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Download by: [Beijing Institute of Technology] Date: 17 November 2016, At: 17:23

International Journal of Control

ISSN: 0020-7179 (Print) 1366-5820 (Online) Journal homepage: http://china.tandfonline.com/loi/tcon20

The active disturbance rejection control of therotating disk–beam system with boundary inputdisturbances

Ya-Ping Guo & Jun-Min Wang

To cite this article: Ya-Ping Guo & Jun-Min Wang (2016) The active disturbance rejection controlof the rotating disk–beam system with boundary input disturbances, International Journal ofControl, 89:11, 2322-2335, DOI: 10.1080/00207179.2016.1155754

To link to this article: http://dx.doi.org/10.1080/00207179.2016.1155754

Accepted author version posted online: 25Feb 2016.Published online: 15 Mar 2016.

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INTERNATIONAL JOURNAL OF CONTROL, VOL. , NO. , –http://dx.doi.org/./..

The active disturbance rejection control of the rotating disk–beam systemwithboundary input disturbances

Ya-Ping Guo and Jun-Min Wang

School of Mathematics and Statistics, Beijing Institute of Technology, Beijing, P.R. China

ARTICLE HISTORYReceived July Accepted February

KEYWORDSRotating disk–beam system;disturbance; ADRC; stable

ABSTRACTThis paper deals with the stabilisation of the rotating disk–beam system, where the control ends ofbeam and disk are suffered from disturbances, respectively. The active disturbance rejection control(ADRC) approach is adopted in investigation. The disturbances are first estimated by the extendedstate observers, and the observer-based feedback control laws are thendesigned to cancel the distur-bances. When the angular velocity of the disk is less than the square root of the first nature frequencyof the beam, it is shown that the feedback control laws are robust to the external disturbances, that is,the vibration of the beam can be suppressedwhile the disk rotatingwith a desired angular velocity inpresence of the disturbances. Finally, the simulation results are provided to illustrate the effectivenessof ADRC approach.

1. Introduction

Many flexible structures in mechanical systems can bemodelled by the Euler–Bernoulli beam equation inter-connected with ordinary differential equation (ODE),such as flexible robots arms (Chen, Chentouf, & Wang,2015), aerospace structures (Baillieul & Levi, 1987) andflexible marine risers Do & Pan (2008), etc. The stabilisa-tion problem of the rotating flexible system has been theobject of a considerable mathematical endeavour (Chen,Chentouf, & Wang, 2014; Chentouf & Wang, 2006, 2015;Laousy, Xu, & Sallet, 1996; Xu & Baillieul, 1993). Indeed,there are two categories of works: in the first one, onlya torque control is exerted (Baillieul & Levi, 1987; Xu& Baillieul, 1993). In turn, in the second one, at leastone boundary control (usually force or moment) is pre-sented in the feedback law in addition of the torquecontrol (Chentouf & Wang, 2006; Laousy et al., 1996).In the practical application, the disturbances (e.g., tem-perature, sound waves and vibration) and uncertainties(e.g., unknown system parameters) have strong adverseeffects on performances of the flexible structures (He &Ge, 2015). To our knowledge, the stabilisation for therotating disk–beam system with the disturbance sufferedfrom the boundary control end is still open. The rotat-ing disk–beam system is nonlinear coupling systems andthe state of the disk and beam is interdependence. Hence,in the presence of the disturbances or uncertainties, thestability analysis of the rotating disk–beam systems isnot easy.

CONTACT Jun-Min Wang [email protected]

Many control approaches have been developed to dealwith the disturbances or uncertainties. These include theslidingmode control for systems with the external distur-bances (Wang, Liu, Ren, &Chen, 2015),modelling uncer-tainties (Qin, Zhong, & Sun, 2013), the adaptive con-trol for systems with unknown parameters (Ge, Zhang,& He, 2011; He, Ge, How, Choo, & Hong, 2011), andthe Lyapunov approach for disturbance (Guo & Kang,2014; He, Sun, & Ge, 2015), to name just a few. Mostof these approaches are the worst-case concern strategyin dealing with disturbance. These control approachesdesigned are usually over-conservative, thus tend to resultin using very large control energy (Feng & Guo, 2014),which is often unnecessary for any particular system.The active disturbance rejection control (ADRC), as anunconventional design approach, is first introduced byHan (2009).

In the past, the ADRC approach has been developedto obtain satisfactory performance of the closed-loop sys-tem, and its applications can be found in lots of literatures(Guo & Jin, 2013a, 2013b; Guo & Zhou, 2015; Xia, Dai,Fu, Li, & Wang, 2014). The ADRC strategy has been wellknown for its performance that estimate or compensatethe system uncertainty or external disturbances. Thus, inthe frame of ADRC, general nonlinear uncertainties orexternal disturbances are permitted (Xia, Shi, Liu, Rees,& Han, 2007). For example, the stabilisation problemof one-dimensional wave equation and Euler–Bernoullibeam equation with nonlinear disturbances have been

© Informa UK Limited, trading as Taylor & Francis Group

INTERNATIONAL JOURNAL OF CONTROL 2323

dealt in Feng and Guo (2014) and Guo and Jin (2013a) bythe ADRC strategy, respectively. TheADRChas been alsoapplied to DC–DC power converts in power electronics(Sun & Gao, 2005), high pointing accuracy and rotationspeed (Li, Yang, & Yang, 2009), motion control (Gao, Hu,& Jiang, 2001), flight system (Huang, Xu, Han, & Lam,2001) and other industrial systems (Xia, Chen, Pu, &Dai,2014). For more detail of the ADRC, we can refer to thereferences in Guo and Zhao (2011)

The objective of this paper is to design the feedbackcontrol, by means of ADRC approach, to achieve stabil-isation for the rotating disk–beam system with bound-ary input disturbances. The disturbances are estimatedthrough extended state observers, and the observer-basedfeedback control laws are then designed for the rotatingdisk–beam system. The stabilisation and well-posednessof the closed-loop system are proved by dividing thesystem into two parts: one is for system (1) with con-trols, and another is for the observer system. For angu-lar velocity of the disk less than a well-defined value,we show that the feedback controllers are robust to theexternal disturbances, that is, regardless of the presenceof the disturbances on the control end, the beam vibra-tions can be suppressed to zero while the disk will rotatewith the desired angular velocity as time goes to infin-ity. Moreover, we relaxed the desired angular velocityto the square root of the first frequency of the beam√

μ1EI/ρ � 3.516√EI/ρ (EI and ρ are, respectively, the

flexural rigidity and themass per unit length of the beam).This improve the desired value 3

√EI/ρ in Laousy et al.

(1996).The rest of this paper is organised as follows. In Sec-

tion 2, the model of the rotating disk–beam system isgiven, along with some results. In Section 3, a feedbackcontrol laws are designed for the disk–beam system viathe ADRC method, where the ADRC is used to estimatethe disturbances. Section 4 is devoted to stability analysisof the closed-loop system, respectively. In Section 5, sim-ulation results are provided to demonstrate the effective-ness of the ADRC. The conclusion is presented in Section6. When angular velocity of the disk is less than a criticalvalue, the stability of the linear beam subsystem is provedin the Appendix.

2. Problem statement and preliminaries

This paper investigates the stabilisation of the rotatingdisk–beam system subject to the boundary disturbances.Assume that a beam (B) clamped at the center of a disk(D) and free at the other end. The disk rotates freelyabout its axis with a non-uniform angular velocity and themotion of the beam is confined to a plane perpendicularto the disk (see Figure 1). Indeed, we can imagine Figure 1

Figure . The disk-beam system.

as a satellite, the beam being its antenna. The purpose ofthis paper is to show that under the effects of externaldisturbance, for any desired angular velocity of the diskless than a well-defined value, the beam vibrations can beforced to decay to zero while the disk will rotate with thedesired angular velocity. The rotating disk–beam systemwith boundary moment control and disturbance as fol-lows (see Figure 1):

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

ρutt + EIuxxxx = ρω2(t )u, x ∈ (0, 1), t > 0,u(0, t ) = ux(0, t ) = 0, t ≥ 0,uxx(1, t ) = U1(t ) + d1(t ), t ≥ 0,uxxx(1, t ) = 0, t ≥ 0,ddt

{(Id + ρ

∫ 1

0u2dx

)ω(t )

}= U2(t ) + d2(t ), t > 0,

(1)

For simplicity and without loss of generality, assume thatEI = ρ = 1 (in fact a simple change of variables will leadto unit physical parameters). Let u(x, t) denote the trans-verse displacement of the beam at time t and position x,ω(t) be the angular velocity of the disk at time t, Id denotethe disk’s moment of inertia, U1(t) be the control inputthrough bend moment, and U2(t) be the control to beapplied on the disk. The functions d1(t) and d2(t) are theexternal disturbances acting on the beam and the disk,respectively, and we have

Assumption : |di(t )|, |di(t )| ≤ M, i = 1, 2,forM > 0 and all t ≥ 0.

When d1(t) and d2(t) are absent, the collocated feedbackcontrols are designed to stabilise system (1) in Laousy

2324 Y.-P. GUO AND J.-M. WANG

et al. (1996):

⎧⎨⎩U1(t ) = −αuxt (1, t ), U2(t )

= −γ(ω(t ) − ω

), α ≥ 0, γ > 0,

uxxx(1, t ) = βut (1, t ), β ≥ 0, α2 + β2 �= 0,

where ω ∈ R is the desired angular velocity of the disk. Itis shown that the system is globally exponentially stablewith the assumption ω <

√9EI/ρ.

Let HnE (0, 1) = {u ∈ Hn(0, 1) : u(0) = ux(0) = 0},

n ∈ N. We consider system (1) in state space

X = H × R, whereH = H2E (0, 1) × L2(0, 1)

with the norm induced by the inner product: for Xi =( fi, gi, ωi) ∈ X, i = 1, 2,

〈X1, X2〉X = ∫ 10

[f ′′1 (x) f ′′

2 (x) − ω2 f1(x) f2(x)+ g1(x)g2(x)

]dx + ω1ω2

which is equivalent to the usual one of H2(0, 1) ×L2(0, 1) × R provided that ω <

√μ1. Here,μ1 is the first

nature frequency of the beam which can be given by thefirst eigenvalue of the self-adjoint positive operator F inL2(0, 1):

Fz = z′′′′, D(F ) = {z ∈ H4

E (0, 1) : z′′(1) = z′′′(1) = 0

}.

Moreover, μ1 satisfies (see Luo & Guo, 1997): 1 +cos(4μ1) cosh(4μ1) = 0 and the numerical simulationtells √

μ1 � 3.516. In addition, for |ω| <√

μ1,

1 + cos(√

|ω|) cosh(√

|ω|) > 0. (2)

We define the unbounded linear operators A and B asfollows:⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

A(y, z) = (z, −yxxxx + ω2y),∀(y, z) ∈ D(A)(⊂ H) → H,

D(A) = {(y, z) ∈ H ∩ (

H4E (0, 1)

×H2E (0, 1)

) ∣∣yxx(1) = 0, yxxx(1) = 0},

B = (0, δ′(x − 1)),

(3)

and define the nonlinear function J in X = H × R: for(φ, ω) ∈ X with φ = (y, z) ∈ H,

J(φ, ω, t ) =(0, (ω2(t ) − ω2)y,

−2ω(t )∫ 10 yzdx +U2(t ) + d2(t )

Id + ∫ 10 y2dx

).

(4)

Then system (1) can be written as

ddt

(φ(t )ω(t )

)=( A 00 1

)(φ(t )ω(t )

)+(B0

)× (U1(t ) + d1(t )) + J(φ(t ), ω(t ), t ),

(5)

By Xu and Baillieul (1993), we know that A generates aC0-semigroup onH. However, it is easy to prove that B isnot admissible to eAt . This is different from those in Fengand Guo (2014) and Guo and Jin (2013a), where the sys-tems are always well-posed. To overcome this difficulty,we design the following control:

U1(t ) = −αuxt (1, t ) +V1(t ), (6)

where V1(t) is a new control variable. Under control (6),system (1) becomes

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ρutt + EIuxxxx = ρω2(t )u, x ∈ (0, 1), t > 0,u(0, t ) = ux(0, t ) = 0, t ≥ 0,uxx(1, t ) = −αuxt (1, t )

+ V1(t ) + d1(t ), t ≥ 0,uxxx(1, t ) = 0, t ≥ 0,ddt

{(Id + ρ

∫ 1

0u2dx

)ω(t )

}= U2(t ) + d2(t ), t > 0.

(7)

Define the unbounded linear operatorsAω as follows:

⎧⎪⎪⎨⎪⎪⎩Aω(y, z) = (z, −yxxxx + ω2y),∀(y, z) ∈ D(Aω )(⊂ H) → H,

D(Aω ) = {(y, z) ∈ H ∩ (

H4E (0, 1)

× H2E (0, 1)

) ∣∣yxx(1) = −αzx(1), yxxx(1) = 0}.

(8)

Then, system (7) can be reformulated into the followingevolution equation in X:

ddt

(φ(t )ω(t )

)= A

(φ(t )ω(t )

)+(B0

)× (V1(t ) + d1(t )) + J(φ(t ), ω(t ), t ),

(9)

where A = diag(Aω, 0). Furthermore, a direct com-putation shows that A∗

ω, the adjoint operator of Aω,

has the form:⎧⎪⎪⎨⎪⎪⎩A∗

ω(ϕ, ψ) = (−ψ, ϕxxxx − ω2ϕ),

∀(ϕ, ψ) ∈ D(A∗ω ),

D(A∗ω ) = {

(ϕ, ψ) ∈ H ∩ (H4

E (0, 1)×H2

E (0, 1)) ∣∣ϕxxx(1) = 0, ϕxx(1) = αψx(1)

}.

(10)

INTERNATIONAL JOURNAL OF CONTROL 2325

2.1 Well-posedness of the system (9)

The well-posedness of (9) is discussed in this subsection.To this end, we need the following results.

Lemma 2.1: Assume that ω <√

μ1 and letAω be definedby (8). Then, A−1

ωexists and is compact on H. Therefore,

σ (Aω ), the spectrum ofAω, consists of isolated eigenvaluesof finite algebraic multiplicity only. Moreover,Aω generatesa C0-semigroup eAωt of contractions inH.

Proof: For given ( f , g) ∈ H, solveAω(u, v ) = ( f , g) toget v = f and u satisfies

{−uxxxx + ω2u = g,u(0) = ux(0) = uxxx(1) = 0, uxx(1) = −α fx(1).

(11)Letm(x) = u(x) + α/2fx(1)x2. Then,m(x) satisfies

{mxxxx − ω2m(x) = −g − ω2α/2 fx(1)x2,m(0) = mx(0) = mxx(1) = mxxx(1) = 0. (12)

Thus, (12) has a unique solution if and only if the homo-geneous equation of (12)

mxxxx − ω2m(x) = 0,m(0) = mx(0) = mxx(1) = mxxx(1) = 0 (13)

only admits a zero solution. Noting that (13) has the fun-damental solution,

m(x) = c1e√

|ω|x + c2e−√

|ω|x + c3ei√

|ω|x + c4e−i√

|ω|x,

we substitute this into the boundary conditions of (13)to get

⎧⎪⎨⎪⎩c1 + c2 + c3 + c4 = 0, c1 − c2 + ic3 − ic4 = 0,c1e

√|ω| + c2e−

√|ω| − c3ei

√|ω| − c4e−i

√|ω| = 0,

c1e√

|ω| − c2e−√

|ω| − c3iei√

|ω| + c4ie−i√

|ω| = 0.(14)

Hence, we have the determinant of the coefficient matrixof (14):

∣∣∣∣∣∣∣∣∣∣∣

1 1 1 11 −1 i −i

e√

|ω| e−√

|ω| −ei√

|ω| −e−i√

|ω|

e√

|ω| −e−√

|ω| −iei√

|ω| ie−i√

|ω|

∣∣∣∣∣∣∣∣∣∣∣= 8i[1 + cosh

√|ω| cos

√|ω|] �= 0,

where we have used ω <√

μ1 and (2). This tells usthat (13) has only zero solution and (12) has the uniquesolution m(x). Thus, there is a unique u(x) = m(x) −

α2 fx(1)x

2 to (11). Consequently, A−1ω

exists and is com-pact onH by the Sobolev embedding theorem. Therefore,σ (Aω ) consists of isolated eigenvalues of finite algebraicmultiplicity only.

Now we show that Aω is dissipative in H. Let X =( f , g) ∈ D(Aω ). Then we have

Re〈AωX,X〉H = −α|gx(1)|2 ≤ 0.

Thereby, Aω is dissipative and Aω generates a C0-semigroup eAωt of contractions on H by the Lumer–Philips theorem (see Pazy, 1983). The proof iscomplete. �

Lemma 2.2: Assume that ω <√

μ1 and let Aω and B bedefined by (8). Then B is admissible to the semigroup eAωt .

Proof: It is noted thatB is admissible for eAωt , or equiva-lently, B∗ is admissible for eA∗

ωt . So, we only need to show

thatB∗ is admissible for eA∗ωt , whereA∗

ωis defined by (10).

Consider the adjoint system

ddt

(u∗

u∗t

)= A∗

ω

(u∗

u∗t

), B∗

(u∗

u∗t

)= −u∗

xt (1),

which yields

⎧⎪⎪⎨⎪⎪⎩u∗tt + u∗

xxxx − ω2u∗ = 0, x ∈ (0, 1), t > 0,u∗(0, t ) = u∗

x(0, t ) = u∗xxx(1, t ) = 0, t ≥ 0,

u∗xx(1, t ) = −αu∗

xt (1, t ), t ≥ 0,y0(t ) = u∗

xt (1, t ), t ≥ 0.

(15)

The energy function of (15) is given by

E∗(t ) = 12

∫ 1

0

[u∗2xx(x, t ) + u∗2

t (x, t ) − ω2u∗2 (x, t )]dx,

ω <√

μ1.

Differentiating E∗(t) with respect to time, we get

E∗(t ) = −αu∗2xt (1, t ) ≤ 0,

whichmeans that E∗(t) is non-increasing in t. Integratingfrom 0 to T with respect to t in the above equation, wehave

∫ T

0u∗2xt (1, t )dt = − 1

α

∫ T

0E∗(t )dt

= 1α

(E∗(0) − E∗(T )

) ≤ 1αE∗(0).

2326 Y.-P. GUO AND J.-M. WANG

On the other hand, for any given (φ, ϕ) ∈ H, solve(A∗

ω− ω)( f , g) = (φ, ϕ), we have

{g = −ω f − φ, fxxxx = ϕ − ωφ,

f (0) = fx(0) = fxxx(1) = 0, fxx(1) = αgx(1),

which has the solution⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

f (x) = 12x2 fxx(1) + x

2

∫ 1

0s2(ϕ(s) − ωφ(s))ds

+16

∫ 1

x(s − x)3(ϕ(s) − ωφ(s))ds

−16

∫ 1

0s3(ϕ(s) − ωφ(s))ds,

g(x) = −ω f (x) − φ(x),

fxx(1) = αgx(1) = − αω

2(1 + αω)

∫ 1

0s2(ψ − ωφ)ds

− α

1 + αωφx(1).

So, ω ∈ σ (A∗ω) and

B∗(A∗ω − ω)−1(φ, ϕ) = ω

2(1 + αω)

∫ 1

0s2(ψ − ωφ)ds

+ 11 + αω

φx(1).

Hence, B∗(A∗ω

− ω)−1 is bounded and thus B∗ is admis-sible for eA∗

ωt . Therefore, B is admissible for eAωt (see

Weiss, 1989). �

We are now in a position to show that (9) is well-posedand deduce the corresponding variation of constant for-mula for mild solutions.Theorem 2.1: Assume that ω <

√μ1 and both di(t) and

di(t ), i=1,2, are bounded measurable. Then for any ini-tial value (u(·, 0), ut (·, 0), ω0) ∈ H × R,V1(t ),U2(t ) ∈L2loc(0, ∞) and di ∈ L2loc(0, ∞), i = 1, 2, the closed-loopsystem (9) admits a unique solution (u(·, t ), ut (·, t ), ω) ∈C(0, ∞;H) which is given by

φ(t ) = eAωtφ0 +∫ t

0eAω (t−s)B (V1(t ) + d1(t )) ds

+∫ t

0eAω (t−s)(ω2(s) − ω2)Pφ(s)ds, (16)

and

ω(t ) = ω0 −∫ t

0

2ω(ξ )∫ 10 uutdx −U2(t ) − d2(t )

Id + ∫ 10 u2dx

dξ,

(17)

where φ(t) = (u, ut) and P(u, v) = (0, u) is the compactoperator onH.

Proof: By Lemma 2.1, Aω generates a C0-semigroupS(t ) = eAωt onH, by (9), so is for A. A trivial verificationshows that (9) is equivalent to the following equation:

(φ(t ), ω(t )) = eAt (φ0, ω0)

+∫ t

0eA(t−s)B (V1(t ) + d1(t )) ds

+∫ t

0eA(t−s)J(φ(s), ω(s), s)ds,

(18)

where B = (B, 0) and eAt = diag(eAωt , I). One can useFréchet derivative to verify that the operator J is continu-ously differentiable inX (Xu & Baillieul, 1993). Since B isadmissible to the semigroup eAωt , B = (B, 0) is admissi-ble to the semigroup eAt . It then follows from the Corol-lary 1.3 of Pazy (1983) that for any initial value (φ0, ω0) ∈H × R and T > 0, the Equation (18) has a unique solu-tion (φ(t ), ω(t )) ∈ C(0,T;H) × R. In fact, the integralEquation (18) is the unique global mild solution of (9).The above Equation (18) can be decomposed into (16)and (17). �

3. Feedback via active disturbance rejectioncontrol

In this section, we use the ADRC approach to attenuatethe disturbances and then design the feedback controlsfor (7).

3.1 Extended state observer of the beam

This subsection is devoted to estimating d1(t). ByTheorem 2.1, the solution of the beam of (9) satisfies

ddt

⟨(uut

),

(φ

ϕ

)⟩H

=⟨(

uut

),A∗

ω

(φ

ϕ

)⟩H

+ (V1(t ) + d1(t ))B∗(

φ

ϕ

)

+⟨(

0(ω2(t ) − ω2) u

),

(φ

ϕ

)⟩H

, (19)

where [φ, ϕ]T ∈ D(A∗ω) is test function. Taking specially

[φ, ϕ]T = [αx2, x2]T. By (10), we have

y(t ) = −2[V1(t ) + d1(t )] + y0(t ) + ω2(t )y(t ), (20)

where

y(t ) =∫ 1

0x2u(x, t )dx, y0(t )

= −2αuxt (1, t ) − 2ux(1, t ). (21)

INTERNATIONAL JOURNAL OF CONTROL 2327

We can design an extended state observer to estimate d1as follows (see Guo & Zhao, 2011):⎧⎪⎪⎪⎨⎪⎪⎪⎩

˙y1(t ) = y2(t ) − 6ε−11 (y1(t ) − y(t )),

˙y2(t ) = −2d1(t ) + ω2(t )y1(t )−(11ε−2

1 + ω2(t ))(y1(t ) − y(t )) − 2V1(t ) + y0(t ),˙d1(t ) = 3ε−3

1 (y1(t ) − y(t )),

where ε1 is the tuning small parameter. The errors are

y1(t ) = y1(t ) − y(t ), y2(t ) = y2(t ) − y(t ),

d1(t ) = −d1(t ) + d1(t ), (22)

which satisfy⎧⎪⎨⎪⎩

˙y1(t ) = y2(t ) − 6ε−11 y1(t ),

˙y2(t ) = 2d1(t ) − 11ε−21 y1(t ),

˙d1(t ) = −3ε−31 y1(t ) + d1(t ).

(23)

Lemma 3.1: Suppose that the disturbance d1(t) and itsderivative d1(t ) are uniformly bounded on [0, �). Then,for any given α > 0, it follows that

|y1(t )| + |y2(t )| + |d1(t )| → 0,as ε1 → 0 uni f ormly in [α, ∞). (24)

Proof: A direct computation gives

ε31

...y 1(t ) + 6ε21 ¨y1(t ) + 11ε1 ˙y1(t ) + 6y1(t ) = 2ε31 d1(t ).

Let t = ε1s and v(s) = y1(ε1s). It then follows from theabove equation that

d3

ds3v(s) + 6

d2

ds2v(s) + 11

dds

v(s) + 6v(s) = 2ε31 d1(ε1s),

which can be rewritten as follows:

d f (s)ds

= A0 f (s) + D(s), (25)

where

A0 =⎛⎝ 0 1 0

0 0 1−6 −11 −6

⎞⎠ , f (s) =

⎛⎝ v(s)

v(s)v(s)

⎞⎠ ,

D(s) =⎛⎝ 0

02ε31 d1(ε1s)

⎞⎠ . (26)

The solution of (25) is

f (s) = eA0s f (0) +∫ s

0eA0(s−τ )D(τ )dτ. (27)

Since A0 is Hurwitz, there exist constants ω, L > 0, suchthat ‖eA0s‖F ≤ Le−ωs, where ‖ · ‖F is the Frobenius norm.From (26) and (27), we get

| f (s)| ≤ |eA0s f (0)| +∣∣∣∣∫ s

0eA0(s−τ )D(τ )dτ

∣∣∣∣≤ | f (0)|Le−ωs +

∫ s

0|D(τ )|Le−ω(s−τ )dτ ≤ | f (0)|Le−ωs

+ 2Lω−1ε31 supt∈[0,∞)

∣∣∣d1(t )∣∣∣ . (28)

Hence,

∣∣∣∣[ ¨y1(t ), ε−11

˙y1(t ), ε−21 y1(t )

]T ∣∣∣∣ ≤ ε−21 L| f (0)|e−ε−1

1 ωt

+ 2ε1ω−1L supt∈[0,∞)

∣∣∣d1(t )∣∣∣ .

Due to bounded of |d(t )|, for any α > 0, we have

∣∣∣ ¨y(t )∣∣∣ + ∣∣∣ε−11

˙y1(t )∣∣∣ + ∣∣ε−2

1 y1(t )∣∣ → 0,

as ε1 → 0 uniformly in [α, ∞). (29)

On the other hand, it follows from (23) that

y2(t ) = ˙y1(t ) + 6ε−11 y1(t ),

d1(t ) = 12

( ¨y1(t ) + 6ε−11

˙y1(t ) + 11ε−21 y1(t )

).

This together with (29) yield (24). The proof iscomplete. �

3.2 Extended state observer of the disk

This subsection mainly estimate the disturbance d2(t).Define

�(t ) =(Id +

∫ 1

0u2dx

)ω(t ). (30)

By the last condition of (7), we get �(t ) = U2(t ) + d2(t ),where �(t) is the state of ODE andU2 is the control. Nowwe are able to design an extended state observer to esti-mate both �(t) and d2(t):

˙�ε2 (t ) = U2(t ) + d2(t ) − ε−1

2 (�ε2 − �),

˙d2(t ) = −ε−22 (�ε2 − �), (31)

2328 Y.-P. GUO AND J.-M. WANG

where ε2 is the tuning small parameter. The errors are

�ε2 (t ) = �ε2 (t ) − �(t ), d2(t ) = −d2(t ) + d2(t ).(32)

Lemma 3.2: Suppose that the disturbance d2(t) and itsderivative d2(t ) are uniformly bounded on [0, �). Then,for any given α0 > 0, it follows that

|�ε2 (t )| + |d2(t )| → 0,as ε2 → 0 uni f ormly in [α0, ∞). (33)

Proof: A simple calculation gives that the errors (32)satisfy

˙�ε2 (t ) = −d2(t ) − ε−1

2 �ε2 (t ),˙d2(t ) = ε−2

2 �ε2 (t ) + d2(t ), (34)

which can be rewritten as

ddt

(�ε2

d2

)= A1

(�ε2

d2

)+ D1d2(t ),

A1 =(−ε−1

2 −1ε−22 0

), D1 =

(01

).

The solution of the above equation is

(�ε2 (t )d2(t )

)= eA1t

(�ε2 (0)d2(0)

)+∫ t

0eA1(t−s)D1d2(s)ds,

(35)

where

eA1t =(

λ1λ2−λ1

eλ1t − λ2λ2−λ1

eλ2t λ1λ2ε−22 (λ2−λ1)

(eλ2t − eλ1t

)ε−22

λ2−λ1

(eλ1t − eλ2t

) − λ2λ2−λ1

eλ1t + λ1λ2−λ1

eλ2t

)

and

eA1tD1 =[

λ1λ2

λ1 − λ2ε2(eλ1t − eλ2t ),

λ2eλ1t − λ1eλ2t

λ1 − λ2

]T.

Here, λ1 and λ2 are two conjugate and negative real parteigenvalues of matrix A1,

λ1 = − 12ε2

+√3

2ε2i, λ2 = − 1

2ε2−

√3

2ε2i.

Therefore, there exists positive constants L, M > 0 suchthat

‖eA1t‖ ≤ Lε−12 e−

12ε2

t, ‖eA1tD1‖ ≤ Me−

12ε2

t. (36)

From (35) and (36), we obtain (33). The proof iscomplete. �

3.3 Feedback control design

Based on the estimates of di(t ), i = 1, 2, we design thefeedback controllers for system (7) as follows:

V1(t ) = −sat(d1(t )

),

U2(t ) = −γ (ω(t ) − ω) − sat(d2(t )

), α > 0, γ > 0,

(37)

where the saturation function sat( · ) defined by

sat(x) :=⎧⎨⎩

M, x ≥ M + 1,−M, x ≤ −M − 1,x, x ∈ (−M − 1, M + 1).

(38)

Here,M is the upper bound of di(t) assumed in Introduc-tion. Under the feedback (37), the closed-loop system of(7) becomes

⎧⎪⎪⎪⎨⎪⎪⎪⎩utt + uxxxx − ω2(t )u = 0, x ∈ (0, 1), t > 0,u(0, t ) = ux(0, t ) = uxxx(1, t ) = 0, t ≥ 0,uxx(1, t ) = −αuxt (1, t ) − sat

(d1(t )

)+ d1(t ), α > 0, t ≥ 0,

(39a)⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

˙y1(t ) = y2(t ) − 6ε−11 (y1(t ) − y(t )),

˙y2(t ) = ω2(t )y1(t ) − (11ε−21 + ω2(t ))(y1(t ) − y(t ))

+2αuxt (1, t ) + 2sat(d1(t )

)− 2d1(t ) + y0(t ),

˙d1(t ) = 3ε−31 {y1(t ) − y(t )},

(39b)

and

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

ddt

�(t ) = −γ (ω(t ) − ω) − sat(d2(t )

)+ d2(t ),

˙�ε2 (t ) = −γ (ω(t ) − ω) − ε−1

2 (�ε2 − �)

− sat(d2(t )

)+ d2(t ),

˙d2(t ) = −ε−22 {�ε2 − �},

(39c)

where y(t) and �(t) are given by (21) and (30), respec-tively. Using the error dynamics defined by (22), (32), it is

INTERNATIONAL JOURNAL OF CONTROL 2329

found that (39) can be rewritten as

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

utt + uxxxx − ω2(t )u = 0,u(0, t ) = ux(0, t ) = uxxx(1, t ) = 0,uxx(1, t ) = −αuxt (1, t )

+ sat(d1(t ) − d1(t )

)+ d1(t ) � −αuxt (1, t ) + h1(t ),

(40a)

⎧⎪⎨⎪⎩

˙y1(t ) = y2(t ) − 6ε−11 y1(t ),

˙y2(t ) = 2d1(t ) − 11ε−21 y1(t ),

˙d1(t ) = −3ε−31 y1(t ) + d1(t ),

(40b)

and

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

d�

dt(t ) = −γ

(ω(t ) − ω

) + sat(d2(t ) − d2(t )

)+ d2(t ) � −γ

(ω(t ) − ω

) + h2(t ),˙�ε2 (t ) = −d2(t ) − ε−1

2 �ε2 (t ),˙d2(t ) = ε−22 �ε2 (t ) + d2(t ).

(40c)From (23) and (34), it is seen that (y1, y2, d1) and

(�, d2) are independent of the ‘u-part’ and ‘�-part’.(y1, y2, d1) and (�, d2) can be arbitrarily small as t→ �,ε1, ε2 → 0 by Lemmas 3.1 and 3.2, respectively. Hence, weonly need to consider the ‘u-part’ and ‘�-part’ of (40),

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

utt + uxxxx − ω2(t )u = 0, x ∈ (0, 1), t > 0,u(0, t ) = ux(0, t ) = uxxx(1, t ) = 0, t ≥ 0,uxx(1, t ) = −αuxt (1, t ) + h1(t ), α > 0, t ≥ 0,ddt

{(Id +

∫ 1

0u2dx

)ω(t )

}= −γ (ω(t ) − ω) + h2(t ), γ > 0.

(41)

4. Stability of the system (41)

In this section, the closed-loop stability of system (41)under the proposed controllers (6) and (37) is established.

Theorem4.1: Assume that ω <√

μ1. The semigroup eAω ,generated by the operator Aω, is exponentially stable inH, i.e., there exist positive constants L0 and β such that‖eAωt‖ ≤ L0e−βt .

Proof: Since it is a tedious work, we put the proof in theAppendix. �Theorem 4.2: Suppose that ω <

√μ1 and both di(t) and

di(t ), i=1,2, are uniformly bounded. For any initial value(u(·, 0), ut (·, 0), ω0) ∈ H × R, then the solution (u( ·, t),ut( ·, t), ω) of system (41) asymptotically tends to equilib-rium point (0, 0, ω) in X, as t → �, ε1, ε2 → 0.

Proof: The proof is divided into two steps. In the firststep, we prove

ω(t ) → ω, as t → ∞, εi → 0, i = 1, 2.

Consider the Lyapunov functionV : X → R+:

V = 12

{Id(ω(t ) − ω)2 + (

ω(t ) − ω)2 ‖u‖2L2(0,1)

+‖u‖2H2E (0,1) + ‖ut‖2L2(0,1) − ω2‖u‖2L2(0,1)

},

where

‖u‖2L2(0,1) =∫ 1

0u2dx, ‖u‖2H2

E (0,1) =∫ 1

0u2xxdx.

Differentiating V(φ(t), ω(t)) along the solution of (41)gives

V = (ω(t ) − ω

)ω(t )

(Id + ‖u‖2L2(0,1)

)+ (

ω(t ) − ω)2 〈u, ut〉L2(0,1)

+ 〈u, ut〉H2E (0,1) + 〈ut , utt〉L2(0,1) − ω2〈u, ut〉L2(0,1)

= − γ(ω(t ) − ω

)2 + (ω(t ) − ω

)h2(t )

− αu2xt (1, t ) + uxt (1, t )h1(t )

≤ −γ(ω(t ) − ω

)2 + γ

2(ω(t ) − ω

)2 + 12γ

h22(t )

− αu2xt (1, t ) + α

2u2xt (1, t ) + 1

2αh21(t )

� −K1 + K2,

where K1 = γ

2

(ω(t ) − ω

)2 + α2 u

2xt (1, t ) and K2 =

12γ h

22(t ) + 1

2αh21(t ). There are two cases:

(1) If K1 ≥ K2, we get V ≤ 0. Hence, each solution isbounded and

∫ ∞0 γ (ω(t ) − ω)2dt converges. This

implies that limt→∞,ε1,ε2→0 ω(t ) = ω accordingto the Barbalat’s lemma.

(2) If K1 < K2, by Lemmas 3.1 and 3.2, we get

limt→∞,ε1,ε2→0

γ

2{ω(t ) − ω

}2 ≤ limt→∞,ε1,ε2→0

K1

≤ limt→∞,ε1,ε2→0

K2 = limt→∞,ε1,ε2→0

12γ

h22(t )

+ 12α

h21(t ) = 0.

2330 Y.-P. GUO AND J.-M. WANG

(a) (b)

0 2 4 6 8 100

5

10

15

20

25

30

35

40

45

t

w

(c)

Figure . The state of open loop system in thepresenceof external disturbances. (a) Displacementu(x, t), (b) velocityut(x, t) and (c) velocityω(t).

In Summary, for any ν > 0, there exist T0 and ε > 0such that

|ω2(t ) − ω2| < ν, for all t ≥ T0, 0 < ε1, ε2 < ε.

(42)

In the second step, we show that φ(t) = (u( ·, t), ut( ·,t)) is asymptotically stable inH, that is,

(u(·, t ), ut (·, t )) → (0, 0), when t → ∞, ε1, ε2 → 0.

By Lemma 3.1 and (40a), for any given ε0 > 0, there existt0 � T0 > 0 and ε∗

0 > 0 such that

|h1(t )| = ∣∣sat(d1(t ) − d1(t )) + d1(t )

∣∣< ε0 for all t > t0 and 0 < ε1, ε2 < ε∗

0 .

We rewrite the solution of (16) as

φ(t ) = eAωtφ0 +∫ t

0eAω (t−s) (ω2(s) − ω2) Pφ(s)ds

+∫ t

0eAω (t−s)Bh1(s)ds

= eAωtφ0 +∫ t

0eAω (t−s) (ω2(s) − ω2) Pφ(s)ds

+ eAω (t−t0)∫ t0

0eAω (t0−s)Bh1(s)ds

+∫ t

t0eAω (t−s)Bh1(s)ds. (43)

The admissibility of B implies that

∥∥∥∥∫ t0

0eAω (t0−s)Bh1(s)ds

∥∥∥∥2

H

≤ Ct0‖h1‖L2(0,t0)Ct0

∥∥∥sat (d1(t ) − d1(t ))

+ d1(t )∥∥∥2L2(0,t0)

� t0Ct0 (2M + 1)2, ∀d1 ∈ L∞(0, ∞), (44)

INTERNATIONAL JOURNAL OF CONTROL 2331

(a) (b)

0 2 4 6 8 10−1

0

1

2

3

4

5

6

7

8

t

w

(c)

Figure . The state of closed loop system in the presence of external disturbance. (a) Displacement u(x, t), (b) velocity ut(x, t) and (c)velocityω(t).

where the constantCt0 is independent of d1. Because eAωt

is exponential stable (see Theorem 4.1) and B is admissi-ble to eAωt with L2loc control, B is admissible to eAωt withL∞loc control. From Proposition 2.5 in Weiss (1989), it fol-

lows that

∥∥∥∥∫ t

t0eAω (t−s)Bh1(s)ds

∥∥∥∥2

H

=∥∥∥∥∫ t

0eAω (t−s)B

(0♦

t0h1)

(s)ds∥∥∥∥2

H≤ C‖h1‖L∞(0,∞)

� C∥∥∥sat(d1(t ) − d1(t )) + d1(t )

∥∥∥L∞(0,∞)

� Cε0,

(45)

where C is a constant and independent of d1, and

(u♦

τv

)={

u(t ), 0 ≤ t ≤ τ,

v(t − τ ), t > τ.

Since ‖eAωt‖ ≤ L0e−βt for some L0, β > 0, inserting (44),(45) into (43), we have

‖φ(t )‖H ≤ L0e−βt‖φ(0)‖H+ L0

∫ t

0e−β(t−s)(ω2(s) − ω2)Pφ(s)ds

+ L0e−β(t−t0)t0Ct0 (2M + 1)2 +Cε0. (46)

Set m(t ) = L0e−βt‖φ(0)‖H + L0e−β(t−t0)t0Ct0 (2M +1)2 +Cε0. It follows from (42) and (46) that

‖φ(t )‖H ≤ m(t ) + νL0∫ t

0e−β(t−s)‖φ(s)‖Hds,

for all t > t0 and 0 < ε1, ε2 < ε∗0 .

2332 Y.-P. GUO AND J.-M. WANG

0 2 4 6 8 10−15

−10

−5

0

5

10

15

20errorestimation of d

1

disturbance d1

Figure . d(t) and d1(t ).

Applying Gronwall’s Lemma to the above inequality, wehave

‖φ(t )‖H≤ m(t ) +

∫ t

0e−β(t−s)m(s) exp

{∫ t

sνL0e−β(t−r)dr

}ds

≤ m(t ) + eνL0∫ t

0e−β(t−s)m(s)ds

= m(t ) + eνL0e−βt∫ t

0eβs

[L0e−βs‖φ(0)‖H

+ L0e−β(s−t0)t0Ct0 (2M + 1)2 +Cε0]ds

= m(t ) + eνL0{L0e−βt t‖φ(0)‖H

+ L0t2e−β(t−t0)Ct0 (2M + 1)2 +C1ε0}, (47)

whereC1 is a positive constant. Passing to the limit as t→� in (47), we finally obtain

limt→∞ ‖φ(t )‖H ≤ Cε0 + eνL0C1ε0.

Since ε0, ν are arbitrary sufficiently small, we deduce thatlimt→∞, ε1,ε2→0 φ(t ) = 0 inH.

Finally, collecting these two steps, we conclude thedesired result that the solution (u( ·, t), ut( ·, t), ω) ofsystem (41) asymptotically tends to equilibrium point(0, 0, ω) as t → �, ε1, ε2 → 0. The proof is complete. �

5. Numerical simulation

In this section, some simulation results are given to illus-trate the effect of our controls design (6) and (37) forthe disk–beam system (1). The numerical results of sys-tem (39) are obtained by the finite difference method.Let Id = 3

2 , γ = 5, α = 2, ω = 3, the initial displace-ment u(x, 0) = 1

20x2 and the initial velocity ut (x, 0) =

0 2 4 6 8 10−60

−40

−20

0

20

40

60errorestimation of d

2

disturbance d2

Figure . d(t) and d2(t ).

120x

2, ω(0)= 8. The grid size isN= 20 for x, and the timestep is dt = 0.0001 for t. The disturbances are chosen asd1(t) = 4cos 4t + 1 and d2(t) = 2sin 2(t − 1).

Figures 2 and 3 show the states of the open-loop andclosed-loop system, respectively. It can be seen that with-out controls both the states of beam and disk are not sta-bilised in Figure 2. Figure 3(a) and 3(b) demonstratesthe displacement and velocity of beam of the closed-loop system (39), respectively. Figure 3(c) shows theangular velocity ω(t) of system (39). In Figure 3, it isshown that after the proposed controllers (6) and (37) areapplied, the beam is stabilised to zero and angular veloc-ity of the disk is a constant. Figures 4 and 5 demonstratethat the estimation errors asymptotically converge to thezero. Furthermore, the peaking value phenomenons areclearly observed for d1(t) and d2(t) in Figures 4 and 5,respectively.

6. Conclusions

This paper concerns with the stability characteristics ofthe ADRC for the rotating disk–beam system with theinput disturbances. By the ADRC, we first estimate thedisturbances and then cancel the disturbances in the feed-back loop. The collocated feedback control is appliedto suppress the rotating beam and the torque control isforced to steer the disk to the desired rotating angularvelocity. In the energy state space, it is shown that thefeedback control laws are robust to the external distur-bances when the angular velocity of the disk is less thanthe square root of the first nature frequency of the beam,and the vibration of the beam can be suppressed while thedisk rotating with a desired angular velocity in presenceof the disturbances. Numerical simulations are presented

INTERNATIONAL JOURNAL OF CONTROL 2333

to show the convergence of both state and approximateddisturbances.

The nonlinear feedback controls and the output feed-back stabilisation for the rotating disk–beam system withthe input disturbances are still open. They will be consid-ered in future.

Acknowledgments

The authors would like to thank anonymous reviewers andeditors for their careful reading and valuable suggestions toimprove the paper.

Disclosure statement

No potential conflict of interest was reported by the authors.

Funding

Thisworkwas supported by theNationalNatural Science Foun-dation of China [grant number 61273130].

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Appendix A. Proof of Theorem 4.1

In this appendix, we present the proof of Theorem 4.1. Itis pointed out that Laousy et al. (1996) have showed thatif the angular velocity ω <

√9EI/ρ (=3 for EI = ρ = 1),

the semigroup eAωt is exponentially stable in H. In thisappendix, we improve this best desired value to √

μ1 �3.516 (EI = ρ = 1) and show that if the angular velocityω <

√μ1, the semigroup eAωt is exponentially stable in

H. Before going on the proof of Theorem 4.1, we recallthe the well-known Huang result (see Huang, 1985):

Lemma A.1: A strongly continuous semigroup of contrac-tions eAt on a Hilbert spaceH is exponentially stable if andonly if

sup{‖(iμI − A)−1‖H; μ ∈ R} < ∞, (A.1)

and {iμ : μ ∈ R} ⊂ ρ(A).

Proof of Theorem 4.1: First, we show {iμ : μ ∈ R} ⊂ρ(Aω ). In Lemma 2.1, it has showedA−1

ωexists. Suppose

that there is a non-zero μ such that iμ ∈ σ (Aω ), that is,there exist� = (y, z) ∈ D(Aω )with ‖�‖H = 1 such thatAω� = iμ�. A simple computation finds

0 = Re〈iμ�, �〉 = Re〈Aω�, �〉 = −αz2x(1),

which together with (8) implies

{yxxxx − (ω2 + μ2)y = 0,y(0) = yx(0) = yxx(1) = yxxx(1) = yx(1) = 0.

A simple argument shows that there is only zero solutionfor the above equation. This is contradiction with iμ ∈σ (Aω ). Hence, there is no eigenvalue on the imaginaryaxis.

Next, we are going to show

sup{‖(iμI − Aω )−1‖H; μ ∈ R} < ∞. (A.2)

Suppose that (A.2) does not hold. Therefore, there exist asequence of real numbers μn → �, and �n = (yn, zn) ∈D(Aω ), satisfying,

⎧⎪⎪⎨⎪⎪⎩

‖(iμnI − Aω )�n‖H → 0, as n → ∞,

‖�n‖H = ‖yn‖∗ + ‖zn‖L2 = 1,

‖yn‖∗ =∫ 1

0y2xxdx − ω2y2dx.

(A.3)

These together with (8) imply that as n → �,

⎧⎪⎪⎨⎪⎪⎩

f n � iμnyn − zn → 0, in H2E (0, 1),

gn � iμnzn + ynxxxx − ω2yn → 0, in L2(0, 1),y(0) = yx(0) = yxxx(1) = 0,yxx(1) = −αzx(1).

(A.4)

From (A.3), we have ‖yn‖∗ and ‖zn‖L2 are uniformlybounded, that is, there exist constants C1, C2 > 0, suchthat

‖yn‖∗ < C1, ‖zn‖L2 < C2. (A.5)

One can check that

‖(iμnI − Aω )�n‖H ≥ α(znx (1))2. (A.6)

By (A.3), we have |znx (1)| → 0, as n → �. By the tracetheorem, this together with the first equation of (A.4)yields that

|μnynx (1)| → 0, as n → ∞. (A.7)

Furthermore, the last boundary condition of (A.4) yields

|ynxx(1)| → 0, as n → ∞. (A.8)

From the second equation of (A.4), we have

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

ynxxx(0) = −∫ 1

0ynxxxxdx

= −∫ 1

0(−iμnzn + gn + ω2yn)dx,

ξ−1n e−ξn |ynxxx(0)| ≤ k1ξ−1

n e−ξn(‖yn‖∗ + ‖gn‖L2 + |μn|‖zn‖L2),

(A.9)

where k1 is a constant. Multiplying the second equationof (A.4) by x and using integration by parts, to yield

∫ 10 (−iμnzn + gn + ω2yn)xdx

= ∫ 10 ynxxxxxdx = ynxx(0) − ynxx(1),

INTERNATIONAL JOURNAL OF CONTROL 2335

and

e−ξn |ynxx(0)| = e−ξn |∫ 1

0(−iμnzn + gn + ω2yn)xdx + ynxx(1)|

≤ k2(e−ξn‖yn‖∗ + e−ξn‖gn‖L2 + e−ξn |μn|‖zn‖L2

)+ e−ξn |ynxx(1)|,

(A.10)

where k2 is a constant. Now, using (A.8)–(A.10) and thefact ‖yn‖∗ and ‖zn‖L2 are uniformly bounded, we get

ξ−1n e−ξn |ynxxx(0)| → 0, e−ξn |ynxx(0)| → 0, as n → ∞.

(A.11)

From the first two equations of (A.4), we get

ynxxxx − (ω2 + μ2n)y

n = gn + iμn f n. (A.12)

Let ξn = (ω2 + μ2n)

14 . Multiplying (A.12) by ξ−1

n eξn(x−1)

and using integration by parts to yield

−ξ−1n ynxxx(0)e

−ξn − ynxx(1) + ynxx(0)e−ξn + ξnynx (1) − ξ 2

n yn(1)

= ξ−1n

∫ 1

0(iμn f n + gn)eξn(x−1)dx. (A.13)

Clearly, as n → �,

ξ−1n

∣∣∣∫ 10 eξn(x−1)gndx

∣∣∣ ≤ k3ξ−1n ‖gn‖L2 → 0,

iμnξ−1n

∣∣∣∫ 10 eξn(x−1) f ndx

∣∣∣ ≤ k4‖ f n‖L2 → 0,

where k3 and k4 are two positive constants. Finally, insert-ing (A.7), (A.8) and (A.11) into (A.13), we have

ξ 2n |yn(1)| =

√μ2n + ω2|yn(1)| → 0, as n → ∞.

(A.14)

Multiply (A.12) by xynx and use integration by parts twice,to yield

− [ynxx(1)]2 − 2ynxx(1)y

nx (1) −

∣∣∣∣√

ω2 + μ2ny

n(1)∣∣∣∣2

+ 3∫ 1

0(ynxx)

2dx +∫ 1

0

∣∣∣∣√

ω2 + μ2ny

n∣∣∣∣2

dx

= 2∫ 1

0(iμn f n + gn)xynxdx.

(A.15)

Applying the Hölder inequality, we get,

2∫ 10 (iμn f n + gn)xynxdx ≤ C3[‖gn‖L2‖yn‖∗

+ ‖ f n‖∗‖μnyn(1)‖L2 + ‖ f n‖∗‖μnyn‖L2 ],

where C3 is a positive constants. The above inequalitytogether with (A.4),(A.5) and (A.14), implies that

2∫ 1

0(iμn f n + gn)xynxdx → 0, as n → ∞. (A.16)

Finally, inserting (A.7), (A.8), (A.14), (A.16) into(A.15), we get

‖yn‖∗ → 0, ‖μnyn‖L2 → 0, ‖zn‖L2≤ ‖μnyn‖L2 + ‖ f n‖L2 → 0, n → ∞, (A.17)

which contradicts (A.3). The proof of Theorem 4.1 fol-lows from the two steps and Lemma A.1. The proof iscomplete. �

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