Calculus 101 to a Nobel Prize in under 50 minutes - School of

Calculus 101 to a Nobel Prize in under 50 minutes

William J. Parnell

Faculty Research Fellow

School of Mathematics, University of Manchester, UK

[email protected]

http://www.maths.man.ac.uk/∼wparnell.

W.J.Parnell, School of Mathematics, University of Manchester. – p.1/51

Scattering of elastic, acoustics, electromagnetic by a single obstacle

Multiple scattering by many obstacles - complicated scattered field andcomplex effects can develop.

Uei(kx−ωt)

Uei(kx−ωt)


Applications

• Composite Material Design


• Biomechanics (imaging, prognosis)


Overview

• Scattering from a single inclusion (Calculus 101)

• Scattering from multiple inclusions

• Some results......

• Anderson Localization (A Nobel Prize)


Calculus 101 - scattering from anisolated inclusion


Scattering from a single, cylindrical void

First, don’t worry about boundary conditions.Let’s just play around with solving the wave equation:

∇2u =1

c2utt

Assume

u = U(r, θ)e−iωt

Urr +1

rUr +

1

r2Uθθ + k2U = 0


Separation of VariablesSeek a separable solution of the form

U(r, θ) = R(r)Θ(θ)

so that, on dividing by RΘ/r2 and rearranging:

r2 Rrr

R+ r

Rr

R+ k2r2 = −

Θθθ

Θ= n2

The Θ problem:

Θθθ + n2Θ = 0

Clearly for fixed n,

Θ(θ) = exp(±inθ)

and also n ∈ Z since solution must be periodic in θ.


TheR problem

We get:

r2Rrr + rRr + (k2r2 − n2)R = 0

which on setting s = kr becomes

s2Rss + sRs + (s2 − n2)R = 0

Frobenius series methods then define a special function.........


TheR problem

We get:

r2Rrr + rRr + (k2r2 − n2)R = 0

which on setting s = kr becomes

s2Rss + sRs + (s2 − n2)R = 0

Frobenius series methods then define a special function.........We define this as Bessel’s function of the first kind and order n:

R(r) = Jn(kr)

which, importantly is bounded at the origin.


Of course the ode is second order so we can generate a secondsolution from the first:

R(r) = Yn(kr)

which is Bessel’s function of the second kind and order n.This solution is singular at the origin.


Hankel functions

Note that we can now define:

H(1)n (kr) = Jn(kr) + iYn(kr)

H(2)n (kr) = Jn(kr) − iYn(kr)

and importantly these have far-field behaviour:

H(1)n (kr)e−iωt ∼ exp(i(kr − ωt))Cn

√

2

πkr

H(2)n (kr)e−iωt ∼ exp(i(kr + ωt))Cn

√

2

πkr

the former is outgoing, the latter is incoming.


Solutions

Hence the following three forms of solutions to the wave equation areusually used, depending upon the domain:

��

��

��

��

U =

∞∑

n=∞

AnJn(kr)einθ

U =∞∑

n=∞

(AnJn(kr) + BnYn(kr))einθ

U =∞∑

n=∞

(AnH(1)n (kr) + BnH(2)

n )einθ


Scattering

Now consider boundary conditions!

Uinc

r = a

What happens when an incidentwave

Uinc = exp(ikx) = exp(ikr cos θ)

is incident on a circular (cylindri-cal) cavity where

∂U

∂r= 0 on r = a


First, we can use the so-called generating function

exp

(

kr

2(t − 1/t)

)

=∞∑

n=−∞

tnJn(kr)

with t = i exp(iθ) to give (Jacobi-Anger identity)

exp(ikr cos θ) =∞∑

n=∞

inJn(kr)einθ

Now write U = Us + Uinc with

Us =∞∑

n=∞

in(AnH(1)n (kr) + BnH(2)

n (kr))einθ

but obviously the scattered field will be outgoing so we choose

Bn = 0


SolutionThe boundary condition

∂U

∂r= 0 on r = a

translates into

∞∑

n=−∞

in (J ′n(kr) + AnH ′

n(kr))∣

∣

∣

r=aeinθ = 0

so that we can satisfy this with An = −J ′n(ka)/H ′

n(ka), i.e.

Us =∞∑

n=∞

−inJ ′

n(ka)

H ′n(ka)

Hn(kr)einθ

so the total field is

U = eikx −∞∑

n=∞

inJ ′

n(ka)

H ′n(ka)

Hn(kr)einθ


Low frequency scattering

Note that is ǫ = ak ≪ 1, we have

Us ∼ A0H0(kr) + A1H1(kr) cos θ

where

A0 =ǫ2π

4iA1 = −

ǫ2π

4i


Multiple Scattering


2D Composite Slab

x1

x2

h−h

D

θinc

O

pspj

rs

rj

θs

θj

θjs

Rjs


Multiple scattering problem

Scale lengthscales on host wavenumber k0.So ǫ = ak0 is radius of fibres..

(∇2 + 1)U = 0

and boundary conditions

∂U

∂r= 0,

on rj = ǫ, j = 1, 2, ..., N .For ease of exposition let’s consider normal incidence, θinc = 0.


Solutions

U = Uinc +N

∑

j=1

∞∑

n=−∞

AjnZnH(1)

n (rj)einθj , in the host phase,

Near jth inclusion, incident wave is (Jacobi-Anger identity)

Uinc(x, y) = eix =∞∑

n=−∞

inJn(rj)einθj

where (rj , θj) are local polar coordinates with origin (pj , qj).Hence solution in vicinity of sth inclusion is


Solutions

U = Uinc +N

∑

j=1

∞∑

n=−∞

AjnZnH(1)

n (rj)einθj , in the host phase,

Near jth inclusion, incident wave is (Jacobi-Anger identity)

Uinc(x, y) = eix =∞∑

n=−∞

inJn(rj)einθj

where (rj , θj) are local polar coordinates with origin (pj , qj).Hence solution in vicinity of sth inclusion is

U(rs, θs) =∞∑

n=−∞

(

inJn(rs)einθs + As

nZnH(1)n (rs)e

inθs

)

+N

∑

j=1,j 6=s

∞∑

n=−∞

AjnZnH(1)

n (rj)einθj


Graf’s addition theorem

Graf’s addition theorem says that in vicinity of jth inclusion:

H(1)n (rj)e

inθj =∞∑

m=−∞

H(1)n−m(|rj − rs|)e

i(n−m)θjsJm(rs)eimθs

and thus


Graf’s addition theorem

Graf’s addition theorem says that in vicinity of jth inclusion:

H(1)n (rj)e

inθj =∞∑

m=−∞

H(1)n−m(|rj − rs|)e


and thus

U(rs, θs) =∞∑

n=−∞

(

inJn(rs)einθs + As

nZnH(1)n (rs)e

inθs

)

+

N∑

j=1,j 6=s

∞∑

n=−∞

AjnZn

∞∑

m=−∞

H(1)n−m(Rjs)e



Application of boundary conditions and orthogonality thus gives

Asm +

N∑

j=1,j 6=s

∞∑

n=−∞

AjnZnHn−m(Rjs)e

i(n−m)θjs = −im

for m ∈ Z.

Various techniques have been developed in order to speed up thesummation of these series in order to calculate An.


"Effective" SH Waves

When ǫ ≪ 1 an effective or homogenized wave propagates through theinhomogeneous medium and it can be shown that this wave is of theform

U = exp(ik∗x)

where

k2∗ =

ρ∗µ∗

ω2

and thus

ρ∗ = (1 − φ)ρ,

µ∗ =(1 − φ)

(1 + φ)µ.


Wave propagation throughinhomogeneous media


Wave Propagation through inhomogeneous media


Anderson Localization - the Nobel Prize


Anderson Localization

• Anderson was awarded the Nobel Prize in 1977.• This work was related to transport of electrons.• Change in behaviour from diffusive regime (in which Ohm’s law

holds) to a localized state where the material behaves as aninsulator. Effect due to interaction of electrons which haveundergone multiple scattering.

• Explained Metal-Insulator transitions.• 1980’s: what about classical wave physics (light, ultrasound, etc.)


I still don’t really understand all of this........

For classical systems, essentially the argument is:• Propagation of light through inhomogeneous media can be

described as a diffusion process• For diffusion of light through a disordered medium, the same

Ohm’s law holds as for diffusion of electrons through a resistor:transmission decreases linearly with system length

• Anderson localization brings diffusion to a halt: transmissiondecreases exponentially rather than linearly with system length

• This transition requires very strong scattering


Localization of light

λ = 1064nm, three powder types.

• Electromagnetic propagation through very strongly scatteringsemiconductor powders.

• Powder is Gallium Arsenide (GaAs).• Very little absorption.


localization of light


Localization of ultrasound


Link with multiple scattering theory

• Clearly there must be one!• However it is still unclear how this effect links in with the theory

described here, it particular the lengthscale effects described.• In particular the sample length effect.


Pre-stressed inhomogeneous media


Problem description

aA

Undeformed Deformed

p∞

p∞

p∞

p∞

p∞

p∞

p∞p∞


Problem description

a

p∞

p∞

p∞

p∞

p∞

p∞

p∞p∞

exp iΓx1


Radial stressOnly 1 equilbrium equation:

dΣrr

dr+

1

r(Σrr − Σθθ) = 0

which can be integrated to give Σrr:

2 4 6 8 10

-5

-4

-3

-2

-1

rΣrr

p∞µ

= 1

p∞µ

= 2

p∞µ

= 4


Incremental deformationSuperpose small amplitude waves on the finite deformation:

u = u + ηu′

where u is the finite displacement, η ≪ 1 and

u′ = (0, 0, w(r, θ)) exp(iωt).


Incremental deformationSuperpose small amplitude waves on the finite deformation:

u = u + ηu′

where u is the finite displacement, η ≪ 1 and

u′ = (0, 0, w(r, θ)) exp(iωt).

Transpires that the modified wave equation is:

(

1 +k

r2

)

∂2w

∂r2+

1

r

(

1 −k

r2

)

∂w

∂r+

1

r2

(

1 −k

r2 + K

)

∂2w

∂θ2+ γ2w = 0.

Can think of this as:

(∇2 + Lk + γ2)w = 0

where Lk = 0 for no pre-stress.W.J.Parnell, School of Mathematics, University of Manchester. – p.45/51

Low Frequency Scattering

Once again assume a low frequency plane incident wave incomingfrom infinity:

winc = eiγr cos θ

so that the scattered wave is described by:

(∇2 + Lk + γ2)ws + I(r, θ) = 0


Matched Asymptotics - Schematic

Scaling lengthscales on either a or γ gives inner and outer problems.

��

��

��

��

OuterOverlap

Inner


Matching

Outer solution:F0(ro) = ǫ2

(

B0H(1)0 (ro) + E0(ro)

)

+ o(r2o)

Inner solution:

f0(ri) = ǫ2(1

4(1 + k) log(r2

i + k) + C0 log ǫ + D0

)

+ o(ǫ2)

Van Dyke’s matching principle says that f(1,1)0 = F

(1,1)0 which gives the

Modified Monopole Scattering Coefficient due to the pre-stress:

B0 =πǫ204i

[

a2L

A2(1 + (L − 1)S1)

(

1 +(A2 − a2)LS1

a2(1 + (L − 1)S1)

)]


Monopole Scattering CoefficientB0

-3 -2 -1 1 2 3

1.0

1.5

2.0

p∞/µ

B0/Γ2

S1 = 1

S1 = 0.9

S1 = 0.8


Dipole Scattering Coefficient:B1:

-3 -2 -1 0 1 2 3

0.5

1.0

1.5

2.0

p∞/µ

|B1|/Γ2

S1 = 1

S1 = 0.9

S1 = 0.8


Effective shear propertiesWe can use this canonical problem in multiple scattering theory inorder to deduce the effective response of a pre-stressedinhomogeneous medium:

0.1 0.2 0.3 0.4

0.5

0.6

0.7

0.8

0.9

1.0

µ∗

φ0

p∞/µ = 1

p∞/µ = 0

p∞/µ = −1


Calculus 101 to a Nobel Prize in under 50 minutes - School of

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