Asymptotic Formulas for the Eigenvalues of the … Formulas for the Eigenvalues of the Timoshenko...

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 InstituteTroy, 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 (wavepropagation in and vibration of solids)

1. INTRODUCTION

Suppose a structural beam is driven by a laterally oscillating sinusoidalforce. 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

2 GEIST AND MCLAUGHLIN

natural frequencies of the beam. Our goal is to address the question: ifa beam’s natural frequencies are known, what can be inferred about itsbending stiffnesses or its mass density? To answer this question we needto know asymptotic formulas for these frequencies. Here we establish suchformulas for beams with variable mass density but otherwise constant beamparameters. We make this assumption as a first step toward solving theproblem with both variable density and variable stiffness. We also makethis assumption because it is consistent with some applications of interestto us. An example of an application consistent with our assumption isan aircraft wing with struts which have been added so that there is anappreciable 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. Thismodel is chosen because it is a more accurate model than the Euler-Bernoulibeam model and because systems of Timoshenko beam models are used tomodel aircraft wings. The mathematical equations that arise are two cou-pled partial differential equations,

(EIψx)x + kAG(wx − ψ) − ρIψtt = 0,

(kAG(wx − ψ))x − ρAwtt = P (x, t).

The dependent variable w = w(x, t) represents the lateral displacementat 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’smodulus, i.e., the modulus of elasticity in tension and compression, and Gis the modulus of elasticity in shear. The non-uniform distribution of shearstress over a cross section depends on cross sectional shape. The coefficientk is introduced to account for this geometry dependent distribution ofshearing stress. I and A represent cross sectional inertia and area, ρ is themass density of the beam per unit length, and P (x, t) is an applied force. Ifwe suppose the beam is anchored so that the so called “free-free” boundaryconditions hold (i.e., shearing forces and moments are assumed to be zeroat each end of the beam), then w and ψ must satisfy the following fourboundary conditions,

wx − ψ|x=0,L = 0, ψx|x=0,L = 0. (1)

EIGENVALUES OF THE TIMOSHENKO BEAM 3

After making a standard separation of variables argument, one finds thatthe Timoshenko differential equations for w and ψ lead to a coupled systemof two second order ordinary differential equations for y(x) and ψ(x),

(EIΨx)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. Wemust 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 impliesthat the values of p2 for which nontrivial solutions to this problem exist;the eigenvalues for this model, are real. Furthermore, it is not difficult toshow that the collection of all eigenvalues for this problem forms a discrete,countable, unbounded set of real non-negative numbers. Moreover, it canbe 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 determineeigenvalues from natural frequency data obtained in an experiment like theone indicated in the opening paragraph.

Suppose from vibration experiments we have determined a set of naturalfrequencies 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 onE, I, kG, A and ρ. This determination is not easy, since the dependenceof eigenvalues on these coefficients is highly nonlinear. Another difficultyarises because the Timoshenko boundary value problem involves two sec-ond order differential equations. When the coefficients in these differentialequations are non-constant, the system of two second order equations can-not be transformed into a single fourth order equation. Therefore, to makeprogress in the case where coefficients are non-constant, the boundary-valueproblem must be handled as a system of equations.

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


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 thisproblem exist if and only if λ = µn, where

µn = n2π2 + Cq −∫ 1

0

q(z)cos(2nπz)dz + αn, (6)

Cq = 2h+ 2H +∫ 1

0

q(z)dz,∞∑n=1

α2n < ∞,

(See Hald-McLaughlin [12, pp. 313-314] as well as Isaacson-Trubowitz[14], Borg [2], and Fulton-Pruess [6, 7]; for other Sturm-Liouville equationswith, effectively, less smooth coefficients, see Coleman-McLaughlin [5].)The importance of equation (6) is that it shows how the eigenvalues forthe Sturm-Liouville problem are related to the coefficient q(z) appearingin (5). Algorithms for reconstructing q from spectral data rely strongly onasymptotic 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 beamparameters which give rise to these eigenvalues? Asymptotic formulas forthe Sturm-Liouville eigenvalues are critical to determining q from spectraldata. We expect that analogous formulas for the Timoshenko eigenvalueswill play a key role in recovering beam parameters like E, kG, or ρ fromsuch data. In this paper, our objective is to determine asymptotic formu-las for the eigenvalues of the Timoshenko beam when free-free boundaryconditions 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-freeTimoshenko beam.

In the next three sections, approximations are derived (accurate to withinan error that isO(1/p)) for the square roots of eigenvalues of free-free beams


with variable density. An important step in deriving these preliminary ap-proximations of the eigenvalues is the use of a transformation (see section1) which changes the Timoshenko differential equations (2) and (3) into anew pair of differential equations, where in the new equations, the coeffi-cient which contains the eigenvalue parameter p2 no longer depends on theindependent variable. The key feature of the transformed system is thatthe largest terms in the new differential equations, and hence the most im-portant terms, are multiplied by coefficients which do not depend on thenew independent variable. As the eigenvalue parameter grows, solutions tothe transformed differential equations approach the solutions of a certainset of constant coefficient differential equations. It is therefore possible toderive an approximate solution (accurate to within an O(1/p) error) to aninitial value problem in which initial conditions are chosen so that the lefttransformed boundary conditions are enforced. In section 2, this initialvalue problem is presented and its approximate solution is derived. Byapplying the two remaining transformed right boundary conditions to theapproximate solution of the initial value problem, a frequency equation isdetermined. In section 3, estimates of square-roots of eigenvalues are madefrom this frequency equation. These estimates appear in Theorem 4.3.

Our approach to deriving the final asymptotic formulas (section 4) isbuilt from the following idea. Suppose ρ0 ≡ ( 1

L

∫ L

0ρ1/2(x)dx)2, and let

ρ(x; t) ≡ ρ0 + tρ(x), where ρ = ρ−ρ0 and t is an auxiliary parameter whichwe allow to vary from 0 to 1. In the Timoshenko differential equations, letρ(x) be replaced by ρ(x; t). Define p2 to be an eigenvalue for a free-freebeam with mass density ρ and constant material and geometric parametersotherwise. 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 goesto ρ(x), and p2 changes continuously in t to an eigenvalue for a beam withvariable density ρ(x). Let L1 =

∫ L

0ρ1/2(x; t)dx, and define µ2 ≡ L2

1p2. 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 theeigenvalue 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 =1L1

∫ 1

0

G(Y , Φ)dt. (8)


The term p2|t=0 is an eigenvalue for a beam where E, I, kG, A andρ = ρ0 are independent of x; i.e. p2|t=0 represents an eigenvalue for auniform beam. Asymptotic formulas for the eigenvalues of free-free andclamped-clamped uniform beams are derived in Geist [8] and published in[10]. From these uniform beam formulas, an asymptotic approximation forthe term p2|t=0 can be made. The final asymptotic eigenvalue formulasfor beams with variable density ρ(x) are obtained by replacing the term1L1

∫ 1

0G(Y , Φ)dt with an asymptotic approximation derived below, and by

replacing p2|t=0 with the appropriate uniform beam eigenvalue formulasgiven in [10].

The function G depends only on E, I, kG, A, ρ and the transformedeigenfunctions Φ and Y . Approximations to the square roots of eigenvaluesgiven in Theorem 4.3 allows us to determine Φ and Y and hence G towithin an error that is O(1/p). Then (8) is used to sharpen our estimatesof the eigenvalues for the Timoshenko beam. From (8) we compute thefinal asymptotic formulas, which are given in Theorem 5.2. Note that theadvantage of this method over say a variational method is that we candetermine more than the first term in the eigenvalue expansion and provea bound in the remainder no matter how large the difference is between ρand ρ0.

2. THE TRANSFORMED PROBLEM

To begin, the free-free Timoshenko boundary value problem is provedequivalent to a certain transformed boundary value problem derived below.This equivalency holds when ρ depends on x; all other beam parametersare assumed constant. A key feature of the transformed problem is thatthe coefficient containing the eigenvalue parameter no longer depends onthe independent variable.

To derive this equivalent problem, a lemma is proved which applies tosingle second order equations. We will use this lemma to prove Theorem2.1, in which the Timoshenko system of differential equations is transformedto 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 =

∫ L

0ρ1/2(x1)dx1 and z(x) = 1

L1

∫ x

0ρ1/2(x2)dx2. Let β,

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

αv′′ + (p2τρ− β)v = f (9)


if and only if V (z) satisfies

Vzz +[(ρ−1/4)xxL2

1

ρ3/4+τL2

1p2

α− βL2

1

αρ

]V =

fL21

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′)2Vzz +(αA2z′)′

AVz + αA′′V,

and provided α(z′)2A �= 0 for x ∈ [0, L], it follows that αv′′+(p2τρ−β)v =f(x) if and only if

Vzz +(A2z′)′

A2(z′)2Vz +

[A′′

A(z′)2+

p2τρ

α(z′)2− β

α(z′)2

]V =

f

α(z′)2A.

Now let A = ρ−1/4(x) and z(x) = 1L1

∫ x

0ρ1/2(x2)dx2. Then (A2z′)′ ≡ 0,

A′′/A(z′)2 = (ρ−1/4)xxL21/ρ

3/4, p2τρ/(α(z′)2) = τL21p

2/α, and β/(α(z′)2) =βL2

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

In the next theorem, new differential equations are determined that arerelated to the Timoshenko equations by the transformation indicated inthe previous lemma. Let L1 and z(x) be defined as they are in Lemma2.1, and let µ2

E = p2L21/E, µ2

kG = p2L21/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 = L2

1[ρ−1/4(x(z))]xx/[ρ3/4(x(z))] and

ρ4 = ρx(x(z))L21/[4ρ

2(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

EIΨxx + (p2Iρ− kAG)Ψ = −kAGyx + F1, (11)

and

kAGyxx + p2Aρy = kAGΨx + F2 (12)

if and only if Φ and Y satisfy


Φzz +p2L2

1

EΦ − L2

1γ

ρΦ + ρ3Φ − γρ4Y +

γL1

ρ1/2Yz =

F1L21

EIρ3/4, (13)

and

Yzz +p2L2

1

kGY + ρ3Y + ρ4Φ − L1

ρ1/2Φz =

F2L21

kAGρ3/4. (14)

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

Theorem 2.1 shows how to transform the original Timoshenko differentialequations into new equations so that in the new equations, coefficientsinvolving p2 are constant with respect to the new independent variablez. Equations (11) - (14) include generic “right hand side” terms so thatTheorem 2.1 is general enough that it applies to differential equations thatarise in the next section. In the case where F1 = F2 ≡ 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 positiveconstants. Then nontrivial Y and Φ are solutions to the boundary valueproblem

Φzz +p2L2

1

EΦ − L2

1γ

ρΦ + ρ3Φ − γρ4Y +

γL1

ρ1/2Yz = 0, (15)

Yzz +p2L2

1

kGY + ρ3Y + ρ4Φ − L1

ρ1/2Φz = 0, (16)

{Yz −

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 nontrivialsolutions to the free-free Timoshenko boundary value problem (2) - (4).The value p2 is an eigenvalue for the free-free Timoshenko boundary valueproblem if and only if there exist nontrivial functions Y and Φ which satisfy


the differential equations (15) and (16) and the boundary conditions givenin (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 (Yz −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 thatfor real c and d, they also satisfy Y (0) = d, Φ(0) = c/α, Yz(0) = c, andΦz(0) = 0. These initial conditions ensure that the boundary conditionson Y and Φ at z = 0 given in (17) are satisfied no matter how c and d arechosen. 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 theboundary conditions at z = 1 are satisfied, then by definition L2

1p2 is an

eigenvalue for the transformed boundary value problem, and by Theorem2.2, 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 derivedwhich are equivalent to the initial value problem discussed above. These in-tegral equations are used to determine approximate solutions to the aboveinitial value problem. The approximate solutions to the initial value prob-lem make possible estimates of the values of p for which nontrivial solutionsexist 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 cosµz +∫ z

0

sin[µ(z − t)]µ

[q(t)w(t) + f(t)]dt. (19)

proof: The elementary proof is based on the well known technique ofvariation of parameters, and is omitted. ✷


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/ρ

1/2(0) and let c and d be arbitrary but fixed constants.Then Φ(z) and Y (z) satisfy the differential equations (15) and (16) andthe initial conditions

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

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

Φ =c

αcosµEz +∫ z

0

sin[µE(z − t)]µE

{(L2

1γ

ρ− ρ3

)Φ + γρ4Y − γL1

ρ1/2Yt

}dt (21)

and

Y =c sin(µkGz)

µkG+

d cos(µkGz) +∫ z

0

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 Yand 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 continuouslydifferentiable. Since yx and Ψx must be continuous, it follows that Yz andΨz are also continuous.

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

(L2

1γ/ρ− ρ3

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

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

shows that the differential equation (15) and initial conditions (20) aresatisfied if and only if integral equation (20) holds. Similarly, since −ρ3

and −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). Ifc and d are allowed to vary over �, the solution to this initial value problemgenerates every solution to differential equations (15) and (16) that satisfies


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

Yz(1) − L1ρ1/2(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 homogeneouslinear system for the arbitrary constants c and d. Values of the eigenvalueparameter p which make possible nontrivial choices for c and d correspondto the square roots of eigenvalues for the free-free Timoshenko beam. Thehomogeneous linear system in c and d has nontrivial solutions if and onlyif the determinant of the corresponding coefficient matrix is zero. Thisdeterminant will define a frequency function. The objective in this sectionis to determine this frequency function from estimates of the coefficients ofc and d in the expressions given in (23).

The following technical fact is used many times in the estimates thatfollow.

Lemma 3.2. Suppose that fz ∈ L∞(0, 1), and that δ is a real constant.Then for z ∈ [0, 1],∣∣∣∣

∫ z

0

sin[µ(z − t) + δ]f(t)dt∣∣∣∣ < ‖f‖∞ + ‖fz‖∞

µ.

proof:∫ z

0

sin[µ(z − t) + δ]f(t)dt =

cos[µ(z − t) + δ]µ

f(t)∣∣∣∣z

0

−∫ z

0

cos[µ(z − t) + δ]µ

ft(t)dt

This implies the result. ✷

Lemma 3.3. Suppose h and δ are real constants and that h is not equalto 0 or L1/

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

when x ∈ [0, L]. Then∣∣∣∣∫ z

0

sin[hµ(z − t) + δ]µ

g(t)Ytdt∣∣∣∣ < O(1/µ)‖Y ‖∞ +O(1/µ)‖Φ‖∞

+O(1/µ)|c| +O(1/µ)|d|.


proof:

Yz = c cos(µkGz) − dµkG sin(µkGz) +

∫ z

0

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

L1

ρ1/2(t)Φt(t)

}dt

⇒∫ z

0


g(t)Ytdt =

∫ z

0

[sin[hµ(z − t) + δ]

µg(t){c cos(µkGt) − dµkG sin(µkGt)

+∫ t

0

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

+sin[hµ(z − t) + δ]

µg(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 thatdg/dz ∈ L∞(0, 1), and Lemma 3.2 can be used to show that the absolutevalue of the first integral on the right of the above equation is boundedabove 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 ofthe 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)

−∫ t

0

[µkG sin[µkG(t− s)]

L1

ρ1/2(s)+ cos[µkG(t− s)

(L1

ρ1/2(s)

)s

]Φds.

Therefore, if we show that

∣∣∣∣∫ z

0


g(t)[∫ t

0

µkG sin[µkG(t− s)]L1

ρ1/2(s)Φ(s)ds

]dt

∣∣∣∣ < O

(1µ

),


then the Lemma follows. After changing the order of integration, the inte-gral above may be rewritten as

∫ z

s=0

∫ z

t=s


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 absolutevalue 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 equalto 0 or L1/

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

when x ∈ [0, L]. Then

∣∣∣∣∫ z

0


g(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 remainfinite as p approaches infinity.

Lemma 3.5. Suppose E �= kG, that ρxx ∈ L∞(0, L), and that ρ(x) > 0for 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 3.3and 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)


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 andd in the functions and (Yz − ΦL1/ρ

1/2(z))/µkG and αΦz/µE .

Lemma 3.6. Suppose E �= kG, that ρxx ∈ L∞(0, L) and ρ(x) > 0 for allx ∈ [0, L], and that α = L1/ρ

1/2(0). Then

∥∥∥∥∥∥Yz(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 Yz/µkG and Φz/µE and can be determinedfrom the integral equations for Y and Φ. The proof follows from the integralequations 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 supposeρxx(x) ∈ L∞(0, L). Then p2 is an eigenvalue for the free-free Timoshenkobeam if and only if

F (p) ≡ sinµkG sinµE + 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).


proof: We seek to determine the values of p2 for which there exist non-trivial functions Φ and Y that solve the transformed differential equationsand all transformed boundary conditions, including those at z = 1. To de-termine all such solutions, we seek nontrivial solutions to the initial valueproblem in (15), (16), and (20) which also solve the transformed boundaryconditions at z = 1. Theorem 2.2 shows that the values of p2 which admitnontrivial solutions when boundary conditions at z = 1 are imposed arethe eigenvalues of the free-free Timoshenko beam.

Lemma 3.6 implies that for any choice of c and d, solutions Y and Φ tothe initial value problem (15), (16), and (20) satisfy

Yz(1) − [L1/ρ1/2(1)] Φ(1)

µkG= δ1,1c+ [sin(µkG) − δ1,2]d (29)

and

αΦz

µE= [− sinµE − δ2,1]c− δ2,2 d (30)

where the δi,j are all O(1/p) functions. Equations (29) and (30) imply thatthe right boundary conditions, i.e., the conditions in (20) when z = 1, maybe satisfied if and only if[

δ1,1 − sinµkG − δ1,2− sinµE − δ2,1 −δ2,2

] (cd

)=

(00

).

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

sinµkG sinµE + 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 thatall 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) whichis 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 shownthat exactly one root of F occurs near isolated roots of sin(µE) · sin(µkG).

Theorem 4.1. Let

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


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

|εp| <c

L1sin−1

{|δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|

}.

(Thus, |εp| 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 now

that ∣∣∣∣∣ 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 + 4c2

,

and hence that,

| sin(pL1/√E)| < |δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|. (33)


Let M be an integer chosen such that |[L1/√E]p −Mπ| ≤ π/2, and let

εp = p− [√E/L1]Mπ. Then from inequality (33) it follows that∣∣∣∣sin

(pL1√E

)∣∣∣∣ =∣∣∣∣sin

(L1εp√E

)∣∣∣∣ = sin(L1|εp|√E

)

<|δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|

⇒ |εp| <√E

L1sin−1

{|δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|

}.

Thus, there is some integer M such that p = εp + [√E/L1]Mπ where

|εp| <√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 p =

√EMπ/L1

or√kGMπ/L1, such that p− p = εp and

|εp| <c

L1sin−1

{|δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|

}. ✷

Theorem 4.2. Suppose p and p are two adjacent roots of the functionsin(µkG) · sin(µE) and that p is a simple root not closer to any other root ofthis 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{p, p}.Let

∆p = sin−1

{c

c

π

2supp>p|δ1,2(p)| + supp>p|δ2,1(p)| + supp>p

|δ|sin(p−1)

}c

L1,

and suppose that |p− p| > 2∆p+ cL1

1p Then there is at least one root of the

characteristic equation (28) in the interval [p−∆p, p+ ∆p]. Furthermore,this interval must contain an odd number of roots of the characteristicequation.

Remark: The quantity ∆p defined above is O(1/p), since all of the δ sappearing in its definition are O(1/p). Thus, the hypothesis above requiresthat p and p be separated by a distance which is O(1/p).


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 p large enough, F (p) must change sign near wheresin(µkG) sin(µE) changes sign, i.e. in the interval [p − ∆p, p + ∆p]. SeeGeist [8, pages 164-168] for a detailed presentation of this proof. ✷

An estimate of ∂F/∂p derived in the next lemma is an important steptoward proving that for each isolated root of sin(µE) · sin(µkG), there isexactly 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

[Yz(1)− L1

ρ1/2(1)Φ(1)

µkG

],

coef.of d

[Yz(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

[Yz(1)− L1

ρ1/2(1)Φ(1)

µkG

],

coef.of d

∂∂p

[Yz(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 coefficientof either c or d. Let “·” denote differentiation with respect to p. Integralequations for Yz, 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, whichsatisfy the initial value problem (15), (16), and (20) may be written asY (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 andΨx are analytic functions of p. See [4, page 37]. This implies that for each


z ∈ [0, 1], Yz, Φz and Φ are analytic functions of p, and so differentiationwith respect to the eigenvalue parameter p is allowed.) Estimates for Yz,Φ, and Φz can be obtained by first showing that the infinity norms of thesefunctions are bounded. Once this is demonstrated, one can generate esti-mates for Yz, Φ, and Φz accurate to within an error that is O(1/p) in thesame way that estimates for Y , Yz, and Φz were produced in section 3.Then, estimates for ∂F/∂p can be calculated using equation (35). For thedetails 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 oneanother.

Theorem 4.3. Suppose p and p are two adjacent roots of the functionsin(µkG)·sin(µE). Suppose p is a simple root of sin(µkG)·sin(µE) not closerto 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{p, p}. Let

∆p = sin−1

{c

c

π

2supp>p|δ1,2(p)| + supp>p|δ2,1(p)| + supp>p

|δ|sin(p−1)

}c

L1,

Γ = supp>p−∆p

{c

L1sin−1

{|δ1,2| + |δ2,1|

2+

√(|δ1,2| + |δ2,1|)2

4+ |δ|

}},(36)

and

r(p) =∂F

∂p− L1√

Ecos(µE) · sin(µkG) − L1√

kGcos(µkG) · sin(µE).

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

|p− p| >

max

{2∆p+

c

L1

1p,

c

L1

π

2

∣∣∣∣tan(L1Γc

)∣∣∣∣ +

√EkG

L21

π

2|r(p)|

cos (ΓL1/c)+ Γ

},

(37)

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

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


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 fromTheorem 4.1 that any root of F nearer to p than it is to any other root ofsin(µkG) sin(µE) must occur in the narrow interval [p−Γ, p+Γ]. Theorem4.2 guarantees that in this interval, there is at least one root of F . Weprove now that |∂F/∂p| > 0 throughout [p−Γ, p+Γ] when the hypothesesof 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 ∈ [p − Γ, p + Γ]. Without loss ofgenerality, suppose p is a zero of sin(µE). Hypothesis (37) implies that

2L21

π√EkG

[|p− p| − Γ] >L1√kG

∣∣∣∣tan(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 ofsin(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

2π

L1√kG

|p− pkG|

≥ 2L21

π√EkG

|p− pkG + p− p|

≥ 2L21

π√EkG

[|p− pkG| − |p− p|]

≥ 2L21

π√EkG

[|p− p| − Γ]. (39)

p ∈ [p− Γ, p+ Γ], p a zero of tan(µE), and ΓL1/√E < π/4 imply that

L1√kG


Γ)∣∣∣∣ +

|r(p)|cos

(L1√E

Γ) =

L1√kG


(p± Γ))∣∣∣∣ +

|r(p)|cos

(L1√E

(p± Γ)) ≥


L1√kG

| tanµE | +|r(p)|

| cosµE |≥ L1√

kG| tanµE || cosµkG| +

r(p)| cosµE |

. (40)

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

L1√E| sinµkG| >

L1√kG

| tanµE cosµkG| +|r(p)|

| cosµE |

⇒∣∣∣∣ L1√E

sinµkG cosµE

∣∣∣∣ >∣∣∣∣ L1√kG

sinµE cosµkG

∣∣∣∣ + |r(p)|

⇒∣∣∣∣∂F∂p (p)

∣∣∣∣ > 0,

as desired.Thus, |∂F/∂p| > 0 for all p ∈ [p − Γ, 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 Theoremfollows when p is a zero of sin(µE). The proof for the case where p is azero of sin(µkG) is identical to the proof above, except that the role of Eand kG are interchanged. ✷

5. THE ASYMPTOTIC FORMULAS

We carry out the following steps in order to sharpen our estimates of theeigenvalues. First, we define a new density function ρ in terms of ρ. Letρ0 = [ 1

L

∫ L

0ρ1/2(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 thatwhen 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, ρ4

and L1 are replaced by z, ρ, ρ3, ρ4, and L1, respectively, where z, ρ3, ρ4 andthe constant L1 are defined as are z, ρ3, ρ4 and L1 except that ρ replacesρ in their definitions. The resulting boundary value problem for Φ andY corresponds to the transformed boundary value problem (see Theorem2.2) one obtains from the original Timoshenko differential equations andboundary conditions when ρ(x) is taken to be ρ(x; t).

It will prove useful to view the transformed differential equations for Φand Y as a single vector equation. Define

U(z) =(

Φ(z)Y (z)

).


Let

B1 =(

1E 00 1

kG

), Q =

(−L2

1γρ + ρ3 −γρ4

ρ4 ρ3

)(41)

S =(

0 γ−1 0

), and µ2 = L2

1p2. (42)

Then the boundary value problem (15-17), where ρ is now replaced by ρ,may be written in vector notation as

Uzz + µ2B1U +QU +L1

ρ1/2SUz =

(00

)(43){

Yz −L1

ρ1/2Φ

}∣∣∣∣∣z=0,1

= 0, and Φz|z=0,1 = 0, (44)

Suppose the transformed eigenvalue parameter, µ2 = L21p

2, is chosen sothat 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(∂∂xEI

∂∂x + (p2Iρ− kAG) kAG ∂

∂x∂∂xkAG

∂∂xkAG

∂∂x + p2Aρ

)u(x) = 80 (45)

[yx − Ψ]|x=0,L = 0, Ψ|x=0,L = 0. (46)

Thus, if L21p

2 = µ2 is an eigenvalue for the transformed boundary valueproblem (43-44) and U(z) is the corresponding eigenfunction, then p2 is aneigenvalue and u(x) = ρ−1/4U(z(x)) is the corresponding eigenfunction forthe 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 tothe auxiliary parameter t, introduced in the definition of ρ. The approachmay be summarized as follows. Let µ2 be an eigenvalue for the trans-formed boundary value problem (43-44) and let U(z) be a correspondingeigenfunction. The differential equation and boundary conditions for U(z)


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 valueproblem. The right hand side of this new boundary value problem willcontain dρµ

2[ρ], which we will show is equal to dµ2/dt. Since the newboundary value problem comes from differentiating the differential equa-tions and boundary conditions (43-44), a boundary value problem witha known solution U(z) ≡/ 0, it follows that a nontrivial solution to thenew, non-homogeneous problem exists, and must be equal to dρµ2[ρ]. Us-ing Theorem 2.1, the new non-homogeneous boundary value problem fordρµ

2[ρ] may be written as a non-homogeneous Timoshenko boundary valueproblem which must also have a nontrivial solution. The fact that thenon-homogeneous Timoshenko boundary value problem has a nontrivialsolution implies that the right hand side of the non-homogeneous problemmust satisfy a certain orthogonality requirement (see Lemma 5.1 below).This orthogonality condition allows us to determine dµ2/dt = dρµ

2[ρ] interms of the transformed eigenfunction U(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 =∫ 1

0

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. Thereforeµ2|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 asµ (and p) get large. Thus, equation (47) provides a means for obtainingsharp estimates for µ2 when t = 1. From these estimates of µ2, we willobtain asymptotic formulas for the free-free Timoshenko eigenvalues whenmass density is ρ(x).

Our next goal then is to calculate an expression for dµ2/dt. Supposeρxx ∈ L∞[0, L]. For these calculations, we first assume that for all such ρ,

i) dρµ2[ρ] exists, that

ii) dρU [ρ], dρ(Uz)[ρ], dρ(Uzz)[ρ] all exist, and that

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


Later, we will give conditions which guarantee that assumptions i), ii) andiii) are valid. As indicated in the discussion above, we formally differentiatethe differential equations and boundary conditions in (43-44) with respectto ρ in the direction of ρ. We use assumptions i), ii), and iii) from above toobtain the differential equation and boundary conditions that dρU [ρ] mustsatisfy. 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]SUz,

and that

[(dρY [ρ])z − (L1/ρ1/2)(dρΦ[ρ]) − (dρ(L1/ρ

1/2)[ρ])Φ]|z=0,1 = 0,

(dρΦ[ρ])|z=0,1 = 0.

When dρµ2[ρ] exists, it follows that

dρµ2[ρ] = lim

ε→0

µ2(ρ+ ερ) − µ2(ρ)ε

= limε→0

µ2(ρ0 + (t+ ε)ρ) − µ2(ρ0 + tρ)ε

=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 00 1/kAG

)(48)

and

F = − ρ3/4

L2B−1

2

{(d

dtµ2

)B1U +

(∂

∂tQ

)U +

(∂

∂t

L1

ρ1/2

)SUz

}. (49)

Now define

V (z) ≡ dρU [ρ] =(dρΦ[ρ]dρY [ρ]

)≡

(Φ(z)Y (z)

).


Then V satisfies

Vzz + µ2B1V +QV +L1

ρ1/2SVz =

L21

ρ3/4B2F ,

{Yz −

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 differentialequation

(∂∂xEI

∂∂x + (p2Iρ− kAG) kAG ∂

∂x∂∂xkAG

∂∂xkAG

∂∂x + p2Aρ

)v(x) = F . (52)

The boundary conditions (50) and (51) may also be rewritten in termsof Ψ and y. By multiplying (50) through by ρ1/4/L1 and noting thatρx(0) = ρx(L) = 0, we find that

{Yz −

L1

ρ1/2Φ − ∂

∂t

(L1

ρ1/2

)Φ

}∣∣∣∣∣z=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) isto 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) mustsatisfy a certain orthogonality condition.


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

EIΨxx + kAG(yx − Ψ) + p2IρΨ = F1, (55)

and

kAG(yx − Ψ)x + p2Aρy = F2, (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 =(F1

F2

)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)) =∫ L

0

(EIΨxxψ + kAG(yx − Ψ)ψ + p2IρΨψ

+ kAG(yx − Ψ)x)y + p2Aρyy)dx

=∫ L

0

(EIΨψxx + kAG(yx − Ψ)ψ − kAG(yx − Ψ)yx

+p2IρΨψ + p2Aρyy)dx + kAG(yx − Ψ)y|Lz=0

=∫ L

0

[EIΨψxx − kAG(yx − ψ)(yx − Ψ) + p2IρΨψ + p2Aρyy]dx

+kAG(yx − Ψ)y|Lx=0

=∫ L

0

{[EIψxx + kAG(yx − ψ) + p2Iρψ]Ψ


+[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 impliesthat

⇒ (F(x), ρ−1/4U(x)) =kAG

L1

∂

∂t

(L1

ρ1/2

)ΦY

∣∣∣∣∣1

z=0

. (57)

Equation (57) holds if and only if

(− ρ

3/4

L2B−1

2

{(d

dtµ2

)B1U +

(∂

∂tQ

)U +

(∂

∂t

L1

ρ1/2

)SUz

}, ρ−1/4U

)

=kAG

L1

∂

∂t

(L1

ρ1/2

)ΦY

∣∣∣∣∣1

z=0

⇔ ddtµ

2 =−(

ρ1/2

L1B−1

2 ( ∂∂tQ)U,U

)−(

ρ1/2

L1

(∂∂t

L1ρ1/2

)B−1

2 SUz,U

)−kAG ∂

∂t

(L1

ρ1/2

)ΦY

∣∣∣1z=0(

ρ1/2

L1B−1

2 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,

ddtµ

2 =−

∫ 1

0UT B−1

2 ( ∂∂tQ)Udz−

∫ 1

0∂∂t

(L1

ρ1/2

)UT B−1

2 SUzdz−kAG ∂∂t

(L1

ρ1/2

)ΦY

∣∣∣1z=0∫ 1

0UT B−1

2 B1Udz

Next we show that the third term in the numerator above, the boundaryterm, may be combined with the second term so that all terms in the


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

∫ 1

0

∂

∂t

(L1

ρ1/2

)UTB−1

2 SUzdz = kAG

∫ 1

0

∂

∂t

(L1

ρ1/2

)(YzΦ − ΦzY )dz

= 2kAG∫ 1

0

∂

∂t

(L1

ρ1/2

)YzΦdz − kAG

∫ 1

0

(ΦY )z∂

∂t

(L1

ρ1/2

)dz

= 2kAG∫ 1

0

∂

∂t

(L1

ρ1/2

)YzΦdz −kAGY Φ

∂

∂t

(L1

ρ1/2

)∣∣∣∣∣1

0

+ kAG

∫ 1

0

ΦY∂2

∂t∂z

(L1

ρ1/2

)dz

⇒ −∫ 1

0

∂

∂t

(L1

ρ1/2

)UTB−1

2 SUzdz − kAGY Φ∂

∂t

(L1

ρ1/2

)∣∣∣∣∣1

0

=

−2kAG∫ 1

0

∂

∂t

(L1

ρ1/2

)YzΦdz − kAG

∫ 1

0

ΦY∂2

∂t∂z

(L1

ρ1/2

)dz.

Therefore,

d

dtµ2 =

−∫ 1

0UT B−1

2 ( ∂∂tQ)Udz−2kAG

∫ 1

0∂∂t

(L1

ρ1/2

)YzΦdz−kAG

∫ 1

0ΦY ∂2

∂t∂z

(L1

ρ1/2

)dz∫ 1

0UT B−1

2 B1Udz

.(58)

The formula above for dµ2/dt holds so long as assumptions i) - iii) hold. Inthe following theorem, conditions are given which guarantee the validity ofthese assumptions, and hence show when dµ2/dt may be calculated usingthe 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)

∫ L

0ρ1/2(x)dx]2, ρ ≡

ρ(x) − ρ0, and ρ(x, t) = ρ0 + tρ(x). Let L1 =∫ L

0ρ1/2(x; t)dx, and let

µ2(t) = L21p

2 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 × 2matrices defined in (41) and in (48).


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 linearscalar equations in which the parameter t appears linearly. This impliesthat for each x, u, ux, and uxx must be analytic in t. See Coddington andLevinson [4, page 37]. This in turn implies that for each z, U(z), Uz(z) ,and Uzz(z) must also be analytic in t when t ∈ [0, 1]. When z = 0 and tis 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 wouldimply U(z) must be identically zero.) From the scalar differential equa-tions for the components of U(z), we conclude that µ2 must be analyticin 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 , dρUz[ρ] = ∂

∂t Uz, and dρUzz[ρ] =∂∂t Uzz. From this we conclude that dρU [ρ], dρUz[ρ], and dρUzz[ρ] all exist,since U , Uz, and Uzz 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 fromthese integral equations that Uzzt and Utzz are continuous functions of zand t. This implies that when ρ(x) ∈ C2[0, L], Uzzt = Utzz. Furthermore,from the integral equations, it follows that Uzzt and Utzz are continuous in ρwith respect to the standard Sobelev norm of order 2. They are continuousmaps which take ρ(x) ∈ H2(0, L) to an element of L2(0, 1). Since C2[0, L] isdense in H2(0, L), it follows that Uzzt = Utzz when ρ(x) ∈ H2(0, L). SinceH2(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 dwhich appear in the integral equations (20) and (21). Theorem 4.1 showsthat if p2 is a large enough eigenvalue, then p must lie near a root ofsin(L1p/

√kG) · sin(L1p/

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

more than one root of sin(L1p/√kG) · sin(L1p/

√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 zeroof sin(L1p/

√kG) or of sin(L1p/

√E). This result will be used to calculate

an estimate of dµ2/dt.In the next Lemma, we show that if p is large enough and satisfies either

(59) or (60) below, then the square root of the nearest eigenvalue to p iswell separated from square roots of other eigenvalues.


Lemma 5.2. Let e ∈ (0, 1) and suppose p is a root of the functionsin(L1p/

√kG) · sin(L1p/

√E) such that either

sin(L1p/√kG) = 0 and | sin(L1p/

√E)| > e (59)

or

sin(L1p/√E) = 0 and | sin(L1p/

√kG)| > e. (60)

Then there exists an M > 0 such that when p > M , exactly one eigenvalueof the free-free Timoshenko beam with density ρ, say p2, is such that itssquare root p lies in the interval

[p− e

√c

2L1, p+ e

√c

2L1

]. Furthermore, |p− 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(L1p/

√kG) · sin(L1p/

√E) as near to p as is any

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

Let Γ be defined as it is in the hypothesis of Theorem 4.3. Let M be chosenlarge enough so that p > M implies that hypothesis (37) of Theorem 4.3is 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 ∈[p− e

√c

2L1, p+ e

√c

2L1

], then

p must be contained in the narrower interval (p− Γ, p+ Γ). On the otherhand, Theorem 4.3 shows that there is exactly one zero of the frequencyfunction in the interval (p−Γ, p+Γ). Together, Theorem 4.1 and Theorem4.3 imply that there is exactly one root of the frequency function in theinterval

[p− e

√c

2L1, p+ e

√c

2L1

]. ✷

Lemma 5.3. Let e ∈ (0, 1), and suppose p is a root of sin(L1p/√kG·sin(L1p/

√E)

such that either (59) or (60) holds. For p large enough, let p2 be the uniqueeigenvalue of the free-free beam with density ρ whose square root p is con-tained in the interval

[p− e

√c

2L1, 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 appearingin (20) and (21)) must satisfy(

cd

)= constant

(O(1/p)

1

).

If (60) holds, (cd

)= constant

(1

O(1/p)

).


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 anestimate for dµ2/dt. We make an estimate dµ2/dt in the case where plies near a root p of sin(L1p/

√kG) · sin(L1p/

√E) which for an arbitrary

but fixed e ∈ (0, 1) satisfies either (59) or (60). A p which meets thiscriterion gives rise to an eigenvalue which we will refer to as being “well-separated” from its neighboring eigenvalues. The estimate of dµ2/dt is usedto calculate asymptotic formulas for these eigenvalues.

In the next lemma, we calculate estimates for one of the terms appearingin the numerator of the right hand side of (58).

Lemma 5.4. Let U and µ2 = L21p

2 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 densityρ. Let e be a fixed but arbitrary number in (0, 1), and let p be a root ofsin(L1p/

√kG) · sin(L1p/

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

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

−2kAG∫ 1

0

∂

∂t

(L1

ρ1/2

)YsΦds =

[( cα

)2

− d2A

I

][kAGEkG − 1

] ∫ 1

0

(L1

ρ1/2

)∂

∂t

(L1

ρ1/2

)ds+O(1/p).

proof: Theorem 2.2 shows that p2 is an eigenvalue for the free-freeTimoshenko beam when L2

1p2 is an eigenvalue for the transformed problem.

Let µE = L1p√E

and µkG = L1p√kG

. Then Theorem 3.1 shows that

Yz = c cos(µkGz) − dµkG sin(µkGz)+


∫ z

0

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+

∫ z

0

sin[µE(z − s)]µE

·{−ρ3(s)Φ(s) +

L21γ

ρ(s)Φ(s) + γρ4(s)Y (s) − γL1

ρ1/2(s)Y (s)

}ds.

Let f(z) ≡ 2kAG ∂∂t

(L1ρ1/2

). Then

∫ 1

0

f(z)Yz(z)Φ(z)dz =

∫ 1

0

f

[c2

αcos(µkGz) cos(µE z) −

cd

αµkG sin(µkGz) cos(µE z)

]dz

−dµkGµE

∫ 1

0

f(z) sin(µkGz)∫ z

0

sin[µE(z−s)]{−ρ3Φ +

L21γ

ρΦ + γρ4Y − L1γ

ρ1/2Ys

}ds dz

∫ 1

0

f(z)c

αcos(µE z)

∫ z

0

cos[µkG(z−s)]{−ρ3Y − ρ4Φ +

L1

ρ1/2Φs

}ds dz+O

(1p

).

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

α must bothbe 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

I1 = −dµkGµE

∫ 1

0

f(z) sin(µkGz)

·[∫ z

0

sin[µE(z − s)]

{−ρ3Φ +

L21γ

ρΦ + γρ4Y − γL1

ρ1/2Ys

}ds

]dz.

By switching the order of integration, we find that

I1 = −dµkGµE

∫ 1

0

∫ 1

s

f(z) sin(µkGz) sin[µE(z − s)]


·{−ρ3(s)Φ(s) +

L21γ

ρ(s)Φ(s) + γρ4(s)Y (s) − γL1

ρ1/2(s)Ys(s)

}dz ds

= −dµkGµE

∫ 1

0

{−ρ3Φ +

L21γ


ρ1/2Ys

}

·[∫ 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 integratethe square bracketed term above by parts with respect to z, and applyLemmas 3.2, 3.3, 3.4, and 3.5 to show that

I1 = −dµkGµE

∫ 1

0

{−ρ3Φ +

L21γ


ρ1/2Ys

}

·f2

[1

µkG + µE− 1µkG − µE

]sin(µkGs)ds+O(1/p)

= −dµkGµE

∫ 1

0

γL1

ρ1/2

f

22µE

µ2kG − µ2

E

sin(µkGs)Ysds+O(1/p).

From the integral equation for Yz, it follows that Yz = −dµkG sin(µkGz) +O(1). Substituting this expression for Yz into the last expression for I1, wefind that

I1 = d2 µkGµE

∫ 1

0

γL1

ρ1/2fµkGµEµ2kG − µ2

E

sin2(µkGs)ds+O(1/p)

= d2kAGA

I

(1

EkG − 1

)∫ 1

0

L1

ρ1/2

∂

∂t

(L1

ρ1/2

)ds+O(1/p).

Using a similar argument as was used to derive the estimate of I1, it ispossible to show that

I2 =∫ 1

0

f(z)c

αcos(µE z)

∫ z

0

cos[µkG(z−s)]{−ρ3Y − ρ4Φ +

L1

ρ1/2Φs

}ds dz

= −[ cα

]2

kAG

(1

EkG − 1

)∫ 1

0

L1

ρ1/2

∂

∂t

(L1

ρ1/2

)ds+O(1/p).


Since ∫ 1

0

2kAG∂

∂t

(L1

ρ1/2

)YsΦds = I1 + I2 +O(1/p),

the lemma follows. ✷

In the next theorem, we make estimates for transformed eigenfunctionsU(z) associated with well separated eigenvalues for the untransformed Tim-oshenko beam with density ρ. We will use these estimates for U to calculatean estimate for dµ2/dt.

Lemma 5.5. Suppose p is an eigenvalue for the free-free Timoshenkobeam with density ρ = ρ0 + tρ(x), where ρ0 = ([1/L]

∫ L

0ρ1/2(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, andA are positive constants. Let p be a root of sin(L1p/

√kG) · sin(L1p/

√E)

as near to p as is any other root of this function. Let e be an arbitrary butfixed constant between 0 and 1. If (59) holds, then for some integer n, p =nπ

√kG/L1, and µ2 = L2

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

an eigenvalue-eigenfunction pair for (43-44). Similarly, if p satisfies (60),then for some integer n, p = nπ

√E/L1 and an eigenvalue-eigenfunction

pair for (43-44) is µ2 = L21p

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

proof: If µE = L1p√E

and µkG = L1p√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

∫ z

0

sin[µE(z − s)]µE

{−ρ3Φ +

L21γ

ρ+ γρ4Y − γL1

ρ1/2Ys

}ds (62)

and

Y =c sin(µkGz)

µkG+ d cos(µkGz)

∫ z

0

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 5.3and Theorem 4.1 we find that p = p+O(1/p) = nπL1/

√kG+O(1/n). This


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 between0 and 1. Suppose p2 is an eigenvalue for the free-free Timoshenko beamwith density ρ(x; t) and that p is a root of the function sin(L1p/

√kG) ·

sin(L1p/√E) nearest p. Let µ2 = L2

1p2. If (59) holds, then p = nπ

√kG

L1for

some integer n, and

dµ2

dt= −kG

∫ 1

0

{(ρ3)t[1 + cos(2nπz)]}dz

− 2kAGI

(EkG − 1

) ∫ 1

0

L1

ρ1/2

∂

∂t

(L1

ρ1/2

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

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

L1for some integer n, and

dµ2

dt= E

∫ 1

0

{[L2

1γ

ρ− ρ3

]t

− cos(2nπz)(ρ3)t

}dz

+2kAG

I(

EkG − 1

) ∫ 1

0

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 thatρ(x) > 0 for all x ∈ [0, L], that ρx(0) = ρx(L) = 0, and that ρxx ∈ L∞[0, L].Let ρ0 = ([1/L]

∫ L

0ρ1/2(x)dx)2, and define

E(p) = sin

(pLρ

1/20√E

)sin

(pLρ

1/20√kG

).

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

pni ≡niπ

√E

Lρ1/20

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


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

Lρ1/20√

kG

)∣∣∣∣∣ > e.

Suppose each pni is large enough that exactly one simple eigenvalue p ofthe Timoshenko beam lies in the interval[

pni−

√c

2Lρ1/20

e, pni+

√c

2Lρ1/20

e

],

where c = min{E, kG}. Then

p2 =n2iπ

2E

L2ρ0− E

L2ρ0

∫ 1

0

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

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

0ρ1/2(s)ds and

CE =A

I

kG

ρ0

[E

E − kG

] ∫ 1

0

(ρ0

ρ(x)− 1

)dz

− E

L2ρ0

∫ 1

0

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(pmiLρ

1/20√

E

)∣∣∣∣∣ > e.

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

p2 of the Timoshenko beam satisfying

p ∈[pmi −

√c

2Lρ1/20

e, pmi +√c

2Lρ1/20

e

].

In this case,

p2 =m2

iπ2kG

L2ρ0− kG

L2ρ0

∫ 1

0

ρ3(x) cos(2miπz)dz + CkG +O(1/mi), (68)


where

CkG = −AI

kG

ρ0

[kG

E − kG

] ∫ 1

0

(ρ0

ρ(x)− 1

)dz

− kG

L2ρ0

∫ 1

0

ρ3(x)dz +A

I

kG

ρ0

(1 +

kG+ E

2(E − kG)

). (69)

proof: To prove formula (66) let µ2 = L21p

2 where p2 is an eigenvalue fora beam with density ρ(x; t). Define p2 = p2|t=1 and p2 = p2|t=0. BecauseL2

1|t=1 = L21|t=0 = L2ρ0 = L2

1, as t changes from 0 to 1, µ2/[L2ρ0] willchange continuously in t from p2, an eigenvalue for a uniform beam withconstant density ρ0, to ρ2, and eigenvalue for a beam with density ρ(x).Furthermore, Lemma 5.6 shows that dµ2/dt satisfies (65). This impliesthat

L2ρ0(p2 − p2) = E

∫ 1

0

{L2

1γ

(1ρ− 1ρ0

)− ρ3 − cos(2nπz)ρ3

}dz

+L2kAG

I(

EkG − 1

) ∫ 1

0

[ρ0

ρ− 1

]dz +O(1/ni) (70)

where L2ρ0p2 = µ2|t=1 is an eigenvalue for the transformed problem with

density ρ(x) and L2ρ0p2 = µ2|t=0 is an eigenvalue for the transformed

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

that

p2 − p2 = E

∫ 1

0

γ

(1ρ− 1ρ0

)dz − E

L2ρ0

∫ 1

0

ρ3dz

+kAG

I(

EkG − 1

) ∫ 1

0

(1ρ− 1ρ0

)dz − E

L2ρ0

∫ 1

0

cos(2niπz)ρ3dz +O

(1ni

)(71)

From Geist-McLaughlin [10], it follows that

p2 =n2iπ

2E

L2ρ0+

(1 − 1

2kG+ E

kG− E

)A

I

kG

ρ0+O

(1ni

).

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 =m2

iπ2kG

L2ρ0+

(1 +

12kG+ E

kG− E

)A

I

kG

ρ0+O

(1mi

). ✷


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Asymptotic Formulas for the Eigenvalues of the … Formulas for the Eigenvalues of the Timoshenko...

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