J Dyn Control Syst (2018) 24:563–576https://doi.org/10.1007/s10883-017-9384-5

Backstepping State Feedback Regulator Designfor an Unstable Reaction-Diffusion PDE with LongTime Delay

Jian-Jun Gu1,2 · Jun-Min Wang1

Received: 14 February 2017 / Revised: 15 August 2017 / Published online: 23 October 2017© Springer Science+Business Media, LLC 2017

Abstract We consider the output regulation of an unstable reaction-diffusion PDE in thepresence of regulator delay and unmatched disturbances, which are generated by an exosys-tem. The systematic design procedure of a backstepping state feedback regulator is firstpresented by mapping the reaction-diffusion PDE cascaded with a transport equation into anerror system, which is shown to be exponentially stable with a prescribed rate in a suitableHilbert space. The regulator design relies on solving regulator equations, and the solvabilitycondition of the regulator equations is then characterized by a transfer function and eigen-values of the exosystem. Finally, the numerical simulations are provided to illustrate theeffect of the regulator.

Keywords Reaction-diffusion PDE · Time delay · Output regulation · Backstepping ·Stability

Mathematics Subject Classification (2010) 93A20 · 93C20 · 93D15

1 Introduction

Control systems with time delays frequently appear in communication and informationtechnologies, such as teleoperated systems [4], computing times in robotics [3], and

� Jian-Jun Gujdgjj@cslg.edu.cn

Jun-Min Wangjmwang@bit.edu.cn

1 School of Mathematics and Statistics, Beijing Institute of Technology, Beijing,People’s Republic of China

2 School of Mathematics and Statistics, Changshu Institute of Technology, Jiangsu,People’s Republic of China

564 Jian-Jun Gu and Jun-Min Wang

communication networks [6]. The presence of time delay usually has a destabilization effectand may deteriorate control system performance. Consequently, the research on time delaysystems is of theoretical importance and practical value. An intensive research activity hasbeen devoted to various control issues for time delay systems, see e.g., stabilization [13, 19,20], observers design [2, 12, 16], and sliding mode control [11]. In [13], the exponentialstability of an unstable reaction-diffusion PDE with long input delay has been establishedby Lyapunov technology, and the decay rate is estimated but not arbitrary. The stabilizationhas been investigated in [19] for a wave equation where the observation signal is subject toa time delay.

Another important control issue is output regulation, that is to design regulators forasymptotic tracking of reference signals and/or rejection of disturbances [5]. When the reg-ulator delay is absent, the regulator has been designed for a diffusive-wave equation in [7,8], parabolic equation in [9], and second order hyperbolic equation in [10]. The kind of reg-ulators are designed so that the original systems are converted into the target systems withsome desirable properties by backstepping approach (see, e.g., [14, 17]).

This paper is concerned with the state feedback regulation of the following unstablereaction-diffusion PDE in the presence of regulator delay and unmatched disturbances:

⎧⎪⎪⎨

⎪⎪⎩

φt (x, t) = φxx(x, t) + c0φ(x, t) + g1(x)d1(t), x ∈ (0, 1), t > 0,

φ(0, t) = qd2(t), t ≥ 0,

φ(1, t) = U(t − τ), t ≥ 0,

yout(t) = C [φ(t)] , t ≥ 0,

(1.1)

where c0 > 0 and , and φ(·, t) is the state variable of the reaction-diffusion equation.U(t) is the input, τ > 0 is time delay in input, and g1 is a continuous function.

Let C1E[0, 1] = {f ∈ C1[0, 1] | f (0) = f (1) = 0}. b ∈ C1

E[0, 1] is assumed to be a

weight function. The output is given by yout(t) = C [φ(t)] = ∫ 10 b(x)φ(x, t)dx denoting the

weighted average of the state over the entire spatial domain [18]. The disturbances di(t) ∈R, i = 1, 2 and the reference signal yref(t) ∈ R, to be tracked by yout(t), are generated bythe following exosystem:

where is diagonalizable in and all the eigenvalues of S are located on theimaginary axis. System (1.2) can model vibration, such assteplike and sinusoidal exogenous signals.

Since time delay is a transfer phenomena, the delay effect can be captured by a transportequation. Let us introduce a new variable u(x, t) = U(t − τx). Then u(x, t) satisfies thetransport equation

τut (x, t) = −ux(x, t), x ∈ (0, 1), t > 0.

Therefore, Eq. 1.1 can be written as the following system of the reaction-diffusion PDEcascaded with a transport equation (see Fig. 1):

⎧⎨

⎩

φt (x, t) = φxx(x, t) + c0φ(x, t) + g1(x)d1(t), x ∈ (0, 1), t > 0,

φ(0, t) = qd2(t), t ≥ 0,

φ(1, t) = u(1, t), t ≥ 0,

(1.3a)

Backstepping State Feedback Regulator Design... 565

Fig. 1 Block diagram for the reaction-diffusion PDE cascaded by transport equation

and {τut (x, t) = −ux(x, t), x ∈ (0, 1), t > 0,

u(0, t) = U(t), t ≥ 0.(1.3b)

In the paper, the output regulation problem is solved by designing a regulator U(t) forEq. 1.3 (or for Eq. 1.1) such that the tracking error satisfies

limt→∞ e(t) = lim

t→∞ [yout(t) − yref(t)] = 0, (1.4)

for all initial values of Eq. 1.3.The main contribution of this paper is to obtain a prescribed decay rate of e(t). For this

purpose, Eq. 1.3a is considered in the state space H−1(0, 1), which is the dual space ofH 1

0 (0, 1) with respect to the pivot space L2(0, 1), and the boundary interconnection requiresthat (1.3b) should be considered in the state space H 1(0, 1). Our regulator design is alsobased on the backstepping approach and error transformations. The error transformationsare found by solving the regulator equations.

We proceed as follows: Section 2 provides the systematic design procedure of a back-stepping regulator by converting (1.3) to a tracking error system, which is shown to beexponentially stable with a prescribed rate in a suitable Hilbert space. Equation 1.4 isobtained by solving the regulator equations. We then present the main results whereinthe solvability condition of the regulator equations and the exponential decay of e(t) areobtained. Section 3 is devoted to the proofs of the main results. In Section 4, numericalsimulations are presented in order to illustrate the effect of the regulator. Finally, someconcluding remarks are given in Section 5.

2 Regulator Design and Main Results

In this section, a backstepping state feedback regulator is designed to transform (1.3) intoa target system, in which the unstable term is moved to the controlled end of the trans-port equation and a dissipative term is added to the parabolic body equation. The errortransformations are then proposed to convert the target system into a tracking error system.Moreover the error system is proven to be exponentially stable in a suitable Hilbert space.This implies (1.4) holds for the designed regulator. Therefore, we divide this section intotwo parts.

2.1 Backstepping Regulator Design

First, we consider the backstepping transformation of the form [13]:

ψ(x, t) = �1[φ(t)](x) = φ(x, t) −∫ x

0k(x, y)φ(y, t)dy, x ∈ [0, 1], (2.1a)

566 Jian-Jun Gu and Jun-Min Wang

andv(x, t) = �2 [u(t), φ(t)] (x) = u(x, t) − ∫ 1

xp(x, y)u(y, t)dy

−∫ 10 γ (x, y)φ(y, t)dy, x ∈ [0, 1], (2.1b)

where the kernels k(x, y), p(x, y), and γ (x, y) satisfy the following equations respectively:⎧⎪⎨

⎪⎩

kxx(x, y) − kyy(x, y) − (c0 + c1) k(x, y) = 0,d

dxk(x, x) = −1

2(c0 + c1) ,

k(x, 0) = 0,

(2.2a)

{px(x, y) + py(x, y) = 0,

p(x, 1) = −τγy(x, 1),(2.2b)

and ⎧⎪⎪⎨

⎪⎪⎩

γx(x, y) + γyy(x, y) + τc0γ (x, y) = 0,

γ (x, 1) = 0,

γ (x, 0) = 0,

γ (1, y) = k(1, y),

(2.2c)

where ddx

k(x, x) = kx(x, x) + ky(x, x). It is straightforward to show that (2.1) maps (1.3)into the target system

⎧⎨

⎩

ψt(x, t) = ψxx(x, t) − c1ψ(x, t) + g�1 (x)w(t), x ∈ (0, 1), t > 0,

ψ(0, t) = qPd2w(t), t ≥ 0,

ψ(1, t) = v(1, t), t ≥ 0,

(2.3a)

and {τvt (x, t) = −vx(x, t) + g�

2 (x)w(t), x ∈ (0, 1), t > 0,

v(0, t) = μww(t), t ≥ 0,(2.3b)

where c1 > 0 is a constant and can be prescribed, and the constant μw is to be determinedin the next subsection. are given by

⎧⎨

⎩

g�1 (x) = �1[g1](x) · Pd1 − ky(x, 0)q · Pd2 ,

g�2 (x) = τγy(x, 0)q · Pd2 − τ

∫ 1

0γ (x, y)g1(y)dy · Pd1 .

The regulator is then obtained by setting x = 0 in Eq. 2.1b and taking into considerationthe boundary conditions of Eqs. 1.3b and 2.3b:

U(t) =∫ 1

0p(0, y)φ(y, t)dy +

∫ 1

0γ (0, y)u(y, t)dy + μww(t), (2.4)

where p(0, y) can be solved from Eq. 2.2b. Since (2.2b) is a simple transport equation,then the solution is given exactly by p(x, y) = −τγy(x, 1). It follows that p(0, y) =−τγy(0, 1). Now we are in a position to find γy(0, 1). To this end, let γ (x, y) = γ (1 −t, y) � γ (t, y), where t = 1 − x, t ∈ [0, 1]. Then (2.2c) can be written as the followinginitial boundary value problems:

⎧⎪⎪⎨

⎪⎪⎩

γt (t, y) − γyy(t, y) − τc0γ (t, y) = 0, t ∈ (0, 1), y ∈ (0, 1),

γ (t, 0) = 0,

γ (t, 1) = 0,

γ (0, y) = k(1, y).

(2.5)

A direct computation gives the solution of Eq. 2.5:

γ (t, y) = 2∞∑

k=1

e(τc0−k2π2)t sin (kπy) ·∫ 1

0sin (kπξ) k(1, ξ)dξ.

Backstepping State Feedback Regulator Design... 567

Consequently, we obtain⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

γ (x, y) = 2∞∑

k=1

e(τc0−k2π2

)(1−x) sin (kπy) ·

∫ 1

0sin (kπξ) k(1, ξ)dξ,

γy(0, 1) = 2∞∑

k=1

(−1)keτc0−k2π2 · (kπ)·∫ 1

0sin (kπξ) k(1, ξ)dξ,

where k(1, y) is determined by Eq. 2.2a, for which the solution has the form [13]

k(x, y) = − (c0 + c1) xI1

[√(c0 + c1)(x2 − y2)

]

√(c0 + c1)(x2 − y2)

with the modified Bessel function I1(·) =∞∑

n=0

(x/2)2n+1

n!(n + 1)! [see 14, P.35].

2.2 Error System and Main Results

To begin with, we consider the error transformations

ϑ(x, t) = ψ(x, t) − m(x)w(t), (2.6a)

ε(x, t) = v(x, t) − n(x)w(t), (2.6b)

where need to satisfy the following equations{

d2xm(x) − c1m(x) − m(x)S + g�

1 (x) = 0,

m(0) = qPd2 ,

and {dxn(x) + τn(x)S − g�

2 (x) = 0,

n(1) = m(1).

Let μw = n(0). Then (2.6) maps (2.3) into the tracking error system:⎧⎨

⎩

ϑt (x, t) = ϑxx(x, t) − c1ϑ(x, t), x ∈ (0, 1), t > 0,

ϑ(0, t) = 0, t ≥ 0,

ϑ(1, t) = ε(1, t), t ≥ 0,

(2.7a)

and {τεt (x, t) = −εx(x, t), x ∈ (0, 1), t > 0,

ε(0, t) = 0, t ≥ 0.(2.7b)

Set C�−11 [m] = Pr . Then e(t) is written in terms of ϑ(x, t):

e(t) = C [φ(t)] − Prw(t)

= C�−11 [ψ(t)] − C�−1

1 [m] w(t)

= C�−11 [ϑ(t)] .

(2.8)

Therefore, if m(x) and n(x) satisfy the regulator equations:⎧⎨

⎩

d2xm(x) − c1m(x) − m(x)S + g�

1 (x) = 0,

m(0) = qPd2 ,

C�−11 [m] = Pr,

(2.9a)

and {dxn(x) + τn(x)S − g�

2 (x) = 0,

n(1) = m(1),(2.9b)

568 Jian-Jun Gu and Jun-Min Wang

then the output regulation of Eq. 1.1 can be obtained from the exponential stability ofEq. 2.7.

We are now ready to investigate the well-posedness and stability of Eq. 2.7. We denoteby ‖ · ‖ the norm of L2(0, 1). Let H 1

E(0, 1) = {g ∈ H 1(0, 1)| g(0) = 0

}with the cor-

responding norm ‖f (·)‖1 = [∫ 10

∣∣f ′(x)

∣∣2 dx]1/2. For each f ∈ H−1(0, 1), we define

‖f ‖−1 = ‖�−1/2f ‖, and �−1 can be given by⎧⎪⎨

⎪⎩

�f (x) = −f ′′(x), D(�) ={f ∈ H 2(0, 1) | f (0) = 0, f (1) = 0

},

(�−1f

)(x) = x

∫ 1

0f (τ)(1 − τ)dτ −

∫ x

0f (τ)(x − τ)dτ.

(2.10)

Moreover, we have the following proposition:

Proposition 1 Let �−1 be defined by Eq. 2.10. Then �−1 is a positive, self-adjointoperator. Therefore, �−1/2 is self-adjoint.

Proof Performing some straightforward computations, we have⟨�−1f, f

⟩

L2(0,1)

=⟨x∫ 1

0 f (τ)(1 − τ)dτ − ∫ x

0 f (τ)(x − τ)dτ, f⟩

L2(0,1)

= ∫ 10 xf (x)dx

∫ 10 f (τ)(1 − τ)dτ − ∫ 1

0 f (x)[∫ x

0 f (τ)(x − τ)dτ]dx

= ∫ 10 f (τ)(1 − τ)dτ

∫ 10 xf (x)dx − ∫ 1

0 xf (x)[∫ x

0 f (τ)dτ]dx

+∫ 10 f (x)

[∫ x

0 τf (τ)dτ]dx

= ∫ 10 f (τ)

[∫ τ

0 xf (x)dx]dτ + ∫ 1

0 f (τ)[∫ 1

τxf (x)dx

]dτ

−∫ 10 τf (τ)

[∫ τ

0 xf (x)dx]dτ − ∫ 1

0 τf (τ)[∫ 1

τxf (x)dx

]dτ

−∫ 10 xf (x)

[∫ x

0 f (τ)dτ]dx + ∫ 1

0 f (x)[∫ x

0 τf (τ)dτ]dx

= ∫ 10 f (τ)(1 − τ)

[∫ τ

0 xf (x)dx]dτ + ∫ 1

0 f (x)(1 − x)[∫ x

0 τf (τ)dτ]dx

= 2∫ 1

0 f (τ)(1 − τ)[∫ τ

0 xf (x)dx]dτ ≥ 0,

which implies that �−1 is a positive operator. On the other hand,⟨�−1f, g

⟩

L2(0,1)

=⟨x∫ 1

0 f (τ)(1 − τ)dτ − ∫ x

0 f (τ)(x − τ)dτ, g⟩

L2(0,1)

= ∫ 10 xg(x)

[∫ 10 f (τ)(1 − τ)dτ

]dx − ∫ 1

0 g(x)[∫ x

0 f (τ)(x − τ)dτ]dx

= ∫ 10 f (x)

[∫ x

0 τg(τ )dτ]dx − ∫ 1

0 xf (x)dx∫ 1

0 τg(τ )dτ

+∫ 10 xf (x)dx

∫ 10 g(τ)dτ − ∫ 1

0 xf (x)[∫ x

0 g(τ)dτ]dx

=⟨f, x

∫ 10 g(τ)(1 − τ)dτ − ∫ x

0 g(τ)(x − τ)dτ⟩

L2(0,1)

= ⟨f, �−1g

⟩

L2(0,1).

Consequently, �−1 is a self-adjoint operator, and �−1/2 is self-adjoint. The proof iscomplete.

Let us consider (2.7a) and (2.7b) in the state space H−1(0, 1) and H 1E(0, 1), respectively.

Define operator A : D(A)(⊂ H−1(0, 1)

) → H−1(0, 1) as follow:

Af = f ′′ − c1f, ∀f ∈ D (A) = H 10 (0, 1). (2.11)

Backstepping State Feedback Regulator Design... 569

Obviously, A = A∗, where A∗ is the adjoint operator of A. Considering Proposition 1 andtaking the inner product on the both sides of Eq. 2.7a with η ∈ D (A∗) renders to

ddt

〈ϑ, η〉H−1(0,1) = ⟨ϑ ′′ − c1ϑ, η

⟩

H−1(0,1)

= ⟨�−1/2

(ϑ ′′ − c1ϑ

), �−1/2η

⟩

L2(0,1)

= ⟨ϑ ′′ − c1ϑ,�−1η

⟩

L2(0,1)

= − (t)[

ddx

(�−1ϑ

)](1) + ⟨

ϑ, η′′ − c1η⟩

H−1(0,1)

= − ⟨δ′(x − 1)�−1 (t), η

⟩

D(A∗)′×D(A∗) + 〈ϑ,A∗η〉H−1(0,1) ,

where δ(·) is the Dirac distribution, and (t) = ε(1, t). Therefore, Eq. 2.7a can be written as⎧⎨

⎩

ϑt (·, t) = Aϑ(·, t) + B (t),

B∗ϑ = −[

d

dx

(�−1ϑ

)]

(1).(2.12)

We have the following lemma concerning A:

Lemma 1 Let A be given by Eq. 2.11. Then A−1 exists and is compact on H−1(0, 1), andconsequently σ(A), the spectrum of A, consists of isolated eigenvalues of finite algebraicmultiplicity only.

Proof For any given f1 ∈ H−1(0, 1), solve f1 = Af to yield

f ′′(x) − c1f (x) = f1(x), f (0) = f (1) = 0,

which has the solution⎧⎪⎪⎨

⎪⎪⎩

f (x) = − sinh(√

c1x)

sinh(√

c1) η(1) + η(x),

η(x) = 1√c1

∫ x

0sinh

[√c1(x − τ)

]f1(τ )dτ.

Therefore, we get the unique solution f ∈ D (A), which implies that A−1 exists and iscompact on H−1(0, 1) by Sobolev embedding theorem [1]. Thus, we finish the proof.

We can now present the main results of the paper, and the proofs will be given inSection 3.

Theorem 1 Let A be given by Eq. 2.11. Then for any ε(·, 0) ∈ H−1(0, 1), and ϑ(·, 0) ∈H 1

E(0, 1), there exist constants M and M > 0 dependent on � with 0 < � < c1 such thatthe following assertions hold:

(1). There exists a unique solution to Eq. 2.7 such that ϑ(·, t) ∈ C([0, ∞) ;H−1(0, 1)

),

ε(·, t) ∈ C([0, ∞) ; H 1

E(0, 1)), and

‖ϑ(·, t)‖−1 + ‖ε(·, t)‖1 ≤ Me−�t[‖ϑ(·, 0)‖−1 + ‖ε(·, 0)‖1

].

(2). The error e(t), given by Eq. 1.4, satisfies

|e(t)| ≤ Me−�t[‖ϑ(·, 0)‖−1 + ‖ε(·, 0)‖1

].

The following corollary is a direct consequence of the above theorem:

Corollary 1 The output regulation problem of Eq. 1.3 is solvable, i.e. Eq. 1.4 holds for theregulator (2.4) if (2.9) is solvable.

570 Jian-Jun Gu and Jun-Min Wang

The next lemma gives a necessary and sufficient condition for the solvability of Eq. 2.9.

Lemma 2 (Cascaded Regulator Equations) Let F(s) = N(s)/D(s) be the transferfunction of Eq. 1.3 and the numerator N(s) = C�−1

1 [β(0, s, ·)], where β(ξ, s, x) =cosh

[√s + c1 (x − ξ)

]. There exists a unique classical solution of Eq. 2.9 if and only if

N(λ) �= 0, ∀λ ∈ σ(S).

Remark 1 The condition in Lemma 2 means that no eigenvalue of the exosystem (1.2) is atransmission zero of Eq. 1.3 [5, P.15].

3 Proofs of the Main Results

This section is devoted to proving Theorem 1 and Lemma 2 stated in the previous section.Therefore, the section is divided into two parts.

3.1 Proof of Theorem 1

We first confirm Assertion (1). Let ε0(x) be the initial state of Eq. 2.7b. Solve ε(x, t) fromEq. 2.7b to obtain

ε(x, t) ={

0, t > τx,

ε0(x − 1

τt), t ≤ τx.

Let us consider the Lyapunov functional

V (t) =∫ 1

0e−c2τxε2

x(x, t)dx,

where the positive constant c2 > c1. Differentiating V (t) with respect to t and integratingby parts yields

V (t) = 2∫ 1

0 e−c2τxεx(x, t)εxt (x, t)dx

= − 2τ

∫ 10 e−c2τxεx(x, t)εxx(x, t)dx

= − e−c2τ

τε2x(1, t) − c2V (t)

≤ −c2V (t).

Consequently, V (t) ≤ −c2V (t). By Holder’s inequality, we have (t) ≤ ‖ε(·, t)‖1 ≤M1e

−c2t ‖ε(·, 0)‖1, where M1 > 0 is a constant depending on c2 .A direct computation shows that λk = −c1 − k2π2 is an eigenvalue of A with the

associated eigenfunction sin . Obviously, {ϑk(x)}∞k=1 forms anorthonormal basis in H−1(0, 1). Therefore, A generates exponentially stable C0-semigroupeAt on H−1(0, 1) with the decay rate � = c1. Consider the adjoint system of Eq. 2.12:

ηt (·, t) = A∗η(·, t), B∗η =∫ 1

0xη(x)dx,

where A∗ = A. It is easy to check that B∗A∗−1 is bounded from H−1(0, 1) intoThen we will show that the control operator B is admissible for the C0-semigroup eAt on

Backstepping State Feedback Regulator Design... 571

H−1(0, 1). For each η(x, 0) = η0(x) ∈ H−1(0, 1), we have η0(x) = ∑∞k=1 ak sin (kπx)

with x ∈ [0, 1], ‖η0‖2−1 = ∑∞k=1 |ak|2 1

k2π2 , and η(x, t) can be written as

η(x, t) =∞∑

k=1

ake−(c1+k2π2)t sin (kπx) , x ∈ [0, 1], t > 0.

Hence,

B∗η(t) = −∫ 10 xη(x, t)dx

= −∞∑

k=1ake

−(c1+k2π2)t∫ 1

0 x sin (kπx) dx

=∞∑

k=1(−1)k

ak

kπe−(c1+k2π2)t ,

and for any T > 0

∫ T

0 |(B∗η) (t)|2 dt = ∫ T

0

∣∣∣∣

∞∑k=1

(−1)kak

kπe−(c1+k2π2)t

∣∣∣∣

2

dt

≤∞∑

k=1

a2k

k2π2 ·∞∑

k=1

∫ T

0 e−2[c1+k2π2]t dt

≤ M2 ‖η(·, 0)‖2−1 ,

where constant M2 > 0 only depends on T . Therefore, B is admissible for the semigroupeAt generated by A on H−1(0, 1). Hence, for any ϑ(·, 0) ∈ H−1(0, 1), there exists a uniquesolution ϑ(x, t) ∈ C

([0, ∞) ;H−1(0, 1)

)to Eq. 2.12 such that

ϑ(·, t) = eAt ϑ(·, 0) +∫ t

0eA(t−s)B (s)ds. (3.1)

Let 0 < � < c1 < c2. Since (t) ≤ M1e−c2t ‖ε(·, 0)‖1, we obtain

∫∞0 e2�t | (t)|2 dt ≤ M2

1

∫∞0 e−2(c2−�)tdt · ‖ε(·, 0)‖2

1≤ M2‖ε(·, 0)‖21,

where M2 = M21

2(c2−�). Hence e�t (t) ∈ L2(0, ∞). Multiply both sides of Eq. 3.1 by e�t

to get

e�tϑ(·, t) = e(A+�)tϑ(·, 0) +∫ t

0e(A+�)(t−s)Be�s (s)ds.

In view of (7.4.1.3) of [15, Theorem 7.4.1.1, on P.654] (for the ε = 0 there), we have∥∥e�tϑ(·, t)∥∥−1 + ∥

∥e�tε(·, t)∥∥1 ≤ M3[‖ϑ(·, 0)‖−1 + ‖e� · ‖L2(0,∞)

]

+M1e−(c2−�)t ‖ε(·, 0)‖1

≤ M3[‖ϑ(·, 0)‖−1 + √

M2 ‖ε(·, 0)‖1]

+M1e−(c2−�)t ‖ε(·, 0)‖1

≤ M[‖ϑ(·, 0)‖−1 + ‖ε(·, 0)‖1

],

where constant M = max{M1 + √

M2M3,M3}. Therefore, we eventually obtain

‖ϑ(·, t)‖−1 ≤ ‖ϑ(·, t)‖−1 + ‖ε(·, t)‖1≤ Me−�t

[‖ϑ(·, 0)‖−1 + ‖ε(·, 0)‖1].

572 Jian-Jun Gu and Jun-Min Wang

Now, we are in a position to prove Assertion (2). Let l(x, y) ∈ C2(T) be the kernel of�−1

1 . From Eq. 2.8, we get

e(t) = C�−11 [ϑ(t)] =

∣∣∣∫ 1

0 b(x)[ϑ(x, t) − ∫ x

0 l(x, y)ϑ(y, t)dy]dx

∣∣∣

≤∣∣∣∫ 1

0 b(x)ϑ(x, t)dx

∣∣∣ + ‖l‖C2(T)

∣∣∣∫ 1

0 b(x)[∫ x

0 ϑ(y, t)dy]dx

∣∣∣

=∣∣∣∫ 1

0 b(x)ϑ(x, t)dx

∣∣∣ + ‖l‖C2(T)

∣∣∣∫ 1

0 ϑ(x, t)[∫ 1

xb(y)dy

]dx

∣∣∣

=∣∣∣∫ 1

0 �1/2b(x)·�−1/2ϑ(x, t)dx

∣∣∣+‖l‖C2(T)

∣∣∣∫ 1

0 �1/2[∫ 1

xb(y)dy

]·�−1/2ϑ(x, t)dx

∣∣∣

≤ Me−�t[‖ϑ(·, 0)‖−1 + ‖ε(·, 0)‖1

],

where M = M ‖b‖C1[0,1] · max{1, ‖l‖C2(T)

}. The proof is complete.

Remark 2 If we consider (2.7a) in the usual state space L2(0, 1), then we find that thecontrol operator B is not admissible for the semigroup eAt . In fact,

d

dt〈ϑ, η〉L2(0,1) = ⟨

ϑ ′′ − c1ϑ, η⟩

L2(0,1)

= − (t)η′(1, t) + ∫ 10 ϑ(x, t)η′′(x, t)dx − c1

∫ 10 ϑ(x, t)η(x, t)dx

= 〈B (t), η〉D(A∗)′×D(A∗) + 〈ϑ,A∗η〉L2(0,1) ,

where B = −δ′(x − 1). Therefore, for any T > 0, we have

∫ T

0 |(B∗η) (t)|2 dt = ∫ T

0 η′(1, t)dt

= ∫ T

0

∣∣∣∣

∞∑k=1

ak(kπ)e−(c1+k2π2)t

∣∣∣∣

2

dt

≤∞∑

k=1a2k ·

∞∑k=1

∫ T

0 (kπ)2e−2[c1+k2π2]t dt

� M ‖η(·, 0)‖2 .

Consequently, the state space of Eq. 2.7a is chosen as H−1(0, 1).

3.2 Proof of Lemma 2

Firstly, a direct computation shows the numerator of the transfer function F(s) : N(s)

= C�−11 [β(0, s, ·)].

Next, we need to decouple (2.9). Since S is diagonalizable in there exists a similar-ity transformation V −1SV = diag

(λv,1, · · · , λv,nw

), where V = [

Vv,1, · · · , Vv,nw

]with

1 2 Postmultiplying the first two equations of Eqs. 2.9a, and2.9b by V , we then obtain the following cascaded ODEs:

{d2xm∗

j (x) − m∗j (x)

(λv,j + c1

) = −g∗1,j (x),

m∗j (0) = qP ∗

d2,j,

(3.2a)

and{

dxn∗j (x) + n∗

j (x)τλv,j = g∗2,j (x),

n∗j (1) = m∗

j (1), j = 1, 2, · · · , nw,(3.2b)

where m∗j (x) = m(x) · Vv,j , n∗

j (x) = n(x) · Vv,j , P ∗d3,j

= Pd3 · Vv,j , and g∗i,j (x) =

g�i (x) · Vv,j , i = 1, 2.

Backstepping State Feedback Regulator Design... 573

The solution to Eq. 3.2a is given by

m∗j (x) = qP ∗

d2,je− j x + 2K1β(0, λv,j , x) − 1

j

∫ x

0 β(ξ, λv,j , x)g∗1,j (ξ)dξ,

where β(ξ, λv,j , x) = sinh[ j (x − ξ)

]with j = √

λv,j + c1 �= 0. The constant K1 needs

to be uniquely determined later. Set P ∗r,j = Pr · Vv,j . Postmultiply C�−1

1 [m] = Pr by V to

yield C�−11

[m∗

j

]= P ∗

r,j . So we obtain

C�−11

[m∗

j

]= qP ∗

d2,j· C�−1

1

[e− j ·] + 2K1C�−1

1

[β(0, λv,j , ·)

]

− 1 jC�−1

1

[∫ ·0 β(ξ, λv,j , ·)g∗

1,j (ξ)dξ]

= P ∗r,j .

(3.3)

K1 can be uniquely solved from Eq. 3.3 if and only if N(λv,j ) �= 0 with λv,j ∈ σ(S):

K1 = 12N(λv,j )

{P ∗

r,j + 1 jC�−1

1

[∫ ·0 β(ξ, λv,j , ·)g∗

1,j (ξ)dξ]

− qP ∗d2,j

· C�−11

[e− j ·]

}.

Therefore,

m∗j (1) = qP ∗

d2,j· e− j + 2K1β(0, λv,j , 1) − 1

j

∫ 10 β(ξ, λv,j , 1)g∗

1,j (ξ)dξ

� K2.

Lastly, combining the Dirichlet boundary condition of Eq. 3.2a, the unique solution toEq. 3.2 under condition (3.3) is given by

m∗j (x) = qP ∗

d2,j· e− j x + 2K1β(0, λv,j , x) − 1

j

∫ x

0β(ξ, λv,j , x)g∗

1,j (ξ)dξ, (3.4a)

and

n∗j (x) = K2e

τλv,j (1−x) −∫ 1

x

eτλv,j (ξ−x)g∗2,j (ξ)dξ, j = 1, 2, · · · , nw. (3.4b)

Hence, the classical solution of Eq. 2.2 is

m(x) = (m∗

1(x),m∗2(x), · · · , m∗

nw(x)

) · V −1,

n(x) = (n∗

1(x), n∗2(x), · · · , n∗

nw(x)

) · V −1,

where m∗j (x) and n∗

j (x), j = 1, 2, · · · , nw , are given by Eq. 3.4. The proof is complete.

4 Simulation Results

In this section, we present numerical simulation results for the output regulation ofEq. 1.1. For simplicity, we set c0 = q = 2, and g1(x) = 1. The output is taken asyout(t) = ∫ 1

0 sin(2πx)φ(x, t)dx. The regulator (2.4) drives yout(t) to track sinusoidal

574 Jian-Jun Gu and Jun-Min Wang

0 1 2 3 4 5 6 7−0.5

0

0.5

t

Ou

tpu

t a

nd

re

fere

nce

sig

na

ly

out(t)

yref

(t)

Fig. 2 The reference “yref(t)” and output “yout(t)” without regulator

reference yref(t), while rejecting the constant disturbances di(t), i = 1, 2. yref(t) and di(t),i = 1, 2, are generated by Eq. 1.2 with

0 1 2 3 4 5 6 7−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

t

Ou

tpu

t a

nd

re

fere

nce

sig

na

l yout

(t)

yref

(t)

Fig. 3 The reference “yref(t)” and output “yout(t)” under regulator

Backstepping State Feedback Regulator Design... 575

where bdiag (X1, X2, · · · , Xn) represents block diagonal matrix with blocks X1, X2,· · · , Xn on the main diagonal, and adiag (a1, a2, · · · , an) denotes anti-diagonal matrix withelements a1, a2, · · · , an on the anti-diagonal.

For numerical computations, the steps of space and time are set as 0.01 and 0.00005,respectively. φ(x, 0) = x5 − 1. When the regulator (2.4) is absent, Fig. 2 shows thatyout(t) does not track yref(t). However, it can be seen in Fig. 3 that yout(t) is forced byregulator (2.4) to track yref(t).

5 Conclusion

In this paper, a state feedback regulator design has been presented for the output regu-lation of an unstable reaction-diffusion PDE with time delay, which was written as thereaction-diffusion PDE cascaded with a transport equation. Using the backstepping anderror transformations, the cascaded system has been converted to an exponentially stableerror system. The error transformations were built on solving regulator equations, and thesolvability condition of the regulator equations has been characterized by a transfer functionand eigenvalues of the exosystem.

Acknowledgements This work was supported by the National Natural Science Foundation of China (GrantNo. 61673061).

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