Stabilization of Timoshenko beam by means of pointwise controls

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ESAIM: Control, Optimisation and Calculus of Variations

DOI: 10.1051/cocv:2003028

STABILIZATION OF TIMOSHENKO BEAM BY MEANS OF POINTWISE CONTROLS

^∗

Gen-Qi Xu

¹

and Siu Pang Yung

²

Abstract. We intend to conduct a fairly complete study on Timoshenko beams with pointwise feedback controls and seek to obtain information about the eigenvalues, eigenfunctions, Riesz-Basis- Property, spectrum-determined-growth-condition, energy decay rate and various stabilities for the beams. One major difficulty of the present problem is the non-simplicity of the eigenvalues. In fact, we shall indicate in this paper situations where the multiplicity of the eigenvalues is at least two. We build all the above-mentioned results from an effective asymptotic analysis on both the eigenvalues and the eigenfunctions, and conclude with the Riesz-Basis-Property and the spectrum-determined-growth- condition. Finally, these results are used to examine the stability effects on the system by the location of the pointwise control relative to the length of the whole beam.

Mathematics Subject Classification. 34K35, 47A65, 93B52, 93C20.

Received January 18, 2002. Revised April 15, 2003.

1. Introduction

The pointwise feedback control stabilization for Euler–Bernoulli beams, or equivalently stabilization of serially connected beams with dissipative joints, has been widely studied and many results have been obtained (see, for example [1–7] and the references therein). As for Timoshenko beams, the most complete physical model for thick beams, relatively few results are on pointwise feedback controls when comparing with the corresponding boundary feedback control problems [8–16]. This may be due to the complicated changes of the eigenfrequencies arisen from the present of the pointwise feedback control. It is the aim of this paper to carry out a thorough study for this problem and we have obtained asymptotic information for the eigenvalues, the eigenfunctions, the Riesz-Basis-Property, the spectrum-determined-growth-condition and their consequences on stabilities. Our approach is simple but effective. A major step is to conduct a complete asymptotic analysis on the eigenvalues and the eigenfunctions. These then form a very efficient building block for us to deduce all the required results.

Keywords and phrases. Timoshenko beam, pointwise feedback control, generalized eigenfunction system, Riesz basis.

∗This research was supported by the Natural Science Foundation of Shan Xi Province in China and a HKRGC grant of code HKU 7133/02P.

1 Department of Mathematics of Shanxi University, TaiYuan 030006, P.R. China.; e-mail:[email protected]

2 Department of Mathematics of the University of Hong Kong, Hong Kong, P.R. China; e-mail:[email protected]

c EDP Sciences, SMAI 2003

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The Timoshenko beam problem with pointwise control that we shall consider is:









ρw(x, t)¨ −κ(w⁰⁰(x, t)−ϕ⁰(x, t)) +u₁(t)δ_ξ = 0, 0< x < `,

I_ρϕ(x, t)¨ −EIϕ⁰⁰(x, t)−κ(w⁰(x, t)−ϕ(x, t)) +u₂(t)δ_ξ = 0, 0< x < `, w(0, t) = 0, ϕ(0, t) = 0,

κ(w⁰(`, t)−ϕ(`, t)) = 0, EIϕ⁰(`, t) = 0,

(1.1)

where w(x, t) is the deflection of the beam from its equilibrium andϕ(x, t) is the rotation angle of a filament of the beam at x. Here and henceforth the dot and the prime denote derivatives with respect to time and space variables respectively. Also,I_ρ, ρ, EI, κ, `,u₁andu₂are respectively the mass moment of inertia, mass density, rigidity coefficient, shear modulus of elasticity, length of the beam, external force and external moment.

We useδ_ξ to denote a Dirac mass concentrated at the pointξ∈(0, `) and consider the following velocity and angular velocity feedback control at pointξ:

u₁(t) =αw(ξ, t),˙

u₂(t) =βϕ(ξ, t),˙ (1.2)

whereαandβ are some positive feedback gain constants that can be tuned.

It is the closed loop system (1.1) and (1.2) that we will conduct our research on. The content of this paper is organized as follows. First, we exhibit some elementary properties of the operatorAdetermined by the closed loop system in Section 2, and then in Section 3, we perform a detail asymptotic analysis for its eigenvalues and eigenfunctions. In Section 4, we apply these asymptotic results to deduce the Riesz basis property for the generalized eigenfunctions ofA. Finally in Section 5, we examine various stabilities of the beams.

2. Basic properties for the operator A of the closed loop system

In this section, we shall establish some basic properties for the closed loop system (1.1, 1.2). To begin, we choose the state spaceHto be:

H:=V₀¹×L²(0, `)×V₀¹×L²(0, `),

where V₀^k :={ϕ∈H^k(0, `)ϕ(0) = 0}, k= 1,2,with H^k(0, `) being the usual Sobolev space of orderk. For Y₁:= [w₁, z₁, ϕ₁, ψ₁]^T andY₂:= [w₂, z₂, ϕ₂, ψ₂]^T ∈ H, in here the superscriptT denotes the transposition, the inner product ofHis defined as

hY₁, Y₂i:=

Z _`

0 κ(w₁⁰(x)−ϕ₁(x))(w⁰₂(x)−ϕ₂(x))dx+ Z _`

0 ρz₁(x)z₂(x)dx +

Z _`

0 EIϕ⁰₁(x)ϕ⁰₂(x)dx+

Z _`

0 I_ρψ₁(x)ψ₂(x)dx and the corresponding norm is denoted byk · k. If we define operatorAin Hby

A





 w

z ϕ ψ





:=





 κ z

ρ(w⁰⁰−ϕ⁰) ψ EI

I_ρϕ⁰⁰+ κ

I_ρ(w⁰−ϕ)







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with domain D(A) :=

(

Y = [w, z, ϕ, ψ]^T ∈ H

w, ϕ∈V₀²(0, ξ)∪H²(ξ, `), z, ψ∈V₀¹(0, `)

κ(w⁰(ξ₊)−w⁰(ξ₋)) =αz(ξ), EI(ϕ⁰(ξ₊)−ϕ⁰(ξ₋)) =βψ(ξ), κ(w⁰(`)−ϕ(`)) = 0, EIϕ⁰(`) = 0

) ,

then the closed loop system (1.1, 1.2) can be written as an evolutionary equation inH:

d

dtY(t) =AY(t), ∀t >0, (2.1)

where

Y(t) := [w(·, t),w(·, t), ϕ(·, t),˙ ϕ(·, t)]˙ ^T.

If (w(t, x), ϕ(x, t)) is a solution pair for (1.1, 1.2), then the energy of the closed loop system is given by E(t) := 1

2kY(t)k²=1 2

Z _`

0

κ|w⁰(x, t)−ϕ(x, t)|²+ρ|w(x, t)|˙ ²+EI|ϕ⁰(x, t)|²+I_ρ|ϕ(x, t)|˙ ² dx.

Theorem 2.1. The operatorAis dissipative with compact resolvents and generates aC₀semigroup of contrac- tions. As a result, the energy of the system (1.1,1.2) is non-increasing in time.

Proof. To show thatAhas compact resolvents, it suffices to show that 0∈ρ(A) because for thenA⁻¹ will be compact due to the Sobolev Embedding theorem and so are the resolvents ofA. So forF := [f₁, f₂, g₁, g₂]^T ∈ H, we need to find a uniqueY := [w, z, ϕ, ψ]^T ∈ D(A) such thatAY =F,i.e.,









 z=f₁

κ

ρ(w⁰⁰−ϕ⁰) =f₂ ψ=g₁

EI

I_ρϕ⁰⁰+ κ

I_ρ(w⁰−ϕ) =g₂ with conditions 





w(0) = 0, ϕ(0) = 0, w(ξ₊) =w(ξ₋), ϕ(ξ₊) =ϕ(ξ₋), κ(w⁰(ξ₊)−w⁰(ξ₋)) =αz(ξ), EI(ϕ⁰(ξ₊)−ϕ⁰(ξ₋)) =βψ(ξ), κ(w⁰(`)−ϕ(`)) = 0, EIϕ⁰(`) = 0.

Solving these equations, we obtain ϕ(x) =− 1

EI

"

I_ρ Z _`

0

k(x, s)g₂(s)ds+ρ Z _`

0

k(x, s)ds Z _`

s

f₂(t)dt

+βg₁(ξ)k(ξ, x) +αf₁(ξ)η(x)

# ,

w(x) = Z _x

0 ϕ(s)ds− ρ κ

Z _`

0 k(x, s)f₂(s)ds−α

κf₁(ξ)k(ξ, x), where

k(x, s) =

s, 0≤s≤x, x, x≤s≤`,

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η(x) =







`ξ−1

2ξ², ξ≤x≤`,

`x−1

2x², 0≤x≤ξ.

Since the Y = [w, f₁, ϕ, g₁]^T given above belongs to D(A) and Y = A⁻¹F, so by the Sobolev Embedding theorem,A⁻¹is compact and so is the resolvent ofA.

Next, for anyY := [w, z, ϕ, ψ]^T ∈ D(A), we have

RehAY, Yi=−α|z(ξ)|²−β|ψ(ξ)|²≤0

from a direct calculation. SoA is dissipative and generates aC₀ semigroup of contractions by Lumer–Phillips

theorem (see [17]) and the last assertion follows accordingly.

Theorem 2.1 tells us that the spectrum of A are all eigenvalues and so our next task is to determine the eigenvalues ofA. Letλ∈ C (C denotes the complex plane) be an eigenvalue ofA, and let Y = [w, z, ϕ, ψ]^T ∈ D(A) be an associated eigenfunction. Then we have

z(x) =λw(x), ψ(x) =λϕ(x), and the function pair (w(x), ϕ(x)) satisfying the following equations

ρλ²w(x)−κ(w⁰⁰(x)−ϕ⁰(x)) = 0, 0< x < `,

I_ρλ²ϕ(x)−EIϕ⁰⁰(x)−κ(w⁰(x)−ϕ(x)) = 0, 0< x < `, (2.2) with conditions 





w(0) = 0, ϕ(0) = 0, w(ξ₊) =w(ξ₋), ϕ(ξ₊) =ϕ(ξ₋), κ(w⁰(ξ₊)−w⁰(ξ₋)) =αλw(ξ), EI(ϕ⁰(ξ₊)−ϕ⁰(ξ₋)) =βλϕ(ξ), κ(w⁰(`)−ϕ(`)) = 0, EIϕ⁰(`) = 0.

(2.3) Let (w_j(λ, x), ϕ_j(λ, x)),j = 1,2,3,4,be the fundamental solutions for (2.2) satisfying

(w₁(λ,0), ϕ₁(λ,0)) = (1,0), (w₂(λ,0), ϕ₂(λ,0)) = (0,1), (w₃⁰(λ,0), ϕ⁰₃(λ,0)) = (1,0), (w⁰₄(λ, x), ϕ⁰₄(λ,0)) = (0,1).

Then any solution of (2.2) can be expressed as

w(λ, x) =w(0)w₁(λ, x) +ϕ(0)w₂(λ, x) +w⁰(0)w₃(λ, x) +ϕ⁰(0)w₄(λ, x),

ϕ(λ, x) =w(0)ϕ₁(λ, x) +ϕ(0)ϕ₂(λ, x) +w⁰(0)ϕ₃(λ, x) +ϕ⁰(0)ϕ₄(λ, x), (2.4) with w(λ,0) = w(0), ϕ(λ,0) = ϕ(0), w⁰(λ,0) = w⁰(0), ϕ⁰(λ,0) = ϕ⁰(0). Employing (2.3), we see that when 0≤x≤ξ,

w(λ, x) =w⁰(0)w₃(λ, x) +ϕ⁰(0)w₄(λ, x), ϕ(λ, x) =w⁰(0)ϕ₃(λ, x) +ϕ⁰(0)ϕ₄(λ, x), whenξ≤x≤`,

w(λ, x) = h

w⁰(0)w₃(λ, ξ) +ϕ⁰(0)w₄(λ, ξ) i

w₁(λ, x−ξ) + h

w⁰(0)ϕ₃(λ, ξ) +ϕ⁰(0)ϕ₄(λ, ξ) i

w₂(λ, x−ξ) +

h w⁰(0)

w⁰₃(λ, ξ) +α

κλw₃(λ, ξ)

+ϕ⁰(0)

w⁰₄(λ, ξ) + α

κλw₄(λ, ξ) i

w₃(λ, x−ξ) +

h w⁰(0)

ϕ⁰₃(λ, ξ) + β

EIλϕ₃(λ, ξ)

+ϕ⁰(0)

ϕ⁰₄(λ, ξ) + β

EIλϕ₄(λ, ξ) i

w₄(λ, x−ξ),

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ϕ(λ, x) = h

w⁰(0)w₃(λ, ξ) +ϕ⁰(0)w₄(λ, ξ) i

ϕ₁(λ, x−ξ) + h

w⁰(0)ϕ₃(λ, ξ) +ϕ⁰(0)ϕ₄(λ, ξ) i

ϕ₂(λ, x−ξ) +

h w⁰(0)

w⁰₃(λ, ξ) +α

κλw₃(λ, ξ)

+ϕ⁰(0)

w⁰₄(λ, ξ) +α

κλw₄(λ, ξ) i

ϕ₃(λ, x−ξ) +

h

w⁰(0)(ϕ⁰₃(λ, ξ) + β

EIλϕ₃(λ, ξ)) +ϕ⁰(0)

ϕ⁰₄(λ, ξ) + β

EIλϕ₄(λ, ξ) i

ϕ₄(λ, x−ξ),

where (w⁰(0), ϕ⁰(0)) is a pair of non-zero solutions of the following linear equations:

w⁰(0)A₁₁(λ) +ϕ⁰(0)A₁₂(λ) = 0,

w⁰(0)A₂₁(λ) +ϕ⁰(0)A₂₂(λ) = 0, (2.5)

with

A₁₁(λ) :=w₃(λ, ξ)[w⁰₁(λ, `−ξ)−ϕ₁(λ, `−ξ)] +ϕ₃(λ, ξ)[w⁰₂(λ, `−ξ)−ϕ₂(λ, `−ξ)]

+[w⁰₃(λ, ξ) +α

κλw₃(λ, ξ)][w⁰₃(λ, `−ξ)−ϕ₃(λ, `−ξ)]

+[ϕ⁰₃(λ, ξ) + β

EIλϕ₃(λ, ξ)][w₄⁰(λ, `−ξ)−ϕ₄(λ, `−ξ)],

A₁₂(λ) :=w₄(λ, ξ)[w⁰₁(λ, `−ξ)−ϕ₁(λ, `−ξ)] +ϕ₄(λ, ξ)[w⁰₂(λ, `−ξ)−ϕ₂(λ, `−ξ)]

+[w⁰₄(λ, ξ) +α

κλw₄(λ, ξ)][w⁰₃(λ, `−ξ)−ϕ₃(λ, `−ξ)]

+[ϕ⁰₄(λ, ξ) + β

EIλϕ₄(λ, ξ)][w₄⁰(λ, `−ξ)−ϕ₄(λ, `−ξ)], A₂₁(λ) :=w₃(λ, ξ)ϕ⁰₁(λ, `−ξ) +ϕ₃(λ, ξ)ϕ⁰₂(λ, `−ξ) + [w⁰₃(λ, ξ) +α

κλw₃(λ, ξ)]ϕ⁰₃(λ, `−ξ) +[ϕ⁰₃(λ, ξ) + β

EIλϕ₃(λ, ξ)]ϕ⁰₄(λ, `−ξ),

A₂₂(λ) :=w₄(λ, ξ)ϕ⁰₁(λ, `−ξ) +ϕ₄(λ, ξ)]ϕ⁰₂(λ, `−ξ) + [w⁰₄(λ, ξ) +α

κλw₄(λ, ξ)]ϕ⁰₃(λ, `−ξ) +[ϕ⁰₄(λ, ξ) + β

EIλϕ₄(λ, ξ)]ϕ⁰₄(λ, `−ξ).

(2.6)

So all the eigenvalues can be found by finding the zeros of the characteristic determinant Γ(λ),

Γ(λ) =

A₁₁(λ)A₁₂(λ) A₂₁(λ)A₂₂(λ)

, (2.7)

of the coefficient matrix in (2.5). Altogether we have the following result.

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Theorem 2.2.

1) Let λ ∈ C. Then λ∈ σ(A) if and only if Γ(λ) = 0. In this case, an eigenfunction corresponding to λ is given by

Y = [w(λ, x), λw(λ, x), ϕ(λ, x), λϕ(λ, x)]^T,

where w(λ, x) andϕ(λ, x) are determined in (2.4). Furthermore, if λ is an eigenvalue for A, then the corre- sponding eigensubspace is of dimension at most two.

2) The spectrum σ(A) ofAdistribute symmetrically with respect to the real axis.

Proof. The first assertion is just a summary of the previous discussion. To prove the second assertion, we use the fact that for any Y ∈ D(A), its conjugate satisfies Y ∈ D(A) and AY = AY. So the second assertion

holds.

3. Asymptotic analysis of the eigenvalues and eigenfunctions of A

Theorem 2.2 gives us a characterization of all the eigenvalues in terms of the zeros of Γ(λ) in (2.7).

We now conduct an asymptotic analysis on Γ(λ) to deduce the corresponding asymptotic behaviour for the eigenvalues. We begin with expanding the eigenfunctions asymptotically, and always assume that

ρ₁6=ρ₂ and denoted ρ²₁:= ρ

κ, ρ²₂:= I_ρ

EI· (3.1)

Assume that λ∈ C with |Imλ| being large enough and −M ≤Reλ≤0, in which M is another large positive constant. Then, by expanding asymptotically inλ, we obtain the following asymptotic expressions:











w₁(λ, x) = coshρ₁λx+O(λ⁻²), ϕ₁(λ, x) = −ρ

EI(ρ²₁−ρ²₂)

sinhρ₁λx

ρ₁λ −sinρ₂λx ρ₂λ

+O(λ⁻³), w₂(λ, x) = ρ²₂

(ρ²₁−ρ²₂)

sinhρ₁λx

ρ₁λ −sinρ₂λx ρ₂λ

+O(λ⁻³), ϕ₂(λ, x) = coshρ₂λx+O(λ⁻²),

w₃(λ, x) = sinhρ₁λx

ρ₁λ +O(λ⁻³), ϕ₃(λ, x) =O(λ⁻²),

w₄(λ, x) =O(λ⁻²), ϕ₄(λ, x) = sinhρ₂λx

ρ₂λ +O(λ⁻³), w⁰₁(λ, x) =ρ₁λsinhρ₁λx+O(λ⁻¹), ϕ⁰₁(λ, x) = −ρ

EI(ρ²₁−ρ²₂)(coshρ₁λx−coshρ₂λx) +O(λ⁻²), w⁰₂(λ, x) = ρ²₂

(ρ²₁−ρ²₂)(coshρ₁λx−coshρ₂λx) +O(λ⁻²), ϕ⁰₂(λ, x) =ρ₂λsinhρ₂λx+O(λ⁻¹),

w⁰₃(λ, x) = coshρ₁λx+O(λ⁻²)), ϕ⁰₃(λ, x) =O(λ⁻¹),

w⁰₄(λ, x) =O(λ⁻¹),

ϕ⁰₄(λ, x) = coshρ₂λx+O(λ⁻²).

(3.2)

Substituting these expressions into Γ(λ) in (2.7), we have:

Γ(λ) =G(λ) +O(λ⁻¹), (3.3)

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where the leading term G(λ) :=

h

coshρ₁λ`+ α

2κρ₁sinhρ₁λ`− α

2κρ₁sinhρ₁λ(`−2ξ) ih

coshρ₂λ`

+ β

EIρ₂sinhρ₂λ`− β

2EIρ₂sinhρ₂λ(`−2ξ) i

.

(3.4)

From expression (3.4), we know thatG(λ) is analytic inC and deduce automatically the following remarks.

Remark. (i) Whenξ=¹₂`, thenG(λ) have no zeros in|Reλ|> M for some large enough positive constantM. It is because whenξ= ¹₂`,G(λ) has a very simple expression:

G(λ) =

coshρ₁λ`+ α

2κρ₁sinhρ₁λ` coshρ₂λ`+ β

EIρ₂sinhρ₂λ`

,

and so G(λ) has no zeros in|Reλ|> M for some large enough positive constantM. (ii) When ξ6= ¹₂` but (1−_2κρ^α₁) = 0 and (1−_2EIρ^β ₂)6= 0, we have

G(λ)e^ρ¹^{λ|`−2ξ|+ρ}²^λ`→ 1

4sign(`−2ξ)

1− β 2EIρ₂

as Reλ→ −∞.

(iii) When (1−_2κρ^α₁)6= 0 and (1−_2EIρ^β ₂)6= 0, we have

G(λ)e^(ρ¹^+ρ²^)λ`→ 1 4

1− α

2κρ₁ 1− β 2EIρ₂

as Reλ→ −∞.

(iv) As a conclusion of (i–iii), we see that, wheneverξ∈(0, `), the zeros ofG(λ) are located asymptotically in a vertical strip of the complex planeC centered at the imaginary axis.

We use these remarks and (3.3, 3.4) to deduce some asymptotic behaviour for the eigenvalues ofA.

Theorem 3.1. With (3.1) and assume that (1−_2κρ^α₁)6= 0 and(1−_2EIρ^β ₂)6= 0. For n∈Z, letζ_n⁽¹⁾ andζ_n⁽²⁾ be the roots of respectively the first and the second equation in







coshρ₁λ`+ α

2κρ₁sinhρ₁λ`− α

2κρ₁sinhρ₁λ(`−2ξ) = 0, coshρ₂λ`+ β

2EIρ₂sinhρ₂λ`− β

2EIρ₂sinhρ₂λ(`−2ξ) = 0.

(3.5)

Let

λ⁽¹⁾_n :=ζ_n⁽¹⁾+η_n⁽¹⁾ and λ⁽²⁾_n :=ζ_n⁽²⁾+η_n⁽²⁾ be the eigenvalues ofA. Then the following assertions are true:

1) when|λ|large enough then −M ≤Reλ≤M and the degree of the zero λ^(j)_n ofΓ(λ)is the same as that of the zero ζ_n^(j) ofG(λ);

2) the eigenvaluesλ^(j)_n ∈σ(A), with sufficiently large modulus, satisfyλ^(j)_n −ζ_n^(j)=O( ¹

ζn^(j))for j= 1,2.

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Proof. The first assertion is just a restatement of the above remarks. For λ^(j)_n :=ζ_n^(j)+η_n^(j) (j = 1,2) to be eigenvalues ofA, we must have Γ(λ^(j)n ) = 0. Thus, for−M ≤Reλ^(j)_n ≤M with|Imλ^(j)n | large enough, we have

G(λ^(j)_n ) =O 1

λ^(j)_n

,

and so by Rouch´e theorem,η_n^(j)=O(1/ζ_n^(j)) forj= 1,2 and this proves the remaining assertions.

We now ready to estimate the asymptotic behaviour of the eigenfunctions that corresponding to the eigen- valuesλ⁽¹⁾_n and λ⁽²⁾_n . For this, we consider





w⁰(0) =A₂₂

λ⁽¹⁾_n

and ϕ⁰(0) =−A₂₁ λ⁽¹⁾_n

, w⁰(0) =A₁₂

λ⁽²⁾_n

and ϕ⁰(0) =−A₁₁ λ⁽²⁾_n

,

(3.6)

respectively, whereA_kj(λ) are given by (2.5, 2.6). Using the estimates in (3.1) and (3.2), we find that A₂₁(λ) =O λ⁻¹

, A₁₂(λ) =O λ⁻¹ ,

A₂₂(λ) = coshρ₂`λ+ β

2EIρ₂sinhρ₂`λ− β

2EIρ₂sinhρ₂λ(`−2ξ) +O λ⁻¹ , A₁₁(λ) = coshρ₁`λ+ α

2κρ₁sinhρ₁`λ− α

2Kρ₁sinhρ₂λ(`−2ξ) +O λ⁻¹ . Let the leading terms be

Ab₂₂(λ) := coshρ₂`λ+ β

2EIρ₂sinhρ₂`λ− β

2EIρ₂sinhρ₂λ(`−2ξ), Ab₁₁(λ) := coshρ₁`λ+ α

2κρ₁sinhρ₁`λ− α

2κρ₁sinhρ₂λ(`−2ξ)

and use the estimates (3.2) in some straight-forward but tedious calculations, we obtain the asymptotic expressions for the eigenfunctions ofA:

Y

λ⁽¹⁾_n

=Ab₂₂

λ⁽¹⁾_n

Ψ

λ⁽¹⁾_n

+Z₁

λ⁽¹⁾_n with||Z₁(λ)||=O(λ⁻¹),

Ψ(λ) :=







sinhρ₁λx ρ₁λ sinhρ₁λx

ρ₁ 0 0





 +χ_[ξ,`]





 α κρ₁

sinhρ₁λξsinhρ₁λ(x−ξ) ρ₁λ

α κρ₁

sinhρ₁λξsinhρ₁λ(x−ξ) ρ₁

0 0







, (3.7)

and

Y

λ⁽²⁾_n

=Ab₁₁

λ⁽²⁾_n

Φ

λ⁽²⁾_n

+Z₂

λ⁽²⁾_n

,

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with||Z₂(λ)||=O(λ⁻¹),

Φ(λ) :=





 0 0 sinhρ₂λx

ρ₂λ sinhρ₂λx

ρ₂





 +χ_[ξ,`]







0 0 β

EIρ₂

sinhρ₂λξsinhρ₂λ(x−ξ) ρ₂λ

β EIρ₂

sinhρ₂λξsinhρ₂λ(x−ξ) ρ₂







. (3.8)

Thus, we can prove the following result.

Theorem 3.2. With (3.1) and assume that(1−_2κρ^α₁)6= 0 and(1−_2EIρ^β ₂)6= 0. Let Ψ(λ),Φ(λ) be defined as in (3.7)–(3.8). Let the zeros ζ_n^(j), λ^(j)_n , j= 1,2, be given as in Theorem 3.1. Then the following assertions are true:

1) an eigenfunction corresponding toλ⁽¹⁾_n is given by Yb

λ⁽¹⁾_n

:= Ψ

ζ_n⁽¹⁾

+Zb₁

λ⁽¹⁾_n

, (3.9)

with ||Zb₁(λ)||=O(λ⁻¹), and an eigenfunction corresponding to λ⁽²⁾_n is given by Yb

λ⁽²⁾_n

:= Φ

ζ_n⁽²⁾

+Zb₂

λ⁽²⁾_n

, (3.10)

with ||Zb₂(λ)||=O(λ⁻¹);

2) for each j= 1,2, the eigenvalues satisfies X

n∈Z

λ^(j)_n ⁻²<∞.

Proof. From the expressions ofAb₁₁andAb₂₂, we first have

n∈Zinf bA₁₁

λ⁽¹⁾_n >0 and inf

n∈Z

bA₂₂

λ⁽²⁾_n >0.

If we take

Yb

λ⁽¹⁾_n

:=

Y

λ⁽¹⁾_n Ab₂₂

λ⁽¹⁾_n

, Yb

λ⁽²⁾_n

:=

Y

λ⁽²⁾_n Ab₁₁

λ⁽²⁾_n

and

Zb₁

λ⁽¹⁾_n

:= Ψ

λ⁽¹⁾_n −Ψ

ζ_n⁽¹⁾

+

Z₁

λ⁽¹⁾_n Ab₂₂

λ⁽¹⁾_n

, Zb₂

λ⁽²⁾_n

:= Φ

λ⁽²⁾_n

−Φ

ζ_n⁽²⁾

+ Z₂

λ⁽²⁾_n

Ab₁₁

λ⁽²⁾_n

, then

Yb

λ⁽¹⁾_n

= Ψ

ζ_n⁽¹⁾

+Zb₁

λ⁽¹⁾_n

, Yb

λ⁽²⁾_n

= Φ

ζ_n⁽²⁾

+Zb₂

λ⁽²⁾_n

.

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Also, Ψ

λ⁽¹⁾_n

−Ψ

ζ_n⁽¹⁾² = Z _`

0 κ

coshρ₁λ⁽¹⁾_n x−coshρ₁ζ_n⁽¹⁾x+χ_[ξ,`] α κρ₁

sinhρ₁λ⁽¹⁾_n ξcoshρ₁λ⁽¹⁾_n (x−ξ)

−sinhρ₁ζ_n⁽¹⁾ξcoshρ₁ζ_n⁽¹⁾(x−ξ) ²

dx +

Z _`

0

ρ ρ²₁

sinhρ₁λ⁽¹⁾_n x−sinhρ₁ζ_n⁽¹⁾x+χ_[ξ,`] α κρ₁

sinhρ₁λ⁽¹⁾_n ξsinhρ₁λ⁽¹⁾_n (x−ξ)

−sinhρ₁ζ_n⁽¹⁾ξsinhρ₁ζ_n⁽¹⁾(x−ξ) ²

dx

=O

λ⁽¹⁾_n −ζ_n⁽¹⁾)²

and Φ

λ⁽²⁾_n

−Φ

ζ_n⁽²⁾²= Z _`

0 κ

sinhρ₂λ⁽²⁾_n x

ρ₂λ⁽²⁾_n −sinhρ₁ζ_n⁽²⁾x ρ₂ζ_n⁽²⁾

+χ_[ξ,`] β EIρ₂

sinhρ₂λ⁽²⁾_n ξsinhρ₂λ⁽²⁾_n (x−ξ) ρ₂λ⁽²⁾_n

−sinhρ₂ζ_n⁽²⁾ξcoshρ₂ζ_n⁽²⁾(x−ξ) ρ₂ζ_n⁽²⁾

₂ dx +

Z _`

0 EI

coshρ₂λ⁽²⁾_n x−coshρ₂ζ_n⁽²⁾x+χ_[ξ,`] β EIρ₂

sinhρ₂λ⁽²⁾_n ξcoshρ₂λ⁽²⁾_n (x−ξ)

−sinhρ₂ζ_n⁽²⁾ξcoshρ₂ζ_n⁽¹⁾(x−ξ) ²

dx +

Z _`

0

I_ρ ρ²₂

sinhρ₂λ⁽¹⁾_n x−sinhρ₁ζ_n⁽²⁾x+χ_[ξ,`] β EIρ₂

sinhρ₁λ⁽²⁾_n ξsinhρ₁λ⁽²⁾_n (x−ξ)

−sinhρ₁ζ_n⁽²⁾ξsinhρ₁ζ_n⁽²⁾(x−ξ) ²

dx

=O

λ⁽²⁾_n −ζ_n⁽²⁾)²

,

so Ψ

λ⁽¹⁾_n

−Ψ

ζ_n⁽¹⁾=O

1/λ⁽¹⁾_n

, Φ

λ⁽²⁾_n −Φ

ζ_n⁽²⁾=O

1/λ⁽²⁾_n )

, and therefore 1) is true because||Zb₁(λ⁽¹⁾_n )||=O(1/λ⁽¹⁾_n ),||Zb₂(λ⁽²⁾_n )||=O(1/λ⁽²⁾_n ).

For 2), since Γ(λ) is an entire function of finite exponential type, so the assertion follows from the standard

result of entire function of finite exponential type (cf.[18]).

We are now in a position to discuss the zeros ofG(λ) in (3.4). For brevity, denote

η :=ρ₁λ`, ζ:=ρ₂λ`, ξ:=s`. (3.11)

Since

G(λ) =Ab₁₁(λ)Ab₂₂(λ) :=g₁(η)g₂(ζ)

with 





g₁(η) :=Ab₁₁(λ) = coshη+ α

2κρ₁sinhη+ α

2κρ₁sinhη(2s−1), g₂(ζ) :=Ab₂₂(λ) = coshζ+ β

2EIρ₂sinhζ+ β

2EIρ₂sinhζ(2s−1),

(3.12)

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so we only need to study the zeros of

g(z) := coshz+Ksinhz+Ksinh(2s−1)z (3.13) where K is a constant and s∈(0,1). Generally speaking, finding the zeros ofg(z) is very difficult. However, we have the following assertion.

Theorem 3.3. Letg(z)be defined by (3.13). Assume thatK6= 1 and s∈(0,1), then the following assertions are true:

1) there exist constantsA, B andR such that for all|x| ≥R, Ae^|x|≤ |g(x+iy)| ≤Be^|x|; 2) letη₀ be the complex number(s) defined by

η₀²= 1

4s(1−s)−K²; (i) if η₀ does not solve the following equation (in unknownν):

ν±√

ν²+K²−1 1 +K

!_(1−2s)

= ν±√

ν²+K²

K , (3.14)

then all the zeros of the functiong(z)are simple and separated;

(ii) if η₀ satisfies equation (3.14), then s must be rational and η₀ is real. Moreover, there exist infinitely many zeros ofg(z)with degree at least two.

Proof. 1) Since g(z) is an entire function of exponential type, so whenK6= 1, we have

|g(x+iy)| → 1 2

(1 +K)e^x, as x→+∞,

|1−K|e^|x|, as x→ −∞. Hence, there existA, B andR such that

Ae^|x|≤ |g(x+iy)| ≤Be^|x|, when |x| ≥R.

2) (i) To consider the multiplicity of the zeros of g(z), we note that

g⁰(z) = sinhz+Kcoshz+ (2s−1)Kcosh(2s−1)z and

g⁰(z) +g(z) = (1 +K)e^z−Ke^(1−2s)z+sK[e^(1−2s)z+ e^−(1−2s)z].

Obviously, if s= ¹₂, g(z) +g⁰(z) = (1 +K)e^z 6= 0, then all zeros ofg(z) are simple and separated. If s6= ¹₂, denote byν the value of coshz+Ksinhz,i.e.,

ν:= coshz+Ksinhz theng(z) = 0 is equivalent to solving the following equations:

coshz+Ksinhz=ν, Ksinh(1−2s)z=ν.

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Since the first equation is equivalent to

e^z= ν±√

ν²+K²−1

1 +K ,

and the second equation is equivalent to

e^(1−2s)z= ν±√

ν²+K²

K ,

so g(z) = 0 has a solution inzis equivalent to the following equation having a solution inν:

ν±√

ν²+K²−1 1 +K

!_(1−2s)

= ν±√

ν²+K²

K ,

which is exactly (3.14). In fact, the solutionz and solutionν are related by

e^z=ν±√

ν²+K²−1 1 +K or

e^(1−2s)z= ν±√

ν²+K²

K ·

Now suppose thatν solves equation (3.14), theng(z) = 0 and g⁰(z) +g(z) =g⁰(z)

=

ν±√

ν²+K²−1 −

ν±√

ν²+K²

+ 2s h∓√

ν²+K² i

=±√

ν²+K²−1±(2s−1)√

ν²+K²

=±h√

ν²+K²−1−p

η₀²+K²−1

±(2s−1) √

ν²+K²−p

η₀²+K² i

(3.15)

because if we set

η₀²= 1

4s(1−s)−K², then

q

η²₀+K²−1±(2s−1) q

η₀²+K²= 0.

Thus, ifν6=η₀ theng⁰(z)6= 0, and hence all zeros of g(z) are simple.

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Let{z_n :n∈Z} be the set of zeros ofg(z), and denote

ν_n:= coshz_n+Ksinhz_n=Ksinh(1−2s)z_n.

Sincez_n lies in the strip−M ≤Rez≤M, so there is a positiveN such that|ν_n| ≤N. Letν be a cluster point of ν_n, then there is a subsequence ν_n_k ofν_n such thatν_n_k → ν. Soν satisfies the limiting equation. On the other hand, if η₀ does not satisfy equation (3.14) then eachz_n is a simple zero ofg(z). Sog⁰(z_n)6= 0. Hence inf_n∈Z|g⁰(z_n)|>0 because if otherwise, there exists a subsequence {z_n_k} such that lim_k→∞g⁰(z_n_k) = 0, then lim_k→∞ν_n_k =ν and (3.15) imply thatν=η₀, which is a contradiction. So the zeros ofg(z) are separated.

2) (ii) If η₀ solves equation (3.14), we know from the previous arguments that there exists a z₀ such that g(z₀) = 0, g⁰(z₀) = 0. Therefore





sinhz₀+Kcoshz₀=±p

η₀²+K²−1 =± ^(2s−1)

2√

s(1−s), Kcosh(2s−1)z₀=∓p

η²₀+K²=∓ ¹

2√

s(1−s)· (3.16)

Writez₀=x₀+iy₀. Notice that the right hands of (3.16) are real and we have (coshx₀+Ksinhx₀) siny₀= 0,

sinh(2s−1)x₀sin(2s−1)y₀= 0,

which imply

siny₀= 0, sin(2s−1)y₀= 0.

Thusy₀=nπand (2s−1)y₀=mπfor somen, m∈Z. This shows thatsis rational. In this case, we also have 0 =g(z₀) = [coshx₀+Ksinhx₀] cosy₀+Ksinh(2s−1)x₀cos(2s−1)y₀,

0 =g⁰(z₀) = [sinhx₀+Kcoshx₀] cosy₀+ (2s−1)Kcosh(2s−1)x₀cos(2s−1)y₀. Sinceg(x)>0 forx≥0, so we havex₀<0 and henceη₀is a real number.

Write s := ^q_p, we can take z_k := z₀ +i2kpπ and we have (2s−1)z_k = (2s−1)z₀ +ik(2q−p)π and g(z_k) =g⁰(z_k) = 0. Sog(z) has infinitely many zeros with degree at least two.

We note down a special case.

Corollary 3.4. If K >1 ands∈

12−¹₂q

1−_K¹2,¹₂ +¹₂ q

1−_K¹2

, then all the zeros ofg(z)are simple and separated.

Proof. Let η₀ be defined as in Theorem 3.3. Ifη₀ satisfies (3.14), then by Theorem 3.3,η₀ is a non-zero real number. This implies that _4s(1−s)¹ −K²>0, which is

s(1−s)< 1 4K² or

|s−1 2|> 1

2 r

1− 1 K²· Therefore

s∈ 0,1 2 −1

2 r

1− 1 K²

![ 1 2+1

2 r

1− 1 K²,1

!

becauseK >1 and the required assertion follows from Theorem 3.3.

Putting all the results together, we have the following conclusion onG(λ).

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Corollary 3.5. With (3.1) and letG(λ) :=Ab₁₁(λ)Ab₂₂(λ)(see (3.11,3.12)). Set ξ:=s`, K₁:= α

2κρ₁, K₂:= β

2EIρ₂, (3.17)

and assume that (1−K₁)6= 0 and(1−K₂)6= 0. Then

1) the zeros of G(λ) lie in a vertical strip centered at the imaginary axis and they are given by the roots of (3.5);

2) whens∈(0,1)is irrational, the zeros ofG(λ) are simple and separated;

3) if s ∈ (0,1) is rational, then the zeros of G(λ) are simple and separated if and only if η₀₁ and η₀₂ defined by

η₀₁:=± s

1

4s(1−s)−K₁², η₀₂:=± s

1

4s(1−s)−K₂² (3.18)

do not solve (3.14).

4. Riesz basis property of the generalized eigenfunction system of A

In this section, we discuss the Riesz basis property of the generalized eigenfunction system ofA. LetK_j and η_0j,j = 1,2, be defined by (3.17) and (3.18) respectively. Here we always assume that (1−K_j)6= 0, j= 1,2, and η_0j, j = 1,2, do not solve equation (3.14). With this assumption, the functions g₁(η) and g₂(ζ) defined in (3.12) always have infinitely many zeros and were denoted by{ζ_n⁽¹⁾:n∈Z} and{ζ_n⁽²⁾:n∈Z} respectively.

According to Theorem 3.2 and the Bari’s theorem (see [18], p. 45, Th. 15), if we can prove the Riesz basis property of the sequences {Ψ(ζn⁽¹⁾) :n∈Z} and {Φ(ζn⁽²⁾) :n∈Z}, then that for the eigenfunctions ofA will also follow. On this regard, we introduce a new norm inH:

kYk²₁:=

Z _`

0

κ|w⁰(x)|²+ρ|z(x)|²+EI|ϕ⁰(x)|²+I_ρ|ψ(x)|²

dx for Y = [w, z, ϕ, ψ]^T. Clearlyk · k1is equivalent to k · k. Let

H1:={Y = [w, z,0,0]^T ∈ H}, H2:={Y = [0,0, ϕ, ψ]^T ∈ H}·

Under the new norm, we have

H=H1⊕ H2,

and we treatg₁ andg₂separately inH₁andH₂ in the following two theorems:

Theorem 4.1. Assume thatK₁ 6= 1 and s∈(0,1) with ξ:=s` such that g₁(η) defined by (3.12) has simple zeros ζ_n⁽¹⁾, n∈Z. Then the sequence{Ψ(ζn⁽¹⁾) :n∈Z} forms a Riesz Basis inH1.

Proof. Let{ζn⁽¹⁾:n∈Z} be the set of zeros ofg₁(η) and Ψ(ζ_n⁽¹⁾) be defined by (3.7). Sinceg₁(iη) is an entire function of the sine type, according to Levin Theorem (cf.[18], p. 172, Th. 10), the set{e^ρ¹^ζ⁽¹⁾ⁿ ^x:n∈Z}forms a Riesz basis forL²[−`, `].

To show that{Ψ(ζ_n⁽¹⁾) :n∈Z}forms a basis inH1, we introduce some more auxiliary space and operator.

LetH :=V₀¹[0, `]×L²[0, `] and define its inner product by hY1, Y₂i:=

Z _`

0 κw⁰₁(x)w₂⁰(x)dx+ Z _`

0 ρz₁(x)z₂(x)dx