Stabilization of reaction–diffusions PDE with delayed distributed actuation

Stabilization of reaction–diffusions PDE with delayed distributed actuation

Systems & Control Letters 133 (2019) 104558 Contents lists available at ScienceDirect Systems & Control Letters journal homepage: www.elsevier.com/l...

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Systems & Control Letters 133 (2019) 104558

Contents lists available at ScienceDirect

Systems & Control Letters journal homepage: www.elsevier.com/locate/sysconle

Stabilization of reaction–diffusions PDE with delayed distributed actuation ∗

Jie Qi a , , Miroslav Krstic b , Shanshan Wang a a

College of Information Science and Technology, Engineering Research Center of Digitized Textile and Fashion Technology Ministry of Education, Donghua University, Shanghai, 201620, China b Department of Mechanical Aerospace Engineering, University of California, San Diego, CA 92093-0411, USA

article

info

Article history: Received 3 February 2019 Received in revised form 27 August 2019 Accepted 8 October 2019 Available online xxxx Keywords: Delayed distributed actuation Backstepping PDE Stabilization

a b s t r a c t This paper pursuits control design for an unstable reaction–diffusion equation with arbitrarily large input delay affecting the in-domain actuator. We introduce a transport PDE which results in an extended spatial domain where the reaction–diffusion PDE and the transport PDE are in cascade. Based on the conception of predictor-feedback, an backstepping integral transformation is introduced to transform the original coupled system to a stable target system. A predictor control which compensates the delay via distributed in-domain actuation is designed. The kernel function weighting the state feedback and the historical control employs a Dirac delta function as its initial condition. Due to singularity of the kernel function, the norm equivalence between the original and the target system is proved before we finally reach the exponential stability in H 1 norm for the coupled system with the delay compensated feedback. A numerical simulation illustrates the effectiveness of the control. © 2019 Elsevier B.V. All rights reserved.

1. Introduction Engineering systems are often subject to transport phenomena which generate the presence of time delay in their actuation path. Typically, input delays arise when modeling a wide class of complex physical systems such as production systems [1], hydraulically actuated systems [2], drill system [3] and even chemical processes. In the last few decades, major contributions intensified the understanding of stability of nonlinear ODEs (Ordinary Differential Equations) contingent on various type of delays that affects their inputs based on predictor-feedback control techniques. The key idea is essentially to define an appropriate time-inversion that enables to derive a nonlinear predictor signal which essentially helps to compensate the effect of the delay when combined with a nominal feedback control law which stabilize the delay-free plant [1,4–10]. More recently, control of PDEs (Partial Differential Equations) with input delays has just attracted attentions recently [11–14]. Based on Lyapunov– Krasovskii functionals method, a stable PDE is controlled by a delayed control operator by applying the Linear Matrix Inequality in [15] and [16] proposes interesting results considering a semilinear case with time-delays. A boundary control to compensate input delay for reaction–diffusion PDE by decomposing the PDE system into one finite-dimensional unstable part and ∗ Corresponding author. E-mail addresses: [email protected] (J. Qi), [email protected] (M. Krstic), [email protected] (S. Wang). https://doi.org/10.1016/j.sysconle.2019.104558 0167-6911/© 2019 Elsevier B.V. All rights reserved.

one stable infinite-dimensional part has been developed [17], and [18] designed a controller based on the PDE backstepping transformation which extends the predictor-based method for an unstable reaction–diffusion PDE prototype system with input delay on boundary. Further results, such as [19] who constructed a boundary controller for a reaction–diffusion PDE with input delay and unmatched disturbance, can be found in the existing literature. For a wave PDE with boundary unknown anti-damping, an adaptive boundary controller is developed to trade off the anti-damping term in [20]. However, the aforementioned results on PDE delays mainly address the control design for boundary input delay that only influence the boundary actuator. Input delays of distributed actuation in domain are much more difficult to compensate compared to the boundary input delay. This is since that the delays in domain affect the entire spatially distributed state and bring a new challenge that one can only overcome by designing a controller that compensate the everywhere delays. In this context few results such as [21], which considers the stabilization of reaction diffusion PDEs with state delay in domain can be found. However, [21] only considers the state delay instead of input delay and just applies boundary control. In the sense, the state delay does not need to be compensated given an appropriate target system is defined. Another interesting approach is proposed in [22] where case of delay varying with time is investigated. In this paper, a controller is designed for an unstable reaction diffusion PDE which subjects to in domain input delays. The

2

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

delays are converted into a transport PDE containing a spatial argument which transforms the time delay into a spatial in domain shift [23]. The resulting system is a cascade of reaction–diffusion and transport PDE that can be stabilized using backstepping technique. An integral transformation which transforms the coupled system into a stable target system in the frame of the PDE backstepping is introduced. Compared to the boundary control for compensating the input delay in [18] which integrates the past control feedback in the delay time interval, we design a distributed in domain control which integrates the past control feedback both in spatial domain and in the delay time interval. Due to the occurrence of in domain input delay, the resulting kernel equation is characterized by a singular initial condition, i.e., the initial condition is denoted by a Dirac delta function. The explicit form of the kernel function is obtained which is formulated by a series of sine orthogonal basis. We apply the Parseval’s theorem to prove the norm equivalence between the coupled system and the stable target system. It is important because the kernel functions of both the transformation and the inverse one contain singular points. Based on the norm equivalence, we prove the exponential stability in H 1 norm of the closed delay compensated system. Finally, the numerical simulation is performed to illustrate the effectiveness of the proposed controller. This paper is organized as follows. Section 2 presents the control design for the system with input delay. The H 1 exponential stability of the closed-loop system under the designed controller is proven in Section 3 and supportive simulation results are provided in Section 4. The paper ends with concluding remarks and future works presented in Section 5. 2. Control design 2.1. Problem description Consider a reaction–diffusion PDE system as follows ut (x, t) = uxx (x, t) + λu(x, t) + U(x, t − D),

(1)

for x ∈ [0, 1], t > 0 and D ∈ R , denoting system (1) delayed by D time unit. The boundary conditions are +

u(0, t) = 0,

u(1, t) = 0.

(2)

Fig. 1. Domain (x, s) of transform PDE v (x, s, t).

For reaction–diffusion PDE (1)–(2), we first represent the delay as a transport PDE, which resulting in the following cascade system ut (x, t) = uxx (x, t) + λu(x, t) + v (x, 0, t), x ∈ (0, 1),

(7)

u(t , 0) = u(t , 1) = 0,

(8)

u(x, 0) = u0 (x),

(9)

vt (x, s, t) = vs (x, s, t),

x ∈ [0, 1], s ∈ [0, D),

(10)

v (x, D, t) = U(x, t),

(11)

v (x, s, 0) = v0 (x, s),

(12)

Extended domain (x, s) is shown in Fig. 1, s resulting from time delay is the spatial argument of the first-order hyperbolic PDE (10) and x relates to spatial derivative in parabolic PDE (7). Note that the solution to (10)–(12) can be expressed as v (x, s, t) = U(x, s + t − D). More precisely, v (x, 0, t) = U(x, t − D) implies that there is no explicit delay in the cascade system, but the delay is transformed into a spatial shift denoted by s and hidden in state v (x, 0, t). In the next subsection a delay compensated controller is designed for the equivalent cascade system (7)–(12). 2.2. Design for the delay-compensated controller We apply the well-known PDE backstepping method to design the controller. First, introduce a stable target system as follows

For D ≡ 0, this is a trivial problem, solvable by many different feedback laws, the nominal one being

ut (x, t) = uxx (x, t) − cu(x, t) + z(x, 0, t), x ∈ (0, 1),

(13)

u(t , 0) = 0,

(14)

U(x, t) = −(c + λ)u(x, t),

u(x, 0) = u0 (x),

c > 0,

(3)

which stabilizes the system (1)–(3) to a zero equilibrium. However, under the occurrence of time delay acting on the in domain actuator, the system (1) subject to the nominal feedback law (3) becomes unstable and a delay compensator is needed to stabilize the system. Such a problem has been considered before for a ODE scalar plant with single input X˙ (t) = λX (t) + U(t − D) t

γ (t − θ )U(θ )dθ

(15)

zt (x, s, t) = zs (x, s, t),

x ∈ [0, 1], s ∈ [0, D],

(16)

z(x, D, t) = 0,

(17)

z(x, s, 0) = z0 (x, s),

(18)

where the transport system z has a mild solution

{

z0 (x, s + t)

0
0

s + t > D.

(4)

z(x, s, t) =

(5)

which means state z(x, s, t) becomes zero after time D. Inspired by predictor-feedback control (5), we introduce the following integrated transformation

for t > 0 and D ∈ R+ . The Predictor feedback for (4) is

[ ∫ U(t) = −(c + λ) γ (D)X (t) +

u(t , 1) = 0,

]

(19)

t −D

where

γ (s) = eλs ,

z(x, s, t) = v (x, s, t) + (c + λ) s ∈ [0, D].

1

γ (x, s, y)u(y, t)dy + (c + λ)

0

(6)

The term in the bracket on the right side of (5) is the predictor which compensates the input delay. And (6) plays the role of state transition. This example is a special case of the general compensator design in the setting of finite dimensional system.



1∫

∫ × 0

s

γ (x, s − r , y)v (y, r , t)drdy,

(20)

0

where kernel function γ (x, s, y) is defined on [0, 1] × [0, D] × [0, 1]. This transformation transforms original system (7)–(12) to target system (13)–(18). Kernel γ (x, s, y) could be derived

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

Fig. 2. Control Kernel. (a) x = y, (b) x =

from the equivalence between the original system and the target system. We first calculate the derivative of z(x, s, t) and get zt (x, s, t) =vt (x, s, t) + (c + λ)

1



γ (x, s, y)ut (y, t)dy

+ (c + λ)

s

γ (x, s − r , y)vt (y, r , t)drdy

=vs (x, s, t) + (c + λ)γ (x, s, 1)ux (1, t) − (c + λ)γ (x, s, 0)ux (0, t) ∫ 1 (γyy (x, s, y) + λγ (x, s, y))u(y, t)dy + (c + λ) 0 ∫ 1 + (c + λ) γ (x, s, y)v (y, 0, t)dy 0 ∫ 1 γ (x, 0, y)v (y, s, t)dy − (c + λ) + (c + λ) 0 ∫ 1 γ (x, s, y)v (y, 0, t)dy × 0 ∫ 1∫ s − (c + λ) γs (x, s − r , y)v (y, r , t)drdy, 0



+ (c + λ)



1

=

(21)

(22)

0

1

(23)

0

we find the kernels equations as follows:

γs (x, s, y) = γyy (x, s, y) + λγ (x, s, y), x ∈ [0, 1], γ (x, s, 0) = 0,

(24)

γ (x, s, 1) = 0,

γ (x, 0, y) = δ (x − y),

0

an sin(nπ x) = f (x),

∞ ∑

2 sin(nπ x) sin(nπ y).

There are singular points in kernel γ (x, s, t). Fig. 2 shows that sum blows up for s = 0, y = x, where parameters are chosen as λ = 10, D = 4 and c = 0. The delay-compensated controller U(x, t) is derived substituting s = D into the transformation (20) and considering the boundary condition (11), (17), which gives 1



U(x, t) = −(c + λ)

− (c + λ)

0



2π 2s

e−n

sin(nπ y) sin(nπ x),

(27)

n=1

where { 2 sin(nπ x), n ≥ 1} forms an orthogonal basis for L2 [0, 1].

(31)

t −D

Substitute (27) into (31), which gives U(x, t)

[

∞ ∑

γ (x, D, y)u(y, t)dy 1∫ t γ (x, t − τ , y)U(y, τ )dτ dy.

0



(26)

Proposition 1. The singular initial value problem (24)–(26) has a unique solution in the sense of distribution

(30)

n=1

(25)

where δ (·) is Dirac delta function. The following proposition presents the solution of γ , which contains singularity.

γ (x, s, y) = 2eλs

(29)

where an = 2 0 sin(nπ y)f (y)dy for n = 1, 2, . . . are the Fourier coefficients of f (x), the initial condition holds. 2 2 2 2 Due to |e−n π s sin(nπ y) sin(nπ x)| ≤ |e−n π s | ≤ n2 π1 2 s for ∀s ∈ (0, D] and n ≥ 1, we deduce that (27) is bounded for 0 < s ≤ D. □

δ (x − y) =

Substituting (21) and (22) into (16) and combining (7), (13) and

y ∈ (0, 1), s ∈ (0, D].

)

sin(nπ y)f (y)dy sin(nπ x)

∑∞

1

γ (x, 0, y)u(y, t)dy,

1

2

∞ ∑

sin(nπ y) sin(nπ x)f (y)dy, 0

Remark 1. As s = 0, it is known γ (x, 0, y) = n=1 2 sin(nπ y) sin(nπ x) from (27). It is well known [24] that the shifting Dirac Delta function where 0 ≤ x, y ≤ 1 can be expanded in Fourier series as

γs (x, s, y)u(y, t)dy



∞ ( ∫ ∑

1

2

∫1

γs (x, s − r , y)v (y, r , t)drdy

z(x, 0, t) = v (x, 0, t) + (c + λ)

(28)

n=1

1

γ (x, 0, y)v (y, s, t)dy.

∫ ∞ ∑ n=1

n=1

0

0

γ (x, 0, y)f (y)dy =

=

s



1 0

0

+ (c + λ)

γ (x, 0, y)u(y, t)dy = u(x, t),



where we use integration by parts. And then we calculate zs (x, s, t) =vs (x, s, t) + (c + λ)

1

for any function f (x) ∈ L2 [0, 1], we have

0



1 . 2

0

0

0

(c) x =

Proof. Substitute the solution (27) back into (24) and (25) and find the equations hold. Knowing that the initial condition (26) are derived from the following equation



0 1∫



1 , 4

3

= −2(c + λ) e

λD

∞ ∑

−n2 π 2 D

e

sin(nπ x)

+

n=1

sin(nπ x)



1



t

] (λ−n2 π 2 )(t −τ )

e 0

sin(nπ y)u(y, t)dy 0

n=1

∞ ∑

1



sin(nπ y)U(y, τ )dτ dy .

t −D

(32)

4

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

Although control term U(y, τ ) appears on the right hand side of the above equation, the control (31) can be calculated in an iteration algorithm as t > D. Control (32) with distributed actuation for constant distributed delays takes some kind of similar form of the control of the ODE system with distributed delay for a discrete input in [6]. But they have two main differences. First, the second term of control (32) which compensates delay integrates both in domain and in delay time interval, while the control in [6] only integrates in the delay time interval. Second, control (32) involves infinite spatial spectrum due to the infinite dimension of PDE, while the control in [6] contains finite inputs.

∑∞

bounded on [0, D] and n=1 sin(nπ x)zn (r , t) = z(x, r , t). Hence (38) is bounded so does (31). 3. Stability Define the L2 norm for function f (x) ∈ L2 [0, 1] and function g(x, s) ∈ L2 ([0, 1] × [0, D]) as follows: 1



2 L2

f (x)dx, ∥g ∥ 2

∥f ∥ = 0

2 L2

1



D



g 2 (x, s)dsdx.

= 0

0

2.3. Inverse transformation

In order to analyze the stability of the system, define the Lyapunov function

Consider the inverse transformation from the target system to the original system

V1 = ∥u∥2L2 + ∥ux ∥2L2 + ∥v∥2L2 + ∥vs ∥2L2 + ∥v (·, 0, t)∥2L2 .

Correspondingly, the Lyapunov function of the target system is

1



v (x, s, t) = z(x, s, t) − (c + λ) η(x, s, y)u(y, t)dy 0 ∫ 1∫ s η(x, s − r , y)z(y, r , t)drdy, − (c + λ)

V2 = ∥u∥2L2 + ∥ux ∥2L2 + A∥z ∥2L2 + A∥zs ∥2L2 + ∥z(·, 0, t)∥2 , (33)

0

0

where η(x, s, y) defined on [0, 1] × [0, D] × [0, 1] are scalar kernel functions. From the equivalence between the target and the original systems, the kernel functions satisfy

ηs (x, s, y) = ηyy (x, s, y) − c η(x, s, y), x, y ∈ [0, 1], s ∈ [0, D], (34)

η(x, s, 0) = 0,

η(x, s, 1) = 0,

(35)

η(x, 0, y) = δ (x − y).

(36)

The solution of kernel equation (34)–(36) is expressed as

η(x, s, y) =

∞ ∑

2 π 2 )s

2e−(c +n

sin(nπ y) sin(nπ x).

(37)

n=1

By using the inverse transformation (33) and combining with (11) and (17), (31) also can be rewritten as 1



η(x, D, y)u(y, t)dy ∫ 1∫ D − (c + λ) η(x, D − r , y)z(y, r , t)drdy.

U(x, t) = −(c + λ)

0

0

(38)

0

Control expressed by (38) also can be regarded as a good realization due to z(y, r , t) = z0 (y, r + t) and z0 (y, r) can be easily obtained from v0 (y, r) through (20). Remark 2. The control U(x, t) (38) whose another form of expression is (31) can be proved to be bounded if the target system (13)–(18) is stable. We also know η(x, s, y) is bounded as 0 < s ≤ D and η(x, 0, y) = −δ (x − y) from the proof of Proposition 1. Let sup0


η(x, D, y)u(y, t)dy ≤ |Mη | 0

|u(y, t)|dy.

(39)

0

Also, in order to state the boundedness of the control, we consider the second term of (38) 1



D



0

η(x, D − r , y)z(y, r , t)drdy D ∞





= 0

e−(c +n

2 π 2 )(D−r)

sin(nπ x)zn (r , t)dr ,

(40)

n=1

∫1

Theorem 1. Consider the system consisting of the plant (7)–(12) and the control law (32). If the initial conditions u0 ∈ H 2 [0, 1], v0 ∈ L2 [0, 1] × H 1 [0, D] are compatible such that u0 (0) = u0 (1) = 0, v0 (x, D) = U(x, 0), then the system is exponential stable, i.e., V1 (t) ≤ Me−α t V1 (0),

where zn (r , t) = 0 z(y, r , t) sin(nπ y)dy is the coefficient of z(y, r , t). According to Abel’s series convergence test, (40) is con2 2 vergent since {e−n π (D−r) } is monotone sequence and uniformly

(43)

where M , α are positive constants. To prove Theorem 1, we first present the stability of the target system in the following Proposition 2. And then we state the norm equivalence between the original system and the target system in the following Proposition 3. Proposition 2. Consider the system (13)–(18), if the initial conditions u0 ∈ H 2 [0, 1], z0 ∈ L2 [0, 1] × H 1 [0, D] are compatible such that u0 (0) = u0 (1) = 0, and z0 (x, D) = 0, then the system is exponential stable, i.e., V2 (t) ≤ M0 e−α0 t V2 (0),

(44)

where M0 , α0 are positive constants.

∫1

Proof. Taking time derivative of 0 u2 (x, t)dx and using the Cauchy and Schwarz Inequality, we get d

1

(∫

dt

u2 (x, t)dx

)

1



u2x dx − (2c − γ )

≤ −2

0

0

+

1

γ

1



u2 dx 0

1



z 2 (x, 0, t)dx,

(45)

0

∫1

where letting 0 < γ < 2c. Take time derivative of 0 u2x (x, t)dx and use the Cauchy and Schwarz Inequality, which gives d dt

0

(42)

where A > 0 is a weighting constant to be chosen in the following analysis. Our main result is the following theorem.

1



(41)

1

(∫

u2x (x 0

)

γ1



1

, t)dx ≤ −2(1 − ) − 2c 2 0 ∫ 1 1 + z 2 (x, 0, t)dx, γ1 0 u2xx dx

1



u2x dx 0

where letting 0 < γ1 − 1 < 2. ∫1∫D Introduce a new Lyapunov function V3 = 0 0 ebs (zs2 (x, s, t) + z 2 (x, s, t))dsdx with constant b > 0, and let V4 = ∥u∥2L2 + ∥ux ∥2L2 + AV3 + ∥z(·, 0, t)∥2 .

(46)

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

V4 is equivalent to V2 in norm, since 1



D



(zs2 + z 2 )dsdx ≤ 0

0

1



≤ ebD



0

(zs2 + z 2 )dsdx.

(47)

0

=4

1



=4

bs

2e (zs zst + zzt )dsdx



zs2 (x, 0, t) + z 2 (x, 0, t) dx

]

[

≤−

≤4

0 1∫

∫ 0

ebs (zs2 (x, s, t) + z 2 (x, s, t))dsdx.

(48)

0

1

(∫

dt

z 2 (x, 0, t)dx

)

≤ γ2

1



0

zs2 (x, 0, t)dx 0

+

1

γ2

z 2 (x, 0, t)dx, for 0 < γ2 . 0

(49) From (45), (46), (48) and (49), we obtain the following inequalV˙ 4 ≤ − (2c − γ )∥u∥2L2 − (2c + 2)∥ux ∥2L2 − 2(1 −

2

)∥uxx ∥2L2

− (A −

1

γ



1

γ1

where we chose γ1 = therefore

− 1 , 2

1

γ2

)∥z(·, 0, t)∥2L2 ,

γ2 =

1 2

and A >

γ

− (A −

1

γ ≤ − α 0 V2 ,

− 1)∥z(·, 0, t)∥

Lemma 1. Let η(x, s, y) be given as (37). The following inequalities are true:

⏐∫ 1 ⏐2 ⏐ ⏐ ⏐ η(x, s, y)f (y)dy⏐⏐ dsdx ≤ ∥f ∥2L2 , ⏐ 0 0 0 ⏐2 ∫ 1 ∫ D ⏐∫ 1 ∫ s ⏐ ⏐ ⏐ ⏐ dsdx ≤ ∥g ∥22 . η (x , s − r , y)g(y , r)drdy ⏐ ⏐ L 0

0

0

(52)

(53)

0

1 0

⏐2 ⏐ η(x, s, y)f (y)dy⏐⏐ dx 0 (∫ 1 ∞ ∑

⏐∫ ⏐ ⏐ ⏐

=

g(y, r) sin(nπ y)dy

dr

0

0

)2

g(y, r) sin(nπ y)dy

dr

g 2 (y, r)dydr .

(56)

0

To obtain the last line, we use Parseval’s theorem for g(y, r) = ∫1 ∑∞ g (r) sin(n π y) and g (r) = 2 g(y , r) sin(nπ y)dy. Theren n n=0 0 fore, the inequality (53) holds. □ Lemma 2. Let η(x, s, y) be given as (37) , the following inequalities are true : 1



D



0

0

1

⏐∫ ⏐ ⏐ ⏐

0

⏐2 ∫ ⏐ ⏐ ηs (x, s, y)f (y)dy⏐ dsdx ≤ C3

1

2

|f ′ (y)| dy, 0

∀ f ∈ HL2 [0, 1], HL2 [0, 1] = {f ∈ H 2 (0, 1), f (0) = f (1) = 0}. (57) 1



D



0

0

⏐∫ s ∫ ⏐ ⏐ ⏐ 0

1 0

⏐2 ⏐ ηs (x, s − r , y)g(y, r)dydr ⏐⏐ dsdx

≤ C4 ∥g ∥2L2 + C5 ∥g(x, 0)∥2L2 + C6 ∥gs (x, s)∥2L2 , ∀ g ∈ L2 [0, 1] × H 1 [0, D].

(58)

where C3 , C4 , C5 and C6 are positive constants.

Proof. From the Parseval’s theorem and (37) , it holds that



)2

1

1



D



Proof. First consider ∀ f ∈ HL2 [0, 1], it holds that

1

e−2(c +n

n=1

2 π 2 )s

f (y) sin(nπ y)dy

4

0

1

∫ )2

ηs (x, s, y)f (y)dy 0

.

(54)

dr

0

0

D

g(y, r) sin(nπ y)dy 0

1

2

0

Before presenting the norm equivalence between the original system and the target system, we first introduce Lemma 1-Lemma 4 as follows.

1

∞ ∫ s( ∫ ∑

)2

1

∫ s (∫ dr 0

2(c + n2 π 2 )



where α0 = min{2c − γ , b, A − γ1 − 1}. From the fact that V4 is equivalent to V2 in norm, we derive the stability result (44) with M0 = ebD . □

2 π 2 )(s−r)

0

n=1



e−2(c +n

∫ s (∫ 2 2 ∞ ∑ 1 − e−2(c +n π )s



(51)

(55)

0

0

∞ ∫ s ∑

n=1

2 L2

= ∥f ∥2L2 ,

1

n=1

=4

V˙ 4 ≤ − (2c − γ )∥u∥2L2 − (2c + 2)∥ux ∥2L2 − Ab∥z ∥2L2 − Ab∥zs ∥2L2

)2

f (y) sin(nπ y)dy

0

n=1

+ 1 with 0 < γ < 2c,

0

⏐∫ 1 ∫ s ⏐2 ⏐ ⏐ ⏐ η(x, s − r , y)g(y, r)drdy⏐⏐ dx ⏐ 0 0 0 )2 ∞ (∫ 1 ∫ s ∑ −(c +n2 π 2 )(s−r) e sin(nπ y)g(y, r)drdy =4

(50) 1

f (y) sin(nπ y)dy

∑∞

≤4

− Ab∥z ∥2L2 − Ab∥zs ∥2L2 − (A − γ2 )∥zs (·, 0, t)∥2L2



2(c + n2 π 2 )

)2

1

where we use the Parseval’s theorem again for f (x) = n=0 bn ∫1 sin(nπ x) and bn = 2 0 f (y) sin(nπ y)dy. For the second inequality, we can use a similar fashion to prove it. Again, we use the Parseval’s theorem, which gives,

ity

γ1

)2

f (y) sin(nπ y)dy

ds 0

∞ (∫ 1 ∑



1



2 π 2 )s

e−2(c +n

n=1

and, moreover d

0

⏐2 ⏐ η(x, s, y)f (y)dy⏐⏐ dsdx (∫ 1

(∫ 2 2 ∞ ∑ 1 − e−2(c +n π )D

D

−b

1

D

n=1

0 1

0

⏐∫ ⏐ ⏐ ⏐

0

n=1

D



0

∞ ∫ ∑

The time derivative of V3 , satisfies the following estimate V˙ 3 =

D



0

D



1



ebs (zs2 + z 2 )dsdx 0 1

0

Substituting (54) into the left side of (52), we have

D



5

1 ∞





2 π 2 )s

(c + n2 π 2 )e−(c +n

= 0

n=1

2 sin(nπ x) sin(nπ y)f (y)dy

6

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

=−

=−

∞ ∑

(c + n2 π 2 )

n=1



2 π 2 )s

2 sin(nπ x)

1



−nπ sin(nπ y)f (y)dy 0

∞ ∑ (c + n2 π 2 )



n=1

e−(c +n

e−(c +n

2 π 2 )s

2 sin(nπ x)

1



where we use the Parseval’s theorem. Third, )2 ∫ 1 ∫ D (∑ ∞ ∫ 1∫ s −(c +n2 π 2 )(s−r) e gr (y, r)2 sin(nπ y)drdy sin(nπ x) dsdx 0

cos(nπ y)f ′ (y)dy, 0

where we use the integration by parts, and (59) is expressed in the form of the Fourier series. Submit the (59) into the left side of (57) and get 1∫

ηs (x, s, y)f (y)dy dsdx ( )2 ∫ D )2 (∫ 1 ∞ ∑ c + n2 π 2 2 2 ≤ cos(nπ y)f ′ (y)dy 4 e−2(c +n π )s ds nπ 0 0 n=1 ) (∫ 1 )2 ( ∞ 2 2 2 −2(c +n2 π 2 )D ∑ c+n π 1−e ′ ≤ 4 cos(nπ y)f (y)dy nπ 2(c + n2 π 2 ) 0 n=1 )2 ∞ ( ∫ 1 c + π2 ∑ ′ ≤ 2 cos(n π y)f (y)dy 2π 2 0 ∫n=11 2 c+π 2 = |f ′ (y)| dy. (60) 2π 2 0 0

0

0

In∫ the last line we use the Parseval’s theorem by considering 1 2 0 cos(nπ y)f ′ (y)dy as Fourier coefficients of the orthogonal basis cos(nπ x) in L2 [0, 1]. For the second inequality, first consider

∫ s∫

=

1

ηs (x, s − r , y)g(y, r)dydr

∫0 1 ∫0 s ∑ ∞ 0

0

(61)

−(c +n2 π 2 )(s−r)

0

(∫

0 1



D

0 D



0

0

2 sin(nπ x) sin(nπ y)g(y, s)dy

)2

δ (x − y)g(y, s)dy

=



1

D



0

g 2 (x, s)dsdx, 0

(62)

1∫

∫ 0

0 D



D

∞ ∑

e−(c +n

0

(

∞ ∑

−2(c +n2 π 2 )s

n=1 D

0

0



2 sin(nπ y)g(y, 0)dy

π2

ds

gr2 (y, r)drdy

e

−(c +n2 π 2 )(s−r)

)2

2 sin(nπ y)

] drdy ds

0

2 2 ∞ ∑ 1 − e−2(c +n π )s ds c + n2 π 2

n=1

D

gr2 (y, r)drdy,

(64)

0

In a same way, we obtain the following two Lemmas. Lemma 3. Let γ (x, s, y) be given as (27). The following inequalities hold 1



0

0 1



D



D

∫ 0

0

⏐∫ ⏐ ⏐ ⏐ ⏐∫ ⏐ ⏐ ⏐

1 0 1

⏐2 ⏐ γ (x, s, y)f (y)dy⏐⏐ dsdx ≤ ∥f ∥2L2 , s

∫ 0

0

(65)

⏐2 ⏐ γ (x, s − r , y)g(y, r)drdy⏐⏐ dsdx ≤ ∥g ∥2L2 .

(66)

Lemma 4. Let γ (x, s, y) be given as (27), the following inequalities are true: 1

D



1

(∫

0

)2 ∫ γs (x, s, y)f (y)dy dsdx ≤ G3

1

2

|f ′ (y)| dy, 0

0

(67)

0

0

0

(68)

where G3 , G4 , G5 and G6 are positive constants. The following proposition states the equivalence between the two Lyapunov functions (41) and (42). Proposition 3. in norm, i.e.,

The Lyapunov function V1 and V2 are equivalence

V1 ≤ α 1 V2 ,

(69)

V2 ≤ α 2 V1 ,

(70)

where α1 and α2 are positive constants. Proof. First consider the L2 norm of v : 1

D



0

0 1



ds

(63)

1

η(x, s, y)u(y, t)dy 0

]2

s



η(x, s − r , y)z(y, r , t)drdy

+ 0

g 2 (y, 0)dy,

0



0

∫ s(

]

s



1

where the Parseval’s theorem and Cauchy–Schwarz inequality are used. Combine (62), (63) and (64), we get (58). □

0

1 0

n=1

v 2 (x, s, t)dsdx ∫ 1∫ D[ ∫ = z(x, s, t) +

0



1

0

0

)2 )

g(y, 0) sin(nπ y)dy

1



∞ ∫ ∑

0

0

0

D



1

g 2 (y, 0)dyds = D



dsdx

)2 )

1

(∫

1



2 sin(nπ y)g(y, 0)dy sin(nπ x)

0

∞ ( ∫ ∑

0



)2

1



e

2



2 π 2 )s

n=1 2 sin(nπ x) sin(nπ y). Also, we have

0

n=1 D

∑∞

n=1

(

=



(

D



[∫

ds

dsdx

dsdx =

0

where we use δ (x − y) =

, r)drdy

gr2 (y

∀ g ∈ L2 [0, 1] × H 1 [0, D].

n=1 1

(∫

0

0

)2

1 ∞



s



)2

gr (y, r)2 sin(nπ y)drdy

≤ G4 ∥g ∥2L2 + G5 ∥g(x, 0)∥2L2 + G6 ∥gs (x, s)∥2L2 ,

where we integrate with respect to r by parts. Now, we need to estimate three integrations above. First, we find



0

2 π 2 )(s−r)

∀ f ∈ HL2 [0, 1], HL2 [0, 1] = {f ∈ H 2 (0, 1), f (0) = 0}. )2 ∫ 1 ∫ D (∫ s ∫ 1 γs (x, s − r , y)g(y, r)dydr dsdx

n=1

0

1

1

e−(c +n 0



0

× 2 sin(nπ x) sin(nπ y)g(y, r)drdy ∫ 1( ∞ ∑ 2 2 = 2 sin(nπ x) (g(y, s)) − e−(c +n π )s g(y, 0) 0 n=1 ) ∫ s 2 2 − e−(c +n π )(s−r) gr (y, r)dr sin(nπ y)dy,



[∫

0 s



0

n=1 D





(c + n2 π 2 )e

1

=

)2

1

(∫ D

0

0

n=1

∑ (∫

=

(59)



0 D ∞



dsdx

0

≤3∥z ∥2L2 + 3

1

∫ 0

D

∫ 0

⏐∫ ⏐ ⏐ ⏐

0

1

⏐2 ⏐ η(x, s, y)u(y, t)dy⏐⏐ dsdx

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

7

Fig. 3. (a) The dynamics of the state u(x, t) without delay compensation. (b) The dynamics of the state u(x, t) with delay compensation. (c) The norm ∥u(x, t)∥2L2 and

∥u(x, t)∥∞ with delay compensation. 1



D



+3 0

0

1

⏐∫ ⏐ ⏐ ⏐

s



0

0

⏐2 ⏐ η(x, s − r , y)z(y, r , t)drdy⏐⏐ dsdx

≤3∥z ∥2L2 + 3C1 ∥u∥2L2 + 3C2 ∥z ∥2L2 ,

the controller for simulation and express it in a semi-discrete form for clear illustration, (71)

where we use Cauchy–Schwarz inequality and Lemma 1. Then, consider the first derivative of vs (x, s, t) to s 1



D



0

0

|vs (x, s, t)| dsdx [ ∫ 1 D ηs (x, s, y)u(y, t)dy zs (x, s, t) +

=



− where n = U(x, t) = −

η(x, 0, y)z(y, s, t)dy

+ 0

1

∫ s∫

ηs (x, s − r , y)z(y, r , t)dydr

+



]2

≤4∥zs ∥2L2 + 4



1



1



1



D



dsdx

D

(∫ s ∫

D

(∫

1

+4 0



0

0

0

0

dsdx

)2 ηs (x, s − r , y)z(y, r , t)drdy dsdx

0 1

+4

ηs (x, s, y)u(y, t)dy

γ (x, D, y)u(y, t)dy + U(x, t)

1

γ (x, 0

D n

(n − i), y)U(y, t −

D n

] (n − i))dy , (73)

. Move U(x, t) to the left side, c+λ

1

[∫

γ (x, D, y)u(y, t)dy.

c+λ+1

0

n−1 ∫ ∑

D

1

γ (x, 0

n

(n − i), y)U(y, t −

D n

] (n − i))dy , (74)

The dynamics for both compensated and uncompensated cases are shown in Fig. 3(a) and (b), respectively. The state diverges in the case without delay compensated control, while the state converge to zero with delay compensated control in Fig. 3(b). The norms of state u(x, t) are also shown in Fig. 3(c), which illustrate the convergence rate of the state.

)2

0

0

0

1

(∫

D

∆s

i=0

0

0

n−1 ∫ ∑ i=0

0

0 1

0

1 0

2

1∫



U(x, t) = −(c + λ)

[∫

)2 η(x, 0, y)z(y, s, t)dy dsdx

5. Conclusions

0

≤4C3 ∥ux (x, t)∥2L2 + 4(C4 + (λ + c)2 )∥z ∥2L2 + 4C5 ∥z(·, 0, t)∥2L2 + 4(1 + C6 )∥zs ∥2L2 ,

(72)

where we use Cauchy–Schwarz inequality and Lemma 2. Since ∥v (x, 0, t)∥2L2 ≤ 2∥z(x, 0, t)∥2L2 + 2(λ + c)2 ∥u∥2L2 and combine the above results, we obtain (69) and using of Lemmas 3 and 4 in a similar way, we can prove (70). □ In the end, combine Propositions 2 and 3, we get the main result in Theorem 1. 4. Simulation In this section, we provide an example to illustrate the effectiveness of the proposed control law for the PDE system (7)–(12). In the numerical example, let reaction coefficient λ = 10, D = 4 and c = 0. The initial conditions are chosen as u0 = 2 cos(π x), v0 = 2 sin(π xs). We discretize the partial differential equation by the finite difference method and integrate with the Crank– Nicolson scheme. The step sizes for discretization of x, s and t are denoted by ∆x, ∆s and ∆t, respectively. Here, we choose ∆x = 0.02, ∆s = 0.08 and ∆t = 0.001, λ = 10, D = 4, c = 8. Since there are blows up in kernel function γ (x, s, y) when x = y and s = 0 shown in Fig. 2, the controller (32) cannot be used in simulation directly. To avoid blows up in simulation, we revise

This paper proposes an extension of the backstepping method for the boundary actuation with delay to the in-domain control with delay. First, a transport PDE also containing the spatial argument is introduced to formulate the time delay, which results the considered system is coupled with the transport PDE. Then, control of the coupled system is designed by the PDE backstepping method. The solution of the kernel equations is expressed by a series in sine basis. H 1 norm stability is proved in the paper. The main difficulty is to prove the norm equivalence between the target and the compensated controlled system which contains singularity. Future research would consider the control design problem with non-constant input delay, such as the delay varying with spatial argument or varying with time. Declaration of competing interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Acknowledgments The work was supported by the National Natural Science Foundation of China 61773112 and China Scholarship Council 201806635015.

8

J. Qi, M. Krstic and S. Wang / Systems & Control Letters 133 (2019) 104558

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