Ponencia en el XVIII CCM

Trapped modes
Scattering problem
Scattering problem for an inﬁnite
nonhomogeneous Timoshenko beam
Hugo Aya1 Ricardo Cano2 Peter Zhevandrov2
1Universidad Distrital
2Universidad de La Sabana
H. Aya, R. Cano and P. Zhevandrov Timoshenko beam

Trapped modes
Scattering problem
Outline
1 Trapped modes
Formulation
Trapped modes
2 Scattering problem
Formulation
T and R coeﬃcients

Trapped modes
Scattering problem
Formulation
Trapped modes
Timoshenko system
Consider the Timoshenko system(Hagedorn & DasGupta 2007):
EIψ − (kGA − ω2
ρI)ψ + kGAy = 0,
−kGAψ + kGAy + ω2
ρAy = 0.
(1)
Here ω is the frequency, A is the area of the cross-section, I is
its second moment, G is the shear modulus, k is the Timoshenko
shear coeﬃcient. The time dependence is assumed to be of the
form 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].

Trapped modes
Scattering problem
Formulation
Trapped modes
Introduce the non-dimensional quantities according to the
formulas
y = L∗y, x = L∗x, ω = ω0ω, ω0 =
kGA
ρ0I
1/2
,
(1) acquires the form



ψ −
γ2
α2
(1 − ω2
)ψ +
γ2
α2
y = −
ω2γ2
α2
f(x)ψ,
−
γ2
α2
ψ +
γ2
α2
y +
ω2γ2
α4
y = −
ω2γ2
α4
f(x)y,
(2)
where
α2
=
r2
g
L2
∗
, γ2
=
kG
E
, rg = I/A.

Trapped modes
Scattering problem
Formulation
Trapped modes
The cut-oﬀ frequency is now ω2 = ω2
0 = 1.
If ≡ 0 in (2), it is well-known that
Ψ ≡
ψ
y
∝ e±ik1,2x
,
where
k2
1,2 =
ω2
2α2
1 + γ2
± (1 − γ2)2 +
4γ2
ω2
(3)

Trapped modes
Scattering problem
Formulation
Trapped modes
We will be interested in the situation when the frequency is
slightly less than the cut-oﬀ frequency, looking for an
eigenfunction of (2) corresponding to
ω2
= 1 − β2
,
where we assume that β > 0 is small, β → 0 as → 0. The
expression under the radical sign in (3) is
1 − γ2 2
+
4γ2
ω2
= 1 + γ2 2
+
4β2γ2
ω2
.
Thus we have
k2
1 > 0, k2
2 < 0.

Trapped modes
Scattering problem
Formulation
Trapped modes
Denote
m2
= k2
1, l2
= −k2
2;
in this way, the four linearly independent solutions of (2) for
= 0 are
Vmeimx
, V−me−imx
, Vlelx
, 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)

Trapped modes
Scattering problem
Formulation
Trapped modes
and
Vm =
1
−ibm
, V−m =
1
ibm
, (6)
Vl =
1
bl
, 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)

Trapped modes
Scattering problem
Formulation
Trapped modes
For small β, we have
l2
= l2
1β2
+ O(β4
), m2
= m2
0 − m2
1β2
+ O(β4
), (10)
where
l2
1 =
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)

Trapped modes
Scattering problem
Formulation
Trapped modes
Integral equation
Our goal is to construct solutions of (2) for > 0 decreasing at
inﬁnity. We convert system (2) into an integral equation with
the aid of the Green matrix of the homogeneous problem(we
note that this scheme was used by Gadyl shin (2002) for the
Schrödinger equation). Let G(x, ξ) be a Green matrix,
G(x, ξ) =
g11(x, ξ) g12(x, ξ)
g21(x, ξ) g22(x, ξ)
, ˆLG = δ(x − ξ)E, (12)
where
ˆL =
ˆL11
ˆL12
ˆL21
ˆL22
, E =
1 0
0 1
,
ˆL11[ψ] = ψ −
γ2
α2
(1 − ω2
)ψ,
ˆL21[ψ] = −
γ2
α2
ψ ,
ˆL12[y] =
γ2
α2
y ,
ˆL22[y] =
γ2
α2
y +
ω2γ2
α4
y.

Trapped modes
Scattering problem
Formulation
Trapped 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) = − ω2
f(x)J G(x, ξ)A(ξ) dξ, (14)
where
J =
γ2
α2
1 0
0 α−2 . (15)

Trapped modes
Scattering problem
Formulation
Trapped modes
Outgoing Green matrix
Since the Green matrix is nonunique due to the existence of
plane waves of the ﬁrst branch, we have to impose some
conditions at x → ±∞. We choose these conditions in such a
way that the Green matrix has only outgoing components at
inﬁnity, that is, describes waves travelling to the right for x > 1
and to the left for x < −1. This Green matrix is often called the
‘outgoing Green matrix’ and hence our G(x, ξ) should have the
following form:
g11(x, ξ)
g21(x, ξ)
=
C1V−le−lx + C2Vmeimx, if x > ξ,
C3Vlelx + C4V−me−imx, if x < ξ,
g12(x, ξ)
g22(x, ξ)
=
C1V−le−lx + C2Vmeimx, if x > ξ,
C3Vlelx + C4V−me−imx, if x < ξ,

Trapped modes
Scattering problem
Formulation
Trapped 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−
=
1
0
,
d
dx
g12(x, ξ)
g22(x, ξ) x→ξ+
−
d
dx
g12(x, ξ)
g22(x, ξ) x→ξ−
=
0
α2/γ2 .
(17)

Trapped modes
Scattering problem
Formulation
Trapped modes
Thus we ﬁnally obtain (t = x − ξ)
G(x, ξ) =
1
2q2
− 1
bl
e−l|t| − i
bm
eim|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.

Trapped modes
Scattering problem
Formulation
Trapped modes
Equation for A(ξ)
The only singular term (as β → 0) of matrix (18) is the
expression
−
1
2q2bl
e−l|t|
= −
α
2γ 1 + γ2
β−1
+ O(1), β → 0,
in the upper left corner. Therefore, it is convenient to introduce
the ‘regularized’ Green matrix
Gr(x, ξ) = G(x, ξ) +
α
2βγ 1 + γ2
1 0
0 0
and the operator ˆT deﬁned by
ˆT [A(x)] = ω2
f(x)J
1
−1
Gr(x, ξ)A(ξ) dξ. (19)

Trapped modes
Scattering problem
Formulation
Trapped modes
It follows from (13),(14) and (15) that
A(x) = − ω2
f(x)J
1
−1
G(x, ξ) A(ξ) dξ
= − ω2
f(x)J
1
−1
Gr(x, ξ)A(ξ) dξ +
2β
γω2B0f(x)
α 1 + γ2
1
0
= − ˆT [A(x)] +
2β
γω2B0f(x)
α 1 + γ2
1
0
,
where
B0 =
1
−1
B(x) dx.
Thus we obtain the following integral equation for A:
1 + ˆT A(x) =
2β
γω2B0f(x)
α 1 + γ2
1
0
. (20)

Trapped modes
Scattering problem
Formulation
Trapped modes
Hence, we have
A =
β
B0
U
V
, (21)
where
U
V
=
ω2γ
2α 1 + γ2
1 + ˆT
−1 f(x)
0
=
ω2γ
2α 1 + γ2
f(x)
0
+O( ).
(22)
Integrating the ﬁrst component of equation (21), multiplying by
β, and dividing by B0, we obtain
β =
1
−1
U(x) dx, (23)
where U is deﬁned by (22) and is analytic in , β in a
neighbourhood of the origin. We have for small
β =
γ
2α 1 + γ2
f(x) dx + O( 2
) (24)

Trapped modes
Scattering problem
Formulation
Trapped modes
Since we have assumed that β > 0, equation (23) has a solution
if
f(x) dx > 0. (25)
Thus we have obtained a solution of (2) given by (13) under the
condition (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 imξ
B(ξ) dξ, D± = e imξ
D(ξ) dξ.

Trapped modes
Scattering problem
Formulation
Trapped modes
Since we are interested in a solution which decreases for large
values of |x|, to achieve this, it is sufficient that the coefficients
of e±imx in (26), (27) be equal to zero. Note that we have a free
parameter m which depends on α and γ. The equality to zero of
the 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 even
and odd, respectively and
B± = B(ξ) cos mξ dξ, D± = i D(ξ) sin mξ dξ, (29)
or
B+ = B−, D+ = −D−, (30)
and hence the two equations in (28) coincide.

Trapped modes
Scattering problem
Formulation
Trapped modes
Thus equations (28) become
B+ + ibmD+ = B− − ibmD− = B(ξ) cos mξ dξ
+ bm D(ξ) sin mξ dξ = 0. (31)
Therefore, to eliminate the oscillating component we have to
satisfy a unique equation for m:
B(ξ) cos mξ dξ + bm D(ξ) sin mξ dξ = 0. (32)
By (22) and the implicit function theorem, if m0 solves
cos mξf(ξ) dξ = 0 and
ξ sin m0ξf(ξ) dξ = 0, (33)
then there exists an analytic in solution of (32),
m = m0 + O( ).

Trapped modes
Scattering problem
Formulation
Trapped modes
Main result
Theorem
Let 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), and
assume that (33) is satisﬁed. Then system (2) possesses, for
ω2 = 1 − β2, an exponentially decreasing solution (trapped mode)
Ψ given by Ψ(x) = G(x, ξ)A(ξ) dξ.

Trapped modes
Scattering problem
Formulation
We still assume that
ω2
= 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 the
conditions of Theorem. Let equation (32) be false,
cos mξf(ξ) dξ = 0. Then the scattering problem has a unique
solution.

Trapped modes
Scattering problem
Formulation
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) = − ω2
f(x)J G(x, ξ)A(ξ)dξ + C1Vmeimx
+ C2V−me−imx
(37)

Trapped modes
Scattering problem
Formulation
From (36) we have for x > 1
Ψ = C1 −
1
2q2
i
bm
B+ − D+ Vmeimx
+C2V−me−imx
+O(e−lx
),
(38)
and for x < −1
Ψ = C2 −
1
2q2
i
bm
B− + D− V−me−imx
+C1Vmeimx
+O(elx
),
(39)
where
B± =
1
−1
e imξ
B(ξ) dξ, D± =
1
−1
e imξ
D(ξ) dξ. (40)

Trapped modes
Scattering problem
Formulation
Reflection and transmission coefficients
Choose C1 = 1, C2 = 0 and denote
T = 1 −
1
2q2
i
bm
B+ − D+ , (41)
R = −
1
2q2
i
bm
B− + D− . (42)
We see that T and R are the reflection and transmission
coefficients.

Trapped modes
Scattering problem
Formulation
From (37) we have
A(x) = − ω2
f(x)J G(x, ξ)A(ξ)dξ + Vmeimx
= − ω2
f(x)J Gr(x, ξ)A(ξ)dξ
+ ω2
f(x)
α
2βγ 1 + γ2
B0
1
0
− JVmeimx
= − ˆT[A(x)] + ω2
f(x)
α
2βγ 1 + γ2
B0
1
0
− JVmeimx
(43)

Trapped modes
Scattering problem
Formulation
Thus
1 + ˆT [A(x)] =
2β
γω2f(x)B0
α 1 + γ2
1
0
− ω2
JVmf(x)eimx
, (44)
and
A(x) =
β
B0
U
V
−
U1
V1
+
U2
V2
(45)
where
U
V
=
γω2
2α 1 + γ2
1 + ˆT
−1 f(x)
0
=
γω2
2α 1 + γ2
f(x)
0
+O( )

Trapped modes
Scattering problem
Formulation
U1
V1
= ω2
1 + ˆT
−1 f(x)eimx
0
U2
V2
= iω2
bmα−2
1 + ˆT
−1 0
f(x)eimx
Integrating the ﬁrst component of equation (45), we obtain
B0 1 −
β
U(x) dx = (U2(x) − U1(x)) dx (46)

Trapped modes
Scattering problem
Formulation
Thus
B0 =
β (U2(x) − U1(x)) dx
β − U(x) dx
. (47)
Therefore, for β = O( ), T = 1 + O( ), R = O( ), but if β
coincides with the real part of the solution of
β − U(x) dx = 0, then we have R = O(1) (the Wood
anomaly).

Ponencia en el XVIII CCM

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Ponencia en el XVIII CCM