 , ( i = 1 , 2 , ⋯ , n 1 ) .
Lemma 2.1 and 2.2 are similar to Lemma 3.1 and 3.2 in [
Theorem 2.1: Define a function N 1 Τ ( s ) = − ∫ 0 1 cosh ( s ( 1 − y ) ) f ( y ) d y satisfies N 1 Τ ( λ i ) Q 1 ξ i ≠ 0 , ( i = 1 , 2 , ⋯ , n 1 ) for all eigen-pairs ( λ i , ξ i ) of S 1 . c 0 , c 1 > 0 are constants, g ( x ) is the unique solution to (2.8). We make P = g ( 0 ) and r 1 ( x ) = − g ( x ) S 1 K 1 , r 2 ( x ) = − g ( x ) 1 K 1 , matrix S 1 − K 1 g ( 0 ) is Hurwitz. Then system (2.7) is well-posed and exponentially stable.
Proof:
We divide system (2.7) into PDE-part and ODE-part, and consider the stability of the solution respectively.
The PDE-part of (2.7) is
{ e t t ( x , t ) = e x x ( x , t ) , e x ( 0 , t ) = c 0 e t ( 0 , t ) + c 1 e ( 0 , t ) , e x ( 1 , t ) = 0 , e ( x , 0 ) = e 0 ( x ) , e t ( x , 0 ) = e 1 ( x ) , (2.9)
Define H 0 = H 1 ( 0 , 1 ) × L 2 ( 0 , 1 ) to be an usual Hilbert space with the following norm induced by the inner product
‖ ( f , g ) Τ ‖ H 0 2 = ∫ 0 1 ( | f ′ ( x ) | 2 d x + | g ( x ) | 2 ) d x + c 0 | f ( 0 ) | 2 , ∀ ( f , g ) ∈ H 0 .
And H = C n 1 × C n 2 × H 0 . We can directly come to the conclusion that (2.9) is exponentially stable from [
‖ ( e ( ⋅ , t ) , e t ( ⋅ , t ) ) Τ ‖ H 0 ≤ M 0 e − w 0 t ‖ ( e 0 ( ⋅ ) , e 1 ( ⋅ ) ) Τ ‖ H 0 , t > 0 , (2.10)
which implies that
| e ( 0 , t ) | ≤ M 0 e − w 0 t ‖ ( e 0 ( ⋅ ) , e 1 ( ⋅ ) ) Τ ‖ H 0 , t > 0.
The ODE-part of (2.7) is governed by
{ v ˜ ˙ 1 ( t ) = ( S 1 − K 1 g ( 0 ) ) v ˜ 1 ( t ) + K 1 e ( 0 , t ) , t > 0 , v ˜ ˙ 2 ( t ) = ( S 2 + K 2 Q 2 ) v ˜ 2 ( t ) , t > 0 , v ˜ 1 ( 0 ) = v ˜ 10 ∈ C n 1 , v ˜ 2 ( 0 ) = v ˜ 20 ∈ C n 2 . (2.11)
There are some positive constants M j , w j > 0 , ( j = 1 , 2 , 3 ) such that the solution v ˜ 1 ( t ) , v ˜ 2 ( t ) of (2.11) has the estimation as
{ ‖ v ˜ 1 ( t ) ‖ = ‖ e [ S 1 − K 1 g ( 0 ) ] t v ˜ 1 ( 0 ) + ∫ 0 t e [ S 1 − K 1 g ( 0 ) ] ( t − s ) K 1 e ( 0 , s ) d s ‖ C n 1 ≤ ‖ e [ S 1 − K 1 g ( 0 ) ] t v ˜ 1 ( 0 ) ‖ C n 1 + ‖ ∫ 0 t e [ S 1 − K 1 g ( 0 ) ] ( t − s ) K 1 e ( 0 , s ) d s ‖ C n 1 ≤ M 1 e − w 1 t ‖ v ˜ 1 ( 0 ) ‖ C n 1 + M 2 e − w 2 t ‖ ( e 0 ( ⋅ ) , e 1 ( ⋅ ) ) Τ ‖ C n 1 , ‖ v ˜ 2 ( t ) ‖ = ‖ e ( S 2 + K 2 Q 2 ) t v ˜ 2 ( 0 ) ‖ C n 2 ≤ M 3 e − w 3 t ‖ v ˜ 2 ( 0 ) ‖ C n 2 . (2.12)
Thus we have
‖ ( v ˜ 1 Τ ( t ) , v ˜ 2 Τ ( t ) , e ( ⋅ , t ) , e t ( ⋅ , t ) ) Τ ‖ H ≤ M 4 e − w 4 t ‖ ( v ˜ 1 Τ ( 0 ) , v ˜ 2 Τ ( 0 ) , e 0 ( ⋅ ) , e 1 ( ⋅ ) ) Τ ‖ H
for any constants M 4 , w 4 > 0 .
Define an invertible bounded operator Α 0 : H → H as
Α 0 ( v ˜ 1 Τ ( t ) , v ˜ 2 Τ ( t ) , e ( x , t ) , e t ( x , t ) ) Τ = ( v ˜ 1 Τ ( t ) , v ˜ 2 Τ ( t ) , e ( x , t ) − g ( x ) v ˜ 1 ( t ) , e t ( x , t ) − g ( x ) [ ( S 1 − K 1 g ( 0 ) ) v ˜ 1 ( t ) + K 1 e ( 0 , t ) ] ) = ( v ˜ 1 Τ ( t ) , v ˜ 2 Τ ( t ) , u ˜ ( x , t ) , u ˜ t ( x , t ) ) .
Then there exists M > 0 which is independent of initial value such that
‖ ( v ˜ 1 Τ ( t ) , v ˜ 2 Τ ( t ) , u ˜ ( ⋅ , t ) , u ˜ t ( ⋅ , t ) ) Τ ‖ H ≤ M ‖ ( v ˜ 1 Τ ( 0 ) , v ˜ 2 Τ ( 0 ) , u ˜ 0 ( x ) , u ˜ 1 ( x ) ) Τ ‖ H . £
We construct a new transformation as
{ v ( t ) = v ( t ) , z ( x , t ) = u ( x , t ) + h ( x ) v ( t ) . (3.1)
Then z ( x , t ) is governed by
{ v ˙ ( t ) = S v ( t ) , t > 0 , z t t ( x , t ) = z x x ( x , t ) , z x ( 0 , t ) = c 0 z t ( 0 , t ) , z x ( 1 , t ) = U ( t ) + δ x ( 1 , t ) + d h ( x ) d x | x = 1 v ( t ) , z ( x , 0 ) = z 0 ( x ) , z t ( x , 0 ) = z 1 ( x ) , (3.2)
where h ( x ) satisfies the BVP as follows
{ d 2 h ( x ) d x 2 = h ( x ) S 2 + f ( x ) P 1 , d h ( x ) d x | x = 0 = c 0 h ( 0 ) S , 1 + c 0 q − 1 h ( 0 ) = P 2 . (3.3)
Moreover, we have
e ( t ) = 1 + c 0 1 − q z ( 0 , t ) − 1 + c 0 1 − q h ( 0 ) v ( t ) − P 2 v ( t ) = 1 + c 0 1 − q z ( 0 , t ) (3.4)
according to the second boundary condition in (3.3).
Lemma 3.1: Assume that S is a diagonalizable matrix, then the regulator equation (3.3) admits a unique solution.
Proof:
Since S is diagonalizable, there exists an invertible matrix A = ( ξ 1 , ξ 2 , ⋯ , ξ n ) such that A − 1 S A = d i a g ( λ 1 , λ 2 , ⋯ , λ n ) , where ξ i is the eigenvector of S corresponding to the eigenvalue λ i , ( i = 1 , 2 , ⋯ , n ) of S. Postmultiplying (3.3) by ξ i , n ODEs are found to be
{ d 2 h i ∗ ( x ) d x 2 = h i ∗ ( x ) λ i 2 + f ( x ) P 1 ∗ , d h i ∗ ( x ) d x | x = 0 = c 0 h i ∗ ( 0 ) λ i , 1 + c 0 q − 1 h i ∗ ( 0 ) = P 2 ∗ . (3.5)
Here h i ∗ ( x ) = h ( x ) ξ i , P 1 ∗ = P 1 ξ i , P 2 ∗ = P 2 ξ i .
Case 1: When λ i = 0 , the BVP (3.5) becomes
{ d 2 h i ∗ ( x ) d x 2 = f ( x ) P 1 ∗ , d h i ∗ ( x ) d x | x = 0 = 0 , 1 + c 0 q − 1 h i ∗ ( 0 ) = P 2 ∗ . (3.6)
A formal simple computation shows that the solution of (3.6) is
h i ∗ ( x ) = h i ∗ ( 0 ) + d h i ∗ ( x ) d x | x = 0 x + ∫ 0 x ( x − y ) f ( y ) d y P 1 ∗ , = q − 1 1 + c 0 P 2 ∗ + ∫ 0 x ( x − y ) f ( y ) d y P 1 ∗ .
Case 2: When λ i ≠ 0 , the BVP (3.5) has a general solution as
h i ∗ ( x ) = m i 1 sinh ( λ i x ) + m i 2 cosh ( λ i x ) + 1 λ i ∫ 0 x sinh ( λ i ( x − y ) ) f ( y ) d y P 1 ∗ . (3.7)
The last two boundary equations in (3.5) conclude that
{ λ i m i 1 = c 0 λ i h i ∗ ( 0 ) = c 0 λ i m i 2 , 1 + c 0 q − 1 m i 2 = P 2 ∗ . (3.8)
Then m i 1 , m i 2 are determined by solving (3.8) as
{ m i 2 = q − 1 1 + c 0 P 2 ∗ , m i 1 = c 0 m i 2 = c ( q − 1 ) 0 1 + c 0 P 2 ∗ .
Thus the unique solution to the regulator Equation (3.3) is obtained and
h ( x ) = [ h 1 ( x ) , h 2 ( x ) , ⋯ , h n ( x ) ] Τ . £
Now we design the output feedback controller for (3.2) as
U ( t ) = − q + c 0 1 + c 0 w t ( 0 , t − 1 ) − c 2 u ^ ( 1 , t ) − [ d h ( x ) d x | x = 1 + c 2 h ( 1 ) ] ( v ^ 1 Τ ( t ) v ^ 2 Τ ( t ) ) Τ , (3.9)
where c 2 > 0 is a constant. Under (3.9), the closed-loop becomes
{ v ˜ ˙ 1 ( t ) = ( S 1 − K 1 g ( 0 ) ) v ˜ 1 ( t ) + K 1 e ( 0 , t ) , t > 0 , v ˜ ˙ 2 ( t ) = ( S 2 + K 2 Q 2 ) v ˜ 2 ( t ) , t > 0 , z t t ( x , t ) = z x x ( x , t ) , z x ( 0 , t ) = c 0 z t ( 0 , t ) , z x ( 1 , t ) = − c 2 z ( 1 , t ) + c 2 u ˜ ( 1 , t ) + [ d h ( x ) d x | x = 1 + c 2 h ( 1 ) ] ( v ˜ 1 Τ ( t ) v ˜ 2 Τ ( t ) ) Τ , v ˜ 1 ( 0 ) = v ˜ 10 ∈ C n 1 , v ˜ 2 ( 0 ) = v ˜ 20 ∈ C n 2 , z ( x , 0 ) = z 0 ( x ) , z t ( x , 0 ) = z 1 ( x ) . (3.10)
Theorem 3.1: For any initial data ( z 0 , z 1 ) ∈ H 0 , μ , α are positive constants, there exists a unique (weak) solution to the PDE-part of (3.10) such that ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) ∈ C ( 0 , ∞ ; H 0 ) . Besides, this solution is exponentially stable in the sense that
‖ ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) Τ ‖ H 0 ≤ μ e − α t ‖ ( z 0 ( ⋅ ) , z 1 ( ⋅ ) ) Τ ‖ H 0 , t > 0. (3.11)
and
lim t → ∞ z ( 0 , t ) = 0 , lim t → ∞ ∫ t − 1 t z t 2 ( 0 , s ) d s = 0. (3.12)
Proof:
Define an operator Α 1 : D ( Α 1 ) → H 0 for the PDE-part of system (3.10) by
{ Α 1 ( f , g ) Τ = ( g , f ″ ) Τ , ∀ ( f , g ) Τ ∈ D ( Α 1 ) , D ( Α 1 ) = { ( f , g ) Τ ∈ H 0 | Α 1 ( f , g ) Τ ∈ H 0 , f ′ ( 0 ) = c 0 g ( 0 ) , f ′ ( 1 ) = − c 2 f ( 1 ) } .
Then the PDE-part of (3.10) can be written as an abstract evolutionary equation in H 0 as follows
d d t ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) Τ = Α 1 ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) Τ + Β ϕ ( t ) , (3.13)
where
{ Β = ( 0 , δ ( x − 1 ) ) Τ ϕ ( t ) = c 2 u ˜ ( 1 , t ) + [ d h ( x ) d x | x = 1 + c 2 h ( 1 ) ] ( v ˜ 1 Τ ( t ) v ˜ 2 Τ ( t ) ) Τ (3.14)
ϕ ( t ) decays to zero exponentially from Theorem 2.1 and the transformation (2.6). It’s well known that Α 1 can generate an exponentially stable C 0 -semigroup by [
It concludes that for any initial value ( z 0 , z 1 ) ∈ H 0 , there exists a unique solution ( z , z t ) ∈ C ( 0 , ∞ ; H 0 ) to the PDE-part of system (3.10), which has the form of
( z ( ⋅ , t ) , z t ( ⋅ , t ) ) Τ = e Α 1 t ( z 0 ( ⋅ ) , z 1 ( ⋅ ) ) Τ + ∫ 0 t e Α 1 ( t − s ) Β ϕ ( s ) d s . (3.15)
The first term on the right side of (3.15) can be estimated as
‖ e Α 1 t ( z ( ⋅ , 0 ) , z t ( ⋅ , 0 ) ) Τ ‖ H 0 ≤ ‖ e Α 1 t ‖ ⋅ ‖ ( z ( ⋅ , 0 ) , z t ( ⋅ , 0 ) ) Τ ‖ H 0 ≤ μ 1 e − α 1 t ‖ ( z 0 ( ⋅ ) , z 1 ( ⋅ ) ) Τ ‖ H 0 → 0 , t → ∞ . (3.16)
The second term on the right side of (3.15) tends to zero exponentially because of the admissibility of B to e Α 1 t and the exponential stability of ϕ ( t ) . From Poincare’s inequality we have
| z ( 0 , t ) | 2 ≤ 2 | z ( 1 , t ) | 2 + 2 ∫ 0 1 | z x ( x , t ) | 2 d x ,
Hence lim t → ∞ z ( 0 , t ) = 0 holds. Next we show that lim t → ∞ ∫ t − 1 t z t 2 ( 0 , s ) d s = 0 .
First define the Lyapunov function as
E z ( t ) = 1 2 ∫ 0 1 ( z t 2 ( x , t ) + z x 2 ( x , t ) ) d x + c 2 2 z 2 ( 1 , t ) , ρ z ( t ) = ∫ 0 1 ( x − 1 ) z t ( x , t ) z x ( x , t ) d x .
Notice the fact that 2 E z ( t ) = ‖ ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) ‖ H 0 2 decays exponentially and | ρ z ( t ) | ≤ E z ( t ) .
Then differentiate ρ z ( t ) along the solution to system (3.2) to give
ρ ˙ z ( t ) = x − 1 2 ( z t 2 ( x , t ) + z x 2 ( x , t ) ) | 0 1 − 1 2 ∫ 0 1 ( z t 2 ( x , t ) + z x 2 ( x , t ) ) d x = 1 2 ( z t 2 ( 0 , t ) + z x 2 ( 0 , t ) ) − 1 2 ∫ 0 1 ( z t 2 ( x , t ) + z x 2 ( x , t ) ) d x . (3.17)
Finally integrating (3.17) from t − 1 to t with respect to t and obtain
1 2 ∫ t − 1 t ( z t 2 ( 0 , s ) + z x 2 ( 0 , s ) ) d s = ρ z ( t ) − ρ z ( t − 1 ) + 1 2 ∫ t − 1 t ∫ 0 1 ( z t 2 ( x , s ) + z x 2 ( x , s ) ) d x d s ≤ E z ( t ) + E z ( t − 1 ) + ∫ t − 1 t E z ( s ) d s , (3.18)
which decays exponentially from the exponential stability of E z ( t ) .
Hence lim t → ∞ ∫ t − 1 t z t 2 ( 0 , s ) d s = 0 holds. £
The closed-loop of system (1.1) corresponding to (1.3), (2.1), (2.4) and (3.9) in the state space Χ = H × H 1 ( 0 , 1 ) × H yields to
{ v ˙ 1 ( t ) = S 1 v 1 ( t ) , t > 0 , v ˙ 2 ( t ) = S 2 v 2 ( t ) , t > 0 , w t t ( x , t ) = w x x ( x , t ) + f ( x ) Q 1 v 1 ( t ) , x ∈ ( 0 , 1 ) , t > 0 , w x ( 0 , t ) = − q w t ( 0 , t ) , w x ( 1 , t ) = u ^ x ( 1 , t ) − q + c 0 1 + c 0 w t ( 0 , t − 1 ) , δ t ( x , t ) = − δ x ( x , t ) , δ ( 0 , t ) = − q + c 0 1 + c 0 w ( 0 , t ) , v ^ ˙ 1 ( t ) = S 1 v ^ 1 ( t ) + K 1 [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] , t > 0 , v ^ ˙ 2 ( t ) = S 2 v ^ 2 ( t ) + K 2 [ Q 2 v ^ 2 ( t ) − r ( t ) ] , t > 0 , u ^ t t ( x , t ) = u ^ x x ( x , t ) − g ( x ) S 1 K 1 [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] + f ( x ) Q 1 v ^ 1 ( t ) − g ( x ) K 1 [ u ^ t ( 0 , t ) − 1 − q 1 + c 0 w t ( 0 , t ) ] , u ^ x ( 0 , t ) = c 0 u ^ t ( 0 , t ) + ( c 1 + c 0 g ( 0 ) K 1 ) [ u ^ ( 0 , t ) − 1 − q 1 + c 0 w ( 0 , t ) ] , u ^ x ( 1 , t ) = − c 2 u ^ ( 1 , t ) − [ d h ( x ) d x x = 1 + c 2 h ( 1 ) ] ( v ^ 1 Τ ( t ) v ^ 2 Τ ( t ) ) Τ , v 10 , v ^ 10 ∈ C n 1 , v 20 , v ^ 20 ∈ C n 2 , δ ( x , 0 ) = δ 0 ( x ) , w ( x , 0 ) = w 0 ( x ) , w t ( x , 0 ) = w 1 ( x ) , u ^ ( x , 0 ) = u ^ 0 ( x ) , u ^ t ( x , 0 ) = u ^ 1 ( x ) . (3.19)
Considering the closed-loop (3.19) in the state space X and define an operator Α 2 for (3.19) which satisfies
{ Α 2 ( φ 1 , φ 2 , f 1 , g 1 , h 1 , ψ 1 , ψ 2 , f 2 , g 2 ) Τ = ( S 1 φ 1 , S 2 φ 2 , g 1 , f ″ 1 + f Q 1 φ 1 , − h ′ 1 , S 1 ψ 1 + K 1 [ f 2 ( 0 ) − 1 − q 1 + c 0 f 1 ( 0 ) ] , S 2 ψ 2 + K 2 Q 2 ( ψ 2 − φ 2 ) , g 2 , f ″ 2 + r 1 [ f 2 ( 0 ) − 1 − q 1 + c 0 f 1 ( 0 ) ] + r 2 [ g 2 ( 0 ) − 1 − q q + c 0 h ′ 1 ( 0 ) ] Τ + f Q 1 ψ 1 ) Τ , ∀ ( φ 1 , φ 2 , f 1 , g 1 , h 1 , ψ 1 , ψ 2 , f 2 , g 2 ) Τ ∈ D ( Α 2 ) , D ( Α 2 ) = { ( φ 1 , φ 2 , f 1 , g 1 , h 1 , ψ 1 , ψ 2 , f 2 , g 2 ) Τ ∈ Χ | Α 2 ( φ 1 , φ 2 , f 1 , g 1 , h 1 , ψ 1 , ψ 2 , f 2 , g 2 ) Τ ∈ Χ , f ′ 1 ( 0 ) = − q g 1 ( 0 ) , f ′ 1 ( 1 ) = f ′ 2 ( 1 ) − h ′ 1 ( 1 ) , h 1 ( 0 ) = − q + c 0 1 + c 0 f 1 ( 0 ) , f ′ 2 ( 0 ) = c 0 g 2 ( 0 ) + ( c 1 + c 0 g ( 0 ) K 1 ) [ f 2 ( 0 ) − 1 − q 1 + c 0 f 1 ( 0 ) ] , f ′ 2 ( 1 ) = − c 2 f 2 ( 1 ) − [ d h ( x ) d x | x = 1 + c 2 h ( 1 ) ] ( ψ 1 Τ ψ 2 Τ ) Τ } .
Then (3.19) can be written as an abstract evolutionary equation in X as
d d t ( v 1 ( t ) , v 2 ( t ) , w ( ⋅ , t ) , w t ( ⋅ , t ) , δ ( ⋅ , t ) , v ^ 1 ( t ) , v ^ 2 ( t ) , u ^ ( ⋅ , t ) , u ^ t ( ⋅ , t ) ) Τ = Α 2 ( v 1 ( t ) , v 2 ( t ) , w ( ⋅ , t ) , w t ( ⋅ , t ) , δ ( ⋅ , t ) , v ^ 1 ( t ) , v ^ 2 ( t ) , u ^ ( ⋅ , t ) , u ^ t ( ⋅ , t ) ) Τ , (4.1)
Now we discuss the closed-loop system (4.1) in X.
Theorem 4.1: Define a function N 1 Τ ( s ) = − ∫ 0 1 cosh ( s ( 1 − y ) ) f ( y ) d y satisfies N 1 Τ ( λ i ) Q 1 ξ i ≠ 0 , ( i = 1 , 2 , ⋯ , n 1 ) for all eigen-pairs ( λ i , ξ i ) of S 1 . c 0 , c 1 > 0 are constants, g ( x ) is the unique solution to BVP (2.8). We make P = g ( 0 ) and r 1 ( x ) = − g ( x ) S 1 K 1 , r 2 ( x ) = − g ( x ) 1 K 1 , matrix S 1 − K 1 g ( 0 ) is Hurwitz. Then for any ( v 1 ( 0 ) , v 2 ( 0 ) , w 0 ( ⋅ ) , w 1 ( ⋅ ) , δ 0 ( ⋅ ) , v ^ 1 ( 0 ) , v ^ 2 ( 0 ) , u ^ 0 ( ⋅ ) , u ^ 1 ( ⋅ ) ) ∈ Χ , system (4.1) admits a unique bounded solution ( v 1 , v 2 , w , w t , δ , v ^ 1 , v ^ 2 , u ^ , u ^ t ) ∈ C ( 0 , ∞ ; Χ ) . Moreover, the tracking error e ( t ) → 0 exponentially when t tends to infinity.
Proof:
Define an invertible bounded operator Α 3 : Χ → Χ by
Α 3 ( v 1 ( t ) , v 2 ( t ) , w ( ⋅ , t ) , w t ( ⋅ , t ) , δ ( ⋅ , t ) , v ^ 1 ( t ) , v ^ 2 ( t ) , u ^ ( ⋅ , t ) , u ^ t ( ⋅ , t ) ) Τ = ( v 1 ( t ) , v 2 ( t ) , w ( ⋅ , t ) + δ ( ⋅ , t ) + h ( x ) v ( t ) , w t ( ⋅ , t ) + δ t ( ⋅ , t ) + h ( x ) v ˙ ( t ) , δ ( ⋅ , t ) , v 1 ( t ) − v ^ 1 ( t ) , v 2 ( t ) − v ^ 2 ( t ) , u ( ⋅ , t ) − u ^ ( ⋅ , t ) , u t ( ⋅ , t ) − u ^ t ( ⋅ , t ) ) Τ = ( v 1 ( t ) , v 2 ( t ) , z ( ⋅ , t ) , z t ( ⋅ , t ) , δ ( ⋅ , t ) , v ˜ 1 ( t ) , v ˜ 2 ( t ) , u ˜ ( ⋅ , t ) , u ˜ t ( ⋅ , t ) ) Τ (4.2)
An equivalent system of (3.19) is found to be
{ v ˙ 1 ( t ) = S 1 v 1 ( t ) , t > 0 , v ˙ 2 ( t ) = S 2 v 2 ( t ) , t > 0 , d ( t ) = Q 1 v 1 ( t ) , r ( t ) = Q 2 v 2 ( t ) , z t t ( x , t ) = z x x ( x , t ) , z x ( 0 , t ) = c 0 z t ( 0 , t ) , z x ( 1 , t ) = − c 2 z ( 1 , t ) + c 2 w ˜ ( 1 , t ) + [ d h ( x ) d x | x = 1 + c 2 h ( 1 ) ] ( v ˜ 1 Τ ( t ) v ˜ 2 Τ ( t ) ) Τ , δ t ( x , t ) = − δ x ( x , t ) , δ ( 0 , t ) = − q + c 0 1 + c 0 w ( 0 , t ) , v ˜ ˙ 1 ( t ) = S 1 v ˜ 1 ( t ) + K 1 u ˜ ( 0 , t ) , t > 0 , v ˜ ˙ 2 ( t ) = ( S 2 + K 2 Q 2 ) v ˜ 2 ( t ) , t > 0 , u ˜ t t ( x , t ) = u ˜ x x ( x , t ) + f ( x ) Q 1 v ˜ 1 ( t ) − g ( x ) S 1 K u ˜ 1 ( 0 , t ) − g ( x ) K 1 u ˜ t ( 0 , t ) , u ˜ x ( 0 , t ) = c 0 u ˜ t ( 0 , t ) + ( c 1 + c 0 g ( 0 ) K 1 ) u ˜ ( 0 , t ) , u ˜ x ( 1 , t ) = 0 , v 10 , v ^ 10 ∈ C n 1 , v 20 , v ˜ 20 ∈ C n 2 , δ ( x , 0 ) = δ 0 ( x ) , z ( x , 0 ) = z 0 ( x ) , z t ( x , 0 ) = z 1 ( x ) , u ˜ ( x , 0 ) = u ˜ 0 ( x ) , u ˜ t ( x , 0 ) = u ˜ 1 ( x ) . (4.3)
We can obtain the solution to the transport system (2.1) explicitly as
δ ( x , t ) = { δ 0 ( x − t ) x ≥ t , q + c 0 q − 1 z ( 0 , t − x ) − q + c 0 q − 1 h ( 0 ) v ( t ) x < t . (4.4)
Let ε 0 = sup t > 0 | q + c 0 q − 1 h ( 0 ) v ( t ) | ∈ R being a designed parameter, then we have the estimation as
| δ ( 0 , t ) | ≤ | q + c 0 q − 1 | | z ( 0 , t ) | + ε 0 ≤ | q + c 0 q − 1 | ‖ ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) ‖ H 0 + ε 0 ,
and
∫ 0 1 ( δ t 2 ( x , t ) + δ x 2 ( x , t ) ) d x ≤ 2 ( q + c 0 q − 1 ) 2 ∫ 0 1 | z t ( 0 , t − x ) | 2 d x + 4 ε 0 q + c 0 q − 1 ∫ 0 1 | z t ( 0 , t − x ) | d x + 2 ε 0 2 = 2 ( q + c 0 q − 1 ) 2 ∫ t − 1 t | z t ( 0 , s ) | 2 d s + 4 ε 0 q + c 0 q − 1 ∫ t − 1 t | z t ( 0 , s ) | d s + 2 ε 0 2 .
Conclusion (3.12) together with ‖ ( z ( ⋅ , t ) , z t ( ⋅ , t ) ) ‖ H 0 exponentially, imply that the solution to (2.1) is bounded as t → ∞ .
According to Theorem 2.1 and the well-posedness and exponential stability of system (3.10), (4.3) admits a unique bounded solution in X and so does the closed-loop (3.19) by the invertible transformation (4.2). As a result, the tracking error
e ( t ) = w ( 0 , t ) − r ( t ) = 1 + c 0 1 − q z ( 0 , t ) → 0
exponentially in the light of (3.12). £
In this paper, the output regulation problem for 1-D anti-stable wave equation is solved. The original system has the anti-damping at the position x = 0 which is anti-collocated with the control, and also subjects to the distributed disturbance with unknown intensity generated by an external system. By proposing an observer-based feedback controller for (1.1), the following objectives are achieved: 1) keep all the states of internal-loop bounded; 2) recover the system state from input and output; 3) regulate the output tracks the given reference signal exponentially.
The author declares no conflicts of interest regarding the publication of this paper.
Li, Z.Y. (2020) Output Feedback Regulation for 1-D Anti-Stable Wave Equation with External System Disturbance. Engineering, 12, 652-665. https://doi.org/10.4236/eng.2020.129046