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Asymptotic Formulas for the Eigenvalues of the Timoshenko

Beam

Bruce Geist 1

Daimler Chrysler Corporation, 800 Chrysler DriveAuburn Hills, MI 48326; [email protected]

and

Joyce R. McLaughlin 1

Department of Mathematical Sciences Rensselaer Polytechnic Institute

Troy, NY 12180; [email protected]

Asymptotic formulas are derived for the eigenvalues of a free-ended Timo-shenko beam which has variable mass density and constant beam parametersotherwise. These asymptotic formulas show how the eigenvalues (and hencehow the natural frequencies) of such a beam depend on the material and geo-metric parameters which appear as coefficients in the Timoshenko differentialequations.

Key Words: eigenvalue asymptotics, Timoshenko beam models

Subject classifications:. 34L (ordinary differential operators), 73D (wave

propagation in and vibration of solids)

1. INTRODUCTION

Suppose a structural beam is driven by a laterally oscillating sinusoidal

force. As the frequency of this applied force is varied, the response varies.

Experimental frequencies for which the response is maximized are called

1Acknowledgment: The work of both authors was completed at Rensselaer Polytech-nic Institute, and was partially supported by funding from the Office of Naval Research,grant number N00014-96-1-0349. The work of the first author was also partially sup-ported by the Department of Education fellowship grant number 6-28069. The workof the second author was also partially supported by the National Science Foundation,

grant number DMS-9802309.

1

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2 GEIST AND MCLAUGHLIN

natural frequencies of the beam. Our goal is to address the question: if

a beam’s natural frequencies are known, what can be inferred about its

bending stiffnesses or its mass density? To answer this question we need

to know asymptotic formulas for these frequencies. Here we establish such

formulas for beams with variable mass density but otherwise constant beam

parameters. We make this assumption as a first step toward solving the

problem with both variable density and variable stiffness. We also make

this assumption because it is consistent with some applications of interest

to us. An example of an application consistent with our assumption is

an aircraft wing with struts which have been added so that there is an

appreciable change in the density and a minimal change in stiffness.

One widely used mathematical model for describing the transverse vibra-

tion of beams was developed by Stephen Timoshenko in the 1920s. This

model is chosen because it is a more accurate model than the Euler-Bernouli

beam model and because systems of Timoshenko beam models are used to

model aircraft wings. The mathematical equations that arise are two cou-pled partial differential equations,

(EI ψx)x + kAG(wx − ψ) − ρIψtt = 0,

(kAG(wx − ψ))x − ρAwtt = P (x, t).

The dependent variable w = w(x, t) represents the lateral displacement

at time t of a cross section located x units from one end of the beam.

ψ = ψ(x, t) is the cross sectional rotation due to bending. E is Young’s

modulus, i.e., the modulus of elasticity in tension and compression, and Gis the modulus of elasticity in shear. The non-uniform distribution of shear

stress over a cross section depends on cross sectional shape. The coefficient

k is introduced to account for this geometry dependent distribution of

shearing stress. I and A represent cross sectional inertia and area, ρ is the

mass density of the beam per unit length, and P (x, t) is an applied force. If

we suppose the beam is anchored so that the so called “free-free” boundary

conditions hold (i.e., shearing forces and moments are assumed to be zero

at each end of the beam), then w and ψ must satisfy the following four

boundary conditions,

wx − ψ|x=0,L = 0, ψx|x=0,L = 0. (1)

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EIGENVALUES OF THE TIMOSHENKO BEAM 3

After making a standard separation of variables argument, one finds that

the Timoshenko differential equations for w and ψ lead to a coupled system

of two second order ordinary differential equations for y(x) and ψ(x),

(EI Ψx)x + kAG(yx − Ψ) + p2IρΨ = 0, (2)

(kAG(yx − Ψ))x + p2Aρy = 0. (3)

Here, p2 is an eigenvalue parameter. The conditions on w and ψ in (1)

imply y and Ψ must satisfy the same free-free boundary conditions. We

must have

yx − Ψ|x=0,L = 0, Ψx|x=0,L = 0. (4)

This boundary value problem for y and Ψ is self-adjoint, which implies

that the values of p2 for which nontrivial solutions to this problem exist;

the eigenvalues for this model, are real. Furthermore, it is not difficult to

show that the collection of all eigenvalues for this problem forms a discrete,

countable, unbounded set of real non-negative numbers. Moreover, it can

be shown that if σ is a natural frequency for a beam, then p2 = (2πσ)2

is one of the beam’s eigenvalues. Therefore, it is possible to determine

eigenvalues from natural frequency data obtained in an experiment like the

one indicated in the opening paragraph.

Suppose from vibration experiments we have determined a set of natural

frequencies for a beam with unknown elastic moduli and mass density, andhave constructed a sequence of eigenvalues from this data. What informa-

tion can the eigenvalues provide about these unknown material parameters?

To address this question, we must determine how eigenvalues depend on

E , I , kG, A and ρ. This determination is not easy, since the dependence

of eigenvalues on these coefficients is highly nonlinear. Another difficulty

arises because the Timoshenko boundary value problem involves two sec-

ond order differential equations. When the coefficients in these differential

equations are non-constant, the system of two second order equations can-

not be transformed into a single fourth order equation. Therefore, to make

progress in the case where coefficients are non-constant, the boundary-value

problem must be handled as a system of equations.

For a simpler, Sturm-Liouville type boundary value problem,

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y(z) + (λ − q(z))y(z) = 0, 0 ≤ z ≤ 1, (5)

y(0) − hy(0) = y(1) + Hy(1) = 0,

it is known that for square integrable q(z), nontrivial solutions y(z) for this

problem exist if and only if λ = µn, where

µn = n2π2 + C q − 10

q(z)cos(2nπz)dz + αn, (6)

C q = 2h + 2H +

10

q(z)dz,∞

n=1α2n < ∞,

(See Hald-McLaughlin [12, pp. 313-314] as well as Isaacson-Trubowitz

[14], Borg [2], and Fulton-Pruess [6, 7]; for other Sturm-Liouville equations

with, effectively, less smooth coefficients, see Coleman-McLaughlin [5].)

The importance of equation (6) is that it shows how the eigenvalues for

the Sturm-Liouville problem are related to the coefficient q(z) appearing

in (5). Algorithms for reconstructing q from spectral data rely strongly on

asymptotic formulas like the one given in (6). (For example, see Hald [11]

and Rundell-Sacks [18].)

Returning now to the Timoshenko beam equations, we ask the question:

what information is contained in the eigenvalues for the Timoshenko beam?

Given a sequence of eigenvalues, can we infer knowledge about the beam

parameters which give rise to these eigenvalues? Asymptotic formulas forthe Sturm-Liouville eigenvalues are critical to determining q from spectral

data. We expect that analogous formulas for the Timoshenko eigenvalues

will play a key role in recovering beam parameters like E , kG, or ρ from

such data. In this paper, our objective is to determine asymptotic formu-

las for the eigenvalues of the Timoshenko beam when free-free boundary

conditions are enforced and when ρ is allowed to vary. We suppose that E ,

kG, A and I are constants and assume ρ is a positive function of x on [0, L]

such that 0 = ρx(0) = ρx(L) and ρxx is in L∞(0, L). Under these assump-

tions, we derive asymptotic formulas for the eigenvalues of the free-free

Timoshenko beam.

In the next three sections, approximations are derived (accurate to within

an error that is O(1/p)) for the square roots of eigenvalues of free-free beams

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with variable density. An important step in deriving these preliminary ap-

proximations of the eigenvalues is the use of a transformation (see section

1) which changes the Timoshenko differential equations (2) and (3) into a

new pair of differential equations, where in the new equations, the coeffi-

cient which contains the eigenvalue parameter p2 no longer depends on the

independent variable. The key feature of the transformed system is that

the largest terms in the new differential equations, and hence the most im-

portant terms, are multiplied by coefficients which do not depend on the

new independent variable. As the eigenvalue parameter grows, solutions to

the transformed differential equations approach the solutions of a certain

set of constant coefficient differential equations. It is therefore possible to

derive an approximate solution (accurate to within an O(1/p) error) to an

initial value problem in which initial conditions are chosen so that the left

transformed boundary conditions are enforced. In section 2, this initial

value problem is presented and its approximate solution is derived. By

applying the two remaining transformed right boundary conditions to theapproximate solution of the initial value problem, a frequency equation is

determined. In section 3, estimates of square-roots of eigenvalues are made

from this frequency equation. These estimates appear in Theorem 4.3.

Our approach to deriving the final asymptotic formulas (section 4) is

built from the following idea. Suppose ρ0 ≡ ( 1L L0

ρ1/2(x)dx)2, and let

ρ(x; t) ≡ ρ0+ tρ(x), where ρ = ρ−ρ0 and t is an auxiliary parameter which

we allow to vary from 0 to 1. In the Timoshenko differential equations, let

ρ(x) be replaced by ρ(x; t). Define ´ p2 to be an eigenvalue for a free-free

beam with mass density ρ and constant material and geometric parameters

otherwise. When t = 0, ρ = ρ0 and ´ p2 is an eigenvalue for a beam where E ,

I , kG, A, and ρ = ρ0 are all independent of x. As t increases to 1, ρ goes

to ρ(x), and ´ p2

changes continuously in t to an eigenvalue for a beam withvariable density ρ(x). Let L1 =

L0

ρ1/2(x; t)dx, and define µ2 ≡ L21´ p2. We

show that there is a function G such that

d(µ2)

dt= G(Y , Φ), (7)

where (Y , Φ) is a transformed eigenfunction pair corresponding to the

eigenvalue ´ p2 of a free-free beam with material parameters E , I , kG, A,

and ρ. Integrating (7) formally with respect to t from 0 to 1, we find that

´ p2

|t=1 − ´ p2

|t=0 =

1

L1 10 G(

ˆY ,

ˆΦ)dt. (8)

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The term ´ p2|t=0 is an eigenvalue for a beam where E , I , kG, A and

ρ = ρ0 are independent of x; i.e. ´ p2|t=0 represents an eigenvalue for a

uniform beam. Asymptotic formulas for the eigenvalues of free-free and

clamped-clamped uniform beams are derived in Geist [8] and published in

[10]. From these uniform beam formulas, an asymptotic approximation for

the term ´ p2|t=0 can be made. The final asymptotic eigenvalue formulas

for beams with variable density ρ(x) are obtained by replacing the term1L1

10

G(Y , Φ)dt with an asymptotic approximation derived below, and by

replacing ´ p2|t=0 with the appropriate uniform beam eigenvalue formulas

given in [10].

The function G depends only on E , I , kG, A, ρ and the transformed

eigenfunctions Φ and Y . Approximations to the square roots of eigenvalues

given in Theorem 4.3 allows us to determine Φ and Y and hence G to

within an error that is O(1/p). Then (8) is used to sharpen our estimates

of the eigenvalues for the Timoshenko beam. From (8) we compute the

final asymptotic formulas, which are given in Theorem 5.2. Note that theadvantage of this method over say a variational method is that we can

determine more than the first term in the eigenvalue expansion and prove

a bound in the remainder no matter how large the difference is between ρ

and ρ0.

2. THE TRANSFORMED PROBLEM

To begin, the free-free Timoshenko boundary value problem is proved

equivalent to a certain transformed boundary value problem derived below.

This equivalency holds when ρ depends on x; all other beam parameters

are assumed constant. A key feature of the transformed problem is that

the coefficient containing the eigenvalue parameter no longer depends onthe independent variable.

To derive this equivalent problem, a lemma is proved which applies to

single second order equations. We will use this lemma to prove Theorem

2.1, in which the Timoshenko system of differential equations is transformed

to a new pair of equations.

Lemma 2.1. Suppose ρ(x) is positive for all x ∈ [0, L] and ρxx(x) ∈L2[0, L]. Let L1 =

L0

ρ1/2(x1)dx1 and z(x) = 1L1

x0

ρ1/2(x2)dx2. Let β ,

α, I , and p2 be constants. Then v(x) = ρ−1/4V (z(x)) satisfies the equation

αv + ( p2τ ρ − β )v = f (9)

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if and only if V (z) satisfies

V zz + (ρ−1/4)xxL2

1

ρ3/4+

τ L21 p

2

α− βL2

1

αρ V =f L2

1

αρ3/4. (10)

proof: Suppose v = A(x)V (z(x)), where A(x) and z(x) are as yet un-

specified smooth functions of x. Then

αv = αA(z)2V zz +(αA2z)

AV z + αAV,

and provided α(z)2A = 0 for x ∈ [0, L], it follows that αv+ ( p2τ ρ−β )v =

f (x) if and only if

V zz +(A2z)

A2(z)2V z +

A

A(z)2+

p2τ ρ

α(z)2− β

α(z)2 V =f

α(z)2A.

Now let A = ρ−1/4(x) and z(x) = 1L1

x0

ρ1/2(x2)dx2. Then (A2z) ≡ 0,

A/A(z)2 = (ρ−1/4)xxL21/ρ3/4, p2τ ρ/(α(z)2) = τ L2

1 p2/α, and β/(α(z)2) =

βL21/αρ, so equation (9) becomes equation (10). ✷

In the next theorem, new differential equations are determined that are

related to the Timoshenko equations by the transformation indicated in

the previous lemma. Let L1 and z(x) be defined as they are in Lemma

2.1, and let µ2E = p2L2

1/E , µ2kG = p2L2

1/kG, and γ = kAG/EI . Since

ρ(x) is positive for all x ∈ [0, L], z(x) is an invertible function. Let x(z)

denote the inverse of z(x), and let ρ3 = L21[ρ−1/4(x(z))]xx/[ρ3/4(x(z))] and

ρ4 = ρx(x(z))L21/[4ρ2(x(z))].

Theorem 2.1. Let E , kG, I , A, and p2

all be constants, let ρ(x) bepositive for all x ∈ [0, L], and let ρxx(x) ∈ L2(0, L). Then

y(x) = ρ−1/4Y (z(x)) and Ψ(x) = ρ−1/4Φ(z(x))

satisfy the equations

EI Ψxx + ( p2Iρ − kAG)Ψ = −kAGyx + F 1, (11)

and

kAGyxx + p2Aρy = kAGΨx + F 2 (12)

if and only if Φ and Y satisfy

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Φzz +p2L2

1

E Φ − L2

1γ

ρΦ + ρ3Φ − γρ4Y +

γL1

ρ1/2Y z =

F 1L21

EI ρ3/4, (13)

and

Y zz +p2L2

1

kGY + ρ3Y + ρ4Φ − L1

ρ1/2Φz =

F 2L21

kAGρ3/4. (14)

proof: This Theorem is an immediate consequence of Lemma 2.1. ✷

Theorem 2.1 shows how to transform the original Timoshenko differential

equations into new equations so that in the new equations, coefficients

involving p2 are constant with respect to the new independent variable

z. Equations (11) - (14) include generic “right hand side” terms so that

Theorem 2.1 is general enough that it applies to differential equations that

arise in the next section. In the case where F 1 = F 2 ≡ 0, equations (11)

and (12) are the homogeneous Timoshenko differential equations (2) and

(3).

Theorem 2.2. Suppose ρ(x) is a positive function of x such that ρxx ∈L2(0, L) and ρx(0) = ρx(L) = 0. Let E , kG, A, I and p2 be positive

constants. Then nontrivial Y and Φ are solutions to the boundary value

problem

Φzz +p2L2

1

E Φ − L2

1γ

ρΦ + ρ3Φ − γρ4Y +

γL1

ρ1/2Y z = 0, (15)

Y zz +p2L2

1

kGY + ρ3Y + ρ4Φ − L1

ρ1/2Φz = 0, (16)

Y z − L1

ρ1/2Φ

z=0,1

= 0, and Φz|z=0,1 = 0, (17)

if and only if y(x) = ρ−1/4Y (z(x)) and Ψ(x) = ρ−1/4Φ(z(x)) are nontrivial

solutions to the free-free Timoshenko boundary value problem (2) - (4).

The value p2 is an eigenvalue for the free-free Timoshenko boundary value

problem if and only if there exist nontrivial functions Y and Φ which satisfy

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the differential equations (15) and (16) and the boundary conditions given

in (17).

proof: This theorem is an easy consequence of Theorem 2.1 and thefact that when ρx|x=0,L = 0, (yx − ψ)|x=0,L = 0 if and only if (Y z −L1Φ/ρ1/2)|z=0,1 = 0 and ψx|x=0,L = 0 if and only if Ψz|z=0,1 = 0. ✷

3. A FREQUENCY EQUATION

Consider the following initial value problem. Let α = L1/ρ1/2(0). Sup-

pose that Y and Φ satisfy differential equations (15) and (16) and that

for real c and d, they also satisfy Y (0) = d, Φ(0) = c/α, Y z(0) = c, and

Φz(0) = 0. These initial conditions ensure that the boundary conditions

on Y and Φ at z = 0 given in (17) are satisfied no matter how c and d are

chosen. They are the least restrictive initial conditions on Y (z) and Φ(z)

which enforce the transformed boundary conditions at the left boundarypoint z = 0. Furthermore, if nontrivial c and d can be chosen so that the

boundary conditions at z = 1 are satisfied, then by definition L21 p

2 is an

eigenvalue for the transformed boundary value problem, and by Theorem

2.2, p2 is an eigenvalue for the free-free Timoshenko boundary value prob-

lem. In the next two lemmas, integral equations for Y and Φ are derived

which are equivalent to the initial value problem discussed above. These in-

tegral equations are used to determine approximate solutions to the above

initial value problem. The approximate solutions to the initial value prob-

lem make possible estimates of the values of p for which nontrivial solutions

exist to the transformed boundary value problem.

Lemma 3.1. Let µ, c and d be fixed constants, and let q(z) and f (z) beintegrable. Then w(z) is the solution to the initial value problem

w + µ2w = q(z)w + f (z),w(0) = c, and w(0) = d,

(18)

if and only if w satisfies the integral equation

w(z) = csin(µz)

µ+ d cos µz +

z0

sin[µ(z − t)]

µ[q(t)w(t) + f (t)]dt. (19)

proof: The elementary proof is based on the well known technique of

variation of parameters, and is omitted. ✷

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Theorem 3.1. Suppose ρ(x) is a positive function of x such that ρxx ∈L2(0, L) and ρx(0) = ρx(L) = 0. Let E , kG, A, I and p2 be positive con-

stants, let α = L1/ρ1/2(0) and let c and d be arbitrary but fixed constants.

Then Φ(z) and Y (z) satisfy the differential equations (15) and (16) and

the initial conditions

Y (0) = d, Φ(0) = c/α, Y z(0) = c, Φz(0) = 0 (20)

if and only if Φ and Y also satisfy the integral equations

Φ =c

αcos µEz + z

0

sin[µE(z − t)]

µE

L21γ

ρ− ρ3

Φ + γρ4Y − γL1

ρ1/2Y t

dt (21)

and

Y =c sin(µkGz)

µkG+

d cos(µkGz) +

z0

sin[µkG(z − t)]

µkG

−ρ3Y − ρ4Φ +

L1

ρ1/2Φt

dt (22)

proof: First observe that since ρxx ∈ L2(0, L), Theorem 2.1 shows that Y

and Φ satisfy (15) and (16) if and only if y(x) = ρ−1/4Y (z(x)) and Ψ(x) =

ρ−1/4Φ(z(x)) satisfy the homogeneous Timoshenko differential equations.

All coefficients in the Timoshenko equations (2) and (3) are continuously

differentiable. Since yx and Ψx must be continuous, it follows that Y z andΨz are also continuous.

Next, apply Lemma 3.1 to the transformed differential equation (15).

Observe that

L21γ/ρ − ρ3

∈ L2(0, 1) and that (from the discussion of

the previous paragraph) (γρ4Y − [γL1/ρ1/2]Y z) ∈ L2(0, 1). Lemma 3.1

shows that the differential equation (15) and initial conditions (20) are

satisfied if and only if integral equation (20) holds. Similarly, since −ρ3and −ρ4Φ + [L1/ρ1/2]Φz are in L2(0, 1), Lemma 3.1 shows that (16) and

(20) hold if and only if (21) holds. ✷

We will use the integral equations of Theorem 3.1 to determine approxi-

mate solutions to the initial value problem given in (15), (16), and (20). If

c and d are allowed to vary over

, the solution to this initial value problem

generates every solution to differential equations (15) and (16) that satisfies

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the left boundary conditions at z = 0. Consider the boundary terms

Y z(1) − L1

ρ1/2(1)Φ(1)

µkGand

αΦz(1)

µE, (23)

where Y and Φ are solutions to the initial value problem (15), (16), and

(20). Setting the above expressions equal to zero leads to a homogeneous

linear system for the arbitrary constants c and d. Values of the eigenvalue

parameter p which make possible nontrivial choices for c and d correspond

to the square roots of eigenvalues for the free-free Timoshenko beam. The

homogeneous linear system in c and d has nontrivial solutions if and only

if the determinant of the corresponding coefficient matrix is zero. This

determinant will define a frequency function. The objective in this section

is to determine this frequency function from estimates of the coefficients of

c and d in the expressions given in (23).

The following technical fact is used many times in the estimates that

follow.

Lemma 3.2. Suppose that f z ∈ L∞(0, 1), and that δ is a real constant.

Then for z ∈ [0, 1], z0

sin[µ(z − t) + δ]f (t)dt

<f ∞ + f z∞

µ.

proof: z0

sin[µ(z − t) + δ]f (t)dt =

cos[µ(z − t) + δ]µ

f (t)z

0

− z0

cos[µ(z − t) + δ]µ

f t(t)dt

This implies the result. ✷

Lemma 3.3. Suppose h and δ are real constants and that h is not equal

to 0 or L1/√

kG. Let dg/dz and ρxx(x(z)) ∈ L∞(0, 1), and let ρ(x) > 0

when x ∈ [0, L]. Then z0

sin[hµ(z − t) + δ]

µg(t)Y tdt

< O(1/µ)Y ∞ + O(1/µ)Φ∞+ O(1/µ)|c| + O(1/µ)|d|.

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proof:

Y z = c cos(µkGz) − dµkG sin(µkGz) +

z0

cos[µkG(z − t)]−ρ3(t)Y (t) − ρ4(t)Φ(t) + L1

ρ1/2(t)Φt(t)

dt

⇒ z0


µg(t)Y tdt =

z0


µg(t){c cos(µkGt) − dµkG sin(µkGt)

+

t0

cos[µkG(t − s)][−ρ3(s)Y (s) − ρ4(s)Φ(s)]ds}

dt

+ sin[hµ(z − t) + δ]µ

g(t)

t

0

cos[µkG(t − s)] L1

ρ1/2(s)Φs(s)ds

dt.

Since by hypothesis hµ = µkG, a double angle formula, the assumption that

dg/dz ∈ L∞(0, 1), and Lemma 3.2 can be used to show that the absolute

value of the first integral on the right of the above equation is bounded

above by

O(1/µ)Y ∞ + O(1/µ)Φ∞ + O(1/µ)|c| + O(1/µ)|d|.

To demonstrate a similar result for the second integral on the right of

the above equation, first note that

t

0

cos[µkG(t − s)]L1

ρ1/2(s)Φs(s)ds =

L1Φ(t)

ρ1/2(t)− L1Φ(0)

ρ1/2(0)cos(µkGt)

− t0

µkG sin[µkG(t − s)]

L1

ρ1/2(s)+ cos[µkG(t − s)

L1

ρ1/2(s)

s

Φds.

Therefore, if we show that

z0

sin[hµ(z

−t) + δ]

µ g(t) t

0 µkG sin[µkG(t − s)]

L1

ρ1/2(s) Φ(s)ds

dt < O

1

µ

,

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then the Lemma follows. After changing the order of integration, the inte-

gral above may be rewritten as

zs=0

zt=s

sin[hµ(z

−t) + δ]

µ g(t)µkG sin[µkG(t − s)]

L1

ρ1/2(s) Φ(s)dtds.

Again, using a double angle formula and the fact that g(z) ∈ L∞(0, 1),

we apply Lemma 3.2 to find that the above integral is bounded in absolute

value by a function of the form O(1/µ)Φ∞. ✷

Lemma 3.4. Suppose h and δ are real constants and that h is not equal

to 0 or L1/√

E . Let dg/dz and ρxx(x(z)) ∈ L∞(0, 1), and let ρ(x) > 0

when x ∈ [0, L]. Then

z0


µg(t)Φtdt

< O(1/µ)Y ∞ + O(1/µ)Φ∞

+ O(1/µ)|c| + O(1/µ)|d|.

proof: The proof of Lemma 3.4 is similar to the proof of Lemma 3.3,

and is therefore omitted. ✷

In the next Theorem, we show that the infinity norms of Y and Φ remain

finite as p approaches infinity.

Lemma 3.5. Suppose E = kG, that ρxx ∈ L∞(0, L), and that ρ(x) > 0

for all x ∈ [0, 1]. Then

Φ

∞ ≤O(1)

|c

|+ O(1/p)

|d

|and

Y ∞ ≤ O(1/p)|c| + O(1)|d|.

proof: The integral equations for Φ and Y together with Lemma 3.3

and Lemma 3.4 imply that

Φ∞ ≤ O(1)|c| + O(1/p)|d| + O(1/p)Φ∞ + O(1/p)Y ∞ (24)

and

Y ∞ ≤ O(1/p)|c| + O(1)|d| + O(1/p)Φ∞ + O(1/p)Y ∞. (25)

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Inequality (24) implies that

(1 − O(1/p))Φ∞ ≤ O(1)|c| + O(1/p)|d| + O(1/p)Y ∞.

For p large enough, this inequality implies that

Φ∞ ≤ O(1)|c| + O(1/p)|d| + O(1/p)Y ∞. (26)

Similarly, from inequality (25), we find that for p large enough

Y ∞ ≤ O(1/p)|c| + O(1)|d| + O(1/p)Φ∞. (27)

The Theorem follows from inequalities (26) and (27). ✷

In the next Theorem, we calculate estimates of the coefficients of c and

d in the functions and (Y z − ΦL1/ρ1/2(z))/µkG and αΦz/µE .

Lemma 3.6. Suppose E = kG, that ρxx

∈L∞

(0, L) and ρ(x) > 0 for all

x ∈ [0, L], and that α = L1/ρ1/2(0). Then

Y z(z) − L1

ρ1/2(z)Φ(z)

µkG− c

cos(µkGz) − ρ1/2(0)ρ1/2(z)

cos(µEz)

µkG+ d sin(µkGz)

∞

and Φz(z)

µE+

c

αsin(µEz)

∞

≤ O(1/p)|c| + O(1/p)|d|.

proof: Integral equations for Y z/µkG and Φz/µE and can be determined

from the integral equations for Y and Φ. The proof follows from the integral

equations for Y and Φ and Lemmas 3.3, 3.4, and 3.5. ✷Lemma 3.6 facilitates the derivation of a frequency equation for the free-

free Timoshenko beam with variable density ρ(x).

Theorem 3.2. Let E , kG, A and I be positive constants such that E =kG. Let ρ(x) > 0 for all x ∈ [0, L], let ρx(0) = ρx(L) = 0, and suppose

ρxx(x) ∈ L∞(0, L). Then p2 is an eigenvalue for the free-free Timoshenko

beam if and only if

F ( p) ≡ sin µkG sin µE + sin(µE)δ1,2 + sin(µkG)δ2,1 + δ = 0, (28)

where the functions δ1,2( p), and δ2,1( p) are O(1/p) and δ( p) is O(1/p2).

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proof: We seek to determine the values of p2 for which there exist non-

trivial functions Φ and Y that solve the transformed differential equations

and all transformed boundary conditions, including those at z = 1. To de-

termine all such solutions, we seek nontrivial solutions to the initial value

problem in (15), (16), and (20) which also solve the transformed boundary

conditions at z = 1. Theorem 2.2 shows that the values of p2 which admit

nontrivial solutions when boundary conditions at z = 1 are imposed are

the eigenvalues of the free-free Timoshenko beam.

Lemma 3.6 implies that for any choice of c and d, solutions Y and Φ to

the initial value problem (15), (16), and (20) satisfy

Y z(1) − [L1/ρ1/2(1)] Φ(1)

µkG= δ1,1c + [sin(µkG) − δ1,2]d (29)

and

αΦz

µE = [− sin µE − δ2,1]c − δ2,2 d (30)

where the δi,j are all O(1/p) functions. Equations (29) and (30) imply that

the right boundary conditions, i.e., the conditions in (20) when z = 1, may

be satisfied if and only if δ1,1 − sin µkG − δ1,2

− sin µE − δ2,1 −δ2,2

cd

=

00

.

This linear system can be non-trivially solved if and only if

sin µkG sin µE + sin(µE)δ1,2 + sin(µkG)δ2,1 + O(1/p2) = 0. ✷

4. ROOTS OF THE FREQUENCY FUNCTION

In this section, four results are presented. Theorem 4.1 shows that

all roots of the frequency equation F must occur near roots of sin(µE) ·sin(µkG). Theorem 4.2 shows that near each root of sin(µE)·sin(µkG) which

is isolated from neighboring roots, there must exist at least one root of F .

In Lemma 4.1, an approximation for ∂F/∂p is calculated. Theorems 4.1,

4.2 and Lemma 4.1 facilitate the proof of Theorem 4.3, in which it is shown

that exactly one root of F occurs near isolated roots of sin(µE) · sin(µkG).

Theorem 4.1. Let

F ( p) ≡ sin µkG sin µE + sin(µE)δ1,2 + sin(µkG)δ2,1 + δ

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where δ1,2 and δ2,1 are O(1/p) and δ is O(1/p2). Let c = max {√E,√

kg},

and suppose p is a root of the frequency function F ( p). Then there exists

a root of the function sin(µkG) sin(µE), say ˆ p, such that p − ˆ p = p, where

| p| <c

L1sin−1

|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

.

(Thus, | p| is O(1/p).)

proof: If p is a root of the frequency equation F , then

sin

pL1√

kG

sin

pL1√

E

= − sin

pL1√

E

δ1,2 − sin

pL1√

kG

δ2,1 − δ.(31)

If either sin( pL1/√

kG) or sin( pL1/√

E ) are zero, then the result follows.

Suppose neither sin( pL1/

√kG) nor sin( pL1/

√E ) is zero. Assume for nowthat sin( pL1/

√E )

sin( pL1/√

kG)

≤ 1, (32)

and divide both sides of equation (31) above by sin( pL1/√

kG). It follows

that

| sin( pL1/√

E )| =

δ1,2 sin( pL1/√

E )

sin( pL1/√

kG)+ δ2,1 +

δ

sin( pL1/√

kG)

≤ |δ1,2| + |δ2,1| + |δ|| sin( pL1/√

E )|

⇒ sin2( pL1/√

E ) < (|δ1,2| + |δ2,1|) | sin( pL1/√

E )| + |δ|.Let s = | sin( pL1/

√E )| , b = |δ1,2| + |δ2,1|, and c = |δ|. Then s > 0 and

s2 − bs − c < 0. This implies that

0 < s <b +

√b2 + 4c

2,

and hence that,

| sin( pL1/√E )| < |δ1,2

|+

|δ2,1

|2 + (|δ1,2| +

|δ2,1

|)2

4 + |δ|. (33)

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Let M be an integer chosen such that |[L1/√

E ] p − Mπ| ≤ π/2, and let

p = p − [√

E/L1]M π. Then from inequality (33) it follows that

sin

pL1

√E

=sin

L1 p

√E

= sinL1

| p

|√E

<|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

⇒ | p| <

√E

L1sin−1

|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

.

Thus, there is some integer M such that p = p + [√

E/L1]Mπ where

| p| <

√E

L1sin−1

|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

.

The argument above was carried out under the assumption that (32) holds.

If instead | sin( pL1/√

E )/ sin( pL1/√

kG)| > 1, the same argument given

above holds in this case provided E and kG are interchanged. In either case,

it follows that there exists a root of sin(µkG) sin(µE), say ˆ p =√

EMπ/L1

or√

kGMπ/L1, such that p − ˆ p = p and

| p| <c

L1sin−1

|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

. ✷

Theorem 4.2. Suppose ˆ p and ˜ p are two adjacent roots of the function

sin(µkG) ·sin(µE) and that ˆ p is a simple root not closer to any other root of

this function than it is to ˜ p. Let c = max{√

E,√

kG}

, c = min{√

E,√

kG}and let ¯ p be the largest root of sin(µkG) sin(µE) which satisfies ¯ p < min{ˆ p, ˜ p}.

Let

∆ p = sin−1

c

c

π

2sup p>¯ p|δ1,2( p)| + sup p>¯ p|δ2,1( p)| + sup p>¯ p

|δ|sin(¯ p−1)

c

L1,

and suppose that |ˆ p − ˜ p| > 2∆ p + cL1

1¯ p Then there is at least one root of the

characteristic equation (28) in the interval [ˆ p − ∆ p, ˆ p + ∆ p]. Furthermore,

this interval must contain an odd number of roots of the characteristic

equation.

Remark: The quantity ∆ p defined above is O(1/¯ p), since all of the δ s

appearing in its definition are O(1/¯ p). Thus, the hypothesis above requires

that ˆ p and ˜ p be separated by a distance which is O(1/¯ p).

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proof sketch: The proof follows from the basic observation that sin(µkG)·sin(µE) is strictly monotonic near its isolated roots, and that F = sin(µkG)·sin(µE) + O(1/p). For ˆ p large enough, F ( p) must change sign near where

sin(µkG) sin(µE) changes sign, i.e. in the interval [ ˆ p−

∆ p, ˆ p + ∆ p]. See

Geist [8, pages 164-168] for a detailed presentation of this proof. ✷

An estimate of ∂F/∂p derived in the next lemma is an important step

toward proving that for each isolated root of sin(µE) · sin(µkG), there is

exactly one zero of F nearby.

Lemma 4.1. Let F be the frequency function defined in (28). Then

∂F

∂p=

L1√E

cos(µE) · sin(µkG) +L1√kG

cos(µkG) · sin(µE) + r( p), (34)

where r( p) = O(1/p).

proof sketch: If δ, δ1,2 and δ2,1 were all known exactly, it might be possi-

ble to calculate ∂F/∂p by direct differentiation. Unfortunately, only orderestimates for the δ s are known; δ, δ1,2 and δ2,1 are not known explicitly.

However, ∂ F /∂p can be estimated using the formula

−∂F

∂p= det

coef.of c

Y z(1)− L1

ρ1/2(1)Φ(1)

µkG

,

coef.of d

Y z(1)− L1

ρ1/2(1)Φ(1)

µkG

coef.of c

∂ ∂p

αΦz(1)µE

,

coef.of d

∂ ∂p

αΦz(1)µE

+ det

coef.of c

∂ ∂p

Y z(1)− L1

ρ1/2(1)Φ(1)

µkG

,

coef.of d

∂ ∂p

Y z(1)− L1

ρ1/2(1)Φ(1)

µkG

coef.of c

αΦz(1)µE

,

coef.of d

αΦz(1)µE

,(35)

where in the expression above each entry in each matrix is a coefficient

of either c or d. Let “·” denote differentiation with respect to p. Integral

equations for Y z, Φ, and Φz can be determined from the integral equa-

tions for Y and Φ. (It follows from Theorem 2.2 that Y and Φ, which

satisfy the initial value problem (15), (16), and (20) may be written as

Y (z) = [ρ(x(z))]1/4y(x(z)) and Φ(z) = [ρ(x(z))]1/4Ψ(x(z)), where y(x)

and Ψ(x) satisfy the original Timoshenko differential equations (2) and

(3) and the initial conditions y(0) = ρ−1/4(0)d, yx(0) = [ρ1/4(0)/L1]c,

Ψ(0) = [ρ1/4(0)/L1]c, and Ψx(0) = 0. For each x

∈[0, L], y, yx, Ψ and

Ψx are analytic functions of p. See [4, page 37]. This implies that for each

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z ∈ [0, 1], Y z, Φz and Φ are analytic functions of p, and so differentiation

with respect to the eigenvalue parameter p is allowed.) Estimates for Y z,

Φ, and Φz can be obtained by first showing that the infinity norms of these

functions are bounded. Once this is demonstrated, one can generate esti-

mates for Y z, Φ, and Φz accurate to within an error that is O(1/p) in the

same way that estimates for Y , Y z, and Φz were produced in section 3.

Then, estimates for ∂F/∂p can be calculated using equation (35). For the

details of this calculation, see Geist [8, pages 173 -189]. ✷

We focus now on proving that for each isolated root of sin(µE) ·sin(µkG),

there is exactly one zero of F nearby. In particular, we demonstrate that

|∂F/∂p| > 0 near roots of sin(µE) · sin(µkG) that are not too close to one

another.

Theorem 4.3. Suppose ˆ p and ˜ p are two adjacent roots of the function

sin(µkG)·sin(µE). Suppose ˆ p is a simple root of sin(µkG)·sin(µE) not closer

to any other root of this function than it is to ˜ p. Let c = max

{

√E,

√kG

},

c = min{√E, √kG} and let ¯ p be the largest root of sin(µkG) · sin(µE) which satisfies ¯ p < min{ˆ p, ˜ p}. Let

∆ p = sin−1

c

c

π

2sup p>¯ p|δ1,2( p)| + sup p>¯ p|δ2,1( p)| + sup p>¯ p

|δ|sin(¯ p−1)

c

L1,

Γ = sup p>¯ p−∆ p

c

L1sin−1

|δ1,2| + |δ2,1|

2+

(|δ1,2| + |δ2,1|)2

4+ |δ|

,(36)

and

r( p) =∂F

∂p −

L1

√E

cos(µE)

·sin(µkG)

−

L1

√kG

cos(µkG)

·sin(µE).

Let ˆ p be large enough so that Γ < cπ/[4L1]. If for p ∈ [ˆ p − Γ, ˆ p + Γ]

|ˆ p − ˜ p| >

max

2∆ p +

c

L1

1

¯ p,

c

L1

π

2

tan

L1Γ

c

+

√EkG

L21

π

2

|r( p)|cos(ΓL1/c)

+ Γ

,

(37)

then there is exactly one root of the characteristic function F in the [ˆ p −Γ, ˆ p + Γ].

Remark: r is O(1/p) and Γ and ∆ p are O(1/¯ p).

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proof: Theorem 4.1 demonstrates that all roots of F occur an O(1/p)

distance from the zeros of sin(µkG) sin(µE). In particular, it follows from

Theorem 4.1 that any root of F nearer to ˆ p than it is to any other root of

sin(µkG) sin(µE) must occur in the narrow interval [ˆ p−

Γ, ˆ p + Γ]. Theorem

4.2 guarantees that in this interval, there is at least one root of F . We

prove now that |∂F/∂p| > 0 throughout [ˆ p − Γ, ˆ p + Γ] when the hypotheses

of this theorem hold.

Lemma 4.1 shows that

∂F

∂p=

L1√E

cos(µE) · sin(µkG) +L1√kG

cos(µkG) · sin(µE) + r( p),

where r( p) is an O(1/p) function. Let p ∈ [ˆ p − Γ, ˆ p + Γ]. Without loss of

generality, suppose ˆ p is a zero of sin(µE). Hypothesis (37) implies that

2L21

π√EkG

[

|ˆ p

−˜ p

| −Γ] >

L1

√kGtan L1

√E

Γ +|r( p)|

cos

L1√ E Γ

. (38)

Let pkG be a zero of sin(µkG) at least as near to p as is any other root of

sin(µE). This implies that |( p − pkG)L1/√

kG| ≤ π/2, and hence that

L1√E

| sin(µkG)| =L1√

E

sin

L1√kG

( p − pkG)

≥ L1√E

2

π

L1√kG

| p − pkG|

≥ 2L21

π√

EkG|ˆ p − pkG + p − ˆ p|

≥2L2

1

π√EkG [|ˆ p − pkG| − | p − ˆ p|]

≥ 2L21

π√

EkG[|ˆ p − ˜ p| − Γ]. (39)

p ∈ [ˆ p − Γ, ˆ p + Γ], ˆ p a zero of tan(µE), and ΓL1/√

E < π/4 imply that

L1√kG

tan

L1√

E Γ

+|r( p)|

cos

L1√ E

Γ =

L1

√kGtan

L1

√E

(ˆ p

±Γ) +

|r( p)|

cos L

1√ E (ˆ p ± Γ) ≥

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L1√kG

| tan µE| +|r( p)|

| cos µE| ≥ L1√kG

| tan µE|| cos µkG| +r( p)

| cos µE | . (40)

Therefore, from (38), (39) and (40) it follows that

L1√E

| sin µkG| >L1√kG

| tan µE cos µkG| +|r( p)|

| cos µE |

⇒ L1√

E sin µkG cos µE

>

L1√kG

sin µE cos µkG

+ |r( p)|

⇒∂F

∂p( p)

> 0,

as desired.

Thus, |∂F/∂p| > 0 for all p ∈ [ˆ p − Γ, ˆ p + Γ], which in turn implies

that there is at most one root of F in this interval. Since it has beenestablished that there is at least one root in this interval, the Theorem

follows when ˆ p is a zero of sin(µE). The proof for the case where ˆ p is a

zero of sin(µkG) is identical to the proof above, except that the role of E

and kG are interchanged. ✷

5. THE ASYMPTOTIC FORMULAS

We carry out the following steps in order to sharpen our estimates of the

eigenvalues. First, we define a new density function ρ in terms of ρ. Let

ρ0 = [ 1L L0

ρ1/2(x)dx]2 and ρ = ρ(x) − ρ0; then define ρ = ρ0 + tρ, where t

is an auxiliary parameter which is allowed to vary from 0 to 1. Note that

when t = 0, ρ is the constant ρ0, and when t = 1, ρ ≡ ρ(x). Let Φ and Y besolutions to the boundary value problem given in (15-17), except z, ρ, ρ3, ρ4and L1 are replaced by z, ρ, ρ3, ρ4, and L1, respectively, where z, ρ3, ρ4 and

the constant L1 are defined as are z, ρ3, ρ4 and L1 except that ρ replaces

ρ in their definitions. The resulting boundary value problem for Φ and

Y corresponds to the transformed boundary value problem (see Theorem

2.2) one obtains from the original Timoshenko differential equations and

boundary conditions when ρ(x) is taken to be ρ(x; t).

It will prove useful to view the transformed differential equations for Φ

and Y as a single vector equation. Define

ˆU (z) =

Φ(z)

Y (z)

.

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Let

B1 = 1E 00 1

kG , Q =

−L2

1γρ + ρ3 −γ ρ4

ρ4 ρ3

(41)

S =

0 γ

−1 0

, and µ2 = L2

1 p2. (42)

Then the boundary value problem (15-17), where ρ is now replaced by ρ,

may be written in vector notation as

U zz + µ2B1U + QU +L1

ρ1/2S U z =

00

(43)

Y z − L1

ρ1/2Φ

z=0,1

= 0, and Φz|z=0,1 = 0, (44)

Suppose the transformed eigenvalue parameter, µ2 = L21 p

2, is chosen so

that a nontrivial solution U (z) exists to the above boundary value problem.

Then by Theorem 2.2, if u is defined as

u(x) ≡

Ψ(x)y(x)

= ρ−1/4U (z(x)) = ρ−1/4(x)

Φ(z(x))

Y (z(x))

,

then u must satisfy

∂ ∂x EI ∂ ∂x + ( p2I ρ − kAG) kAG ∂

∂x∂ ∂xkAG ∂

∂xkAG ∂ ∂x + p2Aρ

u(x) = 0 (45)

[yx − Ψ]|x=0,L = 0, Ψ|x=0,L = 0. (46)

Thus, if L21 p

2 = µ2 is an eigenvalue for the transformed boundary value

problem (43-44) and U (z) is the corresponding eigenfunction, then p2 is an

eigenvalue and u(x) = ρ−1/4U (z(x)) is the corresponding eigenfunction for

the free-free Timoshenko beam with density ρ.

Our method for sharpening the estimates of the eigenvalues of the Tim-

oshenko beam will rely on determining the derivative of µ2 with respect to

the auxiliary parameter t, introduced in the definition of ρ. The approach

may be summarized as follows. Let µ2 be an eigenvalue for the trans-

formed boundary value problem (43-44) and let U (z) be a corresponding

eigenfunction. The differential equation and boundary conditions for U (z)

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are differentiated with respect to ρ in the direction of ρ. This differen-

tiation of (43-44) will lead to a new, non-homogeneous boundary value

problem. The right hand side of this new boundary value problem will

contain dρµ2[ρ], which we will show is equal to dµ2/dt. Since the new

boundary value problem comes from differentiating the differential equa-

tions and boundary conditions (43-44), a boundary value problem with

a known solution U (z) ≡/ 0, it follows that a nontrivial solution to the

new, non-homogeneous problem exists, and must be equal to dρµ2[ρ]. Us-

ing Theorem 2.1, the new non-homogeneous boundary value problem for

dρµ2[ρ] may be written as a non-homogeneous Timoshenko boundary value

problem which must also have a nontrivial solution. The fact that the

non-homogeneous Timoshenko boundary value problem has a nontrivial

solution implies that the right hand side of the non-homogeneous problem

must satisfy a certain orthogonality requirement (see Lemma 5.1 below).

This orthogonality condition allows us to determine dµ2/dt = dρµ2[ρ] in

terms of the transformed eigenfunctionˆU (z); i.e., from the orthogonalitycondition it follows that

d(µ2)

dt= G(U (z(x))).

Integrating both sides of the above equation gives

µ2|t=1 − µ2|t=0 =

10

G(U (z(x)))dt. (47)

Sharp estimates of µ2|t=0 can be obtained by applying formulas from Geist-

McLaughlin [10]. In particular, when t = 0, ρ is ρ0, a constant. Therefore

µ

2

|t=0 may be determined from the formulas for the eigenvalues of theuniform Timoshenko beam. From results in the previous five sections,

G(U (z(x))) can be determined to within an error that converges to zero as

µ (and p) get large. Thus, equation (47) provides a means for obtaining

sharp estimates for µ2 when t = 1. From these estimates of µ2, we will

obtain asymptotic formulas for the free-free Timoshenko eigenvalues when

mass density is ρ(x).

Our next goal then is to calculate an expression for dµ2/dt. Suppose

ρxx ∈ L∞[0, L]. For these calculations, we first assume that for all such ρ,

i) dρµ2[ρ] exists, that

ii) dρU [ρ], dρ(U z)[ρ], dρ(U zz)[ρ] all exist, and that

iii) (dρU [ρ])zz = dρ(U zz)[ρ] and (dρU [ρ])z = dρ(U z)[ρ].

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Later, we will give conditions which guarantee that assumptions i), ii) and

iii) are valid. As indicated in the discussion above, we formally differentiate

the differential equations and boundary conditions in (43-44) with respect

to ρ in the direction of ρ. We use assumptions i), ii), and iii) from above to

obtain the differential equation and boundary conditions that dρU [ρ] must

satisfy. We find that

(dρU [ρ])zz + µ2B1(dρU [ρ]) + Q(dρU [ρ]) +L1

ρ1/2S (dρU [ρ])z =

−(dρµ2[ρ])B1U − (dρQ[ρ])U − dρ(L1/ρ1/2)[ρ]S U z,

and that

[(dρY [ρ])z − (L1/ρ1/2)(dρΦ[ρ]) − (dρ(L1/ρ1/2)[ρ])Φ]|z=0,1 = 0,

(dρΦ[ρ])|z=0,1 = 0.

When dρµ2[ρ] exists, it follows that

dρµ2[ρ] = l i m→0

µ2(ρ + ρ) − µ2(ρ)

= lim→0

µ2(ρ0 + (t + )ρ) − µ2(ρ0 + tρ)

=dµ2

dt.

Similarly, from the definitions of Q and L1/ρ1/2, it follows that dρQ[ρ] =

∂Q/∂t and dρ(L1/ρ1/2)[ρ] = ∂ ∂t (L1/ρ1/2). Define

B2 =

1/EI 0

0 1/kAG

(48)

and

F = − ρ3/4

L2B−12

d

dtµ2

B1U +

∂

∂tQ

U +

∂

∂t

L1

ρ1/2

S U z

. (49)

Now define

¯V (z) ≡ dρ

ˆU [ρ] =

dρΦ[ρ]

dρY [ρ] ≡ Φ(z)

Y (z)

.

8/7/2019 asymtotic



Then V satisfies

V zz + µ2B1V + QV +L1

ρ1/2S V z =

L21

ρ3/4B2F ,

Y z − L1

ρ1/2Φ − ∂

∂t

L1

ρ1/2

z=0,1

= 0, (50)

and

Φz|z=0,1 = 0. (51)

Let v be defined as

v(x) ≡

Ψ(x)y(x)

≡ ρ−1/4(x)V (z(x)).

From Theorem 2.1, we know that v must satisfy the vector differential

equation

∂ ∂x EI ∂ ∂x + ( p2I ρ − kAG) kAG ∂

∂x∂ ∂xkAG ∂

∂xkAG ∂ ∂x + p2Aρ

v(x) = F . (52)

The boundary conditions (50) and (51) may also be rewritten in terms

of Ψ and y. By multiplying (50) through by ρ1/4/L1 and noting that

ρx(0) = ρx(L) = 0, we find that

Y z − L1

ρ1/2Φ − ∂

∂t L1

ρ1/2 Φz=0,1

= 0

⇔

yx(x) − Ψ(x) − ρ1/4

L1

∂

∂t

L1

ρ1/2

Φ(z(x))

x=0,L

= 0 (53)

Similarly, from (51) we have Φz|z=0,1 = 0

⇔ Ψx|x=0,L = 0. (54)

In the next lemma, we show that if the boundary value problem for v(x) is

to have a solution when p2 is an eigenvalue for the boundary value problem

(45)-(46), then the right hand side of the differential equation (52) must

satisfy a certain orthogonality condition.

8/7/2019 asymtotic



Lemma 5.1. Let E , I , kG and A be positive constants, and let ρ be a

positive function such that ρxx ∈ L∞(0, L). Suppose v =

Ψ(x)y(x)

satisfies

the non-homogeneous differential equations

EI Ψxx + kAG(yx − Ψ) + p2I ρΨ = F 1, (55)

and

kAG(yx − Ψ)x + p2Aρy = F 2, (56)

and boundary conditions (53)-(54). Suppose u(x) =

Ψ(x)y(x)

= ρ−1/4U (z(x))

solves the homogeneous Timoshenko boundary value problem given in (45)

and (46). Then F =

F 1F 2

must satisfy

(F (x), u(x)) = (F , ρ−1/4U (z(x))) =kAG

L1

∂

∂t

L1

ρ1/2

ΦY

1

z=0

.

(In the last equation, (·, ·) denotes the standard inner product in L2(0, L).)

proof:

(F (x), u(x)) =

L0

(EI Ψxxψ + kAG(yx − Ψ)ψ + p2I ρΨψ

+ kAG(yx − Ψ)x)y + p2Aρyy)dx

=

L0

(EI Ψψxx + kAG(yx − Ψ)ψ − kAG(yx − Ψ)yx

+ p2I ρΨψ + p2Aρyy)dx + kAG(yx − Ψ)y|Lz=0

=

L0

[EI Ψψxx − kAG(yx − ψ)(yx − Ψ) + p2I ρΨψ + p2Aρyy]dx

+kAG(yx − Ψ)y|Lx=0

= L0 {[EI

ˆψxx + kAG(yx −

ˆψ) + p

2

I ρˆ

ψ]¯Ψ

8/7/2019 asymtotic



+[kAG(yx − ψ)x + p2Aρy]y}dx + kAG(yx − Ψ)y|Lx=0

=ρ1/4

L1

kAG∂

∂t L1

ρ1/2 Φρ−1/4Y

1

z=0

.

⇒ (F (x), u(x)) =kAG

L1

∂

∂t

L1

ρ1/2

ΦY

1

z=0

. ✷

Returning now to (52)-(54), assumptions i) - iii) above imply that

v = ρ−1/4V (z(x)) = ρ−1/4(dρU [ρ](z)) ≡/ 0

is a nontrivial solution to this boundary value problem. Lemma 5.1 implies

that

⇒ (F (x), ρ−1/4U (x)) = kAGL1

∂ ∂t

L1

ρ1/2

ΦY 1

z=0

. (57)

Equation (57) holds if and only if

− ρ3/4

L2B−12

d

dtµ2

B1U +

∂

∂tQ

U +

∂

∂t

L1

ρ1/2

S U z

, ρ−1/4U

=kAG

L1

∂

∂t

L1

ρ1/2

ΦY

1

z=0

⇔ ddtµ2 =

−ρ1/2

L1B−12 ( ∂

∂tQ)U,U −ρ1/2

L1

∂∂t

L1

ρ1/2

B−12 SU z,U

−kAG ∂

∂t

L1

ρ1/2

ΦY

1z=0

ρ1/2

L1B−12 B1U,U

.

The inner products above involve an integration with respect to x. Sinceρ1/2

L1dx = dz, all of the above inner products may be rewritten as integra-

tions with respect to z. Thus,

ddtµ2 =

− 1

0U T B−12 ( ∂

∂tQ)Udz− 1

0

∂∂t

L1

ρ1/2

U T B−12 SU zdz−kAG ∂

∂t

L1

ρ1/2

ΦY

1z=0

1

0U T B−12 B1Udz

Next we show that the third term in the numerator above, the boundary

term, may be combined with the second term so that all terms in the

8/7/2019 asymtotic



numerator are integrals. From the definitions of B2 and S , we find that

1

0

∂

∂t L1

ρ1/2 U T B−12 SU zdz = kAG

1

0

∂

∂t L1

ρ1/2 (Y zΦ − ΦzY )dz

= 2kAG

10

∂

∂t

L1

ρ1/2

Y zΦdz − kAG

10

(ΦY )z∂

∂t

L1

ρ1/2

dz

= 2kAG

10

∂

∂t

L1

ρ1/2

Y zΦdz −kAGY Φ

∂

∂t

L1

ρ1/2

1

0

+ kAG

10

ΦY ∂ 2

∂t∂ z

L1

ρ1/2

dz

⇒ −

1

0

∂

∂t

L1

ρ1/2

U T B−1

2 S U zdz − kAGY Φ∂

∂t

L1

ρ1/2

1

0

=

−2kAG

10

∂

∂t

L1

ρ1/2

Y zΦdz − kAG

10

ΦY ∂ 2

∂t∂ z

L1

ρ1/2

dz.

Therefore,

d

dtµ2 =

−

1

0U T B−12 ( ∂

∂tQ)Udz−2kAG

1

0

∂∂t

L1

ρ1/2Y zΦdz−kAG

1

0ΦY ∂2

∂t∂z

L1

ρ1/2dz

10ˆU T

B

−1

2 B1ˆUdz

.(58)

The formula above for dµ2/dt holds so long as assumptions i) - iii) hold. In

the following theorem, conditions are given which guarantee the validity of

these assumptions, and hence show when dµ2/dt may be calculated using

the last formula given above.

Theorem 5.1. Suppose ρ(x) is a positive function such that ρx(0) =

ρx(L) = 0 and ρxx(x) ∈ L∞(0, L). Let ρ0 = [(1/L) L0

ρ1/2(x)dx]2, ρ ≡ρ(x) − ρ0, and ρ(x, t) = ρ0 + tρ(x). Let L1 =

L0

ρ1/2(x; t)dx, and let

µ2(t) = L21´ p2 be an eigenvalue for the transformed boundary value problem

(43-44). Then dµ2/dt satisfies (7), where U is an eigenfunction of (43-

44) corresponding to the eigenvalue µ2, and Q, B1 and B2 are the 2

×2

matrices defined in (41) and in (48).

8/7/2019 asymtotic



proof: The proof follows from the discussion preceding this theorem,

provided we show assumptions i) - iii) are valid. We observe that u(x) =

ρ−1/4(x)U (z(x)), where u(x) satisfies (45) and (46). The differential equa-

tion for the vector u(x) may be written as a system of four first order linear

scalar equations in which the parameter t appears linearly. This implies

that for each x, u, ux, and uxx must be analytic in t. See Coddington and

Levinson [4, page 37]. This in turn implies that for each z, U (z), U z(z) ,

and U zz(z) must also be analytic in t when t ∈ [0, 1]. When z = 0 and t

is any value between 0 and 1, both components of U (0) cannot simultane-

ously be zero. (If they were both zero, boundary conditions at z = 0 would

imply U (z) must be identically zero.) From the scalar differential equa-

tions for the components of U (z), we conclude that µ2 must be analytic

in t. We have shown already that dρµ2[ρ] = dµ2/dt; hence, assumption i),

that dρµ2[ρ] exits, is valid.

Next, we observe that dρU [ρ] = ∂ ∂t U , dρU z[ρ] = ∂

∂t U z, and dρU zz[ρ] =∂

∂tÛ zz. From this we conclude that dρ

Û [ρ], dρ

Û z[ρ], and dρ

Û zz[ρ] all exist,since U , U z, and U zz are all analytic in t. This proves assumption ii).

To demonstrate that assumption iii) holds under the stated hypotheses,

observe that the components of U satisfy the integral equations (20) and

(21) when ρ is taken to be ρ. When ρxx(x) is continuous, it follows from

these integral equations that U zzt and U tzz are continuous functions of z

and t. This implies that when ρ(x) ∈ C 2[0, L], U zzt = U tzz. Furthermore,

from the integral equations, it follows that U zzt and U tzz are continuous in ρ

with respect to the standard Sobelev norm of order 2. They are continuous

maps which take ρ(x) ∈ H 2(0, L) to an element of L2(0, 1). Since C 2[0, L] is

dense in H 2(0, L), it follows that U zzt = U tzz when ρ(x) ∈ H 2(0, L). Since

H 2(0, L) contains the set of all ρ such that ρxx ∈ L∞(0, L), assumption iii)

must hold when ρxx ∈ L∞(0, L), as desired. ✷In the next three lemmas, our goal is to determine the constants c and d

which appear in the integral equations (20) and (21). Theorem 4.1 shows

that if p2 is a large enough eigenvalue, then p must lie near a root of

sin(L1 p/√

kG) · sin(L1 p/√

E ). We will show that provided p is not near

more than one root of sin(L1 p/√

kG) · sin(L1 p/√

E ), then p must be a

simple eigenvalue, and the vector ( c, d )T must be a multiple of either

( 1, O(1/p) )T or ( O(1/p), 1 )T , depending upon whether p is near a zero

of sin(L1 p/√

kG) or of sin(L1 p/√

E ). This result will be used to calculate

an estimate of dµ2/dt.

In the next Lemma, we show that if ˆ p is large enough and satisfies either

(59) or (60) below, then the square root of the nearest eigenvalue to ˆ p is

well separated from square roots of other eigenvalues.

8/7/2019 asymtotic



Lemma 5.2. Let e ∈ (0, 1) and suppose ˆ p is a root of the function

sin(L1 p/√

kG) · sin(L1 p/√

E ) such that either

sin(L1ˆ p/√

kG) = 0 and

|sin(L1ˆ p/

√E )

|> e (59)

or

sin(L1ˆ p/√

E ) = 0 and | sin(L1ˆ p/√

kG)| > e. (60)

Then there exists an M > 0 such that when ˆ p > M , exactly one eigenvalue

of the free-free Timoshenko beam with density ρ, say ´ p2, is such that its

square root ´ p lies in the interval

ˆ p − e√ c

2L1, ˆ p +

e√ c

2L1

. Furthermore, |´ p− ˆ p| <

O(1/´ p).

proof: It is not difficult to show that when that when (59) or (60) hold,

and when ˜ p is a root of sin(L1 p/

√kG)

·sin(L

1 p/

√E ) as near to ˆ p as is any

other root of this function, then |ˆ p − ˜ p| > e√c/L1, where c = min E,kG.

Let Γ be defined as it is in the hypothesis of Theorem 4.3. Let M be chosen

large enough so that ˆ p > M implies that hypothesis (37) of Theorem 4.3

is satisfied and that |Γ| < e√

c/[2L1]. Theorem 4.1 shows that if there is

a root ´ p of the frequency equation such that ´ p ∈

ˆ p − e√ c

2L1, ˆ p +

e√ c

2L1

, then

´ p must be contained in the narrower interval (ˆ p − Γ, ˆ p + Γ). On the other

hand, Theorem 4.3 shows that there is exactly one zero of the frequency

function in the interval (ˆ p−Γ, ˆ p +Γ). Together, Theorem 4.1 and Theorem

4.3 imply that there is exactly one root of the frequency function in the

interval

ˆ p − e√ c

2L1, ˆ p +

e√ c

2L1

. ✷

Lemma 5.3. Let e ∈ (0, 1), and suppose ˆ p is a root of sin(L1 p/√kG·sin(L1 p/√E )such that either (59) or (60) holds. For ˆ p large enough, let ´ p2 be the unique

eigenvalue of the free-free beam with density ρ whose square root ´ p is con-

tained in the interval

ˆ p − e√ c

2L1, ˆ p +

e√ c

2L1

. If (59) holds, then for the eigen-

value ´ p2, the vector ( c, d )T (where c and d are the constants appearing

in (20) and (21)) must satisfy cd

= constant

O(1/´ p)

1

.

If (60) holds,

c

d

= constant 1

O(1/´ p)

.

8/7/2019 asymtotic



In either case, whether (59) holds or (60) holds, ´ p2 must be a simple eigen-

value.

proof: ´ p2 is an eigenvalue if and only if nontrivial solutions exist to alinear system of the form

00

=

O(1/´ p) − sin( ´ pL1√

kG) − O(1/´ p)

− sin( ´ pL1√ E

) − O(1/´ p) −O(1/´ p)

cd

(61)

(see Theorem 2.15). The conclusion is immediate. ✷

Our next goal is to use formula (58) in Theorem 5.1 to calculate an

estimate for dµ2/dt. We make an estimate dµ2/dt in the case where ´ p

lies near a root ˆ p of sin(L1 p/√

kG) · sin(L1 p/√

E ) which for an arbitrary

but fixed e ∈ (0, 1) satisfies either (59) or (60). A ´ p which meets this

criterion gives rise to an eigenvalue which we will refer to as being “well-

separated” from its neighboring eigenvalues. The estimate of dµ2/dt is used

to calculate asymptotic formulas for these eigenvalues.

In the next lemma, we calculate estimates for one of the terms appearing

in the numerator of the right hand side of (58).

Lemma 5.4. Let U and µ2 = L21´ p2 be an eigenfunction and corre-

sponding eigenvalue for the transformed boundary value problem (43-44).

Then ´ p2 is an eigenvalue for the free-free Timoshenko beam with density

ρ. Let e be a fixed but arbitrary number in (0, 1), and let ˆ p be a root of

sin(L1 p/√

kG) · sin(L1 p/√

E ) as near to ´ p as is any other root of this func-

tion. If ˆ p satisfies either (59) or (60), then

−2kAG

10

∂

∂t

L1

ρ1/2

Y sΦds =

c

α

2− d2

A

I

kAGEkG − 1

10

L1

ρ1/2

∂

∂t

L1

ρ1/2

ds + O(1/´ p).

proof: Theorem 2.2 shows that ´ p2 is an eigenvalue for the free-free

Timoshenko beam when L21´ p2 is an eigenvalue for the transformed problem.

Let µE = L1´ p√ E

and µkG = L1´ p√ kG

. Then Theorem 3.1 shows that

Y z = c cos(µkGz) − dµkG sin(µkGz)+

8/7/2019 asymtotic



z0

cos[µkG(z − s)]

−ρ3(s)Y (s) − ρ4(s)Φ(s) +

L1

ρ1/2(s)Φs(s)

ds

and

Φ =c cos(µE z)

α+

z0

sin[µE(z − s)]

µE

·

−ρ3(s)Φ(s) +L21γ

ρ(s)Φ(s) + γ ρ4(s)Y (s) − γ L1

ρ1/2(s)Y (s)

ds.

Let f (z) ≡ 2kAG ∂ ∂t

L1

ρ1/2

. Then

10

f (z)Y z(z)Φ(z)dz =

10

f

c2

αcos(µkGz)cos(µE z) − cd

αµkG sin(µkGz)cos(µE z)

dz

−dµkG

µE

10

f (z)sin(µkGz)

z0

sin[µE(z−s)]

−ρ3Φ +

L21γ

ρΦ + γ ρ4Y − L1γ

ρ1/2Y s

ds dz

10

f (z)c

αcos(µE z)

z0

cos[µkG(z−s)]

−ρ3Y − ρ4Φ +

L1

ρ1/2Φs

ds dz+O

1

´ p

.

By Lemma 5.3, when ˆ p satisfies (59) or (60), then cdα µkG and c

α must both

be O(1) or smaller. Since µE

= µkG by assumption (otherwise neither (59)

nor (60) could hold), the first integral on the right of the above equationmust be O(1/´ p).

Let

I 1 = −dµkG

µE

10

f (z)sin(µkGz)

· z

0

sin[µE(z − s)]

−ρ3Φ +

L21γ

ρΦ + γ ρ4Y − γ L1

ρ1/2Y s

ds

dz.

By switching the order of integration, we find that

I 1 = −d

µkG

µE 10 1s f (z)sin(µkGz)sin[µE(z − s)]

8/7/2019 asymtotic



·

−ρ3(s)Φ(s) +L21γ

ρ(s)Φ(s) + γ ρ4(s)Y (s) − γ L1

ρ1/2(s)Y s(s)

dz ds

= −d µkG

µE

10

−ρ3Φ + L21γ ρ

Φ + γ ρ4Y − γ L1

ρ1/2Y s

· 1

s

f (z)

2{cos[(µkG − µE)z + µEs] − cos[(µkG + µE)z − µEs]}dz

ds.

Now f ∈ L∞(0, 1) because ρxx ∈ L∞(0, L). Therefore we may integrate

the square bracketed term above by parts with respect to z, and apply

Lemmas 3.2, 3.3, 3.4, and 3.5 to show that

I 1 = −dµkG

µE

10

−ρ3Φ +

L21γ

ρΦ + γ ρ4Y − γ L1

ρ1/2Y s

·f

2

1

µkG + µE− 1

µkG − µE

sin(µkGs)ds + O(1/´ p)

= −dµkG

µE

10

γ L1

ρ1/2f

2

2µE

µ2kG − µ2

E

sin(µkGs)Y sds + O(1/´ p).

From the integral equation for Y z, it follows that Y z = −dµkG sin(µkGz) +

O(1). Substituting this expression for Y z into the last expression for I 1, we

find that

I 1 = d2µkG

µE 1

0

γ L1

ρ1/2f

µkGµE

µ2kG

−µ2E

sin2(µkGs)ds + O(1/´ p)

= d2kAGA

I

1

EkG − 1

10

L1

ρ1/2∂

∂t

L1

ρ1/2

ds + O(1/´ p).

Using a similar argument as was used to derive the estimate of I 1, it is

possible to show that

I 2 =

10

f (z)c

αcos(µE z)

z0

cos[µkG(z−s)]

−ρ3Y − ρ4Φ +

L1

ρ1/2Φs

ds dz

=− c

α2 kAG 1

EkG − 1

1

0

L1

ρ1/2

∂

∂t L1

ρ1/2 ds + O(1/´ p).

8/7/2019 asymtotic



Since 10

2kAG∂

∂t

L1

ρ1/2

Y sΦds = I 1 + I 2 + O(1/´ p),

the lemma follows. ✷

In the next theorem, we make estimates for transformed eigenfunctions

U (z) associated with well separated eigenvalues for the untransformed Tim-

oshenko beam with density ρ. We will use these estimates for U to calculate

an estimate for dµ2/dt.

Lemma 5.5. Suppose ´ p is an eigenvalue for the free-free Timoshenko

beam with density ρ = ρ0 + tρ(x), where ρ0 = ([1/L] L0

ρ1/2(x)dx)2 and

ρ(x) = ρ(x)−ρ0. Suppose ρ(x) is a positive function on [0, L], that ρxx(x) ∈L∞(0, L), and that ρx(0) = ρx(L) = 0. Assume also that E , I , kG, and

A are positive constants. Let ˆ p be a root of sin(L1 p/√

kG) · sin(L1 p/√

E )

as near to ´ p as is any other root of this function. Let e be an arbitrary but

fixed constant between 0 and 1. If (59) holds, then for some integer n, ˆ p =nπ

√kG/L1, and µ2 = L2

1´ p2, U (z) = ( cos(nπz) + O(1/n), O(1/n) )T is

an eigenvalue-eigenfunction pair for (43-44). Similarly, if ˆ p satisfies (60),

then for some integer n, ˆ p = nπ√

E/L1 and an eigenvalue-eigenfunction

pair for (43-44) is µ2 = L21´ p2, U (z) = ( O(1/n), cos(nπz) + O(1/n) )T .

proof: If µE = L1´ p√ E

and µkG = L1´ p√ kG

, then Theorem 3.1 shows that for a

nontrivial choice of the constants c and d and for α = L1/ρ1/2(0),

Φ =c cos(µE z)

α

z0

sin[µE(z − s)]

µE

−ρ3Φ +

L21γ

ρ+ γ ρ4Y − γ L1

ρ1/2Y s

ds (62)

and

Y =c sin(µkGz)

µkG+ d cos(µkGz)

z0

sin[µkG(z − s)]

µkG

−ρ3Y − ρ4Φ +

L1

ρ1/2Φs

ds. (63)

For fixed c and d, Lemmas 3.3, 3.4 and 3.5 show that all terms in (62) and

(63) under the integral sign are O(1/´ p). If (59) holds, then by Lemma 5.3

and Theorem 4.1 we find that ´ p = ˆ p+O(1/ˆ p) = nπL1/√kG+O(1/n). This

8/7/2019 asymtotic



implies that U (z) = ( O(1/n), cos(nπz) + O(1/n) )T . Similarly, when (60)

holds, U (z) = ( cos(nπz) + O(1/n), O(1/n) )T . ✷

The next Lemma gives an estimate of dµ2/dt.

Lemma 5.6. Suppose e is an arbitrary but fixed real number between 0 and 1. Suppose ´ p2 is an eigenvalue for the free-free Timoshenko beam

with density ρ(x; t) and that ˆ p is a root of the function sin(L1 p/√

kG) ·sin(L1 p/

√E ) nearest ´ p. Let µ2 = L2

1´ p2. If (59) holds, then ˆ p = nπ√ kG

L1 for

some integer n, and

dµ2

dt= −kG

10

{(ρ3)t[1 + cos(2nπz)]}dz

− 2kAG

I EkG − 1

10

L1

ρ1/2∂

∂t

L1

ρ1/2

dz + O(1/n). (64)

On the other hand, if (60) holds, then ˆ p = nπ√ EL1

for some integer n, and

dµ2

dt= E

10

L21γ

ρ− ρ3

t

− cos(2nπz)(ρ3)t

dz

+2kAG

I

EkG − 1

10

L1

ρ1/2∂

∂t

L1

ρ1/2

dz + O(1/n). (65)

proof: The proof follows from Theorem 5.1 and Lemmas 5.4 and 5.5. ✷

Finally, for eigenvalues which are well separated from neighboring eigen-

values, we prove the following asymptotic formulas.

Theorem 5.2. Let A, I , E and kG be positive constants. Suppose that

ρ(x) > 0 for all x ∈ [0, L], that ρx(0) = ρx(L) = 0, and that ρxx ∈ L∞[0, L].

Let ρ0 = ([1/L] L0

ρ1/2(x)dx)2, and define

E ( p) = sin

pLρ1/20√

E

sin

pLρ1/20√

kG

.

Let e be a fixed but arbitrary real number between 0 and 1, and let

pni ≡

niπ√

E

Lρ1/20

, i = 1, , 2, . . . ,

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be a sequence of roots of E ( p) such that

sin pniLρ

1/20√

kG > e.

Suppose each pni is large enough that exactly one simple eigenvalue ˇ p of

the Timoshenko beam lies in the interval pni −

√c

2Lρ1/20

e, pni +

√c

2Lρ1/20

e

,

where c = min{E, kG}. Then

ˇ p2 =n2i π2E

L2ρ0− E

L2ρ0

10

ρ3(x) cos(2niπz)dz + C E + O(1/ni), (66)

where x(z) is the inverse of the function z(x) = [1/L1] x0 ρ

1/2

(s)ds and

C E =A

I

kG

ρ0

E

E − kG

10

ρ0

ρ(x)− 1

dz

− E

L2ρ0

10

ρ3(x)dz +A

I

kG

ρ0

1 − kG + E

2(E − kG)

. (67)

Let

pmi ≡ miπ√

kG

Lρ1/20

, i = 1, , 2, . . . ,

be a sequence of roots of E ( p) such that sin

pmi

Lρ1/20√

E

> e.

Suppose each pmiis large enough that there is exactly one simple eigenvalue

ˇ p2 of the Timoshenko beam satisfying

ˇ p ∈ pmi −

√c

2Lρ1/20

e, pmi +

√c

2Lρ1/20

e

.

In this case,

ˇ p2

=

m2i π2kG

L2ρ0 −kG

L2ρ0 10 ρ3(x) cos(2miπz)dz + C kG + O(1/mi), (68)

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where

C kG = −A

I

kG

ρ0

kG

E − kG

10

ρ0

ρ(x)− 1

dz

− kG

L2ρ0

10

ρ3(x)dz +A

I

kG

ρ0

1 +

kG + E

2(E − kG)

. (69)

proof: To prove formula (66) let µ2 = L21´ p2 where ´ p2 is an eigenvalue for

a beam with density ρ(x; t). Define ˇ p2 = ´ p2|t=1 and ¯ p2 = ´ p2|t=0. Because

L21|t=1 = L2

1|t=0 = L2ρ0 = L21, as t changes from 0 to 1, µ2/[L2ρ0] will

change continuously in t from ¯ p2, an eigenvalue for a uniform beam with

constant density ρ0, to ρ2, and eigenvalue for a beam with density ρ(x).

Furthermore, Lemma 5.6 shows that dµ2/dt satisfies (65). This implies

that

L2ρ0

(ˇ p2

−¯ p2) = E 1

0L2

1γ 1

ρ −1

ρ0−

ρ3 −

cos(2nπz)ρ3 dz

+L2kAG

I

EkG − 1

10

ρ0ρ

− 1

dz + O(1/ni) (70)

where L2ρ0ˇ p2 = µ2|t=1 is an eigenvalue for the transformed problem with

density ρ(x) and L2ρ0¯ p2 = µ2|t=0 is an eigenvalue for the transformed

problem with constant density ρ0. Dividing through by L2ρ0 = L21, we find

that

ˇ p2 − ¯ p2 = E

10

γ

1

ρ− 1

ρ0

dz − E

L2ρ0

10

ρ3dz

+ kAGI

EkG − 1

10

1ρ

− 1ρ0

dz − E

L2ρ0

10

cos(2niπz)ρ3dz + O

1ni

(71)

From Geist-McLaughlin [10], it follows that

¯ p2 =n2i π2E

L2ρ0+

1 − 1

2

kG + E

kG − E

A

I

kG

ρ0+ O

1

ni

.

Formula (66) follows from the last equation and (71).

A similar argument proves formulas (68) and (69). For eigenvalues that

are close to the pmis, dµ2/dt satisfies (64) (instead of (65)), and the formula

for ¯ p2 becomes (see [10])

¯ p2

=

m2i π2kG

L2ρ0 +

1 +

1

2

kG + E

kG − E A

I

kG

ρ0 + O 1

mi

.✷

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