Ponencia en el XVIII CCM

Trapped modesScattering problem

Scattering problem for an infinitenonhomogeneous Timoshenko beam

Hugo Aya1 Ricardo Cano2 Peter Zhevandrov2

1Universidad Distrital

2Universidad de La Sabana

H. Aya, R. Cano and P. Zhevandrov Timoshenko beam


Outline

1 Trapped modesFormulationTrapped modes

2 Scattering problemFormulationT and R coefficients



FormulationTrapped modes

Timoshenko system

Consider the Timoshenko system(Hagedorn & DasGupta 2007):{EIψ′′ − (kGA− ω2ρI)ψ + kGAy′ = 0,

−kGAψ′ + kGAy′′ + ω2ρAy = 0.(1)

Here ω is the frequency, A is the area of the cross-section, I isits second moment, G is the shear modulus, k is the Timoshenkoshear coefficient. The time dependence is assumed to be of theform exp(−iωt). We assume that the density ρ has the form

ρ = ρ0

(1 + εf(x)

),−∞ < x <∞, ε� 1,

and f(x) (the perturbation) belongs to C[−1, 1].




Introduce the non-dimensional quantities according to theformulas

y = L∗y, x = L∗x, ω = ω0ω, ω0 =(kGA

ρ0I

)1/2

,

(1) acquires the formψ′′ − γ2

α2(1− ω2)ψ +

γ2

α2y′ = −εω

2γ2

α2f(x)ψ,

−γ2

α2ψ′ +

γ2

α2y′′ +

ω2γ2

α4y = −εω

2γ2

α4f(x)y,

(2)

where

α2 =r2gL2∗, γ2 =

kG

E, rg =

√I/A.




The cut-off frequency is now ω2 = ω20 = 1.

If ε ≡ 0 in (2), it is well-known that

Ψ ≡(ψy

)∝ e±ik1,2x,

where

k21,2 =

ω2

2α2

{1 + γ2 ±

√(1− γ2)2 +

4γ2

ω2

}(3)




We will be interested in the situation when the frequency isslightly less than the cut-off frequency, looking for aneigenfunction of (2) corresponding to

ω2 = 1− β2,

where we assume that β > 0 is small, β → 0 as ε→ 0. Theexpression under the radical sign in (3) is

(1− γ2

)2 +4γ2

ω2=(1 + γ2

)2 +4β2γ2

ω2.

Thus we havek2

1 > 0, k22 < 0.




Denotem2 = k2

1, l2 = −k22;

in this way, the four linearly independent solutions of (2) forε = 0 are

Vmeimx, V−me−imx, Vle

lx, V−le−lx,

where

m2 =ω2

2α2(1 + γ2)

{√1 +

4β2γ2

ω2(1 + γ2)2+ 1

}(4)

l2 =ω2

2α2(1 + γ2)

{√1 +

4β2γ2

ω2(1 + γ2)2− 1

}, (5)




and

Vm =(

1−ibm

), V−m =

(1ibm

), (6)

Vl =(

1bl

), V−l =

(1−bl

)(7)

are the eigenvectors with bl and bm given by

bl = − l2α2/γ2 − 1 + ω2

l=

l

l2 + ω2/α2, (8)

bm =m2α2/γ2 + 1− ω2

m=

m

m2 − ω2/α2. (9)




For small β, we have

l2 = l21β2 +O(β4), m2 = m2

0 −m21β

2 +O(β4), (10)

where

l21 =γ2

α2(1 + γ2), m2

0 =1 + γ2

α2, m2

1 =1 + γ2 + γ4

α2(1 + γ2),

and hence

bl =αγ√1 + γ2

β +O(β3), bm =α√

1 + γ2

γ2+O(β2). (11)




Integral equation

Our goal is to construct solutions of (2) for ε > 0 decreasing atinfinity. We convert system (2) into an integral equation withthe aid of the Green matrix of the homogeneous problem(wenote that this scheme was used by Gadyl′shin (2002) for theSchrödinger equation). Let G(x, ξ) be a Green matrix,

G(x, ξ) =(g11(x, ξ) g12(x, ξ)g21(x, ξ) g22(x, ξ)

), LG = δ(x− ξ)E, (12)

where

L =(L11 L12

L21 L22

), E =

(1 00 1

),

L11[ψ] = ψ′′ − γ2

α2(1− ω2)ψ,

L21[ψ] = −γ2

α2ψ′,

L12[y] =γ2

α2y′,

L22[y] =γ2

α2y′′ +

ω2γ2

α4y.




We look for the solution in the form

Ψ =∫

G(x, ξ)A(ξ) dξ, A(ξ) =(B(ξ)D(ξ)

); (13)

Substituting (13) in (2) and using (12), we obtain

A(x) = −εω2f(x)J∫

G(x, ξ)A(ξ) dξ, (14)

where

J =γ2

α2

(1 00 α−2

). (15)




Outgoing Green matrix

Since the Green matrix is nonunique due to the existence ofplane waves of the first branch, we have to impose someconditions at x→ ±∞. We choose these conditions in such away that the Green matrix has only outgoing components atinfinity, that is, describes waves travelling to the right for x > 1and to the left for x < −1. This Green matrix is often called the‘outgoing Green matrix’ and hence our G(x, ξ) should have thefollowing form:(

g11(x, ξ)g21(x, ξ)

)=

{C1V−le−lx + C2Vme

imx, if x > ξ,

C3Vlelx + C4V−me−imx, if x < ξ,

(g12(x, ξ)g22(x, ξ)

)=

{C1V−le−lx + C2Vme

imx, if x > ξ,

C3Vlelx + C4V−me−imx, if x < ξ,




The Green matrix should satisfy the following conditions:i) Continuity,

G|x→ξ− = G|x→ξ+ ; (16)

ii) Jump condition,d

dx

(g11(x, ξ)g21(x, ξ)

)∣∣∣∣x→ξ+

− d

dx

(g11(x, ξ)g21(x, ξ)

)∣∣∣∣x→ξ−

=(

10

),

d

dx

(g12(x, ξ)g22(x, ξ)

)∣∣∣∣x→ξ+

− d

dx

(g12(x, ξ)g22(x, ξ)

)∣∣∣∣x→ξ−

=(

0α2/γ2

).

(17)




Thus we finally obtain (t = x− ξ)

G(x, ξ) =

12q2

(− 1ble−l|t| − i

bmeim|t| sgn t

(−e−l|t| + eim|t|

)sgn t

(e−l|t| − eim|t|

)ble−l|t| − ibmeim|t|

), (18)

where q2 = l2 +m2.




Equation for A(ξ)

The only singular term (as β → 0) of matrix (18) is theexpression

− 12q2bl

e−l|t| = − α

2γ√

1 + γ2β−1 +O(1), β → 0,

in the upper left corner. Therefore, it is convenient to introducethe ‘regularized’ Green matrix

Gr(x, ξ) = G(x, ξ) +α

2βγ√

1 + γ2

(1 00 0

)and the operator T defined by

T [A(x)] = ω2f(x)J∫ 1

−1Gr(x, ξ)A(ξ) dξ. (19)




It follows from (13),(14) and (15) that

A(x) = −ε ω2 f(x)J∫ 1

−1G(x, ξ) A(ξ) dξ

= −εω2f(x)J∫ 1

−1Gr(x, ξ)A(ξ) dξ +

ε

2βγω2B0f(x)

α√

1 + γ2

(10

)= −ε T [A(x)] +

ε

2βγω2B0f(x)

α√

1 + γ2

(10

),

where

B0 =∫ 1

−1B(x) dx.

Thus we obtain the following integral equation for A:(1 + ε T

)A(x) =

ε

2βγω2B0f(x)

α√

1 + γ2

(10

). (20)




Hence, we have

A =ε

βB0

(UV

), (21)

where(UV

)=

ω2γ

2α√

1 + γ2

(1 + εT

)−1(f(x)

0

)=

ω2γ

2α√

1 + γ2

(f(x)

0

)+O(ε).

(22)Integrating the first component of equation (21), multiplying byβ, and dividing by B0, we obtain

β = ε

∫ 1

−1U(x) dx, (23)

where U is defined by (22) and is analytic in ε, β in aneighbourhood of the origin. We have for small ε

β =εγ

2α√

1 + γ2

∫f(x) dx+O(ε2) (24)




Since we have assumed that β > 0, equation (23) has a solutionif ∫

f(x) dx > 0. (25)

Thus we have obtained a solution of (2) given by (13) under thecondition (25). Our solution contains oscillating terms for|x| > 1. Indeed, we have for x� 1

Ψ =eimx

2q2(B+ + ibmD+)

(−ib−1

m

−1

)+O(e−lx), (26)

and for x� −1

Ψ =e−imx

2q2(B− − ibmD−)

(−ib−1

m

1

)+O(elx), (27)

where

B± =∫e∓imξB(ξ) dξ, D± =

∫e∓imξD(ξ) dξ.




Since we are interested in a solution which decreases for largevalues of |x|, to achieve this, it is sufficient that the coefficientsof e±imx in (26), (27) be equal to zero. Note that we have a freeparameter m which depends on α and γ. The equality to zero ofthe coefficients of eimx and e−imx gives two equations for m,which do not have a solution in general:

B+ + ibmD+ = 0, B− − ibmD− = 0. (28)

To obtain a single equation, assume that the perturbation f(x)is even. This implies that, B(x) and D(x) given by (21) are evenand odd, respectively and

B± =∫B(ξ) cosmξ dξ, D± = ∓i

∫D(ξ) sinmξ dξ, (29)

orB+ = B−, D+ = −D−, (30)

and hence the two equations in (28) coincide.H. Aya, R. Cano and P. Zhevandrov Timoshenko beam



Thus equations (28) become

B+ + ibmD+ = B− − ibmD− =∫B(ξ) cosmξ dξ

+ bm

∫D(ξ) sinmξ dξ = 0. (31)

Therefore, to eliminate the oscillating component we have tosatisfy a unique equation for m:∫

B(ξ) cosmξ dξ + bm

∫D(ξ) sinmξ dξ = 0. (32)

By (22) and the implicit function theorem, if m0 solves∫cosmξf(ξ) dξ = 0 and∫

ξ sinm0ξf(ξ) dξ 6= 0, (33)

then there exists an analytic in ε solution of (32),m = m0 +O(ε).




Main result

TheoremLet f(x) be even and β > 0 and m be solutions of equations(23), (32), where the functions B and D are defined in (21), andassume that (33) is satisfied. Then system (2) possesses, forω2 = 1− β2, an exponentially decreasing solution (trapped mode)Ψ given by Ψ(x) =

∫G(x, ξ)A(ξ) dξ.



FormulationT and R coefficients

We still assume thatω2 = 1− β2.

Scattering problem consists in finding solution of (2) of the form

Ψ|x�−1 = Vmeimx +RV−me−imx +O(elx), (34)

Ψ|x�1 = TVmeimx +O(e−lx). (35)

R is the reflection coefficient.T is the transmission coefficient.This problem does not have a unique solution under theconditions of Theorem. Let equation (32) be false,∫

cosmξf(ξ) dξ 6= 0. Then the scattering problem has a uniquesolution.




Now we look for the solution of (2) in the form

Ψ =∫

G(x, ξ)A(ξ) dξ + C1Vmeimx + C2V−me

−imx, (36)

Substituting (36) in (2) and using (12), we obtain

A(x) = −εω2f(x)J(∫

G(x, ξ)A(ξ)dξ + C1Vmeimx + C2V−me

−imx)

(37)




From (36) we have for x > 1

Ψ =(C1 −

12q2

(i

bmB+ −D+

))Vme

imx+C2V−me−imx+O(e−lx),

(38)and for x < −1

Ψ =(C2 −

12q2

(i

bmB− +D−

))V−me−imx+C1Vme

imx+O(elx),

(39)where

B± =∫ 1

−1e∓imξB(ξ) dξ, D± =

∫ 1

−1e∓imξD(ξ) dξ. (40)




Reflection and transmission coefficients

Choose C1 = 1, C2 = 0 and denote

T = 1− 12q2

(i

bmB+ −D+

), (41)

R = − 12q2

(i

bmB− +D−

). (42)

We see that T and R are the reflection and transmissioncoefficients.




From (37) we have

A(x) = −εω2f(x)J(∫

G(x, ξ)A(ξ)dξ + Vmeimx

)= −εω2f(x)J

∫Gr(x, ξ)A(ξ)dξ

+ εω2f(x)

(α

2βγ√

1 + γ2B0

(10

)− JVmeimx

)

= −εT [A(x)] + εω2f(x)

(α

2βγ√

1 + γ2B0

(10

)− JVmeimx

)(43)




Thus(1 + εT

)[A(x)] =

ε

2βγω2f(x)B0

α√

1 + γ2

(10

)−εω2JVmf(x)eimx, (44)

and

A(x) =ε

βB0

(UV

)− ε(U1

V1

)+ ε

(U2

V2

)(45)

where(UV

)=

γω2

2α√

1 + γ2

(1 + εT

)−1(f(x)

0

)=

γω2

2α√

1 + γ2

(f(x)

0

)+O(ε)




(U1

V1

)= ω2

(1 + εT

)−1(f(x)eimx

0

)(U2

V2

)= iω2bmα

−2(1 + εT

)−1(

0f(x)eimx

)

Integrating the first component of equation (45), we obtain

B0

(1− ε

β

∫U(x) dx

)= ε

(∫(U2(x)− U1(x)) dx

)(46)




Thus

B0 =εβ(∫

(U2(x)− U1(x)) dx)(

β − ε∫U(x) dx

) . (47)

Therefore, for β = O(ε), T = 1 +O(ε), R = O(ε), but if βcoincides with the real part of the solution ofβ − ε

∫U(x) dx = 0, then we have R = O(1) (the Wood

anomaly).


Ponencia en el XVIII CCM

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