J. Math. Anal. Appl. 448 (2017) 1204–1227

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

Exponential stability of an active constrained layer beam

actuated by a voltage source without magnetic effects ✩

Chao Yang, Jun-Min Wang ∗

School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 2 October 2016Available online 25 November 2016Submitted by P. Yao

Keywords:ACL beamRayleigh beamRiesz basisExponential stability

We study the boundary stabilization of an active constrained layer (ACL) beam consisting of a stiff layer, a viscoelastic layer and a piezoelectric layer. The piezoelectric layer is actuated by a voltage source without magnetic effects. The system is modeled as a Rayleigh beam coupled with two wave equations. By using an asymptotic technique, we present the asymptotic expressions for the eigenpairs of the system. We show that the generalized eigenfunctions form a Riesz basis in the state space, and hence the spectrum determined growth condition holds. Finally, the exponential stability of the closed-loop system is established.

© 2016 Elsevier Inc. All rights reserved.

1. Introduction

Active constrained layer (ACL) damping treatments have been one of the main research topics in the vibration suppression of various flexible structures due to their reliability, robustness, low weight, adjusta-bility, and high efficiency [1,2,16,19,22]. The ACL damping treatments integrate both active and passive dampings through constrained layer treatments [18]. The ACL beam is composite material, which usually represents a beam treated with an ACL damping treatment. A typical design of an ACL beam consists of a viscoelastic shear (passive damping) layer sandwiched between a piezoelectric constraining (active damping) layer and the vibrating structure (e.g. an Euler–Bernoulli beam) [17].

Much research of composite beam systems has been carried out in the past decade, and many methods, such as Riesz basis approach, multiplier method, Carleman estimates, Lyapunov’s energy method, and so on, have been used to analyze the systems. For example, the exponential stability and the exact controllability of a sandwich beam system with a boundary control are considered using the Riesz basis approach in [24]. The exact controllability of a Rao–Nakra sandwich beam with boundary controls is studied by the multiplier method in [14]. The analyticity of the solution and the exponential stability of a sandwich beam system are

✩ This work was supported by the National Natural Science Foundation of China with grant number: 61673061.* Corresponding author.

E-mail addresses: chych09@163.com (C. Yang), jmwang@bit.edu.cn (J.-M. Wang).

http://dx.doi.org/10.1016/j.jmaa.2016.11.0670022-247X/© 2016 Elsevier Inc. All rights reserved.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1205

obtained via the Riesz basis approach in [23]. By the method of Carleman estimates, the exact controllability results of a multilayer plate system with clamped (or hinged) and free boundary conditions are obtained in [6,7], respectively. The boundary feedback stabilization of a multilayer Rao–Nakra sandwich beam is investigated via the Riesz basis approach and the compact perturbation method in [12]. Recently, the exact controllability of a multilayer Rao–Nakra sandwich beam with different boundary conditions: clamped, hinged, and clamped-hinged is considered using the multiplier method in [13]. In [15], the exponential stability of a laminated beam with interfacial slip and frictional damping is investigated by the energy method. An ACL beam actuated by charge (or current) is studied in [11], where the exponential stability of the closed-loop system with mechanical feedback controls is considered.

In this paper, we consider an ACL beam consisting of three layers: a viscoelastic layer sandwiched between a stiff (bottom) layer and a piezoelectric (top) layer. The piezoelectric layer is actuated by a voltage source without magnetic effects. The ACL beam is modeled by the following equations [10]:⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

ρϕhϕϕtt − αϕhϕϕxx −Gφ = 0, 0 < x < L, t > 0,

ρψhψψtt − αψhψψxx + Gφ = 0, 0 < x < L, t > 0,

mwtt −K1wxxtt + K2wxxxx −GHφx = 0, 0 < x < L, t > 0,

φ = 1h (−ϕ + ψ + Hwx),

(1.1a)

with the boundary conditions⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

ϕ(0, t) = ψ(0, t) = w(0, t) = wx(0, t) = 0,

αϕhϕϕx(L, t) = u1(t),

αψhψψx(L, t) = u2(t),

K2wxx(L, t) = u3(t),

K1wxtt(L, t) −K2wxxx(L, t) + GHφ(L, t) = 0,

(1.1b)

and the initial data

(ϕ,ϕt, ψ, ψt, w, wt)(x, 0) = (ϕ0, ϕ1, ψ0, ψ1, w0, w1), (1.1c)

where ϕ and ψ are the longitudinal displacements of bottom and top layers respectively, w denotes the transverse displacement of each layer (the transverse displacements of three layers are considered to be equal), and ui (i = 1, 2, 3) are the boundary controls. Moreover, ρϕ, ρψ, hϕ, h, hψ, G, m, H and Ki (i = 1, 2)are positive parameters. For details of (1.1) and its physical analysis, the reader can refer to [9,10].

For brevity in notation, we make the following transformation⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u(x, t) = ϕ(Lx,

√mL4

K2t), v(x, t) = ψ

(Lx,

√mL4

K2t), h1 = 1

h ,

y(x, t) = w(Lx,

√mL4

K2t), αu = αϕmL2

ρϕK2, Gu = GmL4

ρϕhϕK2, K = H

L ,

αv = αψmL2

ρψK2, Gv = GmL4

ρψhψK2, I = K1

mL2 , U1(t) = Lαϕhϕ

u1

(√mL4

K2t),

G1 = GHL3

K2, U2(t) = L

αψhψu2

(√mL4

K2t), U3(t) = L2

K2u3

(√mL4

K2t),

u0(x) = ϕ0(Lx), u1(x) =√

mL4

K2ϕ1(Lx), v0(x) = ψ0(Lx),

v1(x) =√

mL4

K2ψ1(Lx), y0(x) = w0(Lx), y1(x) =

√mL4

K2w1(Lx).

(1.2)

1206 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

Then (1.1) can be written as⎧⎪⎪⎪⎨⎪⎪⎪⎩utt − αuuxx −Guz = 0, 0 < x < 1, t > 0,vtt − αvvxx + Gvz = 0, 0 < x < 1, t > 0,ytt − Iyxxtt + yxxxx −G1zx = 0, 0 < x < 1, t > 0,z = h1(−u + v + Kyx),

with the boundary conditions ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

u(0, t) = v(0, t) = y(0, t) = yx(0, t) = 0,

ux(1, t) = U1(t),

vx(1, t) = U2(t),

yxx(1, t) = U3(t),

Iyxtt(1, t) − yxxx(1, t) + G1z(1, t) = 0,

and the initial data

(u, ut, v, vt, y, yt)(x, 0) = (u0, u1, v0, v1, y0, y1).

We propose the following boundary feedback controls

U1(t) = −l1ut(1, t), U2(t) = −l2vt(1, t), U3(t) = −l3yxt(1, t),

where li (i = 1, 2, 3) are positive constant feedback gains. Then the closed-loop system is⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

utt − αuuxx −Guz = 0, 0 < x < 1, t > 0,

vtt − αvvxx + Gvz = 0, 0 < x < 1, t > 0,

ytt − Iyxxtt + yxxxx −G1zx = 0, 0 < x < 1, t > 0,

z = h1(−u + v + Kyx),

(1.3a)

with the boundary conditions ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

u(0, t) = v(0, t) = y(0, t) = yx(0, t) = 0,

ux(1, t) = −l1ut(1, t),

vx(1, t) = −l2vt(1, t),

yxx(1, t) = −l3yxt(1, t),

Iyxtt(1, t) − yxxx(1, t) + G1z(1, t) = 0,

(1.3b)

and the initial data

(u, ut, v, vt, y, yt)(x, 0) = (u0, u1, v0, v1, y0, y1). (1.3c)

The energy function of the system (1.3) is given by

E(t) = 12

1∫0

(1Gu

u2t + 1

Gvv2t + K

G1y2t + αu

Guu2x + αv

Gvv2x + IK

G1y2xt + K

G1y2xx + 1

h1z2

)dx. (1.4)

It is easy to check that E(t) ≤ 0. So the energy of the system (1.3) is nonincreasing.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1207

Our purpose is to prove that the closed-loop system (1.3) is exponentially stable in the state Hilbert space by Riesz basis approach. To this end, we first prove that the system is well-posed in the state space and the system operator has compact resolvent. Secondly, we show that there are three branches of asymptotic eigenvalues for the system. Finally, we prove that the generalized eigenfunctions of the system form a Riesz basis in the state space and hence the spectrum determined growth condition and the exponential stability hold for the system.

The paper is organized as follows. In section 2, we establish the well-posedness of the system. In sec-tion 3, we give asymptotic estimates of the eigenvalues for the system. Section 4 deals with the asymptotic expansion of the corresponding eigenfunctions. Finally, in section 5, we obtain the main result that the gen-eralized eigenfunctions of the system form a Riesz basis in the state Hilbert space. Therefore, the spectrum determined growth condition and the exponential stability are concluded.

2. Well-posedness of the system

In this section, we transform system (1.3) into an abstract evolution equation and then discuss the well-posedness of the system.

Integrating the third equation of (1.3a) over [x, 1] with respect to the spatial variable yields⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

utt − αuuxx −Guz = 0,vtt − αvvxx + Gvz = 0,1∫

x

yttds + Iyxtt − yxxx + G1z = 0,

z = h1(−u + v + Kyx).

(2.1)

Define an unbounded linear operator A in L2(0, 1) by

Af(x) = If ′(x) +1∫

x

f(s)ds, ∀f ∈ D(A) = H1E(0, 1) := {f ∈ H1(0, 1) : f(0) = 0}. (2.2)

Lemma 2.1. A has a continuous inverse in L2(0, 1) and is given by

A−1g(x) = −

1∫0

g(τ) sinh[√

1I(1 − τ)

]dτ

I cosh√

1I

sinh(√

1Ix

)

+ 1I

x∫0

g(τ) cosh(√

1I(x− τ)

)dτ, ∀g ∈ L2(0, 1). (2.3)

For detailed proof of Lemma 2.1, we refer the reader to [5, Lemma 2.1].Hence, system (1.3) can be written as⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

utt − αuuxx −Guz = 0, 0 < x < 1, t > 0,

vtt − αvvxx + Gvz = 0, 0 < x < 1, t > 0,

ytt −A−1yxxx + G1A−1z = 0, 0 < x < 1, t > 0,

z = h (−u + v + Ky ),

1 x

1208 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

with the boundary conditions

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

u(0, t) = v(0, t) = y(0, t) = yx(0, t) = 0,

ux(1, t) = −l1ut(1, t),

vx(1, t) = −l2vt(1, t),

yxx(1, t) = −l3yxt(1, t).

Let H2E(0, 1) = {f ∈ H2(0, 1) : f(0) = f ′(0) = 0}. We consider system (1.3) in the state Hilbert space

H := H1E(0, 1) × L2(0, 1) ×H1

E(0, 1) × L2(0, 1) ×H2E(0, 1) ×H1

E(0, 1) (2.4)

with inner product induced norm

‖ (u1, u2, v1, v2, y1, y2) ‖2H :=

1∫0

(1Gu

|u2|2 + αu

Gu|u′

1|2 + 1Gv

|v2|2 + αv

Gv|v′1|2 + K

G1|y2|2 + IK

G1|y′2|2 + K

G1|y′′1 |2

+ h1| − u1 + v1 + Ky′1|2)dx, ∀(u1, u2, v1, v2, y1, y2) ∈ H. (2.5)

Define a new variable A : D(A)(⊂ H) → H by

A

⎛⎜⎜⎜⎜⎜⎜⎜⎝

u1u2v1v2y1y2

⎞⎟⎟⎟⎟⎟⎟⎟⎠=

⎛⎜⎜⎜⎜⎜⎜⎜⎝

u2αuu

′′1 + Guh1 (−u1 + v1 + Ky′1)

v2αvv

′′1 −Gvh1 (−u1 + v1 + Ky′1)

y2A−1y′′′1 −G1h1A

−1 (−u1 + v1 + Ky′1)

⎞⎟⎟⎟⎟⎟⎟⎟⎠(2.6)

with

D(A) ={

(u1, u2, v1, v2, y1, y2)T ∈((

H2(0, 1) ×H1E(0, 1)

)2 ×H3(0, 1) ×H2E(0, 1)

)∩H :

u′1(1) = −l1u2(1), v′1(1) = −l2v2(1), y′′1 (1) = −l3y

′2(1)

}. (2.7)

Denote Y (t) := (u(·, t), ut(·, t), v(·, t), vt(·, t), y(·, t), yt(·, t))T , then system (1.3) can be formulated into an evolution equation in H

⎧⎨⎩dY (t)dt

= AY (t), t > 0,

Y (0) := (u0, u1, v0, v1, y0, y1)T .

(2.8)

Lemma 2.2. Let A be defined by (2.6) and (2.7). Then A−1 exists and is compact on H. Hence, the spectrum σ(A) consists entirely of isolated eigenvalues only.

Proof. For any Q = (f1, f2, f3, f4, f5, f6)T ∈ H, we look for Y = (u1, u2, v1, v2, y1, y2)T ∈ D(A) such that AY = Q, which yields

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1209

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αuu′′1 + Guh1 (−u1 + v1 + Ky′1) = f2(x),

αvv′′1 −Gvh1 (−u1 + v1 + Ky′1) = f4(x),

A−1y′′′1 −G1h1A−1 (−u1 + v1 + Ky′1) = f6(x),

u2(x) = f1(x), v2(x) = f3(x), y2(x) = f5(x),

u1(0) = v1(0) = y1(0) = y′1(0) = 0,

u′1(1) = −l1u2(1), v′1(1) = −l2v2(1), y′′1 (1) = −l3y

′2(1).

(2.9)

Let z1(x) := −u1(x) + v1(x) + Ky′1(x). From (2.9), we have{z′′1 (x) = αz1(x) + q1(x), α := Guh1

αu+ Gvh1

αv+ G1Kh1,

z1(0) = 0, z′1(1) = β := l1f1(1) − l2f3(1) − l3Kf ′5(1)

(2.10)

with

q1(x) := − 1αu

f2(x) + 1αv

f4(x) + KAf6(x).

We solve equation (2.10) and obtain

z1(x) =

⎛⎝β −1∫

0

cosh[√

α(1 − τ)]q1(τ)dτ

⎞⎠ sinh (√αx)√

α cosh√α

+ 1√α

x∫0

sinh[√

α(x− τ)]q1(τ)dτ. (2.11)

By the first equation of (2.9), (2.11) and the boundaries u1(0) = 0, u′1(1) = −l1u2(1), we have

u1(x) =x∫

0

(x− τ)q2(τ)dτ −

⎛⎝l1f1(1) +1∫

0

q2(τ)dτ

⎞⎠x, q2(x) := −Guh1

αuz1(x) + 1

αuf2(x). (2.12)

Similarly, v1(x) and y1(x) can be solved as follows:

v1(x) =x∫

0

(x− τ)q3(τ)dτ −

⎛⎝l2f3(1) +1∫

0

q3(τ)dτ

⎞⎠x, q3(x) := Gvh1

αvz1(x) + 1

αvf4(x) (2.13)

and

y1(x) = 12

x∫0

(x− τ)2q4(τ)dτ −

⎛⎝ l3f′5(1)2 + 1

2

1∫0

q4(τ)dτ

⎞⎠x2, q4(x) := G1h1z1(x) + Af6(x). (2.14)

Therefore A−1 exists and is bounded. In light of the Sobolev embedding theorem [8], A−1 is compact. The proof is complete. �Lemma 2.3. Let A be defined by (2.6) and (2.7). Then A generates a C0-semigroup eAt of contractions on H.

Proof. We first prove that A is dissipative in H. For any Y = (u1, u2, v1, v2, y1, y2)T ∈ D(A), we denote

φ := h1 (−u1 + v1 + Ky′1) , ψ := A−1y′′′1 −G1h1A−1 (−u1 + v1 + Ky′1) .

1210 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

Since

〈AY, Y 〉H =⟨(

u2, αuu′′1 + Guφ, v2, αvv

′′1 −Gvφ, y2, ψ

)T

, (u1, u2, v1, v2, y1, y2)T⟩H

= 1Gu

1∫0

(αuu

′′1 + Guφ

)u2dx + αu

Gu

1∫0

u′2u

′1dx + 1

Gv

1∫0

(αvv

′′1 −Gvφ

)v2dx

+ αv

Gv

1∫0

v′2v′1dx + K

G1

1∫0

ψy2dx + IK

G1

1∫0

ψ′y′2dx + K

G1

1∫0

y′′2 y′′1dx

+1∫

0

(−u2 + v2 + Ky′2)φdx

= − αu

l1Gu|u′

1(1)|2 − αv

l2Gv|v′1(1)|2 − αu

Gu

1∫0

(u′

1u′2 − u′

2u′1

)dx

− K

l3G1|y′′1 (1)|2 − αv

Gv

1∫0

(v′1v

′2 − v′2v

′1

)dx− K

G1

1∫0

(y′′1 y

′′2 − y′′2 y

′′1

)dx

+1∫

0

[φ(u2 − v2 −Ky′2

)dx− (u2 − v2 −Ky′2) φ

]dx,

it follows that

Re〈AY, Y 〉H = − αu

l1Gu|u′

1(1)|2 − αv

l2Gv|v′1(1)|2 − K

l3G1|y′′1 (1)|2 ≤ 0.

Hence A is dissipative in H. Therefore, A generates a C0-semigroup eAt of contractions on H by using the Lumer–Phillips theorem. �

Let λ ∈ σ(A) and Yλ := (u1, u2, v1, v2, y1, y2)T ∈ D(A) be an eigenfunction corresponding to λ. From AYλ = λYλ, we obtain u2 = λu1, v2 = λv1, y2 = λy1 and

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αuu′′1 + Guh1 (−u1 + v1 + Ky′1) = λ2u1,

αvv′′1 −Gvh1 (−u1 + v1 + Ky′1) = λ2v1,

A−1y′′′1 −G1h1A−1 (−u1 + v1 + Ky′1) = λ2y1,

u1(0) = 0, u′1(1) = −λl1u1(1),

v1(0) = 0, v′1(1) = −λl2v1(1),

y1(0) = y′1(0) = 0, y′′1 (1) = −λl3y′1(1).

(2.15)

Lemma 2.4. If λ ∈ σ(A), then Reλ < 0.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1211

Proof. From (2.15), we have

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αuu′′1 + Guh1 (−u1 + v1 + Ky′1) = λ2u1,

αvv′′1 −Gvh1 (−u1 + v1 + Ky′1) = λ2v1,

y′′′′1 −G1h1 (−u′1 + v′1 + Ky′′1 ) = λ2Iy′′1 − λ2y1,

u1(0) = 0, u′1(1) = −λl1u1(1),

v1(0) = 0, v′1(1) = −λl2v1(1),

y1(0) = y′1(0) = 0, y′′1 (1) = −λl3y′1(1),

y′′′1 (1) −G1h1[−u1(1) + v1(1) + Ky′1(1)] − λ2Iy′1(1) = 0.

(2.16)

For simplicity, set k1 = 1Guh1

, k2 = 1Gvh1

and k3 = KG1h1

. Multiplying k1u1, the conjugate of k1u1, on both side of the first equation in (2.16) and integrating from 0 to 1 with respect to x, we obtain

λl1αuk1|u1(1)|2 + λ2k1

1∫0

|u1|2dx + αuk1

1∫0

|u′1|2dx−

1∫0

(−u1 + v1 + Ky′1)u1dx = 0. (2.17)

Similarly, multiplying k2v1, the conjugate of k2v1, on both side of the second equation in (2.16) and inte-grating from 0 to 1 with respect to x, we obtain

λl2αvk2|v1(1)|2 + λ2k2

1∫0

|v1|2dx + αvk2

1∫0

|v′1|2dx +1∫

0

(−u1 + v1 + Ky′1) v1dx = 0. (2.18)

In addition, multiplying k3y1, the conjugate of k3y1, on both side of the third equation in (2.16) and integrating from 0 to 1 with respect to x, we obtain

λk3l3|y′1(1)|2 + λ2

⎛⎝k3I

1∫0

|y′1|2dx + k3

1∫0

|y1|2dx

⎞⎠ + k3

1∫0

|y′′1 |2dx +1∫

0

(−u1 + v1 + Ky′1)Ky′1dx = 0.

(2.19)

From (2.17), (2.18) and (2.19), we have

λ[l1αuk1|u1(1)|2 + l2αvk2|v1(1)|2 + l3k3|y′1(1)|2

]+ λ2

⎛⎝k1

1∫0

|u1|2dx + k2

1∫0

|v1|2dx + Ik3

1∫0

|y′1|2dx + k3

1∫0

|y1|2dx

⎞⎠+ αuk1

1∫0

|u′1|2dx + αvk2

1∫0

|v′1|2dx + k3

1∫0

|y′′1 |2dx +1∫

0

| − u1 + v1 + Ky′1|2dx = 0. (2.20)

Let λ = Reλ + iImλ, where Reλ and Imλ are real. We have

Reλ[l1αuk1|u1(1)|2 + l2αvk2|v1(1)|2 + l3k3|y′1(1)|2

]+

[(Reλ)2 − (Imλ)2

]⎛⎝k1

1∫|u1|2dx + k2

1∫|v1|2dx + Ik3

1∫|y′1|2dx + k3

1∫|y1|2dx

⎞⎠

0 0 0 0

1212 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

+ αuk1

1∫0

|u′1|2dx + αvk2

1∫0

|v′1|2dx + k3

1∫0

|y′′1 |2dx +1∫

0

| − u1 + v1 + Ky′1|2dx = 0, (2.21)

and

Imλ[l1αuk1|u1(1)|2 + l2αvk2|v1(1)|2 + l3k3|y′1(1)|2

]+ 2ReλImλ

⎛⎝k1

1∫0

|u1|2dx + k2

1∫0

|v1|2dx + Ik3

1∫0

|y′1|2dx + k3

1∫0

|y1|2dx

⎞⎠ = 0. (2.22)

If Imλ = 0, then Reλ < 0 follows from (2.21). If Imλ �= 0, then Reλ < 0 by (2.22). The proof is complete. �We next formulate the eigenvalue problem for the operator A. For simplicity of notation, we define

r1 :=√

1αu

, r2 :=√

1αv

, c1 := Guh1

αu, c2 := Gvh1

αv, c3 := G1h1. (2.23)

Then (2.15) can be written as

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

r21λ

2u1 − c1Ky′1 − c1v1 + c1u1 − u′′1 = 0,

r22λ

2v1 + c2Ky′1 + c2v1 − c2u1 − v′′1 = 0,

λ2Iy′′1 − λ2y1 + c3Ky′′1 + c3v′1 − c3u

′1 − y′′′′1 = 0,

u1(0) = 0, u′1(1) = −λl1u1(1),

v1(0) = 0, v′1(1) = −λl2v1(1),

y1(0) = y′1(0) = 0, y′′1 (1) = −λl3y′1(1),

y′′′1 (1) = −c3u1(1) + c3v1(1) + (c3K + λ2I)y′1(1).

(2.24)

We use the matrix operator pencil method [20] to solve the equations (2.24). Set

r3 :=√I, u1 := u1, u2 := u′

1, v1 := v1, v2 := v′1, y1 := y1, y2 := y′1, y3 := y′′1 , y4 := y′′′1 (2.25)

and

Φ := (u1, u2, v1, v2, y1, y2, y3, y4)T . (2.26)

Then (2.24) becomes {TD(λ)Φ := Φ′(x) + M(λ)Φ(x) = 0,

TR(λ)Φ := W 0(λ)Φ(0) + W 1(λ)Φ(1) = 0,(2.27)

where

W 0(λ) :=

⎛⎜⎜⎜⎜⎜⎝1 0 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 0

O

⎞⎟⎟⎟⎟⎟⎠ , (2.28)

4×8

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1213

W 1(λ) :=

⎛⎜⎜⎜⎜⎜⎝O4×8

λl1 1 0 0 0 0 0 00 0 λl2 1 0 0 0 00 0 0 0 0 λl3 1 0c3 0 −c3 0 0 −c3K − r2

3λ2 0 1

⎞⎟⎟⎟⎟⎟⎠ , (2.29)

and

M(λ) := D0 − λ2D1 (2.30)

with D0, D1 defined by

D0 :=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 −1 0 0 0 0 0 0−c1 0 c1 0 0 c1K 0 00 0 0 −1 0 0 0 0c2 0 −c2 0 0 −c2K 0 00 0 0 0 0 −1 0 00 0 0 0 0 0 −1 00 0 0 0 0 0 0 −10 c3 0 −c3 0 0 −c3K 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠, (2.31)

D1 :=

⎛⎜⎜⎜⎜⎜⎝r21D11 O2×2 O2×2 O2×2

O2×2 r22D11 O2×2 O2×2

O2×2 O2×2 O2×2 O2×2

O2×2 O2×2 −D11 r23D11

⎞⎟⎟⎟⎟⎟⎠ , D11 :=(

0 01 0

). (2.32)

Theorem 2.1. The characteristic equation (2.15) is equivalent to the first order linear system (2.27). Also λ ∈ σ(A) if and only if (2.27) has a nontrivial solution.

3. Asymptotic behavior of eigenvalues

In this section, we shall give the asymptotic expressions for the eigenvalues of A. We first solve system(2.27) and obtain a fundamental matrix solution by modifying a standard technique of Birkhoff–Langer [3] and Tretter [20] (or [21]) for dealing with the matrix operator pencils. Then we derive the asymptotic expressions of the eigenvalues through the characteristic determinant Δ(λ) of (2.27). A key step is providing an invertible matrix transformation in calculation.

Firstly, we diagonalize the leading term λ2D1 in (2.30). For any 0 �= λ ∈ C, define an invertible matrix in λ by

P (λ) :=

⎛⎜⎝ P1(λ)P2(λ)

P3(λ)

⎞⎟⎠ , P1(λ) :=(

r1λ r1λ

r21λ

2 −r21λ

2

), (3.1)

with

P2(λ) :=(

r2λ r2λ

r22λ

2 −r22λ

2

), P3(λ) :=

⎛⎜⎜⎜⎝1 0 0 00 1

r23

1 −10 0 r3λ r3λ

0 0 r2λ2 −r2λ2

⎞⎟⎟⎟⎠ .

3 3

1214 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

For any λ �= 0, by a simple computation, we obtain

P−1(λ) :=

⎛⎜⎝ P−11 (λ)

P−12 (λ)

P−13 (λ)

⎞⎟⎠ , P−11 (λ) :=

(1

2r1λ1

2r21λ

2

12r1λ − 1

2r21λ

2

), (3.2)

with

P−12 (λ) :=

(1

2r2λ1

2r22λ

2

12r2λ − 1

2r22λ

2

), P−1

3 (λ) :=

⎛⎜⎜⎜⎝1 0 0 00 r2

3 0 − 1λ2

0 0 12r3λ

12r2

3λ2

0 0 12r3λ − 1

2r23λ

2

⎞⎟⎟⎟⎠ .

For simplicity, let ⎧⎪⎪⎨⎪⎪⎩b1 := c1

r1, b2 := c1r2

r21

, b3 := c1K

r21r

23, b4 := c2r1

r22

, b5 := c2r2

,

b6 := c2K

r22r

23, b7 := c3r

21

r23

, b8 := c3r22

r23

, b9 := c3K

r3.

In addition, define

Ψ(x) := P−1(λ)Φ(x), TD(λ) := P−1(λ)TD(λ)P (λ). (3.3)

From (2.27), we have

TD(λ)Ψ = Ψ′(x) − M(λ)Ψ(x) = 0, (3.4)

where

M(λ) = −P−1(λ)M(λ)P (λ) := λM1 + M0 + λ−1M−1 + λ−2M−2, (3.5)

with

M1 := diag(r1,−r1, r2,−r2, 0, 0, r3,−r3), M0 :=(

O5×5 M5×3

M3×5 O3×3

)(3.6)

by

M5×3 :=

⎛⎜⎜⎜⎜⎜⎝0 0 00 0 00 0 00 0 01r23

1 −1

⎞⎟⎟⎟⎟⎟⎠ , M3×5 :=

⎛⎜⎝ c3r21 −c3r

21 −c3r

22 c3r

22 1

− b72

b72

b82 − b8

2 − 12r2

3b72 − b7

2 − b82

b82

12r2

3

⎞⎟⎠ ,

and

M−1 :=(M1

4×4 O4×4

O4×4 M24×4

), M−2 :=

(O4×5 M4×3O4×5 O4×3

)(3.7)

by

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1215

M14×4 :=

⎛⎜⎜⎜⎝b12

b12 − b2

2 − b22

− b12 − b1

2b22

b22

− b42 − b4

2b52

b52

b42

b42 − b5

2 − b52

⎞⎟⎟⎟⎠ ,

M24×4 :=

⎛⎜⎜⎜⎝0 0 0 00 0 −c3r3K −c3r3K

0 0 b92

b92

0 0 − b92 − b9

2

⎞⎟⎟⎟⎠ , M4×3 :=

⎛⎜⎜⎜⎜⎝− b3

2 − b3r23

2b3r

23

2b32

b3r23

2 − b3r23

2b62

b6r23

2 − b6r23

2− b6

2 − b6r23

2b6r

23

2

⎞⎟⎟⎟⎟⎠ .

Secondly, we give an asymptotic expression for the fundamental matrix solution of system (3.4).

Theorem 3.1. Let 0 �= λ ∈ C, and M(λ) given by (3.5) and assume that r1 �= r2 �= r3. For x ∈ [0, 1], set

E(x, λ) := diag(er1λx, e−r1λx, er2λx, e−r2λx, 1, 1, er3λx, e−r3λx

). (3.8)

Then there exists a fundamental matrix solution Ψ(x, λ) for system (3.4), which satisfies

Ψ′(x, λ) = M(λ)Ψ(x, λ) (3.9)

such that, for large enough |λ|,

Ψ(x, λ) =(

Ψ0(x) + Θ(x, λ)λ

)E(x, λ), (3.10)

where

Ψ0(x) :=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 0 0 00 0 0 0 cosh

(1r3x)

1r3

sinh(

1r3x)

0 0

0 0 0 0 r3 sinh(

1r3x)

cosh(

1r3x)

0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠, (3.11)

and

Θ(x, λ) := Ψ1(x) + λ−1Ψ2(x) + · · · (3.12)

with all entries uniformly bounded in [0, 1].

Proof. It is easy to check that

E′(x, λ) = λM1E(x, λ), (3.13)

where M1 is given in (3.6).

1216 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

Now we look for a fundamental matrix solution of (3.9) in the form of

Ψ(x, λ) =(Ψ0(x) + λ−1Ψ1(x) + · · · + λ−nΨn(x) + · · ·

)E(x, λ).

Then the left-hand side of (3.9) is

Ψ′(x, λ) =(Ψ′

0(x) + λ−1Ψ′1(x) + · · · + λ−nΨ′

n(x) + · · ·)E(x, λ)

+ λ(Ψ0(x) + λ−1Ψ1(x) + · · · + λ−nΨn(x) + · · ·

)M1E(x, λ),

and the right-hand side of (3.9) is

M(λ)Ψ(x, λ) =(λM1 + M0 + λ−1M−1 + λ−2M−2

)(Ψ0(x) + λ−1Ψ1(x) + · · · + λ−nΨn(x) + · · ·

)E(x, λ).

According to the coefficients of λ1, λ0, λ−1, . . . , λ−n, . . ., we have

Ψ0(x)M1 − M1Ψ0(x) = 0,

Ψ′0(x) − M0Ψ0(x) + Ψ1(x)M1 − M1Ψ1(x) = 0,

Ψ′1(x) − M0Ψ1(x) − M−1Ψ0(x) + Ψ2(x)M1 − M1Ψ2(x) = 0,

Ψ′2(x) − M0Ψ2(x) − M−1Ψ1(x) − M−2Ψ0(x) + Ψ3(x)M1 − M1Ψ3(x) = 0,

...

Ψ′n(x) − M0Ψn(x) − M−1Ψn−1(x) − M−2Ψn−2(x) + Ψn+1(x)M1

− M1Ψn+1(x) = 0,

...

Using the arguments in [20, p. 135] or [3], we deduce that there is an asymptotic fundamental matrix solution Ψ(x, λ) for system (3.9). It remains to show that the leading order term Ψ0(x) is given by (3.11). Indeed, Ψ0(x) can be determined by the matrix equations

Ψ0(x)M1 − M1Ψ0(x) = 0, (3.14)

and

Ψ′0(x) − M0Ψ0(x) + Ψ1(x)M1 − M1Ψ1(x) = 0, (3.15)

where M1 and M0 are given in (3.6). If Ψ0(x) is known, then one can deduce Ψ1(x) of Θ(x, λ) in (3.12)from (3.15) and

Ψ′1(x) − M0Ψ1(x) − M−1Ψ0(x) + Ψ2(x)M1 − M1Ψ2(x) = 0

with M−1 being given in (3.7). Similarly, we obtain all the terms Ψ1(x), Ψ2(x), . . ., Ψn(x), . . . of Θ(x, λ) in(3.12). So, the proof will be accomplished if we would find Ψ0(x) in (3.10).

Let cij(x) be the (i, j)-entry of the matrix Ψ0(x) with i, j = 1, 2, . . . , 8. Since M1 is diagonal, it follows from (3.14) and r1 �= r2 �= r3 that the entries cij(x) of Ψ0(x) satisfy cij(x) = 0 (i �= j) except c56(x) and

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1217

c65(x). Moreover, the entries cii(x) (i = 1, 2, . . . , 8), c56(x) and c65(x) can be determined by substituting them into (3.15) to obtain ⎧⎪⎪⎨⎪⎪⎩

c′ii(x) = 0, for i = 1, 2, 3, 4, 7, 8,

c′55(x) = 1r23c65(x), c′56(x) = 1

r23c66(x),

c′65(x) = c55(x), c′66(x) = c56(x).

(3.16)

Then (3.11) follows from Ψ0(0) = I. The proof is complete. �The following corollary is the relationship of the solution between system (2.27) and system (3.4).

Corollary 3.1. Let 0 �= λ ∈ C. Suppose that r1 �= r2 �= r3 and Ψ(x, λ) given by (3.10) is a fundamental matrix solution of system (3.4). Then

Φ(x, λ) := P (λ)Ψ(x, λ) (3.17)

is a fundamental matrix solution for the first order linear system (2.27).

Thirdly, we estimate the asymptotic eigenvalues of the system. Note that the eigenvalues of the first order linear system (2.27) are given by the zeros of the characteristic determinant

Δ(λ) := det(TR(λ)Φ

), λ ∈ C, (3.18)

where TR(λ) is given in (2.27) and Φ is a fundamental matrix solution of TD(λ)Φ = 0 [20].In fact, from (2.27), (3.10) and (3.17), we have

TR(λ)Φ = W 0(λ)P (λ)Ψ(0, λ) + W 1(λ)P (λ)Ψ(1, λ). (3.19)

Using (2.28) and (3.1), a simple computation gives

W 0(λ)P (λ) =

⎛⎜⎜⎜⎜⎜⎝r1λ r1λ 0 0 0 0 0 00 0 r2λ r2λ 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1

r23

1 −1O4×8

⎞⎟⎟⎟⎟⎟⎠ .

Similarly,

W 1(λ)P (λ) =

⎛⎜⎜⎜⎜⎜⎝O4×8

r1r4λ2 r1r5λ

2 0 0 0 0 0 00 0 r2r6λ

2 r2r7λ2 0 0 0 0

0 0 0 0 0 r8λ r9λ r10λ

c3r1λ c3r1λ −c3r2λ −c3r2λ 0 −r11 − λ2 −c3K c3K

⎞⎟⎟⎟⎟⎟⎠ ,

where ⎧⎪⎨⎪⎩r4 := l1 + r1, r5 := l1 − r1, r6 := l2 + r2, r7 := l2 − r2,

r8 := l32 , r9 := r3 + l3, r10 := r3 − l3, r11 := c3K

2 .(3.20)

r3 r3

1218 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

For notational simplicity, set [a]1 := a + O(λ−1) with O(λ−m) satisfying

lim|λ|→∞

∣∣λmO(λ−m)∣∣ < ∞, m ∈ Z.

From Theorem 3.1, an asymptotic fundamental matrix solution Ψ(x, λ) of system (3.4) can be given by

Ψ(x, λ) :=(Ψ0(x) + O(λ−1)

)E(x, λ). (3.21)

Since Ψ0(0) = I and E(0, λ) = I, a direct computation yields

W 0(λ)P (λ)Ψ(0, λ) =(

M11 M12O4×4 O4×4

),

where

M11 :=

⎛⎜⎜⎜⎜⎝λ[r1]1 λ[r1]1 O(1) O(1)O(1) O(1) λ[r2]1 λ[r2]1

O(λ−1) O(λ−1) O(λ−1) O(λ−1)O(λ−1) O(λ−1) O(λ−1) O(λ−1)

⎞⎟⎟⎟⎟⎠ ,

and

M12 :=

⎛⎜⎜⎜⎜⎝O(1) O(1) O(1) O(1)O(1) O(1) O(1) O(1)[1]1 O(λ−1) O(λ−1) O(λ−1)

O(λ−1)[

1r23

]1

[1]1 [−1]1

⎞⎟⎟⎟⎟⎠ .

Similarly,

W 1(λ)P (λ)Ψ(1, λ) =(O4×4 O4×4M21 M22

),

where

M21 :=

⎛⎜⎜⎜⎜⎜⎝λ2E1[r1r4]1 λ2E2[r1r5]1 λE3O(1) λE4O(1)

λE1O(1) λE2O(1) λ2E3[r2r6]1 λ2E4[r2r7]1E1O(1) E2O(1) E3O(1) E4O(1)

λE1[m1]1 λE2[m1]1 λE3[m2]1 λE4[m2]1

⎞⎟⎟⎟⎟⎟⎠ ,

and

M22 :=

⎛⎜⎜⎜⎜⎜⎜⎝

λO(1) λO(1) λE5O(1) λE6O(1)

λO(1) λO(1) λE5O(1) λE6O(1)

λ[r8r3 l2

]1

λ[r8 l1

]1

λE5[r9]1 λE6[r10]1

−λ2[r3 l2

]1

−λ2[l1

]1

−λE5O(1) −λE6O(1)

⎞⎟⎟⎟⎟⎟⎟⎠ ,

with

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1219

{E1 := er1λ, E2 := e−r1λ, E3 := er2λ,

E4 := e−r2λ, E5 := er3λ, E6 := e−r3λ,(3.22)

and m1 = c3r1 −O(1), m2 = −c3r2 −O(1), l1 = cosh 1r3, l2 = sinh 1

r3.

Hence, TR(λ)Φ has the following asymptotic expression

W 0(λ)P (λ)Ψ(0, λ) + W 1(λ)P (λ)Ψ(1, λ) =(M11 M12M21 M22

).

Therefore, the asymptotic expansion of the characteristic determinant can be expressed as

Δ(λ) = det(W 0(λ)P (λ)Ψ(0, λ) + W 1(λ)P (λ)Ψ(1, λ)

)= − det

(λ[r1]1 λ[r1]1

λ2E1[r1r4]1 λ2E2[r1r5]1

)× det

(λ[r2]1 λ[r2]1

λ2E3[r2r6]1 λ2E4[r2r7]1

)

× det

⎛⎜⎜⎜⎜⎜⎜⎜⎝

[1]1 O(λ−1) O(λ−1) O(λ−1)

O(λ−1)[

1r23

]1

[1]1 −[1]1

λ[r8 l2r3

]1

λ[r8 l1

]1

λE5[r9]1 λE6[r10]1

−λ2[l2r3

]1

−λ2[l1

]1

−λE5O(1) −λE6O(1)

⎞⎟⎟⎟⎟⎟⎟⎟⎠= λ9r2

1r22 l1Δ1(λ)Δ2(λ)Δ3(λ),

where ⎧⎪⎨⎪⎩Δ1(λ) := r5E2 − r4E1 + O(λ−1),Δ2(λ) := r7E4 − r6E3 + O(λ−1),Δ3(λ) := r10E6 + r9E5 + O(λ−1).

(3.23)

Theorem 3.2. If r1 �= r2 �= r3 and if Δ(λ) be the characteristic determinant of system (2.27), then the following asymptotic expansion for Δ(λ) holds:

Δ(λ) = λ9r21r

22 l1Δ1(λ)Δ2(λ)Δ3(λ) (3.24)

with Δi(λ) being given in (3.23). If li �= ri (i = 1, 2, 3), then there are three branches of asymptotic eigen-values given by (as |n| → ∞ and n ∈ Z)

λjn = μj + r−1j nπi + O(n−1) for j = 1, 2, 3, (3.25)

where

μj :=

⎧⎨⎩1

2rj ln lj−rjlj+rj

, lj > rj1

2rj

(ln rj−lj

lj+rj+ πi

), lj < rj

for j = 1, 2, 3. (3.26)

Moreover, we have, as |n| → ∞,

Reλjn → 12rj

ln∣∣∣∣ lj − rjlj + rj

∣∣∣∣ < 0 for j = 1, 2, 3. (3.27)

Furthermore, the zeros of Δ(λ) are simple when their moduli are sufficiently large.

1220 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

Proof. By Δ(λ) = 0 and (3.24), we obtain that

Δ1(λ)Δ2(λ)Δ3(λ) = 0, (3.28)

and

Δi(λ) = 0 for i = 1, 2, 3.

Let Δ1(λ) = 0, we have

r5E2 − r4E1 + O(λ−1) = 0, (3.29)

that is (from (3.20) and (3.22))

(l1 − r1) e−r1λ − (l1 + r1) er1λ + O(λ−1) = 0. (3.30)

The solutions of the equation

(l1 − r1) e−r1λ − (l1 + r1) er1λ = 0

are given by

λ1n = μ1 + r−11 nπi, n ∈ Z.

By the Rouché’s theorem, the solutions of (3.30) can be expressed as

λ1n = λ1n + O(n−1) = μ1 + r−11 nπi + O(n−1), n ∈ Z and |n| → ∞. (3.31)

Similarly, let Δ2(λ) = 0. Then the equation

(l2 − r2) e−r2λ − (l2 + r2) er2λ + O(λ−1) = 0 (3.32)

has the solutions

λ2n = μ2 + r−12 nπi + O(n−1), n ∈ Z and |n| → ∞. (3.33)

Also, let Δ3(λ) = 0. The equation

(r3 − l3) e−r3λ + (r3 + l3) er3λ + O(λ−1) = 0 (3.34)

has the solutions

λ3n = μ3 + r−13 nπi + O(n−1), n ∈ Z and |n| → ∞. (3.35)

Finally, by r1 �= r2 �= r3, it is easy to see that the last assertion is concluded. �Theorem 3.3. Suppose r1 �= r2 �= r3 and li �= ri (i = 1, 2, 3). Let A be defined by (2.6) and (2.7). Then all eigenvalues of A have the asymptotic expressions given by (3.25). Moreover, all eigenvalues of the system with sufficiently large moduli are simple.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1221

4. Asymptotic behavior of eigenfunctions

In this section, we shall give the asymptotic expressions for the eigenfunctions of A.

Theorem 4.1. Suppose r1 �= r2 �= r3 and li �= ri (i = 1, 2, 3). Let σ(A) := {λ1n, λ2n, λ3n, n ∈ Z} be the eigenvalues of A with λjn (j = 1, 2, 3) being given in (3.25). Then the corresponding eigenfunctions{

(ujn, λjnujn, vjn, λjnvjn, yjn, λjnyjn)T , j = 1, 2, 3, n ∈ Z

}have the following asymptotic expressions for |n| → ∞, n ∈ Z:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u′1n(x) = cosh

[(r1μ1 + nπi)x + O(n−1)

]+ O(n−1),

u′jn(x) = O(n−1) for j = 2, 3,

λ1nu1n(x) = r−11 sinh

[(r1μ1 + nπi)x + O(n−1)

]+ O(n−1),

λjnujn(x) = O(n−1) for j = 2, 3,

v′2n(x) = cosh[(r2μ2 + nπi)x + O(n−1)

]+ O(n−1),

v′jn(x) = O(n−1) for j = 1, 3,

λ2nv2n(x) = r−12 sinh

[(r2μ2 + nπi)x + O(n−1)

]+ O(n−1),

λjnvjn(x) = O(n−1) for j = 1, 3,

y′′3n(x) = cosh[(r3μ3 + nπi)x + O(n−1)

]+ O(n−1),

y′′jn(x) = O(n−1) for j = 1, 2,

λ3ny′3n(x) = r−1

3 sinh[(r3μ3 + nπi)x + O(n−1)

]+ O(n−1),

λjny′jn(x) = O(n−1) for j = 1, 2,

(4.1)

where r1, r2 and r3 are given in (2.23) and (2.25) respectively. Moreover,{(ujn, λjnujn, vjn, λjnvjn, yjn, λjnyjn)T , j = 1, 2, 3, n ∈ Z

}are approximately normalized in H in the sense that there exist positive constants d1 and d2 independent of n such that (j = 1, 2, 3)

d1 ≤∥∥u′

jn

∥∥L2 , ‖λjnujn‖L2 ,

∥∥v′jn∥∥L2 , ‖λjnvjn‖L2 ,∥∥y′′jn∥∥L2 ,

∥∥λjny′jn

∥∥L2 ≤ d2 (4.2)

for all integers n.

Proof. Note that the jth component of Φ(x) in (2.26) with respect to the eigenvalue λ can be obtained by taking the determinant of the matrices which are replaced one of the rows of TR(λ)Φ in (3.19) by eTj Φ so that their determinants are not zero, where ej is the jth column of the identity matrix.

From (3.17), an asymptotic fundamental matrix solution Φ1(x, λ) of system (2.27) can be given by

Φ1(x, λ) := P (λ)Ψ(x, λ),

where Ψ(x, λ) is given in (3.21). A direct computation gives

Φ1(x, λ) =(

Φ11(x, λ) Φ12(x, λ)Φ (x, λ) Φ (x, λ)

),

21 22

1222 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

where

Φ11(x, λ) :=

⎛⎜⎜⎜⎝λ[r1]1er1λx λ[r1]1e−r1λx O(1)er2λx O(1)e−r2λx

λ2 [r21]1 e

r1λx −λ2 [r21]1 e

−r1λx 0 0O(1)er1λx O(1)e−r1λx λ[r2]1er2λx λ[r2]1e−r2λx

0 0 λ2 [r22]1 e

r2λx −λ2 [r22]1 e

−r2λx

⎞⎟⎟⎟⎠ ,

Φ12(x, λ) :=

⎛⎜⎜⎜⎝O(1) O(1) O(1)er3λx O(1)e−r3λx

0 0 0 0O(1) O(1) O(1)er3λx O(1)e−r3λx

0 0 0 0

⎞⎟⎟⎟⎠ ,

Φ21(x, λ) :=

⎛⎜⎜⎜⎝O(λ−1)er1λx O(λ−1)e−r1λx O(λ−1)er2λx O(λ−1)e−r2λx

O(λ−1)er1λx O(λ−1)e−r1λx O(λ−1)er2λx O(λ−1)e−r2λx

O(1)er1λx O(1)e−r1λx O(1)er2λx O(1)e−r2λx

0 0 0 0

⎞⎟⎟⎟⎠ ,

and

Φ22(x, λ) :=

⎛⎜⎜⎜⎜⎜⎝

[l3(x)

]1

[l4(x)r3

]1

O(λ−1)er3λx O(λ−1)e−r3λx[l4(x)r3

]1

[l3(x)r23

]1

[1]1er3λx −[1]1e−r3λx

O(1) O(1) λ[r3]1er3λx λ[r3]1e−r3λx

0 0 λ2 [r23]1 e

r3λx −λ2 [r23]1 e

−r3λx

⎞⎟⎟⎟⎟⎟⎠ ,

with l3(x) := cosh(

1r3x)

and l4(x) := sinh(

1r3x).

Thus, the first component of Φ(x) is given by

u1(x, λ) = − λ−8r−21 r−2

2 l−11 det

(λ[r1]1 λ[r1]1

λ[r1]1er1λx λ[r1]1e−r1λx

)

× det(

λ[r2]1 λ[r2]1λ2E3[r2r6]1 λ2E4[r2r7]1

)

× det

⎛⎜⎜⎜⎜⎜⎜⎝[1]1 O(λ−1) O(λ−1) O(λ−1)

O(λ−1)[

1r23

]1

[1]1 −[1]1

λ[r8 l2r3

]1

λ[r8 l1

]1

λE5[r9]1 λE6[r10]1

−λ2[l2r3

]1

−λ2[l1

]1

−λE5O(1) −λE6O(1)

⎞⎟⎟⎟⎟⎟⎟⎠=

(e−r1λx − er1λx + O(λ−1)

) (r7E4 − r6E3 + O(λ−1)

)×

(r10E6 + r9E5 + O(λ−1)

).

From (3.23), (3.25) and (3.28), it follows that

u1(x, λ) ={

r12(λ)(e−r1λx − er1λx + O(λ−1)

)if λ = λ1n,

O(λ−1) if λ = λ2n or λ3n,(4.3)

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1223

where

r12(λ) :=(r7E4 − r6E3 + O(λ−1)

) (r10E6 + r9E5 + O(λ−1)

). (4.4)

In the same manner we can see that the third component of Φ(x) is given by

v1(x, λ) = − λ−8r−21 r−2

2 l−11 det

(λ[r1]1 λ[r1]1

λ2E1[r1r4]1 λ2E2[r1r5]1

)

× det(

λ[r2]1 λ[r2]1λ[r2]1er2λx λ[r2]1e−r2λx

)

× det

⎛⎜⎜⎜⎜⎜⎜⎝[1]1 O(λ−1) O(λ−1) O(λ−1)

O(λ−1)[

1r23

]1

[1]1 −[1]1

λ[r8 l2r3

]1

λ[r8 l1

]1

λE5[r9]1 λE6[r10]1

−λ2[l2r3

]1

−λ2[l1

]1

−λE5O(1) −λE6O(1)

⎞⎟⎟⎟⎟⎟⎟⎠=

(r5E2 − r4E1 + O(λ−1)

) (e−r2λx − er2λx + O(λ−1)

)×

(r10E6 + r9E5 + O(λ−1)

).

By (3.23), (3.25) and (3.28), we have

v1(x, λ) ={

r13(λ)(e−r2λx − er2λx + O(λ−1)

)if λ = λ2n,

O(λ−1) if λ = λ1n or λ3n,(4.5)

where

r13(λ) :=(r5E2 − r4E1 + O(λ−1)

) (r10E6 + r9E5 + O(λ−1)

). (4.6)

Furthermore, the sixth component of Φ(x) can be given by

y2(x, λ) = − λ−8r−21 r−2

2 l−11 det

(λ[r1]1 λ[r1]1

λ2E1[r1r4]1 λ2E2[r1r5]1

)

× det(

λ[r2]1 λ[r2]1λ2E3[r2r6]1 λ2E4[r2r7]1

)

× det

⎛⎜⎜⎜⎜⎜⎜⎝[1]1 O(λ−1) O(λ−1) O(λ−1)

O(λ−1)[

1r23

]1

[1]1 −[1]1[l4(x)r3

]1

[l3(x)r23

]1

[1]1er3λx −[1]1e−r3λx

−λ2[l2r3

]1

−λ2[l1

]1

−λE5O(1) −λE6O(1)

⎞⎟⎟⎟⎟⎟⎟⎠= −

(r5E2 − r4E1 + O(λ−1)

) (r7E4 − r6E3 + O(λ−1)

)×

(e−r3λx − er3λx + O(λ−1)

).

From (3.23), (3.25) and (3.28), we see that

y2(x, λ) ={

r14(λ)(e−r3λx − er3λx + O(λ−1)

)if λ = λ3n,

O(λ−1) if λ = λ1n or λ2n,(4.7)

1224 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

where

r14(λ) := −(r5E2 − r4E1 + O(λ−1)

) (r7E4 − r6E3 + O(λ−1)

). (4.8)

Together with (3.25), (4.1) can be deduced from (4.3) − (4.8) by setting

un(x) = − u1(x, λ)2r1λr12(λ) , vn(x) = − v1(x, λ)

2r2λr13(λ) , y′n(x) = − y2(x, λ)2r3λr14(λ) . (4.9)

Finally, it follows from (3.25) that⎧⎨⎩∥∥e−rjλjnx

∥∥2L2 = 1−e−2rjμj

2rjμj+ O(n−1) for j = 1, 2, 3,∥∥erjλjnx

∥∥2L2 = e2rjμj−1

2rjμj+ O(n−1) for j = 1, 2, 3,

(4.10)

where μj (j = 1, 2, 3) are given in (3.26). These together with (4.1) yield (4.2). �5. The Riesz basis property and exponential stability of the system

In this section, we prove that the generalized eigenfunctions of A form a Riesz basis for H. Furthermore, we establish the exponential stability of the system (1.3) by Riesz basis approach [4].

Let Yj := (uj , ξj , vj , ζj , yj , ηj)T ∈ H (j = 1, 2), define a new inner product in H by

〈Y1, Y2〉H :=〈u′1, u

′2〉L2 + 〈ξ1, ξ2〉L2 + 〈v′1, v′2〉L2 + 〈ζ1, ζ2〉L2 + 〈y′′1 , y′′2 〉L2 + 〈η′1, η′2〉L2 , (5.1)

and write its induced norm by ‖ · ‖H which is equivalent to (2.5). It is easy to check that H is a Hilbert space under (5.1). For convenience, we introduce another Hilbert space

L :=(L2(0, 1)

)6 (5.2)

with an inner product (for any Xj := (uj , ξj , vj , ζj , yj , ηj)T ∈ L, j = 1, 2

)〈X1, X2〉L :=〈u1, u2〉L2 + 〈ξ1, ξ2〉L2 + 〈v1, v2〉L2 + 〈ζ1, ζ2〉L2 + 〈y1, y2〉L2 + 〈η1, η2〉L2 ,

and define the subspaces of H and L, respectively, by⎧⎪⎨⎪⎩H1 :=

{Y ∈ H | Y = (u, ξ, 0, 0, 0, 0)T

},

H2 :={Y ∈ H | Y = (0, 0, v, ζ, 0, 0)T

},

H3 :={Y ∈ H | Y = (0, 0, 0, 0, y, η)T

},

and ⎧⎪⎨⎪⎩L1 :=

{X ∈ L | X = (u, ξ, 0, 0, 0, 0)T

},

L2 :={X ∈ L | X = (0, 0, v, ζ, 0, 0)T

},

L3 :={X ∈ L | X = (0, 0, 0, 0, y, η)T

}.

Obviously, we have

H = H1 ⊕H2 ⊕H3 and L = L1 ⊕ L2 ⊕ L3, (5.3)

where the sign ⊕ denotes the direct sum in the sense of orthogonality with respect to the inner products 〈·, ·〉H and 〈·, ·〉L in H and L, respectively.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1225

We recall a lemma [25, Lemma 5.2] which is needed in establishing the Riesz basis property of the operator A.

Lemma 5.1. Let {φn(x), n ∈ N} and {1, ψn(x), n ∈ N} be two subsets in L2(0, 1) defined by, respectively,

φn(x) := sinnπx and ψn(x) := cosnπx ∀ x ∈ (0, 1), n ∈ N.

Then {φn(x), n ∈ N} and {1, ψn(x), n ∈ N} are two orthogonal bases in L2(0, 1). Moreover, for any scalars α1, β1 �= 0 ∈ C the vector family{

Ψn := (cosh [(α1 + inπ)x] , β1 sinh [(α1 + inπ)x])T , n ∈ Z

}forms a Riesz basis on the Hilbert space L2(0, 1) × L2(0, 1).

Theorem 5.1. Suppose r1 �= r2 �= r3 and li �= ri (i = 1, 2, 3). Let

Ψjn :=(u′jn, λjnujn, v

′jn, λjnvjn, y

′′jn, λjny

′jn

)T, j = 1, 2, 3, n ∈ Z, (5.4)

where the entries are given as (4.1) corresponding to the eigenvalues λjn. Then {Ψ1n, Ψ2n, Ψ3n, n ∈ Z}forms a Riesz basis in Hilbert space L provided that {Ψjn, j = 1, 2, 3, n ∈ Z} is ω-linearly independent in L.

Proof. Define three vector families by

Φ1n :=(cosh [(r1μ1 + inπ)x] , r−1

1 sinh [(r1μ1 + inπ)x] , 0, 0, 0, 0)T

,

Φ2n :=(0, 0, cosh [(r2μ2 + inπ)x] , r−1

2 sinh [(r2μ2 + inπ)x] , 0, 0)T

,

Φ3n :=(0, 0, 0, 0, cosh [(r3μ3 + inπ)x] , r−1

3 sinh [(r3μ3 + inπ)x])T

.

From Lemma 5.1, we see that the families {Φjn, n ∈ Z} (j = 1, 2, 3) are the Riesz bases for Lj , respectively. From the asymptotic expressions of eigenvalues (3.25) and eigenfunctions (4.1), we conclude that⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

‖u′1n − cosh [(r1μ1 + inπ)x] ‖L2 = O(n−1),

‖r1λ1nu1n − sinh [(r1μ1 + inπ)x] ‖L2 = O(n−1),‖v′2n − cosh [(r2μ2 + inπ)x] ‖L2 = O(n−1),‖r2λ2nv2n − sinh [(r2μ2 + inπ)x] ‖L2 = O(n−1),‖y′′3n − cosh [(r3μ3 + inπ)x] ‖L2 = O(n−1),‖r3λ3ny

′3n − sinh [(r3μ3 + inπ)x] ‖L2 = O(n−1).

(5.5)

Hence, we obtain

‖Ψ1n − Φ1n‖L1 = O(n−1),‖Ψ2n − Φ2n‖L2 = O(n−1),‖Ψ3n − Φ3n‖L3 = O(n−1).

(5.6)

Therefore, by Bari’s theorem [26], {Ψ1n, Ψ2n, Ψ3n, n ∈ Z} forms a Riesz basis for L provided that {Ψjn, j =1, 2, 3, n ∈ Z} is ω-linearly independent in L. �

We now come to the main results of the paper.

1226 C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227

Theorem 5.2. Suppose r1 �= r2 �= r3 and li �= ri (i = 1, 2, 3). Let A be defined by (2.6) and (2.7). Then the generalized eigenfunctions of A form a Riesz basis for H.

Proof. Let Yjn := (ujn, λjnujn, vjn, λjnvjn, yjn, λjnyjn)T (j = 1, 2, 3 and n ∈ Z) be eigenfunctions cor-responding to the eigenvalues λjn (j = 1, 2, 3) in which the entries are given in (4.1). Then {Yjn, j =1, 2, 3, n ∈ Z} is ω-linearly independent in H. The proof will be completed via an isomorphic mapping between two Hilbert spaces H and L that maps Yjn to Ψjn, where {Ψjn, j = 1, 2, 3, n ∈ Z} is given by(5.4).

To this end, for any V := (s1, s2, s3, s4, s5, s6)T ∈ H, we define a linear bounded operator T : H → Lby

T V = (s′1, s2, s′3, s4, s

′′5 , s

′6)T := V .

Since 〈V, Yjn〉H = 〈V , Ψjn〉L, it is easily seen that T is isomorphic and satisfies

‖T V ‖L = ‖V ‖L = ‖V ‖H . (5.7)

In particular, for j = 1, 2, 3 and n ∈ Z,

T Yjn = Ψjn and ‖Yjn‖H = ‖Ψjn‖L.

Moreover, {Yjn, j = 1, 2, 3, n ∈ Z} is ω-linearly independent in H, so is {Ψjn, j = 1, 2, 3, n ∈ Z} in L. Therefore, by Theorem 5.1, {Ψjn, j = 1, 2, 3, n ∈ Z} forms a Riesz basis in L. Hence, {Yjn, j = 1, 2, 3,n ∈ Z} forms a Riesz basis for H. �Theorem 5.3. Suppose r1 �= r2 �= r3 and li �= ri (i = 1, 2, 3). Let A be defined by (2.6) and (2.7), and T (t)a C0-semigroup generated by A in H. Then T (t) is exponentially stable.

Proof. From Theorems 3.3 and 5.2, we see that the spectrum determined growth condition ω(A) = s(A) for T (t) holds, where ω(A) is the growth order of T (t) and s(A) is the spectral bound of A. From Lemma 2.4, Reλ < 0 for all λ ∈ σ(A). This, together with (3.25) and the spectrum determined growth condition, shows that T (t) is an exponentially stable semigroup in H. �References

[1] T. Bailey, A. Gruzen, P. Madden, RCS/Piezoelectric Distributed Actuator Study, Technical Report No. AFAL-TR-88-038, Air Force Astronautics Laboratory produced at the Charles Stark Draper Laboratory, Inc., Cambridge, MA, 1988.

[2] A. Baz, J. Ro, Vibration control of plates with active constrained layer damping, Smart Mater. Struct. 5 (3) (1996) 272–280.

[3] G.D. Birkhoff, R.E. Langer, The boundary problems and developments associated with a system of ordinary linear differ-ential equations of the first order, Proc. Am. Acad. Arts Sci. 58 (2) (1923) 51–128.

[4] B.Z. Guo, Riesz basis approach to the stabilization of a flexible beam with a tip mass, SIAM J. Control Optim. 39 (6) (2001) 1736–1747.

[5] B.Z. Guo, J.M. Wang, S.P. Yung, Boundary stabilization of a flexible manipulator with rotational inertia, Differential Integral Equations 18 (9) (2005) 1013–1038.

[6] S.W. Hansen, O.Y. Imanuvilov, Exact controllability of a multilayer Rao–Nakra plate with clamped boundary conditions, ESAIM Control Optim. Calc. Var. 17 (2011) 1101–1132.

[7] S.W. Hansen, O.Y. Imanuvilov, Exact controllability of a multilayer Rao–Nakra plate with free boundary conditions, Math. Control Relat. Fields 1 (2) (2011) 189–230.

[8] V.G. Maz’ja, Sobolev Spaces, Springer-Verlag, New York, 1985.[9] K.A. Morris, A.Ö. Özer, Modeling and stabilizability of voltage-actuated piezoelectric beams with magnetic effects, SIAM

J. Control Optim. 52 (4) (2014) 2371–2398.[10] A.Ö. Özer, Semigroup well-posedness of a voltage controlled active constrained layered (ACL) beam with magnetic effects,

in: The Proceedings of 2016 American Control Conference, 2016.

C. Yang, J.-M. Wang / J. Math. Anal. Appl. 448 (2017) 1204–1227 1227

[11] A.Ö. Özer, Modeling and stabilization results for a charge or current-actuated active constrained layer (ACL) beam model with the electrostatic assumption, in: Proc. SPIE 9799, Active and Passive Smart Structures and Integrated Systems 2016, 97991F.

[12] A.Ö. Özer, S.W. Hansen, Uniform stabilization of a multilayer Rao–Nakra sandwich beam, Evol. Equ. Control Theory 2 (4) (2013) 695–710.

[13] A.Ö. Özer, S.W. Hansen, Exact boundary controllability results for a multilayer Rao–Nakra sandwich beam, SIAM J. Control Optim. 52 (2) (2014) 1314–1337.

[14] R. Rajaram, Exact boundary controllability results for a Rao–Nakra sandwich beam, Systems Control Lett. 56 (7) (2007) 558–567.

[15] C.A. Raposo, Exponential stability for a structure with interfacial slip and frictional damping, Appl. Math. Lett. 53 (2016) 85–91.

[16] M.C. Ray, J. Oh, A. Baz, Active constrained layer damping of thin cylindrical shells, J. Sound Vib. 240 (5) (2001) 921–935.[17] I.Y. Shen, Hybrid damping through intelligent constrained layer treatments, J. Vib. Acoust. 116 (3) (1994) 341–349.[18] I.Y. Shen, A variational formulation, a work-energy relation and damping mechanisms of active constrained layer treat-

ments, J. Vib. Acoust. 119 (2) (1997) 192–199.[19] R. Stanway, J.A. Rongong, N.D. Sims, Active constrained-layer damping: a state-of-the-art review, Proc. Inst. Mech. Eng.,

Part I, J. Syst. Control Eng. 217 (6) (2003) 437–456.[20] C. Tretter, Spectral problems for systems of differential equations y′+A0y = λA1y with λ-polynomial boundary conditions,

Math. Nachr. 214 (2000) 129–172.[21] C. Tretter, Boundary eigenvalue problems for differential equations Nη = λPη with λ-polynomial boundary conditions,

J. Differential Equations 170 (2001) 408–471.[22] M.A. Trindade, A. Benjeddou, Hybrid active–passive damping treatments using viscoelastic and piezoelectric materials:

review and assessment, J. Vib. Control 8 (2002) 699–745.[23] J.M. Wang, B.Z. Guo, Analyticity and dynamic behavior of a damped three-layer sandwich beam, J. Optim. Theory Appl.

137 (3) (2008) 675–689.[24] J.M. Wang, B.Z. Guo, B. Chentouf, Boundary feedback stabilization of a three-layer sandwich beam: Riesz basis approach,

ESAIM Control Optim. Calc. Var. 12 (1) (2006) 12–34.[25] J.M. Wang, G.Q. Xu, S.P. Yung, Exponential stabilization of laminated beams with structural damping and boundary

feedback controls, SIAM J. Control Optim. 44 (5) (2005) 1575–1597.[26] R.M. Young, An Introduction to Nonharmonic Fourier Series, revised 1st ed., Academic Press, New York, 2001.