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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
Trapped modesScattering problem
Outline
1 Trapped modesFormulationTrapped modes
2 Scattering problemFormulationT and R coefficients
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
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].
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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,
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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
Trapped modesScattering problem
FormulationTrapped modes
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).
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationTrapped modes
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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.
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
(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)
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
Trapped modesScattering problem
FormulationT and R coefficients
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).
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam
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