2004 Conference on Diff. Eqns. and Appl. in Math. Biology, Nanaimo, BC, Canada. Electronic Journal of Differential Equations, Conference 12, 2005, pp. 57–64. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

ON THE ISOSPECTRAL BEAMS

KAZEM GHANBARI

Abstract. The free undamped infinitesimal transverse vibrations of a thin straight beam are modelled by a forth-order differential equation. This paper investigates the families of fourth-order systems which have one spectrum in common, and correspond to four different sets of end-conditions. The analysis is based on the transformation of the beam operator into a fourth-order selfadjoint linear differential operator. This operator is factorized as a product L = H^{∗}H, where H is a second-order differential operator of the form H = D^{2} + rD + s, and H^{∗} is its adjoint operator.

1. Beam Equation

The free undamped infinitesimal transverse vibrations, of frequency ω, of a thin straight beam of length ℓ are governed by the Euler-Bernoulli equation

(EI(x)u′′(x))′′ = ω^{2}ρA(x)u(x), 0 ≤ x ≤ ℓ.  (1.1)

Here E is the Young’s modulus, ρ is the density, both assumed constant, A(x) is the cross-sectional area at x, I(x) is the second moment of this area about the axis through the centroid at right angle to the plane of vibration (the natural axis). Euler-Bernoulli equation (1.1) can be written even in simpler form by the following substitutions. Put

x = ℓs, y(s) = u(x), r(s) = \frac{I(x)}{I(x_{0})},

a(s) = \frac{A(x)}{A(x_{0})}, λ = \frac{ρℓ^{4}ω^{2}A(x_{0})}{EI(x_{0})},

where x_{0} is a fixed point in [0, 1]. Then (1.1) becomes

(r(s)y′′(s))′′ = λa(s)y(s), 0 ≤ s ≤ 1,  (1.2)

where r(s), a(s) > 0 for all s ∈ [0, 1]. For the beam equation (1.2) the most commonly used end-conditions are

(F) y′′ = 0 = (ry′′)′ (Free)

(C) y = 0 = y′ (Clamped)

(P) y = 0 = y′′ (Pinned)

2000 Mathematics Subject Classification. 34B05,34B10.

Key words and phrases. Isospectral, Euler-Bernoulli equation for the vibrating beam, beam operator.

\begin{matrix}c \\ ◯ \end{matrix}2005 Texas State University - San Marcos.

Published April 20, 2005.

(S) y′ = 0 = (ry′′)′ (Sliding).

By using a transformation, we can obtain another form of equation (1.2) which is useful in practice. This transformation is as follows:

z = ∫_{s}^{1} [\frac{a(t)}{r(t)}]^{1/4}dt, b(z) = [\frac{a(s)}{r(s)}]^{1/4}, c^{2}(z) = [r^{3}(s)a(s)]^{1/4}.  (1.3)

We have

\frac{d}{ds} = \frac{d}{dz} \frac{dz}{ds} = −[\frac{a(s)}{r(s)}]^{1/4} \frac{d}{dz} = −b(z)\frac{d}{dz} ,

and

r(s)\frac{d^{2}}{ds^{2}} = r(s)b(z)\frac{d}{dz} [b(z)\frac{d}{dz} ] = c^{2}(z)\frac{d}{dz} [b(z)\frac{d}{dz} ].

Therefore, the (1.2) becomes

b\frac{d}{dz} \{b\frac{d}{dz} [c^{2} \frac{d}{dz} (b\frac{dy}{dz} )]\} = λb^{3}c^{2}y,

which can be written as

(b(c^{2}(by′(z))′)′)′ = λb^{2}c^{2}y(z), 0 ≤ z ≤ L,  (1.4)

where L = ∫_{0}^{1}[\frac{a(t)}{r(t)}]^{1/4}dt. For more details see [1] or [3]. Equation (1.2) can be written in the form v^{(4)} + (Av′)′ + Bv = λv as follows. Put y = \frac{v}{bc}, then

by′ = b[\frac{v′}{bc} − \frac{v}{b^{2}c^{2}} (bc′ + b′c)].

If we put \frac{b′}{b} = β and \frac{c′}{c} = γ, then after some algebraic calculations we find

\begin{aligned}by′ & = \frac{1}{c}[v′ − (β + γ)v] \\ (by′)′ & = \frac{1}{c}[v′′ − (β + γ)v′ − (β′ + γ′)v − γv′ + γ(β + γ)v] \\ (c^{2}(by′)′)′ & = c\{v′′′ − (β + γ)v′′ − [2β′ + 3γ′ + γ^{2}]v′ − (β′′ + γ′′)v \\ & + [γ(β + γ)′ + γ′(β + γ) − γ(β + γ)′ + γ^{2}(β + γ)]v\} \\ (b(c^{2}(by′)′)′)′ & = bc\{v^{(4)} − [(β + γ)′ + 2β′ + 3γ′ + γ^{2} + (β + γ)^{2}]v′′ \\ & − [2γγ′ + β′′ + γ′′ − γ′(β + γ) − γ^{2}(β + γ)]v′ \\ & − [2β′′ + 3γ′′ + 2β′(β + γ) + 3γ′(β + γ) + γ^{2}(β + γ)]v′ \\ & + [γ′(β + γ)′ + γ′′(β + γ) + [γ^{2}(β + γ)]′]v \\ & − [β′′′ + γ′′′ + (β + γ)(β′′ + γ′′)]v \\ & + [γ′(β + γ)^{2} + γ^{2}(β + γ)^{2}]v\}. \end{aligned} (1.5)

It is clear that clamped is the only end-condition that remains invariant under the transformation (3), i.e., v = 0 = v′. By using (5) we can write the equation (4) in v as follows:

v^{(4)} + (Av′)′ + Bv = λv,  (1.6)

where

A = −[3β′ + 4γ′ + γ^{2} + (β + γ)^{2}],  (1.7)
\begin{aligned}B & = −(β + γ)′′′ − β′′(β + γ) + γ′(β + γ)′ + γ′(β + γ)^{2} \\ & + [γ^{2}(β + γ)]′ + γ^{2}(β + γ)^{2}. \end{aligned}  (1.8)

Put θ = β + γ and ϕ = γ′ + γ^{2}, then A and B can be written as follows

\begin{aligned}A & = −3θ′ − θ^{2} − ϕ \\ B & = −θ′′′ − (θ′′ − γ′′)θ + γ′θ′ + γ′θ^{2} + (γ^{2}θ)′ + γ^{2}θ^{2} \\ & = −θ′′′ − θ′′θ + (γ′θ)′ + γ′θ^{2} + (γ^{2}θ)′ + γ^{2}θ^{2} \\ & = −θ′′′ − θ′′θ + [(γ′ + γ^{2})θ]′ + (γ′ + γ^{2})θ^{2} \\ & = −θ′′′ − θ′′θ + (θϕ)′ + θ^{2}ϕ \\ & = (−θ′′ + θϕ)′ + (−θ′′ + θϕ)θ. \end{aligned}  (1.9)

If we substitute β = \frac{b′}{b} and γ = \frac{c′}{c} back in (1.7) and (1.8) then A and B can be written in terms of b, c and their derivatives as follows:

A = 2c′^{2}/c^{2} − 4c′′/c − 3b′′/b + 2b′^{2}/b^{2} − 2b′c′/bc  (1.10)

\begin{aligned}B & = 4b′^{4}/b^{4} − c^{(4)}/c + 4c′^{4}/c^{4} + 4c′′^{2}/c^{2} − b^{(4)}/b \\ & + 3b′′^{2}/b^{2} − 10c′^{2}c′′/c^{3} + 4c′c′′′/c^{2} + 3b′b′′′/b^{2} \\ & − 9b′′b′^{2}/b^{3} − 2b′c′^{3}/bc^{3} + 3b′′b′c′/b^{2}c − c′b′′′/bc \\ & − 2b′^{3}c′/b^{3}c + 4b′c′c′′/bc^{2} + b′′c′′/bc. \end{aligned}  (1.11)

2. Isospectral rods

The linear, free, undamped longitudinal vibrations of a thin elastic rod with variable cross section A = A(x) are governed by the equation

(Au′)′ + λAu = 0.  (2.1)

Gladwell and Morassi [2] found isospectral rods, rods with identical spectrum, by using the Darboux lemma. In this section we apply a different method, namely factorizing the rod operator, to find the the isospectral rods. Then we develop this method to find isospectral beams in the following section. This method first was proposed by Pöchel J and Trubowitz E [7] to Sturm-Liouville operator. The rod equation (2.1) can be written in the Sturm-Liouville form. The cross section function A is positive; write A = a^{2}, y = au. Then

Au′ = a^{2}u′ = ay′ − a′y

so that (2.1) reduces to the Sturm-Liouville form

−y′′ + qy = λy, where q = \frac{a′′}{a} .  (2.2)

Define the rod operator

L = −D^{2} + q.  (2.3)

We want to write the rod operator L as a product of two first order differential operators. Every linear differential operator of the form

H(v) = ∑_{k=0}^{n} p_{k}(x)D^{k}(v)  (2.4)

has the adjoint operator H^{∗} of the form

H^{∗}(u) = ∑_{k=0}^{n} D^{k} [(−1)^{k}p_{k}(x)u] .  (2.5)

See for example Lanczos [6]. Therefore the adjoint operator of H = D + α is H^{∗} = −D + α, where α is a real function. Now consider the rod operator L given by (2.3). We want to write L as a product of the form L = H^{∗}H, where H = D+α. Let

L = −D^{2} + q = −D^{2} + \frac{a′′}{a} = H^{∗}H = (−D + α)(D + α).

Thus

q = α^{2} − α′ = \frac{a′′}{a} .  (2.6)

The operator L = H^{∗}H and L̂ = HH^{∗} are isospectral. Now let

L̂ = HH^{∗} = (D + α)(−D + α) = −D^{2} + q̂ = −D^{2} + \frac{â′′}{â} .

Hence

q̂ = α^{2} + α′.  (2.7)

Substituting the solution α = −\frac{a′}{a} in (2.7) yields

q̂ = 2\frac{a′^{2}}{a^{2}} − \frac{a′′}{a} .  (2.8)

On the other hand

q̂ = \frac{â′′}{â} .

In order to determine the isospectral rods we need the general solution of the following differential equation

\frac{â′′}{â} = 2\frac{a′^{2}}{a^{2}} − \frac{a′′}{a} .  (2.9)

The general solution of the equation (2.9) is

â = \frac{1}{a}\{1 + K ∫_{0}^{x} a^{2}(s)ds\}, where K ∈ ℝ.

This family of isospectral rods coincide with those obtained by Gladwell and Morassi [2].

Remark 2.1. In general if we consider the Sturm-Liouville operator L given by (2.3), using a similar argument we can factor L − μ as follows

L − μ = H^{∗}H = (−D − \frac{g′}{g})(D − \frac{g′}{g}),  (2.10)

where g is a nonzero solution of the Sturm-Liouville equation, i.e.,

−g′′ + (q − μ)g = 0.

If we reverse the factors in (2.10) then we obtain the isospectral Sturm-Liouville operator

L̂ − μ = HH^{∗} = −D^{2} + (q̂ − μ),  (2.11)

where q̂ = q − 2\frac{d^{2}}{dx^{2}} (ln g). For more details see [7].

3. Isospectral beams

According to the beam equation (1.6) we define the beam operator as follows

L = H^{∗}H = D^{4} + D(AD) + B,  (3.1)

where A and B are given by (1.7) and (1.8). Using (2.4) and (2.5) it can be easily checked that the adjoint operator of a general operator of the form H = aD^{2}+bD+c is H^{∗} = aD^{2} +(2a′ −b)D+(a′′ −b′ +c). The idea is to find the isospectral beams by factorizing the beam operator L and then reversing the factors to find the isospectral beams. We therefore suppose that L can be factorized as follows

L = H^{∗}H = [D^{2} − rD + (s − r′)][D^{2} + rD + s]

that is equivalent to the nonlinear system:

{2s + r′ − r^{2} = A, s^{2} + s′′ − (rs)′ = B}.

Define L̂ = HH^{∗}. Then L and L̂ are isospectral and we have

L = H^{∗}H = D^{4} + D(AD) + B, L̂ = HH^{∗} = D^{4} + D(ÂD) + B̂,  (3.2)

where

\begin{matrix}A = 2s + r′ − r^{2} \\ B = s^{2} + s′′ − (rs)′ , \end{matrix}  (3.3)

and

\begin{matrix}Â = 2s − 3r′ − r^{2} \\ B̂ = s^{2} + s′′ − r′′′ − rr′′ + rs′ − sr′. \end{matrix}  (3.4)

Let us call the system (3.3) the principal system because the solutions of (3.3) will produce isospectral beams. It is easy to verify that r = −β − 2γ, and s = γ^{2} + βγ − β′ − γ′ is a solution to the nonlinear system (3.3). Comparing (3.3) and (3.4) we obtain

A − Â = 4r′, B − B̂ = r′′′ + rr′′ − 2rs′.  (3.5)

Substituting r = −β − 2γ, and s = γ^{2} + βγ − β′ − γ′ in (3.5) we find

Â = β′ + 4γ′ − β^{2} − 2γ^{2} − 2βγ,  (3.6)

and

\begin{aligned}B̂ & = γ′′′ + β′′γ + γ′(β + γ)′ − 4βγγ′ − β^{2}γ′ − 4γ′γ^{2} \\ & − 3β′γ^{2} − 2ββ′γ + γ^{2}(β + γ)^{2}. \end{aligned}  (3.7)

It is clear that

Â(β, γ) = A(β, −β − γ),
B̂(β, γ) = B(β, −β − γ).

This implies that β̂ = β and γ̂ = −β − γ. Therefore, b̂ = b and ĉ = \frac{1}{bc}.

Remark 3.1. We can verify this isospectral family as follows. Consider the original beam equation

(b(c^{2}(by′)′)′)′ = λb^{2}c^{2}y.

Put H_{1} = \frac{1}{b^{2}c^{2}} (DbD) and H_{2} = c^{2}(DbD). Then

(b(c^{2}(by′)′)′)′ = λb^{2}c^{2}y ⟺ H_{1}H_{2}(y) = λy.

Therefore H_{2}H_{1}(H_{2}y) = λ(H_{2}y), which can be written as H_{2}H_{1}(v) = λv, where v = H_{2}y.

Remark 3.2. Similar to Remark 2.1, if we factorize the operator L−μ instead the beam operator L given by (3.1) then the corresponding principal system will be

\begin{matrix}2s + r′ − r^{2} = A \\ s^{2} + s′′ − (rs)′ = B − μ = −\frac{g^{(4)}}{g} − \frac{(Ag′)′}{g} \end{matrix}  (3.8)

where g is a nonzero solution of the beam equation, i.e.,

g^{(4)} + (Ag′)′ + (B − μ)g = 0.

In contrast to the principal system (3.3), the system (3.8) does not seem to be easy to solve. The operator L̂ coincide with the beam operator given by Gottlieb [5]. There he finds seven classes of beams isospectral to the standard unit coefficient beam equation

D^{4}v = λv.  (3.9)

Since L and L̂ are isospectral, by using the operator L we can find another seven class of beams isospectral to the standard unit-coefficient beam. Referring to Gottlieb’s notations we define

b = η(δz + ϵ)^{μ}, c = ζ(δz + ϵ)^{ν},  (3.10)

where δ, ϵ, η, ζ are constants, and μ, ν are parameters to be determined from the expressions of A, B in terms of b, c and their derivatives given by (1.10) and (1.11). After some algebraic calculations we find

A = −δ^{2}(δz + ϵ)^{−2}[μ^{2} + 2μν + 2ν^{2} − 4ν − 3μ]  (3.11)
\begin{aligned}B & = δ^{4}(δz + ϵ)^{−4}[ν^{4} − 4ν^{3} + μ^{2}ν^{2} + ν^{2}] \\ & × [2μν^{3} − 2μ^{2} − νμ^{2} − μν − 5μν^{2} + 6(μ + ν)]. \end{aligned}  (3.12)

Using the transformation given by (1.3), we get

\begin{aligned}1 − s & = ∫_{0}^{z} \frac{dz}{b(z)} \\ & = ∫_{0}^{z} \frac{1}{η} (δz + ϵ)^{−μ}dz \\ & = \frac{1}{η(1 − μ)}(δz + ϵ)^{1−μ} − \frac{1}{η(1 − μ)}ϵ^{1−μ}. \end{aligned}  (3.13)

Now put

Z = δz + ϵ, and ξ(s) = ϵ^{1−μ} + η(1 − μ)(1 − s).  (3.14)

Then Z = [ξ(s)]^{χ}, where χ = \frac{1}{1−μ}, and hence

z(s) = δ^{−1} ([ξ(s)]^{χ} − ϵ) .  (3.15)

Thus

b (z(s)) = η[ξ(s)]^{λ}, with λ = \frac{μ}{1 − μ}  (3.16)
c (z(s)) = ζ[ξ(s)]^{κ}, with κ = \frac{ν}{1 − μ}.  (3.17)

Therefore, the seven class of beams isospectral to the standard unit-coefficient beam is corresponding to A = B = 0; so that we can determine these classes by solving the algebraic equations

\begin{matrix}μ^{2} + 2μν + 2ν^{2} − 4ν − 3μ = 0 \\ ν^{4} − 4ν^{3} + μ^{2}ν^{2} + ν^{2} + 2μν^{3} − 2μ^{2} − νμ^{2} − μν − 5μν^{2} + 6(μ + ν) = 0 \end{matrix} (3.18)

which gives the solutions cited in the following table:

\begin{matrix}class & μ & ν & χ & λ & κ \\ (0) & 0 & 0 & 1 & 0 & 0 \\ (1) & 0 & 2 & 1 & 0 & 2 \\ (2) & −1 & 2 & 1/2 & −1/2 & 1 \\ (3) & −1 & 1 & 1/2 & −1/2 & 1/2 \\ (4) & 2 & 1 & −1 & −2 & −1 \\ (5) & 2 & −1 & −1 & −2 & 1 \\ (6) & 3 & −1 & −1/2 & −3/2 & 1/2 \\ (7) & 3 & 0 & −1/2 & −3/2 & 0 \end{matrix}

More isospectral beams. For this purpose, we put θ = β + γ and ϕ = γ′ + γ^{2}. Then (1.9) implies

\begin{matrix}A = A(θ, ϕ) = −3θ′ − θ^{2} + ϕ \\ B = B(θ, ϕ) = (−θ′′ + θϕ)′ + (−θ′′ + θϕ)θ. \end{matrix}  (3.19)

Let Ã = A(θ̃, ϕ̃), and B̃ = B(θ̃, ϕ̃). If A = Ã and B = B̃ then L and L̃ are isospectral, where L is given by (3.1) and

L̃ = D^{4} + D(ÃD) + B̃.  (3.20)

Let θ = θ̃, ϕ = ϕ̃, then A = Ã and B = B̃. Thus we obtain isospectral class as follows

θ = θ̃ ⟺ bc = b̃c̃.  (3.21)

Now the equation ϕ = ϕ̃ implies

\frac{c′′}{c} = \frac{c̃′′}{c̃} ⟺ c′′c̃ − c̃′′c = 0 ⟺ c′c̃ − c̃′c = const.  (3.22)

This equation implies that for arbitrary real constants K_{1}, K_{2} we find

c̃ = c ∫ \frac{K_{1}}{c^{2}} ds + K_{2}c.

Now with a similar way we can find another family isospectral to L̂ which is isospectral to L. Put ϕ = θ^{2} − θ′. After some algebraic computation we can write Â and B̂ as follows

\begin{matrix}Â = 3γ′ − γ^{2} − ϕ \\ B̂ = (γ′′ − γϕ)′ − γ(γ′′ − γϕ). \end{matrix}  (3.23)

If γ = γ̃ and ϕ = ϕ̃ then L̂ and ˜ L̂ are isospectreal. The equation γ = γ̃ implies c = c̃. Now we find b̃ by solving the equation ϕ = ϕ̃. We have

ϕ = ϕ̃ ⟺ (θ̃ − θ)′ = θ̃^{2} − θ^{2} ⟺ \frac{(θ̃ − θ)′}{θ̃ − θ} = θ̃ + θ.

This equation implies that θ̃ − θ = bcb̃c̃ = bb̃c^{2}, which is equivalent to

\frac{b̃′}{b̃} − \frac{b′}{b} = bb̃c^{2} ⟺ b̃′b − b′b̃ = (bb̃)^{2}c^{2} ⟺ (\frac{b̃}{b})′ = b̃^{2}c^{2}.

This equation can be easily solved. In fact, putting u = \frac{b̃}{b}, the above equation becomes

u′ = u^{2}b^{2}c^{2}.

By solving this differential equation, we find

b̃ = \frac{b}{K_{1} − ∫ b^{2}c^{2}ds}, where K_{1} ∈ ℝ.

Suppose

b = c^{2}( ∫ \frac{K_{1}}{c^{2}} ds + K_{2}).

Then we can define another isospectral class as follows: Define b̃ = c^{2} and c̃ = \sqrt{b}. Then bc^{2} = b̃c̃^{2}. Hence r = r̃ where r = β + 2γ. It is clear that s = s̃, thus we have A = Ã and B = B̃ according to the principal system (3.3).

Conclusion. Factoring the beam operator L = D^{4} + D(AD) + B into the product of the form L = HH^{∗}, where H is a second order differential operator of the form H = D^{2} + rD + s and H^{∗} is the adjoint operator of H, and reversing the order of this product we managed to find some 12 classes of isospectral beams. However, determining all classes is linked to nonlinear system (3.3) for which we found particular solution.

Acknowledgement. The author would like to thank anonymous referee for careful reading of the manuscript an giving valuable comments and helpful references.

References

[1] V. Barcilon; Inverse Problem for a vibrating beam in the free-clamped configuration, Phil. Trans. R. Soc. London A 304(1982), 211-251.

[2] G. M. Gladwell and A. Morassi; On the isospectral rods, horns and strings, Inverse Problems, 11 (1995) 533-554.

[3] G. M. L. Gladwell; Inverse Problems in Vibration, Kluwer, Dordrecht 1986.

[4] G. M. L. Gladwell; Isospectral vibrating beams, Proc. R. Soc. London A 458(2002), 2691-2703.

[5] H. P. W. Gottlieb; Isospectral Euler-Bernoulli beams with continous density and rigidity functions, Proceeding of Royal Society of London A 423 (1987) 235-250.

[6] C. Lanczos, 1961 Linear Differential Operators, D. Van Nostrand Company, Ltd, London.

[7] Pöschel J and Trubwitz E; Inverse Spectral Theory (London: Academic), 1987.

[8] G. Subramanian and A. Raman; Isopectral systems for tapered beams, J. Sound and Vibration 198 (1996) 257-266.

Department of Mathematics, Sahand University of Technology, Tabriz, Iran

   E-mail address: kghanbari@sut.ac.ir